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Unformatted text preview: ACM 95b/100b Problem Set 7 Solutions Prepared by: Lei Zhang March 3, 2007 Total: 172 points Include grading section: 2 points Problem 1 (15 points) Show that the second order linear differential equation A ( x ) u 00 + B ( x ) u + C ( x ) u = 0 , A ( x ) 6 = 0 (1) can be transformed to SturmLiouville form d dx p ( x ) du dx + q ( x ) u = 0 via multiplication by an integrating factor h ( x ). Solution 1 We wish to find a function h ( x ) so that we can rewrite (1) as [ hAu ] + hCu = 0 . Expand this using the product rule, and equate derivatives of u with (1) to find the criterion h A + hA = hB. Since we have assumed that A ( x ) 6 = 0 we can rewrite this as h h = ( B A ) A . Integrating we find log( h ) = Z x x B ( s ) A ( s ) ds log( A ) or h ( x ) = 1 A ( x ) exp Z x x B ( s ) A ( s ) ds. Thus we can rewrite (1) in SturmLiouville form where p ( x ) = exp Z x x B ( s ) A ( s ) ds, and q ( x ) = C ( x ) A ( x ) exp Z x x B ( s ) A ( s ) ds. Problem 2 (20 points) Consider the regular SturmLiouville eigenvalue problem d dx p ( x ) du dx + q ( x ) u + r ( x ) u = 0 , a < x < b 1 u ( a ) + 2 u ( a ) = 0 1 u ( b ) + 2 u ( b ) = 0 1 with < a < b < , p ( x ) , p ( x ) , q ( x ) , r ( x ) continuous with p ( x ) > 0, r ( x ) > 0 for a x b , and 1 , 2 , 1 , 2 real with  1  +  2  > 0,  1  +  2  > 0. a) (10 pts) Use the fact that each eigenvalue corresponds to exactly one linearly independent eigenfunc tion to show that the eigenfunctions can be chosen to be real. b) (10 pts) Derive the Rayleigh quotient = p ( x ) u du dx b a + Z b a " p ( x ) du dx 2 q ( x ) u 2 # dx R b a u 2 r ( x ) dx R ( u ) . Note: It can be shown that the smallest eigenvalue, 1 , is the minimum value of the Rayleigh quotient R ( u ) over all continuous functions u ( x ) that satisfy the boundary conditions ( e.g., see Haberman). Hence, it is possible to place an upper bound on the smallest eigenvalue without solving the eigenvalue problem by evaluating R ( u ) for a continuous trial function u ( x ) that satisfies the boundary conditions (but not necessarily the differential equation). Solution 2 a) To show that we can choose real eigenfunctions, assume that we have a complex eigenfunction n ( x ) = U ( x ) + iV ( x ), where U and V are real. The function n ( x ) satisfies the equation so d dx p ( x ) d n dx + q ( x ) n ( x ) + n r ( x ) n ( x ) = 0 Recalling the definition of n , this becomes, pU 00 + p U + qU + n rV + i ( pV 00 + p V + qV + n rV ) = 0 . For the expression to be satisfied, real and imaginary parts must each be zero. Thus, both U ( x ) and V ( x ) must be eigenfunctions. Since each eigenvalue of the SturmLiouville problem corresponds to one linearly independent eigenfunction, U and V must be linearly dependent. Hence, we can choose n = U , a real eigenfunction....
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 Winter '09
 NilesA.Pierce

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