The energy based discontinuous Galerkin method Daniel Appel Applied - PowerPoint PPT Presentation

The energy based discontinuous Galerkin method Daniel Appelö Applied Mathematics University of Colorado , Boulder

Outline • Applications / interests • Why second order? • Why energy based? • Methods Contributors 
 Thomas Hagstrom , Southern Methodist U . Fatemeh Pourahmadian , CU , Boulder . Olof Runborg , Royal Inst . of Technology . Siyang Wang , Chalmers Inst . of Technology . Please Ask Questions Throughout ICERM 2018

Micropolar Elasticity • Classic Hooke’s theory: Continuum description with balance of force and stress . Symmetric tensors . • Voigt ( 1877 ) , Cosserat brothers ( 1909 ): Asymmetric theory with force - stress and couple - stress balance . • Micropolar theory of Eringen - Nowacki: displacements and rotations . • Allows for construction of phononic materials . • Acoustic properties depends on grain - to - grain 
 6 forces . These can be controlled using EM fields . 5 4 Systems of wave equations (dispersive) 3 2 ρ v tt = ( µ + κ ) v xx − κθ x , 1 ρ J θ tt = γθ xx + κ ( v x − 2 θ ) . 0 0 1 2 3 4 5 6 ICERM 2018

Interests in numerical homogenization and beyond • Developing wave solvers that are accurate and geometrically flexible (high order dG with upwind fluxes) . • Methods that couple well with micro - models . 
 In particular micro models described in terms of (discrete) energy balance (Heterogenous Multiscale Models) . • Methods that allow the macro - models to have anisotropy , dispersive and non - linear e ff ects . • Fast solvers that can be used for direct simulation of Maxwell / acoustic cell problems . • Variable order methods that can be used together with multi - level algorithms in UQ , inverse problems and frequency domain solvers . • Taming of the poor CFL constraints of dG . ICERM 2018

Motivation: Why solve in second order form? • “There are three kinds of mathematicians; those who can count and those who cannot… ” . • Maxwell in 3 D has 12 derivative matrices . • Formulated as a second order problem there can be as few as 3 derivative matrices (although for complex materials the saving may be less) . • Many problems in solid mechanics are formulated in terms of energy densities - this fits well with our formulation . • The energy based formulation naturally incorporates upwind fluxes for second order formulations . • The method is parameter free . ICERM 2018

Energy based discontinuous Galerkin method Plan: • What energy are we talking about? • The general formulation… • What the general formulation really is… • SIPDG . • EDG with central flux / Baumann - Oden form . • Full EDG (upwind / alt . / central flux) . • Extension to SBP - SAT finite di ff erences . ICERM 2018

What energy are we talking about? The energy we consider is a non - negative functional of the kinetic and potential energy density . For example: 2 ⏐ ⏐ � 1 ∂ u ⏐ ⏐ E ( t ) = + G ( u , ∇ u , x ) . ⏐ ⏐ 2 ∂ t ⏐ ⏐ Ω ICERM 2018

What energy are we talking about? The energy we consider is a non - negative functional of the kinetic and potential energy density . For example: 2 ⏐ ⏐ � 1 ∂ u ⏐ ⏐ E ( t ) = + G ( u , ∇ u , x ) . ⏐ ⏐ 2 ∂ t ⏐ ⏐ Ω The equations we treat result from the Euler - Lagrange equations derived from the action principle associated with the Lagrangian n be identified as the ⏐ 2 − G − u · f n 1 ⏐ ⏐ ∂ u ⏐ 2 ∂ t ICERM 2018

The Euler - Lagrange equations are � ∂ G ∂ 2 u i � ∂ − ∂ G � ∂ t 2 = + f i , ∂ x k ∂ u i,k ∂ u i k We prefer to evolve “displacement” and “velocity” ∂ u i ∂ t − v i = 0 , ∂ t − � ∂ G � ∂ v i ∂ + ∂ G � = f i , ∂ t − ∂ x k ∂ u i,k ∂ u i k ICERM 2018

The Euler - Lagrange equations are � ∂ G ∂ 2 u i � ∂ − ∂ G � ∂ t 2 = + f i , ∂ x k ∂ u i,k ∂ u i k We prefer to evolve “displacement” and “velocity” ∂ u i ∂ t − v i = 0 , ∂ t − � ∂ G � ∂ v i ∂ + ∂ G � = f i , ∂ t − ∂ x k ∂ u i,k ∂ u i k Seek methods that automagically satisfies � 1 � � d ∂ G 2 | v | 2 + G = � v · f + v i n k , . dt ∂ u i,k ∂ Ω j Ω j Ω j i,k ICERM 2018

B 0 ring testing for velocity ∂ t − � ∂ G � ∂ v i ∂ + ∂ G � = f i , ∂ t − ∂ x k ∂ u i,k ∂ u i k ICERM 2018

Boring testing for velocity ∂ t − � ∂ G � ∂ v i ∂ + ∂ G � = f i , ∂ t − ∂ x k ∂ u i,k ∂ u i k What about the displacement equation? ∂ u i ∂ t − v i = 0 , ICERM 2018

Boring testing for velocity ∂ t − � ∂ G � ∂ v i ∂ + ∂ G � = f i , ∂ t − ∂ x k ∂ u i,k ∂ u i k Note that we are halfway there! � 1 � � ∂ G d 2 | v | 2 + G = � v · f + v i n k , . dt ∂ u i,k Ω j Ω j ∂ Ω j i,k ICERM 2018

Santa’s test Test the displacement with the ∂ u i ∂ t − v i = 0 , variation of the potential energy ICERM 2018

Santa’s test ∂ u i ∂ t − v i = 0 , ICERM 2018

Use solution as test function , integrate by parts , 
 add up to get a Christmas present! ICERM 2018

Use solution as test function , integrate by parts , 
 add up to get a Christmas present! Element - wise energy identity Theorem 1. Suppose U h ( t ) and the fluxes v ∗ , w ∗ are given. Suppose further that (2.2) is linear. Then dU h satisfying Problem 1 is uniquely determined and the energy identity for E h ( t ) = � j E h j ( t ) dt �� � dE h � ∂ G � � � � v h v ∗ i − v h ( u h , ∇ u h , x ) + v h i w ∗ � (2.16) = i f i ( x , t ) + n k , i i,k dt ∂ u i,k ∂ Ω j Ω j i,j i,j k is satisfied. h 1 � ⏐ 2 + G ( u h , ∇ u h , x ) Where E h ⏐ ⏐ v h ⏐ j ( t ) = 2 Ω j ICERM 2018

Accepts upwind and averaged fluxes For the general parametrization We get element - wise contributions to the energy ICERM 2018

Accepts upwind and averaged fluxes For the general parametrization We get element - wise contributions to the energy Two attractive choices are: ∂ G Alternating flux v ∗ i = v h w ∗ ( u h 2 , ∇ u h i,k = 2 , x ) . i, 1 , ∂ u i,k i }} − ζ i Sommerfeld flux i = {{ v h {{ }} − 2 2 [[ D ∇ u i G h ]] , v ∗ i ]] k + {{ ∂ G h i,k = − 1 w ∗ [[ v h }} . 2 ζ i ∂ u i,k ICERM 2018

Examples of equations that fit into the framework σ ji,j − ρ ¨ u i = 0 Micropolar elasticity m ji,j + ε ijk σ jk − ρ J ¨ ϕ i = 0 � � σ ij = λ u k,k δ ij + µ ( u i,j + u j,i ) + κ ( u i,j − u j,i ) / 2 − ε kij ϕ k m ij = αϕ k,k δ ij + β + γ ( ϕ i,j + ϕ j,i ) + β − γ ( ϕ i,j − ϕ j,i ) 2 2 Maxwell Drude material Z t = − Γ e Z + ε 0 ω 2 E t = U , µ 0 ε 0 U t = ∆ E − µ 0 Z , J t = Z , pe U , Y t = − Γ m Y + µ 0 ω 2 H t = V , µ 0 ε 0 V t = ∆ H − ε 0 Y , K t = Y , pm V , E = 1 � � µ 0 U 2 + V 2 � + ( ∇ × E ) 2 + ( ∇ × H ) 2 + Z 2 + ε 0 − G Y 2 � ε 0 µ 0 . ε 0 ω 2 µ 0 ω 2 2 pe pm ICERM 2018

“Ok this is very general , but how do I implement it?!” ICERM 2018

Back to basics SIPDG for wave . u tt − ∇ · ( c ∇ u ) = f, in J × Ω , u = 0 , on J × ∂ Ω , Test as usual , U is the test function . ( u h tt , U ) + a h ( u h , U ) ∀ U ∈ V h , t ∈ J, = ( f, U ) ICERM 2018

Back to basics SIPDG for wave . u tt − ∇ · ( c ∇ u ) = f, in J × Ω , u = 0 , on J × ∂ Ω , Test as usual , U is the test function . ( u h tt , U ) + a h ( u h , U ) ∀ U ∈ V h , t ∈ J, = ( f, U ) Bilinear form: � � � � a h ( u, w ) = c ∇ u · ∇ wdx − [[ u ]] · {{ c ∇ w }} dA K F K ∈ T h F ∈ F h [[ w ]] · {{ c ∇ u }} dA + γ � � � � [[ u ]] · [[ w ]] dA. − h F F F ∈ F h F ∈ F h ICERM 2018

Back to basics SIPDG for wave . u tt − ∇ · ( c ∇ u ) = f, in J × Ω , u = 0 , on J × ∂ Ω , Test as usual , U is the test function . ( u h tt , U ) + a h ( u h , U ) ∀ U ∈ V h , t ∈ J, = ( f, U ) Bilinear form: I . b . p , symmetry , penalty (stabilization) . � � � � a h ( u, w ) = c ∇ u · ∇ wdx − [[ u ]] · {{ c ∇ w }} dA K F K ∈ T h F ∈ F h [[ w ]] · {{ c ∇ u }} dA + γ � � � � [[ u ]] · [[ w ]] dA. − h F F F ∈ F h F ∈ F h Energy conserving - optimal convergence - known bf . s . ICERM 2018

Conservative Energy DG First order in time . u t − v = 0 , in J × Ω , v t − ∇ · ( c ∇ u ) = f, in J × Ω , Velocity equation looks as in SIPDG (b is not the same) . ( v h t , V ) + b h ( u h , V ) = ( f, V ) , U = 0 , ∀ V ∈ V h , t ∈ J, ICERM 2018

Conservative Energy DG First order in time . u t − v = 0 , in J × Ω , v t − ∇ · ( c ∇ u ) = f, in J × Ω , Velocity equation looks as in SIPDG (b is not the same) . ( v h t , V ) + b h ( u h , V ) = ( f, V ) , U = 0 , ∀ V ∈ V h , t ∈ J, We would like this type of estimate! What is missing above? ) = 1 d � ( v h ) 2 + c ∇ u h · ∇ u h dx = ( f, v h ) . � 2 dt K K ∈ T h ICERM 2018

Conservative Energy DG Time for Christmas gi fu s! ( v h t , V ) + b h ( u h , V ) = ( f, V ) , U = 0 , ∀ V ∈ V h , t ∈ J, d h ( u h t , U ) − b h ( U, v h ) = 0 , V = 0 , ∀ U ∈ V h , t ∈ J. ICERM 2018