On a Coupled System of Stochastic It$\widehat{o}$-Differential and the Arbitrary (Fractional) Order Differential Equations with Nonlocal Random and Stochastic Integral Conditions

Abstract

The fractional stochastic differential equations had many applications in interpreting many events and phenomena of life, and the nonlocal conditions describe numerous problems in physics and finance. Here, we are concerned with the combination between the three senses of derivatives, the stochastic It$\widehat{o}$-differential and the fractional and integer orders derivative for the second order stochastic process in two nonlocal problems of a coupled system of two random and stochastic differential equations with two nonlocal stochastic and random integral conditions and a coupled system of two stochastic and random integral conditions. We study the existence of mean square continuous solutions of these two nonlocal problems by using the Schauder fixed point theorem. We discuss the sufficient conditions and the continuous dependence for the unique solution.

1. Introduction

The existence and uniqueness of solutions to stochastic differential equations driven by Brownian motion have been studied by many authors (see [1,2,3] ).

Also the non-local coupled system was studied by some authors (see for example [4,5,6,7,8] and references therein).

The results are important since they cover non-local generalizations of fractional stochastic differential equations (FSDE), more applications are arising in fields such as heat conduction, electromagnetic theory and dynamic system (see for example [9,10]).

Many authors have been interested to study the fractional stochastic differential equations see [11,12,13,14] and investigate their results all the time.

Let $(\Omega ,G,\mu )$ be a probability space see [15].

The motive of this work is to generalize the scope results of A.M.A. El-Sayed [16,17] on the stochastic fractional operators and the solution of non-local coupled systems of stochastic differential equations see [7,16].

Also, we study the existence of solutions of a coupled system of It$\widehat{o}$-differential equation and arbitrary (fractional)orders random differential equation subject to two coupled systems of non-local random and stochastic integral conditions. The effect of random functions and data which ensures the continuous dependence of the solution has been proved.

Let $I=[0,T]$ and $X(t;\omega )=X\left(t\right),t\in I,\omega \in \Omega$ be such that $E\left(X^2\left(t\right)\right)<\infty ,t\in I.$

Let $C=C(I,L_2(\Omega ))$ be the class of all mean square (m.s) continuous stochastic processes on I with norm

$${\parallel X\parallel }_C=\underset{t\in I}{sup}{\parallel X\left(t\right)\parallel }_2,{\parallel X\left(t\right)\parallel }_2={\left(E{X}^2\left(t\right)\right)}^{1/2}.$$

Definition 1.

Let $X\in C(I,L_2(\Omega ))$ and $\mu ,\nu \in (0,1].$ The stochastic integral operator of order ν is defined by

$$I^{\nu }X\left(t\right)={\int }_0^t\frac{{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)}X\left(s\right)ds$$

and the stochastic fractional order derivative is defined by

$$D^{\mu }X\left(t\right)=I^{1-\mu }\frac{dX}{dt}.$$

For the properties of stochastic fractional calculus see [16].

Let $\alpha ,\beta \in (0,1]$ and $T\ge 1.$ Here we prove the existence of solutions $X,Y\in C([0,T],L_2(\Omega ))$ of the two nonlocal fractional coupled system of the two It$\widehat{o}$-deferential and arbitrary orders, differential equations

$$\frac{dX\left(t\right)}{dt}=g_1(t,Y\left(\varphi \left(t\right)\right)),t\in (0,T]$$

(1)

and

$$dY\left(t\right)=g_2(t,D^{\alpha }X\left(t\right)))dW\left(t\right),t\in (0,T]$$

(2)

with the stochastic and random integral conditions

$$X\left(0\right)+{\int }_0^{\tau }{h}_1(s,D^{\beta }X\left(s\right))dW\left(s\right)=X_o,Y\left(0\right)+{\int }_0^{\eta }{h}_2(s,Y\left(s\right))ds=Y_o$$

(3)

and the coupled system of the two stochastic and random non-local integral conditions

$$X\left(0\right)+{\int }_0^{\tau }{h}_1(s,Y\left(s\right))dW\left(s\right)=X_o,Y\left(0\right)+{\int }_0^{\eta }{h}_2(s,X\left(s\right))ds=Y_o$$

(4)

where $X_o$ and $Y_o$ are two second order random variables.

The existence of solutions $X,Y\in C(I,L_2(\Omega ))$ of the problems (1)–(3) and (1)–(2) and (4) are proved. The continuous dependence of the unique solutions $X,Y\in C(I,L_2(\Omega ))$ on $X_o$ and $Y_o,$ $h_1$ and $h_2$ and the solution $Y\in C(I,L_2(\Omega ))$ on $D^{\alpha }X\left(t\right)$ will be studied.

2. Integral Representations of the Solution

Consider the following assumptions:

$\left(a1\right)$

$\varphi :I\to I,\varphi \left(t\right)\le t$ is a continuous on I.

$\left(a2\right)$

$g_i:I\times L_2(\Omega )\to L_2(\Omega )$ are (Caratheodory) measurable in $t\in I,\forall X\in L_2(\Omega )$ and continuous in $X\in L_2(\Omega )\forall t\in I$ and there exist $b_i>0$ and two bounded measurable functions ${\xi }_i:I\to R$ such that

$$\parallel g_i{(t,X)\parallel }_2\le |{\xi }_i\left(t\right)|+b_i{\parallel X\left(t\right)\parallel }_2,i=1,2.$$

(5)

$\left(a3\right)$

$h_i:[0,T]\times L_2(\Omega )\to L_2(\Omega )$ are Caratheodory. There exist $c_i>0$ and two bounded measurable functions ${\sigma }_i:I\to R$ such that

$$\parallel h_i{(t,X)\parallel }_2\le |{\sigma }_i\left(t\right)|+c_i{\parallel X\left(t\right)\parallel }_2,i=1,2.$$

(6)

$\left(a4\right)$

$M=max\{{sup}_{t\in I}|{\xi }_1\left(t\right)|,{sup}_{t\in I}|{\xi }_2\left(t\right)\left|\right\},b=max\{b_1,b_2\}.$

$\left(a5\right)$

$K=max\{{sup}_{t\in I}|{\sigma }_1\left(t\right)|,{sup}_{t\in I}|{\sigma }_2\left(t\right)\left|\right\},c=max\{c_1,c_2\}.$

$\left(a6\right)$

$(b+c)T+b^2{T}^2(2-\alpha )<1.$

Now, operating by $I^{1-\alpha }$ on Equation (1), we obtain

$$D^{\alpha }X\left(t\right))=I^{1-\alpha }\frac{dX}{dt}=I^{1-\alpha }{g}_1(t,Y\left(\varphi \left(t\right)\right)).$$

Let

$$u\left(t\right)=D^{\alpha }X\left(t\right),$$

(7)

then from (1)–(2) we obtain

$$u\left(t\right)=I^{1-\alpha }{g}_1(t,Y\left(\varphi \left(t\right)\right)$$

(8)

and

$$dY\left(t\right)=g_2(t,u\left(t\right))dW\left(t\right).$$

(9)

Then we have the following lemma

Lemma 1.

The solutions of the problems (1)–(3) and (1), (2) and (4) can be given by

$$\begin{array}{ccc}\hfill X\left(t\right)& =& X_0-{\int }_0^{\tau }{h}_1(s,I^{1-\beta }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)+{\int }_0^t{g}_1(s,Y\left(\varphi \left(s\right)\right))ds,\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill Y\left(t\right)& =& Y_0-{\int }_0^{\eta }{h}_2(s,Y\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)\hfill \end{array}$$

(11)

and

$$\begin{array}{ccc}\hfill X\left(t\right)& =& X_o-{\int }_0^{\tau }{h}_1(s,Y\left(s\right))dW\left(s\right)+{\int }_0^t{g}_1(s,Y\left(\varphi \left(s\right)\right))ds,\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill Y\left(t\right)& =& Y_o-{\int }_0^{\eta }{h}_2(s,X\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right),\hfill \end{array}$$

(13)

respectively.

Proof. 

Integrating the Equations (7) and (9) (see [12,13,14,15,16,17,18]) with substitution by (8) and using the non-local conditions (3) and (4) the equivalent between the problem (1)–(3) and the integral representation (10)–(11) and the problem (1), (2) and (4) and the integral representation (12)–(13) can be proved. □

3. Solutions of the Problem (1)–(3)

Theorem 1.

Let the assumptions $\left(a1\right)$–$\left(a6\right)$ be satisfied, then the problem (1)–(3) has at least one solution $X,Y\in C$.

Proof. 

Consider the set $Q_1$ such that

$$Q_1=\{Y\left(t\right)\in L_2{(\Omega ):\parallel Y\left(t\right)\parallel }_2\le r_1\}\subset C(I,L_2(\Omega ))$$

and define the mapping $F_1$ where

$$F_{1}Y\left(t\right)=Y_o-{\int }_0^{\eta }{h}_2(s,Y\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right).$$

(14)

Let $Y\in Q_1,$ then

$$\begin{array}{ccc}\hfill \parallel F_1{Y\left(t\right)\parallel }_2& \le & \parallel Y_o{\parallel }_2+\parallel {\int }_0^{\eta }{h}_2{(s,Y\left(s\right))ds\parallel }_2+{\parallel {\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)\parallel }_2\hfill \\ & \le & \parallel Y_o{\parallel }_2+{\int }_0^{\eta }({\sigma }_2\left(s\right)+c_2{\parallel Y\parallel }_C)ds\hfill \\ & +& \sqrt{{\int }_0^t({\xi }_2\left(s\right)+b_2\parallel I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right))){\parallel }_2^{2}ds}\hfill \\ & \le & \parallel Y_o{\parallel }_2+(K+c{r}_1)T+(M+\frac{b{T}^{1-\alpha }(M+b{r}_1)}{(1-\alpha )\Gamma (1-\alpha )})\sqrt{T}\hfill \\ & \le & \parallel Y_o{\parallel }_2+KT+M\sqrt{T}+\frac{Mb{T}^{\frac{3}{2}-\alpha }}{\Gamma (2-\alpha )}+(cT+\frac{b^2{T}^{\frac{3}{2}-\alpha }}{\Gamma (2-\alpha )})r_1\hfill \\ & \le & \parallel Y_o{\parallel }_2+(K+M)T+Mb{T}^2(2-\alpha )+(cT+b^2{T}^2(2-\alpha ))r_1=r_1\hfill \end{array}$$

where

$$r_1=\frac{(\parallel Y_o{\parallel }_2+(K+M))T+Mb{T}^2(2-\alpha )}{1-cT-b^2{T}^2(2-\alpha )},$$

then $F_1:Q_1\to Q_1$ and the class $\left\{F_{1}Y\left(t\right)\right\},t\in [0,T]$ is uniformly bounded on $Q_1.$ □

Let $t_1,t_2\in [0,T]$ such that $|t_2-t_1|<\delta ,$ then

$$\begin{array}{ccc}\hfill \parallel F_{1}Y\left(t_1\right)-F_{1}Y\left(t_2\right){\parallel }_2& =& \parallel {\int }_0^{t_2}{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)\hfill \\ & -& {\int }_0^{t_1}{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right))){dW\left(s\right)\parallel }_2\hfill \\ & =& \parallel {\int }_{t_1}^{t_2}{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right))){dW\left(s\right)\parallel }_2\hfill \\ & \le & \sqrt{{\int }_{t_1}^{t_2}{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right))){\parallel }_2^{2}ds}\hfill \\ & \le & (M+b\left(\frac{T^{1-\alpha }{(M+b\parallel Y\parallel }_c)}{\Gamma (2-\alpha )}\right))\sqrt{t_2-t_1}.\hfill \end{array}$$

This proves the equicontinuity of the class $\left\{F_{1}Y\left(t\right)\right\},t\in [0,T]$ on $Q_1$.

Let $Y_n\in Q_1,Y_n\to Yw.p.1,$ we get (see [15])

$$\begin{array}{ccc}\hfill {L.i.m}_{n\to \infty }{F}_1{Y}_n& =& {L.i.m}_{n\to \infty }(Y_o-{\int }_0^{\eta }{h}_2(s,Y_n\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y_n\left(\varphi \left(s\right)\right)))dW\left(s\right))\hfill \\ & =& Y_o-{\int }_0^{\eta }{h}_2(s,{L.i.m}_{n\to \infty }{Y}_n\left(s\right))ds\hfill \\ & +& {\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,{L.i.m}_{n\to \infty }{Y}_n\left(\varphi \left(s\right)\right)))dW\left(s\right)\hfill \\ & =& Y_o-{\int }_0^{\eta }{h}_2(s,Y\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)=F_{1}Y.\hfill \end{array}$$

This proves that the operator $F_1:Q_1\to Q_1$ is continuous. Consequently, the closure of $\left\{F_1{Q}_1\right\}$ is compact and (see [15]) and integral Equation (11) has a solution $Y\in C([0,T],L_2(\Omega ))$.

Let the set $Q_2$ be such that

$$Q_2=\{X\left(t\right)\in L_2{(\Omega ):\parallel X\left(t\right)\parallel }_2\le r_2\}\subset C([0,T],L_2(\Omega ))$$

and define the mapping $F_{2}X$ such that

$$F_{2}X\left(t\right)=X_o-{\int }_0^{\tau }{h}_1(s,I^{1-\beta }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)+{\int }_0^t{g}_1(s,Y\left(\varphi \left(s\right)\right))ds.$$

(15)

Let $X\in Q_2$, then

$$\begin{array}{ccc}\hfill \parallel F_2{X\left(t\right)\parallel }_2& \le & \parallel X_o{\parallel }_2+\sqrt{{\int }_0^{\tau }{\parallel h_1(s,I^{1-\beta }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)\parallel }_2^{2}ds}+{\int }_0^t{\parallel g_1(s,Y\left(\varphi \left(s\right)\right))\parallel }_{2}ds\hfill \\ & \le & \parallel X_o{\parallel }_2+\sqrt{{\int }_0^{\tau }\left(\right|{\sigma }_1\left(s\right)|+c_1\parallel I^{1-\beta }{g}_1(s,Y\left(\varphi \left(s\right)\right)){{\parallel }_2)}^{2}ds}+{\int }_0^t\left(\right|{\xi }_1\left(s\right)|+b_1{\parallel Y\parallel }_C)ds\hfill \\ & \le & \parallel X_o{\parallel }_2+(K+\frac{c{T}^{1-\beta }(M+b{r}_1)}{\Gamma (2-\alpha )})\sqrt{T}+(M+b{r}_1)T\hfill \\ & \le & \parallel X_o{\parallel }_2+(K+M)T+\frac{cM{T}^{\frac{3}{2}-\beta }}{\Gamma (2-\alpha )}+(\frac{cb{T}^{\frac{3}{2}-\beta }}{\Gamma (2-\alpha )}+bT)r_1\hfill \\ & \le & \parallel X_o{\parallel }_2+(K+M)T+Mc{T}^2(2-\beta )+(bT+bc{T}^2(2-\beta ))r_1=r_2\hfill \end{array}$$

where

$$r_2={\parallel X_0\parallel }_2+(K+M)T+Mc{T}^2(2-\beta )+(bT+bc{T}^2(2-\beta ))r_1,$$

then the class $\left\{F_{2}X\left(t\right)\right\},t\in [0,T]$ is uniformly bounded.

Let $X\in Q_2;t_1,t_2\in [0,T],$ then

$$\begin{array}{ccc}\hfill \parallel F_{2}X\left(t_2\right)-F_{2}X\left(t_1\right){\parallel }_2& =& \parallel {\int }_{t_1}^{t_2}{g}_1{(s,Y\left(\varphi \left(s\right)\right))ds\parallel }_2\hfill \\ & \le & (M+b{r}_1)(t_2-t_1),\hfill \end{array}$$

then the class $\left\{F_{2}X\left(t\right)\right\},t\in [0,T]$ is equicontinuous.

Let $X_n\in Q_2$ be such that $X_n\to Yw.p.1.$

Using Theorem 1, we get

$$\begin{array}{ccc}\hfill {L.i.m}_{n\to \infty }{F}_2{X}_n& =& {L.i.m}_{n\to \infty }(X_o-{\int }_0^{\tau }{h}_1(s,I^{1-\beta }{g}_1(s,Y_n\left(\varphi \left(s\right)\right)))dW\left(s\right)+{\int }_0^t{g}_1(s,Y_n\left(\varphi \left(s\right)\right))ds)\hfill \\ & =& X_o-{\int }_0^{\tau }{h}_1(s,{L.i.m}_{n\to \infty }{I}^{1-\beta }{g}_1(s,Y_n\left(\varphi \left(s\right)\right)))dW\left(s\right)+{\int }_0^t{g}_1(s,{L.i.m}_{n\to \infty }{Y}_n\left(\varphi \left(s\right)\right))ds\hfill \\ & =& X_o-{\int }_0^{\tau }{h}_1(s,,I^{1-\beta }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)+{\int }_0^t{g}_1(s,Y\left(\varphi \left(s\right)\right))ds=F_{2}X.\hfill \end{array}$$

Applying Schauder Fixed Point Theorem [15], (10) has a solution $X\in C(I,L_2(\Omega )).$

3.1. Uniqueness Theorem

To discuss the uniqueness of the solution Y of (11) consider the assumptions $\left(a^{*}1\right)-\left(a^{*}2\right)$ instead of $\left(a1\right)-\left(a2\right)$ such that

$\left(a^{*}1\right)$

$g_i:I\times L_2(\Omega )\to L_2(\Omega );i=1,2$ are Caratheodory and satisfy second argument Lipschitz condition

$$\parallel g_i(t,u\left(t\right))-g_i{(t,v\left(t\right))\parallel }_2\le b{\parallel u\left(t\right)-v\left(t\right)\parallel }_2.$$

$\left(a^{*}2\right)$

$h_i:I\times L_2(\Omega )\to L_2(\Omega );i=1,2$ Caratheodory and satisfy second argument Lipschitz condition

$$\parallel h_i(t,u\left(t\right))-h_i{(t,v\left(t\right))\parallel }_2\le c{\parallel u\left(t\right)-v\left(t\right)\parallel }_2.$$

It is clear that the assumptions $\left(a^{*}1\right)-\left(a^{*}2\right)$ imply the assumptions $\left(a1\right)-\left(a2\right).$

Theorem 2.

Let $\left(a^{*}1\right)-\left(a^{*}2\right)$ and $\left(a3\right)-\left(a6\right)$ be satisfied, then the solution of (1)–(3) is unique.

Proof. 

Let $Y_1$ and $Y_2$ be two solutions of (11), then

$$\begin{array}{ccc}\hfill \parallel Y_1\left(t\right)-Y_2{\left(t\right)\parallel }_2& \le & {\int }_0^{\eta }{\parallel h_2(s,Y_2\left(s\right))-h_2(s,Y_1\left(s\right))\parallel }_{2}ds\hfill \\ & +& \sqrt{{\int }_0^t{\parallel g_2(s,I^{1-\alpha }{g}_1(s,Y_1\left(\varphi \left(s\right)\right)))-g_2(s,I^{1-\alpha }{g}_1(s,Y_2\left(\varphi \left(s\right)\right)))\parallel }_2^{2}ds}\hfill \\ & \le & cT\parallel Y_1-Y_2{\parallel }_C+\frac{b^2{T}^{\frac{3}{2}-\alpha }}{\Gamma (2-\alpha )}{\parallel Y_1-Y_2\parallel }_C\hfill \\ & \le & (cT+b^2{T}^2(2-\alpha )){\parallel Y_1-Y_2\parallel }_C.\hfill \end{array}$$

and

$$(1-cT-b^2{T}^2(2-\alpha )){\parallel Y_1-Y_2\parallel }_C\le 0.$$

Using $\left(a6\right)$ we can get

$$\parallel Y_1-Y_2{\parallel }_C=0,$$

then the solution of (11) is unique. Consequently, the solution of (10) is unique. □

Combining the results, then we deduce that the solution $X,Y\in C$ of the problem the problem (1)–(3) is unique.

3.2. Continuous Dependence

Theorem 3.

The unique solution of the problem (1)–(3) depends continuously on $X_o,Y_o.$

Proof. 

Let $\widehat{X},\widehat{Y}$ be the solution of

$$\begin{array}{ccc}\hfill \widehat{X}\left(t\right)& =& {\widehat{X}}_o-{\int }_0^{\tau }{h}_1(s,I^{1-\beta }{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right)))dW\left(s\right)+{\int }_0^t{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right))ds,\hfill \\ \hfill \widehat{Y}\left(t\right)& =& {\widehat{Y}}_o-{\int }_0^{\eta }{h}_2(s,\widehat{Y}\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right)))dW\left(s\right).\hfill \end{array}$$

Let

$$\parallel X_o-{\widehat{X}}_o{\parallel }_2\le \delta \mathrm{and}{\parallel Y_o-{\widehat{Y}}_o\parallel }_2\le \delta ,$$

then

$$\begin{array}{ccc}\hfill \parallel Y-\widehat{Y}{\parallel }_C& \le & \parallel Y_o-{\widehat{Y}}_o{\parallel }_2+c\parallel Y-\widehat{Y}{\parallel }_{C}T+\frac{b^2{T}^{\frac{3}{2}-\alpha }}{\Gamma (2-\alpha )})\parallel Y-\widehat{Y}{\parallel }_C\hfill \\ & \le & \parallel Y_o-{\widehat{Y}}_o{\parallel }_2+(cT+b^2{T}^2(2-\alpha )){\parallel Y-\widehat{Y}\parallel }_C,\hfill \end{array}$$

$$(1-cT-b^2{T}^2(2-\alpha ))\parallel Y-\widehat{Y}{\parallel }_C\le {\parallel Y_0-{\widehat{Y}}_0\parallel }_2$$

and

$$\parallel Y-\widehat{Y}{\parallel }_C\le \frac{\delta }{1-cT-b^2{T}^2(2-\alpha )}=ϵ.$$

(16)

In the same way, we have

$$\parallel X-\widehat{X}{\parallel }_C\le \parallel X_o-{\widehat{X}}_o{\parallel }_2+\left[\frac{c{T}^{1-\beta }b{\parallel Y-\widehat{Y}\parallel }_C}{\Gamma (2-\beta )}\right]\sqrt{T}+bT{\parallel Y-\widehat{Y}\parallel }_C,$$

$$\parallel X-\widehat{X}{\parallel }_C\le \parallel X_o-{\widehat{X}}_o{\parallel }_2+[cb{T}^2(2-\beta )+bT]{\parallel Y-\widehat{Y}\parallel }_2$$

and

$$\parallel X-\widehat{X}{\parallel }_C\le \delta \{1+\frac{cb{T}^2(2-\beta )+bT}{1-cT-b^2{T}^2(2-\alpha )}\}=ϵ.$$

(17)

□

Theorem 4.

The solution of (1)–(3) depends continuously on $h_1$ and $h_2$.

Proof. 

Let $\widehat{X},\widehat{Y}$ be the solution of

$$\begin{array}{ccc}\hfill \widehat{X}\left(t\right)& =& X_o-{\int }_0^{\tau }{h}_1^{*}(s,I^{1-\beta }{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right)))dW\left(s\right)+{\int }_0^t{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right))ds,\hfill \\ \hfill \widehat{Y}\left(t\right)& =& Y_o-{\int }_0^{\eta }{h}_2^{*}(s,\widehat{Y}\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right)))dW\left(s\right),\hfill \end{array}$$

If $\parallel h_j^{*}{(s,.)-h(s,.)\parallel }_2\le {\delta }_1,j=1,2$ then

$$\begin{array}{ccc}\hfill \parallel Y\left(t\right)-\widehat{Y}{\left(t\right)\parallel }_2& =& \parallel {\int }_0^{\eta }[h_2^{*}(s,\widehat{Y}\left(s\right))-h_2(s,Y\left(s\right))]ds\hfill \\ & +& {\int }_0^t[g_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))-g_2(s,I^{1-\alpha }{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right)))]{dW\left(s\right)\parallel }_2\hfill \\ & \le & ({\delta }_1+c\parallel Y-\widehat{Y}{\parallel }_C)T+\frac{b^2{T}^{\frac{3}{2}-\alpha }}{\Gamma (2-\alpha )}{\parallel Y-\widehat{Y}\parallel }_C\hfill \\ & \le & {\delta }_{1}T+[cT+b^2{T}^2(2-\alpha )]{\parallel Y-\widehat{Y}\parallel }_C\hfill \end{array}$$

and

$$\parallel Y-\widehat{Y}{\parallel }_C\le \frac{{\delta }_{1}T}{1-cT-b^2{T}^2(2-\alpha )}$$

(18)

and similarly we have

$$\begin{array}{ccc}\hfill \parallel X\left(t\right)-\widehat{X}{\left(t\right)\parallel }_2& \le & \parallel {\int }_0^{\eta }[h_1^{*}(s,I^{1-\beta }{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right)))-h_1(s,I^{1-\beta }{g}_1(s,Y\left(\varphi \left(s\right)\right)))]{dW\left(s\right)\parallel }_2\hfill \\ & +& \parallel {\int }_0^t\left[g_1(s,Y\left(\varphi \left(s\right)\right))\right)-g_1(s,\widehat{Y}\left(\varphi \left(s\right)\right)){]ds\parallel }_2\hfill \\ & \le & \sqrt{{\int }_0^{\tau }{({\delta }_1+\frac{bc{T}^{1-\beta }{\parallel Y-\widehat{Y}\parallel }_C}{\Gamma (2-\beta )})}^{2}ds}+bT{\parallel Y-\widehat{Y}\parallel }_C\hfill \\ & \le & ({\delta }_1+\frac{bc{T}^{1-\beta }{\parallel Y-\widehat{Y}\parallel }_C}{\Gamma (2-\beta )})\sqrt{T}+bTc\hfill \\ & \le & {\delta }_1\sqrt{T}+[bc{T}^2(2-\beta )+bT]{\parallel Y-\widehat{Y}\parallel }_C\hfill \end{array}$$

and

$$\parallel X-\widehat{X}{\parallel }_C\le {\delta }_{1}T[1+\frac{bc{T}^2(2-\beta )+bT}{1-cT-b^2{T}^2(2-\alpha )}]=ϵ.$$

(19)

From (18) and (19) the solution $X,Y$ depends continuously on $h_1,h_2.$ □

Theorem 5.

The solution of (1)–(3) depends continuously on $D^{\alpha }X\left(t\right)=u\left(t\right)$.

Proof. 

Let $\widehat{X},\widehat{Y}$ be the solution of

$$\begin{array}{ccc}\hfill \widehat{X}\left(t\right)& =& X_o-{\int }_0^{\tau }{h}_1(s,I^{\alpha -\beta }\widehat{u}\left(s\right))dW\left(s\right)+I^{\alpha }\widehat{u}\left(t\right),\hfill \\ \hfill \widehat{Y}\left(t\right)& =& Y_o-{\int }_0^{\eta }{h}_2(s,\widehat{Y}\left(s\right))ds+{\int }_0^t{g}_2(s,\widehat{u}\left(s\right)))dW\left(s\right)\hfill \end{array}$$

such that $\parallel \widehat{u\left(t\right)}{-u\left(t\right)\parallel }_2\le {\delta }_2,$ then

$$\begin{array}{ccc}\hfill \parallel X\left(t\right)-\widehat{X}{\left(t\right)\parallel }_2& \le & \parallel {\int }_0^{\eta }[h_1(s,I^{\alpha -\beta }\widehat{u}\left(s\right))-h_1(s,I^{\alpha -\beta }u\left(s\right))]{dW\left(s\right)\parallel }_2+{\parallel I^{\alpha }u\left(t\right)-I^{\alpha }\widehat{u}\left(t\right)\parallel }_2\hfill \\ & \le & \frac{c{T}^{\alpha -\beta }{\delta }_2\sqrt{T}}{\Gamma (\alpha -\beta +1)}+\frac{T^{\alpha }{\delta }_2}{\Gamma (\alpha +1)}\hfill \end{array}$$

and

$$\parallel X-\widehat{X}{\parallel }_C\le T^{\alpha }{\delta }_2[\frac{c{T}^{\frac{1}{2}-\beta }}{\Gamma (\alpha -\beta +1)}+\frac{1}{\Gamma (\alpha +1)}]=ϵ.$$

(20)

By the same way

$$\begin{array}{ccc}\hfill \parallel Y\left(t\right)-\widehat{Y}{\left(t\right)\parallel }_2& \le & \parallel {\int }_0^{\eta }[h_2(s,\widehat{Y}\left(s\right))-h_2(s,Y\left(s\right))]{ds\parallel }_2+{\parallel {\int }_0^t[g_2(s,u\left(s\right))-g_2(s,\widehat{u}\left(s\right))]dW\left(s\right)\parallel }_2\hfill \\ & \le & cT\parallel Y-\widehat{Y}{\parallel }_C)+b\sqrt{T}{\delta }_2=ϵ.\hfill \end{array}$$

Now

$$\parallel Y-\widehat{Y}{\parallel }_C\le \frac{bT{\delta }_2}{(1-cT)}=ϵ$$

(21)

which is complete the proof. □

4. Solutions of the Problem (1)–(2) and (4)

Let $\Lambda =C(I,L_2(\Omega ))\times C(I,L_2(\Omega ))$ be the set of ordered pairs $(X,Y),X,Y\in C$ and

$${\parallel (X,Y)\parallel }_{\Lambda }={\parallel X\parallel }_C+{\parallel Y\parallel }_C.$$

(22)

Define the mapping $P(X,Y)=(P_{1}Y,P_{2}X)$ where $P_{1}Y,P_{2}X$ are given by

$$\begin{array}{ccc}\hfill P_{1}Y\left(t\right)& =& X_0-{\int }_0^{\tau }{h}_1(s,Y\left(s\right))dW\left(s\right)+{\int }_0^t{g}_1(s,Y\left(\varphi \left(s\right)\right))ds,\hfill \end{array}$$

(23)

$$\begin{array}{ccc}\hfill P_{2}X\left(t\right)& =& Y_0-{\int }_0^{\eta }{h}_2(s,X\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right))dW\left(s\right).\hfill \end{array}$$

(24)

Consider the set $Q,$

$$Q_{=}\{(X,Y)\in L_2(\Omega ),(X,Y)\in {\Lambda :\parallel (X,Y)\parallel }_{\Lambda }={\parallel X\left(t\right)\parallel }_2+{\parallel Y\left(t\right)\parallel }_2\}\le r\}.$$

4.1. Existence Theorem

Theorem 6.

Let $T\ge 1$ and $\left(a1\right)-\left(a6\right)$ be satisfied, then (1)–(2) and (4) has a solution $(X,Y)\in \Lambda$.

Proof. 

Let $(X,Y)\in Q,$ then we have

$$\begin{array}{ccc}\hfill \parallel P_1{Y\left(t\right)\parallel }_2& \le & \parallel X_0{\parallel }_2+\parallel {\int }_0^{\eta }{h}_1{(s,Y\left(s\right))dW\left(s\right)\parallel }_2+{\parallel {\int }_0^t{g}_1(s,Y\left(\varphi \left(s\right)\right))ds\parallel }_2\hfill \\ & \le & \parallel X_0{\parallel }_2{+(K+c\parallel Y\parallel }_C)\sqrt{T}{+(M+b\parallel Y\parallel }_C\left)\right)T\hfill \\ & \le & \parallel X_0{\parallel }_2+MT+K\sqrt{T}+(c\sqrt{T}+bT){\parallel Y\parallel }_C\hfill \\ & \le & \parallel X_0{\parallel }_2+(M+K)T+(c+b)T{\parallel Y\parallel }_C,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \parallel P_2{X\left(t\right)\parallel }_2& \le & \parallel Y_0{\parallel }_2+\parallel {\int }_0^{\eta }{h}_2{(s,X\left(s\right))ds\parallel }_2+{\parallel {\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)\parallel }_2\hfill \\ & \le & \parallel Y_0{\parallel }_2={\int }_0^{\eta }{\parallel h_2(s,X\left(s\right))\parallel }_{2}ds+\sqrt{{\int }_0^t{\parallel g_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))\parallel }_2^{2}ds}\hfill \\ & \le & \parallel Y_0{\parallel }_2{+(K+c\parallel X\parallel }_C)T+(M+\frac{b{T}^{1-\alpha }}{\Gamma (2-\alpha )}{(M+b\parallel Y\parallel }_C)\sqrt{T}\hfill \\ & \le & \parallel Y_0{\parallel }_2+KT+M\sqrt{T}+\frac{bM{T}^{\frac{3}{2}-\alpha }}{\Gamma (2-\alpha )}+\frac{b^2{T}^{\frac{3}{2}-\alpha }}{\Gamma (2-\alpha )}{\parallel Y\parallel }_C+Tc{\parallel X\parallel }_C\hfill \\ & \le & \parallel Y_0{\parallel }_2+(K+M)T+Mb{T}^2(2-\alpha )+b^2{T}^2{(2-\alpha )\parallel Y\parallel }_C+Tc{\parallel X\parallel }_C.\hfill \end{array}$$

This implies that

$$\begin{array}{ccc}\hfill {\parallel P(X,Y)\parallel }_X& =& \parallel (P_{1}Y,P_{2}X){\parallel }_{\Lambda }\hfill \\ & =& \parallel P_1{Y\parallel }_C+{\parallel P_{2}X\parallel }_C\hfill \\ & \le & \parallel X_0{\parallel }_2+{\parallel Y_0\parallel }_2+2(K+M)T+Mb{T}^2(2-\alpha )\hfill \\ & +& [(c+b)T+b^2{T}^2(2-\alpha )]{\{\parallel X\parallel }_C+{\parallel Y\parallel }_C\}\hfill \end{array}$$

$$\begin{array}{ccc}& \le & \parallel (X_0,Y_0){\parallel }_{\Lambda }+2(K+M)T+Mb{T}^2(2-\alpha )+[(c+b)T+b^2{T}^2(2-\alpha )]{\parallel (X,Y)\parallel }_{\Lambda }\hfill \\ & \le & \parallel (X_0,Y_0){\parallel }_{\Lambda }+2(K+M)T+Mb{T}^2(2-\alpha )+[(c+b)T+b^2{T}^2(2-\alpha )]r=r\hfill \end{array}$$

where

$$r=\frac{\parallel (X_0,Y_0){\parallel }_{\Lambda }+2(K+M)T+Mb{T}^2(2-\alpha )}{1-cT-bT-b^2{T}^2(2-\alpha )},$$

then the class $\left\{P\right(X,Y\left)\right(t),t=[0,T\left]\right\}$ is uniformly bounded and $P(x,y):Q\to Q.$ □

Let $(X,Y)\in Q,t_1,t_2\in [0,T]$ such that $|t_2-t_1|<\delta ,$ then

$$\begin{array}{ccc}\hfill \parallel P_{1}Y\left(t_2\right)-P_{1}Y\left(t_1\right){\parallel }_2& =& \parallel {\int }_{t_1}^{t_2}{g}_1(s,Y\left({\varphi }_1\left(s\right)\right)){ds\parallel }_2\hfill \\ & \le & {(M+b\parallel Y\parallel }_C)(t_2-t_1),\hfill \\ \hfill and\end{array}$$

(25)

$$\begin{array}{ccc}\hfill \parallel P_{2}X\left(t_2\right)-P_{2}X\left(t_1\right){\parallel }_2& =& \parallel {\int }_{t_1}^{t_2}{g}_2{(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right))dW\left(s\right)\parallel }_2\hfill \\ & \le & (M+b\parallel I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)){\parallel }_2)\sqrt{(t_2-t_1)}.\hfill \end{array}$$

(26)

But

$$\begin{array}{ccc}\hfill P(X\left(t_2\right),Y\left(t_2\right))-P(X\left(t_1\right),Y\left(t_1\right))& =& (P_{1}Y\left(t_2\right),P_{2}X\left(t_2\right))-(P_{1}Y\left(t_1\right),P_{2}Y\left(t_1\right))\hfill \\ & =& ((P_{1}Y\left(t_2\right)-P_{1}Y\left(t_1\right)),(P_{2}X\left(t_2\right)-P_{2}X\left(t_1\right))),\hfill \end{array}$$

then from (25) and (26), $\left\{P\right(X,Y\left)\right(t\left)\right\},t\in [0,T]$ is equicontinuous on $Q$.

Let $(X_n,Y_n)\to (X,Y)w.p.1,\left\{(X_n,Y_n)\right\}\in Q$ then applying stochastic Lebesgue dominated convergence Theorem [15], we can obtain -5.0cm0cm

$$\begin{array}{ccc}\hfill {L.i.m}_{n\to \infty }P(X_n,Y_n)& =& ({L.i.m}_{n\to \infty }{P}_1{Y}_n,{L.i.m}_{n\to \infty }{P}_2{X}_n)\hfill \\ & =& ({L.i.m}_{n\to \infty }\{X_0-{\int }_0^{\tau }{h}_1(s,Y_n\left(s\right))dW\left(s\right)+{\int }_0^t{g}_1(s,Y_n\left(\varphi \left(s\right)\right))ds\},\hfill \\ & & {L.i.m}_{n\to \infty }\{Y_0-{\int }_0^{\eta }{h}_2(s,X_n\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y_n\left(\varphi \left(s\right)\right)))dW\left(s\right)\})\hfill \\ & =& (X_0-{\int }_0^{\tau }{h}_1(s,{L.i.m}_{n\to \infty }{Y}_n\left(s\right))dW\left(s\right)+{\int }_0^t{g}_1(s,{L.i.m}_{n\to \infty }{Y}_n\left(\varphi \left(s\right)\right)))ds,\hfill \\ & & Y_0-{\int }_0^{\eta }{h}_2(s,{L.i.m}_{n\to \infty }{X}_n\left(s\right))ds\hfill \\ & +& {\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,{L.i.m}_{n\to \infty }{Y}_n\left(\varphi \left(s\right)\right))dW\left(s\right))\hfill \\ & =& (X_0-{\int }_0^{\tau }{h}_1(s,Y\left(s\right))dW\left(s\right)+{\int }_0^t{g}_1(s,Y\left(\varphi \left(s\right)\right))ds,\hfill \\ & & Y_0-{\int }_0^{\eta }{h}_2(s,X\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right),)\hfill \\ & =& (P_{1}Y,P_{2}X)=P(x,y).\hfill \end{array}$$

This proves that $P:Q\to Q$ is continuous.

Then the closure of $PQ$ is compact and (see [15]) the problem (1)–(2) and (4) has a solution $(X,Y)\in \Lambda ,X,Y\in C(I,L_2(\Omega )).$

4.2. Uniqueness of the Solution

Theorem 7.

Let $\left(a^{*}1\right)-\left(a^{*}2\right)$ and $\left(a3\right)-\left(a6\right)$ are satisfied. Then the solution of (1)–(2) is unique.

Proof. 

Let $(X_1,Y_1)$ and $(X_2,Y_2)$ be the solution of

$$\begin{array}{ccc}\hfill (x\left(t\right),y\left(t\right))=(X_o& -& {\int }_0^{\tau }{h}_1(s,Y\left(s\right))dW\left(s\right)+{\int }_0^t{g}_1(s,Y\left(\varphi \left(s\right)\right))ds,\hfill \\ \hfill Y_o& -& {\int }_0^{\eta }{h}_2(s,X\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,Y\left(\varphi \left(s\right)\right)))dW\left(s\right)),\hfill \end{array}$$

(27)

then we can get

$$\begin{array}{ccc}\hfill \parallel X_1\left(t\right)-X_2{\left(t\right)\parallel }_2& \le & c\sqrt{T}\parallel Y_1-Y_2{\parallel }_C+bT{\parallel Y_1-Y_2\parallel }_C\hfill \\ & \le & (c\sqrt{T}+bT){\parallel Y_1-Y_2\parallel }_C\hfill \\ & \le & (c+b)T\{\parallel X_1-X_2{\parallel }_C+\parallel Y_1-Y_2{\parallel }_C\}.\hfill \end{array}$$

(28)

Similarly, we can obtain

$$\begin{array}{ccc}\hfill \parallel Y_1\left(t\right)-Y_2{\left(t\right)\parallel }_2& \le & cT\parallel X_1-X_2{\parallel }_C+\frac{b^2{T}^{\frac{3}{2}-\alpha }}{\Gamma (2-\alpha )}{\parallel y_1-y_2\parallel }_C\hfill \\ & \le & cT\parallel X_1-X_2{\parallel }_C+b^2{T}^2(2-\alpha ){\parallel y_1-y_2\parallel }_C\hfill \\ & \le & (cT+b^2{T}^2(2-\alpha ))\{\parallel X_1-X_2{\parallel }_C+\parallel Y_1-Y_2{\parallel }_C\}.\hfill \end{array}$$

(29)

Hence from (28) and (29)

$$\begin{array}{ccc}\hfill \parallel (X_1,Y_1)-(X_2,Y_2){\parallel }_{\Lambda }& =& \parallel (X_1-X_2){\parallel }_C+{\parallel (Y_1-Y_2)\parallel }_C\hfill \\ & \le & ((b+c)T+b^2{T}^2(2-\alpha ))\{\parallel X_1-X_2{\parallel }_C+\parallel Y_1-Y_2{\parallel }_C\}\hfill \\ & \le & ((b+c)T+b^2{T}^2(2-\alpha )){\parallel (X_1,Y_1)-(X_2,Y_2)\parallel }_{\Lambda }.\hfill \end{array}$$

This implies that

$$(1-(b+c)T-b^2{T}^2(2-\alpha )){\parallel (X_1,Y_1)-(X_2,Y_2)\parallel }_{\Lambda }\le 0.$$

then

$$\parallel (X_1,Y_1)-(X_2,Y_2){\parallel }_{\Lambda }=0$$

and $(X_1,Y_1)=(X_2,Y_2).$ □

4.3. Continuous Dependence

Theorem 8.

The solution of the problem (1)–(2) and (4) is continuously dependent on $(X_0,Y_0).$

Proof. 

Let $(\widehat{X},\widehat{Y})$ be the solution of (1)–(2) and (4)

$$\begin{array}{ccc}\hfill \widehat{X}\left(t\right)& =& {\widehat{X}}_o-{\int }_0^{\tau }{h}_1(s,\widehat{Y}\left(s\right))dW\left(s\right)+{\int }_0^t{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right))ds,\hfill \\ \hfill \widehat{Y}\left(t\right)& =& {\widehat{Y}}_o-{\int }_0^{\eta }{h}_2(s,\widehat{X}\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right)))dW\left(s\right))\hfill \end{array}$$

such that $\parallel (X_o,Y_o)-(\widehat{X_o},\widehat{Y_o}){\parallel }_{\Lambda }<{\delta }_3$. Then we have

$$\begin{array}{ccc}\hfill X\left(t\right)-\widehat{X}\left(t\right)& =& X_o-\widehat{X_o}-{\int }_0^{\tau }[h_1(s,\widehat{Y}\left(s\right))-h_1(s,Y\left(s\right))]dW\left(s\right)\hfill \\ & +& {\int }_0^t[g_1(s,Y\left(\varphi \left(s\right)\right))-g_1(s,\widehat{Y}\left(\varphi \left(s\right)\right))]ds\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel X\left(t\right)-\widehat{X}{\left(t\right)\parallel }_2& \le & \parallel X_0-\widehat{X_0}{\parallel }_C+c\sqrt{T}\parallel Y-\widehat{Y}{\parallel }_C+bT{\parallel Y-\widehat{Y}\parallel }_C\hfill \\ & \le & \parallel X_0-\widehat{X_0}{\parallel }_2+(b+c)T\{\parallel X-\widehat{X}{\parallel }_C+\parallel Y-\widehat{Y}{\parallel }_C\}.\hfill \end{array}$$

Similarly, we deduce that

$$\begin{array}{ccc}\hfill \parallel Y\left(t\right)-\widehat{Y}{\left(t\right)\parallel }_2& \le & \parallel Y_0-\widehat{Y_0}{\parallel }_2+cT\parallel X-\widehat{X}{\parallel }_C+b^2{T}^2(2-\alpha )\parallel Y-\widehat{Y}{\parallel }_C\}\hfill \\ & \le & \parallel Y_0-\widehat{Y_0}{\parallel }_2+(cT+b^2{T}^2(2-\alpha ))\{\parallel X-\widehat{X}{\parallel }_C+\parallel Y-\widehat{Y}{\parallel }_C\}\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel (X,Y)-(\widehat{X},\widehat{Y}){\parallel }_{\Lambda }& =& \parallel (X-\widehat{X}{\parallel }_C+\parallel (Y-\widehat{Y}{\parallel }_C\hfill \\ & \le & \parallel X_0-\widehat{X_0}{\parallel }_2+{\parallel Y_0-\widehat{Y_0}\parallel }_2\hfill \\ & +& ((b+c)T+b^2{T}^2(2-\alpha ))\{\parallel X-\widehat{X}{\parallel }_C+\parallel Y-\widehat{Y}{\parallel }_C\}\hfill \\ & \le & \parallel (X_0,Y_0)-(\widehat{X_0},\widehat{Y_0}){\parallel }_{\Lambda }\hfill \\ & +& ((b+c)T+b^2{T}^2(2-\alpha )){\parallel (X,Y)-(\widehat{X},\widehat{Y})\parallel }_{\Lambda }.\hfill \end{array}$$

Hence,

$$\parallel (X,Y)-(\widehat{X},\widehat{Y}){\parallel }_{\Lambda }\le \frac{{\delta }_3}{1-T(b+c)-b^2{T}^2(2-\alpha )}=ϵ.$$

□

Theorem 9.

The solution of (1)–(2) and (4) depends continuously on $h_1$ and $h_2.$

Proof. 

Let $(\widehat{X},\widehat{Y})$ be the solution of (1)–(2) and (4) such that

$$\begin{array}{ccc}\hfill \widehat{X}\left(t\right)& =& X_0-{\int }_0^{\tau }{h}_1^{*}(s,\widehat{Y}\left(s\right))dW\left(s\right)+{\int }_0^t{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right))ds,\hfill \\ \hfill \widehat{Y}\left(t\right)& =& Y_0-{\int }_0^{\eta }{h}_2^{*}(s,\widehat{X}\left(s\right))ds+{\int }_0^t{g}_2(s,I^{1-\alpha }{g}_1(s,\widehat{Y}\left(\varphi \left(s\right)\right)))dW\left(s\right)).\hfill \end{array}$$

Let $\parallel h_i^{*}{(t,u\left(t\right))-h(t,u\left(t\right))\parallel }_2\le {\delta }_4,i=1,2$ then -5.0cm0cm

$$\begin{array}{ccc}\hfill \parallel X\left(t\right)-\widehat{X}{\left(t\right)\parallel }_2& =& \parallel {\int }_0^{\tau }[h_1^{*}(s,\widehat{Y}\left(s\right))-h_1(s,Y\left(s\right))]dW\left(s\right)+{\int }_0^t[g_1(s,Y\left(\varphi \left(s\right)\right))-g_1(s,\widehat{Y}\left(\varphi \left(s\right)\right))]{ds\parallel }_2\hfill \\ & \le & \sqrt{{\int }_0^{\tau }{\parallel h_1^{*}(s,\widehat{Y}\left(s\right))-h_1(s,Y\left(s\right))\parallel }_2^{2}ds}+{\int }_0^t{\parallel g_1(s,Y\left(\varphi \left(s\right)\right))-g_1(s,\widehat{Y}\left(\varphi \left(s\right)\right))\parallel }_{2}ds\hfill \\ & \le & \sqrt{{\int }_0^{\tau }[\parallel h_1^{*}(s,\widehat{Y}\left(s\right))-h_1^{*}(s,Y\left(s\right)){\parallel }_2+\parallel h_1^{*}(s,Y\left(s\right))-h_1(s,Y\left(s\right)){{\parallel }_2]}^{2}ds}\hfill \\ & +& {\int }_0^t{\parallel g_1(s,Y\left(\varphi \left(s\right)\right))-g_1(s,\widehat{Y}\left(\varphi \left(s\right)\right))\parallel }_{2}ds\hfill \\ & \le & \sqrt{{\int }_0^{\tau }(c\parallel Y\left(s\right)-\widehat{Y}\left(s\right){{\parallel }_2+{\delta }_4)}^{2}ds}+{\int }_0^{t}b{\parallel Y\left(s\right)-\widehat{Y}\left(s\right)\parallel }_{2}ds\hfill \\ & \le & (c+b)T\parallel Y-\widehat{Y}{\parallel }_C+{\delta }_{4}T.\hfill \end{array}$$

Similarly we can get

$$\begin{array}{ccc}\hfill \parallel Y-\widehat{Y}{\parallel }_C& \le & cT\parallel X-\widehat{X}{\parallel }_C+\frac{b^2{T}^{\frac{3}{2}-\alpha }}{\Gamma (2-\alpha )}{\parallel Y-\widehat{Y}\parallel }_C+{\delta }_{4}T\hfill \\ & \le & cT\parallel X-\widehat{X}{\parallel }_C+b^2{T}^2(2-\alpha ){\parallel Y-\widehat{Y}\parallel }_C+{\delta }_{4}T\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel (X,Y)-(\widehat{X},\widehat{Y}){\parallel }_{\Lambda }& =& \parallel (X-\widehat{X}{\parallel }_C+\parallel (Y-\widehat{Y}{\parallel }_C\hfill \\ & \le & 2{\delta }_{4}T+cT\parallel X-\widehat{X}{\parallel }_C+(cT+bT+b^2{T}^2(2-\alpha ){\parallel Y-\widehat{Y}\parallel }_C\hfill \\ & \le & 2{\delta }_{4}T+[(b+c)T+b^2{T}^2(2-\alpha )]\{\parallel Y-\widehat{Y}{\parallel }_C+\parallel X-\widehat{X}{\parallel }_C\}\hfill \\ & \le & 2{\delta }_{4}T+[(b+c)T+b^2{T}^2(2-\alpha )]{\parallel (X,Y)-(\widehat{X},\widehat{Y})\parallel }_{\Lambda }.\hfill \end{array}$$

This implies that

$$\parallel (X,Y)-(\widehat{X},\widehat{Y}){\parallel }_{\Lambda }\le \frac{2{\delta }_{4}T}{1-(b+c)T-b^2{T}^2(2-\alpha )}=ϵ$$

which completes the proof. □

Theorem 10.

The solution $(X,Y)$ of (1)–(2) and (4) depends continuously on $u\left(t\right)=D^{\alpha }X\left(t\right)$, $\alpha \in (0,1]$.

Proof. 

Let $(\widehat{X},\widehat{Y})$ be the solution of

$$\begin{array}{ccc}\hfill \widehat{X}\left(t\right)& =& X_o-{\int }_0^{\tau }{h}_1(s,\widehat{Y}\left(s\right))dW\left(s\right)+I^{\alpha }\widehat{u}\left(t\right),\hfill \\ \hfill \widehat{Y}\left(t\right)& =& Y_o-{\int }_0^{\eta }{h}_2(s,\widehat{X}\left(s\right))ds+{\int }_0^t{g}_2(s,\widehat{u}\left(s\right)))dW\left(s\right)\hfill \end{array}$$

such that $\parallel \widehat{u\left(t\right)}{-u\left(t\right)\parallel }_2\le {\delta }_5,$ then

$$\begin{array}{ccc}\hfill \parallel (X,Y)-(\widehat{X},\widehat{Y}){\parallel }_{\Lambda }& =& \parallel (X-\widehat{X}{\parallel }_C+\parallel (Y-\widehat{Y}{\parallel }_C\hfill \\ & \le & \parallel {\int }_0^{\eta }[h_1(s,\widehat{Y}\left(s\right))-h_1(s,Y\left(s\right))]{dW\left(s\right)\parallel }_2+{\parallel I^{\alpha }u\left(t\right)-I^{\alpha }\widehat{u}\left(t\right)\parallel }_2\hfill \\ & +& \parallel {\int }_0^{\eta }[h_2(s,\widehat{X}\left(s\right))-h_2(s,X\left(s\right))]{ds\parallel }_2+{\parallel {\int }_0^t[g_2(s,u\left(s\right))-g_2(s,\widehat{u}\left(s\right))]dW\left(s\right)\parallel }_2\hfill \\ & \le & c\sqrt{T}\parallel Y-\widehat{Y}{\parallel }_C+\frac{T^{\alpha }{\delta }_5}{\Gamma (\alpha +1)}+cT{\parallel X-\widehat{X}\parallel }_C+b\sqrt{T}{\delta }_5\hfill \\ & \le & (\frac{T^{\alpha }}{\Gamma (\alpha +1)}+bT){\delta }_5+cT{\parallel (X,Y)-(\widehat{X},\widehat{Y})\parallel }_{\Lambda }.\hfill \end{array}$$

Now

$$\parallel (X,Y)-(\widehat{X},\widehat{Y}){\parallel }_{\Lambda }\le \frac{{\delta }_5}{1-cT}(\frac{T^{\alpha }}{\Gamma (\alpha +1)}+bT)$$

(30)

which completes the proof. □

5. Example

Consider the coupled system

$$\begin{array}{ccc}\hfill \frac{dX}{dt}\left(t\right)& =& \frac{1-t^2+Y\left(t\right)}{30(1+\parallel Y\left(t\right){\parallel }_2)},t\in (0,1],\hfill \\ \hfill dY\left(t\right)& =& \frac{e^{-t}+D^{\frac{1}{2}}X\left(t\right)}{120(1+\parallel X\left(t\right){\parallel }_2)}dW\left(t\right),t\in (0,1]\hfill \end{array}$$

(31)

subject to

$$\begin{array}{c}\hfill X_0={\int }_0^{\tau }\frac{sX\left(s\right)}{80+s^2}dW\left(s\right),Y_0={\int }_0^{\eta }\frac{Y\left(s\right)}{\sqrt{s+36}}ds\end{array}$$

(32)

where

$$\begin{array}{c}\hfill \parallel g_1{(t,Y\left(t\right))\parallel }_2\le \frac{1}{30}{\parallel Y\left(t\right)\parallel }_2,\parallel g_2{(t,X\left(t\right))\parallel }_2\le \frac{1}{120}{[1+\parallel X\left(t\right)\parallel }_2]\end{array}$$

and

$$\parallel h_1{(t,X\left(t\right)\parallel }_2\le \frac{{\parallel X\left(t\right)\parallel }_2}{80},\parallel h_2{(t,Y\left(t\right)\parallel }_2\le \frac{{\parallel Y\left(t\right)\parallel }_2}{6}.$$

Easily, the coupled system (31) with nonlocal integral conditions (32) satisfies all the assumptions (a1)–(a6) of Theorem 1. with $b=\frac{1}{30},c=\frac{1}{6}.$

6. Conclusions

Here, we have combined between the three senses of derivatives, the stochastic Ito-differential and the fractional and integer orders derivatives for the second order stochastic process in two non-local problems of the coupled system of the two random and stochastic differential Equations (1) and (2) with the stochastic and random integral conditions (3) and the coupled system of the two stochastic and random nonlocal integral conditions (4) where $X_0$ and $Y_0$ are two second order random variables.

The existence of solutions $X,Y\in C(I,L_2(\Omega ))$ of the problems (1)–(3) and (1)–(2) and (4) are proved. The unique solution and the sufficient conditions are discussed. The continuous dependence of the solution $X,Y\in C(I,L_2(\Omega ))$ on the two random variables $X_0$ and $Y_0$ and on the two random functions $h_1$ and $h_2$ and the continuous dependence of the solution $Y\in C(I,L_2(\Omega ))$ on the fractional order derivative $D^{\alpha }X\left(t\right)$ are be studied. An example is given.