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Unformatted text preview: MAE294B/SIO203B: Methods in Applied Mechanics Winter Quarter 2010 http://maecourses.ucsd.edu/mae294b Solution V 1 The L-G solution is y = A [ Q ( x )]- 1 / 4 sin E 1 / 2 Z x p Q ( u ) d u + B [ Q ( x )]- 1 / 4 cos E 1 / 2 Z x p Q ( u ) d u . Differentiating e.g. the sine term gives- 1 4 Q ( x )[ Q ( x )]- 5 / 4 sin E 1 / 2 Z x p Q ( u ) d u + E 1 / 2 [ Q ( x )] 1 / 4 cos E 1 / 2 Z x p Q ( u ) d u . Since E is large, we can neglect the first term. Then in the two boundary conditions the b and b 1 terms can also be neglected. The boundary condition at x = 0 gives A = 0 and the boundary condition at x = 1 gives BQ ( 1 )- 1 / 4 sin E 1 / 2 Z 1 p Q ( u ) d u = . For a non-trivial solution, we obtain E ∼ n π R 1 p Q ( u ) d u ! 2 . For the special case Q = 1, the exact solution is y = A sin ( E 1 / 2 x )+ B cos ( E 1 / 2 x ) . The two boundary conditions can be written as a homogeneous matrix equation, and hence the exact eigenvalue condition is that the determinant a E 1 / 2 (- a 1 E 1 / 2 s + b 1 c )- b ( a 1 E 1 / 2 c + b 1 s ) = vanish, with s = sin E 1 / 2 and c = cos E 1 / 2 . The approximation above corresponds to keeping the O ( E ) term; a a 1 s = 0, so that E ∼ ( n π ) 2 . We can reduce the numbers of parameters from 4 to 2 by dividing by a a 1 6 = 0. Then the eigenvalue condition is- Es + E 1 / 2 ( d 1- d ) c- d d 1 s = . where d = b / a and d 1 = b 1 / a 1 . This equation can be solve numerically starting with the guess E ∼ ( n π ) 2 ....
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This note was uploaded on 03/16/2010 for the course MAE 294b taught by Professor Young,w during the Winter '08 term at UCSD.
- Winter '08