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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 Sturm-Liouville 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 re-write (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 re-write 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 re-write (1) in Sturm-Liouville 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 Sturm-Liouville 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 Sturm-Liouville 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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