# A Generalization for Theorems of Datko and Barbashin Type

Authors: Pham Viet Hai
Journal: Journal of Function Spaces
 (2015)

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## Abstract

The goal of the paper is to give some characterizations for the uniform exponential stability of evolution families by unifying the discrete-time versions of the Barbashin-type theorem and the Datko-type theorem.


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## Body

## 1. Introduction

In operator theory, a bounded linear operator family{U(t,s)}t≥s≥0 (U) is called an evolution family if(i)
U(t,t)=I, the identity;(ii)
U(t,s)=U(t,r)U(r,s), the evolution property;(iii)
for each elementx, the orbit U(·,·)x is continuous. This notion occupies a particularly important role in representing solutions of the nonautonomous linear differential equation(1)u′=Atuon Banach spaces when the operators A(t), t≥0, are linear and unbounded. It is difficult to prove the existence of evolution families when studying on infinite dimensional Banach spaces. In fact, the conditions are obtained in several special cases of (1). We will not continue, in this present paper, the existence problem. Instead, we will assume that the evolution family exists and then study its stability.We say thatU admits a uniform exponential growth if there are two positive numbers M and ω such that, for every t≥s≥0, U(t,s)≤Meω(t-s). When ω<0, U is called uniformly exponentially stable (u.e.s.). During the past decade, an increasing attention is devoted to the stability of evolution families. For example, in 1970, Datko showed that the evolution family with the uniform exponential growth is u.e.s. if, for each element x,(2)sups≥0⁡∫s∞Uτ,sxdτ<∞.In his result, the integral is taken according to the first variable of the evolution family. A similar characterization is named by Barbashin in 1967 when integrating with respect to the second variable:(3)supt≥0⁡∫0tUt,τdτ<∞.Two results become the starting point for the works [1–3], where the discrete versions were established: (i)
(2) ⇔sups⁡∑j=0∞U(j+s,s)x<∞.(ii)
(3) ⇔sups⁡∑j=0nU(n+s,j+s)<∞.Particularly, we can find in [4] the initial studies on unifying the discrete-time versions of the Datko-type theorem and the Barbashin-type theorem. Let us restate the following: the uniform exponential stability is equivalent to the condition(4)supn,m⁡∑j=0nUaj+bn+m,bj+m<∞,where the nondecreasing sequences (an) and (bn) belong to class S; that is, (cn)∈S if supn(cn+1-cn)<∞. Following this idea, [5] provides another characterization:(5)supn,m⁡∑j=0nUaj+m,mUbn+m,bj+m<∞.It is worth mentioning here that conditions (4) and (5) become the Datko-type theorem when bn=0. When an=0, we get the Barbashin-type theorem. Naturally, we raise a question to consider the following assumption:(6)supn,m⁡∑j=0nUajbn+m,a0bj+m<∞.The goal of the paper is to show that condition (6) is equivalent to the uniform exponential stability of the evolution family. The obtained result is an extension of classical theorems due to Barbashin and Datko.


## 2. Preliminaries

We start the paper with some notations. As usual, we denote byR+ and N the set of positive numbers and positive integers, respectively. We write Np for the set of integers j with j≥p. Let (X,·) be a Banach space. The norm on the space of bounded linear operators on X is also denoted by ·. Let Sα denote the set of all nondecreasing positive sequences (bn) with supn∈N(bn+1-bn)≤α. Let us denote by F the set of functionals F defined on the set of positive sequences with the following conditions:(i)
0≤F(s1)≤F(s2) if s1≤s2.(ii)
there isc>0 so that F(αX{n})≥cα for every n∈N and every α≥0.(iii)
limn→∞F(αX{0,…,n})=∞ for every α>0.Notation XA means the characteristic function of a set A. The following lemma is derived from [6, 7].Lemma 1.
IfF∈F and l>0 then limn→∞⁡RFl(n)=∞, where RFl(n)≔infα∈0,l(FαX0,…,n/α2).ByU, we denote the set of all functions g(·,·):T:={(t,s):t≥s≥0}→R+ with the following properties:(i)
g(t,r)≤g(t,s)g(s,r) for all t≥s≥r≥0.(ii)
for eachα>0, there exists Mα>0 such that supt∈[0,α],s≥0g(t+s,s)≤Mα<∞. For simplicity, we will use the symbol “sup” instead of supn,m∈N. In the whole paper, we always assume that the evolution family admits the uniform exponential growth with the constants M and ω. The following lemmas play an important part in the proof.Lemma 2.
Letg∈U. Assume that there exist l,δ>0 and (cn)∈Sδ satisfying(7)limn→∞⁡cn=∞,sup⁡gcn+m,m≤l<∞.Then g is bounded on T.Proof.
In the first step, we demonstrate thatsupt∈R+,m∈Ng(t+m,m)<∞. Indeed, for each t≥0, we have two cases. The first case is t≤c0. In this case, g(t+m,m)≤Mc0. The second case is t≥c0. By limn→∞⁡cn=∞ we can fix n∈N such that t∈[cn,cn+1]. Thus, g(t+m,m)≤g(t+m,cn+m)g(cn+m,m)≤Mδl. Hence, supt∈R+,m∈Ng(t+m,m)≤max⁡{Mc0,Mδl}. Next, we evaluate that supt,s∈R+g(t+s,s)<∞. For each (t,s)∈R+2, there are also two cases.
Ift≤1 then g(t+s,s)≤M1.
Ift≥1 then t+s≥1+[s]>s. We estimate (8)gt+s,s≤gt+s,1+sg1+s,s≤M1gt+s,1+s≤M1max⁡Mc0,Mδl.This implies the desired result.Lemma 3.
Letg∈U. Assume that there are two constants p∈N1 and q∈N and c<1 such that supm∈Nqg(p+m,m)≤c. Then there exist K,ν>0 such that(9)gt+s,s≤Ke-νt,∀t,s∈R+2.Proof.
Step
1. By induction, we now prove that(10)gkp+m,m≤ck,∀m∈Nq,k∈N.For k=0 and k=1, (10) holds. Assume that (10) is true when k=n. Consider k=n+1. We have that(11)gk+1p+m,m≤gk+1p+m,kp+mgkp+m,m≤c·ck=ck+1.Hence, (10) is true.
Step
2. We prove that there exists K1,ν1>0 such that(12)gt+m,m≤K1e-ν1t,∀t≥0,m∈Nq.Indeed, for each t≥0, there exist k∈N and r∈[0,p) such that t=kp+r. We have that(13)gt+m,m≤gt+m,kp+mgkp+m,m≤Mpck=Mpekln⁡c≤K1e-ν1t,where K1:=Mp/c and ν1:=-ln⁡c/p.
Step
3. We prove that there exist K2,ν2>0 such that(14)gt+m,m≤K2e-ν2t,∀t≥0,m∈N.If q=0 then, from the second step, we obtain the desired result. Now we consider the case as q∈N1. For each (m,t)∈N×R+, there are two situations as follows. The first situation is t<q. With this situation, we estimate(15)gt+m,m≤Mq=Mqeν1te-ν1t≤Mqeν1qe-ν1t.The second situation is t≥q. We estimate(16)gt+m,m≤gt+m,q+mgq+m,m≤MqK1e-ν1t-q=K1Mqeν1qe-ν1t.We rewrite(17)gt+m,m≤K1Mqeν1qe-ν1t.From (15) and (17), we can choose ν2:=ν1 and K2:=max⁡{K1Mqeν1q,Mqeν1q}.
Step
4. For each t,s≥0, there are two situations. The first situation is t≤1. Then g(t+s,s)≤M1. The second situation is t≥1. Then t+s≥1+[s]>s. We have that(18)gt+s,s≤gt+s,1+sg1+s,s≤K2e-ν2t+s-1-sM1≤K2e-ν2tM1eν2.This implies the desired result.As a consequence, we have the following.Lemma 4.
If there existl,δ>0, and (cn)∈Sδ satisfying the hypotheses(19)limn→∞⁡cn=∞,sup⁡Ucn+m,m≤l<∞,then U is uniformly bounded.Lemma 5.
If there are two constantsp∈N1 and q∈N, and c<1 such that U(p+m,m)≤c for all m∈Nq then U is uniformly exponentially stable.


## 3. Main Results

Given an evolution familyU, we define the mapping φ(·,·,·):N3→R+ by (20)φm,n,j≔Ucj+anbj+m,ajb0+m,j∈0,…,n,0,j∉0,…,n.The first characterization is given by the following.Theorem 6.
U is u.e.s. if and only if there exist α,β,δ, K,L1,L2,L3>0, (an)∈Sα, (bn)∈Sβ, (cn)∈Sδ, and F∈F such that (1)
(21)sup⁡supj≤nUcn+anbn+m,cj+anbj+m≤L1<∞,sup⁡Uanb0+m,m≤L2<∞,sup⁡Ucn+anbn+m,m≤L3<∞,(2)
sup⁡F(φ(m,n,·))<K.Proof.
Let us prove the necessity. Suppose thatU is uniformly exponentially stable; that is, U(t+s,s)≤Ke-νt. We have that(22)∑j=0nUnj+m,m≤∑j=0nKe-νnj≤∑j=0∞Ke-νj.Thus we only take aj=bj=j, α=β=1, K=∑j=0∞‍Ke-νj, and Lj=K; F(s)=∑n=0∞‍s(n).
Now let us prove the sufficiency. Fix(m,n)∈N2. Let j∈{0,…,n}. Denote l≔L3/L1L2. From the evolution property (23)Ucn+anbn+m,m=Ucn+anbn+m,cj+anbj+m·Ucj+anbj+m,ajb0+mUajb0+m,m,we estimate(24)Ucn+anbn+m,m≤L1L2Ucj+anbj+m,ajb0+m.It follows that(25)φm,n,·≥Ucn+anbn+m,mL1L2X0,…,n.Acting F on two both sides, we have that (26)K≥Fφm,n,·≥FUcn+anbn+m,mL1L2X0,…,n≥Ucn+anbn+m,mL1L22RFln.Hence,(27)RFlnUcn+anbn+m,m≤KL1L2.By Lemma 1, we can fix n1 such that RFl(n1)≥2MeωKL1L2. Now we derive that 2MeωU(cn1+an1bn1+m,m)≤1. From the evolution property (28)Ucn1+an1bn1+1+m,m=Ucn1+an1bn1+1+m,cn1+an1bn1+m·Ucn1+an1bn1+m,m,we estimate (29)Ucn1+an1bn1+1+m,m≤Ucn1+an1bn1+1+m,cn1+an1bn1+m·Ucn1+an1bn1+m,m≤MeωUcn1+an1bn1+m,m≤12.Taking into account that [cn1+an1bn1]+1 does not depend on m. Using Lemma 5, U is uniformly exponentially stable. The proof completes.Our main result is as follows.Theorem 7.
U is u.e.s. if and only if there exist α,β, δ,K>0, (an)∈Sα, (bn)∈Sβ, (cn)∈Sδ, and F∈F such that(30)sup⁡Fφm,n,·<K.Proof.
The necessity is clear. Let us prove the sufficiency. Fix(m,n)∈N2. Let z∈{0,…,n}. It is obvious that(31)φm,n,·≥Ucz+anbz+m,azb0+mXz.Acting F on two both sides of the inequality above, we obtain (32)K≥Fφm,n,·≥FUcz+anbz+m,azb0+mXz≥cUcz+anbz+m,azb0+m.Therefore,(33)Ucz+anbz+m,azb0+m≤Kc.On the other hand, from the evolution property(34)Ucz+anbz+m,m=Ucz+anbz+m,azb0+mUazb0+m,m,we estimate(35)Ucz+anbz+m,m≤Ucz+anbz+m,azb0+mUazb0+m,m≤Meωazb0Ucz+anbz+m,azb0+m≤KMeωazb0c.Now we rewrite the last inequality(36)Ucz+anbz+m,m≤KMeωazb0c.We consider three cases as follows.
Case
1 (
sup
⁡
a
n
=
∞
). In (36), let z=0. We get(37)Uc0+anb0+m,m≤KMeωa0b0c.Using Lemma 5 for the case of the sequence un≔c0+anb0, there exists L>0 such that(38)supt,s≥0⁡Ut+s,s≤L<∞.Applying Theorem 6, U is uniformly exponentially stable.
Case
2 (
A
:
=
sup
⁡
a
n
<
∞
and
sup
⁡
a
n
b
n
=
∞
). With these conditions, it follows from (36) that(39)Ucz+anbz+m,m≤KMeωazb0c≤KMeωAb0c.Let n→∞. Using the continuity of the mapping (t,s)→U(t,s)x, we obtain(40)Ucz+Abz+m,m≤KMeωAb0c.Hence, by Lemma 5 for the case of the sequence un≔cn+Abn, there exists L>0 such that(41)supt,s≥0⁡Ut+s,s≤L<∞.Using Theorem 6, U is uniformly exponentially stable.
Case
3 (
D
≔
sup
⁡
a
n
b
n
<
∞
,
sup
⁡
c
n
=
∞
). In (36), let z=n. We obtain(42)Ucn+anbn+m,m≤KMeωanb0c≤KMeωDc.Let n→∞. Using the continuity of the mapping (t,s)→U(t,s)x, we obtain(43)Ucn+D+m,m≤KMeωDc.Using Lemma 5 for the case of the sequence un≔cn+D, there exists L>0 such that(44)supt,s≥0⁡Ut+s,s≤L<∞.Using Theorem 6, U is uniformly exponentially stable.
Case
4 (
D
≔
sup
⁡
a
n
b
n
<
∞
,
C
≔
sup
⁡
c
n
<
∞
). In this case, we see that(45)Ucn+anbn+m,cj+anbj+m≤Meωcn-cj+anbn-anbj≤MeωC+D,Uanb0+m,m≤Meωanb0≤Meωanbn≤MeωD,Ucn+anbn+m,m≤Meωcn+anbn≤MeωC+D.Now we again apply Theorem 6 to obtain the desired result.From Theorem6, we get the following.Corollary 8.
U is u.e.s. if and only if there exist α,β>0, (aj)∈Sα, and (bj)∈Sβ such that(46)sup⁡∑j=0nUanbj+m,ajb0+m<∞.Proof.
Using Theorem7 for cj=0 and F(s)=∑j=0∞‍s(j), the proof completes.Remark 9.
Note that if we choose in Corollary8  bj=1 (aj=1) then we obtain Barbashin’s theorem (Datko’s theorem).



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*Source: 517348-2015-08-11.xml*