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).