Citation: Mondal Hasan Zahid, Christopher M. Kribs. Ebola: Impact of hospital's admission policy in an overwhelmed scenario[J]. Mathematical Biosciences and Engineering, 2018, 15(6): 1387-1399. doi: 10.3934/mbe.2018063
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Cell-based in vitro assays [27] are efficient methods to study the effect of industrial chemicals on environment or human health. Our work is based on the cytotoxicity profiling project carried by Alberta Centre for Toxicology in which initially 63 chemicals were investigated using the xCELLigence Real-Time Cell Analysis High Troughput (RTCA HT) Assay [26]. We consider a mathematical model represented by stochastic differential equations to study cytotoxicity, i.e. the effect of toxicants on human cells, such as the killing of cells or cellular pathological changes.
The cells were seeded into wells of micro-electronic plates (E-Plates), and the test substances with 11 concentrations (1:3 serial dilution from the stock solution) were dissolved in the cell culture medium [20]. The microelectrode electronic impedance value was converted by a software to Cell Index (
The success of clustering and classification methods depends on providing TCRCs that illustrates the cell population evolution from persistence to extinction. In [1] we consider a model represented by a system of ordinary differential equations to determine an appropriate range for the initial concentration of the toxicant. The model's parameters were estimated based on the data included in the TCRCs [1].
Let
dn(t)dt=βn(t)−γn2(t)−αCo(t)n(t), | (1) |
dCo(t)dt=λ21Ce(t)−η21Co(t), | (2) |
dCe(t)dt=λ22Co(t)n(t)−η22Ce(t)n(t) | (3) |
Here
The deterministic model (1)-(3) is a special case of the class of models proposed in [5], and it is related to the models considered in [7, 11, 15]. However, since we consider an acute dose of toxicant instead of a chronic one, the analysis of the survival/death of the cell population is different from the one done in the previously mentioned papers.
We have noticed that, for the toxicants considered here, the estimated values of the parameters
1. If
limt→∞n(t)=K, limt→∞Co(t)=limt→∞Ce(t)=0. |
2. If
limt→∞n(t)=0, limt→∞Co(t)=C∗eλ21η21, limt→∞Ce(t)=C∗e>βη21αλ21, |
In practice we usually estimate a parameter by an average value plus an error term. To keep the stochastic model as simple as possible, we ignore the relationship between the parameters
˜β=β+error1 ,˜γ=γ+error2 | (4) |
By the central limit theorem, the error terms may be approximated by a normal distribution with zero mean. Thus we replace equation (1) by a stochastic differential equation and, together with equations (2) and (3), we get the stochastic model
dn(t)=n(t)(β−γn(t)−αCo(t))dt+σ1n(t)dB1(t)−σ2n2(t)dB2(t), | (5) |
dCo(t)=(λ21Ce(t)−η21Co(t))dt, | (6) |
dCe(t)=(λ22Co(t)n(t)−η22Ce(t)n(t))dt, | (7) |
Here
Several versions of a stochastic logistic equation similar with (5) were considered in [18], [19], [8], [9], [10] and [21]. The system of stochastic differential equations (5)-(7) is closely related with the stochastic models in a polluted environment considered in [15], [16], and [24]. However, for the models considered in these papers, instead of the equations (6) and (7),
In this paper we extend the methods applied in [15] and [16] to find conditions for extinction, weakly persistence, and weakly stochastically permanence for the model (5)-(7). In addition to this we focus on the ergodic properties when the cell population is strongly persistent. The main contribution of this paper is the proof that
In the next section we prove that there is a unique non-negative solution of system (5)-(7) for any non-negative initial value. In section 3 we investigate the asymptotic behavior, and in section 4 we study the weak convergence of
We have to show that system (5)-(7) has a unique global positive solution in order for the stochastic model to be appropriate. Let
Since equations (6) and (7) are linear in
Co(t)=Co(0)e−η21t+λ21e−η21t∫t0Ce(s)eη21sds | (8) |
Ce(t)=Ce(0)exp(−η22∫t0n(s)ds)+λ22exp(−η22∫t0n(s)ds)∫t0Co(s)n(s)exp(η22∫s0n(l)dl)ds, t≥0. | (9) |
Let's define the differential operator
L=∂∂t+(βn−γn2−αCon)∂∂n+(λ21Ce−η21Co)∂∂Co+(λ22Con−η22Cen)∂∂Ce+12((σ21n2+σ22n4)∂2∂2n) |
For any function
dV(x(t),t)=LV(x(t),t)dt+∂V(x(t),t)∂n(σ1n(t)dB1(t)−σ2n2(t)dB2(t)), | (10) |
where
Theorem 2.1. Let
Proof. The proof is similar with the proof of theorem 3.1 in [29]. Since the coefficients are locally Lipschitz continuous functions, there exists a unique solution on
τm=inf{t∈[0,τe):min{n(t),Ce(t)}≤m−1 or max{n(t),Co(t),Ce(t)}≥m}, | (11) |
where
We show that
We define the
V(x)=Co+α4λ22(Ce−logCe−1)+αCe4λ22+(√n−log√n−1)+n. |
We get
LV(x)=(λ21Ce−η21Co)+α4λ22(1−1Ce)(λ22Con−η22Cen)+α4λ22(λ22Con−η22Cen)+(βn−γn2−αCon)(12√n−12n)+12(σ21n2+σ22n4)(−14n√n+12n2)+(βn−γn2−αCon) |
Omitting some of the negative terms, for any
LV(x)≤λ21Ce+αCon4+αCon4+αCo2−αCon+f(n),≤λ21Ce+αCo2+f(n), |
where
f(n)=−σ22n2√n8+α4λ21η22n+β√n2+γn2+σ214+σ22n24+βn |
Since
Let's define
L˜V(x,t)=−Ce−Ct(1+V(x))+e−CtLV(x)≤0. |
Using Itô's formula (10) for
E[˜V(x(t∧τm),t∧τm)]=˜V(x(0),0)+E[∫t∧τm0L˜V(x(u∧τm),u∧τm)du]≤˜V(x(0),0). |
Notice that for any
E[V(x(τm,ω))IΘm(ω)]≥P(Θm)bm≥ϵbm→∞ |
as
Here we focus on the case when
Lemma 2.2. If
Theorem 2.3. If
limt→∞Co(t)=λ21η21limt→∞Ce(t). |
In this section we assume that
Definition 3.1. The population
Definition 3.2. The population
Definition 3.3. The population
Definition 3.4. The population
Theorem 3.5. a. If
b. If
Proof. The proof is similar with the proof of Theorem 6 in [16]. We start with some preliminary results. By Itô's formula in (5) we have
dlnn(t)=(β−γn(t)−αCo(t)−σ21+σ22n2(t)2)dt+σ1dB1(t)−σ2n(t)dB2(t). |
This means that we have
lnn(t)−lnn(0)=(β−σ212)t−γ∫t0n(s)ds−α∫t0Co(s)ds−σ222∫t0n2(s)ds+σ1B1(t)−σ2∫t0n(s)dB2(s), | (12) |
Notice that the quadratic variation [17] of
⟨M(t),M(t)⟩=σ22∫t0n2(s)ds. |
Now we do the proof for part a. Using the exponential martingale inequality (Theorem 7.4 [17]) and Borel-Cantelli lemma ([22], pp. 102), and proceeding as in the proof of Theorem 6 in [16] we can show that for almost all
sup0≤t≤n(M(t)−12⟨M(t),M(t)⟩)≤2lnn. |
Hence, for all
−σ222∫t0n2(s)ds−σ2∫t0n(s)dB2(s)≤2lnn a.s.. |
Substituting the above inequality in (12) we get
lnn(t)−lnn(0)t≤β−σ212−α∫t0Co(s)dst+σ1B1(t)t+2lnnn−1 a.s., |
for all
lim supt→∞lnn(t)t≤β−σ212−αlim inft→∞∫t0Co(s)dst<0 a.s.. |
Next we prove part b. Suppose that
lim supt→∞lnn(t,ω)t≤0 | (13) |
Moreover, from the law of large numbers for local martingales (Theorem 3.4 in [17]) there exists a set
limt→∞M(t,ω)t=limt→∞B1(t,ω)t=0. |
From (12) we get:
ln(n(t))t=ln(n(0))t+(β−σ212)−α∫t0Co(s)dst−∫t0(γn(s)+σ222n2(s))dst+σ1B1(t)t+M(t,ω)t |
Hence, for any
lim supt→∞lnn(t,ω)t=(β−σ212)−αlim inft→∞∫t0Co(s,ω)dst |
Since we know that
We have the following result regarding the expectation of
Lemma 3.6. There exists a constant
Proof. Using Itô's formula in (5) we get:
d(etn(t))=n(t)et(1+β−αCo(t)−γn(t))dt+σ1n(t)etdB1(t)−σ2n2(t)etdB2(t)≤n(t)et(1+β−γn(t))dt+σ1n(t)etdB1(t)−σ2n2(t)etdB2(t)≤et(1+β)24γdt+σ1n(t)etdB1(t)−σ2n2(t)etdB2(t) | (14) |
Let
ηm=inf{t≥0:n(t)∉(1/m,m)}, | (15) |
for any
E[et∧τmn(t∧τm)]≤n(0)+E[∫t∧τm0es(1+β)24γds]≤n(0)+(1+β)24γ(et−1). |
Letting
E[n(t)]≤n(0)et+(1+β)24γ(1−e−t). |
Thus, there exists a constant
Corollary 1. For any
Proof. For any
P(n(t)>c1(ϵ))≤E[n(t)]c1(ϵ). |
Hence, from Lemma 3.6 we get
lim supt→∞P(n(t)>c1(ϵ))≤lim supt→∞E[n(t)]c1(ϵ)≤ϵ. |
Theorem 3.7. If
Proof. First we show that
By Itô's formula in (5) we get for any real constant c:
d(ectn(t))=ect(1n(t)(c−β+σ21+αCo(t))+γ+σ22n(t))dt−σ1ectn(t)dB1(t)+σ2ectdB2(t) |
Since
d(ectn(t))≤ect(γ+σ22n(t))dt−σ1ectn(t)dB1(t)+σ2ectdB2(t) | (16) |
Taking expectation in (16) and using Lemma 3.6 we get:
E[ec(t∧ηm)n(t∧ηm)]≤1n(0)+E[∫t∧ηm0ecs(γ+σ22n(s))ds]≤1n(0)+(γ+σ22K1)(ect−1)c, |
where
E[1n(t)]≤1n(0)ect+(γ+σ22K1)c(1−e−ct), |
so
Next we show that for any
For any
P(n(t)<c2(ϵ))=P(1n(t)>1c2(ϵ))≤c2(ϵ)E[1n(t)] |
Hence
lim supt→∞P(n(t)<c2(ϵ))≤ϵlim supn→∞E[1/n(t)]/M2≤ϵ. |
Thus
The deterministic system (1)-(3) has a maximum capacity equilibrium point
For stochastic differential equations, invariant and stationary distributions play the same role as fixed points for deterministic differential equations. In general, let
dX(t)=b(X(t))dt+d∑r=1σr(X(t))dBr(t), | (17) |
where
L=l∑i=1bi(x)∂∂xi+12l∑i,j=1Ai,j(x)∂2∂xi∂xj, Ai,j(x)=d∑r=1σr,i(x)σr,j(x). |
Let
Definition 4.1. A stationary distribution [6] for
∫EP(t,x,A)μ(dx)=μ(A), for any t≥0, and any A∈B(E). |
Definition 4.2. The Markov process
It is clear that the stability in distribution implies the existence of a unique stationary measure, but the converse is not always true [2]. We have the following result (see lemma 2.2 in [29] and the references therein).
Lemma 4.3. Suppose that there exists a bounded domain
limt→∞P(t,x,B)=μ(B)Px{limT→∞1T∫T0f(X(t))dt=∫Ef(x)μ(dx)}=1, |
for all
We now study the stochastic system (5)-(7) when
dX(t)=(βX(t)−γX2(t))dt+σ1X(t)dB1(t)−σ2X2(t)dB2(t), | (18) |
dXϵ(t)=(βXϵ(t)−γX2ϵ(t)−αϵXϵ(t))dt+σ1Xϵ(t)dB1(t)−σ2X2ϵ(t)dB2(t), | (19) |
Lemma 4.4. a. For any given initial value
b. For any
c. There exists a constant
Proof. The proofs for a. and b. can be done similarly with the proof of Theorem 2.1, using the
Let
Theorem 4.5. If
limt→∞PX(t,x,B)=μ1(B)Px{limT→∞1T∫T0f(X(t))dt=∫Ef(x)μ1(dx)}=1, |
for all
Proof. We consider the
LV(x)=−σ228x5/2+σ224x2−γ2x3/2+γ2x+(β2−σ218)x1/2+(σ214−β2). |
Since
Let
Let define the processes
dN(t)=(N(t)(σ21−β)+αN(t)Co(t)+γ+σ22N(t))dt−σ1N(t)dB1(t)+σ2dB2(t) a.s., | (20) |
dY(t)=(Y(t)(σ21−β)+γ+σ22Y(t))dt−σ1Y(t)dB1(t)+σ2dB2(t) a.s., | (21) |
dYϵ(t)=(Yϵ(t)(σ21−β+αϵ)+γ+σ22Yϵ(t))dt−σ1Yϵ(t)dB1(t)+σ2dB2(t) a.s.. | (22) |
From the proof of Theorem 3.7 we know that if
Lemma 4.6. If
Proof. The proof is based on the results in Lemma 4.4 and it is similar with the first part of the proof of Theorem 3.7. For completeness we have included it in Appendix C.
We use the processes
Theorem 4.7. Let
Proof. We follow the same idea as in the proof of Theorem 2.4 in [28]. From theorem 4.5 we know that
Firstly, let's notice that
Y(t)≤N(t) and Y(t)≤Yϵ(t) for any t≥0 a.s.. | (23) |
Indeed, if we denote
dξ(t)=(ξ(t)(σ21−β−σ22N(t)Y(t))+αN(t)Co(t))dt−σ1ξ(t)dB1(t) a.s.. |
The solution of the previous linear equation is given by (see chapter 3, [17])
ξ(t)=Φ(t)∫t0αN(s)Co(s)Φ(s)ds a.s., |
where
Φ(t)=exp{−t(β−σ212)−∫t0σ22N(s)Y(s)ds−σ1B1(t)}>0 |
Obviously
Secondly we show that for any
liminft→∞(Yϵ(t)−N(t))≥0 a.s.. | (24) |
From equations (20) and (22) we get
d(Yϵ(t)−N(t))=((Yϵ(t)−N(t))(σ21+αϵ−β−σ22N(t)Yϵ(t))+αN(t)(ϵ−Co(t)))dt−σ1(Yϵ(t)−N(t))dB1(t) a.s.. |
The solution of the linear equation is given by
Yϵ(t)−N(t)=Φ1(t)∫t0αN(s)(ϵ−Co(s))Φ1(s)ds a.s., |
where
0<Φ1(t)=exp{−t(β−αϵ−σ212)−∫t0σ22N(s)Yϵ(s)ds−σ1B1(t)}≤exp{−t(β−αϵ−σ212+σ1B1(t)t)} |
Since
\begin{align} Y_\epsilon(t)-N(t)& = \Phi_1(t)\left(\int_0^T \frac{\alpha N(s) \left(\epsilon-C_o(s)\right)}{\Phi_1(s)}ds+\int_T^t \frac{\alpha N(s) \left(\epsilon-C_o(s)\right)}{\Phi_1(s)}ds\right)\notag\\ &\ge \Phi_1(t)\int_0^T \frac{\alpha N(s) \left(\epsilon-C_o(s)\right)}{\Phi_1(s)}ds\notag \end{align} |
Therefore for any
\liminf\limits_{t\rightarrow\infty}(Y_\epsilon(t)-N(t))\ge \lim\limits_{t\rightarrow\infty} \Phi_1(t)\int_0^T \frac{\alpha N(s) \left(\epsilon-C_o(s)\right)}{\Phi_1(s)}ds = 0 ~~~a.s.. |
Thirdly we prove that
\lim\limits_{\epsilon\rightarrow 0}\lim\limits_{t\rightarrow\infty}E[Y_\epsilon(t)-Y(t)] = 0.\label{fact3} | (25) |
We know from (23) that
\begin{align} &d(Y_\epsilon(t)-Y(t)) = \biggl((Y_\epsilon(t)-Y(t))\left(\sigma_1^2+\alpha\epsilon-\beta-\frac{\sigma_2^2}{Y(t)Y_\epsilon(t)}\right)\notag \\&+\alpha \epsilon Y(t)\biggl)dt-\sigma_1(Y_\epsilon(t)-Y(t))dB_1(t)\notag\\ &\le \biggl((Y_\epsilon(t)-Y(t))\left(\sigma_1^2+\alpha\epsilon-\beta\right)+\alpha \epsilon Y(t)\biggl)dt-\sigma_1(Y_\epsilon(t)-Y(t))dB_1(t)\text{ a.s.}.\notag \end{align} |
From Lemma 4.6 we know that
\begin{align} &E[Y_\epsilon(t)-Y(t)]\le \int_0^t E[Y_\epsilon(s)-Y(s)]\left(\sigma_1^2+\alpha\epsilon-\beta\right)+\alpha \epsilon E[Y(s)]ds\notag\\ &\le \int_0^t E[Y_\epsilon(s)-Y(s)]\left(\sigma_1^2+\alpha\epsilon-\beta\right)ds+t\alpha \epsilon\sup\limits_{t\ge 0}E[Y(t)]\text{ a.s.}.\notag \end{align} |
For any
0\le E[Y_\epsilon(t)-Y(t)]\le\frac{\alpha\epsilon \sup\limits_{t\ge 0}E[Y(t)] }{\beta-\sigma_1^2-\alpha\epsilon}\left(1-\exp(-t(\beta-\sigma_1^2-\alpha\epsilon))\right) |
Taking limits in the previous inequality we get equation (25).
Finally, using (23), (24), and (25) we obtain that
Corollary 2. Let
a. If
\begin{align} &p(x) = \frac{1}{G_1 x^4}\exp\left(-\frac{\beta}{\sigma_2^2}\left(\frac{1}{x}-\frac{\gamma}{\beta}\right)^2\right), x>0\label{dens1} \end{align} | (26) |
\begin{align} &G_1 = \frac{\sigma_2}{2\beta^{5/2}}\left(\Psi\left(\frac{\gamma\sqrt{2\beta}}{\beta\sigma_2}\right)\sqrt{\pi}(\sigma_2^2\beta+2\gamma^2)+ \gamma\sigma_2\beta^{1/2}\exp\left(-\frac{\gamma^2}{\sigma_2^2 \beta}\right)\right)\label{dens2} \end{align} | (27) |
where
b. If
Proof. We know that
a. If
dY(t) = \left(-Y(t)\beta+\gamma+\frac{\sigma_2^2}{Y(t)}\right)dt+\sigma_2dB_2(t)\text{ a.s.}, \label{eqi2new} | (28) |
Let define
\begin{align} q(y)& = \exp\left(-\frac{2}{\sigma_2^2}\int_1^y\left( -\beta u+\frac{\sigma_2^2}{u}+\gamma\right) du\right) = \frac{1}{y^2}\exp\left(-\frac{\beta}{\sigma_2^2}\left(1-\frac{\gamma}{\beta}\right)^2\right)\notag\\ &\exp\left(\frac{\beta}{\sigma_2^2}\left(y-\frac{\gamma}{\beta}\right)^2\right)\notag \end{align} |
It can be easily shown that
\int_0^1 q(y)dy = \infty, ~~ \int_1^\infty q(y)dy = \infty, ~~\int_0^\infty\frac{1}{\sigma_2^2 q(y)}dy = \frac{G_1}{\sigma_2^2}\exp\left(\frac{\beta}{\sigma_2^2}\left(1-\frac{\gamma}{\beta}\right)^2\right), \notag |
where
p_1(x) = \frac{1}{\sigma_2^2q(x)\int_0^\infty\frac{1}{\sigma_2^2 q(y)}dy} = \frac{x^2\exp\left(-\frac{\beta}{\sigma_2^2}\left(x-\frac{\gamma}{\beta}\right)^2\right)}{G_1} |
Thus, by Theorem 4.5,
\begin{align} &\lim\limits_{t\rightarrow \infty}\frac{1}{t}\int_0^tX(u)du = \int_0^\infty xp(x)dx = \frac{\sigma_2}{2\beta^{3/2} G_1}\left( \sigma_2\sqrt{\beta}\exp\biggl(-\frac{\gamma^2}{\sigma_2^2 \beta}\right)\notag\\ &+2\gamma \sqrt{\pi}\Psi\left(\frac{\gamma\sqrt{2\beta}}{\beta\sigma_2}\right)\biggl)\text{ a.s..}\notag \end{align} |
b. If
dY(t) = \left(\gamma-Y(t)(\beta-\sigma_1^2)\right)dt-\sigma_1Y(t)dB_1(t)\text{ a.s.}. |
Proceeding similarly as for a. we can show that
\lim\limits_{t\rightarrow \infty}\frac{1}{t}\int_0^tX(u)du = \left(\frac{2(\beta-\sigma_1^2)}{\sigma_1^2}+1\right)\frac{\sigma_1^2}{2\gamma}\text{ a.s..} |
Notice that if
On the other hand, if
First we illustrate numerically the results obtained in section 3 regarding survival analysis.We consider a cell population exposed to the toxicant monastrol as in the experiments described in [1]. The parameters' values for this toxicant are estimated in [1]:
One of the applications of the mathematical model is for finding the threshold value for
We illustrate this for the model with initial values
Notice also that the results displayed in Figs. 2 and 3 agree with the conclusion of Theorem 2.3. For the stochastic model with
\liminf\limits_{t\rightarrow\infty} n(t, \omega)>0, \lim\limits_{t\rightarrow \infty}C_o(t, \omega) = \lim\limits_{t\rightarrow \infty}C_e(t, \omega) = 0 |
Next we use the same parameters values as stated at the beginning of this section and the initial values
When
We present a stochastic model to study the effect of toxicants on human cells. To account for parameter uncertainties, the model is expressed as a system of coupled ordinary stochastic differential equations. The variables are the cell index
We first prove the positivity of the solutions. Then we investigate the influence of noise on the cell population survival. When the noise variances
Moreover, we prove that when the noise variance
Here we illustrate our results for the toxicant monastrol. We have also considered other toxicants from the experiments described in [1] classified in various clusters [30]. We have noticed that the cluster type does not change the type of stationary distribution, nor has an effect on the behavior of the distributions in response to increased noise variances.
Proof. The proof is similar with the proof of Lemma 3.1 in [1]. We define the stopping time
From (8) with
\begin{align} &0\le C_o(t, \omega) = \lambda_1^2 e^{-\eta_2^2 t}\int_0^t C_e(s, \omega) e^{\eta_1^2 s} ds\le \lambda_1^2 e^{-\eta_2^2 t}C_e(0)\int_0^t e^{\eta_1^2 s} ds\notag\\ & = \frac{\lambda_1^2 C_e(0)}{\eta_1^2}\left(1-e ^{-\eta_1^2t}\right)\le \frac{\lambda_1^2 C_e(0)}{\eta_1^2}\notag \end{align} |
Moreover, on
\begin{align} \frac{dC_e}{dt}\biggl |_{t = \tau}& = \lambda_2^2C_o(\tau)n(\tau)-\eta_2^2 C_e(\tau)n(\tau)\notag\\ &\le C_e(0) n(\tau)\left(\frac{\lambda_1^2 \lambda_2^2}{\eta_1^2}-\eta_2^2\right)<0\notag \end{align} |
Thus we have a contradiction with the definition of
Proof. The proof is similar with the proof of Theorem 3.2 in [1]. Let
If
\int_0^t C_o(s, \omega) n(s, \omega)\exp{\left(\eta_2 ^2\int_0^s n(l, \omega)dl\right )}ds\le \frac{\lambda_1^2 C_e(0)}{\eta_1^2}\exp{\left(\eta_2 ^2|n|_1(\omega)\right )}|n|_1(\omega). |
Thus
\lim\limits_{t\rightarrow\infty} C_e(t, \omega) = C_e(0)\exp(-\eta_2^2|n|_1(\omega))+\lambda_2^2 M(\omega)\exp(-\eta_2^2|n|_1(\omega))<\infty. |
Consequently, there exists
\begin{align} &\int_0^t C_e(s, \omega) e^{\eta_1^2s}ds\ge \int_{T_1(\omega)}^t C_e(s, \omega) e^{\eta_1^2s}ds\notag\\ &\ge C_e(0)\exp(-\eta_2^2|n|_1(\omega))/2\int_{T_1(\omega)}^t e^{\eta_1^2s}ds.\notag \end{align} |
So we can apply L'Hospital's rule in (8), and we get
\lim\limits_{t\rightarrow\infty} C_o(t, \omega) = \frac{\lambda_1^2}{\eta_1^2}\lim\limits_{t\rightarrow\infty} C_e(t, \omega)>0. |
Thus, on
Next, if
0\le \frac{\lambda_2^2}{\eta_2^2}\liminf\limits_{t\rightarrow\infty}C_o(t, \omega)\le \liminf\limits_{t\rightarrow\infty}C_e(t, \omega)\le \limsup\limits_{t\rightarrow\infty}C_e(t, \omega)\le \frac{\lambda_2^2}{\eta_2^2}\limsup\limits_{t\rightarrow\infty}C_o(t, \omega) |
Similarly, from (8) we either get that
0\le \frac{\lambda_1^2}{\eta_1^2}\liminf\limits_{t\rightarrow\infty}C_e(t, \omega)\le \liminf\limits_{t\rightarrow\infty}C_o(t, \omega)\le \limsup\limits_{t\rightarrow\infty}C_o(t, \omega)\le \frac{\lambda_1^2}{\eta_1^2}\limsup\limits_{t\rightarrow\infty}C_e(t, \omega), |
(if
\lim\limits_{t\rightarrow\infty}C_o(t, \omega) = \lim\limits_{t\rightarrow\infty}C_e(t, \omega) = 0, |
because
In conclusion, on
Proof. We choose any
\begin{align} &d(e^{ct}Y(t)) = e^{ct}\biggl(Y(t)(c+\sigma_1^2-\beta)+\gamma +\sigma_2^2 X(t)\biggl)dt-\sigma_1e^{ct}Y(t)dB_1(t)\notag\\ &+\sigma_2e^{ct}dB_2(t)\le e^{ct}\biggl(\gamma +\sigma_2^2 X(t)\biggl)dt-\sigma_1e^{ct}Y(t)dB_1(t)+\sigma_2e^{ct}dB_2(t) \label{eqa1} \end{align} | (29) |
Let
\begin{align} &E\left[e^{c (t\wedge \tau_m)} Y(t\wedge \tau_m)\right]\le \frac{1}{n(0)}+E\left[\int_0^{t\wedge \tau_m} e^{cs}\left(\gamma+\sigma_2^2X(s)\right) ds\right]\notag\\ &\le \frac{1}{n(0)}+\left(\gamma+\sigma_2^2 C_1\right)\frac{(e^{ct}-1)}{c}.\notag \end{align} |
Letting
E\left[Y(t)\right]\le \frac{1}{n(0)e^{ct} }+\frac{\left(\gamma+\sigma_2^2 C_1\right)}{c}(1-e^{-ct}). |
Thus, there exists a constant
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