# A Discontinuous Finite Volume Method for the Darcy-Stokes Equations

Authors: Zhe Yin; Ziwen Jiang; Qiang Xu
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

This paper proposes a discontinuous finite volume method for the Darcy-Stokes equations. An optimal error estimate for the approximation of velocity is obtained in a mesh-dependent norm. First-orderL2-error estimates are derived for the approximations of both velocity and pressure. Some numerical examples verifying the theoretical predictions are presented.


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## Body

## 1. Introduction

The study of discontinuous Galerkin methods has been a very active research field since they were proposed by Reed and Hill [1] in 1973. Discontinuous Galerkin methods use discontinuous functions as finite element approximation and enforce the connections of the approximate solutions between elements by adding some penalty terms. The flexibility of discontinuous functions gives discontinuous Galerkin methods many advantages, such as high parallelizability and localizability. Arnold et al. [2] provided a framework for the analysis of a large class of discontinuous Galerkin methods for second-order elliptic problems.Based on the advantages of using discontinuous functions for approximation in discontinuous Galerkin methods, it is natural to consider using discontinuous functions as trial functions in the finite volume method, which is called the discontinuous finite volume method. Such a method has the flexibility of the discontinuous Galerkin method and the simplicity and conservative properties of the finite volume method. Ye [3] developed a new discontinuous finite volume method and analyzed it for the second-order elliptic problem. Bi and Geng [4] proposed the semidiscrete and the backward Euler fully discrete discontinuous finite volume element methods for the second-order parabolic problems. Ye [5] considered the discontinuous finite volume method for solving the Stokes problems on both triangular and rectangular meshes and derived an optimal order error estimate for the approximation of velocity in a mesh-dependent norm and first-order L2-error estimates for the approximations of both velocity and pressure.The Darcy-Stokes problem is interesting for a variety of reasons. Apart from being a modeling tool in its own right, it also appears, less obviously, in time-stepping methods for Stokes and for high Reynolds number flows (where of course the convective term causes additional difficulties). In [6], the nonconforming Crouzeix-Raviart element is stabilized for the Darcy-Stokes problem with terms motivated by a discontinuous Galerkin approach. In [7], a new stabilized mixed finite element method is presented for the Darcy-Stokes equations.In this paper, we will extend the discontinuous finite volume methods to solve the Darcy-Stokes equations. In our methods, velocity is approximated by discontinuous piecewise linear functions on triangular meshes and by discontinuous piecewise rotated bilinear functions on rectangular meshes. Piecewise constant functions are used as the test functions for velocity in the discontinuous finite volume methods. We obtained an optimal error estimate for the approximation of velocity in a mesh-dependent norm. First-orderL2-error estimates are derived for the approximations of both velocity and pressure. For the sake of simplicity and easy presentation of the main ideas of our method, we restrict ourselves to the model problem.We consider the Darcy-Stokes equations(1.1a)σu-μΔu+∇p=f,x∈Ω,(1.1b)∇·u=0,x∈Ω,(1.1c)u=0,x∈∂Ω,where Ω is a bounded polygonal domain in R2 with boundary ∂Ω. u=(u1,u2) is the velocity, p is the pressure, and f is a given force term. We assume σ=1,μ=1.


## 2. Discontinuous Finite Volume Formulation

Letℛh be a triangular or rectangular partition of Ω. The triangles or rectangles in ℛh are divided into three or four subtriangles by connecting the barycenter of the triangle or the center of the rectangles to their corner nodes, respectively. Then we define the dual partition 𝒯h of the primal partition ℛh to be the union of the triangles shown in Figures 1 and 2 for both triangular and rectangular meshes.Figure 1
ElementT∈𝒯h for triangular mesh.Figure 2
ElementT∈𝒯h for rectangular mesh.LetPk(T) consist of all the polynomials with degree less than or equal to k defined on T. We define the finite dimensional trial function space for velocity on a triangular partition by
(2.1)Vh={v∈L2(Ω)2:v|K∈P1(K)2,∀K∈ℛh}
and on rectangular partition by
(2.2)Vh={v∈L2(Ω)2:v|K∈Q^1(K)2,∀K∈ℛh},
where Q^1 denotes the space of functions of the form a+bx1+cx2+d(x12-x22) on K.LetQh be the finite dimensional space for pressure
(2.3)Qh={q∈L02(Ω):q|K∈P0(K),∀K∈ℛh},
where
(2.4)L02(Ω)={q∈L2(Ω):∫Ω‍qdx=0}.
Define the finite dimensional test function space Wh for velocity associated with the dual partition 𝒯h as
(2.5)Wh={ξ∈L2(Ω)2:ξ|T∈P0(T)2,∀T∈𝒯h}.Multiplying (1.1a) and (1.1b) by ξ∈Wh and q∈Qh, respectively, we have
(2.6)(u,ξ)-∑T∈𝒯h‍∫∂T‍∂u∂n·ξds+∑T∈𝒯h‍∫∂T‍pξ·nds=(f,ξ),∑K∈ℛh‍∫K‍∇·uqdx=0,
where n is the unit outward normal vector on ∂T.LetTj∈𝒯h(j=1,…,t) be the triangles in K∈ℛh, where t=3 for triangular meshes and t=4 for rectangular meshes, as shown as Figures 3 and 4. Then we have
(2.7)∑T∈𝒯h‍∫∂T‍∂u∂n·ξds=∑K∈ℛh‍∑j=1t‍∫Aj+1CAj‍∂u∂n·ξds+∑K∈ℛh‍∫∂K‍∂u∂n·ξds,
where At+1=A1.Figure 3
Triangular partition and its dual.Figure 4
Rectangular partition and its dual.For vectorsv=(v1,v2) and n=(n1,n2), let v⊗n denote the matrix whose ijth component is vi·nj as in [5]. For two matrix valued variables σ and τ, we define σ:τ=∑i,j=12σi,jτi,j. Let Γ=∑K∈ℛh∂K,Γ0=Γ∖∂Ω. Let e be an interior edge shared by two elements K1 and K2 in ℛh. We define the average {·} and jump [·] on e for scalar q, vector w, and matrix τ, respectively. If e∈Γ0,
(2.8){q}=12(q|∂K1+q|∂K2),{w}=12(w|∂K1+w|∂K2),{τ}=12(τ|∂K1+τ|∂K2),[q]=q|∂K1n1+q|∂K2n2,[w]=w|∂K1·n1+w|∂K2·n2,[τ]=τ|∂K1·n1+τ|∂K2·n2,
where n1 and n2 are unit normal vectors on e pointing exterior to K1 and K2, respectively. We also define a matrix valued jump ⟦·⟧ for a vector w as
(2.9)⟦w⟧=w|∂K1⊗n1+w|∂K2⊗n2.
If e∈∂Ω, define
(2.10){q}=q,[w]=w·n,{τ}=τ,⟦w⟧=w⊗n.
A straightforward computation gives
(2.11)∑K∈ℛh‍∫∂K‍qv·nds=∑e∈Γ0‍∫e‍[q]·{v}ds+∑e∈Γ‍∫e‍{q}[v]ds,(2.12)∑K∈ℛh‍∫∂K‍v·τnds=∑e∈Γ0‍∫e‍[τ]·{v}ds+∑e∈Γ‍∫e‍{τ}:⟦v⟧ds.
Let ∫Γ‍qds=∑e∈Γ‍∫e‍qds. Using (2.7), (2.12), and the fact that [∇u]=0 for u∈(H01(Ω)∩H2(Ω))2 on Γ0, (2.7) becomes
(2.13)∑T∈𝒯h‍∫∂T‍∂u∂n·ξds=∑K∈ℛh‍∑j=1t‍∫Aj+1CAj‍∂u∂n·ξds+∫Γ‍⟦ξ⟧:{∇u}ds.
Since [p]=0 for p∈H1(Ω) on Γ0, we also have
(2.14)∑T∈𝒯h‍∫∂T‍pξ·nds=∑K∈ℛh‍∑j=1t‍∫Aj+1CAj‍pξ·nds+∫Γ‍{p}[ξ]ds.
Let V(h)=Vh+(H2(Ω)∩H01(Ω))2. Define a mapping γ:V(h)→Wh,
(2.15)γv|T=1he∫e‍v|Tds,∀T∈𝒯h,
where he is the length of the edge e.We define two norms forV(h) as follows:
(2.16)∥|v∥|12=∥v∥1,h2+∑e∈Γ‍⟦γv⟧e2,∥|v∥|2=∥|v∥|12+∑K∈ℛh‍hK2|v|2,K2,
where ∥v∥1,h2=|v|0,h2+|v|1,h2,  |v|0,h2=∑K∈ℛh‍|v|0,K2,|v|1,h2=∑K∈ℛh‍|v|1,K2, and hK= diameter of K.As in [5], the standard inverse inequality implies that there is a constant C such that
(2.17)∥|v∥|≤C∥|v∥|1,∀v∈Vh.Lemma 2.1.
There exists a positive constantC independent of h such that
(2.18)h∥|v∥|≤C∥v∥,∥v∥≤C∥|v∥|,∀v∈Vh.Proof.
As in [4],
(2.19)h∥|v|∥1,h≤C∥v∥,∥v∥≤C∥|v|∥1,h,∀v∈Vh,
where ∥|v∥|1,h2=|v|1,h2+∑e∈Γ⟦γv⟧e2+∑K∈ℛhhK2|v|2,K2. Since ∥|v∥|1,h≤∥|v∥|, we have ∥v∥≤C∥|v∥|. Note that v∈Vh is a piecewise linear function, and h2∥|v∥|2=h2|v|0,h2+h2|v|1,h2+h2∑e∈Γ⟦γv⟧e2=I1+I2+I3. By Lemma 3.6 in [4], I2≤C∥v∥2,I3≤C∥v∥2, we have h∥|v∥|≤C∥v∥.Lemma 2.2 (see [4]).
The operatorγ is self-adjoint with respect to the L2-inner product, (u,γv)=(v,γu),∀u,v∈Vh. Define ∥|v∥|0=(v,γv)1/2. Then ∥|·∥|0 and ∥·∥ are equivalent; here the equivalence constants are independent of h. And ∥γv∥=∥v∥,∀v∈Vh.Let(2.20)a0(v,ξ)=(v,ξ)-∑K∈ℛh‍∑j=1t‍∫Aj+1CAj‍∂v∂n·ξds-∫Γ‍⟦ξ⟧:{∇v}ds,c(ξ,q)=∑K∈ℛh‍∑j=1t‍∫Aj+1CAj‍qξ·nds+∫Γ‍{q}[ξ]ds,b0(v,q)=∑K∈ℛh‍∫K‍∇·vqdx.
It is clear that the solutions (u,p) of the Darcy-Stokes equations (1.1a)–(1.1c) satisfy the following:
(2.21)a0(u,ξ)+c(ξ,p)=(f,ξ),∀ξ∈Wh,b0(u,q)=0,∀q∈Qh.
Define the following bilinear forms:
(2.22)A0(v,w)=a0(v,γw),∀w,v∈V(h),B0(v,q)=b0(v,q),∀v∈V(h),∀q∈L02(Ω),C(v,q)=c(γv,q),∀v∈V(h),∀q∈L02(Ω).
Then systems (2.21) are equivalent to
(2.23)A0(u,v)+C(v,p)=(f,γv),∀v∈Vh,B0(u,q)=0,∀q∈Qh.We propose two discontinuous finite volume formulations based on modification of the weak formulation (2.23) for Darcy-Stokes problem (1.1a)–(1.1c). Let us introduce the bilinear forms as follows:
(2.24)A1(v,w)=A0(v,w)+α∑e∈Γ‍⟦γv⟧e:⟦γw⟧e,B(v,q)=B0(v,q)-∫Γ‍{q}[γv]ds,
where α>0 is a parameter to be determined later. For the exact solution (u,p) of (1.1a)–(1.1c), we have
(2.25)A0(u,v)=A1(u,v),∀v∈Vh,B0(u,q)=B(u,q),∀q∈Qh.
Therefore, it follows from (2.23) that
(2.26)A1(u,v)+C(v,p)=(f,γv),∀v∈Vh,B(u,q)=0,∀q∈Qh.
The corresponding discontinuous finite volume scheme seeks (uh,ph)∈Vh×Qh, such that
(2.27)A1(uh,v)+C(v,ph)=(f,γv),∀v∈Vh,B(uh,q)=0,∀q∈Qh.Lete be an edge of element K. It is well known (see [2]) that there exists a constant C such that for any function g∈H2(K),
(2.28)∥g∥e2≤C(hK-1∥g∥K2+hK|g|1,K2),(2.29)∥∂g∂n∥e2≤C(hK-1|g|1,K2+hK|g|2,K2),
where C depends only on the minimum angle of K.Let∇hv and ∇h·v be the functions whose restriction to each element ∀K∈ℛh is equal to ∇v and ∇·v, respectively.Lemma 2.3.
Forv,w∈V(h), there exists a positive constant C independent of h such that
(2.30)A1(v,w)≤C∥|v∥|∥|w∥|.Proof.
LetA**(v,w)=(v,γw)+A*(v,w),
(2.31)A*(v,w)=-∑K∈ℛh‍∑j=1t‍∫Aj+1CAj‍∂v∂n·γwds.
By Lemma 3.1 in [5],
(2.32)A*(v,w)=(∇hv,∇hw)+∑K∈ℛh‍∫∂K‍(γw-w)∂v∂nds+∑K∈ℛh‍(Δv,w-γw)K.|A**(v,w)|≤|(v,γw)|+|(∇hv,∇hw)|+|∑K∈ℛh‍∫∂K‍(γw-w)∂v∂nds|+|∑K∈ℛh‍(Δv,w-γw)K|≤C(|v|0,h|w|0,h+|v|1,h|w|1,h+∑K∈ℛh‍(hK-1∥w-γw∥K2+hK|w-γw|1,K2)1/2×(hK-1|v|1,K2+hK|v|2,K2)1/2+∑K∈ℛh‍hK|v|2,K|w|1,K)≤C(|v|0,h|w|0,h+|v|1,h|w|1,h+(∑K∈ℛh‍|w|1,K2)1/2×(|v|1,h+(∑K∈ℛh‍hK2|v|2,K2)1/2)+(∑K∈ℛh‍hK2|v|2,K2)1/2|w|1,h)≤C∥|v∥|∥|w∥|,A1(v,w)=A**(v,w)-∫Γ‍⟦γw⟧:{∇v}ds+α∑e∈Γ‍⟦γve⟧:⟦γwe⟧,≤C(∥|v∥|∥|w∥|+(∑K∈ℛh‍(|v|1,K2+hK2|v|2,K2)1/2)(∑e∈Γ‍⟦γwe2⟧)1/2+α(∑e∈Γ‍⟦γve2⟧)1/2(∑e∈Γ‍⟦γwe2⟧)1/2)≤C∥|v∥|∥|w∥|.Lemma 2.4 (see [5]).
For any(v,q)∈V(h)×Qh, one has
(2.33)C(v,q)=-B(v,q).Lemma 2.5 (see [5]).
For(v,q)∈V(h)×L02(Ω), there exists a positive constant M independent of h such that
(2.34)C(v,q)≤M∥|v∥|(∥q∥+(∑K∈ℛh‍hK2|q|1,K2)1/2).
If (v,q)∈Vh×Qh, then
(2.35)C(v,q)≤M∥|v∥|∥q∥.Lemma 2.6.
For anyv∈Vh, there is a constant C0 independent of h such that for α large enough
(2.36)A1(v,v)≥C0∥|v∥|2.Proof.
Using the proof of Lemmas 3.1 and 3.5 in [5], for v∈Vh,
(2.37)∫Γ‍γv:⟦∇v⟧ds≤C∥|v∥|1(∑e∈Γ‍⟦γv⟧e2)1/2,A*(v,w)=(∇hv,∇hw),∀v,w∈Vh,
we have
(2.38)A1(v,v)=(v,γv)+(∇hv,∇hv)+α∑e∈Γ‍⟦γv⟧e2-∫Γ‍⟦γv⟧:{∇v}ds,≥|v|0,h2+|v|1,h2+α∑e∈Γ‍⟦γv⟧e2-C∥|v∥|1(∑e∈Γ‍⟦γv⟧e2)1/2≥C∥|v∥|12≥C0∥|v∥|2,
when α is large enough.The value ofα depends on the constant in the inverse inequality. Therefore, the value of α for which A1(·,·) is coercive is mesh dependent. We introduce a second discontinuous finite volume scheme which is parameter insensitive. Define a bilinear form as follows:
(2.39)A2(v,w)=A1(v,w)+∫Γ‍⟦γv⟧:{∇w}ds.
Similar to the bilinear form A1(·,·), for the exact solution (u,p) of the Darcy-Stokes problem we have
(2.40)A2(u,v)=A0(u,v),∀v∈Vh.
Consequently, the solution of the Darcy-Stokes problem satisfies the following variational equations:
(2.41)A2(u,v)+C(v,p)=(f,γv),∀v∈Vh,B(u,q)=0,∀q∈Qh.
Our second discontinuous finite volume scheme for (1.1a)–(1.1c) seeks (uh,ph)∈Vh×Qh, such that
(2.42)A2(uh,v)+C(v,ph)=(f,γv),∀v∈Vh,B(uh,q)=0,∀q∈Qh.For any value ofα>0, we have
(2.43)A2(v,v)=(v,γv)+(∇hv,∇hv)+α∑e∈Γ‍⟦γv⟧e2≥C∥|v∥|12≥C0∥|v∥|2,∀v∈Vh.
Similarly, we can prove that
(2.44)A2(v,w)≤C∥|w∥|∥|v∥|,∀v,w∈V(h).
Let A(v,w)=A1(v,w) or A(v,w)=A2(v,w). In the rest of the paper, we assume that the following is true:
(2.45)A(v,v)≥C0∥|v∥|2.
If A(v,w)=A2(v,w), (2.45) holds for any α>0. If A(v,w)=A1(v,w), (2.45) holds for only α large enough.


## 3. Error Estimates

We will derive optimal error estimates for velocity in the norm∥|·∥| and for pressure in the L2-norm. A first-order error estimate for velocity in L2-norm will be obtained.Lete be an interior edge shared by two elements K1 and K2 in ℛh. If ∫e‍v|K1ds=∫e‍v|K2ds, we say that v is continuous on e. We say that v is zero at e∈∂Ω if ∫e‍vds=0. Define a subspace V^h of Vh by
(3.1)V^h={v∈L2(Ω)2:v|K∈Q^1(K)2∀K∈ℛhiscontinuousate∈Γ0andiszeroate∈∂Ω}
for rectangular meshes and by
(3.2)V^h={v∈L2(Ω)2:v|K∈P1(K)2∀K∈ℛhiscontinuousate∈Γ0andiszeroate∈∂Ω}
for triangular mesh.It has been proven in [8, 9] that the following discrete inf-sup condition is satisfied; that is, there exists a positive constant β0 such that
(3.3)supv∈V^h(∇h·v,q)|v|1,h≥β0∥q∥,∀q∈Qh.Lemma 3.1.
The bilinear formB(·,·) satisfies the discrete inf-sup condition
(3.4)supv∈VhB(v,q)∥|v∥|≥β∥q∥,∀q∈Qh,
where β is a positive constant independent of the mesh size h.Proof.
Forv∈V^h⊂Vh and q∈Qh, we have B(v,q)=(∇h·v,q), and ∥|v∥|1=∥v∥1,h. By Poincare-Friedrichs ∥v∥1,h≤C|v|1,h, with (3.3), and (2.17) we get for any q∈Qh(3.5)β0∥q∥≤supv∈V^h(∇·v,q)|v|1,h≤Csupv∈V^hB(v,q)∥|v∥|1≤C1supv∈VhB(v,q)∥|v∥|.
With β=β0/C1, we have proven (3.4).Define an operatorπK:H1(K)→P1(K) or Q^1(K). For all v∈H1(K),
(3.6)∫ei‍πKvds=∫ei‍vds,i=1,…,t,
where ei,i=1,…,t, are the t sides of the element K. t=3 if K is a triangle and t=4 if K is a rectangle. It was proven in [8] that
(3.7)|πKv-v|s,K≤Ch2-s|v|2,K,s=0,1,2.
For all v=(v1,v2)∈H01(Ω)2, define Π1v=(Π1v1,Π1v2)∈Vh by
(3.8)Π1vi|K=πKvi,∀K∈ℛh,i=1,2.
Using the definition of Π1 and integration by parts, we can show that
(3.9)B(v-Π1v,q)=0,∀q∈Qh.
The Cauchy-Schwarz inequality implies
(3.10)⟦γv⟧e2=(1he∫e‍⟦v⟧ds)2≤(1he)2∫e‍⟦v⟧2ds∫e‍ds=∫e‍1he⟦v⟧2ds.
Equations (2.28) and (3.8) imply that
(3.11)∑e∈Γ‍⟦γ(u-Π1u)⟧e2≤C(|u-Π1u|1,h2+∑K∈ℛh‍h-2∥u-Π1u∥K2)≤Ch2∥u∥22.
The definitions of the norm ∥|·∥|, (3.7), and (3.11) give
(3.12)∥|u-Π1u∥|2=|u-Π1u|0,h2+|u-Π1u|1,h2+∑e∈Γ‍⟦γ(u-Π1u)⟧e2+∑K∈ℛh‍h2|u-Π1u|2,K2≤Ch2∥u∥22.Theorem 3.2.
Let(uh,ph)∈Vh×Qh be the solution of (2.27), and let (u,p)∈(H2(Ω)∩H01(Ω))2×(L02(Ω)∩H1(Ω)) be the solution of (1.1a)–(1.1c). Then there exists a constant C independent of h such that
(3.13)∥|u-uh∥|+∥p-ph∥≤Ch(∥u∥2+∥p∥1),(3.14)∥u-uh∥≤Ch(∥u∥2+∥p∥1).Proof.
Letε=u-Π1u,εh=uh-Π1u,η=p-Π2p,ηh=ph-Π2p, where Π2 is L2 projection from L02(Ω)→Qh. Then u-uh=ε-εh,p-ph=η-ηh. Subtracting (2.26) from (2.27) and using Lemma 2.4, we get error equations(3.15a)A(εh,v)-B(v,ηh)=A(ε,v)+C(v,η),∀v∈Vh,(3.15b)B(εh,q)=B(ε,q)=0,∀q∈Qh.By letting v=εh in (3.15a) and q=ηh in (3.15b), the sum of (3.15a) and (3.15b) gives
(3.16)A(εh,εh)=A(ε,εh)+C(εh,η).
Thus, it follows from the coercivity (2.45), the boundedness (2.30), (2.44), and (2.34) that
(3.17)∥|εh∥|2≤C(∥|ε∥|∥|εh∥|+(∥η∥+(∑K∈ℛh‍hK2|η|1,K2)1/2)∥|εh∥|),
which implies the following:
(3.18)∥|εh∥|≤C(∥|ε∥|+∥η∥+(∑K∈ℛh‍hK2|η|1,K2)1/2).
The previous estimate can be rewritten as
(3.19)∥|uh-Π1u∥|≤C(∥|u-Π1u∥|+∥p-Π2p∥+(∑K∈ℛh‍hK2|p-Π2p|1,K2)1/2).
Now using the triangle inequality, (3.7), the definition of Π2, and the inequality mentioned previously, we get
(3.20)∥|u-uh∥|≤C(∥|u-Π1u∥|+∥|uh-Π1u∥|)≤Ch(∥u∥2+∥p∥1),
which completes the estimate for the velocity approximation.
Discrete inf-sup condition (3.4), (3.15a), (3.15b), Lemmas 2.5, 2.4, and inverse inequality give
(3.21)∥ph-Π2p∥≤1βsupv∈VhB(v,Π2p-ph)∥|v∥|1,h=1βsupv∈VhC(v,ph-Π2p)∥|v∥|1,h=1βsupv∈VhC(v,ph-p)+C(v,p-Π2p)∥|v∥|=1βsupv∈VhA(u-uh,v)+C(v,p-Π2p)∥|v∥|≤C(∥|u-uh∥|+∥p-Π2p∥+(∑K∈ℛh‍hK2|p-Π2p|1,K2)1/2)≤Ch(∥u∥2+∥p∥1).
Using the previous inequality and the triangle inequality, we have completed the proof of (3.13).
Using Lemma2.1, (3.12), and (3.13), we have
(3.22)∥uh-Π1uh∥≤C∥|uh-Π1uh∥|≤C(∥|u-uh∥|+∥|u-Π1uh∥|)≤Ch(∥u∥2+∥p∥1).
Equations (3.22) and (3.7) and the triangle inequality imply (3.14). We have completed the proof.


## 4. Numerical Experiments

In this section, we present a numerical example for solving the problems (1.1a)–(1.1c) by using the discontinuous finite volume element method presented with (2.27) and (2.42). Let Ω=(0,1)×(0,1), ℛh be the Delaunay triangulation generated by EasyMesh [10] over Ω with mesh size h as shown in Figure 5. We consider the case of σ=1,μ=1, the exact velocity u1(x,y)=-x2(x-1)2y(y-1)(2y-1), u2(x,y)=-u1(y,x) and the pressure p(x,y)=(x-0.5)(y-0.5). Denote the numerical solution as uh and ph with step hd which is used to generate the mesh data in the EasyMesh input file, and h=max{he:e∈Γ}. For α=2, the numerical results are presented in Tables 1 and 2. It is observed from the tables that the numerical results support our theory.Table 1
Error behavior for scheme (2.27).
h
d
h
|
∥
u
-
u
h
∥
|
|
∥
u
-
u
2
h
∥
|
|
∥
u
-
u
h
∥
|
∥
u
-
u
h
∥
∥
u
-
u
2
h
∥
∥
u
-
u
h
∥
∥
p
-
p
h
∥
∥
p
-
p
2
h
∥
∥
p
-
p
h
∥
1
/
8
1.598
e
-
1
2.082
e
-
2
3.393
e
-
4
1.068
e
-
2
1
/
16
8.372
e
-
2
1.031
e
-
2
2.0
9.649
e
-
5
3.5
5.345
e
-
3
2.0
1
/
32
3.679
e
-
2
5.185
e
-
3
2.0
2.598
e
-
5
3.7
2.650
e
-
3
2.0
1
/
64
1.899
e
-
2
2.611
e
-
3
2.0
6.795
e
-
6
3.8
1.323
e
-
3
2.0
1
/
128
9.413
e
-
3
1.307
e
-
3
2.0
1.730
e
-
6
3.9
6.598
e
-
4
2.0Table 2
Error behavior for scheme (2.42).
h
d
h
|
∥
u
-
u
h
∥
|
|
∥
u
-
u
2
h
∥
|
|
∥
u
-
u
h
∥
|
∥
u
-
u
h
∥
∥
u
-
u
2
h
∥
∥
u
-
u
h
∥
∥
p
-
p
h
∥
∥
p
-
p
2
h
∥
∥
p
-
p
h
∥
1
/
8
1.598
e
-
1
2.071
e
-
2
3.280
e
-
4
1.079
e
-
2
1
/
16
8.372
e
-
2
1.027
e
-
2
2.0
9.204
e
-
5
3.5
5.380
e
-
3
2.0
1
/
32
3.679
e
-
2
5.175
e
-
3
2.0
2.476
e
-
5
3.7
2.659
e
-
3
2.0
1
/
64
1.899
e
-
2
2.608
e
-
3
2.0
6.361
e
-
6
3.8
1.325
e
-
3
2.0
1
/
128
9.413
e
-
3
1.306
e
-
3
2.0
1.613
e
-
6
3.9
6.603
e
-
4
2.0Figure 5
Triangular and its dual partition of(0,1)×(0,1).



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*Source: 761242-2012-12-27.xml*