# The solutions of the heat equation from $0$ datum

Consider the initial value problem $$\partial_t u = \Delta u$$ $$u(0,x) = 0$$ for the heat equation in $$\mathbb R^n$$, where $$u: [0,T] \times \mathbb R^n \to \mathbb R$$ is a smooth solution up to the time $$T$$. Suppose there exists a large constant $$N>0$$ such that $$|u(t,x)| \le N\exp(N|x|^2t^{-1})$$ for any $$x\in \mathbb R^n$$ and $$t\in (0,T]$$.

Can we prove that $$u(t,x)$$ must be identically $$0$$?

Notice that if $$|u(t,x)| \le N\exp(N|x|^2)$$, then the question is true by Tychonoff's theorem.

Have you tried the classical Tychonoff's example, $$u(t,x) = \sum_{k = 0}^\infty \frac{g^{(k)}(t) x^{2k}}{(2k)!},$$ with $$g(t) = e^{-1/t^\alpha}$$ and $$\alpha > 1$$?

As discussed, for example, in these lecture notes, in this case $$|g^{(k)}(t)| \leqslant \frac{k!}{(\theta t)^k} e^{-1/(2 t^\alpha)} \tag{\star}$$ for some $$\theta > 0$$ (edit: see below for a proof), and hence $$|u(t, x)| \leqslant \sum_{k = 0}^\infty \frac{k! x^{2k}}{(\theta t)^k(2k)!} e^{-1 / (2 t^\alpha)} \leqslant \sum_{k = 0}^\infty \frac{x^{2k}}{(2 \theta t)^k k!} e^{-1 / (2 t^\alpha)} = e^{x^2/(2 \theta t)-1/(2t^\alpha)} \leqslant e^{x^2 / (2 \theta t)} .$$ If I am not mistaken, this (counter-)example implies a negative question to your answer.

Edit: For completeness, here is the proof of ($$\star$$). This is, of course, completely standard; I reworked it only to convince myself that there is no error in my answer.

The function $$g(t) = e^{-1/t^\alpha}$$ is holomorphic in the right complex half-plane. By Cauchy's formula (applied to the circle $$\Gamma$$ centred at $$t$$ with radius $$\theta t$$), we have $$g^{(k)}(t) = \frac{k!}{2\pi i} \int_\Gamma \frac{g(z)}{(z - t)^{k + 1}} dz = \frac{k!}{2\pi} \int_0^{2\pi} \frac{g(t + \theta t e^{i s})}{(\theta t e^{i s})^k} ds .$$ Therefore, $$|g^{(k)}(t)| \leqslant \frac{k!}{2\pi (\theta t)^k} \int_0^{2\pi} |g(t + \theta t e^{i s})| ds .$$ Choose $$\theta > 0$$ small enough, so that the image of the disk $$D(1, \theta)$$ under $$z \mapsto z^{-\alpha}$$ is contained in the half-plane $$\Re z > \tfrac{1}{2}$$. Since $$|g(z)| = e^{-\Re z^{-\alpha}}$$, we have $$|g(t + \theta t e^{i s})| = e^{-t^{-\alpha} \Re (1 + \theta e^{i s})^{-\alpha}} \leqslant e^{-t^{-\alpha} / 2} .$$ It follows that $$|g^{(k)}(t)| \leqslant \frac{k!}{(\theta t)^k} e^{-1 / (2 t^\alpha)}$$, as desired.

• Kwasnicki For the Tychonoff argument to work, you need $\alpha >1$. In fact, up to geometrical terms $g^{(k)}$ is bounded above by $k^{k(1+\frac{1}{\alpha})}$ and you have in the series a denominator as $k^{2k}$, so you need $1+\frac{1}{\alpha}<2$. A reference is doi.org/10.1017/S001309150000095X – Bazin Apr 5 '20 at 11:26
• @Bazin: Thanks for spotting the typo, and for the reference! Regarding the estimate of $g^{(k)}$, unfortunately I do not have time now to check the bound given in the lecture notes (which is left as a "good exercise"). – Mateusz Kwaśnicki Apr 5 '20 at 19:46
• @Kwasnicki the good exercise is to prove that your function $g$ belongs to the Gevrey class $G^{1+\frac{1}{\alpha}}$ (this is sharp). Checking the Faa di Bruno theorem is one way to get this, but complex analysis with Cauchy theorem would give you the same answer. – Bazin Apr 9 '20 at 12:23
• @Bazin: Thanks for the comment. I am not sure if I understand correctly, though: you seem to refer to uniform bounds for $g^{(k)}$, while here one needs an estimate that can deteriorate as $t \to 0^+$: a uniform bound for $t^k g^{(k)}(t)$. I believe both estimates follow from Faà di Bruno's formula, but, as said above, I did not attempt to check it. If you did, or if you can improve the answer in any other way, please feel free to edit it. – Mateusz Kwaśnicki Apr 9 '20 at 21:15
• @Kwasnicki you do not need anything else than $\vert g^{(k)}\vert\le C^{k+1} k^{k(1+\frac{1}{\alpha })}$ with $\alpha >1$ for the Tychonoff example to work: you have applying the heat operator a telescoping series and the convergence follows from the estimate $\vert g^{(k)}\vert\frac{x^{2k}}{(2k)!}\lesssim R^{2k}k^{k(-1+\frac{1}{\alpha})}$. – Bazin Apr 10 '20 at 21:07