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.



Uniqueness of solution of the wave equation
Since I haven't been able to [track down Selberg's lecture notes](https://mathoverflow.net/questions/360886/uniqueness-of-solution-of-the-wave-equation/360893#comment909463_360901) since he moved to Bergen, and since the proof of the result I mentioned in [this comment](https://mathoverflow.net/questions/360886/uniqueness-of-solution-of-the-wave-equation/360893#comment909358_360893) 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. 


