Analysis Problem of the Day 81

Today’s problem appeared as Problem 8 on the UCLA Spring 2012 Analysis Qual:

Problem 8. Let $\Omega$ be the subset of the complex plane given by

$\{(x,y) \in \mathbb{R}^2: xy<1, x>0,y>0\}.$

Give an example of an unbounded harmonic function on $\Omega$ that extends continuously to $\partial \Omega$ and vanishes there.

Solution: The region $\Omega$ is bounded by the sets $xy=1 (x,y>0), x=0 (y \geq 0),$ and $y=0 (x \geq 0),$ and is therefore quite complicated to work with, so our goal is to see if we can make it simpler by means of conformal mappings (which preserve harmonic functions). Recall that the map $\phi(z)=z^2$ maps $\phi(x+yi)=x^2-y^2+(2xy)i,$ i.e.

$\phi((x,y)) = (x^2-y^2,2xy)=(w_1,w_2):=w.$

Thus the region $\Gamma$ such that $\phi(\Omega) = \Gamma$ is bounded by the lines $\text{Im } w = 2, \text{Im } w = 0 (\text{Re }w \geq 0),$ and $\text{Im }w=0 (\text{Re }w \leq 0),$ and since $x=y=\frac{\sqrt2}{2}$ maps to $(0,1) \in \Gamma,$ it follows that $\Gamma$ is the horizontal strip $0<\text{Im }w<2.$

Our goal now is to find a harmonic function that vanishes on the edges of the strip $\text{Im }w=0$ and $\text{Im }w=2,$ which reminds us of a periodic function. In particular, let’s suppose the function takes the form $g(w_1,w_2) = f(w_1) \sin (\frac12 \pi w_2),$ since $\sin(0)=\sin(\pi)=0.$ For $f$ to be harmonic, one must have $(-\frac14 \pi^2 f(w_1) + f''(w_1))\sin(\frac12 \pi w_2)=0$ on the interior of $\Gamma,$ i.e. $\frac14 \pi^2 f = f''.$ It is easy to then check that $f(w_1) = e^{\frac12 \pi w_1}$ suffices, so that $g(w_1,w_2) = e^{\frac12 \pi w_1} \sin(\frac12 \pi w_2) = \text{Im } e^{\frac12 \pi w}.$ Thus, the desired harmonic function is $\text{Im } (e^{\frac12 \pi w}) \circ \phi = \text{Im }e^{\frac12 \pi z^2},$ or more explicitly,

$u(x,y) = e^{\frac12 \pi (x^2-y^2)} \sin(\pi xy).$

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