High-order iterative schemes for a nonlinear Kirchhoff–Carrier wave equation associated with the...

Applied Mathematics and Computation 215 (2009) 1908–1925

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Applied Mathematics and Computation

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

The n-order iterative schemes for a nonlinear Kirchhoff–Carrier waveequation associated with the mixed inhomogeneous conditions

Le Xuan Truong a, Le Thi Phuong Ngoc b, Nguyen Thanh Long c,*

a Faculty of Economics and Commerce, Hoa Sen University, 08 Nguyen Van Trang Str., Dist. 1, Ho Chi Minh City, Viet Namb Nhatrang Educational College, 01 Nguyen Chanh Str., Nhatrang City, Viet Namc Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University, 227 Nguyen Van Cu Str., Dist. 5, Ho ChiMinh City, Viet Nam

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

Keywords:Nonlinear Kirchhoff–Carrier wave equationGalerkin methodThe convergence of order N

0096-3003/$ - see front matter � 2009 Elsevier Incdoi:10.1016/j.amc.2009.07.056

* Corresponding author.E-mail addresses: [email protected] (L.X.

[email protected] (N.T. Long).

In this paper, a high-order iterative scheme is established in order to get a convergentsequence at a rate of order NðN P 1Þ to a local unique weak solution of a nonlinear Kirch-hoff–Carrier wave equation associated with mixed nonhomogeneous conditions – theboundary conditions are Dirichlet in one part and Robin in other part of boundary. Onthe other hand, some numerical results were presented.

� 2009 Elsevier Inc. All rights reserved.

1. Introduction

In this paper we consider a nonlinear wave equation with the Kirchhoff–Carrier operator

utt � lðt; kuðtÞk2; kuxðtÞk2Þuxx ¼ f ðuÞ;0 < x < 1;0 < t < T; ð1:1Þ

uxð0; tÞ � huð0; tÞ ¼ gðtÞ;uð1; tÞ ¼ 0; ð1:2Þuðx; 0Þ ¼ ~u0ðxÞ;utðx; 0Þ ¼ ~u1ðxÞ; ð1:3Þ

where l; f ; ~u0; ~u1 are given functions satisfying conditions specified later and h P 0 are given constants. In Eq. (1.1), the non-linear term lðt; kuðtÞk2

; juxðtÞj2Þ depends on the integrals

kuðtÞk2 ¼Z 1

0juðx; tÞj2 dx; kuxðtÞk2 ¼

Z 1

0juxðx; tÞj2 dx: ð1:4Þ

Eq. (1.1) has its origin in the nonlinear vibration of an elastic string (Kirchhoff [7]), for which the associated equation is

qhutt ¼ P0 þEh2L

Z L

0

@u@yðy; tÞ

�� 2 dy

!uxx; ð1:5Þ

here u is the lateral deflection, q is the mass density, h is the cross section, L is the length, E is Young’s modulus and P0 is theinitial axial tension.

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Truong), [email protected], [email protected] (L.T.P. Ngoc), [email protected],

http://dx.doi.org/10.1016/j.amc.2009.07.056

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L.X. Truong et al. / Applied Mathematics and Computation 215 (2009) 1908–1925 1909

In [3], Carrier also established a model of the type

utt ¼ P0 þ P1

Z L

0u2ðy; tÞdy

� �uxx; ð1:6Þ

where P0 and P1 are constants.When f ¼ 0 and l ¼ lðkuxk2Þ is a function depending only on kuxk2

; the Cauchy or mixed problem for (1.1) has been stud-ied by many authors; see Ebihara et al. [5], Pohozaev [22] and the references therein. A survey of the results about the math-ematical aspects of Kirchhoff model can be found in Medeiros et al. [20,21].

In [19] Medeiros has studied the problem (1.1)–(1.3) with f ¼ f ðuÞ ¼ �bu2, where b is a given positive constant, and X is abounded open set of R3. In [6] Hosoya and Yamada have considered (1.1)–(1.3) with f ¼ f ðuÞ ¼ �djujau; where d > 0;a P 0are given constants.

In [10,12] the authors have studied the existence and uniqueness of the equation

utt þ kD2u� lðk 5 uk2ÞDuþ ejut ja�1ut ¼ Fðx; tÞ; x 2 X; t > 0; ð1:7Þ

where k > 0; e > 0;0 < a < 1, are given constants, and X is a bounded open set of Rn.In [11] Long and Diem have studied the linear recursive schemes associated with the nonlinear wave equation

utt � uxx ¼ f ðx; t;u;ux;utÞ; 0 < x < 1;0 < t < T; ð1:8Þ

associated with (1.2), (1.3). Afterwards, this result has been extended in [13] to the nonlinear wave equation with the Kir-chhoff operator

utt � lðkuxðtÞk2Þuxx ¼ f ðx; t;u;ux;utÞ; 0 < x < 1;0 < t < T; ð1:9Þ

associated with (1.3) and the Dirichlet homogeneous boundary condition.In this paper, we associate with Eq. (1.1) a recurrent sequence fumg defined by

@2um

@t2 � lðt; kumðtÞk2; kumxðtÞk2Þumxx ¼XN�1

k¼0

1k!

f ðkÞðum�1Þðum � um�1Þk; ð1:10Þ

0 < x < 1;0 < t < T; with um satisfying (1.2), (1.3). The first term u0 is chosen as u0 � ~u0. If l 2 C1ðR3þÞ; g 2 C3ðRþÞ and

f 2 CNðRÞ, we prove that the sequence fumg converges at a rate of order N to a unique weak solution of the problem(1.1)–(1.3). This result is a relative generalization of [2,4,11,13–18,23]. Finally, we present some numerical results.

2. The n-order iterative schemes

We will omit the definitions of the usual function spaces and denote them by the notationLp ¼ Lpð0;1Þ;Hm ¼ Hmð0;1Þ;Wm;p ¼Wm;pð0;1Þ (see [1]).

Let h�; �i be either the scalar product in L2 or the dual pairing of a continuous linear functional and an element of a functionspace. The notation k � k stands for the norm in L2 and we denote by k � kX the norm in the Banach space X. We call X 0 the dualspace of X. We denote by Lpð0; T; XÞ;1 6 p 61 for the Banach space of real functions u : ð0; TÞ ! X measurable, such that

kukLpð0;T;XÞ ¼Z T

0kuðtÞkp

X dt� �1=p

< þ1 for 1 6 p <1;

and

kukL1ð0;T;XÞ ¼ ess sup0<t<T

kuðtÞkX for p ¼ 1:

Let uðtÞ;u0ðtÞ ¼ utðtÞ ¼ _uðtÞ; u00ðtÞ ¼ uttðtÞ ¼ €uðtÞ; uxðtÞ ¼ 5uðtÞ; uxxðtÞ ¼ DuðtÞ, denote uðx; tÞ; @u@t ðx; tÞ; @

2u@t2 ðx; tÞ; @u

@x ðx; tÞ;@2u@x2 ðx; tÞ, respectively. With l 2 C1ðR3

þÞ;l ¼ lðt; y; zÞ, we put D1l ¼ @l@t ;D2l ¼ @l

@y and D3l ¼ @l@z .

Put

V ¼ fv 2 H1ðXÞ : vð1Þ ¼ 0g; ð2:1Þ

aðu;vÞ ¼Z 1

0uxðxÞvxðxÞdxþ huð0Þvð0Þ: ð2:2Þ

Then V is a closed subspace of H1 and on V three norms kvkH1 ; kvxk and kvkV ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaðv;vÞ

pare equivalent norms.

Then we have the following lemmas, the proofs of which are straightforward and their omitted:

Lemma 2.1. Let h P 0. Then the imbedding V ,!C0ð½0;1�Þ is compact and

kvkC0ð½0;1�Þ 6 kvxk 6 kvkV ;

1ffiffi2p kvkH1 6 kvxk 6 kvkV 6 maxf

ffiffiffihp

;1gkvkH1 ;

(ð2:3Þ

for all v 2 V.

1910 L.X. Truong et al. / Applied Mathematics and Computation 215 (2009) 1908–1925

Lemma 2.2. Let h P 0. Then the symmetric bilinear form að�; �Þ defined by (2.2) is continuous on V � V and coercive on V.

Lemma 2.3. Let h P 0. Then there exists the Hilbert orthonormal base f ~wjg of L2 consisting of the eigenfunctions ~wj correspondingto the eigenvalue kj such that

0 < k1 6 k2 6 . . . 6 kj 6 . . . ; limj!þ1

kj ¼ þ1;

að~wj;vÞ ¼ kjh~wj;vi for all v 2 V ; j ¼ 1;2; . . .

(ð2:4Þ

Furthermore, the sequence f ~wj=ffiffiffiffikj

pg is also the Hilbert orthonormal base of V with respect to the scalar product að�; �Þ.

On the other hand, we also have ~wj satisfying the following boundary value problem.

�D~wj ¼ kj ~wj; in X;

~wjxð0Þ � h ~wjð0Þ ¼ ~wjð1Þ ¼ 0; ~wj 2 C1ðXÞ:

(ð2:5Þ

The proof of Lemma 2.3 can be found in [24, p.87, Theorem 7.7], with H ¼ L2 and V ; að�; �Þ as defined by (2.1), (2.2).

We make the following assumptions:

[H1] h P 0;[ðH2Þ] g 2 C3ðRþÞ;[ðH3Þ] ~u0 2 V \ H2; ~u1 2 V ;

[ðH4Þ] l 2 C1ðR3þÞ, and there exist constants p > 1; l� > 0; li > 0; i 2 f0;1;2;3g, such that

(i) l� 6 lðt; y; zÞ 6 l0ð1þ yp þ zpÞ; for all ðt; y; zÞ 2 R3þ;

(ii) jD1lðt; y; zÞj 6 l1ð1þ yp þ zpÞ; for all ðt; y; zÞ 2 R3þ,

(iii) jD2lðt; y; zÞj 6 l2ð1þ yp�1 þ zpÞ; for all ðt; y; zÞ 2 R3þ,

(iv) jD3lðt; y; zÞj 6 l3ð1þ yp þ zp�1Þ; for all ðt; y; zÞ 2 R3þ,

[ðH5Þ] f 2 CNðRÞ.

In order to consider problem (1.1)–(1.3), we shall reformulate it as a problem with homogeneous boundary conditions asfollows. Let, for x 2 ½0;1� and t P 0,

uðx; tÞ ¼ 11þ h

ðx� 1ÞgðtÞ; ð2:6Þ

vðx; tÞ ¼ uðx; tÞ �uðx; tÞ; ð2:7Þ~v0ðxÞ ¼ ~u0ðxÞ �uðx;0Þ;~v1ðxÞ ¼ ~u1ðxÞ �utðx; 0Þ;

�ð2:8Þ

together with the consistency condition

gð0Þ ¼ ~u0xð0Þ � h~u0ð0Þ: ð2:9Þ

Then problem (1.1)–(1.3) is equivalent to the following initial value problem:

v tt � lðt; kvðtÞ þuðtÞk2; kvxðtÞ þuxðtÞk

2Þvxx ¼ f ðuþ vÞ �utt; 0 < x < 1; 0 < t < T;

vxð0; tÞ � hvð0; tÞ ¼ vð1; tÞ ¼ 0;vðx;0Þ ¼ ~v0ðxÞ;v tðx;0Þ ¼ ~v1ðxÞ:

8><>: ð2:10Þ

The weak formulation of the initial and boundary value problem (2.10) can be made in the following manner:Find vðtÞ defined in the open set ð0; TÞ such that vðtÞ verifies the following variational equation

hv ttðtÞ;wi þ lðt; kvðtÞ þuðtÞk2; kvxðtÞ þuxðtÞk

2ÞaðvðtÞ;wÞ ¼ hf ðuþ vÞ �utt;wi; ð2:11Þ

for all w 2 V , and the initial conditions

vðx;0Þ ¼ ~v0ðxÞ; v tðx;0Þ ¼ ~v1ðxÞ: ð2:12Þ

Given M > 0; T > 0, we put

M1 ¼ M1ðT; gÞ ¼ krukL1ð0;T;L2Þ þ krutkL1ð0;T;L2Þ ¼1

1þ hkgkW1;1ð0;TÞ; ð2:13Þ

K0 ¼ K0ðM; T; f Þ ¼ supfjf ðuÞj : juj 6 M þM1g; ð2:14ÞKi ¼ KiðM; T; f Þ ¼ K0ðM; T; f ðiÞÞ; ð2:15ÞbK i ¼max

06j6iKjðM; T; f Þ; ð2:16Þ

i ¼ 1; 2; . . . ; N.


For each M > 0 and T > 0, we put

WðM; TÞ ¼ fv 2 L1ð0; T; V \ H2Þ : v t 2 L1ð0; T; VÞ;v tt 2 L2ðQ TÞ; kvkL1ð0;T;V\H2Þ 6 M; kv tkL1ð0;T;VÞ 6 M; kv ttkL2ðQT Þ

6 Mg; ð2:17Þ

W1ðM; TÞ ¼ fv 2WðM; TÞ : v tt 2 L1ð0; T; L2Þg; ð2:18Þ

where QT ¼ X� ð0; TÞ.We shall choose the first term v0 ¼ 0. Suppose that

vm�1 2W1ðM; TÞ: ð2:19Þ

We associate with the problem (2.11), (2.12) the following variational problem.Find vm 2W1ðM; TÞ which satisfies the nonlinear variational problem

hv 00mðtÞ;wi þ lmðtÞaðvmðtÞ;wÞ ¼ hFmðtÞ;wi for all w 2 V ; ð2:20Þ

vmð0Þ ¼ ~v0; v 0mð0Þ ¼ ~v1; ð2:21Þ

where

lmðtÞ ¼ lðt; kvmðtÞ þuðtÞk2; k 5 vmðtÞ þ ruðtÞk2Þ; ð2:22Þ

Fmðx; tÞ ¼ fmðx; t;vmÞ ¼ �utt þXN�1

i¼0

1i!

f ðiÞðvm�1 þuÞðvm � vm�1Þi: ð2:23Þ

Then, we have the following theorem.

Theorem 2.4. Let assumptions ðH1Þ � ðH5Þ hold. Then there exist a constant M > 0 depending on ~u0; ~u1;l; g;h and a constantT > 0 depending on ~u0; ~u1;l; g; h; f such that, for v0 � 0, there exists a recurrent sequence fvmg �W1ðM; TÞ defined by (2.20)–(2.23).

Proof. The proof consists of several steps.Step 1: The Galerkin approximation (introduced by Lions [9]). Consider the basis for V as in Lemma 2.4 ðwj ¼ ~wj=

ffiffiffiffikj

pÞ. Put

v ðkÞm ðtÞ ¼Xk

j¼1

cðkÞmj ðtÞwj; ð2:24Þ

where the coefficients cðkÞmj satisfy the system of nonlinear differential equations

€v ðkÞm ðtÞ;wj

D Eþ lðkÞm ðtÞaðv ðkÞm ðtÞ;wjÞ ¼ FðkÞm ðtÞ;wj

D E; 1 6 j 6 k;

v ðkÞm ð0Þ ¼ ~v0k; €v ðkÞm ð0Þ ¼ ~v1k;

8<: ð2:25Þ

where

~v0k ¼Pkj¼1

aðkÞj wj ! ~v0 strongly in H2;

~v1k ¼Pkj¼1

bðkÞj wj ! ~v1 strongly in H1;

8>>>><>>>>: ð2:26Þ

lðkÞm ðtÞ ¼ l t; v ðkÞm ðtÞ þuðtÞ�� 2

; rv ðkÞm ðtÞ þ ruðtÞ�� 2

� �;

FðkÞm ðx; tÞ ¼ fm x; t;v ðkÞm

� ¼ �utt þ

PN�1

i¼0

1i! fðiÞðvm�1 þuÞ v ðkÞm � vm�1

� i:

8>>><>>>: ð2:27Þ

Let us suppose that vm�1 satisfies (2.19). Then it is clear that system (2.25) has a solution v ðkÞm ðtÞ on an interval0 6 t 6 TðkÞm 6 T. The following estimates allow one to take constant TðkÞm ¼ T for all m and k.

Step 2: A priori estimates. Put

SðkÞm ðtÞ ¼ XðkÞm ðtÞ þ Y ðkÞm ðtÞ þR t

0€v ðkÞm ðsÞ�� 2

ds;

sðkÞm ðtÞ ¼ xðkÞm ðtÞ þ yðkÞm ðtÞ þR t


ds;

8><>: ð2:28Þ


where

XðkÞm ðtÞ ¼ _v ðkÞm ðtÞ�� 2

þ lðkÞm ðtÞ v ðkÞm ðtÞ�� 2

V;

Y ðkÞm ðtÞ ¼ _v ðkÞm ðtÞ�� 2

Vþ lðkÞm ðtÞ Dv ðkÞm ðtÞ

�� 2;

xðkÞm ðtÞ ¼ _v ðkÞm ðtÞ�� 2

þ l� v ðkÞm ðtÞ�� 2

V;

yðkÞm ðtÞ ¼ _v ðkÞm ðtÞ�� 2

Vþ l� Dv ðkÞm ðtÞ

�� 2:

8>>>>>>>>>><>>>>>>>>>>:� ð2:29Þ

Then, it follows from (2.24), (2.25), (2.27), (2.28)1, (2.29)1,2, that

SðkÞm ðtÞ ¼ SðkÞm ð0Þ þZ t

0

_lðkÞm ðsÞ v ðkÞm ðsÞ�� 2

V þ Dv ðkÞm ðsÞ�� 2

h idsþ 2

Z t

0FðkÞm ðsÞ; _v ðkÞm ðsÞD E

ds

þ 2Z t

0a FðkÞm ðsÞ; _v ðkÞm ðsÞ�

ds� 2Z t

0FðkÞm ð1; sÞr _v ðkÞm ð1; sÞdsþ

Z t


ds ¼ SðkÞm ð0Þ þX5

j¼1

Ij: ð2:30Þ

We shall estimate, respectively, the following integrals on the right-hand side of (2.30).First integral I1 : we use the following notation:

lðkÞm ðtÞ ¼ l t; v ðkÞm ðtÞ þuðtÞ�� 2

; rv ðkÞm ðtÞ þ ruðtÞ�� 2

� ¼ l v ðkÞm ðtÞ

�:

By (2.27)1, we have

_lðkÞm ðtÞ ¼ D1l½v ðkÞm ðtÞ� þ 2D2l½v ðkÞm ðtÞ�hv ðkÞm ðtÞ þuðtÞ; _v ðkÞm ðtÞ þu0ðtÞi þ 2D3l½v ðkÞm ðtÞ�hrv ðkÞm ðtÞ þ ruðtÞ;r _v ðkÞm ðtÞþ ru0ðtÞi: ð2:31Þ

Put k� ¼ 1þ 1ffiffiffiffil�p . By using the assumption ðH4; ðiiÞ; ðiiiÞ; ðivÞÞ, and the following inequalities

v ðkÞm ðtÞ�� 6 v ðkÞm ðtÞ

�� C0ð½0;1�Þ 6 rv ðkÞm ðtÞ

�� 6 v ðkÞm ðtÞ��

V6

1ffiffiffiffiffiffil�p

ffiffiffiffiffiffiffiffiffiffiffiffisðkÞm ðtÞ

q; ð2:32Þ

_uðkÞm ðtÞ�� 6 ffiffiffiffiffiffiffiffiffiffiffiffi

sðkÞm ðtÞq

; ð2:33Þ

r _v ðkÞm ðtÞ�� 6 _v ðkÞm ðtÞ

�� V 6


q; ð2:34Þ

we deduce from (2.13), (2.31), that

_lðkÞm ðtÞ�� 6 l1 1þ 2 M1 þ k�


q� �2p !

þ 2l2 1þ M1 þ k�


q� �2p�2

þ M1 þ k�


q� �2p" #

M1 þ k�


q� �2

þ 2l3 1þ M1 þ k�


q� �2p

þ M1 þ k�


q� �2p�2" #

M1 þ k�


q� �2

6 ~l1 1þ sðkÞm ðtÞ þ sðkÞm ðtÞ� p þ sðkÞm ðtÞ

� pþ1h i

; ð2:35Þ

where

~l1 ¼ l1 1þ 22pM2p1 þ 22pk2p

�

� þ 2ðl2 þ l3Þ 2M2

1 þ 22p�1M2p1 þ 22pþ1M2pþ2

1 þ 2k2� þ 22pþ2k2pþ1

� þ 22pk2p�1�

� : ð2:36Þ

Using the inequality

sq6 1þ sN0 ; 8 s P 0; 8 q 2 ð0;N0�;N0 ¼maxfN � 1;2pþ 1g; ð2:37Þ

we get from (2.28), (2.29), (2.35), that

I1 ¼Z t

0

_lðkÞm ðsÞ v ðkÞm ðsÞ�� 2

Vþ Dv ðkÞm ðsÞ�� 2

h ids 6 ~l1

Z t

01þ sðkÞm ðsÞ þ sðkÞm ðsÞ

� p þ sðkÞm ðsÞ� pþ1

� 1l�

sðkÞm ðsÞds

64~l1

l�

Z t

01þ sðkÞm ðsÞ

� N0h i

ds 64~l1

l�T þ

Z t

0sðkÞm ðsÞ� N0 ds

� �: ð2:38Þ

We shall now require the following lemma.


Lemma 2.5. We have

FðkÞm ðtÞ�� 6 eK ð1ÞN

XN�1

i¼0

~ai


q� �i

; ð2:39Þ

rFðkÞm ðtÞ�� 6 eK ð2ÞN

XN�1

i¼0

~ai


q� �i

; ð2:40Þ

where eK ð1ÞN ¼ 11þh kg00kL1 þ bK N�1; eK ð2ÞN ¼ 1

1þh kg00kL1 þ M þ 11þh kgkL1

� bK N þ ðN � 1ÞbK N�1; and the constants ~ai; i ¼ 0;1; . . . ;N � 1are defined as follows

~a0 ¼ 1þ 12

XN�1

i¼1

ð2MÞi

i!; ~ai ¼

12i!

ffiffiffiffiffiffi2l�

s !i

; i ¼ 1; . . . ;N � 1: ð2:41Þ

Proof

(i) By (2.3), (2.6), (2.14), (2.15), (2.16), (2.19), (2.27)2, (2.32), we have

FðkÞm ðx; tÞ�� 6 1

1þ hjg00ðtÞj þ K0 þ

XN�1

i¼1

1i!

Ki v ðkÞm

�� þ vm�1j j� i

61

1þ hkg00kL1 þ bK N�1 1þ

XN�1

i¼1

1i!

v ðkÞm ðtÞ��

C0ð½0;1�Þ þ vm�1ðtÞk kC0ð½0;1�Þ

� i" #

61

1þ hkg00kL1 þ bK N�1 1þ

XN�1

i¼1

1i!

1ffiffiffiffiffiffil�p


qþM

� �i" #

6 eK ð1ÞN 1þ 12

XN�1

i¼1

ð2MÞi

i!þXN�1

i¼1

12i!


s !i ffiffiffiffiffiffiffiffiffiffiffiffisðkÞm ðtÞ

q� �i24 35 ¼ eK ð1ÞN

XN�1

i¼0

~ai


q� �i

; ð2:42Þ

with ~ai; i ¼ 0;1; . . . ;N � 1 are defined as (2.41). Hence, (2.39) is proved.(ii) By (2.27)2, we have

rFðkÞm ðx; tÞ ¼�1

1þ hg00ðtÞ þ f 0ðvm�1 þuÞðrvm�1ðtÞ þ ruðtÞÞ

þXN�1

i¼1

1i!

f ðiþ1Þðvm�1 þuÞðrvm�1 þruÞ v ðkÞm � vm�1� i

þXN�1

i¼1

ii!

f ðiÞðvm�1 þuÞ v ðkÞm � vm�1� i�1 rv ðkÞm �rvm�1

� : ð2:43Þ

Using the inequalities (2.3)1, (2.14), (2.15), it follows from (2.43), that

rFðkÞm ðx; tÞ�� 6 1

1þ hkg00kL1 þ K1 jrvm�1ðtÞj þ

11þ h

kgkL1

� �þXN�1

i¼1

1i!

Kiþ1 jrvm�1ðtÞj þ1

1þ hkgkL1

� �v ðkÞm ðtÞ��

V þ vm�1ðtÞk kV

� i

þXN�1

i¼1

ii!

Ki v ðkÞm ðtÞ��

V þ vm�1ðtÞk kV

� i�1rv ðkÞm

�� þ rvm�1j j�

: ð2:44Þ

Hence, we deduce from (2.3), (2.14), (2.15), (2.16), (2.19), (2.32) and (2.44), that

rFðkÞm ðtÞ�� 6 eK ð2ÞN 1þ

XN�1

i¼1

1i!

v ðkÞm ðtÞ��

VþM

� i" #

6 eK ð2ÞN 1þXN�1

i¼1

2i�1

i!1ffiffiffiffiffiffili�

p ffiffiffiffiffiffiffiffiffiffiffiffisðkÞm ðtÞ

q� �i

þMi

!" #6 eK ð2ÞN

XN�1

i¼0

~ai


q� �i

:

ð2:45Þ

Hence, (2.40) is proved. The proof of Lemma 2.5 is complete. h


We now return to the estimates for I2; I3.Second integral I2 : We again use inequality (2.37), and from (2.39), we have

I2 ¼ 2Z t

0FðkÞm ðsÞ; _v ðkÞm ðsÞD E

ds 6 2Z t

0FðkÞm ðsÞ�� _v ðkÞm ðsÞ

�� ds 6 2eK ð1ÞN

XN�1

i¼0

~ai

Z t

01þ sðkÞm ðsÞ

� N0h i

ds

6 2eK ð1ÞN

XN�1

i¼0

~ai T þZ t


� �: ð2:46Þ

Third integral I3: We have

I3 ¼ 2Z t

0a FðkÞm ðsÞ; _v ðkÞm ðsÞ�

ds 6 2Z t

0FðkÞm ðsÞ��

V_v ðkÞm ðsÞ��

V ds 6 2Z t

0FðkÞm ðsÞ��

V

ffiffiffiffiffiffiffiffiffiffiffiffisðkÞm ðsÞ

qds: ð2:47Þ

Using the inequality kvkC0ðXÞ 6ffiffiffi2pkvkH1 for all v 2 H1, we deduce from (2.39), (2.40) that

FðkÞm ðtÞ��

V6

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2hþ 1

pFðkÞm ðtÞ��

H16


pFðkÞm ðtÞ�� þ rFðkÞm ðtÞ

�� h i ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2hþ 1

p eK ð1ÞN þ eK ð2ÞN

� XN�1

i¼0

~ai


q� �i

: ð2:48Þ

Hence, it follows from (2.37), (2.47) and (2.48), that

I3 6 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2hþ 1


� XN�1

i¼0

~ai

Z t

0

ffiffiffiffiffiffiffiffiffiffiffiffisðkÞm ðsÞ

q� �iþ1

ds 6 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2hþ 1


� XN�1

i¼0

~ai

Z t

01þ sðkÞm ðsÞ

� N0h i

ds

6 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2hþ 1


� XN�1

i¼0

~ai T þZ t


� �: ð2:49Þ

Fourth integral: I4 : Note that, by (2.6), (2.27)2, we have FðkÞm ð1; tÞ ¼ f ð0Þ.Hence

I4 ¼ �2Z t

0FðkÞm ð1; sÞr _v ðkÞm ð1; sÞds ¼ �2f ð0Þ

Z t

0r _v ðkÞm ð1; sÞds 6 2jf ð0Þj rv ðkÞm ð1; tÞ

�� þ 2jf ð0Þr~v0kð1Þj: ð2:50Þ

On the other hand

rv ðkÞm ð1; tÞ�� 6 ffiffiffiffiffiffi

2l�

s ffiffiffiffiffiffiffiffiffiffiffiffisðkÞm ðtÞ

q: ð2:51Þ

Hence

jI4j 6 2jf ð0Þj rv ðkÞm ð1; tÞ�� þ 2jf ð0Þr~v0kð1Þj 6 2


sjf ð0Þj


qþ 2jf ð0Þr~v0kð1Þj

612

sðkÞm ðtÞ þ4l�

f 2ð0Þ þ 2jf ð0Þr~v0kð1Þj: ð2:52Þ

Fifth integral I5: Eq. (2.25)1 can be rewritten as follows

€v ðkÞm ðtÞ;wj� �

� lðkÞm ðtÞ Dv ðkÞm ðx; tÞ;wj� �

¼ FðkÞm ðtÞ;wj

D E; 1 6 j 6 k: ð2:53Þ

Hence, it follows after replacing wj with €v ðkÞm ðtÞ and integrating that

I5 ¼Z t


ds 6 2Z t

0FðkÞm ðsÞ�� 2

dsþ 2Z t

0lðkÞm ðsÞ� 2

Dv ðkÞm ðsÞ�� 2

ds ¼ Ið1Þ5 þ Ið2Þ5 : ð2:54Þ

We shall estimate step by step two integrals Ið1Þ5 ; Ið2Þ5 .Estimate Ið1Þ5 :

Using the inequalities (2.37) andPN�1

i¼0 ai

� 26 N

PN�1i¼0 a2

i , for all a0; a1; . . . ; aN�1 2 R, it follows from (2.39), that

Ið1Þ5 ¼ 2Z t

0FðkÞm ðsÞ�� 2

ds 6 2N eK ð1ÞN

� 2Z t

0

XN�1

i¼0

~a2i sðkÞm ðsÞ� i

ds 6 2N eK ð1ÞN

� 2Z t

0

XN�1

i¼0

~a2i 1þ sðkÞm ðsÞ

� N0h i

ds

6 2N eK ð1ÞN

� 2XN�1

i¼0

~a2i T þ

Z t


� �: ð2:55Þ


Estimate Ið2Þ5 :

By using the assumption ðH4; ðiÞÞ; we deduce from (2.13), (2.27)1, (2.32), that

lðkÞm ðtÞ�� 6 l0 1þ v ðkÞm ðtÞ þuðtÞ

�� 2p þ rv ðkÞm ðtÞ þ ruðtÞ�� 2p

h i6 l0 1þ 22p v ðkÞm ðtÞ

�� 2p

V þM2p1

� h i6 l0 1þ 22p 1

lp�

sðkÞm ðtÞ� p þM2p

1

� �� 6 l0 1þ ð2M1Þ2p þ 4

l�

� �p� �1þ sðkÞm ðtÞ

� ph i

: ð2:56Þ

Hence, we obtain from (2.37), (2.56), that

Ið2Þ5 ¼ 2Z t

0lðkÞm ðsÞ� 2

DuðkÞm ðsÞ�� 2

ds 62l2

0

l�1þ ð2M1Þ2p þ 4

l�

� �p� �2 Z t

01þ ðsðkÞm ðsÞÞ

p �2sðkÞm ðsÞds

68l2

0


l�

� �p� �2 Z t

0½1þ sðkÞm ðsÞ

� N0 �ds 68l2

0


l�

� �p� �2

T þZ t

0ðsðkÞm ðsÞÞ

N0

� �ds

¼ ~l5 T þZ t

0ðsðkÞm ðsÞÞ

N0

� �ds: ð2:57Þ

where

~l5 ¼8l2

0


l�

� �p� �2

: ð2:58Þ

It follows from (2.54), (2.55), (2.57), that

I5 6 Kð1ÞN ðT þZ t

0sðkÞm ðsÞ� N0 dsÞ; ð2:59Þ

where

Kð1ÞN ¼ ~l5 þ 2N ~Kð1ÞN

� 2XN�1

i¼0

~a2i : ð2:60Þ

Now, we need an estimate on the term SðkÞm ð0Þ. We have

SðkÞm ð0Þ ¼ k~v1kk2 þ að~v1k; ~v1kÞ þ lð0; k~v0k þuð0Þk2; k 5 ~v0k þruð0Þk2Þ½að~v0k; ~v0kÞ þ kD~v0k2�: ð2:61Þ

By means of the convergences (2.26) we can deduce the existence of a constant M > 0 independent of k and m such that

SðkÞm ð0Þ þ4l�

f 2ð0Þ þ 2jf ð0Þr~v0kð1Þj 6 M2=8: ð2:62Þ

Finally, it follows from (2.30), (2.38), (2.46), (2.49), (2.52), (2.59), (2.62), that

sðkÞm ðtÞ 6 M2=4þ TeC1ðM; TÞ þ eC1ðM; TÞZ t

0sðkÞm ðsÞ� N0 ds; for 0 6 t 6 T ðkÞm 6 T; ð2:63Þ

where

eC1ðM; TÞ ¼ 8~l1

l�þ 2Kð1ÞN þ 4 eK ð1ÞN þ



� � XN�1

i¼0

~ai: ð2:64Þ

Then, we have the following Lemma.

Lemma 2.6. There exists a constant T > 0 independent of k and m such that

sðkÞm ðtÞ 6 M2 8 t 2 ½0; T�; for all k and m: ð2:65Þ

Proof. Put

yðtÞ ¼ M2=4þ TeC1ðM; TÞ þ eC 1ðM; TÞZ t

0sðkÞm ðsÞ� N0 ds;0 6 t 6 T: ð2:66Þ

Clearly

yðtÞ > 0; 0 6 sðkÞm ðtÞ 6 yðtÞ; 0 6 t 6 T;

y0ðtÞ 6 eC1ðM; TÞyN0 ðtÞ; 0 6 t 6 T;

yð0Þ ¼ M2=4þ TeC1ðM; TÞ:

8>><>>: ð2:67Þ


Put ZðtÞ ¼ y1�N0 ðtÞ, after integrating of (2.67)

ZðtÞP ðM2=4þ TeC1ðM; TÞÞ1�N0 � ðN0 � 1ÞeC1ðM; TÞt P ðM2=4þ TeC1ðM; TÞÞ1�N0 � ðN0 � 1ÞeC1ðM; TÞT;8t 2 ½0; T�:ð2:68Þ

Notice that, from (2.64) we have

limT!0þ ½ðM2=4þ TeC1ðM; TÞÞ1�N0 � ðN0 � 1ÞeC1ðM; TÞT� ¼ ðM2=4Þ1�N0 > ðM2Þ1�N0 : ð2:69Þ

Then, from (2.69), we can always choose the constant T > 0 such that

½ðM2=4þ TeC1ðM; TÞÞ1�N0 � ðN0 � 1ÞeC1ðM; TÞT� > ðM2Þ1�N0 : ð2:70Þ

Finally, it follows from (2.67), (2.68) and (2.70), that

0 6 sðkÞm ðtÞ 6 yðtÞ ¼ 1ffiffiffiffiffiffiffiffiffiZðtÞN0�1

p 6 M2;8t 2 ½0; T�: ð2:71Þ

The proof of Lemma 2.6 is complete. h

Remark 2.1. The function XðtÞ ¼ ½ðM2=4þ T ~C1ðM; TÞÞ1�N0 � ðN0 � 1Þ~C1ðM; TÞt�1

1�N0 ; 0 6 t 6 T; is the maximal solution of thefollowing Volterra integral equation with non-decreasing kernel [8].

XðtÞ ¼ M2=4þ TeC1ðM; TÞ þ eC1ðM; TÞZ t

0ðXðsÞÞN0 ds;0 6 t 6 T: ð2:72Þ

By Lemma 2.6, we can take constant TðkÞm ¼ T for all m and k. Therefore, we have

v ðkÞm 2WðM; TÞ for all m and k: ð2:73Þ

From (2.73) we can extract from v ðkÞm

n oa subsequence v ðkjÞ

m

n osuch that

v ðkjÞm ! vm in L1ð0; T; V \ H2Þ weak�;

_v ðkjÞm ! v 0m in L1ð0; T; VÞ weak�;

€v ðkjÞm ! v 00m in L2ðQ TÞ weak;

8>><>>: ð2:74Þ

vm 2WðM; TÞ: ð2:75Þ

We can easily check from (2.25), (2.26), (2.73), (2.74) that vm satisfies (2.20)–(2.23) in L2ð0; TÞ; weak. On the other hand, itfollows from (2.20)1 and vm 2WðM; TÞ that

v 00m ¼ lðt; kvmðtÞ þuðtÞk2; k 5 vmðtÞ þ ruðtÞk2ÞDvm þ Fm 2 L1ð0; T; L2Þ: ð2:76Þ

Hence vm 2W1ðM; TÞ and the proof of Theorem 2.4 is complete.

Theorem 2.7. Let assumptions ðH1Þ—ðH5Þ hold. Then, there exist constants M > 0 and T > 0 such that

(i) The problem (2.10) has a unique weak solution v 2W1ðM; TÞ.(ii) On the other hand, the recurrent sequence fvmg defined by 2.20, 2.21, 2.22, 2.23 converges at a rate of order N to the solu-

tion v strongly in the space W1ðTÞ ¼ fv 2 L1ð0; T; VÞ : v 0 2 L1ð0; T; L2Þg in the sense

kvm � vkL1ð0;T;VÞ þ v 0m � v 0��

L1ð0;T;L2Þ 6 C kvm�1 � vkL1ð0;T;VÞ þ v 0m�1 � v 0��

L1ð0;T;L2Þ

� N;

for all m P 1, where C is a suitable constant.

Furthermore, we have also the estimation

kvm � vkL1ð0;T;VÞ þ kv 0m � v 0kL1ð0;T;L2Þ 6 CTðkTÞNm

; ð2:77Þ

for all m P 1, where CT and kT < 1 are positive constants depending only on T.

Proof. (a) Existence of the solution. First, we note that W1ðTÞ is a Banach space with respect to the norm (see [9]):

kvkW1ðTÞ ¼ kvkL1ð0;T;VÞ þ kv 0kL1ð0;T;L2Þ:


We shall prove that fvmg is a Cauchy sequence in W1ðTÞ. Let wm ¼ vmþ1 � vm. Then wm satisfies the variational problem

w00mðtÞ;w� �

þ lmþ1ðtÞaðwmðtÞ;wÞ ¼ ðlmþ1ðtÞ � lmðtÞÞhDvmðtÞ;wi þ hFmþ1ðtÞ � FmðtÞ;wi 8 w 2 V ;

wmð0Þ ¼ w0mð0Þ ¼ 0;

(ð2:78Þ

where

lmðtÞ ¼ lðt; kvmðtÞ þuðtÞk2; k 5 vmðtÞ þ ruðtÞk2Þ; ð2:79Þ

Fmðx; tÞ ¼ �utt þXN�1

i¼0

1i!

f ðiÞðvm�1 þuÞðvm � vm�1Þi: ð2:80Þ

Taking w ¼ w0m in (2.78)1, after integrating in t we get

rmðtÞ ¼Z t

0l0mþ1ðsÞkwmðsÞk2

V dsþ 2Z t

0ðlmþ1ðsÞ � lmðsÞÞ DvmðsÞ;w0mðsÞ

� �dsþ 2

Z t

0Fmþ1ðsÞ � FmðsÞ;w0mðsÞ� �

ds ¼X3

k¼1

~Jk;

ð2:81Þ

where

rmðtÞ ¼ kw0mðtÞk2 þ lmþ1ðtÞkwmðtÞk2

V P kw0mðtÞk2 þ l�kwmðtÞk2

V ¼ dmðtÞ: ð2:82Þ

We shall estimate all integrals eJk; k ¼ 1;2;3.First, by using the assumption ðH4; ðiiÞ; ðiiiÞ; ðivÞÞ, we deduce from (2.79), that

l0mþ1ðtÞ�� 6 l1ð1þ kvmþ1ðtÞ þuðtÞk2p þ k5 vmþ1ðtÞ þ ruðtÞk2pÞ þ 2l2ð1þ kvmþ1ðtÞ þuðtÞk2p�2

þ k5 vmþ1ðtÞ þ ruðtÞk2pÞ � kvmþ1ðtÞ þuðtÞk v 0mþ1ðtÞ þu0ðtÞ�� þ 2l3ð1þ kvmþ1ðtÞ þuðtÞk2p

þ k5 vmþ1ðtÞ þ ruðtÞk2p�2Þ � k 5 vmþ1ðtÞ þ ruðtÞk rv 0mþ1ðtÞ þ ru0ðtÞ�� : ð2:83Þ

By the following inequalities

kvmþ1ðtÞ þuðtÞk 6 kvmþ1kL1ð0;T;VÞ þM1 6 M þM1 � m1;

k 5 vmþ1ðtÞ þ ruðtÞk 6 kvmþ1kL1ð0;T;VÞ þM1 6 m1;

v 0mþ1ðtÞ þu0ðtÞ�� 6 v 0mþ1

�� L1ð0;T;L2Þ þM1 6 m1;

rv 0mþ1ðtÞ þ ru0ðtÞ�� 6 v 0mþ1

�� L1ð0;T;VÞ þM1 6 m1;

we deduce from (2.83) that:

l0mþ1ðtÞ�� 6 l1 1þ 2m2p

1

� þ 2ðl2 þ l3Þ 1þm2p�2

1 þm2p1

� m2

1 � eM1: ð2:84Þ

On the other hand

lmþ1ðtÞ � lmðtÞ ¼ lðt; kvmþ1ðtÞ þuðtÞk2; k 5 vmþ1ðtÞ þ ruðtÞk2Þ � lðt; kvmðtÞ þuðtÞk2

; k 5 vmðtÞ þ ruðtÞk2Þ

¼ D2l t; kð1Þm ; kð2Þm

� ½kvmþ1ðtÞ þuðtÞk2 � kvmðtÞ þuðtÞk2� þ D3l t; kð1Þm ; kð2Þm

� ½k 5 vmþ1ðtÞ

þ ruðtÞk2 � k5 vmðtÞ þ ruðtÞk2�; ð2:85Þ

where

kð1Þm ¼ kð1Þm ðtÞ ¼ hð1Þm kvmþ1ðtÞ þuðtÞk2 þ ð1� hð1Þm ÞkvmðtÞ þuðtÞk2;

kð2Þm ¼ kð2Þm ðtÞ ¼ hð1Þm k 5 vmþ1ðtÞ þ ruðtÞk2 þ ð1� hð1Þm Þk 5 vmðtÞ þ ruðtÞk2;

(ð2:86Þ

with 0 < hð1Þm < 1.Note that

0 6 kð1Þm 6 maxfkvmðtÞ þuðtÞk2; kvmþ1ðtÞ þuðtÞk2g 6 m2

1: ð2:87Þ

Similarly

0 6 kð2Þm 6 m21: ð2:88Þ


So

D2l t; kð1Þm ; kð2Þm

� �� 6 l2 1þ kð1Þm

� p�1þ kð2Þm

� p� �

6 l2 1þm2p�21 þm2p

1

� � l2m2; ð2:89Þ

D3l t; kð1Þm ; kð2Þm

� �� 6 l3 1þ kð1Þm

� pþ kð2Þm

� p�1� �

6 l3m2: ð2:90Þ

Then, we get

jlmþ1ðtÞ � lmðtÞj 6 2ðl2 þ l3Þm2m1kwmðtÞkV � eM2kwmðtÞkV : ð2:91Þ

Hence, it follows from (2.82), (2.84), (2.91), that

eJ1

�� ¼ Z t

0l0mþ1ðsÞ

�� wmðsÞk2V dsj6

Z t

0jl0mþ1ðsÞjkwmðsÞk2

V ds6 eM11l�

Z t

0dmðsÞds; ð2:92Þ

jeJ2j ¼2jZ t

0ðlmþ1ðsÞ�lmðsÞÞhDvmðsÞ;w0mðsÞids 62

Z t

0

�� lmþ1ðsÞ�lmðsÞjkDvmðsÞk w0mðsÞ�� ds62 eM2M

Z t

0dmðsÞds: ð2:93Þ

Furthermore, by using Taylor’s expansion of the function f ðvm þuÞ around the point vm�1 þu up to order N � 1, weobtain

f ðvm þuÞ � f ðvm�1 þuÞ ¼XN�1

i¼1

1i!

f ðiÞðvm�1 þuÞwim�1 þ

1N!

f ðNÞ kð3Þm

� wN

m�1; ð2:94Þ

where kð3Þm ¼ kð3Þm ðx; tÞ ¼ vm�1 þ h3ðvm � vm�1Þ;0 < h3 < 1.It implies from (2.80), (2.94) that

Fmþ1ðx; tÞ � Fmðx; tÞ ¼XN�1

i¼1

1i!

f ðiÞðvm þuÞwim þ

1N!

f ðNÞ kð3Þm

� wN

m�1: ð2:95Þ

Then

kFmþ1ðtÞ � FmðtÞk 6XN�1

i¼1

Ki

i!kwmðtÞki

V þKN

N!kwm�1ðtÞkN

V 6XN�1

i¼1

Ki

i!kwmðtÞki�1

V kwmðtÞkV þKN

N!kwm�1ðtÞkN

V

6

XN�1

i¼1

Ki

i!ð2MÞi�1ffiffiffiffiffiffil�p

ffiffiffiffiffiffiffiffiffiffiffiffidmðtÞ

qþ KN

N!ffiffiffiffiffiffil�p� N

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidm�1ðtÞ

q� �N

¼ qð1ÞT

ffiffiffiffiffiffiffiffiffiffiffiffidmðtÞ

qþ qð2ÞT

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidm�1ðtÞ

q� �N

; ð2:96Þ

where

qð1ÞT ¼XN�1

i¼1

Ki

i!ð2MÞi�1ffiffiffiffiffiffil�p ; qð2ÞT ¼

KN

N!ð ffiffiffiffiffiffil�p ÞN

: ð2:97Þ

Then we deduce that

eJ3 ¼ 2Z t

0hFmþ1ðsÞ � FmðsÞ;w0mðsÞids 6 2

Z t

0kFmþ1ðsÞ � FmðsÞkkw0mðsÞkds

6 2Z t

0qð1ÞT

ffiffiffiffiffiffiffiffiffiffiffiffidmðsÞ

qþ qð2ÞT

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidm�1ðsÞ

q� �N" # ffiffiffiffiffiffiffiffiffiffiffiffi

dmðsÞq

ds 6 ð2qð1ÞT þ qð2ÞT ÞZ t

0dmðsÞdsþ qð2ÞT

Z T

0dN

m�1ðsÞds: ð2:98Þ

Combining (2.81), (2.92), (2.93) and (2.98), we have

dmðtÞ 6 qð2ÞT

Z T

0dN

m�1ðsÞdsþ qð3ÞT

Z t

0dmðsÞds; ð2:99Þ

where

qð3ÞT ¼eM1

l�þ 2 eM2M þ 2qð1ÞT þ qð2ÞT : ð2:100Þ

By using Gronwall’s lemma, we obtain from (2.99) that

kwmkW1ðTÞ 6 lTkwm�1kNW1ðTÞ; ð2:101Þ


where lT is the constant given by

lT ¼ 1þ 1ffiffiffiffiffiffil�p

� � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiTqð2ÞT ð1þ l�Þ

N exp Tqð3ÞT

� r: ð2:102Þ

Hence, we obtain from (2.101) that

kvm � vmþpkW1ðTÞ 6 ð1� kTÞ�1ðlTÞ�1

N�1ðkTÞNm

; ð2:103Þ

for all m and p where kT ¼ MðlTÞ1

N�1 < 1. It follows that fvmg is a Cauchy sequence in W1ðTÞ: Then there exists v 2W1ðTÞ suchthat vm ! v strongly in W1ðTÞ: Thus, and by applying a similar argument as used in the proof of Theorem 2.4, v 2W1ðM; TÞ isthe unique weak solution of problem (1.10). Passing to the limit as p! þ1 for m fixed, we obtain estimate (2.77) from(2.103). This completes the proof of Theorem 2.7. h

3. Numerical results

Consider the following problem:

v tt � lðt; kvðtÞ þuðtÞk2; kvxðtÞ þuxðtÞk

2Þvxx ¼ f ðuðtÞ þ vÞ �utt þ Fðx; tÞ; ð3:1Þ

0 < x < 1;0 < t < T , with boundary conditions

vxð0; tÞ � vð0; tÞ ¼ vð1; tÞ ¼ 0; ð3:2Þ

and initial conditions

vðx;0Þ ¼ ~v0ðxÞ;v tðx;0Þ ¼ ~v1ðxÞ; ð3:3Þ
where
f ðuÞ ¼ arctgu; ð3:4Þ

uðx; tÞ ¼ 12ðx� 1Þe�t; ð3:5Þ

lðt; y; zÞ ¼ 1þ yþ z; ð3:6Þ~v0ðxÞ ¼ �2x2 þ xþ 1; ~v1ðxÞ ¼ �~v0ðxÞ; ð3:7Þ

Fðx; tÞ ¼ Vexðx; tÞ þ 4e�t 1þ 5930

e�2t

� �þ 1

2ðx� 1Þe�t � f ðVex þuÞ; ð3:8Þ

gðtÞ ¼ e�t; ð3:9Þ
where
Vexðx; tÞ ¼ ð�2x2 þ xþ 1Þe�t: ð3:10Þ

The exact solution of the problem (3.1)–(3.3) with f ;uðx; tÞ;lðt; y; zÞ; ~v0ðxÞ; ~v1ðxÞ; Fðx; tÞ and gðtÞ defined in (3.4)–(3.9)respectively, is the function Vex given in (3.10).

We associate with problem (3.1)–(3.3) a 3-order iterative sequence fvmg defined by

v 00m � lðt; kvmðtÞ þuðtÞk2; kvmxðtÞ þuxðtÞk

2Þvmxx ¼ fmðx; t;vmÞ; ð3:11Þ

0 < x < 1;0 < t < T; with boundary conditions

vmxð0; tÞ � vmð0; tÞ ¼ vmð1; tÞ ¼ 0; ð3:12Þ

and initial conditions

vmðx; 0Þ ¼ ~v0ðxÞ; v 0mðx;0Þ ¼ ~v1ðxÞ; ð3:13Þ

where

fmðx; t;vmÞ ¼ �utt þ Fðx; tÞ þX2

k¼0

1k!

f ðkÞðvm�1 þuÞðvm � vm�1Þk ¼ ½f 0ðvm�1 þuÞ � f 00ðvm�1 þuÞvm�1�vm þ ~f mðx; t;vmÞ;

ð3:14Þ

with

~f mðx; t; vmÞ ¼ f ðvm�1 þuÞ � f 0ðvm�1 þuÞvm�1 þ12

f 00ðvm�1 þuÞ v2m þ v2

m�1

� �utt þ Fðx; tÞ

¼ arctgðvm�1 þuÞ � vm�1

1þ ðvm�1 þuÞ2� vm�1 þuð1þ ðvm�1 þuÞ2Þ2

v2m þ v2

m�1

� �utt þ Fðx; tÞ; m P 1: ð3:15Þ

The first term v0 is chosen as v0 � 0.


To solve problem (3.11)–(3.15) numerically, we consider the differential system for the unknownsUjðtÞ � vmðxj; tÞ;VjðtÞ ¼

dUj

dt ðtÞ, with xj ¼ jDx;Dx ¼ 1N ; j ¼ 0;1; . . . ;N :

dUj

dt ðtÞ ¼ VjðtÞ; j ¼ 0;N � 1;

dV0dt ðtÞ ¼ lðt; kUðtÞ þuðtÞk2

; kUxðtÞ þuxðtÞk2Þ �ð1þDxÞU0ðtÞþU1ðtÞ

ðDxÞ2þ bðmÞ0 ðtÞU0ðtÞ þ ~f mðx0; t;U0ðtÞÞ;

dVj

dt ðtÞ ¼ lðt; kUðtÞ þuðtÞk2; kUxðtÞ þuxðtÞk

2Þ Uj�1ðtÞ�2UjðtÞþUjþ1ðtÞðDxÞ2

þ bðmÞj ðtÞUjðtÞ þ ~f mðxj; t;UjðtÞÞ; j ¼ 1;N � 2;

dVN�1dt ðtÞ ¼ lðt; kUðtÞ þuðtÞk2

; kUxðtÞ þuxðtÞk2Þ UN�2ðtÞ�2UN�1ðtÞ

ðDxÞ2þ bðmÞN�1ðtÞUN�1ðtÞ þ ~f mðxN�1; t;UN�1ðtÞÞ;

Ujð0Þ ¼ ~v0ðxjÞ;Vjð0Þ ¼ ~v1ðxjÞ; j ¼ 0;N � 1;

8>>>>>>>>>>>>><>>>>>>>>>>>>>:ð3:16Þ

and where

UðtÞ ¼ Uðx; tÞ ¼XN�1

j¼0

UjðxÞUjðtÞ; ð3:17Þ

UjðxÞ ¼x�xj�1

Dx ; xj�1 6 x 6 xj;xjþ1�x

Dx ; xj 6 x 6 xjþ1; j ¼ 1;N � 1;

(ð3:18Þ

U0ðxÞ ¼x1�xDx ; 0 6 x 6 x1;

0; x1 6 x 6 1;

�ð3:19Þ

bðmÞj ðtÞ ¼ f 0ðvm�1ðxj; tÞ;uðxj; tÞÞ � f 00ðvm�1ðxj; tÞ;uðxj; tÞÞvm�1ðxj; tÞ

¼ 1

1þ ðvm�1ðxj; tÞ þuðxj; tÞÞ2þ 2ðvm�1ðxj; tÞ þuðxj; tÞÞvm�1ðxj; tÞ½1þ ðvm�1ðxj; tÞ þuðxj; tÞÞ2�2

; j ¼ 0;N � 1;m P 1; ð3:20Þ

~f mðxj; t;UjðtÞÞ ¼ f vm�1ðxj; tÞ þuðxj; tÞ�

� f 0ðvm�1ðxj; tÞ þuðxj; tÞÞvm�1ðxj; tÞ

þ 12

f 00ðvm�1ðxj; tÞ þuðxj; tÞÞ U2j ðtÞ þ v2

m�1ðxj; tÞ�

�uttðxj; tÞ þ Fðxj; tÞ

¼ arctgðvm�1ðxj; tÞ þuðxj; tÞÞ �vm�1ðxj; tÞ

1þ ðvm�1ðxj; tÞ þuðxj; tÞÞ2

� vm�1ðxj; tÞ þuðxj; tÞð1þ ðvm�1ðxj; tÞ þuðxj; tÞÞ2Þ2

U2j ðtÞ þ v2

m�1ðxj; tÞ�

�uttðxj; tÞ þ Fðxj; tÞ; j ¼ 0;N � 1; m P 1: ð3:21Þ

To solve the nonlinear differential system (3.16), we use the following linear recursive scheme generated by the nonlinearterms lðt; kUðtÞ þuðtÞk2

; kUxðtÞ þuxðtÞk2Þ; ~f mðxj; t;UjðtÞÞ; j ¼ 0;N � 1.

dUðnÞj

dt ðtÞ ¼ V ðnÞj ðtÞ; j ¼ 0;N � 1;

dVðnÞ0

dt ðtÞ ¼ lnðtÞ�ð1þDxÞUðnÞ

0ðtÞþUðnÞ

1ðtÞ

ðDxÞ2þ bðmÞ0 ðtÞU

ðnÞ0 ðtÞ þ ~f mðx0; t;U

ðn�1Þ0 ðtÞÞ;

dVðnÞj

dt ðtÞ ¼ lnðtÞUðnÞ

j�1ðtÞ�2UðnÞ

jðtÞþUðnÞ

jþ1ðtÞ

ðDxÞ2þ bðmÞj ðtÞU

ðnÞj ðtÞ þ ~f mðxj; t;U

ðn�1Þj ðtÞÞ; j ¼ 1;N � 2;

dVðnÞN�1dt ðtÞ ¼ lnðtÞ

UðnÞN�2ðtÞ�2UðnÞ

N�1ðtÞðDxÞ2

þ bðmÞN�1ðtÞUðnÞN�1ðtÞ þ ~f mðxN�1; t;U

ðn�1ÞN�1 ðtÞÞ;

UðnÞj ð0Þ ¼ ~v0ðxjÞ; V ðnÞj ð0Þ ¼ ~v1ðxjÞ; j ¼ 0;N � 1;

UðnÞðtÞ ¼ UðnÞðx; tÞ ¼PN�1

j¼0 UjðxÞUðnÞj ðtÞ;

lnðtÞ ¼ l t; kUðn�1ÞðtÞ þuðtÞk2; Uðn�1Þ

x ðtÞ þuxðtÞ�� 2

� �; n P 1;

Uð0Þj ðtÞÞ ¼ V ð0Þj ðtÞ � 0; j ¼ 1;N � 2:

8>>>>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>:

ð3:22Þ


Then system (3.22) is equivalent to:

ð3:23Þ

where

an ¼ anðtÞ ¼ lnðtÞðDxÞ2

; cðmÞj ¼ cðmÞj ðtÞ ¼ �2anðtÞ þ bðmÞj ðtÞ;

~cðmÞ0 ¼ ð1� DxÞan þ cðmÞ0 ;

bðmÞj ¼ bðmÞj ðtÞ ¼ 11þðvm�1ðxj ;tÞþuðxj ;tÞÞ2

þ 2ðvm�1ðxj ;tÞþuðxj ;tÞÞvm�1ðxj ;tÞ½1þðvm�1ðxj ;tÞþuðxj ;tÞÞ2 �2

; j ¼ 0;N � 1; m P 1:

FðnÞj ¼ FðnÞj ðtÞ ¼ ~f m xj; t;Uðn�1Þj ðtÞ

� ¼ arctgðvm�1ðxj; tÞ þuðxj; tÞÞ �

vm�1ðxj ;tÞ1þðvm�1ðxj ;tÞþuðxj ;tÞÞ2

� vm�1ðxj ;tÞþuðxj ;tÞð1þðvm�1ðxj ;tÞþuðxj ;tÞÞ2Þ2

Uðn�1Þj ðtÞ

� 2þ v2

m�1ðxj; tÞ� �

�uttðxj; tÞ þ Fðxj; tÞ; j ¼ 0;N � 1; m P 1:

8>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>:

ð3:24Þ

Rewritten (3.23)

ddt

XðtÞ ¼ AðtÞXðtÞ þ FðtÞ; ð3:25Þ

XðtÞ ¼ UðnÞ0 ðtÞ;UðnÞ1 ðtÞ; . . . ;UðnÞN�1ðtÞ; V ðnÞ0 ðtÞ;V

ðnÞ1 ðtÞ; . . . ;V ðnÞN�1ðtÞ

� T2 R2N;

FðtÞ ¼ 0;0; . . . ;0; FðnÞ0 ; FðnÞ1 ; . . . ; FðnÞN�1

� T2 R2N ;

AðtÞ ¼O EeAðtÞ O

� �;

8>>>>>><>>>>>>:ð3:26Þ

ð3:27Þ

To solve the linear differential system (3.25), we use the following difference scheme

Xiþ1 ¼ ðI þ DtAðtiÞÞXi þ DtFðtiÞ; i ¼ 0;N1 � 1;

X0 ¼ ð~v0ðx0Þ; ~v0ðx1Þ; . . . ; ~v0ðxN�1Þ; ~v1ðx0Þ; ~v1ðx1Þ; . . . ; ~v1ðxN�1ÞÞT ;

(ð3:28Þ

with Xi ¼ XðtiÞ; ti ¼ iDt;Dt ¼ TN1; i ¼ 0;N1.

The 3-order iterative scheme:Let N;N1 be given. We get

xj ¼ jDx;Dx ¼ 1N; j ¼ 0;N;

ti ¼ iDt;Dt ¼ TN1

; i ¼ 0;N1:

Iterative steps with m as follows.

Step 1.First-time, with m ¼ 0 : v0ðxj; tiÞ � 0; i ¼ 0;N1; j ¼ 0;N � 1.


Step 2.Let proceed to the ðm� 1Þ – time, we get vm�1ðxj; tiÞ; i ¼ 0;N1; j ¼ 0;N � 1. Then we shall obtainvmðxj; tiÞ; i ¼ 0;N1; j ¼ 0;N � 1 by steps in turn as follows:

Step 2.1. Iterative steps with n as follows.Step 2.1.1: With n ¼ 0 : Uð0Þj ðtiÞ ¼ 0; i ¼ 0;N1; j ¼ 0;N � 1.
Step 2.1.2: Let proceed to the ðn� 1Þ� time, we get
Uðn�1Þj ðtiÞ; i ¼ 0;N1; j ¼ 0;N � 1:

Then we shall obtain UðnÞj ðtiÞ; i ¼ 0;N1; j ¼ 0;N � 1 by steps in turn as follows:

(i) We calculate the matrix AðtiÞ ¼O EeAðtiÞ O

� �and vector

FðtiÞ ¼ ð0;0; . . . ;0; FðnÞ0 ðtiÞ; FðnÞ1 ðtiÞ; . . . ; FðnÞN�1ðtiÞÞT ;

as in formulas (3.24), (3.26) and (3.27):

an ¼ anðtiÞ ¼ lnðtiÞðDxÞ2

;

cðmÞj ¼ cðmÞj ðtiÞ ¼ �2anðtiÞ þ bðmÞj ðtiÞ; i ¼ 0;N1;

~cðmÞ0 ¼ ~cðmÞ0 ðtiÞ ¼ ð1� DxÞanðtiÞ þ cðmÞ0 ðtiÞ; i ¼ 0;N1;

bðmÞj ¼ bðmÞj ðtiÞ ¼ 11þðvm�1ðxj ;tiÞþuðxj ;tiÞÞ2

þ 2ðvm�1ðxj ;tiÞþuðxj ;tiÞÞvm�1ðxj ;tiÞ½1þðvm�1ðxj ;tiÞþuðxj ;tiÞÞ2 �2

;

j ¼ 0;N � 1; i ¼ 0;N1;m P 1:

FðnÞj ¼ FðnÞj ðtiÞ ¼ ~f mðxj; ti;Uðn�1Þj ðtiÞÞ; j ¼ 0;N � 1; i ¼ 0;N1:

8>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>:

ð3:29Þ

lnðtiÞ ¼ lðti; kUðn�1ÞðtiÞ þuðtiÞk2; kUðn�1Þ

x ðtiÞ þuxðtiÞk2Þ ¼ 1þ 13

e�2ti

þ 16ðDxÞ2 � 1

2Dx� 1

� �e�ti Uðn�1Þ

0 ðtiÞ � ðDxÞ2e�ti Uðn�1ÞN�1 ðtiÞ

� Dxe�tiXN�2

j¼1ð1� jDxÞUðn�1Þ

j ðtiÞ þDx3þ 1

Dx

� �jUðn�1Þ

0 ðtiÞj2 ð3:30Þ

þ 2 1þ 1Dx

� �XN�1

j¼1

jUðn�1Þj ðtiÞj2 þ

Dx3� 2

Dx

� �XN�1

j¼1

jUðn�1Þj�1 ðtiÞjjUðn�1Þ

j ðtiÞj:

(ii) By (3.28), we get Xi; i ¼ 1;N1. Then

Xi ¼ XðtiÞ ¼ UðnÞ0 ðtiÞ;UðnÞ1 ðtiÞ; . . . ;UðnÞN�1ðtiÞ;V ðnÞ0 ðtiÞ;V ðnÞ1 ðtiÞ; . . . ;V ðnÞN�1ðtiÞ� T

;

i ¼ 1;N1:

ð3:31Þ

Let proceed to the n-time (n is large enough) such that

max16i6N1

XN�1

j¼0

UðnÞj ðtiÞ � Uðn�1Þj ðtiÞ

�� < e; ð3:32Þ

where e is given error. Then

UðnÞj ðtiÞ ’ UjðtiÞ ¼ vmðxj; tiÞ:

Step 2.2. We end iterative steps with m when the following condition satisfies:

max06i6N1

XN�1

j¼0

jvmðxj; tiÞ � vm�1ðxj; tiÞj < e; ð3:33Þ

where e is given error as above.In the sequel, we present the numerical results with N ¼ N1 ¼ 10; T ¼ 1

10.

Table 1

m Em

1 8.799999999999999 3.342 0.0078337985131453 0:149047441055927� 10�13

4 0.0


Using Matlap program, we obtain the following table (Table 1) of details of the 3-order iterative scheme, wherem ¼ 1;2;3;4 and (3.22) is proceeded to the five-time, i.e. n ¼ 5. in which

Em �max06i6N1

XN�1

j¼0

jvmðxj; tiÞ � Vexðxj; tiÞj: ð3:35Þ

Remark 3.1. For comparison, we present Figs. 1–3 as below. Figs. 1 and 2, respectively, are the graphical representations forcurved surfaces of the exact solution ðx; tÞ# Vexðx; tÞ, and of the approximate solution ðx; tÞ# v3ðx; tÞ; ðm ¼ 3Þ, where v3ðx; tÞis interpolation polynomial in variables ðx; tÞ, with respect to discontinuous values ðxj; tiÞ; 0 6 i; j 6 10. In Fig. 3, we havedrawn the exact solution x # Vex x; T

2

� by the continuous curve, and we have drawn the approximate solution x # v3 x; T

2

� , by

the discontinuous curve. As in Fig. 3, we can see that two curves are close together.

Fig. 2. Approximate solution ðx; tÞ# v3ðx; tÞ.

Fig. 1. Exact solution ðx; tÞ# Vexðx; tÞ.

Table 2

m Em (1-order) Em (2-order)

1 8.799999999999999 8.7999999999999992 2.352489503627864 0.8555380776969793 0.126320678360079 0.0111092369259884 0.001356809748464 0:1372690722829806 � 10�3 3.36

5 0:2959411667335932 � 10�5 0:3310823283586561 � 10�6

6 0:6765487614579513 � 10�9 0:6819814157843496 � 10�10

7 0.00 0.00

Fig. 3. Exact solution x # Vex x; T2

� and approximate solution x # v3 x; T

2

� , with t ¼ 0:05.


Remark 3.2. With the values of N;N1; T;n are given as above, using the same method, we have established the 1-order and2-order iterative schemes, in which errors Em in (3.35) have been obtained as follows:

Compare Table 1 with Table 2, clearly, Table 1 is much better. The 3-order iterative scheme is much better than the 1-order and 2-order iterative schemes corresponding iterative steps with m, for example with m ¼ 3, error E3 in Table 1 is0:149047441055927 � 10�13, on the other hand, errors E3 in Table 2 are 0:126320678360079;0:11109236925988 � 10�1.

Acknowledgement

The authors wish to express their sincere thanks to the referees for their valuable comments.

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