1 Introduction

The comparison principle plays an important role since it is one of the basic tools to study ODE and PDE. Thus, how to create a new comparison principle is an interesting and important question. In this paper, we shall create a comparison principle with impulsive effect. By means of the comparison principle and monotone iterative method, the existence of the max-minimal solution of second-order impulsive functional integro-differential Eqs. (1.1) is investigated. Also, the iterative sequences of solutions of the system are given. The importance of this method does not need to be particularly pointed out [1–14]. The theorems achieved in this paper are more general and complement many previously known results.

Impulsive differential equations, arising in the mathematical modeling of complex systems and processes, have drawn more and more attention of the research community due to their numerous applications in various fields of science and engineering such as chemistry, physics, biology, medicine, mechanics, etc. (see [15–24]). Boundary value problems (BVP) of differential equations have been investigated for many years. Now, nonlinear boundary conditions have drawn much attention, there exist many articles dealing with the problem for different kinds of boundary value conditions such as multi-point, integral boundary condition, and other conditions (see [25–35]). On the other hand, deviated arguments also play an important role in nonlinear analysis. It should be noticed that such equations appear often in various fields of science and engineering such as mathematical physics, economics, mechanics, etc. (see [36–38]). However, the relevant theory of this type of problem is still at its developing stage, and a great quantity of aspects remain to be explored. For a detailed description, see [39–43].

Here, we use the new result we achieved in the article to investigate the existence theorems of max–min solutions for impulsive systems of the following:

$$ \textstyle\begin{cases} u''(t) = \Upsilon(t,u(t),u(\phi(t)), Xu(t), Yu(t),u(\psi(t,\mu (t)))),\quad t\in J',\\ \triangle u(t_{k})=I_{k}(u(t_{k})),\quad k=1,2,\ldots,m,\\ \triangle u'(t_{k})=I_{k}^{*}(u(t_{k}),u'(t_{k})),\quad k=1,2,\ldots,m,\\ \chi_{1}(u(0),u(T))=0,\qquad\chi_{2}(u'(0),u'(T))=0, \end{cases} $$
(1.1)

where \(t\in J=[0,T](T>0)\), \(\Upsilon\in C(J\times R^{5},R)\), \(I_{k}\in C(R,R)\), \(I_{k}^{*},\chi_{i}(i=1,2)\in C(R\times R,R)\), \(\phi\in C(J,J)\), \(\mu \in C(J,R)\), \(\psi\in C(J\times R,J)\), \(0=t_{0}< t_{1}<\cdots<t_{k}<\cdots<t_{m}<t_{m+1}=T\), \(J'=J\backslash \{t_{1},t_{2},\ldots,t_{m}\}\), and

$$Xu(t)= \int_{0}^{t}k(t,s)u(s)\,ds , \qquad Yu(t)= \int_{0}^{T}h(t,s)u(s)\,ds, $$

\(k(t,s)\in C(D,R^{+})\), \(h(t,s)\in C(J\times J,R^{+})\), \(D=\{(t,s)\in R^{2} |0\leq s\leq t,t\in J\}\), \(R^{+}=[0,+\infty)\). \(\triangle u(t_{k})=u(t_{k}^{+})-u(t_{k}^{-})\), where \(u(t_{k}^{+})\) and \(u(t_{k}^{-})\) denote the right and the left limits of \(u(t)\) at \(t= t_{k}(k=1,2,\ldots,m)\), respectively. \(\triangle u'(t_{k})\) has a similar meaning for \(u'(t)\). Let \(PC(J, R) = \{u: J \to R | u(t)\mbox{ is continuous at }t\neq t_{k}, \mbox{ left continuous at } t=t_{k}\mbox{ and }u(t_{k}^{+})\mbox{ exists}, k=1,2,\ldots,m \}\). In addition, \(PC^{1}(J, R)= \{u\in PC(J, R)| u(t)\mbox{ is continuously differentiable at }t\neq t_{k}, u'(t_{k}^{+})\mbox{ and } u'(t_{k}^{-})\mbox{ exist}, k=1,2,\ldots,m \}\). Obviously, \(PC(J, R)\) and \(PC^{1}(J, R)\) are Banach spaces with respective norms

$$\Vert u \Vert _{PC}=\sup_{t\in J} \bigl\vert u(t) \bigr\vert ,\qquad \Vert u \Vert _{PC^{1}}=\max\bigl\{ \Vert u \Vert _{PC}, \bigl\Vert u' \bigr\Vert _{PC}\bigr\} . $$

2 New comparison principle

Lemma 1

([16])

Let \(s\in[0,T)\), \(c_{k}\ge 0\), \(\phi_{k}\) (\(k=1,2,\ldots,m\)) be constants, and let \(a,b\in PC(J,R)\), \(w\in PC^{1}(J,R)\). If

$$\textstyle\begin{cases} w'(t) \le a(t)w(t)+b(t), \quad t\in[s,T), t\neq t_{k},\\ w(t_{k}^{+})\le c_{k}w(t_{k})+\phi_{k}, \quad t_{k}\in[s,T), \end{cases} $$

then for \(t\in[s,T]\)

$$\begin{aligned} w(t) \le& w\bigl(s^{+}\bigr) \biggl(\prod_{s< t_{k}< t} c_{k}\biggr)\exp\biggl( \int_{s}^{t}a(r)\,dr\biggr)+ \int _{s}^{t}\biggl(\prod _{r< t_{k}< t} c_{k}\biggr)\exp\biggl( \int_{r}^{t}a(\tau)\,d\tau\biggr)b(r)\,dr \\ &{}+\sum_{s< t_{k}< t}\biggl(\prod _{t_{k}< t_{i}< t} c_{k}\biggr)\exp\biggl( \int_{t_{k}}^{t}a(\tau)\,d\tau \biggr)\phi_{k}. \end{aligned}$$

Lemma 2

(New comparison principle)

Assume that \(u\in PC^{1}(J,R)\cap C^{2}(J',R)\) satisfies

$$ \textstyle\begin{cases} u''(t)\le-D_{1}(t)u(t)-D_{2}(t)u(\phi (t))-D_{3}(t)(Xu)(t)-D_{4}(t)(Yu)(t),\quad t\in J',\\ \triangle u(t_{k})\le-L_{k}u(t_{k}),\quad k=1,2,\ldots,m,\\ \triangle u'(t_{k})\le-L_{k}^{*}u'(t_{k}),\quad k=1,2,\ldots,m,\\ u(0)\le\lambda_{1}u(T),\qquad u'(0)\le\lambda_{2}u'(T), \end{cases} $$
(2.1)

where \(D_{i}\in C(J,R^{+})(i=1,2,3,4)\), \(0\le L_{k}<1\), \(0< \lambda_{1},\lambda_{2},L_{k}^{*}<1\), and

$$\begin{aligned} &\lambda_{1}\prod_{k=1}^{m}(1-L_{k})^{2} \prod_{k=1}^{m}\bigl(1-L_{k}^{*} \bigr) \Biggl(1-\lambda _{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr) \\ &\quad \ge \int_{0}^{T}q(t)\,dt \int_{0}^{T}\prod_{s< t_{k}< T}(1-L_{k}) \,ds, \end{aligned}$$
(2.2)

here \(q(t)=D_{1}(t)+D_{2}(t)+D_{3}(t)\int_{0}^{t}k(t,s)\,ds +D_{4}(t)\int_{0}^{T}h(t,s)\,ds\). Then \(u(t)\le0\), \(t\in J\).

Proof

Suppose the contrary. Then, for some \(t\in J\), \(u(t)>0\), thus there exist two cases:

Case a: For some \(\overline{t}\in J\), \(u(\overline{t})>0\), and \(u(t)\ge0\) for \(t\in J\).

Case b: For some \(t^{*},t_{*}\in J\) such that \(u(t^{*})>0\) and \(u(t_{*})<0\).

In Case a, it follows from (2.1) that \(u''(t)\le0\) for \(t\neq t_{k}\) and \(u'(t_{k}^{+})\le(1-L_{k}^{*})u'(t_{k})\). By means of Lemma 1, we have \(u'(t)\le u'(0)\prod_{0< t_{k}< t}(1-L_{k}^{*})\). From this, together with (2.1), we have \(u'(0)\le \lambda_{2}u'(T)\le\lambda_{2}u'(0)\prod_{k=1}^{m}(1-L_{k}^{*})\), which means \(u'(0)\le0\), thus \(u'(t)\le0\). Meanwhile, \(u(t_{k}^{+})\le(1-L_{k})u(t_{k})\le u(t_{k})\). So, \(u(t)\) is nonincreasing in J. Then \(u(0)\le \lambda_{1}u(T)\le\lambda_{1}u(0)\), which is a contradiction.

For Case b, put \(\inf_{t\in J}u(t)=-\gamma\), then \(\gamma>0\), and for some \(i\in\{1,2,\ldots,m\}\), there exists \(t_{*}\in(t_{i},t_{i+1}]\) such that \(u(t_{*})=-\gamma\) or \(u(t_{i}^{+})=-\gamma\). Only consider \(u(t_{*})=-\gamma\), the proof of the case \(u(t_{i}^{+})=-\gamma\) is similar.

By (2.1), we have

$$\textstyle\begin{cases} u''(t)\le\gamma[D_{1}(t)+D_{2}(t)+D_{3}(t)\int_{0}^{t}k(t,s)\,ds +D_{4}(t)\int_{0}^{T}h(t,s)\,ds]\equiv\gamma q(t),\\ u'(t_{k}^{+})\le(1-L_{k}^{*})u'(t_{k}). \end{cases} $$

By Lemma 1, we get

$$ u'(t)\le u'(0)\prod_{0< t_{k}< t} \bigl(1-L_{k}^{*}\bigr)+ \int_{0}^{t}\gamma\prod _{s< t_{k}< t}\bigl(1-L_{k}^{*}\bigr)q(s)\,ds. $$
(2.3)

In (2.3), let \(t=T\), then we have

$$\begin{aligned} u'(0) \le&\lambda_{2}u'(T)\\ \le& \lambda_{2}u'(0)\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)+\lambda_{2} \int_{0}^{T}\gamma\prod _{s< t_{k}< T}\bigl(1-L_{k}^{*}\bigr)q(s)\,ds \\ \le&\lambda_{2}u'(0)\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)+\lambda_{2} \int_{0}^{T}\gamma q(s)\,ds, \end{aligned}$$

which implies

$$ u'(0)\le\lambda_{2} \int_{0}^{T}\gamma q(s)\,ds\Biggl[1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr]^{-1}. $$
(2.4)

From (2.3) and (2.4), we have that

$$\begin{aligned} u'(t) \le&\lambda_{2} \int_{0}^{T}\gamma q(s)\,ds\Biggl[1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr]^{-1}\prod _{0< t_{k}< t}\bigl(1-L_{k}^{*}\bigr)\\ &{} + \int_{0}^{t}\gamma\prod _{s< t_{k}< t}\bigl(1-L_{k}^{*}\bigr)q(s)\,ds \\ \le& \int_{0}^{T}\gamma q(s)\,ds\Biggl\{ \lambda_{2}\prod_{0< t_{k}< t}\bigl(1-L_{k}^{*} \bigr)\Biggl[1-\lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr]^{-1}+1\Biggr\} \\ \le&\lambda_{2}\prod_{0< t_{k}< t} \bigl(1-L_{k}^{*}\bigr) \int_{0}^{T}\gamma q(s)\,ds\\ &{}\times\Biggl\{ \Biggl[1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr]^{-1}+\Biggl[ \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr]^{-1}\Biggr\} \\ \le& \int_{0}^{T}\gamma q(s)\,ds\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1}, \end{aligned}$$

which and \(u(t_{k}^{+})\le(1-L_{k})u(t_{k})\) imply for \(t\in[t_{*},T]\)

$$ \begin{aligned}[b] u(t)\le&{} u(t_{*})\prod _{t_{*}< t_{k}< t}(1-L_{k})+ \int_{t_{*}}^{t}\prod_{s< t_{k}< t}(1-L_{k}) \int _{0}^{T}\gamma q(r)\,dr\\ &{}\times\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1-\lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1}\,ds \\ ={}& u(t_{*})\prod_{t_{*}< t_{k}< t}(1-L_{k})+ \int_{0}^{T}\gamma q(r)\,dr\\ &{}\times\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1-\lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1}\\ &{}\times \int _{t_{*}}^{t}\prod_{s< t_{k}< t}(1-L_{k}) \,ds. \end{aligned} $$
(2.5)

If \(t^{*}>t_{*}\), let \(t=t^{*}\) in (2.5), we have

$$\begin{aligned} 0 < & -\gamma\prod_{t_{*}< t_{k}< t^{*}}(1-L_{k})\\ &{}+\gamma \int_{0}^{T} q(r)\,dr\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1}\\ &{}\times \int_{t_{*}}^{t^{*}}\prod_{s< t_{k}< t^{*}}(1-L_{k}) \,ds, \end{aligned}$$

so,

$$\begin{aligned} \frac{\prod_{k=1}^{m}(1-L_{k})}{\int_{0}^{T}\prod_{s< t_{k}< T}(1-L_{k})\,ds} \le&\frac{\prod_{t_{*}< t_{k}< t^{*}}(1-L_{k})}{\int_{t_{*}}^{t^{*}}\prod_{s< t_{k}< t^{*}}(1-L_{k})\,ds} \\ < & \int_{0}^{T}q(r)\,dr\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1}, \end{aligned}$$

which contradicts (2.2). Hence, \(u(t)\le0\) on J.

If \(t^{*}< t_{*}\), without loss of generality, let \(t_{*}\in(t_{p-1},t_{p}]\) and \(t^{*}\in(t_{q},t_{q+1}]\), \(0\le q\le p-1\), \(p,q\in\{1,2,\ldots,m\}\). By Lemma 1, we have

$$ \begin{aligned}[b] u\bigl(t^{*}\bigr)\le{}& u(0)\prod _{0< t_{k}< t^{*}}(1-L_{k})+ \int_{0}^{T}\gamma q(r)\,dr\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1} \\ &{}\times \int_{0}^{t^{*}}\prod_{s< t_{k}< t^{*}}(1-L_{k}) \,ds \\ ={}&u(0)\prod_{k=1}^{q}(1-L_{k})+ \int_{0}^{T}\gamma q(r)\,dr\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \\ &{}\times\Biggl(1-\lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1} \int _{0}^{t^{*}}\prod_{s< t_{k}< t^{*}}(1-L_{k}) \,ds. \end{aligned} $$
(2.6)

On the other hand,

$$ \begin{aligned}[b] u(0)\le{}&\lambda_{1}u(T) \le \lambda_{1} u(t_{*})\prod_{t_{*}< t_{k}< T}(1-L_{k})+ \lambda_{1} \int_{0}^{T}\gamma q(r)\,dr \\ &{}\times\Biggl[\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr) \Biggl(1-\lambda_{2}\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr)\Biggr) \Biggr]^{-1} \int_{t_{*}}^{T}\prod_{s< t_{k}< T}(1-L_{k}) \,ds \\ ={}&{-}\lambda_{1}\gamma\prod_{k=p}^{m}(1-L_{k})+ \lambda_{1} \int_{0}^{T}\gamma q(r)\,dr \\ &{}\times\Biggl[\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr) \Biggl(1-\lambda_{2}\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr)\Biggr) \Biggr]^{-1} \int_{t_{*}}^{T}\prod_{s< t_{k}< T}(1-L_{k}) \,ds. \end{aligned} $$
(2.7)

From (2.6) and (2.7), we get that

$$\begin{aligned} &\lambda_{1}\prod_{k=p}^{m}(1-L_{k}) \prod_{j=1}^{q}(1-L_{j}) \\ &\quad < \int_{0}^{T}q(r)\,dr\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1} \\ &\qquad {}\times\Biggl[\lambda_{1}\prod_{j=1}^{q}(1-L_{j}) \int_{t_{*}}^{T}\prod_{s< t_{k}< T}(1-L_{k}) \,ds+ \int_{0}^{t^{*}}\prod_{s< t_{k}< t^{*}}(1-L_{k}) \,ds\Biggr]. \end{aligned}$$

By \(\prod_{j=q+1}^{m}(1-L_{j})\) times the above inequality, then

$$\begin{aligned} &\lambda_{1}\prod_{k=p}^{m}(1-L_{k}) \prod_{j=1}^{m}(1-L_{j}) \\ &\quad < \int_{0}^{T}q(r)\,dr\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1}\\ &\qquad {}\times\Biggl[ \lambda_{1}\prod_{j=1}^{m}(1-L_{j}) \int_{t_{*}}^{T}\prod_{s< t_{k}< T}(1-L_{k}) \,ds+\prod_{j=q+1}^{m}(1-L_{j}) \int _{0}^{t^{*}}\prod_{s< t_{k}< t^{*}}(1-L_{k}) \,ds\Biggr] \\ &\quad \le \int_{0}^{T}q(r)\,dr\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1} \\ &\qquad {}\times\biggl[ \int_{t_{*}}^{T}\prod_{s< t_{k}< T}(1-L_{k}) \,ds+ \int_{0}^{t^{*}}\prod_{s< t_{k}< T}(1-L_{k}) \,ds\biggr] \\ &\quad \le \int_{0}^{T}q(r)\,dr\Biggl[\prod _{k=1}^{m}\bigl(1-L_{k}^{*}\bigr) \Biggl(1- \lambda_{2}\prod_{k=1}^{m} \bigl(1-L_{k}^{*}\bigr)\Biggr)\Biggr]^{-1} \\ &\qquad {}\times \int_{0}^{T}\prod_{s< t_{k}< T}(1-L_{k}) \,ds. \end{aligned}$$

So, \(\lambda_{1}\prod_{k=1}^{m}(1-L_{k})^{2}<\int_{0}^{T}q(r)\,dr[\prod_{k=1}^{m}(1-L_{k}^{*})(1-\lambda_{2}\prod_{k=1}^{m}(1-L_{k}^{*}))]^{-1}\int _{0}^{T}\prod_{s<t_{k}<T}(1-L_{k})\,ds\), which contradicts (2.2). Hence, \(u(t)\le0\) on J.

Consider the problem:

$$ \textstyle\begin{cases} u''(t)=\sigma(t)-D_{1}(t)u(t)-D_{2}(t)u(\phi (t))-D_{3}(t)(Xu)(t)-D_{4}(t)(Yu)(t),\quad t\in J',\\ \triangle u(t_{k})=\psi_{k}-L_{k}u(t_{k}),\quad k=1,2,\ldots,m,\\ \triangle u'(t_{k})=\nu_{k}-L_{k}^{*}u'(t_{k}),\quad k=1,2,\ldots,m,\\ u(0)= \lambda_{1}u(T)+m_{1},\qquad u'(0)=\lambda_{2}u'(T)+m_{2}, \end{cases} $$
(2.8)

where \(\sigma\in PC(J,R)\), \(\psi_{k},\nu_{k},m_{1},m_{2}\in R\). □

Lemma 3

\(u(t)\in PC^{1}(J,R)\cap C^{2}(J',R)\) is a solution of the impulsive differential system (2.8) iff \(u(t)\in PC^{1}(J,R)\) is a solution of the impulsive integral system

$$\begin{aligned} u(t)={}&G_{1}+tG_{2} \\ &{}+ \int_{0}^{t}(t-s)\bigl[\sigma(s)-D_{1}(s)u(s)-D_{2}(s)u \bigl(\phi (s)\bigr)-D_{3}(s) (Xu) (s) -D_{4}(s) (Yu) (s)\bigr]\,ds \\ &{}+\sum_{0< t_{k}< t} \bigl\{ \bigl[\psi _{k}-L_{k}u(t_{k}) \bigr]+(t-t_{k})\bigl[\nu_{k}-L_{k}^{*}u'(t_{k}) \bigr]\bigr\} , \end{aligned}$$
(2.9)

where

$$\begin{aligned} G_{1} =&\frac{\lambda_{1}}{1-\lambda_{1}}\biggl\{ \int_{0}^{T}(T-s)\bigl[\sigma (s)-D_{1}(s)u(s)-D_{2}(s)u \bigl(\phi(s)\bigr)-D_{3}(s) (Xu) (s) \\ &{}-D_{4}(s) (Yu) (s)\bigr]\,ds+\sum_{0< t_{k}< T} \bigl\{ \bigl[\psi _{k}-L_{k}u(t_{k}) \bigr]+(T-t_{k})\bigl[\nu_{k}-L_{k}^{*}u'(t_{k}) \bigr]\bigr\} \\ &{}+TG_{2}+m_{1}\biggr\} +m_{1}, \\ G_{2} =&\frac{\lambda_{2}}{1-\lambda_{2}}\biggl\{ \int_{0}^{T}\bigl[\sigma (s)-D_{1}(s)u(s)-D_{2}(s)u \bigl(\phi(s)\bigr)-D_{3}(s) (Xu) (s) \\ &{}-D_{4}(s) (Yu) (s)\bigr]\,ds+\sum_{0< t_{k}< T} \bigl[\nu _{k}-L_{k}^{*}u'(t_{k}) \bigr]+m_{2}\biggr\} +m_{2}. \end{aligned}$$

Lemma 3 is easy, so we omit its proof.

Lemma 4

For \(\sigma\in PC(J,R),~\psi_{k},\nu_{k},m_{1},m_{2}\in R,~0\le L_{k}<1,~0< \lambda_{1},\lambda_{2},L_{k}^{*}<1\) and functions \(M, K, N,L\in C(J,R^{+})\). If

$$ \textstyle\begin{cases} \frac{1}{1-\lambda_{1}}\{\int_{0}^{T}(T-s)p(s)\,ds+\sum_{k=1}^{m}[L_{k}+(T-t_{k})L_{k}^{*}]\}\\ \qquad {}+\frac{\lambda_{2}T}{(1-\lambda_{1})(1-\lambda_{2})}[\int_{0}^{T}p(s)\,ds+\sum_{k=1}^{m}L_{k}^{*}]< 1,\\ \frac{1}{1-\lambda_{2}}[\int_{0}^{T}p(s)\,ds+\sum_{k=1}^{m}L_{k}^{*}]< 1, \end{cases} $$
(2.10)

where \(p(t)=D_{1}(t)+D_{2}(t)+D_{3}(t)\int_{0}^{t}k(t,s)\,ds +D_{4}(t)\int_{0}^{T}h(t,s)\,ds\). (2.8) has a unique solution \(u(t)\in PC^{1}(J,R)\cap C^{2}(J',R)\).

A similar proof can be found in [22] (see Lemma 2.3), so we omit it.

3 Main results

Theorem 1

Assume that condition (2.10) holds. In addition, assume that

\((H_{1})\) There exist \(u_{0}(t)\le v_{0}(t)\in PC^{1}(J,R)\cap C^{2}(J',R)\) such that

$$\textstyle\begin{cases} u_{0}''(t) \le\Upsilon(t,u_{0}(t),u_{0}(\phi(t)), Xu_{0}(t), Yu_{0}(t),u_{0}(\psi (t,\mu(t)))),\\ \triangle u_{0}(t_{k})\le I_{k}(u_{0}(t_{k})),\quad k=1,2,\ldots,m,\\ \triangle u_{0}'(t_{k})\le I_{k}^{*}(u_{0}(t_{k}),u_{0}'(t_{k})),\quad k=1,2,\ldots,m,\\ \chi_{1}(u_{0}(0),u_{0}(T))\le0,\qquad \chi_{2}(u_{0}'(0),u_{0}'(T))\le0, \end{cases} $$

and

$$\textstyle\begin{cases} v_{0}''(t) \ge\Upsilon(t,v_{0}(t),v_{0}(\phi(t)), Xv_{0}(t), Yv_{0}(t),v_{0}(\psi (t,\mu(t)))),\\ \triangle v_{0}(t_{k})\ge I_{k}(v_{0}(t_{k})),\quad k=1,2,\ldots,m,\\ \triangle v_{0}'(t_{k})\ge I_{k}^{*}(v_{0}(t_{k}),v_{0}'(t_{k})),\quad k=1,2,\ldots,m,\\ \chi_{1}(v_{0}(0),v_{0}(T))\ge0,\qquad \chi_{2}(v_{0}'(0),v_{0}'(T))\ge0. \end{cases} $$

\((H_{2})\) Functions \(D_{i}\in C(J,R^{+})(i=1,2,3,4)\), which satisfy (2.2) such that

$$\begin{aligned} \Upsilon(t,u,v,w,z,\xi)-\Upsilon(t,\overline{u},\overline{v},\overline {w}, \overline{z},\overline{\xi}) \ge&-D_{1}(t) (u-\overline{u})-D_{2}(t) (v-\overline{v})\\ &{}-D_{3}(t) (w-\overline {w})-D_{4}(t) (z- \overline{z}), \end{aligned}$$

where \(u_{0}(t)\le\overline{u}\le u\le v_{0}(t)\), \(u_{0}(\phi(t))\le\overline {v}\le v\le v_{0}(\phi(t))\), \(Xu_{0}(t)\le\overline{w}\le w\le Xv_{0}(t)\), \(Yu_{0}(t)\le\overline{z}\le z\le Yv_{0}(t)\), \(u_{0}(\psi(t,\mu(t)))\le\overline{\xi}\le\xi\le v_{0}(\psi(t,\mu (t)))\), \(\forall t\in J\).

\((H_{3})\) There exist constants \(0\le L_{k}<1\), \(0< L_{k}^{*}<1\) (\(k=1,2,\ldots,m\)), and \(0< b_{1}< a_{1}\), \(0< b_{2}< a_{2}\) such that

$$\begin{aligned} &I_{k}(u)-I_{k}(\overline{u})\ge-L_{k}(u- \overline {u}),\\ & I_{k}^{*}\bigl(u,u' \bigr)-I_{k}^{*}\bigl(\overline{u},\overline{u}'\bigr)\ge -L_{k}^{*}\bigl(u'-\overline{u}'\bigr), \\ &\chi_{1}(u,v)-\chi_{1}(\overline{u},\overline{v})\le a_{1}(u-\overline {u})-b_{1}(v-\overline{v}),\\ &\chi_{2}\bigl(u',v'\bigr)-\chi_{2} \bigl(\overline{u}',\overline{v}'\bigr)\le a_{2}\bigl(u'-\overline{u}' \bigr)-b_{2}\bigl(v'-\overline{v}'\bigr), \end{aligned}$$

where \(u_{0}(t_{k})\le\overline{u}\le u\le v_{0}(t_{k})\) (\(k=1,2,\ldots,m\)), \(u_{0}(0)\le\overline{u}\le u\le v_{0}(0)\), \(u_{0}(T)\le\overline{v}\le v\le v_{0}(T)\).

Then the impulsive system (1.1) has the min-maximal solutions \(u^{*},v^{*}\) in \([u_{0},v_{0}]\), respectively. Moreover, there exist monotone iterative sequences \(\{u_{n}(t)\},\{v_{n}(t)\}\subset[u_{0},v_{0}]\) such that \(u_{n}\to u^{*},v_{n}\to v^{*}(n\to\infty)\) uniformly on \(t\in J\), where \(\{u_{n}(t)\},\{v_{n}(t)\}\) satisfy

$$\begin{aligned} &\textstyle\begin{cases} u_{n}''(t)=\Upsilon(t,u_{n-1}(t),u_{n-1}(\phi(t)), Xu_{n-1}(t), Yu_{n-1}(t),u_{n-1}(\psi(t,\mu(t))))\\ \hphantom{u_{n}''(t)=}{}-D_{1}(t)(u_{n}-u_{n-1})(t)-D_{2}(t)(u_{n}-u_{n-1})(\phi (t))-D_{3}(t)X(u_{n}-u_{n-1})(t)\\ \hphantom{u_{n}''(t)=}{}-D_{4}(t)Y(u_{n}-u_{n-1})(t),\quad t\in J',\\ \triangle u_{n}(t_{k})=I_{k}(u_{n-1}(t_{k}))-L_{k}(u_{n}-u_{n-1})(t_{k}),\quad k=1,2,\ldots,m,\\ \triangle u_{n}'(t_{k})=I_{k}^{*}(u_{n-1}(t_{k}),u_{n-1}'(t_{k}))-L_{k}^{*}(u_{n}'-u_{n-1}')(t_{k}),\quad k=1,2,\ldots ,m,\\ u_{n}(0)=u_{n-1}(0)+\lambda_{1}[u_{n}(T)-u_{n-1}(T)]-\frac{1}{a_{1}}\chi _{1}(u_{n-1}(0),u_{n-1}(T)),\quad n=1,2,\ldots,\\ u_{n}'(0)=u_{n-1}'(0)+\lambda_{2}[u_{n}'(T)-u_{n-1}'(T)]-\frac{1}{a_{2}}\chi _{2}(u_{n-1}'(0),u_{n-1}'(T)),\quad n=1,2,\ldots, \end{cases}\displaystyle \end{aligned}$$
(3.1)
$$\begin{aligned} & \textstyle\begin{cases} v_{n}''(t)=\Upsilon(t,v_{n-1}(t),v_{n-1}(\phi(t)), Xv_{n-1}(t), Yv_{n-1}(t),v_{n-1}(\psi(t,\mu(t))))\\ \hphantom{v_{n}''(t)=}{} -D_{1}(t)(v_{n}-v_{n-1})(t)-D_{2}(t)(v_{n}-v_{n-1})(\phi (t))-D_{3}(t)X(v_{n}-v_{n-1})(t)\\ \hphantom{v_{n}''(t)=}{}-D_{4}(t)Y(v_{n}-v_{n-1})(t),\quad t\in J',\\ \triangle v_{n}(t_{k})=I_{k}(v_{n-1}(t_{k}))-L_{k}(v_{n}-v_{n-1})(t_{k}),\quad k=1,2,\ldots,m,\\ \triangle v_{n}'(t_{k})=I_{k}^{*}(v_{n-1}(t_{k}),v_{n-1}'(t_{k}))-L_{k}^{*}(v_{n}'-v_{n-1}')(t_{k}),\quad k=1,2,\ldots ,m,\\ v_{n}(0)=v_{n-1}(0)+\lambda_{1}[v_{n}(T)-v_{n-1}(T)]-\frac{1}{a_{1}}\chi _{1}(v_{n-1}(0),v_{n-1}(T)),\quad n=1,2,\ldots,\\ v_{n}'(0)=v_{n-1}'(0)+\lambda_{2}[v_{n}'(T)-v_{n-1}'(T)]-\frac{1}{a_{2}}\chi _{2}(v_{n-1}'(0),v_{n-1}'(T)),\quad n=1,2,\ldots, \end{cases}\displaystyle \end{aligned}$$
(3.2)

and

$$ u_{0}\le u_{1} \le\cdots\le u_{n}\le\cdots\le u^{*} \le v^{*}\le \cdots\le v_{n}\le\cdots\le v_{1}\le v_{0}, $$
(3.3)

here \(\lambda_{i}=b_{i}/a_{i}(i=1,2)\).

Proof

For any \(u_{n-1},v_{n-1}\in PC^{1}(J,R)\cap C^{2}(J',R)\), it follows from Lemma 4 that (3.1) and (3.2) have unique solutions \(u_{n}\) and \(v_{n}\) in \(PC^{1}(J,R)\cap C^{2}(J',R)\), respectively.

Now, we verify that

$$ u_{n-1}\le u_{n}\le v_{n}\le v_{n-1}, \quad n=1,2,\ldots. $$
(3.4)

Let \(p(t)=u_{0}(t)-u_{1}(t)\), \(q(t)=v_{1}(t)-v_{0}(t)\), \(w(t)=u_{1}(t)-v_{1}(t)\), by (3.1), (3.2) and \((H_{1})-(H_{4})\), we have that

$$\begin{aligned} & \textstyle\begin{cases} p''(t)\le-D_{1}(t)p(t)-D_{2}(t)p(\phi (t))-D_{3}(t)(Xp)(t)-D_{4}(t)(Yp)(t),\quad t\in J',\\ \triangle p(t_{k})\le-L_{k}p(t_{k}),\quad k=1,2,\ldots,m,\\ \triangle p'(t_{k})\le-L_{k}^{*}p'(t_{k}),\quad k=1,2,\ldots,m,\\ p(0)\le\lambda_{1}p(T),\qquad p'(0)\le\lambda_{2}p'(T), \end{cases}\displaystyle \\ & \textstyle\begin{cases} q''(t)\le-D_{1}(t)q(t)-D_{2}(t)q(\phi (t))-D_{3}(t)(Xq)(t)-D_{4}(t)(Yq)(t),\quad t\in J',\\ \triangle q(t_{k})\le-L_{k}q(t_{k}),\quad k=1,2,\ldots,m,\\ \triangle q'(t_{k})\le-L_{k}^{*}q'(t_{k}),\quad k=1,2,\ldots,m,\\ q(0)\le\lambda_{1}q(T),\qquad q'(0)\le\lambda_{2}q'(T), \end{cases}\displaystyle \\ & \textstyle\begin{cases} w''(t)=\Upsilon(t,u_{0}(t),u_{0}(\phi(t)), Xu_{0}(t), Yu_{0}(t),u_{0}(\psi(t,\mu (t))))-D_{1}(t)(u_{1}-u_{0})(t)\\ \hphantom{w''(t)=}{}-D_{2}(t)(u_{1}-u_{0})(\phi (t))-D_{3}(t)X(u_{1}-u_{0})(t)-D_{4}(t)Y(u_{1}-u_{0})(t)\\ \hphantom{w''(t)=}{}-\Upsilon(t,v_{0}(t),v_{0}(\phi(t)), Xv_{0}(t), Yv_{0}(t),v_{0}(\psi (t,\mu(t))))+D_{1}(t)(v_{1}-v_{0})(t)\\ \hphantom{w''(t)=}{}+D_{2}(t)(v_{1}-v_{0})(\phi (t))+D_{3}(t)X(v_{1}-v_{0})(t)+D_{4}(t)Y(v_{1}-v_{0})(t)\\ \hphantom{w''(t)}\le-D_{1}(t)w(t)-D_{2}(t)w(\phi (t))-D_{3}(t)(Xw)(t)-D_{4}(t)(Yw)(t),\quad t\in J',\\ \triangle w(t_{k})=I_{k}(u_{0}(t_{k}))-I_{k}(v_{0}(t_{k}))-L_{k}(u_{1}-u_{0})(t_{k})+L_{k}(v_{1}-v_{0})(t_{k})\\ \hphantom{w(t_{k})}\le-L_{k}w(t_{k}),\quad k=1,2,\ldots,m,\\ \triangle w'(t_{k})=I_{k}^{*}(u_{0}(t_{k}),u_{0}'(t_{k}))-I_{k}^{*}(v_{0}(t_{k}),v_{0}'(t_{k}))-L_{k}^{*}(u_{1}'-u_{0}')(t_{k})+L_{k}^{*}(v_{1}'-v_{0}')(t_{k})\\ \hphantom{w'(t_{k})}\le-L_{k}^{*}w'(t_{k}),\quad k=1,2,\ldots,m,\\ w(0)=u_{0}(0)-v_{0}(0)+\lambda_{1}[u_{1}(T)-u_{0}(T)]-\lambda_{1}[v_{1}(T)-v_{0}(T)]\\ \hphantom{w(0)=}{}-\frac{1}{a_{1}}\chi_{1}(u_{0}(0),u_{0}(T))+\frac{1}{a_{1}}\chi _{1}(v_{0}(0),v_{0}(T))\\ \hphantom{w(0)}\le u_{0}(0)-v_{0}(0)+\lambda_{1}w(T)-\lambda_{1}[u_{0}(T)-v_{0}(T)]\\ \hphantom{w(0)\le}{}+[v_{0}(0)-u_{0}(0)]-\lambda_{1}[v_{0}(T)-u_{0}(T)]\\ \hphantom{w(0)}\le\lambda_{1}w(T),\\ w'(0)\le\lambda_{2}w'(T). \end{cases}\displaystyle \end{aligned}$$

Thus, by means of Lemma 2, we have \(p(t)\le0\), \(q(t)\le0\), \(w(t)\le 0\), \(\forall t\in J\), \(i.e\)., \(u_{0}\le u_{1}\le v_{1}\le v_{0}\).

Assume that \(u_{k-1}\le u_{k}\le v_{k}\le v_{k-1}\) for some \(k\ge1\). Thus, employing the same technique once again, by Lemma 2, one can get \(u_{k}\le u_{k+1}\le v_{k+1}\le v_{k}\). Thus, one can easily show that

$$ u_{0}\le u_{1} \le\cdots\le u_{n}\le\cdots\le v_{n}\le\cdots \le v_{1}\le v_{0},\quad n=1,2, \ldots. $$
(3.5)

Employing the standard arguments, we have

$$\begin{aligned} &\lim_{n\to\infty}u_{n}(t)=u^{*}(t),\\ &\lim_{n\to\infty}v_{n}(t)=v^{*}(t) \end{aligned}$$

uniformly on \(t\in J\), and the limit functions \(u^{*},v^{*}\) satisfy (1.1). Moreover, \(u^{*},v^{*}\in[u_{0},v_{0}]\).

Next, we prove that \(u^{*},v^{*}\) are the min-maximal solutions of impulsive differential system (1.1) in \([u_{0},v_{0}]\). If \(w\in[u_{0},v_{0}]\) is any solution of (1.1). Let \(u_{n-1}(t)\le w(t)\le v_{n-1}(t),\forall t\in J\), for some positive integer n. Put \(p=u_{n}-w\). Then

$$\textstyle\begin{cases} p''(t)=\Upsilon(t,u_{n-1}(t),u_{n-1}(\phi(t)), Xu_{n-1}(t), Yu_{n-1}(t),u_{n-1}(\psi(t,\mu(t))))\\ \hphantom{p''(t)=}{}-D_{1}(t)(u_{n}-u_{n-1})(t)-D_{2}(t)(u_{n}-u_{n-1})(\phi (t))\\ \hphantom{p''(t)=}{}-D_{3}(t)X(u_{n}-u_{n-1})(t)-D_{4}(t)Y(u_{n}-u_{n-1})(t)\\ \hphantom{p''(t)=}{}-\Upsilon(t,w(t),w(\phi(t)), Xw(t), Yw(t),w(\psi(t,\mu(t))))\\ \hphantom{p''(t)}\le D_{1}(t)(w-u_{n-1})(t)+D_{2}(t)(w-u_{n-1})(\phi(t))+D_{3}(t)X(w-u_{n-1})(t)\\ \hphantom{p''(t)=}{}+D_{4}(t)Y(w-u_{n-1})(t)-D_{1}(t)(u_{n}-u_{n-1})(t)-D_{2}(t)(u_{n}-u_{n-1})(\phi (t))\\ \hphantom{p''(t)=}{}-D_{3}(t)X(u_{n}-u_{n-1})(t)-D_{4}(t)Y(u_{n}-u_{n-1})(t)\\ \hphantom{p''(t)}= -D_{1}(t)p(t)-D_{2}(t)p(\phi (t))-D_{3}(t)(Xp)(t)-D_{4}(t)(Yp)(t),\quad t\in J',\\ \triangle p(t_{k})=I_{k}(u_{n-1}(t_{k}))-I_{k}(w(t_{k}))-L_{k}(u_{n}-u_{n-1})(t_{k})\\ \hphantom{\triangle p(t_{k})}\le-L_{k}p(t_{k}),\quad k=1,2,\ldots,m,\\ \triangle p'(t_{k})=I_{k}^{*}(u_{n-1}(t_{k}),u_{n-1}'(t_{k}))-I_{k}^{*}(w(t_{k}),w'(t_{k}))-L_{k}^{*}(u_{n}'-u_{n-1}')(t_{k})\\ \hphantom{\triangle p'(t_{k})}\le-L_{k}^{*}p'(t_{k}),\quad k=1,2,\ldots,m,\\ p(0)=u_{n-1}(0)+\lambda_{1}[u_{n}(T)-u_{n-1}(T)]-\frac{1}{a_{1}}\chi _{1}(u_{n-1}(0),u_{n-1}(T))\\ \hphantom{p(0)=} {}+\frac{1}{a_{1}}\chi_{1}(w(0),w(T))-w(0)\\ \hphantom{p(0)}\le u_{n-1}(0)+\lambda _{1}[u_{n}(T)-u_{n-1}(T)]+[w(0)-u_{n-1}(0)]\\ \hphantom{p(0)=}{}-\lambda _{1}[w(T)-u_{n-1}(T)]-w(0)\\ \hphantom{p(0)}= \lambda_{1}p(T),\\ p'(0)=u_{n-1}'(0)+\lambda_{2}[u_{n}'(T)-u_{n-1}'(T)]-\frac{1}{a_{2}}\chi _{2}(u_{n-1}'(0),u_{n-1}'(T))\\ \hphantom{p(0)=}{}+\frac{1}{a_{2}}\chi_{2}(w(0),w(T))-w(0)\\ \hphantom{p(0)}\le\lambda_{2}p'(T). \end{cases} $$

By Lemma 2, we have \(u_{n}(t)\le w(t),\forall t\in J\). By the same way as above, we can show \(w(t)\le v_{n}(t),\forall t\in J\). That is, \(u_{n}(t)\le w(t)\le v_{n}(t)\), \(\forall t\in J\).

Now, if \(n\to\infty\), then

$$u_{0}(t)\le u^{*}(t)\le w(t)\le v^{*}(t)\le v_{0}(t),\quad \forall t\in J. $$

That is, \(u^{*},v^{*}\) are the min-maximal solutions of (1.1) in \([u_{0},v_{0}]\). □

4 Example

Consider

$$ \textstyle\begin{cases} u''(t)= \frac{1}{100}t^{3}[t-u(t)]+\frac{1}{300}t^{3}[t-u(t^{2})]^{3} +\frac{1}{500}t[t^{3}-\int_{0}^{t}tsu(s)\,ds]^{5}\\ \hphantom{u''(t)= }{}+\frac{1}{700}t^{2}[t^{2}-\int_{0}^{1}t^{2}su(s)\,ds]^{7}-\frac {1}{100}t^{4}e^{-u(\frac{1}{2}t^{3}+\frac{1}{2}t^{2}e^{-t})},\quad t\in [0,1],t\neq\frac{1}{2},\\ \triangle u(\frac{1}{2})=-\frac{1}{4}u(\frac{1}{2}),\\ \triangle u'(\frac{1}{2})=bu(\frac{1}{2})-\frac{3}{8}u'(\frac{1}{2}),\\ \chi_{1}(u(0),u(1))=u^{2}(0)-\frac{1}{2}u(1)=0,\\ \chi_{2}(u'(0),u'(1))=u'(0)-\frac{1}{3}u'(1)=0, \end{cases} $$
(4.1)

where \(0\le b\le\frac{1}{26}\), \(m=1\), \(t_{1}=\frac{1}{2}\), \(\phi(t)=t^{2}\), \(\psi(t,\mu (t))=\frac{1}{2}t^{3}+\frac{1}{2}t^{2}e^{-t}\), \(\forall t\in J\).

Take

$$ u_{0}(t)=0,\quad \forall t\in J, \qquad v_{0}(t)= \textstyle\begin{cases} 1+t+\frac{1}{2}t^{2},\quad t\in[0,\frac{1}{2}],\\ 1+t^{2},\quad t\in(\frac{1}{2},1]. \end{cases} $$
(4.2)

Then

$$ u_{0}(t)\le v_{0}(t),\qquad u_{0}'(t)=0, \qquad v_{0}'(t)= \textstyle\begin{cases} 1+t,\quad t\in[0,\frac{1}{2}],\\ 2t,\quad t\in(\frac{1}{2},1], \end{cases} $$
(4.3)

and

$$\begin{aligned} &\textstyle\begin{cases} u_{0}''(t)=0\le\frac{1}{100}t^{4}+\frac{1}{300}t^{6}+\frac {1}{500}t^{16}+\frac{1}{700}t^{16}-\frac{1}{100}t^{4} \\ \hphantom{u_{0}''(t)}=\Upsilon(t,u_{0}(t),u_{0}(\phi(t)), Xu_{0}(t), Yu_{0}(t),u_{0}(\psi (t,\mu(t)))),\\ \triangle u_{0}(\frac{1}{2})=0= I_{1}(u_{0}(\frac{1}{2})),\\ \triangle u_{0}'(\frac{1}{2})=0= I_{1}^{*}(u_{0}(\frac{1}{2}),u_{0}'(\frac {1}{2})),\\ \chi_{1}(u_{0}(0),u_{0}(1))=u_{0}^{2}(0)-\frac{1}{2}u_{0}(1)=0,\\ \chi_{2}(u_{0}'(0),u_{0}'(1))=u_{0}'(0)-\frac{1}{3}u_{0}'(1)=0, \end{cases}\displaystyle \\ & \textstyle\begin{cases} v_{0}''(t)\ge1>\frac{1}{100}+\frac{1}{300}+\frac{1}{500}+\frac {1}{700}\\ \hphantom{v_{0}''(t)}\ge\Upsilon(t,v_{0}(t),v_{0}(\phi(t)), Xv_{0}(t), Yv_{0}(t),v_{0}(\psi (t,\mu(t)))),\\ \triangle v_{0}(\frac{1}{2})=-\frac{3}{8}\ge-\frac{13}{32}=I_{1}(v_{0}(\frac {1}{2})),\\ \triangle v_{0}'(\frac{1}{2})=-\frac{1}{2}\ge\frac{13b}{8}-\frac {9}{16}=I_{1}^{*}(v_{0}(\frac{1}{2}),v_{0}'(\frac{1}{2})),\\ \chi_{1}(v_{0}(0),v_{0}(1))=v_{0}^{2}(0)-\frac{1}{2}v_{0}(1)=0,\\ \chi_{2}(v_{0}'(0),v_{0}'(1))=v_{0}'(0)-\frac{1}{3}v_{0}'(1)=\frac{1}{3}>0. \end{cases}\displaystyle \end{aligned}$$

Consequently, \(u_{0}\), \(v_{0}\) satisfy \((H_{1})\). Let

$$\begin{aligned} \Upsilon(t,u,v,w,z,\xi) =&\frac{1}{100}t^{3}(t-u)+ \frac{1}{300}t^{3}(t-v)^{3} +\frac{1}{500}t \bigl(t^{3}-w\bigr)^{5}\\ &{}+\frac{1}{700}t^{2} \bigl(t^{2}-z\bigr)^{7}-\frac {1}{100}t^{4}e^{-\xi}, \end{aligned}$$

we have

$$\begin{aligned} &\Upsilon(t,u,v,w,z,\xi)-\Upsilon(t,\overline{u},\overline{v},\overline {w}, \overline{z},\overline{\xi}) \\ &\quad =\frac{1}{100}t^{3}\bigl[(t-u)-(t-\overline{u})\bigr]+ \frac {1}{300}t^{3}\bigl[(t-v)^{3}-(t- \overline{v})^{3}\bigr]+\frac {1}{500}t\bigl[\bigl(t^{3}-w \bigr)^{5}-\bigl(t^{3}-\overline{w}\bigr)^{5}\bigr] \\ &\qquad{} +\frac{1}{700}t^{2}\bigl[\bigl(t^{2}-z \bigr)^{7}-\bigl(t^{2}-\overline{z}\bigr)^{7}\bigr]- \frac {1}{100}t^{4}\bigl(e^{-\xi}-e^{-\overline{\xi}}\bigr) \\ &\quad \ge-\frac{1}{100}t^{3}(u-\overline{u})- \frac{1}{25}t^{3}(v-\overline {v})-\frac{1}{100}t^{13}(w- \overline{w})-\frac{1}{100}t^{14}(z-\overline{z}), \end{aligned}$$

where \(u_{0}(t)\le\overline{u}\le u\le v_{0}(t)\), \(u_{0}(\phi(t))\le\overline {v}\le v\le v_{0}(\phi(t))\), \(Xu_{0}(t)\le\overline{w}\le w\le Xv_{0}(t)\), \(Yu_{0}(t)\le\overline{z}\le z\le Yv_{0}(t)\), \(u_{0}(\psi(t,\mu(t)))\le\overline{\xi}\le\xi\le v_{0}(\psi(t,\mu (t)))\), \(\forall t\in J\). For \(L_{1}=\frac{1}{4}\), \(L_{1}^{*}=\frac{3}{8}\), \(a_{1}=2\), \(b_{1}=\frac {1}{2}\), \(a_{2}=1\), \(b_{2}=\frac{1}{3}\), obviously, \((H_{3})\) and \((H_{4})\) hold. On the other hand, put \(D_{1}(t)=\frac{1}{100}t^{3}\), \(D_{2}(t)=\frac{1}{25}t^{3}\), \(D_{3}(t)=\frac {1}{100}t^{13}\), \(D_{4}(t)=\frac{1}{100}t^{14}\), \(\lambda_{1}=\frac{1}{4}\), \(\lambda_{2}=\frac{1}{3}\), \(L_{1}=\frac {1}{4}\), \(L_{1}^{*}=\frac{3}{8}\), it is easy to see that conditions (2.2) and (2.10) hold. So, \((H_{2})\) also holds.

Thus, Theorem 1 is satisfied. Therefore, our conclusions come from Theorem 1 that (4.1) has the min-maximal solution \(u^{*},v^{*}\in[u_{0},v_{0}]\).