Introduction

Fractional calculus still gaining increasing attention even it is an old mathematical topic, this is due to their multidisciplinary applications such in viscoelasticity, electrochemistry, electromagnetism, rheology...[1]. It is shown that the mean advantages of fractional order derivatives and integrals is their ability to describe memory and hereditary properties of different materials, for more details see [1–3]. Let us note that there exists abundant literature concerning the relation between fractional derivatives and fractional powers of operators [4, 5], fractional partial differential equations [6] and boundary value problems involving fractional calculus [7–13].

The study of systems of fractional differential equations has also known an increasing interest since the behavior of many physical systems can be properly described by fractional-order systems, for example, dielectric polarization, electromagnetic waves, viscoelastic systems, diffusion waves [14]. For some recent results on systems of fractional differential equations and their applications, we refer to [14, 15].

In this work, we consider the following system of fractional differential equations with integral boundary conditions:

$$\begin{aligned} (S):\left\{ \begin{array}{l} \textstyle {}^{c}D_{0^{+}}^{q}u(t)=g(t)f\left( u(t)\right) ,\quad 0<t<1, \\ \textstyle u^{\prime }(0)=0,\quad Eu\left( 0\right) -Bu\left( 1\right) =\int _{0}^{1}h\left( u(s)\right) ds, \end{array} \right. \end{aligned}$$

where \(f,h: {\mathbb {R}} ^{n}\rightarrow {\mathbb {R}} ^{n}\) and \(g:\left[ 0,1\right] \rightarrow {\mathbb {R}} \) are given functions, \(u:\left[ 0,1\right] \rightarrow {\mathbb {R}} ^{n}\) is the unknown function, \(^{c}D_{0^{+}}^{q}\) denotes the Caputo’s fractional derivative, \(1<q<2\). Denote \(u=\left( u_{1},u_{2},\ldots ,u_{n}\right) ^{T}\),

$$\begin{aligned} f\left( u\right)= & {} \left( f_{1}\left( u_{1},u_{2},\ldots ,u_{n}\right) ,\ldots ,f_{n}\left( u_{1},u_{2},\ldots ,u_{n}\right) \right) ^{T}, \\ h\left( u\right)= & {} \left( h_{1}\left( u_{1},u_{2},\ldots ,u_{n}\right) ,\ldots ,h_{n}\left( u_{1},u_{2},\ldots ,u_{n}\right) \right) ^{T}. \end{aligned}$$

Denote by E the identity matrix in \(M_{n}\left( {\mathbb {R}} \right) \) (The vector space of real matrices with n rows and n columns) and \(B=diag\left( \lambda _{1},\ldots \lambda _{n}\right) \in M_{n}\left( {\mathbb {R}} \right) ,\,0<\lambda _{i}<1,\,\forall i\in \left\{ 1,\ldots ,n\right\} \).

Similar boundary value problem for systems of nonlinear fractional differential equations has been widely studied by many authors, but most of them obtained only some existence results without any information about the positivity of solutions, see [1, 5, 8, 16–20].

Boundary value problem with integral boundary conditions is a mathematical model for of various phenomena of physics, ecology, biology, chemistry, etc. Integral conditions come up when values of the function on the boundary is connected to values inside the domain or when direct measurements on the boundary are not possible, see [13, 21–25]

Many methods are used to investigate the existence of solutions for boundary value problems, one can cite fixed point theory, the upper and lower solution method, the variational method...We refer the reader to [7–12, 22–31] for recent developments in this area. On the other hand, fixed point theory is a very powerful mathematical tool in the study of boundary value problems where the existence, uniqueness, positivity and stability knowledge are needed.

This paper is organized as follows: We state in Sect. 2 some background materials and preliminaries. Main results and their proofs are exposed in Sect. 3, which we achieved by an example illustrating the obtained results.

Background Materials and Preliminaries

We recall some basic definitions of the fractional integrals and derivatives. For more details on fractional calculus, see Kilbas et al. [2].

Definition 1

The Riemann-Liouville fractional integral of order \(\alpha \) of a function g is defined by

$$\begin{aligned} I_{a^{+}}^{\alpha }g(t)=\frac{1}{\Gamma \left( \alpha \right) }\int _{a}^{t} \frac{g(s)}{(t-s)^{1-\alpha }}ds, \end{aligned}$$

where \(\Gamma \left( \alpha \right) =\int _{0}^{+\infty }\mathrm{e}^{-t}t^{\alpha -1}dt \) is the Gamma function, \(\alpha >0\).

Definition 2

Let \(q\ge 0,n=[q]+1\). If \(g\in C^{n}[a,b]\), then the Caputo fractional derivative of order q of g defined by

$$\begin{aligned} ^{c}D_{a^{+}}^{q}g(t)=\frac{1}{\Gamma \left( n-q\right) }\int _{a}^{t}\frac{ g^{\left( n\right) }(s)}{(t-s)^{q-n+1}}ds, \end{aligned}$$

exists almost everywhere on \([a,b]\,([q]\) is the entire part of q).

Lemma 3

For \(q>0,\,g\in C(0,1)\), the homogenous fractional differential equation \(^{c}D_{a^{+}}^{q}g(t)=0\) has a solution

$$\begin{aligned} g(t)=c_{1}+c_{2}t+c_{3}t^{2}+\cdots +c_{n}t^{n-1}, \end{aligned}$$

where \(c_{i}\in {\mathbb {R}} ,\,i=0,\ldots ,n\), and \(n=[q]+1\).

Lemma 4

Let \(p,q\ge 0,\,f\in L_{1}[a,b]\). Then \(I_{0^{+}}^{p}I_{0^{+}}^{q}f(t)=I_{0^{+}}^{p+q}f(t)=I_{0^{+}}^{q} I_{0^{+}}^{p}f(t)\) and \(^{c}D_{a^{+}}^{q}I_{0^{+}}^{q}f(t)=f(t)\), for all \(t\in [a,b]\).

Let X be the Banach space of all functions \(u\in C^{n}[0,1]=C[0,1]\times \cdots \times C[0,1]\) with the norm \(\left| \left| .\right| \right| \) defined by \(\left| \left| u\right| \right| =\sum _{i=1}^{i=n}{\max }_{t\in \left[ 0,1\right] }\left| u_{i}\left( t\right) \right| \).

The system (S) can be turned to the following system of n equations with nonlocal conditions:

$$\begin{aligned} (S_{i}):\left\{ \begin{array}{l} \textstyle {}^{c}D_{0^{+}}^{q}u_{i}(t)=g(t)f_{i}\left( u(t)\right) ,\qquad 0<t<1, \\ \textstyle u_{i}^{\prime }(0)=0,\quad u_{i}\left( 0\right) -\lambda _{i}u\left( 1\right) =\int _{0}^{1}h_{i}\left( u(s)\right) ds, \quad i\in \left\{ 1,\ldots ,n\right\} . \end{array} \right. \end{aligned}$$

Lemma 5

If \(0<\lambda _{i}<1,\,\forall i\in \left\{ 1,\ldots ,n\right\} \), then the linear nonhomogeneous problem

$$\begin{aligned} (S_{i}):\left\{ \begin{array}{l} \textstyle {} ^{c}D_{0^{+}}^{q}u_{i}(t)=y(t),\quad 0<t<1, \\ \textstyle u_{i}^{\prime }(0)=0,\quad u_{i}\left( 0\right) -\lambda _{i}u_{i}\left( 1\right) =\int _{0}^{1}z(s)ds, \end{array} \right. \end{aligned}$$

has the following solution

$$\begin{aligned} u_{i}(t)=\frac{1}{\Gamma \left( q\right) }\int _{0}^{1}G_{i}\left( t,s\right) y(s)ds+\frac{1}{1-\lambda _{i}}\int _{0}^{1}z\left( s\right) ds, \end{aligned}$$

where

$$\begin{aligned} \ G_{i}(t,s)=\left\{ \begin{array}{l} \textstyle (t-s)^{q-1}+\frac{\lambda _{i}}{1-\lambda _{i}}\left( 1-s\right) ^{q-1}, \quad s\le t, \\ \textstyle \frac{\lambda _{i}}{1-\lambda _{i}}\left( 1-s\right) ^{q-1},\quad t\le s. \end{array} \right. \end{aligned}$$

Define the integral operator \(T:X\rightarrow X\) by \(Tu=\left( T_{1}u,T_{2}u,\ldots ,T_{n}u\right) \) where

$$\begin{aligned} T_{i}u(t)=\frac{1}{\Gamma \left( q\right) }\int _{0}^{1}g(s)G_{i}\left( t,s\right) f_{i}\left( u\left( s\right) \right) ds+\frac{1}{1-\lambda _{i}} \int _{0}^{1}h_{i}\left( u\left( s\right) \right) ds, \end{aligned}$$

then the system (S) is equivalent to a system of integral equations:

$$\begin{aligned} Tu(t)=\frac{1}{\Gamma \left( q\right) }\int _{0}^{1}g(s)G\left( t,s\right) f\left( u\left( s\right) \right) ds+\int _{0}^{1}H_{\lambda }\left( u\left( s\right) \right) ds, \end{aligned}$$

where the matrix \(G\left( t,s\right) =diag\left( G_{1}\left( t,s\right) ,\ldots ,G_{n}\left( t,s\right) \right) \) and

$$\begin{aligned} \int _{0}^{1}H_{\lambda }\left( u\left( s\right) \right) ds=\left( \frac{1}{ 1-\lambda _{1}}\int _{0}^{1}h_{1}\left( u\left( s\right) \right) ds,\ldots , \frac{1}{1-\lambda _{n}}\int _{0}^{1}h_{n}\left( u\left( s\right) \right) ds\right) ^{T}. \end{aligned}$$

Lemma 6

The function \(u\in X\) is solution of the system (S) if and only if \(T_{i}u(t)=u(t)\), for all \(t\in \left[ 0,1\right] ,\forall i\in \left\{ 1,\ldots ,n\right\} \).

Consequently, existence of solutions for the system (S) can be turned into a fixed point problem in X for the operator T.

Define a positive solution as:

Definition 7

A function u is called positive solution of the system (S) if \(u_{i}(t)\ge 0,\forall t\in \left[ 0,1\right] ,\,\forall i\in \left\{ 1,\ldots ,n\right\} \) and it satisfies the boundary conditions in \((S_{i})\).

Main Results

Let us state the properties of the Green functions:

Lemma 8

For all \(t,s\in [0,1]\) and \(\forall i\in \left\{ 1,\ldots ,n\right\} \), we have the following:

  1. (i)

    \(G_{i}(t,s)\in C\left( [0,1]\times [0,1]\right) ,\,G_{i}(t,s)\ge 0\).

  2. (ii)

    \(\lambda _{i}\gamma _{i}(s)\le G_{i}(t,s)\le \gamma _{i}(s)\), where \( \gamma _{i}(s)=\frac{(1-s)^{q-1}}{1-\lambda _{i}}\).

Now we state the assumptions that will be used to prove the existence of positive solutions:

$$\begin{aligned} \mathbf {(K}_{1}\mathbf {)}&\mathbf {:}0<\lambda _{i}<1,f_{i}\in C\left( {\mathbb {R}}_{+}^{n},{\mathbb {R}}_{+}\right) , \quad \forall i\in \left\{ 1,\ldots ,n\right\} , \\ \text { }g&\in L^{1}\left( \left[ 0,1\right] ,{\mathbb {R}}_{+}\right) ,\text { }\int _{0}^{1}(1-s)^{q-1}g(s)ds>0.\\ \mathbf {(K}_{2})&:h_{i}\in C\left( {\mathbb {R}}_{+}^{n},{\mathbb {R}} _{+}\right) \text {, }h_{i}(v_{1},\ldots ,v_{n})\le l\left( v_{1}+\cdots +v_{n}\right) , \ \\ \forall i&\in \left\{ 1,\ldots ,n\right\} ,\text { }l<\left( \sum _{i=1}^{n}\frac{ 1}{1-\lambda _{i}}\right) ^{-1}. \end{aligned}$$

Lemma 9

Assume that \((K_{1})\) and \((K_{2})\) hold. Then the solution u of the system \(\left( S\right) \) satisfies

$$\begin{aligned} \min _{t\in \left[ 0,1\right] }\sum _{i=1}^{n}u_{i}(t)\ge \left( \min _{1\le i\le n}\lambda _{i}\right) \left\| u\right\| . \end{aligned}$$

Proof

For all \(i\in \left\{ 1,\ldots ,n\right\} \) we have

$$\begin{aligned} u_{i}(t)=\frac{1}{\Gamma \left( q\right) }\int _{0}^{1}G_{i}\left( t,s\right) g(s)f_{i}\left( u\left( s\right) \right) ds+\frac{1}{1-\lambda _{i}} \int _{0}^{1}h_{i}\left( u\left( s\right) \right) ds. \end{aligned}$$

By Lemma 8 we obtain

$$\begin{aligned} u_{i}(t)\le \frac{1}{\Gamma \left( q\right) }\int _{0}^{1}\gamma _{i}\left( s\right) g(s)f_{i}\left( u\left( s\right) \right) ds+\frac{1}{1-\lambda _{i}} \int _{0}^{1}h_{i}\left( u\left( s\right) \right) ds. \end{aligned}$$
(3.1)

Taking the supremum over \(\left[ 0,1\right] \), then summing the obtained inequalities according to i from 1 to n, we get

$$\begin{aligned} \left\| u\right\| \le \frac{1}{\Gamma \left( q\right) } \sum _{i=1}^{n}\left( \int _{0}^{1}\gamma _{i}\left( s\right) g(s)f_{i}\left( u\left( s\right) \right) ds+\frac{1}{1-\lambda _{i}}\int _{0}^{1}h_{i}\left( u\left( s\right) \right) ds\right) . \end{aligned}$$
(3.2)

On the other hand, taking into account the left hand side of inequalities in (ii) of Lemma 8 and the fact that \(0<\lambda _{i}<1\), it yields

$$\begin{aligned} u_{i}(t)\ge & {} \frac{\lambda _{i}}{\Gamma \left( q\right) } \int _{0}^{1}\gamma _{i}\left( s\right) g(s)f_{i}\left( u\left( s\right) \right) ds+\frac{1}{1-\lambda _{i}}\int _{0}^{1}h_{i}\left( u\left( s\right) \right) ds \\\ge & {} \frac{\lambda _{i}}{\Gamma \left( q\right) }\left( \int _{0}^{1}\gamma _{i}\left( s\right) g(s)f_{i}\left( u\left( s\right) \right) ds+\frac{1}{ \left( 1-\lambda _{i}\right) }\int _{0}^{1}h_{i}\left( u\left( s\right) \right) ds\right) \\\ge & {} \left( \min _{1\le i\le n}\lambda _{i}\right) \frac{1}{\Gamma \left( q\right) }\left( \int _{0}^{1}\gamma _{i}\left( s\right) g(s)f_{i}\left( u\left( s\right) \right) ds+\frac{1}{\left( 1-\lambda _{i}\right) } \int _{0}^{1}h_{i}\left( u\left( s\right) \right) ds\right) . \end{aligned}$$

Summing the n obtained inequalities, taking the minimum over t on \(\left[ 0,1\right] \), then applying (3.2) we get the desired result. The proof is complete. \(\square \)

Let us introduce the following notations

$$\begin{aligned} f_{0}^{i}=\underset{\left\| u\right\| \rightarrow 0^{+}}{\lim }\frac{ f_{i}\left( u\right) }{\left\| u\right\| }, \quad f_{\infty }^{i}=\underset{ \left\| u\right\| \rightarrow +\infty }{\lim }\frac{f_{i}\left( u\right) }{\left\| u\right\| }, \quad f_{0}=\sum _{i=0}^{n}f_{0}^{i},f_{\infty }=\sum _{i=0}^{n}f_{\infty }^{i}. \end{aligned}$$

The case \(f_{0}=0\) and \(f_{\infty }=+\infty \) is called superlinear case and the case \(f_{0}=+\infty \) and \(f_{\infty }=0\) is called sublinear case. Let K be the cone:

$$\begin{aligned} K= & {} \Big \{ u=\left( u_{1},\ldots ,u_{2}\right) \in X,u_{i}\left( t\right) \ge 0,\quad \forall t\in \left[ 0,1\right] ,\quad \forall i\in \left\{ 1,\ldots ,n\right\} , \\&\min _{t\in \left[ 0,1\right] }\sum _{i=1}^{n}u_{i}(t)\ge \Big ( \min _{1\le i\le n}\lambda _{i}\Big ) \left\| u\right\| \Big \} . \end{aligned}$$

Lemma 10

The map \(T:X\rightarrow X\) is completely continuous and TK \(\subset \) K.

Proof

By Ascoli-Arzela theorem we prove that \(T_{i}\) is completely continuous mapping for all \(i=1,\ldots ,n\), then T is completely continuous mapping. To prove that TK \(\subset \) K, let us take \(u\in K\), then

$$\begin{aligned} T_{i}u(t)=\frac{1}{\Gamma \left( q\right) }\int _{0}^{1}G_{i}\left( t,s\right) g(s)f_{i}\left( u\left( s\right) \right) ds+\frac{1}{1-\lambda _{i}}\int _{0}^{1}h_{i}\left( u\left( s\right) \right) ds, \end{aligned}$$

arguing as in the proof of Lemma 9, we obtain

$$\begin{aligned} \min _{t\in \left[ 0,1\right] }\sum _{i=1}^{n}Tu(t)\ge \left( \min _{1\le i\le n}\lambda _{i}\right) \left\| Tu\right\| , \end{aligned}$$

thus TK \(\subset \) K. \(\square \)

Now we are ready to state the main result.

Theorem 11

Assume that \((K_{1})\)–\((K_{2})\) hold. Then the fractional boundary value problem (S) has at least one positive solution in superlinear as well as sublinear case.

We recall the well-known Guo-Krasnosel’skii fixed point theorem on cone.

Theorem 12

[32, 33] Let E be a Banach space, and let \(K\subset E\), be a cone. Assume \(\Omega _{1}\) and \(\Omega _{2}\) are open subsets of E with \(0\in \Omega _{1},\,\overline{\Omega _{1}}\subset \Omega _{2}\) and let

$$\begin{aligned} {\mathcal {A}}:K\cap \left( \overline{\Omega _{2}}\backslash \Omega _{1}\right) \rightarrow K, \end{aligned}$$

be a completely continuous operator such that

  1. (i)

    \(\left| \left| {\mathcal {A}}u\right| \right| \le \left| \left| u\right| \right| ,\quad u\in K\cap \partial \Omega _{1}\), and \(\left| \left| {\mathcal {A}}u\right| \right| \ge \left| \left| u\right| \right| ,\quad u\in K\cap \partial \Omega _{2};\) or

  2. (ii)

    \(\left| \left| {\mathcal {A}}u\right| \right| \ge \left| \left| u\right| \right| ,\,u\in K\cap \partial \Omega _{1}\), and \(\left| \left| {\mathcal {A}}u\right| \right| \le \left| \left| u\right| \right| ,\,u\in K\cap \partial \Omega _{2}\).

Then \({\mathcal {A}}\) has a fixed point in \(K\cap \left( \overline{\Omega _{2}}\backslash \Omega _{1}\right) \).

Proof of Theorem 11

Let us consider the superlinear case. From the definition of the limit \(f_{0}=0\), we deduce that \(f_{0}^{i}=0,\, \forall i=1,\ldots ,n\), then for any \(\varepsilon >0\), there exists \(R^{i}>0\), such that if \(0<\left\| u\right\| \le R^{i}\), then \(f_{i}\left( u\right) \le \varepsilon \left\| u\right\| ,\,\forall i=1,\ldots ,n\). Let \(R_{1}=\min \left\{ R^{i},i=1,\ldots ,n,\right\} \) \(\Omega _{1}=\left\{ u\in E,\left\| u\right\| <R_{1}\right\} \), so, for any \(u\in K\cap \partial \Omega _{1}\), we get by using \((K_{1})-(K_{2})\) and Lemma 8

$$\begin{aligned} T_{i}u(t)= & {} \frac{1}{\Gamma \left( q\right) }\int _{0}^{1}G_{i}\left( t,s\right) g(s)f_{i}(u(s))ds+\frac{1}{1-\lambda _{i}} \int _{0}^{1}h_{i}(u(s))ds \\\le & {} \frac{\varepsilon \left\| u\right\| }{\Gamma \left( q\right) } \int _{0}^{1}\gamma _{i}(s)g(s)ds+\frac{l\left\| u\right\| }{1-\lambda _{i}}, \end{aligned}$$

then by taking the maximum over \(t\in \left[ 0,1\right] \) it follows

$$\begin{aligned} \max _{0\le t\le 1}T_{i}u(t)\le \left( \frac{\varepsilon }{\Gamma \left( q\right) }\int _{0}^{1}\gamma _{i}(s)g(s)ds+\frac{l}{1-\lambda _{i}}\right) \left\| u\right\| . \end{aligned}$$
(3.3)

Summing the n obtained inequalities in (3.3) it yields

$$\begin{aligned} \left\| Tu\right\| \le \left\| u\right\| \left( \frac{ \varepsilon }{\Gamma \left( q\right) }\sum _{i=1}^{n}\int _{0}^{1}\gamma _{i}(s)g(s)ds+\sum _{i=1}^{n}\frac{l}{1-\lambda _{i}}\right) . \end{aligned}$$

In view of assumptions \((K_{1})\), one can choose \(\varepsilon \) such that

$$\begin{aligned} \varepsilon \le \frac{\Gamma \left( q\right) \left( 1-l\sum _{i=1}^{n}\frac{1 }{1-\lambda _{i}}\right) }{\sum _{i=1}^{n}\int _{0}^{1}\gamma _{i}(s)g(s)ds}= \frac{\Gamma \left( q\right) \left( 1-l\sum _{i=1}^{n}\frac{1}{1-\lambda _{i}} \right) }{\left( \sum _{i=1}^{n}\frac{1}{1-\lambda _{i}}\right) \left( \int _{0}^{1}(1-s)^{q-1}g(s)ds\right) }, \end{aligned}$$

thus \(\left| \left| Tu\right| \right| \le \left| \left| u\right| \right| \), for \(u\in K\cap \partial \Omega _{1}\).

Next, since \(f_{\infty }=\infty \), then \(f_{\infty }^{i}=\infty ,\,\forall i=1,\ldots ,n\). For any \(M>0\), there exists \(\xi ^{i}>0\), such that \(f_{i}\left( u\right) \ge M\left\| u\right\| \) for \(\left\| u\right\| \ge \zeta ^{i}\). Let \(R=\max \left\{ \zeta ^{i},i=1,\ldots ,n,\right\} \), choose \( R_{2}>\max \left\{ R_{1},\frac{R}{\left( \min _{1\le i\le n}\lambda _{i}\right) }\right\} \) and denote by \(\Omega _{2}=\left\{ u\in E:\left| \left| u\right| \right| <R_{2}\right\} \). Let \( u\in K\cap \partial \Omega _{2}\) then for all \(t\in \left[ 0,1\right] \) we have

$$\begin{aligned} \left\| u\right\| =\sum _{i=1}^{n}\max _{0\le t\le 1}\left| u_{i}(t)\right| \ge \sum _{i=1}^{n}\min _{\left[ 0,1\right] }u_{i}(t)\ge \left( \min _{1\le i\le n}\lambda _{i}\right) \left\| u\right\| =\left( \min _{1\le i\le n}\lambda _{i}\right) R_{2}>R. \end{aligned}$$

Using estimates in Lemma 8 and the fact that \(u\in K\), we obtain for all \( t\in \left[ 0,1\right] \)

$$\begin{aligned} T_{i}u(t)\ge & {} \left\| u\right\| \frac{\lambda _{i}M}{\Gamma \left( q\right) }\int _{0}^{1}\gamma _{i}(s)g(s)ds+\frac{1}{1-\lambda _{i}}\int _{0}^{1}h_{i}(u(s))ds \\\ge & {} \left\| u\right\| \frac{\lambda _{i}M}{\Gamma \left( q\right) } \int _{0}^{1}\gamma _{i}(s)g(s)ds, \end{aligned}$$

thus

$$\begin{aligned} \left\| Tu\right\| \ge \left\| u\right\| \frac{M}{\Gamma \left( q\right) }\sum _{i=1}^{n}\left( \lambda _{i}\int _{0}^{1}\gamma _{i}(s)g(s)ds\right) . \end{aligned}$$
(3.4)

Thanks to assumptions \((K_{1})\) one can choose M such that

$$\begin{aligned} M\ge \frac{\Gamma \left( q\right) }{\sum _{i=1}^{n}\lambda _{i}\int _{0}^{1}\gamma _{i}(s)g(s)ds}=\frac{\Gamma \left( q\right) }{\left( \sum _{i=1}^{n}\frac{\lambda _{i}}{1-\lambda _{i}}\right) \left( \int _{0}^{1}(1-s)^{q-1}g(s)ds\right) }, \end{aligned}$$

then (3.4) implies

$$\begin{aligned} \left| \left| Tu\right| \right| \ge \left| \left| u\right| \right| ,\text { }\forall u\in K\cap \partial \Omega _{2}. \end{aligned}$$

Now we can apply the first statement of Guo-Krasnosel’skii fixed point theorem on cone to conclude that T has a fixed point in\(\ K\cap \left( \overline{\Omega _{2}}\diagdown \Omega _{1}\right) \) such that \(R\le \left| \left| u\right| \right| \le R_{2}\). Applying similar techniques as above, we prove the sublinear case. The proof is complete. \(\square \)

We conclude with an explicit numerical example:

Example 13

Consider the following two-dimensional fractional order system

$$\begin{aligned} (S_{i}):\left\{ \begin{array}{l} \textstyle {} ^{c}\!{D_{0^{+}}}^{\frac{3}{2}}u_{1}(t)=(1-t)\mathrm{e}^{-u_{1}-u_{2}}, \\ \textstyle {} ^{c}\!{D_{0^{+}}}^{\frac{3}{2}}u_{2}(t)=(1-t)\mathrm{e}^{-u_{2}}\quad 0<t<1, \\ \textstyle u_{1}^{\prime }(0)=0,\quad u_{1}\left( 0\right) -\frac{1}{2}u_{1}\left( 1\right) =\int _{0}^{1}\left( \frac{u_{1}\left( s\right) +u_{2}\left( s\right) }{20}\right) ds \\ \textstyle u_{2}^{\prime }(0)=0,\quad u_{2}\left( 0\right) -\frac{1}{4}u_{2}\left( 1\right) =\int _{0}^{1}\frac{\mathrm{e}^{-2}\left( u_{1}\left( s\right) +u_{2}\left( s\right) \right) }{1+u_{1}^{2}\left( s\right) +u_{2}^{2}\left( s\right) }ds. \end{array} \right. \end{aligned}$$

We have \(q=\frac{3}{2},\,\lambda _{1}=\frac{1}{2},\,\lambda _{2}=\frac{1}{4 },\,g(t)=\left( 1-t\right) \in L^{1}\left( \left[ 0,1\right] ,{\mathbb {R}} _{+}\right) ,\,\int _{0}^{1}(1-s)^{q-1}g(s)ds=\frac{2}{5}>0,\,f_{1}\left( u\right) =\mathrm{e}^{-u_{1}-u_{2}},\,f_{2}\left( u\right) =\mathrm{e}^{-u_{2}},\,f_{i}\in C\left( {\mathbb {R}}_{+}^{2},{\mathbb {R}}_{+}\right) ,\,h_{1}\left( u\right) = \frac{1}{20}\left( u_{1}+u_{2}\right) ,\,h_{2}\left( u\right) =\frac{ \mathrm{e}^{-2}\left( u_{1}+u_{2}\right) }{1+u_{1}^{2}+u_{2}^{2}}\le \mathrm{e}^{-2}\left( u_{1}+u_{2}\right) ,\,h_{i}\in C\left( {\mathbb {R}}_{+}^{2},{\mathbb {R}} _{+}\right) \), one can choose \(l=\mathrm{e}^{-2}=0.135\,34\), then \(l<\left( \sum _{i=1}^{2}\frac{1}{1-\lambda _{i}}\right) ^{-1}=0,3\). Since \(\left( K_{1}\right) \) and \(\left( K_{2}\right) \) are satisfied and \( f_{0}^{i}=\infty ,\,f_{\infty }^{i}=0\), then Theorem 11 implies that there exists at least one positive solution in the cone K.