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

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 [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

𝑥(𝑡;𝑤)=+ℎ(𝑡,𝑥(𝑡;𝑤))+^𝑡0𝑘₁(𝑡,𝜏;𝑤)𝑓₁(𝜏,𝑥(𝜏;𝑤))𝑑𝜏_∞0𝑘₂(𝑡,𝜏;𝑤)𝑓₂(𝜏,𝑥(𝜏;𝑤))𝑑𝜏+^𝑡0𝑘₃(𝑡,𝜏;𝑤)𝑓₃(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏),(1.1) 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

Let 𝛽(𝑡;𝑤) be the random process. We will assume that for each 𝑡∈𝑅₊, a minimal 𝜎-algebra 𝐴_𝑡, 𝐴_𝑡⊂𝐴, is such that 𝛽(𝑡;𝑤) is measurable with respect to 𝐴_𝑡. In addition, we will assume that the minimal 𝜎-algebra 𝐴_𝑡 is an increasing family such that

(H1) the random process {𝛽(𝑡;𝑤),𝐴_𝑡∶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∶𝐴_𝑡}=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 𝐴_𝑡 measurable and that 𝐸|𝑥(𝑡;𝑤)|²<∞, for each 𝑡∈𝑅₊. Also we denote

𝐸||𝑥(𝑡;𝑤)||²^1/2=‖𝑥(𝑡;𝑤)‖_𝐿²_{(Ω,𝐴,𝜇)}=_Ω||𝑥(𝑡∶𝑤)||²𝑑𝜇(𝑤)^1/2.(2.1)

Definition 2.1. Denote by 𝐶_𝑐 the linear space of all mean square continuous maps 𝑥(𝑡;𝑤) on 𝑅₊ and define a topology on 𝐶_𝑐 by means of the following family of seminorms. ‖𝑥(𝑡;𝑤)‖_𝑛=sup_{0≤𝑡≤𝑛}𝐸||𝑥(𝑡;𝑤)||²^1/2.(2.2) It is known that such a topology is metrizable and that the metric space 𝐶_𝑐 is complete.

Definition 2.2. Define 𝐶_𝑔⊂𝐶_𝑐 to be the space of all maps 𝑥(𝑡;𝑤) on 𝑅₊ such that 𝐸||𝑥(𝑡;𝑤)||²^1/2≤𝑎𝑔(𝑡),(2.3) where 𝑎>0, a constant and 𝑔(𝑡)>0, a continuous function on 𝑅₊. The norm in the space 𝐶_𝑔 is defined by ‖𝑥(𝑡;𝑤)‖_𝐶𝑔=sup_𝑡≥0𝑔1(𝑡)𝐸||𝑥(𝑡;𝑤)||²^1/2.(2.4)

Definition 2.3. Let 𝐶⊂𝐶_𝑐 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=sup_𝑡≥0𝐸||𝑥(𝑡;𝑤)||²^1/2.(2.5)

Definition 2.4. The pair of Banach spaces (𝐵,𝐷) with 𝐵,𝐷⊂𝐶_𝑐 is called admissible with respect to the operator 𝑇∶𝐶_𝑐→𝐶_𝑐 if 𝑇𝐵⊂𝐷.

Definition 2.5. We will call 𝑥(𝑡;𝑤) a random solution of the stochastic integral equation (1.1) if 𝑥(𝑡;𝑤)∈𝐶_𝑐 for each 𝑡∈𝑅₊ 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 𝐶_𝑔, if every sequence which converges in the topology of 𝐵 converges also in the topology of 𝐶_𝑔.
Finally, let 𝐵,𝐷⊂𝐶_𝑔 be Banach spaces and 𝑇 a linear operator from 𝐶_𝑔 into 𝐶_𝑐. The following lemma is well known [13].

Lemma 2.7. Let 𝑇 be a continuous operator from 𝐶_𝑔 into 𝐶_𝑐. If 𝐵 and 𝐷 are Banach spaces in 𝐶_𝑔 stronger than 𝐶_𝑔 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
𝑇𝑇₁𝑥(𝑡;𝑤)=^𝑡0𝑘₁(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)𝑑𝜏,(2.6)₂𝑇𝑥(𝑡;𝑤)=^∞0𝑘₂(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)∈𝐶_𝑔.
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 𝑡∈𝑅₊ 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 0≤𝜏≤𝑡<∞ 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 0≤𝜏≤𝑡<∞ and continuous as maps from {(𝑡,𝜏)∶0≤𝜏≤𝑡<∞} 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)∶0≤𝜏≤𝑡<∞} into 𝐿_∞(Ω,𝐴,𝜇). Define for 0≤𝜏≤𝑡<∞, ‖‖‖‖||||𝑘𝑘₁(𝑡,𝜏;𝑤)||‖‖=𝜇−esssup_𝑤∈Ω||𝑘₁(𝑡,𝜏;𝑤)₂||,‖‖||𝑘(𝑡,𝜏;𝑤)||‖‖=𝜇−esssup_𝑤∈Ω||𝑘₂(𝑡,𝜏;𝑤)₃||,(𝑡,𝜏;𝑤)||‖‖=𝜇−esssup_𝑤∈Ω||𝑘₃(𝑡,𝜏;𝑤)||.(2.9) The assumptions (𝑎₁)-(𝑎₄) imply that if 𝑥(𝑡;𝑤)∈𝐶_𝑐, then for each 𝑡∈𝑅₊, 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 𝐶_𝑔 into 𝐶_𝑐 provided ^∞0‖‖||𝑘₃(𝑡,𝜏;𝑤)||‖‖²𝑔²(𝜏)𝑑𝜏≤𝑁<∞forsome𝑁>0.(2.12)

Proof. It is easy to show that 𝑇₁,𝑇₂ and 𝑇₃ are linear maps from 𝐶_𝑔 into 𝐶_𝑐. The continuity of 𝑇₁ and 𝑇₂ are also easy to prove [8, 13]. We will prove that 𝑇₃ is continuous.
Let 𝑥(𝑡;𝑤)∈𝐶_𝑔. Then
𝐸||(𝑇₃𝑥)(𝑡;𝑤)||²=𝐸^𝑡0𝑘₃(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)𝑑𝛽(𝜏)²=≤^𝑡0𝐸||𝑘₃(𝑡,𝜏;𝑤)𝑥(𝜏;𝑤)||²𝑑𝐹(𝜏)_𝑡0≤‖𝑥(‖‖𝑡;||𝑘𝑤₃)(‖𝑡,𝜏_2𝐶𝑔;𝑤)||‖‖²0𝑥0200𝑑𝐸||𝑥(𝑡;𝑤)||²𝑑𝐹(𝜏)^𝑡0‖‖||𝑘₃(𝑡,𝜏;𝑤)||‖‖²0𝑥0200𝑑𝑔²(𝜏)𝑑𝐹(𝜏),𝑡<𝑛.(2.13) Hence, on compact intervals [0,𝑛]sup_{0≤𝑡≤𝑛}‖‖(𝑇₃𝑥)(𝑡;𝑤)‖‖_{𝐿2(Ω,𝐴,𝜇)}≤‖𝑥(𝑡;𝑤)‖_𝐶𝑔sup_{0≤𝑡≤𝑛}^𝑡0‖‖||𝑘₃(𝑡,𝜏;𝑤)||‖‖²0𝑥0200𝑑𝑔²(𝜏)𝑑𝐹(𝜏)^1/2≤𝑁₁‖𝑥(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 𝐶_𝑔 into 𝐶_𝑐. 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)‖_𝐶𝑔^∞0‖‖||𝑘₂𝑡₁,𝜏;𝑤−𝑘₂𝑡₂,𝜏;𝑤||‖‖²0𝑥0200𝑑𝑔²(𝜏)𝑑𝜏.(2.15) 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 𝐶_𝑔 into 𝐶_𝑐. 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 𝐶_𝑔 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
‖‖‖‖((𝑇𝑇₁₂𝑥)(𝑡;𝑤)‖‖_𝐷≤𝐾₁‖𝑥(𝑡;𝑤)‖_𝐵,‖‖(𝑇₃𝑥)(𝑡;𝑤)‖‖_𝐷≤𝐾₂‖𝑥(𝑡;𝑤)‖_𝐵,𝑥)(𝑡;𝑤)‖‖_𝐷≤𝐾₃‖𝑥(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,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 𝐶_𝑔, stronger than 𝐶_𝑔, 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 𝑆=𝑥(𝑡;𝑤)∶𝑥(𝑡;𝑤)∈𝐷,‖𝑥(𝑡;𝑤)‖_𝐷≤𝜌(3.1) with values in 𝐵 satisfying ‖‖𝑓₁(𝑡,𝑥(𝑡;𝑤))−𝑓₁(𝑡,𝑦(𝑡;𝑤))‖‖_𝐵≤𝜆₁‖𝑥(𝑡;𝑤)−𝑦(𝑡;𝑤)‖_𝐷(3.2) 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 ‖‖𝑓₃(𝑡,𝑥(𝑡;𝑤))−𝑓₃(𝑡,𝑦(𝑡;𝑤))‖‖_𝐵≤𝜆₃‖𝑥(𝑡;𝑤)−𝑦(𝑡;𝑤)‖_𝐷(3.3) 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 ‖ℎ(𝑡,𝑥(𝑡;𝑤))−ℎ(𝑡,𝑦(𝑡;𝑤))‖_𝐷≤𝛾‖𝑥(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‖ℎ(𝑡,0)‖_𝐷𝛾+𝐾₁𝜆₁+𝐾₃𝜆₃<1,+𝐾₁‖‖𝑓₁(𝑡,0)‖‖_𝐵+𝐾₂‖‖𝑓₂(𝑡,𝑥(𝑡;𝑤))‖‖_𝐵+𝐾₃‖‖𝑓₃(𝑡,0)‖‖_𝐵≤𝜌1−𝛾−𝐾₁𝜆₁−𝐾₃𝜆₃,(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 𝑈₁∶𝑆→𝐷 by 𝑈₁𝑥(𝑡;𝑤)=+ℎ(𝑡,𝑥(𝑡;𝑤))+^𝑡0𝑘₁(𝑡,𝜏;𝑤)𝑓₁(𝜏,𝑥(𝜏;𝑤))𝑑𝜏_𝑡0𝑘₃(𝑡,𝜏;𝑤)𝑓₃(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏).(3.6) We will show that 𝑈₁ is a contraction mapping and that 𝑈₁𝑆⊂𝑆. Let 𝑥(𝑡;𝑤),𝑦(𝑡;𝑤)∈𝑆. Then 𝑈₁𝑥(𝑡;𝑤)−𝑈₁𝑦(𝑡;𝑤)=+ℎ(𝑡,𝑥(𝑡;𝑤))−ℎ(𝑡,𝑦(𝑡;𝑤))_𝑡0+𝑘₁(𝑡,𝜏;𝑤)𝑓₁(𝜏,𝑥(𝜏;𝑤))−𝑓₁(𝜏,𝑦(𝜏;𝑤))𝑑𝜏_𝑡0𝑘₃(𝑡,𝜏;𝑤)𝑓₃(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 ‖‖(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)‖.(3.8) Since (𝛾+𝐾₁𝜆₁+𝐾₃𝜆₃)<1, 𝑈₁ is a contraction operator. Next we show that 𝑈₁𝑆⊂𝑆. From (3.6), we have ‖‖(𝑈₁𝑥)(𝑡;𝑤)‖‖_𝐷=+‖ℎ‖‖‖(𝑡,𝑥(𝑡;𝑤))‖_𝐷+‖‖‖^𝑡0𝑘₁(𝑡,𝜏;𝑤)𝑓₁(𝜏,𝑥(𝜏;𝑤))𝑑𝜏_𝑡0‖‖‖≤‖ℎ(𝑡,0𝑘)₃‖(_𝐷𝑡,𝜏;𝑤)𝑓₃(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏)‖‖‖+𝜆₁+𝛾+𝐾₁𝜆₁+𝐾₃𝜆₃‖𝑥(𝑡;𝑤)‖‖‖𝑓₁(𝑡,0)‖‖_𝐵+𝜆₃‖𝑓(𝑡,0)‖_𝐵.(3.9) Since 𝑥(𝑡;𝑤)∈𝑆, by hypothesis, we have ‖(𝑈₁𝑥)(𝑡;𝑤)‖_𝐷≤𝜌 which implies that 𝑈₁𝑆⊂𝑆.
Let us define the operator 𝑈₂∶𝑆→𝐷 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≤‖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.(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 𝐶_𝑔, stronger than 𝐶_𝑔, 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 𝑆=𝑥(𝑡;𝑤)∶𝑥(𝑡;𝑤)∈𝐷,‖𝑥(𝑡;𝑤)‖_𝐷≤𝜌(3.12) with values in 𝐵 satisfying ‖‖𝑓₁(𝑡,𝑥(𝑡;𝑤))−𝑓₁(𝑡,𝑦(𝑡;𝑤))‖‖_𝐵≤𝜆₁‖𝑥(𝑡;𝑤)−𝑦(𝑡;𝑤)‖_𝐷(3.13) 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 ‖‖𝑓₂(𝑡,𝑥(𝑡;𝑤))−𝑓₂(𝑡,𝑦(𝑡;𝑤))‖‖_𝐵≤𝜆₂‖𝑥(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))‖‖_𝐵≤𝜆₃‖𝑥(𝑡;𝑤)−𝑦(𝑡;𝑤)‖_𝐷(3.15) 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 ‖ℎ(𝑡,𝑥(𝑡;𝑤))−ℎ(𝑡,𝑦(𝑡;𝑤))‖_𝐷≤𝛾‖𝑥(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
𝛾𝛾‖+ℎ(𝐾𝑡,₁0𝜆)₁‖+_𝐷𝐾₂𝜆₂+𝐾₃𝜆₃<1,+𝐾₁‖‖𝑓₁(𝑡,0)‖‖_𝐵+𝐾₂‖‖𝑓₂(𝑡,0)‖‖_𝐵+𝐾₃‖‖𝑓₃(𝑡,0)‖‖_𝐵≤𝜌1−𝛾−𝐾₁𝜆₁−𝐾₂𝜆₂−𝐾₃𝜆₃,(3.17) where 𝐾₁,𝐾₂, and 𝐾₃ are defined by (2.16).

Proof. Define the operator 𝑈∶𝑆→𝐷 as follows (𝑈𝑥)(𝑡;𝑤)=+ℎ(𝑡,𝑥(𝑡;𝑤))+^𝑡0𝑘₁(𝑡,𝜏;𝑤)𝑓₁(𝜏,𝑥(𝜏;𝑤))𝑑𝜏_∞0𝑘₂(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 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
‖(𝑈𝑥)(𝑡;𝑤)−(𝑈𝑦)(𝑡;𝑤)‖_𝐷≤‖ℎ(𝑡,𝑥(𝑡;𝑤))−ℎ(𝑡,𝑦(𝑡;𝑤))‖_𝐷+‖‖‖^𝑡0𝑘₁(𝑡,𝜏;𝑤)𝑓₁(𝜏,𝑥(𝜏;𝑤))−𝑓₁(𝜏,𝑦(𝜏;𝑤))𝑑𝜏‖‖‖_𝐷+‖‖‖^∞0𝑘₂(𝑡,𝜏;𝑤)𝑓₂(𝜏,𝑥(𝜏;𝑤))−𝑓₂(𝜏,𝑦(𝜏;𝑤))𝑑𝜏‖‖‖_𝐷+‖‖‖^𝑡0𝑘₃(𝑡,𝜏;𝑤)𝑓₃(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)‖_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.(3.20) Since (𝛾+𝐾₁𝜆₁+𝐾₂𝜆₂+𝐾₃𝜆₃)<1, 𝑈 is a contraction operator on 𝑆.
We will now show that 𝑈𝑆⊂𝑆. For any 𝑥(𝑡;𝑤)∈𝑆, we have
‖(𝑈𝑥)(𝑡;𝑤)‖_𝐷≤‖ℎ(𝑡,𝑥(𝑡;𝑤))‖_𝐷+‖‖‖^𝑡0𝑘₁(𝑡,𝜏;𝑤)𝑓₁(𝜏,𝑥(𝜏;𝑤)𝑑𝜏‖‖‖_𝐷+‖‖‖^∞0𝑘₂(𝑡,𝜏;𝑤)𝑓₂(𝜏,𝑥(𝜏;𝑤))𝑑𝜏‖‖‖_𝐷+‖‖‖^𝑡0𝑘₃(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.(3.21) Since ‖𝑥(𝑡;𝑤)‖_𝐷≤𝜌, it follows that ‖(𝑈𝑥)(𝑡;𝑤)‖_𝐷≤+𝛾‖𝐾ℎ₁(𝑡,0)‖_𝐷+𝜌𝛾+𝐾₁𝜆₁+𝐾₂𝜆₂+𝐾₃𝜆₃‖‖𝑓₁(𝑡,0)‖‖_𝐵+𝐾₂‖‖𝑓₂(𝑡,0)‖‖_𝐵+𝐾₃‖‖𝑓₃(𝑡,0)‖‖_𝐵.(3.22) Using the condition that 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,(3.23) we have from (3.18) ‖(𝑈𝑥)(𝑡;𝑤)‖_𝐷≤𝜌.(3.24) 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, (𝑈𝑥)(𝑡;𝑤)=+ℎ(𝑡,𝑥(𝑡;𝑤))+^𝑡0𝑘₁(𝑡,𝜏;𝑤)𝑓₁(𝜏,𝑥(𝜏;𝑤))𝑑𝜏_∞0=𝑥(𝑡;𝑤𝑘)₂.(𝑡,𝜏;𝑤)𝑓₂(𝜏,𝑥(𝜏;𝑤))𝑑𝜏+^𝑡0𝑘₃(𝑡,𝜏;𝑤)𝑓₃(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏)(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)𝑓_𝑖(𝑡,𝑥),𝑖=1,2,3 are continuous functions on 𝑅₊×𝑅, such that 𝑓_𝑖(𝑡,0)∈𝐶_𝑔(𝑅₊,𝑅) and |𝑓_𝑖(𝑡,𝑥)−𝑓_𝑖(𝑡,𝑦)|≤𝜆_𝑖𝑔(𝑡)|𝑥−𝑦|, for 𝑥,𝑦∈𝑅 and 0≤𝜆_𝑖<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 solution 𝑥(𝑡;𝑤) of (1.1) such that
‖𝑥(𝑡;𝑤)‖_𝐶≤𝜌(4.2) provided ‖ℎ(𝑡,0)‖,‖𝑓_𝑖(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 (𝐶_𝑔,𝐶_𝑐) 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)|²‖𝑑𝜏+∫^∞0‖|𝑘₂(𝑡,𝜏;𝑤)|²‖𝑑𝜏+∫^𝑡0‖|𝑘₃(𝑡,𝜏;𝑤)|²‖𝑑𝜏<𝐴;(ii)𝑓_𝑖(𝑡,𝑥),𝑖=1,2,3 are continuous functions on 𝑅₊×𝑅, such that 𝑓_𝑖(𝑡,0)∈𝐶_𝑔(𝑅₊,𝑅) and |𝑓_𝑖(𝑡,𝑥)−𝑓_𝑖(𝑡,𝑦)|≤𝜆_𝑖𝑔(𝑡)|𝑥−𝑦|, for 𝑥,𝑦∈𝑅 and 0≤𝜆_𝑖<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 solution 𝑥(𝑡;𝑤) of (1.1) such that
‖𝑥(𝑡;𝑤)‖_𝐶≤𝜌(4.3) provided ‖ℎ(𝑡,0)‖,‖𝑓_𝑖(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)‖|𝑘_𝑖(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,‖ℎ(𝑡,0)‖_𝐶 and ‖𝑓_𝑖(𝑡,0)‖_𝐶𝑔 for 𝑖=1,2,3 small enough.

Proof. We will show that the pair is (𝐶_𝑔,𝐶_𝑐) admissible with respect to the operator 𝑇₂. Let 𝑥(𝑡;𝑤)∈𝐶_𝑔. Then sup_0≤𝑡‖‖(𝑇₂𝑥)(𝑡;𝑤)‖‖_𝐶𝑔≤sup_0≤𝑡^∞0‖‖||𝑘₂(𝑡,𝜏;𝑤)||‖‖²‖𝑥(𝜏;𝑤)‖^2𝐿₂𝑑𝜏^1/2≤‖𝑥(𝑡;𝑤)‖_𝐶𝑔𝐴^∞0𝑔²(𝜏)𝑑𝜏(4.4) which implies that the pair (𝐶_𝑔,𝐶_𝑐) is admissible. Similarly we can show that the pair (𝐶_𝑔,𝐶_𝑐) 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)=ℎ(𝑡,𝑥(𝑡;𝑤))+^𝑛_𝑖=1^𝑡0+𝑎_𝑖(𝑡,𝜏;𝑤)𝑓_𝑖(𝜏,𝑥(𝜏;𝑤))𝑑𝜏_𝑛_𝑖=1^∞0𝑏_𝑖(𝑡,𝜏;𝑤)𝑔_𝑖(𝜏,𝑥(𝜏;𝑤))𝑑𝜏+^𝑛_𝑖=1^𝑡0𝑐_𝑖(𝑡,𝜏;𝑤)𝑘_𝑖(𝜏,𝑥(𝜏;𝑤))𝑑𝛽(𝜏),(4.5) where ℎ,𝑘_𝑖,𝑎_𝑖,𝑏_𝑖,𝑐_𝑖,𝑔_𝑖,𝑓_𝑖, and 𝛽 satisfy appropriate conditions. This general case is treated in a separate paper.

5. Example

Consider the following nonlinear stochastic integral equation:

𝑥(𝑡;𝑤)=14sin𝑥(𝑡;𝑤)+^𝑡0sin𝑡4𝑒^{−𝑠−𝑥2(𝑠;𝑤)}+𝑑𝑠_∞0𝑒^{−𝑡−𝑠}1+||𝑥(𝑠;𝑤)||𝑑𝑠+81^𝑡0ln1+||𝑥(𝑠;𝑤)||𝑑𝛽(𝑠),𝑡∈𝑅₊,(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 [10–13]. The above equation takes the form of (1.1) with

𝑘₁(𝑡,𝑠,𝑤)=sin𝑡4𝑒^−𝑠,𝑘₂(𝑡,𝑠,𝑤)=𝑒^{−𝑡−𝑠},𝑘₃(𝑡,𝑠,𝑤)=1𝑓₁4,ℎ(t,𝑥(𝑡;𝑤))=sin𝑥4(𝑡;𝑤)(𝑠,𝑥(𝑠;𝑤))=𝑒^{−𝑥2(𝑠;𝑤)},𝑓₂(𝑠,𝑥(𝑠;𝑤))=1𝑓1+||𝑥(𝑠;𝑤)₃||,(𝑠,𝑥(𝑠;𝑤))=12ln1+||𝑥(𝑠;𝑤)||.(5.2) Take 𝐵=𝐷=𝐶_𝑔=𝐶_𝑐=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).