1 Introduction and statement of the main result

We investigate the asymptotic structure of global in time solutions u(xt) of the energy-critical semilinear heat equation

$$\begin{aligned} \left\{ \begin{aligned}&u_t=\Delta u +u^5{} & {} \text {in} \quad \Omega \times \mathbb {R}^+,\\&u=0 \quad{} & {} \text {on} \quad \partial \Omega \times \mathbb {R}^+,\\&u(x,0)=u_0(x){} & {} \text {in} \quad \Omega , \end{aligned} \right. \end{aligned}$$
(1.1)

where \(\Omega \subset \mathbb {R}^3\) is a smooth bounded domain and \(u_0\) is a smooth initial datum. The energy associated to the solution u(xt) is

$$\begin{aligned} E(u){:=}\frac{1}{2}\int _{\Omega } \vert \nabla u\vert ^2 \, dx -\frac{1}{6} \int \vert u\vert ^{6} \, dx. \end{aligned}$$

Since classical solutions of (1.1) satisfy

$$\begin{aligned} \frac{d}{dt}E(u(\cdot ,t))=-\int _{\Omega } \vert u_t\vert ^2 \,dx\le 0, \end{aligned}$$

the energy is a Lyapunov functional for (1.1). The stationary equation on the whole space is the Yamabe problem

$$\begin{aligned} \Delta U+U^5=0 {\quad \hbox {in } }{{\mathbb {R}}}^3. \end{aligned}$$

All positive solutions to this equation are given by the Aubin-Talenti bubbles (see [4])

$$\begin{aligned} U_{\mu ,\xi }(x)=\mu ^{-\frac{1}{2}}U\left( \frac{x-\xi }{\mu }\right) , \end{aligned}$$
(1.2)

where \(\mu >0,\xi \in {{\mathbb {R}}}^3\) and

$$\begin{aligned} U(x)=\alpha _3 \frac{1}{\left( 1+\vert x\vert ^2\right) ^{\frac{1}{2}}},\quad \text {where}\quad \alpha _3{:=}3^{\frac{1}{4}}. \end{aligned}$$

Consider the Sobolev embedding \(H_0^1(\Omega )\hookrightarrow L^{p+1}(\Omega )\), which is compact for \(p\in (1,p_S)\), where \(p_S=\frac{n+2}{n-2}\), and the associated constant

$$\begin{aligned} S_p(\Omega ){:=}\inf _{0\ne u \in H_0^1(\Omega )}\frac{\vert \!\vert u\vert \!\vert _{H_0^1(\Omega )}^2}{\vert \!\vert u\vert \!\vert _{L^{p+1}(\Omega )}^2}. \end{aligned}$$

The Aubin-Talenti bubbles achieve the constant \(S_{p_S}({{\mathbb {R}}}^n)\). Thus, the energy \(E(U_{\mu ,\xi })=S_{p_S}({{\mathbb {R}}}^n)\) is invariant with respect to \(\mu ,\xi \). When \(\mu \rightarrow 0\) the Aubin-Talenti bubble becomes singular. This is the reason for the loss of compactness in the Sobolev embedding for \(p=p_S\). Indeed, Struwe proved in [52] that every Palais-Smale sequence associated to the energy functional E looks like

$$\begin{aligned} u_n(x)=u_\infty (x)+\sum _{i=1}^{k}U_{\mu _n^i,\xi _n^i}(x)+o(1) \quad \text {when}\quad n\rightarrow \infty , \end{aligned}$$
(1.3)

up to subsequences, for some \(k\in \mathbb {N}\), where \(u_\infty \in H_0^1(\Omega )\) is a critical point of E and \(\mu _n^i\rightarrow 0\), \(\xi _n^i\in \Omega \). Thus, we say that the compactness is lost by ’bubbling’. When the domain is star-shaped, the Pohozaev identity constrains \(u_\infty \) to vanish.

For classical finite-energy solutions u(xt) the problem (1.1) is well-posed in short time intervals. We refer to the monograph [48] by Quittner and Souplet for an extended review on this problem and more general semilinear parabolic equations. The aim of this paper is exhibiting classical positive finite-energy solutions u(xt) of (1.1) which are globally defined in time and satisfy

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\vert \!\vert u(\cdot ,t)\vert \!\vert _{L^\infty (\Omega )}=\infty . \end{aligned}$$
(1.4)

Given any smooth function \(\varphi (x)\ge 0,\varphi \ne 0\), consider \(\alpha >0\) and \(u_\alpha (x,0){:=}\alpha \varphi (x)\) as initial datum. On one hand, if \(\alpha \) is sufficiently small, then \(u_\alpha (x,t)\) tends uniformly to zero as \(t\rightarrow \infty \). On the other hand, using the eigenfunction method of Kaplan [39], for \(\alpha \) sufficiently large \(u_\alpha (x,t)\) blows-up in finite time. Thus, the threshold number

$$\begin{aligned} \alpha ^*{:=}\sup \left\{ \alpha >0\,:\, \lim \limits _{t\rightarrow \infty }\vert \!\vert u_\alpha (\cdot ,t)\vert \!\vert _\infty =0 \right\} , \end{aligned}$$

is positive. In 1984, the first rigorous proof of the existence in \(L^1\)-weak sense of \(u_{\alpha ^*}(x,t)\) was found by Ni, Sacks and Tavantzis [47]. Du [25] and Suzuki [54] proved, that, for any unbounded sequence of times \(t_n\), \(u_{\alpha ^*}(x,t_n)\) can be decomposed as in (1.3). Thus, when constructing unbounded global solutions for the critical case, it is natural to look for an asymptotic profile as (1.2). Galaktionov and Vázquez [30] proved that, in the radial case \(\Omega =B_1(0)\) with \(\varphi \) radial non-increasing, \(u_{\alpha ^*}(x,t)\) is smooth, global and \(u=u_{\alpha ^*}\) satisfies (1.4). Thus, we naturally wonder what is the asymptotic behavior of global unbounded solutions. Most of the results about the dynamics of threshold solutions in literature concern the radial case. This particular setting allows the construction of specific solutions by means of matched expansions. In [29] Galaktionov and King proved that the threshold behavior of \(u_{\alpha ^*}\) in the radial case is

$$\begin{aligned} \ln \vert \!\vert u_{\alpha ^*}(\cdot ,t)\vert \!\vert _\infty =\left\{ \begin{aligned} \frac{\pi ^2}{4}t(1+o(1)){} & {} \text {if} \quad n=3, \\2\sqrt{t}(1+o(1)){} & {} \text {if} \quad n=4, \end{aligned} \right. \end{aligned}$$
(1.5)

and

$$\begin{aligned} \vert \!\vert u_{\alpha ^*}(\cdot ,t)\vert \!\vert _\infty = \left( \gamma _n t\right) ^{\frac{n-2}{2(n-4)}},\quad \text {if}\quad n\ge 5, \end{aligned}$$

for some explicit constants \(\gamma _n\). Our main theorem is a non-radial extension in dimension 3. The existence of positive non-radial unbounded solutions for the Dirichlet problem in dimension \(n=4\) remains an open problem, which we will consider in a future work. The case of higher dimension \(n\ge 5\) has been already extended to the non-radial case by Cortázar et al. [11]. They found positive multi-spike global solutions which blow-up by bubbling in infinite time. Here, the term multi-spike refers to the fact that the constructed solution is unbounded in a finite number of points in \(\Omega \). Sign-changing solutions which blow-up in infinite time have been discovered by del Pino et al. [19] for \(n\ge 5\), proving stability in case \(n=5,6\).

Our solutions involve the Green function \(G_\gamma \) associated to the elliptic operator

$$\begin{aligned} L_\gamma =-\Delta -\gamma \quad \text {in}\quad \Omega , \end{aligned}$$

where \(\gamma \in [0,\lambda _1)\) and \(\lambda _1\) is the principal Dirichlet eigenvalue. Namely, for all \(y\in \Omega \), \(G_\gamma \) satisfies

$$\begin{aligned}&-\Delta _x G_\gamma (x,y)-\gamma G_\gamma (x,y) =c_3 \delta (x-y) \quad \text {in} \quad \Omega , G_\gamma (x,y)=0 \quad \text {on} \quad \partial \Omega , \end{aligned}$$

where \(\delta (x)\) is the Dirac delta, \(c_3 :=\alpha _3\omega _3\) and the constant \(\omega _3=4\pi \) indicates the area of the unit sphere. The Green function can be decomposed as

$$\begin{aligned} G_\gamma (x,y)=\Gamma (x-y)-H_\gamma (x,y), \end{aligned}$$

where \(\Gamma (x)=\alpha _3\vert x\vert ^{-1}\) and the regular part \(H_\gamma (x,y)\) is defined as the solution, for all \(y \in \Omega \), to

$$\begin{aligned}&\Delta _x H_\gamma (x,y)+\gamma H_\gamma (x,y)=\gamma \frac{\alpha _3}{\vert x-y\vert } \quad \text {in} \quad \Omega ,\nonumber \\&H_\gamma (x,y)=\Gamma (x -y)\quad \text {in} \quad \partial \Omega . \end{aligned}$$

The diagonal \(R_\gamma (x){:=}H_\gamma (x,x)\) is called Robin function associated to \(-\Delta -\gamma \) in \(\Omega \). It turns out (see Lemma 2.1) that for any fixed \(q\in \Omega \) there exists a unique number \(\gamma (q)\in (0,\lambda _1)\) defined by

$$\begin{aligned} \gamma (q){:=}\sup \{ \gamma>0: R_\gamma (q)>0 \}. \end{aligned}$$

Our main theorem shows that, for any \(q\in \Omega \) such that \(3\gamma (q)<\lambda _1\), there exists a global solution to the problem (1.1) which blows-up in infinite time with spike in \(x=q\).

Theorem 1.1

Let \(\Omega \subset {{\mathbb {R}}}^3\) be a bounded smooth domain. Let q be a point in \(\Omega \) such that

$$\begin{aligned} \gamma (q)<\frac{\lambda _1}{3}. \end{aligned}$$
(1.6)

Then, there exist an initial condition \(u_0(x)\in C^1({\bar{\Omega }})\), smooth functions \(\xi (t),\mu (t)\) and \(\theta (x,t)\) such that the solution u(xt) to the problem (1.1) is a positive unbounded global solution with the asymptotic profile

$$\begin{aligned} u(x,t)=\mu (t)^{-\frac{1}{2}} U\left( \frac{x-\xi (t)}{\mu (t)}\right) - \mu (t)^{\frac{1}{2}}\left( H_\gamma (x,\xi (t))+\theta (x,t)\right) \quad \text{ as }\quad t \rightarrow \infty , \end{aligned}$$
(1.7)

where \(\theta \) is a bounded function, and decays uniformly away from the point q. Moreover, the parameters \(\mu (t),\xi (t)\) are smooth functions of time and satisfy

$$\begin{aligned} \ln \left( \frac{1}{\mu (t)}\right) = 2\gamma (q) t(1+o(1)),\quad \xi (t)-q=O(\mu (t)) \quad \text{ as }\quad t\rightarrow \infty . \end{aligned}$$
(1.8)

Furthermore, thanks to the inner-outer gluing scheme, which is based only on elliptic and parabolic estimates, as in [11, 15] we get a codimension-1 stability of the solution stated by Theorem 1.1. In fact, since condition (1.6) is stable under small perturbation of \(q\in \Omega \), the stability result follows exactly as in [11, Proof of Corollary 1.1] (see Remark 7.1 in Sect. 7).

Corollary 1.1

Let u be the solution stated in Theorem 1.1 which blows up at q. Then, there exists a codimension-1 manifold \(\mathcal {M}\) in \(C^{1}({\bar{\Omega }})\) with \(u_0 \in \mathcal {M}\) and such that if \(\tilde{u}_0\in \mathcal {M}\) and it is sufficiently close to \(u_0\), then the solution \({\tilde{u}}\) to (1.1) with initial datum \({\tilde{u}}_0\) is global and blows-up in infinite time with spike in \({\tilde{q}}\) near q and profile (1.7) with \(\ln \vert \!\vert \tilde{u}(\cdot ,t)\vert \!\vert _\infty = \gamma ({\tilde{q}}) t(1+o(1))\) as \(t\rightarrow \infty \).

Condition (1.6) implies that the point q cannot be very close to boundary, since \(\gamma (q)\rightarrow \lambda _1^{-}\) as \(q\rightarrow {\partial } \Omega \) (see Lemma A.2 in Appendix A). Along the proof we need to consider Dirichlet problems of the type

$$\begin{aligned}&u_t =\Delta u+\gamma u+e^{-2\gamma t}f(x) {\quad \hbox {in } }\Omega \times {{\mathbb {R}}}^+,\\&u(x,t)=0 {\quad \hbox {on } } {\partial } \Omega \times {{\mathbb {R}}}^+,\\&u(x,0)=0 {\quad \hbox {in } }\Omega , \end{aligned}$$

for some \(f(x)\in L^p\) with \(p>2\). In order to successfully apply fixed point arguments, we need

$$\begin{aligned} \vert \!\vert u(\cdot ,t)\vert \!\vert _{\infty }\le C e^{-2 \gamma t} \end{aligned}$$

for \(t>1\), which requires condition (1.6). Such assumption (1.6) is useful to get rid of a resonance effect, lastly due to the fact that both the Dirichlet heat kernel \(p_t^{\Omega }(x,y)\) and the parameter \(\mu (t)\) decay exponentially fast. Indeed, the long-term behavior of the Dirichlet heat kernel is

$$\begin{aligned} p_t^{\Omega }(x,y)\sim \phi _1(x)\phi _1(y)e^{-\lambda _1 t} \quad \text{ as }\quad t \rightarrow \infty , \end{aligned}$$

where \(\phi _1\) is the positive eigenfunction of \(-\Delta \) in \(\Omega \) with \(\vert \!\vert \phi _1\vert \!\vert _2=1\). We recall the properties of the Dirichlet heat kernel in Sect. 8. More specifically, we use assumption (1.6) in the following steps of the proof:

  • to get estimates for \(J_1,J_2\) in Lemma 2.2 and Lemma 2.3 respectively;

  • in Lemma 4.1 for solving the outer problem;

  • in Proposition 6.1 for the invertibility theory of the nonlocal operator \(\mathcal {J}\).

The number \(\gamma (q)\) is related to the Brezis-Nirenberg problem. Define

$$\begin{aligned} \mathcal {S}_a(\Omega ){:=}\inf _{u\in H_0^1(\Omega )\setminus \{0\}} \frac{\int _{\Omega }\vert \nabla u\vert ^2\,dx-a \int _{\Omega }\vert u\vert ^2\,dx}{\left( \int _{\Omega }\vert u\vert ^6\,dx\right) ^{\frac{1}{3}}}. \end{aligned}$$

In the celebrated work [2], Brezis and Nirenberg proved the existence of a constant \(\mu _{\text {BN}}\in (0,\lambda _1)\) such that

$$\begin{aligned} {\mu _{\text {BN}}}{:=} \inf \{a>0: \mathcal {S}_a(\Omega )<\mathcal {S}_0\}. \end{aligned}$$

Then, Druet [24] proved

$$\begin{aligned} \min _{q \in \Omega }\gamma (q)= \mu _{\text {BN}}(\Omega ). \end{aligned}$$

Thus, when \(3\mu _{BN}(\Omega )<\lambda _1(\Omega )\) is true, condition (1.6) is satisfied in some open set \(\mathcal {O}\subset \Omega \), and Theorem 1.1 gives the desired solution with blow-up at any fixed point \(q\in \mathcal {O}\).

When we consider the radial case \(\Omega = B_1(0)\) and \(q=0\), an explicit computation gives \(\gamma (0)=\pi ^2/4\), that is consistent with (1.5). In fact, this is the minimum value for \(\gamma (q)\) since Brezis and Nirenberg computed \(\mu _{BN}(B_1)=\pi ^2/4\). By symmetry, we deduce that condition (1.6) is satisfied in the ball \(B_{d^*}\), where \(d^*=\vert q^*\vert \) and \(q^*\) is a point such that \(\gamma (q^*)=\lambda _1/3\).

Also, we can consider smooth perturbation of the ball. Let \(f: \bar{B}_1 \rightarrow {{\mathbb {R}}}^3\) a smooth map and for \(t>0\) define

$$\begin{aligned} \Omega _t:=\{x+t f(x)\,:\, x \in B_1\}. \end{aligned}$$

For small t the domain \(\Omega _t\) is diffeomorphic to the ball. Writing \(\lambda _1\) as Rayleigh quotient and using the definition \(\mu _{BN}\) we can easily see that \(\mu (\Omega _t)=\mu (B_1)+\varepsilon (t)\) and \(\lambda _1(\Omega _t)=\lambda _1(B_1)+{\tilde{\varepsilon }}(t)\) where \(\varepsilon (t),{{\tilde{\varepsilon }}}(t)\rightarrow 0\) as \(t\rightarrow 0\). Thus, for t sufficiently small, the relation \(3\mu _{\text {BN}}(\Omega _t)<\lambda _1(\Omega _t)\) holds, and Theorem 1.1 applies to the domain \(\Omega _t\). This shows that Galaktionov-King’s radial result is stable under small perturbation of the domain.

For the unit cube \( {{\mathcal {C}}}_1\) it is known (see [58, Remark 4.3]) that \(3\mu _{\text {BN}}( {{\mathcal {C}}}_1)< \lambda _1(\mathcal {C}_1)\). Indeed, from \(B_{1/2}\subset \mathcal {C}_1\) and the strict monotonicity of \(\mu _{BN}(\Omega )\) with respect to \(\Omega \) we deduce \(\mu _{\text {BN}}\left( \mathcal {C}_1\right) < \mu _{\text {BN}}\left( B_{1/2}\right) =\pi ^2\). By separation of variables we easily compute \(\lambda _1(\mathcal {C}_1)=3\pi ^2\), thus

$$\begin{aligned} 3 \mu _{\text {BN}}\left( \mathcal {C}_1\right) < 3 \mu _{\text {BN}}\left( B_{1/2}\right) =3 \pi ^2 =\lambda _1( {{\mathcal {C}}}_1). \end{aligned}$$

Hence, a slight modification of Theorem 1.1 applies: since \( {{\mathcal {C}}}_1\) is a Lipschitz domain, by the parabolic regularity theory we get a smooth solution u(xt) in \(\Omega \times {{\mathbb {R}}}^+\) which is Lipschitz continuous in \({\bar{\Omega }} \times [t_0,\infty )\).

Let \(\Omega ^*\) be the ball with the same volume as \(\Omega \). The following estimate holds true:

$$\begin{aligned} \frac{\lambda _1(\Omega ^*)}{4}\le {\mu _{\text {BN}}}(\Omega ) \le \frac{\lambda _1(\Omega ^*)}{4}\min _{x\in \Omega }R_0(x)^2. \end{aligned}$$

The lower bound was proved in [2] by means of a symmetrization argument. Using harmonic transplantation Bandle and Flucher [1] proved the upper bound. Thus, if it happens that we know \(\min _{x\in \Omega } R_0(x)^2<4/3\) we can apply Theorem 1.1 to \(\Omega \). Wang [58] conjectured that \( {\mu _{\text {BN}}}/\lambda _1\in [1/4,4/9)\). In particular, condition \(3\mu _{BN}(\Omega )<\lambda _1(\Omega )\) could be false for "very thin rectangles" (see [58]). The range [1/4, 4/9) is supported by numerical computations made by Budd and Humphries [3].

The main differences with respect to the analogue result [11] in dimension \(n\ge 5\) are the following:

  • the main asymptotic behavior in Theorem 1.1 of the blow-up is dependent on the position of the point \(q\in \Omega .\) As far as we know, this is a completely new phenomenon;

  • since condition (1.6) is not satisfied close to the boundary, we cannot straightforward construct multi-spike solutions in the spirit of [11]. Indeed, roughly speaking, such construction requires spikes relatively far from each other and close to the boundary to suitably bound the interaction between the bubbles.

  • a nonlocal operator controls the dynamic of the parameter \(\mu (t)\). A similar operator has been treated in [15], where the domain \(\Omega ={{\mathbb {R}}}^3\) allows an explicit inversion of the Laplace transform.

The approach developed in this work is inspired by [11, 13, 15]. It is constructive and allows an accurate analysis of the asymptotic dynamics and stability. Let describe the general strategy. The first step consists in choosing a good approximated solution \(u_3\). Here the word ’good’ means that the associated error function

$$\begin{aligned} S[u](x,t){:=}- {\partial } _t u+\Delta u+u^{5} \end{aligned}$$

is sufficiently small in \(\Omega \). Part of the problem consists in understanding what smallness on S[u] is sufficient to find a perturbation \({{\tilde{\phi }}}\) such that

$$\begin{aligned} u=u_3+{{\tilde{\phi }}} \end{aligned}$$

is an exact solution to (1.1). In Sect. 2 we start with the scaled Aubin-Talenti bubble as building block and we modify it to match the boundary at the first order. Then we realize that we need two improvements. The first one is a global correction useful to get solvability conditions for the elliptic linearized operator around the standard bubble

$$\begin{aligned} L[\phi ]{:=}\Delta \phi +5U^4(y)\phi . \end{aligned}$$

Such improvement produces a nonlocal operator which governs the second order term in the expansion of the scaling parameter \(\mu (t)\). This is a low-dimensional effect, lastly due to the fact that

$$\begin{aligned} Z_{n+1}(r){:=}\frac{n-2}{2}U(r)+U'(r)r \notin L^2(\mathbb {R}^n)\quad \text {when} \quad n\in \{3,4\}, \end{aligned}$$

where \(Z_{n+1}\) is the unique (up to multiples) bounded radial function belonging to the kernel of \(L[\phi ]\). Actually, the dimensional restriction in [11] was specially designed to avoid this effect and the presence of the corresponding nonlocal term. Then, by choosing \(\gamma (q)\) as in (1.6) we reduce the error close to \(x=q\); this gives the asymptotic behavior (1.8) of \(\mu (t)\) at the first order. A second correction, local in nature, removes non-radial slow-decay terms and gives the asymptotic for \(\xi \) written in (1.8). At this point we have a sufficiently good ansatz, called \(u_3\), to start the so called inner-outer gluing procedure in Sect. 3: we decompose the problem in a system of nonlinear problems, namely an inner and an outer problem which are weakly coupled thanks to the smallness of \(S[u_3]\). We solve the outer problem in §4, that is a perturbation of the standard heat equation, for suitable parameters \(\mu ,\xi \) and decaying solution \(\phi \) of the inner problem. Then, we look at the inner regime. We can find the inner solution, by fixed point argument, using the adaptation to \(n=3\) of the linear theory for the inner problem developed in [11]. This requires the solvability of orthogonality conditions which, in Sect. 5, we prove to be equivalent to a nonlocal system in the parameters \(\mu ,\xi \). We solve it in Sect. 6 using the invertibility of a nonlocal equation, which we achieve in Sect. 8 by means of a Laplace transform argument combined with asymptotic properties of the heat kernel \(p_t^\Omega (x,y)\). At this point we are ready to find the inner solution \(\phi \) in Sect. 7, which concludes the proof of Theorem 1.1.

Of course, the full problem consists in finding the exact initial datum that evolves in an infinite time blow-up solution. We find the positive initial condition

$$\begin{aligned} u(x,t_0)&= \mu (t_0)^{-1/2}U\left( \frac{x-\xi (t_0)}{\mu (t_0)}\right) -\mu (t_0)^{1/2}H_\gamma (x,t_0)+\mu _0(t_0)^{1/2}J_1(x,t_0)\\\nonumber&\quad +\mu (t_0)^{-1/2}\phi _3\left( \frac{x-\xi (t_0)}{\mu (t_0)},t_0\right) \eta _{l(t_0)} \left( \vert \frac{x-\xi (t_0)}{\mu (t_0)}\vert \right) \\&\quad +\mu _0(t_0)^{1/2}\psi (x,t_0)+\eta _{R(t_0)}\left( \vert \frac{x-\xi (t_0)}{\mu (t_0)}\vert \right) \mu (t_0)^{-1/2}e_0Z_0 \left( \frac{x-\xi (t_0)}{\mu (t_0)}\right) , \end{aligned}$$

for \(t_0\) fixed sufficiently large, where the existence of \(\mu ,\xi ,\phi ,\psi \) and the constant \(e_0\) is a consequence of fixed point arguments, \(\eta _l,l,\eta _R,R\) are defined in (2.5), (2.17) and the functions \(\phi _3,J_1\) solve the problems (2.21) and (2.14). We remark that we do not know if the solution with this initial datum corresponds to a threshold solution in the sense of [47].

We conclude this introduction giving a short bibliographic overview about related problems and recent developments. The rigorous construction of blow-up solutions by bubbling, that is a solution \(u(x,t)\approx U_{\mu (t),\xi (t)}(x)\) with \(\mu \rightarrow 0\) for some special profile U, has been extensively studied in many important problems with criticality. For instance, in the harmonic map flow [13, 49, 50], in the Patlak-Keller-Segel model for chemotaxis [8, 9, 12, 31], in the energy-critical wave equation [26, 36, 40, 41] and energy critical Schrödinger map problem [45].

Concerning the Cauchy problem

$$\begin{aligned}&u_t =\Delta u +u^{\frac{n+2}{n-2}} {\quad \hbox {in } }{{\mathbb {R}}}^n \times {{\mathbb {R}}}^+,\\&u(x,0)=u_0(x) {\quad \hbox {in } }{{\mathbb {R}}}^n, \end{aligned}$$

infinite blow-up positive solutions have been found in dimension \(n=3\) in del Pino et al. [15], with different blow-up rates depending on the space decay of the initial datum. Recently, Wei et al. [59] detected analogue solutions in dimension \(n=4\). These works were inspired by conjectures presented in [27], where Fila and King used matched asymptotic methods to formally analyze the behavior of infinite blow-up solutions in the radial case, also conjecturing that for \(n\ge 5\) such solutions do not exist. However, adding drift terms to the equation, Wang et al. [56] have shown examples of positive initial datum which evolves in multi-spike infinite blow-up by bubbling. For \(n\ge 7\), del Pino et al. [16] proved the existence of sign-changing solutions which blow-up in infinite time in the form of tower of bubbles, that is a supersolution of Aubin-Talenti bubbles at a single point. For the analogue backward problem where \(t\in (-\infty ,0)\), ancient solutions which blow-up in infinite time have been detected by Sun et al. [53] for \(n\ge 7\).

As we have already mentioned, blow-up for the nonlinear heat equation

$$\begin{aligned}&u_t = \Delta u + \vert u\vert ^{p-1}u {\quad \hbox {in } }\Omega \times (0,T), \end{aligned}$$

can also happen in finite time \(T<\infty \). We call it Type I blow-up if the solution satisfies

$$\begin{aligned} \limsup _{t\rightarrow T} (T-t)^{\frac{1}{p-1}}\vert \!\vert u(\cdot ,t)\vert \!\vert _{L^\infty (\Omega )}<\infty , \end{aligned}$$

otherwise, if

$$\begin{aligned} \limsup _{t\rightarrow T} (T-t)^{\frac{1}{p-1}}\vert \!\vert u(\cdot ,t)\vert \!\vert _{L^\infty (\Omega )}=\infty , \end{aligned}$$

we have Type II blow-up. Several works have focused on constructing finite time blow-up solutions for the Cauchy problem. Positive Type II blow-up solutions do not exist in dimension \(n\ge 7\), see Wang and Wei [57], or under radial assumptions in any dimension \(n\ge 3\), see Matano-Merle [44] and the pioneering work by Filippas-Herrero-Velázquez [28]. In dimension \(n \ge 7\), Collot et al. [10] classified the dynamics near the Aubin-Talenti bubble U in the \({\dot{H}}^1\) topology. In particular, they ruled out the Type II scenario for initial conditions \(u_0\) such that \(\vert \!\vert u_0-U\vert \!\vert _{\dot{H}^1({{\mathbb {R}}}^n)}\) is sufficiently small. The existence of positive Type II blow-up in dimensions \(n\in \{3,4,5,6\}\) is an open problem.

Type II blow-up it is still admissible for sign-changing solutions, and in fact examples have been found. Type II blow-up solutions have been constructed by Schweyer [51] in dimension 4 under radial assumption and later by del Pino et al. in the non-radial setting [20] with admissible multi-spike behavior. Also, Type II blow-up solutions have been detected in dimension \(n\in \{3,5,6\}\) in [18, 21, 33, 34] with different blow-up rates. Type II blow-up for the critical heat equation can also happen on curves contained in the boundary of special domains with axial symmetry, see [17].

There have been developments also in the nonlocal generalization of these problems. Concerning the fractional heat equation with critical exponent

$$\begin{aligned} u_t = - \left( -\Delta \right) ^s u + \vert u\vert ^{\frac{4s}{n-2s}}u, \end{aligned}$$

Cai et al. [5] have recently constructed solutions for both the forward and backward Cauchy problem which are sing-changing tower of bubbles at the origin for \(n>6s\), and \(s\in (0,1)\). For \(n \in (4\,s,6\,s)\) and \(s\in (0,1)\) blow-up in finite time has been proved in [7], which is a fractional continuation of the local Type II blow-up cases \(n=4,s=1\) in [51] and \(n=5,s=1\) in [21]. Regarding the associated Dirichlet problem, Musso et al. provided in [46] the existence of positive multi-spike infinite-time blow-up on bounded smooth domains for \(n\in (4s,6s)\) and \(s\in (0,1)\).

2 Approximate solution and estimate of the associated error

In this section we construct an approximate solution to the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u_t = \Delta u + u^5 &{} {\quad \hbox {in } }\Omega \times {{\mathbb {R}}}^+,\\ u=0&{} {\quad \hbox {on } }\partial \Omega \times {{\mathbb {R}}}^+, \end{array}\right. } \end{aligned}$$
(2.1)

and we compute the associated error. Without loss of generality, we construct a solution that blows-up at \(q=0 \in \Omega \). The first approximation \(u_1\) is chosen by selecting a time-scaled version of the stationary solution to the Yamabe problem

$$\begin{aligned} \Delta U + U^5=0 {\quad \hbox {in } }{{\mathbb {R}}}^3, \end{aligned}$$

properly adjusted to be small at the boundary \( {\partial } \Omega \). This is constructed in Sect. 2.1. In order to make the error small at the blow-up point, we need to select a precise first order for the dilation parameter \(\mu (t)\), which matches the radial asymptotic found in [29]. However, we observe in Sect. 2.2 that \(u_1\) is not close enough to an exact solution to make our perturbative scheme rigorous. In Sect. 2.3 we make a global improvement \(u_2\). Such correction involves a nonlocal operator in the lower order term of \(\mu (t)\), similar to a half-fractional Caputo derivative. The last improvement \(u_3\) is only local, and it removes slow-decaying terms in non-radial modes by selecting the first order asymptotic of the translation parameter \(\xi (t)\).

2.1 First global approximation

Our building blocks are the scaled Aubin-Talenti bubbles (1.2) which satisfy

$$\begin{aligned} \Delta U_{\mu ,\xi }+ U_{\mu ,\xi }^5=0 {\quad \hbox {in } }{{\mathbb {R}}}^3. \end{aligned}$$
(2.2)

We look for a solution of the form \(u_1(x,t) \approx U_{\mu (t),\xi (t)}(x)\). We make an ansatz for the parameters \(\mu (t),\xi (t)\). Assuming that \(\mu (t)\rightarrow 0\) and \(\xi \rightarrow 0\in \Omega \) as \(t\rightarrow \infty \), we notice that \(U_{\mu ,\xi }(x)\) is concentrating around \(x=0\) and it is uniformly small away from it. For this reason, we should have

$$\begin{aligned} {\partial } _t u_1- \Delta u_1&=u_1(x,t)^5 \nonumber \\&\approx \delta _0(x-\xi ) \int _{{{\mathbb {R}}}^3} \left( \mu ^{-1/2}U\left( \frac{x-\xi }{\mu }\right) \right) ^5 \,dx\nonumber \\&= \delta _0(x-\xi )\mu ^{1/2} \int _{{{\mathbb {R}}}^3} U(y)^5 \,dy\nonumber \\&= \delta _0(x-\xi ) c_3 \mu ^{1/2}. \end{aligned}$$
(2.3)

Let \(\mu _0(t)\) the first order of \(\mu (t)\), that is

$$\begin{aligned} \mu (t)=\mu _0(t)(1+o(1)) \quad \text{ as }\quad t\rightarrow \infty . \end{aligned}$$

From (2.3) we define the scaled function

$$\begin{aligned}v(x,t){:=}\mu ^{-1/2}u_1(x,t),\end{aligned}$$

which should satisfy

$$\begin{aligned}&v_t \approx \Delta v + \left( -\frac{{\dot{\mu }}}{2\mu } \right) v + c_3 \delta _0(x-\xi ) {\quad \hbox {in } }\Omega \times {{\mathbb {R}}}^+,\nonumber \\&v= 0 {\quad \hbox {on } }\partial \Omega \times {{\mathbb {R}}}^+. \end{aligned}$$
(2.4)

We choose the parameter \(\mu _0(t)\) such that

$$\begin{aligned} -\frac{{\dot{\mu }}_0(t)}{2\mu _0(t)}=\gamma , \end{aligned}$$

for some \(\gamma \in {{\mathbb {R}}}^+\) that will be fixed later. This is equivalent to choose

$$\begin{aligned} \mu _0(t)= b e^{-2\gamma t}, \end{aligned}$$
(2.5)

for some \(b\in {{\mathbb {R}}}^+\). We can fix \(b=1\). Indeed, the equation is translation-invariant in time: we construct, for a sufficiently large initial time \(t_0\), a solution u(xt) in \(\Omega \times [t_0,\infty )\) and we conclude that \(u_0(x,t){:=} u(x,t-t_0)\) is a solution to (2.1) in \(\Omega \times [0,\infty )\). We observe that after shifting the initial time, the main dilation parameter \(\mu _0\) becomes \(\mu _0(t-t_0)=e^{2\gamma t_0}e^{-2\gamma t}\).

With this choice (2.4) reads

$$\begin{aligned}&v_t \approx \Delta v +\gamma v+c_3 \delta _0(x-\xi ) {\quad \hbox {in } }\Omega \times {{\mathbb {R}}}^+,\\&v=0 {\quad \hbox {on } }\partial \Omega \times {{\mathbb {R}}}^+. \end{aligned}$$

Hence, for large time we should have

$$\begin{aligned} v(x,t)\approx G_{\gamma }(x,\xi ), \end{aligned}$$
(2.6)

where \(G_\gamma (x,y)\) is the Green function for the boundary value problem

$$\begin{aligned}&-\Delta _x G_\gamma (x,y)-\gamma G_{\gamma }(x,y)=c_3 \delta (x-y) {\quad \hbox {in } }\Omega ,\nonumber \\&G(\cdot ,y)=0 {\quad \hbox {on } }\partial \Omega . \end{aligned}$$
(2.7)

We write

$$\begin{aligned} G_\gamma (x,y)=\Gamma (x-y)-H_\gamma (x,y), \end{aligned}$$
(2.8)

where

$$\begin{aligned} -\Delta _x \Gamma (x)=c_3 \delta _0(x), \quad \Gamma (x)=\frac{\alpha _3}{\vert x\vert } \end{aligned}$$

is (a multiple of) the fundamental solution of the Laplacian in \({{\mathbb {R}}}^3\) and the regular part \(H_\gamma (x,y)\), for fixed \(y\in \Omega \), satisfies

$$\begin{aligned}&-\Delta _x H_\gamma (x,y)-\gamma H_\gamma (x,y)=-\gamma \Gamma (x-y){\quad \hbox {in } }\Omega ,\nonumber \\&H_\gamma (x,y)=\Gamma (x-y){\quad \hbox {on } }\partial \Omega . \end{aligned}$$
(2.9)

The function \(H_\gamma (\cdot ,y)\in C^{0,1}(\Omega )\) when \(\gamma \in (0,\lambda _1)\). For later purpose, we also write

$$\begin{aligned} H_\gamma (x,y)=\theta _\gamma (x-y)-h_\gamma (x,y), \end{aligned}$$
(2.10)

where

$$\begin{aligned} \theta _\gamma (x):=\alpha _3 \frac{1-\cos (\sqrt{\gamma }\vert x\vert )}{\vert x\vert } \end{aligned}$$
(2.11)

and \(h_\gamma (\cdot ,y)\in C^\infty (\Omega )\) solves

$$\begin{aligned}&\Delta _x h_\gamma (x,y)+\gamma h_\gamma (x,y)=0 {\quad \hbox {in } }\Omega ,\nonumber \\&h_\gamma (x,y)=-\alpha _3 \frac{\cos (\sqrt{\gamma }\vert x-y\vert )}{\vert x-y\vert } {\quad \hbox {on } } {\partial } \Omega . \end{aligned}$$
(2.12)

We also define the Robin function

$$\begin{aligned} R_\gamma (x):=H_\gamma (x,x)=h_\gamma (x,x). \end{aligned}$$

In terms of the original function \(u_1\) the Eq. (2.6) reads as

$$\begin{aligned} u_1(x,t)\approx \mu ^{1/2}\frac{\alpha _3 }{\vert x-\xi \vert } - \mu ^{1/2} H_\gamma (x,\xi ). \end{aligned}$$

We notice that far away from the origin we have

$$\begin{aligned} U_{\mu ,\xi }(x)\approx \mu ^{1/2}\frac{\alpha _3}{\vert x-\xi \vert }. \end{aligned}$$

This formal analysis suggests the ansatz

$$\begin{aligned} u_1(x,t):=U_{\mu ,\xi }(x)- \mu ^{1/2}H_\gamma (x,\xi ). \end{aligned}$$

2.1.1 Dilation parameter \(\mu (t)\)

We write the full dilation parameter in the form

$$\begin{aligned} \mu =\mu _0(t)e^{2\Lambda (t)}, \end{aligned}$$

for some \(\Lambda (t)=o(1)\) as \(t\rightarrow \infty \) to be found, where

$$\begin{aligned} \mu _0(t)=e^{-2\gamma t}. \end{aligned}$$

In this notation we have

$$\begin{aligned} \frac{{\dot{\mu }}(t)}{2\mu (t)}&=\frac{{\dot{\mu }}_0 e^{2\Lambda }}{2\mu _0 e^{2\Lambda }}+\frac{2\dot{\Lambda } \mu _0 e^{2\Lambda }}{2\mu _0 e^{2\Lambda }}\\ {}&=-\gamma +{\dot{\Lambda }}(t), \end{aligned}$$

and

$$\begin{aligned} \Lambda (t)=-\int _{t}^{\infty }{\dot{\Lambda }}(s)\,ds, \end{aligned}$$

where \({\dot{\Lambda }}(s)\) is an integrable function in any \([t_0,\infty )\).

2.2 Error associated to \(u_1\)

The next step consists in computing the error associated to the first ansatz \(u_1\). We define the error operator

$$\begin{aligned} S[u]:=-\partial _t u+ \Delta u+ u^5. \end{aligned}$$

Of course, solving \(S[u]=0\) is equivalent to solve the equation in (2.1). It is well-known that all bounded solutions to the linearized operator

$$\begin{aligned} \Delta _y \phi + 5U(y)^{4}\phi =0 {\quad \hbox {in } }{{\mathbb {R}}}^n, \end{aligned}$$

are linear combinations of the functions \(\{Z_i\}_{i=1}^4\) defined as

$$\begin{aligned} Z_i(y):=\partial _{y_i} U(y), \quad i=1,2,3, \end{aligned}$$

and

$$\begin{aligned} Z_4(y):=\frac{1}{2}U(y)+y\cdot \nabla U(y)=\frac{\alpha _3}{2} \frac{1-\vert y\vert ^2}{\left( 1+\vert y\vert ^2\right) ^{3/2}}. \end{aligned}$$

We define the scaled variable

$$\begin{aligned} y:=y(x,t):=\frac{x-\xi (t)}{\mu (t)}. \end{aligned}$$

Now, we compute \(S[u_1](x,t)\) for \(x\ne \xi (t)\). We have

$$\begin{aligned} \Delta u_1&=\mu ^{-1/2}\Delta _x U\left( \frac{x-\xi }{\mu }\right) - \mu ^{1/2}\Delta _x H_\gamma (x,\xi )\\&=-\mu ^{-5/2}U(y)^5+\mu ^{1/2}\left( \gamma H_\gamma (x,\xi )-\frac{\gamma \alpha _3}{\vert x-\xi \vert }\right) \\&=-\mu ^{-5/2}U(y)^5+\mu ^{1/2}\gamma H_\gamma (x,\xi )-\mu ^{1/2} \frac{\gamma \alpha _3}{\vert x-\xi \vert }, \end{aligned}$$

where we used equations (2.2) and (2.9) for U and \(H_\gamma \) respectively. Using the definition of \(Z_4\), the time-derivative can be written as

$$\begin{aligned} \partial _t u_1=&-\frac{1}{2}\frac{{\dot{\mu }}}{\mu }\mu ^{-1/2}U(y)+\mu ^{-1/2}\nabla _y U(y)\cdot \left[ -\frac{{\dot{\xi }}}{\mu }-\frac{{\dot{\mu }}}{\mu }y\right] \\&-\frac{1}{2}\frac{{\dot{\mu }}}{\mu } \mu ^{1/2}H_\gamma (x,\xi )-\mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_2}H_\gamma (x,\xi )\\ =&-\left( \frac{{\dot{\mu }}}{2\mu }\right) \left[ \mu ^{-1/2}2Z_4(y) +\mu ^{1/2}H_\gamma (x,\xi ) \right] \\ {}&-\mu ^{-3/2}{\dot{\xi }} \cdot \nabla _y U -\mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_2}H_\gamma (x,\xi ). \end{aligned}$$

Hence, the error associated to \(u_1\) is

$$\begin{aligned} S[u_1]=&{\dot{\Lambda }} \left( \mu ^{-1/2}2Z_4(y) +\mu ^{1/2}H_\gamma (x,\xi )\right) -\gamma \mu ^{-1/2}\left( 2Z_4(y)+\frac{ \alpha _3}{\vert y\vert }\right) \nonumber \\ {}&+\mu ^{-3/2}{\dot{\xi }} \cdot \nabla _y U(y)+\mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_2}H_\gamma (x,\xi ) \nonumber \\&-\mu ^{-3/2} 5U(y)^4 H_\gamma (x,\xi )\nonumber \\ {}&+\mu ^{-5/2}\left[ \left( U(y)-\mu H_\gamma (x,\xi )\right) ^5-U(y)^5+\mu 5U(y)^4 H_\gamma (x,\xi )\right] . \end{aligned}$$
(2.13)

2.3 Global improvement

The remaining part of this section concerns the improvement of the natural ansatz \(u_1\). Later in the argument we will divide the error in outer and inner part. We realize that solving the inner-outer system requires a global and a local improvement. Reading Proposition 3.1, which is the linear theory for the inner problem, we see that, to get decay in \(\phi (y,\tau )\) at distance R we need \(a>1\) in the definition of \(\vert \!\vert h\vert \!\vert _{\nu ,2+a}\). This smallness at distance R will make the inner and outer regime weakly decoupled. Our particular h will satisfy \(\vert \!\vert h\vert \!\vert _{\nu ,2+a}<\infty \) with \(a=2\), hence we will use estimate (3.17). Thus, we say that a term is slow-decay in space if it is not controlled by \((1+\vert y\vert )^{-4}\). We can find an exact perturbation with our scheme if we remove such terms. Looking at (2.13) we observe that all the terms in the first two lines are slow-decay. Using the inequality \(\mu (t)\lesssim (1+\vert y\vert )^{-1}\) we can negotiate decay in time with decay in space if needed in other terms. For the moment we can assume \({\dot{\Lambda }}, \Lambda , {\dot{\xi }},\xi \) bounded by some power of \(\mu (t)\). Later we shall specify precise norms for these parameters. Firstly, we decompose

$$\begin{aligned} \mu ^{-3/2} 5U(y)^4 H_\gamma (x,\xi )&=\mu ^{-3/2} 5U(y)^4 \theta _\gamma (x-\xi ) \\&\quad -\mu ^{-3/2}5U(y)^4 h_\gamma (x,\xi ). \end{aligned}$$

We define

$$\begin{aligned} u_2(x,t)=u_1(x,t)+\mu _0^{1/2}J[{{\dot{\Lambda }}}](x,t). \end{aligned}$$

The new error reads as

$$\begin{aligned} S[u_2]&=S[u_1]+\left( - {\partial } _t +\Delta _x\right) \left( \mu _0^{1/2}J[{{\dot{\Lambda }}}](x,t)\right) +u_2^5-u_1^5\\&=S[u_1]+\mu _0^{1/2}\left( - {\partial } _t + \Delta _x + \gamma \right) J[{{\dot{\Lambda }}}]+u_2^5-u_1^5. \end{aligned}$$

Let

$$\begin{aligned} J[{{\dot{\Lambda }}}](x,t):=J_1[{{\dot{\Lambda }}}](x,t)+J_2(x,t). \end{aligned}$$

Plugging \(S[u_1]\) given by (2.13) into \(S[u_2]\) we get

$$\begin{aligned} S[u_2]&= \mu ^{-3/2}{\dot{\xi }} \cdot \nabla _y U(y) +\mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_2} H_\gamma (x,\xi )+\mu ^{-3/2}5U(y)^4h_\gamma (x,\xi )\\&\qquad +\mu _0^{1/2}\left\{ (- {\partial } _t + \Delta _x + \gamma )J_1+\left( \frac{\mu }{\mu _0}\right) ^{\frac{1}{2}}{{\dot{\Lambda }}}\left( \mu ^{-1} 2Z_4\left( y\right) +H_\gamma (x,\xi )\right) \right\} \\&\qquad +\mu _0^{1/2}\left\{ (- {\partial } _t + \Delta _x + \gamma ) J_2- \left( \frac{\mu }{\mu _0}\right) ^{\frac{1}{2}}\right. \\ {}&\left. \qquad \times \left[ \gamma \mu ^{-1} \left( 2Z_4(y)+\frac{\alpha _3}{\vert y\vert }\right) +\mu ^{-2}5U^4 \theta _\gamma (\mu y)\right] \right\} \\&\qquad +\mu ^{-5/2}\left[ \left( U(y)-\mu H_\gamma (x,\xi )+\mu \left( \frac{\mu _0}{\mu }\right) ^{1/2}J[{\dot{\Lambda }}](x,t)\right) ^5\right. \\&\left. \qquad -U(y)^5+\mu 5U(y)^4 H_\gamma (x,\xi )\right] . \end{aligned}$$

We select \(J_1[{\dot{\Lambda }}](x,t)\) such that

$$\begin{aligned} {\partial } _t J_1&= \Delta _x J_1+ \gamma J_1\nonumber \\&\quad +\left( \frac{\mu }{\mu _0}\right) ^{\frac{1}{2}}{{\dot{\Lambda }}}\left( \mu ^{-1} 2Z_4\left( \frac{x-\xi }{\mu }\right) +H_\gamma (x,\xi )\right) {\quad \hbox {in } }\Omega \times [t_0-1,\infty ),\nonumber \\ {}&\quad J_1(x,t)=0 {\quad \hbox {in } } {\partial } \Omega \times [t_0-1,\infty ),\nonumber \\ {}&\quad J_1(x,t_0-1)=0 {\quad \hbox {on } }\Omega , \end{aligned}$$
(2.14)

and

$$\begin{aligned} {\partial } _t J_2 =&\Delta _x J_2+ \gamma J_2-\left( \frac{\mu }{\mu _0}\right) ^{\frac{1}{2}}\bigg [\gamma \left( \mu ^{-1}2Z_4\left( \frac{x-\xi }{\mu }\right) +\frac{\alpha _3}{\vert x-\xi \vert }\right) \nonumber \\&+\mu ^{-2}5U\left( \frac{x-\xi }{\mu }\right) ^4 \theta _\gamma (x-\xi )\bigg ] {\quad \hbox {in } }\Omega \times [t_0,\infty ),\nonumber \\ J_2(x,t)&=0 {\quad \hbox {on } } {\partial } \Omega \times [t_0,\infty ),\nonumber \\ J_2(x,t_0&)=0 {\quad \hbox {in } }\Omega . \end{aligned}$$
(2.15)

The choice of defining \(J_1\) from the time \(t_0-1\), as well as \(\dot{\Lambda }(t)\), will become clear in Sect. 8. For the variable \(\xi (t)\) it is enough to define the extension \(\xi (t)=\xi (t_0)\) for \(t\in [t_0-1,t_0)\). With these choices the error associated to \(u_2\) reads as

$$\begin{aligned} S[u_2]&= \mu ^{-3/2}{\dot{\xi }} \cdot \nabla _y U(y) +\mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_2}H_\gamma (x,\xi )+\mu ^{-3/2}5U(y)^4h_\gamma (x,\xi )\nonumber \\&\quad +\mu ^{-5/2}\left[ \left( U(y)-\mu H_\gamma (x,\xi )+\mu \left( \frac{\mu _0}{\mu }\right) ^{1/2}J[{\dot{\Lambda }}](x,t)\right) ^5\right. \nonumber \\&\left. \quad -U(y)^5+\mu 5U(y)^4 H_\gamma (x,\xi )\right] . \end{aligned}$$
(2.16)

2.3.1 Choice of \(\gamma \)

We observe that with this choice of \(J_2\) we remove the singular term \(\vert x-\xi \vert ^{-1}\) from (2.13). At this point, the main error at \(x=\xi (t)\) is given by the first order of the nonlinear term

$$\begin{aligned} \mu ^{-3/2}5 U(0)^4 R_\gamma (\xi ), \end{aligned}$$

which is, in general, of size \(\mu (t)^{-3/2}\). We realize that we can reduce this error by selecting \(\gamma \) such that \(R_\gamma (0)=0\). The existence and uniqueness of such number is given by the following lemma.

Lemma 2.1

There exists a unique \(\gamma =\gamma ^*(0) \in (0,{\lambda _1})\) such that \(R_{\gamma ^*}(0)=0\).

Proof

We consider the function \(R_\gamma (0)\) as a function of \(\gamma \). Lemma A.2 in [14] shows that

$$\begin{aligned} R_\gamma (0):(0,\lambda _1)\rightarrow (-\infty , R_0(0)) \end{aligned}$$

is smooth in \((0,\lambda _1)\) and \( {\partial } _\gamma R_\gamma (0)<0\). Lemma A.1 in Appendix A shows that \(R_\gamma (0)\rightarrow -\infty \) as \(\gamma \rightarrow \lambda _1^{-}\). By the maximum principle \(H_0(x,y)>0\) for all \(x,y\in \Omega \), hence we have \(R_0(0)>0\) and the intermediate value theorem gives the existence of

$$\begin{aligned} \gamma ^*(0):=\sup \{ \gamma>0 :\, R_\gamma (0)>0 \}. \end{aligned}$$

Finally the monotonicity of \(R_\gamma (0)\) implies the uniqueness of \(\gamma ^*(0)\). \(\square \)

Remark 2.2

(Regularity of \(\gamma ^*(x)\)) Let \(R(\gamma ,x) :=R_{\gamma }(x)\). Since \(R(\gamma ^*(x),x)=0\) and \( {\partial } _\gamma R(\gamma ,x)<0\) for all \(x\in \Omega \), the implicit function theorem implies that \(\gamma ^*(x)\in C^1(\Omega )\) with

$$\begin{aligned} \nabla _x \gamma ^*(x)=-\frac{\nabla _x R(\gamma ,x)}{ {\partial } _\gamma R(\gamma ,x)}. \end{aligned}$$

Remark 2.3

(Radial case) We compute \(\gamma (0)\) in case \(\Omega =B_1(0)\). We look for a radial solution to

$$\begin{aligned}&\Delta H_\gamma +\gamma H_\gamma =\frac{\alpha _3}{\vert x\vert }{\quad \hbox {in } }B_1,\\&H_\gamma (x,0)=\frac{\alpha _3}{\vert x\vert }{\quad \hbox {on } } {\partial } B_1. \end{aligned}$$

We define \(l_0(\vert x\vert ){:=}H_\gamma (x,0)\) for a function \(l_0:[0,1]\rightarrow {{\mathbb {R}}}\). Then \(l_0\) solves

$$\begin{aligned}&{\partial } _{rr} l_0 +\frac{2}{r} {\partial } _r l_0 +\gamma l_0=\gamma \frac{\alpha _3}{r}{\quad \hbox {in } }[0,1],\\&l_0(1)=\alpha _3, \quad l_0(r) \quad \text {bounded at } r=0. \end{aligned}$$

We write \(l_0(r)=\alpha _3l(r)/r\), where l(r) solves

$$\begin{aligned}&{\partial } _{rr}l+\gamma l= \gamma {\quad \hbox {in } }[0,1],\\&l(1)=1, \quad l(r)=O(r)\quad \text {for } r\rightarrow 0. \end{aligned}$$

The solution to this problem is given by

$$\begin{aligned} l(r)=1-\cos (\sqrt{\gamma }r)+\cot (\sqrt{\gamma })\sin (\sqrt{\gamma }r), \end{aligned}$$

and we conclude that

$$\begin{aligned} H_\gamma (r,0)=\alpha _3\left[ \frac{1-\cos (\sqrt{\gamma }r)}{r}+\frac{\sin (\sqrt{\gamma }r)}{r\tan (\sqrt{\gamma })}\right] . \end{aligned}$$

In particular, for \(r=0\) we find

$$\begin{aligned} R_\gamma (0)=H_\gamma (0,0)=\alpha _3\sqrt{\gamma }\cot (\sqrt{\gamma }). \end{aligned}$$

Asking for \(R_\gamma (0)=0\)

$$\begin{aligned} \gamma =\left( \frac{\pi }{2}+k\pi \right) ^2\quad \text {for} \quad k\in \mathbb {N}, \end{aligned}$$

and, recalling that \(\lambda _1(B_1)=\pi ^2\), the unique value in \((0,\lambda _1)\) is

$$\begin{aligned} \gamma ^*=\frac{\pi ^2}{4}, \end{aligned}$$

as predicted in the analysis of Galaktionov and King [29].

For the sake of simplicity we continue to use \(\gamma =\gamma (0)\) to denote the selected number \(\gamma ^*(0)\). Since \(R_\gamma (x)\in C^\infty (\Omega )\) we expand

$$\begin{aligned} R_\gamma (\xi )=R_\gamma (0)+\xi \cdot \nabla _x R_\gamma (0)+\frac{1}{2}\xi ^{\intercal } D_{xx}^2 R_\gamma \left( \xi ^*\right) \xi , \end{aligned}$$

for some \(\xi ^*\in \overline{[0,\xi ]}\). Assuming \(\xi =O(\mu )\) we conclude

$$\begin{aligned} \mu ^{-3/2}5U(0)^4 R_\gamma (\xi )=O\left( \mu ^{-1/2}\right) . \end{aligned}$$

2.4 Local improvement and computation of the final error

In this section we make a further improvement and we obtain the final ansatz. We still need to remove from (2.13) the main order of the terms

$$\begin{aligned} \mu ^{-3/2}{\dot{\xi }} \cdot \nabla _y U + \mu ^{-3/2}5U(y)^4 h_\gamma (x,\xi ). \end{aligned}$$

We define the final ansatz

$$\begin{aligned} u_3(x,t):=u_2(x,t)+\mu (t)^{-1/2}\phi _3\left( \frac{x-\xi (t)}{\mu (t)},t\right) \eta _{l(t)} \left( \vert \frac{x-\xi (t)}{\mu (t)}\vert \right) . \end{aligned}$$

The function \(\eta : [0,\infty )\rightarrow [0,1]\) denotes a smooth cut-off function such that \(\eta (s)\equiv 1\) for \(s<1\) and \({{\,\textrm{supp}\,}}\eta \subset [0,2]\), and we define

$$\begin{aligned} \eta _{l(t)}\left( \vert y\vert \right) :=\eta \left( \frac{\vert y\vert }{l(t)}\right) ,\qquad l(t):=\frac{1}{k_2\mu }, \end{aligned}$$
(2.17)

where \(k_2\) is a constant such that \(B_{\frac{2}{k_2}}(0)\subset \Omega \), to ensure that \({{\,\textrm{supp}\,}}\eta _{l} \Subset \Omega \). Also we define the variable

$$\begin{aligned}z_3(x,t):=\frac{y(x,t)}{l(t)}=\frac{x-\xi (t)}{\mu (t) l(t)}.\end{aligned}$$

We compute

$$\begin{aligned} {\partial } _t \left( \mu ^{-1/2}\phi _3 \eta _{l(t)}\right)&=-\frac{{\dot{\mu }}}{2\mu } \mu ^{-1/2}\phi _3 \eta _{l(t)}+\mu ^{-1/2}\eta _{l(t)}\left[ {\partial } _t \phi _3+\nabla _y \phi _3 \cdot \left( -\frac{{\dot{\mu }}}{\mu }y-\frac{\dot{\xi }}{\mu }\right) \right] \\ {}&\quad +\mu ^{-1/2}\phi _3 {\partial } _t \eta _{l(t)}, \end{aligned}$$

and

$$\begin{aligned} \Delta _x\left( \mu ^{-1/2}\phi _3 \eta _{l}\right) =&\mu ^{-5/2}\eta _{l(t)}\Delta _y \phi _3 + 2\mu ^{-3/2}\nabla _y \phi _3 \cdot \frac{y}{\vert y\vert } \left( \frac{\eta '\left( \vert z_3\vert \right) }{\mu l}\right) \\ {}&+\mu ^{-1/2}\phi _3 \left( \frac{2}{\vert z_3\vert }\frac{\eta '(\vert z_3\vert )}{\mu ^2 l^2}+\frac{\eta ''\left( \vert z_3\vert \right) }{\mu ^2 l^2}\right) . \end{aligned}$$

We define

$$\begin{aligned} \mathcal {N}_3(y,t)&:=\left( U(y)^5- \mu H_\gamma (\mu y+\xi ,\xi )+\mu \left( \frac{\mu _0}{\mu }\right) ^{1/2}J[{\dot{\Lambda }}](\mu y +\xi ,t)+\phi _3(y,t)\eta _l\right) ^5\\&\qquad -U(y)^5 \\ {}&\qquad - 5U(y)^4 \left( -\mu H_\gamma (\mu y +\xi ,\xi )+\mu \left( \frac{\mu _0}{\mu }\right) ^{1/2}J[{\dot{\Lambda }}](\mu y +\xi ,t)+\phi _3 \eta _l\right) \end{aligned}$$

Thus, using (2.16),

$$\begin{aligned} S[u_3]&=- {\partial } _t\left( \mu ^{-1/2}\phi _3 \eta _{l}\right) +\Delta _x \left( \mu ^{-1/2}\phi _3 \eta _{l}\right) +u_3^5-u_2^5+S[u_2]\\&=\mu ^{-3/2}{\dot{\xi }} \cdot \nabla _y U+\mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_2} H_\gamma (x,\xi )+\mu ^{-3/2}5U(y)^4 h_\gamma (x,\xi )\\&\quad +\mu ^{-5/2}\mathcal {N}_3(y,t)+ 5U(y)^4 \mu ^{-3/2} \left( \frac{\mu _0}{\mu }\right) ^{1/2} J(x,t)+\mu ^{-5/2}\phi _3 \eta _l 5U(y)^4\\&\quad -\left( -\frac{{\dot{\mu }}}{2\mu } \right) \mu ^{-1/2}\phi _3 \eta _{l(t)}-\mu ^{-1/2}\eta _{l(t)} \left[ {\partial } _t \phi _3+\nabla _y \phi _3 \cdot \left( -\frac{{\dot{\mu }}}{\mu }y-\frac{{\dot{\xi }}}{\mu }\right) \right] \\&\quad -\mu ^{-1/2}\phi _3 {\partial } _t \eta +\mu ^{-5/2}\eta _{l(t)}\Delta _y \phi _3 + 2\mu ^{-3/2}\nabla _y \phi _3 \cdot \frac{y}{\vert y\vert } \left( \frac{\eta '\left( \vert z_3\vert \right) }{\mu l}\right) \\&\quad +\mu ^{-1/2}\phi _3 \left( \frac{2}{\vert z_3\vert }\frac{\eta '(\vert z_3\vert )}{\mu ^2 l^2}+\frac{\eta ''\left( \vert z_3\vert \right) }{\mu ^2 l^2}\right) . \end{aligned}$$

By Taylor expanding \(h_\gamma (x,\xi )\) centered at \(x=\xi \) we have

$$\begin{aligned} h_\gamma (x,\xi )=R_\gamma (\xi )+ \mu y\cdot \nabla _{x_1}h_\gamma (\xi ,\xi ) + \frac{1}{2}\mu ^2 y^\intercal D_{xx}h_\gamma ({\bar{x}},\xi ) y \end{aligned}$$
(2.18)

for some \({\bar{x}}\in \overline{[\xi ,x]}\). Now, we expand the first terms at \((\xi ,\xi )=(0,0)\). By the Chain Rule we have \(\nabla _{x_1}h_\gamma (x,x)=2 \nabla _x R_\gamma (x)\). Hence, we have

$$\begin{aligned} \nabla _{x_1}h_\gamma (\xi ,\xi )=\frac{1}{2}\nabla _{x}R_\gamma (\xi )=\frac{1}{2}\nabla _{x}R_\gamma (0)+\frac{1}{2}\xi ^\intercal D_{xx}R_\gamma \left( \xi ^{**}\right) , \end{aligned}$$

for some \(\xi ^{**}\in \overline{[0,\xi ]}\). Furthermore, since \(R_\gamma (0)=0\), we have

$$\begin{aligned} R_\gamma (\xi )= \xi \cdot \nabla _{x}R_\gamma (0)+\frac{1}{2}\xi ^\intercal D_{xx}R_\gamma (\xi ^{*})\xi \end{aligned}$$

for some \(\xi ^* \in \overline{[0,\xi ]}\). Plugging these identities in (2.18) we obtain

$$\begin{aligned} h_\gamma (x,\xi )&= \xi \cdot \nabla _{x}R_\gamma (0)+\frac{1}{2}\mu y\cdot \nabla _{x}R_\gamma (0)\nonumber \\ {}&\quad +\frac{1}{2}\xi ^\intercal D_{xx}R_\gamma (\xi ^{*})\xi +\frac{1}{2} \mu y^\intercal D_{xx}R_\gamma \left( \xi ^{**}\right) \xi \nonumber \\ {}&\quad + \frac{1}{2}\mu ^2 y^\intercal D_{xx}h_\gamma ({\bar{x}},\xi )y. \end{aligned}$$
(2.19)

We write

$$\begin{aligned} \xi (t)=\xi _0(t)+\xi _1(t). \end{aligned}$$

Now, we assume the following decay for the parameters \(\xi _1,\dot{\xi }_1, \Lambda ,{\dot{\Lambda }}\):

$$\begin{aligned}&\vert \xi _1(t)\vert +\vert *\vert {{\dot{\xi }}_1(t)}\le C \mu (t)^{1+k},\\&\vert *\vert {\Lambda (t)}\le C\mu (t)^{l_0},\\&\vert *\vert {{\dot{\Lambda }}(t)}\le C \mu (t)^{l_1}, \end{aligned}$$

for some positive constants \(k,l_0,l_1\) to be chosen (in §3.1.3). We write the full error

$$\begin{aligned} S[u_3]=&\mu ^{-3/2}\nabla _y U(y)\cdot \left[ {\dot{\xi }}-\mu ^{-1}\mu _0 {\dot{\xi }}_0 \right] \eta _l\\&+5U(y)^4 \left[ \mu ^{-3/2}h_\gamma (x,\xi )-\mu ^{-5/2}\mu _0 \left( \frac{1}{2}\mu _0 y \cdot \nabla _x R_\gamma (0)\right) \right] \eta _l\\&+\left[ \mu ^{-3/2}\nabla _y U(y)\cdot {\dot{\xi }} +5U(y)^4 \mu ^{-3/2}h_\gamma (x,\xi )\right] (1-\eta _l)\\&+\mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_2} H_\gamma (x,\xi )+\mu ^{-5/2}\mathcal {N}_3(y,t)+ 5U(y)^4 \mu ^{-3/2} \left( \frac{\mu _0}{\mu }\right) ^{1/2} J(x,t) \\&-\left( -\frac{{\dot{\mu }}}{2\mu } \right) \mu ^{-1/2}\phi _3 \eta _{l(t)}-\mu ^{-1/2}\eta _{l(t)} \left[ {\partial } _t \phi _3+\nabla _y \phi _3 \cdot \left( -\frac{{\dot{\mu }}}{\mu }y-\frac{{\dot{\xi }}}{\mu }\right) \right] \\&-\mu ^{-1/2}\phi _3 {\partial } _t \eta +\mu ^{-5/2}\eta _{l(t)}\left[ \Delta _y \phi _3+5U(y)^4\phi _3 +\mathcal {M}[\mu _0,\xi _0]\right] \\ {}&+ 2\mu ^{-3/2}\nabla _y \phi _3 \cdot \frac{y}{\vert y\vert } \left( \frac{\eta '\left( \vert z_3\vert \right) }{\mu l}\right) +\mu ^{-1/2}\phi _3 \left( \frac{2}{\vert z_3\vert }\frac{\eta '(\vert z_3\vert )}{\mu ^2 l^2}+\frac{\eta ''\left( \vert z_3\vert \right) }{\mu ^2 l^2}\right) , \end{aligned}$$

where

$$\begin{aligned} \mathcal {M}[\mu _0,\xi _0]{:=} \mu _0 {\dot{\xi }}_0\cdot \nabla _y U(y)- \frac{5}{2}U(y)^4 \mu _0 \left( \mu _0 y \cdot \nabla _x R_\gamma (0)\right) \end{aligned}$$
(2.20)

For any fixed \(t>t_0\), we select \(\phi _3(x,t)\) as the bounded solution to the elliptic problem

$$\begin{aligned} \Delta _y \phi _3(y,t) + 5U(y)^4 \phi _3(y,t) = - \mathcal {M}[\mu _0,\xi _0](y,t) {\quad \hbox {in } }{{\mathbb {R}}}^3, \end{aligned}$$
(2.21)

with the following orthogonality conditions on the right-hand side:

$$\begin{aligned} \int _{{{\mathbb {R}}}^3} \mathcal {M}[\mu _0,\xi _0](y,t)Z_i(y)\,dy=0 \quad \text {for}\quad t>t_0, \quad \text {and}\quad i=1,2,3,4. \end{aligned}$$
(2.22)

As we shall see in the proof of Lemma 2.4, the conditions (2.22) are essential to have \(\phi _3\) bounded in space and equivalent to choose \(\xi _0(t)\). The condition corresponding to the index \(i=4\) is satisfied by symmetry. When \(i=1,2,3\) the orthogonality condition (2.22) is equivalent to

$$\begin{aligned} \mu _0 {\dot{\xi }}_{0,i} \left( \int _{{{\mathbb {R}}}^3}\vert {\partial } _{y_i} U(y)\vert ^2\,dy\right) -\mu _0^2\left( \int _{{{\mathbb {R}}}^3}5U(y)^4 y_i {\partial } _{y_i}U(y)\,dy\right) \frac{1}{2} {\partial } _{x,i} R_\gamma (0)=0. \end{aligned}$$

Hence, we select \(\xi _{0,i}\) such that

$$\begin{aligned} {\dot{\xi }}_{0,i}(t)= \frac{ {\partial } _{x,i} R_\gamma (0)\left( \int _{{{\mathbb {R}}}^3}5U(y)^4 y_i {\partial } _{y_i}U(y)\,dy\right) }{2\left( \int _{{{\mathbb {R}}}^3}\vert {\partial } _{y_i} U(y)\vert ^2\,dy\right) } \mu _0(t) . \end{aligned}$$

With the condition \(\lim \nolimits _{t\rightarrow \infty }\xi _i(t)=0\) we get

$$\begin{aligned} \xi _{0,i}(t)= \mathfrak {c}_i e^{-2\gamma t},\quad \mathfrak {c}_i= -\frac{\vert *\vert { {\partial } _{x,i} R_\gamma (0)}\vert \int _{{{\mathbb {R}}}^3}5U(y)^4 y_i {\partial } _{y_i}U(y)\,dy\vert }{4\gamma \left( \int _{{{\mathbb {R}}}^3}\vert {\partial } _{y_i} U(y)\vert ^2\,dy\right) }. \end{aligned}$$
(2.23)

Also, we define \(\mathfrak {c}{:=}(\mathfrak {c}_1,\mathfrak {c}_2,\mathfrak {c}_3)\).

Remark 4.1

(No local improvement in the radial case) In case \(\Omega =B_1(0)\), searching \(h_\gamma (r,0)\) solution to (2.12) in the radial form, we see that

$$\begin{aligned} \nabla _x R_\gamma (0)=2\nabla _{x_1} h_\gamma (0,0)=0, \end{aligned}$$

hence conditions (2.22) imply \(\xi _0={0}\), as expected. Thus, the local improvement \(\phi _3\), which in fact involves only non-zero modes, is null in the radial case.

With these choices for \(\phi _3\) and \(\xi _0\) we conclude with the following expression of the error associated to the final ansatz \(u_3\):

$$\begin{aligned} S[u_3]&= \mu ^{-3/2}\nabla _y U(y)\cdot \left[ {\dot{\xi }}_1+\left( 1-\mu ^{-1}\mu _0\right) {\dot{\xi }}_0\right] \eta _l\\&\quad +5U(y)^4 \left[ \mu ^{-3/2}h_\gamma (x,\xi )-\mu ^{-5/2}\mu _0 \left( \frac{1}{2}\mu _0 y \cdot \nabla _x R_\gamma (0)\right) \right] \eta _l\\&\quad +\left[ \mu ^{-3/2}\nabla _y U(y)\cdot {\dot{\xi }} +5U(y)^4 \mu ^{-3/2}h_\gamma (x,\xi )\right] (1-\eta _l)\\&\quad +\mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_2} H_\gamma (x,\xi )+\mu ^{-5/2}\mathcal {N}_3(y,t)+ 5U(y)^4 \mu ^{-3/2} \left( \frac{\mu _0}{\mu }\right) ^{1/2} J(x,t) \\&\quad -\left( -\frac{{\dot{\mu }}}{2\mu }\right) \mu ^{-1/2}\phi _3 \eta _{l(t)}-\mu ^{-1/2}\eta _{l(t)} \left[ {\partial } _t \phi _3+\nabla _y \phi _3 \cdot \left( -\frac{{\dot{\mu }}}{\mu }y-\frac{{\dot{\xi }}}{\mu }\right) \right] \\&\quad -\mu ^{-1/2}\phi _3 {\partial } _t \eta + 2\mu ^{-3/2}\nabla _y \phi _3 \cdot \frac{y}{\vert y\vert } \left( \frac{\eta '\left( \vert z_3\vert \right) }{\mu l}\right) \\&\quad +\mu ^{-1/2}\phi _3 \left( \frac{2}{\vert z_3\vert }\frac{\eta '(\vert z_3\vert )}{\mu ^2 l^2}+\frac{\eta ''\left( \vert z_3\vert \right) }{\mu ^2 l^2}\right) . \end{aligned}$$

2.5 Estimate of the inner and outer error

For later purpose, we split \(S[u_3]\) in inner and outer error. At this stage, it is important to treat the terms involving directly \({{\dot{\Lambda }}}\) as part of the outer error, since, as we shall see, a priori those are the terms with less regularity. Let

$$\begin{aligned} S[u_3]&=S_{\text {in}}+S_{\text {out}}\\&=S_{\text {in}}\eta _{R(t)}(y) + (1-\eta _{R(t)}(y))S_{\text {in}}+S_{\text {out}}, \end{aligned}$$

where we define the inner error

$$\begin{aligned} S_{\text {in}}&:=\mu ^{-3/2} \left( \frac{\mu _0}{\mu }\right) ^{1/2} 5U(y)^4 J(x,t)+\mu ^{-5/2}\mathcal {N}_3 \nonumber \\&\quad +\mu ^{-3/2}\eta _{l}\left( {\dot{\xi }}_1+\left( 1-\mu ^{-1}\mu _0\right) \dot{\xi }_0\right) \cdot \nabla _y U(y)\nonumber \\ {}&\quad +\mu ^{-3/2}\eta _l 5U(y)^4 \left( h_\gamma (x,\xi )-\left( \frac{\mu _0}{\mu }\right) \left( \frac{1}{2}\mu _0 y\cdot \nabla _x R_\gamma (0)\right) \right) , \end{aligned}$$
(2.24)

the outer error

$$\begin{aligned} S_{\text {out}}&:=\mu ^{-3/2}\left[ \nabla _y U(y)\cdot {\dot{\xi }} +5U(y)^4 h_\gamma (x,\xi )\right] (1-\eta _l)\nonumber \\ {}&\quad +\mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_2} H_\gamma (x,\xi )\nonumber \\ {}&\quad -\mu ^{-1/2}\left[ (\gamma -{{\dot{\Lambda }}})\eta _l \left( \phi _3+2y \cdot \nabla _y \phi _3\right) +\eta _l \left( {\partial } _t \phi _3 -\mu ^{-1}{\dot{\xi }} \cdot \nabla _y \phi _3\right) \right. \nonumber \\ {}&\quad \left. + \phi _3 \frac{\eta '(\vert z_3\vert )}{\mu l}{\dot{\xi }}\cdot \frac{z_3}{\vert z_3\vert }\right] \nonumber \\ {}&\quad + 2\mu ^{-3/2}\nabla _y \phi _3 \cdot \frac{y}{\vert y\vert } \left( \frac{\eta '\left( \vert z_3\vert \right) }{\mu l}\right) +\mu ^{-1/2}\phi _3 \left( \frac{2}{\vert z_3\vert }\frac{\eta '(\vert z_3\vert )}{\mu ^2 l^2}+\frac{\eta ''\left( \vert z_3\vert \right) }{\mu ^2 l^2}\right) , \end{aligned}$$
(2.25)

and the radius

$$\begin{aligned} R(t):=\mu (t)^{-\delta }, \end{aligned}$$
(2.26)

for some constant \(\delta >0\) which will be chosen in (2.31) to make both the errors \(S_{\text {in}}\eta _R\) and \(S_{\text {in}}(1-\eta _R)+S_{\text {out}}\) suitably small for a final contraction.

Size of \(S_{\text {in}}\eta _{R}\). We proceed with the estimate of \(S_{\text {in}}\eta _{R}\). More precisely, we need the following conditions on \(\delta ,l_0,l_1,k\):

$$\begin{aligned}&\delta +l_1<1 \end{aligned}$$
(2.27)
$$\begin{aligned}&\delta \in \left( \frac{1-l_1}{2},\frac{1+l_1}{6}\right) \end{aligned}$$
(2.28)
$$\begin{aligned}&l_1\le l_0, \end{aligned}$$
(2.29)
$$\begin{aligned}&k+1\ge 2\delta +l_1, \end{aligned}$$
(2.30)

The condition (2.27) is used to get the estimate in the linear outer problem, and it is due to the fact that both the heat kernel \(p_t^\Omega \) and the parameter \(\mu _0(t)\) have an exponential decay for t large. To make the quadratic term \(U^3 {{\tilde{\phi }}}^2\) smaller than \(S_{\text {in}}\) in the inner problem we need the upper bound in (2.28). The lower bound is necessary to get a positive Hölder exponent in the regularity of \({\dot{\Lambda }}\). The last two conditions (2.29)–(2.30) insure that \(S_{\text {in}}\) is controlled by the first term in (2.24). Thus, we fix the following values satisfying (2.27)–(2.28):

$$\begin{aligned} \delta =\frac{2}{9},\quad l_1=\frac{2}{3}. \end{aligned}$$
(2.31)

Here and in what follows, we write \(a\lesssim b\) if there exists a constant C, independent of \(t_0\), such that \(a\le Cb\). If both the inequalities \(a \lesssim b\) and \(b\lesssim a\) hold we write \(a \sim b\). Using (2.34) and (2.35) we estimate

$$\begin{aligned} \vert \eta _R 5U(y)^4 \left( \frac{\mu _0}{\mu }\right) ^{1/2}\mu ^{-3/2}J[{\dot{\Lambda }}](x,t)\vert&\lesssim \frac{\mu ^{-3/2}}{1+\vert y\vert ^4} \left( \mu ^{l_1}+\frac{\mu }{1+\vert y\vert ^{1-\varepsilon }}\right) ,\\&\lesssim \frac{\mu ^{-3/2+l_1}}{1+\vert y\vert ^4} \end{aligned}$$

and, since we are in the region where \(\eta _R \ne 0\), using (2.38), we obtain

$$\begin{aligned} \vert \eta _R{\mu ^{-5/2}\mathcal {N}_3}\vert&\lesssim \mu ^{-1/2}U(y)^3 \left( \vert H_\gamma (x,\xi )\vert +\vert J(x,t)\vert +\mu ^{-1}\phi _3 \eta _l\right) ^2\\&\lesssim \frac{\mu ^{-1/2}}{1+\vert y\vert ^3} \left( \mu R+\mu ^{l_1}+\frac{\mu }{1+\vert y\vert ^{1-\varepsilon }}+\mu \right) ^2\\&\lesssim \frac{\mu ^{-1/2}}{1+\vert y\vert ^{4}}\mu ^{-\delta } \left( \mu ^{1-\delta }+\mu ^{l_1}\right) ^2\\&\lesssim \frac{\mu ^{-1/2-\delta +2\min \{1-\delta ,l_1\}}}{1+\vert y\vert ^{4}} \end{aligned}$$

Also,

$$\begin{aligned} \vert \eta _R\mu ^{-3/2}\eta _l \left( {\dot{\xi }}_1+(1-\mu ^{-1}\mu _0)\dot{\xi }_0\right) \cdot \nabla _y U\vert&\lesssim \eta _l \frac{\mu ^{-3/2}}{1+\vert y\vert ^{4}}R^{2}\left( \vert *\vert {{\dot{\xi }}_1}+\mu ^2\right) \\&\lesssim \frac{\mu ^{-3/2-2\delta +\min \{1+k,2\}}}{1+\vert y\vert ^{4}} \end{aligned}$$

Now, we estimate the last term of \(S_{\text {in}}\eta _R\) using expansion (2.19) and \(\mu /\mu _0=e^{2\Lambda }\) we get

$$\begin{aligned} \vert \eta _R\mu ^{-3/2}\eta _l U(y)^4 \left( h_\gamma (x,\xi )-\left( \frac{\mu _0}{\mu }\right) \left( \frac{1}{2}\mu _0 y \cdot \nabla _x R_\gamma (0)\right) \right) \vert \lesssim \frac{\mu ^{-3/2+\min \{1,l_0\}}}{1+\vert y\vert ^4}. \end{aligned}$$

Combining these estimates we obtain

$$\begin{aligned} \vert \eta _RS_{\text {in}}\vert \lesssim \frac{1}{1+\vert y\vert ^4}\left[ \mu ^{-1/2-\delta +2\min \{1-\delta ,l_1\}}+\mu ^{-3/2-2\delta +\min \{1+k,2\}}+\mu ^{-3/2+\min \{1,l_0,l_1\}}\right] , \end{aligned}$$

and using the values (2.31) we get

$$\begin{aligned} \vert \eta _RS_{\text {in}}\vert&\lesssim \frac{\mu ^{-\frac{3}{2}+l_1}}{1+\vert y\vert ^4}\lesssim \frac{\mu ^{-5/6}}{1+\vert y\vert ^4} \end{aligned}$$

Size of \(S_{\text {out}}\). For the first term in \(S_{\text {out}}\) we have

$$\begin{aligned}&\vert (1-\eta _l)\nabla _y U \cdot {\dot{\xi }}\vert \lesssim \mu ^{3/2}(1-\eta _l)\\&\vert \mu ^{-3/2}5U^4 h_\gamma (1-\eta _l)\vert \lesssim \mu ^{5/2}(1-\eta _l)\\&\vert \mu ^{1/2}{\dot{\xi }} \cdot \nabla _{x_1} H_\gamma \vert \lesssim \mu ^{3/2} \end{aligned}$$

and using the estimates given by Lemma 2.4 on \(\phi _3,\nabla _y \phi _3\) and \( {\partial } _t \phi _3\) we get

$$\begin{aligned}&\bigg |\mu ^{-1/2}\bigg [(\gamma -{\dot{\Lambda }}) \eta _l \left( \phi _3+2y \cdot \nabla _y \phi _3\right) \\&\quad +\eta _l \left( {\partial } _t \phi _3 -\mu ^{-1}{\dot{\xi }} \cdot \nabla _y \phi _3\right) + \phi _3 \frac{\eta '(\vert z_3\vert )}{\mu l}{\dot{\xi }}\cdot \frac{z_3}{\vert z_3\vert }\bigg ] \bigg |\lesssim \mu ^{3/2} \end{aligned}$$

Finally,

$$\begin{aligned} \vert 2\mu ^{-3/2}\nabla _y \phi _3 \cdot \frac{y}{\vert y\vert } \left( \frac{\eta '\left( \vert z_3\vert \right) }{\mu l}\right) +\mu ^{-1/2}\phi _3 \left( \frac{2}{\vert z_3\vert }\frac{\eta '(\vert z_3\vert )}{\mu ^2 l^2}+\frac{\eta ''\left( \vert z_3\vert \right) }{\mu ^2 l^2}\right) \vert \lesssim \mu ^{3/2}. \end{aligned}$$

We conclude that

$$\begin{aligned} \vert S_{\text {out}}\vert \lesssim \mu ^{\frac{3}{2}}. \end{aligned}$$

Size of \(S_{\text {in}}(1-\eta _R)\). It remains to estimate the size of \(S_{\text {in}}(1-\eta _R)\). We have

$$\begin{aligned} \vert (1-\eta _R)5U^4 \mu ^{-3/2} J[{\dot{\Lambda }}](x,t)\vert&\lesssim \frac{\mu ^{-\frac{3}{2}+l_1+2\delta }}{1+\vert y\vert ^2}(1-\eta _{R}) \end{aligned}$$
(2.32)

Then,

$$\begin{aligned} \vert (1-\eta _R)\mu ^{-5/2}\mathcal {N}_3\vert&\lesssim (1-\eta _R)\mu ^{-1/2}\frac{1}{1+\vert y\vert ^3} \left( \vert H_\gamma (x,\xi )\vert +\vert J(x,t)\vert +\mu ^{-1}\phi _3 \eta _l\right) ^2\\&\lesssim (1-\eta _R)\frac{\mu ^{-1/2}R^{-1}}{1+\vert y\vert ^2}\left( 1+\mu ^{2l_1}+\mu ^2\right) \\&\lesssim (1-\eta _R)\frac{1}{1+\vert y\vert ^2}\mu ^{-1/2+\delta }. \end{aligned}$$

In particular we observe that this is smaller than (2.32), thanks to (2.28). Also,

$$\begin{aligned} \mu ^{-3/2}\eta _l ({\dot{\xi }}_1+(1-\mu ^{-1}\mu _0){\dot{\xi }}_0)\cdot \nabla _y U(y)(1-\eta _R)&\lesssim \frac{\mu ^{\frac{1}{2}+\min \{0,k-\frac{1}{2}\}}}{1+\vert y\vert ^2}(1-\eta _R), \end{aligned}$$

and

$$\begin{aligned}&\vert *\vert {(1-\eta _R)\eta _l 5U(y)^4 \mu ^{-3/2}\left( h_\gamma (x,\xi )-\left( \frac{\mu _0}{\mu }\right) \left( \frac{1}{2}\mu _0 y\cdot \nabla _x R_\gamma (0)\right) \right) }\\&\quad \lesssim \frac{R^{-2}\mu ^{-1/2}}{1+\vert y\vert ^2}(1-\eta _R)\\&\quad \lesssim \frac{\mu ^{2\delta -\frac{1}{2}}}{1+\vert y\vert ^2}(1-\eta _R). \end{aligned}$$

Combining these estimates we find

$$\begin{aligned} \vert S_{\text {in}}(1-\eta _R)\vert&\lesssim \mu ^{-\frac{3}{2}+l_1+2\delta }\frac{1}{1+\vert y\vert ^2}(1-\eta _R(\vert y\vert )). \end{aligned}$$

We conclude that

$$\begin{aligned} \vert S_{\text {in}}(1-\eta _R)+S_{\text {out}}\vert \lesssim \mu ^{-\frac{3}{2}+l_1+2\delta }\frac{1}{1+\vert y\vert ^2}(1-\eta _R)+\mu ^{\frac{3}{2}}. \end{aligned}$$

2.6 Estimates of \(J_1,J_2\) and \(\phi _3\)

The following lemma gives an estimate of \(J_1[{\dot{\Lambda }}](x,t)\) in terms of \({\dot{\Lambda }}\). Observe that

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \left( \frac{\mu (t)}{\mu _0(t)}\right) ^{1/2}\left[ \frac{\mu ^2(t)-\vert x-\xi (t)\vert ^2}{ \left( \mu (t)^2+\vert x-\xi (t)\vert ^2\right) ^{3/2}}+H_\gamma (x,\xi (t))\right] =-\frac{1}{\vert x\vert }+H_\gamma (x,0), \end{aligned}$$

thus, for \(t_0\) large, we will approximate \(J_1\) with \(\mathcal {J}\), that is the solution to

$$\begin{aligned}&{\partial } _t \mathcal {J}= \Delta _x \mathcal {J}+\gamma \mathcal {J}-\dot{\Lambda }(t)G_\gamma (x,0)\quad {\quad \hbox {in } }\Omega \times [t_0-1,\infty ),\nonumber \\ {}&\mathcal {J}(x,t)=0 {\quad \hbox {on } } {\partial } \Omega \times [t_0-1,\infty ),\nonumber \\ {}&\mathcal {J}(x,t_0-1)=0 {\quad \hbox {in } }\Omega . \end{aligned}$$
(2.33)

We define the \(L^\infty \)-weighted space

$$\begin{aligned} X_c{:=}\{ f \in L^\infty (t_0-1,\infty ): \vert \!\vert f\vert \!\vert _{\infty ,c}<\infty \}, \end{aligned}$$

where

$$\begin{aligned} \vert \!\vert f\vert \!\vert _{\infty ,c}{:=}\sup _{t>t_0-1}\vert f(t)\mu _0(t)^{-c}\vert . \end{aligned}$$

Lemma 2.2

(Estimate of \(J_1\)) Suppose \(2 \gamma l_1<\lambda _1-\gamma \) and

$$\begin{aligned} \vert \!\vert {\dot{\Lambda }}\vert \!\vert _{\infty ,l_1}< \infty . \end{aligned}$$

Then we have

$$\begin{aligned} \vert \!\vert J_1(\cdot ,t)\vert \!\vert _{L^\infty (\Omega )}\lesssim \mu _0(t)^{l_1}\vert \!\vert {\dot{\Lambda }}\vert \!\vert _{\infty ,l_1}, \end{aligned}$$
(2.34)

for \(t\ge t_0\).

Since we have selected \(l_1<1\) in (2.31), condition (1.6) guarantees that \(2\gamma l_1<\lambda _1-\gamma \).

Proof

By parabolic comparison, it is enough to prove the bound for \(\mathcal {J}\) defined as the solution to (2.33). Indeed, we have

$$\begin{aligned} \left| {\left( \frac{\mu }{\mu _0}\right) ^{1/2}\dot{\Lambda }(t)\left( \frac{\mu ^2-\vert x-\xi \vert ^2}{\left( \mu ^2+\vert x-\xi \vert ^2\right) ^{3/2}}+H_\gamma (x,\xi )\right) }\right| \lesssim \left| *{{\dot{\Lambda }}(t)}\vert -\frac{1}{\vert x\vert }+H_\gamma (x,0)\vert \right| \end{aligned}$$

We decompose

$$\begin{aligned} \mathcal {J}(x,t)=\sum _{k=1}^{\infty } b_k(t)w_k(x) {\quad \hbox {in } }L^2(\Omega ), \quad \text {for}\quad t\ge t_0-1, \end{aligned}$$

where \(w_k\) is the k-th eigenfunction of \(-\Delta \) on \(\Omega \). Plugging the decomposition into the equation we find

$$\begin{aligned} b_k=c_k \int _{t_0-1}^{t} e^{-\left( \lambda _k-\gamma \right) (t-s)}\dot{\Lambda }(s) \,ds,\quad \text {where}\quad c_k{:=}-\int _\Omega G_\gamma (x,0) w_k(x)\,dx. \end{aligned}$$

In particular, we have

$$\begin{aligned} \vert \!\vert \mathcal {J}(\cdot ,t)\vert \!\vert _{L^2(\Omega )} \le \vert \!\vert G_\gamma (\cdot ,0)\vert \!\vert _{L^2(\Omega )} \int _{t_0-1}^{t} e^{-(\lambda _1-\gamma )(t-s)}\vert *\vert {{\dot{\Lambda }}(s)}\,ds. \end{aligned}$$

Using \(\vert \!\vert {\dot{\Lambda }}\vert \!\vert _{\infty ,l_1}<\infty \) and \(2\gamma l_1<\lambda _1-\gamma \) we obtain

$$\begin{aligned} \vert \!\vert \mathcal {J}(\cdot ,t)\vert \!\vert _{L^2(\Omega )}&\le \vert \!\vert G_\gamma (\cdot ,0)\vert \!\vert _{L^2(\Omega )} \vert \!\vert {\dot{\Lambda }}\vert \!\vert _{\infty ,l_1} e^{-\min \{2\gamma l_1,\lambda _1-\gamma \} (t-(t_0-1))},\\&\lesssim \vert \!\vert {\dot{\Lambda }}\vert \!\vert _{\infty ,{l_1}} e^{-2\gamma l_1 (t-(t_0-1))} \end{aligned}$$

Finally, from standard parabolic estimates, using the \(L^2\)-bound and Eq. (2.33), we get for \(t\ge t_0\)

$$\begin{aligned} \vert \!\vert \mathcal {J}(\cdot ,t)\vert \!\vert _{L^\infty (\Omega ')}\lesssim \vert \!\vert \dot{\Lambda }\vert \!\vert _{\infty ,{l_1}} e^{-2\gamma l_1 t}, \end{aligned}$$

for any \(\Omega ' \Subset \Omega \). By boundary regularity estimates this inequality can be extended to \(\Omega \) thanks to the smoothness of \( {\partial } \Omega \). \(\square \)

Lemma 2.3

(Estimate of \(J_2\)) Let \(J_2(x,t)\) be the unique solution to the problem

$$\begin{aligned} {\partial } _t J_2 =&\Delta _x J_2+ \gamma J_2-\left( \frac{\mu }{\mu _0}\right) ^{\frac{1}{2}}\bigg [\gamma \left( \mu ^{-1}2Z_4\left( \frac{x-\xi }{\mu }\right) +\frac{\alpha _3}{\vert x-\xi \vert }\right) \\&+\mu ^{-2}5U\left( \frac{x-\xi }{\mu }\right) ^4 \theta _\gamma (x-\xi )\bigg ] {\quad \hbox {in } }\Omega \times [t_0,\infty ),\\ J_2(x,t)&=0 {\quad \hbox {on } } {\partial } \Omega \times [t_0,\infty ),\\ J_2(x,t_0&)=0 {\quad \hbox {in } }\Omega . \end{aligned}$$

Suppose that \(3\gamma <\lambda _1\). Then, there exists \(t_0\) large such that

$$\begin{aligned} \vert J_2(\mu y+\xi ,t)\vert \lesssim \mu (t)\frac{1}{1+\vert y\vert ^{1-\varepsilon }}, \end{aligned}$$
(2.35)

for any \(\varepsilon >0\) and for all \((x,t)\in \Omega \times [t_0,\infty )\) where \(y=(x-\xi )/\mu \).

Proof

Firstly, we observe that

$$\begin{aligned} \left| { \frac{1-\vert y\vert ^2}{(1+\vert y\vert ^2)^{3/2}}+\frac{1}{\vert y\vert }}\right| \lesssim \frac{1}{\vert y\vert \left( 1+\vert y\vert ^{2-\varepsilon }\right) }. \end{aligned}$$

Also, by Taylor expanding the function \(\theta _\gamma \) in (2.11) near the origin, we see that

$$\begin{aligned} \vert \mu ^{-2}5U(y)^4\theta _\gamma (\mu y)\vert&\lesssim \frac{\mu ^{-1}}{1+\vert y\vert ^4} \vert y\vert \\&\lesssim \frac{\mu ^{-1}}{\vert y\vert \left( 1+\vert y\vert ^{2-\epsilon }\right) }, \end{aligned}$$

where \(\epsilon >0\) can be taken arbitrarily small. Thus, by parabolic comparison, it is enough to find a supersolution to the problem

$$\begin{aligned}&{\partial } _t u = \Delta _x u + \gamma u + \mu ^{-1} \frac{1}{\vert y(x,t)\vert (1+\vert y(x,t)\vert ^{2-\varepsilon })} {\quad \hbox {in } }{\Omega \times [t_0,\infty )},\\&u(x,t)=0 {\quad \hbox {on } } {\partial } {\Omega \times [t_0,\infty )},\\&u(x,t_0)=0 {\quad \hbox {in } }\Omega . \end{aligned}$$

Let \(v(x,t){:=} \mu (t)^{-1} u(x,t)\), which satisfies

$$\begin{aligned}&{\partial } _t v = \Delta _x v+ (3\gamma -2{{\dot{\Lambda }}})v + \frac{\mu ^{-2}}{\vert y(x,t)\vert \left( 1+\vert y(x,t)\vert ^{2-\varepsilon }\right) } {\quad \hbox {in } }{\Omega \times [t_0,\infty )},\\ {}&v= 0 {\quad \hbox {on } } {\partial } \Omega \times [t_0,\infty ),\\ {}&v(x,t_0)= 0 {\quad \hbox {in } }\Omega . \end{aligned}$$

We look for a supersolution \({{\bar{v}}}\) of the form

$$\begin{aligned} \bar{v}(x,t)=v_0\left( \frac{x-\xi }{\mu }\right) \eta \left( \frac{x-\xi }{C_0}\right) +v_1(x,t). \end{aligned}$$

We need

$$\begin{aligned} {\partial } _t v_1-\Delta _x v_1-(3\gamma -{\dot{\Lambda }})v_1\ge \eta \bigg [&- {\partial } _t v_0+\mu ^{-2}\Delta _y v_0+(3\gamma -{\dot{\Lambda }})v_0\nonumber \\&+\frac{\mu ^{-2}}{\vert y\vert (1+\vert y\vert ^{2-\varepsilon })} \bigg ]\nonumber \\&+(1-\eta )\frac{\mu ^{-2}}{\vert y\vert (1+\vert y\vert ^{2-\varepsilon })}+\left( \Delta _x \eta - {\partial } _t \eta \right) v_0\nonumber \\ {}&+2\mu ^{-1}\nabla _x \eta \cdot \nabla _y v_0, \end{aligned}$$
(2.36)

with \(v_1\ge 0\) on \( {\partial } \Omega \times [t_0,\infty )\) and \(v_0(y(x,t_0))\ge 0\) for \(x\in \Omega \). Without loss of generality let \(\Omega \subset B_1\). Consider the positive radial solution \(v_0(\vert y\vert ,t)\) to

$$\begin{aligned}&\Delta _y v_0 +2\frac{1}{\vert y\vert \left( 1+\vert y\vert ^{2-\varepsilon }\right) }=0 {\quad \hbox {on } }B_{\frac{1}{\mu (t)}},\\&v_0\equiv 0 {\quad \hbox {on } } {\partial } B_{\frac{1}{\mu (t)}}, \end{aligned}$$

given by the formula of variation of parameters

$$\begin{aligned} v_0(\vert y\vert ,t)&=2\omega _3\int _{\vert y\vert }^{\frac{1}{\mu (t)}} \frac{1}{\rho ^2} \int _{0}^{\rho }\frac{s}{1+s^{2-\varepsilon }}\,ds\, d\rho . \end{aligned}$$

From this formula we obtain the following estimates in \((x,t)\in \Omega \times [t_0,\infty )\):

$$\begin{aligned}&\vert v_0(\vert y\vert ,t)\vert +\vert {\partial } _t v_0(\vert y\vert ,t)\vert \lesssim \frac{1}{1+\vert y\vert ^{1-\varepsilon }}, \end{aligned}$$

Thus, if \(\vert x-\xi \vert <C_0\), for \(C_0\) sufficiently small, then

$$\begin{aligned}&- {\partial } _t v_0+\mu ^{-2}\Delta _y v_0+(3\gamma -\dot{\Lambda })v_0+\frac{\mu ^{-2}}{\vert y\vert (1+\vert y\vert ^{2-\varepsilon })} \\ {}&\quad =-\frac{\mu ^{-2}}{\vert y\vert (1+\vert y\vert ^{2-\varepsilon })}+O\left( \frac{1}{1+\vert y\vert ^{1-\varepsilon }}\right) \le 0. \end{aligned}$$

Then, let \(v_1\) be the solution to

$$\begin{aligned} {\partial } _t v_1-\Delta _x v_1-(3\gamma -{\dot{\Lambda }})v_1&=(1-\eta )\frac{\mu ^{-2}}{\vert y\vert (1+\vert y\vert ^{2-\varepsilon })}+(\Delta _x \eta - {\partial } _t \eta )v_0\\ {}&\quad +2\mu ^{-1}\nabla _x \eta \cdot \nabla _y v_0 {\quad \hbox {in } }{\Omega \times [t_0,\infty )}, \end{aligned}$$

with

$$\begin{aligned}&v_1(x,t)=0 {\quad \hbox {on } } {\partial } \Omega \times [t_0,\infty ),\\&v_1(x,t_0)=0 {\quad \hbox {in } }\Omega . \end{aligned}$$

In the right-hand side we have

$$\begin{aligned}&(1-\eta )\frac{\mu ^{-2}}{\vert y\vert (1+\vert y\vert ^{2-\varepsilon })}\lesssim \mu ^{1-\varepsilon }, \\&\vert (\Delta _x- {\partial } _t \eta )v_0\vert \lesssim \mu ^{1-\varepsilon },\\&\vert 2\mu ^{-1}\nabla _x \eta \cdot \nabla _y v_0\vert \lesssim \mu ^{1-\varepsilon }. \end{aligned}$$

Since \(3\gamma -{\dot{\Lambda }}(t)<\lambda _1\) provided that \(t_0\) is sufficiently large, the comparison principle applies and we get \(\vert v_0\vert \lesssim \mu ^{1-\varepsilon }\). Thus, we verified inequality (2.36). Also, we have \(v=v_1\ge 0\) on \( {\partial } \Omega \times [t_0,\infty )\) and \(\eta v_0(y(x,t_0))\ge 0\). Thus, v is a supersolution, and going back to the original function \(u=\mu v\) we get estimate (2.35) for \(J_2\). \(\square \)

Lemma 2.4

(Estimate on \(\phi _3\)) Let \(\mathcal {M}[\xi _0,\mu _0]\) be defined as in (2.20). If the orthogonality conditions (2.22) on \(\mathcal {M}[\xi _0,\mu _0]\) hold, then there exists a bounded solution to the problem

$$\begin{aligned} \Delta _y \phi _3 + 5U(y)^4 \phi _3(y,t)=-\mathcal {M}[\xi _0,\mu _0](y,t) {\quad \hbox {in } }{{\mathbb {R}}}^3. \end{aligned}$$
(2.37)

We have the following estimates on \(\phi _3\) and its derivatives:

$$\begin{aligned} \vert \phi _3(y,t)\vert +(1+\vert y\vert )\vert \nabla _y \phi _3(y,t)\vert +\vert {\partial } _t \phi _3(y,t)\vert \lesssim \mu ^2(t) f(y,t), \end{aligned}$$
(2.38)

where f is a smooth bounded function.

Proof

From the explicit form of the function \(\mathcal {M}\) given in (2.20) we estimate its size by

$$\begin{aligned} \vert \mathcal {M}[\mu _0,\xi _0](y,t)\vert&\le \mu ^2 \frac{1}{1+\vert y\vert ^2}, \end{aligned}$$

and we observe that \(\mathcal {M}\) has only modes \(i=1,2,3\). Thus, we decompose \(\phi _3\) in such modes:

$$\begin{aligned} \phi _3(y)=\sum _{i=1}^3 \phi _{3,i}(r) \vartheta _i(y/r),\quad r{:=}\vert y\vert ,\quad \phi _{3,i}(r){:=}\int _{S^2}\phi _3(r\theta )\vartheta _i(\theta )\, d\theta . \end{aligned}$$

Similarly, we define

$$\begin{aligned}z_i(r){:=}\int _{S^2}Z_i(r\theta )\vartheta _i(\theta ) \,d\theta .\end{aligned}$$

The formula of variation of constants gives

$$\begin{aligned} \phi _{3,i}(r)=z_i(r)\int _0^r \frac{1}{\rho ^2 z_i(\rho )^2}\mathcal {I}_i(\rho ) \,d\rho , \end{aligned}$$

where

$$\begin{aligned} \mathcal {I}_i(\rho ){:=}\int _{0}^{\rho } M_i(s)z_i(s) s^2 \,ds, \end{aligned}$$

and

$$\begin{aligned} M_i(r){:=}\int _{S^2}\mathcal {M}(r\theta )\vartheta _i(\theta ) \,d\theta . \end{aligned}$$

Since

$$\begin{aligned} \vert M_i(r)\vert \lesssim \frac{1}{1+r^2}, \end{aligned}$$

and

$$\begin{aligned} \vert z_i(r)\vert \lesssim \frac{r}{(1+r^3)}, \end{aligned}$$

we deduce

$$\begin{aligned} \vert \mathcal {I}_i(\rho )\vert \lesssim \rho ^4 \quad \text{ as }\quad \rho \rightarrow 0. \end{aligned}$$

Also, by the orthogonality conditions (2.22) we have

$$\begin{aligned} \vert \mathcal {I}_i(\rho )\vert&=\vert \int _{\rho }^{\infty } M_i(s)z_i(s)s^2 \,ds\vert \\&\lesssim \frac{1}{\rho } \quad \text{ as }\quad \rho \rightarrow \infty . \end{aligned}$$

With these estimates we conclude

$$\begin{aligned} \vert \phi _3(r)\vert&\lesssim \frac{r}{1+r^3} \int _{0}^{r}\frac{ \left( 1+\rho ^2\right) ^3}{\rho ^4}\vert \mathcal {I}(\rho )\vert \,d\rho \\&\lesssim \frac{r}{1+r^3} \int _{0}^{r}\frac{\left( 1+\rho ^2\right) ^3}{\rho ^4} \frac{\rho ^4}{1+\rho ^5}\,d\rho \\&\lesssim 1. \end{aligned}$$

Similarly, taking the space and time derivatives of Eq. (2.37), we deduce the bounds on \(\nabla _y \phi _3\) and \( {\partial } _t \phi _3\). \(\square \)

We conclude this section with summarizing the estimates of the error \(S[u_3]\).

Lemma 2.5

Let \(3\gamma <\lambda _1\), \(\mu =\mu _0 e^{2\Lambda }\) and \(\xi =\xi _0+\xi _1\), where \(\mu _0,\xi _0\) are given by (2.5) and (2.23) respectively. Assume

$$\begin{aligned}&\vert \Lambda (t)\vert \lesssim \mu _0(t)^{l_0}, \quad \vert \dot{\Lambda }(t)\vert \lesssim \mu _0(t)^{l_1},\\ {}&R(t)=\mu ^{-\delta },\quad \vert {\dot{\xi }}_1(t)\vert \lesssim \mu _0^{1+k}, \end{aligned}$$

for positive constant \(\delta ,l_0,l_1,k\) satisfying (2.27),(2.28), (2.29) and (2.30). Then, setting \(x=\mu y+\xi \), for \(t_0\) sufficiently large the following estimate on the error function \(S[u_3]\) holds:

$$\begin{aligned} S[u_3](y,t)=S_{\text {in}}(y,t) \eta _{R(t)}\left( \vert y\vert \right) + S_{\text {in}}(y,t) (1-\eta _{R(t)}(\vert y\vert ))+S_{\text {out}}(y,t), \end{aligned}$$

where

$$\begin{aligned}&\vert S_{\text {in}}(y,t)\eta _{R(t)}\vert \lesssim \mu ^{-\frac{3}{2}+l_1}\frac{1}{1+\vert y\vert ^4},\\&\vert S_{\text {out}}(y,t)\vert \lesssim \mu ^{\frac{3}{2}},\\&\vert S_{\text {in}}(y,t)(1-\eta _{R(t)})\vert \lesssim \mu ^{-\frac{3}{2}+l_1+2\delta }\frac{1}{1+\vert y\vert ^2}. \end{aligned}$$

3 The inner-outer scheme

We recall that our final purpose is to find an unbounded global in time solution u to (2.1) of the form

$$\begin{aligned} u=u_3+{{\tilde{\phi }}}, \end{aligned}$$
(3.1)

for a small perturbation \({{\tilde{\phi }}}\). The latter is constructed by means of the inner-outer gluing method. This consists in looking for a perturbation of the form

$$\begin{aligned} \tilde{\phi }(x,t)=\mu _0(t)^{1/2}\psi (x,t)+\eta _{R(t)}\left( \vert y\vert \right) \mu (t)^{-1/2}\phi \left( y,t\right) , \end{aligned}$$
(3.2)

where

$$\begin{aligned} \eta _{R(t)}\left( \vert y\vert \right) =\left( \frac{\vert y\vert }{R(t)}\right) ,\quad y{:=}y(x,t){:=}\frac{x-\xi (t)}{\mu (t)}, \end{aligned}$$

and \(\eta (s)\) is a cut-off function with \({{\,\textrm{supp}\,}}\eta \subset [0,2]\) and \(\eta \equiv 1\) in [0, 1]. We have already chosen \(R=R(t)\) in (2.26). In terms of \({{\tilde{\phi }}}\) the equation reads as

$$\begin{aligned} 0=S[u]&=- {\partial } _t u+ \Delta _x u +u^5\\&= \left( - {\partial } _t u_3+\Delta _x u_3+u_3^5\right) - {\partial } _t {{\tilde{\phi }}} +\Delta _x {{\tilde{\phi }}} +(u_3+{{\tilde{\phi }}})^4-u_3^5\\&=S[u_3]- {\partial } _t{{\tilde{\phi }}}+\Delta _x {{\tilde{\phi }}} + 5u_3^4 {{\tilde{\phi }}} +\mathcal {N}(u_3,{{\tilde{\phi }}}) \end{aligned}$$

where

$$\begin{aligned} \mathcal {N}(u_3,{{\tilde{\phi }}}){:=}(u_3+{{\tilde{\phi }}})^5-u_3^5 -5u_3^4 {{\tilde{\phi }}}. \end{aligned}$$
(3.3)

Hence the problem for \({{\tilde{\phi }}}\) is

$$\begin{aligned}&{\partial } _t {{\tilde{\phi }}}= \Delta _x {{\tilde{\phi }}}+ 5u_3^4 {{\tilde{\phi }}} +S[u_3] +\mathcal {N}(u_3,{\tilde{\phi }}) {\quad \hbox {in } }\Omega \times [t_0,\infty ),\\&{{\tilde{\phi }}} = -u_3 {\quad \hbox {on } } {\partial } \Omega \times [t_0,\infty ). \end{aligned}$$

Now, the main idea is to split the problem for \({{\tilde{\phi }}}\) in a system for \((\psi ,\phi )\), localizing the inner regime. We divide the error in

$$\begin{aligned} S[u_3]=&S_{\text {in}}\eta _R+S_{\text {in}}(1-\eta _R)+S_{\text {out}}, \end{aligned}$$

where \(S_{\text {in}}, S_{\text {out}}\) are defined in (2.24) and (2.25) respectively. Considering \({{\tilde{\phi }}}\) as in (3.2) we compute

$$\begin{aligned} {\partial } _t {{\tilde{\phi }}}&= \frac{{\dot{\mu }}_0}{2\mu _0}\mu _0^{1/2}\psi +\mu _0^{1/2} {\partial } _t \psi + \mu ^{-1/2} \phi {\partial } _t \eta \left( \frac{y(x,t)}{R(t)}\right) -\frac{{\dot{\mu }}}{2\mu }\mu ^{-1/2}\phi \eta _R \\&\quad +\mu ^{-1/2}\left( {\partial } _t \phi + \nabla _y \phi \cdot {\partial } _t y(x,t)\right) \eta _R\\&=-\gamma \mu _0^{1/2}\psi +\mu _0^{1/2} {\partial } _t \psi + \mu ^{-1/2} \phi \left[ \nabla _z \eta \left( \frac{y}{R}\right) \cdot \left( -\frac{\dot{R}}{R}\frac{y}{R}-\frac{{\dot{\mu }}}{\mu }\frac{y}{R} -\frac{{\dot{\xi }}}{\mu R}\right) \right] \\&\quad +\left( -\frac{{\dot{\mu }}}{2\mu }\right) \mu ^{-1/2}\phi \eta _R +\mu ^{-1/2}\eta _R \left( {\partial } _t \phi + \nabla _y \phi \cdot \left( -\frac{{\dot{\mu }}}{\mu }y-\frac{{\dot{\xi }}}{\mu }\right) \right) , \end{aligned}$$

and

$$\begin{aligned} \Delta _x {{\tilde{\phi }}}&= \mu _0^{1/2}\Delta _x \psi + \mu ^{-1/2}\Delta _x \left( \phi (y(x,t),t)\eta _{R(t)}(y(x,t))\right) \\&=\mu _0^{1/2}\Delta _x \psi + \mu ^{-5/2}\eta _R(y)\Delta _y \phi (y,t) + \mu ^{-1/2}\phi \left( \frac{2}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu ^2 R^2}+\frac{\eta ''\left( \vert z\vert \right) }{\mu ^2 R^2}\right) \\&\quad + 2\mu ^{-1/2}\frac{1}{\mu }\nabla _y \phi (y,t) \cdot \frac{z}{\vert z\vert } \frac{\eta '(\vert z\vert )}{\mu R}, \end{aligned}$$

where \(z{:=}y/R\). We split

$$\begin{aligned} 5u_3^4 {{\tilde{\phi }}}= 5u_3^4 \mu _0^{1/2}\psi \eta _R + 5 u_3^4 \mu _0^{1/2}\psi (1-\eta _R)+5u_3^4 \mu ^{-1/2} \phi \eta _R. \end{aligned}$$

Hence, the full equation becomes

$$\begin{aligned}&-\gamma \mu _0^{1/2}\psi +\mu _0^{1/2} {\partial } _t \psi +\mu ^{-1/2}\phi {\partial } _t \eta _R +\eta _R \mu ^{-1/2} {\partial } _t \phi \\ {}&\quad +\eta _R\left\{ (\gamma -{\dot{\Lambda }})\mu ^{-1/2}(\phi +2 \nabla _y \phi \cdot y) -\mu ^{-1/2}\nabla _y \phi \cdot \left( \frac{{\dot{\xi }}}{\mu }\right) \right\} \\&\qquad =\mu _0^{1/2}\Delta _x \psi +\mu ^{-5/2}\eta _R\Delta _y \phi +\mu ^{-1/2}\phi \left( \frac{2}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu ^2 R^2}+\frac{\eta ''\left( \vert z\vert \right) }{\mu ^2 R^2}\right) \\&\qquad \quad +2\mu ^{-1/2}\frac{1}{\mu }\nabla _y \phi \cdot \frac{z}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu R} \\&\qquad \quad +5u_3^4 \mu _0^{1/2}\psi \eta _R + 5u_3^4 \mu _0^{1/2}\psi (1-\eta _R) + 5u_3^4 \mu ^{-1/2}\phi \eta _R\\&\qquad \quad +S_{\text {in}}\eta _R+S_{\text {in}}\left( 1-\eta _R\right) +S_{\text {out}}\\&\qquad \quad +\mathcal {N}(u_3,{{\tilde{\phi }}})(1-\eta _R) +\mathcal {N}(u_3,\tilde{\phi })\eta _R. \end{aligned}$$

We divide the full problem in a system. Firstly, we look for a solution \(\psi \) to

$$\begin{aligned} \mu _0^{1/2} {\partial } _t \psi =&\mu _0^{1/2}\Delta _x \psi +\gamma \mu _0^{1/2}\psi +5u_3^4 \mu _0^{1/2}\psi (1-\eta _R)+\mu ^{-1/2}\phi {\partial } _t \eta _R \nonumber \\ {}&+ \eta _R\left\{ (\gamma -\dot{\Lambda })\mu ^{-1/2}(\phi +2 \nabla _y \phi \cdot y) -\mu ^{-1/2}\nabla _y \phi \cdot \left( \frac{{\dot{\xi }}}{\mu }\right) \right\} \nonumber \\&+\mu ^{-1/2}\phi \left( \frac{2}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu ^2 R^2}+\frac{\eta ''\left( \vert z\vert \right) }{\mu ^2 R^2}\right) +2\mu ^{-1/2}\frac{1}{\mu }\nabla _y \phi \cdot \frac{z}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu R} \nonumber \\&+S_{\text {in}}\left( 1-\eta _R\right) +S_{\text {out}}+\mathcal {N}(u_3,\tilde{\phi })(1-\eta _R), \quad {\quad \hbox {in } }\Omega \times [t_0,\infty )\nonumber \\ \psi (x,t)=&-\mu _0^{-1/2}u_3(x,t)\quad {\quad \hbox {on } } {\partial } \Omega \times [t_0,\infty ). \end{aligned}$$

Thus, after dividing by \(\mu _0^{1/2}\), \(\psi \) solves the outer problem

$$\begin{aligned} {\partial } _t \psi =&\Delta _x \psi +\gamma \psi +5u_3^4 \psi (1-\eta _R)+\mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\phi {\partial } _t \eta _R \nonumber \\ {}&+ \mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\eta _R\left\{ (\gamma -\dot{\Lambda })(\phi +2 \nabla _y \phi \cdot y) -\nabla _y \phi \cdot \left( \frac{{\dot{\xi }}}{\mu }\right) \right\} \nonumber \\&+\mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\left( \phi \left( \frac{2}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu ^2 R^2}+\frac{\eta ''\left( \vert z\vert \right) }{\mu ^2 R^2}\right) +2\frac{\nabla _y \phi }{\mu } \cdot \frac{z}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu R}\right) \nonumber \\&+\mu _0^{-1/2}S_{\text {in}}\left( 1-\eta _R\right) +\mu _0^{-1/2}S_{\text {out}}+\mu _0^{-1/2}\mathcal {N}(u_3,\tilde{\phi })(1-\eta _R), \nonumber \\ {}&\quad {\quad \hbox {in } }\Omega \times [t_0,\infty )\nonumber \\ \psi (x,t)=&-\mu _0^{-1/2}u_3(x,t)\quad {\quad \hbox {on } } {\partial } \Omega \times [t_0,\infty ), \end{aligned}$$
(3.4)

Then, \(\phi \) has to solve the problem

$$\begin{aligned} \mu ^{-1/2} {\partial } _t \phi&= \mu ^{-5/2}\Delta _y \phi + 5u_3^4 \mu ^{-1/2}\phi + 5u_3^4 \mu _0^{1/2}\psi +S_{\text {in}}+\mathcal {N}(u_3,{{\tilde{\phi }}}) \\ {}&\quad {\quad \hbox {in } }B_{2R}(0)\times [t_0,\infty ). \end{aligned}$$

Equivalently, multiplying by \(\mu ^{5/2}\), \(\phi \) solves

$$\begin{aligned} \mu ^2 {\partial } _t \phi =&\Delta _y \phi + 5U^4 \phi +5U^4 \left( \frac{\mu _{0}}{\mu }\right) ^{1/2}\mu \psi (\mu y+\xi ,t)+ B_0\left[ \phi +\mu \psi \right] (\mu y+\xi ,t)\nonumber \\&+\mu ^{5/2}S_{\text {in}}(\mu y+\xi ,t)\nonumber \\&+\mathcal {N}(\mu ^{1/2}u_3,\mu ^{1/2}{{\tilde{\phi }}})(\mu y+\xi ,t) {\quad \hbox {in } }B_{2R}(0)\times [t_0,\infty ), \end{aligned}$$
(3.5)

where \(B_0\) is the linear operator

$$\begin{aligned} B_0[f]{:=}5\left[ \left( U-\mu H_\gamma +\mu J[\dot{\Lambda }]+\mu ^{-1/2}\phi _3(y,t)\eta _3\right) ^4-U^4\right] f, \end{aligned}$$
(3.6)

3.1 General strategy for solving the inner-outer system

We now describe the method we use to solve system (3.4)–(3.5). Firstly, for fixed parameters \(\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }}\) and inner function \(\phi \) in suitable weighted spaces, we solve problem (3.4) in \(\psi =\psi [\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi ]\). This is done in Sect. 4. We insert such \(\psi \) in the inner problem. At this point we need to find \(\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }}\) and \(\phi \). We make the change of variable \(t(\tau )\) defined by the ODE

$$\begin{aligned}&\frac{dt(\tau )}{d\,\tau }=\mu ^2(t(\tau ))\\&t(\tau _0)=t_0, \end{aligned}$$

which explicitly gives

$$\begin{aligned} \tau -\tau _0&=\int _{t_0}^t\frac{ds}{\mu (s)^2}\,ds\end{aligned}$$
(3.7)
$$\begin{aligned}&=\int _{t_0}^t\frac{ds}{\mu _0(s)^2}(1+o(1))\,ds\end{aligned}$$
(3.8)
$$\begin{aligned}&=\frac{1}{4\gamma }\mu _0(t)^{-2}(1+o(1)). \end{aligned}$$
(3.9)

Expressing Eq. (3.5) in the new variables \((y,\tau )\) we get the inner problem

$$\begin{aligned} {\partial } _\tau \phi = \Delta _y \phi + 5U^4 \phi +H[\phi ,\psi ,\Lambda ,\dot{\Lambda },\xi ,{\dot{\xi }}](y,\tau ) {\quad \hbox {in } }B_{2R}\times [\tau _0,\infty ), \end{aligned}$$
(3.10)

where

$$\begin{aligned} H[\phi ,\psi ,\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }}](y,\tau )&:=5U(y)^4 \mu \left( \frac{\mu _0}{\mu }\right) ^{1/2} \psi (\mu y +\xi ,t(\tau ))\nonumber \\ {}&\quad +B_0\left[ \phi +\mu \psi \right] (\mu y+\xi ,t(\tau ))+\mu ^{5/2}S_{\text {in}}(\mu y+\xi ,t(\tau ))\nonumber \\&\quad +\mathcal {N}(\mu ^{1/2}u_3,\mu ^{1/2}{{\tilde{\phi }}})(\mu y+\xi ,t(\tau )). \end{aligned}$$
(3.11)

Let \(Z_0\) be the positive radially symmetric bounded eigenfunction associated to the only negative eigenvalue \(\lambda _0\) of the problem

$$\begin{aligned} -\Delta _y \phi - 5U(y)^4 \phi =\lambda _0\phi \quad \text {for}\quad \phi \in L^\infty ({{\mathbb {R}}}^3). \end{aligned}$$

It is known that \(\lambda _0\) is simple and

$$\begin{aligned} Z_0(y)\sim \frac{e^{-\sqrt{\vert \lambda _0\vert }\vert y\vert }}{\vert y\vert }\quad \text{ as }\quad \vert y\vert \rightarrow \infty . \end{aligned}$$

We solve (3.10) with a multiple of \(Z_0(y)\) as initial datum, namely

$$\begin{aligned} \phi (\tau _0,y)=e_0 Z_0(y) {\quad \hbox {in } }B_{2R}, \end{aligned}$$
(3.12)

for some constant \(e_0=e_0[H]\) to be found. Formally, this initial datum (3.12) allows \(\phi \) to remain small along its trajectory. Indeed, multiplying (3.5) by \(Z_0\) and integrating we obtain

$$\begin{aligned} \mu ^2 {\partial } _t p(t)+\lambda _0 p(t)=q(t), \end{aligned}$$

where

$$\begin{aligned} p(t):=\int _{{{\mathbb {R}}}^3} \phi (y,t)Z_0(y) \,dy,\quad q(t):=\int _{{{\mathbb {R}}}^3}h(y,t)Z_0\,dy. \end{aligned}$$

The general solution p(t) is given by

$$\begin{aligned} p(t)=e^{\vert \lambda _0\vert \int _0^t \mu (s)^{-2}\,ds} \left( p(t_0)+\int _{t_0}^{t}\mu (s)^{-2} q(s) e^{-\vert \lambda _0\vert \mu (s)^{-2}}\,ds\right) . \end{aligned}$$

This shows that in order to get a decaying solution p(t) (and hence \(\phi (y,t)\)), the following initial conditions should hold:

$$\begin{aligned} p(t_0)= \int _{{{\mathbb {R}}}^3}\phi (y,t_0)Z_0(y)\,dy= -\int _{t_0}^{\infty } \mu (s)^{-2} q(s) e^{-\vert \lambda _0\vert \mu (s)^{-2}}\,ds. \end{aligned}$$

This argument formally suggests that, to avoid the instability caused by \(Z_0\), the small initial value \(\phi (y,t_0)\) needs to be constrained along \(Z_0\).

Another important observation is that, in order to solve the problem (3.10)–(3.12) we need to constrain the right-hand side H to be orthogonal to \(\{Z_i\}_{i=1}^4\). Namely we need

$$\begin{aligned} \int _{B_{2R}} H(y,\tau )Z_i(y)\,dy=0 \quad \text {for} \quad \tau \in [\tau _0,\infty ) \quad \text {and}\quad i=1,2,3,4. \end{aligned}$$
(3.13)

Indeed, the elliptic kernel generated by \(\{Z_i\}_{i=1}^4\) is a subset of the kernel of the parabolic operator

$$\begin{aligned} \mu ^2 {\partial } _t \phi = \Delta _y \phi +5 U(y)^4 \phi . \end{aligned}$$

Hence, we expect to have solvability of the inhomogeneous problem (3.10) with suitable space-time decay if the orthogonality conditions (3.13) are satisfied.

As we shall see in Sect. 5, condition (3.13) with index \(i=4\) is equivalent to a nonlocal problem in \(\Lambda \), for fixed \(\phi ,\xi \). Such operator is similar to an half-derivative in the sense of Caputo [6], and we develop an invertibility theory in Sect. 8. In Sect. 5 we solve (3.13) by fixed-point argument and hence we find \(\Lambda ,\xi \). A main ingredient of the full proof is the linear theory for the inner problem developed in [11] and adapted in dimension 3 in [15].

3.1.1 Statement of the linear estimate for the inner problem

We recall the result on the linear theory in dimension 3, proved in [15]. To state the result, we decompose a general function \(h(\cdot ,\tau )\in L^2(B_{2R})\) for any \(\tau \in [\tau _0,\infty )\) in spherical modes. Let \(\{\vartheta _m\}_{m=0}^\infty \) the orthonormal basis of \(L^2(S^2)\) made up of spherical harmonics, namely the eigenfunctions of the problem

$$\begin{aligned} \Delta _{S^2}\vartheta _m +\lambda _m \vartheta _m=0{\quad \hbox {in } }S^2, \end{aligned}$$

where \(0=\lambda _0<\lambda _1=\lambda _2=\lambda _3=2<\lambda _4\le \cdots \). We decompose h into the form

$$\begin{aligned} h(y,\tau )=\sum _{m=1}^{\infty } h_m(\vert y\vert ,\tau )\vartheta _m\left( \frac{y}{\vert y\vert }\right) ,\quad h_j(\vert y\vert ,\tau )=\int _{S^2}h(r\theta ,\tau )\vartheta _m(\theta )\, d \theta . \end{aligned}$$

Furthermore, we write \( h=h^0+h^1+h^\perp \) where

$$\begin{aligned} h^0=h_0\left( \vert y\vert ,\tau \right) ,\quad h^1=\sum _{m=1}^{3}h_m\left( \vert y\vert ,\tau \right) \vartheta _m\left( \frac{y}{\vert y\vert }\right) , \quad h^\perp = \sum _{m=4}^{\infty } h_m\left( \vert y\vert ,\tau \right) \vartheta _m\left( \frac{y}{\vert y\vert }\right) . \end{aligned}$$

We solve the inner problem (3.15) for functions h in the space \(X_{\nu ,2+a}\) defined by

$$\begin{aligned} X_{\nu ,2+a}{:=}\{ h\in L^\infty \left( B_{2R}\times [\tau _0,\infty )\right) : \vert \!\vert h\vert \!\vert _{\nu ,2+a}<\infty \}, \end{aligned}$$
(3.14)

where

$$\begin{aligned} \vert \!\vert h\vert \!\vert _{\nu ,2+a}{:=}\sup _{\tau >\tau _0, y \in B_{2R}}\tau ^{\nu } (1+\vert y\vert ^{2+a})\vert h(y,\tau )\vert . \end{aligned}$$

Proposition 3.1

Let \(\nu ,a\) be positive constants. Then for all sufficiently large \(R>0\) and any \(h(y,\tau )\) with \(\vert \!\vert h\vert \!\vert _{\nu ,2+a}<\infty \) such that

$$\begin{aligned} \int _{B_{2R}} h(y,\tau )Z_j(y)\,dy=0 {\quad \hbox {in } }[\tau _0,\infty ), \quad \text {for}\quad i=1,2,3,4, \end{aligned}$$

there exist \(\phi [h]\) and \(e_0[h]\) which solves

$$\begin{aligned}&{\partial } _\tau \phi = \Delta _ y\phi + 5U(y)^4 \phi +h(y,\tau ){\quad \hbox {in } }B_{2R}\times (\tau _0,\infty )\nonumber \\ {}&\phi (y,\tau _0)=e_0 Z_0(y) {\quad \hbox {in } }B_{2R}. \end{aligned}$$
(3.15)

They define linear operators of h that satisfy the estimates

$$\begin{aligned}&\vert \phi (y,\tau )\vert +(1+\vert y\vert )\vert \nabla _y \phi (y,\tau )\vert \nonumber \\&\quad \lesssim \tau ^{-\nu }\bigg [ \frac{R^2 \theta _0(R,a)}{1+\vert y\vert ^3}\vert \!\vert h^0\vert \!\vert _{\nu ,2+a}+ \frac{R^3 \theta _1(R,a)}{1+\vert y\vert ^4}\vert \!\vert h^1\vert \!\vert _{\nu ,2+a}+\frac{1}{1+\vert y\vert ^a}\vert \!\vert h^{\perp }\vert \!\vert _{\nu ,2+a} \bigg ], \end{aligned}$$
(3.16)

and

$$\begin{aligned} \vert e_0[h]\vert \lesssim \vert \!\vert h\vert \!\vert _{\nu ,2+a}, \end{aligned}$$

where

$$\begin{aligned} \theta _0(R,a) {:=} \left\{ \begin{matrix} 1 &{} \textrm{if } a> 2, \\ \log R &{} \textrm{if } a= 2, \\ R^{2-a} &{} { \mathrm if } a< 2, \end{matrix}\right. , \quad \theta _1(R,a) {:=} \left\{ \begin{matrix} 1 &{} \textrm{if } a> 1, \\ \log R &{} \textrm{if } a= 1, \\ R^{1-a} &{} { \mathrm if } a < 1. \end{matrix}\right. \end{aligned}$$

As we said in Sect. 2.3, in order to make the system for \((\phi ,\psi )\) weakly coupled, \(\phi \) needs to be small at distance \(y\sim R\). For this reason, we need to take \(a>1\) in the statement of Proposition 3.1. This makes clear why we need to improve ansatz \(u_1\) to \(u_3\) in Sect. 2. Since in our problem \(h=H\) as in (3.10) decays as

$$\begin{aligned}\vert h\vert \lesssim \mu ^{1+l_1} \frac{1}{1+\vert y\vert ^{4}}=\tau ^{-\nu } \frac{1}{1+\vert y\vert ^{4}},\end{aligned}$$

where \(\tau \) is given in (3.7), we apply estimate (3.16) with constants

$$\begin{aligned} a=2,\quad \nu =\frac{1+l_1}{2}, \end{aligned}$$

in the simplified form

$$\begin{aligned} \vert \phi \vert +(1+\vert y\vert )\vert \nabla _y \phi (y,\tau )\vert \lesssim \vert \!\vert h\vert \!\vert _{\nu ,4}\tau ^{-\nu }\left[ \frac{R^2 \log (R)}{1+\vert y\vert ^3}+\frac{R^3}{1+\vert y\vert ^4}\right] , \end{aligned}$$
(3.17)

and observe that

$$\begin{aligned} \left[ \frac{R^2 \log (R)}{1+\vert y\vert ^3}+\frac{R^3}{1+\vert y\vert ^4}\right] \lesssim \left\{ \begin{matrix} R^{-1}\log R &{} \textrm{if } \vert y\vert \sim R, \\ R^3 &{} \textrm{if } \vert y\vert \sim 0. \end{matrix}\right. \end{aligned}$$

We look for \(\phi \) in the space of functions

$$\begin{aligned} X_*{:=}\{ \phi (y,t)\in L^\infty (\Omega \times [t_0,\infty )): \vert \!\vert \phi \vert \!\vert _*<\infty \}, \end{aligned}$$

where

$$\begin{aligned} \vert \!\vert \phi \vert \!\vert _{*} :=&\sup _{\tau \in [\tau _0,\infty ), y \in B_{2R} } \tau ^{\nu }\left[ \frac{R^2 \log (R)}{1+\vert y\vert ^3}+\frac{R^3}{1+\vert y\vert ^4}\right] ^{-1}\nonumber \\ {}&\times \left[ \vert \phi (y,\tau )\vert +(1+\vert y\vert )\vert \nabla _y \phi (y,\tau )\vert \right] \nonumber \\ {}&+\sup _{\begin{array}{c} \tau \in [\tau _0,\infty ), y \in B_{2R} \\ \tau _1,\tau _2 \in [\tau ,\tau +1] \end{array}} \tau ^{\nu }\left[ \frac{R^2 \log (R)}{1+\vert y\vert ^3}+\frac{R^3}{1+\vert y\vert ^4}\right] ^{-1}\frac{\vert \phi (y,\tau _1)-\phi (y,\tau _2)\vert }{\vert \tau _1-\tau _2\vert ^{\frac{1}{2}+\varepsilon }}\nonumber \\&+\sup _{\begin{array}{c} \tau \in [\tau _0,\infty ), y \in B_{2R} \\ \tau _1,\tau _2 \in [\tau ,\tau +1] \end{array}}\tau ^{\nu }\left[ \frac{R^2 \log (R)}{1+\vert y\vert ^3}+\frac{R^3}{1+\vert y\vert ^4}\right] ^{-1}\nonumber \\ {}&\times (1+\vert y\vert )\frac{\vert \nabla _y\phi (y,\tau _1)-\nabla _y\phi (y,\tau _2)\vert }{\vert \tau _1-\tau _2\vert ^{\frac{1}{2}+\varepsilon }}, \end{aligned}$$
(3.18)

for \(\varepsilon >0\) fixed small (as in Sect. 3.1.3).

We notice that, by standard parabolic estimates, from (3.17) we also get the bound on the Hölder seminorms in (3.18), thus

$$\begin{aligned} \vert \!\vert \phi \vert \!\vert _*\le C \vert \!\vert h\vert \!\vert _{\nu ,4}. \end{aligned}$$
(3.19)

3.1.2 Spaces for the parameters

We introduce weighted Hölder spaces for the parameters \(\Lambda ,\xi \). Let

$$\begin{aligned} X_{\sharp ,a,b,\sigma }{:=}\{\Lambda \in C(t_0,\infty ): \vert \!\vert \Lambda \vert \!\vert _{\sharp ,a,b,\sigma }<\infty \}, \end{aligned}$$

where

$$\begin{aligned} \vert \!\vert \Lambda \vert \!\vert _{\sharp ,a,b,\sigma }{:=} \sup _{t>t_0}\left\{ \mu (t)^{-a}\vert \!\vert \Lambda \vert \!\vert _{\infty ,[t,t+1]}\right\} +\sup _{t>t_0}\left\{ \mu (t)^{-b}[\Lambda ]_{0,\sigma ,[t,t+1]}\right\} , \end{aligned}$$

and

$$\begin{aligned}&\vert \!\vert \Lambda \vert \!\vert _{\infty ,[t,t+1]}=\sup _{s\in [t,t+1]}\vert \Lambda (s)\vert ,\\&[\Lambda ]_{0,\sigma ,[t,t+1]}{:=}\sup _{\begin{array}{c} s_1,s_2 \in [t,t+1]\\ s_1 \ne s_2 \end{array}} \frac{\vert \Lambda (s_1)-\Lambda (s_2)\vert }{\vert s_1-s_2\vert ^{\sigma }}. \end{aligned}$$

We look for \(\Lambda \) such that

$$\begin{aligned}&\vert \!\vert \Lambda \vert \!\vert _{\sharp ,l_0,\delta _0,\frac{1}{2}+\varepsilon }+\vert \!\vert \dot{\Lambda }\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }<\mathfrak {b}_1, \end{aligned}$$
(3.20)

for some positive constant \(\varepsilon ,\delta _0,\delta _1,l_0,l_1\) to be chosen (see Sect. 3.1.3). We also define \(X_{\sharp ,c,\sigma }{:=}X_{\sharp ,c,c,\sigma }\) and

$$\begin{aligned} \vert \!\vert h\vert \!\vert _{\sharp ,c,\sigma }{:=}\sup _{t>t_0} \mu (t)^{-c} \left[ \vert \!\vert h\vert \!\vert _{\infty ,[t,t+1]}+[h]_{0,\sigma ,[t,t+1]}\right] . \end{aligned}$$

We consider \(\xi _1\) such that

$$\begin{aligned}&\vert \!\vert \xi _1\vert \!\vert _{\sharp ,1+k,\frac{1}{2}+\varepsilon }+\vert \!\vert \dot{\xi }_{1}\vert \!\vert _{\sharp ,1+k,\varepsilon }<\mathfrak {b}_2, \end{aligned}$$
(3.21)

for some \(k>0\) (see Sect. 3.1.3). The positive constants \(\mathfrak {b}_{1},\mathfrak {b}_{2}\) will be selected as small as needed.

3.1.3 Choice of constants

Here we select the constants

$$\begin{aligned} l_0,l_1,\delta _0,\delta _1,\varepsilon ,\delta ,k,\alpha ,\beta ,\sigma ,\kappa , \end{aligned}$$

which are sufficient to find the perturbation \({{\tilde{\phi }}}\) in (3.1) by the inner-outer gluing scheme. Firstly, we indicate where the constants appear in the scheme:

  • \(l_0,l_1,\delta _0,\delta _1,\varepsilon \) appear in the definition (3.20);

  • k is used in the norm (3.21) for \(\xi \);

  • \(\delta \) appears in \(R(t)=\mu ^{-\delta }\), that is the radius of the inner regime;

  • \(\alpha ,\beta \) is used in the norms for the outer problem, see (4.6) and (4.14);

  • \(\sigma >0\) appears in the choice of \(\beta =l_1+\delta +\sigma \) in the outer problem;

  • \(\kappa >0\) is the constant appearing in Proposition 4.1.

We fix the following values:

  • \(\delta =\frac{2}{9}\);

  • \(l_1=k=\frac{2}{3}\);

  • \(l_0=l_1+\frac{\delta }{2}=\frac{7}{9}\);

  • \(\sigma =2\alpha =\varepsilon =\frac{1}{100}\);

  • \(\delta _1=l_1+\delta -\sigma -(1-\delta )(1+\alpha /2)(1+2\varepsilon )\);

  • \(\delta _0 = l_1 + \delta - \sigma - (1-\delta )(1+\alpha /2)2\varepsilon \);

  • \(\beta =\frac{1}{2}+l_1+\delta -\sigma \);

  • \(\kappa = \gamma (\sigma -\alpha \delta )\)

These choices are dictated by the following constraints, based on the estimate of the approximate solution, the linear theory for inner (Proposition 3.1) and outer problem (Lemma 4.1), the characterization of the orthogonality conditions (6.1) and the estimates in Proposition 6.1:

  • \(l_1+\delta <1\) to make \(\beta <3/2\) and apply the outer linear estimate (4.8);

  • we need

    $$\begin{aligned}\delta \in \left( \frac{1-l_1}{2}+{\hat{\varepsilon }},\frac{1+l_1}{6}\right) , \quad \text {where} \, {\hat{\varepsilon }}=\frac{(1+\alpha /2)(1+2\varepsilon )-l_1+\sigma }{1+(1+\alpha /2)(1+2\varepsilon )}-\frac{1-l_1}{2}.\end{aligned}$$

    Up to choosing \(\sigma>\alpha >0\) and \(\varepsilon >0\) small enough, these range is equivalent to (2.27), which, together with the previous restriction, impose a range for \(\delta \) and \(l_1\) leading (for instance) to the choice (2.31);

  • \(l_0\ge l_1\) and \(k+1\ge 2\delta +l_1\) to get \(\mu ^{5/2}S_{\text {in}}\) controlled by the term \(\mu (t) 5U(y)^4 J(x,t)\);

  • \(\sigma>\alpha \delta >0\), \(\varepsilon >0\) and \(\kappa \in (0,2\gamma (\sigma -\alpha \delta ))\). This allows to estimate the \(R^{\alpha }\log R\lesssim e^{-\kappa t}\mu ^{-\sigma }\) when we need to control the term \(\mu ^{-1} \phi \Delta _x \eta _R\) in the outer error;

  • \(k=l_1\). From (5.6) we need \(\vert \xi _1\vert +\vert *\vert {{\dot{\xi }}_1} \lesssim \mu ^{1+l_1}\), thus the choice of k, which is consistent with (2.30);

  • in the outer problem we obtain \(\vert \psi (x,t)\vert \lesssim \frac{\mu ^{l_1-\sigma }R^{-1}}{1+\vert y\vert ^\alpha }\). The nonlocal equation (5.2) and the estimate (2.29) asks for \(\vert \Lambda (t)\vert \lesssim \vert \psi (\xi (t),t)\vert \). Thus, this leads to the a choice of \(l_0\in [l_1,l_1+\delta -\sigma ]\);

  • from estimate (6.3), Eq. (5.2) and the bound on the \(\varepsilon \)-Hölder seminorm of \(\psi \) we get

    $$\begin{aligned}{}[\Lambda ]_{0,\frac{1}{2}+\varepsilon ,[t,t+1]}\lesssim [\psi (\xi (\cdot ),\cdot )]_{0,\varepsilon ,[t,t+1]}\lesssim \frac{\mu ^{l_1+\delta -\sigma }}{(\mu R)^{(1+\frac{\alpha }{2})2\varepsilon }}= \mu ^{\delta _0}, \end{aligned}$$

    which gives \(\delta _0\);

  • similarly, from (4.16) the Hölder estimate on the outer solution gives

    $$\begin{aligned}_{0,\frac{1}{2}+\varepsilon ,[t,t+1]}\lesssim \mu ^{l_1+\delta -\sigma }(\mu R)^{-(1+\frac{\alpha }{2})(1+2\varepsilon )},\end{aligned}$$

    and by Eq. (5.2) and estimate (6.5) we need

    $$\begin{aligned}_{0,\varepsilon ,[t,t+1]}\lesssim [\psi (0,\cdot )]_{0,\frac{1}{2}+\varepsilon ,[t,t+1]}.\end{aligned}$$

    This leads to the choice of \(\delta _1\);

  • after choosing \(\sigma =2\alpha >0\) small so that \(\delta >\sigma \), the constant \(\varepsilon \) is chosen small enough to make \(\delta _1\) positive (any choice of \(\alpha \in (0,\delta /2)\) and \(\varepsilon \) such that \(\delta _1=\delta _1(\alpha ,\varepsilon )>0\) is sufficient).

4 Solving the outer problem

We devote this section to solve the outer problem (3.4)

$$\begin{aligned}&{\partial } _t \psi =\Delta _x \psi +\gamma \psi +V \psi +f[\psi ,\phi ,\Lambda ,\dot{\Lambda },\xi ,{\dot{\xi }}](x,t), {\quad \hbox {in } }\Omega \times [t_0,\infty ),\\&\psi (x,t)=-\mu _0^{-1/2}u_3(x,t)\quad {\quad \hbox {on } } {\partial } \Omega \times [t_0,\infty ),\\&\psi (x,t_0)=\psi _0(x){\quad \hbox {in } }\Omega , \end{aligned}$$

where \(\psi _0(x)\) is any suitable small initial condition,

$$\begin{aligned} f(x,t)=&\mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\phi {\partial } _t \eta _R \nonumber \\ {}&+ \mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\eta _R\left\{ (\gamma -\dot{\Lambda })(\phi +2 \nabla _y \phi \cdot y) -\nabla _y \phi \cdot \left( \frac{{\dot{\xi }}}{\mu }\right) \right\} \nonumber \\&+\mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\left( \phi \left( \frac{2}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu ^2 R^2}+\frac{\eta ''\left( \vert z\vert \right) }{\mu ^2 R^2}\right) +2\frac{\nabla _y \phi }{\mu } \cdot \frac{z}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu R}\right) \nonumber \\&+\mu _0^{-1/2}S_{\text {in}}\left( 1-\eta _R\right) +\mu _0^{-1/2}S_{\text {out}}+\mu _0^{-1/2}\mathcal {N}(u_3,\tilde{\phi })(1-\eta _R) \end{aligned}$$
(4.1)

and potential

$$\begin{aligned} V(x,t)&=5u_3^4 (1-\eta _R), \end{aligned}$$

which, by the definition of \(u_3\), using again the bounds on \(H_\gamma ,J,\phi _3\) and the support of \((1-\eta _R)\), satisfies

$$\begin{aligned} \vert V\vert&\lesssim \mu ^{-2}U(y)^4 \vert \left( 1-\mu \frac{H_\gamma +J+\mu ^{-1}\phi _3 \eta _l}{U}\right) \vert ^4 \nonumber \\ {}&\lesssim (1-\eta _R)\frac{\mu ^{-2}}{1+\vert y\vert ^4}[1+\mu (1+\vert y\vert )]^4\nonumber \\&\lesssim (1-\eta _R)\frac{\mu ^{-2}}{1+\vert y\vert ^4}\nonumber \\&\lesssim \frac{\mu ^{-2}}{1+\vert y\vert ^2}R^{-2}. \end{aligned}$$
(4.2)

Let

$$\begin{aligned} \psi _1(x,t){:=}\mu _0(t)^{1/2}\psi (x,t). \end{aligned}$$

Then, the problem for \(\psi _1\) becomes

$$\begin{aligned}&{\partial } _t \psi _1 = \Delta _x \psi _1+V \psi _1 + F[\psi ,\phi ,\Lambda ,\dot{\Lambda }, \xi ,{\dot{\xi }}](x,t){\quad \hbox {in } }\Omega \times [t_0,\infty ),\nonumber \\&\psi _1(x,t)=g(x,t) {\quad \hbox {on } }\Omega \times [t_0,\infty ),\nonumber \\&\psi _1(x,t_0)=\psi _{1,0}(x) {\quad \hbox {in } }\Omega \end{aligned}$$
(4.3)

where

$$\begin{aligned}&F(x,t){:=}\mu _0(t)^{\frac{1}{2}}f(x,t)\\&g(x,t){:=}-u_3(x,t)\\&\psi _{1,0}(x){:=}\mu _0(t_0)^{\frac{1}{2}}\psi _0(x). \end{aligned}$$

In particular, in the proof of Proposition 4.1 we prove that for any \(\alpha >0\)

$$\begin{aligned}&\vert F(x,t)\vert \lesssim e^{-\kappa t_0}\frac{\mu ^{l_1+\delta -\sigma }\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$
(4.4)

Also, using the definition of \(u_3\), in (4.17) we prove

$$\begin{aligned}&\vert g(x,t)\vert \lesssim \mu ^{\frac{5}{2}}. \end{aligned}$$
(4.5)

Firstly, we consider the linear version of (4.3). Let

$$\begin{aligned} \vert F(x,t)\vert \le \vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2} \frac{\mu ^{\beta }\mu ^{-2}}{1+\vert y\vert ^{\alpha +2}}, \end{aligned}$$
(4.6)

for some \(\beta>0,\alpha >0\), where \(\vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2}\) is the best constant for such inequality. Also, for \(\delta \in (0,1/2)\) and \(\sigma \in (0,1)\) we define the Hölder norms

$$\begin{aligned}&[f]_{0,2\delta ,\delta , \Omega \times [t,t+1]}:=\sup _{\begin{array}{c} x_1\ne x_2 \in \Omega \\ t_1\ne t_2 \in [t,t+1] \end{array}}\frac{\vert f(x_1,t_1)-f(x_2,t_2)\vert }{\vert x_1-x_2\vert ^{2\delta }+\vert t-t_1\vert ^{\delta }}\\&[f(x,\cdot )]_{\sigma ,[t,t+1]}:= \sup _{t_1\ne t_2 \in [t,t+1]}\frac{\vert f(x,t)-f(x,t)\vert }{\vert t_1-t_2\vert ^{\sigma }}\\&[f(\cdot ,t)]_{0,\sigma , \Omega }:=\sup _{x_1\ne x_2 \in \Omega }\frac{\vert f(x_1,t)-f(x_2,t)\vert }{\vert t_1-t_2\vert ^{\sigma }} \end{aligned}$$

Lemma 4.1

Let F such that \(\vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2}<\infty \) for some constants \(\beta <3/2 \) and \(\alpha \in (0,1)\). Furthermore, assume that \(\vert \!\vert e^{as}g(s)\vert \!\vert _{L^{\infty }( {\partial } \Omega \times (t_0,\infty ))}<\infty \) for some \(a>0\) and \(\vert \!\vert h\vert \!\vert _{L^\infty (\Omega )}<\infty \). Let \(\psi _1[F,g,h]\) be the unique solution to

$$\begin{aligned}&{\partial } _t \psi _1 = \Delta _x \psi _1 + V \psi _1 + F(x,t){\quad \hbox {in } }{\Omega \times [t_0,\infty )},\nonumber \\ {}&\psi _1(x,t) = g(x,t) {\quad \hbox {on } } {\partial } {\Omega \times [t_0,\infty )},\nonumber \\&\psi _1(x,t_0)=h(x) {\quad \hbox {in } }\Omega . \end{aligned}$$
(4.7)

Then, for \(b \in (0,\lambda _1)\) and \({\tilde{a}} \in (0,\min \{a,\lambda _1-\varepsilon \}]\) for \(\varepsilon >0\) arbitrary small, we have

$$\begin{aligned} \vert \psi _1(x,t)\vert \lesssim&\vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2} \frac{\mu ^{\beta }}{1+\vert y\vert ^{\alpha }}+e^{-b(t-t_0)}\vert \!\vert h\vert \!\vert _{L^\infty (\Omega )}\nonumber \\&+e^{-{\tilde{a}}(t-t_0)}\vert \!\vert e^{as}g\vert \!\vert _{L^\infty ( {\partial } \Omega \times (t_0,\infty ))} \end{aligned}$$
(4.8)

for all \(x=\mu y+\xi \in \Omega \) and \(t>t_0\). Furthermore, the following local estimate on the gradient holds:

$$\begin{aligned}&\vert \nabla _x\psi _1(x,t)\vert \lesssim \vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2} \frac{ \mu ^{\beta -1}}{1+\vert y\vert ^{\alpha +1}}\quad \text {for}\quad \vert y\vert <R,\nonumber \\ {}&[\nabla _x \psi _1(\cdot ,t)]_{0,2\varepsilon ,B_{\mu R}(\xi )}\lesssim \vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2} {\mu ^{\beta -1-2\varepsilon }} R^{-1-2\varepsilon } \end{aligned}$$
(4.9)

where \(R\le \delta \mu ^{-1}\) for sufficiently small \(\delta >0\). Also, one has

$$\begin{aligned}&\left[ \left( \mu R\right) ^{2+\alpha }\right] ^{\frac{1}{2}+\varepsilon }\sup _{x \in B_{R\mu }(\xi )}{[\psi _1(x,\cdot )]_{0,\frac{1}{2}+\varepsilon ,[t,t+1]}}\nonumber \\&\quad +\left[ \left( \mu R\right) ^{2+\alpha }\right] ^{\varepsilon }\sup _{x\in B_{R\mu (\xi )}}[\psi _1(x,\cdot )]_{0,\varepsilon ,[t,t+1]} \lesssim \vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2}\mu ^{\beta }. \end{aligned}$$
(4.10)

Proof

To prove the result is enough to find a supersolution to the problem

$$\begin{aligned}&{\partial } _t \psi _2 =\Delta \psi _2+F{\quad \hbox {in } }{\Omega \times [t_0,\infty )},\\&\psi _2=g {\quad \hbox {on } } {\partial } {\Omega \times [t_0,\infty )},\\&\psi _2=h {\quad \hbox {in } }\Omega . \end{aligned}$$

We use the notation \(\psi _2=\psi _2[F,g,h]\). Indeed, suppose that \(\bar{\psi }_2\) is a supersolution to this problem. By (4.2) we have

$$\begin{aligned} \vert *\vert {V{\bar{\psi }}_2}\lesssim \frac{\mu ^{\beta -2}}{1+\vert y\vert ^{2+\alpha }}R(t_0)^{-2}, \end{aligned}$$

and hence \(\vert \!\vert V {\bar{\psi }}_2\vert \!\vert _{\beta -2,2+\alpha }<R(t_0)^{-2}\) for \(t_0\) sufficiently large. Thus, we find that a large multiple of \({\bar{\psi }}_2\) is a supersolution of (4.7). Firstly, let \(F,g\equiv 0\) and consider \(\psi _2[0,0,h]\). Let \(v_0(x)\) be the solution to

$$\begin{aligned}&-\Delta _x v_0-b v_0=0 {\quad \hbox {in } }\Omega ,\\&v_0=1{\quad \hbox {on } } {\partial } \Omega , \end{aligned}$$

for \(b \in (0,\lambda _1)\) and define

$$\begin{aligned} {\bar{\psi }}_2 =\vert \!\vert h\vert \!\vert _\infty e^{-b (t-t_0)}v_0(x). \end{aligned}$$

We claim that \({\bar{\psi }}_2\) is a supersolution for \(\psi _2[0,0,h]\). Indeed, we have

$$\begin{aligned}&{\partial } _t {\bar{\psi }}_2 - \Delta _x {\bar{\psi }}_2 =\vert \!\vert h\vert \!\vert _\infty e^{-b(t-t_0)}\left( -b v_0-\Delta _x v_0\right) =0{\quad \hbox {in } }{\Omega \times [t_0,\infty )},\\&{\bar{\psi }}_2(x,t)=\vert \!\vert h\vert \!\vert _\infty e^{-b(t-t_0)}\ge 0 {\quad \hbox {on } } {\partial } {\Omega \times [t_0,\infty )},\\&{\bar{\psi }}_2(x,t_0)=\vert \!\vert h\vert \!\vert _\infty v_0(x)\ge h(x) {\quad \hbox {in } }\Omega , \end{aligned}$$

where the last inequality is a consequence of the maximum principle applied to \(v_0\). Secondly, we look for a supersolution to \(\psi _2[0,g,0]\). Let \(v_1(x)\) to be the solution to

$$\begin{aligned}&-\Delta _x v_1 - {\tilde{a}} v_1 =0 {\quad \hbox {in } }\Omega ,\\&v_1(x)=1 {\quad \hbox {on } } {\partial } \Omega , \end{aligned}$$

where \({\tilde{a}} \in (0,\min \{a,\lambda _1-\varepsilon \}] \) and consider

$$\begin{aligned} {\bar{\psi }}_2(x,t)= \vert \!\vert e^{as}g\vert \!\vert _{L^\infty ( {\partial } \Omega \times (t_0,\infty ))} e^{-{\tilde{a}}(t-t_0)}v_1(x). \end{aligned}$$

We verify that

$$\begin{aligned}&{\partial } _t {\bar{\psi }}_2- \Delta {\bar{\psi }}_2 = \vert \!\vert e^{as}g\vert \!\vert _\infty e^{-{\tilde{a}}(t-t_0)} \left( -{\tilde{a}} v_1-\Delta v_1\right) =0 {\quad \hbox {in } }{\Omega \times [t_0,\infty )},\\&{\bar{\psi }}_2(x,t)=\vert \!\vert e^{as}g\vert \!\vert _\infty e^{-{\tilde{a}}(t-t_0)}\ge g(x,t){\quad \hbox {on } } {\partial } {\Omega \times [t_0,\infty )},\\&{\bar{\psi }}_2(x,t_0)= \vert \!\vert e^{as}g\vert \!\vert _{L^\infty ( {\partial } \Omega \times (t_0,\infty ))} v_1(x)\ge 0 {\quad \hbox {in } }\Omega , \end{aligned}$$

where we used \({\tilde{a}}\le a\) to get the second inequality and \({\tilde{a}}<\lambda _1\) to get the third one by the maximum principle. It remains to find a supersolution for \(\psi _2[F,0,0]\). Let \(\psi _2[F,0,0]=e^{-c(t-t_0)}\psi _3\), where \(c=2\gamma \beta \) so that

$$\begin{aligned} {\partial } _t \psi _3 = \Delta _x \psi _3 + c \psi _3+\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$

We find a bounded \({\bar{\psi }}_3\) supersolution in case \(c<\lambda _1\), that is \(3\gamma <\lambda _1\). Consider

$$\begin{aligned} {\bar{\psi }}_3 = \uppsi _0\left( \frac{x-\xi }{\mu }\right) \eta \left( \frac{x-\xi }{d}\right) +\uppsi _1(x,t). \end{aligned}$$

We need

$$\begin{aligned}&{\partial } _t \uppsi _1-\Delta _x \uppsi _1-c\uppsi _1 \nonumber \\ {}&\quad \ge \eta \left[ - {\partial } _t \uppsi _0+ \mu ^{-2}\Delta _y \uppsi _0 +c \uppsi _0+ \frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}\right] \nonumber \\&\qquad +(1-\eta )\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}+(\Delta _x \eta - {\partial } _t \eta )\uppsi _0+2 \mu ^{-1}\nabla _x \eta \cdot \nabla _y \uppsi _0, \end{aligned}$$
(4.11)

with \({\bar{\psi }}_3(x,t)\ge 0\) on \( {\partial } {\Omega \times [t_0,\infty )}\) and initial datum \(\bar{\psi }_3(x,t_0)\ge 0\). Suppose without loss of generality that \(\Omega \subset B_1\) and take \(\uppsi _0\) as the solution to

$$\begin{aligned}&\Delta _y \uppsi _0=-\frac{2}{1+\vert y\vert ^{2+\alpha }}{\quad \hbox {in } }B_{\mu ^{-1}}\\&\uppsi _0 = 0{\quad \hbox {on } } {\partial } B_{\mu ^{-1}}. \end{aligned}$$

From the variation of parameters formula

$$\begin{aligned} \uppsi _0(\vert y\vert )=2\omega _3\int _{\vert y\vert }^{\mu ^{-1}}\frac{1}{\rho ^2}\int _{0}^{\rho }\frac{s^2}{1+s^{2+\alpha }}\,ds\,d\rho , \end{aligned}$$
(4.12)

we find

$$\begin{aligned}&\vert \uppsi _0\vert \lesssim \frac{1}{1+\vert y\vert ^{\alpha }}, \end{aligned}$$

and

$$\begin{aligned} \vert {\partial } _t \uppsi _0\vert&\lesssim {\partial } _t\left( \frac{1}{\mu }\right) \mu ^2 \int _0^{\mu ^{-1}}\frac{s^2}{1+s^{2+\alpha }}\,ds+ {\partial } _t \left( \vert y(x,t)\vert \right) \frac{1}{\vert y\vert ^2}\int _{0}^{\vert y\vert }\frac{s^2}{1+s^{2+\alpha }}\,ds\\ {}&\lesssim \frac{1}{1+\vert y\vert ^\alpha }+\frac{\vert y\vert ^2}{1+\vert y\vert ^{2+\alpha }}\\&\lesssim \frac{1}{1+\vert y\vert ^{\alpha }} \end{aligned}$$

Also, if \(\vert x-\xi \vert <d\) for d fixed sufficiently small, we obtain

$$\begin{aligned} - {\partial } _t \uppsi _0+ \mu ^{-2}\Delta _y \uppsi _0 +c \uppsi _0+ \frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}=-\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}+O \left( \frac{1}{1+\vert y\vert ^{\alpha }}\right) <0. \end{aligned}$$

Now, we take \(\uppsi _1\) as the solution to

$$\begin{aligned}&{\partial } _t \uppsi _1 -\Delta _x \uppsi _1-c\uppsi _1=(1-\eta ) \frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }} + \left( \Delta _x \eta - {\partial } _t \eta \right) \uppsi _0+2 \mu ^{-1}\nabla _x \eta \cdot \nabla _y \uppsi _0,\\&\uppsi _1=0 {\quad \hbox {on } } {\partial } {\Omega \times [t_0,\infty )},\\&\uppsi _1(x,t_0)=0 {\quad \hbox {in } }\Omega . \end{aligned}$$

We estimate the right-hand side by

$$\begin{aligned}&(1-\eta )\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}\lesssim \mu ^{\alpha },\\&\vert \left( \Delta _x \eta - {\partial } _t \eta \right) \uppsi _0\vert \lesssim \mu ^{\alpha },\\&\vert 2 \mu ^{-1}\nabla _x \eta \cdot \nabla _y \uppsi _0\vert \lesssim \mu ^{\alpha }. \end{aligned}$$

Hence, by comparison principle using \(c<\lambda _1\) we obtain a solution \(\vert *\vert {{\bar{\psi }}_3}\lesssim \mu ^{\alpha }\). Thus, inequality (4.11) is satisfied. Also, \({\bar{\psi }}_3=0\) on \( {\partial } {\Omega \times [t_0,\infty )}\) and \(\psi _3(x,t_0)=\eta \uppsi _0(x,t_0)\ge 0\). We conclude that \({\bar{\psi }}_3\) is a supersolution and the bound (4.8) is proven. Now, we prove the gradient estimate (4.9). Let

$$\begin{aligned}\psi _1(x,t)=:{{\tilde{\psi }}}\left( z(x,t),\tau (t)\right) ,\quad \text {where}\quad z:=\frac{x-\xi (t)}{R(t)\mu (t)}\end{aligned}$$

and \({\dot{\tau }}(t)=(R(t)\mu (t))^{-2}\), that gives \(\tau (t)\sim \mu ^{-2}\). We can take \(\tau (t_0)=2\). The equation for \({{\tilde{\psi }}}\) becomes

$$\begin{aligned} {\partial } _\tau {{\tilde{\psi }}} = \Delta _z {{\tilde{\psi }}} +a(z,\tau ) \cdot \nabla _z {{\tilde{\psi }}} + b(z,\tau ){{\tilde{\psi }}} + {\tilde{F}}(z,\tau ), \end{aligned}$$

where \({\tilde{F}}(z,\tau (t))=(R\mu )^2 F(\mu R z+\xi , \tau (t))\), and the coefficients

$$\begin{aligned}&a(z,\tau ):=(\mu _0 R)\left[ z {\partial } _t(\mu _0)+{\dot{\xi }}\right] ,\quad b(z,\tau ):=(R\mu )^2 V(\mu R z +\xi ,\tau (t))\\&\quad \lesssim \frac{1}{1+R^2\vert z\vert ^2}, \end{aligned}$$

are uniformly bounded. Since \(\vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2}<\infty \), we have

$$\begin{aligned} {\tilde{F}}(z,\tau (t))=(R\mu )^2 F(\mu Rz+\xi , \tau (t))\lesssim \mu ^{\beta }\frac{\vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2}}{1+\vert Rz\vert ^{2+\alpha }}. \end{aligned}$$

We have already proved the \(L^\infty \)-bound

$$\begin{aligned} \vert \!\vert \psi _1\vert \!\vert _{\beta ,\alpha }\lesssim \vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2}. \end{aligned}$$

We apply standard local parabolic estimates for the gradient: let \(\sigma \in (0,1)\) and \(\tau _1\ge \tau (t_0)+2\), then

$$\begin{aligned}&[\nabla _z {{\tilde{\psi }}}_1(\cdot ,\tau _1)]_{0,\sigma ,B_1(0)}+\vert \!\vert \nabla _z {{\tilde{\psi }}}_1(\cdot ,\tau _1)\vert \!\vert _{L^\infty \left( B_1(0)\right) }\\&\quad \lesssim \vert \!\vert {{\tilde{\psi }}}\vert \!\vert _{L^\infty (B_2(0))\times (\tau _1-1,\tau _1)}+\vert \!\vert {\tilde{F}}\vert \!\vert _{L^\infty \left( B_{2}(0)\times (\tau _1-1,\tau _1)\right) }\\&\quad \lesssim \mu (t(\tau _1-1))^{\beta } \vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2}\\&\quad \lesssim \mu (t(\tau _1))^{\beta } \vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2}. \end{aligned}$$

In the original variables, for any \(t\ge t_0+2\) we find

$$\begin{aligned} (R\mu )^{1+\sigma }[\nabla _x \psi _1(\cdot ,t)]_{0,\sigma , B_{\mu R}(\xi )}+R\mu \vert \!\vert \nabla _x \psi _1(\cdot ,t)\vert \!\vert _{L^\infty (B_{\mu R}(\xi ))}\lesssim \mu ^{\beta }\vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2}. \end{aligned}$$
(4.13)

By similar parabolic estimates using \(\vert \!\vert \nabla _x\psi _0\vert \!\vert _\infty <\infty \) we can extend estimate (4.13) up to \(t=t_0\), thus, the proof of (4.9) is complete. Now, we prove estimate (4.10). We consider the Hölder seminorms. We perform the change of variable

$$\begin{aligned} \psi _1(x,t)=\hat{\psi }\left( z,\tau \right) , \end{aligned}$$

where \(z :=(x-\xi )/({R\mu _0})^{1+\frac{\alpha }{2}}\) and \(\tau \) satisfies

$$\begin{aligned} \frac{d\tau }{dt}=\frac{1}{\left( \mu _0(t) R(t)\right) ^{2+\alpha }}, \end{aligned}$$

that is

$$\begin{aligned} \tau -\tau _0&=\int _{t_0}^\infty \frac{\,ds}{\left( \mu _0(t) R(t)\right) ^{2+\alpha }}\,ds\\&=C \left( \mu _0 R\right) ^{-(2+\alpha )}(1+o(1)). \end{aligned}$$

The equation for \({\hat{\psi }}\) is

$$\begin{aligned} {\partial } _\tau {\hat{\psi }} = \Delta _z {\hat{\psi }} +{\hat{a}}(z,\tau )\cdot \nabla _z {\hat{\psi }}+{\hat{b}}(z,\tau ) {\hat{\psi }} + {\hat{f}}(z,\tau ), \end{aligned}$$

where

$$\begin{aligned}&{\hat{a}}(z,\tau )= (\mu _0 R)^{1+\frac{\alpha }{2}} \left[ z {\partial } _t(\mu _0 R)^{1+\frac{\alpha }{2}}+{\dot{\xi }}\right] , \\ {}&{\hat{b}}(z,\tau ) = (\mu _0 R)^{2+\alpha }V((\mu _0 R)^{1+\frac{\alpha }{2}}z+\xi ,t(\tau )),\\&{\hat{F}} = (\mu _0 R)^{2+\alpha } F((\mu _0 R)^{1+\frac{\alpha }{2}}z+\xi ,t(\tau )) \end{aligned}$$

Then, applying local parabolic estimates on \({{\tilde{\psi }}}\), we get

$$\begin{aligned}{}[\psi _1(x,\cdot )]_{0,\frac{1+2\varepsilon }{2},[t,t+1]}=&\sup _{{t_1 \ne t_2 \in [t,t+1]}}\frac{\vert \psi _1(x,t_1)-\psi _1(x,t_2)\vert }{\vert t_1-t_2\vert ^{\frac{1+2\varepsilon }{2}}} \\\lesssim&\sup _{\tau _1,\tau _2 \in [\tau ,\tau +1]}\frac{\vert *\vert {{\hat{\psi }}(z_1,\tau _1)-{\hat{\psi }}(z_1,\tau _2)}}{\vert \tau _1-\tau _2\vert ^{\frac{1+2\varepsilon }{2}}}\frac{\vert \tau _1-\tau _2\vert }{\vert t_1-t_2\vert }^{\frac{1+2\varepsilon }{2}}\\ \lesssim&[{\hat{\psi }}(z,\cdot )]_{0,\frac{1+2\varepsilon }{2},[\tau _1,\tau _2]} \frac{1}{[\left( \mu R\right) ^{2+\alpha }]^{\frac{1+2\varepsilon }{2}}}\\ \lesssim&\vert \!\vert F\vert \!\vert _{\beta -2,\alpha +2} \mu ^{\beta }\frac{1}{[\left( \mu R\right) ^{2+\alpha }]^{\frac{1+2\varepsilon }{2}}}, \end{aligned}$$

where \(\tau _i=\tau (t_i)\) and \(z=z(x,t_i)\) for \(i=1,2\). Similarly, using the Hölder coefficient \((2\varepsilon ,\varepsilon )\), we get

$$\begin{aligned}{}[\psi _1(x,\cdot )]_{ 0,\varepsilon , [t,t+1]}\lesssim \mu ^{\beta }\frac{1}{[\left( \mu R\right) ^{2+\alpha }]^{\varepsilon }}. \end{aligned}$$

\(\square \)

We introduce the following weighted norms for \(\psi \):

$$\begin{aligned}&\vert \!\vert \psi \vert \!\vert _{**}\nonumber \\ {}&:= \sup _{x \in \Omega , t>t_0} \left\{ g_1(x,t)^{-1}\vert \psi (x,t)\vert \right\} +\sup _{t>t_0}\left\{ g_2(t)^{-1}\sup _{x \in B_{\mu R}(\xi )}[\psi (x,\cdot )]_{0,\varepsilon ,[t,t+1]}\right\} \nonumber \\&\quad +\sup _{t>t_0}\sup _{x\in B_{\mu R}(\xi )}\left\{ g_3(x,t)^{-1}\vert \nabla _x \psi (x,t)\vert \right\} +\sup _{t>t_0} g_4(t)^{-1}[\nabla _x \psi ]_{0,2\varepsilon ,\varepsilon ,B_{R\mu }(\xi )\times [t,t+1] }\nonumber \\ {}&\quad +\sup _{t>t_0}\left\{ g_5(t)^{-1}\sup _{x\in B_{\mu R}(\xi )}[\psi (x,\cdot )]_{0,\frac{1}{2}+\varepsilon ,[t,t+1]}\right\} \end{aligned}$$
(4.14)

where

$$\begin{aligned}&g_1(x,t)=\frac{\mu ^{\beta }}{1+\vert y\vert ^{\alpha }},\quad g_2(t)=\mu ^{\beta } \left[ (\mu R)^{2+\alpha }\right] ^{-\varepsilon },\quad g_3(x,t)=\frac{\mu ^{\beta -1}}{1+\vert y\vert ^{\alpha +1}},\\&g_4(t)=\mu ^{\beta }[\mu R]^{-1-2\varepsilon },\quad g_5(t)={\mu ^{\beta }}{\left[ (\mu R)^{2+\alpha }\right] ^{-(\frac{1}{2}+\varepsilon )}} \end{aligned}$$

and define the space of functions

$$\begin{aligned} X_{**}=\{\psi \in L^\infty ({\Omega \times [t_0,\infty )}): \vert \!\vert \psi \vert \!\vert _{**}<\infty \}. \end{aligned}$$

Now, we are ready to solve the outer problem (3.4) for \(\phi \) such that

$$\begin{aligned} \vert \!\vert \phi \vert \!\vert _{*}<{{\textbf {b}}}, \end{aligned}$$
(4.15)

for parameters satisfying (3.20) and (3.21).

Proposition 4.1

Assume that \(\Lambda ,\xi _1,\phi \) satisfy (3.20), (3.21) and (4.15) respectively. Also, suppose \(\psi _0 \in C^2(\bar{\Omega })\) such that

$$\begin{aligned} \vert \!\vert \psi _0\vert \!\vert _{L^\infty }+\vert \!\vert \nabla \psi _0\vert \!\vert <e^{-\kappa t_0}, \end{aligned}$$

for some \(\kappa \in (0,2\gamma (\sigma -\alpha \delta ))\). Then, there exists \(t_0\) large so that problem (3.4) has a unique solution \(\psi =\Psi [\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi ]\) and given \(\alpha >0\), there exists \(C_{**}\) such that

$$\begin{aligned} \vert \!\vert \psi \vert \!\vert _{**}\le e^{-\kappa t_0}C_{**}, \end{aligned}$$
(4.16)

where \(C_{**}=C_{**}({{\textbf {b}}},\mathfrak {b}_1,\mathfrak {b}_2)\) and \({{\textbf {b}}},\mathfrak {b}_1,\mathfrak {b}_2\) are the constants in (4.15), (3.20) and (3.21) respectively.

Proof

Let \(T_1\) the linear operator, defined by Lemma 4.1, such that, given \(\beta <3/2,\alpha >0\) and functions fgh with bounded norms \(\vert \!\vert f\vert \!\vert _{\beta -2,\alpha +2},\vert \!\vert e^{as}g\vert \!\vert _{\infty },\vert \!\vert h\vert \!\vert _\infty \) respectively, T[fgh] is the solution to (4.7).

Let

$$\begin{aligned} \mu _0^{1/2}\psi =\psi _A+\psi _B, \end{aligned}$$

where we define \(\psi _B{:=}T(0,-u_3,\mu _0(t_0)^{1/2}\psi _0)\). From the definition of \(u_3(x,t)\) we expand for \(x\in {\partial } \Omega \) and \(t_0\) large, to get

$$\begin{aligned} u_3(x,t)&=\frac{\alpha _3\mu ^{1/2}}{\left( \mu ^2+\vert x-\xi \vert ^2\right) ^{1/2}}-\mu ^{1/2}\frac{\alpha _3}{\vert x-\xi \vert }\nonumber \\&=\alpha _3 \mu ^{1/2}\vert x-\xi \vert ^{-1} \left[ \left( \frac{\mu ^2+\vert x-\xi \vert ^2}{\vert x-\xi \vert ^2}\right) ^{-1/2}-1\right] \nonumber \\&=\mu ^{5/2}f_B(x,t), \end{aligned}$$
(4.17)

for a smooth bounded function \(f_B(x,t)\) on \( {\partial } \Omega \times [t_0,\infty )\). Hence, Lemma 4.1 gives the bound

$$\begin{aligned} \vert \psi _B\vert \lesssim e^{-b (t-t_0)}\vert \!\vert \psi _0\vert \!\vert _{L^\infty } + e^{-a(t-t_0)}\vert \!\vert e^{as}u_3\vert \!\vert _{L^\infty ( {\partial } \Omega \times [t_0,\infty ))}, \end{aligned}$$

for any \(b<\lambda _1\) and \(a<\min \{5\gamma ,\lambda _1-\varepsilon \}\) for any \(\varepsilon >0\).

Now, we apply the fixed point theorem to find \(\psi _A\) such that \(\psi \) satisfies (4.16). We obtain a solution \(\psi \) if \(\psi _A\) satisfies

$$\begin{aligned} \psi _A=\mathcal {A}(\psi _A),\quad \mathcal {A}(\psi _A):=T[F(\psi _A),0,0], \end{aligned}$$

where \(F(x,t)=\mu _0(t)^{1/2}f(x,t)\) and f is given by (4.1). We look for \(\psi _A\) in

$$\begin{aligned} \mathcal {B}=\{ \psi _A: \quad \vert \!\vert \mu _0^{-1/2}\psi _A\vert \!\vert _{**}<M e^{-\kappa t_0} \}, \end{aligned}$$

where M is a fixed large constant, independent of t and \(t_0\). We prove that \(\mathcal {A}(\psi _A) \in \mathcal {B}\) for any \(\psi _A \in \mathcal {B}\). Firstly, we estimate the \(L^\infty \) norm of \(F(\psi _A)\). From (4.14) we apply Lemma 4.1 with \(\beta =1/2+l_1+\delta <3/2\). We recall that \(F=\mu _0^{1/2}f\) where

$$\begin{aligned} f(x,t)&=\mu _0^{-1/2}\mathcal {N}(u_3,{{\tilde{\phi }}})(1-\eta _R) \\&\quad +\mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\phi {\partial } _t \eta _R \\&\quad + \mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\eta _R\left\{ (\gamma -\dot{\Lambda })(\phi +2 \nabla _y \phi \cdot y) -\nabla _y \phi \cdot \left( \frac{{\dot{\xi }}}{\mu }\right) \right\} \\&\quad +\mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\left( \phi \left( \frac{2}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu ^2 R^2}+\frac{\eta ''\left( \vert z\vert \right) }{\mu ^2 R^2}\right) +2\frac{\nabla _y \phi }{\mu } \cdot \frac{z}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu R}\right) \\&\quad +\mu _0^{-1/2}S_{\text {in}}\left( 1-\eta _R\right) +\mu _0^{-1/2}S_{\text {out}}. \end{aligned}$$

We have \(\eta ''(y/R)\ne 0\) and \(\eta '(y/R)\ne 0\) only if \(\vert y\vert \sim R\), hence we estimate

$$\begin{aligned}&\vert \mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\phi \Delta \eta \vert \nonumber \\&\quad \lesssim \mu ^{-1}\vert \!\vert \phi \vert \!\vert _{*}\mu ^{1+l_1}\left[ \frac{R^2 \log (R)}{1+\vert y\vert ^3}+\frac{R^3}{1+\vert y\vert ^4}\right] \frac{\mu ^{-2}}{1+\vert y\vert ^2}\eta ''\left( \left| {\frac{x-\xi }{\mu R}}\right| \right) \nonumber \\ {}&\quad \lesssim \mu ^{l_1+\delta }\log (R)\frac{\mu ^{-2}}{1+\vert y\vert ^2}\nonumber \\&\quad \lesssim e^{-\kappa t_0}\mu ^{l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$
(4.18)

Using the bound on the gradient given in the definition of \(\vert \!\vert \phi \vert \!\vert _{*}\) we obtain

$$\begin{aligned}&\left| {\mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\left( 2\frac{\nabla _y \phi }{\mu } \cdot \frac{z}{\vert z\vert }\frac{\eta '(\vert z\vert )}{\mu R}\right) }\right| \\&\quad \lesssim \mu ^{-1} \mu ^{1+l_1}\left[ \frac{R^2 \log (R)}{1+\vert y\vert ^4}+\frac{R^3}{1+\vert y\vert ^5}\right] \left( \frac{\vert \eta '(\vert z\vert )\vert }{\mu ^2 R}\right) \\&\quad \lesssim \vert \!\vert \phi \vert \!\vert _{*}\mu ^{l_1+\delta }\log (R)\frac{\mu ^{-2}}{1+\vert y\vert ^2}\\&\quad \lesssim e^{-\kappa t_0}\mu ^{l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$

Similarly, also using the bounds on \({\dot{\Lambda }},{\dot{\xi }}\) we have

$$\begin{aligned}&\left| {\mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\phi {\partial } _t \eta _R}\right| \\&\quad \lesssim \vert \!\vert \phi \vert \!\vert _{*} \mu ^{l_1+\delta }\log (R) \left| {\eta '(\vert z\vert )\frac{z}{\vert z\vert }\cdot \left( -\frac{{\dot{\xi }}}{\mu R}-z \frac{ {\partial } _t(\mu R)}{\mu R}\right) }\right| \\&\quad \lesssim \mu ^{l_1+\delta }\log (R) \frac{\vert \eta '(\vert z\vert )\vert }{\mu ^2 R^2}\\&\quad \lesssim e^{-\kappa t_0}\mu ^{l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$

Also, since \(\delta <1/3\), we have

$$\begin{aligned}&\Bigg \vert {(\mu \mu _0)^{-1/2}\eta _R\left\{ \left( \gamma -{\dot{\Lambda }}\right) (\phi +2 \nabla _y \phi \cdot y) -\nabla _y \phi \cdot \left( \frac{\dot{\xi }}{\mu }\right) \right\} }\Bigg \vert \\&\quad \lesssim \mu ^{-1} \mu ^{1+l_1}\left[ \frac{R^2 \log (R)}{1+\vert y\vert ^4}+\frac{R^3}{1+\vert y\vert ^5}\right] \\&\quad \lesssim \mu ^{l_1}\vert \!\vert \phi \vert \!\vert _* R^3\\&\quad \lesssim \mu ^{l_1+\delta }\frac{\mu ^{-2}}{R^2}\\&\quad \lesssim e^{-\kappa t_0}\mu ^{l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$

Furthermore, using Lemma 2.5 we estimate

$$\begin{aligned} \vert \mu ^{-1/2}S_{\text {out}}\vert&\lesssim \mu \\&\lesssim \mu ^{l_1+\delta }\frac{\mu ^{-2}}{R^2}\\&\lesssim e^{-\kappa t_0}\mu ^{l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$

and

$$\begin{aligned} \vert \mu ^{-1/2}S_{\text {in}}(1-\eta _R)\vert&\lesssim \mu ^{l_1+2\delta } \frac{\mu ^{-2}}{1+\vert y\vert ^2}\\&\lesssim e^{-\kappa t_0}\mu ^{l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$

Finally, since \(\vert \!\vert \mu _0^{-1/2}\psi _A\vert \!\vert _{**}\) is bounded we get

$$\begin{aligned} \vert \mu _0^{-1/2}\mathcal {N}_3(u_3,{{\tilde{\phi }}})(1-\eta _R)\vert \lesssim&\mu ^{-1/2} u_3^3 \left( \mu ^{-1/2}\phi \eta _R +\mu ^{1/2}\psi \right) ^2(1-\eta _R)\nonumber \\ \lesssim&\frac{\mu ^{-2}}{1+\vert y\vert ^3}\left( \mu ^{-1}\vert \phi \vert ^2 \eta _R^2 +\mu _0\vert *\vert {\mu _0^{-1/2}\psi _A}^2\right) (1-\eta _R)\nonumber \\ \lesssim&\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}\bigg [ \mu ^{-1}\vert \!\vert \phi \vert \!\vert _*^2 \mu ^{2(1+l_1)}\left( R^{-1}\log (R)\right) ^2\nonumber \\ {}&\quad \quad \quad \quad \quad +\vert \!\vert \mu _0^{-1/2}\psi _A\vert \!\vert _{**}^2 \mu \frac{\mu ^{2(l_1+\delta -\sigma )}}{R^{2\alpha }} \bigg ]\nonumber \\&\lesssim e^{-\kappa t_0}\mu ^{l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$
(4.19)

Summing up these estimates we conclude that

$$\begin{aligned} \vert f(x,t)\vert \lesssim e^{-\kappa t_0}\mu ^{l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}. \end{aligned}$$

Hence, we have

$$\begin{aligned} \vert F(x,t)\vert \lesssim e^{-\kappa t_0}\mu ^{\frac{1}{2}+l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}, \end{aligned}$$

and Lemma 4.1 gives

$$\begin{aligned} \vert T[F(\psi _A),0,0]\vert \lesssim e^{-\kappa t_0}\mu ^{\frac{1}{2}+l_1+\delta -\sigma }\frac{1}{1+\vert y\vert ^{\alpha }}. \end{aligned}$$

Since \(F\in L^\infty (\Omega \times [t_0,\infty ))\), classic parabolic estimates give \(\psi \in C^{1+{\hat{\sigma }},\frac{1+\hat{\sigma }}{2}}(\Omega \times [t_0,\infty ))\) for any \({\hat{\sigma }}<1\) and from Lemma 4.1 we get

$$\begin{aligned} \vert \!\vert \mu _0^{-1/2}\psi _A\vert \!\vert _{**}\le M e^{-k t_0} \end{aligned}$$
(4.20)

for sufficiently large M (independent of \(t_0\)). This proves \(\mathcal {A}(\psi _A)\in \mathcal {B}\). Now, we claim that the map \(\mathcal {A}(\psi )\) is a contraction, that is: there exists \({{\textbf {c}}}<1\) such that, for any \(\psi _A^{(1)},\psi _A^{(2)} \in \mathcal {B}\),

$$\begin{aligned} \vert \!\vert \mu _0^{-1/2}\mathcal {A}(\psi _A^{(1)})-\mu _0^{-1/2}\mathcal {A}(\psi _A^{(2)})\vert \!\vert _{**}\le {{\textbf {c}}} \vert \!\vert \mu _0^{-1/2}\psi _A^{(1)}-\mu _0^{-1/2}\psi _A^{(2)}\vert \!\vert _{**}. \end{aligned}$$

Since \(\psi \) appears in \(F(\psi )\) only in the nonlinear term \(\mathcal {N}\), we get

$$\begin{aligned} \mathcal {A}(\psi _A^{(1)})-\mathcal {A}(\psi _A^{(2)})=T\bigg [&\mathcal {N}\left( u_3,\psi _A^{(1)}+\psi _{B}+\mu ^{-1/2}\phi \eta _R\right) \\&-\mathcal {N}\left( u_3,\psi _A^{(2)}+ \psi _{B}+\mu ^{-1/2}\phi \eta _R\right) ,0,0\bigg ]. \end{aligned}$$

From definition (3.3) we write

$$\begin{aligned}&\mathcal {N}\left( u_3,\psi _A^{(1)}+\psi _{B}+\mu ^{-1/2}\phi \eta _R\right) - \mathcal {N}\left( u_3,\psi _A^{(2)}+ \psi _{B}+\mu ^{-1/2}\phi \eta _R\right) \\&\quad =\bigg [\left( u_3+\psi _A^{(1)}+\psi _{B}+\mu ^{-1/2}\phi \eta \right) ^5 - (u_3 +\psi _A^{(2)}+\psi _{B} + \mu ^{-1/2}\phi \eta )^5 \nonumber \\&\qquad - 5u_3^{4}(\psi ^{(1)}-\psi ^{(2)})\bigg ]\\&\quad =\bigg [\left( u_3+\psi _A^{(1)}+\psi _{B}+\mu ^{-1/2}\phi \eta \right) ^5- \left( u_3+\psi _A^{(2)}+\psi _{B}+\mu ^{-1/2}\phi \eta \right) ^5 \\ {}&\qquad -5(u_3+\mu ^{-1/2}\phi \eta )^4\mu _0^{1/2} \left( \psi ^{(1)}-\psi ^{(2)}\right) \bigg ]\\&\qquad + 5 \left[ (u_3+\mu ^{-1/2}\phi \eta )^4-u_3^4\right] (\psi ^{(1)}-\psi ^{(2)})\\&\quad =: N_1+N_2. \end{aligned}$$

We estimate

$$\begin{aligned} \vert N_1(x,t)\vert&\lesssim \mu ^{-3/2}U^3\mu _0 \vert \mu _0^{-1/2}(\psi _A^{(1)}-\psi _A^{(2)})\vert ^2\\&\lesssim \mu ^{-1/2}U^3 \vert \mu _0^{-1/2}(\psi _A^{(1)}-\psi _A^{(2)})\vert ^2\\&\lesssim \frac{\mu ^{-1/2}}{1+\vert y\vert ^{3}}\frac{\mu ^{2\left( l_1+\delta -\sigma \right) }}{1+\vert y\vert ^{2\alpha }}\vert \!\vert \mu _0^{-1/2}(\psi _A^{(1)}-\psi _A^{(2)})\vert \!\vert _{**}^2\\&\lesssim e^{-\kappa t_0} \mu ^{\frac{1}{2}+l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}\vert \!\vert \mu _0^{-1/2}(\psi _A^{(1)}-\psi _A^{(2)})\vert \!\vert _{**} \end{aligned}$$

and

$$\begin{aligned} \vert N_2(x,t)\vert&\lesssim u_3^3 \mu _0^{-1/2}\phi \eta \mu _0^{\frac{1}{2}}\vert \mu _0^{-1/2}(\psi _A^{(1)}-\psi _A^{(2)})\vert \\&\lesssim \frac{\mu ^{-3/2}}{1+\vert y\vert ^3} \vert \!\vert \phi \vert \!\vert _* \mu ^{1+l_1}R^3 \frac{\mu ^{l_1+\delta -\sigma }}{1+\vert y\vert ^\alpha }\vert \!\vert \mu _0^{-1/2}(\psi _A^{(1)}-\psi _A^{(2)})\vert \!\vert _{**}\\&\lesssim e^{-\kappa t_0} \mu ^{\frac{1}{2}+l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}\vert \!\vert \mu _0^{1/2}(\psi _A^{(1)}-\psi _A^{(2)})\vert \!\vert _{**} \end{aligned}$$

Finally, using \(\beta =1/2+l_1+\delta -\sigma <3/2\) we apply \(T[\cdot ,0,0]\) to \(F(\psi )\) we obtain

$$\begin{aligned} \vert \mathcal {A}[\psi _A^{(1)}]-\mathcal {A}[\psi _A^{(2)}]\vert \lesssim e^{-\kappa t_0} \mu ^{\frac{1}{2}+l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{\alpha }}\vert \!\vert \mu _0^{-1/2}(\psi ^{(1)}-\psi ^{(2)})\vert \!\vert _{**}. \end{aligned}$$
(4.21)

Arguing as in (4.20), from (4.21) and standard parabolic estimates we obtain

$$\begin{aligned} \vert \!\vert \mu _0^{-1/2}(\mathcal {A}[\psi _A^{(1)}]-\mathcal {A}[\psi _A^{(2)}])\vert \!\vert _{**}\le {{\textbf {c}}} \vert \!\vert \mu _0^{-1/2}(\psi _A^{(1)}-\psi _A^{(2)})\vert \!\vert _{**}, \end{aligned}$$

with \({{\textbf {c}}}<1\) if \(t_0\) is taken sufficiently large. Applying the Banach fixed point theorem we get existence and uniqueness of \(\psi _A\) and hence of \(\psi =\mu _0^{-1/2}(\psi _A+\psi _B)\) with estimate (4.16) that is a consequence of estimates (4.8)–(4.10). \(\square \)

Remark 6.1

(Continuity with respect to the initial condition \(\psi _0\)) Given an initial datum \(\psi _0\) Proposition 4.1 defines a solution \(\psi =\Psi [\psi _0]\) to (3.4), from a small neighborhood of 0 in the \(L^\infty (\Omega )\) space with the \(C^1\)-norm \(\vert \!\vert \psi \vert \!\vert _\infty +\vert \!\vert \nabla \psi _0\vert \!\vert _\infty \) into the Banach space \(L^\infty \) with norm \(\vert \!\vert \psi \vert \!\vert _{**}\) defined in (4.14). In fact, from the proof of Proposition 4.1 and the implicit function theorem, \(\psi _0 \mapsto \Psi [\psi _0]\) is a diffeomorphism and hence

$$\begin{aligned} \vert \!\vert \Psi [\psi _0^1]-\Psi [\psi _0^2]\vert \!\vert _{**}\le c \left[ \vert \!\vert \psi _0^1-\psi _0^2\vert \!\vert _\infty +\vert \!\vert \nabla _x\psi _0^1-\nabla _x\psi _0^2\vert \!\vert _\infty \right] , \end{aligned}$$

for some positive constant c.

The function \(\psi =\Psi [\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi ]\) depends continuously on the parameters \(\Lambda ,\dot{\Lambda },\xi ,{\dot{\xi }},\phi \). To see this we argue similarly to [15, Proposition 4.3]. For example, fix \(\dot{\Lambda },\xi ,{\dot{\xi }},\phi \) and consider

$$\begin{aligned} {\bar{\psi }}{:=}\psi ^{(1)}-\psi ^{(2)}\quad \text {where}\quad \psi ^{(i)}=\Psi [\Lambda _i,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi ],\quad \text {for}\quad i=1,2 \end{aligned}$$

for \(\Lambda _1,\Lambda _2\) satisfying (3.20). Then \({\bar{\psi }}\) solves

$$\begin{aligned} {\partial } _t {\bar{\psi }} = \Delta {\bar{\psi }}+\gamma {\bar{\psi }}+ V[\Lambda _1]\bar{\psi }+\left( V[\Lambda _1]-V[\Lambda _2]\right) \psi ^{(2)}+F[\Lambda _1]-F[\Lambda _2]. \end{aligned}$$

One can easily check each term in F and obtain

$$\begin{aligned} \vert \!\vert F[\Lambda _1]-F[\Lambda _2]\vert \!\vert _{\beta -2,\alpha +2}\le {{\textbf {c}}} \vert \!\vert \Lambda _1-\Lambda _2\vert \!\vert _{l_0,\infty }, \end{aligned}$$

with \({{\textbf {c}}}<1\) if \(t_0\) is large enough. Also, using (4.2) we find that

$$\begin{aligned} \vert \!\vert V[\Lambda _1]-V[\Lambda _2]\vert \!\vert _{\beta -2,\alpha +2}\le {{\textbf {c}}} \vert \!\vert \Lambda _1-\Lambda _2\vert \!\vert _{l_0,\infty }. \end{aligned}$$

Then, arguing as in the proof of (4.8), a multiple of \(\vert \!\vert \Lambda _1-\Lambda _2\vert \!\vert _{\sharp ,l_0,\delta _0,\frac{1}{2}+\varepsilon } \uppsi \), where \(\uppsi \) is the supersolution constructed in Lemma 4.1, is a supersolution for \({\bar{\psi }}\). Similarly, one obtain analogue estimates fixing \(\xi ,{\dot{\Lambda }}, {\dot{\xi }}\). Let us consider all the parameters fixed. We define \(\bar{\psi }:=\psi [\Lambda ,{\dot{\Lambda }},\xi ,\dot{\xi },\phi _1]-\psi [\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi _2]\), which satisfies the equation

$$\begin{aligned} {\partial } _t {\bar{\psi }} = \Delta {\bar{\psi }} +V{\bar{\psi }} + F[\phi _1]-F[\phi _2]. \end{aligned}$$

For instance, we estimate

$$\begin{aligned} \vert \mu ^{-1}\left( \frac{\mu }{\mu _0}\right) ^{1/2}\left( \phi _1-\phi _2\right) {\partial } _t \eta _R \vert \lesssim&\vert \!\vert \phi _1-\phi _2\vert \!\vert _*\mu ^{l_1}R^{-1}\log (R) \mu ^2 R^{2+\alpha }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}\\ \lesssim&\,{{\textbf {c}}}\vert \!\vert \phi _1-\phi _2\vert \!\vert _{*} \mu ^{l_1+\delta -\sigma }\frac{\mu ^{-2}}{1+\vert y\vert ^{2+\alpha }}, \end{aligned}$$

with \({{\textbf {c}}}<1\) when \(t_0\) is fixed large enough, and arguing as in (4.18)-(4.19), we obtain similar estimate on the other terms of \(F[\phi _1]-F[\phi _2]\). Having the \(L^\infty \)-bound, the estimate for the gradient and the Hölder norms of \({\bar{\psi }}\) follow as in the proof of Lemma 4.1. We summarize the continuity of \(\psi [\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi ]\) with respect to the parameters in the following Proposition.

Proposition 4.2

Under the same assumption of Proposition 4.1, the function \(\psi =\Psi [\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi ]\) is continuous with respect to the parameters \(\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi \). Moreover the following estimate holds:

$$\begin{aligned}&\Vert \Psi [\Lambda ^{(1)},{{\dot{\Lambda }}}^{(1)},\xi ^{(1)},{{\dot{\xi }}}^{(1)}, \phi ^{(1)}]-\Psi [\Lambda ^{(2)},{{\dot{\Lambda }}}^{(2)},\xi ^{(2)},{{\dot{\xi }}}^{(2)}, \phi ^{(2)}]\Vert _{**}\\&\quad \le {{\textbf {c}}} \bigg \{\vert \!\vert \Lambda ^{(1)}-\Lambda ^{(2)}\vert \!\vert _{\sharp ,l_0,\delta _0,\frac{1}{2}+\varepsilon } +\vert \!\vert {\dot{\Lambda }}^{(1)}-{\dot{\Lambda }}^{(2)}\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }\\&\qquad +\vert \!\vert \xi _1^{(1)}-\xi _1^{(2)}\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon } +\vert \!\vert {{\dot{\xi }}}_1^{(1)}-{{\dot{\xi }}}_1^{(2)}\vert \!\vert _{\sharp ,1+l_1,\varepsilon } +\vert \!\vert \phi ^{(1)}-\phi ^{(2)}\vert \!\vert _*\bigg \} \end{aligned}$$

where \({{\textbf {c}}}<1\) provided that \(t_0\) is sufficiently large and the constants \(\mathfrak {b}_1,\mathfrak {b}_2\) in (3.20), (3.21) are sufficiently small.

5 Characterization of the orthogonality conditions (3.10)

Given the function \(\psi =\Psi [\Lambda ,{\dot{\Lambda }},\xi ,\dot{\xi },\phi ]\) provided by Proposition 4.1, we plug it in the inner problem for \(\phi \). From the linear theory stated in Proposition 3.1, the inner problem (3.10) with initial datum (3.15) can be solved if the orthogonality conditions

$$\begin{aligned} \int _{B_{2R}} H[\Lambda ,{\dot{\Lambda }},\xi ,\dot{\xi },\phi ](y,t(\tau ))Z_i(y)\,dy=0 \quad \text {for}\quad t>t_0,\quad \text {and}\quad i=1,2,3,4, \end{aligned}$$
(5.1)

are satisfied. The aim of this section is to characterize this set of conditions as an nonlocal system in \(\Lambda ,\xi \) for fixed \(\phi \in X_*\). The next lemma shows that the orthogonality condition with index \(i=4\) is equivalent to a nonlocal equation in the variable \( \Lambda \), for fixed \(\phi ,\xi \).

Lemma 5.1

Assume that \(\Lambda ,\xi ,\phi \) satisfy (3.20), (3.21) and (3.18) respectively. Let \(\psi =\Psi [\Lambda ,\dot{\Lambda },\xi ,{\dot{\xi }},\phi ]\) be the solution to problem (3.4) given by Proposition 4.1. Then, the condition (5.1) with index \(i=4\) is equivalent to

$$\begin{aligned} (1+ a[{\dot{\Lambda }},\xi ](t))\mathcal {J}[\dot{\Lambda }](0,t)=g(t)+G[\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi ](t) \quad \text {for}\quad t \in [t_0,\infty ), \end{aligned}$$
(5.2)

where \(\mathcal {J}\) is the solution to

$$\begin{aligned}&{\partial } _t \mathcal {J}=\Delta \mathcal {J}+\gamma \mathcal {J}-{\dot{\Lambda }}(t) G_\gamma (x,0) {\quad \hbox {in } }\Omega \times [t_0-1,\infty ),\\&\mathcal {J}(x,t)=0 {\quad \hbox {on } } {\partial } \Omega \times [t_0-1,\infty ),\\&\mathcal {J}(x,t_0-1)=0 {\quad \hbox {in } }\Omega . \end{aligned}$$

The function a is smooth, decays as \(t\rightarrow 0\) and \(a[0,0]\equiv 0\). Then, for \(\kappa \in (0,2\gamma (\sigma -\alpha \delta ))\), the following estimates on g and G hold:

$$\begin{aligned} \vert \!\vert g\vert \!\vert _{\sharp ,l_0,\delta _1,\frac{1}{2}+\varepsilon }+ \vert \!\vert g\vert \!\vert _{\sharp ,l_0,\delta _0,\varepsilon }\le C_0 e^{-\kappa t_0} , \end{aligned}$$

and

$$\begin{aligned}&\Vert G[\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi ] \Vert _{\sharp ,l_0,\delta _1,\frac{1}{2}+\varepsilon }+\Vert G[\Lambda ,\dot{\Lambda },\xi ,{\dot{\xi }},\phi ] \Vert _{\sharp ,l_0,\delta _0,\varepsilon }\nonumber \\ {}&\quad \le e^{-\kappa t_0}\big \{\vert \!\vert \Lambda \vert \!\vert _{\sharp ,l_0,\delta _0,\varepsilon }+\vert \!\vert \dot{\Lambda }\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }+\vert \!\vert \xi _1\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }+ \vert \!\vert \dot{\xi }_{1}\vert \!\vert _{\sharp ,1+l_1,\varepsilon }+\vert \!\vert \phi \vert \!\vert _*\big \}. \end{aligned}$$
(5.3)

Furthermore, we have

$$\begin{aligned}&\Vert G[\Lambda ^{(1)},{\dot{\Lambda }}^{(1)},\xi ^{(1)},\dot{\xi }^{(1)},\phi ^{(1)}]-G[\Lambda ^{(2)},\dot{\Lambda }^{(2)},\xi ^{(2)},{\dot{\xi }}^{(2)},\phi ^{(2)}] \Vert _{\sharp ,l_0,\delta _1,\frac{1+2\varepsilon }{2}}\nonumber \\&\qquad \le {\textbf {c}} \big \{ \vert \!\vert \Lambda ^{(1)}- \Lambda ^{(2)}\vert \!\vert _{\sharp ,l_0,\delta _0,\frac{1+2\varepsilon }{2}}+ \vert \!\vert \dot{\Lambda }^{(1)}- \dot{\Lambda }^{(2)}\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }\nonumber \\ {}&\quad +\vert \!\vert \xi _{1}^{(1)}- \xi _{1}^{(2)}\vert \!\vert _{\sharp ,1+l_0,\frac{1}{2}+\varepsilon }+ \vert \!\vert \dot{\xi }_{1}^{(1)}- {\dot{\xi }}_{1}^{(2)}\vert \!\vert _{\sharp ,1+l_0,\varepsilon }+ \vert \!\vert \phi ^{(1)}-\phi ^{(2)}\vert \!\vert _{*}\big \} \end{aligned}$$
(5.4)

with constant \({{\textbf {c}}}<1\) provided that \(t_0\) is fixed sufficiently large and \(\mathfrak {b}_i\) small for \(i=1,2\).

Proof

We recall that

$$\begin{aligned} H[\phi ,\psi ,\mu ,{\dot{\mu }},\xi ,{\dot{\xi }}](y,\tau ){:=}&5U(y)^4 \mu \left( \frac{\mu _0}{\mu }\right) ^{1/2} \psi (\mu y +\xi ,t(\tau ))\nonumber \\ {}&+B_0\left[ \phi +\mu \psi \right] (\mu y+\xi ,t(\tau ))+\mu ^{5/2}S_{\text {in}}(\mu y+\xi ,t(\tau ))\nonumber \\&+\mathcal {N}(\mu ^{1/2}u_3,\mu ^{1/2}{{\tilde{\phi }}})(\mu y+\xi ,t(\tau )). \end{aligned}$$

Hence, (3.13) with index \(i=4\) becomes

$$\begin{aligned} 0=&\mu ^{5/2}\int _{B_{2R}}Z_4(y)S_{\text {in}}(y,t) \,dy+\mu \left( \frac{\mu _0}{\mu }\right) ^{1/2}\int _{B_{2R}} Z_4(y) 5U(y)^4 \psi (\mu y +\xi ,t) \,dy\\&+\int _{B_{2R}}Z_4(y)B_0\left[ \phi +\mu \psi \right] (\mu y+\xi ,t) \,dy\\&+ \int _{B_{2R}}Z_4(y)\mathcal {N}(\mu ^{1/2}u_3,\mu ^{1/2}{{\tilde{\phi }}})(\mu y+\xi ,t(\tau ))\,dy\\ =&: \sum _{j=1}^{4}i_j(t). \end{aligned}$$

We follow the analogue [15, Lemma 5.1] to estimate the terms \(i_j(t)\). Firstly, we analyze \(i_1\). We have

$$\begin{aligned} i_1(t)=&\mu ^{5/2}\int _{B_{2R}} S_{\text {in}}(y,t) Z_4(y)\,dy\\ =&\mu \left( \frac{\mu _0}{\mu }\right) ^{1/2} \int _{B_{2R}}5U(y)^4 J(\mu y+\xi ,t)Z_4(y)\,dy\\&+ \int _{B_{2R}}\mathcal {N}_3(y,t) Z_4(y)\,dy\nonumber \\ {}&+\mu \int _{B_{2R}}Z_4(y)5U(y)^4 h_\gamma (\mu y+\xi ,\xi ) \,dy\\=:&a_1(t)+a_2(t)+a_3(t), \end{aligned}$$

where we used that the integral of \(Z_4(y)U(y)^4 {\partial } _{y_i} U(y)\) on \(B_{2R}(0)\) is null by symmetry for \(i=1,2,3\). Also,

$$\begin{aligned} \mu ^{-1} \left( \frac{\mu _0}{\mu }\right) ^{-1/2} a_1(t)=&\int _{B_{2R}}5U(y)^4 Z_4(y)J[{{\dot{\Lambda }}}](\mu y +\xi ,t) \,dy\\ =&\mathcal {J}[{{\dot{\Lambda }}}](0,t) \int _{B_{2R}}5U(y)^4 Z_4(y) \,dy\\&+\left[ J[{\dot{\Lambda }}](\xi ,t)-\mathcal {J}[{{\dot{\Lambda }}}](0,t)\right] \int _{B_{2R}}5U(y)^4 Z_4(y) \,dy\\&+ \int _{B_{2R}}5U(y)^4 Z_4(y)[J[{{\dot{\Lambda }}}](\mu y +\xi ,t)-J[{{\dot{\Lambda }}}](\xi ,t)] \,dy\\ =:&a_{11}[{\dot{\Lambda }}](t)+a_{12}[{\dot{\Lambda }},\xi ](t)+a_{13}[\dot{\Lambda },\xi ](t). \end{aligned}$$

The main term in the left-hand side of (5.2) is given by

$$\begin{aligned} \mu ^{-1} \left( \frac{\mu _0}{\mu }\right) ^{-1/2} c_1^{-1}(1+O(R^{-2}))^{-1}a_{11}(t)=\mathcal {J}[{\dot{\Lambda }}](0,t), \end{aligned}$$

where \(c_1(1+O(R^{-2}))=\int _{B_{2R}} 5U^4 Z_4 \,dy\). To analyze the terms \(a_{12}\), we decompose \(w[{\dot{\Lambda }}](x,t)=J[\dot{\Lambda }](x,t)-\mathcal {J}[{\dot{\Lambda }}](x,t)\) as a sum of a solution in \({{\mathbb {R}}}^3\) and a smooth one in \(\Omega \) with more decay. Then, using the Duhamel’s formula in \({{\mathbb {R}}}^3\) as in [15, Proof of (7.5)] we deduce

$$\begin{aligned}&\vert \!\vert a_{12}[{\dot{\Lambda }},\xi ]\vert \!\vert _{\sharp ,l_0,\delta _1,\frac{1}{2}+\varepsilon }+\vert \!\vert a_{12}[\dot{\Lambda },\xi ]\vert \!\vert _{\sharp ,l_0,\delta _0,\varepsilon }\lesssim e^{-\kappa t_0}\big \{\vert \!\vert \Lambda \vert \!\vert _{\sharp ,l_0,\delta _0,\frac{1}{2}+\varepsilon }+\vert \!\vert \dot{\Lambda }\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }\nonumber \\ {}&\quad +\vert \!\vert \xi _1\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }\big \}. \end{aligned}$$
(5.5)

We analyze \(a_{13}\) by splitting J as a sum of a solution to the same equation in \({{\mathbb {R}}}^3\) and smooth remainder in \(\Omega \) with more decay, and, proceeding as in [15, Proof of (5.10)], again by Duhamel’s formula in \({{\mathbb {R}}}^3\) we obtain

$$\begin{aligned} \vert J[{{\dot{\Lambda }}}](\mu y +\xi ,t)-J[{{\dot{\Lambda }}}](\xi ,t)\vert = \vert y\mu \vert ^{\sigma } \Pi [{\dot{\Lambda }},\xi ](t)\theta (\vert y\vert ), \end{aligned}$$

for some \(\sigma \in (0,1)\) and bounded smooth function \(\theta \), and \(\Pi [{\dot{\Lambda }},\xi ]\) satisfying the estimate above for \(a_{12}\). After integration, \(a_{13}[{\dot{\Lambda }},\xi ](t)\) satisfies (5.5). Taking into account the behavior of \(J_1,J_2\) and \(\phi _3\) given in (2.34), (2.35) and (2.38) respectively, we have

$$\begin{aligned} a_2&= \int _{B_{2R}} Z_4(y)\mathcal {N}_3(y,t)\,dy\\&= \int _{B_{2R}}Z_4 \left\{ 10 \left( U(y)+s(-\mu H_\gamma +\mu J+\mu ^{-1/2}\phi _3 \eta _l)\right) ^3 \right. \\ {}&\quad \left. \times \left( -\mu H_\gamma +\mu J+\mu ^{-1/2}\phi _3 \eta _l\right) ^2\right\} \,dy\\&=\mu ^2 \int _{B_{2R}} 10 Z_4(y)U(y)^3 Q[\Lambda ,\dot{\Lambda },\xi ](y,t)\,dy, \end{aligned}$$

for some constant \(s\in (0,1)\) and bounded smooth function \(Q[\Lambda ,\xi ](y,t)\) satisfying (5.5).

Finally, Taylor expanding \(h_\gamma (x,\xi )\) at \(x=\xi \), we get

$$\begin{aligned} a_3&=\mu \int _{B_{2R}}Z_4(y)5U(y)^4 h_\gamma (\mu y+\xi ,\xi )\,dy\\&= \mu R_\gamma (\xi ) \int _{B_{2R}} Z_4(y) 5U(y)^4 \,dy\\&\quad +\mu ^3 \int _{B_{2R}}Z_4(y)5U(y)^4 \left( y\cdot D_{xx}h_\gamma (\mu y^*(y)+\xi ,\xi )\cdot y\right) \,dy\\&=\mu ^2 \Pi _2[\Lambda ,\xi ](t), \end{aligned}$$

for some \(y^* \in \overline{[0,y]}\) and a smooth bounded function \(\Pi _2(t)\). The term \(\mu ^{-1/2}\mu _0^{-1/2}i_1(t)=\sum _{i=1}^{3}a_i(t)\) gives the left-hand side in (5.2). Now, we look at \(i_2\). We decompose

$$\begin{aligned}&\mu ^{-\frac{1}{2}}\mu _0^{-\frac{1}{2}}i_2(t) \\ {}&\quad =\int _{B_{2R}}Z_4(y) 5U(y)^4 \psi [\Lambda ,\xi ,{\dot{\Lambda }},{\dot{\xi }},\phi ](\mu y +\xi ,t) \,dy\\&\quad =\psi [0,0,0,0,0](0,t)\int _{B_{2R}}Z_4(y) 5U(y)^4 \,dy\\&\qquad +\int _{B_{2R}}Z_4(y) 5U(y)^4\left\{ \psi [0,0,0,0,0](\mu y+\xi ,t)-\psi [0,0,0,0,0](0,t)\right\} \,dy\\&\qquad +\int _{B_{2R}}Z_4(y) 5U(y)^4 \left\{ \psi [\Lambda ,\xi ,{\dot{\Lambda }},{\dot{\xi }},\phi ](\mu y +\xi ,t) -\psi [0,0,0,0,0](\mu y+\xi ,t)\right\} \,dy\\&\quad =:b_1(t)+b_2[\Lambda ,\xi ](t)+b_3[\Lambda ,\xi ,\dot{\Lambda },{\dot{\xi }},\phi ](t), \end{aligned}$$

The term

$$\begin{aligned} b_1(t)=\psi [0,0,0,0,0](0,t)\int _{B_{2R}} 5U(y)^4 Z_4(y)\,dy, \end{aligned}$$

is independent of parameters and, as a consequence of Proposition 4.1, satisfies the estimate

$$\begin{aligned}&\vert \!\vert b_1\vert \!\vert _{\sharp ,l_0,\delta _0,\varepsilon }+\vert \!\vert b_1\vert \!\vert _{\sharp ,l_0,\delta _1,\frac{1}{2}+\varepsilon }\le C e^{-\kappa t_0}. \end{aligned}$$

Applying the mean value theorem to \(\psi \) and using the gradient estimate we deduce the same bound for \(b_2\). This gives the main term \(b_1(t)+b_2(t)=g(t)\) in the right-hand side of (5.2). We analyze \(b_3(t)\). By Proposition 4.2 applied to

$$\begin{aligned} \psi [ \Lambda ,\xi ,{\dot{\Lambda }},{\dot{\xi }},\phi ]-\psi [0,0,0,0,0] \end{aligned}$$

we obtain

$$\begin{aligned} \vert \!\vert b_3\vert \!\vert _{\sharp ,l_0,\delta _1,\frac{1}{2}+\varepsilon }+\vert \!\vert b_3\vert \!\vert _{\sharp ,l_0,\delta _0,\varepsilon }\lesssim&e^{-\kappa t_0}\bigg \{\vert \!\vert \Lambda \vert \!\vert _{\sharp ,l_0,\delta _0,\frac{1}{2}+\varepsilon }+\vert \!\vert {\dot{\Lambda }}\vert \!\vert _{\sharp , l_1,\delta _1,\varepsilon }\\&+\vert \!\vert {\dot{\xi }}_{1}\vert \!\vert _{\sharp ,1+l_1,\varepsilon }+\vert \!\vert \xi _1\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }+\vert \!\vert \phi \vert \!\vert _*\bigg \}. \end{aligned}$$

Also, again as a consequence of the Lipschitz estimates in \(\psi \) we have for example

$$\begin{aligned} \vert \!\vert b_3[{\dot{\Lambda }}_1]-b_3[\dot{\Lambda }_2]\vert \!\vert _{\sharp ,l_0,\delta _1,\frac{1}{2}+\varepsilon }\le {{\textbf {c}}} \vert \!\vert {\dot{\Lambda }}_1- {\dot{\Lambda }}_2\vert \!\vert _{\sharp , l_1,\delta _1,\varepsilon }, \end{aligned}$$

for any \({\dot{\Lambda }}_1,{\dot{\Lambda }}_2 \in X_{\sharp ,l_1,\delta _1,\varepsilon }\) and fixed \(\Lambda ,\xi ,\dot{\Lambda },{\dot{\xi }}\) in the respective spaces. We analyze \(i_3\). We recall that

$$\begin{aligned} B_0[\phi +\mu \psi ]=5\left[ \left( U-\mu H_\gamma +\mu J[\dot{\Lambda }]+\mu ^{-1/2}\phi _3(y,t)\eta _3\right) ^4-U^4\right] [\phi +\mu \psi ], \end{aligned}$$

which is linear in \(\phi ,\psi \) and satisfies

$$\begin{aligned} \vert B_0[\phi +\mu \psi ](\mu y+\xi ,t)\vert \lesssim \frac{\mu }{1+\vert y\vert ^{3}}\mu \vert \phi +\mu \psi \vert . \end{aligned}$$

It follows that

$$\begin{aligned} \vert i_3(t)\vert \lesssim&\mu {\vert \!\vert \phi \vert \!\vert _* \mu ^{1+l_1}{R^3 }+\vert \!\vert \psi \vert \!\vert _{**}\mu ^2 \mu ^{l_1+\delta -\sigma }}\\ \lesssim&e^{-\kappa t_0}\mu ^{l_0}. \end{aligned}$$

Then, the Hölder bounds on \(\psi \) and \(\phi \) in the respective norms give estimate (5.3) for \(i_3\), and using Proposition 4.2 we also get the Lipschitz property (5.3) for \(i_3\). Finally, we have

$$\begin{aligned}&\vert \mathcal {N}(\mu ^{1/2}u_3,\mu ^{1/2}{{\tilde{\phi }}})(\mu y+\xi ,t(\tau ))\vert \\ {}&\quad \lesssim \frac{1}{1+\vert y\vert ^3} \left( \phi +\mu \psi \right) ^2\\&\quad \lesssim \frac{1}{1+\vert y\vert ^3} \left( \vert \!\vert \phi \vert \!\vert _*^2 \mu ^{2(1+l_1)}R^6+\vert \!\vert \psi \vert \!\vert _{**}^2\mu ^2 \frac{\mu ^{2(l_1+\delta -\sigma )}}{1+\vert y\vert ^{2\alpha }} \right) \\&\quad \lesssim e^{-\kappa t_0}\mu ^{l_0}, \end{aligned}$$

and (5.3)–(5.3) for \(i_4\) follows arguing as for \(i_3\). Summing up the estimates we obtain \(G[\Lambda ,\dot{\Lambda },\xi ,{\dot{\xi }},\phi ](t)=b_3+i_3+i_4\) as in (5.2) with the properties (5.3) and (5.4). \(\square \)

Now, we characterize the conditions

$$\begin{aligned} \int _{B_{2R}} Z_i(y) H[\Lambda ,{\dot{\Lambda }},\xi ,\dot{\xi },\phi ](y,t)\,dy=0, \quad \text {for}\quad t\in (t_0,\infty ) \quad \text {and}\quad i=1,2,3. \end{aligned}$$

This characterization is given in the following lemma, whose proof, similar to the one of Lemma 5.1, is omitted.

Lemma 5.2

The relation (5.1) for \(i=1,2,3\) is equivalent to

$$\begin{aligned} {\dot{\xi }}_{1,i}=\mathfrak {c}_i \mu _0(t)^{1+l_1}+\Theta _i[\Lambda ,\dot{\Lambda },\xi ,\phi ](t) \end{aligned}$$
(5.6)

for smooth bounded function \(\Theta \) which satisfies

$$\begin{aligned}&\Vert \Theta [\Lambda ,{\dot{\Lambda }},\xi ,\phi ] \Vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }+\Vert \Theta [\Lambda ,\dot{\Lambda },\xi ,\phi ] \Vert _{{\sharp ,1+l_1,\varepsilon }}\le e^{-\kappa t_0}\big \{\vert \!\vert \Lambda \vert \!\vert _{\sharp ,l_0,\delta _0,\varepsilon }\nonumber \\&\quad +\vert \!\vert \dot{\Lambda }\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }+\vert \!\vert \xi _1\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }+\vert \!\vert \phi \vert \!\vert _*\big \}. \end{aligned}$$
(5.7)

Furthermore, we have

$$\begin{aligned}&\Vert \Theta [\Lambda ^{(1)},{\dot{\Lambda }}^{(1)},\xi ^{(1)},\dot{\xi }^{(1)},\phi ^{(1)}]-\Theta [\Lambda ^{(2)},\dot{\Lambda }^{(2)},\xi ^{(2)},{\dot{\xi }}^{(2)},\phi ^{(2)}] \Vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }\nonumber \\ {}&\quad \le {{\textbf {c}}} \big \{ \vert \!\vert \Lambda ^{(1)}- \Lambda ^{(2)}\vert \!\vert _{\sharp ,l_0,\delta _0,\frac{1}{2}+\varepsilon }+\vert \!\vert \dot{\Lambda }^{(1)}- \dot{\Lambda }^{(2)}\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }\nonumber \\ {}&\qquad +\vert \!\vert \xi _{1}^{(1)}- \xi _{1}^{(2)}\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }+ \vert \!\vert \dot{\xi }_{1}^{(1)}- {\dot{\xi }}_{1}^{(2)}\vert \!\vert _{\sharp ,1+l_1,\varepsilon }\nonumber \\ {}&\qquad + \vert \!\vert \phi ^{(1)}-\phi ^{(2)}\vert \!\vert _{*}\big \}, \end{aligned}$$
(5.8)

with constant \({{\textbf {c}}}<1\) provided that \(t_0\) is fixed sufficiently large and \(\mathfrak {b}_i\) small for \(i=1,2\).

6 Choice of parameters \(\Lambda ,\xi \)

In the previous section we have proved that if \(\phi \in X_*\) and \(\Lambda ,\xi \) satisfy (3.20) and (3.21) then the system of orthogonality conditions

$$\begin{aligned} \int _{B_{2R}} H[\Lambda ,{\dot{\Lambda }},\xi ,\dot{\xi },\phi ](y,t(\tau ))Z_i(y)\,dy=0 \quad \text {for}\quad t\in [t_0,\infty ) \quad \text {and}\quad i=1,2,3,4, \end{aligned}$$

is equivalent to the nonlocal system in \([t_0,\infty )\)

$$\begin{aligned} {\left\{ \begin{array}{ll} (1+ a[{\dot{\Lambda }},\xi ](t))\mathcal {J}[{\dot{\Lambda }}](0,t)=g(t)+G[\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\phi ](t),&{} \\ {\dot{\xi }}_{1,i}=\mathfrak {c}_i\mu _0(t)^2 \left( 1+\Theta _i[\Lambda ,\dot{\Lambda },\xi ,\phi ]\right) &{} \text {for}\quad i=1,2,3, \end{array}\right. } \end{aligned}$$
(6.1)

with gGa as in Lemma 5.1 and \(\Theta _i\) as in Lemma 5.2. Next, we verify that this system is solvable for \(\Lambda ,\xi \) satisfying (3.20),(3.21) respectively. This relies on the following proposition, proved in Sect. 8, about the solvability of the nonlocal operator \(\mathcal {J}[{\dot{\Lambda }}](0,t)=g(t)\) for g as in (5.2).

Proposition 6.1

Let \(h:[t_0,\infty )\rightarrow {{\mathbb {R}}}\) a function satisfying \(\vert \!\vert h\vert \!\vert _{\sharp ,c_1,c_2,\varepsilon }<\infty \) for some constants \(\varepsilon >0\) and \(c_1,c_2\) such that

$$\begin{aligned} 0<c_2\le c_1<\frac{\lambda _1-\gamma }{2\gamma }. \end{aligned}$$
(6.2)

Then there exists a function \(\Lambda \in C^{\frac{1}{2}+\varepsilon }(t_0-1,\infty )\) satisfying

$$\begin{aligned} \mathcal {J}[{\dot{\Lambda }}](0,t)=h(t) {\quad \hbox {in } }(t_0,\infty ), \end{aligned}$$
(6.3)

where \(\mathcal {J}[{\dot{\Lambda }}]\) satisfies (2.33), and there exists a constant \(C_1\) such that

$$\begin{aligned} \vert \!\vert \Lambda \vert \!\vert _{\sharp ,c_1,c_2,\varepsilon +\frac{1}{2}}\le C_1 \vert \!\vert h\vert \!\vert _{\sharp ,c_1,c_2,\varepsilon }. \end{aligned}$$
(6.4)

Moreover, if \(\vert \!\vert h\vert \!\vert _{\sharp ,c_1,c_2,\frac{1}{2}+\varepsilon } <\infty \) then \(\Lambda \in C^{1,\varepsilon }(t_0-1,\infty )\) and there exists a constant \(C_2\) such that

$$\begin{aligned} \vert \!\vert {\dot{\Lambda }}\vert \!\vert _{\sharp ,c_1,c_2,\varepsilon }\le C_2 \vert \!\vert h\vert \!\vert _{\sharp ,c_1,c_2,\frac{1}{2}+\varepsilon }. \end{aligned}$$
(6.5)

Thus, the linear operators

$$\begin{aligned}&T_1: X_{\sharp ,c_1,c_2,\varepsilon }\rightarrow X_{\sharp ,c_1,c_2,\varepsilon +\frac{1}{2}}\nonumber \\ {}&h(t)\mapsto \Lambda [h](t), \end{aligned}$$
(6.6)

and

$$\begin{aligned}&{\hat{T}}_1{:=}\frac{d}{dt}\circ T_1: X_{\sharp ,c_1,c_2,\frac{1}{2}+\varepsilon }\rightarrow X_{\sharp ,c_1,c_2,\varepsilon }\nonumber \\ {}&h(t)\mapsto {\dot{\Lambda }}[h](t), \end{aligned}$$
(6.7)

are well-defined and continuous.

We are ready to solve the system (6.1) in \(\Lambda ,\xi \) for fixed \(\phi \in X_*\).

Proposition 6.2

Suppose that \(\phi \) satisfies (4.15). Then, there exist \(\Lambda =\Lambda [\phi ](t)\) and \(\xi =\xi [\phi ](t)\) to the nonlinear nonlocal system (6.1) which satisfy (3.20) and (3.21) respectively. Moreover, they satisfy

$$\begin{aligned}&\vert \!\vert \Lambda [\phi _1] - \Lambda [\phi _2]\vert \!\vert _{\sharp ,l_0,\delta _0,\frac{1}{2}+\varepsilon }\le {{\textbf {c}}} \vert \!\vert \phi _1-\phi _2\vert \!\vert _{*},\nonumber \\ {}&\vert \!\vert {\dot{\Lambda }}[\phi _1] - {\dot{\Lambda }}[\phi _2]\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }\le {{\textbf {c}}} \vert \!\vert \phi _1-\phi _2\vert \!\vert _{*},\nonumber \\&\vert \!\vert \xi [\phi _1]-\xi [\phi _2]\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }\le {{\textbf {c}}} \vert \!\vert \phi _1-\phi _2\vert \!\vert _{*},\nonumber \\ {}&\vert \!\vert \dot{\xi }[\phi _1]-{\dot{\xi }}[\phi _2]\vert \!\vert _{\sharp ,1+l_1,\varepsilon }\le {{\textbf {c}}} \vert \!\vert \phi _1-\phi _2\vert \!\vert _{*}, \end{aligned}$$
(6.8)

with constant \({{\textbf {c}}}<1\) provided that \(t_0\) is fixed sufficiently large and \(\mathfrak {b}_i\) small for \(i=1,2\).

Proof

Firstly, we observe that Eq. (5.2) can be rewritten as

$$\begin{aligned} \mathcal {J}[{\dot{\Lambda }}](0,t)=g_1(t)+G_1[{\dot{\Lambda }},\Lambda ,\dot{\xi },\xi ,\phi ](t), \end{aligned}$$

where

$$\begin{aligned} g_1(t)+G_1[{\dot{\Lambda }},\Lambda ,{\dot{\xi }},\xi ]=(1+a[\dot{\Lambda },\xi ])^{-1}[g(t)+G[{\dot{\Lambda }},\Lambda ,\dot{\xi },\xi ,\phi ]](t), \end{aligned}$$
(6.9)

for new functions \(g_1,G_1\) satisfying the same properties of gG in Lemma 5.1. By Proposition 6.1 we reduce the equation for \( \Lambda \) to a fixed point problem

$$\begin{aligned} {\dot{\Lambda }}(t)=\mathcal {F}_1[{\dot{\Lambda }}](t),\quad \mathcal {F}_1[\dot{\Lambda }](t)={\hat{T}}_1\left[ g_1(t)+G_1[\Lambda ,{\dot{\Lambda }},\xi ,\dot{\xi },\phi ]\right] , \end{aligned}$$

where \({\hat{T}}_1\) is defined in (6.7). Let

$$\begin{aligned} {\dot{\Lambda }}_0(t){:=}{\hat{T}}_1[g_1](t) \end{aligned}$$

and define the operator \(\mathcal {L}_1[h] :={\hat{T}}_1 [h-g_1]\). We use the notation

$$\begin{aligned} \mathcal {L}_1[h]=\uplambda [h](t){:=}\dot{\Lambda }[h](t)-{\dot{\Lambda }}_0(t), \end{aligned}$$

for any \(h\in X_{\sharp ,l_1,\delta _1,\frac{1}{2}+\varepsilon }\). Observe that

$$\begin{aligned} \vert *\vert {{\dot{\Lambda }}[h]}&=\vert *\vert {{\dot{\Lambda }}_0}+\vert \uplambda [h]\vert \\&\lesssim \mu ^{l_0}\vert \!\vert g\vert \!\vert _{\sharp ,l_0,\delta _1,\frac{1}{2}+\varepsilon }+\mu ^{l_1}\vert \!\vert h\vert \!\vert _{{\sharp ,l_1,\delta _1,\frac{1}{2}+\varepsilon }}. \end{aligned}$$

Given \(h_j\in X_{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }\) we consider the solution to the ODE

$$\begin{aligned} {\dot{\xi }}_{1,j}=\mathfrak {c}_j\mu _0(t)^{1+l_1}+h_j(t), \end{aligned}$$
(6.10)

given explicitly by

$$\begin{aligned} \xi _{1,j}[h](t)=\mathfrak {c}_j\int _t^\infty \mu _0(s)^{1+l_1} \,ds+\int _t^\infty h(s)\,ds{:=}\Upsilon _{j}+ \int _t^\infty h(s)\,ds. \end{aligned}$$

In particular, we have

$$\begin{aligned}&\vert \xi _{1,j}(t)\vert \lesssim \mu _0(t)^{1+l_1}+\mu _0(t)^{1+l_1} \vert \!\vert h\vert \!\vert _{1+l_1,\infty },&\vert *\vert {{\dot{\xi }}_{1,j}(t)}\lesssim \mu _0(t)^{1+l_1} \vert \!\vert h\vert \!\vert _{1+l_1,\infty } \end{aligned}$$

We define the vector

$$\begin{aligned} \Xi (t) :={\dot{\xi }} -\dot{\Upsilon }=h(t), \end{aligned}$$

where \(h=(h_1,h_2,h_3)\) satisfies \(\vert \!\vert h_i\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }<\infty \) for \(i=1,2,3\). We define

$$\begin{aligned} \vert \!\vert h\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }{:=}\max _{i=1,2,3} \vert \!\vert h_i\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }. \end{aligned}$$

Let \(\mathcal {L}_2\) the linear operator defined as \(\mathcal {L}_2[h]=\Xi \) by relation (6.10) for \(i=1,2,3\). We observe that \((\dot{\Lambda },{\dot{\xi }})\) is a solution to (6.1) if \(( \uplambda ,\Xi )\) satisfies

$$\begin{aligned} (\uplambda ,\Xi )=\mathcal {A}(\uplambda ,\Xi ), \end{aligned}$$

where \(\mathcal {A}\) is the operator

$$\begin{aligned} \mathcal {A}(\uplambda ,\Xi ){:=}\left( \mathcal {A}_1[\uplambda ,\Xi ],\mathcal {A}_2(\uplambda ,\Xi )\right) {:=} \left( {\hat{T}}_1[{\hat{G}}_1[\uplambda ,\Xi ,\phi ]],\mathcal {L}_2[\hat{\Theta }[\uplambda ,\Xi ,\phi ]]\right) , \end{aligned}$$

and

$$\begin{aligned}&{\hat{G}}_1(\uplambda ,\Xi ,\phi ){:=}G_1\left[ \Lambda _0(t)+\int _{t}^{\infty }\uplambda (s)\,ds,{\dot{\Lambda }}_0-\uplambda , \Upsilon +\int _{t}^{\infty }\Xi (s)\,ds,{\dot{\Upsilon }}-\Xi ,\phi \right] ,\\&{\hat{\Theta }} (\uplambda ,\Xi ,\phi ){:=} \Theta \left[ \Lambda _0(t)+\int _{t}^{\infty }\uplambda (s)\,ds,\dot{\Lambda }_0-\uplambda ,\Upsilon (t)+\int _{t}^{\infty }\Xi (s)\,ds,\dot{\Upsilon }-\Xi ,\phi \right] , \end{aligned}$$

with \(G_1\) and \(\Theta \) defined in (6.9) and (5.6). We show that there exists a unique fixed point \(\left( \uplambda ,\Xi \right) =\left( \uplambda [\phi ],\Xi [\phi ]\right) \) in

$$\begin{aligned} \mathcal {B}=\{ (\uplambda ,\Xi ) \in (L^\infty (t_0,\infty ))^4: \vert \!\vert \uplambda \vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }+\vert \!\vert \Xi \vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }\le e^{-\kappa t_0}L \} \end{aligned}$$

for some L fixed large. Indeed, estimates (6.5) and (5.3) give

$$\begin{aligned} \vert \!\vert \mathcal {A}_1[\uplambda ,\Xi ]\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }&\le C_2 \vert \!\vert {\hat{G}}_1[\uplambda ,\Xi ,\phi ]\vert \!\vert _{\sharp ,l_1,\delta _1,\frac{1}{2}+\varepsilon }\\&\le C_2e^{-\kappa t_0}\left\{ \vert \!\vert \uplambda \vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon } +\vert \!\vert \Xi \vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }+\vert \!\vert \phi \vert \!\vert _*\right\} . \end{aligned}$$

Also, from (5.7)

$$\begin{aligned} \vert \!\vert \mathcal {A}_2[\uplambda ,\Xi ]\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }&\le \vert \!\vert \Theta [\uplambda ,\Xi ,\phi ]\vert \!\vert _{\sharp ,1+l_1,\frac{1}{2}+\varepsilon }\\&\le C e^{-\kappa t_0}\left\{ \vert \!\vert \uplambda \vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }+\vert \!\vert \Xi \vert \!\vert _{\sharp ,1+l_1, \frac{1}{2}+\varepsilon }+\vert \!\vert \phi \vert \!\vert _*\right\} . \end{aligned}$$

We have to verify that \(\mathcal {A}\) is a contraction. For instance, we have

$$\begin{aligned} \vert \!\vert \mathcal {A}_1[\uplambda _1,\Xi ]-\mathcal {A}_1[\uplambda _2,\Xi ]\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }&= \vert \!\vert {\hat{T}}_1[{\hat{G}}_1[\uplambda _1,\Theta ,\phi ]-{\hat{G}}_1[\uplambda _2,\Theta ,\phi ]]\vert \!\vert _{\sharp ,l_1,\delta _1,\frac{1}{2}+\varepsilon }\\&\le C_2\vert \!\vert {\hat{G}}_1[\uplambda _1,\Theta ,\phi ]-{\hat{G}}_1[\uplambda _2,\Theta ,\phi ]\vert \!\vert _{\sharp ,l_0,\delta _1,\frac{1}{2}+\varepsilon }\\&\le C_2 {{\textbf {c}}} \vert \!\vert \uplambda _1-\uplambda _2\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon }, \end{aligned}$$

where \(C_2,{{\textbf {c}}}\) is the constant appearing in (6.5) and (5.4) respectively. Since \({{\textbf {c}}}\) can be as small as required provided that \(t_0\) is fixed sufficiently large, we obtain that \(\mathcal {A}_1\) is a contraction map. By means of the Lipschitz property of \({\hat{\Theta }}\) in (5.8) we can estimate \(\mathcal {A}_2[\uplambda _1,\Xi _1]-\mathcal {A}_2[\uplambda _2,\Xi ]\) similarly. Finally, using the estimates on \({\hat{G}},{\hat{\Theta }}\) with respect to \(\Xi \), we get

$$\begin{aligned} \vert \!\vert \mathcal {A}_1(\uplambda _1,\Xi _1)-\mathcal {A}_1(\uplambda _1,\Xi _1)\vert \!\vert _{\sharp , l_1,\delta _1,\varepsilon }\le {{\textbf {c}}} \left[ \vert \!\vert \uplambda _1-\uplambda _2\vert \!\vert _{\sharp ,l_1,\delta _1,\varepsilon } +\vert \!\vert \Xi _1-\Xi _2\vert \!\vert _{\sharp ,l_1,\delta _1,\frac{1}{2}+\varepsilon }\right] . \end{aligned}$$

As a consequence of the Banach fixed point theorem, provided that L and \(t_0\) are fixed large, the map \(\mathcal {A}\) has a unique fixed point \((\uplambda ,\Xi )\) in the space \(\mathcal {B}\). Observe that

$$\begin{aligned} \Lambda [h](t)=-\int _{t}^\infty {\hat{T}}_1[h](s)\,ds=T_1[h], \end{aligned}$$

where \(T_1\) is defined in (6.6), satisfies (3.20) thanks to (6.4). Also, the components of vector \(\xi _1=\int _{t}^\infty \Xi (s)\,ds\) satisfy (3.21). This proves the existence of a solution \((\Lambda ,\xi )\) to the system (6.1) satisfying (3.20)–(3.21). With similar estimates on \(\uplambda [\phi _1]-\uplambda [\phi _2]\) and \(\Xi [\phi _1]-\Xi [\phi _2]\), using (5.4) and (5.8), relations (6.8) follow. \(\square \)

We observe from the proof that \({\hat{T}}_1\), like an half-fractional derivative, loses 1/2-Hölder exponent but we regain it through \(g,G_1\) as a consequence of estimates on \(\psi \) from Proposition 4.1. This is the main reason to put all the terms of \(S[u_3]\) involving directly \({\dot{\mu }}\) in the outer error (2.25). Indeed, to get \({\dot{\Lambda }}\in C^{\varepsilon }\) it is crucial to allow H in (3.11) (and hence \(S_{\text {in}}\) in (2.24)) to depend on \({\dot{\Lambda }}\) only indirectly through \(\psi [\dot{\Lambda }]\) or \(J_1[{\dot{\Lambda }}]\).

Remark 7.1

By remark 4.1 the outer solution \(\psi =\Psi [\psi _0]\) is smooth as a function of the initial datum \(\psi _0\), provided that \(\vert \!\vert \psi _0\vert \!\vert _\infty +\vert \!\vert \nabla \psi _0\vert \!\vert _\infty \) is sufficiently small. Thus, also the parameters \(\Lambda [\psi _0],\xi [\psi _0]\) found in Proposition 6.2 depend smoothly on \(\psi _0\), and from the proof we also obtain

$$\begin{aligned}&\vert \!\vert \Lambda [\psi _0^1]-\Lambda [\psi _0^2]\vert \!\vert _{\infty }\lesssim \vert \!\vert \psi _0^1-\psi _0^2\vert \!\vert _\infty ,\\&\vert \!\vert \xi _1[\psi _0^1]-\xi _1[\psi _0^2]\vert \!\vert _{\infty }\lesssim \vert \!\vert \psi _0^1-\psi _0^2\vert \!\vert _\infty . \end{aligned}$$

7 Final argument: solving the inner problem

This section provides the final step in the proof of Theorem 1.1. At this point, given \(\phi \) satisfying (4.15), we have a solution \(\psi =\Psi [\Lambda [\phi ],\xi [\phi ],\phi ]\) to the outer problem (3.4) and parameters \(\Lambda [\phi ],\xi [\phi ]\) such that the orthogonality conditions (5.1) are satisfied. Thus, to get a solution

$$\begin{aligned} u=u_3+{{\tilde{\phi }}}=u_3+\mu _0^{\frac{1}{2}}\psi +\eta _R \mu ^{-\frac{1}{2}} \phi , \end{aligned}$$

we need to prove the existence of \(\phi \) such that \(\vert \!\vert \phi \vert \!\vert _*<\infty \).

Proof of Theorem 1.1

We make a fixed point argument using the linear estimate (3.17). Proposition 3.1 defines a linear operator \(\mathcal {T}:h \mapsto (\phi [h],e[h])\) which is continuous between the \(L^\infty \)-weighted space described in (3.17). Thus, the solution \(\phi \) to the nonlinear inner problem satisfies

$$\begin{aligned} \phi =\mathcal {A}_{\text {in}}(\phi ),\quad \text {where}\quad \mathcal {A}_{\text {in}}(\phi ){:=}\mathcal {T}(H[\phi ]). \end{aligned}$$
(7.1)

We claim that \(\mathcal {A}_\text {in}\) has a unique fixed point in the space

$$\begin{aligned} \mathcal {B}=\{\phi \in L^\infty (B_{2R}): \vert \!\vert \phi \vert \!\vert _*< {{\textbf {b}}}\}, \end{aligned}$$

for some fixed constant \({{\textbf {b}}}\) large. Firstly, we prove

$$\begin{aligned} \vert H[\Lambda ,\xi ,{\dot{\Lambda }},\dot{\xi }](y,t)\vert \lesssim \frac{\mu ^{1+l_1}}{1+\vert y\vert ^4}. \end{aligned}$$

We recall that

$$\begin{aligned} H[\phi ,\psi ,\mu ,{\dot{\mu }},\xi ,{\dot{\xi }}](y,\tau ){:=}&5U(y)^4 \mu \left( \frac{\mu _0}{\mu }\right) ^{1/2} \psi (\mu y +\xi ,t(\tau ))\\ {}&+B_0\left[ \phi +\mu \psi \right] (\mu y+\xi ,t(\tau ))+\mu ^{5/2}S_{\text {in}}(\mu y+\xi ,t(\tau ))\\&+\mathcal {N}(\mu ^{1/2}u_3,\mu ^{1/2}{{\tilde{\phi }}})(\mu y+\xi ,t(\tau )). \end{aligned}$$

Using the estimate on \(\psi \) given in Proposition 4.1, we have

$$\begin{aligned} \vert 5U(y)^4 \mu \left( \frac{\mu _0}{\mu }\right) ^{1/2} \psi (\mu y +\xi ,t(\tau ))\vert \lesssim e^{-\kappa t_0} \frac{\mu ^{1+l_1+\delta -\sigma }}{1+\vert y\vert ^{4+\alpha }} \end{aligned}$$

and from (3.6) we get

$$\begin{aligned} \vert B_0\left[ \phi +\mu \psi \right] (\mu y+\xi ,t(\tau ))\vert&\lesssim \frac{1}{1+\vert y\vert ^3}\vert \mu H_\gamma +\mu J + \mu ^{-1/2}\phi _3 \eta _3\vert (\phi +\mu \psi )\\&\lesssim \frac{\mu }{1+\vert y\vert ^3} \left( {{\textbf {b}}} e^{-\kappa t_0}\frac{\mu ^{1+l_1}R^3}{1+\vert y\vert ^4} +\mu \vert \!\vert \psi \vert \!\vert _{**}\frac{\mu ^{l_1+\delta -\sigma }}{1+\vert y\vert ^{\alpha }}\right) \\&\lesssim e^{-\kappa t_0}\frac{\mu ^{1+l_1}}{1+\vert y\vert ^4}. \end{aligned}$$

Recalling the estimates on \(\phi \) at \(y\sim 0\) and \(y\sim R\) given by the norm (3.18), using that \(R=\mu ^{- \delta }\) with \(\delta \) satisfying (2.28) we deduce

$$\begin{aligned} \vert \mathcal {N}(\mu ^{1/2}u_3,\mu ^{1/2}{{\tilde{\phi }}})(\mu y+\xi ,t(\tau ))\vert&\lesssim \frac{1}{1+\vert y\vert ^3} \left( \phi +\mu \psi \right) ^2\\&\lesssim \frac{1}{1+\vert y\vert ^3} \left( {{\textbf {b}}} \frac{\mu ^{2(1+l_1)}R^6}{1+\vert y\vert ^8} +\mu ^2 \frac{\mu ^{2(l_1+\delta -\sigma )}}{1+\vert y\vert ^{2\alpha }} \right) \\&\lesssim e^{-\kappa t_0}\frac{\mu ^{1+l_1}}{1+\vert y\vert ^4}. \end{aligned}$$

By Lemma 2.5 we have the main error

$$\begin{aligned} \vert \mu ^{5/2}S_{\text {in}}(\mu y+\xi ,t(\tau ))\vert \lesssim \frac{\mu ^{1+l_1}}{1+\vert y\vert ^4}. \end{aligned}$$

Thanks to the previous section, H satisfies the orthogonality conditions required by Proposition 3.1. Thus, provided that \(t_0\) is large enough, we have

$$\begin{aligned} \vert \!\vert \mathcal {T}[H]\vert \!\vert _*\le C \vert \!\vert H\vert \!\vert _{\nu ,4}< {{\textbf {b}}}, \end{aligned}$$

for \({{\textbf {b}}}\) chosen large, where C is the constant in (3.19). This proves \(\mathcal {A}_{\text {in}}(\phi )\in \mathcal {B}\). Now, we need to prove that for \(\phi ^{(1)},\phi ^{(2)} \in \mathcal {B}\) we have

$$\begin{aligned} \vert *\vert {H[\phi ^{(1)}]-H[\phi ^{(2)}]}\lesssim {{\textbf {c}}} \vert \!\vert \phi ^{(1)}-\phi ^{(2)}\vert \!\vert _{*} \frac{\mu ^{1+l_1}}{1+\vert y\vert ^4}, \end{aligned}$$

for some \({{\textbf {c}}}<1\). This is a consequence of Proposition 4.2 and Proposition 6.2. Indeed, for instance we get

$$\begin{aligned}&5U(y)^4 \mu _0\vert *\vert {e^{\Lambda [\phi ^{(1)}]} \psi [\phi ^{(1)}]- e^{\Lambda [\phi ^{(2)}]} \psi [\phi ^{(2)}]}\\&\quad =5U(y)^4 \mu _0 \vert *\vert {\left[ e^{\Lambda [\phi ^{(1)}]}-e^{\Lambda [\phi ^{(2)}]}\right] \psi [\phi ^{(1)}]+e^{\Lambda [\phi ^{(2)}]} \left[ \psi [\phi ^{(1)}]-\psi [\phi ^{(2)}]\right] }\\&\quad \lesssim {{\textbf {c}}} \vert \!\vert \phi ^{(1)}-\phi ^{(2)}\vert \!\vert _* \frac{\mu ^{1+l_1}}{1+\vert y\vert ^4}, \end{aligned}$$

and similarly we get the same control on the other terms of \(H[\phi ^{(1)}]-H[\phi ^{(2)}]\). Finally, since the operator \(\mathcal {T}:X_{\nu ,4}\rightarrow X_{*}\) is continuous, where \(X_{\nu ,4}\) is defined in (3.14) for \(a=2\), by composition with \(H:X_* \rightarrow X_{\nu ,4}\) we obtain

$$\begin{aligned} \vert \!\vert \mathcal {A}_\text {in}[\phi ^{(1)}]-\mathcal {A}_{\text {in}}[\phi ^{(2)}]\vert \!\vert _*\le {\textbf {c}} \vert \!\vert \phi ^{(1)}-\phi ^{(2)}\vert \!\vert _*, \end{aligned}$$

provided that \(t_0\) is fixed sufficiently large. Thus, \(\mathcal {A}_\text {in}:\mathcal {B}\rightarrow \mathcal {B}\) is a contraction map and by Banach fixed point theorem we obtain the existence and uniqueness of \(\phi \in X_*\) such that (7.1) holds. Finally, we recall that the constant \(e_0=e_0[H]\) in the initial condition \(\phi (y,t_0)=e_0Z_0(y)\) is a linear operator of H. The existence of \(\phi \) immediately defines \(e_0\). This completes the proof of the existence of \(u=u_3 + {{\tilde{\phi }}}\) in Theorem 1.1, with the bubbling profile centered in \(x=0\in \Omega \) and parameters satisfying (1.8). \(\square \)

Remark 8.1

(Continuity of \((\phi ,e_0)\) with respect to \(\psi _0\)) We found the inner perturbation \(\phi \) and its initial datum \(\phi (y,t_0) = e_0Z_0(y)\) based on the existence of the outer solution \(\Psi [\phi ]\) given by Proposition 4.1, which in fact can be found for any initial condition \(\psi _0\in C^1({\bar{\Omega }})\). Furthermore, as a consequence of the continuity of \(\Psi [\psi _0]\) and \(\Lambda [\psi _0],\xi [\psi _0]\) found in Remarks 4.1 and 6.1 we obtain

$$\begin{aligned} \vert *\vert {e_0[\psi _0^1]-e_0[\psi _0^2]}\lesssim \left[ \vert \!\vert \psi _0^1-\psi _0^2\vert \!\vert _{L^\infty (\Omega )}-\vert \!\vert \nabla \psi _0^1 -\nabla \psi _0^2\vert \!\vert _{L^\infty (\Omega )}\right] . \end{aligned}$$

Since we know that \(\Lambda ,{\dot{\Lambda }},\xi ,{\dot{\xi }},\psi \) depends smoothly on \(\psi _0\), by the implicit function theorem, we deduced that map \(\psi _0\mapsto (\phi [\psi _0],e_0[\psi _0])\) is \(C^1\) with respect to \(\psi _0\in C^1({\bar{\Omega }})\). This allows to prove the 1-codimensional stability of Corollary 1.1, under small perturbation. With these ingredients, we can proceed as in [11, Proof of Corollary 1.1].

8 Invertibility theory for the nonlocal linear problem

In this section we prove Proposition 6.1. We deduce the result by Laplace transform method combined with asymptotic estimates of the heat kernel \(p_t^{\Omega }\) associated to \(\Omega \). It turns out that the operator \(\mathcal {J}[{\dot{\Lambda }}]\) is similar to a half-fractional integral of \({\dot{\Lambda }}\). Thus, roughly speaking, we expect the inverse operator to behave as a fractional derivative of order 1/2. In fact, Proposition 6.1 can be seen as a precise statement of this idea.

For later purpose we recall some facts about the Dirichlet heat kernel. For the definition and properties we follow [22, 32]. A function \(p_t^{\Omega }(x,y)\) continuous on \(\bar{\Omega }\times {\bar{\Omega }} \times {{\mathbb {R}}}^+\), \(C^2\) in x and \(C^1\) in t is called Dirichlet heat kernel for the problem

$$\begin{aligned}&{\partial } _t u(x,t)=\Delta u(x,t){\quad \hbox {in } }\Omega \times {{\mathbb {R}}}^+,\\&u(x,t)=0 {\quad \hbox {on } } {\partial } \Omega \times [0,\infty ),\\&u(x,0)=u_0(x ){\quad \hbox {in } }\Omega , \end{aligned}$$

if, for any \(y \in \Omega \), satisfies

$$\begin{aligned}&{\partial } _t p_t^{\Omega }(x,y) = \Delta _x p_t^{\Omega }(x,y) {\quad \hbox {in } }\Omega \times {{\mathbb {R}}}^+,\\&p_t^{\Omega }(x,y)=0 {\quad \hbox {in } } {\partial } \Omega , \end{aligned}$$

and

$$\begin{aligned} \lim \limits _{t\rightarrow 0^+}\int _\Omega p_t^{\Omega }(x,y)u_0(y)\,dy=u_0(x), \end{aligned}$$

uniformly for every function \(u_0 \in C_0({\bar{\Omega }})\). The existence of the Dirichlet heat kernel is a classical result by Levi [43]. It has the following basic properties:

  • \(p_t^{\Omega }(x,y)\ge 0\), \(p_t^{\Omega }(x,y)=p_t^{\Omega }(y,x)\) and \(p_t^{\Omega }(x,y)=0\) if \(x\in {\partial } \Omega \);

  • for any \(y\in \Omega \) the function \(p_t^{\Omega }(x,y)\in C^{\infty }({{\mathbb {R}}}^+\times \Omega )\);

  • it satisfies \( {\partial } _t p_t^{\Omega }(x,y)=\Delta _x p_t^{\Omega }(x,y)\) for \((x,y,t)\in \Omega \times \Omega \times {{\mathbb {R}}}^+\).

Also, from [32, Theorem 10.13] and its proof, the heat kernel \(p_t^{\Omega }(x,y)\) admits the expansion

$$\begin{aligned} p_t^\Omega (x,y)=\sum _{k\ge 1} e^{-\lambda _k t}\phi _k(x)\phi _k(y), \end{aligned}$$
(8.1)

where \(\lambda _k\) is the k-th Dirichlet eigenvalue of \(-\Delta \) on \(\Omega \) and \(\phi _k\) the corresponding eigenfunction and also for \(n\ge 1\) (see [32, Remark 10.15])

$$\begin{aligned} \sum _{k=n}^{\infty }\sup _{x,y \in \Omega }\vert \phi _k(x)\phi _k(y)\vert <\infty \end{aligned}$$
(8.2)

The series (8.1) converges absolutely and uniformly in \([\varepsilon ,\infty ]\times \Omega \times \Omega \) for any \(\varepsilon >0\), as well as in the topology of \(C^\infty ({{\mathbb {R}}}^+\times \Omega \times \Omega )\).

Before starting the proof of Proposition 6.1, we recall an estimate on the short time behavior of the heat kernel \(p_\tau ^\Omega (x,y)\) due to Varadhan [55, Theorem 4.9]. We will use it in the following form as in Hsu [37, Corollary 1.6].

Lemma 8.1

(Short time estimate of \(p_\tau ^{\Omega }\)) Let \(\varepsilon >0\) fixed such that \(B_\varepsilon (0)\subset \Omega \). Then, there exists \(\tau _0>0\) such that, for \(y\in B_{\varepsilon }(0)\) and \(\tau \in (0,\tau _0)\) we have

$$\begin{aligned} p^{{{\mathbb {R}}}^3}_\tau (0,y)(1-e^{-\frac{\delta ^2}{4\tau }}) \le p^\Omega _\tau (0,y), \end{aligned}$$
(8.3)

where \(\delta <\delta _0\) is independent of y and

$$\begin{aligned} \delta _0 :=d( {\partial } \Omega , {\partial } B_{\varepsilon })=\min _{a\in {\partial } \Omega , b \in {\partial } B_\varepsilon (0)} \vert a-b\vert >0. \end{aligned}$$

Proof

Recall the identities in [55, p. 675]

$$\begin{aligned}&\limsup _{\tau \rightarrow 0} 4\tau \log (p^{{{\mathbb {R}}}^3}_{\tau }(x,y)-p^{\Omega }_{\tau }(x,y))\le -d_{ {\partial } \Omega }(x,y)^2, \end{aligned}$$
(8.4)
$$\begin{aligned}&\lim _{\tau \rightarrow 0}4 \tau \log (p^{{{\mathbb {R}}}^3}_\tau (x,y))=- d(x,y)^2, \end{aligned}$$
(8.5)

where

$$\begin{aligned} d_{ {\partial } \Omega }(x,y){:=}\inf _{z \in {\partial } \Omega }\{ d(x,z)+d(z,y) \}. \end{aligned}$$

From (8.4) for \(\tau \in (0,\tau _0)\) we have

$$\begin{aligned} p_{\tau }^{{{\mathbb {R}}}^3}(x,y)- e^{-\frac{ d_{ {\partial } \Omega }^2(x,y)-c(\tau _0)}{4\tau }}\le p_{\tau }^{\Omega }(x,y), \end{aligned}$$

for all \(x,y \in \Omega \), where \(0\le c(\tau _0)=o(1)\) as \(\tau _0\rightarrow 0\). In particular, fix \(x=0\) and consider \(y \in B_{\varepsilon }(0)\subset \Omega \) for a small \(\varepsilon >0\). Then, choosing \(\tau _0\) smaller if needed, we have

$$\begin{aligned} d_{ {\partial } \Omega }^2(0,y)-c(\tau _0)\ge \varepsilon ^2 +\delta _0^2. \end{aligned}$$

Thus for \(y \in B_\varepsilon (0)\)

$$\begin{aligned} e^{-\frac{ d_{ {\partial } \Omega }^2(0,y)-c(\tau _0)}{4\tau }}\le e^{-\frac{\varepsilon ^2+\delta _0^2}{4 \tau }}\le e^{-\frac{d(0,y)^2}{4\tau }}e^{-\frac{\delta _0^2}{4\tau }}, \end{aligned}$$

and (8.5) says

$$\begin{aligned} p_{\tau }^{{{\mathbb {R}}}^3}(0,y)=e^{-\frac{d^2(0,y)}{4\tau }(1+o(1))} \quad \text{ as }\quad \tau \rightarrow 0^+. \end{aligned}$$

Thus, we have for \(\tau <\tau _0\) small and \(y \in B_\varepsilon (0)\)

$$\begin{aligned} p^{{{\mathbb {R}}}^3}_\tau (0,y)(1-e^{-\frac{\delta ^2}{4\tau }}) \le p^\Omega _\tau (0,y), \end{aligned}$$
(8.6)

for any \(\delta <\delta _0\) independent of y. \(\square \)

We mention that the uniform bound (8.6) holds for y ranging in any convex subset of the domain, see [37, p.374–375]. Also, for any \(\tau >0\) and \(x,y \in \Omega \) we have the upper bound

$$\begin{aligned} p^\Omega _\tau (x,y)\le p^{{{\mathbb {R}}}^3}_\tau (x,y), \end{aligned}$$
(8.7)

as a consequence of the maximum principle. Thus, Varadhan’s estimate (8.3) is a precise statement about the idea that for small times the heat kernel “does not feel the boundary”. We refer to Kac [38] and Dodziuk [22] for statements with the same flavor. In the proof of Proposition 6.1 we need the following lemma.

Lemma 8.2

Define the function

$$\begin{aligned} I(\tau ){:=}\int _{\Omega } p_{\tau }^{\Omega }(0,y)G_\gamma (y,0)\,dy, \end{aligned}$$

where \(p_{\tau }^{\Omega }(x,y)\) denotes the Dirichlet heat kernel associated to \(\Omega \) and \(G_\gamma (x,y)\) the Green function of the operator \(-\Delta -\gamma \) on \(\Omega \). Then \(I(\tau )\) has the following asymptotic behavior:

$$\begin{aligned} I(\tau )= {\left\{ \begin{array}{ll} O\left( e^{-\lambda _1 \tau }\right) &{}\text {for} \quad \tau \rightarrow \infty ,\\ c_{1,*} \frac{1}{\sqrt{\tau }}+c_{2,*} \sqrt{\tau }+c_{3,*}\tau +O\left( \tau ^{3/2}\right) &{}\text {for} \quad \tau \rightarrow 0^+, \end{array}\right. } \end{aligned}$$
(8.8)

for some constant \(c_{i,*}\) and \(i=1,2,3\).

Proof

Step 1 (Asymptotic for \(\tau \rightarrow \infty \)). We recall that the heat kernel \(p_\tau ^{\Omega }(x,y)\) admits the series expansion (8.1) which converges absolutely and uniformly in the domain \([\varepsilon ,\infty )\times \Omega \times \Omega \) for any \(\varepsilon >0\), as well as in the topology \(C^\infty ({{\mathbb {R}}}^+\times \Omega \times \Omega )\). By the uniform convergence with respect to \(y\in \Omega \) we obtain for \(\tau >0\)

$$\begin{aligned} I(\tau )&=\int _\Omega \sum _{k=1}^\infty e^{-\lambda _k \tau }\phi _k(0)\phi _k(y)G_\gamma (y,0)\,dy\nonumber \\ {}&= \sum _{k=1}^\infty e^{-\lambda _k \tau }\phi _k(0)\int _\Omega \phi _k(y)G_\gamma (y,0)\,dy. \end{aligned}$$
(8.9)

Multiplying equation (2.7) by \(\phi _k\) and integrating by parts we get

$$\begin{aligned} -\lambda _k \int _\Omega G_\gamma (x,0)\phi _k(x)\,dx&=\int _\Omega G_\gamma (x,0)\Delta \phi _k(x)\\&=\int _\Omega \phi _k(x)\Delta G_\gamma (x,0)\,dx\\&= -\gamma \int _\Omega G_\gamma (x,0)\phi _k(x)\,dx-c_3\int _\Omega \phi _k(x) \delta _0(x)\,dx\\&= -\gamma \int _\Omega G_\gamma (x,0)\phi _k(x)\,dx-c_3 \phi _k(0), \end{aligned}$$

that gives

$$\begin{aligned} \int _\Omega G_\gamma (x,0)\phi _k(x)\,dx=c_3 \frac{\phi _k(0)^2}{\lambda _k-\gamma }. \end{aligned}$$
(8.10)

We plug (8.10) into (8.9). Finally, from (8.2) we obtain the asymptotic behavior (8.8) for \(\tau \rightarrow \infty \).

Step 2 (Asymptotic for \(\tau \rightarrow 0^+\)). Firstly, we split

$$\begin{aligned} I(\tau )=&\int _\Omega p_\tau ^{\Omega }(0,y) \frac{\alpha _3}{\vert y\vert }\,dy+\int _\Omega p_\tau ^{\Omega }(0,y) H_\gamma (y,0)\,dy\\ =&:I_1(\tau )+I_2(\tau ). \end{aligned}$$

We analyze \(I_1(\tau )\). For the region \(B_{\varepsilon }(0)\) we invoke Varadhan’s estimate (8.6) and we obtain

$$\begin{aligned} \int _{B_\varepsilon (0)} \frac{p_\tau ^\Omega (0,y)}{\vert y\vert }\,dy&\ge \int _{B_\varepsilon } \frac{e^{-\frac{\vert y\vert ^2}{4\tau }}}{[4\pi \tau ]^{3/2}}\frac{1}{\vert y\vert }\,dy(1-e^{-\frac{\varepsilon ^2}{\tau }})\\&=4\pi \int _0^\varepsilon \frac{e^{-\frac{\rho ^2}{4\tau }}}{[4\pi \tau ]^{3/2}}\rho \,d\rho (1-e^{-\frac{\varepsilon ^2}{\tau }})\\&=\frac{1}{\sqrt{4\pi \tau }}\int _{0}^{\frac{\varepsilon }{2\sqrt{\tau }}}e^{-r^2}r \,dr(1-e^{-\frac{\varepsilon ^2}{\tau }}) \\&=\frac{1}{\sqrt{4\pi \tau }} \left( \frac{1-e^{-\frac{\varepsilon ^2}{4\tau }}}{2}\right) \left( 1-e^{-\frac{\varepsilon ^2}{\tau }}\right) \\&=\frac{1}{4\sqrt{\pi \tau }}+O\left( \frac{e^{-\frac{c}{\tau }}}{\sqrt{\tau }}\right) \end{aligned}$$

for some \(c>0\), and by (8.7) we have

$$\begin{aligned} \int _{B_\varepsilon (0)} \frac{p_\tau ^\Omega (0,y)}{\vert y\vert }\,dy&\le \int _{B_\varepsilon (0)} \frac{p_\tau ^{{{\mathbb {R}}}^3}(0,y)}{\vert y\vert }\,dy\\&\le \frac{1}{4 \sqrt{\pi \tau }}+O\left( \frac{e^{-\frac{c}{\tau }}}{\sqrt{\tau }}\right) . \end{aligned}$$

From these bounds we conclude

$$\begin{aligned} \int _{B_\varepsilon (0)} \frac{p_\tau ^\Omega (0,y)}{\vert y\vert }\,dy=\frac{1}{4 \sqrt{\pi \tau }}+O\left( \frac{e^{-\frac{c}{\tau }}}{\sqrt{\tau }}\right) . \end{aligned}$$

In the region \(\Omega \setminus B_\varepsilon (0)\) by (8.7) we get

$$\begin{aligned} \int _{\Omega \setminus B_\varepsilon (0)} \frac{p_\tau ^\Omega (0,y)}{\vert y\vert }\,dy&\le \tau ^{-3/2}\int _{\varepsilon }^{1} e^{-\frac{\rho ^2}{cs}}\rho \, d\rho \\&=\tau ^{-1/2} \int _{\frac{\varepsilon }{\sqrt{s}}}^{\frac{1}{\sqrt{s}}} e^{-r^2}r\,dr=O\left( \frac{e^{-\frac{c}{\tau }}}{\sqrt{\tau }}\right) . \end{aligned}$$

We conclude that

$$\begin{aligned} I_1(\tau )&=\alpha _3\int _{\Omega \setminus B_\varepsilon (0)} \frac{p_\tau ^\Omega (0,y)}{\vert y\vert }\,dy+ \alpha _3\int _{B_\varepsilon (0)} \frac{p_\tau ^\Omega (0,y)}{\vert y\vert }\,dy\\&= \frac{c_{1,*}}{\sqrt{\tau }}+O\left( \frac{e^{-\frac{c}{\tau }}}{\sqrt{\tau }}\right) \quad \text{ as }\quad \tau \rightarrow 0^+,\quad \text {with}\quad c_{1,*}=\frac{\alpha _3}{4\sqrt{\pi \tau }}. \end{aligned}$$

Now, we estimate the term \(I_2(\tau )\). We treat it similarly to \(I_1(\tau )\) but we get a lower order term in the expansion since \(H_\gamma (y,0)\) is not singular. We use decomposition (2.10) for \(H_\gamma (y,0)\) and we consider the integral over \(B_{\varepsilon }(0)\). Using the cosine expansion we get

$$\begin{aligned} \theta _\gamma (y,0)=\alpha _3\frac{\gamma }{2}\vert y\vert +O(\vert y\vert ^3). \end{aligned}$$

Thus, we compute the integral associated to the first term with Varadhan’s estimate (8.3) and the upper bound (8.7):

$$\begin{aligned} \int _{B_\varepsilon (0)} p_{\tau }^{\Omega }(y,0)\frac{1-\cos (\sqrt{\gamma }\vert y\vert )}{\vert y\vert } \,dy&= \alpha _3\frac{\gamma }{2}\int _{B_\varepsilon (0)} \frac{e^{-\frac{\vert y\vert ^2}{4\tau }}}{[4\pi \tau ]^{3/2}}\vert y\vert \,dy\left( 1+o\left( e^{-\frac{c}{\tau }}\right) \right) \nonumber \\ {}&=4 \pi \alpha _3\frac{\gamma }{2} \int _{0}^{\varepsilon } \frac{e^{-\frac{\rho ^2}{4\tau }}}{[4\pi \tau ]^{3/2}} \rho ^3 \,d\rho \left( 1+o\left( e^{-\frac{c}{\tau }}\right) \right) \nonumber \\ {}&=4 \pi \alpha _3 \sqrt{\tau }\frac{\gamma }{2}\int _0^{\frac{\varepsilon }{2\sqrt{\tau }}} e^{-r^2}r^{3}\,dr\left( 1+o\left( e^{-\frac{c}{\tau }}\right) \right) \nonumber \\&=c_{2,*} \sqrt{\tau }\left( 1+o\left( e^{-\frac{c}{\tau }}\right) \right) , \end{aligned}$$
(8.11)

for an explicit constant \(c_{2,*}\). The same computation on the remainder \(O\left( \vert y\vert ^3\right) \) gives a term of order \(O\left( \tau ^{3/2}\right) \). Another Taylor expansion at \(y=0\) gives

$$\begin{aligned} h_\gamma (y,0)=\nabla _y h_\gamma (0,0)\cdot y +\frac{1}{2}y\cdot D_{yy}h_\gamma (0,0)\cdot y+O(\vert y\vert ^3), \end{aligned}$$

where \(D_{yy}h_\gamma (0,0)\) denotes the Hessian of \(h_\gamma (\cdot ,0)\) evaluated in \(y=0\). Integrating the first term on \(B_\varepsilon (0)\) against \(p_t^\Omega (0,y)\) and using (8.3)–(8.7) we see by symmetry of the integrand \(p_t^{{{\mathbb {R}}}^3}(0,y)\nabla _y h_\gamma (0,0)\cdot y\) that the integral gives an exponentially decaying term. The second term in (8.12) can be treated similarly to (8.11) and gives a term of order \(c_{3,*}\tau (1+o(1))\) for some explicit constant \(c_{3,*}\). The integral of \(p_{\tau }^\Omega (y,0) H_\gamma (y,0)\) on the complement can be treated as before and gives an exponentially decay term for \(\tau \rightarrow 0\). Thus, we obtain that

$$\begin{aligned} I_2(\tau )=c_{2,*} \sqrt{\tau }+c_{3,*}\tau +O\left( \tau ^{3/2}\right) \quad \text{ as }\quad \tau \rightarrow 0^+. \end{aligned}$$

We conclude that \(I(\tau )=I_1(\tau )+I_2(\tau )\) has the asymptotic (8.8) for \(\tau \rightarrow 0^+\). \(\square \)

We start here the main proof of Proposition 6.1.

8.1 Proof of Proposition 6.1

Firstly, we observe that \(J(0,t_0)=h(t_0)\) is in general not compatible with a null initial condition. For this reason it is natural to solve the problem for \(\mathcal {J}\) starting from \(t=t_0-1\). We look for \(\Lambda (t)\) for \(t \in (t_0-1,\infty )\). The function \(\mathcal {J}\) is a solution to the problem

$$\begin{aligned}&{\partial } _t \mathcal {J}= \Delta _x \mathcal {J}+ \gamma \mathcal {J}- {\dot{\Lambda }}(t) G_\gamma (x,0) {\quad \hbox {in } }\Omega \times (t_0-1,\infty ),\\&\mathcal {J}(x,t)\equiv 0 {\quad \hbox {on } } {\partial } \Omega \times (t_0-1,\infty ), \end{aligned}$$

such that

$$\begin{aligned} \mathcal {J}(0,t)=h^*(t) {\quad \hbox {in } }(t_0,\infty ), \end{aligned}$$

where

$$\begin{aligned} h^*(t)= {\left\{ \begin{array}{ll} h(t)&{}t\in [t_0,\infty ),\\ h_{\text {ext}}(t)&{}t\in [t_0-1,t_0), \end{array}\right. } \end{aligned}$$
(8.12)

and

$$\begin{aligned} h_{\text {ext}}(t)=\eta (t) h(t_0), \end{aligned}$$

where \(\eta \) is a smooth function such that \(\eta (t_0-1)=0\), \(\eta (t_0)=1\) and

$$\begin{aligned} \vert \eta (t_0-\nu )h(t_0)-h(t_0+\nu )\vert \le [h]_{\varepsilon ,[t_0,t_0+1]} \nu ^\varepsilon , \end{aligned}$$

for any \(\nu \le 1\). This choice gives an extension \(h^*(t)\in C^{\varepsilon }\) with

$$\begin{aligned} \vert \!\vert h^*\vert \!\vert _{\sharp ,c_1,c_2,(t_0-1,\infty )}\lesssim \vert \!\vert h\vert \!\vert _{\sharp ,c_1,c_2,(t_0,\infty )}. \end{aligned}$$
(8.13)

Let \(s{:=}t-(t_0-1)\) and for \(s\in (0,\infty )\) define

$$\begin{aligned}&\mathcal {J}_0(x,s):=e^{-\gamma s} \mathcal {J}(x,s+(t_0-1)), \nonumber \\&\beta (s) :=- \Lambda (s+(t_0-1)),\nonumber \\&h_0^*(s) :=h^*(s+(t_0-1)). \end{aligned}$$
(8.14)

The function \(\mathcal {J}_0\) is a solution to

$$\begin{aligned}&{\partial } _s \mathcal {J}_0(x,s)= \Delta _x \mathcal {J}_0 + e^{-\gamma s}\dot{\beta }(s)G_\gamma (x,0) {\quad \hbox {in } }\Omega \times (0,\infty )\\&\mathcal {J}_0(x,s)=0 {\quad \hbox {on } } {\partial } \Omega \times (0,\infty ), \end{aligned}$$

such that

$$\begin{aligned} \mathcal {J}_0[{\dot{\beta }}](0,s)=h_{0}^*(s)e^{-\gamma s} {\quad \hbox {in } }(0,\infty ). \end{aligned}$$
(8.15)

Imposing the initial condition \(\mathcal {J}(x,t_0)\equiv 0\) in \(\Omega \), that is \(\mathcal {J}_0(x,0)\equiv 0\), by Duhamel’s formula we have

$$\begin{aligned} \mathcal {J}_0[{\dot{\beta }}](0,s)=\int _0^s e^{-\gamma (s-\tau )}{\dot{\beta }}(s-\tau ) I(\tau ) \, d\tau , \end{aligned}$$
(8.16)

where

$$\begin{aligned} I(\tau ){:=}\int _\Omega p_\tau ^{\Omega }(0,y) G_\gamma (y,0)\,dy, \end{aligned}$$

and \(p_\tau ^{\Omega }(x,y)\) denotes the heat kernel associated to \(\Omega \). The asymptotic behavior of \(I(\tau )\) is given by Lemma 8.2. We denote the Laplace transform of a function f as

$$\begin{aligned} {\tilde{f}}(\xi ){:=}\int _{0}^{\infty }e^{-\xi s}f(s)\,ds. \end{aligned}$$

We refer to the book [23] by Doetsch for classic properties of the Laplace transform. Applying the Laplace transform to (8.16), using (8.15) and the basic property

$$\begin{aligned} \tilde{\dot{f}}(\xi )=\xi {\tilde{f}}(\xi )-f(0), \end{aligned}$$

we obtain

$$\begin{aligned} {\tilde{h}}_0^*(\xi +\gamma )&= \tilde{{\dot{\beta }}}(\xi +\gamma ) \tilde{I}(\xi )\\&=\left[ (\xi +\gamma ){{\tilde{\beta }}}(\xi +\gamma )-\beta (0)\right] \tilde{I}(\xi ), \end{aligned}$$

and hence

$$\begin{aligned} {\tilde{\beta }}(\xi +\gamma )=\frac{\beta (0)}{\xi +\gamma }+\tilde{h}_0^*(\xi +\gamma ){\tilde{\sigma }}(\xi ), \end{aligned}$$
(8.17)

where

$$\begin{aligned} {{\tilde{\sigma }}}(\xi ){:=}\frac{1}{(\xi +\gamma )\tilde{I}(\xi )}. \end{aligned}$$

By definition we have

$$\begin{aligned} \tilde{I}(\xi )=\int _{0}^{\infty }e^{-\xi s}I(s)\,ds, \end{aligned}$$

that is well defined and analytic in the right-half plane \({\text {Re}}\xi >-\lambda _1\) thanks to Lemma 8.2. By expansion (8.8) we have

$$\begin{aligned} \vert *\vert {e^{-\xi s}I(s)}\lesssim g(s),\quad g(s)= {\left\{ \begin{array}{ll} \frac{1}{\sqrt{s}} &{}\text {for} \quad s \rightarrow 0^+,\\ e^{-(\lambda _1+{\text {Re}}\xi ) s} &{}\text {for} \quad s \rightarrow +\infty , \end{array}\right. } \end{aligned}$$

and g is integrable in \({{\mathbb {R}}}^+\) if \({\text {Re}}{\xi }>-\lambda _1\). Thus, using (8.9), in any half plane \({\text {Re}}\xi \ge c\) where \(c>-\lambda _1\) the dominated convergence theorem applies to get

$$\begin{aligned} {\tilde{I}}(\xi )&= \int _{0}^{\infty } e^{-\xi s} I(s)\,ds\\ {}&= \int _{0}^{\infty } e^{-\xi s} \sum _{k=1}^{\infty } \frac{\phi _k(0)^2}{\lambda _k-\gamma }e^{-\lambda _k s} \,ds\\&= \sum _{k=1}^{\infty }\frac{\phi _k(0)^2}{\lambda _k-\gamma }\int _{0}^{\infty } e^{-\xi s} e^{-\lambda _k s} \,ds\\&= \sum _{k=1}^{\infty }\frac{\phi _k(0)^2}{\lambda _k-\gamma }\frac{1}{\lambda _k+\xi } \end{aligned}$$

At this point we can extend \({\tilde{I}}(\xi )\) analytically from \(\{\xi \in \mathbb {C}: \xi >-\lambda _1\}\) to \(\mathbb {C}{\setminus } \{-\lambda _k\}_{k=1}^\infty \). Let \(\xi =a+ib\) and rewrite the series as

$$\begin{aligned} \tilde{I}(\xi )=&\sum _{k=1}^\infty \frac{\phi _k(0)^2}{\lambda _k-\gamma } \frac{1}{\lambda _k + a+ib}\\ =&\sum _{k=1}^\infty \frac{\phi _k(0)^2}{\lambda _k-\gamma } \frac{\lambda _k+a}{(\lambda _k+a)^2+b^2}-ib \sum _{k=1}^\infty \frac{\phi _k(0)^2}{\lambda _k-\gamma }\frac{1}{(\lambda _k +a)^2+b^2}. \end{aligned}$$

Since the coefficients of the series are positive, \({\tilde{I}} (\xi )=0\) implies \(b=0\). Plugging \(b=0\) into the first series we obtain that a root \(\xi =a\) of \({\tilde{I}}\) satisfies

$$\begin{aligned} \sum _{k=1}^\infty \frac{\phi _k(0)^2}{\lambda _k-\gamma } \frac{1}{\lambda _k+a}=0. \end{aligned}$$

Hence, we deduce that the set of zeros of \({\tilde{I}}\) is given by a sequence \(\{-a_k\}_{k=1}^{\infty }\) where \(a_k\in (\lambda _k,\lambda _{k+1})\). In particular,

$$\begin{aligned} {\tilde{I}}(\xi )\ne 0 \quad \text {for}\quad {\text {Re}}\xi >-\lambda _1. \end{aligned}$$
(8.18)

By standard argument [23, Theorem 33.7] on the Laplace transform, using (8.8), we have

$$\begin{aligned} \tilde{I}(\xi )= c_{1,*}\sqrt{\pi }\xi ^{-1/2}+c_{2,*}\frac{\sqrt{\pi }}{2} \xi ^{-3/2}+c_{3,*}\xi ^{-2}+O(\xi ^{-5/2})\quad \text{ as }\quad \vert \xi \vert \rightarrow \infty , \end{aligned}$$

in the half-plane \({\text {Re}}\xi >-\lambda _1\). Thus, in the same half-plane we have

$$\begin{aligned} {\tilde{\sigma }}(\xi )&=\frac{1}{(\xi +\gamma ){\tilde{I}}(\xi )}\nonumber \\&\quad = d_{1,*}\xi ^{-1/2}+d_{2,*}\xi ^{-3/2}+d_{3,*}\xi ^{-2}+O(\xi ^{-5/2}) \quad \text{ as }\quad \vert \xi \vert \rightarrow \infty . \end{aligned}$$
(8.19)

As a consequence of (8.18), \({{\tilde{\sigma }}}(\xi )\) has a unique singularity at \(\xi =-\gamma \) in the half-plane of convergence. By [23, Theorem 28.3] the function \({{\tilde{\sigma }}}(\xi )\) can be represented as a Laplace transform of a function.Footnote 1 Finally, we compute the inverse Laplace transform by means of the Residue theorem defining the rectangular contour integral \(\mathcal {C}_R\) as in Fig. 1, which is suggested by the proof of [23, Theorem 35.1].

Fig. 1
figure 1

Contour integral \(\mathcal {C}_R\)

For later purpose we observe that, looking at the contour integral \( {{\mathcal {C}}}_R\), the constant \(\mathfrak {a}\in (\gamma ,\lambda _1)\) can be taken arbitrarily close to \(\lambda _1\). An application of the Riemann-Lebesgue Lemma (as in [23, p.237]) implies

$$\begin{aligned}&\lim _{R\rightarrow \infty }\int _{L_{2,R}}e^{\xi \tau }{{\tilde{\sigma }}}(\xi )\, d\xi =0,\\&\lim _{R\rightarrow \infty }\int _{L_{4,R}}e^{\xi \tau }{{\tilde{\sigma }}}(\xi )\, d\xi =0. \end{aligned}$$

Since

$$\begin{aligned} \sigma (\tau )=\lim _{R\rightarrow \infty }\frac{1}{2\pi i}\int _{L_{1,R}}e^{\xi \tau }{{\tilde{\sigma }}}(\xi )\, d\xi \end{aligned}$$

we obtain

$$\begin{aligned} \sigma (t)=\text {Res}\left( e^{\xi t}{{\tilde{\sigma }}}(\xi ),-\gamma \right) e^{-\gamma t}+\lim \limits _{R\rightarrow \infty }\frac{1}{2\pi i} \int _{-\mathfrak {a}-i R}^{-\mathfrak {a}+iR} e^{\xi t}{{\tilde{\sigma }}}(\xi )\, d\xi . \end{aligned}$$
(8.20)

We easily compute

$$\begin{aligned} \text {Res}\left( e^{\xi \tau }{{\tilde{\sigma }}}(\xi ),-\gamma \right)&=\lim \limits _{\xi \rightarrow -\gamma } (\xi +\gamma )\frac{1}{(\xi +\gamma )\tilde{I}(\xi )}=:c_\infty . \end{aligned}$$

Now, we analyze the integral (8.20). We decompose

$$\begin{aligned}&\lim \limits _{R\rightarrow \infty }\int _{-\mathfrak {a}-i R}^{-\mathfrak {a}+iR} e^{\xi \tau }{{\tilde{\sigma }}}(\xi )\, d\xi \\ {}&\quad =i e^{-\mathfrak {a} \tau }\int _{-R}^{R} e^{iy \tau }\bigg [ \tilde{\sigma }(-\mathfrak {a}+iy) - \frac{d_{1,*}}{\sqrt{-\mathfrak {a}+iy}}-\frac{d_{2,*}}{(-\mathfrak {a}+iy)^{3/2}}\\&\qquad - \frac{d_{3,*}}{(-\mathfrak {a}+iy)^2} \bigg ]\,dy+ie^{-\mathfrak {a}\tau }\int _{-iR}^{iR} e^{\xi t} \frac{d_{1,*}}{\sqrt{-\mathfrak {a}+iy}}\,dy\\&\qquad +ie^{-\mathfrak {a}\tau }\int _{-i R}^{iR}\frac{d_{2,*}}{(-\mathfrak {a}+iy)^{3/2}} \,dy+ie^{-\mathfrak {a}\tau }\int _{-i R}^{iR}\frac{d_{3,*}}{(-\mathfrak {a}+iy)^{2}} \,dy\end{aligned}$$

It is easy to see (by means of another contour to avoid the standard branch) that, up to constants, the last three integral are respectively the inverse Laplace transform of \(\xi ^{-1/2},\xi ^{-3/2},\xi ^{-2}\). The integral

$$\begin{aligned} R(\tau ){:=}\int _{-R}^{R} e^{iy \tau }\bigg [ \tilde{\sigma }(-\mathfrak {a}+iy) - \frac{d_{1,*}}{\sqrt{-\mathfrak {a}+iy}}-\frac{d_{2,*}}{(-\mathfrak {a}+iy)^{3/2}} - \frac{d_{3,*}}{(-\mathfrak {a}+iy)^2} \bigg ]\,dy\end{aligned}$$

is absolutely convergent thanks to the second order expansion of \({\tilde{\sigma }}(\xi )\). In fact, obtaining the absolute convergence of \(R(\tau )\) (and \(R'(\tau )\)) is the main reason to use the sharp Varadhan’s estimate on the heat kernel \(p_t^{\Omega }\). Thus, from (8.20) we obtain

$$\begin{aligned} \sigma (\tau )=c_\infty e^{-\gamma \tau }+e^{-\mathfrak {a} \tau } \left[ \frac{C_{1,*}}{\sqrt{\tau }}+C_{2,*}\sqrt{\tau }+C_{3,*}\tau +R(\tau )\right] , \end{aligned}$$

for some constants \(c_\infty , C_{i,*}\) and \(i=1,2,3\), where \(R(\tau )\) is bounded. This gives the asymptotic behavior

$$\begin{aligned} \sigma (\tau )= {\left\{ \begin{array}{ll} c_\infty e^{-\gamma \tau } +O(e^{-\mathfrak {a} \tau }) &{}\text {for} \quad \tau \rightarrow \infty ,\\ \frac{C_{1,*}^{-1}}{\sqrt{\tau }}+c_\infty +O\left( \sqrt{\tau }\right) &{}\text {for} \quad \tau \rightarrow 0^+, \end{array}\right. } \end{aligned}$$

for any \(\mathfrak {a} \in (\gamma ,\lambda _1)\). For later purposes, we observe that \(\sigma (\tau )\) is differentiable. Indeed, differentiating \(R(\tau )\), we still obtain an absolutely convergent integral thanks to the full expansion (8.19), and an application of the dominated convergence theorem gives \(\sigma \in C^1\) with

$$\begin{aligned} \sigma '(\tau )= {\left\{ \begin{array}{ll} -\gamma c_\infty e^{- \gamma \tau } +O(e^{-\mathfrak {a} \tau }) &{}\text {for} \quad \tau \rightarrow \infty ,\\ -(2C_{1,*})^{-1}\tau ^{-3/2}(1+O(\tau )) &{}\text {for} \quad \tau \rightarrow 0^+, \end{array}\right. } \end{aligned}$$

From (8.17), taking the inverse Laplace transform of both sides, we get

$$\begin{aligned} \beta (s)e^{-\gamma s}= \beta (0)e^{-\gamma s} + \int _{0}^{s} e^{-\gamma (s-\tau )}h_0^*(s-\tau ) \sigma (\tau )d\, \tau , \end{aligned}$$

that is

$$\begin{aligned} \beta (s)= \beta (0)+\int _{0}^{s}e^{\gamma \tau }\sigma (\tau )h_0^*(s-\tau )\, d\tau . \end{aligned}$$

Proof of (6.4)

We rewrite this formula as

$$\begin{aligned} \beta (s)&= \beta (0)+c_\infty \int _{0}^{s}h_0^*(\tau )\,d\tau + \int _{0}^{s}h_0^*(\tau )\left[ e^{\gamma (s-\tau )}\sigma (s-\tau )-c_\infty \right] \,d\tau \\&\quad =\left[ \beta (0)+c_\infty \int _0^\infty h_0^*(\tau )\, d\tau \right] -c_\infty \int _{s}^{\infty } h_0^*(\tau )\, d\tau \\&\qquad + \int _{0}^{s}h_0^*(\tau )\left[ e^{\gamma (s-\tau )}\sigma (s-\tau )-c_\infty \right] \,d\tau . \end{aligned}$$

We choose \(\beta (0)=-c_\infty \int _{0}^{\infty }h_0^*(\tau )d\tau \). It remains to estimate

$$\begin{aligned}&\beta _1(s){:=}-c_\infty \int _{s}^{\infty }h_0^*(\tau )\, d\tau ,\\&\beta _2(s){:=} \int _{0}^{s}h_0^*(\tau )\left[ e^{\gamma (s-\tau )}\sigma (s-\tau )-c_\infty \right] \,d\tau . \end{aligned}$$

We recall that the extension \(h_0^*(s)\) has been selected so that (8.13) holds. Here and in what follows, without losing in generality we assume the same value \(c=c_i\) for \(i=1,2\). When we estimate the \(L^\infty \) norm of \(\beta \) we will only use the \(L^\infty \) norm of \(h_0^*\) and hence we get the same \(L^\infty \)-weight constant \(c_1\). Instead, when we estimate the \(C^{1/2+\varepsilon }\) we need both the \(L^\infty \) and \(C^\varepsilon \) norms of \(h_0^*\), thus we will get the same \(C^\varepsilon \)-weight constant \(c_2=\min \{c_1,c_2\}\). Thus, conditionally to \(c_i<(\lambda _1-\gamma )/(2\gamma )\), the weight constant \(c_i\) with \(i=1,2\) for \(\beta \) and \(h_0^*\) are respectively the same. We proceed with the \(L^\infty \) estimate of \(\beta \). We have

$$\begin{aligned} \vert \beta _1(s)\vert&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon }\int _{s}^{\infty }e^{-2\gamma c \tau }\, d\tau \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \mu _0(s)^{c}, \end{aligned}$$

Using hypothesis (6.2) and selecting \(\mathfrak {a}\) close enough to \(\lambda _1\) so that

$$\begin{aligned} c<\mathfrak {a}<\frac{\lambda _1-\gamma }{2\gamma }, \end{aligned}$$
(8.21)

we get

$$\begin{aligned} \vert \beta _2(s)\vert&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \int _{0}^{s} e^{-2 \gamma c\tau }e^{-\mathfrak {a} (s-\tau )}\,ds\\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon }e^{-\min \{2\gamma c,\mathfrak {a}\} s}\\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon }\mu _0(s)^{c}. \end{aligned}$$

Combining the bounds on \(\beta _1\) and \(\beta _2\) we obtain

$$\begin{aligned} \vert \beta (s)\vert \lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp , c,\varepsilon }\mu _0(s)^c. \end{aligned}$$
(8.22)

Now we estimate the \(\left( 1/2+\varepsilon \right) \)-Hölder seminorm. In the following it is enough to assume \(\eta \in (0,1)\). We have

$$\begin{aligned} \vert \beta _1(s)-\beta _1(s-\eta )\vert&\le \vert \int _{s-\eta }^s h_0^*(\tau )\,d \tau \vert \nonumber \\ {}&\le \vert \!\vert h_0^*\vert \!\vert _{\infty ,c}\mu _0(s)^{c}\vert \eta \vert \nonumber \\ {}&\le \vert \!\vert h_0^*\vert \!\vert _{\infty ,c}\mu _0(s)^{c}\vert \eta \vert ^{\frac{1}{2}+\varepsilon } \end{aligned}$$
(8.23)

Let

$$\begin{aligned} l(\tau ){:=}e^{\gamma \tau }\sigma (\tau )-c_\infty . \end{aligned}$$

Following the classical fractional integral estimate of Hardy and Littlewood [35, Theorem 14], we decompose

$$\begin{aligned} \beta _2(s)-\beta _2(s-\eta )&= \int _0^s h_0^*(s-\tau )l(\tau ) \,d\tau - \int _{0}^{s-\eta }h_0^*(s-\eta -\tau )l(\tau )\,d\tau \\&=h_0^*(s)\int _{0}^{s}l(\tau )\,d\tau - \int _{0}^{s} \left[ h_0^*(s)-h_0^*(s-\tau )\right] l(\tau )\,d\tau \\&\quad -h_0^*(s)\int _0^{s-\eta }l(\tau )\,d\tau -\int _{0}^{s-\eta } \left[ h_0^*(s-\eta -\tau )-h_0^*(s)\right] l(\tau )\,d\tau \\&=h_0^*(s)\int _{s-\eta }^{s}l(\tau )\,d\tau -\int _{0}^{\eta } \left[ h_0^*(s)-h_0^*(s-\tau )\right] l(\tau )\,d\tau \\ {}&\quad -\int _{\eta }^{s} \left[ h_0^*(s)-h_0^*(s-\tau )\right] \left( l(\tau )-l(\tau -\eta )\right) \,d\tau \\&=: A_1(s,\eta )+A_2(s,\eta )+A_3(s,\eta ). \end{aligned}$$

For \(s-\eta \in (\eta ,1)\) we have

$$\begin{aligned} \vert A_1\vert&\lesssim \vert h_0^*(s)\vert \int _{s-\eta }^{s}\frac{1}{\sqrt{\tau }}\,d\tau \\&\lesssim \vert h_0^*(s)\vert \left( s^{1/2}-(s-\eta )^{1/2}\right) \\&\lesssim [h_0^*]_{0,\varepsilon ,[s,s+1]} s^{\varepsilon -\frac{1}{2}}\eta \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \mu (s)^c\eta ^{\varepsilon +\frac{1}{2}}. \end{aligned}$$

For \(s-\eta \ge 1\) we get

$$\begin{aligned} \vert A_1\vert&\le \vert h_0^*(s)\vert \int _{s-\eta }^{s} l(\tau )\,d\tau \\&\lesssim \vert h_0^*(s)\vert \int _{s-\eta }^{s}e^{-\mathfrak {a} \tau }\,d\tau \\&\lesssim \vert h_0^*(s)\vert \eta \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon }\mu (s)^c \eta ^{\frac{1}{2}+\varepsilon }. \end{aligned}$$

For \(s-\eta \in (0,\eta )\) we obtain

$$\begin{aligned} \vert A_1\vert \lesssim&\vert h_0^*(s)\vert \int _{s-\eta }^{s} \frac{1}{\sqrt{\tau }}\,d\tau \\ \lesssim&[h_0^*]_{0,\varepsilon ,[s-\eta ,s-\eta +1]} \vert s-\eta \vert ^{\varepsilon } \eta ^{\frac{1}{2}}\\ \lesssim&\vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \mu (s)^c\eta ^{\frac{1}{2}+\varepsilon }. \end{aligned}$$

Now we estimate \(A_2\). We have

$$\begin{aligned} \vert A_2\vert&\le \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \mu (s)^c \int _{0}^{\eta } \vert \tau \vert ^{\varepsilon } \vert l(\tau )\vert \, d\tau \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \mu (s)^c \int _{0}^{\eta } \tau ^{\varepsilon } \frac{1}{\sqrt{\tau }} \,d\tau \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \mu (s)^c \eta ^{\frac{1}{2}+\varepsilon }. \end{aligned}$$

Finally, we estimate \(A_3\). Using the \(L^\infty \) norm of \(h_0^*\) for \(\tau >1\) and \(C^{\varepsilon }\) seminorm for \(\tau <1\) we obtain

$$\begin{aligned} \vert A_3\vert&\lesssim \int _{\eta }^{s}\vert h_0^*(s)-h_0^*(s-\tau )\vert \vert l(\tau )-l(\tau -\eta )\vert \,d\tau \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon }\int _{\eta }^{s} \vert \tau \vert ^{\varepsilon } \vert l(\tau )-l(\tau -\eta )\vert \,d\tau \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \int _{\eta }^{s}\vert \tau \vert ^{\varepsilon } [\tau ^{-1/2}-(\tau -\eta )^{-1/2}] \,d\tau \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \eta \int _{\eta }^{s}\vert \tau \vert ^{\varepsilon } \tau ^{-3/2} \,d\tau \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \eta ^{\frac{1}{2}+\varepsilon } \\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \mu (s-1)^c \eta ^{\frac{1}{2}+\varepsilon }\\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \mu (s)^c \eta ^{\frac{1}{2}+\varepsilon }, \end{aligned}$$

Combining the bounds on \(A_1,A_2,A_3\) and we obtain

$$\begin{aligned} \vert \beta _2(s)-\beta _2(s-\eta )\vert \lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon }\mu (s)^c\vert \eta \vert ^{\frac{1}{2}+\varepsilon }. \end{aligned}$$
(8.24)

Finally, from (8.22), (8.23) and (8.24) we obtain

$$\begin{aligned} \vert \!\vert \beta \vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon } \end{aligned}$$

Going back to the original variable t using (8.14), we obtain

$$\begin{aligned} \vert \!\vert \Lambda \vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\varepsilon }, \end{aligned}$$

and recalling (8.13) the proof of (6.4) is complete. \(\square \)

We proceed to prove the second part of Proposition 6.1: in case \(h\in X_{\sharp ,c,\frac{1}{2}+\varepsilon }\), then \(\Lambda \) is differentiable and \({\dot{\Lambda }} \in X_{\sharp ,c,\varepsilon }\).

Proof of (6.5)

In the same notation of the previous lemma, we need to prove that \(\beta _1(s),\beta _2(s)\) are differentiable and estimate the derivatives. Since

$$\begin{aligned} \beta _1(s){:=}-\int _{s}^{\infty }h_0^*(\tau )\,d\tau , \end{aligned}$$

we clearly have \(\beta _1(s)\in C^{1}(0,\infty )\) and \(\beta _1'(s)=c_\infty h(s)\in X_{{\sharp ,c,\frac{1}{2}+\varepsilon }}\) by hypothesis. To analyze \(\beta _2\), following [35, Theorem 19], we introduce for any \(\epsilon \ge 0\) the function

$$\begin{aligned} \beta _{2,\epsilon }(s)= \int _{0}^{s-\epsilon } h_0^*(\tau ) l(s-\tau )\,d\tau , \end{aligned}$$

so that \(\beta _{2,0}(s)=\beta _2(s)\). Since \(\sigma (\tau )\in C^1\), we can differentiate \(\beta _{2,\epsilon }(s)\) to obtain

$$\begin{aligned} \beta '_{2,\epsilon }(s)&= h_0^*(s-\epsilon )l(\epsilon )+\int _0^{s-\epsilon } h_0^*(\tau ) l'(s-\tau )\, d\tau \\&=-[h_0^*(s)-h_0^*(s-\epsilon )]l(\epsilon )+l(s-\epsilon )h_0^*(s)\\&\quad +\int _0^{s-\epsilon } [h_0^*(\tau )-h_0^*(s)] l'(s-\tau )\, d\tau . \end{aligned}$$

Observe that we can choose the extension \(h_0^*\) such that \(h_0^*(s)=o(s^{1/2})\) for \(s\rightarrow 0\). Since \(h_0^*\in X_{\sharp ,c,\frac{1}{2}+\varepsilon }\), when \(\epsilon \rightarrow 0\) the right-hand side tends uniformly to

$$\begin{aligned} l(s)h_0^*(s)+g(s), \end{aligned}$$

where

$$\begin{aligned} g(s){:=}\int _0^{s} [h_0^*(\tau )-h_0^*(s)]l'(s-\tau )\,ds. \end{aligned}$$

By hypothesis and the choice of the extension we have \(l(s)h_0^*(s)\in X_{\sharp ,c,\frac{1}{2}+\varepsilon }\). Also, the function g(s) is continuous since \(h_0^*(s)\in C^{\frac{1}{2}+\varepsilon }\).

$$\begin{aligned} \beta _2(s_1)-\beta _2(s_2)&=\lim \limits _{\epsilon \rightarrow 0} \left( \beta _{2,\epsilon }(s_1)-\beta _{2,\epsilon }(s_2)\right) \\&=\lim \limits _{\epsilon \rightarrow 0}\int _{s_1}^{s_2} \beta _{2,\epsilon }'(\tau )\,d\tau \\&=\int _{s_1}^{s_2}l(\tau )h_0^*(\tau )+g(\tau )\,d\tau , \end{aligned}$$

hence

$$\begin{aligned} l(s)h_0^*(s)+g(s)=\beta _2'(s). \end{aligned}$$

It remains to prove that \(g(s)\in X_{\sharp ,c,\varepsilon }\). Using the asymptotic of \(\sigma '(t)\) and the assumption (6.2) with \(\mathfrak {a}\) as in (8.21) we have

$$\begin{aligned} \vert g(s)\vert \lesssim&[h]_{0,\frac{1}{2}+\varepsilon ,[s-1,s]}\int _{s-1}^s l'(s-\tau ) \vert s-\tau \vert ^{\frac{1}{2}+\varepsilon } \,d\tau \nonumber \\&+\vert \!\vert h\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }\int _0^{s-1}l'(s-\tau )\mu (\tau )^c \, d\tau \nonumber \\ \lesssim&\vert \!\vert h\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon } \left[ \mu (s)^c \int _{0}^1 \vert w\vert ^{-1+\varepsilon } \, dw +\int _0^{s}e^{-2\gamma c \tau } e^{-\mathfrak {a}(s-\tau )} \right] \, d\tau \nonumber \\ \lesssim&\vert \!\vert h\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }\mu (s)^c. \end{aligned}$$
(8.25)

We write

$$\begin{aligned}&g(s-\eta )-g(s)\\ {}&\quad =\int _0^s \left[ h(s)-h(\tau )\right] l'(s-\tau ) \,d\tau - \int _{0}^{s-\eta } \left[ h(s-\eta )-h(\tau )\right] l'(s-\eta -\tau )\,d\tau \\&\quad = \int _0^{s}\left[ h(s)-h(s-u)\right] l'(u)\,du- \int _\eta ^s [h(s-\eta )-h(s-u)]l'(u-\eta ) \,du\\&\quad = -\int _{\eta }^s[h(s-\eta )-h(s-u)]\left[ l'(u-\eta )-l'(u)\right] \,du\\&\qquad +\int _{\eta }^{s}[h(s)-h(s-\eta )]l'(u)\,du+ \int _{0}^\eta [h(s)-h(s-u)]l'(u)\,du\\&\quad =:B_1(s,\eta )+B_2(s,\eta )+B_3(s,\eta ). \end{aligned}$$

Using again assumption (6.2) we get

$$\begin{aligned} \vert B_1\vert&\lesssim \vert \!\vert h_0^*\vert \!\vert _{0,\frac{1}{2}+\varepsilon ,[s-1,s]} \int _{\eta }^1 \vert u-\eta \vert ^{\frac{1}{2}+\varepsilon }\vert (u-\eta )^{-3/2}-u^{-3/2}\vert \,du\\&\quad +\vert \!\vert h\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon } \int _{1}^s \mu (s-u)\eta \frac{e^{-\mathfrak {a}(u-\eta )}-e^{-\mathfrak {a}u}}{\eta } \ \,du\\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }\mu (s)^c \eta ^{\varepsilon }. \end{aligned}$$

Also

$$\begin{aligned} \vert B_2\vert&\lesssim \vert h_0^*(s)-h_0^*(s-\eta )\vert \eta ^{-1/2}\\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }\mu (s)^c\eta ^{\epsilon }, \end{aligned}$$

and

$$\begin{aligned} \vert B_3\vert&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }\mu (s)^c\int _{0}^{\eta }u^{-1+\varepsilon }\,du\\&\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }\mu (s)^c\eta ^{\varepsilon }. \end{aligned}$$

This proves

$$\begin{aligned} \vert g(s)-g(s-\eta )\vert \lesssim \mu (s)^c\vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon } \vert \eta \vert ^{\varepsilon }. \end{aligned}$$

Combining it with (8.25) we obtain

$$\begin{aligned} \vert \!\vert g\vert \!\vert _{\sharp ,c,\varepsilon }\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }. \end{aligned}$$

Summing up the estimates for \(\beta _1'(s)\) and \(\beta _2'(s)=l(s) h_0^*(s)+g(s)\) we obtain

$$\begin{aligned} \vert \!\vert \beta '(s)\vert \!\vert _{\sharp ,c,\varepsilon }\lesssim \vert \!\vert h_0^*\vert \!\vert _{\sharp ,c,\frac{1}{2}+\varepsilon }. \end{aligned}$$

Finally, in the original variable t, using (8.14) and (8.13), we obtain the bound (6.5). \(\square \)

Remark 2.1

(The initial datum \(J_1(x,t_0)\)) From the proof of Proposition 6.1 we have \(\mathcal {J}(t_0,x)=\int _0^1 h^*(s) I(x,\tau -s)\,ds\) where \(h_0^*\) is an arbitrary smooth function with \(h_0^*(t)=o(t^{1/2})\) for \(t\rightarrow 0\) and \(h_0^*(1)=h(t_0)\), connecting to h(t) at \(t=t_0\) to maintain the \(C^{\varepsilon }\) regularity of h. We observe by estimate (2.2) that

$$\begin{aligned} \vert \!\vert J_1(\cdot ,t_0)\vert \!\vert _{L^\infty (\Omega )}\lesssim \vert \!\vert \mathcal {J}[\dot{\Lambda }](\cdot ,t_0)\vert \!\vert _{L^\infty (\Omega )}\lesssim \vert *\vert {\dot{\Lambda }(t_0)}\lesssim \mu _0(t_0)^{l_1}. \end{aligned}$$

Thus, our initial datum remains positive provided that \(t_0\) is fixed sufficiently large.