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

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In summary: CarlB,Thank you for your detailed explanation. I understand what you did now and will try to do it that way.
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sigmund
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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 [itex]\phi[/itex] that is harmonic in the right half-plane and takes the values [itex]\phi(0,y)=y/\left(1+y^2\right)[/itex] 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 [itex]\phi(z)[/itex] a function [itex]\psi(w)[/itex] harmonic in the unit disk. Show that the values of [itex]\psi(w)[/itex] on the unit circle [itex]w=e^{i\theta}[/itex] must be given by
[tex]
\psi\left(e^{i\theta}\right)=\frac{\sin\theta}{2}~~(1)
[/tex]

*[itex]w=f(z)=\frac{1+z}{1-z}[/itex].
**[itex]z=\frac{w-1}{w+1}[/itex].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 [itex]\psi(w)[/itex], whose values on the unit circle are given by (1). I am stuck here. I know that I have to use the definition of [itex]\psi[/itex]: [itex]\psi=\phi\circ f^{-1}[/itex], 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.
 
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sigmund said:
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.

[tex]f(1,\theta) = \sin(\theta)/2[/tex]

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

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

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
 
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  • #3
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 [Broken].
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.
 
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  • #4
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 [tex]\psi(r,\theta)[/tex] we have that

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

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

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

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

Now [tex]\psi[/tex] 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
 
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1. What is the Dirichlet problem and why is it important in mathematics?

The Dirichlet problem is a mathematical problem that involves finding a function that is harmonic in a given region and satisfies specified boundary conditions. It is important because it has applications in many areas of mathematics and physics, such as potential theory, differential equations, and complex analysis.

2. What is a harmonic function?

A harmonic function is a function that satisfies Laplace's equation, which states that the sum of the second-order partial derivatives of the function is equal to zero. In simpler terms, a harmonic function is one that is smooth and has no sources or sinks in its domain.

3. What is the boundary condition in the Dirichlet problem?

The boundary condition in the Dirichlet problem is a set of conditions that are imposed on the boundary of the given region. These conditions specify the values of the function at the boundary, and the goal is to find a function that satisfies both the boundary condition and Laplace's equation.

4. How is the Dirichlet problem solved?

The Dirichlet problem can be solved using a variety of techniques, including separation of variables, conformal mapping, and complex analysis methods. These methods involve breaking down the problem into simpler equations and then solving them to find the function that satisfies the given conditions.

5. What is the significance of solving the Dirichlet problem in the right half-plane?

Solving the Dirichlet problem in the right half-plane is significant because it allows us to find a harmonic function that is defined in a complex plane with a boundary on the real axis. This is useful in many applications, such as modeling heat flow and electrostatics in a half-plane domain.

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