# Linear transport equation with unbounded coefficients

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.

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 Then the above answer is no longer valid.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

• 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. Jul 18, 2021 at 14:47
• Indeed, your question and your comment are different things. Jul 18, 2021 at 14:51
• hope it is clearer now. Jul 18, 2021 at 15:23
• What is $B$ in the statement? Jul 18, 2021 at 16:01
• @GiorgioMetafune actually, it was supposed to be a constant matrix, but it is not necessary, as we can absorb it in $q$, thanks. Jul 18, 2021 at 16:07

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.

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