Differential Equations Lecture Work Solutions 68

Differential Equations Lecture Work Solutions 68 - 7 c. u...

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7c . u ( r, 0) = 0 u ( r, π/ 2) = 0 u r (1 )=1 Use7atogetthe2ODEs Θ 0 + λ Θ=0 r 2 R 0 + rR 0 λR =0 Θ(0) = Θ ( π/ 2) = 0 λ n = π 2 ± 2 =(2 n ) 2 n =1 , 2 ,... Θ n =s in2 nθ, n =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 n = C n r 2 n + D n r 2 n ,n =1 , 2 ,... Since r 2 n blow up as r 0wehave C n =0 .Thu s u ( r, θ )= X n =1 a n r 2 n sin 2 Apply the inhomogeneous boundary condition u r ( r, θ )= X n =1 2 na n r 2 n 1 sin 2 And at r =1 u r (1 )=
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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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