# Discontinuous Mixed Covolume Methods for Parabolic Problems

Authors: Ailing Zhu; Ziwen Jiang
Journal: The Scientific World Journal
 (2014)

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## Abstract

We present the semidiscrete and the backward Euler fully discrete discontinuous mixed covolume schemes for parabolic problems on triangular meshes. We give the error analysis of the discontinuous mixed covolume schemes and obtain optimal order error estimates in discontinuousHdiv and first-order error estimate in L2.


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## Body

## 1. Introduction

The study of discontinuous Galerkin methods has been a very active research area since its introduction in [1] in 1973. The discontinuous Galerkin method does not require continuity of the approximation functions across the interelement boundary but instead enforces the connection between elements by adding a penalty term. Because of the use of discontinuous functions, discontinuous Galerkin methods have the advantages of a high order of accuracy, high parallelizability, localizability, and easy handling of complicated geometries. Discontinuous Galerkin methods have been used to solve hyperbolic and elliptic equations by many researchers. For example, see [2–10]. In [11], the unified analysis of discontinuous Galerkin methods for elliptic problems was presented. In [12, 13], Ye developed a new discontinuous finite volume method for elliptic and Stokes problems, respectively. The discontinuous finite volume method was used for parabolic equations by Bi and Geng in [14]. In [15], Yang and Jiang extended a new discontinuous mixed covolume method for elliptic problems. In this paper, we consider the semidiscrete and the backward Euler fully discrete discontinuous mixed covolume methods for the second-order parabolic problems and derive the optimal order error estimates in the discontinuous H(div⁡) and first-order L2-error estimates in a mesh-dependent norm.The rest of this paper is organized as follows. In Section2, we introduce some notations and describe the discontinuous mixed covolume schemes for the second-order parabolic problems and give some lemmas which will be used in the convergence analysis. In Section 3, we prove the existence and uniqueness for the semidiscrete and the backward Euler fully discrete discontinuous mixed covolume approximation. A discontinuous mixed covolume elliptic projection is defined in Section 4. Error estimations in both discontinuous H(div⁡) and L2 norms of semidiscrete method and fully discrete method are proved in Sections 5 and 6.Throughout this paper, letterC denotes a generic positive constant independent of the mesh parameter and may stand for different values at its different appearances.


## 2. Discontinuous Mixed Covolume Formulation

In this paper, we consider the following parabolic problems:(1)pt(x,t)-∇·{D(x,t)∇p(x,t)}=f(x,t),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii(x,t)∈Ω×(0,T],p(x,0)=p0(x),x∈Ω,p(x,t)=0,(x,t)∈∂Ω×(0,T],
where Ω∈R2 is a bounded convex polygonal domain with the boundary ∂Ω,  x=(x,y),  p is an unknown function, and D is a symmetric, bounded matrix function which satisfies the following condition: there exist two positive constants α1, α2 such that
(2)α1ξ⊤ξ≤ξ⊤Dξ≤α2ξ⊤ξ,∀ξ=(ξ1,ξ2)∈R2;f is a given function in L2(Ω). Furthermore, we assume that the matrix M=D-1 is locally Lipschitz.Here and in what follows, we will not write the independentx, t for any functions unless it is necessary.Letu=-D∇p, and rewrite (1) as the system of first-order partial differential equations:
(3)Mu+∇p=0,(x,t)∈Ω×(0,T],pt+∇·u=f,(x,t)∈Ω×(0,T],p(x,0)=p0(x),x∈Ω,p(x,t)=0,(x,t)∈∂Ω×(0,T].We will use the standard definitions for the Sobolev spacesHs(K) and their associated inner products (·,·)s,K, norms ∥·∥s,K, and seminorms |·|s,K. The space H0(K) coincides with Ls(K), in which the norm and the inner product are denoted by ∥·∥K and (·,·)K, respectively.Let𝒯h={K} be a triangulation of the domain Ω. As usual, we assume the triangles K to be shape-regular. For a given triangulation 𝒯h, we construct a dual mesh 𝒯h* based upon the primal partition 𝒯h. Each triangle in 𝒯h can be divided into three subtriangles by connecting the barycenter Q of the triangle to their corner nodes Ai(i=1,2,3). Then, we define the dual partition 𝒯h* to be the union of the triangles. Let Pk(T) consist of all the polynomials functions of degree less than or equal to k defined on T. We define the finite-dimensional trial function space for velocity on 𝒯h by
(4)Vh:={v∈L2(Ω)2:v|K∈P1(K)2,∀K∈𝒯h}.
Define the finite-dimensional test function space Wh for velocity associated with the dual partition 𝒯h* as
(5)Wh:={w∈L2(Ω)2:w|T∈P0(T)2,∀T∈𝒯h*}.
Let Hh be the finite-dimensional space for pressure:
(6)Hh:={q∈L2(Ω):q|K∈P0(K),∀K∈𝒯h}.LetΓ denote the union of the boundary of the triangles K of 𝒯h and Γ0:Γ∖∂Ω. The traces of functions in Vh and Hh are double valued on Γ0. Let e be an interior edge shared by two triangles K1 and K2 in Th. Define the normal vectors n1 and n2 on e pointing exterior to K1 and K2, respectively. Next, we introduce some traces operators that we will use in our numerical formulation. We define the average {·} and jump [·] on e for scalar q and vector v, respectively,
(7){q}=12(q|∂K1+q|∂K2),[q]=q|∂K1n1+q|∂K2n2,{v}=12(v|∂K1+v|∂K2),[v]=v|∂K1n1+v|∂K2n2.
If e is an edge on the boundary of Ω, we set
(8){q}=q,[v]=v·n,
where n is the outward unit normal. We do not require either of the quantities [q] or {v} on boundary edges, and we leave them undefined.Multiplying the first and second equations in system (3) by w∈Wh and q∈Hh, respectively, and using the integration by parts formula in the equation, we have
(9)∑T∈𝒯h*‍∫T‍Mu·wdx+∑T∈𝒯h*‍∫∂T‍pw·nds=0,∀w∈Wh,∑K∈𝒯h‍∫K‍ptqdx+∑K∈𝒯h‍∫K‍∇·uqdx=(f,q),∀q∈Hh,
where n is the outward normal vector on ∂T. Let Tj∈𝒯h*(j=1,2,3) be the triangles in K∈𝒯h. Then we have
(10)∑T∈𝒯h*‍∫∂T‍pw·nds=∑K∈𝒯h‍∑j=13‍∫Aj+1QAj‍pw·nds+∑K∈𝒯h‍∫∂K‍pw·nds,∀w∈Wh,
where A4=A1. A straightforward computation gives
(11)∑K∈𝒯h‍∫∂K‍pw·nds=∑e∈Γ0‍∫e‍[p]·{w}ds+∑e∈Γ‍∫e‍{p}·[w]ds,∀w∈Wh.
Let ∫Γ‍pds=∑e∈Γ∫e‍pds. Using the above formula and the fact that [p]=0 for p∈H1(Ω) on Γ0, (10) becomes
(12)∑T∈𝒯h*‍∫∂T‍pw·nds=∑K∈𝒯h‍∑j=13‍∫Aj+1QAj‍pw·nds+∑e∈Γ‍∫e‍{p}[w]ds,∀w∈Wh.
Then, system (9) can be rewritten as follows:
(13)∑T∈𝒯h*‍∫T‍Mu·wdx+∑K∈𝒯h‍∑j=13‍∫Aj+1QAj‍pw·nds+∑e∈Γ‍∫e‍{p}[w]ds=0,∀w∈Wh,∑K∈𝒯h‍∫K‍ptqdx+∑K∈𝒯h‍∫K‍∇·uqdx=(f,q),∀q∈Hh.LetV(h)=Vh+H2(Ω)2. Define a mapping γ:V(h)→Wh as
(14)γv|T=1he∫e‍v|Tds,T∈𝒯h*,
where he is the length of the edge e. For v=(v1,v2)∈V(h), γvi(i=1,2) is defined as
(15)γvi|T=1he∫e‍vi|Tds,T∈𝒯h*,(i=1,2).
Then the system (13) is equivalent to
(16)∑T∈𝒯h*‍∫T‍Mu·γwdx+∑K∈𝒯h‍∑j=13‍∫Aj+1QAj‍pγw·nds+∑e∈Γ‍∫e‍{p}[γw]ds=0,∀w∈V(h),∑K∈𝒯h‍∫K‍ptqdx+∑K∈𝒯h‍∫K‍∇·uqdx=(f,q),∀q∈Hh.
Let
(17)a0(v,w):=∑T∈𝒯h*‍∫T‍Mv·γwdx,b(w,q):=∑K∈𝒯h‍∑j=13‍∫Aj+1QAj‍qγw·nds+∑e∈Γ‍∫e‍{q}[γw]ds,c0(v,q):=∑K∈𝒯h‍∫K‍∇·vqdx.Using the above bilinear forms, it is clear that system (16) can be rewritten as follows:
(18)a0(u,w)+b(w,p)=0,∀w∈V(h),(pt,q)+c0(u,q)=(f,q),∀q∈Hh.In order to define our numerical schemes, we introduce the bilinear forms as follows:(19)a(v,w):=a0(v,w)+α∑e∈Γ‍1he∫e‍M[v][w]ds,c(v,q):=c0(v,q)-∫Γ‍{q}[γv]ds,
where α>0 is a parameter to be determined later. For the exact solution (u,p) of system (3), we have
(20)a0(u,v)=a(u,v),∀v∈Vh,c0(u,q)=c(u,q),∀q∈Hh.
Therefore, it follows from (18) that
(21)a(u,w)+b(w,p)=0,∀w∈Vh,(pt,q)+c(u,q)=(f,q),∀q∈Hh.The discontinuous mixed covolume scheme for (3) reads as follows. Seek (uh,ph)∈Vh×Hh such that
(22)a(uh,w)+b(w,ph)=0,∀w∈Vh,(pht,q)+c(uh,q)=(f,q),∀q∈Hh,
where ph(0)=p~h(0), uh(0)=u~h(0), p~h(0), u~h(0) will be given in Section 4.LetN>0 be a positive integer, let 0=t0<t1<⋯<tj<⋯<tN=T be a subdivision of time. tj=jΔt(0≤j≤N), Δt=T/N. We use the backward Euler difference quotient
(23)∂tphj=phj-phj-1Δt,(j=1,2,…,N)
to approximate the differential quotient (∂phj/∂t)(j=1,2,…,N), in the semidiscrete scheme; then we obtain the backward Euler fully discrete discontinuous mixed covolume scheme for the problem (1): find (uhj,phj)∈Vh×Hh,  (j=1,2,…,N), such that
(24)a(uhj,w)+b(w,phj)=0,∀w∈Vh,(∂tphj,q)+c(uhj,q)=(f,q),∀q∈Hh,(j=1,2,…,N),
where ph0=p~h(0), uh0=u~h(0),p~h(0),u~h(0) will be given in Section 4.We define the following norms forv∈V(h):
(25)|||v|||div⁡2=∥v∥2+∥∇h·v∥2+∑e∈Γ‍1he∫e‍[v]2ds,|||v|||12=∥v∥2+|v|1,h2+∑e∈Γ‍1he∫e‍[v]2ds,|||v|||2=|||v|||12+∑K∈𝒯h‍hK2|v|2,K2,
where ∇h·v is the function whose restriction to each element K∈𝒯h is equal to ∇·v, and |v|1,h2=∑K∈𝒯h|v|1,K2.We will introduce some useful lemmas; for more details, see [6].Lemma 1.
Forv,w∈V(h), one has
(26)a(v,w)≤C|||v|||div⁡|||w|||div⁡.Lemma 2.
For(v,q)∈V(h)×L2(Ω), one has
(27)b(v,q)=-c(v,q).Lemma 3.
For(v,q)∈V(h)×L2(Ω), one has
(28)b(v,q)≤C|||v|||(∥q∥+(∑K∈𝒯hhK2|q|1,K2)1/2);
if (v,q)∈Vh×Hh, then
(29)b(v,q)≤C|||v|||·∥q∥.Lemma 4.
LetZh={vv∈Vh,c(v,q)=0,∀q∈Hh}; for any v∈Zh, there is a constant C0 independent of h such that, for α is large enough,
(30)a(v,v)≥C0|||v|||div⁡2.Lemma 5.
For anyq∈Hh, there is a constant β0 independent of h such that
(31)sup⁡v∈Vhc(v,q)|||v|||≥β0∥q∥.


## 3. Existence and Uniqueness for Discontinuous Mixed Covolume Approximations

In this section, we prove that the discontinuous mixed covolume formulation has a unique solution in the finite element spaceVh×Hh.Theorem 6.
Semidiscrete discontinuous mixed covolume scheme (22) has a unique solution in the space Vh×Hh.Proof.
Only prove that homogenous equation(32)a(uh,w)+b(w,ph)=0,∀w∈Vh,(pht,q)+c(uh,q)=0,∀q∈Hh,ph(0)=0,uh(0)=0
of (22) exists unique zero solution since the number of unknowns is the same as the number of line equations.
By lettingw=uh in the first formula of (32) and q=ph in the second formula of (32), using Lemma 2, the sum of (32) gives
(33)(pht,ph)+a(uh,uh)=0.
Using (1/2)(d/dt)(ph,ph)=(pht,ph) and Lemma 4, we have
(34)12ddt∥ph∥2+C0|||uh|||div⁡2≤0.
Integrating the above formula, we get
(35)∥ph∥2+2C0∫0t‍|||uh|||div⁡2dτ≤0.
Then ∥ph∥=0, |||uh|||div⁡=0. So, ph=0, uh=0, t∈(0,T]. This completes the proof.Theorem 7.
The fully discrete discontinuous mixed covolume method defined in (24) has a unique solution in the finite element space Vh×Hh.Proof.
Only prove that homogenous equation(36)a(uhj,w)+b(w,phj)=0,∀w∈Vh,(∂tphj,q)+c(uhj,q)=0,∀q∈Hh,(j=1,2,…,N),uh0=0,ph0=0
of (24) exists unique zero solution since the number of unknowns is the same as the number of line equations.
By lettingw=uhj in the first formula of (36) and q=phj in the second formula of (36), using Lemma 2, the sum of (36) gives
(37)(∂tphj,phj)+a(uhj,uhj)=0,(j=1,2,…,N).
Using Lemma 4 and
(38)(∂tphj,phj)=1Δt(phj-phj-1,phj)=12Δt[(phj,phj)-(phj-1,phj-1)+(phj-phj-1,phj-phj-1)]=12Δt[∥phj∥2-∥phj-1∥2+∥phj-phj-1∥2]≥12Δt[∥phj∥2-∥phj-1∥2],
we have, from (37),
(39)∥phj∥2-∥phj-1∥2+2C0Δt|||uhj|||div⁡2≤0,(j=1,2,…,N).
Adding the above inequality with j from 1 to i, using ph0=0, we have
(40)∥phi∥2+2C0Δt∑j=1i|||uhj|||div⁡2≤0,(i=1,2,…,N).
Hence, we have ∥phi∥2=0 and |||uhj|||div⁡2=0(i=1,2,…,N); that is, phi=0 and uhi=0(i=1,2,…,N). This completes the proof.


## 4. A Discontinuous Mixed Covolume Elliptic Projection

Define an operatorπK from H1(K) to P1(K) by requiring that, for any ∀v∈H1(K),
(41)∫ei‍πKvds=∫ei‍vds,(i=1,2,3),
where ei(i=1,2,3) are the three sides of the element K∈𝒯h. It was proved in [5] that
(42)|πKv-v|s,K≤h2-s|v|2,K,∀v∈H2(K),(s=0,1,2).For anyv∈H01(Ω)2, define Π1v∈Vh by
(43)(Π1v)i|K=ΠKvi,∀K∈𝒯h,(i=1,2).
Using the definition of Π1 and integration by parts, we can show that
(44)c(v-Π1v,q)=0,∀q∈Hh.
It was proved in [6] that
(45)|||u-Π1u|||div⁡2≤ch2∥u∥22.
Let Π2 be the projection from L02(Ω) to the finite element space Hh.Define a discontinuous mixed covolume elliptic projection by requiring that, findingu~h,p~n:(0,t)→Vh×Hh, such that
(46)a(u-u~h,w)+b(w,p-p~h)=0,∀w∈Vh,c(u-u~h,q)=0,∀q∈Hh.
It was proved in [15] that (46) has a unique solution and the error estimates in Theorem 8.Theorem 8.
Let(u~h,p~h)∈Vh×Hh be the solution of (46) and let (u,p)∈H2(Ω)2×H1(Ω) be the solution of (21). Then there exists a positive constant C independent of h such that
(47)|||u-u~h|||div⁡+∥p-p~h∥≤Ch(∥u∥2+∥p∥1).Theorem 9.
Let(u~h,p~h)∈Vh×Hh be the solution of (46) and let (u,p)∈H2(Ω)2×H1(Ω) be the solution of (21). Then there exists a positive constant C independent of h such that
(48)|||(u-u~h)t|||div⁡+∥(p-p~h)t∥≤Ch(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1).Differentiating each equation of (46) on t and using (44), (45), we can prove this theorem in the same way as [15].


## 5. Error Estimates for Semidiscrete Method

In this section, we will establish the error estimates in theH(div⁡) and L2 norms for the semidiscrete discontinuous mixed covolume method.Theorem 10.
Let(uh,ph)∈Vh×Hh be the solution of (22) and let ph(0)=p~h(0),  (u,p)∈H2(Ω)2×H1(Ω) be the solution of (3). Then there exists a positive constant C independent of h such that
(49)|||u-uh|||div⁡+∥p-ph∥≤Ch[∫0t(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1)dτ+∥u∥2+∥p∥1].Proof.
Letξ=p~h-ph, η=u~h-uh. Subtracting the two equations of (22) from those of (21), respectively, we have
(50)a(u-uh,w)+b(w,p-ph)=0,∀w∈Vh,(pt-pht,q)+c(u-uh,q)=0,∀q∈Hh.
Using (46), we have
(51)a(η,w)+b(w,ξ)=0,∀w∈Vh,(pt-pht,q)+c(η,q)=0,∀q∈Hh.
Differentiating the first equation of (51) on t, we have
(52)a(ηt,w)+b(w,ξt)=0,∀w∈Vh.
By lettingq=ξt in the second formula of (51) and letting w=η in (52), using Lemma 2, the sum of them gives
(53)a(ηt,η)+(ξt,ξt)=((p~h-p)t,ξt).
Using
(54)a(ηt,η)=12ddta(η,η)-12(Mtη,η),
we have
(55)12ddta(η,η)+∥ξt∥2≤C∥ξt∥·∥(p~h-p)t∥+C|||η|||div⁡2.
Multiplying the equation above with 2, integrating them from 0 to t, and using ϵ-inequality, Lemma 4, and (48), we can get
(56)|||η|||div⁡2+2C0(∫0t∥ξt∥dτ)2≤C1h∫0t∥ξt∥·(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1)dτ+C∫0t|||η|||div⁡2dt≤1C0(∫0t∥ξt∥dτ)2+Ch2(∫0t(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1)dτ)2+C∫0t|||η|||div⁡2dt;
then
(57)|||η|||div⁡2+1C0(∫0t∥ξt∥dτ)2≤Ch2(∫0t(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1)dτ)2,
so
(58)|||η|||div⁡≤Ch∫0t(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1)dτ,(59)∫0t‍∥ξt∥dτ≤Ch∫0t‍(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1)dτ;
hence
(60)∥ξ∥≤∫0t‍∥ξt∥dτ≤Ch∫0t‍(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1)dτ.
Now, using the triangle inequalities (47), (58), and (60), we get
(61)|||u-uh|||div⁡+∥p-ph∥≤Ch[∫0t‍(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1)dτ+∥u∥2+∥p∥1].
The proof is complete.


## 6. Error Estimates for Fully Discrete Method

Letξj=p~hj-phj, ζj=p~hj-pj, ηj=u~hj-uhj(j=0,1,…,N); then the error estimates for the backward Euler fully discrete discontinuous mixed covolume method in the H(div⁡) and L2 norms are provided in the next two theorems.Theorem 11.
Let(u,p)∈H2(Ω)2×H1(Ω) be the solution of (3), and let (uhj,phj)∈Vh×Hh(j=1,2,…,N) be the solution of (24) with t=tj(j=1,2,…,N), respectively. If ph0=p~h(0)=p0, uh0=u~h(0)=u0, then there exists a positive constant C independent of h and Δt such that
(62)max⁡0≤i≤N∥pi-phi∥≤CΔt∥ptt∥L∞(L2)+Ch(∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)+∥p∥L∞(H1)∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)).Proof.
Subtracting the two equations of (24) from (21), respectively, with t=tj(j=0,1,…,N), we can get the error equation:
(63)a(uj-uhj,w)+b(w,pj-phj)=0,∀w∈Vh,(ptj-∂tphj,q)+c(uj-uhj,q)=0,iiiiiiii∀q∈Hh,(j=1,2,…,N).
Choosingw=ηj(j=1,2,…,N) and q=ξj(j=1,2,…,N) in the two equations of (63), adding them together, and using Lemma 2, discontinuous mixed covolume elliptic projection, we have
(64)a(ηj,ηj)+(∂tξj,ξj)=(∂tpj-ptj,ξj)+(∂tζj,ξj).
First, we estimate the left item of (64). Using Lemma 4, we have
(65)a(ηj,ηj)≥C0|||ηj|||div⁡2,(∂tξj,ξj)=1Δt(ξj-ξj-1,ξj)=12Δt[(ξj,ξj)-(ξj-1,ξj-1)+(ξj-ξj-1,ξj-ξj-1)]=12Δt[∥ξj∥2-∥ξj-1∥2+∥ξj-ξj-1∥2]>12Δt[∥ξj∥2-∥ξj-1∥2].
Then, we estimate the right item of (64). From
(66)∥∂tpj-ptj∥2=∥1Δt∫tj-1tj‍(tj-1-t)pttdt∥2≤∫Ω‍(1Δt∫tj-1tj‍(tj-1-t)pttdt)2dχ≤1(Δt)2∫Ω‍(∫tj-1tj‍(tj-1-t)2dt∫tj-1tj‍ptt2dt)dχ≤C(Δt)2∥ptt∥L∞(L2)2,
we have
(67)(∂tpj-ptj,ξj)≤C(Δt)2∥ptt∥L∞(L2)2+C∥ξj∥2,∥∂tζj∥=∥ζj-ζj-1Δt∥=∥1Δt∫tj-1tj‍ζtdt∥≤1Δt∫tj-1tj‍∥ζt∥dt≤ChΔt∫tj-1tj‍(∥ut∥2+∥pt∥1+∥u∥2+∥p∥1)dt≤Ch(∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)+∥p∥L∞(H1));
therefore
(68)(∂tζj,ξj)≤Ch2(∥ut∥L∞(H2)2+∥pt∥L∞(H1)2+∥u∥L∞(H2)2+∥p∥L∞(H1)2)2+C∥ξj∥2.
Substituting the estimations above into (64), using ξ0=0, we have
(69)∥ξi∥2+2C0Δt∑j=1i|||ηj|||div⁡2≤C(Δt)2∥ptt∥L∞(L2)2+CΔt∑j=1i∥ξj∥2+Ch2(∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)+∥p∥L∞(H1))2.
By ϵ-inequality and the discrete Gronwall inequality, we have
(70)max⁡0≤i≤N∥ξi∥≤CΔt∥ptt∥L∞(L2)+Ch(∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)+∥p∥L∞(H1)).
From the above formula and (47) and using the triangle inequality, we have
(71)max⁡0≤i≤N∥pi-phi∥≤CΔt∥ptt∥L∞(L2)+Ch(∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)+∥p∥L∞(H1)).
This completes the proof.Theorem 12.
Let(u,p)∈H2(Ω)2×H1(Ω) be the solution of (3), and let (uhj,phj)∈Vh×Hh(j=1,2,…,N) be the solution of (24) with t=tj(j=1,2,…,N), respectively. If ph0=p~h(0)=p0, uh0=u~h(0)=u0, then there exists a positive constant C independent of h and Δt such that
(72)max⁡0≤i≤N|||ui-uhi|||div⁡≤CΔt∥ptt∥L∞(L2)+Ch(∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)+∥p∥L∞(H1)).Proof.
Choosingw=ηj(j=1,2,…,N) and q=∂tξj(j=1,2,…,N) in the two equations of (63), adding them together, and using Lemma 2, discontinuous mixed covolume elliptic projection, we have
(73)a(∂tηj,ηj)+(∂tξj,∂tξj)=(∂tpj-ptj,∂tξj)+(∂tζj,∂tξj).
From
(74)a(∂tηj,ηj)=12∂ta(ηj,ηj)+Δt2a(∂tηj,∂tηj)≥C0|||ηj|||div⁡2
and (66) and (67), we have
(75)|||ηj|||div⁡2≤C(Δt)2∥ptt∥L∞(L2)+Ch2(∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)+∥p∥L∞(H1))2;
therefore
(76)|||ηj|||div⁡≤CΔt∥ptt∥L∞(L2)+Ch(∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)+∥p∥L∞(H1))2.
From the above formula and (47) and using the triangle inequality, we have
(77)max⁡0≤i≤N|||ui-uhi|||div⁡≤CΔt∥ptt∥L∞(L2)+Ch(∥ut∥L∞(H2)+∥pt∥L∞(H1)+∥u∥L∞(H2)+∥p∥L∞(H1)).
The proof is complete.



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*Source: 867863-2014-03-30.xml*