New class of finite element methods: weak Galerkin methods Xiu Ye - - PowerPoint PPT Presentation

new class of finite element methods weak galerkin methods
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New class of finite element methods: weak Galerkin methods Xiu Ye - - PowerPoint PPT Presentation

New class of finite element methods: weak Galerkin methods Xiu Ye University of Arkansas at Little Rock Second order elliptic equation Consider second order elliptic problem: a u = f , in (1) = 0 , on . u (2)


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SLIDE 1

New class of finite element methods: weak Galerkin methods

Xiu Ye

University of Arkansas at Little Rock

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SLIDE 2

Second order elliptic equation

Consider second order elliptic problem:

−∇· a∇u =

f,

in Ω

(1) u

=

0,

  • n ∂Ω.

(2) Testing (1) by v ∈ H1

0(Ω) gives

  • Ω ∇· a∇uvdx =

a∇u ·∇vdx −

  • ∂Ω

a∇u · nvds =

fvdx.

(a∇u, ∇v) = (f, v),

where (f,g) =

  • Ω fgdx.
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SLIDE 3

PDE and its weak form

PDE: find u satisfies

−∇· a∇u =

f,

in Ω

u

=

0,

  • n ∂Ω.

Its weak form: find u ∈ H1

0(Ω) such that

(a∇u, ∇v) = (f, v), ∀v ∈ H1

0(Ω).

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SLIDE 4

Infinity vs finite

Weak form: find u ∈ H1

0(Ω) such that

(a∇u, ∇v) = (f, v), ∀v ∈ H1

0(Ω).

Let Vh ⊂ H1

0(Ω) be a finite dimensional space.

Continuous finite element method: find uh ∈ Vh such that

(a∇uh,∇vh) = (f,vh), ∀vh ∈ Vh,

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SLIDE 5

Continuous finite element method

Find uh ∈ Vh such that

(a∇uh, ∇vh) = (f, vh), ∀vh ∈ Vh.

Let Vh = Span{φ1,··· ,φn} and uh = ∑n

j=1 cjφj, then n

j=1

(a∇φj,∇φi)cj = (f, φi),

i = 1,··· ,n. The equation above is a symmetric and positive definite linear system. Solve it to obtain the finite element solution uh.

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SLIDE 6

Limitations of the continuous finite element methods

  • On approximation functions. Pk only for triangles and Qk for
  • quadrilaterals. Hard to construct high order and special elements

such as C1 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.

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SLIDE 7

Cause and solution

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

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Pros and cons of using discontinuous functions

Pros

  • Flexibility on approximation functions. Polynomial Pk can be used
  • n 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.

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SLIDE 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 C1 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.

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SLIDE 10

Weak functions

Let T be a quadrilateral with ej for j = 1,··· ,4 as its four sides. Define v =

  • v0 ∈ P1(T),

in T 0

vb ∈ P0(e),

  • n e ⊂ ∂T

Define Vh(T) = {v ∈ L2(T) : v = {v0,vb}} = span{φ1,··· ,φ7} where

φj =

  • 1,
  • n ei

0,

  • therwise

j = 1,··· ,4

φ5 =

  • 1,

in T 0

0,

  • n ∂T

φ6 =

  • x,

in T 0

0,

  • n ∂T

φ7 =

  • y,

in T 0

0,

  • n ∂T
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SLIDE 11

Weak derivatives

Define a weak gradient ∇wv ∈ [P0(T)]2 for v = {v0,vb} ∈ Vh(T) on the element T:

(∇wv,q)T = −(v0,∇· q)T +vb,q · n∂T , ∀q ∈ [P0(T)]2.

Let φj = {φj,0,φj,b}, j = 1,··· ,7. The definition of the weak gradient gives that for any q ∈ [P0(T)]2

(∇wφ5,q) = −(φ5,0,∇· q)T +φ5,b,q · n∂T = 0.

We have

∇wφ5 = ∇wφ6 = ∇wφ7 = 0.

Using the definition of ∇w, we can find for j = 1,··· ,4

∇wφj = |ej| |T| nj.

Weak gradient ∇w for all the basis function φj can be found explicitly.

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SLIDE 12

The local stiffness matrix for the WG method

Denote Qb the L2 projection to P0(ej). Qbv0|ej = v0(mj) where mj is the midpoint of ej. Define aT(v,w) = (a∇wv,∇ww)T + h−1Qbv0 − vb,Qbw0 − wb∂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 = (aT(φi,φj)), i,j = 1,··· ,7.

T e1 e2 e3 e4

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Weak Galerkin finite element methods

  • Define weak function v = {v0,vb} such that

v =

  • v0,

in T 0

vb,

  • n ∂T

Define weak Galerkin finite element space Vh = {v = {v0,vb} : v0|T ∈ Pj(T),vb ∈ Pℓ(e),e ⊂ ∂T,vb = 0 on ∂Ω}.

  • Define a weak gradient ∇wv ∈ [Pr(T)]d for v ∈ Vh on each element T:

(∇wv,q)T = −(v0,∇· q)T +vb,q · n∂T, ∀q ∈ [Pr(T)]d.

Weak Galerkin element: (Pj(T),Pℓ(e),[Pr(T)]d). For example:

(P1(T),P0(e),[P0(T)]d).

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SLIDE 14

Weak Galerkin finite element formulation

Define a(uh,vh) = (a∇wuh,∇wvh)+∑

T

h−1

T u0 − ub,v0 − vb∂T.

The WG method: find uh = {u0,ub} ∈ Vh satisfying a(uh,vh) = (f,vh),

∀vh ∈ Vh.

  • Theorem. Let uh be the solution of the WG method associated with local

spaces (Pk(T),Pk(e),[Pk−1(T)]d), then h|||Qhu − uh|||+Qhu − uh ≤ Chk+1uk+1, where Qhu is the L2 projection of u.

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SLIDE 15

Simple formulation: the WG method for the Stokes equations

The weak form of the Stokes equations: find (u,p) ∈ [H1

0(Ω)]d × L2 0(Ω) that

for all (v,q) ∈ [H1

0(Ω)]d × L2 0(Ω)

(∇u,∇v)−(∇· v,p) = (f,v) (∇· u, q) =

0. The weak Galerkin method: find (uh,ph) ∈ Vh × Wh such that for all

(v,q) ∈ Vh × Wh, (∇wuh,∇wv)+ s(uh,v)−(∇w · v,ph) = (f,v) (∇w · uh, q) =

0.

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SLIDE 16

The WG method for the biharmonic equation

The weak form of the Stokes equations: seeking u ∈ H2

0(Ω) satisfying

(∆u,∆v) = (f,v), ∀v ∈ H2

0(Ω),

Weak Galerkin finite element method: seeking uh ∈ Vh satisfying

(∆wuh, ∆wv)+ s(uh, v) = (f, v), ∀v ∈ Vh.

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SLIDE 17

Implementation of the WG method

The WG method: find uh = {u0,ub} ∈ Vh satisfying a(uh,vh) = (f,vh),

∀vh = {v0,vb} ∈ Vh.

Effective implementation of the WG method:

  • 1. Solve u0 as a function of ub from the following local system on element T,

a(uh,vh) = (f,vh),

∀vh = {v0,0} ∈ Vh.

(3)

  • 2. Solve ub from the following global system,

a(uh,vh) = (f,vh),

∀vh = {0,vb} ∈ Vh.

(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.

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SLIDE 18

The recent development: The WG least-squares method

Consider the model problem,

−∇· a∇u =

f,

in Ω

u

=

0,

  • n ∂Ω.

Rewrite the problem as the system of first order equations, q+ a∇u

=

0, in Ω,

∇· q =

f, in Ω, u

=

0,

  • n ∂Ω.

The least-squares method: find (q,u) ∈ H(div;Ω)× H1

0(Ω) such that for any

(σ σ σ,v) ∈ H(div;Ω)× H1

0(Ω),

(q+ a∇u,σ σ σ + a∇v)+(∇· q,∇·σ σ σ) = (f,∇·σ σ σ).

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SLIDE 19

The WG Least-squares method

The least-squares method: find (q,u) ∈ H(div;Ω)× H1

0(Ω) such that for any

(σ σ σ,v) ∈ H(div;Ω)× H1

0(Ω),

(q+ a∇u,σ σ σ + a∇v)+(∇· q,∇·σ σ σ) = (f,∇·σ σ σ).

The WG least-squares method: find (qh,uh) ∈ Σh × Vh such that for any

(σ σ σ,v) ∈ Σh × Vh,

(qh + a∇wuh,σ σ σ + a∇wv)+(∇w · qh,∇w ·σ σ σ)+ s1(uh,v)+ s2(qh,σ σ σ) = (f,∇·σ σ σ).

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SLIDE 20

Define

Dh = {ne : ne is unit and normal to e, e ∈ Eh},

Vh = {v = {v0,vb} : v0|T ∈ Pk+1(T),vb|e ∈ Pk(e),e ∈ ∂T,vb = 0, on ∂Ω},

Σh = {σ σ σ = {σ σ σ 0,σ σ σ b} : σ σ σ 0|T ∈ [Pk(T)]d,σ σ σ b|e = σbne,σb|e ∈ Pk(e), e ∈ ∂T}.

Define s1(w,v)

=

T∈Th

h−1Qbw0 − wb, Qbv0 − vb∂T , s2(t,σ

σ σ) =

T∈Th

h(t0 − tb)· n, (σ

σ σ 0 −σ σ σ b)· n∂T ,

The WG least-squares method: find (qh,uh) ∈ Σh × Vh such that for any

(σ σ σ,v) ∈ Σh × Vh, (qh + a∇wuh,σ σ σ + a∇wv)+(∇w · qh,∇w ·σ σ σ)+ s1(uh,v)+ s2(qh,σ σ σ) = (f,∇·σ σ σ).

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SLIDE 21

We introduce a norm |||·|||V in Vh as

|||v|||2

V = ∑ T∈Th

∇wv2

T + s1(v,v),

and a norm |||·|||Σ in Σh as

|||σ σ σ|||2

Σ = ∑ T∈Th

∇w ·σ σ σ2

T +σ

σ σ 02 + s2(σ σ σ,σ σ σ).

  • Lemma. There is a constant C such that for all (σ

σ σ,v) ∈ Σh × Vh

C(|||σ

σ σ|||2

Σ +|||v|||2 V) ≤ a(v,σ

σ σ;v,σ σ σ).

  • Theorem. Assume the exact solution u ∈ Hk+2(Ω) and q ∈ [Hk+1(Ω)]d.

Then, there exists a constant C such that

|||uh − Qhu|||V +|||qh − Qhq|||Σ ≤ Chk+1(uk+2 +qk+1).

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SLIDE 22

Implementation of the WG least-squares method

The WG least-squares method: find (qh,uh) ∈ Σh × Vh such that for any

(σ σ σ,v) ∈ Σh × Vh,

(qh + a∇wuh,σ σ σ + a∇wv)+(∇w · qh,∇w ·σ σ σ)+ s1(uh,v)+ s2(qh,σ σ σ) = (f,∇·σ σ σ).

Effective implementation of the WG least-squares method:

  • 1. Solve the local systems on each element T ∈ Th for any v = {v0, 0} ∈ Vh(T) and

σ σ σ = {σ σ σ 0,0} ∈ Σh(T),

a(uh,qh;v,σ

σ σ) = (f, ∇w ·σ σ σ)T .

  • 2. Solve a global system,

a(uh,qh;v,σ

σ σ) = 0, ∀v = {0,vb} ∈ Vh,σ σ σ = {0,qb} ∈ Σh,

(5) For the WG least-squares method with k = 0,

# of unknowns = 2×# of interior edges.

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SLIDE 23

Example: Let Ω = (0,1)×(0,1) and the exact solution is given by u = x(1− x)y(1− y).

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

h L2 error

  • rder

L2 error

  • rder

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

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The weak Galerkin method for the elliptic interface problems

Consider a elliptic interface problem,

−∇· A∇u =

f, in Ω, u

=

0,

  • n ∂Ω\Γ,

[[u]]Γ = ψ,

  • n Γ,

[[A∇u · n]]Γ = φ,

  • n Γ,

Vh = {v = {v0,vb} : v0|T ∈ Pk(T),vb|e ∈ Pk(e),e ∈ ∂T,vb = 0,on,∂Ω} The weak Galerkin method: find uh ∈ Vh such that

(A∇wuh,∇wv)+∑

T

h−1u0 − ub,v0 − vb∂T = (f,v0)

+ψ,A∇wv · nΓ −φ,vbΓ −ψ,v0 − vbΓ,∀v ∈ Vh.

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SLIDE 25

Elliptic interface problems: Example 1

Ω = (0,1)2 with Ω1 = [0.2,0.8]2 and Ω2 = Ω/Ω1.

Then the exact solution: u =

  • 5+ 5(x2 + y2),

if (x,y) ∈ Ω1 x2 + y2 + sin(x + y), if (x,y) ∈ Ω2 Permeability: A =

  • 1,

if (x,y) ∈ Ω1 2+ sin(x + y), if (x,y) ∈ Ω2

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SLIDE 26

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

Mesh 1 Mesh 2

0.5 1 0.5 1 −2 2 4 6 8 10 12 2 4 6 8 10

Solution on mesh 1 Solution on mesh 2

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SLIDE 27

Elliptic interface problems: Example 2

The exact solution is u(x,y) =

  • x − y2 + 10 if (x,y) ∈ Ω1

ex cosπy otherwise

−1 −0.5 0.5 1 −1 −0.5 0.5 1 Ω1 Ω2

Figure : The interface Γ in Example 2.

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SLIDE 28

Mesh max{h} Gradient Solution L2 error

  • rder

L2 error

  • rder

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.

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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.