Chem Differential Eq HW Solutions Fall 2011 86

Chem Differential - 86 Chapter 5 Partial Differential Equations in Spherical Coordinates Solutions to Exercises 5.4 5 We apply Theorem 3 and note

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Unformatted text preview: 86 Chapter 5 Partial Differential Equations in Spherical Coordinates Solutions to Exercises 5.4 5. We apply Theorem 3 and note that since f depends only on r and not on θ or φ, the series expansion should also not depend on θ or φ. So all the coefficients in the series are 0 except for the coefficients Aj,0,0, which we will write as Aj for 1 simplicity. Using (16) with m = n = 0, a = 1, f (r, θ, φ) = 1, and Y0,0(θ, φ) = 2√π , we get Aj = 2 2 j1 (α 1 , j ) 2 1 0 2π π 0 0 1 j0 (λ0, j r) √ r2 sin θ dθ dφ dr 2π =2π = = √ 1 2 1 π j1 (α 2 , j ) √ 4π 2 j1 (α 1 , j ) 2 =2 2π π dφ 1 j0 (λ0, j r)r2 dr sin θ dθ 0 0 0 1 j0 (λ0, j r)r2 dr, 0 where λ0, j = α 1 , j , the j th zero of the Bessel function of order 1 . Now 2 2 2 sin x πx J 1/2 ( x ) = (see Example 1, Section 4.7), so the zeros of J1/2 are precisely the zeros of sin x, which are jπ. Hence λ0, j = α 1 , j = jπ. 2 Also, recall that j0 (x) = sin x x (Exercises 38, Section 4.8), so 1 1 j0 (λ0, j r)r2 dr 1 j0 (jπ r)r2 dr = = 0 0 0 sin(jπr) 2 r dr jπ r j +1 = (−1) jπ = = 1 1 sin(jπ r)r dr jπ 0 (−1)j +1 , (jπ)2 where the last integral follows by integration by parts. So, Aj = √ 4 π (−1)j +1 . 2 j1 (jπ) (jπ)2 This can be simplified by using a formula for j1 . Recall from Exercise 38, Section 4.7, sin x − x cos x j1 (x) = . x2 Hence 2 j1 (jπ) = and sin(jπ) − jπ cos(jπ) (jπ)2 2 = − cos(jπ) jπ √ Aj = 4(−1)j +1 π, 2 = (−1)j +1 jπ 2 = 1 , (jπ)2 ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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