On an unsymmetric eigenvalue problem governing free vibrations of - PowerPoint PPT Presentation

On an unsymmetric eigenvalue problem governing free vibrations of fluid-solid structures Markus Stammberger markus.stammberger@tuhh.de This is joint work with Heinrich Voss Hamburg University of Technology TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 1 / 18

Outline Problem definition and properties 1 Numerical methods 2 Numerical Results 3 TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 2 / 18

Problem definition and properties Outline Problem definition and properties 1 Numerical methods 2 Numerical Results 3 TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 3 / 18

Problem definition and properties Problem definition Vibrations of fluid-solid structures can be modelled in terms of solid displacement and fluid pressure and one obtains the classical form of an eigenproblem. Div σ ( u ) + λρ s u = 0 in Ω s , ∆ p + λ c 2 p = 0 in Ω f , σ ( u ) n − p n = 0 on Γ I , ∇ pn − λρ f un = 0 on Γ I , u = 0 on Γ D , ∇ p n = 0 on Γ N , where u : solid displacement p : fluid pressure λ : eigenparameter σ ( u ) : linearized stress tensor ρ s , ρ f : densities of solid and fluid TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 4 / 18

Problem definition and properties Problem definition, cntd. This eigenvalue problem can be given an unsymmetric variational formulation which can be discretized by the Finite-Element method and one obtains the unsymmetric matrix eigenproblem „ M s „ K s « „ x s « « „ x s « C 0 Kx := = λ =: λ Mx , (1) − C T 0 K f x f M f x f where K s , M s ∈ R s × s are symmetric positive definite stiffness and mass matrices of the solid, K f , M f ∈ R f × f are symmetric stiffness and mass matrices of the fluid, where K f is semi positive-definite and M f positive definite, C ∈ R s × f is due to the coupling effects between fluid and solid, x s ∈ R s is the solid displacement vector, and x f ∈ R f the fluid pressure vector. This talks considers the properties of eigenproblem (1) and discusses ways how to use the symmetry of K s , K f , M s , and M f to adapt symmetric eigensolvers to the given problem. TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 5 / 18

Problem definition and properties Problem definition, cntd. This eigenvalue problem can be given an unsymmetric variational formulation which can be discretized by the Finite-Element method and one obtains the unsymmetric matrix eigenproblem „ M s „ K s « „ x s « « „ x s « C 0 Kx := = λ =: λ Mx , (1) − C T 0 K f x f M f x f where K s , M s ∈ R s × s are symmetric positive definite stiffness and mass matrices of the solid, K f , M f ∈ R f × f are symmetric stiffness and mass matrices of the fluid, where K f is semi positive-definite and M f positive definite, C ∈ R s × f is due to the coupling effects between fluid and solid, x s ∈ R s is the solid displacement vector, and x f ∈ R f the fluid pressure vector. This talks considers the properties of eigenproblem (1) and discusses ways how to use the symmetry of K s , K f , M s , and M f to adapt symmetric eigensolvers to the given problem. TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 5 / 18

Problem definition and properties Properties Some properties can easily be derived: Lemma M − 1 M − 1 „ « K s C (1) can be symmetrized by T := s s , i.e. 0 I T T Kx = λ T T Mx is a symmetric eigenvalue problem. (1) has only real non-negative eigenvalues. „ x s « If x := is a right eigenvector of (1) corresponding to the eigenvalue λ , x f „ λ x s « then ˆ x := is a left eigenvector. x f „ K s « 0 Right eigenvectors can be chosen orthonormal with respect to ˜ M := , 0 M f „ M s « 0 left eigenvectors can be chosen orthogonal with respect to ¯ M := . 0 K f Right eigenvectors x and left eigenvectors ˆ x corresponding to distinct eigenvalues satisfy x T Kx = ˆ x T Mx = 0 . ˆ TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 6 / 18

Problem definition and properties Properties Some properties can easily be derived: Lemma M − 1 M − 1 „ « K s C (1) can be symmetrized by T := s s , i.e. 0 I T T Kx = λ T T Mx is a symmetric eigenvalue problem. (1) has only real non-negative eigenvalues. „ x s « If x := is a right eigenvector of (1) corresponding to the eigenvalue λ , x f „ λ x s « then ˆ x := is a left eigenvector. x f „ K s « 0 Right eigenvectors can be chosen orthonormal with respect to ˜ M := , 0 M f „ M s « 0 left eigenvectors can be chosen orthogonal with respect to ¯ M := . 0 K f Right eigenvectors x and left eigenvectors ˆ x corresponding to distinct eigenvalues satisfy x T Kx = ˆ x T Mx = 0 . ˆ TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 6 / 18

Problem definition and properties Properties Some properties can easily be derived: Lemma M − 1 M − 1 „ « K s C (1) can be symmetrized by T := s s , i.e. 0 I T T Kx = λ T T Mx is a symmetric eigenvalue problem. (1) has only real non-negative eigenvalues. „ x s « If x := is a right eigenvector of (1) corresponding to the eigenvalue λ , x f „ λ x s « then ˆ x := is a left eigenvector. x f „ K s « 0 Right eigenvectors can be chosen orthonormal with respect to ˜ M := , 0 M f „ M s « 0 left eigenvectors can be chosen orthogonal with respect to ¯ M := . 0 K f Right eigenvectors x and left eigenvectors ˆ x corresponding to distinct eigenvalues satisfy x T Kx = ˆ x T Mx = 0 . ˆ TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 6 / 18

Problem definition and properties Properties Some properties can easily be derived: Lemma M − 1 M − 1 „ « K s C (1) can be symmetrized by T := s s , i.e. 0 I T T Kx = λ T T Mx is a symmetric eigenvalue problem. (1) has only real non-negative eigenvalues. „ x s « If x := is a right eigenvector of (1) corresponding to the eigenvalue λ , x f „ λ x s « then ˆ x := is a left eigenvector. x f „ K s « 0 Right eigenvectors can be chosen orthonormal with respect to ˜ M := , 0 M f „ M s « 0 left eigenvectors can be chosen orthogonal with respect to ¯ M := . 0 K f Right eigenvectors x and left eigenvectors ˆ x corresponding to distinct eigenvalues satisfy x T Kx = ˆ x T Mx = 0 . ˆ TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 6 / 18

Problem definition and properties Properties Some properties can easily be derived: Lemma M − 1 M − 1 „ « K s C (1) can be symmetrized by T := s s , i.e. 0 I T T Kx = λ T T Mx is a symmetric eigenvalue problem. (1) has only real non-negative eigenvalues. „ x s « If x := is a right eigenvector of (1) corresponding to the eigenvalue λ , x f „ λ x s « then ˆ x := is a left eigenvector. x f „ K s « 0 Right eigenvectors can be chosen orthonormal with respect to ˜ M := , 0 M f „ M s « 0 left eigenvectors can be chosen orthogonal with respect to ¯ M := . 0 K f Right eigenvectors x and left eigenvectors ˆ x corresponding to distinct eigenvalues satisfy x T Kx = ˆ x T Mx = 0 . ˆ TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 6 / 18

Problem definition and properties Properties, cntd. Lemma Let λ j ( A , B ) denote the j smallest eigenvalue of the eigenproblem Ax = λ Bx regarding the multiplicity of eigenvalues. Then it holds that λ j ( K , M ) ≤ λ j ( K s , M s ) , j = 1 , . . . , s λ s + f + 1 − j ( K , M ) ≥ λ s + 1 − j ( K s , M s ) , j = 1 , . . . , s λ j ( K , M ) ≤ λ j ( K f , M f ) , j = 1 , . . . , f λ s + f + 1 − j ( K , M ) ≥ λ f + 1 − j ( K f , M f ) , j = 1 , . . . , f . Proof: Let E s := span { e 1 , . . . , e s } where e j ∈ R s + f denotes the j th unit vector containing a 1 in its j th component and zeros elsewhere. Then it holds that x T T T Kx x T T T Kx λ j ( K , M ) = min max ≤ min max x T T T Mx x T T T Mx dim V = j x ∈ V , x � = 0 dim V = j , V ⊂ E s x ∈ V , x � = 0 y T K s M − 1 K s y s = min max = λ j ( K s , M s ) . y T K s y dim W = j , W ⊂ R s y ∈ W , y � = 0 The second inequality is obtained analogously from the maxmin characterization, and the third and fourth inequalities follow in a similar way. TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 7 / 18

Problem definition and properties An inverse-free Rayleigh functional A Rayleigh quotient for fluid-solid eigenproblems is given immediately by its symmetrized version. As it involves inverse matrices it is numerically less valuable and we are interested in an inverse-free analogon. „ x s « For a given right eigenvector corresponding to the eigenvalue λ it holds x f « T „ K s „ λ x s C « „ x s « x f 0 K f x f λ = « T „ M s „ λ x s « „ x s « 0 − C T x f x f M f This suggests to define a Rayleigh functional p for some general s + f -dimensional vector by the requirement p ( x s , x f ) = p ( x s , x f ) x T s K s x s + p ( x s , x f ) x T s Cx f + x T f K f x f f C T x s + x T p ( x s , x f ) x T s M s x s − x T f M f x f which leads to p ( x s , x f ) 2 x T s M s x s + p ( x s , x f )( x T f M f x f − x T s K s x s − 2 x T s Cx f ) − x T f K f x f = 0 . We therefore choose the unique positive root of this equation as Rayleigh functional. TUHH Markus Stammberger On an unsymmetric eigenvalue problem 12.09.2008 8 / 18