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Let us consider the equation: \begin{align*} (\partial_t - \Delta - b(t,x) \partial_x) u(t,x)& = f(t,x) \\ u(0,x) & = u_0 \end{align*} defined on the whole real line (so in one dimension - but this is only for simplicity) and with a possibly finite time horizon, so say on $[0,T]\times \mathbb{R}.$ Suppose that $b$ and $f$ have polynomial growth, say $$ |u_0(x)|/(1+|x|)^a + |b(t,x)|/ (1 + |x|)^a + |f(t,x)|/(1 + |x|)^a \le C $$ for some $a \ge 0.$

My question is simple: is the solution $u$ to this equation also of polynomial growth?

I am interested in any kind of approach to this question, say through semigroups, operators, Fourier analysis or simple constructions. It seems like a natural question, so I almost expect this to have a well-known answer. We can add any kind of regularity conditions on $b,f$ and $u_0.$ We may even look at the case $ u_0 = 0.$ The only real constraint is the polynomial growth of $b$ and of $f$.

My approach: I have used Schauder estimates in weighted spaces, and this guarantees that the solution is of exponential growth. Also looking at the fundamental solution to this equation has so far not delivered me any better than exponential estimates.

It might even be that the solution can't be expected to grow better than exponential.

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What about $u(t,x) = x e^{t x^2}$, which is a solution of $$\partial_t u = \Delta u + b(t,x) \partial_x u$$ with $$b(t,x) = \frac{x^3 - 6 t x - 4 t^2 x^3}{1 + 2 t x^2}$$ with initial value $u_0(x) = x$?

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  • $\begingroup$ I guess that solves the question quite simply :) $\endgroup$
    – Kore-N
    Commented Sep 21, 2017 at 10:35
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    $\begingroup$ I believe you can use the maximum principle to show that if $u_0$ and $f$ are bounded and if you impose an extra growth condition on the solution, such as $|u(t,x)| \le A \exp(B|x|^2)$, then the solution is indeed bounded. (The extra growth condition is needed to prevent exotic solutions of the heat equation that accumulate energy "at infinity"). However, this will not work without a bound on $f$: if $u_0(x)=0$, $b(t,x)=0$ and $f(t,x)=x$ for $t\in[0,1]$, then $u(1,x)=x$, and so after time $1$ you are back to the unbounded initial value scenario (I mean, $u(1+t,x)$ has unbounded initial data). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 21, 2017 at 12:13
  • $\begingroup$ (Apparently I responded to a comment that is no longer there...) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 21, 2017 at 12:14
  • $\begingroup$ Yes, you responded to a comment I cancelled, because I figured as well that $u_0 = 0$ is not sufficient (just take $v = u-x$ above and it will solve the equation with a polynomial forcing). But your comment is interesting nonetheless. $\endgroup$
    – Kore-N
    Commented Sep 21, 2017 at 12:25
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    $\begingroup$ If $|b(t,x)|\leqslant C(1+|x|)^{1-\epsilon}$, the polynomial bound for the solution is reasonable. If we ignore the Laplacian, we get a transport equation with characteristics $x(t)$ growing at most polynomially (for $\epsilon=0$ we would have exponential growth, while for $\epsilon<0$ – blowup in finite time), so the solution $u(t,x)$ would also have polynomial growth. Adding the Laplace operator should not change this scenario in an essential way. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 21, 2017 at 16:22

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