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A NONLOCAL STOKES SYSTEM WITH VOLUME CONSTRAINTS

QIANG DU AND ZUOQIANG SHI

Abstract. In this paper, we introduce a nonlocal model for linear steady Stokes system with physical no-slip boundary condition. We use the idea of volume constraint to enforce the no-slip boundary condition. We prove that the nonlocal model is well-posed and the solution of the nonlocal system converges to the solution of the original Stokes system when the nonlocality vanishes.

1. Introduction. Recently, nonlocal models and corresponding numerical meth- ods attracts lots of attentions and find many successful applications. In solid mechan- ics, the theory of peridynamics [30] has been shown to be an alternative to conven- tional models of elasticity and fracture mechanics. Many numerical methods have also been developed to compute peridynamic model based on solid mathematical analysis [8, 23, 31, 10, 9, 34]. Nonlocal methods are also successfully applied in image process- ing and data analysis [26, 25, 2, 6, 19, 16, 27, 15, 18, 4, 22, 33]. The idea of integral approximation is also applied to derive numerical scheme for solving PDEs on point cloud.

In this paper, we are trying to extend the nonlocal model in Stokes system in fluid mechanics. A nonlocal model was proposed in [11] for Stokes system with periodic boundary condition. In this paper, we consider the no-slip boundary condition. More precisely, for domain Ω⊂Rn,

∆u(x)− ∇p(x) = f(x), x∈Ω

∇ ·u(x) = 0, x∈Ω.

(1.1)

No-slip boundary condition at the boundary is u= 0, at ∂Ω.

(1.2)

For the pressure, we impose average zero condition Z

p(x)dx= 0.

(1.3)

Comparing to the periodic boundary condition, the no-slip boundary condition is more natural and more often used in real application. However, the theoretical study with no-slip boundary condition is also much more difficult. The first problem is how to enforce no-slip boundary condition in the nonlocal approach. No-slip bound- ary condition is basically Dirichlet type boundary condition. Recently, Du et.al. [8]

proposed volume constraint to deal with the boundary condition in the nonlocal dif- fusion problem. They found that in the nonlocal diffusion problem, since the operator is nonlocal, only enforce the boundary condition on the boundary is not enough, it is necessary to extend the boundary condition to a small region adjacent to the bound- ary. Using this idea, in the nonlocal Stokes system, we extend the no-slip condition to a small layer as shown in Fig. 1. The whole computational domain Ω is decomposed

Department of Applied Physics and Applied Mathematics, Columbia University, New York, NY, 10027, USA,Email: qd2125@columbia.edu

Yau Mathematical Sciences Center, Tsinghua University, Beijing, China, 100084. Email:

zqshi@tsinghua.edu.cn.

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2δ

Vδ δ

Ω = Ωδ∪ Vδ

Fig. 1.Computational domain in non-local Stokes model.

to two parts Ω =Vδ

SΩδ as shown in Fig. 1 anduis enforced to be zero in Vδ, i.e.

(1.4) uδ(x) = 0, x∈ Vδ.

In Ωδ, inspired by the point integral method [29], Stokes equation is approximated by an nonlocal approach,

−Lδuδ(x) +Gδpδ(x) = Z

δ(x,y)f(y)dy, x∈Ωδ, Dδuδ(x)−L¯δpδ(x) = 0, x∈Ω, (1.5)

The integral operators in (2.1) are defined as Lδu(x) = 1

δ2 Z

Rδ(x,y)(u(x)−u(y))dy, (1.6)

Gδp(x) = 1 2δ2

Z

Rδ(x,y)(x−y)p(y)dy, (1.7)

Dδu(x) = 1 2δ2

Z

Rδ(x,y)(x−y)·u(y)dy, (1.8)

δp(x) = Z

δ(x,y)(p(x)−p(y))dy.

(1.9) where

Rδ(x,y) =CδR

kx−yk22

, R¯δ(x,y) =Cδ

kx−yk22

(1.10)

Cδ is a normalization factor such thatR

RnRδ(x,y)dy= 1 and ¯R(r) =R+∞

r R(s)ds.

The kernel function R(r) :R+→R+ is assumed to beC2 smooth and satisfies some mild conditions which are listed in Assumption 1.

Finally, we also need average zero condition for the pressure (1.11)

Z

pδ(x)dx= 0.

(1.4), (2.1) and (1.11) form a complete nonlocal formulation for Stokes equation.

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Nonlocal integral approximation is actually closely related to many numerical schemes of computational fluid dynamics, such as the smoothed particle hydrody- namics (SPH) [14, 20, 21, 24], vortex methods [1, 7] and others [3, 5, 12, 17, 32].

Analysis to the linear steady Stokes equation in this paper could give some new un- derstanding to the theoretical foundation of these methods.

The Stokes system (2.1) is well-known to be a saddle point problem. To preserve the well-posedness, we have to be very careful in the derivation of the nonlocal approx- imation. It is demonstrated that the kernel function has to satisfy some additional conditions for the nonlocal Stokes system with periodic boundary by using Fourier analysis [11]. For the problem we consider in this paper, Fourier transform does not apply. To preserve the well-posedness, we add a relaxation term, ¯Lδpδ(x), in the equation of divergence free. The order of this term isO(δ), so it does not destroy the accuracy of the nonlocal approximation. However, this term is crucial in the analysis of well-posedness.

The rest of the paper is organized as follows. We give the formulating of the nonlocal linear Stokes system in Section 2 and assumptions are also introduced. Then the well-posedness of the nonlocal model is established in Section 3. The vanishing nonlocality limit will be analyzed in Section 4. We will prove the solution of the nonlocal system converges to the solution of the original Stokes system asδ goes to 0. In Section 5, we conclude with a summary and a discussion on future research.

2. Nonlocal Stokes system with volume constraints. Inspired by the vol- ume constraint [8] and the point integral method [29], we consider a nonlocal model of the Stokes system (1.1)-(1.2) as follows:









−Lδuδ(x) +Gδpδ(x) = Z

δ(x,y)f(y)dy, x∈Ωδ, Dδuδ(x)−L¯δpδ(x) = 0, x∈Ω,

uδ(x) = 0, x∈ Vδ, R

pδ(x)dx = 0.

(2.1)

The integral operators have been defined in (1.6)-(1.9). Here, Ωδ and Vδ are subsets of Ω which are defined as

δ={x∈Ω :B(x,2δ)∩∂Ω =∅}, Vδ = Ω\Ωδ. (2.2)

The relation of Ω,∂Ω, Ωδ andVδ are showed in Fig. 1.

Moreover, we assume that the kernel function R(r) satisfies the conditions in Assumption 1.

Assumption1.

Assumptions on the computational domain: Ω, ∂Ωare both compact andC smooth andsatisfies the cone condition.

Assumptions on the kernel function R(r):

(a) (regularity)R∈C1(R+);

(b) (positivity and compact support)R(r)≥0andR(r) = 0for∀r >1;

(c) (nondegeneracy) ∃δ0>0so that R(r)≥δ0 for0≤r≤ 12.

Based on these assumptions, we could prove the well-posedness and the vanishing nonlocality limit of nonlocal Stokes system (2.1).

Remark2.1. If we consider the periodic boundary condition, the well-posedness of the nonlocal system (2.1) is easy to prove.

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Applying Fourier transform on (2.1), we have a linear system:

Aδ(ξ)

δ(ξ) ˆ pδ(ξ)

=

¯λδ(ξ) ˆf(ξ) 0

where

Aδ(ξ) =

λδ(ξ)In ibδ(ξ)

−ibδ(ξ)T cδ(ξ)

i=√

−1,In is ann×n identity matrix and λδ(ξ) = 1

δ2 Z

Rn

Rδ(|s|)(1−cos(ξ·s))ds,

¯λδ(ξ) = Z

Rn

δ(|s|) cos(ξ·s)ds, bδ(ξ) = 1

2 Z

Rn

sRδ(|s|) sin(ξ·s)ds, cδ(ξ) =

Z

Rn

δ(|s|)(1−cos(ξ·s))ds

Notice that λδ(ξ)>0 andcδ(ξ)>0 using the assumption thatRδ ≥0. Then it is easy to verify that the matrixAδ(ξ) is invertible for anyξ6= 0. The well-posedness follows from the nonsingularity ofAδ(ξ).

3. Well-posedness of the nonlocal Stokes system(2.1). In this section, we will prove the well-posedness of the nonlocal Stokes system (2.1). More precisely, we could prove followng theorem:

Theorem 3.1. Suppose Assumption 1 are satisfied. For anyf ∈H−1(Ω), there exits one and only one pair u, p, such that

(a) u∈H1(Ω),p∈L2(Ω). In addition,

kukH1(Ω)+kpkL2(Ω)≤CkfkH1(Ω),

whereC >0 is a constant only depends onand kernel functionR.

(b) u, p verify the nonlocal Stokes system (2.1).

In the proof of the well-posedness, we need several technical lemmas.

Lemma 3.2. ([29]) Ifδis small enough, for any functionu∈L2(M), there exists a constant C >0 independent onδ andu, such that

Z

Z

R

|x−y|2 32δ2

(u(x)−u(y))2dxdy≤C Z

Z

R

|x−y|22

(u(x)−u(y))2dxdy.

Lemma 3.3. ([28]) For any function u∈L2(Ωδ), there exists a constantC >0 only depneds on Ω, such that

1 δ2

Z

δ

Z

δ

Rδ(x,y)(u(x)−u(y))2dxdy+ 1 δ2

Z

δ

u2(x) Z

Vδ

Rδ(x,y)dy

dx≥C Z

δ|∇v|2dx, where

v(x) = 1 wδ(x)

Z

δ

Rδ(x,y)u(y)dy,

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andwδ(x) = Z

Rδ(x,y) dy.

Proof. First, we introduce an extension ofuto Ω,

˜ u(x) =

u(x), x∈Ωδ, 0, x∈ Vδ. It follows that

v(x) = 1 wδ(x)

Z

δ

Rδ(x,y)u(y)dy= 1 wδ(x)

Z

Rδ(x,y)˜u(y)dy Notice that for any x ∈ Ωδ, wδ(x) = 1 and

Z

xRδ(x,y)dy = 0. Then for any x∈Ωδ, we have

∇v(x) = 1 wδ(x)

Z

(∇xRδ(x,y))˜u(y)dy=− 1 wδ(x)

Z

(∇xRδ(x,y))(˜u(x)−u(y))dy˜ This gives that

Z

δ|∇v(x)|2dx= 1

¯ wδ2

Z

δ

Z

(∇xRδ(x,y))(˜u(x)−u(y))dy˜

2

dx

≤ 1

¯ wδ2

Z

δ

Z

|∇xRδ(x,y)|dy Z

|∇xRδ(x,y)|(˜u(x)−u(y))˜ 2dy

dx

≤ C δ2

Z

δ

Z

Rδ(x,y)(˜u(x)−u(y))˜ 2dy

dx

= C δ2

Z

δ

Z

δ

Rδ(x,y)(u(x)−u(y))2dxdy+ C δ2

Z

Vδ

Z

δ

Rδ(x,y)u(y)2dy

! dx which complete the proof.

Lemma 3.4. ([28]) For any function u∈L2(Ωδ), there exists a constantC >0 independent onδ, such that

1 δ2

Z

δ

Z

δ

Rδ(x,y)(u(x)−u(y))2dxdy+ Z

δ

u2(x) Z

Vδ

Rδ(x,y)dy

!

dx≥Ckuk2L2(Ωδ), as long asδ small enough.

Now we can prove the main theorem in this section, Theorem 3.1.

Proof. [Proof of Theorem 3.1:]

First, in the nonlocal Stokes system, we replace the conditionR

pδ(x)dx= 0 by R

δpδ(x)dx= 0 and denote the pressure in the original nonlocal Stokes system as ¯p.

It is obvious that

¯

pδ=pδ− 1

|Ω| Z

pδ(x)dx (3.1)

Multiplying uδ on the first equation of (2.1) and multiplying p on the third equation of (2.1) and adding them together, we can get

1 δ2

Z

Z

Rδ(x,y)|uδ(x)−uδ(y)|2dxdy+

Z

Z

δ(x,y)(pδ(x)−pδ(y))2dxdy

=−2 Z

Z

δ(x,y)f(y)·uδ(x)dxdy (3.2)

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Using this equation, we can get the uniqueness of the solution of the nonlocal Stokes equations (2.1) which also imply the existence of the solution in L2 using standard theory for integral equations.

From (3.2), using Lemma 3.4, there existsC >0, such that kuδk2L2(Ω)≤CkfkH1(Ω)kuδkH1(Ω). (3.3)

In addition, from the first equation of (2.1), uδ has following expression, for any x∈Ωδ,

uδ(x) = 1 wδ(x)

Z

Rδ(x,y)uδ(y)dy+ 1 2wδ(x)

Z

Rδ(x,y)(x−y)pδ(y)dy− δ2 wδ(x)

Z

δ(x,y)f(y)dy (3.4)

Using Lemma 3.3 and (3.2), (3.3), we have k∇

1 wδ(x)

Z

Rδ(x,y)uδ(y)dy

k2L2(Ωδ)

≤C δ2

Z

Z

Rδ(x,y)|uδ(x)−uδ(y)|2dxdy≤CkfkH1(Ω)kuδkH1(Ω) (3.5)

Notice that for anyx∈Ωδ,wδ(x) is a positive constant. Then we have

k∇

1 2wδ(x)

Z

Rδ(x,y)(x−y)pδ(y)dy

k2L2(Ωδ)

(3.6)

≤C Z

δ

Z

xRδ(x,y)(x−y)pδ(y)dy

2

dx+C Z

δ

Z

Rδ(x,y)pδ(y)dy 2

dx

≤C δ2

Z

Z

|Rδ(x,y)||x−y|2pδ(y)dy

2

dx+C Z

Z

Rδ(x,y)pδ(y)dy 2

dx

≤C Z

Z

|Rδ(x,y)|pδ(y)dy 2

dx+C Z

Z

Rδ(x,y)pδ(y)dy 2

dx

≤Ckpk2L2(Ω). where

Rδ(x,y) =CδR

|x−y|22

, R(r) = d drR(r).

In addition, direct calculation gives that k∇

δ2 wδ(x)

Z

δ(x,y)f(y)dy

kL2(Ωδ)≤CkfkH1(Ω) (3.7)

Putting (3.3)-(3.7) together, we obtain

kuδkH1(Ω)≤CkfkH1(Ω)+CkpδkL2(Ω)

(3.8)

Next, we turn to estimate the pressurep. First, considering the problem

∇ ·v(x) =pδ(x), x∈Ωδ. (3.9)

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It is well known (e.g. Section 3.3 of [13]) that if Ωδ satisfies cone condition, there exists at least one solution of (3.9), denoted byv, such that

v∈H01(Ωδ), kvkH1(Ωδ)≤ckpkL2(Ωδ)

(3.10)

Then, we extend v to Ω by assigning the value on Vδ to be 0 and denote the new function also byv. Obviously, we have

v ∈H01(Ωδ)∩H01(Ω), kvkH1(Ω)≤ckpδkL2(Ωδ)

(3.11)

On the other hand, using the second equation of (2.1),∀x∈Ω

¯

wδ(x)pδ(x) = Z

δ(x,y)pδ(y)dy+ 1 2δ2

Z

Rδ(x,y)(x−y)·uδ(y)dy

= Z

δ

δ(x,y)∇ ·v(y)dy+ 1 2δ2

Z

δ

Rδ(x,y)(x−y)·uδ(y)dy + 1

2 Z

Vδ

Rδ(x,y)(x−y)·uδ(y)dy+ Z

Vδ

δ(x,y)pδ(y)dy

=− 1 2δ2

Z

δ

Rδ(x,y)(x−y)·v(y)dy¯ + Z

Vδ

δ(x,y)pδ(y)dy (3.12)

where ¯wδ(x) =R

δ(x,y)dy and ¯v=v−uδ. Then, it follows that

1 2δ2

Z

δ

v(x)¯ Z

Rδ(x,y)(x−y)pδ(y)dy

dx

=− 1 2δ2

Z

pδ(x) Z

δ

Rδ(x,y)(x−y)¯v(y)dy

! dx

= Z

p2δ(x) ¯wδ(x)dx− Z

pδ(x) Z

Vδ

δ(x,y)pδ(y)dy

! dx.

(3.13)

The first term is positive which is a good term. The second term becomes

− Z

pδ(x) Z

Vδ

δ(x,y)pδ(y)dy

! dx (3.14)

= Z

pδ(x) Z

Vδ

δ(x,y)(pδ(x)−pδ(y))dy

! dx−

Z

p2δ(x) Z

Vδ

δ(x,y)dy

! dx.

The second term of (3.14) can be controlled by the first term of (3.13). And the first

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term is bounded by

Z

pδ(x) Z

Vδ

δ(x,y)(pδ(x)−pδ(y))dy

! dx (3.15)

=1 2

Z

Vδ

Z

Vδ

δ(x,y)(pδ(x)−pδ(y))2dydx+ Z

δ

pδ(x) Z

Vδ

δ(x,y)(pδ(x)−pδ(y))dy

! dx

≥1 2

Z

Vδ

Z

Vδ

δ(x,y)(pδ(x)−pδ(y))2dydx+ Z

δ

Z

Vδ

δ(x,y)(pδ(x)−pδ(y))2dy

! dx

− Z

δ

Z

Vδ

δ(x,y)(pδ(x)−pδ(y))pδ(y)dy

! dx

≥1 2

Z

Z

Vδ

δ(x,y)(pδ(x)−pδ(y))2dy

! dx−1

2 Z

Vδ

p2δ(x) Z

δ

δ(x,y)dy

! dx.

Combining (3.13)-(3.15), we get

1 2δ2

Z

δ

¯ v(x)

Z

Rδ(x,y)(x−y)pδ(y)dy

dx≥ Z

δ

p2δ(x) Z

δ

δ(x,y)dy

! dx (3.16)

Now, we are ready to get the estimate ofpδ. Multiplying ¯v on both sides of the first equation of (2.1) and integrating over Ωδ, using the fact that ¯v(x) = 0, x∈ Vδ, we have

− 1 2δ2

Z

Z

Rδ(x,y)(uδ(x)−uδ(y))·(¯v(x)−v(y))dxdy+¯ 1 2δ2

Z

δ

¯ v(x)

Z

Rδ(x,y)(x−y)pδ(y)dy

dx

= Z

δ

¯ v(x)

Z

δ(x,y)f(y)dy

dx (3.17)

Using (3.2), (3.8), (3.11), (3.17) and (3.16), we have 1

2kpk2L2(Ωδ)

≤ 1

2 Z

Z

Rδ(x,y)|uδ(x)−uδ(y)|2dxdy 1/2

1 2δ2

Z

Z

Rδ(x,y)|v(x)¯ −v(y)¯ |2dxdy 1/2

+kv¯kH1(Ωδ)kfkH1(Ω)

≤ 1

2 Z

Z

Rδ(x,y)|uδ(x)−uδ(y)|2dxdy

1/2 1 2δ2

Z

Z

Rδ(x,y)|v(x)−v(y)|2dxdy 1/2

+ 1

2 Z

Z

Rδ(x,y)|uδ(x)−uδ(y)|2dxdy 1/2!

+ (kvkH1(Ωδ)+kuδkH1(Ωδ))kfkH1(Ω)

≤kuδkH1(Ω)kfkH1(Ω)+kuδk1/2H1(Ω)kfk1/2H1(Ω)kvkH1(Ω)+C(kpδkL2(Ωδ)+kfkH1(Ω))kfkH1(Ω)

≤C(kpδkL2(Ω)+kfkH1(Ω))kfkH1(Ω). (3.18)

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On the other hand, using Lemma 3.2, kpkL2(Ω) can be bounded by kpkL2(Ωδ). First, notice that using the nondegeneracy assumption in Assumption 1, it is easy to verify that for anyx∈Ω,

(3.19)

Z

δ

(x,y)dy≥c0>0.

where

(x,y) =Cδ

kx−yk2 4(4δ)2

, Cδ is the normalization factor in (1.10).

kpδk2L2(Ω)≤C Z

Z

δ|pδ(x)|2(x,y)dy

! dx

≤C Z

Z

δ|pδ(x)−pδ(y)|2(x,y)dy

! dx+C

Z

Z

δ|pδ(y)|2(x,y)dy

! dx

≤C Z

Z

|pδ(x)−pδ(y)|2(x,y)dydx+C Z

δ|pδ(y)|2dx

≤C Z

Z

|pδ(x)−pδ(y)|2δ(x,y)dydx+Ckpδk2L2(Ωδ)

Using (3.2), (3.8), (3.18) and (3.20), we have

kpδk2L2(Ω)≤CkuδkH1(Ω)kfkH1(Ω)+Ckpδk2L2(Ωδ)

≤C(kpδkL2(Ω)+kfkH1(Ω))kfkH1(Ω) (3.20)

Together with (3.18), we have

kpδkL2(Ω)≤CkfkH1(Ω). (3.21)

This also gives theH1 estimate ofuδ using (3.8), kuδkH1(Ω)≤CkfkH1(Ω). (3.22)

and using (3.1)

kp¯δkL2(Ω)≤ kpδkL2(Ω)+ 1

|Ω| Z

pδ(x)dx

≤CkfkH1(Ω), (3.23)

here we use the fact that 1

|Ω| Z

pδ(x)dx ≤ 1

p|Ω|kpδkL2(Ω).

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4. Vanishing nonlocality. Besides the well-posedness, another problem we are interested is the limit behavior of the nonlocal Stokes system (2.1) as the nonlocality vanish, i.e. δ→0. In this section, we will answer this question. Under some assump- tions, we can prove that the solution of the nonlocal Stokes system converges to the solution of the Stokes system as δ→ 0. Furthermore, we could give an estimate of the convergence rate. The result is summarized in Theorem 4.2.

Before stating the main theorem, we give several lemmas which will be used in proving the main theorem.

We also need following theorem regarding the nonlocal approximation of the Laplace operator. And the proof can be found in [28].

Theorem 4.1. ([28]) Let u∈H3(Ω) and r(x) =−1

δ2 Z

Rδ(x,y)(u(x)−u(y))dy− Z

δ(x,y)∆u(y)dy, ∀x∈Ωδ.

There exists constants C, T0 depending only onΩ, so that for anyδ≤T0, kr(x)kL2(Ωδ)≤CδkukH3(Ω),

(4.1)

k∇r(x)kL2(Ωδ)≤CkukH3(Ω). (4.2)

This is the main result in this section regarding the convergence of the nonlocal Stokes system as the nonlocality vanish.

Theorem 4.2. Letu(x),p(x)be solution of Stokes system(1.1)anduδ(x), pδ(x) be solution of nonlocal Stokes system (2.1)withf ∈H1(Ω). There exists C >0only depends onandR, such that

ku−uδkH1(Ωδ)+kp−pδkL2(Ω)≤C√

δkfkH1(Ω)

Proof. Let eδ(x) =u(x)−uδ(x) and dδ =p−pδ|Ω1δ|

R

δ(p(x)−pδ(x))dx, theneδ, dδ satisfies

















−1 δ2

Z

Rδ(x,y)(eδ(x)−eδ(y))dy+ 1 2δ2

Z

Rδ(x,y)(x−y)dδ(y)dy = ru(x), x∈Ωδ, eδ(x) = u(x), x∈ Vδ, 1

2 Z

Rδ(x,y)(x−y)·eδ(y)dy− Z

δ(x,y)(dδ(x)−dδ(y))dy = rp(x), x∈Ω, Z

δ

dδ(x)dx = 0, (4.3)

where

ru(x) = Z

δ(x,y)∆u(y)dy+ 1 δ2

Z

Rδ(x,y)(u(x)−u(y))dy, ∀x∈Ωδ (4.4)

rp(x) =− Z

δ(x,y)(p(x)−p(y))dy, ∀x∈Ω.

(4.5)

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First, we focus on the following estimate

1 δ2

Z

δ

eδ(x)· Z

Rδ(x,y)(eδ(x)−eδ(y))dydx (4.6)

=1 δ2

Z

δ

eδ(x)· Z

δ

Rδ(x,y)(eδ(x)−eδ(y))dydx+ 1 δ2

Z

δ

eδ(x)· Z

Vδ

Rδ(x,y)(eδ(x)−eδ(y))dydx

= 1 2δ2

Z

δ

Z

δ

Rδ(x,y)|eδ(x)−eδ(y)|2dxdy+ 1 δ2

Z

δ

eδ(x)· Z

Vδ

Rδ(x,y)(eδ(x)−eδ(y))dydx.

The second term of the right hand side of (4.6) can be calculated as

1 δ2

Z

δ

eδ(x)· Z

Vδ

Rδ(x,y)(eδ(x)−eδ(y))dydx (4.7)

=1 δ2

Z

δ|eδ(x)|2 Z

Vδ

Rδ(x,y)dy

! dx− 1

δ2 Z

δ

eδ(x)· Z

Vδ

Rδ(x,y)u(y)dy

! dx.

Here we use the definition ofeδ and the volume constraint conditionuδ(x) = 0, x∈ Vδ to get thateδ(x) =u(x), x∈ Vδ.

The first term is positive which is good for us. We only need to bound the second term of (4.7) to show that it can be controlled by the first term. First, the second term can be bounded as following

1 δ2

Z

δ

eδ(x)· Z

Vδ

Rδ(x,y)u(y)dy

! dx

(4.8)

≤1 δ2

Z

δ|eδ(x)| Z

Vδ

Rδ(x,y)dy

!1/2

Z

Vδ

Rδ(x,y)|u(y)|2dy

!1/2

dx

≤1 δ2

Z

δ

1

2|eδ(x)|2 Z

Vδ

Rδ(x,y)dy

! dx+ 2

Z

δ

Z

Vδ

Rδ(x,y)|u(y)|2dy

! dx

!

≤ 1 2δ2

Z

δ|eδ(x)|2 Z

Vδ

Rδ(x,y)dy

!

dx+ 2 δ2

Z

Vδ|u(y)|2 Z

δ

Rδ(x,y)dx

! dy

≤ 1 2δ2

Z

δ|eδ(x)|2 Z

Vδ

Rδ(x,y)dy

!

dx+ C δ2

Z

Vδ|u(y)|2dy

≤ 1 2δ2

Z

δ|eδ(x)|2 Z

Vδ

Rδ(x,y)dy

!

dx+Cδkfk2H1(Ω).

Here we use Lemma A.1 in Appendix A to get the last inequality.

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By substituting (4.8), (4.7) in (4.6), we get

1 δ2

Z

δ

eδ(x)· Z

Rδ(x,y)(eδ(x)−eδ(y))dydx (4.9)

≥ 1 2δ2

Z

δ

Z

δ

Rδ(x,y)|eδ(x)−eδ(y)|2dxdy

+ 1 2δ2

Z

δ|eδ(x)|2 Z

Vδ

Rδ(x,y)dy

!

dx−Ckfk2H1(Ω)δ.

This is the key estimate we used to get convergence.

We also need following bound

1 δ2

Z

δ

eδ(x)· Z

Rδ(x,y)(x−y)dδ(y)dy

dx+ 1 δ2

Z

dδ(x) Z

Rδ(x,y)(x−y)·eδ(y)dy

dx (4.10)

= 1 δ2

Z

dδ(x) Z

Vδ

Rδ(x,y)(x−y)·eδ(y)dy

! dx

≤1 δ

Z

Z

Vδ

Rδ(x,y)|dδ(x)||u(y)|dy

! dx

≤1 δ

"

Z

Z

Vδ

Rδ(x,y)|dδ(x)|2dy

! dx

Z

Z

Vδ

Rδ(x,y)|u(y)|2dy

! dx

#1/2

≤C√

δkfkH1(Ω)kdδkL2(Ω).

Multiplying eδ(x), dδ(x) on both sides of the first and third equation of (4.3) and integrating over Ωδ, Ω respectively and adding them together, using (4.9), (4.10), we have

1 δ2

Z

δ

Z

δ

Rδ(x,y)|eδ(x)−eδ(y)|2dxdy+ 1 2δ2

Z

δ|eδ(x)|2 Z

Vδ

Rδ(x,y)dy

! dx (4.11)

+ Z

Z

Rδ(x,y)|dδ(x)−dδ(y)|2dxdy

≤krukL2(Ωδ)keδkL2(Ωδ)+krpkL2(Ω)kdδkL2(Ω)+C√

δkfkH1(Ω)kdδkL2(Ω)+Cδkfk2H1(Ω). To simplify the notation, we denote the right hand side of (4.11) asQ2.

It is well known (e.g. Section 3.3 of [13]) that with the condition thatR

δdδ(x)dx= 0, there exists at least one functionψ, such that

∇ ·ψ(x) =dδ(x), x∈Ωδ, and ψ∈H01(Ωδ), kψkH1(Ωδ)≤ckdδkL2(Ωδ)

(4.12)

Then, we extend ψ to Ω by assigning the value on Vδ to be 0 and denote the new function also byψ. Obviously, we have

ψ∈H01(Ωδ)∩H01(Ω), kψkH1(Ω)≤ckdδkL2(Ωδ)

(4.13)

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Using the third equation of (4.3), we have

¯

wδ(x)dδ(x) = Z

δ(x,y)dδ(y)dy+ 1 2δ2

Z

Rδ(x,y)(x−y)·eδ(y)dy−rp(x)

= Z

δ

δ(x,y)∇ ·ψ(y)dy+ 1 2δ2

Z

δ

Rδ(x,y)(x−y)·eδ(y)dy + 1

2 Z

Vδ

Rδ(x,y)(x−y)·eδ(y)dy+ Z

Vδ

δ(x,y)dδ(y)dy−rp(x)

=− 1 2δ2

Z

δ

Rδ(x,y)(x−y)·ψ(y)dy¯ + 1 2δ2

Z

Vδ

Rδ(x,y)(x−y)·u(y)dy +

Z

Vδ

δ(x,y)dδ(y)dy−rp(x) (4.14)

where ¯wδ(x) =R

δ(x,y)dy and ¯ψ=ψ−eδ. Then, it follows that

1 2δ2

Z

δ

ψ(x)¯ Z

Rδ(x,y)(x−y)dδ(y)dy

dx

=− 1 2δ2

Z

dδ(x) Z

δ

Rδ(x,y)(x−y) ¯ψ(y)dy

! dx

= Z

d2δ(x) ¯wδ(x)dx− Z

dδ(x) Z

Vδ

δ(x,y)dδ(y)dy

! dx

− Z

dδ(x) Z

Vδ

Rδ(x,y)(x−y)·u(y)dy

! dx+

Z

dδ(x)rp(x)dx (4.15)

The first term is positive which is a good term. The second term becomes

− Z

dδ(x) Z

Vδ

δ(x,y)dδ(y)dy

! dx (4.16)

= Z

dδ(x) Z

Vδ

δ(x,y)(dδ(x)−dδ(y))dy

!

dx− 1 2δ2

Z

d2δ(x) Z

Vδ

δ(x,y)dy

! dx.

The second term of (4.16) can be controlled by the first term of (4.15). And the first

(14)

term is bounded by

Z

dδ(x) Z

Vδ

δ(x,y)(dδ(x)−dδ(y))dy

! dx (4.17)

=1 2

Z

Vδ

Z

Vδ

δ(x,y)(dδ(x)−dδ(y))2dydx+ Z

δ

dδ(x) Z

Vδ

δ(x,y)(dδ(x)−dδ(y))dy

! dx

≥1 2

Z

Vδ

Z

Vδ

δ(x,y)(dδ(x)−dδ(y))2dydx+ Z

δ

Z

Vδ

δ(x,y)(dδ(x)−dδ(y))2dy

! dx

− Z

δ

Z

Vδ

δ(x,y)(dδ(x)−dδ(y))dδ(y)dy

! dx

≥1 2

Z

Z

Vδ

δ(x,y)(dδ(x)−dδ(y))2dy

! dx−1

2 Z

Vδ

d2δ(x) Z

δ

δ(x,y)dy

! dx.

Combining (4.15)-(4.17), we get

1 2δ2

Z

δ

ψ(x)¯ Z

Rδ(x,y)(x−y)dδ(y)dy

dx (4.18)

≥ Z

δ

d2δ(x) Z

δ

δ(x,y)dy

!

dx− 1 2δ2

Z

dδ(x) Z

Vδ

Rδ(x,y)(x−y)·u(y)dy

! dx +

Z

dδ(x)rp(x)dx.

In addition, we have

1 2δ2

Z

dδ(x) Z

Vδ

Rδ(x,y)(x−y)·u(y)dy

! dx

(4.19)

≤1 2δ

Z

|dδ(x)| Z

Vδ

Rδ(x,y)|u(y)|dy

! dx

≤1 2δ

"

Z

|dδ(x)|2 Z

Vδ

Rδ(x,y)dy

! dx

Z

Z

Vδ

Rδ(x,y)|u(y)|2dy

! dx

#1/2

≤C√

δkfkH1(Ω)kdδkL2(Ω). and

Z

dδ(x)rp(x)dx (4.20)

= Z

dδ(x) Z

δ(x,y)(p(x)−p(y))dy

dx

≤CδkpkH1(Ω)kdδkL2(Ω)

≤CδkfkH1(Ω)kdδkL2(Ω).

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Multiplying ¯ψon both sides of the first equation of (4.3) and using (4.18), (4.19), (4.20), we have

kdδk2L2(Ωδ)≤1 δ2

Z

δ

Z

δ

Rδ(x,y)(eδ(x)−eδ(y))·( ¯ψ(x)−ψ(y))dxdy¯

+ 1 δ2

Z

δ

ψ(x)¯ · Z

Vδ

Rδ(x,y)(eδ(x)−eδ(y))dy

! dx +kψ¯kL2(Ωδ)krukL2(Ω)+C√

δkfkH1(Ω)kdδkL2(Ω) (4.21)

The first term can be bounded as

1 δ2

Z

δ

Z

δ

Rδ(x,y)(eδ(x)−eδ(y))·( ¯ψ(x)−ψ(y))dxdy¯ (4.22)

≤ 1 δ2

Z

δ

Z

δ

Rδ(x,y)|eδ(x)−eδ(y)|2dxdy

!1/2

1 δ2

Z

δ

Z

δ

Rδ(x,y)|ψ(x)¯ −ψ(y)¯ |2dxdy

!1/2

≤ 1 δ2

Z

δ

Z

δ

Rδ(x,y)|eδ(x)−eδ(y)|2dxdy

!1/2

 1 δ2

Z

δ

Z

δ

Rδ(x,y)|ψ(x)−ψ(y)|2dxdy

!1/2

+ 1

δ2 Z

δ

Z

δ

Rδ(x,y)|eδ(x)−eδ(y)|2dxdy

!1/2

≤Q2+CQkψkH1(Ωδ)≤Q2+CQkdδkL2(Ωδ).

The estimate of the second term of (4.21) is a little involved. First

1 δ2

Z

δ

ψ(x)¯ · Z

Vδ

Rδ(x,y)(eδ(x)−eδ(y))dy

! dx

(4.23)

≤ 1 δ2

Z

δ

Z

Vδ

Rδ(x,y)( ¯ψ(x)−ψ(y))¯ ·(eδ(x)−eδ(y))dy

! dx

+

1 δ2

Z

Vδ

u(x)· Z

δ

Rδ(x,y)(eδ(x)−eδ(y))dy

! dx

 1 δ2

Z

δ

Z

Vδ

Rδ(x,y)|ψ(x)¯ −ψ(y)¯ |2dy

! dx

!1/2

+ 1

δ2 Z

Vδ|u(x)|2 Z

δ

Rδ(x,y)dy

! dx

!1/2

 1

δ2 Z

δ

Z

Vδ

Rδ(x,y)|eδ(x)−eδ(y)|2dy

! dx

!1/2

≤C

kψkH1(Ω)+√

δkfkH1(Ω) 1 δ2

Z

δ

Z

Vδ

Rδ(x,y)|eδ(x)−eδ(y)|2dy

! dx

!1/2

+ C δ2

Z

δ

Z

Vδ

Rδ(x,y)|eδ(x)−eδ(y)|2dy

! dx.

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