2 Imprecise stochastic LotkaVolterra food chain chemostat model
In this section, we introduce the imprecise stochastic LotkaVolterra food chain chemostat model. Let S(t), x(t) and y(t) denote the concentrations of nutrient, prey and predator at time t, respectively. Then a stochastic LotkaVolterra food chain chemostat model takes the following form [19] .
dy(t) = [m 2 x(t)y(t) − Dy(t)]dt + σ 3 y(t)dB 3 (t) + Y γ 3 y(t − ) N (dt, du).
where S 0 is the original input concentration of nutrient and D is the input rate from the feed vessel to the culture vessel, as well as the washout rate from the culture vessel to the receptacle. δ 1 and δ 2 are the yield constants for prey growth on nutrient and predator growth on prey, respectively. B i (t), i = 1, 2, 3 are standard Brownian motions. σ 2 i represent the intensities of B i (t). λ is the characteristic measure of N which is defined on a finite measurable subset Y of (0, +∞) with λ(Y) < ∞. γ i (u) : Y × Ω → R are the bounded and continuous functions satisfying γ i (u) > −1, i = 1, 2, 3. The S(t − ), x(t − ) and y(t − ) are the left limits of S(t), x(t) and y(t), respectively. N denotes the compensated random measure defined by N (dt, du) = N (dt, du)λ(du)dt. We assume B i and N are independent throughout the paper and denote
Before we state the imprecise stochastic LotkaVolterra food chain chemostat model, definitions of Intervalvalued function should recalled (Pal [16] ). Addition
Scalar multiplication: αA = α[a l , a u ] = [αa l , αa u ], where α is a positive real number.
LetD,m 1 ,δ 1 ,σ 1 ,m 2 ,δ 2 ,σ 2 ,σ 3 represent the interval numbers of D, m 1 , δ 1 , σ 1 , m 2 , δ 2 , σ 2 , σ 3 , respectively. The system (2.1) with imprecise parameters becomes:
According to the Theorem 1 in Pal et al [12] and considering the interval-valued function f (p) = (f l ) 1−p (f u ) p for intervalf = [f l , f u ] for p ∈ [0, 1], we can prove that system (2.2) is equivalent to the following system:
Assume that there is a positive constant c such that
For convenience in the following investigation, define
and
.
Assume for each m > 0 there exists L m > 0 such that
Theorem 31 Let Assumptions (H1) and (H2) hold. Then for any given initial value (S(0), x(0), y(0)) ∈ R 3 + , system (2.3) has a unique solution (S(t), x(t), y(t)) ∈ R 3 + for all t ≥ 0 almost surely (a.s.).
Proof. Because the coefficients of system (2.3) are local Lipschitz continuous (Mao [17] ), for any given initial value (S(0), x(0), y(0)) ∈ R 3 + , there is a unique local solution (S(t), x(t), y(t)) on t ∈ [0, τ e ), where τ e is the explosion time (see Mao [17] ). In order to show that the local solution is global, we only need to prove that τ e = ∞ a.s. In this context, choosing a sufficiently large number k 0 ≥ 1 such that (S(0), x(0), y(0)) lie within the interval [ 1 k 0 , k 0 ]. For each integer k ≥ k 0 , we define the stopping time as
where inf ∅ = ∞ (∅ being empty set). By the definition, τ k increases as k → ∞. Set τ ∞ = lim k→∞ τ k . Hence τ ∞ ≤ τ e a.s. If τ ∞ = ∞ a.s. is true, then τ e = ∞ a.s. for all t > 0. In other words, we need to verify τ ∞ = ∞ a.s. If this claim is wrong, then there exist a constant T > 0 and an ǫ ∈ (0, 1) such that
Hence there is an interger k 1 ≥ k 0 such that
Consider the Lyapunov function V :
where A, B are positive constants to be determined later. The nonnegativity of this function can be seen from u − 1 − ln u ≥ 0 for any u > 0. Let k ≥ k 1 and T > 0. Then, for any 0 ≤ t ≤ min τ k , T , the Itǒs formula (Situ [18] ) shows that
where L is a differential operator, and
where K is a positive constant and
Therefore,
Taking the expectations on both sides of (3.2), we obtain that
Let Ω k = {τ k ∧ T } for k ≥ k 1 . Then, by (3.1), we know that P(Ω k ) ≥ ǫ. Noting that for every ω ∈ Ω k , there exist S(τ k , ω), x(τ k , ω) and y(τ k , ω), all of which equal either k or 1
Thus, by (3.3), we konw that
where 1 Ω k (ω) represents the indicator function of Ω k (ω). Setting k → ∞ leads to the contradiction ∞ > V (S(0), x(0), y(0)) + KT = ∞.
Therefore, we have τ ∞ = ∞ a.s. The proof is complete. ⊓ ⊔
When studying dynamical models, two of the most interesting issues are persistence and extinction. In this section, we discuss the extinction of microorganisms in system (2.3) and leave its persistence to the next section.
Before we state the main results of this section, several lemmas (gao et al. [19] ) should be recalled without the proofs and relevant explanations.
(H3) Assume that for some θ > 2,
Lemma 41 Let Assumption (H3) hold. For any initial value (S(0), x(0), y(0)) ∈ R 3 + , there is a unique positive solution (S(t), x(t), y(t)) to system (2.3) and the solution will remain in R 3 + with probability one, i.e., (S(t), x(t), y(t)) ∈ R 3 + for all t ≥ 0 a.s. Then
Lemma 42 For any given initial value (S(0), x(0), y(0)) ∈ R 3 + the solution (S(t), x(t), y(t)) of system (2.3) has the following property
Theorem 41 Let Assumption (H3) hold. For any initial value (S(0), x(0), y(0)) ∈ R 3 + , there is a unique positive solution (S(t), x(t), y(t)) to system (2.3) and the solution will remain in R 3 + with probability one, i.e., (S(t), That is to say, microbes x and y will become extinct exponentially with probability one.
Proof. Integrating from 0 to t on both sides of (2.3), yields
x(s)dB 2 (s)
Clearly, we can derive that
x(s)dB 2 (s)
This together with Lemma 41 implies lim t→∞ φ(t) = 0 a.s.
Applying Ito formula to (2.3) we can conclude that
Integrating (4.3) from 0 to t and then dividing by t on both sides, we obtain 
(4.5)
Moreover, according to (2.5) ,
where Clearly, if R s 0 < 1 holds, then taking the superior limit on both sides of (4.5) and combining with (4.2), (4.6), we know that
Similarly, we get It is now easy to derive from (4.8) and (4.9) that lim t→∞ S t = S 0 a.s.
This completes the proof. ⊓ ⊔ Theorem 42 Let Assumption (H3) hold. For any initial value (S(0), x(0), y(0)) ∈ R 3 + , there is a unique positive solution (S(t), x(t), y(t)) to system (2.3) and the solution will remain in R 3 + with probability one, i.e., (S(t),
Proof. According to (4.5), we know that
This together with (4.2), (4.6) and Zhao et al. (2013) implies
Thus, by (4.8) we get It then follows from (4.5) that
This together with Lemma 42 implies lim inf Therefore, from (4.11 ) and ( 4.13) we have
(4.14)
It follows from (4.1), (4.2), (4.12) and (4.14) that
This completes the proof. ⊓ ⊔
In this section, we establish sufficient conditions for persistence in the mean of system (2.3).
lim inf t→∞ y t > 0 a.s.
Theorem 51 Let Assumption (H3) hold. For any initial value (S(0), x(0), y(0)) ∈ R 3 + , there is a unique positive solution (S(t), x(t), y(t)) to system (2.3) and the solution will remain in R 3 + with probability one, i.e., (S(t), x(t), y(t)) ∈ R 3 + for all t ≥ 0 a.s. If R s 1 > 1, then
Proof. According to (4.8), we know that
This together with (4.5) implies
which yields
].
(5.2)
Taking the inferior limit on both sides of (5.2) and combining with Lemma 42, from (4.2) and (4.6) we have lim inf
Therefore, by the condition R s 1 > 1, we can easily obtain (5.1). ⊓ ⊔