Positive solutions for impulsive fractional differential equations with generalizedperiodic boundary value conditions

Abstract

By constructing Green’s function, we give the natural formulae of solutions forthe following nonlinear impulsive fractional differential equation with generalizedperiodic boundary value conditions:

$$\{\begin{array}{l}{}^{c}D_t^{q}u(t)=f(t,u(t)),t\in J^{\prime }=J\mathrm{\setminus }\{t_1,\dots ,t_m\},J=[0,1],\\ \mathrm{\Delta }u(t_k)=I(u(t_k)),\mathrm{\Delta }{u}^{\prime }(t_k)=J_k(u(t_k)),k=1,\dots ,m,\\ au(0)-bu(1)=0,a{u}^{\prime }(0)-b{u}^{\prime }(1)=0,\end{array}$$

where $1<q<2$ is a real number, ${}^{c}D_t^q$ is the standard Caputo differentiation. We present theproperties of Green’s function. Some sufficient conditions for the existence ofsingle or multiple positive solutions of the above nonlinear fractional differentialequation are established. Our analysis relies on a nonlinear alternative of theSchauder and Guo-Krasnosel’skii fixed point theorem on cones. As applications,some interesting examples are provided to illustrate the main results.

MSC: 34B10, 34B15, 34B37.

1 Introduction

In recent years, the fractional order differential equation has aroused great attentiondue to both the further development of fractional order calculus theory and theimportant applications for the theory of fractional order calculus in the fields ofscience and engineering such as physics, chemistry, aerodynamics, electrodynamics ofcomplex medium, polymer rheology, Bode’s analysis of feedback amplifiers,capacitor theory, electrical circuits, electron-analytical chemistry, biology, controltheory, fitting of experimental data, and so forth. Many papers and books on fractionalcalculus differential equation have appeared recently. One can see [1–17] and the references therein.

In order to describe the dynamics of populations subject to abrupt changes as well asother phenomena such as harvesting, diseases, and so on, some authors have used animpulsive differential system to describe these kinds of phenomena since the lastcentury. For the basic theory on impulsive differential equations, the reader can referto the monographs of Bainov and Simeonov [18], Lakshmikantham et al.[19] and Benchohra et al.[20].

In this article, we consider the following nonlinear impulsive fractional differentialequation with generalized periodic boundary value conditions (for short BVPs (1.1)):

$$\{\begin{array}{l}{}^{c}D_t^{q}u(t)=f(t,u(t)),t\in J^{\prime }=J\mathrm{\setminus }\{t_1,\dots ,t_m\},J=[0,1],\\ \mathrm{\Delta }u(t_k)=I_k(u(t_k)),\mathrm{\Delta }{u}^{\prime }(t_k)=J_k(u(t_k)),k=1,\dots ,m,\\ au(0)-bu(1)=0,a{u}^{\prime }(0)-b{u}^{\prime }(1)=0,\end{array}$$

(1.1)

where a, b are real constants with $a>b>0$. ${}^{c}D_{0^{+}}^q$ is the Caputo fractional derivative of order$1<q<2$. $f:J\times {\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is jointly continuous. $I_k,J_k\in C({\mathbb{R}}^{+},{\mathbb{R}}^{+})$, ${\mathbb{R}}^{+}=[0,+\mathrm{\infty })$. The impulsive point set ${\{t_k\}}_{k=1}^m$ satisfies $0=t_0<t_1<\cdots <t_m<t_{m+1}=1$. $u(t_k^{+})={lim}_{h\to 0^{+}}u(t_k+h)$ and $u(t_k^{-})={lim}_{h\to 0^{-}}u(t_k+h)$ represent the right and left limits of$u(t)$ at the impulsive point $t=t_k$. Let us set $J_0=[0,t_1]$, $J_k=(t_k,t_{k+1}]$, $1\le k\le m$. The goal of this paper is to study the existence ofsingle or multiple positive solutions for the impulsive BVPs (1.1) by a nonlinearalternative of the Schauder and Guo-Krasnosel’skii fixed point theorem oncones.

The rest of the paper is organized as follows. In Section 2, we present some usefuldefinitions, lemmas and the properties of Green’s function. In Section 3, wegive some sufficient conditions for the existence of a single positive solution for BVPs(1.1). In Section 4, some sufficient criteria for the existence of multiplepositive solutions for BVPs (1.1) are obtained. Finally, some examples are provided toillustrate our main results in Section 5.

2 Preliminaries

For the convenience of the reader, we present here the necessary definitions fromfractional calculus theory. These definitions and properties can be found in theliterature.

Definition 2.1 (see [21, 22])

The Riemann-Liouville fractional integral of order $\alpha >0$ of a function $f:(0,+\mathrm{\infty })\to \mathbb{R}$ is given by

$$I_{0+}^{\alpha }f(t)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^t{(t-s)}^{\alpha -1}f(s)ds,$$

provided that the right-hand side is pointwise defined on $(0,+\mathrm{\infty })$.

Definition 2.2 (see [21, 22])

The Caputo fractional derivative of order $\alpha >0$ of a continuous function $f:(0,+\mathrm{\infty })\to \mathbb{R}$ is given by

$${}^{c}D_{0+}^{\alpha }f(t)=\frac{1}{\mathrm{\Gamma }(n-\alpha )}{\int }_0^t\frac{f^{(n)}(s)}{{(t-s)}^{\alpha -n+1}}ds,$$

where $n-1<\alpha \le n$, provided that the right-hand side is pointwise defined on$(0,+\mathrm{\infty })$.

Lemma 2.1 (see [21])

Assume that$u\in C(0,1)\cap L(0,1)$with a Caputo fractional derivative oforder$q>0$that belongs to$u\in C^n[0,1]$, then

$$I_{0+}^q{D}_{0+}^{q}u(t)=u(t)+c_0+c_{1}t+\cdots +c_{n-1}{t}^{n-1}$$

for some$c_i\in \mathbb{R}$, $i=0,1,2,\dots ,n-1$ ($n=-[-q]$) and$[q]$denotes the integer part of the real number q.

Lemma 2.2 (see [23])

Let E be a Banach space. Assumethat$T:E\to E$is a completely continuous operator and theset$V=\{u\in E\mid u=\mu Tu,0<\mu <1\}$is bounded. Then T has a fixedpoint in E.

Lemma 2.3 (Schauder fixed point theorem, see [24])

If U is a close bounded convex subset of a Banachspace E and$T:U\to U$is completely continuous,then T has at least one fixed point in U.

Lemma 2.4 (see [25])

Let E be a Banach space, $P\subseteq E$be a cone, and${\mathrm{\Omega }}_1$, ${\mathrm{\Omega }}_2$be two bounded open balls of E centeredat the origin with$0\in {\mathrm{\Omega }}_1$and${\overline{\mathrm{\Omega }}}_1\subset {\mathrm{\Omega }}_2$. Suppose that$A:P\cap ({\overline{\mathrm{\Omega }}}_2\setminus {\mathrm{\Omega }}_1)\to P$is a completely continuous operator such thateither

• (i) $\parallel Au\parallel \le \parallel u\parallel$, $u\in P\cap \partial {\mathrm{\Omega }}_1$and$\parallel Au\parallel \ge \parallel u\parallel$, $u\in P\cap \partial {\mathrm{\Omega }}_2$, or

• (ii) $\parallel Au\parallel \ge \parallel u\parallel$, $u\in P\cap \partial {\mathrm{\Omega }}_1$and$\parallel Au\parallel \le \parallel u\parallel$, $u\in P\cap \partial {\mathrm{\Omega }}_2$

hold. Then A has at least one fixed pointin$P\cap ({\overline{\mathrm{\Omega }}}_2\setminus {\mathrm{\Omega }}_1)$.

Now we present Green’s function for a system associated with BVPs (1.1).

Lemma 2.5 Given$h\in C(J,{\mathbb{R}}^{+})$and$1<q<2$, the unique solution of

$$\{\begin{array}{l}{}^{c}D_t^{q}u(t)=h(t),t\in J^{\prime },\\ \mathrm{\Delta }u(t_k)=I_k(u(t_k)),\mathrm{\Delta }{u}^{\prime }(t_k)=J_k(u(t_k)),k=1,\dots ,m,\\ au(0)-bu(1)=0,a{u}^{\prime }(0)-b{u}^{\prime }(1)=0,a>b>0,\end{array}$$

(2.1)

is formulated by

$$u(t)={\int }_0^1{G}_1(t,s)h(s)ds+\sum _{i=1}^m{G}_2(t,t_i)J_i(u(t_i))+\sum _{i=1}^m{G}_3(t,t_i)I_i(u(t_i)),t\in J,$$

where

$$G_1(t,s)=\{\begin{array}{ll}\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}+\frac{b{(1-s)}^{q-1}}{(a-b)\mathrm{\Gamma }(q)}+\frac{b(q-1)t{(1-s)}^{q-2}}{(a-b)\mathrm{\Gamma }(q)}+\frac{b^2(q-1){(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q)},& 0\le s\le t\le 1,\\ \frac{b{(1-s)}^{q-1}}{(a-b)\mathrm{\Gamma }(q)}+\frac{b(q-1)t{(1-s)}^{q-2}}{(a-b)\mathrm{\Gamma }(q)}+\frac{b^2(q-1){(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q)},& 0\le t\le s\le 1,\end{array}$$

(2.2)

$$G_2(t,t_i)=\{\begin{array}{ll}\frac{ab}{{(a-b)}^2}+\frac{a(t-t_i)}{a-b},& 0\le t_i<t\le 1,i=1,2,\dots ,m,\\ \frac{ab}{{(a-b)}^2}+\frac{b(t-t_i)}{a-b},& 0\le t\le t_i\le 1,i=1,2,\dots ,m,\end{array}$$

(2.3)

$$G_3(t,t_i)=\{\begin{array}{ll}\frac{a}{a-b},& 0\le t_i<t\le 1,i=1,2,\dots ,m,\\ \frac{b}{a-b},& 0\le t\le t_i\le 1,i=1,2,\dots ,m.\end{array}$$

(2.4)

Proof Let u be a general solution of (2.1) on each interval$(t_k,t_{k+1}]$ ($k=0,1,\dots ,m$). Applying Lemma 2.1, Eq. (2.1) is translated intothe following equivalent integral equation (2.5):

$$u(t)=\frac{1}{\mathrm{\Gamma }(q)}{\int }_0^t{(t-s)}^{q-1}h(s)ds-c_k-d_{k}t,\mathrm{\forall }t\in (t_k,t_{k+1}],$$

(2.5)

where $t_0=0$, $t_{m+1}=1$. Then we have

$$u^{\prime }(t)=\frac{1}{\mathrm{\Gamma }(q-1)}{\int }_0^t{(t-s)}^{q-2}h(s)ds-d_k,t\in (t_k,t_{k+1}].$$

In the light of the generalized periodic boundary value conditions of Eq. (2.1), we get

$$b{\int }_0^1\frac{{(1-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds+a{c}_0-b{c}_m-b{d}_m=0,$$

(2.6)

$$b{\int }_0^1\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds+a{d}_0-b{d}_m=0.$$

(2.7)

Next, using the right impulsive condition of Eq. (2.1), we derive

$$c_{k-1}-c_k=I_k(u(t_k))-J_k(u(t_k))t_k,$$

(2.8)

$$d_{k-1}-d_k=J_k(u(t_k)).$$

(2.9)

By (2.7) and (2.9), we have

$$d_0=\frac{-b}{a-b}{\int }_0^1\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds-\frac{b}{a-b}\sum _{i=1}^m{J}_i(u(t_i)),$$

(2.10)

$$d_m=\frac{-b}{a-b}{\int }_0^1\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds-\frac{a}{a-b}\sum _{i=1}^m{J}_i(u(t_i)).$$

(2.11)

By (2.9) we have

$$\begin{array}{rl}{d}_k& =d_0-\sum _{i=1}^k{J}_i(u(t_i))\\ =\frac{-b}{a-b}{\int }_0^1\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds-\frac{b}{a-b}\sum _{i=1}^m{J}_i(u(t_i))-\sum _{i=1}^k{J}_i(u(t_i)).\end{array}$$

(2.12)

From (2.6), (2.8) and (2.11), we have

$$\begin{array}{rl}{c}_0=& \frac{-b}{a-b}{\int }_0^1\frac{{(1-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds-\frac{b^2}{{(a-b)}^2}{\int }_0^1\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds\\ -\frac{ab}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))-\frac{b}{a-b}\sum _{i=1}^m(I_i(u(t_i))-J_i(u(t_i))t_i).\end{array}$$

(2.13)

According to (2.8), we obtain

$$\begin{array}{rl}{c}_k=& c_0-\sum _{i=1}^k(I_i(u(t_i))-J_i(u(t_i))t_i)\\ =& \frac{-b}{a-b}{\int }_0^1\frac{{(1-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds-\frac{b^2}{{(a-b)}^2}{\int }_0^1\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds\\ -\frac{ab}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))-\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))-\sum _{i=1}^k{I}_i(u(t_i))\\ +\frac{b}{a-b}\sum _{i=1}^m{J}_i(u(t_i))t_i+\sum _{i=1}^k{J}_i(u(t_i))t_i.\end{array}$$

(2.14)

Hence, for $k=1,2,\dots ,m$, (2.12) and (2.14) imply

$$\begin{array}{rl}{c}_k+d_{k}t=& \frac{-b}{a-b}{\int }_0^1\frac{{(1-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds-\frac{(ab-b^2)t+b^2}{{(a-b)}^2}{\int }_0^1\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds\\ -\frac{1}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))[ab+b(a-b)(t-t_i)]-\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ -\sum _{i=1}^k{I}_i(u(t_i))-\sum _{i=1}^k{J}_i(u(t_i))(t-t_i).\end{array}$$

(2.15)

Now substituting (2.10) and (2.13) into (2.5), for $t\in J_0=[0,t_1]$, we obtain

$$\begin{array}{rl}u(t)=& {\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds+\frac{b}{a-b}{\int }_0^1\frac{{(1-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds\\ +\frac{(ab-b^2)t+b^2}{{(a-b)}^2}{\int }_0^1\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds\\ +\frac{1}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))[ab+b(a-b)(t-t_i)]+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ =& {\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds+\frac{b}{a-b}({\int }_0^t+{\int }_t^1)\frac{{(1-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds\\ +\frac{(ab-b^2)t+b^2}{{(a-b)}^2}({\int }_0^t+{\int }_t^1)\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds\\ +\frac{1}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))[ab+b(a-b)(t-t_i)]+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ =& {\int }_0^t[\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}+\frac{b{(1-s)}^{q-1}}{(a-b)\mathrm{\Gamma }(q)}+\frac{[(ab-b^2)t+b^2]{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q-1)}]h(s)ds\\ +{\int }_t^1[\frac{b{(1-s)}^{q-1}}{(a-b)\mathrm{\Gamma }(q)}+\frac{[(ab-b^2)t+b^2]{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q-1)}]h(s)ds\\ +\frac{1}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))[ab+b(a-b)(t-t_i)]+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ =& {\int }_0^1{G}_1(t,s)h(s)ds+\sum _{i=1}^m{G}_2(t,t_i)J_i(u(t_i))+\sum _{i=1}^m{G}_3(t,t_i)I_i(u(t_i)),\end{array}$$

where $G_1(t,s)$, $G_2(t,t_i)$ and $G_3(t,t_i)$ are defined by (2.2)-(2.4).

Substituting (2.15) into (2.5), for $t\in J_k=(t_k,t_{k+1}]$, $k=1,2,\dots ,m$, we have

$$\begin{array}{rl}u(t)=& {\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds+\frac{b}{a-b}{\int }_0^1\frac{{(1-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds\\ +\frac{(ab-b^2)t+b^2}{{(a-b)}^2}{\int }_0^1\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds\\ +\frac{1}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))[ab+b(a-b)(t-t_i)]+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ +\sum _{i=1}^k{I}_i(u(t_i))+\sum _{i=1}^k{J}_i(u(t_i))(t-t_i)\\ =& {\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds+\frac{b}{a-b}({\int }_0^t+{\int }_t^1)\frac{{(1-s)}^{q-1}}{\mathrm{\Gamma }(q)}h(s)ds\\ +\frac{(ab-b^2)t+b^2}{{(a-b)}^2}({\int }_0^t+{\int }_t^1)\frac{{(1-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}h(s)ds\\ +\frac{1}{{(a-b)}^2}(\sum _{i=1}^k+\sum _{i=k+1}^m)J_i(u(t_i))[ab+b(a-b)(t-t_i)]\\ +\frac{b}{a-b}(\sum _{i=1}^k+\sum _{i=k+1}^m)I_i(u(t_i))+\sum _{i=1}^k{I}_i(u(t_i))+\sum _{i=1}^k{J}_i(u(t_i))(t-t_i)\\ =& {\int }_0^t[\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}+\frac{b{(1-s)}^{q-1}}{(a-b)\mathrm{\Gamma }(q)}+\frac{[(ab-b^2)t+b^2]{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q-1)}]h(s)ds\\ +{\int }_t^1[\frac{b{(1-s)}^{q-1}}{(a-b)\mathrm{\Gamma }(q)}+\frac{[(ab-b^2)t+b^2]{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q-1)}]h(s)ds\\ +\sum _{i=1}^k[\frac{ab}{{(a-b)}^2}+\frac{b(a-b)+{(a-b)}^2}{{(a-b)}^2}(t-t_i)]J_i(u(t_i))\\ +\sum _{i=k+1}^m\frac{ab+b(a-b)(t-t_i)}{{(a-b)}^2}{J}_i(u(t_i))+\sum _{i=1}^k[\frac{b}{a-b}+1]I_i(u(t_i))\\ +\sum _{i=k+1}^m\frac{b}{a-b}{I}_i(u(t_i))\\ =& {\int }_0^t[\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}+\frac{b{(1-s)}^{q-1}}{(a-b)\mathrm{\Gamma }(q)}+\frac{[(ab-b^2)t+b^2]{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q-1)}]h(s)ds\\ +{\int }_t^1[\frac{b{(1-s)}^{q-1}}{(a-b)\mathrm{\Gamma }(q)}+\frac{[(ab-b^2)t+b^2]{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q-1)}]h(s)ds\\ +\sum _{i=1}^k\frac{ab+a(a-b)(t-t_i)}{{(a-b)}^2}{J}_i(u(t_i))+\sum _{i=k+1}^m\frac{ab+b(a-b)(t-t_i)}{{(a-b)}^2}{J}_i(u(t_i))\\ +\sum _{i=1}^k\frac{a}{a-b}{I}_i(u(t_i))+\sum _{i=k+1}^m\frac{b}{a-b}{I}_i(u(t_i))\\ =& {\int }_0^1{G}_1(t,s)h(s)ds+\sum _{i=1}^m{G}_2(t,t_i)J_i(u(t_i))+\sum _{i=1}^m{G}_3(t,t_i)I_i(u(t_i)),\end{array}$$

where $G_1(t,s)$, $G_2(t,t_i)$ and $G_3(t,t_i)$ are defined by (2.2)-(2.4). The proof iscomplete. □

Lemma 2.6 Let$0<b<a<+\mathrm{\infty }$, then Green’sfunctions$G_1(t,s)$, $G_2(t,t_i)$and$G_3(t,t_i)$defined by (2.2), (2.3) and (2.4) arecontinuous and satisfy the following:

• (i) $G_1(t,s)\in C(J\times J,{\mathbb{R}}^{+})$, $G_2(t,t_i),G_3(t,t_i)\in C(J\times J,{\mathbb{R}}^{+})$, and$G_1(t,s),G_2(t,t_i),G_3(t,t_i)>0$for all$t,t_i,s\in (0,1)$, where$J=[0,1]$.

• (ii) The functions$G_1(t,s)$, $G_2(t,t_i)$and$G_3(t,t_i)$have the following properties:

  $$\frac{b}{a}M(s)\le G_1(t,s)\le M(s),\mathrm{\forall }t\in J,s\in (0,1),$$

  (2.16)

$$\frac{b^2}{{(a-b)}^2}\le G_2(t,t_i)\le \frac{a^2}{{(a-b)}^2},\mathrm{\forall }t,t_i\in J,$$

(2.17)

$$\frac{b}{a-b}\le G_3(t,t_i)\le \frac{a}{a-b},\mathrm{\forall }t,t_i\in J,$$

(2.18)

where

$$M(s)=\frac{a[(1-s)a-(2-s-q)b]{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q)}>0,s\in [0,1).$$

(2.19)

Proof From the expressions of $G_1(t,s)$, $G_2(t,t_i)$ and $G_3(t,t_i)$, it is obvious that $G_1(t,s),G_2(t,t_i),G_3(t,t_i)\in C(J\times J,{\mathbb{R}}^{+})$ and $G_1(t,s),G_2(t,t_i),G_3(t,t_i)>0$ for all $t,t_i,s\in (0,1)$. Next, we will prove (ii). From the definition of$G_1(t,s)$, we can know that, for given $s\in (0,1)$, $G_1(t,s)$ is increasing with respect to t for$t\in J$. We let

$$\begin{array}{c}{g}_1(t,s)=\frac{{(a-b)}^2{(t-s)}^{q-1}+[(a-b)(1-s)+[(a-b)t+b](q-1)]b{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q)},t\in [s,1],\hfill \\ g_2(t,s)=\frac{[(a-b)(1-s)+[(a-b)t+b](q-1)]b{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q)},t\in [0,s].\hfill \end{array}$$

Hence, we derive

$$\begin{array}{c}\begin{array}{rl}\underset{t\in [0,1]}{min}{G}_1(t,s)& =min\{\underset{t\in [s,1]}{min}{g}_1(t,s),\underset{t\in [0,s]}{min}{g}_2(t,s)\}=min\{g_1(s,s),g_2(0,s)\}=g_2(0,s)\\ =\frac{[(a-b)(1-s)+b(q-1)]b{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q)}\\ =\frac{b[(1-s)a-(2-s-q)b]{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q)}\triangleq m(s),s\in [0,1),\end{array}\hfill \\ \begin{array}{rl}\underset{t\in [0,1]}{max}{G}_1(t,s)& =max\{\underset{t\in [s,1]}{max}{g}_1(t,s),\underset{t\in [0,s]}{max}{g}_2(t,s)\}=max\{g_1(1,s),g_2(s,s)\}=g_1(1,s)\\ =\frac{[(a-b)(1-s)+b(q-1)]a{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q)}\\ =\frac{a[(1-s)a-(2-s-q)b]{(1-s)}^{q-2}}{{(a-b)}^2\mathrm{\Gamma }(q)}\triangleq M(s),s\in [0,1).\end{array}\hfill \end{array}$$

Thus, we have

$$\frac{b}{a}M(s)=m(s)\le G_1(t,s)\le M(s).$$

It is obvious that

$$\frac{b^2}{{(a-b)}^2}=G_2(0,1)\le G_2(t,t_i)\le G_2(1,0)=\frac{a^2}{{(a-b)}^2},\frac{b}{a-b}\le G_3(t,t_i)\le \frac{a}{a-b}.$$

The proof is completed. □

3 Existence of single positive solutions

In this section, we discuss the existence of positive solutions for BVP (1.1).

Let $E=\{u(t):u(t)\in C(J)\}$ denote a real Banach space with the norm$\parallel \cdot \parallel$ defined by $\parallel u\parallel ={max}_{t\in J}|u(t)|$. Let

$$\begin{array}{c}\begin{array}{rl}PC(J)=& \{u\in E\mid u:J\to {\mathbb{R}}^{+},u\in C\left(J^{\prime }\right),u\left(t_k^{-}\right)\text{ and }u\left(t_k^{+}\right)\\ \text{exist with }u\left(t_k^{-}\right)=u(t_k),1\le k\le m\},\end{array}\hfill \\ K=\{u\in PC(J):u(t)\ge \frac{b^2}{a^2}\parallel u\parallel ,t\in J\},\hfill \end{array}$$

(3.1)

$$K_r=\{u\in K:\parallel u\parallel <r\},\partial K_r=\{u\in K:\parallel u\parallel =r\}.$$

(3.2)

Obviously, $PC(J)\subset E$ is a Banach space with the norm $\parallel u\parallel ={max}_{t\in J}|u(t)|$. $K\subset PC(J)$ is a positive cone.

In the following, we need the assumptions and some notations as follows:

(B₁) $0<b<a<1$, $0<{\sigma }_1,{\sigma }_2<+\mathrm{\infty }$, where ${\sigma }_1={\int }_0^{1}M(s)ds$, ${\sigma }_2=\frac{b^3}{a^3}{\int }_0^{1}M(s)ds$.

(B₂) $f\in C(J\times {\mathbb{R}}^{+},{\mathbb{R}}^{+})$ and $f(t,0)=0$ for all $t\in J$.

(B₃) $I_k(u(t_k)),J_k(u(t_k))\in C({\mathbb{R}}^{+},{\mathbb{R}}^{+})$, $k=1,2,\dots ,m$.

Let

$$\begin{array}{r}\overline{N}=max\{\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i)),\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\},\\ f^{\delta }=\underset{u\to \delta }{lim\hspace{0.17em}sup}\underset{t\in J}{max}\frac{f(t,u)}{u},f_{\delta }=\underset{u\to \delta }{lim\hspace{0.17em}inf}\underset{t\in J}{min}\frac{f(t,u)}{u},\end{array}$$

where δ denotes 0 or +∞. In addition, we introduce the followingweight functions:

$$\begin{array}{r}\mathrm{\Phi }(r)=max\{f(t,u(t)):(t,u)\in [0,1]\times [\frac{b^2}{a^2}r,r]\},\\ \varphi (r)=min\{f(t,u(t)):(t,u)\in [0,1]\times [\frac{b^2}{a^2}r,r]\}.\end{array}$$

From Lemma 2.4, we can obtain the following lemma.

Lemma 3.1 Suppose that$f(t,u)$is continuous,then$u\in PC(J)$is a solution of BVPs (1.1) if and onlyif$u\in PC(J)$is a solution of the integral equation

$$u(t)={\int }_0^1{G}_1(t,s)f(s,u(s))ds+\sum _{i=1}^m{G}_2(t,t_i)J_i(u(t_i))+\sum _{i=1}^m{G}_3(t,t_i)I_i(u(t_i)),\mathrm{\forall }t\in J.$$

Define $T:PC(J)\to PC(J)$ to be the operator defined as

$$(Tu)(t)={\int }_0^1{G}_1(t,s)f(s,u(s))ds+\sum _{i=1}^m{G}_2(t,t_i)J_i(u(t_i))+\sum _{i=1}^m{G}_3(t,t_i)I_i(u(t_i)).$$

(3.3)

Then, by Lemma 3.1, the existence of solutions for BVPs (1.1) is translated intothe existence of the fixed point for $u=Tu$, where T is given by (3.3). Thus, the fixed pointof the operator T coincides with the solution of problem (1.1).

Lemma 3.2 Assume that (B₁)-(B₃) hold,then$T:PC(J)\to PC(J)$and$T:K\to K$defined by (3.3) are completelycontinuous.

Proof Firstly, we shall show that $T:PC(J)\to PC(J)$ is completely continuous through three steps.

Step 1. Let $u\in PC(J)$, in view of the nonnegativity and continuity of functions$G_1(t,s)$, $G_2(t,t_i)$, $G_3(t,t_i)$, $f(t,u(t))$, $I_k$, $J_k$ and $a>b>0$, we conclude that $T:PC(J)\to PC(J)$ is continuous.

Step 2. We will prove that T maps bounded sets into bounded sets. Indeed, it isenough to show that for any $r>0$ there exists a positive constant L such that, foreach $u\in {\mathrm{\Omega }}_r=\{u\in PC(J):\parallel u\parallel \le r\}$, $\parallel Tu\parallel \le L$ when $|f(t,u)|\le l_1$, $|J_k|\le l_2$, $|I_k|\le l_3$, where $l_i$ ($i=1,2,3$) are some fixed positive constants. In fact, for each$t\in J_k$, $u\in {\mathrm{\Omega }}_r$, $k=0,1,2,\dots ,m$, by Lemma 2.5, we have

$$\begin{array}{rl}|(Tu)(t)|& \le {\int }_0^1|G_1(t,s)f(s,u(s))|ds+\sum _{i=1}^m|G_2(t,t_i)J_i(u(t_i))|+\sum _{i=1}^m|G_3(t,t_i)I_i(u(t_i))|\\ \le {\sigma }_1{l}_1+\frac{a^{2}m{l}_2}{{(a-b)}^2}+\frac{am{l}_3}{a-b}\triangleq L,\end{array}$$

which imply that $\parallel Tu\parallel \le L$.

Step 3. T is equicontinuous. In fact, since $G_1(t,s)$, $G_2(t,t_i)$, $G_3(t,t_i)$ are continuous on $J\times J$, they are uniformly continuous on $J\times J$. Thus, for fixed $s\in J$ and for any $\epsilon >0$, there exists a constant $\delta >0$ such that for any $t_1,t_2\in J_k$ with $|t_1-t_2|<\delta$, $0\le k\le m$, we have

$$\begin{array}{r}|G_1(t_1,s)-G_2(t_2,s)|<\frac{\epsilon }{3l_1},|G_2(t_1,t_i)-G_2(t_2,t_i)|<\frac{\epsilon }{3m{l}_2},\\ |G_3(t_1,t_i)-G_3(t_2,t_i)|<\frac{\epsilon }{3m{l}_3}.\end{array}$$

Then

$$\begin{array}{r}|Tu(t_2)-Tu(t_1)|\\ =|{\int }_0^1(G_1(t_2,s)-G_1(t_1,s))f(s,u(s))ds+\sum _{i=1}^m(G_2(t_2,t_i)-G_2(t_1,t_i))J_i(u(t_i))\\ +\sum _{i=1}^m(G_3(t_2,t_i)-G_3(t_1,t_i))I_i(u(t_i))|\\ \le l_1{\int }_0^1|G_1(t_2,s)-G_2(t_1,s)|ds+m{l}_2|G_2(t_2,t_i)-G_2(t_1,t_i)|\\ +m{l}_3|G_3(t_2,t_i)-G_3(t_1,t_i)|\\ <\frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon ,\end{array}$$

which means that $T({\mathrm{\Omega }}_r)$ is equicontinuous on all the subintervals$t\in J_k$, $k=0,1,\dots ,m$. Thus, by means of the Arzela-Ascoli theorem, we have that$T:PC(J)\to PC(J)$ is completely continuous.

Next, we will show that $T:K\to K$ is completely continuous. Indeed, for each$t\in J_k$, every $u\in C(J_k,{\mathbb{R}}^{+})$, $k=0,1,2,\dots ,m$, Lemma 2.5 implies that

$$(Tu)(t)\ge \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds+\frac{b^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i)).$$

On the other hand,

$$\parallel Tu\parallel =\underset{t\in J_k}{max}(Tu)(t)\le {\int }_0^{1}M(s)f(s,u(s))ds+\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i)).$$

Thus,

$$\begin{array}{rl}\frac{b^2}{a^2}\parallel Tu\parallel & =\frac{b^2}{a^2}\underset{t\in J_k}{max}(Tu)(t)\\ \le \frac{b^2}{a^2}{\int }_0^{1}M(s)f(s,u(s))ds+\frac{b^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{b^2}{a(a-b)}\sum _{i=1}^m{I}_i(u(t_i))\\ \le (Tu)(t).\end{array}$$

So $(Tu)(t)\ge \frac{b^2}{a^2}\parallel Tu\parallel$ for every $u\in C(J,{\mathbb{R}}^{+})$, which implies $T(K)\subset K$. Similar to the above arguments, we can easily concludethat $T:K\to K$ is a completely continuous operator. The proof iscomplete. □

Theorem 3.1 Assume that (B₁)-(B₃) hold,and suppose that the following assumptions hold:

(A₁) There exists a constant$L_1>0$such that$|f(t,u)-f(t,v)|\le L_1|u-v|$for each$t\in J$and all$u,v\in {\mathbb{R}}^{+}$.

(A₂) There exists a constant$L_2>0$such that$|J_k(u)-J_k(v)|\le L_2|u-v|$for all$u,v\in {\mathbb{R}}^{+}$, $k=1,2,\dots ,m$.

(A₃) There exists a constant$L_3>0$such that$|I_k(u)-I_k(v)|\le L_3|u-v|$for all$u,v\in {\mathbb{R}}^{+}$, $k=1,2,\dots ,m$.

If$\rho ={\sigma }_1{L}_1+\frac{m{a}^2{L}_2}{{(a-b)}^2}+\frac{ma{L}_3}{a-b}<1$, then problem (1.1) has a unique solutionin$K_{\rho }$.

Proof Let the operator $T:K_{\rho }\to K_{\rho }$ be defined by (3.3). For all $u,v\in K_{\rho }$, from Lemma 2.5, we obtain

$$\begin{array}{r}|(Tu)(t)-(Tv)(t)|\\ \le {\int }_0^1{G}_1(t,s)|f(s,u(s))-f(s,v(s))|ds\\ +\sum _{i=1}^m{G}_2(t,t_i)|J_i(u(t_i))-J_i(v(t_i))|+\sum _{i=1}^m{G}_3(t,t_i)|I_i(u(t_i))-I_i(v(t_i))|\\ \le {\sigma }_1{L}_1\parallel u-v\parallel +\frac{m{a}^2{L}_2}{{(a-b)}^2}\parallel u-v\parallel +\frac{ma{L}_3}{a-b}\parallel u-v\parallel =\rho \parallel u-v\parallel ,\end{array}$$

where $\rho ={\sigma }_1{L}_1+\frac{m{a}^2{L}_2}{{(a-b)}^2}+\frac{ma{L}_3}{a-b}<1$. Consequently, T is a contraction mapping.Moreover, from Lemma 3.2, T is completely continuous. Therefore, by theBanach contraction map principle, the operator T has a unique fixed point in$K_{\rho }$ which is the unique positive solution of system (1.1).This completes the proof. □

Theorem 3.2 Assume that (B₁)-(B₃) hold,and suppose that the following assumptions hold:

(A₄) There exists a constant$N_1>0$such that$|f(t,u)|\le N_1$for each$t\in J$and all$u\in {\mathbb{R}}^{+}$.

(A₅) There exists a constant$N_2>0$such that$|J_k(u)|\le N_2$for all$u\in {\mathbb{R}}^{+}$, $k=1,2,\dots ,m$.

(A₆) There exists a constant$N_3>0$such that$|I_k(u)|\le N_3$for all$u\in {\mathbb{R}}^{+}$, $k=1,2,\dots ,m$.

Then BVPs (1.1) have at least one positive solutionin$PC(J)$.

Proof Let $T:PC(J)\to PC(J)$ be cone preserving completely continuous that is definedby (3.3). According to Lemma 2.2, now it remains to show that the set

$$\mathrm{\Omega }=\{u\in PC(J)\mid u=\lambda Tu\text{ for some }0<\lambda <1\}$$

(3.4)

is bounded.

Let $u\in \mathrm{\Omega }$, then $u=\lambda Tu$ for some $0<\lambda <1$. Thus, by Lemma 2.5, for each $t\in J_k$, $k=0,1,\dots ,m$, we have

$$\begin{array}{rl}|u(t)|& =|\lambda Tu|\\ \le {\int }_0^1|G_1(t,s)f(s,u(s))|ds+\sum _{i=1}^m|G_2(t,t_i)J_i(u(t_i))|+\sum _{i=1}^m|G_3(t,t_i)I_i(u(t_i))|\\ \le {\sigma }_1{N}_1+\frac{a^{2}m{N}_2}{{(a-b)}^2}+\frac{am{N}_3}{a-b}.\end{array}$$

Thus, for every $t\in J$, we have $\parallel u(t)\parallel \le {\sigma }_1{N}_1+\frac{a^{2}m{N}_2}{{(a-b)}^2}+\frac{am{N}_3}{a-b}$, which indicates that the set Ω is bounded. Accordingto Lemma 2.2, T has a fixed point $u\in PC(J)$. Therefore, BVPs (1.1) have at least one positivesolution. The proof is complete. □

In the following, we present an existence result when the nonlinearity and the impulsefunctions have sublinear growth.

Theorem 3.3 Assume that (B₁)-(B₃) hold andsuppose that the following assumptions hold:

(A₇) There exist$a_1\in PC(J)$, $b_1>0$and$\alpha \in [0,1)$such that$|f(t,u)|\le a_1(t)+b_1{|u|}^{\alpha }$for each$t\in J$and all$u\in {\mathbb{R}}^{+}$.

(A₈) There exist constants$a_2,b_2>0$and$\alpha \in [0,1)$such that$|J_k(u)|\le a_2+b_2{|u|}^{\alpha }$for all$u\in {\mathbb{R}}^{+}$, $k=1,2,\dots ,m$.

(A₉) There exist constants$a_3,b_3>0$and$\alpha \in [0,1)$such that$|I_k(u)|\le a_3+b_3{|u|}^{\alpha }$for all$u\in {\mathbb{R}}^{+}$, $k=1,2,\dots ,m$.

(A₁₀) $b^{\ast }<1$, $a^{\ast }+b^{\ast }\ge 1$, where$a^{\ast }=p{\sigma }_1+\frac{a^{2}m{a}_2}{{(a-b)}^2}+\frac{am{a}_3}{a-b}$, $b^{\ast }=b_1{\sigma }_1+\frac{a^{2}m{b}_2}{{(a-b)}^2}+\frac{am{b}_3}{a-b}$.

Then BVPs (1.1) have at least one positive solutionin$PC(J)$.

Proof Let $T:PC(J)\to PC(J)$ and Ω be defined by (3.3) and (3.4), respectively.Denote $p={max}_{t\in J}|a_1(t)|$. If $u\in \mathrm{\Omega }$, then for $t\in J$ we have

$$\begin{array}{rl}|u(t)|=& |\lambda Tu|\\ \le & {\int }_0^1|G_1(t,s)(a_1(s)+b_1{|u(s)|}^{\alpha })|ds+\sum _{i=1}^m|G_2(t,t_i)(a_2+b_2{|u|}^{\alpha })|\\ +\sum _{i=1}^m|G_3(t,t_i)(a_3+b_3{|u|}^{\alpha })|\\ \le & p{\int }_0^{1}M(s)ds+b_1{\parallel u\parallel }^{\alpha }{\int }_0^{1}M(s)ds+\frac{a^{2}m(a_2+b_2{\parallel u\parallel }^{\alpha })}{{(a-b)}^2}+\frac{am(a_3+b_3{\parallel u\parallel }^{\alpha })}{a-b}\\ =& a^{\ast }+b^{\ast }{\parallel u\parallel }^{\alpha },\end{array}$$

which imply that $\parallel u\parallel \le a^{\ast }+b^{\ast }{\parallel u\parallel }^{\alpha }$. When $0<\parallel u\parallel \le 1$, then $\parallel u\parallel \le a^{\ast }+b^{\ast }$. When $\parallel u\parallel >1$, then $\parallel u\parallel \le \frac{a^{\ast }}{1-b^{\ast }}$. Taking $C=max\{a^{\ast }+b^{\ast },\frac{a^{\ast }}{1-b^{\ast }},\}$, we have $\parallel u\parallel \le C$ for any solution of (3.4). This shows that the set Ωis bounded. According to Lemma 2.2, T has at least one fixed point in$PC(J)$. Therefore, BVPs (1.1) have at least one positive solutionin $PC(J)$. The proof is complete. □

Theorem 3.4 Assume that (B₁)-(B₃) hold.And suppose that one of the following conditions is satisfied:

(H₁) $f^{\mathrm{\infty }}<\frac{1}{{\sigma }_1}$ (particularly, $f^{\mathrm{\infty }}=0$).

(H₂) There exists a constant$M>0$such that$f(t,u)\le \frac{M}{{\sigma }_1}$for$t\in J$, $u\in [M,+\mathrm{\infty })$.

(H₃) There exists a constant$N>0$such that$\mathrm{\Phi }(N)\le \frac{N}{3{\sigma }_1}$for$t\in J$, $u\in [\frac{b^2}{a^2}N,N]$.

Then BVPs (1.1) have at least one positive solution.

Proof Case 1. Considering $f^{\mathrm{\infty }}<\frac{1}{{\sigma }_1}$, there exists ${\overline{r}}_1>0$ such that $f(t,u)\le (f^{\mathrm{\infty }}+{\epsilon }_1)u$ for all $u\in ({\overline{r}}_1,+\mathrm{\infty })$, $t\in J$, where ${\epsilon }_1$ satisfies ${\sigma }_1(f^{\mathrm{\infty }}+{\epsilon }_1)\le 1$.

Choose $r_1>max\{{\overline{r}}_1,2\overline{N}{(1-{\sigma }_1(f^{\mathrm{\infty }}+{\epsilon }_1))}^{-1}\}$, let $u\in {\mathrm{\Omega }}_1\triangleq K_{r_1}$. We can easily know that ${\mathrm{\Omega }}_1$ is a close bounded convex subset of a Banach space$PC(J)$. Then, for $t\in J$, $u\in {\mathrm{\Omega }}_1$, in view of the nonnegativity and continuity of functions$G_1(t,s)$, $G_2(t,t_i)$, $G_3(t,t_i)$, $f(t,u(t))$, $I_k$, $J_k$ and $a>b>0$, we conclude that $Tu\in P$, $Tu\ge 0$, $t\in J$. By Lemma 2.5, we can obtain the followinginequality:

$$\begin{array}{rl}\frac{b^2}{a^2}\parallel Tu\parallel & =\frac{b^2}{a^2}\underset{t\in J}{max}(Tu)(t)\\ \le \frac{b^2}{a^2}[{\int }_0^{1}M(s)f(s,u(s))ds+\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i))]\\ \le \frac{b^2}{a^2}[{\int }_0^{1}M(s)f(s,u(s))ds+\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i))]\\ \le \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds+\frac{b^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \le (Tu)(t),t\in J.\end{array}$$

Thus $Tu\in K$.

Next, we prove $\parallel Tu\parallel \le r_1$. Indeed, for $t\in J$, $u\in \partial K_{r_1}$, we get

$$\begin{array}{rl}\parallel Tu\parallel & =\underset{t\in J}{max}(Tu)(t)\\ \le {\int }_0^{1}M(s)f(s,u(s))ds+\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \le {\int }_0^{1}M(s)(f^{\mathrm{\infty }}+{\epsilon }_1)u(s)ds+\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \le {\sigma }_1(f^{\mathrm{\infty }}+{\epsilon }_1)\parallel u\parallel +2\overline{N}\\ <{\sigma }_1(f^{\mathrm{\infty }}+{\epsilon }_1)r_1+r_1-{\sigma }_1(f^{\mathrm{\infty }}+{\epsilon }_1)r_1=r_1.\end{array}$$

Therefore, $T({\mathrm{\Omega }}_1)\subset {\mathrm{\Omega }}_1$. From Lemma 3.2, we have that $T:{\mathrm{\Omega }}_1\to {\mathrm{\Omega }}_1$ is completely continuous. Thus BVPs (1.1) have at least apositive solution by Lemma 2.3.

Case 2. Condition (H₂) holds. Let $u\in {\mathrm{\Omega }}_2\triangleq K_d$, where $d>0$ satisfies $d\ge 1+M+{\sigma }_1{max}_{t\in J,u\in [0,M]}f(t,u)+2\overline{N}$. By the ways of Case 1, we can also get$Tu\in K$. Now we prove $\parallel Tu\parallel \le d$. In fact,

$$\begin{array}{rl}\parallel Tu\parallel & =\underset{t\in J}{max}(Tu)(t)\\ \le {\int }_0^{1}M(s)f(s,u(s))ds+\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \le {\int }_{s\in J,u(s)>M}M(s)f(s,u(s))ds+{\int }_{s\in J,0\le u(s)\le M}M(s)f(s,u(s))ds+2\overline{N}\\ \le {\int }_0^{1}M(s)\frac{M}{{\sigma }_1}ds+{\int }_0^{1}M(s)ds\underset{t\in J,u\in [0,M]}{max}f(t,u)+2\overline{N}\\ =M+{\sigma }_1\underset{t\in J,u\in [0,M]}{max}f(t,u)+2\overline{N}<d.\end{array}$$

Therefore, $T({\mathrm{\Omega }}_2)\subset {\mathrm{\Omega }}_2$. From Lemma 3.2 we have that $T:{\mathrm{\Omega }}_2\to {\mathrm{\Omega }}_2$ is completely continuous. Thus BVPs (1.1) have at least apositive solution by Lemma 2.3.

Case 3. Condition (H₃) holds. Let $u\in {\mathrm{\Omega }}_3\triangleq K_N$, where $N>0$ satisfies $N\ge 3\overline{N}$, we get $\frac{b^2}{a^2}\parallel u\parallel \le u(t)\le \parallel u\parallel$. By the ways of Case 1, we can also get$Tu\in K$. Now we prove $\parallel Tu\parallel \le N$. By assumption (H₃), we have

$$f(t,u)\le \mathrm{\Phi }(N)\le \frac{N}{3{\sigma }_1},\mathrm{\forall }t\in J,u\in [\frac{b^2}{a^2}N,N].$$

In view of Lemma 2.6, we have

$$\begin{array}{rl}\parallel Tu\parallel & =\underset{t\in J}{max}(Tu)(t)\\ \le {\int }_0^{1}M(s)f(s,u(s))ds+\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \le {\int }_0^{1}M(s)\frac{N}{3{\sigma }_1}ds+2\overline{N}\\ \le \frac{N}{3}+\frac{2N}{3}=N.\end{array}$$

Therefore, $T({\mathrm{\Omega }}_3)\subset {\mathrm{\Omega }}_3$. From Lemma 3.2 we have $T:{\mathrm{\Omega }}_3\to {\mathrm{\Omega }}_3$ is completely continuous. Thus BVPs (1.1) have at least apositive solution by Lemma 2.3. We complete the proof ofTheorem 3.4. □

4 Existence of multiple positive solutions

In this section, we discuss the multiplicity of positive solutions for BVPs (1.1) by theGuo-Krasnoselskii fixed point theorem.

Theorem 4.1 Assume that (B₁)-(B₃) hold,and suppose that the following two conditions are satisfied:

(H₄) $f_0>\frac{1}{{\sigma }_2}$and$f_{\mathrm{\infty }}>\frac{1}{{\sigma }_2}$ (particularly, $f_0=f_{\mathrm{\infty }}=\mathrm{\infty }$).

(H₅) There exists a constant$c\ge 3\overline{N}$such that$\mathrm{\Phi }(c)<\frac{c}{3{\sigma }_1}$for$t\in J$, $u\in [\frac{b^2}{a^2}c,c]$.

Then for BVPs (1.1) there exist at least two positivesolutions$u_1$, $u_2$, which satisfy

$$0<\parallel u_1\parallel <c<\parallel u_2\parallel .$$

(4.1)

Proof Choose r, R with $0<r<c<R$. Considering $f_0>\frac{1}{{\sigma }_2}$, there exists $r>0$ such that $f(t,u)\ge (f_0-{\epsilon }_2)u$ for $t\in J$, $u\in [0,r]$, where ${\epsilon }_2>0$ satisfies $(f_0-{\epsilon }_2){\sigma }_2\ge 1$. Then, for $u\in \partial K_r$, $t\in J$, we have

$$\begin{array}{rl}\parallel Tu\parallel & =\underset{t\in J}{max}(Tu)(t)\\ \ge \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds+\frac{b^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \ge \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds\ge \frac{b}{a}{\int }_0^{1}M(s)(f_0-{\epsilon }_2)u(s)ds\\ \ge \frac{b}{a}{\int }_0^{1}M(s)(f_0-{\epsilon }_2)\frac{b^2}{a^2}\parallel u\parallel ds\\ =(f_0-{\epsilon }_2){\sigma }_2\parallel u\parallel \ge \parallel u\parallel .\end{array}$$

Therefore,

$$\parallel Tu\parallel \ge \parallel u\parallel ,u\in \partial K_r.$$

(4.2)

Considering $f_{\mathrm{\infty }}>\frac{1}{{\sigma }_2}$, there exists $R>0$ such that $f(t,u)\ge (f_{\mathrm{\infty }}-{\epsilon }_3)u$ for $t\in J$, $u\in [R,\mathrm{\infty })$, where ${\epsilon }_3>0$ satisfies $(f_{\mathrm{\infty }}-{\epsilon }_3){\sigma }_2\ge 1$. Then, for $u\in \partial K_R$, $t\in J$, we have

$$\begin{array}{rl}\parallel Tu\parallel & =\underset{t\in J}{max}(Tu)(t)\\ \ge \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds+\frac{b^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \ge \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds\ge \frac{b}{a}{\int }_0^{1}M(s)(f_{\mathrm{\infty }}-{\epsilon }_3)u(s)ds\\ \ge \frac{b}{a}{\int }_0^{1}M(s)(f_{\mathrm{\infty }}-{\epsilon }_3)\frac{b^2}{a^2}\parallel u\parallel ds=(f_{\mathrm{\infty }}-{\epsilon }_3){\sigma }_2\parallel u\parallel \ge \parallel u\parallel .\end{array}$$

So

$$\parallel Tu\parallel \ge \parallel u\parallel ,u\in \partial K_R.$$

(4.3)

On the other hand, by assumption (H₅), we have

$$f(t,u)\le \mathrm{\Phi }(c)<\frac{c}{3{\sigma }_1},\text{for }t\in J,u\in [\frac{b^2}{a^2}c,c].$$

For $u\in \partial K_c$, where $c>0$ satisfies $c\ge 3\overline{N}$. In view of Lemma 2.6, we have

$$\begin{array}{rl}\parallel Tu\parallel & =\underset{t\in J}{max}(Tu)(t)\\ \le {\int }_0^{1}M(s)f(s,u(s))ds+\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ <{\int }_0^{1}M(s)\frac{c}{3{\sigma }_1}ds+2\overline{N}\le \frac{c}{3}+\frac{2c}{3}=c=\parallel u\parallel .\end{array}$$

Therefore,

$$\parallel Tu\parallel <\parallel u\parallel ,u\in \partial K_c.$$

(4.4)

Thus, applying Lemma 2.4 to (4.2)-(4.4) yields that T has the fixed point$u_1\in K\cap ({\overline{K}}_c\setminus K_r)$ and the fixed point $u_2\in K\cap ({\overline{K}}_R\setminus K_c)$. Thus it follows that problem (1.1) has at least twopositive solutions $u_1$ and $u_2$. Noticing (4.4), we have $\parallel u_1\parallel \ne c$ and $\parallel u_2\parallel \ne c$. Therefore (4.1) holds. The proof iscomplete. □

Theorem 4.2 Assume that (B₁)-(B₃) hold.Further suppose that there exist three positivenumbers${\xi }_i$ ($i=1,2,3$) with$3\overline{N}\le {\xi }_1<{\xi }_2<{\xi }_3$such that one of the following conditions issatisfied:

(H₆) $\varphi ({\xi }_1)\ge \frac{{\xi }_1}{{\sigma }_2}$, $\mathrm{\Phi }({\xi }_2)\le \frac{{\xi }_2}{3{\sigma }_1}$, $\varphi ({\xi }_3)\ge \frac{{\xi }_3}{{\sigma }_2}$.

(H₇) $\mathrm{\Phi }({\xi }_1)\le \frac{{\xi }_1}{3{\sigma }_1}$, $\varphi ({\xi }_2)>\frac{{\xi }_2}{{\sigma }_2}$, $\mathrm{\Phi }({\xi }_3)\le \frac{{\xi }_3}{3{\sigma }_1}$.

Then BVPs (1.1) have at least two positivesolutions$u_1$, $u_2$with

$${\xi }_1\le \parallel u_1\parallel <{\xi }_2<\parallel u_2\parallel \le {\xi }_3.$$

(4.5)

Proof Because the proofs are similar, we prove only case (H₆).Considering $\varphi ({\xi }_1)\ge \frac{{\xi }_1}{{\sigma }_2}$, we have $f(t,u)\ge \varphi ({\xi }_1)\ge \frac{{\xi }_1}{{\sigma }_2}$ for $t\in J$, $u\in [\frac{b^2}{a^2}{\xi }_1,{\xi }_1]$. Then, for $u\in \partial K_{{\xi }_1}$, $t\in J$, we have

$$\begin{array}{rl}|Tu\parallel & =\underset{t\in J}{max}(Tu)(t)\\ \ge \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds+\frac{b^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \ge \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds\ge \frac{b}{a}{\int }_0^{1}M(s)\frac{{\xi }_1}{{\sigma }_2}ds\\ \ge \frac{b^3}{a^3}{\int }_0^{1}M(s)ds\frac{{\xi }_1}{{\sigma }_2}={\xi }_1=\parallel u\parallel .\end{array}$$

Therefore,

$$\parallel Tu\parallel \ge \parallel u\parallel ,u\in \partial K_{{\xi }_1}.$$

(4.6)

Considering $\mathrm{\Phi }({\xi }_2)\le \frac{{\xi }_2}{3{\sigma }_1}$, we have $f(t,u)\le \mathrm{\Phi }({\xi }_2)\le \frac{{\xi }_2}{3{\sigma }_1}$ for $t\in J$, $u\in [\frac{b^2}{a^2}{\xi }_2,{\xi }_2]$. Then, for $u\in \partial K_{{\xi }_2}$, $t\in J$, we derive

$$\begin{array}{rl}\parallel Tu\parallel & =\underset{t\in J}{max}(Tu)(t)\\ \le {\int }_0^{1}M(s)f(s,u(s))ds+\frac{a^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{a}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \le {\int }_0^{1}M(s)\frac{{\xi }_2}{3{\sigma }_1}ds+2\overline{N}\le \frac{{\xi }_2}{3}+\frac{2{\xi }_1}{3}<\frac{{\xi }_2}{3}+\frac{2{\xi }_2}{3}={\xi }_2=\parallel u\parallel .\end{array}$$

So,

$$\parallel Tu\parallel <\parallel u\parallel ,u\in \partial K_{{\xi }_2}.$$

(4.7)

Considering $\varphi ({\xi }_3)\ge \frac{{\xi }_3}{{\sigma }_2}$, we have $f(t,u)\ge \varphi ({\xi }_3)\ge \frac{{\xi }_3}{{\sigma }_2}$ for $t\in J$, $u\in [\frac{b^2}{a^2}{\xi }_3,{\xi }_3]$. Then, for $u\in \partial K_{{\xi }_3}$, $t\in J$, we have

$$\begin{array}{rl}\parallel Tu\parallel & =\underset{t\in J}{max}(Tu)(t)\\ \ge \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds+\frac{b^2}{{(a-b)}^2}\sum _{i=1}^m{J}_i(u(t_i))+\frac{b}{a-b}\sum _{i=1}^m{I}_i(u(t_i))\\ \ge \frac{b}{a}{\int }_0^{1}M(s)f(s,u(s))ds\ge \frac{b}{a}{\int }_0^{1}M(s)\frac{{\xi }_3}{{\sigma }_2}ds\ge \frac{b^3}{a^3}{\int }_0^{1}M(s)ds\frac{{\xi }_3}{{\sigma }_2}={\xi }_3=\parallel u\parallel .\end{array}$$

Therefore,

$$\parallel Tu\parallel \ge \parallel u\parallel ,u\in \partial K_{{\xi }_3}.$$

(4.8)

Thus, applying Lemma 2.4 to (4.6)-(4.8) yields that T has the fixed point$u_1\in K\cap ({\overline{K}}_{{\xi }_2}\setminus K_{{\xi }_1})$ and the fixed point $u_2\in K\cap ({\overline{K}}_{{\xi }_3}\setminus K_{{\xi }_2})$. Thus it follows that BVPs (1.1) have at least twopositive solutions $u_1$ and $u_2$. Noticing (4.7), we have $\parallel u_1\parallel \ne {\xi }_2$ and $\parallel u_2\parallel \ne {\xi }_2$. Therefore (4.5) holds. The proof iscomplete. □

Similar to the above proof, we can obtain the general theorem.

Theorem 4.3 Assume that (B₁)-(B₃) hold.Suppose that there exist$n+1$positive numbers${\xi }_i$ ($i=1,2,\dots ,n+1$) with$3\overline{N}\le {\xi }_1<{\xi }_2<\cdots <{\xi }_{n+1}$such that one of the following conditions issatisfied:

(H₈) $\varphi ({\xi }_{2j-1})>\frac{{\xi }_{2j-1}}{{\sigma }_2}$, $\mathrm{\Phi }({\xi }_{2j})<\frac{{\xi }_{2j}}{3{\sigma }_1}$, $j=1,2,\dots ,[\frac{n+2}{2}]$;

(H₉) $\mathrm{\Phi }({\xi }_{2j-1})<\frac{{\xi }_{2j-1}}{3{\sigma }_1}$, $\varphi ({\xi }_{2j})>\frac{{\xi }_{2j}}{{\sigma }_2}$, $j=1,2,\dots ,[\frac{n+2}{2}]$.

Then BVPs (1.1) have at least n positivesolutions$u_i$ ($i=1,2,\dots ,n$) with

$${\xi }_i<\parallel u_i\parallel <{\xi }_{i+1}.$$

(4.9)

5 Illustrative examples

Example 5.1 Consider the BVPs of impulsive nonlinear fractional orderdifferential equations:

$$\{\begin{array}{l}{}^{c}D_t^{q}u(t)=f(t,u(t)),t\in J,t\ne \frac{1}{2},\\ \mathrm{\Delta }u(\frac{1}{2})=I(u(\frac{1}{2})),\mathrm{\Delta }{u}^{\prime }(\frac{1}{2})=J(u(\frac{1}{2})),\\ au(0)-bu(1)=0,a{u}^{\prime }(0)-b{u}^{\prime }(1)=0.\end{array}$$

(5.1)

If we let $q=\frac{3}{2}$, $a=2$, $b=1$, $f(t,u)=\frac{\mathrm{\Gamma }(\frac{3}{2})cost}{{(t+2\sqrt{5})}^2}\frac{u(t)}{1+u(t)}$, $(t,u)\in [0,1]\times [0,\mathrm{\infty })$, $I(u)=\frac{u}{5+u}$, $J(u)=\frac{u}{10+u}$, $u\in [0,\mathrm{\infty })$.

For $u,v\in [0,\mathrm{\infty })$, $t\in [0,1]$,

$$\begin{array}{c}|f(t,u)-f(t,v)|\le \left|\frac{\mathrm{\Gamma }(\frac{3}{2})cost}{{(t+2\sqrt{5})}^2}\right|\left|\frac{u-v}{(1+u)(1+v)}\right|\le \frac{\mathrm{\Gamma }(\frac{3}{2})}{20}|u-v|,\hfill \\ |I(u)-I(v)|\le \frac{5}{(5+u)(5+v)}|u-v|\le \frac{1}{5}|u-v|,\hfill \\ |J(u)-J(v)|\le \frac{10}{(10+u)(10+v)}|u-v|\le \frac{1}{10}|u-v|.\hfill \end{array}$$

Clearly, $L_1=\frac{\mathrm{\Gamma }(\frac{3}{2})}{20}$, $L_2=\frac{1}{10}$, $L_3=\frac{1}{5}$. Therefore,

$$\rho ={\sigma }_1{L}_1+\frac{m{a}^2{L}_2}{{(a-b)}^2}+\frac{ma{L}_3}{a-b}=\frac{10}{3\mathrm{\Gamma }(\frac{3}{2})}\frac{\mathrm{\Gamma }(\frac{3}{2})}{20}+\frac{2}{5}+\frac{2}{5}=\frac{29}{30}<1.$$

Thus, all the assumptions of Theorem 3.1 are satisfied. Hence, BVPs (5.1) have aunique solution on $[0,1]$.

In addition, in this case, let $N_1=\frac{\mathrm{\Gamma }(\frac{3}{2})}{20}$, $N_2=N_3=1$. It is clear that $|f(t,u)|\le N_1$, $|J_k(u)|\le N_2$, $|I_k(u)|\le N_3$. Thus, BVPs (5.1) have at least one solution on$[0,1]$ by Theorem 3.2.

Example 5.2 Consider the BVPs of impulsive nonlinear fractional orderdifferential equations:

$$\{\begin{array}{l}{}^{c}D_t^{q}u(t))=f(t,u(t),t\in J,t\ne \frac{1}{2},\\ \mathrm{\Delta }u(\frac{1}{2})=I(u(\frac{1}{2})),\mathrm{\Delta }{u}^{\prime }(\frac{1}{2})=J(u(\frac{1}{2})),\\ au(0)-bu(1)=0,a{u}^{\prime }(0)-b{u}^{\prime }(1)=0.\end{array}$$

(5.2)

Let $a=2$, $b=1$, $q=\frac{3}{2}$, $f(t,u)=|\frac{u(t)lnu(t)}{5(1+t^2)}|$, $I(u)=J(u)=\frac{1}{16(1+u)}$. It is easy to see that (H₄) holds. By a simplecomputation, we have

$$\begin{array}{r}{f}_0=\underset{u\to 0}{lim\hspace{0.17em}inf}\underset{t\in [0,1]}{min}\left|\frac{ulnu}{5(1+t^2)u}\right|=\underset{u\to 0}{lim\hspace{0.17em}inf}\frac{|lnu|}{10}=+\mathrm{\infty },\\ f_{\mathrm{\infty }}=\underset{u\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\underset{t\in [0,1]}{min}\left|\frac{ulnu}{5(1+t^2)u}\right|=\underset{u\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{|lnu|}{10}=+\mathrm{\infty }.\end{array}$$

Take $c=1$, it is clear that $3\overline{N}<\frac{3}{4}<c$. For $\frac{1}{4}\le u\le 1$, $f(t,u)=\frac{-ulnu}{5(1+t^2)}$, we can obtain that $f(t,u)$ arrives at maximum at $u=\frac{1}{e}\in [\frac{1}{4},1]$, $t=0$. Thus, we have

$$\mathrm{\Phi }(1)=\underset{t\in [0,1],u\in [\frac{1}{4},1]}{max}f(t,u)=f(0,\frac{1}{e})=\frac{1}{5e}\approx 0.0736<\frac{1}{3{\sigma }_1}=\frac{\sqrt{\pi }}{20}\approx 0.0886.$$

Thus it follows that BVPs (5.2) have at least two positive solutions$u_1$, $u_2$ with $0<\parallel u_1\parallel <1<\parallel u_2\parallel$ by Theorem 4.1.