Unformatted text preview: uaxis, since e x + i π =e x < 0, as follows: f ( x + iπ ) =e x ≤ 1 if x ≥ 0 and1 < f ( x + iπ ) =e x < 0 if x < 0. With these observations, we see that the transformed problem in the uvplane is ∇ 2 U = 0 with boundary values on the uaxis given by U ( u, 0) = 100 if1 < u < 1 and 0 otherwise. The solution in the uvplane follows from Example 5, Section 12.5. We have U ( u, v ) = 100 π ± cot1 ² u1 v ³cot1 ² u + 1 v ³¶ . The solution in the xyplane is φ ( x, y ) = U ◦ f ( z ). To Fnd the formula in terms of ( x, y ), we write z = x + iy , ( u, v ) = f ( z ) = e z = e x cos y + ie x sin y. Thus u = e x cos y and v = e x sin y and so φ ( x, y ) = U ◦ f ( z ) = U ( e x cos y, e x sin y ) = 100 π ± cot1 ² e x cos y1 e x sin y ³cot1 ² e x cos y + 1 e x sin y ³¶ ....
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 Fall '11
 StuartChalk
 Cos, Upper halfplane, cot−1, ParametricPlot Evaluate

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