# A Posteriori Error Estimate for Finite Volume Element Method of the Second-Order Hyperbolic Equations

Authors: Chuanjun Chen; Xin Zhao; Yuanyuan Zhang
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

We establish a posteriori error estimate for finite volume element method of a second-order hyperbolic equation. Residual-type a posteriori error estimator is derived. The computable upper and lower bounds on the error in theH1-norm are established. Numerical experiments are provided to illustrate the performance of the proposed estimator.


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## Body

## 1. Introduction

The finite volume element method is a class of important numerical tools for solving partial differential equations. Due to the local conservation property and some other attractive properties, it is wildly used in many engineering fields, such as heat and mass transfer, fluid mechanics, and petroleum engineering, especially for those arising from conservation laws including mass, momentum, and energy. For the second-order hyperbolic equations, Li et al. [1] have proved the optimal order of convergence in H1-norm. In [2], Kumar et al. have proved optimal order of convergence in L2 and H1-norm for the semidiscrete scheme and quasi-optimal order of convergence in maximum norm.Since the pioneering work of Babuvška and Rheinboldt [3], the adaptive finite element methods based on a posteriori error estimates have become a central theme in scientific and engineering computations. Adaptive algorithm is among the most important means to boost accuracy and efficiency of the finite element discretization. The main idea of adaptive algorithm is to use the error indicator as a guide which shows whether further refinement of meshes is necessary. A computable a posteriori error estimator plays a crucial role in an adaptive procedure. A posteriori error analysis for the finite volume element method has been studied in [4–6] for the second-order elliptic problem, in [7–9] for the convection-diffusion equations, in [10] for the parabolic problems, in [11] for a model distributed optimal problem governed by linear parabolic equations, in [12] for the Stokes problem in two dimensions, and in [13] for the second-order hyperbolic equations.However, to the best of our knowledge, there are few works related to the a posteriori error estimates of the finite volume element method for the second-order hyperbolic problems. The aim of this paper is to establish residual-type a posteriori error estimator of the finite volume element method for the second-order hyperbolic equation. We first construct a computable a posteriori error estimator of the finite volume element method. Then we analyze the residual-type a posteriori error estimates and obtain the computable upper and lower bounds on the error in theH1-norm.The organization of this paper is stated as follows. In Section2, we present the framework of the finite volume element method for the second-order hyperbolic equation. In Section 3, we establish the residual-type a posteriori error estimator of the finite volume element method and derive the upper and lower bounds on the error in the H1-norm. We provide some numerical experiments to illustrate the performance of the error estimator in Section 4.


## 2. Finite Volume Element Formulation

We use standard notation for Sobolev spacesWs,p(Ω) with the norm us,p,Ω [14]. In order to simplify the notation, we denote Ws,2(Ω) by Hs(Ω) and omit the index p=2 and Ω whenever possible.In this paper, we consider the following second-order hyperbolic problem:(1)utt-∇·ax∇u=fx,t,inΩ×0,T,ux,t=0,on⁡∂Ω×0,T,ux,0=u0x,utx,0=v0x,inΩ,where Ω⊂R2 is a polygonal bounded cross section, possessed with a Lipschitz boundary ∂Ω. For simplicity, the right-hand side f is assumed to be measurable and square-integrable on Ω×(0,T] and to be continuous with respect to time. The initial datum u0 and v0 are assumed to be measurable and square-integrable on Ω. a(x,t)=(aij(x,t))i,j=12 is a real-valued smooth matrix function, uniformly symmetric, and positive definite in Ω.The corresponding variational problem is to findu∈H01(Ω), for t>0, satisfying(2)utt,v+au,v=f,v,∀v∈H01Ω,where the bilinear form a(·,·) is defined by (3)au,v=∫Ωax∇u·∇vdx,∀u,v∈H01Ω.Denote byTh the primal quasi-uniform triangulation of Ω with h=max⁡hK, where hK is the diameter of the triangle K∈Th. Let Uh be the standard conforming finite element space of piecewise linear functions, defined on the triangulation Th: (4)Uh=u∈CΩ¯:uKis linear andu∂Ω=0,∀K∈Th. Denote by Th∗ the dual partition which is constructed in the same way as in [1, 15]. Let zK be the barycenter of K. We connect zK with the midpoints of the edges of K by straight line, thus partitioning K into three quadrilaterals Kz, z∈Zh(K), where Zh(K) are the vertices of K. Then with each vertex z∈Zh=∪K∈ThZh(K), we associate a control volume Vz, which consists of the union of the subregions Kz, sharing the vertex z (see Figure 1). Finally, we obtain a group of control volumes covering the domain Ω, which is called the dual partition Th∗ of the triangulation Th. Denote by Zh0 the set of interior vertices of Zh and denote by Eh the set of all interior edges of Th, respectively.Figure 1
(a) The dotted line shows the boundary of the corresponding control volumeVz with z, a common vertex. (b) A triangle K is partitioned into three subregions Kz.
(a)
(b)The partitionTh∗ is regular or quasi-uniform, if there exists a positive constant C>0 such that (5)C-1h2≤meas⁡Vz≤Ch2,∀Vz∈Th∗.The dual partitionTh∗ will also be quasi-uniform [5] if the finite element triangulation Th is quasi-uniform. The test function space Vh is defined by (6)Vh=v∈L2Ω:vVzis constant andv∂Ω=0∀Vz∈Th∗. For any uh∈Uh, we define an interpolation operator Πh:Uh→Vh, such that(7)Πhuh=∑z∈Zh0uhzΨz, where Ψz is the characteristic function of the control volume Vz.According to [16], for each uh∈Uh, there exists a positive constant C independent of h, such that Πh satisfies the following inequality:(8)uh-Πhuh0,K≤ChKuh1,K,∀K∈Th.Introduce the following adjoint elliptic problem:(9)-∇·ax∇u=finΩ,withu=0on∂Ω. Denote by T:L2(Ω)→H2(Ω)⋂‍H01(Ω) the solution operator of this problem, so that(10)aTf,φ=f,φ,∀φ∈H01Ω.Define negative norms by (11)v-s=sup⁡v,φφs;φ∈HsΩ,fors≥0integer. In fact, by Cauchy-Schwarz inequality, we obtain (12)v,φφ1≤vφφ1≤vφ1φ1=v. For our error analysis in the next section, it will be convenient to introduce such a norm defined by (13)v-s=Tsv,v1/2,fors≥0integer. According to Thomée [17], we have the following lemma.Lemma 1.
The normv-s is equivalent to v-s and (Tf,g)=(f,Tg), where s is a nonnegative integer. Particularly, Tv1 is equivalent to v-1 when s=1.
In order to get the fully discrete finite volume element method of (1), we give a partition of the time interval [0,T]: 0=t0<t1<⋯<tN-1<tN=T. Let τn=tn-tn-1, τ=max1≤n≤Nτn, Uhn=Uh(tn), and Uhn,1/2=Uhn+1+Uhn-1/2. With the help of Πh, we obtain the fully discrete finite volume element method of (1): to find Uhn∈Uh, for 1≤n≤N, such that(14)∂t∂¯Uhn,Πhχ+aUhn,1/2,Πhχ=fn,Πhχ,∀χ∈Uh,Uh0=u0,∂¯Uh1=v0,where (15)∂t∂¯Uhn=∂tUhn-∂tUhn-1τn=Uhn+1-Uhn/τn+1-Uhn-Uhn-1/τnτn. By setting v=∂u/∂t=ut and Y=uv, the notation ∇·(a(x)∇)ϕ=∇·(a(x)∇ϕ), (1) can equivalently be written as(16)Yt-01∇·ax∇0Y=F,where F=0f.
LetVhn=∂¯Uhn+1; we define (17)Uτ=t-tn-1τnUhn+1-t-tn-1τnUhn-1,1≤n≤N,Vτ=t-tn-1τnVhn+1-t-tn-1τnVhn-1,1≤n≤N. The residual system, with Yτ=UτVτ, is defined as follows:(18)Y-Yτt-01∇·ax∇0Y-Yτ=PuPvinΩ×0,T,u-Uτ=0on∂Ω×0,T,Y-Yτ·,0=0inΩ,where the quantities Pu in L1(0,T;L2(Ω)) and Pv in L1(0,T;H-1(Ω)) are affine functions on each interval [tn-1,tn],1≤n≤N, that (19)Pu·,t=Vτ-Vhn-1,2≤n≤N,0,n=1. And the quantities Pv are defined as follows.
From the fully discrete algorithm (14), for any φ∈H01(Ω),v∈Uh, we have(20)∂t∂¯Uhn,φ+aUhn,1/2,φ=-fn-∂t∂¯Uhn,φ-Πhv+fn,φ+aUhn,1/2,φ-aUhn,1/2,Πhv.Since (Vτ)t=∂t∂¯Uhn, by (2) and (20), for t∈(tn-1,tn], we get(21)v-Vτt,φ+au-Uhn,1/2,φ=fn-∂t∂¯Uhn,φ-Πhv+f-fn,φ-aUhn,1/2,φ+aUhn,1/2,Πhv.Adding the term a(Uhn,1/2-Uτ,φ) into the two hand sides of (21), we get(22)v-Vτt,φ+au-Uτ,φ=fn-∂t∂¯Uhn,φ-Πhv+f-fn,φ-aUhn,1/2,φ-aUhn,1/2,Πhv+aUhn,1/2-Uτ,φ. So on each interval [tn-1,tn](2≤n≤N), we have(23)Pv,φ=fn-∂t∂¯Uhn,φ-Πhv-aUhn,1/2,φ-aUhn,1/2,Πhv+aUhn,1/2-Uτ,φ+f-fn,φ,∀φ∈H01Ω,v∈Uh.We define(24)Ln,φ=fn-∂t∂¯Uhn,φ-Πhv-aUhn,1/2,φ-aUhn,1/2,Πhv.Then the term Pv on the interval [tn-1,tn](2≤n≤N) can be written as (25)Pv,φ=Ln,φ+aUhn,1/2-Uτ,φ+f-fn,φ,∀φ∈H01Ω,v∈Uh. When t∈[0,t1], (26)Pv·,t=f·,t+∇·ax∇u0+tv0.


## 3. Residual-Type A Posteriori Error Estimates

In this section, we will construct the residual-type a posteriori error estimates of the finite volume element method for (1). We introduce the jump of a vector-valued function across the edge E∈Eh which will be used in the residual-type a posteriori error estimates. Let E be an interior edge shared by elements K+ and K-. Define the unit normal vectors nK+ and nK- on E pointing exterior to K+ and K-, respectively. Let v be a vector-valued function that is smooth inside each of the elements K+ and K-. v+ and v- denote the traces of v on E taken from within the interior of K+ and K-, respectively. Then the jump of v on the edge E is defined by [v]E=v+·nK++v-·nK-. We denote space refinement indicator by ηsn defined by (27)RKn=fn-∂t∂¯Uhn+∇·ax∇Uhn,1/2,REn=-ax∇Uhn,1/2E,ηsn=∑K∈ThhK2RKn0,K2+∑E∈EhhEREn0,E21/2. We define time refinement indicator ηtn as (28)ηtn=τnUhn-Uhn-11+τnVhn-Vhn-1.


### 3.1. Upper Bound

The Scott-Zhang interpolation functionIh:H01(Ω)→Uh is introduced in the following lemma [18].Lemma 2.
For eachφ∈H01(Ω), a positive constant C is independent of hK and hE such that, for any K∈Th, E∈Eh(29)Ihφ1,Ω≤Cφ1,Ω,φ-Ihφ0,K≤ChKφ1,ωK,φ-Ihφ0,E≤ChE1/2φ1,ωE, where ωK=⋃K′⋂‍K≠∅‍K′ and ωE=⋃K⋂‍E≠∅‍K.We also introduce the trace theorem [14].Lemma 3 (trace theorem).
There exists a positive constantC independent of hE such that (30)ω0,E2≤ChE-1ω0,K2+hE∇ω0,K2,∀ω∈H1K,E∈∂K,∀K∈Th. Then we can get the following theorem for the upper bound of the error.Theorem 4.
The following a posteriori error estimate holds between the solutionu of (1) and the solution (Uhn)1≤n≤N of (14), for 2≤m≤N: (31)um-Uhm+∑k=1k=m∫tk-1tkUτ-udt1≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt+C∫0t1f·,t+∇·ax∇u0+tv0dt.Proof.
Taking the inner product of (18) with u-UτT(v-Vτ) and setting (32)Zt=u-Uτ2+v-Vτ-121/2, we obtain, for t∈[tn-1,tn],(33)12dZ2dt=Pu,u-Uτ+Pv,Tv-Vτ≤Puu-Uτ+∇·ax∇Uhn,1/2-Uτ-1Tv-Vτ1+Ln-1Tv-Vτ1+f·,t-fnTv-Vτ≤Puu-Uτ+C∇·ax∇Uhn,1/2-Uτ-1v-Vτ-1+CLn-1v-Vτ-1+Cf·,t-fnv-Vτ-1≤CPu2+Ln-12+f·,t-fn2+∇·ax∇Uhn,1/2-Uτ-121/2Z; hence,(34)dZdt≤CPu2+Ln-12+f·,t-fn2+∇·ax∇Uhn,1/2-Uτ-121/2≤CPu+Ln-1+f·,t-fn+∇·ax∇Uhn,1/2-Uτ-1. Integrating the inequality from tn-1 to tn(2≤n≤N), we have (35)Ztn-Ztn-1≤C∫tn-1tnPu+Ln-1+f·,t-fn+∇·ax∇Uhn,1/2-Uτ-1dt. Using Lemma 1, we obtain(36)∫tn-1tn∇·ax∇Uhn,1/2-Uτ-1dt=∫tn-1tn∇·ax∇Uhn,1/2-Uhn+Uhn-Uτ-1dt≤CUhn-Uhn-11∫tn-1tn1-t-tn-1τndt+Cτn2Uhn+1-Uhn1+Cτn2Uhn-Uhn-11≤CτnUhn+1-Uhn1+CτnUhn-Uhn-11,∫tn-1tnPu·,tdt=Vhn-Vhn-1∫tn-1tnt-tn-1τndt=τn2Vhn-Vhn-1. By the definition of ηtn, we get(37)Ztn-Ztn-1≤Cτnηtn+τnLn-1+∫tn-1tnf·,t-fndt.In order to estimate Ln-1, we choose v=Ihφ in (24); then(38)Ln,φ=fn-∂t∂¯Uhn,φ-Πhv-aUhn,1/2,φ-aUhn,1/2,Πhv=fn-∂t∂¯Uhn,φ-v+fn-∂t∂¯Uhn,v-Πhv-aUhn,1/2,φ-aUhn,1/2,v-aUhn,1/2,v-aUhn,1/2,Πhv≜I1+I2+I3+I4.Using Green’s formula, we have(39)I3=-ax∇Uhn,1/2,∇φ-v=-∑K∈Thax∇Uhn,1/2,∇φ-v=∑K∈Th∇·ax∇Uhn,1/2,φ-v0,K-∑E∈Ehax∇Uhn,1/2E,φ-v0,E. By the definition of RKn,REn, we get(40)I1+I3=∑K∈Thfn-∂t∂¯Uhn+∇·ax∇Uhn,1/2,φ-v0,K-∑E∈Ehax∇Uhn,1/2E,φ-v0,E=∑K∈ThRKn,φ-v0,K+∑E∈EhREn,φ-v0,E. From Cauchy-Schwarz inequality and Lemma 2, we can get (41)I1+I3≤C∑K∈ThhKRKn0,Kφ1,ωK+C∑E∈EhhE1/2REn0,Eφ1,ωE. For I4, since Πhv is a constant in K⋂‍Kz∗,z∈Zh(K),Kz∗∈Th∗, we have(42)∫Kax∇Uhn,1/2·∇vdx=∑z∈ZhK∫K⋂Kz∗ax∇Uhn,1/2·∇v-Πhvdx=-∑z∈ZhK∫K⋂Kz∗∇·ax∇Uhn,1/2·v-Πhvdx+∑z∈ZhK∫∂K⋂Kz∗ax∇Uhn,1/2·nv-Πhvds=-∫K∇·ax∇Uhn,1/2·v-Πhvdx+∫∂Kax∇Uhn,1/2·nv-Πhvds+∑z∈ZhK∫K⋂∂Kz∗ax∇Uhn,1/2·nv-Πhvds. Since a(x)∇Uhn and v are continuous inside each element K∈Th, we have (43)∑z∈ZhK∫K⋂∂Kz∗ax∇Uhn·nvds=0,∑z∈ZhK∫K⋂∂Kz∗ax∇Uhn,1/2·nvds=0. Thus, (44)I4=∑K∈Th∇·ax∇Uhn,1/2,v-Πhv0,K-∑E∈Ehax∇Uhn,1/2E,v-Πhv0,E. Then we get(45)I2+I4=∑K∈Thfn-∂t∂¯Uhn+∇·ax∇Uhn,1/2,v-Πhv0,K-∑E∈Ehax∇Uhn,1/2E,v-Πhv0,E=∑K∈ThRKn,v-Πhv0,K+∑E∈EhREn,v-Πhv0,E. By (8) and Cauchy-Schwarz inequality, we obtain(46)∑K∈ThRKn,v-Πhv0,K≤C∑K∈ThhKRKn0,Kv1,K≤C∑K∈ThhK2RKn0,K21/2∑K∈Thv1,K21/2=C∑K∈ThhK2RKn0,K21/2Ihφ1,Ω≤C∑K∈ThhK2RKn0,K21/2φ1.∑E∈EhREn,v-Πhv0,E≤∑E∈EhREn0,Ev-Πhv0,E≤∑E∈EhhEREn0,E21/2∑E∈EhhE-1v-Πhv0,E21/2. Since Πhv is a piecewise constant function, by Lemma 3 and (8), we get (47)∑E∈EhhE-1v-Πhv0,E2≤C∑E∈EhhE-2v-Πhv0,K2+v1,K2≤Cv12≤Cφ12. Substituting the estimate of I1–I4 into (38) and by the definition of ηsn, we have (48)Ln,φ≤Cηsnφ1; hence (49)Ln,φφ1≤Cηsn,Ln-1≤Cηsn. Substituting the estimation of Ln-1 into (37), we get(50)Ztn-Ztn-1≤Cτnηtn+τnηsn+∫tn-1tnf·,t-fndt.Summing (50) from n=2 to n=m, we obtain(51)Ztm-Zt1≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt.For n=1, we have (52)Zt1-Zt0≤C∫0t1f·,t+∇·ax∇u0+tv0dt. Noting that Z(t0)=Z(0)=0, then(53)Ztm≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt+C∫0t1f·,t+∇·ax∇u0+tv0dt.By the fact that 1/2(a+b)≤a2+b2≤a+b(a,b>0), we have(54)um-Uhm+vm-Vhm-1≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt+C∫0t1f·,t+∇·ax∇u0+tv0dt.In view of the definition of the operator T, we have(55)T∂v∂t+u=Tf·,t,(56)T∂Vτ∂t+Uhm,1/2=Tf·,tm,t∈tm-1,tm.Subtracting (56) from (55), we get (57)T∂v-Vτ∂t+u-Uhm,1/2=Tf·,t-f·,tm,(58)T∂v-Vτ∂t+Tf·,tm-f·,t+Uτ-Uhm,1/2=Uτ-u.Integrating (58) from tm-1 to tm, we obtain(59)Tvm-Vhm-Tvm-1-Vhm-1+∫tm-1tmTf·,tm-f·,tdt+∫tm-1tmUτ-Uhm,1/2dt=∫tm-1tmUτ-udt.Summing (59) from k=1 to k=m, we obtain(60)∑k=1k=m∫tk-1tkUτ-udt=Tvm-Vhm+∑k=1k=m∫tk-1tkTf·,tk-f·,tdt+∑k=1k=m∫tk-1tkUτ-Uhk,1/2dt.Thus, we have(61)∑k=1k=m∫tk-1tkUτ-udt1≤Tvm-Vhm1+∑k=1k=m∫tk-1tkTf·,tk-f·,tdt1+∑k=1k=m∫tk-1tkUτ-Uhk,1/2dt1≤Cvm-Vhm-1+C∑k=1k=m∫tk-1tkf·,tk-f·,t-1dt+∑k=1k=m∫tk-1tkUτ-Uhk1+Uhk-Uhk,1/21dt.Then,(62)∑k=1k=m∫tk-1tkUτ-udt1≤Cvm-Vhm-1+C∑k=1k=m∫tk-1tkf·,tk-f·,tdt+C∑k=1k=mτkUhk-Uhk-11.By (62) and (54), we have(63)um-Uhm+∑k=1k=m∫tk-1tkUτ-udt1≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt+C∫0t1f·,t+∇·ax∇u0+tv0dt.


### 3.2. Lower Bound

In order to derive the local lower bounds on the error, we will introduce some properties of the bubble functions. For each triangleK∈Th, denote by λK,1,λK,2,λK,3 the barycentric coordinates. Define the element-bubble function ψK by (64)ψK=27λK,1λK,2λK,3,inK;ψK=0,inΩ∖K. Assume that K and K′ share the edge E∈Eh. Let the barycentric coordinates with respect to the end points of E be λE,1 and λE,2. Define the edge-bubble function ψE by (65)ψE=4λE,1λE,2,inωE=K∪K′;ψE=0,inΩ∖ωE. For properties of the bubble functions, we have the following lemma [19].Lemma 5.
For each of the elementsK∈Th and E∈Eh, functions ψK and ψE have the following properties: (66)supp⁡ψK⊂K,maxx∈K⁡ψK=1,∫KψKdx=920K~hK2,∇ψK0,K≤ChK-1ψK0,K.ψK∈0,1,supp⁡ψE⊂ωE,maxx∈ωE⁡ψE=1,∫EψEds=23hE,∫ωEψEdx=13ωE~hE2,∇ψE0,ωE≤ChE-1ψE0,ωE,ψE∈0,1. We define the average of RKn on K (RKn¯) and the average of REn on E (REn¯) by (67)RKn¯=1K∫KRKndx,REn¯=1hE∫EREnds. Then we have the following local lower bounds.Theorem 6.
For anyK∈Th,E∈Eh, the following local posteriori lower bounds on the error un-Uhn hold for a positive constant C independent of hK and hE:(68)hKRKn0,K≤Cun-Uhn1,K+hKuttn-∂t∂¯Uhn0,K+Uhn+1-Uhn1,K+Uhn-Uhn-11,K+2hKRKn-RKn¯0,K,(69)hE1/2REn0,E≤Cun-Uhn1,ωE+hEuttn-∂t∂¯Uhn0,ωE+Uhn+1-Uhn1,ωE+Uhn-Uhn-11,ωE+hERKn-RKn¯0,ωE+hE1/2REn-REn¯0,E.Proof.
By triangle inequality, we have(70)RKn0,K≤RKn¯0,K+RKn-RKn¯0,K. By the properties of ψK, the definition of RKn, and Green’s formulation, we have(71)RKn¯0,K2~RKn¯,ψKRKn¯=RKn,ψKRKn¯-RKn-RKn¯,ψKRKn¯=fn-∂t∂¯Uhn+∇·ax∇Uhn,1/2,ψKRKn¯-RKn-RKn¯,ψKRKn¯=fn-uttn,ψKRKn¯+uttn-∂t∂¯Uhn,ψKRKn¯+∇·ax∇Uhn,1/2,ψKRKn¯-RKn-RKn¯,ψKRKn¯=aun,ψKRKn¯-∫Kax∇Uhn,1/2·∇ψKRKn¯dx+uttn-∂t∂¯Uhn,ψKRKn¯-RKn-RKn¯,ψKRKn¯=∫Kax∇un-Uhn,1/2·∇ψKRKn¯dx+uttn-∂t∂¯Uhn,ψKRKn¯-RKn-RKn¯,ψKRKn¯≡P1+P2+P3.For P1, with Cauchy-Schwarz inequality and Lemma 5, we get(72)P1≤Cun-Uhn,1/21,K∇ψKRKn¯0,K=Cun-Uhn,1/21,K∇ψK0,KRKn¯≤ChK-1un-Uhn,1/21,KψK0,KRKn¯=ChK-1un-Uhn,1/21,KψKRKn¯0,K≤ChK-1un-Uhn,1/21,KRKn¯0,K.By Cauchy-Schwarz inequality and Lemma 5, we obtain(73)P2≤uttn-∂t∂¯Uhn0,KψKRKn¯0,K≤Cuttn-∂t∂¯Uhn0,KRKn¯0,K,P3≤RKn-RKn¯0,KψKRKn¯0,K≤RKn-RKn¯0,KRKn¯0,K.Combining (71)–(73), we obtain(74)hKRKn0,K≤Cun-Uhn,1/21,K+hKuttn-∂t∂¯Uhn0,K+2hKRKn-RKn¯0,K≤Cun-Uhn1,K+hKuttn-∂t∂¯Uhn0,K+Uhn+1-Uhn1,K+Uhn-Uhn-11,K+2hKRKn-RKn¯0,K.For (69), by triangle inequality, similarly we have (75)hE1/2REn0,E≤hE1/2REn¯0,E+hE1/2REn-REn¯0,E. By Lemma 5 and Green’s formulation, we get(76)REn¯0,E2~REn¯,ψEREn¯0,E=REn,ψEREn¯0,E+REn¯-REn,ψEREn¯0,E=ax∇Uhn,1/2,∇ψEREn¯0,ωE+∇·ax∇Uhn,1/2,ψEREn¯0,ωE+REn¯-REn,ψEREn¯0,E=∫ωEax∇Uhn,1/2·∇ψEREn¯dx-∫ωEax∇un·∇ψEREn¯dx+∫ωEax∇un·∇ψEREn¯dx+∇·ax∇Uhn,1/2,ψEREn¯0,ωE+REn¯-REn,ψEREn¯0,E=∫ωEax∇Uhn,1/2-un·∇ψEREn¯dx+∇·ax∇Uhn,1/2,ψEREn¯0,ωE+∫ωEfn-∂t∂¯UhnψEREn¯dx+∫ωE∂t∂¯Uhn-uttnψEREn¯dx+REn¯-REn,ψEREn¯0,E=∫ωEax∇Uhn,1/2-un·∇ψEREn¯dx+RKn,ψEREn¯0,ωE+∂t∂¯Uhn-uttn,ψEREn¯+REn¯-REn,ψEREn¯0,E≡O1+O2+O3+O4.Now we will estimate the right-hand terms of (76). By Lemma 5 and the Cauchy-Schwarz inequality, we obtain(77)O1≤CUhn,1/2-un1,ωE∇ψEREn¯0,ωE=CUhn,1/2-un1,ωE∇ψE0,ωEREn¯≤ChE-1Uhn,1/2-un1,ωEψE0,ωEREn¯≤ChE-1/2Uhn,1/2-un1,ωEREn¯0,E,O2≤RKn0,ωEψEREn¯0,ωE=RKn0,ωEψE0,ωEREn¯≤ChERKn0,ωEREn¯≤ChE1/2RKn0,ωEREn¯0,E,O3≤∂t∂¯Uhn-uttn0,ωEψEREn¯0,ωE≤ChE1/2uttn-∂t∂¯Uhn0,ωEREn¯0,E,O4≤REn¯-REn0,EψEREn¯0,E≤REn¯-REn0,EREn¯0,E.Combining (77) with (76), we get(78)REn¯0,E≤ChE-1/2Uhn,1/2-un1,ωE+ChE1/2RKn0,ωE+ChE1/2uttn-∂t∂¯Uhn0,ωE+REn¯-REn0,E. With (74), we obtain (79)hE1/2REn0,E≤Cun-Uhn,1/21,ωE+ChEuttn-∂t∂¯Uhn0,ωE+ChERKn-RKn¯0,ωE+hE1/2REn-REn¯0,E≤Cun-Uhn+Uhn-Uhn,1/21,ωE+hEuttn-∂t∂¯Uhn0,ωE+hERKn-RKn¯0,ωE+hE1/2REn-REn¯0,E≤Cun-Uhn1,ωE+hEuttn-∂t∂¯Uhn0,ωE+Uhn+1-Uhn1,ωE+Uhn-Uhn-11,ωE+hERKn-RKn¯0,ωE+hE1/2REn-REn¯0,E.


## 3.1. Upper Bound

The Scott-Zhang interpolation functionIh:H01(Ω)→Uh is introduced in the following lemma [18].Lemma 2.
For eachφ∈H01(Ω), a positive constant C is independent of hK and hE such that, for any K∈Th, E∈Eh(29)Ihφ1,Ω≤Cφ1,Ω,φ-Ihφ0,K≤ChKφ1,ωK,φ-Ihφ0,E≤ChE1/2φ1,ωE, where ωK=⋃K′⋂‍K≠∅‍K′ and ωE=⋃K⋂‍E≠∅‍K.We also introduce the trace theorem [14].Lemma 3 (trace theorem).
There exists a positive constantC independent of hE such that (30)ω0,E2≤ChE-1ω0,K2+hE∇ω0,K2,∀ω∈H1K,E∈∂K,∀K∈Th. Then we can get the following theorem for the upper bound of the error.Theorem 4.
The following a posteriori error estimate holds between the solutionu of (1) and the solution (Uhn)1≤n≤N of (14), for 2≤m≤N: (31)um-Uhm+∑k=1k=m∫tk-1tkUτ-udt1≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt+C∫0t1f·,t+∇·ax∇u0+tv0dt.Proof.
Taking the inner product of (18) with u-UτT(v-Vτ) and setting (32)Zt=u-Uτ2+v-Vτ-121/2, we obtain, for t∈[tn-1,tn],(33)12dZ2dt=Pu,u-Uτ+Pv,Tv-Vτ≤Puu-Uτ+∇·ax∇Uhn,1/2-Uτ-1Tv-Vτ1+Ln-1Tv-Vτ1+f·,t-fnTv-Vτ≤Puu-Uτ+C∇·ax∇Uhn,1/2-Uτ-1v-Vτ-1+CLn-1v-Vτ-1+Cf·,t-fnv-Vτ-1≤CPu2+Ln-12+f·,t-fn2+∇·ax∇Uhn,1/2-Uτ-121/2Z; hence,(34)dZdt≤CPu2+Ln-12+f·,t-fn2+∇·ax∇Uhn,1/2-Uτ-121/2≤CPu+Ln-1+f·,t-fn+∇·ax∇Uhn,1/2-Uτ-1. Integrating the inequality from tn-1 to tn(2≤n≤N), we have (35)Ztn-Ztn-1≤C∫tn-1tnPu+Ln-1+f·,t-fn+∇·ax∇Uhn,1/2-Uτ-1dt. Using Lemma 1, we obtain(36)∫tn-1tn∇·ax∇Uhn,1/2-Uτ-1dt=∫tn-1tn∇·ax∇Uhn,1/2-Uhn+Uhn-Uτ-1dt≤CUhn-Uhn-11∫tn-1tn1-t-tn-1τndt+Cτn2Uhn+1-Uhn1+Cτn2Uhn-Uhn-11≤CτnUhn+1-Uhn1+CτnUhn-Uhn-11,∫tn-1tnPu·,tdt=Vhn-Vhn-1∫tn-1tnt-tn-1τndt=τn2Vhn-Vhn-1. By the definition of ηtn, we get(37)Ztn-Ztn-1≤Cτnηtn+τnLn-1+∫tn-1tnf·,t-fndt.In order to estimate Ln-1, we choose v=Ihφ in (24); then(38)Ln,φ=fn-∂t∂¯Uhn,φ-Πhv-aUhn,1/2,φ-aUhn,1/2,Πhv=fn-∂t∂¯Uhn,φ-v+fn-∂t∂¯Uhn,v-Πhv-aUhn,1/2,φ-aUhn,1/2,v-aUhn,1/2,v-aUhn,1/2,Πhv≜I1+I2+I3+I4.Using Green’s formula, we have(39)I3=-ax∇Uhn,1/2,∇φ-v=-∑K∈Thax∇Uhn,1/2,∇φ-v=∑K∈Th∇·ax∇Uhn,1/2,φ-v0,K-∑E∈Ehax∇Uhn,1/2E,φ-v0,E. By the definition of RKn,REn, we get(40)I1+I3=∑K∈Thfn-∂t∂¯Uhn+∇·ax∇Uhn,1/2,φ-v0,K-∑E∈Ehax∇Uhn,1/2E,φ-v0,E=∑K∈ThRKn,φ-v0,K+∑E∈EhREn,φ-v0,E. From Cauchy-Schwarz inequality and Lemma 2, we can get (41)I1+I3≤C∑K∈ThhKRKn0,Kφ1,ωK+C∑E∈EhhE1/2REn0,Eφ1,ωE. For I4, since Πhv is a constant in K⋂‍Kz∗,z∈Zh(K),Kz∗∈Th∗, we have(42)∫Kax∇Uhn,1/2·∇vdx=∑z∈ZhK∫K⋂Kz∗ax∇Uhn,1/2·∇v-Πhvdx=-∑z∈ZhK∫K⋂Kz∗∇·ax∇Uhn,1/2·v-Πhvdx+∑z∈ZhK∫∂K⋂Kz∗ax∇Uhn,1/2·nv-Πhvds=-∫K∇·ax∇Uhn,1/2·v-Πhvdx+∫∂Kax∇Uhn,1/2·nv-Πhvds+∑z∈ZhK∫K⋂∂Kz∗ax∇Uhn,1/2·nv-Πhvds. Since a(x)∇Uhn and v are continuous inside each element K∈Th, we have (43)∑z∈ZhK∫K⋂∂Kz∗ax∇Uhn·nvds=0,∑z∈ZhK∫K⋂∂Kz∗ax∇Uhn,1/2·nvds=0. Thus, (44)I4=∑K∈Th∇·ax∇Uhn,1/2,v-Πhv0,K-∑E∈Ehax∇Uhn,1/2E,v-Πhv0,E. Then we get(45)I2+I4=∑K∈Thfn-∂t∂¯Uhn+∇·ax∇Uhn,1/2,v-Πhv0,K-∑E∈Ehax∇Uhn,1/2E,v-Πhv0,E=∑K∈ThRKn,v-Πhv0,K+∑E∈EhREn,v-Πhv0,E. By (8) and Cauchy-Schwarz inequality, we obtain(46)∑K∈ThRKn,v-Πhv0,K≤C∑K∈ThhKRKn0,Kv1,K≤C∑K∈ThhK2RKn0,K21/2∑K∈Thv1,K21/2=C∑K∈ThhK2RKn0,K21/2Ihφ1,Ω≤C∑K∈ThhK2RKn0,K21/2φ1.∑E∈EhREn,v-Πhv0,E≤∑E∈EhREn0,Ev-Πhv0,E≤∑E∈EhhEREn0,E21/2∑E∈EhhE-1v-Πhv0,E21/2. Since Πhv is a piecewise constant function, by Lemma 3 and (8), we get (47)∑E∈EhhE-1v-Πhv0,E2≤C∑E∈EhhE-2v-Πhv0,K2+v1,K2≤Cv12≤Cφ12. Substituting the estimate of I1–I4 into (38) and by the definition of ηsn, we have (48)Ln,φ≤Cηsnφ1; hence (49)Ln,φφ1≤Cηsn,Ln-1≤Cηsn. Substituting the estimation of Ln-1 into (37), we get(50)Ztn-Ztn-1≤Cτnηtn+τnηsn+∫tn-1tnf·,t-fndt.Summing (50) from n=2 to n=m, we obtain(51)Ztm-Zt1≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt.For n=1, we have (52)Zt1-Zt0≤C∫0t1f·,t+∇·ax∇u0+tv0dt. Noting that Z(t0)=Z(0)=0, then(53)Ztm≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt+C∫0t1f·,t+∇·ax∇u0+tv0dt.By the fact that 1/2(a+b)≤a2+b2≤a+b(a,b>0), we have(54)um-Uhm+vm-Vhm-1≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt+C∫0t1f·,t+∇·ax∇u0+tv0dt.In view of the definition of the operator T, we have(55)T∂v∂t+u=Tf·,t,(56)T∂Vτ∂t+Uhm,1/2=Tf·,tm,t∈tm-1,tm.Subtracting (56) from (55), we get (57)T∂v-Vτ∂t+u-Uhm,1/2=Tf·,t-f·,tm,(58)T∂v-Vτ∂t+Tf·,tm-f·,t+Uτ-Uhm,1/2=Uτ-u.Integrating (58) from tm-1 to tm, we obtain(59)Tvm-Vhm-Tvm-1-Vhm-1+∫tm-1tmTf·,tm-f·,tdt+∫tm-1tmUτ-Uhm,1/2dt=∫tm-1tmUτ-udt.Summing (59) from k=1 to k=m, we obtain(60)∑k=1k=m∫tk-1tkUτ-udt=Tvm-Vhm+∑k=1k=m∫tk-1tkTf·,tk-f·,tdt+∑k=1k=m∫tk-1tkUτ-Uhk,1/2dt.Thus, we have(61)∑k=1k=m∫tk-1tkUτ-udt1≤Tvm-Vhm1+∑k=1k=m∫tk-1tkTf·,tk-f·,tdt1+∑k=1k=m∫tk-1tkUτ-Uhk,1/2dt1≤Cvm-Vhm-1+C∑k=1k=m∫tk-1tkf·,tk-f·,t-1dt+∑k=1k=m∫tk-1tkUτ-Uhk1+Uhk-Uhk,1/21dt.Then,(62)∑k=1k=m∫tk-1tkUτ-udt1≤Cvm-Vhm-1+C∑k=1k=m∫tk-1tkf·,tk-f·,tdt+C∑k=1k=mτkUhk-Uhk-11.By (62) and (54), we have(63)um-Uhm+∑k=1k=m∫tk-1tkUτ-udt1≤C∑n=2mτnηtn+ηsn+C∑n=2m∫tn-1tnf·,t-fndt+C∫0t1f·,t+∇·ax∇u0+tv0dt.


## 3.2. Lower Bound

In order to derive the local lower bounds on the error, we will introduce some properties of the bubble functions. For each triangleK∈Th, denote by λK,1,λK,2,λK,3 the barycentric coordinates. Define the element-bubble function ψK by (64)ψK=27λK,1λK,2λK,3,inK;ψK=0,inΩ∖K. Assume that K and K′ share the edge E∈Eh. Let the barycentric coordinates with respect to the end points of E be λE,1 and λE,2. Define the edge-bubble function ψE by (65)ψE=4λE,1λE,2,inωE=K∪K′;ψE=0,inΩ∖ωE. For properties of the bubble functions, we have the following lemma [19].Lemma 5.
For each of the elementsK∈Th and E∈Eh, functions ψK and ψE have the following properties: (66)supp⁡ψK⊂K,maxx∈K⁡ψK=1,∫KψKdx=920K~hK2,∇ψK0,K≤ChK-1ψK0,K.ψK∈0,1,supp⁡ψE⊂ωE,maxx∈ωE⁡ψE=1,∫EψEds=23hE,∫ωEψEdx=13ωE~hE2,∇ψE0,ωE≤ChE-1ψE0,ωE,ψE∈0,1. We define the average of RKn on K (RKn¯) and the average of REn on E (REn¯) by (67)RKn¯=1K∫KRKndx,REn¯=1hE∫EREnds. Then we have the following local lower bounds.Theorem 6.
For anyK∈Th,E∈Eh, the following local posteriori lower bounds on the error un-Uhn hold for a positive constant C independent of hK and hE:(68)hKRKn0,K≤Cun-Uhn1,K+hKuttn-∂t∂¯Uhn0,K+Uhn+1-Uhn1,K+Uhn-Uhn-11,K+2hKRKn-RKn¯0,K,(69)hE1/2REn0,E≤Cun-Uhn1,ωE+hEuttn-∂t∂¯Uhn0,ωE+Uhn+1-Uhn1,ωE+Uhn-Uhn-11,ωE+hERKn-RKn¯0,ωE+hE1/2REn-REn¯0,E.Proof.
By triangle inequality, we have(70)RKn0,K≤RKn¯0,K+RKn-RKn¯0,K. By the properties of ψK, the definition of RKn, and Green’s formulation, we have(71)RKn¯0,K2~RKn¯,ψKRKn¯=RKn,ψKRKn¯-RKn-RKn¯,ψKRKn¯=fn-∂t∂¯Uhn+∇·ax∇Uhn,1/2,ψKRKn¯-RKn-RKn¯,ψKRKn¯=fn-uttn,ψKRKn¯+uttn-∂t∂¯Uhn,ψKRKn¯+∇·ax∇Uhn,1/2,ψKRKn¯-RKn-RKn¯,ψKRKn¯=aun,ψKRKn¯-∫Kax∇Uhn,1/2·∇ψKRKn¯dx+uttn-∂t∂¯Uhn,ψKRKn¯-RKn-RKn¯,ψKRKn¯=∫Kax∇un-Uhn,1/2·∇ψKRKn¯dx+uttn-∂t∂¯Uhn,ψKRKn¯-RKn-RKn¯,ψKRKn¯≡P1+P2+P3.For P1, with Cauchy-Schwarz inequality and Lemma 5, we get(72)P1≤Cun-Uhn,1/21,K∇ψKRKn¯0,K=Cun-Uhn,1/21,K∇ψK0,KRKn¯≤ChK-1un-Uhn,1/21,KψK0,KRKn¯=ChK-1un-Uhn,1/21,KψKRKn¯0,K≤ChK-1un-Uhn,1/21,KRKn¯0,K.By Cauchy-Schwarz inequality and Lemma 5, we obtain(73)P2≤uttn-∂t∂¯Uhn0,KψKRKn¯0,K≤Cuttn-∂t∂¯Uhn0,KRKn¯0,K,P3≤RKn-RKn¯0,KψKRKn¯0,K≤RKn-RKn¯0,KRKn¯0,K.Combining (71)–(73), we obtain(74)hKRKn0,K≤Cun-Uhn,1/21,K+hKuttn-∂t∂¯Uhn0,K+2hKRKn-RKn¯0,K≤Cun-Uhn1,K+hKuttn-∂t∂¯Uhn0,K+Uhn+1-Uhn1,K+Uhn-Uhn-11,K+2hKRKn-RKn¯0,K.For (69), by triangle inequality, similarly we have (75)hE1/2REn0,E≤hE1/2REn¯0,E+hE1/2REn-REn¯0,E. By Lemma 5 and Green’s formulation, we get(76)REn¯0,E2~REn¯,ψEREn¯0,E=REn,ψEREn¯0,E+REn¯-REn,ψEREn¯0,E=ax∇Uhn,1/2,∇ψEREn¯0,ωE+∇·ax∇Uhn,1/2,ψEREn¯0,ωE+REn¯-REn,ψEREn¯0,E=∫ωEax∇Uhn,1/2·∇ψEREn¯dx-∫ωEax∇un·∇ψEREn¯dx+∫ωEax∇un·∇ψEREn¯dx+∇·ax∇Uhn,1/2,ψEREn¯0,ωE+REn¯-REn,ψEREn¯0,E=∫ωEax∇Uhn,1/2-un·∇ψEREn¯dx+∇·ax∇Uhn,1/2,ψEREn¯0,ωE+∫ωEfn-∂t∂¯UhnψEREn¯dx+∫ωE∂t∂¯Uhn-uttnψEREn¯dx+REn¯-REn,ψEREn¯0,E=∫ωEax∇Uhn,1/2-un·∇ψEREn¯dx+RKn,ψEREn¯0,ωE+∂t∂¯Uhn-uttn,ψEREn¯+REn¯-REn,ψEREn¯0,E≡O1+O2+O3+O4.Now we will estimate the right-hand terms of (76). By Lemma 5 and the Cauchy-Schwarz inequality, we obtain(77)O1≤CUhn,1/2-un1,ωE∇ψEREn¯0,ωE=CUhn,1/2-un1,ωE∇ψE0,ωEREn¯≤ChE-1Uhn,1/2-un1,ωEψE0,ωEREn¯≤ChE-1/2Uhn,1/2-un1,ωEREn¯0,E,O2≤RKn0,ωEψEREn¯0,ωE=RKn0,ωEψE0,ωEREn¯≤ChERKn0,ωEREn¯≤ChE1/2RKn0,ωEREn¯0,E,O3≤∂t∂¯Uhn-uttn0,ωEψEREn¯0,ωE≤ChE1/2uttn-∂t∂¯Uhn0,ωEREn¯0,E,O4≤REn¯-REn0,EψEREn¯0,E≤REn¯-REn0,EREn¯0,E.Combining (77) with (76), we get(78)REn¯0,E≤ChE-1/2Uhn,1/2-un1,ωE+ChE1/2RKn0,ωE+ChE1/2uttn-∂t∂¯Uhn0,ωE+REn¯-REn0,E. With (74), we obtain (79)hE1/2REn0,E≤Cun-Uhn,1/21,ωE+ChEuttn-∂t∂¯Uhn0,ωE+ChERKn-RKn¯0,ωE+hE1/2REn-REn¯0,E≤Cun-Uhn+Uhn-Uhn,1/21,ωE+hEuttn-∂t∂¯Uhn0,ωE+hERKn-RKn¯0,ωE+hE1/2REn-REn¯0,E≤Cun-Uhn1,ωE+hEuttn-∂t∂¯Uhn0,ωE+Uhn+1-Uhn1,ωE+Uhn-Uhn-11,ωE+hERKn-RKn¯0,ωE+hE1/2REn-REn¯0,E.


## 4. Numerical Examples

Now we present some numerical examples to show the performance of the proposed error estimator. We consider problem (1) in Ω×[0,T]=[0,1;0,1]×[0,1]. We discretize Ω into N number of rectangles in each direction and then each rectangle is divided into two triangles, resulting in a mesh with size h=2/N. Discretize time by taking time step τn=Δt=h. We consider the following two cases.Case 1.
Consider(80)ax,y=1+sin⁡π4x+sin⁡π4y+e2x+e2y,ux,y,t=x1-xy1-yet.Case 2.
Consider(81)ax,y=ex+y/2,ux,y,t=sin⁡πxsin⁡πyet.Define(82)Om=∑n=2mun-Uhn1,Nm=∑n=2mηtn+ηsn,R=NmOm. We present the results of the above cases when m=N at Tables 1 and 2.Table 1
Error estimates for Case1.
h
u
N
-
U
h
N
0
Rate
u
N
-
U
h
N
1
Rate
O
N
N
N
R
1
/
2
2
1.3839
e
-
02
—
1.6047
e
-
01
—
0.4589
21.3382
46.4952
1
/
2
3
4.0831
e
-
03
1.7610
8.2097
e
-
02
0.9669
0.4417
21.1967
47.9920
1
/
2
4
9.3149
e
-
04
2.1321
4.1279
e
-
02
0.9919
0.4307
21.0618
48.9005
1
/
2
5
2.0574
e
-
04
2.1787
2.0670
e
-
02
0.9979
0.4247
20.9680
49.3734
1
/
2
6
4.6977
e
-
05
2.1308
1.0338
e
-
02
0.9996
0.4215
20.9138
49.6128Table 2
Error estimates for Case2.
h
u
N
-
U
h
N
0
Rate
u
N
-
U
h
N
1
Rate
O
N
N
N
R
1
/
2
2
1.2513
e
-
01
—
2.3532
—
6.6633
66.7292
10.0144
1
/
2
3
3.2994
e
-
02
1.9232
1.1822
0.9931
6.3537
64.6603
10.1768
1
/
2
4
8.1195
e
-
03
2.0227
5.9245
e
-
01
0.9967
6.1806
63.5799
10.2870
1
/
2
5
1.8443
e
-
03
2.1383
2.9641
e
-
01
0.9991
6.0904
63.0179
10.3471
1
/
2
6
4.0436
e
-
04
2.1894
1.4822
e
-
01
0.9999
6.0444
62.7297
10.3782From Tables1 and 2 we can see that the global a posteriori error estimator can predict the exact global error. The error estimator is reliable as evidenced by the ratio R listed on the tables. This list shows that the ratio R is converging to a constant when the mesh size is decreased by half. This shows that the proposed global a posteriori error estimator is robust for predicting the error in the finite volume element method.



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*Source: 510241-2015-12-24.xml*