Uniqueness of solution of the wave equation

Question

Consider the wave equation $$\frac{\partial^2 u}{\partial t^2}-\sum_{i=1}^n\frac{\partial^2 u}{\partial x_i^2}=0$$ with initial conditions $$u|_{t=0}=\frac{\partial u}{\partial t}|_{t=0}=0$$

Does it follow that $u\equiv 0$? If not, are there sufficient extra conditions which guarantee that?

Remarks. (1) For $n=1$ the above uniqueness statement holds. I am particularly interested in $n=3$.

(2) Another version of initial condition which might be of interest for me is $u(x,t)=0$ for any $t\leq 0$ and any $x\in \mathbb{R}^n$.

Sorry if this question is too elementary for this site.

Answer 1

Since I haven't been able to track down Selberg's lecture notes since he moved to Bergen, and since the proof of the result I mentioned in this comment is super-short anyway, let me just include a proof here.

Theorem Let $\psi\in \mathscr{D}'(\mathbb{R}^{n+1})$ be a distributional solution of $\Box \psi = 0$. Suppose further that $\mathrm{supp}(\psi) \cap \{ t \leq 0\} = \emptyset$. Then $\psi \equiv 0$.

Proof: It suffices to show that for every $f\in C^\infty_0(\mathbb{R}^{n+1})$ that $\langle \psi, f\rangle = 0$. Let $T$ be sufficiently large such that $f \equiv 0$ for $t \geq T$. Solve the Cauchy problem $\Box u = f$ with "initial" data $u(T,x) = \partial_t u(T,x) = 0$; this can be done using, e.g. the representation formula. Notice that the representation formula gives $u \equiv 0$ for $t \geq T$. Additionally, the representation formula (finite speed of propagation) shows that for any $S$, on $\{t \geq S\}$, there exists some $R$ such that $u(t,x) \equiv 0$ when $x \geq R$. Let $S < 0$ and choose a smooth cut-off $\chi$ such that $\chi(t) \equiv 1 $ when $t \geq 0$ and $\chi(t) \equiv 0$ when $t \leq S$.

The function $\chi(t) u(t,x) \in C^\infty_0(\mathbb{R}^{n+1})$, and hence

$$ 0 = \langle \Box \psi, \chi u\rangle = \langle \psi, \Box(\chi u) \rangle = \langle \psi, f \rangle + \langle \psi, - u \chi'' - 2 \chi' \partial_t u \rangle.$$

Both $\chi'$ and $\chi''$ are supported on $[S,0]$. So $-u \chi'' - 2 \chi' \partial_t u$ is $C^\infty_0(\mathbb{R}^{1+n})$ with support in $\{ t \leq 0\}$, and hence its pairing against $\psi$ vanishes by assumption.

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The same argument fails for the heat equation due to the failure of finite speed of propagation. Solutions of $\partial_t u - \Delta u = f$ when $f$ has compact support may have spatial tails of size $\exp(-|x|^2)$. You can compensate this by not allowing arbitrary distributions but only distributions "growing no faster than $\exp(|x|^2)$" (interpreted suitably), and get a version of Tychonoff's uniqueness theorem.

Answer 2

The basic customary (and hidden) assumption I'm aware of which allows to conclude that the Cauchy problem $$ \begin{cases} \dfrac{\partial^2 u}{\partial t^2}-\displaystyle\sum_{i=1}^n\dfrac{\partial^2 u}{\partial x_i^2}\equiv\square u(x,t)=0\\ \\ u|_{t=0}=\left.\dfrac{\partial u}{\partial t}\right|_{t=0}=0 \end{cases}\label{cpwe}\tag{CPWE} $$ has the unique null solution is that $u(x,t)$ is Fourier transformable in the sense of distributions, i.e. it is a slowly increasing (or temperate or Schwartz) distribution (see for example references [1], §2.8.1 pp. 148-150 or [2], §5.1-§5.2, pp. 74-78 for the relevant definitions) $$ u(t,x)\in\mathscr{S}^\prime(\Bbb R^n\times\Bbb R) $$ This is a consequence the fact that if we can apply the (partial) Fourier transform $\mathscr{F}_{x\to\xi}:\mathscr{S}^\prime(\Bbb R^n)\to\mathscr{S}^\prime(\Bbb R^n)$ respect to the spatial variable $x\in\Bbb R^n$ to the Cauchy problem \eqref{cpwe}, it becomes the following second order ODE Cauchy problem $$ \begin{cases} \dfrac{\partial^2 \hat{u}_p(\xi,t)}{\partial t^2} + |\xi|^2\hat{u}_p(\xi,t)=0 \\ \hat{u}_p|_{t=0}=\left.\dfrac{\partial\hat{u}_p}{\partial t}\right|_{t=0}=0 \end{cases}\label{1}\tag{1} $$ which admits only (even when considered in $\mathscr{S}^\prime(\Bbb R)$), the (unique) the classical solution $\hat{u}_p(\xi,t)\equiv 0$. The only solution of \eqref{1} induce the uniqueness of the solution ${u}(x,t)\equiv 0$ to the problem \eqref{cpwe} by the isomorphism properties of the Fourier transform (see for example [2], §6.2, pp. 90-92).

Notes

• Tempered distributions are, roughly speaking, distributions "that increase at infinity not faster than a polynomial": this perhaps includes the classes of solutions of the wave equations needed in your research. However, the Fourier transform can be extended to more general classes of generalized functions.
• Strictly speaking, the hypothesis $u(t,x)\in\mathscr{S}^\prime(\Bbb R^n\times\Bbb R)$ is weaker than the strongest one allowed by this method: we could even assume that $u(x,t)\in\mathscr{S}^\prime(\Bbb R^n)\times\mathscr{D}^\prime(\Bbb R)$ as \eqref{1} can be solved in $\mathscr{D}^\prime$ and has the same and unique classical solution $\hat{u}_p(\xi,t)\equiv 0$: see for example [1], §1.5.2, pp. 25-26.
• This method, with the same hypotheses, holds also for the solution of the non homogeneous equation, perhaps answering a question you asked yesterday.

References

[1] Shilov, G. E., Generalized functions and partial differential equations, Mathematics and Its Applications. Vol. 7, (in English) New York-London-Paris: Gordon and Breach Science Publishers, XII, 345 p. (1968), MR0230129, Zbl 0177.36302.

[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.

Answer 3

Yes, as for any strictly hyperbolic equation you do have uniqueness and even much better, well-posedness: you can control the Sobolev norm of $u(t)$ by the Sobolev norms of the initial data $u(0), \dot u(0)$. For instance Theorem 23.2.2 in Hörmander's ALPDO III, gives you that result, but you can also look at Evans' book in the part about the wave equation. If you want a simple practical method, just multiply the equation by $\partial_t u$ and integrate by parts.