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Chem Differential Eq HW Solutions Fall 2011 103

Chem Differential Eq HW Solutions Fall 2011 103 - Fourier...

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Section 6.6 The Biharmonic Operator 103 Exercises 6.6 1. u xxyy = 0, u xxxx = 4!, u yyyy = - 4!, 4 u = 0. 5. Express v in Cartesian coordinates as follows: v = r 2 cos(2 θ )(1 - r 2 ) = r 2 [cos 2 θ - sin 2 θ ](1 - r 2 ) = ( x 2 - y 2 ) ( 1 - ( x 2 + y 2 ) ) . Let u = x 2 - y 2 . Then u is harmonic and so v is biharmonic by Example 1, with A = 1, D = 1, B = C = 0. 7. Write v = r 2 · r n cos and let u = r n cos . Then u is harmonic (use the Laplacian in polar coordinates to check this last assertion) and so v is biharmonic, by Example 1 with A = 1 and B = C = D = 0. 9. Write v = ar 2 ln r + br 2 + c ln r + d = φ + ψ , where φ = [ ar 2 + c ] ln r and ψ = br 2 + d . From Example 1, it follows that ψ is biharmonic. Also, ln r is harmonic (check the Laplacian in polar coordinates) and so, by Example 1, φ is biharmonic. Consequently, v is biharmonic, being the sum of two biharmonic functions. 13. We follow the method of Theorem 1, as illustrated by Example 2. First, solve the Dirichlet problem 2 w = 0, w (1 ) = cos 2 θ , for 0 r < 1, 0 θ 2 π . The solution in this case is w ( r,θ ) = r 2 cos 2 θ . (This is a simple application of the method of Section 4.4, since the boundary function is already given by its
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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 , 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 ) = (1-r 2 ) r 2 cos 2 θ + r 2 cos 2 θ = 2 r 2 cos 2 θ-r 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 ) = (1-r 2 ) v ( r,θ ) =-1 2 (1-r 2 ) ´ a + ∞ ² n =1 r n (cos nθ + b n sin nθ ) µ ....
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