- Email: [email protected]
Critical result on the break-even concentration in a single-species stochastic chemostat model
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.1 (1-10)
J. Math. Anal. Appl. ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Critical result on the break-even concentration in a single-species stochastic chemostat model Dianli Zhao ∗ , Sanling Yuan College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
a r t i c l e
i n f o
Article history: Received 23 July 2015 Available online xxxx Submitted by V. Pozdnyakov Keywords: Stochastic chemostat model Break-even concentration Extinction Stationary distribution
a b s t r a c t is identified as the break-even In this paper, the strictly critical parameter λ > S 0 suffices for extinction concentration for the stochastic chemostat model. λ of the microorganism for any noise intensity σ 2 > 0 is proved firstly, which completely solves the open problem proposed in [10]. Then, by using the Feller = S0, test, we show that the microorganism will go extinct in probability if λ < S 0 , in addition which has not been studied in the known literature. When λ to the previously discussed persistence in mean, the stochastic persistence of the microorganism is studied by use of Chebyshev’s inequality. Besides, the existence of the stationary distribution is proved for the considered model. Numerical simulations are introduced finally to support the obtained results. © 2015 Elsevier Inc. All rights reserved.
1. Introduction A classical chemostat model for the continuous culture of microorganisms can be expressed as the following equations ⎧ mSx ⎪ ⎪ dt, dS = D (Λ − S) − ⎪ ⎨ a+S (1) ⎪ mS ⎪ ⎪ − D xdt, ⎩ dx = a+S where S and x stand for the concentrations of the nutrient and the microorganism at time t respectively. Λ and D are positive constants, which respectively represent the input concentration of the nutrient and mS the common washout rate. a+S denotes the Monod growth functional response, where m > 0 is called the maximal growth rate and a > 0 is the Michaelis–Menten constant. The model (1) has been studied by λm Smith and Waltman [8]. They established the break-even concentration λ, where a+λ = D, and then showed * Corresponding author. E-mail addresses: [email protected] (D. Zhao), [email protected] (S. Yuan). http://dx.doi.org/10.1016/j.jmaa.2015.09.070 0022-247X/© 2015 Elsevier Inc. All rights reserved.
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.2 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
2
that: if λ ≥ Λ for m ≥ D, there is a washout equilibrium E 0 = (Λ, 0) and it is globally asymptotically stable; if λ < Λ, there exists a positive equilibrium E ∗ = (S ∗ , x∗ ) which means the microorganism will be persistent. Recently, Xu and Yuan [10] took notice that the environmental noise always forced the parameter to fluctuate around some average values, by assuming that the maximal growth rate m is perturbed by environmental noise with m → m + σ B˙ (t) ,
where B (t) is a standard Brownian motion on the complete probability space Ω, F, (Ft )t≥0 ,P with the intensity σ 2 > 0, they established the stochastic model: ⎧ mSx 0 σSx ⎪ ⎪ dt − dB(t), − S − dS = D S ⎪ ⎨ a+S a+S ⎪ σSx mS ⎪ ⎪ − D xdt + dB(t), ⎩ dx = a+S a+S
(2)
where S and x stand for the concentrations of the nutrient and the microorganism at time t respectively. S 0 and D are positive constants, which respectively represent the input concentration of the nutrient mS and the common washout rate. a+S denotes the Monod growth functional response where a > 0 is the Michaelis–Menten constant. By defining the break-even concentration for stochastic chemostat model (2) σ 2 (S 0 )2 = Da + as λ m−D 2(a+S 0 )(m−D) , they gave the result as follows:
> S 0 and σ 2 ≤ m a+S (H1) If λ , then the microorganism goes extinct with probability one; S0 < S 0 , then the microorganism is persistent in mean in the chemostat, namely, (H2) If λ 1 lim inf t→∞ t
0
t x (s) ds ≥ 0
σ 2 (S 0 )2 m−D Da S0 − − > 0 a.s. m m − D 2 (a + S 0 ) (m − D)
However, the following important questions are yet to be answered:
m a+S > S 0 is sufficient for extinction of the microorganism regardless of the • When σ 2 > , whether λ S0 size of the noise, which is an open problem proposed by Xu and Yuan in [10]. = S0. • What about the microorganism in the chemostat when λ 0 • When λ < S , except for persistence in mean, if there is any more information that can be given and proved for better understanding how the noise effects on the survival of stochastic chemostat model. 0
> S 0 is sufficient for extinction of the microorganism regardless of the size In this paper, we will (1): show λ = S 0 also implies the extinction of the microorganism; (3): consider the of the noise; (2): try to prove that λ persistence by using the probability which is different from the previous result. On the basis of persistence, we further discuss the existence of stationary distribution. To begin with, let us prepare some known results. Remark 1. In [10], it has been shown that: 2 (B1) for any initial value (S0 , x0 ) ∈ R+ , (2) admits a uniquely positive solution for t ≥ 0 with probability 1; 2 (B2) the model (2) has a positive invariant set Γ = (S, x) ∈ R+ : S + x = S0 .
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.3 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
3
From (B2) in Remark 1, we have only to study the equation: σ S0 − x x m S0 − x − D xdt + dB(t), dx = a + S0 − x a + S0 − x
(3)
with the initial value x(0) = x0 ∈ (0, S 0 ) and then we will mainly focus on discussing the dynamical behavior of this equation (3). It’s easily to see that the microorganism is washed out in chemostat if m ≤ D, so we always assume m > D in the sequel. For more about the biological significance of the deterministic and stochastic chemostat model, one can refer to [10,1,4] and cited therein. When continuous flow bioreactor is concerned, the long time behavior of the established stochastic models has been studied by Zhang and his coauthors, see [2,12] for example. 2. Extinction The following lemma is useful for our proof and we recall it firstly. Consider the one-dimensional timehomogeneous stochastic differential equation; dX = b (X) dt + α (X) dB(t) with X(0) ∈ R+ ,
(4)
satisfying the following conditions: (T1) α2 (X) > 0 for any X ∈ I = (l, r) where −∞ ≤ l < r ≤ ∞; X+ε (T2) For any X ∈ I, ∃ε > 0 such that X−ε 1+|b(x)| α2 (x) dx < ∞. Lemma 2.1. (See e.g. [5,11].) Assume that (T1) and (T2) hold. Let X(t) be a weak solution of (4) in (l, r). x − v 2b(u) du For some fixed constant c ∈ I, the scale function is defined as q (x) = e c α2 (u) dv. If q (l+ ) > −∞ c
and q (r− ) = ∞ hold, then P lim X (t) = l = P sup X (t) < r = 1. t→∞
t≥0
> S 0 holds, then Theorem 2.2. Let x(t) be the solution of system (3) with initial value x0 ∈ (0, S 0 ). If λ
P lim x (t) = 0 = 1. t→∞
Proof. In (3), we denote m S0 − x −D x b (x) = a + S0 − x
and σ S0 − x x . α (x) = a + S0 − x Compute that
x
2 2b (u) du = 2 2 α (u) σ
c
x c
2 = 2 σ
x c
−2 S0 − u u m S0 − u −D du a + S0 − u a + S0 − u 2 mS 0 a + S 0 − D a + S 0 Da2 amS 0 − 2DaS 0 − Da2 + − 2 du (S 0 )2 (S 0 − u) (S 0 )2 u S 0 (S 0 − u)
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.4 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
4
2 =− 2 σ
2 mS 0 a + S 0 − D a + S 0 amS 0 − 2DaS 0 − Da2 0 Da2 ln S − x − ln x + 0 0 (S 0 )2 (S 0 )2 S (S − x)
+ C (c) where C (c) is a constant, and then
q (x) = e−C(c)
x e
mS 0 ln S 0 −v −
amS 0 −2DaS 0 −Da2 (S 0 )2
2 σ2
2 a+S 0 −D a+S 0 (S 0 )2
ln v+
2 Da
S 0 S 0 −v
dv
c
= e−C(c)
x v
− σ22
2 mS 0 a+S 0 −D a+S 0 (S 0 )2
S0 − v
2 σ2
amS 0 −2DaS 0 −Da2 (S 0 )2
eσ
2 2Da
2 S 0 S 0 −v
dv.
(5)
c
By (5), let x → (S 0 )− and s =
1 S 0 −v
2
m a+S D a+S 0 2 − 2 q (S 0 )− ≥ e−C(c) c σ2 (S0 )2 S 0 σ2 S0
0
S
0
S0 − v
2 σ2
amS 0 −2DaS 0 −Da2 (S 0 )2
eσ
2 2Da
2 S 0 S 0 −v
dv
c
≥e
−C(c)
c
2 σ2
D a+S 0 (S 0 )2
2
0 − σ2 2
S
m a+S 0 S0
+∞
− σ22
s
amS 0 −2DaS 0 −Da2 (S 0 )2
−2
2Da2
e σ2 S0 s ds
1 S 0 −c
= ∞. > S 0 implies Note that λ −q 0+ ≤ e−C(c) S 0
2am σ2 S 0
2 σ2
amS 0 −2DaS 0 −Da2 (S 0 )2
S0 − c
2 2DaS 0 +Da2 − σ 2 (S 0 )2
e
< 1, let x → 0+ , we obtain
2 2Da
σ 2 S 0 S 0 −c
c v
− σ22
2 mS 0 a+S 0 −D a+S 0 (S 0 )2
dv
0
=e
−C(c)
2am
S
0 σ2 S 0
S −c 0
− 2 2DaS 2
0 +Da2
σ (S 0 )2
2Da2
e σ2 Λ(Λ−c)
1−
2 σ2
1 amS 0 −2DaS 0 −Da2 (S 0 )2
2
c1− σ2
amS 0 −2DaS 0 −Da2 (S 0 )2
<∞ which means q (0+ ) > −∞. By Lemma 2.1, we get the desired result. The proof is complete. 2 = S 0 . One useful Next, we will show that the microorganism is also washed out in the chemostat for λ lemma is introduced firstly. Lemma 2.3. (See e.g. [9].) Assume that α(X) ≡ 1 and equation (4) admits a non-explosive solution which is unique in the sense of a probability law, denoted by X(t). If the scale functions ρ (x) = x v 2b(u)du x v 0 e dv and θ (x) = 0 e− 0 2b(u)du dv satisfy ρ (−∞) = −∞, ρ (∞) < ∞, θ (−∞) = −∞ and 0 θ (∞) = ∞, then X(t) → −∞ in probability, namely, lim P (X (t) < y) = 1 for any y ∈ R. t→∞
= S 0 , then Theorem 2.4. Suppose that x(t) is the solution of system (3) with initial value x0 ∈ (0, S 0 ). If λ x(t) → 0 as t → ∞ in probability.
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.5 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
Proof. Define X = η(x) =
log x −
1 σS 0
a a+S 0
5
log S 0 − x , clearly, η(x) is a continuous and increasing
function on (0, S 0 ). By the Itô formula, dX = b (X) dt + dB (t) where b (X) =
mS 0 a+S 0
1 σS 0
−
DS 0 a+S 0 −η −1 (X) (a+S 0 )(S 0 −η −1 (X))
−
0 2 2 0 S −η −1 (X) +a η −1 (X) σ 2 a+S 2 2 (a+S 0 )(a+S 0 −η −1 (X))
with x = η −1 (X) since
= S 0 . Compute that by taking u = η (s) λ
x 2b (u) du 0
2 = σS 0
x 0
=
2 2 DS 0 a + S 0 − η −1 (u) σ 2 a + S 0 S 0 − η −1 (u) + a η −1 (u) mS 0 − − du 2 a + S0 (a + S 0 ) (S 0 − η −1 (u)) 2 (a + S 0 ) (a + S 0 − η −1 (u)) η −1
(x)
2 σ 2 (a + S 0 )
2 η −1 (0)
2
= C0 −
σ 2 (a + S 0 )
2 D a + S0 − s σ 2 a + S 0 S 0 − s + as2 a + S 0 − s − ds m− 2 (S 0 − s) 2 (S 0 − v) s S 0 (a + S 0 − s)
A ln S − η
2
0
where C0 is a constant, A = m − D +
x e
u 0
2b(v)du
−1
(x) +
aσ 2 S0 .
σ 2 S 0 + 4a Da2 0 −1 ln a + S − η (x) S 0 (S 0 − η −1 (x)) 2S 0
Then by taking u = η (w) it follows that
du
0 0
=e
+4a
S0S a+S
x 0 2 0 −1 a + S − η (u) C0
[S 0 − η −1 (u)] σ
0
η −1
(x)
η −1 (0)
e
Da2
2
2 S 0 −η −1 (u) σ 2 S 0 a+S 0
du
0
a+S −w
C0
e = 0 σS (a + S 0 )
2A
2 a+S 0 2
−
0
[S 0 − w] σ
+4a
S0S a+S 0 2
e
2A
2 a+S 0 2
−
2 2
2 Da
S 0 −w σ 2 S 0 a+S 0
a a + S0 − 0 dw. w S −w
Since η −1 (0) ∈ 0, S 0 , lim η −1 (x) = 0 and lim η −1 (x) = S 0 , by (6) we get x→−∞
x→∞
0
C0
e ρ (−∞) ≤ − σS 0
+
0S +4a0 2 a + S 0 − η −1 (0) S a+S S0
aeC0 η −1 (0) σS 0 (a + S 0 )
2A 2 σ 2 a+S 0
a + S0
e
−
Da2
2
2 S 0 −η −1 (0) σ 2 S 0 a+S 0
η −1
(0)
1 dw w
0
S 0 +4a
2 S 0 a+S 0
[S 0 − η −1 (0)] σ
2A
2 a+S 0 2
+1
e
−
2 2Da
2 σ 2 S 0 2 a+S 0
= −∞, S 0 +4a
a + S0 eC0 0 S 0 a+S 0 2 a + S ρ (∞) ≤ σS 0 (a + S 0 ) η −1 (0)
S
0
η −1 (0)
e
−
2 2
2 Da
S 0 −w σ 2 S 0 a+S 0
[S 0 − w] σ
2A
2 a+S 0 2
dw < ∞.
(6)
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.6 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
6
Similarly, we obtain that θ (−∞) = −∞ and θ (∞) = ∞. Then by using Lemma 2.3, we complete the proof. 2 > S0 Remark 2. In Theorem 2.2, we show that the microorganism goes extinct with probability one if λ m a+S 0 holds. Comparing with (H1), it’s of interest to see that we do not ask for the assumption σ 2 ≤ , S0 which solves the conjecture proposed in [10]. Besides, Theorem 2.4 gives further study on the behavior = S 0 . Thus, the previously-known result on extinction in [10] is improved of the microorganism under λ significantly. 3. Persistence and stationary distribution In this section, we will discuss the persistence and stationary distribution for the microorganism modeled by (3). Now, we introduce some definitions and lemmas. Definition 3.1. (See e.g. [6].) The microorganism modeled by (3) is said to be 1 t→∞ t
(I) persistence in mean, if lim
t 0
x (s) ds ≥ ζ for some constant ζ > 0;
(II) stochastic persistence in the chemostat, if for any ε ∈ (0, 1), there exist positive constants A1 = A1 (ε) and A2 = A2 (ε) such that for any initial value x0 ∈ R+ , lim inf P (x (t) ≤ A1 ) > 1 − ε and lim inf P (x (t) ≥ A2 ) > 1 − ε. t→∞
t→∞
Lemma 3.2. (See [13].) Let X(t) be a time-homogeneous solution of (4) on E1 (1-dimensional Euclidean space). Assume that: (U.1) In the domain U ⊂ E1 and some neighborhood thereof, the diffusion σ(X) is bounded away from zero; (U.2) If for all X ∈ E1 \U the mean time τX at which a path emerging from X reaches the set U is finite, and sup E (τX ) < ∞ for every compact subset K ⊂ E1 . X∈K
Then the Markov process X(t) has a stationary distribution π(x), and for an integrable function f (·) ⎛ 1 P ⎝ lim t→∞ t
t
∞ f [x (s)] ds =
0
⎞ f (x) π(dx)⎠ = 1.
−∞
< S 0 , then Theorem 3.3. Let x(t) be the solution of system (3) with initial value x0 ∈ (0, S 0 ). If λ (A1) the microorganism x is stochastically persistent in the chemostat; (A2) the model (3) has a stationary distribution, denoted by π(x). Proof. By using the Itô formula to model (3), we have 2 m S 0 − x (t) pσ S 0 − x (t) p (1 + p) σ 2 S 0 − x (t) 1 =− p −D− dB (t) d p dt − p x (t) x (t) a + S 0 − x (t) 2 a + S 0 − x (t) x (t) (a + S 0 − x (t)) 2 pσ S 0 − x (t) mS 0 S0 p (1 + p) σ 2 dB (t) (7) =− p −D− dt + H (t) dt − p x (t) a + S 0 2 a + S0 x (t) (a + S 0 − x (t))
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.7 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
7
where p ∈ (0, 1) is a constant and 2 2 m S 0 − x (t) mS 0 (1 + p) σ 2 p (1 + p) σ 2 S0 S 0 − x (t) + H (t) = p − − x (t) a + S 0 2 a + S0 a + S 0 − x (t) 2 a + S 0 − x (t) 2 2 pmx1−p (t) p (1 + p) σ 2 S0 S 0 − x (t) = − − (a + S 0 ) (a + S 0 − x (t)) 2xp (t) a + S0 a + S 0 − x (t) 1−p
pmS 0 . a (a + S 0 ) mS 0 Define γ = p a+S 0 − D − ≤
(7) by e
γt
(8) (1+p)σ 2 2
S0 a+S 0
2 , and then choose p small enough such that γ > 0. We multiply
and take an integration to show 1 1 = p e−γt + xp (t) x (0)
t
H (s) e−γ(t−s) ds − M (t) ≤
xp
pm(S 0 )1−p 1 + − M (t) (0) γa (a + S 0 )
0
where M (t) =
pσ S 0 −x(s) dB 0 xp (s)(a+S 0 −x(s))
t
E
1 xp (t)
(s). Taking expectations on both sides we get
1 = p e−γt + x (0)
t
E (H (s)) e−γ(t−s) ds − M (t)
0 1−p
≤
pm(S 0 ) 1 + . p x (0) γa (a + S 0 )
In view of the fact that 0 < x(t) ≤ S 0 , by use of Chebyshev’s inequality (see [7]), we have 1 1 1 1 0 =1−P P A2 ≤ x (t) ≤ S = P (A2 ≤ x (t)) = P ≥ p ≥ p Ap2 x (t) Ap2 x (t) 1−p pm(S 0 ) 1 1 ≥ 1 − Ap2 + . ≥ 1 − Ap2 E p x (t) xp (0) γa (a + S 0 ) 0 1−p ) By choosing A2 such that Ap2 xp1(0) + pm(S < ε, we get the desired result. The proof of (A1) is 0 γa(a+S ) complete. Next, we prove (A2). Let k be a sufficiently large number and U be a bounded open subset with a regular boundary such that x:
1 1 ≤ x ≤ S0 − k k
¯ ⊂ 0, S 0 ⊂U ⊂U
¯ is the closure of U . Define a C 2 -function V : R+ → R+ for any p ∈ (0, 1) by where U V (x) =
1 1 . + 0 p px S −x
In view of the Itô formula, we get dV (x (t)) = L
σ S 0 − x (t) 1 σS 0 x (t) +L 0 dt − p − dB (t) , pxp (t) S − x (t) x (t) (a + S 0 − x (t)) (S 0 − x (t)) (a + S 0 − x (t)) 1
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.8 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
8
where the operator LV (t, x) = From (7) and (8), we have
∂ ∂t V
∂ (t, x) + b(x) ∂x V (t, x) +
α2 (x) ∂ 2 2 ∂x2 V
(t, x) by taking (4) as an example.
2 1 mS 0 S0 (1 + p) σ 2 m(S 0 )1−p ≤− p L p − D − + px (t) x (t) a + S 0 2 a + S0 a (a + S 0 ) 1
and
DS 0 mx (t) σ 2 x2 (t) 1 1 =− L 0 2 + S 0 − x (t) a + S 0 − x (t) + D + 2 S − x (t) (S 0 − x (t)) (a + S 0 − x (t)) DS 0 DS 0 σ 2 (S 0 )2 1 mS 0 ≤− − + D + + 2 2 S 0 − x (t) a a2 2(S 0 − x (t)) 2(S 0 − x (t)) 2 DS 0 1 σ2 S 0 m ≤− . 2 + Da + S 0 + Da2 2(S 0 − x (t))
Combining the above equations to show that 2 mS 0 1 (1 + p) σ 2 DS 0 S0 LV (x (t)) ≤ − p −D− − 2 0 0 x (t) a + S 2 a+S 2(S 0 − x (t)) 2 1 m(S 0 )1−p σ2 S 0 m + + + + . a (a + S 0 ) Da S 0 Da2 Let p be small such that
mS 0 a+S 0
−D−
(1+p)σ 2 2
S0 a+S 0
2 > 0, then for sufficiently large k we obtain
LV (x) ≤ −1 for all x ∈ 0, S 0 \U which implies that (U.2) in Lemma 3.2 holds. Besides, one can easily check that (U.1) is satisfied. Thus, the model (3) has a stationary distribution. The proof is complete. 2 Remark 3. From (A1) of Theorem 3.3, the microorganism is stochastically persistent in the chemostat. Obviously, it cannot be deduced from the persistence in mean. Theorem 3.3 also shows us that model (3) admits a stationary distribution denoted by π(x) which satisfies ⎛ 1 P ⎝ lim t→∞ t
t
∞ x (s)ds =
0
⎞ xπ(dx)⎠ = 1.
0
Combining with (H2), we show 1 t→∞ t
t
I = lim
x (s) ds exists a.s. 0
and σ 2 (S 0 )2 m−D Da 0 S − − > 0. I≥ m m − D 2 (a + S 0 ) (m − D) Thus the obtained results in this paper show more information on the dynamics of the microorganism than that given in [10].
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.9 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
9
Fig. 1. Simulations for the SDE model (2) with (a) σ = 0; (b) σ = 0.2; (c) σ = 0.7589; (d) σ = 0.9.
4. Computer simulation and conclusion Computer simulation of the path of S(t), x(t) for the stochastic chemostat model (2) is established by using the classical EM method (see [3]) with step size Δt = 0.001 and initial value (S(0), x(0)) = (0.6, 0.4). We suppose that S 0 = 1.0, a = 0.2, m = 0.6 and D = 0.3. To show the effect of the stochastic perturbations, we choose σ = 0, 0.2, 0.7589, 0.9 separately. In Fig. 1(a), we choose σ = 0, (2) reduces to the corresponding deterministic model. Compute that λ = 0.2 < 1 = S 0 . Then by [8], there is a positive equilibrium which shows the microorganism will be persistent in the chemostat, see Fig. 1(a).
JID:YJMAA
AID:19845 /FLA
10
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.10 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
= 0.2556 < 1 = S 0 . By Theorem 3.3, the microorganism In Fig. 1(b), we choose σ = 0.2. Compute that λ will be persistent in mean and stochastically persistent. = S 0 . By Theorem 2.4, the microorganism goes extinct in probability. In Fig. 1(c), choose σ such that λ m a+S 0
In Fig. 1(d), we choose σ = 0.9. Since σ 2 = 0.81 > 0.72 = holds, by the work in [10], we cannot S0 make sure that the microorganism in the chemostat will be persistent or not. However, by Theorem 2.2, = 1.325 > 1 = S 0 implies that the microorganism will go extinct with probability one. λ for the considered stochastic model is obtained, which In this paper, the strictly critical parameter λ improves the known results significantly; the details can be stated as follows: • Theorem 2.2 shows that the microorganism will go extinct with probability one for any size of the noise m a+S 0 0 2 if λ > S , where we do not ask for σ ≤ . Thus, Theorem 2.2 completely solves the open problem S0 given by Xu and Yuan in [10]. = S 0 in • By using the Feller test, we show that the microorganism will go extinct in probability if λ Theorem 2.4, which has not been studied in the known literature. < S 0 , which • In Theorem 3.3, the stochastic persistence of the microorganism is studied firstly under λ is different with the meaning of persistence in mean. Then we prove that the model has the stationary distribution. In conclusion, the obtained results in this paper present more information on the dynamics than these given in [10]. Acknowledgments This work was supported by NSFC (Nos. 11271260, 11171216), the Hujiang Foundation of China (B14005) and Shanghai Leading Academic Discipline Project (No. XTKX2012). References [1] F. Campillo, M. Jaonnides, I. Larramendy-Valverde, Stochastic modeling of the chemostat, Ecol. Model. 222 (2011) 2676–2689. [2] Z. Chen, T. Zhang, Long time behaviour of a stochastic model for continuous flow bioreactor, J. Math. Chem. 51 (2013) 451–464. [3] D.J. Higham, An algorithmic introduction to numerical simulation of stochastic differential equations, SIAM Rev. 43 (2001) 525–546. [4] L. Imhof, S. Walcher, Exclusion and persistence in deterministic and stochastic chemostat model, J. Differential Equations 217 (2005) 26–63. [5] I. Karatzas, S. Shreve, Brown Motion and Stochastic Calculus, Springer Verlag, New York, 1988. [6] J. Lv, K. Wang, Almost sure permanence of stochastic single species models, J. Math. Anal. Appl. 422 (2015) 675–683. [7] X. Mao, Stochastic Differential Equation and Application, Horwood Publishing, Chichester, 1997. [8] H.L. Smith, P. Waltman, The Theory of the Chemostat, Cambridge University Press, Cambridge, 1995. [9] X. Sun, Y. Wang, Stability analysis of a stochastic logistic model with nonlinear diffusion term, Appl. Math. Model. 32 (2008) 2067–2075. [10] C. Xu, S. Yuan, An analogue of break-even concentration in a simple stochastic chemostat model, Appl. Math. Lett. 48 (2015) 62–68. [11] Q. Yang, D. Jiang, A note on asymptotic behaviors of stochastic population model with Allee effect, Appl. Math. Model. 35 (2011) 4611–4619. [12] T. Zhang, Z. Chen, M. Han, Dynamical analysis of a stochastic model for cascaded continuous flow bioreactors, J. Math. Chem. 52 (2014) 1441–1459. [13] C. Zhu, G. Yin, Asymptotic properties of hybrid diffusion systems, SIAM J. Control Optim. 46 (2007) 1155–1179.
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.1 (1-10)
J. Math. Anal. Appl. ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Critical result on the break-even concentration in a single-species stochastic chemostat model Dianli Zhao ∗ , Sanling Yuan College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
a r t i c l e
i n f o
Article history: Received 23 July 2015 Available online xxxx Submitted by V. Pozdnyakov Keywords: Stochastic chemostat model Break-even concentration Extinction Stationary distribution
a b s t r a c t is identified as the break-even In this paper, the strictly critical parameter λ > S 0 suffices for extinction concentration for the stochastic chemostat model. λ of the microorganism for any noise intensity σ 2 > 0 is proved firstly, which completely solves the open problem proposed in [10]. Then, by using the Feller = S0, test, we show that the microorganism will go extinct in probability if λ < S 0 , in addition which has not been studied in the known literature. When λ to the previously discussed persistence in mean, the stochastic persistence of the microorganism is studied by use of Chebyshev’s inequality. Besides, the existence of the stationary distribution is proved for the considered model. Numerical simulations are introduced finally to support the obtained results. © 2015 Elsevier Inc. All rights reserved.
1. Introduction A classical chemostat model for the continuous culture of microorganisms can be expressed as the following equations ⎧ mSx ⎪ ⎪ dt, dS = D (Λ − S) − ⎪ ⎨ a+S (1) ⎪ mS ⎪ ⎪ − D xdt, ⎩ dx = a+S where S and x stand for the concentrations of the nutrient and the microorganism at time t respectively. Λ and D are positive constants, which respectively represent the input concentration of the nutrient and mS the common washout rate. a+S denotes the Monod growth functional response, where m > 0 is called the maximal growth rate and a > 0 is the Michaelis–Menten constant. The model (1) has been studied by λm Smith and Waltman [8]. They established the break-even concentration λ, where a+λ = D, and then showed * Corresponding author. E-mail addresses: [email protected] (D. Zhao), [email protected] (S. Yuan). http://dx.doi.org/10.1016/j.jmaa.2015.09.070 0022-247X/© 2015 Elsevier Inc. All rights reserved.
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.2 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
2
that: if λ ≥ Λ for m ≥ D, there is a washout equilibrium E 0 = (Λ, 0) and it is globally asymptotically stable; if λ < Λ, there exists a positive equilibrium E ∗ = (S ∗ , x∗ ) which means the microorganism will be persistent. Recently, Xu and Yuan [10] took notice that the environmental noise always forced the parameter to fluctuate around some average values, by assuming that the maximal growth rate m is perturbed by environmental noise with m → m + σ B˙ (t) ,
where B (t) is a standard Brownian motion on the complete probability space Ω, F, (Ft )t≥0 ,P with the intensity σ 2 > 0, they established the stochastic model: ⎧ mSx 0 σSx ⎪ ⎪ dt − dB(t), − S − dS = D S ⎪ ⎨ a+S a+S ⎪ σSx mS ⎪ ⎪ − D xdt + dB(t), ⎩ dx = a+S a+S
(2)
where S and x stand for the concentrations of the nutrient and the microorganism at time t respectively. S 0 and D are positive constants, which respectively represent the input concentration of the nutrient mS and the common washout rate. a+S denotes the Monod growth functional response where a > 0 is the Michaelis–Menten constant. By defining the break-even concentration for stochastic chemostat model (2) σ 2 (S 0 )2 = Da + as λ m−D 2(a+S 0 )(m−D) , they gave the result as follows:
> S 0 and σ 2 ≤ m a+S (H1) If λ , then the microorganism goes extinct with probability one; S0 < S 0 , then the microorganism is persistent in mean in the chemostat, namely, (H2) If λ 1 lim inf t→∞ t
0
t x (s) ds ≥ 0
σ 2 (S 0 )2 m−D Da S0 − − > 0 a.s. m m − D 2 (a + S 0 ) (m − D)
However, the following important questions are yet to be answered:
m a+S > S 0 is sufficient for extinction of the microorganism regardless of the • When σ 2 > , whether λ S0 size of the noise, which is an open problem proposed by Xu and Yuan in [10]. = S0. • What about the microorganism in the chemostat when λ 0 • When λ < S , except for persistence in mean, if there is any more information that can be given and proved for better understanding how the noise effects on the survival of stochastic chemostat model. 0
> S 0 is sufficient for extinction of the microorganism regardless of the size In this paper, we will (1): show λ = S 0 also implies the extinction of the microorganism; (3): consider the of the noise; (2): try to prove that λ persistence by using the probability which is different from the previous result. On the basis of persistence, we further discuss the existence of stationary distribution. To begin with, let us prepare some known results. Remark 1. In [10], it has been shown that: 2 (B1) for any initial value (S0 , x0 ) ∈ R+ , (2) admits a uniquely positive solution for t ≥ 0 with probability 1; 2 (B2) the model (2) has a positive invariant set Γ = (S, x) ∈ R+ : S + x = S0 .
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.3 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
3
From (B2) in Remark 1, we have only to study the equation: σ S0 − x x m S0 − x − D xdt + dB(t), dx = a + S0 − x a + S0 − x
(3)
with the initial value x(0) = x0 ∈ (0, S 0 ) and then we will mainly focus on discussing the dynamical behavior of this equation (3). It’s easily to see that the microorganism is washed out in chemostat if m ≤ D, so we always assume m > D in the sequel. For more about the biological significance of the deterministic and stochastic chemostat model, one can refer to [10,1,4] and cited therein. When continuous flow bioreactor is concerned, the long time behavior of the established stochastic models has been studied by Zhang and his coauthors, see [2,12] for example. 2. Extinction The following lemma is useful for our proof and we recall it firstly. Consider the one-dimensional timehomogeneous stochastic differential equation; dX = b (X) dt + α (X) dB(t) with X(0) ∈ R+ ,
(4)
satisfying the following conditions: (T1) α2 (X) > 0 for any X ∈ I = (l, r) where −∞ ≤ l < r ≤ ∞; X+ε (T2) For any X ∈ I, ∃ε > 0 such that X−ε 1+|b(x)| α2 (x) dx < ∞. Lemma 2.1. (See e.g. [5,11].) Assume that (T1) and (T2) hold. Let X(t) be a weak solution of (4) in (l, r). x − v 2b(u) du For some fixed constant c ∈ I, the scale function is defined as q (x) = e c α2 (u) dv. If q (l+ ) > −∞ c
and q (r− ) = ∞ hold, then P lim X (t) = l = P sup X (t) < r = 1. t→∞
t≥0
> S 0 holds, then Theorem 2.2. Let x(t) be the solution of system (3) with initial value x0 ∈ (0, S 0 ). If λ
P lim x (t) = 0 = 1. t→∞
Proof. In (3), we denote m S0 − x −D x b (x) = a + S0 − x
and σ S0 − x x . α (x) = a + S0 − x Compute that
x
2 2b (u) du = 2 2 α (u) σ
c
x c
2 = 2 σ
x c
−2 S0 − u u m S0 − u −D du a + S0 − u a + S0 − u 2 mS 0 a + S 0 − D a + S 0 Da2 amS 0 − 2DaS 0 − Da2 + − 2 du (S 0 )2 (S 0 − u) (S 0 )2 u S 0 (S 0 − u)
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.4 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
4
2 =− 2 σ
2 mS 0 a + S 0 − D a + S 0 amS 0 − 2DaS 0 − Da2 0 Da2 ln S − x − ln x + 0 0 (S 0 )2 (S 0 )2 S (S − x)
+ C (c) where C (c) is a constant, and then
q (x) = e−C(c)
x e
mS 0 ln S 0 −v −
amS 0 −2DaS 0 −Da2 (S 0 )2
2 σ2
2 a+S 0 −D a+S 0 (S 0 )2
ln v+
2 Da
S 0 S 0 −v
dv
c
= e−C(c)
x v
− σ22
2 mS 0 a+S 0 −D a+S 0 (S 0 )2
S0 − v
2 σ2
amS 0 −2DaS 0 −Da2 (S 0 )2
eσ
2 2Da
2 S 0 S 0 −v
dv.
(5)
c
By (5), let x → (S 0 )− and s =
1 S 0 −v
2
m a+S D a+S 0 2 − 2 q (S 0 )− ≥ e−C(c) c σ2 (S0 )2 S 0 σ2 S0
0
S
0
S0 − v
2 σ2
amS 0 −2DaS 0 −Da2 (S 0 )2
eσ
2 2Da
2 S 0 S 0 −v
dv
c
≥e
−C(c)
c
2 σ2
D a+S 0 (S 0 )2
2
0 − σ2 2
S
m a+S 0 S0
+∞
− σ22
s
amS 0 −2DaS 0 −Da2 (S 0 )2
−2
2Da2
e σ2 S0 s ds
1 S 0 −c
= ∞. > S 0 implies Note that λ −q 0+ ≤ e−C(c) S 0
2am σ2 S 0
2 σ2
amS 0 −2DaS 0 −Da2 (S 0 )2
S0 − c
2 2DaS 0 +Da2 − σ 2 (S 0 )2
e
< 1, let x → 0+ , we obtain
2 2Da
σ 2 S 0 S 0 −c
c v
− σ22
2 mS 0 a+S 0 −D a+S 0 (S 0 )2
dv
0
=e
−C(c)
2am
S
0 σ2 S 0
S −c 0
− 2 2DaS 2
0 +Da2
σ (S 0 )2
2Da2
e σ2 Λ(Λ−c)
1−
2 σ2
1 amS 0 −2DaS 0 −Da2 (S 0 )2
2
c1− σ2
amS 0 −2DaS 0 −Da2 (S 0 )2
<∞ which means q (0+ ) > −∞. By Lemma 2.1, we get the desired result. The proof is complete. 2 = S 0 . One useful Next, we will show that the microorganism is also washed out in the chemostat for λ lemma is introduced firstly. Lemma 2.3. (See e.g. [9].) Assume that α(X) ≡ 1 and equation (4) admits a non-explosive solution which is unique in the sense of a probability law, denoted by X(t). If the scale functions ρ (x) = x v 2b(u)du x v 0 e dv and θ (x) = 0 e− 0 2b(u)du dv satisfy ρ (−∞) = −∞, ρ (∞) < ∞, θ (−∞) = −∞ and 0 θ (∞) = ∞, then X(t) → −∞ in probability, namely, lim P (X (t) < y) = 1 for any y ∈ R. t→∞
= S 0 , then Theorem 2.4. Suppose that x(t) is the solution of system (3) with initial value x0 ∈ (0, S 0 ). If λ x(t) → 0 as t → ∞ in probability.
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.5 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
Proof. Define X = η(x) =
log x −
1 σS 0
a a+S 0
5
log S 0 − x , clearly, η(x) is a continuous and increasing
function on (0, S 0 ). By the Itô formula, dX = b (X) dt + dB (t) where b (X) =
mS 0 a+S 0
1 σS 0
−
DS 0 a+S 0 −η −1 (X) (a+S 0 )(S 0 −η −1 (X))
−
0 2 2 0 S −η −1 (X) +a η −1 (X) σ 2 a+S 2 2 (a+S 0 )(a+S 0 −η −1 (X))
with x = η −1 (X) since
= S 0 . Compute that by taking u = η (s) λ
x 2b (u) du 0
2 = σS 0
x 0
=
2 2 DS 0 a + S 0 − η −1 (u) σ 2 a + S 0 S 0 − η −1 (u) + a η −1 (u) mS 0 − − du 2 a + S0 (a + S 0 ) (S 0 − η −1 (u)) 2 (a + S 0 ) (a + S 0 − η −1 (u)) η −1
(x)
2 σ 2 (a + S 0 )
2 η −1 (0)
2
= C0 −
σ 2 (a + S 0 )
2 D a + S0 − s σ 2 a + S 0 S 0 − s + as2 a + S 0 − s − ds m− 2 (S 0 − s) 2 (S 0 − v) s S 0 (a + S 0 − s)
A ln S − η
2
0
where C0 is a constant, A = m − D +
x e
u 0
2b(v)du
−1
(x) +
aσ 2 S0 .
σ 2 S 0 + 4a Da2 0 −1 ln a + S − η (x) S 0 (S 0 − η −1 (x)) 2S 0
Then by taking u = η (w) it follows that
du
0 0
=e
+4a
S0S a+S
x 0 2 0 −1 a + S − η (u) C0
[S 0 − η −1 (u)] σ
0
η −1
(x)
η −1 (0)
e
Da2
2
2 S 0 −η −1 (u) σ 2 S 0 a+S 0
du
0
a+S −w
C0
e = 0 σS (a + S 0 )
2A
2 a+S 0 2
−
0
[S 0 − w] σ
+4a
S0S a+S 0 2
e
2A
2 a+S 0 2
−
2 2
2 Da
S 0 −w σ 2 S 0 a+S 0
a a + S0 − 0 dw. w S −w
Since η −1 (0) ∈ 0, S 0 , lim η −1 (x) = 0 and lim η −1 (x) = S 0 , by (6) we get x→−∞
x→∞
0
C0
e ρ (−∞) ≤ − σS 0
+
0S +4a0 2 a + S 0 − η −1 (0) S a+S S0
aeC0 η −1 (0) σS 0 (a + S 0 )
2A 2 σ 2 a+S 0
a + S0
e
−
Da2
2
2 S 0 −η −1 (0) σ 2 S 0 a+S 0
η −1
(0)
1 dw w
0
S 0 +4a
2 S 0 a+S 0
[S 0 − η −1 (0)] σ
2A
2 a+S 0 2
+1
e
−
2 2Da
2 σ 2 S 0 2 a+S 0
= −∞, S 0 +4a
a + S0 eC0 0 S 0 a+S 0 2 a + S ρ (∞) ≤ σS 0 (a + S 0 ) η −1 (0)
S
0
η −1 (0)
e
−
2 2
2 Da
S 0 −w σ 2 S 0 a+S 0
[S 0 − w] σ
2A
2 a+S 0 2
dw < ∞.
(6)
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.6 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
6
Similarly, we obtain that θ (−∞) = −∞ and θ (∞) = ∞. Then by using Lemma 2.3, we complete the proof. 2 > S0 Remark 2. In Theorem 2.2, we show that the microorganism goes extinct with probability one if λ m a+S 0 holds. Comparing with (H1), it’s of interest to see that we do not ask for the assumption σ 2 ≤ , S0 which solves the conjecture proposed in [10]. Besides, Theorem 2.4 gives further study on the behavior = S 0 . Thus, the previously-known result on extinction in [10] is improved of the microorganism under λ significantly. 3. Persistence and stationary distribution In this section, we will discuss the persistence and stationary distribution for the microorganism modeled by (3). Now, we introduce some definitions and lemmas. Definition 3.1. (See e.g. [6].) The microorganism modeled by (3) is said to be 1 t→∞ t
(I) persistence in mean, if lim
t 0
x (s) ds ≥ ζ for some constant ζ > 0;
(II) stochastic persistence in the chemostat, if for any ε ∈ (0, 1), there exist positive constants A1 = A1 (ε) and A2 = A2 (ε) such that for any initial value x0 ∈ R+ , lim inf P (x (t) ≤ A1 ) > 1 − ε and lim inf P (x (t) ≥ A2 ) > 1 − ε. t→∞
t→∞
Lemma 3.2. (See [13].) Let X(t) be a time-homogeneous solution of (4) on E1 (1-dimensional Euclidean space). Assume that: (U.1) In the domain U ⊂ E1 and some neighborhood thereof, the diffusion σ(X) is bounded away from zero; (U.2) If for all X ∈ E1 \U the mean time τX at which a path emerging from X reaches the set U is finite, and sup E (τX ) < ∞ for every compact subset K ⊂ E1 . X∈K
Then the Markov process X(t) has a stationary distribution π(x), and for an integrable function f (·) ⎛ 1 P ⎝ lim t→∞ t
t
∞ f [x (s)] ds =
0
⎞ f (x) π(dx)⎠ = 1.
−∞
< S 0 , then Theorem 3.3. Let x(t) be the solution of system (3) with initial value x0 ∈ (0, S 0 ). If λ (A1) the microorganism x is stochastically persistent in the chemostat; (A2) the model (3) has a stationary distribution, denoted by π(x). Proof. By using the Itô formula to model (3), we have 2 m S 0 − x (t) pσ S 0 − x (t) p (1 + p) σ 2 S 0 − x (t) 1 =− p −D− dB (t) d p dt − p x (t) x (t) a + S 0 − x (t) 2 a + S 0 − x (t) x (t) (a + S 0 − x (t)) 2 pσ S 0 − x (t) mS 0 S0 p (1 + p) σ 2 dB (t) (7) =− p −D− dt + H (t) dt − p x (t) a + S 0 2 a + S0 x (t) (a + S 0 − x (t))
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.7 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
7
where p ∈ (0, 1) is a constant and 2 2 m S 0 − x (t) mS 0 (1 + p) σ 2 p (1 + p) σ 2 S0 S 0 − x (t) + H (t) = p − − x (t) a + S 0 2 a + S0 a + S 0 − x (t) 2 a + S 0 − x (t) 2 2 pmx1−p (t) p (1 + p) σ 2 S0 S 0 − x (t) = − − (a + S 0 ) (a + S 0 − x (t)) 2xp (t) a + S0 a + S 0 − x (t) 1−p
pmS 0 . a (a + S 0 ) mS 0 Define γ = p a+S 0 − D − ≤
(7) by e
γt
(8) (1+p)σ 2 2
S0 a+S 0
2 , and then choose p small enough such that γ > 0. We multiply
and take an integration to show 1 1 = p e−γt + xp (t) x (0)
t
H (s) e−γ(t−s) ds − M (t) ≤
xp
pm(S 0 )1−p 1 + − M (t) (0) γa (a + S 0 )
0
where M (t) =
pσ S 0 −x(s) dB 0 xp (s)(a+S 0 −x(s))
t
E
1 xp (t)
(s). Taking expectations on both sides we get
1 = p e−γt + x (0)
t
E (H (s)) e−γ(t−s) ds − M (t)
0 1−p
≤
pm(S 0 ) 1 + . p x (0) γa (a + S 0 )
In view of the fact that 0 < x(t) ≤ S 0 , by use of Chebyshev’s inequality (see [7]), we have 1 1 1 1 0 =1−P P A2 ≤ x (t) ≤ S = P (A2 ≤ x (t)) = P ≥ p ≥ p Ap2 x (t) Ap2 x (t) 1−p pm(S 0 ) 1 1 ≥ 1 − Ap2 + . ≥ 1 − Ap2 E p x (t) xp (0) γa (a + S 0 ) 0 1−p ) By choosing A2 such that Ap2 xp1(0) + pm(S < ε, we get the desired result. The proof of (A1) is 0 γa(a+S ) complete. Next, we prove (A2). Let k be a sufficiently large number and U be a bounded open subset with a regular boundary such that x:
1 1 ≤ x ≤ S0 − k k
¯ ⊂ 0, S 0 ⊂U ⊂U
¯ is the closure of U . Define a C 2 -function V : R+ → R+ for any p ∈ (0, 1) by where U V (x) =
1 1 . + 0 p px S −x
In view of the Itô formula, we get dV (x (t)) = L
σ S 0 − x (t) 1 σS 0 x (t) +L 0 dt − p − dB (t) , pxp (t) S − x (t) x (t) (a + S 0 − x (t)) (S 0 − x (t)) (a + S 0 − x (t)) 1
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.8 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
8
where the operator LV (t, x) = From (7) and (8), we have
∂ ∂t V
∂ (t, x) + b(x) ∂x V (t, x) +
α2 (x) ∂ 2 2 ∂x2 V
(t, x) by taking (4) as an example.
2 1 mS 0 S0 (1 + p) σ 2 m(S 0 )1−p ≤− p L p − D − + px (t) x (t) a + S 0 2 a + S0 a (a + S 0 ) 1
and
DS 0 mx (t) σ 2 x2 (t) 1 1 =− L 0 2 + S 0 − x (t) a + S 0 − x (t) + D + 2 S − x (t) (S 0 − x (t)) (a + S 0 − x (t)) DS 0 DS 0 σ 2 (S 0 )2 1 mS 0 ≤− − + D + + 2 2 S 0 − x (t) a a2 2(S 0 − x (t)) 2(S 0 − x (t)) 2 DS 0 1 σ2 S 0 m ≤− . 2 + Da + S 0 + Da2 2(S 0 − x (t))
Combining the above equations to show that 2 mS 0 1 (1 + p) σ 2 DS 0 S0 LV (x (t)) ≤ − p −D− − 2 0 0 x (t) a + S 2 a+S 2(S 0 − x (t)) 2 1 m(S 0 )1−p σ2 S 0 m + + + + . a (a + S 0 ) Da S 0 Da2 Let p be small such that
mS 0 a+S 0
−D−
(1+p)σ 2 2
S0 a+S 0
2 > 0, then for sufficiently large k we obtain
LV (x) ≤ −1 for all x ∈ 0, S 0 \U which implies that (U.2) in Lemma 3.2 holds. Besides, one can easily check that (U.1) is satisfied. Thus, the model (3) has a stationary distribution. The proof is complete. 2 Remark 3. From (A1) of Theorem 3.3, the microorganism is stochastically persistent in the chemostat. Obviously, it cannot be deduced from the persistence in mean. Theorem 3.3 also shows us that model (3) admits a stationary distribution denoted by π(x) which satisfies ⎛ 1 P ⎝ lim t→∞ t
t
∞ x (s)ds =
0
⎞ xπ(dx)⎠ = 1.
0
Combining with (H2), we show 1 t→∞ t
t
I = lim
x (s) ds exists a.s. 0
and σ 2 (S 0 )2 m−D Da 0 S − − > 0. I≥ m m − D 2 (a + S 0 ) (m − D) Thus the obtained results in this paper show more information on the dynamics of the microorganism than that given in [10].
JID:YJMAA
AID:19845 /FLA
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.9 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
9
Fig. 1. Simulations for the SDE model (2) with (a) σ = 0; (b) σ = 0.2; (c) σ = 0.7589; (d) σ = 0.9.
4. Computer simulation and conclusion Computer simulation of the path of S(t), x(t) for the stochastic chemostat model (2) is established by using the classical EM method (see [3]) with step size Δt = 0.001 and initial value (S(0), x(0)) = (0.6, 0.4). We suppose that S 0 = 1.0, a = 0.2, m = 0.6 and D = 0.3. To show the effect of the stochastic perturbations, we choose σ = 0, 0.2, 0.7589, 0.9 separately. In Fig. 1(a), we choose σ = 0, (2) reduces to the corresponding deterministic model. Compute that λ = 0.2 < 1 = S 0 . Then by [8], there is a positive equilibrium which shows the microorganism will be persistent in the chemostat, see Fig. 1(a).
JID:YJMAA
AID:19845 /FLA
10
Doctopic: Applied Mathematics
[m3L; v1.160; Prn:7/10/2015; 14:09] P.10 (1-10)
D. Zhao, S. Yuan / J. Math. Anal. Appl. ••• (••••) •••–•••
= 0.2556 < 1 = S 0 . By Theorem 3.3, the microorganism In Fig. 1(b), we choose σ = 0.2. Compute that λ will be persistent in mean and stochastically persistent. = S 0 . By Theorem 2.4, the microorganism goes extinct in probability. In Fig. 1(c), choose σ such that λ m a+S 0
In Fig. 1(d), we choose σ = 0.9. Since σ 2 = 0.81 > 0.72 = holds, by the work in [10], we cannot S0 make sure that the microorganism in the chemostat will be persistent or not. However, by Theorem 2.2, = 1.325 > 1 = S 0 implies that the microorganism will go extinct with probability one. λ for the considered stochastic model is obtained, which In this paper, the strictly critical parameter λ improves the known results significantly; the details can be stated as follows: • Theorem 2.2 shows that the microorganism will go extinct with probability one for any size of the noise m a+S 0 0 2 if λ > S , where we do not ask for σ ≤ . Thus, Theorem 2.2 completely solves the open problem S0 given by Xu and Yuan in [10]. = S 0 in • By using the Feller test, we show that the microorganism will go extinct in probability if λ Theorem 2.4, which has not been studied in the known literature. < S 0 , which • In Theorem 3.3, the stochastic persistence of the microorganism is studied firstly under λ is different with the meaning of persistence in mean. Then we prove that the model has the stationary distribution. In conclusion, the obtained results in this paper present more information on the dynamics than these given in [10]. Acknowledgments This work was supported by NSFC (Nos. 11271260, 11171216), the Hujiang Foundation of China (B14005) and Shanghai Leading Academic Discipline Project (No. XTKX2012). References [1] F. Campillo, M. Jaonnides, I. Larramendy-Valverde, Stochastic modeling of the chemostat, Ecol. Model. 222 (2011) 2676–2689. [2] Z. Chen, T. Zhang, Long time behaviour of a stochastic model for continuous flow bioreactor, J. Math. Chem. 51 (2013) 451–464. [3] D.J. Higham, An algorithmic introduction to numerical simulation of stochastic differential equations, SIAM Rev. 43 (2001) 525–546. [4] L. Imhof, S. Walcher, Exclusion and persistence in deterministic and stochastic chemostat model, J. Differential Equations 217 (2005) 26–63. [5] I. Karatzas, S. Shreve, Brown Motion and Stochastic Calculus, Springer Verlag, New York, 1988. [6] J. Lv, K. Wang, Almost sure permanence of stochastic single species models, J. Math. Anal. Appl. 422 (2015) 675–683. [7] X. Mao, Stochastic Differential Equation and Application, Horwood Publishing, Chichester, 1997. [8] H.L. Smith, P. Waltman, The Theory of the Chemostat, Cambridge University Press, Cambridge, 1995. [9] X. Sun, Y. Wang, Stability analysis of a stochastic logistic model with nonlinear diffusion term, Appl. Math. Model. 32 (2008) 2067–2075. [10] C. Xu, S. Yuan, An analogue of break-even concentration in a simple stochastic chemostat model, Appl. Math. Lett. 48 (2015) 62–68. [11] Q. Yang, D. Jiang, A note on asymptotic behaviors of stochastic population model with Allee effect, Appl. Math. Model. 35 (2011) 4611–4619. [12] T. Zhang, Z. Chen, M. Han, Dynamical analysis of a stochastic model for cascaded continuous flow bioreactors, J. Math. Chem. 52 (2014) 1441–1459. [13] C. Zhu, G. Yin, Asymptotic properties of hybrid diffusion systems, SIAM J. Control Optim. 46 (2007) 1155–1179.