The Exact Solutions of Such Coupled Linear Matrix Fractional Differential Equations of Diagonal Unknown Matrices by Using Hadamard Product

(1)

http://dx.doi.org/10.4236/jamp.2016.42049

The Exact Solutions of Such Coupled

Linear Matrix Fractional Differential

Equations of Diagonal Unknown

Matrices by Using Hadamard Product

Zayed Al-Zuhiri1_{, Zeyad Al-Zhour}2_{, Khaled Jaber}1

1_{Department of Mathematics, Faculty of Science and Information Technology, Zarqa University, Zarqa, Jordan} 2_{Department of Basic Sciences and Humanities, College of Engineering, University of Dammam,}

Dammam, Saudi Arabia

Received 6 September 2015; accepted 26 February 2016; published 29 February 2016

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

Abstract

In this paper, we present the general exact solutions of such coupled system of matrix fractional differential equations for diagonal unknown matrices in Caputo sense by using vector extraction operators and Hadamard product. Some illustrated examples are also given to show our new ap-proach.

Keywords

Fractional Operators, Matrix Fractional Differential Equations, Hadamard Product, Vector Extraction Operator

1. Introduction

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solutions of the fractional differential equations have been investigated by many authors see as an example [6]. During the last decades, several methods have been used to solve fractional differential equations, fractional partial differential equations, fractional integro-differential equations and dynamic systems containing fractional derivatives, such as Adomian’s decomposition method [7], variational iteration method [8]-[11], homotopy per-turbation method [12], homotopy analysis method [13], spectral methods [14], and other methods [15].

Recently, Wang [16] studied the synchronized motions in a star network of coupled fractional order systems in which the major element is coupled to each of the non-interacting individual elements and Kilicman and Al- Zhour [17] studied several operational matrices for fractional integration and differentiation and expanded the Kronecker convolution product to the Riemann-Liouville fractional integral of matrices. Al-Zhour [18] intro-duced the exact solution of coupled fractional order systems by using Kronecker structure.

In the present paper, the exact solutions of coupled and uncoupled systems of matrix fractional differential equations for diagonal unknown matrices are presented by using a new attractive method and some illustrated examples are also given to show our new approach.

2. Basic Results and Preliminaries

In this section, we recall some basic results and definitions associated to Hadamard product, Mittage-Leffler function and Caputo fractional derivative that will be used to get our results later.

Definition 2.1. Let A=   aij and B=  bij ∈Mm n, .Then the Hadamard product of A and B is defined by

[19]-[26].

,

ij ij m n

A B B A = =_a b _∈M (2-1)

Definition 2.2. Let A diag a a=

(

11, 22, , ann

)

∈Mn be a diagonal matrix. Then the diagonal extraction

oper-ator of A is defined by [21] [23].

( )

[

]

T

11 22 nn

Vecd A = a a a (2-2)

Theorem 2.3. Let , ,A B Y M∈ n be diagonal matrices. Then

(

)

(

T

)

( )

_.

Vecd AYB = B A Vecd Y (2-3)

Definition 2.4. The one parameter Mittage-Leffler functions and Mittage-Leffler matrix functions of matrix

m

A M∈ are defined, respectively, for p>0 by [18].

( )

₀

₍

₎

Γ 1

k

p k

x E x

pk

∞ = =

+

∑

(2-4)

( )

₀

₍

₎

Γ 1

k

p k

A E A

pk

∞ = =

+

∑

. (2-5)

Note that the Mittage-Leffler matrix function of A M∈ m can be represented by using spectral

decomposi-tion method by [18].

( )

T

( )( )

0 ,

p k k k p k

E A ∞ x y E A λ

=

∑

(2-6)

where x x1, , ,2 xm and y y1, , ,2 ym are the eigenvectors corresponding to the eigenvalue λ λ1, , ,2 λm of A and AT_{, respectively.}

Theorem 2.5. Let A diag a a=

(

11, 22, , amm

)

∈Mm is a diagonal matrix and p>0. Then [18]

( )

(

( )

11 ,

( )

22 , ,

( )

)

.

p p p p mm

E A =diag E a E a E a (2-7)

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( )

_{( )}

(

) (

0

)

1 ( )

( )

1 x _{d .}

n p

n p n

C_{D f x}p _D _{D f x}n _{x t} _f _{t t}

n p

− − − −

= = −

Γ −

∫

(2-8)

Theorem 2.7. The relationship between the Mittage-Leffler function and Caputo derivative are given by:

a) C p

(

( )

p

)

( )

p

p p

D E λx =λE λx (2.9)

b) C p

(

( )

p

)

( )

p

p p

D E Ax =AE Ax (2.10)

3. Main Results

In this section, we present the general exact solutions of the coupled and uncoupled system of fractional diffe-rential equations for diagonal unknown matrices by using the using vector extraction operators and Hadamard product.

Lemma 3.1. Let A M∈ n be a given scalar matrix, c M∈ n,1 be a given scalar vector,f x

( )

∈Mn,1 be a given vector function and y x

( )

∈Mn,1 be an unknown vector function to be solved. Then the exact solution of the following non-homogenous linear fractional system of order 0< <p 1 is given by [18]-[20].

( )

( ) ( )

, 0

C_{D y x}p ₌_{A y x} ₊ _{f x y} ₌_c_(3-1)

is given by:

( )

(

)

1

(

)

( )

0

d .

x

p p

p

p p

y x E Ax c x s − E A x s f s s

= +

_∫

− − (3-2)

Theorem 3.2. Let A=   aij and C M∈ n be given diagonal scalar matrices, U x

( )

∈Mn be a given

di-agonal matrix function and Y x

( )

∈Mn be an unknown diagonal matrix function. Then the general vector

ex-traction solution of the following non-homogeneous matrix fractional differential equation

( )

,

( )

0 , 0 1

C_{D Y x}p ₌_{AY x U x Y}₊ ₌_C _{< <}_p _(3-3)

is given by:

( )

(

)

(

)

(

)

( )

(

)

(

)

(

)

(

)

(

( )

)

1

11

11 0

, ,

, , d .

p

p p

p p nn

x x s

p p

p p nn

Vecd Y x diag E a x E a x Vecd C

diag E a x s E a x s Vecd U s s

−

=

+

_∫

− −

(3-4)

Proof. By using (2-3), then (3.3) can be represented by:

(

)

(

)

( )

(

)

( )

(

11

)

( )

, ,

c p

n

nn

Vecd D Y x Vecd AYI Vecd U

I A Vecd Y Vecd U

diag a a Vecd Y Vecd U

= +

Hence, the vector extraction solution of (3.3) is given by:

( )

(

)

(

)

( )

(

)

(

)

(

)

(

( )

)

(

)

(

)

(

)

( )

(

( )

)

(

( )

)

(

( )

)

11

1

11 0

11

x _p-1 _p _p

p 11 p nn

0

, ,

, , d

, ,

+ x-s

_ò

diag E a x-s ,L,E a x-s Vecd U s ds.

p p

p nn

x

p p p

p nn

p p

p p nn

Vecd Y x E diag a x a x Vecd C

x s E diag a x s a x s Vecd U s s diag E a x E a x Vecd C

−

 

= _ _

 

+ − _ − − _

 

=

∫

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Theorem 3.3. Let A Cij, i∈Mn be given diagonal scalar matrices, and Y xi

( )

∈Mn be an unknown diagonal

matrix functions. Then the general vector extraction solution of the following general system of linear matrix fractional differential equations of order 0< <p 1:

( )

1 11 1 1 1

1 1

, 0 ,

p

n n

i i

p C

nn n

C

n n n

D Y x A Y x A Y x U x

Y C

D Y x A Y x A Y x U x



= + + +



= 



= + + + _

(3-5)

is given by:

( )

(

)

( )

(

)

( )

( )(

)

( )(

)

( )(

)

( )(

)

( )

(

)

( )

(

)

1 ₁₁ ₁ ₁

1

11 1

1

0

1

p p

n

p

p p

n n nn n

p p

n x

p p

n

n nn

Vecd Y x _{diag A x} _{diag A x} _{Vecd C}

E

Vecd Y x diag A x diag A x Vecd C

Vecd U s

diag A x s diag A x s

xs E

Vecd U s

diag A x s diag A x s

−

  _ _{ } _

 ₌    

  _ _{ } _

 

 

 

 ₋ ₋   

   

+   

  

− − 

   

 

∫

d .s

 

(3-6)

Proof. By using (2-3), then (3.5) can be represented by:

( )

(

)

( )

(

)

( )

(

)

( )

(

)

( )

(

)

( )

(

)

1 ₁₁ ₁ 1 1

1

.

C p

n

C p

n nn n n

n

Vecd D Y x _{I A} _{I A} Vecd Y x Vecd U x

I A I A Vecd Y x Vecd U x

Vecd D Y x

  _ _   

  _ __ _{ } _

  =_ _  + 

  _ _   

 

      

 

(3-7)

Now by letting

( )

(

)

( )

(

)

( )

(

)

( )

(

)

1 1

, ,

C p

n n

Vecd D Y x Vecd Y x

D y x y x

Vecd Y x Vecd D Y x

  _ _

   

 

= _{= } 

  _ _

  _ _

 

( )

(

)

( )

(

)

( )

1 ₁₁ ₁ ₁₁ ₁

1 1

, ,

n n

n n nn n nn

Vecd U x _{I A} _{I A} _{diag A} _{diag A}

u x H

Vecd U x I A I A diag A diag A

  _ _ _ _

  _ _ _ _

=  =_ _=_ _

  _ _ _ _

   

 

 

( )

1

n

Vecd C c

Vecd C

 

 

=  

 

 

. Then (3.7) can be written as:

( )

( ) ( ) ( )

, 0 .

C_{D y x}p ₌_{Hy x} ₊_{u x y} ₌_c_(3-8)

Hence by using Lemma 3.1 and simple computations, then we get the solution as in (3-6). Below we will discuss some important special cases of the general system as in Theorem 3.3.

Theorem 3.4. Let , , , , ,A B C D E F M∈ n be given scalar diagonal matrices, U x U x1

( )

, 2

( )

∈Mn be

diagon-al matrix functions, and Y x Y x1

( ) ( )

, 2 ∈Mn be unknown diagonal matrix functions. Then the general solutions

of the following coupled matrix fractional differential equations of order 0< <p 1:

( )

1 1 2 1

1 2

2 1 2 2

, 0 , 0

C p

D Y x AY x BY x U x

Y E Y F

D Y x CY x DY x U x



= + + _

= =



= + + _ (3-9)

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( )

(

)

(

)

(

)

(

)

( )

(

)

(

)

(

)

_{( )}

₍

₎

₍

₎₍

₎

(

)(

)

(

)

(

)

(

)

( )

(

)

(

)(

)

(

)

1 1 1 1 0 1 1 2 2 2 p p p p p p

p p _x

p p p p

p

p p

p

p p

E diagB x E diag AD C x

Vecd Y x E diagA x Vecd E

Vecd F x s E diagA x s

E diagB x s E diag AD C x s

Vecd U s

E diagB x s E diag AD

− − − − −      ₊ ₋    = _ _     ₋ ₋    + + − −      ₋ ₊ ₋ ₋    × _ _    − − − +

∫

(

)

(

)

(

1

)

₍

_{( )}

₎

1

2 d .

2

p C x s

Vecd AD U s− s

  ₋          _    (3.10)

( )

(

)

(

)

(

)

(

)

(

)

( )

(

)

(

)

(

)

_{( )}

(

)

(

)

(

)(

)

(

)(

)

(

)

(

)

(

)

( )

(

)

1 1 2 1 1 1 0 1 1 2 2 2 p p p p p p p p p p x p p p p p p p p

Vecd Y x diag DA E diagA x Vecd E

Vecd F

diag DA x s E diagA x s

Vecd U s

E diag − − − − − −  ₋ ₋    = _ _      ₊ ₋     +_ _ _     + − −  ₋ ₋ ₋ ₋    × _ _    +

∫

(

)(

)

(

)

(

)

(

)

( )

(

)

1 1

2 d .

2

p p

p

B x s E diag AD C x s

Vecd AD U s s

− −   ₋ ₊ ₋ ₋          _    (3.11)

Proof. By multiplying the second equation in (3-9) by _AD−1_{, we get:}

( )

1 1 1 1

2 1 2 2

p C

AD− D Y x ₌AD CY x− ₊AD DY x− ₊AD U x−

( )

(

1

)

1

( )

1

( )

2 1 2 2

p

C_{D AD Y x}− ₌ _{AD CY x}− ₊_{AY x} ₊_{AD U x}−

Then (3-9) can be written as

( )

(

)

( )

1 1 2 1

1 2

1 1 1

2 1 2 2

, 0 , 0

C p

D Y x AY x BY x U x

Y E Y F

D AD Y x− AD CY x− AY x AD U x−



= + + _

= =



= + + _ (3-12)

Now, by using Vecd

( )

. of (3.12), then we get the following equivalent system:

( )

(

)

( )

(

)

(

)

(

( )

)

( )

(

)

( )

(

)

1 1 1

1 1 1 2 2 2 = C p C p

Vecd D Y x I A I B Vecd Y x Vecd U x

I AD C I A Vecd Y x Vecd AD U x

Vecd D AD Y x− − −

  _ __ _ _ _   _ _ +    _ _{ }_ __{ } _   (3.13)

Now by using (3-6), then the solution of (3.13) is given by:

( )

(

)

( )

(

)

( )

(

)

(

)

(

( )

)

( )

(

)

1 1 1 2 1 1 1 1 0 2 d p p x p p p

Vecd Y x _{I A} _{I B} _{Vecd E}

E x

Vecd F I AD C I A

Vecd AD Y x

Vecd U s

I A I B

x s E x s s

I AD C I A Vecd AD U s

(6)

Now we deal with

(

1

)

1 p p

p p

diagA diagB

I A I B

E x E x

diag AD C diagA

I AD C I A− −

      _ _ =      _ _       __ _ _ (3.15) Since

(

1

)

(

1

)

0 0 0 0 0 0 0 0 diagB diagB diagA diagA

diag AD C diag AD C

diagA − − diagA

      ₌                _ _{ } _  Then

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1 1 1 0 0 0 0 0 0 0 0 p p p

p p p p

p p

p p p p

diagA diagB _diagAx diagB x

E x E

diag AD C diagA daigAx diag AD C x

diagB x diagA x

E E

diagA x diag AD C x

− − −      _ _       = _ _+                   = _ _ _ _           But

(

)

(

)

(

)

(

)

(

)

(

)

0 0 0 ₀ p p p

p p _p

p E diagA x diagA x

E

diagA x _E _{diagA x}

 

  _ _

=

 

__ __  

 _{ } _ (3.16)

and

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1 1 1 1 1 0 0 1 2 p p p

p p p p

diagB x E

diagAD C x

E diagB x E diag AD C x E diagB x E diag AD C x E diagB x E diag AD C x E diagB x E diag AD C x

− − − − −     _ _    ₊ ₋ ₋ ₋    =  ₋ ₋ ₊ ₋      (3-17) So,

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1 1 1 1 0 0 0 0

0 ₂ ₂

0

p p

p p p p

p p p p p diagA diagB E x

diag AD C diagA

diagB x diagA x

E E

diagA x diag AD C x

E diagB x E diag AD C x E diagB x E diag AD C x E diagA x

E diagA x E di

− − − −       __ __        _ _   = _ _ _ _           + − − −     =    

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1 1 1 1 1 2 2 2 2 2

p p p p

p p p

p p p p

p p

p p p

p p

agB x E diag AD C x E diagB x E diag AD C x

E diagB x E diagAD C x E diagB x E diagAD C x

E diagA x E diagA x

E diagB x E diagAD C x E di

E diagA x E diagA x

− − − − −         − − + −        ₊ ₋   ₋ ₋              =  ₋ ₋       

(

)

(

)

(

1

)

2

p p

p

agB x E diagAD C x−

(7)

( ) ( )

( )( )

(

)

(

( )( )

)

(

( )( )

)

(

( )( )

)

(

( )( )

)

(

( )( )

)

( )( )

(

)

(

( )( )

)

(

( )( )

)

(

( )( )

)

1

1 1

1

2 2

2 p

p

p p p p

p p

diagA diagB E _{diag AD C} _diagA x s

E diagB x s E diag AD C x s E diagB x s E diag AD C x s

E diagA x s E diagA x s

−

− −

−

  

−

  

_ _ 

 

 ₋ ₊ ₋ ₋   ₋ ₋ ₋ ₋ 

   

− _ _ − _ _

   

=

 ₋ ₋ ₋ ₋ 

 

− _ _ −

 

( )( )

(

)

(

( 1 )( )

)

2

p p

E diagB x s E diag AD C x s−

 

 

 _ _

− + − −

 _ _

 _ _

 

(3.19) Now from (3-13), (3-18) and (3-19), we get

( )

(

)

(

)

(

)

(

)

( )

(

)

(

)

(

)

_{( )}

₍

₎

₍

₎₍

₎

(

)(

)

(

)

(

)

(

)

( )

(

)

(

)(

)

(

)

1

0

1

1 2

2

p p

p p _x

p p p p

p

p p

p

p p

Vecd Y x E diagA x Vecd E

Vecd U s

E diagB x s E diag AD

−

 ₊ ₋ 

 

= _ _

 





 ₋ ₋ 



 

+ _+ − −

 



  

 ₋ ₊ ₋ ₋ 

 

× _ _

 



− − −

+

∫

(

)

(

)

(

1

)

₍

_{( )}

₎

1

2 d

2

p C x s

Vecd AD U s− s



 ₋ 



 



 

  _

  

( )

(

)

(

)

(

)

(

)

(

)

( )

(

)

(

)

(

)

_{( )}

₍

₎

₍

₎₍

₎

(

)(

)

(

)

(

)

(

)

( )

(

)

(

)(

)

(

)

1 2

1

0

1

2

p p

p p _x

p p p p

p

p p

p

p p

Vecd AD Y x

E diagA x Vecd E

Vecd U s

E diagB x s E dia

−

 ₋ ₋ 

 

= _ _

 





 ₊ ₋ 



 

+ + − −



 



  

 ₋ ₋ ₋ ₋ 

 

× _ _

 



− + −

+

∫

(

)

(

)

(

1

)

₍

_{( )}

₎

1

2 d

2

p

g AD C x s

Vecd AD U s s

−



 ₋ 



 



 

  _

  

Since,

( )

(

1

) (

1

)

(

( )

)

(

1

)

(

( )

)

2 2 2

Vecd AD Y x− ₌ I AD Vecd Y x − ₌diag AD Vecd Y x− Then, we get the vector extraction solution as in (3-11).

Corollary 3.5. Let ,E F M∈ n be given scalar diagonal matrices and Y x Y x1

( ) ( )

, 2 ∈Mn be an unknown

diagonal matrix functions. Then the general vector extraction solutions of the following coupled matrix fraction-al differentifraction-al equations of order 0< <p 1:

( )

1 1 2

1 2

2 1 2

, 0 , 0

C p

D Y x Y x Y x

Y E Y F

D Y x Y x Y x



= + _

= =



(8)

are given by:

( )

(

)

( )

(

( )

)

( )

₍

_{( )}

₎

_{( )}

1 ₂ , ,

, , 2

p

p p p p p

p p p p

p

p p p p p

p p p p

E x

Vecd Y x diag E x E x E x E x Vecd E

E x

diag E x E x E x E x Vecd F

= + − + −

+ − − − −

(3.21)

( )

(

)

( )

(

( )

)

( )

₍

_{( )}

₎

_{( )}

2 ₂ , ,

, , .

2

p

p p p p p

p p p p

p

p p p p p

p p p p

E x

Vecd Y x diag E x E x E x E x Vecd E

E x

diag E x E x E x E x Vecd F

= − − − −

+ + − + −

(3.22)

Proof. The proof is straightforward by applying Theorem 3.4 by letting

(

A B C D I= = = = n

)

and by using

the following fact:

( )

p

(

( )

p , ,

( )

p

)

p n p p

E I x =diag E x E x .

4. Illustrated Examples

In the section, we give some illustrated examples to show our new approach as discussed in above section.

Example 4.1. Consider the following matrix linear fractional differential equation:

( )

,

( )

0 , 0 1

C_{D Y x}p ₌_{AY x Y} ₌_C _{< <}_p _(4-1)

where 1 0 , 1 0

0 2 0 3

A=_− _ C=_ _

− −

    and Y x

( )

is diagonal matrix. Then the exact solution of (4-1) by

apply-ing Theorem 3.2 is given by:

( )

(

)

(

( )

(

)

1

, , 2

3

p p

Vecd Y x =diag E −x E − x  _{ }

−  

( )

(

)

1

2 3 2

p p

E x

y x

y x _E _x

 ₋ 

 _{ } _

=

 _{ } _

− −

 _{ } _.

Example 4.2. Consider the following system of order ,0p < <p 1:

1 3 ,1 2 2, 3 2 ,3

C_{D y}p _{= −} _{y D y}C p _{= −}_y C_{D y}p _{= −} _y _(4-2)

where y1

( )

0 =5,y2

( )

0 =2,y3

( )

0 = −4. Then the system (4-2) can be rewritten as:

( )

C_{D Y x}p ₌_{AY x}

1 1

2 2

3 3

0 0 3 0 0 0 0

0 0 0 1 0 0 0

0 0 0 0 2 0 0

C p

D y y

  −  

 ₌ ₋  

    

  _ − _{ } _

 

(4-3)

Now the exact solution of (4-3) by applying Theorem 3.2 is given by:

( )

(

)

(

)

( )

(

)

3 0 0 ₅

0 0 2

4

0 0 2

p p

E x

Vecd Y x E x

E x

 − _{ }

 

= − _{ }

 

 −

 ₋   

 

.

Example 4.3. Consider the following matrix fractional differential equation:

( )

( ) ( )

, 0 , 0 1

(9)

where 1 0 , 1 0 ,

( )

_{( )}

0 0

0 3 0 2

f x

A C U x

g x   −     =_ _ =_ _ _{= } _ −

      and Y x

( )

is diagonal matrix. Then the exact

solu-tion of (4-4) by applying Theorem 3.2 is given by:

( )

(

)

(

( )

)

(

)

1

(

)

(

)

( )

_{( )}

0

1

, , 3

2

, 3 d

p p

x

p p p

p p

Vecd Y x diag E x E x

f s

x s diag E x s E x s s

g s −   = − _{ } −     + − − − − _ _  

∫

( )

(

)

(

)

( )

(

)

(

)

( )

1 0 1 1 2 0 d

2 3 3 d

x p p p p p x p p p p p

E x x s E x s f s s

y x y x

E x x s E x s g s s

− −   − + − − −    _{ } _ =  _{ } _  _{ }₋ ₊ ₋ ₋ _    

∫

.

Example 4.4. Consider the following matrix fractional differential equations of order 0< <p 1:

( )

1 1 2

1 2

2 1 2

2 0 6 0

, 0 , 0

0 4 0 4

C p

D Y x AY x BY x

Y Y

D Y x CY x DY x



= + _ _ _ _ _

= =

    

= + _     (4-5)

where 1

( ) ( )

2

2 0 4 0 1 0 2 0

, , , and ,

0 1 0 3 0 10 0 5

A=_ _ B=_ _ C=_− _ D=_ _ Y x Y x

−

        are diagonal matrices. So

1 1 1 1

1 ₀ ₁ ₀

1 0 1 0

2 _, ₁ _, _,

1 0 0 2 0 5

0 ₅

5

D− AD− AD C− DA−

  _ _   _ _ _₋ _ _ _ =  =_ _ =_ _ =_ _ − − −   __ __        

. (4-6)

Then the exact solution of (4-5) by applying Corollary 3.5 is given by:

( )

(

)

( )

1 4 0

2 0 ₂ ₂

4

0 3 2

0

2 4

0

2 0 ₂ ₆

4

0 3 2

0 2 p p p p p p

p p p

p p

p p p

E x E x

E x

Vecd Y x

E x E x E x

E x E x

E x

E x E x E x

 +      _{ }     = _ __{ }  ₋  ₊ _{ }          −      _{ }     + _ __{ }  ₋  ₋ _{ }        

( )

(

)

( )

( ) ( )

1

2 4 4 2

4 3

p p p

p p

E x E x E x

Vecd Y x

E x E x

 _ − _   =  ₋   

( )

(

)

( )

2 4 0 2 0

1 0 ₂ 2

0 5 ₀ ₃ ₂ 4

0

2 4

0

2 0 ₂ ₆

4

0 3 2

0 2 p p p p p p

p p p

p p

p p p

E x E x

E x

Vecd Y x

E x E x E x

E x E x

E x

E x E x E x

(10)

( )

(

)

( )

( ) ( )

2

2 4 4 2

20 3

p p p

p p

E x E x E x

Vecd Y x

E x E x

 _ + _

 

=

 ₋ ₋ 

 

.

Example 4.5. Consider the following coupled matrix fractional differential equations:

( )

1 1 2

1 2

2 1 2

1 0 3 0

, 0 , 0

0 2 0 1

C p

D Y x Y x Y x

Y Y

D Y x Y x Y x



= +  ₌−  ₌− 

    ₋ 

= + _    . (4-7)

Then the exact solution by applying Corollary 3.5 is given by:

( )

(

)

( )

(

( )

( ) ( )

( )

)

( )

₍

_{( )}

_{( ) ( )}

_{( )}

₎

( )

_{( )}

( )

₍

_{( )}

₎

1

1 ,

2 2

3 ,

1 2

4 2

2

3 2

p

p p p p p

p p p p

p

p p p p p

p p p p

p

p p p

p p

p

p p p

p p

E x

Vecd Y x diag E x E x E x E x

E x

diag E x E x E x E x

E x

E x E x

E x

E x E x

−  

= + − + − _{ }

  −  

+ − − − − _{ }

−  

 

 _− + − _

 

=  

 _ ₊ ₋ _

  

 

( )

(

)

( )

(

( )

( ) ( )

( )

)

( )

₍

_{( )}

_{( ) ( )}

_{( )}

₎

( )

_{( )}

( )

₍

_{( )}

₎

2

1 ,

2 2

3 ,

1 2

4 2

2 _.

3 2

p

p _p _p _p _p

p p p p

p

p p p p p

p p p p

p

p p p

p p

p

p _p _p

p p

E x

Vecd Y x diag E x E x E x E x

E x

diag E x E x E x E x

E x

E x E x

E x

E x E x

−  

= − − − − _{ }

  −  

+ + − + − _{ }

−  

 

 _− − − _

 

=  

  ₋ ₋ 

  

 

5. Conclusion

The general exact solutions of coupled system of matrix fractional differential equations with diagonal matrices coefficients by using vector extraction operators and Hadamard product in Caputo sense are presented with some illustrated examples. How to find the complexity of this method requires further research.

Acknowledgements

The authors express their sincere thanks to referees for very careful reading and helpful suggestion of this paper.

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