Chem Differential Eq HW Solutions Fall 2011 164

Chem Differential Eq HW Solutions Fall 2011 164 - f maps a...

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164 Chapter 12 Green’s Functions and Conformal Mappings boundary. On the outer circular boundary, r = e , and we have φ ( e, θ )= 400 π ± n =0 sinh(2 n +1) (2 n + 1) sinh(2 n +1) sin[(2 n +1) θ ] = 400 π ± n =0 sin[(2 n +1) θ ] (2 n +1) , which is the same series as we found previously; and thus it equals 100 for 0 <θ<π . Now if θ =0o r π (which corresponds to the points on the x -axis), then clearly φ = 0. Hence φ satisFes the boundary conditions, as expected. 25 We have f ( z )= z + z 0 = x + iy + x 0 + iy 0 = x + x 0 + i ( y + y 0 ). Thus
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Unformatted text preview: f maps a point ( x, y ) to the point ( x + x , y + y ). Thus f is a translation by ( x , y ). 37 We have + i = U f + iV f = g ( U + iV ) f. Since V is a harmonic conjugate of U , U + iV is analytic. Thus g f is analytic, being the composition of two analytic functions. Hence + i is analytic and so is a harmonic conjugate of ....
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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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