# Some Stochastic Functional Differential Equations with Infinite Delay: A Result on Existence and Uniqueness of Solutions in a Concrete Fading Memory Space

Authors: Hassane Bouzahir; Brahim Benaid; Chafai Imzegouan
Journal: Chinese Journal of Mathematics
 (2017)

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## Abstract

This paper is devoted to existence and uniqueness of solutions for some stochastic functional differential equations with infinite delay in a fading memory phase space.


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## Body

## 1. Introduction

Let|·| denote the Euclidian norm in Rn. If A is a vector or a matrix, its transpose is denoted by A′ and its trace norm is represented by |A|=(Trace(A′A))1/2. Let a∧b(a∨b) be the minimum (maximum) for a,b∈R.Let(Ω,F,P) be a complete probability space with a filtration {Ft}t≥0 satisfying the usual conditions; that is, it is right continuous and Ft0 contains all P-null sets.M
2
(
(
-
∞
,
T
]
;
R
n
) denotes the family of all Ft-measurable Rn valued processes x(t), t∈(-∞,T] such that E(∫-∞T‍|x(t)|2dt)<∞.Assume thatW(t) is an m-dimensional Brownian motion which is defined on (Ω,F,P); that is, W(t)=W1t,W2t,…,Wmt′.LetCμ={φ∈C(-∞;0];Rn):limθ→-∞eμθφ(θ)existsinRn} denote the family of continuous functions φ defined on (-∞,0] with norm |φ|μ=supθ≤0eμθ|φ(θ)|.Consider then-dimensional stochastic functional differential equation(1)dxt=fxt,tdt+gxt,tdWt,t0≤t≤T,where xt:(-∞,0]→Rn;θ↦xt(θ)=x(t+θ);-∞<θ≤0 can be regarded as a Cμ-value stochastic process, and f:Cμ×[t0,T]→Rn and g:Cμ×[t0,T]→Rn×m are Borel measurable.The initial data of the stochastic process is defined on(-∞,t0]. That is, the initial value xt0=ξ={ξ(θ):-∞<θ≤0} is a Ft0-measurable and Cμ-value random variable such that ξ∈M2(Cμ).Our aim, in this paper, is to study existence and uniqueness of solutions to stochastic functional differential equations with infinite delay of type (1) in a fading memory phase space.


## 2. Preliminary

The theory of partial functional differential equations with delay has attracted widespread attention. However, when the delay is infinite, one of the fundamental tasks is the choice of a suitable phase spaceB. A large variety of phase spaces could be utilized to build an appropriate theory for any class of functional differential equations with infinite delay. One of the reasons for a best choice is to guarantee that the history function t→xt is continuous if x:(-∞,a]→Rn is continuous (where a>0). In general, the selection of the phase space plays an important role in the study of both qualitative and quantitative analysis of solutions. Sometimes, it becomes desirable to approach the problem purely axiomatically. The first axiomatic approach was introduced by Coleman and Mizel in [1]. After this paper, many contributions have been published by various authors until 1978 when Hale and Kato organized the study of functional differential equations with infinite delay in [2]. They assumed that B is a normed linear space of functions mapping (-∞,0] into a Banach space (X,|·|), endowed with a norm |·|B and satisfying the following axioms.(
A
1
) There exist a positive constant H and functions K(·),M(·):[0,+∞)→[0,+∞), with K being continuous and M being locally bounded, such that for any σ∈R and a>0, if x:(-∞,σ+a]→X, xσ∈B, and x(·) is continuous on [σ,σ+a], then for all t in [σ,σ+a], the following conditions hold:(i)
xt∈B,(ii)
|x(t)|≤H|xt|B,(iii)
|xt|B≤K(t-σ)supσ≤s≤t|x(s)|+M(t-σ)|xσ|B.(
A
2
) For the function x(·) in (A1), t→xt is a B-valued continuous function for t in [σ,σ+a].(
A
3
) The space B is complete.Later on, the concept of fading and uniform fading memory spaces has been adopted as the best choice.Forφ∈B,t≥0 and θ≤0, we define the linear operator O(t) by (2)Otφθ=φ0ift+θ≥0,φt+θift+θ<0.(O(t))t≥0 is exactly the solution semigroup associated with the following trivial equation: (3)ddtut=0,u0=φ. We define (4)O0t=OtB0,whereB0≔φ∈B:φ0=0.Let C00 be the set of continuous functions φ:(-∞,0]→X with compact support. We recall the following axiom.(
A
4
) If a uniformly bounded sequence (φn)n≥0 in C00 converges to a function φ compactly on (-∞,0], then φ∈B and |φn-φ|B→0.Definition 1.
(1)
B is called a fading memory space if it satisfies the axioms (A1),(A2),(A3),(A4)and|O0(t)φ|→0 as t→+∞, for all φ∈B0.
(2)
B is called a uniform fading memory space if it satisfies the axioms (A1),(A2),(A3),(A4)and|O0(t)|→0 as t→+∞.Examples. We recall the definitions of some standard examples of phase spaces B.We start first with the phase space ofX-valued bounded continuous functions φ defined on (-∞,0], that is, BC((-∞,0];X) with norm |φ|BC=sup-∞<θ≤0|φ(θ)|.(1) Let(5)BU=φ∈BC-∞,0;X:φis bounded uniformly continuous,where BC is the space of all bounded continuous functions mapping (-∞,0] into X provided with the uniform norm topology.(2) Letμ∈R and (6)Cμ=φ∈C-∞,0;X:limθ→-∞eμθφθexists inX, provided with the norm (7)φμ=sup-∞<θ≤0eμθφθ.(3) For any continuous functiong:(-∞,0]→[0,+∞), we define (8)Cg=φ∈C-∞,0;X:φθgθis bounded on-∞,0,Cg0=φ∈Cg-∞,0;X:limθ→-∞φθgθ=0, endowed with the norm (9)φg=sup-∞<θ≤0φθgθ.Consider the following conditions ong:(g1) sup-∞<θ≤-tg(t+θ)/g(θ) is locally bounded for t≥0,(g2) limθ→-∞g(θ)=+∞,(g3) limt→+∞sup-∞<θ≤-tg(t+θ)/g(θ)=0.Properties of each phase space are summarized in Table1.Table 1
(A1)
(A2)
(A3)
(A4)
Uniform fading memory space
B
C
Yes
No
Yes
No
No
B
U
Yes
Yes
Yes
No
No
C
g
Under (g1)
Under (g1)
Yes
Under (g2)
Under (g3)
C
g
0
Under (g1)
Under (g1)
Yes
Under (g2)
Under (g3)
C
μ
Yes
Yes
Yes
Underμ>0
Underμ>0For other examples, properties, and details about phase spaces, we refer to the book by Hino et al. [3].Fengying and Ke [4] discussed existence and uniqueness of solutions to stochastic functional differential equation with infinite delay in the phase space of bounded continuous functions φ defined on (-∞,0] with values in Rn, that is, BC((-∞,0];Rn) with norm |φ|BC=sup-∞<θ≤0|φ(θ)|.Lemma 2 (page  22 in [3]).
If the phase spaceB satisfies axiom (A4), then BC((-∞,0];Rn) is included in B.


## 3. Existence and Uniqueness

Lemma 3 (see [4]).
Ifp≥2,g∈L2([t0,T];Rn×m) such that E∫t0T‍|g(s)|pds<∞, then(10)E∫t0TgsdWsp≤pp-12p/2Tp-2/2E∫t0Tgspds.Lemma 4 (Borel-Cantelli, page  487 in [5]).
If{En} is a sequence of events and (11)∑n=1∞PEn<∞, then (12)PEni.o.=0, where i.o. is an abbreviation for “infinitively often.”Definitions 1.
R
n-value stochastic process x(t) defined on -∞<t≤T is called the solution of (1) with initial data xt0, if x(t) has the following properties:(i)
x(t) is continuous and {x(t)}t0≤t≤T is Ft-adapted,(ii)
{f(xt,t)}∈L1([t0,T];Rn) and {g(xt,t)}∈L2([t0,T];Rn×m),(iii)
xt0=ξ, for each t0≤t≤T, (13)xt=ξ0+∫t0tfxs,sds+∫t0tgxs,sdWsalmostsurelya.s..x(t) is called unique solution, if any other solution x¯(t) is distinguishable with x(t); that is, (14)Pxt=x¯t,foranyt0≤t≤T=1.
Now, we establish existence and uniqueness of solutions for (1) with initial data xt0. We suppose a uniform Lipschitz condition and a weak linear growth condition.Theorem 5.
Assume that there exist two positive numberK and K¯ such that,
(i) for anyφ,ψ∈Cμ and t∈[t0,T], it follows that(15)fφ,t-fψ,t2∨gφ,t-gψ,t2≤K¯φ-ψμ2,
(ii) for anyt∈[t0,T], it follows that f(0,t),g(0,t)∈L2(Cμ) such that(16)f0,t2∨g0,t2≤K.Then, problem (1), with initial data xt0=ξ∈M2((-∞,0];Rn), has a unique solution x(t). Moreover, x(t)∈M2((-∞,T];Rn).Lemma 6.
Let (15) and (16) hold. If x(t) is the solution of (1) with initial data xt0=ξ, then(17)Esupt0≤t≤Txt2≤Ce6K¯T-t0+1T-t0, where C=3E|ξ|μ2+6K(T-t0+1)(T-t0)+6K¯(T-t0+1)(T-t0)E|ξ|μ2.
Moreover, ifξ∈M2((-∞,0];Rn), then x(t)∈M2((-∞,T];Rn).Proof.
For each numberq≥1, define the stopping time (18)τq=T∧inft∈t0,T:xtμ≥q. Obviously, as q→∞,τq↗T a.s. Let xq(t)=x(t∧τq),t∈[t0,T], and then xq(t) satisfy the following equation: (19)xqt=ξ0+∫t0tfxsq,sIt0,τqsds+∫t0tgxsq,sIt0,τqsdWs. Using the elementary inequality (a+b+c)2≤3(a2+b2+c2), we get(20)xqt2≤3ξμ2+3∫t0tfxsq,sIt0,τqsds2+3∫t0tgxsq,sIt0,τqsdWs2. Taking the expectation on both sides and using the Hölder inequality, Lemma 3, and (15) and (16), we get for all t in [t0,T](21)Exqt2≤3Eξμ2+3E∫t0tfxsq,sIt0,τqsds2+3E∫t0tgxsq,sIt0,τqsdWs2≤3Eξμ2+3t-t0E∫t0tfxsq,s2ds+3E∫t0tgxsq,s2ds≤3Eξμ2+3t-t0E∫t0tfxsq,s-f0,s+f0,s2ds+3E∫t0tgxsq,s-g0,s+g0,s2ds≤3Eξμ2+6t-t0E∫t0tfxsq,s-f0,s2ds+6E∫t0tgxsq,s-g0,s2ds+6t-t0E∫t0tf0,s2ds+6E∫t0tg0,s2ds≤3Eξμ2+6t-t0E∫t0tfxsq,s-f0,s2ds+6E∫t0tgxsq,s-g0,s2ds+6t-t0E∫t0tf0,s2ds+6E∫t0tg0,s2ds+6E∫t0tfxsq,s-f0,s2ds+6t-t0E∫t0tgxsq,s-g0,s2ds+6E∫t0tf0,s2ds+6t-t0E∫t0tg0,s2ds≤3Eξμ2+6t-t0+1E∫t0tfxsq,s-f0,s2ds+6t-t0+1E∫t0tgxsq,s-g0,s2ds+6t-t0+1E∫t0tf0,s2ds+6t-t0+1E∫t0tg0,s2ds≤3Eξμ2+6t-t0+1E∫t0tfxsq,s-f0,s2+gxsq,s-g0,s2+f0,s2+g0,s2ds≤3Eξμ2+6t-t0+1E∫t0tK¯xsqμ2+Kds≤3Eξμ2+6Kt-t0+1t-t0+6K¯t-t0+1E∫t0txsqμ2ds.We have also for each t in [t0,T](22)supt0≤s≤txsqμ2=supt0≤s≤tsup-∞<θ≤0e2μθxqs+θ2=supt0≤s≤tsup-∞<r≤se2μr-sxqr2≤supt0≤s≤te-2μssup-∞<r≤se2μrxqr2≤supt0≤s≤te-2μssup-∞<r≤t0e2μrxqr2∨supt0≤r≤se2μrxqr2≤supt0≤s≤te-2μssup-∞<r-t0≤0e2μrxqr-t0+t02∨supt0≤r≤se2μrxqr2≤supt0≤s≤te-2μssup-∞<r-t0≤0e2μrxt0qr-t02∨supt0≤r≤se2μrxqr2≤supt0≤s≤te-2μse2μt0ξμ2∨supt0≤r≤se2μrxqr2≤supt0≤s≤te-2μse2μt0ξμ2∨supt0≤r≤se2μsxqr2≤e2μt0-sξμ2+supt0≤r≤txqr2≤ξμ2+supt0≤r≤txqr2.Letting t=T, we get(23)Esupt0≤s≤Txqs2≤3Eξμ2+6KT-t0+1T-t0+6K¯T-t0+1E∫t0Tξμ2+supt0≤r≤Txqr2dr≤C+6K¯T-t0+1∫t0TEsupt0≤r≤Txqr2dr, where C=3E|ξ|μ2+6K(T-t0+1)(T-t0)+6K¯(T-t0+1)(T-t0)E|ξ|μ2.
By the Gronwall inequality, we infer(24)Esupt0≤s≤Txqs2≤Ce6K¯T-t0+1T-t0.That is,(25)Esupt0≤s≤Txs∧τq2≤Ce6K¯T-t0+1T-t0. Consequently(26)Esupt0≤s≤τqxs2≤Ce6K¯T-t0+1T-t0. Letting q→∞, that implies the following inequality(27)Esupt0≤s≤Txs2≤Ce6K¯T-t0+1T-t0.Now, to prove the second part or the lemma, suppose that ξ∈M2((-∞,0];Rn). Then(28)Esup-∞≤t≤Txt2=Esup-∞≤t≤t0xt2+Esupt0≤t≤Txt2≤Esup-∞≤t≤t0xt2+Ce6K¯T-t0+1T-t0≤Esup-∞≤t-t0≤0xt-t0+t02+Ce6K¯T-t0+1T-t0≤Esup-∞≤s≤0xt0s2+Ce6K¯T-t0+1T-t0≤Eξ2+Ce6K¯T-t0+1T-t0<∞.
The demonstration of the lemma is complete.Proof of Theorem5.
We begin by checking uniqueness of solution. Letx(t) and x¯(t) be two solutions of (1), by Lemma 6  x(t) and x¯(t)∈M2((-∞,T];Rn). Note that(29)xt-x¯t=∫t0tfxs,s-fx¯s,sds+∫t0tgxs,s-gx¯s,sdWs. By the elementary inequality, (a+b)2≤2(a2+b2), one then gets(30)xt-x¯t2≤2∫t0tfxs,s-fx¯s,sds2+2∫t0tgxs,s-gx¯s,sdWs2. By Hölder inequality, Lemma 3, and (15) and (16), we have(31)Ext-x¯t2≤2t-t0E∫t0tfxs,s-fx¯s,s2ds+2E∫t0tgxs,s-gx¯s,s2ds≤2K¯t-t0E∫t0txs-x¯sμ2ds+2K¯E∫t0txs-x¯sμ2ds≤2K¯t-t0+1E∫t0txs-x¯sμ2ds. From the fact xt0(s)=x¯t0(s)=ξ(s),s∈(-∞,0], and(32)supt0≤s≤txs-x¯sμ2=supt0≤s≤tsup-∞<θ≤0e2μθxs+θ-x¯s+θ2=supt0≤s≤tsup-∞<r≤se2μr-sxr-x¯r2≤supt0≤s≤te-2μssup-∞<r≤se2μrxr-x¯r2≤supt0≤s≤te-2μssup-∞<θ≤0e2μθ+t0xt0θ-x¯t0θ2∨supt0≤r≤se2μrxr-x¯r2≤supt0≤s≤te-2μse2t0μξ-ξμ2∨supt0≤r≤se2μrxr-x¯r2≤supt0≤s≤te-2μssupt0≤r≤se2μsxr-x¯r2≤supt0≤r≤txr-x¯r2. We have(33)Esupt0≤s≤txs-x¯s2≤2K¯t-t0+1∫t0tEsupt0≤r≤sxr-x¯r2ds. Applying the Gronwall inequality yields(34)Ext-x¯t2=0,t0≤t≤T. The above expression means that x(t)=x¯(t) a.s. for t0≤t≤T. Therefore, for all -∞<t≤T,x(t)=x¯(t) a.s., the proof of uniqueness is complete.Next, to check the existence, definext00=ξ and x0(t)=ξ(0) for t0≤t≤T. Let xt0k=ξ,k=1,2,…, and define Picard sequence(35)xkt=ξ0+∫t0tfxsk-1,sds+∫t0tgxsk-1,sdWs,t0≤t≤T.Obviously x0(t)∈M2((-∞,T];Rn). By induction, we can see that xk(t)∈M2((-∞,T];Rn).In fact, by elementary inequality(a+b+c)2≤3(a2+b2+c2)(36)xkt2≤3ξ02+3∫t0tfxsk-1,sds2+3∫t0tgxsk-1,sdWs2. From the Hölder inequality and Lemma 3, we have(37)Exkt2≤3Eξμ2+3t-t0E∫t0tfxsk-1,s-f0,s+f0,s2ds+3E∫t0tgxsk-1,s-g0,s+g0,s2ds. Again the elementary inequality (a+b)2≤2a2+2b2, (22), (15), and (16) imply that(38)Exkt2≤3Eξμ2+3t-t0+1E∫t0t2K¯xsk-1μ2+2Kds≤3Eξμ2+6Kt-t0+1E∫t0tds+6K¯t-t0+1E∫t0txsk-1μ2ds≤C1+6K¯t-t0+1E∫t0txsk-1μ2ds≤C1+6K¯t-t0+1E∫t0tξμ2+supt0≤r≤sxk-1r2ds≤C1+C2+6K¯t-t0+1∫t0tEsupt0≤r≤sxk-1r2ds, where C1=3E|ξ|μ2+6K(t-t0+1)(t-t0) and C2=6K¯(t-t0+1)(t-t0)E|ξ|μ2.Hence, for anyl≥1, one can derive that(39)max1≤k≤l⁡Esupt0≤s≤txks2≤C1+C2+6K¯t-t0+1∫t0tmax1≤k≤l⁡Esupt0≤r≤sxk-1r2ds. Note that(40)max1≤k≤lExk-1s2=maxEξ02,Ex1s2,…,Exl-1s2≤maxEξμ2,Ex1s2,…,Exl-1s2,Exls2=maxEξμ2,max1≤k≤lExks2≤Eξμ2+max1≤k≤lExks2, and then(41)max1≤k≤l⁡Esupt0≤s≤txks2≤C1+C2+6K¯t-t0+1∫t0tEξμ2+max1≤k≤l⁡Esupt0≤r≤sxkr2ds≤C3+6K¯t-t0+1∫t0tmax1≤k≤l⁡Esupt0≤r≤sxkr2ds, where C3=C1+2C2.From the Gronwall inequality, we have(42)max1≤k≤l⁡Exkt2≤C3e6K¯T-t0+1T-t0. Since k is arbitrary(43)Exkt2≤C3e6K¯T-t0+1T-t0t0≤t≤T,k≥1.From the Hölder inequality, Lemma 3, and (15) and (16), as in a similar earlier inequality, one then has(44)Ex1t-x0t2≤2E∫t0tfxs0,sds2+2E∫t0tgxs0,sdWs2≤2t-t0E∫t0tfxs0,s2ds+2E∫t0tgxs0,s2ds≤2t-t0+1E∫t0t2K¯xs0μ2+2Kds≤4Kt-t0+1t-t0+4K¯t-t0+1t-t0Eξμ2. That is,(45)Esupt0≤s≤tx1t-x0t2≤4t-t0+1t-t0K+K¯Eξμ2≔R.By similar arguments as above, we also have(46)Ex2t-x1t2≤2E∫t0tfxs1,s-fxs0,sds2+2E∫t0tgxs1,s-gxs0,sdWs2≤2t-t0E∫t0tfxs1,s-fxs0,s2ds+2E∫t0tgxs1,s-gxs0,s2ds≤2t-t0+1E∫t0tfxs1,s-fxs0,s2+gxs1,s-gxs0,s2ds≤2K¯t-t0+1E∫t0txs1-xs0μ2ds. Then(47)Esupt0≤s≤tx2s-x1s2≤2K¯t-t0+1∫t0tEsupt0≤r≤sx1r-x0r2ds≤2RK¯t-t0+1t-t0=RMt-t0, where M=2K¯(t-t0+1). Similarly(48)Esupt0≤s≤tx3s-x2s2≤M∫t0tEsupt0≤r≤sx2r-x1r2ds≤M∫t0tRMs-t0ds≤RMt-t022. Continue this process to find that (49)Esupt0≤s≤tx4s-x3s2≤M∫t0tEsupt0≤r≤sx3r-x2r2ds≤M∫t0tRMs-t022ds≤RMt-t036. Now we claim that for any k≥0(50)Esupt0≤s≤txk+1s-xks2≤RMt-t0kk!,t0≤t≤T.So, for k=0,1,2,3, inequality (50) holds. We suppose that (50) holds for some k, and check (50) for k+1. In fact(51)Esupt0≤s≤txk+2s-xk+1s2≤2K¯t-t0+1∫t0tEsupt0≤s≤txsk+1-xskμ2ds≤M∫t0tsupt0≤r≤sxk+1r-xkr2ds. From (50)(52)Esupt0≤s≤txk+2s-xk+1s2≤M∫t0tRMs-t0kk!ds≤RMt-t0k+1k+1!, which means that (50) holds for k+1. Therefore, by induction (50) holds for any k≥0.Next to verify{xk(t)} converge to x(t) in L2 with x(t) in M2((-∞,T];Rn) and x(t) is the solution of (1) with initial data xt0. For (50), taking t=T, then(53)Esupt0≤t≤Txk+1t-xkt2≤RMT-t0kk!. By the Chebyshev inequality(54)Psupt0≤t≤Txk+1t-xkt>12k≤R4MT-t0kk!. By using Alembert’s rule, we show that ∑k=0+∞‍R[4M(T-t0)]k/k!converge.That is,∑k=0+∞‍R[4M(T-t0)]k/k!<∞, and by Borel-Cantelli’s lemma, for almost all ω∈Ω, there exists a positive integer k0=k0(ω) such that(55)supt0≤t≤Txk+1t-xkt≤12kask≥k0, and then, {xk(t)} is also a Cauchy sequence in L2. Hence, {xk(t)} converges uniformly and let x(t) be its limit for any t∈(-∞,T]; since xk(t) is continuous on t∈(-∞,T] and Ft adapted, x(t) is also continuous and Ft adapted.So, ask→+∞, xk(t)→x(t) in L2. That is, E|xk(t)-x(t)|2→0 as k→∞.Then from (43)(56)Ext2≤C3e6K¯T-t0+1T-t0∀t0≤t≤T, and therefore(57)E∫-∞Txs2ds=E∫-∞t0xs2ds+E∫t0Txs2ds≤E∫-∞0ξs2ds+∫t0TC3e6K¯T-t0+1T-t0ds<∞. That is, x(t)∈M2((-∞,T];Rn).Now, to show thatx(t) satisfies (1)(58)E∫t0tfxsk,s-fxs,sds+∫t0tgxsk,s-gxs,sdWs2≤2E∫t0tfxsk,s-fxs,sds2+2E∫t0tgxsk,s-gxs,sdWs2≤2t-t0E∫t0tfxsk,s-fxs,s2ds+2E∫t0tgxsk,s-gxs,s2ds≤ME∫t0txsk-xsμ2ds≤M∫t0tEsupt0≤r≤sxkr-xr2ds≤M∫t0TExks-xs2ds. Noting that the sequence {xk}→x(t) means that for any given ε>0 there exists k0 such that k≥k0, for any t∈(-∞,T], one then deduces that (59)Exkt-xt2<ε,∫t0TExkt-xt2ds<T-t0ε, which means that, for any t∈[t0,T], one has(60)∫t0tfxsk,sds⟶∫t0tfxs,sds,∫t0tgxsk,sdWs⟶∫t0tgxs,sdWsinL2. For t0≤t≤T, taking limits on both sides of (35), we deduce that(61)limk→∞xkt=ξ0+limk→∞∫t0tfxsk-1,sds+limk→∞∫t0tgxsk-1,sdWs, and consequently(62)xt=ξ0+∫t0tfxs,sds+∫t0tgxs,sdWst0≤t≤T. Finally, x(t) is the solution of (1), and the demonstration of existence is complete.



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