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Existence of Solutions of Three-Dimensional Fractional Differential Systems

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In this article, we consider the three-dimensional fractional differential system of the form

together with the Neumann boundary conditions,

where are the standard Caputo fractional derivatives, . A new result on the existence of solutions for a class of fractional differential system is obtained by using Mawhin’s coincidence degree theory. Suitable examples are given to illustrate the main results.

1. Introduction

Hu et al. [22] investigated the two-point boundary value problem for fractional differential equations of the following form

where is the Caputo fractional differential operator, , and is continuous.

In [23] , Hu et al. extended the above boundary value problem to the existence of solutions for the following coupled system of fractional differential equations of the form

where are the Caputo fractional derivatives, , and is continuous.

It seems that there has been no work done on the boundary value problem of system involving three nonlinear fractional differential equations. Motivated by the above observation, we investigate the following three-dimensional fractional differential system of the form

(1)

together with the Neumann boundary conditions,

where are the standard Caputo fractional derivatives, , and is continuous.

The main goal of this paper is to establish some new criteria for the existence of solutions of (1). The method is based on Mawhin’s coincidence degree theory. The results in this paper are generalized of the existing ones.

2. Preliminaries

In this section, we give the definitions of fractional derivatives and integrals and some notations which are useful throughout this paper. There are several kinds of definitions of fractional derivatives and integrals. In this paper, we use the Riemann-Liouville left sided definition on the half-axis and the Caputo fractional derivative.

Let X and Y be real Banach spaces and let be a Fred- holm operator with index zero if and ImL is closed in Y and there exist continuous projectors such that

It follows that

is invertible. Here denotes the inverse of.

If is an open bounded subset of X, and, then the map will be called L-compact on, if is bounded and is compact, where I is the identity operator.

Lemma 1. [14] Let be a Fredholm operator with index zero and be L-compact on. Assume that the following condi- tions are satisfied.

1. for every;

2. for every;

3., where is a projection such that.

Then the operator equation has at least one solution in .

Definition 1. [6] The Riemann-Liouville fractional integral of order of a function on the half-axis is given by

provided the right hand side is pointwise defined on

Definition 2. [6] Assume that is -times absolutely continuous function, the Caputo fractional derivative of order of x is given by

where n is the smallest integer greater than or equal to, provided that the right side integral is pointwise defined on.

Lemma 2. [6] Let and. If, then

where, here n is the smallest integer greater

than or equal to.

In this paper, let us take with the norm and with the norm, where. Then we denote with the norm and with the norm. Clearly, both and are Banach spaces.

Define the operators by

where

and

Define the operator by

(2)

where

Let the Nemytski operator be defined as

where is defined by

is defined by

and is defined by

Then Neumann boundary value problem (1) is equivalent to the operator equation

3. Main Results

In this section, we begin with the following theorem on existence of solutions for

Neumann boundary value problem (1).

Theorem 1. Let be continuous. Assume that

(H1) there exist nonnegative functions with

such that for all

where;

(H2) there exists a constant such that for all either

or

(H3) there exists a constant such that for every satisfying either

or

Then Neumann boundary value problem (1) has at least one solution.

Lemma 3. Let L be defined by (2). Then

(3)

and

(4)

Proof. By Lemma 2, has the solution

From the boundary conditions, we have

For, there exists such that. By using the Lemma 2, we get

Then, we have

By the boundary value conditions of (1), we can get that x satisfies

On the other hand, suppose and satisfies Let then and. Hence,. Then we get

Similarly, we have

and

Lemma 4. Let L be defined by (2). Then L is a Fredholm operator of index zero, and are the linear continuous projector opera- tors can be defined as

Further more, the operator can be written by

Proof. Clearly, and. It follows that , we have. By using simple calculation, we get that. Then we have

For, we have

By the definition of, we get

Similarly, we can show that and. Thus, we can get .

Let

,

where . It follows that and, we get. It is clear that

Thus

Hence L is a Fredholm operator of index zero.

From the definitions of P and, we will prove that is the inverse of. Infact, for, we have

(5)

Moreover, for, we have and

which together with the boundary condition yields that

(6)

From (5) and (6), we get.

Lemma 5. Assume is an open bounded subset such that , then N is L-compact on.

Proof. By the continuity of f_{1}, f_{2} and f_{3}, we can get and are bounded. By the Arzela-Ascoli theorem, we will prove that is equicontinuous.

From the continuity of f_{1}, f_{2} and f_{3}, there exist constants such that for all.

Furthermore, for, we have

By

and

Similarly, we can show that

Since and are uniformly continuous on [0, 1], we have is equicontinuous. Thus is compact.

Lemma 6. Assume that hold, then the set

is bounded.

Proof. Let, then. By (4), we get

and

Then, by integral mean value theorem, there exist constants such that and Then we get

From, we get and. Hence we have

(7)

We obtain

(8)

Similarly, we can show that

(9)

and

(10)

By, we get

and

Then

and

So,

(12)

Similarly, we have

(12)

and

(13)

Combining (13) with (12), we get

(14)

Combining (14) with (11), we get

Thus, from and (14), we get

and

From (8), (9) and (10), we have

Hence is bounded.

Lemma 7. Assume that holds, then the set

is bounded.

Proof. For, we have. Then from,

and

From imply that. Thus, we get

Therefore is bounded.

Lemma 8. Assume that the first part of holds, then the set

is bounded.

Proof. For, we have and

(15)

(16)

and

(17)

If, then by, we get. If, then . For, we obtain. Otherwise, if or or, from, one has

or

or

which contradict to (15) or (16) or (17). Hence, is bounded.

Remark 1 Suppose the second part of holds, then the set

is bounded.

Proof of the Theorem 1: Set.

From the Lemma 4 and Lemma 5 we can get L is a Fredholm operator of index zero and N is L-compact on. By Lemma 6 and Lemma 7, we obtain

(1) for every;

(2) for every.

Choose

By Lemma 8 (or Remark 1), we get for. Therefore

Thus, the condition (3) of Lemma 1 is satisfied. By Lemma 1, we obtain has at least one solution in. Hence Neumann boundary value problem (1) has at least one solution. This completes the proof.

4. Examples

In this section, we give two examples to illustrate our main results.

Example 1. Consider the following Neumann boundary value problem of fractional differential equation of the form

(18)

Here. Moreover,

Now let us compute from.

From the above inequality, we get Also,

Here,. Finally,

We get,. And we get, . Choose. Also,

where and. All the condi-

tions of Theorem 1 are satisfied. Hence, boundary value problem (18) has at least one solution.

Example 2. Consider the Neumann boundary value problem of fractional differential equation of the following form

(19)

Here. Moreover,

Now let us compute from.

From the above inequality, we get Also,

Here,. Similarly,

Here,. We get,. Choose.

Also,

where and. Hence all the condi-

tions of Theorem 1 are satisfied. Therefore, boundary value problem (19) has at least one solution.

5. Conclusion

We have investigated some existence results for three-dimensional fractional differential system with Neumann boundary condition. By using Mawhin’s coin- cidence degree theory, we established that the given boundary value problem admits at least one solution. We also presented examples to illustrate the main results.

Acknowledgements

The authors would like to thank the anonymous reviewers for their valuable comments and suggestions to improve the quality of the manuscript.

Cite this paper

*Applied Mathematics*,

**8**, 193-208. doi: 10.4236/am.2017.82016.

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