Existence of positive solutions of higher-order nonlinear neutral equations

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.

1 Introduction

The purpose of this article is to study higher-order neutral nonlinear differential equations of the form

$${[r(t){[x(t)-P_1(t)x(t-\tau )]}^{(n-1)}]}^{\prime }+{(-1)}^n{Q}_1(t)f(x(t-\sigma ))=0,$$

(1)

$${[r(t){[x(t)-P_1(t)x(t-\tau )]}^{(n-1)}]}^{\prime }+{(-1)}^n{\int }_c^d{Q}_2(t,\xi )f(x(t-\xi ))d\xi =0$$

(2)

and

$${[r(t){[x(t)-{\int }_a^b{P}_2(t,\xi )x(t-\xi )d\xi ]}^{(n-1)}]}^{\prime }+{(-1)}^n{\int }_c^d{Q}_2(t,\xi )f(x(t-\xi ))d\xi =0,$$

(3)

where $n⩾2$ is an integer, $\tau >0$, $\sigma ⩾0$, $d>c⩾0$, $b>a⩾0$, r, $P_1\in C([t_0,\mathrm{\infty }),(0,\mathrm{\infty }))$, $P_2\in C([t_0,\mathrm{\infty })\times [a,b],(0,\mathrm{\infty }))$, $Q_1\in C([t_0,\mathrm{\infty }),(0,\mathrm{\infty }))$, $Q_2\in C([t_0,\mathrm{\infty })\times [c,d],(0,\mathrm{\infty }))$, $f\in C(\mathbb{R},\mathbb{R})$, f is a nondecreasing function with $xf(x)>0$, $x\ne 0$.

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

$${[r(t){[x(t)-P(t)x(t-\tau )]}^{\prime }]}^{\prime }+Q(t)f(x(t-\sigma ))=0$$

(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 [2–5], but in this work our results contain not only existence of solutions but also behavior of solutions. For books, we refer the reader to [6–11].

Let ${\rho }_1=max\{\tau ,\sigma \}$. By a solution of (1) we understand a function $x\in C([t_1-{\rho }_1,\mathrm{\infty }),\mathbb{R})$, for some $t_1⩾t_0$, such that $x(t)-P_1(t)x(t-\tau )$ is $n-1$ times continuously differentiable, $r(t){(x(t)-P_1(t)x(t-\tau ))}^{(n-1)}$ is continuously differentiable on $[t_1,\mathrm{\infty })$ and (1) is satisfied for $t⩾t_1$. Similarly, let ${\rho }_2=max\{\tau ,d\}$. By a solution of (2) we understand a function $x\in C([t_1-{\rho }_2,\mathrm{\infty }),\mathbb{R})$, for some $t_1⩾t_0$, such that $x(t)-P_1(t)x(t-\tau )$ is $n-1$ times continuously differentiable, $r(t){(x(t)-P_1(t)x(t-\tau ))}^{(n-1)}$ is continuously differentiable on $[t_1,\mathrm{\infty })$ and (2) is satisfied for $t⩾t_1$. Finally, let ${\rho }_3=max\{b,d\}$. By a solution of (3) we understand a function $x\in C([t_1-{\rho }_3,\mathrm{\infty }),\mathbb{R})$, for some $t_1⩾t_0$, such that $x(t)-{\int }_a^b{P}_2(t,\xi )x(t-\xi )d\xi$ is $n-1$ times continuously differentiable, $r(t){[x(t)-{\int }_a^b{P}_2(t,\xi )x(t-\xi )d\xi ]}^{(n-1)}$ is continuously differentiable on $[t_1,\mathrm{\infty })$ 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

$${\int }_{t_0}^{\mathrm{\infty }}{Q}_1(t)dt=\mathrm{\infty }.$$

(5)

Assume that $0<k_1⩽k_2$ and there exists $\gamma ⩾0$ such that

$$\frac{k_1}{k_2}exp((k_2-k_1){\int }_{t_0-\gamma }^{t_0}{Q}_1(t)dt)⩾1,$$

(6)

$$\begin{array}{c}exp(-k_2{\int }_{t-\tau }^t{Q}_1(s)ds)+exp(k_2{\int }_{t_0-\gamma }^{t-\tau }{Q}_1(s)ds)\hfill \\ \times \frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(exp(-k_1{\int }_{t_0-\gamma }^{u-\sigma }{Q}_1(z)dz))duds\hfill \\ ⩽P_1(t)⩽exp(-k_1{\int }_{t-\tau }^t{Q}_1(s)ds)+exp(k_1{\int }_{t_0-\gamma }^{t-\tau }{Q}_1(s)ds)\hfill \\ \times \frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(exp(-k_2{\int }_{t_0-\gamma }^{u-\sigma }{Q}_1(z)dz))duds,\hfill \\ t⩾t_1⩾t_0+max\{\tau ,\sigma \}.\hfill \end{array}$$

(7)

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

Proof Let Ω be the set of all continuous and bounded functions on $[t_0,\mathrm{\infty })$ with the sup norm. Then Ω is a Banach space. Define a subset A of Ω by

$$A=\{x\in \mathrm{\Omega }:v_1(t)⩽x(t)⩽v_2(t),t⩾t_0\},$$

where $v_1(t)$ and $v_2(t)$ are nonnegative functions such that

$$v_1(t)=exp(-k_2{\int }_{t_0-\gamma }^t{Q}_1(s)ds),v_2(t)=exp(-k_1{\int }_{t_0-\gamma }^t{Q}_1(s)ds),t⩾t_0.$$

(8)

It is clear that A is a bounded, closed and convex subset of Ω. We define the operator $S:A⟶\mathrm{\Omega }$ as

$$(Sx)(t)=\{\begin{array}{l}{P}_1(t)x(t-\tau )-\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(x(u-\sigma ))duds,t⩾t_1,\\ (Sx)(t_1)+v_2(t)-v_2(t_1),t_0⩽t⩽t_1.\end{array}$$

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

$$\begin{array}{rcl}(Sx)(t)& ⩽& P_1(t)v_2(t-\tau )-\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(v_1(u-\sigma ))duds\\ =& P_1(t)exp(-k_1{\int }_{t_0-\gamma }^{t-\tau }{Q}_1(s)ds)\\ -\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(exp(-k_2{\int }_{t_0-\gamma }^{u-\sigma }{Q}_1(z)dz))duds\\ ⩽& v_2(t)\end{array}$$

and

$$\begin{array}{rcl}(Sx)(t)& ⩾& P_1(t)v_1(t-\tau )-\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(v_2(u-\sigma ))duds\\ =& P_1(t)exp(-k_2{\int }_{t_0-\gamma }^{t-\tau }{Q}_1(s)ds)\\ -\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(exp(-k_1{\int }_{t_0-\gamma }^{u-\sigma }{Q}_1(z)dz))duds\\ ⩾& v_1(t).\end{array}$$

For $t\in [t_0,t_1]$ and $x\in A$, we obtain

$$(Sx)(t)=(Sx)(t_1)+v_2(t)-v_2(t_1)⩽v_2(t)$$

and in order to show $(Sx)(t)⩾v_1(t)$, consider

$$H(t)=v_2(t)-v_2(t_1)-v_1(t)+v_1(t_1).$$

By making use of (6), it follows that

$$\begin{array}{rcl}{H}^{\prime }(t)& =& v_2^{\prime }(t)-v_1^{\prime }(t)=-k_1{Q}_1(t)v_2(t)+k_2{Q}_1(t)v_1(t)\\ =& Q_1(t)v_2(t)[-k_1+k_2{v}_1(t)exp(k_1{\int }_{t_0-\gamma }^t{Q}_1(s)ds)]\\ =& Q_1(t)v_2(t)[-k_1+k_{2}exp((k_1-k_2){\int }_{t_0-\gamma }^t{Q}_1(s)ds)]\\ ⩽& Q_1(t)v_2(t)[-k_1+k_{2}exp((k_1-k_2){\int }_{t_0-\gamma }^{t_0}{Q}_1(s)ds)]⩽0,t_0⩽t⩽t_1.\end{array}$$

Since $H(t_1)=0$ and $H^{\prime }(t)⩽0$ for $t\in [t_0,t_1]$, we conclude that

$$H(t)=v_2(t)-v_2(t_1)-v_1(t)+v_1(t_1)⩾0,t_0⩽t⩽t_1.$$

Then $t\in [t_0,t_1]$ and for any $x\in A$,

$$(Sx)(t)=(Sx)(t_1)+v_2(t)-v_2(t_1)⩾v_1(t_1)+v_2(t)-v_2(t_1)⩾v_1(t),t_0⩽t⩽t_1.$$

Hence, S maps A into A.

Second, we show that S is continuous. Let $\{x_i\}$ be a convergent sequence of functions in A such that $x_i(t)\to x(t)$ as $i\to \mathrm{\infty }$. Since A is closed, we have $x\in A$. It is obvious that for $t\in [t_0,t_1]$ and $x\in A$, S is continuous. For $t⩾t_1$,

$$\begin{array}{r}|(S{x}_i)(t)-(Sx)(t)|\\ ⩽P_1(t)|x_i(t-\tau )-x(t-\tau )|\\ +|\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)[f(x_i(u-\sigma ))-f(x(u-\sigma ))]duds|\\ ⩽P_1(t)|x_i(t-\tau )-x(t-\tau )|\\ +\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)|f(x_i(u-\sigma ))-f(x(u-\sigma ))|duds.\end{array}$$

Since $|f(x_i(t-\sigma ))-f(x(t-\sigma ))|\to 0$ as $i\to \mathrm{\infty }$, by making use of the Lebesgue dominated convergence theorem, we see that

$$\underset{t\to \mathrm{\infty }}{lim}\parallel (S{x}_i)(t)-(Sx)(t)\parallel =0$$

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 $\{Sx:x\in A\}$ is uniformly bounded and equicontinuous on $[t_0,\mathrm{\infty })$. Since uniform boundedness of $\{Sx:x\in A\}$ 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 $(Sx)(T)⩽\frac{ϵ}{2}$. For $x\in A$ and $T_2>T_1⩾T$, we have

$$|(Sx)(T_2)-(Sx)(T_1)|⩽|(Sx)(T_2)|+|(Sx)(T_1)|⩽\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ.$$

Note that

$$\begin{array}{rcl}{X}^n-Y^n& =& (X-Y)(X^{n-1}+X^{n-2}Y+\cdots +X{Y}^{n-2}+Y^{n-1})\\ ⩽& n(X-Y)X^{n-1},X>Y>0.\end{array}$$

(9)

For $x\in A$ and $t_1⩽T_1<T_2⩽T$, by using (9) we obtain

$$\begin{array}{r}|(Sx)(T_2)-(Sx)(T_1)|\\ ⩽|P_1(T_2)x(T_2-\tau )-P_1(T_1)x(T_1-\tau )|\\ +\frac{1}{(n-2)!}{\int }_{T_1}^{T_2}\frac{{(s-T_1)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(x(u-\sigma ))duds\\ +\frac{1}{(n-2)!}{\int }_{T_2}^{\mathrm{\infty }}\frac{{(s-T_1)}^{n-2}-{(s-T_2)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(x(u-\sigma ))duds\\ ⩽|P_1(T_2)x(T_2-\tau )-P_1(T_1)x(T_1-\tau )|\\ +\underset{T_1⩽s⩽T_2}{max}\{\frac{1}{(n-2)!}\frac{s^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(x(u-\sigma ))du\}(T_2-T_1)\\ +\frac{1}{(n-3)!}{\int }_{T_2}^{\mathrm{\infty }}\frac{{(s-T_1)}^{n-3}}{r(s)}{\int }_s^{\mathrm{\infty }}{Q}_1(u)f(x(u-\sigma ))duds(T_2-T_1).\end{array}$$

Thus there exits $\delta >0$ such that

$$|(Sx)(T_2)-(Sx)(T_1)|<ϵ\text{if }0<T_2-T_1<\delta .$$

Finally, for $x\in A$ and $t_0⩽T_1<T_2⩽t_1$, there exits $\delta >0$ such that

$$|(Sx)(T_2)-(Sx)(T_1)|=|v_2(T_1)-v_2(T_2)|<ϵ\text{if }0<T_2-T_1<\delta .$$

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

$${\int }_{t_0}^{\mathrm{\infty }}{\tilde{Q}}_2(t)dt=\mathrm{\infty },$$

(10)

where ${\tilde{Q}}_2(t)={\int }_c^d{Q}_2(t,\xi )d\xi$. Assume that $0<k_1⩽k_2$ and there exists $\gamma ⩾0$ such that

$$\begin{array}{c}\frac{k_1}{k_2}exp((k_2-k_1){\int }_{t_0-\gamma }^{t_0}{\tilde{Q}}_2(t)dt)⩾1,\hfill \\ \begin{array}{r}exp(-k_2{\int }_{t-\tau }^t{\tilde{Q}}_2(s)ds)+exp(k_2{\int }_{t_0-\gamma }^{t-\tau }{\tilde{Q}}_2(s)ds)\\ \times \frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(exp(-k_1{\int }_{t_0-\gamma }^{u-\xi }{\tilde{Q}}_2(z)dz))d\xi duds\\ ⩽P_1(t)⩽exp(-k_1{\int }_{t-\tau }^t{\tilde{Q}}_2(s)ds)+exp(k_1{\int }_{t_0-\gamma }^{t-\tau }{\tilde{Q}}_2(s)ds)\\ \times \frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(exp(-k_2{\int }_{t_0-\gamma }^{u-\xi }{\tilde{Q}}_2(z)dz))d\xi duds,\\ t⩾t_1⩾t_0+max\{\tau ,d\}.\end{array}\hfill \end{array}$$

(11)

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

Proof Let Ω be the set of all continuous and bounded functions on $[t_0,\mathrm{\infty })$ with the sup norm. Then Ω is a Banach space. Define a subset A of Ω by

$$A=\{x\in \mathrm{\Omega }:v_1(t)⩽x(t)⩽v_2(t),t⩾t_0\},$$

where $v_1(t)$ and $v_2(t)$ are nonnegative functions such that

$$v_1(t)=exp(-k_2{\int }_{t_0-\gamma }^t{\tilde{Q}}_2(s)ds),v_2(t)=exp(-k_1{\int }_{t_0-\gamma }^t{\tilde{Q}}_2(s)ds),t⩾t_0.$$

It is clear that A is a bounded, closed and convex subset of Ω. We define the operator $S:A⟶\mathrm{\Omega }$ as follows:

$$(Sx)(t)=\{\begin{array}{l}{P}_1(t)x(t-\tau )-\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(x(u-\xi ))d\xi duds,t⩾t_1,\\ (Sx)(t_1)+v_2(t)-v_2(t_1),t_0⩽t⩽t_1.\end{array}$$

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

$$\begin{array}{r}exp(-k_2{\int }_{t-a}^t{\tilde{Q}}_2(s)ds)+exp(k_2{\int }_{t_0-\gamma }^{t-a}{\tilde{Q}}_2(s)ds)\\ \times \frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(exp(-k_1{\int }_{t_0-\gamma }^{u-\xi }{\tilde{Q}}_2(z)dz))d\xi duds\\ ⩽{\tilde{P}}_2(t)⩽exp(-k_1{\int }_{t-b}^t{\tilde{Q}}_2(s)ds)+exp(k_1{\int }_{t_0-\gamma }^{t-b}{\tilde{Q}}_2(s)ds)\\ \times \frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(exp(-k_2{\int }_{t_0-\gamma }^{u-\xi }{\tilde{Q}}_2(z)dz))d\xi duds,\\ t⩾t_1⩾t_0+max\{b,d\},\end{array}$$

(12)

where ${\tilde{P}}_2(t)={\int }_a^b{P}_2(t,\xi )d\xi$. Then (3) has a positive solution which tends to zero.

Proof Let Ω be the set of all continuous and bounded functions on $[t_0,\mathrm{\infty })$ with the sup norm. Then Ω is a Banach space. Define a subset A of Ω by

$$A=\{x\in \mathrm{\Omega }:v_1(t)⩽x(t)⩽v_2(t),t⩾t_0\},$$

where $v_1(t)$ and $v_2(t)$ are nonnegative functions such that

$$v_1(t)=exp(-k_2{\int }_{t_0-\gamma }^t{\tilde{Q}}_2(s)ds),v_2(t)=exp(-k_1{\int }_{t_0-\gamma }^t{\tilde{Q}}_2(s)ds),t⩾t_0.$$

(13)

It is clear that A is a bounded, closed and convex subset of Ω. We define the operator $S:A⟶\mathrm{\Omega }$ as

$$(Sx)(t)=\{\begin{array}{l}{\int }_a^b{P}_2(t,\xi )x(t-\xi )d\xi -\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(x(u-\xi ))d\xi duds,\\ t⩾t_1,\\ (Sx)(t_1)+v_2(t)-v_2(t_1),t_0⩽t⩽t_1.\end{array}$$

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

$$\begin{array}{rcl}(Sx)(t)& ⩽& {\int }_a^b{P}_2(t,\xi )v_2(t-\xi )d\xi -\frac{1}{(n-2)!}\\ \times {\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(v_1(u-\xi ))d\xi duds\\ ⩽& {\tilde{P}}_2(t)exp(-k_1{\int }_{t_0-\gamma }^{t-b}{\tilde{Q}}_2(s)ds)-\frac{1}{(n-2)!}\\ \times {\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(exp(-k_2{\int }_{t_0-\gamma }^{u-\xi }{\tilde{Q}}_2(z)dz))d\xi duds\\ ⩽& v_2(t)\end{array}$$

and

$$\begin{array}{rcl}(Sx)(t)& ⩾& {\int }_a^b{P}_2(t,\xi )v_1(t-\xi )d\xi -\frac{1}{(n-2)!}\\ \times {\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(v_2(u-\xi ))d\xi duds\\ ⩾& {\tilde{P}}_2(t)exp(-k_2{\int }_{t_0-\gamma }^{t-a}{\tilde{Q}}_2(s)ds)-\frac{1}{(n-2)!}\\ \times {\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(exp(-k_1{\int }_{t_0-\gamma }^{u-\xi }{\tilde{Q}}_2(z)dz))d\xi duds\\ ⩾& v_1(t).\end{array}$$

For $t\in [t_0,t_1]$ and $x\in A$, we obtain

$$(Sx)(t)=(Sx)(t_1)+v_2(t)-v_2(t_1)⩽v_2(t)$$

and to show $(Sx)(t)⩾v_1(t)$, consider

$$H(t)=v_2(t)-v_2(t_1)-v_1(t)+v_1(t_1).$$

By making use of (11), it follows that

$$\begin{array}{rcl}{H}^{\prime }(t)& =& v_2^{\prime }(t)-v_1^{\prime }(t)\\ =& -k_1{\tilde{Q}}_2(t)v_2(t)+k_2{\tilde{Q}}_2(t)v_1(t)\\ =& {\tilde{Q}}_2(t)v_2(t)[-k_1+k_2{v}_1(t)exp(k_1{\int }_{t_0-\gamma }^t{\tilde{Q}}_2(s)ds)]\\ ⩽& {\tilde{Q}}_2(t)v_2(t)[-k_1+k_{2}exp((k_1-k_2){\int }_{t_0-\gamma }^{t_0}{\tilde{Q}}_2(s)ds)]⩽0,t_0⩽t⩽t_1.\end{array}$$

Since $H(t_1)=0$ and $H^{\prime }(t)⩽0$ for $t\in [t_0,t_1]$, we conclude that

$$H(t)=v_2(t)-v_2(t_1)-v_1(t)+v_1(t_1)⩾0,t_0⩽t⩽t_1.$$

Then $t\in [t_0,t_1]$ and for any $x\in A$,

$$(Sx)(t)=(Sx)(t_1)+v_2(t)-v_2(t_1)⩾v_1(t_1)+v_2(t)-v_2(t_1)⩾v_1(t),t_0⩽t⩽t_1.$$

Hence, S maps A into A.

Next, we show that S is continuous. Let $\{x_i\}$ be a convergent sequence of functions in A such that $x_i(t)\to x(t)$ as $i\to \mathrm{\infty }$. Since A is closed, we have $x\in A$. It is obvious that for $t\in [t_0,t_1]$ and $x\in A$, S is continuous. For $t⩾t_1$,

$$\begin{array}{r}|(S{x}_i)(t)-(Sx)(t)|\\ ⩽{\int }_a^b{P}_2(t,\xi )|x_i(t-\xi )-x(t-\xi )|d\xi \\ +\frac{1}{(n-2)!}{\int }_t^{\mathrm{\infty }}\frac{{(s-t)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )|f(x_i(u-\xi ))-f(x(u-\xi ))|d\xi duds.\end{array}$$

Since $|f(x_i(t-\xi ))-f(x(t-\xi ))|\to 0$ as $i\to \mathrm{\infty }$ and $\xi \in [c,d]$, by making use of the Lebesgue dominated convergence theorem, we see that

$$\underset{t\to \mathrm{\infty }}{lim}\parallel (S{x}_i)(t)-(Sx)(t)\parallel =0.$$

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 $\{Sx:x\in A\}$ is uniformly bounded and equicontinuous on $[t_0,\mathrm{\infty })$. Since uniform boundedness of $\{Sx:x\in A\}$ 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 $(Sx)(T)⩽\frac{ϵ}{2}$. For $x\in A$ and $T_2>T_1⩾T$, we have

$$|(Sx)(T_2)-(Sx)(T_1)|⩽|(Sx)(T_2)|+|(Sx)(T_1)|⩽\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ.$$

For $x\in A$ and $t_1⩽T_1<T_2⩽T$, by using (9) we obtain

$$\begin{array}{r}|(Sx)(T_2)-(Sx)(T_1)|\\ ⩽{\int }_a^b|P_2(T_2,\xi )x(T_2-\xi )-P_2(T_1,\xi )x(T_1-\xi )|d\xi \\ +\frac{1}{(n-2)!}{\int }_{T_1}^{T_2}\frac{{(s-T_1)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(x(u-\xi ))d\xi duds\\ +\frac{1}{(n-2)!}{\int }_{T_2}^{\mathrm{\infty }}\frac{{(s-T_1)}^{n-2}-{(s-T_2)}^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(x(u-\xi ))d\xi duds\\ ⩽{\int }_a^b|P_2(T_2,\xi )x(T_2-\xi )-P_2(T_1,\xi )x(T_1-\xi )|d\xi \\ +\underset{T_1⩽s⩽T_2}{max}\{\frac{1}{(n-2)!}\frac{s^{n-2}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(x(u-\xi ))d\xi du\}(T_2-T_1)\\ +\frac{1}{(n-3)!}{\int }_{T_2}^{\mathrm{\infty }}\frac{{(s-T_1)}^{n-3}}{r(s)}{\int }_s^{\mathrm{\infty }}{\int }_c^d{Q}_2(u,\xi )f(x(u-\xi ))d\xi duds(T_2-T_1).\end{array}$$

Thus there exits $\delta >0$ such that

$$|(Sx)(T_2)-(Sx)(T_1)|<ϵ\text{if }0<T_2-T_1<\delta .$$

For $x\in A$ and $t_0⩽T_1<T_2⩽t_1$, there exits $\delta >0$ such that

$$|(Sx)(T_2)-(Sx)(T_1)|=|v_2(T_1)-v_2(T_2)|<ϵ\text{if }0<T_2-T_1<\delta .$$

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

$${\left[e^{t/2}{[x(t)-P_1(t)x(t-\frac{3}{2})]}^{(2)}\right]}^{\prime }-qx(t-1)=0,t⩾t_0,$$

(14)

where $q\in (0,\mathrm{\infty })$ and

$$\begin{array}{r}exp(-k_{2}q\tau )+\frac{exp(q[k_2(t+\gamma -\tau -t_0)-k_1(\gamma -\sigma -t_0)])}{k_1}\frac{exp((-q{k}_1-\frac{1}{2})t)}{{(k_{1}q+\frac{1}{2})}^2}\\ ⩽P_1(t)⩽exp(-k_{1}q\tau )+\frac{exp(q[k_1(t+\gamma -\tau -t_0)-k_2(\gamma -\sigma -t_0)])}{k_2}\\ \times \frac{exp((-q{k}_2-\frac{1}{2})t)}{{(k_{2}q+\frac{1}{2})}^2}.\end{array}$$

Note that for $k_1=\frac{2}{3}$, $k_2=1$, $q=1$ and $t_0=\gamma =\frac{13}{2}$, we have

$$\frac{k_1}{k_2}exp((k_2-k_1){\int }_{t_0-\gamma }^{t_0}{Q}_1(t)dt)=\frac{2}{3}exp\left(\frac{1}{3}{\int }_0^{\frac{13}{2}}1dt\right)=5.8194⩾1$$

and

$$exp\left(\frac{-3}{2}\right)+\frac{54}{49}exp\left(\frac{-t-5}{6}\right)⩽P_1(t)⩽exp(-1)+\frac{4}{9}exp\left(\frac{-5t}{6}\right),t⩾8.$$

If $P_1(t)$ 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.