1 Introduction

This paper deals with the classical Keller–Segel problem in \({\mathbb {R}}^2\),

$$\begin{aligned} \left\{ \begin{aligned} u_t =&\Delta u - \nabla \cdot (u \nabla v) \quad {\quad \hbox {in } }{\mathbb {R}}^2\times (0,\infty ),\\ v =&(-\Delta _{{\mathbb {R}}^2})^{-1} u:= \frac{1}{2\pi } \int _{{\mathbb {R}}^2} \, \log \frac{1}{|x-z|}\,u(z,t)\, \textrm{d}z, \\ {}&\qquad \ u(\cdot ,0) = u_0 \quad \hbox {in } {\mathbb {R}}^2, \end{aligned} \right. \end{aligned}$$
(1.1)

which is a well-known model for the dynamics of a population density u(xt) evolving by diffusion with a chemotactic drift. We consider positive solutions which are well defined, unique and smooth up to a maximal time \(0< T \leqq +\infty \). This problem formally preserves mass, in the sense that

$$\begin{aligned} \int _{{\mathbb {R}}^2} u(x,t)dx = \int _{{\mathbb {R}}^2} u_0(x)\,dx =: M \quad \text{ for } \text{ all }\quad t\in (0,T). \end{aligned}$$

An interesting feature of (1.1) is the connection between the second moment of the solution and its mass which is precisely given by

$$\begin{aligned} \frac{d}{dt} \int _{{\mathbb {R}}^2}|x|^2\,u(x,t)\,dx = 4M - \frac{M^2}{2\pi }, \end{aligned}$$

provided that the second moments are finite. If \(M>8\pi \), the negative rate of production of the second moment and the positivity of the solution implies finite blow-up time. If \(M<8\pi \) the solution lives at all times and diffuses to zero with a self similar profile according to [5]. When \(M = 8\pi \) the solution is globally defined in time. If the initial second moment is finite, it is preserved in time, and there is infinite time blow-up for the solution, as was shown in [4].

Globally defined in time solutions of (1.1) are of course its positive finite mass steady states, which consist of the family

$$\begin{aligned} U_{\lambda ,\xi }(x) = \frac{1}{\lambda ^2} U\left( \frac{x-\xi }{\lambda } \right) , \quad U(y) = \frac{8}{(1+|y|^2)^2 }, \quad \lambda >0, \ \xi \in {\mathbb {R}}^2. \end{aligned}$$
(1.2)

We observe that all these steady states have the exact mass \(8\pi \) and infinite second moment

$$\begin{aligned} \int _{{\mathbb {R}}^2} U_{\lambda ,\xi }(x)\, \textrm{d}x\, = \, 8\pi , \quad \int _{{\mathbb {R}}^2}|x|^2\,U_{\lambda ,\xi }(x)\, \textrm{d}x\, = \, +\infty . \end{aligned}$$

As a consequence, if a solution of (1.1) is attracted by the family \((U_{\lambda ,\xi })\), its mass must be larger than \(8\pi \) and if the initial second moment is finite, then blow-up occurs in a singular limit corresponding to \(\lambda \rightarrow 0_+\).

In the critical mass \(M=8\pi \) case, the infinite-time blow-up in (1.1) when the second moment is finite, takes place in the form of a bubble in the form (1.2) with \(\lambda =\lambda (t)\rightarrow 0\) according to [2, 4]. Formal rates and precise profiles were derived in [8, 12] to be

$$\begin{aligned} \lambda (t) \sim \frac{c}{\sqrt{\log t}}\quad \hbox {as } t\rightarrow +\infty . \end{aligned}$$

A radial solution with this rate was built by Ghoul and Masmoudi [26] and its stability within the radial class was established. The framework of the construction in [26] was actually fully nonradial, but for stability a spectral gap inequality only known in the radial case was used. Numerical evidence for this inequality was obtained in [7], and stability for general nonradial perturbation was conjectured in [26]. A related spectral estimate, useful in the analysis of finite time blow-up was found in [15].

In this paper we construct an infinite-time blow-up solution with a different method to that in [26], which in particular leads to a proof of the stability assertion among non-radial functions. The following is our main result:

Theorem 1.1

There exists a nonnegative, radially symmetric function \(u_0^*(x)\) with critical mass \(\int _{{\mathbb {R}}^2} u_0^*(x)\,dx =8\pi \) and finite second moment \(\int _{{\mathbb {R}}^2}|x|^2\, u_0^*(x)\,dx <+\infty \) such that for every \(u_1(x)\) sufficiently close (in suitable sense) to \(u_0^*\) with \(\int _{{\mathbb {R}}^2} u_1\,dx =8\pi \), we have that the solution u(xt) of system (1.1) with initial condition \(u(x,0)= u_1(x) \) has the form

$$\begin{aligned} u(x,t)\, =\, \frac{1}{\lambda (t)^2} U\Bigl (\frac{x-\xi (t)}{\lambda (t)} \Bigr ) (1+ o(1)) , \quad U(y)= \frac{8}{(1+|y|^2)^2} \end{aligned}$$
(1.3)

uniformly on bounded sets of \({\mathbb {R}}^2\), and

$$\begin{aligned} \lambda (t)\ =\ \frac{c}{\sqrt{\log t}}\,(1+o(1)), \quad \xi (t)\rightarrow q \quad \hbox {as } t\rightarrow +\infty , \end{aligned}$$

for some number \(c>0\) and some \(q\in {\mathbb {R}}^2\).

Sufficiently close for the perturbation \(u_1(x):= u_0^*(x) + \varphi (x)\) in this result is measured in the \(C^1\)-weighted norm for some \(\sigma >1\)

$$\begin{aligned} \Vert \varphi \Vert _{*\sigma }:= \Vert (1+|\cdot | ^{4+\sigma }) \varphi \Vert _{L^\infty ({\mathbb {R}}^2)} + \Vert (1+|\cdot |^{5+\sigma }) {\nabla } \varphi (x) \Vert _{L^\infty ({\mathbb {R}}^2)}<+\infty . \end{aligned}$$

The perturbation \(\varphi \) must have zero mass too.

“Uniformly on bounded sets” of \({\mathbb {R}}^2\) in (1.3) means that for any bounded \(K\subset {\mathbb {R}}^2\)

$$\begin{aligned} \lim _{t\rightarrow \infty } \sup _{x\in K} \lambda (t)^2 U\Bigl (\frac{x-\xi (t)}{\lambda (t)} \Bigr )^{-1} \left| u(x,t)- \frac{1}{\lambda (t)^2} U\Bigl (\frac{x-\xi (t)}{\lambda (t)} \Bigr ) \right| =0. \end{aligned}$$

The expansion of u(xt) can be made more precise though, and this is explained along the proof of theorem.

The scaling parameter is rather simple to find at main order from the approximate conservation of second moment, see Sect. 2. The center \(\xi (t)\) actually obeys a relatively simple system of nonlocal ODEs.

We devote the rest of this paper to the proof of Theorem 1.1. Our approach borrows elements of constructions in the works [16,17,18, 21] based on the so-called inner-outer gluing scheme, where a system is derived for an inner equation defined near the blow-up point and expressed in the variable of the blowing-up bubble, and an outer problem that sees the whole picture in the original scale. The result of Theorem 1.1 has already been announced in [20] in connection with [16, 18, 21].

There is a vast literature on chemotaxis in biology and in mathematics. The Patlak–Keller–Segel model [35, 44] is used in mathematical biology to describe the motion of mono-cellular organisms, like Dictyostelium Discoideum, which move randomly but experience a drift in presence of a chemo-attractant. Under certain circumstances, these cells are able to emit the chemo-attractant themselves. Through the chemical signal, they coordinate their motion and eventually aggregate. Such a self-organization scenario is at the basis of many models of chemotaxis and is considered as a fundamental mechanism in biology. Of course, the aggregation induced by the drift competes with the noise associated with the random motion so that aggregation occurs only if the chemical signal is strong enough. A classical survey of the mathematical problems in chemotaxis models can be found in [31, 32]. After a proper adimensionalization, it turns out that all coefficients in the Patlak–Keller–Segel model studied in this paper can be taken equal to 1 and that the only free parameter left is the total mass. For further considerations on chemotaxis, we shall refer to [30] for biological models and to [11] for physics backgrounds.

In many situations of interest, cells are moving on a substrate. The two-dimensional case is therefore of special interest in biology, but also turns out to be particularly interesting from the mathematical point of view as well, because of scaling properties, at least in the simplest versions of the Keller–Segel model. Boundary conditions induce various additional difficulties. In the idealized situation of the Euclidean plane \({\mathbb {R}}^2\), it is known since the early work of W. Jäger and S. Luckhaus [33] that solutions globally exist if the mass M is small and blow-up in finite time if M is large. The blow-up in a bounded domain is studied in [1, 33, 39, 40, 46]. The precise threshold for blow-up, \(M=8\pi \), has been determined in [5, 23], with sufficient conditions for global existence if \(M\leqq 8\pi \) in [5] (also see [22] in the radial case). The key estimate is the boundedness of the free energy, which relies on the logarithmic Hardy–Littlewood–Sobolev inequality established in optimal form in [9]. We refer to [3] for a review of related results. If \(M<8\pi \), diffusion dominates: intermediate asymptotic profiles and exact rates of convergence have been determined in [7]. Also see [25, 41]. In the supercritical case \(M>8\pi \), various formal expansions are known for many years, starting with [27, 28, 49] which were later justified in [38, 45], in the radial case, and in [14], in the non-radially symmetric regime. This latter result is based on the analysis of the spectrum of a linearized operator done in [15], based on the earlier work [19], and relies on a scalar product already considered in [45] and similar to the one used in [6, 7] in the subcritical mass regime. An interesting subproduct of the blow-up mechanism in [29, 45] is that the blow-up takes the form of a concentration in the form of a Dirac distribution with mass exactly \(8\pi \) at blow-up time, as was expected from [24, 29], but it is still an open question to decide whether this is, locally in space, the only mechanism of blow-up.

The critical mass case \(M=8\pi \) is more delicate. If the second moment is infinite, there is a variety of behaviors as observed for instance in [36, 37, 43]. For solutions with finite second moment, blow-up is expected to occur as \(t\rightarrow +\infty \): see [34] for grow-up rates in \({\mathbb {R}}^2\), and [48] for the higher-dimensional radial case. The existence in \({\mathbb {R}}^2\) of a global radial solution and first results of large time asymptotics were established in [2] using cumulated mass functions. In [4], the infinite time blow-up was proved without symmetry assumptions using the free energy and an assumption of boundedness of the second moment. Also see [42, 43] for an existence result under weaker assumptions, and further estimates on the solutions. Asymptotic stability of the family of steady states determined by (1.2) under the mass constraint \(M=8\pi \) has been determined in [10]. The blow-up rate \(\lambda (t)\) and the shape of the limiting profile U were identified in formal asymptotic expansions in [12, 13, 47, 49, 50] and also in [8, Chapter 8]. As already mentioned, a radial solution with rate \(\lambda (t)\sim (\log t)^{-1/2}\) was built and its stability within the radial class was established in [26].

2 Formal Derivation of the Behavior of the Parameters

We consider here a first approximation to a solution u(xt) of (1.1), globally defined in time, such that on bounded sets in x,

$$\begin{aligned} u(x,t) = \frac{1}{\lambda (t)^2}U\left( \frac{x- \xi (t)}{\lambda (t)} \right) (1+ o(1)) \quad \text {as } t\rightarrow +\infty \end{aligned}$$
(2.1)

for certain functions \(0<\lambda (t) \rightarrow 0\) and \(\xi (t) \rightarrow q\in {\mathbb {R}}^2\), where we recall that

$$\begin{aligned} U(y) = \frac{8}{(1+|y|^2)^2}. \end{aligned}$$

We know that (2.1) can only happen in the critical mass, finite second moment case,

$$\begin{aligned} \int _{{\mathbb {R}}^2} u(x,t)dx = 8\pi , \quad \int _{{\mathbb {R}}^2} |x|^2 u(x,t)dx <+\infty , \end{aligned}$$

which according to the results in [4, 12, 26] is consistent with a behavior of the form (2.1). Since the second moment of U is infinite, we do not expect the approximation (2.1) be uniform in \({\mathbb {R}}^2\) but sufficiently far, a faster decay in x should take place as we shall see next. We will find an approximate asymptotic expression for the scaling parameter \(\lambda (t)\) that matches with this behavior.

Let us introduce the function \( \Gamma _0: = (-\Delta )^{-1} U. \) We directly compute

$$\begin{aligned} \Gamma _0 (y) = \log \frac{8}{(1+ |y|^2)^2} \end{aligned}$$

and hence \(\Gamma _0\) solves the Liouville equation

$$\begin{aligned} -\Delta \Gamma _0\, = \, e^{\Gamma _0} = U {\quad \hbox {in } }{\mathbb {R}}^2. \end{aligned}$$

Then \(\nabla \Gamma _0(y) \approx -\frac{4y}{|y|^2} \) for all large y, and hence we get, away from \(x=\xi \),

$$\begin{aligned} -\nabla \cdot ( u\nabla (-\Delta )^{-1} u) \approx 4\nabla u \cdot \frac{x-\xi }{|x-\xi |^2}. \end{aligned}$$

Therefore, defining

$$\begin{aligned} {\mathcal {E}}(u): = \Delta u - \nabla \cdot ( u\nabla (-\Delta )^{-1} u) \end{aligned}$$
(2.2)

and writing in polar coordinates

$$\begin{aligned} u(r,\theta , t) = u(x,t), \quad x= \xi (t) + re^{i\theta }, \end{aligned}$$

we find \( {\mathcal {E}}(u) \approx {\partial } _r^2 u + \frac{5}{r} {\partial } _r u \). Hence, assuming that \({\dot{\xi }} (t)\rightarrow 0 \) sufficiently fast, equation (1.1) approximately reads as

$$\begin{aligned} {\partial } _t u = {\partial } _r^2 u + \frac{5}{r} {\partial } _r u, \end{aligned}$$

which can be idealized as a homogeneous heat equation in \({\mathbb {R}}^6\) for radially symmetric functions. It is therefore reasonable to believe that beyond the self-similar region \(r\gg \sqrt{t}\) the behavior changes into a function of \(r/\sqrt{t}\) with fast decay at \(+\infty \) that yields finiteness of the second moment. To obtain a first global approximation, we simply cut-off the bubble (2.1) beyond the self-similar zone. We introduce a further parameter \(\alpha (t) \) and set

$$\begin{aligned} {{\bar{u}}}(x,t) = \frac{\alpha (t) }{\lambda ^2}U\Bigl ( \frac{x-\xi }{\lambda } \Bigr ) \chi (x,t) , \end{aligned}$$
(2.3)

where

$$\begin{aligned} \chi (x,t)= \chi _0 \Bigl ( \frac{x-\xi }{\sqrt{t}} \Bigr ) \end{aligned}$$
(2.4)

with \(\chi _0\) a smooth radial cut-off function such that

$$\begin{aligned} \chi _0(z) = {\left\{ \begin{array}{ll} 1 \ {} &{} \hbox { if } |z|\leqq 1, \\ 0\ {} &{} \hbox { if } |z|\geqq 2 . \end{array}\right. } \end{aligned}$$
(2.5)

We introduce the parameter \(\alpha (t)\) because the total mass of the actual solution should equal \(8\pi \) for all t. However,

$$\begin{aligned} \frac{1}{\lambda ^2}\int _{{\mathbb {R}}^2} U\Bigl (\frac{x-\xi }{\lambda }\Bigr ) \chi (x,t)\, \textrm{d}x = 8 \pi + 16\pi \Upsilon \frac{\lambda ^2}{t} + O\Bigl ( \frac{\lambda ^4}{t^2}\Bigr ) \end{aligned}$$
(2.6)

as \(t\rightarrow \infty \), where

$$\begin{aligned} \Upsilon = \int _0^\infty ({\tilde{\chi }}_0(s)-1) s^{-3}ds <0, \end{aligned}$$
(2.7)

and \(\chi _0(x) = {\tilde{\chi }}_0(|x|)\). To achieve \( \int _{{\mathbb {R}}^2} {{\bar{u}}}(x,t)\,dx = 8\pi \) we set \(\alpha ={\bar{\alpha }}\) where

$$\begin{aligned} {\bar{\alpha }}(t) = 1 - 2 \Upsilon \frac{\lambda ^2}{t} + O\Bigl ( \frac{\lambda ^4}{t^2}\Bigr ) . \end{aligned}$$

Next we will obtain an approximate value of the scaling parameter \(\lambda (t)\) that is consistent with the existence of a solution \(u(x,t)\approx {{\bar{u}}}(x,t)\) where \({{\bar{u}}}\) is the function in (2.3) with \(\alpha ={\bar{\alpha }}\). Let us consider the “error operator”

$$\begin{aligned} S(u) = - u_t + {\mathcal {E}}(u), \end{aligned}$$
(2.8)

where \({\mathcal {E}}(u) \) is defined in (2.2). We have the following well-known identities, valid for an arbitrary function \(\omega (x)\) of class \(C^2({\mathbb {R}}^2)\) with finite mass and \(D^2 \omega (x) = O(|x|^{-4-\sigma })\) for large |x|:

$$\begin{aligned} \int _{{\mathbb {R}}^2} \mathcal |x|^2{\mathcal {E}}(\omega )\, \textrm{d}x = 4M - \frac{M^2}{2\pi }, \quad M= \int _{{\mathbb {R}}^2} \omega (x) dx \end{aligned}$$
(2.9)

and

$$\begin{aligned} \int _{{\mathbb {R}}^2} x{\mathcal {E}}(\omega )\, \textrm{d}x = 0, \quad \int _{{\mathbb {R}}^2} {\mathcal {E}}(\omega )\, \textrm{d}x = 0. \end{aligned}$$
(2.10)

Let us recall the simple proof of (2.9). Integrating by parts on finite balls with large radii and using the behavior of the boundary terms we get the identities

$$\begin{aligned} \int _{{\mathbb {R}}^2} \mathcal |x|^2\Delta \omega \, \textrm{d}x&= 4M , \nonumber \\ \int _{{\mathbb {R}}^2} |x|^2 \nabla \cdot (\omega \nabla (-\Delta )^{-1})\omega ) \, \textrm{d}x&= - 2 \int _{{\mathbb {R}}^2} x\cdot \omega \nabla (-\Delta )^{-1}\omega \,dx \nonumber \\&= \frac{1}{\pi }\int _{{\mathbb {R}}^2} \int _{{\mathbb {R}}^2} \omega (x)\omega (y) \frac{x\cdot (x-y)}{|x-y|^2} dx\,dy \nonumber \\&= \frac{1}{2\pi } \int _{{\mathbb {R}}^2} \int _{{\mathbb {R}}^2}\omega (x)\omega (y) \frac{(x-y)\cdot (x-y)}{|x-y|^2} dx\,dy \nonumber \\&= \frac{M^2}{2\pi } \end{aligned}$$
(2.11)

and then (2.9) follows. The proof of (2.10) is even simpler. For a solution u(xt) of (1.1) we then get

$$\begin{aligned} \frac{d}{dt} \int _{{\mathbb {R}}^2} u(x,t)|x|^2 dx = 4M - \frac{M^2}{2\pi }, \quad M= \int _{{\mathbb {R}}^2} u(x,t) dx. \end{aligned}$$

In particular, if u(xt) is sufficiently close to \({{\bar{u}}}(x,t)\) and since \(\int _{{\mathbb {R}}^2} {{\bar{u}}}(x,t) dx =8\pi \), we get the approximate validity of the identity

$$\begin{aligned} \frac{d}{dt} \int _{{\mathbb {R}}^2} {{\bar{u}}}(x,t)|x|^2 dx = 0. \end{aligned}$$

This means

$$\begin{aligned} a I(t):= \int _{{\mathbb {R}}^2} \frac{{\bar{\alpha }}}{\lambda ^2 }U\left( \frac{x-\xi }{\lambda }\right) \chi _0 \left( \frac{x-\xi }{\sqrt{t}} \right) |x|^2 dx \, =\, constant. \end{aligned}$$

We readily check that for some constant \(\kappa \)

$$\begin{aligned} I(t) = 16\pi \lambda ^2 \int _0^{\frac{\sqrt{t}}{\lambda }} \frac{ \rho ^3 d\rho }{ (1+\rho ^2)^2 } + \kappa + o(1) = 16\pi \lambda ^2\log \frac{\sqrt{t}}{\lambda }+ \kappa + o(1) \quad \text {as } \lambda \rightarrow 0. \end{aligned}$$

Then we conclude that \(\lambda (t)\) approximately satisfies

$$\begin{aligned} \lambda ^2 \log t \, = \, c^2 \,=\, constant \end{aligned}$$

and hence we get at main order

$$\begin{aligned} \lambda (t) = \frac{c}{\sqrt{\log t}}. \end{aligned}$$

We also notice that the center of mass is preserved for a true solution, thanks to (2.10):

$$\begin{aligned} \frac{d}{dt} \int _{{\mathbb {R}}^2} x u(x,t) dx = 0. \end{aligned}$$

Since the center of mass of \({{\bar{u}}}(x,t)\) is exactly \(\xi (t)\) we then get that approximately

$$\begin{aligned} \xi (t) = constant = q. \end{aligned}$$

3 The Approximations \(u_0\) and \(u_1\)

From now on we to consider the Keller–Segel system starting at a large \(t_0\),

$$\begin{aligned} \left\{ \begin{aligned} u_t =&\Delta u - \nabla \cdot (u \nabla v) \quad {\quad \hbox {in } }{\mathbb {R}}^2\times (t_0,\infty ),\\ v =&(-\Delta _{{\mathbb {R}}^2})^{-1} u:= \frac{1}{2\pi } \int _{{\mathbb {R}}^2} \, \log \frac{1}{|x-z|}\,u(z,t)\, \textrm{d}z, \\ {}&\qquad \ u(\cdot ,t_0) = u_0 \quad \hbox {in } {\mathbb {R}}^2, \end{aligned} \right. \end{aligned}$$
(3.1)

which is equivalent to (1.1). We do this so that some expansions for t large take a simpler form.

In this section we will define a basic approximation to a solution of the Keller–Segel system (3.1). Let us consider parameter functions

$$\begin{aligned} 0<\lambda (t)\rightarrow 0,\quad \xi (t)\rightarrow q, \quad \alpha (t)\rightarrow 1 \quad \hbox {as } t\rightarrow +\infty \end{aligned}$$

that we will later specify. Let us consider the functions

$$\begin{aligned} U (y) = \frac{8}{(1+|y|^2)^2}, \quad \Gamma _0(y) = \log U(y) \end{aligned}$$

and define the approximate solution \(u_0(x,t)\) as

$$\begin{aligned} u_0(x,t)&= \frac{\alpha }{\lambda ^2} U \Bigl ( \frac{x-\xi }{\lambda } \Bigr ) \chi (x,t), \nonumber \\ v_0(x,t)&= (-\Delta _x)^{-1} u_0 = \frac{1}{2\pi }\int _{{\mathbb {R}}^2 } \log \frac{1}{|x-{{\bar{x}}}|} \, u_0({{\bar{x}}},t)\, d{{\bar{x}}}, \end{aligned}$$
(3.2)

where \(\chi \) is the cut-off function (5.3). We consider the error operator

$$\begin{aligned} S(u)&= - {\partial } _t u + {\mathcal {E}}(u ), \end{aligned}$$

where

$$\begin{aligned} {\mathcal {E}}(u ) = \Delta _x u - \nabla _x \cdot ( u \nabla _x v), \quad v = (-\Delta _{x})^{-1} u . \end{aligned}$$

and next measure the error of approximation \(S(u_0)\).

We have

$$\begin{aligned} - {\partial } _t u_0 (x,t)&= - \frac{\dot{\alpha }}{\lambda ^2} U(y) \chi _0(z) + \alpha \frac{{\dot{\lambda }}}{\lambda ^3} Z_0 \chi _0(z) +\frac{\alpha }{\lambda ^3} {\dot{\xi }} \cdot {\nabla } _y U (y) \, \chi _0(z) \nonumber \\&\quad + \frac{\alpha }{\lambda ^2 \sqrt{t}} U(y) {\dot{\xi }} \cdot \nabla _z \chi _0(z) + \frac{\alpha }{2\lambda ^2 t } U(y) \nabla _z\chi _0(z) \cdot z, \nonumber \\ z&= \frac{x-\xi }{\sqrt{t}} \end{aligned}$$
(3.3)

where

$$\begin{aligned} Z_0(y) = 2 U(y) + y\cdot \nabla _y U(y) , \quad y = \frac{x-\xi }{\lambda }. \end{aligned}$$
(3.4)

We also have

$$\begin{aligned} {\mathcal {E}}(u_0)&= \Delta _x u_0 - \nabla _x \cdot ( u_0 \nabla _x v_0 ) \\&= \frac{2\alpha }{\lambda ^3 t^{1/2} } \nabla _z \chi _0(z) \cdot \nabla _y U (y) + \frac{\alpha }{t} \frac{1}{\lambda ^2} \Delta _z \chi _0(z) U(y) - \frac{\alpha }{\lambda ^2 \sqrt{t}} U(y) \nabla _z \chi _0(z)\cdot \nabla _x v_0 \\&\quad + \frac{\alpha \chi _0(z)}{\lambda ^4} \Bigl [ (\chi _0(z) \alpha -1) U^2(y) - \nabla _y U(y) \cdot ( \nabla _y v_0- \nabla _y \Gamma _0 ) \Bigr ]\ . \end{aligned}$$

Let us decompose

$$\begin{aligned} v_0(y) = \alpha \Gamma _0(y) + {\mathcal {R}}(y). \end{aligned}$$
(3.5)

For the term \({\mathcal {R}}\) in (3.5) we directly estimate

$$\begin{aligned} |\nabla _y {\mathcal {R}}(y)| \le {\left\{ \begin{array}{ll} \frac{\lambda ^2}{t} \frac{1}{|y|} &{} |y| \ge \frac{\sqrt{t}}{\lambda }, \\ 0 &{} |y| \le \frac{\sqrt{t}}{\lambda }. \end{array}\right. } \end{aligned}$$
(3.6)

Then

$$\begin{aligned} {\mathcal {E}}(u_0)&= \frac{2\alpha }{\lambda ^3 t^{1/2} } \nabla _z \chi _0(z) \cdot \nabla _y U (y) + \frac{\alpha }{t} \frac{1}{\lambda ^2} \Delta _z \chi _0(z) U(y) - \frac{\alpha }{\lambda ^2 \sqrt{t}} U(y) \nabla _z \chi _0(z) \nabla _x v_0 \\&\quad + \frac{\alpha \chi _0(z)}{\lambda ^4} \Bigl [ (\alpha -1) U^2(y) - (\alpha -1) \nabla _y U(y) \cdot \nabla _y \Gamma _0(y) + \alpha (\chi _0(z)-1)U^2(y) \\&\quad - \nabla _y U(y) \cdot \nabla _y {\mathcal {R}}(y) \Bigr ]. \end{aligned}$$

and thus

$$\begin{aligned} S(u_0)&= - \frac{\dot{\alpha }}{\lambda ^2} U(y) \chi _0(z) + \alpha \frac{{\dot{\lambda }}}{\lambda ^3} Z_0 \chi _0(z) +\frac{\alpha }{\lambda ^3} {\dot{\xi }} \cdot {\nabla } _y U (y) \, \chi _0(z) \nonumber \\&\quad + \frac{\alpha }{\lambda ^2 \sqrt{t}} U(y) {\dot{\xi }} \cdot \nabla _z \chi _0(z) + \frac{\alpha }{2\lambda ^2 t } U(y) \nabla _z\chi _0(z) \cdot z \nonumber \\&\quad + \frac{2\alpha }{\lambda ^3 t^{1/2} } \nabla _z \chi _0(z) \cdot \nabla _y U (y) + \frac{\alpha }{t} \frac{1}{\lambda ^2} \Delta _z \chi _0(z) U(y)\nonumber \\&\quad - \frac{\alpha }{\lambda ^2 \sqrt{t}} U(y) \nabla _z \chi _0(z) \cdot \nabla _x v_0 \nonumber \\&\quad - \frac{\alpha (\alpha -1) \chi _0(z)}{\lambda ^4} \nabla _y \cdot ( U(y) \nabla _y \Gamma _0(y)) \nonumber \\&\quad + \frac{\alpha \chi _0(z)}{\lambda ^4} \Bigl [ \alpha (\chi -1)U^2(y) - \nabla _y U(y) \cdot \nabla _y {\mathcal {R}}(y) \Bigr ]. \end{aligned}$$
(3.7)

For a function \(v(\zeta )\) defined for \(\zeta \in {\mathbb {R}}^2\) consider the operator

$$\begin{aligned} \Delta _6 v(\zeta ) = \Delta v(\zeta ) + 4\frac{\zeta }{|\zeta |^2} \cdot {\nabla } _\zeta v (\zeta ) . \end{aligned}$$
(3.8)

The reason for the notation is that for radial functions \( v= v(r)\), \(r=|\zeta |\), we have

$$\begin{aligned} \Delta _6 v = {\partial } ^2_r v + \frac{5}{r} {\partial } _r v, \end{aligned}$$

which corresponds to Laplace’s operator in \({\mathbb {R}}^6\) on radial functions.

Let \({\tilde{\varphi }}_\lambda (\zeta ,t)\) be the (radial) solution to

$$\begin{aligned} \left\{ \begin{aligned} \partial _t {\tilde{\varphi }}_\lambda&= \Delta _6 {\tilde{\varphi }}_\lambda + E(\zeta ,t) \quad \text {in }{\mathbb {R}}^2 \times ( \frac{t_0}{2},\infty ), \\ {\tilde{\varphi }}_\lambda (\cdot ,\frac{t_0}{2})&= 0 \quad \text {in }{\mathbb {R}}^2, \end{aligned} \right. \end{aligned}$$
(3.9)

given by Duhamel’s formula, where \(E(\zeta ,t)\) is the radial function

$$\begin{aligned} E( \zeta ,t;\lambda ) = \frac{{\dot{\lambda }}}{\lambda ^3} Z_0\Bigl (\frac{\zeta }{\lambda }\Bigr ) \chi _0\Bigl ( \frac{\zeta }{\sqrt{t}}\Bigr ) + \frac{1}{2\lambda ^2 t} U \Bigl (\frac{\zeta }{\lambda }\Bigr ) \nabla _z \chi _0(z) \cdot z + {{\tilde{E}}} (x,t) , \end{aligned}$$
(3.10)

and

$$\begin{aligned} {{\tilde{E}}}(\zeta ,t;\lambda )&= \frac{2}{\lambda ^3 t^{1/2}} \nabla _z \chi _0(z) \cdot \nabla _y U (y) + \frac{1}{\lambda ^2 t} \Delta _z \chi _0(z) U (y) \nonumber \\&\quad - \frac{1}{\lambda ^3 t^{1/2}} U(y) \nabla _z \chi _0(z) \cdot \nabla _y \Gamma _0(y) , \end{aligned}$$
(3.11)

with \(z = \frac{\zeta }{\sqrt{t}}\), \( y = \frac{\zeta }{\lambda } \).

We then define

$$\begin{aligned} \varphi _\lambda (x,t) = {\tilde{\varphi }}_\lambda (x-\xi (t),t) . \end{aligned}$$
(3.12)

The reason to define \(\varphi _\lambda \) for \(t> \frac{t_0}{2}\) is that it gives better properties for the first approximation of \(\lambda \) constructed in Sect. 7. Since \(\lambda (t)\) is defined naturally for \(t>t_0\), we will need to define \(\lambda (t)\) for \(\frac{t_0}{2}<t<t_0\) in an appropriate way (see Proposition 5.1 and Sect. 7). We will write \(\lambda = \lambda _0 + \lambda _1\) where both of these functions are constructed so that they are defined for \(t>\frac{t_0}{2}\). The construction of \(\lambda _0\) is given in Proposition 5.1. In particular \(\lambda _0(t) = \frac{c_0}{\sqrt{\log t}}(1+o(1))\) as \(t\rightarrow \infty \). Note that \(\varphi _\lambda (\cdot ,t_0)\) is not zero.

We define the approximate solution

$$\begin{aligned} u_1 := u_0 + \varphi _\lambda \end{aligned}$$
(3.13)

which depends on the parameter functions \(\alpha (t)\), \(\xi (t)\), \(\lambda (t)\). Correspondingly, we write

$$\begin{aligned} v_1 \,=\ (-\Delta _{x} )^{-1} (u_1)\,. \end{aligned}$$

We will establish in the next sections that a suitable choice of these functions makes it possible to find an actual solution of (3.1) as a lower order perturbation of \(u_1\).

4 The First Error of Approximation

We will assume the following conditions on \(\lambda \), \(\alpha \), \(\xi \)

$$\begin{aligned} \left\{ \begin{aligned}&|\lambda (t)| + t \log (t) |{\dot{\lambda }}(t)| \le \frac{C}{\sqrt{\log (t)}} \\&|{\dot{\xi }}(t) | \le \frac{C}{t^{\gamma }} \\&|\alpha (t)-1|\le \frac{C}{t \log t} , \quad |{\dot{\alpha }}(t)|\le \frac{C}{t^2 \, \log t} , \end{aligned} \right. \end{aligned}$$
(4.1)

where \(\frac{3}{2}<\gamma <2\).

We compute

$$\begin{aligned} S(u_1)&= S(u_0+\varphi _\lambda ) = S(u_0) - \partial _t \varphi _\lambda + {\mathcal {L}}_{u_0}[\varphi _\lambda ] - \nabla \cdot ( \varphi _\lambda \nabla \psi _\lambda ). \end{aligned}$$

where

$$\begin{aligned} {\mathcal {L}}_{u_0}[\varphi ]&= \Delta \varphi - \nabla \cdot ( \varphi \nabla v_0) - \nabla \cdot ( u_0 \nabla \psi ), \\ \psi _\lambda&= (-\Delta )^{-1} \varphi _\lambda , \quad v_0 = ( -\Delta )^{-1} u_0 . \end{aligned}$$

Then

$$\begin{aligned} S(u_1)&= - \frac{\dot{\alpha }}{\lambda ^2} U(y) \chi + (\alpha -1) \frac{{\dot{\lambda }}}{\lambda ^3} Z_0 \chi +\frac{\alpha }{\lambda ^3} {\dot{\xi }} \cdot {\nabla } _y U (y) \, \chi + \frac{\alpha }{\lambda ^2 \sqrt{t}} U(y) {\dot{\xi }} \cdot \nabla \chi _0 \nonumber \\&\quad + \frac{(\alpha -1)}{2 t } \frac{1}{\lambda ^2} U \nabla _z\chi _0 \cdot \frac{x-\xi }{\sqrt{t}} + \frac{2(\alpha -1)}{\lambda ^3 t^{1/2} } \nabla _z \chi _0 \cdot \nabla _y U \nonumber \\&\quad + \frac{(\alpha -1) }{t} \Delta \chi _0 \frac{1}{\lambda ^2} U -\frac{\alpha ^2-1}{\lambda ^3 \sqrt{t}} U \nabla _z \chi _0 \cdot \nabla _y \Gamma _0 - \frac{\alpha }{\lambda ^3 \sqrt{t}} U \nabla _z \chi _0 \cdot \nabla _y {\mathcal {R}} \nonumber \\&\quad - \frac{\alpha (\alpha -1) \chi }{\lambda ^4} \nabla _y \cdot ( U \nabla _y \Gamma _0) + \frac{\alpha ^2 \chi ( 1-\chi )}{\lambda ^4} U^2 - \frac{\alpha \chi }{\lambda ^4} \nabla _y U \cdot \nabla _y {\mathcal {R}} \nonumber \\&\quad + \nabla \varphi _\lambda \cdot {\dot{\xi }} - \frac{4}{r} \partial _r \varphi _\lambda - \nabla \cdot ( \varphi _\lambda \nabla v_0) - \nabla \cdot ( u_0 \nabla \psi _\lambda ) - \nabla \cdot ( \varphi _\lambda \nabla \psi _\lambda ), \end{aligned}$$
(4.2)

where \({\mathcal {R}}\) is defined in the decomposition (3.5).

Lemma 4.1

Let \(\varphi _\lambda \) be defined by (3.12)-(3.9) with \(\lambda \) satisfying (4.1). Then

$$\begin{aligned} |\varphi _\lambda (x,t)| + ( |x-\xi |+\lambda ) |\nabla \varphi _\lambda (x,t)| \le C \frac{1}{t \log t } {\left\{ \begin{array}{ll} \frac{1}{\lambda ^2 + |x-\xi |^2} &{} |x-\xi |\le \sqrt{t} \\ \frac{1}{t} e^{-\frac{|x-\xi |^2}{4t}} &{} |x-\xi |\ge \sqrt{t}. \end{array}\right. } \end{aligned}$$
(4.3)

We also have

$$\begin{aligned} |\nabla \varphi _\lambda (x,t)| \le \frac{C}{t \log t } \frac{|x-\xi |}{(\lambda + |x-\xi |)^4} , \quad |x-\xi | \le \sqrt{t}. \end{aligned}$$
(4.4)

Proof

In terms of the function \({\tilde{\varphi }}_\lambda \) defined in (3.9), with \(r = |x-\xi |\) we claim that

$$\begin{aligned} |{\tilde{\varphi }}_\lambda (r,t)| \le C \frac{1}{t \log t } {\left\{ \begin{array}{ll} \frac{1}{\lambda ^2+r^2} &{} r\le \sqrt{t}, \\ \frac{1}{t} e^{-\frac{r^2}{4t}} &{} r\ge \sqrt{t}. \end{array}\right. } \end{aligned}$$

For the proof of this we use barriers. Consider

$$\begin{aligned} \psi _1(r,t) = \frac{1}{t \log t} \frac{1}{\lambda ^2+r^2} \end{aligned}$$

and note that

$$\begin{aligned} \partial _t \psi _1 - \Bigl (\partial _{rr} +\frac{5}{r}\partial _r\Bigr ) \psi _1 \ge c \frac{\lambda ^{-4} }{t \log t (1+r/\lambda )^4} ,\quad r \le 2 \delta \sqrt{t} \end{aligned}$$

for some \(c>0\), \(\delta >0\).

Let \(\chi _{\delta \sqrt{t}}(r,t ) = {\tilde{\chi }}_0(\frac{r}{\delta \sqrt{t}}) \) where \({\tilde{\chi }}_0\in C^\infty ({\mathbb {R}})\) is such that \({\tilde{\chi }}_0(s) = 1\) for \(s \le 1\) and \({\tilde{\chi }}_0(s) = 0\) for \(s \ge 2\). Consider

$$\begin{aligned} \psi (r,t) = \psi _1 (r,t) \chi _{\delta \sqrt{t}}(r,t) + \frac{C_1}{t^2 \log t} e^{-\frac{r^2}{4t}} . \end{aligned}$$

The function \({{\tilde{E}}}\) (3.11) can be estimated by

$$\begin{aligned} |{{\tilde{E}}}(\zeta ,t) |\le \frac{1}{\lambda ^2 t^3} h_1\Bigl (\frac{\zeta }{\sqrt{t}}\Bigr ) \end{aligned}$$

where \(h_1(z)\) is a smooth function with compact support. Then E (3.10) has the estimate

$$\begin{aligned} |{{\tilde{E}}}(\zeta ,t) | \le C \frac{|\lambda {\dot{\lambda }}|}{(r^2+\lambda ^2)^2} + \frac{1}{\lambda ^2 t^3} h_2\Bigl (\frac{\zeta }{\sqrt{t}}\Bigr ) \end{aligned}$$

where \(h_2(z)\) is a smooth function with compact support.

Then for \(C_1\) sufficiently large

$$\begin{aligned} \partial _t \psi - \Bigl (\partial _{rr} +\frac{5}{r}\partial _r\Bigr ) \psi \ge c |E(r,t)|, \end{aligned}$$

where \(c>0\).

By the comparison principle,

$$\begin{aligned} |{\tilde{\varphi }}_\lambda (r,t) | \le C \psi (r,t), \end{aligned}$$

for some uniform constant C. After a suitable scaling, from standard parabolic estimates we also get

$$\begin{aligned} (\lambda + r) |\nabla _x {\tilde{\varphi }}_\lambda (r,t)| \le C \psi (r,t). \end{aligned}$$

With these two inequalities we obtain (4.3).

To prove (4.4) we change variables \(y = \frac{x-\xi }{\lambda }\) in the equation (3.9) and define

$$\begin{aligned} {\tilde{\varphi }}_\lambda (r, t) = \frac{1}{\lambda ^2} {\hat{\varphi }}_\lambda \Bigl ( \frac{r}{\lambda },t \Bigr ). \end{aligned}$$

We get the equation, after interpreting \(\rho = |y|\), \(y\in {\mathbb {R}}^6\)

$$\begin{aligned} \lambda ^2 \partial _t {\hat{\varphi }} = \Delta _{{\mathbb {R}}^6} {\hat{\varphi }} + \lambda {\dot{\lambda }} ( 2 {\hat{\varphi }}_\lambda + y \cdot \nabla _y {\hat{\varphi }}_\lambda ) + \lambda ^4 E (\lambda y,t), \end{aligned}$$

where E is defined in (3.10). Differentiating with respect to y and using the bound we already have for \(\nabla _y {\hat{\varphi }}_\lambda \) from (4.4), and using standard parabolic estimates, we get

$$\begin{aligned} |D^2_y {\hat{\varphi }}_\lambda (y,t)| \le \frac{C}{t \log t } \frac{1}{(1+|y|)^4} , \quad |y|\le \sqrt{t \log t}. \end{aligned}$$

Using that \(\nabla {\hat{\varphi }}_\lambda (0,t)=0\) we deduce that

$$\begin{aligned} |\nabla _y {\hat{\varphi }}_\lambda (y,t)| \le \frac{C}{t \log t } \frac{|y|}{(1+|y|)^4} , \quad |y|\le \sqrt{t \log t}, \end{aligned}$$

which readily gives (4.4). \(\square \)

Lemma 4.2

Assuming (4.1) we have

$$\begin{aligned} \lambda ^4 |S(u_1)| \chi (x,t) \le C \frac{1}{t \,\log t } \frac{\log (2+|y|)}{1+|y|^6} , \quad y = \frac{x-\xi }{\lambda }, \end{aligned}$$
(4.5)

and

$$\begin{aligned} |S(u_1)| (1-\chi ) \le C \frac{1}{t^4 \log t} e^{-c\frac{|x|^2}{t}}, \end{aligned}$$
(4.6)

for some \(c\in (0,\frac{1}{4})\).

Proof

Let us analyze the terms involving \(\varphi _\lambda \). We estimate, using Lemma 4.1,

$$\begin{aligned} \left| \lambda ^2 U(y) \varphi _\lambda (\xi + \lambda y) \right| \le C \frac{1}{t \log t }\frac{1}{(1+|y|)^6} , \quad |y|\le \sqrt{t \log t}. \end{aligned}$$

Similarly, by (3.5)

$$\begin{aligned} -\frac{4}{r} \partial _r {\tilde{\varphi }}_\lambda - \nabla {\tilde{\varphi }}_\lambda \cdot \nabla v_0&= -\frac{4}{r} \partial _r {\tilde{\varphi }}_\lambda - \nabla {\tilde{\varphi }}_\lambda \cdot \nabla \Gamma _0 - (\alpha -1) \nabla {\tilde{\varphi }}_\lambda \cdot \nabla \Gamma _0 - \nabla {\tilde{\varphi }}_\lambda \cdot \nabla {\mathcal {R}} \nonumber \\&= 4\Bigr ( \frac{r}{r^2+\lambda ^2} - \frac{1}{r}\Bigl ) \partial _r {\tilde{\varphi }}_\lambda - (\alpha -1) \nabla {\tilde{\varphi }}_\lambda \cdot \nabla \Gamma _0 - \nabla {\tilde{\varphi }}_\lambda \cdot \nabla {\mathcal {R}}. \end{aligned}$$
(4.7)

By (4.4)

$$\begin{aligned} \Bigl | \lambda ^4 4\Bigr ( \frac{r}{r^2+\lambda ^2} - \frac{1}{r}\Bigl ) \partial _r \varphi _\lambda \Bigr |&\le \frac{C}{t \log t} \frac{1}{(1+|y|)^6} , \quad |y|\le \sqrt{t \log t}. \end{aligned}$$

The other terms in (4.7) are estimated similarly, using the hypotheses on \(\alpha \) and the estimate on \({\mathcal {R}}\) (3.6), and we get

$$\begin{aligned} \left| -\frac{4}{r} \partial _r {\tilde{\varphi }}_\lambda - \nabla {\tilde{\varphi }}_\lambda \cdot \nabla v_0 \right| \le \frac{C}{t \log t} \frac{1}{(1+|y|)^6} , \quad |y|\le \sqrt{t \log t}. \end{aligned}$$

The terms involving \(\psi _\lambda = (-\Delta )^{-1}\varphi _\lambda \) are estimated using the formula

$$\begin{aligned} \partial _r \psi _\lambda (r,t) = \frac{1}{r} \int _0^r \varphi _\lambda (s,t)sds . \end{aligned}$$

In \( \lambda ^4 S(u_1)\) we have also the term \(- \dot{\alpha }\lambda ^2 U(y) \chi \), which thanks to (4.1) can be estimated as

$$\begin{aligned} \Bigl | \lambda ^2 \dot{\alpha } U(y) \chi \Bigr |&\le \frac{C \lambda ^2}{t^2 \log t} \frac{1}{(1+|y|)^4} \chi (y,t) \le \frac{C }{t \log t} \frac{1}{(1+|y|)^6} \chi (y,t). \end{aligned}$$

The remaining terms are estimated similarly, and we obtain (4.5).

The stated inequality (4.6) follows from the Gaussian decay of \(\varphi _\lambda \) in Lemma 4.1. \(\square \)

5 The Inner–Outer Gluing System

Let us consider the initial approximation

$$\begin{aligned} u_1(x,t) = u_0(x,t) + \varphi _\lambda (x,t) \end{aligned}$$

built in Sect. 3 for a given choice of the parameter functions \(\lambda (t)\), \(\alpha (t)\), \(\xi (t) \) satisfying (4.1). Here \(u_0\) is the function defined in (3.2) and \(\varphi _\lambda \) that in (3.12). We look for a solution of the Keller–Segel equation (3.1) in the form of a small perturbation of \(u_1\), namely

$$\begin{aligned} u (x,t)= u_1 (x,t)+ \Phi (x,t). \end{aligned}$$
(5.1)

We write the perturbation \(\Phi \) as a sum of an “inner” contribution, better expressed in the scale of \(u_0\), and a remote effect that takes into consideration the “outer” regime. Precisely, we write

$$\begin{aligned} \Phi (x,t) = \frac{1}{\lambda ^2} \phi ^i(y,t)\chi (x,t) + \varphi ^o(x,t), \quad y= \frac{x-\xi }{\lambda } , \end{aligned}$$
(5.2)

where \(\chi \) is the smooth cut-off

$$\begin{aligned} \chi (x,t)= \chi _0 \Bigl ( \frac{x-\xi }{\sqrt{t}} \Bigr ) \end{aligned}$$
(5.3)

with \(\chi _0\) a smooth radial cut-off function such that \( \chi _0(z) = 1 \) if \(|z|\leqq 1\), \( \chi _0(z) = 1 \) if \( |z|\geqq 2 \). (The same as defined in (2.4).)

Recall S(u) given by

$$\begin{aligned} S(u) = - \partial _t u + \Delta u - \nabla \cdot (u \nabla v), \quad v = ( -\Delta )^{-1} u, \end{aligned}$$

where the operators act on the original variable x unless otherwise indicated. In the computations that follow we will express the equation

$$\begin{aligned} S(u_1+\Phi ) = 0 \end{aligned}$$

for \(\Phi \) given by (5.2), as a parabolic system in its inner and outer contributions \(\phi ^i\) and \(\varphi ^o\). The coupling in that system will be small if \(\phi ^i(y,t)\) decays sufficiently fast in space and time. That can only be achieved for suitable choices of the parameters \(\alpha , \lambda , \xi \) that yield certain solvability conditions satisfied. The set of all these relations is what we call the inner-outer gluing system. Next we formulate this system. It will be necessary to successively refine its original expression by further decomposing \(\phi ^i\) into two contributions with separate space decay, finally arriving at the equations (5.48), (5.49), (5.50) and (5.52) which are the ones we will actually solve.

Let us observe that

$$\begin{aligned} S(u_1+\Phi )&= S(u_1) - \partial _t \Bigl ( \frac{1}{\lambda ^2} \phi ^i \chi \Bigr ) - \partial _t \varphi ^o + {\mathcal {L}}_{u_1}\Bigr [\frac{1}{\lambda ^2} \phi ^i \chi \Bigl ] + {\mathcal {L}}_{u_1}[\varphi ^o] \\&\quad - \nabla \cdot ( \Phi \nabla (-\Delta )^{-1} \Phi ) , \end{aligned}$$

where

$$\begin{aligned} {\mathcal {L}}_{u_1}[\varphi ]&= \Delta \varphi - \nabla \cdot ( \varphi \nabla v_1) - \nabla \cdot ( u_1 \nabla (-\Delta )^{-1} \varphi ) , \qquad v_1 = ( -\Delta )^{-1} u_1 . \end{aligned}$$

We use the notation

$$\begin{aligned} \psi = \frac{1}{\lambda ^2}(-\Delta )^{-1} \phi ^i, \quad {\hat{\psi }} = \frac{1}{\lambda ^2} (-\Delta )^{-1} ( \phi ^i \chi ), \end{aligned}$$

in the expressions that follow. We expand

$$\begin{aligned} {\mathcal {L}}_{u_1}\left[ \frac{1}{\lambda ^2} \phi ^i \chi \right]&= \chi \frac{1}{\lambda ^2} \Delta \phi ^i + \frac{2}{\lambda ^2}\nabla \chi \cdot \nabla \phi ^i + \frac{1}{\lambda ^2} \phi ^i \Delta \chi - \nabla \cdot \left( \frac{1}{\lambda ^2} \phi ^i \chi \nabla v_1\right) \\&\quad - \nabla \cdot ( u_1 \nabla {\hat{\psi }} ) . \end{aligned}$$

We have

$$\begin{aligned} \nabla \cdot ( u_1 \nabla {\hat{\psi }} )&= \nabla \cdot \left( \frac{\alpha }{\lambda ^2} U \nabla \psi \right) \chi + \nabla \cdot \left( \frac{\alpha }{\lambda ^2} U \nabla ( {\hat{\psi }} -\psi ) \right) \chi + \frac{\alpha }{\lambda ^2} U \nabla \chi \cdot \nabla {\hat{\psi }} \\&\quad + \nabla \cdot ( \varphi _\lambda \nabla \psi ) + \nabla \cdot ( \varphi _\lambda {\nabla } ({\hat{\psi }} -\psi ) ) \end{aligned}$$

and

$$\begin{aligned} \nabla \cdot \left( \frac{1}{\lambda ^2} \phi ^i \chi \nabla v_1\right)&= \nabla \cdot \left( \frac{1}{\lambda ^2} \phi ^i \nabla v_1\right) \chi + \frac{1}{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla v_1 . \end{aligned}$$

Recall the notation

$$\begin{aligned} v_1 = v_0 + \psi _\lambda , \quad v_0 = \frac{\alpha }{\lambda ^2} (-\Delta )^{-1} ( U \chi ) , \quad \psi _\lambda = (-\Delta )^{-1} \varphi _\lambda , \end{aligned}$$

and also (3.5)

$$\begin{aligned} v_0 = \alpha \Gamma _0 + {\mathcal {R}}, \quad {\mathcal {R}} = \frac{\alpha }{\lambda ^2} (-\Delta )^{-1} \bigl ( U (\chi -1) \bigl ) . \end{aligned}$$

Then

$$\begin{aligned} \nabla \cdot \left( \frac{1}{\lambda ^2} \phi ^i \chi \nabla v_1\right)&= \nabla \cdot \left( \frac{1}{\lambda ^2} \phi ^i \nabla v_0\right) \chi + \nabla \cdot \left( \frac{1}{\lambda ^2} \phi ^i \nabla \psi _\lambda \right) \chi + \frac{1}{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla v_0 \\&\quad + \frac{1}{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla \psi _\lambda \\&= \frac{\alpha }{\lambda ^2} \nabla \cdot ( \phi ^i \nabla \Gamma _0) \chi + \nabla \cdot \left( \frac{1}{\lambda ^2} \phi ^i \nabla {\mathcal {R}}\right) \chi + \nabla \cdot \left( \frac{1}{\lambda ^2} \phi ^i \nabla \psi _\lambda \right) \chi \\&\quad + \frac{\alpha }{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla \Gamma _0 + \frac{1}{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla {\mathcal {R}} + \frac{1}{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla \psi _\lambda . \end{aligned}$$

Therefore

$$\begin{aligned} {\mathcal {L}}_{u_1}\left[ \frac{1}{\lambda ^2} \phi ^i \chi \right]&= \chi \frac{1}{\lambda ^2} \Delta \phi ^i + \frac{2}{\lambda ^2}\nabla \chi \cdot \nabla \phi ^i + \frac{1}{\lambda ^2} \phi ^i \Delta \chi \\&\quad - \Bigl [ \nabla \cdot \Bigl ( \frac{\alpha }{\lambda ^2} \phi ^i \nabla \Gamma _0 \Bigr ) \chi + \nabla \cdot \Bigl ( \frac{1}{\lambda ^2} \phi ^i \nabla {\mathcal {R}}\Bigr ) \chi + \nabla \cdot \Bigl ( \frac{1}{\lambda ^2} \phi ^i \nabla \psi _\lambda \Bigr ) \chi \\&\quad + \frac{\alpha }{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla \Gamma _0 + \frac{1}{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla {\mathcal {R}} + \frac{1}{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla \psi _\lambda \Bigr ] \\&\quad - \Bigl [ \nabla \cdot \Bigl ( \frac{\alpha }{\lambda ^2} U \nabla \psi \Bigr ) \chi + \nabla \cdot \Bigl ( \frac{\alpha }{\lambda ^2} U \nabla ( {\hat{\psi }} -\psi ) \Bigr ) \chi + \frac{\alpha }{\lambda ^2} U \nabla \chi \cdot \nabla {\hat{\psi }} \\&\quad \quad + \nabla \cdot ( \varphi _\lambda \nabla \psi ) + \nabla \cdot ( \varphi _\lambda \nabla ({\hat{\psi }} - \psi ) ) \Bigr ] . \end{aligned}$$

Next we expand

$$\begin{aligned} {\mathcal {L}}_{u_1}[\varphi ^o]&= \Delta \varphi ^o - \nabla \cdot ( \varphi ^o \nabla v_1) - \nabla \cdot (u_1 \nabla \psi ^o) , \quad \psi ^{o} = (-\Delta )^{-1} \varphi ^o. \end{aligned}$$

We have

$$\begin{aligned} \nabla \cdot (u_1 \nabla \psi ^o)&= \nabla \cdot \left( \frac{\alpha }{\lambda ^2} U \chi \nabla \psi ^o \right) + \nabla \cdot ( \varphi _\lambda \nabla \psi ^o ) \\&= \nabla \cdot \left( \frac{\alpha }{\lambda ^2} U \nabla \psi ^o \right) \chi + \frac{\alpha }{\lambda ^2} U \nabla \chi \cdot \nabla \psi ^o + \nabla \cdot ( \varphi _\lambda \nabla \psi ^o ) \chi \\&\quad + \nabla \cdot ( \varphi _\lambda \nabla \psi ^o ) (1-\chi ), \end{aligned}$$

and

$$\begin{aligned} \nabla \cdot ( \varphi ^o \nabla v_1)&= \nabla \cdot ( \varphi ^o \nabla v_0) +\nabla \cdot ( \varphi ^o \nabla \psi _\lambda ) \\&= \alpha \nabla \cdot ( \varphi ^o \nabla \Gamma _0) + \nabla \cdot ( \varphi ^o \nabla {\mathcal {R}}) +\nabla \cdot ( \varphi ^o \nabla \psi _\lambda ) \\&= \nabla \varphi ^o \cdot \nabla \Gamma _0 - \frac{1}{\lambda ^2} U \varphi ^o + (\alpha -1) \nabla \cdot ( \varphi ^o \nabla \Gamma _0) \\&\quad + \nabla \cdot ( \varphi ^o \nabla {\mathcal {R}}) + \nabla \cdot ( \varphi ^o \nabla \psi _\lambda ). \end{aligned}$$

Therefore,

$$\begin{aligned} {\mathcal {L}}_{u_1}[\varphi ^o]&= \Delta \varphi ^o - \Bigl [ \nabla \cdot \Bigl ( \frac{\alpha }{\lambda ^2} U \nabla \psi ^o \Bigr ) \chi + \frac{\alpha }{\lambda ^2} U \nabla \chi \cdot \nabla \psi ^o + \nabla \cdot ( \varphi _\lambda \nabla \psi ^o ) \chi \\&\quad + \nabla \cdot ( \varphi _\lambda \nabla \psi ^o ) (1-\chi ) \Bigr ] \\&\quad -\Bigl [ \nabla \varphi ^o \cdot \nabla \Gamma _0 - \frac{1}{\lambda ^2} U \varphi ^o + (\alpha -1) \nabla \cdot ( \varphi ^o \nabla \Gamma _0) \\&\quad + \nabla \cdot ( \varphi ^o \nabla {\mathcal {R}}) + \nabla \cdot ( \varphi ^o \nabla \psi _\lambda ) \Bigr ]. \end{aligned}$$

Based on the previous formulas we formulate the inner equation

$$\begin{aligned} \lambda ^4 \partial _t\left( \frac{1}{\lambda ^2} \phi ^i \right)&= L [\phi ^i] - (\alpha -1) \nabla _y \cdot ( U \nabla _y \psi ) - (\alpha -1) \nabla _y \cdot ( \phi ^i \nabla \Gamma _0) + \lambda ^4 S(u_1) \\&\quad - \lambda ^2 \nabla _y \cdot ( \varphi _\lambda \nabla _y \psi ^o) - \lambda ^2 \nabla _y \cdot ( \varphi ^o \nabla _y \psi _\lambda ) + \lambda ^2 U \varphi ^o - \alpha \nabla _y \cdot ( U \nabla _y \psi ^o) \\&\quad - \lambda ^2 \nabla _y \cdot ( \varphi _\lambda \nabla _y \psi ) - \nabla _y \cdot ( \phi ^i \nabla _y \psi _\lambda ) - (\alpha -1) \lambda ^2 \nabla \cdot ( \varphi ^o \nabla \Gamma _0 ) \\&\quad - \alpha \nabla _y \cdot ( U \nabla _y ( {\hat{\psi }} - \psi ) ) -\lambda ^2 \nabla _y \cdot ( \varphi _\lambda \nabla _y ({\hat{\psi }} - \psi )) \\&\quad - \nabla _y \cdot ( (\phi ^i \chi + \lambda ^2\varphi ^o) \nabla _y ( {\hat{\psi }} + \psi ^o)) , \end{aligned}$$

where

$$\begin{aligned} L[\phi ] = \Delta _y \phi - \nabla _y \cdot ( U \nabla _y \psi ) - \nabla _y \cdot ( \phi \nabla \Gamma _0) . \end{aligned}$$
(5.4)

We slightly modify the inner equation into the form

$$\begin{aligned} \lambda ^2 \partial _t \phi ^i&= L [\phi ^i] + B_0 [\phi ^i] + E_1 {\tilde{\chi }} + F (\phi ^i,\varphi ^o,{{\textbf {p}}}){\tilde{\chi }} \end{aligned}$$
(5.5)

where

$$\begin{aligned} {{\textbf {p}}}&= (\lambda ,\alpha ,\xi ), \nonumber \\ E_1(y,t)&= \lambda ^4 S(u_1({{\textbf {p}}}))(x,t), \quad y = \frac{x-\xi }{\lambda }, \nonumber \\ F(\phi ^i,\varphi ^o,{{\textbf {p}}})&= - \lambda ^2 \nabla _y \cdot ( \varphi _\lambda \nabla _y \psi ^o) - \lambda ^2 \nabla _y \cdot ( \varphi ^o \nabla _y \psi _\lambda ) + \lambda ^2 U \varphi ^o \nonumber \\&\quad - (\alpha -1) \lambda ^2 \nabla _y \cdot ( \varphi ^o \nabla _y \Gamma _0 ) - \alpha \nabla _y \cdot ( U \nabla _y \psi ^o) \nonumber \\&\quad + \lambda {\dot{\xi }} \cdot \nabla _y \phi ^i - \lambda ^2 \nabla _y \cdot ( \varphi _\lambda \nabla _y \psi ) - \nabla _y \cdot ( \phi ^i \nabla _y \psi _\lambda ) \nonumber \\&\quad - (\alpha -1) \nabla _y \cdot ( U \nabla _y \psi ) - (\alpha -1) \nabla _y \cdot ( \phi ^i \nabla _y \Gamma _0) \nonumber \\&\quad - \alpha \nabla _y \cdot ( U \nabla _y ( {\hat{\psi }} - \psi ) ) - \lambda ^2 \nabla _y \cdot ( \varphi _\lambda \nabla _y ( {\hat{\psi }} - \psi ) ) \nonumber \\&\quad - \nabla _y \cdot ( (\phi ^i \chi + \lambda ^2 \varphi ^o) \nabla _y ( {\hat{\psi }} + \psi ^o)) , \qquad {\hat{\psi }} = (-\Delta _y)^{-1} ( \phi ^i \chi ) , \end{aligned}$$
(5.6)
$$\begin{aligned} B_0[\phi ^i]&= \lambda {\dot{\lambda }} ( 2\phi ^i + y \cdot \nabla _{y} \phi ^i ) , \end{aligned}$$
(5.7)

and

$$\begin{aligned} {\tilde{\chi }}(y,t) = \chi _0\Bigl ( \frac{\lambda y}{2\sqrt{t}}\Bigr ) , \end{aligned}$$
(5.8)

with \(\chi _0\) as in (2.5). Similarly we formulate the outer equation as

$$\begin{aligned} \partial _t \varphi ^o&= \Delta \varphi ^o - \nabla \Gamma _0 \cdot \nabla \varphi ^o + G(\phi ^i,\varphi ^o,{\textbf{p}}) \end{aligned}$$
(5.9)

where

$$\begin{aligned} G(\phi ^i,\varphi ^o,{\textbf{p}})&= S(u_1,{{\textbf {p}}} ) (1-\chi ) + \frac{2}{\lambda ^2} \nabla \chi \cdot \nabla \phi ^i + \frac{1}{\lambda ^2} \phi ^i \Delta \chi - \frac{1}{\lambda ^2} \phi ^i \partial _t \chi - \frac{\alpha }{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla \Gamma _0 \nonumber \\&\quad + \frac{1}{\lambda ^2} U \varphi ^o (1-\chi ) - \alpha \lambda ^2 U \nabla \chi \cdot \nabla \psi ^o - \nabla \cdot ( \varphi _\lambda \nabla \psi ^o) (1-\chi ) \nonumber \\&\quad -(\alpha -1) \nabla \cdot (\varphi ^o\nabla \Gamma _0) (1-\chi ) - \nabla \cdot (\varphi ^o \nabla {\mathcal {R}}) - \nabla \cdot (\varphi ^o\nabla \psi _\lambda ) (1-\chi ) \nonumber \\&\quad - \frac{1}{\lambda ^2} \nabla \cdot ( \phi ^i \nabla {\mathcal {R}} ) \chi - \frac{1}{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla {\mathcal {R}} - \frac{1}{\lambda ^2} \phi ^i \nabla \chi \cdot \nabla \psi _\lambda \nonumber \\&\quad - \frac{\alpha }{\lambda ^2} U \nabla \chi \cdot \nabla {\hat{\psi }} - \nabla \cdot ( \varphi _\lambda \nabla ( {\hat{\psi }} - \psi ) )(1-\chi ) \nonumber \\&\quad -\nabla ( \varphi _\lambda \nabla \psi ) (1-\chi ) - \nabla \cdot ( ( \frac{1}{\lambda ^2}\phi ^i\chi + \varphi ^o) \nabla ( {\hat{\psi }} + \psi ^o)) (1-\chi ) . \end{aligned}$$
(5.10)

If \(\phi ^i\), \(\varphi ^o\) is a solution to system (5.5), (5.9), then u given by (5.1), (5.2) satisfies the Keller–Segel system (3.1).

5.1 Choice of \(\lambda _0\) and \(\alpha _0\)

We explain the choice of \(\lambda _0\) in the context of the elliptic equation

$$\begin{aligned} L[\phi ]=h \quad \text {in }{\mathbb {R}}^2, \end{aligned}$$
(5.11)

where h is radial.

Lemma 5.1

Let h(y) be a radial function such that

$$\begin{aligned} \Vert (1+|y|)^{\gamma } h(y) \Vert _{L^\infty ({\mathbb {R}}^2)} < \infty , \end{aligned}$$

for some \(\gamma >4\) and satisfying

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y) dy = 0 \end{aligned}$$
(5.12)
$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y) |y|^2 dy = 0 . \end{aligned}$$
(5.13)

Then there exists a radial solution \(\phi (y) \) of equation (5.11) such that

$$\begin{aligned} |\phi (y) |&\le C \Vert (1+|y|)^{\gamma } h(y) \Vert _{L^\infty ({\mathbb {R}}^2)} \frac{1}{(1+|y|)^{\gamma -2}} , \quad \text {if } \gamma \not =6 \end{aligned}$$
(5.14)
$$\begin{aligned} |\phi (y) |&\le C \Vert (1+|y|)^{\gamma } h(y) \Vert _{L^\infty ({\mathbb {R}}^2)} \frac{\log (1+|y|)}{(1+|y|)^{4}} , \quad \text {if }\gamma =6 , \end{aligned}$$
(5.15)

and

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi (y) dy = 0. \end{aligned}$$
(5.16)

Proof

Defining \(g = \frac{\phi }{U}- (-\Delta )^{-1} \phi \) we obtain the equation

$$\begin{aligned} \nabla \cdot ( U \nabla g ) = h . \end{aligned}$$
(5.17)

Assuming \(\gamma >6\) we choose the radial function g defined by

$$\begin{aligned} g(\rho ) = - \int _\rho ^\infty \frac{1}{r U(r)} \int _0^r h(s) s ds dr, \quad \rho = |y|, \end{aligned}$$

and using (5.12) we get

$$\begin{aligned} |g(\rho ) | \le C \Vert (1+|y|)^{\gamma } h \Vert _{L^\infty ({\mathbb {R}}^2)} \frac{1}{(1+|y|)^{\gamma -6} }. \end{aligned}$$

Now we solve Liouville’s equation

$$\begin{aligned} -\Delta \psi - U \psi = U g \quad \text {in }{\mathbb {R}}^2, \quad \psi (\rho )\rightarrow 0 \quad \text {as }\rho \rightarrow \infty . \end{aligned}$$
(5.18)

Multiplying (5.17) by \(|y|^2\) and using (5.13) we see that

$$\begin{aligned} \int _{{\mathbb {R}}^2} g Z_0 dy = \frac{1}{2} \int _{{\mathbb {R}}^2} h(y) |y|^2 dy= 0, \end{aligned}$$

with \(Z_0\) defined in (3.4). Then by the variations of parameter formula we find that (5.18) has a unique solution \(\psi \), which satisfies

$$\begin{aligned} |\psi (y)| + (1+|y|)|\nabla \psi (y)| \le \Vert (1+|y|)^{\gamma } h \Vert _{L^\infty ({\mathbb {R}}^2)} \frac{1}{(1+|y|)^{\gamma -4} }. \end{aligned}$$
(5.19)

Then we see that \(\phi \) defined by \(\phi = U g + U \psi \) satisfies (5.11), (5.14) and (5.16) because \(\phi = - \Delta \psi \) and \(\psi \) has the decay (5.19).

If \(4<\gamma \le 6\) we do almost the same, except that we define

$$\begin{aligned} g(\rho ) = \int _0^\rho \frac{1}{r U(r)} \int _0^r h(s) s ds dr. \end{aligned}$$

\(\square \)

Remark 5.1

We observe that \(L[Z_0]=0\). This can also be seen in the context of the Lemma 5.1, where \(\phi =Z_0\) which corresponds to g being constant. Indeed, suppose \(g \equiv 1\). Then from (5.18) \(\psi = -1 - \frac{1}{2}z_0\), where \(z_0\) is defined in (9.2). This gives \(\phi = U g + U \psi = -\frac{1}{U}z_0 = -\frac{1}{2}Z_0\). This shows that \(L[Z_0]=0\).

If h doesn’t satisfy the zero second moment condition (5.13), then a solution still exists but with worse decay and non-zero mass. More precisely, if h is radial, \(\Vert (1+|y|)^{\gamma } h(y) \Vert _{L^\infty ({\mathbb {R}}^2)} < \infty \) for some \(\gamma > 6\), and satisfies only (5.12), then one can construct a solution \(\phi \) to (5.11), but any such solution has the estimate

$$\begin{aligned} |\phi (y)|\le C \Vert (1+|y|)^{\gamma } h(y) \Vert _{L^\infty ({\mathbb {R}}^2)} \frac{\log (1+|y|)}{(1+|y|)^{4}}, \end{aligned}$$

so worse decay than the one in (5.14). Moreover, the mass of \(\phi \) becomes

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi = -\int _{{\mathbb {R}}^2} \Delta \psi = - \int _{{\mathbb {R}}^2} g Z_0 = - \frac{1}{2} \int _{{\mathbb {R}}^2} h(y) |y|^2 dy . \end{aligned}$$

For the inner equation (5.5) it is then natural to impose that the first error \(S(u_1) \chi \) satisfies the second moment condition

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1) \chi |y|^2 dy = 0, \quad \text {for all }t>t_0. \end{aligned}$$

The next lemma gives a way of expressing the second moment of \(u_1\).

Lemma 5.2

Let \(u_1\) be defined in (3.13). Then

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1) |x-\xi |^2dx&= 4 \int _{{\mathbb {R}}^2} \varphi _\lambda dx - \alpha \int _{{\mathbb {R}}^2} {{\tilde{E}}}(x-\xi ,t;\lambda ) |x-\xi |^2 dx \nonumber \\&\quad + \int _{{\mathbb {R}}^2} \nabla \varphi _\lambda \, \textrm{d}x \cdot {\dot{\xi }} - \frac{{\dot{\alpha }}}{\lambda ^2} \int _{{\mathbb {R}}^2} U \chi |x-\xi |^2 dx \nonumber \\&\quad - (1-\alpha ) \int _{{\mathbb {R}}^2} E (x-\xi ,t;\lambda ) |x-\xi |^2dx \nonumber \\&\quad + 4 \Bigl ( \int _{{\mathbb {R}}^2} u_0 + \int _{{\mathbb {R}}^2} \varphi _\lambda \Bigr ) \Bigl ( 1 - \frac{1}{8\pi } \int _{{\mathbb {R}}^2} u_0 - \frac{1}{8\pi } \int _{{\mathbb {R}}^2} \varphi _\lambda \Bigr ). \end{aligned}$$
(5.20)

where E, \({{\tilde{E}}}\) are defined in (3.10), (3.11).

Proof of Lemma 5.2

Using (2.11) we see that

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1) |x-\xi |^2dx&= -\int _{{\mathbb {R}}^2} \partial _t u_0 |x-\xi |^2dx -\int _{{\mathbb {R}}^2} \partial _t \varphi _\lambda |x-\xi |^2dx \\&\quad + 4 \Bigl ( \int _{{\mathbb {R}}^2} u_0 + \int _{{\mathbb {R}}^2} \varphi _\lambda \Bigr ) \Bigl ( 1 - \frac{1}{8\pi } \int _{{\mathbb {R}}^2} u_0 - \frac{1}{8\pi } \int _{{\mathbb {R}}^2} \varphi _\lambda \Bigr ) . \end{aligned}$$

But recall that \(\varphi _\lambda (x,t) = {\tilde{\varphi }}_\lambda (x-\xi (t),t)\) where \({\tilde{\varphi }}_\lambda \) satisfies (3.9). Multiplying that equation by \(|\zeta |^2\) and integrating on \({\mathbb {R}}^2\) results in

$$\begin{aligned} \int _{{\mathbb {R}}^2} \partial _t {\tilde{\varphi }}_\lambda |\zeta |^2 \,d\zeta = - 4 \int _{{\mathbb {R}}^2} {\tilde{\varphi }}_\lambda \,d\zeta + \int _{{\mathbb {R}}^2} E(\zeta ,t)|\zeta |^2 \,d\zeta . \end{aligned}$$

Therefore

$$\begin{aligned} \int _{{\mathbb {R}}^2} \partial _t\varphi _\lambda |x-\xi |^2\,dx = -4\int _{{\mathbb {R}}^2} \varphi _\lambda \,dx - \frac{(x-\xi )\cdot {\dot{\xi }}}{|x-\xi |} \int _{{\mathbb {R}}^2} \partial _r {\tilde{\varphi }}_\lambda + \int _{{\mathbb {R}}^2} E(\zeta ,t)|\zeta |^2 \,d\zeta \end{aligned}$$

and then

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1) |x-\xi |^2dx&= -\int _{{\mathbb {R}}^2} \partial _t u_0 |x-\xi |^2dx + 4 \int _{{\mathbb {R}}^2} \varphi _\lambda dx \nonumber \\&\quad + \int _{{\mathbb {R}}^2} \nabla \varphi _\lambda \, \textrm{d}x \cdot {\dot{\xi }} - \int _{{\mathbb {R}}^2} E(x-\xi ,t)|x-\xi |^2dx \nonumber \\&\quad + 4 \Bigl ( \int _{{\mathbb {R}}^2} u_0 + \int _{{\mathbb {R}}^2} \varphi _\lambda \Bigr ) \Bigl ( 1 - \frac{1}{8\pi } \int _{{\mathbb {R}}^2} u_0 - \frac{1}{8\pi } \int _{{\mathbb {R}}^2} \varphi _\lambda \Bigr ) . \end{aligned}$$
(5.21)

But from the formula for \( \partial _t u_0\) (3.3) and the definitions of E and \({{\tilde{E}}}\) (3.10), (3.11) we get

$$\begin{aligned} -\partial _t u_0(x,t) = - \frac{\dot{\alpha }}{\lambda ^2} U(y) \chi _0(z) + \alpha E (x-\xi ,t) - \alpha {{\tilde{E}}}(x-\xi ,t). \end{aligned}$$

Hence

$$\begin{aligned}&\int _{{\mathbb {R}}^2} ( \partial _t u_0 +E(x-\xi ,t))|x-\xi |^2dx \\&\quad = \int _{{\mathbb {R}}^2} ( \partial _t u_0 +\alpha E(x-\xi ))|x-\xi |^2dx + (1-\alpha ) \int _{{\mathbb {R}}^2} E(x-\xi ,t) |x-\xi |^2dx \\&\quad = \frac{{\dot{\alpha }}}{\lambda ^2} \int _{{\mathbb {R}}^2} U \chi |x-\xi |^2 dx + \alpha \int _{{\mathbb {R}}^2} {{\tilde{E}}}(x-\xi ,t) |x-\xi |^2 dx\\&\qquad +(1-\alpha ) \int _{{\mathbb {R}}^2} E(x-\xi ,t) |x-\xi |^2dx . \end{aligned}$$

Replacing this in (5.21) we obtain (5.20). \(\square \)

In the definition (3.13) of \(u_1\) we will stress the dependence on the parameters by writing \({\textbf{p}} = (\lambda ,\alpha ,\xi )\) and \(u_1 = u_1({\textbf{p}})\). At this point we would like to construct \(\lambda _0\) and \(\alpha _0\) so that setting \({\textbf{p}}_0 = ( \lambda _0, \alpha _0, 0 ) \) we have

$$\begin{aligned}&\int _{{\mathbb {R}}^2} u_1({\textbf{p}}_0) dx = 8 \pi , \end{aligned}$$
(5.22)
$$\begin{aligned}&\int _{{\mathbb {R}}^2} S(u_1({\textbf{p}}_0))|x-\xi |^2 dx = O \Bigl ( \frac{1}{t^{\frac{3}{2}+\sigma }}\Bigr ) , \end{aligned}$$
(5.23)

for some \(\sigma >0\). The reason for allowing in (5.23) an error is that it is difficult to solve with right hand side equal to 0 and a remainder of size \(O(t^{-\frac{3}{2}-\sigma })\) with \(\sigma >0\) is sufficiently small to proceed with the rest of the construction.

Assuming that (5.22) holds, we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1) |x-\xi |^2dx&= 4 \int _{{\mathbb {R}}^2} \varphi _\lambda dx - \alpha \int _{{\mathbb {R}}^2} {{\tilde{E}}}(x-\xi ,t;\lambda ) |x-\xi |^2 dx \\&\quad + \int _{{\mathbb {R}}^2} \nabla \varphi _\lambda \, \textrm{d}x \cdot {\dot{\xi }} - \frac{{\dot{\alpha }}}{\lambda ^2} \int _{{\mathbb {R}}^2} U \chi |x-\xi |^2 dx \\&\quad - (1-\alpha ) \int _{{\mathbb {R}}^2} E (x-\xi ,t;\lambda ) |x-\xi |^2dx . \end{aligned}$$

It turns out that the main terms in the expression for \(\int _{{\mathbb {R}}^2} S( u_1 ) |x-\xi |^2 dx\) are the first two. So the equation

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1(\mathbf {p_0})) |x-\xi |^2 dx = 0 \end{aligned}$$

is at main order given by

$$\begin{aligned} 4 \int _{{\mathbb {R}}^2} \varphi _\lambda dx - \int _{{\mathbb {R}}^2} {{\tilde{E}}} |x-\xi |^2dx&= 0 . \end{aligned}$$

It will be shown later that

$$\begin{aligned} \int _{{\mathbb {R}}^2} {{\tilde{E}}} |x-\xi |^2 dx = - 64 \pi \Upsilon \frac{\lambda ^2}{t} + O \Big ( \frac{ \lambda ^4 }{t^2} \Big ) , \end{aligned}$$
(5.24)

see Lemma 7.5, where \(\Upsilon \) is given in (2.7), so that the equation we want to solve becomes at main order,

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _\lambda dx + 16 \pi \Upsilon \frac{\lambda ^2}{t} =0. \end{aligned}$$

In §7 we will show that

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _\lambda dx = -4\pi \int _{t/2}^{t-\lambda ^2} \frac{\lambda {\dot{\lambda }}}{t-s}ds -2\pi \frac{\lambda ^2}{t} - 16 \pi \Upsilon \frac{\lambda ^2}{t} + O \Big ( \frac{ \lambda ^4 \log \log t}{t} \Big ) \end{aligned}$$
(5.25)

see Corollary 7.1. Using (5.25) we see that

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _\lambda dx + 16 \pi \Upsilon \frac{\lambda ^2}{t} = -4 \pi \Bigl [ \int _{t/2}^{t-\lambda ^2} \frac{\lambda {\dot{\lambda }}}{t-s}ds + \frac{\lambda ^2}{2t}\Bigr ] + O \Big ( \frac{ \lambda ^4 \log \log t}{t} \Big ) \end{aligned}$$
(5.26)

so that the equation for \(\lambda \) is at main order

$$\begin{aligned} \int _{t/2}^{t-\lambda ^2} \frac{\lambda {\dot{\lambda }}}{t-s}ds + \frac{\lambda ^2}{2t} = 0. \end{aligned}$$

One can check that \(\lambda ^*(t) = \frac{c_0}{\sqrt{\log t}}\), where \(c_0>0\) is an arbitrary constant, is an approximate solution. Indeed

$$\begin{aligned} \int _{t/2}^{t-(\lambda ^*)^2} \frac{\lambda ^*(s) {\dot{\lambda }}^*(s)}{t-s}ds + \frac{(\lambda ^*)^2}{2t}&\approx \lambda ^*(t) {\dot{\lambda }}^*(t)\int _{t/2}^{t-(\lambda ^*)^2} \frac{ds}{t-s} + \frac{\lambda ^*(t)^2}{2t} \\&\approx \lambda ^*(t) {\dot{\lambda }}^*(t) \log t + \frac{\lambda ^*(t)^2}{2t} \\&= \frac{1}{2} \frac{d}{dt} \bigl [ \lambda ^*(t)^2 \log t \Bigr ]=0. \end{aligned}$$

The error left out in the approximation (5.26) is too big. We give next a result that shows that for an appropriate modification of \(\lambda ^*\) we can achieve a smaller error. Let us write \({{\tilde{E}}}(\lambda )\) the expression defined in (3.11) with the explicit dependence on \(\lambda \).

Proposition 5.1

Let \(c_0>0\) be fixed. For \(t_0>0\) sufficiently large there exists \(\lambda _0:[\frac{t_0}{2},\infty ) \rightarrow (0,\infty ) \) such that

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _{\lambda _0} dx - \frac{1}{4}\int _{{\mathbb {R}}^2} {{\tilde{E}}}(\lambda _0) |x-\xi |^2dx&= O\Bigl ( \frac{1}{t^{\frac{3}{2}+\sigma }}\Bigr ), \quad t >t_0 , \end{aligned}$$
(5.27)

for some \(\sigma >0\). Moreover, for arbitrarily \(\varepsilon >0\) small, \(\lambda _0\) has the expansion

$$\begin{aligned} \lambda _0(t)&= \frac{c_0}{\sqrt{\log t}} + O\Bigl ( \frac{1}{(\log t)^{\frac{3}{2}-\varepsilon }}\Bigr ), \\ {\dot{\lambda }}_0(t)&= -\frac{c_0}{2 t (\log t)^{3/2}} + O\Bigl ( \frac{1}{t (\log t)^{\frac{5}{2} -\varepsilon } } \Bigr ), \\ |\ddot{\lambda }_0(t) |&\le \frac{C}{t^2 (\log t)^{3/2}} , \end{aligned}$$

as \( t \rightarrow \infty \).

We will prove this result in §7.1.

Once \(\lambda _0\) is constructed in Proposition 5.1 we choose \(\alpha _0\) so that (5.22) holds, by imposing

$$\begin{aligned} \alpha _0(t) \int _{{\mathbb {R}}^2} U(y) \chi _0\Bigl (\frac{\lambda _0(t) y}{\sqrt{t}}\Bigr )\,dy + \int _{{\mathbb {R}}^2} \varphi _{\lambda _0}(x,t)\,dx=8\pi , \quad t>t_0. \end{aligned}$$
(5.28)

We note that by (2.6), (5.27) and (5.24) we get

$$\begin{aligned} \alpha _0(t)&= 1+O\Bigl ( \frac{1}{t^{\frac{3}{2}+\sigma }}\Bigr ) \end{aligned}$$

as \(t\rightarrow \infty \). A byproduct of the proof of Proposition 5.1 is that

$$\begin{aligned} \left| \frac{d}{dt} \int _{{\mathbb {R}}^2} \varphi _{\lambda _0}dx\right| \le \frac{C}{t^2} , \end{aligned}$$
(5.29)

and from this and (5.28) we get

$$\begin{aligned} |{\dot{\alpha }}_0(t)|\le \frac{C}{t^2} . \end{aligned}$$
(5.30)

As a corollary of Proposition 5.1 we get:

Corollary 5.1

Let \({{\textbf {p}}}_0=(\lambda _0,\alpha _0,0)\) with \(\alpha _0\) defined by (5.22) and \(\lambda _0\) be given by Proposition 5.1. Then

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1({{\textbf {p}}}_0)) |x-\xi |^2 dx = O \Bigl ( \frac{1}{t^{\frac{3}{2}+\sigma }}\Bigr ) , \end{aligned}$$

for some \(\sigma >0\).

Proof

Using Lemma 5.2 we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1) |x-\xi |^2 dx&=4 \int _{{\mathbb {R}}^2} \varphi _{\lambda _0} dx - \int _{{\mathbb {R}}^2} {{\tilde{E}}}(x-\xi ,t;\lambda _0) |x-\xi |^2 dx \\&\quad - \frac{{\dot{\alpha }}_0}{\lambda ^2} \int _{{\mathbb {R}}^2} U \chi |x-\xi |^2 dx - (1-\alpha _0) \int _{{\mathbb {R}}^2} E(x-\xi ,t;\lambda _0) |x-\xi |^2dx \\&= O \Bigl ( \frac{1}{t^{\frac{3}{2}+\sigma }}\Bigr ) , \end{aligned}$$

for some \(\sigma >0\), since \({\dot{\alpha }}_0(t) = O( \frac{1}{t^2 \log t})\) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} E(x-\xi ,t;\lambda _0) |x-\xi |^2dx = O\Bigl ( \frac{\lambda _0^2}{t}\Bigr ) \end{aligned}$$

by (5.24) and a direct estimate for the remaining terms in E (c.f. (3.10)). \(\square \)

5.2 A further improvement of the approximation

We introduce a correction \(\phi _0^i(y) \), \(y=\frac{x-\xi }{\lambda }\) in the inner approximation to eliminate the radial part of \(S(u_1({{\textbf {p}}})) \) (defined in (4.2)), which we define as

$$\begin{aligned} S_0(u_1({{\textbf {p}}}))&= - \frac{\dot{\alpha }}{\lambda ^2} U(y) \chi + (\alpha -1) \frac{{\dot{\lambda }}}{\lambda ^3} Z_0 \chi + \frac{(\alpha -1)}{2 t } \frac{1}{\lambda ^2} U \nabla _z\chi _0 \cdot \frac{x-\xi }{\sqrt{t}} \nonumber \\&\quad + \frac{2(\alpha -1)}{\lambda ^3 t^{1/2} } \nabla _z \chi _0 \cdot \nabla _y U + \frac{(\alpha -1) }{t} \Delta \chi _0 \frac{1}{\lambda ^2} U \nonumber \\&\quad -\frac{\alpha ^2-1}{\lambda ^3 \sqrt{t}} U \nabla _z \chi _0 \cdot \nabla _y \Gamma _0 - \frac{\alpha }{\lambda ^3 \sqrt{t}} U \nabla _z \chi _0 \cdot \nabla _y {\mathcal {R}} \nonumber \\&\quad - \frac{\alpha (\alpha -1) \chi }{\lambda ^4} \nabla _y \cdot ( U \nabla _y \Gamma _0) + \frac{\alpha ^2 \chi ( 1-\chi )}{\lambda ^4} U^2 - \frac{\alpha \chi }{\lambda ^4} \nabla _y U \cdot \nabla _y {\mathcal {R}}. \nonumber \\&\quad - \frac{4}{r} \partial _r \varphi _\lambda - \nabla \cdot ( \varphi _\lambda \nabla v_0) - \nabla \cdot ( u_0 \nabla \psi _\lambda ) - \nabla \cdot ( \varphi _\lambda \nabla \psi _\lambda ). \end{aligned}$$
(5.31)

With this definition

$$\begin{aligned} S(u_1) = S_0(u_1) + \frac{\alpha }{\lambda ^3} {\dot{\xi }} \cdot {\nabla } _y U (y) \, \chi + \frac{\alpha }{\lambda ^2 \sqrt{t}} U(y) {\dot{\xi }} \cdot \nabla \chi _0 , \end{aligned}$$

and the terms not in \(S(u_1)\) correspond to \( \frac{\alpha }{\lambda ^3} {\dot{\xi }} \cdot {\nabla } _y U (y) \, \chi + \frac{\alpha }{\lambda ^2 \sqrt{t}} U(y)\) which are in mode 1.

Then we want \(\phi _0^i\) to be an appropriate solution to the equation

$$\begin{aligned} L[\phi ^i_0] + \lambda ^4 S_0(u_1({{\textbf {p}}}_0))(x,t) = c_0(t) W_2 \quad \text {in }{\mathbb {R}}^2, \quad x = \xi + \lambda y, \end{aligned}$$
(5.32)

where L is the linear operator (5.4), \(t>t_0\) is regarded as a parameter, \(W_2(y)\) is a fixed smooth radial function with compact support, and

$$\begin{aligned} \int _{{\mathbb {R}}^2} W_2(y)dy=0, \quad \int _{{\mathbb {R}}^2} W_2(y)|y|^2dy=1. \end{aligned}$$
(5.33)

By Lemma 5.2 and Proposition 5.1, the choice \({{\textbf {p}}} = {{\textbf {p}}}_0\) is so that (5.22), (5.23) hold. Since the difference between \(S(u_1)\) and \(S_0(u_1)\) contains terms in mode 1 only, we get from Corollary 5.1

$$\begin{aligned} \int _{{\mathbb {R}}^2} \lambda ^4 S_0(u_1({\textbf{p}}_0))|y|^2 dy&= O \Bigl ( \frac{1}{t^{\frac{3}{2}+\sigma }}\Bigr ) . \end{aligned}$$
(5.34)

In (5.32) we select \(c_0(t)\) such that

$$\begin{aligned} \int _{{\mathbb {R}}^2} [ \lambda ^4S_0(u_1({{\textbf {p}}}_0)) + c_0(t) W_2] |y|^2dy = 0, \quad t>t_0 \end{aligned}$$

and thanks to (5.34) we have

$$\begin{aligned} |c_0(t)|\le \frac{C}{t^{\frac{3}{2}+\sigma }} , \quad t>t_0 . \end{aligned}$$
(5.35)

Note that we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} S_0(u_1({{\textbf {p}}}_0)) d x =0, \end{aligned}$$

which follows from the constant mass in time of \(u_1({\textbf{p}}_0)\) in (5.22) and the form of the operator \(S_0\) (5.31).

We let \(\phi ^i_0\) be the solution to (5.32) constructed in Lemma 5.1. By (5.15) and (4.5)

$$\begin{aligned} |\phi ^i_0(y,t)| \le \frac{C}{t} \frac{\log ( 1+|y|)}{1+|y|^4}, \end{aligned}$$
(5.36)

and

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi ^i_0 (y,t)dy=0, \quad t>t_0. \end{aligned}$$

5.3 Reformulation of the system

In the outer problem (5.9) we would like to separate the effect of the initial condition from the coupling \(G(\phi ^i,\varphi ^o,{\textbf{p}})\).

We take the initial condition in (5.9) to be

$$\begin{aligned} \varphi ^o(\cdot ,t_0)= \varphi _0^*, \end{aligned}$$

and let \(\varphi ^*(x,t)\) denote the solution of

$$\begin{aligned} \left\{ \begin{aligned} \partial _t \varphi ^*&= \Delta \varphi ^* - \nabla _x \Gamma _0 \Big (\frac{x-\xi }{\lambda }\Big ) \cdot \nabla \varphi ^* \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \varphi ^* (\cdot ,t_0)&=\varphi ^*_0 \quad \text {in }{\mathbb {R}}^2. \end{aligned} \right. \end{aligned}$$
(5.37)

The initial condition \(\varphi ^*_0(x)\) will be later used to prove the stability claimed in Theorem 1.1. The topology for \(\varphi _0^*\) will be specified later on.

Note that \(\nabla _x \Gamma _0 (\frac{x-\xi }{\lambda })= -4 \frac{x-\xi }{|x-\xi |^2 +\lambda ^2}\) so that \(\varphi ^* \) is a function of the parameters \(\lambda ,\xi \). Therefore we will write \( \varphi ^*(x,t;{{\textbf {p}}})\) when convenient.

We decompose

$$\begin{aligned} \left\{ \begin{aligned} \phi ^i&= \phi ^i_0 + \phi \\ \varphi ^o&= \varphi ^* + \varphi \\ {{\textbf {p}}}&= {{\textbf {p}}}_0 + {{\textbf {p}}}_1 \end{aligned} \right. \end{aligned}$$
(5.38)

where

$$\begin{aligned} {{\textbf {p}}}_0 = (\lambda _0,\alpha _0,0) , \quad {{\textbf {p}}}_1 = (\lambda _1,\alpha _1,\xi _1) , \end{aligned}$$

with \(\lambda _0\) the function constructed in Proposition 5.1 and \(\alpha _0\) chosen so that (5.22) holds.

We substitute the expressions for \(\phi ^i\), \(\varphi ^o\) and \({{\textbf {p}}}\) in (5.38) into the Eqs. (5.5), (5.9), and are led to the following problem for \(\phi \), \(\varphi \)

$$\begin{aligned}&\left\{ \begin{aligned} \lambda ^2 \partial _t \phi&= L [\phi ] + B_0 [\phi ] + E_2 {\tilde{\chi }}_2+ F_2 (\phi ,\varphi ,{{\textbf {p}}}_1,\varphi ^*_0){\tilde{\chi }} \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \phi (\cdot ,t_0)&= \phi _0 \quad \text {in }{\mathbb {R}}^2 \end{aligned} \right. \end{aligned}$$
(5.39)
$$\begin{aligned}&\left\{ \begin{aligned} \partial _t \varphi&= \Delta \varphi - \nabla _x \Gamma _0\left( \frac{x-\xi }{\lambda }\right) \cdot \nabla \varphi + G_2(\phi ,\varphi ,{\textbf{p}}_1,\varphi ^*_0) \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \varphi (\cdot ,t_0)&= 0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(5.40)

where \({\tilde{\chi }}\) is defined in (5.8),

$$\begin{aligned} E_2&= - \partial _t \phi ^i_0 + B_0[\phi ^i_0] + c_0(t) W_2 \nonumber \\ F_2 (\phi ,\varphi ,{{\textbf {p}}}_1,\varphi ^*_0)&= F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1)\nonumber \\&\quad + \lambda ^4 [ S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ] \nonumber \\&\quad +\lambda \alpha {\dot{\xi }}_1 \cdot {\nabla } _y U (y) \, \chi + \frac{\alpha \lambda ^2}{ \sqrt{t}} U(y) {\dot{\xi }}_1 \cdot \nabla \chi _0 \end{aligned}$$
(5.41)
$$\begin{aligned} G_2(\phi ,\varphi ,{\textbf{p}}_1,\varphi ^*_0)&= G(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) + \lambda ^{-4} E_2 (1-{\tilde{\chi }}_2)\chi \nonumber \\ {\tilde{\chi }}_2(x,t)&= \chi _0\Bigl (\frac{x-\xi }{t^{\frac{1}{2}-\delta }}\Bigr ) , \end{aligned}$$
(5.42)

\(\delta >0\) is a small constant to be fixed later on, and \(\chi _0\) is as in (2.5). We recall that F and G are defined in (5.6) and (5.10). The expressions for \(F_2\) and \(G_2\) depend on the initial condition \(\varphi ^*_0\) through \(\varphi ^*\) (5.37) and \(\phi _0\). The role of \(\phi _0\) will be clarified later on.

By the estimate for \(\ddot{\lambda }_0\) in Proposition 5.1 and (5.35) we get

$$\begin{aligned} |E_2(y,t)| \le \frac{C}{t^2 (\log t)^2} \frac{\log (1+|y|)}{1+|y|^4} + \frac{C}{t^{\frac{3}{2}+\sigma }} |W_2(y)|, \quad |y| \le C \sqrt{t \log t}. \end{aligned}$$
(5.43)

The reason that we introduce the cut-off \({\tilde{\chi }}_2\) is to achieve

$$\begin{aligned} |E_2 {\tilde{\chi }}_2(y,t)|\le \frac{C}{t^\nu (1+|y|)^{6+\sigma }} , \end{aligned}$$

if \(\nu < 1 +2\delta - \frac{\sigma }{2}\). We will choose \(\delta \) and \(\sigma \) positive small numbers such that \(2\delta -\frac{\sigma }{2}>0\) so that we can find \(1<\nu < 1 +2\delta - \frac{\sigma }{2}\).

5.4 Splitting the inner solution \(\phi \)

We perform one more change in the formulation (5.39), (5.40), which consists in decomposing

$$\begin{aligned} \phi = \phi _1 + \phi _2. \end{aligned}$$

The function \(\phi _1\) will solve an equation with part of the right hand side of (5.39), which will be projected so that it satisfies the zero second moment condition.

For any h(yt) with sufficient spatial decay we define

$$\begin{aligned} m_0[h](t) = \int _{{\mathbb {R}}^2} h(y,t) dy ,\quad m_2[h](t) = \int _{{\mathbb {R}}^2} h(y,t) |y|^2 dy , \end{aligned}$$
(5.44)

and

$$\begin{aligned} m_{1,j}[h](t) = \int _{{\mathbb {R}}^2} h(y,t)y_j dy, \quad j=1,2, \end{aligned}$$

which denote the mass, second moment and center of mass of h.

Let \( W_0 \in C^\infty ({\mathbb {R}}^2)\) be radial with compact support such that

$$\begin{aligned} \int _{{\mathbb {R}}^2} W_0 dy =1, \quad \int _{{\mathbb {R}}^2} W_0|y|^2 dy =0. \end{aligned}$$

Let \(W_{1,j}\), \(j=1,2\) be a smooth functions with compact support and with the form \(W_{1,j} (y) = {{\tilde{W}}}(|y|) y_j\) so that

$$\begin{aligned} \int _{{\mathbb {R}}^2} W_{1,j}(y) y_j = 1. \end{aligned}$$
(5.45)

We recall that \(W_2\) defined in (5.33).

Then, \(h - m_0[h]W_0\) has zero mass, \(h-m_2[h]W_2\) has zero second moment, and \(h - m_{1,1}[h] W_{1,1} - m_{1,2}[h]W_{1,2}\) has zero center of mass.

We modify of the operator \(B_0\) appearing in (5.39), and defined in (5.7). The idea is to work with a variant of it, which coincides with it for radial functions, but for functions without radial part it is cutoff outside the region \(|y|\lesssim \frac{\sqrt{t}}{\lambda }\). More precisely, we decompose \(\phi \) in a radial part \([\phi ]_{rad}\) defined by

$$\begin{aligned}{}[\phi ]_{rad}(\rho ,t) = \frac{1}{2\pi } \int _0^{2\pi } \phi (\rho e^{i\theta }, t) d\theta \end{aligned}$$
(5.46)

and a term with no radial mode \(\phi _1 = \phi - [\phi ]_{rad}\). We note that the other linear terms in the equation behave well with this decomposition. Then we define

$$\begin{aligned} B[\phi ] = \lambda {\dot{\lambda }} ( 2 [\phi ]_{rad} + y \cdot \nabla [\phi ]_{rad}) + \lambda {\dot{\lambda }} ( 2\phi _1 + y \cdot \nabla \phi _1) \chi _0\Bigl ( \frac{\lambda y}{5 \sqrt{t}} \Bigr ) \end{aligned}$$
(5.47)

where \(\chi _0\) is a smooth cut-off in \({\mathbb {R}}\) with \(\chi _0(s) = 1\) for \(s\le 1\) and \( \chi _0(s)=1\) for \(s \ge 2\).

With these definitions we introduce the following system for \(\phi _1\), \(\phi _2\), \(\varphi \), \({{\textbf {p}}}_1\),

$$\begin{aligned}&\left\{ \begin{aligned} \lambda ^2 \partial _t \phi _1&= L [\phi _1] + B [\phi _1] + F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \\&\quad - m_0[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] W_0 - m_2[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] W_2 \\&\quad +\sum _{j=1}^2 \mu _j W_{1,j} \qquad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \phi _1(\cdot ,t_0)&= 0\quad \text {in }{\mathbb {R}}^2, \end{aligned} \right. \end{aligned}$$
(5.48)
$$\begin{aligned}&\left\{ \begin{aligned} \lambda ^2 \partial _t \phi _2&= L [\phi _2] + B [\phi _2] + m_2[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] W_2 \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \phi _2(\cdot ,t_0)&= \phi _0 \quad \text {in }{\mathbb {R}}^2, \end{aligned} \right. \end{aligned}$$
(5.49)
$$\begin{aligned}&\left\{ \begin{aligned} \partial _t \varphi&= \Delta \varphi - \nabla \Gamma _0 \cdot \nabla \varphi + G_2(\phi _1+\phi _2,\varphi ,{\textbf{p}}_1,\varphi _0^*) \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \varphi (\cdot ,t_0)&= 0 \quad \text {in }{\mathbb {R}}^2, \end{aligned} \right. \end{aligned}$$
(5.50)

where

$$\begin{aligned} F_3(\phi ,\varphi ,{{\textbf {p}}}_1 ,\varphi _0^*) = E_2 {\tilde{\chi }}_2 + F_2(\phi ,\varphi ,{{\textbf {p}}}_1 ,\varphi _0^*) {\tilde{\chi }}, \end{aligned}$$
(5.51)

In (5.48) \(\mu _j(t)\) are functions so that the right hand side has center of mass equal to zero. A solution \(\phi _1\), \( \phi _2\), \(\varphi \) to (5.48), (5.49) and (5.50) gives a solution to the system (5.39), (5.40) provided \({{\textbf {p}}}_1\) is such that the following equations are satisfied

$$\begin{aligned} \left\{ \begin{aligned} 0&=m_0[ F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) ] (t),\quad \forall t>t_0, \\ 0&=\mu _j(t),\quad \forall t>t_0,\ j=1,2. \end{aligned} \right. \end{aligned}$$
(5.52)

5.5 Mass and second moment

In this section we derive some formulas for the mass and second moment appearing in the right hand side of (5.48).

In the computation of \(m_0[ F_3(\phi ,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) ]\) and \(m_2[ F_3(\phi ,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) ]\), the following formulas will be useful.

Lemma 5.3

We have

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}})) dx&= - \partial _t \int _{{\mathbb {R}}^2} u_0 d x - \partial _t \int _{{\mathbb {R}}^2} \varphi _\lambda dx \\&= - \partial _t \Bigl \{ 8 \pi \alpha \Bigl [ 1+2\Upsilon \frac{\lambda ^2}{t} \Bigr ]+ \alpha e_1 \Bigl ( \frac{\lambda ^2}{t}\Bigr ) + \int _{{\mathbb {R}}^2} \varphi _\lambda dx \Bigr \} \end{aligned}$$

and

$$\begin{aligned} \int _{{\mathbb {R}}^2} ( S(u_1({\textbf{p}}))-S(u_1({\textbf{p}}_0))) dx&= - \partial _t \Bigl \{ \alpha _1 \Bigl [ 8 \pi \Bigl ( 1+2\Upsilon \frac{\lambda ^2}{t}\Bigr )+e_1 \Bigl ( \frac{\lambda ^2}{t}\Bigr ) \Bigr ]\\&\quad + 16 \pi \alpha _0 \Upsilon \frac{\lambda ^2-\lambda _0^2}{t} \\&\quad + \alpha _0 \Bigl ( e_1\Bigl (\frac{\lambda ^2}{t}\Bigr ) -e_1\Bigl (\frac{\lambda _0^2}{t}\Bigr ) \Bigr ) + \int _{{\mathbb {R}}^2} (\varphi _\lambda -\varphi _{\lambda _0}) dx \Bigr \} , \end{aligned}$$

where \(e_1(s)\) is defined by

$$\begin{aligned} \int _{{\mathbb {R}}^2} u_0 dx = 8 \pi \alpha \Bigl [ 1+2\Upsilon \frac{\lambda ^2}{t} \Bigr ] + \alpha e_1\Bigl ( \frac{\lambda ^2}{t}\Bigr ) . \end{aligned}$$
(5.53)

Recall that \(\Upsilon \) is given in (2.7) and note that

$$\begin{aligned} e_1(s) = O(s^2) , \quad \text {as }s\rightarrow 0. \end{aligned}$$

Proof

For this we recall that (c.f. (2.8))

$$\begin{aligned} S(u_1({\textbf{p}})) = - \partial _t u_0 - \partial _t \varphi _\lambda + {\mathcal {E}}(u_0+\varphi _\lambda ) , \end{aligned}$$

so

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}})) dx&= - \partial _t \int _{{\mathbb {R}}^2} u_0 d x - \partial _t \int _{{\mathbb {R}}^2} \varphi _\lambda dx \\&= - \partial _t \Bigl \{ 8 \pi \alpha \Bigl [ 1+2\Upsilon \frac{\lambda ^2}{t} \Bigr ]+ \alpha e_1 \Bigl ( \frac{\lambda ^2}{t}\Bigr ) + \int _{{\mathbb {R}}^2} \varphi _\lambda dx \Bigr \} . \end{aligned}$$

Therefore

$$\begin{aligned} \int _{{\mathbb {R}}^2} ( S(u_1({\textbf{p}}))-S(u_1({\textbf{p}}_0))) dx&= - \partial _t \Bigl \{ \alpha _1 \Bigl [ 8 \pi \Bigl ( 1+2\Upsilon \frac{\lambda ^2}{t}\Bigr )+e_1 \Bigl ( \frac{\lambda ^2}{t}\Bigr ) \Bigr ]\\&\quad + 16 \pi \alpha _0 \Upsilon \frac{\lambda ^2-\lambda _0^2}{t} \\&\quad + \alpha _0 \Bigl ( e_1\Bigl (\frac{\lambda ^2}{t}\Bigr ) -e_1\Bigl (\frac{\lambda _0^2}{t}\Bigr ) \Bigr ) + \int _{{\mathbb {R}}^2} (\varphi _\lambda -\varphi _{\lambda _0}) dx \Bigr \}. \end{aligned}$$

\(\square \)

Lemma 5.4

We have

$$\begin{aligned} \lambda ^4 m_2&[S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ] \\&= - 32\pi \alpha _1 - \frac{{\dot{\alpha }}}{\lambda ^2} \int _{{\mathbb {R}}^2} U(\frac{x-\xi }{\lambda }) \chi _0(\frac{x-\xi }{\lambda }) |x-\xi |^2\,dx\\&\quad + \frac{{\dot{\alpha }}_0}{\lambda _0^2} \int _{{\mathbb {R}}^2} U(\frac{x}{\lambda _0}) \chi (\frac{x}{\lambda _0}) |x|^2\,dx \\&\quad - 4 \Bigl [ \alpha e_1\Bigl (\frac{\lambda ^2}{t}\Bigr ) - \alpha _0 e_1\Bigl (\frac{\lambda _0^2}{t} \Bigr )\Bigr ] - \Bigl [ \alpha e_2\Bigl (\frac{\lambda ^2}{t}\Bigr ) - \alpha _0 e_2\Bigl (\frac{\lambda _0^2}{t} \Bigr )\Bigr ] \\&\quad - \Bigl ( \int _{{\mathbb {R}}^2} (\varphi _{\lambda }-\varphi _{\lambda _0})\,dx\Bigr )^2 \\&\quad -(1-\alpha ) \int _{{\mathbb {R}}^2} E(x-\xi ,t,\lambda )|x-\xi |^2\, \textrm{d}x +(1-\alpha _0) \int _{{\mathbb {R}}^2} E(x,t,\lambda _0)|x|^2\, \textrm{d}x \\&\quad - |\xi |^2 \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}}_0))dx . \end{aligned}$$

Proof

We have defined the second moment \(m_2\) (5.44) integrating with respect to y. Note that

$$\begin{aligned} \lambda ^4 \int _{{\mathbb {R}}^2} f(y) |y|^2dy = \int _{{\mathbb {R}}^2} f\Bigl ( \frac{x-\xi }{\lambda }\Bigr )|x-\xi |^2 dx, \end{aligned}$$

and therefore

$$\begin{aligned}&\lambda ^4 m_2 [S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ] \\&\quad = \lambda ^4 \int _{{\mathbb {R}}^2} S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) (\xi + \lambda y) |y|^2 dy \\&\qquad - \lambda ^4 \int _{{\mathbb {R}}^2} S_0( u_1( {{\textbf {p}}}_0 ) ) (\xi + \lambda y) |y|^2 dy \\&\quad = \int _{{\mathbb {R}}^2} S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) (x) |x-\xi |^2 dy \\&\qquad - \int _{{\mathbb {R}}^2} S_0( u_1( {{\textbf {p}}}_0 ) ) (x) |x-\xi |^2 dy. \end{aligned}$$

We have by Lemma 5.2,

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1) |x-\xi |^2dx&= 4 \int _{{\mathbb {R}}^2} \varphi _\lambda dx - \alpha \int _{{\mathbb {R}}^2} {{\tilde{E}}}(x-\xi ,t;\lambda ) |x-\xi |^2 dx \nonumber \\&\quad + \int _{{\mathbb {R}}^2} \nabla \varphi _\lambda \, \textrm{d}x \cdot {\dot{\xi }} - \frac{{\dot{\alpha }}}{\lambda ^2} \int _{{\mathbb {R}}^2} U \chi |x-\xi |^2 dx \nonumber \\&\quad - (1-\alpha ) \int _{{\mathbb {R}}^2} E (x-\xi ,t;\lambda ) |x-\xi |^2dx \nonumber \\&\quad + 4 \Bigl ( \int _{{\mathbb {R}}^2} u_0 + \int _{{\mathbb {R}}^2} \varphi _\lambda \Bigr ) \Bigl ( 1 - \frac{1}{8\pi } \int _{{\mathbb {R}}^2} u_0 - \frac{1}{8\pi } \int _{{\mathbb {R}}^2} \varphi _\lambda \Bigr ). \end{aligned}$$
(5.54)

where E, \({{\tilde{E}}}\) are defined in (3.10), (3.11). Let

$$\begin{aligned} m = \int _{{\mathbb {R}}^2} (u_0 + \varphi _\lambda ) dx, \quad \delta m = m - 8 \pi . \end{aligned}$$

Since

$$\begin{aligned} \int _{{\mathbb {R}}^2} (u_0 + \varphi _{\lambda _0}) dx = 8 \pi , \end{aligned}$$

by (5.22), we have

$$\begin{aligned} \delta m= \int _{{\mathbb {R}}^2} (\varphi _\lambda - \varphi _{\lambda _0})\,dx. \end{aligned}$$

Replacing m in (5.54) we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}})) |x-\xi |^2dx&= 32 \pi - 4 \int _{{\mathbb {R}}^2} u_0 dx - \frac{1}{2\pi }(\delta m)^2 \nonumber \\&\quad - \alpha \int _{{\mathbb {R}}^2} {{\tilde{E}}}(x-\xi ,t;\lambda ) |x-\xi |^2 dx \nonumber \\&\quad + \int _{{\mathbb {R}}^2} \nabla \varphi _\lambda \, \textrm{d}x \cdot {\dot{\xi }} - \frac{{\dot{\alpha }}}{\lambda ^2} \int _{{\mathbb {R}}^2} U \chi |x-\xi |^2 dx \nonumber \\&\quad - (1-\alpha ) \int _{{\mathbb {R}}^2} E (x-\xi ,t;\lambda ) |x-\xi |^2dx . \end{aligned}$$
(5.55)

Also under (4.1) we have by (5.24):

$$\begin{aligned} \int _{{\mathbb {R}}^2} {{\tilde{E}}} |x-\xi |^2 dx = - 64 \pi \Upsilon \frac{\lambda ^2}{t} +e_2\Bigl ( \frac{\lambda ^2}{t}\Bigr ), \end{aligned}$$
(5.56)

where

$$\begin{aligned} e_2(s) = O(s^2) , \quad \text {as }s\rightarrow 0. \end{aligned}$$

Combining (5.55), (5.53) and (5.56) we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}})) |x-\xi |^2dx&= 32 \pi (1-\alpha ) - \frac{1}{2\pi }(\delta m)^2 - \frac{{\dot{\alpha }}}{\lambda ^2} \int _{{\mathbb {R}}^2} U \chi |x-\xi |^2 dx \\&\quad + \int _{{\mathbb {R}}^2} \nabla \varphi _\lambda \, \textrm{d}x \cdot {\dot{\xi }} - (1-\alpha ) \int _{{\mathbb {R}}^2} E (x-\xi ,t;\lambda ) |x-\xi |^2dx \\&\quad -4 \alpha e_1 \Big ( \frac{ \lambda ^2 }{t} \Big ) - \alpha e_2 \Big ( \frac{ \lambda ^2 }{t} \Big ) . \end{aligned}$$

We can apply this formula to \({\textbf{p}} = {\textbf{p}}_0\) and get

$$\begin{aligned} \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}}_0)) |x|^2dx&= 32 \pi (1-\alpha _0) - \frac{{\dot{\alpha }}_0}{\lambda _0^2} \int _{{\mathbb {R}}^2} U(\frac{x}{\lambda _0}) \chi |x|^2 dx \\&\quad - (1-\alpha _0) \int _{{\mathbb {R}}^2} E (x,t;\lambda _0) |x|^2dx \\&\quad -4 \alpha _0 e_1 \Big ( \frac{ \lambda _0^2 }{t} \Big ) - \alpha _0 e_2 \Big ( \frac{ \lambda _0^2 }{t} \Big ) . \end{aligned}$$

Note that

$$\begin{aligned} \int _{{\mathbb {R}}^2} S_0(u_1({\textbf{p}}_0))|x-\xi |^2dx&= \int _{{\mathbb {R}}^2} S_0(u_1({\textbf{p}}_0))|x|^2dx + |\xi |^2 \int _{{\mathbb {R}}^2} S_0(u_1({\textbf{p}}_0))dx \\&= \int _{{\mathbb {R}}^2} S_0(u_1({\textbf{p}}_0)|x|^2dx + |\xi |^2 \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}}_0)dx \end{aligned}$$

because

$$\begin{aligned} \int _{{\mathbb {R}}^2} S_0(u_1({\textbf{p}}_0) x_j dx =0. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{{\mathbb {R}}^2}&[ S(u_1({\textbf{p}}))-S(u_1({\textbf{p}}_0))] |x-\xi |^2dx \\&= - 32\pi \alpha _1 - \frac{{\dot{\alpha }}}{\lambda ^2} \int _{{\mathbb {R}}^2} U(\frac{x-\xi }{\lambda }) \chi _0(\frac{x-\xi }{\lambda }) |x-\xi |^2\,dx \\&\quad + \frac{{\dot{\alpha }}_0}{\lambda _0^2} \int _{{\mathbb {R}}^2} U(\frac{x}{\lambda _0}) \chi (\frac{x}{\lambda _0}) |x|^2\,dx \\&\quad - 4 \Bigl [ \alpha e_1\Bigl (\frac{\lambda ^2}{t}\Bigr ) - \alpha _0 e_1\Bigl (\frac{\lambda _0^2}{t} \Bigr )\Bigr ] - \Bigl [ \alpha e_2\Bigl (\frac{\lambda ^2}{t}\Bigr ) - \alpha _0 e_2\Bigl (\frac{\lambda _0^2}{t} \Bigr )\Bigr ] \\&\quad - \Bigl ( \int _{{\mathbb {R}}^2} (\varphi _{\lambda }-\varphi _{\lambda _0})\,dx\Bigr )^2 \\&\quad -(1-\alpha ) \int _{{\mathbb {R}}^2} E(x-\xi ,t,\lambda )|x-\xi |^2\, \textrm{d}x +(1-\alpha _0) \int _{{\mathbb {R}}^2} E(x,t,\lambda _0)|x|^2\, \textrm{d}x \\&\quad - |\xi |^2 \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}}_0)dx . \end{aligned}$$

\(\square \)

6 Proof of Theorem 1.1

Next we define norms, which are suitably adapted to the terms in the inner linear problems (5.48), (5.49). Let us write the linearized versions of these problems as

$$\begin{aligned} \left\{ \begin{aligned} \lambda ^2 \partial _t \phi&= L[\phi ] + B[\phi ]+ h(y,t) \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ), \\ \phi (\cdot ,t_0)&= 0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(6.1)

Given positive numbers \(\nu \), p, \(\varepsilon \) and \(m\in {\mathbb {R}}\), we let

$$\begin{aligned} \Vert h\Vert _{0,\nu ,m,p,\varepsilon } =&\inf K \quad \text {such that } \nonumber \\ |h(y,t)|&\le \frac{K}{t^{\nu } (\log t)^{m}} \frac{ 1 }{(1+|y|)^p} {\left\{ \begin{array}{ll} \displaystyle 1 &{} |y|\le \sqrt{ t \log t }, \\ \displaystyle \frac{(t \log t)^{\varepsilon /2}}{|y|^\varepsilon } &{} |y|\ge \sqrt{ t \log t } . \end{array}\right. } \end{aligned}$$
(6.2)

We also defie

$$\begin{aligned} \Vert \phi \Vert _{1,\nu ,m,p,\varepsilon } =&\inf K \quad \text {such that } \\&| \phi (y,t)| + (1+|y|) |\nabla _y \phi (y,t)|\\&\le \frac{K}{t^{\nu } (\log t)^{m}} \frac{ 1 }{(1+|y|)^p} {\left\{ \begin{array}{ll} \displaystyle 1 &{} |y|\le \sqrt{ t \log t }, \\ \displaystyle \frac{(t \log t)^{\varepsilon /2}}{|y|^\varepsilon } &{} |y|\ge \sqrt{ t \log t } . \end{array}\right. } \end{aligned}$$

We develop a solvability theory of problem (6.1) that involves uniform space-time bounds in terms of the above norms. We will establish two results: one in which the solution “loses” one power of t on bounded sets with respect to the time-decay of h, under radial symmetry and the condition of spatial average 0 at all times. Our second result states that for a general h this loss is only \(t^{\frac{1}{2}} \) if in addition the center of mass and second-moment of h are zero at all times.

For the first result we introduce a parameter in the problem in order to get a fast decay of the solution:

$$\begin{aligned} \left\{ \begin{aligned} \lambda ^2 \partial _t \phi&= L[\phi ] + B[\phi ]+ h(y,t) \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ), \\ \phi (\cdot ,t_0)&= c_1 {{\tilde{Z}}}_0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(6.3)

where \({{\tilde{Z}}}_0\) is defined as

$$\begin{aligned} {{\tilde{Z}}}_0(\rho )&= ( Z_0(\rho ) - m_{Z_0} U ) \chi _0\Bigl ( \frac{\rho }{3\lambda (t_0)\sqrt{t_0} }\Bigr ), \end{aligned}$$
(6.4)

where \(m_{Z_0}\) is such that

$$\begin{aligned} \int _{{\mathbb {R}}^2} {{\tilde{Z}}}_0 = 0. \end{aligned}$$

Proposition 6.1

Assume (4.1). Let \(\sigma >0\), \(\varepsilon >0\) with \(\sigma +\varepsilon <2\) and \(1<\nu <\frac{7}{4}\). Let \(0<q<1\). Then there exists a number \(C>0\) such that for \(t_0\) sufficiently large and all radially symmetric \(h=h(|y|,t)\) with \(\Vert h\Vert _{0,\nu ,m,6+\sigma ,\varepsilon }<\infty \) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y,t)dy&= 0 ,\quad \text {for all } t>t_0 , \end{aligned}$$

there exists \(c_1 \in {\mathbb {R}}\) and solution \(\phi (y,t) = {\mathcal {T}}^{i,2}_{{{\textbf {p}}}} [h]\) of problem (6.3) that defines a linear operator of h and satisfies the estimate

$$\begin{aligned} \Vert \phi \Vert _{1,\nu -1,m+q-1,4,2+\sigma +\varepsilon } \le \frac{C}{(\log t_0)^{1-q}} \Vert h\Vert _{0,\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

Moreover \(c_1\) is a linear operator of h and

$$\begin{aligned} |c_1|&\le C \frac{1 }{ t_0^{\nu -1} (\log t_0)^{m}} \Vert h\Vert _{0,\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

We also consider the problem

$$\begin{aligned} \left\{ \begin{aligned} \lambda ^2 \partial _t \phi&= L[\phi ] + B[\phi ]+ h(y,t) + \sum _{j=1}^2 \mu _j(t) W_{1,j} \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ), \\ \phi (\cdot ,t_0)&= 0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(6.5)

where the function \(W_{1,j}\) have been defined in (5.45).

Proposition 6.2

Assume (4.1). Let \(0<\sigma <1\), \(\varepsilon >0\) with \(\sigma +\varepsilon <\frac{3}{2}\) and \(1<\nu < \min ( 1+\frac{\varepsilon }{2},3-\frac{\sigma }{2}, \frac{5}{4})\). Let \(0<q<1\). Then there is C such that for \(t_0\) large the following holds. Suppose that h satisfies \(\Vert h\Vert _{0,\nu ,m,6+\sigma ,\varepsilon }<\infty \) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y,t)dy&=0 , \quad \int _{{\mathbb {R}}^2} h(y,t)|y|^2dy=0, \quad \text {for all } t>t_0 . \end{aligned}$$

Then there exists a solution \(\phi (y,t)\), \(\mu _j(t)\) of problem (6.5) that defines a linear operator of h and satisfies

$$\begin{aligned} \Vert \phi \Vert _{1,\nu -\frac{1}{2} ,m+\frac{q-1}{2},4,2+\sigma +\varepsilon } \le C \Vert h\Vert _{0,\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

The parameters \(\mu _j\) satisfy

$$\begin{aligned} \mu _j(t) = - \int _{{\mathbb {R}}^2} h(y,t) y_jdy + {\tilde{\mu }}_j[h](t) \end{aligned}$$

where \({\tilde{\mu }}_j\) are linear functions of h with

$$\begin{aligned} |{\tilde{\mu }}_j[h]|&\le \frac{C}{t^{\nu +1}(\log t)^{\nu +m+2}}\Vert h \Vert _{\nu ,m,5+\sigma ,\varepsilon }. \end{aligned}$$

We denote this solution by \(\phi = {\mathcal {T}}^{i,1}_{{{\textbf {p}}}} [h]\).

The proof of the Propositions 6.1 and 6.2 is divided into different steps and presented in Sects. 812.

Next we consider the linear outer problem:

$$\begin{aligned} \left\{ \begin{aligned} \partial _t \phi ^o&= L^o [\phi ^o] + g(x,t), \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \phi ^o(\cdot ,t_0)&=\phi ^o_0, \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(6.6)

where

$$\begin{aligned} L^o [\varphi ] :=\Delta _x \varphi - \nabla _x\Big [ \Gamma _0 \Big (\frac{x-\xi (t)}{\lambda (t)}\Big ) \Big ]\cdot \nabla _x \varphi . \end{aligned}$$

For a given function g(xt) we consider the norm \(\Vert g \Vert _{**,o}\) defined as the least \(K\geqq 0\) such that for all \((x,t)\in {\mathbb {R}}^2\times (t_0,\infty )\)

$$\begin{aligned} |g(x,t)| \leqq K\frac{1}{t^a (\log t)^\beta } \frac{1}{ 1+ |\zeta |^b} , \quad \zeta = \frac{x-\xi (t)}{\sqrt{t}}. \end{aligned}$$
(6.7)

Accordingly, we consider for a function \(\phi ^o(x,t)\) the norm \(\Vert \phi \Vert _{*,o}\) defined as the least \(K\geqq 0\) such that

$$\begin{aligned} |\phi ^o(x,t)| + (\lambda +|x-\xi |) | {\nabla } _x \phi ^o (x,t)| \leqq K\frac{1}{t^{a-1} (\log t)^\beta } \frac{1}{ 1+ |\zeta |^{b}} , \quad \zeta = \frac{x-\xi }{\sqrt{t}} \end{aligned}$$
(6.8)

for all \((x,t)\in {\mathbb {R}}^2\times (t_0,\infty )\).

We assume that the parameters \(a,b,\beta \) satisfy the constraints

$$\begin{aligned} 1<a< 4,\quad 2<b< 6, \quad a< 1+ \frac{b}{2}, \quad \quad \beta \in {\mathbb {R}}. \end{aligned}$$
(6.9)

Proposition 6.3

Assume that the parameter functions \({{\textbf {p}}} =(\lambda , \alpha ,\xi )\) satisfy conditions (4.1) and the numbers \(a,b,\beta \) satisfy (6.9). Then there is a constant C so that for \(t_0\) sufficiently large and for \(\Vert g\Vert _{**,o}<\infty \), there exists a solution \(\phi ^o= {\mathcal {T}} ^o_{{{\textbf {p}}}}[g ]\) of (6.6) with \(\phi ^o_0=0\), which defines a linear operator of g and satisfies

$$\begin{aligned} \Vert \phi ^o\Vert _{*,o} \le C \Vert g\Vert _{ **,o} . \end{aligned}$$

For the initial condition \(\phi _0^o\) in (6.6) we consider the norm \(\Vert \varphi _0^o \Vert _{*,b}\) defined as

$$\begin{aligned} \Vert \phi ^o_0 \Vert _{*,b}&= \inf K \quad \text {such that } \nonumber \\&\quad |\phi ^o_0(x)| + (\lambda (t_0) +|x| ) |\nabla _x \phi ^o(x)| \le \frac{K}{( 1+ \frac{|x|}{\sqrt{t_0}} )^b} . \end{aligned}$$
(6.10)

We have an estimate for the solution of (6.6) with \(g=0\) and \(\Vert \phi ^o_0 \Vert _{*,b}<\infty \).

Proposition 6.4

Assume that the parameter functions \({{\textbf {p}}} =(\lambda , \alpha ,\xi )\) satisfy conditions (4.1) and the numbers \(a,b,\beta \) satisfy (6.9). Then there is a constant C so that for \(t_0\) sufficiently large and for \(\Vert \phi ^o_0 \Vert _{*,b}<\infty \) there exists a solution \(\phi ^o\) of (6.6), which defines a linear operator of \( \phi ^o_0\) and satisfies

$$\begin{aligned} \Vert \phi ^o\Vert _{*,o} \le C t_0^{a-1} (\log t_0)^\beta \Vert \phi ^o_0 \Vert _{*,b} . \end{aligned}$$

The proofs of Propositions 6.3 and 6.4 are contained in Sect. 13.

In what follows we work with \({{\textbf {p}}}_1 \) of the form

$$\begin{aligned} {{\textbf {p}}}_1 = (0, \alpha _1,\xi _1), \end{aligned}$$

that is, we take \(\lambda = \lambda _0\), \(\alpha = \alpha _0+\alpha _1\), \(\xi = \xi _1\), where \(\lambda _0\) and \(\alpha _0\) have been fixed in Sect. 5.1, and we write

$$\begin{aligned} {{\textbf {p}}} = {{\textbf {p}}}_0+{{\textbf {p}}}_1. \end{aligned}$$

Next we define suitable operators that allow us to formulate the system of equations (5.48), (5.49), (5.50), and (5.52) as a fixed point problem. We let

$$\begin{aligned} {\mathcal {A}}_{i1}[\phi _1,\phi _2,\varphi , {{\textbf {p}}}_1]&= {\mathcal {T}}^{i,1}_{{{\textbf {p}}}} \Bigl [ F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \\&\quad - m_0[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] W_0\\&\quad - m_2[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] W_2 \\&\quad - m_{1,1}[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] W_{1,1} \\&\quad - m_{1,2}[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] W_{1,2} \Bigr ] \\ {\mathcal {A}}_{i2}[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*]&= {\mathcal {T}}^{i,2}_{{{\textbf {p}}}} \left[ m_2[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1),\varphi _0^*] W_2 \right] \\ A_o[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1, \varphi _0^* ]&= {\mathcal {T}}^o_{{{\textbf {p}}}} [ G_2(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1, \varphi _0^*)] . \end{aligned}$$

Then the equations (5.48), (5.49),(5.50) can be written as

$$\begin{aligned} \phi _1&= {\mathcal {A}}_{i1}[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1, \varphi _0^*] \\ \phi _2&= {\mathcal {A}}_{i2}[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1, \varphi _0^*] \\ \varphi&= {\mathcal {A}}_o[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1, \varphi _0^*. ] \end{aligned}$$

Next we consider the equations (5.52), that is, \(m_0[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] (t)\equiv 0\) and \(\mu _j(t) \equiv 0\). By (5.51) and (5.41)

$$\begin{aligned} m_0[F_3 (\phi ,\varphi ,{{\textbf {p}}}_1, \varphi ^*_0 ) {\tilde{\chi }} ]&= \lambda _0^4 m_0[S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ] +m_0[ E_2 {\tilde{\chi }}_2] \\&\quad + m_0[ F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) {\tilde{\chi }}] \\&\quad + \lambda _0^4 m_0[ ( S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ) ({\tilde{\chi }}-1) ] , \end{aligned}$$

and using Lemma 5.3,

$$\begin{aligned} m_0[F_3(\phi ,\varphi ,{{\textbf {p}}}_1, \varphi ^*_0 ) {\tilde{\chi }} ]&= - \lambda _0^2 \partial _t \Bigl \{ \alpha _1 \Bigl [ 8 \pi \Bigl ( 1+2\Upsilon \frac{\lambda _0^2}{t}\Bigr )+e_1 \Bigl ( \frac{\lambda _0^2}{t}\Bigr ) \Bigr ] \Bigr \} +m_0[ E_2 {\tilde{\chi }}_2] \\&\quad + m_0[ F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) {\tilde{\chi }}] \\&\quad + \lambda _0^4 m_0[ ( S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ) ({\tilde{\chi }}-1) ] . \end{aligned}$$

This motivates the definition

$$\begin{aligned}&{\mathcal {A}}_{\alpha _1} [\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*] \nonumber \\&\quad = - \frac{1}{8\pi ( 1+2\Upsilon \frac{\lambda _0^2}{t}) + e_1(\frac{\lambda _0^2}{t})} \int _t^\infty \frac{1}{\lambda _0^2} \Bigl \{ m_0[ E_2 {\tilde{\chi }}_2] (s) \nonumber \\&\qquad + m_0[ F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) {\tilde{\chi }}](s) \nonumber \\&\qquad + \lambda _0^4 m_0[ ( S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ) ({\tilde{\chi }}-1) ] (s) \Bigr \}ds \end{aligned}$$
(6.11)

Similarly, by (5.51) and (5.41), asking that \(\mu _j\equiv 0\) in (5.48) is equivalent to

$$\begin{aligned} 0&= \lambda _0 \alpha {\dot{\xi }}_{1,j} \int _{{\mathbb {R}}^2} \partial _{y_j}U(y) y_j {\tilde{\chi }} dy + \frac{\alpha \lambda _0^2}{\sqrt{t}} {\dot{\xi }}_{1,j} \int _{{\mathbb {R}}^2} U(y) \partial _{z_j} \chi _0(\frac{\lambda y}{\sqrt{t}}) y_j dy \\&\quad +m_{1,j}[ E_2 {\tilde{\chi }}_2] + m_{1,j}[ F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) {\tilde{\chi }}] + m_{1,j}[B[\phi _1]]. \end{aligned}$$

This motivates the definition

$$\begin{aligned}&{\mathcal {A}}_{\xi _1} [\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*] \nonumber \\&\quad = \int _t^\infty \frac{1}{ \lambda _0 \alpha \int _{{\mathbb {R}}^2} \partial _{y_j}U(y) y_j {\tilde{\chi }} dy} \Bigl \{ \frac{\alpha \lambda _0^2}{\sqrt{t}} {\dot{\xi }}_{1,j} \int _{{\mathbb {R}}^2} U(y) \partial _{z_j} \chi _0(\frac{\lambda y}{\sqrt{t}}) y_j dy \nonumber \\&\qquad +m_{1,j}[ E_2 {\tilde{\chi }}_2](s) + m_{1,j}[ F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) {\tilde{\chi }}](s) +m_{1,j}[B[\phi _1]](s) \Big \}ds \end{aligned}$$
(6.12)

Then we define \({\mathcal {A}}_p\) by

$$\begin{aligned} {\mathcal {A}}_p [\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*] = ( 0, {\mathcal {A}}_{\alpha _1} [\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*], {\mathcal {A}}_{\xi _1} [\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*]). \end{aligned}$$
(6.13)

Then

$$\begin{aligned} {\textbf{p}}_1 = {\mathcal {A}}_p [\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*] \end{aligned}$$

is equivalent to the equations (5.52).

We write

$$\begin{aligned} \vec {\phi } = (\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1), \end{aligned}$$

and

$$\begin{aligned} {\mathcal {A}}[\vec {\phi }] = ({\mathcal {A}}_{i1}[\vec {\phi },\varphi _0^*], {\mathcal {A}}_{i2}[\vec {\phi },\varphi _0^*], {\mathcal {A}}_{o}[\vec {\phi },\varphi _0^*], {\mathcal {A}}_{p}[\vec {\phi },\varphi _0^*] ), \end{aligned}$$

and the objective is to find \(\vec {\phi }\) such that

$$\begin{aligned} \vec {\phi } = {\mathcal {A}}[\vec {\phi }]. \end{aligned}$$

The operator \({\mathcal {A}}\) depends on the initial condition \(\varphi _0^*\) appearing in the parabolic problem (5.37), and we will stress its dependence later on when proving the stability assertion in Theorem 1.1.

We define the spaces on which we will consider the operator \({\mathcal {A}}\) to set up the fixed point problem. For certain choices of constants \(\nu \), q, \(\sigma \), \(\varepsilon \), a, b, \(\beta \), \(\gamma \), \(\Theta \) that we will make precise later, we let

$$\begin{aligned} X_{i}&= \Bigl \{ \, \phi \in L^\infty ( {\mathbb {R}}^2 \times (t_0,\infty ) ) \ | \ \nabla _y \phi \in L^\infty ( {\mathbb {R}}^2 \times (t_0,\infty ) ) , \, \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }<\infty , \\&\quad \int _{{\mathbb {R}}^2} \phi (y,t)dy=0, \ \int _{{\mathbb {R}}^2} \phi (y,t) y dy = 0, \ \ t>t_0 \, \Bigr \} , \\ X_o&= \{ \varphi \in L^\infty ( {\mathbb {R}}^2 \times (t_0,\infty ) ) \ | \ \nabla _y \phi \in L^\infty ( {\mathbb {R}}^2 \times (t_0,\infty ) ) , \, \Vert \varphi \Vert _{*,o}<\infty \, \} , \\ X_p&= \{ \, ( 0,\alpha _1,\xi _1) \in C^1([t_0,\infty ))\ |\ \Vert \alpha _1\Vert _{C^1,\nu +\frac{1}{2},\Theta }<\infty , \ \Vert \xi _1\Vert _{C^1,\gamma ,0}<\infty \, \} \end{aligned}$$

where the norms \(\Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }\) and \(\Vert \varphi \Vert _{*,o}\) are defined in (6.2), (6.7) and \(\Vert \xi _1\Vert _{C^1,\mu ,m}\) is defined by

$$\begin{aligned} \Vert g\Vert _{C^0,\mu ,m}&= \sup _{t\ge t_0} \, t^\mu (\log t)^m |g(t)|.\\ \Vert g \Vert _{C^1,\mu ,m}&= \Vert g \Vert _{C^0,\mu ,m} + \Vert \dot{g}\Vert _{C^0,\mu +1,m}. \end{aligned}$$

for a function \(g \in C^1([t_0, \infty ))\).

We choose in the definition of the outer norm (6.8)

$$\begin{aligned} a = \nu +\frac{5}{2} , \quad 2\nu +3<b<6, \quad \beta < \frac{1+q}{2}. \end{aligned}$$
(6.14)

With these choices we see that (6.9) are satisfied. Also \(\nu \) will be in the range \(1<\nu <\frac{3}{2}\) so the interval for b is not empty in (6.14).

We use the following notation: for \({{\textbf {p}}}_1= (0,\alpha _1,\xi _1)\),

$$\begin{aligned} \Vert {{\textbf {p}}}_1 \Vert _{X_p} = \Vert \alpha _1\Vert _{C^1,\nu +\frac{1}{2},\Theta } + \Vert \xi _1\Vert _{C^1,1+\gamma ,0}, \end{aligned}$$

and for \( \vec {\phi } = ( \phi _1,\phi _2, \varphi , {{\textbf {p}}}_1) \),

$$\begin{aligned} \Vert \vec {\phi } \Vert _X = \Vert \phi _1\Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } + \Vert \phi _2\Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } + \Vert \varphi \Vert _{*,o} + \Vert {{\textbf {p}}}_1 \Vert _{X_p} . \end{aligned}$$
(6.15)

With the above notation, given \(\varphi _0^*\) with \(\Vert \varphi _0^*\Vert _{*,b} \) sufficiently small, we consider the fixed point problem

$$\begin{aligned} \vec {\phi } = {\mathcal {A}}[\vec {\phi }] , \end{aligned}$$
(6.16)

with \(\vec {\phi } \) in a suitable close ball of X. A solution of this fixed point problem yields a solution of the system of Eqs. (5.48), (5.49), (5.50), (5.52), which in turn gives a solution to (3.1).

We claim that for some constant C independent of \(t_0\gg 1\), if \( t_0^{a-1} (\log t_0)^\beta \Vert \varphi _0^* \Vert _{*,b}\le 1 \), and \(\Vert \vec {\phi }\Vert _X\le 1\), then

$$\begin{aligned} \Vert {\mathcal {A}}_{i1}[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*] \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \le \frac{C}{t_0^\vartheta } + C (\log t_0)^{\frac{\sigma }{2}} t_0^{\nu +1+\frac{\sigma }{2}} \Vert \varphi _0^* \Vert _{*,b}, \end{aligned}$$
(6.17)

for some \(\vartheta >0\) small, a constant C independent of \(t_0\), and \(t_0\) sufficiently large.

Indeed, by Proposition 6.2 we have

$$\begin{aligned} \Vert {\mathcal {A}}_{i1}[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*] \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \le C \Vert F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \Vert _{0,\nu ,6+\sigma ,\varepsilon } . \end{aligned}$$

We recall the expansion of \(F_3\) in (5.51). To estimate \( E_2 {\tilde{\chi }}_2\) we use (5.43) to get

$$\begin{aligned} \Vert E_2 {\tilde{\chi }} \Vert _{0,\nu ,0,6+\sigma ,\varepsilon } \le \frac{C }{t_0^{1+2\delta -\frac{\sigma }{2}-\nu } (\log t_0)^{2}} \end{aligned}$$

where \(\delta \), \(\sigma \) are positive small constants and are assumed to satisfy \(2\delta -\frac{\sigma }{2}>0\). Then we take \(\nu \) in the range

$$\begin{aligned} 1<\nu <1+2\delta -\frac{\sigma }{2}, \end{aligned}$$
(6.18)

with \(\nu \) close to 1.

Let us consider the term \( \lambda ^4 [ S_0({{\textbf {p}}}_0 + {{\textbf {p}}}_1) - S_0({{\textbf {p}}}_0)]\) in \(F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \) (c.f. (5.51)). The formula \( \lambda ^4 [ S_0({{\textbf {p}}}_0 + {{\textbf {p}}}_1) - S_0({{\textbf {p}}}_0)]\) (c.f. (5.31)) contains for example the term, evaluated at \(y = \frac{x-\xi _1}{\lambda _0}\),

$$\begin{aligned}&-\lambda _0^2 {\dot{\alpha }} U(y) \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) +\lambda _0^2 {\dot{\alpha }}_0 U\Bigl (\frac{\xi _1+\lambda _0y}{\lambda _0}\Bigr ) \chi _0\Bigl ( \frac{\xi _1+\lambda _0 y }{\sqrt{t}}\Bigr ) \nonumber \\&\quad = -\lambda _0^2 {\dot{\alpha }}_1 U(y) \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) -\lambda _0^2 {\dot{\alpha }}_0 \Big [ U(y)-U\Bigl (\frac{\xi _1+\lambda _0y}{\lambda _0}\Bigr ) \Bigr ] \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) \nonumber \\&\qquad -\lambda _0^2 {\dot{\alpha }}_0 U\Bigl (\frac{\xi _1+\lambda _0y}{\lambda _0}\Bigr ) \Bigl [ \chi _0\Bigl ( \frac{\xi _1+\lambda _0 y }{\sqrt{t}}\Bigr )-\chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) \Bigr ] \end{aligned}$$
(6.19)

But

$$\begin{aligned} \left| -\lambda _0^2 {\dot{\alpha }}_1 U(y) \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) \right|&\le C \frac{1}{t^{\nu +\frac{1-\sigma }{2}} (\log t)^{\Theta -\frac{\sigma }{2}}} \frac{1}{(1+|y|)^{6+\sigma }} \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr )\Vert \alpha _1\Vert _{C^1,\nu +\frac{1}{2},\Theta } , \end{aligned}$$

so

$$\begin{aligned} \Bigl \Vert -\lambda _0^2 {\dot{\alpha }}_1 U(y) \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) \Bigr \Vert _{0,\nu ,6+\sigma ,\varepsilon } \le \frac{C}{t_0^\vartheta } \Vert \alpha _1\Vert _{C^1,\nu +\frac{1}{2},\Theta } , \end{aligned}$$

for some \(\vartheta >0\).

Similarly,

$$\begin{aligned}&\left| -\lambda _0^2 {\dot{\alpha }}_0 \Big [ U(y)-U\Bigl (\frac{\xi _1+\lambda _0y}{\lambda _0}\Bigr ) \Bigr ] \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) \right| \\&\quad \le C \frac{1}{\log t} \frac{1}{t^2 \log t} \frac{1}{(1+|y|)^5} \frac{|\xi _1|}{\lambda _0}\chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) \\ \\&\quad \le C \frac{1}{t^{2+\gamma } (\log t)^{\frac{3}{2}}} \frac{(t\log t)^{\frac{1+\sigma }{2}}}{(1+|y|)^{6+\sigma }} \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) \Vert \xi _1\Vert _{C^1,\gamma ,0} \\&\quad \le C \frac{1}{t^{\frac{3-\sigma }{2}+\gamma } (\log t)^{1-\frac{\sigma }{2}}} \frac{1}{(1+|y|)^{6+\sigma }} \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) \Vert \xi _1\Vert _{C^1,\gamma ,0} \end{aligned}$$

so

$$\begin{aligned} \Bigl \Vert -\lambda _0^2 {\dot{\alpha }}_0 \Big [ U(y)-U\Bigl (\frac{\xi _1+\lambda _0y}{\lambda _0}\Bigr ) \Bigr ] \chi _0\Bigl ( \frac{\lambda _0 y }{\sqrt{t}}\Bigr ) \Bigr \Vert _{0,\nu ,6+\sigma ,\varepsilon } \le \frac{C}{t_0^\vartheta } \Vert \alpha _1\Vert _{C^1,\nu +\frac{1}{2},\Theta } , \end{aligned}$$

for some \(\vartheta >0\). The last term in the expression (6.19) is similar.

The terms in \( \lambda ^4 [ S_0({{\textbf {p}}}_0 + {{\textbf {p}}}_1) - S_0({{\textbf {p}}}_0)]\) that contain the function \(\varphi _{\lambda _0}\) are

$$\begin{aligned}&{\lambda _0}^4 \Bigl [ - \frac{4}{r} \partial _r \varphi _{\lambda _0} - \nabla \cdot ( \varphi _{\lambda _0} \nabla v_0) - \nabla \cdot ( u_0 \nabla \psi _{\lambda _0} ) - \nabla \cdot ( \varphi _{\lambda _0} \nabla \psi _{\lambda _0}) \Bigr ] \\&\quad = 4 \frac{{\lambda _0}^2}{\rho (\rho ^2+1)} \partial _\rho \varphi _{\lambda _0} {-}(\alpha -1) {\lambda _0}^2 \nabla _y \varphi _{\lambda _0} \cdot \nabla _y \Gamma _0 {+} {\lambda _0}^2 \nabla _y \varphi _{\lambda _0} \cdot \nabla _y {\mathcal {R}} {+} 2 {\lambda _0}^2 U \chi \varphi _{\lambda _0} \\&\qquad - \alpha \nabla _y (U \chi ) \cdot \nabla _y \psi _{\lambda _0} - {\lambda _0}^2 \nabla _y ( \varphi _{\lambda _0} \nabla _y \psi _{\lambda _0}) . \end{aligned}$$

In \( {\lambda _0}^4 [ S_0({{\textbf {p}}}_0 + {{\textbf {p}}}_1) - S_0({{\textbf {p}}}_0)]\) these terms appear evaluated at y and then at \(\frac{\xi _1}{\lambda _0}+y\). Using estimates for the the second derivative of \(\varphi _{\lambda _0}\) similar to Lemma 4.1 and assuming

$$\begin{aligned} \sigma<1, \quad \nu <1+\gamma , \end{aligned}$$
(6.20)

we get

$$\begin{aligned} \Vert \lambda ^4 [ S_0({{\textbf {p}}}_0 + {{\textbf {p}}}_1) - S_0({{\textbf {p}}}_0)] \Vert _{0,\nu ,6+\sigma ,\varepsilon } \le C \frac{1}{t_0^\vartheta } \Vert \vec {\phi }\Vert _{X}. \end{aligned}$$

The main term in \(F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \) that depends on the outer solution is \( \lambda ^2 U \varphi ^o \) with \(\varphi ^o = \varphi ^* + \varphi \) defined in (5.38). Then we have

$$\begin{aligned} | \lambda ^2 U \varphi (y,t) {\tilde{\chi }}|&\le \frac{\lambda ^2}{t^{a-1} (\log t)^\beta } \frac{1}{(1+|y|)^4} {\tilde{\chi }} \Vert \varphi \Vert _{*,o} \\&\le C \frac{1}{t^{\nu +\frac{3}{2}} (\log t)^{\beta +1} } \frac{1}{(1+|y|)^{4}} {\tilde{\chi }} \Vert \varphi \Vert _{*,o} \\&\le C \frac{(t \log t)^{1+\frac{\sigma }{2}}}{t^{\nu + \frac{3}{2}} (\log t)^{\beta +1} } \frac{1}{(1+|y|)^{6+\sigma }} {\tilde{\chi }} \Vert \varphi \Vert _{*,o} \\&\le C \frac{1}{t^{\nu + \frac{1-\sigma }{2}} (\log t)^{\beta - \frac{\sigma }{2}} } \frac{1}{(1+|y|)^{6+\sigma }} {\tilde{\chi }} \Vert \varphi \Vert _{*,o}. \end{aligned}$$

Therefore

$$\begin{aligned} \Vert \lambda ^2 U \varphi {\tilde{\chi }} \Vert _{0,\nu ,6+\sigma ,\varepsilon } \le C \frac{1}{t_0^{\frac{1-\sigma }{2}}(\log t_0)^{\beta -\frac{\sigma }{2}}} \Vert \varphi \Vert _{*,o} . \end{aligned}$$

Regarding the function \(\varphi ^*\) (c.f. (5.37)) we note that it has the estimate

$$\begin{aligned} |\varphi ^*(x,t)| \le t_0^{a-1} (\log t_0)^\beta \Vert \varphi _0^* \Vert _{*,b} \frac{1}{t^{a-1} (\log t)^\beta } \frac{1}{ 1+ |\zeta |^{b}} , \quad \zeta = \frac{x-\xi }{\sqrt{t}} \end{aligned}$$
(6.21)

by Proposition 6.4, provided (6.9) holds, and therefore

$$\begin{aligned} \Vert \lambda ^2 U \varphi ^* {\tilde{\chi }} \Vert _{0,\nu ,6+\sigma ,\varepsilon } \le C t_0^{\nu +1+\frac{\sigma }{2}}(\log t_0)^{\frac{\sigma }{2}} \Vert \varphi _0^* \Vert _{*,b} . \end{aligned}$$

Let us analyze some of the terms in \(F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \) that depend on the inner solutions \(\phi _1\) and \(\phi _2\). For instance

$$\begin{aligned} (\alpha -1) \nabla _y \cdot ( \phi _j \nabla _y \Gamma _0) = (\alpha -1) \nabla _y \phi _j \cdot \nabla _y \Gamma _0 - (\alpha -1) \phi _j U . \end{aligned}$$

We have the estimate

$$\begin{aligned} |(\alpha -1) \nabla _y \phi _j \cdot \nabla _y \Gamma _0 {\tilde{\chi }}|&\le \frac{C}{t \log t} \frac{1}{t^{\nu -\frac{1}{2}} (\log t)^{\frac{q-1}{2}}} \frac{1}{(1+|y|)^6} \Vert \phi _j \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \\&\le C \frac{1}{t^{\nu +\frac{1}{2}-\frac{\sigma }{2}} (\log t)^{1+\frac{q-1}{2}-\frac{\sigma }{2}}} \frac{1}{(1+|y|)^{6+\sigma }} \Vert \phi _j \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } , \end{aligned}$$

and we get

$$\begin{aligned} \Vert (\alpha -1) \nabla _y \phi _j \cdot \nabla _y \Gamma _0 {\tilde{\chi }}\Vert _{0,\nu ,6+\sigma ,\varepsilon } \le \frac{C}{t_0^\vartheta }\Vert \phi _j \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } , \end{aligned}$$

for some \(\vartheta >0\).

We also have, writing \(\phi = \phi _1+\phi _2\),

$$\begin{aligned} \Vert \lambda {\dot{\xi }}_1 \nabla \phi {\tilde{\chi }}\Vert _{0,\nu ,6+\sigma ,\varepsilon } \le \frac{C}{t_0^\vartheta } \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \end{aligned}$$

for some \(\vartheta >0\), if

$$\begin{aligned} \gamma > \frac{\sigma }{2}. \end{aligned}$$
(6.22)

Let us estimate the term \( \nabla _y \cdot ( U \nabla _y( {\hat{\psi }} - \psi )) {\tilde{\chi }} \) appearing in (5.6), where \( {\hat{\psi }} = (-\Delta )^{-1} (\lambda ^{-2} \phi ^i \chi )\), \( \psi = (-\Delta )^{-1} (\lambda ^{-2} \phi ^i )\). We recall that \(\phi ^i = \phi ^i_0 + \phi \), c.f. (5.38), and therefore we can decompose \({\hat{\psi }} = {\hat{\psi }}^i_0 + {\hat{\psi }}_1\) where \({\hat{\psi }}^i_0 = ( -\Delta )^{-1}( \lambda ^{-2} \phi ^i_0 \chi )\) and \({\hat{\psi }}_1 = (-\Delta )^{-1} ( \lambda ^{-2} \phi \chi )\). Similarly, we can decompose \(\psi = \psi ^i_0 + \psi _1\) where \(\psi ^i_0 = ( -\Delta )^{-1}( \lambda ^{-2} \phi ^i_0 )\) and \(\psi _1 = (-\Delta )^{-2} ( \lambda ^{-1} \phi )\). By linearity we need to estimate separately \( \nabla _y \cdot ( U \nabla _y( {\hat{\psi }}^i_0 - \psi ^i_0)) \) and \(\nabla _y \cdot ( U \nabla _y( {\hat{\psi }}_1 - \psi _1)) \). Let us consider the latter one. Note that

$$\begin{aligned} {\hat{\psi }}_1 - \psi _1 = (-\Delta )^{-1}[\lambda ^{-2} \phi (1-\chi )]. \end{aligned}$$

From the definition of the norm \(\Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }\)

$$\begin{aligned} |\phi (y,t)|\le \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \frac{1}{t^{\nu -\frac{1}{2}} (\log t)^{\frac{q-1}{2}}} \frac{ 1 }{(1+|y|)^{4}} \end{aligned}$$
(6.23)

and so

$$\begin{aligned} |\nabla _y( {\hat{\psi }}_1-\psi _1)(y,t)|\le & {} C \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \frac{1}{t^{\nu -\frac{1}{2}} (\log t)^{\frac{q-1}{2}}} \frac{1}{(t\log t)^{\frac{3}{2}}},\\{} & {} \quad \text {for } |y|\le 2 \frac{\sqrt{t}}{\lambda }. \end{aligned}$$

Then

$$\begin{aligned} | \nabla _y U \cdot \nabla _y( {\hat{\psi }}_1 - \psi _1)) (y,t)|&\le C \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }\\&\quad \frac{1}{t^{\nu + \frac{1-\sigma }{2}} (\log t)^{\frac{q+1-\sigma }{2}}} \frac{1}{(1+|y|)^{6+\sigma }} , \quad \text {for } |y|\le 2 \frac{\sqrt{t}}{\lambda }. \end{aligned}$$

This and a similar estimate for \(U \phi ( 1-\chi ) \) give

$$\begin{aligned} \Vert \nabla _y \cdot ( U \nabla _y({\hat{\psi }}_1 - \psi _1)) {\tilde{\chi }} \Vert _{0,\nu ,6+\sigma ,\varepsilon } \le \frac{C}{t_0^{\vartheta }} \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \end{aligned}$$

for some \(\vartheta >0\). A similar estimate is obtained for \(\Vert \nabla _y \cdot ( U \nabla _y({\hat{\psi }}^i_0 - \psi ^i_0)) {\tilde{\chi }} \Vert _{0,\nu ,6+\sigma ,\varepsilon }\) using (5.36).

Let us estimate next the term \(\lambda ^2 \nabla _y \cdot (\varphi _\lambda \nabla _y \psi ) {\tilde{\chi }}\), where we recall, \(\psi = (-\Delta )^{-1} (\lambda ^{-2} \phi )\). To do this we use that \(\phi = \phi _1+\phi _2\) has zero mass and center of mass, that is,

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi (y,t)\,dy = \int _{{\mathbb {R}}^2} \phi (y,t)y_j\,dy= 0, \quad t>t_0. \end{aligned}$$

This and the estimate (6.23) imply

$$\begin{aligned} |\psi (y,t)| + (1+|y|) |\nabla _y\psi (y,t)| \le C \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \frac{1}{t^{\nu -\frac{1}{2}} (\log t)^{\frac{q-1}{2}}} \frac{ \log (2+|y|) }{(1+|y|)^2} , \end{aligned}$$

by an argument similar to Remark 9.1. On the other hand, from (4.3)

$$\begin{aligned} |\nabla _y \varphi _\lambda (y,t)|\le \frac{C }{t \log t} \frac{1}{(1+|y|)^3}, \quad |y|\le 2 \frac{\sqrt{t}}{\lambda }. \end{aligned}$$

Therefore

$$\begin{aligned} |\lambda ^2 (\nabla _y \varphi _\lambda \cdot \nabla _y \psi )(y,t)|&\le C\Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \frac{\lambda ^2}{t^{\nu +\frac{1}{2}} (\log t)^{\frac{q+1}{2}}} \frac{ \log (2+|y|) }{(1+|y|)^6} \\&\le C \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \frac{1}{t^{\nu +\frac{1-\sigma }{2}} (\log t)^{\frac{q+1-\sigma }{2}}} \frac{ 1 }{(1+|y|)^{6+\sigma }} , \quad \\&\quad |y|\le 2 \frac{\sqrt{t}}{\lambda }. \end{aligned}$$

From this coupled with a similar estimate for \(\lambda ^2 \varphi _\lambda \phi \) we get

$$\begin{aligned} \Vert \lambda ^2 \nabla _y \cdot (\varphi _\lambda \nabla _y \psi ) {\tilde{\chi }} \Vert _{0,\nu ,6+\sigma ,\varepsilon } \le \frac{C}{t_0^{\vartheta }} \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \end{aligned}$$

for some \(\vartheta >0\).

The remaining terms in \(F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \) are estimated in a similar way and we get the validity (6.17).

Proceeding in the same way we get a Lipschitz bound. Assuming \( t_0^{a-1} (\log t_0)^\beta \Vert \varphi _0^* \Vert _{*,b}\le 1 \), for \(\Vert \vec {\phi }_1\Vert _X \le 1\) and \(\Vert \vec {\phi }_2\Vert _X\le 1\) we have

$$\begin{aligned} \Vert {\mathcal {A}}_{i1}[\vec {\phi }_1,\varphi _0^*]-{\mathcal {A}}_{i1}[\vec {\phi }_2,\varphi _0^*] \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \le \frac{C}{t_0^\vartheta } \Vert \vec {\phi }_1 - \vec {\phi }_2\Vert _{X}, \end{aligned}$$

for some \(\vartheta >0\) small, a constant C independent of \(t_0\), and \(t_0\) sufficiently large. Indeed, the Lipschitz estimate with respect to \(\phi _1\), \(\phi _2\), and \(\varphi \) is direct from the explicit dependence of \(F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \) on these variables, which is either linear or quadratic. The Lipschitz dependence on \(\xi _1\) (where \({{\textbf {p}}}_1 = ( \alpha _1,\xi _1))\) is also direct from the explicit form of \(F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \). The Lipschitz condition with respect to \(\alpha _1\) appears as an explicit dependence on this variable in \(F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) \).

Let us estimate the operator \({\mathcal {A}}_{i2}\). We claim that if \( t_0^{a-1} (\log t_0)^\beta \Vert \varphi _0^* \Vert _{*,b}\le 1 \) and \(\Vert \vec {\phi }\Vert _X\le 1\), then

$$\begin{aligned} \Vert{} & {} {\mathcal {A}}_{i2}[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*] \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }\nonumber \\{} & {} \quad \le C ( \log t_0)^{-\frac{1-q}{2}-\Theta } +C t_0^{a-1} (\log t_0)^{\frac{1-q}{2}} \Vert \varphi _0^*\Vert _{*,b}. \end{aligned}$$
(6.24)

Indeed, we apply Proposition 6.1 to get

$$\begin{aligned}&\Vert {\mathcal {A}}_{i2}[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*] \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \\&\quad \le \frac{C}{(\log t_0)^{1-q}} \left\| m_2[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] W_2 \right\| _{0,\nu +\frac{1}{2},\frac{1-q}{2},6+\sigma ,\varepsilon } \end{aligned}$$

and since \(W_2\) has compact support,

$$\begin{aligned}&\Vert {\mathcal {A}}_{i2}[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*] \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \\&\quad \le \frac{C}{(\log t_0)^{1-q}} \sup _{t>t_0} t^{\nu +\frac{1}{2}} (\log t)^{\frac{1-q}{2}} | m_2[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] (t)|. \end{aligned}$$

Using the definition of \(F_3\) (5.51)

$$\begin{aligned} m_2[ F_3(\phi ,\varphi ,{{\textbf {p}}}_1 ,\varphi _0^*) ] = m_2[ E_2 {\tilde{\chi }}_2] + m_2[F_2(\phi ,\varphi ,{{\textbf {p}}}_1 ,\varphi _0^*) {\tilde{\chi }} ] \end{aligned}$$

We have by (5.43) (assuming \(\sigma <\frac{1}{2}\)),

$$\begin{aligned} |m_2[ E_2 {\tilde{\chi }}_2] (t) | \le \frac{C}{t^{\frac{3+\sigma }{2}}}. \end{aligned}$$

Therefore, asking that

$$\begin{aligned} \nu + \frac{1}{2}<\frac{3+\sigma }{2} \Leftrightarrow \nu < 1+ \frac{\sigma }{2} \end{aligned}$$
(6.25)

we get

$$\begin{aligned} \sup _{t>t_0} t^{\nu +\frac{1}{2}} (\log t)^{\frac{1-q}{2}} |m_2[ E_2 {\tilde{\chi }}_2] (t) | \le \frac{C}{t_0^\vartheta }, \end{aligned}$$

for some \(\vartheta >0\).

By (5.41)

$$\begin{aligned} m_2[F_2 (\phi ,\varphi ,{{\textbf {p}}}_1,\varphi ^*_0) {\tilde{\chi }} ]&= \lambda ^4 m_2[S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ] \\&\quad + m_2[ F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) {\tilde{\chi }}] \\&\quad + \lambda ^4 m_2[ ( S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ) ({\tilde{\chi }}-1) ]. \end{aligned}$$

Of these terms, the largest is the first one. By Lemma 5.4, and since \(\lambda = \lambda _0\), we get

$$\begin{aligned}&\lambda ^4 m_2 [S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ] \nonumber \\&\quad = - 32\pi \alpha _1 - \frac{{\dot{\alpha }}_1}{\lambda _0^2} \int _{{\mathbb {R}}^2} U(\frac{x-\xi }{\lambda _0}) \chi _0(\frac{x-\xi }{\lambda _0}) |x-\xi |^2\,dx \nonumber \\&\qquad + \alpha _1 \int _{{\mathbb {R}}^2} E(x-\xi ,t,\lambda _0)|x-\xi |^2\, \textrm{d}x - |\xi |^2 \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}}_0))dx . \end{aligned}$$
(6.26)

But

$$\begin{aligned} \sup _{t>t_0} t^{\nu +\frac{1}{2}} (\log t)^{\frac{1-q}{2}} |\alpha _1(t)|\le C (\log t_0)^{\frac{1-q}{2}-\Theta } \Vert \alpha _1\Vert _{C^1,\nu +\frac{1}{2},\Theta } , \end{aligned}$$
(6.27)

under the assumption

$$\begin{aligned} \Theta > \frac{1-q}{2}. \end{aligned}$$
(6.28)

The second term in (6.26) is much smaller. For the last term in (6.26) we use Lemma 5.3 and (5.29), (5.30) to get

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}}_0))dx \right| \le \frac{C}{t^2} \end{aligned}$$
(6.29)

and therefore

$$\begin{aligned} |\xi (t)|^2 \left| \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}}_0))dx \right| \le \frac{C}{t^{2+2\gamma }} \Vert \xi _1\Vert _{C^1,\gamma ,0}^2 . \end{aligned}$$

Combining (6.26), (6.27) and (6.29) we get

$$\begin{aligned}&\frac{C}{(\log t_0)^{1-q}} \sup _{t>t_0} t^{\nu +\frac{1}{2}} (\log t)^{\frac{1-q}{2}} \lambda ^4 |m_2[S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) \\&\quad - S_0( u_1( {{\textbf {p}}}_0) ) ] (t)| \le C ( \log t_0)^{-\frac{1-q}{2}-\Theta } \Vert {\textbf{p}}_1\Vert _{X_p}. \end{aligned}$$

Let’s estimate the remaining terms in \(m_2[ F_3(\phi ,\varphi ,{{\textbf {p}}}_1,\varphi _0^*) ]\). Consider

$$\begin{aligned} A(t):= \int _{{\mathbb {R}}^2 } \nabla _y \cdot ( \lambda ^2 \varphi _\lambda \nabla _y (-\Delta _y)^{-1} \phi ) {\tilde{\chi }} |y|^2 dy + \int _{{\mathbb {R}}^2 } \nabla _y \cdot (\phi \nabla _y \psi _\lambda ) {\tilde{\chi }} |y|^2 dy \end{aligned}$$

which appears in the definition of F, where \(\phi = \phi _1 + \phi _2\). Let us recall that \(\psi _\lambda =(-\Delta _x)^{-1}\varphi _\lambda \) and let’s write

$$\begin{aligned} \psi = (-\Delta _y)^{-1}\phi \end{aligned}$$

Integrating by parts,

$$\begin{aligned} A(t)&=\int _{{\mathbb {R}}^2}(\Delta _y\psi _\lambda \nabla _y\psi +\Delta _y\psi \nabla _y\psi _\lambda )\cdot y(2{\tilde{\chi }}+y\cdot \nabla _y {\tilde{\chi }})dy. \end{aligned}$$

Using the Pohozaev type identity

$$\begin{aligned}&\Delta _y\psi _\lambda (\nabla _y\psi \cdot y)+\Delta _y\psi (\nabla _y\psi _\lambda \cdot y) \\&\quad =\nabla _y\cdot [\nabla _y\psi _\lambda (\nabla _y\psi \cdot y)+\nabla _y\psi (\nabla _y\psi _\lambda \cdot y)-y\nabla _y\psi _\lambda \cdot \nabla _y\psi ] \end{aligned}$$

and integrating by parts, we get

$$\begin{aligned} A(t)&=-\int _{{\mathbb {R}}^2}[\nabla _y\psi _\lambda (\nabla _y\psi \cdot y)+\nabla _y\psi (\nabla _y\psi _\lambda \cdot y)\\&\quad -y\nabla _y\psi _\lambda \cdot \nabla _y\psi ]\cdot [2\nabla _y{\tilde{\chi }}+y\Delta _y{\tilde{\chi }}]dy . \end{aligned}$$

Therefore

$$\begin{aligned} |A(t)|&\le C\int _{2 \sqrt{t}/\lambda \le |y| \le 4 \sqrt{t}/\lambda }|\nabla _y\psi |\,\nabla \psi |dy. \end{aligned}$$

Using that \(\psi = (-\Delta )^{-1} \phi \), and

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi (y,t)dy=0,\quad \int _{{\mathbb {R}}^2} \phi (y,t)ydy=0, \end{aligned}$$

we have (see Remark 9.1) for any \(\varrho >0\) small,

$$\begin{aligned} |\nabla \psi (y,t)| \le \frac{C}{1+|y|^{2-\varrho }} \frac{1}{t^{\nu -\frac{1}{2}}(\log t)^{\frac{q-1}{2}}} \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }. \end{aligned}$$

Using that \(\varphi _\lambda \), and \(\psi _\lambda \) are radial and

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} \varphi _\lambda dx \right| \le \frac{C}{t \log t}, \end{aligned}$$

by Lemma 4.1 we have

$$\begin{aligned} |\psi _\lambda (y,t)|\,\leqq \, \frac{C}{t \sqrt{ \log t}} \frac{1}{1+|y|}. \end{aligned}$$

Then

$$\begin{aligned} |A(t)|\le C \frac{1}{t^{\nu +1-\frac{\varrho }{2}} (\log t)^{\frac{q}{2}-1-\frac{\varrho }{2}}}\Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }. \end{aligned}$$

Let us consider the contribution of the term \(\lambda ^2 U \varphi ^*\). Thanks to (6.21)

$$\begin{aligned} \Vert m_2[\lambda ^2 U \varphi ^* {\tilde{\chi }}] W_2 \Vert _{0,\nu +\frac{1}{2},\frac{1-q}{2},6+\sigma ,\varepsilon } \le C t_0^{a-1} (\log t_0)^{\frac{1-q}{2}} \Vert \varphi _0^*\Vert _{*,b}, \end{aligned}$$

under the condition

$$\begin{aligned} \beta >\frac{1-q}{2}. \end{aligned}$$
(6.30)

The other terms in \(m_2\) are estimated in a similar way and we get (6.24).

Similarly we get that if \( t_0^{a-1} (\log t_0)^\beta \Vert \varphi _0^* \Vert _{*,b}\le 1 \), then for \(\Vert \vec {\phi }_1\Vert _X \le 1\) and \(\Vert \vec {\phi }_2\Vert _X\le 1\) we have

$$\begin{aligned} \Vert {\mathcal {A}}_{i2}[\vec {\phi }_1]- {\mathcal {A}}_{i2}[\vec {\phi }_2] \Vert _{0,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \le C ( \log t_0)^{-\frac{1-q}{2}-\Theta } \Vert \vec {\phi }_1 - \vec {\phi }_2\Vert _{X}, \end{aligned}$$

for a constant C independent of \(t_0\), where \(t_0\) sufficiently large.

Let us estimate the operator \( {\mathcal {A}}_o[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*]\). We claim that if \( t_0^{a-1} (\log t_0)^\beta \Vert \varphi _0^* \Vert _{*,b}\le 1 \), then for \(\Vert \vec {\phi }\Vert _X\le 1\),

$$\begin{aligned} \Vert {\mathcal {A}}_o[\vec {\phi },\varphi _0^*] \Vert _{*,o} \le \frac{C}{(\log t_0)^{\frac{q+1}{2}-\beta }} + C t_0^{a-2} (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b}, \end{aligned}$$
(6.31)

and for \(\Vert \vec {\phi }_1\Vert _X\le 1\), \(\Vert \vec {\phi }_2\Vert _X\le 1\) and \( t_0^{a-1} (\log t_0)^\beta \Vert \varphi _0^* \Vert _{*,b}\le 1 \),

$$\begin{aligned} \Vert {\mathcal {A}}_o[\vec {\phi }_1,\varphi _0^*] - {\mathcal {A}}_o[\vec {\phi }_2,\varphi _0^*] \Vert _{*,o} \le \frac{C}{(\log t_0)^{\frac{q+1}{2}-\beta }}\Vert \vec {\phi }_1 - \vec {\phi }_2\Vert _X. \end{aligned}$$

Note that \(\frac{q+1}{2}-\beta >0\) by (6.14).

Indeed, by Proposition 6.3

$$\begin{aligned} \Vert {\mathcal {A}}_o[\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1, \varphi _0^* ] \Vert _{*,o} \le C \Vert G_2(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1, \varphi _0^* ) \Vert _{ **,o} , \end{aligned}$$

where we recall \(G_2\) defined in (5.42).

We start with the term \( \lambda ^{-4} E_2 (1-{\tilde{\chi }}_2)\chi \). Using the estimate (5.43) we get

$$\begin{aligned} \Vert \lambda ^{-4} E_2 (1-{\tilde{\chi }}_2)\chi \Vert _{**,o} \le \frac{C}{t_0^\vartheta } \end{aligned}$$

for some \(\vartheta >0\) provided

$$\begin{aligned} a< 4 (1-\delta ). \end{aligned}$$

We also directly get from (4.6)

$$\begin{aligned} \Vert S(u_1) (1-\chi ) \Vert _{**,o} \le \frac{C}{t_0^\vartheta } \end{aligned}$$

for some \(\vartheta >0\) if \(a<4\).

Regarding the terms in G (c.f. (5.10)) that the depend linearly on \(\phi ^i = \phi ^i_0 + \phi \), we have for \(\Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } < \infty \)

$$\begin{aligned} \Bigl | \frac{1}{\lambda ^2} \phi \Delta \chi \Bigr |(x,t)&\le \frac{C}{\lambda ^2} \frac{1}{t^{\nu -\frac{1}{2}} (\log t)^{\frac{q-1}{2}}} \frac{1}{( |x-\xi |/\lambda |) ^4} \frac{1}{t}\nonumber \\&\Bigl |\Delta _z \chi _0( \frac{x-\xi }{\sqrt{t}}) \Bigr | \Vert \phi \Vert _{1,\nu -\frac{1}{2} ,\frac{q-1}{2},4,2+\sigma +\varepsilon } \nonumber \\&\le C \frac{1}{t^{\nu +\frac{5}{2}} (\log t)^{\frac{q+1}{2}}} \frac{1}{(1+|x-\xi |/\sqrt{t})^b} \Vert \phi \Vert _{1,\nu -\frac{1}{2} ,\frac{q-1}{2},4,2+\sigma +\varepsilon } \end{aligned}$$
(6.32)

which implies

$$\begin{aligned} \Bigl \Vert \frac{1}{\lambda ^2} \phi \Delta \chi \Bigr \Vert _{**,o} \le \frac{C}{(\log t_0)^{\frac{q+1}{2}-\beta }} \Vert \phi \Vert _{1,\nu -\frac{1}{2} ,\frac{q-1}{2},4,2+\sigma +\varepsilon }, \end{aligned}$$

since \(\beta <\frac{q+1}{2}\), which is one of the conditions in (6.14).

We also have, using (5.36),

$$\begin{aligned} \Bigl \Vert \frac{1}{\lambda ^2} \phi _0^i \Delta \chi \Bigr \Vert _{**,o} \le \frac{C}{t_0^\vartheta } \end{aligned}$$

for some \(\vartheta >0\) if

$$\begin{aligned} a<4. \end{aligned}$$

A similar estimate holds for the other terms depending on \(\phi ^i\).

Some of the terms in G that depend on \(\varphi ^o = \varphi ^* + \varphi \) are

$$\begin{aligned} \Bigl | \frac{1}{\lambda ^2} U \varphi ^o (1-\chi ) \Bigr |&\le C \frac{\lambda ^2}{|x-\xi |^4} \frac{1}{t^{a-1} (\log t)^\beta } \frac{1}{(1+|x-\xi |/\sqrt{t})^b} (1-\chi ) \Vert \varphi ^o\Vert _{*,o} \\&\le \frac{C}{t_0 \log t_0} \frac{1}{t^a (\log t)^\beta } \frac{1}{(1+|x-\xi |/\sqrt{t})^b} ( \Vert \varphi ^*\Vert _{*,o} + \Vert \varphi \Vert _{*,o} ) \end{aligned}$$

which implies that

$$\begin{aligned} \Bigl \Vert \frac{1}{\lambda ^2} U \varphi ^o (1-\chi ) \Bigr \Vert _{**,o} \le \frac{C}{t_0 \log t_0} \Vert \varphi \Vert _{*,o} + C t_0^{a-2} (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b}, \end{aligned}$$

by Proposition 6.4. Other terms are estimated in a similar way.

Let us estimate the operator \({\mathcal {A}}_p\), which is defined by the equations (6.13). We claim that if

$$\begin{aligned} (0,{\tilde{\alpha }}_1, {\tilde{\xi }}_1) = {\mathcal {A}}_p [\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1] \end{aligned}$$

and \( t_0^{a-1} (\log t_0)^\beta \Vert \varphi _0^* \Vert _{*,b}\le 1\), \(\Vert \vec {\phi }\Vert _X \le 1\), \(\vec {\phi } = (\phi _1,\phi _2,\varphi ,{\textbf{p}}_1)\), then

$$\begin{aligned} \Vert {\tilde{\alpha }}_1 \Vert _{C^1,\nu +\frac{1}{2},\Theta }&\le C ( \log t_0)^{\Theta -\beta } + C t_0^{a-1} (\log t_0)^{ \Theta } \Vert \varphi _0^*\Vert _{*,b} \nonumber \\ \Vert {\tilde{\xi }}_1 \Vert _{*,\gamma ,0}&\le \frac{C}{t_0^\vartheta } + C t_0^{1+\gamma } (\log t_0)^{\frac{1}{2}}\Vert \varphi _0^*\Vert _{*,b}, \end{aligned}$$
(6.33)

for some \(\vartheta >0\). Similarly, we have the following Lipschitz estimate. If \( t_0^{a-1} (\log t_0)^\beta \Vert \varphi _0^* \Vert _{*,b}\le 1\), then for some \(\vartheta >0\), and for \(\Vert \vec {\phi }_1\Vert _X\le 1\), \(\Vert \vec {\phi }_2\Vert _X\le 1\),

$$\begin{aligned} \Vert {\mathcal {A}}_p[\vec {\phi }_1,\varphi _0^*] - {\mathcal {A}}_p[\vec {\phi }_2,\varphi _0^*] \Vert _{X_p} \le C ( \log t_0)^{\Theta -\beta } \Vert \vec {\phi }_1 - \vec {\phi }_2\Vert _X , \end{aligned}$$
(6.34)

for some \(\vartheta >0\).

Indeed, by (6.11)

$$\begin{aligned} |{\mathcal {A}}_{\alpha _1} [\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*](t)| \le |I_1(t)| + |I_2(t)| + |I_3(t)| \end{aligned}$$

where

$$\begin{aligned} I_1(t)&= \int _t^\infty \frac{1}{\lambda _0^2} m_0[ E_2 {\tilde{\chi }}_2] (s)ds \\ I_2(t)&= \int _t^\infty \frac{1}{\lambda _0^2} m_0[ F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) {\tilde{\chi }}](s)ds \\ I_3(t)&= \int _t^\infty \lambda _0^2 m_0[ ( S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ) ({\tilde{\chi }}-1) ] (s) ds. \end{aligned}$$

Using (5.43) and \(\int _{{\mathbb {R}}^2} E_2 dy=0\) we get

$$\begin{aligned} \left| \frac{1}{\lambda _0^2 } m_0[E_2{\tilde{\chi }}_2](t) \right| \le C\frac{1}{t^{3-2\delta }}. \end{aligned}$$

This gives

$$\begin{aligned} \Vert I_1 \Vert _{C^1,\nu +\frac{1}{2},\Theta } \le C t_0^{\nu -\frac{3}{2}+2\delta }, \end{aligned}$$
(6.35)

under the assumption

$$\begin{aligned} \nu <\frac{3}{2}-2\delta . \end{aligned}$$

The largest contribution in \(I_2\) comes from the term \(\lambda ^2 U \varphi ^o\) in \(F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) \) (c.f. (5.6)). The estimate of this term is

$$\begin{aligned} \left| \frac{1}{\lambda _0^2(t)} \int _{{\mathbb {R}}^2} \lambda _0(t)^2 U(y) \varphi ^o (y,t) dy \right|&\le C \frac{1}{t^{\nu +\frac{3}{2}} (\log t)^\beta } \Vert \varphi ^o\Vert _{*,o} \end{aligned}$$
(6.36)

and so

$$\begin{aligned} \Bigl \Vert \int _{{\mathbb {R}}^2} U(y) \varphi ^o (y,t) dy\Bigr \Vert _{C^1,\nu +\frac{1}{2},\Theta } \le C ( \log t_0)^{\Theta -\beta } \Vert \varphi ^o\Vert _{*,o} , \end{aligned}$$

under the assumption

$$\begin{aligned} \Theta <\beta . \end{aligned}$$
(6.37)

Similar estimates for the remaining terms give

$$\begin{aligned} \Vert I_2 \Vert _{C^1,\nu +\frac{1}{2},\Theta } \le C ( \log t_0)^{\Theta -\beta } \Vert \vec {\phi }\Vert _{X} + C t_0^{a-1} (\log t_0)^{ \Theta } \Vert \varphi _0^*\Vert _{*,b}. \end{aligned}$$
(6.38)

Regarding \(I_3\), using (4.5) we have

$$\begin{aligned} \lambda _0^2 m_0[S_0(u_1({\textbf{p}}) ) ({\tilde{\chi }}-1)]&\le \frac{C}{t^3 \log t }. \end{aligned}$$
(6.39)

Putting together (6.35), (6.38), and (6.39) we get

$$\begin{aligned} \Vert {\mathcal {A}}_{\alpha _1} [\phi _1,\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^* ] \Vert _{C^1,\nu +\frac{1}{2},\Theta } \le C ( \log t_0)^{\Theta -\beta } \Vert \vec {\phi }\Vert _{X} + C t_0^{a-1} (\log t_0)^{ \Theta } \Vert \varphi _0^*\Vert _{*,b} \end{aligned}$$

assuming also that

$$\begin{aligned} \nu <\frac{3}{2}. \end{aligned}$$

The computations leading to (6.33) are very similar, under the assumption

$$\begin{aligned} \gamma < \nu -\frac{1}{2}. \end{aligned}$$
(6.40)

This restriction arises when considering the largest term in the expression (6.12), namely comes from estimating the term \(\lambda _0^2 m_{1,j}[\varphi _{\lambda _0} \phi {\tilde{\chi }}] \) (\(\lambda _0^2 \varphi _{\lambda _0} \phi \) is one of the terms in (5.6))

$$\begin{aligned} \frac{1}{\lambda _0} \lambda _0^2 |m_{1,j}[\varphi _{\lambda _0} \phi {\tilde{\chi }}](t)|&\le C \lambda _0 \int _{{\mathbb {R}}^2} |\varphi _{\lambda _0} \phi y_j| dy \\&\le C \lambda _0 \frac{1}{t (\log t)^2} \frac{1}{t^{\nu -\frac{1}{2}} (\log t)^{\frac{q-1}{2}}} \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \end{aligned}$$

Let us summarize the restrictions on the parameters. We let \(0<q<1\) be fixed. We take

$$\begin{aligned} 0<\delta<\sigma <\min (1,4\delta ), \end{aligned}$$

and

$$\begin{aligned} 1<\nu < \min \Bigl ( 1+2\delta - \frac{\sigma }{2} , \frac{3}{2} , 1+\gamma , 1+\frac{\sigma }{2}\Bigr ). \end{aligned}$$

because of (6.18), (6.20), (6.25). We also need

$$\begin{aligned} \frac{1-q}{2}< \Theta< \beta < \frac{1+q}{2} \end{aligned}$$

by (6.28), (6.37) and by (6.30) and (6.14). We take

$$\begin{aligned} \frac{\sigma }{2}<\gamma < \nu -\frac{1}{2} \end{aligned}$$

by (6.22) and (6.40).

Together with the above inequalities we want also the relations \( \sigma +\varepsilon <2\), \(\nu +\frac{1}{2}<\frac{7}{4}\) for Proposition 6.1 and \(\sigma +\varepsilon <\frac{3}{2}\), \(\nu <\min ( 1 + \frac{\varepsilon }{2}, 3-\frac{\sigma }{2},\frac{5}{4}) \) for Proposition 6.2. The condition (6.9) for Propositions 6.3 and 6.4 hold by (6.14). We see that all these restrictions are satisfied by choosing first \(\delta \), \(\sigma >0\) small so that \(2\delta - \frac{\sigma }{2}>0\). Then we take \(\nu >1\) close to 1, then let \(a=\nu +\frac{5}{2}\) and b satisfying (6.14). Then \(\Theta \), \(\beta \) and \(\gamma \) can be selected. Note that with the above procedure we are getting the restriction \(b>5\).

We already have all elements to solve the fixed point problem (6.16), which we recall

$$\begin{aligned} \vec {\phi } = {\mathcal {A}} [ \vec {\phi } ], \quad \vec {\phi }\in {\mathcal {B}} , \end{aligned}$$

where \({\mathcal {B}}\) is the closed unit ball in the Banach space of functions \(\vec {\phi } \) with \(\Vert \vec {\phi }\Vert _X < +\infty \) and the norm defined in (6.15). Thus

$$\begin{aligned} {\mathcal {B}} = \{ \, \vec {\phi } \in X \, | \, \Vert \vec {\phi }\Vert _X \leqq 1 \, \}. \end{aligned}$$

Let \(\varphi _0^*\) be such that \(t_0^{a-1} (\log t_0)^{\beta } \Vert \varphi _0^*\Vert _{*,b}\le 1\). Estimates (6.17), (6.24), (6.31) and (6.34), imply that, enlarging the parameter \(t_0\) if necessary, \({\mathcal {A}}\) maps \({\mathcal {B}}\) into itself. We also get that \({\mathcal {A}}\) is a contraction mapping on \({\mathcal {B}}\). The contraction mapping principle yields the existence of a unique fixed point in \({\mathcal {B}}\), which then yields the required existence result.

6.1 Stability

Theorem 1.1 gives that if \(\varphi _0^*\) has mass zero and is small so that \(t_0^{\nu +\frac{3}{2}} (\log t_0)^{\beta } \Vert \varphi _0^*\Vert _{*,b}\le 1\), then the function

$$\begin{aligned} u(x,t)&= \frac{\alpha (t)}{\lambda _0(t)^2} \Bigl [ U\Bigl (\frac{x-\xi (t)}{\lambda _0(t)}\Bigr ) + \phi _0^i\Bigl (\frac{x-\xi (t)}{\lambda _0(t)},t\Bigr ) + \phi \Bigl (\frac{x-\xi (t)}{\lambda _0(t)},t\Bigr ) \Bigr ] \chi (x,t) \nonumber \\&\quad + {\tilde{\varphi }}_{\lambda _0}(x-\xi (t),t) + \varphi (x,t) + \varphi ^*(x,t), \end{aligned}$$
(6.41)

solves (3.1) and blows-up in the way described in Theorem 1.1. This follows from the form of the ansatz (3.2), (3.13), (5.1), (5.38), where \(\phi =\phi _1+\phi _2\), and \(\phi _1\), \(\phi _2\), \(\varphi \), \(\varphi ^*\) satisfy respectively the equations (5.48), (5.49), (5.50) and (5.37). The initial value of u is

$$\begin{aligned} u^*(x;\varphi _0^*)&= \frac{\alpha (t_0;\varphi _0^*)}{\lambda _0(t_0)^2} \Big [ U\Bigl ( \frac{x-\xi (t_0;\varphi _0^*)}{\lambda _0(t_0)} \Bigr ) + \phi _0^i \Bigl ( \frac{x-\xi (t_0;\varphi _0^*)}{\lambda _0(t_0)} \Bigr )\\&\qquad \qquad \qquad \qquad + c_1(\varphi _0^*) {{\tilde{Z}}}_0\Bigl ( \frac{x-\xi (t_0)}{\lambda _0(t_0)} \Bigr ) \Big ] \\&\quad \cdot \chi _0 \Bigl ( \frac{x-\xi (t_0;\varphi _0^*)}{\sqrt{t_0}}\Bigr ) + {\tilde{\varphi }}_{\lambda _0}(x-\xi (t_0;\varphi _0^*),t_0) + \varphi _0^*(x). \end{aligned}$$

We recall that \({\tilde{\varphi }}_\lambda \) is defined in (3.9). The function \({\tilde{\varphi }}\) doesn’t depend on \(\xi \) and is radial about the origin.

We let \(u_0^*(x) = u^*(x;0)\). Note that \(u_0^*\) is radial and so its center of mass is zero.

To prove stability, we would like to prove the following intermediate step: given v defined on \({\mathbb {R}}^2\) small, with mass zero and under some additional assumptions to be defined later on, we would like to find \(\varphi _0^*\) with mass zero such that

$$\begin{aligned} u^*(\varphi _0^*)=u_0^*+v. \end{aligned}$$
(6.42)

The equation (6.42) for \(\varphi _0^*\) has the form

$$\begin{aligned}&\frac{\alpha (t_0;\varphi _0^*)}{\lambda _0(t_0)^2} \Big [ U\Bigl ( \frac{x-\xi (t_0;\varphi _0^*)}{\lambda _0(t_0)} \Bigr ) + \phi _0^i \Bigl ( \frac{x-\xi (t_0;\varphi _0^*)}{\lambda _0(t_0)} \Bigr )\nonumber \\&\qquad + c_1(\varphi _0^*) {{\tilde{Z}}}_0\Bigl ( \frac{x-\xi (t_0)}{\lambda _0(t_0)} \Bigr ) \Big ]\cdot \chi _0 \Bigl ( \frac{x-\xi (t_0;\varphi _0^*)}{\sqrt{t_0}}\Bigr ) \nonumber \\&\qquad + {\tilde{\varphi }}_{\lambda _0}(x-\xi (t_0;\varphi _0^*),t_0) + \varphi _0^*(x) \nonumber \\&\quad = \frac{\alpha (t_0;0)}{\lambda _0(t_0)^2} \Big [ U\Bigl ( \frac{x}{\lambda _0(t_0)} \Bigr ) + \phi _0^i \Bigl ( \frac{x}{\lambda _0(t_0)} \Bigr ) + c_1(0) {{\tilde{Z}}}_0\Bigl ( \frac{x}{\lambda _0(t_0)} \Bigr ) \Big ]\cdot \chi _0 \Bigl ( \frac{x}{\sqrt{t_0}}\Bigr ) \nonumber \\&\qquad + {\tilde{\varphi }}_{\lambda _0}(x,t_0) + v . \end{aligned}$$
(6.43)

Computing the mass we find that \(\alpha (t_0;\varphi _0^*)=\alpha (t_0;0)\). Note that \(\lim _{t\rightarrow \infty }\xi (t)=0\) by (6.12). Then the center of mass of \(u(\cdot ,t)\) satisfies

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _{{\mathbb {R}}^2} u(x,t)xdx= 0. \end{aligned}$$

Since the center of mass is preserved

$$\begin{aligned} \int _{{\mathbb {R}}^2} u(x,t_0)xdx= 0. \end{aligned}$$

Let’s assume that the center of mass of v and \(\varphi _0^*\) are both zero. Then, computing the center of mass we find that

$$\begin{aligned} \xi (t_0;\varphi _0^*)=0. \end{aligned}$$
(6.44)

Then the Eq. (6.43) reduces to

$$\begin{aligned} (c_1(\varphi _0^*)-c_1(0) ) \frac{\alpha (t_0;0)}{\lambda _0(t_0)^2} {{\tilde{Z}}}_0\Bigl ( \frac{x}{\lambda _0(t_0)} \Bigr ) + \varphi _0^*(x) = v . \end{aligned}$$
(6.45)

We will prove at the end of this section the following.

Proposition 6.5

There is \(\delta >0\) so that if \(t_0^{\nu +\frac{3}{2}} (\log t_0)^{\beta } \Vert v\Vert _{*,b}\le \delta \), v has mass and center of mass equal to zero, then

$$\begin{aligned} \int _{{\mathbb {R}}^2} v(x)|x|^2dx=0, \end{aligned}$$

is equivalent to

$$\begin{aligned} c_1(v)-c_1(0)=0. \end{aligned}$$

To prove stability we first observe that if \(v:{\mathbb {R}}^2 \rightarrow {\mathbb {R}}\) satisfies \(t_0^{\nu +\frac{3}{2}} (\log t_0)^{\beta } \Vert v\Vert _{*,b}\le \delta \), has mass zero, and

$$\begin{aligned} \int _{{\mathbb {R}}^2} v(x)x_j dx = 0 , \quad \int _{{\mathbb {R}}^2} v(x)|x|^2 dx = 0 , \quad \end{aligned}$$

then \(u_0^*+v = u^*(\varphi _0^*) \) for \(\varphi _0^*=v\), by Proposition 6.5.

Now consider a general v with \(t_0^{\nu +\frac{3}{2}} (\log t_0)^{\beta } \Vert v\Vert _{*,b}\le \delta \) (for a possibly smaller \(\delta >0\)), and mass zero. We want to show that the initial condition \(u_0^*+v\) produces a solution to (3.1) with infinite time blow as described in Theorem 1.1. Consider

$$\begin{aligned} u_{\Lambda ,p}(x) = \frac{1}{\Lambda ^2} \Bigl [ u_0^*\Bigl (\frac{x-p}{\Lambda }\Bigr ) + v\Bigl (\frac{x-p}{\Lambda }\Bigr ) \Bigr ] , \end{aligned}$$

where \(p\in {\mathbb {R}}^2\) and \(\Lambda >0\). Note that \(u_{\Lambda ,p}\) has mass \(8\pi \). Then we select \(\Lambda \) and p such that

$$\begin{aligned} \int _{{\mathbb {R}}^2} u_{\Lambda ,p}(x) x_j dx = 0 , \quad \int _{{\mathbb {R}}^2} u_{\Lambda ,p}(x) |x|^2 d x = \int _{{\mathbb {R}}^2} u_0^*(x)|x|^2dx. \end{aligned}$$

Note that \(|\Lambda ^2-1| \le Ct_0^2 \Vert v\Vert _{*,b}\ll 1\) and \(|p|\le C t_0 \Vert v\Vert _{*,b}\ll 1\). Then we expand

$$\begin{aligned} u_{\Lambda ,p} (x) = u_0^* + w \end{aligned}$$

and w satisfies \(t_0^{\nu +\frac{3}{2}} (\log t_0)^{\beta } \Vert w\Vert _{*,b}\le C \delta \), has mass zero, center of mass zero and second moment equal to 0. By the previous claim, the initial condition \(u_{\Lambda ,p} (x) = u_0^* + w\) is such that the solution to (3.1) blows up as in Theorem 1.1. Then the same is true for the initial condition \(u_0^*+v\) after a scaling and translation in space.

6.2 Proof of Proposition 6.5

Lemma 6.1

Assume that \(t_0^{\nu +\frac{3}{2}} (\log t_0)^{\beta } \Vert v\Vert _{*,b}\le 1\), that v has mass and center of mass equal to zero, and that

$$\begin{aligned} c_1(v)-c_1(0)=0. \end{aligned}$$
(6.46)

Then

$$\begin{aligned} \int _{{\mathbb {R}}^2} v(x)|x|^2dx=0. \end{aligned}$$

Proof

From (6.46), \(\varphi _0^*=v\) solves (6.45), and therefore \(u_0^*+v\) is an initial condition for (3.1) for which the solution blows up in infinite time. The solution u to (3.1) preserves the second moment:

$$\begin{aligned} \int _{{\mathbb {R}}^2} u(x,t)|x|^2dx=\text {const}. \end{aligned}$$

We compute the expansion of \(\int _{{\mathbb {R}}^2} u(x,t)|x|^2dx\) as \(t\rightarrow \infty \), based on the expression (6.41).

Note that \(\lim _{t\rightarrow \infty }\xi (t)=0\) by (6.12). Then

$$\begin{aligned}&\frac{\alpha (t)}{\lambda _0(t)^2} \int _{{\mathbb {R}}^2} U\Bigl (\frac{x-\xi (t)}{\lambda _0(t)}\Bigr ) \chi (x,t) |x|^2 dx \\&\quad = \frac{\alpha (t)}{\lambda _0(t)^2} \int _{{\mathbb {R}}^2} U\Bigl (\frac{x-\xi (t)}{\lambda _0(t)}\Bigr ) \chi (x,t) |x-\xi (t)|^2 dx+o(1), \end{aligned}$$

as \(t\rightarrow \infty \). By explicit computation

$$\begin{aligned} \frac{1}{\lambda _0(t)^2} \int _{{\mathbb {R}}^2} U\Bigl (\frac{x-\xi (t)}{\lambda _0(t)}\Bigr ) \chi (x,t) |x-\xi (t)|^2 dx = 8 \pi \lambda _0^2 \log \Bigl ( \frac{t}{\lambda _0^2}\Bigr )+O(\lambda _0^2) \end{aligned}$$
(6.47)

as \(t\rightarrow \infty \).

Using Lemma 4.1

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _{\lambda _0}(x,t) |x-\xi (t)|^2dx\le \frac{C}{\log (t)} . \end{aligned}$$

Using also Lemma 4.1 to estimate the mass and first moment of \(\varphi _{\lambda _0}\) we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _{\lambda _0}(x,t) |x|^2dx\le \frac{C}{\log (t)} . \end{aligned}$$
(6.48)

Using (6.47), (6.48) and the estimates for \(\phi _0^i\) (5.36), \(\phi =\phi _1+\phi _2\) that arise from \( \Vert \phi \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }<\infty \), and \(\varphi \), \(\varphi ^*\) which arise from \(\Vert \varphi \Vert _{*,o}<\infty \), \(\Vert \varphi ^*\Vert _{*,o}<\infty \), we get that

$$\begin{aligned} \int _{{\mathbb {R}}^2} u(x,t)|x|^2dx =8 \pi \lambda _0^2 \log \Bigl ( \frac{t}{\lambda _0^2}\Bigr )+O(\lambda _0^2) \end{aligned}$$

as \(t\rightarrow \infty \). But \(\lambda _0\) was constructed in Proposition 5.1 with the expansion

$$\begin{aligned} \lambda _0(t)&= \frac{c_0}{\sqrt{\log t}} + O\Bigl ( \frac{1}{(\log t)^{\frac{3}{2}-\varepsilon }}\Bigr ), \end{aligned}$$

as \(t\rightarrow \infty \), where \(c_0>0\) is a constant. Therefore

$$\begin{aligned} \int _{{\mathbb {R}}^2} u(x,t)|x|^2dx =8 \pi c_0^2 \end{aligned}$$

and evaluating at \(t=t_0\) we obtain

$$\begin{aligned} \int _{{\mathbb {R}}^2} u_0^*(x)|x|^2dx+\int _{{\mathbb {R}}^2} v(x)|x|^2dx=8 \pi c_0^2. \end{aligned}$$

We can apply the previous calculation to \(v=0\) and arrive at

$$\begin{aligned} \int _{{\mathbb {R}}^2} u_0^*(x)|x|^2dx=8 \pi c_0^2. \end{aligned}$$

This shows that

$$\begin{aligned} \int _{{\mathbb {R}}^2} v(x)|x|^2dx=0. \end{aligned}$$

\(\square \)

We need an expansion for \(c_1(\varphi _0^*)-c_1(0) \).

Lemma 6.2

Assume that \(t_0^{\nu +\frac{3}{2}} (\log t_0)^{\beta } \Vert \varphi _0^*\Vert _{*,b}\le 1\) and that \(\varphi _0^*\) has mass and center of mass equal to zero. Then

$$\begin{aligned} c_1(\varphi _0^*)-c_1(0) = a_0 \int _{{\mathbb {R}}^2} \varphi _0^*(x)|x|^2dx + R_0(\varphi _0^*) , \end{aligned}$$

where \(a_0\not =0\) and \(R_0\) satisfies

$$\begin{aligned} |R_0(\varphi _0^*)| \le C t_0 \Vert \varphi _0^*\Vert _{*,b} . \end{aligned}$$
(6.49)

Proof

In the following calculations \(\lambda =\lambda _0\).

First we need to estimate the Lipschitz constant of the solutions \(\phi _1\), \(\phi _2\), and \(\varphi \) with respect to \(\varphi _0^*\). We claim that

$$\begin{aligned}&\Vert \phi _1(\varphi _0^*)-\phi _1(0)\Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } +\Vert \phi _2(\varphi _0^*)-\phi _2(0)\Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \nonumber \\&\quad \le C \frac{1}{\log t_0} t_0^{\nu +\frac{1}{2}} (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b} \end{aligned}$$
(6.50)
$$\begin{aligned}&| \varphi (x,t;\varphi _0^*) -\varphi (x,t;0) | \le C \frac{1}{t^{\nu +\frac{3}{2}} (\log t)^{\frac{q+1}{2}}} \frac{1}{(1+|x-\xi |/\sqrt{t})^b} t_0^{\nu +\frac{1}{2}}\nonumber \\&\qquad (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b} \end{aligned}$$
(6.51)
$$\begin{aligned}&|\alpha _1[\varphi _0^*](t)-\alpha _1[0](t)| \le C \frac{1}{t^{\nu +\frac{1}{2}} (\log t)^{\frac{q+1}{2}}} t_0^{\nu +\frac{1}{2}} (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b} . \end{aligned}$$
(6.52)

We discuss briefly the proof of these estimates. One of the main terms in the right hand side of (5.50), written for the difference \(\varphi (\varphi _0^*)-\varphi (0)\) is

$$\begin{aligned}&\Bigl | \frac{1}{\lambda ^2} [ \phi (\varphi _0^*) - \phi (0) ] \Delta \chi \Bigr |(x,t) \\&\quad \le \frac{C}{\lambda ^2} \frac{1}{t^{\nu -\frac{1}{2}} (\log t)^{\frac{q-1}{2}}} \frac{1}{( |x-\xi |/\lambda |) ^4} \frac{1}{t} \Bigl |\Delta _z \chi _0( \frac{x-\xi }{\sqrt{t}}) \Bigr |\\&\quad \Vert \phi (\varphi _0^*) - \phi (0) \Vert _{1,\nu -\frac{1}{2} ,\frac{q-1}{2},4,2+\sigma +\varepsilon } \\&\quad \le C \frac{1}{t^{\nu +\frac{5}{2}} (\log t)^{\frac{q+1}{2}}} \frac{1}{(1+|x-\xi |/\sqrt{t})^b} \Vert \phi (\varphi _0^*) - \phi (0) \Vert _{1,\nu -\frac{1}{2} ,\frac{q-1}{2},4,2+\sigma +\varepsilon } , \end{aligned}$$

which implies

$$\begin{aligned} \Bigl \Vert \frac{1}{\lambda ^2} [ \phi (\varphi _0^*) - \phi (0) ] \Delta \chi \Bigr \Vert _{**,o} \le \frac{C}{(\log t_0)^{\frac{q+1}{2}-\beta }} \Vert \phi (\varphi _0^*) - \phi (0) \Vert _{1,\nu -\frac{1}{2} ,\frac{q-1}{2},4,2+\sigma +\varepsilon }, \end{aligned}$$

since \(\beta <\frac{q+1}{2}\), which is one of the conditions in (6.14). (Here \(\chi \) depends on \(\varphi _0^*\). There is another ther in the difference that depends on \(\chi (\varphi _0^*)-\chi (0)\) and is estimated similarly.) Then

$$\begin{aligned}&| \varphi (x,t;\varphi _0^*) -\varphi (x,t;0) | \nonumber \\&\quad \le C \frac{1}{t^{\nu +\frac{3}{2}} (\log t)^{\frac{q+1}{2}}} \frac{1}{(1+|x-\xi |/\sqrt{t})^b} [ \Vert \phi (\varphi _0^*) - \phi (0) \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }\nonumber \\&\qquad + t_0^{\nu +\frac{1}{2}} (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b} ] . \end{aligned}$$
(6.53)

Considering \(\varphi \) as an operator of \(\phi \) we examine the effect of the therm \(\lambda ^2 U \varphi \). This term appears in the right hand side of (5.48), where the effect is less important, and in the computation of \(\alpha _1\). Estimating the right hand side of (5.50) as in (6.32), using Proposition 6.3 gives that

$$\begin{aligned}&|\alpha _1[\phi (\varphi _0^*](t)-\alpha _1[\phi (0)](t)| \\&\quad \le C \int _t^\infty \int _{{\mathbb {R}}^2} U(y) |\varphi ( \xi + \lambda y,t,;\varphi _0^*)-\varphi ( \xi + \lambda y,t,;0) | dy \\&\quad \le C \frac{1}{t^{\nu +\frac{1}{2}} (\log t)^{\frac{q+1}{2}}} [ \Vert \phi (\varphi _0^*) - \phi (0) \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }\\&\quad + t_0^{\nu +\frac{1}{2}} (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b} ] . \end{aligned}$$

We consider now the effect of \(|\alpha _1[\phi ](t)|\) in the right hand side of (5.49), where thanks to Lemma 5.4 appears mainly as \( \alpha _1(t) W_2(y)\), where \(W_2\) is radial with compact support. Then Proposition 6.1 gives

$$\begin{aligned}&| \phi _2(y,t;\varphi _0^*)-\phi _2(y,t;0) | \\&\quad \le C \frac{1}{(\log t_0)^{1-q}}\frac{1}{t^{\nu -\frac{1}{2}} (\log t)^{\frac{q+1}{2}+q-1} } \frac{1}{(1+|y|)^4} \min \Bigl ( 1 , \frac{(t \log t)^{1/2}}{|y|} \Bigr )^{2+\sigma +\varepsilon } \\&\qquad \cdot [ \Vert \phi (\varphi _0^*) - \phi (0) \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } + t_0^{\nu +\frac{1}{2}} (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b} ] . \end{aligned}$$

Then

$$\begin{aligned}&\Vert \phi _2(\varphi _0^*)-\phi _2(0) \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \\&\quad \le C \frac{1}{\log t_0} [ \Vert \phi (\varphi _0^*) - \phi (0) \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } + t_0^{\nu +\frac{1}{2}} (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b} ] . \end{aligned}$$

The estimate for \(\phi _1\) is actually better, and therefore

$$\begin{aligned}&\Vert \phi _1(\varphi _0^*)-\phi _1(0)\Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } +\Vert \phi _2(\varphi _0^*)-\phi _2(0)\Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \\&\quad \le C \frac{1}{\log t_0} [ \Vert \phi _1(\varphi _0^*)-\phi _1(0) \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon }\\&\quad + \Vert \phi _2(\varphi _0^*)-\phi _2(0) \Vert _{1,\nu -\frac{1}{2},\frac{q-1}{2},4,2+\sigma +\varepsilon } \\&\qquad + t_0^{\nu +\frac{1}{2}} (\log t_0)^{\beta -1} \Vert \varphi _0^*\Vert _{*,b} ]. \end{aligned}$$

This implies (6.50). Replacing this in (6.53) we obtain (6.51), and similarly we get (6.52).

The parameter \(c_1\) appears in the second inner equation in (5.49), which we write as

$$\begin{aligned} \left\{ \begin{aligned} \lambda ^2 \partial _t \phi _2&= L [\phi _2] + B [\phi _2] + h(t) W_2 \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \phi _2(\cdot ,t_0)&= c_1 {{\tilde{Z}}}_0 \quad \text {in }{\mathbb {R}}^2, \end{aligned} \right. \end{aligned}$$
(6.54)

where

$$\begin{aligned} h(t,\varphi _0^*) =m_2[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] (t). \end{aligned}$$

Note that \(\phi _2\) in (6.54) is radial, so the operator B defined (5.47) reduces to \(B[\phi ]=\lambda {\dot{\lambda }} ( 2 \phi + y \cdot \nabla \phi ) = \lambda {\dot{\lambda }} \nabla \cdot (y\phi )\). Multiplying by \(|y|^2\) and integrating on \({\mathbb {R}}^2\) gives

$$\begin{aligned} \lambda ^2 \partial _t \int _{{\mathbb {R}}^2} \phi |y|^2dy +2\lambda {\dot{\lambda }} \int _{{\mathbb {R}}^2} \phi |y|^2dy = h(t). \end{aligned}$$

Then

$$\begin{aligned} \lambda ^2 \int _{{\mathbb {R}}^2} \phi (y,t) |y|^2dy=-\int _t^\infty h(s)ds. \end{aligned}$$

But \(\phi _2(y,t_0)=c_1 {{\tilde{Z}}}_0(y)\) so

$$\begin{aligned} c_1(\varphi _0^*) = -\frac{1}{\lambda (t_0)^2\int _{{\mathbb {R}}^2}{{\tilde{Z}}}_0(y)|y|^2dy}\int _{t_0}^\infty h(s,\varphi _0^*)ds. \end{aligned}$$

In particular

$$\begin{aligned} c_1(\varphi _0^*)-c_1(0) = -\frac{1}{\lambda (t_0)^2\int _{{\mathbb {R}}^2}{{\tilde{Z}}}_0(y)|y|^2dy}\int _{t_0}^\infty [ h(s,\varphi _0^*) - h(s,0)] ds. \end{aligned}$$
(6.55)

The function \(h(t,\varphi _0^*) =m_2[F_3(\phi _1+\phi _2,\varphi ,{{\textbf {p}}}_1,\varphi _0^*)] (t)\) is analyzed near (6.25). We follow the same steps. Using the definition of \(F_3\) (5.51)

$$\begin{aligned} m_2[ F_3(\phi ,\varphi ,{{\textbf {p}}}_1 ,\varphi _0^*) ] = m_2[ E_2 {\tilde{\chi }}_2(\varphi _0^*)] + m_2[F_2(\phi ,\varphi ,{{\textbf {p}}}_1 ,\varphi _0^*) {\tilde{\chi }}(\varphi _0^*) ] . \end{aligned}$$

We note that \({\tilde{\chi }}\), \({\tilde{\chi }}_2\) also depend on \(\varphi _0^*\) because \(\xi \) depends on \(\varphi _0^*\). By (5.41)

$$\begin{aligned} m_2[F_2 (\phi ,\varphi ,{{\textbf {p}}}_1,\varphi ^*_0) {\tilde{\chi }}(\varphi _0^*) ]&=I(\varphi _0^*)+I\!I(\varphi _0^*)+I\!I\!I(\varphi _0^*) \end{aligned}$$

where

$$\begin{aligned}&I(t,\varphi _0^*)=\lambda ^4 m_2[S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) )] \\&I\!I(t,\varphi _0^*)=m_2[ F(\phi ^i_0 + \phi , \varphi ^* + \varphi ,{{\textbf {p}}}_0+{{\textbf {p}}}_1) {\tilde{\chi }}(\varphi _0^*)] \\&I\!I\!I(t,\varphi _0^*)= \lambda ^4 m_2[ ( S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1 ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ) ({\tilde{\chi }}(\varphi _0^*)-1)]. \end{aligned}$$

The main term is \(I(\varphi _0^*)\) and the others are treated as perturbations.

By Lemma 5.4, since \(\lambda = \lambda _0\), we get

$$\begin{aligned} I(t,\varphi _0^*)&= - 32\pi \alpha _1(\varphi _0^*) + I_0(\varphi _0^*) , \end{aligned}$$
(6.56)

where

$$\begin{aligned} I_0(t,\varphi _0^*)&= - \frac{{\dot{\alpha }}_1(\varphi _0^*)}{\lambda _0^2} \int _{{\mathbb {R}}^2} U(\frac{x}{\lambda _0}) \chi _0(\frac{x}{\lambda _0}) |x|^2\,dx + \alpha _1(\varphi _0^*) \int _{{\mathbb {R}}^2} E(x,t,\lambda _0)|x|^2\, \textrm{d}x \\&\quad - |\xi (\varphi _0^*)|^2 \int _{{\mathbb {R}}^2} S(u_1({\textbf{p}}_0))dx . \end{aligned}$$

By (6.11)

$$\begin{aligned} \alpha _1(t,\varphi _0^*) - \alpha _1(t,0) = A_1(t,\varphi _0^*) + A_2(t,\varphi _0^*) \end{aligned}$$
(6.57)

where

$$\begin{aligned} A_1(t,\varphi _0^*)&= -\frac{1}{8\pi ( 1+2\Upsilon \frac{\lambda _0^2}{t}) + e_1(\frac{\lambda _0^2}{t})} \int _t^\infty \frac{1}{\lambda _0^2} \Bigl \{ m_0[ F(\phi ^i_0 \\&\quad + \phi (\varphi _0^*) , \varphi ^* + \varphi (\varphi _0^*) ,{{\textbf {p}}}_0+{{\textbf {p}}}_1(\varphi _0^*)) {\tilde{\chi }}(\varphi _0^*)](s) \\&\quad -m_0[ F(\phi ^i_0 + \phi (0) , \varphi (0) ,{{\textbf {p}}}_0+{{\textbf {p}}}_1(0)) {\tilde{\chi }}(0)](s) \Bigl \}ds \\ A_2(t,\varphi _0^*)&= - \frac{1}{8\pi ( 1+2\Upsilon \frac{\lambda _0^2}{t}) + e_1(\frac{\lambda _0^2}{t})} \int _t^\infty \lambda _0^2 \Bigl \{ m_0[ ( S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1(\varphi _0^*) ) )\\&\quad - S_0( u_1( {{\textbf {p}}}_0) ) ) ({\tilde{\chi }}(\varphi _0^*)-1) ] (s) \\&\quad -m_0[ ( S_0( u_1( {{\textbf {p}}}_0 + {{\textbf {p}}}_1(0) ) ) - S_0( u_1( {{\textbf {p}}}_0) ) ) ({\tilde{\chi }}(0)-1) ] (s) \Bigr \}ds \end{aligned}$$

Let

$$\begin{aligned} {{\tilde{m}}}_0(t,\varphi _0^*)&=\frac{1}{\lambda _0^2} m_0[ F(\phi ^i_0 + \phi (\varphi _0^*) , \varphi ^* + \varphi (\varphi _0^*) ,{{\textbf {p}}}_0+{{\textbf {p}}}_1(\varphi _0^*)) {\tilde{\chi }}(\varphi _0^*)](t) \\&\quad - \int _{{\mathbb {R}}^2} U(y) \varphi ^o(\xi (t,\varphi _0^*)+\lambda y,t,\varphi _0^*) dy \end{aligned}$$

so that

$$\begin{aligned}&m_0[ F(\phi ^i_0 + \phi (\varphi _0^*) , \varphi ^* + \varphi (\varphi _0^*) ,{{\textbf {p}}}_0+{{\textbf {p}}}_1(\varphi _0^*)) {\tilde{\chi }}(\varphi _0^*)](t) \\&\quad =\lambda _0^2(t) \int _{{\mathbb {R}}^2} U(y) \varphi ^o(\xi (t,\varphi _0^*)+\lambda y,t,\varphi _0^*) dy+\lambda _0^2(t){{\tilde{m}}}_0(t,\varphi _0^*) . \end{aligned}$$

By (5.38), \(\varphi ^o=\varphi ^*+\varphi \), where \(\varphi =\varphi (\varphi _0^*)\) solves (5.50) and \(\varphi ^*\) solves (5.37). Therefore

$$\begin{aligned} A_1(t,\varphi _0^*)&= -\frac{1}{8\pi ( 1+2\Upsilon \frac{\lambda _0^2}{t}) + e_1(\frac{\lambda _0^2}{t})} \int _t^\infty \int _{{\mathbb {R}}^2} U(y) \varphi ^*(\xi (s,\varphi _0^*)+\lambda y,s,\varphi _0^*) dy ds \\&\quad + {{\tilde{A}}}_1(t,\varphi _0^*) \end{aligned}$$

where

$$\begin{aligned} {{\tilde{A}}}_1(t,\varphi _0^*)&= -\frac{1}{8\pi ( 1+2\Upsilon \frac{\lambda _0^2}{t}) + e_1(\frac{\lambda _0^2}{t})} \int _t^\infty \int _{{\mathbb {R}}^2} U(y)[\varphi (\xi (s,\varphi _0^*)\\&\quad +\lambda y,s,\varphi _0^*)-\varphi (\xi (s,0)+\lambda y,s,0)] dy ds \\&\quad -\frac{1}{8\pi ( 1+2\Upsilon \frac{\lambda _0^2}{t}) + e_1(\frac{\lambda _0^2}{t})} \int _t^\infty [ {{\tilde{m}}}_0(s,\varphi _0^*)-{{\tilde{m}}}_0(s,0)]ds. \end{aligned}$$

Integrating (5.37) on \({\mathbb {R}}^2\) we find that

$$\begin{aligned} \partial _t \int _{{\mathbb {R}}^2}\varphi ^*(x,t)dx=\lambda (t)^{-2}\int _{{\mathbb {R}}^2}U\Bigl (\frac{x-\xi }{\lambda }\Bigr )\varphi ^*(x,t)\,dx=\int _{{\mathbb {R}}^2}U(y)\varphi ^*(\xi +\lambda y,t)\,dy, \end{aligned}$$

and therefore

$$\begin{aligned} A_1(t,\varphi _0^*)= \frac{1}{8\pi ( 1+2\Upsilon \frac{\lambda _0^2}{t}) + e_1(\frac{\lambda _0^2}{t})} \int _{{\mathbb {R}}^2}\varphi ^*(x,t,\varphi _0^*) dx +{{\tilde{A}}}_1(t,\varphi _0^*). \end{aligned}$$

Then from (6.57)

$$\begin{aligned} \alpha _1(t,\varphi _0^*) - \alpha _1(t,0)&=\frac{1}{8\pi ( 1+2\Upsilon \frac{\lambda _0^2}{t}) + e_1(\frac{\lambda _0^2}{t})} \int _{{\mathbb {R}}^2}\varphi ^*(x,t,\varphi _0^*) dx\\&\quad +{{\tilde{A}}}_1(t,\varphi _0^*)+A_2(t,\varphi _0^*). \end{aligned}$$

Using this and (6.56) we get

$$\begin{aligned} I(t,\varphi _0^*)-I(t,0)&=-\frac{4}{1+2\Upsilon \frac{\lambda _0^2}{t} + \frac{1}{8\pi }e_1(\frac{\lambda _0^2}{t})} \int _{{\mathbb {R}}^2}\varphi ^*(x,t,\varphi _0^*) dx\\&\quad -32\pi {{\tilde{A}}}_1(t,\varphi _0^*)-32\pi A_2(t,\varphi _0^*) \\&\quad +I_0(t,\varphi _0^*)- I_0(t,0). \end{aligned}$$

Hence

$$\begin{aligned} h(t,\varphi _0^*) - h(t,0)&= -4\int _{{\mathbb {R}}^2}\varphi ^*(x,t,\varphi _0^*) dx+{{\tilde{h}}}(t,\varphi _0^*) \end{aligned}$$

where

$$\begin{aligned} {{\tilde{h}}}(\varphi _0^*,t)&= 4 \frac{2\Upsilon \frac{\lambda _0^2}{t} + \frac{1}{8\pi }e_1(\frac{\lambda _0^2}{t})}{1+2\Upsilon \frac{\lambda _0^2}{t} + \frac{1}{8\pi }e_1(\frac{\lambda _0^2}{t})} \int _{{\mathbb {R}}^2}\varphi ^*(x,t,\varphi _0^*) dx\\&\quad -32\pi {{\tilde{A}}}_1(t,\varphi _0^*)-32\pi A_2(t,\varphi _0^*) \\&\quad +I_0(t,\varphi _0^*)- I_0(t,0) +I\!I(t,\varphi _0^*)-I\!I(t,0) +I\!I\!I(t,\varphi _0^*)-I\!I\!I(t,0). \end{aligned}$$

From (6.55) it follows that

$$\begin{aligned} c_1(\varphi _0^*)-c_1(0)&= \frac{4}{\lambda (t_0)^2\int _{{\mathbb {R}}^2}{{\tilde{Z}}}_0(y)|y|^2dy}\int _{t_0}^\infty \int _{{\mathbb {R}}^2}\varphi ^*(x,s,\varphi _0^*) dxds +{{\tilde{c}}}_1(\varphi _0^*) \end{aligned}$$
(6.58)

where

$$\begin{aligned} {{\tilde{c}}}_1(\varphi _0^*)&=-\frac{1}{\lambda (t_0)^2\int _{{\mathbb {R}}^2}{{\tilde{Z}}}_0(y)|y|^2dy}\int _{t_0}^\infty {{\tilde{h}}}(s,\varphi _0^*)ds . \end{aligned}$$

We can relate the integral \(\int _{t_0}^\infty \int _{{\mathbb {R}}^2}\varphi ^*(x,s,\varphi _0^*) dxds\) with the second moment of \(\varphi _0^*\) as follows. We multiply the equation of \(\varphi ^*\) (5.37) by \(|x-\xi (t)|^2\) and integrate on \({\mathbb {R}}^2\) to get

$$\begin{aligned} \partial _t \int _{{\mathbb {R}}^2} \varphi ^*(x,t)|x-\xi (t)|^2dx&=\int _{{\mathbb {R}}^2} \Delta \varphi ^*(x,t) |x-\xi (t)|^2 dx \\&\quad - \int _{{\mathbb {R}}^2} \nabla _x \Gamma _0\Bigl (\frac{x-\xi (t)}{\lambda }\Bigr )\cdot \nabla \varphi ^*(x,t) |x-\xi (t)|^2dx \\&\quad -2 {\dot{\xi }}(t)\cdot \int _{{\mathbb {R}}^2}\varphi ^*(x,t) (x-\xi (t))dx . \end{aligned}$$

But

$$\begin{aligned} \int _{{\mathbb {R}}^2} \Delta \varphi ^* |x-\xi (t)|^2 dx&=4\int _{{\mathbb {R}}^2} \varphi ^*dx \end{aligned}$$

and

$$\begin{aligned}&\int _{{\mathbb {R}}^2} \nabla _x \Gamma _0\Bigl (\frac{x-\xi (t)}{\lambda }\Bigr )\cdot \nabla \varphi ^*(x,t) |x-\xi (t)|^2dx \\&\quad = -\int _{{\mathbb {R}}^2} \varphi ^*(x+\xi (t)) \Bigl [ \Delta _x \Gamma _0\Bigl (\frac{x}{\lambda }\Bigr ) |x|^2 +2 \nabla _x\Gamma _0\Bigl (\frac{x}{\lambda }\Bigr )\cdot x\Bigr ] dx. \end{aligned}$$

Using the explicit expressions for U and \(\Gamma _0\) and writing \(y=\frac{x}{\lambda }\), \(\rho =|y|\), we get

$$\begin{aligned} \Delta _x \Gamma _0\Bigl (\frac{x}{\lambda }\Bigr ) |x|^2 +2 \nabla _x\Gamma _0\Bigl (\frac{x}{\lambda }\Bigr )\cdot x&=-\frac{1}{\lambda ^2} U\Bigl (\frac{x}{\lambda }\Bigr )|x|^2+2\nabla _y \Gamma _0\Bigl (\frac{x}{\lambda }\Bigr )\cdot \frac{x}{\lambda } \\&=- \frac{8 \rho ^2}{(1+\rho ^2)^2} - \frac{8 \rho ^2}{1+\rho ^2} \\&= -\Bigl [ \frac{8 \rho ^2}{(1+\rho ^2)^2} + 8 - \frac{8}{1+\rho ^2}\Bigr ] \\&=-8 + \frac{8}{(1+\rho ^2)^2} . \end{aligned}$$

So

$$\begin{aligned}&\int _{{\mathbb {R}}^2} \varphi ^*(x+\xi (t)) \Bigl [ \Delta _x \Gamma _0\Bigl (\frac{x}{\lambda }\Bigr ) |x|^2 +2 \nabla _x\Gamma _0\Bigl (\frac{x}{\lambda }\Bigr )\cdot x\Bigr ] dx \\&\quad = -8\int _{{\mathbb {R}}^2}\varphi ^*(x,t) dx + \int _{{\mathbb {R}}^2} U\Bigl (\frac{x-\xi (t)}{\lambda }\Bigr )\varphi ^*(x,t)dx \end{aligned}$$

and we find that

$$\begin{aligned} \partial _t \int _{{\mathbb {R}}^2} \varphi ^*(x,t)|x-\xi (t)|^2dx&=-4\int _{{\mathbb {R}}^2} \varphi ^*(x,t)dx+\int _{{\mathbb {R}}^2}U\Bigl (\frac{x-\xi (t)}{\lambda }\Bigr )\varphi ^*(x,t)dx \\&\quad -2 {\dot{\xi }}(t)\int _{{\mathbb {R}}^2}\varphi ^*(x,t) (x-\xi (t))dx . \end{aligned}$$

Integrating and using (6.58) we find that

$$\begin{aligned} c_1(\varphi _0^*)-c_1(0)&= \frac{1}{\lambda (t_0)^2\int _{{\mathbb {R}}^2}{{\tilde{Z}}}_0(y)|y|^2dy} \int _{{\mathbb {R}}^2} \varphi ^*(x,t_0)|x|^2dx+R_0(\varphi _0^*) , \end{aligned}$$

by (6.44), where

$$\begin{aligned} R_0(\varphi _0^*)&= \frac{1}{\lambda (t_0)^2\int _{{\mathbb {R}}^2}{{\tilde{Z}}}_0(y)|y|^2dy} \Bigl [ \int _{t_0}^\infty \int _{{\mathbb {R}}^2}U\Bigl (\frac{x-\xi (s)}{\lambda }\Bigr )\varphi ^*(x,s)dxds \\&\quad -2 \int _{t_0}^\infty {\dot{\xi }}(s)\int _{{\mathbb {R}}^2}\varphi ^*(x,s) (x-\xi (s))dxds \Bigr ] +{{\tilde{c}}}_1(\varphi _0^*). \end{aligned}$$

We claim that \(R_0(\varphi _0^*)\) satisfies (6.49). Indeed, let us look at

$$\begin{aligned} \int _{t_0}^\infty [I_0(s,\varphi _0^*)- I_0(s,0)]ds. \end{aligned}$$

Similarly (6.36), we have

$$\begin{aligned} \left| \frac{{\dot{\alpha }}_1(t,\varphi _0^*)-{\dot{\alpha }}_1(t,0)}{\lambda _0^2} \int _{{\mathbb {R}}^2} U(\frac{x}{\lambda _0}) \chi _0(\frac{x}{\lambda _0}) |x|^2\,dx \right|&\le C \frac{t_0^{\nu +\frac{3}{2}} (\log t_0)^\beta }{t^{\nu +\frac{3}{2}} (\log t)^\beta } \lambda _0^2 \Vert \varphi _0^*\Vert _{*,b} \end{aligned}$$

Similar computations for the other terms of \(I_0\) give

$$\begin{aligned} |I_0(t,\varphi _0^*)- I_0(t,0)| \le C \frac{t_0^{\nu +\frac{3}{2}} (\log t_0)^\beta }{t^{\nu +\frac{3}{2}} (\log t)^\beta } \lambda _0^2 \Vert \varphi _0^*\Vert _{*,b} . \end{aligned}$$

It follows that

$$\begin{aligned} \left| \int _{t_0}^\infty [I_0(s,\varphi _0^*)- I_0(s,0)]ds \right| \le C \frac{t_0}{\log (t_0)}\Vert \varphi _0^*\Vert _{*,b} . \end{aligned}$$

The other terms in \(R_0\) are estimated similarly. \(\square \)

Proof of Proposition 6.5

If \(t_0^{\nu +\frac{3}{2}} (\log t_0)^{\beta } \Vert v\Vert _{*,b}\le 1\) and \(c_1(v)-c_1(0)=0\), then Lemma 6.1 implies that \(\int _{{\mathbb {R}}^2} v(x)|x|^2dx=0\).

To prove the converse, let

$$\begin{aligned} v_1(x) = \frac{1}{(1+\frac{|x|}{\sqrt{t_0}})^b} \end{aligned}$$

so that \(\Vert v_1\Vert _{*,b}=1\) (norm defined in (6.10)). Assuming \(\mu t_0^{\nu +\frac{3}{2}} (\log t_0)^{\beta } \le \delta \) and \(\delta >0\) small, we have by Lemma 6.2

$$\begin{aligned} c_1(v+\mu v_1)-c_1(0) = c \mu t_0^2 + R_0(v+\mu v_1) , \end{aligned}$$

for some constant \(c\not =0\). Note that is \(c_1(\varphi _0^*)\) continuous function of \(\varphi _0^*\), and so is \(R_0(\varphi _0^*)\). By the intermediate value theorem, there is \(\mu = O(t_0 \Vert v\Vert _{*,b})\) such that \(c_1(v+\mu v_1)-c_1(0)=0\). By Lemma 6.1\(\int _{{\mathbb {R}}^2}(v(x) + \mu v_1(x))|x|^2dx=0\), which implies that \(\mu =0\). But then \(c_1(v)-c_1(0)=0\). \(\square \)

7 The mass of \(\varphi _\lambda \)

We devote this section to prove Proposition 5.1. To that purpose, a basic step is to derive a formula for the mass of \(\varphi _\lambda \) defined in (3.12).

Let us write

$$\begin{aligned} \varphi _\lambda = \varphi _\lambda ^{(1)} +\varphi _\lambda ^{(2)} \end{aligned}$$
(7.1)

where \(\varphi _\lambda ^{(1)}\) and \(\varphi _\lambda ^{(2)}\) are the solutions, given by Duhamel’s formula, of the following problems

$$\begin{aligned}&\left\{ \begin{aligned} \partial _t \varphi _\lambda ^{(1)}&= \Delta _6 \varphi _\lambda ^{(1)} + \frac{{\dot{\lambda }}}{\lambda ^3} Z_0(\frac{x}{\lambda }) \chi _0(z) \quad \text {in }{\mathbb {R}}^2 \times ({\textstyle \frac{t_0}{2}},\infty ) \\ \varphi _\lambda ^{(1)}(\cdot ,{\textstyle \frac{t_0}{2}})&= 0 \end{aligned} \right. \end{aligned}$$
(7.2)
$$\begin{aligned}&\left\{ \begin{aligned} \partial _t \varphi _\lambda ^{(2)}&= \Delta _6 \varphi _\lambda ^{(2)} +\frac{1}{2 \lambda ^2 t} U \nabla _z \chi _0(z)\cdot z + {{\tilde{E}}} , \quad \text {in }{\mathbb {R}}^2 \times ({\textstyle \frac{t_0}{2}},\infty ) , \quad z = \frac{x}{\sqrt{t}} , \\ \varphi _\lambda ^{(2)}(\cdot ,{\textstyle \frac{t_0}{2}})&= 0 \end{aligned} \right. \end{aligned}$$
(7.3)

where the operator \(\Delta _6\) is defined in (3.8) and \({{\tilde{E}}}\) in (3.11). We let \(\varphi [p,\lambda ](r,t)\) be the solution of the problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _t \varphi [p,\lambda ]&= \Delta _6 \varphi [p,\lambda ] + \frac{p}{\lambda ^4} Z_0\Bigl (\frac{r}{\lambda }\Bigr ) \chi \Bigr ( \frac{r}{\sqrt{t}}\Bigl ) \quad \text {in } {\mathbb {R}}^2 \times ({\textstyle \frac{t_0}{2}},\infty ), \\ \varphi [p,\lambda ] (\cdot ,{\textstyle \frac{t_0}{2}})&= 0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(7.4)

given by Duhamel’s formula. By definition, we have

$$\begin{aligned} \varphi _{\lambda }^{(1)} = \varphi [\lambda {\dot{\lambda }},\lambda ] . \end{aligned}$$

In definitions (7.2), (7.3), (7.4), the parameter function \(\lambda (t)\) is assumed to be defined for \(t>\frac{t_0}{2}\). In the rest of this section we also assume the validity of the condition stated for \(\lambda \) in (4.1), namely

$$\begin{aligned} |\lambda (t)| + t \log (t) |{\dot{\lambda }}(t)| \le \frac{C}{\sqrt{\log (t)}}, \quad t >\frac{t_0}{2}, \end{aligned}$$
(7.5)

for some fixed constant C. Let us define

$$\begin{aligned} \Vert p\Vert _{\gamma ,m} = \sup _{t\ge t_0/2} \, t^\gamma (\log t)^m |p(t)|. \end{aligned}$$
(7.6)

In what follows we shall only deal with radial functions on \({\mathbb {R}}^2\) and sometimes we will consider them as radial functions on \({\mathbb {R}}^6\). For a fixed constant \(c_0>0\) we let

$$\begin{aligned} \lambda ^*(t) = \frac{c_0}{\sqrt{\log t} } . \end{aligned}$$
(7.7)

The following expansion holds.

Lemma 7.1

Assume that \(\lambda \) satisfies (7.5). Let \(0<\gamma <2\), \(m\in {\mathbb {R}}\) and suppose that \(\Vert p\Vert _{\gamma ,m} <\infty \). Then

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi [p,\lambda ](x)dx = - 4 \pi \int _{t/2}^{t-\lambda (t)^2} \frac{p(s)}{t-s}ds + R[p,\lambda ] \end{aligned}$$

where \(R[p,\lambda ]\) satisfies

$$\begin{aligned} \Vert R[p,\lambda ] \Vert _{\gamma ,m} \le C \Vert p\Vert _{\gamma ,m}. \end{aligned}$$

If \(\lambda _1,\lambda _2\) satisfy

$$\begin{aligned} \Bigl \Vert \frac{\lambda _j}{\lambda ^*} \Bigl \Vert _{L^\infty (t_0/2,\infty ) } <\frac{1}{2}, \quad j=1,2, \end{aligned}$$

then we also have

$$\begin{aligned} \Vert R[p,\lambda ^*+\lambda _1] - R[p,\lambda ^*+\lambda _2] \Vert _{\gamma ,m} \le C \Vert p\Vert _{\gamma ,m} \Bigl \Vert \frac{\lambda _1-\lambda _2}{\lambda ^*} \Bigl \Vert _{L^\infty (t_0/2,\infty ) }. \end{aligned}$$
(7.8)

For the proof of the above result we will need the following calculation.

Lemma 7.2

Let

$$\begin{aligned} f(w) = \frac{1}{(4\pi )^3} \int _{{\mathbb {R}}^6} e^{-\frac{|z|^2}{4}} \frac{1}{|w-z|^4}dz , \quad w \in {\mathbb {R}}^6. \end{aligned}$$

Then

$$\begin{aligned} f(w) = \frac{1}{|w|^4}\Bigl [ 1- e^{ \frac{-|w|^2}{4} }\Bigl ( 1+\frac{|w|^{2}}{4}\Bigr ) \Bigr ] . \end{aligned}$$
(7.9)

Proof

Let \(\varphi _0\) be given by

$$\begin{aligned} \varphi _0(x,t) = \frac{1}{(4\pi )^3} \frac{1}{t^3} \int _{{\mathbb {R}}^6} e^{-\frac{|y|^2}{4t}} \frac{1}{|x-y|^4}dy, \quad x \in {\mathbb {R}}^6, \ t>0, \end{aligned}$$

which solves

$$\begin{aligned} \partial _t \varphi _0&= \Delta _{{\mathbb {R}}^6} \varphi _0 \quad \text {in }{\mathbb {R}}^6\times (0,\infty ) \\ \varphi _0(x,0)&= \frac{1}{|x|^4}. \end{aligned}$$

Then

$$\begin{aligned} f(w) = \varphi _0(w,1). \end{aligned}$$

Write

$$\begin{aligned} \varphi _0(x,t) = \frac{1}{t^2}q\Bigl (\frac{|x|}{\sqrt{t}}\Bigr ) . \end{aligned}$$

Then

$$\begin{aligned} q''(s) + \frac{5}{s}q'(s)+ \frac{s}{2} q'(s) +2 q(s) = 0 \end{aligned}$$

and we want q(s) bounded for \(s\rightarrow 0\), \(q(s) = s^{-4}(1+o(1))\) as \(s\rightarrow \infty \). A calculation using the explicit element in the kernel of the linear operator, \(s^{-4}\), gives

$$\begin{aligned} q(s) = \frac{1}{s^4}\Bigl [ 1-e^{-\frac{s^2}{4}}\Bigl ( 1+\frac{s^2}{4}\Bigr )\Bigr ], \quad s>0, \end{aligned}$$

and then (7.9) follows. \(\square \)

Proof of Lemma 7.1

The solution \(\varphi [p,\lambda ]\) of (7.4) has the formula

$$\begin{aligned} \varphi [p,\lambda ](x,t) = \frac{1}{(4\pi )^3} \int _{t_0/2}^t \frac{p(s)}{\lambda ^4(s)} \frac{1}{(t-s)^3} \int _{{\mathbb {R}}^6} e^{-\frac{|x-y|^2}{4(t-s)}} Z_0\Bigl (\frac{y}{\lambda (s)}\Bigr ) \chi \Bigr ( \frac{y}{\sqrt{s}}\Bigl ) dy ds , \quad x \in {\mathbb {R}}^6. \end{aligned}$$

Writing

$$\begin{aligned} \varphi = \varphi [p,\lambda ] \end{aligned}$$

we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi (x,t)\,dx&= \frac{2}{\pi ^2} \int _{{\mathbb {R}}^6} \varphi (x,t)|x|^{-4}dx \\&= \frac{2}{\pi ^2} \frac{1}{(4\pi )^3} \int _{t_0/2}^t \frac{p(s)}{\lambda ^4(s)} \frac{1}{(t-s)^3} \\&\quad \int _{{\mathbb {R}}^6} \int _{{\mathbb {R}}^6} e^{-\frac{|x-y|^2}{4(t-s)}} |x|^{-4}dx Z_0\Bigl (\frac{y}{ \lambda (s) }\Bigr ) \chi \Bigr ( \frac{y}{\sqrt{s}}\Bigl ) dy ds \\&=\frac{2}{\pi ^2} \frac{1}{(4\pi )^3} \int _{t_0/2}^t \frac{p(s)}{\lambda (s)^4}\\&\quad \int _{{\mathbb {R}}^6} \int _{{\mathbb {R}}^6} e^{-\frac{|z|^2}{4}} \frac{1}{|y-\sqrt{t-s}z|^4} dz Z_0\Bigl (\frac{y}{ \lambda (s) }\Bigr ) \chi \Bigr ( \frac{y}{\sqrt{s}}\Bigl ) dy ds \end{aligned}$$

Using (7.9) we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi (x,t)dx&= \frac{2}{\pi ^2 } \int _{t_0/2}^t \frac{p(s)}{\lambda (s)^4} \int _{{\mathbb {R}}^6} \frac{1}{(t-s)^2} f( (t-s)^{-1/2}|y|) Z_0\Bigl (\frac{y}{ \lambda (s) }\Bigr ) \chi \Bigr ( \frac{y}{\sqrt{s}}\Bigl ) dy ds \\&= 2\pi \int _{t_0/2}^t \frac{p(s)}{\lambda (s)^4} \int _0^\infty \Bigl [ 1-e^{-\frac{r^2}{4(t-s)}} \Bigl ( 1+\frac{r^2}{4(t-s)}\Bigr ) \Bigr ] Z_0\Bigl (\frac{r}{ \lambda (s) }\Bigr ) \chi \Bigr ( \frac{r}{\sqrt{s}}\Bigl ) r dr ds . \end{aligned}$$

Let us notice that

$$\begin{aligned}&\frac{1}{2\pi } \int _{{\mathbb {R}}^2} \varphi (x,t)dx \\&\quad = \int _{t_0/2}^t \frac{p(s)(t-s)}{\lambda (s)^4} \int _0^\infty \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] Z_0\Bigl (\frac{z\sqrt{t-s}}{ \lambda (s) }\Bigr ) \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) z dz ds . \end{aligned}$$

We decompose

$$\begin{aligned} \frac{1}{2\pi } \int _{{\mathbb {R}}^2} \varphi (x,t)dx&= I_1 + I_2 + I_3 \end{aligned}$$

where

$$\begin{aligned} I_1&= \int _{t_0/2}^{t/2} ... \\ I_2&= \int _{t/2}^{t-\lambda (t)^2} ... \\ I_3&= \int _{t-\lambda (t)^2}^t ... \end{aligned}$$

and separately estimate each term. To estimate \(I_1\) we note that for \(s \le t/2\) we have \(\frac{s}{t-s}\le 1\). Assuming that \(\chi (x) = 0 \) for \(x\ge 2\) we obtain

$$\begin{aligned} \int _0^\infty \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] Z_0\Bigl (\frac{z\sqrt{t-s}}{ \lambda (s) }\Bigr ) \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) z dz&= \int _{ 0 } ^{2 \frac{\sqrt{s}}{\sqrt{t-s}} } ... \end{aligned}$$

We estimate for \(s \le t/2\),

$$\begin{aligned}&\Bigl | \int _{0} ^{2 \frac{\sqrt{s}}{\sqrt{t-s}} } \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] Z_0\Bigl (\frac{z\sqrt{t-s}}{ \lambda (s) }\Bigr ) \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) z dz \Bigr | \\&\quad \le C \int _{0} ^{2 \frac{\sqrt{s}}{\sqrt{t-s}} } z^4 \frac{\lambda (s)^4}{(t-s)^2 z^4} z dz \\&\quad \le C \frac{\lambda (s)^4 s}{(t-s)^3 } , \end{aligned}$$

where we have used that \(Z_0(\rho ) \le C / \rho ^4\) and \(1-e^{-\frac{z^2}{4}} ( 1+\frac{z^2}{4}) \le C z^4\). Therefore

$$\begin{aligned} |I_1 |&\le \int _{t_0/2}^{t/2} \frac{|p(s)| s }{(t-s)^2}ds \le \Vert p\Vert _{\gamma ,m} \int _0^{t/2} \frac{ s^{1-\gamma } }{(t-s)^2 (\log s)^m}ds \le \frac{C}{t^\gamma (\log t)^m } \Vert p\Vert _{\gamma ,m}. \end{aligned}$$

Let us analyze \(I_2\). We write

$$\begin{aligned} I_2 = I_{2,*}+ I_{2,a} + I_{2,b} + I_{2,c} + I_{2,d} \end{aligned}$$

where

$$\begin{aligned} I_{2,*}&= -16 \int _{t/2}^{t-\lambda (t)^2} \frac{p(s)(t-s)}{\lambda (s)^4} \int _0^\infty \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] \frac{\lambda (s)^4}{(t-s)^2 z^4} z dz ds \end{aligned}$$

and

$$\begin{aligned} I_{2,a}&= \int _{t/2}^{t-\lambda (t)^2} \frac{p(s)(t-s)}{\lambda (s)^4} \int _0^{\frac{\lambda (s)}{\sqrt{t-s}}} \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] Z_0\Bigl (\frac{z\sqrt{t-s}}{\lambda (s)}\Bigr ) \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) z dz ds \\ I_{2,b}&= 16 \int _{t/2}^{t-\lambda (t)^2} \frac{p(s)(t-s)}{\lambda (s)^4} \int _0^{\frac{\lambda (s)}{\sqrt{t-s}}} \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] \frac{\lambda (s)^4}{(t-s)^2z^4} \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) z dz ds \\ I_{2,c}&= \int _{t/2}^{t-\lambda (t)^2} \frac{p(s)(t-s)}{\lambda (s)^4} \int _{\frac{\lambda (s)}{\sqrt{t-s}}} ^\infty \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] \Bigl [ Z_0\Bigl (\frac{z\sqrt{t-s}}{ \lambda (s) }\Bigr ) + 16\frac{\lambda (s)^4}{(t-s)z^4} \Bigr ] z dz ds \\ I_{2,d}&= \int _{t/2}^{t-\lambda (t)^2} \frac{p(s)(t-s)}{\lambda (s)^4} \int _0^\infty \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] Z_0\Bigl (\frac{z\sqrt{t-s}}{ \lambda (s) }\Bigr ) \Bigl [ \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) -1 \Bigr ] z dz ds \end{aligned}$$

A calculation gives that

$$\begin{aligned} I_{2,*}&= -2 \int _{t/2}^{t-\lambda (t)^2} \frac{p(s)}{t-s} ds . \end{aligned}$$
(7.10)

Next we find a bound for \(I_{2,a}\). Using that \(Z_0\) is a bounded function and \(| 1-e^{-\frac{z^2}{4}} ( 1+\frac{z^2}{4}) |\le C z^4\), we get

$$\begin{aligned}&\Bigl | \int _0^{\frac{\lambda (s)}{\sqrt{t-s}}} \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] Z_0\Bigl (\frac{z\sqrt{t-s}}{\lambda (s)}\Bigr ) \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) z dz \Bigr | \\&\quad \le C \int _0^{\frac{\lambda (s)}{\sqrt{t-s}}} z^5 dz \le C \frac{\lambda (s)^6}{(t-s)^3} . \end{aligned}$$

It follows that

$$\begin{aligned} |I_{2,a}|&\le C \int _{t/2}^{t-\lambda (t)^2} \frac{|p(s)| \lambda (s)^2}{(t-s)^2}ds \le \frac{C}{t^\gamma (\log t)^m} \Vert p\Vert _{\gamma ,m} \int _{t/2}^{t-\lambda (t)^2} \frac{\lambda (s)^2}{(t-s)^2}ds \\&\le \frac{C}{ t^\gamma (\log t)^m} \Vert p\Vert _{\gamma ,m} . \end{aligned}$$

Using that \(| 1-e^{-\frac{z^2}{4}} ( 1+\frac{z^2}{4}) |\le C z^4\), we get

$$\begin{aligned} \Bigl | \int _0^{\frac{\lambda (s)}{\sqrt{t-s}}} \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] \frac{\lambda (s)^4}{(t-s)^2z^4} \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) z dz \Bigr | C&\le \frac{\lambda (s)^4}{(t-s)^2} \int _0^{\frac{\lambda (s)}{\sqrt{t-s}}} z dz \\&\le C \frac{\lambda (s)^6}{(t-s)^3} , \end{aligned}$$

and similarly as before,

$$\begin{aligned} |I_{2,b}|&\le \frac{C}{t^\gamma (\log t)^m} \Vert p\Vert _{\gamma ,m} . \end{aligned}$$

Using that

$$\begin{aligned} Z_0\Bigl (\frac{z\sqrt{t-s}}{\lambda (s)}\Bigr ) = - 16 \frac{\lambda (s)^4}{(t-s)^2 z^4} + O\Bigl ( \frac{\lambda (s)^6}{(t-s)^3 z^6}\Bigr ), \quad \frac{z\sqrt{t-s}}{\lambda (s)}\ge 1, \end{aligned}$$

we get

$$\begin{aligned} |I_{2,c}|&\le C \int _{t/2}^{t-\lambda (t)^2} \frac{p(s)(t-s)}{\lambda (s)^4} \int _{\frac{\lambda (s)}{\sqrt{t-s}}}^\infty \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] \frac{\lambda (s)^6}{(t-s)^3 z^6} \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) z dz \\&\le C \int _{t/2}^{t-\lambda (t)^2} \frac{p(s) \lambda (s)^2}{(t-s)^2} \int _{\frac{ \lambda (s)}{\sqrt{t-s}}}^\infty \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] \frac{1}{ z^5} dz . \end{aligned}$$

But \(\frac{\lambda (s)}{\sqrt{t-s}}\le 2\) in the considered range of s, and then

$$\begin{aligned} |I_{2,c}|&\le C \int _{t/2}^{t-\lambda (t)^2} \frac{|p(s)| \lambda (s)^2 }{(t-s)^2} \log \Bigl ( \frac{ \lambda (s)^2}{t-s} \Bigr ) ds \\&\le \frac{C}{t^\gamma (\log t)^m} \Vert p\Vert _{\gamma ,m} \int _{t/2}^{t-\lambda (t)^2} \frac{\lambda (s)^2 }{(t-s)^2} \log \Bigl ( \frac{ \lambda (s)^2}{t-s} \Bigr ) ds \\&\le \frac{C}{t^\gamma (\log t)^m}\Vert p\Vert _{\gamma ,m}. \end{aligned}$$

Finally, for \(I_{2,d}\),

$$\begin{aligned}&\Bigl | \int _0^\infty \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] Z_0\Bigl (\frac{z\sqrt{t-s}}{\lambda (s)}\Bigr ) \Bigl [ \chi \Bigr ( \frac{z \sqrt{t-s}}{\sqrt{s}}\Bigl ) -1 \Bigr ] z dz \Bigr | \\&\quad \le \int _{2\sqrt{s}/\sqrt{t-s}}^\infty \Bigl [ 1-e^{-\frac{z^2}{4}} \Bigl ( 1+\frac{z^2}{4}\Bigr ) \Bigr ] \frac{\lambda (s)^4}{(t-s)^2 z^4} z dz \\&\quad \le \frac{\lambda (s)^4}{(t-s)^2} \int _{2\sqrt{s}/\sqrt{t-s}}^\infty z^{-3} dz \\&\quad \le C \frac{\lambda (s)^4}{(t-s) s} . \end{aligned}$$

Then

$$\begin{aligned} |I_{2,d}|&\le C \int _{t/2}^{t-\lambda (t)^2} \frac{|p(s)|(t-s)}{\lambda (s)^4} \frac{\lambda (s)^4}{(t-s) s} ds \le \frac{C}{t^\gamma (\log t)^m}\Vert p\Vert _{\gamma ,m}. \end{aligned}$$

Finally we estimate

$$\begin{aligned} |I_3|&= \Bigl | \int _{t-\lambda (t)^2} ^t \frac{p(s)}{\lambda (s)^2} \int _0^\infty \Bigl [ 1-e^{-\frac{\rho ^2 \lambda ^2 }{4(t-s)}} \Bigl ( 1+\frac{\rho ^2 \lambda ^2 }{4(t-s)}\Bigr ) \Bigr ] Z_0(\rho ) \chi \Bigr ( \frac{\lambda \rho }{\sqrt{s}}\Bigl ) \rho d \rho ds \Bigr | \\&\le C \int _{t-\lambda (t)^2} ^t \frac{|p(s)|}{\lambda (s)^2} ds \\&\le \frac{C}{ t^\gamma (\log t)^m} \Vert p\Vert _{\gamma ,m}. \end{aligned}$$

In summary, by (7.10) we have written

$$\begin{aligned} \frac{1}{2\pi } \int _{{\mathbb {R}}^2} \varphi (x,t)dx&= -2 \int _{t/2}^{t- \lambda (t)^2} \frac{p(s)}{t-s} ds + I_1 + I_{2,a} + I_{2,b} + I_{2,c} + I_{2,d} + I_3 , \end{aligned}$$

and each of the expressions \(I_1\), \( I_{2,a}\), \( I_{2,b} \), \( I_{2,c} \), \( I_{2,d} \), \( I_3\) are linear operators of p with the estimate

$$\begin{aligned} \Vert I_j[p] \Vert _{\gamma ,m} \le C \Vert p\Vert _{\gamma ,m}. \end{aligned}$$

The proof of (7.8) follows from the explicit expressions for the terms \(I_j\) in R, and similar estimates as before. \(\square \)

Lemma 7.3

Suppose that \(\lambda \) satisfies (7.5) and \(\varphi _\lambda ^{(2)}\) be given by (7.3). Then

$$\begin{aligned} \varphi _\lambda ^{(2)}(0,t;\lambda )&= -\frac{\lambda (t)^2}{4 t^2} + O\Bigl ( \frac{1}{t^2 (\log t)^2} \Bigr ) , \end{aligned}$$
(7.11)

as \(t\rightarrow \infty \), where \( O( \frac{1}{t^2 (\log t)^2} ) \) is uniform in \(t_0\). With \(\lambda ^*\) given by (7.7), if \(\lambda _1,\lambda _2\) satisfy

$$\begin{aligned} \Bigl \Vert \frac{\lambda _j}{\lambda ^*} \Bigl \Vert _{L^\infty (t_0/2,\infty ) } <\frac{1}{2}, \quad j=1,2, \end{aligned}$$

then we also have

$$\begin{aligned} | \varphi _{\lambda ^*+\lambda _1}^{(2)}(0,t) - \varphi _{\lambda ^*+\lambda _2}^{(2)}(0,t) | \le \frac{C}{t^2 \log t } \Bigl \Vert \frac{\lambda _1-\lambda _2}{\lambda ^*} \Bigr \Vert _{L^\infty (t_0/2,\infty )}. \end{aligned}$$
(7.12)

Proof

For simplicity of notation let us write \(\varphi (x,t;\lambda ) = \varphi _\lambda ^{(2)}(x,t)\). Let us write the right hand side of equation (7.3) in the following form

$$\begin{aligned} E_2(x,t;\lambda )&= - \frac{1}{2\lambda ^2 t} U (y) \nabla _z z_0(z) \cdot z + \frac{2}{\lambda ^3 t^{1/2}} \nabla _z \chi _0(z) \cdot \nabla _y U (y)\\&\quad + \frac{1}{\lambda ^2 t} \Delta _z \chi _0(z) U (y) \\&\quad - \frac{1}{\lambda ^3 t^{1/2}} U(y) \nabla _z \chi _0(z) \cdot \nabla _y \Gamma _0(y) , \quad y = \frac{x}{\lambda }, \ z = \frac{x}{\sqrt{t}}. \end{aligned}$$

To compute \(\varphi (0,t;\lambda )\) let us define the following approximation of it

$$\begin{aligned} {\hat{\varphi }}(r,t) = \lambda ^2 {\tilde{\varphi }}(r,t) , \end{aligned}$$

where \({\tilde{\varphi }}(r,t)\) solves the radial heat equation in dimension 6:

$$\begin{aligned} \left\{ \begin{aligned} {\partial } _t {\tilde{\varphi }}&= {\partial } _r^2 {\tilde{\varphi }}+ \frac{5}{r} {\partial } _r {\tilde{\varphi }} + \frac{1}{t^3} h\Bigl (\frac{r}{\sqrt{t}}\Bigr ), \\ {\tilde{\varphi }}(r,0)&= 0, \end{aligned} \right. \end{aligned}$$
(7.13)

and

$$\begin{aligned} h(\zeta ) = \frac{8}{\zeta ^4} \left[ \chi _0'' - \frac{3}{\zeta } \chi _0'(\zeta )+ \frac{\zeta }{2}\chi _0'(\zeta ) \right] . \end{aligned}$$

The solution \({\tilde{\varphi }}(r,t)\) to problem (7.13) can be expressed in self-similar form as

$$\begin{aligned} {\tilde{\varphi }}(r,t) = \frac{1}{t^2} g(\zeta ), \quad \zeta = \frac{r}{\sqrt{t}}. \end{aligned}$$

We find for g the equation

$$\begin{aligned} g'' + \frac{5}{\zeta }g' + \frac{\zeta }{2} g ' + 2 g + h(\zeta ) = 0, \quad \zeta \in (0,\infty ). \end{aligned}$$
(7.14)

Using that the function \(\frac{1}{\zeta ^4}\) is in the kernel of the homogeneous equation, we find the explicit solution of (7.14),

$$\begin{aligned} g_0(\zeta ) = - \frac{1}{\zeta ^4} \int _0^\zeta x^3 e^{-\frac{1}{4}x^2} \int _0^x h(y) e^{\frac{1}{4}y^2}y\,dy dx. \end{aligned}$$

To find the solution \({\tilde{\varphi }}\) with suitable decay at infinity we let

$$\begin{aligned} g(\zeta ) = g_0(\zeta ) + \frac{1}{8} {{\bar{z}}}(\zeta ) I, \end{aligned}$$
(7.15)

where

$$\begin{aligned} {{\bar{z}}}(\zeta ) = \frac{1}{\zeta ^4} \int _0^\zeta x^3 e^{-\frac{1}{4}x^2}\,dx \end{aligned}$$

is a second solution of the homogeneous equation, linearly independent of \(\frac{1}{\zeta ^4}\) and

$$\begin{aligned} I = \int _0^\infty x^3 e^{-\frac{1}{4}x^2} \int _0^x h(y) e^{\frac{1}{4}y^2}y\,dy dx. \end{aligned}$$

We observe that

$$\begin{aligned} g(\zeta ) = O( e^{- \frac{1}{4} \zeta ^2} )\quad \hbox {as } \zeta \rightarrow +\infty , \end{aligned}$$

which makes the solution (7.15) the only one with decay faster than \(O(\zeta ^{-4})\) as \(\zeta \rightarrow +\infty \). An explicit calculation gives that \( I = -8, \) and therefore

$$\begin{aligned} {\hat{\varphi }} (0,t) = -\frac{\lambda (t)^2}{4 t^2}. \end{aligned}$$
(7.16)

Then, using a barrier for the equation satisfied by \(\varphi (x,t;\lambda ) - {\hat{\varphi }}(x,t)\) we get

$$\begin{aligned} |\varphi (x,t;\lambda ) - {\hat{\varphi }}(x,t)| \le C \frac{1}{t^2 (\log t)^2} e^{-c\frac{|x|^2}{t}}, \end{aligned}$$
(7.17)

for \(t\ge 2\), where \(0<c<\frac{1}{4}\). From (7.16) and (7.17) we obtain (7.11).

The proof of (7.12) is similar. \(\square \)

Lemma 7.4

Suppose that \(\lambda \) satisfies (7.5) and \(\varphi _\lambda ^{(2)}\) be given by (7.3). Then

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _{\lambda }^{(2)}&= -2\pi \frac{\lambda ^2}{t} - 16 \pi \Upsilon \frac{\lambda ^2}{t} + O \Bigl ( \frac{1}{t^2 ( \log t)^2} \Bigr ). \end{aligned}$$
(7.18)

where \(\Upsilon \) is defined in (2.7), that, is, \(\Upsilon = \int _0^\infty (\chi _0(s)-1) s^{-3}ds\).

Proof

Integrating (7.3)

$$\begin{aligned} \frac{d}{dt} \int _{{\mathbb {R}}^2} \varphi _{\lambda }^{(2)}&= - 4 \varphi _{\lambda }^{(2)}(0,t) - \frac{1}{2 \lambda ^2 t} \int _{{\mathbb {R}}^2} U(y) \nabla _z \chi _0(z)\cdot z dx + \int _{{\mathbb {R}}^2} {{\tilde{E}}} dx . \end{aligned}$$

From (7.11)

$$\begin{aligned} \varphi _{\lambda }^{(2)}(0,t)&= -\frac{\lambda (t)^2}{4 t^2} + O\Big ( \frac{1}{t^2 (\log t)^2}\Bigr ) \end{aligned}$$

and we compute

$$\begin{aligned}&- \frac{1}{2 \lambda ^2 t} U(y) \nabla _z \chi _0(z)\cdot z + {{\tilde{E}}} \\&\quad = - \frac{1}{\lambda ^2} U(y) \nabla _z \chi _0(z)\cdot z +\frac{2}{\lambda ^2} \nabla _x \chi \cdot \nabla _x U + \frac{1}{\lambda ^2} \Delta _x \chi U - \frac{1}{\lambda ^2} U \nabla \chi \cdot \nabla \Gamma _0 \\&\quad = \Bigl [ 4 \frac{\lambda ^2}{t^3} \chi _0'(s) \frac{1}{s^3} - 64 \frac{\lambda ^2}{t^3}\chi _0'(s) \frac{1}{s^5} + 8 \frac{\lambda ^2}{t^3} (\chi _0''(s)+ \frac{1}{s}\chi _0'(s)) \frac{1}{s^4} \\&\qquad +32 \frac{\lambda ^2}{t^3}\chi _0'(s) \frac{1}{s^5} \Bigr ] + O\Bigl ( \frac{\lambda ^4}{t^4}\Bigr ) \chi _{\{ 1\le s \le 2 \} } \\&\quad = 8 \frac{ \lambda ^2}{t^3} \frac{1}{s^4} \Big [ \frac{s}{2} \chi _0'(s) - \frac{3}{s} \chi _0'(s) + \chi _0''(s) \Bigr ] + O\Bigl ( \frac{\lambda ^4}{t^4}\Bigr ) \chi _{\{ 1\le s \le 2 \} } \end{aligned}$$

where \(s = \frac{r}{\sqrt{t}}\). Then

$$\begin{aligned}&- \frac{1}{2 \lambda ^2 t} \int _{{\mathbb {R}}^2} U(y) \nabla _z \chi _0(z)\cdot z dx + \int _{{\mathbb {R}}^2} {{\tilde{E}}} dx \\&\quad = 2\pi \frac{8\lambda ^2}{t^2} \int _0^\infty \frac{1}{s^4} \Big [ \frac{s}{2} \chi _0'(s) - \frac{3}{s} \chi _0'(s) + \chi _0''(s) \Bigr ]sds + O\Bigl ( \frac{\lambda ^4}{t^3}\Bigr ) \\&\quad = 16 \pi \frac{\lambda ^2}{t^2} \Bigl [ \int _0^\infty (\chi _0(s)-1)s^{-3}ds + \int _0^\infty (s^{-3}\chi _0')' ds \Bigl ] + O\Bigl ( \frac{\lambda ^4}{t^3}\Bigr ) \\&\quad = 16 \pi \frac{\lambda ^2}{t^2} \Upsilon + O\Bigl ( \frac{\lambda ^4}{t^3}\Bigr ). \end{aligned}$$

Therefore

$$\begin{aligned} \frac{d}{dt} \int _{{\mathbb {R}}^2} \varphi _{\lambda }^{(2)}&= 2\pi \frac{\lambda (t)^2}{t^2} + 16 \pi \Upsilon \frac{\lambda ^2}{t^2} + O \Bigl ( \frac{1}{t^3 ( \log t)^2} \Bigr ) \end{aligned}$$

and integrating we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _{\lambda }^{(2)}&= -2\pi \frac{\lambda ^2}{t} - 16 \pi \Upsilon \frac{\lambda ^2}{t} + O \Bigl ( \frac{1}{t^2 ( \log t)^2} \Bigr ). \end{aligned}$$

This is the desired expansion (7.18). \(\square \)

As a corollary from Lemma 7.1 and Lemma 7.4 we get:

Corollary 7.1

Assume \(\lambda \) satisfies (7.5). Then

$$\begin{aligned}&\int _{{\mathbb {R}}^2} \varphi _\lambda dx = - 4 \pi \int _{t/2}^{t-\lambda (t)^2} \frac{\lambda {\dot{\lambda }} (s)}{t-s}ds -2\pi \frac{\lambda ^2(t)}{t} - 16 \pi \Upsilon \frac{\lambda ^2(t)}{t}\\&+ O \Bigl ( \frac{1}{t^2 ( \log t)^2} \Bigr ) + R[\lambda {\dot{\lambda }},\lambda ], \end{aligned}$$

where R is as in Lemma 7.1.

Lemma 7.5

Let \({{\tilde{E}}}\) be defined by (3.11). Assume that \(\lambda \) satisfies (7.5). Then

$$\begin{aligned} \int _{{\mathbb {R}}^2} {{\tilde{E}}} |x|^2 dx = - 64 \pi \Upsilon \frac{\lambda ^2}{t} + O\Bigl ( \frac{1}{t^2 (\log t)^2} \Bigr ). \end{aligned}$$
(7.19)

Proof

Similarly to the proof of Lemma 7.4 we have

$$\begin{aligned} {{\tilde{E}}}&= 8 \frac{ \lambda ^2}{t^3} \frac{1}{s^4} \Big [ - \frac{3}{s} \chi _0'(s) + \chi _0''(s) \Bigr ] + O\Bigr ( \frac{\lambda ^4}{t^4} \Bigr ) \chi _{ \{ 1 \le s \le 2 \} } , \end{aligned}$$

where \( r = |x|\), \(s = \frac{r}{\sqrt{t}}\), and so

$$\begin{aligned} \int _{{\mathbb {R}}^2} {{\tilde{E}}} |x|^2 dx&= 16 \pi \frac{ \lambda ^2}{t} \int _0^\infty \frac{1}{s^4} \Big [ - \frac{3}{s} \chi _0'(s) + \chi _0''(s) \Bigr ]s^3ds + O\Bigr ( \frac{\lambda ^4}{t^2} \Bigr ) \\&= -64 \pi \frac{ \lambda ^2}{t} \Upsilon + O\Bigr ( \frac{\lambda ^4}{t^2} \Bigr ). \end{aligned}$$

This is (7.19). \(\square \)

Lemma 7.6

Let E be defined by (3.10). Assume that \(\lambda \) satisfies (7.5). Then

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} E |x|^2 dx \right| \le \frac{C}{t\log (t)}. \end{aligned}$$

Proof

We have from (3.10)

$$\begin{aligned} E( \zeta ,t;\lambda ) = \frac{{\dot{\lambda }}}{\lambda ^3} Z_0\Bigl (\frac{\zeta }{\lambda }\Bigr ) \chi _0\Bigl ( \frac{\zeta }{\sqrt{t}}\Bigr ) + \frac{1}{2\lambda ^2 t} U \Bigl (\frac{\zeta }{\lambda }\Bigr ) \nabla _z \chi _0(z) \cdot z + {{\tilde{E}}} (x,t) , \end{aligned}$$

and we have already computed \(\int _{{\mathbb {R}}^2} {{\tilde{E}}} |x|^2 dx\) in (7.19). We have

$$\begin{aligned} \int _{{\mathbb {R}}^2} Z_0\Bigl (\frac{\zeta }{\lambda }\Bigr ) \chi _0\Bigl ( \frac{\zeta }{\sqrt{t}}\Bigr ) |\zeta |^2\,d\zeta&= 2\pi \lambda ^4 \int _0^\infty Z_0(\rho ) \chi _0\Bigl ( \frac{\lambda \rho }{\sqrt{t}}\Bigr )\rho ^3 \, \textrm{d}\rho \\&= O ( \lambda ^4 \log (t) ), \end{aligned}$$

and so

$$\begin{aligned} \left| \frac{{\dot{\lambda }}}{\lambda ^3} \int _{{\mathbb {R}}^2} Z_0\Bigl (\frac{\zeta }{\lambda }\Bigr ) \chi _0\Bigl ( \frac{\zeta }{\sqrt{t}}\Bigr )|\zeta |^2\,d\zeta \right| \le \frac{C}{t \log t}. \end{aligned}$$

\(\square \)

7.1 Proof of Proposition 5.1

Let

$$\begin{aligned} I[ \lambda ] = 4 \int _{{\mathbb {R}}^2} \varphi _{\lambda } dx - \int _{{\mathbb {R}}^2} {{\tilde{E}}}(\lambda ) |x|^2dx. \end{aligned}$$

For the proof we proceed by linearization, that is we look for a function \(\lambda _0\) satisfying

$$\begin{aligned} |I[\lambda _0](t)|\le C \frac{1}{t^{\frac{3}{2}+\sigma }}, \quad t>t_0 \end{aligned}$$

with the expansion

$$\begin{aligned} \lambda _0(t) = \lambda ^*(t) + {\tilde{\lambda }}_0(t) \end{aligned}$$

where \(\lambda ^*\) was defined in (7.7), that is, \(\lambda ^*(t) = \frac{c_0 }{\sqrt{\log t}}\) and \({\tilde{\lambda }}_0(t)\), \(t>\frac{t_0}{2}\), is a correction. Here \(c_0>0\) is a fixed constant.

We claim that

$$\begin{aligned} |I[\lambda ^*](t)| \le C \frac{\log (\log t)}{t (\log t)^2}, \quad t>\frac{t_0}{2} , \end{aligned}$$
(7.20)

with C independent of \(t_0\). In the rest of the proof C will be a constant independent of \(t_0\) (for \(t_0\) large).

Indeed, using the decomposition (7.1) and the notation (7.4) we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _{\lambda ^*} dx = \int _{{\mathbb {R}}^2} \varphi _{\lambda ^*}^{(1)} dx +\int _{{\mathbb {R}}^2} \varphi _{\lambda ^*}^{(2)} dx \end{aligned}$$

and

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _{\lambda ^*}^{(1)} dx = \int _{{\mathbb {R}}^2} \varphi [p^*,\lambda ^*] dx, \quad p^* = \lambda ^*{\dot{\lambda }}^*. \end{aligned}$$

By Lemma 7.1 we have

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} \varphi [p^*,\lambda ^*] dx + 4\pi \int _{t/2}^{t-\lambda ^*(t)^2} \frac{p^*(s)}{t-s}ds \right| \le C \frac{1}{t (\log t)^2} , \quad t>\frac{t_0}{2}. \end{aligned}$$

Therefore

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} \varphi [p^*,\lambda ^*] dx +4\pi \log (t) p^*(t) \right| \le C \frac{\log (\log t)}{t (\log t)^2}, \quad t>\frac{t_0}{2}. \end{aligned}$$

On the other hand, by Lemma 7.4 we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} \varphi _{\lambda ^*}^{(2)}dx&= -2\pi \frac{\lambda ^*(t)^2}{t} - 16 \pi \Upsilon \frac{\lambda ^*(t)^2}{t} + O\Bigl ( \frac{1}{t(\log t)^2} \Bigr ), \end{aligned}$$

and by Lemma 7.5

$$\begin{aligned} \int _{{\mathbb {R}}^2} {{\tilde{E}}}(\lambda ^*) |x|^2 dx = - 64 \pi \Upsilon \frac{\lambda ^*(t)^2}{t} + O\Bigl ( \frac{1}{t^2 (\log t)^2} \Bigr ). \end{aligned}$$

Using the explicit form of \(\lambda ^* \) and the previous formulas we deduce (7.20).

Next let us rewrite slightly the operator \(I [\lambda ]\) as follows. We have

$$\begin{aligned} I[\lambda ] = 4 \int _{{\mathbb {R}}^2} \varphi [\lambda {\dot{\lambda }},\lambda ] dx + 4 \int _{{\mathbb {R}}^2} \varphi _{\lambda }^{(2)} dx - \int _{{\mathbb {R}}^2} {{\tilde{E}}}(\lambda ) |x|^2dx. \end{aligned}$$

Let us define

$$\begin{aligned} R[p,\lambda ] = \int _{{\mathbb {R}}^2} \varphi [p,\lambda ] dx + 4 \pi \int _{t/2}^{t-\lambda ^*(t)^2} \frac{p(s)}{t-s}ds . \end{aligned}$$

This is similar to the decomposition given in Lemma 7.1, but we have changed the interval of integration to \([\frac{t}{2},t-\lambda ^*(t)^2]\). We decompose the integral

$$\begin{aligned} \int _{t/2}^{t-\lambda ^*(t)^2} \frac{p(s)}{t-s}ds&= \int _{t/2}^{t-t^{1-\vartheta }} \frac{p(s)}{t-s}ds + \int _{t-t^{1-\vartheta }}^{t-\lambda ^*(t)^2} \frac{p(s)}{t-s}ds \\&= \int _{t/2}^{t-t^{1-\vartheta }} \frac{p(s)}{t-s}ds + p(t) \int _{t-t^{1-\vartheta }}^{t-\lambda ^*(t)^2} \frac{1}{t-s}ds \\&\quad - \int _{t-t^{1-\vartheta }}^{t-\lambda ^*(t)^2} \frac{p(t)-p(s)}{t-s}ds \end{aligned}$$

where \(0<\vartheta <\frac{1}{2}\) is a fixed constant.

We change variables \(\mu = \lambda ^2\), so that

$$\begin{aligned} I[\lambda ]&= - 8 \pi {\dot{\mu }} (t) ( (1-\vartheta ) \log (t)- 2\log (\lambda ^*(t)) ) - 8 \pi \int _{t/2}^{t-t^{1-\vartheta }} \frac{{\dot{\mu }}(s)}{t-s}ds \\&\quad + 4 \int _{{\mathbb {R}}^2} \varphi _{\sqrt{\mu }}^{(2)}dx + 2 R[ {\dot{\mu }} ,\sqrt{\mu }] - \int _{{\mathbb {R}}^2} {{\tilde{E}}}(\sqrt{\mu }) |x|^2 dx \\&\quad + 8 \pi \int _{t-t^{1-\vartheta }}^{t-\lambda ^*(t)^2} \frac{{\dot{\mu }}(t)-{\dot{\mu }}(s)}{t-s}ds. \end{aligned}$$

Let \(\eta \) be a smooth cut-off such that \(\eta (t) = 0 \) for \(t<\frac{3}{4}t_0\), \(\eta (t) =1 \) for \(t>t_0\). We define

$$\begin{aligned} {{\tilde{I}}} [\mu ]&= - 8 \pi {\dot{\mu }} (t) ( (1-\vartheta ) \log (t)- 2\log (\lambda ^*(t)) ) - 8 \pi \eta (t) \int _{t/2}^{t-t^{1-\vartheta }} \frac{{\dot{\mu }}(s)}{t-s}ds \\&\quad + 4 \eta (t) \int _{{\mathbb {R}}^2} \varphi _{\sqrt{\mu }}^{(2)}dx + 2 \eta (t) R[ {\dot{\mu }} ,\sqrt{\mu }] - \eta (t) \int _{{\mathbb {R}}^2} {{\tilde{E}}}(\sqrt{\mu }) |x|^2 dx \\&\quad + 8 \pi \eta (t) \int _{t-t^{1-\vartheta }}^{t-\lambda ^*(t)^2} \frac{{\dot{\mu }}(t)-{\dot{\mu }}(s)}{t-s}ds \end{aligned}$$

which we write

$$\begin{aligned} {{\tilde{I}}}[\mu ] = \ell [\mu ] + N[\mu ]+ R[\mu ], \end{aligned}$$

where

$$\begin{aligned} \ell [\mu ](t)&= - 8 \pi {\dot{\mu }} (t) ( (1-\vartheta ) \log (t)- 2\log (\lambda ^*(t)) ) - 8 \pi \eta (t) \int _{t/2}^{t-t^{1-\vartheta }} \frac{{\dot{\mu }}(s)}{t-s}ds \\ N[\mu ](t)&= 4 \eta (t) \int _{{\mathbb {R}}^2} \varphi _{\sqrt{\mu }}^{(2)}dx + 2 \eta (t) R[ {\dot{\mu }} ,\sqrt{\mu }] - \eta (t) \int _{{\mathbb {R}}^2} {{\tilde{E}}}(\sqrt{\mu }) |x|^2 dx \\ R[\mu ](t)&= 8 \pi \eta (t) \int _{t-t^{1-\vartheta }}^{t-\lambda ^*(t)^2} \frac{{\dot{\mu }}(t)-{\dot{\mu }}(s)}{t-s} ds. \end{aligned}$$

Note that \(I[\lambda ](t) = {{\tilde{I}}}[\lambda ^2](t)\) for \(t\ge t_0\).

Instead of finding \(\lambda \) such that \(I[\lambda ]=0\) for \(t>t_0\) we are going to construct \(\mu \) such that

$$\begin{aligned} |{{\tilde{I}}}[\mu ](t) |\le \frac{C}{t^{\frac{3}{2}+\sigma }}, \quad t> \frac{t_0}{2}, \end{aligned}$$

for some \(\sigma >0\).

Let \(\mu ^* = (\lambda ^*)^2\) where \(\lambda ^*\) is defined in (7.7). In a first step we will find \(\mu _1\) so that

$$\begin{aligned} \ell [\mu ^*+ \mu _1] + N[\mu ^*+\mu _1] + R[\mu ^*]= 0 , \quad t> \frac{t_0}{2} . \end{aligned}$$
(7.21)

We will look for \(\mu _1\) with \(\Vert \mu _1\Vert _{*,\gamma ,m}<\infty \) where, for a function \(\mu _1 \in C^1([\frac{t_0}{2}, \infty ))\) with \(\lim _{t\rightarrow \infty } \mu _1(t)=0\) we define

$$\begin{aligned} \Vert \mu _1 \Vert _{*,\gamma ,m} = \sup _{t\ge t_0/2} t^\gamma ( \log t)^m |{\dot{\mu }}_1(t)| = \Vert {\dot{\mu }}_1\Vert _{\gamma ,m}. \end{aligned}$$

Equation (7.21) takes the form

$$\begin{aligned} 0&=- 8 \pi {\dot{\mu }}_1 ( (1-\vartheta ) \log (t)- 2\log (\lambda ^*(t)) ) - 8 \pi \eta (t) \int _{t/2}^{t-t^{1-\vartheta }} \frac{{\dot{\mu }}_1(s)}{t-s}ds \nonumber \\&\quad + \eta (t) e_1(t) + \eta (t) F_1[\mu _1](t) , \quad t>\frac{t_0}{2}, \end{aligned}$$
(7.22)

where

$$\begin{aligned} e_1(t) = {{\tilde{I}}}[\mu ^*] \end{aligned}$$

and \(F_1\) is an operator with the following properties:

$$\begin{aligned} \Vert F_1[{\tilde{\mu }}_1] \Vert _{\gamma ,m}&\le C \Vert {\tilde{\mu }}_1 \Vert _{*,\gamma ,m} , \end{aligned}$$
(7.23)
$$\begin{aligned} \Vert F_1[{\tilde{\mu }}_1] -F_1[{\tilde{\mu }}_2] \Vert _{\gamma ,m}&\le C \Vert {\tilde{\mu }}_1 -{\tilde{\mu }}_2\Vert _{*,\gamma ,m} , \end{aligned}$$
(7.24)

for \({\tilde{\mu }}_j\) satisfying \(\Vert {\tilde{\mu }}_j\Vert _{*,\gamma ,m} \le 1\), with \(0<\gamma <2\), \(m\in {\mathbb {R}}\), where \(\Vert \ \Vert _{\gamma ,m}\) is defined in (7.6). From (7.20) we find

$$\begin{aligned} |e_1(t) |\le C \frac{\log ( \log t)}{t (\log t)^2}, \quad t > \frac{t_0}{2}. \end{aligned}$$

Now we apply the contraction mapping principle to the Eq. (7.22) written in the form

$$\begin{aligned} {\dot{\mu }}_1&= - \eta (t) I_r[{\dot{\mu }}_1] + \frac{1}{8\pi ( (1-\vartheta ) \log (t)- 2\log (\lambda ^*(t)))} \eta (t) \bigl [ e_1(t)\nonumber \\&\quad + F[\mu _1](t) \bigr ] , \ t>\frac{t_0}{2}, \end{aligned}$$
(7.25)

where

$$\begin{aligned} I_r[{\dot{\mu }}_1] = \frac{1}{( (1-\vartheta ) \log (t)- 2\log (\lambda ^*(t))} \int _{t/2}^{t-t^{1-\vartheta }} \frac{{\dot{\mu }}_1(s)}{t-s}ds . \end{aligned}$$

We directly check that

$$\begin{aligned} \Vert I_r[{\dot{\mu }}] \Vert _{\gamma ,m} \le \frac{\vartheta }{1-\vartheta } \Vert {\dot{\mu }}_1\Vert _{\gamma ,m} . \end{aligned}$$

Let X be the space \(X = \{ \mu _1 \in C^1([ \frac{t_0}{2},\infty ) ) \, | \, \lim _{t\rightarrow \infty } \mu _1(t) = 0 \}\) with the norm \(\Vert \mu _1 \Vert _X = \Vert \mu _1 \Vert _{*,1,3-\varepsilon }\), where \(0<\varepsilon <1\). It follows that if \(\vartheta <\frac{1}{2}\) the equation (7.25) has a unique solution \(\mu _1 \) in the ball \({{\overline{B}}}_1(0)\) of X.

Therefore we have found \(\mu _1\) with \(\Vert \mu _1 \Vert _{*,1,3-\varepsilon } \le 1\) so that \(\mu = \mu ^* + \mu _1\) satisfies

$$\begin{aligned} {{\tilde{I}}}[\mu ] = - 8 \pi \eta (t) \int _{t-t^{1-\vartheta }}^{t-\lambda ^*(t)^2} \frac{{\dot{\mu }}_1(t)-{\dot{\mu }}_1(s)}{t-s}ds. \end{aligned}$$
(7.26)

To estimate this remainder we then need a bound for \(\ddot{\mu }\). Differentiating with respect to t in the decompositions used in Lemmas 7.17.47.5 we obtain

$$\begin{aligned} |\dot{e}_1(t)|\le C \frac{\log (\log t)}{t^2 (\log t)^2} , \quad t > \frac{t_0}{2}. \end{aligned}$$

Differentiating in t equation (7.25) and using the contraction mapping principle we get that for any \(\varepsilon >0\) small

$$\begin{aligned} |\ddot{\mu }_1(t)| \le \frac{C}{t^{2-\varepsilon }}. \end{aligned}$$

Using this we find that the remainder (7.26) has the estimate

$$\begin{aligned} \left| \int _{t-t^{1-\vartheta }}^{t-\lambda ^*(t)^2} \frac{{\dot{\mu }}(t)-{\dot{\mu }}(s)}{t-s}ds \right|&\le \frac{C}{t^{1+\vartheta -\varepsilon } }, \quad t>\frac{t_0}{2}, \end{aligned}$$

where \(\mu = \mu ^* + \mu _1\).

Next we introduce another correction \(\mu _2\) to improve the decay of the remainder. We consider \(\mu = \mu ^* + \mu _1 + \mu _2\) and we consider the following equation for \(\mu _2\):

$$\begin{aligned} \ell [\mu ^*+\mu _1+\mu _2] + N[\mu ^*+\mu _1+\mu _2] + R[\mu ^*+\mu _1] = 0 , \quad t > \frac{t_0}{2}. \end{aligned}$$

Similarly as before, this equation can be written as

$$\begin{aligned} 0&=- 8 \pi {\dot{\mu }}_2 ( (1-\vartheta ) \log (t)- 2\log (\lambda ^*(t)) ) - 8 \pi \eta (t) \int _{t/2}^{t-t^{1-\vartheta }} \frac{{\dot{\mu }}_2(s)}{t-s}ds \nonumber \\&\quad + \eta (t) e_2(t) + \eta (t) F_2[\mu _2](t) , \quad t>\frac{t_0}{2}, \end{aligned}$$
(7.27)

where \(F_2\) satisfies the same estimate (7.23) (7.24), and \(e_2\) has the estimate

$$\begin{aligned} |e_2(t) |\le \frac{C}{t^{1+\vartheta -\varepsilon }} , \quad t > \frac{t_0}{2}. \end{aligned}$$

Using again the contraction mapping principle we find a solution \(\mu _2\) of (7.27) with \( \Vert \mu _2\Vert _{*,1+\vartheta -\varepsilon ,1} \le 1\). Then for \(\mu = \mu ^* + \mu _1 + \mu _2\)

$$\begin{aligned} {{\tilde{I}}} [\mu ](t) = - 8 \pi \eta (t) \int _{t-t^{1-\vartheta }}^{t-\lambda ^*(t)^2} \frac{{\dot{\mu }}_2(t)-{\dot{\mu }}_2(s)}{t-s}ds. \end{aligned}$$

To estimate this remainder we need the following bound for \(\ddot{\mu }_2\)

$$\begin{aligned} |\ddot{\mu }_2(t) |\le \frac{C}{t^{2+\vartheta -\varepsilon }} \end{aligned}$$
(7.28)

which is obtained from an estimate for \(\dot{e}_2\), differentiating with respect to t equation (7.27). The estimate for \(\dot{e}_2\) is obtained from an analogous estimate for \(\frac{d^3 \mu _1}{dt^3}\).

From (7.28) we find

$$\begin{aligned} |{{\tilde{I}}} [\mu ](t)| \le \frac{C}{t^{1+2\vartheta -\varepsilon }} \quad t>\frac{t_0}{2}, \end{aligned}$$

where we recall that \(0<\vartheta <\frac{1}{2}\) is arbitrary.

Thus letting \(\lambda _0 = \sqrt{\mu }\), \(\mu = \mu ^* + \mu _1 + \mu _2\) we obtain

$$\begin{aligned} |I[\lambda _0] |\le \frac{C}{t^{1+2\vartheta -\varepsilon }} \quad t>t_0. \end{aligned}$$

Choosing \(\vartheta >\frac{1}{4}\) and \(\varepsilon >0\) small, we obtain the properties stated in Proposition 5.1. \(\square \)

8 Inner Linear Theory

In this section we consider the problem

$$\begin{aligned} \left\{ \begin{aligned} \lambda ^2 \partial _t \phi&= L[\phi ] + B[\phi ]+ h(y,t) +\sum _{j=1}^2 \mu _j(t) W_{1,j} \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \phi (\cdot ,t_0)&= 0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(8.1)

that appears in the inner equations (5.48) and (5.49), where, we recall

$$\begin{aligned} L[\phi ]&= \nabla \cdot \Bigl [ U \nabla \Bigl ( \frac{\phi }{U}- (-\Delta )^{-1}\phi \Bigr ) \Bigr ] , \nonumber \\ (-\Delta )^{-1}\phi (y, t)&= \frac{1}{2\pi } \int _{{\mathbb {R}}^2} \log \Bigl (\frac{1}{|y-z|}\Bigr ) \phi (z,t )dz. \end{aligned}$$
(8.2)

Slightly more general than the operator B defined in (5.47) we will consider

$$\begin{aligned} B[\phi ] = \zeta _1(t) [\phi ]_{rad} + \zeta _2(t) y \cdot \nabla [\phi ]_{rad} + ( \zeta _1(t) \phi _1 + \zeta _2(t) y \cdot \nabla \phi _1) \chi _0\Bigl ( \frac{\lambda y}{5 \sqrt{t}} \Bigr ), \end{aligned}$$

where \([\phi ]_{rad}\) is the radial part of \(\phi \) (defined in (5.46)) and \(\phi _1 = \phi - [\phi ]_{rad}\), and where \(\chi _0\) is the smooth cut-off function defined in (2.5). In the sequel we will keep the same notation for B.

In what follows we will analyze the linear initial value problem (8.1) where we assume that the functions \(\lambda (t)\), \(\zeta _i(t)\) are continuous, \(t_0>1\) and that for some positive numbers c, C we have

$$\begin{aligned} \frac{c}{\sqrt{\log t}}\,\le & {} \, \lambda (t) \, \le \, \frac{C}{\sqrt{\log t}}\quad \text {for all } t>t_0, \\ |\zeta _i(t)|\leqq & {} \frac{C}{t\log ^2 t }\quad \text {for all } t>t_0. \end{aligned}$$

We change the time variable into

$$\begin{aligned} \tau = \tau _0 + \int _{t_0}^t \frac{1}{\lambda (s)^2}ds , \end{aligned}$$

where \(\tau _0 = t_0 \log t_0\). Then

$$\begin{aligned} {{\tilde{c}}}_1 t \log t \, \le \, \tau \, \le \, {{\tilde{c}}}_2 t \log t \end{aligned}$$

for some \({{\tilde{c}}}_1,{{\tilde{c}}}_2>0\). Identifying \(\phi (y,t)\) and h(yt) with \(\phi (y,\tau )\) and \(h(y,\tau )\) we rewrite (8.1) as

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L[\phi ] + B[\phi ] + h +\sum _{j=1}^2 \mu _j(\tau ) W_{1,j} \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \phi (\cdot ,\tau _0)&= 0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(8.3)

We consider problem (8.3) for functions \(h(y,\tau )\) that have fast decay in space. More precisely, we assume that for all \(T>0\) there is \(C_T\) such that

$$\begin{aligned} |h(y,\tau )| \le \frac{C_T}{1+|y|^6} \quad \text {for all } (y,\tau ) \in {\mathbb {R}}^2 \times (\tau _0,T). \end{aligned}$$

In this case, by a solution \(\phi (y,\tau )\) of (8.3) we understand a continuous function \(\phi (y,\tau )\), of class \(C^1\) in y, such that for any \(T>\tau _0\) there exists a \(C_T>0\) with

$$\begin{aligned} |\phi (y,\tau )|+ (1+|y|)|\nabla _y\phi (y,\tau )|\, \le \, \frac{C_T}{1+|y|^6} \quad \text {for all } (y,\tau ) \in {\mathbb {R}}^2 \times (\tau _0,T), \end{aligned}$$
(8.4)

and satisfies the integral equation

$$\begin{aligned} \phi (y,\tau )&= \int _{\tau _0}^\tau \int _{{\mathbb {R}}^2} G(y-z,\tau -s)\, [ -\nabla \phi \nabla \Gamma _0 - \nabla U \nabla (-\Delta )^{-1} \phi \nonumber \\&\quad + 2 U \phi + B[\phi ] + h + \sum _{j=1}^2\mu _j(s) W_{1,j}](z,s)\, \textrm{d}z \,\textrm{d}s , \end{aligned}$$
(8.5)

where \((-\Delta )^{-1}\phi \) is defined in (8.2) and \(G(y,\tau )\) is the two-dimensional heat kernel,

$$\begin{aligned} G(y,\tau ) = \frac{1}{4 \pi \tau } e^{-\frac{|y|^2}{4 \tau }}. \end{aligned}$$

From the formula

$$\begin{aligned} \nabla (-\Delta )^{-1} h (y) = - \frac{1}{2\pi } \int _{{\mathbb {R}}^2} \frac{y-z}{|y-z|^2}h(z)dz \end{aligned}$$

we see that if \(|\phi (y)|\le \frac{C}{1+|y|^6}\) then

$$\begin{aligned} |\nabla (-\Delta )^{-1} \phi (y)|\le \frac{C}{1+|y|} \Vert (1+|y|^6) \phi \Vert _{L^\infty ({\mathbb {R}}^2)}. \end{aligned}$$

Using this estimate, existence and uniqueness of a solution of (8.5) satisfying (8.4) are standard. For a short time \(T>\tau _0\) this is established by a contraction mapping argument in an appropriate \(L^\infty \)-weighted space. Then a direct linear continuation procedure applies.

A first natural condition to impose on h in (8.3) is that

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y,\tau )dy=0\quad \text {for all }\tau >\tau _0, \end{aligned}$$

in order to achieve that the solution has also zero mass at all times.

We want to find solutions to (8.3) that have fast decay in space and time. For this we need to assume fast space-time decay of the right hand side, which we do by working with the following class of norms.

Given positive numbers \(\nu \), p, \(\varepsilon \) and \(m\in {\mathbb {R}}\), we let \(\Vert h\Vert _{\nu ,m,p,\varepsilon }\) denote the least \(K\geqq 0\) such that for all \(\tau >\tau _0\) and for all \(y\in {\mathbb {R}}^2\)

$$\begin{aligned} |h(y,\tau )| \ \le \ \, \frac{K}{\tau ^{\nu } (\log \tau )^{m}} \frac{ 1 }{(1+|y|)^p} {\left\{ \begin{array}{ll} \displaystyle 1 &{} |y|\le \sqrt{ \tau }, \\ \displaystyle \frac{\tau ^{\varepsilon /2}}{|y|^\varepsilon } &{} |y|\ge \sqrt{\tau } . \end{array}\right. } \end{aligned}$$
(8.6)

This is similar to the norm introduced in (6.2) but defined using \(\tau \) instead of t. We will give the results in Sects. 912 using the norm (8.6).

Still, fast decay of the right hand side doesn’t imply fast decay of the solution. For example, consider Eq. (8.1) without the operator B and without the \(\mu _j\), that is,

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L[\phi ] + h(y,t) \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \phi (\cdot ,\tau _0)&= 0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(8.7)

and suppose that h has compact support in space and time, and that \(\phi \) has sufficient space-time decay. Then, multiplying (8.7) by \(|y|^2\) and integrating in \({\mathbb {R}}^2 \times (\tau _0,\infty ) \) gives

$$\begin{aligned} \int _{\tau _0}^\infty \int _{{\mathbb {R}}^2} h (y,\tau ) |y|^2 dy d\tau = 0, \end{aligned}$$

because if \(\phi \) is a regular function with fast decay, then

$$\begin{aligned} \int _{{\mathbb {R}}^2} L[\phi ]|y|^2 dy = 0, \end{aligned}$$

see Remark 9.2 below. It is then necessary to impose a condition on h, or to adjust a parameter in the problem in order to get a fast decay of the solution. We develop here the theory by adjusting the parameter \(c_1\) in the equation below

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L[\phi ] + B[\phi ]+ h(y,t) \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \phi (\cdot ,t_0)&= c_1 {{\tilde{Z}}}_0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(8.8)

where \({{\tilde{Z}}}_0\) is defined in (6.4).

Proposition 8.1

Let \(\sigma >0\), \(\varepsilon >0\) with \(\sigma +\varepsilon <2\) and \(1<\nu <\frac{7}{4}\). Let \(0<q<1\). Then there exists a number \(C>0\) such that for \(t_0\) sufficiently large and all radially symmetric \(h=h(|y|,\tau )\) with \(\Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }<\infty \) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y,\tau )dy&= 0 ,\quad \text {for all } \tau >\tau _0 , \end{aligned}$$

there exists \(c_1 \in {\mathbb {R}}\) and solution \(\phi (y,\tau ) = {\mathcal {T}}^{i,2}_{{{\textbf {p}}}} [h]\) of problem (8.8) that defines a linear operator of h and satisfies the estimate

$$\begin{aligned} \Vert \phi \Vert _{\nu -1,m+q,4,2+\sigma +\varepsilon } \le \frac{C}{(\log \tau _0)^{1-q}} \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

Moreover \(c_1\) is a linear operator of h and

$$\begin{aligned} |c_1|&\le C \frac{1 }{ \tau _0^{\nu -1} (\log \tau _0)^{m+1}} \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

We have stated this result only in the radial setting, because this is what is needed, but there is a version of it in the non-radial case.

The next result is for the problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L[\phi ] + B[\phi ]+ h(y,\tau ) +\sum _{j=1}^2 \mu _j W_{1,j} \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \phi (\cdot ,\tau _0)&= 0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(8.9)

and holds without the radial symmetry assumption.

Proposition 8.2

Let \(0<\sigma <1\), \(\varepsilon >0\) with \(\sigma +\varepsilon <\frac{3}{2}\) and \(1<\nu < \min ( 1+\frac{\varepsilon }{2},3-\frac{\sigma }{2}, \frac{5}{4})\). Let \(0<q<1\). Then there is C such that for \(\tau _0\) large the following holds. Suppose that h satisfies \(\Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }<\infty \) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y,\tau )dy&=0 , \quad \int _{{\mathbb {R}}^2} h(y,\tau )|y|^2dy=0, \quad \text {for all } \tau >\tau _0 . \end{aligned}$$

Then there exists a solution \(\phi (y,\tau )\), \(\mu _j\) of problem (8.9) that defines a linear operator of h and satisfies

$$\begin{aligned} \Vert \phi \Vert _{\nu -\frac{1}{2} ,m+\frac{q}{2},4,2+\sigma +\varepsilon } \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

The parameters \(\mu _j\) satisfy

$$\begin{aligned} \mu _j(\tau ) = - \int _{{\mathbb {R}}^2} h(y,\tau ) y_jdy + {\tilde{\mu }}_j[h](\tau ) \end{aligned}$$

where \({\tilde{\mu }}_j\) are linear functions of h with

$$\begin{aligned} | {\tilde{\mu }}_j[h](\tau )|\le C \frac{1}{\tau ^{\nu +1}(\log \tau )^{m+1}} \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

We denote this solution by \(\phi = {\mathcal {T}}^{i,1}_{{{\textbf {p}}}} [h]\).

Propositions 6.1 and 6.2 given in Sect. 6 are direct corollaries of Propositions 8.1 and  8.2. The only changes are due to the change in the time variable, because \(\tau \sim t \log t\), and the fact that the norms for the solutions in Propositions 6.1 and 6.2 include a gradient term. The estimate for the gradient follows from the weighted \(L^\infty \) estimate, scaling and standard parabolic estimates.

The proofs of Propositions 8.1 and  8.2 are contained in Sects. 912. They are based on an energy inequality obtained by multiplying the equation by a suitable test function, and using an inequality for a quadratic form. Section 9 contains some preliminaries on this quadratic form.

In Proposition 10.1, we obtain an additive decomposition of the solution \(\phi (y,\tau )\) of (8.8) into a part with a relatively slow space decay that loses \(\tau ^{1/2}\) with respect to the time decay of the right hand side, and a term along \(Z_0(y)\) that loses an entire power of \(\tau \). This is the key element for the proof of Proposition 8.1 in Sect. 10 (p.80).

Then the proof of Proposition 8.2 in the radial case uses Proposition 10.1 after formally applying the operator \(L^{-1}\) to the original equation and performing a concentration procedure that improves the space decay of the resulting error. This is done on Sect. 11, and we give there a proof of Proposition 8.2 in the case of radial functions.

The proof of Proposition 8.2 in the general case is in Sect. 12 (p.96).. The idea is that the decomposition obtained in Proposition 10.1 for solutions with no radial mode does not contain the term along \(Z_0\), which allows us to obtain a much better estimate.

9 Preliminaries for the Linear Theory

A central ingredient in obtaining good estimates for the linearized parabolic operator associated to the inner problem is the analysis of the quadratic form

$$\begin{aligned} \phi \mapsto \int _{{\mathbb {R}}^2} g\phi ,\quad g = \frac{\phi }{U} - (-\Delta )^{-1} \phi . \end{aligned}$$
(9.1)

This quadratic form arises when considering the linearized Keller–Segel problem (8.1). Indeed, \(L[\phi ] = \nabla \cdot ( U \nabla g) \) and it is natural to test the Eq. (8.1) with g, since

$$\begin{aligned} \int _{{\mathbb {R}}^2} L[\phi ] g = \int _{{\mathbb {R}}^2} \nabla \cdot ( U \nabla g) g = - \int _{{\mathbb {R}}^2} U |\nabla g|^2. \end{aligned}$$

But from the time derivative we get \(\lambda ^2 \int _{{\mathbb {R}}^2} \partial _t \phi g \), which leads to (9.1).

We observe that g has degeneracy directions. Indeed, if \(\psi = (-\Delta )^{-1} \phi \) then

$$\begin{aligned} \Delta \psi + U(y)\psi = -Ug {\quad \hbox {in } }{\mathbb {R}}^2. \end{aligned}$$

The operator \(\Delta \psi + U(y) \psi \) is classical. It corresponds to linearizing the Liouville equation

$$\begin{aligned} \Delta v + e^{v} =0\quad \text {in }{\mathbb {R}}^2 \end{aligned}$$

around the solution \(\Gamma _0 = \log U\). It is well known that the bounded kernel of this linearization is spanned by the generators of rigid motions, namely dilation and translations of the equation, which are precisely the functions \(z_0,z_1,z_2\) defined by

$$\begin{aligned} \left\{ \begin{aligned} z_0(y)&= \nabla \Gamma _0(y) \cdot y + 2 \\ z_j(y)&= \partial _{y_j} \Gamma _0(y), \quad j=1,2 . \end{aligned} \right. \end{aligned}$$
(9.2)

Note that g is precisely annihilated at the linear combinations of these functions. In the rest of this section we will state and prove several estimates that take into account this issue, which will be crucial later on.

The quadratic form (9.1) can be naturally transformed into a similar one in \(S^2\) by stereographic projection \(\Pi :S^2\setminus \{(0,0,1)\}\rightarrow {\mathbb {R}}^2\)

$$\begin{aligned} \Pi (y_1,y_2,y_3) = \Bigl ( \frac{y_1}{1-y_3},\frac{y_2}{1-y_3}\Bigr ). \end{aligned}$$

For \(\varphi :{\mathbb {R}}^2 \rightarrow {\mathbb {R}}\) we write

$$\begin{aligned} {\tilde{\varphi }} = \varphi \circ \Pi , \quad {\tilde{\varphi }}: S^2 \setminus \{(0,0,1)\}\rightarrow {\mathbb {R}}. \end{aligned}$$

Then we have the following formulas:

$$\begin{aligned}&\int _{S^2} {\tilde{\varphi }} = \frac{1}{2}\int _{{\mathbb {R}}^2} \varphi U \\&\int _{S^2} {{\tilde{U}}} |\nabla _{S^2} {\tilde{\varphi }}|^2 = \int _{{\mathbb {R}}^2} U |\nabla _{{\mathbb {R}}^2} \varphi |^2 \\&\frac{1}{2} {{\tilde{U}}} \Delta _{S^2} {\tilde{\varphi }} = ( \Delta _{{\mathbb {R}}^2} \varphi ) \circ \Pi . \end{aligned}$$

9.1 The Liouville equation

Here we consider the linearized Liouville equation

$$\begin{aligned} \Delta \psi + U \psi + h=0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned}$$
(9.3)

The stereographic projection transforms the linearized Liouville equation (9.3) into

$$\begin{aligned} \Delta _{S^2} {\tilde{\psi }} +2 {\tilde{\psi }} + 2 {{\tilde{h}}} = 0 \end{aligned}$$
(9.4)

in \(S^2{\setminus }\{P\}\), \(P = (0,0,1)\), where \({\tilde{\psi }} = \psi \circ \Pi \), \({{\tilde{h}}} = (U^{-1} h) \circ \Pi \).

The functions in (9.2) are transformed through the stereographic projection into constant multiples of the coordinate functions

$$\begin{aligned} {{\tilde{z}}}_j(\omega ) = c_j \omega _j,\quad j=1,2, \quad {{\tilde{z}}}_0(\omega ) = c_0 \omega _3, \quad \omega = (\omega _1,\omega _2,\omega _3) \in S^2. \end{aligned}$$

By standard elliptic theory, if \({{\tilde{h}}} \in L^p(S^2)\), \(p>2\), then exists a solution \({\tilde{\psi }}_0 \in W^{2,p}(S^2)\) to (9.4) in \(S^2\) if and only if \({{\tilde{h}}}\) satisfies

$$\begin{aligned} \int _{S^2} {{\tilde{h}}} {{\tilde{z}}}_j = 0 , \quad j=1,2,3. \end{aligned}$$

This solution is unique if we normalize it such that

$$\begin{aligned} \int _{S^2} {\tilde{\psi }}_0 {{\tilde{z}}}_j = 0, \quad j=1,2,3, \end{aligned}$$

and then satisfies the estimate

$$\begin{aligned} \Vert {\tilde{\psi }}_0 \Vert _{C^{1,\alpha }(S^2)} \le C \Vert {{\tilde{h}}}\Vert _{L^p(S^2)} \end{aligned}$$

where \(\alpha = 1- \frac{2}{p}\). By subtracting off a suitable linear combination of the functions \({{\tilde{z}}}_j\), \(j=0,1,2\) we obtain the unique solution \({\tilde{\psi }}_1\) to (9.4) in \(S^2\) satisfying

$$\begin{aligned} {\tilde{\psi }}_1(P) = 0, \quad \nabla _{S^2} {\tilde{\psi }}_1(P) = 0 . \end{aligned}$$
(9.5)

For this solution we also have the estimate

$$\begin{aligned} \Vert {\tilde{\psi }}_1 \Vert _{C^{1,\alpha }(S^2)} \le C \Vert {{\tilde{h}}}\Vert _{L^p(S^2)}. \end{aligned}$$
(9.6)

Lemma 9.1

Let \(0<\sigma <1\). Then there is C such that if \(\psi \) satisfies (9.3) and \(\psi (y)\rightarrow 0\) as \(|y|\rightarrow \infty \) with h satisfying \(\Vert (1+|y|)^{3+\sigma }h\Vert _{L^\infty ({\mathbb {R}}^2)}<+\infty \) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} (U \psi + h(y)) dy= 0, \quad \int _{{\mathbb {R}}^2} (U\psi + h(y)) y_j \, \textrm{d}y = 0, \quad j=1,2 , \end{aligned}$$
(9.7)

then

$$\begin{aligned} \Vert (1+|y|)^{1+\sigma }\psi \Vert _{L^\infty ({\mathbb {R}}^2)}\, \le \, C\Vert (1+|y|)^{3+\sigma }h\Vert _{L^\infty ({\mathbb {R}}^2)}. \end{aligned}$$

Remark 9.1

Let \(h:{\mathbb {R}}^2\rightarrow {\mathbb {R}}\) satisfy \(\Vert (1+|y|)^{2+\sigma }h\Vert _{L^\infty ({\mathbb {R}}^2)}<+\infty \) where \(0<\sigma <1\). If

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y) dy = 0 \end{aligned}$$

then

$$\begin{aligned} |(-\Delta )^{-1} h(y)| \le \frac{C}{(1+|y|)^\sigma } \Vert (1+|y|)^{2+\sigma }h\Vert _{L^\infty ({\mathbb {R}}^2)}. \end{aligned}$$

If \(h:{\mathbb {R}}^2\rightarrow {\mathbb {R}}\) satisfy \(\Vert (1+|y|)^{3+\sigma }h\Vert _{L^\infty ({\mathbb {R}}^2)}<+\infty \) where \(0<\sigma <1\) and in addition to mass zero we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y)y_jdy=0, \quad j=1,2, \end{aligned}$$

then

$$\begin{aligned} |(-\Delta )^{-1} h(y)| \le \frac{C}{(1+|y|)^{1+\sigma }} \Vert (1+|y|)^{3+\sigma }h\Vert _{L^\infty ({\mathbb {R}}^2)}. \end{aligned}$$

The first claim is standard. For the second, write

$$\begin{aligned} (-\Delta )^{-1} h(x) = \frac{1}{2\pi } \int _{{\mathbb {R}}^2} \left( \log |x| - \log |x-y| + \frac{y\cdot x}{|x|^2} \right) h(y)dy, \end{aligned}$$

and estimate the integral after splitting it into the regions \(|y|<\frac{|x|}{2}\) and its complement.

Proof of Lemma 9.1

We claim that \(\psi = (-\Delta )^{-1} (U\psi +h)\). Indeed the function \(\psi - (-\Delta )^{-1} (U\psi +h)\) is harmonic in \({\mathbb {R}}^2\) and decays to 0 at infinity, and therefore it is equal to 0. The assumptions (9.7) and Remark 9.1 imply that

$$\begin{aligned} \Vert (1+|y|)^{1+\sigma }\psi \Vert _{L^\infty ({\mathbb {R}}^2)}\, \le \, C\Vert (1+|y|)^{3+\sigma }h\Vert _{L^\infty ({\mathbb {R}}^2)} + C\Vert \psi \Vert _{L^\infty ({\mathbb {R}}^2)}. \end{aligned}$$
(9.8)

Let \({\tilde{\psi }} = \psi \circ \Pi \), so that it satisfies (9.4) in \(S^2\setminus \{P\}\) with \({{\tilde{h}}} = (U^{-1} h)\circ \Pi \). Note that \({{\tilde{h}}} \in L^p(S^2)\) for some \(p>2\). More precisely,

$$\begin{aligned} \Vert {{\tilde{h}}} \Vert _{L^p(S^2)} \le C \Vert (1+|y|)^{3+\sigma } h\Vert _{L^\infty ( {\mathbb {R}}^2)} , \end{aligned}$$
(9.9)

with \(p<\frac{2}{1-\sigma }\). The singularity at P is removable and thus \({\tilde{\psi }}\) satisfies (9.4) in \(S^2\). By elliptic regularity \({\tilde{\psi }} \in C^{1,\alpha }(S^2)\) for some \(\alpha >0\). Since \(\psi \) decays at infinity, \({\tilde{\psi }}(P)=0\). By (9.8) we have also \(\nabla _{S^2} {\tilde{\psi }}(P)= 0\).

We let \({\tilde{\psi }}_1\) denote the solution to (9.4) satisfying (9.5). The solution to (9.4) in \(S^2\) satisfying (9.5) is unique, so that we have \({\tilde{\psi }} = {\tilde{\psi }}_1\) and by estimate (9.6), (9.9) and (9.8) we obtain

$$\begin{aligned} \Vert (1+|y|)^{1+\sigma }\psi \Vert _{L^\infty ({\mathbb {R}}^2)}\, \le \, C\Vert (1+|y|)^{3+\sigma }h\Vert _{L^\infty ({\mathbb {R}}^2)}. \end{aligned}$$

\(\square \)

9.2 A quadratic form

Here we discuss properties of the quadratic form (9.1). For this we consider a function \(\phi :{\mathbb {R}}^2 \rightarrow {\mathbb {R}}\) with sufficient decay, in the form

$$\begin{aligned} |\phi (y)|\le \frac{1}{(1+|y|)^{2+\sigma }} , \end{aligned}$$
(9.10)

with \(0<\sigma <1\), and zero mass,

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi \,dy&=0 . \end{aligned}$$
(9.11)

We recall g defined in (9.1) \(g = \frac{\phi }{U} - (-\Delta )^{-1}\phi \), and use the notation

$$\begin{aligned} \psi = (-\Delta )^{-1}\phi \end{aligned}$$

so that

$$\begin{aligned} -\Delta \psi - U \psi = U g \quad \text {in }{\mathbb {R}}^2. \end{aligned}$$

We next introduce a normalized version of g, namely \(g^\perp \) defined by

$$\begin{aligned} g^\perp = g + a, \end{aligned}$$

where \(a \in {\mathbb {R}}\) is chosen so that

$$\begin{aligned} \int _{{\mathbb {R}}^2} g^\perp U d y = 0 . \end{aligned}$$

As shown in Lemma 9.3 below, the quadratic form \(\int _{{\mathbb {R}}^2} \phi g\) is equivalent to \(\int _{{\mathbb {R}}^2} U(g^\perp )^2 \).

It will be convenient to work with functions \(\phi ^\perp \), \(\psi ^\perp \), which are analogues of \(\phi \), \(\psi \) but associated to \(g^\perp \). In particular, we want a choice of \(\psi ^\perp \) such that

$$\begin{aligned} -\Delta \psi ^\perp - U \psi ^\perp = U g^\perp , \quad \psi ^\perp (y) \rightarrow 0\quad \text {as }|y|\rightarrow \infty . \end{aligned}$$
(9.12)

Let \(\psi _0=1+\frac{1}{2}z_0\), where \(z_0\) is defined in (9.2), and observe that

$$\begin{aligned} -\Delta \psi _0 - U \psi _0 = -U, \quad \psi _0(y) \rightarrow 0\quad \text {as }|y|\rightarrow \infty . \end{aligned}$$

Then \(\psi ^\perp \) defined by

$$\begin{aligned} \psi ^\perp = \psi -a \Bigl (1+\frac{1}{2}z_0\Bigl ) = \psi - a \psi _0, \end{aligned}$$

indeed satisfies (9.12).

Define

$$\begin{aligned} \phi ^\perp = U( g^\perp + \psi ^\perp ), \end{aligned}$$

and obtain the relations

$$\begin{aligned} \phi = \phi ^\perp + \frac{a}{2} U z_0 , \quad -\Delta \psi ^\perp = \phi ^\perp , \quad \int _{{\mathbb {R}}^2} \phi ^\perp =0. \end{aligned}$$

We note that \(\phi - \phi ^\perp = \frac{a}{2} U z_0\) is a constant times \(Z_0 = U z_0\), which is in the kernel of the operator L.

Lemma 9.2

If \(\phi :{\mathbb {R}}^2 \rightarrow {\mathbb {R}}\) satisfies (9.10) and (9.11), then

$$\begin{aligned} \int _{{\mathbb {R}}^2} g U z_j = \int _{{\mathbb {R}}^2} g^\perp U z_j = 0, \quad j=0,1,2, \end{aligned}$$

where \(z_j\) are the functions defined in (9.2).

Proof

By the definition of \(\psi \) and from (9.10), (9.11) we have

$$\begin{aligned} |\psi (y)| + (1+|y|) |\nabla \psi (y)| \le \frac{C}{(1+|y|)^{\sigma }} , \end{aligned}$$

and hence also

$$\begin{aligned} |\psi ^\perp (y)| + (1+|y|) |\nabla \psi ^\perp (y)| \le \frac{C}{(1+|y|)^{\sigma }}. \end{aligned}$$
(9.13)

We multiply (9.12) by \(z_j\), integrate in the ball \(B_R(0)\) and let \(R\rightarrow \infty \). Since \(z_j\) is in the kernel of \(\Delta +U\) we just have to check that

$$\begin{aligned} \int _{\partial B_R} \Bigl ( \frac{\partial \psi ^\perp }{\partial \nu } z_j - \psi ^\perp \frac{\partial z_j}{\partial \nu }\Bigr ) \rightarrow 0 , \quad \text {as }R\rightarrow \infty , \end{aligned}$$

where \(\nu \) is the exterior normal vector to \(\partial B_R\). This follows from (9.13), and the explicit bounds

$$\begin{aligned}&|z_0(y)| \le C, \quad |z_j(y)| \le \frac{C}{(1+|y|)}, \quad j=1,2, \\&|\nabla z_j(y)| \le \frac{C}{(1+|y|)^{2}}. \end{aligned}$$

\(\square \)

A consequence of the previous lemma is the following.

Remark 9.2

Suppose that \(\phi :{\mathbb {R}}^2 \rightarrow {\mathbb {R}}\) satisfies (9.10) and (9.11). Then

$$\begin{aligned} \int _{{\mathbb {R}}^2} L[\phi ]|y|^2dy=0. \end{aligned}$$

Indeed, integrating on \(B_R\), with the notation \(g = \frac{\phi }{U}-(-\Delta )^{-1}\phi \),

$$\begin{aligned} \int _{B_R} L[\phi ]|y|^2dy&= \int _{B_R} \nabla \cdot ( U \nabla g ) |y|^2dy \\&= -2 \int _{B_R} U \nabla g \cdot y dy + R^2 \int _{\partial B_R} U \nabla g \cdot \nu dS(y) \\&= 2 \int _{B_R} g Z_0 dy - 2 \int _{\partial B_R} U g y \cdot \nu dy + R^2 \int _{\partial B_R} U \nabla g \cdot \nu dS(y) . \end{aligned}$$

By (9.10) and (9.11), \(g(y) = O ( |y|^{2-\sigma } )\), \(\nabla g(y) = O ( |y|^{1-\sigma } )\) as \(|y|\rightarrow \infty \). Therefore the boundary terms tend to 0 as \(R\rightarrow \infty \), and we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} L[\phi ]|y|^2dy = 2 \int _{{\mathbb {R}}^2} g Z_0 dy =0, \end{aligned}$$

by Lemma 9.2.

Lemma 9.3

There are constants \(c_1>0\), \(c_2>0\) such that if \(\phi :{\mathbb {R}}^2 \rightarrow {\mathbb {R}}\) satisfies

$$\begin{aligned} |\phi (y)|\le \frac{1}{(1+|y|)^{3+\sigma }} , \quad 0<\sigma <1 \end{aligned}$$

and (9.11), then

$$\begin{aligned} c_1 \int _{{\mathbb {R}}^2} U (g^\perp )^2 \le \int _{{\mathbb {R}}^2} \phi g^\perp \le c_2 \int _{{\mathbb {R}}^2} U (g^\perp )^2 . \end{aligned}$$
(9.14)

Proof

By Lemma 9.2

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi g&= \int _{{\mathbb {R}}^2} ( \phi ^\perp + \frac{a}{2} U z_0) g = \int _{{\mathbb {R}}^2} \phi ^\perp (g^\perp +a) = \int _{{\mathbb {R}}^2} \phi ^\perp g^\perp \\&= \int _{{\mathbb {R}}^2} U( g^\perp + \psi ^\perp ) g^\perp . \end{aligned}$$

Let \({{\tilde{g}}}^\perp = g^\perp \circ \Pi \), \({\tilde{\psi }}^\perp = \psi ^\perp \circ \Pi \) and write (9.12) as

$$\begin{aligned} -\Delta _{S^2} {\tilde{\psi }}^\perp - 2 {\tilde{\psi }}^\perp = 2 {{\tilde{g}}}^\perp , \quad \text {in }S^2. \end{aligned}$$
(9.15)

We also get

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^2} \phi g&= \int _{S^2} [ ({{\tilde{g}}}^\perp )^2 + {\tilde{\psi }}^\perp {{\tilde{g}}}^\perp ]. \end{aligned}$$

Multiplying (9.15) by \({\tilde{\psi }}^\perp \) we find that

$$\begin{aligned} \int _{S^2} {{\tilde{g}}}^\perp {\tilde{\psi }}^\perp = \frac{1}{2} \int _{S^2}|\nabla _{S^2} {\tilde{\psi }}^\perp |^2 - \int _{S^2}({\tilde{\psi }}^\perp )^2 \end{aligned}$$

and hence

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^2} \phi g = \int _{S^2} ({{\tilde{g}}}^\perp )^2 +\frac{1}{2} \int _{S^2} |\nabla _{S^2} {\tilde{\psi }}^\perp |^2 - \int _{S^2}({\tilde{\psi }}^\perp )^2 . \end{aligned}$$

We recall that the eigenvalues of \(-\Delta \) on \(S^2\) are given by \( \{ k ( k+1) \ | \ k \ge 0 \} \). The eigenvalue 0 has a constant eigenfunction and the eigenvalue 2 has eigenspace spanned by the coordinate functions \(\pi _i(x_1,x_2,x_3) = x_i\), for \((x_1,x_2,x_3) \in S^2\) and \(i=1,2,3\). Let \((\lambda _j)_{j\ge 0}\) denote all eigenvalues, repeated according to multiplicity, with \(\lambda _0 = 0\), \(\lambda _1=\lambda _2=\lambda _3=2\), and let \((e_j)_{j\ge 0}\) denote the corresponding eigenfunctions so that they form an orthonormal system in \(L^2(S^2)\), and \(e_1,e_2,e_3\) are multiples of the coordinate functions \(\pi _1,\pi _2,\pi _3\). We decompose \({\tilde{\psi }}\) and \({{\tilde{g}}}\) to get that

$$\begin{aligned} {\tilde{\psi }}^\perp = \sum _{j=0}^\infty {\tilde{\psi }}^\perp _j e_j, \quad {{\tilde{g}}}^\perp = \sum _{j=0}^\infty {{\tilde{g}}}^\perp _j e_j, \end{aligned}$$
(9.16)

where

$$\begin{aligned} {\tilde{\psi }}^\perp _j = \langle {\tilde{\psi }}^\perp ,e_j\rangle _{L^2(S^2)}, \quad {{\tilde{g}}}^\perp _j = \langle {{\tilde{g}}}^\perp ,e_j\rangle _{L^2(S^2)}. \end{aligned}$$

Then

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^2} \phi g&= \sum _{j=0}^\infty ( {{\tilde{g}}}^\perp _j)^2 +\frac{1}{2} \sum _{j=0}^\infty (\lambda _j -2)({\tilde{\psi }}^\perp _j)^2 \\&= \sum _{j=0}^\infty ( {{\tilde{g}}}^\perp _j)^2 - ({\tilde{\psi }}^\perp _0)^2 +\frac{1}{2} \sum _{j=4}^\infty (\lambda _j -2)({\tilde{\psi }}^\perp _j)^2 . \end{aligned}$$

Equation (9.15) gives us that

$$\begin{aligned} (\lambda _j-2){\tilde{\psi }}^\perp _j = 2{{\tilde{g}}}^\perp _j, \end{aligned}$$
(9.17)

and then

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^2} \phi g&= \sum _{j=1}^\infty ( {{\tilde{g}}}^\perp _j)^2 + \sum _{j=4}^\infty \frac{2}{\lambda _j-2} ({{\tilde{g}}}^\perp _j)^2 . \end{aligned}$$

By Lemma 9.2\({{\tilde{g}}}^\perp _1 = {{\tilde{g}}}^\perp _2 = {{\tilde{g}}}^\perp _3 = 0\). Therefore

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^2} \phi g&= \sum _{j=4}^\infty \frac{\lambda _j}{\lambda _j-2} ({{\tilde{g}}}^\perp _j)^2 \end{aligned}$$
(9.18)

and

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^2} (g^\perp )^2 U = \sum _{j=4}^\infty ({{\tilde{g}}}^\perp _j)^2. \end{aligned}$$

This proves (9.14). \(\square \)

Lemma 9.4

There exist positive constants \(c_1\), \(c_2\) such that if \(\phi :{\mathbb {R}}^2 \rightarrow {\mathbb {R}}\) is radially symmetric and satisfies \((1+|y|)^{3+\sigma } \phi \in L^\infty ({\mathbb {R}}^2)\) with \(0<\sigma <1\), and

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi (y)dy=0,\quad \end{aligned}$$

then

$$\begin{aligned} c_1 \int _{{\mathbb {R}}^2} U (g^\perp )^2 \,&\le \, \int _{{\mathbb {R}}^2} U^{-1} (\phi ^\perp )^2\, \le \, c_2 \int _{{\mathbb {R}}^2} U (g^\perp )^2 , \end{aligned}$$
(9.19)
$$\begin{aligned} \int _{{\mathbb {R}}^2} U (\psi ^\perp )^2\,&\le \, c_2 \int _{{\mathbb {R}}^2} U (g^\perp )^2 . \end{aligned}$$
(9.20)

Proof

Using the same notation as in the proof of Lemma 9.3, we have

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^2} U^{-1}(\phi ^\perp )^2&=\frac{1}{2}\int _{{\mathbb {R}}^2} U[(\psi ^\perp )^2+2\psi ^\perp g^\perp + (g^\perp )^2] \\&=\int _{S^2}[ ({\tilde{\psi }}^\perp )^2 + 2 {\tilde{\psi }}^\perp {{\tilde{g}}}^\perp + ({{\tilde{g}}}^\perp )^2] \\&= \sum _{j=0}^\infty [({\tilde{\psi }}^\perp _j)^2 +2{\tilde{\psi }}^\perp _j {{\tilde{g}}}^\perp _j +({{\tilde{g}}}^\perp _j)^2] . \end{aligned}$$

As in the previous proof, \({{\tilde{g}}}^\perp _j=0\) for \(j=0,1,2,3\). Using (9.17) we get

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^2} U^{-1}(\phi ^\perp )^2&= \sum _{j=0}^3 ({\tilde{\psi }}^\perp _j)^2 +\sum _{j=4}^\infty \frac{\lambda _j^2}{(\lambda _j-2)^2} ({{\tilde{g}}}^\perp _j)^2 . \end{aligned}$$

This formula already gives

$$\begin{aligned} \int _{{\mathbb {R}}^2} U(g^\perp )^2 \le C \int _{{\mathbb {R}}^2} U^{-1}(\phi ^\perp )^2. \end{aligned}$$

We observe that \({\tilde{\psi }}^\perp _1={\tilde{\psi }}^\perp _2=0\) by radial symmetry. We also have \({\tilde{\psi }}^\perp _0=0\), by (9.17). Let

$$\begin{aligned} {\hat{\psi }} = \sum _{j=4}^\infty {\tilde{\psi }}^\perp _j e_j \end{aligned}$$

and note that it satisfies

$$\begin{aligned} -\Delta _{S^2} {\hat{\psi }} -2 {\hat{\psi }} = 2 {{\tilde{g}}}^\perp \quad \text {in } S^2. \end{aligned}$$

By (9.17),

$$\begin{aligned} \Vert {\hat{\psi }}\Vert _{L^2(S^2)} \le C \Vert {{\tilde{g}}}^\perp \Vert _{L^2 (S^2)}, \end{aligned}$$

and from elliptic estimates

$$\begin{aligned} \Vert {\hat{\psi }}\Vert _{C^\alpha (S^2)} \le C \Vert {{\tilde{g}}}^\perp \Vert _{L^2 (S^2)}, \end{aligned}$$
(9.21)

for any \(0<\alpha <1\). Since \((1+|y|)^{3+\sigma }\phi \in L^\infty ( {\mathbb {R}}^2 )\) and \(\phi \) has total mass 0, we have \((1+|y|)^{1+\sigma }\psi \in L^\infty ( {\mathbb {R}}^2 )\) (here the functions are radial) and also \((1+|y|)^{1+\sigma }\psi ^\perp \in L^\infty ( {\mathbb {R}}^2 )\). It follows that \({\tilde{\psi }}^\perp (P)\) = 0 where \(P= (0,0,1)\). Since \({\tilde{\psi }}^\perp \) and \({\hat{\psi }}\) differ by a constant times \(\pi _3\) we have

$$\begin{aligned} {\tilde{\psi }}^\perp = {\hat{\psi }} - \frac{{\hat{\psi }}(P)}{\pi _3(P)} \pi _3, \end{aligned}$$

where \(\pi _3(x_1,x_2,x_3)=x_3\). This implies, by (9.21),

$$\begin{aligned} \Vert {\tilde{\psi }}^\perp \Vert _{L^2(S^2)}&\le C \Vert {\hat{\psi }} \Vert _{L^2(S^2)} + C|{\hat{\psi }}(P)| \le C \Vert {{\tilde{g}}}^\perp \Vert _{L^2(S^2)}. \end{aligned}$$

This proves the other inequality in (9.19) and (9.20). \(\square \)

Lemma 9.5

Suppose that \(\phi = \phi (y,t)\), \(y\in {\mathbb {R}}^2\), \(t>0\) is a function satisfying

$$\begin{aligned} |\phi (y,t)|\le \frac{1}{(1+|y|)^{2+\sigma }}, \end{aligned}$$

with \(0<\sigma <1\),

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi (y,t)\,dy&=0 ,\quad \forall t>0, \end{aligned}$$

and that \(\phi \) is differentiable with respect to t and \(\phi _t\) satisfies also

$$\begin{aligned} |\phi _t(y,t)|\le \frac{1}{(1+|y|)^{2+\sigma }} . \end{aligned}$$

Then

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi _t g = \frac{1}{2} \partial _t \int _{{\mathbb {R}}^2} \phi g \end{aligned}$$

where for each t, g(yt) is defined as

$$\begin{aligned} g = \frac{\phi }{U}-(-\Delta ^{-1})\phi + c (t) \end{aligned}$$

and \(c(t)\in {\mathbb {R}}\) is chosen so that

$$\begin{aligned} \int _{{\mathbb {R}}^2} g(y,t) U (y)\,dy= 0. \end{aligned}$$

Proof

Using the notation of the previous lemma, we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi _t g&= \int _{{\mathbb {R}}^2} U( g_t + \psi _t ) g = 2 \int _{S^2} ( {{\tilde{g}}}_t {{\tilde{g}}} + {\tilde{\psi }}_t {{\tilde{g}}}) . \end{aligned}$$

We have

$$\begin{aligned} -\Delta _{S^2} {\tilde{\psi }} - 2 {\tilde{\psi }} = 2 {{\tilde{g}}} , \quad \text {in }S^2. \end{aligned}$$

And differentiating in t we get

$$\begin{aligned} -\Delta _{S^2} {\tilde{\psi }}_t - 2 {\tilde{\psi }}_t = 2 {{\tilde{g}}}_t , \quad \text {in }S^2. \end{aligned}$$
(9.22)

Multiplying by \({{\tilde{g}}}\) and integrating we find that

$$\begin{aligned} \int _{S^2} {\tilde{\psi }}_t {{\tilde{g}}} = - \frac{1}{2}\int _{S^2} \Delta {\tilde{\psi }}_t {{\tilde{g}}} -\int _{S^2} {{\tilde{g}}}_t {{\tilde{g}}} . \end{aligned}$$

Thus

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi _t g = - \int _{S^2} \Delta {\tilde{\psi }}_t {{\tilde{g}}} \end{aligned}$$

Decompose as in (9.16) and find that

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi _t g = \sum _{j=0}^\infty \lambda _j ({\tilde{\psi }}_j)_t {{\tilde{g}}}_j \end{aligned}$$

But from (9.22)

$$\begin{aligned} (\lambda _j-2) ({\tilde{\psi }}_j)_t = 2({{\tilde{g}}}_j)_t. \end{aligned}$$

We note that \({{\tilde{g}}}_j=0\) for \(j=0,1,2,3\). Indeed, this is true for \(j=0\) by the assumption \(\int _{{\mathbb {R}}^2} g U=0\). By Lemma 9.2 this is true also for \(j=1,2,3\). Then

$$\begin{aligned} \frac{1}{2} \int _{{\mathbb {R}}^2} \phi _t g = \sum _{j=4}^\infty \frac{\lambda _j}{\lambda _j-2} ({{\tilde{g}}}_j)_t {{\tilde{g}}}_j \end{aligned}$$

and the desired conclusion follows from (9.18). \(\square \)

9.3 A Poincaré inequality

Lemma 9.6

Let \(B_R(0)\subset {\mathbb {R}}^2\) be the open ball centered at 0 of radius R. There exists \(C>0\) such that, for any \(R >0\) large and any \(g\in H^1(B_R)\) with \(\int _{B_R} g\,U\,dx =0\) we have

$$\begin{aligned} \frac{C}{ R^2} \int _{B_R} g^2 U\le \int _{B_R} |\nabla g|^2 U. \end{aligned}$$

Proof

Using a Fourier decomposition we only need to consider the radial case, that is, we claim that if g(r) satisfies

$$\begin{aligned} \int _0^R g(r) \frac{r}{(1+r^2)^2}dr = 0, \end{aligned}$$
(9.23)

then there is C such that for all R large

$$\begin{aligned} \int _0^R g(r)^2 \frac{r}{(1+r^2)^2}dr \le CR^2 \int _0^R g'(r)^2 \frac{r}{(1+r^2)^2}dr . \end{aligned}$$

Let \(0<\delta <1\) to be fixed later on. From (9.23) we have

$$\begin{aligned} \int _\delta ^Rg(r) \frac{r}{(1+r^2)^2}dr = - \int _0^\delta g(r) \frac{r}{(1+r^2)^2}dr . \end{aligned}$$

But

$$\begin{aligned} \int _\delta ^Rg(r) \frac{r}{(1+r^2)^2}dr&= -\frac{1}{2} \int _\delta ^R g(r) \frac{d}{dr}\Bigl ( \frac{1}{1+r^2}\Bigr )dr \\&= - \frac{1}{2}\frac{g(R)}{1+R^2} + \frac{1}{2}\frac{g(\delta )}{1+\delta ^2} + \frac{1}{2} \int _\delta ^R g'(r) \frac{1}{1+r^2} dr . \end{aligned}$$

Therefore

$$\begin{aligned} \frac{1}{2}\frac{|g(\delta )|}{1+\delta ^2}&\le \frac{1}{2}\frac{|g(R)|}{1+R^2} + \frac{1}{2} \int _\delta ^R |g'(r)| \frac{1}{1+r^2} dr + \int _0^\delta |g(r)| \frac{r}{(1+r^2)^2}dr . \end{aligned}$$

By the Cauchy-Schwarz inequality

$$\begin{aligned} \int _\delta ^R |g'(r)| \frac{1}{1+r^2} dr \le \Bigl ( \int _\delta ^R g'(r)^2 \frac{r}{(1+r^2)^2}dr\Bigr )^{1/2} ( \log R - \log \delta )^{1/2} \end{aligned}$$
$$\begin{aligned} \int _0^\delta |g(r)| \frac{r}{(1+r^2)^2}dr \le \delta \Bigl ( \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr \Bigr )^{1/2} . \end{aligned}$$

Hence

$$\begin{aligned} g(\delta )^2&\le 2 \frac{g(R)^2}{R^4} + 2 ( \log R - \log \delta ) \int _\delta ^R g'(r)^2 \frac{r}{(1+r^2)^2}dr\nonumber \\&\qquad + 4 \delta ^2 \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr . \end{aligned}$$
(9.24)

We compute now

$$\begin{aligned} \int _\delta ^R g(r)^2 \frac{r}{(1+r^2)^2}dr&= -\frac{1}{2} \int _\delta ^R g(r)^2 \frac{d}{dr}\Bigl ( \frac{1}{1+r^2}\Bigr )dr \\&= - \frac{1}{2}\frac{g(R)^2}{1+R^2} + \frac{1}{2}\frac{g(\delta )^2}{1+\delta ^2} + \int _\delta ^R g(r) g'(r) \frac{1}{1+r^2} dr . \end{aligned}$$

Using (9.24) and the Cauchy-Schwartz inequality we get

$$\begin{aligned} \int _\delta ^R g(r)^2 \frac{r}{(1+r^2)^2}dr&\le - \frac{1}{2}\frac{g(R)^2}{1+R^2} + \frac{g(R)^2}{R^4}\\&\quad + ( \log R - \log \delta ) \int _\delta ^R g'(r)^2 \frac{r}{(1+r^2)^2}dr \\&\quad + 2 \delta ^2 \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr \\&\quad + A R^2 \int _\delta ^R g'(r)^2 \frac{r}{(1+r^2)^2} dr \\&\quad + \frac{1}{A R^2} \int _\delta ^R g(r)^2 \frac{1}{r} dr . \end{aligned}$$

But \(\frac{1}{AR^2 r} \le \frac{1}{2}\frac{r}{(1+r^2)^2}\) for \(r\in [\delta ,R]\) if \(A = 4 (1+\frac{1}{\delta ^2})\) and \(R\ge 1\). Choosing \(A = 4 (1+\frac{1}{\delta ^2})\) and \(R\ge 2\) we have

$$\begin{aligned} \int _\delta ^R g(r)^2 \frac{r}{(1+r^2)^2}dr&\le [ 2AR^2 + 2 ( \log R - \log \delta ) ] \int _\delta ^R g'(r)^2 \frac{r}{(1+r^2)^2}dr \nonumber \\&\quad + 4 \delta ^2 \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr \end{aligned}$$
(9.25)

With \(\delta >0\) still to be chosen we get from (9.24) for \(0<x<\delta \)

$$\begin{aligned} g(x)^2&\le 2 \frac{g(R)^2}{R^4} + 2 ( \log R - \log x) \int _0^R g'(r)^2 \frac{r}{(1+r^2)^2}dr \\&\quad + 4 x^2 \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr . \end{aligned}$$

Integrating we get

$$\begin{aligned} \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr&\le \delta ^2 \frac{g(R)^2}{R^4} + 2 \log R \int _0^R g'(r)^2 \frac{r}{(1+r^2)^2}dr\nonumber \\&\quad + \delta ^4 \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr . \end{aligned}$$
(9.26)

Using the condition (9.23) we obtain

$$\begin{aligned} \int _0^Rg(r) \frac{r}{(1+r^2)^2}dr&= \frac{1}{2} \int _0^R g(r) \frac{d}{dr}\Bigl ( \frac{r^2}{1+r^2}\Bigr )dr \\&= \frac{1}{2} g(R)\frac{R^2}{1+R^2} - \frac{1}{2} \int _0^R g'(r) \frac{r^2}{1+r^2}dr . \end{aligned}$$

Then

$$\begin{aligned} g(R)^2 \le 4 R^4 \int _0^R g'(r)^2 \frac{r}{(1+r^2)^2}dr . \end{aligned}$$

Using this combined with (9.26) we get

$$\begin{aligned} \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr&\le \delta ^2 4 \int _0^R g'(r)^2 \frac{r}{(1+r^2)^2}dr\\&\quad + 2 \log R \int _0^R g'(r)^2 \frac{r}{(1+r^2)^2}dr \\&\quad + \delta ^4 \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr . \end{aligned}$$

Taking \(\delta =\frac{1}{2}\) (this fixes A) gives

$$\begin{aligned} \int _0^\delta g(r)^2 \frac{r}{(1+r^2)^2}dr&\le 4 (\log R +1) \int _0^R g'(r)^2 \frac{r}{(1+r^2)^2}dr. \end{aligned}$$

Combining this with (9.25) we get

$$\begin{aligned} \int _0^R g(r)^2 \frac{r}{(1+r^2)^2}dr \le C R^2 \int _0^R g'(r)^2 \frac{r}{(1+r^2)^2}dr. \end{aligned}$$

\(\square \)

10 Linear Theory: A Decomposition

Here we consider

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L [\phi ] + B[\phi ]+ h, \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \phi (\cdot ,\tau _0)&= \phi _0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(10.1)

The results of this section are going to be used later only in the case of radial functions, so we make this assumption here. We write in the rest of this section \(\phi = \phi ( y,\tau ) = \phi (\rho ,\tau )\), where \(y\in {\mathbb {R}}^2\), \(\rho = |y|\).

The operator B is assumed to be one of the following two:

$$\begin{aligned} B[\phi ] = \zeta (\tau ) ( 2 \phi + y \cdot \nabla \phi ) = \zeta (\tau ) \nabla \cdot ( y \phi ) , \end{aligned}$$
(10.2)

or

$$\begin{aligned} B[\phi ] = \zeta (\tau ) y \cdot \nabla \phi , \end{aligned}$$
(10.3)

where

$$\begin{aligned} \zeta (\tau ) = - \frac{\zeta _0 }{\tau \log \tau }+O\Big (\frac{1}{\tau (\log \tau )^{1+\sigma _0}} \Bigr ),\quad \text {as }\tau \rightarrow \infty , \end{aligned}$$

for some constants \(\zeta _0>0\), \(0<\sigma _0<1\).

We assume that \(\Vert h\Vert _{**}<\infty \) where

$$\begin{aligned} \Vert h\Vert _{**}&= \inf K , \quad \text {such that } \\ |h(y,\tau )|&\le K \frac{ 1 }{ \tau ^\nu (\log \tau )^m } \frac{ 1 }{(1+|y|)^{6+\sigma } } \min \Bigl ( 1, \frac{ \tau ^{\varepsilon /2}}{|y|^\varepsilon } \Bigr ) , \quad \tau >\tau _0, \ y\in {\mathbb {R}}^2, \end{aligned}$$

where \(\nu >1\), \(\varepsilon >0\), \(\sigma >0\), \(m\in {\mathbb {R}}\). This is the same norm as in (8.6).

We also assume that h has zero mass

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y,\tau )dy=0\quad \text {for all }\tau >\tau _0, \end{aligned}$$
(10.4)

and the same for the initial condition

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi _0 dy = 0. \end{aligned}$$
(10.5)

It follows from the equation (10.1), (10.4), and (10.5) that the solution \(\phi \) to (10.1) defined in §8 satisfies

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi (y,\tau )dy=0\quad \text {for all }\tau >\tau _0 . \end{aligned}$$

We recall the decomposition of \(\phi \) introduced in §9.2. Given \(\phi :{\mathbb {R}}^2\rightarrow {\mathbb {R}}\) with sufficient decay and mass zero, we let \(g = \frac{\phi }{U}- (-\Delta ^{-1})\phi \), and define a so that \(\int _{{\mathbb {R}}^2} ( g + a ) U dy= 0\). Then define \(g^\perp = g + a \), \(\psi ^\perp = \psi - a( 1+\frac{1}{2}z_0)\), and

$$\begin{aligned} \phi ^\perp = \phi - \frac{a}{2}Z_0 . \end{aligned}$$
(10.6)

Actually a is directly computed by

$$\begin{aligned} a = - \frac{1}{8 \pi }\int _{{\mathbb {R}}^2} U g = \frac{1}{8 \pi }\int _{{\mathbb {R}}^2} U (-\Delta )^{-1} \phi = \frac{1}{8 \pi }\int _{{\mathbb {R}}^2}\Gamma _0\phi . \end{aligned}$$
(10.7)

In the time dependent situation \(a = a(\tau )\) and all functions depend on \(y\in {\mathbb {R}}^2\) and \(\tau \).

A difficulty to obtain estimates is the presence of a kernel in the linear operator if \(B=0\), since \(Z_0\) satisfies \(L[Z_0]=0\). It can be proved that the solution \(\phi \) of (10.1) with zero initial condition and \(\Vert h\Vert _{**}<\infty \) has the bound

$$\begin{aligned} \sup _{y} |\phi (y,\tau )|\le C \Bigl ( \frac{\log \tau _0}{\log \tau } \Bigr )^{2\zeta _0-\sigma _0} \Vert h\Vert _{**} , \end{aligned}$$

and probably this estimate cannot be improved much. Also \(\phi \) has a some decay at spatial infinity and in particular it has finite second moment

$$\begin{aligned} \int _{{\mathbb {R}}^2} |\phi (y,\tau )| \, |y|^2\,dy<\infty , \quad \tau >\tau _0. \end{aligned}$$

Therefore \(Z_0\) doesn’t describe well the class of solution we want to consider, even for the case \(B=0\), in which \(\zeta (\tau ) \equiv 0\).

A better candidate to describe the solutions \(\phi \) of (10.1) with zero initial condition and \(\Vert h\Vert _{**}<\infty \) is obtained by considering the initial value problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau Z_B&= L [Z_B] + B[Z_B], \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ Z_B(\cdot ,\tau _0)&= {{\tilde{Z}}}_0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(10.8)

where \( {{\tilde{Z}}}_0\) is defined in (6.4). Note that since \(Z_0\) has mass zero and decays like \(1/\rho ^4\) we have \( m_{Z_0} = O ( \frac{1}{\tau _0} ) \).

We will then consider the problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L [\phi ] + B[\phi ]+ h, \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \phi (\cdot ,\tau _0)&= c_1 {{\tilde{Z}}}_0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(10.9)

for radial functions \(\phi \), h, \(\phi _0\), where \(c_1\in {\mathbb {R}}\) is a parameter. We assume that \(\Vert h\Vert _{**}<\infty \).

Proposition 10.1

Let us assume that \(1<\nu <\frac{7}{4}\). Then there is \(C>0\) such that for any \(\tau _0\) sufficiently large the following holds. Suppose that \(\Vert h\Vert _{**}<\infty \) is radially symmetric and satisfies the zero mass condition (10.4). Then there exists \(c_1\) such that the solution \(\phi = \phi ^\perp + \frac{a}{2}Z_0\) of (10.9) satisfies

$$\begin{aligned} |a(\tau ) |&\le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}}\Vert h\Vert _{**}, \nonumber \\ |\phi ^\perp (\rho ,\tau )|&\le C f(\tau ) R(\tau ) \frac{1}{1+|y|^2} \Vert h\Vert _{**}. \end{aligned}$$
(10.10)

where \(R(\tau )>0\) is defined by

$$\begin{aligned} R(\tau )^2 = \frac{ \tau }{(\log \tau )^{q}}, \end{aligned}$$
(10.11)

where \(0< q<1\), and

$$\begin{aligned} f(\tau ) = \frac{1}{\tau ^\nu (\log \tau )^m} . \end{aligned}$$
(10.12)

Moreover \(c_1\) is a linear function of h and satisfies

$$\begin{aligned} |c_1|&\le C \frac{f(\tau _0)R(\tau _0)^2 }{(\log \tau _0)^{1-q}} \Vert h\Vert _{**}. \end{aligned}$$

We always decompose \(\phi \) as in (10.6):

$$\begin{aligned} \phi = \phi ^\perp + \frac{a(\tau )}{2} Z_0 \end{aligned}$$

and write

$$\begin{aligned} g = \frac{\phi }{U}- (-\Delta )^{-1} \phi , \quad g^\perp = \frac{\phi ^\perp }{U}- (-\Delta )^{-1} \phi ^\perp . \end{aligned}$$

Let us denote

$$\begin{aligned} \omega (\tau ) = \Bigl ( \int _{{\mathbb {R}}^2 \setminus B_{R(\tau )}(0)} U g(\tau )^2 \Bigr )^{1/2}. \end{aligned}$$
(10.13)

The strategy for the proof of Proposition 10.1 is contained in the following lemmas. The first one is an a-priori estimate for the solution, assuming that \(a(T_2)=0\) for some \(T_2\).

Lemma 10.1

There is C such that for \(\tau _0\) large the following holds. Suppose that \(\Vert h\Vert _{**}<\infty \) is radially symmetric and satisfies the zero mass condition (10.4) and consider (10.9). Let \(\phi ^\perp \), a be the decomposition (10.6). Suppose that for some \(c_1\in {\mathbb {R}}\) there is \(T_2>\tau _0\) is such that

$$\begin{aligned} a(T_2)=0. \end{aligned}$$

Then

$$\begin{aligned}&|a(\tau )| \le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}}\Vert h\Vert _{**}, \quad \tau \in [\tau _0,T_2] \end{aligned}$$
(10.14)
$$\begin{aligned}&|\omega (\tau )| \le C \frac{f(\tau ) R(\tau )}{(\log \tau _0)^{1-q}}\Vert h\Vert _{**}, \quad \tau \in [\tau _0,T_2] \end{aligned}$$
(10.15)
$$\begin{aligned}&|c_1| \le C \frac{f(\tau _0)R(\tau _0)^2 }{(\log \tau _0)^{1-q}} \Vert h\Vert _{**} . \end{aligned}$$
(10.16)

The constant C is independent of \(T_2\) and \(c_1\).

There is a variant of the previous lemma, where the hypothesis \(a(T_2)=0\) is replaced by an assumption about its time decay.

Lemma 10.2

There is C such that for \(\tau _0\) large the following holds. Suppose that \(\Vert h\Vert _{**}<\infty \) is radially symmetric and satisfies the zero mass condition (10.4) and consider (10.9). Let \(\phi ^\perp \), a be the decomposition (10.6). Suppose that for some \(c_1\in {\mathbb {R}}\),

$$\begin{aligned} \frac{a}{f R^2} \in L^\infty (\tau _0,\infty ). \end{aligned}$$

Then

$$\begin{aligned}&|a(\tau )| \le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}}\Vert h\Vert _{**}, \quad \tau >\tau _0 \end{aligned}$$
(10.17)
$$\begin{aligned}&|\omega (\tau )| \le C \frac{f(\tau ) R(\tau )}{(\log \tau _0)^{1-q}}\Vert h\Vert _{**}, \quad \tau >\tau _0, \end{aligned}$$
(10.18)
$$\begin{aligned}&|c_1| \le C \frac{f(\tau _0)R(\tau _0)^2 }{(\log \tau _0)^{1-q}} \Vert h\Vert _{**} . \end{aligned}$$
(10.19)

Lemma 10.3

Let \(Z_B\) be the solution to (10.8) and write it as \(Z_B = Z_B^\perp + \frac{a_Z}{2}Z_0\) according to the decomposition (10.6). Then \(a_Z(\tau )\not =0\) for all \(\tau \ge \tau _0\).

Lemma 10.4

There is C such that for \(\tau _0\) large the following holds. Suppose that \(\Vert h\Vert _{**}<\infty \) is radially symmetric and satisfies the zero mass condition (10.4). Then there is a unique \(c_1 \in {\mathbb {R}}\) such that the solution \(\phi = \phi ^\perp + \frac{a}{2}Z_0\) of (10.9) (as in (10.6)) satisfies (10.17), (10.18) and (10.19).

In the first results we do some computations and obtain some estimates, which are used as technical steps in the main argument.

The next lemma is a calculation to help us deal with the term B when we multiply the equation by a suitable test function. It holds for operators more general than B as in (10.2) and (10.3). Let

$$\begin{aligned} {{\tilde{B}}}[\phi ] = \zeta _1(\tau ) \phi + \zeta _2(\tau ) y\cdot \nabla \phi , \end{aligned}$$

with \(\zeta _1(\tau )\), \(\zeta _2(\tau )\) satisfying

$$\begin{aligned} |\zeta _i(\tau )| \leqq \frac{C}{\tau \log \tau }\quad \text {for all } \tau >\tau _0. \end{aligned}$$
(10.20)

Lemma 10.5

We have

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} {{\tilde{B}}}[\phi ] g^\perp \right| \le \frac{C}{\tau \log \tau } \int _{{\mathbb {R}}^2} U (g^\perp )^2 dy +C \frac{|a(\tau )|}{\tau \log \tau } \Vert \nabla g^\perp U^{\frac{1}{2}}\Vert _{L^2} . \end{aligned}$$
(10.21)

Proof

We have

$$\begin{aligned} \int _{{\mathbb {R}}^2} {{\tilde{B}}}[\phi ] g^\perp dy = \int _{{\mathbb {R}}^2} [ \zeta _1(\tau ) \phi + \zeta _2(\tau ) y\cdot \nabla \phi ] g^\perp dy. \end{aligned}$$

By Lemma 9.3 and the hypothesis (10.20) we have

$$\begin{aligned} \left| \zeta _1(\tau ) \int _{{\mathbb {R}}^2} \phi g^\perp dy \right|&\le \frac{C}{\tau \log \tau } \int _{{\mathbb {R}}^2} U (g^\perp )^2 dy. \end{aligned}$$
(10.22)

Let us write

$$\begin{aligned} \int _{{\mathbb {R}}^2} y \cdot \nabla \phi (y) g^\perp (y)dy&= \int _{{\mathbb {R}}^2} y \cdot \nabla \phi ^\perp (y) g^\perp (y)dy + \frac{a(\tau )}{2} \int _{{\mathbb {R}}^2} y \cdot \nabla Z_0(y) g^\perp (y)dy. \end{aligned}$$

We claim that

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} y \cdot \nabla \phi ^\perp (y) g^\perp (y)dy\right| \le C \int _{{\mathbb {R}}^2} (g^\perp )^2 Udy. \end{aligned}$$
(10.23)

Indeed, we write

$$\begin{aligned} \int _{{\mathbb {R}}^2} y \cdot \nabla \phi ^\perp (y) g^\perp (y)dy&= \int _{{\mathbb {R}}^2} y \cdot \nabla ( U g^\perp ) g^\perp (y)dy + \int _{{\mathbb {R}}^2} y \cdot \nabla ( U \psi ^\perp ) g^\perp (y)dy . \end{aligned}$$
(10.24)

But

$$\begin{aligned} \int _{{\mathbb {R}}^2} y \cdot \nabla ( U g^\perp ) g^\perp (y)dy&= \int _{{\mathbb {R}}^2} y \cdot \nabla U (g^\perp )^2(y)dy + \int _{{\mathbb {R}}^2} U y \cdot \nabla g^\perp g^\perp (y)dy \\&= \int _{{\mathbb {R}}^2} y \cdot \nabla U (g^\perp )^2(y)dy + \frac{1}{2} \int _{{\mathbb {R}}^2} U y \cdot \nabla [ (g^\perp )^2 ](y)dy \\&= \frac{1}{2}\int _{{\mathbb {R}}^2} y \cdot \nabla U (g^\perp )^2(y)dy - \int _{{\mathbb {R}}^2} U (g^\perp )^2 (y)dy , \end{aligned}$$

and so

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} y \cdot \nabla ( U g^\perp ) g^\perp (y)dy \right| \le C \int _{{\mathbb {R}}^2} (g^\perp )^2 U dy. \end{aligned}$$
(10.25)

The second term in (10.24) is:

$$\begin{aligned} \int _{{\mathbb {R}}^2} y \cdot \nabla ( U \psi ^\perp ) g^\perp (y)dy&= \int _{{\mathbb {R}}^2} (y \cdot \nabla U) \psi ^\perp g^\perp (y)dy + \int _{{\mathbb {R}}^2} U (y \cdot \nabla \psi ^\perp ) g^\perp (y)dy . \end{aligned}$$

We estimate the first term above

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} (y \cdot \nabla U) \psi ^\perp g^\perp (y)dy \right|&\le C \Bigl ( \int _{{\mathbb {R}}^2} (\psi ^\perp )^2 Udy\Bigr )^{1/2} \Bigl ( \int _{{\mathbb {R}}^2} (g^\perp )^2 Udy\Bigr )^{1/2} \nonumber \\&\le C \int _{{\mathbb {R}}^2} (g^\perp )^2 Udy, \end{aligned}$$
(10.26)

by (9.20). To estimate \(\int _{{\mathbb {R}}^2} U (y \cdot \nabla \psi ^\perp ) g^\perp (y)dy \) we write it using radial symmetry:

$$\begin{aligned} \int _{{\mathbb {R}}^2} U (y \cdot \nabla \psi ^\perp ) g^\perp (y)dy = 2 \pi \int _0^\infty U(\rho ) (\psi ^\perp )'(\rho ) g^\perp (\rho ) \rho ^2 d\rho . \end{aligned}$$

We use that \(\psi ^\perp \) satisfies

$$\begin{aligned} -\Delta \psi ^\perp - U \psi ^\perp = U g^\perp \quad \text {in }{\mathbb {R}}^2, \quad \psi ^\perp (\rho ,\tau )\rightarrow 0 \quad \text {as }\rho \rightarrow \infty . \end{aligned}$$

Then, by the variations of parameters formula, since that \(\int _{{\mathbb {R}}^2} U g^\perp z_0 dy=0\), we have

$$\begin{aligned} (\psi ^\perp )'(\rho ) = z_0'(\rho ) \int _\rho ^\infty U(r) g^\perp (r) {{\bar{z}}}_0(r) r \, \textrm{d} r + {{\bar{z}}}_0'(\rho ) \int _0^\rho U(r) g^\perp (r) z_0(r) r \, \textrm{d} r , \end{aligned}$$

where \({{\bar{z}}}_0\) is a second linear independent function in the kernel of \(\Delta +U\) satisfying

$$\begin{aligned} |{{\bar{z}}}_0(\rho ) | \le C( |\log \rho | + 1) . \end{aligned}$$

We then compute

$$\begin{aligned} \int _0^\infty U(\rho ) (\psi ^\perp )'(\rho ) g^\perp (\rho ) \rho ^2 d\rho = I_1 + I_2 \end{aligned}$$

where

$$\begin{aligned} I_1&= \int _0^\infty \int _\rho ^\infty U(\rho ) U(r) z_0'(\rho ) {{\bar{z}}}_0(r) g^\perp (r) g^\perp (\rho ) \rho ^2 r d r d\rho \\ I_2&= - \int _0^\infty \int _\rho ^\infty U(\rho )U(r) {{\bar{z}}}_0'(\rho ) z_0(r) g^\perp (r) g^\perp (\rho ) \rho ^2 r d r d\rho . \end{aligned}$$

We directly check that

$$\begin{aligned} |I_1|+|I_2| \le C \int _{{\mathbb {R}}^2} (g^\perp )^2 Udy. \end{aligned}$$

From this we get that

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} U (y \cdot \nabla \psi ^\perp ) g^\perp (y)dy \right| \le C \int _{{\mathbb {R}}^2} (g^\perp )^2 Udy. \end{aligned}$$
(10.27)

Combining (10.24), (10.25), (10.26), (10.27) we obtain (10.23).

Next we claim that

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} y \cdot \nabla Z_0(y) g^\perp (y)dy \right|&\le C \Vert \nabla g^\perp U^{\frac{1}{2}}\Vert _{L^2}. \end{aligned}$$
(10.28)

Indeed, write

$$\begin{aligned} y \cdot \nabla Z_0&= \nabla \cdot ( y Z_0) - 2 Z_0 = \nabla \cdot ( y Z_0 - 2 \nabla z_0) - 4 Z_0 \end{aligned}$$

where \(z_0\) is defined in (9.2) and satisfies the linearized Liouville equation \(\Delta z_0 + U z_0=0\). We have used here that \(Z_0 = U z_0\). So

$$\begin{aligned} \int _{{\mathbb {R}}^2} y \cdot \nabla Z_0(y) g^\perp (y)dy = - \int _{{\mathbb {R}}^2} ( y Z_0 - 2 \nabla z_0) \nabla g^\perp dy - 4 \int _{{\mathbb {R}}^2} g^\perp Z_0 dy . \end{aligned}$$

But \(\int _{{\mathbb {R}}^2} Z_0 g^\perp dy=\int _{{\mathbb {R}}^2} U z_0 g^\perp dy=0\) by Lemma 9.2, and \(|y Z_0 - 2 \nabla z_0| \le \frac{C}{|y|^4}\), so

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} y \cdot \nabla Z_0(y) g^\perp (y)dy \right|&\le C \Bigl ( \int _{{\mathbb {R}}^2} \frac{1}{(1+|y|)^4}|\nabla g^\perp |^2 dy\Bigr )^{\frac{1}{2}} \le C \Vert \nabla g^\perp U^{\frac{1}{2}}\Vert _{L^2}. \end{aligned}$$

This proves (10.28).

From (10.22), (10.23) and (10.28) we conclude the validity of (10.21).

\(\square \)

In the next lemma we get an estimate for \(\int _{{\mathbb {R}}^2} \phi g^\perp \), but with right hand side that depends on the solution.

Lemma 10.6

We make the same assumptions of Proposition 10.1. Let f be given by (10.12), \(\omega \) be defined in (10.13) and let \(R:[\tau _0,\infty ) \rightarrow (0,\infty )\) be continuous. There is \(c>0\), \(\varepsilon >0\) and \(C>0\) such that for \(\tau _0\) sufficiently large, if

$$\begin{aligned} \sup _{\tau \ge \tau _0}\frac{R^2(\tau )}{\tau \log \tau } \le \varepsilon \end{aligned}$$
(10.29)

then

$$\begin{aligned} \partial _\tau \int _{{\mathbb {R}}^2} \phi g^\perp + \frac{c}{R^2} \int _{{\mathbb {R}}^2} \phi g^\perp \le C f(\tau )^2 \Vert h\Vert _{**}^2 + C \frac{a(\tau )^2}{R^4} + C \frac{\omega (\tau )^2}{R^2}, \end{aligned}$$

for some constant \(c>0\).

Proof

Equation (10.9) can be written in the form

$$\begin{aligned} \partial _\tau \phi = \nabla \cdot ( U \nabla g^\perp ) + B[\phi ] + h , \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) . \end{aligned}$$

We multiply this equation by \(g^\perp \) and integrate on \({\mathbb {R}}^2\), using Lemma 9.5:

$$\begin{aligned} \frac{1}{2} \partial _\tau \int _{{\mathbb {R}}^2} \phi g^\perp + \int _{{\mathbb {R}}^2 } U |\nabla g^\perp |^2 = \int _{{\mathbb {R}}^2} B[\phi ] g^\perp + \int _{{\mathbb {R}}^2} h g^\perp . \end{aligned}$$
(10.30)

Let \(H = (-\Delta )^{-1}h\), and observe that, since h is radial and \(\int _{{\mathbb {R}}^2} h dy = 0\),

$$\begin{aligned} |\nabla H ( \rho ,\tau )|&= \left| \frac{1}{\rho }\int _{\rho }^\infty h(s,\tau ) s ds \right| \le C f(\tau ) \Vert h\Vert _{**} \frac{1}{ (1+\rho )^{5+\sigma }} . \end{aligned}$$

It follows that

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} h g^\perp \right|&= \left| \int _{{\mathbb {R}}^2} \nabla \cdot \nabla H g^\perp \right| = \left| \int _{{\mathbb {R}}^2} \nabla H \cdot \nabla g^\perp \right| \\&\le \frac{1}{2}\int _{{\mathbb {R}}^2} U |\nabla g^\perp |^2 + C \int _{{\mathbb {R}}^2} |\nabla H |^2 U^{-1} \\&\le \frac{1}{2}\int _{{\mathbb {R}}^2} U |\nabla g^\perp |^2 + C f(\tau )^2 \Vert h\Vert _{**}^2 . \end{aligned}$$

This combined with (10.30) gives

$$\begin{aligned} \frac{1}{2} \partial _\tau \int _{{\mathbb {R}}^2} \phi g^\perp + \frac{1}{2}\int _{{\mathbb {R}}^2 } U |\nabla g^\perp |^2 \le \left| \int _{{\mathbb {R}}^2} B[\phi ] g^\perp \right| + C f(\tau )^2 \Vert h\Vert _{**}^2. \end{aligned}$$
(10.31)

We use the inequality in Lemma 9.6 to get

$$\begin{aligned} \frac{c}{R^2} \int _{B_{R}} (g^\perp - {{\bar{g}}}^\perp _{R})^2 U \le \int _{{\mathbb {R}}^2 } U |\nabla g^\perp |^2 , \end{aligned}$$
(10.32)

for some \(c>0\), where

$$\begin{aligned} {{\bar{g}}}^\perp _{R} = \frac{1}{\int _{B_{R}} U }\int _{B_{R}} g^\perp U . \end{aligned}$$

From

$$\begin{aligned} \int _{B_R} (g^\perp )^2 U = \int _{B_R} (g^\perp - {{\bar{g}}}^\perp _R)^2 U + 2 \int _{B_R} g^\perp {{\bar{g}}}^\perp _R U -\int _{B_R} ( {{\bar{g}}}^\perp _R)^2 U \end{aligned}$$

we get

$$\begin{aligned} \int _{B_R} (g^\perp )^2 U \le 2 \int _{B_R} (g^\perp - {{\bar{g}}}^\perp _R)^2 U + C ( {{\bar{g}}}^\perp _R)^2 . \end{aligned}$$

so, using (10.32),

$$\begin{aligned} \frac{c}{R^2}\int _{B_R} (g^\perp )^2 U&\le \int _{{\mathbb {R}}^2 } U |\nabla g^\perp |^2 +C \frac{1}{R^2} ( {{\bar{g}}}^\perp _R)^2 , \end{aligned}$$

for a new \(c>0\). This implies

$$\begin{aligned} \frac{c}{R^2}\int _{{\mathbb {R}}^2} (g^\perp )^2 U&\le \int _{{\mathbb {R}}^2 } U |\nabla g^\perp |^2 +C \frac{1}{R^2} ( {{\bar{g}}}^\perp _R)^2 + C \frac{1}{R^2} \int _{{\mathbb {R}}^2\setminus B_R} U(g^\perp )^2. \end{aligned}$$

Using that \(g^\perp = g + a\) we get

$$\begin{aligned} \frac{c}{R^2}\int _{{\mathbb {R}}^2} (g^\perp )^2 U&\le \int _{{\mathbb {R}}^2 } U |\nabla g^\perp |^2 +C \frac{1}{R^2} ( {{\bar{g}}}^\perp _R)^2 + C \frac{\omega ^2}{R^2} + C \frac{a^2}{R^4}. \end{aligned}$$
(10.33)

But

$$\begin{aligned} \int _{{\mathbb {R}}^2} g^\perp U d y = 0 \end{aligned}$$

and this implies

$$\begin{aligned} {{\bar{g}}}^\perp _{R}&= - \frac{1}{\int _{B_{R}} U } \int _{{\mathbb {R}}^2\setminus B_{R}} g^\perp U \end{aligned}$$

so

$$\begin{aligned} ({{\bar{g}}}_{R}^\perp )^2&\le \frac{C}{R^2}\int _{{\mathbb {R}}^2\setminus B_{R}} (g^\perp )^2 U \le \frac{C a^2}{R^4} + \frac{C}{R^2 }\int _{{\mathbb {R}}^2\setminus B_{R}} g^2 U . \end{aligned}$$

This combined with (10.33) gives

$$\begin{aligned} \frac{c}{R^2} \int _{{\mathbb {R}}^2} (g^\perp )^2 U&\le \int _{{\mathbb {R}}^2 } U |\nabla g^\perp |^2 +C \frac{a^2}{R^4} + C \frac{\omega ^2}{R^2}. \end{aligned}$$

We use this together with (10.31) to obtain (for a new \(c>0\))

$$\begin{aligned}&\frac{1}{2} \partial _\tau \int _{{\mathbb {R}}^2} \phi g^\perp + \frac{c}{R^2}\int _{{\mathbb {R}}^2} (g^\perp )^2 U + \frac{1}{4}\int _{{\mathbb {R}}^2 } U |\nabla g^\perp |^2 \nonumber \\&\quad \le \left| \int _{{\mathbb {R}}^2} B[\phi ] g^\perp \right| + C f(\tau )^2 \Vert h\Vert _{**}^2 + C \frac{a^2}{R^4} + C \frac{\omega ^2}{R^2 }. \end{aligned}$$
(10.34)

We obtain from Lemma 10.5 and the assumption (10.29) that

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} B[\phi ] g^\perp \right|&\le \frac{C}{\tau \log \tau } \int _{{\mathbb {R}}^2} (g^\perp )^2 Udy +C \frac{|a(\tau )|}{\tau \log \tau } \Vert \nabla g^\perp U^{\frac{1}{2}}\Vert _{L^2} \nonumber \\&\le \frac{C}{\tau \log \tau } \int _{{\mathbb {R}}^2} (g^\perp )^2 Udy + \frac{|a(\tau )|^2}{R^4} + C\frac{R^4}{\tau ^2 (\log \tau )^2}\Vert \nabla g^\perp U^{\frac{1}{2}}\Vert _{L^2}^2. \nonumber \\&= C \frac{\varepsilon }{R^2 } \int _{{\mathbb {R}}^2} (g^\perp )^2 Udy + \frac{|a(\tau )|^2}{R^4} + C \varepsilon ^2 \Vert \nabla g^\perp U^{\frac{1}{2}}\Vert _{L^2}^2. \end{aligned}$$
(10.35)

Taking \(\varepsilon >0\) small, and combining (10.34) and (10.35) we get

$$\begin{aligned} \partial _\tau \int _{{\mathbb {R}}^2} \phi g^\perp + \frac{1}{R^2} \int _{{\mathbb {R}}^2} (g^\perp )^2 U \le C f(\tau )^2 \Vert h\Vert _{**}^2 +\frac{C a^2}{R^4} + C \frac{\omega ^2}{R^2}. \end{aligned}$$

By Lemma 9.3 we obtain

$$\begin{aligned} \partial _\tau \int _{{\mathbb {R}}^2} \phi g^\perp + \frac{c}{R^2} \int _{{\mathbb {R}}^2} \phi g^\perp \le C f(\tau )^2 \Vert h\Vert _{**}^2 + C \frac{a^2}{R^4} + C \frac{\omega ^2}{R^2}, \end{aligned}$$

for some constant \(c>0\), which is the desired conclusion. \(\square \)

The next lemma provides a pointwise estimate for \(g = \frac{\phi }{U}- (-\Delta ^{-1})\phi \) assuming a certain bound for \(\Vert U^{1/2} g \Vert _{L^2}\).

Lemma 10.7

Assume \(\nu >0\). Let \(\phi \) be the solution to (10.9) as in §8. Suppose that \(\tau _1\ge \tau _0\) and

$$\begin{aligned} \Vert g (\tau ) U^{\frac{1}{2}} \Vert _{L^2({\mathbb {R}}^2)} \le K_1 f_1(\tau ) ,\quad \tau \in [\tau _0,\tau _1] , \end{aligned}$$
(10.36)

where \(K_1 \ge 0 \) and

$$\begin{aligned} f_1(\tau ) = \frac{(\log \tau )^\mu }{\tau ^{\nu -1}}, \end{aligned}$$

where \(\mu \in {\mathbb {R}}\). Then

$$\begin{aligned} |U(y) g (y,\tau )| \le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)}+ \frac{|c_1|}{f_1(\tau _0)} \Bigr ) f_1(\tau ) \frac{1}{ (1+|y|)^{2}} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$

Proof

We define

$$\begin{aligned} g_0 = U g, \end{aligned}$$

and obtain from (10.1) the equation

$$\begin{aligned} \partial _\tau g_0 = U \partial _\tau g&= \partial _\tau \phi - U (-\Delta ^{-1}) \partial _\tau \phi \nonumber \\&= \nabla \cdot \Bigl [ U \nabla \Bigl ( \frac{g_0}{U} \Bigr ) \Bigr ] - U (-\Delta )^{-1} \left[ \nabla \cdot ( U \nabla g) \right] + h-U(-\Delta )^{-1} h \nonumber \\&\quad + B[g_0] + B[ U \psi [g_0] ] - U (-\Delta )^{-1} ( B [ g_0 + U \psi [g_0]] ) , \end{aligned}$$
(10.37)

where we regard \(\psi [g_0]\) as the operator that maps \(g_0\) to the unique radial solution to

$$\begin{aligned} -\Delta \psi - U \psi = g_0 \quad \text {in }{\mathbb {R}}^2, \quad \psi (\rho ,\tau ) \rightarrow 0 \quad \text {as }\rho \rightarrow \infty . \end{aligned}$$
(10.38)

We note that this problem has indeed a solution since \(\int _{{\mathbb {R}}^2} g_0 z_0 dy = 0\) by Lemma 9.2, which is unique by imposing \(\psi (\rho ,\tau ) \rightarrow 0\) as \( \rho \rightarrow \infty \) in the radial setting. This solution is given by the variations of parameters formula

$$\begin{aligned} \psi (\rho ,\tau ) = z_0(\rho ) \int _\rho ^\infty g_0(r,\tau ) {{\bar{z}}}_0(r) r \, \textrm{d} r + {{\bar{z}}}_0(\rho ) \int _0^\rho g_0(r,\tau ) z_0(r) r \, \textrm{d} r , \end{aligned}$$

where \({{\bar{z}}}_0\) is a second linear independent function in the kernel of \(\Delta +U\) satisfying \( |{{\bar{z}}}_0(\rho ) | \le C( |\log \rho | + 1) \).

We compute

$$\begin{aligned} \nabla \cdot (U\nabla g)&=\Delta g U + \nabla U \cdot \nabla g =\Delta (g U) - \nabla U \cdot \nabla g - g \Delta U, \end{aligned}$$

and hence

$$\begin{aligned} (-\Delta )^{-1}[ \nabla \cdot (U\nabla g)]&= - g U - (-\Delta )^{-1} \left[ \nabla U \cdot \nabla g + g \Delta U \right] \\&= - g U - v \end{aligned}$$

where

$$\begin{aligned} v:= (-\Delta )^{-1} ( \nabla \cdot ( g_0 \nabla \Gamma _0 ) ). \end{aligned}$$
(10.39)

We write (10.37) as

$$\begin{aligned} \partial _\tau g_0&= \Delta g_0 - \nabla g_0 \cdot \nabla \Gamma _0 + 2 U g_0 + B[g_0] + {{\tilde{h}}} \end{aligned}$$
(10.40)

where

$$\begin{aligned} {{\tilde{h}}} = U v + B[ U \psi [g_0] ] - U (-\Delta )^{-1} ( B [ g_0 + U \psi [g_0]] )+ h-U(-\Delta )^{-1} h . \end{aligned}$$
(10.41)

Note that since we are working with radial functions, we can integrate (10.39) explicitly and obtain

$$\begin{aligned} v(\rho ,\tau ) = \int _\rho ^\infty g_0(s,\tau ) \Gamma _0'(s) ds . \end{aligned}$$
(10.42)

We claim that for any \(y \in {\mathbb {R}}^2\):

$$\begin{aligned} \Vert {{\tilde{h}}} \Vert _{ L^p (B_1(y))} \le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)} \Bigr ) f_1(\tau ) \frac{1}{(1+|y|)^{4-\frac{4}{p}}} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$
(10.43)

Indeed, let us start with

$$\begin{aligned} \int _0^\infty |v(\rho )|^p U(\rho ) \rho d \rho&\le \int _0^\infty \Bigl ( \int _\rho ^\infty U(s) g(s)^2 s ds\Bigr )^{p/2}\nonumber \\&\quad \Bigl ( \int _\rho ^\infty U(s) \frac{\Gamma _0'(s)^2}{s} ds\Bigr )^{p/2} U(\rho ) \rho d\rho \nonumber \\&\le C \Vert g U^{\frac{1}{2}}\Vert _{L^2({\mathbb {R}}^2)}^p, \end{aligned}$$
(10.44)

which follows from (10.42) and Hölder’s inequality

Let us write \(\psi =\psi [g_0]\) and \({\tilde{\psi }} = \psi \circ \Pi \), where \(\Pi \) is the stereographic projection. Writing (10.38) in \(S^2\) and using standard \(L^p\) theory we find that for any \(p>2\)

$$\begin{aligned} \Vert {\tilde{\psi }} \Vert _{L^\infty (S^2)}+\Vert \nabla _{S^2} {\tilde{\psi }} \Vert _{L^p(S^2)} \le C \Vert g U^{\frac{1}{2}}\Vert _{L^2({\mathbb {R}}^2)}, \end{aligned}$$

which implies

$$\begin{aligned} \Vert \psi \Vert _{L^\infty ({\mathbb {R}}^2) } + \left( \int _{{\mathbb {R}}^2} |\nabla \psi |^p U^{1-\frac{p}{2}} \right) ^{\frac{1}{p}} \le C \Vert g U^{\frac{1}{2}}\Vert _{L^2({\mathbb {R}}^2)}. \end{aligned}$$
(10.45)

Let \(y\in {\mathbb {R}}^2 \). From (10.36) we see that

$$\begin{aligned} \Vert g_0(\cdot ,\tau ) \Vert _{L^2(B_1(y))} \le C K_1 f_1(\tau ) \frac{1 }{(1+|y|)^{2}} ,\quad \tau \in [\tau _0,\tau _1], \end{aligned}$$

and from (10.36) and (10.44) we have

$$\begin{aligned} \Vert U v (\cdot ,\tau ) \Vert _{L^p(B_1(y))}\le C K_1 f_1(\tau ) \frac{1}{ (1+|y|)^{4-\frac{4}{p}}} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$
(10.46)

Similarly, inequalities (10.45) and (10.36) imply

$$\begin{aligned} \Vert B[U \psi [g_0]] \Vert _{L^p(B_1(y))} \le C K_1 f_1(\tau ) \frac{1}{\tau \log \tau } \frac{1}{ (1+|y|)^{4}} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$
(10.47)

Let’s estimate

$$\begin{aligned} (-\Delta )^{-1} ( B [ g_0 + U \psi [g_0]] )&= \zeta _1(\tau ) (-\Delta )^{-1}( y\cdot \nabla (g_0 + U \psi [g_0]) )\\&\quad + \zeta _2(\tau ) (-\Delta )^{-1}(g_0 + U \psi [g_0]). \end{aligned}$$

Note that \(\psi = (-\Delta )^{-1}\phi = (-\Delta )^{-1}(g_0 + U \psi ) \). But we can estimate \(\psi \) from

$$\begin{aligned} \psi (\rho ) = z_0(\rho )\int _\rho ^\infty \frac{1}{z_0(r)^2 r} \int _r^\infty g_0(s) z_0(s) s ds ,\quad \rho >1 . \end{aligned}$$
(10.48)

Then (10.36) yields

$$\begin{aligned} |\psi (\rho ,\tau ) |&\le \frac{C}{1+\rho } \Vert g U^{\frac{1}{2}}\Vert _{L^2( {\mathbb {R}}^2) } \le C K_1 f_1(\tau ) \frac{1 }{1+\rho } ,\quad \tau \in [\tau _0,\tau _1], \end{aligned}$$
(10.49)

and so

$$\begin{aligned} | U (-\Delta )^{-1} ( g_0 + U \psi [g_0] ) | \le C K_1 f_1(\tau ) \frac{1 }{1+|y|^5} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$
(10.50)

Concerning the term \((-\Delta )^{-1} ( y \cdot \nabla ( g_0 + U \psi ))\), we notice that if we let \(w = g_0 + U \psi \), then \(\int _{{\mathbb {R}}^2} y \cdot \nabla w = 0\), and

$$\begin{aligned} (-\Delta )^{-1} (y \cdot \nabla w)(\rho ) = \int _\rho ^\infty r w(r,\tau )dr - 2 \psi (\rho ,\tau ). \end{aligned}$$

Using (10.36) and (10.49) we get

$$\begin{aligned} \left| ((-\Delta )^{-1} (y \cdot \nabla (g_0 + U \psi )))(\rho ,\tau ) \right| \le C K_1 f_1(\tau ) \frac{1 }{1+\rho } ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$

From this and (10.50) we find that

$$\begin{aligned} |U (-\Delta )^{-1} ( B [ g_0 {+} U \psi [g_0]] )(y,\tau )| {\le } C K_1 f_1(\tau ) \frac{1 }{ \tau \log \tau (1{+}|y|)^5} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$
(10.51)

Finally the estimates

$$\begin{aligned} \Vert h\Vert _{L^p(B_1(y))} + \Vert U (-\Delta )^{-1} h\Vert _{L^p(B_1(y))} \le C \frac{\Vert h\Vert _{**}}{R(\tau _0)} f_1(\tau ) \frac{1}{(1+|y|)^{4-\frac{4}{p}}} \end{aligned}$$
(10.52)

are directly obtained.

Combining (10.46), (10.47), (10.51), and (10.52) we deduce (10.43).

From equation (10.40), the estimate (10.43), standard parabolic \(L^p\) estimates restricted to \(B_1(y)\times ( \max (\tau -1,\tau _0),\tau )\) and embedding into Hölder spaces, we deduce that

$$\begin{aligned} |g_0 (y,\tau )| \le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)}+ \frac{|c_1|}{f_1(\tau _0)} \Bigr ) f_1(\tau ) \frac{1}{ (1+|y|)^{2}} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$
(10.53)

This is the desired conclusion. We also get from (10.53):

$$\begin{aligned} |v (y,\tau )| \le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)}+ \frac{|c_1|}{f_1(\tau _0)} \Bigr ) f_1(\tau ) \frac{1}{ (1+|y|)^{2}} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$
(10.54)

\(\square \)

In some of the proofs below the following barrier will be useful. Consider the equation

$$\begin{aligned} \begin{aligned}&\partial _\tau \phi = \Delta _{{\mathbb {R}}^6} \phi + h \quad \text {in } (\tau _0,\infty )\times {\mathbb {R}}^6 \\&\phi (\tau _0,\cdot ) = 0 \end{aligned} \end{aligned}$$
(10.55)

where \(\Delta _{{\mathbb {R}}^6}\) is the laplacian in \({\mathbb {R}}^6\). Suppose that h has the estimate

$$\begin{aligned} |h (y,\tau ) | \le \frac{1}{\tau ^{\gamma +1} } \frac{1}{(1+|y|/\sqrt{\tau })^b} \end{aligned}$$

for some \(\gamma ,b\in {\mathbb {R}}\).

If \(\gamma <3\) and \(\gamma <\frac{b}{2}\) then there is a barrier satisfying

$$\begin{aligned} C_1 \frac{1}{\tau ^{\gamma } } \frac{1}{ (1+|y|/\sqrt{\tau })^b} \le \phi (y,\tau ) \le C_2 \frac{1}{\tau ^{\gamma } } \frac{1}{ (1+|y|/\sqrt{\tau })^b}. \end{aligned}$$

Indeed, we can consider all functions to be radial and write \(\rho = |y|\), \(y\in {\mathbb {R}}^6\). Let

$$\begin{aligned} {\bar{\phi }}(\rho ,\tau ) = \frac{1}{\tau ^{\gamma } } g\Bigl ( \frac{\rho }{\sqrt{\tau }}\Bigr ) , \quad \zeta = \frac{\rho }{\sqrt{\tau }}. \end{aligned}$$
(10.56)

Then

$$\begin{aligned} \partial _\tau {\bar{\phi }} - \Bigl (\partial _{\rho \rho } + \frac{5}{\rho } \partial _\rho \Bigr ) {\bar{\phi }} = - \frac{1}{\tau ^{\gamma +1}} \Bigl [ g''(\zeta ) + \frac{5}{\zeta } g'(\zeta ) +\frac{\zeta }{2}g'(\zeta ) + \gamma g(\zeta ) \Bigr ]. \end{aligned}$$

Let \(g_1(\zeta ) = \frac{1}{(1+\zeta ^2)^{b/2}}\). Since \(\gamma <\frac{b}{2}\) we have

$$\begin{aligned} -\Bigl [ g_1''(\zeta ) + \frac{5}{\zeta } g_1'(\zeta ) +\frac{\zeta }{2}g_1'(\zeta ) + \gamma g_1(\zeta ) \Bigr ] \ge \frac{c}{\zeta ^b}, \quad \zeta \ge M, \end{aligned}$$

for some \(c, M>0\). Let \(g_0(\zeta ) = e^{-\frac{\zeta ^2}{4}}\) be the Gaussian kernel, which satisfies

$$\begin{aligned} g_0''(\zeta ) + \frac{5}{\zeta } g_0'(\zeta ) +\frac{\zeta }{2}g_0'(\zeta ) + 3 g_0(\zeta ) = 0. \end{aligned}$$

Let \(g = C_1 g_0 + g_1\). Since \(\gamma <3\), we can find \(C_1\) large so that

$$\begin{aligned} -\Bigl [ g''(\zeta ) + \frac{5}{\zeta } g'(\zeta ) +\frac{\zeta }{2}g'(\zeta ) + \gamma g(\zeta ) \Bigr ] \ge \frac{c}{1+\zeta ^b}, \quad \zeta >0. \end{aligned}$$

Then \({\bar{\phi }}\) defined by (10.56) with \(g = C_1 g_0 + g_1\) is a supersolution to (10.55).

In the next lemma we improve the spatial decay of \(g = \frac{\phi }{U}- (-\Delta ^{-1})\phi \).

Lemma 10.8

Assume \(1<\nu <\frac{7}{4}\). Let \(\phi \) be the solution to (10.9) as in §8. Suppose that \(\tau _1\ge \tau _0\) and

$$\begin{aligned} \Vert g (\tau ) U^{\frac{1}{2}} \Vert _{L^2({\mathbb {R}}^2)} \le K_1 f_1(\tau ) ,\quad \tau \in [\tau _0,\tau _1] , \end{aligned}$$

where \(K_1\ge 0\) and

$$\begin{aligned} f_1(\tau ) = \frac{(\log \tau )^\mu }{\tau ^{\nu -1}}, \end{aligned}$$

where \(\mu \in {\mathbb {R}}\). Then

$$\begin{aligned} |U (\rho ) g(\rho ,\tau ) |\le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)}+ \frac{|c_1|}{f_1(\tau _0)} \Bigr ) f_1(\tau ) \frac{1}{(1+\rho )^{4}} , \quad \tau \in [\tau _0,\tau _1] . \end{aligned}$$
(10.57)

Proof

We us the same notation as in Lemma 10.7 and consider (10.40) for \(g_0 = U g\) with \({{\tilde{h}}}\) defined in (10.41). We are going to use barriers to estimate \(g_0\).

We claim that \({{\tilde{h}}}\) satisfies

$$\begin{aligned} |{{\tilde{h}}}(y,\tau )| \le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)} \Bigr ) f_1(\tau ) \Bigl ( \frac{1}{(1+|y|)^6} +\frac{1}{\tau \log \tau (1+|y|)^5} \Bigr ) ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$
(10.58)

Indeed, from (10.53) and (10.54) we find that

$$\begin{aligned} | - U v + h-U(-\Delta )^{-1} h | \le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)}\Bigr ) f_1(\tau ) \frac{1 }{ (1+|y|)^{6}} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$
(10.59)

To estimate \(B[U \psi [g_0]]\) we use (10.49) a similar estimate for \(\partial _\rho \psi \), and the assumptions on \(\zeta _1\), \(\zeta _2\) in (10.20), to obtain

$$\begin{aligned} |B[U\psi [g_0]|\le C K_1 f_1(\tau ) \frac{1 }{\tau \log \tau } \frac{1}{1+|y|^5} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$

This, (10.59) and (10.51) prove (10.58).

To get better spatial decay we construct a barrier and apply the maximum principle to equation (10.40) in \(({\mathbb {R}}^2 {\setminus } B_{R_0} (0) ) \times (\tau _0,\tau _1)\), where \(R_0\) is a fixed large constant. Several of constants C below depend on \(R_0\) but we will not keep track of the explicit dependence.

The linear operator for \(g_0\) in (10.40), acting on radial functions with \(\rho = |y|\), is given by

$$\begin{aligned}&\partial _\tau g_0 -[ \Delta g_0 - \nabla g_0 \cdot \nabla \Gamma _0 + B[g_0] + 2 U g_0] \\&\quad = \partial _\tau g_0 - \partial _{\rho \rho } g_0 - \frac{1}{\rho } \partial _\rho g_0 - \frac{4 \rho }{1+\rho ^2} \partial _\rho g_0 + O\Bigl ( \frac{1}{1+\rho ^4} \Bigr ) g_0 \\&\qquad + O\Bigl ( \frac{1}{\tau \log \tau } \Bigr ) g_0 + O\Bigl ( \frac{1}{\tau \log \tau } \Bigr ) \rho \partial _\rho g_0 . \end{aligned}$$

The main part outside of a ball \(B_{R_0}(0)\) with \(R_0\) big is given by \(\partial _\tau - \partial _{\rho \rho } - \frac{5}{\rho }\partial _\rho \).

By (10.58) we need to construct \({{\bar{g}}}_1\) such that

$$\begin{aligned} \partial _\tau {{\bar{g}}}_1 -[ \Delta {{\bar{g}}}_1 - \nabla {{\bar{g}}}_1 \cdot \nabla \Gamma _0 + B[{{\bar{g}}}_1] +2 U {{\bar{g}}}_1 ] \ge h_1 \end{aligned}$$

where

$$\begin{aligned} h_1(\rho ,\tau )&= f_1(\tau ) \Bigl ( \frac{1}{(1+\rho )^6} +\frac{1}{\tau \log \tau (1+\rho )^5} \Bigr ) . \end{aligned}$$

To construct \({{\bar{g}}}_1\), let \(0<\vartheta <1\), and let \({{\tilde{g}}}_1(\rho )\) be radial and solve

$$\begin{aligned} -\Delta _6 {{\tilde{g}}}_1 = \frac{1}{1+\rho ^{6-\vartheta }} \quad \text {in }{\mathbb {R}}^6, \end{aligned}$$

such that \( {{\tilde{g}}}_1(\rho ) (1+\rho ^{4-\vartheta })\) is bounded below and above by positive constants. Let

$$\begin{aligned} {{\bar{g}}}_1(\rho ,\tau )&= f_1(\tau ) {{\tilde{g}}}_1(\rho ) \chi _0\Bigl (\frac{\rho }{\delta \sqrt{\tau }}\Bigr ) + C_1 \frac{ f_1(\tau ) }{\tau ^{2-\vartheta /2} (1+\rho /\sqrt{\tau })^{5}} + C_2 \frac{ f_1(\tau ) }{\tau ^{2-\vartheta /2}} e^{-\frac{\rho ^2}{4\tau } } , \end{aligned}$$

For appropriate \(\delta >0\), \(C_1\), and \(C_2\), the function \( {{\bar{g}}}_1(\rho ,\tau ) \) is a supersolution in \(({\mathbb {R}}^2 {\setminus } B_{R_0} (0) ) \times (\tau _0,\tau _1)\) for the right hand side \(h_1\). More precisely, writing \(M = R_0^{2-\vartheta } ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)} + \frac{1}{f_1(\tau _0) } |c_1| )\), we have

$$\begin{aligned} \left( \partial _\tau -[ \Delta - \nabla (\cdot ) \cdot \nabla \Gamma _0 + B ] \right) M {{\bar{g}}}_1 \ge |{{\tilde{h}}}|, \quad \text {in }({\mathbb {R}}^2 \setminus B_{R_0} (0) ) \times (\tau _0,\tau _1), \end{aligned}$$
$$\begin{aligned} M {{\bar{g}}}_1 \ge |g_0| , \quad \text {on } \rho = R_0, \ \tau \in (\tau _0,\tau _1), \end{aligned}$$

because of Lemma 10.7, and

$$\begin{aligned} M {{\bar{g}}}_1 (\tau _0) \ge \left| c_1 U g_{{{\tilde{Z}}}_0} \right| , \quad \text {in }{\mathbb {R}}^2, \end{aligned}$$

where

$$\begin{aligned} g_{{{\tilde{Z}}}_0} = \frac{{{\tilde{Z}}}_0}{U} - (-\Delta )^{-1} {{\tilde{Z}}}_0, \end{aligned}$$

is the function g associated to \({{\tilde{Z}}}_0\) defined in (6.4). We note that \(|U g_{{{\tilde{Z}}}_0}(\rho )|\le C \frac{1}{1+\rho ^4}\) and is supported on \(\rho \le 2\sqrt{\tau _0}\). Here we are using that \(\nu <\frac{3}{2}+\frac{\vartheta }{2}\).

Using the maximum principle we get

$$\begin{aligned} |g_0 (y,\tau )| \le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)}+ \frac{|c_1|}{f_1(\tau _0)} \Bigr ) f_1(\tau ) \frac{1}{(1+\rho )^{4-\vartheta }} ,\quad \tau \in [\tau _0,\tau _1]. \end{aligned}$$

The constant C here depends on \(R_0\), but \(R_0\) is fixed and we will not keep track of the dependence of C on \(R_0\).

By (10.42) and (10.48) we have

$$\begin{aligned} |{{\tilde{h}}} (y,\tau )| \le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)}+ \frac{|c_1|}{f_1(\tau _0)} \Bigr ) f_1(\tau )\Bigl ( \frac{1}{(1+\rho )^{6+\sigma }} +\frac{1}{\tau \log \tau ( 1+\rho )^{6-\vartheta }} \Bigr ) . \end{aligned}$$

We can now repeat the argument with a new barrier. Consider \({{\tilde{g}}}_2(\rho )\) the radial solution to

$$\begin{aligned} \begin{aligned}&-\Delta _6 {{\tilde{g}}}_2 = \frac{1}{1+\rho ^{6+\sigma }} \quad \text {in }{\mathbb {R}}^6, \quad c_1 \frac{1}{1+\rho ^4} \le {{\tilde{g}}}_2(\rho ) \le c_2 \frac{1}{1+\rho ^4} , \end{aligned} \end{aligned}$$
(10.60)

where \(c_1\), \(c_2>0\). Let

$$\begin{aligned} {{\bar{g}}}_2(\rho ,\tau )&= f_1(\tau ) {{\tilde{g}}}_2(\rho ) \chi _0\Bigl (\frac{\rho }{\delta \sqrt{\tau }}\Bigr ) + C_1 \frac{f_1(\tau )}{\tau ^{2} (1+\rho /\sqrt{\tau })^{6-\vartheta }} + C_2 \frac{ f_1(\tau )}{\tau ^{2}} e^{-\frac{\rho ^2}{4\tau } } . \end{aligned}$$

For appropriate constants \(\delta \), \(C_1\), \(C_2\), and assuming that \(\nu <2-\frac{\vartheta }{2}\) we get a suitable supersolution and we obtain

$$\begin{aligned} |g_0 (y,\tau )| \le C \Bigl ( K_1 +\frac{\Vert h\Vert _{**}}{R(\tau _0)}+ \frac{|c_1|}{f_1(\tau _0)} \Bigr ) f_1(\tau ) \frac{1}{(1+\rho )^4}. \end{aligned}$$

This proves (10.57).

The restriction on \(\nu \) were \(\nu < \frac{3}{2}+\frac{\vartheta }{2}\) and \(\nu <2-\frac{\vartheta }{2}\). Choosing \(\vartheta =\frac{1}{4}\) we find that for \(\nu <\frac{7}{4}\) both barriers work. \(\square \)

The next result is a technical step used in several places.

Lemma 10.9

Let \(\phi :{\mathbb {R}}^2 \rightarrow {\mathbb {R}}\) be radial such that \(\int _{{\mathbb {R}}^2} \phi = 0\) and \(|\phi (y)|\le \frac{C}{(1+|y|)^{2+\sigma }}\) for some \(\sigma >0\). Let \(g = \frac{\phi }{U}-(-\Delta )^{-1} \phi \) and assume that \(\Vert g\Vert _{L^\infty } <\infty \). Then

$$\begin{aligned} |\phi (y) |\le C \frac{\Vert g\Vert _{L^\infty }}{(1+|y|)^4}. \end{aligned}$$
(10.61)

Proof

Let \(\psi = (-\Delta )^{-1}\phi \). Since \(\psi \) satisfies

$$\begin{aligned} -\Delta \psi - U \psi = U g \quad \text {in }{\mathbb {R}}^2, \quad \psi (\rho )\rightarrow 0 \quad \text {as }\rho \rightarrow \infty , \end{aligned}$$

we have necessarily

$$\begin{aligned} \int _{{\mathbb {R}}^2} U g z_0 dy=0 . \end{aligned}$$

We have the variations of parameters formula

$$\begin{aligned} \psi (\rho ) = z_0(\rho ) \int _\rho ^\infty \frac{1}{z_0(r)^2 r} \int _r^\infty U g (s,\tau ) z_0(s) s \, \textrm{d}s \, \textrm{d}r , \quad \rho >1 . \end{aligned}$$
(10.62)

From (10.62) we find

$$\begin{aligned} |\psi (\rho ,\tau ) | \le C \Vert g\Vert _{L^\infty }. \end{aligned}$$

This and the formula \(\phi = U g + U \psi \) gives (10.61). \(\square \)

Next we give a proof of Proof of Lemma 10.1, but first we point some estimates of \({{\tilde{Z}}}_0\) defined in (6.4). Using the general decomposition (10.6), we write

$$\begin{aligned} {{\tilde{Z}}}_0 = {{\tilde{Z}}}_0^\perp + \frac{{{\tilde{a}}}_0}{2} Z_0 . \end{aligned}$$

By (10.7)

$$\begin{aligned} {{\tilde{a}}}_0 =\frac{1}{8\pi }\int _{{\mathbb {R}}^2} \Gamma _0 {{\tilde{Z}}}_0 = 2 + O\left( \frac{\log \tau _0}{\tau _0}\right) . \end{aligned}$$

Hence \({{\tilde{Z}}}_0^\perp \) satisfies

$$\begin{aligned} {{\tilde{Z}}}_0^\perp (\rho )&= {{\tilde{Z}}}_0(\rho ) - \frac{{{\tilde{a}}}_0}{2}Z_0(\rho )\\&= ( Z_0(\rho ) - m_{Z_0} U(\rho ) ) \chi _3 - \Bigl ( 1 + O \Bigl (\frac{\log \tau _0}{\tau _0}\Bigr )\Bigr ) Z_0(\rho )\\&= Z_0(\rho ) ( \chi _3-1) +O\Bigl ( \frac{\log \tau _0}{\tau _0} \frac{1}{1+\rho ^4} \Bigr ) \end{aligned}$$

where

$$\begin{aligned} \chi _3(\rho )=\chi _0\Bigl ( \frac{\rho }{3\sqrt{\tau _0}}\Bigr ). \end{aligned}$$

Let \({{\tilde{g}}}_0=\frac{{{\tilde{Z}}}_0}{U}-(-\Delta _y)^{-1}{{\tilde{Z}}}_0\) and \({{\tilde{g}}}_0^\perp = {{\tilde{g}}}_0+{{\tilde{a}}}_0\). Note that since \(Z_0\) has mass zero and decays like \(1/\rho ^4\) we have \( m_{Z_0} = O ( \frac{1}{\tau _0} ) \). We claim that

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2}{{\tilde{Z}}}_0 g_0^\perp dy\right| \le C \frac{\log \tau _0}{\tau ^2} . \end{aligned}$$
(10.63)

Indeed, let us use the notation

$$\begin{aligned} {\tilde{\psi }}_0&=(-\Delta _y)^{-1} {{\tilde{Z}}}_0 \end{aligned}$$

so that

$$\begin{aligned} {{\tilde{g}}}_0 =\frac{{{\tilde{Z}}}_0}{U}-{\tilde{\psi }}_0. \end{aligned}$$

Let us write

$$\begin{aligned} {{\tilde{Z}}}_0&= Z_0 + h , \end{aligned}$$

where

$$\begin{aligned} h=Z_0(\chi _3-1) - m_{Z_0} U \chi _3 . \end{aligned}$$

Since \(\Delta z_0 + U z_0 = 0\) and \(\lim _{\rho \rightarrow \infty } z_0(\rho ) = -2\) we have \((-\Delta )^{-1} Z_0 = z_0+2\). Therefore

$$\begin{aligned} {\tilde{\psi }}_0=(-\Delta )^{-1}{{\tilde{Z}}}_0=z_0+2 + (-\Delta )^{-1}h. \end{aligned}$$

Since the mass of \({{\tilde{Z}}}_0\) is zero

$$\begin{aligned} \int _{{\mathbb {R}}^2} {{\tilde{Z}}}_0 {{\tilde{g}}}_0^\perp&=\int _{{\mathbb {R}}^2} {{\tilde{Z}}}_0 {{\tilde{g}}}_0 \\&= \int _{{\mathbb {R}}^2}(Z_0+h)\Bigl ( \frac{Z_0+h}{U}-{\tilde{\psi }}_0\Bigr )dy\\&= \int _{{\mathbb {R}}^2}(Z_0+h)\Bigl ( \frac{Z_0+h}{U}-z_0-2-(-\Delta )^{-1}h\Bigr )dy\\&= \int _{{\mathbb {R}}^2} \frac{Z_0^2+2Z_0h+h^2}{U}dy -\int _{{\mathbb {R}}^2}Z_0 (z_0+2+(-\Delta )^{-1}h)dy \\&\quad -\int _{{\mathbb {R}}^2}h (z_0+2+(-\Delta )^{-1}h)dy \end{aligned}$$

But \(Z_0=Uz_0\) and the mass of h is zero, so

$$\begin{aligned} \int _{{\mathbb {R}}^2} {{\tilde{Z}}}_0 g^\perp&= \int _{{\mathbb {R}}^2} z_0hdy +\int _{{\mathbb {R}}^2} \frac{h^2}{U}dy -\int _{{\mathbb {R}}^2}Z_0 (-\Delta )^{-1}hdy -\int _{{\mathbb {R}}^2}h (-\Delta )^{-1}h dy \\&=\int _{{\mathbb {R}}^2} \frac{h^2}{U}dy -\int _{{\mathbb {R}}^2}h (-\Delta )^{-1}h dy, \end{aligned}$$

because, integrating by parts,

$$\begin{aligned} \int _{{\mathbb {R}}^2}Z_0 (-\Delta )^{-1}h dy =\int _{{\mathbb {R}}^2} (-\Delta z_0) (-\Delta )^{-1}h dy =\int _{{\mathbb {R}}^2} z_0 h. \end{aligned}$$

By direct computation

$$\begin{aligned} |(-\Delta _y)^{-1}h(\rho )| \le C\log (\tau _0){\left\{ \begin{array}{ll} \frac{1}{\rho ^2}&{}\rho \ge \sqrt{\tau _0} \\ \frac{1}{\tau _0}&{}\rho \le \sqrt{\tau _0}. \end{array}\right. } \end{aligned}$$

With this inequality we estimate

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} \frac{h^2}{U}dy\right| \le \frac{C}{\tau ^2} \end{aligned}$$

and

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2}h (-\Delta )^{-1}h dy\right| \le C\frac{\log \tau _0}{\tau ^2}. \end{aligned}$$

This proves (10.63).

Proof of Lemma 10.1

We let R be defined by (10.11). We multiply equation (10.9) by \(g^\perp \) and integrate in \({\mathbb {R}}^2\). Using Lemmas 10.6 and 9.3 we get

$$\begin{aligned} \partial _\tau \int _{{\mathbb {R}}^2} \phi g^\perp + \frac{c}{R^2} \int _{{\mathbb {R}}^2} \phi g^\perp \le C f(\tau )^2 \Vert h\Vert _{**}^2 +\frac{C a^2}{R^4} + \frac{C}{R^2 } \omega (\tau )^2 , \end{aligned}$$
(10.64)

for some \(c>0\), where

$$\begin{aligned} \omega (\tau ) = \Bigl ( \int _{{\mathbb {R}}^2\setminus B_{R(\tau )}} g^2 U \Bigr )^{1/2}. \end{aligned}$$

Let us write

$$\begin{aligned} \Vert \varphi \Vert _{\infty ,T_2} = \Vert \varphi \Vert _{L^\infty (\tau _0,T_2)} , \end{aligned}$$

and note that

$$\begin{aligned} \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2}^2<\infty , \quad \Bigl \Vert \frac{\omega }{Rf}\Bigr \Vert _{\infty ,T_2}<\infty . \end{aligned}$$

The following inequalities are valid for \(\tau _0<\tau <T_2\). From (10.64) we get

$$\begin{aligned} \partial _\tau \int _{{\mathbb {R}}^2} \phi g^\perp + \frac{c}{R^2} \int _{{\mathbb {R}}^2} \phi g^\perp \le C f(\tau )^2 \Bigl ( \Vert h\Vert _{**}^2 + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2}^2 + \Bigl \Vert \frac{\omega }{Rf}\Bigr \Vert _{\infty ,T_2}^2 \Bigr ) . \end{aligned}$$

By Gronwall’s inequality and Lemma 9.3 we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} (g^\perp )^2 U&\le C f(\tau )^2 R(\tau )^2 \Bigl ( \Vert h\Vert _{**}^2 + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2}^2 + \Bigl \Vert \frac{\omega }{Rf}\Bigr \Vert _{\infty ,T_2}^2 + c_1^2 D(\tau _0)^2 \Bigr ) \end{aligned}$$
(10.65)

where

$$\begin{aligned} D(\tau _0) = \frac{1}{f(\tau _0) R(\tau _0)} \frac{\sqrt{\log \tau _0}}{\tau _0}, \end{aligned}$$

and we have used (10.63).

From (10.65) we find

$$\begin{aligned}&\int _{{\mathbb {R}}^2} g^2 U \le C f(\tau )^2 R(\tau )^4 \Bigl ( \frac{1}{R(\tau _0)^2} \Vert h\Vert _{**}^2 + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{ \infty ,T_2 }^2\nonumber \\&\quad + \frac{1}{R(\tau _0)^2}\Bigl \Vert \frac{\omega }{R f}\Bigr \Vert _{ \infty ,T_2 }^2 + c_1^2 \frac{D(\tau _0)^2}{R(\tau _0)^2} \Bigr ) \end{aligned}$$
(10.66)

Using Lemma 10.8 we get

$$\begin{aligned} |U g|&\le C f(\tau ) R(\tau )^2 \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2}\nonumber \\&\quad + \frac{1}{R(\tau _0)} \Bigl \Vert \frac{\omega }{R f}\Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) \frac{1}{(1+\rho )^{4}} , \end{aligned}$$
(10.67)

where we have used that for \(\tau _0\) large, \( \frac{D(\tau _0)}{R(\tau _0)}< \frac{1}{f(\tau _0) R(\tau _0)^2} \).

We use this to estimate

$$\begin{aligned} \int _{{\mathbb {R}}^2\setminus B_{R}} g^2 U&\le C f(\tau )^2 R(\tau )^2 \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2f } \Bigr \Vert _{ \infty ,T_2 } \\&\quad + \frac{1}{R(\tau _0)}\Bigl \Vert \frac{\omega }{R f}\Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr )^2, \end{aligned}$$

which implies

$$\begin{aligned} \frac{\omega (\tau ) }{R(\tau )f(\tau )} \le C \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2f } \Bigr \Vert _{ \infty ,T_2} + \frac{1}{R(\tau _0)}\Bigl \Vert \frac{\omega }{R f}\Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ). \end{aligned}$$

We deduce that

$$\begin{aligned} \Bigl \Vert \frac{\omega }{Rf}\Bigr \Vert _{\infty ,T_2} \le C \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2f } \Bigr \Vert _{ \infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ). \end{aligned}$$
(10.68)

Combining this inequality with (10.66) we obtain

$$\begin{aligned} \int _{{\mathbb {R}}^2} g^2 U \le C f(\tau )^2 R(\tau )^4 \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr )^2 , \end{aligned}$$
(10.69)

and with (10.65) we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} (g^\perp )^2 U \le C f(\tau )^2 R(\tau )^2 \Bigl ( \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr )^2 . \end{aligned}$$
(10.70)

Going back to (10.67) we find

$$\begin{aligned} |U g (\rho ,\tau )|\le Cf(\tau ) R(\tau )^2 \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty } + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) \frac{1}{1+\rho ^4} . \end{aligned}$$
(10.71)

Using Lemma 10.9 we also obtain

$$\begin{aligned} |\phi (\rho ,\tau )|\le C f(\tau ) R(\tau )^2 \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) \frac{1}{1+\rho ^4} . \end{aligned}$$
(10.72)

We multiply the equation satisfied by \(\phi \) (10.9) by \(|y|^2 \chi _0(\frac{y}{R})\), and integrate on \({\mathbb {R}}^2\)

$$\begin{aligned} \partial _\tau \int _{{\mathbb {R}}^2} \phi |y|^2 \chi _0\Bigl (\frac{y}{R}\Bigr ) dy&=\int _{{\mathbb {R}}^2} (L[\phi ] + h) |y|^2 \chi _0(\frac{y}{R}) dy +\int _{{\mathbb {R}}^2} B[\phi ] |y|^2 \chi _0(\frac{y}{R}) dy \nonumber \\&\quad - \frac{R'(\tau )}{R}\int _{{\mathbb {R}}^2} \phi |y|^2 \nabla \chi _0(\frac{y}{R}) \cdot \frac{y}{R} dy, \end{aligned}$$
(10.73)

where \(R' = \frac{d R}{d \tau }\).

We integrate (10.73) from \(\tau \) to \(T_2\), use the decomposition (10.6) and that \(a(T_2)=0\) to get

$$\begin{aligned} |a(\tau ) | \log \tau&\le \left| \int _\tau ^{T_2} \int _{{\mathbb {R}}^2} (L[\phi (s)] + h) |y|^2 \chi _0(\frac{y}{R(s)}) dy ds\right| \nonumber \\&\quad + \left| \int _\tau ^{T_2} \int _{{\mathbb {R}}^2} B[\phi (s)] |y|^2 \chi _0(\frac{y}{R(s)}) dy ds\right| \nonumber \\&\quad + \left| \int _\tau ^{T_2} \frac{R'(s)}{R(s)} \int _{{\mathbb {R}}^2} \phi (s) |y|^2 \nabla \chi _0(\frac{y}{R(s)}) \cdot \frac{y}{R(s)} dy ds\right| \nonumber \\&\quad + \left| \int _{{\mathbb {R}}^2} \phi ^\perp (T_2) |y|^2 \chi _0\Bigl (\frac{y}{R(T_2)}\Bigr ) dy\right| +\left| \int _{{\mathbb {R}}^2} \phi ^\perp (\tau ) |y|^2 \chi _0\Bigl (\frac{y}{R(\tau )}\Bigr ) dy\right| . \end{aligned}$$
(10.74)

By Lemma 9.4 and (10.70)

$$\begin{aligned} \int _{B_{2R(\tau )}} |\phi (\tau )^\perp | |y|^2 dy&\le C R(\tau ) \Bigl ( \int _{{\mathbb {R}}^2} (\phi ^\perp (\tau ) )^2 U^{-1} \Bigr )^{1/2} \nonumber \\&\le C R(\tau ) \Bigl ( \int _{{\mathbb {R}}^2} (g^\perp (\tau ))^2 U \Bigr )^{1/2} \nonumber \\&\le C f(\tau ) R(\tau )^2 \Bigl ( \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) . \end{aligned}$$
(10.75)

Analogously,

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} \phi ^\perp (T_2) |y|^2 \chi _0\Bigl (\frac{y}{R(T_2)}\Bigr ) dy \right|&\le C f(T_2) R(T_2)^2 \Bigl ( \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) \nonumber \\&\le C f(\tau ) R(\tau )^2 \Bigl ( \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) . \end{aligned}$$
(10.76)

Integrating by parts

$$\begin{aligned} \left| \int _\tau ^{T_2} \int _{{\mathbb {R}}^2} B[\phi (s)] |y|^2 \chi _0(\frac{y}{R(s)}) dy ds\right|&\le C \int _\tau ^{T_2} \frac{1}{s \log s} \int _{{\mathbb {R}}^2} |\phi (y,s)| |y|^2 \chi _0(\frac{y}{R(s)}) dy ds \nonumber \\&\quad + C \int _\tau ^{T_2} \frac{1}{s \log s } \int _{{\mathbb {R}}^2} |\phi (y,s)| |y|^2 |\nabla \chi _0(\frac{y}{R(s)})| dy ds. \end{aligned}$$
(10.77)

Let’s estimate, using (10.72)

$$\begin{aligned}&\int _\tau ^{T_2} \frac{1}{s \log s} \int _{{\mathbb {R}}^2} |\phi (y,s)| |y|^2 \chi _0(\frac{y}{R(s)}) dy ds \\&\quad \le C\int _\tau ^{T_2} \frac{1}{s} f(s) R(s)^2 ds \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) \\&\quad \le C f(\tau ) R(\tau )^2 \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ). \end{aligned}$$

The second term in (10.77) is even smaller, and we deduce that

$$\begin{aligned} \left| \int _\tau ^{T_2} \int _{{\mathbb {R}}^2} B[\phi (s)] |y|^2 \chi _0(\frac{y}{R(s)}) dy ds\right|&\le C f(\tau ) R(\tau )^2 \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} \nonumber \\&\quad + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ). \end{aligned}$$
(10.78)

From (10.72) we also get

$$\begin{aligned}&\left| \int _\tau ^{T_2} \frac{R'(s)}{R(s)} \int _{{\mathbb {R}}^2} \phi (s) |y|^2 \nabla \chi _0(\frac{y}{R(s)}) \cdot \frac{y}{R(s)} dy ds\right| \nonumber \\&\quad \le C\int _\tau ^{T_2} \frac{R'(s)}{R(s)} f(s) R(s)^2 ds \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) \nonumber \\&\quad \le C f(\tau ) R(\tau )^2 \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ). \end{aligned}$$
(10.79)

Next we look at

$$\begin{aligned} \int _{{\mathbb {R}}^2} L[\phi ] |y|^2 \chi _0(\frac{y}{R}) dy&= - 2 \int _{{\mathbb {R}}^2} U \nabla g \cdot y \chi _0(\frac{y}{R}) dy - \frac{1}{R}\int _{{\mathbb {R}}^2} U |y|^2 \nabla g\cdot \nabla \chi _0(\frac{y}{R}) dy \nonumber \\&= 2 \int _{{\mathbb {R}}^2} g Z_0 \chi _0(\frac{y}{R}) dy +\frac{4}{R}\int _{{\mathbb {R}}^2} g U y \cdot \nabla \chi _0(\frac{y}{R}) dy \nonumber \\&\quad +\frac{1}{R}\int _{{\mathbb {R}}^2} g |y|^2 \nabla U \cdot \nabla \chi _0(\frac{y}{R}) dy +\frac{1}{R^2}\int _{{\mathbb {R}}^2} g |y|^2 U \Delta \chi _0(\frac{y}{R}) dy . \end{aligned}$$
(10.80)

We have \(\int _{{\mathbb {R}}^2} g Z_0 = 0\) by Lemma 9.2 and therefore, using (10.71), we find that

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} g Z_0 \chi _0(\frac{y}{R(\tau )}) dy \right|&\le \left| \int _{{\mathbb {R}}^2\setminus B_{R(\tau )}(0) } g Z_0 dy\right| \\&\le f(\tau ) \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty } + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) . \end{aligned}$$

The remaining terms in (10.80) are estimated using (10.69) or (10.71) and we get

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} L[\phi ] |y|^2 \chi _0(\frac{y}{R}) dy \right| \le C f(\tau ) \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) . \end{aligned}$$

Therefore

$$\begin{aligned}&\left| \int _\tau ^{T_2} \int _{{\mathbb {R}}^2} L[\phi (s)] |y|^2 \chi _0(\frac{y}{R(s)}) dy ds\right| \le C f(\tau ) R(\tau )^2 (\log \tau )^q\nonumber \\&\quad \Bigl ( \frac{1}{R(\tau _0)} \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) . \end{aligned}$$
(10.81)

Finally

$$\begin{aligned} \left| \int _\tau ^{T_2} \int _{{\mathbb {R}}^2} h |y|^2 \chi _0(\frac{y}{R(s)}) dy ds\right|&\le C f(\tau ) R(\tau )^2 (\log \tau )^q \Vert h\Vert _{**} . \end{aligned}$$
(10.82)

From (10.74), (10.75), (10.76), (10.78), (10.79), (10.81), and (10.82) we get

$$\begin{aligned} |a(\tau ) | \log \tau \le C f(\tau ) R(\tau )^2 (\log \tau )^q \Bigl ( \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) . \end{aligned}$$
(10.83)

Assuming \(\tau _0\) large, we deduce that

$$\begin{aligned} \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} \le \frac{C}{(\log \tau _0)^{1-q}} \Bigl ( \Vert h\Vert _{**} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ). \end{aligned}$$
(10.84)

Note that \(a(\tau _0)\) and \(c_1\) are related. Indeed, the initial condition is \(\phi _0 = c_1 {{\tilde{Z}}}_0 = \phi _0^\perp + \frac{a(\tau _0)}{2} Z_0\) with

$$\begin{aligned} a(\tau _0) = \frac{c_1}{8\pi } \int _{{\mathbb {R}}^2} {{\tilde{Z}}}_0 \Gamma _0, \end{aligned}$$

by (10.7). We note that \( \int _{{\mathbb {R}}^2} {{\tilde{Z}}}_0 \Gamma _0 = 16 \pi + O(\frac{\log \tau _0}{\tau _0})\). So by (10.84)

$$\begin{aligned} |c_1|\le C |a(\tau _0)| \le C \frac{f(\tau _0)R(\tau _0)^2 }{(\log \tau _0)^{1-q}}\Vert h\Vert _{**} + C \frac{1 }{(\log \tau _0)^{1-q}} |c_1|. \end{aligned}$$

For \(\tau _0\) large, we deduce that

$$\begin{aligned} |c_1|\le C |a(\tau _0)| \le C \frac{f(\tau _0)R(\tau _0)^2 }{(\log \tau _0)^{1-q}} \Vert h\Vert _{**}. \end{aligned}$$
(10.85)

This proves (10.16). Replacing this in (10.84) we get

$$\begin{aligned} \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} \le \frac{C}{(\log \tau _0)^{1-q}} \Vert h\Vert _{**} , \end{aligned}$$
(10.86)

which proves (10.14). Combining (10.68), (10.85) and (10.86) we obtain (10.15).

Finally, we also obtain from (10.72)

$$\begin{aligned} |\phi (\rho ,\tau )|\le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}} \frac{1}{1+\rho ^4} \Vert h\Vert _{**}. \end{aligned}$$
(10.87)

\(\square \)

Proof of Lemma 10.2

The proof is a slight modification of the one of Lemma 10.1. Using the same notation as in that proof, integrating (10.73) from \(\tau \) to \(T_2>\tau \) yields

$$\begin{aligned}&\int _{{\mathbb {R}}^2} \phi (T_2) |y|^2 \chi _0\Bigl (\frac{y}{R(T_2)}\Bigr ) dy - \int _{{\mathbb {R}}^2} \phi (\tau ) |y|^2 \chi _0\Bigl (\frac{y}{R(\tau )}\Bigr ) dy \\&\quad = \int _\tau ^{T_2} \int _{{\mathbb {R}}^2} (L[\phi (s)] + h) |y|^2 \chi _0(\frac{y}{R(s)}) dy ds \\&\qquad + \int _\tau ^{T_2} \int _{{\mathbb {R}}^2} B[\phi (s)] |y|^2 \chi _0(\frac{y}{R}) dy ds \\&\qquad - \int _\tau ^{T_2} \frac{R'(s)}{R(s)}\int _{{\mathbb {R}}^2} \phi (s) |y|^2 \nabla \chi _0(\frac{y}{R(s)}) \cdot \frac{y}{R(s)} dy ds, \end{aligned}$$

Similarly to (10.83) we obtain

$$\begin{aligned} |a(\tau ) | \log \tau&\le C f(\tau ) R(\tau )^2 (\log \tau )^q \Bigl ( \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) \nonumber \\&\quad + C |a(T_2)| \log (T_2). \end{aligned}$$
(10.88)

The assumption \(\frac{a}{f R^2} \in L^\infty (\tau _0,\infty )\) implies that

$$\begin{aligned} \lim _{\tau \rightarrow \infty } a(\tau ) \log \tau =0. \end{aligned}$$

Letting \(T_2\rightarrow \infty \) in (10.88) we obtain

$$\begin{aligned} |a(\tau ) | \log \tau \le C f(\tau ) R(\tau )^2 (\log \tau )^q \Bigl ( \Vert h\Vert _{**} + \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{L^\infty (\tau _0,\infty ) } + |c_1| \frac{1}{f(\tau _0) R(\tau _0)^2} \Bigr ) . \end{aligned}$$

Then the same argument as in Lemma 10.1 gives the estimates for a, \(\omega \) and \(c_1\). \(\square \)

Proof of Lemma 10.3

Here \(Z_B\) is the solution to (10.8). Assume to the contrary that there is some \(T_2>\tau _0\) such that

$$\begin{aligned} a_Z(T_2)=0. \end{aligned}$$

We follow the same computations as in the proof of Lemma 10.1 with \(h=0\) and \(c_1=1\). By the inequality (10.84) in the proof of Lemma 10.1

$$\begin{aligned} \Bigl \Vert \frac{a}{R^2 f} \Bigr \Vert _{\infty ,T_2} \le \frac{C}{(\log \tau _0)^{1-q}} \frac{1}{f(\tau _0) R(\tau _0)^2} \end{aligned}$$

which implies

$$\begin{aligned} |a(\tau _0)|\le \frac{C}{(\log \tau _0)^{1-q}}. \end{aligned}$$
(10.89)

But by (10.7)

$$\begin{aligned} a(\tau _0) = \frac{1}{8\pi }\int _{{\mathbb {R}}^2}\Gamma _0 {{\tilde{Z}}}_0 = 2 + O\left( \frac{\log \tau _0}{\tau _0}\right) \end{aligned}$$

which contradicts (10.89). \(\square \)

Proof of Lemma 10.4

We let \(T_n \) be a sequence such that \(T_n\rightarrow \infty \) as \(n\rightarrow \infty \). Let \({\bar{\phi }}\) be the solution to (10.1) with initial condition equal to 0. This solution exists but for the moment we don’t have any control of its asymptotic behavior as \(\tau \rightarrow \infty \). Let \({\bar{\phi }}^\perp \), \({{\bar{a}}}(\tau )\) be the decomposition (10.6) of \({\bar{\phi }}\). Let \(Z_B^\perp \), \(a_Z(\tau )\) be the decomposition (10.6) of \(Z_B\). Using Lemma 10.3 there is \(c_n \in {\mathbb {R}}\) such that

$$\begin{aligned} {{\bar{a}}}(T_n) + c_n a_Z(T_n) = 0. \end{aligned}$$

Let us define

$$\begin{aligned} \phi _n = {\bar{\phi }} + c_n Z_B, \end{aligned}$$

and let

$$\begin{aligned} \phi _n = \phi _n^\perp + \frac{a_n}{2}Z_0 \end{aligned}$$

be the decomposition (10.6) of \(\phi _n\). Then by Lemma 10.1 we have

$$\begin{aligned}&|a_n(\tau )| \le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}}\Vert h\Vert _{**}, \quad \tau \in [\tau _0,T_n]\\&|\omega _n(\tau )| \le C \frac{f(\tau ) R(\tau )}{(\log \tau _0)^{1-q}}\Vert h\Vert _{**}, \quad \tau \in [\tau _0,T_n]\\&|c_n| \le C \frac{f(\tau _0)R(\tau _0)^2 }{(\log \tau _0)^{1-q}} \Vert h\Vert _{**}. \end{aligned}$$

Moreover, we also have the uniform estimate

$$\begin{aligned} |\phi _n(\rho ,\tau )|\le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}} \frac{1}{1+\rho ^4} \Vert h\Vert _{**} \end{aligned}$$

for \(\tau \in [\tau _0,T_n]\) from (10.87).

By using standard parabolic estimates, passing to a subsequence we may assume that \(c_n\rightarrow c_1\) and \(\phi _n\rightarrow \phi \) locally uniformly in space-time, and that \(\phi \) is a solution of (10.9) for some \(c_1\) such that

$$\begin{aligned} |c_1|&\le C \frac{f(\tau _0)R(\tau _0)^2 }{(\log \tau _0)^{1-q}} \Vert h\Vert _{**}. \end{aligned}$$

Moreover \(\phi \) satisfies

$$\begin{aligned} |\phi (\rho ,\tau )|\le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}} \frac{1}{1+\rho ^4} \Vert h\Vert _{**} \end{aligned}$$

and writing the decomposition (10.6) as \(\phi = \phi ^\perp + \frac{a}{2}Z_0\) we have

$$\begin{aligned} |a(\tau )|&\le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}}\Vert h\Vert _{**}. \end{aligned}$$

We also get

$$\begin{aligned} |\omega _n(\tau )|&\le C \frac{f(\tau ) R(\tau )}{(\log \tau _0)^{1-q}}\Vert h\Vert _{**}, \end{aligned}$$

where \(\omega \) is defined in (10.13).

The uniqueness of \(c_1\) is a consequence of Lemma 10.2. \(\square \)

Proof of Proposition 10.1

We have already constructed \(\phi \) and \(c_1\) in Lemma 10.4, we have the uniqueness of \(\phi \) and the estimates for a and \(c_1\) in Lemma 10.2.

We only need to prove the estimate for \(\phi ^\perp \) stated in (10.10). By the construction of \(\phi \) in Lemma 10.4 and (10.70), (10.85) and (10.86), we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} (g^\perp )^2 U \le C f(\tau )^2 R(\tau )^2 \Vert h\Vert _{**}^2 , \quad \tau >\tau _0. \end{aligned}$$
(10.90)

We claim that from this inequality we have

$$\begin{aligned} U |g^\perp (y,\tau )|&\le C f(\tau ) R(\tau ) \frac{1}{(1+|y|)^{2}} \Vert h\Vert _{**}. , \quad \tau >\tau _0. \end{aligned}$$

The proof of this estimate is similar to that of (10.57) in Lemma 10.8.

Indeed, we define

$$\begin{aligned} g_0^\perp = U g^\perp \end{aligned}$$

and obtain the equation

$$\begin{aligned} \partial _\tau g_0^\perp&= \nabla \cdot \Bigl ( U \nabla \Bigl (\frac{g_0^\perp }{U}\Bigr )\Bigr ) - U ( -\Delta )^{-1} \nabla \cdot \Bigl ( U \nabla \Bigl (\frac{g_0^\perp }{U}\Bigr )\Bigr ) \nonumber \\&\quad + h - U(-\Delta )^{-1} h \nonumber \\&\quad + B[g_0^\perp ]-U(-\Delta )^{-1} B[g_0^\perp ] + B[U \psi [g_0^\perp ]-U(-\Delta )^{-1} B[U \psi [g_0^\perp ] \nonumber \\&\quad + a'(\tau ) U + \frac{a}{2}B[Z_0]-\frac{a}{2}U(-\Delta )^{-1} B[Z_0] . \end{aligned}$$
(10.91)

Here the notation \(\psi [g_0^\perp ]\) is the one introduced in the proof of Lemma 10.7 in (10.38).

To get an estimate for the solution we need an estimate for \(a'(\tau )\). Since \(g^\perp = g + a\) and \(\int _{{\mathbb {R}}^2} U g^\perp =0\) we have

$$\begin{aligned} a(\tau ) =-\frac{1}{8\pi } \int _{{\mathbb {R}}^2} U g(\tau ) dy=-\frac{1}{8\pi } \int _{{\mathbb {R}}^2} g_0(\tau ) dy. \end{aligned}$$

But integrating (10.37) we find

$$\begin{aligned} \partial _\tau \int _{{\mathbb {R}}^2} g_0(\tau ) dy&= - \int _{{\mathbb {R}}^2} U (-\Delta )^{-1} \Bigl ( \nabla \cdot ( U \nabla \frac{g_0}{U}) \Bigr ) d y -\int _{{\mathbb {R}}^2} U(-\Delta )^{-1} h dy \\&\quad - \int _{{\mathbb {R}}^2} U (-\Delta )^{-1} ( B [ g_0 + U \psi [g_0]] ) dy, \end{aligned}$$

which gives the expression

$$\begin{aligned} a'(\tau )&= \frac{1}{8\pi } \int _{{\mathbb {R}}^2} U (-\Delta )^{-1} \Bigl ( \nabla \cdot ( U \nabla \frac{g_0}{U}) \Bigr ) dy + \frac{1}{8\pi }\int _{{\mathbb {R}}^2} U(-\Delta )^{-1} h dy \\&\quad + \frac{1}{8\pi }\int _{{\mathbb {R}}^2} U (-\Delta )^{-1} ( B [ g_0 + U \psi [g_0]] ) dy. \end{aligned}$$

We claim that

$$\begin{aligned} |a'(\tau ) |\le C f(\tau ) R(\tau ) \Vert h\Vert _{**}. \end{aligned}$$
(10.92)

Indeed, we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} U (-\Delta )^{-1} \Bigl ( \nabla \cdot ( U \nabla \frac{g_0}{U}) \Bigr ) dy&= \int _{{\mathbb {R}}^2} \Gamma _0 \nabla \cdot ( U \nabla g^\perp ) dy =-\int _{{\mathbb {R}}^2} \nabla U \cdot \nabla g^\perp dy \\&= \int _{{\mathbb {R}}^2} \Delta U g^\perp . \end{aligned}$$

Then, by (10.90)

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} U (-\Delta )^{-1} \Bigl ( \nabla \cdot ( U \nabla \frac{g_0}{U}) \Bigr ) dy \right|&\le C \Bigl ( \int _{{\mathbb {R}}^2} (g^\perp )^2 U \Bigr )^{1/2} \\&\le C f(\tau ) R(\tau ) \Vert h\Vert _{**}. \end{aligned}$$

We also have, for the case of the operator (10.2),

$$\begin{aligned} \int _{{\mathbb {R}}^2} U (-\Delta )^{-1}( B[g_0 ]) \, \textrm{d}y&= \int _{{\mathbb {R}}^2} \Gamma _0 B[g_0 ] = \zeta (\tau ) \int _{{\mathbb {R}}^2} \Gamma _0 \nabla \cdot (y g_0) dy \\&= - \zeta (\tau ) \int _{{\mathbb {R}}^2} \nabla \Gamma _0 \cdot y U g dy \end{aligned}$$

But by construction and (10.69), (10.85) and (10.86), we get

$$\begin{aligned} \Bigl ( \int _{{\mathbb {R}}^2} g^2 U \Bigr )^{1/2} \le \frac{C}{(\log \tau _0)^{1-q}} f(\tau ) R(\tau )^2 \Vert h\Vert _{**}. \end{aligned}$$
(10.93)

so, using (10.93)

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} U (-\Delta )^{-1}( B[g_0 ]) \, \textrm{d}y \right|&\le \frac{C}{\tau \log \tau } \Bigr ( \int _{{\mathbb {R}}^2} U g^2\Bigl )^{1/2} \\&\le \frac{C}{\tau \log \tau } \frac{1}{(\log \tau _0)^{1-q}} f(\tau ) R(\tau )^2 \Vert h\Vert _{**} \\&\le C f(\tau ) \Vert h\Vert _{**} \\&\le C f(\tau ) R(\tau ) \Vert h\Vert _{**}. \end{aligned}$$

The last term is estimated similarly and we get (10.92).

Repeating the argument in of Lemma 10.7 we obtain from (10.90)

$$\begin{aligned} |g_0^\perp (y,\tau )| \le C f(\tau ) R(\tau )\Vert h\Vert _{**} \frac{1}{(1+|y|)^{2}} . \end{aligned}$$

An argument similar to Lemma 10.9 gives

$$\begin{aligned} |\phi ^\perp (\rho ,\tau )|\le C f(\tau ) R(\tau ) \frac{1}{(1+|y|)^2} \Vert h\Vert _{**}. \end{aligned}$$

\(\square \)

We have an estimate for \(\phi ^\perp \) stronger than (10.10) under a stricter assumption on \(\nu \).

Lemma 10.10

Let us assume that \(1<\nu <\frac{3}{2}\). Under the same assumption of Proposition 10.1 let \(\phi = \phi ^\perp + \frac{a}{2}Z_0\) be the solution of (10.9). Then

$$\begin{aligned} |\phi ^\perp (y,\tau ) |&\le C R(\tau )f(\tau )\Vert h\Vert _{**} {\left\{ \begin{array}{ll} \frac{1}{(1+|y|)^{2}} &{} |y|\le \sqrt{\tau } \\ \frac{\tau }{|y|^4} &{} |y|\ge \sqrt{\tau }, \end{array}\right. } \end{aligned}$$

Proof

We write (10.91) as

$$\begin{aligned} \partial _\tau g_0^\perp&= \Delta g_0 ^\perp - \nabla g_0^\perp \cdot \nabla \Gamma _0 + 2 U g_0^\perp + B[g_0^\perp ] + {{\tilde{h}}}_1 \end{aligned}$$
(10.94)

where

$$\begin{aligned} {{\tilde{h}}}_1&= - U (-\Delta )^{-1}( \nabla \cdot ( g_0^\perp \nabla \Gamma _0 ) ) \\&\quad -U(-\Delta )^{-1} B[g_0^\perp ] + B[U \psi [g_0^\perp ]]-U(-\Delta )^{-1} B[U \psi [g_0^\perp ]] \\&\quad + a'(\tau ) U + \frac{a}{2}B[Z_0]-\frac{a}{2}U(-\Delta )^{-1} B[Z_0] \\&\quad + h - U (-\Delta )^{-1} h. \end{aligned}$$

Then, similarly to (10.58), we have

$$\begin{aligned} |{{\tilde{h}}}_1(y,\tau )| \le C f(\tau ) R(\tau ) \Vert h\Vert _{**} \frac{1}{(1+|y|)^4} . \end{aligned}$$

Let

$$\begin{aligned} {{\bar{g}}}^\perp (\rho ,\tau )&= f(\tau ) R(\tau ) {{\tilde{g}}}_3(\rho ) \chi _0\Bigl (\frac{\rho }{\delta \sqrt{\tau }}\Bigr ) +A_1 \frac{f(\tau ) R(\tau ) }{\tau } \frac{1}{(1+\rho /\sqrt{\tau })^{4}}\\&\quad +A_2 \frac{f(\tau ) R(\tau ) }{\tau }e^{-\frac{\rho ^2}{4\tau } } \end{aligned}$$

where \(-\Delta _6 {{\tilde{g}}}_3 = \frac{1}{1+\rho ^4}\) with \({{\tilde{g}}}_2(\rho )\rightarrow 0\) as \(\rho \rightarrow \infty \). If \(\nu <\frac{3}{2}\), for appropriate positive constants \(\delta \), \(A_1\), \(A_2\), and C, the function \(C \Vert h\Vert _{**} {{\bar{g}}}^\perp \) is supersolution to (10.94) in \(\{ (y,\tau ) | \tau> \tau _0, \ |y|> R_0 \}\). We deduce that

$$\begin{aligned} |g_0^\perp (y,\tau )| \le C f(\tau ) R(\tau ) \Vert h\Vert _{**} \frac{\min (1,\frac{\tau }{|y|^2})}{(1+|y|)^{2}} . \end{aligned}$$

An argument similar to Lemma 10.9 gives

$$\begin{aligned} |\phi ^\perp (\rho ,\tau )|\le C f(\tau ) R(\tau ) \Vert h\Vert _{**} {\left\{ \begin{array}{ll} \frac{1}{1+|y|^2} &{} |y |\le \sqrt{\tau }\\ \frac{\tau }{|y|^4}&{} |y |\ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$

\(\square \)

Proof of Proposition 8.1

By Proposition 10.1 there is \(c_1\) such that the solution \(\phi \) to (10.9) has the properties stated in Proposition 10.1. We recall that by (10.87) \(\phi \) satisfies

$$\begin{aligned} |\phi (\rho ,\tau )|\le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}} \frac{1}{1+\rho ^4} \Vert h\Vert _{**} . \end{aligned}$$
(10.95)

We will construct a barrier to estimate \(\phi \) for \(|y|\ge R_0\), where \(R_0\) is a large constant. We consider the equation (10.9) in \({\mathbb {R}}^2 \setminus B_{R_0}(0)\) written in the form

$$\begin{aligned} \partial _\tau \phi = \Delta \phi - 4 \nabla \Gamma _0 \nabla \phi + 2 U \phi + B[\phi ] + {{\bar{h}}} , \end{aligned}$$
(10.96)

where

$$\begin{aligned} {{\bar{h}}} = - \nabla U \nabla \psi + h. \end{aligned}$$

Since \(\psi = ( -\Delta )^{-1}\phi \), from (10.95) we get

$$\begin{aligned} | \nabla \psi (\rho ,\tau )|\le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}} \frac{1}{1+\rho ^3} \Vert h\Vert _{**} . \end{aligned}$$

This gives

$$\begin{aligned} |\nabla U \cdot \nabla \psi | \le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}} \frac{1}{1+\rho ^8} \Vert h\Vert _{**} . \end{aligned}$$
(10.97)

By (10.97) and the definition of the norm \( \Vert h\Vert _{**}\),

$$\begin{aligned} |{{\bar{h}}}(y,\tau ) |\le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}} \frac{1}{(1+\rho )^{6+ \sigma }} \min \Bigl ( 1, \frac{\tau ^{\varepsilon /2}}{\rho ^\varepsilon }\Bigr )\Vert h\Vert _{**}, \end{aligned}$$

where we have used that \(\sigma +\varepsilon <2\). Let \({{\tilde{g}}}_2\) be defined by (10.60) and let

$$\begin{aligned} {\bar{\phi }} (\rho ,\tau )&= f(\tau ) R(\tau )^2 {{\tilde{g}}}_2(\rho ) \chi _0\Bigl (\frac{\rho }{\delta \sqrt{\tau }}\Bigr ) +A_1 \frac{f(\tau ) R(\tau )^2 }{\tau ^2 } \frac{1}{(1+\rho /\sqrt{\tau })^{6+\sigma +\varepsilon }} \\&\quad +A_2 \frac{f(\tau ) R(\tau )^2 }{\tau ^2 }e^{-\frac{\rho ^2}{4\tau } } . \end{aligned}$$

Then for suitable positive constants \(\delta \), \(A_1\), \(A_2\), and C, the function \(C (\log \tau _0)^{q-1} \Vert h\Vert _{**} {\bar{\phi }} \) is a supersolution to (10.96) in \(\{ (y,\tau ) | \tau> \tau _0, \ |y|> R_0 \}\). For this we need \(\nu <2\). Moreover \(|\phi (\rho ,\tau )| \le C {\bar{\phi }}(\rho ,\tau ) (\log \tau _0)^{q-1} \Vert h\Vert _{**}\) at \(\rho = R_0\) by (10.95). By the maximum principle

$$\begin{aligned} |\phi (y,\tau ) |\le C {\bar{\phi }}(y,\tau ) (\log \tau _0)^{q-1} \Vert h\Vert _{**}, \quad |y|>R_0. \end{aligned}$$

This gives the explicit bound

$$\begin{aligned} |\phi (\rho ,\tau ) | \le C \frac{f(\tau ) R(\tau )^2}{(\log \tau _0)^{1-q}} \frac{1}{(1+\rho ^4) } \min \Bigl ( 1 , \frac{\tau ^{1/2}}{\rho }\Bigr )^{2+\sigma +\varepsilon } \Vert h\Vert _{**} \end{aligned}$$

\(\square \)

We include here some results that will be useful later. Let

$$\begin{aligned} {{\hat{Z}}}_0 = L[ {{\tilde{Z}}}_0]. \end{aligned}$$

Lemma 10.11

The function \({{\hat{Z}}}_0\) satisfies

$$\begin{aligned} |{{\hat{Z}}}_0(\rho )| \le C \frac{1}{\tau _0 ( 1 + \rho )^4}, \end{aligned}$$
(10.98)

and is supported on \(\rho \le 2 \tau _0\).

Proof

Let \( \psi = (-\Delta )^{-1} {{\tilde{Z}}}_0\) and \(g = \frac{{{\tilde{Z}}}_0}{U} - \psi \). By (6.4) and using that \(Z_0 = U z_0\), \(z_0\) defined in (9.2),

$$\begin{aligned} g&= \frac{(Z_0-m_{Z_0} U ) \chi }{U} - \psi = z_0 \chi - m_{Z_0} \chi - \psi , \end{aligned}$$

where \(\chi (\rho ) = \chi _0(\frac{\rho }{\sqrt{\tau _0}})\). Note that \({{\tilde{Z}}}_0\) has mass zero and support in \(B_{2\sqrt{\tau _0}}\). It follows that \(\psi \) has also support contained in \(B_{2\sqrt{\tau _0}}\) and then g has support contained in \(B_{2\sqrt{\tau _0}}\). Therefore \({{\hat{Z}}}_0 = L[{{\tilde{Z}}}_0] = \nabla \cdot ( U \nabla g)\) has also support contained in \(B_{2\sqrt{\tau _0}}\).

To get an estimate for \({{\hat{Z}}}_0\) let us write

$$\begin{aligned} \psi&= (-\Delta )^{-1} (Z_0-m_{Z_0} U ) \chi ) = (-\Delta )^{-1} Z_0 + \psi _1, \end{aligned}$$

where

$$\begin{aligned} \psi _1 = (-\Delta )^{-1} (Z_0(\chi -1)-m_{Z_0} U \chi ). \end{aligned}$$

Since \(\Delta z_0 + U z_0 = 0\) and \(\lim _{\rho \rightarrow \infty } z_0(\rho ) = -2\) we have \((-\Delta )^{-1} Z_0 = z_0+2\). So

$$\begin{aligned} \psi = z_0 + 2 + \psi _1 \end{aligned}$$

Hence

$$\begin{aligned} g = z_0 ( \chi -1 ) - 2 - m_{Z_0} \chi - \psi _1 \end{aligned}$$

and so

$$\begin{aligned} {{\hat{Z}}}_0&= L[{{\tilde{Z}}}_0] = \nabla \cdot ( U \nabla g) \nonumber \\&= \nabla \cdot ( U ( \nabla z_0(\chi -1) + z_0 \nabla \chi - m_{Z_0} \nabla \chi - \nabla \psi _1)). \end{aligned}$$
(10.99)

Using radial symmetry and \(m_{Z_0} = O ( \frac{1}{\tau _0})\) we get

$$\begin{aligned} |\nabla \psi _1(\rho ) | \le C \frac{1}{\tau _0 (1+\rho )}. \end{aligned}$$

From this and (10.99) we get (10.98). \(\square \)

Consider the initial value problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi _1&= L[\phi _1] + B[\phi _1] \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \phi _1(\cdot ,\tau _0)&= {{\hat{Z}}}_0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(10.100)

Lemma 10.12

Let \(0<\gamma <2\). Let \(1<\nu _0<\frac{7}{4}\)

$$\begin{aligned} f_0(\tau ) = \frac{1}{\tau ^{\nu _0}} . \end{aligned}$$

and let \(R(\tau ) \) be as in (10.11). Then the solution \(\phi _1\) of (10.100) satisfies

$$\begin{aligned} | \phi _1(\rho ,\tau ) | \le C \frac{f_0(\tau ) R(\tau )^2}{\tau _0 f_0(\tau _0) R(\tau _0)^2} \frac{1}{(1+\rho ^4) } \min \Bigl ( 1 , \frac{\tau ^{1/2}}{\rho }\Bigr )^{2+\gamma } . \end{aligned}$$

Proof

A suitable modification in the proof of Proposition 10.1 gives the following result. Consider

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L[\phi ] + B[\phi ] \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \phi (\cdot ,t_0)&= \phi _0 + c_1 {{\tilde{Z}}}_0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(10.101)

Then there is \(C>0\) such that for any \(\tau _0\) sufficiently large the following holds. Suppose that \(\phi _0\) is a radial function with zero mass in \({\mathbb {R}}^2\), supported in \(B_{2\sqrt{\tau _0}}(0)\), and such that

$$\begin{aligned} |\phi _0(\rho )| \le M \frac{1}{1+\rho ^4}. \end{aligned}$$

Then there exists \(c_1\) such that the solution \(\phi \) of (10.101) satisfies

$$\begin{aligned} |\phi (\rho ,\tau ) | \le C M \frac{f_0(\tau ) R(\tau )^2}{f_0(\tau _0) R(\tau _0)^2} \frac{1}{(1+\rho ^4) } \min \Bigl ( 1 , \frac{\tau ^{1/2}}{\rho }\Bigr )^{2+\gamma } . \end{aligned}$$

Moreover \(c_1\) is a linear function of \(\phi _0\) and satisfies

$$\begin{aligned} |c_1|&\le C M \frac{1 }{(\log \tau _0)^{1-q}} . \end{aligned}$$

Let us apply this statement to \(\phi _0 = L[{{\tilde{Z}}}_0]\), which is radial, with mass zero, support in \(B_{2\sqrt{\tau _0}}(0)\), and satisfies

$$\begin{aligned} |\phi _0(\rho )|\le \frac{1}{\tau _0} \frac{1}{1+\rho ^4} , \end{aligned}$$

by Lemma 10.11. Then there exists \(c_1\) such that the solution \({\tilde{\phi }}\) to (10.101) with \(\phi _0 = L[{{\tilde{Z}}}_0]\) satisfies

$$\begin{aligned} |{\tilde{\phi }}(\rho ,\tau ) | \le C \frac{f_0(\tau ) R(\tau )^2}{\tau _0 f_0(\tau _0) R(\tau _0)^2} \frac{1}{(1+\rho ^4) } \min \Bigl ( 1 , \frac{\tau ^{1/2}}{\rho }\Bigr )^{2+\gamma } . \end{aligned}$$
(10.102)

We claim that \(c_1=0\). To prove this, we multiply (10.101) by \(|y|^2\) and integrate on \({\mathbb {R}}^2 \times (\tau _0,\infty )\). Let’s work with

$$\begin{aligned} B[\phi ] = \zeta (\tau ) \nabla \cdot (y \phi ) . \end{aligned}$$

The case of the operator (10.3) is similar. Then we get

$$\begin{aligned} \partial _\tau \int _{{\mathbb {R}}^2} {\tilde{\phi }}(y,\tau )|y|^2d y = -2 \zeta (\tau )\int _{{\mathbb {R}}^2} {\tilde{\phi }}(y,\tau )|y|^2d y , \end{aligned}$$

because \(\int _{{\mathbb {R}}^2} L[\phi ]|y|^2dy=0\), see Remark 9.2. Integrating

$$\begin{aligned} \int _{{\mathbb {R}}^2} {\tilde{\phi }}(y,\tau )|y|^2d y&= e^{-2\int _{\tau _0}^\tau \zeta } \int _{{\mathbb {R}}^2} {\tilde{\phi }}(y,\tau _0)|y|^2d y = c_1 e^{-2\int _{\tau _0}^\tau \zeta } \int _{{\mathbb {R}}^2} {{\tilde{Z}}}_0 (y)|y|^2 dy, \end{aligned}$$

because \(\int _{{\mathbb {R}}^2} L[{{\tilde{Z}}}_0] |y|^2 dy = 0\). Using the asymptotic expansion of \(\zeta \) one gets

$$\begin{aligned} e^{-2\int _{\tau _0}^\tau \zeta }\rightarrow \infty , \quad \text {as }\tau \rightarrow \infty . \end{aligned}$$

But the bound (10.102) implies that

$$\begin{aligned} \lim _{\tau \rightarrow \infty } \int _{{\mathbb {R}}^2} {\tilde{\phi }}(y,\tau )|y|^2d y = 0. \end{aligned}$$

This only can happen if \(c_1=0\).

We deduce that \(\phi _1\) defined in (10.100) coincides with \({\tilde{\phi }}\), and then (10.102) holds for \(\phi _1\). \(\square \)

11 Linear Estimate with Second Moment (Radial)

We will prove in this section Proposition 8.2 in the radial case \(h(\rho ,\tau )\). In this case \(\mu _j\equiv 0\).

Proposition 11.1

Let \(0<\sigma <1\), \(\varepsilon >0\) with \(\sigma +\varepsilon <2\) and \(1<\nu < \min ( 1+\frac{\varepsilon }{2},3-\frac{\sigma }{2}, \frac{5}{4})\). Let \(0<q<1\). Then there is C such that for \(\tau _0\) large the following holds. Suppose that h satisfies \(\Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }<\infty \) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y,\tau )dy&=0 , \quad \int _{{\mathbb {R}}^2} h(y,\tau )|y|^2dy=0. \end{aligned}$$

Then the solution \(\phi (y,\tau )\) of problem (8.9) satisfies

$$\begin{aligned} \Vert \phi \Vert _{\nu -\frac{1}{2} ,m+\frac{q}{2},4,2+\sigma +\varepsilon } \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

To describe the idea of the proof more easily let us consider for a moment the equation (8.9) without B:

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L[\phi ] + h(y,t) \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \phi (\cdot ,\tau _0)&= 0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(11.1)

The idea is to formally apply a suitable left inverse \(L^{-1}\) of L to (11.1) (to be defined later on in Lemma 11.1). If we call \(\Phi = L^{-1} \phi \), \(H = L^{-1}h\), then we would like to solve

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \Phi&= L[\Phi ] + H(y,t) \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \Phi (\cdot ,\tau _0)&= 0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(11.2)

In order to get good properties of H, in this step we have already used that h satisfies the second moment condition. At this point we would like to apply Proposition 10.1, which gives a decomposition

$$\begin{aligned} \Phi = \Phi ^\perp + \frac{a(\tau )}{2}Z_0. \end{aligned}$$

Note that \(\Phi ^\perp \) decays in time like \(1/\tau ^{\nu -1/2}\) and so \(\phi = L \Phi \) also decays in time like \(1/\tau ^{\nu -1/2}\), which is better than the estimate provided by Proposition 8.1. It turns out that H decays in space like \(1/\rho ^{4+\sigma }\) so we can’t apply directly Proposition 10.1 to (11.2). What we do is concentrate H by solving first a nicer problem. We write \(\Phi = \Phi _1 + \Phi _2\) where \(\Phi _1\) is asked to solve

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \Phi _1&= L_0[\Phi _1] + H(y,t) \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \Phi _1(\cdot ,\tau _0)&= 0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$

where

$$\begin{aligned} L_0[\phi ] = \nabla \cdot \Bigl ( U \nabla \Bigl (\frac{\phi }{U} \Bigr ) \Bigr ) = \Delta \phi - \nabla \phi \cdot \nabla \Gamma _0 + U \phi . \end{aligned}$$
(11.3)

Lemma 11.2 below deals with \(\Phi _1\). Then the problem for \(\Phi _2\) becomes

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \Phi _2&= L_0[\Phi _2] + L[\Phi _1]-L_0[\Phi _1] \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \Phi _1(\cdot ,\tau _0)&= 0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$

It turns out that the right hand side in this equation has better spatial decay and we can apply Proposition 10.1.

In the next lemmas we give some preliminary results, and the proof of Proposition 11.1 is given at the end of this section.

We define the inverse of L that we use. For \(h:{\mathbb {R}}^2\rightarrow {\mathbb {R}}\) define \(\Vert h\Vert _{\tau ,6+\sigma ,\varepsilon }\) as the smallest K such that

$$\begin{aligned} |h(y)| \le \frac{ K }{(1+|y|)^{6+\sigma }} {\left\{ \begin{array}{ll} 1 &{} |y|\le \sqrt{\tau } \\ \frac{\tau ^{\varepsilon /2}}{|y|^\varepsilon } &{} |y|\ge \sqrt{\tau } , \end{array}\right. } \end{aligned}$$

which depends on \(\tau \), treated as parameter here, \(\sigma \), and \(\varepsilon \).

Lemma 11.1

Let \(\sigma ,\varepsilon >0\). Let \(h=h(\rho )\) be radial and satisfy \(\Vert h\Vert _{\tau ,6+\sigma ,\varepsilon }<\infty \) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} h dy = \int _{{\mathbb {R}}^2} h|y|^2 dy = 0. \end{aligned}$$

Then there exists H radially symmetric such that \(L[H] = h\) in \({\mathbb {R}}^2\) and satisfies

$$\begin{aligned} \Vert H \Vert _{\tau ,4+\sigma ,\varepsilon } \le C \Vert h\Vert _{\tau ,6+\sigma ,\varepsilon } \end{aligned}$$
(11.4)

Moreover, H defines a linear operator of h and satisfies

$$\begin{aligned} \int _{{\mathbb {R}}^2} H d y = 0. \end{aligned}$$
(11.5)

Proof

Write the equation \(L[H] = h\) as

$$\begin{aligned} \nabla \cdot ( U \nabla g ) = h \end{aligned}$$

where \(g = \frac{H}{U}- (-\Delta )^{-1} H\). We choose g as

$$\begin{aligned} g(\rho ) = -\int _\rho ^\infty \frac{1}{r U(r)} \int _0^r h(s) s ds dr. \end{aligned}$$

Using that \(\int _{{\mathbb {R}}^2} h = 0\) we check that

$$\begin{aligned} |g(\rho ) | \le C\Vert h\Vert _{\tau ,6+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{(1+\rho )^\sigma } &{} \rho \le \sqrt{\tau }\\ \frac{ \tau ^{\varepsilon /2}}{\rho ^{\sigma +\varepsilon }} &{} \rho \ge \sqrt{\tau }\end{array}\right. } \end{aligned}$$

Now we solve Liouville’s equation

$$\begin{aligned} -\Delta \psi - U \psi = U g \quad \text {in }{\mathbb {R}}^2, \quad \psi (\rho )\rightarrow 0 \quad \text {as }\rho \rightarrow \infty , \end{aligned}$$

Since \(\int _{{\mathbb {R}}^2} h |y|^2 dy=0\) we check that

$$\begin{aligned} \int _{{\mathbb {R}}^2} g Z_0 dy = 0. \end{aligned}$$

Then we can use the variations of parameter formula, and get

$$\begin{aligned} |\psi (\rho )|\le C\Vert h\Vert _{\tau ,6+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{(1+\rho )^{2+\sigma }} &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{\varepsilon /2}}{\rho ^{2+\sigma +\varepsilon }} &{} \rho \ge \sqrt{\tau }\end{array}\right. } \end{aligned}$$

Then define \(H = U ( g + \psi )\), which is the desired solution, and note that it satisfies (11.4). Property (11.5) follows from \(H = - \Delta \psi \) and the decay of \(\psi \). \(\square \)

To take into account the operator B we define

$$\begin{aligned} \Lambda [ \phi ] = y \cdot \nabla \phi , \end{aligned}$$

and compute

$$\begin{aligned} \Lambda \circ L [\Phi ] - L \circ \Lambda [\Phi ]&= \nabla \cdot ( \Phi U y ) -2 L[\Phi ] - \nabla \cdot ( (y \cdot \nabla U+ 2 U) \nabla (-\Delta )^{-1}\Phi ). \end{aligned}$$
(11.6)

Indeed, write \(\Psi = (-\Delta )^{-1} \Phi \). Then

$$\begin{aligned} L \Phi = \Delta \Phi - \nabla \Gamma _0 \cdot \nabla \Phi - \nabla U \cdot \nabla \Psi + 2 U \Phi . \end{aligned}$$
(11.7)

By direct computation

$$\begin{aligned}&\Lambda \Delta \Phi = \Delta \Lambda \Phi - 2 \Delta \Phi \end{aligned}$$
(11.8)
$$\begin{aligned}&\Lambda ( \nabla \Gamma _0 \cdot \nabla \Phi ) = \nabla (\Lambda \Gamma _0) \cdot \nabla \Phi + \nabla \Gamma _0 \cdot \nabla ( \Lambda \Phi ) - 2 \nabla \Gamma _0 \cdot \nabla \Phi \end{aligned}$$
(11.9)
$$\begin{aligned}&\Lambda ( \nabla U \cdot \nabla \Psi ) = \nabla (\Lambda U) \cdot \nabla \Psi + \nabla U \cdot \nabla ( \Lambda \Psi ) - 2 \nabla U \cdot \nabla \Psi , \end{aligned}$$
(11.10)

but \(-\Delta \Psi = \Phi \), and therefore

$$\begin{aligned} - \Delta ( \Lambda \Psi ) + 2 \Delta \Psi = \Lambda \Phi . \end{aligned}$$

Applying \((-\Delta )^{-1}\) gives

$$\begin{aligned} \Lambda \Psi = ( -\Delta )^{-1}( \Lambda \Phi ) + 2 \Psi . \end{aligned}$$

Substituting this into (11.10) we obtain

$$\begin{aligned} \Lambda ( \nabla U \cdot \nabla \Psi )&= \nabla (\Lambda U) \cdot \nabla \Psi + \nabla U \cdot \nabla [ ( -\Delta )^{-1}( \Lambda \Phi ) + 2 \Psi ] - 2 \nabla U \cdot \nabla \Psi \nonumber \\&= \nabla (\Lambda U) \cdot \nabla \Psi + \nabla U \cdot \nabla [ ( -\Delta )^{-1}( \Lambda \Phi ) ]. \end{aligned}$$
(11.11)

Combining (11.7), (11.8), (11.9), (11.11) we find that

$$\begin{aligned}&\Lambda L \Phi = L \Lambda \Phi - 2 L \Phi + 4 U \Phi - 2\nabla U \cdot \nabla \Psi - \nabla (\Lambda \Gamma _0)\cdot \nabla \Phi \\&\quad - \nabla (\Lambda U) \cdot \nabla \Psi + 2 \Lambda (U) \Phi . \end{aligned}$$

But

$$\begin{aligned} - 2\nabla U \cdot \nabla \Psi - \nabla (\Lambda U) \cdot \nabla \Psi&= - \nabla Z_0 \cdot \nabla \Psi \\&= - \nabla \cdot (Z_0 \nabla \Psi ) - Z_0 \Phi , \end{aligned}$$

so that

$$\begin{aligned} \Lambda L \Phi = L \Lambda \Phi - 2 L \Phi + 4 U \Phi - \nabla (\Lambda \Gamma _0)\cdot \nabla \Phi + 2 \Lambda (U) \Phi - \nabla \cdot (Z_0 \nabla \Psi ) - Z_0 \Phi . \end{aligned}$$

Using that

$$\begin{aligned} 2 \Lambda (U) \Phi - Z_0 \Phi&= -2 U \Phi + \Lambda (U) \Phi \end{aligned}$$

we then obtain

$$\begin{aligned} \Lambda L \Phi = L \Lambda \Phi - 2 L \Phi + 2 U \Phi - \nabla (\Lambda \Gamma _0)\cdot \nabla \Phi + \Lambda (U) \Phi - \nabla \cdot (Z_0 \nabla \Psi ) \end{aligned}$$

Let’s consider the terms \(2 U \Phi - \nabla (\Lambda \Gamma _0)\cdot \nabla \Phi + \Lambda (U) \Phi \). Noting that \( \nabla (\Lambda \Gamma _0) = \nabla ( y \cdot \nabla \Gamma +2) = \nabla z_0\) and that \(Z_0 = 2 U + \Lambda (U)\), we can write

$$\begin{aligned} 2 U \Phi - \nabla (\Lambda \Gamma _0)\cdot \nabla \Phi + \Lambda (U) \Phi&= 2 U \Phi - \nabla z_0 \cdot \nabla \Phi + \Lambda (U) \Phi \\&= Z_0 \Phi - \nabla \cdot ( \nabla z_0 \Phi ) + \Delta z_0 \Phi . \end{aligned}$$

But \(\Delta z_0 + Z_0=0\), so

$$\begin{aligned} \Lambda L \Phi = L \Lambda \Phi - 2 L \Phi - \nabla \cdot ( \nabla z_0 \Phi ) - \nabla \cdot (Z_0 \nabla \Psi ) . \end{aligned}$$

We can again write \(\nabla z_0 = \nabla ( y \cdot \nabla \Gamma _0)\) and using the radial symmetry of the functions \(\Gamma _0\), \(z_0\) and the notation \(\rho = |y|\)

$$\begin{aligned} \nabla z_0&= \frac{y}{\rho } \partial _\rho z_0 = \frac{y}{\rho } \partial _\rho ( \rho \partial _\rho \Gamma _0) = y \Delta \Gamma _0 = - y U. \end{aligned}$$

Then

$$\begin{aligned} \Lambda L \Phi = L \Lambda \Phi - 2 L \Phi + \nabla \cdot ( y U \Phi ) - \nabla \cdot (Z_0 \nabla \Psi ) . \end{aligned}$$

This proves (11.6).

Formula (11.6) leads us to consider the following equation for \(\Phi = L^{-1}[\phi ]\):

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \Phi&= L [\Phi ] + {{\tilde{B}}}[\Phi ] + \zeta _1(\tau ) A[\Phi ]+ H \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \Phi (\cdot ,\tau _0)&=0 \end{aligned} \right. \end{aligned}$$
(11.12)

where

$$\begin{aligned} A[\Phi ] = L^{-1}[ \nabla \cdot ( \Phi U y ) - \nabla \cdot ( Z_0 \nabla (-\Delta )^{-1}\Phi ) ] , \end{aligned}$$

\(Z_0(y)= 2 U(y) + y \cdot \nabla _y U(y)\), and \({{\tilde{B}}}\) has the same form as B:

$$\begin{aligned} {{\tilde{B}}}[\Phi ] = {\tilde{\zeta }}_1(\tau ) y \cdot \nabla \Phi + {\tilde{\zeta }}_2(\tau ) \Phi \end{aligned}$$

with \({\tilde{\zeta }}_1(\tau )\), \({\tilde{\zeta }}_2(\tau )\) satisfying

$$\begin{aligned} |{\tilde{\zeta }}_i(\tau )| \leqq \frac{C}{\tau \log \tau }\quad \text {for all } \tau >\tau _0. \end{aligned}$$
(11.13)

and \(\zeta _1\) satisfies the same restriction, that is, (10.20).

The next lemma allows us to reduce to an equation like (11.12) but with a right hand side with more spatial decay.

Lemma 11.2

Let \(\sigma >0\), \(\varepsilon >0\) and \(1<\nu < \min ( 1+\frac{\varepsilon }{2},3-\frac{\sigma }{2})\). Let \(H(y,\tau )\) be radial in y and satisfy

$$\begin{aligned} \int _{{\mathbb {R}}^2} H(\cdot ,\tau )=0 \end{aligned}$$
(11.14)

and \(\Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon }<\infty \). Then there exists \(H_1\) and \(\Phi _1\) such that

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \Phi _1&= L [\Phi _1]+ {{\tilde{B}}}[\Phi _1] + H - H_1 , \quad \text {in } {\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \Phi _1(\cdot ,\tau _0)&= 0\quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$

Moreover \(\Phi _1\) and \(H_1\) are linear operators of H and satisfy

$$\begin{aligned}&|\Phi _1(\rho ,\tau )| \le \frac{C}{\tau ^\nu (\log \tau )^m } \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{ (1+\rho )^{2+\sigma }} &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{1+\varepsilon /2}}{ (1+\rho )^{4+\sigma +\varepsilon } } &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$
(11.15)
$$\begin{aligned}&|H_1(\rho ,\tau )| \le \frac{C}{\tau ^\nu (\log \tau )^m } \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{ (1+\rho )^{6+\sigma }} &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{\varepsilon /2}}{ (1+\rho )^{6+\sigma +\varepsilon } } &{} \rho \ge \sqrt{\tau }. \end{array}\right. }, \end{aligned}$$
(11.16)
$$\begin{aligned}&\int _{{\mathbb {R}}^2} \Phi _1 dy =0 \end{aligned}$$
(11.17)
$$\begin{aligned}&\int _{{\mathbb {R}}^2} H_1(\cdot ,\tau )=0. \end{aligned}$$
(11.18)

Proof

Write the operator L as

$$\begin{aligned} L[\phi ] = L_0[\phi ] - \nabla \cdot ( U \nabla (-\Delta )^{-1} \phi ) \end{aligned}$$

where \(L_0\) is defined in (11.3). Consider the problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \Phi _1&= L_0 [\Phi _1] +{{\tilde{B}}}[\Phi _1] +H , \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \Phi _1(\cdot ,\tau _0)&= 0\quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$

The idea is to formally apply \(L_0^{-1}\) to this equation. Similarly to the proof of (11.6) we compute

$$\begin{aligned} \Lambda \circ L_0 [\Phi ] - L_0 \circ \Lambda [\Phi ]&= \nabla \cdot ( \Phi U y ) - 2 L_0 [ \Phi ]. \end{aligned}$$

This leads us to consider the problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau {\tilde{\Phi }}&= L_0 [{\tilde{\Phi }} ] + B_1[{\tilde{\Phi }}] +{{\tilde{H}}} , \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ {\tilde{\Phi }}(\cdot ,\tau _0)&= 0\quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(11.19)

where \({{\tilde{H}}}\) is a radial function satisfying

$$\begin{aligned} L_0[{{\tilde{H}}}] = H \quad \text {in }{\mathbb {R}}^2 \end{aligned}$$

and

$$\begin{aligned} B_1[{\tilde{\Phi }}] = {\hat{\zeta }}_1(\tau ) y \cdot \nabla {\tilde{\Phi }} +{\hat{\zeta }}_2(\tau ) {\tilde{\Phi }} \end{aligned}$$

with

$$\begin{aligned} {\hat{\zeta }}_1(\tau ) = {\tilde{\zeta }}_1(\tau ) = O\Bigl ( \frac{1}{\tau \log \tau }\Bigr ) , \quad {\hat{\zeta }}_2(\tau ) = {\tilde{\zeta }}_2(\tau )-2{\tilde{\zeta }}_1 (\tau ) = O\Bigl ( \frac{1}{\tau \log \tau }\Bigr ) , \end{aligned}$$
(11.20)

by (11.13).

We claim that there is a choice of \({{\tilde{H}}}\), which defines a linear operator of H, and satisfies

$$\begin{aligned} |{{\tilde{H}}}| + (1+\rho )|\nabla {{\tilde{H}}}| \le C\frac{1}{\tau ^\nu (\log \tau )^m} \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{(1+\rho )^{2+\sigma }} &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{\varepsilon /2}}{(1+\rho )^{2+\sigma +\varepsilon }} &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$
(11.21)

Indeed, the equation \(L_0[{{\tilde{H}}}] = H\) for radial functions has the form

$$\begin{aligned} \partial _\rho \Bigl ( \rho U \partial _\rho \Bigl ( \frac{{{\tilde{H}}}}{U} \Bigr ) \Bigr ) = \rho H. \end{aligned}$$

We select the solution

$$\begin{aligned} {{\tilde{H}}}(\rho ,\tau ) = U(\rho ) \int _0^\rho \frac{1}{r U(r)} \int _0^r H(s,\tau ) s ds dr . \end{aligned}$$

Using (11.14) we get (11.21).

Instead of (11.19) we consider

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau {\tilde{\Phi }}_1&= \Delta _{{\mathbb {R}}^2} {\tilde{\Phi }}_1 - \nabla \Gamma _0 \cdot \nabla {\tilde{\Phi }}_1 + B_1[{\tilde{\Phi }}_1] +{{\tilde{H}}} , \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ {\tilde{\Phi }}_1(\cdot ,\tau _0)&= 0\quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(11.22)

We then have the following estimate for \({\tilde{\Phi }}_1\):

$$\begin{aligned} |{\tilde{\Phi }}_1|\le \frac{C}{\tau ^\nu (\log \tau )^m } \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{ (1+\rho )^\sigma } &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{1+\varepsilon /2}}{ (1+\rho )^{2+\sigma +\varepsilon } } &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$
(11.23)

For the proof of this we construct a barrier. First we find a solution to

$$\begin{aligned} \Delta _{{\mathbb {R}}^2} \phi _1 - \nabla \Gamma _0\cdot \nabla \phi _1 + \frac{1}{(1+\rho )^{2+\sigma }}&= 0 \quad \text {in } {\mathbb {R}}^2, \\ \phi _1 (\rho )&\rightarrow 0 \quad \text {as } \rho \rightarrow \infty . \end{aligned}$$

The equation may be integrated explicitly, noting that

$$\begin{aligned} \Delta _{{\mathbb {R}}^2} \phi - \nabla \Gamma _0 \cdot \nabla \phi = \phi _{\rho \rho } + \Bigl ( \frac{1}{\rho }+ \frac{4\rho }{1+\rho ^2}\Bigr ) \phi _\rho \end{aligned}$$

and that the constants are in the kernel of this operator. We then have

$$\begin{aligned} \phi _1 (\rho ) = \int _\rho ^\infty \frac{1}{r(1+r^2)^2}\int _0^r \frac{1}{(1+s)^{2+\sigma }} s (1+s^2)^2 ds dr \end{aligned}$$

and this implies

$$\begin{aligned} | \phi _1 (\rho ) | + (1+\rho ) | \phi _1 '(\rho ) | \le \frac{C}{(1+\rho )^\sigma } \end{aligned}$$

Let

$$\begin{aligned} \chi (\rho ,\tau ) = \chi _0(\frac{\rho }{\delta \sqrt{\tau }}) , \end{aligned}$$

where \(\chi _0\in C^\infty ({\mathbb {R}})\), \(\chi _0(s)=1\) for \(s\le 1\) and \(\chi _0(s)=0\) for \(s\ge 2\). Define \({\tilde{\phi }}_1 = \frac{1}{\tau ^\nu (\log \tau )^m} \phi _1\chi \). We have

$$\begin{aligned}&(\partial _\tau - \Delta _{{\mathbb {R}}^2} + \nabla \Gamma _0\cdot \nabla ) {\tilde{\phi }}_1 \\&\quad \ge \frac{1}{\tau ^\nu (\log \tau )^m(1+\rho )^{2+\sigma }} \chi - \frac{C_1}{\tau ^{\nu +\sigma /2+1}(\log \tau )^m} \chi _{\{\delta \sqrt{\tau }\le \rho \le 2 \delta \sqrt{\tau }\}} , \end{aligned}$$

for some \(C_1>0\), \(\delta >0\) (assuming \(\tau _0\) large). Now consider

$$\begin{aligned}&\phi _2(\rho ,\tau ) = \frac{1}{\tau ^{\nu +\sigma /2}(\log \tau )^m}\frac{1}{(1+\rho /\sqrt{\tau })^{2+\sigma +\varepsilon }} ,\\&\quad \qquad \phi _3(\rho ,\tau ) = \frac{1}{\tau ^{\nu +\sigma /2}(\log \tau )^m} e^{-\frac{\rho ^2}{4 \tau }}. \end{aligned}$$

A computation, using (11.20), shows that

$$\begin{aligned} {\bar{\phi }} =A_1 {\tilde{\phi }}_1 + A_2 \phi _2 + A_3 \phi _3 \end{aligned}$$

satisfies

$$\begin{aligned} (\partial _\tau - \Delta _{{\mathbb {R}}^2} + \nabla \Gamma _0\cdot \nabla + B_1 ) {\bar{\phi }} \ge \frac{c}{\tau ^\nu (\log \tau )^m} {\left\{ \begin{array}{ll} \frac{1}{(1+\rho )^{2+\sigma }} &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{\varepsilon /2}}{(1+\rho )^{2+\sigma +\varepsilon }} &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$

for some \(c>0\). This step needs \(\nu -1<\frac{\varepsilon }{2}\) and \(\nu +\frac{\sigma }{2}<3\). By comparison, we find that \({\tilde{\Phi }}_1\) satisfies (11.23).

The solution \({\tilde{\Phi }}_1\) of (11.22) satisfies

$$\begin{aligned} \partial _\tau {\tilde{\Phi }}_1&= L_0[ {\tilde{\Phi }}_1 ] - U {\tilde{\Phi }}_1 + B_1[{\tilde{\Phi }}_1] +{{\tilde{H}}} \end{aligned}$$

Applying \(L_0\) to this equation we find that

$$\begin{aligned} \Phi _1 = L_0[ {\tilde{\Phi }}_1] \end{aligned}$$

satisfies

$$\begin{aligned} \partial _\tau \Phi _1&= L [\Phi _1]+ {{\tilde{B}}}[\Phi _1]+ H - H_1 \end{aligned}$$

with

$$\begin{aligned} H_1 = - \nabla \cdot ( U \nabla \Psi _1)+L_0[ U {\tilde{\Phi }}_1] + {\tilde{\zeta }}_1 \nabla \cdot ( {\tilde{\Phi }}_1 U y ) , \quad \Psi _1 = (-\Delta )^{-1} \Phi _1. \end{aligned}$$
(11.24)

Let us verify that \(\Phi _1\) and \(H_1\) satisfy the conditions stated in (11.15), (11.16), (11.18). Indeed, from standard parabolic estimates and (11.23) we have

$$\begin{aligned} |\nabla {\tilde{\Phi }}_1|\le \frac{C}{\tau ^\nu (\log \tau )^m } \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{ (1+\rho )^{1+\sigma }} &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{1+\varepsilon /2}}{ (1+\rho )^{3+\sigma +\varepsilon } } &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$
(11.25)

Differentiating in \(y_j\), \(j=1,2\) the equation (11.22) and using standard parabolic estimates, together with (11.21) and (11.25), we obtain

$$\begin{aligned} | D^2 {\tilde{\Phi }}_1| \le \frac{C}{\tau ^\nu (\log \tau )^m } \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{ (1+\rho )^{2+\sigma }} &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{1+\varepsilon /2}}{ (1+\rho )^{4+\sigma +\varepsilon } } &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$
(11.26)

The definition \(\Phi _1 =L_0[{\tilde{\Phi }}_1]\) and the estimates (11.23), (11.25), (11.26) give the estimate (11.15).

We compute

$$\begin{aligned} H_1 = -\nabla U \cdot \nabla \Psi _1 + U \Phi _1 + \nabla U \cdot \nabla {\tilde{\Phi }}_1 + U \Delta {\tilde{\Phi }}_1 +{\tilde{\zeta }}_1 \nabla \cdot ( {\tilde{\Phi }}_1 U y ) . \end{aligned}$$

Note that \(\int _{{\mathbb {R}}^2} \Phi _1(\cdot ,\tau )=0\). So, by a direct radial computation of \(\Psi _1 = (-\Delta )^{-1} \Phi _1\) and (11.15) we obtain

$$\begin{aligned} |\nabla \Psi _1(\rho ,\tau )| \le \frac{C}{\tau ^\nu (\log \tau )^m } \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{ (1+\rho )^{1+\sigma }} &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{1+\varepsilon /2}}{ (1+\rho )^{3+\sigma +\varepsilon } } &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$

This estimate and the ones already obtained for \({\tilde{\Phi }}_1\) (11.25), (11.26) and for \(\Phi _1\) (11.15) yields

$$\begin{aligned} |H_1(\rho ,\tau )| \le \frac{C}{\tau ^\nu (\log \tau )^m } \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{ (1+\rho )^{6+\sigma }} &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{\varepsilon /2}}{ (1+\rho )^{6+\sigma +\varepsilon } } &{} \rho \ge \sqrt{\tau }. \end{array}\right. }, \end{aligned}$$

which is the desired estimate (11.16).

Finally, the zero mass condition (11.18) follows from the form of \(H_1\) (11.24) and its decay. The mass condition for \(\Phi _1\) (11.17) follows from \(\Phi _1 =L_0[{\tilde{\Phi }}_1]\) and the decay of \({\tilde{\Phi }}_1\) (11.23) and (11.25). \(\square \)

Next we would like to obtain a result similar to Proposition 10.1 for the problem (11.12). In order to simplify this step, we will modify this equation by allowing a parameter in the initial condition. This technical obstruction will be removed in the proof of Proposition 11.1. Thus we consider

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \Phi&= L [\Phi ] + {{\tilde{B}}}[\Phi ] + \zeta _1(\tau ) A[\Phi ]+ H \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \Phi (\cdot ,\tau _0)&= c_1 {{\tilde{Z}}}_0 , \end{aligned} \right. \end{aligned}$$
(11.27)

where \({{\tilde{Z}}}_0\) is defined in (6.4).

The next result allows us to say that if in equation (11.27) the right hand side has fast decay, then we can decompose the solution similarly as in Proposition 10.1. This result is an extension of that proposition to an equation that has the extra operator A in it, which is treated as a perturbation.

Lemma 11.3

Let \(0<\sigma <1\), \(\varepsilon >0\), \(\sigma +\varepsilon <2\), \(1<\nu <\min (1+\frac{\varepsilon }{2},3-\frac{\sigma }{2},\frac{3}{2})\). Let \(0<q<1\). Then there is \(C>0\) such that for \(\tau _0\) sufficiently large and for H radially symmetric with \(\Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon }<\infty \) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} H(y,\tau )dy=0 \quad \text {for all } \tau >\tau _0 \end{aligned}$$

the solution \(\Phi \) to (11.27) can be decomposed as \(\Phi = \Phi _0 + \frac{a(\tau )}{2} Z_0\) with the estimates

$$\begin{aligned} |\Phi _0(\rho ,\tau ) |&\le C \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} } \min \Bigl ( \frac{1}{(1+|y|)^{2}} , \frac{\tau }{|y|^4} \Bigr ) \\ |a(\tau )|&\le C \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu -1} (\log \tau )^{m+q} } . \end{aligned}$$

Moreover \(\Phi _0\) and a are linear operators of H.

Proof of Lemma 11.3

We will treat the operator A as a perturbation and therefore consider

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \Phi&= L [\Phi ] + {{\tilde{B}}}[\Phi ]+ H \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \Phi (\cdot ,\tau _0)&=c_1 {{\tilde{Z}}}_0 . \end{aligned} \right. \end{aligned}$$
(11.28)

Let \(\Phi _1\), \(H_1\) be the functions constructed in Lemma 11.2. Setting \(\Phi = \Phi _1 + \Phi _2\), (11.28) is equivalent to the following equation for \(\Phi _2\)

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \Phi _2&= L [\Phi _2] + {{\tilde{B}}} [\Phi _2] + H_1, \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \Phi _2(\cdot ,\tau _0)&= c_1 {{\tilde{Z}}}_0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(11.29)

We now apply Proposition 10.1 to (11.29). We have that \(\Vert H_1\Vert _{\nu ,m,6+\sigma ,\varepsilon }<\infty \) by (11.16), \(H_1\) is radial and satisfies the zero mass condition (11.18). By Proposition 10.1 and Lemma 10.10 there exists \(c_1\) such that the solution \(\Phi _2 \) of (11.29) satisfies

$$\begin{aligned} \Phi _2(y,\tau ) = \Phi _2^\perp (y,\tau ) + \frac{a(\tau )}{2} Z_0(y) , \end{aligned}$$

with the estimates

$$\begin{aligned} |\Phi _2^\perp (y,\tau ) |&\le C \frac{ \Vert H_1 \Vert _{\nu ,m,6+\sigma ,\varepsilon }}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} } {\left\{ \begin{array}{ll} \displaystyle \frac{1}{(1+|y|)^{2}} &{} |y|\le \sqrt{\tau }\\ \displaystyle \frac{\tau }{|y|^4} &{} |y|\ge \sqrt{\tau }, \end{array}\right. } \end{aligned}$$
(11.30)
$$\begin{aligned} |a(\tau ) |&\le C \frac{\Vert H_1\Vert _{\nu ,m,6+\sigma ,\varepsilon }}{\tau ^{\nu -1} (\log \tau )^{m+q}} . \end{aligned}$$
(11.31)

(We are ignoring the factor \(\frac{1}{(\log \tau _0)^{1-q}}\) in the estimate of \(a(\tau )\).) We also know that \(c_1\) is a linear function of \(H_1\) and satisfies

$$\begin{aligned} |c_1|&\le C \frac{\Vert H_1\Vert _{\nu ,m,6+\sigma ,\varepsilon }}{\tau _0^{\nu -1} (\log \tau _0)^{m+1}} . \end{aligned}$$

Combining (11.15) and (11.30) we conclude that \(\Phi \), the solution to (11.28), can be decomposed as

$$\begin{aligned} \Phi = \Phi _0 + \frac{a(\tau )}{2} Z_0 \end{aligned}$$

where \(\Phi _0(y,\tau ) = \Phi _1 + \Phi _2^\perp \) is radial and satisfies

$$\begin{aligned} |\Phi _0(y,\tau )| \le C \frac{\Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } }{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} } \min \Bigl ( \frac{1}{(1+|y|)^{2}} , \frac{\tau }{|y|^4} \Bigr ) \end{aligned}$$

and \(a(\tau )\) satisfies, combining (11.16) and (11.31),

$$\begin{aligned} |a(\tau ) |&\le C \frac{1}{\tau ^{\nu -1} (\log \tau )^{m+q}} \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } . \end{aligned}$$

We summarize the previous finding as follows. Given H radial satisfying \(\int _{{\mathbb {R}}^2} H(\cdot ,\tau )=0\) for \(\tau >\tau _0\) and \(\Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } <\infty \), let us denote \(T_0(H) =\Phi _0 = \Phi _1 + \Phi _2^\perp \) and \(T_a(H) = a(\tau ) \) so that the solution \(\Phi \) of (11.28), is \(\Phi = \Phi _0 + \frac{a(\tau )}{2} Z_0 = T_0[H] + \frac{1}{2} T_a[H] Z_0\). Then \(T_0\), \(T_a\) are linear and have the estimates

$$\begin{aligned} \Vert T_0[H] \Vert _0&\le C\Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } \end{aligned}$$
(11.32)
$$\begin{aligned} \Vert T_a[H] \Vert _a&\le C \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon } , \end{aligned}$$
(11.33)

where

$$\begin{aligned} \Vert \Phi _0\Vert _{0}&= \sup _{ \tau>\tau _0 , \, y \in {\mathbb {R}}^2} \tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} \frac{1}{\min \Bigl ( \frac{1}{(1+|y|)^{2}} , \frac{\tau }{|y|^4} \Bigr ) } |\Phi _0(y,\tau )| \\ \Vert a\Vert _{a}&= \sup _{ \tau >\tau _0 } \tau ^{\nu -1} (\log \tau )^{m+q} |a(\tau )| . \end{aligned}$$

Moreover \(c_1\) is a linear function of H and satisfies

$$\begin{aligned} |c_1|&\le C \frac{\Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon }}{\tau _0^{\nu -1} (\log \tau _0)^{m+1}} . \end{aligned}$$

We will apply these estimates to treat problem (11.27), which can be written as the fixed point problem

$$\begin{aligned} \Phi _0&= T_0 [ H + \zeta _1 A[ \Phi _0 + a Z_0] ] \\ a&= T_a [ H + \zeta _1 A[ \Phi _0 + a Z_0] ] \end{aligned}$$

By (11.32) and (11.33)

$$\begin{aligned} \Vert T_0[ \zeta _1A[ \Phi _0 {+} a Z_0 ] ] \Vert _0 {+}\Vert T_a[ \zeta _1A[ \Phi _0 {+} a Z_0 ] ] \Vert _a \le C \Vert \zeta _1 A[ \Phi _0 {+} a Z_0 ] \Vert _{\nu ,m,4+\sigma ,\varepsilon }. \end{aligned}$$

We claim that

$$\begin{aligned} \Vert \zeta _1 A[ \Phi _0 ] \Vert _{\nu ,m,4+\sigma ,\varepsilon } \le C \tau _0^{-\vartheta } \Vert \Phi _0\Vert _0 , \end{aligned}$$
(11.34)

for some \(\vartheta >0\), where C is independent of \(\tau _0\), and

$$\begin{aligned} \Vert \zeta _1 A[ a Z_0 ] \Vert _{\nu ,m,4+\sigma ,\varepsilon } \le \frac{C}{(\log \tau _0)^{1+q}}\Vert a\Vert _a . \end{aligned}$$
(11.35)

Assume for the moment that (11.34), (11.35) hold. The we see that

$$\begin{aligned} \Vert \Phi _0\Vert _0 + \Vert a\Vert _a \le \frac{C}{(\log \tau _0)^{1+q}} (\Vert \Phi _0\Vert _0 + \Vert a\Vert _a) + C \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon }. \end{aligned}$$

For \(\tau _0\) large this gives

$$\begin{aligned} \Vert \Phi _0\Vert _0 + \Vert a\Vert _a \le C \Vert H\Vert _{\nu ,m,4+\sigma ,\varepsilon }, \end{aligned}$$

which is the desired result.

For the proof of estimates (11.34), (11.35) we will need the following property. If \(\Phi \) satisfies \(|\Phi (y)|\le \frac{1}{(1+|y|)^{2+\kappa }}\) for some \(\kappa >0\) and \(\int _{{\mathbb {R}}^2}\Phi dy=0\), then

$$\begin{aligned} \int _{{\mathbb {R}}^2} \nabla \cdot [ \Phi U y - Z_0 \nabla \Psi ]|y|^2 dy=0 , \quad \Psi = (-\Delta )^{-1}\Phi . \end{aligned}$$
(11.36)

Indeed,

$$\begin{aligned} \int _{{\mathbb {R}}^2} \nabla \cdot ( \Phi U y ) |y|^2dy&= - 2 \int _{{\mathbb {R}}^2} \Phi U |y|^2dy = 2 \int _{{\mathbb {R}}^2} \Delta \Psi U |y|^2dy \\&= -2 \int _{{\mathbb {R}}^2} \nabla \Psi \cdot \nabla (U |y|^2) dy \\&= -2 \int _{{\mathbb {R}}^2} \nabla \Psi \cdot y Z_0 dy \end{aligned}$$

and

$$\begin{aligned} \int _{{\mathbb {R}}^2} \nabla \cdot ( Z_0 \nabla \Psi ) |y|^2 dy&= -2 \int _{{\mathbb {R}}^2} Z_0 \nabla \Psi \cdot y dy. \end{aligned}$$

To prove (11.34), let us write \(\Psi _0 = (-\Delta )^{-1} \Phi _0\). Then

$$\begin{aligned} A[\Phi _0] = L^{-1}[ \nabla \cdot ( \Phi _0 U y - Z_0 \nabla \Psi _0) ] . \end{aligned}$$

Using the definition of \(L^{-1}\) given in Lemma 11.1 we have that

$$\begin{aligned} L^{-1}[ \nabla \cdot ( \Phi U y ) - \nabla \cdot ( Z_0 \nabla (-\Delta )^{-1}\Phi ) ] = U g + U \psi \end{aligned}$$

where

$$\begin{aligned} g(\rho ,\tau ) = -\int _\rho ^\infty \Bigl [ \Phi _0(s,\tau ) s -\frac{Z_0(s)}{U(s)} \partial _\rho \Psi _0(s,\tau ) \Bigr ]ds , \end{aligned}$$
(11.37)

and \(\psi \) is the decaying solution to the Liouville equation

$$\begin{aligned} -\Delta \psi - U \psi = U g . \end{aligned}$$

From the definition \(\Psi _0 = (-\Delta )^{-1} \Phi _0\) and using that \(\int _{{\mathbb {R}}^2} \Phi _0 dy=0\) we have

$$\begin{aligned} \partial _\rho \Psi _0(\rho ,\tau )=\frac{1}{\rho }\int _\rho ^\infty \Phi _0(s,\tau )sds \end{aligned}$$

which gives the estimate

$$\begin{aligned} | \partial _\rho \Psi _0(\rho ,\tau )|\le C \Vert \Phi _0\Vert _0 \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}}} {\left\{ \begin{array}{ll} \frac{\log ( \frac{2\sqrt{\tau }}{1+\rho })}{1+\rho } &{} \rho \le \sqrt{\tau }, \\ \frac{\tau }{\rho ^3} &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$

Then formula (11.37) gives

$$\begin{aligned} |g(\rho ,\tau ) |&\le C \Vert \Phi _0\Vert _0 \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}}} {\left\{ \begin{array}{ll} \log ^2( \frac{2\sqrt{\tau }}{1+\rho }) &{} \rho \le \sqrt{\tau }, \\ \frac{\tau }{\rho ^2} &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \\&\le C \Vert \Phi _0\Vert _0 \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}-2}} \min \Bigl (1,\frac{\tau }{\rho ^2}\Bigr ). \end{aligned}$$

We note that by (11.36) we have \(\int _{{\mathbb {R}}^2} U g z_0 dy = 0\). Then, \(\psi \) has the estimate

$$\begin{aligned} |\psi (\rho ,\tau ) |\le C \Vert \Phi _0\Vert _0 \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}-2}} \frac{1}{(1+\rho )^2} \min \Bigl (1,\frac{\tau }{\rho ^2}\Bigr ). \end{aligned}$$

It follows that \(A[ \Phi _0] = U g + U \psi \) satisfies

$$\begin{aligned} |A[ \Phi _0] (\rho ,\tau )| \le C \Vert \Phi _0\Vert _0 \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}-2}} \frac{1}{(1+\rho )^4} \min \Bigl (1,\frac{\tau }{\rho ^2}\Bigr ). \end{aligned}$$

From this inequality we obtain (11.34).

The proof of (11.35) is similar. This time \(A[a Z_0] = U g_1 + U \psi _1\) where

$$\begin{aligned} g_1(\rho ,\tau ) = -a(\tau ) \int _\rho ^\infty \Bigl [ Z_0(s) s -\frac{Z_0(s)}{U(s)} z_0'(s) \Bigr ]ds , \end{aligned}$$

and \(\psi _1\) is the radial decaying solution to

$$\begin{aligned} -\Delta \psi _1 - U \psi _1 = U g_1 . \end{aligned}$$

We then obtain that

$$\begin{aligned} | A[aZ_0](\rho ,\tau ) |\le C \Vert a\Vert _a \frac{1}{\tau ^{\nu -1} (\log \tau )^{m+q}} \frac{1}{(1+\rho )^6} . \end{aligned}$$

From this estimate we deduce (11.35). \(\square \)

Before proving Proposition 11.1 as stated, we obtain a version of it for the problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L[\phi ] + B[\phi ]+ h(y,\tau ) \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \phi (\cdot ,\tau _0)&= c_1 {{\hat{Z}}}_0 \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(11.38)

where

$$\begin{aligned} {{\hat{Z}}}_0 = L[ {{\tilde{Z}}}_0]. \end{aligned}$$

Lemma 11.4

Let \(0<\sigma <1\), \(\varepsilon >0\), \(\sigma +\varepsilon <2\) and \(1<\nu < \min ( 1+\frac{\varepsilon }{2},3-\frac{\sigma }{2}, \frac{3}{2})\). Let \(0<q<1\). Then there is C such that for \(\tau _0\) large the following holds. Suppose that h is radially symmetric, satisfies \(\Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }<\infty \) and

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y,\tau )dy&=0, \quad \int _{{\mathbb {R}}^2} h(y,\tau )|y|^2dy=0,\quad \tau >\tau _0. \end{aligned}$$

Then there exist \(c_1\in {\mathbb {R}}\) and a solution \(\phi (y,\tau )\) of problem (11.38) that define linear operators of h and satisfy

$$\begin{aligned} \Vert \phi \Vert _{\nu -\frac{1}{2} ,m+\frac{q-1}{2},4,2+\sigma +\varepsilon }&\le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \\ |c_1|&\le C \frac{1 }{ \tau _0^{\nu -1} (\log \tau _0)^{m+1}} \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

Proof

Consider equation (11.27), where H is the function constructed in Lemma 11.1. By Lemma 11.3, there is \(c_1\) such that the solution \(\Phi \) of (11.27) can be decomposed as \(\Phi = \Phi _0 + \frac{a(\tau )}{2} Z_0\), where \(\Phi _0\) and a satisfy the estimates stated in that proposition. In combination with (11.4) we find

$$\begin{aligned} |\Phi _0(\rho ,\tau ) |&\le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} } \min \Bigl ( \frac{1}{(1+|y|)^{2}} , \frac{\tau }{|y|^4} \Bigr ) \nonumber \\ |a(\tau )|&\le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu -1} (\log \tau )^{m+q} } . \end{aligned}$$
(11.39)
$$\begin{aligned} |c_1|&\le C \frac{1 }{ \tau _0^{\nu -1} (\log \tau _0)^{m+1}} \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$
(11.40)

Moreover \(\Phi _0\), a, \(c_1\) are linear operators of H.

From standard parabolic estimates and (11.39) we obtain

$$\begin{aligned} |\nabla \Phi _0(\rho ,\tau ) |&\le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }\frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} } \min \Bigl ( \frac{1}{(1+|y|)^{3}} , \frac{\tau }{|y|^5} \Bigr ). \end{aligned}$$
(11.41)

We consider the equation for \(\Phi _0 = \Phi - \frac{a(\tau )}{2}Z_0\), obtained from (11.27), and differentiate with respect to \(y_j\), \(j=1,2\). Using standard parabolic estimates, together with (11.39), (11.41), and the bound for \(a'(\tau )\) in (10.92), we obtain

$$\begin{aligned} | D^2 \Phi _0(\rho ,\tau )| \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} } \min \Bigl ( \frac{1}{(1+|y|)^{4}} , \frac{\tau }{|y|^6} \Bigr ). \end{aligned}$$
(11.42)

Let us define \(\phi = L[\Phi ]\). Then \(\phi \) satisfies (11.38) because \(L[Z_0]=0\) and thanks to (11.39), (11.41), (11.42) we find

$$\begin{aligned} |\phi (\rho ,\tau )| \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} } \min \Bigl ( \frac{1}{(1+|y|)^{4}} , \frac{\tau }{|y|^6} \Bigr ). \end{aligned}$$
(11.43)

In the rest of the proof we show that

$$\begin{aligned} |\phi (\rho ,\tau )| \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} } \frac{1}{(1+\rho )^4} {\left\{ \begin{array}{ll} 1 &{} \rho \le \sqrt{\tau }\\ \frac{\tau ^{1+\sigma /2+\varepsilon /2}}{\rho ^{2+\sigma + \varepsilon }} &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$

For this we consider the equation (11.38) written in the form

$$\begin{aligned} \partial _\tau \phi = \Delta \phi - \nabla \Gamma _0 \nabla \phi + 2 U \phi + B[\phi ] + {{\bar{h}}} , \end{aligned}$$
(11.44)

where

$$\begin{aligned} {{\bar{h}}} = - \nabla U \nabla \psi + h. \end{aligned}$$

Using (11.43) and the radial formula for \(\psi = (-\Delta )^{-1} \phi \), we get

$$\begin{aligned} |\nabla \psi (y,\tau )| \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu -\frac{1}{2}} (\log \tau )^{m+\frac{q}{2}} } {\left\{ \begin{array}{ll} \frac{1}{(1+\rho )^3} &{} \rho \le \sqrt{\tau }\\ \frac{\tau }{\rho ^{5}} &{} \rho \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$

This estimate and the definition of the norm \( \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }\), give

$$\begin{aligned} |{{\bar{h}}}(y,\tau ) |\le C \frac{1}{\tau ^{\nu -\frac{1}{2}} ( \log \tau )^{m+\frac{q}{2}} } \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } {\left\{ \begin{array}{ll} \frac{1}{(1+|y|)^{6+\sigma }} &{} |y| \le \sqrt{\tau }\\ \frac{1}{\tau ^{3+\sigma /2}(\frac{|y|}{\sqrt{\tau }})^{6+\sigma +\varepsilon }} &{} |y| \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$

We now construct a barrier very similar to the proof of Proposition 8.1

$$\begin{aligned} {\bar{\phi }} (\rho ,\tau )&= A_1 \frac{1}{\tau ^{\nu -\frac{1}{2}} ( \log \tau )^{m+\frac{q}{2}} } {{\tilde{g}}}_2(\rho ) \chi _0\Bigl (\frac{\rho }{\sqrt{\tau }}\Bigr )\\&\quad +A_2 \frac{1}{\tau ^{\nu +\frac{3}{2}} (\log \tau )^{m+\frac{q}{2}} } \frac{1}{(1+\rho /\sqrt{\tau })^{6+\sigma +\varepsilon }} \\&\quad +A_3 \frac{1}{\tau ^{\nu +\frac{3}{2}} (\log \tau )^{m+\frac{q}{2}} } e^{-\frac{\rho ^2}{4\tau } } , \end{aligned}$$

where \({{\tilde{g}}}_2\) is the function (10.60). We consider (11.44) in \(\{ \, (y,\tau ) \ | \ \tau> \tau _0, \ |y|> R_0\, \}\) where \(R_0>0\) is a large constant. For suitable constants \(A_1\), \(A_2\), \(A_3\), C the function \(C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } {\bar{\phi }} \) is a supersolution. This computation requires \(\nu <\frac{3}{2}\).

Moreover \(\phi (y,\tau ) \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } {\bar{\phi }}(y,\tau )\) at \(|y| = R_0\). The initial conditions also compare well. Indeed, by Lemma 10.11 and (11.40)

$$\begin{aligned} |\phi (\rho ,\tau _0)|&= c_1 |{{\hat{Z}}}_0(\rho )| \le C \frac{1 }{ \tau _0^{\nu -1} (\log \tau _0)^{m+1}} \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \frac{1}{\tau _0} \frac{1}{1+\rho ^6} , \end{aligned}$$

and this is supported on \(\rho \le 2 \sqrt{\tau }_0\), so

$$\begin{aligned} |\phi (\rho ,\tau _0)| \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } {\bar{\phi }}(y,\tau ). \end{aligned}$$

By the maximum principle

$$\begin{aligned} |\phi (y,\tau ) |\le C {\bar{\phi }}(y,\tau ) \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } , \quad |y|> R_0. \end{aligned}$$

This finishes the proof. \(\square \)

Proof of Proposition 11.1

Let \({\hat{\phi }}\), \(c_1\) be the solution to (11.38) constructed in Lemma 11.4. Let \(\phi _1\) be the solution to (10.100). By Lemma 10.12\(\phi _1\) satisfies

$$\begin{aligned} | \phi _1(\rho ,\tau ) | \le C \frac{\tau _0^{\nu _0-1} R(\tau )^2}{ \tau ^{\nu _0} R(\tau _0)^2} \frac{1}{(1+\rho ^4) } \min \Bigl ( 1 , \frac{\tau ^{1/2}}{\rho }\Bigr )^{2+\sigma +\varepsilon } , \end{aligned}$$
(11.45)

where \(1<\nu _0<\frac{7}{4}\). Then the solution \(\phi \) to (8.9) that we construct is given by

$$\begin{aligned} \phi = {\hat{\phi }} - c_1 \phi _1 . \end{aligned}$$

To get the desired estimate on \(\phi \) we need to estimate \(|c_1 \phi _1|\). Let f be given by (10.12). By (11.40) and (11.45)

$$\begin{aligned} |c_1 \phi _1(\rho ,\tau )|&\le C \frac{1 }{ \tau _0^{\nu -1} (\log \tau _0)^{m+1}} \frac{\tau _0^{\nu _0-1} R(\tau )^2}{ \tau ^{\nu _0} R(\tau _0)^2} \frac{1}{(1+\rho ^4) } \min \Bigl ( 1 , \frac{\tau ^{1/2}}{\rho }\Bigr )^{2+\sigma +\varepsilon } \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \\&\le C \frac{1}{\log \tau _0 R(\tau _0)} f(\tau ) R(\tau ) \frac{1}{(1+\rho ^4) } \min \Bigl ( 1 , \frac{\tau ^{1/2}}{\rho }\Bigr )^{2+\sigma +\varepsilon } \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \\&\le C f(\tau ) R(\tau ) \frac{1}{(1+\rho ^4) } \min \Bigl ( 1 , \frac{\tau ^{1/2}}{\rho }\Bigr )^{2+\sigma +\varepsilon } \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon } \end{aligned}$$

provided \(\frac{1}{2}+\nu -\nu _0<0\). However \(\nu _0\) can be taken close to \(\frac{7}{4}\), so we obtain the result by assuming \(\nu < \frac{5}{4}\) in addition to the other constraints needed in Lemma 11.4, namely \(1<\nu < \min ( 1+\frac{\varepsilon }{2},3-\frac{\sigma }{2}, \frac{3}{2})\). \(\square \)

12 Linear Estimate with Second Moment (General)

A convenient property of problem (8.3) is that it can be split into Fourier modes. If we decompose

$$\begin{aligned} h(y,\tau )&= h_0(|y|,\tau ) + h_1(y,\tau ) , \quad h_0(\rho ,\tau ) = \frac{1}{2\pi } \int _0^{2\pi } h(\rho e^{i\theta }, \tau ) d\theta \end{aligned}$$
(12.1)
$$\begin{aligned} \phi (y,\tau )&= \phi _0(|y|,\tau ) + \phi _1(y,\tau ) , \quad \phi _0(\rho ,\tau ) = \frac{1}{2\pi } \int _0^{2\pi } \phi (\rho e^{i\theta }, \tau ) d\theta , \end{aligned}$$
(12.2)

then \(\phi \) solves (8.3) if and only if \(\phi _i\) solves (8.3) where h is replaced with \(h_i\), for \(i=0,1\). If \(h=h_1\) we say that h has no radial mode.

For the proof Proposition 8.2 in the general case we will consider in a first step the equation (8.3) but without the operator B, namely,

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L [\phi ] + h, \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ), \\ \phi (\cdot ,\tau _0)&= 0\quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(12.3)

for functions with no radial mode, as explained at the beginning of Sect. 11. Later on, we will consider equation (8.3) for functions with no radial mode, where we will treat the operator \(B[\phi ]\) as a perturbation term that can be assimilated to the right hand side.

The main step in the proof is the following estimate, valid when the functions involved have no radial mode:

Proposition 12.1

Let \(0<\sigma <1\), \(0<\varepsilon <2\), \(0<\nu <\min ( 1 + \frac{\varepsilon }{2}, \frac{3}{2}-\frac{\sigma }{2})\), \(m\in {\mathbb {R}}\). Then there is a \(C>0\) such that for any \(\tau _0\) sufficiently large the following holds. Suppose that \(h(y,\tau )\) has no radial mode and satisfies \(\Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon }<\infty \),

$$\begin{aligned} \int _{{\mathbb {R}}^2} h(y,\tau ) y_j dy=0\quad \text {for all }\tau >\tau _0, \quad j=1,2. \end{aligned}$$
(12.4)

Then the solution \(\phi (y,\tau )\) of (12.3) satisfies

$$\begin{aligned} |\phi (y,\tau ) | \le C \frac{\Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon }}{\tau ^\nu (\log \tau )^{m} } {\left\{ \begin{array}{ll} \frac{1}{(1+|y|)^{3+\sigma }} , &{} |y|\le \sqrt{\tau } . \\ \frac{\tau ^{1+\frac{\varepsilon }{2}}}{|y|^{5+\sigma +\varepsilon }} , &{} |y|\ge \sqrt{\tau } . \end{array}\right. } \end{aligned}$$
(12.5)

Proof

Since \(h(y,\tau )\) has no radial mode, all functions involved in the proof have also this property. We use the notation from §9.2, particular \(g = \frac{\phi }{U}-(-\Delta )^{-1}\phi \), \(g^\perp = g - a\) with \(a(\tau )\in {\mathbb {R}}\) such that

$$\begin{aligned} \int _{{\mathbb {R}}^2} g^\perp (y,\tau ) U dy =0. \end{aligned}$$

But

$$\begin{aligned} \int _{{\mathbb {R}}^2} g (y,\tau ) U dy =0 \end{aligned}$$

because g has no radial mode, so that \(a(\tau ) = 0\), \(g^\perp = g\), \(\phi ^\perp = \phi \). Then the proof proceeds as the proof of Proposition 10.1 with some simplifications, since there is no need to estimate a.

We write (12.3) as

$$\begin{aligned} \partial _\tau \phi = \nabla \cdot ( U \nabla g^\perp ) + h , \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) . \end{aligned}$$

We multiply this equation by g and integrate in \({\mathbb {R}}^2\).

Let \(R>0\) be a large fixed constant and let

$$\begin{aligned} f(\tau ) = \frac{1}{\tau ^\nu (\log \tau )^m}. \end{aligned}$$

Let \(T_2>\tau _0\) and let

$$\begin{aligned} \Vert \varphi \Vert _{\infty ,T_2} = \sup _{\tau \in [\tau _0,T_2]} |\varphi (\tau )|. \end{aligned}$$

The following estimates are valid for \(\tau \in [\tau _0,T_2]\). As in the proof of Proposition 10.1 we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} g^2 U \le Cf(\tau )^2 R^2 \Bigl ( \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon }^2 + \Bigl \Vert \frac{\omega }{f R}\Bigr \Vert _{\infty ,T_2}^2 \Bigr ) , \end{aligned}$$
(12.6)

where

$$\begin{aligned} \omega (\tau ) = \Bigl ( \int _{{\mathbb {R}}^2\setminus B_{R}} g(\tau )^2 U \Bigr )^{1/2}. \end{aligned}$$

Similarly as in Lemma 10.8, from (12.6) we get

$$\begin{aligned} |U g(y,\tau )| \le Cf(\tau ) R \Bigl ( \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } + \Bigl \Vert \frac{\omega }{f R}\Bigr \Vert _{\infty ,T_2} \Bigr ) \frac{ 1 }{(1+|y|)^{3+\sigma }} . \end{aligned}$$
(12.7)

The proof is presented below. We use this to estimate

$$\begin{aligned} \omega (\tau ) = \Bigl ( \int _{{\mathbb {R}}^2\setminus B_{R}} g^2 U \Bigr )^{1/2} \le C f(\tau ) R^{1-\sigma } \Bigl ( \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } + \Bigl \Vert \frac{\omega }{f R}\Bigr \Vert _{\infty ,T_2} \Bigr ), \end{aligned}$$

which implies

$$\begin{aligned} \frac{\omega (\tau ) }{f(\tau ) R} \le C R^{-\sigma }\Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } + C R^{-\sigma } \Bigl \Vert \frac{\omega }{f R}\Bigr \Vert _{\infty ,T_2}. \end{aligned}$$

We deduce that

$$\begin{aligned} \Bigl \Vert \frac{\omega }{f R}\Bigr \Vert _{\infty ,T_2} \le C R^{-\sigma } \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } , \end{aligned}$$

by choosing R as a large constant.

Now we let \(T_2\rightarrow \infty \) and find

$$\begin{aligned} \omega (\tau ) \le C f(\tau ) R \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon }^2 , \quad \tau >\tau _0. \end{aligned}$$
(12.8)

The inequalities that follow hold for \(\tau >\tau _0\).

Combining (12.8) with (12.6) we obtain

$$\begin{aligned} \int _{{\mathbb {R}}^2} g^2 U \le Cf(\tau )^2 R^2 \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon }^2 , \quad \tau >\tau _0. \end{aligned}$$

and using (12.7) we also get

$$\begin{aligned} |U g(y,\tau )| \le C f(\tau ) R \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } \frac{ 1 }{(1+|y|)^{3+\sigma }} . \end{aligned}$$

Let \(\psi = (-\Delta )^{-1}\phi \) so that \( \phi = U g + U \psi \). Using Lemma 9.1 and the previous estimate we obtain

$$\begin{aligned} |\psi (y,\tau )| + (1+|y|)|\nabla \psi (y,\tau )| \le C \frac{R }{\tau ^{\nu } (\log \tau )^{m}} \frac{ 1 }{(1+|y|)^{1+\sigma }} \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } . \end{aligned}$$
(12.9)

We consider the equation (12.3) in \({\mathbb {R}}^2 {\setminus } B_{R}(0)\) written in the form

$$\begin{aligned} \partial _\tau \phi = \Delta \phi - \nabla \Gamma _0 \nabla \phi + 2 U \phi + {{\bar{h}}} , \end{aligned}$$

where

$$\begin{aligned} {{\bar{h}}} = - \nabla U \nabla \psi + h. \end{aligned}$$

By (12.9) and the definition of the norm \( \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon }\),

$$\begin{aligned} |{{\bar{h}}}(y,\tau ) |\le C \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu } ( \log \tau )^{m} } \frac{1}{(1+|y|)^{5+\sigma }} {\left\{ \begin{array}{ll} 1 &{} |y| \le \sqrt{\tau }\\ \frac{\tau ^{\varepsilon /2}}{|y|^\varepsilon } &{} |y| \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$

Here we are using \(\varepsilon <2\). Using barriers as in the proof of Lemma 10.8 we get

$$\begin{aligned} |\phi (y,\tau ) | \le C \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } \frac{1}{\tau ^{\nu } ( \log \tau )^{m} } \frac{1}{(1+|y|)^{3+\sigma }} {\left\{ \begin{array}{ll} 1 &{} |y| \le \sqrt{\tau }\\ \frac{\tau ^{1+\varepsilon /2}}{|y|^{2+\varepsilon }} &{} |y| \ge \sqrt{\tau }. \end{array}\right. } \end{aligned}$$

(For this we need \( \nu < 1+\frac{\varepsilon }{2}\), \(\nu + \frac{\sigma }{2} < \frac{3}{2}\).) This proves (12.5). \(\square \)

Proof of (12.7)

We define

$$\begin{aligned} g_0 = U g, \end{aligned}$$

which satisfies the equation

$$\begin{aligned} \partial _\tau g_0&= \Delta g_0 - \nabla g_0 \cdot \nabla \Gamma _0 +2 U g_0 + {{\tilde{h}}} \end{aligned}$$
(12.10)

where

$$\begin{aligned} {{\tilde{h}}} = U v + h-U(-\Delta )^{-1} h \end{aligned}$$

and

$$\begin{aligned} v:= (-\Delta )^{-1} ( \nabla \cdot ( g_0 \nabla \Gamma _0 ) ). \end{aligned}$$

As in the proof of Lemma 10.7 we obtain

$$\begin{aligned} |g_0 (y,\tau )| \le C \frac{R }{\tau ^\nu (\log \tau )^m (1+|y|)^{2}} K, \end{aligned}$$
(12.11)

where

$$\begin{aligned} K = \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } + \Bigl \Vert \frac{\omega }{f R}\Bigr \Vert _{\infty ,T_2}. \end{aligned}$$

Applying parabolic estimates to (12.10) and a scaling argument we find

$$\begin{aligned} | \nabla g_0 (y,\tau )| \le C \frac{R K }{\tau ^\nu (\log \tau )^m (1+|y|)^3} . \end{aligned}$$
(12.12)

Using (12.11), (12.12) and \(g_0 = g U\) we get that

$$\begin{aligned} | \nabla U \cdot \nabla g + g \Delta U |\le C \frac{R K}{\tau ^\nu (\log \tau )^m (1+|y|)^4 } . \end{aligned}$$

We observe that for \(i=1,2\)

$$\begin{aligned} \int _{{\mathbb {R}}^2} \nabla (U \nabla g) y_i \,dy = 0 . \end{aligned}$$
(12.13)

Indeed,

$$\begin{aligned} \int _{{\mathbb {R}}^2} \nabla (U \nabla g) y_i \,dy&= -\int _{{\mathbb {R}}^2} U \nabla g e_i =\int _{{\mathbb {R}}^2} g \nabla U e_i . \end{aligned}$$

But from \(g = \frac{\phi }{U} - \psi \), \( \psi = (-\Delta )^{-1}\phi \) we have

$$\begin{aligned} -\Delta \psi - U \psi = U g = g_0. \end{aligned}$$

Multiplying this equation by \(z_i = \nabla \Gamma _0 e_i\) defined in (9.2) and integrating we get

$$\begin{aligned} \int _{{\mathbb {R}}^2} g U \nabla \Gamma _0 e_i =0 , \end{aligned}$$

which is the desired claim (12.13). We note that

$$\begin{aligned} \left\{ \begin{aligned} -\Delta v&= \nabla U \cdot \nabla g + g \Delta U = \nabla \cdot ( g \nabla U) \quad \text {in }{\mathbb {R}}^2. \\ v(y)&\rightarrow 0 \quad \text {as }|y|\rightarrow \infty . \end{aligned} \right. \end{aligned}$$

Now we can apply Remark 9.1 and deduce that for any \(\vartheta \in (0,1)\) there is C such that

$$\begin{aligned} |v(y,\tau ) | \le C \frac{R K}{\tau ^\nu (\log \tau )^m (1+|y|)^{2-\vartheta } } . \end{aligned}$$
(12.14)

We next estimate \({{\tilde{h}}}\). From Remark 9.1 and the assumptions on h, in particular (12.4), we have

$$\begin{aligned} |((-\Delta )^{-1}h)(y,\tau )| \le C \frac{\Vert h\Vert }{\tau ^\nu (\log \tau )^m( 1+|y|)^{2-\vartheta }} , \end{aligned}$$
(12.15)

for any \(\vartheta \in (0,1)\). Also from (12.11) we have

$$\begin{aligned} |U g_0 (y,\tau )| \le C \frac{R }{\tau ^\nu (\log \tau )^m (1+|y|)^{6}} K . \end{aligned}$$

Therefore, from (12.15), (12.11), (12.14) we find that, for any \(\vartheta >0\),

$$\begin{aligned} |{{\tilde{h}}}(y,\tau )|&\le C \frac{R K}{\tau ^{\nu } (\log \tau )^{m}} \Bigl [ \frac{ 1 }{(1+|y|)^{5+\sigma }} \min \Bigl ( 1 , \frac{\tau ^{\varepsilon /2}}{\rho ^\varepsilon } \Bigr ) +\frac{ 1 }{(1+|y|)^{6-\vartheta }} \Bigr ] . \end{aligned}$$

We now use a barrier as in the proof of Lemma 10.8, in a domain of the form \(({\mathbb {R}}^2 {\setminus } B_{R_0} ) \times (\tau _0,\infty )\) where \(R_0\) is a large constant. We let \({{\tilde{g}}}(y)\) be the radial decaying solution to \(-\Delta _6 {{\tilde{g}}} = \frac{1}{(1+|y|)^{5+\sigma }}\) and

$$\begin{aligned}&{{\bar{g}}}(y,\tau ) = \frac{1}{\tau ^\nu (\log \tau )^m} {{\tilde{g}}}(y) \chi _0\Bigl ( \frac{y}{\delta \sqrt{\tau }}\Bigr ) + C_1 \frac{1}{\tau ^{\nu +\frac{3}{2}+\frac{\sigma }{2}} (\log \tau )^m}\\&\quad \Bigl [ \frac{1}{(1+|y|/\sqrt{\tau })^{\mu }} + C_2 e^{-\frac{|y|^2}{4\tau }} \Bigr ] \end{aligned}$$

where

$$\begin{aligned} \mu = \min ( 5 + \sigma + \varepsilon , 6 -\vartheta ). \end{aligned}$$

We assume that \(\nu <\frac{3}{2}-\frac{\sigma }{2}-\frac{\vartheta }{2}\), \( \nu < 1+\frac{\varepsilon }{2}\), \(\nu + \frac{\sigma }{2} < \frac{3}{2}\), and \(\sigma +\vartheta <1\). Since \(\vartheta >0\) is arbitrary we only need \(\nu <\frac{3}{2}-\frac{\sigma }{2}\), \( \nu < 1+\frac{\varepsilon }{2}\) and \(\sigma <1\). Then, for an appropriate choice of \(C_1\), \(C_2\), the function \(R K {{\bar{g}}}(y,\tau )\) is a supersolution. By the maximum principle

$$\begin{aligned} |g_0(y,\tau )|&\le C R K {{\bar{g}}}(y,\tau ). \end{aligned}$$

This proves the desired estimate (12.7). \(\square \)

Next we consider equation (8.3), which we recall,

$$\begin{aligned} \left\{ \begin{aligned} \partial _\tau \phi&= L[\phi ] + B[\phi ] + h +\sum _{j=1}^2 \mu _j(\tau ) W_{1,j} \quad \text {in }{\mathbb {R}}^2 \times (\tau _0,\infty ) \\ \phi (\cdot ,\tau _0)&= 0 \quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(12.16)

For \( \phi \) with no radial mode we can write

$$\begin{aligned} B[\phi ] = ( \zeta _1(t) \phi + \zeta _2(t) y \cdot \nabla \phi ) \chi _0\Bigl ( \frac{\lambda y}{5 \sqrt{t}} \Bigr ). \end{aligned}$$

Corollary 12.1

Let \(0<\sigma <1\), \(0<\varepsilon <2\), \(1<\nu <\min ( 1 + \frac{\varepsilon }{2}, \frac{3}{2}-\frac{\sigma }{2})\), \(m\in {\mathbb {R}}\). Then there is a \(C>0\) such that for any \(\tau _0\) sufficiently large the following holds. Suppose that \(h(y,\tau )\) has no radial mode and satisfies \(\Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon }<\infty \). Then there is a solution \(\phi (y,\tau )\), \(\mu _j\) of (12.16) that is a linear operator of h and satisfies

$$\begin{aligned} \Vert \phi \Vert _{\nu ,m,3+\sigma ,2+\varepsilon }&\le C \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } \end{aligned}$$
(12.17)
$$\begin{aligned} \mu _j(\tau )&= - \int _{{\mathbb {R}}^2} h(y,\tau ) y_jdy + {\tilde{\mu }}_j[h](\tau ) \nonumber \\ |{\tilde{\mu }}_j[h]|&\le \frac{C}{\tau ^{\nu +1+\sigma }(\log \tau )^{m+1}}\Vert h \Vert _{\nu ,m,5+\sigma ,\varepsilon }. \end{aligned}$$
(12.18)

Proof

Using Proposition 12.1, there is a linear operator T so that given h with \( \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon }<\infty \), with no radial mode, and satisfying the condition (12.4) associates the solution \(\phi \) of (12.3). Then the solution \(\phi \) of (12.16) can be written as

$$\begin{aligned} \phi = T \Bigl [B[\phi ] + h+\sum _{j=1}^2\mu _j(\tau )W_{1,j}\Bigr ], \end{aligned}$$

where \(\mu _j\) is chosen so that

$$\begin{aligned} \int _{{\mathbb {R}}^2} ( B[\phi ] + h)y_jdy+\mu _j(\tau )=0,\quad \forall \tau >\tau _0. \end{aligned}$$
(12.19)

The estimate (12.5) implies

$$\begin{aligned} \Vert \phi \Vert _{\nu ,m,3+\sigma ,2+\varepsilon } \le \Vert B[\phi ] + h \Vert _{\nu ,m,5+\sigma ,\varepsilon }+\sup _{\tau >\tau _0} \tau ^\nu (\log \tau )^m\sum _{j=1}^2|\mu _j(\tau )|. \end{aligned}$$

Using standard parabolic estimates we also get

$$\begin{aligned} \Vert |y| \nabla \phi \Vert _{\nu ,m,3+\sigma ,2+\varepsilon } \le \Vert B[\phi ] + h \Vert _{\nu ,m,5+\sigma ,\varepsilon } +\sup _{\tau >\tau _0} \tau ^\nu (\log \tau )^m\sum _{j=1}^2|\mu _j(\tau )|. \end{aligned}$$

To estimate \(\mu _j\) note that multiplying (12.16) by \(y_j\) and integrating we get that

$$\begin{aligned} \int _{{\mathbb {R}}^2} \phi y_j dy =0,\quad \forall \tau >\tau _0. \end{aligned}$$

Therefore

$$\begin{aligned} \left| \int _{{\mathbb {R}}^2} B(\phi ) dy\right| \le \frac{C}{\tau ^{\nu +1+\sigma }(\log \tau )^{m+1}} \Vert \phi \Vert _{\nu ,m,3+\sigma ,2+\varepsilon }, \end{aligned}$$

and from the definition (12.19)

$$\begin{aligned} \sup _{\tau >\tau _0} \tau ^\nu (\log \tau )^m\sum _{j=1}^2|\mu _j(\tau )| \le C \Vert h\Vert _{\nu ,m,5+\sigma ,\varepsilon } + \frac{C}{\tau _0^{1+\sigma }\log \tau _0} \Vert \phi \Vert _{\nu ,m,3+\sigma ,2+\varepsilon }. \end{aligned}$$

We also have that

$$\begin{aligned} \Vert B[\phi ] \Vert _{\nu ,m,5+\sigma ,\varepsilon } \le \frac{C}{\log \tau _0} \Vert |\phi | + |y||\nabla \phi | \Vert _{\nu ,m,3+\sigma ,2+\varepsilon } . \end{aligned}$$

Then for \(\tau _0\) large we deduce the estimate (12.17).

Finally, from (12.19) we get (12.18) with \({\tilde{\mu }}_j\) a linear operator of h satisfying

$$\begin{aligned} |{\tilde{\mu }}_j[h]|&= \left| \int _{{\mathbb {R}}^2} B(\phi ) dy\right| \le \frac{C}{\tau ^{\nu +1+\sigma }(\log \tau )^{m+1}} \Vert \phi \Vert _{\nu ,m,3+\sigma ,2+\varepsilon } \\&\le \frac{C}{\tau ^{\nu +1+\sigma }(\log \tau )^{m+1}}\Vert h \Vert _{\nu ,m,5+\sigma ,\varepsilon }. \end{aligned}$$

\(\square \)

We are now in a position to prove Proposition 8.2 in the general case.

Proof of Proposition 8.2

We decompose \(h = h_0 + h_1\) and \(\phi = \phi _0 + \phi _1\) as in (12.1), (12.2). We apply Proposition 11.1 to get

$$\begin{aligned} \Vert \phi _0 \Vert _{\nu -\frac{1}{2} ,m+\frac{q}{2},4,2+\sigma +\varepsilon } \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

To estimate \(\phi _1\) we use Corollary 12.1. First we select \(0<\vartheta <\frac{1}{2}\). Then note that

$$\begin{aligned} \Vert h_1\Vert _{\nu ,m,6-\vartheta ,\sigma +\varepsilon +\vartheta } \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

Then Corollary 12.1 gives a solution \(\phi _1\) of (12.16) such that

$$\begin{aligned} \Vert \phi _1\Vert _{\nu ,m,3+{\bar{\sigma }},2+{\bar{\varepsilon }}} \le C \Vert h_1\Vert _{\nu ,m,5+{\bar{\sigma }},{\bar{\varepsilon }}} . \end{aligned}$$

We take \({\bar{\sigma }}=1-\vartheta \) and \({\bar{\varepsilon }}=\varepsilon +\sigma +\vartheta \) and get

$$\begin{aligned} \Vert \phi _1\Vert _{\nu ,m,4-\vartheta ,2+\sigma +\varepsilon +\vartheta } \le C \Vert h_1\Vert _{\nu ,m,6-\vartheta ,\sigma +\varepsilon +\vartheta } \end{aligned}$$

and

$$\begin{aligned} \mu _j(\tau ) = - \int _{{\mathbb {R}}^2} h(y,\tau ) y_jdy + {\tilde{\mu }}_j[h](\tau ) \end{aligned}$$

because \(h_0\) is radial, with

$$\begin{aligned} |{\tilde{\mu }}_j[h]|&\le \frac{C}{\tau ^{\nu +2-\vartheta }(\log \tau )^{m+1}} \Vert h_1\Vert _{\nu ,m,6-\vartheta ,\sigma +\varepsilon +\vartheta }. \end{aligned}$$

But

$$\begin{aligned} \Vert \phi _1 \Vert _{\nu -\frac{1}{2} ,m+\frac{q}{2},4,2+\sigma +\varepsilon }&\le \Vert \phi _1 \Vert _{\nu ,m+\frac{q}{2},4-\vartheta ,2+\sigma +\varepsilon +\vartheta } \end{aligned}$$

and hence

$$\begin{aligned} \Vert \phi _1 \Vert _{\nu -\frac{1}{2} ,m+\frac{q}{2},4,2+\sigma +\varepsilon } \le C \Vert h\Vert _{\nu ,m,6+\sigma ,\varepsilon }. \end{aligned}$$

To apply Corollary 12.1 we need \(1<\nu <1+\frac{\varepsilon }{2}\) and \(\nu <1+\frac{\vartheta }{2}\). Given \(1<\nu < \min ( 1+\frac{\varepsilon }{2},3-\frac{\sigma }{2}, \frac{5}{4})\) we can select \(\vartheta \in (0,\frac{1}{2})\) such that \(\nu <1+\frac{\vartheta }{2}\) and then proceed. This concludes the proof. \(\square \)

13 The Outer Problem

We consider the linear outer problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _t \phi ^o&= L^o [\phi ^o] + g(x,t), \quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \phi ^o(\cdot ,t_0)&=0, \quad \text {in }{\mathbb {R}}^2 , \end{aligned} \right. \end{aligned}$$
(13.1)

where

$$\begin{aligned} L^o [\varphi ] :=\Delta _x \varphi - \nabla _x\Big [ \Gamma _0 \Big (\frac{x-\xi (t)}{\lambda (t)}\Big ) \Big ]\cdot \nabla _x \varphi = \Delta _x \varphi + 4 \frac{(x-\xi )}{|x-\xi |^2+\lambda ^2}\cdot \nabla _x\varphi . \end{aligned}$$

For \(g:{\mathbb {R}}^2 \times (t_0,\infty )\rightarrow {\mathbb {R}}\) we consider the norm \(\Vert g \Vert _{**,o}\) defined as the least K such that for all \((x,t)\in {\mathbb {R}}^2\times (t_0,\infty )\)

$$\begin{aligned} |g(x,t)| \leqq K\frac{1}{(t-t_0+A)^a (\log t)^\beta } \frac{1}{ 1+ |\zeta |^b} , \quad \zeta = \frac{x-\xi (t)}{\sqrt{t-t_0+A}}, \end{aligned}$$

where \(A>0\) is a constant.

We also define the norm \(\Vert \phi \Vert _{*,o}\) as the least K such that

$$\begin{aligned}&|\phi ^o(x,t)| + (\lambda +|x-\xi |) | {\nabla } _x \phi ^o (x,t)|\\&\quad \le K\frac{1}{(t-t_0+A)^{a-1} (\log t)^\beta } \frac{1}{ 1+ |\zeta |^{b}} , \quad \zeta = \frac{x-\xi }{\sqrt{t-t_0+A}} \end{aligned}$$

for all \((x,t)\in {\mathbb {R}}^2\times (t_0,\infty )\).

We assume that the parameters ab satisfy the constraints

$$\begin{aligned} 1<a< 4,\quad 2<b< 6, \quad a< 1+ \frac{b}{2}. \end{aligned}$$
(13.2)

There is no restriction on \(\beta \).

We recall from (4.1) that we are assuming that

$$\begin{aligned} | {\dot{\lambda }}(t)| \le \frac{C}{t ( \log t)^{3/2}} , \quad t>t_0, \end{aligned}$$
(13.3)

and

$$\begin{aligned} |{\dot{\xi }}(t)|\le \frac{C}{t^{\frac{3}{2}+\sigma }},\quad t>t_0, \end{aligned}$$
(13.4)

where \(0<\sigma <\frac{1}{2}\).

Proposition 13.1

Assume that ab satisfy (13.2), \(\frac{A}{\lambda (t_0)^2}\) is sufficiently large, and \(\lambda , \xi \) satisfy (13.3), (13.4). Then there is a constant C so that for \(t_0\) sufficiently large and for \(\Vert g\Vert _{**,o}<\infty \) there exists a solution \(\phi ^o= {\mathcal {T}} ^o_{{{\textbf {p}}}}[g]\) of (13.1), which defines a linear operator of g and satisfies

$$\begin{aligned} \Vert \phi ^o\Vert _{*,o} \le C \Vert g\Vert _{ **,o} . \end{aligned}$$

Proposition 6.3 in Sect. 6 follows from Proposition 13.1 with \(A = t_0\).

Lemma 13.1

Let \(2<\beta <6\) and h(r) satisfy

$$\begin{aligned} |h(r)|\le \frac{\lambda ^{-2}}{(r/\lambda +1)^\beta } =\frac{\lambda ^{\beta -2}}{(r+\lambda )^\beta } , \end{aligned}$$
(13.5)

where \(\lambda >0\). Then there is a unique bounded radial function \(\varphi (r)\) satisfying

$$\begin{aligned} L^o[\varphi ] + h = 0 \quad \text {in }{\mathbb {R}}^2. \end{aligned}$$

Moreover \(\varphi \) satisfies

$$\begin{aligned} |\varphi (r) | + (\lambda +r) |\partial _r \varphi (r) | \le \frac{C}{(1+r/\lambda )^{\beta -2}} = C \frac{\lambda ^{\beta -2}}{(r+\lambda )^{\beta -2}} \end{aligned}$$
(13.6)

Proof

The equation for \(\varphi \) is given by

$$\begin{aligned} \partial _{rr} \varphi (r) + \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \varphi (r) + h(r) = 0 , \quad r>0. \end{aligned}$$

We change variables \(\rho = \frac{r}{\lambda }\) and let \(\varphi (r) = {\bar{\varphi }}(\frac{r}{\lambda })\). Then we need to solve

$$\begin{aligned} \partial _{\rho \rho } {\bar{\varphi }} + \Bigl (\frac{1}{\rho }+ \frac{4\rho }{1 + \rho ^2} \Bigr ) \partial _{\rho } {\bar{\varphi }} + {{\bar{h}}}(\rho ) = 0, \quad \rho >0, \end{aligned}$$

where

$$\begin{aligned} {{\bar{h}}}(\rho ) = \lambda ^2 h(\lambda \rho ). \end{aligned}$$

By (13.5)

$$\begin{aligned} |{{\bar{h}}}(\rho )|\le \frac{1}{(1+\rho )^\beta }. \end{aligned}$$

The bounded solution is given by

$$\begin{aligned} {\bar{\varphi }}(\rho ) =\int _\rho ^\infty \frac{1}{v(1+v^2)^2} \int _0^v {{\bar{h}}}(s) s (1+s^2)^2 \, \textrm{d} s \, \textrm{d} v. \end{aligned}$$

By direct computation we get

$$\begin{aligned} |{\bar{\varphi }} (\rho )| + (1+\rho ) |\partial _\rho {\bar{\varphi }}(\rho )| \le \frac{C}{(1+\rho )^{\beta -2}}, \end{aligned}$$

and this implies (13.6). \(\square \)

Proof of Proposition 13.1

To find a pointwise estimate for the solution \(\phi ^o\) we construct a barrier.

Using polar coordinates \(x-\xi (t) = r e^{i\theta }\), \(L^o\) can be written as:

$$\begin{aligned} L^o [\varphi ] = \partial _{rr} \varphi + \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \varphi + \frac{1}{r^2}\partial _{\theta \theta }\varphi . \end{aligned}$$

First we construct a function \({\tilde{\psi }}(r,t)\) such that

$$\begin{aligned} \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \bigr ]{\tilde{\psi }} \ge \frac{1}{(t-t_0+A)^a (\log t)^\beta } \frac{1}{(1+r/\sqrt{t-t_0+A})^b}. \end{aligned}$$

Let

$$\begin{aligned} \psi _1(r,t) = \frac{1}{(t-t_0+A)^{a-1} ( \log t)^\beta } \Bigl [ \frac{1}{(1+\frac{r^2}{t-t_0+A})^{b/2}} + C_1 e^{-\frac{r^2}{4(t-t_0+A)}}\Bigr ]. \end{aligned}$$

Choosing a large constant \(C_1\), \(\psi _1\) satisfies

$$\begin{aligned}&\partial _t \psi _1 - \partial _{rr} \psi _1 - \frac{5}{r} \partial _r \psi _1\\&\quad \ge c \frac{1}{(t-t_0+A)^a ( \log t)^\beta } \frac{1}{(1+ \frac{r}{\sqrt{t-t_0+A}})^b},\quad \text {for } r>0, \ t>t_0, \end{aligned}$$

where \(c>0\). Here we require \(a<4\) and \( a < 1+ \frac{b}{2}\), which are part of the conditions (13.2). Then

$$\begin{aligned} \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \bigr ]\psi _1&= \Bigl [ \partial _t -\partial _{rr} - \frac{5}{r} \partial _r \Bigr ]\psi _1 + 4\frac{\lambda ^2}{r(r^2+\lambda ^2)} \partial _r \psi _1 \nonumber \\&\ge c \frac{1}{(t-t_0+A)^a ( \log t)^\beta } \frac{1}{(1+ \frac{r}{\sqrt{t-t_0+A}})^b} \nonumber \\&\quad - 4 \frac{ \lambda ^2}{r(r^2+\lambda ^2)} |\partial _r \psi _1|, \end{aligned}$$
(13.7)

but

$$\begin{aligned} \partial _r \psi _1 = \frac{r}{(t-t_0+A)^a (\log t)^\beta } \Bigl [ -\frac{b}{(1+\frac{r^2}{t-t_0+A})^{b/2+1}} - \frac{C_1}{2} e^{-\frac{r^2}{4(t-t_0+A)}}\Bigr ], \end{aligned}$$

and so

$$\begin{aligned} \frac{\lambda ^2}{r(r^2+\lambda ^2)} |\partial _r \psi _1|&\le C \frac{ \lambda ^2}{r^2+\lambda ^2} \frac{1}{(t-t_0+A)^a (\log t)^\beta } \frac{1}{(1+\frac{r^2}{t-t_0+A})^{b/2+1}} . \end{aligned}$$
(13.8)

We note that for \(r\le \sqrt{t-t_0+A}\) we have

$$\begin{aligned}&\frac{\lambda ^2}{r(r^2+\lambda ^2)} |\partial _r \psi _1| \le \frac{ \lambda ^2}{r^2+\lambda ^2} \frac{1}{(t-t_0+A)^a (\log t)^\beta }\nonumber \\&\le C \frac{ \lambda ^2}{(r^2+\lambda ^2)^2} \frac{1}{(t-t_0+A)^{a-1} (\log t)^\beta } , \end{aligned}$$
(13.9)

where we have used that \(A\ge \lambda (t)^2\).

Let \({\tilde{\psi }}_2(r;\lambda )\) be the bounded solution of

$$\begin{aligned} - \Bigl [\partial _{rr} + \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \Bigr ]{\tilde{\psi }}_2 = \frac{\lambda ^2}{(r^2+\lambda ^2)^2} , \quad r>0, \end{aligned}$$

given by Lemma 13.1. Then \({\tilde{\psi }}_2\) can be written as

$$\begin{aligned} {\tilde{\psi }}_2(r;\lambda ) = {\bar{\psi }}_2\Bigl (\frac{r}{\lambda }\Bigr ) , \end{aligned}$$

for a function \({\bar{\psi }}_2\) satisfying

$$\begin{aligned} |{\bar{\psi }}_2(\rho ) | + (1+\rho ) |{\bar{\psi }}_2'(\rho )|\le \frac{C}{1+\rho ^2}. \end{aligned}$$
(13.10)

Let

$$\begin{aligned} \psi _2(r,t) = \frac{1}{(t-t_0+A)t^{a-1} (\log t)^\beta } {\tilde{\psi }}_2(r;\lambda (t)). \end{aligned}$$

Then, using (13.10) and (13.3), we get

$$\begin{aligned}&\Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \bigr ]\psi _2 \\&\quad = \frac{1}{(t-t_0+A)^{a-1}(\log t)^\beta }\frac{\lambda ^2}{(r^2+\lambda ^2)^2}\\&\qquad \Bigl [ 1 - \Bigl ( \frac{a-1}{t-t_0+A} + \frac{\beta }{t\log t}\Bigr ) {\tilde{\psi }}_2(r)\frac{(r^2+\lambda ^2)^2}{\lambda ^2} - \frac{{\dot{\lambda }}}{\lambda } {\bar{\psi }}_2'\Bigl (\frac{r}{\lambda }\Bigr )\frac{r}{\lambda }\frac{(r^2+\lambda ^2)^2}{\lambda ^2}\Bigr ] \\&\quad \ge \frac{1}{(t-t_0+A)^{a-1}(\log t)^\beta }\frac{\lambda ^2}{(r^2+\lambda ^2)^2} \Bigl [ 1 - C \frac{r^2+\lambda ^2}{t-t_0+A}\Bigr ] . \end{aligned}$$

Therefore there is \(\delta >0\) (fixed independent of \(t_0\)) such that for all \(t_0\) large,

$$\begin{aligned}&\Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \bigr ]\psi _2 \nonumber \\&\quad \ge \frac{1}{2} \frac{1}{(t-t_0+A)^{a-1}(\log t)^\beta }\frac{\lambda ^2}{(r^2+\lambda ^2)^2}, \quad \text {for } r\le 2\delta \sqrt{t}. \end{aligned}$$
(13.11)

Let \(\chi _0 \in C^\infty ({\mathbb {R}})\) be such that \(\chi _0(s) = 1 \) if \(s\le 1\) and \(\chi _0(s)=0\) if \(s\ge 2 \) and define

$$\begin{aligned} \chi _{\delta } (r,t) = \chi _0\Bigl (\frac{r}{\delta \sqrt{t-t_0+A}}\Bigr ) . \end{aligned}$$

We consider

$$\begin{aligned} {\tilde{\psi }}= \psi _1 + M \psi _2 \chi _{\delta } , \end{aligned}$$

where \(M>0\) is a constant to be fixed later. We compute, using (13.7)

$$\begin{aligned} \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \bigr ]{\tilde{\psi }}&\ge c \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+\frac{r}{\sqrt{t-t_0+A}})^b} \nonumber \\&\quad - 4 \frac{ \lambda ^2}{r(r^2+\lambda ^2)} |\partial _r \psi _1| \nonumber \\&\quad + M \chi _{\delta } \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \Bigr ]\nonumber \\&\quad {\tilde{\psi }}_2 + R(r,t), \end{aligned}$$
(13.12)

where

$$\begin{aligned} R&= M \Bigl [ \psi _2 \partial _t \chi _{\delta } -2 \partial _r {\tilde{\psi }}_2 \partial _r \chi _{\delta } - {\tilde{\psi }}_2\Bigl ( \partial _{rr}\chi _{\delta } + \frac{1}{r}\partial _r \chi _{\delta }+4\frac{r}{r^2+\lambda ^2}\partial _r \chi _{\delta } \Bigr )\Bigr ] . \end{aligned}$$

We have, by (13.10),

$$\begin{aligned} |R(r,t)|\le C_2 M \lambda ^2 \frac{1}{(t-t_0+A)^{a+1} (\log t)^\beta } , \end{aligned}$$
(13.13)

where \(C_2\) is independent of M (although it depends on \(\delta \)), and is supported on \(\delta \sqrt{t-t_0+A} \le r \le 2 \delta \sqrt{t-t_0+A}\).

We claim that there is \(M>0\) and \({{\tilde{c}}}>0\) so that for all \(t_0\) sufficiently large

$$\begin{aligned} \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \Bigr ]{\tilde{\psi }}&\ge {{\tilde{c}}} \frac{1}{t^a ( \log t)^\beta (1+r/\sqrt{t})^b} , \end{aligned}$$
(13.14)

for all \(r>0\), \(t>t_0\).

Indeed, if \(r\le \delta \sqrt{t-t_0+A}\), then from (13.12), (13.7), (13.11) and (13.9) we get

$$\begin{aligned} \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \bigr ]{\tilde{\psi }}&\ge c \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+\frac{r}{\sqrt{t-t_0+A}})^b} \nonumber \\&\quad - C \frac{ \lambda ^2}{(r^2+\lambda ^2)} \frac{1}{(t-t_0+A)^{a-1} (\log t)^\beta } \nonumber \\&\quad + M \chi _{\delta } \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \Bigr ]{\tilde{\psi }}_2 \nonumber \\&\ge c \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+\frac{r}{\sqrt{t-t_0+A}})^b} , \end{aligned}$$
(13.15)

if \(M\ge C\). Here we fix \(M=C\).

If \( \delta \sqrt{t-t_0+A} \le r\le 2\delta \sqrt{t-t_0+A}\), then by (13.12), (13.7), (13.9) and (13.13) we get

$$\begin{aligned} \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \bigr ]{\tilde{\psi }}&\ge c \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+\frac{r}{\sqrt{t-t_0+A}})^b} \\&\quad - C_2 M \lambda ^2 \frac{1}{(t-t_0+A)^{a+1} (\log t)^\beta } \\&= \frac{1}{(t-t_0+A)^a ( \log t)^\beta } \Bigl ( \frac{c}{3^ b} - \frac{C_2 M \lambda ^2}{t-t_0+A} \Bigr ) \end{aligned}$$

By taking \(\frac{A}{\lambda (t_0)^2}\) large, we get

$$\begin{aligned} \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \bigr ]{\tilde{\psi }}&\ge \frac{c}{2} \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+ \frac{r}{\sqrt{t-t_0+A}})^b} , \end{aligned}$$
(13.16)

for \( \delta \sqrt{t-t_0+A} \le r\le 2\delta \sqrt{t-t_0+A}\).

If \(r\ge 2\delta \sqrt{t-t_0+A}\), by (13.12) and (13.8)

$$\begin{aligned} \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \bigr ]\psi _1&\ge c \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+\frac{r}{\sqrt{t-t_0+A}})^b} \nonumber \\&\quad -C \frac{ \lambda ^2}{r^2+\lambda ^2} \frac{1}{(t-t_0+A)^a (\log t)^\beta } \frac{1}{(1+\frac{r^2}{t-t_0+A})^{b/2+1}} \nonumber \\&\ge \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+\frac{r}{\sqrt{t-t_0+A}})^b} \Bigl [c - C \frac{\lambda ^2}{t-t_0+A} \Bigr ] \nonumber \\&\ge \frac{c}{2} \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+\frac{r}{\sqrt{t-t_0+A}})^b} \end{aligned}$$
(13.17)

if \(\frac{A}{\lambda (t_0)^2}\) is sufficiently large.

Combining (13.15), (13.16) and (13.17) we deduce the estimate (13.14).

Let

$$\begin{aligned} \psi (x,t) = {\tilde{\psi }}(|x-\xi |,t). \end{aligned}$$

Then, by (13.14),

$$\begin{aligned} (\partial _t - L^ o)[\psi ]&= \Bigl [ \partial _t - \partial _{rr} - \Bigl (\frac{1}{r}+ \frac{4r}{\lambda ^2 + r^2} \Bigr ) \partial _{r} \Bigr ]{\tilde{\psi }} -\partial _r {\tilde{\psi }} \frac{(x-\xi )\cdot {\dot{\xi }}}{|x-\xi |} \\&\ge {{\tilde{c}}} \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+\frac{r}{\sqrt{t-t_0+A}})^b} - |{\dot{\xi }}| \, |\partial _r {\tilde{\psi }}|, \end{aligned}$$

but

$$\begin{aligned} |\partial _r \psi |&\le C\frac{1}{(t-t_0+A)^{a-1/2}(\log t)^\beta } \frac{1}{(1+ \frac{r}{\sqrt{t-t_0+A}})^{b+1}}\\&\quad + C\frac{1}{(t-t_0+A)^{a-1}(\log t)^\beta } \frac{1}{\lambda } \frac{1}{(1+r/\lambda )^3} \chi _{\delta }(r,t) \\&\quad + C\frac{1}{\delta (t-t_0+A)^{a-1/2}(\log t)^\beta } \frac{1}{(1+r/\lambda )^2} \chi _0'\Bigl ( \frac{r}{\delta \sqrt{t-t_0+A}} \Bigr ) . \end{aligned}$$

Using (13.4) we see that if \(t_0\) is sufficiently large,

$$\begin{aligned} (\partial _t - L^ o)[\psi ]&\ge \frac{{{\tilde{c}}}}{2} \frac{1}{(t-t_0+A)^a ( \log t)^\beta (1+\frac{r}{\sqrt{t-t_0+A}})^b} . \end{aligned}$$

\(\square \)

A direct consequence of the proof of Proposition 13.1 (using the same barriers) is the following, for the initial value problem

$$\begin{aligned} \left\{ \begin{aligned} \partial _t \phi ^o = L^o [\phi ^o] ,&\quad \text {in }{\mathbb {R}}^2 \times (t_0,\infty ) \\ \phi ^o(\cdot ,t_0)= \phi ^o_0,&\quad \text {in }{\mathbb {R}}^2 . \end{aligned} \right. \end{aligned}$$
(13.18)

Consider the norm

$$\begin{aligned} \Vert \phi ^o_0 \Vert _{*,b} =&\inf K \quad \text {such that } \\&|\phi ^o_0(x)| \le \frac{K}{( 1+ \frac{|x-\xi (0)|}{\sqrt{t-t_0+A}} )^b} \end{aligned}$$

where \(b \in (2,6)\), \(A>0\).

Proposition 13.2

Assume that ab satisfy (13.2), \(\frac{A}{\lambda (t_0)^2}\) is sufficiently large, and \(\lambda , \xi \) satisfy (13.3), (13.4). Then there is a constant C so that for \(t_0\) sufficiently large and for \(\Vert \phi ^o_0 \Vert _{*,b}<\infty \) there exists a solution \(\phi ^o\) of (13.18), which defines a linear operator of \( \phi ^o_0\) and satisfies

$$\begin{aligned} \Vert \phi ^o\Vert _{*,o} \le C A^{a-1} (\log t_0)^\beta \Vert \phi ^o_0 \Vert _{*,b} . \end{aligned}$$