Chem Differential Eq HW Solutions Fall 2011 66

Chem Differential Eq HW Solutions Fall 2011 66 - 66 Chapter...

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Unformatted text preview: 66 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates which shows that (1 − α2 )A0,n = a0,n is the nth Bessel coefficient of the Bessel mn series expansion of order 0 of the function 1. This series is computed in Example 1, Section 4.8. We have ∞ 1= 2 J0 (α0,nr) 0 < r < 1. α J (α0,n) n=1 0,n 1 Hence (1 − α2 )A0,n = mn 2 α0,nJ1 (α0,n) and so ∞ u(r, θ) = n=1 A0,n = ⇒ 2 (1 − α2 )α0,nJ1(α0,n) mn ; 2 (1 − J0(α0,nr). α2 )α0,nJ1 (α0,n) mn 9. Let r 0 h( r ) = if 0 < r < 1/2, if 1/2 < r < 1. Then the equation becomes 2u = f (r, θ), where f (r, θ) = h(r) sin θ. We proceed as in the previous exercise and try ∞ ∞ ∞ u(r, θ) = ∞ Jm (λmn r)(Amn cos mθ + Bmn sin mθ) = m=0 n=1 φmn(r, θ), m=0 n=1 where φmn (r, θ) = Jm (λmn r)(Amn cos mθ + Bmn sin mθ). We plug this solution into the equation, use the fact that 2(φmn ) = −λ2 φmn = −α2 φmn , and get mn mn ∞ ∞ 2 φmn(r, θ) = h(r) sin θ m=0 n=1 ∞ ∞ 2 ⇒ (φmn (r, θ)) = h(r) sin θ m=0 n=1 ∞∞ −α2 φmn(r, θ) = h(r) sin θ. mn ⇒ m=0 n=1 We recognize this expansion as the expansion of the function h(r) sin θ in terms of the functions φmn . Because the right side is proportional to sin θ, it follows that all Amn and Bmn are zero, except B1,n. So ∞ −α2nB1,n J1 (α1nr) = h(r) sin θ, 1 sin θ n=1 which shows that −α2nB1,n is the nth Bessel coefficient of the Bessel series expan1 sion of order 1 of the function h(r): −α2nB1,n 1 = = = = 2 J2 (α1,n)2 1/2 r2J1 (α1,nr) dr 0 2 α3,nJ2(α1,n)2 1 2 α3,nJ2(α1,n)2 1 α1,n /2 s2 J1(s) ds 0 s2 J2 (s) J2 (α1,n/2) . 2α1,nJ2(α1,n)2 α1,n /2 0 ...
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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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