17
XFEM project An Extended Finite Element Method for a Two-Phase Stokes problem P. Lederer, C. Pfeiler, C. Wintersteiger Advisor: Dr. C. Lehrenfeld August 5, 2015 Contents 1 Problem description 2 1.1 Physics ......................................... 2 1.1.1 Bubble ...................................... 3 1.1.2 Curvature .................................... 3 1.2 Weak formulation .................................... 5 2 Pressure enrichment 5 2.1 Pressure enrichment Q x h ................................ 5 2.2 Ghost penalty ...................................... 7 3 Velocity enrichment 8 3.1 Velocity enrichment V x h ................................ 8 3.2 Nitsche XFEM formulation .............................. 9 3.3 The weighted average ................................. 10 3.4 Choice of λ ....................................... 10 4 Numerical examples 11 4.1 First example ...................................... 11 4.2 Second example ..................................... 14 5 Conclusion 16 1

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XFEM project

An Extended Finite Element Method for a

Two-Phase Stokes problem

P. Lederer, C. Pfeiler, C. WintersteigerAdvisor: Dr. C. Lehrenfeld

August 5, 2015

Contents

1 Problem description 21.1 Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Bubble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.2 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Weak formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Pressure enrichment 52.1 Pressure enrichment Qxh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Ghost penalty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Velocity enrichment 83.1 Velocity enrichment Vx

h . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Nitsche XFEM formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.3 The weighted average . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.4 Choice of λ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Numerical examples 114.1 First example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.2 Second example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

5 Conclusion 16

1

The theory of this work is based on the notes [Leh15] of the XFEM - lecture in the summerterm 2015 held by Dr. C. Lehrenfeld.

In this project we consider an approximation of a two-phase stokes problem where the interfaceis not resolved by the triangulation. In section 1 we discuss the problem and the physicalintepretation of the occuring equations by presenting two model problems. Depending on theparameters, the solution can have weak and strong discontinuities. Therefore new approximationspaces are designed in the section 2 and 3. In the last section we present numerical examples,where an exact solution exists and analyze the convergence of the new methods.

1 Problem description

We consider an open domain Ω ⊂ R2, which is separated into two subdomains Ω1 and Ω2 bythe interface Γ = Ω1 ∩ Ω2. We assume that one phase is completely surrounded by the other,i.e. ∂Ω ∩ Γ = ∅.Let Thh>0 be a family of shape regular meshes of Ω. We only consider simplicial meshesTh. The elements of such a mesh Th are simplices, i.e. triangles in 2D, and it holds h =max hT |T ∈ Th , where hT := diam(T ) is the local element size. In general the interface Γdoes not coincide with the boundaries of the elements of Th. We call the interface unfitted w.r.t.the mesh Th. We assume that the resolution close to the interface is sufficiently high, such thatΓ can be resolved by the triangulation, in the sense that, if Γ ∩ T =: ΓT 6= ∅, then ΓT can berepresented as the graph of a function on a planar cross-section of T . Further the resolutionclose to the interface should be sufficiently high, such that |ΓT |h ≤ cΓ|T | for all T cut by Γ, andwith a constant cΓ > 0 independent of h and the cut position of Γ.

We consider the two-phase Stokes problem on the domain Ω, which results as

−div (µiD(u)) +∇p = ρig in Ωi, i = 1, 2, (1a)

div (u) = 0 in Ωi, i = 1, 2, (1b)

JuK = 0 on Γ, (1c)

J−µD(u) · n + pnK = f on Γ, (1d)

u = 0 on ∂Ω, (1e)

with the domainwise constant density ρ, domainwise constant viscosity µ, given right hand sideg ∈ [L2(Ω)]2 and surface tension force f ∈ [L2(Γ)]2. D(u) denotes the symmetric gradientD(u) := ∇u + (∇u)>.

1.1 Physics

Before we introduce two examples of a two phase Stokes problem, we first take a closer lookat the equations (1a) - (1e) and the occuring material parameters ρ, µ. Equation (1a) is the socalled momentum conservation equation and can be derived by balancing the momentums in andon a control volume under the assumptions of a Newtonian fluid. Another important fact for thederivation of (1a) is that we consider a creeping flow, which means that the advective inertialforces are small compared to the viscous forces. The second equation (1b) can be derived fromthe conservation of mass and (1e) is a given boundary condition. The new aspects compared toan one-phase stokes problem, are the equations (1c) and (1d), which will be called the interfaceconditions. We are looking for continuous velocity fields what leads to (1c). Equation (1d)describes the momentum balance at the interface. The stress from the domains Ω1 and Ω2

2

Ω1 :µ1 = 1

ρig =

(00

)Ω2 :µ2 = 100

ρig =

(0−1

)

Γ :

f =

(00

)

Figure 1: A bubble in a surrounding fluid

balances with the force f on the interface. A typical f would be the surface tension force.To classify the fluid in Ωi we have two material parameters: The dynamic viscosity µi whichdescribes the resistance due to deformation by the stress tensor and the density ρi. To completeour equations we consider the gravitational force ρig as right hand side, which acts as a pushingforce on the fluid domains.

1.1.1 Bubble

In the first example (Figure 1) we want to simulate a bubble with a very small density ρ1 (forexample air) surrounded by a fluid with a higher density ρ2 (e.g. water). Also the viscositiesare chosen differently with µ1 = 1 and µ2 = 100. The force on the interface Γ is set to zero. Dueto the different viscosities we expect a jump in the pressure to satisfy the interface condition(1d). In Figure 2 and 3 the velocity and pressure (turned 180 degrees) fields are displayed. Weobserve, that the gravitational force drags the mass of the outer fluid down and therefore thelighter bubble rises up and the displayed velocity field is produced. Far away from the bubblethe pressure is essentially the hydrostatic pressure and increases almost linear. Closer to thebubble (Ω2), we see that the pressure is still increasing towards the bottom, but due to theinteraction with the bubble the increase is no longer linear. Furthermore, we see that we get adiscontinuity across the interface Γ. Only at two points (cf. Figure 3) the pressure is continuousas the normal component of D(u) vanishes there and thus equation (1e) demands JpK = 0.Note that the pressure gradient (from bottom to top) is not a driving force in this example!

1.1.2 Curvature

In the second example (Figure 4) we demonstrate how the surface tension force affects thevelocity field. The surface tension can be modeled as σκnΓ, where κ is the mean curvaturein each point on the interface and σ is a material parameter which controls the influence ofthe curvature. The curvature κ has different impacts depending on the substance system. Forinstance the resulting surface tension force is higher for an air bubble than for an oil bubble

3

Figure 2: Velocity plot for example 1.1.1

J−µD(u) · n + pnK =0

Figure 3: Pressure plot for example 1.1.1

4

Ω1 :µ1 = 1

ρig =

(00

)Ω2 :µ2 = 1

ρig =

(00

)

Γ :f = σκnΓ

Figure 4: Rounded square and considered paramters for example 1.1.2

(with the same gemoetry). When we look at this example (see Figure 4 for the consideredgeometrical configuration and the parameters. Note: ρ and µ are globally constant), one cansee that the curvature is much higher in the corners as on the straight sides (where it is actualzero). In Figure 5 we see how this affects the velocity. In the corners the flow is directedtowards the center of the bubble and on the straight sides outwards. This goes hand in handwith the physical expectations, as the bubble tends to a spherical shape if the surface tensionis the dominant force.

1.2 Weak formulation

We multiply (1a) by v ∈ V = [H1(Ω)]2 and (1b) by q ∈ Q = L2(Ω). Partial integration andthe interface condition (1d), then lead to the weak formulation. Approximating the spaces Vand Q by Vh and Qh leads to the discrete problem: Find (u, p) ∈ Vh ×Qh, such that

1

2

∑i=1,2

µi (D(u),D(v))Ωi− (p,div(v))Ωi

=∑i=1,2

ρi (g,v)Ωi+ (f ,v)Γ ∀v ∈ Vh, (2a)

∑i=1,2

(q,div(u))Ωi= 0 ∀q ∈ Qh, (2b)

where (a, b)Ωi:=∫

Ωia · b dx. The choice of the finite element spaces Vh and Qh with respect

to (1c) and (1e) will be discussed in the following sections.b

2 Pressure enrichment

2.1 Pressure enrichment Qxh

A popular choice for the velocity-pressure space Vh ×Qh is the Taylor-Hood element, which isLBB-stable. One chooses globally continuous piecewise polynomials of degree k for the velocity,

5

Figure 5: Velocity plot for example 1.1.2

and globally continuous piecewise polynomials of degree k − 1 for the pressure

Vh :=v ∈ [C(Ω)]2

∣∣∣v|T ∈ [Pk(T )]2 for all T ∈ Th,

Qh :=q ∈ C(Ω)

∣∣∣ q|T ∈ Pk−1(T ) for all T ∈ Th.

In this work we consider the element with k = 2. However, due to the surface tension force f atthe interface Γ, the pressure solution p will have jump discontinuities which can not be resolvedby Qh very well. In particular it holds

infph∈Qh

‖ph − p‖L2(Ω) . h1/2 ‖p‖H1(Ω1,2) , (3)

where ‖.‖2Hk(Ω1,2) := ‖.‖2Hk(Ω1) +‖.‖2Hk(Ω2). This estimate is sharp, even if p is arbitrarily smoothon the subdomains Ω1,Ω2, as long as it has a jump across the interface Γ. The situation cannot be improved by choosing piecewise polynomials of higher degree for Qh.Therefore we introduce the enriched space QΓ

h := Qh ⊕Qxh, where Qxh is chosen as follows:Define J := 1, . . . , n, where n = dim(Qh), and let (ϕi)i∈J be the nodal basis in Qh. LetJΓ := j ∈ J |Γ ∩ suppϕj 6= ∅ be the index set of those basis functions in Qh which havetheir support intersected by Γ. Let χ2 be the characteristic function with χ2(x) = 0 for x ∈ Ω1

and χ2(x) = 1 for x ∈ Ω2. Using this, for j ∈ JΓ we introduce the so-called enrichmentfunction Φj(x) := |χ2(x) − χ2(xj)|, where xj is the vertex with index j corresponding to ϕj .We introduce the new basis functions ϕΓ

j := Φjϕj , j ∈ JΓ. Now we can define

Qxh := spanϕΓj | j ∈ JΓ

.

6

With the velocity-pressure space Vh×QΓh, jump discontinuities of the pressure at the interface

Γ can be resolved, and so (3) can be improved to

infph∈Qh

‖ph − p‖L2(Ω) . hk ‖p‖Hk+1(Ω1,2) . (4)

(a) ph ∈ Qh (b) ph ∈ QΓh

Figure 6: Solution without and with pressure space enrichment using triangulation and interfaceposition as in Figure 7a.

Enriching the pressure space leads to a new problem. The discrete ”inf-sup” condition obtainedby the LBB-stable Taylor-Hood element is no longer given. In particular, with the enriched spaceQΓh, the discretization error and the conditioning of the linear system are no longer independent

of the position of the interface Γ, c.f. Figure 6/7 .

x

y

interface Γ

(a) Almost no bad cut positions

x

y

interface Γ

(b) Many bad cut positions

Figure 7: Triangulation and interface positions

2.2 Ghost penalty

To fix the problems observed, we introduce the so-called ghost penalty stabilization: Let Fh bethe set of facets corresponding to the elements of Th. Define ΩΓ

i,h := Ωi∩⋃T ∈ Th |T ∩ Γ 6= ∅

7

and let FΓi,h :=

F ∈ Fh

∣∣∣F ∩ ΩΓi,h 6= ∅

be the set of facets which have some part in ΩΓ

i,h.

Then, the ghost penalty term can be defined as

J(p, q) := −γ∑i=1,2

∑F∈FΓ

i,h

h3 (JEi∇p · nK , JEi∇q · nK)F , (5)

for (p, q) ∈ QΓh × QΓ

h, where Ei is the canonical extension operator (on the discrete spaceQΓh).The stabilization parameter γ is O(1). In particular, the term J(p, q) is independent of

the cut position within the elements which results in the robustness of the method, c.f. Figure8. The problem without the ghost penalty stabilization was, that the enriched space QΓ

h wasto big to fulfill the LBB condition. With the new stabilization the LBB condition is still notvalid, but by bounding the gradients close to the interface, one can show the discrete“inf-sup”condition of the big system.

(a) without ghost penalty (b) with ghost penalty

Figure 8: ph ∈ QΓh using triangulation and interface position as in Figure 7b.

3 Velocity enrichment

3.1 Velocity enrichment Vxh

In the past section we enriched the pressure space and obtained QΓh, s.t. jumps in the pressure

solution at the interface Γ can be resolved. Now, due to (1d), if µ1 6= µ2, the velocity solutionu will have weak discontinuities, i.e. kinks at the interface Γ. Those discontinuities can not beresolved very well by the approximation space Vh. In particular, for a function u which has akink at the interface Γ, we have the approximation error estimate

infvh∈Vh

‖vh − u‖L2(Ω) . h3/2 ‖u‖H2(Ω1,2) , (6)

which again, is sharp, even if u is arbitrarily smooth on the subdomains Ω1,Ω2. Increasing thepiecewise polynomial degree in Vh again will not cure the situation. To achieve better results,we introduce the enriched velocity approximation space VΓ

h := Vh ⊕ Vxh. The space Vx

h isconstructed from Vh similarly as Qxh has been constructed from Qh. The only difference is thatthe polynomial degree of Vh is one degree higher than the degree of Qh and thus we can notrefer to a nodal basis for the construction of Vx

h: Therefore let again J := 1, . . . , n, wheren = dim(Vh) and (ϕi)i∈J a basis of Vh. Now define JΓ := j ∈ J | suppϕj ∩ Γ 6= ∅ the index

8

set of basis functions which are non trivial on Ω1 and non trivial on Ω2. Let χ2 again be theindicator function with χ2(x) = 0 for x ∈ Ω1 and χ2(x) = 1 for x ∈ Ω2. Now define ϕΓ

j := χ2ϕjfor j ∈ JΓ and

Vxh := span

ϕΓj | j ∈ JΓ

.

Note, that while it is irrelevant for the theoretical construction of VΓh whether we choose ϕΓ

j :=

χ2ϕj or ϕΓj := (1 − χ2)ϕj as new basis function for j ∈ JΓ, it does make a difference for the

arising linear system and the computational structure. For details on the right choice we referto the lecture notes [Leh15].

3.2 Nitsche XFEM formulation

We start with a few simple calculations, the result of which facilitate the introduction of theconsidered Nitche discretization. Consider the term (f ,v)Γ in (2a) and equation (1d),

(f ,v)Γ =

∫Γ− Jσ(u, p)n · vK ds ,

where σi(u, p) := −µiD(ui) + Ipi. Since n1 = −n2, we get

Jσn · vK = σ1n1 · v1 + σ2n2 · v2

= σ1n1 · v2 + σ2n2 · v2 + σ1n1 · (v1 − v2)

= (σ1 − σ2)n1︸ ︷︷ ︸=f

·v2 + σ1n1 · (v1 − v2) =: A

but a similar computation shows

Jσn · vK = σ1n1 · v1 + σ2n2 · v1 + σ2n2 · (v2 − v1)

= (σ1 − σ2)n1︸ ︷︷ ︸=f

·v1 + σ2n1 · (v1 − v2) =: B

Now since A = B it holds for all 0 ≤ κ1, κ2 ≤ 1 with κ1 + κ2 = 1, that A = κ1A + κ2B = B.This observation shows

Jσn · vK = κ1A+ κ2B

= f · (κ1v2 + κ2v1) + (κ1σ1 + κ2σ2)n1 · (v1 − v2)

= f · (κ1v2 + κ2v1) + σ(u, p)n1 · JvK ,

where the weighted average is defined as σ := κ1σ1 + κ2σ2. We obtain

(f ,v)Γ = Nc((u, p), (v, q)) + (f , κ1v2 + κ2v1)Γ ,

where Nc((u, p), (v, q)) :=∫

Γ σ(u, p)n1 · JvK ds. We add the symmetrical counterpartNc((v, q), (u, p)) to preserve symmetry of the bilinear form. This is consistent, due to JuK = 0on Γ for the solution u.By introducing the enriched space VΓ

h we now can resolve kinks and jumps in the velocity. Asthe velocity has to be continuous, we add another symmetric (stabilization) term

Ns(u,v) :=

∫Γ

λ µh

JuK · JvK ds ,

9

which penalizes jumps in the velocity. Note that Ns(u,v) = 0 for the solution u. The parameterλ has to be chosen accordingly, which will be discussed in section 3.4. We define the Nitscheterm

N((u, p), (v, q)) := Nc((u, p), (v, q)) +Nc((v, q), (u, p)) +Ns(u,v) (7)

and arrive at the Nitsche XFEM-formulation (with ghost penalty stabilization):Find (u, p) ∈ VΓ

h ×QΓh, such that

1

2

∑i=1,2

µi (D(u),D(v))Ωi− (p,div(v))Ωi

+N((u, p), (v, q))

+∑i=1,2

(q,div(u))Ωi+ J(p, q)

=∑i=1,2

ρi (g,v)Ωi+ (f , κ1v2 + κ2v1)Γ .

(8)

for all (v, q) ∈ VΓh ×QΓ

h.This formulation is consistent for any valid choice of κ1, κ2. However, the stability of the methoddoes depend on the precise choice of κ1, κ2.

3.3 The weighted average

We choose the Heaviside weighting for κ1, κ2. This means κi|T = κT,i, where for T ∈ Th, let

κT,1 :=

1 if |Ω1 ∩ T | > |Ω2 ∩ T |0 else,

κT,2 := 1− κT,2 .(9)

In particular this choice of weights guarantees, that for i = 1, 2, v ∈ VΓh and all T ∈ Th

κ2T,i

∫ΓT

h ‖D(v)‖2 ds ≤ ctr

∫Ti

‖D(v)‖2 dx (10)

holds, with a constant ctr > 0 only depending on the polynomial degree k = 2, cκ and cΓ. Thisis important for the stability of the method. cκ is a fixed constant such that

κ2i ≤ cκ

|Ti||T |

holds for all T ∈ Th, i = 1, 2. Since we chose the Heaviside weighting, this holds with cκ := 2.

3.4 Choice of λ

For a sufficiently large λ stability of the discretization can be ensured. The drawback of choosingλ unnecessarily large is an increase in the condition number. However it is sufficient to chooseλ > cλ := max 1, 4ctr (for details see [Leh15]).

In Figure 9 the differences between using and not using a velocity enrichment can be observed.Without V x

h the kink can not be approximated very well.

10

(a) (uh, ph) ∈ Vh ×QΓh (b) (uh, ph) ∈ VΓ

h ×QΓh

Figure 9: Approximation of the velocity kink

4 Numerical examples

The results in the following section were calculated using the software package Netgen/NGSolveand the extension ngsxfem, which was presented in the lecture. For the new methods for thetwo-phase stokes problem we implemented some further extensions in the ngsxfem package.

To observe the expected convergence rates, we consider an example given in [KGR15] wherethe exact solution is given by

u(x, y) = α(r)e−r2

(−yx

)where r =

√x2 + y2

α(r) =

µ−1

1 if r < rΓ

µ−12 + (µ−1

1 − µ−12 )er

2−r2Γ if r ≥ rΓ

p(x, y) = x3 +

c if x ∈ Ω1

0 else,

where Ω := (−1, 1)2, Ω1 := B2/3 is a ball with radius rΓ = 2/3 and a given force on the interfacef = cnΓ. Note that the coefficient α(r) is continuous and either constant (if µ1 = µ2) or notconstant (if the viscosities are different), which will produce a kink in the velocity. The forcef = cnΓ will only act on the pressure and will produce a discontinuity across the interface. Bythat we mean that the pressure can be decomposed p = p1 + p2 with p2 = χ2c and p1 thesolution of the problem without surface tension force.

4.1 First example

In the first example we choose µ1 = µ2 = 1 and c = 0.5 (cf. Figure 10).

11

(a) velocity (b) pressure

Figure 10: Solutions of the first example

Due to (3) the L2-error of the pressure only converges with order 1/2 using no enrichment. Ifwe use the pressure space QΓ

h, we observe a quadradic convergence (using piecewise linears forthe pressure + enrichment) (cf. Figure 11).

0 1 2 3 410−4

10−3

10−2

10−1

100

Qh × Vh

QΓh × Vh

QΓh × Vh + Ghost

QΓh × V Γ

h

QΓh × V Γ

h + Ghost

h1/2

h2

Figure 11: pressure error ‖p− ph‖L2(Ω)

As expected we lose one order of convergence in the H1-error and can see that without theghost penalty, the gradient of the pressure can not be controlled (cf. Figure 12).

0 1 2 3 4

10−1

100

101

Qh × Vh

QΓh × Vh

QΓh × Vh + Ghost

QΓh × V Γ

h

QΓh × V Γ

h + Ghost

h1

Figure 12: pressure error ‖p− ph‖H1(Ω)

12

Although we have no kinks in the velocity, we see, using no enrichment Vxh, that the L2-error

of the velocity only converges with order 3/2 as the L2 pressure error converges only with order1/2, consider interface condition (1d) (cf. Figure 13). If the pressure space is enriched, weobserve the optimal order 3 independent of the choice for the velocity space.

0 1 2 3 410−7

10−6

10−5

10−4

10−3

10−2

10−1

Qh × Vh

QΓh × Vh

QΓh × Vh + Ghost

QΓh × V Γ

h

QΓh × V Γ

h + Ghost

h3/2

h3

Figure 13: velocity error ‖u− uh‖L2(Ω)

In Figure 14 we see one order less than in Figure 13, as expected for the H1 norm.

0 1 2 3 4

10−4

10−3

10−2

10−1

Qh × Vh

QΓh × Vh

QΓh × Vh + Ghost

QΓh × V Γ

h

QΓh × V Γ

h + Ghost

h2

Figure 14: velocity error ‖u− uh‖H1(Ω)

13

4.2 Second example

In the second example we choose µ1 = 1, µ2 = 10 and c = 0.5 (cf. Figure 15).

(a) velocity (b) pressure

Figure 15: Solutions of the second example

Similar as in Figure 11 we observe the convergence rate 1/2 without enrichment, but also inthe case QΓ

h ×Vh (and also with the ghost penalty) the rate is not optimal, due to the couplingbetween the pressure and the velocity in equation (1d). Note, that in this example we have akink in the velocity across the interface, which means that a bad approximation for the velocityimpacts the pressure approximation (cf. Figure 17 and Figure 16).

Figure 16: Pressure with and without velocity enrichment

14

0 1 2 3 410−4

10−3

10−2

10−1

100

Qh × Vh

QΓh × Vh

QΓh × Vh + Ghost

QΓh × V Γ

h

QΓh × V Γ

h + Ghost

h1/2

h2

Figure 17: pressure error ‖p− ph‖L2(Ω)

The H1-error shows the same situation as in Figure 17 but we see that the ghost penalty is (alsoin the case of a velocity enrichment) important to get an optimal order as the ghost penaltypenalizes large gradients in the pressure close to the interface (cf. Figure 18).

0 1 2 3 4

10−1

100

101

Qh × Vh

QΓh × Vh

QΓh × Vh + Ghost

QΓh × V Γ

h

QΓh × V Γ

h + Ghost

h1

Figure 18: pressure error ‖p− ph‖H1(Ω)

15

In Figure 19 and 20 the L2- and H1-error of the velocity are shown. As expected due to (6),if no velocity enrichment is used, we get no better convergence rate than 3/2 respectively 1/2.Including an enrichment for the velocity and the pressure, we get the optimal rates.

0 1 2 3 410−7

10−6

10−5

10−4

10−3

10−2

10−1

100

Qh × Vh

QΓh × Vh

QΓh × Vh + Ghost

QΓh × V Γ

h

QΓh × V Γ

h + Ghost

h3/2

h3

Figure 19: velocity error ‖u− uh‖L2(Ω)

0 1 2 3 4

10−4

10−3

10−2

10−1

100

Qh × Vh

QΓh × Vh

QΓh × Vh + Ghost

QΓh × V Γ

h

QΓh × V Γ

h + Ghost

h1/2

h2

Figure 20: velocity error ‖u− uh‖H1(Ω)

5 Conclusion

With the combination of the already known ghost penalty stabilization for the pressure enrich-ment and the new derived Nitsche formulation for the velocity enrichment one is now able toresolve occuring jumps and/or weak discontinuities. The numerical examples show that the op-timal convergence rates can be observed. An open question is a proof for the discrete“inf-sup”condition although we did not observe problems in our numerical examples what indicates thatthe condition should be valid.Another open question is how the new method can be extended to time dependent problems, asfor example the method of lines is not straight forward for velocities with weak discontinuities.

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References

[KGR15] Matthias Kirchhart, Sven Gross, and Arnold Reusken. Analysis of an xfem discretiza-tion for stokes interface problems. 2015. Institut fur Geometrie und praktische Mathe-matik (https://www.igpm.rwth-aachen.de).

[Leh15] Christoph Lehrenfeld. Extended finite element methods for interface problems, lecturenotes. 2015. TU Wien.

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