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.


1 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.

  • $\begingroup$ I think I understand. A continuous function on a bounded interval must achieve a maximum. In this case, the maximum is achieved at $(x,\tau)$, so it must be that $u$ is increasing with respect to $t$ at the point $(x,\tau)$. I'm curious if there is a specific theorem in calculus which guarantees that $u_t(x,\tau)\geq 0$. $\endgroup$
    – Axion004
    May 21, 2019 at 0:01
  • $\begingroup$ I understand that 100%. I understand your answer about 75% - I was trying to find something from calculus that would show that $u$ is non-decreasing in $t$ at $(x,\tau)$. $\endgroup$
    – Axion004
    May 21, 2019 at 16:22
  • $\begingroup$ @Axion004 sorry, I should have been more careful. That's probably not true but currently taking me too much effort to produce a correct counterexample. Its not needed for the proof, though. $\endgroup$ May 21, 2019 at 18:42
  • $\begingroup$ That makes sense (even without the counterexample). Thanks for editing your response. $\endgroup$
    – Axion004
    May 21, 2019 at 19:08
  • $\begingroup$ You need to have it wobble like a (smoothed) zig-zag, where the gradients go to 0 but theres infinitely many minima and maxima as you approach $\tau$. You can write a piecewise linear function that "basically" does this, and then modify it to be $C^1$...but its a pain to write it out properly. $\endgroup$ May 21, 2019 at 19:17

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