Chem Differential Eq HW Solutions Fall 2011 56

Chem Differential Eq HW Solutions Fall 2011 56 - 56 Chapter...

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Unformatted text preview: 56 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates Solutions to Exercises 4.3 1. The condition g(r, θ) = 0 implies that a∗ = 0 = b∗ . Since f (r, θ) is mn mn proportional to sin 2θ, only b2,n will be nonzero, among all the amn and bmn . This is similar to the situation in Example 2. For n = 1, 2, . . ., we have b2,n = 2 πJ3 (α2,n)2 1 2π (1 − r2)r2 sin 2θJ2 (α2,nr) sin 2θr dθ dr 0 0 =π = = = 2 πJ3 (α2,n)2 1 2π sin2 2θ dθ(1 − r2)r3 J2 (α2,nr) dr 0 1 0 2 (1 − r2 )r3J2 (α2,nr) dr J3 (α2,n)2 0 2 2 4J4 (α2,n) J4 (α2,n) = 2 , J3 (α2,n)2 α2,n α2,nJ3 (α2,n)2 2 where the last integral is evaluated with the help of formula (15), Section 4.3. We can get rid of the expression involving J4 by using the identity Jp−1 (x) + Jp+1 (x) = 2p Jp ( x ) . x With p = 3 and x = α2,n, we get =0 J2 (α2,n) +J4 (α2,n) = 6 α2,n J3 (α2,n) ⇒ J4(α2,n) = 6 α2,n J3 (α2,n). So b2,n = 24 . α3,nJ3(α2,n) 2 Thus ∞ u(r, θ, t) = 24 sin 2θ J2 (α2,nr) cos(α2,nt). α3,nJ3 (α2,n) n=1 2 5. We have amn = bmn = 0. Also, all a∗ and b∗ are zero except b∗,n. We have mn mn 2 b∗,n = 2 2 πα2,nJ3(α2,n)2 1 2π (1 − r2)r2 sin 2θJ2 (α2,nr) sin 2θr dθ dr. 0 0 The integral was computed in Exercise 1. Using the computations of Exercise 1, we find 24 b∗,n = 4 . 2 α2,nJ3(α2,n) hus ∞ u(r, θ, t) = 24 sin 2θ J2(α2,nr) sin(α2,nt). α4,nJ3 (α2,n) n=1 2 9. (a) For l = 0 and all k ≥ 0, the formula follows from (7), Section 4.8, with p = k: rk+1 Jk (r) dr = rk+1Jk+1 (r) + C. (b) Assume that the formula is true for l (and all k ≥ 0). Integrate by parts, using ...
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