# Analysis of a System for Linear Fractional Differential Equations

Authors: Fang Wang; Zhen-hai Liu; Ping Wang
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

The main purpose of this paper is to obtain the unique solution of the constant coefficient homogeneous linear fractional differential equationsDt0qX(t)=PX(t),X(a)=B and the constant coefficient nonhomogeneous linear fractional differential equations Dt0qX(t)=PX(t)+D,X(a)=B if P is a diagonal matrix and X(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T] and prove the existence and uniqueness of these two kinds of equations for any P∈L(Rm) and X(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T]. Then we give two examples to demonstrate the main results.


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## Body

## 1. Introduction

System of fractional differential equations has gained a lot of interest because of the challenges it offers compared to the study of system of ordinary differential equations. Numerous applications of this system in different areas of physics, engineering, and biological sciences have been presented in [1–3]. The differential equations involving the Riemman-Liouville differential operators of fractional order 0<q<1 appear to be more important in modeling several physical phenomena and therefore seem to deserve an independent study of their theory parallel to the well-known theory of ordinary differential equations. The existence and uniqueness of solution for fractional differential equations with any X(t)∈C[t0,T]×C[t0,T]×⋯×C[t0,T] have been studied in many papers, see [4–28]. In [4] Daftradar-Gejji and Babakhani have studied the existence and uniqueness of
(1.1)D0q(X(t)-X0)=PX(t),
where D0q denotes the standard Riemman-Liouville fractional derivative, 0<q<1,X(t)=(x1(t),x2(t),…,xm(t))T, X(0)=X0=(x10,x20,…,xm0)T, P∈L(Rm) which is an m dimensional linear space. They have obtained that the system (1.1) has a unique solution defined on [0,T] if P∈L(Rm) and X(t)∈C[t0,T]×C[t0,T]×⋯×C[t0,T]. In [17]  Belmekki et al. have studied the existence of periodic solution for some linear fractional differential equation in C1-q[0,1]. In [21] Ahmad and Nieto have studied the Riemann-Liouwille fractional differential equations with fractional boundary conditions. In comparison with the earlier results of this type we get more general assumptions. We assume X(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T] instead of X(t)∈C[t0,T]×C[t0,T]×⋯×C[t0,T] and consider the following system of fractional differential equations:
(1.2)Dt0qX(t)=PX(t),X(a)=B,Dt0qX(t)=PX(t)+D,X(a)=B,
where D0q denotes the standard Riemman-Liouville fractional derivative, 0<q<1, P∈L(Rm),
(1.3)X(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T],(a,B)∈(t0,T]×Rm and D is a constant vector. We completely generalize the results in [4] and obtain the new results if X(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T]. Furthermore, we also obtain some results of the unique solution of the homogeneous and nonhomogeneous initial value problems with the classical Mittag-Leffler special function [5] which is similar to the ordinary differential equations. Now we introduce the first Mittag-Leffler function eq(t-t0) defined by
(1.4)eq(t-t0)=∑k=1+∞‍(t-t0)kq-1Γ(kq).
The function eq(t-t0) belongs to C1-q[t0,T]. Indeed, taking the norm in C1-q[t0,T], we have
(1.5)∥eq(t-t0)∥1-q≤∑k=1+∞‍(T-t0)(k-1)qΓ(kq)<+∞.
The formula remains valid for q→1-. In this case, e1(t-t0)=exp(t-t0). Then we introduce the second Mittag-Leffler function Eq(t-t0) defined by
(1.6)Eq(t-t0)=∑k=1+∞‍(t-t0)kqΓ(kq+1).
The formula remains also valid for q→1-. In this case, E1(t-t0)=exp(t-t0)-1.The paper is organized as follows. In Section2 we recall the definitions of fractional integral and derivative and related basic properties and preliminary results used in the text. In Section 3 we obtain the unique solution of the constant coefficient homogeneous and nonhomogeneous linear fractional differential equations for P being the diagonal matrix. In Section 4 we prove the existence and uniqueness of these two kinds of equations for any P∈L(Rm). In Section 5 we give some specific examples to illustrate the results.


## 2. Definitions and Preliminary Results

Let us denote byC[t0,T] the space of all continuous real functions defined on [t0,T], which turns out to be a Banach space with the norm
(2.1)∥x∥=maxt∈[t0,T]|x(t)|.We define similarly another Banach spaceC1-q[t0,T], in which function x(t) is continuous on (t0,T] and (t-t0)1-qx(t) is continuous on [t0,T] with the norm:
(2.2)∥x∥1-q=maxt∈[t0,T](t-t0)1-q|x(t)|.L
[
t
0
,
T
] is the space of real functions defined on [t0,T] which are Lebesgue integrable on [t0,T].ObviouslyC1-q[t0,T]⊂L(t0,T).The definitions and results of the fractional calculus reported below are not exhaustive but rather oriented to the subject of this paper. For the proofs, which are omitted, we refer the reader to [6] or other texts on basic fractional calculus.Definition 2.1 (see [6]).
The fractional primitive of orderq>0 of function x(t)∈C1-q[t0,T] is given by
(2.3)It0qx(t)=1Γ(q)∫t0t‍(t-s)q-1x(s)ds.
From [17] we know It0qx(t) exists for all q>0, when x∈C1-q[t0,T];  consider  also that when x∈C[t0,T] then It0qx(t)∈C[t0,T] and moreover
(2.4)It0qx(t0)=0.Definition 2.2 (see [6]).
The fractional derivative of order0<q<1 of a function x(t)∈C1-q[t0,T] is given by
(2.5)Dt0qx(t)=1Γ(1-q)ddx∫t0t‍(t-s)-qx(s)ds.
We haveDt0qIt0qx(t)=x(t) for all x(t)∈C1-q[t0,T].Lemma 2.3 (see [6]).
Let0<q<1. If one assumes x(t)∈C1-q[t0,T], then the fractional differential equation
(2.6)Dt0qx(t)=0
has x(t)=c(t-t0)q-1,c∈R, as solutions.From this lemma we can obtain the following law of composition.Lemma 2.4 (see [6]).
Assume thatx(t)∈C1-q[t0,T] with a fractional derivative of order 0<q<1 that belongs to C1-q[t0,T]. Then
(2.7)It0qDt0qx(t)=x(t)+c(t-t0)q-1,
for some c∈R. When the function x is in C[t0,T], then c=0.Lemma 2.5 (see [6]).
LetU be a nonempty closed subset of a Banach space E, and let αn≥0 for every n and such ∑n=0∞‍αn converges. Moreover, let the mapping A:U→U satisfy the inequality
(2.8)∥Anu-Anv∥≤αn∥u-v∥,
for every n∈N and any u,v∈U. Then, A has a uniquely defined fixed point u*. Furthermore, for any u0∈U, the sequence (Anu0)n=1∞ converges to this fixed point u*.Lemma 2.6 (see [12]).
LetP∈L(Rm)  and  have real eigenvalues λ1,λ2,…,λr. Then there exists a basis of Rm in which the matrix representation of P assumes Jordan form, that is, the matrix of P is made of diagonal blocks of the form diag(J1,J2,…,Jr), where each Ji consists of diagonal blocks of the form
(2.9)(λi0⋯001λi⋯0001⋯00⋮⋮⋱⋮⋮00⋯1λi).Lemma 2.7 (see [12]).
LetP∈L(Rm)  and  have complex eigenvalues μj=αj+iβj,j=1,2,…,r, with multiplicity. Then there exists a basis of Rm, where P has matrix form diag(J^1,J^2,…,J^r), where each J^i consists of diagonal blocks of the type
(2.10)(D0⋯00I2D⋯000I2⋯00⋮⋮⋱⋮⋮00⋯I2D),D=(αi-βiβiαi),I2=(1001).Lemma 2.8 (see [12]).
LetP∈L(Rm). Then Rm has a basis giving P a matrix representation composed of diagonal blocks of type Ji and/or matrices J^i, where Ji and J^i are as defined in the preceding lemmas.Now, we will introduce Lemma2.9 to prove the following Theorem 4.4 in Section 4.Lemma 2.9.
Let0<q<1. Assume that x(t) and f(t) belong to C1-q[t0,T]. Then For the initial value problem
(2.11)Dt0qx(t)=λx(t)+f(t),x(a)=b
has a unique solution x(t)∈C1-q[t0,T] provided t0<a<a0, where a0 is a suitable constant depending on t0, q, and λ.Proof.
The initial value problem (2.11) will be solved in two steps.(1) Local existence.Our problem is equivalent to the problem of determination of fixed points of the following operator:(2.12)Ax(t)=c(t-t0)q-1+1Γ(q)∫t0t‍(t-s)q-1(λx(s)+f(s))ds,with(2.13)c=(b-1Γ(q)∫t0a‍(a-s)q-1(λx(s)+f(s))ds)(a-t0)1-q.It is immediate to verify thatA:C1-q[t0,T]→C1-q[t0,T] is also well defined. Indeed,
(2.14)|(t-t0)1-qAx(t)|≤|λ||1Γ(q)∫t0t‍(t-s)q-1(s-t0)q-1(s-t0)1-qx(s)ds|+|1Γ(q)∫t0t‍(t-s)q-1(s-t0)q-1(s-t0)1-qf(s)ds|≤|c|+|λ|∥x∥1-q|Iq(t-t0)q-1|+∥f∥1-q|Iq(t-t0)q-1|≤|c|+|λ|∥x∥1-qΓ(q)Γ(2q)(t-t0)q+∥f∥1-qΓ(q)Γ(2q)(t-t0)q,forx(t) and f(t) belong to ∈C1-q[t0,T].Then we can also proveA is a contraction operator. Indeed,
(2.15)(t-t0)1-q|Ax(t)-Ay(t)|≤|λ|Γ(q)Γ(2q)(a-t0)q∥x-y∥1-q+|λ|Γ(q)Γ(2q)(t-t0)q∥x-y∥1-q≤|λ|Γ(q)Γ(2q)(a-t0)q∥x-y∥1-q+|λ|Γ(q)Γ(2q)(T-t0)q∥x-y∥1-q,for allx(t),y(t)∈C1-q[t0,T]. Let us assume
(2.16)|λ|Γ(q)Γ(2q)(a-t0)q<12,that is,(2.17)a<a0=(Γ(2q)2|λ|Γ(q))1/q+t0.TakingT-a>0 sufficiently small, we also have
(2.18)|λ|Γ(q)Γ(2q)(T-t0)q<12,
and then
(2.19)∥Ax(t)-Ay(t)∥1-q≤L∥x-y∥1-q,withL<1. Therefore A is a contraction operator. This shows that initial problem (2.11) has a unique solution.(2) Continuation of solution.Since we know the value ofx(t) on (t0,a], then we can compute
(2.20)c*=(b-1Γ(q)∫t0a‍(a-s)q-1(λx(s)+f(s))ds)(a-t0)1-q.We can solve the integral problem(2.21)y(t)=c*(t-t0)q-1+1Γ(q)∫t0t‍(t-s)q-1(λy(s)+f(s))ds,obtaining a unique solutiony(t)∈C1-q[t0,T] for all T>t0. Now x(t) and y(t) agree on (t0,a]. Thus the solution admits y(t) as its continuation. Hence the  proof of  Lemma 2.9  is complete.


## 3. Initial Value Problem: Continuous Solutions on(t0,T]

We open this section with some basic examples, concerning the case when the solutions inC1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T] are submitted to an initial condition.Theorem 3.1.
Let0<q<1. For all (a,B)∈(t0,T]×Rm the initial value problem
(3.1)Dt0qX(t)=0,X(a)=B
admits
(3.2)X(t)=B(a-t0)1-q(t-t0)q-1,
as unique solution in C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T].Proof.
According to Lemma2.4, the initial value problem (3.1) is equivalent to the following equations:
(3.3)X(t)=C(t-t0)q-1,C=B(a-t0)1-q.
Hence the  proof of Theorem 3.1  is complete.Theorem 3.2.
Let0<q<1. Assume
(3.4)F(t)=(f1(t),f2(t),…,fm(t))T∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T].
Then for all (a,B)∈(t0,T]×Rm the initial value problem
(3.5)Dt0qX(t)=F(t),X(a)=B
has a unique solution in
(3.6)C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T],
given by
(3.7)X(t)=(x1(t),x2(t),…,xm(t))T,
with
(3.8)xi(t)=(bi-1Γ(q)∫t0a‍(a-s)q-1fi(s)ds)(a-t0)1-q(t-t0)q-1+1Γ(q)∫t0t‍(t-s)q-1fi(s)ds,(i=1,2,…,m).Proof.
According to Lemma2.4, the initial value problem (3.5) is equivalent to the following equations:
(3.9)X(t)=C(t-t0)q-1+1Γ(q)∫t0t‍(t-s)q-1F(s)ds,C=(B-1Γ(q)∫t0a‍(a-s)q-1F(s)ds)(a-t0)1-q.
Hence the  proof of  Theorem 3.2  is complete.The result remains true even ifq→1-. In this case, (3.5) is reduced to the ordinary differential equations
(3.10)X′(t)=F(t),X(a)=B,
which have a unique solution in
(3.11)C[t0,T]×C[t0,T]×⋯×C[t0,T],
given by
(3.12)X(t)=(x1(t),x2(t),…,xm(t))T,
with
(3.13)xi(t)=bi-∫t0a‍fi(s)ds+∫t0t‍fi(s)ds,(i=1,2,…,m).Theorem 3.3.
Let0<q<1. For all (a,B)∈(t0,T]×Rm the initial value problem
(3.14)Dt0qX(t)=PX(t),X(a)=B,
where
(3.15)P=(λ10⋯000λ2⋯0000⋯00⋮⋮⋱⋮⋮00⋯0λm)
has a unique solution in
(3.16)C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T],
given by
(3.17)X(t)=(x1(t),x2(t),…,xm(t))T,
with
(3.18)xi(t)=bieq-1(λi1/q(a-t0))eq(λi1/q(t-t0)),(i=1,2,…,m).Proof.
We can write (3.27) in the following form:
(3.19)Dt0qx1(t)=λ1x1(t),Dt0qx2(t)=λ2x2(t),⋮Dt0qxm(t)=λmxm(t),x1(a)=b1,…,xm(a)=bm.
According to Lemma 2.4,(3.20)Dt0qxi(t)=λixi(t)
is equivalent to the following equations:
(3.21)xi(t)=ci(t-t0)q-1+It0q(λixi(t)),
for some ci∈R. From (3.21) we obtain, by iteration,
(3.22)xi(t)=ciΓ(q)((t-t0)q-1Γ(q)+λi(t-t0)2q-1Γ(2q)+⋯+λin-1(t-t0)nq-1Γ(nq))+λinIt0nqxi(t).
Letting n→+∞, ∥λinIt0nqxi(t)∥1-q→0 if xi(t)∈C1-q[t0,T]. Indeed,
(3.23)∥λinIt0nqxi(t)∥1-q≤|λin|∥xi(t)∥1-qΓ(q)Γ((n+1)q)(t-t0)nq.
On the other hand,(3.24)eq(t-t0)=∑k=1+∞‍(t-t0)kq-1Γ(kq),
then we can obtain
(3.25)xi(t)=ciΓ(q)λi(1/q-1)eq(λi1/q(t-t0)).
Since xi(a)=bi,
(3.26)xi(t)=bieq-1(λi1/q(a-t0))eq(λi1/q(t-t0)),(i=1,2,…,m).
Hence the  proof of  Theorem 3.3  is complete.
The result remains valid even ifq→1-. In this case,
(3.27)X′(t)=PX(t),X(a)=B
has a unique solution in
(3.28)C[t0,T]×C[t0,T]×⋯×C[t0,T],
given by
(3.29)X(t)=(x1(t),x2(t),…,xm(t))T,
with
(3.30)xi(t)=biexp(-λi(a-t0))exp(λi(t-t0)),(i=1,2,…,m).Theorem 3.4.
Let0<q<1. For all (a,B)∈(t0,T]×Rm the initial value problem
(3.31)Dt0qX(t)=PX(t)+D,X(a)=B,
where
(3.32)P=(λ10⋯000λ2⋯0000⋯00⋮⋮⋱⋮⋮00⋯0λm),
and D=(d1,d2,…,dm) has a unique solution in
(3.33)C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T],
given by
(3.34)X(t)=(x1(t),x2(t),…,xm(t))T,
with
(3.35)xi(t)=(bi-diλi-1Eq(λi1/q(a-t0)))eq-1(λi1/q(a-t0))eq(λi1/q(t-t0))+diλi-1Eq(λi1/q(t-t0))(i=1,2,…,m).Proof.
We can write (3.44) in the following form:(3.36)Dt0qx1(t)=λ1x1(t)+d1,Dt0qx2(t)=λ2x2(t)+d2,⋮Dt0qxm(t)=λmxm(t)+dmx1(a)=b1,…,xm(a)=bm.
According to Lemma2.4, the equation
(3.37)Dt0qxi(t)=λixi(t)+di
is equivalent to the following equations:
(3.38)xi(t)=ci(t-t0)q-1+It0q(λixi(t))+It0q(di),
for some ci∈R. From (3.38) we obtain, by iteration,
(3.39)It0qλixi(t)=It0q(ciλi(t-t0)q-1)+It02q(λi2xi(t))+It02q(λidi),xi(t)=ci(t-t0)q-1+It0q(λici(t-t0)q-1)+It02q(λi2xi(t))+It0q(di)+It02q(λidi),It0qxi(t)=It0q(ci(t-t0)q-1)+It02q(λixi(t))+It02q(di),xi(t)=ciΓ(q)((t-t0)q-1Γ(q)+λi(t-t0)2q-1Γ(2q)+⋯+λin-1(t-t0)nq-1Γ(nq))+di((t-t0)qΓ(q+1)+λi(t-t0)2qΓ(2q+1)+⋯+λin-1(t-t0)nqΓ(nq+1))+λinIt0nqxi(t).
Lettingn→+∞, ∥λinIt0nqxi(t)∥1-q→0 if xi(t)∈C1-q[t0,T]. Indeed,
(3.40)∥λinIt0nqxi(t)∥1-q=maxt∈[t0,T](t-t0)1-q|λin1Γ(q)∫t0t‍(t-s)nq-1xi(s)ds|≤|λin|∥xi(t)∥1-qΓ(q)Γ((n+1)q)(t-t0)nq.
On the other hand,(3.41)eq(t-t0)=∑k=1+∞‍(t-t0)kq-1Γ(kq),Eq(t-t0)=∑k=1+∞‍(t-t0)kqΓ(kq+1).
Then we can obtain(3.42)xi(t)=ciΓ(q)λi(1/q-1)eq(λi1/q(t-t0))+diλi-1Eq(λi1/q(t-t0)).
We know thatdiλi-1Eq(λi1/q(t-t0)) is satisfied for the fractional nonhomogeneous linear differential equation Dt0qxi(t)=λ1xi(t)+di. So we can also deduce that the general solution of the fractional nonhomogeneous linear differential equation is equal to the general solution of the corresponding homogeneous linear differential equation plus the special solution of the nonhomogeneous linear differential equation. If X(a)=B, xi(a)=bi, then
(3.43)xi(t)=(bi-diλi-1Eq(λi1/q(a-t0)))eq-1(λi1/q(a-t0))eq(λi1/q(t-t0))+diλi-1Eq(λi1/q(t-t0))(i=1,2,…,m).
Hence the  proof of  Theorem 3.4  is complete.The result remains valid even ifq→1-. In this case,
(3.44)X′(t)=PX(t)+D,X(a)=B,
where
(3.45)P=(λ10⋯000λ2⋯0000⋯00⋮⋮⋱⋮⋮00⋯0λm),
and D=(d1,d2,…,dm) has a unique solution in
(3.46)C[t0,T]×C[t0,T]×⋯×C[t0,T],
given by
(3.47)X(t)=(x1(t),x2(t),…,xm(t))T,
with
(3.48)xi(t)=(bi-diλi-1E1(λi(a-t0)))e1-1(λi(a-t0))e1(λi(t-t0))+diλi-1E1(λi(t-t0))=(bi+diλi-1exp(-λ(a-t0)))exp(λi(t-t0))-dλ-1,(i=1,2,…,m).


## 4. Existence and Uniqueness of the Solution

In Section3 we have obtained the unique solution of the constant coefficient homogeneous and nonhomogeneous linear fractional differential equations for P being the diagonal matrix. In the present section we will prove the existence and uniqueness of these two kinds of equations for any P∈L(Rm).Theorem 4.1.
Let0<q<1 and P∈L(Rm). If the matrix P has distinct real eigenvalues, then for all (a,B)∈(t0,T]×Rm the initial value problem
(4.1)Dt0qX(t)=PX(t),X(a)=B
has the unique solution X(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T].Proof.
Since the matrixP has distinct real eigenvalues, there exists an invertible matrix Q such that
(4.2)Q-1PQ=(λ10⋯000λ2⋯0000⋯00⋮⋮⋱⋮⋮00⋯0λm),
where λ1,λ2,…,λm are the eigenvalues of the matrix P. If we define Y(t)=Q-1X(t),
(4.3)Dt0qY(t)=Dt0qQ-1X(t)=Q-1Dt0qX(t)=Q-1PX(t)=RY(t),
with R=diag(λ1,λ2,…,λm). From the above Theorem 3.3 we know the initial value problem
(4.4)Dt0qY(t)=RY(t),Y(a)=Q-1B
has a unique solution Y(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T] defined on [t0,T]. Then X(t)=QY(t) uniquely solves the equations (4.1), where t∈[t0,T]. Hence the  proof of  Theorem 4.1  is complete.Theorem 4.2.
Let0<q<1. For all (a,B)∈(t0,T]×R2 the initial value problem
(4.5)Dt0qX(t)=PX(t),X(a)=B,
where
(4.6)P=(α-ββα)
has the unique solution X(t)∈C1-q[t0,T]×C1-q[t0,T] defined on [t0,T].Proof.
Let us define(4.7)Z(t)=x1(t)+ix2(t),μ=α+iβ.
We can find that (4.5) is equivalent to the following equation
(4.8)Dt0qZ(t)=μZ(t),Z(a)=x1(a)+x2(a)=b1+ib2.
Obviously, Z(t)∈C1-q[t0,T] if x1(t) and x2(t) belong to C1-q[t0,T]. From the above Theorem 3.3 in Section 3, we know the complex Equation (4.7) has a unique solution defined on [t0,T]. Hence the  proof of  Theorem 4.2  is complete.Theorem 4.3.
Let0<q<1 and P∈R2. If P has eigenvalues α±iβ, for all (a,B)∈(t0,T]×R2 the initial value problem
(4.9)Dt0qX(t)=PX(t),X(a)=B
has a unique solution X(t)∈C1-q[t0,T]×C1-q[t0,T].Proof.
SinceP has eigenvalues α±iβ, there exists an invertible matrix Q such that P=QSQ-1 where
(4.10)S=(α-ββα).
Define
(4.11)Y(t)=Q-1X(t),
then
(4.12)Dt0qY(t)=Dt0qQ-1X(t)=Q-1Dt0qX(t)=Q-1PX(t)=SY(t).
From the above Theorem 4.2, we know the initial value problem
(4.13)Dt0qY(t)=SY(t),Y(a)=Q-1B
has a unique solution defined on [t0,T]. Hence the  proof of  result  is complete.Theorem 4.4.
Let0<q<1 and P∈Rm be an elementary Jordan matrix:
(4.14)(λ0⋯001λ⋯0001⋯00⋮⋮⋱⋮⋮00⋯1λ).
The initial value problem
(4.15)Dt0qX(t)=PX(t),X(a)=B
has a unique solution X(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T] provided (a,B)∈(t0,a0]×Rm, where a0 is a suitable constant depending on t0, q, and λ.Proof.
From the (4.15), we can write the equations in the following form:
(4.16)Dt0qx1(t)=λx1(t),Dt0qx2(t)=x1(t)+λx2(t),⋮Dt0qxm(t)=xm-1(t)+λxm(t),x1(a)=b1,…,xm(a)=bm.
Consider the first equation(4.17)Dt0qx1(t)=λx1(t),x1(a)=b1.
We can obtain the solution of this equation(4.18)x1(t)=b1eq-1(λ1/q(a-t0))eq(λ1/q(t-t0)).
Consider the second equation(4.19)Dt0qx2(t)=x1(t)+λx2(t),x2(a)=b2,
where now x1(t)∈C1-q[t0,T] is a known function. Since x1(t),x2(t)∈C1-q[t0,T], according to Lemma 2.9, (4.19) has a unique solution in C1-q[t0,T]. Now x1(t) and x2(t) are known functions which will be substituted in
(4.20)Dt0qx3(t)=x2(t)+λx3(t),x3(a)=b3
and so on. Thus the system of equations given in (4.15) has unique solution in C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T].Theorem 4.5.
Let0<q<1 and P∈L(Rm). The initial value problem
(4.21)Dt0qX(t)=PX(t),X(a)=B
has the unique solution X(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T] provided (a,B)∈(t0,a0]×Rm, where a0 is a suitable constant depending on t0, q, and λ.Proof.
In view of Lemma2.8, there exists an invertible matrix Q such that Q-1PQ is composed of diagonal blocks of the type Ji and J^i, as defined in the preceding Lemmas 2.7 and 2.8. Let B=Q-1PQ and Y(t)=Q-1X(t). Consider the initial value problem:
(4.22)Dt0qY(t)=Dt0qQ-1X(t)=Q-1Dt0qX(t)=Q-1PX(t)=BY(t),Y(a)=Q-1X(a)=Q-1B.
Then in view of Theorems 4.1–4.5, (4.22) has a unique solution: Y(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T]. Therefore (4.21) has a unique solution Q-1Y(t)∈C1-q[t0,T]×C1-q[t0,T]×⋯×C1-q[t0,T].Remark 4.6.
All the above results are valid forq→1-. Moreover, we can also discuss the case if a=t0, in this case, we cannot consider the usual initial condition x(t0)=b, but limt→t0(t-t0)1-qx(t)=b. We can also obtain some similar results by the same method, So we did not give the detailed process and conclusion in this paper.


## 5. Illustrative Examples

In this section, we give some specific examples to illustrate the above results.Example 5.1.
Consider the following system, where0<q<1, t∈[t0,T], (a,B)∈(t0,T]×R3,B=(b1,b2,b3),(5.1)Dt0qx1(t)=3x1(t)-x2(t)+x3(t),Dt0qx2(t)=-x1(t)+5x2(t)-x3(t),Dt0qx3(t)=x1(t)-x2(t)+3x3(t),x1(a)=b1,x2(a)=b2,x3(a)=b3.
Here
(5.2)P=(3-11-15-11-13),
having the eigenvalues 2, 3, and 6. Choose the eigenvectors  g1=(1,0,-1)T,g2=(1,1,1)T, and  g3=(1,-2,1)T. Then
(5.3)(200030006)=Q-1(3-11-15-11-13)Q,
where
(5.4)Q-1=(120-1213131316-1316).
Define theY=Q-1X. Then the system of equation in Y is decoupled, namely,
(5.5)Dt0qy1(t)=2y1(t),Dt0qy2(t)=3y2(t),Dt0qy3(t)=6y3(t),y1(a)=12b1-12b3,y2(a)=13b1+13b2+13b3,y3(a)=16b1-13b2+16b3.
In view of (3.30), we can obtain
(5.6)y1(t)=(12b1-12b3)eq-1(21/q(a-t0))eq(21/q(t-t0)),y2(t)=(13b1+13b2+13b3)eq-1(31/q(a-t0))eq(31/q(t-t0)),y3(t)=(16b1-13b2+16b3)eq-1(61/q(a-t0))eq(61/q(t-t0)).
Hence(5.7)x1(t)=y1(t)+y2(t)+y3(t),x2(t)=y2(t)-2y3(t),x3(t)=-y1(t)+y2(t)+y3(t).Example 5.2.
Consider the following system, where0<q<1, t∈[t0,T], (a,B)∈(t0,T]×R2,B=(b1,b2),
(5.8)Dt0qx1(t)=-2x1(t)-x2(t),Dt0qx2(t)=13x1(t)+4x2(t),x1(a)=b1,x2(a)=b2.
Here
(5.9)P=(-2-1134)
having the eigenvalues 1±2i. Choose the  eigenvectors   g1=(1,-3-2i)T,  and  g2=(1,-3+2i)T, Then
(5.10)(1+2i001-2i)=Q-1(-2-1134)Q,
where
(5.11)Q=(11-3-2i-3+2i),Q-1=(2+3i4i42-3i4-i4).
Define the Y=Q-1X. Then the system of equation in Y is decoupled, namely,
(5.12)Dt0qy1(t)=(1+2i)y1(t),y1(a)=12b1+(34b1+14b2)iDt0qy2(t)=(1-2i)y2(t),y2(a)=12b1-(34b1+14b2)i.
In view of (3.30), we can obtain
(5.13)y1(t)=(12b1+(34b1+14b2)i)eq-1((1+2i)1/q(a-t0))eq((1+2i)1/q(t-t0))y2(t)=(12b1-(34b1+14b2)i)eq-1((1-2i)1/q(a-t0))eq((1-2i)1/q(t-t0)).
Hence
(5.14)x1(t)=y1(t)+y2(t)x2(t)=-3(y1(t)+y2(t))-2i(y1(t)-y2(t)).



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*Source: 193061-2012-09-09.xml*