The section focuses on parameter dependence of roughness for the general exponential dichotomy on time scales under the sufficiently small linear perturbation. We consider the linear perturbed system
(3.1)
where , is an open subset of a Banach space (the parameter space). In the rest of the section, we let be the evolution operator associated to system (3.1) for each .
To obtain our conclusion, we let
and assume that the following conditions hold:
-
(a1) there exist a positive constant c and a function such that
and
where ;
-
(a2) there exist positive constants , such that
-
(a3) and for each fixed ;
-
(a4) is a decreasing function and is an increasing function.
Now we state our main result in this section.
Theorem 3.1 Assume that system (2.1) admits a general exponential dichotomy on a time scale with and conditions (a1)-(a4) hold with sufficiently small c. Then system (3.1) also admits a general exponential dichotomy on the time scale , i.e., for each , there exist projections such that
(3.2)
and
(3.3)
(3.4)
where are the complementary projections of ,
(3.5)
Moreover, if Y is a finite-dimensional space and is Lipschitz continuous for the parameter λ, then the stable subspace and the unstable subspace are Lipschitz continuous for the parameter λ.
The proof of Theorem 3.1 is nontrivial and is achieved in several steps:
-
(i)
construct some bounded solutions of perturbed system (3.1) (Lemmas 3.1, 3.2);
-
(ii)
semigroup properties of the bounded solutions of system (3.1) (Lemma 3.3);
-
(iii)
construction of the projections in (3.2) (Lemmas 3.4, 3.5 and (3.15));
-
(iv)
norm bounds for the evolution operator (Lemmas 3.6, 3.7, 3.8);
-
(v)
and are Lipschitz continuous for the parameter λ (Lemma 3.9).
We set
where
It is not difficult to show that and are both Banach spaces.
Lemma 3.1 For each , there exists a unique bounded solution satisfying
(3.6)
and is Lipschitz continuous for the parameter λ.
Proof Direct calculation shows that satisfying (3.6) is a solution of (3.1). For each , we define an operator on by
By (2.2), (2.3), (a1) and (a2), we have
for and
for . For , we get
Then
and
that is,
(3.7)
This implies that . Similarly, for each , we get
If c is sufficiently small, then is a contraction and for each , there exists a unique such that and (3.6) holds.
Next we show that is Lipschitz continuous for the parameter λ. For any , there exist bounded solutions satisfying (3.6). It follows from (a2) that
for and
for . Moreover, we also have
for . It follows from (2.2), (2.3), (a1) and (a2) that
for and
for . For , we get
Then
The proof of the lemma is complete. □
Similarly, we have the following lemma.
Lemma 3.2 For each , there exists a unique bounded solution satisfying
(3.8)
and is Lipschitz continuous for the parameter λ.
Lemma 3.3 For each , the bounded solutions and of system (3.1) satisfy
(3.9)
Proof From (3.6), we get
for , and let . We define the operator by
for any and each , where is obtained from replacing s by l. Obviously, . Carrying out similar arguments to the proof of Lemma 3.1, we have
and
for any . This means that there exists a unique such that . On the other hand, we also note that and . Therefore, . Similarly, the second identity of (3.9) holds. □
Now we construct the projections for each . We first set
(3.10)
Lemma 3.4 For each , we have
-
(b1) , are projections for each and ;
-
(b2) , , ;
-
(b3) , , , ;
-
(b4) , .
Proof It follows from Lemma 3.3 that (b1) and (b2) hold. By (3.6) and (3.8), we get
(3.11)
(3.12)
which imply that (b3) holds. By Lemma 3.1 and Lemma 3.2, we have and since satisfies identity (3.6) with and satisfies identity (3.8) with . Therefore, (b4) holds. □
Lemma 3.5 For each , is invertible.
Proof For each , combining (b3) and (b4) together gives
It follows from (3.11) and (3.12) that
Note that
(3.13)
and
(3.14)
Then
and
Then
which means that is invertible if c is sufficiently small. □
We set
(3.15)
for each and . It is not difficult to show that , , . Then , are projections and (3.2) is valid.
Lemma 3.6
We have
(3.16)
(3.17)
Proof We first prove that for each , if is a bounded solution of (3.1), then
(3.18)
A straightforward calculation shows that
(3.19)
(3.20)
and
It follows from (3.20), (2.2) and (2.3) that
(3.21)
and
We note that
for and
for . Then
when letting in (3.21). Consequently,
which means that (3.18) holds.
For each , we let , , . It is clear that and are solutions of (3.1) with the same initial value . Then
is a bounded solution of (3.1) with the initial value since is bounded for . It follows from (3.18) that
By (2.2), (2.3), (a1) and (a2), we get
and
for . Then
(3.22)
For , we have
and
For , we get
and
Then
(3.23)
for . Combining (3.22) and (3.23) together gives (3.16). Similarly, (3.17) holds. □
Lemma 3.7
We have
(3.24)
Proof According to Lemma 3.4, we get
Let , , . By using (3.15), we obtain
On the other hand,
Then and since is invertible. It follows from Lemma 3.6 that
and
which yield the desired inequalities. □
Lemma 3.8 For each , we have
(3.25)
Proof For each and any , we set
It follows from Lemma 3.7 that and are bounded solutions of (3.1). Combining (3.6) and (3.8) together gives
and
Taking leads to
and