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

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 $$0<\tau (in the interior) and $$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.

I think your sketch has hidden the problem. You do the proof for sets of the form $$\Omega \times[0,T')$$, $$T'. Being a continuous function, $$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.
• 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 $u_t(x,\tau)\geq 0$. – Axion004 May 21 at 0:01
• 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 $(x,\tau)$. – Axion004 May 21 at 16:22
• 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. – Calvin Khor May 21 at 19:17