An extended problem

Exercises 3.4 An extended problem

Consider the open set

\begin{equation*} \Omega = B_1(0) \cap \{x_2 \gt 0\} \subset \R^2. \end{equation*}

Using the techniques from Chapter 3, what can we say about the problem

\begin{align} \left\{ \begin{alignedat}{2} \Delta u \amp= x_2 \amp\quad\amp \ina \Omega \\ u \amp= 0 \amp\quad\amp \ona \partial \Omega \end{alignedat} \right.\tag{✶} \end{align}

for \(u \in C^2(\Omega) \cap C^1(\overline\Omega)\text{?}\)

1. Weak maximum principle.

Can you apply the weak maximum principle to (✶)? If so, what is the conclusion?

Solution.

As \(\Omega\) is bounded, \(\Delta\) is uniformly elliptic with no zeroth-order terms, and \(\Delta u = x_2 \ge 0\) in \(\Omega\text{,}\) we can indeed apply the weak maximum principle to get that

\begin{equation*} \max_{\overline \Omega} u = \max_{\partial\Omega} u = 0 \text{.} \end{equation*}

2. Uniqueness.

Can you apply our result on uniqueness to (✶)? If so, what is the conclusion?

Solution.

As \(\Omega\) is bounded and \(\Delta\) is uniformly elliptic with no zeroth-order terms, Corollary 3.5 indeed implies that \(u\) has at most one solution \(u \in C^2(\Omega) \cap C^0(\overline\Omega)\text{.}\)

3. Strong maximum principle.

Does the Strong maximum principle apply to (✶)? If so, what is the conclusion?

Solution.

As \(\Omega\) is bounded and \(\Delta\) is uniformly elliptic with no zeroth-order terms, we can indeed apply Strong maximum principle. By our earlier application of the weak maximum principle, \(\max_{\overline\Omega} u = 0\text{,}\) and so the conclusion is that either \(u \lt 0\) in \(\Omega\) or else \(u \equiv 0\text{.}\) The latter is not possible, since it would imply that \(\Delta u \equiv 0\text{,}\) contradicting (✶). We therefore deduce that \(u \lt 0\) in \(\Omega\text{.}\)

4. Hopf lemma.

At which points \(p \in \partial\Omega\) does the Hopf lemma apply to (✶)? What is the conclusion? Pick some explicit examples and write the conclusion in terms of \(\partial_1 u\) and \(\partial_2 u\text{.}\)

Solution.

As in the previous problem, we know that \(\Omega\) is connected, \(\Delta\) is uniformly elliptic with no zeroth order terms, and \(\Delta u = x_2 \ge 0\) in \(\Omega\text{.}\) Moreover, we are given that \(u \in C^2(\Omega) \cap C^1(\overline\Omega)\text{.}\) By the above problems, we know that \(u\) achieves \(\sup_\Omega u = 0\) at every point \(p \in \partial\Omega\text{,}\) and moreover that \(u\) is not constant. Therefore we can apply the Hopf lemma at every point of \(p \in \partial\Omega\) where \(\Omega\) satisfies the interior ball property, obtaining \(\partial u/\partial n \gt 0\) there. Drawing a picture of \(\Omega\text{,}\) we see that this is every point \(p \in \partial\Omega\) aside from the corner points \((\pm1,0)\text{.}\)

As an example, consider the origin \((0,0)\text{.}\) There the outward pointing normal is clearly \((0,-1)\text{,}\) and so we get

\begin{equation*} 0 \lt \frac{\partial u}{\partial n}(0,0) = (0,-1) \cdot \nabla u(0,0) = -\partial_2 u(0,0), \end{equation*}

i.e. that \(\partial_2 u(0,0) \lt 0\text{.}\)

5. Comparison.

Find an explicit function \(v\) which solves the first line of (✶), and use this and the comparison principle to prove an upper or lower bound on \(u\text{.}\) What can you say at this point about \(u(0,\frac 12)\text{?}\)

Solution.

Integrating twice with respect to \(x_2\text{,}\) we see that \(v=x_2^3/6\) is such a function. As \(x_2 \le 1\) on \(\partial \Omega\text{,}\) we see that \(v \le 1/6\) there. Thus we have \(u = 0 \ge v - 1/6\) on \(\partial\Omega\) and \(\Delta u = \Delta (v-1/6)\) in \(\Omega\text{.}\) Again, \(\Omega\) is bounded and \(L\) is uniformly elliptic with no zeroth-order coefficient, and so we can apply the comparison principle to get that \(u \ge v - 1/6\) in \(\Omega\text{.}\) In particular, \(u(0,\frac 12) \ge v(0,\frac 12) - 1/6 = -7/48\text{.}\)

This is just one example; there are many other comparison functions that could have been chosen.

6. Symmetry.

Show that if \(u(x_1,x_2)\) solves (✶) then so does its reflection \(\tilde u(x_1,x_2) = u(-x_1,x_2)\text{.}\) What does Corollary 3.5 now imply?

Solution.

Using the chain rule we can check that if \(u\) solves (✶) then so does \(\tilde u\text{.}\) Uniqueness then implies \(\tilde u \equiv u\text{,}\) i.e. that \(u\) is even in \(x_1\text{.}\)

These questions are continued further in Section 4.6.

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