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At a recent study session my professor and peers gave an intuitive set-map illustration of how a harmonic function on a domain $D$ has a harmonic conjugate function on $D$ (see below).


The problem was posed thusly:

"Let $D = \{a \lt |z| \lt b \}$ \ $(-b , -a)$, an annulus slit along the negative real axis. [Demonstrate] that any harmonic function on $D$ has a harmonic conjugate on $D$." - Gamelin

In the image above

$\hspace{2in}$ $u: A \rightarrow \mathbb{R}$ (annulus to reals)
$\hspace{2in}$ $f: A \rightarrow S$ (annulus to rectangle)
$\hspace{2in}$ $\tilde{u}: S \rightarrow \mathbb{R}$ (rectangle to reals)
$\hspace{2in}$ $\tilde{v}: S \rightarrow \mathbb{R}$ (rectangle to reals)
$\hspace{2in}$ $v: A \rightarrow \mathbb{R}$ (annulus to reals),

where the "~" denotes complex conjugacy. What I'd like to understand is how these data can be used to piece together the existence of a harmonic conjugate within $D$.


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Is this the purdue math club? – Eric Haengel Feb 21 '13 at 21:06
You mean the image? No, this is not the Purdue Math Club. – Trancot Feb 21 '13 at 21:38
Ok, sorry it just looked a lot like it haha. – Eric Haengel Feb 21 '13 at 21:39
@RobertIsrael: OK, so for example, if $f =$ Log$(z)$, whose inverse is $f^{-1} = e^z$, then because $f^{-1}$ is differentiable which then means $u \circ f^{-1}$ is harmonic (see 2.5.3 above), then $u$ has a harmonic conjugate on $D$, correct? – Trancot Feb 22 '13 at 19:52
up vote 1 down vote accepted

Conformal maps preserve harmonicity, i.e. if $f$ is a one-to-one conformal map from $A$ onto $S$, then $u$ is a harmonic function on $A$ if and only if $u \circ f^{-1}$ is harmonic on $S$. If you already know that a harmonic function on $S$ has a harmonic conjugate, say $w$, then $w \circ f$ is a harmonic function on $A$, and this will be the harmonic conjugate of $u$.

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