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https://doi.org/10.14317/jami.2019.053

AN EXISTENCE OF THE SOLUTION TO NEUTRAL STOCHASTIC FUNCTIONAL DIFFERENTIAL EQUATIONS

UNDER SPECIAL CONDITIONS

YOUNG-HO KIM

Abstract. In this paper, we show the existence of solution of the neutral stochastic functional differential equations under non-Lipschitz condition, a weakened linear growth condition and a contractive condition. Further- more, in order to obtain the existence of solution to the equation we used the Picard sequence.

AMS Mathematics Subject Classification : 60H05, 60H10, 60H20.

Key words and phrases : Gronwall’s inequality, Bihari’s inequality, non- Lipschitz condition, Existence and Uniqueness, Neutral stochastic func- tional differential equations.

1. Introduction

Together with the development of science systems, usually using the ordinary differential equations to describe the trajectory of systems with phenomenon of time delay is have difference in real measured trajectory. Moreover, we can not ignore the effect of the science systems with time delay.

So, we need an another class of stochastic equations depending on past and present values but that involves derivatives with delays as well as the function itself. Such equations historically have been referred to as neutral stochastic functional differential equations, or neutral stochastic differential delay equations (see, [2], [3], [4], [6], [8], [9]). Such equations are more difficult to motivate but often arise in the study of two or more simple electrodynamics or oscillatory systems with some interconnections between them.

For example, In studying the collision problem in electrodynamics, Diver [1]

considered the system of neutral type

˙

x(t) = f1(x(t), x(δ(t))) + f2(x(t), x(δ(t))) ˙x(δ(t)),

Received August 6, 2018. Revised October 5, 2018. Accepted October 10, 2018.

This research was supported by Changwon National University in 2017-2018 2019 Korean SIGCAM and KSCAMc .

53

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where δ(t) ≤ t. Generally, a neutral functional differential equation has the form d

dt[x(t) − D(xt)] = f (xt, t).

Taking into account stochastic perturbations, we are led to a neutral stochastic functional differential equation

d[x(t) − D(xt)] = f (xt, t)dt + g(xt, t)dB(t) (1) Neutral stochastic functional differential equations(NSDEs) are known to model problems from several areas of science and engineering. For instance, in 2007, Mao [8] published the stochastic differential equations and applications, in 2010, Li and Fu [7] considered the stability analysis of stochastic functional differential equations with infinite delay and its application to recurrent neural networks, in 2013, Wei et al. [10] considered the existence and uniqueness of the solution to following neutral stochastic functional differential equations with in- finite delay. Also, Kim [5] considered the solution to following neutral stochastic functional differential equations with infinite delay

d[x(t) − G(t, xt)] = f (t, xt)dt + g(t, xt)dB(t), (2) where xt= {x(t + θ) : −∞ < θ ≤ 0}.

Motivated by [5], [10], one of the objectives of this paper is to get one proof to existence theorem for given NSDEs. The other objective of this paper is to estimate on how fast the Picard iterations xn(t) convergence the unique solution x(t) of the neutral NSDEs.

2. Preliminary

Let | · | denote Euclidean norm in Rn. If A is a vector or a matrix, its trans- pose is denoted by AT; if A is a matrix, its trace norm is represented by |A| = ptrace(ATA). And BC((−∞, 0]; Rd) denote the family of bounded continuous Rd-value functions ϕ defined on (−∞, 0] with norm kϕk = sup−∞<θ≤0|ϕ(θ)|.

M2((−∞, T ]; Rd) denote the family of all Rd-valued measurable Ft-adapted pro- cess ψ(t) = ψ(t, w), t ∈ (−∞, T ] such that ERT

−∞|ψ(t)|2dt < ∞.

Let t0 be a positive constant and (Ω, F , P ), throughout this paper unless otherwise specified, be a complete probability space with a filtration {Ft}t≥t0

satisfying the usual conditions (i.e. it is right continuous and Ft0 contains all P -null sets).

Let B(t) is a m-dimensional Brownian motion defined on complete probability space, that is B(t) = (B1(t), B2(t), ..., Bm(t))T.

For 0 ≤ t0≤ T < ∞, let f : [t0, T ] × BC((−∞, 0]; Rd) → Rd and g : [t0, T ] × BC((−∞, 0]; Rd) → Rd×m are Borel measurable mapping and G : [t0, T ] × BC((−∞, 0]; Rd) → Rdbe continuous mapping.

With all the above preparation, consider the following d-dimensional neutral SFDEs:

d[x(t) − G(t, xt)] = f (t, xt)dt + g(t, xt)dB(t), t0≤ t ≤ T, (3)

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where xt= {x(t + θ) : −∞ < θ ≤ 0} can be considered as a BC((−∞, 0]; Rd)- value stochastic process. The initial value of the system (3)

xt0 = ξ = {ξ(θ) : −∞ < θ ≤ 0} (4) is an Ft0measurable, BC((−∞, 0]; Rd)-value random variable such that ξ ∈ M2((−∞, 0]; Rd).

To be more precise, we give the definition of the solution of the equation (3) with initial data (4).

Definition 2.1. [10] Rd-value stochastic process x(t) defined on −∞ < t ≤ T is called the solution of (3) with initial data (4), if x(t) has the following properties:

(i) x(t) is continuous and {x(t)}t0≤t≤T is Ft-adapted;

(ii) {f (t, xt)} ∈ L1([t0, T ]; Rd) and {g(t, xt)} ∈ L2([t0, T ]; Rd×m) ; (iii) xt0= ξ, for each t0≤ t ≤ T,

x(t) = ξ(0) + G(t, xt) − G(t0, ξ) + Z t

t0

f (s, xs)ds + Z t

t0

g(s, xs)dB(s) a.s. (5) The x(t) is called as a unique solution, if any other solution x(t) is distinguishable with x(t), that is

P {x(t) = x(t), for any − ∞ < t ≤ T } = 1.

The following lemmas are known as special name for stochastic integrals which was appear in [8] and will play an important role in next section.

Lemma 2.2. (H¨older’s inequality) [8] If 1p+1q = 1 for any p, q > 1, f ∈ Lp, and g ∈ Lq, then f g ∈ L1 andRb

a f g dx ≤ (Rb

a|f |p dx)1/p(Rb

a |g|q dx)1/q.

Lemma 2.3. (Gronwall’s inequality) [8] Let u(t) and b(t) be nonnegative con- tinuous functions for t ≥ α, and let u(t) ≤ a +Rt

αb(s)u(s)ds, t ≥ α, where a ≥ 0 is a constant. Then

u(t) ≤ a exp

Z t α

b(s)ds



, t ≥ α.

Lemma 2.4. (Bihari’s inequality) [8] Let T ≥ 0 and u0≥ 0, let u(t) and b(t) be continuous functions on [0, T ]. Let κ(·) : R+ → R+ is a concave nondecreasing function such that κ(0) = 0, κ(u) > 0 for all u > 0. If u(t) ≤ a+Rt

αb(s)κ(u(s))ds, for all 0 ≤ t ≤ T. Then

u(t) ≤ G−1



G(u0) + Z T

t

v(s)ds



for all t ∈ [0, T ] such that G(u0) +RT

t v(s)ds ∈ Dom(G−1), where G(r) = Rr

1 1

k(s)ds, r ≥ 0 and G−1 is the inverse function of G.

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Lemma 2.5. (Moment inequality) [8] If p ≥ 2, g ∈ M2([0, T ]; Rd×m) such that ERT

0 |g(s)|pds < ∞, then E

 sup

0≤t≤T

Z t 0

g(s) dB(s)

p

 p3 2(p − 1)

p2 Tp−22 E

Z T 0

|g(s)|pds.

In order to attain the solution of equation (3) with initial data (4), we propose the following assumptions:

(H1) (non-uniform Lipschitz condition) For any ϕ, ψ ∈ BC((−∞, 0]; Rd) and t ∈ [t0, T ], we assume that

|f (t, ϕ) − f (t, ψ)|2∨ |g(t, ϕ) − g(t, ψ)|2≤ κ(kϕ − ψk2),

where κ(·) : R+ → R+ is a concave continuous nondecreasing function such that κ(0) = 0, κ(u) > 0 for all u > 0 andR

0+ 1

k(u)du = ∞.

(H2) (weakened linear growth condition) For any t ∈ [t0, T ], it follows that f (t, 0), g(t, 0) ∈ L2such that

|f (t, 0)|2∨ |g(t, 0)|2≤ K, where K is a positive constant.

(H3) (contractive condition) Assuming that there exists a positive number K0 such that 0 < K0 < 1 and for any ϕ, ψ ∈ BC((−∞, 0]; Rd) and t ∈ [t0, T ], it follows that

|G(t, ϕ) − G(t, ψ)| ≤ K0(kϕ − ψk).

3. Main results

In order to obtain the existence of solutions to neutral SFDEs, let x0t0 = ξ and x0(t) = ξ(0), for t0 ≤ t ≤ T. For each n = 1, 2, . . . , set xnt0 = ξ and define the following Picard sequence

xn(t) − ξ(0)

= G(t, xn−1t ) − G(t0, xn−1t0 ) + Z t

t0

f (s, xn−1s )ds + Z t

t0

g(s, xn−1s )dB(s). (6) Now we give the existence theorem to the solution of equation (3) with initial data (4) by approximate solutions by means of Picard sequence.

Theorem 3.1. Assume that (H1)-(H3) hold. Then, there exists a unique so- lution to the neutral SFDEs (3) with initial data (4). Moreover, the solution belongs to M2((−∞, T ]; Rd).

We prepare a lemma in order to prove this theorem.

Lemma 3.2. Let the assumption (H1) and (H3) hold. If x(t) is a solution of equation (2.1) with initial data (2.2), then

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E sup

−∞<t≤T

|x(s ∧ τn)|2



c1(α + K)(T − t0) + 4 (1 −√

K0)(1 − K0)Ekξk2



exp(c1β(T − t0)), where α and β are positive constants, c1= 6(T − t0+ 4)/(1 −√

K0)(1 − K0). In particular, x(t) belong to M2((−∞, T ]; Rd).

Proof. For each number n ≥ 1, define the stopping time τn= T ∧ inf{t ∈ [t0, T ] : kx(t)k ≥ n}.

Obviously, as n → ∞, τn ↑ T a.s. Let xn(t) = x(t ∧ τn), t ∈ (−∞, T ]. Then, for t0≤ t ≤ T, xn(t) satisfy the following equation

xn(t) = G(t, xnt) − G(t0, xnt0) + Jn(t), where

Jn(t) = ξ(0) + Z t

t0

f (s, xns)I[t0n](s) ds + Z t

t0

g(s, xns)I[t0n](s) dB(s).

Applying the elementary inequality (a+b)2aα2+1−αb2 when a, b > 0, 0 < α < 1, we have

|xn(t)|2 ≤ 1 K0

|G(t, xnt) − G(t0, xnt0)|2+ 1 1 − K0

|Jn(t)|2

≤ p

K0kxntk2+ K0 1 −√

K0

kξk2+ 1 1 − K0

|Jn(t)|2,

where condition (H3) has also been used. Taking the expectation on both sides, one sees that

E sup

t0<s≤t

|xn(s)|2

≤p K0E

sup

−∞<s≤t

|xn(s)|2

+ K0 1 −√

K0

Ekξk2+ 1 1 − K0

E sup

t0≤s≤t

|Jn(s)|2 . Noting that sup−∞<s≤t|xn(s)|2≤ kξk2+ supt

0≤s≤t|xn(s)|2, we get E

sup

−∞<s≤t

|xn(s)|2

≤p K0E

sup

−∞<s≤t

|xn(s)|2

+ K0

1 −√

K0Ekξk2+ 1 1 − K0E

sup

t0≤s≤t

|Jn(s)|2 . Consequently

E sup

−∞<s≤t

|xn(s)|2

≤ 1

(1 −√

K0)2Ekξk2+ 1 (1 −√

K0)(1 − K0)E sup

t0≤s≤t

|Jn(s)|2 . (7)

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On the other hand, by the elementary inequality (a + b + c)2≤ 3a2+ 3b2+ 3c2, one can show that

|Jn(s)|2≤ 3



Ekξk2+

Z t t0

|f (s, xns)|2ds

2

+

Z s t0

g(r, xnr) dB(r)

2 . By H¨older’s inequality and Lemma (2.5), one can show that

E sup

t0≤s≤t

|Jn(s)|2

≤ 3



Ekξk2+ (T − t0) Z t

t0

E|f (s, xns)|2ds + 4 Z s

t0

E|g(s, xns)|2ds

 . By the condition (H2), one can show that

E sup

t0≤s≤t

|Jn(s)|2

≤ 3Ekξk2+ 6(T − t0+ 4) Z t

t0

E(κ(kxnsk2) + K)ds.

Since κ(·) is concave and κ(0) = 0, we can find a positive constants α and β such that κ(u) ≤ α + βu for all u ≥ 0. Therefore, we have

E sup

t0≤s≤t

|Jn(s)|2

≤ 3Ekξk2+ 6(T − t0+ 4)(α + K)(T − t0) + 6β(T − t0+ 4) Z t

t0

Ekxnsk2ds.

Substituting this into (7) yields that E

sup

−∞<s≤t

|xn(s)|2

≤ c1(α + K)(T − t0) + 4 (1 −√

K0)(1 − K0)Ekξk2+ c1β Z t

t0

Ekxnsk2ds, where c1= 6(T − t0+ 4)/(1 −√

K0)(1 − K0). Therefore, we have E

sup

−∞<s≤t

|xn(s)|2

≤ c1(α + K)(T − t0) + 4 (1 −√

K0)(1 − K0)Ekξk2 (8) +c1β

Z t t0

sup

−∞<r≤s

E|xn(r)|2dr.

The Gronwall’s inequality then yields that E

sup

−∞<s≤t

|xn(s)|2



c1(α + K)(T − t0) + 4 (1 −√

K0)(1 − K0)Ekξk2



exp(c1β(T − t0)).

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For all n = 0, 1, 2, . . . , we deduce that E

sup

−∞<s≤t

|x(s ∧ τn)|2



c1(α + K)(T − t0) + 4 (1 −√

K0)(1 − K0)Ekξk2



exp(c1β(T − t0)).

Consequently the required inequality follows by letting n → ∞.  Proof of Theorem 3.1. To check the uniqueness, let x(t) and x(t) be any two solutions of (3) with initial data (4). By Lemma 3.2, x(t), x(t) ∈ M2((−∞, T ]; Rd). Note that

x(t) − x(t) = G(t, xt) − G(t, xt) + J (t), where J (t) =Rt

t0[f (s, xs) − f (s, xs)]ds +Rt

t0[g(s, xs) − g(s, xs)]dB(s). One then gets

|x(t) − x(t)|2≤ 1 K0

|G(t, xt) − G(t, xt)|2+ 1 1 − K0

|J (t)|2, where 0 < K0< 1. We derive that

|x(t) − x(t)|2≤ K0kxt− xtk2+ 1 1 − K0

|J (s)|2. Therefore

E sup

t0≤s≤t

|x(s) − x(s)|2

≤ K0E

supt0≤s≤t|x(s) − x(s)|2 +(1−K1

0)E

supt0≤s≤t|J (s)|2 . Consequently

E sup

t0≤s≤t

|x(t) − x(t)|2

≤ 1

(1 − K0)2E sup

t0≤s≤t

|J (s)|2 . On the other hand, one can show that

E sup

t0≤s≤t

|J (s)|2

≤ 2



(T − t0)E Z t

t0

|f (s, xs) − f (s, xs)|2ds + 4E Z t

t0

|g(s, xs) − g(s, xs)|2ds



≤ 2(T − t0+ 4) Z t

t0

Eκ(kxs− xsk2) ds.

For any ε > 0, by the Jensen Inequality of the continuous function κ, this yields that

E sup

t0≤s≤t

|J (s)|2

≤ ε + 2(T − t0+ 4) Z t

t0

κ(E sup

t0<r≤s

|x(r) − x(r)|2) ds.

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By the Bihari’s inequality, this yields that for sufficiently small ε > 0 E

sup

t0≤s≤t

|J (s)|2

≤ G−1(G(ε) + 2(T − t0+ 4)(T − t0)), where G(r) =Rr

0 1

κ(u)du on r > 0 and G−1(·) is the inverse function of G(·). By assumptionR

0+ 1

κ(u)du = ∞ and the definition of κ(·) on see that lim↓0G() =

∞ and then lim

↓0G−1(G(ε) + 2(T − t0+ 4)(T − t0)) = 0.

Therefore, by letting  → 0, we have E supt0≤s≤t|J (s)|2= 0. This implies that E

sup

t0<s≤t

|x(t) − x(t)|2

= 0.

Hence, we get x(t) = x(t) for t0≤ t ≤ T a.s. The uniqueness has been proved.

Now we check the existence of the solution using the Picard sequence (6).

Obviously, from the Picard iterations, we have x0(t) ∈ M2([t0, T ] : Rd). More- over, one can show the boundedness of the sequence {xn(t), n ≥ 0} that xn(t) ∈ M2((−∞, T ] : Rd), in fact

xn(t) = G(t, xn−1t ) − G(t0, xn−1t0 ) + Jn−1(t), where

Jn−1(t) = ξ(0) + Z t

t0

f (s, xn−1s )ds + Z t

t0

g(s, xn−1s )dB(s).

Applying the elementary inequality (a+b)2aα2+1−αb2 when a, b > 0, 0 < α < 1, we have

|xn(t)|2 ≤ 1 K0

|G(t, xn−1t ) − G(t0, ξ)|2+ 1 1 − K0

|Jn−1(t)|2

≤ p

K0kxn−1t k2+ K0 1 −√

K0

kξk2+ 1 1 − K0

|Jn−1(t)|2, where condition (H3) has also been used. Taking the expectation on both sides, one sees that

E sup

t0≤s≤t

|xn(s)|2

−p

K0E sup

t0≤s≤t

|xn−1(s)|2

≤ K0

1 −√

K0Ekξk2+ 1 1 − K0E

sup

t0≤s≤t

|Jn−1(s)|2

. (9)

On the other hand, by elementary inequality, H¨older’s inequality and moment inequality, one can show that

E sup

t0≤s≤t

|Jn−1(s)|2

≤ 3



Ekξk2+ (T − t0)E Z t

t0

|f (s, xn−1s )|2ds + 4E Z t

t0

|g(s, xn−1s )|2ds



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≤ 3Ekξk2+ 6(T − t0+ 4)E Z t

t0

(κ(kxn−1s k2) + K) ds.

Since κ(·) is concave and κ(0) = 0, we can find a positive constants α and β such that κ(u) ≤ α + βu for all u ≥ 0. Therefore, we have

E sup

t0≤s≤t

|Jn−1(s)|2

≤ 3Ekξk2+ γ1+ 6β(T − t0+ 4) Z t

t0

Ekxn−1s k2ds, where γ1= 6(T − t0+ 4)(T − t0)(α + K). Substituting this into (9) yields that

E sup

t0≤s≤t

|xn(s)|2

≤ c2+p

K0E sup

t0≤s≤t

|xn−1(s)|2+6β(T − t0+ 4) 1 − K0

Z t t0

E sup

t0≤r≤s

|xn−1(r)|2dr, where c2 = 1−Kγ1

0 +4+6β(T −t0+4)(T −t0)

(1−

K0)(1−K0) Ekξk2. It also follows note that for any k ≥ 1,

max

1≤n≤kE

sup |xn−1(u)|2

= maxn

Ekξk2, E(sup |x1(u)|2), . . . , E(sup |xk−1(u)|2)o

≤ maxn

Ekξk2, E(sup |x1(u)|2), . . . , E(sup |xk−1(u)|2), E(sup |xk(u)|2)o

≤ Ekξk2+ max

1≤n≤kE(sup |xn(u)|2).

Therefore, one can derive that max

1≤n≤kE sup

t0≤s≤t

|xn(s)|2

≤ c3+ 6β(T − t0+ 4) (1 −√

K0)(1 − K0) Z t

t0

max

1≤n≤kE sup

t0≤r≤s

|xn(r)|2 dr, where c3 = 1−c2K

0 + 1+6β(T −t(1−K0+4)(T −t0)

0)(1−K0) Ekξk2. By Gronwall’s inequality, we have

max

1≤n≤kE sup

t0≤s≤t

|xn(s)|2

≤ c3exp6β(T − t0+ 4)(T − t0) (1 −√

K0)(1 − K0)

 . Since k is arbitrary, for all n = 0, 1, 2, . . . , we deduce that

E sup

t0≤s≤t

|xn(s)|2

≤ c3exp6β(T − t0+ 4)(T − t0) (1 −√

K0)(1 − K0)

 , which shows the boundedness of the sequence {xn(t), n ≥ 0}.

Next, we check that the sequence {xn(t)} is Cauchy sequence. For all n ≥ 0 and t0≤ t ≤ T, we have

xn+1(t) − xn(t) = G(t, xnt) − G(t, xn−1t )

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+ Z t

t0

[f (s, xns) − f (s, xn−1s )]ds + Z t

t0

[g(s, xns) − g(s, xn−1s )]dB(s).

Using an elementary inequality (u + v)21αu2+1−α1 v2and the condition (H3), we derive that

E sup

t0<s≤t

|xn+1(s) − xn(s)|2

≤ K0E sup

t0<s≤t

|xn(s) − xn−1(s)|2

+2(T − t0+ 4) 1 − K0

E Z t

t0

κ sup

t0≤r≤s

|xn(r) − xn−1(r)|2 ds.

This yields that max

1≤n≤kE sup

t0<s≤t

|xn+1(s) − xn(s)|2

≤ 2(T − t0+ 4) (1 − K0)2

Z t t0

κ max

1≤n≤kE sup

t0≤u≤s

|xn+1(u) − xn(u)|2

ds.

Let Z(t) = lim supn,m→∞max1≤n≤kE

supt0≤s≤t|xn+1(s) − xn(s)|2

, we get

Z(t) ≤  +2(T − t0+ 4) (1 − K0)2

Z t t0

κ(Z(s)) ds.

By Bihari’s inequality, we get Z(t) = 0. This shows the sequence {xn(t), n ≥ 0}

is a Cauchy sequence in L2. Hence, as n → ∞, xn(t) → x(t), that is E|xn(t) − x(t)|2 → 0. Therefore, we obtain that x(t) ∈ M2((−∞, T ]; Rd). Now to show that x(t) satisfy (5).

E

Z t t0

[f (s, xns) − f (s, xs)]ds + Z t

t0

[g(s, xns) − g(s, xs)] dB(s)

2

≤ 2



(T − t0)E Z t

t0

|f (s, xns) − f (s, xs)|2ds + 4E Z t

t0

|g(s, xns) − g(s, xs)|2ds



≤ 2(T − t0+ 4) Z t

t0

κ E

sup

t0≤u≤s

|xn(u) − x(u)|2

ds.

Noting that sequence xn(t) is uniformly converge on (−∞, T ], it means that E

sup

t0≤u≤s

|xn(u) − x(u)|2

→ 0

as n → ∞, further κ(E(supt0≤u≤s|xn(u) − x(u)|2)) → 0 as n → ∞. Hence, taking limits on both sides in the Picard sequence, we obtain that

x(t) = ξ(0) + G(t, xt) − G(t0, xt0) + Z t

t0

f (s, xs)ds + Z t

t0

g(s, xs)dB(s).

The above expression demonstrates that x(t) is a solution of equation (3) satis- fying the initial condition (4). So far, the existence of theorem is complete. 

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Remark 3.1. In the proof of Theorem 3.1 we have shown that the Picard iterations xn(t) converge to the unique solution x(t) of equation (3). In the next study, we should gives an estimate on the difference between xn(t) and x(t) under some special condition, and it clearly shows that one can use the Picard iteration procedure to obtain the approximate solutions to equations (3).

References

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2. T.E. Govindan, Stability of mild solution of stochastic evolution equations with variable delay, Stochastic Anal. Appl. 21 (2003), 1059-1077.

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4. Y.-H. Kim, An estimate on the solutions for stochastic functional differential equations , J. Appl. Math. and Informatics 29 (2011) no.5-5, 1549-1556.

5. Y.-H. Kim, A note on the solutions of neutral SFDEs with infinite delay, J. inequalities and Applications 181 (2013) no.1, 1-11.

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8. X. Mao, Stochastic Differential Equations and Applications, Horwood Publication Chich- ester, UK (2007).

9. X. Mao, Y. Shen, and C. Yuan, Almost surely asymptotic stability of neutral stochastic differential delay equations with Markovian switching, Stochastic process. Appl. 118 (2008), 1385-1406.

10. F. Wei and Y. Cai, Existence, uniqueness and stability of the solution to neutral sto- chastic functional differential equations with infinite delay under non-Lipschitz conditions, Advances in Difference Equations 151 (2013), 1-12.

Young-Ho Kim received M.Sc. from Chung-Ang University and Ph.D at Chung-Ang University. Since 1997 he has been at Changwon National University. His research interests include Stochastic differential Equations and Stochastic Analysis.

Department of Mathematics, Changwon National University, Changwon Gyeongsangnam- do 51140, Korea.

e-mail: yhkim@changwon.ac.kr

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