Chem Differential Eq HW Solutions Fall 2011 74

Chem Differential Eq HW Solutions Fall 2011 74 - 74 Chapter...

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Unformatted text preview: 74 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates Solutions to Exercises 4.8 1. (a) Using the series definition of the Bessel function, (7), Section 4.7, we have d −p [x Jp (x)] = dx ∞ d dx = k =0 ∞ = x 2 2k (−1)k dx p k !Γ(k + p + 1) dx 2 2 2k k =0 ∞ 2p (k − m=0 (−1)m + p + 2) −x−p (−1)m m!Γ(m + p + 2) m=0 (−1)k 2k 1 + p + 1) 2 2p k!Γ(k x 2 2k −1 2k −1 2m+1 x 2 ∞ = k =0 (−1) x − 1)!Γ(k + p + 1) 2 2p m!Γ(m ∞ = k k =0 ∞ = (−1)k + p + 1) 2p k!Γ(k (set m = k − 1) x 2 2m+p+1 = −x−p Jp+1 (x). To prove (7), use (1): dp [x Jp (x)] = xp Jp−1 (x) dx ⇒ xp Jp−1 (x) dx = xp Jp (x) + C. Now replace p by p + 1 and get xp+1 Jp (x) dx = xp+1 Jp+1 (x) + C, which is (7). Similarly, starting with (2), d −p [x Jp (x)] = −x−p Jp+1 (x) dx ⇒− ⇒ x−pJp+1 (x) dx = x−p Jp (x) + C x−p Jp+1 (x) dx = −x−p Jp (x) + C. Now replace p by p − 1 and get x−p+1 Jp (x) dx = −x−p+1 Jp−1(x) + C, which is (8). (b) To prove (4), carry out the differentiation in (2) to obtain x−p Jp (x) − px−p−1Jp (x) = −x−p Jp+1 (x) ⇒ xJp (x) − pJp (x) = −xJp+1 (x), upon multiplying through by xp+1 . To prove (5), add (3) and (4) and then divide by x to obtain Jp−1 (x) − Jp+1 (x) = 2Jp (x). To prove (6), subtract (4) from (3) then divide by x. 5. J1(x) dx = −J0 (x) + C , by (8) with p = 1. 9. J3 (x) dx = x2[x−2J3 (x)] dx x2 = u, x−2J3(x) dx = dv, 2x dx = du, v = −x−2 J2 (x) = −J2 (x) + 2 x−1 J2(x) dx = −J2 (x) − 2x−1 J1(x) + C 2 2 J1(x) − J1(x) + C (use (6) with p = 1) x x 4 = J0 (x) − J1(x) + C. x = J0 ( x ) − ...
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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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