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Here we consider a system of two nonlinear differential equations that is destined to unify different known mathematical models, in particular, very often investigated models of predator–prey type [47, 53, 60, 65, 72, 82, 83, 94, 107, 108, 112, 113, 127, 128, 153, 180, 235, 249, 267, 283, 288, 304, 305, 311, 314, 317, 321, 325]. The system under consideration is exposed to stochastic perturbations and is linearized in a neighborhood of the positive point of equilibrium. Asymptotic mean-square stability conditions for the trivial solution of the constructed linear system are at the same time sufficient conditions for the stability in probability of the positive equilibrium point of the initial nonlinear system by stochastic perturbations.

10.1 System Under Consideration

Consider the system of two nonlinear differential equations

(10.1)

Here x i (t), i=1,2, is the value of the process x i at time t, and x it =x i (t+s), s≤0, is a trajectory of the process x i to the point of time t.

Put, for example,

(10.2)

where K i (s) and R i (s), i=0,1,2, are nondecreasing functions such that

$$ \begin{gathered} K_i=\int ^\infty_0dK_i(s)<\infty, \qquad R_i=\int^\infty_0dR_i(s)< \infty , \\ \hat{K}_i=\int^\infty_0s\,dK_i(s)< \infty, \qquad\hat{R}_i=\int^\infty_0s\,dR_i(s)< \infty, \end{gathered} $$
(10.3)

and all integrals are understood in the Stieltjes sense.

In the case (10.2)–(10.3) system (10.1) takes the form

$$ \begin{aligned} \dot{x}_1(t)=&x_1(t) \biggl(a-\int^\infty_0f_0 \bigl(x_1(t-s)\bigr)\,dK_0(s) \biggr)-\prod ^2_{i=1}\int^\infty_0f_i \bigl(x_i(t-s)\bigr)\,dK_i(s), \\ \dot{x}_2(t)=&-x_2(t) \biggl(b+\int^\infty_0g_0 \bigl(x_1(t-s)\bigr)\,dR_0(s) \biggr)+\prod ^2_{i=1}\int^\infty_0g_i \bigl(x_i(t-s)\bigr)\,dR_i(s). \end{aligned} $$
(10.4)

Systems of type (10.4) are investigated in some biological problems. Put here, for example,

$$ \begin{gathered} f_0(x)=f_1(x)=f_2(x)=g_1(x)=g_2(x)=x, \\ g_0(x)=0, \qquad dK_1(s)=\delta(s)\,ds,\qquad dR_0(s)=0 \end{gathered} $$
(10.5)

(δ(s) is Dirac’s function). If a and b are positive constants, x 1(t) and x 2(t) are respectively the densities of prey and predator populations, then (10.4) is transformed to the mathematical predator–prey model [267] with distributed delay

$$ \begin{aligned} \dot{x}_1(t)&=x_1(t) \biggl(a-\int^\infty_0x_1(t-s)\,dK_0(s)- \int^\infty _0x_2(t-s)\,dK_2(s) \biggr), \\ \dot{x}_2(t)&=-bx_2(t)+\int^\infty_0x_1(t-s)\,dR_1(s) \int^\infty_0x_2(t-s)\,dR_2(s). \end{aligned} $$
(10.6)

Putting in (10.6)

$$ \begin{gathered} dK_0(s)=a_1 \delta(s)\,ds, \qquad dK_2(s)=a_2\delta(s)\,ds, \\ dR_1(s)=b_1\delta(s-h_1)\,ds, \qquad dR_2(s)=\delta(s-h_2)\,ds, \end{gathered} $$
(10.7)

we obtain the known predator–prey mathematical model with fixed delays

$$ \begin{aligned} \dot{x}_1(t)&=x_1(t) \bigl(a-a_1x_1(t)-a_2x_2(t) \bigr), \\ \dot{x}_2(t)&=-bx_2(t)+b_1x_1(t-h_1)x_2(t-h_2). \end{aligned} $$
(10.8)

If here h 1=h 2=0, we have the classical Lotka–Volterra model

$$\begin{aligned} \dot{x}_1(t)&=x_1(t) \bigl(a-a_1x_1(t)-a_2x_2(t) \bigr), \\ \dot{x}_2(t)&=x_2(t) \bigl(-b+b_1x_1(t) \bigr). \end{aligned} $$

Many authors [15, 19, 23, 50, 69, 70, 116, 306, 309] consider the so-called ratio-dependent predator–prey models with delays of type

$$ \begin{aligned} \dot{x}_1(t)&=x_1(t) \biggl(a-\int^\infty_0x_1(t-s)\,dK_0(s) \biggr)\\ &\quad {}-\int^\infty_0{x_1^k(t-s)x_2(t)\over x_1^k(t-s)+a_2x_2^k(t-s)}\,dK_1(s), \\ \dot{x}_2(t)&=-bx_2(t)+\int^\infty_0 {x_1^m(t-s)x_2(t)\over x_1^m(t-s)+b_2x_2^m(t-s)}\,dR_1(s). \end{aligned} $$
(10.9)

Here it is supposed that m and k are positive constants.

System (10.9) follows from (10.1) if

$$ \begin{aligned} F_0(x_{1t},x_{2t})&= \int^\infty_0x_1(t-s)\,dK_0(s), \qquad G_0(x_{1t},x_{2t})=0, \\ F_1(x_{1t},x_{2t})&=\int^\infty_0f \bigl(x_1(t-s),x_2(t-s)\bigr)x_2(t)\,dK_1(s), \\ G_1(x_{1t},x_{2t})&=\int^\infty_0g \bigl(x_1(t-s),x_2(t-s)\bigr)x_2(t)\,dR_1(s), \\ f(x_1,x_2)&={x_1^k\over x_1^k+a_2x_2^k},\qquad g(x_1,x_2)={x_1^m\over x_1^m+b_2x_2^m}. \end{aligned} $$
(10.10)

Putting in (10.9), for example,

$$ \begin{gathered} dK_0(s)=a_0 \delta(s)\,ds,\qquad dK_1(s)=a_1\delta(s)\,ds, \\ dR_1(s)=b_1\delta(s-h)\,ds,\qquad k=m=1, \end{gathered} $$
(10.11)

we obtain the system

$$ \begin{aligned} \dot{x}_1(t)&=x_1(t) \biggl(a-a_0x_1(t)-{a_1x_2(t)\over x_1(t)+a_2x_2(t)} \biggr), \\ \dot{x}_2(t)&=x_2(t) \biggl(-b+{b_1x_1(t-h)\over x_1(t-h)+b_2x_2(t-h)} \biggr), \end{aligned} $$
(10.12)

which was considered in [23, 50].

10.2 Equilibrium Points, Stochastic Perturbations, Centering, and Linearization

10.2.1 Equilibrium Points

Let in system (10.1) F i =F i (ϕ,ψ) and G i =G i (ϕ,ψ), i=0,1, be functionals defined on H×H, where H is a set of functions ϕ=ϕ(s), s≤0, with the norm ∥ϕ∥=sup s≤0|ϕ(s)|, the functionals F i and G i are nonnegative for nonnegative functions ϕ and ψ. Let us suppose also that system (10.1) has a positive equilibrium point \((x_{1}^{*},x_{2}^{*})\). This point is obtained from the conditions \(\dot{x}_{1}(t)\equiv0\), \(\dot{x}_{2}(t)\equiv0\) and is defined by the system of algebraic equations

$$ \begin{aligned} x^*_1 \bigl(a-F_0\bigl(x^*_1,x^*_2\bigr) \bigr)&=F_1\bigl(x^*_1,x^*_2\bigr), \\ x^*_2\bigl(b+G_0\bigl(x_1^*,x^*_2 \bigr)\bigr)&=G_1\bigl(x^*_1,x^*_2\bigr). \end{aligned} $$
(10.13)

From (10.13) it follows that system (10.1) has a positive solution by the condition

$$ a>F_0\bigl(x^*_1,x^*_2 \bigr) $$
(10.14)

only. For example, if \(a>K_{0}f_{0}(x^{*}_{1})\), a positive equilibrium point of system (10.4) is defined by the system of algebraic equations

$$ \begin{aligned} x^*_1 \bigl(a-K_0f_0\bigl(x^*_1\bigr) \bigr)&=K_1K_2f_1\bigl(x^*_1 \bigr)f_2\bigl(x^*_2\bigr), \\ x^*_2\bigl(b+R_0g_0\bigl(x_1^* \bigr)\bigr)&=R_1R_2g_1\bigl(x^*_1 \bigr)g_2\bigl(x^*_2\bigr). \end{aligned} $$
(10.15)

In particular, from (10.5), (10.14), (10.15) it follows that system (10.6) has a positive equilibrium point

$$ x^*_1={b\over R_1R_2}, \qquad x^*_2={a-K_0x^*_1\over K_2}={a-(R_1R_2)^{-1}K_0b\over K_2}, $$
(10.16)

provided that a>(R 1 R 2)−1 K 0 b. For system (10.8), from (10.7), (10.16) we obtain

$$ x^*_1={b\over b_1}, \qquad x^*_2={A\over a_2}, \quad A=a-b{a_1\over b_1}>0. $$
(10.17)

From (10.13), (10.10) it follows that the positive equilibrium point for system (10.9) is

$$x_1^*={A\over K_0},\qquad x_2^*= {A\over BK_0},\quad A=a-{K_1\over B+a_2B^{1-k}}>0,\ B= \biggl( {bb_2\over R_1-b} \biggr)^{1\over m}>0. $$

In particular, by (10.11), for system (10.12), it is

$$ x_1^*={A\over a_0},\qquad x_2^*={A\over Ba_0},\quad A=a-{a_1\over B+a_2}>0, \ B={bb_2\over b_1-b}>0. $$
(10.18)

10.2.2 Stochastic Perturbations and Centering

Similarly to Sect. 9.2, we will assume that system (10.1) is exposed to stochastic perturbations that are of white noise type and are directly proportional to the deviations of the system state (x 1(t),x 2(t)) from the equilibrium point \((x^{*}_{1}, x^{*}_{2})\) and influence \(\dot{x}_{1}(t)\), \(\dot{x}_{2}(t)\), respectively. In this way system (10.1) is transformed to the form

$$ \begin{aligned} \dot{x}_1(t)&=x_1(t) \bigl(a-F_0(x_{1t},x_{2t}) \bigr)-F_1(x_{1t},x_{2t})+ \sigma_1\bigl(x_1(t)-x^*_1\bigr)\dot{w}_1(t), \\ \dot{x}_2(t)&=-x_2(t) \bigl(b+G_0(x_{1t},x_{2t}) \bigr)+G_1(x_{1t},x_{2t})+\sigma_2 \bigl(x_2(t)-x^*_2\bigr)\dot{w}_2(t). \end{aligned} $$
(10.19)

Here σ 1, σ 2 are constants, and w 1(t), w 2(t) are independent standard Wiener processes.

Centering system (10.19) at the positive point of equilibrium via the new variables \(y_{1}=x_{1}-x^{*}_{1}\), \(y_{2}=x_{2}-x^{*}_{2}\), we obtain

$$ \begin{aligned} \dot{y}_1(t)&= \bigl(y_1(t)+x^*_1\bigr) \bigl(a-F_0 \bigl(y_{1t}+x_1^*,y_{2t}+x_2^*\bigr) \bigr) \\ &\quad {}-F_1\bigl(y_{1t}+x_1^*,y_{2t}+x_2^* \bigr)+\sigma_1y_1(t)\dot{w}_1(t), \\ \dot{y}_2(t)&=-\bigl(y_2(t)+x^*_2\bigr) \bigl(b+G_0\bigl(y_{1t}+x_1^*,y_{2t}+x_2^* \bigr) \bigr) \\ &\quad {}+G_1\bigl(y_{1t}+x_1^*,y_{2t}+x_2^* \bigr)+\sigma_2y_2(t)\dot{w}_2(t). \end{aligned} $$
(10.20)

It is clear that the stability of equilibrium point \((x_{1}^{*},x_{2}^{*})\) of system (10.19) is equivalent to the stability of the trivial solution of system (10.20).

For system (10.4), the representations (10.19) and (10.20) respectively take the forms

(10.21)

and

(10.22)

In particular, for system (10.6), from (10.21), (10.22) by (10.5), (10.16) we obtain

(10.23)

and

$$ \begin{aligned} \dot{y}_1(t)=&- \bigl(y_1(t)+x^*_1\bigr) \biggl(\int ^\infty_0y_1(t-s)\,dK_0(s)+ \int^\infty_0y_2(t-s)\,dK_2(s) \biggr) \\ &+\sigma_1y_1(t)\dot{w}_1(t), \\ \dot{y}_2(t)=&-by_2(t)+R_2x^*_2 \int^\infty_0y_1(t-s)\,dR_1(s)\\ &+R_1x^*_1 \int^\infty_0y_2(t-s)\,dR_2(s) \\ &+\prod^2_{i=1}\int^\infty_0y_i(t-s)\,dR_i(s)+ \sigma_2y_2(t)\dot{w}_2(t). \end{aligned} $$
(10.24)

For (10.8), systems (10.23) and (10.24) take respectively the forms

$$ \begin{aligned} \dot{x}_1(t)&=x_1(t) \bigl(a-a_1x_1(t)-a_2x_2(t) \bigr)+\sigma_1\bigl(x_1(t)-x^*_1\bigr)\dot{w}_1(t), \\ \dot{x}_2(t)&=-bx_2(t)+b_1x_1(t-h_1)x_2(t-h_2)+ \sigma _2\bigl(x_1(t)-x^*_2\bigr)\dot{w}_2(t) \end{aligned} $$
(10.25)

and

$$ \begin{aligned} \dot{y}_1(t)=&- \bigl(y_1(t)+x^*_1\bigr) \bigl(a_1y_1(t)+a_2y_2(t) \bigr)+\sigma_1y_1(t)\dot{w}_1(t), \\ \dot{y}_2(t)=&-by_2(t)+b_1 \bigl(x^*_2y_1(t-h_1)+x^*_1y_2(t-h_2) \bigr) \\ &+b_1y_1(t-h_1)y_2(t-h_2)+ \sigma_2y_2(t)\dot{w}_2(t). \end{aligned} $$
(10.26)

10.2.3 Linearization

Along with the considered nonlinear system, we will use the linear part of this system. Let us suppose that the functionals in (10.19) have the representations (10.2) with differentiable functions f i (x), g i (x), i=0,1,2. Using for all these functions the representation

$$f\bigl(z+x^*\bigr)=f_0+f_1z+o(z),\quad f_0=f\bigl(x^*\bigr),\ f_1={df\over dx} \bigl(x^*\bigr), $$

and neglecting o(z), we obtain the linear part (process (z 1(t),z 2(t))) of system (10.22)

$$ \begin{aligned} \dot{z}_1(t)&=(a-K_0f_{00})z_1(t)- \int^\infty_0z_1(t-s)\,dK(s) \\ &\quad {}-K_1f_{10}f_{21}\int^\infty_0z_2(t-s)\,dK_2(s)+ \sigma_1z_1(t)\dot{w}_1(t), \\ \dot{z}_2(t)&=-(b+R_0g_{00})z_2(t)+ \int^\infty_0z_1(t-s)\,dR(s) \\ &\quad {}+R_1g_{10}g_{21}\int^\infty_0z_2(t-s)\,dR_2(s)+ \sigma_2z_2(t)\dot{w}_2(t), \end{aligned} $$
(10.27)

where

$$ \begin{aligned} dK(s)&=K_2f_{20}f_{11}\,dK_1(s)+f_{01}x^*_1\,dK_0(s), \\ dR(s)&=R_2g_{20}g_{11}\,dR_1(s)-g_{01}x^*_2\,dR_0(s). \end{aligned} $$
(10.28)

Below we will speak about system (10.27) as about the linear part corresponding to system (10.22) or, for brevity, as about the linear part of system (10.22).

In particular, by conditions (10.5), (10.16), and (10.28) from (10.27) we obtain the linear part of system (10.24)

$$ \begin{aligned} \dot{z}_1(t)=&-x_1^* \biggl( \int^\infty_0z_1(t-s)\,dK_0(s)+ \int^\infty _0z_2(t-s)\,dK_2(s) \biggr)+\sigma_1z_1(t)\dot{w}_1(t), \\ \dot{z}_2(t)=&-bz_2(t)+R_2x_2^* \int^\infty_0z_1(t-s)\,dR_1(s)+R_1x_1^* \int^\infty_0z_2(t-s)\,dR_2(s) \\ &+\sigma_2z_2(t)\dot{w}_2(t). \end{aligned} $$
(10.29)

From (10.26) or, via (10.7), from (10.29) we have the linear part of system (10.26)

$$ \begin{aligned} \dot{z}_1(t)&=-x^*_1 \bigl(a_1z_1(t)+a_2z_2(t)\bigr)+ \sigma_1z_1(t)\dot{w}_1(t), \\ \dot{z}_2(t)&=-bz_2(t)+b_1 \bigl(x^*_2z_1(t-h_1)+x^*_1z_2(t-h_2) \bigr)+\sigma _2z_2(t)\dot{w}_2(t). \end{aligned} $$
(10.30)

As it is shown in Sect. 5.3, if the order of nonlinearity of the system under consideration is higher than one, then a sufficient condition for the asymptotic mean-square stability of the linear part of the considered nonlinear system is also a sufficient condition for the stability in probability of the initial system. So, below we will obtain sufficient conditions for the asymptotic mean-square stability of the linear part of considered nonlinear systems.

10.3 Stability of Equilibrium Point

Obtain now sufficient conditions for the asymptotic mean-square stability of the trivial solution of system (10.27) as the linear part of (10.22). The obtained conditions will be at the same time sufficient conditions for the stability in probability of the equilibrium point of (10.21).

Following the procedure of constructing Lyapunov functionals (Sect. 2.2.2), rewrite (10.27) in the form

$$ \begin{aligned} \dot{Z}_1(t)&=a_{11}z_1(t)+a_{12}z_2(t)+ \sigma_1z_1(t)\dot{w}_1(t), \\ \dot{Z}_2(t)&=a_{21}z_1(t)+a_{22}z_2(t)+ \sigma_2z_2(t)\dot{w}_2(t), \end{aligned} $$
(10.31)

where

$$ \begin{aligned} Z_1(t)&=z_1(t)- \int^\infty_0\int^t_{t-s}z_1( \theta)\,d\theta \,dK(s)-K_1f_{10}f_{21}\int ^\infty_0\int^t_{t-s}z_2( \theta)\,d\theta\, dK_2(s), \\ Z_2(t)&=z_2(t)+\int^\infty_0 \int^t_{t-s}z_1(\theta)\,d\theta\, dR(s)+R_1g_{10}g_{21}\int^\infty_0 \int^t_{t-s}z_2(\theta)\,d\theta\, dR_2(s), \end{aligned} $$
(10.32)

and, by (10.15), (10.28),

$$ \begin{aligned} a_{11}&=a-K-K_0f_{00}=K_1K_2f_{20} \biggl({f_{10}\over x_1^*}-f_{11} \biggr)-K_0f_{01}x_1^*, \\ a_{12}&=-K_1K_2f_{10}f_{21}, \qquad a_{21}=R=R_1R_2g_{20}g_{11}-R_0g_{01}x_2^*, \\ a_{22}&=R_1R_2g_{10}g_{21}-b-R_0g_{00}=-R_1R_2g_{10} \biggl({g_{20}\over x_2^*}-g_{21} \biggr). \end{aligned} $$
(10.33)

System (10.31), (10.32) is a system of stochastic differential equations of neutral type, so, following (2.10), we have to suppose that

$$\begin{aligned} \int^\infty_0s\,dK(s)+K_1|f_{10}f_{21}| \int^\infty_0s\,dK_2(s)&<1, \\ \int^\infty_0s\,dR(s)+R_1|g_{10}g_{21}| \int^\infty_0s\,dR_2(s)&<1, \end{aligned} $$

or, by (10.3), (10.28), that

$$ \begin{aligned} |f_{01}|x^*_1 \hat{K}_0&+K_2|f_{20}f_{11}|\hat{K}_1+K_1|f_{10}f_{21}|\hat{K}_2<1, \\ |g_{01}|x^*_2\hat{R}_0&+R_2|g_{20}g_{11}| \hat{R}_1+R_1|g_{10}g_{21}|\hat{R}_2<1. \end{aligned} $$
(10.34)

10.3.1 First Way of Constructing a Lyapunov Functional

Let \(\hat{A}=\|a_{ij}\|\) be the matrix with the elements defined by (10.33), and P=∥p ij ∥ be the matrix with the elements defined by (1.29) for some q>0. Represent p 11, p 22 in the form

$$ p_{ii}={1\over2}\bigl(qp_{ii}^{(0)}+p_{ii}^{(1)} \bigr),\quad i=1,2, $$
(10.35)

where

$$ \begin{gathered} p_{11}^{(0)}= {a^2_{22}+\det(\hat{A})\over|\operatorname{Tr}(\hat{A})|\det(\hat{A})},\qquad p_{11}^{(1)}= {a^2_{21}\over|\operatorname{Tr}(\hat{A})|\det(\hat{A})}, \\ p_{22}^{(0)}={a^2_{12}\over|\operatorname{Tr}(\hat{A})|\det(\hat{A})},\qquad p_{22}^{(1)}={a^2_{11}+\det(\hat{A})\over|\operatorname{Tr}(\hat{A})|\det(\hat{A})}, \end{gathered} $$
(10.36)

and put

$$ d\mu_{ij}(s)=qd\mu^{(0)}_{ij}(s)+d \mu^{(1)}_{ij}(s),\quad i,j=1,2, $$
(10.37)

where

$$ \begin{gathered} d\mu^{(0)}_{11}=\,dK(s)- {a_{12}\over|\operatorname{Tr}(\hat{A})|}\,dR(s),\qquad d\mu ^{(1)}_{11}= {a_{21}\over|\operatorname{Tr}(\hat{A})|}\,dR(s), \\ d\mu^{(0)}_{12}=K_1f_{10}f_{21}\,dK_2(s)- {a_{12}\over|\operatorname{Tr}(\hat{A})|}R_1g_{10}g_{21}\,dR_2(s), \\ d\mu^{(1)}_{12}={a_{21}\over|\operatorname{Tr}(\hat{A})|}R_1g_{10}g_{21}\,dR_2(s), \\ d\mu^{(0)}_{21}=-{a_{12}\over|\operatorname{Tr}(\hat{A})|}\,dK(s),\qquad d\mu ^{(1)}_{21}={a_{21}\over|\operatorname{Tr}(\hat{A})|}\,dK(s)-dR(s), \\ d\mu^{(0)}_{22}=-{a_{12}\over|\operatorname{Tr}(\hat{A})|}K_1f_{10}f_{21}\,dK_2(s), \\ d\mu^{(1)}_{22}={a_{21}\over|\operatorname{Tr}(\hat{A})|}K_1f_{10}f_{21}\,dK_2(s)-R_1g_{10}g_{21}\,dR_2(s), \end{gathered} $$
(10.38)

and dK(s), dR(s) are defined by (10.28).

Put also

$$ \begin{gathered} \delta_i= {1\over2}\sigma^2_i, \quad i=1,2, \\ \nu^{(m)}_{ij}=\int^\infty_0s\bigl|d \mu^{(m)}_{ij}(s)\bigr|,\quad i,j=1,2,\ m=0,1, \end{gathered} $$
(10.39)

and

$$ \begin{gathered} A_1=1- \nu^{(0)}_{11}-p^{(0)}_{11} \delta_1, \qquad A_2=1-\nu ^{(1)}_{22}-p^{(1)}_{22} \delta_2, \\ B_1=\nu^{(1)}_{11}+p^{(1)}_{11} \delta_1,\qquad B_2=\nu ^{(0)}_{22}+p^{(0)}_{22} \delta_2, \\ C_1=\nu^{(1)}_{12}+\nu^{(1)}_{21}, \qquad C_2=\nu^{(0)}_{12}+\nu^{(0)}_{21}. \end{gathered} $$
(10.40)

Theorem 10.1

If A 1>0, A 2>0, and conditions (10.34) and

$$ \sqrt{(A_1C_1+B_1C_2) (A_2C_2+B_2C_1)}+B_1B_2<A_1A_2 $$
(10.41)

hold, then the trivial solution of system (10.27) is asymptotically mean-square stable and the equilibrium point of system (10.21) is stable in probability.

Proof

We will consider now system (10.31)–(10.33) and suppose that the trivial solution of the appropriate auxiliary system without delays of type (2.60) with a ij , i,j=1,2, defined by (10.33) is asymptotically mean-square stable, and so conditions (2.62) hold.

Consider the functional

$$ V_1(t)=p_{11}Z_1^2(t)+2p_{12}Z_1(t)Z_2(t)+p_{22}Z_2^2(t) $$
(10.42)

with p ij , i,j=1,2, defined by (1.29). Let L be the generator of system (10.31). Then, by (10.31), (10.42),

(10.43)

Putting

$$ \rho={a_{21}-a_{12}q\over|\operatorname{Tr}(\hat{A})|} $$
(10.44)

and using (1.29), (2.62), we obtain

$$\begin{aligned} 2(p_{11}a_{11}+p_{12}a_{21})&=-q, \qquad 2(p_{12}a_{12}+p_{22}a_{22})=-1, \\ 2(p_{12}a_{11}+p_{22}a_{21})&= {-(a_{12}a_{22}q+a_{21}a_{11})a_{11}+(a^2_{11}+\det (\hat{A})+a^2_{12}q)a_{21}\over|\operatorname{Tr}(\hat{A})|\det(\hat{A})} \\ &={\det(\hat{A})a_{21}-(a_{11}a_{22}-a_{12}a_{21})a_{12}q\over|\operatorname{Tr}(\hat{A})|\det(\hat{A})}=\rho, \\ 2(p_{11}a_{12}+p_{12}a_{22})&= {((a^2_{22}+\det(\hat{A}))q+ a^2_{21})a_{12}-(a_{12}a_{22}q+a_{21}a_{11})a_{22}\over|\operatorname{Tr}(\hat{A})|\det (\hat{A})} \\ &={\det(A)a_{12}q-a_{21}(a_{11}a_{22}-a_{21}a_{12})\over |\operatorname{Tr}(\hat{A})|\det(\hat{A})}=-\rho. \end{aligned} $$

So, (10.43) takes the form

(10.45)

Substituting (10.32) into (10.45), we have

By (10.37), (10.38), (10.44), it can be written in the form

(10.46)

Using (10.35), (10.37), (10.39) and some positive number γ from (10.46), we obtain

From this by (10.39) we have

(10.47)

where

$$\begin{aligned} dF_i(s)&={1\over2} \bigl(qdF_i^{(0)}(s)+dF_i^{(1)}(s) \bigr),\quad i=1,2, \\ dF^{(0)}_1(s)&=\bigl|d\mu^{(0)}_{11}(s)\bigr|+ \gamma^{-1}\bigl|d\mu^{(0)}_{21}(s)\bigr|, \\ dF^{(1)}_1(s)&=\bigl|d\mu^{(1)}_{11}(s)\bigr|+ \gamma^{-1}\bigl|d\mu^{(1)}_{21}(s)\bigr|, \\ dF^{(0)}_2(s)&=\bigl|\gamma d\mu^{(0)}_{12}(s)\bigr|+\bigl|d \mu^{(0)}_{22}(s)\bigr|, \\ dF^{(1)}_2(s)&=\bigl|\gamma d\mu^{(1)}_{12}(s)\bigr|+\bigl|d \mu^{(1)}_{22}(s)\bigr|. \end{aligned} $$

Note that for the functional

$$V_2(t)=\sum^2_{i=1}\int ^\infty_0\int^t_{t-s}( \theta-t+s)z^2_i(\theta )\,d\theta\, dF_i(s), $$

we have

$$ LV_2(t)=\hat{F}_1z^2_1(t)+ \hat{F}_2z^2_2(t)-\sum ^2_{i=1}\int^\infty_0 \int^t_{t-s}z^2_i(\theta)\,d \theta\, dF_i(s), $$
(10.48)

where

$$\begin{aligned} \hat{F}_1&={1\over2} \bigl[q \bigl(\nu^{(0)}_{11}+\gamma^{-1}\nu ^{(0)}_{21} \bigr)+\nu^{(1)}_{11}+ \gamma^{-1}\nu^{(1)}_{21} \bigr], \\ \hat{F}_2&={1\over2} \bigl[q \bigl(\gamma \nu^{(0)}_{12}+\nu ^{(0)}_{22} \bigr)+\gamma \nu^{(1)}_{12}+\nu^{(1)}_{22} \bigr]. \end{aligned} $$

From (10.47), (10.48), for the functional V=V 1+V 2, by (10.40) we obtain

(10.49)

By Theorem 2.1, if there exist positive numbers q and γ such that

$$ \begin{aligned} &q \biggl(-A_1+ {\gamma^{-1}\over2}C_2 \biggr)+B_1+ {\gamma^{-1}\over 2}C_1<0, \\ & {-}A_2+{\gamma\over2}C_1+q \biggl(B_2+{\gamma\over2}C_2 \biggr)<0, \end{aligned} $$
(10.50)

then the trivial solution of system (10.27) is asymptotically mean-square stable.

Rewrite (10.50) in the form

$$ \biggl(B_1+{\gamma^{-1}\over2}C_1 \biggr) \biggl(A_1-{\gamma^{-1}\over 2}C_2 \biggr)^{-1}< q< \biggl(A_2-{\gamma\over2}C_1 \biggr) \biggl(B_2+{\gamma\over2}C_2 \biggr)^{-1}. $$
(10.51)

So, if

$$ \biggl(B_1+{\gamma^{-1}\over2}C_1 \biggr) \biggl(A_1-{\gamma^{-1}\over 2}C_2 \biggr)^{-1}< \biggl(A_2-{\gamma\over2}C_1 \biggr) \biggl(B_2+{\gamma\over2}C_2 \biggr)^{-1}, $$
(10.52)

then there exists q>0 such that (10.51) holds.

Rewriting (10.52) in the form

$${\gamma\over2}(A_1C_1+B_1C_2)+ {\gamma^{-1}\over2}(A_2C_2+B_2C_1)<A_1A_2-B_1B_2 $$

and calculating the infimum of the left-hand part of the obtained inequality with respect to γ>0, we obtain (10.41). So, if (10.41) holds, then there exist positive numbers q and γ such that (10.50) holds, and therefore the trivial solution of system (10.27) is asymptotically mean-square stable. The proof is completed. □

Put now

$$ D_1={a_1\over b_1}-Ah_1- {\delta_1\over b},\qquad D_2=1-bh_2- {a_1\delta _2\over Ab_1}, $$
(10.53)

and note that the first condition (10.34) for system (10.30) is a trivial one and the second condition takes the form \(Aa_{2}^{-1}b_{1}h_{1}+bh_{2}<1\) or, via the representation (10.17) for A,

$$ b_1h_1a+(a_2h_2-a_1h_1)b<a_2. $$
(10.54)

Corollary 10.1

If D 1>0, D 2>0, and conditions (10.54) and

$$ \begin{gathered} \sqrt{A(D_1h_1+h_2) (\delta_2h_1+D_2bh_2)}+ {\delta_2\over b}<D_1D_2 \end{gathered} $$
(10.55)

hold, then the trivial solution of system (10.30) is asymptotically mean-square stable, and the equilibrium point of system (10.25) is stable in probability.

Proof

Calculating for (10.30) the parameters (10.33), (10.36), (10.38), (10.39), (10.40), we obtain

$$\begin{gathered} a_{11}=-{a_1b\over b_1},\qquad a_{12}=-{a_2b\over b_1},\qquad a_{21}= {Ab_1\over a_2},\qquad a_{22}=0, \\ p_{11}^{(0)}={b_1\over a_1b},\qquad p_{11}^{(1)}={Ab_1^3\over a_1a^2_2b^2},\qquad p_{22}^{(0)}={a^2_2\over Aa_1b_1},\qquad p_{22}^{(1)}={a_1\over Ab_1}+{b_1\over a_1b}, \\ \nu^{(0)}_{11}={Ab_1h_1\over a_1},\qquad\nu ^{(1)}_{11}={A^2b_1^3h_1\over a_1a_2^2b},\qquad \nu^{(0)}_{12}={a_2bh_2\over a_1},\qquad \nu^{(1)}_{12}={Ab^2_1h_2\over a_1a_2}, \\ \nu^{(0)}_{21}=0,\qquad\nu^{(1)}_{21}= {Ab_1h_1\over a_2},\qquad\nu ^{(0)}_{22}=0,\qquad \nu^{(1)}_{22}=bh_2, \\ A_1={b_1\over a_1}D_1,\qquad A_2=D_2-{b_1\delta_2\over a_1b}, \\ B_1={Ab_1^3\over a_1a_2^2b} \biggl(Ah_1+ {\delta_1\over b} \biggr), \qquad B_2={a^2_2\delta_2\over Aa_1b_1}, \\ C_1={Ab_1\over a_1a_2}(a_1h_1+b_1h_2), \qquad C_2={a_2bh_2\over a_1}. \end{gathered} $$

From this it follows that

$$ \begin{aligned} A_1C_1+B_1C_2&= {Ab_1^2\over a_1a_2}(D_1h_1+h_2), \\ A_2C_2+B_2C_1&= {a_2\over a_1}(\delta_2h_1+D_2bh_2), \\ A_1A_2-B_1B_2&= {b_1\over a_1} \biggl(D_1D_2- {\delta_2\over b} \biggr). \end{aligned} $$
(10.56)

From the representations for a ij , i,j=1,2, it follows also that conditions (2.62) hold. Substituting (10.56) into (10.41), we obtain (10.55). The proof is completed. □

Remark 10.1

Note that condition (10.55) does not depend on a 2. The dependence on a 2 is included in condition (10.54).

Remark 10.2

By the absence of the delays, i.e., by h 1=h 2=0, condition (10.54) is trivial, and condition (10.55) can be written in the form

$$\delta_1<b{a_1\over b_1},\qquad\delta_2< {Ab_1(a_1b-b_1\delta_1)\over Ab^2_1+a_1(a_1b-b_1\delta_1)}. $$

The same conditions can be obtained immediately from Corollary 2.3.

10.3.2 Second Way of Constructing a Lyapunov Functional

Let us consider another way of constructing of a Lyapunov functional for system (10.30).

Theorem 10.2

If D 1>0, D 2>0, and conditions (10.54) and

$$ \Bigl(\sqrt{Abh_2^2+4 \delta_2b^{-1}D_1}+\sqrt{Ab}h_2 \Bigr) \Bigl(\sqrt {Abh_1^2+4D_2}+ \sqrt{Ab}h_1 \Bigr)<4D_1D_2 $$
(10.57)

hold, where A and D 1, D 2 are defined by (10.17) and (10.53), respectively, then the trivial solution of system (10.30) is asymptotically mean-square stable, and the equilibrium point of system (10.25) is stable in probability.

Proof

Using (10.17), rewrite system (10.30) in the form

$$ \begin{aligned} \dot{z}_1(t)=&- {a_1b\over b_1}z_1(t)-{a_2b\over b_1}z_2(t)+ \sigma _1z_1(t)\dot{w}_1(t), \\ \dot{Z}_2(t)=&{Ab_1\over a_2}z_1(t)+ \sigma_2z_2(t)\dot{w}_2(t), \end{aligned} $$
(10.58)

where

$$ \begin{aligned} Z_2(t)=&z_2(t)+ {Ab_1\over a_2}J_1(z_{1t})+bJ_2(z_{2t}), \\ J_i(z_{it})=&\int^t_{t-h_i}z_i(s)\,ds, \quad i=1,2. \end{aligned} $$
(10.59)

Consider now the functional

$$ V_1(t)=z_1^2(t)+2\mu z_1(t)Z_2(t)+\gamma Z_2^2(t), $$
(10.60)

where the parameters μ and γ will be chosen below. Then by (10.60), (10.58) we have

(10.61)

Defining now γ by the equality

$$ \gamma{Ab_1\over a_2}=\mu{a_1b\over b_1}+ {a_2b\over b_1}, $$
(10.62)

from (10.61) we obtain

$$\begin{aligned} LV_1(t)=&-2 \biggl( {a_1b\over b_1}-\mu{Ab_1\over a_2}-\delta_1 \biggr)z^2_1(t)-2 \biggl(\mu{a_2b\over b_1}- \gamma\delta_2 \biggr)z^2_2(t) \\ &+2{a_2b\over b_1}z_1(t) \biggl({Ab_1\over a_2}J_1(z_{1t})+bJ_2(z_{2t}) \biggr) \\ &-2\mu{a_2b\over b_1}z_2(t) \biggl({Ab_1\over a_2}J_1(z_{1t})+bJ_2(z_{2t}) \biggr). \end{aligned} $$

By (10.59) from this, for some positive γ 1, γ 2, we have

(10.63)

By the representations (10.53) for D 1, D 2 and (10.62) for γ inequality (10.63) can be written in the form

Put now

Then

and as a result, for the functional V=V 1+V 2, we obtain

By Theorem 2.1, if

$$ \begin{aligned} &-2bD_1+2\mu {Ab_1\over a_2}+\gamma_1{a_2b^2h_2\over b_1}+\gamma _2^{-1}\mu Abh_1<0, \\ &-2\mu{a_2b\over b_1}D_2+2{a^2_2b\over Ab_1^2} \delta_2+\gamma ^{-1}_1{a_2b^2h_2\over b_1}+ \gamma_2\mu Abh_1<0, \end{aligned} $$
(10.64)

then the trivial solution of system (10.30) is asymptotically mean-square stable.

Rewrite (10.64) in the form

$$ {2{a^2_2b\over Ab_1^2}\delta_2+\gamma^{-1}_1{a_2b^2h_2\over b_1}\over 2{a_2b\over b_1}D_2-\gamma_2Abh_1} <\mu<{2bD_1-\gamma_1{a_2b^2h_2\over b_1}\over2{Ab_1\over a_2}+\gamma _2^{-1}Abh_1}. $$
(10.65)

So, if the inequality

$$ {2{a^2_2b\over Ab_1^2}\delta_2+\gamma^{-1}_1{a_2b^2h_2\over b_1}\over 2{a_2b\over b_1}D_2-\gamma_2Abh_1} <{2bD_1-\gamma_1{a_2b^2h_2\over b_1}\over2{Ab_1\over a_2}+\gamma_2^{-1}Abh_1} $$
(10.66)

holds, then there exists μ such that (10.65) holds too.

It is easy to check that from (10.65), (10.62) the condition μ 2<γ follows, which ensures the positivity of the functional (10.60).

Representing (10.66) in the form

$${{a^2_2\over Ab_1^2}\delta_2+\gamma^{-1}_1{a_2bh_2\over2b_1}\over D_1-\gamma_1{a_2bh_2\over2b_1}}\times {{Ab^2_1\over a^2_2b}+\gamma_2^{-1}{Ab_1h_1\over2a_2}\over D_2-\gamma _2{Ab_1h_1\over2a_2}}<1 $$

and using Lemma 2.4 twice, we obtain (10.57) □

Remark 10.3

Note that the representation (10.31)–(10.33) for system (10.30) coincides with (10.58), (10.59). So, conditions (10.55) and (10.57) are equivalent and give the same stability region. For simplicity, let us check this statement by the condition h 1=0. Indeed, in this case from (10.57) we have

$$Abh_2^2+4\delta_2b^{-1}D_1< (2D_1\sqrt{D_2}-\sqrt{Ab}h_2 )^2=4D_1^2D_2-4D_1h_2 \sqrt{AbD_2}+Abh_2^2 $$

or \(\delta_{2}b^{-1}<D_{1}D_{2}-h_{2}\sqrt{AbD_{2}}\), which is equivalent to (10.55) by h 1=0. Similarly, it is easy to get that (10.55) coincides with (10.57) by the condition h 2=0 or by the condition δ 2=0. In the general case the necessary transformation is bulky enough.

The regions of stability in probability for a positive point of equilibrium of system (10.25), obtained by condition (10.55) (or (10.57)), are shown in the space of the parameters (a,b) for a 1=0.6, b 1=1 and different values of the other parameters: in Fig. 10.1 for h 1=0, h 2=0, δ 1=0, δ 2=0, in Fig. 10.2 for h 1=0, h 2=0, δ 1=0.2, δ 2=0.3, in Fig. 10.3 for a 2=0.6, h 1=0.1, h 2=0.15, δ 1=0, δ 2=0, and in Fig. 10.4 for a 2=0.07, h 1=0.01, h 2=0.15, δ 1=0.05, δ 2=0.1.

Fig. 10.1
figure 1

Region of stability in probability for (10.25): a 1=0.6, b 1=1, h 1=0, h 2=0, δ 1=0, δ 2=0

Full size image
Fig. 10.2
figure 2

Region of stability in probability for (10.25): a 1=0.6, b 1=1, h 1=0, h 2=0, δ 1=0.2, δ 2=0.3

Full size image
Fig. 10.3
figure 3

Region of stability in probability for (10.25): a 1=0.6, a 2=0.6, b 1=1, h 1=0.1, h 2=0.15, δ 1=0, δ 2=0

Full size image
Fig. 10.4
figure 4

Region of stability in probability for (10.25): a 1=0.6, a 2=0.07, b 1=1, h 1=0.01, h 2=0.15, δ 1=0.05, δ 2=0.1

Full size image

The equation of the straight line in Figs. 10.1 and 10.2 is ab 1=ba 1, which corresponds to the condition A=0. In Figs. 10.3 and 10.4 the straight line 1 also corresponds to this equation and the straight line 2 is defined by the equation b 1 h 1 a+(a 2 h 2−a 1 h 1)b=a 2, which follows from condition (10.54).

Note that the stability of the positive equilibrium point of the difference analogue of system (10.25) is investigated in [278].

10.3.3 Stability of the Equilibrium Point of Ratio-Dependent Predator–Prey Model

Consider now system (10.9) with stochastic perturbations, i.e.,

(10.67)

System (10.9) was obtained from (10.1) by conditions (10.10). So, by (10.13), (10.14) the positive equilibrium point \((x^{*}_{1},x^{*}_{2})\) of system (10.9) (and also (10.67)) is defined by the conditions

$$ \begin{gathered} x_1^* \bigl(a-K_0x_1^*\bigr)=K_1f \bigl(x^*_1,x^*_2\bigr)x_2^*, \\ b=R_1g\bigl(x^*_1,x^*_2\bigr),\qquad a>K_0x_1^*. \end{gathered} $$
(10.68)

Suppose that the functions f(x 1,x 2) and g(x 1,x 2) in (10.10) are differentiable and can be represented in the form

$$\begin{gathered} f\bigl(y_1+x^*_1,y_2+x^*_2 \bigr)=f_0+f_1y_1-f_2y_2+o(y_1,y_2), \\ g\bigl(y_1+x^*_1,y_2+x^*_2 \bigr)=g_0+g_1y_1-g_2y_2+o(y_1,y_2), \end{gathered} $$

where \(\lim_{|y|\to0}{o(y_{1},y_{2})\over|y|}=0\) for \(|y|=\sqrt {y_{1}^{2}+y^{2}_{2}}\), and

$$\begin{gathered} f_0=f\bigl(x^*_1,x^*_2 \bigr),\qquad f_1=x_2^* \hat{f},\qquad f_2=x_1^* \hat{f},\quad\hat{f}={ka_2(x_1^*x_2^*)^{k-1}\over ((x_1^*)^k+a_2(x_2^*)^k)^2}, \\[-2pt] g_0=g\bigl(x^*_1,x^*_2\bigr),\qquad g_1=x_2^* \hat{g},\qquad g_2=x_1^* \hat{g}, \quad\hat{g}={mb_2(x_1^*x_2^*)^{m-1}\over((x_1^*)^m+b_2(x_2^*)^m)^2}. \end{gathered} $$

So, the functionals F 0(x 1t ,x 2t ), F 1(x 1t ,x 2t ), G 1(x 1t ,x 2t ) in (10.10) have the representations

$$ \begin{aligned} F_0 \bigl(y_{1t}+x^*_1,y_{2t}+x^*_2 \bigr)&=K_0x^*_1+\int^\infty_0y_1(t-s)\,dK_0(s), \\[-2pt] F_1\bigl(y_{1t}+x^*_1,y_{2t}+x^*_2 \bigr)&=K_1f_0x^*_2+f_1x^*_2 \int^\infty _0y_1(t-s)\,dK_1(s)+K_1f_0y_2(t) \\[-2pt] &\quad {}-f_2x^*_2\int^\infty_0y_2(t-s)\,dK_1(s)+o(y_1,y_2), \\ G_1\bigl(y_{1t}+x^*_1,y_{2t}+x^*_2 \bigr)&=R_1g_0x^*_2+g_1x^*_2 \int^\infty _0y_1(t-s)\,dR_1(s)+R_1g_0y_2(t) \\[-2pt] &\quad -g_2x^*_2\int^\infty_0y_2(t-s)\,dR_1(s)+o(y_1,y_2). \end{aligned} $$
(10.69)

By (10.68), (10.69) the linear part of system (10.67) has the form

(10.70)

where \(dK(s)=x^{*}_{1}\,dK_{0}(s)+f_{1}x^{*}_{2}\,dK_{1}(s)\). Rewrite system (10.70) in the form (10.31) with

(10.71)

Further investigation is similar to the previous sections.

For short, consider system (10.67) by conditions (10.11). The point of equilibrium in this case is defined by (10.18). From (10.11), (10.18), (10.70) and (10.71) it follows that system (10.67) and the linear part of this system respectively take the forms

$$ \begin{aligned} \dot{x}_1(t)&=x_1(t) \biggl(a-a_0x_1(t)-{a_1x_2(t)\over x_1(t)+a_2x_2(t)} \biggr)+ \sigma_1\bigl(x_1(t)-x^*_1\bigr)\dot{w}_1(t), \\ \dot{x}_2(t)&=x_2(t) \biggl(-b+{b_1x_1(t-h)\over x_1(t-h)+b_2x_2(t-h)} \biggr)+\sigma_2\bigl(x_1(t)-x^*_2\bigr)\dot{w}_2(t) \end{aligned} $$
(10.72)

and

$$ \begin{aligned} \dot{z}_1(t)&=a_{11}z_1(t)+a_{12}z_2(t)+ \sigma_1z_1(t)\dot{w}_1(t), \\ \dot{z}_2(t)&=a_{21}z_1(t-h)+a_{22}z_2(t-h)+ \sigma_2z_2(t)\dot{w}_2(t), \end{aligned} $$
(10.73)

where

(10.74)

Let \(\hat{A}=\|a_{ij}\|\) be the matrix with the elements defined by (10.74). Suppose that

$$ \begin{gathered} b\in(0,b_1), \\ a>\left \{ \begin{array}{l@{\quad}l} a_1\alpha& \text{if}\ a_1\alpha^2\le b_1b_2\beta^2, \\[4pt] a_1\alpha+B(a_1\alpha^2-b_1b_2\beta^2) & \text{if} \ a_1\alpha ^2>b_1b_2\beta^2. \end{array} \right . \end{gathered} $$
(10.75)

By conditions (10.75) conditions (2.62) for the matrix \(\hat{A}\) hold. Indeed,

$$ \operatorname{Tr}(\hat{A})=B\bigl(a_1\alpha^2-b_1b_2 \beta^2\bigr)-A<0,\qquad\det(\hat{A})=ABb_1b_2 \beta^2>0. $$
(10.76)

Let P=∥p ij ∥ be the matrix with the elements defined by (1.29) for some q>0 and represented in the form (10.35), (10.36). Using (10.35), (10.36), (10.44), (10.76), put

$$ \rho=\rho^{(0)}q+\rho^{(1)}, \quad \rho^{(0)}=-{a_{12}\over|\operatorname{Tr}(\hat{A})|},\ \rho^{(1)}= {a_{21}\over|\operatorname{Tr}(\hat{A})|}, $$
(10.77)

and

$$ \begin{gathered} A_1=1-p_{11}^{(0)} \delta_1-\rho^{(0)}|a_{21}|h,\qquad A_2=1-p_{22}^{(1)}\delta_2-|a_{22}|h, \\ B_1=p_{11}^{(1)}\delta_1+ \rho^{(1)}|a_{21}|h,\qquad B_2=p_{22}^{(0)} \delta_2, \quad\delta_i={1\over2} \sigma_i^2,\ i=1,2, \\ C_1=\bigl(|a_{21}|+\rho^{(1)}|a_{22}| \bigr)h,\qquad C_2=\rho^{(0)}|a_{22}|h. \end{gathered} $$
(10.78)

Rewrite system (10.73) in the form

$$ \begin{aligned} \dot{z}_1(t)&=a_{11}z_1(t)+a_{12}z_2(t)+ \sigma_1z_1(t)\dot{w}_1(t), \\ \dot{Z}_2(t)&=a_{21}z_1(t)+a_{22}z_2(t)+ \sigma_2z_2(t)\dot{w}_2(t), \end{aligned} $$
(10.79)

where

$$ \begin{gathered} Z_2(t)=z_2(t)+ \int^t_{t-h}\bigl(a_{21}z_1(s)+a_{22}z_2(s) \bigr)\,ds, \end{gathered} $$
(10.80)

and following condition (2.10), suppose that the parameters a 21 and a 22 in (10.74) satisfy the condition \(h\sqrt {a^{2}_{21}+a^{2}_{22}}<1\) or, via (10.74), \(b_{1}b_{2}\beta^{2}h\sqrt {1+B^{2}}<1\), which is equivalent to

$$ (b_1-b)\sqrt{(b_1-b)^2+b^2b^2_2}< {b_1b_2\over h}. $$
(10.81)

Theorem 10.3

Let conditions (10.75), (10.81) hold. If A 1>0, A 2>0, and

$$ \sqrt{(A_1C_1+B_1C_2) (A_2C_2+B_2C_1)}+B_1B_2<A_1A_2, $$
(10.82)

then the trivial solution of system (10.73) is asymptotically mean-square stable, and the equilibrium point of system (10.72) is stable in probability.

Proof

Consider the functional

$$V_1(t)=p_{11}z_1^2(t)+2p_{12}z_1(t)Z_2(t)+p_{22}Z_2^2(t) $$

with p ij , i,j=1,2, defined by (1.29). Let L be the generator of system (10.79). Then, using (10.77), similarly to (10.45), for system (10.79), we obtain

(10.83)

Substituting (10.80) into (10.83) and using some positive γ, we obtain

Putting

for the functional V=V 1+V 2, we have

(10.84)

Using the representations (10.35), (10.36), (10.77), (10.78), we can rewrite (10.84) in the form

which coincides with (10.49). So, from this (10.82) follows, which coincides with (10.41). The proof is completed. □

The regions of stability in probability for a positive point of equilibrium of system (10.72), obtained by conditions (10.81), (10.82), are shown in the space of the parameters (a,b) for a 0=0.3, a 1=5 a 2=0.5, b 1=6, b 2=2, h=0.4 and different values of δ 1, δ 2: in Fig. 10.5 for δ 1=1.5, δ 2=0.05, in Fig. 10.6 for δ 1=1, δ 2=0.55.

Fig. 10.5
figure 5

Region of stability in probability for (10.69): a 0=0.3, a 1=5, a 2=0.5, b 1=6, b 2=2, h=0.4, δ 1=1.5, δ 2=0.05

Full size image
Fig. 10.6
figure 6

Region of stability in probability for (10.69): a 0=0.3, a 1=5, a 2=0.5, b 1=6, b 2=2, h=0.4, δ 1=1, δ 2=0.55

Full size image

In the both figures the thick line shows the stability region given by conditions (10.75) that corresponds to the values of the parameters h=δ 1=δ 2=0.