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Unformatted text preview: Solutions to Homework Set 4 1) Linear Differential operators : a) Integrating by parts gives us h u  Lv i w = Z b a wu * i d dx v dx = [ iwu * v ] b a + Z b a w i w d dx wu * v dx [ Q ] b a + h L u  v i w . Therefore the formal adjoint is L = i w d dx w i d dx + i (ln w ) , and the boundary term is Q [ u, v ] = iwu * v. b) We have that d dx [ u * v (3) ( u ) * v 00 + ( u 00 ) * v ( u (3) ) * v ] = u * v (4) ( u (4) ) * v and so Z 1 u * v (4) dx = [ u * v (3) ( u ) * v 00 + ( u 00 ) * v ( u 3) ) * v ] 1 + Z 1 ( u (4) ) * v dx. The formal adjoint M is therefore d 4 /dx 4 , which is the same as M . The operator is therefore formally selfadjoint. Is it truly selfadjoint? We are told that D ( M ) is defined by requiring v and v (3) to be zero at both ends, but we are told nothing about v 00 and v . To make the integratedout term vanish we therefore need to impose u = 0 and u 00 = 0 at x = 0 , 1. Thus D ( M ) = { u, u (4) L 2 [0 , 1] : u (0) = u (1) = u 00 (0) = u 00 (1) = 0 } . These are not the same boundary conditions as those imposed on M , and so M is not truly selfadjoint. 2) SturmLiouville forms : The equation p y 00 + p 1 y + p 2 y = 0 becomes Ly = 1 w ( wp y ) + p 2 y = 0 , provided we take w ( x ) = 1 p exp Z x p 1 ( ) p ( ) d . 1 a) We apply the general method, and so compute Z x p 1 p 2 d = Z x  1 2 ( + + 2) 1 2 d = ln[(1 + x ) +1 (1 x ) +1 ] . Therefore w = (1 + x ) (1 x ) , and Ly = (1 + x ) (1 x ) d dx (1 + x ) +1 (1 x ) +1 dy dx . When n is an integer, the equation d dx (1 + x ) +1 (1 x ) +1 dy dx + n ( n + + + 1)(1 + x ) (1 x ) y = 0 has polynomial solutions y = P ( , ) n ( x ). These are the Jacobi polynomials. b) Just set = = 1 / 2. c) We find w = 1 x 2 , and Lu = (1 x 2 ) 1 d dx (1 x 2 ) du dx m 2 1 x 2 u. This is the differential operator appearing in Legendres equation ....
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This homework help was uploaded on 01/29/2008 for the course PHYS 598 taught by Professor Stone during the Fall '07 term at University of Illinois at Urbana–Champaign.
 Fall '07
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