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Having found the solution for the Dirichlet problem in the region $A=\{x+iy: 0\leq y\leq 1\}$ such that $u(x,0)=0$ and $u(x,1)=1$ to be $u(x,y)=y$, I am asked to find, using conformal maps, the solution in $B=\{z:r_1\leq|z| \leq r_2 \}$ such that $u(z)=0$ on the internal disc and $u(z)=1$ on the external one.

Now, I could find a conformal map from $A$ onto $B$ to be $z \rightarrow e^{i((z-i)\log r_1- z\log r_2)}$

But I think I need a map from $B$ onto $A$ instead and this one is obviously not invertible..

Once I find this conformal map I would be done as the solution wound simply be the composition of the solution in the strip and the conformal map.

EDIT1: I can find a solution quite easily which is $u(x,y)= \frac{1}{\log\frac{r_2}{r1}}\log(\frac{\sqrt{x^2+y^2}}{r_1})$ but I would like to use the conformal map method!

EDIT2: continuing on mrs's hint: having found the solution, we have that if there is a conformal map $f$ from $B$ onto $A$ then $f(x+iy)=u_1(x,y)+i\ u_2(x,y)$ where $u_2=u$ the solution we found, then we use C-R equations to work out the harmonic conjugate of $u$ and we find that $f(x,y)=\frac{1}{\log\frac{r_2}{r1}}(\frac{r_2}{r_1}\tan^{-1}\frac{x}{y}+ i\log\frac{\sqrt{x^2+y^2}}{r_1})$

EDIT3: the $f$ I found, sadly, has two problems: it is possibly not holomorphic when $y=0$ and the image of the annulus under it is only a rectangle...

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Is this from Ahlfors? I remember doing this once upon a time, and I think that's where it was from. – mixedmath Jan 1 '13 at 19:49
Hint: Find a radial function that solves the problem. – mrf Jan 1 '13 at 19:53
It is from one of my problem sheets, probably my professor took it from there! – Moritzplatz Jan 1 '13 at 19:56
@mrf I edited my question, thank you :) – Moritzplatz Jan 1 '13 at 20:09
To your edit3: the mapping does not have to be one to one. It is just necessary mapped function is unique on the annulus. – Fabian Jan 1 '13 at 21:32
up vote 2 down vote accepted

It is not important that the mapping is one to one. Being a solution of the Dirichlet problem is a local property, independent whether the mapping involves multiple sheets. Also the fact that the mapping is not holomorphic at in the image at $r=0$ is important as this is outside the region $B$. What is important however is to check whether the mapped function is unique, i.e., whether the function value on the different sheets give the same result.

The mapping $f\colon A\mapsto B$ is given by your expression $$f(z) = e^{(i z+1 )\log r_1- iz\log r_2} = r_1 \left(\frac{r_1}{r_2}\right)^{iz}.$$ As you have noticed the inverse function is multivalued $$f^{-1}(w)= i\frac{ \log(w/r_1)}{\log(r_2/r_1)} $$ but this will not hinder us from mapping the solution $u$ from $A$ to $B$.

Starting with $u(x,y)=y$ defined on $A$, we map it via $f$ to $B$ to find the solution $v= \mathop{\rm Im} f^{-1}(z)$ on $B$; more explicitly $$v(r=|w|) = \frac{ \log( r/r_1)}{\log (r_2/r_1)}.$$ Note that even though the inverse was initially multivalued the solution $v$ is uniquely defined and in fact independent of $\text{arg} w$!

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thank you very much, I was scared by the multivaluedness of the inverse... so the point is that even if it is multivalued, the image of B is A? – Moritzplatz Jan 1 '13 at 22:27
@Moritzplatz: yes. but in fact it covers $A$ multiple times. So you have to be careful that the answer does not depend on which of the cover (=sheets) you are. This is equivalent to asking that your answer is analytic on $A$ without any branch cut. – Fabian Jan 1 '13 at 22:28
oh thanks i get it now! – Moritzplatz Jan 1 '13 at 22:47
You are welcome. – Fabian Jan 1 '13 at 22:48

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