Existence of nonoscillatory solutions of second-order nonlinear neutral differential equations with distributed deviating arguments

Abstract

Some sufficient conditions are provided for the existence of nonoscillatory solutions of nonlinear second-order neutral differential equations with distributed deviating arguments. The main tool for proving our results is the Banach contraction principle. Two examples are given to illustrate the effectiveness of our results.

Keywords:

• Neutral differential equation
• fixed point
• second order
• nonoscillatory solution

1991 Mathematics Subject Classifications:

• 34K11
• 34K40

1. Introduction

The purpose of this article is to study the second-order neutral nonlinear differential equations with distributed deviating arguments of the form (1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) and (2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) where $g\in C\left(\right[t_0,\mathrm{\infty }),\mathbb{R})$, $\tau >0$, $r\in C\left(\right[t_0,\mathrm{\infty }),(0,\mathrm{\infty }\left)\right)$, $p\in C\left(\right[t_0,\mathrm{\infty }),\mathbb{R})$, $P\in C\left(\right[t_0,\mathrm{\infty })\times [a,b],\mathbb{R})$ for $0<a<b,$ and ${\sigma }_i\in C\left(\right[t_0,\mathrm{\infty })\times [a_i,b_i],\mathbb{R})$ with $\underset{t\to \mathrm{\infty }}{lim}{\sigma }_i(t,\xi )=\mathrm{\infty }$ for $\xi \in [a_i,b_i]$, $a_i\ge 0,$ $i=1,2.$

Throughout this study, we assume that $f_i\in C\left(\right[t_0,\mathrm{\infty })\times \mathbb{R},\mathbb{R})$ is nondecreasing in the second variable, $i=1,2,$ $x{f}_i(t,x)>0$ for $x\ne 0$, $i=1,2,$ and satisfies (3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3)

where $q_i\in C\left(\right[t_0,\mathrm{\infty }),(0,\mathrm{\infty }\left)\right)$, $i=1,2,$ and $[e,f]$ (0<e<f or e<f<0) is any closed interval. Furthermore, suppose that (4) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{q_i\left(u\right)}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty },i=1,2,\end{array}$(4) (5) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|f_i(u,d)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\text{for some}\mathrm{d}\ne 0,\\ & i=1,2,\end{array}$(5) (6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) hold.

In recent years, there have been many studies concerning the oscillatory and nonoscillatory behaviour of neutral differential equations (see [1–16] and references cited therein). For some studies of the qualitative analysis of Volterra integro-differential equations and a variable delay system of differential equations of second order, we refer the reader to [17, 18] and references cited therein. For example, in 2010, Candan and Dahiya [3] considered the existence of nonoscillatory solutions of first and second-order neutral equations of the form $\begin{array}{rl}& \frac{{\mathrm{d}}^k}{\mathrm{d}{t}^k}\left[x\left(t\right)+P\left(t\right)x(t-\tau )\right]+{\int }_a^b{q}_1(t,\xi )x(t-\xi )\mathrm{d}\xi \\ & -{\int }_c^d{q}_2(t,\mu )x(t-\mu )\mathrm{d}\mu =0,\end{array}$

in 2016, Candan [4] investigated nonoscillatory solutions of higher order neutral differential equations of form $\begin{array}{rl}& [r\left(t\right)[[x\left(t\right)-{\int }_a^b{p}_2(t,\xi )x(t-\xi )\mathrm{d}\xi {]}^{(n-1)}{]}^{\gamma }{]}^{\prime }\\ & +(-1{)}^n{\int }_c^d{Q}_2(x,\xi \left)G\right(x(t-\xi ))\mathrm{d}\xi =0,\end{array}$ and in 2019, Çına et al. [8] studied the existence of nonoscillatory solutions of a nonlinear second-order neutral differential equation with forcing term $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }\\ & +f_1(t,x({\sigma }_1\left(t\right)\left)\right)-f_2(t,x({\sigma }_2\left(t\right)\left)\right)=g\left(t\right).\end{array}$ Motivated by the above-mentioned studies, the aim of this paper is to give some sufficient conditions for the existence of nonoscillatory solutions of (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ) and (2(2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) ), more general than the latter equation, by using the Banach contraction principle.

Let $T_0=min\{t_1-\tau ,\underset{t\ge t_1}{inf}\underset{\xi \in [a_1,b_1]}{min}{\sigma }_1(t,\xi ),\underset{t\ge t_1}{inf}\underset{\xi \in [a_2,b_2]}{min}{\sigma }_2(t,\xi \left)\right\}$ for $t_1\ge t_0$. By a solution of Equation (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ), we mean a function $x\in C\left(\right[T_0,\mathrm{\infty }),\mathbb{R})$ in the sense that both $x\left(t\right)-p\left(t\right)x(t-\tau )$ and $r\left(t\right)\left(x\right(t)-p(t\left)x\right(t-\tau ){)}^{\prime }$ are continuously differentiable on $[t_1,\mathrm{\infty }]$ and such that Equation (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ) is satisfied for $t\ge t_1$.

Let $T_1=min\{t_1-b,\underset{t\ge t_1}{inf}\underset{\xi \in [a_1,b_1]}{min}{\sigma }_1(t,\xi ),\underset{t\ge t_1}{inf}\underset{\xi \in [a_2,b_2]}{min}{\sigma }_2(t,\xi \left)\right\}$ for $t_1\ge t_0$. By a solution of Equation (2(2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) ), we mean a function $x\in C\left(\right[T_1,\mathrm{\infty }),\mathbb{R})$ in the sense that both $x\left(t\right)-{\int }_a^{b}p(t,\xi )x(t-\xi )\mathrm{d}\xi$ and $r\left(t\right)\left(x\right(t)-{\int }_a^{b}p(t,\xi \left)x\right(t-\xi )\mathrm{d}\xi {)}^{\prime }$ are continuously differentiable on $[t_1,\mathrm{\infty }]$ and such that Equation (2(2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) ) is satisfied for $t\ge t_1$.

As is customary, a solution of (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ) or (2(2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) ) is said to be oscillatory if it has arbitrarily large zeros. Otherwise, the solution is called nonoscillatory.

2. Main results

We suppose throughout this paper that X is the set of all continuous and bounded functions on $[t_0,\mathrm{\infty })$ with the norm $\parallel x\parallel =\underset{t\ge t_0}{sup}|x\left(t\right)|<\mathrm{\infty }$.

Theorem 2.1

Assume that (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ) hold and $0\le p\left(t\right)\le p<1$. Then (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ) has a bounded nonoscillatory solution.

Proof.

Suppose (5(5) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|f_i(u,d)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\text{for some}\mathrm{d}\ne 0,\\ & i=1,2,\end{array}$(5) ) holds with d>0. A similar argument holds for d<0. Let $N_2=d$. Set $A=\{x\in X:N_1\le x(t)\le N_2,t\ge t_0\},$ where $N_1$ and $N_2$ are positive constants such that $N_1<(1-p)N_2.$ It is clear that A is a closed, bounded and convex subset of X. In view of (4(4) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{q_i\left(u\right)}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty },i=1,2,\end{array}$(4) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ), we can choose a $t_1>t_0$ sufficiently large such that $t-\tau \ge t_0$, ${\sigma }_i(t,\xi )\ge t_0$, $\xi \in [a_i,b_i]$, i = 1, 2 for $t\ge t_1$ and (7) $\begin{array}{rl}& p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)q_1\right(u)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s\le {\theta }_1<1,\end{array}$(7) where ${\theta }_1$ is a constant, (8) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)f_1\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -N_1,\end{array}$(8) and (9) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_2-a_2\left)f_2\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le (1-p)N_2-\alpha ,\end{array}$(9) where $\alpha \in (N_1,(1-p\left)N_2\right).$ Define a mapping $S:A⟶X$ as follows: $\begin{array}{r}\left(Sx\right)\left(t\right)=\left\{\begin{array}{l}\alpha +p\left(t\right)x(t-\tau )-{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\\ \times \left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ -\left.{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s,\\ t\ge t_1\\ \left(Sx\right)\left(t_1\right),t_0\le t\le t_1.\end{array}\right.\end{array}$ It is clear that Sx is continuous. For every $x\in A$ and $t\ge t_1$, by (9(9) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_2-a_2\left)f_2\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le (1-p)N_2-\alpha ,\end{array}$(9) ), we get $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\alpha +p\left(t\right)x(t-\tau )\\ & -{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)]\mathrm{d}u\mathrm{d}s\\ & \le \alpha +p{N}_2+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_2-a_2\left)f_2\right(u,d)\\ & +|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le N_2\end{array}$ and taking (8(8) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)f_1\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -N_1,\end{array}$(8) ) into account, we have $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\alpha +p\left(t\right)x(t-\tau )-{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\\ & \times [{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)]\mathrm{d}u\mathrm{d}s\\ & \ge \alpha -{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)f_1\right(u,d)\\ & +|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \ge N_1.\end{array}$ Thus $SA\subset A$. Now we show that S is a contraction mapping on A. In fact, for $x,y\in A$ and $t\ge t_1$, in view of (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) ) and (7(7) $\begin{array}{rl}& p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)q_1\right(u)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s\le {\theta }_1<1,\end{array}$(7) ), we have $\begin{array}{rl}& |\left(Sx\right)\left(t\right)-\left(Sy\right)\left(t\right)|\le p|x(t-\tau )-y(t-\tau )|\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}|f_1(u,x({\sigma }_1(u,\xi )\left)\right)\\ & -f_1(u,y({\sigma }_1(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}|f_2(u,x({\sigma }_2(u,\xi )\left)\right)\\ & -f_2(u,y({\sigma }_2(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & \le p|x(t-\tau )-y(t-\tau )|\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}{q}_1\left(u\right)|x\left({\sigma }_1\right(u,\xi \left)\right)\\ & -y\left({\sigma }_1\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}{q}_2\left(u\right)|x\left({\sigma }_2\right(u,\xi \left)\right)\\ & -y\left({\sigma }_2\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & \le \parallel x-y\parallel \left[p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)q_1\right(u)\right.\\ & +\left.(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s\phantom{{\int }_{t_1}^{\mathrm{\infty }}}\right]\\ & \le {\theta }_1\parallel x-y\parallel .\end{array}$ This implies that $\parallel Sx-Sy\parallel \le {\theta }_1\parallel x-y\parallel .$ Since ${\theta }_1<1$, S is a contraction mapping on A. Consequently, S has the unique fixed point $x\in A$ such that $Sx=x$, which is obviously a positive solution of (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ). This completes the proof.

Theorem 2.2

Assume that (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ) hold and $1<p_1\le p\left(t\right)\le p_2<\mathrm{\infty }$. Then (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ) has a bounded nonoscillatory solution.

Proof.

Suppose (5(5) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|f_i(u,d)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\text{for some}\mathrm{d}\ne 0,\\ & i=1,2,\end{array}$(5) ) holds with d>0, the case d<0 can be treated similarly. Let $N_4=d$. Set $A=\{x\in X:N_3\le x(t)\le N_4,t\ge t_0\},$ where $N_3$ and $N_4$ are positive constants such that $p_2{N}_3<(p_1-1)N_4.$ It is obvious that A is a closed, bounded and convex subset of X. Because of (4(4) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{q_i\left(u\right)}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty },i=1,2,\end{array}$(4) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ), we can choose a $t_1>t_0$ sufficiently large such that ${\sigma }_i(t+\tau ,\xi )\ge t_0$, $\xi \in [a_i,b_i],$ i = 1, 2 for $t\ge t_1$ and (10) $\begin{array}{rl}& \frac{1}{p_1}[1+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)q_1\right(u)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s]\le {\theta }_2<1,\end{array}$(10) where ${\theta }_2$ is a constant, (11) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)f_1\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le (p_1-1)N_4-\alpha \end{array}$(11) and (12) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_2-a_2\left)f_2\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -p_2{N}_3,\end{array}$(12) where $\alpha \in (p_2{N}_3,(p_1-1\left)N_4\right).$ Consider the mapping $S:A⟶X$ defined by $\begin{array}{r}\left(Sx\right)\left(t\right)=\left\{\begin{array}{l}\frac{1}{p(t+\tau )}\left[\alpha +x(t+\tau )+{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\right.\\ \times \left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ -\left.\left.{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s\right],\\ t\ge t_1\\ \left(Sx\right)\left(t_1\right),t_0\le t\le t_1.\end{array}\right.\end{array}$It is obvious that Sx is continuous. For every $x\in A$ and $t\ge t_1$, by (11(11) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)f_1\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le (p_1-1)N_4-\alpha \end{array}$(11) ), we have $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\frac{1}{p(t+\tau )}[\alpha +x(t+\tau )+{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\\ & \times [{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)]\mathrm{d}u\mathrm{d}s]\\ & \le \frac{1}{p_1}[\alpha +N_4+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)f_1\right(u,d)\\ & +|g\left(u\right)|]\mathrm{d}u\mathrm{d}s]\\ & \le N_4\end{array}$ and from (12(12) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_2-a_2\left)f_2\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -p_2{N}_3,\end{array}$(12) ), we obtain $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\frac{1}{p(t+\tau )}\left[\alpha +x(t+\tau )+{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\right.\\ & \times \left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ & -\left.\left.{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s\right]\\ & \ge \frac{1}{p_2}\left[\alpha -{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_2-a_2\left)f_2\right(u,d)\right.\\ & +\left.|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\phantom{{\int }_s^{\mathrm{\infty }}}\right]\\ & \ge N_3.\end{array}$ Thus $SA\subset A$. Finally, for $x,y\in A$ and $t\ge t_1$, in view of (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) ) and (10(10) $\begin{array}{rl}& \frac{1}{p_1}[1+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)q_1\right(u)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s]\le {\theta }_2<1,\end{array}$(10) ), we have $\begin{array}{rl}& |\left(Sx\right)\left(t\right)-\left(Sy\right)\left(t\right)|\le \frac{1}{p(t+\tau )}[|x(t+\tau )-y(t+\tau )|\\ & +{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}|f_1(u,x({\sigma }_1(u,\xi )\left)\right)\\ & -f_1(u,y({\sigma }_1(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}|f_2(u,x({\sigma }_2(u,\xi )\left)\right)\\ & -f_2(u,y({\sigma }_2(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s]\\ & \le \frac{1}{p_1}[|x(t+\tau )-y(t+\tau )|\\ & +{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}{q}_1\left(u\right)|x\left({\sigma }_1\right(u,\xi \left)\right)\\ & -y\left({\sigma }_1\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}{q}_2\left(u\right)|x\left({\sigma }_2\right(u,\xi \left)\right)\\ & -y\left({\sigma }_2\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s]\\ & \le \frac{\parallel x-y\parallel }{p_1}[1+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times \left[\right(b_1-a_1\left)q_1\right(u)+(b_2-a_2\left)q_2\right(u\left)\right]\mathrm{d}u\mathrm{d}s]\\ & \le {\theta }_2\parallel x-y\parallel .\end{array}$ This implies that $\parallel Sx-Sy\parallel \le {\theta }_2\parallel x-y\parallel$ with ${\theta }_2<1$. Hence, S is a contraction mapping on A. Consequently, S has the unique fixed point $x\in A$ such that $Sx=x$, which is obviously a positive solution of (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ). Thus, the theorem is proved.

Theorem 2.3

Assume that (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ) hold and $-1<-p\le p\left(t\right)\le 0$. Then (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ) has a bounded nonoscillatory solution.

Proof.

Suppose (5(5) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|f_i(u,d)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\text{for some}\mathrm{d}\ne 0,\\ & i=1,2,\end{array}$(5) ) holds with d>0. A similar argument holds for d<0. Let $N_6=d$. Set $A=\{x\in X:N_5\le x(t)\le N_6,t\ge t_0\},$ where $N_5$ and $N_6$ are positive constants such that $N_5+p{N}_6<N_6.$ It is obvious that A is a closed, bounded and convex subset of X. In view of (4(4) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{q_i\left(u\right)}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty },i=1,2,\end{array}$(4) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ), we can choose a $t_1>t_0$ sufficiently large such that $t-\tau \ge t_0$, ${\sigma }_i(t,\xi )\ge t_0$, $\xi \in [a_i,b_i]$, i = 1, 2 for $t\ge t_1$ and (13) $\begin{array}{rl}& p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)q_1\right(u)+(b_2-a_2\left)q_2\right(u\left)\right]\\ & \times \mathrm{d}u\mathrm{d}s\le {\theta }_3<1,\end{array}$(13) where ${\theta }_3$ is a constant, (14) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)f_1\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -N_5-p{N}_6,\end{array}$(14) and (15) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_2-a_2\left)f_2\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le N_6-\alpha ,\end{array}$(15) where $\alpha \in (N_5+p{N}_6,N_6).$ Define a mapping $S:A⟶X$ as follows: $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\left\{\begin{array}{l}\alpha +p\left(t\right)x(t-\tau )-{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}[{\int }_{a_1}^{b_1}\\ \times f_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \\ -{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)]\mathrm{d}u\mathrm{d}s,\\ t\ge t_1\\ \left(Sx\right)\left(t_1\right),t_0\le t\le t_1.\end{array}\right.\end{array}$ It is clear that Sx is continuous. For every $x\in A$ and $t\ge t_1$, from (15(15) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_2-a_2\left)f_2\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le N_6-\alpha ,\end{array}$(15) ), we have $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\alpha +p\left(t\right)x(t-\tau )-{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\\ & \times [{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)]\mathrm{d}u\mathrm{d}s\\ & \le \alpha +{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_2-a_2\left)f_2\right(u,d)\\ & +|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le N_6\end{array}$ and by using (14(14) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)f_1\right(u,d)+|g(u\left)|\right]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -N_5-p{N}_6,\end{array}$(14) ), we have $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\alpha +p\left(t\right)x(t-\tau )-{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\\ & \times [{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)]\mathrm{d}u\mathrm{d}s\\ & \ge \alpha -p{N}_6-{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)f_1\right(u,d)\\ & +|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \ge N_5.\end{array}$ Thus $SA\subset A$. We shall show that S is a contraction mapping on A. In fact, for $x,y\in A$ and $t\ge t_1$, in view of (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) ) and (13(13) $\begin{array}{rl}& p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)q_1\right(u)+(b_2-a_2\left)q_2\right(u\left)\right]\\ & \times \mathrm{d}u\mathrm{d}s\le {\theta }_3<1,\end{array}$(13) ), we have $\begin{array}{rl}& |\left(Sx\right)\left(t\right)-\left(Sy\right)\left(t\right)|\le p|x(t-\tau )-y(t-\tau )|\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}|f_1(u,x({\sigma }_1(u,\xi )\left)\right)\\ & -f_1(u,y({\sigma }_1(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}|f_2(u,x({\sigma }_2(u,\xi )\left)\right)\\ & -f_2(u,y({\sigma }_2(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & \le p|x(t-\tau )-y(t-\tau )|\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}{q}_1\left(u\right)|x\left({\sigma }_1\right(u,\xi \left)\right)\\ & -y\left({\sigma }_1\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}{q}_2\left(u\right)|x\left({\sigma }_2\right(u,\xi \left)\right)\\ & -y\left({\sigma }_2\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & \le \parallel x-y\parallel [p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times \left[\right(b_1-a_1\left)q_1\right(u)+(b_2-a_2\left)q_2\right(u\left)\right]\mathrm{d}u\mathrm{d}s]\\ & \le {\theta }_3\parallel x-y\parallel .\end{array}$ This implies that $\parallel Sx-Sy\parallel \le {\theta }_3\parallel x-y\parallel .$ Since ${\theta }_3<1$, S is a contraction mapping on A. Consequently, S has the unique fixed point $x\in A$ such that $Sx=x$, which is obviously a positive solution of (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ). This completes the proof.

Theorem 2.4

Assume that (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ) hold and $-\mathrm{\infty }<-p_1\le p\left(t\right)\le -p_2<-1$. Then (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ) has a bounded nonoscillatory solution.

Proof.

Suppose (5(5) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|f_i(u,d)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\text{for some}\mathrm{d}\ne 0,\\ & i=1,2,\end{array}$(5) ) holds with d>0, the case d<0 can be treated similarly. Let $N_8=d$. Set $A=\{x\in X:N_7\le x(t)\le N_8,t\ge t_0\},$ where $N_7$ and $N_8$ are positive constants such that $p_1{N}_7+N_8<p_2{N}_8.$ It is obvious that A is a closed, bounded and convex subset of X. In view of (4(4) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{q_i\left(u\right)}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty },i=1,2,\end{array}$(4) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ), there exists a $t_1>t_0$ sufficiently large such that ${\sigma }_i(t+\tau ,\xi )\ge t_0$, $\xi \in [a_i,b_i],$ i = 1, 2 for $t\ge t_1$ and (16) $\begin{array}{rl}& \frac{1}{p_2}[1+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)q_1\left(u\right)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s]\le {\theta }_4<1,\end{array}$(16) where ${\theta }_4$ is a constant, (17) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)f_1(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -p_1{N}_7-N_8\end{array}$(17) and (18) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_2-a_2)f_2(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le p_2{N}_8-\alpha ,\end{array}$(18) where $\alpha \in (p_1{N}_7+N_8,p_2{N}_8).$ Consider the mapping $S:A⟶X$ defined by $\begin{array}{r}\left(Sx\right)\left(t\right)=\left\{\begin{array}{l}-\frac{1}{p(t+\tau )}\left[\alpha -x(t+\tau )-{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\right.\\ \times \left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ \left.-{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)]\mathrm{d}u\mathrm{d}s\right],\\ t\ge t_1\\ \left(Sx\right)\left(t_1\right),t_0\le t\le t_1.\end{array}\right.\end{array}$ It is obvious that Sx is continuous. For every $x\in A$ and $t\ge t_1$, from (18(18) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_2-a_2)f_2(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le p_2{N}_8-\alpha ,\end{array}$(18) ), we have $\begin{array}{rl}\left(Sx\right)\left(t\right)& =-\frac{1}{p(t+\tau )}[\alpha -x(t+\tau )\\ & -{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ & -\left.\left.{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s\right]\\ & \le \frac{1}{p_2}[\alpha +{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times [(b_2-a_2)f_2(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s]\\ & \le N_8\end{array}$ and from (17(17) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)f_1(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -p_1{N}_7-N_8\end{array}$(17) ), we obtain $\begin{array}{rl}\left(Sx\right)\left(t\right)& =-\frac{1}{p(t+\tau )}\left[\alpha -x(t+\tau )-{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\right.\\ & \times \left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ & -\left.\left.{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s\right]\\ & \ge \frac{1}{p_1}[\alpha -N_8-{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times [(b_1-a_1)f_1(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s]\\ & \ge N_7.\end{array}$ Thus $SA\subset A$. In fact, for $x,y\in A$ and $t\ge t_1$, by using (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) ) and  (16(16) $\begin{array}{rl}& \frac{1}{p_2}[1+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)q_1\left(u\right)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s]\le {\theta }_4<1,\end{array}$(16) ), we have $\begin{array}{rl}& |\left(Sx\right)\left(t\right)-\left(Sy\right)\left(t\right)|\le \frac{1}{|p(t+\tau )|}[|x(t+\tau )-y(t+\tau )|\\ & +{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}|f_1(u,x({\sigma }_1(u,\xi )\left)\right)\\ & -f_1(u,y({\sigma }_1(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}|f_2(u,x({\sigma }_2(u,\xi )\left)\right)\\ & -f_2(u,y({\sigma }_2(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s]\\ & \le \frac{1}{p_2}[|x(t+\tau )-y(t+\tau )|\\ & +{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}{q}_1\left(u\right)|x\left({\sigma }_1\right(u,\xi \left)\right)\\ & -y\left({\sigma }_1\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_{t+\tau }^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}{q}_2\left(u\right)|x\left({\sigma }_2\right(u,\xi \left)\right)\\ & -y\left({\sigma }_2\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s]\\ & \le \frac{\parallel x-y\parallel }{p_2}[1+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times [(b_1-a_1)q_1\left(u\right)+(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s]\\ & \le {\theta }_4\parallel x-y\parallel .\end{array}$ This shows that $\parallel Sx-Sy\parallel \le {\theta }_4\parallel x-y\parallel$ with ${\theta }_4<1$. This implies that S is a contraction mapping on A. Thus, S has the unique fixed point $x\in A$ such that $Sx=x$, which is clearly a positive solution of (1(1) $\begin{array}{rl}& {\left(r\left(t\right){\left(x\left(t\right)-p\left(t\right)x(t-\tau )\right)}^{\prime }\right)}^{\prime }+{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right)\end{array}$(1) ). Hence the proof is complete.

Theorem 2.5

Assume that (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ) hold, $P(t,\xi )\ge 0$ and ${\int }_a^{b}P(t,\xi )\mathrm{d}\xi \le p<1$. Then (2(2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) ) has a bounded nonoscillatory solution.

Proof.

Suppose (5(5) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|f_i(u,d)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\text{for some}\mathrm{d}\ne 0,\\ & i=1,2,\end{array}$(5) ) holds with d>0. A similar argument holds for d<0. Let $N_{10}=d$. Set $A=\{x\in X:N_9\le x(t)\le N_{10},t\ge t_0\},$ where $N_9$ and $N_{10}$ are positive constants such that $N_9<(1-p)N_{10}.$ It is obvious that A is a closed, bounded and convex subset of X. Because of (4(4) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{q_i\left(u\right)}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty },i=1,2,\end{array}$(4) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ), we can take a $t_1>t_0$ sufficiently large such that $t-b\ge t_0$, ${\sigma }_i(t,\xi )\ge t_0$, $\xi \in [a_i,b_i]$, i = 1, 2 for $t\ge t_1$ and (19) $\begin{array}{rl}& p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)q_1\left(u\right)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s\le {\theta }_5<1,\end{array}$(19) where ${\theta }_5$ is a constant, (20) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)f_1(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -N_9,\end{array}$(20) and (21) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_2-a_2)f_2(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le (1-p)N_{10}-\alpha ,\end{array}$(21) where $\alpha \in (N_9,(1-p\left)N_{10}\right).$ Define a mapping $S:A⟶X$ as follows: $\begin{array}{r}\left(Sx\right)\left(t\right)=\left\{\begin{array}{l}\alpha +{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi \\ -{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \\ -{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)]\mathrm{d}u\mathrm{d}s,\\ t\ge t_1\\ \left(Sx\right)\left(t_1\right),t_0\le t\le t_1.\end{array}\right.\end{array}$ It is clear that Sx is continuous. For every $x\in A$ and $t\ge t_1$, from (21(21) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_2-a_2)f_2(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le (1-p)N_{10}-\alpha ,\end{array}$(21) ), we have $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\alpha +{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi \\ & -{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ & -\left.{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s\\ & \le \alpha +p{N}_{10}+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times [(b_2-a_2)f_2(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le N_{10}\end{array}$ and by using (20(20) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)f_1(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -N_9,\end{array}$(20) ), we get $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\alpha +{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi \\ & -{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ & -\left.{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s\\ & \ge \alpha -{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times [(b_1-a_1)f_1(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \ge N_9.\end{array}$ Thus $SA\subset A$. Now we show that S is a contraction mapping on A. In fact, for $x,y\in A$ and $t\ge t_1$, from (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) ) and (19(19) $\begin{array}{rl}& p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)q_1\left(u\right)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s\le {\theta }_5<1,\end{array}$(19) ), we have $\begin{array}{rl}& |\left(Sx\right)\left(t\right)-\left(Sy\right)\left(t\right)|\le {\int }_a^{b}P(t,\xi )|x(t-\xi )-y(t-\xi )|\mathrm{d}\xi \\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}|f_1(u,x({\sigma }_1(u,\xi )\left)\right)\\ & -f_1(u,y({\sigma }_1(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}|f_2(u,x({\sigma }_2(u,\xi )\left)\right)\\ & -f_2(u,y({\sigma }_2(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & \le {\int }_a^{b}P(t,\xi )|x(t-\xi )-y(t-\xi )|\mathrm{d}\xi \\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}{q}_1\left(u\right)|x\left({\sigma }_1\right(u,\xi \left)\right)\\ & -y\left({\sigma }_1\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}{q}_2\left(u\right)|x\left({\sigma }_2\right(u,\xi \left)\right)\\ & -y\left({\sigma }_2\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & \le \parallel x-y\parallel [p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times [(b_1-a_1)q_1\left(u\right)+(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s]\\ & \le {\theta }_5\parallel x-y\parallel .\end{array}$ This implies that $\parallel Sx-Sy\parallel \le {\theta }_5\parallel x-y\parallel .$ Since ${\theta }_5<1$, S is a contraction mapping on A. Consequently, S has the unique fixed point $x\in A$ such that $Sx=x$, which is obviously a positive solution of (2(2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) ). This completes the proof.

Theorem 2.6

Assume that (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ) hold, $P(t,\xi )\le 0$ and $-1<-p\le {\int }_a^{b}P(t,\xi )\mathrm{d}\xi$. Then (2(2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) ) has a bounded nonoscillatory solution.

Proof.

Suppose (5(5) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|f_i(u,d)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\text{for some}\mathrm{d}\ne 0,\\ & i=1,2,\end{array}$(5) ) holds with d>0, the case d<0 can be treated similarly. Let $N_{12}=d$. Set $A=\{x\in X:N_{11}\le x(t)\le N_{12},t\ge t_0\},$ where $N_{11}$ and $N_{12}$ are positive constants such that $p{N}_{12}+N_{11}<N_{12}.$ It is clear that A is a closed, bounded and convex subset of X. By (4(4) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{q_i\left(u\right)}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty },i=1,2,\end{array}$(4) )–(6(6) $\begin{array}{rl}& {\int }_{t_0}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{|g\left(u\right)|}{r\left(s\right)}\mathrm{d}u\mathrm{d}s<\mathrm{\infty }\end{array}$(6) ), we can take a $t_1>t_0$ sufficiently large such that $t-b\ge t_0$, ${\sigma }_i(t,\xi )\ge t_0$, $\xi \in [a_i,b_i]$, i = 1, 2 for $t\ge t_1$ and (22) $\begin{array}{rl}& p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)q_1\left(u\right)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s\le {\theta }_6<1,\end{array}$(22) where ${\theta }_6$ is a constant, (23) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)f_1(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -p{N}_{12}-N_{11},\end{array}$(23) and (24) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_2-a_2)f_2(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le N_{12}-\alpha ,\end{array}$(24) where $\alpha \in (p{N}_{12}+N_{11},N_{12}).$ Consider the mapping $S:A⟶X$ defined by $\begin{array}{r}\left(Sx\right)\left(t\right)=\left\{\begin{array}{l}\alpha +{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi -{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\\ \times \left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ \left.-{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s,\\ t\ge t_1\\ \left(Sx\right)\left(t_1\right),t_0\le t\le t_1.\end{array}\right.\end{array}$ It is obvious that Sx is continuous. For every $x\in A$ and $t\ge t_1$, by (24(24) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_2-a_2)f_2(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le N_{12}-\alpha ,\end{array}$(24) ), we have $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\alpha +{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi \\ & -{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ & -\left.{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s\\ & \le \alpha +{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times [(b_2-a_2)f_2(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le N_{12}\end{array}$ and taking (23(23) $\begin{array}{rl}& {\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)f_1(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \le \alpha -p{N}_{12}-N_{11},\end{array}$(23) ) into account, we get $\begin{array}{rl}\left(Sx\right)\left(t\right)& =\alpha +{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi -{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}\\ & \times \left[{\int }_{a_1}^{b_1}{f}_1(u,x({\sigma }_1(u,\xi )\left)\right)\mathrm{d}\xi \right.\\ & -\left.{\int }_{a_2}^{b_2}{f}_2(u,x({\sigma }_2(u,\xi )\left)\right)\mathrm{d}\xi -g\left(u\right)\right]\mathrm{d}u\mathrm{d}s\\ & \ge \alpha -p{N}_{12}-{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\\ & \times [(b_1-a_1)f_1(u,d)+|g\left(u\right)|]\mathrm{d}u\mathrm{d}s\\ & \ge N_{11}.\end{array}$ Thus $SA\subset A$. Now we show that S is a contraction mapping on A. In fact, for $x,y\in A$ and $t\ge t_1$, in view of (3(3) $\begin{array}{rl}|f_i(t,x)-f_i(t,y)|& \le q_i\left(t\right)|x-y|\text{for}t\in [t_0,\mathrm{\infty })\\ \text{and}x,y\in [e,f],\end{array}$(3) ) and (22(22) $\begin{array}{rl}& p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}[(b_1-a_1)q_1\left(u\right)\\ & +(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s\le {\theta }_6<1,\end{array}$(22) ), we have $\begin{array}{rl}& |\left(Sx\right)\left(t\right)-\left(Sy\right)\left(t\right)|\le {\int }_a^b(-P(t,\xi \left)\right)|x(t-\xi )\\ & -y(t-\xi )|\mathrm{d}\xi +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}\\ & \times |f_1(u,x({\sigma }_1(u,\xi )\left)\right)-f_1(u,y({\sigma }_1(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}|f_2(u,x({\sigma }_2(u,\xi )\left)\right)\\ & -f_2(u,y({\sigma }_2(u,\xi )\left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & \le {\int }_a^b(-P(t,\xi \left)\right)|x(t-\xi )-y(t-\xi )|\mathrm{d}\xi \\ & +{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_1}^{b_1}{q}_1\left(u\right)|x\left({\sigma }_1\right(u,\xi \left)\right)\\ & -y\left({\sigma }_1\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s+{\int }_t^{\mathrm{\infty }}\frac{1}{r\left(s\right)}{\int }_s^{\mathrm{\infty }}{\int }_{a_2}^{b_2}{q}_2\left(u\right)\\ & \times |x\left({\sigma }_2\right(u,\xi \left)\right)-y\left({\sigma }_2\right(u,\xi \left)\right)|\mathrm{d}\xi \mathrm{d}u\mathrm{d}s\\ & \le \parallel x-y\parallel \left[p+{\int }_{t_1}^{\mathrm{\infty }}{\int }_s^{\mathrm{\infty }}\frac{1}{r\left(s\right)}\left[\right(b_1-a_1\left)q_1\right(u)\right.\\ & +\left.(b_2-a_2)q_2\left(u\right)]\mathrm{d}u\mathrm{d}s\phantom{{\int }_{t_1}^{\mathrm{\infty }}}\right]\\ & \le {\theta }_6\parallel x-y\parallel .\end{array}$ This implies that $\parallel Sx-Sy\parallel \le {\theta }_6\parallel x-y\parallel$ with ${\theta }_6<1,$ and S is a contraction mapping on A. Consequently, S has the unique fixed point $x\in A$ such that $Sx=x$, which is obviously a positive solution of (2(2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) ). Thus the theorem is proved.

Example 2.7

For t>4, consider the equation (25) $\begin{array}{rl}& \left({\mathrm{e}}^t\right(x\left(t\right)-{\mathrm{e}}^{-t}x(t-2){)}^{\prime }{)}^{\prime }+{\int }_0^{2}2{\mathrm{e}}^{-t+2}x(t-2\xi )\mathrm{d}\xi \\ & -{\int }_2^4{\mathrm{e}}^{-t+2}x(t-\xi )\mathrm{d}\xi \\ & =2{\mathrm{e}}^{2-t}-{\mathrm{e}}^{2-2t}+{\mathrm{e}}^{4-2t}.\end{array}$(25) Note that $r\left(t\right)={\mathrm{e}}^t$, $p\left(t\right)={\mathrm{e}}^{-t}$, $\tau =2$, ${\sigma }_1(t,\xi )=t-2\xi$, ${\sigma }_2(t,\xi )=t-\xi$, $f_1(t,x)=2{\mathrm{e}}^{-t+2}x$, $f_2(t,x)={\mathrm{e}}^{-t+2}x$ and $g\left(t\right)=2{\mathrm{e}}^{2-t}-{\mathrm{e}}^{2-2t}+{\mathrm{e}}^{4-2t}.$ We can check that all the conditions of Theorem 2.1 are satisfied. We note that $x\left(t\right)={\mathrm{e}}^{-t}+2$ is a nonoscillatory solution of (25(25) $\begin{array}{rl}& \left({\mathrm{e}}^t\right(x\left(t\right)-{\mathrm{e}}^{-t}x(t-2){)}^{\prime }{)}^{\prime }+{\int }_0^{2}2{\mathrm{e}}^{-t+2}x(t-2\xi )\mathrm{d}\xi \\ & -{\int }_2^4{\mathrm{e}}^{-t+2}x(t-\xi )\mathrm{d}\xi \\ & =2{\mathrm{e}}^{2-t}-{\mathrm{e}}^{2-2t}+{\mathrm{e}}^{4-2t}.\end{array}$(25) ).

Example 2.8

For t>4, consider the equation (26) $\begin{array}{rl}& \left({\mathrm{e}}^{t/2}\right(x\left(t\right)-{\int }_3^4{\mathrm{e}}^{-t-2\xi }x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_1^{2}3{\mathrm{e}}^{-t}x(t-2\xi )\mathrm{d}\xi \\ & -{\int }_2^{4}2{\mathrm{e}}^{-t}x(t-\xi )\mathrm{d}\xi =\frac{1}{4}{\mathrm{e}}^{-t/2-8}-\frac{1}{4}{\mathrm{e}}^{-t/2-6}\\ & +3{\mathrm{e}}^{-3t/2}-\frac{15}{2}{\mathrm{e}}^{-5t/2}-{\mathrm{e}}^{-t}+\frac{1}{4}{\mathrm{e}}^{4-3t}-\frac{1}{4}{\mathrm{e}}^{8-3t}.\end{array}$(26) Equation (26(26) $\begin{array}{rl}& \left({\mathrm{e}}^{t/2}\right(x\left(t\right)-{\int }_3^4{\mathrm{e}}^{-t-2\xi }x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_1^{2}3{\mathrm{e}}^{-t}x(t-2\xi )\mathrm{d}\xi \\ & -{\int }_2^{4}2{\mathrm{e}}^{-t}x(t-\xi )\mathrm{d}\xi =\frac{1}{4}{\mathrm{e}}^{-t/2-8}-\frac{1}{4}{\mathrm{e}}^{-t/2-6}\\ & +3{\mathrm{e}}^{-3t/2}-\frac{15}{2}{\mathrm{e}}^{-5t/2}-{\mathrm{e}}^{-t}+\frac{1}{4}{\mathrm{e}}^{4-3t}-\frac{1}{4}{\mathrm{e}}^{8-3t}.\end{array}$(26) ) is a special case of (2(2) $\begin{array}{rl}& (r\left(t\right)(x\left(t\right)-{\int }_a^{b}P(t,\xi )x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_{a_1}^{b_1}{f}_1(t,x({\sigma }_1(t,\xi )\left)\right)\mathrm{d}\xi \\ & -{\int }_{a_2}^{b_2}{f}_2(t,x({\sigma }_2(t,\xi )\left)\right)\mathrm{d}\xi =g\left(t\right),\end{array}$(2) ) with $r\left(t\right)={\mathrm{e}}^{t/2}$, $P(t,\xi )={\mathrm{e}}^{-t-2\xi }$, ${\sigma }_1(t,\xi )=t-2\xi$, ${\sigma }_2(t,\xi )=t-\xi$, $f_1(t,x)=3{\mathrm{e}}^{-t}x$, $f_2(t,x)=2{\mathrm{e}}^{-t}x$ and $g\left(t\right)=\frac{1}{4}{\mathrm{e}}^{-t/2-8}-\frac{1}{4}{\mathrm{e}}^{-t/2-6}+3{\mathrm{e}}^{-3t/2}-\frac{15}{2}{\mathrm{e}}^{-5t/2}-{\mathrm{e}}^{-t}+\frac{1}{4}{\mathrm{e}}^{4-3t}-\frac{1}{4}{\mathrm{e}}^{8-3t}.$ The conditions of Theorem 2.5 are clearly satisfied. It is obvious that $x\left(t\right)={\mathrm{e}}^{-2t}+1$ is a nonoscillatory solution of (26(26) $\begin{array}{rl}& \left({\mathrm{e}}^{t/2}\right(x\left(t\right)-{\int }_3^4{\mathrm{e}}^{-t-2\xi }x(t-\xi )\mathrm{d}\xi {)}^{\prime }{)}^{\prime }\\ & +{\int }_1^{2}3{\mathrm{e}}^{-t}x(t-2\xi )\mathrm{d}\xi \\ & -{\int }_2^{4}2{\mathrm{e}}^{-t}x(t-\xi )\mathrm{d}\xi =\frac{1}{4}{\mathrm{e}}^{-t/2-8}-\frac{1}{4}{\mathrm{e}}^{-t/2-6}\\ & +3{\mathrm{e}}^{-3t/2}-\frac{15}{2}{\mathrm{e}}^{-5t/2}-{\mathrm{e}}^{-t}+\frac{1}{4}{\mathrm{e}}^{4-3t}-\frac{1}{4}{\mathrm{e}}^{8-3t}.\end{array}$(26) ).

Disclosure statement

No potential conflict of interest was reported by the authors.

ORCID

M. Tamer Şenel http://orcid.org/0000-0003-1915-5697

T. Candan http://orcid.org/0000-0002-6603-3732

B. Çına http://orcid.org/0000-0003-1294-0983