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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.
 Fall '11
 StuartChalk

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