Question about one step in the proof of the weak maximum principle for the heat equation

Question

I'm confused about one step in the proof of the weak maximum principle for the heat equation in McOwen.

Theorem (Weak Maximum Principle): Let $u\in C^{2;1}(U)\cap C(\overline{U})$ satisfy $\Delta u \geq u_t$ in $U$. Then $u$ achieves its maximum on the parabolic boundary of $u$:

$$\max_{(x,t)\in~\overline{U}}u(x,t)=\max_{(x,t)\in~\Gamma} u(x,t)$$

Sketch of first part of proof:

Assume $u_t < \Delta u$.

By contradiction, suppose that u has a maximum point at $(x,\tau)$ for some $0<\tau<T$ (in the interior) and $x\in\Omega$.

Then, at $(x,\tau)$,

$$u_t(x,\tau)\geq 0$$

since $\tau$ is a maximum. By the second derivative test from calculus,

$$\Delta u(x,\tau) \leq 0$$

So this means that $u_t \geq \Delta u$ in direct contradiction to our hypothesis that $u_t < \Delta u$. Therefore,

$$\max_{(x,t)\in~\overline{U}}u(x,t)=\max_{(x,t)\in~\Gamma} u(x,t)$$

Question: Why can we conclude that $u_t(x,\tau)\geq 0$ at $(x,\tau)$? Since the point $(x,\tau)$ is a maximum, shouldn't it be that $u_t(x,\tau)= 0$? I may be forgetting something that I learned in calculus.

Answer 1

I think your sketch has hidden the problem. You do the proof for sets of the form $\Omega \times[0,T')$, $T'<T$. Being a continuous function, $u$ certainly achieves some maximum on $\bar\Omega \times [0,T']$, and its possible that the maximum is achieved at time $T'$ (which is not necessarily in the parabolic boundary of $\Omega \times[0,T')$). In this case you get an inequality on the derivative by $$u(x,\tau)- u(x,\tau-\epsilon)\ge 0 \implies u_t(x,\tau)= \lim_{\epsilon\to 0}\frac{u(x,\tau)- u(x,\tau-\epsilon)}\epsilon\ge 0 $$

I previously said that $u$ would be non-increasing at the maximum, but I don't think this is true for an arbitrary $C^1$ function, and its not needed for the proof.