Regularity is equivalent to the exponential dichotomy

Một phần của tài liệu almost periodic functions and differential equations (Trang 183 - 186)

7 Statement of the principle of the stationary point

Theorem 1. Regularity is equivalent to the exponential dichotomy

In one direction this assertion has already been proved in § 1; it remains to prove that regularity implies exponential dichotomy on J. Our proof is constructed so that regularity is really only used in the last stage, until then our arguments are based only on correctness.

Suppose that u(t) (t to) is a solution of Lu =0. We call it uni- formly stable to the right if

t s to)

liu(t)11. ( (8)

and uniformly exponentially stable to the right if

/1 exP (—ci(t —s)) s to). (9)

In a similar way we speak of uniform stability and exponential uniform stability to the left if a solution u (t) is defined for t to.

Lemma 1. If L is correct, then uniform stability (to the right or left) implies uniform exponential stability (to the right or left), and the constants l and c 1 depend only on 1 in (8) and k in (3).

Proof. To be definite we consider stability to the right. We show

that we can find a number T = T(1, k) such that

iiu(s + liu(s)11/2 2"; s to). (10) For suppose there is a solution z(t) with the estimate (8) and an

interval A = [a, bic [to, co) such that iz(a)11= 1 and liz(b)11---- From

Regularity and exponential dichotomy 171 (8) we obtain immediately

4(01= 1, Ilz(t)11--- 1/21 (t E A). (11)

Let 0 be such that supp 4) c A and OM = 1 for t E [a +s, b — e], and let

f= 0 (t)z(t)Ilz MIL' , t

u(t) = z(t) f co o (s)liz(s)11 - ' ds.

By using (11) it is easy to calculate that iluli c ...--. 2 -11 -2(4 —2E). Since

liflic = 1 and Lu =1' from (3) we obtain that A --.21c1 2 , which proves (10). Then this estimate and (8) imply (11) if we take / 1 = 21 and c l = T-1 ln 2. Thus, Lemma 1 is proved.

We introduce the following manifolds:

Ni (to) = {u (to) e B : Lu = 0, sup lits(t)11< 001,

t- -- to

N2(to) = 114 (to) E B :Lu =0, sup Ilu(t)11< col. (13)

t --.. to

First of all we note that under our conditions the solution of the

homogeneous equation can have an infinite number of extensions

to the whole line. In (13) it is required that such an extension is bounded on (-00, to); correctness implies the uniqueness of this

extension. We prove the inequalities

liti(t)11--- Iliti(to)li (t % to, ti (to) E N 140))) (14) liti (Oil -- liiti (t 011 (t to, U (to) E N2(to)). (15) To this end let 0 be such that:

supp 0 c {to, 00), OW = 1

(t---- to+ 1), 10'(t)i ---- 2. (16)

If u (to) e Ni(to), then for v = Ou we have Lv = (Yu =f (t En. There- fore, from (3) and (1) we obtain

sup ilu (t)II -_ II v li c .--.. k lifli c --. 2k exp collu (to)ii, tt 0+1

which together with (1) gives (14). Next, if u(to) E N2(t0), then by extending the solution to the whole line and setting v = (1— (1))u, we obtain (15) by a similar method.

The estimates (14) and (15) indicate that the manifolds Ni (to) and N2(t0) are closed; by Lemma 1, these estimates imply the correspond- ing exponential estimates. Consequently, it remains to prove that

the subspaces Ni(to) and N2(t0) are mutually complementary.

(12)

172 Linear equations in a Banach space

Let x (to) be arbitrary and x(t) (t-- --- to) be a solution of the

homogeneous equation. We put f= O'x(t) (t En, where 0 satisfies

the conditions (16). By the regularity property, the equation Lu =f has a unique solution u EC ; the function v (t)= 0 (t)x (t) (t eJ) also satisfies this equation. We set z(t)= v(t)-u(t) and y(t)= x(t)z(t).

Then obviously Lz = 0 (t EJ) and Ly = 0 (t...-- to). It is clear from the

definition of 0 that

sup Ilz(t)ii<œ), sup ilY(t)11< co,

t ,-5to t?--t0+1

that is, z(to) e N2(t0) and y(to) e Ni(to). Thus we have obtained the

decomposition x(t0)= y(t0)+ z(t0), which proves that N1 (t0) and N2(t0) are mutually complementary. Theorem 1 is proved.

We consider a family of operators L a satisfying the inequality (1) uniformly with respect to a. We say that the La are uniformly regular if the inverse operators La-1 : C --> C are uniformly bounded. Further- more, we say that the La have the uniform exponential dichotomy

on J if the constants in conditions (d2)-(d4) are bounded above and

separated from zero.

The preceding arguments prove that uniform regularity is equivalent to the uniform exponential dichotomy on J.

Finally, we remark that a regular operator is automatically strongly

correct. In fact, we can prove more than this, namely, the existence of a bounded inverse L-1:mi(B._.> ) C (B). If f

E M i (B), then we set u (t) = f_c°,,o G(t, s)f(s) ds. In view of (6) we have

ilti(t)11= i f OE) exp (-c i lsplifis +Oil ds.

By dividing the axis of integration into unit intervals we find that

1 00

ilu (t)II= 21 sup (f Ilf(s + t)11 ds) E exp (-CM)

tel I) m=0

= killfIlmi. (17)

3 Theorems on regularity

In many cases it is desirable to deduce regularity from simpler properties, such as correctness or weak regularity. We prefer to start from correctness since firstly it arises in a number of problems

on stability, and secondly, weak regularity can be reduced to correct- ness by duality arguments.

Below we distinguish two basic situations in which correctness implies regularity.

Theorems on regularity 173 1. We say that the compactness condition holds if the operators U(t, to) (t > to) are compact. The simplest examples show that the compactness condition is not enough, we need certain conditions on the dependence of A(t) on t e J.

A function 6 j -> R (R is a metric space) is called Poisson stable (has the P-property) if e(t + 4,)-* e(t) (t e J) for some sequence tm -> —00.

If the operator A(t) were bounded, then the P-property would mean the Poisson stability of the function A(t) :J -> Horn (B, B). In general, the P-property must be formulated in terms of a solving operator, namely, there exists a sequence tm -> —co such that

U(t + tm, s + tm )x -> U(t, s)x (t -.- s, x E B).

Một phần của tài liệu almost periodic functions and differential equations (Trang 183 - 186)

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