# Existence of Nonoscillatory Solutions of First-Order Neutral Differential Equations

Authors: Božena Dorociaková; Anna Najmanová; Rudolf Olach
Journal: Abstract and Applied Analysis
 (2011)

---

## Abstract

This paper contains some sufficient conditions for the existence of
positive solutions which are bounded below and above by positive functions
for the first-order nonlinear neutral differential equations. These equations
can also support the existence of positive solutions approaching zero at infinity


---

## Body

## 1. Introduction

This paper is concerned with the existence of a positive solution of the neutral differential equations of the form(1.1)ddt[x(t)-a(t)x(t-τ)]=p(t)f(x(t-σ)),t≥t0,
where τ>0,σ≥0,a∈C([t0,∞),(0,∞)),p∈C(R,(0,∞)),f∈C(R,R),f is nondecreasing function, and xf(x)>0,x≠0.By a solution of (1.1) we mean a function x∈C([t1-m,∞),R),m=max{τ,σ}, for some t1≥t0, such that x(t)-a(t)x(t-τ) is continuously differentiable on [t1,∞) and such that (1.1) is satisfied for t≥t1.The problem of the existence of solutions of neutral differential equations has been studied by several authors in the recent years. For related results we refer the reader to [1–11] and the references cited therein. However there is no conception which guarantees the existence of positive solutions which are bounded below and above by positive functions. In this paper we have  presented some conception. The method also supports the existence of positive solutions approaching zero at infinity.As much as we know, for (1.1) in the literature, there is no result for the existence of solutions which are bounded by positive functions. Only the existence of solutions which are bounded by constants is treated, for example, in [6, 10, 11]. It seems that conditions of theorems are rather complicated, but cannot be simpler due to Corollaries 2.3, 2.6, and 3.2.The following fixed point theorem will be used to prove the main results in the next section.Lemma 1.1 ([see [6, 10] Krasnoselskii's fixed point theorem]).
LetX be a Banach space, let Ω be a bounded closed convex subset of X, and let S1,S2 be maps of Ω into X such that S1x+S2y∈Ω for every pair x,y∈Ω. If S1 is contractive and S2 is completely continuous, then the equation
(1.2)S1x+S2x=x
has a solution in Ω.


## 2. The Existence of Positive Solution

In this section we will consider the existence of a positive solution for (1.1). The next theorem gives us the sufficient conditions for the existence of a positive solution which is bounded by two positive functions.Theorem 2.1.
Suppose that there exist bounded functionsu,v∈C1([t0,∞),(0,∞)), constant c>0 and t1≥t0+m such that
(2.1)u(t)≤v(t),t≥t0,(2.2)v(t)-v(t1)-u(t)+u(t1)≥0,t0≤t≤t1,(2.3)1u(t-τ)(u(t)+∫t∞p(s)f(v(s-σ))ds)≤a(t)≤1v(t-τ)(v(t)+∫t∞p(s)f(u(s-σ))ds)≤c<1,t≥t1.
Then (1.1) has a positive solution which is bounded by functions u,v.Proof.
LetC([t0,∞),R) be the set of all continuous bounded functions with the norm ∥x∥=supt≥t0|x(t)|. Then C([t0,∞),R) is a Banach space. We define a closed, bounded, and convex subset Ω of C([t0,∞),R) as follows:
(2.4)Ω={x=x(t)∈C([t0,∞),R):u(t)≤x(t)≤v(t),t≥t0}.
We now define two maps S1 and S2:Ω→C([t0,∞),R) as follows:
(2.5)(S1x)(t)={a(t)x(t-τ),t≥t1,(S1x)(t1),t0≤t≤t1,(S2x)(t)={-∫t∞p(s)f(x(s-σ))ds,t≥t1,(S2x)(t1)+v(t)-v(t1),t0≤t≤t1.
We will show that for any x,y∈Ω we have S1x+S2y∈Ω. For every x,y∈Ω and t≥t1, we obtain
(2.6)(S1x)(t)+(S2y)(t)≤a(t)v(t-τ)-∫t∞p(s)f(u(s-σ))ds≤v(t).
For t∈[t0,t1], we have
(2.7)(S1x)(t)+(S2y)(t)=(S1x)(t1)+(S2y)(t1)+v(t)-v(t1)≤v(t1)+v(t)-v(t1)=v(t).
Furthermore, for t≥t1, we get
(2.8)(S1x)(t)+(S2y)(t)≥a(t)u(t-τ)-∫t∞p(s)f(v(s-σ))ds≥u(t).
Let t∈[t0,t1]. With regard to (2.2), we get
(2.9)v(t)-v(t1)+u(t1)≥u(t),t0≤t≤t1.
Then for t∈[t0,t1] and any x,y∈Ω, we obtain
(2.10)(S1x)(t)+(S2y)(t)=(S1x)(t1)+(S2y)(t1)+v(t)-v(t1)≥u(t1)+v(t)-v(t1)≥u(t).
Thus we have proved that S1x+S2y∈Ω for any x,y∈Ω.
We will show thatS1 is a contraction mapping on Ω. For x,y∈Ω and t≥t1 we have(2.11)|(S1x)(t)-(S1y)(t)|=|a(t)∥x(t-τ)-y(t-τ)|≤c‖x-y‖.
This implies that
(2.12)‖S1x-S1y‖≤c‖x-y‖.
Also for t∈[t0,t1], the previous inequality is valid. We conclude that S1 is a contraction mapping on Ω.
We now show thatS2 is completely continuous. First we will show that S2 is continuous. Let xk=xk(t)∈Ω be such that xk(t)→x(t) as k→∞. Because Ω is closed, x=x(t)∈Ω. For t≥t1 we have(2.13)|(S2xk)(t)-(S2x)(t)|≤|∫t∞p(s)[f(xk(s-σ))-f(x(s-σ))]ds|≤∫t1∞p(s)|f(xk(s-σ))-f(x(s-σ))|ds.
According to (2.8), we get
(2.14)∫t1∞p(s)f(v(s-σ))ds<∞.
Since |f(xk(s-σ))-f(x(s-σ))|→0 as k→∞, by applying the Lebesgue dominated convergence theorem, we obtain
(2.15)lim⁡k→∞‖(S2xk)(t)-(S2x)(t)‖=0.
This means that S2 is continuous.
We now show thatS2Ω is relatively compact. It is sufficient to show by the Arzela-Ascoli theorem that the family of functions {S2x:x∈Ω} is uniformly bounded and equicontinuous on [t0,∞). The uniform boundedness follows from the definition of Ω. For the equicontinuity we only need to show, according to Levitans result [7], that for any given ε>0 the interval [t0,∞) can be decomposed into finite subintervals in such a way that on each subinterval all functions of the family have a change of amplitude less than ɛ. Then with regard to condition (2.14), for x∈Ω and any ɛ>0, we take t*≥t1 large enough so that(2.16)∫t*∞p(s)f(x(s-σ))ds<ε2.
Then, for x∈Ω,T2>T1≥t*, we have
(2.17)|(S2x)(T2)-(S2x)(T1)|≤∫T2∞p(s)f(x(s-σ))ds+∫T1∞p(s)f(x(s-σ))ds<ε2+ε2=ε.
For x∈Ω and t1≤T1<T2≤t*, we get
(2.18)|(S2x)(T2)-(S2x)(T1)|≤∫T1T2p(s)f(x(s-σ))ds≤max⁡t1≤s≤t*{p(s)f(x(s-σ))}(T2-T1).
Thus there exists δ1=ε/M, where M=maxt1≤s≤t*{p(s)f(x(s-σ))}, such that
(2.19)|(S2x)(T2)-(S2x)(T1)|<εif0<T2-T1<δ1.
Finally for any x∈Ω,t0≤T1<T2≤t1, there exists a δ2>0 such that
(2.20)|(S2x)(T2)-(S2x)(T1)|=|v(T1)-v(T2)|=|∫T1T2v′(s)ds|≤max⁡t0≤s≤t1{|v′(s)|}(T2-T1)<εif0<T2-T1<δ2.
Then {S2x:x∈Ω} is uniformly bounded and equicontinuous on [t0,∞), and hence S2Ω is relatively compact subset of C([t0,∞),R). By Lemma 1.1 there is an x0∈Ω such that S1x0+S2x0=x0. We conclude that x0(t) is a positive solution of (1.1). The proof is complete.Corollary 2.2.
Suppose that there exist functionsu,v∈C1([t0,∞),(0,∞)), constant c>0 and t1≥t0+m such that (2.1), (2.3) hold and
(2.21)v′(t)-u′(t)≤0,t0≤t≤t1.
Then (1.1) has a positive solution which is bounded by the functions u,v.Proof.
We only need to prove that condition (2.21) implies (2.2). Let t∈[t0,t1] and set
(2.22)H(t)=v(t)-v(t1)-u(t)+u(t1).
Then with regard to (2.21), it follows that
(2.23)H′(t)=v′(t)-u′(t)≤0,t0≤t≤t1.
Since H(t1)=0 and H′(t)≤0 for t∈[t0,t1], this implies that
(2.24)H(t)=v(t)-v(t1)-u(t)+u(t1)≥0,t0≤t≤t1.
Thus all conditions of Theorem 2.1 are satisfied.Corollary 2.3.
Suppose that there exists a functionv∈C1([t0,∞),(0,∞)), constant c>0 and t1≥t0+m such that
(2.25)a(t)=1v(t-τ)(v(t)+∫t∞p(s)f(v(s-σ))ds)≤c<1,t≥t1.
Then (1.1) has a solution x(t)=v(t),t≥t1.Proof.
We putu(t)=v(t) and apply Theorem 2.1.Theorem 2.4.
Suppose that there exist functionsu,v∈C1([t0,∞),(0,∞)), constant c>0 and t1≥t0+m such that (2.1), (2.2), and (2.3) hold and
(2.26)lim⁡t→∞v(t)=0.
Then (1.1) has a positive solution which is bounded by the functions u,v and tends to zero.Proof.
The proof is similar to that of Theorem2.1 and we omit it.Corollary 2.5.
Suppose that there exist functionsu,v∈C1([t0,∞),(0,∞)), constant c>0 and t1≥t0+m such that (2.1), (2.3), (2.21), and (2.26) hold. Then (1.1) has a positive solution which is bounded by the functions u,v and tends to zero.Proof.
The proof is similar to that of Corollary2.2, and we omitted it.Corollary 2.6.
Suppose that there exists a functionv∈C1([t0,∞),(0,∞)), constant c>0 and t1≥t0+m such that (2.25), (2.26) hold. Then (1.1) has a solution x(t)=v(t),t≥t1 which tends to zero.Proof.
We putu(t)=v(t) and apply Theorem 2.4.


## 3. Applications and Examples

In this section we give some applications of the theorems above.Theorem 3.1.
Suppose that(3.1)∫t0∞p(t)dt=∞,0<k1≤k2 and there exist constants c>0,γ≥0,t1≥t0+m such that
(3.2)k1k2exp⁡((k2-k1)∫t0-γt0p(t)dt)≥1,(3.3)exp⁡(-k2∫t-τtp(s)ds)+exp⁡(k2∫t0-γt-τp(s)ds)×∫t∞p(s)f(exp⁡(-k1∫t0-γs-σp(ξ)dξ))ds≤a(t)≤exp⁡(-k1∫t-τtp(s)ds)+exp⁡(k1∫t0-γt-τp(s)ds)×∫t∞p(s)f(exp⁡(-k2∫t0-γs-σp(ξ)dξ))ds≤c<1,t≥t1.
Then (1.1) has a positive solution which tends to zero.Proof.
We set(3.4)u(t)=exp⁡(-k2∫t0-γtp(s)ds),v(t)=exp⁡(-k1∫t0-γtp(s)ds),t≥t0.
We will show that the conditions of Corollary 2.5 are satisfied. With regard to (2.21), for t∈[t0,t1], we get
(3.5)v′(t)-u′(t)=-k1p(t)v(t)+k2p(t)u(t)=p(t)v(t)[-k1+k2u(t)exp⁡(k1∫t0-γtp(s)ds)]=p(t)v(t)[-k1+k2exp⁡((k1-k2)∫t0-γtp(s)ds)]≤p(t)v(t)[-k1+k2exp⁡((k1-k2)∫t0-γt0p(s)ds)]≤0.
Other conditions of Corollary 2.5 are also satisfied. The proof is complete.Corollary 3.2.
Suppose thatk>0,c>0,t1≥t0+m, (3.1) holds, and
(3.6)a(t)=exp⁡(-k∫t-τtp(s)ds)+exp⁡(k∫t0t-τp(s)ds)×∫t∞p(s)f(exp⁡(-k∫t0s-σp(ξ)dξ))ds≤c<1,t≥t1.
Then (1.1) has a solution
(3.7)x(t)=exp⁡(-k∫t0tp(s)ds),t≥t1,
which tends to zero.Proof.
We putk1=k2=k,γ=0 and apply Theorem 3.1.Example 3.3.
Consider the nonlinear neutral differential equation(3.8)[x(t)-a(t)x(t-2)]′=px3(t-1),t≥t0,
where p∈(0,∞). We will show that the conditions of Theorem 3.1 are satisfied. Condition (3.1) obviously holds and (3.2) has a form
(3.9)k1k2exp⁡((k2-k1)pγ)≥1,0<k1≤k2,γ≥0. For function a(t), we obtain
(3.10)exp⁡(-2pk2)+13k1exp⁡(p[k2(γ-t0-2)-3k1(γ-t0-1)+(k2-3k1)t])≤a(t)≤exp⁡(-2pk1)+13k2exp⁡(p[k1(γ-t0-2)-3k2(γ-t0-1)+(k1-3k2)t]),t≥t0.
For p=1,k1=1,k2=2,γ=1,t0=1, condition (3.9) is satisfied and
(3.11)e-4+13ee-t≤a(t)≤e-2+e46e-5t,t≥t1≥3.
If the function a(t) satisfies (3.11), then (3.8) has a solution which is bounded by the functions u(t)=exp(-2t),v(t)=exp(-t),t≥3.
For example ifp=1,k1=k2=1.5,γ=1,t0=1, from (3.11) we obtain(3.12)a(t)=e-3+e1.54.5e-3t,
and the equation
(3.13)[x(t)-(e-3+e1.54.5e-3t)x(t-2)]′=x3(t-1),t≥3,
has the solution x(t)=exp(-1.5t) which is bounded by the function u(t) and v(t).



---

*Source: 346745-2011-06-22.xml*