Open Access

Monotone iterative procedure and systems of a finite number of nonlinear fractional differential equations

Advances in Difference Equations20152015:167

https://doi.org/10.1186/s13662-015-0504-9

Received: 7 January 2015

Accepted: 17 May 2015

Published: 3 June 2015

Abstract

The aim of the paper is to present a nontrivial and natural extension of the comparison result and the monotone iterative procedure based on upper and lower solutions, which were recently established in (Wang et al. in Appl. Math. Lett. 25:1019-1024, 2012), to the case of any finite number of nonlinear fractional differential equations.

Keywords

monotone iterative proceduresystem of fractional differential equationsupper and lower solution

MSC

26A3334A0834B15

1 Introduction

Fractional derivatives and integrals are used for a better description of material properties. In the literature we can find many interesting papers concerning this theory; see e.g., [113]. The study of systems involving fractional differential/integral equations is also important as such systems occur in various problems of applied nature; for example, see [1422]. Some basic theory of fractional differential equations involving the Riemann-Liouville differential operator can be found in [2325].

In the paper we consider the following system of nonlinear fractional differential equations:
$$ \left \{ \begin{array}{l} D^{\alpha}u_{1}(t) = f_{1}(t,u_{1}(t),u_{2}(t),\ldots,u_{n}(t)),\quad t \in(0,T], \\ D^{\alpha}u_{2}(t) = f_{2}(t,u_{1}(t),u_{2}(t),\ldots,u_{n}(t)),\quad t \in(0,T], \\ \ldots, \\ D^{\alpha}u_{n}(t) = f_{n}(t,u_{1}(t),u_{2}(t),\ldots,u_{n}(t)), \quad t \in(0,T], \\ t^{1-\alpha}u_{1}(t)|_{t=0}=x^{1}_{0}, \qquad t^{1-\alpha}u_{2}(t)|_{t=0}=x^{2}_{0},\qquad \ldots , \qquad t^{1-\alpha}u_{n}(t)|_{t=0}=x^{n}_{0}, \end{array} \right . $$
(1.1)
where \(D^{\alpha}\) is the standard Riemann-Liouville fractional derivative of order α, \(0\leq\alpha\leq1\), \(T>0\), \(f^{i} \in C([0,T]\times\mathbb{R}^{n},\mathbb{R})\), \(1\leq i \leq n\), and \(x^{1}_{0},\ldots,x^{n}_{0} \in\mathbb{R}\) satisfy
$$ \sum_{i=2}^{n}x^{i}_{0} - x^{1}_{0} \geq0. $$
(1.2)
We investigate system (1.1) with respect to the existence of a solution via the method of upper and lower solutions. There is also presented the concept of an iterative procedure, where the appropriately constructed sequences are convergent to the extreme solution. The paper is a continuation of the investigations in [10] of Wang et al., where the authors examined system (1.1) in the case \(n=2\). After proving the main results we state, for convenience of the reader, the introduced techniques in the case of three nonlinear fractional differential equations and also present a concrete example.

2 Preliminaries

First, let us recall the needed notations and crucial results which will be needed in the next sections of the article.

Denote by \(C_{1-\alpha}([0,T])\) the family of all functions \(u \in C((0,T])\) such that \(t^{1-\alpha}u \in C([0,T])\). A basic theorem concerning the existence of the result and its uniqueness for the linear fractional equation is as follows.

Lemma 2.1

([23])

Let \(0<\alpha\leq1\), \(M \in\mathbb{R}\), and \(\sigma\in C_{1-\alpha }([0,T])\) be fixed. Then the linear initial value problem
$$ \left \{ \begin{array}{l} D^{\alpha}u(t) = \sigma(t) - Mu(t), \quad t \in(0,T], \\ t^{1-\alpha}u(t)|_{t=0} = u_{0}, \end{array} \right .\quad i,j \in \mathbb{N}, $$
(2.1)
has a unique solution, given by the following formula:
$$u(t)=\Gamma(\alpha)u_{0}t^{\alpha-1}E_{\alpha,\alpha} \bigl(-Mt^{\alpha}\bigr)+\int_{0}^{t}(t-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(t-s)^{\alpha}\bigr)\sigma(s)\, ds, $$
where \(E_{\alpha,\beta}\) is the Mittag-Leffler function, i.e. the function of the form
$$E_{\alpha,\beta}(z)=\sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma(\alpha k + \beta)}, \quad \alpha, \beta> 0, z \in\mathbb{R}. $$

The comparison result for the initial value problem (2.1) due to Wang et al. is as follows.

Lemma 2.2

([10])

Let \(0<\alpha\leq1\) and \(M\in\mathbb{R}\) be given. Then, if \(w\in C_{1-\alpha}([0,T])\) satisfies
$$ \left \{ \begin{array}{l} D^{\alpha}w(t) + Mw(t) \geq0, \quad t \in(0,T], \\ t^{1-\alpha}w(t)|_{t=0} \geq0, \end{array} \right . $$
then \(w(t)\geq0\) for all \(t\in(0,T]\).

The same authors also proved the following result, which will be needed in the sequel.

Lemma 2.3

([10])

Let \(0<\alpha\leq1\), \(M\in\mathbb{R}\), and \(N\geq0\) be given. Assume that \(u,v \in C_{1-\alpha}([0,T])\) satisfy
$$ \left \{ \begin{array}{l} D^{\alpha}u(t) \geq-Mu(t) + Nv(t),\quad t \in(0,T], \\ D^{\alpha}v(t) \geq-Mv(t) + Nu(t), \quad t \in(0,T], \\ t^{1-\alpha}u(t)|_{t=0} \geq0,\qquad t^{1-\alpha}v(t)|_{t=0} \geq0. \end{array} \right . $$
Then \(u(t)\geq0\), \(v(t)\geq0\) for all \(t\in(0,T]\).

3 The results

In the sequel we will use the following notation:
$$ \delta_{ij}= \left \{ \begin{array}{l@{\quad}l} 1 &\text{if } i=j, \\ -1 &\text{if } i\neq j, \end{array} \right . \quad i,j \in\mathbb{N}. $$
\(C_{1-\alpha}([0,T])^{n}\) denotes \(C_{1-\alpha}([0,T])\times C_{1-\alpha }([0,T])\times\cdots\times C_{1-\alpha}([0,T])\) (n times).

Lemma 3.1

Let \(0<\alpha\leq1\) be fixed, \(M_{i} \in\mathbb{R}\), \(\sigma_{i} \in C_{1-\alpha}([0,T])\), \(i=1,2,\ldots, n\). Then the linear problem of n equations
$$ \left \{ \textstyle\begin{array}{l} D^{\alpha}u_{1}(t) = \sigma_{1}(t) - M_{1}u_{1}(t) - \sum_{i,j=2}^{n}M_{j}\delta _{ji}u_{i}(t),\quad t \in(0,T], \\ D^{\alpha}u_{j}(t)= \sigma_{j}(t)+(M_{j}-\sum_{i=1}^{n}M_{i})u_{j}(t) \\ \hphantom{D^{\alpha}u_{j}(t)=}{}-M_{j}(\sum_{i=1}^{n}u_{i}(t)-u_{j}(t)),\quad t\in(0,T], 2\leq j \leq n, \\ t^{1-\alpha}u_{i}(t)|_{t=0} = x^{i}_{0},\quad 1\leq i \leq n, \end{array} \right . $$
(3.1)
has a unique solution in \(C_{1-\alpha}([0,T])^{n}\).

Proof

First observe that for any \(p_{1}, p_{2},\ldots, p_{n} \in C_{1-\alpha}([0,T])\) the system
$$ \left \{ \begin{array}{l} u_{1} + u_{2} + \cdots+ u_{n} = p_{1}, \\ u_{1} - u_{2} + \cdots+ u_{n} = p_{2}, \\ \ldots, \\ u_{1} + u_{2} + \cdots- u_{n} = p_{n} \end{array} \right . $$
(3.2)
has exactly one solution, which is a consequence of the fact that
$$ \det \left [ \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 1 & 1 & 1 & \cdots& 1\\ 1 &-1 & 1 & \cdots& 1\\ 1 & 1 & -1 & \cdots& 1\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ 1 & 1 & 1 & \cdots& -1 \end{array} \right ]_{n\times n} = (-2)^{n-1} \neq0. $$
Next, observe that system (3.1) can be transformed to system (3.2), where \(p_{1}, p_{2}, \ldots, p_{n}\) solve the following n problems:
$$\begin{aligned}& \left \{ \begin{array}{l} D^{\alpha}p_{1}(t) =(\sigma_{1}(t)+\sigma_{2}(t)+\cdots+\sigma _{n}(t))-(M_{1}+M_{2}+\cdots+M_{n})p_{1}(t), \\ t^{1-\alpha}p_{1}(t)|_{t=0} =x^{1}_{0}+x^{2}_{0}+\cdots+x^{n}_{0}, \end{array} \right . \\& \left \{ \begin{array}{l} D^{\alpha}p_{2}(t) =(\sigma_{1}(t)-\sigma_{2}(t)+\cdots+\sigma _{n}(t))-(M_{1}-M_{2}+\cdots+M_{n})p_{2}(t), \\ t^{1-\alpha}p_{2}(t)|_{t=0} =x^{1}_{0}-x^{2}_{0}+\cdots+x^{n}_{0}, \end{array} \right . \\& \vdots \\& \left \{ \begin{array}{l} D^{\alpha}p_{n}(t) =(\sigma_{1}(t)+\sigma_{2}(t)+\cdots-\sigma _{n}(t))-(M_{1}+M_{2}+\cdots-M_{n})p_{n}(t), \\ t^{1-\alpha}p_{n}(t)|_{t=0} =x^{1}_{0}+x^{2}_{0}+\cdots-x^{n}_{0}. \end{array} \right . \end{aligned}$$
Finally, observe that the solutions of the above equations are unique due to Lemma 2.1, which ends the proof. □

Now we can state and proof the comparison result for system (3.1).

Theorem 3.1

Let \(0<\alpha\leq1\), \(M_{1} \in\mathbb{R}\), \(M_{2}, \ldots, M_{n} \geq0\), and let \(u_{1},\ldots, u_{n} \in C_{1-\alpha }([0,T])\) satisfy
$$ \left \{ \textstyle\begin{array}{l} D^{\alpha}u_{1}(t) \geq- M_{1}u_{1}(t) + \sum_{i,j=2}^{n}M_{j}\delta _{ji}u_{i}(t),\quad t \in(0,T], \\ D^{\alpha}u_{s}(t) \geq-M_{1}u_{s}(t)+ (\sum_{i=2}^{n}M_{i}-M_{s} )u_{s}(t) \\ \hphantom{D^{\alpha}u_{s}(t) \geq}{}+M_{s} (\sum_{i=1}^{n}u_{i}(t)-u_{s}(t) ), \quad 2 \leq s \leq n, t \in(0,T], \\ t^{1-\alpha}u_{s}(t)|_{t=0} \geq0,\quad 1\leq s \leq n. \end{array} \right . $$
(3.3)
Then
$$\begin{aligned}& \sum_{i=1}^{n}u_{i}(t) \geq0,\quad t\in(0,T], \end{aligned}$$
(3.4)
$$\begin{aligned}& u_{s}(t)\geq0, \quad t\in(0,T], 2 \leq s \leq n, \end{aligned}$$
(3.5)
$$\begin{aligned}& -u_{s}(t)+\sum_{i=1}^{n}u_{i}(t) \geq0, \quad t\in(0,T], 2 \leq s \leq n. \end{aligned}$$
(3.6)

Proof

Put \(r(t)=\sum_{s=1}^{n}u_{s}(t)\). Using (3.3) we obtain
$$\begin{aligned} D^{\alpha}r(t) =&\sum_{s=1}^{n}D^{\alpha}u_{s}(t) \\ \geq&- M_{1}u_{1}(t) + \sum_{i,j=2}^{n}M_{j} \delta_{ji}u_{i}(t)-M_{1} \sum _{s=2}^{n}u_{s}(t) - 2\sum _{s=2}^{n}M_{s} u_{s}(t) \\ &{}+ \sum _{s=2}^{n}\sum_{i=2}^{n}M_{i}u_{s}(t) + \sum_{s=2}^{n}\sum _{i=1}^{n}M_{s} u_{i}(t) \\ =&- M_{1}r(t) + \sum_{i,j=2}^{n} \bigl(M_{j}\delta_{ji}u_{i}(t)+M_{i}u_{j}(t) \bigr) -2 \sum_{s=2}^{n}M_{s}u_{s}(t)+ \sum_{s=2}^{n}M_{s}r(t). \end{aligned}$$
Observe that
$$ \sum_{i,j=2, i\neq j}^{n} \bigl(M_{j}\delta_{ji}u_{i}(t)+M_{i}u_{j}(t) \bigr)=0. $$
(3.7)
Hence, we obtain
$$\begin{aligned} \begin{aligned}[b] D^{\alpha}r(t)\geq{}&- \Biggl(M_{1} - \sum _{s=2}^{n}M_{s} \Biggr)r(t)+\sum _{i,j=2, i\neq j}^{n} \bigl(M_{j} \delta_{ji}u_{i}(t)+M_{i}u_{j}(t) \bigr) \\ &{}+\sum_{i,j=2, i=j}^{n} \bigl(M_{j} \delta_{ji}u_{i}(t)+M_{i}u_{j}(t) \bigr) -2 \sum_{s=2}^{n}M_{s}u_{s}(t) \\ ={}&- \Biggl(M_{1} - \sum_{s=2}^{n}M_{s} \Biggr)r(t)+2 \sum_{i=2}^{n}M_{i}u_{i}(t)-2 \sum_{s=2}^{n}M_{s}u_{s}(t) \\ ={}&- \Biggl(M_{1} - \sum_{s=2}^{n}M_{s} \Biggr)r(t). \end{aligned} \end{aligned}$$
(3.8)
Moreover, observe that
$$ t^{1-\alpha}r(t)=\sum_{s=1}^{n}t^{1-\alpha}u_{s}(t) \geq0. $$
(3.9)
Applying (3.8) and (3.9) to Lemma 2.2 we get (3.4).
Now, consider any \(2\leq s\leq n\) and denote
$$r_{s}(t)=\sum_{i=1}^{n}u_{i}(t)-u_{s}(t), \quad t \in(0,T]. $$
By (3.3) we have
$$\begin{aligned} D^{\alpha}r_{s}(t) =&\sum_{i=1}^{n}D^{\alpha}u_{i}(t)-D^{\alpha }u_{s}(t)=D^{\alpha}u_{1}(t) +\sum_{i=2}^{n}D^{\alpha}u_{i}(t)-D^{\alpha }u_{s}(t) \\ \geq&- M_{1}u_{1}(t) + \sum_{i,j=2}^{n}M_{j} \delta_{ji}u_{i}(t)-\sum_{i=2}^{n}M_{1}u_{i}(t)+M_{1}u_{s}(t) \\ &{}+\sum_{i=2}^{n} \Biggl(\sum _{j=2}^{n}M_{j}-M_{i} \Biggr)u_{i}(t)- \Biggl(\sum_{j=2}^{n}M_{j}-M_{s} \Biggr)u_{s}(t) \\ &{}+\sum_{i=2}^{n}M_{i} \Biggl( \sum_{j=1}^{n}u_{j}(t)-u_{i}(t) \Biggr)-M_{s} \Biggl(\sum_{j=1}^{n}u_{j}(t)-u_{s}(t) \Biggr) \\ =&-M_{1}u_{1}(t)+\sum_{i,j=2}^{n} \bigl(M_{j}\delta_{ji}u_{i}(t)+M_{j}u_{i}(t) \bigr)-M_{1}\sum_{i=2}^{n}u_{i}(t)+M_{1}u_{s}(t)+M_{s}u_{s}(t) \\ &{}-2\sum_{i=2}^{n}M_{i}u_{i}(t)-u_{s}(t) \sum_{j=2}^{n}M_{j}+\sum _{i=2}^{n}\sum_{j=1}^{n}M_{i}u_{j}(t)-M_{s}r_{s}(t). \end{aligned}$$
Again, using (3.7), we obtain
$$\begin{aligned} D^{\alpha}r_{s}(t) \geq&-M_{1}\sum _{i=1}^{n}u_{i}(t)+M_{1}u_{s}(t)+M_{s}u_{s}(t)-u_{s}(t) \sum_{j=2}^{n}M_{j} \\ &{}+ \sum_{i=2}^{n}M_{i}r_{s}(t)+u_{s}(t) \sum_{i=2}^{n}M_{i}-M_{s}r_{s}(t) \\ =&- \Biggl(M_{1}-\sum_{i=2}^{n}M_{i}+M_{s} \Biggr)r_{s}(t)+M_{s}u_{s}(t). \end{aligned}$$
(3.10)
Moreover, observe that (3.3) implies
$$ D^{\alpha}u_{s}(t)\geq- \Biggl(M_{1} - \sum_{i=2}^{n}M_{i} + M_{s} \Biggr)u_{s}(t)+M_{s}r_{s}(t). $$
(3.11)
Finally, note that (3.10) and (3.11) applied to Lemma 2.3 give (3.5) and (3.6). □

Now, we are in a position to enunciate the main result.

Theorem 3.2

Suppose that there exist \(u^{1}_{0},u^{2}_{0},\ldots, u^{n}_{0} \in C_{1-\alpha }([0,T])\), \(u^{1}_{0}\leq\sum_{i=2}^{n}u_{0}^{i}\), satisfying
$$ \left \{ \begin{array}{l} D^{\alpha}u^{1}_{0}(t) \leq f_{1}(t,u^{1}_{0}(t),u^{2}_{0}(t),\ldots, u^{n}_{0}(t)), \quad t\in(0,T], \\ D^{\alpha}u^{s}_{0}(t) \geq f_{s}(t,u^{1}_{0}(t),u^{2}_{0}(t),\ldots, u^{n}_{0}(t)), \quad t\in(0,T], 2\leq s \leq n, \\ t^{1-\alpha}u^{1}_{0}(t)|_{t=0} \leq x^{1}_{0}, \\ t^{1-\alpha}u^{s}_{0}(t)|_{t=0} \geq x^{s}_{0},\quad 2\leq s \leq n, \end{array} \right . $$
(3.12)
and there exist \(M_{1}\in\mathbb{R}\), \(M_{2},\ldots, M_{n} >0\) such that
  1. (i)
    $$ f_{1}(t,\alpha_{1},\ldots,\alpha_{n})-f_{1}(t, \beta_{1},\ldots,\beta_{n}) \geq- M_{1}( \alpha_{1}-\beta_{1}) - \sum_{i,j=2}^{n}M_{j} \delta_{ji}(\alpha_{i} - \beta_{i}), $$
    (3.13)
     
  2. (ii)
    $$\begin{aligned}& f_{s}(t,\alpha_{1},\ldots,\alpha_{n}) -f_{s}(t,\beta_{1},\ldots,\beta_{n}) \\& \quad \geq \Biggl(-M_{1}+\sum_{i=2}^{n}M_{i}-M_{s} \Biggr) (\alpha_{s}-\beta_{s}) -M_{s}\Biggl( \alpha_{1}-\beta_{1}+\alpha_{s}-\beta_{s} -\sum_{i=2}^{n}(\alpha_{i}- \beta_{i})\Biggr), \end{aligned}$$
    where \(\alpha_{i}, \beta_{i} \in\mathbb{R}\), \(1\leq i \leq n\) satisfy for all \(t \in[0,T]\) and \(2\leq s \leq n\),
    $$\begin{aligned}& u_{0}^{1}(t)-\Biggl(\sum_{i=2}^{n}u_{0}^{i}(t)-u_{0}^{s}(t) \Biggr) \leq\beta_{1} - \Biggl(\sum_{i=2}^{n} \beta_{i} - \beta_{s}\Biggr) \leq\alpha_{1} - \Biggl(\sum_{i=2}^{n}\alpha_{i} - \alpha_{s}\Biggr)\leq u_{0}^{s}(t), \\& u_{0}^{1}(t)-\Biggl(\sum_{i=2}^{n}u_{0}^{i}(t)-u_{0}^{s}(t) \Biggr)\leq\alpha_{s} \leq\beta_{s} \leq u^{s}_{0}(t), \end{aligned}$$
     
  3. (iii)
    $$\begin{aligned}& \sum_{s=2}^{n}f_{s} \bigl(t,u^{1}(t),u^{2}(t),\ldots, u^{n}(t)\bigr) -f_{1}\bigl(t,u^{1}(t),u^{2}(t),\ldots, u^{n}(t)\bigr) \\& \quad \geq\Biggl(-M_{1}+\sum_{s=2}^{n}M_{s} \Biggr) \Biggl(\sum_{s=2}^{n}u^{s}(t)-u^{1}(t) \Biggr), \end{aligned}$$
    (3.14)
    where
    $$u_{0}^{1}-\Biggl(\sum_{i=2}^{n}u_{0}^{i}-u_{0}^{s} \Biggr) \leq u^{1} - \Biggl(\sum_{i=2}^{n}u^{i} - u^{s}\Biggr) \leq u^{s} \leq u_{0}^{s}, \quad 2\leq s \leq n. $$
     
Then there exists a solution \((\bar{u}^{1},\bar{u}^{2},\ldots,\bar{u}^{n})\) of system (1.1) such that
$$(n-1)u_{0}^{1}-(n-2)\sum_{i=2}^{n}u^{i}_{0} \leq\bar{u}^{1}\leq\sum_{i=2}^{n}u^{i}_{0}, \qquad u_{0}^{1} - \sum_{i=2}^{n}u_{0}^{i}+u_{0}^{s} \leq\bar{u}^{s} \leq u_{0}^{s},\quad 2\leq s\leq n. $$
Moreover, there exist iterative sequences \((u^{1}_{k}), (u^{2}_{k}),\ldots, (u^{n}_{k})\) such that \(u^{i}_{k}\to\bar{u}^{i}\), \(k \to\infty\), \(i=1,2,\ldots ,n\), uniformly on compact subsets of \((0,T]\).

Proof

Let us first consider the linear system of the form
$$ \left \{ \textstyle\begin{array}{l} D^{\alpha}u^{1}(t) = f_{1}(t,u^{1}_{0}(t),u^{2}_{0}(t),\ldots, u^{n}_{0}(t))+ M_{1}u_{0}^{1}(t) + \sum_{i,j=2}^{n}M_{j}\delta_{ji}u^{i}_{0}(t) \\ \hphantom{D^{\alpha}u^{1}(t) =}{}- M_{1}u^{1}(t) - \sum_{i,j=2}^{n}M_{j}\delta_{ji}u^{i}(t), \quad t\in(0,T], \\ D^{\alpha}u^{s}(t)= f_{s}(t,u^{1}_{0}(t),u^{2}_{0}(t),\ldots, u^{n}_{0}(t))+M_{1}u^{s}_{0}(t) + (\sum_{i=2}^{n}M_{i}-M_{s})u^{s}(t) \\ \hphantom{D^{\alpha}u^{s}(t)=}{}+ M_{s}(u_{0}^{1}(t)+\sum_{i=2}^{n}u^{i}(t)-u^{s}(t))-M_{1}u^{s}(t)-(\sum_{i=2}^{n}M_{i}-M_{s})u^{s}_{0}(t) \\ \hphantom{D^{\alpha}u^{s}(t)=}{}-M_{s}(u^{1}(t)+\sum_{i=2}^{n}u^{i}_{0}(t)-u^{s}_{0}(t)),\quad t\in(0,T], 2\leq s \leq n, \\ t^{1-\alpha}u^{s}(t)|_{t=0} = x^{s}_{0},\quad 1\leq s \leq n, \end{array} \right . $$
(3.15)
where \(u^{1}, u^{2},\ldots,u^{n} \in C_{1-\alpha}([0,T])\). Due to Lemma 3.1 there exists a system of solutions \((u^{1}_{1}, u^{2}_{1},\ldots, u^{n}_{1}) \in C([0,T])^{n}\) for system (3.15). Using induction we obtain the sequence \((u^{1}_{k}, u^{2}_{k},\ldots, u^{n}_{k}) \in C([0,T])^{n}\), \(k \in\mathbb{N}\), satisfying
$$ \left \{ \textstyle\begin{array}{l} D^{\alpha}u^{1}_{k}(t) = f_{1}(t,u^{1}_{k-1}(t),u^{2}_{k-1}(t),\ldots, u^{n}_{k-1}(t))+ M_{1}u_{k-1}^{1}(t) + \sum_{i,j=2}^{n}M_{j}\delta _{ji}u^{i}_{k-1}(t) \\ \hphantom{D^{\alpha}u^{1}_{k}(t) =}{}- M_{1}u^{1}_{k}(t) - \sum_{i,j=2}^{n}M_{j}\delta_{ji}u^{i}_{k}(t), \quad t\in(0,T], \\ D^{\alpha}u^{s}_{k}(t)= f_{s}(t,u^{1}_{k-1}(t),u^{2}_{k-1}(t),\ldots, u^{n}_{k-1}(t))+M_{1}u^{s}_{k-1}(t) \\ \hphantom{D^{\alpha}u^{s}_{k}(t)=}{}+ (\sum_{i=2}^{n}M_{i}-M_{s})u^{s}_{k}(t)+ M_{s}(u_{k-1}^{1}(t)+\sum_{i=2}^{n}u^{i}_{k}(t)-u^{s}_{k}(t)) \\ \hphantom{D^{\alpha}u^{s}_{k}(t)=}{}-M_{1}u^{s}_{k}(t)-(\sum_{i=2}^{n}M_{i}-M_{s})u^{s}_{k-1}(t) \\ \hphantom{D^{\alpha}u^{s}_{k}(t)=}{}-M_{s}(u^{1}_{k}(t)+\sum_{i=2}^{n}u^{i}_{k-1}(t)-u^{s}_{k-1}(t)),\quad t\in(0,T], 2\leq s \leq n, \\ t^{1-\alpha}u^{s}_{k}(t)|_{t=0} = x^{s}_{0}, \quad 1\leq s \leq n. \end{array} \right . $$
(3.16)
Now, put \(p^{1}_{1}=u^{1}_{1}-u^{1}_{0}\), \(p^{s}_{1}=u^{s}_{0}-u^{s}_{1}\), \(2\leq s \leq n\). From (3.12) and (3.15), for all \(t\in(0,T]\), we obtain
$$\begin{aligned}& D^{\alpha}p^{1}_{1}(t) = D^{\alpha}u_{1}^{1}(t) -D^{\alpha}u^{1}_{0}(t) \\& \hphantom{D^{\alpha}p^{1}_{1}(t)}=f_{1} \bigl(t,u^{1}_{0}(t),u^{2}_{0}(t), \ldots, u^{n}_{0}(t)\bigr)+ M_{1}u_{0}^{1}(t) + \sum_{i,j=2}^{n}M_{j} \delta_{ji}u^{i}_{0}(t) \\& \hphantom{D^{\alpha}p^{1}_{1}(t) = }{} - M_{1}u^{1}_{1}(t) - \sum _{i,j=2}^{n}M_{j} \delta_{ji}u^{i}_{1}(t)-D^{\alpha }u^{1}_{0}(t) \\& \hphantom{D^{\alpha}p^{1}_{1}(t)}\geq-M_{1}p_{1}^{1}(t) + \sum _{i,j=2}^{n}M_{j}\delta_{ji}p_{1}^{i}(t), \\& D^{\alpha}p^{s}_{1}(t) =D^{\alpha}u^{s}_{0}(t)-D^{\alpha}u^{s}_{1}(t) \\& \hphantom{D^{\alpha}p^{s}_{1}(t)}=D^{\alpha}u^{s}_{0}(t)-f_{s} \bigl(t,u^{1}_{0}(t),u^{2}_{0}(t), \ldots, u^{n}_{0}(t)\bigr)-M_{1}u^{s}_{0}(t) \\& \hphantom{D^{\alpha}p^{s}_{1}(t) =}{} - \Biggl(\sum_{i=2}^{n}M_{i}-M_{s} \Biggr)u^{s}_{1}(t)- M_{s}\Biggl(u_{0}^{1}(t)+ \sum_{i=2}^{n}u^{i}_{1}(t)-u^{s}_{1}(t) \Biggr)+M_{1}u^{s}_{1}(t) \\& \hphantom{D^{\alpha}p^{s}_{1}(t) =}{} +\Biggl(\sum_{i=2}^{n}M_{i}-M_{s} \Biggr)u^{s}_{0}(t) +M_{s}\Biggl(u^{1}_{1}(t)+ \sum_{i=2}^{n}u^{i}_{0}(t)-u^{s}_{0}(t) \Biggr) \\& \hphantom{D^{\alpha}p^{s}_{1}(t)} \geq-M_{1}p_{1}^{s}(t) + \Biggl( \sum_{i=2}^{n}M_{i}-M_{s} \Biggr)p_{1}^{s}(t) +M_{s}\Biggl(\sum _{i=1}^{n}p^{i}_{1}(t)-p_{1}^{s}(t) \Biggr)\quad \text{for all } 2\leq s \leq n, \\& t^{1-\alpha}p_{1}^{1}(t)|_{t=0} = t^{1-\alpha}u_{1}^{1}(t)|_{t=0}-t^{1-\alpha }u^{1}_{0}(t)|_{t=0} \geq x^{1}_{0}-x^{1}_{0}=0, \\& t^{1-\alpha}p^{s}_{1}(t)|_{t=0}=t^{1-\alpha}u^{s}_{0}(t)|_{t=0}-t^{1-\alpha }u^{s}_{1}(t)|_{t=0} \geq x^{s}_{0}-x^{s}_{0}=0, \quad 2\leq s \leq n. \end{aligned}$$
Hence, using Theorem 3.1, we have
$$ u_{1}^{s} \leq u_{0}^{s}, \quad 2\leq s \leq n $$
(3.17)
and
$$ u^{1}_{1}-u^{1}_{0}+\sum _{i=2}^{n}\bigl(u^{i}_{0}-u^{i}_{1} \bigr)\geq u_{0}^{s}-u_{1}^{s},\quad 2 \leq s \leq n. $$
(3.18)
Consider now \(q_{1}=\sum_{i=2}^{n}u^{i}_{1}-u^{1}_{1}\). Using (3.14) and (3.15) we have
$$\begin{aligned} D^{\alpha}q_{1}(t) =& \sum_{s=2}^{n}u^{s}_{1}(t)-u^{1}_{1}(t)= \sum_{s=2}^{n}D^{\alpha}u^{s}_{1}(t)-D^{\alpha}u^{1}_{1}(t) \\ =&\sum_{s=2}^{n}f_{s} \bigl(t,u^{1}_{0}(t),u^{2}_{0}(t), \ldots, u^{n}_{0}(t)\bigr)+\sum_{s=2}^{n}M_{1}u^{s}_{0}(t) \\ &{}+ \sum_{s=2}^{n}\Biggl(\sum _{i=2}^{n}M_{i}-M_{s} \Biggr)u^{s}_{1}(t)+ \sum_{s=2}^{n}M_{s} \Biggl(u_{0}^{1}(t)+\sum_{i=2}^{n}u^{i}_{1}(t)-u^{s}_{1}(t) \Biggr) \\ &{}-\sum_{s=2}^{n}M_{1}u^{s}_{1}(t)- \sum_{s=2}^{n}\Biggl(\sum _{i=2}^{n}M_{i}-M_{s} \Biggr)u^{s}_{0}(t) \\ &{}-\sum_{s=2}^{n}M_{s} \Biggl(u^{1}_{1}(t)+\sum_{i=2}^{n}u^{i}_{0}(t)-u^{s}_{0}(t) \Biggr) -f_{1}\bigl(t,u^{1}_{0}(t),u^{2}_{0}(t), \ldots, u^{n}_{0}(t)\bigr) \\ &{}- M_{1}u_{0}^{1}(t) - \sum _{i,j=2}^{n}M_{j}\delta_{ji}u^{i}_{0}(t) + M_{1}u^{1}_{1}(t) + \sum _{i,j=2}^{n}M_{j}\delta_{ji}u^{i}_{1}(t) \\ =& \sum_{s=2}^{n}f_{s} \bigl(t,u^{1}_{0}(t),u^{2}_{0}(t), \ldots, u^{n}_{0}(t)\bigr) -f_{1} \bigl(t,u^{1}_{0}(t),u^{2}_{0}(t), \ldots, u^{n}_{0}(t)\bigr) \\ &{}-\Biggl(M_{1}-\sum_{s=2}^{n}M_{s} \Biggr)q_{1}(t)+\Biggl(M_{1}-\sum _{s=2}^{n}M_{s}\Biggr) \Biggl(\sum _{s=2}^{n}u_{0}^{s}(t)-u_{0}^{1}(t) \Biggr) \\ \geq&-\Biggl(M_{1}-\sum_{s=2}^{n}M_{s} \Biggr)q_{1}(t). \end{aligned}$$
Moreover, (1.2) implies
$$t^{1-\alpha}q_{1}(t)|_{t=0}=\sum _{i=2}^{n}t^{1-\alpha }u^{i}_{1}(t)|_{t=0}-t^{1-\alpha}u^{1}_{1}(t)|_{t=0}= \sum_{i=2}^{n}x^{i}_{0}-x^{1}_{0} \geq0. $$
Now, from Lemma 2.2 we conclude
$$ u^{1}_{1}(t)\leq\sum _{i=2}^{n}u^{i}_{1}(t) \quad \text{for all } t\in[0,T]. $$
(3.19)
Combining (3.17) and (3.18) with (3.19) we obtain for all \(2\leq s \leq n\) the inequalities
$$u^{1}_{0}-\Biggl(\sum_{i=2}^{n}u^{i}_{0}-u^{s}_{0} \Biggr)\leq u^{1}_{1} - \Biggl(\sum _{i=2}^{n}u_{1}^{i}-u_{1}^{s} \Biggr)\leq u_{1}^{s}\leq u_{0}^{s}. $$
Let \(2\leq s\leq n\) be fixed and suppose now that for some \(k \in \mathbb{N}\) the following inequalities hold:
$$ u^{1}_{k-1}-\Biggl(\sum _{i=2}^{n}u^{i}_{k-1}-u^{s}_{k-1} \Biggr)\leq u^{1}_{k} - \Biggl(\sum _{i=2}^{n}u_{k}^{i}-u_{k}^{s} \Biggr)\leq u_{k}^{s}\leq u_{k-1}^{s}. $$
(3.20)
Denote \(p^{1}_{k+1}=u^{1}_{k+1}-u^{1}_{k}\), \(p^{s}_{k+1}=u^{s}_{k}-u^{s}_{k+1}\), \(2\leq s \leq n\). From (3.13), (3.16), and (3.20) we obtain
$$\begin{aligned}& D^{\alpha}p^{1}_{k+1}(t) =D^{\alpha}u^{1}_{k+1}(t)-D^{\alpha}u^{1}_{k}(t) \\ & \hphantom{D^{\alpha}p^{1}_{k+1}(t)} = f_{1}\bigl(t,u^{1}_{k}(t),u^{2}_{k}(t), \ldots, u^{n}_{k}(t)\bigr)+ M_{1}u_{k}^{1}(t) + \sum_{i,j=2}^{n}M_{j} \delta_{ji}u^{i}_{k}(t) - M_{1}u^{1}_{k+1}(t) \\ & \hphantom{D^{\alpha}p^{1}_{k+1}(t) =}{} - \sum_{i,j=2}^{n}M_{j} \delta _{ji}u^{i}_{k+1}(t)-f_{1} \bigl(t,u^{1}_{k-1}(t),u^{2}_{k-1}(t), \ldots, u^{n}_{k-1}(t)\bigr)-M_{1}u_{k-1}^{1}(t) \\ & \hphantom{D^{\alpha}p^{1}_{k+1}(t) =}{} -\sum_{i,j=2}^{n}M_{j} \delta_{ji}u^{i}_{k-1}(t)+ M_{1}u^{1}_{k}(t)+ \sum_{i,j=2}^{n}M_{j} \delta_{ji}u^{i}_{k}(t) \\ & \hphantom{D^{\alpha}p^{1}_{k+1}(t)} \geq-M_{1}\bigl(u^{1}_{k1}(t)-u^{1}_{k-1}(t) \bigr)-\sum_{i,j=2}^{n}M_{j}\delta _{ji}\bigl(u^{i}_{k}(t)-u^{i}_{k-1}(t) \bigr) + M_{1}u_{k}^{1}(t) \\ & \hphantom{D^{\alpha}p^{1}_{k+1}(t) =}{} + \sum_{i,j=2}^{n}M_{j} \delta_{ji}u^{i}_{k}(t) - M_{1}u^{1}_{k+1}(t) - \sum_{i,j=2}^{n}M_{j} \delta_{ji}u^{i}_{k+1}(t)-M_{1}u_{k-1}^{1}(t) \\ & \hphantom{D^{\alpha}p^{1}_{k+1}(t) =}{} -\sum_{i,j=2}^{n}M_{j} \delta_{ji}u^{i}_{k-1}(t)+ M_{1}u^{1}_{k}(t)+ \sum_{i,j=2}^{n}M_{j} \delta_{ji}u^{i}_{k}(t) \\ & \hphantom{D^{\alpha}p^{1}_{k+1}(t)}=-M_{1}p^{1}_{k+1}(t)+ \sum_{i,j=2}^{n}M_{j} \delta_{ji}p^{i}_{k+1}(t), \\ & D^{\alpha}p^{s}_{k+1}(t) =D^{\alpha}u^{s}_{k}(t)-D^{\alpha}u^{s}_{k+1}(t) \\ & \hphantom{D^{\alpha}p^{s}_{k+1}(t)} \geq\Biggl(-M_{1}+\sum_{i=2}^{n}M_{i}-M_{s} \Biggr) \bigl(u^{s}_{k-1}(t)-u^{s}_{k}(t) \bigr)-M_{s}\Biggl(u^{1}_{k-1}(t)-u^{1}_{k}(t) \\ & \hphantom{D^{\alpha}p^{s}_{k+1}(t) =}{}+u^{s}_{k-1}(t)-u^{s}_{k}(t) -\sum_{i=2}^{n} \bigl(u^{i}_{k-1}-u^{i}_{k}\bigr)\Biggr) + M_{1}u^{s}_{k-1}(t) + \Biggl(\sum _{i=2}^{n}M_{i}-M_{s} \Biggr)u^{s}_{k}(t) \\ & \hphantom{D^{\alpha}p^{s}_{k+1}(t) =}{} + M_{s}\Biggl(u_{k-1}^{1}(t)+\sum _{i=2}^{n}u^{i}_{k}(t)-u^{s}_{k}(t) \Biggr)-M_{1}u^{s}_{k}(t)-\Biggl(\sum _{i=2}^{n}M_{i}-M_{s} \Biggr)u^{s}_{k-1}(t) \\ & \hphantom{D^{\alpha}p^{s}_{k+1}(t) =}{} -M_{s}\Biggl(u^{1}_{k}(t)+\sum _{i=2}^{n}u^{i}_{k-1}(t)-u^{s}_{k-1}(t) \Biggr) -M_{1}u^{s}_{k}(t) - \Biggl(\sum _{i=2}^{n}M_{i}-M_{s} \Biggr)u^{s}_{k+1}(t) \\ & \hphantom{D^{\alpha}p^{s}_{k+1}(t) =}{} - M_{s}\Biggl(u_{k}^{1}(t)+\sum _{i=2}^{n}u^{i}_{k+1}(t)-u^{s}_{k+1}(t) \Biggr) +M_{1}u^{s}_{k+1}(t)+\Biggl(\sum _{i=2}^{n}M_{i}-M_{s} \Biggr)u^{s}_{k}(t) \\ & \hphantom{D^{\alpha}p^{s}_{k+1}(t) =}{} +M_{s}\Biggl(u^{1}_{k+1}(t)+\sum _{i=2}^{n}u^{i}_{k}(t)-u^{s}_{k}(t) \Biggr) \\ & \hphantom{D^{\alpha}p^{s}_{k+1}(t)} =-M_{1}p^{s}_{k+1}(t)+\Biggl(\sum _{i=2}^{n}M_{i}-M_{s} \Biggr)p^{s}_{k+1}(t)+M_{s}\Biggl(\sum _{i=1}^{n}p^{i}_{k+1}(t)-p^{s}_{k+1}(t) \Biggr). \end{aligned}$$
Also observe that \(t^{1-\alpha}p^{1}_{k+1}(t)|_{t=0}=t^{1-\alpha }p^{s}_{k+1}(t)|_{t=0}=0\), which, together with the above, due to Theorem 3.1, gives
$$\begin{aligned}& u_{k+1}^{s}\leq u_{k}^{s}, \quad 2\leq s \leq n, \end{aligned}$$
(3.21)
$$\begin{aligned}& u_{k}^{s} - u_{k+1}^{s} \leq\sum_{i=2}^{n}\bigl(u_{k}^{i} - u_{k+1}^{i}\bigr)+u_{k+1}^{1}-u_{k}^{1}. \end{aligned}$$
(3.22)
Consider now \(q_{k}=\sum_{i=2}^{n}u^{i}_{k}-u^{1}_{k}\). Using the same arguments as with \(q_{1}\) we obtain
$$ D^{\alpha}q_{k}(t)\geq-\Biggl(M_{1}-\sum _{s=2}^{n}M_{s}\Biggr)q_{k}(t) $$
and
$$t^{1-\alpha}q_{k}(t)|_{t=0}\geq0, $$
which, due to Lemma 2.2, gives
$$ u_{k}^{1}\leq\sum _{s=2}^{n}u^{i}_{k}. $$
(3.23)
Summarizing, by (3.21)-(3.23) and induction, we obtain the following inequalities describing the sequences \((u_{k}^{s})_{k \in\mathbb {N}\cup\{0\}}\):
$$\begin{aligned} u^{1}_{0}-\Biggl(\sum _{i=2}^{n}u^{i}_{0}-u^{s}_{0} \Biggr) \leq& u^{1}_{1}-\Biggl(\sum _{i=2}^{n}u^{i}_{1}-u^{s}_{1} \Biggr) \\ \leq&\cdots\leq u^{1}_{k}-\Biggl(\sum _{i=2}^{n}u^{i}_{k}-u^{s}_{k} \Biggr) \\ \leq& u_{k}^{s}\leq\cdots \leq u_{1}^{s} \leq u_{0}^{s}, \end{aligned}$$
(3.24)
where \(2\leq s \leq n\). The inequalities (3.24) imply
$$\lim_{k\to\infty}u_{k}^{s} (t)= \bar{u}^{s}(t),\quad s=2,\ldots, n. $$
Observe that
$$u^{1}_{0}-\Biggl(\sum_{i=2}^{n}u^{i}_{0}-u^{s}_{0} \Biggr) \leq\bar{u}^{s} \leq u_{0}^{s},\quad s=2, \ldots, n. $$
In order to show that the sequence \((u_{k}^{1})\) is convergent observe first that from (3.24) there exists a function \(x^{*}\) such that
$$\lim_{k\to\infty}\Biggl(u^{1}_{k}(t)-\sum _{i=2}^{n-1}u^{i}_{k}(t) \Biggr)=x^{*}(t). $$
Hence, putting \(\bar{u}^{1}=x^{*}+\sum_{s=2}^{n-1}\bar{u}^{s}\), we have
$$\begin{aligned} \lim_{k\to\infty}\bigl(u^{1}_{k}(t) - \bar{u}^{1}(t)\bigr)&=\lim_{k\to\infty }\Biggl(u^{1}_{k}(t)-x^{*}(t)- \sum_{s=2}^{n-1}\bar{u}^{s}(t)+\sum _{s=2}^{n-1}u_{k}^{s}(t)- \sum_{s=2}^{n-1}u_{k}^{s}(t) \Biggr) \\ &=\lim_{k\to\infty}\Biggl(u^{1}_{k}(t)-\sum _{s=2}^{n-1}u_{k}^{s}(t)-x^{*}(t)+ \sum_{s=2}^{n-1}\bigl(u_{k}^{s}(t)- \bar{u}^{s}(t)\bigr)\Biggr) \\ &=\lim_{k\to\infty}\Biggl(u^{1}_{k}(t)-\sum _{s=2}^{n-1}u_{k}^{s}(t)-x^{*}(t) \Biggr)+\sum_{s=2}^{n-1}\lim _{k\to\infty }\bigl(u_{k}^{s}(t)- \bar{u}^{s}(t)\bigr)=0. \end{aligned}$$
In order to show the uniform convergence of sequences \((u^{2}_{k}), (u^{3}_{k}),\ldots, (u^{n}_{k})\), observe that from (3.24) and from the fact that \(u^{s}_{k} \to\bar{u}^{s}\), \(s=2,3,\ldots,n\), we have
$$\bar{u}^{s} \leq u_{k}^{s} \leq\cdots\leq u_{1}^{s} \leq u_{0}^{s} \quad \text{for all } k \in\mathbb{N}. $$
Then, the uniform convergence of sequences \((u_{k}^{s})\), \(s=2,3,\ldots, n\), on a compact subset of \((0,T]\) is a straightforward consequence of Dini’s theorem, which states that if a monotone sequence of continuous functions is convergent on a compact set, then it converges uniformly.
Showing a uniform convergence of \((u_{k}^{1})\) requires some observations. Take any \(2\leq s \leq n\) and denote
$$h_{k} = u^{1}_{k} - \Biggl(\sum _{i=2}^{n}u_{k}^{i} - u_{k}^{s}\Biggr), \quad k \in\mathbb{N}\cup\{0\}. $$
From (3.24) and the convergence of \((u_{k}^{1}), \ldots, (u_{k}^{n})\) we have
$$h_{0} \leq h_{1} \leq\cdots\leq h_{k} \leq \bar{u}^{1} - \Biggl(\sum_{i=2}^{n} \bar {u}^{i} - \bar{u}^{s}\Biggr). $$
Applying again Dini’s result we get the uniform convergence of \((h_{k})\) on every compact subset of \((0,T]\). Finally note that
$$u_{k}^{1} = h_{k} + \Biggl(\sum _{i=2}^{n}u_{k}^{i} - u_{k}^{s}\Biggr), \quad k \in\mathbb{N}, $$
and thus \((u^{1}_{k})\) is uniformly convergent on a compact subset of \((0,T]\) to \(\bar{u}^{1}\) as a linear combination of sequences uniformly convergent.
Moreover, observe that the limit functions satisfy the properties
$$\begin{aligned}& (n-1)u_{0}^{1}-(n-2)\sum_{i=2}^{n}u^{i}_{0} \leq\bar{u}^{1}\leq\sum_{i=2}^{n}u^{i}_{0}, \\& u_{0}^{1} - \sum_{i=2}^{n}u_{0}^{i}+u_{0}^{s} \leq\bar{u}^{s} \leq u_{0}^{s}, \quad 2\leq s\leq n. \end{aligned}$$
Taking k to ∞ in (3.16) we see that \((\bar{u}^{1},\bar {u}^{2},\ldots,\bar{u}^{n})\) is a system of solutions of system (1.1). Also observe that from (3.24) we have the following relations between the limit functions:
$$\bar{u}^{1}-\Biggl(\sum_{i=2}^{n} \bar{u}^{i}-\bar{u}^{s}\Biggr)\leq\bar{u}^{s}, \quad 2\leq s \leq n, $$
which ends the proof. □

Remark 3.1

Observe that using the same methods as in the proof of Theorem 3.2 we can see that \((\bar{u}^{1},\bar{u}^{2},\ldots,\bar{u}^{n})\) is an extremal solution of system (1.1) in the sense that if \((u^{1},\ldots,u^{n})\) were any other solution such that
$$ u_{0}^{1}-\Biggl(\sum_{i=2}^{n}u_{0}^{i}-u_{0}^{s} \Biggr)\leq u^{1}-\Biggl(\sum_{i=2}^{n}u^{i}-u^{s} \Biggr)\leq u_{0}^{s}, \qquad u_{0}^{1}- \Biggl(\sum_{i=2}^{n}u_{0}^{i}-u_{0}^{s} \Biggr)\leq u^{s} \leq u_{0}^{s} $$
for any \(2\leq s \leq n\), then we would have
$$\bar{u}^{1}-\Biggl(\sum_{i=2}^{n} \bar{u}^{i}-\bar{u}^{s}\Biggr)\leq u^{1} - \Biggl( \sum_{i=2}^{n}u^{i} - u^{s}\Biggr),\quad u^{s} \leq\bar{u}^{s}, 2\leq s \leq n. $$

4 The system of three fractional differential equations

In order to see the nature of the iterative procedure introduced in the proof of Theorem 3.2, we consider the case \(n=3\).

Corollary 4.1

If there exist \(u_{0},v_{0},w_{0} \in C_{1-\alpha}([0,T])\), \(u_{0}\leq v_{0} + w_{0}\) such that
$$ \left \{ \begin{array}{l} D^{\alpha}u_{0}(t) \leq f(t,u_{0}(t),v_{0}(t),w_{0}(t)), \quad t\in(0,T], \\ D^{\alpha}v_{0}(t) \geq g(t,u_{0}(t),v_{0}(t),w_{0}(t)),\quad t\in(0,T], \\ D^{\alpha}w_{0}(t) \geq h(t,u_{0}(t),v_{0}(t),w_{0}(t)),\quad t\in(0,T], \\ t^{1-\alpha}u_{0}(t)|_{t=0} \leq x_{0}, \\ t^{1-\alpha}v_{0}(t)|_{t=0} \geq y_{0}, \\ t^{1-\alpha}w_{0}(t)|_{t=0} \geq z_{0}, \end{array} \right . $$
(4.1)
and there exist \(M\in\mathbb{R}\), \(N,S \geq0\) satisfying
$$\begin{aligned}& f(t,\alpha_{1},\alpha_{2},\alpha_{3})-f(t, \beta_{1},\beta_{2},\beta_{n}) \geq- M( \alpha_{1}-\beta_{1}) + (-N+S) (\alpha_{2}- \beta_{2})+(N-S) (\alpha _{3}-\beta_{3}), \\& g(t,\alpha_{1},\alpha_{2},\alpha_{3})-g(t, \beta_{1},\beta_{2},\beta_{3}) \geq-N( \alpha_{1}-\beta_{1})+(-M+S) (\alpha_{2}- \beta_{2})+N(\alpha_{3}-\beta _{3}), \\& h(t,\alpha_{1},\alpha_{2},\alpha_{3})-h(t, \beta_{1},\beta_{2},\beta_{3}) \geq-S( \alpha_{1}-\beta_{1})+S(\alpha_{2}- \beta_{2})+(-M+N) (\alpha_{3}-\beta_{3}), \end{aligned}$$
where \(\alpha_{i}, \beta_{i} \in\mathbb{R}\), \(1\leq i \leq3\) satisfy, for all \(t \in[0,T]\),
$$\begin{aligned}& u_{0}(t)-w_{0}(t)\leq\beta_{1}- \beta_{3} \leq\alpha_{1}-\alpha_{3}\leq v_{0}(t),\qquad u_{0}(t)-w_{0}(t)\leq \alpha_{2} \leq\beta_{2} \leq v_{0}(t), \\& u_{0}(t)-v_{0}(t)\leq\beta_{1}- \beta_{2} \leq\alpha_{1}-\alpha_{2}\leq w_{0}(t),\qquad u_{0}(t)-v_{0}(t)\leq \alpha_{3} \leq\beta_{3} \leq w_{0}(t) \end{aligned}$$
and
$$ (g+h-f) (t,u,v,w)\geq(-M+N+S) (v+w-u), $$
(4.2)
where
$$\begin{aligned}& u_{0}(t)-w_{0}(t)\leq u - w \leq v \leq v_{0}(t), \\& u_{0}(t)-v_{0}(t)\leq u - v \leq w \leq w_{0}(t). \end{aligned}$$
Then there exists a solution
$$\bigl(u^{*},v^{*},w^{*}\bigr)\in[2u_{0}-v_{0}-w_{0},v_{0}+w_{0}] \times[u_{0}-w_{0},v_{0}] \times[u_{0}-v_{0},w_{0}] $$
of (4.1) and the sequences \((u_{n}) \subseteq [2u_{0}-v_{0}-w_{0},v_{0}+w_{0}]\), \((v_{n}) \subseteq[u_{0}-w_{0},v_{0}] \), \((w_{n}) \subseteq [u_{0}-v_{0},w_{0}]\) such that \(u_{n} \to u^{*}\), \(v_{n} \to v^{*}\), \(w_{n} \to w^{*}\) uniformly on compact subsets of \((0,T]\). Moreover, the following inequalities hold:
$$\begin{aligned}& u_{0} - v_{0} \leq u_{1} - v_{1}\leq \cdots\leq u_{n}-v_{n}\leq\cdots\leq u^{*}-v^{*} \leq w^{*} \leq \cdots\leq w_{n}\leq\cdots\leq w_{1} \leq w_{0}, \\& u_{0} - w_{0} \leq u_{1} - w_{1}\leq \cdots\leq u_{n}-w_{n}\leq\cdots\leq u^{*}-w^{*} \leq v^{*} \leq \cdots\leq v_{n}\leq\cdots\leq v_{1} \leq v_{0}. \end{aligned}$$

4.1 Example

Consider the nonlinear problem of the form
$$ \left \{ \begin{array}{l} D^{0.5}u(t) = \Gamma(1.5)^{-1}v(t)-\Gamma (1.5)^{-1}w(t)+(v(t)-t)^{2}+2(t-w(t)+u(t))^{2}, \\ D^{0.5}v(t) = \Gamma(1.5)^{-1}v(t)+(v(t)-t)^{2}+(t-w(t)+u(t))^{2}, \\ D^{0.5}w(t) = -\Gamma(1.5)^{-1}u(t)+\Gamma (1.5)^{-1}v(t)+(t-w(t)+u(t))^{2}, \\ t^{0.5}u(t)|_{t=0} = t^{0.5}v(t)|_{t=0} = t^{0.5}w(t)|_{t=0} = 0, \end{array} \right . $$
(4.3)
where \(t\in[0,1]\). Taking
$$\begin{aligned}& f(t,u,v,w) = \Gamma(1.5)^{-1}v-\Gamma(1.5)^{-1}w+(v-t)^{2}+2(t-w+u)^{2}, \\& g(t,u,v,w) = \Gamma(1.5)^{-1}v+(v-t)^{2}+(t-w+u)^{2}, \\& h(t,u,v,w) = -\Gamma(1.5)^{-1}u+\Gamma(1.5)^{-1}v+(t-w+u)^{2} \end{aligned}$$
and
$$u_{0}(t)=0, \qquad v_{0}(t)=w_{0}(t)=t, \quad t \in[0,1], $$
we obtain, for all \(t\in[0,1]\),
$$\begin{aligned}& D^{0.5}u_{0}(t)=0 = f\bigl(t,u_{0}(t),v_{0}(t),w_{0}(t) \bigr), \\& D^{0.5}v_{0}(t)=\frac{\sqrt{t}}{\Gamma(1.5)} \geq\frac{t}{\Gamma(1.5)} = g \bigl(t,u_{0}(t),v_{0}(t),w_{0}(t)\bigr), \\& D^{0.5}w_{0}(t)=\frac{\sqrt{t}}{\Gamma(1.5)} \geq0 = h \bigl(t,u_{0}(t),v_{0}(t),w_{0}(t)\bigr). \end{aligned}$$
Next, for all \(\alpha_{i}, \beta_{i} \in\mathbb{R}\), \(1\leq i \leq3\) such that
$$\begin{aligned} \begin{aligned} &-t\leq\beta_{1} - \beta_{3} \leq\alpha_{1}- \alpha_{3}\leq t,\qquad -t\leq\alpha _{2}\leq \beta_{2} \leq t, \\ &-t\leq\beta_{1} - \beta_{2} \leq\alpha_{1}- \alpha_{2}\leq t, \qquad -t\leq\alpha _{3}\leq \beta_{3} \leq t, \end{aligned} \end{aligned}$$
one can calculate that
$$\begin{aligned}& f(t,\alpha_{1},\alpha_{2},\alpha_{3})-f(t, \beta_{1},\beta_{2},\beta_{3})\geq\Gamma (1.5)^{-1}(\alpha_{2}-\beta_{2})- \Gamma(1.5)^{-1}(\alpha_{3}-\beta_{3}), \\& g(t,\alpha_{1},\alpha_{2},\alpha_{3})-g(t, \beta_{1},\beta_{2},\beta_{3})\geq\Gamma (1.5)^{-1}(\alpha_{2}-\beta_{2}), \\& h(t,\alpha_{1},\alpha_{2},\alpha_{3})-h(t, \beta_{1},\beta_{2},\beta_{3})\geq - \Gamma(1.5)^{-1}(\alpha_{1}-\beta_{1})+ \Gamma(1.5)^{-1}(\alpha_{2}-\beta_{2}). \end{aligned}$$
Therefore it is sufficient to take in Corollary 4.1 \(M=0\), \(N=0\), \(S=\Gamma(1.5)^{-1}\). Finally observe that condition (4.2) also holds. Thus, the system of fractional differential equations (4.3) has a solution \((u^{*},v^{*},w^{*}) \in[-2t,2t]\times[-t,t] \times[-t,t]\).
Now, using the proof of Theorem 3.2 and Lemma 3.1, we can derive the iterative procedure \((u_{k},v_{k},w_{k})\) convergent to the solution \((u^{*},v^{*},w^{*})\). First observe that the sequences \((u_{k})\), \((v_{k})\), \((w_{k})\) satisfy the following system of linear equations:
$$\begin{aligned}& D^{0.5}u_{k}=f(t,u_{k-1},v_{k-1},w_{k-1})- \Gamma(1.5)^{-1} v_{k-1}+\Gamma(1.5)^{-1}w_{k-1}+ \Gamma(1.5)^{-1}v_{k}-\Gamma (1.5)^{-1}w_{k}, \\& D^{0.5}v_{k}=g(t,u_{k-1},v_{k-1},w_{k-1})- \Gamma(1.5)^{-1}v_{k-1}+\Gamma (1.5)^{-1}v_{k}, \\& D^{0.5}w_{k}=h(t,u_{k-1},v_{k-1},w_{k-1})+ \Gamma(1.5)^{-1}u_{k-1} -\Gamma(1.5)^{-1}v_{k-1}- \Gamma(1.5)^{-1}u_{k}+\Gamma(1.5)^{-1}v_{k}, \\& t^{0.5}u_{k}(t)|_{t=0}=t^{0.5}v_{k}(t)|_{t=0}=t^{0.5}w_{k}(t)|_{t=0}=0, \end{aligned}$$
which can be equivalently transformed to the system
$$ \left \{ \begin{array}{l} u_{k}+v_{k}+w_{k}=p_{k}, \\ u_{k}-v_{k}+w_{k}=q_{k}, \\ u_{k}+v_{k}-w_{k}=r_{k}, \end{array} \right . $$
where \(p_{k}\), \(q_{k}\), \(r_{k}\) are the solutions of the following systems:
$$\begin{aligned}& \left \{ \begin{array}{l} D^{0.5}p_{k} = (f+g+h)(t,u_{k-1},v_{k-1},w_{k-1}) \\ \hphantom{D^{0.5}p_{k} =}{}+\Gamma (1.5)^{-1}u_{k-1}-3\Gamma(1.5)^{-1}v_{k-1}+\Gamma (1.5)^{-1}w_{k-1}-\Gamma(1.5)^{-1}p_{k}, \\ t^{0.5}p_{k}(t)|_{t=0} = 0, \end{array} \right . \\& \left \{ \begin{array}{l} D^{0.5}q_{k} = (f-g+h)(t,u_{k-1},v_{k-1},w_{k-1}) \\ \hphantom{D^{0.5}q_{k} =}{}+\Gamma (1.5)^{-1}u_{k-1}-\Gamma(1.5)^{-1}v_{k-1}+\Gamma(1.5)^{-1}w_{k-1}-\Gamma (1.5)^{-1}q_{k}, \\ t^{0.5}q_{k}(t)|_{t=0} = 0, \end{array} \right . \\& \left \{ \begin{array}{l} D^{0.5}r_{k} = (f+g-h)(t,u_{k-1},v_{k-1},w_{k-1}) \\ \hphantom{D^{0.5}r_{k} =}{}-\Gamma (1.5)^{-1}u_{k-1}-\Gamma(1.5)^{-1}v_{k-1}+\Gamma(1.5)^{-1}w_{k-1}+\Gamma (1.5)^{-1}r_{k}, \\ t^{0.5}r_{k}(t)|_{t=0} = 0. \end{array} \right . \end{aligned}$$
The solutions of the above systems, due to Lemma 2.1, are given by the formulas
$$\begin{aligned}& p_{k}(t)= \int_{0}^{t} (t-s)^{-0.5}E_{0.5,0.5}\bigl(-\Gamma (1.5)^{-1}(t-s)^{0.5} \bigr) \bigl((f+g+h) \bigl(s,u_{k-1}(s),v_{k-1}(s),w_{k-1}(s) \bigr) \\& \hphantom{p_{k}(t)=}{} +\Gamma(1.5)^{-1}u_{k-1}(s)-3 \Gamma(1.5)^{-1}v_{k-1}(s)+\Gamma (1.5)^{-1}w_{k-1}(s) \bigr)\, ds, \\& q_{k}(t)= \int_{0}^{t} (t-s)^{-0.5}E_{0.5,0.5}\bigl(-\Gamma (1.5)^{-1}(t-s)^{0.5} \bigr) \bigl((f-g+h) \bigl(s,u_{k-1}(s),v_{k-1}(s),w_{k-1}(s) \bigr) \\& \hphantom{q_{k}(t)=}{} +\Gamma(1.5)^{-1}u_{k-1}(s)- \Gamma(1.5)^{-1}v_{k-1}(s)+\Gamma (1.5)^{-1}w_{k-1}(s) \bigr)\, ds, \\& r_{k}(t)= \int_{0}^{t} (t-s)^{-0.5}E_{0.5,0.5}\bigl(\Gamma (1.5)^{-1}(t-s)^{0.5} \bigr) \bigl((f+g-h) \bigl(s,u_{k-1}(s),v_{k-1}(s),w_{k-1}(s) \bigr) \\& \hphantom{r_{k}(t)=}{} -\Gamma(1.5)^{-1}u_{k-1}(s)- \Gamma(1.5)^{-1}v_{k-1}(s)+\Gamma (1.5)^{-1}w_{k-1}(s) \bigr)\, ds. \end{aligned}$$
In consequence, the iterative sequences are of the form
$$\begin{aligned}& u_{k}(t)=\frac{1}{2}(q_{k}+r_{k}) \\& \hphantom{u_{k}(t)}=\frac{1}{2}\int_{0}^{t}(t-s)^{-0.5} \bigl[E_{0.5,0.5}\bigl(-\Gamma (1.5)^{-1}(t-s)^{0.5} \bigr) \bigl((f-g+h) \bigl(s,u_{k-1}(s),v_{k-1}(s),w_{k-1}(s) \bigr) \\& \hphantom{u_{k}(t)=}{} +\Gamma(1.5)^{-1}u_{k-1}(s)- \Gamma(1.5)^{-1}v_{k-1}(s)+\Gamma (1.5)^{-1}w_{k-1}(s) \bigr) \\& \hphantom{u_{k}(t)=}{} +E_{\alpha,\alpha}\bigl(\Gamma{1.5}^{-1}(t-s)^{\alpha } \bigr) \bigl((f+g-h) \bigl(s,u_{k-1}(s),v_{k-1}(s),w_{k-1}(s) \bigr) \\& \hphantom{u_{k}(t)=}{} -\Gamma(1.5)^{-1}u_{k-1}(s)- \Gamma(1.5)^{-1}v_{k-1}(s)+\Gamma (1.5)^{-1}w_{k-1}(s) \bigr) \bigr]\, ds, \\& v_{k}(t)=\frac{1}{2}(p_{k}-q_{k}) \\& \hphantom{v_{k}(t)}=\frac{1}{2}\int_{0}^{t}(t-s)^{-0.5} \bigl[E_{0.5,0.5}\bigl(-\Gamma (1.5)^{-1}(t-s)^{0.5} \bigr) \bigl((f+g+h) \bigl(s,u_{k-1}(s),v_{k-1}(s),w_{k-1}(s) \bigr) \\& \hphantom{v_{k}(t)=}{} +\Gamma(1.5)^{-1}u_{k-1}(s)-3 \Gamma(1.5)^{-1}v_{k-1}(s)+\Gamma (1.5)^{-1}w_{k-1}(s) \bigr) \\& \hphantom{v_{k}(t)=}{} -E_{0.5,0.5}\bigl(-\Gamma (1.5)^{-1}(t-s)^{0.5} \bigr) \bigl((f-g+h) \bigl(s,u_{k-1}(s),v_{k-1}(s),w_{k-1}(s) \bigr) \\& \hphantom{v_{k}(t)=}{} +\Gamma(1.5)^{-1}u_{k-1}(s)- \Gamma(1.5)^{-1}v_{k-1}(s)+\Gamma (1.5)^{-1}w_{k-1}(s) \bigr) \bigr]\, ds, \\& w_{k}(t)=\frac{1}{2}(p_{k}-r_{k}) \\& \hphantom{w_{k}(t)}=\frac{1}{2}\int_{0}^{t}(t-s)^{-0.5} \bigl[E_{0.5,0.5}\bigl(-\Gamma (1.5)^{-1}(t-s)^{0.5} \bigr) \bigl((f+g+h) \bigl(s,u_{k-1}(s),v_{k-1}(s),w_{k-1}(s) \bigr) \\& \hphantom{w_{k}(t)=}{} +\Gamma(1.5)^{-1}u_{k-1}(s)-3 \Gamma(1.5)^{-1}v_{k-1}(s)+\Gamma (1.5)^{-1}w_{k-1}(s) \bigr) \\& \hphantom{w_{k}(t)=}{} -E_{0.5,0.5}\bigl(\Gamma (1.5)^{-1}(t-s)^{0.5} \bigr) \bigl((f+g-h) \bigl(s,u_{k-1}(s),v_{k-1}(s),w_{k-1}(s) \bigr) \\& \hphantom{w_{k}(t)=}{} -\Gamma(1.5)^{-1}u_{k-1}(s)- \Gamma(1.5)^{-1}v_{k-1}(s)+\Gamma (1.5)^{-1}w_{k-1}(s) \bigr) \bigr]\, ds. \end{aligned}$$

Declarations

Acknowledgements

The author is very grateful to the reviewers for the remarks, which improved the final version of the manuscript. This article was financially supported by University of Łódź as a part of donation for the research activities aimed at the development of young scientists, grant no. 545/1117.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Nonlinear Analysis, Faculty of Mathematics and Computer Science, University of Łódź, Łódź, Poland

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