1 Model formulation

Gopalsamy proposed a general nonautonomous Lotka–Volterra model with infinite delay for n-interacting species in [1], which was described by an n-dimensional ordinary differential equation

$$ \frac{\mathrm{d}x_{i}(t)}{\mathrm{d}t} = x_{i}(t) \Biggl(b_{i}+ \sum_{j=1}^{n}a_{ij}x_{j}(t) +\sum_{j=1}^{n}b_{ij}x_{j}(t- \tau_{ij}) + \sum_{j=1}^{n}c_{ij} \int_{-\infty}^{t}k_{ij}(t-s)x_{j}(s) \,\mathrm{d}s \Biggr), $$
(1)

where \(i=1,2,\ldots,n\), and the sufficient conditions for the global attractivity of a positive solution of model (1) were obtained therewith. Motivated by model (1), the researchers investigated the modified models [2–26] and obtained some good results regarding n-interacting species in the recent literatures. We have known that the intrinsic growth rates and the carrying capacity of the species often vary when one of the below factors changes, for instance here, the situation of nutrition supply, adequacy of food resources and changes of climate as well. Therefore, the ecosystems governed by the deterministic models would inevitably be affected by the surrounding environmental noises. It is very natural to consider the stochastic ecosystems with the continuous-time Markovian chains \(r(t)\) (\(t\geq 0\)), which take values in a finite state space \(\mathbb{S}=\{1, 2, \ldots, m\}\). Let the Markovian chain \(r(t)\) be generated by the generator \(\Gamma=(q_{ij})_{m\times m}\)

$$ \mathrm{P}\bigl\{ r(t+\Delta t)=j|r(t)=i\bigr\} = \textstyle\begin{cases} q_{ij}\Delta t+\mathrm{o}(\Delta t), & \mbox{if }i\neq j, \\ 1+q_{ij}\Delta t+\mathrm{o}(\Delta t), & \mbox{if }i=j, \end{cases} $$
(2)

where \(\Delta t>0\), and \(q_{ij}\geq0\) is the transition rate from state i to state j if \(i\neq j\); while \(q_{ii}=-\sum_{j\neq i}q_{ij}\).

In this paper, we modify model (1) further and propose a more general nonautonomous Lotka–Volterra model with Markovian switching

$$\begin{aligned} \frac{\mathrm{d}x_{i}(t)}{\mathrm{d}t} =&x_{i}(t) \Biggl(b_{i}\bigl(r(t)\bigr)+ \sum_{j=1}^{n}a_{ij}\bigl(r(t) \bigr)x_{j}(t) +\sum_{j=1}^{n}b_{ij} \bigl(r(t)\bigr)x_{j}(t-\tau_{ij}) \\ &{}+ \sum_{j=1}^{n}c_{ij} \bigl(r(t)\bigr) \int_{-\infty }^{t}k_{ij}(t-s)x_{j}(s) \,\mathrm{d}s \Biggr),\quad t\geq 0, \end{aligned}$$
(3)

where \(i=1,2,\ldots,n\) and \(r(t)\in\mathbb{S}\). Assuming that initially, \(r(t)=k\in\mathbb{S}\), then (3) obeys the following differential equation:

$$\begin{aligned} \frac{\mathrm{d}x_{i}(t)}{\mathrm{d}t} =& x_{i}(t) \Biggl(b_{i}(k)+ \sum _{j=1}^{n}a_{ij}(k)x_{j}(t) +\sum_{j=1}^{n}b_{ij}(k)x_{j}(t- \tau_{ij}) \\ &{} + \sum_{j=1}^{n}c_{ij}(k) \int_{-\infty}^{t}k_{ij}(t-s)x_{j}(s) \,\mathrm{d}s \Biggr), \end{aligned}$$
(4)

until \(r(t)\) jumps to another state, say \(l\in\mathbb{S}\), thus (3) follows the differential equation with state l before \(r(t)\) jumps to a new state, that is

$$\begin{aligned} \frac{\mathrm{d}x_{i}(t)}{\mathrm{d}t} =& x_{i}(t) \Biggl(b_{i}(l)+ \sum _{j=1}^{n}a_{ij}(l)x_{j}(t) +\sum_{j=1}^{n}b_{ij}(l)x_{j}(t- \tau_{ij}) \\ &{} + \sum_{j=1}^{n}c_{ij}(l) \int_{-\infty}^{t}k_{ij}(t-s)x_{j}(s) \,\mathrm{d}s \Biggr). \end{aligned}$$
(5)

Now, we consider the perturbations of intrinsic growth rates and interacting rates between species, that is to say, \(b_{i}(\varsigma)\) and \(a_{ij}(\varsigma)\), \(b_{ij}(\varsigma)\), \(c_{ij}(\varsigma)\) will be disturbed by the white noises in the form of

$$\begin{aligned}& b_{i}(\varsigma)\rightarrow b_{i}(\varsigma)+ \sigma_{i}(\varsigma)\dot{B}_{i}(t),\qquad a_{ij}( \varsigma )\rightarrow a_{ij}(\varsigma)+\alpha_{ij}( \varsigma)\dot{B}_{i}(t), \end{aligned}$$
(6)
$$\begin{aligned}& b_{ij}(\varsigma)\rightarrow b_{ij}(\varsigma)+ \beta_{ij}(\varsigma)\dot{B}_{i}(t),\qquad c_{ij}( \varsigma)\rightarrow c_{ij}(\varsigma)+\gamma_{ij}(\varsigma) \dot{B}_{i}(t), \end{aligned}$$
(7)

where \(\dot{B}_{i}(t)\) are the white noises and \(\sigma_{i}(\varsigma)\) represent the intensities of the white noises in regime \(\varsigma\in\mathbb{S}\) for \(i=1, 2, \ldots, n\). Next, we are about to investigate the dynamical properties of the following stochastic Lotka–Volterra model under Markovian switching:

$$\begin{aligned} \begin{aligned}[b] \mathrm{d}x_{i}(t)&= x_{i}(t) \Biggl(b_{i} \bigl(r(t)\bigr)+\sum_{j=1}^{n}a_{ij} \bigl(r(t)\bigr)x_{j}(t) +\sum_{j=1}^{n}b_{ij} \bigl(r(t)\bigr)x_{j}(t-\tau_{ij}) \\ &\quad {}+ \sum_{j=1}^{n}c_{ij} \bigl(r(t)\bigr) \int_{-\infty }^{t}k_{ij}(t-s)x_{j}(s) \,\mathrm{d}s \Biggr)\,\mathrm{d}t \\ &\quad {}+x_{i}(t) \Biggl(\sigma_{i}\bigl(r(t)\bigr) + \sum _{j=1}^{n}\alpha_{ij}\bigl(r(t) \bigr)x_{j}(t)+\sum_{j=1}^{n}\beta _{ij}\bigl(r(t)\bigr)x_{j}(t-\tau_{ij}) \\ &\quad {}+\sum_{j=1}^{n}\gamma_{ij} \bigl(r(t)\bigr) \int_{-\infty }^{t}k_{ij}(t-s)x_{j}(s) \,\mathrm{d}s \Biggr)\,\mathrm{d}B_{i}(t),\quad i=1,2, \ldots,n. \end{aligned} \end{aligned}$$
(8)

The initial conditions of model (8) are supposed to be given as follows:

$$\begin{aligned} (\mathrm{H}1)&\quad x_{i}(\theta)=\varphi_{i}(\theta)>0, \qquad \sup_{-\infty< \theta\leq 0} \bigl\vert \varphi_{i}(\theta) \bigr\vert < \infty, \end{aligned}$$
(9)
$$\begin{aligned} (\mathrm{H}2)&\quad \lambda>0,\quad \int_{0}^{\infty} k_{ij}(s)e^{\lambda s}\, \mathrm{d}s=\bar{k}_{ij}< \infty, \end{aligned}$$
(10)

where \(\varphi_{i}\) (\(i=1, 2, \ldots, n\)) are the continuous functions which are defined on the interval \((-\infty, 0]\) with its Euclidian norm, and \(k_{ij}(s)\) denotes the weight function with the property \(\int_{0}^{\infty} k_{ij}(s)\,\mathrm{d}s\leq1\).

If \(\sigma_{i}(r)=\beta_{ij}(r)=\gamma_{ij}(r)=0\), then model (8) turns to be a stochastic Lotka–Volterra model with infinite delay, where the stochastically ultimate boundedness and pth (\(0< p\leq2\)) moment in time average of the solution were considered by Wan and Zhou [20]. If \(b_{ij}(r)=c_{ij}(r)=0\), \(\sigma_{i}(r)=\beta_{ij}(r)=\gamma_{ij}(r)=0\), model (8) becomes a stochastic Lotka–Volterra model with Markovian switching, where the sufficient conditions of the stochastic permanence and extinction were presented by Liu et al. [16]. If \(b_{ij}(r)=c_{ij}(r)=0\), \(\alpha_{i}(r)=\beta_{ij}(r)=\gamma_{ij}(r)=0\), Wu et al. [13] discussed the ergodic property of a positive recurrence. Besides, we prefer to mention the related work herewith. For example, Hu and Wang [17] investigated the asymptotic stability in distribution and stochastic boundedness of the solution. Liu and Shen [18] showed the persistence in the mean, extinction, partial permanence, and partial extinction. In addition, Zhu and Yin [19] derived the stochastic boundedness and positive recurrence of the solution.

The framework of this paper will go as follows. We will show that the existence and uniqueness of the positive global solution always holds with probability one for any positive initial value in the next section. Later, the stochastically ultimate boundedness will be derived when constructing a proper function therewith. Consequently, the moment estimation of the solution will be investigated for the stochastic Lotka–Volterra model (8) with Markovian chains in this paper.

2 Existence and uniqueness of global solution

To proceed with our main results in this section, we denote \(\mathbb{R}_{+}:=[0, \infty)\), and we will show that model (8) admits a unique and global solution and the solution will remain in \(\mathbb{R}_{+}^{n}\) almost surely.

Theorem 2.1

For any \(b_{i}(r(t)), a_{ij}(r(t)), b_{ij}(r(t)), c_{ij}(r(t))\in\mathbb{R}_{+}\) (\(i,j=1,2,\ldots, n\)), model (8) admits a unique solution \(x(t)\), and the solution remains in \(\mathbb{R}_{+}^{n}\) with probability one.

Proof

According to Theorem 3.1 in [27], the coefficients of model (8) clearly satisfy the local Lipschitz condition, but do not satisfy the linear growth condition. To show that the solution of model (8) is a global solution, we only need to prove that the explosion time \(\tau_{e}=\infty\) holds almost surely. Let \(m_{0}>1\) be sufficiently large such that

$$ \frac{1}{m_{0}}\leq\min_{t\geq0} \bigl\vert x(t) \bigr\vert \leq\max_{t\geq0} \bigl\vert x(t) \bigr\vert \leq m_{0}, $$
(11)

where \(x(t)=(x_{1}(t), x_{2}(t), \ldots, x_{n}(t))^{\mathrm{T}}\). We define the stopping time

$$ \tau_{m}=\inf \biggl\{ t\in[0,\tau_{e}): x_{i}(t)\notin \biggl(\frac{1}{m}, m \biggr), \mbox{for some } i=1,2,\ldots,n \biggr\} ,\quad m\geq m_{0}. $$
(12)

As usual, ∅ denotes the empty set, we set \(\inf\emptyset=\infty\). Clearly, \(\tau_{m}\) increases when m tends to infinity. We set

$$ \tau_{\infty}=\lim_{m\rightarrow\infty}\tau_{m}, $$
(13)

thus \(\tau_{\infty}\leq\tau_{e}\) by the definition of stopping time. Now, we define two \(C^{2}\)-functions in order to check that \(\tau_{\infty}=\infty\) almost surely:

$$\begin{aligned}& V_{1}\bigl(x(t)\bigr)= \sum_{i=1}^{n} \bigl(x_{i}^{0.5}(t)-1-0.5\ln x_{i}(t)\bigr), \end{aligned}$$
(14)
$$\begin{aligned}& \begin{aligned}[b] V_{2}\bigl(x(t)\bigr) &= \sum _{i=1}^{n}\sum_{j=1}^{n} \biggl( \frac{1}{2n} \int_{t-\tau_{ij}}^{t}x_{j}^{2}(s)\, \mathrm{d}s +\frac{1}{2n} \int_{0}^{\infty}k_{ij}(s) \int_{t-s}^{t}x_{j}^{2}(u)\, \mathrm{d}u\,\mathrm{d}s \\ &\quad {} +\frac{|\beta(r(\tau_{k}))|^{2}}{4(1-l_{1})(1-l_{2})l_{3}} \int_{t-\tau_{ij}}^{t}x_{j}^{2}(s)\, \mathrm{d}s \\ &\quad {} +\frac{|\gamma(r(\tau_{k}))|^{2}}{4(1-l_{1})(1-l_{2})(1-l_{3})} \int_{0}^{\infty} k_{ij}(s) \int_{t-s}^{t} x_{j}^{2}(u)\, \mathrm{d}u\,\mathrm{d}s \biggr), \end{aligned} \end{aligned}$$
(15)

where \(0< l_{i}<1\) (\(i=1, 2, 3\)), and \(|\sigma(r(\tau_{k}))|\) means the norm of vector \((\sigma_{1}(r(\tau_{k})), \ldots, \sigma_{n}(r(\tau_{k})))\), \(|\beta(r(\tau_{k}))|\) denotes the norm of the matrix \((b_{ij}(r(\tau_{k})))_{n\times n}\), so the same to norms \(|\alpha(r(\tau_{k}))|\) and \(|\gamma(r(\tau_{k}))|\). For the sake of simplicity, model (8) could be rewritten in the following form:

$$ \mathrm{d}x_{i}(t)=x_{i}(t)f_{i}(t)\, \mathrm{d}t+x_{i}(t)g_{i}(t)\,\mathrm{d}B_{i}(t), \quad i=1,2,\ldots,n. $$
(16)

We define the same differential operator \(\mathcal{L}\) associated with equation (8) by

$$ \mathcal{L}=\frac{\partial}{\partial t}+\sum_{i=1}^{n}x_{i}(t)f_{i}(t) \frac{\partial}{\partial x_{i}(t)}+\frac{1}{2}\sum_{i=1}^{n}x_{i}^{2}(t)g_{i}^{2}(t) \frac{\partial ^{2}}{\partial x_{i}^{2}(t)}, $$
(17)

then the differential operator \(\mathcal{L}\) acts on \(V_{1}(x(t))\), generalized Itô’s formula gives

$$\begin{aligned} \mathrm{d}V_{1}\bigl(x(t)\bigr) =& 0.5 \sum _{i=1}^{n}\bigl(x_{i}^{0.5}(t)-1 \bigr)f_{i}(t)\,\mathrm{d}t +0.25\sum_{i=1}^{n} \bigl(-0.5x_{i}^{0.5}(t)+1\bigr)g_{i}^{2}(t) \,\mathrm{d}t \\ &{}+ 0.5 \sum_{i=1}^{n} \bigl(x_{i}^{0.5}(t)-1\bigr)g_{i}(t)\, \mathrm{d}B_{i}(t). \end{aligned}$$
(18)

The elementary inequality \(ab\leq\frac{n}{4}a^{2}+\frac{1}{n}b^{2}\) (\(a>0\), \(b>0\)) gives that

$$\begin{aligned}& \sum_{i=1}^{n}\bigl(x_{i}^{0.5}(t)-1 \bigr)f_{i}(t) \\& \quad = \sum_{i=1}^{n}\bigl(x_{i}^{0.5}(t)-1 \bigr) \Biggl( b_{i}\bigl(r(\tau_{k})\bigr)+ \sum _{j=1}^{n} a_{ij}\bigl(r( \tau_{k})\bigr)x_{j}(t) \\& \qquad {} + \sum_{j=1}^{n} b_{ij} \bigl(r(\tau_{k})\bigr)x_{j}(t-\tau_{ij}) +\sum _{j=1}^{n}c_{ij}\bigl(r( \tau_{k})\bigr) \int_{0}^{\infty }k_{ij}(s)x_{j}(t-s) \,\mathrm{d}s \Biggr) \\& \quad \leq \sum_{i=1}^{n} \bigl(x_{i}^{0.5}(t)-1\bigr)b_{i}\bigl(r( \tau_{k})\bigr)+ \sum_{i=1}^{n} \sum_{j=1}^{n} \biggl[ 0.25n \bigl(x_{i}^{0.5}(t)-1\bigr)^{2} \bigl(a_{ij}^{2}\bigl(r(\tau_{k})\bigr) \\& \qquad {}+b_{ij}^{2}\bigl(r(\tau_{k}) \bigr)+c_{ij}^{2}\bigl(r(\tau_{k})\bigr) \bigr) + \frac{1}{n}x_{j}^{2}(t)+ \frac{1}{n}x_{j}^{2}(t- \tau_{ij}) \\& \qquad {}+ \frac{1}{n} \biggl( \int_{0}^{\infty}k_{ij}(s)x_{j}(t-s) \,\mathrm{d}s \biggr)^{2} \biggr]. \end{aligned}$$
(19)

The inequalities \((u+v)^{2}\leq \frac{u^{2}}{l_{i}}+\frac{v^{2}}{1-l_{i}}\) for \(0< l_{i}<1\) and \((\sum_{i=1}^{n}a_{i}b_{i})^{2}\leq\sum_{i=1}^{n} a_{i}^{2}\sum_{i=1}^{n} b_{i}^{2}\) yield that

$$\begin{aligned} g_{i}^{2}(t) \leq& \frac{1}{l_{1}} \sigma_{i}^{2}\bigl(r(\tau_{k})\bigr) + \frac{1}{l_{2}(1-l_{1})}\sum_{j=1}^{n} \alpha_{ij}^{2}\bigl(r(\tau _{k})\bigr)\sum _{j=1}^{n} x_{j}^{2}(t) \\ &{}+ \frac{1}{l_{3}(1-l_{1})(1-l_{2})}\sum_{j=1}^{n} \beta_{ij}^{2}\bigl(r(\tau _{k})\bigr)\sum _{j=1}^{n} x_{j}^{2}(t- \tau_{ij}) \\ &{}+ \frac{1}{(1-l_{1})(1-l_{2})(1-l_{3})}\sum_{j=1}^{n} \gamma_{ij}^{2}\bigl(r(\tau_{k})\bigr)\sum _{j=1}^{n} \biggl( \int_{0}^{\infty }k_{ij}(s)x_{j}(t-s) \,\mathrm{d}s \biggr)^{2}, \end{aligned}$$
(20)

which implies that

$$\begin{aligned}& \sum_{i=1}^{n}\bigl(-0.5x_{i}^{0.5}(t)+1 \bigr)g_{i}^{2}(t) \\& \quad < \sum_{i=1}^{n}g_{i}^{2}(t) \\& \quad \leq \sum_{i=1}^{n}\frac{\sigma_{i}^{2}(r(\tau_{k}))}{l_{1}} +\sum_{i=1}^{n} \Biggl(\sum _{j=1}^{n} \frac{\alpha_{ij}^{2}(r(\tau_{k}))}{l_{2}(1-l_{1})}\sum _{j=1}^{n}x_{j}^{2}(t) \Biggr) \\& \qquad {}+ \sum_{i=1}^{n} \Biggl(\sum _{j=1}^{n}\frac{\beta_{ij}^{2}(r(\tau _{k}))}{(1-l_{1})(1-l_{2})l_{3}} \sum _{j=1}^{n}x_{j}^{2}(t- \tau_{ij}) \Biggr) \\& \qquad {}+ \sum_{i=1}^{n} \Biggl[\sum _{j=1}^{n}\frac{\gamma_{ij}^{2}(r(\tau _{k}))}{(1-l_{1})(1-l_{2})(1-l_{3})} \sum _{j=1}^{n} \biggl( \int_{0}^{\infty}k_{ij}(s)x_{j}(t-s) \,\mathrm{d}s \biggr)^{2} \Biggr] \\& \quad \leq \frac{|\sigma(r(\tau_{k}))|^{2}}{l_{1}} + \frac{|\alpha(r(\tau_{k}))|^{2}}{(1-l_{1})l_{2}} \bigl\vert x(t) \bigr\vert ^{2} + \frac{|\beta(r(\tau_{k}))|^{2}}{(1-l_{1})(1-l_{2})l_{3}} \sum_{i=1}^{n} \sum_{j=1}^{n}x_{j}^{2}(t- \tau_{ij}) \\& \qquad {}+ \frac{|\gamma(r(\tau_{k}))|^{2}}{(1-l_{1})(1-l_{2})(1-l_{3})} \sum_{i=1}^{n} \sum_{j=1}^{n} \biggl( \int_{0}^{\infty} k_{ij}(s)x_{j}(t-s) \,\mathrm{d}s \biggr)^{2}. \end{aligned}$$
(21)

By a similar argument, we derive that

$$\begin{aligned} \mathrm{d}V_{2}\bigl(x(t)\bigr) =& \Biggl[-\frac{1}{2n}\sum _{i=1}^{n}\sum_{j=1}^{n}x_{j}^{2}(t- \tau_{ij}) +\frac{1}{2n}\sum_{i=1}^{n} \sum_{j=1}^{n} \int_{0}^{\infty}k_{ij}(s) \bigl(x_{j}^{2}(t)-x_{j}^{2}(t-s) \bigr)\,\mathrm{d}s \\ &{}+0.5 \bigl\vert x(t) \bigr\vert ^{2} +\frac{|\beta(r(\tau_{k}))|^{2}}{4(1-l_{1})(1-l_{2})l_{3}} \sum _{i=1}^{n}\sum_{j=1}^{n} \bigl(x_{j}^{2}(t)-x^{2}_{j}(t- \tau_{ij})\bigr) \\ &{}+\frac{|\gamma(r(\tau_{k}))|^{2}}{4(1-l_{1})(1-l_{2})l_{3}}\sum_{i=1}^{n}\sum _{j=1}^{n} \int_{0}^{\infty}k_{ij}(s) \bigl(x_{j}^{2}(t)-x^{2}_{j}(t- \tau_{ij})\bigr)\,\mathrm{d}s \Biggr]\,\mathrm{d}t, \end{aligned}$$
(22)

where

$$\begin{aligned} \biggl( \int_{0}^{\infty}k_{ij}(s)x_{j}(t-s) \,\mathrm{d}s \biggr)^{2} \leq& \int_{0}^{\infty} \bigl(\sqrt{k_{ij}(s)} \bigr)^{2}\,\mathrm{d}s\cdot \int_{0}^{\infty} \bigl(\sqrt{k_{ij}(s)}x_{j}(t-s) \bigr)^{2}\,\mathrm{d}s \\ \leq& \int_{0}^{\infty}k_{ij}(s)x_{j}^{2}(t-s) \,\mathrm{d}s, \end{aligned}$$
(23)

together with expressions (18), (22), and (23), which gives that

$$\begin{aligned}& \mathrm{d}\bigl(V_{1}\bigl(x(t)\bigr)+V_{2}\bigl(x(t) \bigr)\bigr) \\& \quad \leq \Biggl(0.5 \sum_{i=1}^{n} \bigl(x_{i}^{0.5}(t)-1\bigr)b_{i}\bigl(r( \tau_{k})\bigr)+ 0.125\sum_{i=1}^{n} \sum_{j=1}^{n}n\bigl(x_{i}^{0.5}(t)-1 \bigr)^{2} \bigl[a_{ij}^{2}\bigl(r( \tau_{k})\bigr) \\& \qquad {}+ b_{ij}^{2}\bigl(r(\tau_{k}) \bigr)+c_{ij}^{2}\bigl(r(\tau_{k})\bigr) \bigr]+1.5 \bigl\vert x(t) \bigr\vert ^{2} +\frac{1}{4l_{1}} \bigl\vert \sigma\bigl(r(\tau_{k})\bigr) \bigr\vert ^{2} \\& \qquad {}+ \frac{1}{4l_{2}(1-l_{1})} \bigl\vert \alpha\bigl(r(\tau_{k}) \bigr) \bigr\vert ^{2} \bigl\vert x(t) \bigr\vert ^{2} +\frac{n}{4l_{3}(1-l_{1})(1-l_{2})} \bigl\vert \beta\bigl(r(\tau_{k})\bigr) \bigr\vert ^{2} \bigl\vert x(t) \bigr\vert ^{2} \\& \qquad {} +\frac{n}{4(1-l_{1})(1-l_{2})(1-l_{3})} \bigl\vert \gamma\bigl(r(\tau _{k}) \bigr) \bigr\vert ^{2} \bigl\vert x(t) \bigr\vert ^{2} \Biggr)\,\mathrm{d}t \\& \qquad {}+ 0.5 \sum_{i=1}^{n} \bigl(x_{i}^{0.5}(t)-1\bigr)g_{i}(t)\, \mathrm{d}B_{i}(t) \\& \quad :=F\bigl(x(t)\bigr)\,\mathrm{d}t+0.5 \sum_{i=1}^{n} \bigl(x_{i}^{0.5}(t)-1\bigr)g_{i}(t)\, \mathrm{d}B_{i}(t), \end{aligned}$$
(24)

where \(F(x(t))\) is a bounded constant due to the boundedness of all \(x(t)\) lying within the interval \((\frac{1}{m}, m)\), that is,

$$ \sup_{\frac{1}{m}< x(t)< m} F\bigl(x(t)\bigr)\leq\hat{F}< \infty. $$
(25)

Inequality (24) thus becomes

$$ \mathrm{d}\bigl(V_{1}\bigl(x(t)\bigr)+V_{2} \bigl(x(t)\bigr)\bigr)\leq \hat{F}\,\mathrm{d}t+0.5 \sum _{i=1}^{n}\bigl(x_{i}^{0.5}(t)-1 \bigr)g_{i}(t)\,\mathrm{d}B_{i}(t). $$
(26)

Integrating both sides of (26) from 0 to \(\tau_{m}\wedge T\) and taking expectations imply that

$$\begin{aligned} \mathbb{E}V_{1}\bigl(x(\tau_{m}\wedge T)\bigr) < & \mathbb{E}\bigl[V_{1}\bigl(x(\tau_{m}\wedge T) \bigr)+V_{2}\bigl(x(\tau_{m}\wedge T)\bigr)\bigr] \\ \leq& V_{1}\bigl(x(0)\bigr)+V_{2}\bigl(x(0)\bigr)+ \hat{F}T, \end{aligned}$$
(27)

where T is an arbitrary positive constant. For every \(w\in\Omega_{m}=\{\tau_{m}\leq T\}\), there exists some i such that \(x_{i}(\tau_{m}, w)\) equals either m or \(\frac{1}{m}\), then

$$ V_{1}\bigl(x(\tau_{m}\wedge T)\bigr)\geq\min \biggl\{ \sqrt{m}-1-0.5\ln m, \sqrt{\frac{1}{m}}-1+0.5\ln m \biggr\} , $$
(28)

which leads to

$$\begin{aligned}& \mathbb{P}\{\tau_{m}\leq T\}\min \biggl\{ \sqrt{m}-1-0.5\ln{m}, \sqrt{ \frac{1}{m}}-1+0.5\ln{m} \biggr\} \\& \quad \leq\mathbb{E}\bigl[\textrm{1}_{\Omega_{m}}(\omega)V_{1}\bigl(x( \tau_{m}\wedge T)\bigr)\bigr]\leq V_{1}\bigl(x(0) \bigr)+V_{2}\bigl(x(0)\bigr)+\hat{F}T, \end{aligned}$$
(29)

where \(\textrm{1}_{\Omega_{m}}(\omega)\) is the indicator function of \(\Omega_{m}\). We therefore get that

$$ \lim_{m\rightarrow\infty}\mathbb{P}\{\tau_{m}\leq T\}=0, $$
(30)

when m tends to infinity. From the arbitrariness of a positive number T, the following assertion holds:

$$ \mathbb{P} \{\tau_{\infty}=\infty\}=1. $$
(31)

The proof is complete. □

3 Stochastically ultimate boundedness

The moment estimation of the solution to model (8) will be investigated as the first part of this section. Then the stochastically ultimate boundedness of the solution follows later in Theorem 3.2.

Lemma 3.1

If condition (H2) holds, then there exists positive constants p, λ such that

$$ \limsup_{t\rightarrow\infty}\mathbb{E} \bigl\vert x(t) \bigr\vert ^{p}\leq \frac{\bar{Q}}{\lambda}n^{\frac{p}{2}}, $$
(32)

where Q̄ is a positive constant and \(0< p<1\).

Proof

Let us define a \(C^{2}\)-function

$$ V_{3}\bigl(x(t)\bigr)=\sum_{i=1}^{n}x_{i}^{p}(t). $$
(33)

For any given \(\lambda>0\), applying Itô’s formula to \(e^{\lambda t}V_{3}(x(t))\) and taking expectation, we derive that

$$ e^{\lambda t}\mathbb{E}\bigl(V_{3}\bigl(x(t)\bigr)\bigr)= \mathbb{E}V_{3}\bigl(x(0)\bigr)+\mathbb{E} \int_{0}^{t}e^{\lambda s}\bigl[ \mathcal{L}V_{3}\bigl(x(s)\bigr)+\lambda V_{3}\bigl(x(s) \bigr)\bigr]\,\mathrm{d}s, $$
(34)

where

$$\begin{aligned} \mathcal{L}V_{3}\bigl(x(t)\bigr) =& p\sum _{i=1}^{n}x_{i}^{p}(t)f_{i}(t) +\frac{p(p-1)}{2}\sum_{i=1}^{n}x_{i}^{p}(t)g_{i}^{2}(t) \\ \leq& \sum_{i=1}^{n}px_{i}^{p}(t)b_{i} \bigl(r(\tau_{k})\bigr) \\ &{}+ \frac{np^{2}}{4}\sum _{i=1}^{n}\sum_{j=1}^{n}x_{i}^{2p}(t) \bigl[a_{ij}^{2}\bigl(r(\tau_{k}) \bigr)+b_{ij}^{2}\bigl(r(\tau_{k})\bigr) +c_{ij}^{2}\bigl(r(\tau_{k})\bigr) \bigr] \\ &{}+ \frac{1}{n}\sum_{i=1}^{n}\sum _{j=1}^{n} \biggl( \int_{0}^{\infty}k_{ij}(s)x_{j}(t-s) \,\mathrm{d}s \biggr)^{2} + \bigl\vert x(t) \bigr\vert ^{2}+ \frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^{n}x_{j}^{2}(t- \tau _{ij}) \\ &{}- \frac{p(1-p)}{2}\sum_{i=1}^{n}x_{i}^{p}(t) \biggl[\sigma_{i}^{2}\bigl(r(\tau _{k})\bigr)+ \alpha_{ii}^{2}\bigl(r(\tau_{k}) \bigr)x_{i}^{2} +\beta_{ii}^{2}\bigl(r( \tau_{k})\bigr)x_{i}^{2}(t-\tau_{ii}) \\ &{}+\gamma_{ii}^{2}\bigl(r(\tau_{k})\bigr) \biggl( \int_{0}^{\infty}k_{ii}(s)x_{i}(t-s) \,\mathrm{d}s \biggr)^{2} \biggr]. \end{aligned}$$
(35)

Again we define

$$\begin{aligned} V_{4}\bigl(x(t)\bigr) =& \frac{1}{n}\sum _{i=1}^{n}\sum_{j=1}^{n} \int_{t-\tau _{ij}}^{t}e^{\lambda(s+\tau_{ij})}x_{j}^{2}(s) \,\mathrm{d}s \\ &{}+ \frac{1}{n}\sum_{i=1}^{n}\sum _{j=1}^{n} \int_{0}^{\infty }k_{ij}(s) \int_{t-s}^{t}e^{\lambda(u+s)}x_{j}^{2}(u) \,\mathrm{d}u\,\mathrm{d}s, \end{aligned}$$
(36)

according to the same argument, we then derive that

$$\begin{aligned} \mathrm{d}V_{4}\bigl(x(t)\bigr) \leq& \Biggl(e^{\lambda(t+\tau)} \bigl\vert x(t) \bigr\vert ^{2}- \frac{1}{n}\sum _{i=1}^{n}\sum_{j=1}^{n}e^{\lambda t}x_{j}^{2}(t- \tau_{ij}) \\ &{}+e^{\lambda{t}}\bar{k} \bigl\vert x(t) \bigr\vert ^{2}- \frac{1}{n}\sum_{i=1}^{n}\sum _{j=1}^{n}e^{\lambda t} \int_{0}^{\infty}k_{ij}(s)x_{j}^{2}(t-s) \,\mathrm{d}s \Biggr)\,\mathrm{d}t, \end{aligned}$$
(37)

where

$$ \tau=\max_{1\leq i,j\leq n}\tau_{ij},\qquad \bar{k}=\max _{1\leq i,j\leq n}\bar{k}_{ij}. $$
(38)

Further, we combine (23), (35), and (37) to get that

$$\begin{aligned}& \mathrm{d}\bigl(e^{\lambda t}V_{3}\bigl(x(t)\bigr)+V_{4} \bigl(x(t)\bigr)\bigr) \\& \quad =\mathrm{d}V_{4}\bigl(x(t)\bigr)+\lambda e^{\lambda t}V_{3} \bigl(x(t)\bigr)\,\mathrm{d}t+e^{\lambda t}\mathcal{L}V_{3} \bigl(x(t)\bigr)\,\mathrm{d}t+e^{\lambda t} \sum_{i=1}^{n}px_{i}^{p}(t)g_{i}(t) \,\mathrm{d}B_{i}(t) \\& \quad \leq e^{\lambda t} \Biggl\{ \lambda V_{3}\bigl(x(t)\bigr)+ \bigl(e^{\lambda\tau}+\bar{k}+1\bigr) \bigl\vert x(t) \bigr\vert ^{2} + \sum_{i=1}^{n}px_{i}^{p}(t)b_{i} \bigl(r(\tau_{k})\bigr) \\& \qquad {}+ \frac{np^{2}}{4}\sum_{i=1}^{n} \sum_{j=1}^{n}x_{i}^{2p}(t) \bigl[a_{ij}^{2}\bigl(r(\tau_{k})\bigr) +b_{ij}^{2}\bigl(r(\tau_{k})\bigr)+c_{ij}^{2} \bigl(r(\tau_{k})\bigr) \bigr] \\& \qquad {}- \frac{p(1-p)}{2}\sum_{i=1}^{n}x_{i}^{p}(t) \biggl[\sigma_{i}^{2}\bigl(r(\tau_{k})\bigr)+ \alpha_{ii}^{2}\bigl(r(\tau_{k}) \bigr)x_{i}^{2}(t) \\& \qquad {} +\beta_{ii}^{2}\bigl(r(\tau_{k}) \bigr)x_{i}^{2}(t-\tau_{ii})+\gamma _{ii}^{2}\bigl(r(\tau_{k})\bigr) \\& \qquad {}\times \biggl( \int_{0}^{\infty}k_{ij}(s)x_{i}(t-s) \,\mathrm{d}s \biggr)^{2} \biggr] \Biggr\} \,\mathrm{d}t+e^{\lambda t} \sum_{i=1}^{n}px_{i}^{p}(t)g_{i}(t) \,\mathrm{d}B_{i}(t) \\& \quad < e^{\lambda t}Q\bigl(x(t)\bigr)\,\mathrm{d}t+e^{\lambda t} \sum _{i=1}^{n}px_{i}^{p}(t)g_{i}(t) \,\mathrm{d}B_{i}(t), \end{aligned}$$
(39)

where the term with \(p(1-p)\) is positive, and the boundedness \(\bar{Q} =\sup_{x(t)\in\mathbb{R}_{+}^{n}}Q(x(t))< \infty\) is derived due to the boundedness of \(x(t)\in\mathbb{R}_{+}^{n}\). Integrating both sides of (39) and taking expectation from 0 to t therefore give

$$\begin{aligned} e^{\lambda t}\mathbb{E}V_{3}\bigl(x(t)\bigr) \leq& V_{3}\bigl(x(0)\bigr)+V_{4}\bigl(x(0)\bigr)+\mathbb{E} \int_{0}^{t}e^{\lambda s}Q\bigl(x(s)\bigr)\, \mathrm{d}s \\ \leq& V_{3}\bigl(x(0)\bigr)+V_{4}\bigl(x(0)\bigr)+ \frac{\bar{Q}e^{\lambda t}}{\lambda}, \end{aligned}$$
(40)

which implies that

$$ \limsup_{t\rightarrow\infty}\mathbb{E}V_{3}\bigl(x(t)\bigr)\leq \frac{\bar {Q}}{\lambda}. $$
(41)

The inequality \(|x(t)|^{2}\leq n\max_{i}x_{i}^{2}(t)\) admits the following inequality:

$$ \bigl\vert x(t) \bigr\vert ^{p}\leq n^{\frac{p}{2}}\max _{ i}x_{i}^{p}(t)\leq n^{\frac{p}{2}} \sum_{i=1}^{n}x_{i}^{p}(t)=n^{\frac{p}{2}}V_{3} \bigl(x(t)\bigr), $$
(42)

which yields that

$$ \limsup_{t\rightarrow\infty}\mathbb{E} \bigl\vert x(t) \bigr\vert ^{p}\leq \frac{\bar{Q}}{\lambda}n^{\frac{p}{2}}. $$
(43)

The proof is complete. □

Theorem 3.2

If condition (H2) holds, then the following property is valid for any arbitrary positive constant \(\varepsilon\in(0,1)\):

$$ \limsup_{t\rightarrow\infty}\mathbb{P}\bigl\{ \bigl\vert x(t) \bigr\vert \leq \zeta\bigr\} \geq1-\varepsilon, $$
(44)

that is, the solution \(x(t)\) of model (8) is stochastically ultimately bounded.

Proof

We take \(p=\frac{1}{2}\) in (43), then there exists a positive constant \(\frac{\bar{Q}}{\lambda}n^{\frac{1}{4}}\) such that

$$ \limsup_{t\rightarrow\infty}\mathbb{E} \bigl\vert x(t) \bigr\vert ^{\frac{1}{2}}\leq \frac{\bar{Q}}{\lambda}n^{\frac{1}{4}}. $$
(45)

For any positive constant ε, let

$$ \zeta=\frac{\bar{Q}^{2}}{\lambda^{2}\varepsilon^{2}}n^{\frac{1}{2}}, $$
(46)

the Chebyshev’s inequality gives

$$ \limsup_{t\rightarrow\infty}\mathbb{P}\bigl\{ \bigl\vert x(t) \bigr\vert \geq \zeta\bigr\} \leq \limsup_{t\rightarrow\infty}\frac{\mathbb{E}|x(t)|^{\frac{1}{2}}}{\zeta ^{\frac{1}{2}}}\leq \varepsilon, $$
(47)

which yields that

$$ \limsup_{t\rightarrow\infty}\mathbb{P}\bigl\{ \bigl\vert x(t) \bigr\vert < \zeta\bigr\} \geq1-\varepsilon. $$
(48)

The proof is complete. □

4 Moment estimation of the solution

We are about to discuss other properties of the solution to model (8) in this section, for instance, the absolute value of the solution will grow slower than time t when t goes to infinity. Moreover, the pth moment in time average of the solution to model (8) is discussed in Theorem 4.2.

Theorem 4.1

If condition (H2) holds, then the solution of model (8) has the property that

$$ \limsup_{t\rightarrow\infty}\frac{\ln|x(t)|}{\ln t}\leq1 \quad \textit{a.s.} $$
(49)

Proof

In order to prove this theorem, we need to define two \(C^{2}\)-functions:

$$\begin{aligned}& V_{5}\bigl(x(t)\bigr)=\sum_{i=1}^{n}x_{i}(t), \end{aligned}$$
(50)
$$\begin{aligned}& \begin{aligned}[b] V_{6}\bigl(x(t)\bigr) &= \sum _{i=1}^{n}\sum_{j=1}^{n} \bigl\vert b_{ij}\bigl(r(\tau_{k})\bigr) \bigr\vert \int_{t-\tau_{ij}}^{t}e^{\lambda(s+\tau _{ij})}x_{j}(s)\, \mathrm{d}s \\ &\quad {} +\sum_{i=1}^{n}\sum _{j=1}^{n} \bigl\vert c_{ij}\bigl(r( \tau_{k})\bigr) \bigr\vert \int_{0}^{\infty}k_{ij}(s) \int_{t-s}^{t}e^{\lambda(u+s)}x_{j}(u)\, \mathrm{d}u\,\mathrm{d}s. \end{aligned} \end{aligned}$$
(51)

Generalized Itô’s formula yields

$$\begin{aligned}& \mathrm{d}V_{5}\bigl(x(t)\bigr)= \sum_{i=1}^{n}x_{i}(t)f_{i}(t) \,\mathrm{d}t +\sum_{i=1}^{n}x_{i}(t)g_{i}(t) \,\mathrm{d}B_{i}(t), \end{aligned}$$
(52)
$$\begin{aligned}& \begin{aligned}[b] \mathrm{d}V_{6}\bigl(x(t)\bigr) &= \Biggl(\sum _{i=1}^{n}\sum_{j=1}^{n} \bigl\vert b_{ij}\bigl(r(\tau_{k})\bigr) \bigr\vert e^{\lambda (t+\tau_{ij})}x_{j}(t) \\ &\quad {}- \sum_{i=1}^{n}\sum _{j=1}^{n} \bigl\vert b_{ij}\bigl(r( \tau_{k})\bigr) \bigr\vert e^{\lambda t}x_{j}(t- \tau_{ij}) \\ &\quad {}+ \sum_{i=1}^{n}\sum _{j=1}^{n} \bigl\vert c_{ij}\bigl(r( \tau_{k})\bigr) \bigr\vert \int_{0}^{\infty }k_{ij}(s)e^{\lambda (t+s)}x_{j}(s) \,\mathrm{d}s \\ &\quad {}- \sum_{i=1}^{n}\sum _{j=1}^{n} \bigl\vert c_{ij}\bigl(r( \tau_{k})\bigr) \bigr\vert \int_{0}^{\infty }k_{ij}(s)e^{\lambda t}x_{j}(t-s) \,\mathrm{d}s \Biggr)\,\mathrm{d}t. \end{aligned} \end{aligned}$$
(53)

For any given \(\lambda>0\), expressions (52) and (53) imply that

$$\begin{aligned}& \mathrm{d}\bigl(e^{\lambda t}\ln V_{5}\bigl(x(t) \bigr)+V_{6}\bigl(x(t)\bigr)\bigr) \\& \quad =\mathrm{d}V_{6}\bigl(x(t)\bigr)+e^{\lambda t} \Biggl( \lambda\ln V_{5}\bigl(x(t)\bigr)+ \sum_{i=1}^{n}x_{i}(t)f_{i}(t)V_{5}^{-1} \bigl(x(t)\bigr) - \frac{|x^{\mathrm{T}}(t)g(t)|^{2}}{2V_{5}^{2}(x(t))} \Biggr)\,\mathrm{d}t \\& \qquad {} + \frac{e^{\lambda t}|x^{\mathrm{T}}(t)g(t)|}{V_{5}(x(t))}\,\mathrm{d}B(t). \end{aligned}$$
(54)

We denote

$$ M^{\ast}(t)= \int_{0}^{t}\frac{e^{\lambda s}|x^{\mathrm{T}}(t)g(t)|}{V_{5}(x(t))}\,\mathrm{d}B(s), $$
(55)

then the strong law of large numbers (see Theorem 3.4 in [27]) for a local martingale \(M^{\ast}(t)\) yields that

$$ \lim_{t\rightarrow\infty}\frac{M^{\ast}(t)}{t}=0 \quad \mbox{a.s.} $$
(56)

and

$$ \bigl\langle M^{\ast}(t),M^{\ast}(t)\bigr\rangle = \int_{0}^{t}\frac{e^{2\lambda s}|x^{\mathrm{T}}(t)g(t)|^{2}}{V_{5}^{2}(x(t))}\,\mathrm{d}s. $$
(57)

The exponential martingale inequality (see Theorem 7.4 in [27]) shows that when choosing

$$ T=v, \qquad \alpha=\frac{\eta}{e^{\lambda k}},\qquad \beta=\frac{ve^{\lambda k\ln k}}{\eta}, $$
(58)

and letting \(v>1\), \(0<\eta<1\), we get

$$ \mathbb{P} \biggl\{ \sup_{0\leq t\leq v} \biggl(M^{\ast}(t)- \frac{\eta}{2e^{\lambda k}}\bigl\langle M^{\ast}(t),M^{\ast}(t)\bigr\rangle \biggr)>\frac{ve^{\lambda k}}{\eta}\ln k \biggr\} \leq\frac{1}{k^{v}}. $$
(59)

By the Borel–Cantelli lemma, for almost all \(w\in\Omega\), there exists a random integer \(v_{0}=v_{0}(w)\) such that

$$ M^{\ast}(t)\leq\frac{\eta}{2e^{\lambda k}}\bigl\langle M^{\ast}(t), M^{\ast}(t)\bigr\rangle +\frac{ve^{\lambda k}}{\eta}\ln k,\quad v\geq v_{0}(\omega). $$
(60)

Therefore, we have

$$\begin{aligned}& \mathrm{d}\bigl(e^{\lambda t}\ln V_{5}\bigl(x(t) \bigr)+V_{6}\bigl(x(t)\bigr)\bigr) \\& \quad \leq e^{\lambda t} \Biggl[ \sum_{i=1}^{n} \sum_{j=1}^{n} \bigl\vert b_{ij} \bigl(r(\tau_{k})\bigr) \bigr\vert e^{\lambda \tau_{ij}}x_{j}(t)- \frac{ \vert x^{\mathrm{T}}(t)g(t) \vert ^{2}}{2V_{5}^{2}(x(t))} \\& \qquad {}+ \sum_{i=1}^{n}\sum _{j=1}^{n} \bigl\vert c_{ij}\bigl(r( \tau_{k})\bigr) \bigr\vert \int_{0}^{\infty }k_{ij}(s)e^{\lambda s}x_{j}(s) \,\mathrm{d}s \\& \qquad {}+ \lambda\ln V_{5}\bigl(x(t)\bigr)+\sum _{i=1}^{n} \bigl\vert b_{i}\bigl(r( \tau_{k})\bigr) \bigr\vert + \sum_{i=1}^{n} \sum_{j=1}^{n} \bigl\vert a_{ij} \bigl(r(\tau_{k})\bigr) \bigr\vert x_{j}(t) \Biggr]\, \mathrm{d}t \\& \qquad {}+ \frac{e^{\lambda t} \vert x^{\mathrm{T}}(t)g(t) \vert }{V_{5}(x(t))}\,\mathrm{d}B(t) \\& \quad \leq e^{\lambda t} \biggl[\sqrt{n} \bigl\vert x(t) \bigr\vert (a+b+c)+\lambda\ln V_{5}\bigl(x(t)\bigr)+\bar{b} - \frac{ \vert x^{\mathrm{T}}(t)g(t) \vert ^{2}}{2V_{5}^{2}(x(t))} \biggr]\,\mathrm{d}t \\& \qquad {}+ \frac{e^{\lambda t} \vert x^{\mathrm{T}}(t)g(t) \vert }{V_{5}(x(t))}\,\mathrm{d}B(t), \end{aligned}$$
(61)

where

$$\begin{aligned}& a=\max_{j}\sum_{i=1}^{n} \bigl\vert a_{ij}\bigl(r(\tau_{k})\bigr) \bigr\vert , \qquad b=\max_{j}\sum_{i=1}^{n} \bigl\vert b_{ij}\bigl(r(\tau_{k})\bigr) \bigr\vert e^{\lambda\tau_{ij}}, \end{aligned}$$
(62)
$$\begin{aligned}& c=\max_{j}\sum_{i=1}^{n} \bigl\vert c_{ij}\bigl(r(\tau_{k})\bigr) \bigr\vert \bar{k}_{ij},\qquad \bar{b}=\sum_{i=1}^{n} \bigl\vert b_{i}\bigl(r(\tau_{k})\bigr) \bigr\vert . \end{aligned}$$
(63)

For all \(0\leq t\leq\tau_{k}\leq k\) and \(k\geq k_{0}(w)\), it follows \(e^{t}\leq e^{k}\) and \(V_{5}^{2}(x(t))\leq n|x(t)|^{2}\), we then get

$$\begin{aligned} e^{\lambda t}\ln V_{5}\bigl(x(t)\bigr) \leq& e^{\lambda t}\ln V_{5}\bigl(x(t)\bigr)+V_{6}\bigl(x(t)\bigr) \\ =&\ln V_{5}\bigl(x(0)\bigr)+V_{6}\bigl(x(0)\bigr)+ \int_{0}^{t}\,\mathrm{d}\bigl(e^{\lambda s}\ln V_{5}\bigl(x(s)\bigr)+V_{6}\bigl(x(s)\bigr)\bigr) \\ \leq&C_{0}+ \int_{0}^{t}e^{\lambda s}\bigl[\bigl((a+b+c) \sqrt{n}+\lambda\ln\sqrt{n}\bigr) \bigl\vert x(s) \bigr\vert +\bar{b}\bigr]\, \mathrm{d}s \\ &{}- \int_{0}^{t}(1-\eta)\frac{e^{\lambda s} \vert x^{\mathrm{T}}(s)g(x(s)) \vert ^{2}}{2n \vert x(s) \vert ^{2}}\,\mathrm{d}s+ \frac{ve^{\lambda k}}{\eta}\ln k \\ \leq&C_{0}+ \int_{0}^{t}e^{\lambda s}\bigl[\bigl((a+b+c) \sqrt{n}+\lambda\ln\sqrt{n}\bigr) \bigl\vert x(s) \bigr\vert +\bar{b}\bigr]\, \mathrm{d}s +\frac{ve^{\lambda k}}{\eta}\ln k \\ :=&C_{0}+ \int_{0}^{t}e^{\lambda s}U\bigl(x(s)\bigr)\, \mathrm{d}s+\frac{ve^{\lambda k}}{\eta}\ln k \\ < & C_{0}+C_{1}\lambda^{-1} \bigl(e^{\lambda t}-1\bigr)+\frac{ve^{\lambda k}}{\eta}\ln k \end{aligned}$$
(64)

due to the boundedness of x in \(\mathbb{R}_{+}^{n}\). For \(0\leq k-1\leq t\leq k\) and \(k\geq k_{0}(w)\), expression (64) thus gives

$$ \frac{\ln V_{5}(x(t))}{\ln t}\leq\frac{C_{0}+\eta^{-1}ve^{\lambda k}\ln k+\lambda^{-1}C_{1}(e^{\lambda t}-1)}{e^{\lambda t}\ln(k-1)}, $$
(65)

taking superior limit on both sides of (64) as t tends to infinity, which then becomes

$$ \limsup_{t\rightarrow\infty}\frac{\ln V_{5}(x(t))}{\ln t}\leq\frac{ve^{\lambda}}{\eta} \quad \mbox{a.s.} $$
(66)

Letting \(v\rightarrow1\), \(\eta\rightarrow1\), and \(\lambda\rightarrow0\) yields

$$ \limsup_{t\rightarrow\infty}\frac{\ln V_{5}(x(t))}{\ln t}\leq1 \quad \mbox{a.s.} $$
(67)

Noting that \(|x(t)|^{2}\leq n\max_{i}x_{i}^{2}(t)\), which leads to

$$ \frac{V_{5}^{2}(x(t))}{n}\leq \bigl\vert x(t) \bigr\vert ^{2}\leq nV_{5}^{2}\bigl(x(t)\bigr), $$
(68)

then we have that

$$ \limsup_{t\rightarrow\infty}\frac{\ln|x(t)|}{\ln t}\leq1\quad \mbox{a.s.} $$
(69)

The proof is complete. □

Theorem 4.2

If condition (H1) is valid, for any positive constant p, then there exists a positive constant C such that the following property holds:

$$ \limsup_{T\rightarrow\infty}\frac{1}{T} \int_{0}^{T}\mathbb {E} \bigl\vert x(t) \bigr\vert ^{p}\,\mathrm{d}t\leq C. $$
(70)

Proof

Let

$$ H\bigl(x(t)\bigr)=F\bigl(x(t)\bigr)+ \bigl\vert x(t) \bigr\vert ^{p}, $$
(71)

where \(F(x(t))\) is the same as in (24). Then there exists a positive constant C such that

$$ \sup_{x(t)\in\mathbb{R}_{+}^{n}}H\bigl(x(t)\bigr)\leq C. $$
(72)

Integrating both sides of (18) from 0 to T and taking expectation give

$$ V_{1}\bigl(x(T)\bigr)+V_{2}\bigl(x(T)\bigr)-V_{1} \bigl(x(0)\bigr)-V_{2}\bigl(x(0)\bigr)\leq \int_{0}^{T}F\bigl(x(t)\bigr)\,\mathrm{d}t, $$
(73)

the relationship \(|x(t)|^{p}=H(x(t))-F(x(t))\) implies that

$$ \int_{0}^{T}\mathbb{E} \bigl\vert x(t) \bigr\vert ^{p}\,\mathrm{d}t \leq CT+ V_{1}\bigl(x(0) \bigr)+V_{2}\bigl(x(0)\bigr)-V_{1}\bigl(x(T) \bigr)-V_{2}\bigl(x(T)\bigr). $$
(74)

Letting \(T\rightarrow\infty\) yields that

$$ \limsup_{t\rightarrow\infty}\frac{1}{T} \int_{0}^{T} \mathbb{E} \bigl\vert x(t) \bigr\vert ^{p}\,\mathrm{d}t\leq C. $$
(75)

The proof is complete. □

5 Example and conclusion

We take the weight function \(k_{ij}(t)=e^{-(\lambda+1)t}\) in (8), then (H2) is always valid because \(\int_{0}^{\infty}k_{ij}(s)e^{\lambda s}\,\mathrm{d}s=1<\infty\). Therefore the existence and uniqueness of the positive solution is derived with probability one as time approaches infinity. Moreover, the pth moment of the solution is controlled by a bounded constant as presented in Lemma 3.1, and the solution is stochastically ultimately bounded as proved in Theorem 3.2. In addition, the absolute value of the solution will not explode in a long run, and the pth moment in the mean always keeps a constant no matter how large the time scale is.