Soliton Resolution for Critical Co-rotational Wave Maps and Radial Cubic Wave Equation

Abstract

In this paper we prove the soliton resolution conjecture for all times, for all solutions in the energy space, of the co-rotational wave map equation. To our knowledge this is the first such result for all initial data in the energy space for a wave-type equation. We also prove the corresponding results for radial solutions, which remain bounded in the energy norm, of the cubic (energy-critical) nonlinear wave equation in space dimension 4.

Appendices

Appendix A. Radiation Term for the Free Wave Equation

In this appendix we consider the linear inhomogeneous wave equation in any space dimension $N\ge 3$:

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t^2u-\Delta u=f,\quad (t,x)\in {\mathbb {R}}\times {\mathbb {R}}^N\\&{u}_{\restriction t=0}=(u_0,u_1), \end{aligned}\right. \end{aligned}$$

(A.1)

with radial data.

Proposition A.1

Let

$$\begin{aligned} (u_0,u_1)\in {\mathcal {H}}=({\dot{H}}^1\times L^2)_{{{\,\mathrm{rad}\,}}}({\mathbb {R}}^N), \quad f\in L^1({\mathbb {R}},L^2_{{{\,\mathrm{rad}\,}}}({\mathbb {R}}^N)), \end{aligned}$$

and u be the corresponding solution of (A.1). Then there exists $G\in L^2({\mathbb {R}})$ such that

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _0^{+\,\infty } \left| r^{\frac{N-1}{2}} \partial _r u(t,r)-G(r-t)\right| ^2dr&=0 \end{aligned}$$

(A.2)

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _0^{+\,\infty } \left| r^{\frac{N-1}{2}}\partial _t u(t,r)+G(r-t)\right| ^2dr&=0 \end{aligned}$$

(A.3)

If furthermore $(u_0,u_1)\in {\dot{H}}^2\times {\dot{H}}^1$ and $\partial _t f\in L^1L^2$, then

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _0^{+\,\infty } \left| r^{\frac{N-1}{2}} \partial _{r}\partial _t u(t,r)+G'(r-t)\right| ^2dr&=0 \end{aligned}$$

(A.4)

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _0^{+\,\infty } \left| r^{\frac{N-1}{2}}\partial _t^2 u(t,r)-G'(r-t)\right| ^2dr&=0 \end{aligned}$$

(A.5)

Proof

The fact that (A.2) and (A.3) hold in the case $f=0$ is well-known (see [23, Theorem 2.1]) and goes back at least to the work of Friedlander [33].

To prove (A.2) and (A.3) in the general case, one can reduce to the case $f=0$, by recalling that if u is a solution of (A.1) with $f\in L^1({\mathbb {R}},L^2({\mathbb {R}}^N))$, there exists a finite energy solution $v_L$ of the free wave equation $(\partial _t^2-\Delta )v_L=0$ such that

$$\begin{aligned} \lim _{t\rightarrow \infty } \left\| {u}(t)-{v}_L(t)\right\| _{{\mathcal {H}}}=0. \end{aligned}$$

(A.6)

This follows immediately from the Duhamel formulation of (A.1), which yields that (A.6) holds with

$$\begin{aligned} v_L(t)=&\cos (t\sqrt{-\Delta })\left( u_0-\int _0^{+\,\infty } \frac{\sin (s\sqrt{-\Delta })}{\sqrt{-\Delta }}f(s)ds\right) \\&+\frac{\sin (t\sqrt{-\Delta })}{\sqrt{-\Delta }}\left( u_1+\int _0^{+\,\infty }\cos \left( s\sqrt{-\Delta }\right) f(s)ds \right) . \end{aligned}$$

We next assume that $u_0\in {\dot{H}}^2$, $u_1\in {\dot{H}}^1$, $\partial _t f\in L^1({\mathbb {R}},L^2)$ and prove (A.4), (A.5). We first note that $\partial _t u$ is solution of

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t^2\partial _tu-\Delta \partial _tu=\partial _t f, \quad (t,x)\in {\mathbb {R}}\times {\mathbb {R}}^N\\&{u}_{\restriction t=0}=(u_1,\Delta u_0+f(0))\in {\mathcal {H}}, \end{aligned}\right. \end{aligned}$$

(A.7)

(where f(0) is well defined and in $L^2({\mathbb {R}}^N)$ by an elementary trace lemma). By the considerations above, there exists $H\in L^2({\mathbb {R}})$ such that

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _0^{+\,\infty } \left| r^{\frac{N-1}{2}} \partial _r \partial _tu(t,r)-H(r-t)\right| ^2dr&=0 \end{aligned}$$

(A.8)

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _0^{+\,\infty } \left| r^{\frac{N-1}{2}}\partial _t^2 u(t,r)+H(r-t)\right| ^2dr&=0, \end{aligned}$$

(A.9)

and we are reduced to prove that $H=-G'$.

Let $\chi \in C^{\infty }({\mathbb {R}})$ such that $\chi (\eta )=0$ if $\eta \le \frac{1}{2}$ and $\chi (\eta )=1$ if $\eta \ge 1$. Since

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _0^1 |G(r-t)|^2dr=0, \end{aligned}$$

the equality (A.3) implies

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _{{\mathbb {R}}} \left| \chi (r)r^{\frac{N-1}{2}} \partial _tu(t,r)+G(r-t)\right| ^2dr=0, \end{aligned}$$

that is

$$\begin{aligned} \lim _{\tau \rightarrow \infty } \int _{{\mathbb {R}}} \left| (\eta +\tau )^{\frac{N-1}{2}} \chi (\eta +\tau )\partial _tu(\tau ,\eta +\tau )+G(\eta )\right| ^2d\eta =0. \end{aligned}$$

In other words, the family of functions

$$\begin{aligned} \eta \mapsto (\eta +\tau )^{\frac{N-1}{2}}\chi (\eta +\tau )(\partial _t u)(\tau ,\eta +\tau ) \end{aligned}$$

converge to $-G$ in $L^2({\mathbb {R}})$ as $\tau$ goes to infinity. Taking the derivative in $\eta$ and using that

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _{0}^{+\,\infty }|\partial _t u(t,r)|^2r^{N-3}dr=0 \end{aligned}$$

(which follows from the fact that there exists a finite energy solution $w_L$ of $\partial _t^2w_L-\Delta w_L=0$ such that $\lim _{t\rightarrow \infty } \Vert \partial _tu(t)-w_L(t)\Vert _{{\dot{H}}^1}=0,$ and a classical dispersive estimate, see e.g. [23, Theorem 2.1]) we obtain that the family of functions

$$\begin{aligned} \eta \mapsto (\eta +\tau )^{\frac{N-1}{2}}\chi (\eta +\tau )(\partial _t\partial _r u)(\tau ,\eta +\tau ) \end{aligned}$$

converge to $-G'$ in ${\mathcal {D}}'({\mathbb {R}})$ as $\tau \rightarrow \infty$. However by (A.8), we obtain that this family converge to H in $L^2({\mathbb {R}})$ as $\tau \rightarrow \infty$. By uniqueness of the distributional limit, $H=-G'$ concluding the proof. $\quad \square$

Remark A.2

We recall that for all $G\in L^2({\mathbb {R}})$, there exists a unique finite energy solution u of (A.1) such that (A.2) and (A.3) hold: see again [23, Theorem 2.1].

Appendix B. Pseudo-Ortogonality and Channels of Energy

1.1 B.1. Linear Wave Equation

In this subsection, we prove (6.10) and (6.11)

Since $\int _{\rho _n<|x|<\rho _n'}\ldots =\int _{\rho _n<|x|}\ldots -\int _{\rho _n'<|x|}$, it is sufficient to treat the case where $\rho _n'=\infty$ for all n. Rescaling $u_n$, we also assume

$$\begin{aligned} \forall n,\quad \lambda _{j,n}=1. \end{aligned}$$

Pseudo-orthogonality between two profiles. We first prove (6.10). We will use the pseudo-orthogonality (6.1) of the scaling and time translation parameters. We fix $j\ne k$. Let

$$\begin{aligned} \varepsilon _n:=\,&\int _{|x|>\rho _n}\nabla _{t,x}U^j_n(s_n,x)\cdot \nabla _{t,x}U^k_n(s_n,x)dx\\ =\,&\int _{|x|>\rho _n}\nabla _{t,x}U^j\left( s_n-t_{j,n},x\right) \cdot \frac{1}{\lambda _{k,n}^2}\nabla _{t,x}U^k \left( \frac{s_n-t_{k,n}}{\lambda _{k,n}},\frac{x}{\lambda _{k,n}}\right) dx. \end{aligned}$$

Arguing by contradiction, we can extract subsequences and assume that the following limits exist in ${\mathbb {R}}\cup \{\pm \infty \}$.

$$\begin{aligned} \lim _{n\rightarrow \infty }s_n-t_{j,n}=T_j,\quad \lim _{n\rightarrow \infty } \frac{s_n-t_{k,n}}{\lambda _{k,n}}=T_k. \end{aligned}$$

Case 1: $T_j\in {\mathbb {R}}$, $T_k\in {\mathbb {R}}$. In this case we have

$$\begin{aligned} \varepsilon _n=o_n(1)+\int _{|x|>\rho _n} \nabla _{t,x}U^j(T_j,x)\cdot \frac{1}{\lambda _{k,n}^2}\nabla _{t,x} U^k\left( T_k,\frac{x}{\lambda _{k,n}} \right) dx. \end{aligned}$$

Since (6.1) implies $\lim _n\lambda _{k,n}\in \{0,\infty \}$, we obtain right away $\lim _{n}\varepsilon _n=0$.

Case 2. $T_j=+\,\infty$, $T_k\in {\mathbb {R}}$. We note that in this case, there exists $g^j\in L^2({\mathbb {R}})$ such that

$$\begin{aligned} \varepsilon _n&= o_n(1)+\int _{\rho _n}^{\infty } g^j (r-s_n+t_{j,n})\frac{1}{\lambda _{k,n}^2} \partial _tU^k \left( T_k,\frac{r}{\lambda _{k,n}}\right) r^{3/2}dr\nonumber \\&\quad -\int _{\rho _n}^{\infty } g^j(r-s_n+t_{j,n})\frac{1}{\lambda _{k,n}^2} \partial _rU^k\left( T_k,\frac{r}{\lambda _{k,n}}\right) r^{3/2}dr. \end{aligned}$$

(B.1)

Indeed if $\lim _{n\rightarrow \infty } t_{j,n}\in \{\pm \infty \}$, $U^j$ is a solution of the free wave equation (3.1), and (B.1) follows from the asymptotic behaviour for these solutions, recalled in Proposition A.1 above. If $t_{j,n}=0$ for all n, then by the definition of $R_n$ and the assumptions $\rho _n\ge R_n+|s_n|$, there exists a solution $V^j_{L}$ of (3.1) such that

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _{|x|>\rho _n} \left| \nabla _{t,x}U^j(s_n-t_{j,n},x)-\nabla _{t,x}V^j_L (s_n-t_{j,n},x)\right| ^2dx=0, \end{aligned}$$

and the claim follows again from Proposition A.1. Letting $h^k=r^{3/2} \partial _tU^k(T_k,r)\in L^2(0,\infty )$, we see that

$$\begin{aligned}&\int _{\rho _n}^{\infty } g^j(r-s_n+t_{j,n})\frac{1}{\lambda _{k,n}^2} \partial _tU^k\left( T_k,\frac{r}{\lambda _{k,n}}\right) r^{3/2}dr \\&\quad = \int _{\rho _n}^{\infty } g^j(r-s_n+t_{j,n})h^k\left( \frac{r}{\lambda _{k,n}}\right) dr, \end{aligned}$$

which tends to 0 as n tends to infinity because $\lim _ns_n-t_{j,n}=\infty$. Since the term in the second line of (B.1) can be treated in the exact same way, we obtain $\lim _{n\rightarrow \infty }\varepsilon _n=0$.

The proof in the cases $T_j=-\infty$, $T_k\in {\mathbb {R}}$, and $T_j\in {\mathbb {R}}$, $T_k=\pm \infty$ are very close and we omit them.

Case 3. $T_j=+\,\infty$ and $T_k=+\,\infty$. Similarly to Case 2, there exist $g^j \in L^2({\mathbb {R}})$ and $g^k\in L^2({\mathbb {R}})$ such that

$$\begin{aligned} \varepsilon _n=o_n(1)+2\int _{\rho _n}^{\infty } g^j(r-s_n+t_{j,n})\frac{1}{\lambda _{k,n}^2}g^k\left( \frac{r-s_n+t_{k,n}}{\lambda _{k,n}} \right) dr, \end{aligned}$$

and the fact that $\lim _{n\rightarrow \infty }\varepsilon _n=0$ follows easily from the pseudo-orthogonality property (6.1).

The proof in the case $T_j=-\infty$, $T_k=-\infty$ is the same and we omit it.

Case 4. $T_j=+\,\infty$ and $T_k=-\infty$. Using Proposition A.1 as in cases 2 and 3, we obtain that the contributions of the integrals of $\partial _tU^j_n\partial _tU^k_n$ and $\partial _rU^j_n\partial _rU^k_n$ in the definition of $\varepsilon _n$ are opposite, and thus

$$\begin{aligned} \lim _{n\rightarrow \infty }\varepsilon _n=0. \end{aligned}$$

The proof in the case $T_j=-\infty$, $T_k=+\,\infty$ is exactly the same.

Pseudo-orthogonality between a profile and the dispersive remainder.

We next prove (6.11). We assume again that $T_j=\lim _{n\rightarrow \infty } s_n-t_{j,n}$ exists in ${\mathbb {R}}\cup \{\pm \infty \}$.

Case 1. $T_j\in {\mathbb {R}}$. Then

$$\begin{aligned} \varepsilon _n=o_n(1)+\int _{|x|>\rho _n} \nabla _{t,x}U^j(T_j,x)\nabla _{t,x}w^J_n(s_n,x)dx. \end{aligned}$$

If $\lim _{n\rightarrow \infty }\rho _n=+\,\infty$, we deduce immediately that $\lim _{n\rightarrow \infty }\varepsilon _n=0$. If not, we can assume $\lim _{n\rightarrow \infty }\rho _n=\rho _{\infty }\in [0,\infty )$, and we obtain

$$\begin{aligned} \varepsilon _n=o_n(1)+\int _{|x|>\rho _{\infty }} \nabla _{t,x}U^j(T_j,x)\nabla _{t,x}w^J_n(s_n,x)dx. \end{aligned}$$

By (6.10), we can take J arbitrarily large. By the definitions of the profiles, and (6.5) we obtain, for J large enough,

(B.2)

As a consequence, $\lim _n\varepsilon _n=0$.

Case 2. $T_j=+\,\infty$. Arguing as in Case 2 in the proof of (6.10), we obtain that there exists $g^j\in L^2({\mathbb {R}})$ such that

$$\begin{aligned} \varepsilon _n=o_n(1)+\int _{\rho _n}^{\infty } g^j(r+t_{j,n}-s_n)\Big (\partial _tw_n^J(s_n,r) -\partial _rw_n^J(s_n,r)\Big )r^{3/2}dr. \end{aligned}$$

If $\lim _{n}\rho _n+t_{j,n}-s_n=+\,\infty$, we obtain right away that $\varepsilon _n=o_n(1)$.

If $\lim _{n}\rho _n+t_{j,n}-s_n=-\infty$, then

$$\begin{aligned} \int _{0}^{\rho _n}\left| g^j(r+t_{j,n}-s_n)\right| ^2dr\le \int _{-\infty }^{\rho _n+t_{j,n}-s_n} |g(\eta )|^2\underset{n\rightarrow \infty }{\longrightarrow }0, \end{aligned}$$

and thus

$$\begin{aligned} \varepsilon _n=\,&o_n(1)+\int _{0}^{\infty } g^j(r+t_{j,n}-s_n) \Big (\partial _tw_n^J(s_n,r)-\partial _rw_n^J(s_n,r)\Big )r^{3/2}dr\\ =\,&o_n(1)+\int \nabla _{t,x}V^j_L(s_n-t_{j,n})\cdot \nabla _{t,x}w_n^J(s_n,x)dx, \end{aligned}$$

where $V^j_L$ is the radial solution of the linear wave equation such that

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _0^{\infty }|r^{3/2} \partial _tV^j_L(t,r)-g^j(r-t)|^2dr+\int _0^{\infty }|r^{3/2} \partial _rV^j_L(t,r)+g^j(r-t)|^2dr=0. \end{aligned}$$

By conservation of the energy for the free wave equation (3.1),

$$\begin{aligned} \varepsilon _n=o_n(1)+\int \nabla _{t,x}V^j_L(0)\cdot \nabla _{t,x}w_n^J(t_{j,n},x)dx, \end{aligned}$$

and thus by (6.5), $\varepsilon _n=o_n(1)$.

Finally, if $\lim _{n}\rho _n+t_{j,n}-s_n=c\in {\mathbb {R}}$, we see that

$$\begin{aligned}&\varepsilon _n=o_n(1)+\int _{0}^{\infty } g^j_c(r+t_{j,n}-s_n) \Big (\partial _tw_n^J(s_n,r)-\partial _rw_n^J(s_n,r)\Big )r^{3/2}dr\\&\quad =o_n(1)+\int \nabla _{t,x}V^j_L(s_n-t_{j,n})\cdot \nabla _{t,x}w_n^J(s_n,x)dx, \end{aligned}$$

where $g_c(\eta )=1\!\!1_{\eta >c}g(\eta )$, and $V^j_L$ is the radial solution of the linear wave equation such that

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _0^{\infty }|r^{3/2} \partial _tV^j_L(t,r)-g^j_c(r-t)|^2dr+\int _0^{\infty }|r^{3/2} \partial _rV^j_L(t,r)+g^j_c(r-t)|^2dr=0. \end{aligned}$$

The same proof as before yields $\varepsilon _n=o_n(1)$.

Since the case $T_j=-\infty$ can be treated exactly in the same way, the proof of (6.11) is complete.

1.2 B.2. Pythagorean expansion of the energy for wave maps

In this subsection we prove (7.21). In view of Claim 7.2, it is sufficient to prove

$$\begin{aligned}&j\ne k\Longrightarrow \lim _{n\rightarrow \infty }\int _0^{\infty } \partial _{r}\Psi _{0,n}^j\partial _r\Psi _{0,n}^krdr+\int _0^{\infty } \Psi _{1,n}^j\Psi _{1,n}^krdr=0 \end{aligned}$$

(B.3)

$$\begin{aligned}&j\ne k\Longrightarrow \lim _{n\rightarrow \infty }\int _0^{\infty } \left| \sin \Psi _{0,n}^j\sin \Psi _{0,n}^k\right| \frac{dr}{r}=0 \end{aligned}$$

(B.4)

$$\begin{aligned}&j\le J\Longrightarrow \lim _{n\rightarrow \infty }\int _0^{\infty } \partial _{r}\Psi _{0,n}^j\partial _r\omega _{0,n}^Jrdr+\int _0^{\infty } \Psi _{1,n}^j\omega _{1,n}^Jrdr=0\end{aligned}$$

(B.5)

$$\begin{aligned}&j\le J\Longrightarrow \lim _{n\rightarrow \infty }\int _0^{\infty }\left| \sin \Psi _{0,n}^j\sin \omega _{0,n}^J\right| \frac{dr}{r}=0. \end{aligned}$$

(B.6)

Recall that if $j\in {\mathcal {J}}_L$, i.e. $\lim _{n\rightarrow \infty } -t_{j,n}/\lambda _{j,n}\in \{\pm \infty \}$, then

$$\begin{aligned} \left( \Psi ^j_{0,n}(r),\Psi ^j_{1,n}(r)\right) =\left( \Psi ^j_L\left( \frac{-t_{j,n}}{\lambda _{j,n}},\frac{r}{\lambda _{j,n}} \right) ,\frac{1}{\lambda _{j,n}}\partial _t \Psi _L^j\left( \frac{-t_{j,n}}{\lambda _{j,n}},\frac{r}{\lambda _{j,n}} \right) \right) , \end{aligned}$$

where $\Psi ^j_L$ is a solution of the linear equation (7.6) with initial data in ${\mathbf {H}}$. Thus $U^j=\frac{1}{r}\Psi ^j$ is a radial solution of the $1+4$ dimensional wave equation. By standard dispersive estimates for linear wave equations,

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _0^{\infty } \left| \Psi _{0,n}^j(r)\right| ^2\frac{dr}{r}=\lim _{n\rightarrow \infty } \int _0^{\infty }\left| U^j\left( \frac{-t_{j,n}}{\lambda _{j,n}},r \right) \right| ^2rdr=0. \end{aligned}$$

(B.7)

This yields (B.4) when $j\in {\mathcal {J}}_L$ or $k\in {\mathcal {J}}_k$, and (B.6) when $j\in {\mathcal {J}}_L$.

Next, we assume $j\in {\mathcal {J}}_L$ and $k\in {\mathcal {J}}_L$ and see that by (B.7) letting $U^j_n(t,r)=\frac{1}{\lambda _{j,n}}U^j\left( \frac{t-t_{j,n}}{\lambda _{j,n}},\frac{r}{\lambda _{j,n}} \right)$ and defining similarly $U^k_n$, we have

$$\begin{aligned}&\lim _{n\rightarrow \infty }\int _0^{\infty } \partial _{r}\Psi _{0,n}^j \partial _r\Psi _{0,n}^krdr+\int _0^{\infty } \Psi _{1,n}^j\Psi _{1,n}^krdr\\&\quad = \int _{0}^{\infty } \partial _rU_n^j(0,r)\partial _tU_n^j(0,r)r^3dr+ \int _{0}^{\infty } \partial _rU_n^j(0,r)\partial _rU_n^j(0,r)r^3dr+o_n(1), \end{aligned}$$

and (B.3) follows from (6.10).

It remains to prove (B.3) when $j\in {\mathcal {J}}_C$ or $k\in {\mathcal {J}}_C$, (B.4) when $j\in {\mathcal {J}}_C$ and $k\in {\mathcal {J}}_C$, and (B.5), (B.6) when $j\in {\mathcal {J}}_C$.

Proof of (B.3) when $j\in {\mathcal {J}}_C$, $k\in {\mathcal {J}}_L$ Assume to fix ideas $\lim _{n\rightarrow \infty } -t_{j,n}/\lambda _{j,n}=+\,\infty$. By the asymptotic formulas (A.2) and (A.3), there exists $G^k\in L^2({\mathbb {R}})$ such that

$$\begin{aligned} \lim _{t\rightarrow +\,\infty } \int _0^{\infty }\left| r^{3/2}\partial _tU^k(t,r)+G^k(r-t)\right| dr+\int _0^{\infty } \left| r^{3/2}\partial _rU^k(t,r)-G^k(r-t)\right| dr. \end{aligned}$$

Using also that $\lim _{t\rightarrow \infty }\int _0^{\infty } |U^k(t,r)|^2r dr=0$, we deduce

$$\begin{aligned} \lim _{t\rightarrow +\,\infty } \int _0^{\infty }\left| r^{1/2}\partial _t\Psi ^k(t,r)+G^k(r-t)\right| dr+\int _0^{\infty } \left| r^{1/2}\partial _r\Psi ^k(t,r)-G^k(r-t)\right| dr. \end{aligned}$$

Thus

$$\begin{aligned}&\left| \int _{0}^{\infty }\partial _r\Psi _{0,n}^j(r) \partial _r\Psi _{0,n}^k(r)rdr\right| \\&\quad =\left| \int _{0}^{\infty } \frac{1}{\lambda _{j,n}}\partial _r\Psi _{0}^j \left( \frac{r}{\lambda _{j,n}}\right) G^k \left( \frac{t_{k,n}+r}{\lambda _{k,n}}\right) \frac{r^{1/2}}{\lambda _{k,n}^{1/2}}dr \right| \\&\quad =\left| \int _{0}^{\infty }\partial _r\Psi _{0}^j\left( s\right) \frac{\lambda _{k,n}^{1/2}}{\lambda _{j,n}^{1/2}} G^k\left( \frac{t_{k,n}+\lambda _{j,n}s}{\lambda _{k,n}}\right) s^{1/2}ds \right| , \end{aligned}$$

which goes to 0 when n goes to infinity (this is immediate when $G^k$ is continuous and compactly supported, and follows by density in the general case). By a very similar computation,

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{0}^{\infty }\Psi _{1,n}^j(r)\Psi _{1,n}^k(r)rdr=0, \end{aligned}$$

which concludes the proof of (B.3) in this case.

Proof of (B.3) and (B.4) when $j,k\in {\mathcal {J}}_C$. We have

$$\begin{aligned} \int _0^{\infty } \partial _r\Psi _{0,n}^j\partial _r\Psi _{0,n}^k rdr=\,&\int _0^{\infty } \frac{1}{\lambda _{j,n}}\partial _r\Psi _0^j \left( \frac{r}{\lambda _{j,n}} \right) \frac{1}{\lambda _{k,n}} \partial _r\Psi _0^k\left( \frac{r}{\lambda _{k,n}} \right) rdr\\ =\,&\int _0^{\mu _n}\ldots +\int _{\mu _n}^{\infty }\ldots , \end{aligned}$$

where $\mu _n=\sqrt{\lambda _{j,n}\lambda _{k,n}}$. Using Cauchy–Schwarz on each of the integrals, we see that it goes to 0 when n goes to $\infty$, since $\lim _n\frac{\lambda _{j,n}}{\lambda _{k,n}} +\frac{\lambda _{k,n}}{\lambda _{j,n}}=\infty$. By the same proof for the term involving $\Psi _{1,n}^j$ and $\Psi _{1,n}^k$, we obtain (B.3).

The proof of (B.4) when $j,k \in {\mathcal {J}}_C$ is the same.

Proof of (B.6) when $j\in {\mathcal {J}}_C$ We have

$$\begin{aligned} \int _0^{\infty }\left| \sin \Psi _{0,n}^j(r) \sin \omega _{0,n}^J(r)\right| \frac{dr}{r}=\int _0^{\infty }\left| \sin \Psi _{0}^j\left( r \right) \sin \omega _{0,n}^J(\lambda _{j,n}r)\right| \frac{dr}{r}. \end{aligned}$$

Let $\varepsilon >0$. Then

$$\begin{aligned}&\int _{\varepsilon }^{\varepsilon ^{-1}}\left| \sin \Psi _{0}^j(r) \sin \omega _{0,n}^J(\lambda _{j,n}r)\right| \frac{dr}{r}\\&\quad \le \left( \int _{\varepsilon }^{\varepsilon ^{-1}}\left| \sin \Psi _{0}^j(r)\right| ^{3/2}\frac{dr}{r^{3/2}}\right) ^{2/3}\left( \int _{\varepsilon }^{\varepsilon ^{-1}} \left| \omega _{0,n}^J(\lambda _{j,n}r)\right| ^3dr \right) ^{1/3}. \end{aligned}$$

Let $w_n^J(r)=\frac{1}{r}\omega _n^J(\lambda _{j,n}r)$. Then (7.23) implies that $\lambda _{j,n}w_n^J(\lambda _{j,n}\cdot )$, considered as a radial function on ${\mathbb {R}}^4$, converges weakly to 0 in ${\dot{H}}^1({\mathbb {R}}^4)$. This implies that $\lambda _{j,n}w_n^J(\lambda _{j,n}\cdot )$ converges strongly in $L^3_{{{\,\mathrm{loc}\,}}}({\mathbb {R}}^4)$, and thus

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _{\varepsilon }^{\varepsilon ^{-1}} \left| \omega _{0,n}^J(\lambda _{j,n}r)\right| ^3dr =0. \end{aligned}$$

As a consequence, for all $\varepsilon$

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{\varepsilon }^{\varepsilon ^{-1}}\left| \sin \Psi _{0}^j(r) \sin \omega _{0,n}^J(\lambda _{j,n}r)\right| \frac{dr}{r}=0, \end{aligned}$$

and (B.6) follows since $\int _{0}^{\infty } \sin ^2\Psi _{0}^j(r)\frac{dr}{r}$ is finite and $\int _0^{\infty }\sin ^2\omega _{0,n}^J(\lambda _{j,n}r)\frac{dr}{r}$ is uniformly bounded.

Appendix C. Boundedness of Integral Operators on $L^pL^q$ Spaces

In this appendix, we consider, as in the core of the article, functions defined for $t\in {\mathbb {R}}$, $x\in {\mathbb {R}}^4$ that are radial in the space variable. As before we use the notation

$$\begin{aligned} \Vert u\Vert _{(L^pL^q)(R)}:=\left\| 1\!\!1_{\{|x|>R+|t|\}} u\right\| _{L^p({\mathbb {R}},L^q({\mathbb {R}}^4))}. \end{aligned}$$

Lemma C.1

Let $R\ge 0$. For $w\in (L^2L^8)(R)$, define

$$\begin{aligned} (Aw)(t,r)=\frac{1}{r^2}\int _{0}^t w(\tau ,r)d\tau . \end{aligned}$$

Then $Aw \in (L^2L^{8/3})(R)$ and

$$\begin{aligned} \left\| Aw\right\| _{(L^2L^{8/3})(R)} \lesssim \Vert w\Vert _{(L^2L^8)(R)}, \end{aligned}$$

where the implicit constant is independent of $R\ge 0$.

Proof

Step 1. We first claim that if $h\in (L^2L^8)(R)$, then $\frac{|t|}{r^2}h\in (L^2L^{8/3})(R)$. Indeed

$$\begin{aligned}&\left( \int _{R+|t|}^{\infty } |h(t,r)|^{8/3} \left( \frac{|t|}{r^2} \right) ^{8/3}r^3dr \right) ^{3/8}\\&\quad \lesssim |t|\left( \int _{R+|t|} h(t,r)^8r^3dr \right) ^{1/8} \left( \int _{R+|t|}^{\infty } \frac{r^3}{r^8}dr \right) ^{1/4}\\&\quad \lesssim \frac{|t|}{R+|t|}\left( \int _{R+|t|}^{\infty } h(t,r)^8r^3dr\right) ^{1/8}. \end{aligned}$$

Now $\frac{|t|}{R+|t|}\le 1$ and $t\mapsto \left( \int _{R+|t|}^{\infty } h(t,r)^8r^3dr\right) ^{1/8}$ is in $L^2$ in time, concluding this step.

Step 2. In this step, we consider a positive function f in $L^2({\mathbb {R}},L^8({\mathbb {R}}^4))$, and define

$$\begin{aligned} M(f)(t,x)=\sup _{I\ni t} \frac{1}{|I|} \int _{I}f(\tau ,x)d\tau , \end{aligned}$$

where the supremum is taken over all finite intervals I that contain t, and we prove 1

$$\begin{aligned} \left\| M(f)\right\| _{L^2L^8}\lesssim \Vert f\Vert _{L^2L^8}. \end{aligned}$$

(C.1)

Indeed, if $f\in L^2L^2$ then

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

(C.2)

by Fubini and Hardy–Littlewood maximal theorem in the time variable. Furthermore, if $f\in L^2L^{\infty }$, one can prove:

$$\begin{aligned} \Vert M(f)\Vert _{L^2L^{\infty }}\lesssim \Vert f\Vert _{L^2L^{\infty }}. \end{aligned}$$

(C.3)

Indeed,

$$\begin{aligned} M(f)(t,x)\lesssim \sup _{I\ni t}\frac{1}{|I|} \int _{I}\Vert f(\tau )\Vert _{L^{\infty }_x}d\tau \end{aligned}$$

and thus

$$\begin{aligned} \Vert Mf(t)\Vert _{L^{\infty }_x}\le \sup _{I\ni t}\frac{1}{|I|} \int _{I}\Vert f(\tau )\Vert _{L^{\infty }_x}d\tau . \end{aligned}$$

Using Hardy–Littlewood maximum theorem in the t variable, we obtain (C.3). Interpolation gives (C.1).

Step 3. Let $w\in (L^2L^8)(R)$. Let

$$\begin{aligned} f(t,r)=|w(t,r)|1\!\!1_{\{r>t+R\}}. \end{aligned}$$

Then, for $r>|t|+R$, we have

$$\begin{aligned} \frac{1}{|t|}\left| \int _{0}^{t}w(\tau ,r)d\tau \right| \le \frac{1}{|t|}\int _{-|t|}^{|t|}1\!\!1_{r>\tau +R}|w(\tau ,r)|d\tau \le 2M(f)(t,r), \end{aligned}$$

since $r>|t|+R\ge |\tau |+R$ for $0\le |\tau |\le |t|$. As a consequence, for $r>|t|+R$,

$$\begin{aligned} \left| A(w)(t,r)\right| \lesssim \frac{|t|}{r^2}M(f)(t,r). \end{aligned}$$

By Step 1 with $h(t,r)=M(f)(t,r)$, then Step 2, we have

$$\begin{aligned} \left\| A(w)\right\| _{(L^2L^{8/3})(R)}\lesssim \Vert M(f)\Vert _{(L^2L^8)(R)}\lesssim \Vert w\Vert _{(L^2L^8)(R)}. \end{aligned}$$

$\square$

Lemma C.2

Let $R\ge 1$, $\lambda >1/R$. For $w\in (L^3L^6)(R)$, we define

$$\begin{aligned} (Bw)(t,r)= \frac{t}{r^4\log (r\lambda )^{1/2}}\int _0^{t} w(\tau ,r)d\tau . \end{aligned}$$

Then B is bounded from $(L^3L^6)(R)$ to $(L^1L^2)(R)$ and

$$\begin{aligned} \Vert Bw\Vert _{(L^1L^2)(R)}\lesssim \frac{1}{\log (\lambda R)^{1/3}}\Vert w\Vert _{(L^3L^6)(R)} \end{aligned}$$

Proof

We just work with $t>0$. By Minkowski in time,

$$\begin{aligned} \left( \int _{R+t}^{\infty } \left( \int _0^t \frac{w(\tau ,r) t}{r^4\log (\lambda r)}d\tau \right) ^2r^3dr \right) ^{1/2}\le \int _{0}^t \left\| w(\tau ,r)\frac{t}{r^4\log (\lambda r)}\right\| _{L^2(R+t)}d\tau . \end{aligned}$$

Furthermore, by Hölder

$$\begin{aligned} \left\| w(\tau ,r)\frac{t}{r^4\log (\lambda r)}\right\| _{L^2(R+t)} =\,&t\left( \int _{R+t}^{\infty } |w(\tau ,r)|^2\frac{r^3}{r^8\log ^2 (\lambda r)}dr \right) ^{1/2}\\ \le \,&t\left( \int _{R+t}^{\infty }|w(\tau ,r)|^6r^3dr \right) ^{1/6} \left( \int _{R+t}^{\infty }\frac{r^3}{r^{12}\log ^3(\lambda r)}dr \right) ^{1/3}\\ \lesssim \,&\frac{t}{(R+t)^{8/3}\log (\lambda (R+t))} \int _0^t \left( \int _{R+t}^{\infty } |w(\tau ,r)|^6r^3dr \right) ^{1/6}d\tau \end{aligned}$$

Integrating, we obtain

$$\begin{aligned}&\left\| 1\!\!1_{t>0}B(w)\right\| _{(L^1L^2)(R)}\\&\quad \lesssim \int _0^{\infty } \frac{t}{(R+t)^{8/3}\log (\lambda (R+t))} \int _0^t \left( \int _{R+t}^{\infty } |w(\tau ,r)|^6r^3dr \right) ^{1/6}d\tau dt \end{aligned}$$

By Fubini,

$$\begin{aligned}&\left\| 1\!\!1_{t>0}B(w)\right\| _{(L^1L^2)(R)}\\&\quad \lesssim \int _0^{\infty } \left( \int _{R+\tau }^{\infty } |w(\tau ,r)|^6r^3dr \right) ^{1/6} \int _{\tau }^{\infty } \frac{t}{(R+t)^{8/3} \log (\lambda (R+t))} dt\,d\tau \\&\quad \lesssim \int _{0}^{\infty } \left( \int _{R+\tau }^{\infty } |w(\tau ,r)|^6r^3dr \right) ^{1/6} \int _{\tau }^{\infty }\frac{dt}{(R+t)^{5/3}\log (\lambda (R+t))} d \tau \\&\quad \lesssim \int _{0}^{\infty } \left( \int _{R+\tau }^{\infty } |w(\tau ,r)|^6r^3dr \right) ^{1/6} \frac{d\tau }{(R+\tau )^{2/3}\log (\lambda (R+\tau ))}\\&\quad \lesssim \left( \int _0^{\infty } \left( \int _{R+\tau }^{\infty } |w(\tau ,r)|^{6}r^3dr \right) ^{1/2}d\tau \right) ^{1/3}\left( \int _0^{\infty }\frac{d\tau }{(R+\tau )\log (\lambda (R+\tau ))^{3/2}}d\tau \right) ^{2/3}, \end{aligned}$$

where at the last line we have used Hölder’s inequality. The desired conclusion follows easily. $\quad \square$