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

and Applied Analysis 3 for all uh x ∈ Uh, where φi x is the basis function associated with the nodes xi i 1, 2, . . . , r − 1 , φi x ⎧ ⎪ ⎪⎨ ⎪ ⎪⎩ 1 − xi − x hi , x ∈ Ii, 1 − x − xi hi 1 , x ∈ Ii 1, 0, x / ∈ Ii ∪ Ii 1. 2.1 It is easy to know that the derivative of uh with respect to x is uhx x uh xi − uh xi−1 hi , xi−1 ≤ x ≤ xi, i 1, 2, . . . , r. 2.2 The test function space Vh ⊂ L2 I associated with the dual partition T ∗ h 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 I∗ j , that is, ψj x { 1, x ∈ I∗ j , 0, x / ∈ I∗ j , j 1, 2, . . . , r − 1. 2.3 Then for any vh x ∈ Vh can be expressed as vh x ∑r−1 j 1 vh xj ψj x . Obviously, Uh ⊂ H1 0 I , Vh ⊂ L2 I , dimUh dimVh r − 1. 2.4 Meanwhile,Uh ⊂W1,∞ 0 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 functions u x, t and f ux x, t : u u x, tn , uj u ( xj , t ) , unj u ( xj , tn ) , ∂tu n u 1 − u τ , ∂ttu n u 1 − 2u un−1 τ2 , u 1/2 u 1 u 2 , un,1/4 u 1 2u un−1 4 , f1/2 ux f ( u 1 x ) f ux 2 , f1/4 ux f ( u 1 x ) 2f ux f ( un−1 x ) 4 , f 1/4 ∗ ux 3f ux f ( un−1 x ) 4 . 2.5 4 Abstract and Applied Analysis Then we can get ∂ttu n ∂tu n − ∂tun−1 τ , un,1/4 u 1/2 un−1/2 2 , 1 2 ( ∂tu n ∂tun−1 ) 1 τ ( u 1/2 − un−1/2 ) . 2.6 Let u be the solution of 1.1 . Integrating 1.1 a over the dual element I∗ j ∈ T ∗ h , we obtain ∫xj 1/2 xj−1/2 uttdx uxt ( xj−1/2 ) − uxt ( xj 1/2 ) f ( ux ( xj−1/2 )) − fux ( xj 1/2 )) 0, 2.7 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 utt, vh a∗ ut, vh b∗ ( f u , vh ) 0, ∀vh ∈ Vh, t ∈ 0, T , u 0 u0, ut 0 u1, 2.8 where for any arbitrary w ∑r−1 j 1 wjψj ∈ Vh, the bilinear forms a∗ v,w , b∗ f v , w are defined by


1.1
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 u 0 x and u t x, 0 u 1 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,

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 T h be a partition for the interval I 0, 1 , with its nodes 0 x 0 < x 1 < • • • < x r 1.The length of the element I i x i−1 , x i is denoted by h i x i − x i−1 , i 1, 2, . . ., r, h max 1≤i≤r h i is maximum of h i .We suppose T h is regular, that is, there exists a positive constant μ > 0 such that h i ≥ μh, i 1, 2, . . ., r.For the definition of the finite volume scheme, the dual partition T * h of T h is needed, which is 0 The dual elements are denoted by I * 0 x 0 , x 1/2 , I * j x j−1/2 , x j 1/2 , j 1, 2, . . ., r − 1, I * r x r−1/2 , x r , where x j−1/2 1/2 x j−1 x j , j 1, 2, . . ., r.Secondly, we define the piecewise linear trial function space U h over the partition T h , satisfying U h ⊂ H 1 0 I , where H 1 0 I is the Sobolev space on I. Then u h x r−1 i 1 u h x i ϕ i x Abstract and Applied Analysis 3 for all u h x ∈ U h , where ϕ i x is the basis function associated with the nodes x i i 1, 2, . . ., r − 1 ,

2.1
It is easy to know that the derivative of u h with respect to x is The test function space V h ⊂ L 2 I associated with the dual partition T * h is defined as the set of all piecewise constants with v h 0 v h 1 0 for all v h x ∈ V h .We may choose the basis function ψ j x of V h in such a way that ψ j x is the characteristic function of I * j , that is, x ∈ I * j , 0, x / ∈ I * j , j 1, 2, . . ., r − 1.

2.3
Then for any v h x ∈ V h can be expressed as v h x r−1 j 1 v h x j ψ j x .Obviously,
We introduce some notations for functions u x, t and f u x x, t :

2.5
Then we can get

2.6
Let u be the solution of 1.1 .Integrating 1.1 a over the dual element I * j ∈ T * h , we obtain where j 0, 1, . . ., r, x −1/2 x 0 , x r 1/2 x r , and 0 < t ≤ T .The problem 2.7 can be rewritten in a variational form.For any arbitrary v h ∈ V h , we multiply the integral relation in 2.7 by v h x j and sum over all j 0, 1, . . ., r to obtain where for any arbitrary w r−1

2.9
Since f is a nonlinear function, we will consider the following linear finite volume scheme: find

2.10
where Π h u 0 and Π h u 1 are the interpolation projection of u 0 and u 1 onto the trial function space U h , respectively, and the interpolation operator Π h is defined as

2.11
We also need to introduce the interpolation Π * h : 12 By Sobolev's interpolation theory, we know that

2.13
In this paper we adopt the standard notation W m,p I for Sobolev space on I with norm • m,p and seminorm |•| m,p .In order to simplify the notations, we denote W m,2 I by H m I and skip the index p 2 when possible, that is, • .We denote by L q 0, T; W m,p I the Banach space of all L q integrable functions from 0, T into W m,p I with the norm • L q W m,p T 0 • q m,p dt 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.
. ., N be the solution of 2.10 .According to u 0 h Π h u 0 and u 1 h Π h u 0 τΠ h u 1 , u 0 hj u 0 x j , u 1 hj u 0 x j τu 1 x j , j 1, 2, . . ., r − 1 are known.Hence the existence and uniqueness of the solution u n h n 0, 1, 2, . . ., N of scheme 2.10 are equivalent to the existence and uniqueness of {u n hi } r−1 i 1 , n 2, . . ., N.

2.14
Using 2.10 we obtain hj Abstract and Applied Analysis that is,

2.16
Noticing u n h0 u n hr 0, the coefficient matrix of the system 2.16 is strictly diagonally dominant matrix.So, when {u n−1 hi } r−1 i 1 and {u n−2 hi } r−1 i 1 are known, the solution {u n hi } r−1 i 1 is existent and unique.Combining the above conditions, the solution {u n hi } r−1 i 1 n 2, . . ., N is existent and unique when {u 0 hi } r−1 i 1 and {u 1 hi } r−1 i 1 are known.This completes the proof.

Error Estimates
In this section, we will prove the optimal error estimates in the L 2 and H 1 norms as well as the superconvergence error estimates in the H 1 norm.This needs some assumptions about the data.
The initial functions u 0 and u 1 satisfy Noting u h0 u hr v h0 v hr 0 and 11, Lemma 3.1 we have

3.4
Hence we have On the other hand, using 11, Lemma 3.3 , we also have

3.7
For error estimate, we will use the generalized adjoint finite volume element projection u 11 of the solution Let u and u denote the solutions of 1.1 and 3.8 , respectively.Under the assumptions H 1 , H 2 , and H 3 together with 11, Theorems 3.4-3.5 and the one-dimensional imbedding theorem in Sobolev space, then we can obtain the following.
Next, we give the corresponding error estimates.Choosing t t n−1 , t n , t n 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
Similarly, taking t t n−1 , t n , t n 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 Subtracting 3.11 from 2.10 and using 3.12 , we get the error equation

3.14
Then the error equation 3.13 can be rewritten as

3.15
Theorem 3.2.Let {u n h } N n 0 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 h 0 , such that when h < h 0 3.17 Proof.Firstly, we estimate |θ 1 | 1 .Applying the Taylor formula, and noting the definitions of u 1 h and u 0 in 2.10 and 3.8 , respectively, we know

3.20
Notice that u 1 u t 0 u 0 t , 2.13 and 3.9 , there exists a positive constant h 0 such that, when h < h 0 we have

3.23
This completes the proof of 3.17 .
Theorem 3.3.Let u, {u n h } N n 0 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 < h 0 , one has

3.26
For any

3.27
Let where

3.29
By 3.28 , we have

3.30
Abstract and Applied Analysis 13 On the other hand

3.32
Combining the above five equalities with 3.26 , we can get

3.33
Using the ε-inequality, we find Abstract and Applied Analysis

3.34
Now we estimate I 1 .Noting 2.2 , we have u n hx x dx.

3.36
So

3.37
When τ is sufficiently small, we can take τ suitable such that c 1 τ 2 ≤ 1/16, from the above equalities we have t n u t dt and 3.9 , we obtain

3.39
Abstract and Applied Analysis 15 Then

3.40
Substituting the estimate of I 1 into 3.34 , we have

3.42
where ξ n j lies between u n 1 x x j−1/2 sθ n 1 x x j−1/2 and u n x x j−1/2 sθ n x x j−1/2 , by 2.13 , 3.9 , and the inverse estimate, we obtain

3.44
From the assumption H 1 , we find that 1 0 f ξ n j ds is bounded: this leads to the boundedness of

3.45
Substituting the estimates of B 2 , B 3 , B 4 , and B 5 into 3.33 , we can derive 3.24 .This completes the proof of the theorem.
Theorem 3.4.Let u, {u n h } N n 0 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
Proof.Taking v h Π * h ∂ t θ n in the error equation 3.15 and making a simple calculation yield

3.47
Form 3.4 , we derive that

3.49
By 3.7 , we get

3.50
Applying Taylor's formula with integral-type remainder, we have

3.52
Note that Π * h ∂ t θ n ≤ c ∂ t θ n , we get

18
Abstract and Applied Analysis

3.53
For Π * h v r−1 j 1 v j ψ j x ∈ V h , v 0 v r 0, we have

3.54
Noting that

3.55
we see that

3.58
Multiplying the above inequation by τ and then summing it with respect to n from 1 to n, we have

3.59
Now we make the induction hypothesis:

3.60
Substituting 3.60 into 3.59 and noting 3.6 and the assumption H 1 , we conclude Abstract and Applied Analysis

3.62
By the discrete Gronwall's lemma, it follows that

3.67
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 that θ 0 0 and 2.13 , by Theorems 3.2 and 3.4, Lemma 3.1, and the triangle inequality, we obtain the following.

3.69
Remark 3.6.The optimal order error estimates in L 2 , H 1 norms and superconvergence error estimates in H 1 norm of the solution of the finite volume scheme 2.10 are achieved in Theorem 3.5.
For simplicity, we choose a constant spatial step h and a constant time step τ.The corresponding nodes are denoted by x i ih, i 0, 1, . . ., 2r and the time levels by t j 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 u h , we get the discrete norms u − u h 0,h and |u − u h | 1,h , which are given in Tables 1 and 2.
From Tables 1 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.
Lemma 3.1.Let u 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
Similarly, letting h, τ be sufficiently small such that h, τ satisfy c 4

Table 3 :
The orders of u − u h 0,h and |u − u h | 1,h .