Energy partition for the linear radial wave equation

Abstract

We consider the radial free wave equation in all dimensions and derive asymptotic formulas for the space partition of the energy, as time goes to infinity. We show that the exterior energy estimate, which Duyckaerts et al. obtained in odd dimensions (Duyckaerts et al., J Eur Math Soc 13:533–599, 2011; J Eur Math Soc, 2013) fails in even dimensions. Positive results for restricted classes of data are obtained.

1 Introduction

In this paper we consider solutions to the linear wave equation

$$\begin{aligned} \Box u=0,\quad u(0)=f,\; \partial _t u(0)=g \end{aligned}$$

(1)

where $(f,g)\in (\dot{H}^1\times L^2)(\mathbb R ^d)$ are radial. Denote by $u(t) = S(t)(f,g)$ the solution to this wave equation (1) with initial data $(f,g)$ at time 0.

The origin of our work lies in the exterior energy estimates obtained by Duyckaerts et al. [5, 6] which state that for $d\ge 3$ and odd one has either one of the following estimates (even in the nonradial setting):

$$\begin{aligned} \forall \; t&\ge 0, \quad \int _{|x|\ge t} |\nabla _{t,x} S(t)(f,g)(x)|^2\, dx \ge \frac{1}{2} \int _\mathbb{R ^d} (|\nabla f(x)|^2 + |g(x)|^2)\, dx, \quad \hbox {or}\nonumber \\ \forall \; t&\le 0, \quad \int _{|x|\ge -t} |\nabla _{t,x} S(t)(f,g)(x)|^2 \, dx \ge \frac{1}{2} \int _\mathbb{R ^d} (|\nabla f(x)|^2 + |g(x)|^2)\, dx, \end{aligned}$$

(2)

where

$$\begin{aligned} |\nabla _{t,x} S(t)(f,g)|^2 = |\nabla u(t)|^2 + |\partial _t u(t)|^2 \end{aligned}$$

is the linear energy density (see [6, Proposition 2.7]). No result of this type was established there for even dimensions, and the method of proof used in odd dimensions does not apply in even dimensions.

In this paper we show that (2) fails in even dimensions. To be specific, there does not exist a positive constant which can be substituted on the right-hand side for $\frac{1}{2}$ and so that the resulting inequality will hold for all $(f,g)$.This will be based on a computation of the asymptotic exterior energy

$$\begin{aligned} \lim _{t\rightarrow \pm \infty } \int _{|x|\ge |t|} |\nabla _{t,x} S(t)(f,g)(x)|^2\, dx =: \lim _{t\rightarrow \pm \infty } \Vert \nabla _{t,x} S(t)(f,g)(x) \Vert _{L^2(|x| \ge |t|)}^2. \end{aligned}$$

Note that the exterior energy is decreasing in $|t|$, whence (2) reduces to the computation of these limits. Since the propagator $S(t)$ is difficult to work with on the “physical side”, we employ the Fourier transform in this computation. To state our asymptotic result, we introduce the Hankel transform $H$ and the Hilbert transform ${\fancyscript{ H}}$ on the half-line $(0,\infty )$:

$$\begin{aligned} (H\varphi )(\rho ):=\int _0^\infty \frac{\varphi (\sigma )}{\rho +\sigma }\, d\sigma , \quad \hbox {and} \quad ({\fancyscript{ H}}\varphi )(\rho ):=\int _0^\infty \frac{\varphi (\sigma )}{\rho -\sigma }\, d\sigma \end{aligned}$$

where the second integral is to be taken in the principal value sense. Both these operators are bounded and self-adjoint (anti-selfadjoint, respectively) on $L^2((0,\infty ),d\rho )$, with norm $\pi$. Furthermore, $H$ is a positive operator since it is of the form $H=\fancyscript{L}^2$ where $\fancyscript{L}$ is the Laplace transform, see for example Lax [7, Section 16.3.3] for details. This positivity is important for our purposes. In even dimensions, we find the following expression for the asymptotic exterior energy in terms of $H$ and ${\fancyscript{ H}}$. In the following two theorems, we use the notation

$$\begin{aligned} \langle f,g\rangle := \int _0^\infty f(x) \overline{g(x)}\, dx \end{aligned}$$

for two functions $f,g$ on the half-line $(0,\infty )$.

Theorem 1

Let $d$ be even, $(f,g)\in \dot{H}^1\times L^2(\mathbb R ^d)$ be radial as above, and denote by $\hat{f},\hat{g}$ their Fourier transforms in $\mathbb R ^d$. Then for some constant $C(d)>0$ one has for the solution $u$ of (1)

$$\begin{aligned}&\lim _{t\rightarrow \pm \infty } C(d) \Vert \nabla _{t,x} S(t)(f,g) \Vert _{L^2(|x| \ge |t|)}^2 \!=\! \frac{\pi }{2} \int (\rho ^2|\hat{f}(\rho )|^2\!+\!|\hat{g}(\rho )|^2)\rho ^{d-1}\, d\rho \nonumber \\&\quad +\frac{(-1)^{\frac{d}{2}} }{2} \left( \left\langle H(\rho ^{\frac{d+1}{2}} \hat{f}), \rho ^{\frac{d+1}{2}} \hat{f} \right\rangle \!-\! \left\langle H( \rho ^{\frac{d-1}{2}} \hat{g}), \rho ^{\frac{d-1}{2}} \hat{g} \right\rangle \right) \!\pm \! \mathrm Re \left\langle \rho ^{\frac{d+1}{2}} \hat{f}, {\fancyscript{ H}}( \rho ^{\frac{d-1}{2}} \hat{g}) \right\rangle .\nonumber \\ \end{aligned}$$

(3)

The constant $C(d)$ is explicit, see below.

This immediately implies that for $d \equiv 2 \mod 4$, there can be no exterior energy estimate for the initial value problem with data $(f,0)$, whereas there is such an estimate for data of the form $(0,g)$.

Corollary 2

Let $d\ge 2$ be even. Let $\Box u =0,\,u(0)=f\in \dot{H}^1(\mathbb R ^d)$ be radial, $\partial _t u(0)=0$. Then for all $t\ge 0$, and provided $d\equiv 0 \mod 4$,

$$\begin{aligned} \Vert \nabla _{t,x} S(t)(f,0) \Vert _{L^2(r\ge t)}^{2} \ge c(d)\, \Vert \nabla f \Vert _{L^2}^{2} \end{aligned}$$

(4)

where $c(d)>0$ is an absolute constant that only depends on the dimension. If $d\equiv 2 \mod 4$ then there can be no estimate of the form (4) for all $t\ge 0$: more precisely, there exists as sequence $f_n \in \dot{H}^1(\mathbb {R} ^d)$ such that

$$\begin{aligned} \lim _{t \rightarrow \pm \infty } \Vert \nabla _{x,t} S(t) (f_n,0) \Vert _{L^2(r \ge t)}^2 = o(\Vert \nabla f_n \Vert _{L^2}^2) \quad \hbox {as} \quad n \rightarrow +\infty . \end{aligned}$$

For the dual initial value problem $\Box u =0,\,u(0)=0,\,\partial _t u(0)=g\in L^2(\mathbb R ^d)$ radial, one has for all $t \ge 0$

$$\begin{aligned} \Vert \nabla _{t,x} S(t)(0,g) \Vert _{L^2(r\ge t)}^{2} \ge c(d)\, \Vert g\Vert _2^{2} \end{aligned}$$

if $d\equiv 0 \mod 4$, whereas it fails (in the same sense as above) if $d\equiv 2 \mod 4$.

This is in sharp contrast with the asymptotics for odd dimensions:

Theorem 3

Let $d$ be odd, $(f,g)\in (\dot{H}^1\times L^2)(\mathbb R ^d)$ be radial, and denote by $\hat{f},\hat{g}$ their Fourier transforms in $\mathbb R ^d$. Then for some constant $C(d)>0$ one has for the solution $u$ of (1)

$$\begin{aligned}&\lim _{t\rightarrow \pm \infty } C(d) \Vert \nabla _{t,x} S(t)(f,g)(x) \Vert _{L^2(|x| \ge |t|)}^2 =\frac{\pi }{2} \int (\rho ^2|\hat{f}(\rho )|^2+|\hat{g}(\rho )|^2)\rho ^{d-1}\, d\rho \nonumber \\&\quad \pm \left( (-1)^{\frac{d-1}{2}} \mathrm Re \left\langle H\left( \rho ^{\frac{d+1}{2}} \hat{f}\right) , \rho ^{\frac{d-1}{2}} \hat{g} \right\rangle + \mathrm Re \left\langle \rho ^{\frac{d+1}{2}} \hat{f}, {\fancyscript{H}} \left( \rho ^{\frac{d-1}{2}}\hat{g}\right) \right\rangle \right) . \end{aligned}$$

(5)

From this one immediately deduces (2) up to constants. We prove Theorems 1, 3 in Sect. 2. The failure of (2) presents a serious obstruction for the extension of the nonlinear machinery developed in [5, 6] to even dimensions. However, see [3, 4] for an application of the exterior energy estimate in four dimensions restricted to data $(f,0)$ in the context of equivariant wave maps.

In order to salvage some aspect of (2) in even dimensions, we show in Sect. 3 that at least a delayed exterior energy estimate holds. This is natural in view of two facts:

• energy equipartition,

• at least one of the Cauchy data $(f,0)$ or $(0,g)$ is favorable in each even dimension.

The equipartition property here refers to the fact that after some time, which of course depends on the solution, the energy will split more or less evenly between $\nabla u$ and $\partial _t u$. “Delayed” refers to lifting the forward (say) light-cone upwards by a certain amount. Equivalently, it means calculating the energy over $|x|\ge t-T$ instead of $|x|\ge t$ for some $T>0$. Figure 1 shows the distinction between an exterior region both without and with a time delay.

Fig. 1

Exterior region without and with time delay

The choice of this $T$ is a delicate matter and depends on the data $(f,g)$. The following proposition expresses our main quantitative energy evacuation result. In odd dimensions, results of this nature are obtained via the sharp Huygens principle and are simpler to obtain. The novelty here lies again with even dimensions.

Proposition 4

For all $\varepsilon >0$ and for all $(f,g) \in ( \dot{H}^1 \times L^2)(\mathbb R ^d)$ radial there exists $T=T(\varepsilon ,f,g,d)>0$ such that

$$\begin{aligned} \Vert \nabla _{t,x} S(t)(f,g) \Vert _{L^2(|x| \le t - T)}^2 \le \varepsilon \Vert (f,g) \Vert _{\dot{H}^1 \times L^2}^2 \end{aligned}$$

(6)

for all $t\ge T$.

In combination with finite propagation speed, Proposition 4 implies the following result on the concentration of energy near the light-cone. Such statements are well-known in odd dimensions, see [5, Lemma 4.1] for the three-dimensional version.

Theorem 5

Let $(f,g) \in (\dot{H}^1 \times L^2)(\mathbb R ^{d})$ be radial. Then we have the following vanishing of the energy away from the forward light-cone $\{|x|=t\ge 0\}$:

$$\begin{aligned} \lim _{T \rightarrow +\infty } \limsup _{t \rightarrow +\infty } \Vert \nabla _{t,x} S(t)(f,g) \Vert _{L^2(||x|-t| \ge T )} =0. \end{aligned}$$

Finally, in Sect. 4 we present various results connected with the profile decomposition of Bahouri–Gérard [2], in particular the Pythagorean expansion of the linear energy with sharp cut-offs (Corollary 8). These are even-dimensional versions of tools that are of essential importance to the nonlinear theory developed in [5, 6].

Although these results are technical in nature we include them here since the methods needed to establish them in even dimensions are similar to those used earlier in the paper. The key new point is a bilinear convergence result, see Lemma 6, which follows from some representation formulas and computations analogous to those of Proposition 4.

See our followup work [3, 4] with Andrew Lawrie for concrete applications of these results.

2 Asymptotic representation of the exterior energy

Before going into the detailed computations, let us first give an outline of the proofs of Theorems 1, 3 and Proposition 4. Denote by $u$ the solution to the linear wave equation (1) with initial data $(f,g)$ at time 0:

$$\begin{aligned} u(t) = S(t)(f,g). \end{aligned}$$

The starting point is a representation formula of $u(t)$ through Bessel functions. More precisely $u(t)$ is given by

$$\begin{aligned} u(t) = \cos (t|\nabla |)f+\frac{\sin (t|\nabla |)}{|\nabla |} g. \end{aligned}$$

Let $\hat{f},\hat{g}$ be the Fourier transforms in $\mathbb R ^d$:

$$\begin{aligned} \hat{f}(\xi )=\int _\mathbb{R ^d} e^{-ix\cdot \xi } f(x) \, dx,\qquad f(x)=(2\pi )^{-d} \int _\mathbb{R ^d} e^{ix\cdot \xi } \hat{f}(\xi )\, d\xi . \end{aligned}$$

For radial functions, $\hat{f}$ is again radial. Recall that

$$\begin{aligned} \widehat{\sigma _\mathbb{{S} ^{d-1} }} (\xi ) = (2\pi )^{\frac{d}{2}} |\xi |^{-\nu } J_\nu (|\xi |),\quad \nu : = \frac{d-2}{2}\ge 0, \end{aligned}$$

where $J_\nu$ is the Bessel function of the first type of order $\nu$. It is characterized as being the solution of

$$\begin{aligned} x^2 J_{\nu }''(x) + x J_\nu '(x) + (x^2 - \nu ^2) J_\nu (x) =0 \end{aligned}$$

(7)

which is regular at $x=0$ (unique up to a multiplicative constant). The inversion formula takes the form

$$\begin{aligned} f(r) = (2\pi )^{-\frac{d}{2}} \int _0^\infty \hat{f}(\rho ) J_\nu (r\rho ) (r\rho )^{-\nu } \rho ^{d-1}\, d\rho . \end{aligned}$$

The Plancherel identity takes the form $\Vert \hat{f}\Vert _2^2=(2\pi )^d \Vert f\Vert _2^2$. For the solution $u(t,r)$ this means that

$$\begin{aligned} u(t,r) \!&= \! (2\pi )^{-\frac{d}{2}} \int _0^\infty \left( \cos (t\rho ) \hat{f}(\rho ) \!+\! \frac{\sin (t \rho )}{\rho } \hat{g}(\rho ) \right) J_{\nu }(r \rho ) (r\rho )^{-\nu } \rho ^{d-1} \, d\rho , \qquad \end{aligned}$$

(8)

$$\begin{aligned} \partial _t u(t,r) \!&= \! (2\pi )^{-\frac{d}{2}} \int _0^\infty \left( -\sin (t\rho ) \rho \hat{f}(\rho ) \!+\! \cos (t \rho ) \hat{g}(\rho ) \right) J_{\nu }(r \rho ) (r\rho )^{-\nu } \rho ^{d-1} \, d\rho .\,\,\qquad \end{aligned}$$

(9)

We shall invoke the standard asymptotics for the Bessel functions, see [1],

$$\begin{aligned} J_\nu (x)&= \sqrt{\frac{2}{\pi x} } \left[ (1+ \omega _2(x)) \cos (x-\tau ) + \omega _1(x) \sin (x-\tau ) \right] , \nonumber \\ J_\nu '(x)&= \sqrt{\frac{2}{\pi x} } \left[ \tilde{\omega }_1(x) \cos (x-\tau ) - (1+\tilde{\omega }_2(x)) \sin (x-\tau ) \right] . \end{aligned}$$

(10)

with phase-shift $\tau = (d-1)\frac{\pi }{4}$, and with the bounds (for $n \ge 0,\,x \ge 1$)

$$\begin{aligned} | \omega _1^{(n)}(x)| + |\tilde{\omega }_1^{(n)}(x)| \le C_n\, x^{-1-n}, \quad | \omega _2^{(n)}(x)| + |\tilde{\omega }_2^{(n)}(x)| \le C_n \, x^{-2-n}. \end{aligned}$$

(11)

Using the representations (8) and (9), we have explicit formulas for $\Vert \nabla _{x,t} u \Vert _{L^2(|x| \ge |t|)}$ or the delayed quantity. Then we expand these formulas and using the asymptotics of the Bessel functions; by a delicate inspection of all terms involved, we derive the desired expansion as $t \rightarrow +\infty$. We shall make frequent use of the monotonicity of the energy on outer cones, i.e., the fact that

$$\begin{aligned} \Vert (u,\partial _t u)(t) \Vert _{\dot{H}^1 \times L^2(|x| \ge t-T)} =: \Vert \nabla _{t,x} u(t) \Vert _{L^2(|x| \ge t-T)} \le \Vert \nabla _{t,x} u(s) \Vert _{L^2(|x|\ge s-T)} \end{aligned}$$

for all $T \le s \le t$. Also, given a set $\fancyscript{S}\subset \mathbb {R} ^d$ (possibly depending on time), define the localized energy functional on $\fancyscript{S}$ as

$$\begin{aligned} \Vert (u,v) \Vert _{\dot{H}^1 \times L^2(\fancyscript{S})}^2 := \int _{x \in \fancyscript{S}} \frac{1}{2}(| v|^2 + |\nabla u|^2)(x)\, dx. \end{aligned}$$

We now proceed with the details. The goal in the remaining of this section is to prove the expression (3) for the asymptotic exterior energy in even dimensions, as well as the exterior energy estimate on the region $\{|x|>|t|\}$. We shall also contrast this to the analogous known results in odd dimensions.

Throughout this section, the delay $T=0$.

Proof

(Proof of Theorem 1) First, notice that it suffices to let $f,g$ be Schwartz functions by energy bounds, and we may assume that $\hat{f}(\rho )$ and $\hat{g}(\rho )$ are supported on $0<\rho _{*}<\rho <\rho ^{*}<\infty$. We begin with the kinetic part of the outer energy, viz.

$$\begin{aligned}&(2\pi )^d \frac{1}{2} \Vert \partial _t u(t) \Vert _{L^2(|x| \ge t)}^2 = (2\pi )^d|\mathbb {S} ^{d-1}|\int _{t}^\infty \frac{1}{2} |\partial _t u(t,r)|^2 \, r^{d-1} \, dr \nonumber \\&\quad = (2\pi )^d |\mathbb {S} ^{d-1}| \lim _{\varepsilon \rightarrow 0+} \int _{t}^\infty \frac{1}{2} |\partial _t u(t,r)|^2 r^{d-1} e^{-\varepsilon r} \, dr\nonumber \\&\quad = \lim _{\varepsilon \rightarrow 0+}\int _t^\infty \int \int \frac{1}{2} \left( -\sin (t\rho _1) \rho _1 \hat{f}(\rho _1) + \cos (t \rho _1) \hat{g}(\rho _1) \right) \nonumber \\&\qquad \cdot \left( -\sin (t\rho _2) \rho _2 \overline{ \hat{f}(\rho _2)} + \cos (t \rho _2) \overline{\hat{g}(\rho _2)} \right) \nonumber \\&\qquad \cdot J_{\nu }(r \rho _1) J_{\nu }(r \rho _2) (r^2 \rho _1 \rho _2)^{-\nu } (\rho _1 \rho _2)^{d-1} \, d\rho _1 d\rho _2\, r^{d-1}\, e^{-\varepsilon r} \, dr. \end{aligned}$$

(12)

For each $\varepsilon >0$ fixed, the integrals here are absolutely convergent. In view of the asymptotic expansion of the Bessel functions as stated above, the leading term for (12) is given by the following expression, with $\displaystyle \mu =\nu +\frac{1}{2}=\frac{d-1}{2}$:

$$\begin{aligned}&{\frac{1}{\pi }} \lim _{\varepsilon \rightarrow 0+}\int \limits _t^\infty \int \limits _0^\infty \int \limits _0^\infty \left( -\sin (t\rho _1) \rho _1 \hat{f}(\rho _1) + \cos (t \rho _1) \hat{g}(\rho _1) \right) \nonumber \\&\quad \cdot \left( -\sin (t\rho _2) \rho _2 \overline{\hat{f}(\rho _2)} + \cos (t \rho _2) \overline{ \hat{g}(\rho _2)} \right) \nonumber \\&\quad \cdot \cos (r \rho _1 -\tau ) \cos (r \rho _2-\tau ) (\rho _1 \rho _2)^\mu \, d\rho _1 d\rho _2 \, e^{-\varepsilon r}\, dr. \end{aligned}$$

(13)

We shall show later that this indeed captures the correct asymptotic behavior of the exterior kinetic energy. To be specific, and adding in the contribution by $\partial _r u$ we make the following claim:

$$\begin{aligned}&(2\pi )^d |\mathbb {S} ^{d-1}|^{-1} \left( \Vert \partial _t u(t) \Vert _{L^2(|x| \ge t)}^2 + \Vert \partial _r u(t) \Vert _{L^2(|x| \ge t)}^2\right) \nonumber \\&\quad = \frac{2}{\pi }\lim _{\varepsilon \rightarrow 0+}\int _t^\infty \int \int \left( -\sin (t\rho _1) \rho _1 \hat{f}(\rho _1) + \cos (t \rho _1) \hat{g}(\rho _1) \right) \nonumber \\&\qquad \cdot \left( -\sin (t\rho _2) \rho _2 \overline{ \hat{f}(\rho _2)} + \cos (t \rho _2) \overline{\hat{g}(\rho _2)} \right) \nonumber \\&\qquad \cdot \cos (r \rho _1-\tau ) \cos (r \rho _2-\tau ) (r^2 \rho _1 \rho _2)^{\mu } \, d\rho _1 d\rho _2\, \, e^{-\varepsilon r} \, dr+\nonumber \\&\qquad + \frac{2}{\pi }\lim _{\varepsilon \rightarrow 0+}\int _t^\infty \int \int \left( \cos (t\rho _1) \rho _1 \hat{f}(\rho _1) + \sin (t \rho _1) \hat{g}(\rho _1) \right) \nonumber \\&\qquad \cdot \left( \cos (t\rho _2) \rho _2 \overline{ \hat{f}(\rho _2)} + \sin (t \rho _2) \overline{\hat{g}(\rho _2)} \right) \nonumber \\&\qquad \cdot \sin (r \rho _1-\tau ) \sin (r \rho _2-\tau ) (\rho _1 \rho _2)^{\mu } \, d\rho _1 d\rho _2\, \, e^{-\varepsilon r} \, dr + o(1) \end{aligned}$$

(14)

where $o(1)$ is with respect to $t\rightarrow \pm \infty$. We now proceed to extract (3) from the integrals. In order to carry out the $r$-integration in (14), we use (note $\displaystyle 2\tau \in (\mathbb Z +\frac{1}{2})\pi$)

$$\begin{aligned} \cos (r \rho _1\!-\!\tau ) \cos (r \rho _2-\tau )\!&= \!\frac{1}{2}[\cos (r(\rho _1\!+\!\rho _2)-2\tau ) \!+\! \cos (r(\rho _1\!-\!\rho _2))]\\&= \frac{1}{2}[(-1)^\nu \sin (r(\rho _1\!+\!\rho _2)) \!+\! \cos (r(\rho _1-\rho _2))]\\&\times \sin (r \rho _1\!-\!\tau ) \sin (r \rho _2-\tau )\\&= \frac{1}{2}[-\cos (r(\rho _1\!+\!\rho _2)-2\tau ) \!+\! \cos (r(\rho _1\!-\!\rho _2))] \\&= \frac{1}{2}[-(-1)^\nu \sin (r(\rho _1\!+\!\rho _2)) \!+\! \cos (r(\rho _1\!-\!\rho _2))] \end{aligned}$$

(recall $\displaystyle \nu =\frac{d-2}{2}$). In what follows, we slightly abuse notation by writing $\hat{f'}(\rho ):=\rho \hat{f}(\rho )$.

For any smooth compactly supported functions $\phi , \psi$ on $(0,\infty )$, one has for every $t\in \mathbb R$

$$\begin{aligned}&\lim _{\varepsilon \rightarrow 0+} \int _t^\infty \int \int \cos (r(\rho _1 - \rho _2)) \phi (\rho _1) \psi (\rho _2) \, e^{-\varepsilon r}\,dr \, d\rho _1 d\rho _2\nonumber \\&\quad = \pi \int \phi (\rho ) \psi (\rho ) d\rho - \int \int \frac{\sin (t(\rho _1 - \rho _2))}{\rho _1-\rho _2} \phi (\rho _1) \psi (\rho _2) \, d\rho _1 d\rho _2 \end{aligned}$$

(15)

$$\begin{aligned}&\lim _{\varepsilon \rightarrow 0+} \int _t^\infty \int \int \sin (r(\rho _1 + \rho _2)) \phi (\rho _1) \psi (\rho _2) e^{-\varepsilon r}\,dr\, d\rho _1 d\rho _2 \nonumber \\&= \int \int \frac{\cos (t(\rho _1 + \rho _2))}{\rho _1+\rho _2} \phi (\rho _1) \psi (\rho _2) \, d\rho _1 d\rho _2. \end{aligned}$$

(16)

To prove (15) we note that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0+} \int _t^\infty \cos (a r) e^{-\varepsilon r}\, dr&= \lim _{\varepsilon \rightarrow 0+} \frac{1}{2} \left( -\frac{e^{t(ia-\varepsilon )} }{ia-\varepsilon } + \frac{e^{-t(ia+\varepsilon )} }{ia+\varepsilon }\right) = \pi \delta _{0}(a) - \frac{\sin (ta)}{a} \end{aligned}$$

where the limit is to be taken in the distributional sense. For (16) the argument is essentially the same.

Carrying out the $r$-integration using (15), (16) and ignoring constant prefactors yields:

$$\begin{aligned}&\int \int \Bigg [ \cos (t(\rho _1-\rho _2)) (\hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} \!+\! \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} ) \!-\!\sin (t(\rho _1\!-\!\rho _2)) \cdot \\&\quad \cdot (\hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} \!-\! \hat{g}(\rho _1) \overline{ \hat{f'}(\rho _2)} ) \Bigg ] \left( \pi \delta _0(\rho _1\!-\!\rho _2) \!-\! \frac{\sin (t(\rho _1\!-\!\rho _2))}{\rho _1\!-\!\rho _2} \right) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \\&\quad + (-1)^{\frac{d}{2}} \int \int \Bigg [\cos (t(\rho _1\!+\!\rho _2)) \left( \hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} \!-\! \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} \right) \!+\! \sin (t(\rho _1\!+\!\rho _2))\cdot \\&\quad \cdot (\hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} \!+\! \hat{g}(\rho _1) \overline{ \hat{f'}(\rho _2)} ) \Bigg ] \frac{\cos (t(\rho _1+\rho _2))}{\rho _1\!+\!\rho _2} (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \end{aligned}$$

which further simplifies to (integration extending over $(0,\infty )$)

$$\begin{aligned}&\pi \int _0^\infty (|\hat{f'}(\rho )|^2\!+\!|\hat{g}(\rho )|^2)\rho ^{d-1}\, d\rho \nonumber \\&\quad -\frac{1}{2} \int \int \frac{\sin (2t(\rho _1\!-\!\rho _2)) }{\rho _1\!-\!\rho _2} \left( \hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} \!+ \!\hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} \right) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \nonumber \\&\quad + \int \int \frac{\sin ^2(t(\rho _1\!-\!\rho _2)) }{\rho _1\!-\!\rho _2} (\hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} \!-\! \hat{g}(\rho _1) \overline{ \hat{f'}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \nonumber \\&\quad +(-1)^{\frac{d}{2}} \int \int \frac{\cos ^2(t(\rho _1\!+\!\rho _2)) }{\rho _1+\rho _2} (\hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} \!-\! \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \nonumber \\&\quad + \frac{(-1)^\frac{d}{2}}{2} \int \int \frac{\sin (2t(\rho _1\!+\!\rho _2)) }{\rho _1\!+\!\rho _2} (\hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} \!+\! \hat{g}(\rho _1) \overline{ \hat{f'}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2.\nonumber \\ \end{aligned}$$

(17)

It remains to determine the limit as $t\rightarrow \infty$. First, recall that for any $a>0$ (with $\fancyscript{F}$ denoting the Fourier transform on $\mathbb R$)

$$\begin{aligned} \fancyscript{F}\left[ \frac{\sin (ax)}{x}\right] (\xi )=\pi \chi _{(-a,a)}(\xi ),\quad \fancyscript{F}\left[ \frac{\cos (ax)}{x}\right] (\xi ) = \pi i [-\chi _{(-\infty ,-a)}+\chi _{(a,\infty )}].\nonumber \\ \end{aligned}$$

(18)

The integral on the second line is of the form

$$\begin{aligned} \int _\mathbb{R ^{2}} \frac{\sin (2t(\rho _1-\rho _2))}{\rho _1-\rho _2} \phi (\rho _{1}) \overline{\phi (\rho _{2})}\, d\rho _{1}d\rho _{2} = \frac{1}{2} \int _{-2t}^{2t} |\hat{\phi }(\xi )|^{2}\, d\xi \end{aligned}$$

(19)

with $\phi (\rho )=\hat{f}(\rho )\rho ^{\mu }$. As $t\rightarrow \infty$, this approaches

$$\begin{aligned} \frac{1}{2}\Vert \hat{\phi }\Vert _2^2=\pi \Vert \phi \Vert _2^2. \end{aligned}$$

Hence, the first and second lines in (17) approach

$$\begin{aligned} \frac{\pi }{2}\int _0^\infty (|\hat{f'}(\rho )|^2+|\hat{g}(\rho )|^2)\rho ^{d-1}\, d\rho \qquad \hbox { as }t\rightarrow \infty . \end{aligned}$$

Integration by parts shows that the fifth line vanishes in the limit as $t\rightarrow \infty$ (the data are Schwartz). For the expressions on the third and fourth lines, respectively, we use

$$\begin{aligned} \cos ^2(x)=\frac{1}{2} + \frac{1}{2} \cos (2x),\qquad \sin ^2(x)=\frac{1}{2}-\frac{1}{2} \cos (2x) \end{aligned}$$

and (18), (19) to deduce that in the limit of (17) as $t\rightarrow \infty$ equals

$$\begin{aligned}&\frac{\pi }{2} \int (|\hat{f'}(\rho )|^2+|\hat{g}(\rho )|^2)\rho ^{d-1}\, d\rho \\&\quad +\frac{(-1)^{\frac{d}{2}} }{2} \int \int \frac{1 }{\rho _1+\rho _2} (\hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} - \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \\&\quad + \mathrm Re \int \int \frac{1 }{\rho _1-\rho _2} \hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2, \end{aligned}$$

up to a constant prefactor, and with integration extending over $(0,\infty )$. This is exactly what (3) claims.

It remains to verify the claimed dominance of the leading order terms of the Bessel expansion, see (14). For simplicity, we restrict ourselves to the kinetic energy (12). Subtracting (13) from (12) yields, with $\omega _j$ as in (11),

$$\begin{aligned}&{\frac{2}{\pi }} \lim _{\varepsilon \rightarrow 0+}\int \limits _t^\infty \int \limits _0^\infty \int \limits _0^\infty \left( -\sin (t\rho _1) \rho _1 \hat{f}(\rho _1) + \cos (t \rho _1) \hat{g}(\rho _1) \right) \\&\quad \cdot \left( -\sin (t\rho _2) \rho _2 \overline{\hat{f}(\rho _2)} + \cos (t\rho _2) \overline{ \hat{g}(\rho _2)} \right) \\&\quad \cdot \Bigg [ \left( \omega _2(r\rho _1) + \omega _2(r\rho _2) + \omega _2(r\rho _1) \omega _2(r\rho _2) \right) \cos (r \rho _1 -\tau ) \cos (r \rho _2-\tau ) \\&\quad +\, \omega _1(r\rho _1) (1+\omega _2(r\rho _2)) \sin (r \rho _1 -\tau ) \cos (r \rho _2-\tau ) \\&\quad +\, \omega _1(r\rho _2) (1+\omega _2(r\rho _1)) \sin (r \rho _2 -\tau ) \cos (r \rho _1-\tau ) \\&\quad +\, \omega _1(r\rho _1) \omega _1(r\rho _2) \sin (r\rho _1-\tau )\sin (r\rho _2-\tau )\Bigg ] (\rho _1 \rho _2)^\mu \, d\rho _1 d\rho _2 \, e^{-\varepsilon r}\, dr. \end{aligned}$$

All terms here are treated in a similar fashion. As a representative example, consider for all $\varepsilon >0$ the error term

$$\begin{aligned} E_1(\varepsilon )&:= \int \limits _t^\infty \int \limits _0^\infty \int \limits _0^\infty \sin ((t+T) \rho _1) \sin ((t+T) \rho _2) \cos (r\rho _1 - \tau ) \sin (r\rho _2-\tau ) \omega _1(r\rho _2) \\&\cdot \hat{f}(\rho _1) \overline{\hat{f}(\rho _2)} (\rho _1 \rho _2)^{\mu +1} e^{-\varepsilon r} \, d\rho _1 d\rho _2\,dr, \end{aligned}$$

As before, we write

$$\begin{aligned} \cos (r\rho _1 - \tau ) \sin (r\rho _2-\tau ) = -\frac{1}{2} \left[ (-1)^\nu \cos (r (\rho _1+\rho _2)) + \sin (r(\rho _1- \rho _2)) \right] , \end{aligned}$$

expand the trigonometric functions on the right-hand side into complex exponentials, and perform an integration by parts in the $r$ variable as follows: for any $\sigma \in \mathbb R$ and dropping the subscripts on $\omega ,\rho$ for simplicity, one has

$$\begin{aligned} \int _t^\infty e^{-[\varepsilon \mp i\sigma ]r} \, \omega (r\rho )\, dr = \frac{e^{-[\varepsilon \mp i\sigma ]t}}{\varepsilon \mp i\sigma } \omega (t\rho ) + \int _t^\infty \frac{e^{-[\varepsilon \mp i\sigma ]r}}{\varepsilon \mp i\sigma } \omega '(r\rho )\rho \, dr. \end{aligned}$$

(20)

We apply this with $\sigma =\rho _1+\rho _2$ and $\sigma =\rho _1-\rho _2$ to the fully expanded form of $E_1(\varepsilon )$ as explained above. In both cases one has the uniform bounds

$$\begin{aligned} \sup _{\varepsilon >0}\left\| \int _{-\infty }^\infty \frac{\phi (\rho _2)}{(\rho _1 \pm \rho _2) \pm i\varepsilon } \, d\rho _2\right\| _{L^2(\rho _1)} \le C\Vert \phi \Vert _2. \end{aligned}$$

In order to use this, we distribute the exponential factors as well as all weights over the functions $\hat{f}(\rho _1)$ and $\overline{\hat{f}(\rho _2)}$, respectively. For the first term on the right-hand side of (20) we then obtain an estimate $O(t^{-1})$ from the decay of the weight $\omega$, whereas for the integral in (20) we obtain a $O(r^{-2})$-bound via

$$\begin{aligned} \sup _{\rho >0} |\omega '(r\rho )\rho ^2| \le C\, r^{-2} \end{aligned}$$

which then leads to the final bound

$$\begin{aligned} \int _t^\infty O(r^{-2})\, dr = O(t^{-1}). \end{aligned}$$

The $O$-here are uniform in $\varepsilon >0$. Note that various $\rho$-factors which are introduced by the $\omega$-weights are harmless due to our standing assumption that $0<\rho _*<\rho <\rho ^*$.

All error terms fall under this scheme. In fact, those involving two $\omega$-factors yield a $O(t^{-2})$-estimate. This concludes the proof.

As an immediate corollary one obtains the exterior energy estimate in even dimensions.

Proof

(Proof of Corollary 2) Denote the left-hand side of (4) by $E(t)$. Since $E(t)$ is decreasing, it suffices to consider the limit as $t\rightarrow \infty$. Let us fix dimensions $d=4,8,12,\ldots$ and data $(f,0)$. Then (3) implies that

$$\begin{aligned} \lim _{t\rightarrow \infty } C(d) \Vert \nabla _{t,x} S(t)(f,0) \Vert _{L^2(r\ge t)}^{2} = \frac{\pi }{2} \Vert \nabla f\Vert _2^2 + \frac{1}{2} \left\langle H\left( \rho ^{\frac{d+1}{2}} \hat{f}\right) , \rho ^{\frac{d+1}{2}} \hat{f}\right\rangle .\qquad \quad \end{aligned}$$

(21)

It is well-known that the Hankel transform $H$ is a positive operator on $L^2((0,\infty ))$, since $H=\fancyscript{L}^2$ where $\fancyscript{L}$ is the Laplace transform which is self-adjoint. See for example [7, Section 16.3.3].

The failure of the estimate for $d=2,6,10, \ldots$ and data $(f,0)$ follows just as easily since the operator norm of $H$ on $L^2$ equals $\pi$. More precisely, ${1\!\!1}_{a,b}/\sqrt{ \cdot }$ where $b/a \rightarrow +\infty$ is an explicit extremizing family for $\fancyscript{L}$. Let $f_n = {1\!\!1}_{1,n}/\sqrt{ \cdot }$, then

$$\begin{aligned} \left\langle H( \rho ^{\frac{d+1}{2}} \hat{f}), \rho ^{\frac{d+1}{2}} \hat{f}\right\rangle \!=\! \left\| L \left( \rho ^{\frac{d+1}{2}} \hat{f}_n\right) \right\| _{L^2((0,\infty ))}^2 \!\rightarrow \!\pi \! \left\| \rho ^{\frac{d+1}{2}} \hat{f}_n \right\| _{L^2((0,\infty ))}^2 \quad \hbox {as} \quad n\!\rightarrow \!\infty . \end{aligned}$$

Then with data $(f_n,0)$

$$\begin{aligned}&\lim _{t\rightarrow \pm \infty } C(d) \Vert \nabla _{t,x} S(t)(f_n,0)(x) \Vert ^2_{L^2(|x|\ge |t|)} = o\left( \Vert \rho ^{\frac{d+1}{2}} \hat{f}_n \Vert _{L^2((0,\infty ))}^2\right) \\&\quad =o\left( \Vert f_n\Vert _{\dot{H}^1(\mathbb R ^d)}^2\right) . \end{aligned}$$

The Cauchy problem with data $(0,g)$ is treated analogously.

For the sake of completeness, we contrast the even-dimensional case of Theorem 1 with the odd-dimensional one of Theorem 3. The asymptotic calculations are completely analogous to the ones above, with the dimension entering only (in an essential way) through the phase-shift $\displaystyle \tau =\frac{d-1}{4}\pi$ in the expansions of the Bessel functions for large arguments. The key feature being that $2\tau$ is an integer if $d$ is odd, and a half-integer otherwise.

Proof

(Proof of Theorem 3) We begin by computing the asymptotic form of the exterior energy as in even dimensions, say for $t\ge 0$. With all Fourier transforms being those in $\mathbb R ^d$, one has

$$\begin{aligned}&(2\pi )^d \left( \Vert \partial _t u(t) \Vert _{L^2(|x| \ge t)}^2 + \Vert \partial _r u(t) \Vert _{L^2(|x| \ge t)}^2\right) \nonumber \\&= \frac{2}{\pi }\lim _{\varepsilon \rightarrow 0+}\int _{t}^\infty \int \int \left( -\sin (t\rho _1) \rho _1 \hat{f}(\rho _1) + \cos (t \rho _1) \hat{g}(\rho _1) \right) \nonumber \\&\quad \cdot \left( -\sin (t\rho _2) \rho _2 \overline{ \hat{f}(\rho _2)} + \cos (t \rho _2) \overline{\hat{g}(\rho _2)} \right) \nonumber \\&\quad \cdot \cos (r \rho _1-\tau ) \cos (r \rho _2-\tau ) (r^2 \rho _1 \rho _2)^{\mu } \, d\rho _1 d\rho _2\, \, e^{-\varepsilon r} \, dr+\nonumber \\&\quad + \frac{2}{\pi }\lim _{\varepsilon \rightarrow 0+}\int _{t}^\infty \int \int \left( \cos (t\rho _1) \rho _1 \hat{f}(\rho _1) + \sin (t \rho _1) \hat{g}(\rho _1) \right) \nonumber \\&\quad \cdot \left( \cos (t\rho _2) \rho _2 \overline{ \hat{f}(\rho _2)} + \sin (t \rho _2) \overline{\hat{g}(\rho _2)} \right) \nonumber \\&\quad \cdot \sin (r \rho _1-\tau ) \sin (r \rho _2-\tau ) (\rho _1 \rho _2)^{\mu } \, d\rho _1 d\rho _2\, \, e^{-\varepsilon r} \, dr + o(1) \end{aligned}$$

(22)

where the $o(1)$ is for $t\rightarrow \infty$. Here $\displaystyle \tau =\frac{d-1}{4}\pi ,\,\displaystyle \mu =\frac{d-1}{2}$. Moreover, we used the asymptotic expansions of the Bessel functions (10), and we absorbed all error terms in the $o(1)$, which is justified by the exact same reasoning as in the proof of (3). In order to carry out the $r$-integration, we use (note $2\tau \in \mathbb Z \pi$)

$$\begin{aligned} \cos (r \rho _1-\tau ) \cos (r \rho _2-\tau )&= \frac{1}{2}[\cos (r(\rho _1+\rho _2)-2\tau ) + \cos (r(\rho _1-\rho _2))]\\&= \frac{1}{2}[(-1)^\mu \cos (r(\rho _1+\rho _2)) + \cos (r(\rho _1-\rho _2))] \\ \sin (r \rho _1-\tau ) \sin (r \rho _2-\tau )&= \frac{1}{2}[-\cos (r(\rho _1+\rho _2)-2\tau ) + \cos (r(\rho _1-\rho _2))] \\&= \frac{1}{2}[-(-1)^\mu \cos (r(\rho _1+\rho _2)) + \cos (r(\rho _1-\rho _2))]. \end{aligned}$$

In what follows, we slightly abuse notation by writing $\hat{f'}(\rho ):=\rho \hat{f}(\rho )$. Carrying out the $r$-integration using (15), (16) and applying trigonometric identities yields (ignoring constant prefactors):

$$\begin{aligned}&\int \int \Bigg [ \cos (t(\rho _1\!-\!\rho _2)) (\hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} \!+\! \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} )\!-\!\sin (t(\rho _1-\rho _2)) \\&\quad (\hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} \!-\! \hat{g}(\rho _1) \overline{ \hat{f'}(\rho _2)} ) \Bigg ] \left( \pi \delta _0(\rho _1\!-\!\rho _2) \!-\! \frac{\sin (t(\rho _1\!-\!\rho _2))}{\rho _1\!-\!\rho _2} \right) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \\&\quad + (-1)^\mu \int \int \Bigg [\cos (t(\rho _1\!+\!\rho _2)) (\hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} \!-\! \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} ) \\&\quad + \sin (t(\rho _1\!+\!\rho _2)) (\hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)}\!+\! \hat{g}(\rho _1) \overline{ \hat{f'}(\rho _2)} ) \Bigg ] \frac{\sin (t(\rho _1\!+\!\rho _2))}{\rho _1\!+\!\rho _2} (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \end{aligned}$$

which further simplifies to (integration extending over $(0,\infty )$)

$$\begin{aligned}&\pi \int _0^\infty (|\hat{f'}(\rho )|^2+|\hat{g}(\rho )|^2)\rho ^{d-1}\, d\rho \\&\quad -\frac{1}{2} \int \int \frac{\sin (2t(\rho _1-\rho _2)) }{\rho _1-\rho _2} (\hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} + \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \\&\quad +\frac{1}{2} \int \int \frac{1 - \cos (2t(\rho _1-\rho _2)) }{\rho _1-\rho _2} (\hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} - \hat{g}(\rho _1) \overline{ \hat{f'}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \\&\quad +\frac{(-1)^\mu }{2} \int \int \frac{\sin (2t(\rho _1+\rho _2)) }{\rho _1+\rho _2} (\hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} - \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \\&\quad + \frac{(-1)^\mu }{2} \int \int \frac{1 - \cos (2t(\rho _1\!+\!\rho _2)) }{\rho _1\!+\!\rho _2} (\hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} \!+\! \hat{g}(\rho _1) \overline{ \hat{f'}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2. \end{aligned}$$

We may now pass to the limit as $t\rightarrow \infty$. The terms involving $\sin (2t(\rho _1+\rho _2))$ and $\cos (2t(\rho _1+\rho _2))$ in the fourth and fifth lines, respectively, vanish in the limit as $t\rightarrow \infty$ as can be seen by integration by parts (we may again assume that the data are Schwartz). The asymptotic form of the terms involving $\sin (2t(\rho _1-\rho _2))$ and $\cos (2t(\rho _1-\rho _2))$ in the second and third lines, respectively, follows from (18):

$$\begin{aligned}&\lim _{t\rightarrow \infty } \int \int \frac{\sin (2t(\rho _1-\rho _2)) }{\rho _1-\rho _2} (\hat{f'}(\rho _1) \overline{ \hat{f'}(\rho _2)} + \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2\\&\quad = \pi \int (|\hat{f'}(\rho )|^2+|\hat{g}(\rho )|^2)\rho ^{d-1}\, d\rho \end{aligned}$$

and

$$\begin{aligned} \lim _{t\rightarrow \infty } \int \int \frac{ \cos (2t(\rho _1-\rho _2)) }{\rho _1-\rho _2} (\hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} - \hat{g}(\rho _1) \overline{ \hat{f'}(\rho _2)} ) (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 =0. \end{aligned}$$

In conclusion, we obtain the following asymptotic expression for the left-hand side of (14) for $d$ odd as $t\rightarrow \pm \infty$:

$$\begin{aligned}&\frac{\pi }{2}\int (|\hat{f'}(\rho )|^2+|\hat{g}(\rho )|^2)\rho ^{d-1}\, d\rho \nonumber \\&\quad \pm \, \mathrm Re \int \int \left[ \frac{1 }{\rho _1-\rho _2} + (-1)^\mu \frac{1 }{\rho _1+\rho _2} \right] \hat{f'}(\rho _1) \overline{\hat{g}(\rho _2)} (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2\qquad \end{aligned}$$

(23)

up to a constant prefactor, and with integration extending over $(0,\infty )$. This is exactly (5).

In order to deduce (2) from (23), one choses the direction of time so as the make the second line of (23) nonnegative.

3 Delayed exterior energy and energy concentration

We now turn to a delayed version of the exterior energy bound. We will rely on the radial Fourier formalism from the proof of Theorem 1 without further mention.

Proof

(Proof of Proposition 4) Denote by $u(t,x) = S(t)(f,g)$ the solution of the wave equation (1) as above. We first remark that by conservation of energy (6) is equivalent to the following:

$$\begin{aligned} \Vert (f,g) \Vert _{\dot{H}^1 \times L^2}^2 - \Vert \nabla _{t,x} u \Vert _{L^2( |x| \ge t -T)}^2 \le \varepsilon \Vert (f,g) \Vert _{\dot{H}^1 \times L^2}^2 \end{aligned}$$

(24)

for all $t\ge T$ where $T=T(\varepsilon ,f,g,d)$. Due to the fact that

$$\begin{aligned} t\mapsto \Vert \nabla _{t,x} u(t)\Vert _{L^2(|x| \ge t -T)} \end{aligned}$$

is monotone decreasing, we see that (24) is a consequence of the following bound

$$\begin{aligned} \limsup _{t \rightarrow +\infty } \left[ \Vert (f,g) \Vert _{\dot{H}^1 \times L^2}^2 - \Vert \nabla _{t,x} u \Vert _{L^2( |x| \ge t -T)}^2 \right] \le \varepsilon \Vert (f,g) \Vert _{\dot{H}^1 \times L^2}^2 \end{aligned}$$

(25)

which we now prove. Moreover, it suffices to let $f,g$ be Schwartz functions by energy bounds, and we may assume that $\hat{f}(\rho )$ and $\hat{g}(\rho )$ are supported on $0<\rho _{*}<\rho <\rho ^{*}<\infty$. We begin with the kinetic part of the outer energy, viz.

$$\begin{aligned}&(2\pi )^d \frac{1}{2} \Vert \partial _t u(t+T) \Vert _{L^2(|x| \ge t)}^2 = (2\pi )^d|\mathbb {S} ^{d-1}|\int _{t}^\infty \frac{1}{2} |\partial _t u(t+T,r)|^2 \, r^{d-1} \, dr \nonumber \\&\quad = (2\pi )^d |\mathbb {S} ^{d-1}| \lim _{\varepsilon \rightarrow 0+} \int _{t}^\infty \frac{1}{2} |\partial _t u(t+T,r)|^2 r^{d-1} e^{-\varepsilon r} \, dr \nonumber \\&\quad = \lim _{\varepsilon \rightarrow 0+}\int _t^\infty \int \int \frac{1}{2} \left( -\sin ((t+T)\rho _1) \rho _1 \hat{f}(\rho _1) + \cos ((t+T) \rho _1) \hat{g}(\rho _1) \right) \nonumber \\&\qquad \cdot \left( -\sin ((t+T)\rho _2) \rho _2 \overline{ \hat{f}(\rho _2)} + \cos ((t+T) \rho _2) \overline{\hat{g}(\rho _2)} \right) \nonumber \\&\qquad \cdot J_{\nu }(r \rho _1) J_{\nu }(r \rho _2) (r^2 \rho _1 \rho _2)^{-\nu } (\rho _1 \rho _2)^{d-1} \, d\rho _1 d\rho _2\, r^{d-1}\, e^{-\varepsilon r} \, dr. \end{aligned}$$

(26)

For each $\varepsilon >0$ fixed, the integrals here are absolutely convergent. In view of the asymptotic expansion of the Bessel functions (10), the leading term for (12) is given by the following expression, where $\mu =\nu +\frac{1}{2}=\frac{d-1}{2}$:

$$\begin{aligned} I(T,t)&:= {\frac{1}{\pi }} \lim _{\varepsilon \rightarrow 0+}\int \limits _t^\infty \int \limits _0^\infty \int \limits _0^\infty \left( -\sin ((t+T)\rho _1) \rho _1 \hat{f}(\rho _1) + \cos ((t+T) \rho _1) \hat{g}(\rho _1) \right) \nonumber \\&\cdot \left( -\sin ((t+T)\rho _2) \rho _2 \overline{\hat{f}(\rho _2)} + \cos ((t+T) \rho _2) \overline{ \hat{g}(\rho _2)} \right) \nonumber \\&\cdot \cos (r \rho _1 -\tau ) \cos (r \rho _2-\tau ) (\rho _1 \rho _2)^\mu \, d\rho _1 d\rho _2 \, e^{-\varepsilon r}\, dr. \end{aligned}$$

(27)

We now proceed to estimate $I(T,t)$, and then show later that the higher order corrections to the Bessel asymptotics contribute terms that vanish as $t\rightarrow \infty$. To be more precise, we shall show at the end of the proof that

$$\begin{aligned}&\forall \, T,t \ge 0, \quad \frac{(2\pi )^d}{2 } \Vert \partial _t u(t+T) \Vert _{L^2(|x| \ge t)}^2 = I(T,t) + O(t^{-1}) \hbox { as }t\rightarrow \infty .\quad \end{aligned}$$

(28)

First, we expand $I$ as follows: With $\mu :=\nu + \frac{1}{2}=\frac{d-1}{2}$,

$$\begin{aligned} I(T,t) \!&= \! {\frac{1}{2\pi }} \lim _{\varepsilon \rightarrow 0+} \int \limits _t^\infty \int \limits _0^\infty \int \limits _0^\infty \Bigg ( \sin ((t\!+\!T)\rho _1)\sin ((t\!+\!T)\rho _2) \rho _1 \hat{f}(\rho _1) \rho _2 \overline{\hat{f}(\rho _2)} \nonumber \\&- \sin ((t\!+\!T)\rho _1) \cos ((t\!+\!T)\rho _2) \rho _1 \hat{f}(\rho _1) \overline{\hat{g}(\rho _2) } \nonumber \\&- \sin ((t\!+\!T)\rho _2) \cos ((t\!+\!T)\rho _1) \rho _2 \overline{\hat{f}(\rho _2)} \hat{g}(\rho _1) \nonumber \\&+ \cos ((t\!+\!T)\rho _1) \cos ((t\!+\!T)\rho _2) \hat{g}(\rho _1) \overline{\hat{g}(\rho _2) } \Bigg ) \nonumber \\&\cdot \Bigg [ (-1)^\nu \sin (r (\rho _1 \!+\! \rho _2)) \!+\! \cos (r (\rho _1 \!- \!\rho _2)) \Bigg ] (\rho _1 \rho _2)^{\mu }\, d\rho _1 d\rho _2\, e^{-\varepsilon r}\, dr.\qquad \qquad \end{aligned}$$

(29)

Inserting (15) and (16) into (29) yields

$$\begin{aligned} I(T,t) = I_1(t+T) + {(-1)^\nu } I_2(T,t) - I_3(T,t), \end{aligned}$$

where

$$\begin{aligned} I_1(s)&= \frac{1}{2} \mathrm Re \int \limits _0^\infty \left( \sin ^2 (s\rho ) \rho ^2 |\hat{f}(\rho )|^2 - 2 \sin (2s\rho ) \rho \hat{f}(\rho ) \overline{\hat{g}(\rho )}\right. \\&\left. + \cos ^2 (s\rho ) |\hat{g}(\rho )|^2 \right) \, \rho ^{d-1} d\rho , \\ I_2(T,t) \!&= \! \frac{1}{4\pi } \int \limits _0^\infty \!\int \limits _0^\infty \!\Bigg (\! \left( \cos ((t\!+\!T)(\rho _1\!-\!\rho _2))\! -\! \cos ((t\!+\!T)(\rho _1\!+\!\rho _2)) \right) \rho _1 \hat{f}(\rho _1) \rho _2 \overline{ \hat{f}(\rho _2) }\\&- \left( \sin ( (t\!+\!T) (\rho _1\!+\!\rho _2) ) \!+\! \sin ((t\!+\!T)(\rho _1\!-\!\rho _2))\right) \rho _1 \hat{f}(\rho _1) \overline{ \hat{g}(\rho _2) } \\&- \left( \sin ( (t\!+\!T) (\rho _1\!+\!\rho _2) ) \!-\! \sin ((t\!+\!T)(\rho _1\!-\!\rho _2))\right) \hat{g}(\rho _1) \rho _2 \overline{ \hat{f}(\rho _2) } \\&+ \left( \cos ((t+T)(\rho _1-\rho _2)) + \cos ((t+T)(\rho _1+\rho _2)) \right) \hat{g}(\rho _1) \overline{ \hat{g}(\rho _2) } \Bigg ) \\&\cdot \frac{\cos (t (\rho _1 + \rho _2))}{\rho _1+\rho _2} (\rho _1 \rho _2)^{\mu } d\rho _1 d\rho _2, \\ I_3(T,t) \!&= \! \frac{1}{4\pi } \!\int \limits _0^\infty \!\int \limits _0^\infty \!\Bigg (\! \left( \cos ((t\!+\!T)(\rho _1\!-\!\rho _2)) \!-\! \cos ((t\!+\!T)(\rho _1\!+\!\rho _2)) \right) \rho _1 \hat{f}(\rho _1) \rho _2 \overline{ \hat{f}(\rho _2) }\\&- \left( \sin ( (t+T) (\rho _1+\rho _2) ) + \sin ((t+T)(\rho _1-\rho _2))\right) \rho _1 \hat{f}(\rho _1) \overline{ \hat{g}(\rho _2) } \\&- \left( \sin ( (t+T) (\rho _1+\rho _2) ) - \sin ((t+T)(\rho _1-\rho _2)) \right) \hat{g}(\rho _1) \rho _2 \overline{ \hat{f}(\rho _2) } \\&+ \left( \cos ((t+T)(\rho _1-\rho _2)) + \cos ((t+T)(\rho _1+\rho _2)) \right) \hat{g}(\rho _1)\overline{ \hat{g}(\rho _2) } \Bigg ) \\&\cdot \frac{\sin (t (\rho _1 - \rho _2)) }{\rho _1-\rho _2} (\rho _1 \rho _2)^{\mu } d\rho _1 d\rho _2. \end{aligned}$$

Passing to the limit $s\rightarrow \infty$ (by Riemann–Lebesgue or using the Schwartz property of the integrand) yields

$$\begin{aligned} \displaystyle I_1(s)&\longrightarrow \frac{1}{4} \int _0^\infty (\rho ^2 |\hat{f}(\rho )|^2 + |\hat{g}(\rho )|^2) \rho ^{d-1}\, d\rho =: I_1(\infty )\\ I_1(\infty )&= \frac{(2\pi )^d}{4|\mathbb {S} ^{d-1}|} ( \Vert f \Vert ^2_{\dot{H}^1} + \Vert g \Vert _{L^2}^2) \end{aligned}$$

where the second line uses the Plancherel formula $\Vert \hat{g}\Vert _2^2=(2\pi )^d\Vert g\Vert _2^2$ in $L^2(\mathbb R ^d)$. Next, for the terms containing the Hankel-transform kernel $\displaystyle \frac{1}{\rho _{1}+\rho _{2}}$ we claim that the following representation holds:

$$\begin{aligned} I_2(T,t)&= \frac{1}{8\pi } \int \int \frac{\cos (T(\rho _1+\rho _2))}{\rho _1+\rho _2} \left( - \rho _1 \hat{f}(\rho _1) \rho _2 \overline{ \hat{f}(\rho _2) } + \hat{g}(\rho _1) \overline{ \hat{g}(\rho _2) } \right) (\rho _1 \rho _2)^\mu \, d\rho _1 d\rho _2 \nonumber \\&- \frac{1}{4\pi }\mathrm Re \int \int \frac{\sin (T(\rho _1+\rho _2))}{\rho _1+\rho _2} \rho _1 \hat{f}(\rho _1) \overline{ \hat{g}(\rho _2) } (\rho _1 \rho _2)^\mu \, d\rho _1 d\rho _2+ \widetilde{I}_2(T,t),\qquad \end{aligned}$$

(30)

where

$$\begin{aligned} \forall \, T\!,t\!\ge \! 0, \quad |\widetilde{I}_2(T,t)| \!\le \! c(d) (\Vert f \Vert _{\dot{H}^1}^2\!+\! \Vert g \Vert _{L^2}^2), \quad \forall \, T \!\ge \! 0, \quad \widetilde{I}_2(T,t) \!\rightarrow \! 0 \hbox { as } t \!\rightarrow \! + \infty . \end{aligned}$$

Moreover, the convergence as $t\rightarrow \infty$ here holds uniformly in $T\ge 0$.

To verify this claim, notice first that as $\rho _1, \rho _2 \in [\rho _*,\rho ^*]$ for some $\rho _*,\rho ^* >0$, the denominator $\displaystyle \frac{1}{\rho _1+\rho _2}$ does not create any singularity. We simplify the trigonometric terms as follows, denoting by $z$ either $\sin$ or $\cos$ (which can change from one line to the next):

$$\begin{aligned}&z ((t+T)(\rho _1-\rho _2)) \cos (t (\rho _1 + \rho _2)) \\&\quad =\frac{1}{2} \left[ z (2t \rho _1 + T(\rho _1-\rho _2)) + z( -2t \rho _2 + T(\rho _1 - \rho _2))\right] . \end{aligned}$$

Note that there is no complete cancellation of $t$ in this process. Hence for all terms of the type

$$\begin{aligned} I_1(t,T) := \int \int z (2 t\rho _1 + T (\rho _1 - \rho _2)) \frac{h_1(\rho _1) h_2(\rho _2)}{\rho _1 + \rho _2} (\rho _1 \rho _2)^\mu \, d\rho _1 d\rho _2 \end{aligned}$$

(and symmetrically in $\rho _1$ and $\rho _2$), we see that for all $T$ one has $I_1(t,T) = o_t(1)$ as $t \rightarrow +\infty$, and uniformly in $T\ge 0$. The uniformity is established as follows: by the support and smoothness properties of $h_1$,

$$\begin{aligned}&\int \int e^{\pm i \left[ 2 t\rho _1 + T (\rho _1 - \rho _2)\right] } \, \frac{h_1(\rho _1) h_2(\rho _2)}{\rho _1 + \rho _2} (\rho _1 \rho _2)^\mu \, d\rho _1 d\rho _2 \\&\quad = \int \int e^{\pm i (2 t+T)\rho _1} \, \frac{h_1(\rho _1)}{\rho _1 + \rho _2} (\rho _1 \rho _2)^\mu \, d\rho _1\, e^{\mp iT \rho _2 } h_2(\rho _2) \, d\rho _2 \\&\quad = \mp \frac{1}{i(2t+T)} \int \int e^{\pm i (2 t+T)\rho _1} \,\partial _{\rho _1} \left( \frac{h_1(\rho _1)\rho _1^\mu }{\rho _1 + \rho _2}\right) \, d\rho _1\, \rho _2^\mu e^{\mp iT \rho _2 } h_2(\rho _2)\, d\rho _2 \end{aligned}$$

as desired. For the terms with $z((t+T)(\rho _1+\rho _2))$ we use the identity

$$\begin{aligned} \sin ((t\!+\!T)(\rho _1 \!+\! \rho _2)) \cos (t(\rho _1 \!+\! \rho _2)) \!=\! \frac{1}{2} \left[ \sin ((2t\!+\!T)(\rho _1 \!+\! \rho _2)) \!+\! \sin (T(\rho _1\!+\!\rho _2)) \right] . \end{aligned}$$

The first term yields a contribution of $o_t(1)$ as before whence

$$\begin{aligned}&\int \int \sin ((t\!+\!T)(\rho _1\!+\!\rho _2)) \left( \rho _1 \hat{f}(\rho _1) \overline{ \hat{g}(\rho _2) } \!+\! \hat{g}(\rho _1) \rho _2 \overline{ \hat{f}(\rho _2) }\right) \frac{\cos (t (\rho _1 \!+\! \rho _2))}{\rho _1\!+\!\rho _2} \\&\quad \times (\rho _1 \rho _2)^{\mu } d\rho _1 d\rho _2 \!=\! \mathrm Re \int \int \frac{\sin (T(\rho _1\!+\!\rho _2))}{\rho _1\!+\!\rho _2} \rho _1 \hat{f}(\rho _1) \overline{ \hat{g}(\rho _2)} (\rho _1 \rho _2)^{\mu } d\rho _1 d\rho _2 \!+\! o_t(1) \end{aligned}$$

as $t\rightarrow \infty$, uniformly in $T\ge 0$. We also used the symmetry here to reduce to one pair of functions. In the same way,

$$\begin{aligned} \cos ((t\!+\!T)(\rho _1 \!+\! \rho _2)) \cos (t(\rho _1 \!+\! \rho _2)) \!=\! \frac{1}{2} \left[ \cos ((2t\!+\!T)(\rho _1 \!+\! \rho _2)) \!+\! \cos (T(\rho _1\!+\!\rho _2))\right] . \end{aligned}$$

The first term makes a contribution of $o_t(1)$, again uniformly in $T\ge 0$, and thus

$$\begin{aligned}&\int \int \cos ((t+T)(\rho _1+\rho _2)) \left( -\rho _1 \hat{f}(\rho _1) \rho _2 \overline{ \hat{f}(\rho _2) } + \hat{g}(\rho _1) \overline{ \hat{g}(\rho _2) } \right) \\&\quad \times \frac{\cos (t (\rho _1 + \rho _2))}{\rho _1+\rho _2} (\rho _1 \rho _2)^{\mu } \, d\rho _1 d\rho _2 \!=\! \frac{1}{2} \int \int \frac{\cos (T(\rho _1+\rho _2))}{\rho _1+\rho _2} \\&\quad \times \left( -\rho _1 \hat{f}(\rho _1) \rho _2 \overline{ \hat{f}(\rho _2)}+\, \hat{g}(\rho _1) \overline{\hat{g}(\rho _2) }\right) (\rho _1 \rho _2)^{\mu } \, d\rho _1 d\rho _2 + o_t(1) \end{aligned}$$

and we have proved (30).

It remains to deal with the $I_3$-term. Let $\hat{h}_1, \hat{ h}_2$ denote any of the functions

$$\begin{aligned} {1\!\!1}_{[0,+\infty )}(\rho ) \rho ^{\mu +1} \hat{f}(\rho ) \hbox {\ \ or\ \ } {1\!\!1}_{[0,+\infty )}(\rho ) \rho ^{\mu } {\hat{g}}(\rho ). \end{aligned}$$

Here $\hat{h}_j$ are the one-dimensional Fourier transforms. We write the trigonometric factors in exponential form: all the terms are of the type

$$\begin{aligned}&\int \int e^{i(t+T)(\rho _1 \pm \rho _2)} \frac{\sin (t(\rho _1-\rho _2))}{\rho _1-\rho _2} \hat{h}_1(\rho _1) \overline{\hat{h}_2 (\rho _2) } \, d\rho _1 d\rho _2 \nonumber \\&\quad = \frac{1}{2}\int (\widehat{{1\!\!1}_{[-t,t]}} * (e^{i(t+T)\rho _1} \hat{h}_1))(\rho _2) \, \overline{ e^{\mp i (t+T) \rho _2} \hat{h}_2(\rho _2)}\, d\rho _2 \nonumber \\&\quad = \frac{1}{4\pi } \int _{-t}^t h_1(r + (t+T)) \overline{ h_2(r \mp (t+T)) }\, dr \end{aligned}$$

(31)

where we used Plancherel on the last line. Via Cauchy–Schwarz we can bound these terms by

$$\begin{aligned} \Vert h_1 \Vert _{L^2(|x| \ge T)} \Vert h_2 \Vert _{L^2(|x| \ge T)}. \end{aligned}$$

(32)

Due to the distinction between the Fourier transform on the line and in $\mathbb R ^d$ we cannot simply express the previous expression by one involving the energy of $(f,g)$ over $\{|x|>T\}$. However, it is clear that (32) can be made arbitrarily small by taking $T\ge T_*$.

In summary, we arrive at the following preliminary conclusion:

Given $\varepsilon >0$ there exists $T_*=T_*(\varepsilon ,f,g)$ such that the following holds: for any $T \ge T_*$

$$\begin{aligned}&I(T,t) \ge I_1(\infty )(1-\varepsilon ) + \frac{(-1)^\nu }{8\pi } \int \int \frac{\cos (T(\rho _1\!+\!\rho _2))}{\rho _1+\rho _2} \nonumber \\&\quad \times \left( - \rho _1 \hat{f}(\rho _1) \rho _2 \overline{\hat{f}(\rho _2)} \!+\! \hat{g}(\rho _1) \overline{\hat{g}(\rho _2)} \right) (\rho _1 \rho _2)^{\mu }\, d\rho _1 d\rho _2 \nonumber \\&\quad - \frac{(-1)^\nu }{4\pi } \mathrm Re \int \int \frac{\sin (T(\rho _1+\rho _2))}{\rho _1\!+\!\rho _2} \rho _1 \hat{f}(\rho _1) \overline{\hat{g}(\rho _2) } (\rho _1 \rho _2)^{\mu } \, d\rho _1 d\rho _2\!+\! \widetilde{I}(T,t),\nonumber \\ \end{aligned}$$

(33)

where

$$\begin{aligned} \forall \, t\ge 0, \quad |\widetilde{I}(T,t)| \le c_1(d) (\Vert f \Vert _{\dot{H}^1}^2+ \Vert g \Vert _{L^2}^2), \quad \hbox {and} \quad \widetilde{I}(T,t) \rightarrow 0 \hbox { as } t \rightarrow +\infty . \end{aligned}$$

The constants here do not depend on $T$, and the vanishing of $\tilde{I}$ as $t\rightarrow \infty$ holds uniformly in $T\ge 0$.

To proceed we first note that

$$\begin{aligned} \frac{1}{T} \int _0^T \left( \Vert f \Vert _{\dot{H}^1(|x| \ge \tau )}^2 + \Vert g \Vert _{L^2(|x| \ge \tau )}^2 \right) \, d\tau \longrightarrow 0 \end{aligned}$$

(34)

as $T\rightarrow \infty$. The double integrals in (33) will be dealt with by randomizing $T$, in other words, by taking averages in $T$. This process becomes degenerate for small frequencies $\rho _1, \rho _2$. However, by the uncertainty principle (which amounts to an application of Bernstein’s inequality), these small frequencies occur only with small probability and can therefore be ignored.

To be specific, we rely on the following simple fact: let $h \in L^2(\mathbb R ^d)$ be such that $\displaystyle \Vert h \Vert _{L^2(|x| \ge R)} \le \delta \Vert h \Vert _{L^2}$. Then, with $\hat{h}$ being the Fourier transform in $\mathbb R ^d$,

$$\begin{aligned} \Vert \hat{h} \Vert _{L^2(|\xi | \le \rho )} \le c(d) ((R\rho )^{\frac{d}{2}} + \delta ) \Vert h \Vert _{L^2}. \end{aligned}$$

(35)

To prove this property, let $h_1 := h {1\!\!1}_{[|x| \le R]},\,h_2 := h -h_1$. Then $\Vert \hat{h}_2 \Vert _{L^2} \le c(d) \delta \Vert h \Vert _{L^2}$ and

$$\begin{aligned} \Vert \hat{h}_1 \Vert _{L^\infty } \le \Vert h_1 \Vert _{L^1} \le c(d)R^{\frac{d}{2}} \Vert h_1 \Vert _{L^2} \le c(d)R^{\frac{d}{2}} \Vert h \Vert _{L^2}. \end{aligned}$$

Now, by Cauchy–Schwarz,

$$\begin{aligned} \Vert \hat{h}_1 \Vert _{L^2(|\xi | \le \rho )} \le \sqrt{| \{ |\xi | \le \rho \}|}\, \Vert \hat{h}_1 \Vert _{L^\infty } \le c(d) (R\rho )^{\frac{d}{2}} \Vert h \Vert _{L^2}. \end{aligned}$$

As $\hat{h} = \hat{h}_1 + \hat{h}_2$, (35) follows.

We apply (35) to establish the following “randomized estimate” on the double integrals in (33). We formulate it as a general principle:

Given $\delta > 0$ and any $h_1,h_2 \in L^2(\mathbb R ^d)$ radial, there exists $T^*= T^*(\delta ,h_1,h_2)$ such that for all $T \ge T^*$,

$$\begin{aligned} \left| \frac{1}{T} \int _0^T \int \int \frac{e^{i\tau (\rho _1+\rho _2)}}{\rho _1 +\rho _2} \hat{h}_1(\rho _1) \overline{ \hat{h}_2(\rho _2)}\, (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 d\tau \right| \le c(d) \delta ^2 \Vert h_1 \Vert _{L^2} \Vert h_2 \Vert _{L^2}.\nonumber \\ \end{aligned}$$

(36)

With $T, \rho > 0$ to be determined later, we split the integral into two parts:

$$\begin{aligned} I_{\le \rho } (t,T)&:= \frac{1}{T} \int _0^T \int \int _{\rho _1 + \rho _2 \le \rho } \frac{e^{i\tau (\rho _1+\rho _2)}}{\rho _1 +\rho _2} \hat{h}_1(\rho _1) \overline{ \hat{h}_2(\rho _2)}\, (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 d\tau , \\ I_{\ge \rho } (t,T)&:= \frac{1}{T} \int _0^T \int \int _{\rho _1 + \rho _2 \ge \rho } \frac{e^{i\tau (\rho _1+\rho _2)}}{\rho _1 +\rho _2} \hat{h}_1(\rho _1) \overline{ \hat{h}_2(\rho _2)}\, (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 d\tau \end{aligned}$$

where it is understood that $\rho _1,\rho _2>0$. Then with $R$ as in (35)

$$\begin{aligned} |I_{\le \rho } (t,T)|&\le \frac{1}{T} \int _0^T \int \int \frac{|\hat{h}_1(\rho _1)| {1\!\!1}_{[\rho _1 \le \rho ]} |\hat{h}_2(\rho _2)| {1\!\!1}_{[\rho _2 \le \rho ]}}{\rho _1+\rho _2} \, (\rho _1 \rho _2)^{\mu } \, d\rho _1 d\rho _2 d\tau \\&\le \Vert H ( |\hat{h}_1(\rho _1)|\rho _1^\mu {1\!\!1}_{[\rho _1 \le \rho ]}) \Vert _{L^2} \Vert \hat{h}_2(\rho _2)\rho _2^\mu \Vert _{L^2(|\rho _2| \le \rho )} \\&\le c(d)^2 ((R\rho )^{\frac{d}{2}} + \delta )^2 \Vert h_1 \Vert _{L^2} \Vert h_2 \Vert _{L^2}. \end{aligned}$$

where we used $L^2$-boundedness of the Hankel transform $\displaystyle (Hf)(r) := \textstyle \int \nolimits _0^\infty \frac{f(s)}{r+s}\, ds$ and (35) to pass to the final estimate. For the second term, we integrate first in $\tau$

$$\begin{aligned} |I_{\ge \rho }(t,T)|&\le \frac{2}{T} \int \int _{\rho _1 + \rho _2 \ge \rho } \frac{|\hat{h}_1(\rho _1) \hat{h}_2(\rho _2)|}{(\rho _1+\rho _2)^2} \, (\rho _1\rho _2)^\mu \, d\rho _1 d\rho _2 \\&\le \frac{2}{\rho T} \langle H(|\hat{h}_1|\rho _1^\mu ) , |\hat{h}_2|\rho _2^\mu \,\rangle \le \frac{C(d)}{\rho T} \Vert h_1 \Vert _{L^2} \Vert h_2 \Vert _{L^2} \end{aligned}$$

where $\langle \cdot ,\cdot \rangle$ is the $L^2(0,\infty )$-pairing. Taking first $R$ large (depending on $\delta , h_1,h_2$), then $\rho$ small, and finally $T$ large implies (36).

It is now a simple matter to finish the estimation of the principal term. Indeed, fix a small $\varepsilon >0$ (to be determined later) and let $T^*, T_* \ge 0$ be sufficiently large. Then for all $T \ge \max (T^*,T_*)$ and $t \ge 0$, we obtain the following lower bound on (33):

$$\begin{aligned} \frac{1}{T} \int _0^{T} I(\tau ,t) \, d\tau \ge (1 - \varepsilon ) I_1(\infty ) - \frac{1}{T} \int _0^T |\widetilde{I}(\tau ,t)| \, d\tau . \end{aligned}$$

By the asymptotic behavior of $\widetilde{I}(\tau ,t)$ we see that given $\varepsilon >0$ there exists $T_{0}$, depending on $\varepsilon , f ,g$ and $d$, such that

$$\begin{aligned} \limsup _{t\rightarrow \infty } \frac{1}{T} \int _0^{T} I(\tau ,t) \, d\tau \ge (1- \varepsilon ) I_1(\infty ). \end{aligned}$$

Recall that so far we have only dealt with the kinetic part of the energy, i.e., the one given by $\partial _{t}u$. The other contribution coming from $\partial _r u(t,r)$ equals

$$\begin{aligned}&(2\pi )^d |\mathbb {S} ^{d-1}|^{-1} \frac{1}{2} \Vert \partial _r u(t+T) \Vert _{L^2(|x| \ge t)}^2 = (2\pi )^d\int _{t}^\infty \frac{1}{2} |\partial _r u(t+T,r)|^2 \, r^{d-1} \, dr \\&\quad = (2\pi )^d \lim _{\varepsilon \rightarrow 0+} \int _{t}^\infty \frac{1}{2} |\partial _r u(t+T,r)|^2 r^{d-1} e^{-\varepsilon r} \, dr \\&\quad = \lim _{\varepsilon \rightarrow 0+}\int _t^\infty \int \int \frac{1}{2} \left( \cos ((t+T)\rho _1) \rho _1 \hat{f}(\rho _1) + \sin ((t+T) \rho _1) \hat{g}(\rho _1) \right) \\&\qquad \cdot \left( \cos ((t+T)\rho _2) \rho _2 \overline{ \hat{f}(\rho _2)} + \sin ((t+T) \rho _2) \overline{\hat{g}(\rho _2)} \right) \\&\qquad \cdot J'_{\nu }(r \rho _1) J'_{\nu }(r \rho _2) (r^2 \rho _1 \rho _2)^{-\nu } (\rho _1 \rho _2)^{d-1} \, d\rho _1 d\rho _2\, r^{d-1}\, e^{-\varepsilon r} \, dr + o_t(1) \end{aligned}$$

as $t\rightarrow \infty$. The final term here results from the derivatives in $r$ falling on the $r^{-2\nu }$ weight outside of the Bessel functions, see below for the treatment of such error terms. Plugging in the asymptotics from (10), and performing the same type of arguments as before now yields

$$\begin{aligned} \limsup _{t\rightarrow \infty } \frac{1}{T} \int _0^{T} \Vert \nabla _{t,x} u(t) \Vert _{L^2(|x|>t-\tau )}^2 \, d\tau \ge (1 - c_2(d) (\delta ^2+\varepsilon )) \Vert (f,g) \Vert _{\dot{H}^1 \times L^2}^2 \end{aligned}$$

for all $T\ge T_{0}$. We also used the Plancherel identity $\Vert \hat{f}\Vert _2^2=(2\pi )^d \Vert f\Vert _2^2$. By the monotonicity of the exterior energy, we can take $T=T_{0}$ which leads to the desired result.

It remains to verify the dominance of the leading order terms of the Bessel expansion as expressed by (28). This is very similar to the corresponding argument in the proof of Theorem 1. Indeed, subtracting (27) from (26) yields, with $\omega _j$ as in (11),

$$\begin{aligned} I(T,t)&:= {\frac{2}{\pi }} \lim _{\varepsilon \rightarrow 0+}\int \limits _t^\infty \int \limits _0^\infty \int \limits _0^\infty \left( -\sin ((t+T)\rho _1) \rho _1 \hat{f}(\rho _1) + \cos ((t+T) \rho _1) \hat{g}(\rho _1) \right) \\&\cdot \left( -\sin ((t+T)\rho _2) \rho _2 \overline{\hat{f}(\rho _2)} + \cos ((t+T) \rho _2) \overline{ \hat{g}(\rho _2)} \right) \\&\cdot \Bigg [ \left( \omega _2(r\rho _1) + \omega _2(r\rho _2) + \omega _2(r\rho _1) \omega _2(r\rho _2) \right) \cos (r \rho _1 -\tau ) \cos (r \rho _2-\tau ) \\&+ \omega _1(r\rho _1) (1+\omega _2(r\rho _2)) \sin (r \rho _1 -\tau ) \cos (r \rho _2-\tau ) \\&+ \omega _1(r\rho _2) (1+\omega _2(r\rho _1)) \sin (r \rho _2 -\tau ) \cos (r \rho _1-\tau ) \\&+ \omega _1(r\rho _1) \omega _1(r\rho _2) \sin (r\rho _1-\tau )\sin (r\rho _2-\tau )\Bigg ] (\rho _1 \rho _2)^\mu \, d\rho _1 d\rho _2 \, e^{-\varepsilon r}\, dr. \end{aligned}$$

All terms here are treated in a similar fashion. As one example, consider for all $\varepsilon >0$ the error term

$$\begin{aligned} E_1(\varepsilon )&:= \int \limits _t^\infty \int \limits _0^\infty \int \limits _0^\infty \sin ((t+T) \rho _1) \sin ((t+T) \rho _2) \cos (r\rho _1 - \tau ) \sin (r\rho _2-\tau ) \omega _1(r\rho _2) \\&\cdot \hat{f}(\rho _1) \overline{\hat{f}(\rho _2)} (\rho _1 \rho _2)^{\mu +1} e^{-\varepsilon r} \, d\rho _1 d\rho _2\,dr, \end{aligned}$$

As before, we write

$$\begin{aligned} \cos (r\rho _1 - \tau ) \sin (r\rho _2-\tau ) = -\frac{1}{2} \left[ (-1)^\nu \cos (r (\rho _1+\rho _2)) + \sin (r(\rho _1- \rho _2)) \right] , \end{aligned}$$

expand the trigonometric functions on the right-hand side into complex exponentials, and perform an integration by parts in the $r$ variable as in (20). We apply this with $\sigma =\rho _1+\rho _2$ and $\sigma =\rho _1-\rho _2$ to the fully expanded form of $E_1(\varepsilon )$ as explained above. In both cases one has the uniform bounds

$$\begin{aligned} \sup _{\varepsilon >0} \left\| \int _{-\infty }^\infty \frac{\phi (\rho _2)}{(\rho _1 \pm \rho _2) \pm i\varepsilon } \, d\rho _2\right\| _{L^2(\rho _1)} \le C\Vert \phi \Vert _2. \end{aligned}$$

In order to use this, we distribute the exponential factors as well as all weights over the functions $\hat{f}(\rho _1)$ and $\overline{\hat{f}(\rho _2)}$, respectively. For the first term on the right-hand side of (20) we then obtain an estimate $O(t^{-1})$ from the decay of the weight $\omega$, whereas for the integral in (20) we obtain a $O(r^{-2})$-bound via

$$\begin{aligned} \sup _{\rho >0} |\omega '(r\rho )\rho ^2| \le C\, r^{-2} \end{aligned}$$

which then leads to the final bound

$$\begin{aligned} \int _t^\infty O(r^{-2})\, dr = O(t^{-1}). \end{aligned}$$

The $O$-here are uniform in $\varepsilon >0$. Note that various $\rho$-factors which are introduced by the $\omega$-weights are harmless due to our standing assumption that $0<\rho _*<\rho <\rho ^*$.

Proof

(Proof of Theorem 5) This is an immediate consequence of Proposition 4 and the monotonicity of the energy on the region $\{|x|\ge t+T\}$.

4 Concentration compactness decompositions

The section collects some admittedly more technical results which, however, are of crucial importance in the implementation of certain nonlinear arguments in our followup work [3, 4].

4.1 A bilinear convergence property

We begin with the following result is useful when considering energy splitting in a concentration-compactness decomposition. The issue here is to localize such a splitting to the exterior of balls. We show that such a localization does not affect a Pythagoras-type property of the energy. Technically speaking, the main issue here is the inclusion of the cut-off $\{|x|>r_n\}$ or $\{|x|<r_n\}$ in (37). We present this material here since it rests on the exact same considerations as the exterior energy estimates from above.

Lemma 6

Let ${\varvec{ w}}_n=(w_{n,0},w_{n,1})$ be a bounded sequence of radial functions in $\dot{H}^1 \times L^2$. Let $t_n, r_n$ be two sequences $(r_n \ge 0)$. Assume that $\nabla _{x,t} S(-t_n) {\varvec{ w}}_n \rightharpoonup 0$ in $L^2$ as $n \rightarrow \infty$. Then for any ${\varvec{ U}} =(u_0,u_1) \in (\dot{H}^1\times L^2)(\mathbb R ^d)$, one has

$$\begin{aligned}&\int _{|x|> r_n} \nabla _{x,t} S(t_n) {\varvec{U} }\cdot (\nabla _x w_{n,0}, w_{n,1}) \, dx \rightarrow 0 \quad \hbox {as} \quad n \rightarrow +\infty , \end{aligned}$$

(37)

$$\begin{aligned}&\int _{|x| < r_n} \nabla _{x,t} S(t_n) {\varvec{ U}} \cdot (\nabla _x w_{n,0}, w_{n,1}) \, dx \rightarrow 0 \quad \hbox {as} \quad n \rightarrow +\infty . \end{aligned}$$

(38)

Proof

By conservation of the linear energy, one has

$$\begin{aligned}&\int _\mathbb{{R} ^d} \nabla _{x,t} S(t_n) {\varvec{ U}} \cdot (\nabla _x w_{n,0}, w_{n,1}) \, dx \nonumber \\&\quad = \int _\mathbb{{R} ^d} \nabla _{x,t} {\varvec{ U}} \cdot S(-t_n)(\nabla _x w_{n,0}, w_{n,1}) \, dx \rightarrow 0 \quad \hbox {as} \quad n \rightarrow +\infty . \end{aligned}$$

(39)

Hence, (38) and (37) are equivalent.

By unitarity of the evolution we may assume that $u$ is a Schwartz function with Fourier support away from the origin. Also it suffices to show the claim assuming that the sequences

$$\begin{aligned} (t_n)_n, \quad (r_n)_n, \quad (t_n - r_n)_n \quad \hbox {and} \quad (t_n+r_n)_n \quad \hbox {have a limit in } {\overline{\mathbb{{R} }}}. \end{aligned}$$

If $t_n$ has a finite limit, then $S(t_n) {\varvec{ U}}$ converges strongly in $L^2$ and $(\nabla _x w_{n,0}, w_{n,1})$ converges weakly in $L^2$. Now recall the following simple fact: if $f_n \rightharpoonup f$ weakly in $L^2$, and $\alpha _n \rightarrow \alpha \in \overline{\mathbb{{R} }}$, the dominated convergence theorem shows that

$$\begin{aligned} {1\!\!1}_{|x| \ge \alpha _n} f_n \rightharpoonup {1\!\!1}_{(\alpha ,+\infty )} f \quad \hbox {weakly in } L^2. \end{aligned}$$

Applying this to $\alpha _n=r_n$ and $f_n = (\nabla _x w_{n,0}, w_{n,1})$ yields the result in this case.

We now turn to the case when $\lim t_n \in \{ \pm \infty \}$. We have shown above that the sequence $\nabla _{x,t} S(t_n) {\varvec{ U}}$ asymptotically concentrates its $L^2$ mass where $||x|-|t_n|| \le R$. In particular,

$$\begin{aligned} \int _{|x| \le |t_n|/2} |\nabla _{x,t} S(t_n) {\varvec{ U}}|^2 \rightarrow 0. \end{aligned}$$

If $r_n$ is bounded, it then transpires that

$$\begin{aligned} \int _{|x|\le r_n} \nabla _{x,t} S(t_n) {\varvec{ U}} \cdot (\nabla _x w_{n,0}, w_{n,1}) \, dx \rightarrow 0, \end{aligned}$$

and we are done with this case.

It remains to treat the case where both $(t_n)_n$ and $(r_n)_n$ have infinite limits. We proceed as in the proof of Proposition 4, using the Fourier representation and the Bessel functions $J_\nu$ with $\displaystyle \nu =\frac{d-2}{2}$. Retaining only the leading orders in the expansions of these functions the dominant contribution to (37) is given by

$$\begin{aligned}&\int _{r_n}^\infty \int _0^\infty \left( \cos (t_n \rho )\rho \widehat{u_0}(\rho )+\sin (t_n\rho )\widehat{u_1}(\rho )\right) \sin (r\rho -\tau )(r\rho )^{-\nu -\frac{1}{2}}\rho ^{d-1}\,d\rho \\&\quad \int _0^\infty \widehat{w_{n,0}}(\sigma )\sigma \sin (r\sigma -\tau )(r\sigma )^{-\nu -\frac{1}{2}} \sigma ^{d-1}\,d\sigma \, e^{-\varepsilon r} r^{d-1}\, dr \\&\quad + \int _{r_n}^\infty \int _0^\infty \left( -\sin (t_n\rho ) \rho \widehat{u_0}(\rho )+\cos (t_n\rho )\widehat{u_1}(\rho )\right) \cos (r\rho -\tau )(r\rho )^{-\nu -\frac{1}{2}}\rho ^{d-1}\,d\rho \\&\qquad \int _0^\infty \widehat{w_{n,1}}(\sigma )\cos (r\sigma -\tau )(r\sigma )^{-\nu -\frac{1}{2}} \sigma ^{d-1}\,d\sigma \, e^{-\varepsilon r} r^{d-1}\, dr \end{aligned}$$

in the limit $\varepsilon \rightarrow 0^+$. Carrying out the $r$-integration and passing to the limit yields the expression

$$\begin{aligned}&\int _0^\infty \int _0^\infty \left( \cos (t_n \rho )\rho \widehat{u_0}(\rho )+\sin (t_n\rho )\widehat{u_1}(\rho )\right) \rho ^{\frac{d-1}{2}} \, \sigma \widehat{w_{n,0}}(\sigma ) \sigma ^{\frac{d-1}{2}} \nonumber \\&\quad \left( \pi \delta _0(\rho -\sigma ) - \frac{\sin (r_n (\rho -\sigma ))}{\rho -\sigma } - (-1)^\nu \frac{\cos (r_n(\rho +\sigma ))}{\rho +\sigma }\right) \,d\rho d\sigma \nonumber \\&\quad + \int _0^\infty \int _0^\infty \left( - \sin (t_n \rho )\rho \widehat{u_0}(\rho )+\cos (t_n\rho )\widehat{u_1}(\rho )\right) \rho ^{\frac{d-1}{2}} \, \widehat{w_{n,1}}(\sigma ) \sigma ^{\frac{d-1}{2}} \nonumber \\&\quad \left( \pi \delta _0(\rho -\sigma ) - \frac{\sin (r_n(\rho -\sigma ))}{\rho -\sigma } + (-1)^\nu \frac{\cos (r_n(\rho +\sigma ))}{\rho +\sigma }\right) \,d\rho d\sigma . \end{aligned}$$

(40)

The $\delta _0$ make the following contribution to (40):

$$\begin{aligned}&\int _0^\infty \rho \widehat{u_0}(\rho ) \left[ \cos (t_n\rho ) \rho \widehat{w_{n,0}}(\rho ) -\sin (t_n\rho ) \widehat{w_{n,1}}(\rho ) \right] \rho ^{d-1}\, d\rho \\&\quad + \int _0^\infty \widehat{u_1}(\rho ) \left[ \sin (t_n\rho ) \rho \widehat{w_{n,0}}(\rho ) + \cos (t_n\rho ) \widehat{w_{n,1}}(\rho ) \right] \rho ^{d-1}\, d\rho \end{aligned}$$

which tends to $0$ by the assumption on $w_n$. Next, we extract the terms involving the Hilbert transform kernel from (40) (ignoring multiplicative constants):

$$\begin{aligned}&\int _0^\infty \int _0^\infty \Bigg \{ \rho \widehat{u_0}(\rho ) \left[ \cos (t_n\rho ) \sigma \widehat{w_{n,0}}(\sigma ) -\sin (t_n\rho ) \widehat{w_{n,1}}(\sigma ) \right] \nonumber \\&\qquad \qquad + \widehat{u_1}(\rho ) \left[ \sin (t_n\rho ) \sigma \widehat{w_{n,0}}(\sigma ) + \cos (t_n\rho ) \widehat{w_{n,1}}(\sigma ) \right] \Bigg \} \frac{\sin (r_n(\rho -\sigma ))}{\rho -\sigma }\, (\rho \sigma )^{\frac{d-1}{2}}\, d\rho d\sigma .\nonumber \\ \end{aligned}$$

(41)

Using simple trigonometry, the terms involving $u_0$ can be transformed into the following expression:

$$\begin{aligned}&\int _0^\infty \int _0^\infty \rho \widehat{u_0}(\rho ) \Bigg \{\left[ \sin ((t_n + r_n)(\rho -\sigma )) - \sin ((t_n-r_n)(\rho -\sigma ))\right] \cos (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma ) \nonumber \\&\qquad + \left[ \cos ((t_n + r_n)(\rho -\sigma )) - \cos ((r_n-t_n)(\rho -\sigma ))\right] \sin (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma ) \nonumber \\&\qquad - \left[ \cos ((t_n - r_n)(\rho -\sigma )) - \cos ((t_n + r_n)(\rho -\sigma ))\right] \cos (t_n \sigma ) \widehat{w_{n,1}}(\sigma ) \nonumber \\&\qquad - \left[ - \sin ((t_n - r_n)(\rho -\sigma )) + \sin ((t_n+r_n)(\rho -\sigma ))\right] \sin (t_n \sigma ) \widehat{w_{n,1}}(\sigma ) \Bigg \}\nonumber \\&\qquad \times \frac{(\rho \sigma )^{\frac{d-1}{2}}}{\rho -\sigma } \, d\rho d\sigma =\int _0^\infty \int _0^\infty \rho \widehat{u_0}(\rho ) \Bigg \{ \left[ \sin ((t_n+r_n)(\rho -\sigma ))\right. \nonumber \\&\qquad \left. + \sin ((r_n - t_n)(\rho -\sigma ))\right] \left( \cos (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma )-\sin (t_n \sigma ) \widehat{w_{n,1}}(\sigma ) \right) \nonumber \\&\qquad + \left[ \cos ((t_n+r_n)(\rho -\sigma )) - \cos ((r_n-t_n)(\rho -\sigma ))\right] \nonumber \\&\qquad \times \left( \sin (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma ) + \cos (t_n \sigma ) \widehat{w_{n,1}}(\sigma )\right) \Bigg \} \frac{(\rho \sigma )^{\frac{d-1}{2}}}{\rho -\sigma } \, d\rho d\sigma . \end{aligned}$$

(42)

Define, with $\fancyscript{F}_1$ the Fourier transform on $\mathbb R$,

$$\begin{aligned} \tilde{u}_0 := \fancyscript{F}_1^{-1}\left( {1\!\!1}_\mathbb{R ^+} \rho \widehat{u_0}(\rho )\rho ^{\frac{d-1}{2}}\right) \in L^2(\mathbb R ). \end{aligned}$$

Then with some constant $c$,

$$\begin{aligned} \int _0^\infty e^{\pm iB_n(\rho -\sigma )} \frac{\rho \widehat{u_0}(\rho )}{\rho -\sigma }\, \rho ^{\frac{d-1}{2}} \,d\rho = c{\fancyscript{F}}_1(\mathrm{sign}({\cdot }\pm B_n)\tilde{u}_0)(\sigma ). \end{aligned}$$

If $B_n$ has a limit in $\mathbb {R}$ or $\pm \infty$, then this converges strongly in $L^2(\mathbb R )$: in our case, $B_n$ is $t_n+r_n$ or $t_n-r_n$. Thus, (42) can be reduced to the form $\langle v_n, \tilde{v}_n\rangle \rightarrow 0$ where $v_n$ converges strongly in $L^2$ and $\tilde{v}_n\rightharpoonup 0$ weakly in $L^2$ as $n\rightarrow \infty$. Analogously, the terms involving $u_1$ in (41) are reduced to the following expressions:

$$\begin{aligned}&\int _0^\infty \int _0^\infty \widehat{u_1}(\rho ) \Bigg \{\left[ \sin ((t_n+r_n)(\rho -\sigma )) - \sin ((t_n-r_n)(\rho -\sigma ))\right] \\&\quad \left( \sin (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma ) + \cos (t_n \sigma ) \widehat{w_{n,1}}(\sigma ) \right) \\&\quad - \left[ \cos ((t_n+r_n)(\rho -\sigma )) - \cos ((t_n-r_n)(\rho -\sigma ))\right] \\&\quad \left( \cos (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma )-\sin (t_n \sigma ) \widehat{w_{n,1}}(\sigma )\right) \Bigg \} \frac{(\rho \sigma )^{\frac{d-1}{2}}}{\rho -\sigma } \, d\rho d\sigma \end{aligned}$$

which converges to zero by the same reason.

It remains to handle the terms in (40) involving the Hankel kernel $\displaystyle \frac{1}{\rho +\sigma }$. Using the same type of trigonometric identities as above the terms involving the Hankel kernel as well as $u_0$ are transformed into the following ones:

$$\begin{aligned}&\int _0^\infty \int _0^\infty \rho \widehat{u_0}(\rho ) \Bigg \{\left[ \sin ((t_n+r_n)(\rho +\sigma )) + \sin ((t_n-r_n)(\rho +\sigma ))\right] \\&\quad (\sin (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma ) + \cos (t_n \sigma ) \widehat{w_{n,1}}(\sigma )) \\&\quad + \left[ \cos ((t_n+r_n)(\rho +\sigma )) + \cos ((t_n-r_n)(\rho +\sigma ))\right] \\&\quad (\cos (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma )-\sin (t_n \sigma ) \widehat{w_{n,1}}(\sigma ) ) \Bigg \} \frac{(\rho \sigma )^{\frac{d-1}{2}}}{\rho +\sigma } \, d\rho d\sigma . \end{aligned}$$

We proceed as in the case of the Hilbert transform, considering

$$\begin{aligned} \breve{u}_0 := \fancyscript{F}_1^{-1}\left( {1\!\!1}_\mathbb{R ^-} \rho \widehat{u_0}(\rho )\rho ^{\frac{d-1}{2}}\right) \end{aligned}$$

instead of $\tilde{u}_0$, and noticing

$$\begin{aligned} \int _0^\infty e^{\pm iB_n(\rho +\sigma )} \frac{\rho \widehat{u_0}(\rho )}{\rho +\sigma }\, \rho ^{\frac{d-1}{2}} \,d\rho = c{\fancyscript{F}}_1(\mathrm{sign}({\cdot }\mp B_n)\breve{u}_0)(\sigma ). \end{aligned}$$

We argue analogously for the terms involving the Hankel kernel as well as $u_1$, which are of the form

$$\begin{aligned}&\int _0^\infty \int _0^\infty \widehat{u_1}(\rho ) \Bigg \{ \left[ \sin ((t_n+r_n)(\rho +\sigma )) + \sin ((t_n-r_n)(\rho +\sigma ))\right] \\&\quad (\cos (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma ) - \sin (t_n \sigma ) \widehat{w_{n,1}}(\sigma )) \\&\quad - \left[ \cos ((t_n+r_n)(\rho +\sigma )) + \cos ((t_n-r_n)(\rho +\sigma ))\right] \\&\quad (\sin (t_n \sigma ) \sigma \widehat{w_{n,0}}(\sigma )+\cos (t_n \sigma ) \widehat{w_{n,1}}(\sigma ) ) \Bigg \} \frac{(\rho \sigma )^{\frac{d-1}{2}}}{\rho +\sigma } \, d\rho d\sigma . \end{aligned}$$

By inspection, these also vanish in the limit $n\rightarrow \infty$.

It remains to deal with the errors resulting from the lower orders in (10). In contrast to the leading order, no use is going to be made of the weak convergence assumption on $w_n$. Indeed, just by means of $L^2$-estimation and the gain of (at least) one power stemming from the $\omega _j$ and $\tilde{\omega }_j$ factors in (10), one obtains a $O(r_n^{-1})$ bound on all of the contributions of these terms to the left-hand side of (37) (recall our assumption $\rho >\rho _*>0$, and the same for $\sigma$). To be more specific, the error terms are of the form

$$\begin{aligned} \int _{r_n}^\infty \left( U_{n,0}(r) w_{n,1}'(r) + U_{n,1}(r) w_{n,0}'(r) + U_{n,1}(r) w_{n,1}'(r) \right) r^{d-1}\, dr \end{aligned}$$

(43)

where

$$\begin{aligned} U_{n,1}(r)&= \int _0^\infty \left( \cos (t_n\rho )\widehat{u_0}(\rho )\! +\! \frac{\sin (t_n\rho )}{\rho } \widehat{u_1}(\rho )\right) \left( \tilde{\omega }_1(r\rho )\cos (r\rho -\tau ) \right. \\&\left. - \tilde{\omega }_2(r\rho )\sin (r\rho -\tau )\right) (r\rho )^{-\nu -\frac{1}{2}} \rho ^{d}\, d\rho , \\ w_{n,1}(r)&= \int _0^\infty \widehat{w_n}(\sigma ) \left( \tilde{\omega }_1(r\sigma )\cos (r\sigma -\tau ) - \tilde{\omega }_2(r\sigma )\sin (r\sigma -\tau )\right) (r\sigma )^{-\nu -\frac{1}{2}}\sigma ^d\, d\sigma . \end{aligned}$$

Let us consider the first term in (43):

$$\begin{aligned}&\int _{r_n}^\infty U_{n,0}(r) w_{n,1}'(r) r^{d-1}\, dr\nonumber \\&\quad = \lim _{\varepsilon \rightarrow 0+} \int _{t_n+A}^\infty \int _0^\infty \!\left( \!\cos (t_n\rho )\widehat{u_0}(\rho ) \!+\! \frac{\sin (t_n\rho )}{\rho } \widehat{u_1}(\rho )\right) \sin (r\rho \!-\!\tau )(r\rho )^{-\nu \!-\!\frac{1}{2}} \rho ^{d}\, d\rho \nonumber \\&\qquad \int _0^\infty \! \widehat{w_n}(\sigma )\left( \tilde{\omega }_1(r\sigma ) \cos (r\sigma \!-\!\tau ) \!-\!\tilde{\omega }_2(r\sigma )\sin (r\sigma \!-\!\tau )\right) (r\sigma )^{-\nu \!-\!\frac{1}{2}}\sigma ^d\, d\sigma \, e^{-\varepsilon r} r^{d-1}\, dr \nonumber \\ \end{aligned}$$

(44)

In view of (20) the $r$-integral here is of the form

$$\begin{aligned} \int _t^\infty e^{-[\varepsilon \mp i\tau ]r} \, \omega (r\sigma )\, dr = \frac{e^{-[\varepsilon \mp i\tau ]t}}{\varepsilon \mp i\tau } \omega (t\sigma ) + \int _t^\infty \frac{e^{-[\varepsilon \mp i\tau ]r}}{\varepsilon \mp i\tau } \omega '(r\sigma )\sigma \, dr \end{aligned}$$

(45)

for all $t>0, \sigma >0$. Inserting the boundary term on the right-hand side of (45) into (43) yields expressions of the form, for $j=1,2$,

$$\begin{aligned} \int _0^\infty \int _0^\infty \frac{e^{\pm i t_n\rho } e^{-(\varepsilon \pm i (\rho \pm \sigma ))(t_n+A)} }{\varepsilon \pm i (\rho \pm \sigma ) } \tilde{\omega }_j(r_n \sigma ) \widehat{u_0}(\rho )\widehat{w_n}(\sigma ) (\sigma \rho )^{\frac{d+1}{2}}\, d\rho d\sigma \end{aligned}$$

where the signs are chosen independently of each other. By the $L^2$-boundedness of the Hilbert, respectively, Hankel transforms and the fact that $\sigma >\rho _*>0$, we conclude that uniformly in $\varepsilon >0$ this expression is $O(t_n^{-1})$. Similarly, the integral on the right-hand side of (45) yields

$$\begin{aligned} \int _{r_n}^\infty \left[ \int _0^\infty \int _0^\infty \frac{e^{\pm i t_n\rho } e^{-(\varepsilon \pm i (\rho \pm \sigma ))r} }{\varepsilon \pm i (\rho \pm \sigma ) } \tilde{\omega }_j'(r\sigma ) \sigma \, \widehat{u_0}(\rho )\widehat{w_n}(\sigma ) (\sigma \rho )^{\frac{d+1}{2}}\, d\rho d\sigma \right] \, dr. \end{aligned}$$

Again by $L^2$-boundedness the expression in brackets is $O(r^{-2})$ uniformly in $\varepsilon >0$ and $n$. Integrating this in $r>r_n$ then yields $O(r_n^{-1})$ as before. This shows that the entire first term on the right-hand side of (43) is $O(r_n^{-1})$. The second and third terms satisfy the same bound and we are done. $\square$

4.2 Energy partition for profile decompositions

We first recall the notion of a profile decomposition which originates in this form in [2]. It plays a fundamental role in the analysis of nonlinear equations at large energies. See for example [3–6].

We denote $S$ the Strichartz space $L^{\frac{2(d+1)}{d-2}}_{t,x}(\mathbb {R} ^d)$, and its associated norm

$$\begin{aligned} \Vert f \Vert _{S} = \Vert f \Vert _{L_t^{\frac{2(d+1)}{d-2}}(\mathbb {R} , L_x^{\frac{2(d+1)}{d-2}}(\mathbb {R} ^d))}. \end{aligned}$$

Definition 7

We say that a sequence $(u_{0,n},u_{1,n}) \subset \dot{H}^1 \times L^2$ admits a profile decomposition $(U_\mathrm{L }^j, \partial _t U_\mathrm{L }^j)_{j \in \mathbb {N} } \subset \dot{H}^1 \times L^2$ (solutions to the linear wave equation (1)), with parameters $(\lambda _{j,n},t_{j,n})$, and remainder $w_n^J$ (also solutions to the linear wave equation (1)) if there holds

$$\begin{aligned}&{\left\{ \begin{array}{ll} \displaystyle u_{0,n} = \sum _{j=1}^J\frac{1}{\lambda _{j,n}^{d/2-1}} U_\mathrm{L }^j \left( - \frac{t_{j,n}}{\lambda _{j,n}}, \frac{x}{\lambda _{j,n}} \right) + w_{n}^J(0,x), \\ \displaystyle u_{1,n} = \sum _{j=1}^J \frac{1}{\lambda _{j,n}^{d/2}} \partial _t U_\mathrm{L }^j \left( - \frac{t_{j,n}}{\lambda _{j,n}} \frac{x}{\lambda _{j,n}} \right) + \partial _t w_{n}^J(0,x), \end{array}\right. } \end{aligned}$$

(46)

where $\lim _{J \rightarrow +\infty } \limsup _{n \rightarrow +\infty } \Vert w^J_n \Vert _{S} =0,$

and the parameters are pseudo-orthogonal, that is for all $i \ne j$,

$$\begin{aligned} \left| \ln \frac{\lambda _{j,n}}{\lambda _{i,n}} \right| + \frac{|t_{j,n} - t_{i,n}|}{\lambda _{j,n}} \rightarrow +\infty \quad \hbox {as} \quad n \rightarrow + \infty . \end{aligned}$$

The following result is the Pythagorean expansion of the truncated energy.

Corollary 8

Let $\{(u_{0,n},u_{1,n})\}$ be a bounded sequence in $\dot{H}^1 \times L^2$, and assume it admits a profile decomposition (46) with profiles $({\varvec{ U}}_\mathrm{L }^j)_{j \in \mathbb {N} }$, parameters $(\lambda _{j,n},t_{j,n})$, and remainder $w_n^J$. Let $t_n, r_n$ be two sequences. Then we have the Pythagorean expansion:

$$\begin{aligned}&\int _{|x| \ge r_n} \left( | \nabla _x u_{0,n}(x)|^2 + | u_{1,n}(x) |^2 \right) \, dx \\&\quad = \sum _{j=1}^J \int _{|x| \ge r_n} \frac{1}{\lambda _{j,n}^d} \left| \nabla _{x,t} U_\mathrm{L } \left( - \frac{t_{j,n}}{\lambda _{j,n}}, \frac{x}{\lambda _{j,n}} \right) \right| ^2 \, dx + \int _{|x| \ge r_n} | \nabla _{x,t} w_{n}(0,x)|^2 \, dx + o_{n}(1). \end{aligned}$$

Proof

It suffices to prove that the cross terms go to $0$, i.e.

$$\begin{aligned} \forall i&\ne j, \quad \int _{|x| \ge r_n} \frac{1}{\lambda _{i,n}^{d/2}} \nabla _{x,t} U_\mathrm{L }^i \left( - \frac{t_{i,n}}{\lambda _{i,n}}, \frac{x}{\lambda _{i,n}} \right) \frac{1}{\lambda _{j,n}^{d/2}} \nabla _{x,t} U_\mathrm{L }^j \left( - \frac{t_{j,n}}{\lambda _{j,n}}, \frac{x}{\lambda _{j,n}} \right) \, dx \rightarrow 0, \\ \forall j&\le J, \quad \int _{|x| \ge r_n} \frac{1}{\lambda _{j,n}^{d/2}} \nabla _{x,t} U_\mathrm{L }^j \left( - \frac{t_{j,n}}{\lambda _{j,n}}, \frac{x}{\lambda _{j,n}} \right) \nabla _{x,t} w_{n}^J(0,x) \, dx \rightarrow 0. \end{aligned}$$

After scaling, this takes the expression

$$\begin{aligned} \forall i&\ne j, \quad \int _{|x| \ge r_n/\lambda _{i,n}} \nabla _{x,t} U_\mathrm{L }^i \left( - \frac{t_{i,n}}{\lambda _{i,n}}, x \right) \frac{\lambda _{i,n}^{d/2}}{\lambda _{j,n}^{d/2}} \nabla _{x,t} U_\mathrm{L }^j \left( - \frac{t_{j,n}}{\lambda _{j,n}}, \frac{\lambda _{i,n}}{\lambda _{j,n}} x \right) \, dx \rightarrow 0, \\ \forall j&\le J, \quad \int _{|x| \ge r_n/\lambda _{j,n}} \nabla _{x,t} U_\mathrm{L }^j \left( - \frac{t_{j,n}}{\lambda _{j,n}}, x \right) \lambda _{j,n}^{d/2} \nabla _{x,t} w_{n}^J(0,\lambda _{j,n}^{d/2} x) \, dx \rightarrow 0. \end{aligned}$$

In both cases we will use Lemma 6: we have to check the weak convergence.

In the first case, we have

$$\begin{aligned}&\nabla _{x,t} S\left( \frac{t_{i,n}}{\lambda _{i,n}} \right) \left( \frac{\lambda _{i,n}^{d/2-1}}{\lambda _{j,n}^{d/2-1}} U_\mathrm{L }^j \left( - \frac{t_{j,n}}{\lambda _{j,n}}, \frac{\lambda _{i,n}}{\lambda _{j,n}} x \right) , \frac{\lambda _{i,n}^{d/2}}{\lambda _{j,n}^{d/2}} \partial _t U_\mathrm{L }^j \left( - \frac{t_{j,n}}{\lambda _{j,n}}, \frac{\lambda _{i,n}}{\lambda _{j,n}} x \right) \right) \\&\quad = \frac{\lambda _{i,n}^{d/2}}{\lambda _{j,n}^{d/2}} \nabla _{x,t} U_\mathrm{L }^j \left( \frac{t_{i,n} - t_{j,n}}{\lambda _{j,n}}, \frac{\lambda _{i,n}}{\lambda _{j,n}} x \right) . \end{aligned}$$

From pseudo-orthogonality, it is clear that this last expression tends weakly to 0 in $L^2$. Let us focus on the second, then

$$\begin{aligned} \nabla _{x,t} S\left( \frac{t_{j,n}}{\lambda _{j,n}} \right) (\lambda _{j,n}^{d/2-1} w_{n,0}^J(0,\lambda _{j,n} x), \lambda _{j,n}^{d/2} w_{n,1}^J(0,\lambda _{j,n} x)) = \lambda _{j,n}^{d/2} \nabla _{x,t} w_n \left( t_{j,n}, \lambda _{j,n} x \right) . \end{aligned}$$

But by construction of a profile decomposition, for $j \le J$, recall that

$$\begin{aligned} \lambda _{j,n}^{d/2} \nabla _{x,t} w \left( t_{j,n}, \lambda _{j,n} x \right) \rightharpoonup 0 \quad \hbox {weakly in } L^2 \quad \hbox {as} \quad n \rightarrow +\infty . \end{aligned}$$

4.3 Asymptotic vanishing of Strichartz norms

The goal here is to prove a technical statement on the stability of the asymptotic vanishing of global Strichartz norms under spatial cutoffs, see Lemma 11 below. In odd dimensions, this was established in [5, Claim 2.11]. This statement will play an important role in the applications of this paper to wave maps, see [3, 4].

Lemma 9

[5, Lemma 4.1] Let $v$ be a solution to the linear wave equation (1), and $(t_n) \subset \mathbb {R} ,\,(\lambda _n) \subset \mathbb {R} ^*_+$ be two sequences. Define the sequence

$$\begin{aligned} v_n(t,x) = \frac{1}{\lambda _n^{d/2-1}} v \left( \frac{t}{\lambda _n}, \frac{x}{\lambda _n} \right) . \end{aligned}$$

Assume that $\displaystyle \frac{t_n}{\lambda _n} \rightarrow \ell \in {\overline{\mathbb{{R} }}}$. Then

$$\begin{aligned}&\hbox {If } \ell \in \{ \pm \infty \}, \quad \limsup \nolimits _{n \rightarrow \infty } \Vert \nabla _{x,t} v_n(t_n) \Vert ^2_{L^2(||x| - |t_n|| \ge R \lambda _n)} \rightarrow 0 \quad \hbox {as} \quad R \rightarrow +\infty , \\&\hbox {If } \ell \in \mathbb {R} , \quad \limsup \nolimits _{n \rightarrow \infty } \Vert \nabla _{x,t} v_n(t_n) \Vert ^2_{L^2(|\ln (x/\lambda _n)| \ge \ln R)} \rightarrow 0 \quad \hbox {as} \quad R \rightarrow +\infty . \end{aligned}$$

Proof

First consider the case $\ell \in \mathbb {R}$, then notice that

$$\begin{aligned} \Vert \nabla _{x,t} v_n(t_n) \Vert _{L^2( |\ln (|x|/\lambda _n)| \ge \ln R)} = \Vert \nabla _{x,t} v(\ell ) \Vert _{L^2(|\ln |x|| \ge \ln R)} + o_n(1), \end{aligned}$$

from where the result follows. In the case $|\ell | = +\infty$, then

$$\begin{aligned} \Vert \nabla _{x,t} v_n(t_n) \Vert _{L^2(||x| - |t_n|| \ge R \lambda _n)} = \Vert \nabla _{x,t} v(t_n/\lambda _n) \Vert _{L^2(||x| - |t_n/\lambda _n|| \ge R)}, \end{aligned}$$

and the result follows from Theorem 5.

Lemma 10

[5, Claim A.1] Let $(u,\partial _t u)$ and $(w_n, \partial _t w_n)$ be solutions to the linear wave equation (1) bounded in $\dot{H}^1 \times L^2$, and let $(\lambda _n)_n,\,(\mu _n),\,(t_n)_n,\,(s_n)_n$ be sequences of real numbers (with $\lambda _n, \mu _n >0$). Assume that

$$\begin{aligned} \lambda _n^{d/2} \nabla _{x,t} w_n (t_n, \lambda _n x) \rightharpoonup (0,0) \quad \hbox {weakly in} \quad L^2. \end{aligned}$$

(47)

If $\varphi$ is either a radial, compactly supported smooth function such that $\varphi =1$ (or $\varphi \equiv 1$) in a neighbourhood of $0$, we have

$$\begin{aligned}&\int \varphi \left( \frac{ x }{\mu _n} \right) \nabla _{x,t} w_n(s_n,x) \cdot \frac{1}{\lambda _n^{d/2}} \nabla _{x,t} u \left( \frac{s_n - t_n}{\lambda _n}, \frac{x}{\lambda _n} \right) \, dx \rightarrow 0, \end{aligned}$$

(48)

$$\begin{aligned}&\hbox {and} \quad \int (1- \varphi )\left( \frac{x}{\mu _n} \right) \nabla _{x,t} w_n(s_n,x) \cdot \frac{1}{\lambda _n^{d/2}} \nabla _{x,t} u \left( \frac{s_n - t_n}{\lambda _n}, \frac{x}{\lambda _n} \right) \, dx \rightarrow 0.\quad \end{aligned}$$

(49)

as $n\rightarrow \infty$.

Proof

By conservation of the linear energy for solutions to (1), and we have

$$\begin{aligned}&\int \nabla _{x,t} w_n(s_n,x) \cdot \frac{1}{\lambda _n^{d/2}} \nabla _{x,t} u \left( \frac{s_n - t_n}{\lambda _n}, \frac{x}{\lambda _n} \right) \, dx \nonumber \\&\quad = \int \lambda _n^{d/2} \nabla _{x,t} w_n(t_n,x) \cdot \nabla _{x,t} u \left( \frac{s_n-t_n}{\lambda _n}, x \right) \, dx \nonumber \\&\quad = \int \lambda _n^{d/2} \nabla _{x,t} w_n(t_n, \lambda _n x) \cdot \nabla _{x,t} u \left( 0, x \right) \, dx \rightarrow 0. \end{aligned}$$

(50)

where we used weak convergence (47). This settles the case $\varphi \equiv 1$. Also this shows that is suffices to prove (49). For this we will use Lemma 6. Writing $\displaystyle \varphi (z) = - \int {1\!\!1}_{[y \ge z]} \varphi '(y) dy$, we have

$$\begin{aligned}&\int \varphi \left( \frac{ x }{\mu _n} \right) \nabla _{x,t} w_n(s_n,x) \cdot \frac{1}{\lambda _n^{d/2}} \nabla _{x,t} u \left( \frac{s_n - t_n}{\lambda _n}, \frac{x}{\lambda _n} \right) \, dx = -\int _0^\infty \varphi '(y) F_n(y) dy, \\&\hbox {where} \quad F_n(y) = \int {1\!\!1}_{[x \le \mu _n y]} \nabla _{x,t} w_n(s_n,x) \cdot \frac{1}{\lambda _n^{d/2}} \nabla _{x,t} u \left( \frac{s_n - t_n}{\lambda _n}, \frac{x}{\lambda _n} \right) \, dx. \end{aligned}$$

Unscaling, we see that

$$\begin{aligned} F_n(y) = \int {1\!\!1}_{[x \le \mu _n y/\lambda _n]} \lambda _{n}^{d/2} \nabla _{x,t} w_n(s_n,\lambda _n x) \cdot \nabla _{x,t} u \left( \frac{s_n - t_n}{\lambda _n}, x \right) \, dx. \end{aligned}$$

Now we compute

$$\begin{aligned}&\nabla _{x,t} S \left( \frac{t_n-s_n}{\lambda _n} \right) \left( \lambda _{n}^{d/2-1} w_n(s_n,\lambda _n x), \lambda _{n}^{d/2} \partial _t w_n(s_n,\lambda _n x) \right) \nonumber \\&\quad = \lambda _{n}^{d/2} \nabla _{x,t} w_n(t_n, \lambda x) \rightharpoonup 0 \quad \hbox {in } L^2, \end{aligned}$$

(51)

by hypothesis. Hence (37) ensures that for all $y,\,F_n(y) \rightarrow 0$ as $n \rightarrow +\infty$. Furthermore, it is clear that

$$\begin{aligned} |F_n(y)| \le \Vert (u,\partial _t u) \Vert _{\dot{H}^1 \times L^2} \Vert (w_n, \partial _t w_n) \Vert _{\dot{H}^1 \times L^2} \le M. \end{aligned}$$

Hence for all $n,\,| \varphi '(y) F_n(y) | \le M |\varphi '(y)|$. As $\varphi ' \in L^1$, the Theorem of dominated convergence applies and

$$\begin{aligned} \int _0^\infty \varphi '(y) F_n(y) dy \rightarrow 0. \end{aligned}$$

We are now in a position to derive the aforementioned stability result for the asymptotic vanishing of the Strichartz norms.

Lemma 11

[5, Claim 2.11] Let $w_n$ be a sequence of radial solutions to the linear wave equation (1) with bounded energy and such that

$$\begin{aligned} \Vert w_n \Vert _{S} \rightarrow 0 \quad \hbox {as} \quad n \rightarrow + \infty . \end{aligned}$$

Let $(w_{0,n},w_{1,n})$ be the initial data of $w_n,\,\chi \in {\fancyscript{D}}(\mathbb {R} ^d)$ radial and such that $\chi = 1$ around the origin, and $\lambda _n$ be a sequence of positive numbers. Consider the solution $v_n$ to (1) with truncated data

$$\begin{aligned} (v_{0,n}, v_{1,n}) := ( \varphi (|\cdot |/\lambda _n) w_{0,n}, \varphi (|\cdot | / \lambda _n) w_{1,n}), \end{aligned}$$

where $\varphi =\chi$ or $\varphi = 1-\chi$. Then

$$\begin{aligned} \Vert v_n \Vert _{S} \rightarrow 0 \quad \hbox {as} \quad n \rightarrow + \infty . \end{aligned}$$

Proof

It suffice to consider the case $\varphi = \chi$. By scaling invariance, we can assume that $\lambda _n =1$ for all $n$. Notice that convergence to 0 in the Strichartz space $S$ is equivalent to having trivial profile decomposition, more precisely, one has the following:

Let $(u_n,\partial _t u_n)$ be a sequence of solution to (1). Then $\Vert u_n \Vert _{S(\mathbb {R} )} \rightarrow 0$ if and only if for any sequence $(t_n) \subset \mathbb {R} ,\,(\mu _n) \subset (0,+\infty )$,

$$\begin{aligned} \mu _n^{d/2} \nabla _{x,t} u_n (-\mu _n t_n, \mu _n x) \rightharpoonup 0 \quad \hbox {weakly in } L^2. \end{aligned}$$

This a consequence of the construction of a profile decomposition, see [2] for further details.

Hence, let $(t_n) \subset \mathbb {R} ,\,(\mu _n) \subset (0,+\infty )$ be two sequences, and $(u_0,u_1) \in \dot{H}^1 \times L^2$. By density, we can assume that $(u_0,u_1)$ are radial, smooth and compactly support outside 0, say in $\{ x \mid |x| \in [\rho _*,\rho ^*] \}$ for some $\rho _*, \rho ^* >0$. Define $(u,\partial _t u)$ be the solution to (1) with initial data $(u_0,u_1)$. It suffices to prove that

$$\begin{aligned} \int \frac{1}{\mu _n^{d/2}} \nabla _{x,t} v_n \left( - \frac{t_n}{\mu _n}, \frac{x}{\mu _n} \right) \nabla _{x,t} u(0,x) \, dx \rightarrow 0. \end{aligned}$$

Now we compute,

$$\begin{aligned}&\int \frac{1}{\mu _n^{d/2}} \nabla _{x,t} v_n \left( - \frac{t_n}{\mu _n}, \frac{x}{\mu _n} \right) \nabla _{x,t} u(x) \, dx\\&\quad = \int \nabla _{x,t} v_n \left( 0, x \right) \mu _n^{d/2} \nabla _{x,t} u(-t_n, \mu _n x) \, dx \\&\quad = \int \varphi \left( x \right) \nabla _{x} w_n \left( 0, x \right) \mu _n^{d/2} \nabla _{x} u(t_n,\mu _n x) \, dx \\&\qquad + \int \varphi \left( x \right) \partial _t w_n \left( 0, x \right) \mu _n^{d/2} \partial _t u(t_n,\mu _n x) \, dx \\&\qquad + \int \nabla _x \varphi \left( x \right) w_n \left( 0, x \right) \mu _n^{d/2} \nabla _{x} u(t_n,\mu _n x) \, dx. \end{aligned}$$

The claim together with (48), (49) shows that the first two terms of the right-hand side converge to 0. Hence we are left to prove that

$$\begin{aligned} I_n := \int \nabla _x \varphi \left( x \right) w_n \left( 0, x \right) \mu _n^{d/2} \nabla _{x} u(t_n,\mu _n x) \, dx \rightarrow 0. \end{aligned}$$

It suffices to prove this for subsequences, hence we can assume that $t_n,\,\mu _n$ and $t_n/\mu _n$ have a limit in $\overline{\mathbb{{R} }}$. The claim ensures that $w_n(0,x) \rightharpoonup 0$ in $\dot{H}^1$. Recall that $\nabla _x \varphi$ has compact support away from 0: due to Hardy’s inequality, we deduce

$$\begin{aligned} \frac{1}{|x|} w_n(0,x) \rightharpoonup 0 \quad \hbox {in } L^2\hbox {-weak}, \quad \hbox {and then} \quad \nabla _x \varphi \left( x \right) w_n \left( 0, x \right) \rightharpoonup 0 \quad \hbox {in } L^2\hbox {-weak}. \end{aligned}$$

In particular $\Vert w_n(0,x)/|x| \Vert _{L^2}, \Vert \nabla _x \varphi \left( x \right) w_n \left( 0, x \right) \Vert _{L^2}$ are bounded.

First assume that $\displaystyle t_n \rightarrow \tau \in \mathbb {R}$. By Lemma 9, we see that $\displaystyle \mu _n^{d/2} \nabla _{x,t} u \left( t_n, \frac{x}{\mu _n} \right)$ concentrates $L^2$ mass on annuli of the form

$$\begin{aligned} \{ x \mid \mu _n/R \le |x| \le \mu _n R \}. \end{aligned}$$

Hence if $\mu _n \rightarrow +\infty$ or if $\mu _n \rightarrow 0$, as $\nabla _x \varphi$ has compact support away from 0, we see that $I_n \rightarrow 0$.

If $\mu _n \rightarrow \mu \in (0,+\infty )$, then

$$\begin{aligned} \mu _n^{d/2} \nabla _{x} u(t_n,\mu _n x) \rightarrow \mu ^{d/2} \nabla _x u(\tau , \mu x) \quad \hbox {strongly in } L^2. \end{aligned}$$

As $\nabla _x \varphi \left( x \right) w_n \left( 0, x \right) \rightharpoonup 0$ in $L^2$-weak, we deduce that $I_n \rightarrow 0$.

We now turn to the case when $\displaystyle |t_n| \rightarrow +\infty$. Then Lemma 9 shows that

$$\begin{aligned} \limsup _{n \rightarrow +\infty } \Vert \mu _n^{d/2} \nabla _{x,t} u \left( t_n, \mu _n x \right) \Vert _{L^2( | |x| - |t_n|/\mu _n| \ge R/\mu _n)} \rightarrow 0 \quad \hbox {as} \quad R \rightarrow +\infty . \end{aligned}$$

(52)

If $\displaystyle \frac{|t_n|}{\mu _n} \rightarrow + \infty$, then for all $R,\,{{\mathrm{\mathrm{Supp}}}}(\nabla _x \varphi ) \subset \{ | |x| - |t_n|/\mu _n| \ge \mu _n R \}$ when $n$ is large enough: hence we see that $I_n \rightarrow 0$.

Otherwise, $\displaystyle \frac{t_n}{\mu _n} \rightarrow \ell \in \mathbb {R}$. Using (52) and $\mu _n \rightarrow +\infty$, we see that

$$\begin{aligned} \Big \Vert \mu _n^{d/2} \nabla _{x,t} u \left( t_n, \frac{x}{\mu _n} \right) \Big \Vert _{L^2( | |x| - \ell | \ge \mu _n^{-1/2})} \rightarrow 0. \end{aligned}$$

Hence, separating the integral in $I_n$ between the regions $\{ |x| - \ell | \ge \mu _n^{-1/2} \}$ and its complement, and writing $\nabla _x \varphi \left( x \right) = |x| \nabla _x \varphi (x) \frac{1}{|x|}$, we have

$$\begin{aligned} I_n&= \int _{| |x| - \ell | \le \mu _n^{-1/2}} |x| \nabla _x \varphi \left( x \right) \frac{1}{|x|} w_n \left( 0, x \right) \mu _n^{d/2} \nabla _{x} u(t_n,\mu _n x) dx + o(1) \\&= |\ell | \nabla _x \varphi \left( \ell \right) \int _{| |x| - \ell | \le \mu _n^{-1/2}} \frac{1}{|x|} w_n \left( 0, x \right) \mu _n^{d/2} \nabla _{x} u(t_n,\mu _n x) dx + o(1). \end{aligned}$$

(we used the continuity of $|x| \nabla _x \varphi \left( x \right)$ at $\ell$ on the last line, and $1/\sqrt{\mu _n} \rightarrow 0$).

Now as $\displaystyle \frac{1}{|x|} w_n \left( 0, x \right) \rightharpoonup 0$ in $L^2$, we deduce that the last integral converges to 0, and $I_n \rightarrow 0$. This completes the proof.