# On the Existence and Uniqueness of the Solution of Linear Fractional Differential-Algebraic System

Authors: Zaiyong Feng; Ning Chen
Journal: Mathematical Problems in Engineering
 (2016)

---

## Abstract

The existence and uniqueness of the solution of a new kind of system—linear fractional differential-algebraic equations (LFDAE)—are investigated. Fractional derivatives involved are under the Caputo definition. By using the tool of matrix pair, the LFDAE in which coefficients matrices are both square matrices have unique solution under the condition that coefficients matrices make up a regular matrix pair. With the help of equivalent transformation and Kronecker canonical form of the coefficients matrices, the sufficient condition for existence and uniqueness of the solution of the LFDAE in which coefficients matrices are both not square matrices is proposed later. Two examples are given to justify the obtained theorems in the end.


---

## Body

## 1. Introduction

The dynamical behavior of a mechanical system was usually modeled via differential-algebraic equations (DAE) whose general form appears asF
t
,
y
,
y
˙
=
0, including both differential and algebraic equations to describe the corresponding constraints, for example, by Newton’s laws of motion or by position constraints such as the movement on a given surface. On the other hand, researchers had effectively solved engineering problems with fractional differential equations (FDE), which involves fractional derivatives y
α in the model [1–6].Recently, some investigators tried using fractional differential-algebraic equations (FDAE), which denote the combination of DAE and FDE, in dealing with the studied system. The general form of FDAE appears asF
t
,
y
,
y
α
=
0, where F, y, y
α could be vectors if necessary. Till now, the majority of these attempts concentrated on the algorithms for solving the FDAE [7–10], while fundamental problems such as the existence and uniqueness of the solution were neglected.Existence and uniqueness of the solution of the model are of great significance, since the existence of the solution guarantees the practicability of the model, while the uniqueness of the solution guarantees the validity of the obtained solution.In this paper, the existence and uniqueness of the solution of linear fractional differential-algebraic systems are discussed to lay the groundwork for the further studies and applications.


## 2. Linear Fractional Differential-Algebraic System

Consider the initial value problem for linear fractional differential-algebraic equations (LFDAE) in the form(1)
E
D
t
α
t
0
y
=
A
y
+
z
,
y
t
=
t
0
=
0
,where 0
<
α
<
1, y
t
∈
R
n, z
t
∈
R
m, y
t
,
z
t
∈
L
1
0
,
T, 0
<
t
<
T
<
+
∞, m
≤
n. E
,
A
∈
R
m
×
n are constant coefficients matrices. D
t
α
t
0 denotes Caputo’s fractional derivative operator [11–13], D
t
α
t
0
y
=
(
1
/
Γ
1
-
α
)
∫
t
0
t
(
y
′
ξ
/
t
-
ξ
α
)
d
ξ, since Caputo’s fractional derivative allows us to couple the fractional differential equations with initial conditions in the traditional form y
t
=
t
0
=
y
0
=
0. And it is linked to other fractional definitions under certain conditions [11]. For convenience, we write system (1) as (E
,
A) in short.For givenz
t
∈
R
m, whether there exists unique solution y
t
∈
R
n for LFDAE (1) is a problem of great significance in application and it is our main concern as well.System (1) is also valid when y
t
=
t
0
≠
0. In this case, by a simple coordinate transformation, we can obtain E
D
t
α
t
0
y
1
=
A
y
1
+
z
1
,
y
1
t
=
t
0
=
0, where y
1
=
y
-
y
0
,
A
=
a
i
j
m
×
n
,
z
1
i
=
z
i
+
∑
j
=
1
n
a
i
j
y
0
j
,
i
=
1,2
,
…
,
m
.It is worth pointing out that algebraic constraintsg
(
y
,
z
)
=
0 are represented while rows in E whose elements are all 0 appear.


## 3. Equivalent Transformation of LFDAE

To discuss the theorems on the existence and uniqueness of the solution of the LFDAE introduced in Section2, we give definition of equivalent matrix pair [14] as a necessary preparation.Definition 1.
Two pairs of matrices(
E
1
,
A
1
) and (
E
2
,
A
2
), i
=
1
,
2, are called (strongly) equivalent if there exist nonsingular matrices P
1
∈
R
m
×
m and P
2
∈
R
n
×
n such that E
2
=
P
1
E
1
P
2, A
2
=
P
1
A
1
P
2. If this is the case, one writes (E
1
,
A
1)~(E
2
,
A
2).
Multiplying nonsingular matrixP
1
∈
R
m
×
m on the left and setting y
=
P
2
y
~, P
2
∈
R
n
×
n is a nonsingular matrix, LFDAE (1), where E
,
A
∈
R
m
×
n
m
≠
n, y
t
∈
R
n, z
t
∈
R
m is transformed into(2)
E
~
D
t
α
t
0
y
~
=
A
~
y
~
+
z
~
,
y
~
t
=
t
0
=
0
,
E
~
=
P
1
E
P
2
,
A
~
=
P
1
A
P
2
,
z
t
~
=
P
1
z
twhich is again a linear fractional differential-algebraic equation with constant coefficients. System (2) could be written as E
~
,
A
~ and (
E
,
A
)~(
E
~
,
A
~
). Because of the linearity of the operator D
t
α
t
0 and the nonsingularity of square matrices P
1, P
2, it is evident that system (2) has the same solution property as system (1). Thus, by equivalent transformation, we can consider the transformed system (2) instead of (1) with respect to solvability and related questions.


## 4. Regular Matrix Pair and the Solution of LFDAE

To simplify the problem, we now focus on the LFDAE in whichE, A are both square matrices; that is, E
,
A
∈
C
n
×
n. Recall Definition 2 [14].Definition 2.
LetE
,
A
∈
C
m
×
n, the matrix pair (
E
,
A
) is called regular matrix pair if m
=
n, and the characteristic polynomial p
(
λ
)
=
det
⁡
(
λ
E
-
A
) is not the zero polynomial. A matrix pair which is not regular is called singular.
Lemma3 allows us to gain an equivalent but more simple form of the matrix pair (E
,
A)—Weierstrass canonical form [14].Lemma 3.
LetE
,
A
∈
C
n
×
n and (
E
,
A
) be regular. Then, one has the Weierstrass canonical form(3)
E
,
A
~
I
0
0
N
,
J
0
0
I
,where J is a matrix in Jordan canonical form and N is a nilpotent matrix also in Jordan canonical form. Moreover, it is allowed that one or the other block is not present.
Letl be the index of nilpotent matrix N
∈
C
k
×
k; that is, N
l
=
0 and N
l
-
1
≠
0 (in fact, l is always less than k); we have Theorem 4 which is fundamental on the solvability of LFDAE (1).Theorem 4.
Letl be the index of nilpotent matrix N
∈
C
k
×
k; z
t is differentiable enough; that is, D
t
0
t
i
α
z
t
i
=
0,1
,
2
,
…
,
l is well defined, where D
t
i
α
t
0
=
D
t
α
t
0
D
t
α
t
0
⋯
D
t
α
t
0
︷
i denotes sequential fractional derivatives [11, 15–17]. Then linear fractional differential-algebraic equation∗
N
D
t
α
t
0
y
t
=
y
t
+
z
thas unique solution y
t
=
-
∑
i
=
0
l
-
1
N
i
D
t
i
α
t
0
z
t.Proof.
Firstly, we prove solutions of∗ have the form of y
t
=
-
∑
i
=
0
l
-
1
N
i
D
t
i
α
t
0
z
t.
SinceD
t
0
t
α is a linear operator maps function y
t into its derivative of order α, moving the left hand term in ∗ to the right, we have(4)
I
-
N
D
t
0
t
α
y
t
+
z
t
=
0
.
BecauseN is a matrix consisting of constant number, it is commutable with operator D
t
0
t
α; using Neumann series and taking l as the index of nilpotent matrix N into account, we obtain(5)
y
t
=
-
I
-
N
D
t
0
t
α
-
1
z
t
=
-
∑
i
=
0
+
∞
N
D
t
0
t
α
i
z
t
=
-
∑
i
=
0
l
-
1
N
i
D
t
0
t
i
α
z
t
.
Secondly, we prove functiony
t
=
-
∑
i
=
0
l
-
1
N
i
D
t
0
t
i
α
z
t is the solution of ∗.
Substitutingy
t
=
-
∑
i
=
0
l
-
1
N
i
D
t
0
t
i
α
z
t into ∗ yields(6)
N
D
t
0
t
α
y
t
-
y
t
-
z
t
=
-
N
D
t
0
t
α
∑
i
=
0
l
-
1
N
i
D
t
0
t
i
α
z
t
-
-
∑
i
=
0
l
-
1
N
i
D
t
0
t
i
α
z
t
-
z
t
=
-
∑
i
=
1
l
-
1
N
i
D
t
0
t
i
α
z
t
-
-
∑
i
=
0
l
-
1
N
i
D
t
0
t
i
α
z
t
-
z
t
,
N
l
=
0
=
z
t
-
z
t
=
0
.
Obviously,y
(
t
) is really the solution of ∗.
So equationN
D
t
0
t
α
y
t
=
y
t
+
z
t has the unique solution y
t
=
-
∑
i
=
0
l
-
1
N
i
D
t
0
t
i
α
z
t.Now let us elaborate on Theorem5 below.Theorem 5.
LetE
,
A
∈
R
n
×
n
,
y
t
;
z
t
∈
R
n
,
0
<
t
<
T
<
+
∞ and y
t
,
z
t
∈
L
1
0
,
T
,
z
t is differentiable enough; then there exists unique solution for the linear fractional differential-algebraic equations(7)
E
D
t
0
t
α
y
=
A
y
+
z
,
y
t
=
t
0
=
0if the matrix pair (
E
,
A
) is regular.Proof.
As claimed in Section3, system (7) has the same solution property as system (8),(8)
E
¯
D
t
α
t
0
y
¯
=
A
¯
y
¯
+
z
¯
,
y
¯
t
=
t
0
=
0
,
E
¯
=
P
1
E
P
2
∈
R
n
×
n
,
A
¯
=
P
1
A
P
2
∈
R
n
×
n
,
z
¯
=
P
1
z
.
We now discuss the solvability of system (8). Using Lemma 3, E
¯
=
P
1
E
P
2, A
¯
=
P
1
A
P
2 in system (8) could be obtained in the following Weierstrass canonical form:(9)
E
¯
=
I
0
0
N
,
A
¯
=
J
0
0
I
.Hence system (8) appears as (10)
I
0
0
N
D
t
α
t
0
y
¯
=
J
0
0
I
y
¯
+
z
¯
,
y
¯
t
=
t
0
=
0
.By setting y
¯
=
y
1
¯
,
y
2
¯, system (8) is separated into two subsystems:(11)
I
D
t
0
t
α
y
1
¯
=
J
y
1
¯
+
z
1
¯
y
1
¯
t
=
t
0
=
0
,
(12)
N
D
t
α
t
0
y
2
¯
=
I
y
2
¯
+
z
2
¯
y
2
¯
t
=
t
0
=
0
.
Subsystem (11) is a normal fractional differential system, which has unique solution with given initial value [11]. As discussed in Theorem 4, subsystem (12) has unique solution y
2
t
¯
=
-
∑
i
=
0
l
-
1
N
i
D
t
0
t
i
α
z
2
t
¯.
Hence, system (8) has unique solution; accordingly system (7) has unique solution as a result. And the proof of Theorem 5 is completed.Remark 6.
Since the solutiony
t in Theorem 4 is obtained without specifying the initial value of y
t, initial value y
2
¯
t
=
t
0
=
0 is consistent when D
t
0
t
i
α
z
2
t
0
¯
=
0
i
=
1,2
,
…
,
l
-
1, where l is the index of nilpotent matrix N.Remark 7.
As derivative of fractional order is the generalization of derivative of integer order, the condition “the matrix pair(
E
,
A
) is regular” plays the same role as in theory of differential-algebraic equation [18].


## 5. Kronecker Canonical Form and the Solution of LFDAE

We have investigated the existence and uniqueness of the solution of LFDAE (1) in which E
,
A
∈
R
n
×
n are both square matrices. Nevertheless, the general form of LFDAE (1), where E
,
A
∈
R
m
×
n
m
≠
n are both not square matrices, is frequently modeled in mechanical systems. We go on to investigate this case by tools of equivalent transformation and the Kronecker canonical form of the system.Kronecker canonical form which is referred to asE
~, A
~ in Lemma 8 plays an important role in analyzing the existence and uniqueness of the solution of LFDAE. In 1974, Gantmacher pointed out Lemma 8 [19, 20].Lemma 8.
LetA
∈
R
m
×
n; then there exist nonsingular matrices P
∈
R
m
×
m and Q
∈
R
n
×
n, such that (E
,
A)~(E
~
,
A
~) (P
E
Q
=
E
~
,
P
A
Q
=
A
~), where E
~, A
~ take the next form:(13)
E
~
=
P
E
Q
=
diag
⁡
0
n
0
×
n
0
,
L
1
,
L
2
,
…
,
L
p
,
L
1
~
,
L
2
~
,
…
,
L
q
~
,
I
,
N
,
A
~
=
P
A
Q
=
diag
⁡
0
n
0
×
n
0
,
J
1
,
J
2
,
…
,
J
p
,
J
1
~
,
J
2
~
,
…
,
J
q
~
,
A
1
,
I
,
L
i
=
1
0
1
0
⋱
⋱
1
0
,
J
i
=
0
1
0
1
⋱
⋱
0
1
∈
R
n
i
×
n
i
+
1
,
L
j
~
=
1
0
1
⋱
⋱
1
0
,
J
j
~
=
0
1
0
⋱
⋱
0
1
∈
R
n
j
~
+
1
×
n
j
~
i
=
1,2
,
…
,
p
,
j
=
1,2
,
…
,
q
,
N
=
d
i
a
g
N
1
,
N
2
,
…
,
N
l
∈
R
h
×
h
,
N
s
=
0
1
0
1
⋱
⋱
0
1
0
∈
R
k
s
×
k
s
,
A
1
,
I
∈
R
g
×
g
.
Dimension of each matrix satisfies(14)
n
0
+
∑
i
n
i
+
∑
j
n
j
~
+
1
+
∑
s
k
s
+
g
=
m
;
n
0
+
∑
i
n
i
+
1
+
∑
j
n
j
~
+
∑
s
k
s
+
g
=
n
;
∑
s
k
s
=
h
.According to Lemma8, setting P
1
=
P, P
2
=
Q
,
E
~
,
A
~ in system (2) transformed from system (1) have the Kronecker canonical form through equivalent transformation: y
=
Q
y
~, z
(
t
)
~
=
P
z
t.Considering the structure ofE
~, A
~, we now divide the vector y
(
t
)
~ and vector z
(
t
)
~ as below:(15)
y
t
~
=
y
0
T
;
y
L
1
T
,
y
L
2
T
,
…
,
y
L
p
T
;
y
L
~
1
T
,
y
L
~
2
T
,
…
,
y
L
~
q
T
;
y
I
T
;
y
N
1
T
,
y
N
2
T
,
…
,
y
N
l
T
⊤
;
z
t
~
=
z
0
T
;
z
L
1
T
,
z
L
2
T
,
…
,
z
L
p
T
;
z
L
~
1
T
,
z
L
~
2
T
,
…
,
z
L
~
q
T
;
z
I
T
;
z
N
1
T
,
z
N
2
T
,
…
,
z
N
l
T
⊤
.Hence, LFDAE (2) could be transformed into the following equivalent equations:(16)
0
n
0
×
n
0
D
t
α
t
0
y
0
t
=
z
0
t
,
(17)
L
i
D
t
α
t
0
y
L
i
t
=
J
i
y
L
i
t
+
z
L
i
t
,
i
=
1,2
,
…
,
p
;
(18)
L
j
~
D
t
0
t
α
y
L
~
j
t
=
J
j
~
y
L
~
j
t
+
z
L
~
j
t
,
j
=
1,2
,
…
,
q
;
(19)
N
s
D
t
α
y
N
s
t
=
y
N
s
t
+
z
N
s
t
,
s
=
1,2
,
…
,
l
,and the equation(20)
D
t
α
y
I
t
=
A
1
y
I
t
+
z
I
t
.Now, let us discuss (16)–(20) to investigate the existence and uniqueness of the solution of LFDAE (2) in which E
~, A
~ take the Kronecker canonical form.With regard to (16), it is easy to see that the equation is solvable if and only if all components of z
0
t are zeros; that is, z
0
t
=
0. In this case, (16) either has no solution while z
0
t
≠
0 or has solutions of infinite number (as long as z
0
t
=
0, any function that is differentiable of order α could be considered as the solution of (16)). So there exists no unique solution for (16).With regard to (17), considering the particular form of L
i, J
i, (17) appears as(21)
1
0
1
0
⋱
⋱
1
0
D
t
α
t
0
y
L
i
t
=
0
1
0
1
⋱
⋱
0
1
y
L
i
t
+
z
L
i
t
,
i
=
1,2
,
…
,
p
,
L
i
,
J
i
∈
R
n
i
×
n
i
+
1
.Denoting components ofy
L
i
t as y
L
i
1
,
y
L
i
2
,
…
,
y
L
i
n
i
+
1
T and z
L
i
t as z
L
i
1
,
z
L
i
2
,
…
,
z
L
i
n
i
T, the above equation is precisely equal to (22)
D
t
α
t
0
y
L
i
1
t
=
y
L
i
2
t
+
z
L
i
1
t
,
D
t
α
t
0
y
L
i
2
t
=
y
L
i
3
t
+
z
L
i
2
t
,
⋮
D
t
α
t
0
y
L
i
n
i
t
=
y
L
i
n
i
+
1
t
+
z
L
i
n
i
t
,
i
=
1,2
,
…
,
p
.We further have(23)
y
L
i
2
t
=
D
t
α
t
0
y
L
i
1
t
-
z
L
i
1
t
,
y
L
i
3
t
=
D
t
α
t
0
y
L
i
2
t
-
z
L
i
2
t
,
⋮
y
L
i
n
i
+
1
t
=
D
t
α
t
0
y
L
i
n
i
t
-
z
L
i
n
i
t
,
i
=
1,2
,
…
,
p
.Sincey
L
i
1
t in (23) is free of all restrictions, any function differentiable of order α could be considered as y
L
i
1
t; thus (17) has solutions of infinite number.With regard to (18), considering the particular form of matrices L
j
~, J
j
~, (18) appears as(24)
1
0
1
⋱
⋱
1
0
D
t
α
t
0
y
L
j
~
t
=
0
1
0
⋱
⋱
0
1
y
L
j
~
t
+
z
L
j
~
t
,
j
=
1,2
,
…
,
q
,
L
j
~
,
J
j
~
∈
R
n
j
~
+
1
×
n
j
~
.Denoting components ofy
L
j
~
t as y
L
j
~
1
,
y
L
j
~
2
,
…
,
y
L
j
~
n
j
~
T and z
L
j
~
t as z
L
j
~
1
,
z
L
j
~
2
,
…
,
z
L
j
~
n
j
~
+
1
T, the above equation is precisely equal to(25)
D
t
α
t
0
y
L
j
~
1
t
=
z
L
j
~
1
t
,
D
t
α
t
0
y
L
j
~
2
t
=
y
L
j
~
1
t
+
z
L
j
~
2
t
,
⋮
D
t
α
t
0
y
L
j
~
n
j
~
t
=
y
L
j
~
n
j
~
-
1
t
+
z
L
j
~
n
j
~
t
,
j
=
1,2
,
…
,
q
,and an additional formula (26)
y
L
j
~
n
j
~
t
+
z
L
j
~
n
j
~
+
1
t
=
0from which we get26
′
y
L
j
~
n
j
~
t
=
-
z
L
j
~
n
j
~
+
1
t
.Thus (25) could be rewritten in a recursive form of reverse order:25
′
D
t
α
t
0
y
L
j
~
1
t
=
z
L
j
~
1
t
,
y
L
j
~
1
t
=
D
t
α
t
0
y
L
j
~
2
t
-
z
L
j
~
2
t
,
⋮
y
L
j
~
n
j
~
-
1
t
=
D
t
α
t
0
y
L
j
~
n
j
~
t
-
z
L
j
~
n
j
~
t
,
j
=
1,2
,
…
,
q
.Obviously, to any given differential enoughz
L
j
~
t, (25) has unique solution y
L
j
~
t. But the obtained y
L
j
~
1
t hardly satisfy the equation D
t
α
t
0
y
L
j
~
1
t
=
z
L
j
~
1
t in (25) except very special situation. Therefore, (18) has no feasible solution for any given z
L
j
~
t.With regard to (19), considering the particular form of matrices N
s, (19) appears as(27)
0
1
0
1
⋱
⋱
0
1
0
D
t
α
y
N
s
t
=
y
N
s
t
+
z
N
s
t
,
s
=
1,2
,
…
,
l
;
N
s
∈
R
k
s
×
k
s
.Denoting components ofy
N
s
t as y
N
s
1
,
y
N
s
2
,
…
,
y
N
s
k
s
T and z
N
s
t as z
N
s
1
,
z
N
s
2
,
…
,
z
N
s
k
s
T, the above equation is equal to(28)
D
t
α
t
0
y
N
s
2
t
=
y
N
s
1
t
+
z
N
s
1
t
,
D
t
α
t
0
y
N
s
3
t
=
y
N
s
2
t
+
z
N
s
2
t
,
⋮
D
t
α
t
0
y
N
s
k
s
t
=
y
N
s
k
s
-
1
t
+
z
N
s
k
s
-
1
t
,
y
N
s
k
s
t
+
z
N
s
k
s
t
=
0
,
s
=
1,2
,
…
,
l
.We further obtain the equivalence of (28) as below:(29)
y
N
s
1
t
=
D
t
α
t
0
y
N
s
2
t
-
z
N
s
1
t
,
y
N
s
2
t
=
D
t
α
t
0
y
N
s
3
t
-
z
N
s
2
t
,
⋮
y
N
s
k
s
-
1
t
=
D
t
α
t
0
y
N
s
k
s
t
-
z
N
s
k
s
-
1
t
,
y
N
s
k
s
t
=
-
z
N
s
k
s
t
,
s
=
1,2
,
…
,
l
.Obviously, to any givenz
N
s
t, solution y
N
s
t could be derived from (29) if z
t is differentiable enough; that is, D
t
i
α
t
0
z is well defined. Thus, (19) always has unique solution towards any given suitable vector z
N
s
t.With regard to (20), where E
=
I
∈
R
g
×
g, A
1
∈
R
g
×
g are square matrices, because of Theorem 5, (20) has unique solution if the matrix pair (
I
,
A
1
) is regular.Bringing the analysis above together, LFDAE (2) has unique solution if (16) to (18) do not appear in their Kronecker canonical form, while matrix pair (
I
,
A
1
) is regular. The above-mentioned summarization brings us to Theorem 9.Theorem 9.
LetE
,
A
∈
R
m
×
n. The linear fractional differential-algebraic equations(30)
E
D
t
α
t
0
y
=
A
y
+
b
z
y
t
=
t
0
=
0in which y
t
∈
R
n, z
t
∈
R
m, 0
<
t
<
T
<
+
∞ satisfy the following:(1)
y
t
,
z
t
∈
L
1
0
,
T;
(2)
D
t
i
α
t
0
z
i
=
0,1
,
2
,
…
,
n is well defined, and D
t
i
α
t
0
z
0
=
0
,
i
=
0,1
,
2
,
…
,
n.
Then, the system has unique solution if there exist nonsingular matricesP
∈
R
m
×
m and Q
∈
R
n
×
n, such that (E
,
A)~(E
~
,
A
~), where E
~
=
diag
⁡
I
,
N, A
~
=
diag
⁡
A
1
,
I, and matrix pair (I
,
A
1) is regular.Remark 10.
Theorem9 is also suitable when other fractional derivatives are involved. Since D
t
0
t
α in the sequential fractional derivatives in Theorem 4 can mean the Riemann-Liouville, the Grünwald-Letnikov, the Caputo, or even any other definition of fractional derivatives [11], Theorem 5 is applicable to these fractional derivatives. Meanwhile, analysis in Section 5 is still valid for other definitions of fractional derivatives, so it is the same as Theorem 9 as a result.


## 6. Examples

Now let us take some examples on the theorems represented before.Example 1.
LFDAE in whichE
=
N is a nilpotent matrix and A
=
I is as below (example of Theorem 4): (31)
0
1
2
0
0
3
0
0
0
D
t
0.5
0
y
1
t
D
t
0.5
0
y
2
t
D
t
0.5
0
y
3
t
=
1
0
0
0
1
0
0
0
1
y
1
t
y
2
t
y
3
t
+
t
2.5
t
2
t
1.5
,
y
0
=
0
.
In the example,E
=
0
1
2
0
0
3
0
0
0
=
N is a nilpotent matrix with index l
=
3, A
=
1
0
0
0
1
0
0
0
1
=
I. So according to Theorem 4, the unique solution of Example 1 is (32)
y
t
=
-
∑
i
=
0
l
-
1
N
i
0
D
t
i
×
0.5
z
t
=
-
z
t
+
N
D
0
t
0.5
z
t
+
N
2
0
D
t
2
×
0.5
z
t
=
-
t
2.5
t
2
t
1.5
+
0
1
2
0
0
3
0
0
0
D
0
t
0.5
t
2.5
t
2
t
1.5
+
0
1
2
0
0
3
0
0
0
2
D
0
t
2
×
0.5
t
2.5
t
2
t
1.5
=
-
t
2.5
t
2
t
1.5
+
0
1
2
0
0
3
0
0
0
Γ
3.5
Γ
3
t
2
Γ
3
Γ
2.5
t
1.5
Γ
2.5
Γ
2
t
+
0
0
3
0
0
0
0
0
0
Γ
3.5
Γ
2.5
t
1.5
Γ
3
Γ
2
t
Γ
2.5
Γ
1.5
t
0.5
=
-
t
2.5
+
Γ
3
Γ
2.5
t
1.5
+
2
Γ
2.5
Γ
2
t
+
3
Γ
2.5
Γ
1.5
t
0.5
t
2
+
3
Γ
2.5
Γ
2
t
t
1.5
.
Figure1 shows the solution of Example 1. The correctness and the initial value consistency of the solution could be verified by computation.Figure 1
Solution of Example1.Example 2.
LFDAE with Kronecker canonical form is as below (example of Theorem5):a
2
1
2
5
3
2
2
6
0
0
0
0
0
0
1
2
D
t
0.5
0
y
1
t
D
t
0.5
0
y
2
t
D
t
0.5
0
y
3
t
D
t
0.5
0
y
4
t
=
4
3
5
11
5
4
5
13
0
0
1
2
0
0
3
5
y
1
t
y
2
t
y
3
t
y
4
t
+
2
t
2
t
t
2
twith the initial value y
0
=
0.
In this example,E
=
2
1
2
5
3
2
2
6
0
0
0
0
0
0
1
2, A
=
4
3
5
11
5
4
5
13
0
0
1
2
0
0
3
5; there exist two nonsingular matrices P and Q:(33)
P
=
1
0
3
-
2
-
1
1
-
2
1
0
0
-
2
1
0
0
1
0
,
Q
=
1
-
1
0
1
-
1
2
1
-
3
0
0
2
-
1
0
0
-
1
1
,satisfying (34)
E
~
=
P
E
Q
=
1
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
,
A
~
=
P
A
Q
=
1
2
0
0
0
1
0
0
0
0
1
0
0
0
0
1
,
Z
~
=
P
Z
=
t
0
0
t
.
Obviously,E
~
=
diag
⁡
I
,
N, A
~
=
diag
⁡
A
1
,
I; we further get λ
I
-
A
1
=
λ
-
1
-
2
0
λ
-
1
=
λ
-
1
2, which is not a zero polynomial; hence the matrix pair (I
,
A
1) is regular.
From Theorem9, the fractional system has unique solution. Now let us find the solution.
By settingy
t
=
Q
y
t
~, that is, y
t
~
=
Q
-
1
y
t, multiplying P on the left to a, we obtain the equivalent transformation of a in the next form:b
1
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
D
t
0.5
0
y
1
t
~
D
t
0.5
0
y
2
t
~
D
t
0.5
0
y
3
t
~
D
t
0.5
0
y
4
t
~
=
1
2
0
0
0
1
0
0
0
0
1
0
0
0
0
1
y
1
t
~
y
2
t
~
y
3
t
~
y
4
t
~
+
t
0
0
t
,
y
~
0
=
0
.
It is easy to obtain the unique solution forb by Laplace transform [11]: (35)
y
1
t
~
=
t
3
/
2
E
1
/
2
,
5
/
2
t
=
∑
k
=
0
+
∞
t
k
+
3
/
2
Γ
k
+
5
/
2
,
y
2
t
~
=
0
,
y
3
t
~
=
-
2
π
t
,
y
4
t
~
=
-
t
.
Accordingly, the solution for LFDAEa is (36)
y
1
t
=
∑
k
=
0
+
∞
t
k
+
3
/
2
Γ
k
+
5
/
2
-
t
,
y
2
t
=
-
∑
k
=
0
+
∞
t
k
+
3
/
2
Γ
k
+
5
/
2
-
2
π
t
+
3
t
,
y
3
t
=
-
4
π
t
+
t
,
y
4
t
=
2
π
t
-
t
.And the solution satisfies y
0
=
0. The solution of Example 2 is shown in Figure 2.Figure 2
Solution of Example2.



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*Source: 1674725-2016-04-11.xml*