Wave Equations with Moving Potentials

Abstract

In this paper, we study the some reversed Strichartz estimates along general time-like trajectories for wave equations in ${\mathbb {R}}^{3}$. Some applications of the reversed Strichartz estimates and the structure of wave operators to the wave equation with one potential are also discussed. These techniques are useful to analyze the stability problem of traveling solitons.

Appendices

Appendix A: Pointwise Decay

For the sake of completeness, in this appendix, we provide the proof of dispersive estimates for the free wave equation in ${\mathbb {R}}^{3}$ based on the idea of reversed Strichartz estimates.

Theorem A.1

In ${\mathbb {R}}^{3}$, suppose $f\in L^{2},\,\nabla f\in L^{1}$ and $g\in L^{2},\,\Delta g\in L^{1}$. Then one has the following estimates:

$$\begin{aligned}&\left\| \frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f\right\| _{L_{x}^{\infty }}\lesssim \frac{1}{\left| t\right| }\left\| \nabla f\right\| _{L_{x}^{1}}, \end{aligned}$$

(A.1)

$$\begin{aligned}&\left\| \cos \left( t\sqrt{-\Delta }\right) g\right\| _{L_{x}^{\infty }}\lesssim \frac{1}{\left| t\right| }\left\| \Delta g\right\| _{L_{x}^{1}}. \end{aligned}$$

(A.2)

Remark

Note that the second estimate is slightly different from the estimates commonly used in the literature. For example, in Krieger–Schlag [KS] one needs the $L^{1}$ norm of $D^{2}g$ instead of $\Delta g$.

Proof

First of all, we consider

$$\begin{aligned} \frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f. \end{aligned}$$

(A.3)

In ${\mathbb {R}}^{3}$, one has

$$\begin{aligned} \frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f=\frac{1}{4\pi t}\int _{\left| x-y\right| =t}f(y)\,dy. \end{aligned}$$

(A.4)

Without loss of generality, we assume $t\ge 0$.

Multiplying t and integrating, we obtain

$$\begin{aligned} \int _{0}^{\infty }\left| t\frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f\right| dt\lesssim & {} \int _{0}^{\infty }\int _{{\mathbb {S}}^{2}}\left| f(x+r\omega )\right| r^{2}\,d\omega dr\nonumber \\\lesssim & {} \left\| f\right\| _{L_{x}^{1}}. \end{aligned}$$

(A.5)

Therefore,

$$\begin{aligned} \left\| t\frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f\right\| _{L_{x}^{\infty }L_{t}^{1}}\lesssim \left\| f\right\| _{L_{x}^{1}}. \end{aligned}$$

(A.6)

Notice that, from the estimate above, we also have

$$\begin{aligned} \left\| \int _{t}^{\infty }\frac{\sin \left( s\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f\,ds\right\| _{L_{x}^{\infty }}\lesssim \frac{1}{\left| t\right| }\left\| f\right\| _{L_{x}^{1}}. \end{aligned}$$

(A.7)

Replacing f with $\Delta f$, it implies that

$$\begin{aligned} \left\| \int _{t}^{\infty }\sqrt{-\Delta }\sin \left( s\sqrt{-\Delta }\right) f\,ds\right\| _{L_{x}^{\infty }}\lesssim \frac{1}{\left| t\right| }\left\| \Delta f\right\| _{L_{x}^{1}}. \end{aligned}$$

(A.8)

On the other hand,

$$\begin{aligned} \int _{0}^{\infty }\left| t\cos \left( t\sqrt{-\Delta }\right) f\right| dt\lesssim & {} \int _{0}^{\infty }\int _{{\mathbb {S}}^{2}}\left| rf\left( x+r\omega \right) d\omega +r^{2}\partial _{r}f\left( x+r\omega \right) \right| \,d\omega dr\nonumber \\\lesssim & {} \left\| \nabla f\right\| _{L_{x}^{1}} \end{aligned}$$

(A.9)

where in the last inequality, we applied integration by parts in r in the first term of the RHS of the first line.

Therefore,

$$\begin{aligned} \left\| t\cos \left( t\sqrt{-\Delta }\right) f\right\| _{L_{x}^{\infty }L_{t}^{1}}\lesssim \left\| \nabla f\right\| _{L_{x}^{1}}. \end{aligned}$$

(A.10)

Hence

$$\begin{aligned} \left\| \int _{t}^{\infty }\cos \left( s\sqrt{-\Delta }\right) f\,ds\right\| _{L_{x}^{\infty }}\lesssim \frac{1}{\left| t\right| }\left\| \nabla f\right\| _{L_{x}^{1}}. \end{aligned}$$

(A.11)

Finally, we check

$$\begin{aligned} \frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f=\int _{t}^{\infty }\cos \left( s\sqrt{-\Delta }\right) f\,ds, \end{aligned}$$

(A.12)

and

$$\begin{aligned} \cos \left( t\sqrt{-\Delta }\right) g=\int _{t}^{\infty }\sqrt{-\Delta }\sin \left( s\sqrt{-\Delta }\right) g\,ds. \end{aligned}$$

(A.13)

It suffices to show expressions hold for tast functions. Let $f,\,g,\,h$ be any test functions. Define

$$\begin{aligned} Ag=\cos \left( t\sqrt{-\Delta }\right) g-\int _{t}^{\infty }\sqrt{-\Delta }\sin \left( s\sqrt{-\Delta }\right) g\,ds \end{aligned}$$

(A.14)

and

$$\begin{aligned} Bf=\frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f-\int _{t}^{\infty }\cos \left( s\sqrt{-\Delta }\right) f\,ds. \end{aligned}$$

(A.15)

It is easy to check that $A,\,B$ are independent of t by taking the time derivative of the above expressions.

To see $A=B=0$, for A, one observes that

$$\begin{aligned} \left\langle \cos \left( t\sqrt{-\Delta }\right) g,\,h\right\rangle \rightarrow 0 \end{aligned}$$

(A.16)

and

$$\begin{aligned} \left\| \int _{t}^{\infty }\frac{\sin \left( s\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f\,ds\right\| _{L_{x}^{\infty }}\lesssim \frac{1}{\left| t\right| }\left\| f\right\| _{L_{x}^{1}}. \end{aligned}$$

(A.17)

Therefore,

$$\begin{aligned} \left\langle Ag,\,h\right\rangle \rightarrow 0,\quad t\rightarrow \infty . \end{aligned}$$

(A.18)

Since A is independent of t, one concludes that

$$\begin{aligned} \left\langle Ag,\,h\right\rangle =0 \end{aligned}$$

(A.19)

for any pair of test functions and hence

$$\begin{aligned} A=0. \end{aligned}$$

(A.20)

Similarly, we get

$$\begin{aligned} B=0. \end{aligned}$$

(A.21)

Therefore by our calculations above, we can obtain the dispersive estimates for the free wave equation,

$$\begin{aligned} \left\| \frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f\right\| _{L_{x}^{\infty }}= & {} \left\| \int _{t}^{\infty }\cos \left( s\sqrt{-\Delta }\right) f\,ds\right\| _{L_{x}^{\infty }}\nonumber \\\lesssim & {} \frac{1}{\left| t\right| }\left\| \nabla f\right\| _{L_{x}^{1}}, \end{aligned}$$

(A.22)

and

$$\begin{aligned} \left\| \cos \left( t\sqrt{-\Delta }\right) g\right\| _{L_{x}^{\infty }}= & {} \left\| \int _{t}^{\infty }\sqrt{-\Delta }\sin \left( s\sqrt{-\Delta }\right) g\,ds\right\| _{L_{x}^{\infty }}\nonumber \\\lesssim & {} \frac{1}{\left| t\right| }\left\| \Delta g\right\| _{L_{x}^{1}}. \end{aligned}$$

(A.23)

The theorem is proved. $\quad \square$

Appendix B: Local Energy Decay

We derive the local energy decay estimate for the free wave equation by the Fourier method.

Recall the coarea formula: for a a real-valued Lipschitz function u and a $L^{1}$ function g then

$$\begin{aligned} \int _{{\mathbb {R}}^{n}}g(x)\left| \nabla u(x)\right| dx=\int _{{\mathbb {R}}}\int _{\left\{ u(x)=t\right\} }g(x)\,d\sigma (x)dt, \end{aligned}$$

(B.1)

where $\sigma$ is the surface measure.

Lemma B.1

For $F\in C_{0}^{\infty }$, $\phi$ smooth and non-degenerate,i.e. $\left| \nabla \phi (x)\right| \ne 0$, one has

$$\begin{aligned} \int _{{\mathbb {R}}}\int _{{\mathbb {R}}^{n}}e^{i\lambda \phi (x)}F(x)\,dxd\lambda =\left( 2\pi \right) ^{n}\int _{\left\{ \phi =0\right\} }\frac{F(x)}{\left| \nabla \phi (x)\right| }\,d\sigma (x). \end{aligned}$$

(B.2)

Proof

From (B.1),

$$\begin{aligned} \int _{{\mathbb {R}}}\int _{{\mathbb {R}}^{n}}e^{i\lambda \phi (x)}F(x)\,dxd\lambda =\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}e^{i\lambda y}\int _{\left\{ \phi =y\right\} }\frac{F(x)}{\left| \nabla \phi (x)\right| }\,d\sigma (x)dyd\lambda . \end{aligned}$$

(B.3)

Denote $\int _{\left\{ \phi =y\right\} }\frac{F(x)}{\left| \nabla \phi (x)\right| }\,d\sigma (x)=g(y)$, then

$$\begin{aligned} \int _{{\mathbb {R}}}\int _{{\mathbb {R}}^{n}}e^{i\lambda \phi (x)}F(x)\,dxd\lambda= & {} \int _{{\mathbb {R}}}\int _{{\mathbb {R}}}e^{i\lambda y}g(y)\,dyd\lambda \nonumber \\= & {} \left( 2\pi \right) ^{\frac{n}{2}}\int _{{\mathbb {R}}}{\hat{g}}(\lambda )\,d\lambda \nonumber \\= & {} \left( 2\pi \right) ^{n}g(0)\nonumber \\= & {} \int _{\left\{ \phi =0\right\} }\frac{F(x)}{\left| \nabla \phi (x)\right| }\,d\sigma (x). \end{aligned}$$

(B.4)

We are done. $\quad \square$

It suffices to consider the half wave evolution,

$$\begin{aligned} e^{it\sqrt{-\Delta }}f. \end{aligned}$$

(B.5)

Theorem B.2

(Local energy decay). Let $\chi \ge 0$ be a smooth cut-off function such that ${\hat{\chi }}$ has compact support. Then

$$\begin{aligned} \left\| \chi (x)e^{it\sqrt{-\Delta }}f\right\| _{L_{t,x}^{2}}\lesssim \left\| f\right\| _{L_{x}^{2}}. \end{aligned}$$

(B.6)

Proof

Consider

$$\begin{aligned}&\int _{{\mathbb {R}}}\int _{{\mathbb {R}}^{n}}\left| e^{it\sqrt{-\Delta }}f\right| ^{2}(x)\chi (x)\,dxdt\nonumber \\&\quad = \int _{{\mathbb {R}}}\left\langle e^{it\sqrt{-\Delta }}f,\chi (x)e^{it\sqrt{-\Delta }}f\right\rangle _{L^{2}}dt\nonumber \\&\quad = \int _{{\mathbb {R}}}\left\langle e^{it\left| \xi \right| }{\hat{f}}\left( \xi \right) ,\left[ e^{it\left| \xi \right| }{\hat{f}}\left( \xi \right) \right] *{\hat{\chi }}\left( \xi \right) \right\rangle _{L^{2}}dt\nonumber \\&= \quad \int _{{\mathbb {R}}}\int _{{\mathbb {R}}^{n}}\int _{{\mathbb {R}}^{n}}e^{it\left( \left| \xi \right| -\left| \eta \right| \right) }{\hat{\chi }}\left( \xi -\eta \right) {\hat{f}}\left( \xi \right) {\hat{f}}\left( \eta \right) \,d\eta d\xi dt. \end{aligned}$$

(B.7)

Applying Lemma B.1 with $\phi \left( \xi ,\eta \right) =\left| \xi \right| -\left| \eta \right|$, the surface $\left\{ \phi =0\right\}$ becomes $\left\{ \left| \xi \right| =\left| \eta \right| \right\}$ and $\left| \nabla \phi \right| =\sqrt{2}$. It follows that

$$\begin{aligned} \int _{{\mathbb {R}}}\int _{{\mathbb {R}}^{n}}\left| e^{it\sqrt{-\Delta }}f\right| ^{2}(x)\chi (x)\,dxdt\simeq & {} \int _{\left| \xi \right| =\left| \eta \right| }{\hat{\chi }}\left( \xi -\eta \right) {\hat{f}}\left( \xi \right) {\hat{f}}\left( \eta \right) \,d\sigma \nonumber \\\lesssim & {} \int _{\left| \xi \right| =\left| \eta \right| }\left| {\hat{\chi }}\left( \xi -\eta \right) \right| \left[ \left| {\hat{f}}\left( \xi \right) \right| ^{2}+\left| {\hat{f}}\left( \eta \right) \right| ^{2}\right] \,d\sigma \nonumber \\\lesssim & {} \int _{{\mathbb {R}}^{n}}\left| {\hat{f}}\left( \xi \right) \right| ^{2}\int _{\left| \xi \right| =\left| \eta \right| }\left| {\hat{\chi }}\left( \xi -\eta \right) \right| \,d\sigma d\xi \nonumber \\\lesssim & {} \sup _{\xi }\left| K\left( \xi \right) \right| \int _{{\mathbb {R}}^{n}}\left| {\hat{f}}\left( \xi \right) \right| ^{2}d\xi \nonumber \\\lesssim & {} \int _{{\mathbb {R}}^{n}}\left| f(x)\right| ^{2}dx. \end{aligned}$$

(B.8)

It reduces to show that

$$\begin{aligned} K(\xi )=\int _{\left| \xi \right| =\left| \eta \right| }\left| {\hat{\chi }}\left( \xi -\eta \right) \right| \,d\sigma \end{aligned}$$

(B.9)

is bounded uniformly in $\xi$. Since ${\hat{\chi }}\left( \xi \right)$ decays fast, we have

$$\begin{aligned} \left| {\hat{\chi }}(\xi )\right| \lesssim \left\langle \xi \right\rangle ^{-N} \end{aligned}$$

and

$$\begin{aligned} \left| {\hat{\chi }}\left( \xi \right) \right| \lesssim \left| \xi \right| ^{1-\epsilon -n}, \end{aligned}$$

where as usual, $\left\langle \xi \right\rangle =\left( 1+\left| \xi \right| ^{2}\right) ^{\frac{1}{2}}$.

Note that

$$\begin{aligned} K(\xi )= & {} \int _{\left| \xi \right| =\left| \eta \right| }\left| {\hat{\chi }}\left( \xi -\eta \right) \right| \,d\sigma \nonumber \\= & {} \int _{\left| \zeta -\xi \right| =\left| \xi \right| }\left| {\hat{\chi }}\left( \zeta \right) \right| \,d\sigma \nonumber \\\lesssim & {} \int _{\left| \zeta -\xi \right| =\left| \xi \right| ,\left| \zeta \right| <1}\left| {\hat{\chi }}\left( \zeta \right) \right| \,d\sigma \nonumber \\&+\int _{\left| \zeta -\xi \right| =\left| \xi \right| ,\left| \zeta \right| >1}\left| {\hat{\chi }}\left( \zeta \right) \right| \,d\sigma \nonumber \\\lesssim & {} C(n) \end{aligned}$$

(B.10)

which is uniformly bounded in $\xi$ and only depends on n.

Therefore, we can conclude

$$\begin{aligned} \left\| \chi (x)e^{it\sqrt{-\Delta }}f\right\| _{L_{t,x}^{2}}\lesssim \left\| f\right\| _{L_{x}^{2}}. \end{aligned}$$

(B.11)

We are done. $\quad \square$

With dyadic decomposition and weights, one has a global version of the result above:

Corollary B.3

$\forall \epsilon >0$, one has

$$\begin{aligned} \left\| \left( 1+\left| x\right| \right) ^{-\frac{1}{2}-\epsilon }e^{it\sqrt{-\Delta }}f\right\| _{L_{t,x}^{2}}\lesssim _{\epsilon }\left\| f\right\| _{L_{x}^{2}}. \end{aligned}$$

(B.12)

Proof

Let $\chi (x)$ from Theorem B.2 be a smooth version of $1_{B_{1}(0)}$, the indicator function of the unit ball. It follows that

$$\begin{aligned}&\left\| \chi \left( 2^{-j}x\right) e^{it\sqrt{-\Delta }}f\right\| _{L_{t,x}^{2}}\nonumber \\&\quad =2^{\frac{jn}{2}}2^{\frac{j}{2}}\left\| \chi \left( x\right) \left( e^{it\sqrt{-\Delta }}f\right) \left( 2^{j}t,2^{j}x\right) \right\| _{L_{t,x}^{2}}\nonumber \\&\quad =2^{\frac{jn}{2}}2^{\frac{j}{2}}\left\| \chi \left( x\right) \left( e^{it\sqrt{-\Delta }}f\left( 2^{j}\cdot \right) \right) \right\| _{L_{t,x}^{2}}\nonumber \\&\quad \lesssim 2^{\frac{jn}{2}}2^{\frac{j}{2}}\left\| f\left( 2^{j}\cdot \right) \right\| _{L_{x}^{2}}\nonumber \\&\quad \lesssim 2^{\frac{j}{2}}\left\| f\right\| _{L_{x}^{2}}. \end{aligned}$$

(B.13)

Notice that

$$\begin{aligned} \left( 1+\left| x\right| \right) ^{-\frac{1}{2}-\epsilon }\simeq \sum _{j\ge 0}2^{-j\left( \frac{1}{2}+\epsilon \right) }\chi \left( 2^{-j}x\right) \end{aligned}$$

(B.14)

then with our computations above, we can conclude that

$$\begin{aligned} \left\| \sum _{j\ge 0}2^{-j\left( \frac{1}{2}+\epsilon \right) }\chi \left( 2^{-j}x\right) e^{it\sqrt{-\Delta }}f\right\| _{L_{t,x}^{2}}\lesssim _{\epsilon }\left\| f\right\| _{L_{x}^{2}}, \end{aligned}$$

(B.15)

and hence

$$\begin{aligned} \left\| \left( 1+\left| x\right| \right) ^{-\frac{1}{2}-\epsilon }e^{it\sqrt{-\Delta }}f\right\| _{L_{t,x}^{2}}\lesssim _{\epsilon }\left\| f\right\| _{L_{x}^{2}}. \end{aligned}$$

(B.16)

The corollary is proved. $\quad \square$

Appendix C: Global Existence

In this appendix, we discuss the global existence of solutions to the wave equation with time-dependent potentials. Lorentz transformations are important tools in our analysis. Lorentz transformations are rotations of space-time, therefore, a priori, one needs to show the global existence of solutions to wave equations with time-dependent potentials.

Theorem C.1

Assume $V(x,t)\in L_{t,x}^{\infty }$. Then for each $\left( g,\,f\right) \in H^{1}\left( {\mathbb {R}}^{3}\right) \times L^{2}\left( {\mathbb {R}}^{3}\right) ,$ there is a unique solution $\left( u,\,u_{t}\right) \in C\left( {\mathbb {R}},\,H^{1}\left( {\mathbb {R}}^{3}\right) \right) \times C\left( {\mathbb {R}},\,L^{2}\left( {\mathbb {R}}^{3}\right) \right)$ to

$$\begin{aligned} \partial _{tt}u-\Delta u+V(x,t)u=0 \end{aligned}$$

(C.1)

with initial data

$$\begin{aligned} u(x,0)=g,\quad \partial _{t}u(x,0)=f. \end{aligned}$$

(C.2)

Proof

By Duhamel’s formula, we might write the solution as

$$\begin{aligned} u=\frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f+\cos \left( t\sqrt{-\Delta }\right) g+\int _{0}^{t}\frac{\sin \left( \left( t-s\right) \sqrt{-\Delta }\right) }{\sqrt{-\Delta }}V(\cdot ,s)u(s)\,ds.\nonumber \\ \end{aligned}$$

(C.3)

Starting from the local existence, we try to construct the solution in

$$\begin{aligned} X=C\left( [0,\,T],\,H^{1}\left( {\mathbb {R}}^{3}\right) \right) \times C\left( [0,\,T),\,L^{2}\left( {\mathbb {R}}^{3}\right) \right) \end{aligned}$$

(C.4)

with $T\le 1$. One can view u as the fixed-point of the map

$$\begin{aligned} S(h)(t)=\frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f+\cos \left( t\sqrt{-\Delta }\right) g+\int _{0}^{t}\frac{\sin \left( \left( t-s\right) \sqrt{-\Delta }\right) }{\sqrt{-\Delta }}V(\cdot ,s)h(s)\,ds.\nonumber \\ \end{aligned}$$

(C.5)

Let

$$\begin{aligned} R=2\left\| \frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f+\cos \left( t\sqrt{-\Delta }\right) g\right\| _{X}. \end{aligned}$$

(C.6)

We will show when T is small enough, S will be a contraction map in $B_{X}(0,R)$.

Clearly,

$$\begin{aligned} \left\| S(h)(t)\right\| _{X}\le & {} \left\| \frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f+\cos \left( t\sqrt{-\Delta }\right) g\right\| _{X}\nonumber \\&+\,\left\| \int _{0}^{t}\frac{\sin \left( \left( t-s\right) \sqrt{-\Delta }\right) }{\sqrt{-\Delta }}V(\cdot ,s)h(s)\,ds\right\| _{X}. \end{aligned}$$

(C.7)

By direct calculations,

$$\begin{aligned} \left\| \int _{0}^{t}\frac{\sin \left( \left( t-s\right) \sqrt{-\Delta }\right) }{\sqrt{-\Delta }}V(\cdot ,s)h(s)\,ds\right\| _{L_{x}^{2}}\le & {} T^{2}\left\| V\left( \cdot ,t\right) h(t)\right\| _{L^{2}}, \end{aligned}$$

(C.8)

$$\begin{aligned} \left\| \int _{0}^{t}\frac{\sin \left( \left( t-s\right) \sqrt{-\Delta }\right) }{\sqrt{-\Delta }}V(\cdot ,s)h(s)\,ds\right\| _{{\dot{H}}_{x}^{1}}\le & {} T\left\| V\left( \cdot ,t\right) h(t)\right\| _{L^{2}}, \end{aligned}$$

(C.9)

and

$$\begin{aligned} \left\| \partial _{t}\left( \int _{0}^{t}\frac{\sin \left( \left( t-s\right) \sqrt{-\Delta }\right) }{\sqrt{-\Delta }}V(\cdot ,s)h(s)\,ds\right) \right\| _{L_{x}^{2}}\le T\left\| V\left( \cdot ,t\right) h(t)\right\| _{L^{2}}.\nonumber \\ \end{aligned}$$

(C.10)

Therefore, we can pick $T\left\| V\right\| _{L_{t,x}^{\infty }}<\frac{1}{10}$, we have

$$\begin{aligned} \left\| S(h)(t)\right\| _{X}\le \left\| \frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f+\cos \left( t\sqrt{-\Delta }\right) g\right\| _{X}+\frac{1}{2}\left\| h\right\| _{X}. \end{aligned}$$

(C.11)

Hence, S maps $B_{X}\left( 0,R\right)$ into itself.

Next we show S is a contraction. The calculations are straightforward.

$$\begin{aligned} \left\| S(h_{1}-h_{2})(t)\right\| _{X}\le \left\| \int _{0}^{t}\frac{\sin \left( \left( t-s\right) \sqrt{-\Delta }\right) }{\sqrt{-\Delta }}V(\cdot ,s)\left( h_{1}(s)-h_{2}(s)\right) \,ds\right\| _{X}.\nonumber \\ \end{aligned}$$

(C.12)

The the same arguments as above give

$$\begin{aligned} \left\| S(h_{1}-h_{2})(t)\right\| _{X}\le \frac{1}{2}\left\| (h_{1}-h_{2})(t)\right\| _{X}. \end{aligned}$$

(C.13)

Therefore, by fixed point theorem, there is $u\in X$ such that

$$\begin{aligned} u=S(u), \end{aligned}$$

(C.14)

in other words, there exist $u\in C\left( [0,\,T],\,H^{1}\left( {\mathbb {R}}^{3}\right) \right) \times C\left( [0,\,T),\,L^{2}\left( {\mathbb {R}}^{3}\right) \right)$ such that

$$\begin{aligned} u=\frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f+\cos \left( t\sqrt{-\Delta }\right) g+\int _{0}^{t}\frac{\sin \left( \left( t-s\right) \sqrt{-\Delta }\right) }{\sqrt{-\Delta }}V(\cdot ,s)u(s)\,ds.\nonumber \\ \end{aligned}$$

(C.15)

We notice that the choice of T is independent of the size of the initial data. Then we can repeat the argument above with $\left( u(T),\partial _{t}u(T)\right)$ as initial condition to construct the solution from T to 2T. Iterating this process, one can easily construct the solution $\left( u,\,u_{t}\right) \in C\left( {\mathbb {R}},\,H^{1}\left( {\mathbb {R}}^{3}\right) \right) \times C\left( {\mathbb {R}},\,L^{2}\left( {\mathbb {R}}^{3}\right) \right)$.

Finally, we notice the uniqueness of the solution follows from Grönwall’s inequality. Suppose one has two solutions $u_{1}$ and $u_{2}$ to our equation with the same data, then

$$\begin{aligned} \left\| u_{1}-u_{2}\right\| _{H^{1}\times L^{2}}(t)\le \int _{0}^{t}\left( t-s\right) \left\| u_{1}-u_{2}\right\| (s)\,ds. \end{aligned}$$

(C.16)

Applying Grönwall’s inequality over [0, T], we obtain

$$\begin{aligned} \left\| u_{1}-u_{2}\right\| _{X}=0, \end{aligned}$$

(C.17)

which means $u_{1}\equiv u_{2}$ on [0, T]. Then by the same iteration argument as above, we can conclude that in $C\left( {\mathbb {R}},\,H^{1}\left( {\mathbb {R}}^{3}\right) \right) \times C\left( {\mathbb {R}},\,L^{2}\left( {\mathbb {R}}^{3}\right) \right)$

$$\begin{aligned} u_{1}\equiv u_{2}. \end{aligned}$$

(C.18)

Therefore, one obtains the uniqueness.

The theorem is proved. $\quad \square$

In our setting, $V(x,t)=V\left( x-\vec {v}(t)\right)$ satisfies the assumption of Theorem C.1, therefore we have the global existence and uniqueness.

Corollary C.2

For each $\left( g,\,f\right) \in H^{1}\left( {\mathbb {R}}^{3}\right) \times L^{2}\left( {\mathbb {R}}^{3}\right) ,$ there is a unique global solution $\left( u,\,u_{t}\right) \in C\left( {\mathbb {R}},\,H^{1}\left( {\mathbb {R}}^{3}\right) \right) \times C\left( {\mathbb {R}},\,L^{2}\left( {\mathbb {R}}^{3}\right) \right)$ to the wave equation

$$\begin{aligned} \partial _{tt}u-\Delta u+V\left( x-\vec {v}(t)\right) u=0 \end{aligned}$$

(C.19)

with initial data

$$\begin{aligned} u(x,0)=g,\,\partial _{t}u(x,0)=f. \end{aligned}$$

(C.20)

Remark

The theorem above also applies to the charge transfer model in [GC2]:

$$\begin{aligned} \partial _{tt}u-\Delta u+\sum _{i=1}^{m}\sum _{j=1}^{m}V_{v_{j}}\left( x-\vec {v}_{j}t\right) u=0. \end{aligned}$$

Appendix D: Revered Strichartz Estimates

In this appendix, we present an alternative approach to the homogeneous endpoint reversed Strichartz estimates based on the Fourier transformation.

We only consider $\frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f=\frac{1}{2}\frac{e^{it\sqrt{-\Delta }}}{\sqrt{-\Delta }}f-\frac{1}{2}\frac{e^{-it\sqrt{-\Delta }}}{\sqrt{-\Delta }}f$. We can further reduce to consider

$$\begin{aligned} \frac{e^{it\sqrt{-\Delta }}}{\sqrt{-\Delta }}f. \end{aligned}$$

(D.1)

With Fourier transform and polar coordinates $\xi =\lambda \omega$, we have

$$\begin{aligned} \frac{e^{it\sqrt{-\Delta }}}{\sqrt{-\Delta }}f= & {} \int _{0}^{\infty }\int _{{\mathbb {S}}^{2}}\frac{e^{2\pi it\lambda }}{\lambda }e^{2\pi i\lambda \left( \omega \cdot x\right) }\lambda ^{2}{\hat{f}}\left( \lambda \omega \right) d\omega d\lambda \nonumber \\= & {} \int _{{\mathbb {R}}}e^{2\pi it\lambda }\left( \chi _{[0,\infty )}(\lambda )\int _{{\mathbb {S}}^{2}}e^{2\pi i\lambda \left( \omega \cdot x\right) }\lambda {\hat{f}}\left( \lambda \omega \right) d\omega \right) d\lambda \nonumber \\= & {} \int _{{\mathbb {R}}}e^{2\pi it\lambda }G(x,\lambda )d\lambda \end{aligned}$$

(D.2)

where

$$\begin{aligned} G(x,\lambda )=\chi _{[0,\infty )}(\lambda )\int _{{\mathbb {S}}^{2}}e^{2\pi i\lambda \left( \omega \cdot x\right) }\lambda {\hat{f}}\left( \lambda \omega \right) d\omega . \end{aligned}$$

(D.3)

By Plancherel’s Theorem, we know for fixed x,

$$\begin{aligned} \left\| \frac{e^{it\sqrt{-\Delta }}}{\sqrt{-\Delta }}f\right\| _{L_{t}^{2}}= & {} \left\| G(x,\lambda )\right\| _{L_{\lambda }^{2}}. \end{aligned}$$

(D.4)

$$\begin{aligned} G^{2}(x,\lambda )= & {} \left( \chi _{[0,\infty )}(\lambda )\int _{{\mathbb {S}}^{2}}e^{2\pi i\lambda \left( \omega \cdot x\right) }\lambda {\hat{f}}\left( \lambda \omega \right) d\omega \right) ^{2}\nonumber \\\lesssim & {} \chi _{[0,\infty )}(\lambda )\int _{{\mathbb {S}}^{2}}\lambda ^{2}\left| {\hat{f}}\left( \lambda \omega \right) \right| ^{2}d\omega \end{aligned}$$

(D.5)

$$\begin{aligned} \left\| \frac{e^{it\sqrt{-\Delta }}}{\sqrt{-\Delta }}f\right\| _{L_{t}^{2}}^{2}\lesssim & {} \int _{0}^{\infty }\int _{{\mathbb {S}}^{2}}\lambda ^{2}\left| {\hat{f}}\left( \lambda \omega \right) \right| ^{2}d\omega d\lambda \nonumber \\\lesssim & {} \int \left| {\hat{f}}\left( \xi \right) \right| ^{2}d\xi \nonumber \\= & {} \int \left| f\left( x\right) \right| ^{2}dx. \end{aligned}$$

(D.6)

Therefore,

$$\begin{aligned} \left\| \frac{\sin \left( t\sqrt{-\Delta }\right) }{\sqrt{-\Delta }}f\right\| _{L_{x}^{\infty }L_{t}^{2}}\lesssim \left\| f\right\| _{L^{2}} \end{aligned}$$

(D.7)

as desired.

Remark

The two dimension version was obtained in [Oh] and is mentioned in [B]:

$$\begin{aligned} \left\| e^{it\sqrt{-\Delta }}f\right\| _{L_{x}^{\infty }L_{t}^{2}}\lesssim \left\| f\right\| _{{\dot{B}}_{2,1}^{1/2}}. \end{aligned}$$

(D.8)