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ASE 365 - Lecture 35

# ASE 365 - Lecture 35 - u u M K 2 ϖ = problem eigenvalue...

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Unformatted text preview: u u M K 2 ϖ = : problem eigenvalue Algebraic i i i i i th M K i u u u 2 2 ϖ ϖ = : satisfies } , { eigenpair j j j j j th M K j u u u 2 2 ϖ ϖ = : satisfies } , { eigenpair j T i j j T i T i i T j i i T j T j M K M K u u u u u u u u u u 2 2 ϖ ϖ = = : : : by Multiply j T i T i T j i T j T i T j K K K K K K u u u u u u u u = = = ) ( , & scalar is Since i T j j i M u u ) ( 2 2 ϖ ϖ- = . and both to" respect with " orthogonal are and that say We have also must we then true, is this If have must we , if Therefore, K M K M j i i T j i T j j i u u u u u u . . 2 2 = = ≠ ϖ ϖ Orthogonality of eigenfunctions ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 dx AU U dx U EA U U EA U dx U EA U dx AU U dx U EA U U EA U dx U EA U s L r s s L r L s r s L r r L s r r L s L r s r L s ρ ϖ ρ ϖ ∫ ∫ ∫ ∫ ∫ ∫ = ′ ′ + ′- = ′ ′- = ′ ′ + ′- = ′ ′- 2 2 : integrate Multiply, . , satisfy eigenpairs and The s s s r r r th th AU U EA AU U EA s r ρ ϖ ρ ϖ 2 2 = ′ ′- = ′ ′- ∫- = L s r s r dx U AU 2 2 ) ( ρ ϖ ϖ ( 29 ( 29 . , If 2 2 = ′ ′- = ′ ′- → = ≠ ∫ ∫ ∫ dx U EA U dx U EA U dx U AU r L s s L r s L r s r ρ ϖ ϖ ( 29 ( 29 . and then , let we If rs r r rs r L s r L r s rs r L s r L r r m k dx U EA U dx U EA U m dx U AU dx AU m δ ϖ δ δ ρ ρ 2 2 = = ′ ′- = ′ ′- = ≡ ∫ ∫ ∫ ∫ : Example U(x) s s s r r r AU U EA AU U EA ρ ϖ ρ ϖ 2 2 ) ( ) ( = ′ ′- = ′ ′- dx AU U dx U EA U U EA U dx U EA U dx AU U dx U EA U U EA U dx U EA U s L r s s L r L s r s L r r L s r r L s L r s r L s ρ ϖ ρ ϖ ∫ ∫ ∫ ∫ ∫ ∫ = ′ ′ + ′- = ′ ′- = ′ ′ + ′- = ′ ′- 2 2 ) ( ) ( L L L L L U k U m U EA u k u m u EA- = ′ →-- = ′ 2 ) ( ϖ : BC k m dx AU U dx U EA U U k U U m U dx AU U dx U EA U U k U U m U s L r s s L r L s r L s s r r L s r r L s L r s L r r s ρ ϖ...
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ASE 365 - Lecture 35 - u u M K 2 ϖ = problem eigenvalue...

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