# Existence of Positive Solutions of Neutral Differential Equations

Authors: B. Dorociaková; M. Kubjatková; R. Olach
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

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


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## Body

## 1. Introduction

This paper is concerned with the existence of a positive solution of the neutral differential equations of the form:(1.1)dndtn[x(t)-a(t)x(t-τ)]=(-1)n+1p(t)f(x(t-σ)),t≥t0,
where n>0 is an integer, τ>0, σ≥0, a∈C([t0,∞),(0,∞)), p∈C(R,(0,∞)), f∈C(R,R), f is a nondecreasing function and xf(x)>0, x≠0.By a solution of (1.1) we mean a function x∈C([t1-τ,∞),R) for some t1≥t0, such that x(t)-a(t)x(t-τ) is n-times 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 and discussed by several authors in the recent years. For related results we refer the reader to [1–17] 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. Maybe it is due to the technical difficulties arising in the analysis of the problem. In this paper we presented some conception. The method also supports the existence of positive solutions which approaching zero at infinity. Some examples illustrating the results.The existence and asymptotic behavior of solutions of the nonlinear neutral differential equations and systems have been also solved in [1–7, 12, 15].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 and discussed, for example, in [10, 15, 17]. It seems that conditions of theorems are rather complicate, but cannot be simpler due to Corollaries 2.4, 2.8, and 3.3.The following fixed point theorem will be used to prove the main results in the next section.Lemma 1.1 (see [7, 10, 12] 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 a 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) which is bounded by two positive functions. We will use the notation m=max{τ,σ}.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)+1(n-1)!∫t∞(s-t)n-1p(s)f(v(s-σ))ds)≤a(t)≤1v(t-τ)(v(t)+1(n-1)!∫t∞(s-t)n-1p(s)f(u(s-σ))ds)≤c<1,t≥t1.
Then (1.1) has a positive solution which is bounded by the 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)={-1(n-1)!∫t∞(s-t)n-1p(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)-1(n-1)!∫t∞(s-t)n-1p(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)-1(n-1)!∫t∞(s-t)n-1p(s)f(v(s-σ))ds≥u(t).
Finally let t∈[t0,t1] and 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 get
(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 inequality above 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)|≤1(n-1)!|∫t∞(s-t)n-1p(s)[f(xk(s-σ))-f(x(s-σ))]ds|≤1(n-1)!∫t1∞(s-t1)n-1p(s)|f(xk(s-σ))-f(x(s-σ))|ds.
According to (2.8) we get
(2.14)∫t1∞(s-t1)n-1p(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 that
(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 Levitan result [8], 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 change of amplitude less than ε. With regard to the condition (2.14), for x∈Ω and any ε>0 we take t*≥t1 large enough so that(2.16)1(n-1)!∫t*∞(s-t1)n-1p(s)f(x(s-σ))ds<ε2.
Then for x∈Ω,T2>T1≥t* we have
(2.17)|(S2x)(T2)-(S2x)(T1)|≤1(n-1)!∫T2∞(s-t1)n-1p(s)f(x(s-σ))ds+1(n-1)!∫T1∞(s-t1)n-1p(s)f(x(s-σ))ds<ε2+ε2=ε.
For x∈Ω, t1≤T1<T2≤t* and n≥2 we get
(2.18)|(S2x)(T2)-(S2x)(T1)|=1(n-1)!|∫T1∞(s-T1)n-1p(s)f(x(s-σ))ds-∫T2∞(s-T2)n-1p(s)f(x(s-σ))ds|=1(n-1)!|∫T1T2(s-T1)n-1p(s)f(x(s-σ))ds+∫T2∞(s-T1)n-1p(s)f(x(s-σ))ds-∫T2∞(s-T2)n-1p(s)f(x(s-σ))ds|≤1(n-1)!∫T1T2sn-1p(s)f(x(s-σ))ds+1(n-1)!∫T2∞[(s-T1)n-1-(s-T2)n-1]p(s)f(x(s-σ))ds≤max⁡t1≤s≤t*{1(n-1)!sn-1p(s)f(x(s-σ))}(T2-T1)+1(n-1)!∫T2∞[(s-T1)-(s-T2)]×[(s-T1)n-2+(s-T1)n-3(s-T2)+⋯+(s-T1)(s-T2)n-3+(s-T2)n-2]p(s)f(x(s-σ))ds≤max⁡t1≤s≤t*{1(n-1)!sn-1p(s)f(x(s-σ))}(T2-T1)+1(n-2)!∫T2∞(T2-T1)(s-T1)n-2p(s)f(x(s-σ))ds.
With regard to the condition (2.14) we have that
(2.19)1(n-2)!∫T2∞(s-T1)n-2p(s)f(x(s-σ))ds<B,B>0.
Then we obtain
(2.20)|(S2x)(T2)-(S2x)(T1)|<(max⁡t1≤s≤t*{1(n-1)!sn-1p(s)f(x(s-σ))}+B)(T2-T1).
Thus there exists a δ1=ε/(M+B), where
(2.21)M=max⁡t1≤s≤t*{1(n-1)!sn-1p(s)f(x(s-σ))},
such that
(2.22)|(S2x)(T2)-(S2x)(T1)|<εif0<T2-T1<δ1.
For n=1 we proceed by the similar way as above. Finally for any x∈Ω, t0≤T1<T2≤t1 there exists a δ2>0 such that
(2.23)|(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 all conditions of Theorem2.1 are satisfied and(2.24)lim⁡t→∞v(t)=0.
Then (1.1) has a positive solution which tends to zero.Corollary 2.3.
Suppose that there exist bounded functionsu,v∈C1([t0,∞),(0,∞)), constant  c>0 and  t1≥t0+m such that (2.1), (2.3) hold and(2.25)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 (2.25) implies (2.2). Let t∈[t0,t1] and set
(2.26)H(t)=v(t)-v(t1)-u(t)+u(t1).
Then with regard to (2.25), it follows that 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
(2.27)H(t)=v(t)-v(t1)-u(t)+u(t1)≥0,t0≤t≤t1.
Thus all conditions of Theorem 2.1 are satisfied.Corollary 2.4.
Suppose that there exists a bounded functionv∈C1([t0,∞),(0,∞)), constant  c>0 and  t1≥t0+m such that(2.28)a(t)=1v(t-τ)(v(t)+1(n-1)!∫t∞(s-t)n-1p(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 2.5.
Suppose thatp is bounded and there exist bounded functions  u,v∈C1([t0,∞),(0,∞)), constant  c>0 and  t1≥t0+m such that (2.1), (2.2) hold and
(2.29)1u(t-τ)(u(t)-1(n-1)!∫t1t(t-s)n-1p(s)f(u(s-σ))ds)≤a(t)≤1v(t-τ)(v(t)-1(n-1)!∫t1t(t-s)n-1p(s)f(v(s-σ))ds)≤c<1,t≥t1,
if n is odd,
(2.30)1u(t-τ)(u(t)+1(n-1)!∫t1t(t-s)n-1p(s)f(v(s-σ))ds)≤a(t)≤1v(t-τ)(v(t)+1(n-1)!∫t1t(t-s)n-1p(s)f(u(s-σ))ds)≤c<1,t≥t1,
if n is even, and
(2.31)∫t1t(t-s)n-2p(s)f(v(s-σ))ds≤K,t≥t1,K>0,n≥2.
Then (1.1) has a positive solution which is bounded by the functions u, v.Proof.
LetC([t0,∞),R) be the set as in the proof of Theorem 2.1. We define a closed, bounded, and convex subset Ω of C([t0,∞),R) as in the proof of Theorem 2.1. We define two maps S1 and S2:Ω→C([t0,∞),R) as follows:
(2.32)(S1x)(t)={a(t)x(t-τ),t≥t1,(S1x)(t1),t0≤t≤t1,(S2x)(t)={(-1)n+1(n-1)!∫t1t(t-s)n-1p(s)f(x(s-σ))ds,t≥t1,(S2x)(t1)+v(t)-v(t1),t0≤t≤t1.
We shall show that for any x,y∈Ω we have S1x+S2y∈Ω. For n odd, every x,y∈Ω and t≥t1 we obtain
(2.33)(S1x)(t)+(S2y)(t)≤a(t)v(t-τ)+1(n-1)!∫t1t(t-s)n-1p(s)f(v(s-σ))ds≤v(t).
For t∈[t0,t1], we have
(2.34)(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
(2.35)(S1x)(t)+(S2y)(t)≥a(t)u(t-τ)+1(n-1)!∫t1t(t-s)n-1p(s)f(u(s-σ))ds≥u(t).
Let t∈[t0,t1] and according to (2.2) we have
(2.36)v(t)-v(t1)+u(t1)≥u(t).
Then for t∈[t0,t1] and any x,y∈Ω we get
(2.37)(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.
Forn even by the similar way as above we can prove that S1x+S2y∈Ω for any x,y∈Ω.
As in the proof of Theorem2.1, we can show 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
(2.38)|(S2xk)(t)-(S2x)(t)|≤1(n-1)!∫t1t(t-s)n-1p(s)|f(xk(s-σ))-f(x(s-σ))|ds.
According to (2.33) there exists a positive constant M such that
(2.39)∫t1t(t-s)n-1p(s)f(v(s-σ))ds≤Mfort≥t1.
The inequality above also holds for n even.
Since|f(xk(s-σ))-f(x(s-σ))|→0 as k→∞, by applying the Lebesgue dominated convergence theorem we obtain that(2.40)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 n≥2 and with regard to (2.31) we have(2.41)|ddt(S2x)(t)|=1(n-2)!∫t1t(t-s)n-2p(s)f(x(s-σ))ds≤1(n-2)!∫t1t(t-s)n-2p(s)f(v(s-σ))≤M1,
and for n=1 we obtain
(2.42)|ddt(S2x)(t)|=p(t)f(v(t-σ))≤M2,
for t≥t1, M2>0 and |(d/dt)(S2x)(t)|=|v′(t)|≤M3 for t0≤t≤t1, M3>0, which shows the equicontinuity of the family S2Ω, (cf. [7, page 265]). Hence S2Ω is relatively compact and therefore S2 is completely continuous. By Lemma 1.1, there is x0∈Ω such that S1x0+S2x0=x0. Thus x0(t) is a positive solution of (1.1). The proof is complete.Corollary 2.6.
Suppose that all conditions of Theorem2.5 are satisfied and(2.43)lim⁡t→∞v(t)=0.
Then (1.1) has a positive solution which tends to zero.Corollary 2.7.
Suppose thatp is bounded and there exist bounded functions  u,v∈C1([t0,∞),(0,∞)), constant  c>0 and  t1≥t0+m such that (2.1), (2.29), (2.30), (2.31) hold and(2.44)v′(t)-u′(t)≤0,t0≤t≤t1.
Then (1.1) has a positive solution which is bounded by the functions u,v.Proof.
The proof is similar to that of Corollary2.3 and we omit it.Corollary 2.8.
Suppose thatp is bounded and there exists a bounded function  v∈C1([t0,∞),(0,∞)), constant  c>0 and  t1≥t0+m such that (2.31) holds and(2.45)a(t)=1v(t-τ)(v(t)+(-1)n(n-1)!∫t1t(t-s)n-1p(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 2.5.


## 3. Applications and Examples

In this section, we give some applications of the theorems above.Theorem 3.1.
Suppose that0<k1≤k2 and there exist  γ≥0,  c>0,  t1≥t0+m such that
(3.1)k1k2exp⁡((k2-k1)∫t0-γt0p(t)dt)≥1,(3.2)exp⁡(-k2∫t-τtp(s)ds)+1(n-1)!exp⁡(k2∫t0-γt-τp(s)ds)×∫t∞(s-t)n-1p(s)f(exp⁡(-k1∫t0-γs-σp(ξ)dξ))ds≤a(t)≤exp⁡(-k1∫t-τtp(s)ds)+1(n-1)!exp⁡(k1∫t0-γt-τp(s)ds)×∫t∞(s-t)n-1p(s)f(exp⁡(-k2∫t0-γs-σp(ξ)dξ))ds≤c<1,t≥t1.
Then (1.1) has a positive solution which is bounded by two exponential functions.Proof.
We set(3.3)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 2.3 are satisfied. With regard to (3.1) for t∈[t0,t1] we get
(3.4)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 2.3 are also satisfied. The proof is complete.Corollary 3.2.
Suppose that all conditions of Theorem3.1 are satisfied and(3.5)∫t0∞p(t)dt=∞.
Then (1.1) has a positive solution which tends to zero.Corollary 3.3.
Suppose thatk>0,c>0,t1≥t0+m and
(3.6)a(t)=exp⁡(-k∫t-τtp(s)ds)+1(n-1)!exp⁡(k∫t0t-τp(s)ds)×∫t∞(s-t)n-1p(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.Proof.
We putk1=k2=k,γ=0 and apply Theorem 3.1.Example 3.4.
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. The condition (3.1) 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, the 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.Example 3.5.
Consider the nonlinear differential equation:(3.12)[x(t)-a(t)x(t-π)]′=p(t)f(x(t-π)),t≥0,
where f(x)=x,x>0,p(t)=0.8exp(π-t+0.05cost),t≥0, and
(3.13)e0.1cos⁡t(e0.1cos⁡t+0.8(eπ-t-1))≤a(t)≤e0.1cos⁡t(e0.1cos⁡t+0.8b(eπ-t-1))<1,
for t≥π,b∈[1,2]. Set
(3.14)u(t)=e0.1cos⁡t,v(t)=be0.1cos⁡t,t≥0.
Then we have
(3.15)v′(t)-u′(t)=-0.1be0.1cos⁡tsin⁡t+0.1e0.1cos⁡tsin⁡t=-0.1(b-1)e0.1cos⁡tsin⁡t≤0fort∈[0,π].
By Corollary 2.7, (3.12) has a solution which is bounded by the functions e0.1cost and be0.1cost,t≥π. If
(3.16)a(t)=e0.1cos⁡t(e0.1cos⁡t+0.8(eπ-t-1))fort≥π,
then (3.12) has the positive periodic solution x(t)=u(t)=e0.1cost,t≥π.



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