Solve Dirichlet Problem: Find Function Harmonic in Right Half-Plane

[sigmund]

I have the following problem [From: E. B. Saff & A. D. Snider: Fundamentals of Complex Analysis -- with Applications to Engineering and Science, pp. 375-376]:

Consider the problem of finding a function \phi that is harmonic in the right half-plane and takes the values \phi(0,y)=y/\left(1+y^2\right) on the imaginary axis.
According to the text the mappings (7)* and (8)** provide a correspondence between the right half-plane and the unit disk. (Of course, one should interchange the roles of z and w in the formulas). Thus the w-plane inherits from \phi(z) a function \psi(w) harmonic in the unit disk. Show that the values of \psi(w) on the unit circle w=e^{i\theta} must be given by
<br /> \psi\left(e^{i\theta}\right)=\frac{\sin\theta}{2}~~(1)<br />

*w=f(z)=\frac{1+z}{1-z}.
**z=\frac{w-1}{w+1}.I know that (*) maps the unit circle onto the the imaginary axis and its interior onto the right half-plane. Furthermore, (**) maps the imaginary axis onto the unit circle, because (**) is the inverse of (*).
Now I have to find a function \psi(w), whose values on the unit circle are given by (1). I am stuck here. I know that I have to use the definition of \psi: \psi=\phi\circ f^{-1}, but I am not sure how to apply it for the actual problem.
Hopefully some of you could give me some hints. I do not ask for a solution to the problem, because that will not help me in future problems of this problem. The important thing is to understand the principle behind the solution procedure.

 

[CarlB]

Now I have to find a function , whose values on the unit circle are given by (1). I am stuck here. I know that I have to use the definition of : , but I am not sure how to apply it for the actual problem.

Ooooooo! This is a good one!

Here's the direction I'd go in. BUT it doesn't look easy.

f(1,\theta) = \sin(\theta)/2

\frac{df}{d\theta} = \cos(\theta)/2

\left( \frac{d}{d\theta} + i\right) f(z) = z/2

From there, you have to fix the LHS to make it analytic. Of course you only have specified one of the derivatives with respect to r and theta and this is the freedom you need.

Good luck. Do report how it goes.

Carl

 

[sigmund]

CarlB,
Thank you for answering. I did not figure out the problem myself, but my teacher has provided a solution for this problem. See page 3 of the following, where the problem has been solved: http://www2.mat.dtu.dk/education/01141/S/7homework05.pdf .
Actually, I did not really figure out, what you did. Could you therefore, in detail, explain it once more? It would be nice if you could comment on my teachers solution too.

 

[CarlB]

Looking back, I see that I should have used something other than "f" as you were already using it and this is more than confusing. I should have used \psi.
Let me try again...

given \psi(r,\theta) we have that

\psi(1,\theta) = \sin(\theta)/2 and so

\frac{d\psi(1,\theta)}{d\theta} = \cos(\theta)/2

You can combine these together to give an analytic thing on the right hand side:

(\frac{d}{d\theta} + i)\psi = \cos(\theta)/2 + i\sin(\theta)/2 = z/2.

Now \psi is an analytic function. You know the derivative of psi in the theta direction but you don't know the derivative of psi in the r direction. That's okay cause analytic functions have a relation between their derivatives in r and theta (or between their derivatives in x and y when writing z=x+iy). So you should be able to convert the problem into an analytic differential equation and solve that. But it gets sticky.

The instructor used a nice trick. I suggest doing it his way.

Carl