The Convergence and Boundedness of Solutions to SFDEs with the G-Framework

Abstract

Generally, stochastic functional differential equations (SFDEs) pose a challenge as they often lack explicit exact solutions. Consequently, it becomes necessary to seek certain favorable conditions under which numerical solutions can converge towards the exact solutions. This article aims to delve into the convergence analysis of solutions for stochastic functional differential equations by employing the framework of G-Brownian motion. To establish the goal, we find a set of useful monotone type conditions and work within the space ${\mathbb{C}}_r((-\infty ,0];R^n)$. The investigation conducted in this article confirms the mean square boundedness of solutions. Furthermore, this study enables us to compute both ${\mathbb{L}}_G^2$ and exponential estimates.

1. Introduction

Stochastic functional differential equations (SFDEs) find applications in various fields of engineering and science, such as neural networks [1], financial assets [2,3,4], population dynamics [5,6], and gene expression [7]. These dynamic systems are also used in the study of turbulent flow analysis and modeling [8]. A vast amount of literature exists on moment estimates, convergence, stability, and existence of solutions for SFDEs [9,10,11,12,13]. The study of SFDEs driven by G-Brownian motion is relatively new, dating back to the invention of G-Brownian theory in 2006 [14]. In [2,15], the classical Lipschitz condition and linear growth condition were used to establish the existence-uniqueness theorem for SFDEs in the space $BC((-\infty ,0];R^n)$. The study of SFDEs under the G-framework with non-Lipschitz conditions and mean square stability was investigated in [16]. The work in [17,18,19] provides insights into pth moment estimates, the Cauchy–Maruyama approximation scheme, and exponential estimates for solutions to SFDEs within the framework of G-Brownian motion. Asymptotic estimates were studied in [20], while SFDEs under the G-Lévy processes were investigated in [21].

In this article, we introduce some useful monotone type conditions to study SFDEs under the G-framework within the space ${\mathbb{C}}_r((-\infty ,0];R^n)$. Our findings contribute to the growing body of research on SFDEs driven by G-Brownian motion and deepen our understanding of the role of G-framework in stochastic analysis. We study the convergence of solutions for a SFDEs using the framework of G-Brownian motion. Our analysis results in the mean square boundedness of solutions and allows us to compute both ${\mathbb{L}}_G^2$ and exponential estimates. Consider a matrix A; its transpose is denoted by $A^T$. Let $\mathbb{C}((-\infty ,0];R^n)$ denotes the set of continuous mappings from $(-\infty ,0]$ to $R^n$. Define the space ${\mathbb{C}}_r((-\infty ,0];R^n)$, $r>0$ as

$${\mathbb{C}}_r((-\infty ,0];R^n)=\{\alpha \in C((-\infty ,0];R^n):\underset{\sigma \to -\infty }{lim}{e}^{r\sigma }\alpha \left(\sigma \right)\mathrm{exists}\mathrm{in}{R}^n\},$$

Associated with norm ${\parallel \alpha \parallel }_r={sup}_{-\infty <\sigma \le 0}{e}^{r\sigma }\left|\alpha \left(\sigma \right)\right|<\infty$, the space ${\mathbb{C}}_r((-\infty ,0];R^n)$ is a Banach space of bounded continuous mappings. For each $0<r_1\le r_2<\infty$, ${\mathbb{C}}_{r_1}\subseteq {\mathbb{C}}_{r_2}$ [22,23]. Represent the $\sigma$-algebra of ${\mathbb{C}}_r$ by $\mathcal{B}\left({\mathbb{C}}_r\right)$ and ${\mathbb{C}}_r^0=\{\alpha \in {\mathbb{C}}_r:{lim}_{\sigma \to -\infty }{e}^{r\sigma }\alpha \left(\sigma \right)=0\}$. Let ${\mathbb{L}}^2\left({\mathbb{C}}_r\right)$ denote the space of all $\mathcal{F}$-measurable stochastic processes $\alpha$ taking values in ${\mathbb{C}}_r$, such that $\widehat{\mathbb{E}}{\left|\alpha \right|}_r^2<\infty$. Similarly, let ${\mathbb{L}}^2\left({\mathbb{C}}_r^0\right)$ denote the space of all $\mathcal{F}$-measurable stochastic processes $\alpha$ taking values in ${\mathbb{C}}_r^0$, such that ${\mathbb{E}\left|\alpha \right|}_r^2<\infty$. Let $(\mathsf{\Omega },\mathcal{F},P)$ be a complete probability space, where $\mathcal{F}$ is a sigma-algebra of subsets. The natural filtration ${\mathcal{F}}_t$ on $(\mathsf{\Omega },\mathcal{F},P)$ is defined as the sigma-algebra, denoted by ${\mathcal{F}}_t=\sigma \mathcal{B}\left(v\right):0\le v\le t$, where $\mathcal{B}\left(v\right)$ represents the Borel sigma-algebra of ${\mathbb{C}}_r$. We use $\mathcal{P}$ to represent the set of all probability measures on $({\mathbb{C}}_r,\mathcal{B}\left({\mathbb{C}}_r\right))$. Additionally, ${\mathbb{L}}_b\left({\mathbb{C}}_r\right)$ denotes the collection of continuous bounded functionals on ${\mathbb{C}}_r$. Finally, let ${\mathsf{\Lambda }}_0$ be the collection of probability measures on $(-\infty ,0]$ satisfying ${\int }_{-\infty }^0\mu \left(d\sigma \right)=1$ for every $\mu \in {\mathsf{\Lambda }}_0$. We define

$${\mathsf{\Lambda }}_k=\{\mu \in {\mathsf{\Lambda }}_0:{\mu }^{\left(k\right)}={\int }_{-\infty }^0{e}^{-k\sigma }\mu \left(d\sigma \right)<\infty \},$$

(1)

where ${\mathsf{\Lambda }}_{k_0}\subset {\mathsf{\Lambda }}_k\subset {\mathsf{\Lambda }}_0$ for any $k\in (0,k_0)$ [23]. Let $\kappa :{\mathbb{C}}_r((-\infty ,0];R^n)\to R^n$, $\eta :{\mathbb{C}}_r((-\infty ,0];R^n)\to R^n$ and $\gamma :{\mathbb{C}}_r((-\infty ,0];R^n)\to R^n$ be Borel measurable. Consider the SFDEs driven by G-Brownian motion of the form

$$dz\left(t\right)=\kappa \left(z_t\right)dt+\eta \left(z_t\right)d\langle B,B\rangle \left(t\right)+\gamma \left(z_t\right)dB\left(t\right),$$

(2)

on $t\ge 0$ where $z_t=\{z(t+\theta ):-\infty <\theta \le 0\}$. Equation (2) has the starting value $z_0=\zeta \in {\mathbb{C}}_r((-\infty ,0];R^n)$. Let $\langle B,B\rangle \left(t\right)$ denote the quadratic variation process of the G-Brownian motion $B\left(t\right)$ defined on a complete probability space $(\mathsf{\Omega },\mathcal{F},\mathbf{P})$, where $B\left(t\right)$ is a one-dimensional process under the filtration ${\mathcal{F}}_{t\ge 0}$ satisfying the usual conditions. This paper presents the analysis of solutions for equations of the type (2). The remaining paper is arranged as follows. Section 2 presents the basic results. In Section 3, some useful lemmas are given. Section 4 investigates the mean square boundedness and convergence of solutions. In Section 5, we first study the ${\mathbb{L}}_G^2$ estimate and then derive the exponential estimate. Section 6 contains conclusions.

2. Basic Notions and Results

This section presents some fundamental concepts and results that we utilize in the following research work of this article [14,24,25]. The following two basic lemmas can be utilized in forthcoming sections of this paper [9].

Lemma 1.

Let $c,l\ge 0$ and $\delta \in (0,1).$ Then

$$\begin{array}{c}\hfill {(c+l)}^2\le \frac{c^2}{\delta }+\frac{l^2}{1-\delta }.\end{array}$$

Lemma 2.

Let $p\ge 2$ and $\widehat{\delta },c,l>0.$ Then,

(i) 

$c^{p-1}l\le \frac{(p-1)\widehat{\delta }{c}^p}{p}+\frac{l^p}{p{\widehat{\delta }}^{p-1}}.$

(ii) 

$c^{p-2}{l}^2\le \frac{(p-2)\widehat{\delta }{c}^p}{p}+\frac{2l^p}{p{\widehat{\delta }}^{\frac{p-2}{2}}}.$

Let $\mathcal{H}$ be a space of real mappings defined on a non-empty set $\mathsf{\Omega }$.

Definition 1.

∀$x,y\in \mathcal{H}$, a functional $\widehat{\mathbb{E}}:\mathcal{H}\to R$ assuring the below given features are called a G-expectation

1.

$\widehat{\mathbb{E}}\left[y\right]\ge \widehat{\mathbb{E}}\left[x\right]$ whenever $y\ge x$.

2.

$\widehat{\mathbb{E}}\left[m_1\right]=m_1$, for any $m_1\in R$.

3.

$\widehat{\mathbb{E}}\left[m_{2}y\right]=m_2\widehat{\mathbb{E}}\left[y\right]$, for any $m_2\in R^{+}$.

4.

$\widehat{\mathbb{E}}[y+x]\le \widehat{\mathbb{E}}\left[y\right]+\widehat{\mathbb{E}}\left[x\right]$.

Suppose that $\mathsf{\Omega }$ is the space of $R^n$-valued continuous paths ${\left(w\left(t\right)\right)}_{t\ge 0}$ such that $w\left(0\right)=0$ associated with the norm

$$\mathbb{D}(w^1,w^2)=\sum _{j=1}^{\infty }{2}^{-j}\left(\underset{t\in [0,j]}{max}|w^1\left(t\right)-w^2\left(t\right)|\wedge 1\right).$$

Choose ${\mathsf{\Omega }}_T=\{\omega \wedge T:\omega \in \mathsf{\Omega }\}$ [14]. Assume the canonical process $B\left(t\right)=B(t,w)$ where $t\ge 0$ and $w\in \mathsf{\Omega }$. Let $\lambda \in C_{b.Lip}\left(R^{n\times d}\right)$ and $t_1,t_2,\dots ,t_n\in [0,T]$, then

$${\mathbb{L}}_{ip}^0\left({\mathsf{\Omega }}_T\right)=\left\{\lambda (B\left(t_1\right),B\left(t_2\right),\dots ,B\left(t_n\right)):n\ge 1\right\}.$$

Notice that ${\mathbb{L}}_{ip}^0\left({\mathsf{\Omega }}_t\right)\subseteq {\mathbb{L}}_{ip}^0\left({\mathsf{\Omega }}_T\right)$, ${\mathbb{L}}_{ip}^0\left(\mathsf{\Omega }\right)={\cup }_{m=1}^{\infty }{\mathbb{L}}_{ip}^0\left({\mathsf{\Omega }}_m\right)$ and the completion of ${\mathbb{L}}_{ip}^0\left(\mathsf{\Omega }\right)$ associated with $\widehat{\mathbb{E}}{\left[\right|.|}^p{]}^{\frac{1}{p}}$, $p\ge 1$ is ${\mathbb{L}}_G^p\left(\mathsf{\Omega }\right)$. Related to ${\left\{B\left(t\right)\right\}}_{t\ge 0}$, we can express the filtration as ${\mathcal{F}}_t=\sigma \{B\left(u\right),0\le u\le t\}$ where $\mathcal{F}={\left\{{\mathcal{F}}_t\right\}}_{t\ge 0}$. Let $0\le t_0\le t_1\le \dots \le t_N<\infty$ and ${▵}_T=\{t_0,t_1,\dots ,t_N\}$ is a partition of $[0,T].$ Let $p\ge 1$, then ${\mathbb{M}}_G^{p,0}(0,T)$ is given by

$${\mathbb{M}}_G^{p,0}(0,T)=\{{\rho }_t\left(w\right)=\sum _{o=0}^{N-1}{\vartheta }_j\left(w\right)I_{[t_o,t_{o+1}]}\left(t\right):{\vartheta }_o\in {\mathbb{L}}_G^p\left({\mathsf{\Omega }}_{t_o}\right),o=0,1,\dots ,N-1\},$$

The space ${\mathbb{M}}_G^p(0,T)$ is the completion of ${\mathbb{M}}_G^{p,0}(0,T)$ under the norm $\parallel \rho \parallel =$${\left\{{\int }_0^T\mathbb{E}\left[\right|{\rho }_s{|}^p]ds\right\}}^{1/p},$$p\ge 1.$

Definition 2.

The G-Brownian motion is an n-dimensional stochastic process ${\left\{B\left(t\right)\right\}}_{t\ge 0}$ fulfilling the following characteristics:

(i) 

$B\left(0\right)=0.$

(ii) 

$B(t+v)-B\left(t\right)$ is G-normally distributed and independent of $B\left(t_1\right),B\left(t_2\right),\cdots \cdots B\left(t_n\right)$ for any $n\in \mathbb{N}$ and $0\le t_1\le t_2\le ,\cdots ,\le t_n\le t$.

Definition 3.

Let $\rho \left(t\right)\in {\mathbb{M}}_G^{2,0}(0,T)$. The G-Itô integral $\mathcal{I}\left(\rho \right)$ is defined as the stochastic integral of a function ρ with respect to G-Brownian motion given by

$$\begin{array}{c}\hfill \mathcal{I}\left(\rho \right)={\int }_0^T\rho \left(s\right)d{B}^a\left(s\right)=\sum _{k=0}^{N-1}{\xi }_k\left(B^a\left(t_{k+1}\right)-B^a\left(t_k\right)\right),\end{array}$$

One can extend $\mathcal{I}:{\mathbb{M}}_G^{2,0}(0,T)\mapsto {\mathbb{L}}_G^2\left({\mathcal{F}}_T\right)$ to $\mathcal{I}:{\mathbb{M}}_G^2(0,T)\mapsto {\mathbb{L}}_G^2\left({\mathcal{F}}_T\right)$, where for $\rho \in {\mathbb{M}}_G^2(0,T)$ we have

$$\begin{array}{c}\hfill {\int }_0^T\rho \left(s\right)d{B}^a\left(s\right)=\mathcal{I}\left(\rho \right).\end{array}$$

Definition 4.

Let $\langle B^a\rangle \left(0\right)=0$. The G-quadratic variation process ${\left\{\langle B^a\rangle \left(t\right)\right\}}_{t\ge 0}$ is given as follows:

$$\begin{array}{c}\hfill \begin{array}{cc}& \langle B^a\rangle \left(t\right)=\underset{N\to \infty }{lim}\sum _{j=0}^{N-1}{\left(B^a\left(t_{j+1}^N\right)-B^a\left(t_j^N\right)\right)}^2={B^a\left(t\right)}^2-2{\int }_0^t{B}^a\left(s\right)d{B}^a\left(s\right).\hfill \end{array}\end{array}$$

Consider a function ${\mathcal{K}}_{0,T}:{\mathbb{M}}_G^{0,1}(0,T)\mapsto {\mathbb{L}}_G^2\left({\mathcal{F}}_T\right)$ given as

$$\begin{array}{c}\hfill {\mathcal{K}}_{0,T}\left(\rho \right)={\int }_0^T\rho \left(s\right)d\langle B^a\rangle \left(s\right)=\sum _{i=0}^{N-1}{\xi }_i\left({\langle B^a\rangle }_{(}{t}_{i+1})-\langle B^a\rangle \left(t_i\right)\right).\end{array}$$

One can extend ${\mathcal{K}}_{0,T}$ to ${\mathbb{M}}_G^1(0,T)$. For $\rho \in {\mathbb{M}}_G^1(0,T)$, and it is given by

$$\begin{array}{c}\hfill {\int }_0^T\rho \left(s\right)d\langle B^a\rangle \left(s\right)={\mathcal{K}}_{0,T}\left(\rho \right).\end{array}$$

Lemma 3

([14]). Assume that $\gamma \in {\mathbb{M}}_G^p(0,T)$ and $p\ge 2$. Then

$$\widehat{\mathbb{E}}\left[\underset{0\le t\le T}{sup}|{\int }_0^t\gamma \left(s\right)dB\left(s\right){|}^p\right]\le a_3\widehat{\mathbb{E}}{\left[{\int }_0^t{\left|\gamma \left(s\right)\right|}^{2}ds\right]}^{\frac{p}{2}},$$

where $a_3\in (0,\infty )$ is a p dependent constant.

Lemma 4

([14]). Assume that $\gamma \in {\mathbb{M}}_G^p(0,T)$, $p\ge 1$. Then,

$$\widehat{\mathbb{E}}\left[\underset{0\le t\le T}{sup}|{\int }_0^t\gamma \left(s\right)d\langle B,B\rangle \left(s\right){|}^p\right]\le a_2\widehat{\mathbb{E}}{\left[{\int }_0^t{\left|\gamma \left(s\right)\right|}^{2}ds\right]}^{\frac{p}{2}},$$

where $a_2\in (0,\infty )$ depends on p.

Lemma 5

([26]). Let $p\ge 0$, $z\in {\mathbb{L}}_G^p$. For each $m_1>0$,

$$\nu \left(\right|z|>m_1)\le \frac{\widehat{\mathbb{E}}[{\left|z\right|}^p]}{m_1},$$

where $\widehat{\mathbb{E}}{\left|z\right|}^p<\infty$.

3. Some Useful Results

In this section, we introduce and discuss some important assumptions and establish two lemmas. We consider the following hypotheses:

(H) Let $a_i>0$, $i=1,2,\dots ,5$ and $\alpha \left(\sigma \right)-\beta \left(\sigma \right)=\mathsf{\Psi }\left(\sigma \right)$. $Forany\alpha ,\beta \in {\mathbb{C}}_r((-\infty ,0];R^n)$ and for any probability measure ${\mu }_1,{\mu }_2,{\mu }_3\in {\mathsf{\Lambda }}_{2r}$ the following inequalities hold

$${\left[\mathsf{\Psi }\left(0\right)\right]}^T[\kappa \left(\alpha \right)-\kappa \left(\beta \right)]\le -a_1{\left|\mathsf{\Psi }\left(0\right)\right|}^2+a_2{\int }_{-\infty }^0{\left|\mathsf{\Psi }\left(\sigma \right)\right|}^2{\mu }_1\left(d\sigma \right),$$

(3)

$${\left[\mathsf{\Psi }\left(0\right)\right]}^T[\eta \left(\alpha \right)-\eta \left(\beta \right)]\le -a_3{\left|\mathsf{\Psi }\left(0\right)\right|}^2+a_4{\int }_{-\infty }^0{\left|\mathsf{\Psi }\left(\sigma \right)\right|}^2{\mu }_2\left(d\sigma \right),$$

(4)

$${|\gamma \left(\alpha \right)-\gamma \left(\beta \right)|}^2\le a_5{\int }_{-\infty }^0{\left|\mathsf{\Psi }\left(\sigma \right)\right|}^2{\mu }_3\left(d\sigma \right).$$

(5)

Lemma 6.

Let $a<pr$ and $p\ge 1$. Then,

$$\begin{array}{cc}\hfill \parallel z_t{\parallel }_r^p& \le e^{-at}{\parallel \zeta \parallel }_r^p+\underset{0<u\le t}{sup}{\left|z\left(u\right)\right|}^p,\hfill \end{array}$$

where $\zeta \in {\mathbb{C}}_r((-\infty ,0];R^n)$.

Proof. 

Suppose $pr>a$. We can deduce the following by employing the definition of norm ${\parallel ·\parallel }_r$:

$$\begin{array}{cc}\hfill \parallel z_t{\parallel }_r^p& ={\left[\underset{-\infty <\sigma \le 0}{sup}{e}^{r\sigma }\left|z(t+\sigma )\right|\right]}^p\le \underset{-\infty <\sigma \le 0}{sup}{e}^{a\sigma }{\left|z(t+\sigma )\right|}^p\hfill \\ & \le \underset{0<u\le t}{sup}{e}^{-a(t-u)}{\left|z\left(u\right)\right|}^p+\underset{-\infty <u\le 0}{sup}{e}^{-a(t-u)}{\left|z\left(u\right)\right|}^p\hfill \\ & =e^{-at}{\parallel \zeta \parallel }_r^p+e^{-at}\underset{0<u\le t}{sup}{e}^{as}{\left|z\left(u\right)\right|}^p\hfill \\ & \le \underset{0<u\le t}{sup}{\left|z\left(u\right)\right|}^p+e^{-at}{\parallel \zeta \parallel }_r^p.\hfill \end{array}$$

The proof stands completed. □

Throughout this paper, we let that for any $p\ge 1$, $a<pr$.

Lemma 7.

Let $a<pr$, $p\ge 2$ and ${\mu }_i\in {\mathsf{\Lambda }}_k$, ∀$i\in N$. Then,

$${\int }_0^t{\int }_{-\infty }^0{\left|z(\sigma +s)\right|}^p{\mu }_i\left(d\sigma \right)ds\le \frac{{\mu }_i^{\left(pr\right)}}{pr}{\parallel \zeta \parallel }_r^p+{\int }_0^t{\left|z\left(s\right)\right|}^{p}ds,$$

(6)

$${\int }_0^t{\int }_{-\infty }^0{e}^{as}{\left|z(\sigma +s)\right|}^p{\mu }_i\left(d\sigma \right)ds\le \frac{{\mu }_i^{\left(pr\right)}}{pr-a}{\parallel \zeta \parallel }_r^p+{\mu }_i^{\left(pr\right)}{\int }_0^t{e}^{as}{\left|z\left(s\right)\right|}^{p}ds,$$

(7)

where $\zeta \in {\mathbb{C}}_r((-\infty ,0];R^n)$.

Proof. 

As for each $i\in Z^{+}$, ${\mu }_i\in {\mathsf{\Lambda }}_{pr}$ and $\zeta \in C_r((-\infty ,0];R^n)$, by using the Fubini theorem and the definition of norm, it follows that

$$\begin{array}{cc}\hfill & {\int }_0^t{\int }_{-\infty }^0{\left|z(\sigma +s)\right|}^p{\mu }_i\left(d\sigma \right)ds\hfill \\ & ={\int }_0^t\left[{\int }_{-\infty }^{-s}{e}^{pr(\sigma +s)}{\left|z(\sigma +s)\right|}^p{e}^{-pr(s+\sigma )}{\mu }_i\left(d\sigma \right)+{\int }_{-s}^0{\left|z(\sigma +s)\right|}^p{\mu }_i\left(d\sigma \right)\right]ds\hfill \\ & \le {\parallel \zeta \parallel }_r^p{\int }_0^t{e}^{-prs}ds{\int }_{-\infty }^0{e}^{-pr\sigma }{\mu }_i\left(d\sigma \right)+{\int }_{-\infty }^0{\mu }_i\left(d\sigma \right){\int }_0^t{\left|z\left(s\right)\right|}^{p}ds.\hfill \end{array}$$

Observing that ${\int }_{-\infty }^0{\mu }_i\left(d\sigma \right)=1$ and ${\int }_{-\infty }^0{e}^{-pr\sigma }{\mu }_i\left(d\sigma \right)={\mu }_i^{\left(pr\right)}$, $i\in N$, it follows that

$${\int }_0^t{\int }_{-\infty }^0{\left|z(s+\sigma )\right|}^p{\mu }_i\left(d\sigma \right)ds\le \frac{{\mu }_i^{\left(pr\right)}}{pr}{\parallel \zeta \parallel }_r^p+{\int }_0^t{\left|z\left(s\right)\right|}^{p}ds.$$

The proof of (6) is complete. Using similar arguments as used above we determine

$$\begin{array}{cc}\hfill & {\int }_0^t{\int }_{-\infty }^0{e}^{as}{\left|z(\sigma +s)\right|}^p{\mu }_i\left(d\sigma \right)ds\hfill \\ & ={\int }_0^t{e}^{as}ds\left[{\int }_{-\infty }^{-s}{\left|z(\sigma +s)\right|}^p{\mu }_i\left(d\sigma \right)+{\int }_{-s}^0{\left|z(\sigma +s)\right|}^p{\mu }_i\left(d\sigma \right)\right]\hfill \\ & ={\int }_0^t{e}^{as}ds{\int }_{-\infty }^{-s}{\left|z(\sigma +s)\right|}^p{\mu }_i\left(d\sigma \right)+{\int }_{-t}^0{\mu }_i\left(d\sigma \right){\int }_{-\sigma }^t{e}^{as}{\left|z(\sigma +s)\right|}^{p}ds\hfill \\ & \le {\int }_0^t{e}^{as}ds{\int }_{-\infty }^{-s}{e}^{pr(\sigma +s)}{\left|z(s+\sigma )\right|}^p{e}^{-pr(\sigma +s)}{\mu }_i\left(d\sigma \right)+{\int }_{-\infty }^0{\mu }_i\left(d\sigma \right){\int }_0^t{e}^{a(s-\sigma )}{\left|z\left(s\right)\right|}^{p}ds\hfill \\ & \le {\parallel \zeta \parallel }_r^p{\int }_0^t{e}^{-(pr-a)s}ds{\int }_{-\infty }^0{e}^{-pr\sigma }{\mu }_i\left(d\sigma \right)+{\int }_{-\infty }^0{e}^{-a\sigma }{\mu }_i\left(d\sigma \right){\int }_0^t{e}^{as}{\left|z\left(s\right)\right|}^{p}ds.\hfill \end{array}$$

With reference to Equation (1), and taking note that $pr>a$, we can conclude that

$${\int }_0^t{\int }_{-\infty }^0{e}^{as}{\left|z(\sigma +s)\right|}^p{\mu }_i\left(d\sigma \right)ds\le \frac{{\mu }_i^{\left(pr\right)}}{pr-a}{\parallel \zeta \parallel }_r^p+{\mu }_i^{\left(pr\right)}{\int }_0^t{e}^{as}{\left|z\left(s\right)\right|}^{p}ds.$$

The proof of (7) is complete. □

4. Convergence and Mean Square Boundedness

Firstly, let us derive the mean square boundedness for the solutions to Equation (2).

Theorem 1.

Let the inequalities (3)–(5) be satisfied. Assume Equation (2) with initial condition $\zeta \in C_r((-\infty ,0];R^n)$ has just one solution $z\left(t\right)$. Let $a_i$, $i=1,2,\dots ,5$ assure $2a_1>2a_2{\mu }_1^{\left(2r\right)}+2b_1{a}_4{\mu }_2^{\left(2r\right)}+b_1{a}_5{\mu }_3^{\left(2r\right)}-2b_1{a}_3$. Then there is $a\in (0,(2a_1+2b_1{a}_3-2a_2{\mu }_1^{\left(2r\right)}-2b_1{a}_4{\mu }_2^{\left(2r\right)}-b_1{a}_5{\mu }_3^{\left(2r\right)})\wedge 2r)$ so that

$$\widehat{\mathbb{E}}{\left[\right|z\left(t\right)|}^2]\le c_1+c_2{e}^{-at},$$

(8)

where

$$c_1=\frac{1}{a}\left(\frac{1}{\delta }{\left|\kappa \left(0\right)\right|}^2+\frac{b_1}{{\delta }_1}{\left|\eta \left(0\right)\right|}^2+\frac{b_1}{{\delta }_2}{\left|\gamma \left(0\right)\right|}^2\right)$$

and

$$c_2=\widehat{\mathbb{E}}{\left|z\left(0\right)\right|}^2+\frac{2a_2{\mu }_1^{\left(2r\right)}}{2r-a}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+\frac{2b_1{a}_4{\mu }_2^{\left(2r\right)}}{2r-a}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+\frac{b_1{a}_5{\mu }_3^{\left(2r\right)}}{(2r-a)(1-{\delta }_2)}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2.$$

The values of δ,${\delta }_1$ and ${\delta }_2$ are sufficiently small so that

$$2a_1-\delta -a-b_1{\delta }_1+2b_1{a}_3-2a_2{\mu }_1^{\left(2r\right)}-2b_1{a}_4{\mu }_2^{\left(2r\right)}-\frac{b_1{a}_5}{1-{\delta }_2}{\mu }_3^{\left(2r\right)}>0.$$

Proof. 

By using the G-Itô formula, G-Itô integral and Lemma 4, it follows

$$\begin{array}{cc}\hfill \widehat{\mathbb{E}}[e^{at}{\left|z\left(t\right)\right|}^2]& \le \widehat{\mathbb{E}}{\left|z\left(0\right)\right|}^2+\widehat{\mathbb{E}}{\int }_0^t{e}^{as}\left[{a\left|z\left(s\right)\right|}^2+2z^T\left(s\right)\kappa \left(z_s\right)\right]ds\hfill \\ & +b_1\widehat{\mathbb{E}}{\int }_0^t{e}^{as}\left[2z^T\left(s\right)\eta \left(z_s\right)+{\left|\gamma \left(z_s\right)\right|}^2\right]ds.\hfill \end{array}$$

(9)

Utilizing (3), (4) and Lemma 2, we determine

$$\begin{array}{cc}\hfill z^T\left(t\right)\kappa \left(z_t\right)& \le (\frac{\delta }{2}-a_1)|z\left(t\right)|{}^2+\frac{1}{2\delta }|\kappa \left(0\right)|{}^2+a_2{\int }_{-\infty }^0{\left|z(t+\sigma )\right|}^2{\mu }_1\left(d\sigma \right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill z^T\left(t\right)\eta \left(z_t\right)& \le (\frac{{\delta }_1}{2}-a_3)|z\left(t\right)|{}^2+\frac{1}{2{\delta }_1}|\eta \left(0\right)|{}^2+a_4{\int }_{-\infty }^0{\left|z(t+\sigma )\right|}^2{\mu }_2\left(d\sigma \right).\hfill \\ \end{array}$$

It follows from Lemma 1 and the condition given in (5) that

$$\begin{array}{cc}\hfill |\gamma \left(z_t\right){|}^2& \le \frac{1}{{\delta }_2}{\left|\gamma \left(0\right)\right|}^2+\frac{a_5}{1-{\delta }_2}{\int }_{-\infty }^0{\left|z(t+\sigma )\right|}^2{\mu }_3\left(d\sigma \right).\hfill \end{array}$$

Using the above inequalities, (9) becomes

$$\begin{array}{cc}\hfill \widehat{\mathbb{E}}[e^{at}{\left|z\left(t\right)\right|}^2]& \le \widehat{\mathbb{E}}{\left|z\left(0\right)\right|}^2+\frac{1}{a}\left(\frac{1}{\delta }{\left|\kappa \left(0\right)\right|}^2+\frac{b_1}{{\delta }_1}{\left|\eta \left(0\right)\right|}^2+\frac{b_1}{{\delta }_2}{\left|\gamma \left(0\right)\right|}^2\right)(e^{at}-1)\hfill \\ & +(\delta +a-2a_1+b_1{\delta }_1-2b_1{a}_3)\widehat{\mathbb{E}}{\int }_0^t{e}^{as}{\left|z\left(s\right)\right|}^{2}ds\hfill \\ & +2a_2\widehat{\mathbb{E}}{\int }_0^t{e}^{as}{\int }_{-\infty }^0{\left|z(s+\sigma )\right|}^2{\mu }_1\left(d\sigma \right)ds\hfill \\ & +2b_1{a}_4\widehat{\mathbb{E}}{\int }_0^t{e}^{as}{\int }_{-\infty }^0{\left|z(s+\sigma )\right|}^2{\mu }_2\left(d\sigma \right)ds\hfill \\ & +b_1\frac{a_5}{1-{\delta }_2}\widehat{\mathbb{E}}{\int }_0^t{e}^{as}{\int }_{-\infty }^0{\left|z(s+\sigma )\right|}^2{\mu }_3\left(d\sigma \right)ds.\hfill \end{array}$$

(10)

In view of Lemma 7, it follows

$${\int }_0^t{\int }_{-\infty }^0{e}^{as}{\left|z(s+\sigma )\right|}^2{\mu }_i\left(d\sigma \right)ds\le \frac{1}{2r-a}{\parallel \zeta \parallel }_r^2{\mu }_i^{\left(2r\right)}+{\mu }_i^{\left(2r\right)}{\int }_0^t{e}^{as}{\left|z\left(s\right)\right|}^{2}ds.$$

(11)

Substituting (11) in (10), we derive

$$\begin{array}{cc}\hfill & \widehat{\mathbb{E}}[e^{at}{\left|z\left(t\right)\right|}^2]\le \widehat{\mathbb{E}}{\left|z\left(0\right)\right|}^2+\frac{2a_2{\mu }_1^{\left(2r\right)}}{2r-a}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+\frac{2b_1{a}_4{\mu }_2^{\left(2r\right)}}{2-ra}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+\frac{b_1{a}_5{\mu }_3^{\left(2r\right)}}{(2r-a)(1-{\delta }_2)}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2\hfill \\ & +\frac{1}{\lambda }\left(\frac{1}{\delta }{\left|\kappa \left(0\right)\right|}^2+\frac{b_1}{{\delta }_1}{\left|\eta \left(0\right)\right|}^2+\frac{b_1}{{\delta }_2}{\left|\gamma \left(0\right)\right|}^2\right)(e^{at}-1)\hfill \\ & -(2a_1-\delta -a-b_1{\delta }_1+2b_1{a}_3-2a_2{\mu }_1^{\left(2r\right)}-2b_1{a}_4{\mu }_2^{\left(2r\right)}-\frac{b_1{a}_5}{1-{\delta }_2}{\mu }_3^{\left(2r\right)})\widehat{\mathbb{E}}{\int }_0^t{e}^{as}{\left|z\left(s\right)\right|}^{2}ds.\hfill \end{array}$$

As $2a_1>2a_2{\mu }_1^{\left(2r\right)}+2b_1{a}_4{\mu }_2^{\left(2r\right)}+b_1{a}_5{\mu }_3^{\left(2r\right)}-2b_1{a}_3$ and $a\in (0,(2a_1+2b_1{a}_3-2a_2{\mu }_1^{\left(2r\right)}-2b_1{a}_4{\mu }_2^{\left(2r\right)}-b_1{a}_5{\mu }_3^{\left(2r\right)})\wedge 2r)$. Selecting $\delta$, ${\delta }_1$ and ${\delta }_2$ sufficiently small so that

$$2a_1-\delta -a-b_1{\delta }_1+2b_1{a}_3-2a_2{\mu }_1^{\left(2r\right)}-2b_1{a}_4{\mu }_2^{\left(2r\right)}-\frac{b_1{a}_5}{1-{\delta }_2}{\mu }_3^{\left(2r\right)}>0,$$

we obtain the desired result:

$$\widehat{\mathbb{E}}{\left[\right|z\left(t\right)|}^2]\le c_1+c_2{e}^{-at},$$

where

$$c_1=\frac{1}{a}\left(\frac{1}{\delta }{\left|\kappa \left(0\right)\right|}^2+\frac{b_1}{{\delta }_1}{\left|\eta \left(0\right)\right|}^2+\frac{b_1}{{\delta }_2}{\left|\gamma \left(0\right)\right|}^2\right)$$

and

$$c_2=\widehat{\mathbb{E}}|z_0|{}^2+\frac{2a_2{\mu }_1^{\left(2q\right)}}{2r-a}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+\frac{2b_1{a}_4{\mu }_2^{\left(2r\right)}}{2r-a}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+\frac{b_1{a}_5{\mu }_3^{\left(2r\right)}}{(2r-a)(1-{\delta }_2)}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2.$$

□

Theorem 1 describes that Equation (2) has a mean square bounded solution. The following Theorem 2 expresses that any two distinct solutions of Equation (2) are convergent.

Theorem 2.

Assuming that all hypotheses of Theorem 1 are satisfied, let $z\left(t\right)$ and $y\left(t\right)$ be two solutions of Equation (2) associated with initial values ζ and ξ, respectively. Then, we have

$$\widehat{\mathbb{E}}{\left[\right|z\left(t\right)-y\left(t\right)|}^2]\le c_3\widehat{\mathbb{E}}{|\xi -\zeta |}_r^2{e}^{-at},$$

(12)

where $c_3=1+\frac{1}{2r-a}(2a_2{\mu }_1^{\left(2r\right)}+2b_1{a}_4{\mu }_2^{\left(2r\right)}+b_1{a}_5{\mu }_3^{\left(2r\right)})$.

Proof. 

Define $\varphi \left(t\right)=z\left(t\right)-y\left(t\right)$, $\widehat{\gamma }\left(t\right)=\gamma \left(z_t\right)-\gamma \left(y_t\right)$, $\widehat{\kappa }\left(t\right)=\kappa \left(z_t\right)-\kappa \left(y_t\right)$, and $\widehat{\eta }\left(t\right)=\eta \left(z_t\right)-\eta \left(y_t\right)$. Utilizing the G-Itô integral, Lemma 4, and G-Itô formula, it follows that

$$\begin{array}{cc}\hfill \widehat{\mathbb{E}}[e^{at}{\left|\varphi \left(t\right)\right|}^2]& \le \widehat{\mathbb{E}}{|\xi \left(0\right)-\zeta \left(0\right)|}^2+\widehat{\mathbb{E}}{\int }_0^t{e}^{as}{\left[a\right|\varphi \left(s\right)|}^2+2{\varphi }^T\left(s\right)\widehat{\kappa }\left(s\right)]ds\hfill \\ & +b_1\widehat{\mathbb{E}}{\int }_0^t{e}^{\varphi s}[2{\varphi }^T\left(s\right)\widehat{\eta }\left(s\right)+|\widehat{\gamma }\left(s\right){|}^2]ds.\hfill \end{array}$$

(13)

From hypothesis H, we derive

$$\begin{array}{cc}\hfill & {\varphi }^T\left(t\right)\widehat{\kappa }\left(t\right)\le -a_1{\left|\varphi \left(t\right)\right|}^2+a_2{\int }_{-\infty }^0\varphi (\sigma +t){\mu }_1\left(d\sigma \right),\hfill \\ & {\varphi }^T\left(t\right)\widehat{\eta }\left(t\right)\le -a_3{\left|\varphi \left(t\right)\right|}^2+a_4{\int }_{-\infty }^0\varphi (\sigma +t){\mu }_2\left(d\sigma \right)\hfill \\ \end{array}$$

and

$$|\widehat{\gamma }{\left(t\right)|}^2\le a_5{\int }_{-\infty }^0\varphi (t+\sigma ){\mu }_3\left(d\sigma \right).$$

Utilizing the above inequalities, (13) becomes

$$\begin{array}{cc}\hfill \widehat{\mathbb{E}}[e^{at}{\left|\varphi \left(t\right)\right|}^2]& \le \widehat{\mathbb{E}}{|\xi \left(0\right)-\zeta \left(0\right)|}^2+(a-2a_1-2b_1{a}_3)\widehat{\mathbb{E}}{\int }_0^t{e}^{as}{\left|\varphi \left(s\right)\right|}^{2}ds\hfill \\ & +2a_2\widehat{\mathbb{E}}{\int }_0^t{\int }_{-\infty }^0{e}^{as}\varphi (s+\sigma ){\mu }_1\left(d\sigma \right)ds+2b_1{a}_4\widehat{\mathbb{E}}{\int }_0^t{\int }_{-\infty }^0{e}^{as}\varphi (s+\sigma ){\mu }_2\left(d\sigma \right)ds\hfill \\ & +k_1{a}_5\widehat{\mathbb{E}}{\int }_0^t{\int }_{-\infty }^0{e}^{as}\varphi (s+\sigma ){\mu }_3\left(d\sigma \right)ds.\hfill \end{array}$$

(14)

From Lemma 7 for $i=1,2,3$, it follows

$$\begin{array}{cc}\hfill & {\int }_0^t{\int }_{-\infty }^0{e}^{as}{\left|\varphi (s+\sigma )\right|}^2{\mu }_i\left(d\sigma \right)ds\le \frac{1}{2q-a}{\parallel \zeta -\xi \parallel }_r^2{\mu }_i^{\left(2r\right)}+{\mu }_i^{\left(2r\right)}{\int }_0^t{e}^{as}{\left|\varphi \left(s\right)\right|}^{2}ds.\hfill \end{array}$$

(15)

Plugging (15) in (14), we determine

$$\begin{array}{cc}\hfill e^{at}\widehat{\mathbb{E}}{\left|\varphi \left(t\right)\right|}^2& \le \widehat{\mathbb{E}}{|\xi \left(0\right)-\zeta \left(0\right)|}^2+\frac{1}{2r-a}[2a_2{\mu }_1^{\left(2r\right)}+2b_1{a}_4{\mu }_2^{\left(2r\right)}+b_1{a}_5{\mu }_3^{\left(2r\right)}]\widehat{\mathbb{E}}{\parallel \zeta -\xi \parallel }_r^2\hfill \\ & -(2a_1+2b_1{a}_3-a-2a_2{\mu }_1^{\left(2r\right)}-2k_1{a}_4{\mu }_2^{\left(2r\right)}-b_1{a}_5{\mu }_3^{\left(2r\right)})\widehat{\mathbb{E}}{\int }_0^t{e}^{as}{\left|\varphi \left(s\right)\right|}^{2}ds.\hfill \end{array}$$

As $2a_1>2a_2{\mu }_1^{\left(2r\right)}+2b_1{a}_4{\mu }_2^{\left(2r\right)}+b_1{a}_5{\mu }_3^{\left(2r\right)}-2b_1{a}_3$ and $a\in (0,(2a_1+2b_1{a}_3-2a_2{\mu }_1^{\left(2r\right)}-2b_1{a}_4{\mu }_2^{\left(2r\right)}-b_1{a}_5{\mu }_3^{\left(2r\right)})\wedge 2r)$ it follows that

$$\begin{array}{cc}\hfill \widehat{\mathbb{E}}{\left|\varphi \left(t\right)\right|}^2& \le [1+\frac{1}{2r-a}(2a_2{\mu }_1^{\left(2r\right)}+2b_1{a}_4{\mu }_2^{\left(2r\right)}+b_1{a}_5{\mu }_3^{\left(2r\right)})]\widehat{\mathbb{E}}{\parallel \zeta -\xi \parallel }_r^2{e}^{-at},\hfill \end{array}$$

consequently, we derive the following required result

$$\begin{array}{cc}\hfill \widehat{\mathbb{E}}{|z\left(t\right)-y\left(t\right)|}^2& \le c_3\widehat{\mathbb{E}}{\parallel \zeta -\xi \parallel }_r^2{e}^{-at},\hfill \end{array}$$

where $c_3=1+\frac{1}{2r-a}(2a_2{\mu }_1^{\left(2r\right)}+2b_1{a}_4{\mu }_2^{\left(2r\right)}+b_1{a}_5{\mu }_3^{\left(2r\right)})$. □

If $\kappa \left(0\right)=\eta \left(0\right)=\gamma \left(0\right)=0$, then from Theorem 2 we can obtain that the trivial solution of Equation (2) is mean square exponentially stable.

Example 1.

Consider $z\left(t\right)$ and $y\left(t\right)$ as two solutions of the equation

$$dz\left(t\right)=z_{t}dt+sin\left(z_t\right)d\langle B\rangle \left(t\right)+z_{t}dB\left(t\right)$$

with initial values ζ and ξ, respectively. Define $\varphi \left(t\right)=z\left(t\right)-y\left(t\right)$, $\widehat{\eta }\left(t\right)=\widehat{\gamma }\left(t\right)=z_t-y_t$ and $\widehat{\kappa }\left(t\right)=sin\left(z_t\right)-sin\left(y_t\right)$. Under the given hypothesis one can easily derive that $y\left(t\right)$ is mean square convergent to $z\left(t\right)$.

5. The Exponential Estimate

Firstly, let us determine the ${\mathcal{L}}_G^2$ estimates. Let Equation (2) with initial condition $\zeta \in C_r((-\infty ,0];R^n)$ has just one solution $z\left(t\right)$ on $t\in [0,\infty )$.

Theorem 3.

Assume that the hypothesis H holds and ${E\parallel \zeta \parallel }_r^2<\infty$. For every $t\ge 0$,

$$\widehat{\mathbb{E}}\left[\underset{-\infty <s\le t}{sup}{\left|z\left(t\right)\right|}^2\right]\le [\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+m_1]e^{m_{2}t},$$

where $m_1=c+\frac{2}{r}[r+a_2{\mu }_1^{\left(2r\right)}+b_1(a_5{\mu }_3^{\left(2r\right)}+{\mu }_2^{\left(2r\right)})+2b_3{a}_5{\mu }_3^{\left(2r\right)}]\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2$, $c={2\left[\right|\kappa \left(0\right)|}^2+b_1{\left(\right|\eta \left(0\right)|}^2+{2\left|\gamma \left(0\right)\right|}^2)+4b_3{\left|\gamma \left(0\right)\right|}^2]T$ and $m_2=2[2a_2-2a_1+1+b_1(2a_5-2a_3+3)+4b_3{a}_5]$.

Proof. 

Using the G-It$\widehat{o}$ formula and properties of the G-expectation, it follows that

$$\begin{array}{cc}\hfill \widehat{\mathbb{E}}\left[\underset{0\le s\le t}{sup}{\left|z\left(t\right)\right|}^2\right]& \le \widehat{\mathbb{E}}{\left|z\left(0\right)\right|}^2+2\widehat{\mathbb{E}}\left[\underset{0\le s\le t}{sup}{\int }_0^t{z}^T\left(s\right)\kappa \left(z_s\right)ds\right]\hfill \\ & +\widehat{\mathbb{E}}\left[\underset{0\le s\le t}{sup}{\int }_0^t(2z^T\left(s\right)\eta \left(z_s\right)+|\gamma \left(z_s\right){|}^2)d\langle B,B\rangle \left(s\right)\right]\hfill \\ & +2\widehat{\mathbb{E}}\left[\underset{0\le s\le t}{sup}{\int }_0^t{z}^T\left(s\right)\gamma \left(z_s\right)dB\left(s\right)\right].\hfill \end{array}$$

(16)

From our assumption H, $2a_1{a}_2\le {\sum }_{i=1}^2{a}_i^2$ and ${\left({\sum }_{i=1}^2{a}_i\right)}^2\le 2{\sum }_{i=1}^2{a}_i^2$, it follows that

$$\begin{array}{cc}\hfill z^T\left(t\right)\kappa \left(z_t\right)& \le -(a_1-\frac{1}{2})|z\left(t\right){|}^2+\frac{1}{2}|\kappa \left(0\right){|}^2+a_2{\int }_{-\infty }^0{\left|z(t+\sigma )\right|}^2{\mu }_1\left(d\sigma \right),\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill z^T\left(t\right)\eta \left(z_t\right)& \le -(a_3-\frac{1}{2})|z\left(t\right){|}^2+\frac{1}{2}|\eta \left(0\right){|}^2+a_4{\int }_{-\infty }^0{\left|z(t+\sigma )\right|}^2{\mu }_2\left(d\sigma \right),\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill |\gamma \left(z_t\right){|}^2& \le {2\left|\gamma \left(0\right)\right|}^2+2a_5{\int }_{-\infty }^0{\left|z(t+\sigma )\right|}^2{\mu }_3\left(d\sigma \right).\hfill \end{array}$$

(19)

In view of (17) and (11), we determine that

$$\begin{array}{cc}& 2\widehat{\mathbb{E}}\left[\underset{0<s\le t}{sup}{\int }_0^t{z}^T\left(s\right)\kappa \left(z_s\right)ds\right]\le {\left|\kappa \left(0\right)\right|}^{2}T+\frac{a_2}{r}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2{\mu }_1^{\left(2r\right)}+(2a_2-2a_1+1)\widehat{\mathbb{E}}{\int }_0^t{\left|z\left(s\right)\right|}^{2}ds.\hfill \end{array}$$

Utilizing (18), (19), (11) and Lemma 4, it follows that

$$\begin{array}{cc}\hfill & \widehat{\mathbb{E}}\left[\underset{0\le s\le t}{sup}{\int }_0^t(2z^T\left(s\right)\eta \left(z_s\right)+|\gamma \left(z_s\right){|}^2)d\langle B,B\rangle \left(s\right)\right]\le b_1\widehat{\mathbb{E}}\left[{\int }_0^t(2z^T\left(s\right)\eta \left(z_s\right)+|\gamma \left(z_s\right){|}^2)\right]ds\hfill \\ & \le b_1{\left[\right|\eta \left(0\right)|}^2+{2\left|\gamma \left(0\right)\right|}^2]T+b_1\frac{1}{r}(a_5{\mu }_3^{\left(2r\right)}+{\mu }_2^{\left(2r\right)})\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2\hfill \\ & +b_1(2a_5-2a_3+3)\widehat{\mathbb{E}}{\int }_0^t{\left|z\left(s\right)\right|}^{2}ds.\hfill \end{array}$$

The inequality $a_1{a}_2\le \frac{1}{2}{\sum }_{i=1}^2{a}_i$, Lemma 3 and (19) give

$$\begin{array}{cc}\hfill & 2\widehat{\mathbb{E}}\left[\underset{0<s\le t}{sup}{\int }_0^t{z}^T\left(s\right)\gamma \left(z_s\right)dB\left(t\right)\right]\le 2b_2\widehat{\mathbb{E}}{\left[{\int }_0^t{\left|z^T\left(s\right)\gamma \left(z_s\right)\right|}^{2}ds\right]}^{\frac{1}{2}}\hfill \\ & \le \frac{1}{2}\widehat{\mathbb{E}}\left[\underset{0<s\le t}{sup}{\left|z\left(s\right)\right|}^2\right]+4b_2^2{\left|\gamma \left(0\right)\right|}^{2}T+4b_2^2{a}_5\widehat{\mathbb{E}}{\int }_0^t{\int }_{-\infty }^0{\left|z(s+\sigma )\right|}^2{\mu }_3\left(d\sigma \right)ds.\hfill \end{array}$$

Using Lemma 7, we derive

$$\begin{array}{cc}\hfill & 2\widehat{\mathbb{E}}\left[\underset{0<s\le t}{sup}{\int }_0^t{z}^T\left(s\right)\gamma \left(z_s\right)dB\left(s\right)\right]\hfill \\ & \le \frac{1}{2}\widehat{\mathbb{E}}\left[\underset{0<s\le t}{sup}{\left|z\left(s\right)\right|}^2\right]+4b_3{\left|\gamma \left(0\right)\right|}^{2}T+\frac{4b_3{a}_5}{2r}{\mu }_3^{\left(2r\right)}\widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+4b_3{a}_5\widehat{\mathbb{E}}{\int }_0^t{\left|z\left(s\right)\right|}^{2}ds,\hfill \end{array}$$

(20)

where $b_3=b_2^2$. By substituting the aforementioned inequalities into Equation (16), and letting $m_1=c+\frac{2}{r}[r+a_2{\mu }_1^{\left(2r\right)}+b_1(a_5{\mu }_3^{\left(2r\right)}+{\mu }_2^{\left(2r\right)})+2b_3{a}_5{\mu }_3^{\left(2r\right)}]E{\parallel \zeta \parallel }_r^2$, $m_2=2[2a_2-2a_1+1+b_1(2a_5-2a_3+3)+4b_3{a}_5]$, we can evaluate the result:

$$\begin{array}{cc}\hfill \widehat{\mathbb{E}}[\underset{0\le s\le t}{sup}{\left|z\left(t\right)\right|}^2]& \le m_1+m_2{\int }_0^t\widehat{\mathbb{E}}\left[\underset{0\le s\le t}{sup}{\left|z\left(s\right)\right|}^2\right]ds,\hfill \end{array}$$

(21)

where $c=2\left[{\left|\kappa \left(0\right)\right|}^2+b_1{\left(\right|\eta \left(0\right)|}^2+{2\left|\gamma \left(0\right)\right|}^2)+4b_3{\left|\gamma \left(0\right)\right|}^2\right]T$. By observing that

$$\widehat{\mathbb{E}}[\underset{-\infty <s\le t}{sup}{\left|z\left(s\right)\right|}^2]\le \widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+\widehat{\mathbb{E}}[\underset{0\le s\le t}{sup}{\left|z\left(s\right)\right|}^2],$$

it follows that

$$\begin{array}{cc}\hfill \widehat{\mathbb{E}}\left[\underset{-\infty <s\le t}{sup}{\left|z\left(s\right)\right|}^2\right]& \le \widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+m_1+m_2{\int }_0^t\widehat{\mathbb{E}}\left[\underset{0\le s\le t}{sup}{\left|z\left(s\right)\right|}^2\right]ds\hfill \\ & \le \widehat{\mathbb{E}}{\parallel \zeta \parallel }_r^2+m_1+m_2{\int }_0^t\widehat{\mathbb{E}}\left[\underset{-\infty <s\le t}{sup}{\left|z\left(s\right)\right|}^2\right]ds.\hfill \end{array}$$

Finally, the required result is obtained by using the Grownwall inequality. □

Theorem 4.

Under the conditions of Theorem 3, it follows that

$$\underset{t\to \infty }{lim}sup\frac{1}{t}log\left|z\left(t\right)\right|\le \alpha ,$$

where $\alpha =2a_2-2a_1+1+b_1(2a_5-2a_3+3)+4b_3{a}_5$ and $a_1,a_2,\cdots$ are positive constants.

Proof. 

Assuming that $m_1=m+\frac{2}{r}[r+a_2{\mu }_1^{\left(2r\right)}+b_1(a_5{\mu }_3^{\left(2r\right)}+{\mu }_2^{\left(2r\right)})+2b_3{a}_5{\mu }_3^{\left(2r\right)}]E{\parallel \zeta \parallel }_r^2$ and $m_2=2[2a_2-2a_1+1+b_1(2a_5-2a_3+3)+4b_3{a}_5]$ then from the inequality (21), we can conclude that

$$\widehat{\mathbb{E}}\left[\underset{0\le s\le t}{sup}{\left|z\left(s\right)\right|}^2\right]\le m_1{e}^{m_{2}t},$$

(22)

where $m=2\left[{\left|\kappa \left(0\right)\right|}^2+b_1{\left(\right|\eta \left(0\right)|}^2+{2\left|\gamma \left(0\right)\right|}^2)+4b_3{\left|\gamma \left(0\right)\right|}^2\right]T$. For each $q=1,2,3,\cdots ,$ from (22) it follows that

$$\widehat{\mathbb{E}}\left[\underset{q-1\le t\le q}{sup}{\left|z\left(t\right)\right|}^2\right]\le m_1{e}^{m_{2}q}.$$

By utilizing Lemma 5 for every given $\delta >0$, we obtain

$$\begin{array}{cc}\hfill \nu \left\{w:\underset{q-1\le t\le q}{sup}{\left|z\left(t\right)\right|}^2>e^{(m_2+\delta )q}\right\}& \le \frac{\widehat{\mathbb{E}}\left[{sup}_{q-1\le t\le q}{\left|z\left(t\right)\right|}^2\right]}{e^{(m_2+\delta )q}}\hfill \\ & \le \frac{m_1{e}^{m_{2}q}}{e^{(m_2+\delta )q}}\hfill \\ & =m_1{e}^{-\delta q}.\hfill \end{array}$$

But the Borel–Cantelli lemma gives that for almost every $w\in \mathsf{\Omega }$ there is a random number $q_0=q_0\left(w\right)\in Z$ in a manner that when $q\ge q_0,$ then

$$\underset{q-1\le t\le q}{sup}{\left|z\left(t\right)\right|}^2\le e^{(m_2+\delta )q},$$

which implies

$$\begin{array}{cc}\hfill \underset{t\to \infty }{lim}sup\frac{1}{t}log\left|z\left(t\right)\right|& \le \frac{m_2+\delta }{2}\hfill \\ & =2a_2-2a_1+1+b_1(2a_5-2a_3+3)+4b_3{a}_5+\frac{\delta }{2},\hfill \end{array}$$

as $\delta$ is arbitrary and letting $\alpha =2a_2-2a_1+1+b_1(2a_5-2a_3+3)+4b_3{a}_5$, we can conclude that

$$\alpha \ge \underset{t\to \infty }{lim}sup\frac{1}{t}log\left|z\left(t\right)\right|.$$

The proof stands completed. □

The lemma above expresses that the second moment of the Lyapunov exponent, as defined in [27] as ${lim}_{t\to \infty }sup\frac{1}{t}log\left|z\left(t\right)\right|$, is bounded above by $\alpha$.

6. Conclusions

Several stochastic functional differential equations (SFDEs) in financial mathematics do not hold the standard Lipschitz assumption such as the Cox–Ingersoll–Ross, Heston and Ait–Sahalia models. In this article, some useful monotone-type conditions have been introduced. We have proved that any two solutions of SFDEs in the G-framework under distinct initial conditions are convergent. The solutions are mean square bounded. The ${\mathcal{L}}_G^2$ and exponential estimates have been calculated. We anticipate that the findings presented in this article will offer valuable insights into the analysis of equations, even when not under the constraints of standard assumptions. This contribution is poised to have a substantial positive impact on the examination of various unresolved inquiries, including the investigation into the existence, uniqueness, convergence and stability of solutions for backward and forward stochastic dynamic systems driven by G-Brownian motion with conditions of a monotone nature.