# Oscillation Criteria for a Class of Second-Order Nonlinear Differential Equations with Damping Term

Authors: Zigen Ouyang; Jichao Zhong; Shuliang Zou
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

A class of 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
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.


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## Body

## 1. Introduction

Consider the following second-order nonlinear differential equations with damping term:(1.1)(r(t)|x′(t)|σ-1x′(t))′+p(t)|x′(t)|σ-1x′(t)+q(t)f(x(t))=0,t≥t0,
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 x≠0. We restrict our attention to those solutions x(t) of (1.1) which exist on some half line [tx,∞) and satisfy sup{|x(t)|:t≥T}>0 for any T≥tx.Recently, there are many authors who have investigated the oscillation for second-order differential equations [1–9], Li [10] and Zhao investigated oscillation criteria for the following equation:
(1.2)(r(t)(x′(t))σ)′+p(t)(x′(t))σ+q(t)f(x(t))=0,t≥t0,
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 [10] are too complex.More recently, Rogovchenko and Tuncay [11] have obtained oscillation criteria of the following:
(1.3)(r(t)x′(t))′+p(t)x′(t)+q(t)f(x(t))=0,t≥t0.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 [10].A solutionx(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 Section2, 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 [2, 3]).
Let the functionK(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
(2.1)u(t)≥(≤)h(t)+∫t0tK(t,s,u(s))ds,t≥t0,
and v*(t) is the minimal (maximal) solution of
(2.2)v(t)=h(t)+∫t0tK(t,s,v(s))ds,t≥t0.
Then u(t)≥(≤)v*(t) for all t≥t0.Now, we give our main results.Theorem 2.2.
Assume thatf′(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
(2.3)∫t0∞q(t)ρ(t)dt=∞,(2.4)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 t≥t0.
We consider the following three cases.
Case 1.
Suppose thatx′(t) is oscillatory. Then there exists t1≥t0 such that x′(t1)=0. From (1.1), we have
(2.5)(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
(2.6)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 thatx′(t)<0. From (1.1), we obtain
(2.7)-(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 t1≥t0, such that
(2.8)r(t)(-x′(t))σ≥M,t≥t1,
it follows
(2.9)x(t)≤-∫t1∞1r1/σ(t)M1/σdt+x(t1),t≥t1,
which means that limt→∞x(t)=-∞, this contradicts the assumption that x(t)>0.Case 3.
Suppose thatx′(t)>0. Let w(t)=ρ(t)r(t)(x′(t))σ, then
(2.10)w′(t)=(r(t)(x′(t))σ)′ρ(t)+r(t)(x′(t))σρ′(t),t≥t0,
in view of (1.1), we obtain
(2.11)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)),t≥t0,
noticing that
(2.12)(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)),t≥t0,
integrating the above from t0 to t, we get
(2.13)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
(2.14)0≤limt→∞w(t)f(x(t))=-∞,
this is a contradiction, the proof is complete.Remark 2.3.
If we replacep(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 thatf′(x)≥0 holds. Suppose also that
(2.15)ρ0(t)=exp(∫t0tp(s)r(s)ds),(2.16)∫t0∞dt(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))2≥0 for t≥t0 and
(2.17)w′(t)=(r(t)|x′(t)|σ-1x′(t))′ρ0(t)+r(t)|x′(t)|σ-1x′(t)ρ0′(t),t≥t0,
in view of (1.1) and (2.15), we obtain
(2.18)w′(t)f(x(t))=-q(t)ρ0(t),t≥t0,
since
(2.19)(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)),t≥t0,
integrating the above from t0 to t, we have
(2.20)-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,t≥t0.
In view of (2.3), there exists a constant m>0 and t1≥t0 such that
(2.21)-w(t0)f(x(t0))+∫t0tq(s)ρ0(s)ds+∫t0t1w(s)x′(s)f′(x(s))f2(x(s))ds≥m,
which means that
(2.22)-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
(2.23)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
(2.24)u(t)≥mf(x(t))+∫t1tf(x(t))f′(x(s))(-x′(s))f2(x(s))u(s)ds.
Define
(2.25)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
(2.26)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
(2.27)u(t)≥v(t),t≥t0.
Dividing both sides of (2.26) by f(x(t)) and deriving both sides of (2.26), it follows
(2.28)(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,
(2.29)(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
(2.30)v′(t)≡0.
So v(t)=v(t1)=mf(x(t1)), t≥t0. From (2.27), we obtain
(2.31)-x′(t)≥(mf(x(t1)))1/σ1(ρ0(t)r(t))1/σ,t≥t1.
Integrating both sides of the above from t1 to t, we have
(2.32)-x(t)+x(t1)≥(mf(x(t1)))1/σ∫t1tds(ρ0(s)r(s))1/σ.
Letting t→∞ in (2.32), and using (2.16), it follows that limt→∞x(t)≤-∞, which contradicts to that x(t) is eventually positive. The proof is complete.In the following, we always suppose thatH(t)∈C2(R;R) and it satisfies the following two conditions:(H1)
H(t)>0 for t≥t0, H(t) is a bounded function;(H2)
H′(t)=h(t), h(t) is a bounded function.Theorem 2.5.
Assume thatf′(x)≥0, ∫t0∞(1/r1/σ(t))dt=∞ hold, and
(2.33)p(t)≤0,q(t)>0,
or
(2.34)p(t)≤0,q(t)≤0,limt→∞p(t)q(t)=M>0.
Suppose further that there exists a function H(t) that satisfies  (H1),  (H2), and such that
(2.35)∫t0∞H(t)φ(t)dt=∞,(2.36)lim supt→∞v(t)r(t)<∞,
where
(2.37)φ(t)=v(t)(q(t)-p(t)h(t)-(r(t)h(t))′),(2.38)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 t≥t0.
Define(2.39)u(t)=v(t)r(t)(|x′(t)|σ-1x′(t)f(x(t))+h(t)).
Deriving (2.39), we get
(2.40)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
(2.41)φ(t)H(t)≤-H(t)u′(t)-h(t)u(t).
We consider the following three cases.
Case 1 (u(t) 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
(2.42)∫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
(2.43)limtn→∞∫t0tnH(t)φ(t)dt≤H(t0)u(t0),
which contradicts (2.35).Case 2 (u(t) is eventually positive).
Integrating both sides of (2.41) from t0 to ∞, we obtain
(2.44)∫t0∞H(t)φ(t)dt≤H(t0)u(t0)-limt→∞H(t)u(t)≤H(t0)u(t0),
which also contradicts to (2.35).Case 3 (u(t) is eventually negative).
Iflim supt→∞u(t)>-∞, then there exists a sequence {t¯n}, n=1,2,…, that satisfies {t¯n}→∞ as n→∞ and such that limt¯n→∞u(t¯n)=lim supt→∞u(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
(2.45)∫t0t¯nH(t)φ(t)dt≤H(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
(2.46)∞=limt¯n→∞∫t0t¯nH(t)φ(t)dt≤H(t0)u(t0)-limt¯n→∞H(t¯n)u(t¯n)≤H(t0)u(t0)-M1M2<∞,
which is obviously a contradiction.
Iflim supt→∞u(t)=-∞, limt→∞u(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
(2.47)p(t)≤0,q(t)>0,
or
(2.48)p(t)≤0,q(t)≤0,limt→∞p(t)q(t)=M>0.
Suppose further that there exists a function H(t) that satisfies  (H1),  (H2) and such that
(2.49)∫t0∞H(t)φ¯(t)dt=∞,
where
(2.50)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 t≥t1. Define
(2.51)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
(2.52)p(t)≤0,q(t)>0,
or
(2.53)p(t)≤0,q(t)≤0,limt→∞p(t)q(t)=M>0.
Suppose further that there exists a function H(t) that satisfies  (H1), (H2) and such that
(2.54)∫t0∞H(t)ϕ(t)dt=∞,
where
(2.55)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 t≥t0.
Define(2.56)u(t)=v(t)r(t)(|x′(t)|σ-1x′(t)x(t)+h(t)).
Noting that xf(x)≥0 for x≠0, so f(x)/x≥0 for x≠0. Deriving (2.56), we obtain
(2.57)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
(2.58)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
(2.59)p(t)≤0,q(t)>0,
or
(2.60)p(t)≤0,q(t)≤0,limt→∞p(t)q(t)=M>0.
Suppose further that there exists a function H(t) satisfies  (H1), (H2) and such that
(2.61)∫t0∞H(t)ϕ¯(t)dt=∞,
where
(2.62)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 t≥t0.
Define(2.63)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. Examples

Example 3.1.
Consider the following delay differential equation:(3.1)(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 [10].
By takingρ(t)=1/t2, then
(3.2)∫t0∞q(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:(3.3)(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 takingH(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 supt→∞v(t)r(t)=lim supt→∞t0<∞, φ(t)=v(t)(q(t)-p(t)h(t)-(r(t)h(t))′)=t0, and ∫t0∞H(t)φ(t)dt=∫t0∞t0Cdt=∞.
From Theorem2.5 or Theorem 2.6, it follows that (3.3) is oscillatory.
In fact,x(t)=sint is such an oscillatory solution.



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