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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 ≡ =∇ · ( p ∇ u ) + qu in a bounded Ω ⊂ 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 = λmu for m ( x ) > 0 in Ω. Prove the orthogonality of eigenfunctions corresponding to unequal eigenvalues with respect to innerproduct h ., . i defined by h u, v i = Z Ω muvd x Solution: We note that ( A u, v ) = Z Ω v * {∇ · ( p ∇ u ) + qu } d x = Z ∂ Ω p ∂u ∂n d x + Z Ω { p ∇ u · ∇ v * + qu } d x = Z ∂ Ω p ∂u ∂n v * u ∂v * ∂n + Z Ω u {∇ · ( p ∇ v * ) + qv * } d x For Dirichlet BC, the integral term over ∂ Ω 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 ∂ Ω p ∂u ∂n v * u ∂v * ∂n = Z ∂ Ω 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 ∂ Ω pu * ∂u ∂n d x + Z Ω { p ∇ u · ∇ u * + qu } d x For Dirichlet condition, the boundary term drops out, and we have ( u, A u ) = Z Ω p ∇ u  2 + q  u  2 d x > for arbitrary u ∈ 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 ∂ Ω a  u  2 d x + Z Ω p ∇ u  2 + q  u  2 d x > 1 for arbitary nonzero u ∈ C 2 , iff a ≥ 0 and q > 0. Thus, under these conditions A is positive. If we have only a ≥ 0 and q ≥ 0, then we will obtain A semipositive in all three cases. If u is an eigenvector of A corresponding to eigenvalue λ , it follows that λ < u, u > = ( A u, u ) = ( u, A u ) = λ * ( u, mu ) = λ * < u, u > This implies that λ is real. Further, if u , v are eigenvectors corresponding to two unequal eigenvalues λ and μ , it follows that λ < u, v > = ( A u, v ) = ( u, A v ) = μ < u, v >, since both eigenvalues λ and μ must be real. Therefore, ( λ μ ) < 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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This note was uploaded on 07/26/2011 for the course MATH 716 taught by Professor Tanveer,s during the Spring '08 term at Ohio State.
 Spring '08
 Tanveer,S
 Math

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