Research
Open Access
Existence results for a kind of fourth-order impulsive integral boundary value problems
- Meng Sun1 and
- Yepeng Xing1Email author
Received: 10 December 2015
Accepted: 29 March 2016
Published: 14 April 2016
Abstract
In this paper we investigate the existence of solutions to a kind of fourth-order impulsive differential equations with integral boundary value conditions. By employing the Schauder fixed point theorem, we obtain sufficient conditions which ensure the system has at lease one solution. Also by using the contraction mapping theorem we get the uniqueness result. Finally an example is given to illustrate the effectiveness of our results.
Keywords
fourth order impulsive integral boundary conditions1 Introduction
Fourth-order boundary value problems have attached much attention from many authors; for example, see Sun and Wang [1], Yao [2], O’Regan [3], Yang [4], Zhang [5], Gupta [6], Agarwal [7], Bonanno and Bella [8], and Han and Xu [9]. In particular, we would like to mention some results as follows. In [10], Zhang and Liu studied the following fourth-order four-point boundary value problem: where \(0<\xi, \eta<1\), \(0\leq a< b<1\). By using the upper and lower solutions method, fixed point theorems, and the properties of the Green’s functions \(G(t,s)\) and \(H(t,s)\), the authors gave sufficient conditions for the existence of one positive solution.
$$\left \{ \textstyle\begin{array}{l} (\phi_{p}(x''(t)))''=\omega(t)f(t,x(t)),\quad t\in[0,1], \\ x(0)=0,\qquad x(1)=ax(\xi), \\ x''(0)=0, \qquad x''(1)=bx''(\eta), \end{array}\displaystyle \right . $$
Zhou and Zhang [11] employed a new existence theory to study the fourth-order p-Laplacian elasticity problems: where \(a,b> 0\), \(J=[0,1]\), \(\phi_{m}(s)\) is an m-Laplace operator, i.e.
\(\phi_{m}(s)=|s|^{m-2}s\), \(m>1\), \((\phi_{m})^{-1}=\phi_{m^{*}}\), \(\frac{1}{m}+\frac {1}{m^{*}}=1\), \(F:[0,1]\times R \times R \rightarrow R\) is continuous. In their paper, a new technique for dealing with the bending term of the fourth-order p-Laplacian elasticity problems was introduced and several new and more general results were obtained for the existence of at least single, double, or triple positive solutions.
$$\left \{ \textstyle\begin{array}{l} (\phi_{m}(y''))''=F(t,y,y''),\quad 0< t< 1, \\ ay(0)-by'(0)=\int^{1}_{0}g(s)y(s)\,ds, \\ ay(1)+by'(1)=\int^{1}_{0}g(s)y(s)\,ds, \\ \phi_{m}(y''(0))=\phi_{m}(y''(1))=\int^{1}_{0}h(t)\phi_{m}(y''(t))\,dt, \end{array}\displaystyle \right . $$
Feng [12] studied a fourth-order boundary value problem with impulses and integral boundary conditions, By using a suitably constructed cone and fixed point theory for cones, the existence of multiple positive solutions was established. Some papers considered the existence, multiplicity, and nonexistence of positive solutions for fourth-order impulsive differential equations with one-dimensional m-Laplacian; for example, see [13–17]. Most recently Feng and Qiu [18], studied a fourth-order impulse integral boundary value problem with one-dimensional m-Laplacian and deviating arguments:
$$\left \{ \textstyle\begin{array}{l} (\phi_{p}(y''(t)))''=f(t,y(t)),\quad t\in[0,1], t\neq t_{k}, k=1,2,\ldots,n, \\ \Delta y'|_{t=t_{k}}=- I_{k}(y(t_{k})), \quad k=1,2,\ldots,n, \\ y(0)=y(1)=\int^{1}_{0}g(s)y(s)\,ds, \\ \phi_{p}(y''(0))=\phi_{p}(y''(1))=\int^{1}_{0}h(s)\phi_{p}(y''(s))\,ds. \end{array}\displaystyle \right . $$
$$\left \{ \textstyle\begin{array}{l} (\phi_{m}(y''))''=\lambda\omega(t)f(t,y(\alpha(t))), \quad t\in{J}, t\neq t_{k}, k=1,2,\ldots,n, \\ \Delta y'|_{t=t_{k}}=-\mu I_{k}(t_{k},y(t_{k})),\quad k=1,2,\ldots,n, \\ ay(0)-by'(0)=\int^{1}_{0}g(s)y(s)\,ds, \\ ay(1)+by'(1)=\int^{1}_{0}g(s)y(s)\,ds, \\ \phi_{m}(y''(0))=\phi_{m}(y''(1))=\int^{1}_{0}h(t)\phi_{m}(y''(t))\,dt. \end{array}\displaystyle \right . $$
We see that in the above system the right-hand side function f has nothing to do with the term \(y'\), the jumping function \(I_{k}\) does not contain the term \(y'(t_{k})\). What is more, there is no restriction on the impulses for state function, i.e.
\(\Delta y|_{t=t_{k}}\) does not appear. Definitely for more extensive applications, we would better consider the following boundary value problem: where \(a,b> 0\), \(J=[0,1]\), \(\phi_{m}(s)\) is an m-Laplace operator, i.e.
\(\phi_{m}(s)=|s|^{m-2}s\), \(m>1\), \((\phi_{m})^{-1}=\phi_{m^{*}}\), \(\frac{1}{m}+\frac {1}{m^{*}}=1\), \(0=t_{0}< t_{1}< t_{2}<\cdots<t_{k}<\cdots<t_{m}<t_{m+1}=1\), \(f\in {C[J\times R^{n}\times R^{n},R^{n}]}\), \(I_{k}\in{C[R^{n},R^{n}]}\), \(\bar{I}_{k}\in{C[R^{n}\times R^{n},R^{n}]}\), \(\Delta y|_{t=t_{k}}=y(t^{+}_{k})-y(t^{-}_{k})\), here \(y(t^{+}_{k})\) and \(y(t^{-}_{k})\) represent the right-hand limit and left-hand limit of \(y(t)\) at \(t=t_{k}\), respectively. \(\Delta y'|_{t=t_{k}}\) has a similar meaning for \(y'(t)\). In addition, f, g, and h satisfy the following conditions.
$$ \left \{ \textstyle\begin{array}{l} (\phi_{m}(y''))''=f(t,y,y'), \quad t\in{J}, t\neq t_{k}, k=1,2,\ldots,m, \\ \Delta y|_{t=t_{k}}=I_{k}(y(t_{k})),\quad k=1,2,\ldots,m, \\ \Delta y'|_{t=t_{k}}=\bar{I}_{k}(y(t_{k}),y'(t_{k})),\quad k=1,2,\ldots,m, \\ ay(0)-by'(0)=\int^{1}_{0}g(s)y(s)\,ds, \\ ay(1)+by'(1)=\int^{1}_{0}g(s)y(s)\,ds, \\ \phi_{m}(y''(0))=\phi_{m}(y''(1))=\int^{1}_{0}h(t)\phi_{m}(y''(t))\,dt, \end{array}\displaystyle \right . $$
(1.1)
- (H1)
\(f\in{C[J\times R^{n}\times R^{n},R^{n}]}\), \(\Delta y|_{t=t_{k}}=y(t^{+}_{k})-y(t^{-}_{k})\), where \(y(t^{+}_{k})\) and \(y(t^{-}_{k})\) represent the right-hand limit and left-hand limit of \(y(t)\) at \(t=t_{k}\), respectively;
- (H2)
\(I_{k}\in{C[R^{n},R^{n}]}\), \(\bar{I}_{k}\in{C[R^{n}\times R^{n},R^{n}]}\);
- (H3)\(g,h\in{L^{1}[0,1]}\) are nonnegative and \(\xi\in[0,a)\), \(\upsilon\in [0,1)\) where$$ \xi= \int^{1}_{0}g(t)\,dt, \qquad \upsilon= \int^{1}_{0}h(t)\,dt. $$(1.2)
The remainder of the paper is organized as below. In Section 2, we give the expression of the solution to BVP (1.2). For this purpose, we do some computation and estimation of the Green’s function. In Section 3, we show the existence and uniqueness of solutions to BVP (1.2) by the Schauder fixed point theorem and contraction mapping theorem. Section 4 gives an example to illustrate our main result.
2 Preliminaries and lemmas
We shall reduce problem (1.1) to an integral equation. To this aim, first, by means of the transformation we convert problem (1.1) into and
$$\phi_{m}\bigl(y''(t)\bigr)=-x(t), $$
$$ \left \{ \textstyle\begin{array}{l} x''(t)+f(t,y,y')=0,\quad t\in{J}, \\ x(0)=x(1)=\int^{1}_{0}h(t)x(t)\,dt \end{array}\displaystyle \right . $$
(2.1)
$$ \left \{ \textstyle\begin{array}{l} y''(t)=-\phi_{m^{*}}(x(t)),\quad t\in{J}, t\neq t_{k}, \\ \Delta y|_{t=t_{k}}=I_{k}(y(t_{k})),\quad k=1,2,\ldots,m, \\ \Delta y'|_{t=t_{k}}=\bar{I}_{k}(y(t_{k}),y'(t_{k})),\quad k=1,2,\ldots,m, \\ ay(0)-by'(0)=\int^{1}_{0}g(s)y(s)\,ds, \\ ay(1)+by'(1)=\int^{1}_{0}g(s)y(s)\,ds. \end{array}\displaystyle \right . $$
(2.2)
Lemma 2.1
If (H1), (H2), and (H3) hold, then problem (2.1) has a unique solution
\(x(t)\), which is given by
where
$$x(t)= \int^{1}_{0}H(t,s)f\bigl(s,y,y'\bigr) \,ds, $$
$$\begin{aligned}& H(t,s)=G(t,s)+\frac{1}{1-v} \int^{1}_{0}G(s,\tau)h(\tau)\,d\tau, \end{aligned}$$
(2.3)
$$\begin{aligned}& G(t,s)= \left \{ \textstyle\begin{array}{l@{\quad}l} t(1-s),& 0\leq t\leq s\leq1, \\ s(1-t),& 0\leq s\leq t\leq1. \end{array}\displaystyle \right . \end{aligned}$$
(2.4)
Proof
Integrating (2.1) from 0 to t we get Integrating it again, we have From (2.1) we know that \(x(0)=x(1)=\int^{1}_{0}h(t)x(t)\,dt\). Letting \(t=1\) we then obtain Hence Thus we get In order to get the expression of \(x(t)\), different from [3], we multiply both sides of (2.5) with function \(h(t)\) and then integrating it from 0 to 1, we have and Hence, Finally, we obtain Thus where This completes the proof. □
$$\begin{aligned}& x'(t)-x'(0)=- \int^{t}_{0}f\bigl(t,y,y'\bigr)\,dt, \\& x'(t)=x'(0)- \int^{t}_{0}f\bigl(t,y,y'\bigr)\,dt. \end{aligned}$$
$$\begin{aligned}& x(t)-x(0)=x'(0)t- \int^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds, \\& x(t)=x(0)+x'(0)t- \int^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds. \end{aligned}$$
$$x(1)=x(0)+x'(0)- \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds. $$
$$x'(0)= \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds. $$
$$ x(t)= \int^{1}_{0}h(t)x(t)\,dt+t \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds- \int ^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds. $$
(2.5)
$$\begin{aligned} \int^{1}_{0}h(t)x(t)\,dt =& \int^{1}_{0}h(s)\,ds \int^{1}_{0}h(t)x(t)\,dt+ \int ^{1}_{0}th(t)\,dt \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ &{}- \int^{1}_{0}h(t)\,dt \int ^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds \end{aligned}$$
$$\begin{aligned} (1-\upsilon) \int^{1}_{0}h(t)x(t)\,dt =& \int^{1}_{0}th(t)\,dt \int ^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ &{}- \int^{1}_{0}h(t)\,dt \int^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds. \end{aligned}$$
$$\begin{aligned} \int^{1}_{0}h(t)x(t)\,dt =&\frac{1}{1-\upsilon}\biggl( \int^{1}_{0}th(t)\,dt \int ^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ &{}- \int^{1}_{0}h(t)\,dt \int ^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds\biggr). \end{aligned}$$
(2.6)
$$\begin{aligned} x(t) =&\frac{1}{1-\upsilon}\biggl( \int^{1}_{0}th(t)\,dt \int ^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{0}h(t)\,dt \int ^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds\biggr) \\ &+t \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ =&\frac{1}{1-\upsilon}\biggl( \int^{1}_{0} \int ^{1}_{0}th(t) (1-s)f\bigl(s,y,y' \bigr)\,dt\,ds- \int^{1}_{0} \int ^{t}_{0}h(t) (t-s)f\bigl(s,y,y' \bigr)\,ds\,dt\biggr) \\ &{}+ \int^{1}_{0}t(1-s)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ =& \int^{1}_{0}\frac{1}{1-\upsilon}\biggl( \int^{1}_{0}t(1-s)h(t)\,dt- \int ^{t}_{0}(t-s)h(t)\,dt\biggr)f\bigl(s,y,y' \bigr)\,ds \\ &{}+ \int^{1}_{0}t(1-s)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ =& \int^{1}_{0}\frac{1}{1-\upsilon}\biggl( \int^{1}_{0}t(1-s)h(t)\,dt- \int ^{t}_{0}(t-s)h(t)\,dt\biggr)f\bigl(s,y,y' \bigr)\,ds \\ &{}+ \int^{t}_{0}s(1-t)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{t}t(1-s)f\bigl(s,y,y'\bigr) \,ds. \end{aligned}$$
$$x(t)= \int^{1}_{0}H(t,s)f\bigl(s,y,y'\bigr) \,ds, $$
$$\begin{aligned}& H(t,s)=G(t,s)+\frac{1}{1-v} \int^{1}_{0}G(s,\tau)h(\tau)\,d\tau, \\& G(t,s)= \left \{ \textstyle\begin{array}{l@{\quad}l} t(1-s),& 0\leq t\leq s\leq1, \\ s(1-t),& 0\leq s\leq t\leq1. \end{array}\displaystyle \right . \end{aligned}$$
Let \(e(t)=t(1-t)\). Then from (2.3) and (2.4) we can prove that \(H(t,s)\) and \(G(t,s)\) have the following properties.
Lemma 2.2
Let
\(G(t,s)\)
and
\(H(t,s)\)
be given as in Lemma
2.1. Assume that (H3) holds. Then we have
where
$$\begin{aligned}& H(t,s)>0, \qquad G(t,s)>0,\quad \forall t,s\in(0,1), \\& H(t,s)\geq0,\qquad G(t,s)\geq0,\quad \forall t,s\in{J}, \\& e(t)e(s)\leq G(t,s)\leq G(t,t)=t(1-t)=e(t)\leq\bar{e}= \max_{t\in {J}}e(t)= \frac{1}{4},\quad \forall t,s\in{J}, \\& \rho e(s)\leq H(t,s)\leq\gamma s(1-s)=\gamma e(s)\leq\frac{1}{4}\gamma , \quad \forall t,s\in{J}, \end{aligned}$$
$$\gamma=\frac{1}{1-\upsilon}, \qquad \rho=\frac{\int^{1}_{0}e(\tau)h(\tau)\,d\tau }{1-\upsilon}. $$
Lemma 2.3
If (H1), (H2), and (H3) hold, then problem (2.2) has a unique solution
\(y(t)\)
expressed in the form
where
and
\(d=a(a+2b)\).
$$\begin{aligned} y(t) =& \int^{1}_{0}H_{1}(t,s)\phi_{m^{*}} \bigl(x(s)\bigr)\,ds-\sum^{m}_{k=1}H_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-\sum^{m}_{k=1}H_{2}(t)I_{k} \bigl(y(t_{k})\bigr), \end{aligned}$$
$$\begin{aligned}& H_{1}(t,s)=G_{1}(t,s)+\frac{1}{a-\xi} \int^{1}_{0}G_{1}(s,\tau)g(\tau )\,d\tau, \end{aligned}$$
(2.7)
$$\begin{aligned}& G_{1}(t,s)=\frac{1}{d} \left \{ \textstyle\begin{array}{l@{\quad}l} (b+as)(b+a(1-t)), &0\leq s\leq t\leq1, \\ (b+at)(b+a(1-s)), &0\leq t\leq s\leq1, \end{array}\displaystyle \right . \end{aligned}$$
(2.8)
$$\begin{aligned}& H_{2}(t)=\frac{at-a-b}{a+2b}+\frac{1}{a-\xi} \int^{1}_{0}\frac {at-a-b}{a+2b}g(t) \,dt, \end{aligned}$$
(2.9)
Proof
First, we assume that \(t\in I_{i}\), \(I_{i}=(t_{i},t_{i+1})\) (\(i=0,1,2,\ldots,m\)). Integrating both sides of (2.2) from \(t_{i}\) to \(t^{-}_{i+1}\), we get Adding the above equations, we find Similarly, we can get Let \(t=1\) in (2.10) and (2.11). We obtain It is easy to get which implies Then we have the following equations: Obviously, Substituting (2.10), (2.11) in (2.9), we have In a similar way as the proof of (2.6) in Lemma 2.1, we get Hence, we finally get Hence where Then the proof is completed. □
$$\begin{aligned}& y'\bigl(t^{-}_{1}\bigr)-y'(0)=- \int^{t_{1}}_{0}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt, \\& y'\bigl(t^{-}_{2}\bigr)-y' \bigl(t^{+}_{1}\bigr)=- \int^{t_{2}}_{t_{1}}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt, \\& \ldots, \\& y'(t)-y'\bigl(t^{+}_{i}\bigr)=- \int^{t}_{t_{i}}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt. \end{aligned}$$
$$\begin{aligned}& y'(t)-y'(0)-\sum_{k:t_{k}< t} \bigl(y'\bigl(t^{+}_{k}\bigr)-y' \bigl(t^{-}_{k}\bigr)\bigr)=- \int ^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt, \\& y'(t)=y'(0)+\sum_{k:t_{k}< t} \bigl(y'\bigl(t^{+}_{k}\bigr)-y' \bigl(t^{-}_{k}\bigr)\bigr)- \int ^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt, \\& y'(t)=y'(0)+\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)- \int ^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt. \end{aligned}$$
(2.10)
$$\begin{aligned} y(t) =&y(0)+y'(0)t- \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(t)\bigr) \,ds \\ &{}+\sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+\sum_{k:t_{k}< t}I_{k}\bigl(y(t_{k}) \bigr). \end{aligned}$$
(2.11)
$$\left \{ \textstyle\begin{array}{l} by'(1)=by'(0)+b\sum_{k:t_{k}< t}\bar{I}_{k}(y(t_{k}),y'(t_{k}))-b\int ^{t}_{0}\phi_{m^{*}}(x(t))\,dt, \\ ay(1)=ay(0)+ay'(0)-a\int^{1}_{0}(1-s)\phi_{m^{*}}(x(s))\,ds \\ \hphantom{ay(1)={}}{}+a \sum_{k:t_{k}< t}(1-t_{k})\bar{I}_{k}(y(t_{k}),y'(t_{k}))+a\sum_{k:t_{k}< t}I_{k}(y(t_{k})). \end{array}\displaystyle \right . $$
$$\begin{aligned} ay(1)+by'(1) =&ay(0)+(a+b)y'(0)-a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds \\ &{}+a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}+b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \\ =& \int^{1}_{0}g(s)y(s)\,ds, \end{aligned}$$
$$\begin{aligned} ay(0)+(a+b)y'(0) =&a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds \\ &{}-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a\sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt+ \int^{1}_{0}g(s)y(s)\,ds. \end{aligned}$$
$$\left \{ \textstyle\begin{array}{l} ay(0)+(a+b)y'(0) \\ \quad =a\int^{1}_{0}(1-s)\phi_{m^{*}}(x(s))\,ds-a\sum_{k:t_{k}< t}(1-t_{k})\bar{I}_{k}(y(t_{k}),y'(t_{k})) \\ \qquad {}-a \sum_{k:t_{k}< t}I_{k}(y(t_{k}))-b\sum_{k:t_{k}< t}\bar{I}_{k}(y(t_{k}),y'(t_{k})) \\ \qquad {}-b\int^{t}_{0}\phi_{m^{*}}(x(t))\,dt+\int^{1}_{0}g(s)y(s)\,ds, \\ ay(0)-by'(0)=\int^{1}_{0}g(s)y(s)\,ds. \end{array}\displaystyle \right . $$
$$\begin{aligned}& y'(0) =\frac{1}{a+2b}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(t)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\& \hphantom{y'(0) ={}}{} -a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr], \\& y(0) =\frac{b}{a(a+2b)}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\& \hphantom{y(0) ={}}{} -a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\& \hphantom{y(0) ={}}{} +\frac{1}{a} \int^{1}_{0}g(s)y(s)\,ds. \end{aligned}$$
$$\begin{aligned} y(t) =&\frac{b}{a(a+2b)}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}+\frac{1}{a} \int^{1}_{0}g(s)y(s)\,ds+\frac{t}{a+2b}\biggl[a \int ^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s) \bigr)\,ds \\ &{}-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}- \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(t)\bigr) \,ds+ \sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+\sum_{k:t_{k}< t}I_{k}\bigl(y(t_{k}) \bigr). \end{aligned}$$
$$\begin{aligned} \int^{1}_{0}g(s)y(s)\,ds =&\frac{a}{a-\xi}\biggl\{ \frac{b}{a(a+2b)}\biggl[ \int ^{1}_{0}g(t)\, dt a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(t)\bigr) \,ds \\ &{}-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}+\frac{t}{a+2b} \int^{1}_{0}g(t)\,dt\biggl[a \int^{1}_{0}(1-s)\phi _{m^{*}}\bigl(x(s) \bigr)\,ds \\ &{}-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}- \int^{1}_{0}g(t)\,dt \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds \\ &{}+ \sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \int^{1}_{0}g(t)\,dt + \int^{1}_{0}g(t) \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)\,dt\biggr\} . \end{aligned}$$
$$\begin{aligned} y(t) =&\frac{b}{a(a+2b)}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}+\frac{a}{a-\xi}\biggl\{ \frac{b}{a(a+2b)}\biggl[ \int^{1}_{0}g(t)\, dt a \int ^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s) \bigr)\,ds \\ &{}- a\sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}+\frac{t}{a+2b} \int^{1}_{0}g(t)\,dt\biggl[a \int^{1}_{0}(1-s)\phi _{m^{*}}\bigl(x(s) \bigr)\,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}- \int^{1}_{0}g(t)\,dt \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds+ \sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \int^{1}_{0}g(t)\,dt \\ &{}+ \int^{1}_{0}g(t)\,dt \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)\biggr\} \\ &{}+\frac{t}{a+2b}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}- \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds+ \sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+\sum_{k:t_{k}< t}I_{k}\bigl(y(t_{k}) \bigr). \end{aligned}$$
$$y(t)= \int^{1}_{0}H_{1}(t,s)\phi_{m^{*}} \bigl(x(s)\bigr)\,ds-\sum^{m}_{k=1}H_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H_{2}(t)I_{k} \bigl(y(t_{k})\bigr), $$
$$\begin{aligned}& H_{1}(t,s)=G_{1}(t,s)+\frac{1}{a-\xi} \int^{1}_{0}G_{1}(s,\tau)g(\tau )\,d\tau, \\& G_{1}(t,s)=\frac{1}{d} \left \{ \textstyle\begin{array}{l@{\quad}l} (b+as)(b+a(1-t)),& 0\leq s\leq t\leq1, \\ (b+at)(b+a(1-s)),& 0\leq t\leq s\leq1, \end{array}\displaystyle \right . \\& H_{2}(t)=\frac{at-a-b}{a+2b}+\frac{1}{a-\xi} \int^{1}_{0}\frac {at-a-b}{a+2b}g(t)\,dt. \end{aligned}$$
It follows from (2.7), (2.8), and (2.9) that \(H_{1}(t,s)\), \(G_{1}(t,s)\), and \(H_{2}(t)\) have the following properties.
Lemma 2.4
Suppose (H3) holds and assume that
\(G_{1}(t,s)\)
and
\(H_{1}(t,s)\)
are given as in Lemma
2.3. Then we have
where
$$\begin{aligned}& \frac{1}{d}b^{2}\leq G_{1}(t,s)\leq G_{1}(s,s)\leq\frac{(a+b)^{2}}{d},\quad \forall t,s\in{J}, \end{aligned}$$
(2.12)
$$\begin{aligned}& \rho_{1}\leq H_{1}(t,s)\leq\frac{a}{a-\xi}G_{1}(s,s) \leq\rho_{2},\quad \forall t,s\in{J}, \end{aligned}$$
(2.13)
$$\begin{aligned}& H_{2}(t)\leq\rho_{2}, \end{aligned}$$
(2.14)
$$\delta=\frac{b}{a+b}, \qquad \rho_{1}=\frac{b^{2}\gamma^{1}}{a+2b},\qquad \rho _{2}=\frac{a(a+b)^{2}}{(a-\xi)d}. $$
Proof
Clearly, it follows from the definition of \(G_{1}(t,s)\) that (2.12) holds. Now we show that (2.13) and (2.14) are true.
In fact, for \(t\in[\zeta,1]\) and \(s\in J\), we have Consequently, This completes the proof. □
$$\begin{aligned}& H_{1}(t,s)=G_{1}(t,s)+\frac{1}{a-\xi} \int^{1}_{0}G_{1}(s,\tau)g(\tau )\,d\tau \\& \hphantom{H_{1}(t,s)}\leq G_{1}(s,s)+\frac{\xi}{a-\xi}G_{1}(s,s)\leq \frac {a(a+b)^{2}}{(a-\xi)d}=\rho_{2}, \\& H_{2}(t)=\frac{at-a-b}{a+2b}+\frac{1}{a-\xi} \int^{1}_{0}\frac {at-a-b}{a+2b}g(t)\,dt\leq \frac{at}{a+2b}+\frac{\xi}{a-\xi}\frac{at}{a+2b}, \\& \frac{at}{a+2b}\leq\frac{a}{a+2b}\leq\frac{(a+b)^{2}}{a+2b}. \end{aligned}$$
$$ H_{2}(t)\leq\frac{(a+b)^{2}}{a+2b}+\frac{\xi}{a-\xi}\frac {(a+b)^{2}}{a+2b}= \frac{a(a+b)^{2}}{(a-\xi)d}=\rho_{2}. $$
(2.15)
Combining Lemma 2.1 with Lemma 2.3, we can get directly the following result.
Lemma 2.5
Assume that (H1)-(H3) hold. Then
\(y(t)\)
has the following form:
$$\begin{aligned} y(t) =& \int^{1}_{0}H_{1}(t,s)\phi_{m^{*}} \biggl( \int^{1}_{0}H(s,\tau)f\bigl(\tau ,y( \tau),y'(\tau)\bigr)\,d\tau\biggr)\,ds \\ &{}-\sum ^{m}_{k=1}H_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H_{2}(t)I_{k} \bigl(y(t_{k})\bigr). \end{aligned}$$
Proof
The conclusion is so straightforward that we omit it here. □
We next give some notations and a fixed point theorem which will be used to prove our main results. Let Clearly, \(\mathit{PC}^{1}[J,R^{n}]\) is a Banach space with the norm \(\|y\|_{\mathit{PC}^{1}}=\max\{\|y\|_{\mathit{PC}^{1}},\|y'\|_{\mathit{PC}^{1}}\}\).
$$\begin{aligned} \mathit{PC}^{1}\bigl[J,R^{n}\bigr] =&\bigl\{ y:J\rightarrow R^{n}\mid y'(t) \mbox{ is continuous when } t\neq t_{k},y\bigl(t^{+}_{k}\bigr),y \bigl(t^{+}_{-}\bigr),y'\bigl(t^{+}_{k} \bigr),y'\bigl(t^{-}_{k}\bigr) \\ &{}\mbox{exist and } y\bigl(t^{-}_{k}\bigr)=y(t_{k}),k=1,2, \ldots,m \bigr\} . \end{aligned}$$
Lemma 2.6
[19]
\(H\subset \mathit{PC}^{1}[J,R^{n}]\) is a relatively compact set if and only if \(\forall y\in H\), y and \(y'\) are uniformly bounded in J and equi-continuous on \(J_{k}\) (\(k=0, 1, 2, \ldots, m\)).
Definition 2.1
A function \(y\in \mathit{PC}^{1}\) is said to be a solution of (1.1) if it satisfies every equation in system (1.1).
Lemma 2.7
(Schauder fixed point theorem)
If K is a nonempty convex subset of a Banach space V and T is a continuous mapping of K into itself such that \(T(K)\) is contained in a compact subset of K, then T has a fixed point.
Definition 2.2
Define an operator \(A:\mathit{PC}^{1}[J,R^{n}]\rightarrow \mathit{PC}^{1}[J,R^{n}]\) by
$$\begin{aligned} (Ay) (t) =& \int^{1}_{0}H_{1}(t,s)\phi_{m^{*}} \biggl( \int^{1}_{0}H(s,\alpha )f\bigl(\alpha,y,y' \bigr)\, d\alpha \biggr)\,ds \\ &{}-\sum^{m}_{k=1}H_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H_{2}(t)I_{k} \bigl(y(t_{k})\bigr). \end{aligned}$$
(2.16)
Lemma 2.8
Assume that (H1)-(H3) hold. Then \(y(t)\in{J}\) is a fixed point of A if and only if \(y(t)\) is a solution of problem (1.1).
Lemma 2.9
The operator \(A: \mathit{PC}^{1}[J,R^{n}]\rightarrow \mathit{PC}^{1}[J,R^{n}]\) is completely continuous.
Proof
According to (2.16) we have From (2.16) and (2.17) we know that \(A:\mathit{PC}^{1}[J,R^{n}]\rightarrow \mathit{PC}^{1}[J,R^{n}]\) is continuous. For any bounded set \(S\in \mathit{PC}^{1}[J,R^{n}]\), and any function \(y(t)\in S\), we see that \((Ay)(t)\) and \((Ay)'(t)\) are uniformly bounded and equi-continuous on \(J_{k}\) (\(k=0,1,2,\ldots,m\)). Hence, according to Lemma 2.6 we see that \(A(S)\) is a relatively compact set, therefore A is a completely continuous operator. □
$$\begin{aligned} (Ay)'(t) =& \int^{1}_{0}H'_{1t}(t,s) \phi_{m^{*}} \biggl( \int^{1}_{0}H(s,\tau )f\bigl(\tau,y,y' \bigr)\,d\tau \biggr)\,ds \\ &{}-\sum^{m}_{k=1}H'_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H'_{2}(t)I_{k} \bigl(y(t_{k})\bigr). \end{aligned}$$
(2.17)
3 Main results
Let
$$\begin{aligned}& \beta= \varlimsup_{\|y\|+\|y'\|\rightarrow\infty} \biggl(\frac{\| f(t,y,y')\|}{\phi_{m^{*}}(\|y\|+\|y'\|)} \biggr), \\& \beta_{k}= \varlimsup_{\|y\|\rightarrow\infty} \biggl(\frac{\|I_{k}(y)\| }{\|y\|} \biggr), \\& \bar{\beta}_{k}= \varlimsup_{\|y\|+\|y'\|\rightarrow\infty} \biggl(\frac{\| \bar{I}_{k}(y,y')\|}{\|y\|+\|y'\|} \biggr)\quad (k=1,2,\ldots,m). \end{aligned}$$
Theorem 3.1
Assume that (H1)-(H3) hold. Let
\(\eta= \max \{\eta _{1},\eta_{2}\}<1\), then (1.1) has at least one solution, where
$$\begin{aligned}& \eta_{1}=2\rho_{2}\phi_{m^{*}}\biggl( \frac{\gamma\beta}{4}\biggr)+2m\rho_{2}\bar{\beta}_{k}+m \rho_{2}\beta_{k}, \\& \eta_{2}=2\rho_{3}\phi_{m^{*}}\biggl( \frac{\gamma\beta}{4}\biggr)+2m\rho_{3}\bar{\beta}_{k}+m \rho_{2}\beta_{k}. \end{aligned}$$
Proof
From the definition of β, there exists \(N>0\), s.t. Similarly, we get This together with (2.14), (2.15) implies that where \(\eta_{1}=2\rho_{2}\phi_{m^{*}}(\frac{\gamma\beta}{4})+2m\rho _{2}\bar{\beta}_{k}+m\rho_{2}\beta_{k}\).
$$\bigl\Vert f\bigl(t,y,y'\bigr)\bigr\Vert \leq\beta \phi_{m^{*}}\bigl(\Vert y\Vert +\bigl\Vert y'\bigr\Vert \bigr),\quad \forall t\in{J}, \phi _{m^{*}}\bigl(\Vert y\Vert + \bigl\Vert y'\bigr\Vert \bigr)\geq N. $$
$$\begin{aligned}& \bigl\Vert I_{k}(y)\bigr\Vert \leq\beta_{k}\Vert y \Vert , \\& \bigl\Vert \bar{I}_{k}\bigl(y,y'\bigr)\bigr\Vert \leq\bar{\beta}_{k}\bigl(\Vert y\Vert +\bigl\Vert y' \bigr\Vert \bigr),\quad \forall y,y'\in R^{n}\ (k=1,2, \ldots,m). \end{aligned}$$
$$\begin{aligned} \bigl\Vert (Ay) (t)\bigr\Vert =&\Biggl\Vert \int^{1}_{0}H_{1}(t,s)\phi_{m^{*}} \biggl( \int ^{1}_{0}H(s,\tau)f\bigl(\tau,y,y' \bigr)\,d\tau \biggr)\,ds \\ &{}- \sum^{m}_{k=1}H_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H_{2}(t)I_{k} \bigl(y(t_{k})\bigr)\Biggr\Vert \\ \leq&\biggl\Vert \rho_{2}\phi_{m^{*}}\biggl( \int^{1}_{0}H(s,\tau)f\bigl(\tau,y( \tau),y'(\tau )\bigr)\,d\tau\biggr)\biggr\Vert \\ &{}+m \rho_{2}\bigl(\bar{\beta}_{k}\bigl(\Vert y\Vert +\bigl\Vert y'\bigr\Vert \bigr)\bigr)+m\rho_{2} \beta_{k}\Vert y\Vert \\ \leq&\rho_{2}\phi_{m^{*}}\biggl(\frac{\gamma}{4}\beta \phi_{m}\bigl(\Vert y\Vert +\bigl\Vert y'\bigr\Vert \bigr)\biggr)+2m\rho_{2}\bar{\beta}_{k}\Vert y \Vert +m\rho_{2}\beta_{k}\Vert y\Vert \\ \leq&2\rho_{2}\phi_{m^{*}}\biggl(\frac{\gamma\beta}{4}\biggr) \Vert y\Vert +2m\rho_{2}\bar{\beta}_{k}\Vert y\Vert +m\rho_{2}\beta_{k}\Vert y\Vert \\ \leq& \biggl(2\rho_{2}\phi_{m^{*}}\biggl(\frac{\gamma\beta}{4} \biggr)+2m\rho_{2}\bar{\beta}_{k}+m\rho_{2} \beta_{k} \biggr)\Vert y\Vert \\ \leq&\eta_{1}\Vert y\Vert \\ \leq&\Vert y\Vert , \end{aligned}$$
From (2.7), (2.8), and (2.9), we have and which together with (2.17) imply where \(\eta_{2}=2\rho_{3}\phi_{m^{*}}(\frac{\gamma\beta}{4})+2m\rho _{3}\bar{\beta}_{k}+m\rho_{2}\beta_{k}\).
$$\begin{aligned}& H'_{1t}(t,s)=G'_{1t}(t,s)= \frac{1}{d} \left \{ \textstyle\begin{array}{l@{\quad}l} -a(b+as), &0\leq s\leq t\leq1, \\ a(b+a(1-s)),& 0\leq t\leq s\leq1, \end{array}\displaystyle \right . \\& \max_{t,s\in{J},t\neq s}\bigl\vert H'_{1t}(t,s) \bigr\vert =\max_{t,s\in{J},t\neq s}\bigl\vert G'_{1t}(t,s) \bigr\vert \leq\frac{1}{d}a(a+b)=\frac{a+b}{a+2b}=\rho_{3}, \end{aligned}$$
$$H'_{2}(t)=\frac{a}{a+2b}\leq\rho_{2}, $$
$$\begin{aligned} \bigl\Vert (Ay)'(t)\bigr\Vert =&\Biggl\Vert \int^{1}_{0}H'_{1t}(t,s) \phi_{m^{*}} \biggl( \int ^{1}_{0}H(s,\tau)f\bigl(\tau,y,y' \bigr)\,d\tau \biggr)\,ds \\ &{}- \sum^{m}_{k=1}H'_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H'_{2}(t)I_{k} \bigl(y(t_{k})\bigr)\Biggr\Vert \\ \leq&\rho_{3}\phi_{m^{*}}\biggl(\frac{\gamma}{4}\beta \phi_{m}\bigl(\Vert y\Vert +\bigl\Vert y'\bigr\Vert \bigr)\biggr)+m\rho_{2}\bigl(\bar{\beta}_{k}\bigl( \Vert y\Vert +\bigl\Vert y'\bigr\Vert \bigr)\bigr)+m \rho_{2}\beta_{k}\Vert y\Vert \\ \leq&\rho_{3}\phi_{m^{*}}\biggl(\frac{\gamma\beta}{4}\biggr) \bigl(\Vert y\Vert +\bigl\Vert y'\bigr\Vert \bigr)+2m\rho _{3}\bar{\beta}_{k}\Vert y\Vert +m\rho_{2} \beta_{k}\Vert y\Vert \\ \leq& \biggl(2\rho_{3}\phi_{m^{*}}\biggl(\frac{\gamma\beta}{4} \biggr)+2m\rho_{3}\bar{\beta}_{k}+m\rho_{2} \beta_{k} \biggr)\Vert y\Vert \\ \leq&\eta_{2}\Vert y\Vert \\ \leq&\Vert y\Vert , \end{aligned}$$
On the other hand, according to Lemma 2.9, we know operator A is a completely continuous operator. Together with Lemma 2.6 (Schauder fixed point theorem), we know A has a fixed point in \(\mathit{PC}^{1}[J,R^{n}]\). □
Theorem 3.2
Assume that (H1)-(H3) hold. If there exist nonnegative real numbers
α, \(\alpha_{k}\), \(\bar{\alpha}_{k}\), s.t.
and
\(\xi= \max\{\xi_{1},\xi_{2}\}<1\), then (1.1) has a unique solution, where
$$\begin{aligned}& \bigl\Vert \phi_{m^{*}}(x)-\phi_{m^{*}}(y)\bigr\Vert \leq \alpha\bigl(\Vert x-y\Vert \bigr), \\& \bigl\Vert \bar{I}_{k}\bigl(x,x'\bigr)- \bar{I}_{k}\bigl(y,y'\bigr)\bigr\Vert \leq\bar{\alpha}_{k}\bigl(\Vert x-y\Vert +\bigl\Vert x'-y' \bigr\Vert \bigr), \\& \bigl\Vert I_{k}(x)-I_{k}(y)\bigr\Vert \leq \alpha_{k}\Vert x-y\Vert , \end{aligned}$$
$$\begin{aligned}& \xi_{1}=\rho_{2}(2\alpha+2m\bar{\alpha}_{k}+m \alpha_{k}), \\& \xi_{2}=2\rho_{3}\alpha+2m\rho_{3}\bar{\alpha}_{k}+m\rho_{2}. \end{aligned}$$
Proof
From (2.16) and (2.17), similar to the proof of Theorem 3.1, we get Computing straightforwardly we have
$$\begin{aligned} (Ax) (t)-(Ay) (t) =& \int^{1}_{0}H_{1}(t,s) \bigl( \phi_{m^{*}}(x)-\phi _{m^{*}}(y)\bigr)+\sum ^{m}_{k=1} H_{1}(t,t_{k}) \bigl( \bar{I}_{k}\bigl(y,y'\bigr)-\bar{I}_{k} \bigl(x,x'\bigr)\bigr) \\ &{}+\sum^{m}_{k=1} H_{2}(t) \bigl(I_{k}(y)-I_{k}(x)\bigr). \end{aligned}$$
$$\begin{aligned} \bigl\Vert (Ax) (t)-(Ay) (t)\bigr\Vert \leq&\rho_{2}\alpha\bigl( \Vert x-y\Vert \bigr)+ \sum^{m}_{k=1} \rho _{2}\bar{\alpha}_{k}\bigl(\Vert x-y\Vert +\bigl\Vert x'-y'\bigr\Vert \bigr) \\ &{}+\sum ^{m}_{k=1}\rho_{2}\alpha _{k} \bigl(\Vert x-y\Vert \bigr) \\ \leq&\rho_{2}(\alpha+2m\bar{\alpha}_{k}+m \alpha_{k})\Vert x-y\Vert _{\mathit{PC}^{1}} \\ =&\xi_{1}\Vert x-y\Vert _{\mathit{PC}^{1}} \\ \leq&\xi \Vert x-y\Vert _{\mathit{PC}^{1}}. \end{aligned}$$
Also we obtain It follows from \(\xi<1\) that A has a unique fixed point and therefore (1.1) has a unique solution. □
$$\begin{aligned}& (Ax)'(t)-(Ay)'(t)= \int^{1}_{0}H'_{1t}(t,s) \bigl(\phi_{m^{*}}(x)-\phi _{m^{*}}(y)\bigr) \\& \hphantom{(Ax)'(t)-(Ay)'(t)={}}{}+\sum ^{m}_{k=1}H'_{1}(t,t_{k}) \bigl(\bar{I}_{k}\bigl(y,y'\bigr)-\bar{I}_{k} \bigl(x,x'\bigr)\bigr) \\& \hphantom{(Ax)'(t)-(Ay)'(t)={}}{}+\sum^{m}_{k=1} H'_{2}(t) \bigl(I_{k}(y)-I_{k}(x) \bigr), \\& \bigl\Vert (Ax)'(t)-(Ay)'(t)\bigr\Vert \leq \rho_{3}\alpha \Vert x-y\Vert +m\rho_{3}\bar{\alpha}_{k}\bigl(\Vert x-y\Vert +\bigl\Vert x'-y' \bigr\Vert \bigr)+m\rho_{2}\alpha_{k}\Vert x-y\Vert \\& \hphantom{\bigl\Vert (Ax)'(t)-(Ay)'(t)\bigr\Vert } \leq(\rho_{3}\alpha+2m\rho_{3}\bar{\alpha}_{k}+m\rho_{2}\alpha_{k})\Vert x-y\Vert _{\mathit{PC}^{1}} \\& \hphantom{\bigl\Vert (Ax)'(t)-(Ay)'(t)\bigr\Vert } =\xi_{2}\Vert x-y\Vert _{\mathit{PC}^{1}} \\& \hphantom{\bigl\Vert (Ax)'(t)-(Ay)'(t)\bigr\Vert } \leq\xi \Vert x-y\Vert _{\mathit{PC}^{1}}. \end{aligned}$$
4 Example
In this section, we will illustrate the main results by a simple example.
Let \(n=1\), \(t_{1}=\frac{1}{2}\), \(a=b=1\), \(I_{1}(y(t_{1}))=\bar{I}_{1}(y(t_{1}, y'(t_{1})))=\frac{1}{2}\), \(f(t,y,y')=\sqrt[3]{t-y+y'}-\frac {1}{42}y'-3\ln(1+y^{2})\), and \(g(s)=\frac{1}{3}\), \(h(t)=\frac{1}{2}\), \(m=3\) in \(\phi_{m}\). Then equation (1.1) turns to the following equation: Following Theorem 3.1, we have the following result.
$$ \left \{ \textstyle\begin{array}{l} (\phi_{3}(y''))''=\sqrt[3]{t-y+y'}-\frac{1}{41}y'-3\ln(1+y^{2}),\quad t\in {J}, t\neq\frac{1}{2}, \\ \Delta y|_{t=\frac{1}{2}}=\frac{1}{2}, \\ \Delta y'|_{t=\frac{1}{2}}=\frac{1}{2}, \\ y(0)-y'(0)=\int^{1}_{0}\frac{1}{3}y(s)\,ds, \\ y(1)+y'(1)=\int^{1}_{0}\frac{1}{3}y(s)\,ds, \\ \phi_{3}(y''(0))=\phi_{3}(y''(1))=\int^{1}_{0}\frac{1}{2}\phi_{3}(y''(t))\,dt. \end{array}\displaystyle \right . $$
(4.1)
Theorem 4.1
The problem (4.1) has at least one positive solution.
Proof
Obviously, \(f(t,y,y')\in{C[J\times R^{n}\times R^{n}, R^{n}]}\), \(I_{1}(y(t_{1}))\in {C[R^{n}, R^{n}]}\), \(\bar{I}_{1}(y(t_{1}, y'(t_{1})))\in C[R^{n}\times R^{n},R^{n}]\). From (4.1), we get so we get \(\beta\leq\frac{1}{41}\), \(\beta_{1}=0\), \(\bar{\beta}_{1}=0\), \(\rho _{2}=2\), \(\rho_{3}=\frac{2}{3}\), \(\gamma=2\), \(\phi_{m^{*}}(\frac{\gamma\beta }{4})\leq\sqrt{\frac{1}{82}}\).
$$\begin{aligned}& \bigl\Vert f\bigl(t,y,y'\bigr)\bigr\Vert \leq\sqrt[3]{t+\Vert y\Vert +\bigl\Vert y'\bigr\Vert }+\frac{1}{41}\bigl\Vert y'\bigr\Vert +3\ln\bigl(1+\Vert y\Vert ^{2} \bigr), \\& \bigl\Vert I_{1}(y)\bigr\Vert =\frac{1}{2},\qquad \bigl\Vert \bar{I}_{1}\bigl(t,y,y'\bigr)\bigr\Vert = \frac{1}{2},\quad \forall t\in{J}, y,y' \in R^{n}, \end{aligned}$$
Hence, \(\eta_{1}\leq\sqrt{\frac{8}{41}}\leq1\), \(\eta_{2}\leq\sqrt{\frac {8}{369}}\leq1\), \(\eta\leq\sqrt{\frac{8}{41}}\leq1\). From Theorem 3.1, we get the result.
This completes the proof. □
Declarations
Acknowledgements
The authors express their sincere thanks to the anonymous reviews for their valuable suggestions and corrections for improving the quality of the paper. This work was supported by NNSF of China No. 11431008 and NNSF of China No. 11271261.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
(1)
Department of Mathematics, Shanghai Normal University
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