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Boundary Value Problems

Volume 2010, Article ID 781750,16pages doi:10.1155/2010/781750

Research Article

A Linear Difference Scheme for Dissipative

Symmetric Regularized Long Wave Equations with

Damping Term

Jinsong Hu,

1

Youcai Xu,

2

and Bing Hu

2

1School of Mathematics and Computer Engineering, Xihua University, Chengdu 610039, China

2School of Mathematics, Sichuan University, Chengdu 610064, China

Correspondence should be addressed to Youcai Xu,xyc@scu.edu.cn

Received 24 August 2010; Accepted 14 November 2010

Academic Editor: V. Shakhmurov

Copyrightq2010 Jinsong Hu et al. 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.

We study the initial-boundary problem of dissipative symmetric regularized long wave equations with damping term by finite difference method. A linear three-level implicit finite difference scheme is designed. Existence and uniqueness of numerical solutions are derived. It is proved that the finite difference scheme is of second-order convergence and unconditionally stable by the discrete energy method. Numerical simulations verify that the method is accurate and efficient.

1. Introduction

A symmetric version of regularized long wave equationSRLWE,

utρxuuxuxxt0,

ρtux0,

1.1

has been proposed to model the propagation of weakly nonlinear ion acoustic and space charge waves1 . The sec2solitary wave solutions are

ux, t 3

v21 v sec

21 2

v21

v2 xvt,

ρx, t 3

v21 v2 sec

21 2

v21

v2 xvt.

(2)

The four invariants and some numerical results have been obtained in 1 , where vis the velocity,v2>1. Obviously, eliminatingρfrom1.1, we get a class of SRLWE:

uttuxx

1 2

u2

xtuxxtt0. 1.3

Equation1.3is explicitly symmetric in thex andt derivatives and is very similar to the regularized long wave equation that describes shallow water waves and plasma drift waves

2, 3 . The SRLW equation also arises in many other areas of mathematical physics 4–6 . Numerical investigation indicates that interactions of solitary waves are inelastic7 ; thus, the solitary wave of the SRLWE is not a solution. Research on the wellposedness for its solution and numerical methods has aroused more and more interest. In8 , Guo studied the existence, uniqueness, and regularity of the numerical solutions for the periodic initial value problem of generalized SRLW by the spectral method. In9 , Zheng et al. presented a Fourier pseudospectral method with a restraint operator for the SRLWEs and proved its stability and obtained the optimum error estimates. There are other methods such as pseudospectral method, finite difference method for the initial-boundary value problem of SRLWEssee9–

15 .

In applications, the viscous damping effect is inevitable, and it plays the same important role as the dispersive effect. Therefore, it is more significant to study the dissipative symmetric regularized long wave equations with the damping term

utρxυuxxuuxuxxt0, 1.4

ρtuxγρ0, 1.5

whereυ, γare positive constants,υ >0 is the dissipative coefficient, andγ >0 is the damping coefficient. Equations 1.4-1.5are a reasonable model to render essential phenomena of nonlinear ion acoustic wave motion when dissipation is considered. Existence, uniqueness, and wellposedness of global solutions to 1.4-1.5 are presented see 16–20 . But it is difficult to find the analytical solution to 1.4-1.5, which makes numerical solution important.

To authors’ knowledge, the finite difference method to dissipative SRLWEs with

damping term1.4-1.5has not been studied till now. In this paper, we propose linear three level implicit finite difference scheme for1.4-1.5with

ux,0 u0x, ρx,0 ρ0x, xxL, xR , 1.6

and the boundary conditions

uxL, t uxR, t 0, ρxL, t ρxR, t 0, t∈0, T . 1.7

We show that this difference scheme is uniquely solvable, convergent, and stable in both theoretical and numerical senses.

Lemma 1.1. Suppose thatu0H1,ρ0L2, the solution of1.41.7satisfiesu

L2 ≤C,uxL2 ≤

(3)

Proof. Let

Et u2L2ux

2

L2ρ

2

L2

xR

xL

u2dx xR

xL

ux2dx

xR

xL

ρ2dx, t∈0, T . 1.8

Multiplying1.4byuand integrating overxL, xR , we have

xR

xL

uutuρxυuuxxu2uxuuxxt

dx0. 1.9

According to

xR

xL

uutdx

1 2

d dt

xR

xL u2dx,

xR

xL

uρxdxuρ|xxRLxR

xL

ρduxR

xL

uxρdx,

xR

xL

uuxxdxuux|xxRL xR

xL

uxdu

xR

xL

ux2dx,

xR

xL

u2uxdx 1

3u 3|xR

xL 0,

xR

xL

uuxxtdxuuxt|xxRL xR

xL

uxtdu

1 2

d dt

xR

xL

ux2dx,

1.10

we get

d dt

xR

xL

u2u2x

dx−2

xR

xL

uxρdx2υ

xR

xL

ux2dx0, 1.11

Then, multiplying1.5byρand integrating overxL, xR , we have

xR

xL

ρρtρuxγρ2

dx0. 1.12

By

xR

xL

ρρtdx

1 2

d dt

xR

xL

ρ2dx, 1.13

we get

d dt

xR

xL

ρ2dx2 xR

xL

uxρdx2γ

xR

xL

(4)

Adding1.14to1.11, we obtain

d dt

xR

xL

u2u22

dx−2υ xR

xL

ux2dx−2γ

xR

xL

ρ2dx≤0. 1.15

SoEtis decreasing with respect tot, which implies thatEt u2

L2ux

2

L2ρ

2

L2≤E0,

t ∈ 0, T . Then, it indicates that uL2 ≤ C,uxL2 ≤ C, andρL2 ≤ C. It is followed from

Sobolev inequality thatuL∞ ≤C.

2. Finite Difference Scheme and Its Error Estimation

Lethandτbe the uniform step size in the spatial and temporal direction, respectively. Denote xj xLjhj 0,1,2, . . . , J,tn nτn 0,1,2, . . . , N,N T/τ ,unjuxj, tn,ρnj

ρxj, tn, andZ0h{u uj|u0uJ 0, j0,1,2, . . . , J}. We define the difference operators

as follows:

un

j

x

unj1unj

h ,

un

j

x

unjunj1

h ,

un

j

x

unj1unj1

2h ,

un

j

t

unj1−unj−1

2τ ,

unj

un1

j unj−1

2 , u

n, vnhJ−1 j0

unjvjn, un2un, un, un max 0≤jJ−1

unj.

2.1

Then, the average three-implicit finite difference scheme for the solution of1.4–1.7is as follow:

unj

t

unj

xxt

ρjn

xυ

unj

xx

1 3

unjunj

x

unjunj

x

0, 2.2

ρjn

t

unj

xγρ n

j 0, 2.3

u0j u0xj

, ρ0j ρ0xj

, 0≤jJ, 2.4

un

0 unJ 0, ρn0 ρnJ 0, 1≤nN. 2.5

Lemma 2.1. Summation by parts follows [12,21] that for any two discrete functionsu, vZ0

h

uj

x, vj

uj,

vj

x

, vj,

uj

xx

vj

x,

uj

x

(5)

Lemma 2.2discrete Sobolev’s inequality 12,21 . There exist two constantsC1 andC2 such that

unC1unC2unx. 2.7

Lemma 2.3discrete Gronwall inequality12,21 . Suppose thatwk,ρkare nonnegative functions and ρkis nondecreasing. IfC >0and

wkρk

k−1

l0

wl. 2.8

ThenwkρkeCτk.

Theorem 2.4. Ifu0H1,ρ0L2, then the solution of 2.22.5satisfies

unC, unxC, ρnC, un∞≤C n1,2, . . . , N. 2.9

Proof. Taking an inner product of 2.2 with 2unj i.e., ujn1 unj−1 and considering the

boundary condition2.5andLemma 2.1, we obtain

1 2τ

un12−un−12

1

2τ

unx1

2

unx−1

2

ρnj

x,2u n j

υunj

xx,2u n j

P,2unj

0,

2.10

whereP 1/3unjunjx unju n

jx . Since

ρn

j

x,2u n j

ρn j,2

unj

x

,

unj

xx,2u n j

−2unx

2 ,

P,2unj

2

3h

J−1

j0

unj

unj

x

unjunj

x

unj

1

12

J−1

j0

ujnunj11unj11−unj11−unj11unj1unj11unj11−ujn1unj11unj11

×un1

j unj−1

1

12

J−1

j0

un j unj1

un1

j1 unj−11

un1

j unj−1

−1 12

J−1

j0

unj unj1unj1ujn−1unj11unj110,

(6)

we obtain

1 2τ

un12−un−12

1

2τ

unx1

2

unx−1

2

ρjn,2

unj

x

2υunx

2

0. 2.12

Taking an inner product of2.3with 2ρnj i.e., ρn1

j ρnj−1, we obtain

1 2τ

ρn12ρn−12

un j

x,2ρ n j

2γρnj2 0. 2.13

Adding2.12to2.13, we have

un12−un−12unx1

2

unx−1

2

ρn12−ρn−12

2τρn j,2

unj

x

un

j

x,2ρ n j

−4υτunx

2

4γτρnj2.

2.14

Since

ρnj,2

unj

x

ρnj,

unj1

x

unj−1

x

ρn21 2

unx1

2

unx−1

2 ,

unj

x,2ρ n j

unj

x, ρ n1

j ρjn−1

unx2

1 2

ρn12ρn−12

.

2.15

Equation2.14can be changed to

un12−un−12unx1

2

unx−1

2

ρn12−ρn−12

Cτunx1

2

unx2unx−1

2

ρn12ρn2ρn−12

.

2.16

LetAnun12un2un1

x 2unx2ρn12ρn2, and2.16is changed to

AnAn−1≤CτAnAn−1. 2.17

Ifτis sufficiently small which satisfies 1−Cτ >0, then

AnAn−1≤CτAn−1. 2.18

Summing up2.18from 1 ton, we have

AnA0n−1

l0

(7)

FromLemma 2.3, we obtainAn C, which implies that,un C,un

xC, andρnC.

ByLemma 2.2, we obtainun ∞≤C.

Theorem 2.5. Assume thatu0H2,ρ0H1, the solution of dierence scheme2.22.5satisfies: ρn

xC, unxxC, unx∞≤C, ρn∞≤C n1,2, . . . , N. 2.20

Proof. Differentiating backward2.2–2.5with respect tox, we obtain

unj

xt

unj

xxxt

ρnj

xxυ

unj

xxx

1 3

unjunj

x

unjunj

x

x 0, 2.21

ρnj

xt

unj

xxγ

ρnj

x0, 2.22

u0j

xu0,x

xj

, ρ0j

xρ0,x

xj

, 0≤jJ, 2.23

un0xunJ

x0,

ρn0xρnJ

x0, 0≤nN. 2.24

Computing the inner product of2.21with 2unx i.e., uxn1unx−1and considering2.24and

Lemma 2.1, we obtain

1 2τ

unx1

2

unx−1

2

1

2τ

unxx1

2

unxx−1

2

ρnxx,2unx

υunxxx,2unx

R,2unx

0,

2.25

whereR 1/3unjunjx unju n

jx x. It follows fromTheorem 2.4that

unjC j 0,1,2, . . . , J. 2.26

By the Schwarz inequality andLemma 2.1, we get

R,2unx

2

3

unjunj

x

unjunj

x

x, u n x −2 3

unjunj

x

unjunj

x, u n xx

−2 3h

J−1

j0

unj

unj

x

unjunj

x

unj

xx

≤ 2 3Ch

J−1

j0 unj

x

·unj

xx

Cunx

2 unxx

2

Cunx1

2

unx−1

2

unxx1

2

unxx−1

2 .

(8)

Noting that

ρnxx,2unx

−2unxx, ρxn

ρnx, unxx1unxx1

ρxn2

1 2

unxx1

2

unxx−1

2 ,

unxxx,2unx

−2unxx

2 ,

2.28

it follows from2.25that

unx1

2

unx−1

2

unxx1

2

unxx−1

2

≤ −4υτunxx

2

Cτunx1

2

unx−1

2

unxx1

2

unxx−1

2

ρnx

2 .

2.29

Computing the inner product of2.22with 2ρnxi.e.,ρn1

x ρnx−1and considering2.24and

Lemma 2.1, we obtain

1 2τ

ρnx1

2

ρxn−1

2

unxx, ρnx2γρnx20. 2.30

Since

unxx,2ρnxunxx, ρnx1ρnx−1

unxx2

1 2

ρxn1

2

ρnx−1

2

, 2.31

then2.30is changed to

ρxn1−ρxn−1≤ −4γτρnx

2

unxx2ρxn1

2

ρnx−1

2

. 2.32

Adding2.29to2.32, we have

unx1

2

unx−1

2

unxx1

2

unxx−1

2

ρnx1

2

ρxn−1

2

≤ −4υτunxx

2

4γτρnx2

Cτun1

x

2

un−1

x

2

un

xx2unxx1

2

un−1

xx

2

ρn1

x

2

ρn x

2 ρn−1

x

2

Cτunx1

2

unx−1

2

unxx2unxx1

2

unxx−1

2

ρnx1

2

ρxn

2 ρnx−1

2 .

2.33

LetingBnun1

x 2unx2uxxn12unxx2ρxn12ρnx2, we obtainBnBn−1≤CτBn

Bn−1. Choosing suitableτwhich is small enough to satisfy 1Cτ >0, we get

(9)

Summing up2.34from 1 ton, we have

BnB0

n−1

l0

Bl. 2.35

By Lemma 2.3, we get Bn C, which implies that ρn

xC,unxxC. It follows from

Theorem 2.4andLemma 2.2thatun

x∞≤C,ρn∞≤C.

3. Solvability

Theorem 3.1. The solutionunof2.22.5is unique.

Proof. Using the mathematical induction, clearly,u0,ρ0 are uniquely determined by initial conditions2.4. then select appropriate second-order methodssuch as the C-N Schemes and calculate u1 and ρ1 i.e. u0, ρ0, and u1, ρ1 are uniquely determined. Assume that u0, u1, . . . , unandρ0, ρ1, . . . , ρnare the only solution, now considerun1andρn1in2.2and

2.3:

1 2τu

n1

j

1 2τ

unj1

xx

υ 2

unj1

xx

1 6

unjunj1

x

unjunj1

x

0, 3.1

1 2τρ

n1

j

γ 2ρ

n1

j 0. 3.2

Taking an inner product of3.1withun1, we have

1 2τ

un12 1 2τ

unx1

2

υ

2 unx1

2

1

6h

J−1

j0

unjunj1

x

unjunj1

x

unj10. 3.3

Since

1 6h

J−1

j0

unjunj1

x

unjunj1

x

unj1

1

12

J−1

j0

ujnunj11−ujn11unj1unj11−unj1unj11unj1

1

12

J−1

j0

unjunj1unj11ujn1unj1ujn11− 1 12

J−1

j0

unj1ujn11unj1unjunj11unj10,

3.4

then it holds

1 2τ

un12

1 2τ

υ 2

unx1

2

(10)

Taking an inner product of3.2withρn1and adding to3.5, we have

1 2τ

un12

1 2τ

υ 2

unx1

2

1 2τ

γ 2

ρn120, 3.6

which implies that3.1-3.2have only zero solution. So the solutionun1

j andρnj1of2.2–

2.5is unique.

4. Convergence and Stability

Let vx, t and ∅x, t be the solution of problem 1.4–1.7; that is, vjn uxj, tn, ∅nj

ρxj, tn, then the truncation of the difference scheme2.2–2.5is

rjn

vnj

t

vnj

xxt

n

j

xυ

vnj

xx

1 3

vnj

vnj

x

vjnvnj

x

, 4.1

snjnj

t

vnj

n

j. 4.2

Making use of Taylor expansion, it holds|rjn||snj|2h2ifh, τ → 0.

Theorem 4.1. Assume thatu0H1,ρ0 L2, then the solutionunandρnin the senses of norms · ∞and · L2, respectively, to the difference scheme2.22.5converges to the solution of problem 1.41.7and the order of convergence isOτ2h2.

Proof. Subtracting2.2from4.1subtracting2.3from4.2, and lettingenj vjnunj,ηnj

n

jρnj, we have

rjnenj

t

enj

xxt

ηjn

xυ

enj

xxQ, 4.3

snj ηjn

t

ejn

xγη n

j, 4.4

where

Q 1 3

vjnvnj

xu n j

unj

x

1

3

vjnvnj

x

unjunj

x

. 4.5

Computing the inner product of4.3with 2en, we get

en12−en−12exn1

2

enx−1

2

−4υτenx

2

2τrjn,2enj

ηnj

x,2e n j

Q,2enj

.

(11)

According to

Q,2enj

−2 3h

J−1

j0

vnjvnj

xu n j

unj

x

enj

2 3h

J−1

j0

vnjvnj

x

unjunj

x

enj

−2 3h

J−1

j0

vnjenj

xe n j

unj

x

enj

2 3h

J−1

j0

vnjvnjunju n j

enj

x

−2 3h

J−1

j0

vnjenj

xe n j

unj

x

enj

2 3h

J−1

j0

enjvnj unje n j

enj

x,

4.7

it follow fromLemma 1.1, Theorems2.4, and2.5that

vnjC, vnjC,

unj

x

C, unjC j0,1,2, . . . , J. 4.8

By the Schwarz inequality, we obtain

Q,2en≤ 2 3Ch

J−1

j0

enj

x en j ·enj

2 3Ch

J−1

j0 en

jenj

·enj

x

Cenx

2

en2en2

Cen12en2en−12enx1

2

enx−1

2 .

4.9

Since

rjn,2enj

rjn, enj1ejn−1≤ rn21 2

en12en−12

,

ηnj

x,2e n j

ηnj,2enj

x

ηn21 2

exn1

2

enx−1

2 ,

4.10

it follows from4.9–4.10and4.6that

en12−en−12enx1

2

enx−1

2

≤2τrnCτen12en2en−12enx1

2

enx−1

2

η2

.

(12)

Computing the inner product of4.4with 2ηn, we obtain

ηn12−ηn−122τsnj,2ηjn−2τenj

x,2η n j

−2γτηn2

Cτηn12ηn−12enx2

2τsn2.

4.12

Adding4.12to4.11, we have

en12−en−12enx1

2

exn−1

2

ηn12 ηn−12

≤2τrn22τsn2Cτen12en2en−12enx1

2

en−1

x

2

en

x2ηn1

2

ηn2 ηn−12

.

4.13

Leting

Dnen2en12en

x2enx1

2

ηn2

ηn12, 4.14

we get

DnDn−1≤2τrn22τsn2CτDn1Dn. 4.15

Ifτis sufficiently small which satisfies 1−Cτ >0, then

DnDn−1CτDn−1rn2sn2. 4.16

Summing up4.16from 1 ton, we have

DnD0

n

l1

rl2

n

l1

sl2

n−1

l0

Dl. 4.17

Select appropriate second-order methodssuch as the C-N Schemes, and calculateu1 and ρ1, which satisfies

(13)

Noticing that

τ

n

l1

rl2max

1≤ln

rl2T·Oτ2h22,

τ

n

l1

sl2≤max 1≤ln

sl2≤T·2h22,

4.19

we then have

Dn2h22

n−1

l0

Dl. 4.20

ByLemma 2.3, we get

DnOτ2h22. 4.21

This yields

enOτ2h2, en xO

τ2h2, ηnOτ2h2. 4.22

ByLemma 2.2, we have

en ∞≤O

τ2h2. 4.23

Similarly toTheorem 4.1, we can prove the result as follows.

Theorem 4.2. Under the conditions ofTheorem 4.1, the solutionunandρnof2.22.5is stable in

the senses of norm · ∞and · L2, respectively.

5. Numerical Simulations

Since the three-implicit finite difference scheme can not start by itself, we need to select other two-level schemes such as the C-N Schemeto get u1, ρ1. Then, reusing initial valueu0, ρ0, we can work outu2, ρ2, u3, ρ3, . . .. Iterative numerical calculation is not required, for this scheme is linear, so it saves computing time.

Whent0, the damping does not have an effect and the dissipative will not appear. So the initial conditions of1.4–1.7are same as those of1.1:

u0x 5 2sec

2

√ 5

6 x, ρ0x

5 3sec

2

√ 5

(14)

Table 1:The error ratios in the sense ofl∞at various time steps.

τh0.1 τh0.05 τh0.025

μ

t0.2 5.783531e−4 1.366490e−4 3.178799e−5 t0.4 9.505742e−4 2.237941e−4 5.198658e−5 t0.6 1.159542e−3 2.724234e−4 6.320922e−5 t0.8 1.246682e−3 2.925785e−4 6.789465e−5 t1.0 1.248960e−3 2.936257e−4 6.817804e−5

ρ

t0.2 1.292902e−3 3.176066e−4 7.553391e−5 t0.4 2.182523e−3 5.367686e−4 1.277456e−4 t0.6 2.182523e−3 6.760967e−4 1.610159e−4 t0.8 3.046673e−3 7.521463e−4 1.792741e−4 t1.0 3.154536e−3 7.796078e−4 1.859421e−4

0 0.5 1 1.5 2 2.5

t=0 t=0.5 t=1

−20 −15 −10 −5 0 5 10 15 20

Figure 1:Whenτh0.05, the wave graph ofuat various times.

LetxL −20,xR 20,T 1.0, andυ γ 1. Since we do not know the exact solution of

1.4-1.5, an error estimates method in21 is used: a comparison between the numerical solutions on a coarse mesh and those on a refine mesh is made. We consider the solution on meshτ h1/160 as the reference solution. InTable 1, we give the ratios in the sense ofl∞ at various time steps.

Whenτ h0.05, a wave figure comparison ofuandρat various time steps is as in Figures1and2.

(15)

−0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

t=0 t=0.5 t=1

−20 −15 −10 −5 0 5 10 15 20

Figure 2:Whenτh0.05, the wave graph ofρat various times.

Acknowledgments

The work of Jinsong Hu was supported by the research fund of key disciplinary of application

mathematics of Xihua University Grant no. XZD0910-09-1. The work of Youcai Xu was

supported by the Youth Research Foundation of Sichuan Universityno. 2009SCU11113.

References

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Figure

Figure 1: When τ � h � 0.05, the wave graph of u at various times.
Figure 2: When τ � h � 0.05, the wave graph of ρ at various times.

References

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