solutionhw6 - Solutions to Homework Set 6 1 Flexible rod...

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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 0 u * v (4) dx = [ u * v (3) - ( u 0 ) * v 00 + ( u 00 ) * v 0 - ( u (3) ) * v ] 1 0 + Z 1 0 ( 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 self-adjoint. 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 ) = δ ( x - y ) we see that we need G , G 0 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 , 0 < x < y < 1, c ( x - 1) + d ( x - 1) 3 , 0 < y < x < 1 . The jump condition, the continuity at x = y of G 00 , G 0 , 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 ] , 0 < x < y < 1, 1 6 [ y ( y - 1)( y + 1)( x - 1) + y ( x - 1) 3 ] , 0 < y < x < 1 . e) To see that that the Green function does indeed obey G ( x, y ) = G ( y, x ) it helps to rearrange the second alternative so as to expose the powers of y : G ( x, y ) = ( 1 6 [ y ( y - 1)( y - 2) x + ( y - 1) x 3 ] , 0 < x < y < 1, 1 6 [ x ( x - 1)( x - 2) y + ( x - 1) y 3 ] , 0 < y < x < 1 .
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