Solution to HW3 No. 12 (MTH847 Fall 2017)

Question 12 on HW3 asked

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.