Chem Differential Eq HW Solutions Fall 2011 163

Chem Differential Eq HW Solutions Fall 2011 163 - Section...

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Section 12.6 Solving Dirichlet Problems with Conformal Mappings 163 u 4 ( x, y ), where u 4 ( x, y )= ± n =1 D n sinh nx sin ny ; D n = 2 π sinh n ² π 0 100 sin ny dy = - 200 sinh n (cos - 1) = 200 sinh n (1 - ( - 1) n ); u 3 ( x, y ± n =1 C n sinh[ n (1 - x )] sin ny ; C n = 2 π sinh n ² π 0 100 sin ny dy = 200 sinh n (1 - ( - 1) n ) . Thus (back to the variables u and v ) U ( u, v u 3 + u 4 = ± n =1 200 sinh n (1 - ( - 1) n ) sinh nu sin nv + ± n =1 200 sinh n (1 - ( - 1) n ) sinh[ n (1 - u )] sin nv = 200 π ± n =1 sin nv n sinh n (1 - ( - 1) n ) ³ sinh nu + sinh[ n (1 - u )] ´ = 400 π ± n =0 sinh[(2 n +1) u ] + sinh[(2 n + 1)(1 - u )] (2 n + 1) sinh(2 n sin[(2 n v ] . To get the solution in the xy -plane, substitute u = 1 2 ln( x 2 + y 2 ) and v = cot - 1 ( y x ) . The solution takes on a neater form if we use polar coordinates and substitute x 2 + y 2 = r 2 and θ = cot - 1 ( y x ) . Then φ ( x, y U ( 1 2 ln( x 2 + y 2 ) , cot - 1 µ y x U (ln r, θ ) = 400 π ± n =0 sinh[(2 n + 1) ln
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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