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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 [email protected] Mathematical Sciences Deptartment, WPI
FEM for an Interface Problem 1
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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
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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
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Interface Problem
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+
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Illustration of interface
Illustration of Ω, Γ.
−1 −0.5 0 0.5 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Ω
Γ
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Equivalent Formulation
For simplicity we assume here that α ≡ 0
−∆u = f + F in Ω ⊂ R2
u = 0 on ∂Ω
F (x) =
∫ A
0
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
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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)
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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
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Variational Formulation for Interface Problem
Find u ∈ H10 (Ω) such that∫
Ω
∇u · ∇vdx =
∫Ω
fvdx+
∫Γ
βvds
for all v ∈ H10 (Ω).
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The Natural Method
Find uh ∈ Vh such that;∫Ω
∇uh · ∇v dx =
∫Ω
f v dx+
∫Γ
βv ds ∀v ∈ Vh
Ex: Vh is the conforming piecewise polynomials of degree k
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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
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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
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1Error Plot
×10-3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
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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+)
)
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Numerical Tests
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.0510
−3
10−2
10−1
100
d
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1Error Plot
×10-3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
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.
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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
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Poisson Interface Problem
Goal
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 =
∫Ω
f vdx+
∫Γ
βv ds−∑T∈T Γ
h
∫T
∇wuT · ∇v dx
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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−)
)
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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]
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Interface Problem The Natural Method Poisson Interface Problem Stokes Interface Problem High-Contrast Transmission Problem Concluding Remarks & Future Work References
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
Marcus Sarkis [email protected] Mathematical Sciences Deptartment, WPI
FEM for an Interface Problem 19
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Interface Problem The Natural Method Poisson Interface Problem Stokes Interface Problem High-Contrast Transmission Problem Concluding Remarks & Future Work References
Figure: Illustration of notation. T± = T ∩ Ω±
Marcus Sarkis [email protected] Mathematical Sciences Deptartment, WPI
FEM for an Interface Problem 20
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Without Knowing u
For each x`,Ti ∈ Γη = an+ bt[
Dk−`η u(x`,Ti )
]= a
[Dk−`n u(x`,Ti )
]+ b
[Dk−`t u(x`,Ti )
][D`ηu(x`,Ti )
]=∑j=0
(l
j
)ajb`−j
[DjnD
`−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
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Existence and Uniqueness
Lemma
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
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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±)
Lemma
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
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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
1
h
(‖u+‖Ck+1(Ω+) + ‖u−‖Ck+1(Ω−)
)where C > 0 is a constant independent of the mesh
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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
h ‖eh‖L2 r ‖eh‖L∞ r ‖∇eh‖L2 r ‖∇eh‖L∞ r
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
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Numerical ExamplesConsider a exact solution of problem with α non zero
u(x) =
x2
1 − x22 if r ≤ 2/3
0 if r > 2/3x ∈ [−1, 1]2, r = ‖x‖2
h ‖eh‖L2 r ‖eh‖L∞ r ‖∇eh‖L2 r ‖∇eh‖L∞ r
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
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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
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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
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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
h
∫Tq∇ ·wu
T dx
for all (v, q) ∈ V h ×Mh.
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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
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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−)
)
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Numerical Examples
Consider a exact solution of Stokes problem on Ω = [−1, 1]
u(x, y) =
(0
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
3.5e-1 1.27e-2 1.22e-2 1.55e-1 1.92e-11.8e-1 3.42e-3 1.89 3.42e-3 1.83 7.86e-2 0.98 1.08e-1 0.838.8e-2 8.84e-4 1.95 9.67e-4 1.82 3.93e-2 1.00 5.71e-2 0.924.4e-2 2.25e-4 1.98 2.60e-4 1.89 1.96e-2 1.00 3.06e-2 0.902.2e-2 5.66e-5 1.99 6.84e-5 1.93 9.79e-3 1.00 1.58e-2 0.95
Table: Errors and orders of convergence for velocity, on a uniform mesh
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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
00.5
1
−1
−0.5
0
0.5
1
−3
−2
−1
0
1
2
3
ph
Figure: Plot of the approximate pressure by our method, on a uniform grid.
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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
π9
if r > 1/3
h ‖euh ‖L2 r ‖euh ‖L∞ r ‖∇euh ‖L2 r ‖∇euh ‖L∞ r
2.5e-01 3.02e-2 3.99e-2 5.16e-1 8.10e-11.3e-01 8.48e-3 1.83 1.79e-2 1.16 2.79e-1 0.89 5.48e-1 0.566.3e-02 2.03e-3 2.06 5.35e-3 1.74 1.36e-1 1.03 3.35e-1 0.713.1e-02 5.09e-4 2.00 1.68e-3 1.67 6.84e-2 0.99 2.06e-1 0.701.6e-02 1.26e-4 2.02 4.22e-4 1.99 3.36e-2 1.03 1.05e-1 0.97
Table: Errors and orders of convergence for velocity, on structured meshes.
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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
1
0.5
0
-0.5
-1-1
-0.5
0
0.5
-1
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
1
First component of the velocity (uh)1 (left) and pressure ph (right).
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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
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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+
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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
h
∫e
(ρ∇hv · n
[w] +
ρ∇hw · n
[v])
+∑
e∈EΓh
(γ
|e−|
∫e−
ρ−
[w] · [v] +γ
|e+|
∫e+
ρ+
[w] · [v]
)
∑e∈EΓ
h
(|e−|
∫e−
ρ−
[∇hv · n] [∇hw] + |e+|∫e+
ρ+
[∇hv · n] [∇hw · n]
)
Here we denote by ∇hv the functions whose restriction to each T± with T ∈ Th is ∇v
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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+)
)
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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
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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.