Chem Differential Eq HW Solutions Fall 2011 82

Chem Differential Eq HW Solutions Fall 2011 82 - 82 Chapter...

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Unformatted text preview: 82 Chapter 5 Partial Differential Equations in Spherical Coordinates So ∞ u(r, θ) = 60 + 20 (4n + 3) n=0 5. Solution We have (−1)n (2n)! r2n+1P2n+1(cos θ). + 1) 22n(n!)2(n ∞ An rnPn (cos θ), u(r, θ) = n=0 with An = = = 2n + 1 2 2n + 1 2 2n + 1 2 π f (θ)Pn (cos θ) sin θ dθ 0 π 2 cos θ Pn(cos θ) sin θ dθ 0 1 x Pn(x) dx, 0 where, as in Exercise 3, we made the change of variables x = cos θ. At this point, we have to appeal to Exercise 11, Section 5.6, for the evaluation of this integral. (The cases n = 0 and 1 can be done by referring to the explicit formulas for the Pn, but we may as well at this point use the full result of Exercise 11, Section 5.6.) We have 1 1 1 1 xP0(x) dx = ; xP1(x) dx = ; 20 3 0 1 x P2n(x) dx = 0 and (−1)n+1 (2n − 2)! ; 22n((n − 1)!)2n(n + 1) n = 1, 2, . . . ; 1 x P2n+1(x) dx = 0; n = 1, 2, . . . . 0 Thus, A0 = 11 1 =; 22 4 A1 = 31 1 =; 23 2 A2n+1 = 0, n = 1, 2, 3, . . . ; and for n = 1, 2, . . ., A2n = 2(2n) + 1 (−1)n+1 (2n − 2)! (−1)n+1 (2n − 2)! = (4n + 1) 2n+1 . 2 22n((n − 1)!)2n(n + 1) 2 ((n − 1)!)2n(n + 1) So ∞ u(r, θ) = 11 (−1)n+1 (4n + 1)(2n − 2)! 2n + r cos θ + r P2n(cos θ). 42 22n+1((n − 1)!)2 n(n + 1) n=1 ...
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