Applied Mathematics and Computation 204 (2008) 311–316

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Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate/amc

Crank–Nicolson finite difference scheme for the Rosenau–Burgers equation

Bing Hu a, Youcai Xu a,*, Jinsong Hu b

a School of Mathematics, Sichuan University, Chengdu 610064, PR Chinab School of Mathematics and Computer Engineering, Xihua University, Chengdu 610039, PR China

a r t i c l e i n f o a b s t r a c t

Keywords:Rosenau–Burgers equationC–N schemeConvergenceStability

0096-3003/$ - see front matter � 2008 Elsevier Incdoi:10.1016/j.amc.2008.06.051

* Corresponding author.E-mail address: xyc@scu.edu.cn (Y. Xu).

In this paper, a Crank–Nicolson finite difference scheme for the numerical solution of theinitial-boundary value problem of Rosenau–Burgers equation is proposed. Existence anduniqueness of numerical solutions are derived. It is proved that the finite difference schemeis convergent in the order of oðs2 þ h2Þ and stable. Numerical simulations show that themethod is efficient.

� 2008 Elsevier Inc. All rights reserved.

1. Introduction

In the study of the dynamics of dense discrete systems, the case of wave–wave and wave–wall interactions cannot bedescribed using the well-known KDV equation. To overcome this shortcoming of the KDV equation, Rosenau [1,2] proposedthe so-called Rosenau equation

ut þ uxxxxt þ ux þ uux ¼ 0; x 2 ½0; L�; t 2 ½0; T�: ð1:1Þ

The existence and the uniqueness of the solution for (1.1) was proved by Park [3]. But it is difficult to find the analyticalsolution for (1.1). Since then, much work has been done on the numerical method for (1.1) ([4–7] and also the refer-ences therein). On the other hand, for the further consideration of the nonlinear wave, the viscous term �uxx needs to beincluded

ut þ uxxxxt � uxx þ ux þ uux ¼ 0: ð1:2Þ

This equation is usually called the Rosenau–Burgers equation. A great of work has been devoted to the Cauchy problem of theRosenau–Burgers equation [8–11]. But the numerical method to the initial-boundary value problem of Rosenau–Burgersequation has not been studied till now. In this paper, we propose a Crank–Nicolson finite difference scheme for the Rose-nau–Burgers equation (1.2) with the boundary conditions

uð0; tÞ ¼ uðL; tÞ ¼ 0; uxxð0; tÞ ¼ uxxðL; tÞ ¼ 0; t 2 ½0; T� ð1:3Þ

and an initial condition

uðx;0Þ ¼ u0ðxÞ; x 2 ½0; L�: ð1:4Þ

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

. All rights reserved.

312 B. Hu et al. / Applied Mathematics and Computation 204 (2008) 311–316

The rest of this paper is organized as follows: in Section 2, we will describe a Crank–Nicolson finite difference scheme forthe Rosenau–Burgers equation and discuss the estimate for the difference solution. In Section 3, we will show that thescheme is uniquely solvable. Then, in Section 4, we will prove the convergence and stability for the difference scheme. Finallysome numerical tests are given in Section 5 to verify our theoretical analysis.

2. Finite difference scheme and estimate for the difference solution

Let h and s be the uniform step size in the spatial and temporal direction, respectively. Denote xj ¼ jhð0 6 j 6 JÞ;tn ¼ nsð0 6 n 6 NÞ;un

j � uðjh;nsÞ and Z0h ¼ fu ¼ ðujÞju0 ¼ uJ ¼ 0; j ¼ 0;1;2; . . . ; Jg. Throughout this paper, we will denote C

as a generic constant independent of step sizes h and s. We define the difference operators as follows:

ðunj Þx ¼

unjþ1 � un

j

h; ðun

j Þ�x ¼un

j � unj�1

h; ðun

j Þx̂ ¼un

jþ1 � unj�1

2h;

ðunj Þt ¼

unþ1j � un

j

s; ðun

j Þx�x ¼ ðunj Þ�xx ¼

unjþ1 � 2un

j þ unj�1

h2 ; unþ1

2j ¼

unþ1j þ un

j

2;

ðun; vnÞ ¼ hXJ�1

j¼0

unj vn

j ; kunk2 ¼ ðun; unÞ; kunk1 ¼ max06j6J�1

kunj k:

We propose a Crank–Nicolson finite difference scheme for the solution of (1.2)–(1.4)

ðunj Þt þ ðun

j Þxx�x�xt � unþ1

2j

� �x�xþ u

nþ12

j

� �x̂þ 1

3u

nþ12

jþ1 þ unþ1

2j þ u

nþ12

j�1

� �� u

nþ12

j

� �x̂¼ 0; ð2:1Þ

u0j ¼ u0ðxjÞ; 0 6 j 6 J � 1; ð2:2Þ

un0 ¼ un

J ¼ 0; ðun0Þx�x ¼ ðun

J Þ�xx ¼ 0: ð2:3Þ

Lemma 2.1. It follows from summation by parts that for any two mesh functions u; v 2 Z0h:

ððujÞx; vjÞ ¼ �ðuj; ðvjÞ�xÞ; ðvj; ðujÞx�xÞ ¼ �ððvjÞx; ðujÞxÞ:

Then we have

ððvjÞx;ujÞ ¼ �ðvj; ðujÞ�xÞ; ð2:4Þðuj; ðujÞx�xÞ ¼ �ððujÞx; ðujÞxÞ ¼ �kuxk2

: ð2:5Þ

If ðun0Þx�x ¼ ðun

J Þx�x ¼ 0, which implies

ðuj; ðujÞxx�x�xÞ ¼ kuxxk2: ð2:6Þ

Lemma 2.2 (Discrete Sobolev’s inequality [12]). There exist two constants C1 and C2 such that

kunk1 6 C1kunk þ C2kunxk:

Theorem 2.1. If u0 2 H20½0; L�, then the solution un of (2.1)–(2.3) satisfies kunk 6 C; kun

xk 6 C, which yieldkunk1 6 Cðn ¼ 1;2; . . . ;NÞ.

Proof. Taking an inner product of (2.1) with 2unþ12 (i.e. unþ1 þ un), we obtain

1sðkunþ1k2 � kunk2Þ þ ðun

j Þxx�x�xt;2unþ1

2j

� �� u

nþ12

j

� �x�x;2u

nþ12

j

� �þ u

nþ12

j

� �x̂;2u

nþ12

j

� �þ P;2u

nþ12

j

� �¼ 0; ð2:7Þ

where

P ¼ 13

unþ1

2jþ1 þ u

nþ12

j þ unþ1

2j�1

� �� u

nþ12

j

� �x̂: �

Considering the boundary condition (2.3) and Lemma 2.1, we get

ðunj Þxx�x�xt;2u

nþ12

j

� �¼ 1

skunþ1

xx k2 � kun

xxk2

� �;

unþ1

2j

� �x�x;2u

nþ12

j

� �¼ �2kunþ1

2x k2

;

unþ1

2j

� �x̂;2u

nþ12

j

� �¼ 0;

ð2:8Þ

B. Hu et al. / Applied Mathematics and Computation 204 (2008) 311–316 313

P;2unþ1

2j

� �¼ 2

3hXJ

j¼1

unþ1

2jþ1 þ u

nþ12

j þ unþ1

2j�1

� �� u

nþ12

j

� �x̂u

nþ12

j ¼ 13

XJ

j¼1

unþ1

2jþ1 þ u

nþ12

j þ unþ1

2j�1

� �� u

nþ12

jþ1 � unþ1

2j�1

� �u

nþ12

j

¼ 13

XJ

j¼1

unþ1

2jþ1 þ u

nþ12

j

� �u

nþ12

jþ1 unþ1

2j � 1

3

XJ

j¼1

unþ1

2j þ u

nþ12

j�1

� �u

nþ12

j�1 unþ1

2j ¼ 0: ð2:9Þ

Thus, (2.7) yields

1sðkunþ1k2 � kunk2Þ þ 1

sðkunþ1

xx k2 � kun

xxk2Þ ¼ �2kunþ1

2x k2

6 0:

We have

kunk2 þ kunxxk

26 kun�1k2 þ kun�1

xx k26 � � � 6 ku0k2 þ ku0

xxk26 C:

Then

kunk 6 C; kunxxk 6 C:

Using (2.5) and Cauchy–Schwarz inequality, we derive

kunxk

26 kunk � kun

xxk 612ðkunk2 þ kun

xxk2Þ 6 C: ð2:10Þ

Making use of Lemma 2.2, the theorem follows.

3. Solvability

The following Brouwer fixed point theorem will be needed in order to show the existence of solution for (2.1)–(2.3). Forthe proof, see [13].

Lemma 3.1 (Brouwer fixed point theorem). Let H be a finite dimensional inner product space, suppose that g : H! H iscontinuous and there exists an a > 0 such that ðgðxÞ; xÞ > 0 for all x 2 H with kxk ¼ a. Then there exists x� 2 H such that gðx�Þ ¼ 0and kx�k 6 a.

Theorem 3.1. There exists un 2 Z0h which satisfies the difference scheme (2.1)–(2.3).

Proof. In order to prove the theorem by the mathematical induction, we assume that u0; u1; . . . ;un which satisfy (2.1)–(2.3)exist for n 6 N � 1. Next prove there also exists unþ1 which satisfies (2.1)–(2.3).

Let g be a operator on Z0h defined by

gðvÞ ¼ 2v� 2un þ 2vxx�x�x � 2unxx�x�x � svx�x þ svx̂ þ

s3ðvjþ1 þ vj þ vj�1Þvx̂: ð3:1Þ

Then g is obviously continuous. Taking an inner product of (3.1) with v, and using

ðvx̂; vÞ ¼ 0; ððvjþ1 þ vj þ vj�1Þvx̂; vÞ ¼ 0;

we have

ðgðvÞ; vÞ ¼ 2kvk2 � 2ðun; vÞ þ 2kvxxk2 � 2ðunxx�x�x; vÞ � sðvx�x; vÞ ¼ 2kvk2 � 2ðun; vÞ þ 2kvxxk2 � 2ðun

xx; vxxÞ þ skvxk2

P 2kvk2 � 2kunk � kvk þ 2kvxxk2 � 2kunxxk � kvxxk þ skvxk2

P 2kvk2 � kunk2 � kvk2 þ 2kvxxk2 � kunxxk

2 � kvxxk2 þ skvxk2 ¼ kvk2 � kunk2 � kunxxk

2 þ kvxxk2 þ skvxk2

P kvk2 � kunk2 � kunxxk

2:

Hence, it is obvious that ðgðvÞ; vÞ > 0 for all v 2 Z0h with kvk2 ¼ kunk2 þ kun

xxk2 þ 1. It follows from Lemma 3.1 that there

exists v� 2 Z0h such that gðv�Þ ¼ 0. If we take unþ1 ¼ 2v� � un, then unþ1 satisfies (2.1)–(2.3). This completes the proof. h

4. Convergence and stability

Let vðx; tÞ be the solution of problem (1.2)–(1.4), vnj ¼ vðjh;nsÞ, then the truncation of the difference scheme (2.1)–(2.3)

is

rnj ¼ ðvn

j Þt þ ðvnj Þxx�x�xt � v

nþ12

j

� �x�xþ v

nþ12

j

� �x̂þ 1

3v

nþ12

jþ1 þ vnþ1

2j þ v

nþ12

j�1

� �� v

nþ12

j

� �x̂: ð4:1Þ

Making use of Taylor expansion, we know that rnj ¼ oðs2 þ h2Þ holds if h; s! 0.

314 B. Hu et al. / Applied Mathematics and Computation 204 (2008) 311–316

Lemma 4.1 (Discrete Gronwall inequality [12]). Suppose wðkÞ;qðkÞ are nonnegative function and qðkÞ is nondecreasing. IfC > 0, and

wðkÞ 6 qðkÞ þ CsXk�1

l¼0

wðlÞ 8k;

then

wðkÞ 6 qðkÞeCsk 8k:

Theorem 4.1. Suppose u0 2 H20½0; L�, then the solution un of the difference scheme (2.1)–(2.3) converges to the solution vðx; tÞ of

problem (1.2), (1.3), (1.4) in norm k � k1 and the rate of convergence is oðs2 þ h2Þ.

Proof. Subtracting (2.1) from (4.1) and letting enj ¼ vn

j � unj , we have

rnj ¼ ðen

j Þt þ ðenj Þxx�x�xt � e

nþ12

j

� �x�xþ e

nþ12

j

� �x̂þ 1

3v

nþ12

jþ1 þ vnþ1

2j þ v

nþ12

j�1

� �� v

nþ12

j

� �x̂� 1

3u

nþ12

jþ1 þ unþ1

2j þ u

nþ12

j�1

� �� u

nþ12

j

� �x̂: �

ð4:2Þ

Let

Q ¼ 13

vnþ1

2jþ1 þ v

nþ12

j þ vnþ1

2j�1

� �� v

nþ12

j

� �x̂� 1

3u

nþ12

jþ1 þ unþ1

2j þ u

nþ12

j�1

� �� u

nþ12

j

� �x̂:

Computing the inner product of (4.2) with 2enþ12, and using e

nþ12

j

� �x̂;2e

nþ12

j

� �¼ 0, we obtain

rnj ;2enþ1

2

� �¼ 1

sðkenþ1k2 � kenk2Þ þ 1

sðkenþ1

xx k2 � ken

xxk2Þ þ 2kenþ1

2x k2 þ Q ;2enþ1

2

� �:

That is

kenþ1k2 � kenk2 þ kenþ1xx k

2 � kenxxk

2 ¼ �2skenþ12

x k2 þ s rnj ;2enþ1

2

� �þ s �Q ;2enþ1

2

� �: ð4:3Þ

Similar to the proof of (2.9), we have

hXJ

j¼1

ðenþ12

jþ1 þ enþ1

2j þ e

nþ12

j�1 Þ � enþ1

2j

� �x̂e

nþ12

j ¼ 0:

Hence

�Q ;2enþ12

� �¼ �2

3hXJ

j¼1

vnþ1

2jþ1 þ v

nþ12

j þ vnþ1

2j�1

� �� v

nþ12

j

� �x̂e

nþ12

j þ 23

hXJ

j¼1

unþ1

2jþ1 þ u

nþ12

j þ unþ1

2j�1

� �� u

nþ12

j

� �x̂e

nþ12

j

¼ �23

hXJ

j¼1

enþ1

2jþ1 þ e

nþ12

j þ enþ1

2j�1

� �� e

nþ12

j

� �x̂e

nþ12

j � 23

hXJ

j¼1

enþ1

2jþ1 þ e

nþ12

j þ enþ1

2j�1

� �� u

nþ12

j

� �x̂e

nþ12

j

� 23

hXJ

j¼1

unþ1

2jþ1 þ u

nþ12

j þ unþ1

2j�1

� �� e

nþ12

j

� �x̂e

nþ12

j

¼ �23

hXJ

j¼1

enþ1

2jþ1 þ e

nþ12

j þ enþ1

2j�1

� �� u

nþ12

j

� �x̂e

nþ12

j � 23

hXJ

j¼1

unþ1

2jþ1 þ u

nþ12

j þ unþ1

2j�1

� �� e

nþ12

j

� �x̂e

nþ12

j :

By Theorem 2.1 and using Cauchy-Schwarz inequality, we can get

�Q ;2enþ12

� �6

23

ChXJ

j¼1

enþ1

2jþ1

��� ���þ enþ1

2j

��� ���þ enþ1

2j�1

��� ���� �e

nþ12

j

��� ���þ 23

ChXJ

j¼1

enþ1

2j

� �x̂

��� ��� � enþ1

2j

��� ��� 6 C enþ12

��� ���2þ e

nþ12

x̂

��� ���2� �

6 C kenþ1k2 þ kenk2 þ kenþ1x k2 þ ken

xk2

� �: ð4:4Þ

Noting that

rnj ;2enþ1

2

� �¼ ðrn

j ; enþ1 þ enÞ 6 krnk2 þ 1

2ðkenþ1k2 þ kenk2Þ; ð4:5Þ

from (4.3) and (4.4) we have

kenþ1k2 � kenk2 þ kenþ1xx k

2 � kenxxk

26 Cs kenþ1k2 þ kenk2 þ kenþ1

x k2 þ kenxk

2� �

þ skrnk2: ð4:6Þ

B. Hu et al. / Applied Mathematics and Computation 204 (2008) 311–316 315

Similar to (2.10) we can prove

kenþ1x k2

612ðkenþ1k2 þ kenþ1

xx k2Þ;

kenxk

26

12ðkenk2 þ ken

xxk2Þ:

ð4:7Þ

(4.6) can be changed into

kenþ1k2 þ kenþ1xx k

2 � ðkenk2 þ kenxxk

2Þ 6 Csðkenþ1k2 þ kenk2 þ kenþ1xx k

2 þ kenxxk

2Þ þ skrnk2: ð4:8Þ

Let Bn ¼ kenk2 þ kenxxk

2. Then (4.8) can be rewritten as follows:

ð1� CsÞðBnþ1 � BnÞ 6 2CsBn þ skrnk2:

If s is sufficiently small which satisfies 1� Cs ¼ d > 0, then

Bnþ1 � Bn6 CsBn þ Cskrnk2

: ð4:9Þ

Summing up (4.9) from 0 to n� 1, we have

Bn6 B0 þ Cs

Xn�1

l¼0

krlk2 þ CsXn�1

l¼0

Bl:

Noticing

sXn�1

l¼0

krlk26 ns max

06l6n�1krlk2

6 T � oðs2 þ h2Þ2; e0 ¼ 0;

we get B0 ¼ oðs2 þ h2Þ2. Hence we obtain

Bn6 oðs2 þ h2Þ2 þ Cs

Xn�1

l¼0

Bl:

From Lemma 4.1, we get

Bn6 oðs2 þ h2Þ2:

That is

kenk 6 oðs2 þ h2Þ; kenxxk 6 oðs2 þ h2Þ:

From (4.7), we have kenxk 6 oðs2 þ h2Þ. Using Lemma 2.2, we get kenk1 6 oðs2 þ h2Þ.

Finally, we can similarly prove results as follows:

Theorem 4.2. Under the conditions of Theorem 4.1, the solution un of (2.1)–(2.3) is stable for initial data in norm k � k1.

Theorem 4.3. The solution un of (2.1)–(2.3) is unique.

5. Numerical simulations

Consider the Rosenau–Burgers

ut þ uxxxxt � uxx þ ux þ uux ¼ 0; ðx; tÞ 2 ½0;1� � ½0;1�; ð5:1Þ

with an initial condition

uðx;0Þ ¼ u0ðxÞ ¼ x4ð1� xÞ4; x 2 ½0;1� ð5:2Þ

and the boundary conditions

uð0; tÞ ¼ uð1; tÞ ¼ 0; uxxð0; tÞ ¼ uxxð1; tÞ ¼ 0; t 2 ½0;1�: ð5:3Þ

We discretize the Eqs. (5.1)–(5.3) using a finite difference method as (2.1)–(2.3). Then we obtain a nonlinear system of dif-ference equations

ðunj Þt þ ðun

j Þxx�x�xt � unþ1

2j

� �x�xþ u

nþ12

j

� �x̂þ 1

3ðunþ1

2jþ1 þ u

nþ12

j þ unþ1

2j�1 Þ � u

nþ12

j

� �x̂¼ 0:

It is easy to solve the system of finite difference equations by Newton iterative algorithm.

Table 1The ratios of square error at various time step

h ¼ 0:2 h ¼ 0:1 h ¼ 0:05

t ¼ 0:1 3.85032e�2 1.65089e�2 5.69725e�3t ¼ 0:2 3.85008e�2 1.65036e�2 5.69210e�3t ¼ 0:3 3.84983e�2 1.64982e�2 5.68696e�3t ¼ 0:4 3.84959e�2 1.16492e�2 5.68183e�3t ¼ 0:5 3.84935e�2 1.64875e�2 5.67671e�3t ¼ 0:6 3.84911e�2 1.64822e�2 5.67159e�3t ¼ 0:7 3.84887e�2 1.64768e�2 5.66648e�3t ¼ 0:8 3.84863e�2 1.64751e�2 5.66138e�3t ¼ 0:9 3.84839e�2 1.64661e�2 5.65629e�3t ¼ 1 3.84815e�2 1.64608e�2 5.65121e�3

316 B. Hu et al. / Applied Mathematics and Computation 204 (2008) 311–316

Since we do not know the exact solution of (5.1) and (5.2), a comparison between the numerical solutions on a coarsemesh and those on a refine mesh is made. In order to obtain the error estimates we consider the solution on mesh h ¼ 1

160as the reference solution. In Table 1, we give ratios of square error at time step for various step sizes of h with fixed s ¼ 0:05.

From the numerical results, we get that the finite difference scheme of this paper is efficient.

Acknowledgement

The authors thank the referees for their valuable comments which improved this article.

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