Oscillatory and non-oscillatory behaviour of second-order neutral delay differential equations

Oscillatory and non-oscillatory behaviour of second-order neutral delay differential equations

Applied Mathematics and Computation 135 (2003) 333–344 www.elsevier.com/locate/amc Oscillatory and non-oscillatory behaviour of second-order neutral ...

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Applied Mathematics and Computation 135 (2003) 333–344 www.elsevier.com/locate/amc

Oscillatory and non-oscillatory behaviour of second-order neutral delay differential equations Sabah Hafez Abdallah Department of Mathematics, University College for Women, Ain Shams University, P.O. Box 9019, Nasr City, Cairo 11765, Egypt

Abstract The oscillatory and non-oscillatory behaviour of solutions of the neutral delay differential equation: d ½P ðtÞ½xðtÞ þ hðtÞxðt  sÞ0 0 þ qðtÞxa ðt  rÞ ¼ 0; 0 ¼ ; dt where p; h; q 2 Cð½t0 ; 1Þ; RÞ and s and r are non-negative real numbers, are studied. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: Neutral delay differential equation; Oscillatory solutions

1. Introduction In this paper we are concerned with establishing criteria for the oscillatory behaviour of the solution of the second-order nonlinear neutral delay differential equation: ½P ðtÞ½xðtÞ þ hðtÞxðt  sÞ0 0 þ qðtÞxa ðt  rÞ ¼ 0;

t P t0 ;

ð1:1Þ

where q : ½t0 ; 1Þ ! R is continuous and does not vanish eventually, P : ½t0 ; 1Þ ! R is positive continuously differentiable and s and r are nonnegative real numbers, xa ðt  rÞ : R ! R is continuous such that uxa ðuÞ > 0 for u 6¼ 0 and ðd=duÞxa ðuÞ P 0. Let u 2 Cð½t0  m; t0 ; RÞ, where m ¼ maxfs; rg. By a solution of Eq. (1.1) with initial function u at t0 we mean a function 0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 0 1 ) 0 0 3 3 5 - 6

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x 2 Cð½t0  m; 1Þ; RÞ such that xðtÞ ¼ uðtÞ for t  m 6 t 6 t0 and x satisfies Eq. (1.1) for t P t0 . Second-order neutral delay differential equations have applications in problems dealing with vibrating masses attached to an elastic and also appear, as the Euler equation, in some vibrational problems (see [1–8]). Definition 1.1. A solution of Eq. (1.1) is called oscillatory if it has arbitrary large zeros and it is non-oscillatory if it is eventually positive or negative. In this paper we consider two cases q P 0 and q changes sign for all large t, to give sufficient conditions in order that every solution of Eq. (1.1) is oscillatory and to investigate the asymptotic nature of non-oscillatory solutions of Eq. (1.1). With respect to their asymptotic behaviour, all the solutions of (1.1) may be a priori divided into the following classes: M þ ¼ {x ¼ xðtÞ solution of (1.1): there exists tx P t0 : xðtÞx0 ðtÞ P 0 for t P tk }; M  ¼ {x ¼ xðtÞ is solution of (1.1): there exists tx P t0 : xðtÞx0 ðtÞ 6 0 for t P tk }; Os ¼ {x ¼ xðtÞ is solution of (1.1): there exists ftn g, tn ! 1 : xðtn Þ ¼ 0}; Wos ¼ {x ¼ xðtÞ solution of (1.1): xðtÞ 6¼ 0 for t sufficiently large and for all a > 0 9ta1 , ta2 such that x0 ðta1 Þ  x0 ðta2 Þ < 0}. 2. Existence and oscillation of solutions First we examine the existence of solutions of Eq. (1.1) in the class M þ . Theorem 2.1. Assume ðiÞ hðtÞ P 0 and non-decreasing for all t P t0 ; Z t ðiiÞ Lim sup qðsÞ ds ¼ 1 t!1

ð2:1Þ ð2:2Þ

t0

hold. Then for Eq. (1.1) we have M þ ¼ u. Proof. Suppose that Eq. (1.1) has a solution x 2 M þ . There is no loss of generality in assuming that there exists t1 P t0 such that xðtÞ > 0, x0 ðtÞ P 0, xðt  mÞ > 0, x0 ðt  mÞ P 0, for all t P t1 (the proof is similar if xðtÞ < 0, xðt  mÞ < 0 for all large t). Let zðtÞ ¼ xðtÞ þ hðtÞxðt  sÞ: By condition (i) we see that zðtÞ > 0 for t P t1 ; for some t1 P T :

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335

And from the given equation (1.1) it follows that 0

½P ðtÞz0 ðtÞ ¼ qðtÞxa ðt  rÞ 6 0

ð2:3Þ

for t P t1 :

Therefore, pðtÞz0 ðtÞ is a non-increasing function of t. We claim that z0 ðtÞ P 0 for t P t1 : Otherwise, z0 ðtÞ < 0

for t P t1 :

Integrating Eq. (2.3) from t1 to t we obtain Z t Z t 0 ½P ðsÞz0 ðsÞ ds ¼  qðsÞ ds; a t1 x ðs  rÞ t1 Z t Z t P ðtÞz0 ðtÞ P ðt1 Þz0 ðt1 Þ P ðsÞz0 ðsÞx0 ðs  rÞ  þa ds ¼  qðsÞ ds xa ðt  rÞ xa ðt1  rÞ xaþ1 ðs  rÞ t1 t1

ð2:4Þ

or P ðtÞz0 ðtÞ P ðt1 Þz0 ðt1 Þ  6 xa ðt  rÞ xa ðt1  rÞ

Z

t

qðsÞ ds: t1

From (2.2) we obtain Lim inf t!1

P ðtÞz0 ðtÞ ¼ 1; xa ðt  rÞ

which contradicts the assumption z0 ðtÞ P 0 for all large t.



Example 1 [9]. Consider the neutral differential equation " 0 #0 t1 3ðt  2Þ xðt  1Þ t xðtÞ þ xðt  2Þ ¼ 0; t P 2: þ 2 t t ðt  3Þ We see that: P ðtÞ ¼ t P 2 is positive continuously differentiable; hðtÞ ¼

t1 is positive continuously differentiable; t

s ¼ 1 and r ¼ 2 are non-negative real numbers. Let u 2 Cð½t0  m; t0 ; RÞ, where m ¼ maxf1; 2g ¼ 2; u 2 Cð½0; 2; RÞ;

qðsÞ ¼

3ðs  2Þ ; s2 ðs  3Þ

ðIÞ

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S.H. Abdallah / Appl. Math. Comput. 135 (2003) 333–344

Z

t

Lim sup t!1

t0

3ðs  2Þ ds ¼ Lim sup t!1 s2 ðs  3Þ

Z t t0

 1 1  ln s  þ lnðs  3Þ ¼ 0; 3 s

i.e., All conditions of Theorem 2.1 are satisfied. In fact (I) has the solution xðtÞ ¼ ðt  1Þ=t 2 M þ . In the following theorem we consider the existence of solution in the class M þ subject to the following: 1 < h1 6 hðtÞ 6 0;

ð2:5Þ

qðtÞ P 0

ð2:6Þ

Z

t

Lim t!1

for all t P t0 ;

qðtÞ dt ¼ 1

ð2:7Þ

1 ds ¼ 1; P ðsÞ

ð2:8Þ

t0

and Z Lim t!1

t0

t

Theorem 2.2. Assume that conditions (2.5)–(2.8) hold. Then for Eq. (1.1) we have M þ ¼ u. Proof. As in the proof of Theorem 2.1, suppose that Eq. (1.1) has a solution x 2 M þ . Without loss of generality we assume that there exists t1 > t0 such that xðtÞ > 0, xðt  mÞ > 0, x0 ðtÞ P 0, x0 ðt  mÞ P 0 for all t P t1 (the proof is similar if xðtÞ < 0, xðt  mÞ < 0 for all large t). Let zðtÞ ¼ xðtÞ þ hðtÞxðt  sÞ: We claim that zðtÞ > 0 for all t P t1 : For, if zðtÞ 6 0; then xðtÞ < h1 xðt  sÞ; and therefore xðt þ sÞ < h1 xðtÞ:

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By induction, xðt þ 2sÞ < h1 xðt þ sÞ; 2

xðt þ 2sÞ < h1 ðhÞxðtÞ i:e: xðt þ 2sÞ < ðh1 Þ xðtÞ; .. . We obtain n

xðt þ nsÞ < ðh1 Þ xðtÞ; which implies that xðtÞ ! 0 as n ! 1, which is a contradiction. From Eq. (1.1) we obtain 0

½pðtÞz0 ðtÞ ¼ qðtÞxa ðt  rÞ;

t P t1 ;

from condition (2.6) and x ðt  rÞ > 0 we have pðtÞz0 ðtÞ is non-increasing on ½t1 ; 1Þ. Now suppose that pðtÞz0 ðtÞ < 0 for t P t1 . Then there exists t2 > t1 such that a

P ðtÞz0 ðtÞ 6 pðt2 Þz0 ðt2 Þ < 0 z0 ðtÞ 6

for all t P t2 ;

P ðt2 Þz0 ðt2 Þ : P ðtÞ

Integrating this equation from t2 ! t we obtain Z t P ðt2 Þz0 ðt2 Þ ds: zðtÞ  zðt2 Þ 6 P ðsÞ t2 This implies that zðtÞ ! 1 as t ! 1, which is a contradiction. Thus P ðtÞz0 ðtÞ P 0. Now proceeding as in the proof of Theorem 2.1, we obtain, by using (2.7) Lim t!1

P ðtÞz0 ðtÞ ¼ 1; xa ðt  rÞ

which contradicts the assumption z0 ðtÞ > 0 for all large t. This completes the proof of the theorem.  Example 2. Consider the neutral differential 0 0

½t2 ðxðtÞ 2xðt  1ÞÞ  þ

2t xðt  2Þ ¼ 0; t2

t P 3:

ðIIÞ

We see that: P ðtÞ ¼ t2 is positive continuously differentiable and s ¼ 1, r ¼ 2 are non-negative real numbers, hðtÞ ¼ 2 according to Theorem 2.2, we see that hðtÞ ¼ 2 < 0;

qðtÞ ¼

2t >0 t2

for t > 3

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S.H. Abdallah / Appl. Math. Comput. 135 (2003) 333–344

Z

t

qðtÞ dt ¼ Lim

Lim t!1

Z

t!1

t0

t t0

2t dt ¼ 2 Lim t!1 t2

Z t t0

 2 1þ dt t2

t

¼ 2 Lim ½ t þ 2 lnðt  2Þ ¼ 1; t!1 t

0

Z

t

Lim t!1

t0

1 ds ¼ Lim t!1 pðsÞ

Z t0

t

 t   1 1 1 1 1 þ ds ¼ Lim  ¼ Lim  ¼ : 2 t!1 t!1 s s t t0 t0 t0

All conditions of Theorem 2.2 are satisfied except the condition (2.8). In fact Eq. (II) has the solution xðtÞ ¼ t, t > 3 x0 ðtÞ ¼ 1;

xðtÞ  x0 ðtÞ ¼ tð1Þ > 0

for t > 3:

Hence xðtÞ ¼ t 2 M þ . Next, we examine the problem of the existence of solution of Eq. (1.1) in the class M  . Theorem 2.3. Assume that s 6 r. If the function 1=xa ðuÞ is locally integrable on ð0; cÞ and ðc; 0Þ for some c > 0, that is Z c Z 0 du du < 1; > 1; ð2:9Þ a ðuÞ a ðuÞ x x 0 c xa is submultiplicative;

ð2:10Þ

hðtÞ P 0 and non-increasing for all t P t0

ð2:11Þ

and Z

t

Lim sup t!1

T

1 xa ½1 þ hðsÞpðsÞ

Z

s

qðsÞ ds ds ¼ 1

for all T P t0 ;

ð2:12Þ

T

then for Eq. (1.1) we have M  ¼ u. Proof. Suppose that Eq. (1.1) has a solution x 2 M  . There is no loss of generality in assuming that there exists t1 P t0 such that xðtÞ > 0, x0 ðtÞ 6 0, xðt  mÞ > 0, x0 ðt  mÞ 6 0 for all t P t1 (the proof is similar if xðtÞ < 0 and x0 ðtÞ P 0 for all large t). Following the same steps as in the proof of Theorem 2.2, let zðtÞ ¼ xðtÞ þ hðtÞxðt  sÞ: Then by using (2.11) we see that zðtÞ > 0 and

z0 ðtÞ 6 0

for all t P t1 :

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From (2.4) we obtain P ðtÞz0 ðtÞ P ðt1 Þz0 ðt1 Þ ¼ a xa ðt  rÞ xa ðt1  rÞ Z t 6 qðsÞ ds

Z

t t1

P ðsÞz0 ðsÞx0 ðs  rÞ ds  xaþ1 ðs  rÞ

Z

t

qðsÞ ds t1

t1

or z0 ðtÞ 1 6 a x ðt  rÞ P ðtÞ

Z

t

qðsÞ ds;

ð2:13Þ

t P t1 :

t1

Since x is non-increasing and s 6 r, we have zðtÞ 6 xðt  rÞ þ hðtÞxðt  rÞ 6 ½1 þ hðtÞxðt  rÞ; and by using (2.10), we then obtain xa ðzðtÞÞ 6 xa ½1 þ hðtÞxa ½t  r;

ð2:14Þ

a

xa ½t  rÞ P

x ðzðtÞÞ ; þ hðtÞ

xa ½1

1 xa ½1 þ hðtÞ 6 : xa ðt  rÞ xa ½zðtÞ Substituting in Eq. (2.13) we get Z t z0 ðtÞ z0 ðtÞxa ½1 þ hðtÞ 1 6 6  qðsÞ ds; xa ðt  rÞ xa ½zðtÞ P ðtÞ t1 Z t z0 ðtÞ 1 6 qðsÞ ds; t P t1 : xa ½zðtÞ P ðtÞxa ½1 þ hðtÞ t1

t P t1 :

Integrating this equation from t1 to t we obtain Z t Z t Z s z0 ðtÞ 1 ds 6  qðsÞ ds ds; a a t1 x ðzðsÞÞ t1 x ½1 þ hðsÞP ðsÞ t1 let zðsÞ ¼ u. Then z0 ðsÞds ¼ du Z zðtÞ Z t Z s du 1 6  qðsÞ ds ds; a a zðt1 Þ x ðuÞ t1 x ½1 þ hðsÞP ðsÞ t1 or Z

zðt1 Þ

zðtÞ

du P a x ðuÞ

Z t1

t

1 a x ½1 þ hðsÞP ðsÞ

by using (2.12), which implies that

Z

s

qðsÞ ds ds t1

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S.H. Abdallah / Appl. Math. Comput. 135 (2003) 333–344

Z

zðt1 Þ

Lim sup t!1

zðtÞ

du xa ðuÞ

¼1

which contradicts (2.9). The proof is now complete.

ð2:15Þ 

Next we establish sufficient conditions under which Eq. (1.1) has no weakly oscillatory solution. Theorem 2.4. Let the assumption (2.6) be satisfied. If hðtÞ  h P 0

for all t P t0 ;

ð2:16Þ

then for Eq. (1.1) we have Wos ¼ u. Proof. Let x be a weakly oscillatory solution of (1.1). There is no loss of generality in assuming that there exists t1 P t0 such that xðtÞ > 0, xðt  mÞ > 0 for all t P t0 (the proof is similar if xðtÞ < 0, xðt  mÞ < 0 for all large t). Let zðtÞ ¼ xðtÞ þ hðtÞxðt  sÞ. Then by using (2.16), zðtÞ > 0 and weakly oscillatory and Eq. (1.1) takes the form: 0

½P ðtÞz0 ðtÞ þ qðtÞxa ðt  aÞ ¼ 0: Putting F ðtÞ ¼ P ðtÞz0 ðtÞ we get for t P t0 , F 0 ðtÞ ¼ ½P ðtÞz0 ðtÞ0 ; F 0 ðtÞ ¼ qðtÞxa ðt  rÞ 6 0 and so F is non-increasing which gives a contradiction, since F is an oscillatory function.  Theorem 2.5. Let conditions (2.7), (2.8) and (2.16) hold. Then every solution of Eq. (1.1) is either oscillatory or weakly oscillatory. Proof. From Theorem 2.1 it follows that for Eq. (1.1) M þ ¼ u. In order to complete the proof it suffices to show that for Eq. (1.1) M  ¼ u. Let x be a solution of class M  of Eq. (1.1). Assume without loss of generality that there exists t1 P t0 such that xðtÞ > 0, xðt  mÞ > 0, x0 ðtÞ 6 0, x0 ðt  mÞ 6 0 for all t P t1 (the proof is similar if xðtÞ < 0 and x0 ðtÞ P 0 for all large t). Let zðtÞ ¼ xðtÞ þ hðtÞxðt  sÞ. Then by using (2.16), we have zðtÞ > 0, and z0 ðtÞ 6 0 for all t P t1 . Proceeding as in the proof of Theorem 2.1, we obtain for ðt P t1 Þ

S.H. Abdallah / Appl. Math. Comput. 135 (2003) 333–344

P ðtÞz0 ðtÞ P ðt1 Þz0 ðt1 Þ  þa xa ðt  rÞ xa ðt1  rÞ Z t qðsÞ ds: ¼

Z

t t1

341

P ðsÞz0 ðsÞxa1 ðs  rÞx0 ðs  rÞ ds x2a ðs  rÞ

t1

Let, W ðtÞ ¼

P ðtÞz0 ðtÞ ; xa ðt  rÞ

W ðt1 Þ ¼

P ðt1 Þz0 ðt1 Þ < 0; xa ðt1  rÞ

 xa1 ðs  rÞx0 ðs  rÞ W ðtÞ ¼ W ðt1 Þ þ a W ðsÞ ds xa ðs  rÞ t1   Z t Z t axa1 ðs  rÞx0 ðs  rÞ qðsÞ ds 6 W ðt1 Þ þ W ðsÞ  ds: xa ðs  rÞ t1 t1 Z

t



By using Gronwall’s inequality we get W ðtÞ 6

W ðt1 Þxa ðt1  rÞ : xa ðt  rÞ

Thus P ðtÞz0 ðtÞ W ðt1 Þxa ðt1  rÞ 6 ; xa ðt  rÞ xa ðt  rÞ P ðtÞz0 ðtÞ 6 W ðt1 Þxa ðt1  rÞ ¼ k z0 ðtÞ 6

ðk < 0Þ;

k : P ðtÞ

Thus for all large t, we have Z t ds ; zðtÞ  zðt1 Þ 6 k P ðsÞ t1 which because of (2.8) implies Lim zðtÞ ¼ 1: t!1

This is a contradiction and the proof is complete.



2.1. Behaviour of solutions in the classes M  and M þ In this section, we study the asymptotic behaviour of the eventually monotone solution of Eq. (1.1).

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Theorem 2.6. If assumptions (2.10)–(2.12) are satisfied then for every solution x 2 M  we have Lim xðtÞ ¼ 0: t!1

Proof. The assertion follows from the same argument as given in the proof of Theorem 2.3 taking into account (2.15) which implies Lim zðtÞ ¼ 0 and t!1

z0 ðtÞ P xðtÞ

for all t P t1 :

Then Lim xðtÞ ¼ 0: t!1

This completes the proof.



Finally, we examine the asymptotic behaviour of solutions in the class M þ . The following theorem holds. Theorem 2.7. If the assumption (2.1) holds with hðtÞ bounded and Z t Z s 1 ds ds ¼ þ1 for all T P t; Lim sup qðsÞ t!1 P ðsÞ T T

ð2:17Þ

is satisfied, then every solution in the class M þ is unbounded. Proof. Let x be a solution of Eq. (1.1) such that x 2 M þ . There is no loss of generality in assuming that there exists t1 P t0 such that xðtÞ > 0, x0 ðtÞ P 0, xðt  mÞ > 0, x0 ðt  mÞ P 0 for all t P t1 (the proof is similar if xðtÞ < 0, x0 ðtÞ < 0 for all large t). Let zðxÞ ¼ xðtÞ þ hðtÞxðt  sÞ. Then in view of (2.1) zðtÞ P 0 and z0 ðtÞ > 0 for all t P t1 . Consider the function Z t P ðtÞz0 ðtÞ 1 ds; W ðtÞ ¼  a x ðt  rÞ t1 P ðsÞ   0 Z t ½P ðtÞz0 ðtÞ ds 1 0 0 W ðtÞ ¼  a  P ðtÞz ðtÞ a x ðt  rÞP ðtÞ x ðt  rÞ t1 P ðsÞ Z t 0 a1 0 aP ðtÞz ðtÞx ðt  rÞx ðt  rÞ ds þ x2a ðt  rÞ P ðsÞ t1 from Eq. (I) we get Z W 0 ðtÞ ¼ qðtÞ t1

t

ds z0 ðtÞ aP ðtÞz0 ðtÞxa1 ðt  rÞx0  a þ P ðsÞ x ðt  rÞ x2a ðt  rÞ

Z

t t1

ds ; P ðsÞ

S.H. Abdallah / Appl. Math. Comput. 135 (2003) 333–344

W 0 ðtÞ P qðtÞ

Z

t t1

343

ds z0 ðtÞ  a : P ðsÞ x ðt  rÞ

Integrating this equation from t1 ! t we get Z t Z t Z s ds z0 ðsÞ ds  W ðtÞ P ds: qðsÞ a t1 t1 P ðsÞ t1 x ðs  rÞ

ð2:18Þ

As the function ðz0 ðsÞ=xa ðt  rÞÞ is positive for t > t1 then Z t z0 ðsÞ Lim ds exists: t!1 t xa ðs  rÞ 1 Assume that Z t Lim t!1

t1

z0 ðsÞ ds ¼ k < 1: xa ðs  rÞ

Taking into account (2.17), and from (2.18) we get Lim sup W ðtÞ ¼ 1 t!1

which gives a contradiction, since W is negative for all values of t P t1 . Thus Z t z0 ðsÞ Lim ds ¼ 1: ð2:19Þ a t!1 t x ðs  rÞ 1 Now for all values of t P t1 , we have xa ðt  rÞ P xa ðt1  rÞ; i.e 1 1 6 ; xa ðt  rÞ xa ðt1  rÞ and consequently Z t Z z0 ðsÞ 1 t 0 1 ds 6 z ðsÞ ds ¼ ½zðtÞ zðt1 Þ: a ðs  rÞ x c c t1 t1 From (2.19) we get Lim zðtÞ ¼ 1: t!1

Since zðtÞ ¼ xðtÞ þ hðtÞxðt  sÞ and x is a non-negative we have zðtÞ 6 ½1 þ hðtÞxðtÞ:

ð2:20Þ

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S.H. Abdallah / Appl. Math. Comput. 135 (2003) 333–344

Thus from (2.20) we get Lim xðtÞ ¼ 1; t!1

which completes the proof.



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