# Solution to HW3 No. 12 (MTH847 Fall 2017)

Let $$u\in C^2((0,T)\times \Omega)\cap C^0(\overline{[0,T]\times \Omega})$$, where $$\Omega$$ is open and bounded, and suppose $$u$$ satisfies $$\partial_t u - \triangle u \leq -\epsilon < 0$$ for some constant $$\epsilon > 0$$.

• Prove that $$u$$ cannot attain a local maximum on the set $$(0,T] \times \Omega$$.
• Prove, without using the mean value property, that if $$v \in C^2((0,T)\times \Omega)\cap C^0(\overline{[0,T]\times \Omega})$$ solves the heat equation, then $$v$$ satisfies the (weak) maximum principle.
(Hint: apply the first part to the function $$u = v - \epsilon t$$.)

### Solution to the first part

The first step is to show that for functions $$u$$ under the assumption, any point $$(t_0, x_0) \in (0,T)\times \Omega$$ cannot be a local maximum. We argue by contradiction: suppose $$(t_0, x_0)$$ is a local maximum, then we know that $$\partial_t u(t_0,x_0) = 0$$ and $$\nabla^2 u(t_0,x_0)$$ is a negative semi-definite matrix. This implies necessarily $$\partial_t u(t_0,x_0) - \triangle u(t_0,x_0) \geq 0$$, which contradicts the assertion that the same quantity is strictly negative.

A similar argument holds if $$u \in C^2((0,T] \times \Omega)$$, except that if the local maximum is attained at $$t = T$$ instead of having $$\partial_t u = 0$$ there, we have $$\partial_t u \geq 0$$ there.

(This second step is a modification of an argument shown to me by Connor Mooney.) Next, suppose the local maximum is achieved at $$(T,x_0)$$ for some $$x_0 \in \Omega$$. By possibly shrinking the domain $$\Omega$$ and also the time interval, we can assume without loss of generality that in fact $$(T,x_0)$$ achieves the global maximum. Consider the function $w(t,x) = u(t,x) - \frac{\epsilon}{8} (|x - x_0|^2) + \frac{\epsilon}{4} (t - T) .$ We have that $\partial_t w - \triangle w = \partial_t u - \triangle u + \frac{\epsilon}{2} \leq - \frac{\epsilon}{2}.$ On the other hand, we also have that $w |_{[0,T]\times \partial\Omega} < u(T,x_0), \qquad w|_{{0} \times \Omega} \leq u(T,x_0) - \frac{T\epsilon}{4}.$ This means that by continuity we can restrict to the set $$[0,T-\delta]$$ and have that the maximum of $$w$$ is attained away from the parabolic boundary. Now, since $$w \in C^2((0,T-\delta] \times \Omega)$$, by the derivative argument (as in the first step) we obtain a contradiction.

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