Chem Differential Eq HW Solutions Fall 2011 60

# Chem Differential Eq HW Solutions Fall 2011 60 - 60 Chapter...

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Unformatted text preview: 60 Chapter 4 Partial Diﬀerential Equations in Polar and Cylindrical Coordinates Thus bn = = = = = = = Hence 2 πn 1 πn π /4 sin θ sin 4nθ dθ 0 π /4 [− cos[(4n + 1)θ] + cos[(4n − 1)θ]] dθ 0 1 sin[(4n + 1)θ] sin[(4n − 1)θ] − + πn 4n + 1 4n − 1 π /4 0 sin[(4n + 1) π sin[(4n − 1) π 1 4 4 − + πn 4n + 1 4n − 1 cos(nπ) sin π ] cos(nπ) sin π 1 4 4 − − πn 4n + 1 4n − 1 √ (−1)n 2 −1 1 − πn 2 4n + 1 4n − 1 √ (−1)n+1 2 4 . 2−1 π 16n √∞ 42 (−1)n+1 4n u(r, θ) = r sin 4nθ. π n=1 16n2 − 1 17. Since u satsiﬁes Laplace’s equation in the disk, the separation of variables method and the fact that u is 2π-periodic in θ imply that u is given by the series (4), where the coeﬃcients are to be determined from the Neumann boundary condition. From ∞ rn u(r, θ) = a0 + an cos nθ + bn sin nθ , a n=1 it follows that ∞ ur (r, θ) = n n=1 rn−1 an an cos nθ + bn sin nθ . Using the boundary condition ur (a, θ) = f (θ), we obtain ∞ f (θ ) = n an cos nθ + bn sin nθ . a n=1 In this Fourier series expansion, the n = 0 term must be 0. But the n = 0 term is given by 2π 1 f (θ) dθ, 2π 0 thus the compatibility condition 2π f (θ) dθ = 0 0 must hold. Once this condition is satsiﬁed, we determine the coeﬃcients an and bn by using the Euler formulas, as follows: n 1 an = a π 2π f (θ) cos nθ dθ 0 ...
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