# solh5 - Solution to Set 5: Math 716 1. Show that the...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 inner-product. 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 inner-product 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 semi-positive 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 Gram-Schmidt orgonalization procedure to obtain or- thogonality. Thus all eigenvectors are orthogonal....
View Full Document

## 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.

### Page1 / 6

solh5 - Solution to Set 5: Math 716 1. Show that the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online