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Question. Assume that $U\in C^1(\,[0, \infty)\to \mathcal{S}'(\mathbb{R}^n)\,)$ is a solution to the following tempered distributional Cauchy problem $$\tag{CP}\begin{cases} \frac{ d U}{dt} = f \cdot U(t), & t>0 \\ U(0) = 0 \end{cases} $$ where $f\in C^\infty(\mathbb{R}^n)$ is a smooth function not depending on $t$. Is it true that $U(t)=0$ at all times $t>0$?

The background for this question comes from a passage in the book on PDEs by Michael E. Taylor. The author claims that the problem \begin{equation} \begin{cases} \frac{\partial u}{\partial t} -\Delta u =0, & t>0,\ x\in \mathbb{R}^n \\ u(0, x)=u_0 \end{cases} \end{equation} has a unique tempered distributional solution if $u_0$ is a tempered distribution. (This is not entirely trivial, as uniqueness fails if one imposes no growth conditions on $u$). The proof via Fourier transform is essentially reduced to the statement that the Cauchy problem $$ \begin{cases} \frac{\partial \hat{u}}{\partial t}(t, \xi) + \lvert \xi\rvert^2 \hat{u}(t, \xi)=0, & t > 0 \\ \hat{u}(0, \xi)=0\\ \hat{u}\in C^1(\, [0, \infty)_t \to \mathcal{S}'(\mathbb{R}^n_\xi)\, ) \end{cases} $$ has the unique solution $\hat{u}(t, \xi)\equiv 0$. This does not seem obvious to me and that's why I am posing this question.

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  • $\begingroup$ What is $S'(\mathbb { R ^ n })$? $\endgroup$
    – Ellya
    Apr 5, 2014 at 22:43
  • $\begingroup$ @ellya: The space of tempered distributions on $\mathbb{R}^n$. It is the dual space of the Schwartz space $\mathcal{S}$, which is a Fréchet space with topology given by the seminorms $$\lvert f\rvert_{\alpha, \beta}=\sup_{x\in \mathbb{R}^n}\lvert x^\alpha\partial^\beta f(x)\rvert.$$ Here's the related Wikipedia page. $\endgroup$ Apr 6, 2014 at 7:57

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I think this is a great question. This is the direction I would try. So I presume that you are able to prove this if $f$ is compactly supported (since multiplication by $f$ would then be a continuous operation from $\mathcal S \to \mathcal S$). Secondly, I presume that there is a theorem that says if $\mu, \nu \in \mathcal S'$, and $\mu(\phi) = \nu(\phi)$ for all $\phi \in \mathcal S$ which are compactly supported, then $\mu = \nu$. I think that whatever proof is used to show this would have to be an important part of my argument.

So pick a bump function $\rho \in \mathcal S$ such that $\rho(x) = 1$ if $|x| \le 1$, and $\rho(x) = 0$ if $|x| \ge 2$, and $\rho(x) \in [0,1]$ for all $x \in \mathbb R^n$.

Then for each positive integer $m$, let $U_m$ be the solution to the equation $$ \frac d{dt} U_m = \rho(\cdot/m) f \cdot U_m ,\quad U_m(0) = 0$$ and show that $U_m \equiv 0$. Then show that $U(t,\cdot) = U_m(t,\cdot)$ on $B(0,m)$. And then conclude $U = \lim_{m\to\infty} U_m = 0$.

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  • $\begingroup$ I am honoured by your commentary on my question. I am going to think a bit about your answer. Could you please elaborate a little on your remark "...since multiplication by $f$ would then be a continuous operation from $\mathcal{S}$ into $\mathcal{S}$"." ? I don't understand why this property should help me in proving uniqueness in that case. $\endgroup$ Apr 6, 2014 at 15:20
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    $\begingroup$ I was thinking of rewriting it as an integral equation: $|U(t)(\phi)| = \int_0^t U(s)(f \phi) \, ds$, then on the left take the supremum over all $\phi$ in the unit ball of one of the semi-norms $\|\cdot\|$ that defines $\mathcal S$, and bounding the right hand side by the integral of a similar quantity, and then applying Gronwall's inequality. $\endgroup$ Apr 6, 2014 at 15:34
  • $\begingroup$ I have still not tried to follow your approach. However, I have had a chat with my PhD director, and he suggested a (IMHO) very nice and clever way to prove the result. I have published it as a separate answer. My guess is that combining his argument and yours one can finally obtain a complete proof of the result, but I still haven't checked all details. $\endgroup$ Apr 7, 2014 at 15:34
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Here's a partial solution suggested to me by my thesis director. It is partial because it needs some additional assumption on $f$. Either $f$ must be real valued and each of its derivatives must be bounded above, or $f$ must be complex valued and each of its derivatives must be bounded. $^{[1]}$ In particular, this method of proof works when $f$ is the symbol of the Laplacian and also when $f$ has compact support.

This partial result could then be plugged into Stephen Montgomery-Smith's great answer to eliminate the additional assumption on $f$. I haven't still checked the details, though.

Let us consider our Cauchy problem together with a suitable dual problem: $$\tag{CP} \begin{cases} \partial_t U - f\cdot U = 0, & U \in C^1(\, [0, +\infty) \to \mathcal{S}'\, ) \\ U(0) = 0 \end{cases} $$ and $$ \tag{CP'} \begin{cases} \partial_t \phi + f\cdot \phi = 0, & t\in(0, T) \\ \phi(T)=\psi \in \mathcal{S} \end{cases} $$ Here $T > 0$ and $\psi\in \mathcal{S}$ are arbitrary. Our task is to prove that $$ \langle U(T), \psi\rangle=0. $$ Now the dual problem (CP') has the unique classical solution $$ \phi(t, \xi) = e^{( T - t )f(\xi)} \psi(\xi). $$ Our assumptions on $f$ imply that $\phi \in C^1(\,[0, +\infty)\to \mathcal{S}\, )$. We may therefore use $\phi(t)$ as a test function for $U(t)$. The resulting pairing is constant, because $$ \begin{split} \frac{d}{dt} \langle U(t), \phi(t) \rangle & = \langle f U (t), \phi(t) \rangle \\ & = \langle U(t), f\phi(t)\rangle \\ & = -\frac{d}{dt} \langle U(t), \phi(t) \rangle. \end{split} $$ We infer that $$ \langle U(T), \psi\rangle = \langle U(T), \phi(T)\rangle=\langle U(0), \phi(0)\rangle = 0. $$ Since $T$ and $\psi$ are arbitrary, we conclude that $U\equiv 0$.


$^{[1]}$ Actually, we need the slightly less stringent requirement that $e^{\lambda f}$ is a slowly increasing function for any $\lambda>0$. For more information one may look into this French Wikipedia page.

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  • $\begingroup$ From the way you had worded your problem, I had assumed you had already done it for $f(x) = |x|^2$. So I was considering cases like $f(x) = e^{|x|^2}$. Although in this latter case, even though you get uniqueness, you don't get existence. $\endgroup$ Apr 7, 2014 at 16:10

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