Differential Equations Lecture Work Solutions 63

# Differential Equations Lecture Work Solutions 63 - 6 a. urr...

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6a. u rr + 1 r u r + 1 r 2 u θθ = 0 outside circle u ( a, θ )=ln2+4co s3 θ u ( r, θ )= R ( r )Θ( θ ) r 2 R 0 + rR 0 λR =0 Θ 0 + λ Θ=0 Θ(0) = Θ (2 π ) λ =0 R 0 = α 0 + β 0 ln r Θ 0 (0) = Θ 0 (2 π ) λ = n 2 R n = α n r n + β n r n Since we are solving λ< 0 trivial solution outside the circle λ =0 Θ 0 ( θ )=1 ln r →∞ as r →∞ λ> 0 λ n = n 2 r n →∞ as r →∞ Θ n = A n cos + B n sin thus R 0 = α 0 R n = β n r n u ( r, θ )= a 0 α 0 · 1 | {z } = a 0 / 2 + X n =1 a n ( A n cos + B n sin ) β n r n u ( r, θ )= a 0 / 2+ X n =1 ( a n cos + b n sin ) r n Use the boundary condition: u (
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## This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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