Differential Equations Lecture Work Solutions 67

Differential Equations Lecture Work Solutions 67 - 7 b. u...

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7b . u θ ( r, 0) = 0 u θ ( r, π/ 2) = 0 u r (1 )= g ( θ ) Use7atogetthe2ODEs Θ 0 + λ Θ=0 r 2 R 0 + rR 0 λR =0 Θ 0 (0) = Θ 0 ( π/ 2) = 0 λ n = π 2 ± 2 =(2 n ) 2 n =0 , 1 , 2 ,... Θ n =co s2 nθ, n =0 , 1 , 2 ,... Now substitute the eigenvalues in the R equation r 2 R 0 + rR 0 (2 n ) 2 R =0 The solution is R 0 = C 0 ln r + D 0 ,n =0 R n = C n r 2 n + D n r 2 n ,n =1 , 2 ,... Since ln r and r 2 n blow up as r 0wehave C 0 = C n =0 .Thu s u ( r, θ )= a 0 + X n =1 a n r 2 n cos 2 Apply the inhomogeneous boundary condition u r ( r, θ )= X n =1 2 na n r 2 n 1 cos 2 And at r =1 u r (1 )= X n =1 2 na n cos 2 = g ( θ
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