# Existence for Eventually Positive Solutions of High-Order Nonlinear Neutral Differential Equations with Distributed Delay

Authors: Huanhuan Zhao; Youjun Liu; Jurang Yan
Journal: Discrete Dynamics in Nature and Society
 (2012)

---

## Abstract

We consider the existence for eventually positive solutions of high-order nonlinear neutral differential equations with distributed delay. We useLebesgue's dominated convergence theorem to obtain new necessary and sufficient condition for the existence of eventually positive solutions.


---

## Body

## 1. Introduction and Preliminary

In this paper, we consider the high-order nonlinear neutral differential equation:(1.1)[r(t)x(t)-∫abp(t,τ)x(t-τ)dτ](n)+∫cdq(t,σ)f(x(t-σ))dσ=0,t≥t0,
and the associated inequality:(1.2)[r(t)x(t)-∫abp(t,τ)x(t-τ)dτ](n)+∫cdq(t,σ)f(x(t-σ))dσ≤0,t≥t0,(1)
wheren is a positive integer, 0<a<b,0<c<d,(2)
p∈C([t0,∞)×[a,b],R+),(3)
r∈C([t0,∞),R+),r(t)>0, q∈C([t0,∞)×[c,d],R+),(4)
f(u) are continuously nondecreasing real function with respect to u defined on R such that uf(u)>0, for u>0, for n is positive odd integer; f(u) are continuously decreasing real function with respect to u defined on R such that uf(u)>0, for u>0, for n is positive even integer.Recently, there has been a lot of activities concerning the existence of eventually positive solutions for nonlinear neutral differential equations. See [1–8]. In [1], Liu et al. have studied the even-order neutral differential equation:(1.3)[a(t)x(t)-∑i=1mbi(t)x(t-τi)](n)-∑j=1lpj(t)fj(t,x(t-σj))=0,t≥0,
and the associated differential inequality:(1.4)[a(t)x(t)-∑i=1mbi(t)x(t-τi)](n)-∑j=1lpj(t)fj(t,x(t-σj))≥0,t≥0.
They have obtained that the existences of eventually positive solutions of (1.3) and (1.4) are equivalent. In [2], Ouyang et al. has studied the odd-order neutral differential equation:(1.5)[a(t)x(t)-∑i=1mbi(t)x(t-τi)](n)+∑j=1lpj(t)fj(x(t-σj))=0,t≥0,
and the associated differential inequality:(1.6)[a(t)x(t)-∑i=1mbi(t)x(t-τi)](n)+∑j=1lpj(t)fj(x(t-σj))≤0,t≥0.
He has obtained that the existences of eventually positive solutions of (1.5) and (1.6) are equivalent.As usual, a solution of (1.1) is a continuous function x(t) defined on [-μ,∞) such that y(t):=r(t)x(t)-∫abp(t,τ)x(t-τ)dτ is n times differentiable and (1.1) holds for all n≥0. Such a solution x(t) is called an eventually positive solution if there is T≥t0, such that x(t)>0, for t≥T. Here, μ=max{b,d}.Lemma 1.1.
Assumesupt≥t0r(t)<∞, and ∫abp(t,τ)dτ/r(t)≤M, let x(t) is an eventually bounded positive solution of inequality (1.2), and set
(1.7)y(t)=r(t)x(t)-∫abp(t,τ)x(t-τ)dτ.
If n is a positive odd integer, then eventually,
(1.8)y(n)(t)≤0,(-1)k+1y(k)(t)<0,(k=0,1,…,n-1),lim⁡t→∞y(k)(t)=0,(k=1,2,…,n-1).
If n is a positive even integer, then eventually,
(1.9)y(n)(t)≤0,(-1)ky(k)(t)<0,(k=0,1,…,n-1),lim⁡t→∞y(k)(t)=0,(k=1,2,…,n-1).Proof.
We have the following cases.
Case 1 (Ifn is a positive odd integer).
Becausex(t) is bounded, and supt≥t0r(t)<∞,∫abp(t,τ)dτ/r(t)≤M, thus y(t) is bounded. From (1.2), we have y(n)(t)≤-∫cdq(t,σ)f(x(t-σ))dσ≤0. Assume y(n-1)(t)≤0, since y(n)(t)≤0, y(n-1)(t) decreases, set limt→∞y(n-1)(t)=L, then -∞≤L<0.
Thus,(1.10)y(n-2)(t)-y(n-2)(t0)=∫t0ty(n-1)(t)dt≤(L+ɛ)(t-t0)⟶-∞,(t⟶∞),
that is, limt→∞y(n-2)(t)=-∞. Simile, limt→∞y(k)(t)=-∞,(k=0,1,…,n-2), this is a contradiction and y(t) is bounded, therefore, y(n-1)(t)>0.
Again, since it decreases, setlimt→∞y(n-1)(t)=L, thus L≥0. Next, we proof L=0. Assume L>0, then,
(1.11)y(n-2)(t)-y(n-2)(t0)=∫t0ty(n-1)(t)dt≥(L-ɛ)(t-t0)⟶∞,(t⟶∞),
then, limt→∞y(n-2)(t)=∞. Simile, limt→∞y(k)(t)=∞,(k=0,1,…,n-2), this is a contradiction and y(t) is bounded, therefore, L=0. That is, limt→∞y(n-1)(t)=0. Similarly, we obtain
(1.12)(-1)k+1y(k)(t)<0,(k=0,1,…,n-1),lim⁡t→∞y(k)(t)=0,(k=1,2,…,n-1).Case 2 (Ifn is a positive even integer).
The proof of Case2 is similar to that of part Case 1, therefore, it is omitted. We obtain
(1.13)(-1)ky(k)(t)<0,(k=0,1,…,n-1),lim⁡t→∞y(k)(t)=0,(k=1,2,…,n-1).
The proof is complete.Lemma 1.2 (see [3, page 21]).
Letη∈(-∞,0),τ∈(0,∞),t0∈R, and suppose that a function x(t)∈C([t0-τ,∞),R) satisfies the inequality:
(1.14)x(t)≤η+max⁡t-τ≤s≤tx(s),t≥t0.
Then, x(t) cannot be a nonnegative function.Lemma 1.3.
Suppose thatsupt≥t0r(t)<∞, and ∫abp(t,τ)dτ/a(t)≤1. Let x(t) be an eventually positive solution of (1.2) and set
(1.15)y(t)=r(t)x(t)-∫abp(t,τ)x(t-τ)dτ,
then eventually
(1.16)y(t)>0.Proof.
From (1.2) and (1.7), eventually, we have
(1.17)y(n)(t)≤-∫cdq(t,σ)f(x(t-σ))dσ≤0,
and the hypotheses on q(t,σ),f(x), and x(t) yield that y(n) is not eventually zero. Thus, y(i)(t)(i=0,l,…,n-1) is eventually nonzero. Hence, if y(t)>0 does not hold, then eventually y(t)<0,y′(t)<0, or y(t)<0,y′(t)>0.Case 1.
y(t)<0,y′(t)<0, then there exists t1>0 such that y(t)≤y(t1)<0 for t>t1. Then, α:=-y(t1)/supt≥t0r(t)>0. In view of (1.3) and ∫abp(t,τ)dτ/a(t)≤1, we obtain
(1.18)x(t)=y(t)r(t)+1r(t)∫abp(t,τ)x(t-τ)dτ≤-α+1r(t)∫abp(t,τ)dτmax⁡t-b≤s≤tx(s)≤-α+max⁡t-b≤s≤tx(s).
According to Lemma 1.2, we obtain x(t)<0. This is a contradiction and so y(t) is eventually positive.Case 2.
y(t)<0,y′(t)>0, one argues that y′′(t)<0 and repeats the argument to obtain that y(n)(t)>0 which contradicts the offset after y(n)(t)<0.
The proof is complete.


## 2. Comparison Theory of Existence for Eventually Positive Solution

Theorem 2.1.
Assume all conditions of Lemma1.1 hold, n is a positive odd integer. And(H1)
p(t,τ)+q(t,σ)>0, for sufficiently large t.Then (1.1) has an eventually bounded positive solution if and only if inequality (1.2) has an eventually bounded positive solution.Proof.
It is clear that an eventually bounded positive solution of (1.1) is also an eventually bounded positive solution of (1.2). So, it suffices to prove that if (1.2) has an eventually bounded positive solution x(t), for t>t0, then so does (1.1). Set
(2.1)y(t)=r(t)x(t)-∫abp(t,τ)x(t-τ)dτ.
It follows from Lemma 1.1 and (1.2) that eventually,
(2.2)y(n)(t)≤0,(-1)k+1y(k)(t)>0,(k=0,1,…,n-1),lim⁡t→∞yk(t)=0,(k=1,2,…,n-1).
By using (1.8) and integrating (1.2) from t to ∞, we obtain
(2.3)y(n-1)(∞)-y(n-1)(t)≤-∫t∞[∫cdq(t,σ)f(x(t-σ))dσ]ds,y(n-1)(t)≥∫t∞[∫cdq(s,σ)f(x(s-σ))dσ]ds.
By repeating the same procedure n times and by using (1.8), we are led to the inequality:
(2.4)y(t)≥∫t∞dtn∫tn∞dtn-1∫tn-1∞dtn-2⋯∫t2∞[∫cdq(s,σ)f(x(s-σ))dσ]ds,t≥t0.
Using Tonelli′s theorem, we reverse the order of integration and obtain
(2.5)=∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(x(s-σ))dσ]ds,t≥t0.
That is,
(2.6)x(t)≥1r(t)∫abp(t,τ)x(t-τ)dτ+1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(x(s-σ))dσ]ds,t≥t0.
Let T≥t0 be such that (2.6) hold, and x(t-μ)>0,t≥T. Now, we consider the set of functions
(2.7)Ω={z∈C([T-μ,∞),R+):0≤z(t)≤1,t≥T-μ}
and define an operator S on Ω as follows:
(2.8)(Sz)(t)={t-T+μμ(Sz)(T)+(1-t-T+μμ),T-μ≤t<T,1x(t){1r(t)∫abp(t,τ)z(t-τ)x(t-τ)dτ+1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(z(s-σ)x(s-σ))dσ]ds},t≥T.
Then, it follows from Lebesgue′s dominated convergence theorem that S is continuous. By using (2.6), it is easy to see that S maps Ω into itself, and for any z∈Ω, we have (Sz)(t)>0, for T-μ≤t<T. Next, we define the sequence zk(t)∈Ω(2.9)z0(t)≡1,t≥T-μ,zk+1(t)=(Szk)(t),t≥T-μ,k=0,1,….
Then, by using (2.6) and a simple induction, we can easily see that
(2.10)0≤zk+1(t)≤zk(t)≤1,t≥T-μ.
Set limk→∞zk(t)=z(t),t≥T-μ then z(t) satisfies
(2.11)z(t)={t-T+μμ(Sz)(T)+(1-t-T+μμ),T-μ≤t<T,1x(t){1r(t)∫abp(t,τ)z(t-τ)x(t-τ)dτ+1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(z(s-σ)x(s-σ))dσ]ds},t≥T.
Again, set ω(t)=z(t)x(t), then ω(t) satisfies ω(t)>0,T-μ≤t≤T, and
(2.12)ω(t)=1r(t)∫abp(t,τ)ω(t-τ)dτ+1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(ω(s-σ))dσ]ds,t≥T.
Thus, ω(t) is a positive solution of (1.1) for t≥T.
The following:(2.13)ω(t)>0,t≥T-μ.
Assume that there exists t*≥T-μ, such that ω(t)>0, for T-μ≤t≤t*, and ω(t*)=0. Then,
(2.14)0=ω(t*)=1r(t*)∫abp(t*,τ)ω(t*-τ)dτ+1r(t*)∫t*∞(s-t*)n-1(n-1)![∫cdq(s,σ)f(ω(s-σ))dσ]ds,t*≥T,
which implies
(2.15)p(t*,τ)=0,q(s,σ)f(ω(s-σ))≡0,
which contradicts (H1). Thus, ω(t) is an eventually bounded positive solution of (1.1).
The proof is complete.Theorem 2.2.
Assume all conditions of Lemma1.3 hold, n is a positive odd integer. And(H1)
p(t,τ)+q(t,σ)>0, for sufficiently large t.Then, (1.1) has an eventually positive solution if and only if inequality (1.2) has an eventually positive solution.Proof.
It is clear that an eventually positive solution of (1.1) is also an eventually positive solution of (1.2). So, it suffices to prove that if (1.2) has an eventually positive solution x(t), for t>t0, then so does (1.1). Set
(2.16)y(t)=r(t)x(t)-∫abp(t,τ)x(t-τ)dτ.
It follows from Lemma1.3 and (1.2) that eventually, y(n)(t)≤0,y(t)>0, which implies that there exists a nonnegative even integer n*≤n-1, such that eventually
(2.17)y(k)(t)>0,(k=0,1,…,n*),(-1)(k)y(k)(t)>0,(k=n*,…,n-1).
We consider the following possible cases.
Case 1 (n*=0).
Sincen is a positive integer, n-1 is an even integer, we can easily see that there exists a T′>0, such y(n-1)(t)>0, and yn(t)≤0,t≥T′,y(n-1)(∞)≥0.
By using (2.17) and integrating (1.2) from t to ∞, we obtain
(2.18)y(n-1)(∞)-y(n-1)(t)≤-∫t∞[∫cdq(s,σ)f(x(s-σ))dσ]ds,y(n-1)(t)≥∫t∞[∫cdq(s,σ)f(x(s-σ))dσ]ds.
By repeating the same procedure n times and by using (2.31), we are led to the inequality:
(2.19)y(t)≥∫t∞dtn∫tn∞dtn-1∫tn-1∞dtn-2⋯∫t2∞[∫cdq(s,σ)f(x(s-σ))dσ]ds,t≥t0.
Using Tonelli′s theorem, we reverse the order of integration and obtain
(2.20)=∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(x(s-σ))dσ]ds,t≥t0.
That is,
(2.21)x(t)≥1r(t)∫abp(t,τ)x(t-τ)dτ+1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(t,σ)f(x(t-σ))dσ]ds,t≥t0.
Let T≥t0 be such that (2.21) hold, and x(t-μ)>0,t≥T. Now, we consider the set of functions
(2.22)Ω={z∈C([T-μ,∞),R+):0≤z(t)≤1,t≥T-μ},
and define an operator S on Ω as follows:
(2.23)(Sz)(t)={t-T+μμ(Sz)(T)+(1-t-T+μμ),T-μ≤t<T,1x(t){1r(t)∫abp(t,τ)z(t-τ)x(t-τ)dτ+1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(z(s-σ)x(s-σ))dσ]ds},t≥T.
Then, it follows from Lebesgue′s dominated convergence theorem that S is continuous. By using (2.21), it is easy to see that S maps Ω into itself, and for any z∈Ω, we have (Sz)(t)>0, for T-μ≤t<T. Next, we define the sequence zk(t)∈Ω(2.24)z0(t)≡1,t≥T-μ,zk+1(t)=(Szk)(t),t≥T-μ,k=0,1,….
Then, by using (2.21) and a simple induction, we can easily see that
(2.25)0≤zk+1(t)≤zk(t)≤1,t≥T-μ.
Set limk→∞zk(t)=z(t),t≥T-μ, then z(t) satisfies:
(2.26)z(t)={t-T+μμ(Sz)(T)+(1-t-T+μμ),T-μ≤t<T,1x(t){1r(t)∫abp(t,τ)z(t-τ)x(t-τ)dτ+1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(z(s-σ)x(s-σ))dσ]ds},t≥T.
Again, set ω(t)=z(t)x(t), then ω(t) satisfies ω(t)>0,T-μ≤t≤T, and
(2.27)ω(t)=1r(t)∫abp(t,τ)ω(t-τ)dτ+1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(ω(s-σ))dσ]ds,t≥T.
Thus, ω(t) is a positive solution of (1.1) for t≥T.
Consider the following:(2.28)ω(t)>0,t≥T-μ.
Assume that there exists t*≥T-μ, such that ω(t)>0, for T-μ≤t≤t*, and ω(t*)=0. Then,
(2.29)0=ω(t*)=1r(t*)∫abp(t*,τ)ω(t*-τ)dτ+1r(t*)∫t*∞(s-t*)n-1(n-1)![∫cdq(s,σ)f(ω(s-σ))dσ]ds,t*≥T,
which implies
(2.30)p(t*,τ)=0,q(s,σ)f(ω(s-σ))≡0,
which contradicts (H1). Thus, ω(t) is an eventually positive solution of equation.
Case 2 (2≤n*≤n-1).
By using (2.17) and integrating (1.2) from t to ∞, we obtain
(2.31)y(n*)(t)≥∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(x(s-σ))dσ]ds,t≥t0.
Let T≥0 be such that (2.17) and (H1) hold. Integrating (2.31) from T to t and using (2.17), we have
(2.32)x(t)≥1r(t)∫abp(t,τ)x(t-τ)dτ+1r(t)∫Tt(t-s)n*-1(n*-1)!∫t∞(u-s)n-n*-1(n-n*-1)![∫cdq(s,σ)f(x(s-σ))dσ]duds,t≥t0.
Using a method similar to the proof of Case 1 yields that (1.1) also has an eventually positive solution.
The proof is complete.Theorem 2.3.
Assume all conditions of Lemma1.1 hold, n is a positive even integer, and(H1)
p(t,τ)+q(t,σ)>0, for sufficiently large t.Then, (1.1) has an eventually positive solution if and only if inequality (1.2) has an eventually positive solution.Proof.
It is clear that an eventually bounded positive solution of (1.1) is also an eventually bounded positive solution of (1.2). So, it suffices to prove that if (1.2) has an eventually bounded positive solution x(t), for t>t0, then so does (1.1). Set
(2.33)y(t)=r(t)x(t)-∫abp(t,τ)x(t-τ)dτ.
It follows from Lemma 1.1 and (1.2) that eventually,
(2.34)y(n)(t)≤0,(-1)ky(k)(t)<0,(k=0,1,…,n-1),lim⁡t→∞yk(t)=0,(k=1,2,…,n-1).
By using (1.9) and integrating (1.2) from t to ∞, we obtain
(2.35)y(n-1)(∞)-y(n-1)(t)≤-∫t∞[∫cdq(s,σ)f(x(s-σ))dσ]ds,y(n-1)(t)≥∫t∞[∫cdq(s,σ)f(x(s-σ))dσ]ds.
By repeating the same procedure n times and by using (1.9), we are led to the inequality:
(2.36)y(t)≤-∫t∞dtn∫tn∞dtn-1∫tn-1∞dtn-2⋯∫t2∞[∫cdq(s,σ)f(x(s-σ))dσ]ds,t≥t0.
Using Tonelli′s theorem, we reverse the order of integration and obtain
(2.37)=-∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(x(s-σ))dσ]ds,t≥t0.
That is,
(2.38)x(t)≤1r(t)∫abp(t,τ)ω(t-τ)dτ-∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(x(s-σ))dσ]ds,t≥t0.
Let T≥0 be such that (2.38) hold, and x(t-μ)>0,t≥T. Now, we consider the set of functions:
(2.39)Ω={z∈C([T-μ,∞),R):0<z(t)≤1,t≥T-μ},
and define an operator S on Ω as follows:
(2.40)(Sz)(t)={t-T+μμ(Sz)(T)+(1-t-T+μμ),T-μ≤t<T,x(t)1r(t)∫abp(t,τ)x(t-τ)z(t-τ)dτ-1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(x(s-σ)z(s-σ))dσ]ds,t≥T.
Then, it follows from Lebesgue′s dominated convergence theorem that S is continuous. By using (2.38), it is easy to see that S maps Ω into itself and, for any z∈Ω, we have (Sz)(t)<0, for T-μ≤t<T. Next, we define the sequence zk(t)∈Ω:
(2.41)z0(t)≡1,t≥T-μ,zk+1(t)=(Szk)(t),t≥T-μ,k=0,1,….
Then, by using (2.38) and a simple induction, we can easily see that
(2.42)0<zk+1(t)≤zk(t)≤1,t≥T-μ.
Set limk→∞zk(t)=z(t),t≥T-μ, then z(t) satisfies
(2.43)z(t)={t-T+μμ(Sz)(T)+(1-t-T+μμ),T-μ≤t<T,x(t)1r(t)∫abp(t,τ)x(t-τ)z(t-τ)dτ-1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(x(s-σ)z(s-σ))dσ]ds,t≥T.
Again, set ω(t)=x(t)/z(t), then ω(t) satisfies ω(t)>0,T-μ≤t≤T, and
(2.44)ω(t)=1r(t)∫abp(t,τ)ω(t-τ)dτ-1r(t)∫t∞(s-t)n-1(n-1)![∫cdq(s,σ)f(ω(s-σ))dσ]ds,t≥T.
Thus, ω(t) is a positive solution of (1.1) for t≥T.
Consider the following(2.45)ω(t)>0,t≥T-μ.
Assume that there exists t*≥T-μ, such that ω(t)>0, for T-μ≤t≤t*, and ω(t*)=0. Then,
(2.46)0=ω(t*)=1r(t*)∫abp(t*,τ)ω(t*-τ)dτ-1r(t*)∫t*∞(s-t*)n-1(n-1)![∫cdq(s,σ)f(ω(s-σ))dσ]ds,t*≥T,
which implies
(2.47)p(t*,τ)=0,q(s,σ)f(ω(s-σ))≡0,
which contradicts (H1). Thus, ω(t) is an eventually positive solution of (1.1).
The proof is complete.Example 2.4.
Consider high-order neutral differential equation with distributed delay(2.48)(x(t)-∫12e-tx(t-τ)dτ)(n)+12e-1-3e-2∫12σx(t-σ)dσ=0.
Here, r(t)≡1,p(t-τ)=e-t,q(t,σ)=σ/2e-1-3e-2,a=c=1,b=d=2. It is easy to see that
(2.49)r(t)≡1<∞,∫12e-tdτ1=e-t<1,t>0.
From Theorem 2.2, we have that (1.1) has an eventually positive solution if and only if inequality (1.2) has an eventually positive solution. In fact, x(t)=et is a positive solution of (1.1).



---

*Source: 618058-2012-05-13.xml*