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Unformatted text preview: Solutions to Homework Set 6 1) Flexible rod again : This is a somewhat tedious, but ultimately rewarding exercise. a) Look back at your solution for homework set 4, where you showed that 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. From this we see that taking v = v 00 = 0 at the endpoints as the boundary conditions for L requires u = u 00 = 0 at the endpoints for the as the boundary conditions for L . The operator is therefore selfadjoint. b) There are no zero modes for L . The solution is therefore unique, and F has no condi tions to satisfy c) From 4 x G ( x, y ) = ( x y ) we see that we need G , G and G 00 to be continuous at x = y , and G 000 must obey the jump condition G 000 ( y + , y ) G 000 ( y , y ) = 1 . The Green function, considered as a function of x , must also satisfy the boundary conditions at x = 0 , 1. d) We satisfy the boundary conditions at the ends by writing G ( x, y ) = ( ax + bx 2 , < x < y < 1, c ( x 1) + d ( x 1) 3 , < y < x < 1 . The jump condition, the continuity at x = y of G 00 , G , and G (in that order) then read: 6 d 6 b = 1 , 6 d ( y 1) = 6 by, c + 3 d ( y 1) 2 = a + 3 by 2 , c ( y 1) + d ( y 1) 3 = ay + by 3 . The first two equations quickly give d = y/ 6 and b = ( y 1) / 6. Rather more effort produces c = y ( y 1)( y + 1) / 6 and a = y ( y 1)( y 2) / 6. Thus G ( x, y ) = ( 1 6 [ y ( y 1)( y 2) x + ( y 1) x 3 ] , < x < y < 1, 1 6 [ y ( y 1)( y + 1)( x 1) + y ( x 1) 3 ] , < y < x < 1 ....
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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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