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Consider the PDE

$$\partial_t f(x,t) = \langle q(x), \nabla \rangle f(t,x) + p(x),$$

with Schwartz initial data $f(0,x) = f_0(x) \in \mathscr S(\mathbb R^n).$

I am wondering then if $q$ and all its derivatives are polynomially bounded and $p$ is Schwartz, too:

Does there exist a solution to this equation that decays faster than any polynomial in space $x$ at any fixed time $t>0$?

This sounds plausible to me, but I am not sure how one argues for such an equation. I assume it must be a classical question.

As there was apparently some confusion about the meaning of this question, let me ask it again:

Fix a time $t>0$, then as a function of $x$, does the solution decay faster than any polynomial? This seems to be true in your case for example, as it is just a translation of a Schwartz function.

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2 Answers 2

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No. If e.g. $n=1$, $p=0$, and $Bq(x)=1$ for all $x$, then $f(t,x)=f_0(t+x)$, which does not decay along the lines $\{(t,x)\colon t+x=c\}$ for real $c$.


The OP has changed the question, now looking for decay only in $x$, faster than any polynomial, for each $t>0$. Then the above answer is no longer valid.

However, then the answer is still no, in general; here, we just need to change the space variable. E.g., let $n=1$ and $Bq(x)=x^2+1$ for all $x$. Then \begin{equation} f(t,x)=f_0(\tan(t+\tan^{-1}x)), \end{equation} which is not even defined at any point $(t,x)$ such that $t+\tan^{-1}x=\pi/2$. For each $t\notin\pi\mathbb Z$, the solution $f$ will explode to $\pm\infty$ at all the points of the form $x=(-1)^k\cot t$ for $k\in\mathbb Z$.

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  • $\begingroup$ sorry, perhaps the question was not clear enough: What is meant is: Fix a time $t>0$, then as a function of $x$, does the solution decay faster than any polynomial? This seems to be true in your case for example, as it is just a translation of a Schwartz function. $\endgroup$ Jul 18, 2021 at 14:47
  • $\begingroup$ Indeed, your question and your comment are different things. $\endgroup$ Jul 18, 2021 at 14:51
  • $\begingroup$ hope it is clearer now. $\endgroup$ Jul 18, 2021 at 15:23
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    $\begingroup$ What is $B$ in the statement? $\endgroup$ Jul 18, 2021 at 16:01
  • $\begingroup$ @GiorgioMetafune actually, it was supposed to be a constant matrix, but it is not necessary, as we can absorb it in $q$, thanks. $\endgroup$ Jul 18, 2021 at 16:07
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No. Consider the case $p=0$. In that case, solutions are constant along characteristics. But polynomial growth does not preclude characteristics from diverging to infinity in finite time.

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  • $\begingroup$ thank you. What could be a natural assumption on $q$ such that the statement in the question would be true? $\endgroup$ Jul 18, 2021 at 17:21

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