# A New Estimation of the Growth Bound of a Periodic Evolution Family on Banach Spaces

Authors: Constantin Buse; Aftab Khan; Gul Rahmat; Afshan Tabassum
Journal: Journal of Function Spaces and Applications
 (2013)

---

## Abstract

Let𝒰={U(t,s)}t≥s≥0 be a strongly continuous and q-periodic evolution family acting on a complex Banach space X. We prove that if 0<p<∞,x∈X, and sups≥0sup||x||≤1∫0∞‍∥U(t+s,s)x∥pdt:=Cp<∞, then the growth bound of the family 𝒰 is less than or equal to -1/(pCp).


---

## Body

## 1. Introduction

Knowing the rate of convergence of an iterative process, we can control the speed of its convergence. This helps us to obtain the limit of the process, with the desired accuracy, in a determined time. On the other hand, in the study of certain dynamical systems or differential equations, we meet many timesX-valued functions f, defined on ℝ+:=[0,∞), having the property that the map t↦e-ωt∥f(t)∥ is bounded on ℝ+, for a suitable real number ω. Here, and in the following, X stands for a real or complex Banach space. Clearly, every ω′≥ω has the same property. The infimum of all real numbers ω which verify the given boundedness condition is called exponential growth of the function f. In this note, we obtain an estimate of the exponential growth of a periodic evolution family that satisfies an integral condition, originally given by Datko. The theoretical result allows us to estimate the L2-norm of the solution of a periodic Cauchy problem with time-varying coefficients, which is naturally led by the one-dimensional heat equation.The classical theorem of Datko [1] states that a strongly continuous evolution family 𝒰={U(t,s)}t≥s≥0, acting on a real or complex Banach space X, is uniformly exponentially stable (i.e., there are two positive constants N and ν) such that
(1)∥U(t,s)∥≤Ne-ν(t-s),∀t≥s≥0,
if, for some (and then for all) 1≤p<∞, one has
(2)sups≥0∫0∞∥U(t+s,s)x∥pdt<∞,∀x∈X.
As usual, the norm of X and of ℒ(X) is denoted by ∥·∥. Here, ℒ(X) denotes the Banach algebra of all the bounded linear operators acting on X. By σ(T), we denote the spectrum of the linear operator T, and when it is bounded, its spectral radius is defined by
(3)r(T):=sup{|z|:z∈σ(T)}=limn→∞∥Tn∥1/n.
We use classical notations for the set of real numbers, complex numbers, and integer numbers. The set of all nonnegative integer numbers will be denoted by ℤ+.It is known that ifT={T(t)}t≥0 is a strongly continuous semigroup on a complex Banach space X and if 1≤p<∞ and Cp>0 are such that
(4)∫0∞∥T(t)x∥pdt≤Cp∥x∥p,∀x∈X,
then ω0(T)≤-1/(pCp). See [2], where the case of semigroups acting on Hilbert spaces is analyzed, and see [3, pages 81-82] for the general case.In this note, we extend this result to periodic strongly continuous evolution families acting on real or complex Banach spaces. Moreover, we prove the result withp∈(0,∞) instead of 1≤p<∞.As an application of our theoretical results, we provide a simple example, which apparently cannot be treated using the results described above for semigroups.A family𝒰:={U(t,s)}t≥s≥0⊂ℒ(X) is called a strongly continuous and  q-periodic (for some q≥1) evolution family if it satisfies the following.(1)
U(t,t)=I for all t≥0.(2)
U(t,r)U(r,s)=U(t,s) for all t≥r≥s≥0.(3)
U(t+q,s+q)=U(t,s) for all t≥s≥0.(4)
The mapt→U(t,s)x:[s,∞)→X is continuous for every x∈X and is called exponentially bounded if there exist ω∈ℝ and Mω≥1, such that
(5)∥U(t,s)∥≤Mωeω(t-s)fort≥s≥0.The growth bound ω0(𝒰) of an exponentially bounded evolution family 𝒰 is the infimum of all ω∈ℝ for which there exists Mω≥1 such that (5) is fulfilled. It is known [4] that
(6)ω0(𝒰)=limt→∞ln∥U(t,0)∥t.
The family 𝒰 is uniformly exponentially stable if its growth bound is negative. A q-periodic and strongly continuous evolution family 𝒰={U(t,s):t≥s≥0} is uniformly exponentially stable if and only if ω0(𝒰) is negative or, equivalently, if the spectral radius of the monodromy operator U(q,0) is less than one. The next result from the abstract theory of operators, originally given by Müller [5], will be useful in what follows.Lemma 1.
LetX be a complex Banach space, and let V∈ℒ(X). If the spectral radius of  V is greater than or equal to 1, then, for all 0<ε<1 and any sequence (an) with an→0 (as n→∞) and ∥(an)∥∞≤1, there exists a unit vector u0∈X, such that
(7)∥Vnu0∥≥(1-ε)·|an|,∀n∈ℤ+.Moreover, this result remains valid for real Banach spaces provided that the operatorV is power bounded; that is, sup{∥Vn∥:n∈ℤ+} is finite. See [6] for further details.


## 2. Preliminary Results

The following two lemmas will be useful in the proof of Theorem5 below. Its proof is essentially contained in [6, Theorem 1.2]. Because it has an interest in itself, we state it as a separate statement and infer its proof for the sake of completeness.Lemma 2.
Letξ:[0,∞)→[0,1] be a nonincreasing function such that limt→∞ξ(t)=0. For each δ>0, there exists a nonincreasing function ρ:[0,∞)→[0,1], having the properties (1)
ρ(t)≥ξ(t),  for all t≥0,(2)
limt→∞ρ(t)=0,(3)
ρ(t+s)≥(1+δ)-1ρ(t)ρ(s),  for all t,s≥0.Proof.
Let0=K0<K1<K2… such that 0≤ξ(t)≤(1+δ)-n, when t≥Kn. Next, choose the integers 0=R0<R1<R2… in such a way that Rn≥Kn for all n and Rn+Rm≤Rn+m for all m,n≥0. For example, we can take Rn=K0+K1+K2+⋯+Kn.
Define the functionρ:[0,∞)→[0,1] by
(8)ρ(t):=(1+δ)-nwhenRn≤t<Rn+1.
Clearly, ρ is nonincreasing, 0≤ξ(t)≤ρ(t)≤1, and limt→∞ρ(t)=0.
We claim thatρ(t+s)≥(1+δ)-1ρ(t)ρ(s) for all t,s≥0. Indeed, choose the integers jt and js such that Rjt≤t<Rjt+1, Rjs≤s<Rjs+1. Then, ρ(t)=(1+δ)-jt and ρ(s)=(1+δ)-js. Now,
(9)t+s<Rjt+1+Rjs+1≤Rjt+js+2
yields
(10)ρ(t+s)≥(1+δ)-jt-js-1=(1+δ)-1ρ(t)ρ(s).Lemma 3.
Let𝒰={U(t,s)}t≥s≥0 be a strongly continuous and q-periodic evolution family on a complex Banach space X such that r(U(q,0))≥1. Then, for any 0<ϵ<1 and every nonincreasing function ξ:[0,∞)→[0,1], with limt→∞ξ(t)=0, there exist t0≥0 and a unit vector x∈X, such that
(11)∥U(t+t0,t0)x∥≥(1-ϵ)ξ(t)∀t≥0.Proof.
Step  1. Define β:[0,∞)→[0,1] by
(12)β(t):={ξ(0),if0≤t<q,ξ(t-q),ift≥q.
The map β(t) is nonincreasing, and limt→∞β(t)=0.
Sincer(U(q,0))≥1, there exists a norm one vector x0∈X, such that
(13)∥Uk(q,0)x0∥≥12β(k)∀k∈ℤ+.
This follows by Lemma 1 with an=β(n), ϵ=1/2, and V=U(q,0). For t≥0, let N be an integer number, such that Nq≤t<Nq+q. We infer
(14)∥U(Nq+q,0)x0∥=∥U(Nq+q,t)U(t,0)x0∥≤M∥U(t,0)x0∥,
where M=Mωeqω. Thus,
(15)∥U(t,0)x0∥≥1M∥U(Nq+q,0)x0∥≥12Mβ((N+1))≥12Mβ((N+1)q)=12Mξ(Nq)≥12Mξ(t).Step  2. Let us choose δ>0 such that (1-δ)/(1+δ)≥1-ϵ. By Lemma 2, we can find a nonincreasing function ρ:[0,∞)→[0,1], such that ρ(t)≥ξ(t), limt→∞ρ(t)=0, and ρ(t+s)≥(1+δ)-1ρ(t)ρ(s), for all t,s≥0.
In view of the first step, applied to the functionρ, we may choose a unit vector y0∈X, such that
(16)∥U(t,0)y0∥≥12Mρ(t)∀t≥0.
Denote
(17)η:=inft≥0∥U(t,0)y0∥ρ(t).
Then, η≥1/(2M), and for all t≥0, we have that ∥U(t,0)y0∥≥ηρ(t). Thus,
(18)supt≥0ηρ(t)∥U(t,0)y0∥=1.
Now, let us choose t0≥0, such that
(19)1-δ≤ηρ(t0)∥U(t0,0)y0∥,
and let
(20)x:=U(t0,0)y0∥U(t0,0)y0∥.
We infer
(21)∥U(t+t0,t0)x∥=∥U(t+t0,t0)U(t0,0)y0∥∥U(t0,0)y0∥=∥U(t+t0,0)y0∥∥U(t0,0)y0∥≥ηρ(t+t0)∥U(t0,0)y0∥≥ηρ(t)ρ(t0)(1+δ)-1∥U(t0,0)y0∥≥1-δ1+δξ(t)≥(1-ϵ)ξ(t),
which completes the proof.Theorem 4.
Let𝒰={U(t,s)}t≥s≥0 be a strongly continuous and q-periodic evolution family on a complex Banach space X, and let p∈(0,∞). The following three statements are equivalent.(i)
∫0∞‍∥U(t,0)x∥pdt≤Cp<∞, for all unit vectors x∈X.(ii)
For everys≥0, one has
(22)sup∥x∥≤1(∫0∞‍∥U(t+s,s)x∥pdt):=Cp(s)<∞.(iii)
The growth bound,ω0(𝒰), is negative.The equivalence of (i) and (iii) is given, for example, in [7, Theorem 3.3], and (ii) implying (i) is obvious. We have only to prove that (i) implies (ii). To do this, let s=nq+r, with n∈ℤ+ and r∈[0,q). Let ω<0 and Mω≥1 such that
(23)∥U(t,s)∥≤Mωeω(t-s)∀t≥s≥0.
Then,
(24)∫0∞‍∥U(t+s,s)x∥pdt=∫0∞‍∥U(t+r,r)x∥pdt=∫0q-r‍∥U(t+r,r)x∥pdt+∫q-r∞‍∥U(t+r,r)x∥pdt≤qMωpepqω+∫q-r∞‍∥U(t+r-q,0)U(q,r)x∥pdt=qMωpepqω+∥U(q,r)x∥p×∫0∞‍∥U(t,0)U(q,r)x∥U(q,r)x∥∥pdt≤qMωpepqω+CpMωpepqω≤Mωpepqω(q+Cp):=Kp.


## 3. The Result and Its Consequences

Theorem 5.
Let𝒰={U(t,s)}t≥s≥0 be a strongly continuous and q-periodic evolution family acting on a complex Banach space X. Suppose that the integral condition (22) is fulfilled for any s≥0 and some (and then for all) 0<p<∞. Then, there exists t0≥0 such that ω0(𝒰)≤-1/(pCp(t0)).Proof.
Letω=ω0(𝒰). The family
(25)ℬ:={B(t,s):=e-ω(t-s)U(t,s)}t≥s≥0
is a q-periodic evolution, family and ω0(ℬ)=0. Let ξ(t):=e-δt with δ>0, and let 0<ϵ<1 be fixed. By applying Lemma 3 to the q-periodic evolution family ℬ, there exist t0≥0 and a norm one vector x∈X, such that
(26)∥B(t+t0,t0)x∥≥(1-ϵ)e-δt,∀t≥0,
or equivalently
(27)∥U(t+t0,t0)x∥≥(1-ϵ)e(ω-δ)t,∀t≥0.
Hence, via Theorem 4,
(28)∫0∞‍(1-ϵ)pe(ω-δ)ptdt≤∫0∞‍∥U(t+t0,t0)x∥pdt≤Cp(t0).
That is, -(1-ϵ)p/(ω-δ)p≤Cp(t0). Since ϵ>0 and δ>0 were arbitrary, the last inequality yields -1/pω≤Cp(t0) that is, ω0(𝒰)≤-1/(pCp(t0)).Corollary 6.
Let𝒰={U(t,s)}t≥s≥0 be a strongly continuous and q-periodic evolution family acting on a complex Banach space X. If 0<p<∞ and C>0 are such that
(29)sups≥0∫0∞‍∥U(t+s,s)x∥pdt≤C∥x∥p,∀x∈X,
then ω0(𝒰)≤-1/(pC).Corollary 7.
Let0<p<∞ and T={T(t)}t≥0 be a C0-semigroup on a Banach space X. If
(30)sup∥x∥≤1∫0∞‍∥T(t)x∥pdt=Cp<∞,
then
(31)ω0(T)≤-1pCp.Theorem 8.
Let𝒰={U(t,s)}t≥s≥0 be a strongly continuous and q-periodic evolution family acting on a real or complex Banach space X. Assume that ω0:=ω0(𝒰) is negative, and, in addition, the family
(32)𝒰0:={e-ω0(t-s)U(t,s):t≥s≥0}
is uniformly bounded, and let 0<p<∞. Then, there exists t0≥0 such that ω0(𝒰)≤-1/(pCp(t0)), where Cp(t0) is defined in (22).Actually, under the addition boundedness assumption, the monodromy operator associated with the family𝒰0, that is, e-qU(q,0), is power bounded, and we can use [6, Theorem 2.1].


## 4. Some Examples and Remarks

Now we present two examples which illustrate the theoretical result.Example 1.
Leta:ℝ→ℂ be a continuous and q-periodic function, and let us define U(t,s):=e∫st‍a(τ)dτ. Clearly, the family 𝒰:={U(t,s):t≥s} is a norm continuous q-periodic evolution family on X:=ℂ. If
(33)sups∈ℝ∫0∞‍(e∫ss+t‍Re[a(τ)]dτ)pdt=Cp<∞,
Then, for every ν>-1/(pCp) and every t≥s, one has
(34)e∫st‍Re[a(τ)]dτ≤Mνeν(t-s).Example 2.
Let us chooseH:=L2([0,π],ℂ) to be the state space. Endowed with the usual inner product and norm, it becomes a complex Hilbert space, and, in addition, the one parameter family {T(t)}t≥0, given by
(35)(T(t)x)(ξ)=2π∑n=1∞‍e-tn2cn(x)sinnξ,ξ∈[0,π],t≥0,
where cn(x):=∫0π‍x(s)sin(ns)ds, is a strongly continuous semigroup on H. The semigroup T={T(t)}t≥0 is generated by the linear operator A given by Ax=x¨, and the maximal domain of A is the set D(A) of all x∈H such that x and x˙ are absolutely continuous, x¨∈H, and x(0)=x(π)=0. Moreover, T(t) is a self-adjoint operator for every t≥0 [8, Example 1.3, pages 178, 198]. Consider the nonautonomous Cauchy problem
(36)∂z(t,ξ)∂t=f(t)∂2z(t,ξ)∂2ξ,t>0,ξ∈[0,π]z(t,0)=z(t,π)=0,t≥0z(0,ξ)=x(ξ),
where x(·) is a given function in H, and f:ℝ+→[1,∞) is a function having the following properties.(1)
f(·) is q-periodic for some q≥1.(2)
There existα∈(0,1] and c>0 such that
(37)|f(t)-f(s)|≤c|t-s|α∀t,s≥0.
LetF(t)=∫0t‍f(t)dt. As is well known, the solution z(·) of the above Cauchy problem satisfies the evolution condition
(38)z(t)=U(t,s)z(s),
where U(t,s)=T(F(t)-F(s)). See [9, Example 2.9b].
We can estimate a positive constantν having the property that the map t↦evt∥z(t)∥ is bounded on ℝ+.
Indeed,(39)∫0∞∥U(t,0)x∥2dt=2π∫0∞‍∑n=1∞‍cn2(x)e-2n2F(t)dt=2π∑n=1∞‍cn2(x)∫0∞‍e-2n2F(t)dt=∥x∥22∫0∞‍e-2n2F(t)dt≤∥x∥22∫0∞‍e-2F(t)dt.
On the other hand
(40)∫0∞‍e-2F(t)dt=∑k=0∞‍∫kq(k+1)q‍e-2F(t)dt=∑k=0∞‍∫0q‍e-2F(kq+r)dr=∑k=0∞‍e-2kF(q)∫0q‍e-2F(r)dr≤q∑k=0∞‍e-2kF(q)=qe2F(q)e2F(q)-1:=C2.
Hence,(41)∫0∞∥U(t,0)x∥2dt≤C2∥x∥22.
Now by Theorem8, we get ω0(𝒰)≤-1/(2K2), with K2 being given in (24).Remark 9.
Even in the case of Hilbert spaces, the inequality(42)ω0(T)≤-1pCp
could be sharp.
Indeed, letH:=l2(ℤ+,ℂ) be the Hilbert space of all ℂ-valued sequences x=(xn) verifying
(43)∥x∥2:=(∑n=1∞‍|xn|2)1/2<∞.
Letε>0 be fixed, and set
(44)(T(t)x)n:=e-εtxn,n∈ℤ+,t≥0.
Obviously, the family T={T(t)}t≥0 is a strongly continuous semigroup on l2(ℤ+,ℂ), and
(45)2ω0(T)·(sup∥x∥2≤1∫0∞∥T(t)x∥22dt)=-1.Remark 10.
It seems that the theoretical result of this paper cannot be applied to wave equations. We try to justify this statement in the following.
LetA be a closed operator on a complex Banach space X, and let x∈X. Consider the Cauchy Problem
(46)P2(x,0){u′′(t)=Au(t)(t≥0)u(0)=xu′(0)=0.
Recall that a classical solution ofP2(x,0) is a function u∈C2(ℝ+,X) such that u(t)∈D(A) for all t≥0, and P2(x,0) holds. A function u∈C(ℝ+,X) is called a mild solution for P2(x,0) if
(47)∫0t‍(t-s)u(s)ds∈D(A),u(t)=x+A∫0t‍(t-s)u(s)ds
for all t≥0. Any C2(ℝ+,X)-valued mild solution of P2(x,0) is a classical solution. The mild solution of P2(x,0) leads to the notion of cosine function. See [10] or [11, pages 206–221], for further details.
Recall that a strongly continuous functioncos:ℝ+→ℒ(X) is called cosine function if(1)
cos(0)=I, where I is the identity operator on X,(2)
2cos(t)cos(s)=cos(t+s)+cos(t-s) for all t≥0 and s≥0.It is well known that any cosine function is exponentially bounded; that is, there are ω∈ℝ and M>0 such that ∥cos(t)∥≤Meωt for all t≥0. Moreover, the uniform growth bound of any cosine function is nonnegative [11, Proposition 3.14.6]. Any cosine function has a generator. More exactly, there exist a positive ω and a linear operator A such that (ω2,∞)⊂ρ(A) and
(48)λR(λ2,A)=∫0∞e-λtcos(t)dt(λ>ω).
See [11, Proposition 3.14.4]. On the other hand, if A is a generator of a cosine function, the mild solution of the Cauchy Problem P2(x,0) is given by u(t)=cos(t)x, t≥0. Finally, we remind the reader that the result of this paper refers to exponentially stable evolution families, and then it cannot be applied to cosine functions.



---

*Source: 260920-2013-10-22.xml*