sol7 - MAE294B/SIO203B Methods in Applied Mechanics Winter...

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MAE294B/SIO203B: Methods in Applied Mechanics Winter Quarter 2010 http://maecourses.ucsd.edu/mae294b Solution VII 1 The function h ( t ) = - t 4 ( 1 - t ) 2 has maxima at the endpoints 0 and 1 of the interval. The contribution from t = 1 is the standard local maximum times 1 / 2. The contribution from t = 0 is different since h ( 0 ) = 0 and the prefactor t vanishes. One can proceed simply by expanding locally and writing I Z 0 t e - xt 4 d t + Z 1 - e - x ( 1 - t ) 2 d t = 3 4 π x . 2 The function h = - t 3 + 3 t 2 - 2 t is a cubic with a maximum at t 0 = 1 + 1 / 3 in the range of integration. Expanding gives the exact result h ( t ) = h ( t 0 )+ 1 2 h ( t 0 )( t - t 0 ) 2 + 1 6 h ( t 0 )( t - t 0 ) 3 = h ( t 0 ) - 3 ( t - t 0 ) 2 - ( t - t 0 ) 3 . Substitute in: I = Z 0 e x [ h ( t 0 ) - 3 ( t - t 0 ) 2 ] 1 - x ( t - t 0 ) 3 + x 2 2 ( t - t 0 ) 6 + ··· d t . The usual change of variable combined with extending the range gives I e 2 x / 3 3 π 1 / 2 3 - 1 / 4 x - 1 / 2 1 + 5 16 3 x + O ( x - 2 ) . 3 The function h ( x , t ) = xt 2 - t 3 - t 2 / x 2 has potential maxima on the range of integration at 0 or at zeroes of x - 3 t 2 - 2 t / x 2 provided these give maxima. Rather than solving this equation exactly,
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