An extended problem

Exercises 4.6 An extended problem

We continue with the example from Exercises 3.4. 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 4, 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 \(u \in C^2(\Omega) \cap C^0(\overline\Omega)\text{?}\)

1. Getting a harmonic function.

Find an explicit function \(v\) which solves the first line of (✶) and is also odd in \(x_2\text{.}\) What problem does the difference \(w = u-v\) solve?

Solution.

We’ve shown on Exercises 3.4 that \(v=x_2^3/6\) is such a function. The difference \(w=u-v\) then solves

\begin{align} \left\{ \begin{alignedat}{2} \Delta w \amp= 0 \amp\quad\amp \ina \Omega, \\ w \amp= -v \amp\quad\amp \ona \partial \Omega. \end{alignedat} \right.\tag{✶✶} \end{align}

2. Existence.

By applying Theorem 4.17 to the function \(w\) from the previous problem, show that (✶) has a unique solution \(u \in C^2(\Omega) \cap C^0(\overline\Omega)\text{.}\)

Solution.

Drawing a picture, we visually confirm that \(\Omega\) satisfies the exterior ball property. As it is also bounded and connected, and \(-v \in C^0(\partial\Omega)\text{,}\) Theorem 4.17 guarantees that (✶✶) has a unique solution \(w \in C^2(\Omega) \cap C^0(\overline \Omega)\text{.}\) Indeed, by Exercise 4.2.2 we have \(w \in C^\infty(\Omega)\text{.}\) Thus (✶) has a unique solution \(u = w + v \in C^\infty(\Omega) \cap C^0(\overline\Omega)\text{.}\)

3. Extension.

Argue that the function \(w\) from the previous problem satisfies the hypotheses of Exercise 4.3.1. What is the conclusion?

Solution.

Our domain \(\Omega\) is precisely the type of domain considered in Exercise 4.3.1. Moreover, \(w\) is a harmonic function with \(w=-v = -x_2^3/6 = 0\) on \(\partial \Omega \cap \{x_2 = 0\}\text{.}\) Thus we can apply the result from this exercise, and conclude that the function

\begin{align*} W(x_1,x_2) = \begin{cases} w(x_1,x_2) \amp x_2 \ge 0, \\ -w(x_1,-x_2) \amp x_2 \lt 0 \end{cases} \end{align*}

is harmonic in \(B_1(0)\text{.}\) Indeed, we have that \(W \in C^2(B_1(0)) \cap C_0(\overline{B_1(0)})\) satisfies

\begin{align} \left\{ \begin{alignedat}{2} \Delta W \amp= 0 \amp\quad\amp \ina B_1(0), \\ W \amp= -x_2^3/6 \amp\quad\amp \ona \partial B_1(0). \end{alignedat} \right.\tag{✶✶✶} \end{align}

4. Explicit formula.

The previous problem gives you a harmonic function in the ball \(B_1(0)\text{.}\) Argue using Lemma 4.7 that this harmonic function is in fact a polynomial, and use the proof of Lemma 4.7 to find an explicit formula for this polynomial. Finally, use this to find an explicit formula for the unique solution of (✶), and then check your answers to previous questions against this formula.

Solution.

Looking at (✶✶✶) we see that it is precisely the type of problem covered by Lemma 4.7. Thus the unique solution \(W\) is a polynomial, which we can for instance calculate using the technique from the proof of Lemma 4.7. We ultimately find that the corresponding \(u\) is

\begin{equation*} u = - \tfrac 18 (1-\abs x^2)x_2 \text{.} \end{equation*}

Using this explicit formula we can check, for instance, that \(u(0,\frac 12) = -3/64 \in (-7/48,0)\) and that \(\partial_2 u(0,0) = -1/8 \lt 0\text{.}\)

5. Generalising.

What happens to the preceding arguments (including the arguments in Exercises 3.4) if we replace the right hand side \(x_2\) in (✶) with, say,

\(x_2^2\text{,}\)
\(x_1^3 + x_2^3\text{,}\)
\(\sin x_2\text{,}\) or
\(\sinh x_2\text{?}\)

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