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# Existence results for nonlinear fractional differential equations involving different Riemann-Liouville fractional derivatives

- Guotao Wang
^{1}, - Sanyang Liu
^{1}Email author, - Dumitru Baleanu
^{2, 3, 4}and - Lihong Zhang
^{5}

**2013**:280

https://doi.org/10.1186/1687-1847-2013-280

© Wang et al.; licensee Springer 2013

**Received:**9 May 2013**Accepted:**15 August 2013**Published:**4 October 2013

## Abstract

By applying an iterative technique, a necessary and sufficient condition is obtained for the existence of the unique solution of nonlinear fractional differential equations involving two Riemann-Liouville derivatives of different fractional orders. Finally, an example is also given to illustrate the availability of our main results.

## Keywords

- different fractional-order
- nonlinear fractional differential equations
- Riemann-Liouville derivative
- monotone iterative technique

## 1 Introduction

where $t\in J=[0,T]$ ($0<T<\mathrm{\infty}$), $f\in C(J\times {\mathbb{R}}^{3},\mathbb{R})$, *D* is the standard Riemann-Liouville fractional derivative, $1<\alpha \le 2$, $0<\beta \le 1$ and $0<\alpha -\beta \le 1$. It is worthwhile to indicate that the nonlinear term *f* involves the unknown function’s Riemann-Liouville fractional derivatives with different orders.

The method of upper and lower solutions coupled with the monotone iterative technique is an interesting and powerful mechanism. The importance and advantage of the method needs no special emphasis [6, 7]. There have appeared some papers dealing with the existence of the solution of nonlinear Riemann-Liouville-type fractional differential equations [8–18] or nonlinear Caputo-type fractional differential equations [19–22] by using the method. For example, by employing the method of lower and upper solutions combined with the monotone iterative technique, Lakshmikanthan and Vatsala [13], McRae [14] and Zhang [17] successfully investigated the initial value problems of Riemann-Liouville fractional differential equation ${D}^{\alpha}u(t)=f(t,u(t))$, where $0<\alpha \le 1$.

However, in the existing literature [8–18], only one case when $\alpha \in (0,1]$ is considered. The research, involving Riemann-Liouville fractional derivative of order $1<\alpha \le 2$, proceeds slowly and there appear some new difficulties in employing the monotone iterative method. To overcome these difficulties, we apply a substitution ${D}^{\alpha}u(t)=y(t)$. Note that the technique has been discussed for fractional problems in papers [10, 11]. To the best of our knowledge, it is the first paper, in which the monotone iterative method is applied to nonlinear Riemann-Liouville-type fractional differential equations, involving two different fractional derivatives ${D}^{\alpha}$ and ${D}^{\beta}$.

We organize the rest of this paper as follows. In Section 2, by using the monotone iterative technique and the method of upper and lower solutions, the minimal and maximal solutions of an equivalent problem of (1.1) are investigated and two explicit monotone iterative sequences, converging to the corresponding minimal and maximal solution, are given. In addition, the uniqueness of the solution for fractional differential equations (1.1) is discussed. In Section 3, an example is given to illustrate our results.

## 2 Existence results

**Lemma 2.1**

*For a given function*$y\in C(J,\mathbb{R})$,

*the following problem*

*has a unique solution* $u(t)={I}^{\alpha}y(t)$, *where* *I* *is the fractional integral and* ${I}^{\alpha}y(t)={\int}_{0}^{t}\frac{{(t-s)}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )}y(s)\phantom{\rule{0.2em}{0ex}}ds$, $1<\alpha \le 2$, $0<\beta \le 1$ *and* $0<\alpha -\beta \le 1$.

*Proof*One can reduce equation ${D}^{\alpha}u(t)=y(t)$ to an equivalent integral equation

for some ${c}_{1},{c}_{2}\in \mathbb{R}$.

By the condition ${D}^{\beta}u(0)=0$, it follows that ${c}_{1}=0$. Therefore, we have $u(t)={I}^{\alpha}y(t)$.

Conversely, by a direct computation, we can get ${D}^{\alpha}u(t)=y(t)$ and ${D}^{\beta}u(t)={I}^{\alpha -\beta}y(t)$. It is easy to verify $u(t)={I}^{\alpha}y(t)$ satisfies (2.1).

This completes the proof. □

where $y(t)={D}^{\alpha}u(t)$, $\mathrm{\forall}t\in J$ and ${I}^{\alpha}$, ${I}^{\alpha -\beta}$ are the standard fractional integrals.

Now, we list for convenience the following condition:

_{1}) There exist ${y}_{0},{z}_{0}\in C(J,\mathbb{R})$ satisfying ${y}_{0}\le {z}_{0}$ such that

_{2}) There exists a function $M\in C(J,(-1,+\mathrm{\infty}))$ such that

where ${y}_{0}\le v\le u\le {z}_{0}$, $\mathrm{\forall}t\in J$.

_{3}) There exist functions $N,K,L\in C(J,[0,+\mathrm{\infty}))$ such that

where ${y}_{0}\le v\le u\le {z}_{0}$, $\mathrm{\forall}t\in J$.

**Theorem 2.1**

*Assume that*(H

_{1})

*and*(H

_{2})

*hold*.

*Then problem*(2.5)

*has the minimal and maximal solution*${y}^{\ast}$, ${z}^{\ast}$

*in the ordered interval*$[{y}_{0},{z}_{0}]$.

*Moreover*,

*there exist explicit monotone iterative sequences*$\{{y}_{n}\},\{{z}_{n}\}\subset [{y}_{0},{z}_{0}]$

*such that*${lim}_{n\to \mathrm{\infty}}{y}_{n}(t)={y}^{\ast}(t)$

*and*${lim}_{n\to \mathrm{\infty}}{z}_{n}(t)={z}^{\ast}(t)$,

*where*${y}_{n}(t)$, ${z}_{n}(t)$

*are defined as*

*and*

*Proof*Define an operator $Q:[{y}_{0},{z}_{0}]\to C(J,\mathbb{R})$ by $x=Q\eta $, where

*x*is the unique solution of the corresponding linear problem corresponding to $\eta \in [{y}_{0},{z}_{0}]$ and

*Q*has the following properties:

_{1}) and the definition of

*Q*, we know that

Thus, we can obtain $p(t)\ge 0$, $\mathrm{\forall}t\in J$. That is, ${y}_{0}\le Q{y}_{0}$. Similarly, we can prove that $Q{z}_{0}\le {z}_{0}$. Then, (a) holds.

_{2}), we have

Hence, we have $q(t)\ge 0$, $\mathrm{\forall}t\in J$. That is, $Q{h}_{2}\ge Q{h}_{1}$. Then, (b) holds.

Employing the same arguments used in Ref. [17], we see that $\{{y}_{n}\}$, $\{{z}_{n}\}$ converge to their limit functions ${y}^{\ast}$, ${z}^{\ast}$, respectively. That is, ${lim}_{n\to \mathrm{\infty}}{y}_{n}(t)={y}^{\ast}(t)$ and ${lim}_{n\to \mathrm{\infty}}{z}_{n}(t)={z}^{\ast}(t)$. Moreover, ${y}^{\ast}(t)$, ${z}^{\ast}(t)$ are solutions of (2.5) in $[{y}_{0},{z}_{0}]$. (2.7) is true.

Thus, taking limit in (2.12) as $n\to +\mathrm{\infty}$, we have ${y}^{\ast}\le w\le {z}^{\ast}$. That is, ${y}^{\ast}$, ${z}^{\ast}$ are the minimal and maximal solution of (2.5) in the ordered interval $[{y}_{0},{z}_{0}]$, respectively.

This completes the proof. □

**Theorem 2.2**

*Let*$N(t)\ge -M(t)$.

*Assume conditions*(H

_{1})-(H

_{3})

*hold*.

*If*

*then problem* (2.5) *has a unique solution* $x(t)\in [{y}_{0},{z}_{0}]$.

*Proof*By Theorem 2.1, we have proved that ${y}^{\ast}$, ${z}^{\ast}$ are the minimal and maximal solution of (2.5) and

Now, we are going to show that problem (2.5) has a unique solution *x*, *i.e.*, ${y}^{\ast}(t)={z}^{\ast}(t)=x(t)$.

_{3}), we have

which implies that ${max}_{t\in J}p(t)\le 0$. Since $p(t)\ge 0$, then it holds $p(t)=0$. That is, ${y}^{\ast}(t)={z}^{\ast}(t)$. Therefore, problem (2.5) has a unique solution $x\in [{y}_{0},{z}_{0}]$. □

Let $x(t)$ be the unique solution of (2.5). Noting that $x\in [{y}_{0},{z}_{0}]$ and $u(t)={I}^{\alpha}x(t)$, we can easily obtain the following theorem.

**Theorem 2.3** *Let all conditions of Theorem * 2.2 *hold*. *Then problem* (1.1) *has a unique solution* $u\in [{I}^{\alpha}{y}_{0},{I}^{\alpha}{z}_{0}]$, $\mathrm{\forall}t\in J$.

## 3 Example

where $t\in [0,1]$.

Hence, condition (H_{1}) holds.

Then, all conditions of Theorem 2.3 are satisfied. In consequence, the problem (3.1) has a unique solution ${u}^{\ast}\in [0,\frac{4{t}^{\frac{3}{2}}}{3\sqrt{\pi}}]$.

## Declarations

### Acknowledgements

The authors would like to thank the referees for their useful comments and remarks. This work is supported by the NNSF of China (No.61373174) and the Natural Science Foundation for Young Scientists of Shanxi Province, China (No. 2012021002-3).

## Authors’ Affiliations

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