Higher-0rder Finite Element Methods for Elliptic Problems ...€¦ · Outline 1 Interface Problem 2...

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Interface Problem The Natural Method Poisson Interface Problem Stokes Interface Problem High-Contrast Transmission Problem Concluding Remarks & Future Work References

Higher-0rder Finite Element Methods

for Elliptic Problems with Interfaces

Marcus Sarkis

Mathematical Sciences Deptartment, WPI

May 12, 2015. Hydraulic Fracturing IMA Workshop

Joint work with Johnny Guzman and Manuel Sanchez (Brown)

Marcus Sarkis msarkis@wpi.edu Mathematical Sciences Deptartment, WPI

FEM for an Interface Problem 1

Outline

1 Interface Problem

2 The Natural Method

3 Poisson Interface Problem

4 Stokes Interface Problem

5 High-Contrast Transmission Problem

6 Concluding Remarks & Future Work

7 References

Outline

1 Interface Problem

7 References

Interface Problem

−∆u± = f in Ω±

u = 0 on ∂Ω

[u] = α on Γ

[∇u · n] = β on Γ

We denote

[u] = u+ − u−

[∇u · n] = ∇u− · n− +∇u+ · n+

Illustration of interface

Illustration of Ω, Γ.

−1 −0.5 0 0.5 1

−0.8

−0.6

−0.4

−0.2

Equivalent Formulation

For simplicity we assume here that α ≡ 0

−∆u = f + F in Ω ⊂ R2

u = 0 on ∂Ω

F (x) =

β(s)δ(x−X(s))ds ∀x ∈ Ω

• X : [0, A)→ Γ is the arch-length parametrization of Γ

• δ is a two-dimensional Dirac function

• This could be thought of as Peskin’s Formulation

Previous Work

Some Finite Difference methods

• IBM Peskin (77)• IIM LeVeque, Li (94)• Beale, A. Layton (96)• Mori (98)• Marquez, Nave, Rosales (11)

Some Finite Element Methods

• Boffi, Gastaldi (03)• Gong, Li, Li (07)• He, Lin, Lin (11)• Adjerid, Ben-Romd, Lin (14)

Outline

1 Interface Problem

7 References

Variational Formulation for Interface Problem

Find u ∈ H10 (Ω) such that∫

∇u · ∇vdx =

for all v ∈ H10 (Ω).

The Natural Method

Find uh ∈ Vh such that;∫Ω

∇uh · ∇v dx =

f v dx+

βv ds ∀v ∈ Vh

Ex: Vh is the conforming piecewise polynomials of degree k

Numerical Example

Exact solution of the interface problem in Ω = [−1, 1]2

u(x) =

1 if r ≤ R1− log( r

R) if r > R

where r = ‖x‖2 and R = 1/3

Then, the data are given by f± = 0, α = 0 and β = 1R

Vh =v ∈ H1

0 (Ω) : v|T ∈ P1(T ) ∀T ∈ Th

Numerical Example

h ‖eh‖L2 r ‖eh‖L∞ r ‖∇eh‖L2 r ‖∇eh‖L∞ r

3.5e-1 6.74e-3 2.01e-2 1.31e-1 5.05e-11.8e-1 2.98e-3 1.18 1.57e-2 0.36 1.17e-1 0.17 4.95e-1 0.038.8e-2 1.21e-3 1.30 9.98e-3 0.65 1.15e-1 0.02 9.11e-1 -0.884.4e-2 4.39e-4 1.46 5.21e-3 0.94 8.41e-2 0.45 9.06e-1 0.01

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

1Error Plot

×10-3

Question

How far do we need to be from the interface to recover anoptimal estimate?

Let z ∈ Ω and d = dist(z,Γ)

|∇(Ihu−uh)(z)| ≤ Ch(1+log(1/h)h

d2)(‖u‖C2(Ω−) + ‖u‖C2(Ω+)

Numerical Tests

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.0510

10−2

10−1

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

1Error Plot

×10-3

Semi-log plot of gradient error for the natural method with h = .0028.|∇eNh (dT )| (red) for every triangle T and curve 2h + log(1/h)(h/d)2 (blue).

The distance d in the x-axis varies from 0 to√h.

Outline

1 Interface Problem

7 References

Poisson Interface Problem

Recover the high accuracy of the natural method

Vh =v ∈ H1

0 (Ω) : v|T ∈ Pk(T ) ∀T ∈ Th

The set T Γh = T ∈ Th : T ∩ Γ 6= ∅

Find uh ∈ Vh such that for all v ∈ Vh the following holds∫Ω

∇uh · ∇v dx =

f vdx+

βv ds−∑T∈T Γ

∇wuT · ∇v dx

Main Result

Ih : Lagrange interpolation operator onto the Vh

Theorem

If u± ∈ Ck+1(Ω±

) , f |Ω± ∈ Ck−1(Ω±

), β smooth, then

‖∇(Ihu− uh)‖L∞(Ω) ≤ C hk(‖u+‖Ck+1(Ω+) + ‖u−‖Ck+1(Ω−)

What do we need?

We will construct wuT , for T ∈ T Γ

h , such that satisfies

‖∇(Ihu+ wuT − u)‖L∞(T±) ≤ Chk

where T± = T ∩ Ω±

• P1(T ) conforming correction, [GSS 14]

• Pk(T ) nonconforming correction, [GSS 15a]

Construction of wuT

Consider the local space for T ∈ T Γh

Sk(T ) =w ∈ L2(T ) : w|T± ∈ Pk(T±)

For each T ∈ T Γh , let wu

T ∈ Sk(T ) such that[Dk−`η wu

T (x`,Ti )]

=[Dk−`η u(x`,Ti )

]for 0 ≤ i ≤ ` and 0 ≤ ` ≤ k

IhwuT = 0

Figure: Illustration of notation. T± = T ∩ Ω±

Without Knowing u

For each x`,Ti ∈ Γη = an+ bt[

Dk−`η u(x`,Ti )

[Dk−`n u(x`,Ti )

[Dk−`t u(x`,Ti )

][D`ηu(x`,Ti )

]=∑j=0

)ajb`−j

`−jt u(x`,Ti )

]The RHS obtained from normal and tangential derivatives of f andtangential derivaties of α and β. Derived from the equations

−∆u = f, [u] = α, [Dnu] = β

Existence and Uniqueness

Given data ci,` for 0 ≤ i ≤ ` and 0 ≤ ` ≤ k. There exist aunique function in w ∈ Sk(T ) such that[

Dk−`η w(x`,Ti )

]= ci,` for 0 ≤ i ≤ ` and 0 ≤ ` ≤ k

Ihw = 0

• Note that is a square system of (k + 1)(k + 2) equations• Explict construction using

p(s, r) = pk(s) + rpk−1(s) + r2pk−2(s) + . . . rkpo

Local Estimates

Existence of extension

Assuming u± ∈ Ck+1(Ω±) and Γ smooth there exist extensionsu±E ∈ Ck+1(Ω) such that

u±E =u± in Ω±

‖u±E‖Ck+1(Ω) ≤C ‖u±‖Ck+1(Ω±)

Suppose that u our solution satisfies u± ∈ Ck+1(Ω±) and wuT is

the correction function. Then, for j = 0, 1, we have

hjT ‖Dj(u−Ihu−wu

T )‖L∞(T±) ≤ C hk+1T

(‖u+E‖Ck+1(TE) + ‖u−E‖Ck+1(TE)

)TE denotes the patch of elements associated to T

Main Result

Theorem

Suppose that Ω is convex and that the family of meshes Thh>0

are quasi-uniform and shape regular, then

‖∇(Ihu− uh)‖L∞(Ω) ≤ C hk(‖u+‖Ck+1(Ω+) + ‖u−‖Ck+1(Ω−)

)‖Ihu− uh‖L∞(Ω) ≤ Chk+1 log

(‖u+‖Ck+1(Ω+) + ‖u−‖Ck+1(Ω−)

)where C > 0 is a constant independent of the mesh

Numerical Example

Exact solution of the interface problem in Ω = [−1, 1]2

u(x) =

1, if r ≤ 1/31− log(3r) if r > 1/3

1.8e-1 8.87e-5 3.97e-4 3.80e-3 2.53e-29.0e-2 9.73e-6 3.29 7.46e-5 2.49 9.04e-4 2.14 7.43e-3 1.824.7e-2 1.11e-6 3.33 1.06e-5 3.00 2.15e-4 2.21 2.58e-3 1.632.4e-2 1.30e-7 3.15 1.42e-6 2.95 5.06e-5 2.13 7.34e-4 1.841.2e-2 1.59e-8 3.14 2.24e-7 2.76 1.27e-5 2.07 2.16e-4 1.836.1e-3 1.96e-9 3.04 3.15e-8 2.85 3.15e-6 2.02 5.55e-5 1.98

Vh =v ∈ H1

0 (Ω) : v|T ∈ P2(T ) ∀T ∈ Th

Numerical ExamplesConsider a exact solution of problem with α non zero

u(x) =

1 − x22 if r ≤ 2/3

0 if r > 2/3x ∈ [−1, 1]2, r = ‖x‖2

2.5e-1 7.55e-4 2.19e-3 2.18e-2 1.05e-11.2e-1 5.41e-5 3.80 2.22e-4 3.31 2.56e-3 3.09 1.96e-2 2.426.2e-2 4.37e-6 3.63 3.60e-5 2.62 4.83e-4 2.40 5.78e-3 1.763.1e-2 4.41e-7 3.31 5.11e-6 2.82 8.11e-5 2.57 1.53e-3 1.921.6e-2 3.38e-8 3.70 6.99e-7 3.07 1.45e-5 2.48 4.35e-4 1.90

Outline

1 Interface Problem

7 References

Stokes Interface Problem

Stokes Problem

−∆u+∇p = f in Ω

∇ · u = 0 in Ω

u = 0 on ∂Ω

[u] = α on Γ

[Dnu− pn] = β on Γ

Finite Element Method

Find (uh, ph) ∈ V h ×Mh such that

∫Ω∇uh : ∇v dx−

∫Ωph∇ · vdx =

∫Ωf · vdx +

∫Γβ · vds

−∑

T∈T Γh

(∫Tw

pT∇ · vdx +

∫T∇wu

T : ∇v dx)

∫Ωq∇ · uhdx =−

∑T∈T Γ

∫Tq∇ ·wu

for all (v, q) ∈ V h ×Mh.

Corrections

Velocity

For each T ∈ T Γh , let wuT ∈ Sk(T ) such that[

Dk−`η wuT (x`,Ti )

]=[Dk−`η u(x`,Ti )

]for 0 ≤ i ≤ ` and 0 ≤ ` ≤ k

Ih(wuT ) = 0

Pressure

For each T ∈ T Γh , let wp

T ∈ Sk−1(T ) such that[Dk−`η wp

T (x`,Ti )]

=[Dk−`η p(x`,Ti )

]for 0 ≤ i ≤ ` and 0 ≤ ` ≤ k − 1

Jh(wpT ) = 0

Main Result

Theorem [GSS15b]

There exists a constant C > 0 such that where k is the integersatisfying an assumption for V h and Mh

‖∇(Ihu− uh)‖L∞(Ω) + ‖Jh(p)− ph‖L∞(Ω) ≤ Chk(‖u+‖Ck+1(Ω+) + ‖u−‖Ck+1(Ω−) + ‖p+‖Ck(Ω+) + ‖p−‖Ck(Ω−)

Numerical Examples

Consider a exact solution of Stokes problem on Ω = [−1, 1]

u(x, y) =

u2(x, y)

), u2(x, y) =

2/3 + x if x ≤ 1/34/3− x if x > 1/3

p(x, y) =

x2 + y2 + 1/3 if x ≤ 1/3x2 + y2 − 8/3 if x > 1/3

(x, y) ∈ Ω

In this case the interface is Γ = (x, y) ∈ Ω : x = 1/3h ‖euh ‖L2 r ‖euh ‖L∞ r ‖∇euh ‖L2 r ‖∇euh ‖L∞ r

Table: Errors and orders of convergence for velocity, on a uniform mesh

Numerical Examples

h ‖eph‖L2 r ‖ep

h‖L∞ r

3.5e-1 6.01e-2 9.30e-21.8e-1 2.22e-2 1.44 6.07e-2 0.628.8e-2 7.45e-3 1.58 3.52e-2 0.794.4e-2 2.34e-3 1.67 1.92e-2 0.882.2e-2 7.09e-4 1.73 1.00e-2 0.93

Table: Errors and orders of convergence for pressure, on a uniform mesh

−1−0.5

−0.5

Figure: Plot of the approximate pressure by our method, on a uniform grid.

Numerical Examples

Exact solution of Stokes interface problem on Ω = [−1, 1]

u(x, y) =

u1(x, y) =

3y if r ≤ 1/34y3r− y if r > 1/3

u2(x, y) =

−3, if r ≤ 1/3x− 4x

3rif r > 1/3

, p(x, y) =

4− π

9if r ≤ 1/3

if r > 1/3

h ‖euh ‖L2 r ‖euh ‖L∞ r ‖∇euh ‖L2 r ‖∇euh ‖L∞ r

Table: Errors and orders of convergence for velocity, on structured meshes.

Numerical Examples

h ‖eph‖L2 r ‖eph‖L∞ r2.5e-1 1.39e-1 1.84e-11.3e-1 3.39e-2 2.04 7.71e-2 1.266.3e-2 1.32e-2 1.36 4.29e-2 0.853.1e-2 3.79e-3 1.80 2.36e-2 0.861.6e-2 1.46e-3 1.38 1.27e-2 0.90

First component of the velocity (uh)1 (left) and pressure ph (right).

Outline

1 Interface Problem

7 References

High-Contrast Coefficients

Interface Problem

−ρ±∆u± = f± in Ω±

u = 0 on ∂Ω

[u] = 0 on Γ

[ρ∇u · n] = 0 on Γ

Denote

[u] = u+ − u−

[ρ∇u · n] = ρ−∇u− · n− + ρ+∇u+ · n+

Discontinuous Galerkin

Find uh ∈ Vh such that

ah(uh, vh) = (f, vh) for all vh ∈ Vh,Bilinear Form

ah(w, v) :=

∫Ωρ∇hw · ∇hv −

∑e∈EΓ

(ρ∇hv · n

ρ∇hw · n

e∈EΓh

|e−|

∫e−

[w] · [v] +γ

[w] · [v]

∑e∈EΓ

(|e−|

∫e−

[∇hv · n] [∇hw] + |e+|∫e+

[∇hv · n] [∇hw · n]

Here we denote by ∇hv the functions whose restriction to each T± with T ∈ Th is ∇v

Main Result

Theorem

The error estimate that we prove is of the form

‖u− uh‖V ≤ C h(√

ρ−‖u‖H2(Ω−) +√ρ+‖u‖H2(Ω+)

Summary & Future Work

Summarizing

Analysis of the natural method

Higher-order method for Poisson interface problem

Higher-order method for Stokes interface problem

Second-order high constrast problems

Future Work

Fracturing problems

Time-evolving problems

References

GSS 14 J. Guzman, M. Sanchez-Uribe and S. On the accuracy offinite element approximations to a class of interfaceproblems. Math. Comp. Accepted, 2014

GSS 15a J. Guzman, M. Sanchez-Uribe and S. Higher-order finiteelements methods for elliptic problems with interfaces.Submitted 2015.

GSS 15b J. Guzman, M. Sanchez-Uribe and S. A finite elementmethod for high-contrast interface problems with errorestimates independent of contrast. To be submitted 2015.

Higher-0rder Finite Element Methods for Elliptic Problems ...€¦ · Outline 1 Interface Problem 2...

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