Semigroups on Frechet spaces and equations with infinite delays

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Semigroups on Frechet spaces and equations with infinite delays


Department of Mathematics, SSN College of Engineering, Old Mahabalipuram Road, Kalavakkam 603 110, India

MS received 28 February 2005

Abstract. In this paper, we show existence and uniqueness of a solution to a functional differential equation with infinite delay. We choose an appropriate Frechet space so as to cover a large class of functions to be used as initial functions to obtain existence and uniqueness of solutions.

Keywords. Functional differential equation; infinite delay; semigroup; Frechet space.

1. Introduction and preliminaries

In this paper we study linear functional differential equations with infinite delay. Consider x(t)=ax(t)+


bix(tτi), t ≥0

x(θ)=φ(θ), θ(−∞,0] (1.1)

wherea∈R,{bi}i=1is an arbitrary sequence of real numbers,{τi}i=1is a strictly increas- ing sequence of strictly positive reals such that limi→∞τi = ∞andφ:(−∞,0]−→Ris continuous.

For the special case{bi}i=1l1, (1.1) can be uniquely solved for any givenφ ∈BC (−∞,0], the space of all bounded real-valued continuous functions. The proof of this is indicated in Example1.2. Denote this solution byxφ. Consider the family of operators St, t≥0 on BC(−∞,0] defined as

[Stφ](θ)=xφ(t+θ), if t+θ >0

=φ(t+θ), if t+θ≤0. (1.2)

It is elementary to see that{St}is not a strongly continuous semigroup on BC(−∞,0].

We prove this as follows: IfSt is a semigroup, we must have


Given > 0, we can findδ > 0 such that Stφφ for |t| ≤ δ. Now, let δ =min(1, δ). Considerθ1, θ2(−∞,−1] with 0 ≤θ2θ1δ. Lett =θ2θ1. Note that 0≤tδand





|φ(θ2)φ(θ1)| ≤ |φ(t+θ1)φ(θ1|

≤ |(Stφ)(θ1)φ(θ1)|



Thus, we have shown thatφis uniformly continuous on(−∞,−1]. By the uniform con- tinuity ofφon [−1,0], uniform continuity ofφon(−∞,0] follows. But this is a contra- diction as there are bounded continuous functions which are not uniformly continuous.

On the space BUC(−∞,0], the space of bounded uniformly continuous functions, the family of operatorsStdefined by (1.2) do form a semigroup but this space is properly con- tained in BC(−∞,0]. In the literature, certain Banach spaces, which contain BC(−∞,0], which are classes of functions satisfying certain growth conditions, are used in the con- text of infinite delay equations. Refer [5, 6]. Our approach is to find a Frechet space that contains BC(−∞,0] on whichSt’s form a semigroup.

Instead of constructing a weight function which is related tobiandτi, we obtain a family of semi-norms for the initial functionφthat enable us to get estimates for the solution and also to capture a Frechet spaceFsuch that (1.2) defines a strongly continuous semigroup on F. We do not make any explicit summability assumption on the sequence{bi}i=1, but the spaceF heavily depends on the properties of{bi}i=1. If{bi} ∈l1, then BC(−∞,0]⊂F but ifbi =1i, BC(−∞,0] is not contained inF.

The basic theory of finite delay differential equations is covered in [7]. [8] and [1] are some basic references for Banach phase spaces related to infinite delay equations. Consider the following examples:

Example 1.1. LetX= BC[0,∞)andA:D(A)=BC1[0,∞)−→BC[0,∞)be defined as [Aϕ](x)=ϕ(x). DefineSt:X−→X, t≥0, as(Stϕ)(x)=ϕ(t+x). It is easy to see thatAsatisfies the following conditions:

(a) the resolvents(λIA)1u=

0 eλτSτudτ exist forλ >0,whereIis the identity operator onX,

(b) (λIA)11λ and thatSt satisfy the following conditions:

(i) S0=I, (ii) St+s =StSs.

Despite the above observationsSt is not strongly semigroup onXsince the con- dition

(iii) limt0Stϕ=ϕ, ϕX does not hold.

The proof that (iii) does not hold is similar to the proof of the analogous assertion in the case of the infinite delay equations. The Hille–Yosida theorem is not applicable precisely becauseAis not densely defined. One way to overcome this difficulty is to construct a smaller Banach space called the Hille–Yosida space for the operatorA[10] on which the restriction ofAgenerates a semigroup. In this construction, the condition (a) plays a crucial role.

But consider the operatorA:D(A)=C1[0,∞)−→C[0,)defined as [Aϕ](x)= ϕ(x).Ais an ‘extension’ ofAthat does generate the semigroupSt: C[0,)−→C[0,)


defined as [St]ϕ(x) =ϕ(t+x)on the Frechet space C[0,). Thus,it is clear that by considering the Hille–Yosida space,a lot of useful information is lost.

Example 1.2. We now indicate the proof of assertion that for {bi} ∈ l1, (1.1) can be uniquely solved for any givenφ∈BC(−∞,0].

Fort∈[0, τ1], t−τi(−∞,0] and hence forφ∈BC(−∞,0], φ(t−τi)is meaningful.

DefineziC[0, τ1] aszi = biφ(tτi). Nowzi = |bi|supt[0,τ1]|φ(tτi)| ≤

|bi|φ= |bi|φ. Since{bi}is inl1, the series

i=1ziconverges. Hence

i=1zi converges in C[0, τ1]. Thus,

i=1φ(tτi)C[0, τ1]. Now, consider the ordinary differential equation



biφ(tτi), t ∈[0, τ1],


whose solution exists and is unique.

Now, assume the existence of a unique functionyk:(−∞, kτ1]−→Rsuch thatyk(θ)= φ(θ)forθ(−∞,0] and whose restriction to [0, kτ1] is a solution to



bix(tτi), t[0, kτ1],


As before, the expression

i=1biyk(tτi)defines a continuous function on [kτ1, (k+ 1)τ1]. Now, consider



biyk(tτi), t ∈[kτ1, (k+1)τ1], x(kτ1)=yk(kτ1)

which has a unique solution z. Define yk+1 as yk+1(s) = yk(s), s(−∞, kτ1] and yk+1(s)=z(s), s∈[kτ1, (k+1)τ1]. Thus, we obtain a solution to (1.1) in(−∞, (k+1)τ1].

By induction and patching up of solutions, we get a unique solution to (1.1) on the whole of(−∞,).

LetX=BC(−∞,0] and{bi}i=1l1. Define A:D(A)=

ϕ ∈BC1(−∞,0]:ϕ(0)=






D(A)is not dense inXand henceAdoes not generate a semigroup. Motivated by Exam- ple 1.1, we look for a Frechet spaceF that contains BC(−∞,0] and an ‘extension’Ato Asuch thatAgenerates a semigroup onF.


Remark 1.3. The general theory of semigroups on Frechet spaces is very complicated. For example,even a bounded linear operator on a Frechet space need not generate a semigroup [12]. In [11],a generalisation of the Hille–Yosida theorem for a closed and unbounded operator in a locally convex space is proved. But the hypotheses of this theorem are not easily verified in many concrete cases. By proving various estimates for the solutionxof (1.1),we are able to capture a Frechet spaceF on which the solution of (1.1) gives rise to a semigroup.

Refer to [3], [13] and [2] for applications of semigroups on locally convex spaces to PDE’s.

We need the following definitions and results in the next section.


(i) A topological vector spaceXis said to be a Frechet space if its topology is generated by a family of countable semi-norms{qi}i=1andX is complete with respect to the family{qi}i=1.

(ii) A linear mapS:XX is said to be bounded if for everyi ∈ N, there are finitely many indicesj1, j2, . . . , jmand a constantCsuch that for allφX,

qi(Sφ)Cmax(qj1(φ), qj2(φ), . . . , qjm(φ)).

The basic theory of Frechet spaces and the proof of the following proposition can be found in [14].


A linear mapS:XXis continuous if and only if it is bounded.


A family of bounded linear operators{St:t ≥0}onXis said to be a strongly continuous semigroup if the properties (i), (ii) and (iii) of Example 1.1 hold.


Let l1(X) denote the Banach space of all sequences {xi}i=1 of elements of a Banach X such that

i=1xi < ∞. Let C([0, T];l1)be the Banach space of all continuous functionsh: [0, T]−→l1. The Banach spacel1(C([0, T]))is isometrically embedded in C([0, T];l1).

Proof. LetGl1(C([0, T])). DefineG: [0, T]−→l1(R)as [G(s)](i)=G(i)(s). It is given that


sup{|G(i)(s)|:s∈[0, T]}<.

It is easy to check thatGis a bounded function and that sup


|[G(s)](i)|:s∈[0, T]



sup{|G(i)(s)|:s∈[0, T]}.


It remains to be shown that



|[G(t)](i)−[G(t0)](i)| =0.

By the hypothesis for eachi, limtt0|[G(t)](i)G(t0)](i)| =0 and for a given >0, there existsK∈Nsuch that


sup{|G(i)(s)|:s∈[0, T]}< /3.

Further, there existsδ >0 such that|tt0|< δimplies that K


|[G(t)](i)−[G(t0)](i)|< /3.

Hence for allt, t0in [0, T] with|tt0|< δ,


|[G(t)][(i)−[G(t0)][(i)| = K i=1




|[G(t)][(i)−[G(t0)][(i)| ≤/3



|[G(t)][(i)| +



/3+/3+/3=. The result is proved.

The following definitions and statements on Frechet space valued Riemann integral are found in [9].

Theorem 1.8. LetX be a Frechet space and letu: [a, b] −→ X be continuous. The integralb

a u(t)dtXcan be defined uniquely which has the following properties:

(i) for every continuous linear functionalx:X−→R, x


a u(t)dt = b

a x(u(t))dt, (ii) for every seminormqk,

qk b

a u(t)dtb

aqk(u(t))dt, (iii) b

a u(t)dt+c

b u(t)dt =c



(iv) b

a[u(t)+v(t)]dt =b

a u(t)dt+b

a v(t)dt, (v) cb

a u(t)dt =b

a cu(t)dt.


LetJbe a sub interval ofR. A functionu:J −→Xis said to be differentiable att0J if there existsyXwith the following property: for every >0 andk∈N, there exists δ >0 such that for allh∈Rwith|t0+h|< δandt0+hJ, we have



hy < .

Further, the elementyis denoted byu(t0).


LetJbe a sub interval ofR. A functionu:J −→Xis said to be continuously differentiable onJ, ifuis differentiable at every pointt0J and the function mappingt tou(t)is continuous onJ. The class of all such functions is denoted by C1(J;X).

Fundamental theorem of integral calculus

Ifu: [a, b]−→Xis continuously differentiable, then u(a)u(b)=


a u(t)dt.

A Frechet phase space

For a givenϕC((−∞,0])define the family of seminorms{.k:k∈N}as ϕk =sup{|ϕ(θ)|:θ∈[−k,0]}.

Let{bi}i=1and{τi}i=1be as in the beginning of this section. Givenk∈N, definen(k)∈N as the smallest positive integer such thatτi1for allin(k).

DefineF as

F = {ϕC(−∞,0]:pk(ϕ) <∞ for all k∈N}, where the seminormspkare defined as follows:



sup{|biϕ(sτi)|:s∈[0, kτ1]}.


The spaceF equipped with the topology generated by the family of seminorms{ · k:k∈ N} ∪ {pk:k∈N}is a Frechet space.


Proof. Letφj be a Cauchy sequence inF. Clearly, there existsφC(−∞,0] such that φjconverges toφuniformly on every compact set of(−∞,0]. Consider the Banach space l1(C([0, kτ1])). For everyj ∈N, defineGjl1(C([0,kτ1]))as

G(i)j (s)=bn(k)+(i1)φj(sτn(k)+(i1)), i∈N.

By the hypothesis,Gjis a Cauchy sequence inl1(C([0, kτ1]))and hence converges to some Gl1(C([0, kτ1])). For fixediands, limj→∞G(i)j (s) =G(i)(s)and henceG(i)(s)= bn(k)+(i1)φ(sτn(k)+(i1)). This implies thatpk(φ) <∞andφj converges toφinF. The proof is complete.

Now we state the definition of a mild solution of an abstract Cauchy problem in Frechet spaces from [4].

LetXbe a Frechet space whose topology is given by the family of seminorms{qk:k∈N}

andD(A)be a subset ofX. For a closed operatorA:D(A)−→XandφX, consider the abstract Cauchy problem

du dt =Au,

u(0)=φ. (1.3)


A functionu(·)C([0,),D(A))C1([0,), X)that satisfies (1.3) is said to be a solution of Problem (1.3). A functionu(·)C([0,), X)is said to be a mild solution of (1.3), ifv(t)=t

0u(s)dsD(A), for allt ≥0, and d(v(·))

dt (t)=A(v(t))+φ, t ≥0. (1.4)

2. Main results

LetF be as in the previous section. DefineD(A)= {ϕF:ϕF andϕ(0)=}and A:D(A)−→F as=ϕwhere





We say that a functionx:R−→Ris a solution to (1.1) if the following hold:

(i) xis continuous andx(θ)=φ(θ)for allθ(−∞,0].

(ii) The restriction ofxto [0,∞)is continuously differentiable.

(iii) x(t)=ax(t)+

i=1bix(tτi)for allt≥0.

Remark 2.2. Note that our definition of the solution does not imply thatxis differentiable from the left att =0.


Theorem 2.3. LetF, LandAbe as in the beginning of this section. ThenAgenerates a strongly continuous semigroup{St:t ≥0}of bounded linear operators onF. Further, for a givenφF,the mapx:R→Rdefined as

x(t)=φ(t), t(−∞,0], x(t)=[Stφ](0), t(0,) is a unique solution to(1.1).

Besides, fixingφF and definingu: [0,)F as u(t) = Stφ, u(·)is a mild solution to the abstract Cauchy problem (1.3).

We need the following lemmas to prove Theorem 2.3 and we actually define the semi- group via the solution to (1.1).

Lemma 2.4. LetφF. The problem(1.1)has a unique solutionx:R→R.

Further,for eachk∈ Nthere exists a finite subset of{.k:k∈ N} ∪ {pk:k∈ N}

and a constantCk ≥0 such that

sup{|x(s)|:s∈[0, kτ1]} ≤Ckmax{q(φ):q∈ }. (2.1) Proof. Considert ∈[0, τ1]. Clearly,tτitτ1for alli∈N. Thus fort∈[0, τ1], t− ti ≤0. Definey1: [0, τ1]→Ras the unique solution of the initial value problem.




x(0)=φ(0). (2.2)

Note that asφF, t

i=1biφ(tτi)defines a continuous function on [0, τ1]. We have

y1(t)=φ(0)eat+eat t






Definex1:(−∞, τ1]→Ras

x1(s)=φ(s), s(−∞,0]

=y1(s), s∈[0, τ1].

In the remaining part of the proof we shall assume thata=0. The estimates fora =0 are easier to obtain. Clearly,

sup{|x1(t)|:t ∈[0, τ1]} ≤





eas−1 a


(2.3) Here, note that forr > 0,eara1 >0 for alla =0. Now we claim that for eachk ∈ N, there exists a functionxk:(−∞, kτ1]→Rwith the following properties:


(i) For eacht ∈[0, kτ1],



converges and this summation defines a continuous function on [0, kτ1].

(ii) xkis the unique solution to x(t)=ax(t)+


bix(tτi), for t ∈[0, kτ1]

x(θ)=φ(θ), for t(−∞,0]. (2.4)

(iii) There exists a finite subset of{ · k:k∈N} ∪ {pk:k∈N}and a constantCk≥0 such that

sup{|xk(s)|:s∈[0, kτ1]} ≤Ckmax{q(φ):q ∈ }. We prove this by induction onk. The casek=1 is already proved.

Assuming that our claim is true for arbitraryk∈ N, we show that the claim is true for k+1. DefineIk =[(k−1)τ1, kτ1] andxkIkas

xkIk =sup{|xk(s)|:s∈[(k−1)τ1, kτ1]}.

ForsIk+1andi∈N, sτi(−∞, kτ1] and so we can consider the summation



n(k+1)1 i=1




The first summation involves only finitely many terms. For s ∈ [kτ1, (k +1)τ1] and in(k+1), s−τi < s(k+1)τ1≤0 and hence





So, forsIk+1, the summation

i=1bixk(sτi)defines a continuous function onIk+1. For a givenk∈N− {1}, definem(k)as the smallest positive integer such that

m(k) <min{(k−1)τ1τi:i=1,2, . . . , n(k)−1}.

The following estimates follow from the definition of the seminorms and the integersm(k):




:s∈[kτ1, (k+1)τ1]

pk+1(φ). (2.5)

Moreover, sup

n(k+1)1 i=n


:s∈[kτ1, (k+1)τ1]

n(k+1)1 i=1

|bi| ×max{φm(k+1),xkIk}. (2.6)


Defineyk+1: [kτ1, (k+1)τ1]→Ras yk+1(t)=xk(kτ1)ea(t1)+eat

t 1





From (2.5) and (2.6), we get the estimate sup



:s∈[kτ1, (k+1)τ1]



|bi| ×max{φm(k+1),xkIk} +pk+1(φ)

. (2.7)

Definingxk+1:(−∞, (k+1)τ1]→Ras xk+1(s)=xk(s), s(−∞, kτ1]

=yk+1(s), s(kτ1, (k+1)τ1],

our claims (i) and (ii) are proved. Now, we proceed to prove (iii) fork+1. From the definition ofyk+1and the estimate (2.7), we get, fork∈N,

sup{|xk+1(s)|:s∈[kτ1, (k+1)τ1]} = xk+1Ik+1

=sup{|yk+1(s)|:s∈[kτ1, (k+1)τ1]}



xkIk +


ea(k+1)s−1 a




|bi| ×max{φm(k+1),xkIk} +pk+1(φ)

. (2.8)


xkIk ≤sup{|xk(s)|:s∈[0, kτ1]}

the assertion (iii) fork+1 follows from the estimate (2.8) and hence the assertion (iii) fork.

The solution to (1.1) is obtained by patching the functionsxk. Uniqueness ofx now follows.

Lemma 2.5. LetϕF andx:R→Rbe a continuous function such thatx(θ)=ϕ(θ) for allθ(−∞,0]. Defineu: [0,)C((−∞,0])as [u(t)](θ) = x(t+θ). Then uC([0,);F ).

Proof. Fixt ∈ [0, jτ1]. It is trivial to check thatu(t)C((−∞,0]). We also have the estimate

u(t)k =sup{θ∈[−k,0]:|[u(t)](θ)|}

≤max(ϕk,sup{|x(t)|:t ∈[0, jτ1}). (2.9)


Now we show thatpk(u(t)) <∞for eachk ∈N. Lets ∈[0, kτ1] andt ∈[0, jτ1]. It is clear thatn(k+j)n(k)and we have the following assertions:

in(k)(sτi)≤0 and

n(k)i < n(k+j)⇒ −(k+j)τ1t+sτi1. Thus, for everyKn(k+j), we have

K i=n(k)


n(k+j)1 i=n(k)


+ K i=n(k+j)



n(k+j)1 i=n(k)


+ K



Now choosing a positive integerm(k+j)τ1, we have the estimate pk(u(t))

n(k+j) i=1

|bi|max(sup{|x(t)|:t ∈[0, jτi}m)+p(k+j)(φ).

(2.10) Equations (2.9) and (2.10) show thatu(t)F.

Next, we show the continuity ofuat an arbitraryt0 ∈ [0,∞). Let > 0 be given.

By the uniform continuity ofxon [−k−1+t0, t0+1], there existsδ >0 such that for p, q∈[−k−1+t0, t0+1] with|pq|< δ,|x(p)x(q)|< . Takeδ =min(1, δ) and considerswith|t0s| < δ. Clearly, for anyθ ∈ [−k,0], t0+θands+θ both belong to [−k−1+t0, t0+1] and hence

sup{|x(t0+θ)x(s+θ)|:θ ∈[−k,0]}< . That is, given >0, there is aδ>0 such that


whenever|st0|< δ.

Next, we claim that for a given >0 andt0 ∈ [0, jτ1], there existsδ > 0 such that pk(u(t0)u(t)) < whenevert0∈[0, jτ1] satisfies|tt0|< δ.


By Proposition 1.6, and the definition ofF, for a givenϕF, Gk: [0, kτ1]→ l1(R) defined as [Gk(s)](i)=bn(k)+(i1)ϕ(sτn(k)+(i1))is a continuous function. Consider

K i=n(k)


n(k+j)1 i=n(k)


+ K i=n(k+j)



n(k+j)1 i=n(k)


= K i=n(k+j)


and hence i=n(k)


n(k+j)1 i=n(k)





By the uniform continuity ofGn(k+j)on [0, (k+j)τ1], given >0, there exists aδ1>0 such that|pq|< δ1implies that


|bi||[φ(p−τi)φ(qτi)]|< /2.

So, fort0andt ∈[0, jτ1] with|t0t|< δ1, sup


|bi||[φ(p−τi)φ(qτi)]|:s∈[0, kτ1]

< /2.

Since the expressionn(k+j)1

i=n(k) bi[x(t0+sτi)x(t+sτi)] involves evaluation of xover a compact set, given >0, there existsδ2>0 such that

n(k+j)1 i=n(k)

sup{|bi||[x(t0+sτi)x(t+sτi)]|}< /2

whenever t0, t ∈ [0, jτ1] and |t0t| < δ2. Our claim is now proved by taking δ = min(δ1, δ2).

Proof of Theorem 2.3. By Lemma 2.4, for a givenφF, we have a solutionxφ:R→R to (1.1).


DefineSt:FF as

[Stφ](θ)=xφ(t+θ), t+θ >0

=φ(t+θ), t+θ≤0.

Lemma 2.5 shows thatStφF for allt ≥0 and the mapu(t)=Stφis a continuous function from [0,∞)intoF. From the definition ofSt one can verify linearity of eachSt and the propertySt+s =StSs. We need only to check that eachSt is a bounded linear map onF. This follows from the estimates (2.9), 2.10) and (2.1)

Next, we have

[u(t)](θ)=φ(t+θ), t+θ≤0 [u(t)](θ)=φ(0)+



L(u(s))ds, t +θ >0 (2.11)

and using (2.11), it is easy to see thatuis indeed a mild solution to (1.3).

The Banach space Cg defined below was the phase space used in [4] for the study of infinite delay equations of which (1.1) is a special case. An interesting observation is that F contains Cg.

Example 2.6. Letg:(−∞,0] → [1,∞)be a continuous non-increasing function such that


big(τi) <. (2.12)

Consider the space Cgof allϕC(−∞,0]→Rsuch that ϕgdef

=sup |ϕ(θ)|

g(θ) :θ(−∞,0]

<. Then CgF.

Proof. Consider|biϕ(sτi)| = |bi||ϕ(sg(sττii))|g(sτi). Fors∈[0, kτ1] andin(k),0≥ sτi ≥ −τi and hence

sup{|biϕ(sτi)|:s∈[0, kτ1]} ≤g)|bi|g(τi).

Now, the assertion follows from (2.12).


Major part of this work was done while the author was with the Department of Mathemat- ics, Indian Institute of Science, Bangalore. The author acknowledges the referee whose comments greatly improved the presentation of the paper and who pointed out many mis- prints.



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