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Unformatted text preview: Solution to Set 5: Math 716 1. Show that the partial differential operator A defined by: A u â‰¡ =âˆ‡ Â· ( p âˆ‡ u ) + qu in a bounded Î© âŠ‚ R n for p ( x ) > 0 is symmetric, with respect to the usual L 2 innerproduct. What condition on q ( x ) makes A positive. Suppose we consider the eigenfunctions u satisfying A u = Î»mu for m ( x ) > 0 in Î©. Prove the orthogonality of eigenfunctions corresponding to unequal eigenvalues with respect to innerproduct h ., . i defined by h u, v i = Z Î© muvd x Solution: We note that ( A u, v ) = Z Î© v * {âˆ‡ Â· ( p âˆ‡ u ) + qu } d x = Z âˆ‚ Î© p âˆ‚u âˆ‚n d x + Z Î© { p âˆ‡ u Â· âˆ‡ v * + qu } d x = Z âˆ‚ Î© p âˆ‚u âˆ‚n v * u âˆ‚v * âˆ‚n + Z Î© u {âˆ‡ Â· ( p âˆ‡ v * ) + qv * } d x For Dirichlet BC, the integral term over âˆ‚ Î© clearly drops out, and we clearly obtain from the above expression, ( A u, v ) = ( u, A v ). For Robin boundary condition (which includes Neumann as a special case (when a = 0), we obtain Z âˆ‚ Î© p âˆ‚u âˆ‚n v * u âˆ‚v * âˆ‚n = Z âˆ‚ Î© p { auv * uav * } = 0 Therefore, in all cases, the boundary term drops out, we get ( A u, v ) = ( u, A v ), implying that A is a symmetric operator in every case. From the above calculation, we obtain ( u, A u ) = Z âˆ‚ Î© pu * âˆ‚u âˆ‚n d x + Z Î© { p âˆ‡ u Â· âˆ‡ u * + qu } d x For Dirichlet condition, the boundary term drops out, and we have ( u, A u ) = Z Î© p âˆ‡ u  2 + q  u  2 d x > for arbitrary u âˆˆ C 2 iff q ( x ) > 0 almost everywhere ( i.e. it can be zero at most on a set of measure 0). For Robin BC (including Neumann as a special case), we get ( u, A u ) = Z âˆ‚ Î© a  u  2 d x + Z Î© p âˆ‡ u  2 + q  u  2 d x > 1 for arbitary nonzero u âˆˆ C 2 , iff a â‰¥ 0 and q > 0. Thus, under these conditions A is positive. If we have only a â‰¥ 0 and q â‰¥ 0, then we will obtain A semipositive in all three cases. If u is an eigenvector of A corresponding to eigenvalue Î» , it follows that Î» < u, u > = ( A u, u ) = ( u, A u ) = Î» * ( u, mu ) = Î» * < u, u > This implies that Î» is real. Further, if u , v are eigenvectors corresponding to two unequal eigenvalues Î» and Î¼ , it follows that Î» < u, v > = ( A u, v ) = ( u, A v ) = Î¼ < u, v >, since both eigenvalues Î» and Î¼ must be real. Therefore, ( Î» Î¼ ) < u, v > = 0 implying < u, v > For equal eigenvalues, we can use GramSchmidt orgonalization procedure to obtain or thogonality. Thus all eigenvectors are orthogonal....
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 Spring '08
 Tanveer,S
 Math, dx, Eigenfunction, AU, Dirichlet boundary condition, Neumann boundary condition, cj Jm

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