• Research
• Open Access

# Existence of solutions for a coupled system of fractional differential equations at resonance

Boundary Value Problems20122012:98

https://doi.org/10.1186/1687-2770-2012-98

• Accepted: 22 August 2012
• Published:

## Abstract

In this paper, by using the coincidence degree theory, we study the existence of solutions for a coupled system of fractional differential equations at resonance. A new result on the existence of solutions for a fractional boundary value problem is obtained.

MSC:34B15.

## Keywords

• fractional differential equation
• boundary value problem
• coincidence degree theory
• resonance

## 1 Introduction

In recent years, the fractional differential equations have received more and more attention. The fractional derivative has been occurring in many physical applications such as a non-Markovian diffusion process with memory , charge transport in amorphous semiconductors , propagations of mechanical waves in viscoelastic media , etc. Phenomena in electromagnetics, acoustics, viscoelasticity, electrochemistry and material science are also described by differential equations of fractional order (see ).

Recently, boundary value problems for fractional differential equations have been studied in many papers (see ). Moreover, the existence of solutions to a coupled systems of fractional differential equations have been studied by many authors (see ). But the existence of solutions for a coupled system of fractional differential equations at resonance are seldom considered. Motivated by all the works above, in this paper, we consider the following boundary value problem (BVP for short) for a coupled system of fractional differential equations given by
$\left\{\begin{array}{c}{D}_{{0}^{+}}^{\alpha }u\left(t\right)=f\left(t,v\left(t\right),{v}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right),\hfill \\ {D}_{{0}^{+}}^{\beta }v\left(t\right)=g\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right),\hfill \\ u\left(0\right)=v\left(0\right)=0,\phantom{\rule{2em}{0ex}}{u}^{\prime }\left(0\right)={u}^{\prime }\left(1\right),\phantom{\rule{2em}{0ex}}{v}^{\prime }\left(0\right)={v}^{\prime }\left(1\right),\hfill \end{array}$
(1.1)

where ${D}_{{0}^{+}}^{\alpha }$, ${D}_{{0}^{+}}^{\beta }$ are the standard Caputo fractional derivatives, $1<\alpha \le 2$, $1<\beta \le 2$ and $f,g:\left[0,1\right]×{\mathbb{R}}^{2}\to \mathbb{R}$ is continuous.

The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions and lemmas. In Section 3, we establish a theorem on the existence of solutions for BVP (1.1) under nonlinear growth restriction of f and g, basing on the coincidence degree theory due to Mawhin (see ). Finally, in Section 4, an example is given to illustrate the main result.

## 2 Preliminaries

In this section, we will introduce some notations, definitions and preliminary facts which are used throughout this paper.

Let X and Y be real Banach spaces, and let $L:domL\subset X\to Y$ be a Fredholm operator with index zero, and $P:X\to X$, $Q:Y\to Y$ be projectors such that
$\begin{array}{c}ImP=KerL,\phantom{\rule{2em}{0ex}}KerQ=ImL,\hfill \\ X=KerL\oplus KerP,\phantom{\rule{2em}{0ex}}Y=ImL\oplus ImQ.\hfill \end{array}$
It follows that
$L{|}_{domL\cap KerP}:domL\cap KerP\to ImL$

is invertible. We denote the inverse by ${K}_{P}$.

If Ω is an open bounded subset of X, and $domL\cap \overline{\mathrm{\Omega }}\ne \mathrm{\varnothing }$, the map $N:X\to Y$ will be called L-compact on $\overline{\mathrm{\Omega }}$ if $QN\left(\overline{\mathrm{\Omega }}\right)$ is bounded and ${K}_{P}\left(I-Q\right)N:\overline{\mathrm{\Omega }}\to X$ is compact, where I is an identity operator.

Lemma 2.1 

Let $L:domL\subset X\to Y$ be a Fredholm operator of index zero and $N:X\to Y$ L-compact on $\overline{\mathrm{\Omega }}$. Assume that the following conditions are satisfied:
1. (1)

$Lx\ne \lambda Nx$ for every $\left(x,\lambda \right)\in \left[\left(domL\setminus KerL\right)\right]\cap \partial \mathrm{\Omega }×\left(0,1\right)$;

2. (2)

$Nx\notin ImL$ for every $x\in KerL\cap \partial \mathrm{\Omega }$;

3. (3)

$deg\left(QN{|}_{KerL},KerL\cap \mathrm{\Omega },0\right)\ne 0$, where $Q:Y\to Y$ is a projection such that $ImL=KerQ$.

Then the equation $Lx=Nx$ has at least one solution in $domL\cap \overline{\mathrm{\Omega }}$.

Definition 2.1 The Riemann-Liouville fractional integral operator of order $\alpha >0$ of a function x is given by
${I}_{{0}^{+}}^{\alpha }x\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -1}x\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,$

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

Definition 2.2 The Riemann-Liouville fractional derivative of order $\alpha >0$ of a function x is given by
${}^{R}D_{{0}^{+}}^{\alpha }x\left(t\right)=\frac{{d}^{n}}{d{t}^{n}}{I}_{{0}^{+}}^{n-\alpha }x\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(n-\alpha \right)}\frac{{d}^{n}}{d{t}^{n}}{\int }_{0}^{t}{\left(t-s\right)}^{n-\alpha -1}x\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,$

where n is the smallest integer greater than or equal to α, provided that the right-hand side integral is pointwise defined on $\left(0,+\mathrm{\infty }\right)$.

Definition 2.3 The Caputo fractional derivative of order $\alpha >0$ of a function x is given by
${D}_{{0}^{+}}^{\alpha }x\left(t\right){=}^{R}{D}_{{0}^{+}}^{\alpha }\left[x\left(t\right)-\sum _{k=0}^{n-1}\frac{{x}^{\left(k\right)}\left(0\right)}{k!}{t}^{k}\right],$

where n is the smallest integer greater than or equal to α, provided that the right-hand side integral is pointwise defined on $\left(0,+\mathrm{\infty }\right)$.

Lemma 2.2 

Assume that $x\in C\left(0,1\right)\cap L\left(0,1\right)$ with a Caputo fractional derivative of order $\alpha >0$ that belongs to $C\left(0,1\right)\cap L\left(0,1\right)$. Then
${I}_{{0}^{+}}^{\alpha }{D}_{{0}^{+}}^{\alpha }x\left(t\right)=x\left(t\right)+{c}_{0}+{c}_{1}t+{c}_{2}{t}^{2}+\cdots +{c}_{n-1}{t}^{n-1},$

where ${c}_{i}\in \mathbb{R}$, $i=0,1,2,\dots ,n-1$, here n is the smallest integer greater than or equal to α.

Lemma 2.3 

Assume that $\alpha >0$ and $x\in C\left[0,1\right]$. Then
${D}_{{0}^{+}}^{\alpha }{I}_{{0}^{+}}^{\alpha }x\left(t\right)=x\left(t\right).$

In this paper, we denote $X={C}^{1}\left[0,1\right]$ with the norm ${\parallel x\parallel }_{X}=max\left\{{\parallel x\parallel }_{\mathrm{\infty }},{\parallel {x}^{\prime }\parallel }_{\mathrm{\infty }}\right\}$ and $Y=C\left[0,1\right]$ with the norm ${\parallel y\parallel }_{Y}={\parallel y\parallel }_{\mathrm{\infty }}$, where ${\parallel x\parallel }_{\mathrm{\infty }}={max}_{t\in \left[0,1\right]}|x\left(t\right)|$. Then we denote $\overline{X}=X×X$ with the norm ${\parallel \left(u,v\right)\parallel }_{\overline{X}}=max\left\{{\parallel u\parallel }_{X},{\parallel v\parallel }_{X}\right\}$ and $\overline{Y}=Y×Y$ with the norm ${\parallel \left(x,y\right)\parallel }_{\overline{Y}}=max\left\{{\parallel x\parallel }_{Y},{\parallel y\parallel }_{Y}\right\}$. Obviously, both $\overline{X}$ and $\overline{Y}$ are Banach spaces.

Define the operator ${L}_{1}:domL\subset X\to Y$ by
${L}_{1}u={D}_{{0}^{+}}^{\alpha }u,$
where
$dom{L}_{1}=\left\{u\in X|{D}_{{0}^{+}}^{\alpha }u\left(t\right)\in Y,u\left(0\right)=0,{u}^{\prime }\left(0\right)={u}^{\prime }\left(1\right)\right\}.$
Define the operator ${L}_{2}:dom{L}_{2}\subset X\to Y$ by
${L}_{2}v={D}_{{0}^{+}}^{\beta }v,$
where
$dom{L}_{2}=\left\{v\in X|{D}_{{0}^{+}}^{\beta }v\left(t\right)\in Y,v\left(0\right)=0,{v}^{\prime }\left(0\right)={v}^{\prime }\left(1\right)\right\}.$
Define the operator $L:domL\subset \overline{X}\to \overline{Y}$ by
$L\left(u,v\right)=\left({L}_{1}u,{L}_{2}v\right),$
(2.1)
where
$domL=\left\{\left(u,v\right)\in \overline{X}|u\in dom{L}_{1},v\in dom{L}_{2}\right\}.$
Let $N:\overline{X}\to \overline{Y}$ be the Nemytski operator
$N\left(u,v\right)=\left({N}_{1}v,{N}_{2}u\right),$
where ${N}_{1}:Y\to X$
${N}_{1}v\left(t\right)=f\left(t,v\left(t\right),{v}^{\prime }\left(t\right)\right)$
and ${N}_{2}:Y\to X$
${N}_{2}u\left(t\right)=g\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right).$
Then BVP (1.1) is equivalent to the operator equation
$L\left(u,v\right)=N\left(u,v\right),\phantom{\rule{1em}{0ex}}\left(u,v\right)\in domL.$

## 3 Main result

In this section, a theorem on the existence of solutions for BVP (1.1) will be given.

Theorem 3.1 Let $f,g:\left[0,1\right]×{\mathbb{R}}^{2}\to \mathbb{R}$ be continuous. Assume that

(${H}_{1}$) there exist nonnegative functions ${p}_{i},{q}_{i},{r}_{i}\in C\left[0,1\right]$ ($i=1,2$) with
$\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)-4\left({Q}_{1}+{R}_{1}\right)\left({Q}_{2}+{R}_{2}\right)}{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)}>0$
such that for all $\left(u,v\right)\in {\mathbb{R}}^{2}$, $t\in \left[0,1\right]$
$|f\left(t,u,v\right)|\le {p}_{1}\left(t\right)+{q}_{1}\left(t\right)|u|+{r}_{1}\left(t\right)|v|,$
and
$|g\left(t,u,v\right)|\le {p}_{2}\left(t\right)+{q}_{2}\left(t\right)|u|+{r}_{2}\left(t\right)|v|,$

where ${P}_{i}={\parallel {p}_{i}\parallel }_{\mathrm{\infty }}$, ${Q}_{i}={\parallel {q}_{i}\parallel }_{\mathrm{\infty }}$, ${R}_{i}={\parallel {r}_{i}\parallel }_{\mathrm{\infty }}$ ($i=1,2$);

(${H}_{2}$) there exists a constant $B>0$ such that for $\mathrm{\forall }t\in \left[0,1\right]$, $|u|>B$, $v\in \mathbb{R}$ either
$uf\left(t,u,v\right)>0,\phantom{\rule{2em}{0ex}}ug\left(t,u,v\right)>0,$
or
$uf\left(t,u,v\right)<0,\phantom{\rule{2em}{0ex}}ug\left(t,u,v\right)<0;$
(${H}_{3}$) there exists a constant $D>0$ such that for every ${c}_{1},{c}_{2}\in \mathbb{R}$ satisfying $min\left\{{c}_{1},{c}_{2}\right\}>D$ either
${c}_{1}{N}_{1}\left({c}_{2}t\right)>0,\phantom{\rule{2em}{0ex}}{c}_{2}{N}_{2}\left({c}_{1}t\right)>0,$
or
${c}_{1}{N}_{1}\left({c}_{2}t\right)<0,\phantom{\rule{2em}{0ex}}{c}_{2}{N}_{2}\left({c}_{1}t\right)<0.$

Then BVP (1.1) has at least one solution.

Now, we begin with some lemmas below.

Proof By Lemma 2.2, ${L}_{1}u={D}_{{0}^{+}}^{\alpha }u\left(t\right)=0$ has the solution
$u\left(t\right)={c}_{0}+{c}_{1}t,\phantom{\rule{1em}{0ex}}{c}_{0},{c}_{1}\in \mathbb{R}.$
Combining it with the boundary value conditions of BVP (1.1), one has
$Ker{L}_{1}=\left\{u\in X|u={c}_{1}t,{c}_{1}\in \mathbb{R}\right\}.$
For $x\in Im{L}_{1}$, there exists $u\in dom{L}_{1}$ such that $x={L}_{1}u\in Y$. By Lemma 2.2, we have
$u\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -1}x\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+{c}_{0}+{c}_{1}t.$
Then, we have
${u}^{\prime }\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}x\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+{c}_{1}.$
By the conditions of BVP (1.1), we can get that x satisfies
${\int }_{0}^{1}{\left(1-s\right)}^{\alpha -2}x\left(s\right)\phantom{\rule{0.2em}{0ex}}ds=0.$
On the other hand, suppose $x\in Y$ and satisfies ${\int }_{0}^{1}{\left(1-s\right)}^{\alpha -2}x\left(s\right)\phantom{\rule{0.2em}{0ex}}ds=0$. Let $u\left(t\right)={I}_{{0}^{+}}^{\alpha }x\left(t\right)$, then $u\in dom{L}_{1}$. By Lemma 2.3, we have ${D}_{{0}^{+}}^{\alpha }u\left(t\right)=x\left(t\right)$ so that $x\in Im{L}_{1}$. Then we have
$Im{L}_{1}=\left\{x\in Y|{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -2}x\left(s\right)\phantom{\rule{0.2em}{0ex}}ds=0\right\}.$
Similarly, we can get
$\begin{array}{c}Ker{L}_{2}=\left\{v\in X|v={c}_{2}t,{c}_{2}t\in \mathbb{R}\right\},\hfill \\ Im{L}_{2}=\left\{y\in Y|{\int }_{0}^{1}{\left(1-s\right)}^{\beta -2}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds=0\right\}.\hfill \end{array}$

Then, the proof is complete. □

Lemma 3.2 Let L be defined by (2.1), then L is a Fredholm operator of index zero, and the linear continuous projector operators $P:\overline{X}\to \overline{X}$ and $Q:\overline{Y}\to \overline{Y}$ can be defined as
$\begin{array}{c}P\left(u,v\right)=\left({P}_{1}u,{P}_{2}v\right)=\left({u}^{\prime }\left(0\right)t,{v}^{\prime }\left(0\right)t\right),\hfill \\ Q\left(x,y\right)=\left({Q}_{1}x,{Q}_{2}y\right)=\left(\left(\alpha -1\right){\int }_{0}^{1}{\left(1-s\right)}^{\alpha -2}x\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\left(\beta -1\right){\int }_{0}^{1}{\left(1-s\right)}^{\beta -2}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right).\hfill \end{array}$
Furthermore, the operator ${K}_{P}:ImL\to domL\cap KerP$ can be written by
${K}_{P}\left(x,y\right)=\left({I}_{{0}^{+}}^{\alpha }x\left(t\right),{I}_{{0}^{+}}^{\beta }y\left(t\right)\right).$
Proof Obviously, $ImP=KerL$ and ${P}^{2}\left(u,v\right)=P\left(u,v\right)$. It follows from $\left(u,v\right)=\left(\left(u,v\right)-P\left(u,v\right)\right)+P\left(u,v\right)$ that $\overline{X}=KerP+KerL$. By simple calculation, we can get that $KerL\cap KerP=\left\{\left(0,0\right)\right\}$. Then we get
$\overline{X}=KerL\oplus KerP.$
For $\left(x,y\right)\in \overline{Y}$, we have
${Q}^{2}\left(x,y\right)=Q\left({Q}_{1}x,{Q}_{2}y\right)=\left({Q}_{1}^{2}x,{Q}_{2}^{2}y\right).$
By the definition of ${Q}_{1}$, we can get
${Q}_{1}^{2}x={Q}_{1}x\cdot \left(\alpha -1\right){\int }_{0}^{1}{\left(1-s\right)}^{\alpha -2}\phantom{\rule{0.2em}{0ex}}ds={Q}_{1}x.$

Similar proof can show that ${Q}_{2}^{2}y={Q}_{2}y$. Thus, we have ${Q}^{2}\left(x,y\right)=Q\left(x,y\right)$.

Let $\left(x,y\right)=\left(\left(x,y\right)-Q\left(x,y\right)\right)+Q\left(x,y\right)$, where $\left(x,y\right)-Q\left(x,y\right)\in KerQ=ImL,Q\left(x,y\right)\in ImQ$. It follows from $KerQ=ImL$ and ${Q}^{2}\left(x,y\right)=Q\left(x,y\right)$ that $ImQ\cap ImL=\left\{\left(0,0\right)\right\}$. Then, we have
$\overline{Y}=ImL\oplus ImQ.$
Thus
$dimKerL=dimImQ=codimImL.$

This means that L is a Fredholm operator of index zero.

Now, we will prove that ${K}_{P}$ is the inverse of $L{|}_{domL\cap KerP}$. By Lemma 2.3, for $\left(x,y\right)\in ImL$, we have
$L{K}_{P}\left(x,y\right)=\left({D}_{{0}^{+}}^{\alpha }\left({I}_{{0}^{+}}^{\alpha }x\right),{D}_{{0}^{+}}^{\beta }\left({I}_{{0}^{+}}^{\beta }y\right)\right)=\left(x,y\right).$
(3.3)
Moreover, for $\left(u,v\right)\in domL\cap KerP$, we have ${u}^{\prime }\left(0\right)={v}^{\prime }\left(0\right)=0$ and
$\begin{array}{rcl}{K}_{P}L\left(u,v\right)& =& \left({I}_{{0}^{+}}^{\alpha }{D}_{{0}^{+}}^{\alpha }u\left(t\right),{I}_{{0}^{+}}^{\beta }{D}_{{0}^{+}}^{\beta }v\left(t\right)\right)\\ =& \left(u\left(t\right)+{c}_{0}+{c}_{1}t,v\left(t\right)+{d}_{0}+{d}_{1}t\right),\phantom{\rule{1em}{0ex}}{c}_{0},{c}_{1},{d}_{0},{d}_{1}\in \mathbb{R},\end{array}$
which, together with $u\left(0\right)=v\left(0\right)=0$, yields that
${K}_{P}L\left(u,v\right)=\left(u,v\right).$
(3.4)

Combining (3.3) with (3.4), we know that ${K}_{P}={\left(L{|}_{domL\cap KerP}\right)}^{-1}$. The proof is complete. □

Lemma 3.3 Assume $\mathrm{\Omega }\subset \overline{X}$ is an open bounded subset such that $domL\cap \overline{\mathrm{\Omega }}\ne \mathrm{\varnothing }$, then N is L-compact on $\overline{\mathrm{\Omega }}$.

Proof By the continuity of f and g, we can get that $QN\left(\overline{\mathrm{\Omega }}\right)$ and ${K}_{P}\left(I-Q\right)N\left(\overline{\mathrm{\Omega }}\right)$ are bounded. So, in view of the Arzelá-Ascoli theorem, we need only prove that ${K}_{P}\left(I-Q\right)N\left(\overline{\mathrm{\Omega }}\right)\subset \overline{X}$ is equicontinuous.

From the continuity of f and g, there exists a constant ${A}_{i}>0$, $i=1,2$, such that $\mathrm{\forall }\left(u,v\right)\in \overline{\mathrm{\Omega }}$
$|\left(I-{Q}_{1}\right){N}_{1}v|\le {A}_{1},\phantom{\rule{2em}{0ex}}|\left(I-{Q}_{2}\right){N}_{2}u|\le {A}_{2}.$
Furthermore, for $0\le {t}_{1}<{t}_{2}\le 1$, $\left(u,v\right)\in \overline{\mathrm{\Omega }}$, we have
$\begin{array}{c}|{K}_{P}\left(I-Q\right)N\left(u\left({t}_{2}\right),v\left({t}_{2}\right)\right)-\left({K}_{P}\left(I-Q\right)N\left(u\left({t}_{1}\right),v\left({t}_{1}\right)\right)\right)|\hfill \\ \phantom{\rule{1em}{0ex}}=|\left({I}_{{0}^{+}}^{\alpha }\left(I-{Q}_{1}\right){N}_{1}v\left({t}_{2}\right),{I}_{{0}^{+}}^{\beta }\left(I-{Q}_{2}\right){N}_{2}u\left({t}_{2}\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\left({I}_{{0}^{+}}^{\alpha }\left(I-{Q}_{1}\right){N}_{1}v\left({t}_{1}\right),{I}_{{0}^{+}}^{\beta }\left(I-{Q}_{2}\right){N}_{2}u\left({t}_{1}\right)\right)|\hfill \\ \phantom{\rule{1em}{0ex}}=|\left({I}_{{0}^{+}}^{\alpha }\left(I-{Q}_{1}\right){N}_{1}v\left({t}_{2}\right)-{I}_{{0}^{+}}^{\alpha }\left(I-{Q}_{1}\right){N}_{1}v\left({t}_{1}\right),\hfill \\ \phantom{\rule{2em}{0ex}}{I}_{{0}^{+}}^{\beta }\left(I-{Q}_{2}\right){N}_{2}u\left({t}_{2}\right)-{I}_{{0}^{+}}^{\beta }\left(I-{Q}_{2}\right){N}_{2}u\left({t}_{1}\right)\right)|.\hfill \end{array}$
By
$\begin{array}{c}|{I}_{{0}^{+}}^{\alpha }\left(I-{Q}_{1}\right){N}_{1}v\left({t}_{2}\right)-{I}_{{0}^{+}}^{\alpha }\left(I-{Q}_{1}\right){N}_{1}v\left({t}_{1}\right)|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}|{\int }_{0}^{{t}_{2}}{\left({t}_{2}-s\right)}^{\alpha -1}\left(I-{Q}_{1}\right){N}_{1}v\left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{{t}_{1}}{\left({t}_{1}-s\right)}^{\alpha -1}\left(I-{Q}_{1}\right){N}_{1}v\left(s\right)\phantom{\rule{0.2em}{0ex}}ds|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{{A}_{1}}{\mathrm{\Gamma }\left(\alpha \right)}\left[{\int }_{0}^{{t}_{1}}{\left({t}_{2}-s\right)}^{\alpha -1}-{\left({t}_{1}-s\right)}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{t}_{1}}^{{t}_{2}}{\left({t}_{2}-s\right)}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}ds\right]\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{A}_{1}}{\mathrm{\Gamma }\left(\alpha +1\right)}\left({t}_{2}^{\alpha }-{t}_{1}^{\alpha }\right)\hfill \end{array}$
and
$\begin{array}{c}|{\left({I}_{{0}^{+}}^{\alpha }\left(I-{Q}_{1}\right){N}_{1}v\right)}^{\prime }\left({t}_{2}\right)-{\left({I}_{{0}^{+}}^{\alpha }\left(I-{Q}_{1}\right){N}_{1}v\right)}^{\prime }\left({t}_{1}\right)|\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{\alpha -1}{\mathrm{\Gamma }\left(\alpha \right)}|{\int }_{0}^{{t}_{2}}{\left({t}_{2}-s\right)}^{\alpha -2}\left(I-{Q}_{1}\right){N}_{1}v\left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{{t}_{1}}{\left({t}_{1}-s\right)}^{\alpha -2}\left(I-{Q}_{1}\right){N}_{1}v\left(s\right)\phantom{\rule{0.2em}{0ex}}ds|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{{A}_{1}}{\mathrm{\Gamma }\left(\alpha -1\right)}\left[{\int }_{0}^{{t}_{1}}{\left({t}_{1}-s\right)}^{\alpha -2}-{\left({t}_{2}-s\right)}^{\alpha -2}\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{t}_{1}}^{{t}_{2}}{\left({t}_{2}-s\right)}^{\alpha -2}\phantom{\rule{0.2em}{0ex}}ds\right]\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{{A}_{1}}{\mathrm{\Gamma }\left(\alpha \right)}\left[{t}_{2}^{\alpha -1}-{t}_{1}^{\alpha -1}+2{\left({t}_{2}-{t}_{1}\right)}^{\alpha -1}\right].\hfill \end{array}$
Similar proof can show that
$\begin{array}{c}|{I}_{{0}^{+}}^{\beta }\left(I-{Q}_{2}\right){N}_{2}u\left({t}_{2}\right)-{I}_{{0}^{+}}^{\beta }\left(I-{Q}_{2}\right){N}_{2}u\left({t}_{1}\right)|\le \frac{{A}_{2}}{\mathrm{\Gamma }\left(\beta +1\right)}\left({t}_{2}^{\beta }-{t}_{1}^{\beta }\right),\hfill \\ |{\left({I}_{{0}^{+}}^{\beta }\left(I-{Q}_{2}\right){N}_{2}u\right)}^{\prime }\left({t}_{2}\right)-{\left({I}_{{0}^{+}}^{\beta }\left(I-{Q}_{2}\right){N}_{2}u\right)}^{\prime }\left({t}_{1}\right)|\le \frac{{A}_{2}}{\mathrm{\Gamma }\left(\beta \right)}\left[{t}_{2}^{\beta -1}-{t}_{1}^{\beta -1}+2{\left({t}_{2}-{t}_{1}\right)}^{\beta -1}\right].\hfill \end{array}$

Since ${t}^{\alpha }$, ${t}^{\alpha -1}$, ${t}^{\beta }$ and ${t}^{\beta -1}$ are uniformly continuous on $\left[0,1\right]$, we can get that ${K}_{P}\left(I-Q\right)N\left(\overline{\mathrm{\Omega }}\right)\subset \overline{X}$ is equicontinuous.

Thus, we get that ${K}_{P}\left(I-Q\right)N:\overline{\mathrm{\Omega }}\to \overline{X}$ is compact. The proof is complete. □

Lemma 3.4 Suppose (${H}_{1}$), (${H}_{2}$) hold, then the set
${\mathrm{\Omega }}_{1}=\left\{\left(u,v\right)\in domL\setminus KerL|L\left(u,v\right)=\lambda N\left(u,v\right),\lambda \in \left(0,1\right)\right\}$

is bounded.

Proof Take $\left(u,v\right)\in {\mathrm{\Omega }}_{1}$, then $N\left(u,v\right)\in ImL$. By (3.2), we have
$\begin{array}{c}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -2}f\left(s,v\left(s\right),{v}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds=0,\hfill \\ {\int }_{0}^{1}{\left(1-s\right)}^{\beta -2}g\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds=0.\hfill \end{array}$

Then, by the integral mean value theorem, there exist constants $\xi ,\eta \in \left(0,1\right)$ such that $f\left(\xi ,v\left(\xi \right),{v}^{\prime }\left(\xi \right)\right)=0$ and $g\left(\eta ,u\left(\eta \right),{u}^{\prime }\left(\eta \right)\right)=0$. So, from (${H}_{2}$), we get $|v\left(\xi \right)|\le B$ and $|u\left(\eta \right)|\le B$.

From $\left(u,v\right)\in domL$, we get $u\left(0\right)=v\left(0\right)=0$, then
By $L\left(u,v\right)=\lambda N\left(u,v\right)$ and $\left(u,v\right)\in domL$, we have
$u\left(t\right)=\frac{\lambda }{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -1}f\left(s,v\left(s\right),{v}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{u}^{\prime }\left(0\right)t$
and
$v\left(t\right)=\frac{\lambda }{\mathrm{\Gamma }\left(\beta \right)}{\int }_{0}^{t}{\left(t-s\right)}^{\beta -1}g\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{v}^{\prime }\left(0\right)t.$
Then we get
${u}^{\prime }\left(t\right)=\frac{\lambda }{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}f\left(s,v\left(s\right),{v}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{u}^{\prime }\left(0\right)$
and
${v}^{\prime }\left(t\right)=\frac{\lambda }{\mathrm{\Gamma }\left(\beta -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\beta -2}g\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{v}^{\prime }\left(0\right).$
Take $t=\eta$, we get
${u}^{\prime }\left(\eta \right)=\frac{\lambda }{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{\eta }{\left(\eta -s\right)}^{\alpha -2}f\left(s,v\left(s\right),{v}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{u}^{\prime }\left(0\right).$
Together with $|{u}^{\prime }\left(\eta \right)|\le B$, (${H}_{1}$) and (3.6), we have
$\begin{array}{rcl}|{u}^{\prime }\left(0\right)|& \le & |{u}^{\prime }\left(\eta \right)|+\frac{\lambda }{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{\eta }{\left(\eta -s\right)}^{\alpha -2}|f\left(s,v\left(s\right),{v}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \le & B+\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{\eta }{\left(\eta -s\right)}^{\alpha -2}\left[{p}_{1}\left(s\right)+{q}_{1}\left(s\right)|v\left(s\right)|+{r}_{1}\left(s\right)|{v}^{\prime }\left(s\right)|\right]\phantom{\rule{0.2em}{0ex}}ds\\ \le & B+\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{\eta }{\left(\eta -s\right)}^{\alpha -2}\left[{P}_{1}+{Q}_{1}{\parallel v\parallel }_{\mathrm{\infty }}+{R}_{1}{\parallel {v}^{\prime }\parallel }_{\mathrm{\infty }}\right]\phantom{\rule{0.2em}{0ex}}ds\\ \le & B+\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{\eta }{\left(\eta -s\right)}^{\alpha -2}\left[{P}_{1}+\left({Q}_{1}+{R}_{1}\right){\parallel {v}^{\prime }\parallel }_{\mathrm{\infty }}\right]\phantom{\rule{0.2em}{0ex}}ds\\ \le & B+\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}\left[{P}_{1}+\left({Q}_{1}+{R}_{1}\right){\parallel {v}^{\prime }\parallel }_{\mathrm{\infty }}\right].\end{array}$
So, we have
$\begin{array}{rcl}{\parallel {u}^{\prime }\parallel }_{\mathrm{\infty }}& \le & \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}|f\left(s,v\left(s\right),{v}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds+|{u}^{\prime }\left(0\right)|\\ \le & \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\left[{p}_{1}\left(s\right)+{q}_{1}\left(s\right)|v\left(s\right)|+{r}_{1}\left(s\right)|{v}^{\prime }\left(s\right)|\right]\phantom{\rule{0.2em}{0ex}}ds+|{u}^{\prime }\left(0\right)|\\ \le & \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}\left[{P}_{1}+\left({Q}_{1}+{R}_{1}\right){\parallel {v}^{\prime }\parallel }_{\mathrm{\infty }}\right]{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\phantom{\rule{0.2em}{0ex}}ds+|{u}^{\prime }\left(0\right)|\\ \le & B+\frac{2}{\mathrm{\Gamma }\left(\alpha \right)}\left[{P}_{1}+\left({Q}_{1}+{R}_{1}\right){\parallel {v}^{\prime }\parallel }_{\mathrm{\infty }}\right].\end{array}$
(3.7)
Similarly, we can get
${\parallel {v}^{\prime }\parallel }_{\mathrm{\infty }}\le B+\frac{2}{\mathrm{\Gamma }\left(\beta \right)}\left[{P}_{2}+\left({Q}_{2}+{R}_{2}\right){\parallel {u}^{\prime }\parallel }_{\mathrm{\infty }}\right].$
(3.8)
Together with (3.7) and (3.8), we have
${\parallel {u}^{\prime }\parallel }_{\mathrm{\infty }}\le B+\frac{2}{\mathrm{\Gamma }\left(\alpha \right)}\left\{{P}_{1}+\left({Q}_{1}+{R}_{1}\right)\left[B+\frac{2}{\mathrm{\Gamma }\left(\beta \right)}\left({P}_{2}+\left({Q}_{2}+{R}_{2}\right){\parallel {u}^{\prime }\parallel }_{\mathrm{\infty }}\right)\right]\right\}.$
Thus, from $\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)-4\left({Q}_{1}+{R}_{1}\right)\left({Q}_{2}+{R}_{2}\right)}{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)}>0$, we obtain that
${\parallel {u}^{\prime }\parallel }_{\mathrm{\infty }}\le \frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)B+2\mathrm{\Gamma }\left(\beta \right)\left[{P}_{1}+\left({Q}_{1}+{R}_{1}\right)B\right]+4{P}_{2}\left({Q}_{1}+{R}_{1}\right)}{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)-4\left({Q}_{1}+{R}_{1}\right)\left({Q}_{2}+{R}_{2}\right)}:={M}_{1}$
and
${\parallel {v}^{\prime }\parallel }_{\mathrm{\infty }}\le \frac{1}{\mathrm{\Gamma }\left(\beta \right)}\left[{P}_{2}+{Q}_{2}B+\left({Q}_{2}+{R}_{2}\right){M}_{1}\right]:={M}_{2}.$
Together with (3.5) and (3.6), we get
${\parallel \left(u,v\right)\parallel }_{\overline{X}}\le max\left\{{M}_{1},{M}_{2}\right\}:=M.$

So ${\mathrm{\Omega }}_{1}$ is bounded. The proof is complete. □

Lemma 3.5 Suppose (${H}_{3}$) holds, then the set
${\mathrm{\Omega }}_{2}=\left\{\left(u,v\right)|\left(u,v\right)\in KerL,N\left(u,v\right)\in ImL\right\}$

is bounded.

Proof For $\left(u,v\right)\in {\mathrm{\Omega }}_{2}$, we have $\left(u,v\right)=\left({c}_{1}t,{c}_{2}t\right)$, ${c}_{1},{c}_{2}\in \mathbb{R}$. Then from $N\left(u,v\right)\in ImL$, we get
$\begin{array}{c}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -2}f\left(s,{c}_{2}s,{c}_{2}\right)\phantom{\rule{0.2em}{0ex}}ds=0,\hfill \\ {\int }_{0}^{1}{\left(1-s\right)}^{\beta -2}g\left(s,{c}_{1}s,{c}_{1}\right)\phantom{\rule{0.2em}{0ex}}ds=0,\hfill \end{array}$
which, together with (${H}_{3}$), implies $|{c}_{1}|,|{c}_{2}|\le D$. Thus, we have
${\parallel \left(u,v\right)\parallel }_{\overline{X}}\le D.$

Hence, ${\mathrm{\Omega }}_{2}$ is bounded. The proof is complete. □

Lemma 3.6 Suppose the first part of (${H}_{3}$) holds, then the set
${\mathrm{\Omega }}_{3}=\left\{\left(u,v\right)\in KerL|\lambda \left(u,v\right)+\left(1-\lambda \right)QN\left(u,v\right)=\left(0,0\right),\lambda \in \left[0,1\right]\right\}$

is bounded.

Proof For $\left(u,v\right)\in {\mathrm{\Omega }}_{3}$, we have $\left(u,v\right)=\left({c}_{1}t,{c}_{2}t\right)$, ${c}_{1},{c}_{2}\in \mathbb{R}$ and
If $\lambda =0$, then $|{c}_{1}|,|{c}_{2}|\le D$ because of the first part of (${H}_{3}$). If $\lambda =1$, then ${c}_{1}={c}_{2}=0$. For $\lambda \in \left(0,1\right]$, we can obtain $|{c}_{1}|,|{c}_{2}|\le D$. Otherwise, if $|{c}_{1}|$ or $|{c}_{2}|>D$, in view of the first part of (${H}_{3}$), one has
$\lambda {c}_{1}^{2}t+\left(1-\lambda \right)\left(\alpha -1\right){\int }_{0}^{1}{\left(1-s\right)}^{\alpha -2}{c}_{1}f\left(s,{c}_{2}s,{c}_{2}\right)\phantom{\rule{0.2em}{0ex}}ds>0,$
or
$\lambda {c}_{2}^{2}t+\left(1-\lambda \right)\left(\beta -1\right){\int }_{0}^{1}{\left(1-s\right)}^{\beta -2}{c}_{2}g\left(s,{c}_{1}s,{c}_{1}\right)\phantom{\rule{0.2em}{0ex}}ds>0,$

which contradicts (3.9) or (3.10). Therefore, ${\mathrm{\Omega }}_{3}$ is bounded. The proof is complete. □

Remark 3.1 If the second part of (${H}_{3}$) holds, then the set
${\mathrm{\Omega }}_{3}^{\prime }=\left\{\left(u,v\right)\in KerL|-\lambda \left(u,v\right)+\left(1-\lambda \right)QN\left(u,v\right)=\left(0,0\right),\lambda \in \left[0,1\right]\right\}$

is bounded.

Proof of Theorem 3.1 Set $\mathrm{\Omega }=\left\{\left(u,v\right)\in \overline{X}|{\parallel \left(u,v\right)\parallel }_{\overline{X}}. It follows from Lemma 3.2 and 3.3 that L is a Fredholm operator of index zero and N is L-compact on $\overline{\mathrm{\Omega }}$. By Lemma 3.4 and 3.5, we get that the following two conditions are satisfied:
1. (1)

$L\left(u,v\right)\ne \lambda N\left(u,v\right)$ for every $\left(\left(u,v\right),\lambda \right)\in \left[\left(domL\setminus KerL\right)\cap \partial \mathrm{\Omega }\right]×\left(0,1\right)$;

2. (2)

$Nx\notin ImL$ for every $\left(u,v\right)\in KerL\cap \partial \mathrm{\Omega }$.

Take
$H\left(\left(u,v\right),\lambda \right)=±\lambda \left(u,v\right)+\left(1-\lambda \right)QN\left(u,v\right).$
According to Lemma 3.6 (or Remark 3.1), we know that $H\left(\left(u,v\right),\lambda \right)\ne 0$ for $\left(u,v\right)\in KerL\cap \partial \mathrm{\Omega }$. Therefore,
$\begin{array}{rcl}deg\left(QN{|}_{KerL},\mathrm{\Omega }\cap KerL,\left(0,0\right)\right)& =& deg\left(H\left(\cdot ,0\right),\mathrm{\Omega }\cap KerL,\left(0,0\right)\right)\\ =& deg\left(H\left(\cdot ,1\right),\mathrm{\Omega }\cap KerL,\left(0,0\right)\right)\\ =& deg\left(±I,\mathrm{\Omega }\cap KerL,\left(0,0\right)\right)\ne 0.\end{array}$

So, the condition (3) of Lemma 2.1 is satisfied. By Lemma 2.1, we can get that $L\left(u,v\right)=N\left(u,v\right)$ has at least one solution in $domL\cap \overline{\mathrm{\Omega }}$. Therefore, BVP (1.1) has at least one solution. The proof is complete. □

## 4 Example

Example 4.1 Consider the following BVP:
$\left\{\begin{array}{c}{D}_{{0}^{+}}^{\frac{3}{2}}u\left(t\right)=\frac{1}{16}\left[v\left(t\right)-10\right]+\frac{{t}^{2}}{16}{e}^{-|{v}^{\prime }\left(t\right)|},\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],\hfill \\ {D}_{{0}^{+}}^{\frac{5}{4}}v\left(t\right)=\frac{1}{12}\left[u\left(t\right)-8\right]+\frac{{t}^{3}}{12}{sin}^{2}\left({u}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],\hfill \\ u\left(0\right)=v\left(0\right)=0,\phantom{\rule{2em}{0ex}}{u}^{\prime }\left(0\right)={u}^{\prime }\left(1\right),\phantom{\rule{2em}{0ex}}{v}^{\prime }\left(0\right)={v}^{\prime }\left(1\right).\hfill \end{array}$
(4.1)

Choose ${p}_{1}\left(t\right)=\frac{11}{16}$, ${p}_{2}\left(t\right)=\frac{3}{4}$, ${q}_{1}\left(t\right)=\frac{1}{16}$, ${q}_{2}\left(t\right)=\frac{1}{12}$, ${r}_{1}\left(t\right)={r}_{2}\left(t\right)=0$, $B=D=10$.

By simple calculation, we can get that (${H}_{1}$), (${H}_{2}$) and the first part of (${H}_{3}$) hold.

By Theorem 3.1, we obtain that BVP (4.1) has at least one solution.

## Declarations

### Acknowledgements

The authors would like to thank the referees very much for their helpful comments and suggestions. This research was supported by the Fundamental Research Funds for the Central Universities (2010LKSX09) and the National Natural Science Foundation of China (11271364).

## Authors’ Affiliations

(1)
Department of Mathematics, China University of Mining and Technology, Xuzhou, 221008, People’s Republic of China

## References 