Annali di Matematica https://doi.org/10.1007/s10231-018-0758-4

Persistent competition models on two complementary nutrients with density-dependent consumption rates Fethi Borsali1 · Karim Yadi1

Received: 2 February 2018 / Accepted: 5 May 2018 © Fondazione Annali di Matematica Pura ed Applicata and Springer-Verlag GmbH Germany, part of Springer Nature 2018

Abstract We give a result of uniform persistence for a theoretical competition model of 3 species of microorganisms on 2 complementary nutrients in a chemostat with densitydependent growth functions. The proof consists of giving first sufficient conditions of persistence, which is by the way available for more than 3 species. Then, in the case of 2 species, we show under additional assumptions, and with the use of the so-called Hofbauer–Hutson theorem, that it is possible to obtain a uniform persistence around a positive equilibrium point. Finally, the main result is established from what preceeds and by means of Thieme–Zhao theorem. These results are illustrated by some numerical simulations. Keywords Dynamical systems · Global stability · Uniform persistence · Chemostat · Complementary nutrients · Density dependence Mathematics Subject Classifications 34C11 · 37N25 · 92B05

1 Introduction The chemostat is a laboratory apparatus used for continuous cultivation of microorganisms in which the effect of limiting nutrient on the population of microorganisms can be investigated. A theoretical approach to the description of the competition of two or more populations of microorganisms on a limiting nutrient in a chemostat has been developed in the book of Smith and Waltman [18] (see also Harmand et al. [8], Ajbar and Alhumaizi [1] and the references therein). Our interest goes to models where growth functions are obtained by the combination of two limiting nutrients. The different classifications of limiting nutrients were

B

Fethi Borsali [email protected] Karim Yadi [email protected]

1

Laboratoire Systèmes Dynamiques et Applications, University of Tlemcen, BP 119, 13000 Tlemcen, Algeria

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introduced by Rapport [17] and Covich [6]. We mainly distinguish between two types, the socalled complementary nutrients, that is nutrients of different essential substances which are independently required for growth, and substitutable nutrients, that is alternative sources of an essential substance. Note that the first type is called complementary by Leon and Tumpson [12] and essential by Tilman [20]. Leon and Tumpson found the conditions of local stability of a positive equilibrium point and hence conditions of coexistence of the two populations. Hsu et al. [10] made a global study when the growth functions are of Michaelis–Menten type and obtained the classical results of the two dimensional Lotka–Volterra competition model. The same model with general monotonic and non-monotonic growth functions was studied by Butler and Wolkowicz [5]. Li [13] examined a model generalizing that of Butler and Wolkowicz with several populations of microorganisms and for a class of general growth functions. He proved global stability of a positive equilibrium point under some conditions. Mazenc and Malisoff [16] considered a more general model with M resources and N species and different removal rates. When M = N , they obtained a conditional existence of a positive equilibrium point by Brouwer degree theory. For particular families of the growth functions they proved the global asymptotic stability of such a point with the use of Lyapunov functions. Recently, Borsali and Yadi [4] studied the same model of two microorganisms where they added terms of interspecific interaction. They proved the permanence of the model and the existence of a positive equilibrium point under some conditions. On the other hand, many models where the growth function depends not only on the substrate but also on the microorganism, have been considered. In this context, Arditi and Ginzburg [2,3] developed predator-prey models with functional responses depending on the prey and the predator. They suggested a class of functional responses which take into account the interference between the predators and depending on the amount of prey consumed by one predator that is on the ratio of the amount of consumed prey and the predator per unit time : the consumption rate decreases proportionally with the abundance of predators. The models with this type of functional response are called ratio-dependent. This approach has also been applied in competitive models in the chemostat. More exactly, in the following chemostat model of competition of n species “xi ” on a single nutrient “s”, n    1 s˙ = s ◦ − s D − μi (s, x) xi , yi i=1

(1)

x˙i = (μi (s, x) − D) xi , i ∈ {1, 2, . . . , n} , the growth function, denoted by μi (s, x), depends on the substrate and microorganisms. These models are rather called density-dependent. Lobry and Mazenc [15] examined the following general density-dependent competition model called intra-specific competition model : s˙ = f (s) −

n  1

y i=1 i

μi (s, xi ) xi ,

(2)

x˙i = (μi (s, xi ) − di ) xi , i ∈ {1, 2, . . . , n} . Here, xi is the biomass of the i th population in the chemostat, s is the concentration of the nutrient, s ◦ is the input concentration of the resource s, D is the dilution rate, yi is the growth yield constant, and di is the removal rate of the i th species. The function f is a positive function and of class C 1 . The growth function μi is supposed to vanish at s = 0, to be increasing with respect to s and decreasing with respect to xi . This assumption is natural. It ensures that the growth function is zero in the absence of the nutrient, increases with

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the amount of substrate, and decreases with the abundance of the microorganism. Under some assumptions, they showed that the model (2) has an interior equilibrium point globally asymptotically stable and therefore the coexistence of all the species of microorganisms. The demonstration is based on the construction of a Lyapunov function. Another study of the delay density-dependent model with the growth yield constants depending on the substrate was conducted by Xu and Li [14]. They also have the same result as that of Lobry and Mazenc [15]. The objective of the present paper is to study a density-dependent competition model of several microorganisms on two complementary nutrients. The model is written : n    1 s˙ = s ◦ − s D − μi (s, r, xi ) xi , ysi

  r˙ = r ◦ − r D −

i=1 n  i=1

1 μi (s, r, xi ) xi , yri

(3)

x˙i = (μi (s, r, xi ) − D) xi , i ∈ {1, 2, . . . , n} , where the functions μi (s, r, xi ) := min ( pi (s, xi ) , qi (r, xi )) satisfy the following assumptions : H1. The functions s  → pi (s, xi ) and r  → qi (r, xi ) are positive, continuously differentiable, increasing and such that pi (0, xi ) = qi (0, xi ) = 0. H2. The functions xi  → pi (s, xi ) and xi  → qi (r, xi ) are positive, continuously differentiable and decreasing. Since pi and qi are C 1 , the functions μi are locally Lipschitz. In particular, the assumptions H1 and H2 insure the existence and uniqueness property of solutions of (3) for prescribed initial conditions in Rn+2 + . Moreover, the break-even concentrations will be defined as being the respective values λsi and λri of s and r such that pi (λsi , 0) = qi (λri , 0) = D. This work should be seen as a contribution to the mathematical study of theoretical chemostat models in at least two aspects. The first one lies in the framework of multinutrients models, for which the references are less abundant. The second one concerns the densitydependence notion for which we believe that it will lead to new interesting questions. Even if this article in inspired by works in [15], the mathematical treatment turns to be more complex, except perhaps for the persistence result of Theorem 2.5 bellow. The aim of this work being theoretical, we do not give the connection with any realistic biological experiment. In Sect. 2, we first show that the positive solutions of (3) are positively bounded and give conditions of washing out. We then prove the persistence of the model under some assumptions (Theorem 2.5). Section 3 is devoted to the study of the same model limited to two microorganisms and for which we obtain a uniform persistence result under some assumptions (Theorem 3.3). Moreover, we deduce that there exists a positive equilibrium point which is globally asymptotically stable (Theorem 3.4). Finally, in Sect. 4, thanks to the previous results we prove that our model (3) with three microorganisms is actually uniformly persistent and that there exists at least a positive equilibrium point (Theorem 4.1). Some numerical simulations are presented in this article to illustrate the results. Note that one can make the conjecture that the equilibrium point for the last case is unique and attractive.

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2 Persistence in case of several species of microorganisms 2.1 Elementary properties The first two propositions recall the usual properties of dissipativity (i.e. the solutions are positively bounded), invariance and conditions of washing out. The proof of the first proposition is classical and is omitted. Proposition 2.1 Under assumptions H1 and H2, the system (3) is dissipative and the positive orthant Rn+2 + is positively invariant. More exactly, there exists t0 > 0 such for all t > t0 one has s(t) < s ◦ , r (t) < r ◦ and xi (t) < min(ysi s ◦ , yri r ◦ ), i ∈ {1, 2, ..., n}. Proposition 2.2 Suppose that H1 and H2 are satisfied. If s ◦ < λsi or r ◦ < λri then, for all positive initial conditions, one has lim s (t) = s ◦ , lim r (t) = r ◦ and lim xi (t) = 0.

t→+∞

t→+∞

t→+∞

Proof If s ◦ < λsi or r ◦ < λri for all i ∈ {1, 2, . . . , n}, then there exists a real constant ε > 0 such that s ◦ + ε < λsi or r ◦ + ε < λri . From Proposition 2.1, we deduce that ∃t0 > 0 : ∀t > t0 s(t) < s ◦ + ε, r (t) < r ◦ + ε. On the other hand, from the constant variational formula, we have :  t  xi (t) = xi (0) exp (μi (s (τ ) , r (τ ) , xi (τ )) −D) dτ , 0  t0 = xi (0) exp (μi (s (τ ) , r (τ ) , xi (τ )) −D) dτ 0  t + (μi (s (τ ) , r (τ ) , xi (τ )) −D) dτ . t0

Hence,

 xi (t) = C x i (0) exp

t

 (μi (s (τ ) , r (τ ) , xi (τ )) −D) dτ ,

t0

where

 C := exp

t0

 (μi (s (τ ) , r (τ ) , xi (τ )) −D) dτ

> 0.

0

By assumption H2, we obtain :  xi (t) < C x i (0) exp  < C x i (0) exp

t  t0 t 

   μi s ◦ + ε, r ◦ + ε, xi (τ ) −D dτ     μi s ◦ + ε, r ◦ + ε, 0 −D dτ .

t0

Finally, we have : xi (t) < C x i (0) exp (Ai (t − t0 )) ,

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where Ai := μi (s ◦ + ε, r ◦ + ε, 0) −D. The real constant Ai is negative since s ◦ + ε < λsi or r ◦ + ε < λri for all i ∈ {1, 2, . . . , n}. Finally, the solutions of the model (with positive initial conditions) are positive, then : lim xi (t) = 0.

t→+∞

Now, let us set 1 := s +

n  1 xi , ysi i=1

n  1 2 := r + xi , yri i=1

which transforms the system (3) in : ˙ 1 = D(s ◦ − 1 ),  ˙ 2 = D(r ◦ − 2 ), 



  n n   1 1 xi , 2 − xi ,x j − D x j , j = 1..n. x˙ j = μ j 1 − ysi yri i=1

(4)

i=1

Hence, the solutions of the two first equations are simply :   1 (t) = s ◦ + 1 (0) − s ◦ e−Dt ,   2 (t) = r ◦ + 2 (0) − r ◦ e−Dt . Finally, we deduce : lim s (t) = s ◦ and lim r (t) = r ◦ .

t→+∞

t→+∞

Remark 2.3 In what follows, we suppose that μi (s ◦ , r ◦ , 0) > D (i.e. s ◦ > λsi and r ◦ > λri ).

2.2 A persistence result Let us additionally assume that, for all i ∈ {1, 2, .., n}, H3. There exists a real number h ∈]0, 1] such that pi (s, xi ) ≤ pi (s, 0) qi (r, xi ) ≤ qi (r, 0)

1 (1 + xi )h 1 (1 + xi )h

, .

H4. The functions s  → pi (s, 0) and r  → qi (r, 0) are, respectively, bounded by two real constants m si > 0 and m ri > 0. Note that, the functions pi and qi being C 1 , there exist two real constants K si > 0 and K ri > 0 such that pi (s, 0) ≤ K si s and qi (r, 0) ≤ K ri r.

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H5. The following inequalities, where Mi := min (m si , m ri ) > D, are satisfied : ⎛ ⎞  1 n  h 2M 1 i ◦ s > D+ K si ⎝ − 1⎠ , y D i=1 si ⎛ ⎞  1 n  h 2M 1 i K ri ⎝ − 1⎠ . r◦ > D + yri D i=1

H6. The functions μi satisfy μi (D, D, 0) > D. Note that the assumption μi (s ◦ , r ◦ , 0) := min ( pi (s ◦ , 0) , qi (r ◦ , 0)) > D is a consequence of assumptions H5 and H6. Remark 2.4 We give an example of a density-dependent growth function which is often used in biology and which satisfies the assumptions H3 and H4, namely the Beddington–DeAngelis type growth function : p (s, x) =

ms , s + k + ax

where m, k, and a are positive constants. Note that this function satisfies obviously the assumptions H1 and H2. One has : p (s, 0) =

ms . s+k

Assumption H4 is clearly satisfied by p (s, 0). We show that : p (s, x) = p (s, 0) with A(s) :=

A(s) , A(s) + x

s+k . Now, if A(s) ≤ 1, it is obvious that : a A(s) 1 ≤ . A(s) + x 1+x

Hence, we obtain : p (s, x) ≤ p (s, 0)

1 , 1+x

which is the condition H3 with h = 1. On the other hand, knowing that s < s ◦ we show that if A(s) > 1 then : A(s) C ≤ , A(s) + x 1+x where C := A(s ◦ ) > 1. Hence, p (s, x) ≤ p (s, 0)

C . 1+x

Finally, since C > 1, the two inequalities imply : p (s, x) ≤ p (s, 0)

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1 . 1+x

Persistent competition models on two complementary nutrients…

Now, we can state and prove the main theorem of this section, which gives sufficient conditions of persistence of the model (3). Theorem 2.5 If assumptions H1 to H6 are satisfied, then the model (3) is persistent. Proof According to assumption H4, one has μi (s, r, 0) ≤ Mi . From the equations of xi in (3) and assumption H3, one deduces that   Mi x˙i ≤ − D xi . (1 + xi )h If

Mi

(1 + xi ) Hence,

h

≤

D D then x˙i ≤ − xi < 0. In other words, if xi ≥ 2 2  ∃t1 > 0 such that ∀t ≥ t1 xi (t) <

 Note that

2Mi D

2Mi D



2Mi D

1 h

− 1 then x˙i < 0.

1 h

− 1.

1 h

> 1 since Mi > D. In the other hand, for all t ≥ t1 , one has



 1 n  2Mi h 1 s˙ > s − s D − μi (s, r, xi ) −1 , y D i=1 si 

 1 n    ◦ 2Mi h 1 μi (s, r, xi ) −1 . r˙ > r − r D − yri D 



◦

i=1

According to assumption H2, one can see that μi (s, r, xi ) ≤ μi (s, r, 0). Hence, 

 1 n  2Mi h 1 s˙ > s − s D − μi (s, r, 0) −1 , y D i=1 si 

 1 n   ◦  2Mi h 1 r˙ > r − r D − μi (s, r, 0) −1 . yri D 



◦

i=1

Moreover, from H4, we know that μi (s, r, 0) ≤ pi (s, 0) ≤ K si s, μi (s, r, 0) ≤ qi (r, 0) ≤ K ri r. Therefore, 

 1 n  2Mi h 1 K si − 1 s, s˙ > s − s D − y D i=1 si 

 1 n    ◦ 2Mi h 1 K ri − 1 r. r˙ > r − r D − yri D 

◦



i=1

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F. Borsali, K. Yadi

If s, r ∈]0, D], then 



 1 n  2Mi h 1 s˙ > s − D − K si − 1 s, y D i=1 si  

 1 n  2Mi h 1 ◦ K ri − 1 r. r˙ > r − D − yri D ◦

i=1

According to assumption H5, one concludes that s˙ > 0 and r˙ > 0 for all s, r ∈]0, D]. Hence,



∃t2 > t1 : ∀t ≥ t2 , s(t) > D and ∃ t2 > t1 : ∀t ≥ t2 , r (t) > D, then

∀t ≥ max(t 2 , t2 ), x˙i = (μi (s, r, xi ) − D) xi > (μi (D, D, xi ) − D) xi . Now, by continuity of the function μi with respect to xi , for all εi > 0, there exists δ > 0 such that |μi (D, D, xi ) − μi (D, D, 0)| ≤ εi for all xi ≤ δ. Hence, for all xi ≤ δ one has μi (D, D, xi ) − D ≥ μi (D, D, 0) − εi − D. Ai Let Ai := μi (D, D, 0) − D which is positive according to H6. If we chose εi = , 2 then there exists δ > 0 such that, for all xi ≤ δ, μi (D, D, xi ) − D ≥

Ai , 2

which implies that ∀xi ≤ δ, x˙i >

Ai xi > 0. 2

Finally,  

∃t3 > max t2 , t2 : ∀t ≥ t3 , xi (t) > δ, which establishes the persistence of the model (3).



3 Global asymptotic stability in case of two species Consider the model 2    1 μi (s, r, xi ) xi , s˙ = s ◦ − s D − ysi i=1

2    1 r˙ = r ◦ − r D − μi (s, r, xi ) xi , yri

(5)

i=1

x˙1 = (μ1 (s, r, x1 ) − D) x1 , x˙2 = (μ2 (s, r, x2 ) − D) x2 , which is the system (3) with n = 2 and suppose that assumptions H1 and H2 are still satisfied.

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Persistent competition models on two complementary nutrients…

3.1 Uniform persistence In the following Lemma, we omit the subscribe i for convenience.  Lemma 3.1 The equation μ ]0, min(ys

s◦,

yr

r ◦ )[.

s◦

1 1 − x, r ◦ − x, x ys yr

 − D = 0 has a unique solution in

Proof Define :   1 1 F (x) := μ s ◦ − x, r ◦ − x, x − D. ys yr One has :     F (0) = μ s ◦ , r ◦ , 0 − D > 0 and F min(ys s ◦ , yr r ◦ ) = −D < 0. We check now that the function F is decreasing. Recall that both functions p and q are increasing with respect to their first argument and decreasing with respect to the second one. Let x1 < x2 .       1 1 1 p s ◦ − x1 , x1 > p s ◦ − x2 , x1 > p s ◦ − x2 , x2 ys ys ys      1 1  min p s ◦ − x2 , x2 , q r ◦ − x2 , x2 , ys yr then

    1 1 1 p s ◦ − x1 , x1 > μ s ◦ − x2 , r ◦ − x2 , x2 . ys ys yr

Likewise,       1 1 1 q r ◦ − x1 , x1 > q r ◦ − x2 , x1 > q r ◦ − x2 , x2 yr yr yr      1 1 ◦ ◦  min p s − x2 , x2 , q r − x2 , x2 , ys yr then

    1 1 1 q r ◦ − x1 , x1 > μ s ◦ − x2 , r ◦ − x2 , x2 . yr ys yr

These two inequalities imply     1 1 1 1 μ s ◦ − x1 , r ◦ − x1 , x1 > μ s ◦ − x2 , r ◦ − x2 , x2 . ys yr ys yr We deduce that F (x1 ) > F (x2 ). Hence, the equation F (x) = 0 has a unique solution in ]0, min(ys s ◦ , yr r ◦ )[.

We can make the additional following assumption :

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 H7. The unique solution x˜i of the equation μi s ◦ − y1si xi , r ◦ − ]0, min(ysi s ◦ , yri r ◦ )[ verifies :   x˜2 ◦ x˜2 ,r − , 0 − D > 0, μ1 s ◦ − ys2 yr 2   x ˜ x˜1 1 ◦ ◦ ,r − , 0 − D > 0. μ2 s − ys1 yr 1

1 yri x i , x i



− D = 0 in

Remark 3.2 The assumption μi (s ◦ , r ◦ , 0) := min ( pi (s ◦ , 0) , qi (r ◦ , 0)) > D is a consequence of the assumption H7 (see Remark 2.3). Concerning the function μi satisfying hypothesis H7, we reconsider here the example of Beddington–DeAngelis growth function and show that the solutions x˜i can be explicitly determined. This can be done also for the so-called Contois growth function p (s, x) = ms/(s + kx), where m, k > 0, but, unfortunately, assumption H4 is not satisfied in this case. For convenience, we will consider the equation : μ(s ◦ − αx, r ◦ − βx, x) − D = 0, where α = 1/ys and β = 1/yr . Let us set : F (x) := μ(s ◦ − αx, r ◦ − βx, x) − D, with μ (s, r, x) := min ( p (s, x) , q (r, x)) . Consider then

p (s, x) =

ms mr , , q (r, x) =

s + k + ax r + k + a x

where m, m , k, k , a and a are positive constants. If F (x) = p (s ◦ − αx, x) − D, then : F (x) = 0 ⇔ (α (m − D) + a D)x = (m − D) s ◦ − k D (m − D) s ◦ − k D ⇔x= > 0. α (m − D) + a D The positiveness of x is a consequence of the condition of non-washing out (Remark 2.3). The case F (x) = q (r ◦ − βx, x) − D is similar. We can state the following result giving sufficient conditions of permanence of the system (5). The proof will use a corollary 2 of the paper [7] cited in the appendix. This corollary is a consequence of Hofbauer’s and Hutson’s theorem [9,11]. Theorem 3.3 If assumptions H1, H2 and H7 are satisfied, then the model (5) is permanent. Proof According to Proposition 2.1, the system (5) is dissipative and the following bounded set   C := (s, r, x1 , x2 ) ∈ R4+ : s ≤ s ◦ , r ≤ r ◦ , xi ≤ min(ysi s ◦ , yri r ◦ )

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Persistent competition models on two complementary nutrients…

is an attracting one, in the sense that all trajectories eventually lie in it. By setting : 1 := s +

2  1 xi , ysi i=1

2 := r +

2  1 xi , yri i=1

the system (5) becomes : ˙ 1 = D(s ◦ − 1 ),  ˙ 2 = D(r ◦ − 2 ), 



  2 2   1 1 x˙1 = μ1 1 − xi , 2 − xi ,x1 − D x1 , y y i=1 si i=1 ri



  2 2   1 1 x˙2 = μ2 1 − xi , 2 − xi ,x2 − D x2 . ysi yri i=1

(6)

i=1

The solutions of the two first equations are :   1 (t) = s ◦ + 1 (0) − s ◦ e−Dt ,   2 (t) = r ◦ + 2 (0) − r ◦ e−Dt . The couple ( 1 (t) , 2 (t)) converges exponentially to (s ◦ , r ◦ ). Hence, the ω-limit set of all positive trajectories of the model (5) lies in the simplex :   2 2   1 1 ◦ ◦

:= (s, r, x1 , x2 ) ∈ C : s + xi = s , r + xi = r . ysi yri i=1

i=1

The system (6) reduced on is given by :



  2 2   1 1 ◦ ◦ x˙1 = μ1 s − xi , r − xi ,x1 − D x1 , y y i=1 si i=1 ri



  2 2   1 1 ◦ ◦ xi , r − xi ,x2 − D x2 , x˙2 = μ2 s − ysi yri i=1

and is defined on :



X := (x1 , x2 ) ∈

R2+

(7)

i=1

 2 2   1 1 ◦ ◦ : xi ≤ s and xi ≤ r . ysi yri i=1

i=1

X is obviously a closed set. Let us show that it is positively invariant. Suppose that 2 2 1 1 xi (0) = s ◦ and i=1 xi (0) = r ◦ . One can see that : i=1 ysi yri  

2

2   d  1 d  1   ◦ xi (t)  = −Ds < 0, xi (t)  = −Dr ◦ < 0.   dt ysi dt yri i=1

t=0

i=1

t=0

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F. Borsali, K. Yadi

Hence, for all t > 0 : 2 2   1 1 xi (t) < s ◦ and xi (t) < r ◦ . ysi yri i=1

i=1

On the other hand, one has : 

t

xi (t) = xi (0) exp

μi

0

   2 2   1 1 ◦ xi (τ ), r − xi (τ ), xi (τ ) − D dτ . s − ysi yri ◦

i=1

i=1

Hence, xi (t) ≥ 0 when xi (0) ≥ 0. 2 ∩ X = [0, min(y s ◦ , y r ◦ )] ∪ [0, min(y s ◦ , y r ◦ )]. The set Now, let set S := ∂ R+ s1 r1 s2 r2 S is obviously compact and positively invariant. Define the function P : X → R+ by P (x1 , x2 ) := x1 x2 , which clearly vanishes on S. We compute the derivative of P at (x1 , x2 ) ∈ X \S in the direction of the field F defined by the reduced system (7). P˙ (x1 , x2 ) := grad P (x1 , x2 ) , F (x1 , x2 ) =

2  ∂ P (x1 , x2 ) j=1

∂x j

F j (x1 , x2 )

 2

  2 2    1 1 ◦ ◦ = x1 x2 μi s − xi , r − xi ,xi − 2D . ysi yri i=1

i=1

i=1

Let ψ : X −→ R defined by : ψ (x1 , x2 ) :=

2  i=1

μi

2 2   1 1 xi , r ◦ − xi ,xi s − ysi yri ◦

i=1

 − 2D.

i=1

One has : P˙ (x1 , x2 ) = P (x1 , x2 ) ψ (x1 , x2 ) ∀ (x1 , x2 ) ∈ X \S. The function ψ is bounded below by −2D. The condition (i) of Corollary A.1 of the appendix is then verified. For the condition (ii), it is sufficient to check that ψ (E k ) > 0 where E k is an equilibrium point in S. Indeed, the unidimensional set S is positively invariant and its ω-limit set, denoted by (S), is formed by the equilibrium points E k in S. Let us determine these equilibrium points. If (x1 (0), x2 (0)) ∈ S such that x1 (0) > 0 and x2 (0) = 0, then x2 (t) ≡ 0 and the system (7) is written :     1 1 ◦ ◦ x˙1 = g1 s − x1 , r − x1 , x1 − D x1 . ys1 yr 1   1 1 ◦ ◦ The nullclines are g1 s − x1 , r − x1 , x1 − D = 0 or x1 = 0. If x1 = 0, then ys1 yr 1 E 0 (0, 0) is an equilibrium point in S. If x1  = 0, then according to assumption H7, the first equation has a unique solution x˜1 in ]0, min(ys1 s ◦ , yr 1 r ◦ )[. Hence, E 1 (x˜1 , 0) is an equilibrium point in S. Analogously, if (x1 (0), x2 (0)) ∈ S such that x1 (0) = 0 and x2 (0) > 0, then there exists another equilibrium point E 2 (0, x˜2 ) for x˜2 ∈]0, min(ys2 s ◦ , yr 2 r ◦ ). Hence,

(S) = {E 0 , E 1 , E 2 }. From assumption H7, one has :

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Persistent competition models on two complementary nutrients…

ψ (E 0 ) =

2    μi s ◦ , r ◦ , 0 − 2D > 0, i=1

 x˜1 ◦ ,r − ψ (E 1 ) = μ2 s ◦ − ys1  x˜2 ◦ ,r − ψ (E 2 ) = μ1 s ◦ − ys2

 x˜1 , 0 − D > 0, yr 1  x˜2 , 0 − D > 0. yr 2

The condition (ii) of Corollary A.1 is then satisfied which implies that S is a uniform repeller and that the model (7) is uniformly persistent. Finally, since lim  1 (t) = s ◦ and lim  2 (t) = r ◦ , the whole system (5) is uniformly persistent.

t→+∞

t→+∞



3.2 Global asymptotic stability Uniform persistence and properties of the planar systems lead to the following result of existence of a G.A.S. equilibrium point. Theorem 3.4 Under assumptions H1, H2 and H7, the system (5) has a positive G.A.S. equilibrium point. Proof First, we determine the trivial equilibrium points of the reduced system (7) and their nature. The Jacobian matrix at an arbitrary point E(x1 , x2 ) is given by :  μ1 − D + x1 a1 x1 b12 J (E) = , x2 b21 μ2 − D + x2 a2 with the following notations :

 2 2   1 1 ◦ ◦ xi , r − xi ,x j , μ j := μ j s − y y i=1 si i=1 ri



 2 2   ∂pj 1 1 1 ∂pj ◦ ◦ xi , x j + xi , x j or a j := − s − s − ys j ∂s y ∂x j y i=1 si i=1 si 



2 2   ∂q j 1 ∂q j 1 1 ◦ ◦ xi , x j + xi , x j , r − a j := − r − yr j ∂r y ∂x j y i=1 si i=1 si

 2  1 1 ∂pj ◦ b jk := − xi , x j or s − ysk ∂s y i=1 si 

2  1 ∂q j 1 ◦ xi , x j with k ∈ {1, 2} , k  = j. b jk := − r − yr k ∂r ysi i=1

The nullclines equations are given by :  



2 2   1 1 ◦ ◦ xi , r − xi ,x1 − D x1 = 0, μ1 s − y y i=1 si i=1 ri



  2 2   1 1 ◦ ◦ xi , r − xi ,x2 − D x2 = 0. μ2 s − ysi yri i=1

i=1

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F. Borsali, K. Yadi

1st case : x1 = x2 = 0. The Jacobian matrix at the equilibrium point E 0 (0, 0) is simply :  μ1 (s ◦ , r ◦ , 0) − D 0 , J (E 0 ) = 0 μ2 (s ◦ , r ◦ , 0) − D the eigenvalues of which are vi0 := μi (s ◦ , r ◦ , 0) − D, i ∈ {1, 2}. The origin E 0 unstable, since μi (s ◦ , r ◦ ) > D. 2nd case : x1  = 0, x2 = 0. According to assumption H7, the equation :   1 1 ◦ ◦ μ1 s − x1 , r − x1 , x1 − D = 0, ys1 yr 1 has a unique solution x˜1 in ]0, min(ys1 s ◦ , yr 1 r ◦ )[. The Jacobian matrix evaluated at the equilibrium point E 1 (x˜1 , 0) is given by :  x˜ a˜ x˜1 b˜12 , J (E 1 ) = 1 1 0 μ˜ 2 − D where



◦

μ˜ 2 := μ2 s − 1 ∂ p1 ys1 ∂s 1 ∂q1 a˜ 1 := − yr 1 ∂r 1 ∂ p1 b˜12 := − ys2 ∂s 1 ∂q1 b˜12 := − yr 2 ∂r a˜ 1 := −

 1 1 ◦ x˜1 , r − x˜1 ,0 , ys1 yr 1     ∂ p1 ◦ 1 1 s − s◦ − x˜1 , x˜1 + x˜1 , x˜1 or ys1 ∂ x1 ys1     1 1 ∂q 1 x˜1 , x˜1 + x˜1 , x˜1 , r◦ − r◦ − yr 1 ∂ x1 yr 1   1 x˜1 , x˜1 or s◦ − ys1   1 r◦ − x˜1 , x˜1 . yr 1

The eigenvalues of J (E 1 ) are v11 := x˜1 a˜ 1 and v21 :=μ˜ 2 − D. According to assumption H2, the value v11 is negative while, according to assumption H7, the value v21 is positive. The equilibrium point E 1 is unstable (saddle point). 3r d case : x1 = 0, x2  = 0. This case is analogous to the second one. The obtained equilibrium point, also a saddle point, is E 2 (0, x˜2 ) where x˜2 is the unique solution of :   1 1 μ2 s ◦ − x2 , r ◦ − x2 , x2 − D = 0, ys2 yr 2 in ]0, min(ys2 s ◦ , yr 2 r ◦ )[. Now, let us show that the model (7) has no limit cycle with the use of Dulac–Bendixson Criterion [18]. Let A be the simply connected attracting domain of the system (7):   A := (x1 , x2 ) ∈ R2+ : ε ≤ xi ≤ min(ysi s ◦ , yri r ◦ ) , where ε is a positive real number such that lim inf xi (t) > ε. Such a uniform bound exists t→+∞

according the Theorem 3.3. Let B : A −→ R the C 1 function defined by B (x1 , x2 ) := 1/x1 x2 and denote by F := (F1 , F2 ) the vector field of (7) :

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Persistent competition models on two complementary nutrients…





  2 2   1 1 ◦ F1 (x1 , x2 ) := μ1 s − xi , r − xi ,x1 − D x1 , y y i=1 si i=1 ri



  2 2   1 1 ◦ ◦ xi , r − xi ,x2 − D x2 . F2 (x1 , x2 ) := μ2 s − ysi yri ◦

i=1

i=1

The divergence of the field B.F is then given by : div ((B.F) (x1 , x2 )) :=

∂ ∂ (B (x1 , x2 ) F1 (x1 , x2 )) + (B (x1 , x2 ) F2 (x1 , x2 )) . ∂ x1 ∂ x2

Simple calculations lead to ∂ 1 1 ∂ (B (x1 , x2 ) F1 (x1 , x2 )) = a1 and (B (x1 , x2 ) F2 (x1 , x2 )) = a2 , ∂ x1 x2 ∂ x2 x1 where 1 ∂pj a j := − ys j ∂s 1 ∂q j a j := − yr j ∂r



2  1 xi , x j s − ysi



◦



i=1

2  1 xi , x j r − ysi ◦

i=1



∂pj + ∂x j ∂q j + ∂x j



2  1 xi , x j s − ysi ◦



 or

i=1

 2  1 xi , x j . r − ysi ◦

i=1

Finally, div ((B.F) (x1 , x2 )) =

1 1 a1 + a2 . x2 x1

According to assumption H2, the quantity div ((B.F) (x1 , x2 )) is negative for all (x1 , x2 ) in A. Hence, according to Dulac–Bendixson Criterion and Poincaré–Bendixson theorem, the model (7) has no limit cycle and admits at least a positive equilibrium point in A. Let E(x1∗ , x2∗ ) be such a point and let us evaluate the Jacobian matrix at it :  ∗ ∗ ∗ ∗ x1 a1 x1 b12 J (E) = ∗ x ∗a∗ , x2∗ b21 2 2 where : a ∗j

1 ∂pj := − ys j ∂s

a ∗j := −

1 ∂q j yr j ∂r

b∗jk := −

1 ∂pj ysk ∂s

b∗jk := −

1 ∂q j yr k ∂r





 2 2   ∂pj 1 ∗ ∗ 1 ∗ ∗ ◦ x ,x + x ,x s − s − or y i j ∂x j y i j i=1 si i=1 si 



2 2   ∂q 1 1 j x ∗, x ∗ + x ∗, x ∗ , r◦ − r◦ − y i j ∂x j y i j i=1 si i=1 si 

2  1 ∗ ∗ ◦ x ,x or s − y i j i=1 si

 2  1 ∗ ∗ ◦ x ,x r − with k ∈ {1, 2} , k  = j. ysi i j ◦

i=1

The trace T r (J (E)) = x1∗ a1∗ + x2∗ a2∗ of J (E) is clearly negative. The determinant is given ∗ . If, for example, at the equilibrium point, μ = p ( by det (J (E)) = x1∗ x2∗ a1∗ a2∗ −b∗12 b21 j j the case μ j = q j is similar), one has :

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F. Borsali, K. Yadi

det (J (E)) =

∂ p1 ∂ p2 1 ∂ p1 ∂ p2 1 ∂ p2 ∂ p1 − − > 0. ∂ x1 ∂ x2 ys1 ∂s ∂ x2 ys2 ∂s ∂ x1

The equilibrium point E is then locally asymptotically stable. Therefore, knowing that there is no limit cycle, and that the equilibrium E is an arbitrary one, which is locally asymptotically stable, one can conclude that E is unique and G.A.S.. Return back to the whole system (6). In the proof of Theorem 3.3, we showed that the equilibrium point E sr (s ◦ , r ◦ ) is G.A.S. for the first two equations of the model (6). Moreover, the solutions of (6) are positively bounded. Then, one can Principle Lemma  use the Separation  [21] to conclude that the positive equilibrium point E ∗ s ◦ , r ◦ , x1∗ , x2∗ of the model (6), then   2 1 2 1 ∗ ∗ ∗ ∗ ◦ ◦ x ,r − x , x1 , x2 of the also the positive equilibrium point s − i=1 ysi i i=1 yri i initial model (5), is G.A.S.



4 Uniform persistence of three species of microorganisms Reconsider the model (3) with n = 3. 3    1 μi (s, r, xi ) xi , s˙ = s ◦ − s D − ysi i=1

3    1 r˙ = r ◦ − r D − μi (s, r, xi ) xi , yri

(8)

i=1

x˙i = (μi (s, r, xi ) − D) xi , i ∈ {1, 2, 3} . We show in this section that we are able to say more than for the general case even if we do not prove the existence of a G.A.S. point. We assume that hypotheses H1 to H7 for all i ∈ {1, 2, 3} are fulfilled, where H7 should be understood as follows : the unique   1 1 solution x˜i of the equation μi s ◦ − xi , r ◦ − xi , xi −D = 0 in ]0, min(ysi s ◦ , yri r ◦ )[ ysi yri satisfies :  μi

 x˜ j ◦ x˜ j ,r − , 0 − D > 0, i, j ∈ {1, 2, 3}, i  = j. s − ys j yr j ◦

We can state and show that : Theorem 4.1 Under assumptions H1–H7, the model (8) is permanent. Proof 1. Reduction : According to Proposition 2.1, the system (3) is dissipative and the following set in an attracting one :   C := (s, r, x1 , x2 , x3 ) ∈ R5+ : s ≤ s ◦ , r ≤ r ◦ , xi ≤ min(ysi s ◦ , yri r ◦ ) . If we set 1 := s +

123

3

3 1 1 xi and 2 := r + xi , the system (8) becomes : i=1 ysi i=1 yri

Persistent competition models on two complementary nutrients…

˙ 1 = D(s ◦ − 1 ),  ˙ 2 = D(r ◦ − 2 ), 



  3 3   1 1 x˙1 = μ1 1 − xi , 2 − xi , x1 − D x1 , y y i=1 si i=1 ri



  3 3   1 1 xi , 2 − xi , x2 − D x2 , x˙2 = μ2 1 − y y i=1 si i=1 ri



  3 3   1 1 x˙3 = μ3 1 − xi , 2 − xi , x3 − D x3 , ysi yri i=1

(9)

i=1

the ω-limit set of which lies in the simplex : 

 3 3   1 1 ◦ ◦

:= (s, r, x1 , x2 , x3 ) ∈ C : s + xi = s , r + xi = r . ysi yri i=1

i=1

On the system (9) is reduced to :



  3 3   1 1 ◦ xi , r − xi , x1 − D x1 , x˙1 = μ1 s − y y i=1 si i=1 ri  



3 3   1 1 ◦ ◦ xi , r − xi , x2 − D x2 , x˙2 = μ2 s − y y i=1 si i=1 ri  



3 3   1 1 ◦ ◦ x˙3 = μ3 s − xi , r − xi , x3 − D x3 . ysi yri ◦

i=1

(10)

i=1

The system (10) is defined on :  X := (x1 , x2 , x3 ) ∈

R3+

 3 3   1 1 ◦ ◦ : xi ≤ s and xi ≤ r . ysi yri i=1

i=1

2. Boundary equilibria: In order to use Thieme–Zhao’s theorem, we set X 1 := int X and X 2 := ∂ X . Hence X = X 1 ∪ X 2 , X 1 ∩ X 2 = ∅. The set X 2 is a closed subset of X . Moreover, X 1 is positively invariant for (10) since, for all positive initial condition, one has : 

t

xi (t) = xi (0) exp

μi

0

   3 3   1 1 ◦ s − xi (τ ), r − xi (τ ), xi (τ ) − D dτ > 0. ysi yri ◦

i=1

i=1

Now, let us determine the set M of the equilibria of the system (10) in X 2 . First, we claim that any solution of this system starting in X 1 converges to a point of M. Indeed, in the proof of Theorem 2.5, we have already showed that, under assumptions H1–H6, there exists a positive real number δ and a positive instant t0 such that, for all t ≥ t0 and all i ∈ {1, 2, . . . , n}, one has xi (t) > δ. Hence, lim inf xi (t) > 0 for all component xi (t) of a positive solution of the t→+∞

system (10 ). Now, the jacobian matrix of (10) at an arbitrary point E(x1 , x2 , x3 ) is given by :

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F. Borsali, K. Yadi

⎡

⎤ x1 b12 x1 b13 μ1 − D + x1 a1 ⎦, x2 b21 μ2 − D + x2 a2 x2 b23 J (E) = ⎣ x3 b31 x3 b32 μ3 − D + x3 a3 where :



 3 3   1 1 ◦ xi , r − xi ,x j , μ j := μ j s − y y i=1 si i=1 ri



 3 3   ∂pj 1 1 1 ∂pj ◦ ◦ xi , x j + xi , x j or a j := − s − s − ys j ∂s y ∂x j y i=1 si i=1 si



 3 3   ∂q j 1 1 1 ∂q j ◦ ◦ xi , x j + xi , x j , r − r − a j := − yr j ∂r y ∂x j y i=1 si i=1 si 

3  1 1 ∂pj ◦ xi , x j or b jk := − s − ysk ∂s y i=1 si 

3  1 1 ∂q j ◦ xi , x j with k ∈ {1, 2, 3} , k  = j. b jk := − r − yr k ∂r ysi ◦

i=1

The nullclines being :



  3 3   1 1 ◦ xi , r − xi , x1 − D x1 = 0, μ1 s − y y i=1 si i=1 ri  



3 3   1 1 ◦ ◦ xi , r − xi , x2 − D x2 = 0, μ2 s − y y i=1 si i=1 ri



  3 3   1 1 ◦ ◦ xi , r − xi , x3 − D x3 = 0, μ3 s − ysi yri ◦

i=1

i=1

we distinguish the following cases : If x1 = x2 = x3 = 0, the origin E 0 (0, 0, 0) ∈ X 2 is an equilibrium point at which the jacobian matrix is : ⎡ ⎤ μ1 (s ◦ , r ◦ , 0) − D 0 0 ⎦. 0 μ2 (s ◦ , r ◦ , 0) − D 0 J (E 0 ) = ⎣ 0 0 μ3 (s ◦ , r ◦ , 0) − D Since the eigenvalues are all positive, E 0 is unstable. If x1  = 0, x2 = x3 = 0, according to assumption H7, the nullcline   1 1 μ1 s ◦ − x1 , r ◦ − x1 , x1 − D = 0, ys1 yr 1 has a unique solution x˜1 in ]0, min(ys1 s ◦ , yr 1 r ◦ )[. The jacobian matrix of the system (10 ) at the boundary equilibrium point E 1 (x˜1 , 0, 0) ∈ X 2 is : ⎡ ⎤ x˜1 a˜ 1 x˜1 b˜12 x˜1 b˜13 ⎦, J (E 1 ) = ⎣ 0 μ˜ 2 − D 0 0 0 μ˜ 3 − D

123

Persistent competition models on two complementary nutrients…

where

  1 1 μ˜ j := μ j s ◦ − x˜ , r ◦ − x˜1 , 0 , j ∈ {2, 3} , ys1 yr 1  1    1 ∂ p1 ◦ 1 1 ∂ p1 ◦ x˜1 , x˜1 + x˜1 , x˜1 or a˜ 1 := − s − s − ys1 ∂s ys1 ∂ x1 ys1     ∂q1 ◦ 1 ∂q1 ◦ 1 1 r − r − x˜1 , x˜1 + x˜1 , x˜1 , a˜ 1 := − yr 1 ∂r yr 1 ∂ x1 yr 1   ∂ p 1 1 1 x˜1 , x˜1 or s◦ − b˜1k := − ysk ∂s ys1   1 ∂q1 ◦ 1 r − x˜1 , x˜1 , k ∈ {2, 3} . b˜1k := − yr k ∂r yr 1

The eigenvalues of J (E 1 ) are v11 := x˜1 a˜ 1 , v21 := μ˜ 2 − D and v31 := μ˜ 3 − D. The first one is negative according to H2, the others are positives according to H7. The equilibrium E 1 is then a saddle point. The stable separatrix lies on the x1 −axis. The other two cases where either x2  = 0 or x3  = 0 is analogous and determines two other saddle points E 2 (0, x˜2 , 0) and E 3 (0, 0, x˜3 ) in X 2 for which the stable separatrices are, respectively, the x 2 −axis and the x3 −axis. If x1  = 0, x2  = 0, x3 = 0, the nontrivial nullclines are :

 2 2   1 1 ◦ ◦ μ1 s − xi , r − xi ,x1 − D = 0, y y i=1 si i=1 ri

 2 2   1 1 ◦ ◦ xi , r − xi ,x2 − D = 0. μ2 s − ysi yri i=1

i=1

In the proof of Theorem 3.4 we showed that the model (7) has a unique positive equilibrium which is globally asymptotically stable. Hence, the system (10) has an equilibrium point E 12 (x 1 , x 2 , 0) in X 2 , at which the jacobian matrix is : ⎡ ⎤ x 1 a¯ 1 x 1 b¯12 x 1 b¯13 J (E 12 ) = ⎣ x 2 b¯21 x 2 a¯ 2 x 2 b¯23 ⎦ , 0 0 μ¯ 3 − D where :



 2 2   1 1 ◦ μ¯ j := μ j s − x¯i , r − x¯i ,x¯ j , y y i=1 si i=1 ri



 2 2   ∂pj 1 1 1 ∂pj ◦ ◦ x¯i , x¯ j + x¯i , x¯ j or a¯ j := − s − s − ys j ∂s y ∂x j y i=1 si i=1 si



 2 2   ∂q j 1 1 1 ∂q j ◦ ◦ x¯i , x¯ j + x¯i , x¯ j , r − r − a¯ j := − yr j ∂r y ∂x j y i=1 si i=1 si

 2  ∂pj 1 ◦ ¯b jk := − 1 x¯i , x¯ j or s − ysk ∂s y i=1 si

 2  ∂q j 1 1 ◦ x¯i , x¯ j with k ∈ {1, 2, 3} , k  = j. r − b¯ jk := − yr k ∂r ysi ◦

i=1

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F. Borsali, K. Yadi

The characteristic equation of J (E 12 ) is given by :   (μ3 − D − v) v 2 + Av + B = 0, where A := −x 1 a 1 − x 2 a 2 ,   B := x 1 x 2 a 1 a 2 − b¯12 b¯21 . One has : 1 ∂ p1 ∂ p2 1 ∂ p1 ∂ p2 ∂ p 1 ∂ p2 − + ys1 ys2 ∂s ∂s ∂ x1 ∂ x2 ys1 ∂s ∂ x2 1 ∂ p2 ∂ p1 1 ∂ p1 1 ∂ p2 > 0, − − ys2 ∂s ∂ x1 ys2 ∂s ys1 ∂s

a 1 a 2 − b¯12 b¯21 =

or 1 ∂q1 ∂q2 1 ∂q1 ∂q2 ∂q1 ∂q2 − + yr 1 yr 2 ∂r ∂r ∂ x1 ∂ x2 yr 1 ∂r ∂ x2 1 ∂q2 ∂q1 1 ∂q1 1 ∂q2 − − > 0. yr 2 ∂r ∂ x1 yr 2 ∂r yr 1 ∂r

a 1 a 2 − b¯12 b¯21 =

Hence, the coefficients A and B are positive. According to Routh Criterion (see for example [18]), the equation v 2 + Av + B = 0 has two eigenvalues with negative real parts. The persistence result of Theorem 2.5 implies that the other eigenvalue μ3 − D must be positive. Hence, E 12 is a saddle point in X 2 . The stable manifold is the x1 x 2 -plane. If x1  = 0, x3  = 0, x2 = 0, or x2  = 0, x3  = 0, x1 = 0, one obtains two other saddle points E 13 and E 23 in X 2 the stable manifolds of which are, respectively, the x1 x 3 -plane and the x2 x 3 -plane. 3. Thieme–Zhao’s conditions: Now, having the set of the equilibria of the system (10) in X , namely M = {E 0 , E 1 , E 2 , E 3 , E 12 , E 13 , E 23 } , we begin to show that the three hypotheses of Thieme–Zhao’s theorem ([19,22], see also the Appendix) are verified. Condition T1 :If (x1 (0), x2 (0) , x3 (0))  = E 0 ∈ M, we distinguish the following cases : If x1 (0)  = 0, x2 (0) = x3 (0) = 0, since xi (t) ≡ 0 for i ∈ {2, 3}, the system (10) simply becomes :     1 1 ◦ ◦ x˙1 = μ1 s − x1 , r − x1 , x1 − D x1 . ys1 yr 1 It admits two equilibria, namely the origin which is clearly unstable and x˜1 in ]0, min(ys1 s ◦ , yr 1 r ◦ )[ for which the global asymptotic stability is obvious according to what precedes and corresponds to the point E 1 . The cases x1 (0) = 0, x2 (0)  = 0, x3 (0) = 0 and x1 (0) = 0, x2 (0) = 0, x3 (0)  = 0 are similar. It means that all solutions starting in X 2 stay in X 2 and converge to E 1 ,E 2 or E 3 .

123

Persistent competition models on two complementary nutrients…

If x1 (0)  = 0, x2 (0)  = 0, x3 (0) = 0, the system (10) becomes :



  2 2   1 1 ◦ ◦ x˙1 = μ1 s − xi , r − xi ,x1 − D x1 , y y i=1 si i=1 ri



  2 2   1 1 ◦ ◦ xi , r − xi ,x2 − D x2 . x˙2 = μ2 s − ysi yri i=1

i=1

In the proof of Theorem 3.4, we already showed that this system has a unique positive equilibrium point which is G.A.S. and denoted by (x 1 , x 2 ). Hence, the solutions of (10) such that x1 (0)  = 0, x2 (0)  = 0, x3 (0) = 0 converge to the corresponding equilibria E 12 (x 1 , x 2 , 0) ∈ M. The cases x1 (0)  = 0, x3 (0)  = 0, x2 (0) = 0 and x2 (0)  = 0, x3 (0)  = 0, x1 (0) = 0 are analogous and the considered solutions converge, respectively, to E 13 and E 23 . Condition T2 : The equilibrium points of M are hyperbolic. The origin is unstable, and the others are saddle points. Therefore, the singleton of each point of M is an isolated invariant set. Moreover, the persistence of the system (3) as showed in Sect. 2 means that the set M is a repeller for X 1 . Condition T3 : Recall that the stable manifolds of E 0 , E i (i ∈ {1, 2, 3}) and E i j (i, j ∈{1, 2, 3} {E 0 }, the axes (O x i ) and the planes  with i < j) are, respectively,  Pi j := xi , x j ∈ R2+ : xi  = 0 and x j  = 0 . Thus, the only chains occurring in X 2 are : E 0 → E i , E 1 → E 12 , E 2 → E 23 , E 3 → E 13 , E 0 → E i j , E 1 → E 13 , E 2 → E 12 , E 3 → E 23 . There is no cycle (closed chain) in X 2 . In virtue of Theorem [19] and [22], X 2 is a uniform repeller for X 1 . Actually, X 1 being a convex set, there exists at least an equilibrium point in X 1 . Hence, the system (10) is uniformly persistent.



5 Numerical simulations We give some examples of a model of two and three microorganisms in competition on two complementary nutrients. First, the growth functions pi , qi are of Beddington–DeAngelis type, then we present a case where the growth functions are of the form pi (s)θi (xi ), qi (s)θi (xi ), where pi and qi are increasing and θi is decreasing1 . The parameters used in these examples do not necessarily have a biological meaning and the simulations are given just as an illustration of the result. Simulation 1 Let the right-hand side functions of the model (5) be given by : 4s 5r , q1 (r, x1 ) = , 2x1 + s + 2 2x1 + r + 1 5s 6r p2 (s, x2 ) = , q2 (r, x2 ) = , x2 + s + 2 2x2 + r + 4 p1 (s, x1 ) =

while the parameters are s ◦ = 10, r ◦ = 20, D = 1, ys1 = 2, ys2 = 3, yr 1 = 1 and yr2 = 7. One can check that the assumptions of the theorem are all satisfied. The numerical simulations have been performed with the use of Maple 18. In 1 This multiplicative form is justified for example by Zwietering at al. [23].

123

F. Borsali, K. Yadi Fig. 1 Eight trajectories of Simulation 1 in the space (x1 , x2 ) with the initial data given in the text

Fig. 1, we represent eight trajectories with initial conditions (s(0), r (0), x1 (0), x2 (0)) = (2.5, 3.5, 15, 25), (2.5, 3.5, 20, 10), (2.5, 3.5, 0.1, 15), (2.5, 3.5, 6, 27), (2.5, 3.5, 15, 0.1), (2.5, 3.5, 1.75, 0.1), (2.5, 3.5, 0.1, 0.1) and (2.5, 3.5, 0.1, 30) in the phase space (x 1 , x2 ). These numerical simulations show that the solutions converge to the positive equilibrium point, in accordance with the uniform persistence and the global asymptotic stability property of Theorems 3.3 and 3.4. Simulation 2 We make the following choice of the functions and the parameters occurring in the model (8) 4s 5r , q1 (r, x1 ) = , 2x1 + s + 2 2x1 + r + 1 5s 6r , q2 (r, x2 ) = , p2 (s, x2 ) = x2 + s + 2 2x2 + r + 4 4s 7r p3 (s, x3 ) = , q3 (r, x3 ) = , x3 + s + 2 x3 + r + 2 p1 (s, x1 ) =

and s ◦ = 30, r ◦ = 35, D = 1, ys1 = 5, ys2 = 8, ys3 = 5, yr 1 = 8, yr 2 = 4 and yr 3 = 9. Six trajectories are represented in Fig. 2 , namely those with initial conditions (s(0), r (0), x1 (0), x2 (0), x3 (0)) = (2.5, 3.5, 30, 85, 1), (2.5, 3.5, 30, 5, 80), (2.5, 3.5, 30, 30, 25) , (2.5, 3.5, 0.5, 0.5, 0.5), (2.5, 3.5, 5, 30, 70) and (2.5, 3.5, 10, 90, 30). The uniform persistence of the three species of microorganisms is confirmed, and the solutions seem to converge to a positive equilibrium point. Simulation 3 Now, the right-hand side functions of (5) are given by : 4s 5r 1 , q1 (r ) = , θ1 (x1 ) = , 2+s 2+r 1 + 1.5x1 5s 6r 1 , p2 (s) = , q2 (r ) = , θ2 (x2 ) = 1+s 2+r 1 + 2x2 p1 (s) =

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Persistent competition models on two complementary nutrients… Fig. 2 Six trajectories of Simulation 2 in the space (x1 , x2 , x3 ) with the initial data given in the text

Fig. 3 Eight trajectories of Simulation 3 in the space (x1 , x2 ) with the initial data given in the text

with s ◦ = 10, r ◦ = 20, D = 1, ys1 = 2, ys2 = 3, yr 1 = 1 and yr2 = 7. In Fig. 3, we represent eight trajectories with initial conditions (s(0), r (0), x1 (0), x2 (0)) = (2.5, 3.5, 5, 5), (2.5, 3.5, 5, 15), (2.5, 3.5, 0.1, 15), (2.5, 3.5, 2.5, 15), (2.5, 3.5, 5, 0.1), (2.5, 3.5, 1.75, 0.1), (2.5, 3.5, 0.1, 0.1) and (2.5, 3.5, 0.1, 6) in the phase space (x 1 , x2 ). Simulation 4 We consider the model (8) with the following functions : 4s 5r 1 , , q1 (r ) = , θ1 (x1 ) = 2+s 2+r 1 + x1 5s 6r 1 p2 (s) = , q2 (r ) = , θ2 (x2 ) = , 1+s 2+r 1 + x2 4s 7r 1 . p3 (s) = , q3 (r ) = , θ3 (x3 ) = 1+s 1+r 1 + x3 p1 (s) =

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F. Borsali, K. Yadi Fig. 4 Six trajectories of Simulation 4 in the space (x1 , x2 , x3 ) with the initial data given in the text

The numerical simulation is performed with the parameters s ◦ = 30, r ◦ = 35, D = 1, ys1 = 5, ys2 = 8, ys3 = 5, yr 1 = 8, yr 2 = 4 and yr 3 = 9. Six trajectories are represented in Fig. 4 , namely those with initial conditions (s(0), r (0), x1 (0), x2 (0), x3 (0)) = (2.5, 3.5, 5, 0.5, 1), (2.5, 3.5, 5, 5, 11), (2.5, 3.5, 5, 1, 5), (2.5, 3.5, 0.5, 15, 3), (2.5, 3.5, 0.5, 5, 7) and (2.5, 3.5, 0.5, 15, 10). Acknowledgements The authors would like to thank the anonymous referee for these helpful comments and recommendations.

A Appendix Corollary A.1 (Hofbauer–Hutson theorem) [7] Let X ⊆ Rn+ be closed, F : X −→ Rn be locally Lipschitz and consider the ordinary differential system x˙ = F(x)

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Assume that X is positively invariant for the system (12). Let S be a compact set and P ∈ C(X, R+ ) ∩ C 1 (X \S, R) be such that P (x) = 0 ⇔ x ∈ S. Moreover, suppose that there exist a lower semicontinuous function ψ : X −→ R, bounded below, and an α ∈ [0, 1] such that : (i) P˙ (x) ≥ [P (x)]α ψ(x) ∀x ∈ X \S, T (ii) ∀x ∈  sup 0 ψ(x (t))dt > 0, T ≥0

¯ where  denotes S or, whenever S is positively invariant, (S). Then S is a uniform repeller. By (S) we denote the whole ω-limit of the orbits on S. Theorem A.2 (Thieme–Zhao) [19,22] Let X ⊆ Rn+ be closed, F : X −→ Rn be locally Lipschitz and consider the ordinary differential system x˙ = F(x)

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Persistent competition models on two complementary nutrients…

Assume that X is positively invariant for the system (12) which is supposed to be dissipative on X . Let X = X 1 ∪ X 2 , X 1 ∩ X 2 = ∅, with X 2 being a closed subset of Rn and X 1 positively invariant. Let M be a finite set of equilibria of (12) in X 2 . Assume that : T1. Every solution that starts in X 2 and stays in X 2 for all forward times converges to one of the equilibria in M, T2. Every equilibrium in M is an isolated invariant set in X and a weak repeller for X 1 , T3. M is acyclic in X 2 . Then X 2 is a uniform strong repeller for X 1 . Moreover, if X 1 is convex, there exists at least one equilibrium point in X 1 .

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