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Existence of Solutions of Three-Dimensional Fractional Differential Systems

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DOI: 10.4236/am.2017.82016    780 Downloads   970 Views  

ABSTRACT

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.

u ( 0 ) = u ( 1 ) = 0 , v ( 0 ) = v ( 1 ) = 0 , w ( 0 ) = w ( 1 ) = 0,

where D 0 + α , D 0 + β , D 0 + γ are the standard Caputo fractional derivatives, 1 < α , β , γ 2 . 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.

Keywords:

1. Introduction

Fractional calculus is a very effective tool in the modeling of many phenomena like control of dynamical systems, porous media, electro chemistry, viscoelasticity, electromagnetic and so on. The fractional theory and its applications are mentioned by many papers and monographs, we refer [1] - [9] . For nonlinear fractional boundary value problem, many fixed point theorems were applied to investigate the existence of solutions as in references [10] [11] [12] [13] . On the other hand, there is another effective approach, Mawhin’s coincidence theory, which proves to be very useful for determining the existence of solutions for fractional order differential equations. In recent years, boundary value problems for fractional differential equations at resonance have been studied in many papers (see [14] - [21] ). The main motivation for investigating the fractional boundary value problem arises from fractional advection-dispersion equation.

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

D 0 + α x ( t ) = f ( t , x ( t ) , x ( t ) ) , t [ 0 , 1 ] , x ( 0 ) = 0 , x ( 0 ) = x ( 1 ) ,

where D 0 + α is the Caputo fractional differential operator, 1 < α 2 , and f : [ 0,1 ] × R 2 R 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

D 0 + α u ( t ) = f ( t , v ( t ) , v ( t ) ) , t ( 0 , 1 ) , D 0 + β v ( t ) = g ( t , u ( t ) , u ( t ) ) , t ( 0 , 1 ) , u ( 0 ) = v ( 0 ) = 0 , u ( 0 ) = u ( 1 ) , v ( 0 ) = v ( 1 ) .

where D 0 + α , D 0 + β are the Caputo fractional derivatives, 1 < α 2 , 1 < β 2 , and f , g : [ 0,1 ] × R 2 R 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

D 0 + α u ( t ) = f 1 ( t , v ( t ) , v ( t ) ) , t ( 0 , 1 ) , D 0 + β v ( t ) = f 2 ( t , w ( t ) , w ( t ) ) , t ( 0 , 1 ) , D 0 + γ w ( t ) = f 3 ( t , u ( t ) , u ( t ) ) , t ( 0 , 1 ) , (1)

together with the Neumann boundary conditions,

u ( 0 ) = u ( 1 ) = 0 , v ( 0 ) = v ( 1 ) = 0 , w ( 0 ) = w ( 1 ) = 0 ,

where D 0 + α , D 0 + β , D 0 + γ are the standard Caputo fractional derivatives, 1 < α , β , γ Math_19#, and f 1 , f 2 , f 3 : [ 0 , 1 ] × R × R R 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 R + and the Caputo fractional derivative.

Let X and Y be real Banach spaces and let L : dom L X Y be a Fred- holm operator with index zero if dim Ker L = codim Im L < and ImL is closed in Y and there exist continuous projectors P : X X , Q : Y Y such that

Im P = Ker L , Ker Q = Im L , X = Ker L Ker P , Y = Im L Im Q .

It follows that

L | dom L Ker P : dom L Ker P Im L

is invertible. Here K P denotes the inverse of L | dom L Ker P .

If Ω is an open bounded subset of X, and dom L Ω ¯ ϕ , then the map N : X Y will be called L-compact on Ω ¯ , if Q N ( Ω ¯ ) is bounded and K P ( I Q ) N : Ω ¯ X is compact, where I is the identity operator.

Lemma 1. [14] Let L : dom L X Y be a Fredholm operator with index zero and N : X Y be L-compact on Ω ¯ . Assume that the following condi- tions are satisfied.

1. L x λ N x for every ( x , λ ) [ ( dom L \ Ker L ) Ω ] × ( 0 , 1 ) ;

2. N x Im L for every x Ker L Ω ;

3. deg ( Q N | Ker L , Ker L Ω , 0 ) 0 , where Q : Y Y is a projection such that Im L = Ker Q .

Then the operator equation L x = N x has at least one solution in dom L Ω ¯ .

Definition 1. [6] The Riemann-Liouville fractional integral of order α > 0 of a function y : R + R on the half-axis R + is given by

( I 0 + α y ) ( t ) : = 1 Γ ( α ) 0 t ( t v ) α 1 y ( v ) d v for t > 0

provided the right hand side is pointwise defined on R + .

Definition 2. [6] Assume that x ( t ) is ( n 1 ) -times absolutely continuous function, the Caputo fractional derivative of order α > 0 of x is given by

( D 0 + α x ) ( t ) : = I 0 + n α d n x ( t ) d t n = 1 Γ ( n α ) 0 t ( t v ) n α 1 x ( n ) ( v ) d v for t > 0

where n is the smallest integer greater than or equal to α , provided that the right side integral is pointwise defined on ( 0, + ) .

Lemma 2. [6] Let α > 0 and n = [ α ] . If x ( n 1 ) A C [ 0,1 ] , then

I 0 + α D 0 + α x ( t ) : = x ( t ) + c 0 + c 1 t + c 2 t 2 + + c n 1 t n 1

where c i = x ( i ) ( 0 ) i ! R , i = 0 , 1 , 2 , , n 1 , here n is the smallest integer greater

than or equal to α .

In this paper, let us take X = C 1 [ 0 , 1 ] with the norm x X = max { x , x } and Y = C [ 0 , 1 ] with the norm y Y = y , where x = max t [ 0 , 1 ] | x ( t ) | . Then we denote X ¯ = X × X × X with the norm ( u , v , w ) X ¯ = max { u X , v X , w X } and Y ¯ = Y × Y × Y with the norm ( x , y , z ) Y ¯ = max { x Y , y Y , z Y } . Clearly, both X ¯ and Y ¯ are Banach spaces.

Define the operators L i : dom L X Y , ( i = 1 , 2 , 3 ) by

L 1 u = D 0 + α u , L 2 v = D 0 + β v and L 3 w = D 0 + γ w ,

where

dom L 1 = { u X | D 0 + α u ( t ) Y , u ( 0 ) = u ( 1 ) = 0 } ,

dom L 2 = { v X | D 0 + β v ( t ) Y , v ( 0 ) = v ( 1 ) = 0 } ,

and

dom L 3 = { w X | D 0 + γ w ( t ) Y , w ( 0 ) = w ( 1 ) = 0 } .

Define the operator L : dom L X ¯ Y ¯ by

L ( u , v , w ) = ( L 1 u , L 2 v , L 3 w ) , (2)

where dom L = { ( u , v , w ) X ¯ | u dom L 1 , v dom L 2 , w dom L 3 } .

Let the Nemytski operator N : X ¯ Y ¯ be defined as

N ( u , v , w ) = ( N 1 v , N 2 w , N 3 u ) ,

where N 1 : X Y is defined by

N 1 v ( t ) = f 1 ( t , v ( t ) , v ( t ) ) ,

N 2 : X Y is defined by

N 2 w ( t ) = f 2 ( t , w ( t ) , w ( t ) ) ,

and N 3 : X Y is defined by

N 3 u ( t ) = f 3 ( t , u ( t ) , u ( t ) ) .

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

L ( u , v , w ) = N ( u , v , w ) , ( u , v , w ) dom L .

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 f 1 , f 2 , f 3 : [ 0,1 ] × R × R R be continuous. Assume that

(H1) there exist nonnegative functions a i , b i , c i C [ 0 , 1 ] , ( i = 1 , 2 , 3 ) with

Γ ( α ) Γ ( β ) Γ ( γ ) ( B 1 + C 1 ) ( B 2 + C 2 ) ( B 3 + C 3 ) Γ ( α ) Γ ( β ) Γ ( γ ) > 0

such that for all ( u , v ) R 2 , t [ 0 , 1 ] ,

| f i ( t , u , v ) | a i ( t ) + b i ( t ) | u | + c i ( t ) | v | for i = 1 , 2 , 3 ,

where A i = a i , B i = b i , C i = c i , ( i = 1 , 2 , 3 ) ;

(H2) there exists a constant M > 0 such that for all t [ 0 , 1 ] , | u | > M , v R either

u f 1 ( t , u , v ) > 0 , u f 2 ( t , u , v ) > 0 , u f 3 ( t , u , v ) > 0

or

u f 1 ( t , u , v ) < 0 , u f 2 ( t , u , v ) < 0 , u f 3 ( t , u , v ) < 0 ;

(H3) there exists a constant M * > 0 such that for every m 1 , m 2 , m 3 R satisfying min { m 1 , m 2 , m 3 } > M * either

m 1 N 1 ( m 2 ) > 0 , m 2 N 2 ( m 3 ) > 0 , m 3 N 3 ( m 1 ) > 0

or

m 1 N 1 ( m 2 ) < 0 , m 2 N 2 ( m 3 ) < 0 , m 3 N 3 ( m 1 ) < 0.

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

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

Ker L = ( Ker L 1 , Ker L 2 , Ker L 3 ) = { ( u , v , w ) X ¯ | ( u , v , w ) = ( u ( 0 ) , v ( 0 ) , w ( 0 ) ) } , (3)

and

Im L = ( Im L 1 , Im L 2 , Im L 3 ) = { ( x , y , z ) Y ¯ | 0 1 ( 1 s ) α 2 x ( s ) d s = 0 1 ( 1 s ) β 2 y ( s ) d s = 0 1 ( 1 s ) γ 2 z ( s ) d s = 0 } . (4)

Proof. By Lemma 2, L 1 u = D 0 + α u ( t ) = 0 has the solution

u ( t ) = u ( 0 ) + u ( 0 ) t .

From the boundary conditions, we have

Ker L 1 = { u X | u = u ( 0 ) } .

For x Im L 1 , there exists u dom L 1 such that x = L 1 u Y . By using the Lemma 2, we get

u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 x ( s ) d s + u ( 0 ) + u ( 0 ) t .

Then, we have

u ( t ) = 1 Γ ( α 1 ) 0 t ( t s ) α 2 x ( s ) d s + u ( 0 ) .

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

0 1 ( 1 s ) α 2 x ( s ) d s = 0.

On the other hand, suppose x Y and satisfies 0 1 ( 1 s ) α 2 x ( s ) d s = 0. Let u ( t ) = I 0 + α x ( t ) , then u dom L 1 and D 0 + α u ( t ) = x ( t ) . Hence, x Im L 1 . Then we get

Im L 1 = { x Y | 0 1 ( 1 s ) α 2 x ( s ) d s = 0 } .

Similarly, we have

Ker L 2 = { v X | v = v ( 0 ) } , Im L 2 = { y Y | 0 1 ( 1 s ) β 2 y ( s ) d s = 0 } ,

and

Ker L 3 = { w X | w = w ( 0 ) } , Im L 3 = { z Y | 0 1 ( 1 s ) γ 2 z ( s ) d s = 0 } .

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

P ( u , v , w ) = ( P 1 u , P 2 v , P 3 w ) = ( u ( 0 ) , v ( 0 ) , w ( 0 ) ) ,

Q ( x , y , z ) = ( Q 1 x , Q 2 y , Q 3 z ) = ( ( α 1 ) 0 1 ( 1 s ) α 2 x ( s ) d s , ( β 1 ) 0 1 ( 1 s ) β 2 y ( s ) d s , ( γ 1 ) 0 1 ( 1 s ) γ 2 z ( s ) d s ) .

Further more, the operator K P : Im L dom L Ker P can be written by

K P ( x , y , z ) = ( I 0 + α x ( t ) , I 0 + β y ( t ) , I 0 + γ z ( t ) ) .

Proof. Clearly, Im P = Ker L and P 2 ( u , v , w ) = P ( u , v , w ) . It follows that ( u , v , w ) = ( ( u , v , w ) P ( u , v , w ) ) + P ( u , v , w ) , we have X ¯ = Ker P + Ker L . By using simple calculation, we get that Ker L Ker P = { ( 0 , 0 , 0 ) } . Then we have

X ¯ = Ker P Ker L .

For ( x , y , z ) Y ¯ , we have

Q 2 ( x , y , z ) = Q ( Q 1 x , Q 2 y , Q 3 z ) = ( Q 1 2 x , Q 2 2 y , Q 3 2 z )

By the definition of Q 1 , we get

Q 1 2 x = Q 1 x ( α 1 ) 0 1 ( 1 s ) α 2 x ( s ) d s = Q 1 x .

Similarly, we can show that Q 2 2 y = Q 2 y and Q 3 2 z = Q 3 z . Thus, we can get Q 2 ( x , y , z ) = Q ( x , y , z ) .

Let

( x , y , z ) = ( ( x , y , z ) Q ( x , y , z ) ) + Q ( x , y , z ) ,

where ( x , y , z ) Q ( x , y , z ) Ker Q , Q ( x , y , z ) Im Q . It follows that Ker Q = Im L and Q 2 ( x , y , z ) = Q ( x , y , z ) , we get Im Q Im L = { ( 0 , 0 , 0 ) } . It is clear that

Y ¯ = Im L Im Q .

Thus

dim Ker L = dim Im Q = codim Im L .

Hence L is a Fredholm operator of index zero.

From the definitions of P and K P , we will prove that K P is the inverse of L | dom L Ker P . Infact, for ( x , y , z ) Im L , we have

L K P ( x , y , z ) = ( D 0 + α ( I 0 + α x ) , D 0 + β ( I 0 + β y ) , D 0 + γ ( I 0 + γ z ) ) = ( x , y , z ) . (5)

Moreover, for ( u , v , w ) dom L Ker P , we have u ( 0 ) = v ( 0 ) = w ( 0 ) = 0 and

K P L ( u , v , w ) = ( I 0 + α ( D 0 + α u ( t ) ) , I 0 + β ( D 0 + β v ( t ) ) , I 0 + γ ( D 0 + γ w ( t ) ) ) = ( u ( t ) u ( 0 ) u ( 0 ) t , v ( t ) v ( 0 ) v ( 0 ) t , w ( t ) w ( 0 ) w ( 0 ) t ) ,

which together with the boundary condition u ( 0 ) = v ( 0 ) = w ( 0 ) = 0 yields that

K P L ( u , v , w ) = ( u , v , w ) . (6)

From (5) and (6), we get K P = ( L | dom L Ker P ) 1 .

Lemma 5. Assume Ω X ¯ is an open bounded subset such that dom L Ω ¯ ϕ , then N is L-compact on Ω ¯ .

Proof. By the continuity of f1, f2 and f3, we can get Q N ( Ω ¯ ) and K P ( I Q ) N ( Ω ¯ ) are bounded. By the Arzela-Ascoli theorem, we will prove that K P ( I Q ) N ( Ω ¯ ) X ¯ is equicontinuous.

From the continuity of f1, f2 and f3, there exist constants M i > 0 , ( i = 1 , 2 , 3 ) such that for all ( u , v , w ) Ω ¯ .

| ( I Q 1 ) N 1 v | M 1 , | ( I Q 2 ) N 2 w | M 2 , | ( I Q 3 ) N 3 u | M 3 .

Furthermore, for 0 t 1 < t 2 1 , ( u , v , w ) Ω ¯ , we have

| ( K P ( I Q ) N ( u ( t 2 ) , v ( t 2 ) , w ( t 2 ) ) ) ( K P ( I Q ) N ( u ( t 1 ) , v ( t 1 ) , w ( t 1 ) ) ) | = | K P ( I Q ) ( N 1 v ( t 2 ) , N 2 w ( t 2 ) , N 3 u ( t 2 ) ) K P ( I Q ) ( N 1 v ( t 1 ) , N 2 w ( t 1 ) , N 3 u ( t 1 ) ) | = | ( I 0 + α ( I Q 1 ) N 1 v ( t 2 ) , I 0 + β ( I Q 2 ) N 2 w ( t 2 ) , I 0 + γ ( I Q 3 ) N 3 u ( t 2 ) ) ( I 0 + α ( I Q 1 ) N 1 v ( t 1 ) , I 0 + β ( I Q 2 ) N 2 w ( t 1 ) , I 0 + γ ( I Q 3 ) N 3 u ( t 1 ) ) | = | I 0 + α ( I Q 1 ) N 1 v ( t 2 ) I 0 + α ( I Q 1 ) N 1 v ( t 1 ) , I 0 + β ( I Q 2 ) N 2 w ( t 2 ) I 0 + β ( I Q 2 ) N 2 w ( t 1 ) , I 0 + γ ( I Q 3 ) N 3 u ( t 2 ) I 0 + γ ( I Q 3 ) N 3 u ( t 1 ) | .

By

| I 0 + α ( I Q 1 ) N 1 v ( t 2 ) I 0 + α ( I Q 1 ) N 1 v ( t 1 ) | 1 Γ ( α ) | 0 t 2 ( t 2 s ) α 1 ( I Q 1 ) N 1 v ( s ) d s 0 t 1 ( t 1 s ) α 1 ( I Q 1 ) N 1 v ( s ) d s | M 1 Γ ( α ) [ 0 t 1 ( ( t 2 s ) α 1 ( t 1 s ) α 1 ) d s + t 1 t 2 ( t 2 s ) α 1 d s ] = M 1 Γ ( α + 1 ) ( t 2 α t 1 α )

and

| ( I 0 + α ( I Q 1 ) N 1 v ) ( t 2 ) ( I 0 + α ( I Q 1 ) N 1 v ) ( t 1 ) | α 1 Γ ( α ) | 0 t 2 ( t 2 s ) α 2 ( I Q 1 ) N 1 v ( s ) d s 0 t 1 ( t 1 s ) α 2 ( I Q 1 ) N 1 v ( s ) d s | M 1 Γ ( α 1 ) [ 0 t 1 ( ( t 2 s ) α 2 ( t 1 s ) α 2 ) d s + t 1 t 2 ( t 2 s ) α 2 d s ] = M 1 Γ ( α ) ( t 2 α 1 t 1 α 1 ) .

Similarly, we can show that

| I 0 + β ( I Q 2 ) N 2 w ( t 2 ) I 0 + β ( I Q 2 ) N 2 w ( t 1 ) | M 2 Γ ( β + 1 ) ( t 2 β t 1 β ) ,

| ( I 0 + β ( I Q 2 ) N 2 w ) ( t 2 ) ( I 0 + β ( I Q 2 ) N 2 w ) ( t 1 ) | M 2 Γ ( β + 1 ) ( t 2 β 1 t 1 β 1 ) ,

| I 0 + γ ( I Q 3 ) N 3 u ( t 2 ) I 0 + γ ( I Q 3 ) N 3 u ( t 1 ) | M 3 Γ ( γ + 1 ) ( t 2 γ t 1 γ ) ,

| ( I 0 + γ ( I Q 3 ) N 3 u ) ( t 2 ) ( I 0 + γ ( I Q 3 ) N 3 u ) ( t 1 ) | M 3 Γ ( γ + 1 ) ( t 2 γ 1 t 1 γ 1 ) .

Since t α , t α 1 , t β , t β 1 , t γ and t γ 1 are uniformly continuous on [0, 1], we have K P ( I Q ) N ( Ω ¯ ) X is equicontinuous. Thus K P ( I Q ) N : Ω ¯ X is compact.

Lemma 6. Assume that ( H 1 ) , ( H 2 ) hold, then the set

Ω 1 = { ( u , v , w ) dom L \ Ker L | L ( u , v , w ) = λ N ( u , v , w ) , λ ( 0 , 1 ) }

is bounded.

Proof. Let ( u , v , w ) Ω 1 , then N ( u , v , w ) Im L . By (4), we get

0 1 ( 1 s ) α 2 f 1 ( s , v ( s ) , v ( s ) ) d s = 0,

0 1 ( 1 s ) β 2 f 2 ( s , w ( s ) , w ( s ) ) d s = 0

and

0 1 ( 1 s ) γ 2 f 3 ( s , u ( s ) , u ( s ) ) d s = 0.

Then, by integral mean value theorem, there exist constants ξ , η , ζ ( 0,1 ) such that f 1 ( ξ , v ( ξ ) , v ( ξ ) ) = 0 , f 2 ( η , w ( ( η ) , w ( η ) ) ) = 0 and f 3 ( ζ , u ( ζ ) , u ( ζ ) ) = 0. Then we get

v ( ξ ) f 1 ( ξ , v ( ξ ) , v ( ξ ) ) = 0 , w ( η ) f 2 ( η , w ( η ) , w ( η ) ) = 0 and u ( ζ ) f 3 ( ζ , u ( ζ ) , u ( ζ ) ) = 0.

From ( H 2 ) , we get | v ( ξ ) | M , | w ( η ) | M and | u ( ζ ) | M . Hence we have

| u ( t ) | = | u ( ζ ) + ζ t u ( s ) d s | M + u . (7)

We obtain

u M + u . (8)

Similarly, we can show that

v M + v (9)

and

w M + w . (10)

By L ( u , v , w ) = λ N ( u , v , w ) , we get

u ( t ) = λ Γ ( α ) 0 t ( t s ) α 1 f 1 ( s , v ( s ) , v ( s ) ) d s + u ( 0 ) ,

v ( t ) = λ Γ ( β ) 0 t ( t s ) β 1 f 2 ( s , w ( s ) , w ( s ) ) d s + v ( 0 )

and

w ( t ) = λ Γ ( γ ) 0 t ( t s ) γ 1 f 3 ( s , u ( s ) , u ( s ) ) d s + w ( 0 ) .

Then

u ( t ) = λ Γ ( α 1 ) 0 t ( t s ) α 2 f 1 ( s , v ( s ) , v ( s ) ) d s ,

v ( t ) = λ Γ ( β 1 ) 0 t ( t s ) β 2 f 2 ( s , w ( s ) , w ( s ) ) d s

and

w ( t ) = λ Γ ( γ 1 ) 0 t ( t s ) γ 2 f 3 ( s , u ( s ) , u ( s ) ) d s .

So,

u = 1 Γ ( α 1 ) 0 t ( t s ) α 2 | f 1 ( s , v ( s ) , v ( s ) ) | d s 1 Γ ( α 1 ) 0 t ( t s ) α 2 [ a 1 ( s ) + b 1 ( s ) | v ( s ) | + c 1 ( s ) | v ( s ) | ] d s 1 Γ ( α 1 ) [ A 1 + B 1 M + ( B 1 + C 1 ) v ] 0 t ( t s ) α 2 d s 1 Γ ( α ) [ A 1 + B 1 M + ( B 1 + C 1 ) v ] . (12)

Similarly, we have

v 1 Γ ( β ) [ A 2 + B 2 M + ( B 2 + C 2 ) w ] (12)

and

w 1 Γ ( γ ) [ A 3 + B 3 M + ( B 3 + C 3 ) u ] . (13)

Combining (13) with (12), we get

v 1 Γ ( β ) Γ ( γ ) [ ( A 2 + B 2 M ) Γ ( γ ) + ( B 2 + C 2 ) ( A 3 + B 3 M ) + ( B 2 + C 2 ) ( B 3 + C 3 ) u ] . (14)

Combining (14) with (11), we get

u 1 Γ ( α ) Γ ( β ) Γ ( γ ) [ ( A 1 + B 1 M ) Γ ( β ) Γ ( γ ) + ( B 1 + C 1 ) ( A 2 + B 2 M ) Γ ( γ ) + ( B 1 + C 1 ) ( B 2 + C 2 ) ( A 3 + B 3 M ) + ( B 1 + C 1 ) ( B 2 + C 2 ) ( B 3 + C 3 ) u ] .

Thus, from Γ ( α ) Γ ( β ) Γ ( γ ) ( B 1 + C 1 ) ( B 2 + C 2 ) ( B 3 + C 3 ) Γ ( α ) Γ ( β ) Γ ( γ ) > 0 and (14), we get

u 1 Γ ( α ) Γ ( β ) Γ ( γ ) ( B 1 + C 1 ) ( B 2 + C 2 ) ( B 3 + C 3 ) [ ( A 1 + B 1 M ) Γ ( β ) Γ ( γ ) + ( B 1 + C 1 ) ( A 2 + B 2 M ) Γ ( γ ) + ( B 1 + C 1 ) ( B 2 + C 2 ) ( A 3 + B 3 M ) ] : = K 1

v 1 Γ ( β ) Γ ( γ ) [ ( A 2 + B 2 M ) Γ ( γ ) + ( B 2 + C 2 ) ( A 3 + B 3 M ) + ( B 2 + C 2 ) ( B 3 + C 3 ) K 1 ] : = K 2

and

w 1 Γ ( γ ) [ A 3 + B 3 M + ( B 3 + C 3 ) K 1 ] : = K 3 .

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

( u , v , w ) X ¯ max { K 1 + M , K 2 + M , K 3 + M } : = K .

Hence Ω 1 is bounded.

Lemma 7. Assume that ( H 3 ) holds, then the set

Ω 2 = { ( u , v , w ) | ( u , v , w ) Ker L , N ( u , v , w ) Im L }

is bounded.

Proof. For ( u , v , w ) Ω 2 , we have ( u , v , w ) = ( m 1 , m 2 , m 3 ) , m 1 , m 2 , m 3 R . Then from N ( u , v , w ) Im L ,

0 1 ( 1 s ) α 2 f 1 ( s , m 2 , 0 ) d s = 0 ,

0 1 ( 1 s ) β 2 f 2 ( s , m 3 , 0 ) d s = 0

and

0 1 ( 1 s ) γ 2 f 3 ( s , m 1 , 0 ) d s = 0.

From ( H 3 ) imply that | m 1 | , | m 2 | , | m 3 | M * . Thus, we get

( u , v , w ) X ¯ M * .

Therefore Ω 2 is bounded.

Lemma 8. Assume that the first part of ( H 3 ) holds, then the set

Ω 3 = { ( u , v , w ) Ker L | λ ( u , v , w ) + ( 1 λ ) Q N ( u , v , w ) = ( 0 , 0 , 0 ) , λ [ 0 , 1 ] }

is bounded.

Proof. For ( u , v , w ) Ω 3 , we have ( u , v , w ) = ( m 1 , m 2 , m 3 ) , m 1 , m 2 , m 3 R and

λ m 1 + ( 1 λ ) ( α 1 ) 0 1 ( 1 s ) α 2 f 1 ( s , m 2 ,0 ) d s = 0 , (15)

λ m 2 + ( 1 λ ) ( β 1 ) 0 1 ( 1 s ) β 2 f 2 ( s , m 3 , 0 ) d s = 0 (16)

and

λ m 3 + ( 1 λ ) ( γ 1 ) 0 1 ( 1 s ) γ 2 f 3 ( s , m 1 ,0 ) d s = 0. (17)

If λ = 0 , then by ( H 3 ) , we get | m 1 | , | m 2 | , | m 3 | M * . If λ = 1 , then m 1 = m 2 = m 3 = 0 . For λ ( 0,1 ] , we obtain | m 1 | , | m 2 | , | m 3 | M * . Otherwise, if | m 1 | or | m 2 | or | m 3 | > M * , from ( H 3 ) , one has

λ m 1 2 + ( 1 λ ) ( α 1 ) 0 1 ( 1 s ) α 2 m 1 f 1 ( s , m 2 , 0 ) d s > 0

or

λ m 2 2 + ( 1 λ ) ( β 1 ) 0 1 ( 1 s ) β 2 m 2 f 2 ( s , m 3 , 0 ) d s > 0

or

λ m 3 2 + ( 1 λ ) ( γ 1 ) 0 1 ( 1 s ) γ 2 m 3 f 3 ( s , m 1 ,0 ) d s > 0,

which contradict to (15) or (16) or (17). Hence, Ω 3 is bounded.

Remark 1 Suppose the second part of ( H 3 ) holds, then the set

Ω 3 = { ( u , v , w ) Ker L | λ ( u , v , w ) + ( 1 λ ) Q N ( u , v , w ) = ( 0 , 0 , 0 ) , λ [ 0 , 1 ] }

is bounded.

Proof of the Theorem 1: Set Ω = { ( u , v , w ) X ¯ | ( u , v , w ) X ¯ < max { K , M * } + 1 } .

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

(1) L ( u , v , w ) λ N ( u , v , w ) for every ( ( u , v , w ) , λ ) [ ( dom L \ Ker L ) Ω ] × ( 0 , 1 ) ;

(2) N x Im L for every ( u , v , w ) Ker L Ω .

Choose

H ( ( u , v , w ) , λ ) = ± λ ( u , v , w ) + ( 1 λ ) Q N ( u , v , w ) .

By Lemma 8 (or Remark 1), we get H ( ( u , v , w ) , λ ) 0 for ( u , v , w ) Ker L Ω . Therefore

deg ( Q N | Ker L , Ker L Ω , 0 ) = deg ( H ( . , 0 ) , Ker L Ω , 0 ) = deg ( H ( . , 1 ) , Ker L Ω , 0 ) = deg ( ± I , Ker L Ω , 0 ) 0.

Thus, the condition (3) of Lemma 1 is satisfied. By Lemma 1, we obtain L ( u , v , w ) = N ( u , v , w ) has at least one solution in dom L Ω ¯ . 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

D 0 + 5 4 u ( t ) = 1 8 ( v ( t ) 6 ) + t 2 8 ( 1 + v ( t ) ) 1 2 , t ( 0 , 1 ) , D 0 + 3 2 v ( t ) = 1 6 ( w ( t ) 4 ) + t 3 6 cos 2 w ( t ) , t ( 0 , 1 ) , D 0 + 7 4 w ( t ) = 1 10 ( u ( t ) 8 ) + t 3 10 sin 2 u ( t ) , t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = 0 , v ( 0 ) = v ( 1 ) = 0 , w ( 0 ) = w ( 1 ) = 0. (18)

Here α = 5 4 , β = 3 2 , γ = 7 4 . Moreover,

f 1 ( t , v ( t ) , v ( t ) ) = 1 8 ( v ( t ) 6 ) + t 2 8 ( 1 + v ( t ) ) 1 2 ,

f 2 ( t , w ( t ) , w ( t ) ) = 1 6 ( w ( t ) 4 ) + t 3 6 cos 2 w ( t ) ,

f 3 ( t , u ( t ) , u ( t ) ) = 1 10 ( u ( t ) 8 ) + t 3 10 sin 2 u ( t ) .

Now let us compute a 1 ( t ) , b 1 ( t ) , c 1 ( t ) from f 1 ( t , v ( t ) , v ( t ) ) .

f 1 ( t , v ( t ) , v ( t ) ) = 1 8 ( v ( t ) 6 ) + t 2 8 ( 1 + v ( t ) ) 1 2 = 1 8 ( v ( t ) 6 ) + t 2 8 ( 1 + 1 2 v ( t ) + ) 1 8 ( v ( t ) 6 ) + t 2 8 .

| f 1 ( t , v ( t ) , v ( t ) ) | 1 8 | v ( t ) | + 7 8 .

From the above inequality, we get a 1 ( t ) = 7 8 , b 1 ( t ) = 1 8 , c 1 ( t ) = 0. Also,

f 2 ( t , w ( t ) , w ( t ) ) = 1 6 ( w ( t ) 4 ) + t 3 6 cos 2 w ( t )

| f 2 ( t , w ( t ) , w ( t ) ) | 1 6 | w ( t ) | + 5 6 .

Here, a 2 ( t ) = 5 6 , b 2 ( t ) = 1 6 , c 2 ( t ) = 0 . Finally,

f 3 ( t , u ( t ) , u ( t ) ) = 1 10 ( u ( t ) 8 ) + t 3 10 sin 2 u ( t ) .

| f 3 ( t , u ( t ) , u ( t ) ) | 1 10 | u ( t ) | + 8 10 .

We get, a 3 ( t ) = 4 5 , b 3 ( t ) = 1 10 , c 3 ( t ) = 0 . And we get, B 1 ( t ) = 1 8 , B 2 ( t ) = 1 6 , B 3 ( t ) = 1 10 , C 1 ( t ) = C 2 ( t ) = C 3 ( t ) = 0 . Choose M = M * = 8 . Also,

Γ ( α ) Γ ( β ) Γ ( γ ) ( B 1 + C 1 ) ( B 2 + C 2 ) ( B 3 + C 3 ) Γ ( α ) Γ ( β ) Γ ( γ ) = Γ ( 5 4 ) Γ ( 3 2 ) Γ ( 7 4 ) ( B 1 B 2 B 3 ) Γ ( 5 4 ) Γ ( 3 2 ) Γ ( 7 4 ) 0.73605543 > 0 ,

where Γ ( 1 4 ) 3.625 , Γ ( 3 4 ) 1.225416702 and Γ ( 1 2 ) = π . 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

D 0 + 4 3 u ( t ) = 1 17 ( v ( t ) 15 ) + t 4 20 log ( 1 + v ( t ) ) , t ( 0 , 1 ) , D 0 + 5 4 v ( t ) = 1 9 ( w ( t ) 7 ) + t 6 25 ( 1 + w ( t ) ) 1 3 , t ( 0 , 1 ) , D 0 + 3 2 w ( t ) = 1 13 ( u ( t ) 11 ) + t 7 15 arctan u ( t ) , t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = 0 , v ( 0 ) = v ( 1 ) = 0 , w ( 0 ) = w ( 1 ) = 0. (19)

Here α = 4 3 , β = 5 4 , γ = 3 2 . Moreover,

f 1 ( t , v ( t ) , v ( t ) ) = 1 17 ( v ( t ) 15 ) + t 4 20 log ( 1 + v ( t ) ) ,

f 2 ( t , w ( t ) , w ( t ) ) = 1 9 ( w ( t ) 7 ) + t 6 25 ( 1 + w ( t ) ) 1 3 ,

f 3 ( t , u ( t ) , u ( t ) ) = 1 13 ( u ( t ) 11 ) + t 7 15 arctan u ( t ) .

Now let us compute a 1 ( t ) , b 1 ( t ) , c 1 ( t ) from f 1 ( t , v ( t ) , v ( t ) ) .

f 1 ( t , v ( t ) , v ( t ) ) = 1 17 ( v ( t ) 15 ) + t 4 20 log ( 1 + v ( t ) ) = 1 17 ( v ( t ) 15 ) + t 4 20 ( v ( t ) ( v ( t ) ) 2 2 ! + )

| f 1 ( t , v ( t ) , v ( t ) ) | 1 17 | v ( t ) | + 15 17 + 1 20 | v ( t ) |

From the above inequality, we get a 1 ( t ) = 15 17 , b 1 ( t ) = 1 17 , c 1 ( t ) = 1 20 . Also,

f 2 ( t , w ( t ) , w ( t ) ) = 1 9 ( w ( t ) 7 ) + t 6 25 ( 1 + w ( t ) ) 1 3 = 1 9 ( w ( t ) 7 ) + t 6 25 ( 1 + 1 3 w ( t ) + )

| f 2 ( t , w ( t ) , w ( t ) ) | 1 9 | w ( t ) | + 7 9 + 1 25 ( 1 + 1 3 | w ( t ) | + ) 1 9 | w ( t ) | + 184 225 + 1 75 | w ( t ) | .

Here, a 2 ( t ) = 184 225 , b 2 ( t ) = 1 9 , c 2 ( t ) = 1 75 . Similarly,

f 3 ( t , u ( t ) , u ( t ) ) = 1 13 ( u ( t ) 11 ) + t 7 10 arctan u ( t ) . = 1 13 ( u ( t ) 11 ) + t 7 15 ( u ( t ) ( u ( t ) 3 ) 3 + ) .

| f 3 ( t , u ( t ) , u ( t ) ) | 1 13 | u ( t ) | + 11 13 + 1 15 u ( t ) .

Here, a 3 ( t ) = 11 13 , b 3 ( t ) = 1 13 , c 3 ( t ) = 1 15 . We get, B 1 ( t ) = 1 17 , B 2 ( t ) = 1 9 , B 3 ( t ) = 1 13 , C 1 ( t ) = 1 20 , C 2 ( t ) = 1 75 , C 3 ( t ) = 1 15 . Choose M = M * = 15 .

Also,

Γ ( α ) Γ ( β ) Γ ( γ ) ( B 1 + C 1 ) ( B 2 + C 2 ) ( B 3 + C 3 ) Γ ( α ) Γ ( β ) Γ ( γ ) = Γ ( 4 3 ) Γ ( 5 4 ) Γ ( 3 2 ) ( ( 1 17 + 1 20 ) ( 1 9 + 1 75 ) ( 1 13 + 1 15 ) ) Γ ( 4 3 ) Γ ( 5 4 ) Γ ( 3 2 ) 0.96872 > 0 ,

where Γ ( 1 4 ) 3.625 , Γ ( 1 3 ) 0.1924 and Γ ( 1 2 ) = π . 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

Sadhasivam, V. , Kavitha, J. and Deepa, M. (2017) Existence of Solutions of Three-Dimensional Fractional Differential Systems. Applied Mathematics, 8, 193-208. doi: 10.4236/am.2017.82016.

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