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,
1Youcai Xu,
2and Bing Hu
21School 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,
utρxuux−uxxt0,
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
v2−1 v sec
21 2
v2−1
v2 x−vt,
ρx, t 3
v2−1 v2 sec
21 2
v2−1
v2 x−vt.
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:
utt−uxx
1 2
u2
xt−uxxtt0. 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
utρx−υuxxuux−uxxt0, 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, x∈xL, 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 thatu0∈H1,ρ0∈L2, the solution of1.4–1.7satisfiesu
L2 ≤C,uxL2 ≤
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
uutuρx−υuuxxu2ux−uuxxt
dx0. 1.9
According to
xR
xL
uutdx
1 2
d dt
xR
xL u2dx,
xR
xL
uρxdxuρ|xxRL− xR
xL
ρdu− xR
xL
uxρdx,
− xR
xL
uuxxdx−uux|xxRL xR
xL
uxdu
xR
xL
ux2dx,
xR
xL
u2uxdx 1
3u 3|xR
xL 0,
− xR
xL
uuxxtdx−uuxt|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
Adding1.14to1.11, we obtain
d dt
xR
xL
u2u2xρ2
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 ,unj ≈ uxj, tn,ρnj ≈
ρxj, tn, andZ0h{u uj|u0uJ 0, j0,1,2, . . . , J}. We define the difference operators
as follows:
un
j
x
unj1−unj
h ,
un
j
x
unj −unj−1
h ,
un
j
x
unj1−unj−1
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≤j≤J−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
x−υ
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≤j≤J, 2.4
un
0 unJ 0, ρn0 ρnJ 0, 1≤n≤N. 2.5
Lemma 2.1. Summation by parts follows [12,21] that for any two discrete functionsu, v∈Z0
h
uj
x, vj
−uj,
vj
x
, vj,
uj
xx
−vj
x,
uj
x
Lemma 2.2discrete Sobolev’s inequality 12,21 . There exist two constantsC1 andC2 such that
un∞≤C1unC2unx. 2.7
Lemma 2.3discrete Gronwall inequality12,21 . Suppose thatwk,ρkare nonnegative functions and ρkis nondecreasing. IfC >0and
wk≤ρk Cτ
k−1
l0
wl. 2.8
Thenwk≤ρkeCτk.
Theorem 2.4. Ifu0∈H1,ρ0∈L2, then the solution of 2.2–2.5satisfies
un ≤C, unx ≤C, ρn≤C, 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
ujnunj11unj−11−unj−11−unj−−11unj1unj11unj−11−ujn−1unj−11unj−−11
×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 unj−1unj1ujn−1unj−11unj−−110,
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
ρn12−ρn−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
An−An−1≤CτAnAn−1. 2.17
Ifτis sufficiently small which satisfies 1−Cτ >0, then
An−An−1≤CτAn−1. 2.18
Summing up2.18from 1 ton, we have
An≤A0Cτn−1
l0
FromLemma 2.3, we obtainAn ≤ C, which implies that,un ≤ C,un
x ≤ C, andρn ≤ C.
ByLemma 2.2, we obtainun ∞≤C.
Theorem 2.5. Assume thatu0∈H2,ρ0∈H1, the solution of difference scheme2.2–2.5satisfies: ρn
x≤C, unxx ≤C, unx∞≤C, ρn∞≤C n1,2, . . . , N. 2.20
Proof. Differentiating backward2.2–2.5with respect tox, we obtain
unj
xt−
unj
xxxt
ρnj
xx−υ
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≤j≤J, 2.23
un0xunJ
x0,
ρn0xρnJ
x0, 0≤n≤N. 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
unj≤C 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 .
Noting that
ρnxx,2unx
−2unxx, ρxn
−ρnx, unxx1unx−x1
≤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 Cτ
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 obtainBn−Bn−1≤CτBn
Bn−1. Choosing suitableτwhich is small enough to satisfy 1−Cτ >0, we get
Summing up2.34from 1 ton, we have
Bn≤B0Cτ
n−1
l0
Bl. 2.35
By Lemma 2.3, we get Bn ≤ C, which implies that ρn
x ≤ C,unxx ≤ C. It follows from
Theorem 2.4andLemma 2.2thatun
x∞≤C,ρn∞≤C.
3. Solvability
Theorem 3.1. The solutionunof2.2–2.5is 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−ujn−11unj1unj11−unj−1unj−11unj1
1
12
J−1
j0
unjunj1unj11ujn1unj1ujn11− 1 12
J−1
j0
unj−1ujn−11unj1unjunj−11unj10,
3.4
then it holds
1 2τ
un12
1 2τ
υ 2
unx1
2
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
x−υ
vnj
xx
1 3
vnj
vnj
x
vjnvnj
x
, 4.1
snj ∅nj
t
vnj
xγ∅ n
j. 4.2
Making use of Taylor expansion, it holds|rjn||snj|Oτ2h2ifh, τ → 0.
Theorem 4.1. Assume thatu0 ∈ H1,ρ0 ∈L2, then the solutionunandρnin the senses of norms · ∞and · L2, respectively, to the difference scheme2.2–2.5converges to the solution of problem 1.4–1.7and the order of convergence isOτ2h2.
Proof. Subtracting2.2from4.1subtracting2.3from4.2, and lettingenj vjn−unj,ηnj
∅n
j −ρnj, we have
rjnenj
t−
enj
xxt
ηjn
x−υ
enj
xxQ, 4.3
snj ηjn
t
ejn
xγη n
j, 4.4
where
Q 1 3
vjnvnj
x−u 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
.
According to
−Q,2enj
−2 3h
J−1
j0
vnjvnj
x−u 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
vnjvnj −unju 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
vnj≤C, vnj≤C,
unj
x
≤C, unj≤C 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
.
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
Dn−Dn−1≤2τrn22τsn2CτDn1Dn. 4.15
Ifτis sufficiently small which satisfies 1−Cτ >0, then
Dn−Dn−1≤CτDn−1Cτrn2Cτsn2. 4.16
Summing up4.16from 1 ton, we have
Dn≤D0Cτ
n
l1
rl2Cτ
n
l1
sl2Cτ
n−1
l0
Dl. 4.17
Select appropriate second-order methodssuch as the C-N Schemes, and calculateu1 and ρ1, which satisfies
Noticing that
τ
n
l1
rl2≤nτmax
1≤l≤n
rl2≤T·Oτ2h22,
τ
n
l1
sl2≤nτmax 1≤l≤n
sl2≤T·Oτ2h22,
4.19
we then have
Dn≤Oτ2h22Cτ
n−1
l0
Dl. 4.20
ByLemma 2.3, we get
Dn≤Oτ2h22. 4.21
This yields
en ≤Oτ2h2, en x ≤O
τ2h2, ηn≤Oτ2h2. 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 solutionunandρnof2.2–2.5is 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
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.
−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.
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