International Journal of Mathematics and Mathematical Sciences

International Journal of Mathematics and Mathematical Sciences / 2010 / Article

Research Article | Open Access

Volume 2010 |Article ID 603819 | 16 pages | https://doi.org/10.1155/2010/603819

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

Academic Editor: Jewgeni Dshalalow
Received13 Aug 2009
Accepted19 Jan 2010
Published08 Mar 2010

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.

1. Introduction

Random or stochastic integral equations are important in the study of many physical phenomena in life sciences, engineering, and technology [113]. 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, 1440]. 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, 4147]), (ii)random contractor method, ([17, 21, 35, 4752]), (iii)measure of noncompactness method, ([11, 5361]).

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, 6163]. 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

𝑥(𝑡;𝑤)=(𝑡,𝑥(𝑡;𝑤))+𝑡0𝑘1(𝑡,𝜏;𝑤)𝑓1(+𝜏,𝑥(𝜏;𝑤))𝑑𝜏0𝑘2(𝑡,𝜏;𝑤)𝑓2(𝜏,𝑥(𝜏;𝑤))𝑑𝜏+𝑡0𝑘3(𝑡,𝜏;𝑤)𝑓3(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏),(1.1) where 𝑡𝑅+,𝛽(𝑡) is a stochastic process and

(a-i)𝑤Ω, 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 𝑡𝑅+, the nonnegative real numbers, (a-iii)(𝑡,𝑥) is a scalar function defined for 𝑡𝑅+ and 𝑥𝑅, the real line, (a-iv)𝑘1(𝑡,𝜏;𝑤) and 𝑘3(𝑡,𝜏;𝑤) are stochastic kernels defined for 𝑡 and 𝜏 satisfying 0𝜏𝑡<, (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 𝑡𝑅+ and 𝑥𝑅, 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, 4146].

2. Preliminaries

Let 𝛽(𝑡;𝑤) be the random process. We will assume that for each 𝑡𝑅+, a minimal 𝜎-algebra 𝐴𝑡, 𝐴𝑡𝐴, 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),𝐴𝑡𝑡𝑅+} 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)|2𝐴𝑡}=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 that 𝑥(𝑡;𝑤) is 𝐴𝑡 measurable and that 𝐸|𝑥(𝑡;𝑤)|2<, for each 𝑡𝑅+. Also we denote

𝐸||||𝑥(𝑡;𝑤)21/2=𝑥(𝑡;𝑤)𝐿2(Ω,𝐴,𝜇)=Ω||||𝑥(𝑡𝑤)2𝑑𝜇(𝑤)1/2.(2.1)

Definition 2.1. Denote by 𝐶𝑐 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. 𝑥(𝑡;𝑤)𝑛=sup0𝑡𝑛𝐸||||𝑥(𝑡;𝑤)21/2.(2.2) It is known that such a topology is metrizable and that the metric space 𝐶𝑐 is complete.

Definition 2.2. Define 𝐶𝑔𝐶𝑐 to be the space of all maps 𝑥(𝑡;𝑤) on 𝑅+ such that 𝐸||||𝑥(𝑡;𝑤)21/2𝑎𝑔(𝑡),(2.3) where 𝑎>0, a constant and 𝑔(𝑡)>0, a continuous function on 𝑅+. The norm in the space 𝐶𝑔 is defined by 𝑥(𝑡;𝑤)𝐶𝑔=sup𝑡01𝐸||||𝑔(𝑡)𝑥(𝑡;𝑤)21/2.(2.4)

Definition 2.3. Let 𝐶𝐶𝑐 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 𝑥(𝑡;𝑤)𝐶=sup𝑡0𝐸||||𝑥(𝑡;𝑤)21/2.(2.5)

Definition 2.4. The pair of Banach spaces (𝐵,𝐷) with 𝐵,𝐷𝐶𝑐 is called admissible with respect to the operator 𝑇𝐶𝑐𝐶𝑐 if 𝑇𝐵𝐷.

Definition 2.5. We will call 𝑥(𝑡;𝑤) a random solution of the stochastic integral equation (1.1) if 𝑥(𝑡;𝑤)𝐶𝑐 for each 𝑡𝑅+ and satisfies equation (1.1) 𝜇-a.e., for all 𝑡>0.

Definition 2.6. The Banach space 𝐵 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, let 𝐵,𝐷𝐶𝑔 be Banach spaces and 𝑇 a linear operator from 𝐶𝑔 into 𝐶𝑐. The following lemma is well known [13].

Lemma 2.7. Let 𝑇 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
𝑇1𝑥(𝑡;𝑤)=𝑡0𝑘1(𝑇𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)𝑑𝜏,(2.6)2𝑥(𝑡;𝑤)=0𝑘2𝑇(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)𝑑𝜏,(2.7)3𝑥(𝑡;𝑤)=𝑡0𝑘3(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)𝑑𝛽(𝜏),(2.8) for 𝑥(𝑡;𝑤)𝐶𝑔.
We state the following assumptions for our use.
(𝑎1)The functions 𝑓1(𝑡,𝑥(𝑡;𝑤)),𝑓2(𝑡,𝑥(𝑡;𝑤)), and 𝑓3(𝑡,𝑥(𝑡;𝑤)) are continuous functions of 𝑡𝑅+ with values in 𝐿2(Ω,𝐴,𝜇).(𝑎2)For each 𝑡 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 0𝜏𝑡< has values in the space 𝐿(Ω,𝐴,𝜇).(𝑎3)The stochastic kernels 𝑘1(𝑡,𝜏;𝑤) and 𝑘3(𝑡,𝜏;𝑤) are essentially a bounded function with respect to 𝜇 for every 𝑡 and 𝜏 such that 0𝜏𝑡< and continuous as maps from {(𝑡,𝜏)0𝜏𝑡<} into 𝐿(Ω,𝐴,𝜇).(𝑎4)The stochastic kernel 𝑘2(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)0𝜏𝑡<} into 𝐿(Ω,𝐴,𝜇). Define for 0𝜏𝑡<, ||𝑘1||(𝑡,𝜏;𝑤)=𝜇esssup𝑤Ω||𝑘1||,||𝑘(𝑡,𝜏;𝑤)2||(𝑡,𝜏;𝑤)=𝜇esssup𝑤Ω||𝑘2||,||𝑘(𝑡,𝜏;𝑤)3||(𝑡,𝜏;𝑤)=𝜇esssup𝑤Ω||𝑘3||.(𝑡,𝜏;𝑤)(2.9) The assumptions (𝑎1)-(𝑎4) imply that if 𝑥(𝑡;𝑤)𝐶𝑐, then for each 𝑡𝑅+, 𝐸||𝑘3||(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)2||𝑘3||(𝑡,𝜏;𝑤)2𝐸||||𝑥(𝑡;𝑤)2.(2.10) Because of the continuity assumptions on |𝑘3(𝑡,𝜏;𝑤)| and 𝐸|𝑥(𝜏;𝑤)|2 it follows from the above inequality that 𝑡0𝐸||𝑘3||(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)2𝑑𝐹(𝜏)<,(2.11) 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 0||𝑘3||(𝑡,𝜏;𝑤)2𝑔2(𝜏)𝑑𝜏𝑁<forsome𝑁>0.(2.12)

Proof. It is easy to show that 𝑇1,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.
Let 𝑥(𝑡;𝑤)𝐶𝑔. Then
𝐸||(𝑇3||𝑥)(𝑡;𝑤)2=𝐸𝑡0𝑘3(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)𝑑𝛽(𝜏)2=𝑡0𝐸||𝑘3||(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)2𝑑𝐹(𝜏)𝑡0||𝑘3||(𝑡,𝜏;𝑤)2||||0𝑥0200𝑑𝐸𝑥(𝑡;𝑤)2𝑑𝐹(𝜏)𝑥(𝑡;𝑤)2𝐶𝑔𝑡0||𝑘3||(𝑡,𝜏;𝑤)20𝑥0200𝑑𝑔2(𝜏)𝑑𝐹(𝜏),𝑡<𝑛.(2.13) Hence, on compact intervals [0,𝑛]sup0𝑡𝑛(𝑇3𝑥)(𝑡;𝑤)𝐿2(Ω,𝐴,𝜇)𝑥(𝑡;𝑤)𝐶𝑔sup0𝑡𝑛𝑡0||𝑘3||(𝑡,𝜏;𝑤)20𝑥0200𝑑𝑔2(𝜏)𝑑𝐹(𝜏)1/2𝑁1𝑥(𝑡;𝑤)𝐶𝑔,(2.14) where 𝑁1 is a constant depends upon 𝑛. This proves the continuity of 𝑇3. The linearity of 𝑇3 is obvious.
To show that 𝑇2 maps 𝐶𝑔 into 𝐶𝑐. Let 𝑦(𝑡;𝑤)=0𝑘2(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)𝑑𝜏. Then
𝑦(𝑡1;𝑤)𝑦(𝑡2;𝑤)𝐿2(Ω,𝐴,𝜇)=𝑥(𝑡;𝑤)𝐶𝑔0||𝑘2𝑡1,𝜏;𝑤𝑘2𝑡2||,𝜏;𝑤20𝑥0200𝑑𝑔2(𝜏)𝑑𝜏.(2.15) The right-hand side of the above inequality goes to zero as 𝑡2𝑡1, 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 operators 𝑇1,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
(𝑇1𝑥)(𝑡;𝑤)𝐷𝐾1𝑥(𝑡;𝑤)𝐵,(𝑇2𝑥)(𝑡;𝑤)𝐷𝐾2𝑥(𝑡;𝑤)𝐵,(𝑇3𝑥)(𝑡;𝑤)𝐷𝐾3𝑥(𝑡;𝑤)𝐵.(2.16) The constants 𝐾1,𝐾2,𝐾3 are the bounds of the operator 𝑇1,𝑇2,𝑇3.

Theorem 2.9 (Krasnoselskii Theorem). Let 𝑆 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,𝑦𝑆, (ii)𝑈1 is a contraction operator on 𝑆, (iii)𝑈2 is completely continuous. Then there is at least one point 𝑥𝑆 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)|2𝑔2(𝜏)𝑑𝜏𝑁< for some 𝑁>0; (iii)𝑥(𝑡;𝑤)𝑓1(𝑡,𝑥(𝑡;𝑤)) is a continuous map from 𝑥𝑆=(𝑡;𝑤)𝑥(𝑡;𝑤)𝐷,𝑥(𝑡;𝑤)𝐷𝜌(3.1) with values in 𝐵 satisfying 𝑓1(𝑡,𝑥(𝑡;𝑤))𝑓1(𝑡,𝑦(𝑡;𝑤))𝐵𝜆1𝑥(𝑡;𝑤)𝑦(𝑡;𝑤)𝐷(3.2) for 𝑥(𝑡;𝑤),𝑦(𝑡;𝑤)𝑆 and 𝜆10 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 𝑓3(𝑡,𝑥(𝑡;𝑤))𝑓3(𝑡,𝑦(𝑡;𝑤))𝐵𝜆3𝑥(𝑡;𝑤)𝑦(𝑡;𝑤)𝐷(3.3) for 𝑥(𝑡;𝑤),𝑦(𝑡;𝑤)𝑆 and 𝜆3 a constant;(vi)𝑥(𝑡;𝑤)(𝑡,𝑥(𝑡;𝑤)) is a continuous map from 𝑆 into 𝐷 such that (𝑡,𝑥(𝑡;𝑤))(𝑡,𝑦(𝑡;𝑤))𝐷𝛾𝑥(𝑡;𝑤)𝑦(𝑡;𝑤)𝐷(3.4) 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
𝛾+𝐾1𝜆1+𝐾3𝜆3𝛾<1,(𝑡,0)𝐷+𝐾1𝑓1(𝑡,0)𝐵+𝐾2𝑓2(𝑡,𝑥(𝑡;𝑤))𝐵+𝐾3𝑓3(𝑡,0)𝐵𝜌1𝛾𝐾1𝜆1𝐾3𝜆3,(3.5) where 𝐾1,𝐾2, and 𝐾3 are defined by (2.16).

Proof. The set 𝑆 closed, bounded, and convex in 𝐷. Let 𝑥(𝑡;𝑤),𝑦(𝑡;𝑤)𝑆. Then define the operator 𝑈1𝑆𝐷 by 𝑈1𝑥(𝑡;𝑤)=(𝑡,𝑥(𝑡;𝑤))+𝑡0𝑘1(𝑡,𝜏;𝑤)𝑓1(+𝜏,𝑥(𝜏;𝑤))𝑑𝜏𝑡0𝑘3(𝑡,𝜏;𝑤)𝑓3(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏).(3.6) We will show that 𝑈1 is a contraction mapping and that 𝑈1𝑆𝑆. Let 𝑥(𝑡;𝑤),𝑦(𝑡;𝑤)𝑆. Then 𝑈1𝑥𝑈(𝑡;𝑤)1𝑦+(𝑡;𝑤)=(𝑡,𝑥(𝑡;𝑤))(𝑡,𝑦(𝑡;𝑤))𝑡0𝑘1𝑓(𝑡,𝜏;𝑤)1(𝜏,𝑥(𝜏;𝑤))𝑓1+(𝜏,𝑦(𝜏;𝑤))𝑑𝜏𝑡0𝑘3𝑓(𝑡,𝜏;𝑤)3(𝜏,𝑥(𝜏;𝑤))𝑓3(𝜏,𝑦(𝜏;𝑤))𝑑𝛽(𝜏).(3.7) 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 (𝑈1𝑈𝑥)(𝑡;𝑤)1𝑦(𝑡;𝑤)𝐷(𝑡,𝑥(𝑡;𝑤))(𝑡,𝑦(𝑡;𝑤))𝐷+𝐾1𝑓1(𝜏,𝑥(𝜏;𝑤))𝑓1(𝜏,𝑦(𝜏;𝑤))𝐵+𝐾3𝑓3(𝜏,𝑥(𝜏;𝑤))𝑓3(𝜏,𝑦(𝜏;𝑤))𝐵𝛾+𝐾1𝜆1+𝐾3𝜆3𝑥(𝑡;𝑤)𝑦(𝑡;𝑤).(3.8) Since (𝛾+𝐾1𝜆1+𝐾3𝜆3)<1, 𝑈1 is a contraction operator. Next we show that 𝑈1𝑆𝑆. From (3.6), we have (𝑈1𝑥)(𝑡;𝑤)𝐷=(𝑡,𝑥(𝑡;𝑤))𝐷+𝑡0𝑘1(𝑡,𝜏;𝑤)𝑓1(+𝜏,𝑥(𝜏;𝑤))𝑑𝜏𝑡0𝑘3(𝑡,𝜏;𝑤)𝑓3(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏)(𝑡,0)𝐷+𝛾+𝐾1𝜆1+𝐾3𝜆3𝑥(𝑡;𝑤)+𝜆1𝑓1(𝑡,0)𝐵+𝜆3𝑓(𝑡,0)𝐵.(3.9) Since 𝑥(𝑡;𝑤)𝑆, by hypothesis, we have (𝑈1𝑥)(𝑡;𝑤)𝐷𝜌 which implies that 𝑈1𝑆𝑆.
Let us define the operator 𝑈2𝑆𝐷 as
𝑈2𝑥(𝑡;𝑤)=0𝑘2(𝑡,𝜏;𝑤)𝑓2(𝜏,𝑥(𝜏;𝑤))𝑑𝜏.(3.10) 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 (𝑈1𝑥)(𝑡;𝑤)+(𝑈1𝑦)(𝑡;𝑤)𝐷(𝑡,𝑥(𝑡;𝑤))𝐷+𝐾1𝑓1(𝜏,𝑥(𝜏;𝑤))𝐵+𝐾2𝑓2(𝜏,𝑦(𝜏;𝑤))𝐵+𝐾3𝑓3(𝜏,𝑥(𝜏;𝑤))𝐵(𝑡,0)𝐷+𝛾+𝐾1𝜆1+𝐾3𝜆3𝜌+𝐾1𝑓1(𝑡,0)𝐵+𝐾2𝑓2(𝑡,𝑥(𝑡;𝑤)𝐵+𝐾3𝑓3(𝑡,0)𝐵𝜌.(3.11) 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)|2𝑔2(𝜏)𝑑𝜏𝑁< for some 𝑁>0; (iii)𝑥(𝑡;𝑤)𝑓1(𝑡,𝑥(𝑡;𝑤)) is a continuous map from 𝑥𝑆=(𝑡;𝑤)𝑥(𝑡;𝑤)𝐷,𝑥(𝑡;𝑤)𝐷𝜌(3.12) with values in 𝐵 satisfying 𝑓1(𝑡,𝑥(𝑡;𝑤))𝑓1(𝑡,𝑦(𝑡;𝑤))𝐵𝜆1𝑥(𝑡;𝑤)𝑦(𝑡;𝑤)𝐷(3.13) for 𝑥(𝑡;𝑤),𝑦(𝑡;𝑤)𝑆 and 𝜆10 a constant;(iv)𝑥(𝑡;𝑤)𝑓2(𝑡,𝑥(𝑡;𝑤)) is a continuous map from 𝑆 with values in 𝐵 satisfying 𝑓2(𝑡,𝑥(𝑡;𝑤))𝑓2(𝑡,𝑦(𝑡;𝑤))𝐵𝜆2𝑥(𝑡;𝑤)𝑦(𝑡;𝑤)𝐷(3.14) 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 𝑓3(𝑡,𝑥(𝑡;𝑤))𝑓3(𝑡,𝑦(𝑡;𝑤))𝐵𝜆3𝑥(𝑡;𝑤)𝑦(𝑡;𝑤)𝐷(3.15) for 𝑥(𝑡;𝑤),𝑦(𝑡;𝑤)𝑆 and 𝜆3 a constant;(vi)𝑥(𝑡;𝑤)(𝑡,𝑥(𝑡;𝑤)) is a continuous map from 𝑆 into 𝐷 such that (𝑡,𝑥(𝑡;𝑤))(𝑡,𝑦(𝑡;𝑤))𝐷𝛾𝑥(𝑡;𝑤)𝑦(𝑡;𝑤)𝐷(3.16) 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
𝛾+𝐾1𝜆1+𝐾2𝜆2+𝐾3𝜆3𝛾<1,(𝑡,0)𝐷+𝐾1𝑓1(𝑡,0)𝐵+𝐾2𝑓2(𝑡,0)𝐵+𝐾3𝑓3(𝑡,0)𝐵𝜌1𝛾𝐾1𝜆1𝐾2𝜆2𝐾3𝜆3,(3.17) where 𝐾1,𝐾2, and 𝐾3 are defined by (2.16).

Proof. Define the operator 𝑈𝑆𝐷 as follows (𝑈𝑥)(𝑡;𝑤)=(𝑡,𝑥(𝑡;𝑤))+𝑡0𝑘1(𝑡,𝜏;𝑤)𝑓1(+𝜏,𝑥(𝜏;𝑤))𝑑𝜏0𝑘2(𝑡,𝜏;𝑤)𝑓2(𝜏,𝑥(𝜏;𝑤))𝑑𝜏+𝑡0𝑘3(𝑡,𝜏;𝑤)𝑓3(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏).(3.18)
We will show that 𝑈 is a contraction operator on 𝑆 and that 𝑈𝑆𝑆. Let 𝑥(𝑡;𝑤),𝑦(𝑡;𝑤)𝑆. Then (𝑈𝑥)(𝑡;𝑤)(𝑈𝑦)(𝑡;𝑤)𝐷 as 𝑈𝑆𝐷 and 𝐷 is a Banach space. Also
(𝑈𝑥)(𝑡;𝑤)(𝑈𝑦)(𝑡;𝑤)𝐷(𝑡,𝑥(𝑡;𝑤))(𝑡,𝑦(𝑡;𝑤))𝐷+𝑡0𝑘1𝑓(𝑡,𝜏;𝑤)1(𝜏,𝑥(𝜏;𝑤))𝑓1(𝜏,𝑦(𝜏;𝑤))𝑑𝜏𝐷+0𝑘2𝑓(𝑡,𝜏;𝑤)2(𝜏,𝑥(𝜏;𝑤))𝑓2(𝜏,𝑦(𝜏;𝑤))𝑑𝜏𝐷+𝑡0𝑘3𝑓(𝑡,𝜏;𝑤)3(𝜏,𝑥(𝜏;𝑤))𝑓3(𝜏,𝑦(𝜏;𝑤))𝑑𝛽(𝜏)𝐷.(3.19) Thus, in view of (2.16), we have (𝑈𝑥)(𝑡;𝑤)(𝑈𝑦)(𝑡;𝑤)𝐷𝛾𝑥(𝑡;𝑤)𝑦(𝑡;𝑤)𝐷+𝐾1𝑓1(𝑡,𝑥(𝑡;𝑤)𝑓1(𝑡,𝑦(𝑡;𝑤)𝐵+𝐾2𝑓2(𝑡,𝑥(𝑡;𝑤))𝑓2(𝑡,𝑦(𝑡;𝑤))𝐵+𝐾3𝑓3(𝑡,𝑥(𝑡;𝑤))𝑓3(𝑡,𝑦(𝑡;𝑤))𝐵𝛾+𝐾1𝜆1+𝐾2𝜆2+𝐾3𝜆3𝑥(𝑡;𝑤)𝑦(𝑡;𝑤)𝐷.(3.20) Since (𝛾+𝐾1𝜆1+𝐾2𝜆2+𝐾3𝜆3)<1, 𝑈 is a contraction operator on 𝑆.
We will now show that 𝑈𝑆𝑆. For any 𝑥(𝑡;𝑤)𝑆, we have
(𝑈𝑥)(𝑡;𝑤)𝐷(𝑡,𝑥(𝑡;𝑤))𝐷+𝑡0𝑘1(𝑡,𝜏;𝑤)𝑓1(𝜏,𝑥(𝜏;𝑤)𝑑𝜏𝐷+0𝑘2(𝑡,𝜏;𝑤)𝑓2(𝜏,𝑥(𝜏;𝑤))𝑑𝜏𝐷+𝑡0𝑘3(𝑡,𝜏;𝑤)𝑓3(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏)𝐷(𝑡,𝑥(𝑡;𝑤))𝐷+𝐾1𝑓1(𝑡,𝑥(𝑡;𝑤))𝐵+𝐾2𝑓2(𝑡,𝑥(𝑡;𝑤))𝐵+𝐾3𝑓3(𝑡,𝑥(𝑡;𝑤))𝐵𝛾𝑥(𝑡;𝑤)𝐷+𝛾(𝑡,0)𝐷+𝜆1𝐾1𝑥(𝑡;𝑤)𝐷+𝐾1𝑓1(𝑡,0)𝐵+𝜆2𝐾2𝑥(𝑡;𝑤)𝐷+𝐾2𝑓2(𝑡,0)𝐵+𝜆3𝐾3𝑥(𝑡;𝑤)𝐷+𝐾3𝑓3(𝑡,0)𝐵.(3.21) Since 𝑥(𝑡;𝑤)𝐷𝜌, it follows that (𝑈𝑥)(𝑡;𝑤)𝐷𝛾(𝑡,0)𝐷+𝜌𝛾+𝐾1𝜆1+𝐾2𝜆2+𝐾3𝜆3+𝐾1𝑓1(𝑡,0)𝐵+𝐾2𝑓2(𝑡,0)𝐵+𝐾3𝑓3(𝑡,0)𝐵.(3.22) Using the condition that 𝛾(𝑡,0)𝐷+𝐾1𝑓1(𝑡,0)𝐵+𝐾2𝑓2(𝑡,0)𝐵+𝐾3𝑓3(𝑡,0)𝐵𝜌1𝛾𝐾1𝜆1𝐾2𝜆2𝐾3𝜆3,(3.23) we have from (3.18) (𝑈𝑥)(𝑡;𝑤)𝐷𝜌.(3.24) Hence (𝑈𝑥)(𝑡;𝑤)𝑆 for all 𝑥(𝑡;𝑤)𝑆 or 𝑈𝑆𝑆. 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, (𝑈𝑥)(𝑡;𝑤)=(𝑡,𝑥(𝑡;𝑤))+𝑡0𝑘1(𝑡,𝜏;𝑤)𝑓1(+𝜏,𝑥(𝜏;𝑤))𝑑𝜏0𝑘2(𝑡,𝜏;𝑤)𝑓2(𝜏,𝑥(𝜏;𝑤))𝑑𝜏+𝑡0𝑘3(𝑡,𝜏;𝑤)𝑓3(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏)=𝑥(𝑡;𝑤).(3.25)

4. Applications

In this section we will give some application of Theorem 3.2.

Theorem 4.1. Suppose the stochastic integral equation (1.1) satisfies the following conditions: (i)there exists a constant 𝐴>0 and a continuous function 𝑔(𝑡), such that 𝑡0||𝑘1(||𝑡,𝜏;𝑤)2𝑔2(𝜏)𝑑𝜏+0||𝑘2(||𝑡,𝜏;𝑤)2𝑔2(𝜏)𝑑𝜏+𝑡0||𝑘3(||𝑡,𝜏;𝑤)2𝑔2(𝜏)𝑑𝜏<𝐴;(4.1)(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)|𝑥𝑦|, for 𝑥,𝑦𝑅 and 0𝜆𝑖<1,𝑖=1,2,3;(iii)(𝑡,𝑥) is a continuous functions on 𝑅+×𝑅, such that |(𝑡,𝑥)(𝑡,𝑦)|𝛾|𝑥𝑦|, for 𝑥,𝑦𝑅 and 0𝛾<1.
Then there exists a unique random solution 𝑥(𝑡;𝑤) of (1.1) such that
𝑥(𝑡;𝑤)𝐶𝜌(4.2) provided (𝑡,0),𝑓𝑖(𝑡,0)𝐶𝑔,𝑖=1,2,3 are small enough.

Proof. It is easy to show that the hypothesis of Theorem 3.2 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)𝑡0|𝑘1(𝑡,𝜏;𝑤)|2𝑑𝜏+0|𝑘2(𝑡,𝜏;𝑤)|2𝑑𝜏+𝑡0|𝑘3(𝑡,𝜏;𝑤)|2𝑑𝜏<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)|𝑥𝑦|, for 𝑥,𝑦𝑅 and 0𝜆𝑖<1,𝑖=1,2,3;(iii)(𝑡,𝑥) is a continuous functions on 𝑅+×𝑅, such that |(𝑡,𝑥)(𝑡,𝑦)|𝛾|𝑥𝑦|, for 𝑥,𝑦𝑅 and 0𝛾<1.
Then there exists a unique random solution 𝑥(𝑡;𝑤) of (1.1) such that
𝑥(𝑡;𝑤)𝐶𝜌(4.3) provided (𝑡,0),𝑓𝑖(𝑡,0)𝐶𝑔,𝑖=1,2,3 are small enough.

Proof. Take 𝑔(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 𝑡0𝑔2(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,(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 sup0𝑡(𝑇2𝑥)(𝑡;𝑤)𝐶𝑔sup0𝑡0||𝑘2||(𝑡,𝜏;𝑤)2𝑥(𝜏;𝑤)2𝐿2𝑑𝜏1/2𝑥(𝑡;𝑤)𝐶𝑔𝐴0𝑔2(𝜏)𝑑𝜏(4.4) 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 𝑥(𝑡;𝑤)=(𝑡,𝑥(𝑡;𝑤))+𝑛𝑖=1𝑡0𝑎𝑖(𝑡,𝜏;𝑤)𝑓𝑖+(𝜏,𝑥(𝜏;𝑤))𝑑𝜏𝑛𝑖=10𝑏𝑖(𝑡,𝜏;𝑤)𝑔𝑖(𝜏,𝑥(𝜏;𝑤))𝑑𝜏+𝑛𝑖=1𝑡0𝑐𝑖(𝑡,𝜏;𝑤)𝑘𝑖(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏),(4.5) where ,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:

1𝑥(𝑡;𝑤)=4sin𝑥(𝑡;𝑤)+𝑡0sin𝑡4𝑒𝑠𝑥2(𝑠;𝑤)+𝑑𝑠0𝑒𝑡𝑠||||11+𝑥(𝑠;𝑤)𝑑𝑠+8𝑡0||||ln1+𝑥(𝑠;𝑤)𝑑𝛽(𝑠),𝑡𝑅+,(5.1) where 𝛽(𝑡) is a stochastic process. This equation is a particular case of general stochastic integral equation occurring in mathematical biology and chemotherapy [1013]. The above equation takes the form of (1.1) with

𝑘1(𝑡,𝑠,𝑤)=sin𝑡4𝑒𝑠,𝑘2(𝑡,𝑠,𝑤)=𝑒𝑡𝑠,𝑘3(1𝑡,𝑠,𝑤)=4,(t,𝑥(𝑡;𝑤))=sin𝑥(𝑡;𝑤)4𝑓1(𝑠,𝑥(𝑠;𝑤))=𝑒𝑥2(𝑠;𝑤),𝑓21(𝑠,𝑥(𝑠;𝑤))=||𝑥||,𝑓1+(𝑠;𝑤)31(𝑠,𝑥(𝑠;𝑤))=2||||.ln1+𝑥(𝑠;𝑤)(5.2) 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 𝜌10, 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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