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

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Unformatted text preview: 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 ) 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, ) = 1, (b) u r ( a, ) = . Solution 1 For both parts of this problem, we first partially solve the PDE 2 u + u = 0 . (1) If we let u ( r, ) f ( r ) g ( ) , (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 d 2 + g = 0- g ( ) = a m sin( m ) + b m cos( m ) , = m 2 , m = 0 , 1 , 2 ,... (3) (using the condition of 2 periodicity in and handling the special case of = 0 correctly). This implies that for 6 = 0 d dr r df dr + r- m 2 r f = 0- f ( r ) = J m r (4) and for = 0 d dr r df dr - m 2 r f = 0- f ( r ) = r m (5) (in both cases of in order to have f ( r ) finite at r = 0). 1 (a) For this part, let v = u- 1, then v satisfies homogenous boundary condition, and 2 v = r cos(2 ), do eigenfunction expansion for v , v ( a, ) = 0- f ( a ) = J m a = 0 (6) holds for 6 = 0 (there is no nontrivial solution for = 0). Thus, = 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 > . So, the function v mn = A mn J m p mn r cos( m ) + B mn J m p mn r sin( m ) (8) solves the PDE system 2 v mn + mn v mn = 0 (9) with v mn ( a, ) = 0. We therefore write v as the series v ( r, ) = X m =0 X n =1 A mn J m p mn r cos( m ) + X m =1 X n =1 B mn J m p mn r sin( m ) , (10) which implies that 2 v =- X m =0 X n =1 mn A mn J m p mn r cos( m )- X m =1 X n =1 mn B mn J m p mn r sin( m ) . (11) We also write a series expansion for r cos(2 ): r cos(2 ) = X m =0 X n =1 C mn J m p mn r cos( m ) + X m =1 X n =1 D mn J m p mn r sin( m ) , (12) where (by orthogonality of the eigenfunctions) C mn = R a r 2 J 2 ( 2 n r ) dr R a J 2 2 ( 2 n r ) rdr m = 2 m 6 = 2 (13) and D mn = 0 (14) So, 2 u = r cos(2 ) (15) 2 implies that- X m =0 X n =1 mn A mn J m p mn r cos( m )- X m =1 X n =1 mn B mn J m p mn r sin( m ) = X m =0 X n =1 C mn J m p mn r cos( m ) + X m =1 X n =1 D mn J m p mn r sin( m ) ....
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This note was uploaded on 01/08/2011 for the course ACM 95c taught by Professor Nilesa.pierce during the Spring '09 term at Caltech.

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

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