# Existence of Solutions of Nonlinear Stochastic Volterra Fredholm Integral Equations of Mixed Type

Authors: K. Balachandran; J.-H. Kim
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

We establish sufficient conditions for the existence and
uniqueness of random solutions of nonlinear Volterra-Fredholm stochastic integral
equations of mixed type by using admissibility theory and fixed point theorems. The
results obtained in this paper generalize the results of several papers.


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## Body

## 1. Introduction

Random or stochastic integral equations are important in the study of many physical phenomena in life sciences, engineering, and technology [1–13]. Currently there are two basic versions of stochastic integral equations being studied by mathematical statisticians and probabilists namely, those integral equations involving Ito-Doob type of stochastic integrals and those which can be formed as probabilistic analogues of classical deterministic integral equations whose formulation involves the usual Lebesgue integral. Equations of the later category have been studied extensively by several authors [4, 10, 14–40]. Many papers have been appeared on the problem of existence of solutions of nonlinear random integral equations and the results are established by applying various fixed point techniques. These methods are broadly classified into three categories:(i)
admissibility theory, ([2, 7, 24, 27, 41–47]),(ii)
random contractor method, ([17, 21, 35, 47–52]),(iii)
measure of noncompactness method, ([11, 53–61]).All these methods are effectively used to study the existence of solutions for stochastic integral equations. Further asymptotic behaviour and stability of solutions of stochastic integral equations are discussed in the papers [33, 42, 50, 54, 55, 59, 61–63]. In this paper we will study the existence of random solutions of nonlinear stochastic integral equations of mixed type.Consider a nonlinear stochastic integral equation of the form(1.1)x(t;w)=h(t,x(t;w))+∫0tk1(t,τ;w)f1(τ,x(τ;w))dτ+∫0∞k2(t,τ;w)f2(τ,x(τ;w))dτ+∫0tk3(t,τ;w)f3(τ,x(τ;w))dβ(τ),
where t∈R+,β(t) is a stochastic process and(a-i)
w∈Ω, the supporting set of the complete probability measure space (Ω,A,μ), with the σ-algebra A and probability measure μ,(a-ii)
x(t;w) is the unknown random function for t∈R+, the nonnegative real numbers,(a-iii)
h(t,x) is a scalar function defined for t∈R+ and x∈R, the real line,(a-iv)
k1(t,τ;w) and k3(t,τ;w) are stochastic kernels defined for t and τ satisfying 0≤τ≤t<∞,(a-v)
k2(t,τ;w) is the stochastic kernel defined for t and τ in R+,(a-vi)
f1(t,x),f2(t,x),f3(t,x)  are scalar functions defined for t∈R+ and x∈R, the real line.The first and the second part of the stochastic integral (1.1) are to be understood as an ordinary Lebesque integral with probabilistic characterization, while the third part is an Ito-Doob stochastic integral. Our aim is to investigate the existence as well as uniqueness of random solutions of the stochastic integral equation (1.1) by making use of “admissibility theory” that was first introduced by Tsokos [40] and fixed point theorems due to Krasnoselskii and Banach. The results generalize the previous results of [2, 7, 24, 27, 41–46].


## 2. Preliminaries

Letβ(t;w) be the random process. We will assume that for each t∈R+, a minimal σ-algebra At, At⊂A, is such that β(t;w) is measurable with respect to At. In addition, we will assume that the minimal σ-algebra At is an increasing family such that(H1)
the random process{β(t;w),At:t∈R+} is a real martingale(H2)
there is a real continuous nondecreasing function,F(t), such that for s<t we have E{|β(t;w)-β(s;w)|2}=E{|β(t;w)-β(s;w)|2:At}=F(t)-F(s)μ- a.e. where E denotes the expected value of the random process.In the definitions that follow, we will assume thatx(t;w) is At measurable and that E|x(t;w)|2<∞, for each t∈R+. Also we denote(2.1){E|x(t;w)|2}1/2=∥x(t;w)∥L2(Ω,A,μ)=(∫Ω|x(t:w)|2dμ(w))1/2.Definition 2.1.
Denote byCc the linear space of all mean square continuous maps x(t;w) on R+ and define a topology on Cc by means of the following family of seminorms.
(2.2)∥x(t;w)∥n=sup0≤t≤n{E|x(t;w)|2}1/2.
It is known that such a topology is metrizable and that the metric space Cc is complete.Definition 2.2.
DefineCg⊂Cc to be the space of all maps x(t;w) on R+ such that
(2.3){E|x(t;w)|2}1/2≤ag(t),
where a>0, a constant and g(t)>0, a continuous function on R+. The norm in the space Cg is defined by
(2.4)∥x(t;w)∥Cg=supt≥0{1g(t){E|x(t;w)|2}1/2}.Definition 2.3.
LetC⊂Cc be the space of maps x(t;w) on R+ with {E|x(t;w)|2}1/2<M, for some M>0. The norm in space C is defined by
(2.5)∥x(t;w)∥C=supt≥0{E|x(t;w)|2}1/2.Definition 2.4.
The pair of Banach spaces(B,D) with B,D⊂Cc is called admissible with respect to the operator T:Cc→Cc if TB⊂D.Definition 2.5.
We will callx(t;w) a random solution of the stochastic integral equation (1.1) if x(t;w)∈Cc for each t∈R+ and satisfies equation (1.1) μ-a.e., for all t>0.Definition 2.6.
The Banach spaceB is said to be stronger than Cg, if every sequence which converges in the topology of B converges also in the topology of Cg.
Finally, letB,D⊂Cg be Banach spaces and T a linear operator from Cg into Cc. The following lemma is well known [13].Lemma 2.7.
LetT be a continuous operator from Cg into Cc. If B and D are Banach spaces in Cg stronger than Cg and if the pair (B,D) is admissible with respect to T, then T is a continuous operator from B into D.
Let us define the operators
(2.6)(T1x)(t;w)=∫0tk1(t,τ;w)x(τ;w)dτ,(2.7)(T2x)(t;w)=∫0∞k2(t,τ;w)x(τ;w)dτ,(2.8)(T3x)(t;w)=∫0tk3(t,τ;w)x(τ;w)dβ(τ),
for x(t;w)∈Cg.
We state   the following assumptions for our use.
(a1)
The functionsf1(t,x(t;w)),f2(t,x(t;w)), and f3(t,x(t;w)) are continuous functions of t∈R+ with values in L2(Ω,A,μ).(a2)
For eacht and τ in R+,k2(t,τ;w) has values in the space L∞(Ω,A,μ) and the functions k1(t,τ;w) and k3(t,τ;w) for each t and τ such that 0≤τ≤t<∞ has values in the space L∞(Ω,A,μ).(a3)
The stochastic kernelsk1(t,τ;w) and k3(t,τ;w) are essentially a bounded function with respect to μ for every t and τ such that 0≤τ≤t<∞ and continuous as maps from {(t,τ):0≤τ≤t<∞} into L∞(Ω,A,μ).(a4)
The stochastic kernelk2(t,τ;w) is essentially a bounded function with respect to μ for every t and τ in R+ and continuous as maps from {(t,τ):0≤τ≤t<∞} into L∞(Ω,A,μ).
Define for 0≤τ≤t<∞,
(2.9)∥|k1(t,τ;w)|∥=μ-esssupw∈Ω|k1(t,τ;w)|,∥|k2(t,τ;w)|∥=μ-esssupw∈Ω|k2(t,τ;w)|,∥|k3(t,τ;w)|∥=μ-esssupw∈Ω|k3(t,τ;w)|.
The  assumptions (a1)–(a4) imply that if x(t;w)∈Cc, then for each t∈R+,
(2.10)E|k3(t,τ;w)x(τ;w)|2≤∥|k3(t,τ;w)|∥2E|x(t;w)|2.
Because of the continuity assumptions on |k3(t,τ;w)| and E|x(τ;w)|2 it follows from the above inequality that
(2.11)∫0tE|k3(t,τ;w)x(τ;w)|2dF(τ)<∞,
which together with (H1) and (H2) implies that the integral in (2.8) is well defined.Lemma 2.8.
Under the assumptions(a1)–(a4), (H1) and (H2), T1,T2, and T3 are continuous linear operators from Cg into Cc provided
(2.12)∫0∞∥|k3(t,τ;w)|∥2g2(τ)dτ≤N<∞forsomeN>0.Proof.
It is easy to show thatT1,T2 and T3 are linear maps from Cg into Cc. The continuity of T1 and T2 are also easy to prove [8, 13]. We will prove that T3 is continuous.
Letx(t;w)∈Cg. Then
(2.13)E|(T3x)(t;w)|2=E{∫0tk3(t,τ;w)x(τ;w)dβ(τ)}2=∫0tE|k3(t,τ;w)x(τ;w)|2dF(τ)≤∫0t∥|k3(t,τ;w)|∥2‍E|x(t;w)|2dF(τ)≤∥x(t;w)∥Cg2∫0t∥|k3(t,τ;w)|∥2‍g2(τ)dF(τ),t<n.
Hence, on compact intervals [0,n](2.14)sup0≤t≤n∥(T3x)(t;w)∥L2(Ω,A,μ)≤∥x(t;w)∥Cg{sup0≤t≤n[∫0t∥|k3(t,τ;w)|∥2‍g2(τ)dF(τ)]1/2}≤N1∥x(t;w)∥Cg,
where N1 is a constant depends upon n. This proves the continuity of T3. The linearity of T3 is obvious.
To show thatT2 maps Cg into Cc. Let y(t;w)=∫0∞k2(t,τ;w)x(τ;w)dτ. Then
(2.15)∥y(t1;w)-y(t2;w)∥L2(Ω,A,μ)=∥x(t;w)∥Cg∫0∞∥|k2(t1,τ;w)-k2(t2,τ;w)|∥2‍g2(τ)dτ.
The right-hand side of the above inequality goes to zero as t2→t1, since k2(t,τ;w)g(τ)∈L2(Ω,A,μ). Thus, this proves that T2 maps Cg into Cc. The proof of the continuity of T2 is similar to that of T3.
Let the operatorsT1,T2, and T3 be as defined in (2.6), (2.7), and (2.8) and let the assumptions of Lemma 2.8 hold. Then it follows from Lemma 2.7 that, if B and D are Banach spaces stronger than Cg and the pair (B,D) is admissible with respect to the operators T1,T2 and T3, then T1,T2, and T3 are continuous from B into D. Thus, there exist positive constants K1,K2, and K3 such that
(2.16)∥(T1x)(t;w)∥D≤K1∥x(t;w)∥B,∥(T2x)(t;w)∥D≤K2∥x(t;w)∥B,∥(T3x)(t;w)∥D≤K3∥x(t;w)∥B.
The constants K1,K2,K3 are the bounds of the operator T1,T2,T3.Theorem 2.9 (Krasnoselskii Theorem).
LetS be a closed, bounded and convex subset of a Banach space X and let U1 and U2 be operators on S satisfying the following conditions:  (i)
U1(x)+U2(y)∈S whenever x,y∈S,(ii)
U1 is a contraction operator on S,(iii)
U2 is completely continuous.Then there is at least one point x*∈S such that U1(x*)+U2(x*)=x*.


## 3. Main Results

In this section we will prove the main result of this paper.Theorem 3.1.
For the stochastic integral equation (1.1) assume the following conditions  (i)
B and D are Banach spaces in Cg, stronger than Cg, such that (B,D) is admissible with respect to the operators T1,T2, and T3 defined by (2.6), (2.7), and (2.8);(ii)
∫0∞∥|k2(t,τ;w)|2∥g2(τ)dτ≤N<∞ for some N>0;(iii)
x(t;w)→f1(t,x(t;w)) is a continuous map from
(3.1)S={x(t;w):x(t;w)∈D,∥x(t;w)∥D≤ρ}
with values in B satisfying
(3.2)∥f1(t,x(t;w))-f1(t,y(t;w))∥B≤λ1∥x(t;w)-y(t;w)∥D
for x(t;w),y(t;w)∈S and λ1≥0 a constant;(iv)
x(t;w)→f2(t,x(t;w)) is a completely continuous map from S into B;(v)
x(t;w)→f3(t,x(t;w)) is a continuous map from S with values in B satisfying
(3.3)∥f3(t,x(t;w))-f3(t,y(t;w))∥B≤λ3∥x(t;w)-y(t;w)∥D
for x(t;w),y(t;w)∈S and λ3 a constant;(vi)
x(t;w)→h(t,x(t;w)) is a continuous map from S into D such that
(3.4)∥h(t,x(t;w))-h(t,y(t;w))∥D≤γ∥x(t;w)-y(t;w)∥D
for x(t;w),y(t;w)∈S and γ>0 a constant.
Then there exists a unique random solution of (1.1) in S provided
(3.5)γ+K1λ1+K3λ3<1,γ∥h(t,0)∥D+K1∥f1(t,0)∥B+K2∥f2(t,x(t;w))∥B+K3∥f3(t,0)∥B≤ρ(1-γ-K1λ1-K3λ3),
where K1,K2, and K3 are defined by (2.16).Proof.
The setS closed, bounded, and convex in D. Let x(t;w),y(t;w)∈S. Then define the operator U1:S→D by
(3.6)(U1x)(t;w)=h(t,x(t;w))+∫0tk1(t,τ;w)f1(τ,x(τ;w))dτ+∫0tk3(t,τ;w)f3(τ,x(τ;w))dβ(τ).
We will show that U1 is a contraction mapping and that U1S⊂S. Let x(t;w),y(t;w)∈S. Then
(3.7)(U1x)(t;w)-(U1y)(t;w)=h(t,x(t;w))-h(t,y(t;w))+∫0tk1(t,τ;w)[f1(τ,x(τ;w))-f1(τ,y(τ;w))]dτ+∫0tk3(t,τ;w)[f3(τ,x(τ;w))-f3(τ,y(τ;w))]dβ(τ).
From our assumption it is clear that (U1x)(t;w)-(U1y)(t;w)∈D and f1(τ,x(τ;w))-f1(τ,y(τ;w)),f3(τ,x(τ;w))-f3(τ,y(τ;w))∈B. Furthermore
(3.8)∥(U1x)(t;w)-(U1y)(t;w)∥D≤∥h(t,x(t;w))-h(t,y(t;w))∥D+K1∥f1(τ,x(τ;w))-f1(τ,y(τ;w))∥B+K3∥f3(τ,x(τ;w))-f3(τ,y(τ;w))∥B≤(γ+K1λ1+K3λ3)∥x(t;w)-y(t;w)∥.
Since (γ+K1λ1+K3λ3)<1, U1 is a contraction operator. Next we show that U1S⊂S. From (3.6), we have
(3.9)∥(U1x)(t;w)∥D=∥h(t,x(t;w))∥D+∥∫0tk1(t,τ;w)f1(τ,x(τ;w))dτ∥+∥∫0tk3(t,τ;w)f3(τ,x(τ;w))dβ(τ)∥≤∥h(t,0)∥D+((γ+K1λ1+K3λ3))∥x(t;w)∥+λ1∥f1(t,0)∥B+λ3∥f(t,0)∥B.
Since x(t;w)∈S, by hypothesis, we have ∥(U1x)(t;w)∥D≤ρ which implies that U1S⊂S.
Let us define the operatorU2:S→D as
(3.10)(U2x)(t;w)=∫0∞k2(t,τ;w)f2(τ,x(τ;w))dτ.
It is clear that U2 is composition of continuous map T2 and completely continuous map f2. Hence U2 is completely continuous. Furthermore, if x(t;w),y(t;w)∈S, we have
(3.11)∥(U1x)(t;w)+(U1y)(t;w)∥D≤∥h(t,x(t;w))∥D+K1∥f1(τ,x(τ;w))∥B+K2∥f2(τ,y(τ;w))∥B+K3∥f3(τ,x(τ;w))∥B≤∥h(t,0)∥D+(γ+K1λ1+K3λ3)ρ+K1∥f1(t,0)∥B+K2∥f2(t,x(t;w)∥B+K3∥f3(t,0)∥B≤ρ.
This shows that if x(t;w),y(t;w)∈S, then (U1x)(t;w)+(U2y)(t;w)∈S. Hence, applying Krasnoselskii's fixed point theorem, we can conclude that there exists a random solution of (1.1) in the set S.
We will now consider the case under which the stochastic integral equation (1.1) possesses a unique solution. This will be achieved by using the Banach contraction mapping principle.Theorem 3.2.
For the stochastic integral equation (1.1) assume the following conditions  (i)
B and D are Banach spaces in Cg, stronger than Cg, such that (B,D) is admissible with respect to the operators T1,T2 and T3 defined by (2.6), (2.7), and (2.8);(ii)
∫0∞∥|k2(t,τ;w)|2∥g2(τ)dτ≤N<∞ for some N>0;(iii)
x(t;w)→f1(t,x(t;w)) is a continuous map from
(3.12)S={x(t;w):x(t;w)∈D,∥x(t;w)∥D≤ρ}
with values in B satisfying
(3.13)∥f1(t,x(t;w))-f1(t,y(t;w))∥B≤λ1∥x(t;w)-y(t;w)∥D
for x(t;w),y(t;w)∈S and λ1≥0 a constant;(iv)
x(t;w)→f2(t,x(t;w)) is a continuous map from S with values in B satisfying
(3.14)∥f2(t,x(t;w))-f2(t,y(t;w))∥B≤λ2∥x(t;w)-y(t;w)∥D
for x(t;w),y(t;w)∈S and λ2≥ a constant;(v)
x(t;w)→f3(t,x(t;w)) is a continuous map from S with values in B satisfying
(3.15)∥f3(t,x(t;w))-f3(t,y(t;w))∥B≤λ3∥x(t;w)-y(t;w)∥D
for x(t;w),y(t;w)∈S and λ3 a constant;(vi)
x(t;w)→h(t,x(t;w)) is a continuous map from S into D such that
(3.16)∥h(t,x(t;w))-h(t,y(t;w))∥D≤γ∥x(t;w)-y(t;w)∥D
for x(t;w),y(t;w)∈S and γ>0 a constant.
Then there exists a unique random solution of (1.1) in S provided
(3.17)γ+K1λ1+K2λ2+K3λ3<1,γ∥h(t,0)∥D+K1∥f1(t,0)∥B+K2∥f2(t,0)∥B+K3∥f3(t,0)∥B≤ρ(1-γ-K1λ1-K2λ2-K3λ3),
where K1,K2, and K3 are defined by (2.16).Proof.
Define the operatorU:S→D as follows
(3.18)(Ux)(t;w)=h(t,x(t;w))+∫0tk1(t,τ;w)f1(τ,x(τ;w))dτ+∫0∞k2(t,τ;w)f2(τ,x(τ;w))dτ+∫0tk3(t,τ;w)f3(τ,x(τ;w))dβ(τ).
We will show thatU is a contraction operator on S and that US⊂S. Let x(t;w),y(t;w)∈S. Then (Ux)(t;w)-(Uy)(t;w)∈D as US⊂D and D is a Banach space. Also
(3.19)∥(Ux)(t;w)-(Uy)(t;w)∥D≤∥h(t,x(t;w))-h(t,y(t;w))∥D+∥∫0tk1(t,τ;w)[f1(τ,x(τ;w))-f1(τ,y(τ;w))]dτ∥D+∥∫0∞k2(t,τ;w)[f2(τ,x(τ;w))-f2(τ,y(τ;w))]dτ∥D+∥∫0tk3(t,τ;w)[f3(τ,x(τ;w))-f3(τ,y(τ;w))]dβ(τ)∥D.
Thus, in view of (2.16), we have
(3.20)∥(Ux)(t;w)-(Uy)(t;w)∥D≤γ∥x(t;w)-y(t;w)∥D+K1∥f1(t,x(t;w)-f1(t,y(t;w)∥B+K2∥f2(t,x(t;w))-f2(t,y(t;w))∥B+K3∥f3(t,x(t;w))-f3(t,y(t;w))∥B≤(γ+K1λ1+K2λ2+K3λ3)∥x(t;w)-y(t;w)∥D.
Since (γ+K1λ1+K2λ2+K3λ3)<1, U is a contraction operator on S.
We will now show thatUS⊂S. For any x(t;w)∈S, we have
(3.21)∥(Ux)(t;w)∥D≤∥h(t,x(t;w))∥D+∥∫0tk1(t,τ;w)f1(τ,x(τ;w)dτ∥D+∥∫0∞k2(t,τ;w)f2(τ,x(τ;w))dτ∥D+∥∫0tk3(t,τ;w)f3(τ,x(τ;w))dβ(τ)∥D≤∥h(t,x(t;w))∥D+K1∥f1(t,x(t;w))∥B+K2∥f2(t,x(t;w))∥B+K3∥f3(t,x(t;w))∥B≤γ∥x(t;w)∥D+γ∥h(t,0)∥D+λ1K1∥x(t;w)∥D+K1∥f1(t,0)∥B+λ2K2∥x(t;w)∥D+K2∥f2(t,0)∥B+λ3K3∥x(t;w)∥D+K3∥f3(t,0)∥B.
Since ∥x(t;w)∥D≤ρ, it follows that
(3.22)∥(Ux)(t;w)∥D≤γ∥h(t,0)∥D+ρ(γ+K1λ1+K2λ2+K3λ3)+K1∥f1(t,0)∥B+K2∥f2(t,0)∥B+K3∥f3(t,0)∥B.
Using the condition that
(3.23)γ∥h(t,0)∥D+K1∥f1(t,0)∥B+K2∥f2(t,0)∥B+K3∥f3(t,0)∥B≤ρ(1-γ-K1λ1-K2λ2-K3λ3),
we have from (3.18)
(3.24)∥(Ux)(t;w)∥D≤ρ.
Hence (Ux)(t;w)∈S for all x(t;w)∈S or US⊂S. Thus the condition of Banach's fixed point theorem is satisfied and hence there exists a fixed point x(t;w)∈S such that (Ux)(t;w)=x(t;w). That is,
(3.25)(Ux)(t;w)=h(t,x(t;w))+∫0tk1(t,τ;w)f1(τ,x(τ;w))dτ+∫0∞k2(t,τ;w)f2(τ,x(τ;w))dτ+∫0tk3(t,τ;w)f3(τ,x(τ;w))dβ(τ)=x(t;w).


## 4. Applications

In this section we will give some application of Theorem3.2.Theorem 4.1.
Suppose the stochastic integral equation (1.1) satisfies the following conditions:  (i)
there exists a constantA>0 and a continuous function g(t), such that
(4.1)∫0t∥|k1(t,τ;w)|∥2g2(τ)dτ+∫0∞∥|k2(t,τ;w)|∥2g2(τ)dτ+∫0t∥|k3(t,τ;w)|∥2g2(τ)dτ<A;(ii)
fi(t,x),i=1,2,3 are continuous functions on R+×R, such that fi(t,0)∈Cg(R+,R) and |fi(t,x)-fi(t,y)|≤λig(t)|x-y|, for x,y∈R and 0≤λi<1,i=1,2,3;(iii)
h(t,x) is a continuous functions on R+×R, such that |h(t,x)-h(t,y)|≤γ|x-y|, for x,y∈R and 0≤γ<1.
Then there exists a unique random solutionx(t;w) of (1.1) such that
(4.2)∥x(t;w)∥C≤ρ
provided ∥h(t,0)∥,∥fi(t,0)∥Cg,i=1,2,3 are small enough.Proof.
It is easy to show that the hypothesis of Theorem3.2 are satisfied by simply showing the pair of spaces (Cg,Cc) is admissible with respect to the operators T1,T2, and T3. This follows from Lemma 2.8.Corollary 4.2.
Suppose the stochastic integral equation (1.1) satisfies the following conditions:  (i)
∫0t∥|k1(t,τ;w)|2∥dτ+∫0∞∥|k2(t,τ;w)|2∥dτ+∫0t∥|k3(t,τ;w)|2∥dτ<A;(ii)
fi(t,x),i=1,2,3 are continuous functions on R+×R, such that fi(t,0)∈Cg(R+,R) and |fi(t,x)-fi(t,y)|≤λig(t)|x-y|, for x,y∈R and 0≤λi<1,i=1,2,3;(iii)
h(t,x) is a continuous functions on R+×R, such that |h(t,x)-h(t,y)|≤γ|x-y|, for x,y∈R and 0≤γ<1.
Then there exists a unique random solutionx(t;w) of (1.1) such that
(4.3)∥x(t;w)∥C≤ρ
provided ∥h(t,0)∥,∥fi(t,0)∥Cg,i=1,2,3 are small enough.Proof.
Takeg(t)=1 in Theorem 4.1.Corollary 4.3.
Suppose the stochastic integral equation (1.1) satisfies the following conditions:  (i)
∥|ki(t,τ;w)|∥≤A, i=1,2,3 and ∫0tg2(τ)τ<∞;(ii)
same as conditions(iv),(v),and(vi) in Theorem 3.2.
Then there exists a unique random solution of (1.1) provided γ,∥h(t,0)∥C and ∥fi(t,0)∥Cg for  i=1,2,3 small enough.Proof.
We will show that the pair is(Cg,Cc) admissible with respect to the operator T2. Let x(t;w)∈Cg. Then
(4.4)sup0≤t∥(T2x)(t;w)∥Cg≤sup0≤t{∫0∞∥|k2(t,τ;w)|∥2∥x(τ;w)∥L22dτ}1/2≤∥x(t;w)∥CgA∫0∞g2(τ)dτ
which implies that the pair (Cg,Cc) is admissible. Similarly we can show that the pair (Cg,Cc) is admissible with respect to the operators T1,T3. It is easy to check the other conditions of Theorem 3.2 and hence there exists a unique random solution of equation of the stochastic integral equation (1.1).Remark 4.4.
Using the same argument one can establish the existence of a unique random solution of the following general stochastic integral equation(4.5)x(t;w)=h(t,x(t;w))+∑i=1n∫0tai(t,τ;w)fi(τ,x(τ;w))dτ+∑i=1n∫0∞bi(t,τ;w)gi(τ,x(τ;w))dτ+∑i=1n∫0tci(t,τ;w)ki(τ,x(τ;w))dβ(τ),
where h,ki,ai,bi,ci,gi,fi, and β satisfy appropriate conditions. This general case is treated in a separate paper.


## 5. Example

Consider the following nonlinear stochastic integral equation:(5.1)x(t;w)=14sinx(t;w)+∫0tsint4e-s-x2(s;w)ds+∫0∞e-t-s1+|x(s;w)|ds+18∫0tln(1+|x(s;w)|)dβ(s),t∈R+,
where β(t) is a stochastic process. This equation is a particular case of general stochastic integral equation occurring in mathematical biology and chemotherapy [10–13]. The above equation takes the form of (1.1) with(5.2)k1(t,s,w)=sint4e-s,k2(t,s,w)=e-t-s,k3(t,s,w)=14,h(t,x(t;w))=sinx(t;w)4f1(s,x(s;w))=e-x2(s;w),f2(s,x(s;w))=11+|x(s;w)|,f3(s,x(s;w))=12ln(1+|x(s;w)|).
Take B=D=Cg=Cc=C and g(t)=1. It is easy to see that γ=1/4, K1=K3=1/4, K2=1, λ1=1, λ2=1/4, and λ3=1/2. Further γ+K1λ1+K2λ2+K3λ3=7/8<1 and by taking ρ≥10, the other condition of Theorem 3.2 is satisfied. It is clear that (5.1) satisfies assumptions (i) to (vi) of Theorem 3.2. Hence there exists a unique random solution for (5.1).



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*Source: 603819-2010-03-08.xml*