On Existence and Continuity Results of Solution for Multi-time Scale Fractional Stochastic Differential Equation

Abstract

In this paper, we study four critical aspects of a new class of multi-time scale nonlinear fractional stochastic differential equations with fractional integral in the sense of Riemann Liouville. A primary goal of this paper is to investigate the existence, uniqueness, Ulam-Hyers stability, and continuity of the solutions under sufficient assumptions using the Banach contraction theorem. At the end of the paper, a specific example is provided to demonstrate the efficiency and effectiveness of the new results of this paper.

Appendices

Appendix A: Proof of Theorem 3.3

In order to seek the existence and uniqueness of the solution to the system (1.2) in ${\mathcal {S}}$ for $T>0$, we use the Banach fixed point theorem with the norm $|{\mathcal {F}}(\tau )|=\underset{0<\tau \le T}{sup} E |{\mathcal {F}}(\tau )|^2<\infty$. Define the operator ${\mathcal {J}}X(\tau )$ as follows:

$$\begin{aligned} {\mathcal {J}}X(\tau )= & {} x_0+\frac{1}{\Gamma (\beta )}\int _{0}^{\tau }(\tau -\mu )^{\beta -1}{\mathcal {F}}(X_\mu ,\mu )d\mu \\{} & {} +\frac{1}{\Gamma (\beta )}\int _{0}^{\tau }(\tau -\mu )^{\beta -1}{\mathcal {U}}(X_\mu ,\mu )dB_\mu \\{} & {} +\frac{\Gamma (\alpha +1)}{\Gamma (\alpha +\beta -1)}\int _{0}^{\tau }(\tau -\mu )^{\alpha +\beta -2}\,\,{\mathcal {G}}(X_\mu ,\mu )d\mu . \end{aligned}$$

By Theorem 3.2, the operator ${\mathcal {J}}:{\mathcal {S}}\rightarrow {\mathcal {S}}$ is well defined. For $X_\tau ,Y_\tau \in {\mathcal {S}}$ with $x_0=y_0,$ we have

$$\begin{aligned} E |{\mathcal {J}}X_\tau -{\mathcal {J}}Y_\tau |^2\le & {} \frac{3}{\Gamma ^2(\beta )} E |\int _{0}^{\tau }(\tau -\mu )^{\beta -1}({\mathcal {F}}(X_\mu ,\mu )-{\mathcal {F}}(Y_\mu ,\mu ))d\mu |^2\nonumber \\{} & {} +\frac{3}{\Gamma ^2(\beta )} E |\int _{0}^{\tau }(\tau -\mu )^{\beta -1}({\mathcal {U}}(X_\mu ,\mu )-{\mathcal {U}}(Y_\mu ,\mu ))dB_\mu |^2\nonumber \\{} & {} +\frac{3\Gamma ^2(\alpha +1)}{\Gamma ^2(\alpha +\beta -1)} E |\int _{0}^{\tau }(\tau -\mu )^{\alpha +\beta -2}\,\,{\mathcal {G}}(X_\mu ,\mu )d\mu |^2\nonumber \\\le & {} \frac{3T{\mathcal {K}}}{\Gamma ^2(\beta )}\int _{0}^{\tau }(\tau -\mu )^{2\beta -2} E |X_\mu -Y_\mu |^2d\mu \nonumber \\{} & {} +\frac{3{\mathcal {K}}}{\Gamma ^2(\beta )}\int _{0}^{\tau }(\tau -\mu )^{2\beta -2} E |X_\mu -Y_\mu |^2d\mu \nonumber \\{} & {} +\frac{3T{\mathcal {K}}\Gamma ^2(\alpha +1)}{\Gamma ^2(\alpha +\beta -1)}\int _{0}^{\tau }(\tau -\mu )^{2\alpha +2\beta -4} E |X_\mu -Y_\mu |^2d\mu \nonumber \\\le & {} 3\lambda _1\int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }_1-1} E |X_\mu -Y_\mu |^2d\mu \nonumber \\{} & {} +3\lambda _2\int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }_2-1} E |X_\mu -Y_\mu |^2d\mu . \end{aligned}$$

(8.1)

Since, $\alpha \in [\frac{1}{2},1), \beta \in (\frac{1}{2},1],$ hence, $0<\bar{\beta }_1<1, -1<\bar{\beta }_2<1.$ Thus, we discuss the following two cases: Case 1: Let $\bar{\beta }_1,\bar{\beta }_2\in (0,1).$ In this case, the two curves $\frac{\tau ^{\bar{\beta }_1}}{\bar{\beta }_1},\frac{\tau ^{\bar{\beta }_2}}{\bar{\beta }_2},$ where $0<\tau \le T<+\infty ,$ have the intersection $T_*=(\frac{\bar{\beta }_1}{\bar{\beta }_2})^{\frac{1}{2-2\alpha }}, \frac{1}{2-2\alpha }\ge 1.$ Thus, we split [0, T] into two intervals $[0,T_*]$ and $[T_*,T],$ where $0<T_*<T<+\infty .$ Now, if $\tau \in [0,T_*],$ one can choose $\bar{\beta }=\min \{\bar{\beta }_1,\bar{\beta }_2\}$ such that, for $\lambda =\max \{\lambda _1,\lambda _2\},$ we have

$$\begin{aligned} \lambda _1\int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }_1-1} E |X_\mu -Y_\mu |^2d\mu \le \lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |X_\mu -Y_\mu |^2d\mu ,\\ \lambda _2\int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }_2-1} E |X_\mu -Y_\mu |^2d\mu \le \lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |X_\mu -Y_\mu |^2d\mu . \end{aligned}$$

Consequently, (8.1) can be written as

$$\begin{aligned} E |{\mathcal {J}}X_\tau -{\mathcal {J}}Y_\tau |^2\le & {} 6\lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |X_\mu -Y_\mu |^2d\mu \nonumber \\\le & {} 6\lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1}\underset{0\le \mu \le T_*}{sup} E |X_\mu -Y_\mu |^2d\mu \nonumber \\\le & {} \frac{6\lambda T_*^{\bar{\beta }}}{\bar{\beta }}|X_\tau -Y_\tau |. \end{aligned}$$

(8.2)

Then, we directly get

$$\begin{aligned} E |{\mathcal {J}}^2X_\tau -{\mathcal {J}}^2Y_\tau |^2\le & {} 6\lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |{\mathcal {J}}X_\mu -{\mathcal {J}}Y_\mu |^2d\mu \\\le & {} 6\lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1}\frac{6\lambda \,\mu ^{\bar{\beta }}}{\bar{\beta }}|X_\mu -Y_\mu | d\mu \\\le & {} \frac{(6\lambda )^2}{\bar{\beta }}|X_\tau -Y_\tau |\int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1}\mu ^{\bar{\beta }}d\mu . \end{aligned}$$

To evaluate the integral on the right hand side, let $\mu =\tau z$. Then,

$$\begin{aligned} \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1}\mu ^{\bar{\beta }}d\mu =\tau ^{2\bar{\beta }}\int _{0}^{1}(1-z)^{\bar{\beta }-1} z^{\bar{\beta }}dz=\frac{\tau ^{2\bar{\beta }}\,\Gamma (\bar{\beta })\Gamma (\bar{\beta }+1)}{\Gamma (2\bar{\beta }+1)}, \end{aligned}$$

where $\int _{0}^{1}(1-z)^{\bar{\beta }-1} z^{\bar{\alpha }-1}dz=\frac{\Gamma (\bar{\beta })\Gamma (\bar{\alpha })}{\Gamma (\bar{\alpha }+\bar{\beta })},$ which leads to

$$\begin{aligned} E |{\mathcal {J}}^2X_\tau -{\mathcal {J}}^2Y_\tau |^2\le \frac{(6\lambda \,\Gamma (\bar{\beta }))^2\,T_*^{2\bar{\beta }}}{\Gamma (2\bar{\beta }+1)}|X_\tau -Y_\tau |. \end{aligned}$$

(8.3)

Repeating the same procedure, we get

$$\begin{aligned} E |{\mathcal {J}}^3X_\tau -{\mathcal {J}}^3Y_\tau |^2\le \frac{(6\lambda \,\Gamma (\bar{\beta }))^3\,T_*^{3\bar{\beta }}}{\Gamma (3\bar{\beta }+1)}|X_\tau -Y_\tau |. \end{aligned}$$

(8.4)

From (8.2–8.4), for $m\ge 1, m\in \mathbb {N},$ we have

$$\begin{aligned} E |{\mathcal {J}}^mX_\tau -{\mathcal {J}}^mY_\tau |^2\le \frac{(6\lambda \,\Gamma (\bar{\beta }))^m\,T_*^{m\bar{\beta }}}{\Gamma (m\bar{\beta }+1)}|X_\tau -Y_\tau |. \end{aligned}$$

(8.5)

However, we have

$$\begin{aligned} \underset{0<\tau \le T_*}{sup} E |{\mathcal {J}}^mX_\tau -{\mathcal {J}}^mY_\tau |^2\le {\mathfrak {C}}_1\underset{0<\tau \le T_*}{sup} E |X_\tau -Y_\tau |^2, \end{aligned}$$

where ${\mathfrak {C}}_1=\frac{(6\lambda \,\Gamma (\bar{\beta }))^m\,T_*^{m\bar{\beta }}}{\Gamma (m\bar{\beta }+1)}.$ This means that

$$\begin{aligned} |{\mathcal {J}}^mX_\tau -{\mathcal {J}}^mY_\tau |\le {\mathfrak {C}}_1|X_\tau -Y_\tau |. \end{aligned}$$

(8.6)

Next, if $\tau \in [T_*,T],$ one can choose $\bar{\beta }=\max \{\bar{\beta }_1,\bar{\beta }_2\}$ such that, for $\lambda =\max \{\lambda _1,\lambda _2\},$ we have

$$\begin{aligned} \lambda _1\int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }_1-1} E |X_\mu -Y_\mu |^2d\mu \le \lambda \int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |X_\mu -Y_\mu |^2d\mu ,\\ \lambda _2\int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }_2-1} E |X_\mu -Y_\mu |^2d\mu \le \lambda \int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |X_\mu -Y_\mu |^2d\mu . \end{aligned}$$

Consequently, (8.1) can be written as

$$\begin{aligned} E |{\mathcal {J}}X_\tau -{\mathcal {J}}Y_\tau |^2\le & {} 6\lambda \int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |X_\mu -Y_\mu |^2d\mu \\\le & {} 6\lambda \int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }-1}\underset{T_*\le \mu \le T}{sup} E |X_\mu -Y_\mu |^2d\mu \\\le & {} \frac{6\lambda (T-T_*)^{\bar{\beta }}}{\bar{\beta }}|X_\tau -Y_\tau |. \end{aligned}$$

Then, we directly get

$$\begin{aligned} E |{\mathcal {J}}^2X_\tau -{\mathcal {J}}^2Y_\tau |^2\le & {} 6\lambda \int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |{\mathcal {J}}X_\mu -{\mathcal {J}}Y_\mu |^2d\mu \\\le & {} 6\lambda \int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }-1}\frac{6\lambda \,(\mu -T_*)^{\bar{\beta }}}{\bar{\beta }}|X_\mu -Y_\mu |d\mu \\\le & {} \frac{(6\lambda )^2}{\bar{\beta }}|X_\tau -Y_\tau |\int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }-1}(\mu -T_*)^{\bar{\beta }}d\mu . \end{aligned}$$

In order to evaluate the integral on the right hand side, let $\mu -T_*=(\tau -T_*) z$. Then, we have

$$\begin{aligned} \int _{T_*}^{\tau }(\tau -\mu )^{\bar{\beta }-1}(\mu -T_*)^{\bar{\beta }}d\mu= & {} (\tau -T_*)^{2\bar{\beta }}\int _{0}^{1}(1-z)^{\bar{\beta }-1} z^{\bar{\beta }}dz\\= & {} \frac{(\tau -T_*)^{2\bar{\beta }}\,\Gamma (\bar{\beta })\Gamma (\bar{\beta }+1)}{\Gamma (2\bar{\beta }+1)}, \end{aligned}$$

which leads to

$$\begin{aligned} E |{\mathcal {J}}^2X_\tau -{\mathcal {J}}^2Y_\tau |^2\le \frac{(6\lambda \,\Gamma (\bar{\beta }))^2\,(T-T_*)^{2\bar{\beta }}}{\Gamma (2\bar{\beta }+1)}|X_\tau -Y_\tau |. \end{aligned}$$

Repeating the same calculations, for $m\ge 1, m\in \mathbb {N},$ following is valid:

$$\begin{aligned} |{\mathcal {J}}^mX_\tau -{\mathcal {J}}^mY_\tau |\le {\mathfrak {C}}_2|X_\tau -Y_\tau |, \end{aligned}$$

(8.7)

where, ${\mathfrak {C}}_2=\frac{(6\lambda \,\Gamma (\bar{\beta }))^m\,(T-T_*)^{m\bar{\beta }}}{\Gamma (m\bar{\beta }+1)}.$ Case 2: Let $\bar{\beta }_2\in (-1,0], \bar{\beta }_1\in (0,1).$ In this case, there is no an intersection point between the curves. Thus, for $\tau \in [0,T],$ where $T>0$ is arbitrary, we have

$$\begin{aligned} \lambda _1\int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }_1-1} E |X_\mu -Y_\mu |^2d\mu \le \lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |X_\mu -Y_\mu |^2d\mu ,\\ \lambda _2\int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }_2-1} E |X_\mu -Y_\mu |^2d\mu \le \lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |X_\mu -Y_\mu |^2d\mu , \end{aligned}$$

where $\bar{\beta }=\bar{\beta }_1,$ and $\lambda =\max \{\lambda _1,\lambda _2\}.$ Consequently, (8.1) can be written as

$$\begin{aligned} E |{\mathcal {J}}X_\tau -{\mathcal {J}}Y_\tau |^2\le & {} 6\lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1} E |X_\mu -Y_\mu |^2d\mu \\\le & {} 6\lambda \int _{0}^{\tau }(\tau -\mu )^{\bar{\beta }-1}\underset{0\le \mu \le T}{sup} E |X_\mu -Y_\mu |^2d\mu \\\le & {} \frac{6\lambda T^{\bar{\beta }}}{\bar{\beta }}|X_\tau -Y_\tau |. \end{aligned}$$

Similarly, for $m\ge 1, m\in \mathbb {N},$ we can obtain

$$\begin{aligned} |{\mathcal {J}}^mX_\tau -{\mathcal {J}}^mY_\tau |\le {\mathfrak {C}}_3|X_\tau -Y_\tau |, \end{aligned}$$

(8.8)

where ${\mathfrak {C}}_3=\frac{(6\lambda \,\Gamma (\bar{\beta }))^m\,T^{m\bar{\beta }}}{\Gamma (m\bar{\beta }+1)}.$ From (8.6), (8.7) and (8.8), as long as the chosen $\bar{\beta }>0,$ we find ${\mathfrak {C}}_i\rightarrow 0$ as $m\rightarrow \infty ,$ for $i\in \mathbb {N}_1^3.$ Consequently, for sufficiently large m, we have ${\mathfrak {C}}_i<1,$ for any $i\in \mathbb {N}_1^3,$ which proves that ${\mathcal {J}}$ is contraction mapping on ${\mathcal {S}}.$ In view of Theorem 2.5, the operator ${\mathcal {J}}$ has fixed point in ${\mathcal {S}},$ which means that the system (1.2) has a unique mild solution, and this terminates the proof.

Appendix B: Proof of Theorem 5.1

In view of Theorem 3.3, (5.1) and (5.2) have unique solutions, which are given by

$$\begin{aligned} X^{\beta _1}(\tau )= & {} x_0+\frac{1}{\Gamma (\beta _1)}\int _{0}^{\tau }(\tau -\mu )^{\beta _1-1}{\mathcal {F}}(X^{\beta _1}_\mu ,\mu )d\mu \\{} & {} +\frac{1}{\Gamma (\beta _1)}\int _{0}^{\tau }(\tau -\mu )^{\beta _1-1}{\mathcal {U}}(X^{\beta _1}_\mu ,\mu )dB_\mu \\{} & {} +\frac{\Gamma (\alpha +1)}{\Gamma (\alpha +\beta _1-1)}\int _{0}^{\tau }(\tau -\mu )^{\alpha +\beta _1-2}\,\,{\mathcal {G}}(X^{\beta _1}_\mu ,\mu )d\mu ,\\ X^{\beta _2}(\tau )= & {} x_0+\frac{1}{\Gamma (\beta _2)}\int _{0}^{\tau }(\tau -\mu )^{\beta _2-1}{\mathcal {F}}(X^{\beta _2}_\mu ,\mu )d\mu \\{} & {} +\frac{1}{\Gamma (\beta _2)}\int _{0}^{\tau }(\tau -\mu )^{\beta _2-1}{\mathcal {U}}(X^{\beta _2}_\mu ,\mu )dB_\mu \\{} & {} +\frac{\Gamma (\alpha +1)}{\Gamma (\alpha +\beta _2-1)}\int _{0}^{\tau }(\tau -\mu )^{\alpha +\beta _2-2}\,\,{\mathcal {G}}(X^{\beta _2}_\mu ,\mu )d\mu , \end{aligned}$$

respectively. Then, we have

$$\begin{aligned} E |X_\tau ^{\beta _1}-X_\tau ^{\beta _2}|^2\le & {} 3 E \big |\int _{0}^{\tau }\frac{(\tau -\mu )^{\beta _1-1}}{\Gamma (\beta _1)}{\mathcal {F}}(X_\mu ^{\beta _1},\mu )d\mu \nonumber \\{} & {} -\int _{0}^{\tau }\frac{(\tau -\mu )^{\beta _2-1}}{\Gamma (\beta _2)}{\mathcal {F}}(X_\mu ^{\beta _2},\mu )d\mu \big |^2\nonumber \\{} & {} +3 E \big |\int _{0}^{\tau }\frac{(\tau -\mu )^{\beta _1-1}}{\Gamma (\beta _1)}{\mathcal {U}}(X^{\beta _1}_\mu ,\mu )dB\nonumber \\{} & {} -\int _{0}^{\tau }\frac{(\tau -\mu )^{\beta _2-1}}{\Gamma (\beta _2)}{\mathcal {U}}(X^{\beta _2}_\mu ,\mu )dB\big |^2\nonumber \\{} & {} +3 E \big |\int _{0}^{\tau }\frac{\Gamma (\alpha +1)\,(\tau -\mu )^{\alpha +\beta _1-2}}{\Gamma (\alpha +\beta _1-1)}{\mathcal {G}}(X^{\beta _1}_\mu ,\mu )d\mu \nonumber \\{} & {} -\int _{0}^{\tau }\frac{\Gamma (\alpha +1)\,(\tau -\mu )^{\alpha +\beta _2-2}}{\Gamma (\alpha +\beta _2-1)}{\mathcal {G}}(X^{\beta _2}_\mu ,\mu )d\mu \big |^2\nonumber \\:= & {} 3({\mathcal {I}}_1+{\mathcal {I}}_2+{\mathcal {I}}_3), \end{aligned}$$

(8.9)

where

$$\begin{aligned} {\mathcal {I}}_1= & {} E \big |\int _{0}^{\tau }\frac{(\tau -\mu )^{\beta _1-1}}{\Gamma (\beta _1)}{\mathcal {F}}(X_\mu ^{\beta _1},\mu )d\mu -\int _{0}^{\tau }\frac{(\tau -\mu )^{\beta _2-1}}{\Gamma (\beta _2)}{\mathcal {F}}(X_\mu ^{\beta _2},\mu )d\mu \big |^2,\\ {\mathcal {I}}_2= & {} E \big |\int _{0}^{\tau }\frac{(\tau -\mu )^{\beta _1-1}}{\Gamma (\beta _1)}{\mathcal {U}}(X^{\beta _1}_\mu ,\mu )dB-\int _{0}^{\tau }\frac{(\tau -\mu )^{\beta _2-1}}{\Gamma (\beta _2)}{\mathcal {U}}(X^{\beta _2}_\mu ,\mu )dB\big |^2,\\ {\mathcal {I}}_3= & {} E \big |\int _{0}^{\tau }\frac{\Gamma (\alpha +1)\,(\tau -\mu )^{\alpha +\beta _1-2}}{\Gamma (\alpha +\beta _1-1)}{\mathcal {G}}(X^{\beta _1}_\mu ,\mu )d\mu \\{} & {} -\int _{0}^{\tau }\frac{\Gamma (\alpha +1)\,(\tau -\mu )^{\alpha +\beta _2-2}}{\Gamma (\alpha +\beta _2-1)}{\mathcal {G}}(X^{\beta _2}_\mu ,\mu )d\mu \big |^2. \end{aligned}$$

We start estimating ${\mathcal {I}}_1$ as follows:

$$\begin{aligned} {\mathcal {I}}_1= & {} E \big |\int _{0}^{\tau }\left( \frac{(\tau -\mu )^{\beta _1-1}}{\Gamma (\beta _1)}{\mathcal {F}}(X^{\beta _1}_\mu ,\mu ) -\frac{(\tau -\mu )^{\beta _1-1}}{\Gamma (\beta _1)}{\mathcal {F}}(X^{\beta _2}_\mu ,\mu )\right. \\{} & {} \left. +\frac{(\tau -\mu )^{\beta _1-1}}{\Gamma (\beta _1)}{\mathcal {F}}(X^{\beta _2}_\mu ,\mu ) -\frac{(\tau -\mu )^{\beta _2-1}}{\Gamma (\beta _2)}{\mathcal {F}}(X^{\beta _2}_\mu ,\mu )\right) d\mu \big |^2\\\le & {} \frac{T}{\Gamma ^2(\beta _1)}\int _{0}^{\tau }(\tau -\mu )^{2\beta _1-2} E |{\mathcal {F}}(X_\mu ^{\beta _1},\mu )-{\mathcal {F}}(X_\mu ^{\beta _2},\mu )|^2d\mu \\{} & {} +T\int _{0}^{\tau }\left( \frac{(\tau -\mu )^{\beta _1-1}}{\Gamma (\beta _1)}-\frac{(\tau -\mu )^{\beta _2-1}}{\Gamma (\beta _2)}\right) ^2d\mu \int _{0}^{\tau } E |{\mathcal {F}}(X_\mu ^{\beta _2},\mu )|^2d\mu \\\le & {} \frac{T{\mathcal {K}} }{\Gamma ^2(\beta _1)}\int _{0}^{\tau }(\tau -\mu )^{2\beta _1-2} E |X_\mu ^{\beta _1}-X_\mu ^{\beta _2}|^2d\mu \\{} & {} +T\,{\mathcal {K}}\int _{0}^{\tau }\left( \frac{(\tau -\mu )^{2\beta _1-2}}{\Gamma ^2(\beta _1)} -\frac{2(\tau -\mu )^{\beta _1+\beta _1-2}}{\Gamma ^(\beta _1)\Gamma (\beta _2)}+\frac{(\tau -\mu )^{2\beta _2-2}}{\Gamma ^2(\beta _2)}\right) d\mu \\{} & {} \int _{0}^{\tau }(1+ E |X_\mu ^{\beta _2}|^2)d\mu \\\le & {} \frac{T{\mathcal {K}} }{\Gamma ^2(\beta _1)}\int _{0}^{\tau }(\tau -\mu )^{2\beta _1-2} E |X_\mu ^{\beta _1}-X_\mu ^{\beta _2}|^2d\mu \\{} & {} +{\mathcal {K}}\,C\left( \frac{T^{2\beta _1}}{(2\beta _1-1)\Gamma ^2(\beta _1)}-\frac{2T^{\beta _1+\beta _2}}{(\beta _1+\beta _2-1)\Gamma (\beta _1)\Gamma (\beta _2)}+\frac{T^{2\beta _2}}{(2\beta _2-1)\Gamma ^2(\beta _2)}\right) . \end{aligned}$$

We define a continuous function ${\mathfrak {B}}_1(\beta _1, \beta _2)$ as follows:

$$\begin{aligned} {\mathfrak {B}}_1(\beta _1, \beta _2)= & {} {\mathcal {K}}\,C\left( \frac{T^{2\beta _1}}{(2\beta _1-1)\Gamma ^2(\beta _1)}-\frac{2T^{\beta _1+\beta _2}}{(\beta _1+\beta _2-1)\Gamma (\beta _1)\Gamma (\beta _2)}\right. \\{} & {} \left. +\frac{T^{2\beta _2}}{(2\beta _2-1)\Gamma ^2(\beta _2)}\right) \end{aligned}$$

such that ${\mathfrak {B}}_1(\beta _1, \beta _2)\rightarrow 0$ as $\beta _1\rightarrow \beta _2.$ Then, we have the following inequality:

$$\begin{aligned} {\mathcal {I}}_1\le \frac{T{\mathcal {K}} }{\Gamma ^2(\beta _1)}\int _{0}^{\tau }(\tau -\mu )^{2\beta _1-2} E |X_\mu ^{\beta _1}-X_\mu ^{\beta _2}|^2d\mu +{\mathfrak {B}}_1(\beta _1, \beta _2). \end{aligned}$$

(8.10)

For estimating ${\mathcal {I}}_2$ and ${\mathcal {I}}_3,$ we follow the same procedure and define ${\mathfrak {B}}_2(\beta _1, \beta _2)$ and ${\mathfrak {B}}_3(\beta _1, \beta _2)$ as follows:

$$\begin{aligned} {\mathfrak {B}}_2(\beta _1, \beta _2)= & {} {\mathcal {K}}\,C\left( \frac{T^{2\beta _1-1}}{(2\beta _1-1)\Gamma ^2(\beta _1)}-\frac{2T^{\beta _1+\beta _2-1}}{(\beta _1+\beta _2-1)\Gamma (\beta _1)\Gamma (\beta _2)}+\frac{T^{2\beta _2-1}}{(2\beta _2-1)\Gamma ^2(\beta _2)}\right) ,\\ {\mathfrak {B}}_3(\beta _1, \beta _2)= & {} {\mathcal {K}}\,C\left( \frac{\Gamma ^2(\alpha +1)T^{2\alpha +2\beta _1-2}}{(2\alpha +2\beta _1-3)\Gamma ^2(\alpha +\beta _1-1)}+\frac{\Gamma ^2(\alpha +1)T^{2\alpha +2\beta _2-2}}{(2\alpha +2\beta _2-3)\Gamma ^2(\alpha +\beta _2-1)}\right. \\{} & {} \left. -\frac{2\Gamma ^2(\alpha +1)T^{2\alpha +\beta _1+\beta _2-2}}{(2\alpha +\beta _1+\beta _2-3)\Gamma (\alpha +\beta _1-1)\Gamma (\alpha +\beta _2-1)}\right) \end{aligned}$$

such that ${\mathfrak {B}}_2(\beta _1, \beta _2),{\mathfrak {B}}_3(\beta _1, \beta _2)\rightarrow 0$ as $\beta _1\rightarrow \beta _2.$ Then, we get

$$\begin{aligned} {\mathcal {I}}_2\le & {} \frac{{\mathcal {K}} }{\Gamma ^2(\beta _1)}\int _{0}^{\tau }(\tau -\mu )^{2\beta _1-2} E |X_\mu ^{\beta _1}-X_\mu ^{\beta _2}|^2d\mu + {\mathfrak {B}}_2(\beta _1, \beta _2), \end{aligned}$$

(8.11)

$$\begin{aligned} {\mathcal {I}}_3\le & {} \frac{T{\mathcal {K}} \Gamma ^2(\alpha +1)}{\Gamma ^2(\alpha +\beta _1-1)}\int _{0}^{\tau }(\tau -\mu )^{2\alpha +2\beta _1-4} E |X_\mu ^{\beta _1}-X_\mu ^{\beta _2}|^2d\mu \nonumber \\{} & {} + {\mathfrak {B}}_3(\beta _1, \beta _2). \end{aligned}$$

(8.12)

Substituting (8.10–8.12) into (8.9), we get

$$\begin{aligned} E |X_\mu ^{\beta _1}-X_\mu ^{\beta _2}|^2\le & {} {\mathfrak {B}}(\beta _1, \beta _2)+{\bar{r}}_1\int _{0}^{\tau }(\tau -\mu )^{\varpi _1-1} E |X_\mu ^{\beta _1}-X_\mu ^{\beta _2}|^2d\mu \nonumber \\{} & {} +{\bar{r}}_2\int _{0}^{\tau }(\tau -\mu )^{\varpi _2-1} E |X_\mu ^{\beta _1}-X_\mu ^{\beta _2}|^2d\mu , \end{aligned}$$

(8.13)

where ${\mathfrak {B}}(\beta _1, \beta _2)=\sum _{i=1}^{3}3 {\mathfrak {B}}_i(\beta _1, \beta _2), r_1=\frac{3{\mathcal {K}}(T+1) }{\Gamma ^2(\beta _1)}$ and $r_2=\frac{3T{\mathcal {K}} \Gamma ^2(\alpha +1)}{\Gamma ^2(\alpha +\beta _1-1)}.$ In addition, $\varpi _1=2\beta _1-1$ and $\varpi _2=2\alpha +2\beta _1-3.$

Applying Theorem 2.4, we obtain

$$\begin{aligned}{} & {} E |X_\tau ^{\beta _1}-X_\tau ^{\beta _2}|^2\nonumber \\{} & {} \le {\mathfrak {B}}(\beta _1, \beta _2)\left[ 1 +\sum _{m=1}^{\infty }\sum _{\ell =0}^{m}C_m^\ell \frac{({\bar{r}}_1\Gamma (\varpi _1))^{\ell }({\bar{r}}_2\Gamma (\varpi _2))^{m-\ell }\, T^{\ell \varpi _1+(m-\ell )\varpi _2)}}{\Gamma (\ell \varpi _1+(m-\ell )\varpi _2+1)}\right] .\nonumber \\ \end{aligned}$$

(8.14)

Based on (8.14), we notice that ${\mathfrak {B}}(\beta _1, \beta _2)\rightarrow 0$ as $\beta _1\rightarrow \beta _2.$ Therefore, $E |X_\tau ^{\beta _1}-X_\tau ^{\beta _2}|^2\rightarrow 0$ as $\beta _1\rightarrow \beta _2,$ and this shows that $X_\tau ^{\beta _1}\rightarrow X_\tau ^{\beta _2}$ as $\beta _1\rightarrow \beta _2.$ Hence, the proof is complete.