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# This amounts to two conditions condition 1 ad bc 0

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Unformatted text preview: ) Use the result of part (a) to solve some-dimensional versions: z Solution (a) Solving it directly would be a nightmare—in fact none the usual methods would be at all tractable. Therefore, we use the appendix to transform this region into a simpler one, and we find that the map w = z + 1/z maps this into the upper-half place taking the boundary of the above region onto the x-axis. and gives us the following region in the wplane: u=a u=b Now, we can solve the Dirichlet problem for the w-region: It is radially symmetric, and Dirichlet's problem in radial coordinates is: 1 1 Ô2u = urr + ur + 2 uøø = 0 r r 23 So, any linear function in ø will work, like u = b + (a-b)ø/π. In complex notation, this is U(w) = b + (a-b)Arg(w)/π Now notice that Arg(w) is the imaginary part of the analytic function f(z) = Ln z (which is another reason that it is satisfies Laplace’s equation). So, let us take (a-b) F(w) = b + Ln(w) π Since w = z + 1/z, we have (a-b) F(z) = b + Ln(z + 1/z) π its imaginary part is a function of x and y that satisfies the original equation. (b) If u on the boundary is independent of z, then the same solution...
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## This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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