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Math Help - Harmonic Functions and Conformal Mappings

  1. #1
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    Harmonic Functions and Conformal Mappings

    Hello there,

    I am trying to find harmonic functions with set boundary conditions for closed regions by using conformal mapping.

    For example, I am looking for a harmonic function u(x,y) that is defined and continuous on D − {0}, where D {0 <=arg z <=3pi/2}, (3/4 plane) such that
    u(x, 0) = 1 for x > 0, and u(0, y) = 0 for y < 0.

    So, I decided to perform a conformal mapping z--> 2/3log(z), which will map the 3/4 plane to an infinite strip. How can I utilize this information to find the desired harmonic function? I thought that if I could somehow apply the boundary conditions given and find them on the infinite strip, I could then find a harmonic function and then push back the harmonic function to the 3/4 plane. However, this is all talk and I am not sure how to go about accomplishing this.

    I appreciate any help that can be provided. Thanks!
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  2. #2
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    Quote Originally Posted by ComplexXavier View Post
    I am trying to find harmonic functions with set boundary conditions for closed regions by using conformal mapping.

    For example, I am looking for a harmonic function u(x,y) that is defined and continuous on D − {0}, where D {0 <= argz <= 3pi/2}, (3/4 plane) such that u(x, 0) = 1 for x > 0, and u(0, y) = 0 for y < 0.

    So, I decided to perform a conformal mapping z--> 2/3log(z), which will map the 3/4 plane to an infinite strip. How can I utilize this information to find the desired harmonic function? I thought that if I could somehow apply the boundary conditions given and find them on the infinite strip, I could then find a harmonic function and then push back the harmonic function to the 3/4 plane.
    You are thinking along the right lines there. Suppose that z\mapsto f(z) is a conformal map taking D\setminus\{0\} to a strip 0\leqslant \text{Im }z\leqslant k. Next, you want to find a harmonic function taking the value 0 everywhere on the lower boundary of the strip, and the value 1 everywhere on the upper boundary. The easiest way to do this is to use the fact that the real part of an analytic function is harmonic. The analytic function g(z) = -iz/k has exactly the properties that you want. So define u(x,y) = \text{Re }g(f(x+iy)).
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