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Asymptotic behavior of a third-order nonlinear neutral delay differential equation

Abstract

The objective of this paper is to study asymptotic nature of a class of third-order neutral delay differential equations. By using a generalized Riccati substitution and the integral averaging technique, a new Philos-type criterion is obtained which ensures that every solution of the studied equation is either oscillatory or converges to zero. An illustrative example is included.

MSC:34K11.

1 Introduction

In this work, we study the oscillation and asymptotic behavior of a third-order nonlinear neutral differential equation with variable delay arguments

( r ( t ) [ x ( t ) + P ( t ) x ( t τ ( t ) ) ] ) + i = 1 m Q i (t) f i ( x ( t σ i ( t ) ) ) =0,t t 0 ,
(1)

where m1 is an integer and t 0 >0. We assume that the following hypotheses are satisfied.

(A1) r C 1 ([ t 0 ,),(0,)), P,τ, Q i , σ i C([ t 0 ,),[0,)), f i C(R,R), and u f i (u)>0 for u0, i=1,2,,m;

(A2) r (t)0, t 0 r 1 (t)dt=, and 0P(t) p 0 <1;

(A3) lim t (tτ(t))= lim t (t σ i (t))=, i=1,2,,m;

(A4) there exist constants α i >0 such that f i (u)/u α i for u0 and i=1,2,,m.

Throughout, we define

z(t):=x(t)+P(t)x ( t τ ( t ) ) .
(2)

By a solution of equation (1), we mean a function xC([ T x ,),R), T x t 0 , which has the properties z C 2 ([ T x ,),R), r z C 1 ([ T x ,),R), and satisfies (1) on [ T x ,). We consider only those solutions x of (1) which satisfy assumption sup{|x(t)|:tT}>0 for all T T x . We assume that (1) possesses such solutions. A solution of (1) is called oscillatory if it has arbitrarily large zeros on [ T x ,); otherwise, it is termed nonoscillatory.

As is well known, the third-order differential equations are derived from many different areas of applied mathematics and physics, for instance, deflection of buckling beam with a fixed or variable cross-section, three-layer beam, electromagnetic waves, gravity-driven flows, etc. In recent years, the oscillation theory of third-order differential equations has received a great deal of attention since it has been widely applied in research of physical sciences, mechanics, radio technology, lossless high-speed computer network, control system, life sciences, and population growth.

Numerous research activities are concerned with the oscillation of solutions to different functional differential equations, for some related contributions, we refer the reader to [116] and the references cited therein. In the following, we provide some background details regarding the study of various classes of neutral differential equations. Baculíková and Džurina [3] studied a second-order neutral differential equation

( r ( t ) [ x ( t ) + p ( t ) x ( τ ( t ) ) ] ) +q(t)x ( σ ( t ) ) =0.

Agarwal et al. [1], Grace et al. [7], and Zhang et al. [16] considered a third-order nonlinear differential equation

( a ( t ) ( b ( t ) x ( t ) ) ) +q(t) x γ ( σ ( t ) ) =0.

Baculíková and Džurina [4], Candan [5, 6], Karpuz [8], Li and Rogovchenko [9], Li and Thandapani [10], and Li et al. [11, 12] investigated a class of third-order neutral differential equations

( r ( t ) [ x ( t ) + p ( t ) x ( τ ( t ) ) ] ) +q(t)x ( σ ( t ) ) =0.
(3)

Define τ ˜ (t):=tτ(t) and σ ˜ i (t):=t σ i (t), i=1,2,,m. It follows from conditions (A1) and (A3) that τ ˜ (t)t, σ ˜ i (t)t, and lim t τ ˜ (t)= lim t σ ˜ i (t)=, i=1,2,,m. Hence, equation (3) is a special case of (1). As a matter of fact, equation (1) reduces to the form of (3) if m=1 and f 1 (u)=u.

There are two techniques in the study of oscillation of third-order neutral differential equations. One of them is comparison method which is used to reduce the third-order neutral differential equations to the first-order differential equations or inequalities; see, e.g., [810]. Another technique is the Riccati technique; see, e.g., [46, 1012]. In this paper, using a generalized Riccati substitution which differs from those reported in [46, 1012], a new asymptotic criterion for (1) is presented. In what follows, all functional inequalities are tacitly supposed to hold for all sufficiently large t.

2 Some lemmas

Lemma 1 Assume that conditions (A1)-(A4) hold and x is a positive solution of (1). Then there are only the following two possible cases for z defined by (2):

  1. (I)

    z(t)>0, z (t)>0, z (t)>0, z (t)0, and ( r ( t ) z ( t ) ) 0;

  2. (II)

    z(t)>0, z (t)<0, z (t)>0, z (t)0, and ( r ( t ) z ( t ) ) 0,

for tT, where T t 0 is sufficiently large.

Proof The proof is similar to that of Baculíková and Džurina [[4], Lemma 1], and hence is omitted. □

Lemma 2 Assume that conditions (A1)-(A4) hold and let x be a positive solution of (1) and corresponding z satisfy case (II) in Lemma  1. If

t 0 v 1 r ( u ) ( i = 1 m u Q i ( s ) d s ) dudv=,
(4)

then lim t x(t)= lim t z(t)=0.

Proof Suppose that x is a positive solution of (1). Since z(t)>0 and z (t)<0, there exists a finite constant l0 such that lim t z(t)=l0. We shall prove that l=0. Assume now that l>0. Then for any ε>0, there exists a t 1 T such that l+ε>z(t)>l for t t 1 . Choose 0<ε<l(1 p 0 )/ p 0 . It is not hard to find that

x ( t ) = z ( t ) P ( t ) x ( t τ ( t ) ) > l P ( t ) x ( t τ ( t ) ) > l p 0 z ( t τ ( t ) ) > l p 0 ( l + ε ) : = N ( l + ε ) > N z ( t ) ,
(5)

where N:=(l p 0 (l+ε))/(l+ε)>0. Using (1) and (5), we conclude that

0 = ( r ( t ) z ( t ) ) + i = 1 m Q i ( t ) f i ( x ( t σ i ( t ) ) ) ( r ( t ) z ( t ) ) + i = 1 m α i Q i ( t ) x ( t σ i ( t ) ) ( r ( t ) z ( t ) ) + N i = 1 m α i Q i ( t ) z ( t σ i ( t ) ) ( r ( t ) z ( t ) ) + N i = 1 m α i Q i ( t ) z ( t ) .
(6)

Integrating (6) from t to ∞, we obtain

0r(t) z (t)+N i = 1 m α i t Q i (s)z(s)ds.

Noting that z(t)>l, we get

0 z (t)+ l N r ( t ) i = 1 m α i t Q i (s)ds.
(7)

Integrating (7) from t to ∞, we have

0 z (t)+lN t 1 r ( u ) ( i = 1 m α i u Q i ( s ) d s ) du.
(8)

Integrating (8) from t 1 to ∞, we deduce that

t 1 v 1 r ( u ) ( i = 1 m α i u Q i ( s ) d s ) dudv z ( t 1 ) l N ,

which contradicts (4). Hence, l=0 and lim t z(t)=0. Then it follows from 0x(t)z(t) that lim t x(t)=0. The proof is complete. □

Lemma 3 (See [[4], Lemma 3])

Assume that u(t)>0, u (t)>0, and u (t)0 for t t 0 . If σC([ t 0 ,),[0,)), σ(t)t, and lim t σ(t)=, then for every α(0,1), there exists a T α t 0 such that u(σ(t))/σ(t)αu(t)/t for t T α .

Remark 1 If u satisfies conditions of Lemma 3, then u(t σ i (t))/u(t)α(t σ i (t))/t for i=1,2,,m when using conditions (A1) and (A3).

Lemma 4 (See [[4], Lemma 4])

Assume that u(t)>0, u (t)>0, u (t)>0, and u (t)0 for t t 0 . Then for each β(0,1), there exists a T β t 0 such that u(t)βt u (t)/2 for t T β .

Remark 2 If u satisfies conditions of Lemma 4, then u(t σ i (t))/ u (t σ i (t))β(t σ i (t))/2 for i=1,2,,m when using condition (A3).

3 Main results

We use the integral averaging technique to establish a Philos-type (see Philos [13]) criterion for (1). Let

D:= { ( t , s ) : t s t 0 } and D 0 := { ( t , s ) : t > s t 0 } .

We say that a function HC(D,R) belongs to the class X if

  1. (i)

    H(t,t)=0, t t 0 , H(t,s)>0, (t,s) D 0 ;

  2. (ii)

    H has a nonpositive continuous partial derivative H/s on D 0 with respect to the second variable, and there exist functions ρ C 1 ([ t 0 ,),(0,)), δ C 1 ([ t 0 ,),R), and hC( D 0 ,R) such that

    H ( t , s ) s + ( 2 δ ( s ) + ρ ( s ) ρ ( s ) ) H(t,s)=h(t,s) H ( t , s ) .
    (9)

Theorem 1 Assume that conditions (A1)-(A4) and (4) are satisfied. If

lim sup t 1 H ( t , t 0 ) t 0 t [ H ( t , s ) G ( s ) 1 4 ρ ( s ) r ( s ) h 2 ( t , s ) ] ds=
(10)

holds for some α(0,1), β(0,1), and for some HX, where

G(t):=ρ(t) [ α β ( 1 p 0 ) 2 i = 1 m α i Q i ( t ) ( t σ i ( t ) ) 2 t + r ( t ) δ 2 ( t ) ( r ( t ) δ ( t ) ) ] ,
(11)

then every solution x of (1) is either oscillatory or satisfies lim t x(t)=0.

Proof Suppose to the contrary and assume that (1) has a nonoscillatory solution x. Without loss of generality, we can assume that there exists a t 1 t 0 such that x(t)>0, x(tτ(t))>0, and x(t σ i (t))>0 for t t 1 and i=1,2,,m. By Lemma 1, we observe that z satisfies either (I) or (II) for tT, where T t 1 is large enough. We consider each of the two cases separately.

Assume first that case (I) holds. It follows from z (t)>0 that

x ( t ) = z ( t ) P ( t ) x ( t τ ( t ) ) z ( t ) p 0 x ( t τ ( t ) ) z ( t ) p 0 z ( t τ ( t ) ) ( 1 p 0 ) z ( t ) .
(12)

Using (1) and (12), we deduce that

( r ( t ) z ( t ) ) = i = 1 m Q i ( t ) f i ( x ( t σ i ( t ) ) ) i = 1 m α i Q i ( t ) x ( t σ i ( t ) ) ( 1 p 0 ) i = 1 m α i Q i ( t ) z ( t σ i ( t ) ) .
(13)

Define a generalized Riccati substitution by

ω(t):=ρ(t) [ r ( t ) z ( t ) z ( t ) + r ( t ) δ ( t ) ] .
(14)

Then we have

ω = ρ [ r z z + r δ ] + ρ [ r z z + r δ ] = ρ ρ ω + ρ ( r δ ) + ρ ( r z z ) = ρ ρ ω + ρ ( r δ ) + ρ ( r z ) z ρ r ( z z ) 2 .
(15)

By virtue of (14), we conclude that

( z z ) 2 = [ ω ρ r δ ] 2 = ( ω ρ r ) 2 + δ 2 2 ω δ ρ r .
(16)

Substituting (13) and (16) into (15), we obtain

ω = ρ ( r z ) z + ρ ρ ω + ρ ( r δ ) ρ r [ ω 2 ρ 2 r 2 + δ 2 2 ω δ ρ r ] = ρ ( r z ) z ρ [ r δ 2 ( r δ ) ] + ( ρ ρ + 2 δ ) ω ω 2 r ρ ( 1 p 0 ) ρ i = 1 m α i Q i z ( t σ i ( t ) ) z ( t ) ρ [ r δ 2 ( r δ ) ] + ( ρ ρ + 2 δ ) ω ω 2 r ρ .
(17)

It follows from Remarks 1 and 2 that, for any α(0,1) and β(0,1),

z ( t σ i ( t ) ) z ( t ) = z ( t σ i ( t ) ) z ( t σ i ( t ) ) z ( t σ i ( t ) ) z ( t ) α β 2 ( t σ i ( t ) ) 2 t ,
(18)

i=1,2,,m. Combining (17) and (18), we get

ω ( t ) α β ( 1 p 0 ) 2 ρ ( t ) i = 1 m α i Q i ( t ) ( t σ i ( t ) ) 2 t ρ ( t ) [ r ( t ) δ 2 ( t ) ( r ( t ) δ ( t ) ) ] + ( ρ ( t ) ρ ( t ) + 2 δ ( t ) ) ω ( t ) ω 2 ( t ) r ( t ) ρ ( t ) = G ( t ) + A ( t ) ω ( t ) B ( t ) ω 2 ( t ) ,

where G is defined as in (11), A(t):=( ρ (t)/ρ(t))+2δ(t), and B(t):=1/(r(t)ρ(t)). Replacing in the latter inequality t with s, multiplying both sides by H(t,s) and integrating with respect to s from some T 1 ( T 1 T) to t, we derive from H(t,t)=0 and (9) that

T 1 t H ( t , s ) G ( s ) d s T 1 t H ( t , s ) [ ω ( s ) + A ( s ) ω ( s ) B ( s ) ω 2 ( s ) ] d s = H ( t , T 1 ) ω ( T 1 ) + T 1 t [ ( H ( t , s ) s + A ( s ) H ( t , s ) ) ω ( s ) H ( t , s ) B ( s ) ω 2 ( s ) ] d s = H ( t , T 1 ) ω ( T 1 ) T 1 t [ h ( t , s ) H ( t , s ) ω ( s ) + H ( t , s ) B ( s ) ω 2 ( s ) ] d s = H ( t , T 1 ) ω ( T 1 ) T 1 t ( H ( t , s ) B ( s ) ω ( s ) + h ( t , s ) 2 B ( s ) ) 2 d s + T 1 t h 2 ( t , s ) 4 B ( s ) d s H ( t , T 1 ) ω ( T 1 ) + T 1 t h 2 ( t , s ) 4 B ( s ) d s ,

and hence

lim sup t 1 H ( t , T 1 ) T 1 t [ H ( t , s ) G ( s ) 1 4 ρ ( s ) r ( s ) h 2 ( t , s ) ] dsω( T 1 ),

which contradicts condition (10).

Assume now that case (II) holds. By virtue of Lemma 2, lim t x(t)=0. This completes the proof. □

Corollary 1 The conclusion of Theorem  1 remains intact if condition (10) is replaced by the assumptions

lim sup t 1 H ( t , t 0 ) t 0 t H(t,s)G(s)ds=

and

lim sup t 1 H ( t , t 0 ) t 0 t ρ(s)r(s) h 2 (t,s)ds<.

As an application of Theorem 1, we provide the following example.

Example 1 For t1, consider a third-order neutral delay differential equation

( x ( t ) + 1 3 x ( t 3 ) ) + t 3 x ( t 4 ) +4 t 3 x ( t 2 ) =0.
(19)

Let ρ(t)=t, δ(t)=0, and H(t,s)= ( t s ) 2 . It is not difficult to verify that all assumptions of Theorem 1 are satisfied. Hence, every solution x of (19) is either oscillatory or satisfies lim t x(t)=0. As a matter of fact, one such solution is x(t)= t 1 .

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Acknowledgements

The authors are grateful to the editors and two anonymous referees for a very thorough reading of the manuscript and for pointing out several inaccuracies. This research is supported by NNSF of P.R. China (Grant No. 61403061) and the AMEP of Linyi University, P.R. China.

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Jiang, Y., Li, T. Asymptotic behavior of a third-order nonlinear neutral delay differential equation. J Inequal Appl 2014, 512 (2014). https://doi.org/10.1186/1029-242X-2014-512

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