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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.

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  • $\begingroup$ Isn't this just about Cauchy boundary conditions?en.wikipedia.org/wiki/Cauchy_boundary_condition And other links from there.The second case is obviously trying to trick the system. Physically, $u$ could hardly have a derivative, but mathematically the time derivative could still suddenly be non-zero. $\endgroup$
    – user155396
    May 20 '20 at 18:59
  • $\begingroup$ Version 2 follows from Version 1: In version 2 at $t = -1$ your assumptions imply that $u(x,-1)$ and $\partial_t u(x,-1)$ both vanish identically. $\endgroup$ May 20 '20 at 19:25
  • $\begingroup$ 2 does not follow from 1 if the wave equation only applies 'outside' the region, i.e. $t>0$ - as one would assume. $\endgroup$
    – user155396
    May 21 '20 at 14:03
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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.


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.

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  • $\begingroup$ Nicely written! $\endgroup$ May 22 '20 at 7:55
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    $\begingroup$ +1 Very nice proof. It seems to me that the technique you used in the proof can be almost verbatim extended to the case of hyperbolic equations of $C^\infty$-smooth coefficients: while my answer relies on an explicit representation which cannot be easily obtained for variable coefficient equations (Hadamard docet), you need only to know that a compactly supported initial datum will propagate with a finite speed and thus remain compactly supported. $\endgroup$ May 23 '20 at 19:12
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    $\begingroup$ @DanieleTampieri: I believe you are correct. $\endgroup$ May 25 '20 at 3:45
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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.

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    $\begingroup$ @MKO I am not aware of any counterexample. The references (even in the updated version of the answer) deal not specifically to the theory of hyperbolic PDEs. Also ,to my memory, while counterexample based on the growth of the solutions are known for parabolic PDEs even to non specialist, this not so for hyperbolic PDEs. $\endgroup$ May 21 '20 at 6:04
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    $\begingroup$ Uniqueness holds under minimal assumption, i.e., any solution $u\in C^1([0,T];D'(R^n))$ which satisfies the Cauchy problem in the sense of distributions of $D'(R^{n+1})$ with zero initial data, vanishes identically. The proof was written in some detail in a set of lecture notes by Sigmund Selberg on Nonlinear Wave Equations, it should be possible to find them around (or ask him directly) $\endgroup$ May 21 '20 at 7:17
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    $\begingroup$ Thus the behaviour at spatial infinity of the solution has no influence whatsoever on uniqueness $\endgroup$ May 21 '20 at 7:20
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    $\begingroup$ If $u$ is a $C^2$ solution of the wave equation with zero initial data then it vanishes identically. No additional conditions are required. This is proved by classical local energy estimates $\endgroup$ May 21 '20 at 10:14
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    $\begingroup$ Well, Evans, as stated before. Section 2.4. $\endgroup$ May 21 '20 at 13:44
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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.

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  • $\begingroup$ Thanks. It seems here you assume that some integrals converge (e.g. Sobolev norm is finite). This is interesting, but I am not sure this condition is satisfied in my situation. $\endgroup$
    – makt
    May 20 '20 at 19:30
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    $\begingroup$ @MKO: the result also holds for distributional solutions. It is convenient to state it using your second version: if $\psi$ is a distribution on $\mathbb{R}^{1+n}$ that solves the linear wave equation in the distribution sense, and if the support of $\psi$ is disjoint from $\{t \leq 0\}$, then in fact $\psi$ is identically zero. $\endgroup$ May 20 '20 at 19:36
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    $\begingroup$ @MKO: furthermore, the energy estimates can also be spatially localized due to finite speed of propagation; so if your worries are "growth at spatial infinity", you don't need to. See e.g. section 3 of my notes. $\endgroup$ May 20 '20 at 19:37

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