# The Backward Euler Fully Discrete Finite Volume Method for the Problem of Purely Longitudinal Motion of a Homogeneous Bar

Authors: Ziwen Jiang; Deren Xie
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We present a linear backward Euler fully discrete finite volume method for the initial-boundary-value problem of purely longitudinal motion of a homogeneous bar and an give optimal order error estimates inL2 and H1 norms. Furthermore, we obtain the superconvergence error estimate of the generalized projection of the solution u in H1 norm. Numerical experiment illustrates the convergence and stability of this scheme.


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## Body

## 1. Introduction

We consider the following mixed boundary-initial value problem:(1.1)(a)utt=uxxt+f(ux)x,(x,t)∈(0,1)×[0,T],(b)u(x,0)=u0(x),ut(x,0)=u1(x),x∈(0,1),(c)u(0,t)=u(1,t)=0,t∈[0,T].
This problem (1.1) arises when one considers the purely longitudinal motion of a homogeneous bar [1]. The displacement of the cross-section of the bar at time t is denoted by u(x,t). When both ends of the bar are fixed, u(0,t)=u(1,t)=0, t∈[0,T], that is the boundary condition. u(x,0)=u0(x) and ut(x,0)=u1(x) are the initial data.In theoretical analysis, the problem (1.1) was first treated by Greenberg et al. [2], by assuming that the function f was monotonic, that is,
(1.2)f′(s)>0,∀s∈(-∞,+∞),
and that the initial data was smooth, specifically
(1.3)u0∈C4([0,1]),u1∈C2([0,1]).
Under these assumptions they showed the existence of a unique smooth solution which decays the zero solution as t→∞. (See also Greenberg [3] and Greenberg and MacCamy [4].) Andrews [1] made the hypotheses that u0∈W01,∞(0,1), u1∈H01(0,1), the function f is locally Lipschitz continuous, and there exists a constant a0>0 such that (f(s1)-f(s2))(s1-s2)>0 whenever |s1-s2|≥a0; then he proved the existence of a unique global weak solution. Under the hypotheses that u0,u1∈H2(0,1)∩H01(0,1), and there exists a constant a1>0 such that f'(s)≥a1, s∈(-∞,+∞), Y. Liu and D. Liu [5] proved the existence of a unique global strong solution of the problem (1.1).Numerical simulation methods for the problem (1.1) are recently studied by several authors ([6–11]). In [6], Gao et al. studied a finite difference method of the problem (1.1) in the domain [0,1]×[0,T](T>0) and proved the convergence of the method by using discrete functional analysis and prior estimate. In [7, 8], Jiang et al. studied two finite element methods of (1.1) and obtained the optimal error estimates of this finite element scheme in L2 and H1 norms. In [9], Z. Jiang  and Y. Jiang, and in [10], Jiang and Li, studied a mixed finite element method and a expanded mixed finite element method, respectively, and obtained the optimal error estimates of these schemes. However, few work on finite volume methods of (1.1) was found (see [11]). As we know, finite volume methods (also called generalized difference methods) were proposed in eighties last century and developed very quickly. Now this kind of method becomes one of the main numerical methods for solving differential equations, for example, convection diffusion equations [12–14] and Navier-Stokes equations [15].In this paper we want to make further study of finite volume methods for the problem (1.1). First in Section 2, we derive a finite volume weak form of (1.1) in the case that f(s) is nonlinear, then propose a linear backward Euler fully discrete finite volume scheme of (1.1). Existence and uniqueness of the solution of this scheme are proved. Next, in Section 3, we give optimal error estimates in L2 and H1 norms and superconvergence in H1 norm by using new defined projections. Numerical experiments and computational results are presented in Section 4, which confirm our theoretical analysis.


## 2. The Linear Backward Euler Fully Discrete Finite Volume Scheme

In this section, we construct the finite volume method of the problem (1.1) and prove the existence and uniqueness of the solution of this finite volume scheme.Firstly, letTh be a partition for the interval I=[0,1], with its nodes 0=x0<x1<⋯<xr=1. The length of the element Ii=[xi-1,xi] is denoted by hi=xi-xi-1, i=1,2,…,r,  h=max1≤i≤rhi is maximum of hi. We suppose Th is regular, that is, there exists a positive constant μ>0 such that hi≥μh, i=1,2,…,r. For the definition of the finite volume scheme, the dual partition Th* of Th is needed, which is 0=x0<x1/2<x3/2<⋯<xr-1/2<xr=1. The dual elements are denoted by I0*=[x0,x1/2], Ij*=[xj-1/2,xj+1/2], j=1,2,…,r-1, Ir*=[xr-1/2,xr], where xj-1/2=(1/2)(xj-1+xj), j=1,2,…,r.Secondly, we define the piecewise linear trial function spaceUh over the partition Th, satisfying Uh⊂H01(I), where H01(I) is the Sobolev space on I. Then uh(x)=∑i=1r-1uh(xi)φi(x) for all uh(x)∈Uh, where φi(x) is the basis function associated with the nodes xi(i=1,2,…,r-1),
(2.1)φi(x)={1-xi-xhi,x∈Ii,1-x-xihi+1,x∈Ii+1,0,x∉Ii∪Ii+1.
It is easy to know that the derivative of uh with respect to x is
(2.2)uhx(x)=uh(xi)-uh(xi-1)hi,xi-1≤x≤xi,i=1,2,…,r.
The test function space Vh⊂L2(I) associated with the dual partition Th* is defined as the set of all piecewise constants with vh(0)=vh(1)=0 for all vh(x)∈Vh. We may choose the basis function ψj(x) of Vh in such a way that ψj(x) is the characteristic function of Ij*, that is,
(2.3)ψj(x)={1,x∈Ij*,0,x∉Ij*,j=1,2,…,r-1.
Then for any vh(x)∈Vh can be expressed as vh(x)=∑j=1r-1vh(xj)ψj(x).Obviously,(2.4)Uh⊂H01(I),Vh⊂L2(I),dimUh=dimVh=r-1.
Meanwhile, Uh⊂W01,∞(I).Thirdly, for the time interval[0,T], we give an isometric partition and denote the nodes ti=iτ, i=0,1,…,N, τ=T/N.We introduce some notations for functionsu(x,t) and f(ux(x,t)):
(2.5)un=u(x,tn),uj=u(xj,t),ujn=u(xj,tn),∂tun=un+1-unτ,∂tt2un=un+1-2un+un-1τ2,un+1/2=un+1+un2,un,1/4=un+1+2un+un-14,f1/2(uxn)=f(uxn+1)+f(uxn)2,f1/4(uxn)=f(uxn+1)+2f(uxn)+f(uxn-1)4,f(1/4)*(uxn)=3f(uxn)+f(uxn-1)4.
Then we can get
(2.6)∂tt2un=∂tun-∂tun-1τ,un,1/4=un+1/2+un-1/22,12(∂tun+∂tun-1)=1τ(un+1/2-un-1/2).Letu be the solution of (1.1). Integrating (1.1) a over the dual element Ij*∈Th*, we obtain
(2.7)∫xj-1/2xj+1/2‍uttdx+uxt(xj-1/2)-uxt(xj+1/2)+f(ux(xj-1/2))-f(ux(xj+1/2))=0,
where j=0,1,…,r, x-1/2=x0, xr+1/2=xr, and 0<t≤T. The problem (2.7) can be rewritten in a variational form. For any arbitrary vh∈Vh, we multiply the integral relation in (2.7) by vh(xj) and sum over all j=0,1,…,r to obtain
(2.8)(utt,vh)+a*(ut,vh)+b*(f(u),vh)=0,∀vh∈Vh,t∈(0,T],u(0)=u0,ut(0)=u1,
where for any arbitrary w=∑j=1r-1wjψj∈Vh, the bilinear forms a*(v,w), b*(f(v),w) are defined by
(2.9)a*(v,w)=∑j=1r-1‍wja*(v,ψj),a*(v,ψj)=vx(xj-1/2)-vx(xj+1/2),b*(f(v),w)=∑j=1r-1‍wjb*(f(v),ψj),b*(f(v),ψj)=f(vx(xj-1/2))-f(vx(xj+1/2)).Sincef is a nonlinear function, we will consider the following linear finite volume scheme: find uhn∈Uh, n=0,1,2,…,N such that
(2.10)(∂tt2uhn,vh)+a*(∂tuhn,vh)+b*(f(1/4)*(uhn),vh)=0,∀vh∈Vh,n=1,…,N-1,uh0=Πhu0,uh1=Πhu0+τΠhu1,
where Πhu0 and Πhu1 are the interpolation projection of u0 and u1 onto the trial function space Uh, respectively, and the interpolation operator Πh is defined as Πh:H01(I)→Uh(2.11)Πhw=∑i=1r-1‍wiφi,∀w∈H01(I).
We also need to introduce the interpolation Πh*:H01(I)→Vh, defined by,
(2.12)Πh*w=∑i=1r-1‍wiψi,∀w∈H01(I).
By Sobolev’s interpolation theory, we know that
(2.13)|w-Πhw|m,p≤ch2-m|w|2,p,m=0,1,1≤p≤∞.
In this paper we adopt the standard notation Wm,p(I) for Sobolev space on I with norm ∥·∥m,p and seminorm |·|m,p. In order to simplify the notations, we denote Wm,2(I) by Hm(I) and skip the index p=2 when possible, that is, |·|m,p=|·|m, ∥·∥m,p=∥·∥m, ∥·∥0=∥·∥. We denote by Lq(0,T;Wm,p(I)) the Banach space of all Lq integrable functions from (0,T) into Wm,p(I) with the norm ∥·∥Lq(Wm,p)=(∫0T‍∥·∥m,pqdt)1/q for q∈[1,∞) and the standard modification for q=∞. After all these denotations, we give the existence and uniqueness of the solution of the finite volume scheme (2.10).Theorem 2.1.
The solution of the finite volume scheme (2.10) is existent and unique.Proof.
Letuhn=∑i=1r-1uhinφi(x)∈Uh (n=0,1,2,…,N) be the solution of (2.10). According to uh0=Πhu0 and uh1=Πhu0+τΠhu1, uhj0=u0(xj), uhj1=u0(xj)+τu1(xj), j=1,2,…,r-1 are known. Hence the existence and uniqueness of the solution uhn(n=0,1,2,…,N) of scheme (2.10) are equivalent to the existence and uniqueness of {uhin}i=1r-1, n=2,…,N.Forn=2,…,N, choosing vh=ψj, j=1,2,…,r-1. By (2.2) we have
(2.14)(∂tt2uhn-1,vh)=∫01‍∂tt2uhn-1ψj(x)dx=∫xj-1/2xj+1/2‍∂tt2uhn-1dx=1τ2∫xj-1/2xj+1/2‍(uhn-2uhn-1+uhn-2)dx=1τ2∫xj-1/2xj+1/2‍∑i=1r-1‍(uhin-2uhin-1+uhin-2)φi(x)dx=1τ2{(uhj-1n-2uhj-1n-1+uhj-1n-2)∫xj-1/2xj+1/2‍φj-1(x)dx+(uhjn-2uhjn-1+uhjn-2)∫xj-1/2xj+1/2‍φj(x)dx+(uhj+1n-2uhj+1n-1+uhj+1n-2)∫xj-1/2xj+1/2‍φj+1(x)dx}=1τ2{18hj(uhj-1n-2uhj-1n-1+uhj-1n-2)+38(hj+hj+1)(uhjn-2uhjn-1+uhjn-2)+18hj+1(uhj+1n-2uhj+1n-1+uhj+1n-2)},a*(∂tuhn-1,vh)=a*(∂tuhn-1,ψj)=∂tuhxn-1(xj-1/2)-∂tuhxn-1(xj+1/2)=∂tuhjn-1-uhj-1n-1hj-∂tuhj+1n-1-uhjn-1hj+1=1τhj(uhjn-uhj-1n-uhjn-1+uhj-1n-1)-1τhj+1(uhj+1n-uhjn-uhj+1n-1+uhjn-1),b*(f(1/4)*(uhn-1),vh)=b*(f(1/4)*(uhn-1),ψj)=f(1/4)*(uhxn-1(xj-1/2))-f(1/4)*(uhxn-1(xj+1/2))=f(1/4)*(uhjn-1-uhj-1n-1hj)-f(1/4)*(uhj+1n-1-uhjn-1hj+1)=34f(uhjn-1-uhj-1n-1hj)+14f(uhjn-2-uhj-1n-2hj)-34f(uhj+1n-1-uhjn-1hj+1)-14f(uhj+1n-2-uhjn-2hj+1).
Using (2.10) we obtain
(2.15)1τ2[18hj(uhj-1n-2uhj-1n-1+uhj-1n-2)+38(hj+hj+1)(uhjn-2uhjn-1+uhjn-2)+18hj+1(uhj+1n-2uhj+1n-1+uhj+1n-2)]+1τhj(uhjn-uhj-1n-uhjn-1+uhj-1n-1)-1τhj+1(uhj+1n-uhjn-uhj+1n-1+uhjn-1)+34f(uhjn-1-uhj-1n-1hj)+14f(uhjn-2-uhj-1n-2hj)-34f(uhj+1n-1-uhjn-1hj+1)-14f(uhj+1n-2-uhjn-2hj+1)=0,
that is,
(2.16)(hj8-τhj)uhj-1n+[38(hj+hj+1)+τ(1hj+1hj+1)]uhjn+(hj+18-τhj+1)uhj+1n=(hj4-τhj)uhj-1n-1+[34(hj+hj+1)+τ(1hj+1hj+1)]uhjn-1+(hj+14-τhj+1)uhj+1n-1+3τ24[f(uhj+1n-1-uhjn-1hj+1)-f(uhjn-1-uhj-1n-1hj)]-18hjuhj-1n-2-38(hj+hj+1)uhjn-2-18hj+1uhj+1n-2+τ24[f(uhj+1n-2-uhjn-2hj+1)-f(uhjn-2-uhj-1n-2hj)],j=1,2,…,r-1.
Noticing uh0n=uhrn=0, the coefficient matrix of the system (2.16) is strictly diagonally dominant matrix. So, when {uhin-1}i=1r-1 and {uhin-2}i=1r-1 are known, the solution {uhin}i=1r-1 is existent and unique. Combining the above conditions, the solution {uhin}i=1r-1(n=2,…,N) is existent and unique when {uhi0}i=1r-1 and {uhi1}i=1r-1 are known. This completes the proof.


## 3. Error Estimates

In this section, we will prove the optimal error estimates in theL2 and H1 norms as well as the superconvergence error estimates in the H1 norm. This needs some assumptions about the data.(H1)
The nonlinear functionf(s) satisfies
(3.1)0<l≤f′(s)≤L,max{|f′′(s)|,|f′′′(s)|}≤M,s∈(-∞,+∞).(H2)
The initial functionsu0 and u1 satisfy
(3.2)u0∈H01(I)∩H2(I),u1∈H01(I)∩W3,1(I).(H3)
The solutionu of (1.1) satisfies
(3.3)u∈L∞(0,T;W3,1(I)),ut∈L∞(0,T;W2,∞(I)∩W3,1(I)),utt∈L∞(0,T;W3,1(I)),utttt∈L2(0,T;L2(I)).For anyuh,vh∈Uh, uh=∑i=1r-1uhiφi(x), vh=∑i=1r-1vhiφi(x), where uhi=uh(xi), vhi=vh(xi), i=1,2,…,r-1. Using the definition of the interpolation operator Πh* we know that Πh*uh=∑i=1r-1uhiφi(x), Πh*vh=∑i=1r-1vhiφi(x)∈Vh. Noting uh0=uhr=vh0=vhr=0 and [11, Lemma 3.1] we have
(3.4)(uh,Πh*vh)=(vh,Πh*uh)=18∑i=1r-1‍[hiuhi-1vhi+3(hi+hi+1)uhivhi+hi+1uhi+1vhi].
Hence
(3.5)(uh,Πh*uh)=18∑i=1r-1‍[hi(uhi+uhi+1)2+2(hi+hi+1)uhi2]+h1uh12.
So, when uh≠0, we can get (uh,Πh*uh)>0. Let |∥uh∥|0=(uh,Πh*uh)1/2. By [11, Lemma 3.2], we have
(3.6)23|∥uh∥|0≤∥uh∥≤43|∥uh∥|0,∀uh∈Uh.
On the other hand, using [11, Lemma 3.3], we also have
(3.7)a*(uh,Πh*uh)=∑i=1r‍1hi(uhi-uhi-1)2=|uh|12.
For error estimate, we will use the generalized adjoint finite volume element projection u~[11] of the solution u of (1.1), that is, u~:[0,T]→Uh satisfies
(3.8)a*(ut-u~t,vh)+b*(f(u)-f(u~),vh)=0,∀vh∈Vh,0<t≤T,u~(0)=Πhu0.
Let u and u~ denote the solutions of (1.1) and (3.8), respectively. Under the assumptions (H1), (H2), and (H3) together with [11, Theorems 3.4-3.5] and the one-dimensional imbedding theorem in Sobolev space, then we can obtain the following.Lemma 3.1.
Letu and u~ be the solutions of (1.1) and (3.8), respectively. If the assumptions (H 1), (H 2), and (H 3) hold, then one has
(3.9)|Πhu-u~|1+|u-u~|1+|ut-u~t|1+|utt-u~tt|1≤ch,t∈[0,T],(3.10)∥u-u~∥+∥ut-u~t∥+∥utt-u~tt∥≤ch2,t∈[0,T].Next, we give the corresponding error estimates. Choosingt=tn-1,tn,tn+1 in the first equation of (2.8), then multiplying the three equations by 1/4, 1/2, 1/4, respectively, and adding them, we have
(3.11)(uttn,1/4,vh)+a*(utn,1/4,vh)+b*(f1/4(un),vh)=0,∀vh∈Vh,n=1,2,…,N-1.
Similarly, taking t=tn-1,tn,tn+1 in the first equation of (3.8), then multiplying the three equations by 1/4, 1/2, 1/4, respectively, and adding them, we obtain
(3.12)a*(utn,1/4-u~tn,1/4,vh)+b*(f1/4(un)-f1/4(u~n),vh)=0,∀vh∈Vh,n=1,2,…,N-1.
Subtracting (3.11) from (2.10) and using (3.12), we get the error equation
(3.13)(∂tt2uhn-uttn,1/4,vh)+a*(∂tuhn-u~tn,1/4,vh)+b*(f(1/4)*(uhn)-f1/4(u~n),vh)=0,∀vh∈Vh,
where n=1,2,…,N-1. Let
(3.14)uhn-un=θn+ρn,θn=uhn-u~n,ρn=u~n-un,n=1,2,…,N-1.
Then the error equation (3.13) can be rewritten as
(3.15)(∂tt2θn,vh)+a*(∂tθn,vh)+b*(f(1/4)*(uhn)-f1/4(u~n),vh)=-(∂tt2ρn,vh)+(uttn,1/4-∂tt2uhn,vh)+a*(ρtn,1/4-∂tρn,vh)+a*(utn,1/4-∂tun,vh),∀vh∈Vh,n=1,2,…,N-1.Theorem 3.2.
Let{uhn}n=0N and u~ be the solutions of (2.10) and (3.8), respectively. If the assumptions (H 1), (H 2), and (H 3) hold, then there exists a positive constant h0, such that when h<h0(3.16)|uh1-u~1|1=|θ1|1≤c(h2+τ2),(3.17)∥∂t(uh0-u~0)∥=|∂tθ0|≤c(h2+τ).Proof.
Firstly, we estimate|θ1|1. Applying the Taylor formula,
(3.18)u~1=u~(τ)=u~(0)+τu~t(0)+12τ2u~tt(ξ)=u~0+τu~t0+12τ2u~tt(ξ),ξ∈(0,τ),
and noting the definitions of uh1 and u~0 in (2.10) and (3.8), respectively, we know
(3.19)uh1-u~1=θ1=τ(Πhu1-u~t0)-12τ2u~tt(ξ),ξ∈(0,τ),
so
(3.20)|θ1|12=|τ(Πhu1-u~t0)-12τ2u~tt(ξ)|12≤4τ2(|Πhu1-u1|12+|u1-u~t0|12)+12τ4|u~tt(ξ)|12.
Notice that u1=ut(0)=ut0, (2.13) and (3.9), there exists a positive constant h0 such that, when h<h0 we have
(3.21)|θ1|12≤cτ2h2+12τ4(|u~tt(ξ)|1+ch)2≤c(τ2h2+τ4)≤c(h4+τ4).
This implies (3.16) holds. Secondly, noting θ0=0 and (3.19), we obtain
(3.22)∂tθ0=θ1-θ0τ=Πhu1-u~t0-12τu~tt(ξ),ξ∈(0,τ),
hence, by (2.16), (3.10), and h<h0 we have
(3.23)∥∂tθ0∥2=∥(Πhu1-u1)+(u1-u~t0)-12τu~tt(ξ)∥2≤4∥Πhu1-u1∥2+4∥u1-u~t0∥+12τ2∥u~tt(ξ)∥2≤ch4+12τ2(∥utt(ξ)∥+ch2)2≤c(h4+τ2).
This completes the proof of (3.17).Theorem 3.3.
Letu, {uhn}n=0N and u~ be the solutions of (1.1), (2.10), and (3.8), respectively. Assume that (H 1), (H 2), and (H 3) hold; then for any n=1,2,…,N-1, when τ is sufficiently small and h<h0, one has
(3.24)b*(f(1/4)*(uhn)-f1/4(u~n),Πh*∂tθn)≥14τ∑j=1r-1‍1hj[aj-1/2n+1(θjn+1-θj-1n+1)2-aj-1/2n(θjn-θj-1n)2]-34|∂tθn|12-c[(1+|θxn|0,∞2)|θn+1|12+|θn|12+|θn-1|12]-cτ2,
where
(3.25)aj-1/2k=∫01‍f′(u~xk(xj-1/2)-sθxk(xj-1/2))ds,k=n,n+1,j=1,2,…,r-1.Proof.
Expanding the termb*(f(1/4)*(uhn)-f1/4(u~n),Πh*∂tθn), we have
(3.26)b*(f(1/4)*(uhn)-f1/4(u~n),Πh*∂tθn)=b*(f1/4(uhn)-f1/4(u~n),Πh*∂tθn)+b*(14[f(uhn)-f(uhn+1)],Πh*∂tθn)=12τ[b*(f1/2(uhn)-f1/2(u~n),Πh*θn+1)-b*(f1/2(uhn-1)-f1/2(u~n-1),Πh*θn)]-12τ[b*(f1/2(uhn)-f1/2(u~n),Πh*θn)-b*(f1/2(uhn-1)-f1/2(u~n-1),Πh*θn+1)]+14τ[b*(f(uhn)-f(uhn+1),Πh*θn+1)-b*(f(uhn)-f(uhn+1),Πh*θn)].
For any vh∈Vh, since vh=∑i=1r-1vhiψi(x) and vh0=vhr=0, then
(3.27)b*(f(u),vh)=∑i=1r-1‍vhi[f(ux(xi-1/2))-f(ux(xi+1/2))]=∑i=1r‍f(ux(xi-1/2))(vhi-vhi-1).
Let θi(xj)=θji, thus Πh*θi=∑j=1r-1θjiψj(x)∈Vh. Applying transformation of variable, we know
(3.28)b*(f1/2(uhm)-f1/2(u~m),Πh*θi)=∑j=1r‍(θji-θj-1i)[f1/2(uhxm(xj-1/2))-f1/2(u~xm(xj-1/2))]=12∑j=1r‍(θji-θj-1i){[f(uhxm+1(xj-1/2))-f(u~xm+1(xj-1/2))]12∑j=1r‍(θji-θj-1i)+[f(uhxm(xj-1/2))-f(u~xm(xj-1/2))]uhxm+1}=12∑j=1r‍(θji-θj-1i)[aj-1/2m+1θxm+1(xj-1/2)+aj-1/2mθxm(xj-1/2)]=12∑j=1r‍1hj[aj-1/2m+1(θji-θj-1i)(θjm+1-θj-1m+1)12∑j=1r‍1hj+aj-1/2m(θji-θj-1i)(θjm-θj-1m)],
where
(3.29)aj-1/2k=∫01‍f′(u~xk(xj-1/2)+sθxk(xj-1/2))ds,θxk(xj-1/2)=1hj(θjk-θj-1k),k=m,m+1.
By (3.28), we have
(3.30)b*(f1/2(uhn)-f1/2(u~n),Πh*θn+1)=12∑j=1r‍1hj[aj-1/2n+1(θjn+1-θj-1n+1)212∑j=1r‍1hj+aj-1/2n(θjn+1-θj-1n+1)(θjn-θj-1n)aj-1/2n+1(θjn+1-θj-1n+1)2],b*(f1/2(uhn-1)-f1/2(u~n-1),Πh*θn)=12∑j=1r‍1hj[aj-1/2n(θjn-θj-1n)212∑j=1r‍1hj+aj-1/2n-1(θjn-θj-1n)(θjn-1-θj-1n-1)aj-1/2n(θjn-θj-1n)2],b*(f1/2(uhn)-f1/2(u~n),Πh*θn)=12∑j=1r‍1hj[aj-1/2n(θjn-θj-1n)2aj-1/2n+1(θjn+1-θj-1n+1)(θjn-θj-1n)12∑j=1r‍1hj+aj-1/2n(θjn-θj-1n)2aj-1/2n+1(θjn+1-θj-1n+1)(θjn-θj-1n)],b*(f1/2(uhn-1)-f1/2(u~n-1),Πh*θn+1)=12∑j=1r‍1hj[aj-1/2n(θjn+1-θj-1n+1)(θjn-θj-1n)12∑j=1r‍1hj+aj-1/2n-1(θjn+1-θj-1n+1)(θjn-1-θj-1n-1)].
On the other hand
(3.31)b*(f(uhn)-f(uhn+1),Πh*θn+1)-b*(f(uhn)-f(uhn+1),Πh*θn)=∑j=1r‍[f(uhxn(xj-1/2))-f(uhxn+1(xj-1/2))][(θjn+1-θj-1n+1)-(θjn-θj-1n)]=∑j=1r‍aj-(1/2)*n(uhxn(xj-1/2)-uhxn+1(xj-1/2))(θjn+1-θjn-(θj-1n+1-θj-1n))=-τ2∑j=1r‍aj-1/2*n∂tuhxn(xj-1/2)(∂tθjn-∂tθj-1n),
where
(3.32)aj-1/2*n=∫01‍f′(uhxn(xj-1/2))+s(uhxn(xj-1/2)-uhxn+1(xj-1/2))ds.
Combining the above five equalities with (3.26), we can get
(3.33)b*(f(1/4)*(uhn)-f1/4(u~n),Πh*∂tθn)=14τ∑j=1r‍1hj[aj-1/2n+1(θjn+1-θj-1n+1)2-aj-(1/2)n(θjn-θj-1n)2]+14τ∑j=1r‍1hjaj-1/2n(θjn-θj-1n)[θjn+1-θj-1n+1-(θjn-θj-1n)]+14τ∑j=1r‍1hjaj-1/2n-1(θjn-1-θj-1n-1)[θjn+1-θj-1n+1-(θjn-θj-1n)]-14τ∑j=1r‍1hj(aj-1/2n+1-aj-1/2n)(θjn+1-θj-1n+1)(θjn-θj-1n)-τ4∑j=1r‍aj-1/2*n(∂tθjn-∂tθj-1n)∂tuhxn(xj-1/2)=B1+B2+B3+B4+B5.
Using the ε-inequality, we find
(3.34)-B2=14τ∑j=1r‍1hjaj-1/2n(θjn-θj-1n)[θjn-θj-1n-(θjn+1-θj-1n+1)]=-14∑j=1r‍1hjaj-1/2n(θjn-θj-1n)(∂tθjn-∂tθj-1n)≤c∑j=1r‍(θjn-θj-1n)2hj+116∑j=1r‍(∂tθjn-∂tθj-1n)2hj=c|θn|12+116|∂tθn|12,-B3=14τ∑j=1r‍1hjaj-1/2n-1(θjn-1-θj-1n-1)[θjn-θj-1n-(θjn+1-θj-1n+1)]=-14∑j=1r‍1hjaj-1/2n-1(θjn-1-θj-1n-1)(∂tθjn-∂tθj-1n)≤c|θn-1|12+116|∂tθn|12,-B5=τ4∑j=1r‍aj-1/2*n(∂tθjn-∂tθj-1n)∂tuhxn(xj-1/2)≤116|∂tθn|12+cτ2∑j=1r‍hj(∂tuhxn(xj-1/2))2=116|∂tθn|12+I1.
Now we estimate I1. Noting (2.2), we have
(3.35)uhxn(xj-1/2)=uhjn-uhj-1nhj=1hj∫xj-1/2xj‍uhxn(x)dx.
Furthermore,
(3.36)(∂tuhxn(xj-1/2))2=(1hj∫xj-1/2xj‍∂tuhxn(x)dx)2≤1hj∫xj-1xj‍(∂tuhxn(x))2dx.
So
(3.37)I1≤cτ2∑j=1r‍∫xj-1xj‍(∂tuhxn(x))2dx=cτ2|∂tuhn|12≤cτ2|∂tθn|12+cτ2|∂tu~n|12≤c1τ2|∂tθn|12+cτ2|∂tu~n|12.
When τ is sufficiently small, we can take τ suitable such that c1τ2≤1/16, from the above equalities we have
(3.38)I1≤116|∂tθn|12+cτ2|∂tu~n|12.
Moreover, in view of ∂tu~n=(1/τ)∫tntn+1‍u~tdt and (3.9), we obtain
(3.39)|∂tu~n|12=∫01‍(∂tu~xn(x))2dx=1τ2∫01‍(∫tntn+1‍u~tdt)2dx≤1τ∫01‍∫tntn+1‍(u~t)2dtdx=1τ∫tntn+1‍|u~t|12dt≤1τ∫tntn+1‍(ch+|ut|1)2dt≤c.
Then
(3.40)I1≤116|∂tθn|12+cτ2.
Substituting the estimate of I1 into (3.34), we have
(3.41)-B5≤18|∂tθn|12+cτ2,-B4=14τ∑j=1r‍1hj(aj-1/2n+1-aj-1/2n)(θjn+1-θj-1n+1)(θjn-θj-1n).
Since
(3.42)aj-1/2n+1-aj-1/2n=∫01‍[f′(u~xn+1(xj-1/2)+sθxn+1(xj-1/2))-f′(u~xn(xj-1/2)+sθxn(xj-1/2))f′(u~xn+1(xj-(1/2))+sθxn+1(xj-(1/2)))]ds=∫01‍f′′(ξjn)[u~xn+1(xj-1/2)-u~xn(xj-1/2)+s(θxn+1(xj-1/2)-θxn(xj-1/2))]ds=τ∫01‍f′′(ξjn)ds∂tu~xn(xj-1/2)+τ∫01‍f′′(ξjn)sds∂tθxn(xj-1/2),
where ξjn lies between u~xn+1(xj-1/2)+sθxn+1(xj-1/2) and u~xn(xj-1/2)+sθxn(xj-1/2), by (2.13), (3.9), and the inverse estimate, we obtain
(3.43)|u~t|1,∞≤|ut|1,∞+|ut-u~t|1,∞≤|ut|1,∞+|ut-Πhut|1,∞+|Πhut-u~t|1,∞≤|ut|1,∞+ch|ut|2,∞+ch-1/2|Πhut-u~t|1≤|ut|1,∞+ch|ut|2,∞+ch1/2.
Thus, when h<h0 we have ∥u~xt∥L∞(L∞)≤c, then
(3.44)|∂tu~xn(xj-1/2)|=|u~xt(xj-1/2,tn+sτ)|≤∥u~xt∥L∞(L∞)≤c.
From the assumption (H1), we find that ∫01‍f′′(ξjn)ds is bounded: this leads to the boundedness of ∫01‍f′′(ξjn)ds∂tu~xn(xj-1/2) and ∫01‍f′′(ξjn)sds. So
(3.45)-B4=14∑j=1r‍1hj[∫01‍f′′(ξjn)ds∂tu~xn(xj-1/2)+∫01‍f′′(ξjn)sds∂tθxn(xj-1/2)]×(θjn+1-θj-1n+1)(θjn-θj-1n)=14∑j=1r‍1hj[∫01‍f′′(ξjn)ds∂tu~xn(xj-1/2)](θjn+1-θj-1n+1)(θjn-θj-1n)+14∑j=1r‍1hj(∫01‍f′′(ξjn)sds)(∂tθjn-∂tθj-1n)(θjn+1-θj-1n+1)θxn(xj-1/2)≤c|θn+1|12+c|θn|12+c|θxn|0,∞|θn+1|1|∂tθn|1≤c[(1+|θxn|0,∞)|θn+1|12+|θn|12]+12|∂tθn|12.
Substituting the estimates of B2, B3, B4, and B5 into (3.33), we can derive (3.24). This completes the proof of the theorem.Theorem 3.4.
Letu, {uhn}n=0N and u~ be the solutions of (1.1), (2.10), and (3.8), respectively. Assume that (H 1), (H 2), and (H 3) hold; then for any n=1,2,…,N-1, when h and τ are sufficiently small, one has
(3.46)max1≤n≤N-1∥∂tθn∥+|θn|1≤c(h2+τ).Proof.
Takingvh=Πh*∂tθn in the error equation (3.15) and making a simple calculation yield
(3.47)(∂tt2θn,Πh*∂tθn)+a*(∂tθn,Πh*∂tθn)+b*(f(1/4)*(uhn)-f1/4(u~n),Πh*∂tθn)=-(∂tt2ρn,Πh*∂tθn)+(uttn,1/4-∂tt2un,Πh*∂tθn)+a*(ρtn,1/4-∂tρn,Πh*∂tθn)+a*(utn,1/4-∂tun,Πh*∂tθn).
Form (3.4), we derive that
(3.48)(∂tθn,Πh*∂tθn)-2(∂tθn-1,Πh*∂tθn)+(∂tθn-1,Πh*∂tθn-1)=n(∂tθn-∂tθn-1,Πh*(∂tθn-∂tθn-1))≥0.
Then
(3.49)(∂tt2θn,Πh*∂tθn)=1τ(∂tθn-∂tθn-1,Πh*∂tθn)=12τ{(∂tθn,Πh*∂tθn)-(∂tθn-1,Πh*∂tθn-1)+[(∂tθn,Πh*∂tθn)-2(∂tθn-1,Πh*∂tθn)+(∂tθn-1,Πh*∂tθn-1)]}≥12τ{|∥∂tθn∥|02-|∥∂tθn-1∥|02}.
By (3.7), we get
(3.50)a*(∂tθn,Πh*∂tθn)=|∂tθn|12.
Applying Taylor’s formula with integral-type remainder,
(3.51)f(t)=f(tn)+f′(tn)(t-tn)+⋯+f(k)(tn)k!(t-tn)k+1k!∫tnt‍f(k+1)(s)(t-s)kds,
we have
(3.52)∂tt2ρn=1τ2[∫tntn+1‍(tn+1-s)ρtt(s)ds+∫tntn-1‍(tn-1-s)ρtt(s)ds]=1τ2∫-ττ‍(τ-|s|)ρtt(tn+s)ds=1τ2∫tn-1tn+1‍(τ-|tn-s|)ρtt(s)ds,uttn,1/4-∂tt2un=∫-ττ‍(τ-|s|)[14-(τ-|s|)26τ2]utttt(tn+s)ds=∫tn-1tn+1‍(τ-|tn-s|)[14-(τ-|tn-s|)26τ2]utttt(s)ds,utn,1/4-∂tun=14[∫tntn+1‍utt(s)ds+∫tntn-1‍utt(s)ds]-∫tntn+1‍utt(s)tn+1-sτds,ρtn,1/4-∂tρn=14[∫tntn+1‍ρtt(s)ds+∫tntn-1‍ρtt(s)ds]-∫tntn+1‍ρtt(s)tn+1-sτds.
Note that ∥Πh*∂tθn∥≤c∥∂tθn∥, we get
(3.53)(∂tt2ρn,Πh*∂tθn)≤∥∂tt2ρn∥∥Πh*∂tθn∥≤c∥∂tt2ρn∥2+c∥∂tθn∥2≤cτ-1∫tn-1tn+1‍∥ρtt∥2dt+c∥∂tθn∥2,(uttn,1/4-∂tt2un,Πh*∂tθn)≤∥uttn,1/4-∂tt2un∥∥Πh*∂tθn∥≤c∥uttn,1/4-∂tt2un∥2+∥∂tθn∥2≤cτ3∫tn-1tn+1‍∥utttt∥2dt+c∥∂tθn∥2.
For Πh*v=∑j=1r-1vjψj(x)∈Vh, v0=vr=0, we have
(3.54)a*(u,Πh*v)=∑j=1r-1‍vj(ux(xj-1/2)-ux(xj+1/2))=∑j=1r‍ux(xj-1/2)(vj-vj-1)≤(∑j=1r‍(vj-vj-1)2hj)1/2(∑j=1r‍hjux2(xj-1/2))1/2.
Noting that
(3.55)ux(xj-1/2)=ux(x)-∫xj-1/2x‍uxx(s)ds,x∈[xj-1,xj],ux2(xj-1/2)=(ux(x)-∫xj-1/2x‍uxx(s)ds)2≤2ux2(x)+2(∫xj-1/2x‍uxx(s)ds)2≤2ux2(x)+2(∫xj-1xj‍|uxx(x)|dx)2≤2ux2(x)+2hj(∫xj-1xj‍uxx2(x)dx),
we see that
(3.56)∑j=1r‍hjux2(xj-1/2)=∑j=1r‍∫xj-1xj‍ux2(xj-1/2)dx≤2∑j=1r‍∫xj-1xj‍ux2(x)dx+2∑j=1r‍hj2∫xj-1xj‍uxx2(x)dx≤2|u|12+2h2|u|22,
so
(3.57)a*(utn,1/4-∂tun,Πh*∂tθn)≤2(|utn,1/4-∂tun|12+h2|utn,1/4-∂tun|22)1/2|∂tθn|1≤c(|utn,1/4-∂tun|12+h2|utn,1/4-∂tun|22)+18|∂tθn|12≤cτ∫tn-1tn+1‍(|utt|12+h2|utt|22)dt+18|∂tθn|12,a*(ρtn,1/4-∂tρn,Πh*∂tθn)≤cτ∫tn-1tn+1‍(|ρtt|12+h2|ρtt|22)dt+18|∂tθn|12=cτ∫tn-1tn+1‍(|ρtt|12+h2|utt|22)dt+18|∂tθn|12.
Substituting (3.24), (3.49)–(3.57) into (3.47) and using the Lemma 3.1, we obtain
(3.58)12τ{|∥∂tθn∥|02-|∥∂tθn-1∥|02}+14τ∑j=1r‍1hj[aj-1/2n+1(θjn+1-θj-1n+1)2-aj-1/2n(θjn-θj-1n)2]≤c∥∂tθn∥2+c[(1+|θxn|0,∞2)|θn+1|12+|θn|12+|θn-1|12]+c(τ2+τ4+τ2h2+h4).
Multiplying the above inequation by τ and then summing it with respect to n from 1 to n, we have
(3.59)12|∥∂tθn∥|02+14∑j=1r‍1hjaj-1/2n+1(θjn+1-θj-1n+1)2≤12|∥∂tθ0∥|02+14∑j=1r‍1hjaj-1/2n+1(θj1-θj-11)2+cτ∑i=1n‍[∥∂tθi∥2+(1+|θxn|0,∞2)|θi+1|12+|θi|12+|θi-1|12]+c(τ2+h4).
Now we make the induction hypothesis:
(3.60)max1≤k≤n|θxk|0,∞≤1.
Substituting (3.60) into (3.59) and noting (3.6) and the assumption (H1), we conclude
(3.61)c2(∥∂tθn∥2+|θn+1|12)≤1234∥∂tθn∥2+14l|θn+1|12≤12|∥∂tθn∥|02+14∑j=1r‍1hjaj-1/2n+1(θjn+1-θj-1n+1)2≤1232∥∂tθ0∥2+14l|θ1|12+cτ∑i=1n‍∥∂tθi∥2+cτ∑i=0n+1‍|θi|12+c(τ2+h4)≤c3(∥∂tθ0∥2+|θ1|12+τ2+h4+τ∑i=1n‍∥∂tθi∥2+τ∑i=0n+1‍|θi|12).
Taking τ is sufficiently small, such that c3τ<c2 and c1τ2≤1/16 (these ensure that Theorem 3.3. holds), combining the above equalities and Theorem 3.2. we have
(3.62)∥∂tθn∥2+|θn+1|12≤c(τ2+h4)+c3τc2(∑i=1n‍∥∂tθi∥2+∑i=0n+1‍|θi|12).
By the discrete Gronwall’s lemma, it follows that
(3.63)∥∂tθn∥2+|θn+1|12≤c(τ2+h4),n=1,2,…,N-1.
To complete our proof we must verify that (3.60) holds. In fact, by considering the inverse hypothesis |θn|1,∞≤c4h-1/2|θn|1, n=1,2,…,N-1 and (3.16), we have
(3.64)|θx1|0,∞=|θ1|1,∞≤c4h-1/2c(τ2+h2)≤c5h-1/2(τ2+h2).
Thus, we can choose h, τ such that h, τ satisfy c5h-1/2(τ2+h2)≤1 for h, τ are sufficiently small, that is, (3.60) holds for n=1. From the proof of (3.63), as h, τ are sufficiently small and satisfy
(3.65)c1τ2≤116,c3τ<c2,c5h-1/2(τ2+h2)≤1,
then (3.63) holds for n=1. Thus
(3.66)|θ2|1≤c6(h2+τ).
Hence
(3.67)|θx2|0,∞=|θ2|1,∞≤c4h-1/2|θ2|1≤c4c6h-1/2(h2+τ).
Similarly, letting h, τ be sufficiently small such that h, τ satisfy c4c6h-1/2(h2+τ)≤1, then we get |θx2|0,∞≤1. Furthermore, setting h, τ to satisfy
(3.68)c1τ2≤116,c3τ<c2,c5h-1/2(τ2+h2)≤1,c4c6h-1/2(h2+τ)≤1,
we conclude that (3.60) holds for n=2. Using the above facts, we find that (3.63) holds for n=2 as h, τ are sufficiently small and satisfy (3.68). By the induction argument for n=1,2,…,N-1, we deduce that (3.60) holds as h, τ satisfy (3.68). Then (3.63) holds for n=1,2,…,N-1, which implies (3.46) is valid. This completes the proof of the theorem.Note thatθ0=0 and (2.13), by Theorems 3.2 and 3.4, Lemma 3.1, and the triangle inequality, we obtain the following.Theorem 3.5.
Under the assumption of Theorem3.4, for h, τ are sufficiently small, one has
(3.69)max0≤n≤N∥uhn-un∥≤c(h2+τ),max0≤n≤N|uhn-un|1≤c(h+τ),max0≤n≤N|uhn-u~n|1≤c(h2+τ).Remark 3.6.
The optimal order error estimates inL2, H1 norms and superconvergence error estimates in H1 norm of the solution of the finite volume scheme (2.10) are achieved in Theorem 3.5.


## 4. Numerical Experiments

To test the finite volume scheme (2.10) and the theoretical analysis, we do some experiments by considering the following problem:
(4.1)(a)utt=uxxt+π2-1π2uxx,(x,t)∈(0,1)×(0,1],(b)u(x,0)=sin(πx),ut(x,0)=-sin(πx),x∈(0,1),(c)u(0,t)=u(1,t)=0,t∈[0,1].
The exact solution of the problem (4.1) is u=e-tsin(πx), (x,t)∈[0,1]×[0,1].For simplicity, we choose a constant spatial steph and a constant time step τ. The corresponding nodes are denoted by xi=ih, i=0,1,…,2r and the time levels by tj=jτ, j=0,1,…,2N.The discrete forms∥u∥0,h and |u|1,h of forms ∥u∥ and |u|1 are calculated by the compound Simpson formula. Here we choose r=N; this yields h=τ. The program is written in MATLAB and run in Windows XP. By the calculated approximate solution uh, we get the discrete norms ∥u-uh∥0,h and |u-uh|1,h, which are given in Tables 1 and 2.Table 1
∥u-uh∥0,h.
h
t
=
0.1
t
=
0.2
t
=
0.3
t
=
0.4
t
=
0.5
t
=
0.6
t
=
0.7
t
=
0.8
t
=
0.9
1
/
20
0.1025
0.0922
0.0831
0.0742
0.0671
0.0600
0.0548
0.0490
0.0447
1
/
40
0.0520
0.0469
0.0424
0.0374
0.0346
0.0309
0.0279
0.0253
0.0229
1
/
80
0.0261
0.0235
0.0212
0.0191
0.0173
0.0156
0.0141
0.0128
0.0116
1
/
160
0.0131
0.0116
0.0106
0.0096
0.0087
0.0078
0.0071
0.0064
0.0058Table 2
|u-uh|1,h.
h
t
=
0.1
t
=
0.2
t
=
0.3
t
=
0.4
t
=
0.5
t
=
0.6
t
=
0.7
t
=
0.8
t
=
0.9
1
/
20
0.3510
0.3165
0.2852
0.2569
0.2315
0.2088
0.1884
0.1702
0.1539
1
/
40
0.1736
0.1567
0.1413
0.1274
0.1150
0.1038
0.0938
0.0849
0.0768
1
/
80
0.0863
0.0779
0.0703
0.0634
0.0573
0.0518
0.0468
0.0424
0.0384
1
/
160
0.0430
0.0388
0.0351
0.0316
0.0286
0.0258
0.0234
0.0212
0.0192From Tables1 and 2, we can see that the scheme (2.10) is indeed efficient. For h=τ, we get the first-order optimal convergence under the discrete norm ∥u∥0,h and the discrete norm |u|1,h, respectively (see Table 3). Meanwhile, the scheme (2.10) is stable.Table 3
The orders of∥u-uh∥0,h and |u-uh|1,h.
Time
Order
For
∥
u
-
u
h
∥
0
,
h
Order
For
|
u
-
u
h
|
1
,
h
t
0.05
/
0.025
0.025
/
0.0125
0.0125
/
0.00625
0.05
/
0.025
0.025
/
0.0125
0.0125
/
0.00625
0.1
0.9790
0.9947
1.0000
1.0157
1.0083
1.0050
0.2
0.9752
0.9969
1.0185
1.0142
1.0083
1.0056
0.3
0.9708
1.0000
1.0000
1.0132
1.0072
1.0021
0.4
0.9884
0.9695
0.9925
1.0118
1.0068
1.0046
0.5
0.9555
1.0000
0.9917
1.0094
1.0050
1.0025
0.6
0.9574
0.9861
1.0000
1.0083
1.0028
1.0056
0.7
0.9739
0.9846
0.9898
1.0061
1.0031
1.0000
0.8
0.9536
0.9830
1.0000
1.0034
1.0017
1.0000
0.9
0.9649
0.9812
1.0000
1.0028
1.0000
1.0000



---

*Source: 475801-2012-12-20.xml*