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ps6sol

# ps6sol - ACM 95/100c Problem Set 6 Solutions Lei Zhang May...

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ACM 95/100c Problem Set 6 Solutions Lei Zhang May 7, 2006 Each problem is worth 10 points - each part of a multi-part problem is weighted equally. Problem 1 (Collaboration allowed) Solve (using an appropriate two dimensional eigenfunction expansion) the Poisson equation 2 u = r cos(2 θ ) inside a circle of radius a subject to each set of boundary conditions given below. In case a solution may not exist indicate under what conditions it does. (a) u ( a, θ ) = 1, (b) u r ( a, θ ) = α . Solution 1 For both parts of this problem, we first partially solve the PDE 2 u + λu = 0 . (1) If we let u ( r, θ ) f ( r ) g ( θ ) , (2) then by plugging into the PDE and separating terms (as has been done before in the class notes and many times in the solutions for problem sets), we get d 2 g 2 + μg = 0 -→ g ( θ ) = a m sin ( ) + b m cos ( ) , μ = m 2 , m = 0 , 1 , 2 , . . . (3) (using the condition of 2 π periodicity in θ and handling the special case of μ = 0 correctly). This implies that for λ 6 = 0 d dr r df dr + λr - m 2 r f = 0 -→ f ( r ) = J m λr · (4) and for λ = 0 d dr r df dr - m 2 r f = 0 -→ f ( r ) = r m (5) (in both cases of λ in order to have f ( r ) finite at r = 0). 1

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(a) For this part, let v = u - 1, then v satisfies homogenous boundary condition, and 2 v = r cos(2 θ ), do eigenfunction expansion for v , v ( a, θ ) = 0 -→ f ( a ) = J m λa · = 0 (6) holds for λ 6 = 0 (there is no nontrivial solution for λ = 0). Thus, λ = λ mn = j m,n a 2 , m = 0 , 1 , 2 , . . . , n = 1 , 2 , 3 , . . . , (7) where j m,n is the n th zero of J m ( z ) , z > 0 . So, the function v mn = A mn J m p λ mn r · cos ( ) + B mn J m p λ mn r · sin ( ) (8) solves the PDE system 2 v mn + λ mn v mn = 0 (9) with v mn ( a, θ ) = 0. We therefore write v as the series v ( r, θ ) = X m =0 X n =1 A mn J m p λ mn r · cos ( ) + X m =1 X n =1 B mn J m p λ mn r · sin ( ) , (10) which implies that 2 v = - X m =0 X n =1 λ mn A mn J m p λ mn r · cos ( ) - X m =1 X n =1 λ mn B mn J m p λ mn r · sin ( ) . (11) We also write a series expansion for r cos(2 θ ): r cos(2 θ ) = X m =0 X n =1 C mn J m p λ mn r · cos ( ) + X m =1 X n =1 D mn J m p λ mn r · sin ( ) , (12) where (by orthogonality of the eigenfunctions) C mn = R a 0 r 2 J 2 ( λ 2 n r ) dr R a 0 J 2 2 ( λ 2 n r ) rdr m = 2 0 m 6 = 2 (13) and D mn = 0 (14) So, 2 u = r cos(2 θ ) (15) 2
implies that - X m =0 X n =1 λ mn A mn J m p λ mn r · cos ( ) - X m =1 X n =1 λ mn B mn J m p λ mn r · sin ( ) = X m =0 X n =1 C mn J m p λ mn r · cos ( ) + X m =1 X n =1 D mn J m p λ mn r · sin ( ) . (16) By orthogonality, - λ mn A mn = C mn (17) and - λ mn B mn = D mn . (18) We conclude that A mn = - R a 0 r 2 J 2 ( λ 2 n r ) dr λ 2 n R a 0 J 2 2 ( λ 2 n r ) rdr m = 2 0 m 6 = 2 (19) and B mn = 0 . (20) (b) We can use the eigenfunctions in the first part, u has expansion, u ( r, θ ) = X m =0 A m 0 r m cos( ) + X m =1 B m 0 r m sin( ) + X m =0 X n =1 A mn J m p λ mn r · cos ( ) + X m =1 X n =1 B mn J m p λ mn r · sin ( ) (21) apply the boundary condition u r ( a, θ ) = α (suppose α is constant), by orthogonality condition, we have X n =1 p λ 0 n A 0 n J 0 0 p λ 0 n a · = α (22) A m 0

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• Spring '09
• NilesA.Pierce
• Sin, Boundary value problem, Boundary conditions, Dirichlet boundary condition, Green's function, Amn Jm

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