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Differential Equations Lecture Work Solutions 69

# Differential Equations Lecture Work Solutions 69 - 8a 1 1...

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8a. u rr + 1 r u r + 1 r 2 u θθ =0 u ( a, θ )= f ( θ ) u ( b, θ )= g ( θ ) r 2 R 0 + rR 0 λR =0 Θ 0 + λ Θ=0 Θ(0) = Θ(2 π ) Θ 0 (0) = Θ 0 (2 π ) The eigenvalues and eigenfunctions can be found in the summary of chapter 4 λ 0 =0 Θ 0 =1 fo r n =0 λ n = n 2 Θ n =co s and sin for n =1 , 2 ,... Use these eigenvalues in the R equation and we get the following solutions: R 0 = A 0 + B 0 ln rn =0 R n = A n r n + B n r n n =1 , 2 ,... Since r = 0 is outside the domain and r is Fnite, we have no reason to throw away any of the 4 parameters A 0 ,A n ,B 0 ,B n . Thus the solution u ( r, θ )=( A 0 + B 0 ln r ) | {z } R 0 · 1 |{z} Θ 0 · a 0 + X n =1 ( A n r n + B n r n ) | {z } R n ( a n cos + b n sin ) |
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