Hindawi Publishing Corporation Computational and Mathematical Methods in Medicine Volume 2016, Article ID 5218163, 14 pages http://dx.doi.org/10.1155/2016/5218163
Research Article The Dynamical Behaviors in a Stochastic SIS Epidemic Model with Nonlinear Incidence Ramziya Rifhat, Qing Ge, and Zhidong Teng College of Mathematics and Systems Science, Xinjiang University, Urumqi 830046, China Correspondence should be addressed to Zhidong Teng; zhidong
[email protected] Received 8 February 2016; Accepted 22 May 2016 Academic Editor: Chuangyin Dang Copyright Β© 2016 Ramziya Rifhat et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. A stochastic SIS-type epidemic model with general nonlinear incidence and disease-induced mortality is investigated. It is proved Μ 0 < 1 and together with Μ 0 . That is, when π
that the dynamical behaviors of the model are determined by a certain threshold value π
Μ an additional condition, the disease is extinct with probability one, and when π
0 > 1, the disease is permanent in the mean in probability, and when there is not disease-related death, the disease oscillates stochastically about a positive number. Furthermore, Μ 0 > 1, the model admits positive recurrence and a unique stationary distribution. Particularly, the effects of the intensities when π
of stochastic perturbation for the dynamical behaviors of the model are discussed in detail, and the dynamical behaviors for the stochastic SIS epidemic model with standard incidence are established. Finally, the numerical simulations are presented to illustrate the proposed open problems.
1. Introduction Our real life is full of randomness and stochasticity. Therefore, using stochastic dynamical models can gain more real benefits. Particularly, stochastic dynamical models can provide us with some additional degrees of realism in comparison to their deterministic counterparts. There are different possible approaches which result in different effects on the epidemic dynamical systems to include random perturbations in the models. In particular, the following three approaches are seen most often. The first one is parameters perturbation; the second one is the environmental noise that is proportional to the variables; and the last one is the robustness of the positive equilibrium of the deterministic models. In recent years, various types of stochastic epidemic dynamical models are established and investigated widely. The main research subjects include the existence and uniqueness of positive solution with any positive initial value in probability mean, the persistence and extinction of the disease in probability mean, the asymptotical behaviors around the disease-free equilibrium and the endemic equilibrium of the deterministic models, and the existence of the stationary distribution as well as ergodicity. Many important results
have been established in many literatures, for example, [1β16] and the references cited therein. Particularly, for stochastic SI type epidemic models, in [6], Gray et al. constructed a stochastic SIS epidemic model with constant population size where the authors not only obtained the existence of the unique global positive solution with any positive initial value, but also established the threshold value conditions; that is, the disease dies out or persists. Furthermore, in the case of the persistence, the authors also showed the existence of a stationary distribution and finally computed the mean value and variance of the stationary distribution. However, from articles [1β16] and the references cited therein, we see that there are still many important problems which are not studied completely and impactfully. For example, see the following. (1) The stochastic epidemic models with general nonlinear incidence are not investigated. Up to now, only some special cases of nonlinear incidence, for example, saturated incidence rate, are considered. But, we all know that the nonlinear incidence rate in the theory of mathematical epidemiology is very important.
2
Computational and Mathematical Methods in Medicine (2) For the stochastic epidemic models with the standard incidence, up to now, we do not find any interesting researches. (3) The conditions obtained on the existence of unique stationary distribution are very rigorous. Whether there is a unique stationary distribution only when the model is permanent in the mean with probability one is still an open problem.
Motivated by the above work, in this paper, we consider the following deterministic SIS epidemic model with nonlinear incidence rate and disease-induced mortality: ππ (π‘) = Ξ β π½π (π (π‘) , πΌ (π‘)) + πΎπΌ (π‘) β ππ (π‘) , ππ‘ ππΌ (π‘) = π½π (π (π‘) , πΌ (π‘)) β (π + πΎ + πΌ) πΌ (π‘) . ππ‘
(1)
In model (1), π and πΌ denote the susceptible and infectious individuals, Ξ denotes the recruitment rate of the susceptible, π is the natural death rate of π and πΌ, πΌ is the disease-related death rate, the transmission of the infection is governed by a nonlinear incidence rate π½π(π, πΌ), where π½ denotes the transmission coefficient between compartments π and πΌ, π(π, πΌ) is a continuously differentiable function of π and πΌ, and πΎ denotes the per capita disease contact rate. Now, we assume that the random effects of the environment make the transmission coefficient π½ of disease in deterministic model (1) generate random disturbance. That Μ is, π½ β π½ + ππ΅(π‘), where π΅(π‘) is a one-dimensional standard Brownian motion defined on some probability space. Thus, model (1) will become into the following stochastic SIS epidemic model with nonlinear incidence rate: ππ (π‘) = [Ξ β π½π (π (π‘) , πΌ (π‘)) + πΎπΌ (π‘) β ππ (π‘)] ππ‘ β ππ (π (π‘) , πΌ (π‘)) ππ΅ (π‘) , ππΌ (π‘) = [π½π (π (π‘) , πΌ (π‘)) β (π + πΎ + πΌ) πΌ (π‘)] ππ‘
(2)
+ ππ (π (π‘) , πΌ (π‘)) ππ΅ (π‘) . In this paper, we investigate the dynamical behaviors of model (2). By using the Lyapunov function method, ItΛoβs formula, and the theory of stochastic analysis [17, 18], we will establish a series of new interesting criteria on the extinction of the disease, permanence in the mean of the model with probability one. The stochastic oscillation of the disease about a positive number in the case where there is not diseaserelated death is also obtained. Further, we study the positive recurrence and the existence of stationary distribution for model (2), and a new criterion is established. Particularly, the effects of the intensities of stochastic perturbation for the dynamical behaviors of the model are discussed in detail. For some special cases of nonlinear incidence π(π, πΌ), for example, π(π, πΌ) = ππΌ/π (standard incidence) and π(π, πΌ) = β(π)π(πΌ), many idiographic criteria on the extinction, permanence, and stationary distribution are established. Lastly, some affirmative answers for the open problems which are proposed in this paper also are given by the numerical
examples (the numerical simulation method can be found in [19]). The organization of this paper is as follows. In Section 2, the preliminaries are given, and some useful lemmas are introduced. In Section 3, the sufficient conditions are established which ensure that the disease dies out with probability one. In Section 4, we establish the sufficient conditions which ensure that the disease in model (2) is permanent in the mean with probability one, and when there is not diseaserelated death the disease oscillates stochastically about a positive number. In Section 5, the existence on the unique stationary distribution of model (2) is proved. In Section 6, the numerical simulations are carried out to illustrate some open problems. Lastly, a brief discussion is given in the end to conclude this work.
2. Preliminaries Denote π
+2 = {(π₯1 , π₯2 ) : π₯1 > 0, π₯2 > 0}, π
+0 = [0, β), and π
+ = (0, β). Throughout this paper, we assume that model (2) is defined on a complete probability space (Ξ©, {πΉπ‘ }π‘β₯0 , π) with a filtration {πΉπ‘ }π‘β₯0 satisfying the usual conditions; that is, {πΉπ‘ }π‘β₯0 is right continuous and πΉ0 contains all π-null sets. In model (2), π and πΌ denote the susceptible and infected fractions of the population, respectively, and π = π + πΌ is the total size of the population among whom the disease is spreading; the parameters Ξ, π, π½, and πΎ are given as in model (1); the transmission of the infection is governed by a nonlinear incidence rate π½ππ(πΌ); π΅(π‘) denotes onedimensional standard Brownian motion defined on the above probability space; and π represents the intensity of the Brownian motion π΅(π‘). Throughout this paper, we always assume the following. (H) π(π, πΌ) is two-order continuously differentiable for any π β₯ 0, πΌ β₯ 0, and π + πΌ > 0. For each fixed πΌ β₯ 0, π(π, πΌ) is increasing for π > 0 and for each fixed π β₯ 0, π(π, πΌ)/πΌ is decreasing for πΌ > 0. π(π, 0) = π(0, πΌ) = 0 for any π > 0 and πΌ > 0, and ππ(π0 , 0)/ππΌ > 0, where π0 = Ξ/π. Particularly, when π(π, πΌ) = β(π)π(πΌ), then assumption (H) becomes in the following form: (Hβ ) β(π) and π(πΌ) are continuously differentiable for π β₯ 0 and πΌ β₯ 0, β(π) is increasing for π β₯ 0, and π(πΌ)/πΌ is decreasing for πΌ > 0. Remark 1. From (H), by simple calculating, we can obtain that for any π > 0 and πΌ > 0, 0 β€ π(π, πΌ) β€ (ππ(π, 0)/ππΌ)πΌ, and for any π2 > π1 > 0, ππ(π2 , 0)/ππΌ β₯ ππ(π1 , 0)/ππΌ. Remark 2. When π(π, πΌ) = ππΌ/π (standard incidence), where π = π+πΌ, π(π, πΌ) = ππΌ/(1+π1 πΌ+π2 π) (Beddington-DeAngelis incidence) with constants π1 β₯ 0 and π2 β₯ 0, and π(π, πΌ) = ππΌ/(1 + ππΌ2 ) with constant π β₯ 0, then (H) is satisfied. Now, we give the following result for function π(π, πΌ).
Computational and Mathematical Methods in Medicine
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Lemma 3. For any constants π > π > 0, let π· = {(π, πΌ) : π > 0, πΌ > 0, π β€ π + πΌ β€ π}. Then, π (π, πΌ) π (π, πΌ) , } < β, (π,πΌ)βπ· π πΌ σ΅¨σ΅¨ 1 ππ (π, πΌ) π (π, πΌ) σ΅¨σ΅¨ σ΅¨σ΅¨ 1 ππ (π, πΌ) σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ max {σ΅¨σ΅¨σ΅¨ β σ΅¨,σ΅¨ σ΅¨} < β. (π,πΌ)βπ· σ΅¨σ΅¨ πΌ ππΌ πΌ2 σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ πΌ ππ σ΅¨σ΅¨σ΅¨ max {
lim
π (π, πΌ) ππ (0, πΌ) = , π ππ
π+πΎ+πΌ
, (8)
2
Μ 0 = π
0 β π
π2 (ππ (π0 , 0) /ππΌ) 2 (π + πΎ + πΌ)
.
We have that π
0 is the basic reproduction number of deterministic model (1). On the extinction of the disease in probability for model (2) we have the following result. Theorem 5. Assume that one of the following conditions holds: (b) π2 > π½2 /2(π + πΎ + πΌ). Then disease πΌ in model (2) is extinct with probability one. That is, for any initial value (π(0), πΌ(0)) β π
+2 , solution (π(π‘), πΌ(π‘)) of model (2) has limπ‘ββ πΌ(π‘) = 0 a.s.
πΌ > 0, πΌ = 0, (π, πΌ) β π·, (6)
1 ππ (π, πΌ) { , πΌ > 0, { { { πΌ ππΌ πΊ (π, πΌ) = { 2 { π π (π, 0) { { , πΌ = 0, { ππΌππ
π½ (ππ (π0 , 0) /ππΌ)
Μ 0 < 1; (a) π2 β€ π½/(ππ(π0 , 0)/ππΌ) and π
Hence, conclusion (3) holds. Define the functions 1 ππ (π, πΌ) π (π, πΌ) { , β { { { πΌ ππΌ πΌ2 π» (π, πΌ) = { 2 { 1 π π (π, 0) { { , { 2 ππΌ2
π
0 =
(4)
(5)
π (π, πΌ) ππ (π, 0) lim = . πΌβ0 π ππΌ
Define the constants
(3)
The proof of Lemma 3 is simple. In fact, from (H), we have
πβ0
3. Extinction of the Disease
ππ (π0 + π, 0) π (π (π‘) , πΌ (π‘)) β (0, ]. πΌ (π‘) ππΌ
(π, πΌ) β π·.
1 ππ (π, πΌ) π2 π (π, 0) lim = . πΌβ0 πΌ ππ ππΌππ
(9)
With ItΛoβs formula (see [17, 18]), we have π log πΌ (π‘) = [π½
Using the LβHospital principle, from (H), we have 1 ππ (π, πΌ) π (π, πΌ) 1 π2 π (π, 0) lim ( ) = , β πΌβ0 πΌ ππΌ πΌ2 2 ππΌ2
Proof. By Lemma 4 we have (π(π‘), πΌ(π‘)) β π
+2 a.s. for all π‘ β₯ 0 and lim supπ‘ββ (π(π‘)+πΌ(π‘)) β€ π0 . For any π > 0 there is π0 > 0 such that π(π‘) + πΌ(π‘) < π0 + π for all π‘ β₯ π0 . Hence, for any π‘ β₯ π0 ,
β (7)
This shows that π»(π, πΌ) and πΊ(π, πΌ) are continuous for (π, πΌ) β π·. Therefore, conclusion (4) also is true. Next, on the existence of global positive solutions and the ultimate boundedness of solutions for model (2) with probability one, we have the result as follows. Lemma 4. For any initial value (π(0), πΌ(0)) β π
+2 , model (2) has a unique solution (π(π‘), πΌ(π‘)) defined on π‘ β π
+0 satisfying (π(π‘), πΌ(π‘)) β π
+2 for all π‘ β₯ 0 with probability one. Furthermore, when πΌ > 0 then π0 β€ lim inf π‘ββ π(π‘) β€ lim supπ‘ββ π(π‘) β€ π0 , and when πΌ = 0 then limπ‘ββ π(π‘) = π0 , where π(π‘) = π(π‘) + πΌ(π‘) and π0 = Ξ/(π + πΌ). Lemma 4 can be proved by using the method which is given in [6]. We hence omit it here.
β
π (π (π‘) , πΌ (π‘)) β (π + πΎ + πΌ) πΌ (π‘)
π2 π (π (π‘) , πΌ (π‘)) 2 ) ] ππ‘ + π ( 2 πΌ (π‘)
(10)
π (π (π‘) , πΌ (π‘)) ππ΅ (π‘) . πΌ (π‘)
Hence, for any π > 0, log πΌ (π‘) log πΌ (0) π½ + π π‘ π (π (π ) , πΌ (π )) ππ β€ + β« π‘ π‘ π‘ πΌ (π ) 0 β (π + πΎ + πΌ) β
π2 1 π‘ π (π (π ) , πΌ (π )) 2 ) ππ β« ( 2 π‘ 0 πΌ (π )
+
π π‘ π (π (π ) , πΌ (π )) ππ΅ (π ) . β« π‘ 0 πΌ (π )
(11)
Define a function π (π’) = (π½ + π) π’ β
π2 2 π’ β (π + πΎ + πΌ) . 2
(12)
4
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When π = 0, π(π’) is monotone increasing for π’ β π
+ , and when π > 0, π(π’) is monotone increasing for π’ β [0, (π½ + π)/π2 ) and monotone decreasing for π’ β [(π½ + π)/π2 , β). If condition (a) holds, then when π = 0, from (9), we directly have π(
ππ (π0 + π, 0) π (π (π‘) , πΌ (π‘)) ) β€ π( ) πΌ (π‘) ππΌ
βπ‘ β₯ π0 . (13)
When π > 0, since ππ(π0 , 0)/ππΌ β€ π½/π2 , we can choose π > 0 such that π β€ π and ππ(π0 + π, 0)/ππΌ < (π½ + π)/π2 . From (9) we also have inequality (13). Hence, when π‘ β₯ π0 , π (π (π ) , πΌ (π )) log πΌ (π‘) log πΌ (0) 1 π‘ ) ππ β€ + β« π( π‘ π‘ π‘ 0 πΌ (π ) + β€
π π‘ π (π (π ) , πΌ (π )) ππ΅ (π ) β« π‘ 0 πΌ (π )
π (π (π ) , πΌ (π )) log πΌ (0) 1 π0 ) ππ + β« π( π‘ π‘ 0 πΌ (π )
Now, we give a further discussion for conditions (a) and (b) of Theorem 5 by using the intensity π of stochastic perturbation and basic reproduction number π
0 of deterministic model (1). Μ 0 < 1, and it is easy When π
0 β€ 1, then, for any π > 0, π
to prove that one of the conditions (a) and (b) of Theorem 5 holds. Therefore, for any π > 0, the conclusions of Theorem 5 Μ 0 = 1 we have hold. Let 1 < π
0 β€ 2. From π
πβπ=
β2 (π + πΎ + πΌ) (π
0 β 1) ππ (π0 , 0) /ππΌ
.
(20)
π½ β2 (π + πΎ + πΌ)
, (21)
ππ (π0 + π, 0) log πΌ (π‘) lim sup β€ π( ) π‘ ππΌ π‘ββ
(15)
a.s.
From the arbitrariness of π and π, we further obtain 0
ππ (π , 0) 1 2 ππ (π , 0) log πΌ (π‘) lim sup β€π½ β π ( ) π‘ ππΌ 2 ππΌ π‘ββ
2
(16)
β (π + πΎ + πΌ)
If condition (b) holds, then, since π > 0, π(π’) has maximum value (π½ + π)2 /2π2 β (π + πΎ + πΌ) at π’ = (π½ + π)/π2 , and for any π‘ β₯ 0, we have 2
(π½ + π) π (π (π‘) , πΌ (π‘)) β (π + πΎ + πΌ) , )β€ πΌ (π‘) 2π2
(17)
2
π (π (π ) , πΌ (π )) π ππ΅ (π ) . β« π‘ 0 πΌ (π )
Since π1 β€ π2 , we easily prove that when π > π one of the conditions (a) and (b) of Theorem 5 holds. Therefore, for any π > π, the conclusions of Theorem 5 hold. When π
0 > 2, we have π1 > π2 and π1 β₯ π β₯ π2 . Hence, condition (a) in Theorem 5 does not hold. We only can obtain that for any π > π1 the conclusions of Theorem 5 hold. Summarizing the above discussions we have the following result as a corollary of Theorem 5.
(a) π
0 β€ 1 and π > 0; (b) 1 < π
0 β€ 2 and π > π; (c) π
0 > 2 and π > π1 . Then disease πΌ in model (2) is extinct with probability one. Corollary 7. Let π(π, πΌ) = ππΌ/π (standard incidence). Assume that one of the following conditions holds:
which implies log πΌ (π‘) log πΌ (0) (π½ + π) β (π + πΎ + πΌ) β€ + π‘ π‘ 2π2
π½ π2 = β . 0 ππ (π , 0) /ππΌ
Corollary 6. Assume that one of the following conditions holds:
Μ 0 β 1) < 0 a.s. = (π + πΎ + πΌ) (π
+
(19)
a.s.
From (16) and (19) we finally have limπ‘ββ πΌ(π‘) = 0 a.s. This completes the proof.
π1 =
By the large number theorem for martingales (see [17] or Lemma A.1 given in [9]), we obtain
π‘
π‘ββ
π½2 log πΌ (π‘) β€ 2 β (π + πΎ + πΌ) < 0 π‘ 2π
Denote
π π‘ π (π (π ) , πΌ (π )) + β« ππ΅ (π ) . π‘ 0 πΌ (π )
π½π (
lim sup
(14)
ππ (π0 + π, 0) 1 + π( ) (π‘ β π0 ) π‘ ππΌ
0
With the large number theorem for martingales and arbitrariness of π, we obtain
Μ 0 = π½/(π + πΎ + πΌ) β π2 /2(π + πΎ + πΌ) < 1; (a) π2 β€ π½ and π
(18)
(b) π2 > π½2 /2(π + πΎ + πΌ). Then disease πΌ in model (2) is extinct with probability one.
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Corollary 8. Let π(π, πΌ) = β(π)π(πΌ). Assume that (Hβ ) holds and one of the following conditions holds: Μ 0 = π½β(π0 )πσΈ (0)/(π + πΎ + πΌ) β (a) π2 β€ π½/β(π0 )πσΈ (0) and π
2 2 0 σΈ π (β(π )π (0)) /2(π + πΎ + πΌ) < 1; (b) π2 > π½2 /2(π + πΎ + πΌ). Then disease πΌ in model (2) is extinct with probability one. Remark 9. It is easy to see that in Theorem 5 the conditions π
0 > 2 and π β€ π β€ π1 are not included. Therefore, an interesting conjecture for model (2) is proposed; that is, if the above condition holds, then the disease still dies out with probability one. In Section 6, we will give an affirmative answer by using the numerical simulations; see Example 1. Μ0 = Remark 10. In the above discussions, we see that case π
1 has not been considered. An interesting open problem Μ 0 = 1 the disease in model (2) also is is whether when π
extinct with probability one. A numerical example is given in Section 6; see Example 2.
4. Permanence of the Disease On the permanence of the disease in the mean with probability one for model (2), we establish the following results. Μ 0 > 1, then disease πΌ in model (2) is Theorem 11. If π
permanent in the mean with probability one. That is, there is a constant ππΌ > 0 such that, for any initial value (π(0), πΌ(0)) β π
+2 , solution (π(π‘), πΌ(π‘)) of model (2) satisfies π‘
1 lim inf β« πΌ (π ) ππ β₯ ππΌ π‘ββ π‘ 0
π.π .
(22)
Μ 0 > 1, we choose a small enough constant π > 0 Proof. From π
such that 0
π½
ππ (π , 0) ππΌ
0
2
ππ (π + π, 0) 1 β (π + πΎ + πΌ) β π2 ( ) 2 ππΌ
(23)
> 0. By Lemma 4, it is clear that, for any initial value (π(0), πΌ(0)) β π
+2 , solution (π(π‘), πΌ(π‘)) of model (2) satisfies π‘ lim supπ‘ββ (1/π‘) β«0 πΌ(π )ππ β€ π0 and for above π > 0 there is π0 > 0 such that π0 β π β€ π(π‘) + πΌ(π‘) β€ π0 + π a.s. for all π‘ β₯ π0 . Denote the set π·π = {(π, πΌ) : π0 β π β€ π + πΌ β€ π0 + π}. Since ππ(π‘) = (Ξ β ππ(π‘) β πΌπΌ(π‘))ππ‘, we obtain for any π‘ > π0 π‘
β« (π (π ) β π0 ) ππ = β π0
π+πΌ π‘ β« πΌ (π ) ππ π π0
π (π0 ) β π (π‘) + . π
From (10), for any π‘ β₯ π0 , ππ (π0 , 0)
0
ππΌ
0 π (π (π ) , πΌ (π )) ππ (π , 0) β + ] ππ β (π + πΎ πΌ (π ) ππΌ
(25)
2
π‘
π (π (π ) , πΌ (π )) 1 ) ππ + πΌ) π‘ β π2 β« ( 2 πΌ (π ) 0 + πβ«
π‘
0
π (π (π ) , πΌ (π )) ππ΅ (π ) . πΌ (π )
Since π(π, πΌ)/πΌ for π > 0 and πΌ > 0 is continuously differentiable, limπΌβ0 (π(π, πΌ)/πΌ) = ππ(π, 0)/ππΌ exists for any π > 0, and set π·π is convex and connected, by the Lagrange mean value theorem when π‘ β₯ π0 we have 0 π (π (π‘) , πΌ (π‘)) ππ (π , 0) β πΌ (π‘) ππΌ
=(
1 ππ (π (π‘) , π (π‘)) π (π (π‘) , π (π‘)) ) πΌ (π‘) β π (π‘) ππΌ π2 (π‘)
+
(26)
1 ππ (π (π‘) , π (π‘)) (π (π‘) β π0 ) , π (π‘) ππ
where (π(π‘), π(π‘)) β π·π . Let constants σ΅¨σ΅¨ 1 ππ (π, πΌ) π (π, πΌ) σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ π1 = max {σ΅¨σ΅¨σ΅¨ β σ΅¨} , (π,πΌ)βπ·π σ΅¨σ΅¨ πΌ ππΌ πΌ2 σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ 1 ππ (π, πΌ) σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ π2 = max {σ΅¨σ΅¨σ΅¨ σ΅¨} . (π,πΌ)βπ·π σ΅¨σ΅¨ πΌ ππ σ΅¨σ΅¨σ΅¨
(27)
From Lemma 3 we have 0 < π1 , π2 < β. For any π‘ β₯ π0 , we have 1 ππ (π (π‘) , π (π‘)) π (π (π‘) , π (π‘)) β₯ βπ1 β π (π‘) ππΌ π2 (π‘) 1 ππ (π (π‘) , π (π‘)) β€ π2 π (π‘) ππ
a.s., (28) a.s.
From (25) and Remark 1 we further have log πΌ (π‘) = log πΌ (0) + π½ β«
π0
0
+π½ (24)
π‘
log πΌ (π‘) = log πΌ (0) + π½ β« [
ππ (π0 , 0) ππΌ π‘
+ π½ β« [( π0
π (π (π ) , πΌ (π )) ππ πΌ (π )
(π‘ β π0 )
1 ππ (π (π‘) , π (π‘)) π (π‘) ππΌ
6
Computational and Mathematical Methods in Medicine β
π (π (π‘) , π (π‘)) 1 ) πΌ (π ) + 2 π (π‘) π (π‘)
π=π½
ππ (π0 , 0) ππΌ
β (π + πΎ + πΌ) 2
ππ (π0 + π, 0) 1 β π2 ( ) , 2 ππΌ
ππ (π (π‘) , π (π‘)) β
(π (π ) β π0 )] ππ β (π + πΎ ππ π‘ π (π (π ) , πΌ (π )) 2 1 + πΌ) π‘ β π2 β« ( ) ππ‘ 2 πΌ (π ) 0
π0 = π½ (π1 + π2
π+πΌ ). π (30)
π‘
π (π (π ) , πΌ (π )) + πβ« ππ΅ (π ) β₯ log πΌ (0) πΌ (π ) 0 + π½β«
π0
0
By the large number theorem for martingales and Lemma 4, limπ‘ββ (π»(π‘)/π‘) = 0 a.s. Therefore, from Lemma 5.2 given in π‘ [16], we finally obtain lim inf π‘ββ (1/π‘) β«0 πΌ(π )ππ β₯ π/π0 a.s. This completes the proof.
0
ππ (π , 0) π (π (π ) , πΌ (π )) ππ + π½ (π‘ β π0 ) πΌ (π ) ππΌ π‘
π‘
π0
π0
Μ 0 > 1 is equivalent to Remark 12. From (20), we have that π
π < π. Therefore, Theorem 11 also can be rewritten by using intensity π of stochastic perturbation in the following form: if π < π, then disease πΌ in model (2) is permanent in the mean with probability one.
β π½π1 β« πΌ (π ) ππ + π½π2 β« (π (π ) β π0 ) ππ 2
ππ (π0 + π, 0) 1 β (π + πΎ + πΌ) π‘ β π2 ( ) π‘ 2 ππΌ + πβ«
π‘
0
+ π½β«
π0
0
0
ππ (π , 0) π (π (π ) , πΌ (π )) ππ‘ + π½ (π‘ β π0 ) πΌ (π ) ππΌ π‘
β π½π1 β« πΌ (π ) ππ β π½π2 π0
+ π½π2 β
π ( + πβ«
ππΌ
π‘
π+πΌ π‘ β« πΌ (π ) ππ π π0
Theorem 14. Susceptible π in model (2) also is permanent in the mean with probability one. That is, there is a constant ππ > 0 such that, for any initial value (π(0), πΌ(0)) β π
+2 , solution (π(π‘), πΌ(π‘)) of model (2) satisfies
1 1 (π (π0 ) β π (π‘)) β (π + πΎ + πΌ) π‘ β π 2
ππ (π0 + π, 0)
2
Remark 13. Combining Corollary 6 and Remark 12 we can obtain that when 1 < π
0 β€ 2, number π is a threshold value. When 0 < π < π, the disease πΌ in model (2) is permanent in the mean and when π > π, the disease πΌ is extinct with probability one. However, when π
0 > 2, then the alike results are not established. Therefore, it yet is an interesting open problem.
π (π (π ) , πΌ (π )) ππ΅ (π ) = log πΌ (0) πΌ (π )
2
) π‘
π (π (π ) , πΌ (π )) ππ΅ (π ) = π» (π‘) + ππ‘ πΌ (π )
0 π‘
β π0 β« π (π ) ππ , 0
(29) where π» (π‘) = log πΌ (0) + π½ β«
π0
0
π (π (π ) , πΌ (π )) ππ πΌ (π )
0
βπ½
ππ (π , 0) ππΌ
π0
+ π½ (π1 + π2 + π½π2 + πβ«
π‘
0
π0 π+πΌ ) β« πΌ (π ) ππ π 0
1 (π (π0 ) β π (π‘)) π π (π (π‘) , πΌ (π )) ππ΅ (π ) , πΌ (π )
1 π‘ (31) lim inf β« π (π ) ππ β₯ ππ π.π . π‘ββ π‘ 0 Proof. By Lemma 4 we easily see that, for any initial value (π(0), πΌ(0)) β π
+2 , solution (π(π‘), πΌ(π‘)) of model (2) satisfies π‘ lim supπ‘ββ (1/π‘) β«0 π(π )ππ β€ π0 and for any small enough constant π > 0 there is π0 > 0 such that π0 β π β€ π(π‘) + πΌ(π‘) β€ π0 + π for all π‘ β₯ π0 . Hence, by Lemma 3, when π‘ β₯ π0 we have π(π(π‘), πΌ(π‘)) β€ ππ π(π‘), where ππ = maxπ·π {π(π, πΌ)/π} < β. Integrating the first equation of model (2) we obtain for any π‘ β₯ π0 π (π‘) β π (0) π‘ =Ξβ
1 π‘ β« [π½π (π (π ) , πΌ (π )) + ππ (π ) β πΎπΌ (π )] ππ π‘ 0
π π‘ β« π (π (π ) , πΌ (π )) ππ΅ (π ) π‘ 0 1 π0 β₯ Ξ β β« [π½π (π (π ) , πΌ (π )) + ππ (π )] ππ π‘ 0 1 π‘ β β« [π½ππ + π] π (π ) ππ π‘ π0 β
β
π π‘ β« π (π (π ) , πΌ (π )) ππ΅ (π ) . π‘ 0
(32)
Computational and Mathematical Methods in Medicine
7
Therefore, with the large number theorem for martingales, we finally have 1 π‘ Ξ lim inf β« π (π ) ππ β₯ . π‘ββ π‘ 0 π½ππ + π
(33)
This completes the proof.
Proof. From Lemma 4, we know that limπ‘ββ (π(π‘)+πΌ(π‘)) = π0 . Without loss of generality, we assume that π(π‘) + πΌ(π‘) β‘ π0 for all π‘ β₯ 0. From (10), for any π‘ β₯ 0, π‘
log πΌ (π‘) = log πΌ (0) + β« [π½ 0 [
π (π0 β πΌ (π ) , πΌ (π )) πΌ (π ) 2
0 π2 π (π β πΌ (π ) , πΌ (π )) ] ) ππ β (π + πΎ) β ( 2 πΌ (π ) ]
As consequences of Theorems 11 and 14, we have the following corollaries. Corollary 15. Let π(π, πΌ) = ππΌ/π (standard incidence). If Μ 0 = (π½β(1/2)π2 )/(π+πΎ+πΌ) > 1, then model (2) is permanent π
in the mean with probability one.
π‘
+β« π 0
(38)
π (π (π‘) , πΌ (π )) ππ΅ (π ) . πΌ (π )
Define a function π’(πΌ) = π(π0 β πΌ, πΌ)/πΌ. Then, for any π‘ β₯ 0, π‘
Corollary 16. Let π(π, πΌ) = β(π)π(πΌ). Assume that (Hβ ) holds Μ 0 = π½β(π0 )πσΈ (0)/(π + πΎ + πΌ) β π2 (β(π0 )πσΈ (0))2 /2(π + and π
πΎ + πΌ) > 1; then model (2) is permanent in the mean with probability one.
log πΌ (π‘) = log πΌ (0) + β« π (π’ (πΌ (π ))) ππ
We further have the result on the weak permanence of model (2) in probability.
where function π(π’) = π½π’β(π2 /2)π’2 β(π+πΎ). With condition Μ 0 > 1 we have π(0) = β(π + πΎ) < 0 and π
Μ 0 > 1. Then there is a constant Corollary 17. Assume that π
π > 0 such that, for any initial value (π(0), πΌ(0)) β π
+2 , solution (π(π‘), πΌ(π‘)) of model (2) satisfies
0
(39)
π‘
π (π (π ) , πΌ (π )) +β« π ππ΅ (π ) , πΌ (π ) 0
π(
ππ (π0 , 0) ππΌ
0
2
0
ππ (π , 0) π2 ππ (π , 0) )=β ( ) +π½ 2 ππΌ ππΌ (40) β (π + πΎ) > 0.
lim sup πΌ (π‘) β₯ π, π‘ββ
lim sup π (π‘) β₯ π
(34)
π‘ββ
a.s. Now, we discuss special case: πΌ = 0 for model (2); that is, there is not disease-related death in model (2). We can establish the following more precise results on the weak permanence of the disease in probability compared to the conclusion given in Corollary 17. Μ 0 > 1, then, for any Theorem 18. Let πΌ = 0 in model (2). If π
initial value (π(0), πΌ(0)) β π
+2 , solution (π(π‘), πΌ(π‘)) of model (2) satisfies lim sup πΌ (π‘) β₯ π π.π .,
(35)
lim inf πΌ (π‘) β€ π π.π .,
(36)
π‘ββ
π‘ββ
where π > 0 satisfies the equation π+πΎ , π = 0, { { π (π β π, π) { { π½ ={ 2 (π + πΎ) { π , π > 0. { { 2 β 2π2 (π + πΎ) β π½ + π½ { 0
(37)
Hence, π(π’) = 0 has a positive root π in (0, ππ(π0 , 0)/ππΌ) which is π+πΎ , π = 0, { { { { π½ π={ 2 (π + πΎ) { , π > 0. { { 2 β 2π2 (π + πΎ) β π½ + π½ {
(41)
Since π’(πΌ) is monotone decreasing for πΌ β (0, π0 ), π’(π0 ) = 0, and lim+ π’ (πΌ) = lim+
πΌβ0
πΌβ0
π (π0 β πΌ, πΌ) πΌ
=
ππ (π0 , 0) ππΌ
,
(42)
there is a unique π β (0, π0 ) such that π’(π) = π(π0 βπ, π)/π = π and π(π’(π)) = π(π) = 0. When π > 0 and π½/π2 < ππ(π0 , 0)/ππΌ, since function π(π’) has maximum value π(π½/π2 ) at π’ = π½/π2 and π(π½/π2 ) > π(ππ(π0 , 0)/ππΌ), there is a unique ΜπΌ, such that π’(ΜπΌ) = π½/π2 . From π β (0, ππ(π0 , 0)/ππΌ) and π(π) = 0 we have π < π½/π2 . Hence, 0 < ΜπΌ < π < π0 . From the above discussion we obtain that π(π’(πΌ)) > 0 is strictly increasing on πΌ β (0, ΜπΌ), π(π’(πΌ)) > 0 is strictly decreasing on πΌ β (ΜπΌ, π), and π(π’(πΌ)) < 0 is strictly decreasing on πΌ β (π, π0 ). When π2 β€ π½/(ππ(π0 , 0)/ππΌ), similarly to the above discussion, we can obtain that π(π’(πΌ)) > 0 is strictly decreasing
8
Computational and Mathematical Methods in Medicine
on πΌ β (0, π) and π(π’(πΌ)) < 0 is strictly decreasing on πΌ β (π, π0 ). Now, we firstly prove that (35) is true. If it is not true, then there is an enough small π0 β (0, 1) such that π(Ξ©1 ) > π0 , where Ξ©1 = {lim supπ‘ββ πΌ(π‘) < π}. Hence, for every π β Ξ©1 , there is a constant π1 = π1 (π) β₯ π0 such that πΌ (π‘) β€ π β π0
βπ‘ β₯ π1 .
(43)
With the above discussion we know that π(π’(πΌ(π‘))) β₯ π(π’(π β π0 )) > 0 for all π‘ β₯ π1 . From (39) we further obtain for any π‘ β₯ π1 π1
log πΌ (π‘) β₯ log πΌ (0) + β« π (π’ (πΌ (π ))) ππ 0
+ π (π’ (π β π0 )) (π‘ β π1 ) π‘
+β« π 0
(44)
a more less positive π than number π such that any solution (π(π‘), πΌ(π‘)) of model (2) with initial value (π(0), πΌ(0)) β π
+2 , lim inf π‘ββ πΌ(π‘) β₯ π a.s. In Section 6, we will give an affirmative answer by using the numerical simulations; see Example 3. From Theorem 18, we easily see that number π will arise from the change when the noise intensity π changes. Therefore, it is very interesting and important to discuss how number π changes along with the change of π. We have the following result. Theorem 20. Assume that πΌ = 0 in model (2) and Μ 0 > 1. Let number π be given in Theorem 18 and π
0 = π
π½(ππ(π0 , 0)/ππΌ)/(π + πΎ). Then one has the following. (a) π as the function of π is defined for
π (π (π ) , πΌ (π )) ππ΅ (π ) . πΌ (π )
0 π1 , where Ξ©2 = {lim inf π‘ββ πΌ(π‘) > π}. Hence, for every π β Ξ©2 , there is π2 = π2 (π) β₯ π0 such that πΌ (π‘) β₯ π + π1
βπ‘ β₯ π2 .
ππ (π0 , 0) /ππΌ
(47)
Μ. fl π
Μ ). (b) π is monotone decreasing for π β (0, π (c) limπβ0 π = πΌβ , where (πβ , πΌβ ) is the endemic equilibrium of deterministic model (1). (d) If 1 β€ π
0 β€ 2, then limπβΜπ π = 0, and if π
0 > 2, then limπβΜπ π = π2 , where π2 satisfies
(45)
With the above discussion we have π(π’(πΌ(π‘))) β€ π(π’(π+π1 )) < 0 for all π‘ β₯ π2 . Together with (39), we further obtain for any π‘ β₯ π2
β2 (π + πΎ) (π
0 β 1)
π (π0 β π2 , π2 ) π2
=
ππ (π0 , 0) /ππΌ π
0 β 1
.
(48)
Proof. Since
π2
log πΌ (π‘) = log πΌ (0) + β« π (π’ (πΌ (π ))) ππ
π (π0 β π, π)
0
π
π‘
+ β« π (π’ (πΌ (π ))) ππ +β« π 0
π (π (π ) , πΌ (π )) ππ΅ (π ) πΌ (π )
(46) π=
π2
β€ log πΌ (0) + β« π (π’ (πΌ (π ))) ππ 0
+ π (π’ (π + π1 )) (π‘ β π2 ) π‘
+β« π 0
(49)
by the inverse function theorem we obtain that π as the function of π is defined for π β (0, ππ(π0 , 0)/ππΌ). From
π2 π‘
= π,
π (π (π ) , πΌ (π )) ππ΅ (π ) . πΌ (π )
With the large number theorem for martingales, we have lim supπ‘ββ (log πΌ(π‘)/π‘) β€ π(π’(π + π1 )) < 0, which implies πΌ(π‘) β 0 as π‘ β β. This leads to a contradiction with (45). This completes the proof. Μ 0 > 1 and πΌ = Remark 19. Theorem 18 indicates that if π
0, then any solution (π(π‘), πΌ(π‘)) of model (2) with initial value (π(0), πΌ(0)) β π
+2 oscillates about a positive number π. Therefore, an interesting open problem is whether there is
π½ β βπ½2 β 2π2 (π + πΎ) π2
(50)
,
Μ. we can obtain that π β (0, ππ(π0 , 0)/ππΌ) when 0 < π < π Μ. Therefore, π as a function of π is defined for 0 < π < π Computing the derivative of π with respect to π, we have 2 (π + πΎ) ππ β2π½ = 3 + ππ π πβπ½2 β 2π2 (π + πΎ) + =
2βπ½2 β 2π2 (π + πΎ)
(51)
π3
2π½2 β 2π2 (π + πΎ) β 2π½βπ½2 β 2π2 (π + πΎ) π3 βπ½2 β 2π2 (π + πΎ)
.
Computational and Mathematical Methods in Medicine
9 decreasing as π increases. Thus, both limπβ0 π and limπβΜπ π exist. Let limπβ0 π = π1 and limπβΜπ π = π2 . We have
Since 2
[2π½2 β 2π2 (π + πΎ)] β (2π½βπ½2 β 2π2 (π + πΎ))
2
lim π = lim
(52)
πβ0
πβ0
2
4
= 4π (π + πΎ) > 0, we have ππ/ππ > 0. From the definition of π, we easily see that π is monotone decreasing for π. From (49) and (H), we obtain that ππ/ππ exists and is continuous for π. Since (π/ππ)(π(π0 β π, π)/π) < 0, we have ππ/ππ < 0. Therefore, ππ/ππ = (ππ/ππ)(ππ/ππ) < 0. It follows that π is monotone
lim π =
π½ β βπ½2 β 2Μ π2 (π + πΎ) Μ2 π
πβΜ π
=
lim
π
πβΜ π
=
ππ (π0 , 0) ππΌ
πβΜ π
.
(55)
=
(ππ (π0 , 0) /ππΌ) (π + πΎ) π½ (ππ (π0 , 0) /ππΌ) β (π + πΎ) ππ (π0 , 0) /ππΌ π
0 β 1
lim
π (π β π, π) π
π2
π½ (ππ (π0 , 0) /ππΌ)
π (π0 β π, π) π
0
=
ππ (π , 0) /ππΌ π
0 β 1
.
=
ππ (π0 , 0) /ππΌ (π
0 β 1)
.
= >
(57)
β (π + πΎ)
>
π+πΎ ; π½
(59)
(ππ (π0 , 0) /ππΌ) (π + πΎ) π½ (ππ (π0 , 0) /ππΌ) β (π + πΎ) π+πΎ = π½
π (π0 β πΌβ , πΌβ ) πΌβ
(60)
,
where (πβ , πΌβ ) is the endemic equilibrium of deterministic model (1). Hence,
Therefore, we have limπβΜπ π = π2 , and π2 satisfies π (π0 β π2 , π2 )
(ππ (π0 , 0) /ππΌ) (π + πΎ)
namely,
, πβΜ π
πβΜ π
(54)
Remark 21. When π
0 > 2, then from (56) we obtain
πβΜ π
which implies
.
of model (2) with initial value (π(0), πΌ(0)) β π
+2 , which implies that π = 0. Therefore, when π
0 > 2, we can propose an interesting open problem: whether there is a critical value π, π1 ) such that when π β (0, πβ ) we have the fact that πβ β (Μ π is monotonically decreasing and π > 0 and when π > πβ we have π = 0.
lim π =
(56)
(53)
Hence, limπβ0 (π(π0 β π, π)/π) = limπβ0 π = (π + πΎ)/π½. This shows that π(π0 β π1 , π1 )/π1 = (π + πΎ)/π½. Let (πβ , πΌβ ) be the endemic equilibrium of deterministic model (1); then we have π(π0 β πΌβ , πΌβ )/πΌβ = (π + πΎ)/π½. Hence, π1 = πΌβ . This shows that limπβ0 π = πΌβ . On the other hand, we have
lim
0
π+πΎ . π½
2 (π½ (ππ (π0 , 0) /ππΌ) β (π + πΎ))
This shows that limπβΜπ π = 0. If π
0 > 2, then we have from (54) lim π =
π½ + βπ½2 β 2π2 (π + πΎ)
=
σ΅¨ σ΅¨ (ππ (π0 , 0) /ππΌ) (π½ (ππ (π0 , 0) /ππΌ) β σ΅¨σ΅¨σ΅¨σ΅¨π½ (ππ (π0 , 0) /ππΌ) β 2 (π + πΎ)σ΅¨σ΅¨σ΅¨σ΅¨)
If 1 β€ π
0 β€ 2, then from (54) we obtain limπβΜπ π = ππ(π0 , 0)/ππΌ. Hence, π (S0 β π, π)
2 (π + πΎ)
(58)
This completes the proof. Conclusion (b) of Theorem 20 shows that when πΌ = 0 in model (2), number π monotonically decreases when π Μ ), and when π = 0, π has a maximum value increases in (0, π πΌβ by Conclusion (c). Therefore, 0 < π < πΌβ when π > 0. If Μ , π has a minimum value 0 and 1 β€ π
0 β€ 2, then when π = π Μ , π has a minimum value π2 > 0 by if π
0 > 2 then when π = π Conclusion (d). Μ = π from It is clear that when in model (2) πΌ = 0 then π (20). On the other hand, from Conclusion (c) of Corollary 7, we see that if π
0 > 2 then when π > π1 , where π1 is given in (21), we have limπ‘ββ πΌ(π‘) = 0 a.s. for any solution (π(π‘), πΌ(π‘))
π (π0 β π2 , π2 ) π2
>
π (π0 β πΌβ , πΌβ ) πΌβ
.
(61)
Consequently, 0 < π2 < πΌβ . Remark 22. When π(π, πΌ) = ππΌ, we easily validate that Theorems 20 and 24 degenerate into Theorems 5.1 and 5.4 which are given in [19], respectively. Therefore, Theorems 18 and 20 are the considerable extension of Theorems 5.1 and 5.4 in general nonlinear incidence cases, respectively. Remark 23. For the case πΌ > 0 in model (2), an interesting Μ 0 > 1 whether we and important open problem is when π
also can establish similar results as Theorems 18 and 20. Furthermore, as an improvement of the results obtained in
10
Computational and Mathematical Methods in Medicine
Corollary 17 we also propose another open problem: only Μ 0 > 1 we also can establish the permanence of the when π
disease with probability one; that is, there is a constant π > 0 such that, for any solution (π(π‘), πΌ(π‘)) of model (2) with initial value (π(0), πΌ(0)) β π
+2 , one has limπ‘ββ πΌ(π‘) β₯ π, a.s. In Section 6, we will give an affirmative answer by using the numerical simulations; see Example 3.
and 0 < V < 1 is a constant. Computing πΏΞ¨1 , by Remark 1, we have πΏΞ¨1 = βπΌβ(V+1) (π½π (π, πΌ) β (π + πΌ + πΎ) πΌ) + β
π2 πΌβ(V+2) π2 (π, πΌ) β€ πΌβV (π + πΌ + πΎ 0
5. Stationary Distribution Μ 0 > 1 model From Theorems 11 and 14 we obtain that when π
(2) is permanent in the mean with probability one. However, Μ 0 > 1 model (2) also has a stationary distribution. We when π
have an affirmative answer as follows. Μ 0 > 1, then model (2) is positive recurrent Theorem 24. If π
and has a unique stationary distribution. Proof. Here, the method given in the proof of Theorem 5.1 in [17] is improved and developed. By Lemma 4 and Remark 9 we only need to give the proof in region Ξ, where Ξ = {(π, πΌ) : π β₯ 0, πΌ β₯ 0, π0 β€ π + πΌ β€ π0 }. Let (π(π‘), πΌ(π‘)) be any solution of model (1) with (π(0), πΌ(0)) β Ξ a.s. for all π‘ β₯ 0. Let π > 0 be a large enough constant, and let
1 (1 + V) 2
2
(66)
0
ππ (π , 0) ππ (π , 0) 1 + (1 + V) π2 ( ) βπ½ ) 2 ππΌ ππΌ + πΌβV π½ (
ππ (π0 , 0) ππΌ
β
π (π, πΌ) ). πΌ
Applying the Lagrange mean value theorem, we have ππ (π0 , 0)
β
ππΌ
π (π, πΌ) 1 ππ (π, π) 0 = (π β π) πΌ π ππ +(
(67) π (π, π) 1 ππ (π, π) β )πΌ 2 π π ππΌ
β€ π1 (π0 β π) + π2 πΌ + π3 π
, where (π, π) β Ξ and
π· = {(π, πΌ) β Ξ :
1 1 1 1 < π < π0 β , < πΌ < π0 β } . (62) π π π π
When (π, πΌ) β Ξ \ π·, then either 0 < π < 1/π or 0 < πΌ < 1/π. The diffusion matrix for model (56) is 2 2
2 2
π π (π, πΌ) βπ π (π, πΌ) π΄ (π, πΌ) = ( 2 2 ). βπ π (π, πΌ) π2 π2 (π, πΌ) 2 2
2
(π,πΌ)βΞ
1 ππ (π, πΌ) }, πΌ ππ
(63)
0
(64)
(68)
π (π, πΌ) 1 ππ (π, πΌ) π2 = max { β }. (π,πΌ)βΞ πΌ2 πΌ ππΌ By Lemma 3, we have 0 β€ π1 ; π2 < β. We hence have 0
2
ππ (π , 0) 1 πΏΞ¨1 β€ πΌ (π + πΌ + πΎ + (1 + V) π2 ( ) 2 ππΌ βV
For any (π, πΌ) β π· we have π π (π, πΌ) β₯ π (π(1/π, π β 1/π)/(ππ0 β 1))2 . Choose a Lyapunov function as follows: π (π, πΌ) = Ξ¨1 (πΌ) + Ξ¨2 (π, πΌ) + Ξ¨3 (π) ,
π1 = max {
βπ½
ππ (π0 , 0) ππΌ
(69) ) + π½π1 (π0 β π) πΌβV + π½π2 πΌ1βV .
Computing πΏΞ¨2 , by Remark 1, we have 1 πΏΞ¨2 = β πΌβV (Ξ β ππ β π½π (π, πΌ) + πΎπΌ) β πΌβ(V+1) (π0 V β π) (π½π (π, πΌ) β (π + πΌ + πΎ) πΌ) +
1 (1 + V) 2
1 β
π2 π2 (π, πΌ) πΌβ(V+2) (π0 β π) β πΌβ(V+1) π2 π2 (π, πΌ) 2
where
1 = β πΌβV (π (π0 β π) β π½π (π, πΌ) + πΎπΌ) V
1 Ξ¨1 (πΌ) = πΌβV , V 1 Ξ¨2 (π, πΌ) = πΌβV (π0 β π) , V 1 Ξ¨3 (π) = , π
(65)
β πΌβV (π0 β π) (π½ β
π2 (
π (π, πΌ) 1 β (π + πΌ + πΎ)) + (1 + V) πΌ 2
π (π, πΌ) 2 βV 0 π (π, πΌ) 2 1βV ) πΌ (π β π) β π2 ( ) πΌ πΌ πΌ
Computational and Mathematical Methods in Medicine = πΌβV (π0 β π) (β +
11
π π (π, πΌ) +π+πΌ+πΎβπ½ V πΌ
and when V > 0 is small enough, it follows that 0
2
ππ (π , 0) 1 ) π + πΌ + πΎ + (1 + V) π2 ( 2 ππΌ
2
π (π, πΌ) π½ π (π, πΌ) 1 ) ) + πΌ1βV ( (1 + V) π2 ( 2 πΌ V πΌ
π (π, πΌ) 2 1 πΏ β π2 ( ) ) β πΌβV+1 β€ πΌβV (π0 β π) 2 πΌ V
βπ½
ππ (π0 , 0) ππΌ
< 0,
(74) 0
β
(β
+
ππ (π , 0) π 1 β + π + πΌ + πΎ + (1 + V) π2 ( ) V 2 ππΌ
π +π+πΌ+πΎ V
1 (1 + V) π2 ( 2
ππ (π0 , 0) ππΌ
2
) )+
+ π½π1 < 0;
0 π½ ππ (π , 0)
V
we finally obtain that when π > 0 is large enough
ππΌ
πΏπ < β1 a.s. β (π, πΌ) β Ξ \ π·.
2
π (π, πΌ) 1 πΏ β
πΌ1βV β π2 ( ) πΌ1βV β πΌβV+1 . 2 πΌ V Computing πΏΞ¨3 , we have
πΎ πΎ Ξ 1 β 2 πΌ β€ β 2 + (π + π½π0 + π2 π02 ) β 2 πΌ π π π π
This completes the proof.
(71)
where, by Lemma 3, π0 = maxΞ {π(π, πΌ)/π} < β. From the above calculations, we obtain that for any (π, πΌ) β Ξ \ π· 0
ππ (π0 , 0) ππΌ
) + πΌβV (π0 β π) (β 0
2
2
(72)
π½ ππ (π , 0) 1 Ξ )β 2 + (π + π½π0 V ππΌ 2π 2π
2
+ π2 π02 ) . Since 0
2
0
ππ (π , 0) ππ (π , 0) 1 π + πΌ + πΎ + π2 ( 1; then model (2) is positive recurrent and has a unique stationary distribution. Combining Corollary 6, Theorem 11, Remark 12, Theorem 24, and Remark 26, we can finally establish the following summarization result by using intensity π of stochastic perturbation and basic reproduction number π
0 of deterministic model (1).
0
β
(π½π2 +
Particularly, for some special cases of nonlinear incidence π(π, πΌ), we have the following idiographic results on the stationary distribution as the consequences of Theorem 24. Corollary 27. Let π(π, πΌ) = ππΌ/π (standard incidence). If Μ 0 = (π½ β (1/2)π2 )/(π + πΎ + πΌ) > 1, then model (2) is positive π
recurrent and has a unique stationary distribution.
π +π+πΌ+πΎ V
ππ (π , 0) 1βV 1 ) + π½π1 ) + (π0 ) + (1 + V) π2 ( 2 ππΌ
Remark 25. Comparing Theorem 24 with Theorem 6.2 given in [19], we see that Theorem 6.2 is extended and improved to the general stochastic SIS epidemic model (2). Μ 0 > 1 is equivalent to π < π, we also have Remark 26. Since π
that if π < π, then model (2) is positive recurrent and has a unique stationary distribution.
2 πΎ Ξ 1 β€β 2 + (π + π½π0 + π2 π02 ) β 2 πΌ, 2π 2Ξ π
ππ (π , 0) 1 ) πΏπ β€ πΌ (π + πΌ + πΎ + (1 + V) π2 ( 2 ππΌ
(76)
= 1.
1 1 (Ξ β ππ β π½π (π, πΌ) + πΎπΌ) + 3 π2 π2 (π, πΌ) π2 π
π (π, πΌ) 2 1 π (π, πΌ) 1 Ξ π β€β 2 + +π½ + π2 ( ) π π π π π π
βπ½
1 π β« (π (π‘) , πΌ (π‘)) ππ‘ = β« (π, πΌ) π (π (π, πΌ))} πββ π 0 Ξ
π { lim
βV
(75)
From Theorem 2.2, given in [10], we know that model (2) has a unique stationary distribution π such that (70)
πΏΞ¨3 = β
2
(73)
Corollary 29. (a) Let π
0 β€ 1. Then for any π > 0 the disease in model (2) is extinct with probability one. (b) Let 1 < π
0 β€ 2. Then for any 0 < π < π model (2) is permanent in the mean with probability one and has a unique stationary distribution, and for any π > π the disease in model (2) is extinct with probability one.
12
Computational and Mathematical Methods in Medicine 2
1.8 1.6
1.5
1.4 1.2
1 I(t)
I(t)
1
0.5
0.8 0.6 0.4
0
0.2 0
β0.5
0
50
100
150 Time T
200
250
300
Stochastic Deterministic
β0.2
0
50
100
150 Time T
200
250
300
Stochastic Deterministic (a)
(b)
Figure 1: (a) is trajectories of the solution πΌ(π‘) with the initial value πΌ(0) = 0.5 and (b) with the initial value πΌ(0) = 0.06.
(c) Let π
0 > 2. Then for any 0 < π < π model (2) is permanent in the mean with probability one and has a unique stationary distribution, and for any π > π1 , where π1 is given in (20), the disease in model (2) is extinct with probability one.
6. Numerical Simulations In this section we analyze the stochastic behavior of model (2) by means of the numerical simulations in order to make readers understand our results more better. The numerical simulation method can be found in [19]. Throughout the following numerical simulations, we choose π(π, πΌ) = ππΌ/(1 + ππΌ), where π > 0 is a constant. The corresponding discretization system of model (2) is given as follows: ππ+1 = ππ + [Ξ β + πΌπ+1
π½ππ πΌπ + πΎπΌπ β πππ ] Ξπ‘ 1 + πΌπΌπ
ππ πΌπ 1 [πππ βΞπ‘ + π2 (ππ2 β 1) Ξπ‘] , 1 + πΌπΌπ 2
π½π πΌ = πΌπ + [ π π β (π + πΎ) πΌπ ] Ξπ‘ 1 + πΌπΌπ +
(77)
ππ πΌπ 1 [πππ βΞπ‘ + π2 (ππ2 β 1) Ξπ‘] , 1 + πΌπΌπ 2
where ππ (π = 1, 2, . . .) are the Gaussian random variables which follow standard normal distribution π(0, 1). Example 1. In model (2) we choose Ξ = 2000, π½ = 0.60, π = 11, πΎ = 13, π = 0.075, and πΌ = 2. Μ 0 = 0.6715 < 1, By computing we have π
0 = 4.195 > 2, π
2 2 2 π½/π β π = β0.0023 < 0, and π β π½ /2(π + πΎ) = β0.0019 < 0 which is the case of Remark 9. From the numerical 0
simulations, we see that the disease will die out (see Figure 1). An affirmative answer is given for the open problem proposed in Remark 9. Example 2. In model (2), choose Ξ = 2000, π½ = 0.9, π = 30, πΎ = 12, and π = 0.09. Μ 0 = 1. From the numerical By computing we have π
simulations given in Figure 2 we know that the disease will die out. Therefore, an affirmative answer is given for the open problem proposed in Remark 10. Example 3. In model (2) choose Ξ = 2000, π½ = 0.5, π = 30, πΎ = 20, π = 0.02, and πΌ = 2. Μ 0 = 1.200, π
0 = 1.2500, and π = 0.1037. We have π
The numerical simulations are found in Figure 3. We can see that solution πΌ(π‘) of model (2) oscillates up and down at π, which further show that the conclusions of Theorems 14 and 18 are true. At the same time, this example also shows that the disease in model (2) is permanent with probability one. Therefore, an affirmative answer is given for the open problems proposed in Remarks 19 and 23.
7. Discussion In this paper we investigated a class of stochastic SIS epidemic models with nonlinear incidence rate, which include the standard incidence, Beddington-DeAngelis incidence, and nonlinear incidence β(π)π(πΌ). A series of criteria in the probability mean on the extinction of the disease, the persistence and permanence in the mean of the disease, and the existence of the stationary distribution are established. Furthermore, the numerical examples are carried out to illustrate the proposed open problems in this paper.
Computational and Mathematical Methods in Medicine
13
0.7
0.8
0.6
0.7 0.6 0.5
0.4
I(t)
I(t)
0.5
0.3
0.3
0.2
0.2
0.1 0
0.4
0.1 0
50
100 Time T
150
0
200
0
Deterministic Stochastic
50
100 Time T
150
200
Deterministic Stochastic (a)
(b)
Figure 2: (a) is trajectories of the solution πΌ(π‘) with the initial value πΌ(0) = 0.5 and (b) with the initial value πΌ(0) = 0.06. 0.5
0.35
0.45
0.3
0.4
0.25
0.35 I(t)
I(t)
0.3 0.25
0.15
0.2 0.15
0.1
0.1
0.05
0.05 0
0.2
0
50
100
150
200
0
0
50
Time T
100 Time T
150
200
Stochastic Deterministic π
Stochastic Deterministic π (a)
(b)
Figure 3: (a) is trajectories of the solution πΌ(π‘) with the initial value πΌ(0) = 0.5 and (b) with the initial value πΌ(0) = 0.06.
It is easily seen that the research given in [6] for the stochastic SIS epidemic model with bilinear incidence is extended to the model with general nonlinear incidence and disease-induced mortality. Particularly, we see that stochastic SIS epidemic model with standard incidence is investigated for the first time. The researches given in this paper show that stochastic model (2) has more rich dynamical properties than the corresponding deterministic model (1). Particularly, stochastic model (2) has no endemic equilibrium. Thus, this can bring more difficulty for us to investigate model (2), but, on
the other hand, this also makes model (2) have more rich researchful subjects than deterministic model (1). We can discuss not only the extinction, persistence, and permanence in the mean of disease in probability, but also the existence and uniqueness of stationary distribution, the asymptotical behaviors of solutions of stochastic model (2) around the equilibrium of deterministic model (1), and so forth. In addition, we easily see that when intensity π > 0 of Μ 0 . This shows that the stochastic perturbation, then π
0 > π
Μ when π
0 > 1 we still can have π
0 < 1. Therefore, there is a very interesting and important phenomenon; that is, for
14 deterministic model (1) the disease is permanent, but for the corresponding stochastic model (2) the disease is extinct with probability one; see Conclusion (c) of Corollary 29.
Competing Interests The authors declare that they have no competing interests.
Acknowledgments This research is supported by the Doctorial Subjects Foundation of The Ministry of Education of China (Grant no. 2013651110001) and the National Natural Science Foundation of China (Grants nos. 11271312, 11401512, and 11261056).
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