Chem Differential Eq HW Solutions Fall 2011 78

Chem Differential Eq HW Solutions Fall 2011 78 - 78 Chapter...

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Unformatted text preview: 78 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates On the left, we have the desired integral times (b2 − a2 ) and, on the right, we have c Jp (bc)aJp (ac) − bJp (ac)Jp (bc) − c aJp (0)Jp (0) − bJp (0)Jp (0) . Since Jp (0) = 0 if p > 0 and J0(x) = −J1 (x), it follows that Jp (0)Jp (0) − Jp (0)Jp (0) = 0 for all p > 0. Hence the integral is equal to c Jp (ax) Jp (bx)x dx = I= 0 c aJp (bc)Jp (ac) − bJp (ac)Jp (bc) . b2 − a2 Now using the formula Jp ( x ) = 1 Jp−1(x) − Jp+1 (x) , 2 we obtain I= c aJp (bc) Jp−1 (ac) − Jp+1 (ac) − bJp (ac) Jp−1 (bc) − Jp+1 (bc) . 2(b2 − a2) Simplify with the help of the formula Jp+1 (x) = 2p Jp (x) − Jp−1 (x) x and you get I = = c 2p aJp (bc) Jp−1 (ac) − ( Jp (ac) − Jp−1(ac)) 2(b2 − a2) ac 2p −bJp (ac) Jp−1 (bc) − ( Jp (bc) − Jp−1 (bc)) bc c aJp (bc)Jp−1(ac) − bJp (ac)Jp−1 (bc) , b2 − a2 as claimed. Note that this formula implies the orthogonality of Bessel functions. In fact its proof mirrors the proof of orthogonality from Section 4.8. ...
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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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