A Primal-Dual Weak Galerkin Finite Element Method for Second Order Elliptic Equations in Non-Divergence Form

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1 A Primal-Dual Weak Galerkin Finite Element Method for Second Order Elliptic Equations in Non-Divergence Form Chunmei Wang Visiting Assistant Professor School of Mathematics Georgia Institute of Technology Collaborated with: Junping Wang (NSF) Supported by NSF Grant DMS Jiangsu Provincial Foundation Award BK October 26, 2015

2 Talk Outline Model Problem and Background Primal-Dual Weak Galerkin Finite Element Method Error Estimates Numerical Tests

3 Model Problem Second Order Elliptic Equations in Non-Divergence Form Find u = u(x) satisfying u Ω = 0, such that d a ij iju 2 = f, in Ω. i,j=1 Assume a(x) = (a ij (x)) d d L (Ω).

4 Background of Second Order Elliptic Equations in Non-Divergence Form probability and stochastic processes nonlinear PDEs in conjunction with linearization techniques such as the Newton s iterative method a(x) is hardly smooth nor even continuous. a(x) is merely essentially bounded in the application to Hamilton-Jacobi-Bellman equations. For nonlinear PDEs discretized by discontinuous finite elements, their linearization involves at most piecewise smooth coefficients.

5 Variational Equation Assume the model problem has a unique strong solution in X = H 2 (Ω) H0 1 (Ω) satisfying u 2 C f 0. Variational Equation: Find u X such that b(u, w) = (f, w) w Y = L 2 (Ω). b(u, w) = (Lu, w) Lu = d i,j=1 a ij ij 2u b(, ) satisfies the inf-sup condition b(v, σ) sup Λ σ Y, σ Y. v X,v 0 v X

6 Weak Hessian W (K) = {v = {v 0, v b, v g } : v 0 L 2 (K), v b L 2 ( K), v g [L 2 ( K)] d }. Weak Second Order Partial Derivative The weak second order partial derivative of v W (K) is defined as a bounded linear functional 2 ij,w v on H2 (K) so that its action on each ϕ H 2 (K) is given by 2 ij,w v, ϕ K := (v 0, 2 jiϕ) K v b n i, j ϕ K + v gi, ϕn j K. weak Hessian of v W (K): 2 w,k v = { 2 ij,w v } d d

7 Discrete Weak Hessian Discrete Weak Second Order Partial Derivative A discrete weak second order partial derivative of v W (K), denoted by ij,w,r,k 2 v, is defined as the unique polynomial satisfying ( 2 ij,w,r,k v, ϕ) K = (v 0, 2 jiϕ) K v b n i, j ϕ K + v gi, ϕn j K, ϕ P r (K). discrete weak Hessian of v W (K): 2 w,r,k v = { 2 ij,w,r,k v} d d

8 Weak Finite Element Spaces W k (T ) = {{v 0, v b, v g } P k (T ) P k (e) [P k 1 (e)] d } W h,k = { } {v 0, v b, v g } : {v 0, v b, v g } T W k (T ), T T h Wh,k 0 = {{v 0, v b, v g } W h,k, v b e = 0, e Ω} } S h,k = {σ : σ T S k (T ), T T h P k 2 (T ) S k (T ) P k 1 (T ) The choice of S k (T ) = P k 2 (T ) has the least degrees of freedom, but the resulting numerical solution may not be as accurate as the case of S k (T ) = P k 1 (T ).

9 Primal-Dual Weak Galerkin Finite Element Algorithms Find u h W 0 h,k satisfying b h (v, σ) = (f, σ), σ S h,k. b h (v, σ) = d T T h i,j=1 (a ij ij,w 2 v, σ) T. The problem is not well-posed unless an inf-sup condition is satisfied. Constrained optimization problem: Find u h Wh,k 0 such that ( ) 1 u h = argmin v W 0 h,k,b h (v,σ)=(f,σ), σ S h,k 2 s h(v, v). Stablizer s h (v, v) = T h 3 T v 0 v b, v 0 v b T +h 1 T v 0 v g, v 0 v g T measures the continuity of v Wh,k 0 in the sense that v is a classical conforming element if and only if s h (v, v) = 0.

10 Primal-Dual Weak Galerkin Finite Element Algorithms Primal-Dual Weak Galerkin Algorithm Find (u h ; λ h ) Wh,k 0 S h,k satisfying s h (u h, v) + b h (v, λ h ) = 0, v Wh,k 0, b h (u h, σ) = (f, σ), σ S h,k.

11 L 2 Projections Q 0 : L 2 projection onto P k (T ) Q b : L 2 projection onto P k (e) Q g = (Q g1, Q g2,..., Q gd ): L 2 projection onto [P k 1 (e)] d Q h : L 2 projection onto W h,k, such that on each element T, Q h w = {Q 0 w, Q b w, Q g ( w)} Q h : L 2 projection onto S h,k

12 Inf-Sup condition ( Brezzi and Babuska ) Inf-Sup Condition Assume that the coefficient matrix a = {a ij } d d is uniformly piecewise continuous in Ω with respect to the finite element partition T h. For any σ S h,k, there exists v σ W 0 h,k satisfying b h (v σ, σ) 1 2 σ 2 0, v σ 2 2,h C σ 2 0, provided that the meshsize h < h 0 for a sufficiently small, but fixed parameter h 0 > 0.

13 Error Estimates Error Estimates in a Discrete H 2 Norm Assume that the coefficient functions a ij are uniformly piecewise continuous in Ω. Assume that the exact solution u is sufficiently regular such that u H k+1 (Ω). Let (u h ; ρ h ) W 0 h,k S h,k be primal-dual WG solution. There exists a constant C such that u h Q h u 2,h + λ h Q h λ 0 Ch k 1 u k+1, provided that the meshsize h < h 0 holds true for a sufficiently small, but fixed h 0 > 0. v 2 2,h = T T h d i,j=1 Q h(a ij 2 ij v 0) 2 T + s h(v, v)

14 Error Estimates Error Estimates in H 1 Let u h = {u 0, u b, u g } Wh,k 0 be the primal-dual WG solution. Assume that a ij C 1 (Ω), and the exact solution u H k+1 (Ω). There exists a constant C such that T Th u 0 u 2 T 1 2 Ch k u k+1, provided that the meshsize h is sufficiently small and the dual problem has the H 1 -regularity with the a priori estimate w 1 C θ 1.

15 Error Estimates Error Estimates in L 2 Assume a ij C 1 (Ω) [ Π T Th W 2, (T ) ]. In addition, assume that the dual problem has H 2 -regularity with the a priori estimate w 2 C θ 0, and P 1 (T ) S k (T ) for all T T h. Then, we have u 0 u 0 Ch k+1 u k+1, u b Q b u L 2 Ch k+1 u k+1, u g Q b u L 2 Ch k u k+1. provided that the meshsize h is sufficiently small. ( e b L 2 = h T e b 2 T T T h ) 1 2 ( e g L 2 = h T e g 2 T T T h ) 1 2

16 Error Estimates Remark In the case of P 1 (T ) S 2 (T ), we have the following sub-optimal order error estimate u 0 u 0 Ch k u k+1 provided that (1) a ij C 1 (Ω), (2) the meshsize h is sufficiently small, and (3) the dual problem has the H 1 -regularity with the a priori estimate w 1 C θ 1.

17 Numerical Tests the exact solution u = sin(x 1 ) sin(x 2 ) a 11 = 3, a 12 = a 21 = 1, a 22 = 2 λ h is piecewise linear Table: numerical error and convergence order for domain Ω = (0, 1) 2 1/h e 0 0 order e g L 2 order λ h 0 order e e e e e e

18 Numerical Tests Table: numerical error and convergence order for L-shaped domain with vertexes (0,0), (2,0), (1,1), (1,2), and (0,2). 2/h e 0 0 order e g L 2 order λ h 0 order e e e e e

19 Numerical Tests the exact solution u = sin(x 1 )sin(x 2 ) Ω = ( 1, 1) 2 a11 = 1 + x 1, a12 = a21 = 0.5 x 1 x 2 1 3, a22 = 1 + x 2 Table: numerical error and convergence order ( λ h is piecewise linear) 2/h e 0 0 order e g L 2 order λ h 0 order e e e e e

20 Numerical Tests Table: numerical error and convergence order (λ h is piecewise constant) 2/h e 0 0 order e g L 2 order λ h 0 order e e e

21 Numerical Tests Consider 2 (1 + δ ij ) x i x j x i x j 2 iju = f, in Ω, i,j=1 u = 0, on Ω. Ω = ( 1, 1) 2 the exact solution u(x 1, x 2 ) = (x 1 e 1 x1 x 1 )(x 2 e 1 x2 x 2 ).

22 Numerical Tests Table: Numerical error and convergence order (λ h is piecewise linear). 2/h e 0 0 order e g L 2 order λ h 0 order

23 Numerical Tests figure for u

24 Numerical Tests figure for rho h

25 Numerical Tests Table: Numerical error and convergence order (λ h is piecewise constant). 2/h e 0 0 order e g L 2 order λ h 0 order e e

26 Numerical Tests figure for rho h

27 Numerical Tests Consider 2 (δ ij + x ix j x 2 ) 2 iju = (2α 2 α) x α 2, in Ω. i,j=1 the exact solution u = x α (α > 1) Ω = (0, 1) 2 α = 1.6 Table: numerical error and convergence order(λ h is piecewise linear) 1/h e 0 0 order e g L 2 order λ h 0 order e e e

28 Numerical Tests figure for u

29 Numerical Tests Table: numerical error and convergence order (λ h is piecewise constant) 1/h e 0 0 order e g L 2 order λ h 0 order e e e

30 Thank you very much for your attention! Chunmei Wang Visiting Assistant Professor School of Mathematics Georgia Institute of Technology Homepage: cwang462/