# Existence of Nonoscillatory Solutions for a Third-Order Nonlinear Neutral Delay Differential Equation

Authors: Zeqing Liu; Lin Chen; Shin Min Kang; Sun Young Cho
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

The aim of this paper is to study the solvability of a third-order nonlinear neutral delay differential equation of the form{α(t)[β(t)(x(t)+p(t)x(t−τ))′]′}′+f(t,x(σ1(t)),x(σ2(t)),…,x(σn(t)))=0, t≥t0. By using the Krasnoselskii's fixed point theorem and the Schauder's fixed point theorem, we demonstrate the existence of uncountably many bounded nonoscillatory solutions for the above differential equation. Several nontrivial examples are given to illustrate our results.


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## Body

## 1. Introduction and Preliminaries

In recent years, the study of the oscillation, nonoscillation, asymptotic behaviors and existence of solutions for various kinds of first- and second-order neutral delay differential equations and systems of differential equations have attracted much attention, for example, see [1–12] and the references therein. Dorociaková and Olach [2] discussed the existence of nonoscillatory solutions and asymptotic behaviors for the first-order delay differential equation (1.1)x′(t)+p(t)x(t)+q(t)x(τ(t))=0,t≥0.
Elbert [3] and Huang [5] established a few oscillation and nonoscillation criteria for the second-order linear differential equation (1.2)x′′(t)+q(t)x(t)=0,t≥0,
where q∈C([0,+∞),ℝ+). Tang and Liu [10] studied the existence of bounded oscillation for the second-order linear delay differential equation of unstable type (1.3)x′′(t)=p(t)x(t-τ),t≥t0,
where τ>0,p∈C([t0,+∞),ℝ+)  and  p(t)≢0 on any interval of length τ. In view of the Banach fixed point theorem, Kulenović and Hadžiomerspahić [7] deduced the existence of a nonoscillatory solution for the second-order linear neutral delay differential equation with positive and negative coefficients (1.4)[x(t)+cx(t-τ)]′′+Q1(t)x(t-σ1)-Q2(t)x(t-σ2)=0,t≥t0,
where c∈ℝ∖{-1,1},τ>0,σ1,σ2∈[0,+∞),Q1,Q2∈C([t0,+∞),ℝ+). Qin et al. [9] and Yang et al. [11] developed several oscillation criteria for the second-order differential equation (1.5)[r(t)(x(t)+p(t)x(t-τ))′]′+q(t)f(x(t-δ))=0,t≥t0,
where τ and δ are nonnegative constants, r,p,q∈C([t0,+∞),ℝ), and f∈C(ℝ,ℝ). By utilizing the Krasnoselskii’s fixed point theorem, Zhou [12] discussed the existence of nonoscillatory solutions of the second-order nonlinear neutral differential equation (1.6)[r(t)(x(t)+p(t)x(t-τ))′]′+∑i=1mQi(t)f(xi(t-σi))=0,t≥t0,
where m≥1 is an integer, τ>0,σi≥0,r,p,Qi∈C([t0,+∞),ℝ), and fi∈C(ℝ,ℝ) for i∈{1,2,…,m}.However, the existence of nonoscillatory solutions of third-order neutral differential equations received much less attention, moreover, the results in [7, 11, 12] only figured out the existence of a nonoscillatory solution of (1.3)–(1.5), respectively.Motivated by the papers mentioned above, in this paper, we investigate the following third-order nonlinear neutral delay differential equation(1.7){α(t)[β(t)(x(t)+p(t)x(t-τ))′]′}′+f(t,x(σ1(t)),x(σ2(t)),…,x(σn(t)))=0,t≥t0,
where n≥1 is an integer, τ>0,α,β∈C([t0,+∞),ℝ+∖{0}),p∈C([t0,+∞),ℝ), and f∈C([t0,+∞)×ℝn,ℝ). By applying the Krasnoselskii's fixed point theorem and the Schauder's fixed point theorem, we obtain some sufficient conditions for the existence of uncountably many bounded nonoscillatory solutions of (1.7).Throughout this paper, we assume thatℝ=(-∞,+∞),ℝ+=[0,+∞),C([t0,+∞),ℝ) denotes the Banach space of all continuous and bounded functions on [t0,+∞) with the norm ∥x∥=supt≥t0|x(t)| for each x∈C([t0,+∞),ℝ) and
(1.8)A(N,M)={x∈C([t0,+∞),R):N≤x(t)≤M,t≥t0}forM>N>0.
It is easy to see that A(N,M) is a bounded closed and convex subset of C([t0,+∞),ℝ).By a solution of (1.7), we mean a function x∈C([t1-τ,+∞),ℝ) for some t1≥t0 such that x(t)+p(t)x(t-τ),β(t)(x(t)+p(t)x(t-τ))′ and α(t)[β(t)(x(t)+p(t)x(t-τ))′]′are continuously differentiable in [t1,+∞) and such that (1.7) is satisfied for t≥t1. As is customary, a solution of (1.7) is called  oscillatory if it has arbitrarily large zeros, and otherwise it is said to be nonoscillatory.Definition 1.1 (see [6]).
A familyF of functions in C([t0,+∞),ℝ) isequicontinuous on [t0,+∞) if for any ε>0, the interval [t0,+∞) can be decomposed into a finite number of subintervals I1,I2,…,In such that
(1.9)|f(x)-f(y)|≤ε,∀f∈F,x,y∈Ii,i∈{1,2,…,n}.Lemma 1.2 (see Krasnoselskii’s fixed point theorem, [4]).
LetX be a Banach space. Let Ω be a bounded closed convex subset of X and S1 and S2 mappings from Ω into X such that S1x+S2y∈Ω for every pair x,y∈Ω. If S1 is a contraction and S2 is completely continuous, then the equation S1x+S2x=x has at least one solution in Ω.Lemma 1.3 (see Schauder’s fixed point theorem, [4]).
LetΩ be a nonempty closed convex subset of a Banach space X. Let S:Ω→Ω be a continuous mapping such that SΩ is a relatively compact subset of X. Then S has at least one fixed point in Ω.


## 2. Main Results

Now we study those conditions under which (1.7) possesses uncountably many bounded nonoscillatory solutions.Theorem 2.1.
Assume that there exist constantsM,N,c1,c2,T0 and a function h∈C([t0,+∞),ℝ+) satisfying
(2.1)min⁡{c1,c2}≥0,c1+c2<1,0<N<(1-c1-c2)M,(2.2)-c2≤p(t)≤c1,t≥T0>t0;(2.3)|f(t,u1,u2,…,un)|≤h(t),t∈[t0,+∞),ui∈[N,M],i∈{1,…,n};(2.4)∫t0+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<+∞.
Then  (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Proof.
Setk∈(c1M+N,(1-c2)M). From (2.4), we pick up T>T0 such that
(2.5)∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<min⁡{(1-c2)M-k,k-c1M-N}.
Define two mappings S1k and S2k:A(N,M)→C([t0,+∞),ℝ) by
(2.6)(S1kx)(t)={k-p(t)x(t-τ),t≥T,(S1kx)(T),t0≤t<T,(2.7)(S2kx)(t)={∫t+∞∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv,t≥T,(S2kx)(T),t0≤t<T
for x∈A(N,M).
Firstly, we prove thatS1k(A(N,M))+S2k(A(N,M))⊆A(N,M). By (2.1)–(2.3) and (2.5)–(2.7), we get that
(2.8)(S1kx)(t)+(S2ky)(t)≤k+c2M+∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤k+c2M+min⁡{(1-c2)M-k,k-c1M-N}≤M,x,y∈A(N,M),t≥T,(S1kx)(t)+(S2ky)(t)≥k-c1M-∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≥k-c1M-min⁡{(1-c2)M-k,k-c1M-N}≥N,x,y∈A(N,M),t≥T,
which infer that S1k(A(N,M))+S2k(A(N,M))⊆A(N,M) for any x,y∈A(N,M).
Secondly, we show thatS1k is a contraction mapping. By (2.1), (2.2), and (2.6), we deduce that
(2.9)|(S1kx)(t)-(S1ky)(t)|≤|p(t)||x(t-τ)-y(t-τ)|≤(c1+c2)‖x-y‖
for x,y∈A(N,M) and t≥T, which gives that
(2.10)‖S1kx-S1ky‖≤(c1+c2)‖x-y‖.
Thirdly, we show thatS2k is completely continuous. Let x∈A(N,M) and {xm}m≥1⊆A(N,M) with limm→+∞xm=x. By (2.7), we obtain that
(2.11)|(S2kxm)(t)-(S2kx)(t)|≤∫T+∞∫v+∞∫s+∞1α(s)β(v)|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(u)),…,x(σn(u)))|dudsdv,t≥T,m≥1.
Using (2.3) and (2.4), we conclude that
(2.12)|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(u)),…,x(σn(u)))|≤2h(u),u∈[T,+∞),m≥1,∫s+∞|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(u)),…,x(σn(u)))|du≤2∫s+∞h(u)du,s∈[T,∞),m≥1,∫v+∞∫s+∞1α(s)|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(u)),…,x(σn(u)))|duds≤2∫v+∞∫s+∞h(u)α(s)duds,v,s∈[T,∞),m≥1.
In light of (2.11)–(2.12),
(2.13)|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(u)),…,x(σn(u)))|⟶0asm⟶+∞,
and the Lebesgue dominated convergence theorem, we conclude that
(2.14)lim⁡m→+∞‖S2kxm-S2kx‖=0,
which means that S2k is continuous in A(N,M).
Now we show thatS2k is completely continuous. By virtue of (2.3), (2.4), and (2.7), we get that
(2.15)‖S2kx‖≤∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv,x∈A(N,M).
That is, S2k(A(N,M)) is uniform bounded. It follows from (2.4) that for each ε>0, there exists T*>T such that
(2.16)∫T*+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<ε2.
In view of (2.3), (2.7), and (2.16), we infer that
(2.17)|(S2kx)(t2)-(S2kx)(t1)|≤∫t2+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv+∫t1+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<ε,x∈A(N,M),t2>t1≥T*.
From (2.3) and (2.7), we get that
(2.18)|(S2kx)(t2)-(S2kx)(t1)|≤∫t1t2∫v+∞∫s+∞h(u)α(s)β(v)dudsdv,x∈A(N,M),T≤t1≤t2≤T*,
which together with (2.4) ensures that there exists δ>0 satisfying
(2.19)|(S2kx)(t2)-(S2kx)(t1)|<ε,x∈A(N,M),t1,t2∈[T,T*]with|t2-t1|<δ.
Clearly,
(2.20)|(S2kx)(t2)-(S2kx)(t1)|=0<ε,x∈A(N,M),t0≤t1≤t2≤T.
That is, S2k(A(N,M)) is equicontinuous on [t0,+∞). Consequently, S2k is completely continuous. By Lemma 1.2, there is x0∈A(N,M) such that S1x0+S2x0=x0, which is a bounded nonoscillatory solution of (1.7).
Lastly, we demonstrate that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M). Let k1,k2∈(c1M+N,(1-c2)M) and k1≠k2. For each j∈{1,2}, we choose Tj>T0 and the mappings Sjk1 and Sjk2 satisfying (2.5)–(2.7) with k and T replaced by kj and Tj, respectively, and some T3>max{T1,T2} such that
(2.21)∫T3+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<|k1-k2|2.
Obviously, there are x,y∈A(N,M) such that S1k1x+S2k1x=x and S1k2y+S2k2y=y, respectively. That is, x and y are two bounded nonoscillatory solutions of (1.7) in A(N,M). In order to prove that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M), we prove only that x≠y. In fact, by (2.6) and (2.7), we gain that for t≥T3(2.22)x(t)=k1-p(t)x(t-τ)+∫t+∞∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv,y(t)=k2-p(t)y(t-τ)+∫t+∞∫v+∞∫s+∞1α(s)β(v)f(u,y(σ1(u)),…,y(σn(u)))dudsdv,
which together with (2.1)–(2.3) imply that
(2.23)|x(t)-y(t)|≥|k1-k2|-|p(t)||x(t-τ)-y(t-τ)|-∫t+∞∫v+∞∫s+∞1α(s)β(v)×|f(u,x(σ1(u)),…,x(σn(u)))-f(u,y(σ1(u)),…,y(σn(u)))|dudsdv≥|k1-k2|-(c1+c2)‖x-y‖-2∫T3+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv,t≥T3,
which means that
(2.24)‖x-y‖≥11+c1+c2(|k1-k2|-2∫T3+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv)>0,
by (2.1) and (2.21). That is, x≠y. This completes the proof.Theorem 2.2.
Assume that there exist constantsM,N,c1,c2,T0 and a function h∈C([t0,+∞),ℝ+) satisfying (2.3), (2.4) and
(2.25)0<(1-c2)N<(1-c1)M;(2.26)0≤c2≤p(t)≤c1<1,t≥T0>t0.
Then (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Proof.
Setk∈(c1M+N,M+c2N). It follows from (2.4) that there exists T>T0 such that
(2.27)∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<min⁡{M+c2N-k,k-c1M-N}.
Define two mappings S1k and S2k:A(N,M)→C([t0,+∞),ℝ) by (2.6) and (2.7), respectively. By virtue of (2.3), (2.6), (2.7), (2.26), and (2.27), we obtain that
(2.28)(S1kx)(t)+(S2ky)(t)≤k-p(t)x(t-τ)+∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤k-c2N+min⁡{M+c2N-k,k-c1M-N}≤M,x,y∈A(N,M),t≥T,(2.29)(S1kx)(t)+(S2ky)(t)≥k-p(t)x(t-τ)-∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≥k-c1M-min⁡{M+c2N-k,k-c1M-N}≥N,x,y∈A(N,M),t≥T,
which yield that S1k(A(N,M))+S2k(A(N,M))⊆A(N,M).
By a similar argument used in the proof of Theorem2.1, we gain that S1k is a contraction mapping S2k is completely continuous, and (1.7) possesses uncountably many nonoscillatory solutions. This completes the proof.Theorem 2.3.
Assume that there exist constantsM,N,c1,c2,T0 and a function h∈C([t0,+∞),ℝ+) satisfying  (2.3), (2.4), (2.25), and
(2.30)-1<-c1≤p(t)≤-c2≤0,t≥T0>t0.
Then (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Proof.
Setk∈((1-c2)N,(1-c1)M). It follows from (2.4) that there exists T>T0 satisfying
(2.31)∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<min⁡{(1-c1)M-k,k-(1-c2)N}.
Let S1k and S2k:A(N,M)→C([t0,+∞),ℝ) be defined by (2.6) and (2.7), respectively. In view of (2.3), (2.6), (2.7), (2.30), and (2.31), we obtain that
(2.32)(S1x)(t)+(S2y)(t)≤k-p(t)x(t-τ)+∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤k+c1M+min⁡{(1-c1)M-k,k-(1-c2)N}≤M,x,y∈A(N,M),t≥T,(S1x)(t)+(S2y)(t)≥k-p(t)x(t-τ)-∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≥k+c2N-min⁡{(1-c1)M-k,k-(1-c2)N}≥N,x,y∈A(N,M),t≥T,
which mean that S1k(A(N,M))+S2k(A(N,M))⊆A(N,M).
The rest of the proof is similar to the proof of Theorem2.1 and is omitted. This completes the proof.Theorem 2.4.
Assume that there exist constantsM and N with M>N>0 and a function h∈C([t0,+∞),ℝ+) satisfying (2.3) and (2.4). If p(t)=1 for each t∈[t0,+∞), then (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Proof.
Setk∈(N,M). It follows from (2.4) that there exists T≥t0+τ satisfying
(2.33)∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<min⁡{M-k,k-N}.
Define a mapping Sk:A(N,M)→C([t0,+∞),ℝ) by
(2.34)(Skx)(t)={k+∑i=1∞∫t+(2i-1)τt+2iτ∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv,t≥T,(Skx)(T),t0≤t<T
for x∈A(N,M).
Firstly, we prove thatSk(A(N,M))⊆A(N,M). By virtue of (2.3), (2.33) and (2.34), we obtain that
(2.35)(Skx)(t)≤k+∑i=1∞∫t+(2i-1)τt+2iτ∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤k+∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤M,x∈A(N,M),t≥T,(Skx)(t)≥k-∑i=1∞∫t+(2i-1)τt+2iτ∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≥k-∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≥N,x∈A(N,M),t≥T,
which imply that Sk(A(N,M))⊆A(N,M).
Secondly, we show thatSk is continuous in A(N,M). Let x∈A(N,M) and {xm}m≥1⊆A(N,M) with limm→+∞xm=x. By (2.34), we get that
(2.36)|(Skxm)(t)-(Skx)(t)|≤∑i=1∞∫t+(2i-1)τt+2iτ∫v+∞∫s+∞1α(s)β(v)|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(u)),…,x(σn(u)))|dudsdv≤∫T+∞∫v+∞∫s+∞1α(s)β(v)|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(u)),…,x(σn(u)))|dudsdv,t≥T,m≥1.
In view of (2.12), (2.13), (2.36), and the Lebesgue dominated convergence theorem, we deduce that
(2.37)lim⁡m→+∞‖Skxm-Skx‖=0,
which means that Sk is continuous in A(N,M).
Thirdly, we show thatSk(A(N,M)) is relatively compact. From (2.3), (2.33), and (2.34), we gain that
(2.38)‖Skx‖≤k+∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤2M,x∈A(N,M),
which means that Sk(A(N,M)) is uniform bounded.
Letε>0. It follows from (2.4) that there exist T̅>T*>T such that
(2.39)∫T*+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<ε2,(2.40)∫T̅+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<ε4.
By (2.3), (2.34), and (2.39), we deduce that
(2.41)|(Skx)(t2)-(Skx)(t1)|≤∑i=1∞(∫t2+(2i-1)τt2+2iτ+∫t1+(2i-1)τt1+2iτ)∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤2∫T*+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<ε,x∈A(N,M),t2>t1≥T*.
Choose an integer z≥1 with T+(2z+1)τ≥T̅, and put
(2.42)A=max⁡{∫v+∞∫s+∞h(u)α(s)β(v)duds:v∈[T+τ,T*+(2z-1)τ]},B=max⁡{∫v+∞∫s+∞h(u)α(s)β(v)duds:v∈[T+2τ,T*+2zτ]},δ=min⁡{ε1+4Az,ε1+4Bz}.
Equation (2.4) means that max{A,B}<+∞. It follows from (2.3), (2.34), and (2.40)–(2.42) that for x∈A(N,M),t1,t2∈[T,T*]witht1≤t2≤t1+δ(2.43)|(Skx)(t2)-(Skx)(t1)|≤|∑i=1z(∫t2+(2i-1)τt2+2iτ-∫t1+(2i-1)τt1+2iτ)×∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv∫t1+(2i-1)τt1+2iτ|+∑i=z+1∞(∫t2+(2i-1)τt2+2iτ+∫t1+(2i-1)τt1+2iτ)∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤|∑i=1z(∫t2+(2i-1)τt1+(2i-1)τ+∫t1+(2i-1)τt2+2iτ+∫t1+2iτt1+(2i-1)τ)×∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv∫t1+2iτt1+(2i-1)τ|+2∫T̅+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤∑i=1z(∫t1+(2i-1)τt2+(2i-1)τ+∫t1+2iτt2+2iτ)∫v+∞∫s+∞h(u)α(s)β(v)dudsdv+ε2≤Az(t2-t1)+Bz(t2-t1)+ε2<ε.
It is not difficult to verify that
(2.44)|(Skx)(t2)-(Skx)(t1)|=0<ε,x∈A(N,M),t0≤t1≤t2≤T.
Therefore Sk(A(N,M)) is equicontinuous on [t0,+∞), and consequently Sk is relatively compact. By Lemma 1.3, there is x0∈A(N,M) such that Sx0=x0, which together with (2.34) yields that for t≥T+τ(2.45)x0(t)=k+∑i=1∞∫t+(2i-1)τt+2iτ∫v+∞∫s+∞1α(s)β(v)f(u,x0(σ1(u)),…,x0(σn(u)))dudsdv,x0(t-τ)=k+∑i=1∞∫t+2(i-1)τt+(2i-1)τ∫v+∞∫s+∞1α(s)β(v)f(u,x0(σ1(u)),…,x0(σn(u)))dudsdv,
which mean that
(2.46)x0(t)+x0(t-τ)=2k+∫t+∞∫v+∞∫s+∞1α(s)β(v)f(u,x0(σ1(u)),…,x0(σn(u)))dudsdv.
It is easy to show that x0 is a bounded nonoscillatory solution of (1.7).
Finally, we demonstrate that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M). Let k1,k2∈(N,M) and k1≠k2. For each j∈{1,2}, we pick up a positive integer Tj≥t0+τ and the mapping Skj satisfying (2.33) and (2.34), where k and T are replaced by kj and Tj, respectively, and some T3>max{T1,T2} satisfying (2.21). Clearly, there are x and y∈A(N,M) such that Sk1x=x and Sk2y=y, respectively. That is, x and y are bounded nonoscillatory solutions of (1.7) in A(N,M). By (2.34) we get that for t≥T3(2.47)x(t)=k1+∑i=1∞∫t+(2i-1)τt+2iτ∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv,y(t)=k2+∑i=1∞∫t+(2i-1)τt+2iτ∫v+∞∫s+∞1α(s)β(v)f(u,y(σ1(u)),…,y(σn(u)))dudsdv,
which together with (2.3) and (2.21) yield that
(2.48)‖x-y‖≥|k1-k2|-2∫T3+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv>0,
which implies that x≠y. This completes the proof.Theorem 2.5.
Assume that there exist constantsM and N with M>N>0 and a function h∈C([t0,+∞),ℝ+) satisfying (2.3) and
(2.49)∑i=1∞∫t0+iτ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<+∞.
If p(t)=-1 for each t∈[t0,+∞), then (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Proof.
Setk∈(N,M). It follows from (2.49) that there exists T>t0 satisfying
(2.50)∑i=1∞∫T+iτ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<min⁡{M-k,k-N}.
Define a mapping Sk:A(N,M)→C([t0,+∞),ℝ) by
(2.51)(Skx)(t)={k-∑i=1∞∫t+iτ+∞∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv,t≥T,(Skx)(T),t0≤t<T
for x∈A(N,M).
Firstly, we prove thatSk(A(N,M))⊆A(N,M). By (2.3), (2.50), and (2.51), we obtain that
(2.52)(Skx)(t)≤k+∑i=1∞∫t+iτ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤M,x∈A(N,M),t≥T,(Skx)(t)≥k-∑i=1∞∫t+iτ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≥N,x∈A(N,M),t≥T,
which imply that S(A(N,M))⊆A(N,M).
Secondly, we show thatSk is continuous in A(N,M). Let x∈A(N,M) and {xm}m≥1⊆A(N,M) with limm→+∞xm=x. By (2.51), we get that
(2.53)|(Skxm)(t)-(Skx)(t)|≤∑i=1∞∫T+iτ+∞∫v+∞∫s+∞1α(s)β(v)|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(u)),…,x(σn(u)))|dudsdv,t≥T,m≥1.
Equation (2.53) together with (2.3), (2.49) and the Lebesgue dominated convergence theorem yields that
(2.54)lim⁡m→+∞‖Skxm-Skx‖=0,
that is, Sk is continuous in A(N,M).
Thirdly, we show thatSk(A(N,M)) is relatively compact. From (2.3), (2.50), and (2.51), we obtain that
(2.55)‖Skx‖≤k+∑i=1∞∫T+iτ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤2M,x∈A(N,M),
which means that Sk(A(N,M)) is uniform bounded. It follows from (2.49) that, for each ε>0, there exists T*>T such that
(2.56)∑i=1∞∫T*+iτ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<ε2.
Notice that (2.3), (2.51), and (2.56) yield that
(2.57)|(Skx)(t2)-(Skx)(t1)|≤2∑i=1∞∫T*+iτ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<ε,x∈A(N,M),t2>t1≥T*.
Choose a positive integer z satisfying T+zτ≥T*. Put
(2.58)D=max⁡{∫v+∞∫s+∞h(u)α(s)β(v)duds:v∈[T+τ,T*+zτ]},δ=ε1+2Dz.
By (2.3), (2.51), and (2.58), we gain that
(2.59)|(Skx)(t2)-(Skx)(t1)|=|∑i=1∞(∫t2+iτ+∞-∫t1+iτ+∞)∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv|≤∑i=1∞∫t1+iτt2+iτ∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤∑i=1z∫t1+iτt2+iτ∫v+∞∫s+∞h(u)α(s)β(v)dudsdv+∑i=z+1∞∫t1+iτt2+iτ∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤Dz(t2-t1)+∑i=1∞∫T*+iτ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<ε,x∈A(N,M),t1,t2∈[T,T*]witht1≤t2≤t1+δ.
Clearly,
(2.60)|(Skx)(t2)-(Skx)(t1)|=0<ε,x∈A(N,M),t0≤t1≤t2≤T.
That is, Sk(A(N,M)) is equicontinuous on [t0,+∞), and Sk is relatively compact. The rest argument is similar to the proof of Theorem 2.4 and is omitted. This completes the proof.Theorem 2.6.
Assume that there exist constantsM,N,c1,c2,T0 and a function h∈C([t0,+∞),ℝ+) satisfying (2.3), (2.4), and
(2.61)1<c2<c1<c22,0<c12-c2c1N<c22-c1c2M;(2.62)c2≤p(t)≤c1,t≥T0≥t0+τ.
Then (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Proof.
Setk∈(c1N+(c1/c2)M,c2M+(c2/c1)N). It follows from (2.4) that there exists T>T0 such that
(2.63)∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<min⁡{c2M+c2c1N-k,c2c1k-M-c2N}.
Define two mappings S1k and S2k:A(N,M)→C([t0,+∞),ℝ) by
(2.64)(S1kx)(t)={kp(t+τ)-x(t+τ)p(t+τ),t≥T,(S1kx)(T),t0≤t<T,(2.65)(S2kx)(t)={1p(t+τ)∫t+τ+∞∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv,t≥T,(S2kx)(T),t0≤t<T
for x∈A(N,M).
Firstly, we show thatS1k(A(N,M))+S2k(A(N,M))⊆A(N,M). From (2.3), (2.61)–(2.65), we get that
(2.66)(S1kx)(t)+(S2ky)(t)≤kc2-Nc1+1c2min⁡{c2M+c2c1N-k,c2c1k-M-c2N}≤M,x,y∈A(N,M),t≥T,(S1kx)(t)+(S2ky)(t)≥kc1-Mc2-1c2min⁡{c2M+c2c1N-k,c2c1k-M-c2N}≥N,x,y∈A(N,M),t≥T,
That is, S1k(A(N,M))+S2k(A(N,M))⊆A(N,M).
Secondly, by (2.61), (2.62), and (2.64), we conclude that
(2.67)‖S1kx-S1ky‖≤1c2‖x-y‖,x,y∈A(N,M),
which implies that S1k is a contraction mapping in A(N,M).
Thirdly, we show thatS2k is completely continuous. Let x∈A(N,M) and {xm}m≥1⊆A(N,M) be such that xm→x as m→+∞. By (2.61), (2.62), and (2.65), we gain that
(2.68)|(S2kxm)(t)-(S2kx)(t)|≤1p(t+τ)∫t+τ+∞∫v+∞∫s+∞1α(s)β(v)|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(t)),…,x(σn(t)))|dudsdv≤1c2∫T+∞∫v+∞∫s+∞1α(s)β(v)|f(u,xm(σ1(u)),…,xm(σn(u)))-f(u,x(σ1(u)),…,x(σn(u)))|dudsdv,t≥T,m≥1.
In view of (2.12), (2.13), (2.68), and the Lebesgue dominated convergence theorem, we obtain that
(2.69)lim⁡m→+∞‖S2kxm-S2kx‖=0,
that is, S2k is continuous in A(N,M).
For eachε>0, it follows from (2.4) that there exists T*≥T satisfying
(2.70)∫T*+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<min⁡{c2ε2,c22ε4c1}.
From (2.3), (2.61), (2.62), (2.65), and (2.70), we gain that
(2.71)|(S2kx)(t2)-(S2kx)(t1)|≤1c2∫t2+τ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv+1c2∫t1+τ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤2c2∫T*+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<ε,x∈A(N,M),t2>t1≥T*.
By (2.3), (2.61), (2.62), (2.65), and (2.70), we obtain that, for x∈A(N,M),T≤t1≤t2≤T*,
(2.72)|(S2kx)(t2)-(S2kx)(t1)|≤1p(t2+τ)∫t1+τt2+τ∫v+∞∫s+∞h(u)α(s)β(v)dudsdv+|1p(t2+τ)-1p(t1+τ)|∫t1+τ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv≤1c2∫t1+τt2+τ∫v+∞∫s+∞h(u)α(s)β(v)dudsdv+2c1c22∫T*+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv,
which means that there exists δ>0 such that
(2.73)|(S2kx)(t2)-(S2kx)(t1)|<ε,x∈A(N,M),t1,t2∈[T,T*]with|t2-t1|<δ.
It is easy to verify that
(2.74)|(S2kx)(t2)-(S2kx)(t1)|=0<ε,x∈A(N,M),t0≤t1≤t2≤T.
That is, S2k(A(N,M)) is equicontinuous on [t0,+∞), and S2k is completely continuous. By Lemma 1.2, there is x0∈A(N,M) such that S1kx0+S2kx0=x0, which is a bounded nonoscillatory solution of (1.7).
Lastly, we demonstrate that (1.7) possesses uncountably many nonoscillatory solutions. Let k1,k2∈(c1N+(c1/c2)M,c2M+(c2/c1)N) and k1≠k2. For each j∈{1,2}, we choose Tj>t0 and contraction mappings Sjk1 and Sjk2 satisfying (2.63)–(2.65) with k and T replaced by kj and Tj, respectively, and some T3>max{T1,T2} satisfying
(2.75)∫T3+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<c2|k1-k2|2c1.
Obviously, there are x,y∈A(N,M) such that S1k1x+S2k1x=x and S1k2y+S2k2y=y, respectively, and x and y are two bounded nonoscillatory solutions of (1.7) in A(N,M). By (2.64) and (2.65), we gain that for t≥T3(2.76)x(t)=k1p(t+τ)-x(t+τ)p(t+τ)+1p(t+τ)∫t+τ+∞∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv,y(t)=k2p(t+τ)-y(t+τ)p(t+τ)+1p(t+τ)∫t+τ+∞∫v+∞∫s+∞1α(s)β(v)f(u,y(σ1(u)),…,y(σn(u)))dudsdv,
which together with (2.3), (2.61), and (2.62) mean that
(2.77)|x(t)-y(t)|≥1p(t+τ)|k1-k2|-1p(t+τ)|x(t-τ)-y(t-τ)|-1p(t+τ)∫t+τ+∞∫v+∞∫s+∞1α(s)β(v)×|f(u,x(σ1(u)),…,x(σn(u)))-f(u,y(σ1(u)),…,y(σn(u)))|dudsdv≥1c1|k1-k2|-1c2‖x-y‖-2c2∫T3+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv,t≥T3,
which together with (2.75) yields that
(2.78)‖x-y‖≥c21+c2(1c1|k1-k2|-2c2∫T3+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv)>0,
that is, x≠y. This completes the proof.Theorem 2.7.
Assume that there exist constantsM,N,c1,c2,T0 and a function h∈C([t0,+∞),ℝ+) satisfying (2.3), (2.4), and
(2.79)0<(c1-1)N<(c2-1)M;(2.80)-∞<-c1≤p(t)≤-c2<-1,t≥T0≥t0+τ.
Then (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Proof.
Setk∈((c1-1)N,(c2-1)M). It follows from (2.4) that there exists T>T0 such that
(2.81)∫T+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv<min⁡{(c2-1)M-k,c2(k+N)c1-c2N}.
Define two mappings S1k and S2k:A(N,M)→C([t0,+∞),ℝ) by
(2.82)(S1kx)(t)={-kp(t+τ)-x(t+τ)p(t+τ),t≥T,(S1kx)(T),t0≤t<T,(S2kx)(t)={1p(t+τ)∫t+τ+∞∫v+∞∫s+∞1α(s)β(v)f(u,x(σ1(u)),…,x(σn(u)))dudsdv,t≥T,(S2kx)(T),t0≤t<T
for x∈A(N,M). From (2.3), (2.80)–(2.82), we obtain that
(2.83)(S1kx)(t)+(S2ky)(t)≤kc2+Mc2+1c2min⁡{(c2-1)M-k,c2(k+N)c1-c2N}≤M,x,y∈A(N,M),t≥T,(S1kx)(t)+(S2ky)(t)≥kc1+Nc1-1c2min⁡{(c2-1)M-k,c2(k+N)c1-c2N},≥N,x,y∈A(N,M),t≥T
which implies that S1k(A(N,M))+S2k(A(N,M))⊆A(N,M). The rest of the proof is similar to the proof of Theorem 2.6 and is omitted. This completes the proof.


## 3. Examples

In this section we construct seven examples as applications of the results presented in Section2.Example 3.1.
Consider the following third-order nonlinear neutral delay differential equation:(3.1){t3et[tln⁡(1+t)(x(t)+3sin(t2-2)+17x(t-τ))′]′}′+1et(x2(2t21+3t)+x2(t-1t))=0,t≥t0,
where τ>0 and t0>0 are fixed. Let n=2, and let M and N be two positive constants with M>7N and
(3.2)α(t)=t3et,β(t)=tln⁡(1+t),p(t)=3sin(t2-2)+17,c1=47,c2=27,σ1(t)=2t21+3t,σ2(t)=t-1t,h(t)=2M2et,f(t,u,v)=u2+v2et,(t,u,v)∈[t0,+∞)×R2.
Obviously, (2.1)–(2.3) hold. On the other hand,
(3.3)∫t0+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv=M2∫t0+∞ln⁡(1+v)v3dv<+∞.
That is, (2.4) holds. Thus Theorem 2.1 means that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Example 3.2.
Consider the following third-order nonlinear neutral delay differential equation:(3.4){(2t+1)3t[(4t2+4t+5)(x(t)+2t21+3t2x(t-τ))′]′}′+2x(t+1)t2=0,t≥t0,
where τ>0 and t0>0 are fixed. Let n=1, and let M and N be two positive constants with (3t02+1)M>3(t02+1)N and
(3.5)α(t)=(2t+1)3t,β(t)=4t2+4t+5,p(t)=2t21+3t2,c1=23,c2=2t021+3t02,σ1(t)=t+1,h(t)=2Mt2,f(t,u)=2ut2,(t,u)∈[t0,+∞)×R.
Obviously, (2.3), (2.25), and (2.26) hold. Moreover,
(3.6)∫t0+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv=2M16[12(2t0+1)-12(π2-arctan(2t0+1))]<+∞.
Hence (2.4) holds. Thus Theorem 2.2 ensures that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Example 3.3.
Consider the following third-order nonlinear neutral delay differential equation:(3.7){et-t2[(t4+1)1/3(x(t)+-3t-15t+2x(t-τ))′]′}′+tx(t3)et2(1+x2(t2+2t))=0,t≥0,
where τ>0 is fixed and t0=0. Let n=2, and let M and N be two positive constants with 4M>5N and
(3.8)α(t)=et-t2,β(t)=(t4+1)1/3,p(t)=-3t-15t+2,c1=35,c2=12,σ1(t)=t(t+2),σ2(t)=t3,h(t)=Mt(1+N2)et2,f(t,u,v)=tvet2(1+u2),(t,u,v)∈[t0,+∞)×R2.
Clearly, (2.3), (2.25), and (2.30) hold, and
(3.9)∫t0+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv=M2(1+N2)∫t0+∞e-v(t4+1)-1/3dv<+∞,
hence (2.4) holds. Therefore Theorem 2.3 guarantees that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Example 3.4.
Consider the following third-order nonlinear neutral delay differential equation:(3.10){t3+1t2arctant[1t(x(t)+x(t-τ))′]′}′+x2(t2-t)(x(t2-t)+3)t4=0,t≥t0,
where τ>0 and t0>0 are fixed. Let n=1, and let M and N be two positive constants with M>N and
(3.11)α(t)=t3+1t2arctant,β(t)=1t,p(t)=1,σ1(t)=t2-t,h(t)=M2(M+3)t4,f(t,u)=u2(u+3)t4,(t,u)∈[t0,+∞)×R.
Obviously, (2.3) holds. Furthermore,
(3.12)∫t0+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv=M2(M+3)3∫t0+∞varctan⁡vv3+1dv<+∞,
that is, (2.4) holds. Thus Theorem 2.4 ensures that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Example 3.5.
Consider the following third-order nonlinear neutral delay differential equation:(3.13){t3[1t(x(t)-x(t-τ))′]′}′+x2(t+(1/t)+ln⁡(1+t))t3=0,t≥t0,
where τ>0 and t0>0 are fixed. Let n=1, and let M and N be two positive constants with M>N and
(3.14)α(t)=t3,β(t)=1t,p(t)=-1,σ1(t)=t+1t+ln⁡(1+t),h(t)=M2t3,f(t,u)=u2t3,(t,u)∈[t0,+∞)×R.
Obviously, (2.3) holds. Notice that
(3.15)∑i=1+∞∫t0+iτ+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv=M216∑i=1+∞1(t0+iτ)2<+∞,
that is, (2.49) holds. Hence Theorem 2.5 implies that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Example 3.6.
Consider the following third-order nonlinear neutral delay differential equation:(3.16){t2et2[(1+ln⁡2t)(x(t)+(5π2+arctan⁡t2/3)x(t-τ))′]′}′+tet2(x2(t2+1)+1)=0,t≥t0,
where τ>0 and t0>0 are fixed. Let n=1, and let M and N be two positive constants with (5/2)(9π-(5/2))N<3((25π/4)-3)M and
(3.17)α(t)=t2et2,β(t)=1+ln⁡2t,c1=3π,c2=5π2,p(t)=5π2+arctan⁡t2/3,σ1(t)=t2+1,h(t)=t(N2+1)et2,f(t,u)=t(u2+1)et2,(t,u)∈[t0,+∞)×R.
Clearly, (2.3), (2.61), and (2.62) hold, and
(3.18)∫t0+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv=12(N2+1)(π2-arctan(ln⁡⁡t0))<+∞,
that is, (2.4) holds also. Thus Theorem 2.6 ensures that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).Example 3.7.
Consider the following third-order nonlinear neutral delay differential equation(3.19){1t[t2+1et2(x(t)+(sin(t-1t)+4cos⁡(ln⁡t)-7)x(t-τ))′]′}′+tx2(t2+1)et2=0,t≥t0,
where τ>0 and t0>0 are fixed. Let n=1, and let M and N be two positive constants with M>11N and
(3.20)α(t)=1t,β(t)=t2+1et2,c1=12,c2=2,p(t)=sin(t-1t)+4cos⁡(ln⁡t)-7,σ1(t)=t2+1,h(t)=tM2et2,f(t,u)=tu2et2,(t,u)∈[t0,+∞)×R.
Obviously, (2.3), (2.79), and (2.80) hold. Note that
(3.21)∫t0+∞∫v+∞∫s+∞h(u)α(s)β(v)dudsdv=M24(π2-arctan⁡t0)<+∞,
that is, (2.4) holds. It follows from Theorem 2.7 that (1.7) possesses uncountably many bounded nonoscillatory solutions in A(N,M).



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*Source: 693890-2011-08-15.xml*