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Unformatted text preview: Fourier series.) We now consider a second Dirichlet problem on the unit disk with boundary values v (1 , ) = 1 2 ( w r (1 , )g ( ) ) . Since g ( ) = 0 and w r ( r, ) = 2 r cos 2 , it follows that v (1 , ) = cos 2 . The solution of th Dirichlet problem in v is v ( r, ) = r 2 cos 2 . Thus the solution of biharmonic problem is u ( r, ) = (1r 2 ) r 2 cos 2 + r 2 cos 2 = 2 r 2 cos 2 r 4 cos 2 . This can be veried directly by plugging into the biharmonic equation and the boundary conditions. 17. u (1 , 0) = 0 implies that w = 0 and so v (1 , ) =g ( ) 2 . So v ( r, ) =1 2 a + n =1 r n (cos n + b n sin n ) , where a n and b n are the Fourier coecients of g . Finally, u ( r, ) = (1r 2 ) v ( r, ) =1 2 (1r 2 ) a + n =1 r n (cos n + b n sin n ) ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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