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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2011, Article ID 979705, 27 pages
doi:10.1155/2011/979705
Research Article
Lyapunov Stability of Quasilinear Implicit Dynamic
Equations on Time Scales
N. H. Du,
1
N. C. Liem,
1
C. J. Chyan,
2
and S. W. Lin
2
1
Department of Mathematics, Mechanics and Informatics, Vietnam National University, 334 Nguyen Trai,
Hanoi, Vietnam
2
Department of Mathematics, Ta mkang University, 151 Ying Chuang Road, Tamsui, Taipei County
25317, Taiwan
Correspondence should be addressed to N. H. Du,
Received 29 September 2010; Accepted 4 February 2011
Academic Editor: Stevo Stevic
Copyright q 2011 N. H. Du et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
This paper studies the stability of the solution x ≡ 0 for a class of quasilinear implicit dynamic
equationsontimescalesoftheformA
t
x
Δ
ft, x. We deal with an index concept to study the
solvability and use Lyapunov functions as a tool to approach the stability problem.
1. Introduction
The stability theory of quasilinear differential-algebraic equations DAEs for short
A
t
x
t
f
t, x
t
,x
t
,f
t, 0, 0
0 ∀t ∈
, 1.1
with A
.
being a given m × m-matrix function, has been an intensively discussed field in
both theory and practice. This problem can be seen in many real problems, such as in
electric circuits, chemical reactions, and vehicle systems. M
¨
arz in 1 has dealt with the
question whether the zero-solution of 1.1 is asymptotically stable in the Lyapunov sense
with ft, x
t,xt Bxtgt, x
t,xt,withA being constant a nd small perturbation
g.
Together with the theory of DAEs, there has been a great interest in singular difference
equation SDEalso referred to as descriptor systems, implicit difference equations
A
n
x
n 1
f
n, x
n 1
,x
n
,n∈ . 1.2
2 Journal of Inequalities and Applications
This model a ppears in many practical areas, such as the Leontiev dynamic model of
multisector economy, the Leslie population growth model, and singular discrete optimal
control problems. On the other hand, SDEs occur in a natural way of using discretization
techniques for solving DAEs and partial differential-algebraic equations, and so forth, which
have already attracted much attention from researchers cf. 2–4.Whenfn, xn1,xn
B
n
xngn, xn 1,xn,in5, the authors considered the solvability of Cauchy problem
for 1.2; the question of stability of the zero-solution of 1.2 has been considered in 6 where
the nonlinear perturbation gn, xn 1,xn is small a nd does not depend on xn 1.
Further, in recent years, to unify the presentation of continuous and discrete analysis,
a new theory was born and is more and more extensively concerned, that is, the theory of the
analysis on time scales. The most popular examples of time scales are
and .Using
“language” of time scales, we rewrite 1.1 and 1.2 under a unified form
A
t
x
Δ
t
f
t, x
Δ
t
,x
t
, 1.3
with t in time scale
and Δ being the derivative operator on .When , 1.3 is 1.1;if
, we have a similar equation to 1.2 if it is rewritten under the form A
n
xn1−xn
−A
n
xnfn, xn 1,xn; n ∈ .
The purpose of this paper is to answer the question whether results of stability for
1.1 and 1.2 can be extended and unified for the implicit dynamic equations of the form
1.3. The main tool to study the stability of this implicit dynamic equation is a generalized
direct Lyapunov method, and the results of this paper can be considered as a generalization
of 1.1 and 1.2.
The organization of this paper is as follows. In Section 2, we present shortly some
basic notions of the analysis on time scales and give the solvability of Cauchy problem for
quasilinear implicit dynamic equations
A
t
x
Δ
B
t
x f
t, x
,
1.4
with small perturbation ft, x and for quasilinear implicit dynamic equations of the style
A
t
x
Δ
f
t, x
,
1.5
with the assumption of differentiability for ft, x. The main results of this paper are
established in Section 3 where we deal with the stability of 1.5. The technique we use in
this section is somewhat similar to the one in 6–8. However, we need some improvements
because of the complicated structure of every time scale.
2. Nonlinear Implicit Dynamic Equations on Time Scales
2.1. Some Basic Notations of the Theory of the Analysis on Time Scales
A time scale is a nonempty closed subset of the real numbers , and we usually denote it
by the symbol
. We assume throughout that a time scale is endowed with the topology
inherited from the real numbe rs with the standard topology. We define the forward jump
operator and the backward jump operator σ, ρ :
→ by σtinf{s ∈ : s>t}
Journal of Inequalities and Applications 3
supplemented by inf ∅ sup
and ρtsup{s ∈ : s<t} supplemented by
sup ∅ inf
.Thegraininess μ : →
∪{0} is given by μtσt − t.Apointt ∈
is said to be right-dense if σtt, right-scattered if σt >t, left-dense if ρtt, left-scattered if
ρt <t,andisolated if t is right-scattered and left-scattered. For every a, b ∈
,bya, b,we
mean the set {t ∈
: a t b}.Theset
k
is defined to be if does not have a left-scattered
maximum; otherwise, it is
without this left-scattered maximum. Let f be a function defined
on
,valuedin
m
. We say that f is delta differentiable or simply: differentiable at t ∈
k
provided there exists a vector f
Δ
t ∈
m
, c alled the derivative of f,suchthatforall>0
there is a neighborhood V around t with fσt − fs − f
Δ
tσt − s |σt − s|
for all s ∈ V .Iff is differentiable for every t ∈
k
,thenf is said to be differentiable on
.If , then delta derivative is f
t from continuous calculus; if ,thedelta
derivative is the forward difference, Δf, from discrete calculus. A function f defined on
is rd-continuous if it is continuous at every right-dense point and if the left-sided limit exists
at every left-dense point. The set of all rd-continuous functions from
to a Banach space
X is denoted by C
rd
,X. A matrix function f from to
m×m
is said to be regressive if
detI μtft
/
0forallt ∈
k
, and denote R the set of regressive functions from to
m×m
.Moreover,denoteR
the set of positively regressive functions from to ,thatis,theset
{f :
→ :1 μtft > 0 ∀t ∈ }.
Theorem 2.1 see 9–11. Let t ∈
and let A
t
be a rd-continuous m × m-matrix function and q
t
rd-continuous function. Then, for any t
0
∈
k
, the initial value problem (IVP)
x
Δ
A
t
x q
t
,x
t
0
x
0
2.1
has a unique solution x· defined on t
t
0
. Further, if A
t
is regressive, this solution exists on t ∈ .
The solution of the corresponding matrix-valued IVP X
Δ
A
t
X, XsI always
exists for t
s,evenA
t
is not regressive. In this case, Φ
A
t, s is defined only with t s
see 12, 13 and is called the Cauchy operator of the dynamic equation 2.1. If we suppose
further that A
t
is regressive, the Cauchy operator Φ
A
t, s is defined for all s, t ∈ .
We now recall the chain rule for multivariable functions on time scales, this result has
been proved in 14.LetV :
×
m
→ and g : →
m
be continuously differentiable.
Then V ·,g· :
→ is delta differentiable and there holds
V
Δ
t, g
t
V
Δ
t
t, g
t
1
0
V
x
σ
t
,g
t
hμ
t
g
Δ
t
,g
Δ
t
dh
V
Δ
t
t, g
σ
t
1
0
V
x
t, g
t
hμ
t
g
Δ
t
,g
Δ
t
dh,
2.2
where V
x
is the deriva tive in the second variable of the function V V t, x in normal
meaning and ·, · is the scalar product.
We refer to 12, 15 for more information on the analysis on time scales.
4 Journal of Inequalities and Applications
2.2. Linear Equations with Small Nonlinear Perturbation
Let be a time scale. We consider a class of nonlinear equations of the form
A
t
x
Δ
B
t
x f
t, x
.
2.3
The homogeneous linear implicit dynamic equations LIDEs associated to 2.3 are
A
t
x
Δ
B
t
x,
2.4
where A
.
,B
.
∈ C
rd
k
,
m×m
and ft, x is rd-continuous in t, x ∈ ×
m
.Inthecasewhere
the matrices A
t
are invertible for every t ∈ , we can multiply both sides of 2.3 by A
−1
t
to
obtain an ordinary dynamic equation
x
Δ
A
−1
t
B
t
x A
−1
t
f
t, x
,t∈ ,
2.5
which has been well studied. If there is at least a t such that A
t
is singular, we cannot solve
explicitly the leading term x
Δ
. In fact, we are concerned with a so-called ill-posed problem
where the solutions of Cauchy problem may e xist only on a submanifold or even they do not
exist. One of the ways to solve this equation is to impose some further assumptions stated
under the form of indices of the equation.
We introduce the so-called index-1 of 2.4. Suppose that rank A
t
r for all t ∈
and let T
t
∈ GL
m
such that T
t
|
ker A
t
is an isomorphism between ker A
t
and ker A
ρt
;
T
.
∈ C
rd
k
,
m×m
.LetQ
t
be a projector onto ker A
t
satisfying Q
.
∈ C
rd
k
,
m×m
.Wecan
find such operators T
t
and Q
t
by the following way: let matrix A
t
possess a singular value
decomposition
A
t
U
t
Σ
t
V
t
, 2.6
where U
t
, V
t
are orthogonal matrices and Σ
t
is a diagonal matrix with singular values σ
1
t
σ
2
t
··· σ
r
t
> 0 on its main diagonal. Since A
.
∈ C
rd
k
,
m×m
, on the above decomposition
of A
t
, we can choose the matrix V
t
to be in C
rd
k
,
m×m
see 16. Hence, by putting Q
t
V
t
diagO, I
m−r
V
t
and T
t
V
ρt
V
−1
t
,weobtainQ
t
and V
t
as the requirement.
Let
S
t
{
x ∈
m
,B
t
x ∈ imA
t
}
, 2.7
and P
t
: I − Q
t
.
Under these notations, we have the following Lemma.
Lemma 2.2. The following assertions are equivalent
i kerA
ρt
∩ S
t
{0};
ii the matrix G
t
A
t
− B
t
T
t
Q
t
is nonsingular;
iii
m
kerA
ρt
⊕ S
t
, for all t ∈ .
Journal of Inequalities and Applications 5
Proof. i⇒ii Let t ∈
and x ∈
m
such that A
t
− B
t
T
t
Q
t
x 0 ⇔ B
t
T
t
Q
t
xAx.This
equation implies T
t
Q
t
x ∈ S
t
. Since ker A
ρt
∩ S
t
{0} and T
t
Q
t
x ∈ ker A
ρt
, it follows that
T
t
Q
t
x 0. Hence, Q
t
x 0 which implies A
t
x 0. This means that x ∈ ker A
t
.Thus,x Q
t
x
0, that is, the matrix G
t
A
t
− B
t
T
t
Q
t
is nonsingular.
ii⇒iii It is obvious that x I T
t
Q
t
G
−1
t
B
t
x−T
t
Q
t
G
−1
t
B
t
x.WeseethatT
t
Q
t
G
−1
t
B
t
x ∈
ker A
ρt
and B
t
I T
t
Q
t
G
−1
t
B
t
x B
t
x − A
t
− B
t
T
t
Q
t
G
−1
t
B
t
x A
t
G
−1
t
B
t
x A
t
G
−1
t
B
t
x ∈ imA
t
.
Thus, I T
t
Q
t
G
−1
t
B
t
x ∈ S
t
and we have
m
S
t
ker A
ρt
.
Let x ∈ ker A
ρt
∩ S
t
,thatis,x ∈ S
t
and x ∈ ker A
ρt
.Sincex ∈ S
t
,thereisaz ∈
m
such that B
t
x A
t
z A
t
P
t
z and since x ∈ ker A
ρt
, T
−1
t
x ∈ ker A
t
. Therefore, T
−1
t
x Q
t
T
−1
t
x.
Hence, A
t
− B
t
T
t
Q
t
T
−1
t
x −A
t
− B
t
T
t
Q
t
P
t
z which follows that T
−1
t
x −P
t
z.Thus,T
−1
t
x 0
and then x 0. So, we have that iii. iii⇒i is obvious.
Lemma 2.2 is proved.
Lemma 2.3. Suppose that the matrix G
t
is nonsingular. Then, there hold the following assertions:
1
P
t
G
−1
t
A
t
, 2.8
2
Q
t
−G
−1
t
B
t
T
t
Q
t
, 2.9
3
Q
t
: −T
t
Q
t
G
−1
t
B
t
is the projector onto ker A
ρt
along S
t
, 2.10
4
a
P
t
G
t
−1
B
t
P
t
G
−1
t
B
t
P
ρt
, 2.11
b
Q
t
G
t
−1
B
t
Q
t
G
−1
t
B
t
P
ρt
− T
t
−1
Q
ρt
, 2.12
5
T
t
Q
t
G
−1
t
does not depend on the choice of T
t
and Q
t
. 2.13
Proof. 1 Noting that G
t
P
t
A
t
− B
t
T
t
Q
t
P
t
A
t
P
t
A
t
,weget2.8.
2 From B
t
T
t
Q
t
A
t
− G
t
, it follows G
−1
t
B
t
T
t
Q
t
P
t
− I −Q
t
. Thus, we have 2.9.
3
Q
2
t
T
t
Q
t
G
−1
t
B
t
T
t
Q
t
G
−1
t
B
t
2.9
−T
t
Q
t
Q
t
G
−1
t
B
t
−T
t
Q
t
G
−1
t
B
t
Q
t
and A
ρt
Q
t
−A
ρt
T
t
Q
t
G
−1
t
B
t
0. This means that
Q
t
is a projector onto ker A
ρt
. From the proof of iii,
Lemma 2.2,weseethat
Q
t
is the projector onto ker A
ρt
along S
t
.
4 Since T
−1
t
Q
ρt
x ∈ ker A
t
for any x,
P
t
G
−1
t
B
t
Q
ρt
P
t
G
−1
t
B
t
T
t
T
−1
t
Q
ρt
−P
t
G
−1
t
A
t
− B
t
T
t
Q
t
Q
t
T
−1
t
Q
ρt
0. 2.14
Therefore, P
t
G
−1
t
B
t
P
t
G
−1
t
B
t
P
ρt
so we have 2.11. Finally,
Q
t
G
−1
t
B
t
Q
t
G
−1
t
B
t
P
ρt
Q
t
G
−1
t
B
t
T
t
Q
t
T
−1
t
Q
ρt
Q
t
G
−1
t
B
t
P
ρt
− Q
t
G
−1
t
A
t
− B
t
T
t
Q
t
Q
t
T
−1
t
Q
ρt
Q
t
G
−1
t
B
t
P
ρt
− Q
t
T
−1
t
Q
ρt
Q
t
G
−1
t
B
t
P
ρt
− T
−1
t
Q
ρt
.
2.15
Thus, we get 2.12.
6 Journal of Inequalities and Applications
5 Let T
t
be another linear transformation from
m
onto
m
satisfying T
t
|
ker A
t
to be
an isomorphism from ker A
t
onto ker A
ρt
and Q
t
a projector onto ker A
t
.DenoteG
t
A
t
−
B
t
T
t
Q
t
. It is easy to see that
T
t
Q
t
G
−1
t
G
t
T
t
Q
t
G
−1
t
A
t
− B
t
T
t
Q
t
T
t
Q
t
P
t
− T
t
Q
t
G
−1
t
B
t
T
t
Q
t
T
t
Q
t
T
t
Q
t
T
t
Q
t
. 2.16
Therefore, T
t
Q
t
G
−1
t
T
t
Q
t
G
−1
t
. The proof of Lemma 2.3 is complete.
Definition 2.4. The LIDE 2.4 is said to be index-1 if for all t ∈ , the following conditions
hold:
i rank A
t
r constant 1 r m − 1,
ii kerA
ρt
∩ S
t
{0}.
Now, we add the following assumptions.
Hypothesis 2.5. 1 The homogeneous LIDE 2.4 is of index-1.
2 ft, x is rd-continuous and satisfies the Lipschitz condition,
f
t, w
− f
t, w
L
t
w − w
, ∀w, w
∈
m
, 2.17
where
γ
t
: L
t
T
t
Q
t
G
−1
t
< 1 ∀t ∈
k
. 2.18
Remark 2.6. By the item 2.13 of Lemma 2.3, the condition 2.18 is independent from the
choice of T
t
and Q
t
.
We assume further that we can choose the projector function Q
t
onto ker A
t
such that
Q
ρt
Q
t
for all right-dense and left-scattered t; Q
ρt
is differentiable at every t ∈
k
and
Q
ρt
Δ
is rd-continuous. For each t ∈
k
,wehaveP
ρt
xt
Δ
P
ρσt
x
Δ
tP
ρt
Δ
xt.
Therefore,
A
t
x
Δ
t
A
t
P
t
x
Δ
t
A
t
P
ρt
x
t
Δ
−
P
ρt
Δ
x
t
, 2.19
and the implicit equation 2.3 canberewrittenas
A
t
P
ρt
x
Δ
A
t
P
ρt
Δ
B
t
x f
t, x
,t∈
k
. 2.20
Thus, we should look for solutions of 2.3 from the space C
1
N
:
C
1
N
k
,
m
x
·
∈ C
rd
k
,
m
: P
ρt
x
t
is differentiable at every t ∈
k
. 2.21
Note that C
1
N
does not depend on the choice of the projector function sinc e the relations
P
t
P
t
P
t
and P
t
P
t
P
t
are true for each two projectors P
t
and P
t
along the space ker A
t
.
Journal of Inequalities and Applications 7
We now describe shortly the decomposition technique for 2.3 as follows.
Since 2.3 has index-1 and by virtue of Lemma 2.2, we see that the matrices G
t
are
nonsingular for all t ∈
k
. Multiplying 2.3 by P
t
G
−1
t
and Q
t
G
−1
t
, respectively, it yield s
P
t
x
Δ
P
t
G
−1
t
B
t
x P
t
G
−1
t
f
t, x
,
0 Q
t
G
−1
t
B
t
x Q
t
G
−1
t
f
t, x
.
2.22
Therefore, by using the results of Lemma 2.3,weget
P
ρt
x
Δ
P
ρt
Δ
I T
t
Q
t
G
−1
t
B
t
P
ρt
x P
t
G
−1
t
B
t
P
ρt
x
P
ρt
Δ
T
t
Q
t
G
−1
t
P
t
G
−1
t
f
t, x
,
Q
ρt
x T
t
Q
t
G
−1
t
B
t
P
ρt
x T
t
Q
t
G
−1
t
f
t, x
.
2.23
By denoting u P
ρt
x, v Q
ρt
x, 2.23 becomes a dynamic equation on time scale
u
Δ
P
ρt
Δ
I T
t
Q
t
G
−1
t
B
t
u P
t
G
−1
t
B
t
u
P
ρt
Δ
T
t
Q
t
G
−1
t
P
t
G
−1
t
f
t, u v
, 2.24
and an algebraic relation
v T
t
Q
t
G
−1
t
B
t
u T
t
Q
t
G
−1
t
f
t, u v
.
2.25
For fixed u ∈
m
and t ∈
k
, we consider a mapping C
t
:imQ
ρt
→ im Q
ρt
given by
C
t
v
: T
t
Q
t
G
−1
t
B
t
u T
t
Q
t
G
−1
t
f
t, u v
.
2.26
We see that
C
t
v
− C
t
v
T
t
Q
t
G
−1
t
f
t, u v
− f
t, u v
γ
t
v − v
, 2.27
for any v, v
∈ im Q
ρt
.Sinceγ
t
< 1, C
t
is a contractive mapping. Hence, by the fixed point
theorem, there exists a mapping g
t
:imP
ρt
→ im Q
ρt
satisfying
g
t
u
T
t
Q
t
G
−1
t
B
t
u T
t
Q
t
G
−1
t
f
t, u g
t
u
,
2.28
and it is easy to see that g
t
u is rd-continuous in t.
Moreover,
g
t
u
− g
t
u
T
t
Q
t
G
−1
t
B
t
u − u
T
t
Q
t
G
−1
t
f
t, u g
t
u
− f
t, u
g
t
u
T
t
Q
t
G
−1
t
B
t
u − u
L
t
T
t
Q
t
G
−1
t
u − u
g
t
u
− g
t
u
.
2.29
8 Journal of Inequalities and Applications
This deduces
g
t
u
− g
t
u
γ
t
1 − γ
t
−1
L
−1
t
L
t
B
t
u − u
.
2.30
Thus, g
t
is Lipschitz continuous with the Lipschitz constant δ
t
: γ
t
1 − γ
t
−1
L
−1
t
L
t
B
t
.
Substituting g
t
into 2.24,weobtain
u
Δ
P
ρt
Δ
I T
t
Q
t
G
−1
t
B
t
u P
t
G
−1
t
B
t
u
P
ρt
Δ
T
t
Q
t
G
−1
t
P
t
G
−1
t
f
t, u g
t
u
.
2.31
It is easy to see that the right-hand side of 2.31 satisfies the Lipschitz condition with the
Lipschitz constant
ω
t
P
ρt
Δ
I T
t
Q
t
G
−1
t
B
t
P
t
G
−1
t
B
t
L
t
1 δ
t
P
ρt
Δ
T
t
Q
t
G
−1
t
P
t
G
−1
t
. 2.32
Applying the global existence theorem see 12,weseethat2.31, with the initial
condition ut
0
P
ρt
0
x
0
has a unique solution utut; t
0
,x
0
, t t
0
.
Thus, we get the following theorem.
Theorem 2.7. Let Hypothesis 2.5 and the assumptions on the projector Q
t
be satisfied. Then, 2.3
with the initial condition
P
ρt
0
x
t
0
− x
0
0 2.33
has a unique solution. This solution is expressed by
x
t
x
t; t
0
,x
0
u
t; t
0
,x
0
g
t
u
t; t
0
,x
0
,t
t
0
,t∈
k
,
2.34
where utut; t
0
,x
0
is the solution of 2.31 with ut
0
P
ρt
0
x
0
.
We now describe the solution space of the implicit dynamic equation 2.3.Denote
Ł
t
x ∈
m
: Q
ρt
x T
t
Q
t
G
−1
t
B
t
P
ρt
x T
t
Q
t
G
−1
t
f
t, x
,
Ω
t
x ∈
m
: B
t
x f
t, x
∈ im A
t
.
2.35
Lemma 2.8. There hold the following statements:
iŁ
t
Ω
t
,
ii If ft, 00 for all t ∈
then Ω
t
∩ ker A
ρt
{0}.
Proof. i Let y ∈ Ł
t
,thatis,Q
ρt
y T
t
Q
t
G
−1
t
B
t
P
ρt
y T
t
Q
t
G
−1
t
ft, y.Wehave
y P
ρt
y Q
ρt
y
I T
t
Q
t
G
−1
t
B
t
P
ρt
y T
t
Q
t
G
−1
t
f
t, y
. 2.36
Journal of Inequalities and Applications 9
Hence,
B
t
y f
t, y
B
t
I T
t
Q
t
G
−1
t
B
t
P
ρt
y
I B
t
T
t
Q
t
G
−1
t
f
t, y
I B
t
T
t
Q
t
G
−1
t
B
t
P
ρt
y
I B
t
T
t
Q
t
G
−1
t
f
t, y
I B
t
T
t
Q
t
G
−1
t
B
t
P
ρt
y f
t, y
.
2.37
From
I B
t
T
t
Q
t
G
−1
t
I
A
t
− G
t
G
−1
t
A
t
G
−1
t
,
2.38
it yields
B
t
y f
t, y
A
t
G
−1
t
B
t
P
ρt
y f
t, y
∈ im A
t
⇒ y ∈ Ω
t
.
2.39
Conversely, suppose that y ∈ Ω
t
, that is, there exists z ∈
m
such that B
t
y ft, yA
t
z.We
have to prove
Q
ρt
y T
t
Q
t
G
−1
t
B
t
P
ρt
y T
t
Q
t
G
−1
t
f
t, y
,
2.40
or equivalently,
y T
t
Q
t
G
−1
t
f
t, y
T
t
Q
t
G
−1
t
B
t
P
ρt
y P
ρt
y.
2.41
Indeed,
T
t
Q
t
G
−1
t
f
t, y
T
t
Q
t
G
−1
t
B
t
P
ρt
y P
ρt
y
T
t
Q
t
G
−1
t
f
t, y
T
t
Q
t
G
−1
t
B
t
y − T
t
Q
t
G
−1
t
B
t
Q
ρt
y P
ρt
y
T
t
Q
t
G
−1
t
f
t, y
B
t
y
− T
t
Q
t
G
−1
t
B
t
Q
ρt
y P
ρt
y
T
t
Q
t
G
−1
t
A
t
z − T
t
Q
t
G
−1
t
B
t
Q
ρt
y P
ρt
y
T
t
Q
t
P
t
z − T
t
Q
t
G
−1
t
B
t
Q
ρt
y P
ρt
y
−T
t
Q
t
G
−1
t
B
t
Q
ρt
y P
ρt
y Q
ρt
y P
ρt
y y,
2.42
where we have already used a result of Lemma 2.3 that
Q −T
t
Q
t
G
−1
t
B
t
is a projector onto
ker A
ρt
.SoŁ
t
Ω
t
.
ii Let y ∈ Ω
t
∩ ker A
ρt
.Theny ∈ Ω
t
and P
ρt
y 0. Since Ω
t
Ł
t
,wehavey ∈ Ł
t
.
This means that Q
ρt
y T
t
Q
t
G
−1
t
B
t
P
ρt
y T
t
Q
t
G
−1
t
ft, yT
t
Q
t
G
−1
t
ft, Q
ρt
y.Fromthe
assumption ft, 00, it follows that Q
ρt
y L
t
T
t
Q
t
G
−1
t
Q
ρt
y γ
t
Q
ρt
y.Thefact
γ
t
< 1 implies that Q
ρt
y 0. Thus y P
ρt
y Q
ρt
y 0. The lemma is proved.
10 Journal of Inequalities and Applications
Remark 2.9. 1 By virtue of Lemma 2.8, we find out that the solution space Ł
t
is independent
fromthechoiceofprojectorQ
t
and operator T
t
.
2 Since G
−1
ρt
0
A
ρt
0
P
ρt
0
and A
ρt
0
P
ρt
0
A
ρt
0
, the initial condition 2.33 is
equivalent to the condition A
ρt
0
xt
0
A
ρt
0
x
0
. This implies that the initial condition is
not also dependent on choice of projectors.
3 Noting that if xt is a solution of 2.3 with the initial condition 2.33,thenxt ∈
Ł
t
for all t t
0
.Conversely,letx
0
∈ Ł
t
Ω
t
and let xs; t, x
0
, s t, be the solution of
2.3 satisfying the initial condition P
ρt
xt; t, x
0
− x
0
0. We see that xt; t, x
0
P
ρt
x
g
t
P
ρt
xP
ρt
x
0
g
t
P
ρt
x
0
x
0
. This means that there exists a solution of 2.3 passing
x
0
∈ Ł
t
.
2.3. Quasilinear Implicit Dynamic Equations
Now we consider a quasilinear implicit dynamic equation of the form
A
t
x
Δ
f
t, x
,
2.43
with A
.
∈ C
rd
k
,
m×m
and f : ×
m
→
m
assumed to be continuously differentiable in
the variable x and continuous in t, x.
Suppose that rank A
t
r for all t ∈ . We keep all assumptions on the projector Q
t
and
operator T
t
stated in Section 2.2.
Equation 2.43 is said to be of index-1 if the matrix
G
t
: A
t
− f
x
t, x
T
t
Q
t
2.44
is invertible for every t ∈
and x ∈
m
.
Denote
S
t, x
z ∈
m
,f
x
t, x
z ∈ imA
t
;kerA
t
N
t
. 2.45
Further introduce the set
Ω
t
x ∈
m
,f
t, x
∈ imA
t
, 2.46
containing all solutions of 2.43.ThesubspaceSt, x manifests its geometrical meaning
S
t, x
T
x
Ω
t
for x ∈ Ω
t
, 2.47
where T
x
is the tangent space of Ω
t
at the point x.
Suppose that 2.43 is of index-1. Then, by Lemma 2.2, this condition is equivalent to
one of the following conditions:
Journal of Inequalities and Applications 11
1 St, x ⊕ N
ρt
m
,
2 St, x ∩ N
ρt
{0}.
3 Let B
t
∈
m×m
be a matrix such that the matrix G
t
A
t
− B
t
T
t
Q
t
is invertible we
can choose B
t
f
x
t, 0,e.g.. From the relation
G
t
A
t
− B
t
T
t
Q
t
B
t
T
t
Q
t
− f
x
t, x
T
t
Q
t
G
t
B
t
− f
x
t, x
T
t
Q
t
I
B
t
− f
x
t, x
T
t
Q
t
G
−1
t
G
t
,
2.48
it follows that
I
B
t
− f
x
t, x
T
t
Q
t
G
−1
t
2.49
is invertible.
Lemma 2.10. Suppose that the bounded linear operator triplet:
: X → Y, : Y → Z, : Z →
X is given, where X, Y, Z are Banach spaces. Then the operator I −
is invertible if and only if
I −
is invertible.
Proof . See 17, Lemma 1.
By virtue of 2.49 and Lemma 2.10,wegetthat
I T
t
Q
t
G
−1
t
B
t
− f
x
t, x
is invertible.
2.50
Now we come to split 2.43. Multiplying both sides of 2.43 by P
t
G
−1
t
and Q
t
G
−1
t
,
respectively, and putting u P
ρt
x, v Q
ρt
x,weobtain
u
Δ
P
ρt
Δ
u v
P
t
G
−1
t
f
t, u v
,
0 T
t
Q
t
G
−1
t
f
t, u v
.
2.51
Consider the function
k
t, u, v
: T
t
Q
t
G
−1
t
f
t, u v
.
2.52
We see that
∂k
∂v
t, u, v
h T
t
Q
t
G
−1
t
f
x
t, u v
h,
2.53
where h ∈ Q
ρt
m
.
12 Journal of Inequalities and Applications
Let h ∈ Q
ρt
m
be a vector satisfying T
t
Q
t
G
−1
t
f
x
t, u vh 0. Paying attention to
T
t
Q
t
G
−1
t
B
t
h −h,wehave
−T
t
Q
t
G
−1
t
f
x
t, u v
h
I T
t
Q
t
G
−1
t
B
t
− f
x
t, u v
h. 2.54
Therefore, by 2.50 we get h 0. This means that ∂k/∂vt, u, v|
Q
ρt
m
is an isomorphism
of Q
ρt
m
. By the implicit function theorem, equation kt, u, v0hasauniquesolutionv
g
t
u.Moreover,thefunctionv g
t
u is continuous in t, u and continuously differentiable
in u. Its derivative is
∂g
t
u
∂u
−T
t
Q
t
G
−1
t
f
x
t, u g
t
u
|
Q
ρt
m
−1
T
t
Q
t
G
−1
t
f
x
t, u g
t
u
|
P
ρt
m
.
2.55
Then, by substituting v g
t
u into the first equation of 2.51 we come to
u
Δ
P
ρt
Δ
u g
t
u
P
t
G
−1
t
f
t, u g
t
u
.
2.56
It is obvious that the ordinary dynamic equation 2.56 with the initial condition
u
t
0
P
ρt
0
x
0
2.57
is locally uniquely solvable and the solution xt; t
0
,x
0
of 2.43 with the initial condition
2.33 can be expressed by xt; t
0
,x
0
ut; t
0
,x
0
g
t
ut; t
0
,x
0
.
Now suppose further that ft, x satisfies the Lipschitz condition in x and we can find
amatrixB
t
such that
T
t
Q
t
G
−1
t
f
x
t, x
|
Q
ρt
m
−1
T
t
Q
t
G
−1
t
f
x
t, x
|
P
ρt
m
2.58
is bounded for all t ∈
and x ∈
m
. Then, the right-hand side of 2.56 also satisfies the
Lipschitz condition. Thus, from the global existence theorem see 12, 2.56 with the initial
condition 2.57 has a unique solution defined on t
0
, sup .
Therefore, we have the following theorem.
Theorem 2.11. Given an index-1 quasilinear implicit dynamic equation 2.43, then there holds the
following.
(1) Equation 2.43 is locally solvable, that is, for any t
0
∈
k
, x
0
∈
m
, there exists a unique
solution xt; t
0
,x
0
of 2.43,definedont
0
,b with some b ∈ ,b>t
0
, satisfying the initial
condition 2.33.
(2) Moreover, if ft, x satisfies the Lipschitz condition in x and we can find a matrix B
t
such
that
T
t
Q
t
G
−1
t
f
x
t, x
|
Q
ρt
m
−1
T
t
Q
t
G
−1
t
f
x
t, x
|
P
ρt
m
2.59
Journal of Inequalities and Applications 13
is bounded, then this solution is defined on t
0
, sup and we have the expression
x
t; t
0
,x
0
u
t; t
0
,x
0
g
t
u
t; t
0
,x
0
,t
t
0
, 2.60
where ut; t
0
,x
0
is the solution of 2.56 with ut
0
P
ρt
0
x
0
.
Remark 2.12. 1 We note that the expression T
t
Q
t
G
−1
t
B
t
depends only on choosing the
matrix B
t
.
2 The assumption that T
t
Q
t
G
−1
t
f
x
t, x|
Q
ρt
m
−1
T
t
Q
t
G
−1
t
f
x
t, x|
P
ρt
m
is bounded for
amatrixfunctionB
t
seems to be too strong. In Section 3, we show a condition for the global
solvability via Lyapunov functions.
3 If x
0
∈ Ω
t
,thereexistsz ∈
m
satisfying A
t
z ft, x
0
.Hence,T
t
Q
t
G
−1
t
ft, x
0
0.
Therefore, by the same argument as in Section 2 .2, we can prove that for every x
0
∈ Ω
t
,there
is a unique solution passing through x
0
.
3. Stability Theorems of Implicit Dynamic Equations
For the reason of our purpose, in this section we suppose that is an upper unbounded time
scale, that is, sup
∞.Forafixedτ ∈ ,denote
τ
{t ∈ ,t τ}.
Consider an implicit dynamic equation of the form
A
t
x
Δ
f
t, x
,t∈
τ
,
3.1
where A
.
∈ C
rd
k
τ
,
m×m
and f·, · ∈ C
rd
τ
×
m
,
m
.
First, we suppose that for each t
0
∈
k
τ
, 3.1 with the initial condition
A
ρt
0
x
t
0
− x
0
0 3.2
has a unique solution defined on
t
0
. The condition ensuring the existence of a unique
solution can be refered to Section 2. We denote the solution with the initial condition 3.2
by xtxt; t
0
,x
0
. Remember that we look for the solution of 3.1 in the space C
1
N
k
τ
,
m
.
Let ft, 00forallt ∈
τ
, which follows that 3.1 has the trivial solution x ≡ 0.
We mention again that Ω
t
{x ∈
m
,ft, x ∈ im A
t
}.Notingthatifxtxt; t
0
,x
0
is the solution of 3.1 and 3.2 then xt ∈ Ω
t
for all t ∈
t
0
.
Definition 3.1. The trivial solution x ≡ 0of3.1 is said to be
1 A-stable resp., P -stable if, for each >0andt
0
∈
k
τ
, there exists a positive δ
δt
0
, such that A
ρt
0
x
0
<δresp., P
ρt
0
x
0
<δ implies xt; t
0
,x
0
<for all
t
t
0
,
2 A-uniformly resp., P-uniformly stable if it is A-stable resp., P -stable and the
number δ mentioned in the part 1. of this definition is independent of t
0
,
3 A-asymptotically resp., P-asymptotically stable if it is stable and for each t
0
∈
k
τ
, there exist positive δ δt
0
such that the inequality A
ρt
0
x
0
<δ
resp., P
ρt
0
x
0
<δ implies lim
t →∞
xt; t
0
,x
0
0. If δ is independent of
t
0
, then the corresponding stability is A-uniformly asymptotically P-uniformly
asymptotically stable,
14 Journal of Inequalities and Applications
4 A-uniformly globally asymptotically resp., P-uniformly globally asymptotically
stable if for any δ
0
> 0 there exist functions δ·, T· such that A
ρt
0
x
0
<δ
resp., P
ρt
0
x
0
<δ implies xt; t
0
,x
0
<for a ll t t
0
and if A
ρt
0
x
0
<δ
0
resp., P
ρt
0
x
0
<δ
0
then xt; t
0
,x
0
<for all t t
0
T,
5 P-exponentially stable if there is positive constant α with −α ∈R
such that for
every t
0
∈
k
τ
there exists an N Nt
0
1, the solution of 3.1 with the initial
condition P
ρt
0
xt
0
− x
0
0satisfiesxt; t
0
,x
0
NP
ρt
0
x
0
e
−α
t, t
0
,t
t
0
,t∈
τ
. If the constant N can be chosen independent of t
0
, then this solution is
called P-uniformly exponentially stable.
Remark 3.2. From G
−1
t
A
t
P
t
and A
t
A
t
P
t
, the notions of A-stable and P-stable as well as A-
asymptotically stable and P-asymptotically stable are equivalent. Therefore, in the following
theorems we will omit the prefixes A and P when talking about stability and asymptotical
stability. However, the concept of A-uniform stability implies P-uniform stability if the
matrices A
t
are uniformly bounded and P-uniform stability implies A-uniform stability if
the matrices G
t
are uniformly bounded.
Denote
:
φ ∈ C
0,a
,
,φ
0
0,φis strictly increasing; a>0
, 3.3
and
φ is the domain of definition of φ.
Proposition 3.3. The trivial solution x ≡ 0 of 3.1 is A-uniformly (resp., P -uniformly) stable if and
only if there exists a function ϕ ∈
such that for each t
0
∈
k
τ
and any solution xt; t
0
,x
0
of 3.1
the inequality
x
t; t
0
,x
0
ϕ
A
ρt
0
x
0
,
resp.,
x
t; t
0
,x
0
ϕ
P
ρt
0
x
0
∀t
t
0
, 3.4
holds, provided A
ρt
0
x
0
∈ ϕ (resp., P
ρt
0
x
0
∈ ϕ).
Proof. We only need to prove the proposition for the A-uniformly stable case.
Sufficiency. Suppose there exists a function ϕ ∈
satisfying 3.4 for each >0; we
take δ δ > 0suchthatϕδ <,thatis,ϕ
−1
>δ.Ifxt; t
0
,x
0
is an arbitrary solution of
3.1 and A
ρt
0
x
0
<δ,thenxt; t
0
,x
0
ϕA
ρt
0
x
0
<ϕδ <,for all t t
0
.
Necessity. Suppose that the trivial solution x ≡ 0of3.1 is A-uniformly stable, that is,
for each >0thereexistsδ δ > 0suchthatforeacht
0
∈
k
τ
the inequality A
ρt
0
x
0
<δ
implies xt; t
0
,x
0
<,forallt t
0
. For the sake of simplicity in computation, we choose
δ <.Denote
γ
sup
δ
: δ
has such a property
. 3.5
It is clear that γ is an increasing positive function in .Further,γ
and by definition,
there holds
A
ρt
0
x
0
<γ
then
x
t; t
0
,x
0
< ∀t
t
0
. 3.6
Journal of Inequalities and Applications 15
By putting
β
:
1
0
γ
t
dt, 3.7
it is seen that
β ∈
, 0 <β
<γ
. 3.8
Let ϕ : 0, sup β →
be the inverse function of β. It is clear that ϕ also belongs to .
For t
t
0
,wedenote
t
xt; t
0
,x
0
.If
t
0, then xt; t
0
,x
0
t
0
ϕA
ρt
0
x
0
∀t t
0
by ϕ ∈ remember that xt; t
0
,x
0
0 does not imply that x·; t
0
,x
0
≡
0. Consider the case where
t
> 0. If A
ρt
0
x
0
<β
t
, then by the relations 3.6 and 3.8
we have xs; t
0
,x
0
<
t
, ∀s t
0
.Inparticular,xt; t
0
,x
0
<
t
which is a contradiction.
Thus A
ρt
0
x
0
β
t
, this implies xt; t
0
,x
0
t
ϕA
ρt
0
x
0
, ∀t t
0
,provided
sup β>A
ρt
0
x
0
.
The proposition is proved.
Similarly, we have the following proposition.
Proposition 3.4. The trivial solution x ≡ 0 of 3.1 is A-stable (resp., P -stable) if and only if for each
t
0
∈
k
τ
and any solution xt; t
0
,x
0
of 3.1 there exists a function ϕ
t
0
∈ such that there holds the
following:
x
t; t
0
,x
0
ϕ
t
0
A
ρt
0
x
0
resp.,
x
t; t
0
,x
0
ϕ
t
0
P
ρt
0
x
0
∀t
t
0
, 3.9
provided A
ρt
0
x
0
∈ ϕ
t
0
(resp., P
ρt
0
x
0
∈ ϕ
t
0
).
In order to use the Lyapunov function technique related to 3.1, we suppose that
A
ρt
∈ C
1
rd
k
τ
,
m×m
.Byusing2.3, we can define the derivative of the function V :
τ
×
m
→
along every solution curve as follows:
V
Δ
3.10
t, A
ρt
x
V
Δ
t
t, A
ρt
x
1
0
V
x
σ
t
,A
ρt
x hμ
t
A
ρt
x
Δ
,
A
ρt
x
Δ
dh.
3.10
Remark 3.5. Note that when the function V is independent of t and even if the vector field
associated with the implicit dynamic equation 3.1 is autonomous, the derivative V
Δ
3.10
may
depend on t.
Theorem 3.6. Assume that there exist a constant c>0, −c ∈R
and a function V :
τ
×
m
→
being rd-continuous and a function ψ ∈ , ψ defined on 0, ∞ satisfying
1 ψx
V t, A
ρt
x for all x ∈ Ω
t
and t ∈
τ
,
2 V
Δ
3.10
t, A
ρt
x c/1 − cμtV t, A
ρt
x, for any x ∈ Ω
t
and t ∈
k
τ
.
16 Journal of Inequalities and Applications
Assume further that 3.1 is locally solvable. Then, 3.1 is globally solvable, that is, every solution
with the initial condition 3.2 is defined on
t
0
.
Proof. Denote
W
t, x
V
t, x
e
−c
t, t
0
. 3.11
By the condition 2,wehave
W
Δ
3.10
t, A
ρt
x
V
Δ
3.10
t, A
ρt
x
e
−c
σ
t
,t
0
− cV
t, A
ρt
x
e
−c
t, t
0
c
1 − cμ
t
V
t, A
ρt
x
1 − cμ
t
e
−c
t, t
0
− cV
t, A
ρt
x
e
−c
t, t
0
0.
3.12
Therefore, for all t
t
0
W
t, A
ρt
x
t
− W
t
0
,A
ρt
0
x
t
0
t
t
0
W
Δ
3.10
τ, A
ρτ
x
τ
Δτ
0.
3.13
From the condition 1, it follows that
e
−c
t, t
0
ψ
x
t
W
t, A
ρt
x
t
W
t
0
,A
ρt
0
x
t
0
V
t
0
,A
ρt
0
x
t
0
3.14
or
x
t
ψ
−1
V
t
0
,A
ρt
0
x
t
0
e
−c
t, t
0
ψ
−1
V
t
0
,A
ρt
0
x
t
0
e
c/1−cμt
t, t
0
.
3.15
The last inequality says that the solution xt can be lengthened on
t
0
,thatis,3.1 is globally
solvable.
Theorem 3.7. Assume that there exist a function V :
τ
×
m
→
being rd-continuous and
a function ψ ∈
, ψ defined on 0, ∞ satisfying the conditions
1 V t, 0 ≡ 0 for all t ∈
τ
,
2 ψx
V t, A
ρt
x for all x ∈ Ω
t
and t ∈
τ
,
3 V
Δ
3.10
t, A
ρt
x 0 for any x ∈ Ω
t
and t ∈
k
τ
.
Assume further that 3.1 is locally solvable. Then the trivial solution of 3.1 is stable.
Proof. By virtue of Theorem 3.6 and the conditions 2 and 3, it follows that 3.1 is globally
solvable. Suppose on the contrary that the trivial solution x ≡ 0of3.1 is not stable. Then,
there exists an
0
> 0suchthatforallδ>0thereexistsasolutionxt of 3.1 satisfying
A
ρt
0
xt
0
<δand xt
1
; t
0
,xt
0
0
for some t
1
t
0
.Put
1
ψ
0
.
Journal of Inequalities and Applications 17
By the assumption that V t
0
, 00andV t, x is rd-continuous, we can find δ
0
> 0
such that if y <δ
0
then V t
0
,y <
1
. With given δ
0
> 0, let xt be a solution of 3.1 such
that A
ρt
0
xt
0
<δ
0
and xt
1
; t
0
,xt
0
0
for some t
1
t
0
.
Since xt ∈ Ω
t
and by the condition 3 ,
t
1
t
0
V
Δ
3.10
t, A
ρt
x
t
Δt V
t
1
,A
ρt
1
x
t
1
− V
t
0
,A
ρt
0
x
t
0
0.
3.16
Therefore, V t
1
,A
ρt
1
xt
1
V t
0
,A
ρt
0
xt
0
<
1
.Further,xt
1
∈ Ω
t
1
and by the condition
2 we have V t
1
,A
ρt
1
xt
1
ψxt
1
ψ
0
1
. This is a contradiction. The theorem
is proved.
Theorem 3.8. Assume that there exist a function V :
τ
×
m
→
being rd-continuous and
functions ψ, φ ∈
, ψ defined on 0, ∞, δ ∈ C
rd
t
0
, ∞, 0, ∞ such that
t
t
0
δ
s
Δs −→ ∞ as t −→ ∞ ,
3.17
satisfying the conditions
1 lim
x → 0
V t, x0 uniformly in t ∈
τ
,
2 ψx
V t, A
ρt
x for all x ∈ Ω
t
and t ∈
τ
,
3 V
Δ
3.10
t, A
ρt
x −δtφA
ρt
x for any x ∈ Ω
t
and t ∈
k
τ
.
Further, 3.1 is locally solv able. Then the trivial solution of 3.1 is asymptotically stable.
Proof. Also from Theorem 3.6 and the conditions 2 and 3, it implies that 3.1 is globally
solvable.
And since V
Δ
3.10
t, A
ρt
x −δtφA
ρt
x 0, the trivial solution of 3.1 is
stable by Theorem 3.7. Consider a bounded solution xt of 3.1. First, we show that
lim inf
t →∞
V t, A
ρt
xt 0. Assume on the contrary that inf
t∈
t
0
V t, A
ρt
xt > 0. From
the condition 1, it follows that inf
t∈
t
0
A
ρt
xt : r>0. By the condition 3,wehave
V
t, A
ρt
x
t
V
t
0
,A
ρt
0
x
t
0
t
t
0
V
Δ
3.10
s, A
ρs
x
s
Δs
V
t
0
,A
ρt
0
x
t
0
−
t
t
0
δ
s
φ
A
ρs
x
s
Δs V
t
0
,x
t
0
− φ
r
t
t
0
δ
s
Δs −→ − ∞ ,
3.18
as t →∞, which gets a contradiction.
Thus, inf
t∈
t
0
V t, A
ρt
xt 0. Further, from the condition 3 for any s t we get
V
t, A
ρt
x
t
− V
s, A
ρs
x
s
t
s
V
Δ
3.10
τ, A
ρτ
x
τ
Δτ 0.
3.19
18 Journal of Inequalities and Applications
This means that V t, A
ρt
xt is a decreasing function. Consequently,
lim
t →∞
V
t, A
ρt
x
t
inf
t∈
t
0
V
t, A
ρt
x
t
0,
3.20
which follows that lim
t →∞
xt 0 by the condition 2.
Theorem 3.9. Suppose that there exist a function a ∈ , a defined on 0, ∞, and a function V ∈
C
rd
τ
×
m
,
such that
1 lim
x → 0
V t, x0 uniformly in t ∈
τ
and ax V t, A
ρt
x for all x ∈ Ω
t
and
t ∈
τ
,
2 V
Δ
3.10
t, A
ρt
x 0, for any x ∈ Ω
t
and t ∈
k
τ
.
Assume further that 3.1 is locally solvable. Then, the trivial solution of 3.1 is A-uniformly stable.
Proof. The proof is similar to the one of Theorem 3.7 with a remark that since lim
x → 0
V t, x
0 uniformly in t ∈
τ
,wecanfindδ
0
> 0suchthatify <δ
0
then sup
t∈
τ
V t, y <
1
.
The proof is complete.
Remark 3.10. The conclusion of Theorem 3.9 is still true if the condition 1 is replaced by
“there exist two functions a, b ∈
, a defined on 0, ∞ and a function V ∈ C
rd
τ
×
m
,
such
that ax
V t, A
ρt
x bA
ρt
x for all x ∈ Ω
t
and t ∈
τ
”.
We present a theorem of uniform global asymptotical stability.
Theorem 3.11. If there exist functions a, b, c ∈
, a defined on 0, ∞, and a function V ∈ C
rd
τ
×
m
,
satisfying
1 ax
V t, A
ρt
x bA
ρt
x for all x ∈ Ω
t
and t ∈
τ
,
2 V
Δ
3.10
t, A
ρt
x −cA
ρt
x for any x ∈ Ω
t
and t ∈
k
τ
.
Assume further that 3.1 is locally solvable. Then, the trivial solution of 3.1 is A-uniformly globally
asymptotically stable.
Proof. Let δ
0
> 0 be given. Define δmin{b
−1
a,δ
0
} and
T
max
t∈
μ
t
2b
δ
0
c
δ
. 3.21
T is not necessary in
.
Let xt be a solution of 3.1 with A
ρt
0
xt
0
<δ. From the condition 2,wesee
that
V
t, A
ρt
x
t
− V
t
0
,A
ρt
0
x
t
0
t
t
0
V
Δ
3.10
s, A
ρs
x
s
Δs
0.
3.22
Journal of Inequalities and Applications 19
Therefore,
a
x
t
V
t, A
ρt
x
t
V
t
0
,A
ρt
0
x
t
0
b
A
ρt
0
x
t
0
<b
δ
a
. 3.23
Hence, xt <for all t
t
0
.
Because the trivial solution of 3.1 is A-uniformly stable, we only need to show that
there exists a t
∗
∈ t
0
,t
0
T such that A
ρt
∗
xt
∗
<δ. Assume that such a t
∗
does not
exist, that is A
ρt
xt δ for all t ∈ t
0
,t
0
T. From the condition 2,weget
V
t
0
T
,A
ρt
0
T
x
t
0
T
t
0
T
t
0
c
A
ρs
x
s
Δs
V
t
0
,A
ρt
0
x
t
0
b
A
ρt
0
x
t
0
b
δ
0
.
3.24
Since V
0,
c
δ
T
b
δ
⇒ T
b
δ
0
c
δ
, 3.25
which contradicts the definition of T in 3.21. The proof is complete.
When A
ρt
is not differentiable, one supposes that there exists a Δ-differentiable
projector Q
t
onto ker A
t
and Q
ρt
Δ
is rd-continuous on
k
τ
;moreover,Q
ρt
Q
t
for all
t ∈
τ
ls
rd
.LetP
t
I − Q
t
.
We choose matrix functions T
t
,B
t
∈ C
rd
k
τ
,
m×m
such that T
t
|
ker A
t
is an isomorphism
between ker A
t
and ker A
ρt
and the matrix G
t
A
t
− B
t
T
t
Q
t
is invertible. Define
V
Δ
3.26
t, P
ρt
x
V
Δ
t
t, P
ρt
x
1
0
V
x
σ
t
,P
ρt
x hμ
t
P
ρt
x
Δ
,
P
ρt
x
Δ
dh,
3.26
where P
ρt
x
Δ
P
ρt
Δ
x P
t
G
−1
t
ft, xsee 2.51.
From now on we remain following the above assumptions on the operators Q
t
,T
t
,B
t
whenever V
Δ
3.26
t, P
ρt
x is mentioned.
BythesameargumentasTheorem 3.6, we have the following theorem.
Theorem 3.12. Assume that there exist a constant c>0, −c ∈R
and a function V :
τ
×
m
→
being rd-continuous and a function ψ ∈ , ψ defined on 0, ∞ satisfying
1 ψx
V t, P
ρt
x for all x ∈ Ω
t
and t ∈
τ
,
2 V
Δ
3.26
t, P
ρt
x c/1 − cμtV t, P
ρt
x, for any x ∈ Ω
t
and t ∈
k
τ
.
Assume further that 3.1 is locally solvable. Then, 3.1 is globally solvable.
20 Journal of Inequalities and Applications
Theorem 3.13. Assume that 3.1 is locally solvable. Then, the trivial solution x ≡ 0 of 3.1 is stable
if there exist a function V :
τ
×
m
→
being rd-continuous and a function ψ ∈ , ψ defined on
0, ∞ such that
1 V t, 0 ≡ 0 for all t ∈
τ
,
2 V t, P
ρt
y ψy for all y ∈ Ω
t
and t ∈
τ
,
3 V
Δ
3.26
t, P
ρt
x 0 for all x ∈ Ω
t
and t ∈
k
τ
.
Proof. Assume that there is a function V satisfying the assertions 1, 2,and3 but the
trivial solution x ≡ 0of3.1 is not stable. Then, there exist a positive
0
> 0andat
0
∈
k
τ
such that ∀δ>0; there exists a solution xtxt; t
0
,x
0
of 3.1 satisfying P
ρt
0
x
0
<δ
and xt
1
; t
0
,x
0
0
,forsomet
1
t
0
.Let
1
ψ
0
.SinceV t
0
, 00, it is possible to find
a δ δ
0
,t
0
> 0 satisfying V t
0
,P
ρt
0
z <
1
when P
ρt
0
z <δ,z∈
m
.Considerthe
solution xt satisfying P
ρt
0
x
0
<δand xt
1
; t
0
,x
0
0
for a t
1
t
0
.
From the assumption 3, it follows that
t
1
t
0
V
Δ
3.26
t, P
ρt
x
t
Δt V
t
1
,P
ρt
1
x
t
1
− V
t
0
,P
ρt
0
x
0
0.
3.27
This implies
V
t
0
,P
ρt
0
x
0
V
t
1
,P
ρt
1
x
t
1
ψ
x
t
1
ψ
0
1
. 3.28
We get a contradiction because
1
>Vt
0
,P
ρt
0
x
0
when P
ρt
0
x
0
<δ.
The proof of the theorem is complete.
Theorem 3 .14. Assume that 3.1 is locally solvable. If there exist two functions a, b ∈ , a defined
on 0, ∞ and a function V :
τ
×
m
→
being rd-continuous such that
1 ax
V t, P
ρt
x bP
ρt
x for all x ∈ Ω
t
and t ∈
τ
,
2 V
Δ
3.26
t, P
ρt
x 0 for all x ∈ Ω
t
and t ∈
k
τ
,
then the trivial solution of 3.1 is P-uniformly stable.
Proof. The proof is similar to the one of Theorem 3.9.
Theorem 3.15. If there exist functions a, b, c ∈ , a defined on 0, ∞ and a function V ∈ C
rd
τ
×
m
,
satisfying
1 ax
V t, P
ρt
x bP
ρt
x for all x ∈ Ω
t
and t ∈
τ
,
2 V
Δ
3.10
t, P
ρt
x −cP
ρt
x for any x ∈ Ω
t
and t ∈
k
τ
.
Assume further that 3.1 is locally solvable. Then, the trivial solution of 3.1 is P -uniformly globally
asymptotically stable.
Proof. Similarly to the proof of Theorem 3.11.
It is difficult to establish the inverse theorem for Theorems from 3.7 to 3.15,thatis,if
the trivial solution of 3.1 is sta ble, there exists a function V satisfying the assertions in the
Journal of Inequalities and Applications 21
above theorems. However, if the structure of the time scale
is rather simple we have the
following theorem.
Theorem 3.16. Suppose that
τ
contains no right-dense points and the trivial solution x ≡ 0 of 3.1
is P-uniformly stable. Then, there exists a function V :
τ
× U →
being rd-continuous satisfying
the conditions (1), (2), and (3) of Theorem 3.13,whereU is an open neighborhood of 0 in
m
.
Proof. Suppose the trivial solution of 3.1 is P-uniformly stable. Due to Proposition 3.3,there
exist functions ϕ ∈
such that for any solution xt; t
0
,x
0
of 3.1,wehave
x
t; t
0
,x
0
ϕ
P
ρt
0
x
0
∀t
t
0
, 3.29
provided P
ρt
0
x
0
∈ ϕ.
Let
ϕ0,a and U {x : x <a}.Foranyz ∈
m
satisfying P
ρt
0
z <aand
t ∈
τ
,weput
V
t, z
: sup
s t
x
s; t, z
,
3.30
where xs; t, z is the unique solution of 3.1 satisfying the initial condition P
ρt
xtP
ρt
z.
It is seen that V is defined for all z satisfying P
ρt
0
z∈ ϕ, V t, 0 ≡ 0, and V t, x ∈
C
rd
τ
×
m
,
.
Let y ∈ Ω
t
. By the definition, Vt, P
ρt
ysup
s t
xs; t, P
ρt
y xt; t, P
ρt
y.
From 2.60, xs; t, P
ρt
yus; t, P
ρt
ygs, us; t, P
ρt
y for all s ∈
t
.Inparticular,
xt; t, P
ρt
yP
ρt
y gt, P
ρt
yy.Thus,V t, P
ρt
y y∀y ∈ Ω
t
,t∈
τ
.Hence,we
have the assertion 2 of the theorem.
Due to the unique solvability of 3.1,wehavexs; t, P
ρt
yxs; σt,xσt,t,
P
ρt
y with s σt. Therefore, V t, P
ρt
ysup
s t
xs; t, P
ρt
y and
V
σ
t
,P
ρσt
x
σ
t
,t,P
ρt
y
sup
s σt
x
s; σ
t
,x
σ
t
,t,P
ρt
y
sup
s σt
x
s; t, P
ρt
y
V
t, P
ρt
y
.
3.31
This implies
V
Δ
3.26
t, P
ρt
y
t
V
σ
t
,P
ρσt
x
σ
t
,t,P
ρt
y
− V
t, P
ρt
y
μ
t
0.
3.32
The proof is complete.
Now we give an example on using Lyapunov functions to test the stability of
equations. The following result finds out that the stability of a linear equation will be ensured
if nonlinear perturbations are sufficiently small Lipschitz.
Consider a nonlinear equation of the form 2.3
Ax
Δ
Bx f
t, x
,
3.33
22 Journal of Inequalities and Applications
where A and B are constant matrices with ind A, B1, ft, 00 ∀t ∈
,andft, x
satisfing the Lipschitz condition
f
t, x
− f
t, y
<L
x − y
, 3.34
where L is sufficiently small. Let Q be defined by 2.9 with T
t
I and G A − BQ, P
I − Q.ByTheorem 2.7, we see that there exists a unique solution satisfying the condition
Pxt
0
− x
0
0foranyx
0
∈
m
.
Besides, also consider the homogeneous equation associated to 3.33
Ax
Δ
Bx,
3.35
and suppose this equation has index-1. As in Section 2, multiplying 3.33 by PG
−1
we get
Px
Δ
Mx PG
−1
f
t, x
,
3.36
where M PG
−1
B PG
−1
BP.
Note that the general solution of 3.35 is
x
t; t
0
,x
0
e
M
t, t
0
Px
t
0
exp
tM
⎛
⎝
s∈I
t,t
0
I μ
s
M
exp
−μ
s
M
⎞
⎠
Px
t
0
,t
t
0
,
3.37
in there I
t,t
0
is denoted the set of right-scattered points of the interval t
0
,t.
Denote σA, B{λ :detλA−B0}. It is easy to show that the trivial solution x ≡ 0
of 3.35 is P-uniformly exponentially stable if and only if σA, B ⊂ S,whereS is the domain
of uniform exponential stability of
. On the exponential stable domain of a time scale, we
can refer to 10, 18, 19. By the definition of exponential stability, it implies that the graininess
function of the time scale
is upper bounded. Let μ
∗
sup
t∈
μt.
We denote the set
U
⎧
⎪
⎨
⎪
⎩
λ :
λ
1
μ
∗
1
μ
∗
if μ
∗
/
0
{
λ :
λ<0
}
if μ
∗
0,
3.38
and suppose σA, B ⊂ U.SinceU ⊂ S, this condition implies that 3.35 is P-uniformly
exponentially stable.
If μ
∗
/
0, define
H μ
∗
∞
k0
I μ
∗
M
n
P
FP
I μ
∗
M
n
Q
FQ, 3.39
where the matrix F is supposed to be symmetric positive definite. It is clear that H is
symmetric positive definite.
Journal of Inequalities and Applications 23
Since σA, B ⊂ U, the above series is convergent. Further, for any k
0wehave
I μ
∗
M
k1
P
FP
I μ
∗
M
k1
−
I μ
∗
M
k
P
FP
I μ
∗
M
k
I μ
∗
M
k1
P
FP
I μ
∗
M
k1
−
I μ
∗
M
k
I μ
∗
M
k1
−
I μ
∗
M
k
P
FP
I μ
∗
M
k
I μ
∗
M
μ
∗
I μ
∗
M
k
P
FP
×
I μ
∗
M
k
M μ
∗
M
I μ
∗
M
k
P
FP
I μ
∗
M
k
.
3.40
Thus,
I μ
∗
M
n1
P
FP
I μ
∗
M
n1
− P
FP
I μ
∗
M
n
k0
μ
∗
I μ
∗
M
k
P
FP
I μ
∗
M
k
M
μ
∗
M
n
k0
I μ
∗
M
k
P
FP
I μ
∗
M
k
.
3.41
Letting n →∞and paying attention to lim
n →∞
I μ
∗
M
n
P
FPI μ
∗
M
n
0, we obtain
−P
FP
I μ
∗
M
HM M
H HM M
H μ
∗
M
HM. 3.42
In the case where μ
∗
0andF is symmetric positive definite, by putting
H
∞
0
exp
tM
P
FP exp
tM
dt Q
FQ,
3.43
we can examine easily that the matrix H also satisfies 3.42, H is symmetric and positive
definite.
Theorem 3.17. Suppose that σA, B ⊂ U and the homogeneous equation 3.35 is of index-1 and
the constant L is sufficiently small. Then, the trivial solution x ≡ 0 of 3.33 is P-uniformly globally
asymptotically stable.
24 Journal of Inequalities and Applications
Proof. Let H be a symmetric and positive definite constant matrix satisfying 3.42.
Consider the Lyapunov function V x : x
Hx. The derivative of V along the solution of
3.33 is
V
Δ
3.26
Px
Px
Δ
H
Px
σ
Px
H
Px
Δ
Mx PG
−1
f
t, x
T
H
Px μ
t
Mx PG
−1
f
t, x
Px
H
Mx PG
−1
f
t, x
Mx PG
−1
f
t, x
HPx μ
t
Mx PG
−1
f
t, x
T
H
Mx PG
−1
f
t, x
Px
H
Mx PG
−1
f
t, x
Mx PG
−1
f
t, x
HPx μ
∗
Mx PG
−1
f
t, x
H
Mx PG
−1
f
t, x
Px
H
Mx PG
−1
f
t, x
Px
M
H HM μ
∗
M
HM
Px
PG
−1
f
t, x
× H
Px μ
∗
Mx μ
∗
PG
−1
f
t, x
Px
H
I μ
∗
M
PG
−1
f
t, x
−
Px
P
FPPx
PG
−1
f
t, x
H
Px μ
∗
Mx μ
∗
PG
−1
f
t, x
Px
H
I μ
∗
M
PG
−1
f
t, x
−
Px
FPx
PG
−1
f
t, x
H
Px μ
∗
MPx μ
∗
PG
−1
f
t, x
Px
H
I μ
∗
M
PG
−1
f
t, x
.
3.44
From the Lipschitz condition and 2.25, it is seen that Qx
KPx where K QG
−1
B
LQG
−1
/1 − LQG
−1
. Therefore,
f
t, x
L
1 K
Px
. 3.45
Combining this inequality and the above appreciation, we see that when L is sufficiently
small there exists β>0suchthat
V
Δ
3.26
Px
−β
Px
2
.
3.46
By Theorem 3.15, 3.33 is P-uniformly globally asymptotically stable.
Journal of Inequalities and Applications 25
Example 3.18. Let
∪
∞
k0
2k, 2k 1 and consider
A
t
x
Δ
B
t
x f
t, x
,
3.47
with
A
t
t 1
10
00
,B
t
−t − 20
0 −t − 1
,f
t, x
sin x
1
t 1
0, 1
. 3.48
We have ker A
t
span{0, 1
},rankA
t
1forallt ∈ .ItiseasytoverifythatQ
t
00
01
is
the canonical projector onto ker A
t
, P
t
I − Q
t
10
00
. Let us choose T
t
I.Weseethat
G
t
A
t
− B
t
T
t
Q
t
t 1
10
01
. 3.49
Since t
0, det G
t
t 1
2
/
0, 3.47 has index-1.
It is obvious that ft, w
1
− ft, w
2
1/t 1w
1
− w
2
, ∀w
1
,w
2
∈
2
.Further,
γ
t
L
t
T
t
Q
t
G
−1
t
1/t1
2
< 1, for all t ∈ . Thus, according to Theorem 2.7 for each t
0
∈ ,
3.47 with the initial condition P
ρt
0
xt
0
P
ρt
0
x
0
has the unique solution.
It is easy to compute, G
−1
t
1/t 1
10
01
, T
t
Q
t
G
−1
t
B
t
P
ρt
x 0, 0
, T
t
Q
t
G
−1
t
ft, x
sin x
1
/t1
2
0, 1
,wherex x
1
,x
2
,P
t
G
−1
t
B
t
−1/t1
t20
00
,andP
t
G
−1
t
ft, x
0, 0
.
Therefore, utP
ρt
xt satisfies u
Δ
−1/t 1
t20
00
u. M oreover, we have
Ł
t
x
x
1
,x
2
∈
2
,x
2
sin x
1
t 1
2
. 3.50
Let the Lyapunov function be V t, x : 2x,t∈
,x∈
2
.
Put x x
1
,x
2
∈ Ł
t
,wehaveV t, P
ρt
x2P
ρt
x 2|x
1
| and
x
x
2
1
x
2
2
1/2
x
2
1
sin
2
x
1
t 1
4
1/2
x
2
1
sin
2
x
1
1/2
2
|
x
1
|
.
3.51
Hence,
x
V
t, P
ρt
x
2
P
ρt
x
, ∀x ∈ Ł
t
,t∈ . 3.52
We have for any solution xt of 3.47 and t ∈
noting that t 0,
