Globally bounded solutions of a system of nonlinear functional differential equations with iterated deviating argument

Applied Mathematics and Computation 219 (2012) 2180–2185

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Globally bounded solutions of a system of nonlinear functional differential equations with iterated deviating argument Stevo Stevic´ Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, 11000 Beograd, Serbia

a r t i c l e

i n f o

Keywords: Functional differential equation Globally bounded solutions Bounded first derivative Iterated deviating argument

a b s t r a c t Some sufficient conditions which guarantee the existence of a bounded solution together with its first derivative on the whole real line of a system of nonlinear functional differential equations with an iterated deviating argument dependent of unknown function, are given. Ó 2012 Elsevier Inc. All rights reserved.

1. Introduction Various questions of the theory of nonlinear differential or nonlinear functional differential equations, not solved (or partially solved) with respect to the highest-order derivatives, have been considered for a long time, see, for example, [1–7,9–17,31] (see also the references therein). One of the systems of functional differential equations whose particular cases have been studied considerably is the following:

x0 ðtÞ ¼ AxðtÞ þ Fðt; xðtÞ; xðf ðt; xðtÞÞÞ; x0 ðgðtÞÞÞ;

ð1Þ

where t 2 R; A ¼ diagðA1 ; A2 Þ; A1 and A2 are real constant matrices of dimensions p  p and q  q, respectively, p þ q ¼ N, and Fðt; x; y; zÞ; f ðt; xÞ, and gðtÞ are real-valued functions continuous for t 2 R and x; y; z 2 RN . Some iteration methods for approximating fixed points which are iterations of some iterative processes are introduced in our papers [18–22] (for related ideas in theory of difference equations see also [23–29]). This idea naturally suggests investigation of equations with continuous arguments, whose deviations of an argument depend on an unknown function which depend also of the function. These deviations are called iterated deviations, and we introduced them in [30] for the case of functional equations. Our aim is to present some sufficient conditions that guarantee the existence of bounded solutions together with its first derivative of the following system of nonlinear functional differential equations on R

x0 ðtÞ ¼ AxðtÞ þ Fðt; xðtÞ; xðv 1 ðtÞÞ; x0 ðgðtÞÞÞ;

ð2Þ

v 1 ðtÞ ¼ u1 ðt; xðu2 ðt; . . . xðuk ðt; xðtÞÞÞ . . .ÞÞÞ:

ð3Þ

where

A = diag(A1 ; A2 ), A1 and A2 are real constant matrices of dimensions p  p and q  q, respectively, p þ q ¼ N, F : R  RN  RN  RN ! RN ; uj : R  RN ! R, j ¼ 1; k; g : R ! R, and xðtÞ is an unknown function. For related results in the literature, see, for example, [2,8,10,12,14,15]. E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2012.08.063

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2. Main result In this section we prove the main result in this paper which is incorporated in the following theorem: Theorem 1. Suppose that the following conditions are satisfied: (i) the following inequalities hold

jeA1 t j 6 N1 ea1 t

and jeA2 t j 6 N2 ea2 t ;

t 2 Rþ ;

ð4Þ

where a1 , a2 ; N 1 , and N 2 are positive constants; (ii) functions Fðt; x; y; zÞ; uj ðt; xÞ; j ¼ 1; k; gðtÞ, are continuous for t 2 R; x; y; z 2 RN , and

supjFðt; 0; 0; 0Þj ¼ M f < 1; t2R

(iii) functions Fðt; x; y; zÞ; uj ðt; xÞ; j ¼ 1; k, satisfy the following Lipschitz conditions

jFðt; x1 ; y1 ; z1 Þ  Fðt; x2 ; y2 ; z2 Þj 6 L1 jx1  x2 j þ L2 jy1  y2 j þ L3 jz1  z2 j; juj ðt; x1 Þ  uj ðt; x2 Þj 6 lj jx1  x2 j;

ð5Þ

j ¼ 1; k;

ð6Þ N

N

N

where Li ; lj ; i 2 f1; 2; 3g, j ¼ 1; k, are positive constants, ðt; x1 ; y1 ; z1 Þ, ðt; x2 ; y2 ; z2 Þ 2 R  R  R  R , and ðt; x1 Þ; ðt; x2 Þ 2 R  RN ; (iv) there is a q 2 ð0; 1Þ such that

jAj þ

Q 3 k K j ji¼1 li 1X L2 X Lj þ < 1; q j¼1 q j¼1 ð1  qÞj

ð7Þ

! Q  3 k X X K j ji¼1 li N1 N2 6 q; Lj þ L2 þ j a1 a2 j¼1 j¼1 ð1  qÞ where

ð8Þ

       N1 N2 N1 N2 ; Mf jAj þ1 : K ¼ max M f þ þ

a1

a2

a1

a2

Then, for sufficiently small Li ; lj ; i 2 f1; 2; 3g, j ¼ 1; k, system of equations (2) has a solution continuously differentiable and bounded for t 2 R together with its first derivative. Proof. Consider the system of integral equations

xðtÞ ¼

Z

þ1

Gðt  sÞFðs; xðsÞ; xðv 1 ðsÞÞ; x0 ðgðsÞÞÞds;

ð9Þ

1

where

v 1 ðtÞ is defined in (3) and (

GðtÞ ¼

diagðeA1 t ; 0Þ;

t>0

diagð0; eA2 t Þ; t < 0:

By definition it is easy to see that function GðtÞ satisfies the following conditions: (a) if E denote the N  N identity matrix, then

Gðþ0Þ  Gð0Þ ¼ E; (b) for every t – 0

G0 ðtÞ ¼ AGðtÞ; from which it follows that every solution of system (9) is a solution of system (2). We will find a solution of system (9) by the method of successive approximations and also show that it is unique. Let sequence ðxn ðtÞÞn2N0 be defined as follows:

x0 ðtÞ ¼ 0; xnþ1 ðtÞ ¼

Z

x00 ðtÞ ¼ 0; þ1

1

ð10Þ

0 Gðt  sÞFðs; xn ðsÞ; xn ðv ð1Þ n ðsÞÞ; xn ðgðsÞÞÞds;

0 x0nþ1 ðtÞ ¼ Axnþ1 ðtÞ þ Fðt; xn ðtÞ; xn ðv ð1Þ n ðtÞÞ; xn ðgðtÞÞÞ;

n 2 N0 ;

ð11Þ

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where

v ðjÞ n ðtÞ ¼ uj ðt; xn ðujþ1 ðt; . . . xn ðuk ðt; xn ðtÞÞÞ . . .ÞÞÞ; We may regard that We show that

v

ðkþ1Þ ðtÞ n

j ¼ 1; k:

ð12Þ

¼ t.

jxnþ1 ðtÞ  xn ðtÞj 6 Kqn ;

ð13Þ

jx0nþ1 ðtÞ  x0n ðtÞj 6 Kqn ;

ð14Þ

and

for all t 2 R and n 2 N0 , where q 2 ð0; 1Þ is such that (7) and (8) hold. We have

 Z t Z þ1  Gðt  sÞFðs; 0; 0; 0Þds 6 N1 ea1 ðtsÞ jFðs; 0; 0; 0Þjds þ N2 ea2 ðstÞ jFðs; 0; 0; 0Þjds 1 1 t   N1 N2 : 6 Mf þ

Z  jx1 ðtÞ  x0 ðtÞj ¼ 

þ1

a1

a2

From this it follows that

    N1 N2 þ1 : jx01 ðtÞ  x00 ðtÞj ¼ jAx1 ðtÞ þ Fðt; 0; 0; 0Þj 6 jAjjx1 ðtÞj þ jFðt; 0; 0; 0Þj 6 M f jAj þ

a1

a2

Hence, (13) and (14) hold for n ¼ 0. If (13) and (14) hold for an n  1, then

jx0n ðtÞj 6

n1 n1 X X jx0iþ1 ðtÞ  x0i ðtÞj 6 Kqi 6 i¼0

i¼0

K : 1q

ð15Þ

Using (15) and the inductive hypothesis we also have that Z þ1  Z þ1   ð1Þ 0 0  jxnþ1 ðtÞ  xn ðtÞj ¼  Gðt  sÞFðs; xn ðsÞ; xn ðv ð1Þ ð s ÞÞ; x ðgð s ÞÞÞd s  Gðt  s ÞFð s ; x ð s Þ; x ð v ð s ÞÞ; x ðgð s ÞÞÞd s n1 n1 n n n1 n1  1 1 Z þ1 Z þ1 ð1Þ 6 L1 jGðt  sÞjjxn ðsÞ  xn1 ðsÞjds þ L2 jGðt  sÞjjxn ðv ð1Þ n ðsÞÞ  xn ðv n1 ðsÞÞjds 1 1 Z þ1 Z þ1 ð1Þ ð1Þ þ L2 jGðt  sÞjjxn ðv n1 ðsÞÞ  xn1 ðv n1 ðsÞÞjds þ L3 jGðt  sÞjjx0n ðgðsÞÞ  x0n1 ðgðsÞÞjds 1 1 ! Z þ1 3 X KL2 ð1Þ ð1Þ 6 jGðt  sÞj Kqn1 Lj þ ð16Þ jv ðsÞ  v n1 ðsÞj ds: 1q n 1 j¼1 By using condition (6) and inequality (15) we obtain ð1Þ

ð2Þ

ð2Þ

ð2Þ jv nð1Þ ðsÞ  v n1 ðsÞj ¼ ju1 ðt; xn ðv ð2Þ n ðsÞÞÞ  u1 ðt; xn1 ðv n1 ðsÞÞÞj 6 l1 jxn ðv n ðsÞÞ  xn1 ðv n1 ðsÞÞj ð2Þ

6 l1 ðjxn ðv nð2Þ ðsÞÞ  xn1 ðv nð2Þ ðsÞÞj þ jxn1 ðv nð2Þ ðsÞÞ  xn1 ðv n1 ðsÞÞjÞ   K ð2Þ jv ð2Þ 6 l1 Kqn1 þ n ðsÞ  v n1 ðsÞj : 1q

ð17Þ

Assume that we have proved the inequality ð1Þ

jv nð1Þ ðsÞ  v n1 ðsÞj 6 Kqn1

m2 X j¼0

Kj

Qj

i¼0 liþ1 j

ð1  qÞ

þ

K m1

Qm2 i¼0

liþ1

ð1  qÞm1

ðmÞ

jv ðmÞ n ðsÞ  v n1 ðsÞj;

ð18Þ

for some 2 6 m 6 k. We have ðmÞ

ðmþ1Þ

ðmþ1Þ

jv nðmÞ ðsÞ  v n1 ðsÞj ¼ jum ðt; xn ðv ðmþ1Þ ðsÞÞÞ  um ðt; xn1 ðv n1 ðsÞÞÞj 6 lm jxn ðv nðmþ1Þ ðsÞÞ  xn1 ðv n1 ðsÞÞj n ðmþ1Þ

6 lm ðjxn ðv nðmþ1Þ ðsÞÞ  xn1 ðv nðmþ1Þ ðsÞÞj þ jxn1 ðv nðmþ1Þ ðsÞÞ  xn1 ðv n1 ðsÞÞjÞ   K ðmþ1Þ ð s Þ  v ð s Þj : jv ðmþ1Þ 6 lm Kqn1 þ n1 1q n

ð19Þ

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Using (19) in (18) we get ð1Þ

jv nð1Þ ðsÞ  v n1 ðsÞj 6 Kqn1

m2 j Qj X K i¼0 liþ1

þ

K m1

Qm2

i¼0 liþ1 lm m1

 Kqn1 þ

 K ðmþ1Þ s s jv ðmþ1Þ ð Þ  v ð Þj n1 1q n

ð1  qÞ ð1  qÞj Q Qm1 j j m m1 XK K ðmþ1Þ i¼0 liþ1 i¼0 liþ1 þ jv nðmþ1Þ ðsÞ  v n1 ðsÞj: ¼ Kqn1 j ð1  qÞm j¼0 ð1  qÞ j¼0

ð20Þ

From (17), (20) and the induction we have that (18) holds for 2 6 m 6 k þ 1. If we take m ¼ k þ 1 in (18) we obtain ð1Þ

jv nð1Þ ðsÞ  v n1 ðsÞj 6 Kqn1

k1 j Qj X K i¼0 liþ1 j¼0

ð1  qÞ

j

þ

Kk

Qk1

i¼0 liþ1 k

ð1  qÞ

ðkþ1Þ

jv nðkþ1Þ ðsÞ  v n1 ðsÞj ¼ Kqn1

k1 j Qj X K i¼0 liþ1 j¼0

ð1  qÞj

:

ð21Þ

Using (21) in (16) and then employing condition (8), we get

! 3 k1 j Qj X X K i¼0 liþ1 KL2 n1 ds jxnþ1 ðtÞ  xn ðtÞj 6 jGðt  sÞj Kq Lj þ Kq j 1q 1 j¼1 j¼0 ð1  qÞ !  Z Q  Z þ1 3 k t X X K j ji¼1 li a1 ðtsÞ a2 ðstÞ  N 6 Kqn1 Lj þ L2 e d s þ N e d s 1 2 j 1 t j¼1 j¼1 ð1  qÞ ! Q   j 3 k X X K j i¼1 li N1 N2 6 Kqn : 6 Kqn1 Lj þ L2 þ j a1 a2 j¼1 j¼1 ð1  qÞ Z

þ1

n1

ð22Þ

Using (5), (21) and (7), we get ð1Þ

0 0 jx0nþ1 ðtÞ  x0n ðtÞj ¼ jAðxnþ1 ðtÞ  xn ðtÞÞ þ Fðt; xn ðtÞ; xn ðv ð1Þ n ðtÞÞ; xn ðgðtÞÞÞ  Fðt; xn1 ðtÞ; xn1 ðv n1 ðtÞÞ; xn1 ðgðtÞÞÞj ð1Þ

6 jAjjxnþ1 ðtÞ  xn ðtÞj þ L1 jxn ðtÞ  xn1 ðtÞj þ L2 jxn ðv nð1Þ ðtÞÞ  xn ðv n1 ðtÞÞj ð1Þ

ð1Þ

þ L2 jxn ðv n1 ðtÞÞ  xn1 ðv n1 ðtÞÞj þ L3 jx0n ðgðtÞÞ  x0n1 ðgðtÞÞj Q 3 k X X K j ji¼1 li 6 jAjKqn þ Kqn1 Lj þ L2 Kqn1 6 Kqn : j j¼1 j¼1 ð1  qÞ

ð23Þ

Hence, (13) and (14) hold for every t 2 R and n 2 N0 . Using (13) and (14) and the fact 0 < q < 1, we see that the vector series

x0 ðtÞ þ

1 X ðxnþ1 ðtÞ  xn ðtÞÞ n¼0

and

x00 ðtÞ þ

1 X ðx0nþ1 ðtÞ  x0n ðtÞÞ n¼0

converge uniformly on R, which implies that the sequences ðxn ðtÞÞn2N0 and ðx0n ðtÞÞn2N0 converge to a continuously differentiable function, say xðtÞ, and its derivative respectively, which is a solution of system (9). Moreover, by letting n ! 1 in the following inequality

jxn ðtÞj 6

n1 n1 X X jxiþ1 ðtÞ  xi ðtÞj 6 Kqi 6 i¼0

i¼0

K ; 1q

ð24Þ

we get that xðtÞ is bounded on R, while letting n ! 1 in inequality (15) we get that its first derivative is bounded too, and both functions are bounded by K=ð1  qÞ. Now we prove that solution xðtÞ is unique. Assume to the contrary that there is another continuously differentiable solution zðtÞ on R, which is bounded together with its first derivative and such that xðtÞ – zðtÞ. Let

  kx  zk1 ¼ max supjxðtÞ  zðtÞj; supjx0 ðtÞ  z0 ðtÞj : t2R

t2R

Let

v j ðtÞ ¼ uj ðt; xðujþ1 ðt; . . . xðuk ðt; xðtÞÞÞ . . .ÞÞÞ uj ðtÞ ¼ uj ðt; zðujþ1 ðt; . . . zðuk ðt; zðtÞÞÞ . . .ÞÞÞ; j ¼ 1; k. We may regard that

v kþ1 ðtÞ ¼ ukþ1 ðtÞ ¼ t. Then we have

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 Z þ1 Z þ1   jxðtÞ  zðtÞj ¼  Gðt  sÞFðs; xðsÞ; xðv 1 ðsÞÞ; x0 ðgðsÞÞÞds  Gðt  sÞFðs; zðsÞ; zðu1 ðsÞÞ; z0 ðgðsÞÞÞds 1 Z 1 þ1 6 jGðt  sÞjðL1 jxðsÞ  zðsÞj þ L2 jxðv 1 ðsÞÞ  zðu1 ðsÞÞj þ L3 jx0 ðgðsÞÞ  z0 ðgðsÞÞjds Z1 þ1 6 jGðt  sÞjðL1 jxðsÞ  zðsÞj þ L2 jxðv 1 ðsÞÞ  xðu1 ðsÞÞjÞds 1 Z þ1 þ jGðt  sÞjðL2 jxðu1 ðsÞÞ  zðu1 ðsÞÞj þ L3 jx0 ðgðsÞÞ  z0 ðgðsÞÞjÞds 1 ! Z þ1 3 X KL2 6 jGðt  sÞj kx  zk1 Lj þ jv 1 ðsÞ  u1 ðsÞj ds: 1q 1 j¼1

ð25Þ

We have

jv 1 ðsÞ  u1 ðsÞj ¼ ju1 ðs; xðv 2 ðsÞÞÞ  u1 ðs; zðu2 ðsÞÞÞj 6 l1 jxðv 2 ðsÞÞ  zðu2 ðsÞÞj  6 l1 ðjxðv 2 ðsÞÞ  xðu2 ðsÞÞj þ jxðu2 ðsÞÞ  zðu2 ðsÞÞjÞ 6 l1 kx  zk1 þ

 K jv 2 ðsÞ  u2 ðsÞj : 1q

ð26Þ

Assume that we have proved the inequality

jv 1 ðsÞ  u1 ðsÞj 6 kx  zk1

m2 X j¼0

Kj

Qj

i¼0 liþ1 j

ð1  qÞ

þ

K m1

Qm2 i¼0

liþ1

ð1  qÞm1

jv m ðsÞ  um ðsÞj;

ð27Þ

for some 2 6 m 6 k. We have

jv m ðsÞ  um ðsÞj ¼ jum ðt; xðv mþ1 ðsÞÞÞ  um ðt; zðv mþ1 ðsÞÞÞj 6 lm jxðv mþ1 ðsÞÞ  zðumþ1 ðsÞÞj 6 lm ðjxðv mþ1 ðsÞÞ  xðumþ1 ðsÞÞj þ jxðumþ1 ðsÞÞ  zðumþ1 ðsÞÞjÞ   K jv mþ1 ðsÞ  umþ1 ðsÞj : 6 lm kx  zk1 þ 1q

ð28Þ

Using (28) in (27) we get

jv 1 ðsÞ  u1 ðsÞj 6 kx  zk1

m 2 X j¼0

¼ kx  zk1

Kj

Qj

i¼0 liþ1 j

ð1  qÞ

m1 j Qj X K i¼0 liþ1 j¼0

ð1  qÞj

þ

þ

K m1

Qm2

i¼0 liþ1 lm m1

ð1  qÞ

Q K m m1 i¼0 liþ1 m ð1  qÞ

 kx  zk1 þ

K jv mþ1 ðsÞ  umþ1 ðsÞj 1q

jv mþ1 ðsÞ  umþ1 ðsÞj:



ð29Þ

From (26), (29) and the induction we have that (27) holds for 2 6 m 6 k þ 1. For m ¼ k þ 1 we get

jv 1 ðsÞ  u1 ðsÞj 6 kx  zk1

k1 j Qj X K i¼0 liþ1 j¼0

ð1  qÞj

:

ð30Þ

Using (30) into (25) we get

jxðtÞ  zðtÞj 6 kx  zk1 6 kx  zk1

Z

! Q 3 k X X K j ji¼1 li Lj þ L2 ds j 1 j¼1 j¼1 ð1  qÞ ! Q  3 k X X K j ji¼1 li N1 N2 6 qkx  zk1 : Lj þ L2 þ j a1 a2 j¼1 j¼1 ð1  qÞ þ1

jGðt  sÞj

ð31Þ

From (6) and (30) we have that

jx0 ðtÞ  z0 ðtÞj ¼ jAðxðtÞ  zðtÞÞ þ ðFðt; xðtÞ; xðv 1 ðtÞÞ; x0 ðgðtÞÞÞ  Fðt; zðtÞ; zðu1 ðtÞÞ; z0 ðgðtÞÞÞj 6 jAjjxðtÞ  zðtÞj þ L1 jxðtÞ  zðtÞj þ L2 jxðv 1 ðtÞÞ  xðu1 ðtÞÞj þ L2 jxðu1 ðtÞÞ  zðu1 ðtÞÞj þ L3 jx0 ðgðtÞÞ  z0 ðgðtÞÞj 1 0 j jQ ! K l i 3 3 k C B X X X KL2 B i¼1 C 6 kx  zk1 jAj þ Lj þ Lj þ L2 jv 1 ðtÞ  u1 ðtÞj 6 kx  zk1 BjAj þ C 6 qkx  zk1 : j A @ 1q j¼1 j¼1 j¼1 ð1  qÞ ð32Þ

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From (31) and (32) it follows that

kx  zk1 6 qkx  zk1 which along with q < 1, implies xðtÞ  zðtÞ, which is a contradiction, finishing the proof of the theorem. h Remark 1. If jAj < 1, then it is easy to see that for sufficiently small Li ; lj ; i 2 f1; 2; 3g j ¼ 1; k, there is a q 2 ð0; 1Þ such that conditions (7) and (8) hold. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31]

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