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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 = 0 or π (which corresponds to the points on the x -axis), then clearly φ = 0. Hence φ satisfies 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
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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 φ + iψ = U ◦ f + iV ◦ f = g ² ³´ µ ( 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 φ + iψ is analytic and so ψ is a harmonic conjugate of φ ....
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