# Nonoscillatory Solutions of Second-Order Differential Equations without Monotonicity Assumptions

Authors: Lianwen Wang; Rhonda McKee
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

The continuability, boundedness, monotonicity, and asymptotic properties of nonoscillatory solutions for a class of second-order nonlinear differential equations[p(t)h(x(t))f(x′(t))]′=q(t)g(x(t)) are discussed without monotonicity assumption for function g. It is proved that all solutions can be extended to infinity, are eventually monotonic, and can be classified into disjoint classes that are fully characterized in terms of several integral conditions. Moreover, necessary and sufficient conditions for the existence of solutions in each class and for the boundedness of all solutions are established.


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## Body

## 1. Introduction

This paper studies the continuability, boundedness, monotonicity, and asymptotic properties of nonoscillatory solutions for a class of second-order nonlinear differential equations(1.1)[p(t)h(x(t))f(x′(t))]′=q(t)g(x(t)),t≥a.
Some special cases of (1.1) such as half-linear equation
(1.2)[p(t)Φp(x′(t))]′=q(t)Φp(x(t)),
where Φp(r)=|r|p-2r,  p>1, is the so-called p-Laplacian operator, Emden-Fowler equation
(1.3)[p(t)Φp(x′(t))]′=q(t)Φβ(x(t)),
and differential equation
(1.4)[p(t)h(x(t))x′(t)]′=q(t)g(x(t))
have been extensively discussed in the literature; see, for example, [1–15] and references cited therein. Equation (1.1) with general nonlinear function f(r) is investigated in [16–18]. It is worth to point out that g(r) is assumed to be monotonic in most cited papers, but [2, 6] explain that this assumption does not hold in some applications. The aim of this paper is to investigate the continuability, boundedness, monotonicity, and asymptotic properties of nonoscillatory solutions of (1.1) without monotonic assumption for g. Some techniques and ideas have been used by the authors in [17].By solution of (1.1), we mean a differentiable function x such that p(t)h(x(t))f(x′(t)) is differentiable and satisfies (1.1) on the maximum existence interval [a,αx), αx≤∞. A solution x of (1.1) is said to be eventually monotonic if there exists a tx≥a such that x is monotonic on [tx,αx). In this paper, we consider only solutions that are not eventually identically equal to zero.Throughout the paper, we always assume that(H)p(t),q(t):[a,∞)→ℝarecontinuousandp(t)>0andq(t)>0;h(r):ℝ→ℝiscontinuousandh(r)>0;g(r):ℝ→ℝiscontinuousandrg(r)>0forr≠0;f(r):ℝ→ℝiscontinuous,increasing,andrf(r)>0forr≠0.(H1) There exists a constantM1>0 such that
(1.5)|f-1(uv)|≤M1|f-1(u)||f-1(v)|,∀u,v∈R.Remark 1.1.
(H1) holds forp-Laplacian operator; indeed,
(1.6)f-1(uv)=f-1(u)f-1(v).
However, there are nonlinear functions f that satisfy (H1) but not (1.6); see [17].The paper is organized as follows: Section1 briefly addresses the background and the motivation of the paper. Continuability, classification, and boundedness of solutions are discussed in Section 2. Sections 3 and 4 deal with the existence of class A and class B solutions, respectively. Finally, several remarks are provided in Section 5 to compare our results with existing ones.


## 2. Continuability, Classification, and Boundedness of Solutions

In this section we discuss continuability, classification, and boundedness of solutions of (1.1). First of all, we cite a result from [17] that will be used later on.Lemma 2.1.
Ifx is a solution of (1.1) with maximal existence interval [a,αx), αx≤∞, then x is eventually monotonic. Moreover, if x is bounded on all finite subinterval of [a,αx), then αx=∞.Remark 2.2.
From Lemma2.1 all solutions of (1.1) except eventually trivial solutions can be classified into two classes
(2.1)A={xisdefinedon⁡[a,αx):x(t)x′(t)>0inaleftneighborhoodofαx},B={xisdefinedon⁡[a,∞):x(t)x′(t)<0fort≥a}.Next theorem establishes the continuability for all solutions of (1.1), in other words, all solutions can be extended to [a,∞).Theorem 2.3.
Assume the following assmputions hold.(H2)
There exists a real numberm>0 and a continuous function G(r):ℝ→ℝ  such that G(r) is increasing and |g(r)|≤|G(r)| for |r|≥m, and rG(r)>0 for r≠0;(H3)
There exists a real numberr0>0 such that
(2.2)∫r0∞drf-1(z(r))=∞,∫-∞-r0drf-1(z(r))=-∞,
where z(r)=G(r)/h(r). Then all solutions of (1.1) can be extended to [a,∞).Proof.
The proof is similar to that of Theorem2.3 [17]. We point out that as in the proof of Theorem 2.3 [17], for a class A solution x, we have
(2.3)f(x′(t))≤G(x(t))p(t)h(x(t))(p(d)h(x(d))f(x′(d))G(x(d))+∫dtq(s)ds),∫x(t1)x(t)drf-1(z(r))≤M12f-1(k)∫t1tf-1(1p(s)∫dsq(σ)dσ)ds.Remark 2.4.
The functiong(r)=r+sinr is not monotonic. Clearly, |g(r)|≤2|r|, so g is bounded by an increasing function G(r)=2r. Therefore, the existing results which require the monotonic condition for g would not apply, but Theorem 2.3 does.From Remark2.2 and Theorem 2.3, all solutions of (1.1) can be classified further into four disjoint classes
(2.4)Ab={x∈A:lim⁡t→∞|x(t)|=l<∞},A∞={x∈A:lim⁡t→∞|x(t)|=∞},Bb={x∈B:lim⁡t→∞x(t)=l≠0},B0={x∈B:lim⁡t→∞x(t)=0}.We will show that the existence of solutions in each class and the boundedness of all solutions are fully characterized by means of convergence or divergence of the following integrals:(2.5)J1=∫a∞f-1(1p(t)∫at‍q(s)ds)dt,J2=∫a∞f-1(-1p(t)∫at‍q(s)ds)dt,J3=∫a∞f-1(1p(t)∫t∞q(s)ds)dt,J4=∫a∞f-1(-1p(t)∫t∞q(s)ds)dt,J5=∫a∞f-1(1p(t))dt.Theorem 2.5.
Let (H2) and (H3) hold. Then all positive (negative) solutions of (1.1) are bounded if and only if J1<∞(J2>-∞).Proof.
We consider positive solutions only since the case of negative solutions can be handled similarly.
Necessity. Let x be a positive bounded class A solution. Then x(t)>0 and x′(t)>0 for t≥b>a and limt→∞x(t)=l∈(0,∞). By the Extreme Value Theorem, we have L1∶=minx(b)≤r≤lg(r)>0. Hence
(2.6)p(t)h(x(t))f(x′(t))=p(b)h(x(b))f(x′(b))+∫btq(s)g(x(s))ds≥L1∫btq(s)ds.
Since x is continuous and bounded and h(r) is continuous, then h(x(t)) is bounded. Let h(x(t))≤K for t∈[a,∞). Then
(2.7)Kp(t)f(x′(t))≥p(t)h(x(t))f(x′(t))≥L1∫btq(s)ds,KL1f(x′(t))≥1p(t)∫btq(s)ds.
By (H1), we have
(2.8)f-1(1p(t)∫btq(s)ds)≤f-1(KL1f(x′(t)))≤M1f-1(KL1)x′(t).
Integrating from b to t and letting t→∞, we have
(2.9)J1=∫b∞f-1(1p(t)∫bt‍q(s)ds)dt≤M1f-1(KL1)(l-x(b))<∞.
Sufficiency. We will prove by contradiction. Let x be a unbounded class A solution. Then x(t)>0 and x′(t)>0 on [b,∞), and there exists a real number d≥b such that x(t)≥m for d≤t<∞. Similar to the proof of Theorem 2.3, we have
(2.10)∫x(t1)x(t)drf-1(z(r))≤M12f-1(k)∫t1tf-1(1p(s)∫ds‍q(σ)dσ)ds.
Letting t→∞ and noting that x(∞)=∞, we have
(2.11)∫x(t1)∞drf-1(z(r))≤M12f-1(k)∫t1∞‍f-1(1p(s)∫bs‍q(σ)dσ)ds≤M12f-1(k)J1<∞,
a contradiction to (H3). Therefore, x is bounded.Corollary 2.6.
Let (H2) and (H3) hold. If (1.1) has a positive (negative) bounded class A solution, then all positive (negative) solutions are bounded. On the other hand, if (1.1) has an unbounded positive (negative) class A solution, then all positive (negative) solutions are unbounded.


## 3. Class A Solutions

In this section, we consider the existence of classAb and class A∞ solutions of (1.1). The necessary and sufficient conditions for the existence of class Ab solutions and the sufficient conditions for the existence of class A∞ solutions are provided.Theorem 3.1.
Equation (1.1) has both positive and negative class A solutions.Proof.
Similar to the proof of Theorem3.1 in [17].Theorem 3.2.
Equation (1.1) has a positive (negative) Ab solution if and only if J1<∞(J2>-∞).Proof.
Necessity. Without loss of generality, we assume that x is a positive Ab solution. In this case, there exists a b≥a such that x(t)>0 and x′(t)>0 for t≥b. Note that x(∞)∶=limt→∞x(t)<∞, we have
(3.1)m1∶=min⁡x(b)≤r≤x(∞)g(r)>0,c1∶=max⁡b≤t<∞h(x(t))≤max⁡x(b)≤r≤x(∞)⁡h(r)<∞.
Then
(3.2)p(t)h(x(t))f(x′(t))=p(b)h(x(b))f(x′(b))+∫btq(s)g(x(s))ds≥m1∫btq(s)ds,
and hence
(3.3)1p(t)∫btq(s)ds≤1m1h(x(t))f(x′(t)).
Taking f-1 on both sides and applying (H1) imply that
(3.4)f-1(1p(t)∫btq(s)ds)≤f-1(1m1h(x(t))f(x′(t)))≤M1f-1(c1m1)x′(t).
Therefore
(3.5)J1=∫b∞f-1(1p(t)∫bt‍q(s)ds)ds≤M1f-1(c1m1)(x(∞))-(x(b))<∞.
Sufficiency. Define
(3.6)m2=max⁡1≤r≤2g(r)>0,c2=min⁡1≤r≤2h(r)>0.
Since J1<∞, we may select a d≥a such that
(3.7)∫d∞f-1(1p(t)∫dt‍q(s)ds)dt≤1M1f-1(m2/c2).
Let CB[d,∞) be the Banach space of all bounded and continuous functions defined on [d,∞) endowed with the supremum norm, and let X={x∈CB[d,∞):1≤x(t)≤2,t≥d}. Clearly, X is a bounded convex subset of CB[d,∞). Define a mapping F1:X→CB[d,∞) by
(3.8)(F1x)(t)=1+∫dt‍f-1(1p(s)h(x(s))∫ds‍q(σ)g(x(σ))dσ)ds.
In order to apply Schauder's fixed-point theorem to show that F1 has a fixed point in X, we need to prove that F1 maps into X and is continuous, and F1(X) is precompact in CB[d,∞).
Letx∈X. Considering (3.7), we have
(3.9)1≤(F1x)(t)≤1+M1f-1(m2c2)∫dtf-1(1p(s)∫ds‍q(σ)dσ)ds≤1+M1f-1(m2c2)∫d∞‍f-1(1p(t)∫dt‍q(s)ds)dt≤2.
Hence, F1 maps X into X.
Now, we show that ifxn, x*∈X and ∥xn-x*∥→0 as n→∞, then ∥F1xn-F1x*∥→0. Indeed, for any fixed s∈[d,∞), since xn(s)→x*(s) as n→∞, we have
(3.10)|f-1(1p(s)h(xn(s))∫dsq(σ)g(xn(σ))dσ)f-1-f-1(1p(s)h(x*(s))∫dsq(σ)g(x*(σ))dσ)|⟶0asn⟶∞.
Note that
(3.11)|f-1(1p(s)h(xn(s))∫ds‍q(σ)g(xn(σ))dσ)-f-1(1p(s)h(x*(s))∫dsq(σ)g(x*(σ))dσ)|f-1≤|f-1(1p(s)h(xn(s))∫ds‍q(σ)g(xn(σ))dσ)|+|f-1(1p(s)h(x*(s))∫dsq(σ)g(x*(σ))dσ)|f-1≤2M1f-1(m2c2)f-1(1p(s)∫dsq(σ)dσ)≔F(s),
and that
(3.12)∫d∞F(s)ds=∫d∞2M1f-1(m2c2)f-1(1p(s)∫ds‍q(σ)dσ)ds=2M1f-1(m2c2)J1.
By Lebesgue’s dominated convergence theorem and considering (3.11) and (3.12) we have
(3.13)‖F1xn-F1x*‖F1xn≤sup⁡b≤t<∞∫dt‍|f-1(1p(s)h(xn(s))∫ds‍q(σ)g(xn(σ))dσ)-f-1(1p(s)h(x*(s))∫ds‍q(σ)g(x*(σ))dσ)|dsF1xn≤∫d∞‍|f-1(1p(s)h(xn(s))∫dsq(σ)g(xn(σ))dσ)-f-1(1p(s)h(x*(s))∫ds‍q(σ)g(x*(σ))dσ)|ds⟶0
as n→∞. Therefore, F1 is continuous in X.
Finally, we show the precompactness ofF1(X) in CB[d,∞), which means that for any sequence xn∈X, F1xn has a convergent subsequence in CB[d,∞). This can be proved by showing that F1xn has a convergent subsequence in C[b1,b2] for any compact subinterval [b1,b2] of [b,∞) as well as the diagonal rule. In fact, F1xn is uniformly bounded on [b1,b2]. Since
(3.14)(F1xn)′(t)=f-1(1p(t)h(xn(t))∫dt‍q(s)g(xn(s))ds)≤M1f-1(m2c2)f-1(1p(t)∫dt‍q(t)ds).
By the Mean Value Theorem, we have
(3.15)|(F1xn)(t1)-(F1xn)(t2)|=|(F1xn)′(ξ)(t1-t2)|≤M1f-1(m2c2)max⁡b1≤t≤b2f-1(1p(t)∫dt‍q(s)ds)|t1-t2|.
Then F1xn is uniformly bounded and equicontinuous in C[b1,b2]. So F1xn has a convergent subsequence in C[b1,b2] by Arzelà-Ascoli Theorem.
Now all conditions of Schauder's fixed-point theorem are satisfied, soF1 has a fixed point x¯ in X, that is,
(3.16)x¯(t)=1+∫dtf-1(1p(s)h(x¯(s))∫ds‍q(σ)g(x¯(σ))dσ)ds.
It is easy to verify that [p(t)h(x¯(t))f(x¯′(t))]′=q(t)g(x¯(t)). Hence, x¯ is a positive Ab solution of (1.1). The proof is complete.Theorem 3.3.
Let (H2) and (H3) hold. Then(a)
A∞=∅ if and only if J1<∞ and J2>-∞.(b)
Equation (1.1) has a positive (negative) A∞ solution if J1=∞(J2=-∞).Proof.
By Theorem2.5 all solutions of (1.1) are bounded if and only if J1<∞ and J2>-∞, so part (a) follows.
IfJ1=∞, there is no positive Ab solution of (1.1) from Theorem 3.2. Therefore, Theorem 3.1 guarantees the existence of a positive A∞ solution of (1.1). Similarly, there exists a negative A∞ solution of (1.1) if J2=-∞.


## 4. Class B Solutions

In this section the existence of classB, Bb, and B0 solutions are discussed. We assume that (1.1) has a unique solution for any initial conditions x(a)=x0≠0 and x′(a)=x1.Theorem 4.1.
Assume the following assumptions hold.(H2a)
There exists a continuous functionG(r):ℝ→ℝ  such that G is increasing, rG(r)>0 for r≠0 and |g(r)|≤|G(r)|;(H4)
There existsr0>0 such that
(4.1)∫0±r0drf-1(z(r))=∞.Then (1.1) has (a)
both positive and negative solutions in class B;(b)
no solution which is eventually identically equal to zero.Proof.
(a) We prove that classB has a positive solution, the case of having a negative solution is similar. Assume x0>0. The solution of (1.1) with initial conditions x(a)=x0 and x′(a)=c, denoted by x(t):=x(t,c), has the form
(4.2)x(t)=x0+∫atf-1(p(a)h(x0)f(c)p(s)h(x(s))+1p(s)h(x(s))∫asq(σ)g(x(σ))dσ)ds.
Define two sets U and L as
(4.3)U={c∈R:thereexistssomet¯≥asuchthatx′(t¯,c)>0},L={c∈R:thereexistssomet¯≥asuchthatx(t¯,c)<0}.
Then U∩L=∅. Clearly, U≠∅. We claim that U is open. Indeed, if c0∈U, there exists t¯>a such that x′(t¯,c0)>0. For any c∈ℝ, we have
(4.4)p(t¯)h(x(t¯,c0))f(x′(t¯,c0))-p(t¯)h(x(t¯,c))f(x′(t¯,c))=p(a)h(x0)f(c0)-p(a)h(x0)f(c)+∫at¯q(s)(g(x(s,c0))-g(x(s,c)))ds.
Since (1.1) has a unique solution for any initial conditions x(a)≠0, x′(a), this solution is continuously dependent on initial data. If c→c0, we have g(x(s,c))-g(x(s,c0))→0 uniformly for s on [a,t¯]. Hence, x′(t¯,c)>0 for all c that are close to c0, this proves the openness of U.
Next we show thatL≠∅. Define
(4.5)M2≔min⁡0≤r≤x0h(r)>0,M3:=min⁡a≤t≤a+1p(t)>0.
Let
(4.6)c<f-1(M2M3f-1(-x0)-G(x0)∫aa+1q(s)dsp(a)h(x0))<0.
If there exists b∈(a,a+1] such that x(b,c)<0, then c∈L and L≠∅. Otherwise, x(t,c)≥0 on [a,a+1]. In this case, we claim x′(t,c)<0 on [a,a+1]. If this is not true, since x′(a,c)=c<0, there exists t1∈(a,a+1] such that x′(t1,c)=0 and x′(t,c)<0 for t∈[a,t1). Taking into account (4.6) we have
(4.7)0=p(t)h(x(t1,c))f(x′(t1,c))=p(a)h(x0)f(c)+∫at1‍q(s)g(x(s,c))ds≤p(a)h(x0)f(c)+G(x0)∫aa+1‍q(s)ds<0.
This is a contradiction and hence x′(t,c)<0 on [a,a+1]. Notice that
(4.8)x(a+1,c)=x0+∫aa+1f-1(p(a)h(x0)f(c)p(t)h(x(t))+1p(t)h(x(t))∫atq(s)g(x(s))ds)dt≤x0+∫aa+1f-1(p(a)h(x0)f(c)+G(x0)∫aa+1‍q(s)dsM2M3)dt<0,
we know c∈L. Clearly, L is open, then ℝ-(U∪L)≠∅. Take c∈ℝ-(U∪L), x(t,c) is a nonincreasing nonnegative solution on [a,∞). We will show that x(t,c)>0 on [a,∞). If not, there exists t0>a such that x(t0)=0 and x(t)=0 for t≥t0 and x′(t0)=0. Note that for t∈[a,t0] we have
(4.9)x′(t)=f-1(-1p(t)h(x(t))∫tt0q(s)g(x(s))ds)≥f-1(-G(x(t))p(t)h(x(t))∫tt0‍q(s)ds)≥M1f-1(z(x(t)))f-1(-1p(t)∫tt0‍q(s)ds).
Dividing both sides by f-1(z(x(t))) and integrating from a to t0, we have
(4.10)∫at0‍x′(t)f-1(z(x(t)))dt≥M1∫at0f-1(-1p(t)∫tt0‍q(s)ds)dt.
That is
(4.11)∫0x0‍1f-1(z(r))dr≤-M1∫at0f-1(-1p(t)∫tt0‍q(s)ds)dt<∞,
a contradiction to (H4). Therefore, x(t)>0 for t≥a and x∈B.
The proof of part (b) follows from the end part of the proof of part (a).Theorem 4.2.
Equation (1.1) has a positive (negative) Bb solution if and only if J4>-∞(J3<∞).Proof.
Necessity. We assume that x is a positive Bb solution. The case of negative Bb solution is similar. In this case, we have x(t)>0 and x′(t)<0 for t≥a. Let
(4.12)m1=min⁡x(∞)≤r≤x(a)g(r)>0,c1=max⁡x(∞)≤r≤x(a)h(r)>0
and note that p(t)h(x(t))f(x′(t))<0, (p(t)h(x(t))f(x′(t)))′>0. Then
(4.13)lim⁡t→∞p(t)h(x(t))f(x′(t))=B≤0.
Integrating both sides of (1.1) from t to ∞ implies that
(4.14)m1∫t∞‍q(s)ds≤∫t∞‍q(s)g(x(s))ds=B-(p(t)h(x(t))f(x′(t)))≤-p(t)h(x(t))f(x′(t)).
Hence,
(4.15)f-1(-1p(t)∫t∞‍q(s)ds)≥M1f-1(c1m1)x′(t).
Again, integrating both sides of the above inequality we have
(4.16)J4=∫a∞‍f-1(-1p(t)∫t∞‍q(s)ds)dt≥M1f-1(c1m1)(x(∞)-x(a))>-∞.
Sufficiency. Let
(4.17)m2=max⁡1≤r≤2g(r)>0,c2=min⁡1≤r≤2h(r)>0.
Since J4>-∞ we choose d>a such that
(4.18)∫d∞‍f-1(-1p(t)∫t∞‍q(s)ds)dt≥-1M1f-1(m2/c2).
Let X and CB[d,∞) as defined in Theorem 3.2. Define F2:X→CB[d,∞) by
(4.19)(F2x)(t)=1-∫t∞‍f-1(-1p(s)h(x(s))∫s∞‍q(σ)g(x(σ))dσ)ds.
For any x∈X, we have
(4.20)1≤(F2x)(t)≤1-∫d∞‍M1f-1(m2c2)f-1(-1p(s)∫s∞‍q(σ)dσ)ds≤2.
This proves that F2 maps X into X. Similar to the proof of Theorem 3.2, we are able to show that F2 is continuous in X, and F2(X) is precompact in CB[d,∞). Then F2 has a fixed-point x¯(t) in X by Schauder's fixed-point theorem, that is,
(4.21)x¯(t)=1-∫t∞‍f-1(-1p(s)h(x¯(s))∫s∞‍q(σ)g(x¯(σ))dσ)ds.
It is easy to verify that x¯(t) is a positive Bb solution of (1.1). The proof is complete.Theorem 4.3.
Let (H2a) and (H4) hold and letJ5=∞. Then (1.1) has a positive (negative) B0 solution if and only if J4=-∞ (J4=∞).Proof.
We prove the assertion for positive solutions without loss of generality.
Necessity. Assume x(t) is a positive B0 solution. Then x(t)>0 and x′(t)<0 for t≥a, x(∞)=0, and limt→∞p(t)h(x(t))f(x′(t))=L∈(-∞,0]. We claim L=0. In fact, if L<0, since p(t)h(x(t))f(x′(t)) is negative and increasing on [a,∞), then p(t)h(x(t))f(x′(t))≤L and
(4.22)x′(t)≤f-1(Lc1p(t))≤M1f-1(Lc1)f-1(1p(t)),
where c1=max0≤r≤x(a)h(r)>0.
Integrating both sides froma to ∞ and noting that x(∞)=0, we have
(4.23)x(a)≥-M1f-1(Lc1)∫a∞‍f-1(1p(t))dt,
a contradiction to J5=∞ and hence L=0.
Integrating both sides of (1.1) from t to ∞ we have
(4.24)p(t)h(x(t))f(x′(t))=-∫t∞‍q(s)g(x(s))ds.
Then
(4.25)x′(t)=f-1(-1p(t)h(x(t))∫t∞‍q(s)g(x(s))ds)≥M1f-1(z(x(t)))f-1(-1p(t)∫t∞‍q(s)ds).
Hence,
(4.26)x′(t)f-1(z(x(t)))≥M1f-1(-1p(t)∫t∞‍q(s)ds).
Integrating both sides of the above inequality from a to ∞ implies that
(4.27)∫0x(a)‍drf-1(z(r))≤-M1∫a∞‍f-1(-1p(t)∫t∞‍q(s)ds)dt.
Therefore, J4=-∞ from (H4).
Sufficiency. By Theorem 4.1 (1.1) has a positive class B solution x, either x∈Bb or x∈B0. Note that J4=-∞ implies that x∉Bb from Theorem 4.2. So x∈B0. The proof is complete.


## 5. Remarks

In this section, we present several remarks about comparison of our results with the existing ones in the literature.Theorems2.3 and 2.5 improve [10, Theorem 1] since (H3) reduces to (iii) of [10] if f(r)=r and the differentiability of p(·) and h(·) is not required. Theorems 2.3, 2.5, and 4.2 complement and generalize [2, Theorem 8]. Moreover, under (H2), Theorems 2.3, 2.5, and 4.2 improve [2, Theorem 8] since (H3) improves (22) of [2]; see the discussion in [16]. Theorem 2.5 generalizes [13, Theorem 3.9]. Theorems 2.3, 3.1, and 4.2 generalize [16, Theorem 1]. Theorem 3.2 generalizes [2, Theorem 3], [16, Theorem 3], and [18, Theorem 2.1]. Theorem 3.3 generalizes [18, Theorem 2.2]. Theorem 4.1 generalizes [13, Theorem 2.1] and improves [3, Theorem 6] under (H2a) since (hp) in [3] is replaced by a weaker condition (H4). Theorem 4.2 generalizes [2, Theorem 1], [16, Theorem 5], and [18, Theorem 3.1]. Theorem 4.3 generalize [16, Theorem 6] and [18, Theorem 3.2].



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*Source: 313725-2012-07-16.xml*