Unformatted text preview: Fourier series.) We now consider a second Dirichlet problem on the unit disk with boundary values v (1 , θ ) = 1 2 ( w r (1 ,θ )g ( θ ) ) . Since g ( θ ) = 0 and w r ( r, θ ) = 2 r cos 2 θ , it follows that v (1 , θ ) = cos 2 θ . The solution of th Dirichlet problem in v is v ( r, θ ) = r 2 cos 2 θ . Thus the solution of biharmonic problem is u ( r, θ ) = (1r 2 ) r 2 cos 2 θ + r 2 cos 2 θ = 2 r 2 cos 2 θr 4 cos 2 θ. This can be veri±ed directly by plugging into the biharmonic equation and the boundary conditions. 17. u (1 , 0) = 0 implies that w = 0 and so v (1 , θ ) =g ( θ ) 2 . So v ( r, θ ) =1 2 ± a + ∞ ² n =1 r n (cos nθ + b n sin nθ ) ³ , where a n and b n are the Fourier coeﬃcients of g . Finally, u ( r,θ ) = (1r 2 ) v ( r,θ ) =1 2 (1r 2 ) ´ a + ∞ ² n =1 r n (cos nθ + b n sin nθ ) µ ....
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 Fall '11
 StuartChalk
 Cartesian Coordinate System, Polar coordinate system, Biharmonic equation, Elliptic partial differential equations, Biharmonic Operator

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