Partial Differentials Notes_Part_11

Partial Differentials Notes_Part_11 - We have thus found...

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Unformatted text preview: We have thus found the series solution hatwide u ( x ) = summationdisplay k = 0 u 2 k x m +2 k = summationdisplay k = 0 ( 1) k x m +2 k 2 2 k k ( k 1) 3 2 ( r + k )( r + k 1) ( r + 2)( r + 1) . (12 . 93) So far, we not paid attention to the precise values of the indices r = m . In order to continue the recurrence, we need to ensure that the denominators in (12.92) are never 0. Since n > 0, a vanishing denominator will appear whenever 2 r + n = 0, and so r = 1 2 n is either a negative integer 1 , 2 , 3 , . . . , or half-integer, 1 2 , 3 2 , 5 2 , . . . . This will occur when the order m = r = 1 2 n is either an integer or a half-integer. Indeed, these are precisely the situations when the two indices, namely r 1 = m and r 2 = m , differ by an integer, r 2 r 1 = n , and so we are in the tricky case ( iii ) of the Frobenius method. There is, in fact, a major difference between the integral and the half integral cases. Recall that the odd coefficients u 2 k +1 = 0 in the Frobenius series automatically vanish, and so we only have to worry about the recurrence relation (12.92) for even values of n . When n = 2 k , the factor 2 r + n = 2( r + k ) = 0 vanishes only when r = k is a negative integer; the half integral values do not, in fact cause problems. Therefore, if the order m 0 is not a non-negative integer, then the Bessel equation of order m admits two linearly independent Frobenius solutions, given by the expansions (12.93) with exponents r = + m and r = m . On the other hand, if m is an integer, there is only one Frobenius solution, namely the expansion (12.93) for the positive indexsolution, namely the expansion (12....
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This note was uploaded on 02/10/2012 for the course MATH 5587 taught by Professor Olver during the Fall '10 term at University of Central Florida.

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Partial Differentials Notes_Part_11 - We have thus found...

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