SSM-Ch18 - Chap. 18 Complex Analysis and Potential Theory...

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Chap . 18 Complex Analysis and Potential Theory Complex analysis offers help in solving problems involving the Laplace equation in two variables exactly or approximately, where the latter case means that a dependence on a third variable can be neglected in the region of space considered. This relation to potential theory is the main reason of the importance of complex analysis in applied mathematics. The chapter first discusses special problems from several areas and then general properties of harmonic functions (solutions of Laplace’s equation that have continuous second partial derivatives). Sec . 18 . 1 Electrostatic Fields This section explains the advantage of working with complex potentials, whose real and imaginary parts both have a physical meaning. Problem Set 18 . 1 . Page 753 3 . Real and complex potential . Use w u u U iv u z 2 u x 2 u y 2 U 2 ixy , thus u u x 2 u y 2 , v u 2 xy . This function w u z 2 maps the axes y u 0 and x u 0 onto the real w -axis v u 0. It also maps xy u 1 onto v u 2 xy u 2. Hence it maps the given region onto the horizontal strip of width 2 between v u 0 and v u 2. The potential in this strip is u u u , v U u av U b with a and b determined from the boundary values, u u u ,0 U u b u 110 u u u ,2 U u 2 a U b u 2 a U 110 u 60, a u u 25. Hence the potential in the w -plane is u u u , v U u 110 u 25 v . This is the real part of the complex potential F u w U u 110 U 25 iw u 110 U 25 i u u U iv U u 110 u 25 v U 25 iu . In the z -plane you thus have the complex potential F u z U u 110 U 25 iz 2 u 110 u 50 xy U 25 u x 2 u y 2 U i and thus the real potential u u x , y U u 110 u 50 xy . The curves u u const are the equipotential lines (a conducting boundary is always an equipotential line!). The hyperbolas ± u x 2 u y 2 u const are the lines of force and are orthogonal to the equipotential lines. 5 . Cylinders . Example 2 tells you that u u r U u a ln r U b with the constants a and b to be determined from the boundary conditions. Use that ln 1 2 u u ln 2. Calculate u u 0.5 U u a ln 0.5 U b u u a ln 2 U b u u 110 u u 2.0 U u a ln 2.0 U b u 110. By addition, b u 0. Subtract and use b u 0 to obtain u u 2.0 U u u u 0.5 U u 2 a ln 2 u 220, hence a ln 2 u 110, a u 110/ln 2. This gives
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160 Complex Analysis Part D u u r U u 110 ln 2 ln r and F u z U u 110 ln 2 Ln z . 11 . Two source lines . The equipotential lines in Example 7 are | u z u c U / u z U c U | u k u const u k and c real U . Hence | z u c | u k | z U c |. Squaring gives | z u c | 2 u K | z U c | 2 u K u k 2 U . Writing this in terms of the real and imaginary parts and taking all the terms to the left, you obtain u x u c U 2 U y 2 u K uu x U c U 2 U y 2 U u 0. Writing out the squares gives x 2 u 2 cx U c 2 U y 2 u K u x 2 U 2 cx U c 2 U y 2 U u 0. (A) For k u 1, hence K u 1, most terms cancel, and you are left with u 4 cx u 0, hence x u 0 (because c U 0). This is the y -axis. Then | z u c | 2 u | z U c | 2 u y 2 U c 2 , | z u c |/| z U c | u 1, Ln 1 u 0. Hence this shows that the
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.

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SSM-Ch18 - Chap. 18 Complex Analysis and Potential Theory...

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