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.



Linear transport equation with unbounded coefficients
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 $c$.




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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 Then the above answer is no longer valid.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 $k\in\mathbb Z$.


