New class of finite element methods: weak Galerkin methods

Transcription

1 New class of finite element methods: weak Galerkin methods Xiu Ye University of Arkansas at Little Rock

2 Second order elliptic equation Consider second order elliptic problem: a u = f, in Ω (1) u = 0, on Ω. (2) Testing (1) by v H0 1 (Ω) gives a uvdx = a u vdx a u nvds = fvdx. Ω Ω Ω Ω where (f,g) = Ω fgdx. (a u, v) = (f, v),

3 PDE and its weak form PDE: find u satisfies a u = f, in Ω u = 0, on Ω. Its weak form: find u H0 1 (Ω) such that (a u, v) = (f, v), v H 1 0(Ω).

4 Infinity vs finite Weak form: find u H0 1 (Ω) such that (a u, v) = (f, v), v H 1 0(Ω). Let V h H0 1 (Ω) be a finite dimensional space. Continuous finite element method: find u h V h such that (a u h, v h ) = (f,v h ), v h V h,

5 Continuous finite element method Find u h V h such that (a u h, v h ) = (f, v h ), v h V h. Let V h = Span{φ 1,,φ n } and u h = n j=1 c jφ j, then n j=1 (a φ j, φ i )c j = (f, φ i ), i = 1,,n. The equation above is a symmetric and positive definite linear system. Solve it to obtain the finite element solution u h.

6 Limitations of the continuous finite element methods On approximation functions. P k only for triangles and Q k for quadrilaterals. Hard to construct high order and special elements such as C 1 conforming element. On mesh generation. Only triangular or quadrilateral meshes can be used in 2D. Hybrid meshes or meshes with hanging nodes are not allowed. Not compatible to hp adaptive technique.

7 Cause and solution Cause: Continuity requirement of approximating functions cross element boundaries. Solution: Use discontinuous approximations.

8 Pros and cons of using discontinuous functions Pros Flexibility on approximation functions. Polynomial P k can be used on any polygonal element. Easy to construct high order element. Flexibility on mesh generation. Hybrid meshes or meshes with hanging nodes are allowed. Compatible to hp adaptive technique. Cons There are more unknowns. Complexity in finite element formulations due to enforcing connections of numerical solutions between element boundaries.

9 Weak Galerkin finite element methods Weak Galerkin (WG) methods use discontinuous approximations. The WG methods keep the advantages: Flexible in approximations. Avoid construction of special elements such as C 1 conforming elements. Flexible in mesh generation. Hybrid meshes or meshes with hanging nodes can be used. and minimize the disadvantages: Simple formulations. Comparable number of unknowns to the continuous finite element methods if implemented appropriately.

10 Weak functions Let T be a quadrilateral with e j for j = 1,,4 as its four sides. Define { v0 P 1 (T ), in T 0 v = v b P 0 (e), on e T Define V h (T ) = {v L 2 (T ) : v = {v 0,v b }} = span{φ 1,,φ 7 } where { 1, on ei φ j = j = 1,,4 0, otherwise { { { 1, in T 0 x, in T 0 y, in T 0 φ 5 = φ 6 = φ 7 = 0, on T 0, on T 0, on T

11 Weak derivatives Define a weak gradient w v [P 0 (T )] 2 for v = {v 0,v b } V h (T ) on the element T : ( w v,q) T = (v 0, q) T + v b,q n T, q [P 0 (T )] 2. Let φ j = {φ j,0,φ j,b }, j = 1,,7. The definition of the weak gradient gives that for any q [P 0 (T )] 2 ( w φ 5,q) = (φ 5,0, q) T + φ 5,b,q n T = 0. We have w φ 5 = w φ 6 = w φ 7 = 0. Using the definition of w, we can find for j = 1,,4 w φ j = e j T n j. Weak gradient w for all the basis function φ j can be found explicitly.

12 The local stiffness matrix for the WG method Denote Q b the L 2 projection to P 0 (e j ). Q b v 0 ej = v 0 (m j ) where m j is the midpoint of e j. Define a T (v,w) = (a w v, w w) T + h 1 Q b v 0 v b,q b w 0 w b T. The local stiffness matrix A for the WG method on the element T for second order elliptic problem is a 7 7 matrix A = (a T (φ i,φ j )), i,j = 1,,7. e 4 e 1 T e 3 e 2

13 Weak Galerkin finite element methods Define weak function v = {v 0,v b } such that { v0, in T 0 v = v b, Define weak Galerkin finite element space on T V h = {v = {v 0,v b } : v 0 T P j (T ),v b P l (e),e T,v b = 0 on Ω}. Define a weak gradient w v [P r (T )] d for v V h on each element T : ( w v,q) T = (v 0, q) T + v b,q n T, q [P r (T )] d. Weak Galerkin element: (P j (T ),P l (e),[p r (T )] d ). For example: (P 1 (T ),P 0 (e),[p 0 (T )] d ).

14 Weak Galerkin finite element formulation Define a(u h,v h ) = (a w u h, w v h ) + h 1 T u 0 u b,v 0 v b T. T The WG method: find u h = {u 0,u b } V h satisfying a(u h,v h ) = (f,v h ), v h V h. Theorem. Let u h be the solution of the WG method associated with local spaces (P k (T ),P k (e),[p k 1 (T )] d ), then h Q h u u h + Q h u u h Ch k+1 u k+1, where Q h u is the L 2 projection of u.

15 Simple formulation: the WG method for the Stokes equations The weak form of the Stokes equations: find (u,p) [H0 1(Ω)]d L 2 0 (Ω) that for all (v,q) [H0 1(Ω)]d L 2 0 (Ω) ( u, v) ( v,p) = (f,v) ( u, q) = 0. The weak Galerkin method: find (u h,p h ) V h W h such that for all (v,q) V h W h, ( w u h, w v) + s(u h,v) ( w v,p h ) = (f,v) ( w u h, q) = 0.

16 The WG method for the biharmonic equation The weak form of the Stokes equations: seeking u H0 2 (Ω) satisfying ( u, v) = (f,v), v H 2 0(Ω), Weak Galerkin finite element method: seeking u h V h satisfying ( w u h, w v) + s(u h, v) = (f, v), v V h.

17 Implementation of the WG method The WG method: find u h = {u 0,u b } V h satisfying a(u h,v h ) = (f,v h ), v h = {v 0,v b } V h. Effective implementation of the WG method: 1. Solve u 0 as a function of u b from the following local system on element T, a(u h,v h ) = (f,v h ), v h = {v 0,0} V h. (3) 2. Solve u b from the following global system, a(u h,v h ) = (f,v h ), v h = {0,v b } V h. (4) Theorem. The global system (4) is symmetric and positive definite. For the lowest order WG method, the number of unknowns of (4) is # of unknowns = # of interior edges.

18 The recent development: The WG least-squares method Consider the model problem, a u = f, in Ω u = 0, on Ω. Rewrite the problem as the system of first order equations, q + a u = 0, in Ω, q = f, in Ω, u = 0, on Ω. The least-squares method: find (q,u) H(div;Ω) H0 1 (Ω) such that for any (σ,v) H(div;Ω) H0 1(Ω), (q + a u, σ + a v) + ( q, σ) = (f, σ).

19 The WG Least-squares method The least-squares method: find (q,u) H(div;Ω) H0 1 (Ω) such that for any (σ,v) H(div;Ω) H0 1(Ω), (q + a u, σ + a v) + ( q, σ) = (f, σ). The WG least-squares method: find (q h,u h ) Σ h V h such that for any (σ,v) Σ h V h, (q h + a w u h, σ + a w v) + ( w q h, w σ) + s 1 (u h,v) + s 2 (q h, σ) = (f, σ).

20 Define D h = {n e : n e is unit and normal to e, e E h }, V h = {v = {v 0,v b } : v 0 T P k+1 (T ),v b e P k (e),e T,v b = 0, on Ω}, Σ h = {σ = {σ 0, σ b } : σ 0 T [P k (T )] d, σ b e = σ b n e,σ b e P k (e), e T }. Define s 1 (w,v) = T T h h 1 Q b w 0 w b, Q b v 0 v b T, s 2 (t, σ) = T T h h (t 0 t b ) n, (σ 0 σ b ) n T, The WG least-squares method: find (q h,u h ) Σ h V h such that for any (σ,v) Σ h V h, (q h + a w u h, σ + a w v) + ( w q h, w σ) + s 1 (u h,v) + s 2 (q h, σ) = (f, σ).

21 We introduce a norm V in V h as and a norm Σ in Σ h as v 2 V = T T h w v 2 T + s 1 (v,v), σ 2 Σ = T T h w σ 2 T + σ s 2 (σ, σ). Lemma. There is a constant C such that for all (σ,v) Σ h V h C( σ 2 Σ + v 2 V ) a(v, σ;v, σ). Theorem. Assume the exact solution u H k+2 (Ω) and q [H k+1 (Ω)] d. Then, there exists a constant C such that u h Q h u V + q h Q h q Σ Ch k+1 ( u k+2 + q k+1 ).

22 Implementation of the WG least-squares method The WG least-squares method: find (q h,u h ) Σ h V h such that for any (σ,v) Σ h V h, (q h + a w u h, σ + a w v) + ( w q h, w σ) + s 1 (u h,v) + s 2 (q h, σ) = (f, σ). Effective implementation of the WG least-squares method: 1. Solve the local systems on each element T T h for any v = {v 0, 0} V h (T ) and σ = {σ 0,0} Σ h (T ), 2. Solve a global system, a(u h,q h ;v, σ) = (f, w σ) T. a(u h,q h ;v, σ) = 0, v = {0,v b } V h, σ = {0,q b } Σ h, (5) For the WG least-squares method with k = 0, # of unknowns = 2 # of interior edges.

23 Example: Let Ω = (0,1) (0,1) and the exact solution is given by u = x(1 x)y(1 y) h L 2 error order L 2 error order e e e e e e e e e e e e e e e e e e

24 The weak Galerkin method for the elliptic interface problems Consider a elliptic interface problem, A u = f, in Ω, u = 0, on Ω \ Γ, [[u]] Γ = ψ, on Γ, [[A u n]] Γ = φ, on Γ, V h = {v = {v 0,v b } : v 0 T P k (T ),v b e P k (e),e T,v b = 0,on, Ω} The weak Galerkin method: find u h V h such that (A w u h, w v) + T h 1 u 0 u b,v 0 v b T = (f,v 0 ) + ψ,a w v n Γ φ,v b Γ ψ,v 0 v b Γ, v V h.

25 Elliptic interface problems: Example 1 Ω = (0,1) 2 with Ω 1 = [0.2,0.8] 2 and Ω 2 = Ω/Ω 1. Then the exact solution: { 5 + 5(x 2 + y 2 ), if (x,y) Ω 1 u = x 2 + y 2 + sin(x + y), if (x,y) Ω 2 Permeability: A = { 1, if (x,y) Ω sin(x + y), if (x,y) Ω 2

26 Mesh 1 Mesh Solution on mesh 1 Solution on mesh 2

27 Elliptic interface problems: Example 2 The exact solution is u(x,y) = { x y if (x,y) Ω 1 e x cosπy otherwise Ω Ω Figure : The interface Γ in Example 2.

28 Mesh max{h} Gradient Solution L 2 error order L 2 error order Level e e e-01 Level e e e Level e e e Level e e e Level e e e Figure : The WG approximation of Example 2 on mesh level 5. Left: Numerical solution; Right: Exact solution.

29 Summary The weak Galerkin finite element methods represent advanced methodology for handling discontinuous functions in finite element procedure. The weak Galerkin finite element methods have the flexibility of using discontinuous elements and the simplicity of using continuous elements.