The Weak Galerkin Methods and Applications Lin Mu, Junping Wang and Xiu Ye University of Arkansas at Little Rock
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),
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).
Infinite 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,
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.
Construction of V h V h = {v H 1 0(Ω); v T P k (T ), T T h }.
Continuous finite element method Find u h V h H0 1 (Ω) such that (a u h, v h ) = (f, v h ), v h V h. Theorem. If u H k+1 (Ω) H0 1 (Ω), then (u u h ) C inf v V h (u v) Ch k u k+1.
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.
Cause and solution Cause: Continuity requirement of approximating functions cross element boundaries. Solution: Use discontinuous approximations.
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.
Discontinuous Galerkin method Find u h V h such that for all v h V h T (a u h, v h ) T ((a{ u h },[v h ]) e σ(a{ v h },[u h ]) e ) where α is a penalty parameter. e + α h 1 ([u h ], [v h ]) e = (f,v h ), e
Local discontinuous Galerkin method Find q h V h, u h W h such that (a 1 q h,v) + ( v,u h ) Th û h,v n Th = 0, v V h (q h, w) Th ˆq h n,w Th = (f,w), w W h, where û h = {u h } β [u h ], ˆq h = {q h } + β[q h ] α[u h ].
Mixed Hybrid method and the HDG method Mixed hybrid finite element method: find q h V h, u h W h and û h M h such that (a 1 q h,v) ( v,u h ) Th + û h,v n Th = 0, v V h ( q h,w) Th = (f,w), w W h, µ,q h n Th = 0, µ M h. HDG method: find q h V h, u h W h and û h M h such that (a 1 q h,v) ( v,u h ) Th + û h,v n Th = 0, v V h ( q h,w) Th + τ u h û h,w Th = (f,w), w W h µ,q h n Th + τ u h û h, µ Th = 0, µ M h.
References D. Arnold, F. Brezzi, B. Cockburn and D. Marini, Unified analysis of discontinuous Galerkin methods for elliptic problems,siam J. Numer. Anal., 39 (2002), 1749-1779. B. Cockburn and C.-W. Shu, The local discontinuous Galerkin method for time-dependent convection-diffusion systems, SIAM J. Numer. Anal., 35 (1998), pp. 2440 2463. B. M. Fraejis de Veubeke, Displacement and equilibrium models in the finite element method, in Stress Analysis, O. Zienkiewicz and G. Holister, eds., Wiley, New York, 1965. B. Cockburn, J. Gopalakrishnan, and R. Lazarov, Unified hybridization of discontinuous Galerkin, mixed, and conforming Galerkin methods for second order elliptic problems, SIAM J. Nu- mer. Anal. 47 (2009), 1319-136.
Weak Galerkin finite element methods Weak Galerkin (WG) methods use discontinuous approximations. The WG methods keep the advantages: Flexible in approximations. Flexible in mesh generation. Minimize the disadvantages: Simple formulations: ( w u h, w v) + s(u h,v) = (f,v). Comparable number of unknowns to the continuous finite element methods if implemented appropriately.
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 ).
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.
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
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.
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
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.
Application 1: 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.
Application 2: 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.
Application 3: Brinkman equations Brinkman equations: where 0 ε 2 1. Stokes equations: Darcy equations: ε 2 u + u + p = f,in Ω, u = g, in Ω u + p = f, in Ω, u = 0, in Ω u + p = 0, in Ω, u = g, in Ω
Challenge of algorithm design for the Brinkman equations The main challenge for solving Brinkman equations is in the construction of numerical schemes that are stable for both the Darcy flow and the Stokes flow. Stokes stable elements such as Crouzeix-Raviart element, MINI element and Taylor-Hood element do not work well for the Darcy flow (small ε). Darcy stable elements such as RT elements and BDM elements do not work well for the Stokes flow (large ε).
WG methods for the Brinkman equations The weak form of the Brinkman equations: find (u,p) H 1 0 (Ω)d L 2 0 (Ω) s.t. for (v,q) H1 0 (Ω)d L 2 0 (Ω) ε 2 ( u, v) + (u,v) ( v,p) = (f,v), ( u,q) = (g,q). Weak Galerkin finite element methods for the Brinkman equations: find (u h,p h ) V h W h s.t. for (v,q) V h W h ε 2 ( w u h, w v) + (u 0,v 0 ) + s(u h,v) ( w v,p h ) = (f,v), ( w u h,q) = (g,q), where s(v,w) = h 1 K v 0 v b,w 0 w b K. K T h
WG Methods for Brinkman problems Example 1. Let Ω = (0,1) 2, u = curl(sin 2 (πx)sin 2 (πy)), and p = sin(πx). Table : Numerical Convergence Test for Crouzeix Raviart element, Raviart Thomas element, and WG element. ε 1 2 2 2 4 2 8 0 Crouzeix-Raviart rate H 1,velocity 0.98 0.97 0.74 0.03-0.03 rate L 2,pressure 1.00 0.93 0.98 0.12-0.03
WG Methods for Brinkman problems Example 1. Let Ω = (0,1) 2, u = curl(sin 2 (πx)sin 2 (πy)), and p = sin(πx). Table : Numerical Convergence Test for Crouzeix Raviart element, Raviart Thomas element, and WG element. ε 1 2 2 2 4 2 8 0 Crouzeix-Raviart rate H 1,velocity 0.98 0.97 0.74 0.03-0.03 rate L 2,pressure 1.00 0.93 0.98 0.12-0.03 Raviart-Thomas rate H 1,velocity -0.07-0.07 0.28 0.97 0.97 rate L 2,pressure -0.04 0.08 0.86 1.01 1.01
WG Methods for Brinkman problems Example 1. Let Ω = (0,1) 2, u = curl(sin 2 (πx)sin 2 (πy)), and p = sin(πx). Table : Numerical Convergence Test for Crouzeix Raviart element, Raviart Thomas element, and WG element. ε 1 2 2 2 4 2 8 0 Crouzeix-Raviart rate H 1,velocity 0.98 0.97 0.74 0.03-0.03 rate L 2,pressure 1.00 0.93 0.98 0.12-0.03 Raviart-Thomas rate H 1,velocity -0.07-0.07 0.28 0.97 0.97 rate L 2,pressure -0.04 0.08 0.86 1.01 1.01 WG-FEM rate H 1,velocity 1.00 0.99 0.98 0.97 0.97 rate L 2, velocity 2.00 2.00 1.96 1.91 1.91 rate L 2, pressure 1.00 1.00 0.99 0.98 0.98
Application 4: 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.
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 1 2 + sin(x + y), if (x,y) Ω 2
1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 0.2 0.4 0.6 0.8 1 0 0 0.2 0.4 0.6 0.8 1 Mesh 1 Mesh 2 12 10 8 6 4 2 10 8 6 4 0 2 1 0.5 0 0.5 1 2 0 Solution on mesh 1 Solution on mesh 2
Elliptic interface problems: Example 2 The exact solution is u(x,y) = { x y 2 + 10 if (x,y) Ω 1 e x cosπy otherwise 1 0.5 0 Ω 1 0.5 Ω 2 1 1 0.5 0 0.5 1 Figure : The interface Γ in Example 2.
Mesh max{h} Gradient Solution L 2 error order L 2 error order Level 1 4.7778e-01 1.8483e+00 5.6857e-01 Level 2 2.3889e-01 7.5111e-01 1.2991 1.4087e-01 2.0130 Level 3 1.1944e-01 3.4091e-01 1.1396 3.5107e-02 2.0044 Level 4 5.9720e-02 1.6283e-01 1.0660 8.7630e-03 2.0023 Level 5 2.9860e-02 7.9658e-02 1.0315 2.1891e-03 2.0011 Figure : The WG approximation of Example 2 on mesh level 5. Left: Numerical solution; Right: Exact solution.
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).
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).
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).
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 0 2 + 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 ).
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.
Example: Let Ω = (0,1) (0,1) and the exact solution is given by u = x(1 x)y(1 y). 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 0.2 0.4 0.6 0.8 1 0 0 0.2 0.4 0.6 0.8 1 h L 2 error order L 2 error order 3.2327e-01 6.1187e-02 5.4044e-03 1.6178e-01 2.1245e-02 1.5281 1.5135e-03 1.8386 8.0929e-02 8.4355e-03 1.3335 4.0547e-04 1.9015 4.0847e-02 3.8201e-03 1.1586 1.0480e-04 1.9788 2.0610e-02 1.8519e-03 1.0585 2.6515e-05 2.0091 1.0351e-02 9.1819e-04 1.0187 6.6586e-06 2.0064
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.