*2.2. Uniqueness Theorem*

Replace the assumptions (*A*1) and (*A*2) by (*A*∗1) and (*A*∗2), respectively, such that (*A*∗1) The functions *fi* : [0, *T*] × *L*2(Ω) → *L*2(Ω), *i* = 1, 2 are measurable in *t* ∈ [0, *T*] for all *x* ∈ *L*2(Ω) and satisfy the Lipschitz condition with respect to the second argument

$$||f\_i(t, u) - f\_i(t, v)||\_2 \le b||u - v||\_2.$$

(*A*∗2) The functions *hi* : [0, *T*] × *L*2(Ω) → *L*2(Ω), *i* = 1, 2 are measurable in *t* ∈ [0, *T*] for all *x* ∈ *L*2(Ω) and satisfy the Lipschitz condition with respect to the second argument

$$||h\_i(t, \mathfrak{u}) - h\_i(t, \mathfrak{v})||\_2 \le c||\mathfrak{u} - \mathfrak{v}||\_2.$$

**Remark 1.** *Let the assumptions* (*A*∗1) *and* (*A*∗2) *be satisfied, then we can obtain*

$$\|\|f\_i(t,\mu)\|\|\_2 - \|f\_i(t,0)\|\|\_2 \le \|\|f\_i(t,\mu) - f\_i(t,0)\|\|\_2 \le b\|\|\mu\|\|\_{2\prime}$$

$$\|\|f\_i(t,\mu)\|\|\_2 \le \|\|f\_i(t,0)\|\|\_2 + b\|\|\mu\|\|\_2 \le M + b\|\|\mu\|\|\_2$$

*and*

$$\|\|h\_i(t,\mu)\|\|\_2 \le \|\|h\_i(t,0)\|\|\_2 + c\|\|\mu\|\|\_2 \le K + c\|\|\mu\|\|\_2.$$

**Theorem 2.** *Let the assumptions* (*A*∗1) − (*A*∗2) *and* (*A*3) − (*A*5) *be satisfied, then the solution of problem (1)–(3) is unique.*

**Proof.** Let (*x*1, *y*1) and (*x*2, *y*2) be two solutions of the problem (1)–(3), then

$$\begin{aligned} \left(\mathbf{x}\_i(t), y\_i(t)\right) &= \left(\mathbf{x}\_0 \quad -\int\_0^\tau h\_1(s, y(s))dW(s) + \int\_0^t f\_1(s, y(\phi\_1(s)))ds, \\ y\_0 & - \int\_0^\eta h\_2(s, \mathbf{x}(s))ds + \int\_0^t f\_2(s, \mathbf{x}(\phi\_2(s)))dW(s)), \quad i = 1, 2 \end{aligned} \tag{16}$$

where

$$\begin{split} \|\mathbf{x}\_{1}(t) - \mathbf{x}\_{2}(t)\|\_{2} &\leq \quad \|\int\_{0}^{\tau} [h\_{1}(s, y\_{2}(s)) - h\_{1}(s, y\_{1}(s))] d\mathcal{W}(s)\|\_{2} + \|\int\_{0}^{t} (f\_{1}(s, y\_{1}) - f\_{1}(s, y\_{2})) ds\|\_{2} \\ &\leq \quad \sqrt{\int\_{0}^{\tau} c^{2} \|y\_{2} - y\_{1}\|\_{\mathcal{C}}^{2} ds} + Tb\|y\_{1} - y\_{2}\|\_{\mathcal{C}} \leq T\sqrt{c} \|y\_{1} - y\_{2}\|\_{\mathcal{C}} + Tb\|y\_{1} - y\_{2}\|\_{\mathcal{C}} \\ &\leq \quad T(b + c) \|y\_{1} - y\_{2}\|\_{\mathcal{C}} \\ &\leq \quad T(b + c) \max\{\|x\_{1} - x\_{2}\|\_{\mathcal{C}}, \|y\_{1} - y\_{2}\|\_{\mathcal{C}}\} \end{split}$$

and

$$\begin{split} \|y\_1(t) - y\_2(t)\|\_2 &\le \int\_0^\eta \|h\_2(s, \mathbf{x}\_2(s)) - h\_2(s, \mathbf{x}\_1(s))\|\_2 ds + \sqrt{\int\_0^t b^2 \|\mathbf{x}\_1(s) - \mathbf{x}\_2(s)\|\_{\mathcal{L}}} ds \\ &\le \quad \sqrt{T} b \|\mathbf{x}\_1 - \mathbf{x}\_2\|\_{\mathcal{C}} + cT \|\mathbf{x}\_2 - \mathbf{x}\_1\|\_{\mathcal{C}}} \\ &\le \quad T(b + c) \|\mathbf{x}\_1 - \mathbf{x}\_2\|\_{\mathcal{C}} \\ &\le \quad T(b + c) \max\{\|\mathbf{x}\_1 - \mathbf{x}\_2\|\_{\mathcal{C}}, \|\mathbf{y}\_1 - \mathbf{y}\_2\|\_{\mathcal{C}}\}. \end{split}$$

Hence,

$$\begin{aligned} \| (\mathbf{x}\_1, y\_1) - (\mathbf{x}\_2, y\_2) \|\_{X} &= \| (\mathbf{x}\_1 - \mathbf{x}\_2), (y\_1, y\_2) \|\_{X} \\ &= \max \{ \| (\mathbf{x}\_1 - \mathbf{x}\_2) \|\_{\mathcal{C}}, \| (y\_1, y\_2) \|\_{\mathcal{C}} \} \\ &\leq \| T(b + c) \max \{ \| \mathbf{x}\_1 - \mathbf{x}\_2 \|\_{\mathcal{C}}, \| y\_2 - y\_1 \|\_{\mathcal{C}} \} \\ &\leq \| T(b + c) \| (\mathbf{x}\_1, y\_1) - (\mathbf{x}\_2, y\_2) \|\_{X} . \end{aligned}$$

This implies that

$$\|(1 - T(b + c))\| \|(\mathfrak{x}\_1, y\_1) - (\mathfrak{x}\_2, y\_2)\|\_X \le 0$$

and

$$\|(\mathfrak{x}\_1, \mathfrak{y}\_1) - (\mathfrak{x}\_2, \mathfrak{y}\_2)\|\_X = 0,$$

then (*x*1, *y*1)=(*x*2, *y*2) and the solution of the problem (1)–(3) is unique.

*2.3. Continuous Dependence*

**Theorem 3.** *Let the assumptions of Theorem 2 be satisfied. Then the solution (16) of the problem (1)–(3) depends continuously on the two random data* (*x*0, *y*0).

**Proof.** Let (*x*ˆ, *y*ˆ) be the solution of the coupled system

$$\begin{aligned} \mathfrak{X}(t) &= \mathfrak{X}\_0 - \int\_0^\tau h\_1(s, \mathfrak{Y}(s)) dW(s) + \int\_0^t f\_1(s, \mathfrak{Y}(\phi\_1(s))) ds \\ \mathfrak{Y}(t) &= \mathfrak{Y}\_0 - \int\_0^\eta h\_2(s, \mathfrak{X}(s)) ds + \int\_0^t f\_2(s, \mathfrak{X}(\phi\_2(s))) dW(s), \end{aligned}$$

such that (*x*0, *y*0) − (*x*ˆ0, *y*ˆ0)*<sup>X</sup>* < *δ*1, then

$$\begin{array}{rcl} \|\mathbf{x} - \widehat{\mathbf{x}}\|\|\_{\mathcal{C}} & \leq & \|\|\mathbf{x}\_{0} - \widehat{\mathbf{x}}\_{0}\|\|\_{\mathcal{C}} + T(b + c) \|\|\mathbf{y} - \widehat{\mathbf{y}}\|\|\_{\mathcal{C}} \\ & \leq & \delta\_{1} + T(b + c) \|\|\mathbf{y} - \widehat{\mathbf{y}}\|\|\_{\mathcal{C}} \\ & \leq & \delta\_{1} + T(b + c) \max\{\|\|\mathbf{x} - \widehat{\mathbf{x}}\|\|\_{\mathcal{C}}, \|\|\mathbf{y} - \widehat{\mathbf{y}}\|\|\_{\mathcal{C}}\} \\ \|\|\mathbf{y} - \widehat{\mathbf{y}}\|\|\_{\mathcal{C}} & \leq & \|\|\mathbf{y}\_{0} - \widehat{\mathbf{y}}\_{0}\|\|\_{\mathcal{C}} + T(b + c) \|\|\mathbf{x} - \widehat{\mathbf{x}}\|\|\_{\mathcal{C}'} \\ & \leq & \delta\_{1} + T(b + c) \|\|\mathbf{x} - \widehat{\mathbf{x}}\|\|\_{\mathcal{C}} \\ & \leq & \delta\_{1} + T(b + c) \max\{\|\|\mathbf{x} - \widehat{\mathbf{x}}\|\|\_{\mathcal{C}'}, \|\|\mathbf{y} - \widehat{\mathbf{y}}\|\|\_{\mathcal{C}}\}. \end{array}$$

Then

$$\begin{array}{rcl} \|(\mathfrak{x},\mathfrak{y})-(\mathfrak{x},\mathfrak{y})\|\_{X} &=& \|(\mathfrak{x}-\mathfrak{x},\mathfrak{y}-\mathfrak{y})\|\_{X} \\ &=& \max\{\|\mathfrak{x}-\mathfrak{x}\|\_{\mathbb{C}^{\prime}}\|\mathfrak{y}-\mathfrak{y}\|\_{\mathbb{C}}\} \\ &\leq& \delta\_{1}+T(b+c)\max\{\|\mathfrak{x}-\mathfrak{x}\|\_{\mathbb{C}^{\prime}}\|\mathfrak{y}-\mathfrak{y}\|\_{\mathbb{C}}\} \\ &\leq& \delta\_{1}+T(b+c)\|(\mathfrak{x},\mathfrak{y})-(\mathfrak{x},\mathfrak{y})\|\_{X}. \end{array}$$

This implies that

$$\|\left(\left(x,y\right)-\left(\hat{x},\hat{y}\right)\|\right)\_X \le \frac{\delta\_1}{1-T(b+c)} = \epsilon$$

which completes the proof.

**Theorem 4.** *The solution (16) of the problem (1)–(3) depends continuously on the two random functions h*<sup>1</sup> *and h*2*.*

**Proof.** Let (*x*ˆ, *y*ˆ) be the solutions of the coupled system

$$\begin{aligned} \dot{\mathfrak{X}}(t) &= \quad \mathfrak{x}\_0 - \int\_0^\tau h\_1^\*(s, \dot{\mathfrak{Y}}(s)) dW(s) + \int\_0^t f\_1(s, \dot{\mathfrak{Y}}(\Phi\_1(s))) ds, \\ \dot{\mathfrak{Y}}(t) &= \quad \mathfrak{y}\_0 - \int\_0^\eta h\_2^\*(s, \mathfrak{X}(s)) ds + \int\_0^t f\_2(s, \mathfrak{X}(\Phi\_2(s))) dW(s) \end{aligned}$$

such that *h*<sup>∗</sup> *<sup>i</sup>* (*s*, .) − *h*(*s*, .)<sup>2</sup> ≤ *δ*2, *i* = 1, 2, then

*x*(*t*) − *x*ˆ(*t*)<sup>2</sup> = *τ* 0 [*h*∗ <sup>1</sup> (*s*, *<sup>y</sup>*ˆ(*s*)) <sup>−</sup> *<sup>h</sup>*1(*s*, *<sup>y</sup>*(*s*))]*dW*(*s*) + *<sup>t</sup>* 0 [ *f*1(*s*, *y*(*φ*1(*s*))) − *f*1(*s*, *y*ˆ(*φ*1(*s*)))]*ds*<sup>2</sup> ≤ *<sup>τ</sup>* 0 *h*<sup>∗</sup> <sup>1</sup> (*s*, *<sup>y</sup>*ˆ(*s*)) − *<sup>h</sup>*1(*s*, *<sup>y</sup>*(*s*))<sup>2</sup> <sup>2</sup>*ds* + *t* 0 *f*1(*s*, *y*(*φ*1(*s*))) − *f*1(*s*, *y*ˆ(*φ*1(*s*)))2*ds* ≤ *<sup>τ</sup>* 0 [*h*<sup>∗</sup> <sup>1</sup> (*s*, *y*ˆ(*s*)) − *h*<sup>∗</sup> <sup>1</sup> (*s*, *y*(*s*))<sup>2</sup> + *h*<sup>∗</sup> <sup>1</sup> (*s*, *y*(*s*)) − *h*1(*s*, *y*(*s*))2]2*ds* + *t* 0 *f*1(*s*, *y*(*φ*1(*s*))) − *f*1(*s*, *y*ˆ(*φ*1(*s*)))2*ds* ≤ *<sup>τ</sup>* 0 (*cy*(*s*) − *y*ˆ(*s*)<sup>2</sup> + *δ*2)2*ds* + *t* 0 *by*(*s*) − *y*ˆ(*s*)2*ds* ≤ (*c* √ *T* + *bT*)*y* − *y*ˆ*<sup>C</sup>* + *δ*<sup>2</sup> √ *T* ≤ *T*(*b* + *c*) max{*x* − *x*ˆ*C*, *y* − *y*ˆ*C*} + *δ*2*T*

Similarly we can obtain

$$\begin{split} \|y(t) - \hat{y}(t)\|\_{2} &= \|\int\_{0}^{\eta} [h\_{2}^{\*}(s, \mathfrak{x}(s)) - h\_{2}(s, \mathfrak{x}(s))] ds + \int\_{0}^{t} [f\_{2}(s, \mathfrak{x}(\phi\_{2}(s))) - f\_{2}(s, \mathfrak{x}(\phi\_{2}(s)))] dW(s))\|\_{2} \\ &\leq \|cT + b\sqrt{T}\| \|\mathbf{x} - \mathfrak{x}\|\_{\mathbb{C}} + \delta\_{2}T \\ &\leq \|T(b + c)\| \|\mathbf{x} - \mathfrak{x}\|\_{\mathbb{C}} + \delta\_{2}T \\ &\leq \|T(b + c)\max\{\|\mathbf{x} - \hat{\mathfrak{x}}\|\_{\mathbb{C}}, \|\mathbf{y} - \hat{\mathfrak{y}}\|\_{\mathbb{C}}\} + \delta\_{2}T \\ &\leq \|T(b + c)\max\{\|\mathbf{x} - \hat{\mathfrak{x}}\|\_{\mathbb{C}}, \|\mathbf{y} - \hat{\mathfrak{y}}\|\_{\mathbb{C}}\} + \delta\_{2}T. \end{split}$$

Now

$$\begin{array}{rcl} \|(\mathfrak{x},\mathfrak{y}) - (\mathfrak{x},\mathfrak{y})\|\_{X} &=& \max\{\|\mathfrak{x} - \mathfrak{x}\|\_{\mathbb{C}\mathscr{I}}, \|\mathfrak{y} - \mathfrak{y}\|\_{\mathbb{C}}\} \\ &\leq& T(b+c)\max\{ (\|\mathfrak{x} - \mathfrak{x}\|\_{\mathbb{C}\mathscr{I}}, \|\mathfrak{y} - \mathfrak{y}\|\_{\mathbb{C}}) + \delta\_{2}T \} \\ &\leq& T(c+b) \|(\mathfrak{x},\mathfrak{y}) - (\mathfrak{x},\mathfrak{y})\|\_{X} + \delta\_{2}T. \end{array}$$

This implies that

$$\|\|(x, y) - (\pounds, \mathfrak{H})\|\|\_{\mathcal{X}} \le \frac{\delta\_2 T}{1 - T(b + c)} = \epsilon$$

which completes the proof.
