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Abstract and Applied Analysis

Volume 2012 (2012), Article ID 475801, 23 pages

http://dx.doi.org/10.1155/2012/475801

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

School of Mathematical Sciences, Shandong Normal University, Jinan, Shandong 250014, China

Received 28 September 2012; Accepted 29 November 2012

Academic Editor: Xiaodi Li

Copyright © 2012 Ziwen Jiang and Deren Xie. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### 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 in and norms. Furthermore, we obtain the superconvergence error estimate of the generalized projection of the solution in norm. Numerical experiment illustrates the convergence and stability of this scheme.

#### 1. Introduction

We consider the following mixed boundary-initial value problem: 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 is denoted by . When both ends of the bar are fixed, , , that is the boundary condition. and are the initial data.

In theoretical analysis, the problem (1.1) was first treated by Greenberg et al. [2], by assuming that the function was monotonic, that is, and that the initial data was smooth, specifically Under these assumptions they showed the existence of a unique smooth solution which decays the zero solution as . (See also Greenberg [3] and Greenberg and MacCamy [4].) Andrews [1] made the hypotheses that , , the function is locally Lipschitz continuous, and there exists a constant such that whenever ; then he proved the existence of a unique global weak solution. Under the hypotheses that , and there exists a constant such that , , 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 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 and 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 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 and norms and superconvergence in 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, let be a partition for the interval , with its nodes . The length of the element is denoted by , , is maximum of . We suppose is regular, that is, there exists a positive constant such that , . For the definition of the finite volume scheme, the dual partition of is needed, which is . The dual elements are denoted by , , , , where , .

Secondly, we define the piecewise linear trial function space over the partition , satisfying , where is the Sobolev space on . Then for all , where is the basis function associated with the nodes , It is easy to know that the derivative of with respect to is The test function space associated with the dual partition is defined as the set of all piecewise constants with for all . We may choose the basis function of in such a way that is the characteristic function of , that is, Then for any can be expressed as .

Obviously, Meanwhile, .

Thirdly, for the time interval , we give an isometric partition and denote the nodes , , .

We introduce some notations for functions and : Then we can get

Let be the solution of (1.1). Integrating (1.1) a over the dual element , we obtain where , , , and . The problem (2.7) can be rewritten in a variational form. For any arbitrary , we multiply the integral relation in (2.7) by and sum over all to obtain where for any arbitrary , the bilinear forms , are defined by

Since is a nonlinear function, we will consider the following linear finite volume scheme: find , such that where and are the interpolation projection of and onto the trial function space , respectively, and the interpolation operator is defined as We also need to introduce the interpolation , defined by, By Sobolev’s interpolation theory, we know that In this paper we adopt the standard notation for Sobolev space on with norm and seminorm . In order to simplify the notations, we denote by and skip the index when possible, that is, , , . We denote by the Banach space of all integrable functions from into with the norm for and the standard modification for . 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. *Let () be the solution of (2.10). According to and , , , are known. Hence the existence and uniqueness of the solution of scheme (2.10) are equivalent to the existence and uniqueness of , .

For , choosing , . By (2.2) we have Using (2.10) we obtain that is, Noticing , the coefficient matrix of the system (2.16) is strictly diagonally dominant matrix. So, when and are known, the solution is existent and unique. Combining the above conditions, the solution is existent and unique when and are known. This completes the proof.

#### 3. Error Estimates

In this section, we will prove the optimal error estimates in the and norms as well as the superconvergence error estimates in the norm. This needs some assumptions about the data. The nonlinear function satisfies The initial functions and satisfy The solution of (1.1) satisfies

For any , , , where , , . Using the definition of the interpolation operator we know that , . Noting and [11, Lemma 3.1] we have Hence So, when , we can get . Let . By [11, Lemma 3.2], we have On the other hand, using [11, Lemma 3.3], we also have For error estimate, we will use the generalized adjoint finite volume element projection of the solution of (1.1), that is, satisfies Let and denote the solutions of (1.1) and (3.8), respectively. Under the assumptions , , and together with [11, Theorems ] and the one-dimensional imbedding theorem in Sobolev space, then we can obtain the following.

Lemma 3.1. *Let and be the solutions of (1.1) and (3.8), respectively. If the assumptions (H _{1}), (H _{2}), and (H _{3}) hold, then one has
*

Next, we give the corresponding error estimates. Choosing in the first equation of (2.8), then multiplying the three equations by , , , respectively, and adding them, we have Similarly, taking in the first equation of (3.8), then multiplying the three equations by , , , respectively, and adding them, we obtain Subtracting (3.11) from (2.10) and using (3.12), we get the error equation where . Let Then the error equation (3.13) can be rewritten as

Theorem 3.2. *Let and 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 , such that when *

*Proof. *Firstly, we estimate . Applying the Taylor formula,
and noting the definitions of and in (2.10) and (3.8), respectively, we know
so
Notice that , (2.13) and (3.9), there exists a positive constant such that, when we have
This implies (3.16) holds. Secondly, noting and (3.19), we obtain
hence, by (2.16), (3.10), and we have
This completes the proof of (3.17).

Theorem 3.3. *Let , and 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 , when is sufficiently small and , one has
*

*where*

*Proof. *Expanding the term , we have
For any , since and , then
Let , thus . Applying transformation of variable, we know
where
By (3.28), we have
On the other hand
where
Combining the above five equalities with (3.26), we can get
Using the -inequality, we find
Now we estimate . Noting (2.2), we have
Furthermore,
So
When is sufficiently small, we can take suitable such that , from the above equalities we have
Moreover, in view of and (3.9), we obtain
Then
Substituting the estimate of into (3.34), we have
Since
where lies between and , by (2.13), (3.9), and the inverse estimate, we obtain
Thus, when we have , then
From the assumption , we find that is bounded: this leads to the boundedness of and . So
Substituting the estimates of , , , and into (3.33), we can derive (3.24). This completes the proof of the theorem.

Theorem 3.4. *Let , and 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 , when and are sufficiently small, one has
*

*Proof. *Taking in the error equation (3.15) and making a simple calculation yield
Form (3.4), we derive that