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# Existence of positive solutions of higher-order nonlinear neutral equations

Journal of Inequalities and Applications20132013:573

https://doi.org/10.1186/1029-242X-2013-573

• Accepted: 11 November 2013
• Published:

## Abstract

In this work, we consider the existence of positive solutions of higher-order nonlinear neutral differential equations. In the special case, our results include some well-known results. In order to obtain new sufficient conditions for the existence of a positive solution, we use Schauder’s fixed point theorem.

## Keywords

• neutral equations
• fixed point
• higher-order
• positive solution

## 1 Introduction

${\left[r\left(t\right){\left[x\left(t\right)-{P}_{1}\left(t\right)x\left(t-\tau \right)\right]}^{\left(n-1\right)}\right]}^{\prime }+{\left(-1\right)}^{n}{Q}_{1}\left(t\right)f\left(x\left(t-\sigma \right)\right)=0,$
(1)
${\left[r\left(t\right){\left[x\left(t\right)-{P}_{1}\left(t\right)x\left(t-\tau \right)\right]}^{\left(n-1\right)}\right]}^{\prime }+{\left(-1\right)}^{n}{\int }_{c}^{d}{Q}_{2}\left(t,\xi \right)f\left(x\left(t-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi =0$
(2)
and
${\left[r\left(t\right){\left[x\left(t\right)-{\int }_{a}^{b}{P}_{2}\left(t,\xi \right)x\left(t-\xi \right)\phantom{\rule{0.2em}{0ex}}d\xi \right]}^{\left(n-1\right)}\right]}^{\prime }+{\left(-1\right)}^{n}{\int }_{c}^{d}{Q}_{2}\left(t,\xi \right)f\left(x\left(t-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi =0,$
(3)

where $n⩾2$ is an integer, $\tau >0$, $\sigma ⩾0$, $d>c⩾0$, $b>a⩾0$, r, ${P}_{1}\in C\left(\left[{t}_{0},\mathrm{\infty }\right),\left(0,\mathrm{\infty }\right)\right)$, ${P}_{2}\in C\left(\left[{t}_{0},\mathrm{\infty }\right)×\left[a,b\right],\left(0,\mathrm{\infty }\right)\right)$, ${Q}_{1}\in C\left(\left[{t}_{0},\mathrm{\infty }\right),\left(0,\mathrm{\infty }\right)\right)$, ${Q}_{2}\in C\left(\left[{t}_{0},\mathrm{\infty }\right)×\left[c,d\right],\left(0,\mathrm{\infty }\right)\right)$, $f\in C\left(\mathbb{R},\mathbb{R}\right)$, f is a nondecreasing function with $xf\left(x\right)>0$, $x\ne 0$.

The motivation for the present work was the recent work of Culáková et al. [1] in which the second-order neutral nonlinear differential equation of the form
${\left[r\left(t\right){\left[x\left(t\right)-P\left(t\right)x\left(t-\tau \right)\right]}^{\prime }\right]}^{\prime }+Q\left(t\right)f\left(x\left(t-\sigma \right)\right)=0$
(4)

was considered. Note that when $n=2$ in (1), we obtain (4). Thus, our results contain the results established in [1] for (1). The results for (2) and (3) are completely new.

Existence of nonoscillatory or positive solutions of higher-order neutral differential equations was investigated in [25], but in this work our results contain not only existence of solutions but also behavior of solutions. For books, we refer the reader to [611].

Let ${\rho }_{1}=max\left\{\tau ,\sigma \right\}$. By a solution of (1) we understand a function $x\in C\left(\left[{t}_{1}-{\rho }_{1},\mathrm{\infty }\right),\mathbb{R}\right)$, for some ${t}_{1}⩾{t}_{0}$, such that $x\left(t\right)-{P}_{1}\left(t\right)x\left(t-\tau \right)$ is $n-1$ times continuously differentiable, $r\left(t\right){\left(x\left(t\right)-{P}_{1}\left(t\right)x\left(t-\tau \right)\right)}^{\left(n-1\right)}$ is continuously differentiable on $\left[{t}_{1},\mathrm{\infty }\right)$ and (1) is satisfied for $t⩾{t}_{1}$. Similarly, let ${\rho }_{2}=max\left\{\tau ,d\right\}$. By a solution of (2) we understand a function $x\in C\left(\left[{t}_{1}-{\rho }_{2},\mathrm{\infty }\right),\mathbb{R}\right)$, for some ${t}_{1}⩾{t}_{0}$, such that $x\left(t\right)-{P}_{1}\left(t\right)x\left(t-\tau \right)$ is $n-1$ times continuously differentiable, $r\left(t\right){\left(x\left(t\right)-{P}_{1}\left(t\right)x\left(t-\tau \right)\right)}^{\left(n-1\right)}$ is continuously differentiable on $\left[{t}_{1},\mathrm{\infty }\right)$ and (2) is satisfied for $t⩾{t}_{1}$. Finally, let ${\rho }_{3}=max\left\{b,d\right\}$. By a solution of (3) we understand a function $x\in C\left(\left[{t}_{1}-{\rho }_{3},\mathrm{\infty }\right),\mathbb{R}\right)$, for some ${t}_{1}⩾{t}_{0}$, such that $x\left(t\right)-{\int }_{a}^{b}{P}_{2}\left(t,\xi \right)x\left(t-\xi \right)\phantom{\rule{0.2em}{0ex}}d\xi$ is $n-1$ times continuously differentiable, $r\left(t\right){\left[x\left(t\right)-{\int }_{a}^{b}{P}_{2}\left(t,\xi \right)x\left(t-\xi \right)\phantom{\rule{0.2em}{0ex}}d\xi \right]}^{\left(n-1\right)}$ is continuously differentiable on $\left[{t}_{1},\mathrm{\infty }\right)$ and (3) is satisfied for $t⩾{t}_{1}$.

The following fixed point theorem will be used in proofs.

Theorem 1 (Schauder’s fixed point theorem [9])

Let A be a closed, convex and nonempty subset of a Banach space Ω. Let $S:A\to A$ be a continuous mapping such that SA is a relatively compact subset of Ω. Then S has at least one fixed point in A. That is, there exists $x\in A$ such that $Sx=x$.

## 2 Main results

Theorem 2 Let
${\int }_{{t}_{0}}^{\mathrm{\infty }}{Q}_{1}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=\mathrm{\infty }.$
(5)
Assume that $0<{k}_{1}⩽{k}_{2}$ and there exists $\gamma ⩾0$ such that
$\frac{{k}_{1}}{{k}_{2}}exp\left(\left({k}_{2}-{k}_{1}\right){\int }_{{t}_{0}-\gamma }^{{t}_{0}}{Q}_{1}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)⩾1,$
(6)
$\begin{array}{c}exp\left(-{k}_{2}{\int }_{t-\tau }^{t}{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+exp\left({k}_{2}{\int }_{{t}_{0}-\gamma }^{t-\tau }{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left(exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{u-\sigma }{Q}_{1}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\hfill \\ \phantom{\rule{1em}{0ex}}⩽{P}_{1}\left(t\right)⩽exp\left(-{k}_{1}{\int }_{t-\tau }^{t}{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+exp\left({k}_{1}{\int }_{{t}_{0}-\gamma }^{t-\tau }{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left(exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{u-\sigma }{Q}_{1}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds,\hfill \\ \phantom{\rule{2em}{0ex}}t⩾{t}_{1}⩾{t}_{0}+max\left\{\tau ,\sigma \right\}.\hfill \end{array}$
(7)

Then (1) has a positive solution which tends to zero.

Proof Let Ω be the set of all continuous and bounded functions on $\left[{t}_{0},\mathrm{\infty }\right)$ with the sup norm. Then Ω is a Banach space. Define a subset A of Ω by
$A=\left\{x\in \mathrm{\Omega }:{v}_{1}\left(t\right)⩽x\left(t\right)⩽{v}_{2}\left(t\right),t⩾{t}_{0}\right\},$
where ${v}_{1}\left(t\right)$ and ${v}_{2}\left(t\right)$ are nonnegative functions such that
${v}_{1}\left(t\right)=exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{t}{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right),\phantom{\rule{2em}{0ex}}{v}_{2}\left(t\right)=exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{t}{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right),\phantom{\rule{1em}{0ex}}t⩾{t}_{0}.$
(8)
It is clear that A is a bounded, closed and convex subset of Ω. We define the operator $S:A⟶\mathrm{\Omega }$ as
$\left(Sx\right)\left(t\right)=\left\{\begin{array}{l}{P}_{1}\left(t\right)x\left(t-\tau \right)-\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left(x\left(u-\sigma \right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t⩾{t}_{1},\\ \left(Sx\right)\left({t}_{1}\right)+{v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right),\phantom{\rule{1em}{0ex}}{t}_{0}⩽t⩽{t}_{1}.\end{array}$

We show that S satisfies the assumptions of Schauder’s fixed point theorem.

First, S maps A into A. For $t⩾{t}_{1}$ and $x\in A$, using (7) and (8), we have
$\begin{array}{rcl}\left(Sx\right)\left(t\right)& ⩽& {P}_{1}\left(t\right){v}_{2}\left(t-\tau \right)-\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left({v}_{1}\left(u-\sigma \right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ =& {P}_{1}\left(t\right)exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{t-\tau }{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\\ -\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left(exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{u-\sigma }{Q}_{1}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ ⩽& {v}_{2}\left(t\right)\end{array}$
and
$\begin{array}{rcl}\left(Sx\right)\left(t\right)& ⩾& {P}_{1}\left(t\right){v}_{1}\left(t-\tau \right)-\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left({v}_{2}\left(u-\sigma \right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ =& {P}_{1}\left(t\right)exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{t-\tau }{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\\ -\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left(exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{u-\sigma }{Q}_{1}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ ⩾& {v}_{1}\left(t\right).\end{array}$
For $t\in \left[{t}_{0},{t}_{1}\right]$ and $x\in A$, we obtain
$\left(Sx\right)\left(t\right)=\left(Sx\right)\left({t}_{1}\right)+{v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)⩽{v}_{2}\left(t\right)$
and in order to show $\left(Sx\right)\left(t\right)⩾{v}_{1}\left(t\right)$, consider
$H\left(t\right)={v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)-{v}_{1}\left(t\right)+{v}_{1}\left({t}_{1}\right).$
By making use of (6), it follows that
$\begin{array}{rcl}{H}^{\prime }\left(t\right)& =& {v}_{2}^{\prime }\left(t\right)-{v}_{1}^{\prime }\left(t\right)=-{k}_{1}{Q}_{1}\left(t\right){v}_{2}\left(t\right)+{k}_{2}{Q}_{1}\left(t\right){v}_{1}\left(t\right)\\ =& {Q}_{1}\left(t\right){v}_{2}\left(t\right)\left[-{k}_{1}+{k}_{2}{v}_{1}\left(t\right)exp\left({k}_{1}{\int }_{{t}_{0}-\gamma }^{t}{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\right]\\ =& {Q}_{1}\left(t\right){v}_{2}\left(t\right)\left[-{k}_{1}+{k}_{2}exp\left(\left({k}_{1}-{k}_{2}\right){\int }_{{t}_{0}-\gamma }^{t}{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\right]\\ ⩽& {Q}_{1}\left(t\right){v}_{2}\left(t\right)\left[-{k}_{1}+{k}_{2}exp\left(\left({k}_{1}-{k}_{2}\right){\int }_{{t}_{0}-\gamma }^{{t}_{0}}{Q}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\right]⩽0,\phantom{\rule{1em}{0ex}}{t}_{0}⩽t⩽{t}_{1}.\end{array}$
Since $H\left({t}_{1}\right)=0$ and ${H}^{\prime }\left(t\right)⩽0$ for $t\in \left[{t}_{0},{t}_{1}\right]$, we conclude that
$H\left(t\right)={v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)-{v}_{1}\left(t\right)+{v}_{1}\left({t}_{1}\right)⩾0,\phantom{\rule{1em}{0ex}}{t}_{0}⩽t⩽{t}_{1}.$
Then $t\in \left[{t}_{0},{t}_{1}\right]$ and for any $x\in A$,
$\left(Sx\right)\left(t\right)=\left(Sx\right)\left({t}_{1}\right)+{v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)⩾{v}_{1}\left({t}_{1}\right)+{v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)⩾{v}_{1}\left(t\right),\phantom{\rule{1em}{0ex}}{t}_{0}⩽t⩽{t}_{1}.$

Hence, S maps A into A.

Second, we show that S is continuous. Let $\left\{{x}_{i}\right\}$ be a convergent sequence of functions in A such that ${x}_{i}\left(t\right)\to x\left(t\right)$ as $i\to \mathrm{\infty }$. Since A is closed, we have $x\in A$. It is obvious that for $t\in \left[{t}_{0},{t}_{1}\right]$ and $x\in A$, S is continuous. For $t⩾{t}_{1}$,
$\begin{array}{r}|\left(S{x}_{i}\right)\left(t\right)-\left(Sx\right)\left(t\right)|\\ \phantom{\rule{1em}{0ex}}⩽{P}_{1}\left(t\right)|{x}_{i}\left(t-\tau \right)-x\left(t-\tau \right)|\\ \phantom{\rule{2em}{0ex}}+|\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)\left[f\left({x}_{i}\left(u-\sigma \right)\right)-f\left(x\left(u-\sigma \right)\right)\right]\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}⩽{P}_{1}\left(t\right)|{x}_{i}\left(t-\tau \right)-x\left(t-\tau \right)|\\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)|f\left({x}_{i}\left(u-\sigma \right)\right)-f\left(x\left(u-\sigma \right)\right)|\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds.\end{array}$
Since $|f\left({x}_{i}\left(t-\sigma \right)\right)-f\left(x\left(t-\sigma \right)\right)|\to 0$ as $i\to \mathrm{\infty }$, by making use of the Lebesgue dominated convergence theorem, we see that
$\underset{t\to \mathrm{\infty }}{lim}\parallel \left(S{x}_{i}\right)\left(t\right)-\left(Sx\right)\left(t\right)\parallel =0$

and therefore S is continuous.

Third, we show that SA is relatively compact. In order to prove that SA is relatively compact, it suffices to show that the family of functions $\left\{Sx:x\in A\right\}$ is uniformly bounded and equicontinuous on $\left[{t}_{0},\mathrm{\infty }\right)$. Since uniform boundedness of $\left\{Sx:x\in A\right\}$ is obvious, we need only to show equicontinuity. For $x\in A$ and any $ϵ>0$, we take $T⩾{t}_{1}$ large enough such that $\left(Sx\right)\left(T\right)⩽\frac{ϵ}{2}$. For $x\in A$ and ${T}_{2}>{T}_{1}⩾T$, we have
$|\left(Sx\right)\left({T}_{2}\right)-\left(Sx\right)\left({T}_{1}\right)|⩽|\left(Sx\right)\left({T}_{2}\right)|+|\left(Sx\right)\left({T}_{1}\right)|⩽\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ.$
Note that
$\begin{array}{rcl}{X}^{n}-{Y}^{n}& =& \left(X-Y\right)\left({X}^{n-1}+{X}^{n-2}Y+\cdots +X{Y}^{n-2}+{Y}^{n-1}\right)\\ ⩽& n\left(X-Y\right){X}^{n-1},\phantom{\rule{1em}{0ex}}X>Y>0.\end{array}$
(9)
For $x\in A$ and ${t}_{1}⩽{T}_{1}<{T}_{2}⩽T$, by using (9) we obtain
$\begin{array}{r}|\left(Sx\right)\left({T}_{2}\right)-\left(Sx\right)\left({T}_{1}\right)|\\ \phantom{\rule{1em}{0ex}}⩽|{P}_{1}\left({T}_{2}\right)x\left({T}_{2}-\tau \right)-{P}_{1}\left({T}_{1}\right)x\left({T}_{1}-\tau \right)|\\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left(n-2\right)!}{\int }_{{T}_{1}}^{{T}_{2}}\frac{{\left(s-{T}_{1}\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left(x\left(u-\sigma \right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left(n-2\right)!}{\int }_{{T}_{2}}^{\mathrm{\infty }}\frac{{\left(s-{T}_{1}\right)}^{n-2}-{\left(s-{T}_{2}\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left(x\left(u-\sigma \right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}⩽|{P}_{1}\left({T}_{2}\right)x\left({T}_{2}-\tau \right)-{P}_{1}\left({T}_{1}\right)x\left({T}_{1}-\tau \right)|\\ \phantom{\rule{2em}{0ex}}+\underset{{T}_{1}⩽s⩽{T}_{2}}{max}\left\{\frac{1}{\left(n-2\right)!}\frac{{s}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left(x\left(u-\sigma \right)\right)\phantom{\rule{0.2em}{0ex}}du\right\}\left({T}_{2}-{T}_{1}\right)\\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left(n-3\right)!}{\int }_{{T}_{2}}^{\mathrm{\infty }}\frac{{\left(s-{T}_{1}\right)}^{n-3}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{Q}_{1}\left(u\right)f\left(x\left(u-\sigma \right)\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\left({T}_{2}-{T}_{1}\right).\end{array}$
Thus there exits $\delta >0$ such that
Finally, for $x\in A$ and ${t}_{0}⩽{T}_{1}<{T}_{2}⩽{t}_{1}$, there exits $\delta >0$ such that

Therefore SA is relatively compact. In view of Schauder’s fixed point theorem, we can conclude that there exists $x\in A$ such that $Sx=x$. That is, x is a positive solution of (1) which tends to zero. The proof is complete. □

Theorem 3 Let
${\int }_{{t}_{0}}^{\mathrm{\infty }}{\stackrel{˜}{Q}}_{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=\mathrm{\infty },$
(10)
where ${\stackrel{˜}{Q}}_{2}\left(t\right)={\int }_{c}^{d}{Q}_{2}\left(t,\xi \right)\phantom{\rule{0.2em}{0ex}}d\xi$. Assume that $0<{k}_{1}⩽{k}_{2}$ and there exists $\gamma ⩾0$ such that
$\begin{array}{c}\frac{{k}_{1}}{{k}_{2}}exp\left(\left({k}_{2}-{k}_{1}\right){\int }_{{t}_{0}-\gamma }^{{t}_{0}}{\stackrel{˜}{Q}}_{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)⩾1,\hfill \\ \begin{array}{r}exp\left(-{k}_{2}{\int }_{t-\tau }^{t}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+exp\left({k}_{2}{\int }_{{t}_{0}-\gamma }^{t-\tau }{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\\ \phantom{\rule{2em}{0ex}}×\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{u-\xi }{\stackrel{˜}{Q}}_{2}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}⩽{P}_{1}\left(t\right)⩽exp\left(-{k}_{1}{\int }_{t-\tau }^{t}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+exp\left({k}_{1}{\int }_{{t}_{0}-\gamma }^{t-\tau }{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\\ \phantom{\rule{2em}{0ex}}×\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{u-\xi }{\stackrel{˜}{Q}}_{2}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds,\\ \phantom{\rule{2em}{0ex}}t⩾{t}_{1}⩾{t}_{0}+max\left\{\tau ,d\right\}.\end{array}\hfill \end{array}$
(11)

Then (2) has a positive solution which tends to zero.

Proof Let Ω be the set of all continuous and bounded functions on $\left[{t}_{0},\mathrm{\infty }\right)$ with the sup norm. Then Ω is a Banach space. Define a subset A of Ω by
$A=\left\{x\in \mathrm{\Omega }:{v}_{1}\left(t\right)⩽x\left(t\right)⩽{v}_{2}\left(t\right),t⩾{t}_{0}\right\},$
where ${v}_{1}\left(t\right)$ and ${v}_{2}\left(t\right)$ are nonnegative functions such that
${v}_{1}\left(t\right)=exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{t}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right),\phantom{\rule{2em}{0ex}}{v}_{2}\left(t\right)=exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{t}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right),\phantom{\rule{1em}{0ex}}t⩾{t}_{0}.$
It is clear that A is a bounded, closed and convex subset of Ω. We define the operator $S:A⟶\mathrm{\Omega }$ as follows:
$\left(Sx\right)\left(t\right)=\left\{\begin{array}{l}{P}_{1}\left(t\right)x\left(t-\tau \right)-\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(x\left(u-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t⩾{t}_{1},\\ \left(Sx\right)\left({t}_{1}\right)+{v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right),\phantom{\rule{1em}{0ex}}{t}_{0}⩽t⩽{t}_{1}.\end{array}$

Since the remaining part of the proof is similar to those in the proof of Theorem 2, it is omitted. Thus the theorem is proved. □

Theorem 4 Suppose that (10) and (11) hold. In addition, assume that
$\begin{array}{r}exp\left(-{k}_{2}{\int }_{t-a}^{t}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+exp\left({k}_{2}{\int }_{{t}_{0}-\gamma }^{t-a}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\\ \phantom{\rule{2em}{0ex}}×\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{u-\xi }{\stackrel{˜}{Q}}_{2}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}⩽{\stackrel{˜}{P}}_{2}\left(t\right)⩽exp\left(-{k}_{1}{\int }_{t-b}^{t}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+exp\left({k}_{1}{\int }_{{t}_{0}-\gamma }^{t-b}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\\ \phantom{\rule{2em}{0ex}}×\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{u-\xi }{\stackrel{˜}{Q}}_{2}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds,\\ \phantom{\rule{2em}{0ex}}t⩾{t}_{1}⩾{t}_{0}+max\left\{b,d\right\},\end{array}$
(12)

where ${\stackrel{˜}{P}}_{2}\left(t\right)={\int }_{a}^{b}{P}_{2}\left(t,\xi \right)\phantom{\rule{0.2em}{0ex}}d\xi$. Then (3) has a positive solution which tends to zero.

Proof Let Ω be the set of all continuous and bounded functions on $\left[{t}_{0},\mathrm{\infty }\right)$ with the sup norm. Then Ω is a Banach space. Define a subset A of Ω by
$A=\left\{x\in \mathrm{\Omega }:{v}_{1}\left(t\right)⩽x\left(t\right)⩽{v}_{2}\left(t\right),t⩾{t}_{0}\right\},$
where ${v}_{1}\left(t\right)$ and ${v}_{2}\left(t\right)$ are nonnegative functions such that
${v}_{1}\left(t\right)=exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{t}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right),\phantom{\rule{2em}{0ex}}{v}_{2}\left(t\right)=exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{t}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right),\phantom{\rule{1em}{0ex}}t⩾{t}_{0}.$
(13)
It is clear that A is a bounded, closed and convex subset of Ω. We define the operator $S:A⟶\mathrm{\Omega }$ as
$\left(Sx\right)\left(t\right)=\left\{\begin{array}{l}{\int }_{a}^{b}{P}_{2}\left(t,\xi \right)x\left(t-\xi \right)\phantom{\rule{0.2em}{0ex}}d\xi -\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(x\left(u-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds,\\ \phantom{\rule{1em}{0ex}}t⩾{t}_{1},\\ \left(Sx\right)\left({t}_{1}\right)+{v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right),\phantom{\rule{1em}{0ex}}{t}_{0}⩽t⩽{t}_{1}.\end{array}$

We show that S satisfies the assumptions of Schauder’s fixed point theorem.

First of all, S maps A into A. For $t⩾{t}_{1}$ and $x\in A$, using (12), (13), the decreasing nature of ${v}_{2}$ and ${v}_{1}$, we have
$\begin{array}{rcl}\left(Sx\right)\left(t\right)& ⩽& {\int }_{a}^{b}{P}_{2}\left(t,\xi \right){v}_{2}\left(t-\xi \right)\phantom{\rule{0.2em}{0ex}}d\xi -\frac{1}{\left(n-2\right)!}\\ ×{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left({v}_{1}\left(u-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ ⩽& {\stackrel{˜}{P}}_{2}\left(t\right)exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{t-b}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)-\frac{1}{\left(n-2\right)!}\\ ×{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{u-\xi }{\stackrel{˜}{Q}}_{2}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ ⩽& {v}_{2}\left(t\right)\end{array}$
and
$\begin{array}{rcl}\left(Sx\right)\left(t\right)& ⩾& {\int }_{a}^{b}{P}_{2}\left(t,\xi \right){v}_{1}\left(t-\xi \right)\phantom{\rule{0.2em}{0ex}}d\xi -\frac{1}{\left(n-2\right)!}\\ ×{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left({v}_{2}\left(u-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ ⩾& {\stackrel{˜}{P}}_{2}\left(t\right)exp\left(-{k}_{2}{\int }_{{t}_{0}-\gamma }^{t-a}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)-\frac{1}{\left(n-2\right)!}\\ ×{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(exp\left(-{k}_{1}{\int }_{{t}_{0}-\gamma }^{u-\xi }{\stackrel{˜}{Q}}_{2}\left(z\right)\phantom{\rule{0.2em}{0ex}}dz\right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ ⩾& {v}_{1}\left(t\right).\end{array}$
For $t\in \left[{t}_{0},{t}_{1}\right]$ and $x\in A$, we obtain
$\left(Sx\right)\left(t\right)=\left(Sx\right)\left({t}_{1}\right)+{v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)⩽{v}_{2}\left(t\right)$
and to show $\left(Sx\right)\left(t\right)⩾{v}_{1}\left(t\right)$, consider
$H\left(t\right)={v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)-{v}_{1}\left(t\right)+{v}_{1}\left({t}_{1}\right).$
By making use of (11), it follows that
$\begin{array}{rcl}{H}^{\prime }\left(t\right)& =& {v}_{2}^{\prime }\left(t\right)-{v}_{1}^{\prime }\left(t\right)\\ =& -{k}_{1}{\stackrel{˜}{Q}}_{2}\left(t\right){v}_{2}\left(t\right)+{k}_{2}{\stackrel{˜}{Q}}_{2}\left(t\right){v}_{1}\left(t\right)\\ =& {\stackrel{˜}{Q}}_{2}\left(t\right){v}_{2}\left(t\right)\left[-{k}_{1}+{k}_{2}{v}_{1}\left(t\right)exp\left({k}_{1}{\int }_{{t}_{0}-\gamma }^{t}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\right]\\ ⩽& {\stackrel{˜}{Q}}_{2}\left(t\right){v}_{2}\left(t\right)\left[-{k}_{1}+{k}_{2}exp\left(\left({k}_{1}-{k}_{2}\right){\int }_{{t}_{0}-\gamma }^{{t}_{0}}{\stackrel{˜}{Q}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\right]⩽0,\phantom{\rule{1em}{0ex}}{t}_{0}⩽t⩽{t}_{1}.\end{array}$
Since $H\left({t}_{1}\right)=0$ and ${H}^{\prime }\left(t\right)⩽0$ for $t\in \left[{t}_{0},{t}_{1}\right]$, we conclude that
$H\left(t\right)={v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)-{v}_{1}\left(t\right)+{v}_{1}\left({t}_{1}\right)⩾0,\phantom{\rule{1em}{0ex}}{t}_{0}⩽t⩽{t}_{1}.$
Then $t\in \left[{t}_{0},{t}_{1}\right]$ and for any $x\in A$,
$\left(Sx\right)\left(t\right)=\left(Sx\right)\left({t}_{1}\right)+{v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)⩾{v}_{1}\left({t}_{1}\right)+{v}_{2}\left(t\right)-{v}_{2}\left({t}_{1}\right)⩾{v}_{1}\left(t\right),\phantom{\rule{1em}{0ex}}{t}_{0}⩽t⩽{t}_{1}.$

Hence, S maps A into A.

Next, we show that S is continuous. Let $\left\{{x}_{i}\right\}$ be a convergent sequence of functions in A such that ${x}_{i}\left(t\right)\to x\left(t\right)$ as $i\to \mathrm{\infty }$. Since A is closed, we have $x\in A$. It is obvious that for $t\in \left[{t}_{0},{t}_{1}\right]$ and $x\in A$, S is continuous. For $t⩾{t}_{1}$,
$\begin{array}{r}|\left(S{x}_{i}\right)\left(t\right)-\left(Sx\right)\left(t\right)|\\ \phantom{\rule{1em}{0ex}}⩽{\int }_{a}^{b}{P}_{2}\left(t,\xi \right)|{x}_{i}\left(t-\xi \right)-x\left(t-\xi \right)|\phantom{\rule{0.2em}{0ex}}d\xi \\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left(n-2\right)!}{\int }_{t}^{\mathrm{\infty }}\frac{{\left(s-t\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)|f\left({x}_{i}\left(u-\xi \right)\right)-f\left(x\left(u-\xi \right)\right)|\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds.\end{array}$
Since $|f\left({x}_{i}\left(t-\xi \right)\right)-f\left(x\left(t-\xi \right)\right)|\to 0$ as $i\to \mathrm{\infty }$ and $\xi \in \left[c,d\right]$, by making use of the Lebesgue dominated convergence theorem, we see that
$\underset{t\to \mathrm{\infty }}{lim}\parallel \left(S{x}_{i}\right)\left(t\right)-\left(Sx\right)\left(t\right)\parallel =0.$

Thus S is continuous.

Finally, we show that SA is relatively compact. In order to prove that SA is relatively compact, it suffices to show that the family of functions $\left\{Sx:x\in A\right\}$ is uniformly bounded and equicontinuous on $\left[{t}_{0},\mathrm{\infty }\right)$. Since uniform boundedness of $\left\{Sx:x\in A\right\}$ is obvious, we need only to show equicontinuity. For $x\in A$ and any $ϵ>0$, we take $T⩾{t}_{1}$ large enough such that $\left(Sx\right)\left(T\right)⩽\frac{ϵ}{2}$. For $x\in A$ and ${T}_{2}>{T}_{1}⩾T$, we have
$|\left(Sx\right)\left({T}_{2}\right)-\left(Sx\right)\left({T}_{1}\right)|⩽|\left(Sx\right)\left({T}_{2}\right)|+|\left(Sx\right)\left({T}_{1}\right)|⩽\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ.$
For $x\in A$ and ${t}_{1}⩽{T}_{1}<{T}_{2}⩽T$, by using (9) we obtain
$\begin{array}{r}|\left(Sx\right)\left({T}_{2}\right)-\left(Sx\right)\left({T}_{1}\right)|\\ \phantom{\rule{1em}{0ex}}⩽{\int }_{a}^{b}|{P}_{2}\left({T}_{2},\xi \right)x\left({T}_{2}-\xi \right)-{P}_{2}\left({T}_{1},\xi \right)x\left({T}_{1}-\xi \right)|\phantom{\rule{0.2em}{0ex}}d\xi \\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left(n-2\right)!}{\int }_{{T}_{1}}^{{T}_{2}}\frac{{\left(s-{T}_{1}\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(x\left(u-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left(n-2\right)!}{\int }_{{T}_{2}}^{\mathrm{\infty }}\frac{{\left(s-{T}_{1}\right)}^{n-2}-{\left(s-{T}_{2}\right)}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(x\left(u-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}⩽{\int }_{a}^{b}|{P}_{2}\left({T}_{2},\xi \right)x\left({T}_{2}-\xi \right)-{P}_{2}\left({T}_{1},\xi \right)x\left({T}_{1}-\xi \right)|\phantom{\rule{0.2em}{0ex}}d\xi \\ \phantom{\rule{2em}{0ex}}+\underset{{T}_{1}⩽s⩽{T}_{2}}{max}\left\{\frac{1}{\left(n-2\right)!}\frac{{s}^{n-2}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(x\left(u-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\right\}\left({T}_{2}-{T}_{1}\right)\\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left(n-3\right)!}{\int }_{{T}_{2}}^{\mathrm{\infty }}\frac{{\left(s-{T}_{1}\right)}^{n-3}}{r\left(s\right)}{\int }_{s}^{\mathrm{\infty }}{\int }_{c}^{d}{Q}_{2}\left(u,\xi \right)f\left(x\left(u-\xi \right)\right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\left({T}_{2}-{T}_{1}\right).\end{array}$
Thus there exits $\delta >0$ such that
For $x\in A$ and ${t}_{0}⩽{T}_{1}<{T}_{2}⩽{t}_{1}$, there exits $\delta >0$ such that

Therefore SA is relatively compact. In view of Schauder’s fixed point theorem, we can conclude that there exists $x\in A$ such that $Sx=x$. That is, x is a positive solution of (1) which tends to zero. The proof is complete. □

Example 1 Consider the neutral differential equation
${\left[{e}^{t/2}{\left[x\left(t\right)-{P}_{1}\left(t\right)x\left(t-\frac{3}{2}\right)\right]}^{\left(2\right)}\right]}^{\prime }-qx\left(t-1\right)=0,\phantom{\rule{1em}{0ex}}t⩾{t}_{0},$
(14)
where $q\in \left(0,\mathrm{\infty }\right)$ and
$\begin{array}{r}exp\left(-{k}_{2}q\tau \right)+\frac{exp\left(q\left[{k}_{2}\left(t+\gamma -\tau -{t}_{0}\right)-{k}_{1}\left(\gamma -\sigma -{t}_{0}\right)\right]\right)}{{k}_{1}}\frac{exp\left(\left(-q{k}_{1}-\frac{1}{2}\right)t\right)}{{\left({k}_{1}q+\frac{1}{2}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}⩽{P}_{1}\left(t\right)⩽exp\left(-{k}_{1}q\tau \right)+\frac{exp\left(q\left[{k}_{1}\left(t+\gamma -\tau -{t}_{0}\right)-{k}_{2}\left(\gamma -\sigma -{t}_{0}\right)\right]\right)}{{k}_{2}}\\ \phantom{\rule{2em}{0ex}}×\frac{exp\left(\left(-q{k}_{2}-\frac{1}{2}\right)t\right)}{{\left({k}_{2}q+\frac{1}{2}\right)}^{2}}.\end{array}$
Note that for ${k}_{1}=\frac{2}{3}$, ${k}_{2}=1$, $q=1$ and ${t}_{0}=\gamma =\frac{13}{2}$, we have
$\frac{{k}_{1}}{{k}_{2}}exp\left(\left({k}_{2}-{k}_{1}\right){\int }_{{t}_{0}-\gamma }^{{t}_{0}}{Q}_{1}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\frac{2}{3}exp\left(\frac{1}{3}{\int }_{0}^{\frac{13}{2}}1\phantom{\rule{0.2em}{0ex}}dt\right)=5.8194⩾1$
and
$exp\left(\frac{-3}{2}\right)+\frac{54}{49}exp\left(\frac{-t-5}{6}\right)⩽{P}_{1}\left(t\right)⩽exp\left(-1\right)+\frac{4}{9}exp\left(\frac{-5t}{6}\right),\phantom{\rule{1em}{0ex}}t⩾8.$

If ${P}_{1}\left(t\right)$ fulfils the last inequality above, a straightforward verification yields that the conditions of Theorem 2 are satisfied and therefore (14) has a positive solution which tends to zero.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Arts and Sciences, Niğde University, Niğde, 51200, Turkey

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