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Unformatted text preview: 5.3.8 Multiplying the equation by Ï† yields: 0 = Ï† d 2 Ï† dx 2 + ( Î» x 2 ) Ï† 2 = d dx Ï† dÏ† dx dÏ† dx 2 + Î»Ï† 2 x 2 Ï† 2 . Integrating this identity over the interval 0 â‰¤ x â‰¤ 1 allows us to conclude: Ï† dÏ† dx 1 Z 1 " dÏ† dx 2 + x 2 Ï† 2 # dx = Î» Z 1 Ï† 2 dx. If Ï† is an eigenfunction, and hence not identically zero, we can now divide by R 1 Ï† 2 dx to obtain the standard Rayleigh quotient Î» = Ï† dÏ† dx 1 + R 1 h ( dÏ†/dx ) 2 + x 2 Ï† 2 i dx R 1 Ï† 2 dx The first derivatives of Ï† vanish at the boundary which means Î» = R 1 h ( dÏ†/dx ) 2 + x 2 Ï† 2 i dx R 1 Ï† 2 dx â‰¥ . If Î» = 0, we would then have 0 = Z 1 dÏ† dx 2 dx + Z 1 x 2 Ï† 2 dx â‰¥ since each term in the integrand is nonnegative. This implies that each integral vanishes, and that each integrand vanishes. In particular, x 2 Ï† 2 ( x ) = 0 for all 0 â‰¤ x â‰¤ 1, which implies that Ï† ( x ) = 0 for all < x â‰¤ 1 and hence Ï† (0) = 0 by continuity. Thus Ï† is the trivial function and hence Î» = 0 cannot be an eigenvalue.an eigenvalue....
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 Spring '10
 sim
 Flux, Boundary value problem, dx, Stokes' theorem

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