# Proof of weak maximum principle for heat equation

I am having trouble with the last step of the proof of the weak maximum principle for the heat equation, as found in

th. 2.4 pp. 36-38.

Here is a sketch of the proof.

Let $w_t - D\Delta w = q \leq 0$ on the space-time cylinder $Q_T$ with parabolic boundary $\partial_p Q_T$ and let $0 < \varepsilon < T$.

Defining the auxiliary function $u = w - \varepsilon t \leq w$, we have $$u_t - D\Delta u = q-\varepsilon < 0$$ and we prove that $$\max_{\overline{Q}_{T-\varepsilon}} u = \max_{\partial_p Q_{T-\varepsilon}} u$$ by a calculus argument: the Hessian is negative semidefinite and the time derivative is non-negative at the maximum, leading to a contradiction.

Moreover we prove that $$\max_{\overline{Q}_{T-\varepsilon}} w \leq \max_{\partial_p Q_T} w +\varepsilon T$$ essentially by subset properties w.r.t. maxima and the definition of $u$.

So far, so good.

Here comes the tricky part, at least for me. It goes

Since $w$ is continuous in $\overline{Q}_{T-\varepsilon}$ we deduce that $$\max_{\overline{Q}_{T-\varepsilon}} w \to \max_{\overline{Q}_T} w \quad\text{as}\; \varepsilon\to 0$$

therefore in the limit above we get $$\max_{\overline{Q}_T} w \leq \max_{\partial_p Q_T} w$$ which concludes the proof since the opposite inequality is true for subsets.

• $w$ is not only continuous, but also uniformly continuous on the compact $\overline{Q}_T$ and all its subsets, by Heine-Cantor theorem, but I cannot figure out how to use this property with the maxima.
• I have also explored the possibility that they meant 'uniformly convergent' instead of 'continuous', but this seems only applicable to other proof strategies, in which only $\overline{Q}_T$ and not $\overline{Q}_{T-\varepsilon}$ is used, like for instance in Evans (1998) ch. 7 th. 8. For that line of reasoning, I am having trouble applying the triangle inequality to prove that $$\max_{\overline{Q}_T} u \to \max_{\overline{Q}_T} w \quad\text{as}\; \varepsilon\to 0$$ although this is slightly off topic.

• For $\varepsilon\to 0$ the sets $\overline{Q}_{T-\varepsilon}$ are increasing, therefore we can build a corresponding sequence of maxima which is increasing and so converges to its supremum over $\varepsilon$, but here again how can I prove that $$\sup_{\varepsilon>0} \max_{\overline{Q}_{T-\varepsilon}} w = \max_{\overline{Q}_T} w$$

Any help will be greatly appreciated.

• I have a question about the end of the proof by Salsa, could you help? Jun 11, 2021 at 19:58
• @soundwave I am no expert, but if I can help I will. Feel free to ask. Jun 17, 2021 at 18:37
• Thanks for reply, before the "My comments" section you wrote "which concludes the proof since the opposite inequality is true for subsets.", but I don't understand what it means, could you elaborate more? Many thanks (in Salsa's book it is just written "which concludes the proof" without any further explanation) Jun 19, 2021 at 6:56
• Another question: why is the time derivative $u_t$ non-negative at the maximum? Jun 19, 2021 at 23:17
• Suppose $\max_{\overline {Q}_{T-\varepsilon}} u=u(x_0,t_0)\notin \partial_p Q_{T-\varepsilon}$. If $t_0<T-\varepsilon$ then $u_t(x_0,t_0)=0$ because it is a maximum on an internal point. If $t_0=T-\varepsilon$ then $u_t(x_0,t_0)\geq 0$ because the maximum is on the boundary and $u$ is non decreasing w.r.t. time. So $u_t(x_0,t_0)\geq 0$ Jun 23, 2021 at 19:39

$\def\Q#1{\bar Q_{#1}}\def\M#1{\max_{\Q{#1}}w}$Compactness and continuity makes this fairly easy. We know that the maximum over $\Q T$ is attained at some point $(x_0,t_0)$. If $t_0 < T$ then $\M {T - \epsilon} = \M{T}$ for all $\epsilon \in [0,T-t_0]$, so we are done; so we just need to consider the case $t_0 = T$.
Given an arbitrary $\epsilon > 0$, the continuity of $w$ provides a $\delta > 0$ such that $|w(x,t) - \M T| < \epsilon$ whenever $|(x,t) - (x_0,T)| < \delta$. Thus if we take a time $t > T - \delta$ we can conclude $$\M t \ge w(x_0,t) >\M T - \epsilon.$$ On the other hand we have $\M t \le \M T$ since $\Q t \subset \Q T$; so we have shown that $\M t \to \M T$ as $t \nearrow T$.
• Thanks, the second part of your proof is clear; re the first part, I do not understand why for $t_0<T$ the maxima are equal. Is it because the corresponding sets are increasing w.r.t. time? Aug 18, 2017 at 15:28
• @GiovanniMariani: If the maximum over $[0,T]$ is obtained at $t_0 < T$, then of course the maximum over a smaller $[0,t]$ (with $t \ge t_0$) will again be obtained at the same point. Aug 19, 2017 at 5:31
• @AntonyCarapetis I see, it comes from the fact that $(x_0,t_0)\in \overline{Q}_{t_0}$ then $\max_{\overline{Q}_T} w=w(x_0,t_0)\leq \max_{\overline{Q}_{t_0}} w \leq \max_{\overline{Q}_T} w$ Aug 19, 2017 at 9:39