AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation89705810.1155/2009/897058897058Research ArticleOscillation Criteria for a Class of Second-Order Nonlinear Differential Equations with Damping TermOuyangZigen1, 2ZhongJichao1, 2ZouShuliang1ZhouYong1Center of Nuclear Energy Economy and ManagementUniversity of South ChinaHengyang 421001Chinanhu.edu.cn2School of Mathematics and PhysicsUniversity of South ChinaHengyang 421001Chinanhu.edu.cn200911102009200931082009290920092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

A class of second-order nonlinear differential equations with damping term (r(t)|x(t)|σ1x(t))+p(t)|x(t)|σ1x(t)+q(t)f(x(t))=0 are investigated in this paper. By using a new method, we obtain some new sufficient conditions for the oscillation of the above equation, and some references are extended in this paper. Examples are inserted to illustrate this result.

1. Introduction

Consider the following second-order nonlinear differential equations with damping term: (r(t)|x(t)|σ-1x(t))+p(t)|x(t)|σ-1x(t)+q(t)f(x(t))=0,tt0, where r(t)C1([t0,);R+), p(t),q(t)C([t0,);R), σ is a positive constant, and f is a continuous real-valued function on the real line R and satisfies xf(x)0 for x0. We restrict our attention to those solutions x(t) of (1.1) which exist on some half line [tx,) and satisfy sup{|x(t)|:tT}>0 for any Ttx.

Recently, there are many authors who have investigated the oscillation for second-order differential equations , Li  and Zhao investigated oscillation criteria for the following equation: (r(t)(x(t))σ)+p(t)(x(t))σ+q(t)f(x(t))=0,tt0, where σ is a quotient of odd positive integer. It is obvious that (1.2) is a special case of (1.1). In fact, the conditions of Theorem 3.2 in  are too complex.

More recently, Rogovchenko and Tuncay  have obtained oscillation criteria of the following: (r(t)x(t))+p(t)x(t)+q(t)f(x(t))=0,tt0.

Motivated by the above discussions, we investigate the oscillation of (1.1) in this paper; our oscillatory conditions and the proof of the main results are more simple than those of Theorem 3.2 in .

A solution x(t) of (1.1) is oscillatory if and only if it has arbitrarily large zeros; otherwise, it is nonoscillatory. Equation (1.1) is oscillatory if and only if every solution of (1.1) is oscillatory.

The paper is arranged as follows. In Section 2, we will establish our main results. Finally, examples are given to illustrate our results.

2. Main Results

To obtain our results, we introduce a lemma as follows.

Lemma 2.1 (see [<xref ref-type="bibr" rid="B2">2</xref>, <xref ref-type="bibr" rid="B3">3</xref>]).

Let the function K(t,s,x):R×R×R+R be such that for each fixed t, s, the function K(t,s,·) is nondecreasing. Further, let h(t) be a given function and u(t) satisfies that u(t)()h(t)+t0tK(t,s,u(s))ds,tt0, and v*(t) is the minimal (maximal) solution of v(t)=h(t)+t0tK(t,s,v(s))ds,tt0. Then u(t)()v*(t) for all tt0.

Now, we give our main results.

Theorem 2.2.

Assume that f(x)0, p(t)0, q(t)>0, and t0(1/r1/σ(t))dt= hold. Suppose that there exists a positive function ρ(t) such that t0q(t)ρ(t)dt=,p(t)ρ(t)r(t)ρ(t). Then every solution of (1.1) is oscillatory.

Proof.

Assume that (1.1) has a nonoscillatory solution x(t). Without loss of generality, suppose that it is an eventually positive solution (if it is an eventually negative solution, the proof is similar), that is, x(t)>0 for all tt0.

We consider the following three cases.

Case 1.

Suppose that x(t) is oscillatory. Then there exists t1t0 such that x(t1)=0. From (1.1), we have (r(t)|x(t)|σ-1x(t)exp(t0tp(s)r(s)ds))=(r(t)|x(t)|σ-1x(t))exp(t0tp(s)r(s)ds)+p(t)|x(t)|σ-1x(t)exp(t0tp(s)r(s)ds)=-q(t)f(x(t))exp(t0tp(s)r(s)ds)<0, which means that r(t)|x(t)|σ-1x(t)exp(t0tp(s)r(s)ds)<r(t1)|x(t1)|σ-1x(t1)exp(t0t1p(s)r(s)ds)=0,t>t1, it follows that x(t)<0 for all t>t1, which contradicts to the assumption that x(t) is oscillatory.

Case 2.

Suppose that x(t)<0. From (1.1), we obtain -(r(t)|x(t)|σ-1x(t))=(r(t)(-x(t))σ)=-p(t)(-x(t))σ+q(t)f(x(t))0, then there exists an M>0 and a t1t0, such that r(t)(-x(t))σM,tt1, it follows x(t)-t11r1/σ(t)M1/σdt+x(t1),tt1, which means that limtx(t)=-, this contradicts the assumption that x(t)>0.

Case 3.

Suppose that x(t)>0. Let w(t)=ρ(t)r(t)(x(t))σ, then w(t)=(r(t)(x(t))σ)ρ(t)+r(t)(x(t))σρ(t),tt0, in view of (1.1), we obtain w(t)f(x(t))=-q(t)ρ(t)-ρ(t)p(t)(x(t))σf(x(t))+ρ(t)r(t)(x(t))σf(x(t)),tt0, noticing that (w(t)f(x(t)))=w(t)f(x(t))-w(t)f(x(t))x(t)f2(x(t))=-q(t)ρ(t)-ρ(t)p(t)(x(t))σf(x(t))+ρ(t)r(t)(x(t))σf(x(t))-w(t)f(x(t))x(t)f2(x(t)),tt0, integrating the above from t0 to t, we get w(t)f(x(t))=w(t0)f(x(t0))-t0t(q(s)ρ(s)+(ρ(s)p(s)-ρ(s)r(s))(x(s))σf(x(s))+w(s)x(s)f(x(s))f2(x(s)))ds. Using (2.3), (2.4), and x(t)>0, we have 0limtw(t)f(x(t))=-, this is a contradiction, the proof is complete.

Remark 2.3.

If we replace p(t)0, q(t)>0 by p(t)0, q(t)0, limt(p(t)/q(t))=M>0, Theorem 2.2 holds also.

Theorem 2.4.

Assume that f(x)0 holds. Suppose also that ρ0(t)=exp(t0tp(s)r(s)ds),t0dt(ρ0(t)r(t))1/σ=, and ρ0(t) such that (2.3) holds. Then every solution of (1.1) is oscillatory.

Proof.

To the contrary, (1.1) has a nonoscillatory solution x(t). Without loss of generality, we assume that x(t) is an eventually positive solution. Let w(t)=ρ0(t)r(t)|x(t)|σ-1x(t), then w(t)x(t)=ρ0(t)r(t)|x(t)|σ-1(x(t))20 for tt0 and w(t)=(r(t)|x(t)|σ-1x(t))ρ0(t)+r(t)|x(t)|σ-1x(t)ρ0(t),tt0, in view of (1.1) and (2.15), we obtain w(t)f(x(t))=-q(t)ρ0(t),tt0, since (w(t)f(x(t)))=w(t)f(x(t))-w(t)f(x(t))x(t)f2(x(t))=-q(t)ρ0(t)-w(t)f(x(t))x(t)f2(x(t)),tt0, integrating the above from t0 to t, we have -w(t)f(x(t))=-w(t0)f(x(t0))+t0tq(s)ρ0(s)ds+t0tw(s)x(s)f(x(s))f2(x(s))ds,tt0. In view of (2.3), there exists a constant m>0 and t1t0 such that -w(t0)f(x(t0))+t0tq(s)ρ0(s)ds+t0t1w(s)x(s)f(x(s))f2(x(s))dsm, which means that -w(t)f(x(t))m+t1tw(s)x(s)f(x(s))f2(x(s))ds. Because that x(t) is positive, then (2.22) implies -w(t)>0, or equivalently x(t)<0. Let u(t)=-w(t)=-ρ0(t)r(t)|x(t)|σ-1x(t)=ρ0(t)r(t)(-x(t))σ, thus (2.22) can be changed as u(t)mf(x(t))+t1tf(x(t))f(x(s))(-x(s))f2(x(s))u(s)ds. Define K(t,s,u)=f(x(t))f(x(s))(-x(s))f2(x(s))u. Then, for any fixed t and s, K(t,s,u) is nondecreasing in u. Let v(t) be the minimal solution of the equation v(t)=mf(x(t))+t1tf(x(t))f(x(s))(-x(s))f2(x(s))v(s)ds. Applying Lemma 2.1, we obtain u(t)v(t),tt0. Dividing both sides of (2.26) by f(x(t)) and deriving both sides of (2.26), it follows (v(t)f(x(t)))=(m+t1tf(x(s))(-x(s))f2(x(s))v(s)ds)=f(x(t))(-x(t))f2(x(t))v(t). On the other hand, (v(t)f(x(t)))=v(t)f(x(t))-f(x(t))x(t)f2(x(t))v(t). Combining (2.28) and (2.29), it means v(t)0. So v(t)=v(t1)=mf(x(t1)), tt0. From (2.27), we obtain -x(t)(mf(x(t1)))1/σ1(ρ0(t)r(t))1/σ,tt1. Integrating both sides of the above from t1 to t, we have -x(t)+x(t1)(mf(x(t1)))1/σt1tds(ρ0(s)r(s))1/σ. Letting t in (2.32), and using (2.16), it follows that limtx(t)-, which contradicts to that x(t) is eventually positive. The proof is complete.

In the following, we always suppose that H(t)C2(R;R) and it satisfies the following two conditions:

H(t)>0 for tt0, H(t) is a bounded function;

H(t)=h(t), h(t) is a bounded function.

Theorem 2.5.

Assume that f(x)0, t0(1/r1/σ(t))dt= hold, and p(t)0,q(t)>0, or p(t)0,q(t)0,limtp(t)q(t)=M>0. Suppose further that there exists a function H(t) that satisfies (H1), (H2), and such that t0H(t)φ(t)dt=,lim suptv(t)r(t)<, where φ(t)=v(t)(q(t)-p(t)h(t)-(r(t)h(t))),v(t)=exp(t0t(p(s)r(s)-h(s)H(s))ds). Then every solution of (1.1) is oscillatory.

Proof.

For the sake of contradiction, (1.1) has a nonoscillatory solution x(t). Without loss of generality, we may assume that x(t)>0 for all tt0.

Define u(t)=v(t)r(t)(|x(t)|σ-1x(t)f(x(t))+h(t)). Deriving (2.39), we get u(t)=(p(t)r(t)-h(t)H(t))u(t)+v(t)[-p(t)|x(t)|σ-1x(t)f(x(t))-q(t)-r(t)|x(t)|σ-1(x(t))2f(x(t))f2(x(t))+(r(t)h(t))]-h(t)H(t)u(t)+p(t)v(t)h(t)-v(t)q(t)+v(t)(r(t)h(t))=-h(t)H(t)u(t)-φ(t). Multiplying (2.40) by H(t), it follows φ(t)H(t)-H(t)u(t)-h(t)u(t). We consider the following three cases.

Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M143"><mml:mi>u</mml:mi><mml:mo stretchy="false">(</mml:mo><mml:mi>t</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula> is oscillatory).

Then there exists a sequence {tn}, n=1,2,,tn as n and such that u(tn)=0, n=1,2,. Integrating both sides of (2.41) from t0 to tn, we obtain t0tnH(t)φ(t)dt-t0tnH(t)u(t)dt-t0tnh(t)u(t)dt=-H(t)u(t)|t0tn-t0tn(-H(t)u(t)+h(t)u(t))dt=H(t0)u(t0)-H(tn)u(tn)=H(t0)u(t0), that is limtnt0tnH(t)φ(t)dtH(t0)u(t0), which contradicts (2.35).

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M153"><mml:mi>u</mml:mi><mml:mo stretchy="false">(</mml:mo><mml:mi>t</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula> is eventually positive).

Integrating both sides of (2.41) from t0 to , we obtain t0H(t)φ(t)dtH(t0)u(t0)-limtH(t)u(t)H(t0)u(t0), which also contradicts to (2.35).

Case 3 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M157"><mml:mi>u</mml:mi><mml:mo stretchy="false">(</mml:mo><mml:mi>t</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula> is eventually negative).

If lim suptu(t)>-, then there exists a sequence {t¯n}, n=1,2,, that satisfies {t¯n} as n and such that limt¯nu(t¯n)=lim suptu(t)=M1>-. Because H(t) is a bounded function, then there exists a M2>0 such that H(t¯n)M2, n=1,2,. According to (2.41), we obtain t0t¯nH(t)φ(t)dtH(t0)u(t0)-H(t¯n)u(t¯n)H(t0)u(t0)-M2u(t¯n). Using (2.35) and taking limit as t¯n, it is easy to show that =limt¯nt0t¯nH(t)φ(t)dtH(t0)u(t0)-limt¯nH(t¯n)u(t¯n)H(t0)u(t0)-M1M2<, which is obviously a contradiction.

If lim suptu(t)=-, limtu(t)=-. From the definition of h(t), combining (2.36) and (2.39), it follows that x(t)<0 and limt(|x(t)|σ-1x(t)/f(x(t)))=-, which means that limt((-x(t))σ/f(x(t)))=. Owing to p(t)0, q(t)0, or p(t)0, q(t)0 and limt(p(t)/q(t))=M>0, using the similar method of the proof of Case 2 in Theorem 2.2, we will derive a contradiction. Then the proof is complete.

Theorem 2.6.

Assume that (2.36) holds, f(x)0, t0(1/r1/σ(t))dt=, and p(t)0,q(t)>0, or p(t)0,q(t)0,limtp(t)q(t)=M>0. Suppose further that there exists a function H(t) that satisfies (H1), (H2) and such that t0H(t)φ¯(t)dt=, where φ¯(t)=v(t)(q(t)+p(t)h(t)+(r(t)h(t))), and v(t) is defined in (2.38). Then every solution of (1.1) is oscillatory.

Proof.

For the sake of contradiction, (1.1) has a nonoscillatory solution. Without loss of generality, we may assume that (1.1) has an eventually positive solution (if it has an eventually negative solution, the proof is similar), then there exists a t1>t0 such that x(t)>0 for all tt1. Define u(t)=v(t)r(t)(|x(t)|σ-1x(t)f(x(t))-h(t)). The rest of the proof is similar to Theorem 2.5. The proof is complete.

Theorem 2.7.

Assume that (2.36) holds, f(x)0,t0(1/r1/σ(t))dt=, and p(t)0,q(t)>0, or p(t)0,q(t)0,limtp(t)q(t)=M>0. Suppose further that there exists a function H(t) that satisfies (H1), (H2) and such that t0H(t)ϕ(t)dt=, where ϕ(t)=v(t)(-p(t)h(t)-(r(t)h(t))), where v(t) is defined in (2.38). Then every solution of (1.1) is oscillatory.

Proof.

For the sake of contradiction, (1.1) has a nonoscillatory solution x(t). Without loss of generality, we may assume that x(t)>0 for all tt0.

Define u(t)=v(t)r(t)(|x(t)|σ-1x(t)x(t)+h(t)). Noting that xf(x)0 for x0, so f(x)/x0 for x0. Deriving (2.56), we obtain u(t)=(p(t)r(t)-h(t)H(t))u(t)+v(t)[-p(t)|x(t)|σ-1x(t)x(t)-q(t)f(x(t))x(t)-r(t)|x(t)|σ-1(x(t))2x2(t)+(r(t)h(t))]-h(t)H(t)u(t)+p(t)v(t)h(t)+v(t)(r(t)h(t))=-h(t)H(t)u(t)-ϕ(t). Multiplying (2.57) by H(t), we get H(t)ϕ(t)-H(t)u(t)-h(t)u(t). The rest of the proof is similar to Theorem 2.5; the proof is complete.

Theorem 2.8.

Assume that (2.36) holds, f(x)0,t0(1/r1/σ(t))dt=, and p(t)0,q(t)>0, or p(t)0,q(t)0,limtp(t)q(t)=M>0. Suppose further that there exists a function H(t) satisfies (H1), (H2) and such that t0H(t)ϕ¯(t)dt=, where ϕ¯(t)=v(t)(p(t)h(t)+(r(t)h(t))), where v(t) is defined in (2.38). Then every solution of (1.1) is oscillatory.

Proof.

For the sake of contradiction, (1.1) has a nonoscillatory solution x(t). Without loss of generality, we may assume that x(t)>0 for all tt0.

Define u(t)=v(t)r(t)(|x(t)|σ-1x(t)x(t)-h(t)). The rest of the proof is similar to Theorem 2.5; the proof is complete.

3. ExamplesExample 3.1.

Consider the following delay differential equation: (tx(t))-2x(t)+(t+34t)x(t)=0. It is obvious that σ=1, q(t)=(t+3/4t)>0, p(t)=-2<0, r(t)=t, and t0(1/t)dt=. It is difficult to distinguish whether every solution of (3.1) is oscillatory by Theorem  3.2 of .

By taking ρ(t)=1/t2, then t0q(t)ρ(t)dt=t0(t+34t)1t2dt=,p(t)ρ(t)=-2t2=r(t)ρ(t). From Theorem 2.2 or Theorem 2.4, it is easy to show that (3.1) is oscillatory.

In fact, x(t)=t1/2cost is such an oscillatory solution.

Example 3.2.

Consider the following differential equation: (tx(t))-x(t)+tx(t)=0. It is obvious that σ=1, r(t)=t, p(t)=-1<0, q(t)=t>0, and t0(1/r(t))dt=t0(1/t)dt=.

We are taking H(t)=C>0, h(t)=0. By a simple calculation, it is easy to show that v(t)=exp(t0t(p(s)/r(s)-h(s)/H(s))ds)=exp(t0t(-1/s-0/C)ds)=t0/t, lim suptv(t)r(t)=lim suptt0<, φ(t)=v(t)(q(t)-p(t)h(t)-(r(t)h(t)))=t0, and t0H(t)φ(t)dt=t0t0Cdt=.

From Theorem 2.5 or Theorem 2.6, it follows that (3.3) is oscillatory.

In fact, x(t)=sint is such an oscillatory solution.

Acknowledgments

This work was supported by the Natural Science Foundation of Hunan Province under Grant no. 07JJ3130, the Doctor Foundation of University of South China under Grant no. 5-XQD-2006-9, and the Subject Lead Foundation of University of South China under Grant no. 2007XQD13.

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