1 Introduction

Let (Mg) and (Nh) be Riemannian manifolds with metrics g and h, respectively. A map \(U: M\rightarrow N\) is called harmonic if it is a critical point of the functional

$$\begin{aligned} \mathcal S(U):=\int _M g^{jk}\partial _j U^a\partial _k U^b h_{ab}\circ U, \end{aligned}$$

where we employ Einstein’s summation convention throughout. Note that \(\mathcal S(U)\) is a natural generalization of the Dirichlet energy. The Euler–Lagrange equations associated to \(\mathcal S\) are

$$\begin{aligned} \Delta _M U^a-g^{jk}\Gamma ^a_{bc}(U)\partial _j U^b\partial _k U^c =0, \end{aligned}$$

where \(\Gamma ^a_{bc}\) are the Christoffel symbols on the target manifold N and

$$\begin{aligned} \Delta _M=\frac{1}{\sqrt{\det g}}\partial _j \left( \sqrt{\det g}\,g^{jk}\partial _k\right) \end{aligned}$$

is the Laplace–Beltrami operator on M. The study of harmonic maps is a classical subject in geometric analysis, see e.g. [18, 19, 26,27,28, 34,35,36,37, 39]. The basic mathematical questions concern the existence and, ideally, the classification of harmonic maps. A standard tool in this respect is the associated heat flow, i.e., one considers a one-parameter family \(\{U_t: t\ge 0\}\) of maps from M to N that evolve according to the heat equation

$$\begin{aligned} \partial _t U_t^a=\Delta _M U_t^a-g^{jk}\Gamma ^a_{bc}(U_t)\partial _j U_t^b\partial _k U_t^c. \end{aligned}$$

The idea then is to take an arbitrary map \(U_0: M\rightarrow N\) as initial data at \(t=0\) and due to the regularizing effects of the heat flow, the solution \(U_t\) is expected to converge to an equilibrium as \(t\rightarrow \infty \). In other words, the heat flow is supposed to deform arbitrary maps into harmonic ones. Indeed, this strategy works well under certain curvature assumptions as is demonstrated in the classical paper [27]. In the general case, however, the flow tends to form singularities (or “blow up”) in finite time [1, 2, 7, 8, 10, 20, 21, 25, 30,31,32,33, 43, 44]. This is a severe obstruction which can only be overcome if one is able to continue the flow past the singularity in a well-defined manner. Such a construction is a challenging endeavor which presupposes a detailed understanding of possible blowup scenarios. Naturally, one is mainly interested in blowup behavior that is stable under small perturbations of the initial data.

In this paper we are interested in singularity formation in the heat flow of harmonic maps \(U: \mathbb {S}^d\rightarrow \mathbb {S}^d\). As it turns out, the blowup is a local phenomenon and the curvature of the base manifold is irrelevant for the asymptotic behavior near the singularity. Consequently, we may equally well consider maps \(U: \mathbb {R}^d\rightarrow \mathbb {S}^d\), cf. [21, 38]. Furthermore, we restrict ourselves to the case \(d=3\) and assume corotational symmetry. That is to say, we choose standard spherical coordinates \((r,\theta ,\varphi )\) on \(\mathbb {R}^3\), hyperspherical coordinates on \(\mathbb {S}^3\), and make the ansatz \(U(r,\theta ,\varphi )=(u(r),\theta ,\varphi )\) for the map \(U: \mathbb {R}^3\rightarrow \mathbb {S}^3\). Under this symmetry reduction, the Euler–Lagrange equations associated to the functional \(\mathcal S\) reduce to a single nonlinear ordinary differential equation for u which reads

$$\begin{aligned} u''(r)+\frac{2}{r}u'(r)-\frac{\sin (2u(r))}{r^2}=0,\qquad r\ge 0. \end{aligned}$$

In order to obtain the associated heat flow, we introduce an artificial time dependence and consider the Cauchy problem for the equation

$$\begin{aligned} \partial _t u(r,t)-\partial _r^2 u(r,t)-\frac{2}{r}\partial _r u(r,t)+\frac{\sin (2u(r,t))}{r^2}=0. \end{aligned}$$
(1.1)

Our main result shows the existence of a stable self-similar blowup scenario for Eq. (1.1). For the precise formulation we introduce the following function space.

Definition 1.1

Let

$$\begin{aligned} \tilde{Y}:=\{h\in C^\infty _c([0,\infty )): h^{(2k)}(0)=0 \text{ for } \text{ all } k\in \mathbb {N}_0\} \end{aligned}$$

and set

$$\begin{aligned} \Vert h\Vert _Y:=\Vert |\cdot |^{-1}h(|\cdot |)\Vert _{\dot{H}^2(\mathbb {R}^5)}+\Vert |\cdot |^{-1}h(|\cdot |)\Vert _{\dot{H}^4(\mathbb {R}^5)}. \end{aligned}$$

The Banach space Y is defined as the completion of \(\tilde{Y}\) with respect to \(\Vert \cdot \Vert _Y\).

Theorem 1.2

There exists an \(f_0 \in C^\infty ([0,\infty ))\cap Y\) with \(f_0>0\) on \((0,\infty )\) such that, for any \(T_0>0\) and \(t\in [0,T_0)\),

$$\begin{aligned} u^*_{T_0}(r,t):=f_0\left( \frac{r}{\sqrt{T_0-t}} \right) \end{aligned}$$

is a solution to Eq. (1.1). Furthermore, there exist \(\delta ,M,\omega _0>0\) such that the following holds. For any \(h\in Y\) satisfying \(\Vert h\Vert _Y \le \frac{\delta }{M^2}\), there exists a \(T_h \in [T_0-\frac{\delta }{M},T_0+\frac{\delta }{M}]\) such that Eq. (1.1) with initial data \(u(r,0)=u^*_{T_0}(r,0)+h(r)\) has a unique solution \(u_h\) that blows up at \(t=T_h\) and converges to \(u^*_{T_h}\) in the sense that

$$\begin{aligned} \frac{\Vert u_h(\cdot ,t)-u^*_{T_h}(\cdot ,t)\Vert _Y}{\Vert u_{T_h}^*(\cdot , t)\Vert _Y} \le \delta (T_h-t)^{\omega _0} \end{aligned}$$

for all \(t\in [0,T_h)\). In particular, the class \(\{u^*_{T_0}: T_0>0\}\) of self-similar solutions is nonlinearly asymptotically stable under small perturbations of the initial data.

Some remarks are in order.

  • The map \(U: \mathbb {R}^3\rightarrow \mathbb {S}^3\) has values on the sphere and thus, there is no blowup in \(L^\infty \). However, the self-similar solution \(u_{T_0}^*\) blows up in Y. Indeed, a simple scaling argument shows

    $$\begin{aligned} \Vert u_{T_0}^*(\cdot ,t)\Vert _Y\simeq (T_0-t)^{-\frac{5}{4}} \end{aligned}$$

    for \(t\in [0,T_0)\).

  • The blowup profile \(f_0\) is constructed in the companion paper [3] by a novel computer-assisted (but rigorous) method. It is not known in closed form. Furthermore, \(f_0\) is not the only self-similar profile. In fact, there exist infinitely many self-similar solutions to Eq. (1.1), see [20]. To the knowledge of the authors, Theorem 1.2 is the first result on stable blowup with a nonunique blowup profile that is not known explicitly.

  • The norm \(\Vert \cdot \Vert _Y\) might look odd at first glance since it is based on homogeneous Sobolev spaces on \(\mathbb {R}^5\) whereas Eq. (1.1) is posed on \(\mathbb {R}^3\). However, if one sets \(u(r,t)=rv(r,t)\), Eq. (1.1) transforms into a radial heat equation on \(\mathbb {R}^5\) for the function v. In addition, this transformation regularizes the nonlinearity at the center, see below. In this sense, the effective dimension of the problem is 5 and it is natural to work with radial functions on \(\mathbb {R}^5\).

  • In the formulation of Theorem 1.2 we do not specify the precise solution concept we are using. We will study Eq. (1.1) in similarity coordinates by semigroup theory which yields a canonical notion of strong solution (which is actually called “mild solution” in semigroup theory). Since Eq. (1.1) is parabolic, smoothing effects will kick in immediately and turn strong solutions into classical ones.

  • For obvious reasons, self-similar solutions of the form \(f(\frac{r}{\sqrt{T_0-t}})\) are called shrinkers. Since Eq. (1.1) is not time-reversible, there is another, independent class of self-similar solutions, so-called expanders, which take the form \(f(\frac{r}{\sqrt{t-T_0}})\). The latter have also attracted considerable interest, in particular in connection with the question of unique continuation beyond blowup [2, 22, 23], but they play no role in the present paper.

1.1 Related results

The analysis of harmonic maps is a vast subject that is impossible to review in this paper. We restrict ourselves to a brief discussion of recent blowup results that are directly related to our work and refer the reader to the monographs and survey articles [18, 19, 26, 28, 35, 39] for the general background.

As already indicated, self-similar solutions for the corotational heat flow of harmonic maps \(U:\mathbb {R}^d\rightarrow \mathbb {S}^d\) for \(d\in \{3,4,5,6\}\) are constructed in [20, 21]. Expanding self-similar solutions are studied in [23]. For \(d\ge 7\), there are no self-similar shrinkers [5] and the blowup is of a more complicated nature [1, 4]. The case \(d=2\) is of special interest since it is energy-critical and blowup takes place via shrinking of a soliton [32, 33, 44]. The unique continuation beyond blowup is investigated in [2, 22]. Needless to say, there are similar results for closely related problems like the Yang–Mills heat flow or the nonlinear heat equation, see the discussion in [17] for a brief overview. Of particular interest in this context is the recent paper [9] which also considers self-similar blowup for a nonlinear heat equation with a blowup profile that is not known in closed form. In contrast to our result, however, the blowup studied in [9] is highly unstable and the necessary spectral properties can be obtained by a perturbative argument.

1.2 Outline of the proof

The proof of Theorem 1.2 proceeds by a perturbative construction around the blowup solution \(u_{T_0}^*\). We would like to emphasize that this is a robust approach that uses no structure other than the spectral stability of the self-similar profile \(f_0\) which is established in [3]. As a consequence, our method provides a universal framework for studying self-similar blowup in general parabolic evolution equations. We briefly outline the main steps.

  • We consider Eq. (1.1) with initial data \(u(r,0)=u_{T_0}^*(r,0)+h(r)\). By time translation invariance we may assume \(T_0=1\) and we introduce similarity coordinates \(s=-\log (T-t)+\log T\), \(y=\frac{r}{\sqrt{T-t}}\) which go back to [24]. Here, \(T>0\) is a free parameter which will be adjusted later. Then we rescale the dependent variable u in a suitable manner to obtain the evolution equation

    $$\begin{aligned} \partial _s \tilde{w}-\partial _y^2 \tilde{w}-\tfrac{4}{y}\partial _y \tilde{w}+\tfrac{y}{2}\partial _y \tilde{w}-\tfrac{2}{y^2}\tilde{w}+\tfrac{1}{2} \tilde{w}+\tfrac{\sin (y\tilde{w})}{y^3}=0, \end{aligned}$$

    where \(\tilde{w}=\tilde{w}(y,s)\), with initial data \(\tilde{w}(y,0)=f_0(\sqrt{T}y)/y+h(\sqrt{T} y)/y\). This equation has the static solution \(\tilde{w}(y,s)=f_0(y)/y\). To study its stability, we make the ansatz \(\tilde{w}(y,s)=f_0(y)/y+w(y,s)\) which leads to an evolution equation of the form

    $$\begin{aligned} \left\{ \begin{array}{l} \partial _s w(\cdot ,s)=\hat{\mathcal L}w(\cdot ,s)+\mathcal N(w(\cdot ,s)) \\ w(y,0)=f_0(\sqrt{T} y)/y-f_0(y)/y+h(\sqrt{T}y)/y \end{array} \right. \end{aligned}$$
    (1.2)

    for the perturbation w. The linear operator \(\hat{\mathcal L}\) is given by

    $$\begin{aligned} \hat{\mathcal L}=\partial _y^2 +\tfrac{4}{y}\partial _y -\tfrac{y}{2}\partial _y-\tfrac{1}{2} -V_0(y) \end{aligned}$$

    with the potential \(V_0(y)=\tfrac{2\cos (2f_0(y))-2}{y^2}\) and \(\mathcal N\) denotes the nonlinear remainder. In the spirit of standard local well-posedness theory we now try to solve Eq. (1.2) by treating the nonlinear terms perturbatively. Consequently, we first have to understand the linearized equation that arises from (1.2) by dropping the nonlinear terms.

  • The operator \(\hat{\mathcal L}\), interpreted as an operator acting on radial functions on \(\mathbb {R}^5\), has a self-adjoint extension \(\mathcal L\) on \(L^2_\sigma (\mathbb {R}^5)\) with the weight \(\sigma (x)=e^{-|x|^2/4}\). Here we encounter the fundamental problem in studying self-similar blowup for parabolic equations: In order to apply self-adjoint spectral theory, it seems necessary to study the evolution in Sobolev spaces with exponentially decaying weights. This, however, is impossible since one cannot control nonlinear terms in such spaces.

    There are (at least) two ways around this issue. First, one can study the evolution in unweighted Sobolev spaces and rely on nonself-adjoint spectral theory. This approach was chosen in [17] for the study of the Yang–Mills heat flow. In this paper we follow a different strategy which is based on the simple observation that in a certain sense the problem splits into a self-adjoint part on a compact domain, where the exponentially decaying weight is irrelevant, and a nonself-adjoint part on an unbounded domain which, however, is easy since the potential term is negligible there. We remark that this is not a new discovery but a well-known phenomenon in parabolic problems, see e.g. [6, 9, 29, 40]. Somewhat paradoxically, we can therefore study the linearized evolution on unweighted spaces by using self-adjoint spectral theory in a weighted space.

    More precisely, we consider the semigroup \(e^{s\mathcal L}\) on \(L^2_\sigma (\mathbb {R}^5)\) generated by the self-adjoint operator \(\mathcal L\). From [3] we know that \(\mathcal L\) has precisely one nonnegative eigenvalue \(\lambda =1\) with eigenfunction \(\psi _1\). As usual, this instability is related to the freedom in choosing the parameter T in the similarity coordinates. From self-adjoint spectral theory we obtain the weighted decay estimate

    $$\begin{aligned} \Vert e^{s\mathcal L}f\Vert _{L^2_\sigma }\lesssim e^{-c_0 s}\Vert f\Vert _{L^2_\sigma } \end{aligned}$$

    for some constant \(c_0>0\), provided \(f\perp \psi _1\). Similar bounds hold for higher Sobolev spaces with weights. As a matter of fact, also on unweighted homogeneous Sobolev spaces of sufficiently high degree we have decay, but a priori only for the free operator \(\mathcal L_0=\mathcal L-V_0\). Indeed, an integration by parts shows

    $$\begin{aligned} (\Delta \mathcal L_0 f | \Delta f)_{L^2}\le -\tfrac{1}{4} \Vert f\Vert _{L^2}^2 \end{aligned}$$

    on the unweighted \(L^2\). Similar bounds hold for higher derivatives. Consequently, by combining the unweighted bounds, the weighted decay, and the smallness of \(V_0(y)\) for large y, we derive the unweighted decay

    $$\begin{aligned} \Vert e^{s\mathcal L}f\Vert _X\lesssim e^{-\omega _0 s}\Vert f\Vert _X,\qquad f\perp \psi _1 \end{aligned}$$

    for some \(\omega _0>0\), where \(X=\dot{H}^2(\mathbb {R}^5)\cap \dot{H}^4(\mathbb {R}^5)\).

  • From now on we follow the argument introduced in our earlier works [11,12,13,14,15,16] on self-similar blowup for wave-type equations. We first show that the nonlinearity is locally Lipschitz on X. This is not hard but requires at least some work due to the removable singularity of the nonlinearity at the center. Then we employ Duhamel’s principle to rewrite Eq. (1.2) as

    $$\begin{aligned} \phi (s)=e^{s\mathcal L}\mathcal U(h,T)+\int _0^s e^{(s-s')\mathcal L}\mathcal N(\phi (s'))\mathrm {d}s' \end{aligned}$$
    (1.3)

    where \(\phi (s)(y)=w(y,s)\) and \(\mathcal U(h,T)\) is an abbreviation for the initial data. In general, Eq. (1.3) does not have a global solution due to the unstable eigenvalue \(1\in \sigma (\mathcal L)\). We deal with this issue by employing the Lyapunov–Perron method. That is to say, we first suppress the instability by subtracting a correction term and instead of Eq. (1.3), we consider the modified equation

    $$\begin{aligned} \phi (s)=e^{s\mathcal L}\mathcal U(h,T)+\int _0^s e^{(s-s')\mathcal L}\mathcal N(\phi (s'))\mathrm {d}s' -e^s\mathcal C(\phi ,\mathcal U(h,T)) \end{aligned}$$
    (1.4)

    with

    $$\begin{aligned} \mathcal C(\phi ,\mathcal U(h,T))=\mathcal P \mathcal U(h,T)+\int _0^\infty e^{-s'}\mathcal P\mathcal N(\phi (s'))\mathrm {d}s'. \end{aligned}$$

    Here, \(\mathcal P\) is the orthogonal projection on the unstable subspace \(\langle \psi _1\rangle \). By a fixed point argument we show that for any small h and T close to 1, Eq. (1.4) has a global (in s) solution \(\phi _{h,T}\) that decays like the stable linear flow, i.e., \(\Vert \phi _{h,T}(s)\Vert \lesssim e^{-\omega _0 s}\). In the final step we prove that for any small h, there exists a \(T_h\) close to 1 which makes the correction term vanish. In other words, \(\phi _{h,T_h}\) is a solution to the original equation (1.3).

2 Preliminary transformations

The basic evolution equation is

$$\begin{aligned} \partial _t u(r,t)-\partial _r^2 u(r,t)-\frac{2}{r}\partial _r u(r,t)+\frac{\sin (2u(r,t))}{r^2}=0 \end{aligned}$$
(2.1)

where \(r\ge 0\). For any \(T_0>0\), we have the self-similar solution

$$\begin{aligned} u_{T_0}^*(r,t)=f_0\left( \frac{r}{\sqrt{T_0-t}}\right) \end{aligned}$$

with \(f_0\) constructed in [3]. Our goal is to study the evolution of small initial perturbations of \(u_{T_0}^*\). By time translation invariance, we may restrict ourselves to \(T_0=1\). Consequently, we consider the Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t u(r,t)-\partial _r^2 u(r,t)-\frac{2}{r}\partial _r u(r,t)+\frac{\sin (2u(r,t))}{r^2}=0 \\ u(r,0)=u_1^*(r,0)+h(r)=f_0(r)+h(r) \end{array} \right. \end{aligned}$$
(2.2)

where h is a free function. In order to regularize the nonlinearity, it is useful to change variables according to \(u(r,t)=rv(r,t)\). This yields

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t v(r,t)-\partial _r^2 v(r,t)-\frac{4}{r}\partial _r v(r,t)-\frac{2}{r^2}v(r,t)+\frac{\sin (2rv(r,t))}{r^3}=0 \\ v(r,0)=f_0(r)/r+h(r)/r. \end{array} \right. \end{aligned}$$
(2.3)

Accordingly, we write \(u_{T_0}^*(r,t)=rv_{T_0}^*(r,t)\) for the self-similar solution. Now we switch to similarity coordinates \(s=-\log (T-t)+\log T\), \(y=\frac{r}{\sqrt{T-t}}\) and define the new dependent variable \(\tilde{w}\) by

$$\begin{aligned} \tilde{w}(y,s):=\sqrt{T} e^{-s/2}v\left( \sqrt{T} y e^{-s/2}, T(1-e^{-s})\right) , \end{aligned}$$

or, equivalently,

$$\begin{aligned} v(r,t)=\frac{1}{\sqrt{T-t}}\tilde{w}\left( \frac{r}{\sqrt{T-t}}, -\log (T-t)+\log T\right) . \end{aligned}$$

Here, \(T>0\) is a free parameter that will be needed to account for the time translation invariance of the problem which introduces an artificial instability. Eq. (2.3) transforms into

$$\begin{aligned} \left\{ \begin{array}{l} \partial _s \tilde{w}(y,s)-\partial _y^2 \tilde{w}(y,s)-\frac{4}{y}\partial _y \tilde{w}(y,s)+\frac{y}{2}\partial _y \tilde{w}(y,s)-\frac{2}{y^2}\tilde{w}(y,s)+\frac{1}{2} \tilde{w}(y,s)+\frac{\sin (2y\tilde{w}(y,s))}{y^3}=0 \\ \tilde{w}(y,0)=f_0(\sqrt{T} y)/y+h(\sqrt{T} y)/y. \end{array} \right. \end{aligned}$$
(2.4)

Observe that the only trace of the parameter T is in the initial data. Furthermore, by construction,

$$\begin{aligned} \tilde{w}_T^*(y,s):&=\sqrt{T} e^{-s/2}v_T^*\left( \sqrt{T}ye^{-s/2},T(1-e^{-s})\right) = \frac{1}{y}u_T^*\left( \sqrt{T}ye^{-s/2},T(1-e^{-s})\right) \\&=f_0(y)/y \end{aligned}$$

is a static solution to Eq. (2.4). By making the ansatz \(\tilde{w}(y,s)=f_0(y)/y+w(y,s)\), we rewrite Eq. (2.4) as

$$\begin{aligned} \left\{ \begin{array}{l} \partial _s w(y,s)=\hat{\mathcal L} w(y,s)+\hat{\mathcal N}(w(y,s)) \\ w(y,0)=f_0(\sqrt{T} y)/y-f_0(y)/y+h(\sqrt{T}y)/y \end{array} \right. \end{aligned}$$
(2.5)

with the linear operator \(\hat{\mathcal L}\) defined by

$$\begin{aligned} \begin{aligned} \hat{\mathcal L}&=\partial _y^2 +\frac{4}{y}\partial _y -\frac{y}{2}\partial _y-\frac{1}{2} -\frac{2\cos (2f_0(y))-2}{y^2} \end{aligned} \end{aligned}$$
(2.6)

and the nonlinearity

$$\begin{aligned} \hat{\mathcal N}(w(y,s))=-\frac{1}{y^3}\left[ \sin (2f_0(y)+2yw(y,s))-\sin (2f_0(y))-2y\cos (2f_0(y))w(y,s) \right] . \end{aligned}$$
(2.7)

3 The linearized evolution

In this section we study the linearized equation, i.e., we drop the nonlinearity in Eq. (2.5) and focus on

$$\begin{aligned} \partial _s w(y,s)=\hat{\mathcal L}w(y,s). \end{aligned}$$
(3.1)

Furthermore, we do not specify the initial data explicitly because their specific form is irrelevant for the linear theory.

Note that the operator \(\hat{\mathcal L}\) contains the 5-dimensional radial Laplacian and for the rest of this paper we actually find it convenient to switch to 5-dimensional notation. To this end, we define the operator

$$\begin{aligned} \Lambda f(x):=\tfrac{1}{2} x\nabla f(x)+\tfrac{1}{2} f(x) \end{aligned}$$

acting on functions \(f: \mathbb {R}^5\rightarrow \mathbb {R}\). In the following, the variable x is used to denote an element of \(\mathbb {R}^5\). In this spirit we define the potential \(V_0: \mathbb {R}^5\rightarrow \mathbb {R}\) by

$$\begin{aligned} V_0(x):=-\frac{2\cos (2f_0(|x|))-2}{|x|^2}. \end{aligned}$$

By [3], \(f_0\) is oddFootnote 1 and thus, \(V_0\in C^\infty (\mathbb {R}^5)\), see [45]. Now we define a differential operator \(\tilde{\mathcal L}\) by

$$\begin{aligned} \tilde{\mathcal L} f:=\Delta f-\Lambda f+V_0 f \end{aligned}$$

where throughout, \(\Delta \) denotes the Laplacian on \(\mathbb {R}^5\). Then we have

$$\begin{aligned} \tilde{\mathcal L}f(x)=\tilde{f}''(|x|)+\frac{4}{|x|}\tilde{f}'(|x|) -\frac{|x|}{2}\tilde{f}'(|x|)-\frac{1}{2} \tilde{f}(|x|)+V_0(x)\tilde{f}(|x|) \end{aligned}$$

for all radial functions \(f:\mathbb {R}^5\rightarrow \mathbb {R}\) with \(f(x)=\tilde{f}(|x|)\). Consequently, the linearized equation (3.1) can be written as

$$\begin{aligned} \partial _s \phi (s)=\tilde{\mathcal L}\phi (s) \end{aligned}$$
(3.2)

where \(\phi (s)(x)=w(|x|,s)\). Formally, the solution of Eq. (3.2) is given by \(\phi (s)=e^{s\tilde{\mathcal L}}\phi (0)\). In the following, we make this rigorous.

3.1 Basic semigroup theory

As usual, for \(\Omega \subset \mathbb {R}^d\) open and \(w: \Omega \rightarrow [0,\infty )\) a weight function, we write

$$\begin{aligned} (f|g)_{L^2_w(\Omega )}:=\int _\Omega f(x)g(x)w(x)\mathrm {d}x,\qquad \Vert f\Vert _{L^2_w(\Omega )}:=\sqrt{(f|f)_{L^2_w(\Omega )}} \end{aligned}$$

and denote by \(L^2_w(\Omega )\) the completion of \(C^\infty _c(\Omega )\) with respect to \(\Vert \cdot \Vert _{L^2_w(\Omega )}\).

We promote \(\tilde{\mathcal L}\) to an unbounded linear operator on the Hilbert space

$$\begin{aligned} H:=\{f\in L^2_\sigma (\mathbb {R}^5): f \text{ radial }\} \end{aligned}$$

with \(\sigma (x)=e^{-|x|^2/4}\), by specifying the domain \(\mathcal D(\tilde{\mathcal L}):=\{f\in C^\infty _c(\mathbb {R}^5): f \text{ radial }\}\).

Proposition 3.1

The operator \(\tilde{\mathcal L}: \mathcal D(\tilde{\mathcal L})\subset H\rightarrow H\) is essentially self-adjoint and the spectrum of its closure \(\mathcal L\) satisfies \(\sigma (\mathcal L)\cap [0,\infty )=\{1\}\). The spectral point 1 is a simple eigenvalue and \(\mathcal L\) generates a strongly continuous one-parameter semigroup \(e^{s\mathcal L}\) on H. The function \(\psi _1(x):=f_0'(|x|)/\Vert f_0'(|\cdot |)\Vert _{L^2_\sigma (\mathbb {R}^5)}\) is an eigenfunction of \(\mathcal L\) with eigenvalue 1. Moreover, there exists a constant \(c_0>0\) such that

$$\begin{aligned} \Vert e^{s\mathcal L}f\Vert _{L^2_\sigma (\mathbb {R}^5)}\le e^{-c_0 s}\Vert f\Vert _{L^2_\sigma (\mathbb {R}^5)} \end{aligned}$$

for all \(f\in H\) satisfying \((f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}=0\) and all \(s\ge 0\).

Proof

Via \(\tilde{f}\mapsto |\mathbb {S}^4|^{-1/2}\tilde{f}(|\cdot |): L^2_\rho (0,\infty )\rightarrow H\) with the weight \(\rho (y)=y^4 e^{-y^2/4}\), \(\tilde{\mathcal L}\) is unitarily equivalent to the Sturm–Liouville operator

$$\begin{aligned} \mathcal T \tilde{f}(y):=\frac{1}{\rho (y)}\frac{\mathrm {d}}{\mathrm {d}y}\left[ \rho (y)\tilde{f}'(y)\right] -\frac{1}{2} \tilde{f}(y)-\frac{2\cos (2f_0(y))-2}{y^2} \tilde{f}(y) \end{aligned}$$

with domain \(\mathcal D(\mathcal T):=\{\tilde{f}\in C^\infty _c([0,\infty )): \tilde{f}^{(2k)}(0)=0 \text{ for } \text{ all } k\in \mathbb {N}_0\}\). The equation

$$\begin{aligned} \frac{1}{\rho (y)}\frac{\mathrm {d}}{\mathrm {d}y}\left[ \rho (y)\tilde{f}'(y)\right] =0 \end{aligned}$$
(3.3)

has the explicit solution

$$\begin{aligned} \tilde{f}_1(y)=\int _1^y \rho (s)^{-1}\mathrm {d}s = \int _1^y s^{-4}e^{s^2/4}\mathrm {d}s . \end{aligned}$$

For \(y\in (0,1]\) we have

$$\begin{aligned} |\tilde{f}_1(y)|=\int _y^1 s^{-4}e^{s^2/4}\mathrm {d}s\ge \int _y^1 s^{-4}\mathrm {d}s=\tfrac{1}{3} y^{-3}-\tfrac{1}{3} \end{aligned}$$

and thus, \(\tilde{f}_1\notin L^2_\rho (0,1)\). Similarly, for \(y\ge 1\),

$$\begin{aligned} \tilde{f}_1(y)=2\int _1^y s^{-5}\partial _s e^{s^2/4}\mathrm {d}s\ge 2y^{-5}(e^{y^2/4}-e^{1/4}) \end{aligned}$$

which implies \(\tilde{f}_1\notin L^2_\rho (1,\infty )\). By the Weyl alternative, the Sturm–Liouville operator defined by (3.3) is in the limit-point case at both endpoints and the Kato–Rellich theorem implies that \(\mathcal T\) (and hence \(\tilde{\mathcal L}\)) is essentially self-adjoint, see e.g. [42].

In fact, by \(\tilde{f}\mapsto |\mathbb {S}^4|^{-1/2}\tilde{f}(|\cdot |)/|\cdot |: L^2_{\tilde{\rho }}(0,\infty )\rightarrow H\) with \(\tilde{\rho }(y)=y^2 e^{-y^2/4}\), \(\mathcal L\) is unitarily equivalent to the operator \(-\mathcal A_0\) studied in [3]. Consequently, from [3] we obtain \(\sigma (\mathcal L)\cap [0,\infty )=\{1\}\) with 1 a simple eigenvalue. The corresponding normalized eigenfunction is given by \(\psi _1(x)=f_0'(|x|)/\Vert f_0(|\cdot |)\Vert _{L^2_\sigma (\mathbb {R}^5)}\), see [3]. Since \(0\notin \sigma (\mathcal L)\), we obtain \(-c_0:=\sup \sigma (\mathcal L)\setminus \{1\}<0\) and the self-adjointness of \(\mathcal L\) implies the bound

$$\begin{aligned} (\mathcal L f | f)_{L^2_\sigma (\mathbb {R}^5)}\le -c_0 \Vert f\Vert _{L^2_\sigma (\mathbb {R}^5)}^2 \end{aligned}$$
(3.4)

for all \(f\in \mathcal D(\mathcal L)\) with \((f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}=0\). From this, the stated bound on the semigroup \(e^{s\mathcal L}\) follows. \(\square \)

3.2 Estimates in local Sobolev norms

We upgrade the \(L^2_\sigma \) bound on \(e^{s\mathcal L}\) to a local \(H^4\) bound. In the following we use

$$\begin{aligned} \Vert f\Vert _{G(\mathcal L)}:=\Vert \mathcal Lf\Vert _{L^2_\sigma (\mathbb {R}^5)}+\Vert f\Vert _{L^2_\sigma (\mathbb {R}^5)} \end{aligned}$$

for \(f\in \mathcal D(\mathcal L)\) to denote the graph norm of \(\mathcal L\). Furthermore, the letter C (possibly with subscripts to indicate dependencies) denotes a positive constant that might change its value at each occurrence and \(c_0>0\) is the constant from Proposition 3.1. Finally, for \(R>0\) we set

$$\begin{aligned} \mathbb {B}^5_R:=\{x\in \mathbb {R}^5: |x|<R\}. \end{aligned}$$

Lemma 3.2

Let \(f\in \mathcal D(\mathcal L)\) and \(R\ge 1\). Then \(\nabla f, \Delta f\in L^2(\mathbb {B}^5_R)\) and we have the bound

$$\begin{aligned} \Vert \Delta f\Vert _{L^2(\mathbb {B}_R^5)}+\Vert \nabla f\Vert _{L^2\left( \mathbb {B}^5_R\right) }\le C_R \Vert f\Vert _{G(\mathcal L)} \end{aligned}$$

for all \(R\ge 1\) and all \(f\in \mathcal D(\mathcal L)\).

Proof

Let \(f\in C^\infty _c(\mathbb {R}^5)\) and \(R\ge 1\). An integration by parts yields

$$\begin{aligned} (\mathcal L f | f)_{L^2_\sigma (\mathbb {R}^5)}\le -\Vert \nabla f\Vert _{L^2_\sigma (\mathbb {R}^5)}^2+C\Vert f\Vert _{L^2_\sigma (\mathbb {R}^5)}^2 \end{aligned}$$

and we infer

$$\begin{aligned} \Vert \nabla f\Vert _{L^2\left( \mathbb {B}^5_R\right) }\le C_R \Vert \nabla f\Vert _{L^2_\sigma (\mathbb {R}^5)}\le C_R \Vert f\Vert _{G(\mathcal L)}. \end{aligned}$$
(3.5)

Now let \(f\in \mathcal D(\mathcal L)\). Since \(C^\infty _c(\mathbb {R}^5)\) is a core for \(\mathcal L\), there exists a sequence \((f_n)\subset C^\infty _c(\mathbb {R}^5)\) such that \(f_n \rightarrow f\) in the graph norm \(\Vert \cdot \Vert _{G(\mathcal L)}\). Consequently, Eq. (3.5) shows that \((\partial _j f_n)\) is Cauchy in \(L^2(\mathbb {B}^5_R)\) for any \(j\in \{1,2,\dots ,5\}\). We set \(g_j:=\lim _{n\rightarrow \infty }\partial _j f_n \in L^2(\mathbb {B}^5_R)\). By dominated convergence we infer

$$\begin{aligned} \int _{\mathbb {B}^5_R}g_j \varphi =\lim _{n\rightarrow \infty }\int _{\mathbb {B}^5_R}\partial _j f_n \varphi =-\lim _{n\rightarrow \infty }\int _{\mathbb {B}^5_R}f_n\partial _j\varphi =-\int _{\mathbb {B}^5_R}f\partial _j \varphi \end{aligned}$$

for any \(\varphi \in C^\infty _c(\mathbb {B}^5_R)\). Consequently, \(\partial _j f=g_j\) in the weak sense and this shows \(\nabla f\in L^2(\mathbb {B}^5_R)\) with the bound (3.5).

Let \(f\in C^\infty _c(\mathbb {R}^5)\). Then we have

$$\begin{aligned} \Vert \mathcal L f\Vert _{L^2(\mathbb {B}_R^5)}^2&=(\Delta f-\Lambda f+V_0f | \Delta f-\Lambda f+V_0 f)_{L^2(\mathbb {B}_R^5)} \\&=\Vert \Delta f\Vert _{L^2(\mathbb {B}_R^5)}^2+2(\Delta f | -\Lambda f+V_0 f)_{L^2(\mathbb {B}_R^5)} +\Vert -\Lambda f+V_0 f\Vert _{L^2(\mathbb {B}_R^5)}^2 \end{aligned}$$

which yields the bound

$$\begin{aligned} \Vert \Delta f\Vert _{L^2(\mathbb {B}_R^5)}&\lesssim \Vert \mathcal L f\Vert _{L^2(\mathbb {B}_R^5)} +\Vert \Lambda f\Vert _{L^2\left( \mathbb {B}^5_R\right) }+\Vert V_0 f\Vert _{L^2(\mathbb {B}_R^5)} \\&\le C_R\Vert f\Vert _{G(\mathcal L)} +C_R \Vert \nabla f\Vert _{L^2(\mathbb {B}_R^5)} \\&\le C_R\Vert f\Vert _{G(\mathcal L)} \end{aligned}$$

by Eq. (3.5). Consequently, a density argument as above finishes the proof. \(\square \)

In order to control the full Sobolev norm

$$\begin{aligned} \Vert f\Vert _{H^k\left( \mathbb {B}^5_R\right) }=\sum _{|\alpha |\le k}\Vert \partial ^\alpha f\Vert _{L^2(\mathbb {B}_R^5)} \end{aligned}$$

for \(k=2\), we need two technical results which are completely elementary since we restrict ourselves to radial functions. First, we have a trace lemma.

Lemma 3.3

Let \(R\ge 1\). Then we have the bound

$$\begin{aligned} \Vert \nabla f\Vert _{L^\infty (\partial \mathbb {B}^5_R)}\le C_R \Vert \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) } \end{aligned}$$

for all radial \(f\in C^2(\overline{\mathbb {B}^5_R})\).

Proof

By assumption, there exists a function \(\tilde{f}\in C^2([0,R])\) such that \(f(x)=\tilde{f}(|x|)\). The fundamental theorem of calculus yields

$$\begin{aligned} |R^4 \tilde{f}'(R)|=\left| \int _0^R \partial _r [r^4 \tilde{f}'(r)]\mathrm {d}r\right| \le C_R \left( \int _0^R \left| \tfrac{1}{r^4}\partial _r [r^4 \tilde{f}'(r)]\right| ^2 r^8 \mathrm {d}r \right) ^{1/2}\le C_R\Vert \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) } \end{aligned}$$

and thus,

$$\begin{aligned} |\partial _j f(x)|=\left| \tfrac{x_j}{|x|}\tilde{f}'(|x|)\right| \le C_R \Vert \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) } \end{aligned}$$

for all \(x\in \partial \mathbb {B}^5_R\) and \(j\in \{1,2,\dots ,5\}\). \(\square \)

Next, by an extension argument and Fourier analysis, we easily get control on mixed derivatives. Here and in the following, \(\mathcal F\) is the Fourier transform

$$\begin{aligned} (\mathcal F f)(\xi ):=\int _{\mathbb {R}^5}e^{-\mathrm {i}\,\xi x}f(x)\mathrm {d}x. \end{aligned}$$

Lemma 3.4

Let \(R\ge 1\). Then we have the bound

$$\begin{aligned} \Vert \partial _j\partial _k f\Vert _{L^2\left( \mathbb {B}^5_R\right) }\le C_R\left( \Vert \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) }+ \Vert \nabla f\Vert _{L^2\left( \mathbb {B}^5_R\right) }+\Vert f\Vert _{L^2\left( \mathbb {B}^5_R\right) }\right) \end{aligned}$$

for all radial \(f\in C^2(\overline{\mathbb {B}^5_R})\) and all \(j,k\in \{1,2,\dots ,5\}\).

Proof

Let \(\tilde{f}\in C^2([0,R])\) such that \(f(x)=\tilde{f}(|x|)\). We define an extension \(\tilde{\mathcal E}\tilde{f}\) of \(\tilde{f}\) by

$$\begin{aligned} \tilde{\mathcal E} \tilde{f}(r):=\left\{ \begin{array}{ll}\tilde{f}(r) &{}\quad \text{ for } \;r\in [0,R] \\ \tilde{f}(2R-r)-2\tilde{f}'(R)(R-r) &{}\quad \text{ for } \;r\in (R,2R) \end{array} \right. . \end{aligned}$$

Then we have

$$\begin{aligned} \lim _{r\rightarrow R+}\tilde{\mathcal E}\tilde{f}(r)&=\tilde{f}(R) \\ \lim _{r\rightarrow R+}(\tilde{\mathcal E}\tilde{f})'(r)&=\lim _{r\rightarrow R+}\big [-\tilde{f}'(2R-r)+2\tilde{f}'(R)\big ]=\tilde{f}'(R) \\ \lim _{r\rightarrow R+}(\tilde{\mathcal E}\tilde{f})''(r)&=\lim _{r\rightarrow R+}\tilde{f}''(2R-r)=\tilde{f}''(R) \end{aligned}$$

and thus, \(\tilde{\mathcal E}\tilde{f}\in C^2([0,2R))\). Furthermore,

$$\begin{aligned} \int _R^{\frac{3}{2} R} |\tilde{\mathcal E}\tilde{f}(r)|^2 r^4 \mathrm {d}r&=\int _{\frac{1}{2} R}^R |\tilde{\mathcal E}\tilde{f}(2R-r)|^2(2R-r)^4 \mathrm {d}r \le C_R \int _{\frac{1}{2} R}^R |\tilde{\mathcal E}\tilde{f}(2R-r)|^2 r^4 \mathrm {d}r \\&\le C_R \int _{\frac{1}{2} R}^R |\tilde{f}(r)|^2 r^4 \mathrm {d}r+C_R |\tilde{f}'(R)|^2. \end{aligned}$$

Analogously, we obtain

$$\begin{aligned} \Vert |\cdot |^2(\tilde{\mathcal E} \tilde{f})^{(k)}\Vert _{L^2(R,\frac{3}{2} R)}\le C_R \Vert |\cdot |^2 \tilde{f}^{(k)}\Vert _{L^2(\frac{1}{2}R,R)}+C_R |\tilde{f}'(R)| \end{aligned}$$
(3.6)

for any \(k\in \{0,1,2\}\).

Now let \(\chi : \mathbb {R}^5 \rightarrow [0,1]\) be a smooth cut-off that satisfies \(\chi (x)=1\) for \(|x|\le 1\) and \(\chi (x)=0\) for \(|x|\ge \frac{3}{2}\) and set

$$\begin{aligned} \mathcal E f(x):=\chi (\tfrac{x}{R})\tilde{\mathcal E}\tilde{f}(|x|). \end{aligned}$$

Then \(\mathcal E f\in C_c^2(\mathbb {R}^5)\) with \({\text {supp}}(\mathcal E f)\subset \overline{\mathbb {B}^5_{\frac{3}{2}R}}\) and \(\mathcal Ef=f\) on \(\mathbb {B}^5_R\). From Eq. (3.6) and Lemma 3.3 we obtain the bound

$$\begin{aligned} \Vert \Delta \mathcal E f\Vert _{L^2(\mathbb {R}^5)}&\simeq \Vert \Delta \mathcal E f\Vert _{L^2\left( \mathbb {B}^5_R\right) } +\Vert \Delta \mathcal E f\Vert _{L^2(\mathbb {B}^5_{\frac{3}{2}R}\setminus \mathbb {B}^5_R)} \\&\le C_R \left( \Vert \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) }+\Vert \nabla f\Vert _{L^2\left( \mathbb {B}^5_R\right) }+\Vert f\Vert _{L^2\left( \mathbb {B}^5_R\right) }\right) . \end{aligned}$$

Consequently, the estimate

$$\begin{aligned} \Vert \partial _j\partial _k f\Vert _{L^2\left( \mathbb {B}^5_R\right) }\le \Vert \partial _j\partial _k \mathcal E f\Vert _{L^2(\mathbb {R}^5)} \lesssim \Vert |\cdot |^2 \mathcal F\mathcal E f\Vert _{L^2(\mathbb {R}^5)} \simeq \Vert \Delta \mathcal E f\Vert _{L^2(\mathbb {R}^5)} \end{aligned}$$

finishes the proof. \(\square \)

Now we can control the linear evolution on the Sobolev space \(H^2(\mathbb {B}^5_R)\).

Corollary 3.5

Let \(f\in \mathcal D(\mathcal L)\) and \(R\ge 1\). Then \(e^{s\mathcal L}f \in H^2(\mathbb {B}^5_R)\) for all \(s\ge 0\) and we have the bound

$$\begin{aligned} \Vert e^{s\mathcal L}f\Vert _{H^2\left( \mathbb {B}^5_R\right) } \le C_R e^{-c_0 s}\Vert f\Vert _{G(\mathcal L)} \end{aligned}$$

for all \(s\ge 0\) and all \(f\in \mathcal D(\mathcal L)\) satisfying \((f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}=0\).

Proof

By Lemma 3.4 it suffices to control \(\nabla e^{s\mathcal L}f\) and \(\Delta e^{s\mathcal L}f\). Since \(\mathcal D(\mathcal L)\) is invariant under \(e^{s\mathcal L}\), Lemma 3.2 implies \(\nabla e^{s\mathcal L}f, \Delta e^{s\mathcal L}f \in L^2(\mathbb {B}^5_R)\). Consequently, from Lemma 3.2 and Proposition 3.1 we infer

$$\begin{aligned} \Vert \Delta e^{s\mathcal L}f\Vert _{L^2\left( \mathbb {B}^5_R\right) }+\Vert \nabla e^{s\mathcal L}f\Vert _{L^2\left( \mathbb {B}^5_R\right) }&\le C_R \left( \Vert \mathcal L e^{s\mathcal L}f\Vert _{L^2_\sigma (\mathbb {R}^5)}+\Vert e^{s\mathcal L}f\Vert _{L^2_\sigma (\mathbb {R}^5)} \right) \\&=C_R \left( \Vert e^{s\mathcal L}\mathcal L f\Vert _{L^2_\sigma (\mathbb {R}^5)}+\Vert e^{s\mathcal L}f\Vert _{L^2_\sigma (\mathbb {R}^5)} \right) \\&\le C_R e^{-c_0 s}\Vert f\Vert _{G(\mathcal L)} \end{aligned}$$

since

$$\begin{aligned} (\mathcal L f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}=(f|\mathcal L\psi _1)_{L^2_\sigma (\mathbb {R}^5)} =(f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}=0. \end{aligned}$$

\(\square \)

Next, we improve the above by two derivatives.

Lemma 3.6

Let \(f\in \mathcal D(\mathcal L^2)\) and \(R\ge 1\). Then \(\nabla \Delta f, \Delta ^2 f\in L^2\left( \mathbb {B}^5_R\right) \) and we have the bound

$$\begin{aligned} \Vert \Delta ^2 f\Vert _{L^2(\mathbb {B}_R^5)}+\Vert \nabla \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) }\le C_R \Vert f\Vert _{G(\mathcal L^2)} \end{aligned}$$

for all \(R\ge 1\) and all \(f\in \mathcal D(\mathcal L^2)\).

Proof

Let \(f\in C^\infty _c(\mathbb {R}^5)\) and \(R\ge 1\). From Lemma 3.2 we have the bound

$$\begin{aligned} \Vert \nabla \mathcal L f\Vert _{L^2\left( \mathbb {B}^5_R\right) }\le C_R \Vert f\Vert _{G(\mathcal L^2)}. \end{aligned}$$

Expanding the square yields

$$\begin{aligned} \Vert \nabla \mathcal L f\Vert _{L^2\left( \mathbb {B}^5_R\right) }^2&=\Vert \nabla (\Delta -\Lambda +V_0)f\Vert _{L^2\left( \mathbb {B}^5_R\right) }^2 \\&=\Vert \nabla \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) }^2+2(\nabla \Delta f | \nabla (-\Lambda +V_0)f)_{L^2\left( \mathbb {B}^5_R\right) }\\&\quad +\,\Vert \nabla (-\Lambda +V_0)f\Vert _{L^2\left( \mathbb {B}^5_R\right) }^2 \end{aligned}$$

and thus,

$$\begin{aligned} \Vert \nabla \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) }&\lesssim \Vert \nabla \mathcal Lf\Vert _{L^2\left( \mathbb {B}^5_R\right) } +\Vert f\Vert _{H^2\left( \mathbb {B}^5_R\right) }\le C_R \Vert f\Vert _{G(\mathcal L^2)} \end{aligned}$$

by Lemmas 3.2 and 3.4.

For \(\Delta ^2 f\) we expand \(\Vert \Delta \mathcal L f\Vert _{L^2(\mathbb {B}^5_R)}^2\) and use Lemma 3.2 together with the bound on \(\nabla \Delta f\) to obtain

$$\begin{aligned} \Vert \Delta ^2 f\Vert _{L^2\left( \mathbb {B}^5_R\right) } \lesssim \Vert \Delta \mathcal L f\Vert _{L^2\left( \mathbb {B}^5_R\right) }^2 +\Vert \nabla \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) }^2 +C_R\Vert f\Vert _{G(\mathcal L)}^2\le C_R \Vert f\Vert _{G(\mathcal L^2)}^2. \end{aligned}$$

\(\square \)

Corollary 3.7

Let \(f\in \mathcal D(\mathcal L^2)\) and \(R\ge 1\). Then \(e^{s\mathcal L}f \in H^4(\mathbb {B}^5_R)\) for all \(s\ge 0\) and we have the bound

$$\begin{aligned} \Vert e^{s\mathcal L}f\Vert _{H^4\left( \mathbb {B}^5_R\right) } \le C_R e^{-c_0 s}\Vert f\Vert _{G(\mathcal L^2)} \end{aligned}$$

for all \(s\ge 0\) and all \(f\in \mathcal D(\mathcal L^2)\) satisfying \((f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}=0\).

Proof

It suffices to note that \(\mathcal D(\mathcal L^2)\) is invariant under \(e^{s\mathcal L}\) so that Lemma 3.6 can be applied to \(e^{s\mathcal L}f\). Proposition 3.1 and Lemma 3.4 then yields the statement. \(\square \)

3.3 Estimates in unweighted global Sobolev norms

Next, we prove bounds in \(\dot{H}^2(\mathbb {R}^5)\) and \(\dot{H}^4(\mathbb {R}^5)\). The intersection \(\dot{H}^2(\mathbb {R}^5)\cap \dot{H}^4(\mathbb {R}^5)\) will be our main space where we study the evolution. First, we have to ensure that unweighted Sobolev spaces are invariant under \(e^{s\mathcal L}\).

Lemma 3.8

Let \(k\in \mathbb {N}_0\) and \(f\in H^k(\mathbb {R}^5)\). Then we have \(e^{s\mathcal L}f\in H^k(\mathbb {R}^5)\) for all \(s\ge 0\) and \(e^{s\mathcal L}\) is a strongly continuous semigroup on \(H^k(\mathbb {R}^5)\).

Proof

We denote by \(\mathcal L_0=\mathcal L-V_0+\frac{1}{2}\) the principal part of \(\mathcal L\). The operator \(\mathcal L_0: \mathcal D(\mathcal L)\subset L^2_\sigma (\mathbb {R}^5)\rightarrow L^2_\sigma (\mathbb {R}^5)\) is self-adjoint and it generates the semigroup \(e^{s\mathcal L_0}\). As a matter of fact, \(e^{s\mathcal L_0}\) can be given explicitly and we have

$$\begin{aligned}{}[e^{s\mathcal L_0}f](x)=(K_s * f)(e^{-s/2} x) \end{aligned}$$

where

$$\begin{aligned} K_s(x)=[\pi \alpha (s)]^{-5/2}e^{-|x|^2/\alpha (s)},\qquad \alpha (s)=4(1-e^{-s}). \end{aligned}$$

This is easily verified by an explicit computation. Since \(K_s\in L^1(\mathbb {R}^5)\) for any \(s> 0\), dominated convergence and Young’s inequality immediately imply the invariance of \(H^k(\mathbb {R}^5)\) under \(e^{s\mathcal L_0}\). By rescaling we infer

$$\begin{aligned} e^{s\mathcal L_0}f(x)-f(x)&=[\pi \alpha (s)]^{-\frac{5}{2}}\int _{\mathbb {R}^5} e^{-|x'|^2/\alpha (s)}\big [f(e^{-s/2}x-x')-f(x)\big ]\mathrm {d}x' \\&=\pi ^{-\frac{5}{2}}\int _{\mathbb {R}^5}\big [f(e^{-s/2}x-\alpha (s)^\frac{1}{2} x')-f(x)\big ] e^{-|x'|^2}\mathrm {d}x' \end{aligned}$$

and Minkowski’s inequality yields

$$\begin{aligned} \Vert e^{s\mathcal L_0}f-f\Vert _{L^2(\mathbb {R}^5)}\lesssim \int _{\mathbb {R}^5}\big \Vert f(e^{-s/2}(\cdot )-\alpha (s)^\frac{1}{2} x')-f\big \Vert _{L^2(\mathbb {R}^5)}e^{-|x'|^2}\mathrm {d}x'. \end{aligned}$$

Since scaling and translation are continuous operations on \(L^2(\mathbb {R}^5)\), we infer

$$\begin{aligned} \Vert f(e^{-s/2}(\cdot )-\alpha (s)^\frac{1}{2} x')-f\big \Vert _{L^2(\mathbb {R}^5)}\rightarrow 0 \end{aligned}$$

as \(s\rightarrow 0+\) for any fixed \(x'\in \mathbb {R}^5\). Consequently, by dominated convergence, we obtain

$$\begin{aligned} \Vert e^{s\mathcal L_0}f-f\Vert _{L^2(\mathbb {R}^5)}\rightarrow 0 \end{aligned}$$

as \(s\rightarrow 0+\). The same argument yields \(\Vert e^{s\mathcal L_0}f-f\Vert _{H^k(\mathbb {R}^5)}\rightarrow 0\) as \(s\rightarrow 0+\). We conclude that \(e^{s\mathcal L_0}\) is strongly continuous on \(H^k(\mathbb {R}^5)\). Evidently, the map \(f\mapsto V_0 f\) is bounded on \(H^k(\mathbb {R}^5)\) and thus, by the bounded perturbation theorem, \(e^{s\mathcal L}\) is a strongly continuous semigroup on \(H^k(\mathbb {R}^5)\). \(\square \)

Lemma 3.9

Let \(f\in \mathcal D(\mathcal L)\cap \dot{H}^2(\mathbb {R}^5)\). Then \(e^{s\mathcal L}f\in \dot{H}^2(\mathbb {R}^5)\) for all \(s\ge 0\) and there exists a constant \(c_1>0\) such that

$$\begin{aligned} \Vert e^{s\mathcal L}f\Vert _{\dot{H}^2(\mathbb {R}^5)}\lesssim e^{-c_1 s}\left( \Vert f\Vert _{G(\mathcal L)}+\Vert f\Vert _{\dot{H}^2(\mathbb {R}^5)} \right) \end{aligned}$$

for all \(s\ge 0\) and all \(f\in \mathcal D(\mathcal L)\cap \dot{H}^2(\mathbb {R}^5)\) satisfying \((f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}=0\).

Proof

Let \(f\in C^\infty _c(\mathbb {R}^5)\) and note the commutator relation \([\Delta , \Lambda ]f=\Delta f\) which yields

$$\begin{aligned} \Delta \mathcal L f&=\Delta ^2 f-\Delta \Lambda f+\Delta (V_0 f) =\Delta ^2 f-\Lambda \Delta f-\Delta f+\Delta (V_0 f). \end{aligned}$$

Consequently, with \((-\Lambda \Delta f|\Delta f)_{L^2(\mathbb {R}^5)}= \frac{3}{4} \Vert \Delta f\Vert _{L^2(\mathbb {R}^5)}^2\) we obtain

$$\begin{aligned} (\Delta \mathcal L f | \Delta f)_{L^2(\mathbb {R}^5)}=&-\Vert \nabla \Delta f\Vert _{L^2(\mathbb {R}^5)}^2 \nonumber \\&-(\Lambda \Delta f |\Delta f)_{L^2(\mathbb {R}^5)}-\Vert \Delta f\Vert _{L^2(\mathbb {R}^5)}^2 +(\Delta (V_0 f)|\Delta f)_{L^2(\mathbb {R}^5)} \nonumber \\ \le&-\Vert \nabla \Delta f\Vert _{L^2(\mathbb {R}^5)}^2-\tfrac{1}{4} \Vert \Delta f\Vert _{L^2(\mathbb {R}^5)}^2 +(\Delta (V_0 f)|\Delta f)_{L^2(\mathbb {R}^5)}. \end{aligned}$$
(3.7)

Now we claim the estimate

$$\begin{aligned} |(\Delta (V_0 f)|\Delta f)_{L^2(\mathbb {R}^5)}|\le C_R \Vert f\Vert _{G(\mathcal L)}^2+ \tfrac{C}{R^2}\Vert \Delta f\Vert _{L^2(\mathbb {R}^5)}^2 \end{aligned}$$
(3.8)

for all \(R\ge 1\). To prove this, we note that \(\Delta (V_0 f)=\Delta V_0 f+2\nabla V_0 \nabla f+V_0\Delta f\) and estimate each of these terms individually. Clearly,

$$\begin{aligned} |(V_0 \Delta f | \Delta f)_{L^2(\mathbb {R}^5)}|&\lesssim \Vert |V_0|^\frac{1}{2} \Delta f\Vert _{L^2(\mathbb {R}^5)}^2 = \Vert |V_0|^\frac{1}{2} \Delta f\Vert _{L^2\left( \mathbb {B}^5_R\right) }^2+ \Vert |V_0|^\frac{1}{2} \Delta f\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5_R)}^2 \\&\le C_R \Vert f\Vert _{G(\mathcal L)}^2+\tfrac{C}{R^2}\Vert \Delta f\Vert _{L^2(\mathbb {R}^5)}^2 \end{aligned}$$

where we have used Lemma 3.2 and the decay \(|V_0(x)|\lesssim \langle x\rangle ^{-2}\). Next,

$$\begin{aligned} |(\nabla V_0 \nabla f | \Delta f)_{L^2(\mathbb {R}^5)}|&\lesssim \Vert |\nabla V_0|^\frac{2}{3} \nabla f\Vert _{L^2(\mathbb {R}^5)}^2+\Vert |\nabla V_0|^\frac{1}{3} \Delta f\Vert _{L^2(\mathbb {R}^5)}^2. \end{aligned}$$

Thanks to the decay \(|\nabla V_0(x)|\lesssim \langle x\rangle ^{-3}\), the last term can be estimated as before. For the first term we use the decay of \(\nabla V_0\), Lemma 3.2, and Hardy’s inequality to estimate

$$\begin{aligned} \Vert |\nabla V_0|^\frac{2}{3} \nabla f\Vert _{L^2(\mathbb {R}^5)}&\lesssim \Vert \langle \cdot \rangle ^{-2}\nabla f\Vert _{L^2(\mathbb {R}^5)} \simeq \Vert \nabla f\Vert _{L^2\left( \mathbb {B}^5_R\right) }+\Vert |\cdot |^{-2}\nabla f\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5_R)} \\&\le C_R \Vert f\Vert _{G(\mathcal L)}+\tfrac{C}{R}\Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5_R)} \\&\le C_R \Vert f\Vert _{G(\mathcal L)}+\tfrac{C}{R}\Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\mathbb {R}^5)} \\&\le C_R \Vert f\Vert _{G(\mathcal L)}+\tfrac{C}{R}\Vert \Delta f\Vert _{L^2(\mathbb {R}^5)}. \end{aligned}$$

In view of the decay \(|\Delta V_0(x)|\lesssim \langle x \rangle ^{-4}\), the term \((\Delta V_0 f|\Delta f)_{L^2(\mathbb {R}^5)}\) can be estimated analogously. This proves Eq. (3.8).

Having Eq. (3.8) at our disposal, we obtain from Eq. (3.7) the bound

$$\begin{aligned} (\Delta \mathcal L f | \Delta f)_{L^2(\mathbb {R}^5)}\le \left( -\tfrac{1}{4}+\tfrac{C}{R^2}\right) \Vert \Delta f\Vert _{L^2(\mathbb {R}^5)}^2+C_R \Vert f\Vert _{G(\mathcal L)}^2. \end{aligned}$$
(3.9)

By approximation, Eq. (3.9) extends to all \(f\in \mathcal D (\mathcal L)\) satisfying \(\mathcal L f, f \in H^2(\mathbb {R}^5)\). From Lemma 3.8 we know that \(\mathcal L e^{s\mathcal L}f, e^{s\mathcal L}f\in H^2(\mathbb {R}^5)\) and Eq. (3.9) yields

$$\begin{aligned} \tfrac{1}{2} \partial _s \Vert \Delta e^{s\mathcal L}f\Vert _{L^2(\mathbb {R}^5)}^2&= (\partial _s \Delta e^{s\mathcal L}f | \Delta e^{s\mathcal L}f )_{L^2(\mathbb {R}^5)}=(\Delta \mathcal L e^{s\mathcal L}f | \Delta e^{s\mathcal L}f)_{L^2(\mathbb {R}^5)} \\&\le \left( -\tfrac{1}{4}+\tfrac{C}{R^2}\right) \Vert \Delta e^{s\mathcal L}f\Vert _{L^2(\mathbb {R}^5)}^2 +C_R \Vert e^{s\mathcal L}f\Vert _{G(\mathcal L)}^2 \\&\le -\tfrac{1}{8} \Vert \Delta e^{s\mathcal L}f\Vert _{L^2(\mathbb {R}^5)}^2 + C_R e^{-2c_0 s}\Vert f\Vert _{G(\mathcal L)}^2 \end{aligned}$$

by choosing \(R\ge 1\) sufficiently large. From now on R is fixed and hence, \(C_R=C\). Upon setting \(c_1=\frac{1}{2}\min \{c_0,\frac{1}{8}\}>0\), we infer

$$\begin{aligned} \tfrac{1}{2} \partial _s \Vert \Delta e^{s\mathcal L}f\Vert _{L^2(\mathbb {R}^5)}^2\le -2c_1 \Vert \Delta e^{s\mathcal L}f\Vert _{L^2(\mathbb {R}^5)}^2 + C e^{-4c_1 s}\Vert f\Vert _{G(\mathcal L)}^2 \end{aligned}$$

and this inequality may be rewritten as

$$\begin{aligned} \tfrac{1}{2} \partial _s \left[ e^{4c_1 s}\Vert \Delta e^{s\mathcal L}f\Vert _{L^2(\mathbb {R}^5)}^2 \right] \le C \Vert f\Vert _{G(\mathcal L)}^2. \end{aligned}$$

Consequently, integration yields the bound

$$\begin{aligned} \Vert \Delta e^{s\mathcal L}f\Vert _{L^2(\mathbb {R}^5)}^2&\lesssim \langle s\rangle e^{-4c_1 s} \left( \Vert f\Vert _{G(\mathcal L)}^2+\Vert f\Vert _{\dot{H}^2(\mathbb {R}^5)}^2 \right) \\&\lesssim e^{-2c_1 s} \left( \Vert f\Vert _{G(\mathcal L)}^2+\Vert f\Vert _{\dot{H}^2(\mathbb {R}^5)}^2 \right) . \end{aligned}$$

By a density argument, this bound holds for all \(f\in \mathcal D(\mathcal L)\cap \dot{H}^2(\mathbb {R}^5)\). \(\square \)

It is now straightforward to upgrade to \(\dot{H}^4\).

Lemma 3.10

Let \(f\in \mathcal D(\mathcal L^2)\cap \dot{H}^2(\mathbb {R}^5)\cap \dot{H}^4(\mathbb {R}^5)\). Then \(e^{s\mathcal L}f\in \dot{H}^4(\mathbb {R}^5)\) for all \(s\ge 0\) and there exists a constant \(c_1>0\) such that

$$\begin{aligned} \Vert e^{s\mathcal L}f\Vert _{\dot{H}^4(\mathbb {R}^5)}\lesssim e^{-c_1 s}\left( \Vert f\Vert _{G(\mathcal L^2)}+\Vert f\Vert _{\dot{H}^2(\mathbb {R}^5)}+\Vert f\Vert _{\dot{H}^4(\mathbb {R}^5)} \right) \end{aligned}$$

for all \(s\ge 0\) and all \(f\in \mathcal D(\mathcal L^2)\cap \dot{H}^2(\mathbb {R}^5)\cap \dot{H}^4(\mathbb {R}^5)\) satisfying \((f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}=0\).

Proof

Let \(f\in C^\infty _c(\mathbb {R}^5)\). By applying the commutator relation \([\Delta ,\Lambda ]f=\Delta f\) twice, we obtain the estimate

$$\begin{aligned} (\Delta ^2\mathcal Lf |\Delta ^2 f)_{L^2(\mathbb {R}^5)}\le -\tfrac{5}{4} \Vert \Delta ^2 f\Vert _{L^2(\mathbb {R}^5)}^2 +(\Delta ^2 (V_0 f)|\Delta ^2 f)_{L^2(\mathbb {R}^5)}, \end{aligned}$$

cf. Eq. (3.7). Consequently, it suffices to follow the logic in the proof of Lemma 3.9 and apply Lemma 3.6. \(\square \)

3.4 Control of the linearized flow

Finally, we arrive at the main result on the linearized flow. First, we define the main Sobolev space we will be working with and prove an elementary embedding result.

Definition 3.11

The Banach space X is defined as the completion of all radial functions in \(C^\infty _c(\mathbb {R}^5)\) with respect to the norm

$$\begin{aligned} \Vert f\Vert _X=\Vert f\Vert _{\dot{H}^2(\mathbb {R}^5)\cap \dot{H}^4(\mathbb {R}^5)}=\Vert \Delta f\Vert _{L^2(\mathbb {R}^5)}+\Vert \Delta ^2 f\Vert _{L^2(\mathbb {R}^5)}. \end{aligned}$$

Lemma 3.12

Let \(s\in [0,\frac{3}{2})\). Then we have the bound

$$\begin{aligned} \Vert |\nabla |^s f\Vert _{L^\infty (\mathbb {R}^5)}\lesssim \Vert f\Vert _X \end{aligned}$$

for all \(f\in C^\infty _c(\mathbb {R}^5)\).

Proof

We readily estimate

$$\begin{aligned} \Vert |\nabla |^s f\Vert _{L^\infty (\mathbb {R}^5)}&\lesssim \Vert |\cdot |^s \mathcal F f\Vert _{L^1(\mathbb {R}^5)}\simeq \Vert |\cdot |^s \mathcal F f\Vert _{L^1(\mathbb {B}^5)}+\Vert |\cdot |^s\mathcal F f\Vert _{L^1(\mathbb {R}^5\setminus \mathbb {B}^5)} \\&\lesssim \Vert |\cdot |^{-2}\Vert _{L^2(\mathbb {B}^5)}\Vert |\cdot |^2 \mathcal F f\Vert _{L^2(\mathbb {R}^5)} +\Vert |\cdot |^{-4+s}\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5)}\Vert |\cdot |^4\mathcal F f\Vert _{L^2(\mathbb {R}^5)} \\&\lesssim \Vert f\Vert _{\dot{H}^2(\mathbb {R}^5)}+\Vert f\Vert _{\dot{H}^4(\mathbb {R}^5)}. \end{aligned}$$

\(\square \)

Now we can prove the following simple but useful embedding theorem.

Lemma 3.13

We have the continuous embeddings

$$\begin{aligned} H^4_{\mathrm {rad}}(\mathbb {R}^5)\hookrightarrow X \hookrightarrow C^1(\mathbb {R}^5)\cap W^{1,\infty }(\mathbb {R}^5) \end{aligned}$$

where \(H^4_\mathrm {rad}(\mathbb {R}^5)=\{f\in H^4(\mathbb {R}^5): f \text{ radial }\}\).

Proof

Let \(f\in H^4_{\mathrm {rad}}(\mathbb {R}^5)\). Then there exists a sequence \((f_n)_{n\in \mathbb {N}}\subset C^\infty _c(\mathbb {R}^5)\) of radial functions such that \(f_n\rightarrow f\) with respect to \(\Vert \cdot \Vert _{H^4(\mathbb {R}^5)}\). This implies that \((f_n)_{n\in \mathbb {N}}\) is Cauchy with respect to \(\Vert \cdot \Vert _X\) and thus, there exists a limiting element \(\hat{f} \in X\) such that \(f_n\rightarrow \hat{f}\) in X. We define a map \(\iota : H^4_\mathrm {rad}(\mathbb {R}^5)\rightarrow X\) by setting \(\iota (f):=\hat{f}\). Obviously, \(\iota \) is linear. We claim that \(\iota \) is injective. Indeed, if \(\iota (f)=0\), there exists a sequence \((f_n)_{n\in \mathbb {N}}\subset C^\infty _c(\mathbb {R}^5)\) that converges to f in \(H^4_{\mathrm {rad}}(\mathbb {R}^5)\) and to 0 in X. By Lemma 3.12 we see that \(\lim _{n\rightarrow \infty }\Vert f_n\Vert _{L^\infty (\mathbb {R}^5)}=0\). In particular, \(f_n\rightharpoonup 0\) in \(L^2(\mathbb {R}^5)\). On the other hand, \(f_n\rightarrow f\) in \(H^4_\mathrm {rad}(\mathbb {R}^5)\) implies \(f_n\rightharpoonup f\) in \(L^2(\mathbb {R}^5)\) and the uniqueness of weak limits shows that \(f=0\). Clearly, we have \(\Vert \iota (f)\Vert _X\lesssim \Vert f\Vert _{H^4(\mathbb {R}^5)}\) and thus, \(\iota : H^4_\mathrm {rad}(\mathbb {R}^5)\rightarrow X\) is a continuous embedding.

The second assertion is proved similarly. Indeed, given \(f\in X\) we find a sequence \((f_n)_{n\in \mathbb {N}}\subset C^\infty _c(\mathbb {R}^5)\) such that \(f_n\rightarrow f\) in X. By Lemma 3.12, \((f_n)_{n\in \mathbb {N}}\) is Cauchy in \(W^{1,\infty }(\mathbb {R}^5)\) and therefore converges to a limiting function \(\hat{f}\in C^1(\mathbb {R}^5)\cap W^{1,\infty }(\mathbb {R}^5)\). Using this, we define an inclusion map \(\iota : X\rightarrow C^1(\mathbb {R}^5)\cap W^{1,\infty }(\mathbb {R}^5)\) by setting \(\iota (f):=\hat{f}\). It remains to show that \(\iota \) is injective. If \(\iota (f)=0\), it follows that there exists a sequence \((f_n)_{n\in \mathbb {N}}\subset C^\infty _c(\mathbb {R}^5)\) that converges to f in X and to 0 in \(L^\infty (\mathbb {R}^5)\). Consequently,

$$\begin{aligned} \left| \int _{\mathbb {R}^5} \Delta f_n \varphi \right| = \left| \int _{\mathbb {R}^5} f_n \Delta \varphi \right| \lesssim \Vert f_n\Vert _{L^\infty (\mathbb {R}^5)}\rightarrow 0 \end{aligned}$$

for any \(\varphi \in C_c^\infty (\mathbb {R}^5)\) and thus, \(\Delta f_n \rightharpoonup 0\) in \(L^2(\mathbb {R}^5)\). Analogously, we obtain \(\Delta ^2 f_n\rightharpoonup 0\) in \(L^2(\mathbb {R}^5)\). By the uniqueness of weak limits we therefore have \(\lim _{n\rightarrow \infty }\Vert f_n\Vert _X=0\) and this shows \(f=0\). \(\square \)

Theorem 3.14

The Sobolev space X is invariant under \(e^{s\mathcal L}\) and there exists a constant \(\omega _0>0\) such that

$$\begin{aligned} \Vert e^{s\mathcal L}f\Vert _X\lesssim e^{-\omega _0 s}\Vert f\Vert _X \end{aligned}$$

for all \(s\ge 0\) and all \(f\in X\) satisfying \((f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}=0\).

Proof

By Lemma 3.13, \(\mathcal D(\mathcal L^2)\cap H^4(\mathbb {R}^5)\hookrightarrow X\). Since the former space is invariant under \(e^{s\mathcal L}\), see Lemma 3.8, it follows that \(e^{s\mathcal L} f\in X\) for all \(s\ge 0\) and all \(f\in C^\infty _c(\mathbb {R}^5)\). Consequently, in view of Lemmas 3.9 and 3.10, and a density argument, it suffices to prove the bound

$$\begin{aligned} \Vert f\Vert _{G(\mathcal L^2)}\lesssim \Vert f\Vert _X \end{aligned}$$

for all \(f\in C^\infty _c(\mathbb {R}^5)\). Thanks to the strong decay of the weight \(\sigma (x)=e^{-|x|^2/4}\), we immediately obtain

$$\begin{aligned} \Vert f\Vert _{G(\mathcal L^2)}&\lesssim \Vert |\cdot |^{-2}f\Vert _{L^2(\mathbb {R}^5)}+\Vert |\cdot |^{-1} \nabla f\Vert _{L^2(\mathbb {R}^5)}+\Vert \Delta f\Vert _{L^2(\mathbb {R}^5)} \\&\quad +\,\Vert |\cdot |^{-1}\nabla \Delta f\Vert _{L^2(\mathbb {R}^5)} +\Vert \Delta ^2 f\Vert _{L^2(\mathbb {R}^5)} \\&\lesssim \Vert f\Vert _X \end{aligned}$$

by Hardy’s inequality. \(\square \)

4 The nonlinear evolution

Now we turn to the full nonlinear problem Eq. (2.5). As before with the linear operator, we switch to 5-dimensional notation and define the nonlinearity \(\mathcal N\), acting on functions \(f:\mathbb {R}^5\rightarrow \mathbb {R}\), by

$$\begin{aligned} \mathcal N(f)(x):=-\frac{1}{|x|^3}\left[ \sin (2f_0(|x|)+2|x|f(x))-\sin (2f_0(|x|))-2|x|\cos (2f_0(|x|))f(x)\right] . \end{aligned}$$

With this convention, Eq. (2.5) can be written as

$$\begin{aligned} \left\{ \begin{array}{l} \partial _s \phi (s)=\mathcal L\phi (s)+\mathcal N(\phi (s)) \\ \phi (0)=\mathcal U(h,T) \end{array} \right. \end{aligned}$$
(4.1)

where

$$\begin{aligned} \mathcal U(h,T)(x):=f_0(\sqrt{T} |x|)/|x|-f_0(|x|)/|x|+h(\sqrt{T} |x|)/|x| \end{aligned}$$

and \(\phi : [0,\infty )\rightarrow X\).

So far, this is purely formal. In what follows we first prove basic embedding theorems and then some Moser-type inequalities. These will allow us to show that the nonlinearity is locally Lipschitz on X. Next, we study mapping properties of the “initial data operator” \(\mathcal U\) and finally, we implement an infinite-dimensional version of the Lyapunov–Perron method to prove global existence for Eq. (4.1).

4.1 Further properties of the space X

Corollary 4.1

(Algebra property) We have the bound

$$\begin{aligned} \Vert fg\Vert _X\lesssim \Vert f\Vert _X \Vert g\Vert _X \end{aligned}$$

for all \(f,g\in X\). As a consequence, X is a Banach algebra.

Proof

This is a straightforward consequence of the Leibniz rule, the Gagliardo–Nirenberg inequality (see e.g. [41]), and Lemma 3.12. \(\square \)

Next, we prove weighted \(L^\infty \) bounds outside of balls. As opposed to Lemma 3.12 and Corollary 4.1, the restriction to radial functions is crucial here.

Lemma 4.2

We have the bounds

$$\begin{aligned} \Vert |\cdot |^\frac{3}{2} f\Vert _{L^\infty (\mathbb {R}^5\setminus \mathbb {B}^5)}&\lesssim \Vert f\Vert _{\dot{H}^1(\mathbb {R}^5)} \\ \Vert |\cdot |^\frac{1}{2} f\Vert _{L^\infty (\mathbb {R}^5\setminus \mathbb {B}^5)}&\lesssim \Vert f\Vert _{\dot{H}^2(\mathbb {R}^5)} \end{aligned}$$

for all radial \(f\in C^\infty _c(\mathbb {R}^5)\).

Proof

Let \(f\in C^\infty _c(\mathbb {R}^5)\) be radial and write \(f(x)=\tilde{f}(|x|)\). The fundamental theorem of calculus yields

$$\begin{aligned} \tilde{f}(r)=-\int _r^\infty \tilde{f}'(s)\mathrm {d}s=-\int _r^\infty s^{-2}\tilde{f}'(s)s^2 \mathrm {d}s \end{aligned}$$

and thus, by Cauchy–Schwarz,

$$\begin{aligned} |\tilde{f}(r)|\le \Vert |\cdot |^2 \tilde{f}'\Vert _{L^2(1,\infty )}\left( \int _r^\infty s^{-4}\mathrm {d}s\right) ^{1/2}\lesssim r^{-\frac{3}{2}} \Vert |\cdot |^2 \tilde{f}'\Vert _{L^2(1,\infty )}\lesssim r^{-\frac{3}{2}}\Vert \nabla f\Vert _{L^2(\mathbb {R}^5)} \end{aligned}$$

for all \(r\ge 1\). This implies the first assertion.

For the second statement we proceed similarly and use

$$\begin{aligned} \tilde{f}(r)=\int _r^\infty \int _s^\infty \tilde{f}''(t)\mathrm {d}t = \int _r^\infty \int _s^\infty t^{-2}\tilde{f}''(t)t^2\mathrm {d}t \end{aligned}$$

to obtain the bound

$$\begin{aligned} |\tilde{f}(r)|\le \Vert |\cdot |^2 \tilde{f}''\Vert _{L^2(1,\infty )} \int _r^\infty \left( \int _s^\infty t^{-4}\mathrm {d}t \right) ^{1/2}\mathrm {d}s \lesssim r^{-\frac{1}{2}}\Vert |\cdot |^2 \tilde{f}''\Vert _{L^2(1,\infty )} \end{aligned}$$

for all \(r\ge 1\). Now note that

$$\begin{aligned} \tilde{f}'(|x|)=\tfrac{x^j}{|x|}\partial _j f(x) \end{aligned}$$

and thus, by Hardy’s inequality, we infer

$$\begin{aligned} r^\frac{1}{2} |\tilde{f}(r)|\lesssim \Vert \Delta f\Vert _{L^2(\mathbb {R}^5)}+\Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\mathbb {R}^5)} \lesssim \Vert \Delta f\Vert _{L^2(\mathbb {R}^5)} \end{aligned}$$

for all \(r\ge 1\), which is the desired result. \(\square \)

4.2 Nonlinear estimates

For \(\delta >0\) we set

$$\begin{aligned} X_\delta :=\{f\in X: \Vert f\Vert _X\le \delta \}. \end{aligned}$$

The goal of this section is to prove that the nonlinearity \(\mathcal N\) is locally Lipschitz on X. The key results in this respect are the following Moser-type inequalities. First, we focus on large radii where we need to assume a decay property.

Proposition 4.3

Let \(\Phi \in C^4(\mathbb {R}\times \mathbb {R}^5)\) and suppose

$$\begin{aligned} |\partial ^\alpha \Phi (v,x)|\lesssim \langle x\rangle ^{-1} \end{aligned}$$

for all \((v,x)\in \mathbb {R}\times \mathbb {R}^5\setminus \mathbb {B}^5\) and all multi-indices \(\alpha \in \mathbb {N}_0^6\) with \(|\alpha |\le 4\). For \(f: \mathbb {R}^5\rightarrow \mathbb {R}\) set

$$\begin{aligned} \mathcal M(f)(x):=f(x)^2 \Phi (|x|f(x),x). \end{aligned}$$

Then we have the bound

$$\begin{aligned} \Vert \mathcal M(f)-\mathcal M(g)\Vert _{\dot{H}^2(\mathbb {R}^5\setminus \mathbb {B}^5)\cap \dot{H}^4(\mathbb {R}^5\setminus \mathbb {B}^5)}\lesssim (\Vert f\Vert _X+\Vert g\Vert _X)\Vert f-g\Vert _X \end{aligned}$$

for all \(f,g \in X_1 \cap C^\infty _c(\mathbb {R}^5)\).

Proof

Let \(f,g \in X_1\cap C^\infty _c(\mathbb {R}^5)\) and set \(\mathcal I(f)(x):=\Phi (|x|f(x),x)\). Then we have

$$\begin{aligned} \mathcal M(f)-\mathcal M(g)=f^2 \mathcal I(f)-g^2\mathcal I(g)=(f^2-g^2)\mathcal I(f)+g^2[\mathcal I(f)-\mathcal I(g)] \end{aligned}$$

and

$$\begin{aligned} \mathcal I(f)(x)-\mathcal I(g)(x)&=\int _0^1 \partial _t \Phi \big (|x|g(x)+t|x|(f(x)-g(x)),x\big )\mathrm {d}t \\&=[f(x)-g(x)]|x|\int _0^1 \partial _1 \Phi \big (|x|g(x)+t|x|(f(x)-g(x)),x\big )\mathrm {d}t \\&=:[f(x)-g(x)]\mathcal J(f,g)(x). \end{aligned}$$

Consequently, it suffices to prove

$$\begin{aligned} \Vert gh\mathcal I(f)\Vert _{\dot{H}^2(\Omega )\cap \dot{H}^4(\Omega )}\lesssim \Vert g\Vert _X\Vert h\Vert _X,\qquad \Vert g^2h\mathcal J(f,g)\Vert _{\dot{H}^2(\Omega )\cap \dot{H}^4(\Omega )}\lesssim \Vert g\Vert _X^2\Vert h\Vert _X \end{aligned}$$

for all radial \(f,g,h\in X_1\cap C^\infty _c(\mathbb {R}^5)\), where \(\Omega :=\mathbb {R}^5\setminus \mathbb {B}^5\).

We start with the estimate for \(\mathcal I(f)\). By the chain rule,

$$\begin{aligned} \langle \cdot \rangle |\nabla \mathcal I(f)|&\lesssim |\nabla F|+1 \\ \langle \cdot \rangle |\Delta \mathcal I(f)|&\lesssim |\Delta F|+|\nabla F|^2+|\nabla F|+1 \\ \langle \cdot \rangle |\nabla \Delta \mathcal I(f)|&\lesssim |\nabla \Delta F| +|\Delta F||\nabla F|+|\Delta F|+|\nabla F|^3+|\nabla F|^2+|\nabla F|+1 \\ \langle \cdot \rangle |\Delta ^2 \mathcal I(f)|&\lesssim |\Delta ^2 F| +|\nabla \Delta F||\nabla F| +|\nabla \Delta F| +|\Delta F|^2+|\Delta F||\nabla F|^2 \\&\quad +\,|\Delta F||\nabla F| +|\Delta F|+|\nabla F|^4+|\nabla F|^3+|\nabla F|^2+|\nabla F|+1, \end{aligned}$$

where \(F(x)=|x|f(x)\). The strategy is to use Lemma 4.2 to absorb the growing weight in F. We consider \(\Vert gh\Delta \mathcal I(f)\Vert _{L^2(\Omega )}\) and estimate

$$\begin{aligned} \Vert \langle \cdot \rangle ^{-1}gh\Delta F\Vert _{L^2(\Omega )}&\lesssim \Vert gh\Delta f\Vert _{L^2(\Omega )} +\Vert |\cdot |^{-1}gh\nabla f\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-2}g h f\Vert _{L^2(\Omega )} \\&\lesssim \Vert g\Vert _X\Vert h\Vert _X\Vert f\Vert _X \\ \Vert \langle \cdot \rangle ^{-1}gh |\nabla F|^2\Vert _{L^2(\Omega )}&\lesssim \Vert |\cdot | gh|\nabla f|^2\Vert _{L^2(\Omega )} +\Vert |\cdot |^{-1}ghf^2\Vert _{L^2(\Omega )} \\&\lesssim \Vert |\cdot |^\frac{1}{2} g\Vert _{L^\infty (\Omega )} \Vert h\Vert _{L^\infty (\Omega )} \Vert |\cdot |^\frac{3}{2} \nabla f\Vert _{L^\infty (\Omega )} \Vert |\cdot |^{-1}\nabla f \Vert _{L^2(\Omega )} \\&\quad +\,\Vert |\cdot |^\frac{1}{2} g\Vert _{L^\infty (\Omega )}\Vert |\cdot |^\frac{1}{2} h\Vert _{L^\infty (\Omega )} \Vert f\Vert _{L^\infty (\Omega )}\Vert |\cdot |^{-2}f\Vert _{L^2(\Omega )} \\&\lesssim \Vert g\Vert _X\Vert h\Vert _X\Vert f\Vert _X^2 \\ \Vert \langle \cdot \rangle ^{-1}gh \nabla F\Vert _{L^2(\Omega )}&\lesssim \Vert gh\nabla f\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-1}ghf\Vert _{L^2(\Omega )} \\&\lesssim \Vert |\cdot |^\frac{1}{2} g\Vert _{L^\infty (\Omega )}\Vert |\cdot |^\frac{1}{2} h\Vert _{L^\infty (\Omega )} \left( \Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-2}f\Vert _{L^2(\Omega )} \right) \\&\lesssim \Vert g\Vert _X\Vert h\Vert _X\Vert f\Vert _X \end{aligned}$$

by Lemma 4.2 and Hardy’s inequality. This yields \(\Vert \Delta [gh \mathcal I(f)]\Vert _{L^2(\Omega )}\lesssim \Vert g\Vert _X\Vert h\Vert _X\).

Next, we estimate \(\Vert gh\Delta ^2 \mathcal I(f)\Vert _{L^2(\Omega )}\). The easy terms are

$$\begin{aligned} \Vert \langle \cdot \rangle ^{-1}\Delta ^2 F\Vert _{L^2(\Omega )}&\lesssim \sum _{k=0}^4 \Vert |\cdot |^{-4+k}\nabla ^k f\Vert _{L^2(\Omega )}\lesssim \Vert f\Vert _X \\ \Vert \langle \cdot \rangle ^{-1}\nabla \Delta F\nabla F\Vert _{L^2(\Omega )}&\lesssim \sum _{k=0}^3 \Vert |\cdot |^{-2+k}\nabla ^k f\nabla f\Vert _{L^2(\Omega )} +\sum _{k=0}^3 \Vert |\cdot |^{-3+k}\nabla ^k ff\Vert _{L^2(\Omega )} \\&\lesssim \left( \Vert |\cdot |^\frac{3}{2} \nabla f\Vert _{L^\infty (\Omega )}+\Vert f\Vert _{L^\infty (\Omega )}\right) \sum _{k=0}^3 \Vert |\cdot |^{-3+k}\nabla ^k f\Vert _{L^2(\Omega )} \\&\lesssim \Vert f\Vert _X^2 \\ \Vert \langle \cdot \rangle ^{-1}\nabla \Delta F\Vert _{L^2(\Omega )}&\lesssim \sum _{k=0}^3 \Vert |\cdot |^{-3+k}\nabla ^k f\Vert _{L^2(\Omega )}\lesssim \Vert f\Vert _X \end{aligned}$$

as well as

$$\begin{aligned}&\Vert \langle \cdot \rangle ^{-1}\Delta F|\nabla F|^2\Vert _{L^2(\Omega )}\\&\quad \lesssim \Vert |\cdot |^2 \Delta f|\nabla f|^2\Vert _{L^2(\Omega )} +\Vert |\cdot | |\nabla f|^3\Vert _{L^2(\Omega )}+\Vert f|\nabla f|^2\Vert _{L^2(\Omega )} \\&\qquad +\,\Vert |\cdot | \Delta f f^2\Vert _{L^2(\Omega )} +\Vert \nabla f f^2\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-1}f^3\Vert _{L^2(\Omega )} \\&\quad \lesssim \Vert |\cdot |^\frac{3}{2} \nabla f\Vert _{L^\infty (\Omega )}^2 \left( \Vert \Delta f\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\Omega )} +\Vert |\cdot |^{-2}f\Vert _{L^2(\Omega )} \right) \\&\qquad +\,\Vert |\cdot |^\frac{1}{2} f\Vert _{L^\infty (\Omega )}^2\left( \Vert \Delta f\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-2}f\Vert _{L^2(\Omega )}\right) \\&\quad \lesssim \Vert f\Vert _X^3 \end{aligned}$$

and

$$\begin{aligned} \Vert \langle \cdot \rangle ^{-1}\Delta F\nabla F\Vert _{L^2(\Omega )}&\lesssim \sum _{k=0}^2\Vert |\cdot |^{-1+k}\nabla ^k f\nabla f\Vert _{L^2(\Omega )} +\sum _{k=0}^2 \Vert |\cdot |^{-2+k}\nabla ^k f f\Vert _{L^2(\Omega )} \\&\lesssim \left( \Vert |\cdot |^\frac{3}{2} \nabla f\Vert _{L^\infty (\Omega )}+\Vert f\Vert _{L^\infty (\Omega )}\right) \sum _{k=0}^2 \Vert |\cdot |^{-2+k}\nabla ^k f\Vert _{L^2(\Omega )} \\&\lesssim \Vert f\Vert _X^2 \\ \Vert \langle \cdot \rangle ^{-1}|\nabla F|^4\Vert _{L^2(\Omega )}&\lesssim \Vert |\cdot |^3|\nabla f|^4\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-1}f^4\Vert _{L^2(\Omega )} \\&\lesssim \Vert |\cdot |^\frac{3}{2} \nabla f\Vert _{L^\infty (\Omega )}^3 \Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\Omega )}+\Vert |\cdot |^\frac{1}{2} f\Vert _{L^\infty (\Omega )}^3 \Vert |\cdot |^{-2}f\Vert _{L^2(\Omega )} \\&\lesssim \Vert f\Vert _X^4. \end{aligned}$$

Analogously, we estimate

$$\begin{aligned} \Vert \langle \cdot \rangle ^{-1}|\nabla F|^3\Vert _{L^2(\Omega )}&\lesssim \Vert |\cdot |^2 |\nabla f|^3\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-1}f^3\Vert _{L^2(\Omega )} \\&\lesssim \Vert |\cdot |^\frac{3}{2} \nabla f\Vert _{L^\infty (\Omega )}^2 \Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\Omega )}+\Vert |\cdot |^\frac{1}{2} f\Vert _{L^\infty (\Omega )}^2 \Vert |\cdot |^{-2}f\Vert _{L^2(\Omega )} \\&\lesssim \Vert f\Vert _X^2. \end{aligned}$$

It remains to control the most delicate term, \(|\Delta F|^2\). For this one we use Hardy and Lemma 4.2 to obtain

$$\begin{aligned} \Vert \langle \cdot \rangle ^{-1}gh|\Delta F|^2\Vert _{L^2(\Omega )}&\lesssim \Vert |\cdot |gh|\Delta f|^2\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-1}gh|\nabla f|^2\Vert _{L^2(\Omega )} +\Vert |\cdot |^{-3}ghf^2\Vert _{L^2(\Omega )} \\&\lesssim \Vert |\cdot |^\frac{1}{2} g\Vert _{L^\infty (\Omega )} \Vert |\cdot |^\frac{1}{2} h\Vert _{L^\infty (\Omega )} \\&\quad \times \, \left( \Vert |\Delta f|^2\Vert _{L^2(\Omega )}+\Vert |\cdot |^{-1}|\nabla f|^2\Vert _{L^2(\Omega )} +\Vert |\cdot |^{-2}f^2\Vert _{L^2(\Omega )} \right) \\&\lesssim \Vert g\Vert _X\Vert h\Vert _X\Vert f\Vert _X^2 . \end{aligned}$$

The above estimates easily imply \(\Vert \Delta ^2[gh \mathcal I(f)]\Vert _{L^2(\Omega )}\lesssim \Vert g\Vert _X\Vert h\Vert _X\). Putting everything together, we arrive at the desired \(\Vert gh\mathcal I(f)\Vert _{\dot{H}^2(\Omega )\cap \dot{H}^4(\Omega )}\lesssim \Vert g\Vert _X\Vert h\Vert _X\). The bound on \(\mathcal J(f,g)\) is proved in the exact same way. \(\square \)

The next bound controls the nonlinearity near the center. Here the issue is to handle powers of \(|\cdot |^{-1}\) that arise by differentiation.

Lemma 4.4

Let \(\Phi \in C^4(\mathbb {R})\) and suppose \(\Phi '(0)=0\). For \(f:\mathbb {R}^5\rightarrow \mathbb {R}\) set

$$\begin{aligned} \mathcal M(f)(x)=f(x)^2\Phi (|x|f(x)) \end{aligned}$$

Then we have

$$\begin{aligned} \Vert \mathcal M(f)-\mathcal M(g)\Vert _{\dot{H}^2(\mathbb {B}^5)\cap \dot{H}^4(\mathbb {B}^5)}\lesssim (\Vert f\Vert _X+\Vert g\Vert _X)\Vert f-g\Vert _X \end{aligned}$$

for all \(f,g\in X_1 \cap C^\infty _c(\mathbb {R}^5)\).

Proof

As in the proof of Proposition 4.3, we write \(\mathcal I(f)(x)=\Phi (|x|f(x))\) and \(\mathcal I(f)-\mathcal I(g)=(f-g)\mathcal J(f,g)\) with

$$\begin{aligned} \mathcal J(f,g)(x)=|x|\int _0^1 \Phi '\big (|x|g(x)+t|x|(f(x)-g(x))\big ) \mathrm {d}t \end{aligned}$$

and it suffices to show

$$\begin{aligned} \Vert \mathcal I(f)\Vert _{\dot{H}^2(\mathbb {B}^5)\cap \dot{H}^4(\mathbb {B}^5)}+\Vert \mathcal J(f,g)\Vert _{\dot{H}^2(\mathbb {B}^5) \cap \dot{H}^4(\mathbb {B}^5)}\lesssim 1 \end{aligned}$$

for all \(f,g\in X_1\cap C^\infty _c(\mathbb {R}^5)\). We begin with the bound on \(\mathcal I(f)\). By the chain rule we infer

$$\begin{aligned} |\mathcal I(f)|&\lesssim 1 \\ |\nabla \mathcal I(f)|&\lesssim |\Phi '\circ F||\nabla F| \\ |\Delta \mathcal I(f)|&\lesssim |\Phi '\circ F||\Delta F|+|\nabla F|^2 \\ |\nabla \Delta \mathcal I(f)|&\lesssim |\Phi '\circ F||\nabla \Delta F| +|\Delta F||\nabla F|+|\nabla F|^3 \\ |\Delta ^2 \mathcal I(f)|&\lesssim |\Phi '\circ F||\Delta ^2 F|+|\nabla \Delta F||\nabla F|+|\Delta F|^2+ |\Delta F||\nabla F|^2+|\nabla F|^4 \end{aligned}$$

on the ball \(\mathbb {B}^5\), where \(F(x):=|x|f(x)\) and we have used the fact that \(\Vert F\Vert _{L^\infty (\mathbb {B}^5)}\lesssim 1\) which follows from Lemma 3.12. We consider \(\Vert \Delta \mathcal I(f)\Vert _{L^2(\mathbb {B}^5)}\) and estimate

$$\begin{aligned} \Vert (\Phi '\circ F) \Delta F\Vert _{L^2(\mathbb {B}^5)}&\lesssim \Vert \Delta f\Vert _{L^2(\mathbb {B}^5)} +\Vert \nabla f\Vert _{L^2(\mathbb {B}^5)}+\Vert |\cdot |^{-1}f\Vert _{L^2(\mathbb {B}^5)} \\&\lesssim \Vert \Delta f\Vert _{L^2(\mathbb {R}^5)} +\Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\mathbb {R}^5)} +\Vert |\cdot |^{-2}f\Vert _{L^2(\mathbb {R}^5)} \\&\lesssim \Vert f\Vert _X \\ \Vert |\nabla F|^2\Vert _{L^2(\mathbb {B}^5)}&\lesssim \Vert |\nabla f|^2\Vert _{L^2(\mathbb {B}^5)}+\Vert f^2\Vert _{L^2(\mathbb {B}^5)} \\&\lesssim \Vert \nabla f\Vert _{L^\infty (\mathbb {R}^5)}\Vert |\cdot |^{-1}\nabla f\Vert _{L^2(\mathbb {R}^5)} +\Vert f\Vert _{L^\infty (\mathbb {R}^5)}\Vert |\cdot |^{-2}f\Vert _{L^2(\mathbb {R}^5)} \\&\lesssim \Vert f\Vert _X^2 \end{aligned}$$

by Lemma 3.12 and Hardy’s inequality. This yields the desired \(\Vert \Delta \mathcal I(f)\Vert _{L^2(\mathbb {B}^5)}\lesssim 1\).

Next, we estimate \(\Vert \Delta ^2 \mathcal I(f)\Vert _{L^2(\mathbb {B}^5)}\). The most delicate term is \(|\Phi '\circ F||\Delta ^2 F|\) where we absorb one singular factor \(|\cdot |^{-1}\) by exploiting the assumption \(\Phi '(0)=0\). More precisely,

$$\begin{aligned} \Vert (\Phi '\circ F) \Delta ^2 F\Vert _{L^2(\mathbb {B}^5)}&\lesssim \Vert |\cdot |^{-1}\Phi '\circ F\Vert _{L^\infty (\mathbb {B}^5)}\sum _{k=0}^4 \Vert \cdot |^{-2+k}\nabla ^k f\Vert _{L^2(\mathbb {B}^5)} \\&\lesssim \Vert f\Vert _X^2 \end{aligned}$$

by Hardy’s inequality, the bound

$$\begin{aligned} |(\Phi '\circ F)(x)|=|\Phi '(|x|f(x))|\lesssim |x||f(x)|, \end{aligned}$$

which follows from \(\Phi '(0)=0\), and Lemma 3.12. Furthermore, by Gagliardo–Nirenberg (see e.g. [41], p. 8, Proposition 3.1),

$$\begin{aligned} \Vert |\Delta F|^2\Vert _{L^2(\mathbb {B}^5)}&\lesssim \Vert (\chi \Delta F)^2\Vert _{L^2(\mathbb {R}^5)} \lesssim \Vert \nabla (\chi \nabla F)\nabla (\chi \nabla F)\Vert _{L^2(\mathbb {B}^5)} +\Vert (\nabla \chi \nabla F)^2\Vert _{L^2(\mathbb {R}^5)} \\&\lesssim \Vert \nabla (\chi \nabla F)\Vert _{L^4(\mathbb {R}^5)}^2 +\Vert (\nabla \chi \nabla F)^2\Vert _{L^2(\mathbb {R}^5)} \\&\lesssim \Vert \chi \nabla F\Vert _{L^\infty (\mathbb {R}^5)}\Vert \Delta (\chi \nabla F)\Vert _{L^2(\mathbb {R}^5)} +\Vert \nabla \chi \nabla F\Vert _{L^\infty (\mathbb {R}^5)}^2 \\&\lesssim \Vert f\Vert _X^2, \end{aligned}$$

where \(\chi : \mathbb {R}^5\rightarrow [0,1]\) is a smooth cut-off satisfying \(\chi (x)=1\) for \(|x|\le 1\) and \(\chi (x)=0\) for \(|x|\ge 2\). The remaining terms are readily estimated as

$$\begin{aligned} \Vert \nabla \Delta F\nabla F\Vert _{L^2(\mathbb {B}^5)}&\lesssim \Vert \nabla F\Vert _{L^\infty (\mathbb {B}^5)} \Vert \nabla \Delta F\Vert _{L^2(\mathbb {B}^5)}\lesssim \Vert f\Vert _X^2 \\ \Vert \Delta F|\nabla F|^2\Vert _{L^2(\mathbb {B}^5)}&\lesssim \Vert \nabla F\Vert _{L^\infty (\mathbb {B}^5)}^2 \Vert \Delta F\Vert _{L^2(\mathbb {B}^5)}\lesssim \Vert f\Vert _X^3 \end{aligned}$$

This shows \(\Vert \Delta ^2\mathcal I(f)\Vert _{L^2(\mathbb {B}^5)}\lesssim 1\). The proof of the bound on \(\mathcal J(f,g)\) is identical. \(\square \)

In fact, we need a slightly more general form of Lemma 4.4.

Corollary 4.5

Let \(\Phi \in C^4(\mathbb {R}^2)\) and suppose \(\partial _1\Phi (0,0)=\partial _2\Phi (0,0)=0\). Set

$$\begin{aligned} \mathcal M(f)(x)=f(x)^2\Phi (|x|f(x),|x|\varphi _0(x)) \end{aligned}$$

where \(\varphi _0\in C^\infty _c(\mathbb {R}^5)\) is a fixed function. Then we have

$$\begin{aligned} \Vert \mathcal M(f)-\mathcal M(g)\Vert _{\dot{H}^2(\mathbb {B}^5)\cap \dot{H}^4(\mathbb {B}^5)}\lesssim (\Vert f\Vert _X+\Vert g\Vert _X)\Vert f-g\Vert _X \end{aligned}$$

for all \(f\in X_1 \cap C^\infty _c(\mathbb {R}^5)\).

Proof

This is a straightforward generalization of Lemma 4.4. \(\square \)

We are now in a position to prove that the nonlinearity \(\mathcal N\) is locally Lipschitz on X.

Lemma 4.6

We have the bound

$$\begin{aligned} \Vert \mathcal N(f)-\mathcal N(g)\Vert \lesssim (\Vert f\Vert _X+\Vert g\Vert _X)\Vert f-g\Vert _X \end{aligned}$$

for all \(f,g \in X_1\).

Proof

By a density argument it suffices to consider \(f,g\in X_1\cap C^\infty _c(\mathbb {R}^5)\). Recall that \(\mathcal N(f)(x)=N(f(x),|x|)\) with

$$\begin{aligned} N(u,y)=-\frac{1}{y^3}\left[ \sin (2f_0(y)+2yu)-\sin (2f_0(y))-2y\cos (2f_0(y))u \right] . \end{aligned}$$

Note that

$$\begin{aligned} \partial _1 N(u,y)&=-\frac{2}{y^2}\left[ \cos (2f_0(y)+2yu)-\cos (2f_0(y)) \right] \\ \partial _1^2 N(u,y)&=\frac{4}{y}\sin (2f_0(y)+2yu) \\ \partial _1^3 N(u,y)&=8\cos (2f_0(y)+2yu) . \end{aligned}$$

Evidently, \(N(0,y)=\partial _1 N(0,y)=0\) and the fundamental theorem of calculus yields

$$\begin{aligned} N(u,y)&=\int _0^1 \partial _{t_1}N(t_1u,y)\mathrm {d}t_1 = u\int _0^1 \partial _1 N(t_1 u,y)\mathrm {d}t_1 =u^2 \int _0^1 t_1 \int _0^1 \partial _1^2 N(t_2t_1 u,y)\mathrm {d}t_2 \mathrm {d}t_1 \\&=\frac{4u^2}{y}\int _0^1 t_1 \int _0^1 \sin (2f_0(y)+2t_2t_1yu)\mathrm {d}t_2 \mathrm {d}t_1 \\&=8u^2 \frac{f_0(y)}{y}\int _0^1 t_1\int _0^1 \int _0^1 \cos (2t_3f_0(y)+2t_3t_2t_1yu)\mathrm {d}t_3 \mathrm {d}t_2 \mathrm {d}t_1 \\&\quad +\,8u^3\int _0^1 t_1^2\int _0^1 t_2\int _0^1 \cos (2t_3f_0(y)+2t_3t_2t_1yu)\mathrm {d}t_3 \mathrm {d}t_2 \mathrm {d}t_1. \end{aligned}$$

We define \(\Phi _1, \tilde{\Phi }_1: \mathbb {R}^2\rightarrow \mathbb {R}\) and \(\Phi _2: \mathbb {R}\times \mathbb {R}^5\rightarrow \mathbb {R}\) by

$$\begin{aligned} \Phi _1(v,v_0)&:=8\int _0^1 t_1\int _0^1 \int _0^1 \cos (2t_3v_0+2t_3t_2t_1v)\mathrm {d}t_3 \mathrm {d}t_2 \mathrm {d}t_1 \\ \tilde{\Phi }_1(v,v_0)&:=8\int _0^1 t_1^2\int _0^1 t_2\int _0^1 \cos (2t_3v_0+2t_3t_2t_1v)\mathrm {d}t_3 \mathrm {d}t_2 \mathrm {d}t_1 \\ \Phi _2(v,x)&:=\frac{4}{|x|}\int _0^1 t_1 \int _0^1 \sin (2f_0(|x|)+2t_2t_1v)\mathrm {d}t_2 \mathrm {d}t_1 \end{aligned}$$

and set

$$\begin{aligned} \mathcal M_1(f)(x)&:=f(x)^2 \Phi _1(|x|f(x),|x|\varphi _0(x)) \\ \tilde{\mathcal M}_1(f)(x)&:=f(x)^2 \tilde{\Phi }_1(|x|f(x),|x|\varphi _0(x)) \\ \mathcal M_2(f)(x)&:=f(x)^2 \Phi _2(|x|f(x),x) \end{aligned}$$

where \(\varphi _0(x):=\chi (x)f_0(|x|)/|x|\) with \(\chi :\mathbb {R}^5\rightarrow [0,1]\) the usual smooth cut-off satisfying \(\chi (x)=1\) for \(|x|\le 1\) and \(\chi (x)=0\) for \(|x|\ge 2\). By [3], \(f_0\) is odd and thus, \(\varphi _0\in C^\infty _c(\mathbb {R}^5)\). This yields the representations

$$\begin{aligned} \mathcal N(f)(x)=\varphi _0(x)\mathcal M_1 f(x)+f(x)\tilde{\mathcal M}_1 (f)(x) \end{aligned}$$

for \(|x|\le 1\) and

$$\begin{aligned} \mathcal N(f)(x)=\mathcal M_2(f)(x) \end{aligned}$$

for \(|x|\ge \frac{1}{2}\). Evidently, we have

$$\begin{aligned} \partial _1 \Phi _1(0,0)=\partial _2\Phi _1(0,0)=\partial _1 \tilde{\Phi }_1(0,0)=\partial _2\tilde{\Phi }_1(0,0)=0 \end{aligned}$$

and

$$\begin{aligned} |\partial ^\alpha \Phi _2(v,x)|\le C_\alpha \langle x\rangle ^{-1} \end{aligned}$$

for all \((v,x)\in \mathbb {R}\times \mathbb {R}^5\setminus \mathbb {B}^5\) and all multi-indices \(\alpha \in \mathbb {N}_0^6\). As a consequence, Corollary 4.5 and Proposition 4.3 apply to \(\mathcal M_1\), \(\tilde{\mathcal M}_1\), and \(\mathcal M_2\), respectively, and Corollary 4.1 yields the claim. \(\square \)

4.3 The initial data operator

Now we consider the initial data operator

$$\begin{aligned} \mathcal U(h,T)(x):=f_0(\sqrt{T} |x|)/|x|-f_0(|x|)/|x|+h(\sqrt{T} |x|)/|x| . \end{aligned}$$

Recall that

$$\begin{aligned} \tilde{Y}=\{h\in C^\infty _c([0,\infty )): h^{(2k)}(0)=0 \text{ for } \text{ all } k\in \mathbb {N}_0\} \end{aligned}$$

and Y was defined as the completion of \(\tilde{Y}\) with respect to the norm

$$\begin{aligned} \Vert h\Vert _Y=\Vert |\cdot |^{-1}h(|\cdot |)\Vert _X. \end{aligned}$$

First, we need to make sure that \(\mathcal U(h,T)\) has values in X. The following more general result will be helpful in this respect.

Lemma 4.7

Let \(f\in C^\infty (\mathbb {R}^5)\) and assume the bounds

$$\begin{aligned} |\nabla ^k f(x)|\lesssim \langle x\rangle ^{-1-k} \end{aligned}$$

for all \(x\in \mathbb {R}^5\) and \(k\in \{0,1,2,3,4\}\). Then \(f\in X\).

Proof

Let \(\chi : \mathbb {R}^5\rightarrow [0,1]\) be the usual smooth cut-off satisfying \(\chi (x)=1\) for \(|x|\le 1\) and \(\chi (x)=0\) for \(|x|\ge 2\). For \(n\in \mathbb {N}\) we set \(\chi _n(x):=\chi (x/n)\). Then \(\chi _n f\in C^\infty _c(\mathbb {R}^5)\) for any \(n\in \mathbb {N}\) and \(\chi _nf -f=0\) on \(\mathbb {B}^5_n\). Thus, thanks to the decay \(|f(x)|\lesssim \langle x \rangle ^{-1}\),

$$\begin{aligned} \Vert \chi _n f-f\Vert _{L^\infty (\mathbb {R}^5)}&\le \Vert \chi _n f-f\Vert _{L^\infty (\mathbb {B}^5_n)} +\Vert \chi _n f-f\Vert _{L^\infty (\mathbb {R}^5\setminus \mathbb {B}^5_n)}\lesssim \Vert f\Vert _{L^\infty (\mathbb {R}^5\setminus \mathbb {B}^5_n)} \\&\lesssim n^{-1} \end{aligned}$$

and we see that \(\chi _n f\rightarrow f\) in \(L^\infty (\mathbb {R}^5)\). Now let \(m\le n\) and note that \(\chi _n f-\chi _m f=0\) on the ball \(\mathbb {B}^5_m\). Furthermore,

$$\begin{aligned} \Vert \chi _n f\Vert _{\dot{H}^2(\mathbb {R}^5\setminus \mathbb {B}^5_m)}&\simeq \Vert \Delta \chi _n f\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5_m)} +\Vert \nabla \chi _n \nabla f\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5_m)} +\Vert \chi _n \Delta f\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5_m)} \\&\lesssim \Vert |\cdot |^{-1}\Delta \chi _n\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5_m)} +\Vert |\cdot |^{-2}\nabla \chi _n\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5_m)} +\Vert |\cdot |^{-3}\chi _n\Vert _{L^2(\mathbb {R}^5\setminus \mathbb {B}^5_m)} \\&\lesssim m^{-\frac{1}{2}} \end{aligned}$$

and similarly for \(\Vert \chi _n f\Vert _{\dot{H}^4(\mathbb {R}^5\setminus \mathbb {B}^5_m)}\). In summary, we find

$$\begin{aligned} \Vert \chi _mf -\chi _n f\Vert _X\lesssim m^{-\frac{1}{2}}+n^{-\frac{1}{2}} \end{aligned}$$

for all \(n,m\in \mathbb {N}\) and thus, \((\chi _n f)_{n\in \mathbb {N}}\) is Cauchy in X. This shows \(f\in X\), cf. Lemma 3.13. \(\square \)

Corollary 4.8

The function \(x\mapsto f_0(|x|)/|x|: \mathbb {R}^5\rightarrow \mathbb {R}\) belongs to X.

Proof

By [3], \(f_0: [0,\infty )\rightarrow \mathbb {R}\) is smooth, odd, and satisfies the bounds \(|f_0^{(k)}(y)|\lesssim y^{-k}\) for all \(y\ge 1\) and \(k\in \{0,1,2,3,4\}\). Consequently, the function \(x\mapsto f_0(|x|)/|x|: \mathbb {R}^5\rightarrow \mathbb {R}\) verifies the hypotheses of Lemma 4.7. \(\square \)

Lemma 4.9

The map \(\mathcal U: Y\times [\frac{1}{2},\frac{3}{2}]\rightarrow X\) is well-defined and continuous. Furthermore, we have the bound

$$\begin{aligned} \Vert \mathcal U(h,T)\Vert _X\lesssim \Vert h\Vert _Y+|T-1| \end{aligned}$$

for all \((h,T)\in Y\times [\frac{1}{2},\frac{3}{2}]\).

Proof

Corollary 4.8 and the very definition of Y show that \(\mathcal U\) has values in X. Hence, \(\mathcal U\) is well-defined. For brevity we set \(I:=[\frac{1}{2},\frac{3}{2}]\), \(\varphi _0(x):=f_0(|x|)/|x|\), and \(H(x):=h(|x|)/|x|\). Since \(\Vert h\Vert _Y=\Vert H\Vert _X\), we have to show that the maps \(T \mapsto \varphi _0(\sqrt{T}(\cdot )): I\rightarrow X\) and \((H,T)\mapsto H(\sqrt{T}(\cdot )): X\times I\rightarrow X\) are continuous. The fundamental theorem of calculus yields

$$\begin{aligned} \varphi _0(\sqrt{T_1} x)-\varphi _0(\sqrt{T_2} x)&=\int _0^1 \partial _t \varphi _0\left( \sqrt{T_2} x+t\left( \sqrt{T_1}-\sqrt{T_2}\right) x\right) \mathrm {d}t \\&=\left( \sqrt{T_1}-\sqrt{T_2}\right) \int _0^1 x\nabla \varphi _0\left( \sqrt{T_2}x+t\left( \sqrt{T_1}-\sqrt{T_2}\right) x\right) \mathrm {d}t \\&=:\left( \sqrt{T_1}-\sqrt{T_2}\right) \psi _{T_1,T_2}(x). \end{aligned}$$

Obviously, \(\psi _{T_1,T_2}\in C^\infty (\mathbb {R}^5)\) and \(|\nabla ^k \psi _{T_1,T_2}(x)|\lesssim \langle x\rangle ^{-1-k}\) for all \(x\in \mathbb {R}^5\) and \(k\in \{0,1,2,3,4\}\). Consequently, \(\psi _{T_1,T_2}\in X\) by Lemma 4.7 and we infer

$$\begin{aligned} \left\| \varphi _0(\sqrt{T_1}(\cdot ))-\varphi _0(\sqrt{T_2}(\cdot ))\right\| _X\le \left| \sqrt{T_1}-\sqrt{T_2}\right| \Vert \psi _{T_1,T_2}\Vert _X\lesssim \left| \sqrt{T_1}-\sqrt{T_2}\right| \end{aligned}$$

for all \(T_1,T_2\in I\).

Now let \(\epsilon >0\) and choose \(H_1, H_2 \in X\) such that \(\Vert H_1-H_2\Vert _X\le \epsilon /100\). Furthermore, choose \(\tilde{H}_1 \in C^\infty _c(\mathbb {R}^5)\) such that \(\Vert H_1-\tilde{H}_1\Vert _X\le \epsilon /100\). Then we have

$$\begin{aligned} \left\| H_1(\sqrt{T_1}(\cdot ))-H_2(\sqrt{T_2}(\cdot ))\right\| _X&\le \left\| H_1(\sqrt{T_1}(\cdot ))-H_1(\sqrt{T_2}(\cdot ))\right\| _X \\&\quad +\,\left\| H_1(\sqrt{T_2}(\cdot ))-H_2(\sqrt{T_2}(\cdot ))\right\| _X \\&\le \left\| H_1(\sqrt{T_1}(\cdot ))-H_1(\sqrt{T_2}(\cdot ))\right\| _X +\tfrac{\epsilon }{4} \end{aligned}$$

and

$$\begin{aligned} \left\| H_1(\sqrt{T_1}(\cdot ))-H_1(\sqrt{T_2}(\cdot ))\right\| _X&\le \left\| H_1(\sqrt{T_1}(\cdot ))-\tilde{H}_1(\sqrt{T_1}(\cdot ))\right\| _X \\&\quad +\,\left\| \tilde{H}_1(\sqrt{T_1}(\cdot ))-\tilde{H}_1(\sqrt{T_2}(\cdot ))\right\| _X\\&\quad +\,\left\| \tilde{H}_1(\sqrt{T_2}(\cdot ))-H_1(\sqrt{T_2}(\cdot ))\right\| _X \\&\le \tfrac{\epsilon }{2}+C_\epsilon \left| \sqrt{T_1}-\sqrt{T_2}\right| \end{aligned}$$

for all \(T_1,T_2\in I\) again by the fundamental theorem of calculus. Consequently, we may choose \(|T_1-T_2|\) so small that

$$\begin{aligned} \left\| H_1(\sqrt{T_1}(\cdot ))-H_2(\sqrt{T_2}(\cdot ))\right\| _X<\epsilon . \end{aligned}$$

This proves the continuity of \(\mathcal U\). Finally, from the above it is obvious that

$$\begin{aligned} \Vert \mathcal U(h,T)\Vert _X\lesssim \Vert h\Vert _Y+|\sqrt{T}-1|\lesssim \Vert h\Vert _Y+|T-1|. \end{aligned}$$

\(\square \)

4.4 Global existence for the modified equation

Now we turn to the solution of Eq. (4.1). As an intermediate step we consider the Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{l} \partial _s \phi (s)=\mathcal L\phi (s)+\mathcal N(\phi (s)) \\ \phi (0)=f \end{array} \right. \end{aligned}$$
(4.2)

for given small \(f\in X\). We employ Duhamel’s principle to obtain the weak formulation

$$\begin{aligned} \phi (s)=e^{s\mathcal L}f+\int _0^s e^{(s-s')\mathcal L}\mathcal N(\phi (s'))\mathrm {d}s'. \end{aligned}$$
(4.3)

As a matter of fact, this equation does not have global solutions for arbitrary f due to the unstable subspace of the semigroup \(e^{s\mathcal L}\). Thus, we modify Eq. (4.3) by adding a correction term that stabilizes the evolution. In order to obtain this term, we formally project the evolution to the unstable subspace. That is to say, we define the projection operator \(\mathcal P: H\rightarrow H\) by

$$\begin{aligned} \mathcal Pf:=(f|\psi _1)_{L^2_\sigma (\mathbb {R}^5)}\psi _1. \end{aligned}$$

Note that by [3], \(\psi _1(x)=f_0'(|x|)/\Vert f_0'(|\cdot |)\Vert _{L^2_\sigma (\mathbb {R}^5)}\) satisfies the assumptions of Lemma 4.7 and thus, \(\mathcal P\) has values in X. Furthermore, by Lemma 3.12, \(\mathcal P|_X\) is a bounded projection on X. Applying \(\mathcal P\) to Eq. (4.3) and using the fact that \(\mathcal Pe^{s\mathcal L}f=e^s \mathcal Pf\), we obtain (at least formally)

$$\begin{aligned} \mathcal P\phi (s)=e^s \mathcal Pf+e^s \int _0^s e^{-s'}\mathcal P\mathcal N(\phi (s'))\mathrm {d}s'. \end{aligned}$$

This suggests to subtract the term \(e^s \mathcal C(\Phi , f)\), where

$$\begin{aligned} \mathcal C(\phi , f):=\mathcal Pf+\int _0^\infty e^{-s'}\mathcal P\mathcal N(\phi (s'))\mathrm {d}s'. \end{aligned}$$

In order to put this on a sound functional analytic footing, we introduce the Banach space

$$\begin{aligned} \mathcal X:=\left\{ \phi \in C([0,\infty ),X): \Vert \phi \Vert _{\mathcal X}<\infty \right\} \end{aligned}$$

with the norm

$$\begin{aligned} \Vert \phi \Vert _{\mathcal X}:=\sup _{s>0} e^{\omega _0 s}\Vert \phi (s)\Vert _X, \end{aligned}$$

where \(\omega _0>0\) is the constant from Theorem 3.14. Furthermore, for \(\delta >0\), we set

$$\begin{aligned} \mathcal X_\delta :=\{\phi \in \mathcal X: \Vert \phi \Vert _{\mathcal X}\le \delta \}. \end{aligned}$$

Now we define \(\mathcal K: \mathcal X\times X\rightarrow \mathcal X\) by

$$\begin{aligned} \mathcal K(\phi , f)(s):=e^{s\mathcal L}f+\int _0^s e^{(s-s')\mathcal L}\mathcal N(\phi (s'))\mathrm {d}s' -e^s \mathcal C(\phi ,f) \end{aligned}$$

and show that \(\mathcal K(\cdot ,f)\) is a contraction on \(\mathcal X_\delta \), provided \(f\in X\) is sufficiently small.

Lemma 4.10

There exists a constant \(c>0\) such that \(\mathcal K(\phi ,f)\in \mathcal X_\delta \) for all \(\phi \in \mathcal X_\delta \) and all \(f\in X_{\delta /c}\), provided \(\delta >0\) is sufficiently small.

Proof

By definition, we have

$$\begin{aligned} \mathcal P\mathcal K(\phi ,f)(s)=-e^s\int _s^\infty e^{-s'}\mathcal P\mathcal N(\phi (s'))\mathrm {d}s' \end{aligned}$$

and thus, by Lemma 4.6,

$$\begin{aligned} \Vert \mathcal P\mathcal K(\phi ,f)(s)\Vert _X&\lesssim e^s \int _s^\infty e^{-s'}\Vert \phi (s')\Vert _X^2 \mathrm {d}s'\lesssim e^{-2\omega _0 s}\Vert \phi \Vert _{\mathcal X}^2 \lesssim \delta ^2 e^{-2\omega _0 s}. \end{aligned}$$

Similarly,

$$\begin{aligned} (1-\mathcal P)\mathcal K(\phi ,f)(s)=e^{s\mathcal L}(1-\mathcal P)f+\int _0^s e^{(s-s')\mathcal L} (1-\mathcal P)\mathcal N(\phi (s'))\mathrm {d}s' \end{aligned}$$

and thus,

$$\begin{aligned} \Vert (1-\mathcal P)\mathcal K(\phi ,f)(s)\Vert _X&\lesssim e^{-\omega _0 s}\Vert f\Vert _X+\int _0^s e^{-\omega _0 (s-s')}\Vert \phi (s')\Vert _X^2 \mathrm {d}s' \\&\lesssim \tfrac{\delta }{c}e^{-\omega _0 s}+\Vert \phi \Vert _{\mathcal X}^2 e^{-\omega _0 s}\int _0^s e^{-\omega _0 s'}\mathrm {d}s' \\&\lesssim \tfrac{\delta }{c}e^{-\omega _0 s}+\delta ^2 e^{-\omega _0 s} \end{aligned}$$

by Theorem 3.14 and Lemma 4.6. In summary, this yields \(\Vert \mathcal K(\phi ,f)\Vert _{\mathcal X}\lesssim \frac{\delta }{c}+\delta ^2\) and by choosing \(c>0\) large enough and \(\delta >0\) small enough, we obtain \(\Vert \mathcal K(\phi ,f)\Vert _{\mathcal X}\le \delta \). \(\square \)

Lemma 4.11

Let \(\delta >0\) be sufficiently small. Then we have the bound

$$\begin{aligned} \Vert \mathcal K(\phi ,f)-\mathcal K(\psi ,f)\Vert _{\mathcal X}\le \tfrac{1}{2} \Vert \phi -\psi \Vert _{\mathcal X} \end{aligned}$$

for all \(\phi ,\psi \in \mathcal X_\delta \) and all \(f\in X\).

Proof

We have

$$\begin{aligned} \mathcal P\mathcal K(\phi ,f)(s)-\mathcal P\mathcal K(\psi ,f)(s)=-e^s\int _s^\infty e^{-s'} \mathcal P\big [ \mathcal N(\phi (s'))-\mathcal N(\psi (s')) \big ]\mathrm {d}s' \end{aligned}$$

and Lemma 4.6 yields

$$\begin{aligned} \Vert \mathcal P\mathcal K(\phi ,f)(s)-\mathcal P\mathcal K(\psi ,f)(s)\Vert _X\lesssim \delta e^{-2\omega _0 s}\Vert \phi -\psi \Vert _{\mathcal X}, \end{aligned}$$

cf. the proof of Lemma 4.10. Similarly,

$$\begin{aligned} (1-\mathcal P)\mathcal K(\phi ,f)(s)-(1-\mathcal P)\mathcal K(\psi ,f)(s)= \int _0^s e^{(s-s')\mathcal L}(1-\mathcal P)\big [ \mathcal N(\phi (s'))-\mathcal N(\psi (s')) \big ]\mathrm {d}s' \end{aligned}$$

and thus,

$$\begin{aligned} \Vert (1-\mathcal P)\mathcal K(\phi ,f)(s)-(1-\mathcal P)\mathcal K(\psi ,f)(s)\Vert _X\lesssim \delta e^{-\omega _0 s}\Vert \phi -\psi \Vert _{\mathcal X}. \end{aligned}$$

In summary, we infer

$$\begin{aligned} \Vert \mathcal K(\phi ,f)-\mathcal K(\psi ,f)\Vert _{\mathcal X}\lesssim \delta \Vert \phi -\psi \Vert _{\mathcal X} \end{aligned}$$

and by choosing \(\delta >0\) sufficiently small, we arrive at the claim. \(\square \)

Based on the above, it is now easy to construct a global solution to the modified equation

$$\begin{aligned} \phi (s)=e^{s\mathcal L}\mathcal U(h,T)+\int _0^s e^{(s-s')\mathcal L}\mathcal N(\phi (s'))\mathrm {d}s'-e^s\mathcal C(\phi ,\mathcal U(h,T)). \end{aligned}$$
(4.4)

Corollary 4.12

Let \(M>0\) be sufficiently large and \(\delta >0\) sufficiently small. Then, for every \(h\in Y\) and every \(T>0\) satisfying

$$\begin{aligned} \Vert h\Vert _Y+|T-1|\le \tfrac{\delta }{M}, \end{aligned}$$

there exists a unique \(\phi _{h,T}\in \mathcal X_\delta \) such that

$$\begin{aligned} \phi _{h,T}=\mathcal K(\phi _{h,T},\mathcal U(h,T)). \end{aligned}$$

In particular, \(\phi _{h,T}\) is a solution to Eq. (4.4). Furthermore, the solution map \((h,T)\mapsto \phi _{h,T}\) is continuous.

Proof

By Lemma 4.9, we can achieve

$$\begin{aligned} \Vert \mathcal U(h,T)\Vert _X\le \tfrac{\delta }{c} \end{aligned}$$

for any given \(c>0\) by choosing M sufficiently large. Thus, the existence and uniqueness of \(\phi _{h,T}\) is a consequence of Lemmas 4.10 and 4.11, and the contraction mapping principle.

For the continuity of the solution map we note that

$$\begin{aligned} \Vert \phi _{h_1,T_1}-\phi _{h_2,T_2}\Vert _{\mathcal X}&=\Vert \mathcal K(\phi _{h_1,T_1},\mathcal U(h_1,T_1)) -\mathcal K(\phi _{h_2,T_2},\mathcal U(h_2,T_2))\Vert _{\mathcal X} \\&\le \Vert \mathcal K(\phi _{h_1,T_1},\mathcal U(h_1,T_1)) -\mathcal K(\phi _{h_2,T_2},\mathcal U(h_1,T_1))\Vert _{\mathcal X} \\&\quad +\,\Vert \mathcal K(\phi _{h_2,T_2},\mathcal U(h_1,T_1)) -\mathcal K(\phi _{h_2,T_2},\mathcal U(h_2,T_2))\Vert _{\mathcal X} \\&\le \tfrac{1}{2} \Vert \phi _{h_1,T_1}-\phi _{h_2,T_2}\Vert _{\mathcal X}+C\Vert \mathcal U(h_1,T_1)-\mathcal U(h_2,T_2)\Vert _X \end{aligned}$$

by Lemma 4.11 since

$$\begin{aligned} \Vert \mathcal K&(\phi _{h_2,T_2},\mathcal U(h_1,T_1))(s) -\mathcal K(\phi _{h_2,T_2},\mathcal U(h_2,T_2))(s)\Vert _X \\&=\Vert e^{s\mathcal L}(1-\mathcal P)[\mathcal U(h_1,T_1)-\mathcal U(h_2,T_2)]\Vert _X \\&\lesssim e^{-\omega _0 s}\Vert \mathcal U(h_1,T_1)-\mathcal U(h_2,T_2)\Vert _X \end{aligned}$$

by Theorem 3.14. Consequently, Lemma 4.9 finishes the proof. \(\square \)

Corollary 4.12 provides us with a solution to the modified Eq. (4.4). Thus, in order to obtain a (mild) solution to Eq. (4.1), we have to get rid of the correction term \(\mathcal C(\phi ,\mathcal U(h,T))\). So far, h and T can be chosen freely, subject to the smallness conditions in Corollary 4.12. In the last step of the construction we now show that for any small \(h\in Y\) there exists in fact a \(T_h>0\) such that \(\mathcal C(\phi _{h,T_h}, \mathcal U(h,T_h))=0\).

Lemma 4.13

Let \(M>0\) be sufficiently large and \(\delta >0\) sufficiently small. Then, for every \(h\in Y\) satisfying \(\Vert h\Vert _Y\le \frac{\delta }{M^2}\), there exists a \(T_h \in [1-\frac{\delta }{M}, 1+\frac{\delta }{M}]\) such that

$$\begin{aligned} \mathcal C(\phi _{h,T_h},\mathcal U(h,T_h))=0. \end{aligned}$$

Proof

For brevity we set \(I_{M,\delta }=[1-\frac{\delta }{M}, 1+\frac{\delta }{M}]\). The map \(\mathcal C\) has values in \(\langle \psi _1\rangle \) and thus, it suffices to consider the real-valued function \(F_h: I_{M,\delta }\rightarrow \mathbb {R}\) given by

$$\begin{aligned} F_h(T):=\big (\mathcal C(\phi _{h,T},\mathcal U(h,T)) \big | \psi _1 \big )_{L^2_\sigma (\mathbb {R}^5)}. \end{aligned}$$

By Corollary 4.12, \(F_h\) is continuous. Furthermore, by noting that

$$\begin{aligned} \left. \partial _T \frac{f_0\left( \sqrt{T}|x|\right) }{|x|}\right| _{T=1}=\tfrac{1}{2} f_0'(|x|)=\tfrac{1}{2} \Vert f_0(|\cdot |)\Vert _{L^2_\sigma (\mathbb {R}^5)}\psi _1(x) \end{aligned}$$

we obtain by a Taylor expansion the representation

$$\begin{aligned} \mathcal U(h,T)(x)=\tfrac{1}{2}(T-1)\Vert f_0(|\cdot |)\Vert _{L^2_\sigma (\mathbb {R}^5)}\psi _1(x)+(T-1)^2 f_T(x) + h(\sqrt{T}|x|)/|x| \end{aligned}$$

where \(T\mapsto f_T: I_{M,\delta }\rightarrow X\) is continuous and \(\Vert f_T\Vert _X\lesssim 1\) for all \(T\in I_{M,\delta }\). This yields

$$\begin{aligned} F_h(T)&=\big (\mathcal C(\phi _{h,T},\mathcal U(h,T)) \big | \psi _1 \big )_{L^2_\sigma (\mathbb {R}^5)} \\&=\big (\mathcal P\mathcal U(h,T)\big |\psi _1\big )_{L^2_\sigma (\mathbb {R}^5)} +\int _0^\infty e^{-s'}\big (\mathcal P\mathcal N(\phi _{h,T}(s'))\big | \psi _1\big )_{L^2_\sigma (\mathbb {R}^5)}\mathrm {d}s' \\&=\tfrac{1}{2} (T-1)\Vert f_0(|\cdot |)\Vert _{L^2_\sigma (\mathbb {R}^5)}+O(\tfrac{\delta }{M^2} T^0)+O(\delta ^2 T^0). \end{aligned}$$

Consequently, by setting \(\tilde{F}_h(T)=2\Vert f_0(|\cdot |)\Vert _{L^2_\sigma (\mathbb {R}^5)}^{-1}F_h(T)\), we infer

$$\begin{aligned} \tilde{F}_h(T)=T-1+O(\tfrac{\delta }{M^2}T^0)+O(\delta ^2T^0) \end{aligned}$$

and \(\tilde{F}_h(T)=0\) is equivalent to \(T-1=G(T)\) for a continuous function \(G: I_{M,\delta }\rightarrow \mathbb {R}\) that satisfies

$$\begin{aligned} |G(T)|\le C\tfrac{\delta }{M^2}+C\delta ^2 \end{aligned}$$

for all \(T\in I_{M,\delta }\). By choosing \(M>0\) sufficiently large and \(\delta >0\) sufficiently small, we can achieve \(|G(T)|\le \frac{\delta }{M}\) for all \(T\in I_{M,\delta }\) and thus, \(1+G\) is a continuous self-map of the interval \(I_{M,\delta }\) which necessarily has a fixed point \(T_h\in I_{M,\delta }\). \(\square \)

4.5 Proof of Theorem 1.2

Without loss of generality we set \(T_0=1\). Lemma 4.13 yields a strong solution \(w_h(y,s)=\phi _{h,T_h}(s)(ye_1)\) of Eq. (2.5) with \(T=T_h\) and

$$\begin{aligned} \Vert w_h(|\cdot |,s)\Vert _X\le \delta e^{-\omega _0 s}. \end{aligned}$$

By construction, see Sect. 2,

$$\begin{aligned} u_h(r,t):=f_0\left( \frac{r}{\sqrt{T_h-t}}\right) +\frac{r}{\sqrt{T_h-t}}w_h\left( \frac{r}{\sqrt{T_h-t}}, -\log (T_h-t)+\log T_h\right) \end{aligned}$$

is a solution of Eq. (1.1) with initial data \(u_h(r,0)=u_1^*(0,r)+h(r)\). By scaling, we infer

$$\begin{aligned} \Vert u_h(\cdot ,t)-u_{T_h}^*(\cdot ,t)\Vert _Y&=(T_h-t)^{-\frac{1}{2}}\left\| w_h\left( \frac{|\cdot |}{\sqrt{T_h-t}}, -\log (T_h-t)+\log T_h\right) \right\| _X \\&\lesssim (T_h-t)^{-\frac{5}{4}}\left\| w_h(|\cdot |,-\log (T_h-t)+\log T_h)\right\| _X \\&\lesssim \delta (T_h-t)^{-\frac{5}{4}+\omega _0} \end{aligned}$$

for all \(t\in [0,T_h)\). Furthermore, from Corollary 4.8 it follows that \(f_0\) belongs to Y and the blowup speed of \(u_{T_h}^*\) in Y is

$$\begin{aligned} \Vert u_{T_h}^*(\cdot ,t)\Vert _Y&=\left\| |\cdot |^{-1}f_0\left( \frac{|\cdot |}{\sqrt{T_h-t}}\right) \right\| _X \\&=(T_h-t)^{-\frac{1}{4}}\Vert |\cdot |^{-1}f_0(|\cdot |)\Vert _{\dot{H}^2(\mathbb {R}^5)} +(T_h-t)^{-\frac{5}{4}}\Vert |\cdot |^{-1}f_0(|\cdot |)\Vert _{\dot{H}^4(\mathbb {R}^5)} \\&\simeq (T_h-t)^{-\frac{5}{4}}. \end{aligned}$$

Consequently, the statement of Theorem 1.2 follows by choosing M sufficiently large.