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Global asymptotic behavior of a chemostat model with delayed response in growth
Applied Mathematics and Computation 147 (2004) 147–161 www.elsevier.com/locate/amc
Global asymptotic behavior of a chemostat model with delayed response in growth H.M. El-Owaidy a, A.A. Moniem a
b,*
Department of Mathematics, Faculty of Science, AL-Azhar University, Nasr City, Cairo 11884, Egypt b Akhbar EL-Yom Academy, 6 October City, Giza 12573, Egypt
Abstract This paper studies the global asymptotic behavior of solutions of a competition model between n species in the chemostat. The model incorporates time delays that allow the competing organisms to be stored the nutrient (so that the free nutrient concentration does not reflect the nutrient available for growth) and allow growth response functions to be nonmonotone. By introducing three auxiliary functions and using a lemma, it is shown that at most one competitor survives, and the substrate and the surviving competitor (if two exist) approach limiting values. Ó 2002 Elsevier Inc. All rights reserved.
1. Introduction Competition modeling is one of the more challenging aspects of mathematical biology. The chemostat is a basic piece of laboratory apparatus, yet it has begun to occupy an increasingly central role in ecological studies. The ecological environment created by a chemostat is one of the few completely controlled experimental systems for testing microbial growth and competition. As a tool in biotechnology, the chemostat plays an important role in bio processing. See the monograph of Smith and Waltman [6] for a detailed description of a chemostat and for various mathematical methods for analyzing
*
Corresponding author. E-mail address: [email protected] (A.A. Moniem).
0096-3003/$ - see front matter Ó 2002 Elsevier Inc. All rights reserved. doi:10.1016/S0096-3003(02)00658-6
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chemostat models. It is quite natural then that it should be used as a model for studying delays problems. Delays occur naturally in biological system by two obvious sources of delays: delays due to the cell cycle; and delays due to the possibility the organism stores the nutrient. Delays appear in a chemostat model in Bush and Cook [1]. They have investigated a model of growth of one organism in the chemostat with a delay in the intrinsic growth rate of the microorganism but with no delay in the substrate equation. (As a consequence there is a delay in the growth equation but no delay in the consumption term in the nutrient equation.) This approach was extended to the competitive situation in [5]. Moreover, in [4], Freedman et al. proposed and studied a model of growth of one organism in the chemostat with a time delay that defines as the time that elapses between the uptake of nutrient by cells and the incorporation of this nutrient as biomass. Ellermeyer [8] extended the results of [4] to the model for competition between two microorganisms in the chemostat. Under the assumption that uptake functions are monotone increasing, the sufficient conditions one established in [8] under which at least one species is persistent. In [7], El-Owaidy and Ismail, improved on and extended the results of [8,9] by introducing three auxiliary functions (which improve the proofs of the main results of [9]) and the uptake functions was relaxed to nonmonotone functions (the necessity for this can be found in [3]). They showed that at most one competitor survives, and that the substrate and the surviving competitor (if two exist) approach limiting values, provided that some parameters are distinct. The purpose of this paper is to extend the number of competing species to be an n microorganism. We introduce three extended auxiliary functions to extend the proofs of the main results of [7]. This paper is organized as follow. In Section 2, we present some preliminary results about the model (2.1). Our main results are stated and proven in Section 3. Finally, we conclude with a brief discussion in Section 4.
2. The model We consider, as a model of an n microbial populations competing for a single nutrient in a chemostat with delays in uptake conversion, the following system of retarded functional differential equations: 9 8 n X dsðtÞ > > 0 > > ¼ Dðs sðtÞÞ Pi ðsðtÞÞxi ðtÞ; = < dt i¼1 ð2:1Þ > > > ; : dxi ¼ Dxi ðtÞ þ eDsi Pi ðsðt si ÞÞxi ðt si Þ: > dt In these equations, xi ðtÞ, sðtÞ and s0 denote the concentration of the ith population of microorganisms at time t, the concentration of substrate at time t
H.M. El-Owaidy, A.A. Moniem / Appl. Math. Comput. 147 (2004) 147–161
149
and the concentration of substrate in the feed bottle, respectively. Also, D denotes the input rate from the feed bottle containing the substrate and the washout rate of substrate, microorganisms and byproducts from culture vessel; Pi ðsÞ is the function that represents the rate of conversion of nutrient to biomass (i.e. the density-dependent uptakes of nutrient by population xi ). There is a delay of time si in converting nutrient into xi . Through our analysis of the system (2.1), we assume that si are positive constants with s ¼ maxfs1 ; s2 ; . . . ; sn g, i ¼ 1; 2; . . . ; n. Also, we assume that D and s0 are positive constants. In [3], we assume that the Pi satisfy the following: ii(i) i(ii) (iii) (iv)
Pi : R þ ! R þ , Pi is continuously differential, Pi ð0Þ ¼ 0, There exist unique, positive, extended real numbers ki and li with ki < li such that Pi ðsÞ < D
if s 62 ½ki ; li and Pi ðsÞ > D
if s 2 ðki ; li Þ
ð2:2Þ
Here ki and li represent the break-even concentration of the substrate for the ith competitor. The interpretation of (iv) is that species xi increases when ki < s < li and decreases if 0 6 s 6 ki or if li < s < 1 (if li 6¼ 1). In particular, if 0 < ki < li < 1 then xi grows at moderate nutrient concentration (ki < s < li ) and declines at low concentration (s < ki ) and at high concentration (li < s). Remark. A further assumption of a generic character is that all the finite ki and li is distinct from each other and from s0 (which is biologically reasonable). nþ1 Let Rþ ¼ fðs; x1 ; x2 ; . . . ; xn Þ : s P 0; xi P 0; i ¼ 1; 2; . . . ; ng and let C þ ð½s; nþ1 nþ1 0 ; Rþ Þ denote the space of continuous functions from ½s; 0 into Rþ . Also, we assume that the initial conditions for the system (2.1) take the following form:
8 9 < sðhÞ ¼ U0 ðhÞ; xi ðhÞ ¼ Ui ðhÞ; i ¼ 1; 2; . . . ; n; = nþ1 ðU ; U ; . . . ; Un Þ 2 C þ ð½s; 0 ; Rþ Þ; : 0 1 ; Ui ð0Þ > 0; i ¼ 1; 2; . . . ; n:
ð2:3Þ
One can easily show that solutions of the system (2.1) corresponding to the inequality (2.2) are defined for all t P 0 and remain positive for all t P 0. Such solutions will be called positive solutions. The following lemma shows that positive solutions of the system (2.1) are bounded for all positive time.
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Lemma 2.1. If ðsðtÞ; x1 ðtÞ; x2 ðtÞ; . . . ; xn ðtÞÞ is a positive solution of system (2.1), then sðtÞ þ
n X
eDsi xi ðt þ si Þ ¼ s0 þ eðtÞ for t P 0;
i¼1
where eðtÞ ! 0 exponentially as t ! þ1. Pn Proof. Let ZðtÞ ¼ sðtÞ þ i¼1 eDsi xi ðt þ si Þ. From (2.1) we obtain Z 0 ðtÞ ¼ Dðs0 ZðtÞÞ, from which we obtain ZðtÞ ¼ s0 þ ðZ0 s0 ÞeDt , where Z0 ¼ Zð0Þ. This completes the proof. We shall need the following two lemmas: Lemma 2.2 [2]. Let f ðtÞ 2 C 2 ½t0 ; 1Þ, f ðtÞ P 0 and L > 0. i(i) If f 0 ðtÞ P 0, f ðtÞ is bounded and f 00 ðtÞ < L for all t P t0 , then f 0 ðtÞ ! 0 as t ! þ1. (ii) If f 0 ðtÞ 6 0 and f 00 ðtÞ P L > 1 for all t P t0 , then f 0 ðtÞ ! 0 as t ! þ1. Lemma 2.3 [9]. Let f ðtÞ 2 C 3 ½t0 ; 1Þ and 0 6 f ðtÞ 6 L for all t P t0 . Suppose that f1 ¼ lim inf t!þ1 f ðtÞ and f 1 ¼ lim supt!þ1 f ðtÞ. If f ðtÞ is not eventually monotone, then there exist sequences ftn g and ftn g, tn ! þ1, tn ! þ1 as n ! þ1, such that f ðtn Þ ! f 1 and f ðtn Þ ! f1 as n ! þ1 and f 0 ðtn Þ ¼ f 0 ðtn Þ ¼ 0 for all n.
3. Main results The principal result of this paper will be stated and proved in this current section. Theorem 3.1. If the break-even concentration of the substrate for the ith competitor is greater than or equal to the concentration of substrate in the feed bottle; then limt!þ1 xi ðtÞ ¼ 0 for all positive solutions of the system (2.1). Proof. Let ðsðtÞ; x1 ; x2 ; . . . ; xn Þ be a positive solution of system (2.1) and let 1 x1 i ¼ lim supt!þ1 xi ðtÞ. We consider that xi > 0 to derive a concentration by considering two cases. Case 1. Suppose that xi ðtÞ is eventually monotone. Then xi ðtÞ ! x1 as i t ! þ1. By Lemma 2.2, we see that x0i ðtÞ ! 0 as t ! þ1. For fixed i, Lemma 2.1 implies that
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x0i ðtÞ ¼ Dxi ðtÞ þe
Dsi
" 0
Dsi
Pi s þ eðt si Þ e xi ðtÞ
n X
151
# e
Dsj
xj ðt þ sj si Þ xi ðt si Þ:
j¼1 j6¼i
From this we have " 0
Dsi
Pi s þ eðt si Þ e xi ðtÞ
n X
# e
Dsj
xj ðt þ sj si Þ ¼
j¼1 j6¼i
½x0i ðtÞ þ Dxi ðtÞ eDsi : xi ðt si Þ
It follows that when t ! þ1, we obtain: " 0
Dsi
Pi s þ eðt si Þ e xi ðtÞ
n X
# e
Dsj
xj ðt þ sj si Þ ! DeDsi :
j¼1 j6¼i
Note that Pi ðsÞ ! DeDsi if s ! ki or li . It follows from the boundedness of sðtÞ and continuity of Pi ðsÞ that: " 0
lim
t!þ1
Dsi
s þ eðt si Þ e xi ðtÞ
n X
# e
Dsj
xj ðt þ sj si Þ ¼ ki or li :
j¼1 j6¼i
Assume that the left hand side of the above equation equal to ki . Since xj ðtÞ are positive for all j 6¼ i, j ¼ 1; 2; . . . ; n, we have s0 eDsi x1 i P ki , which yields Dsi 0 1 x1 6 e ðs k Þ 6 0, which contradict x > 0. If we replace li by ki , we find i i i that similar argument leads to a contradiction. Case 2. Suppose that xi ðtÞ is not eventually monotone. Then, Lemma 2.3 implies that there exists a sequence ftk g, tk ! þ1 as k ! þ1, such that x0i ðtk Þ ¼ 0, k ¼ 1; 2; . . . ; xi ðtk Þ ! x1 i as k ! þ1. From second equation of the system (2.1) and by Lemma 2.1, we get " 0
Dsi
Pi s þ eðtk si Þ e xi ðtk Þ
n X
# e
Dsj
xj ðtk þ sj si Þ ¼
j¼1 j6¼1
Dxi ðtk ÞeDsi : xi ðtk si Þ
Recall that lim supt!þ1 xi ðtÞ ¼ x1 i . For any g > 0 there is an N ðgÞ > 0 such that xi ðtk si Þ < x1 þ g for k > N . As consequence, i " 0
Dsi
Pi s þ eðtk si Þ e xi ðtk Þ
n X j¼1 j6¼i
# e
Dsj
xj ðtk þ sj si Þ P
Dxi ðtk ÞeDsi x1 i þg
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for k > N . It follows that: 2 6 lim inf Pi 4s0 þ eðtk si Þ eDsi xi ðtk Þ
n X
tk !þ1
Since g > 0 is arbitrarily fixed. We have 2 6 lim inf Pi 4s0 þ eðtk si Þ eDsi xi ðtk Þ
j¼1 j6¼i
n X
tk !þ1
j¼1 j6¼i
3 Dx1 eDsi 7 : eDsj xj ðtk þ sj si Þ5 P 1i xi þ g
3 7 eDsj xj ðtk þ sj si Þ5 P DeDsi :
Note that Pi ðsÞ < DeDsi when s 2 ð0; ki Þ. We have 2 3 n X 6 7 lim inf 4s0 þ eðtk si Þ eDsi xi ðtk Þ eDsj xj ðtk þ sj si Þ5 P ki ; tk !þ1
j¼1 j6¼i
1 Dsi 0 from which we obtain s0 eDsi x1 ðs ki Þ 6 0. We i P ki , which yields xi 6 e are led to a contradiction, and therefore, the proof of Theorem 3.1 is completed.
Due to Theorem 3.1, we shall assume that the populations are labeled so that ki < kj < s0 for i < j. Set bi ¼ eDsi ðs0 ki Þ, i ¼ 1; 2; . . . ; n. By the same type of arguments as those in the proof of Theorem 3.1, one can obtain the following lemma. Lemma 3.1. Let ki < kj < s0 for i < j and i; j ¼ 1; 2; . . . ; n. Set bi ¼ eDsi ðs0 ki Þ, for fixed i; then any positive solution ðsðtÞ; x1 ; x2 ; . . . ; xn Þ of the system (2.1) satisfies lim supt!þ1 xi ðtÞ 6 bi , i ¼ 1; 2; . . . ; n. For mathematical reasons, we shall assume that s0 < l in the following theorem. Theorem 3.2. If ki < kj < s0 < l for i < j; then any positive solution of the system (2.1) satisfies limt!þ1 sðtÞ ¼ ki , limt!þ1 xi ðtÞ ¼ bi and limt!þ1 xj ðtÞ ¼ 0, i; j ¼ 1; 2; . . . ; n. To prove this theorem, we need the following lemma. Lemma 3.2. If ki < kj < s0 < l for i < j; then lim inf t!þ1 xi ðtÞ P 0 for all positive solutions of system (2.1), i; j ¼ 1; 2; . . . ; n.
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153
Proof. Fix n > 0 small enough such that kj 2n > ki . By Lemmas 2.1 and 3.1, we see that there is a t0 > 0 such that 2 3 n X 60 7 eDsj xj ðt þ sj si Þ5 P kj n for t P t0 : ð3:1Þ 4s þ eðt si Þ j¼1 j6¼i
Also, let d > 0 be small enough such that n > deDsi :
ð3:2Þ
Suppose lim inf t!þ1 xi ðtÞ ¼ 0. Then there is a sufficiently large t1 such that xi ðt1 Þ < d. Set m ¼ mint2½t1 si ;t1 xi ðtÞ. Since t1 is large enough and xi ðtÞ is positive when t is large, we must have m > 0. Set t2 ¼ supft : t P t1 ; xi ðuÞ P m for all u 2 ½t1 si ; t g. Since xi ðuÞ P m for all u 2 ½t1 si ; t1 , the set ft : t P t1 ; xi ðuÞ P m for all u 2 ½t1 si ; t g is empty. Now we have two cases for t2 . Case 1. If t2 is infinite, then x0i ðtÞ P m > 0 for all t P t1 si . This would contradict lim inf t!þ1 xi ðtÞ ¼ 0. Case 2. If t2 is finite, it is clear that x0i 6 ðt2 Þ 6 0. However, x0i ðt2 Þ ¼ Dxi ðt2 Þ 2 6 þ eDsi Pi 4s0 þ eðt2 si Þ eDsi xi ðt2 Þ x0i ðt2 Þ P xi ðt2 Þ 2
n X j¼1 j6¼i
3 7 eDsj xj ðt2 þ sj si Þ5xi ðt2 si Þ;
2
6 6 4 D þ eDsi Pi 4s0 þ eðt2 si Þ eDsi xi ðt2 Þ
By (3.1) and (3.2), we have 2 60 4s þ eðt2 si Þ eDsi d
n X j¼1 j6¼i
n X j¼1 j6¼i
33 77 eDsj xj ðt2 þ sj si Þ55:
3 7 eDsj xj ðt2 þ sj si Þ5 > ki :
It follows from s0 < l that: 2 2 6 6 4 D þ eDsi Pi 4s0 þ eðt2 si Þ eDsi xi ðt2 Þ
n X j¼1 j6¼i
33 77 eDsj xj ðt2 þ sj si Þ55 > 0:
Consequently, x0i ðt2 Þ > 0. We are led to a contradiction. The proof of Lemma 3.2 is completed.
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Proof of Theorem 3.2. Let ðsðtÞ; x1 ; x2 ; . . . ; xn Þ be a positive solution of the system (2.1). We need to prove that limt!þ1 sðtÞ ¼ ki , limt!þ1 xi ðtÞ ¼ bi and limt!þ1 xj ðtÞ ¼ 0 for i < j and i; j ¼ 1; 2; . . . ; n. In view of Lemma 2.1, it suffices to show that limt!þ1 xi ðtÞ ¼ bi and limt!þ1 xj ðtÞ ¼ 0 for i < j. We get x0i ðtÞ ¼ Dxi ðtÞ
2
6 þ eDsi Pi 4s0 þ eðt si Þ eDsi xi ðtÞ
n X j¼1 j6¼i
3 7 eDsj xj ðt þ sj si Þ5xi ðt si Þ: ð3:3Þ
Let we define three auxiliary functions fi ðxj Þ, gi ðxk Þ and hi ðxL Þ with k < L < j, n Y eDsm ; fi ðxj Þ ¼ s0 ki xj eDsj
i; j ¼ 1; 2; . . . ; n;
m¼1 m6¼j
n Y gi ðxk Þ ¼ s0 ki xk eDsk eDsm ;
i; k ¼ 1; 2; . . . ; n;
m¼1 m6¼k
n Y hi ðxL Þ ¼ s0 ki xL eDsL eDsm ;
i; L ¼ 1; 2; . . . ; n:
m¼1 m6¼L
By the definition of fi ðxj Þ, gi ðxk Þ and hi ðxL Þ; we can see that if common sm ¼ 0 between gi ðxk Þ and hi ðxL Þ except sk ; sL 6¼ 0 in hi ðxL Þ and in gi ðxk Þ, respectively, then i(i) xL ¼ gi ðxk Þ are inverse functions of xk ¼ hi ðxL Þ for each corresponding i, (ii) gL ð0Þ ¼ bL . Also, if common sm ¼ 0 between fi ðxj Þ and hi ðxL Þ except sj 6¼ 0 in hi ðxL Þ and sL 6¼ 0 in fi ðxj Þ, we obtain i(i) xj ¼ hi ðxL Þ are inverse functions of xL ¼ fi ðxj Þ for each corresponding i, (ii) hj ð0Þ ¼ bj . Finally, for common sm ¼ 0 between fi ðxj Þ and gi ðxk Þ except sj 6¼ 0 in gi ðxk Þ and sk 6¼ 0 in fi ðxj Þ, we get i(i) xj ¼ gi ðxk Þ are inverse functions of xk ¼ fi ðxj Þ for each corresponding i, (ii) fk ð0Þ ¼ bk .
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155
Let ek ¼ fk ðbk Þ, b0 ¼ gk ð0Þ. Since kk < kL < kj < s0 , we have bj > 0, bL > 0, bk > 0 and b0 > bL . Notice that fi ðxj Þ is strictly decreasing, we see that 0 < ek < bk . Set eL ¼ fL ð0Þ. It easy to see that eL < bk . Let we consider that lim inf t!þ1 xk ðtÞ ¼ p. Then p > 0 by Lemma 3.2. At this point we need the following. Claim. If p P eL , then bk lim xm ðtÞ ¼ 0 t!þ1
if m ¼ k ; if m ¼ 6 k
k; m ¼ 1; 2; . . . ; n:
Proof. Set lim supt!þ1 xL ðtÞ ¼ qL > 0 and lim supt!þ1 xi ðtÞ ¼ r ¼ 0, L; i ¼ 1; 2; . . . ; n. Case 1. Assume that xL ðtÞ and xi ðtÞ are eventually monotone and converge to qL and ri , respectively, as t ! þ1. Then by Lemmas 2.1 and 2.2, x0i ðtÞ ! 0 as t ! þ1 and x0L ðtÞ ! 0 as t ! þ1. From the equation of system (3.3), replace i by L, one obtains 2 3 n X eDsi xi ðt þ si sL Þ5 PL 4s0 þ eðt sL Þ eDsL xL ðtÞ eDsk xk ðt þ sk sL Þ i¼1 i6¼k6¼L
¼
½x0L ðtÞ þ DxL ðtÞ eDsL ; xL ðt sL Þ
which leads to 2 lim inf PL 4s0 þ eðt sL Þ eDsL xL ðtÞ eDsk xk ðt þ sk sL Þ t!þ1
n X
3 eDsi xi ðt þ si sL Þ5 P DeDsL ;
i¼1 i6¼k6¼L
consequently, 2 lim inf 4s0 þ eðt sL Þ eDsL xL ðtÞ eDsk xk ðt þ sk sL Þ t!þ1
n X i¼1 i6¼k6¼L
3 eDsi xi ðt þ si sL Þ5 P kL :
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H.M. El-Owaidy, A.A. Moniem / Appl. Math. Comput. 147 (2004) 147–161
Thus, we have 2
3 n X eDsi ri 5 P kL : lim inf 4s0 eDsL qL eDsk p t!þ1
i¼1 i6¼k6¼L
As a consequence, 2
3 n X eDsi ri 5 qL 6 eDsL 4s0 kL eDsk p i¼1 i6¼k6¼L
6 eDsL ½s0 kL eDsk eL ¼ gL ðeL Þ ¼ 0; which contradicts qL > 0. Case 2. Assume that xL ðtÞ and xi ðtÞ are not eventually monotone. Then by Lemma 2.3, there exists a sequence ftn g, tn ! þ1 as n ! þ1, such that x0L ðtn Þ ¼ x0i ðtn Þ ¼ 0 and 8 9 < qL > 0 if m ¼ L = xi ðtn Þ ¼ ri ¼ 0 if i ¼ m ; : ; p > 0 if m ¼ k
i 6¼ k 6¼ L; i; k; L ¼ 1; 2; . . . ; n:
Thus, from the equation of system (3.3) and i is replaced by L, we have 2 PL 4s0 þ eðtn sL Þ eDsL xL ðtn Þ eDsk xk ðtn þ sk sL Þ
n X
3 eDsi xi ðtn þ si sL Þ5 ¼
i¼1 i6¼k6¼L
DxL ðtn ÞeDsL : xL ðtn sL Þ
For any g > 0, there is an N ðgÞ > 0 such that xL ðtn sL Þ < qL þ g when N ðgÞ < n and lim supt!þ1 xL ðtn Þ ¼ qL . As a consequence, if N ðgÞ < n and as tn ! þ1, we get 2 lim inf PL 4s0 þ eðtn sL Þ eDsL xL ðtn Þ eDsk xk ðtn þ sk sL Þ tn !þ1
n X i¼1 i6¼k6¼L
3 eDsi xi ðtn þ si sL Þ5 P
DqL eDsL : qL þ g
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157
Since g > 0 is arbitrarily fixed, we have 2 lim inf PL 4s0 þ eðtn sL Þ eDsL xL ðtn Þ eDsk xk ðtn þ sk sL Þ tn !þ1
n X
3 eDsi xi ðtn þ si sL Þ5 P DeDsL :
i¼1 i6¼k6¼L
Furthermore, we can note that PL ðsÞ < DeDsL for 0 6 s 6 kL . We conclude that, for any f > 0, there is an N ðfÞ > 0 such that 2 lim inf 4s0 þ eðtn sL Þ eDsL xL ðtn Þ eDsk xk ðtn þ sk sL Þ tn !þ1
n X
3 eDsi xi ðtn þ si sL Þ5 P kL f;
i¼1 i6¼k6¼L
for n > N ðfÞ. One obtains 2 4s0 qL eDsL peDsk
n X
3 ri eDsi 5 P kL f;
i¼1 i6¼k6¼L
from which, we get 2 qL 6 eDsL 4s0 kL þ f peDsk
n X
3 ri eDsi 5:
i¼1 i6¼k6¼L
Since ri ¼ 0 and f > 0 is arbitrarily fixed, we have qL 6 eDsL ½s0 kL eL eDsk ¼ gL ðeL Þ ¼ 0, which contradicts qL > 0. Thus, in all cases, we must have qL ¼ 0, and therefore, p if m ¼ k lim xm ðtÞ ¼ ; m; k ¼ 1; 2; . . . ; n: 0 if m 6¼ k t!þ1 Now, we want to show that limt!þ1 xk ðtÞ ¼ p ¼ bk . Since lim supt!þ1 xk ðtÞbk by Lemma 3.1. Therefore, we must show that lim inf t!þ1 xk ðtÞ ¼ p ¼ bk . Suppose that eL 6 p < bk is in the following two possible cases. Case 1. Assume that xk ðtÞ is eventually monotone and converge to p as t ! þ1. Then, by Lemmas 2.1 and 2.2, x0k ðtÞ ! 0 as t ! þ1. From the equation of system (3.3), replace i by k, one obtains
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2 Pk 4s0 þ eðt sk Þ eDsk xk ðtÞ
n X
3 eDsi xi ðt þ si sk Þ5
i¼1 i6¼k
¼
½x0k ðtÞ þ Dxk ðtÞ eDsk : xk ðt sk Þ
From the result of the claim and limt!þ1 x0k ðtÞ ¼ 0, one obtains Pk ½s0 peDsk ¼ DeDsk , which, together with s0 < l, leads to ½s0 peDsk ¼ kk . Consequently, p ¼ ½s0 kk eDsk ¼ fk ð0Þ ¼ bk , contrary to p < bk . Case 2. Assume that xk ðtÞ is not eventually monotone. Then, by Lemma 2.3, there exists a sequence ftn g, tn ! þ1 as n ! þ1, such that x0k ðtn Þ ¼ 0 and xk ðtn Þ ! p as n ! þ1. Thus, from the equation of system (3.3) and i is replaced by k, we have 2 3 n X Dxk ðtn ÞeDsk : Pk 4s0 þ eðtn sk Þ eDsk xk ðtn Þ eDsi xi ðtn þ si sk Þ5 ¼ xk ðtn sk Þ i¼1 i6¼k
Since lim inf t!þ1 xk ðtÞ ¼ p, it follows that: 2 lim sup Pk 4s0 þ eðtn sk Þ eDsk xk ðtn Þ tn !þ1
n X
3 eDsi xi ðtn þ si sk Þ5 6 DeDsk ;
i¼1 i6¼k
which yields Pk ½s0 peDsk 6 DeDsk . Since s0 < l, we have ½s0 peDsk 6 kk . Hence, p P ½s0 kk eDsk ¼ fk ð0Þ ¼ bk , which contradicts p < bk . Summarizing the discussions of the last two cases, we conclude that p ¼ bk , and hence limt!þ1 xk ðtÞ ¼ bk . This completes the proof of the claim. To complete the proof of Theorem 3.2, it suffices to show that lim inf t!þ1 xk ðtÞ ¼ p ¼ bk > eL , k < L. First we verify the following assertion. Assertion 1. If ek ¼ fk ðbk Þ, then lim inf t!þ1 xk ðtÞ ¼ p ¼ bk > ek . Proof. Suppose that p < ek is in the following two possible cases. Case 1. Assume that xk ðtÞ is eventually monotone and converge to p as t ! þ1. Then, x0k ðtÞ ! 0 as t ! þ1 From the equation of system (3.3), replace i by k and k < j, one has 2 Pk 4s0 þ eðt sk Þ eDsk xk ðtÞ eDsj xj ðt þ sj sk Þ
n X i¼1 i6¼k6¼j
3 eDsi xi ðt þ si sk Þ5 ¼
½x0k ðtÞ þ Dxk ðtÞ eDsk ; xk ðt sk Þ
H.M. El-Owaidy, A.A. Moniem / Appl. Math. Comput. 147 (2004) 147–161
159
from which, we get 2 lim 4s0 þ eðt sk Þ eDsk xk ðtÞ eDsj xj ðt þ sj sk Þ
t!þ1
n X
3 eDsi xi ðt þ si sk Þ5 ¼ kk ;
i¼1 i6¼k6¼j
which leads to 2 4s0 þ eðt sk Þ eDsk lim xk ðtÞ eDsj lim xj ðtÞ t!þ1
t!þ1
n X i¼1 i6¼k6¼j
3 eDsi lim xi ðtÞ5 ¼ kk : t!þ1
By using Lemma 3.1, we have 2 4s0 bj eDsj
n X
3 bi eDsi kk 5eDsk 6 p:
i¼1 i6¼k6¼j
Setting bi ¼ 0 for i 6¼ j, this reduces to p P ½s0 kk bj eDsj eDsk :
ð3:4Þ
Since kk < kj , we have p P ½s0 kk bk eDsj eDsk ¼ fk ðbk Þ ¼ ek , contrary to p < ek . Case 2. Assume that xk ðtÞ is not eventually monotone. Then, there exists a sequence ftn g, tn ! þ1 as n ! þ1, such that x0k ðtn Þ ¼ 0 and xk ðtn Þ ! p as n ! þ1. Thus, from the equation of system (3.3) and i is replaced by k, k < j, we have 2 Pk 4s0 þ eðtn sk Þ eDsk xk ðtn Þ eDsj xj ðt þ sj sk Þ
n X i¼1 i6¼k6¼j
3 eDsi xi ðtn þ si sk Þ5 ¼
Dxk ðtn ÞeDsk ; xk ðtn sk Þ
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from which, we get 2 lim sup Pk 4s0 þ eðtn sk Þ eDsk xk ðtn Þ eDsj xj ðt þ sj sk Þ tn !þ1
n X
3 eDsi xi ðtn þ si sk Þ5 6 DeDsk :
i¼1 i6¼k6¼j
Thus, for any g > 0, there is an N ðgÞ > 0 such that 2 lim sup4s0 þ eðtn sk Þ eDsk xk ðtn Þ eDsj xj ðt þ sj sk Þ tn !þ1
n X
3 eDsi xi ðtn þ si sk Þ5 6 kk þ g;
i¼1 i6¼k6¼j
for n > N ðgÞ. By using Lemma 3.1 and since g > 0 is arbitrarily fixed, we have 2 3 n X 0 Ds Ds 4s kk bj e j bi e i 5eDsk 6 p: i¼1 i6¼k6¼j
Setting bi ¼ 0, it leads to p P ½s0 kk bj eDsj eDsk . As a consequence of the Eq. (3.4) in the first case, we obtain p > ek , contrary to p < ek . Thus, we proved that in all cases p > ek . Now, we must show that eL < lim inf t!þ1 xk ðtÞ ¼ p ¼ bk . Suppose that p < eL . Thus, there are two cases to be considered. Case 1. From the Eq. (3.4), we have p P eDsk ½s0 kk bj eDsj . Putting bj ¼ 0, say by the claim, and for all k < L, we have p P eDsk ½s0 kk > eDsk ½s0 kL ¼ fL ð0Þ ¼ eL , which contradicts p < eL . Case 2. Furthermore, for all kL > kk , from the Eq. (3.4) and the claim, we have p > eDsk ½s0 kL ¼ fL ð0Þ ¼ eL , which contradicts p < eL . Thus in all cases, p > eL . Therefore, the proof of Theorem 3.2 is completed. Remark. Theorem 3.2 implies that limt!þ1 sðtÞ ¼ kk , limt!þ1 xk ðtÞ ¼ eDsk ½s0 kk and limt!þ1 xm ðtÞ ¼ 0, 8m 6¼ k. This indicates that Theorem 3.2 is generalization of the results of [7] where three populations are considered. Assertion 2. If kL > kk , then eL > ek . Proof. It easy to show that bk ¼ ek ½1 eDðsk sj Þ 1 ¼ eL þ ðkL kk ÞeDsk ;
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161
which leads to eL þ ðkL kk ÞeDsk ¼ ek þ bk eDsj : 0
Since s ¼ kL þ bL e therefore we have
DsL
¼ kk þ bk e
ðkL kk ÞeDsk < bk eDsj :
Dsk
ð3:5Þ
, from which, one obtains kL kk 6 bk eDsk , ð3:6Þ
By (3.5) and (3.6), we have ek þ bk eDsj < eL þ bk eDsj , which implies that ek < eL . 4. Discussion In this paper, a chemostat model with delayed response in growth is considered. The model incorporates time delays that allow the competing organisms to be stored the nutrient (so that the free nutrient concentration does not reflect the nutrient available for growth) and allow growth response functions to be nonmonotone. The main results of El-Owaidy and Ismail are generalized. Our results show that at most one competitor survives, and the substrate and the surviving competitor (if two exist) approaches limiting values, provided that some parameters are distinct. Thus the principal of competitive exclusion remains valid for that model. This indicates that this kind of time delay does not influence the competitive outcome of the organisms. References [1] A.W. Bush, A.E. Cook, The effect of time delay and growth rate inhibition in the bacterial treatment of wastewater, J. Theor. Biol. 63 (1975) 385–395. [2] G.B. Hsu, S. Hubbell, P. Waltman, A mathematical theory for single-nutrient competition in continuous cultures of microorganisms, SIAM J. Appl. Math. 32 (1977) 366–383. [3] G.J. Butler, G.S. Wolkowicz, A mathematical model of the chemostat with a general class of functions describing nutrient uptake, SIAM J. Appl. Math. 45 (1985) 138–151. [4] H.I. Freedman, J.W.H. So, P. Waltman, Chemostat competition with time delays, in: R. Vichnevetsky, P. Borne, J. Vignes (Eds.), Proceeding of IMACS (1988): 12th World Congress on Scientific Computation: Modeling and Simulation of System Paris, Gerfidn Cite Scientifique, France, 1988, pp. 102–104. [5] H.I. Freedman, J.W.H. So, P. Waltman, Coexistence in a model of competition in the chemostat incorporating discrete delays, SIAM J. Appl. Math. 49 (1989) 859–870. [6] H.L. Smith, P. Waltman, The Theory of Chemostat, Cambridge University Press, Cambridge, UK, 1995. [7] H.M. El-Owaidy, M. Ismail, Asymptotic behavior of the chemostat model with delayed response in growth, Chaos, Solitons Fract. 13 (2002) 787–795. [8] S.F. Ellermeyer, Competition in the chemostat: global asymptotic behavior of a model with delayed response in growth, SIAM J. Appl. Math. 54 (1994) 456–465. [9] W. Wendi, M. Zhien, Converges in the chemostat model with delayed response in growth, J. Syst. Sci. Math. Sci. 12 (1999) 23–32.
Global asymptotic behavior of a chemostat model with delayed response in growth H.M. El-Owaidy a, A.A. Moniem a
b,*
Department of Mathematics, Faculty of Science, AL-Azhar University, Nasr City, Cairo 11884, Egypt b Akhbar EL-Yom Academy, 6 October City, Giza 12573, Egypt
Abstract This paper studies the global asymptotic behavior of solutions of a competition model between n species in the chemostat. The model incorporates time delays that allow the competing organisms to be stored the nutrient (so that the free nutrient concentration does not reflect the nutrient available for growth) and allow growth response functions to be nonmonotone. By introducing three auxiliary functions and using a lemma, it is shown that at most one competitor survives, and the substrate and the surviving competitor (if two exist) approach limiting values. Ó 2002 Elsevier Inc. All rights reserved.
1. Introduction Competition modeling is one of the more challenging aspects of mathematical biology. The chemostat is a basic piece of laboratory apparatus, yet it has begun to occupy an increasingly central role in ecological studies. The ecological environment created by a chemostat is one of the few completely controlled experimental systems for testing microbial growth and competition. As a tool in biotechnology, the chemostat plays an important role in bio processing. See the monograph of Smith and Waltman [6] for a detailed description of a chemostat and for various mathematical methods for analyzing
*
Corresponding author. E-mail address: [email protected] (A.A. Moniem).
0096-3003/$ - see front matter Ó 2002 Elsevier Inc. All rights reserved. doi:10.1016/S0096-3003(02)00658-6
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chemostat models. It is quite natural then that it should be used as a model for studying delays problems. Delays occur naturally in biological system by two obvious sources of delays: delays due to the cell cycle; and delays due to the possibility the organism stores the nutrient. Delays appear in a chemostat model in Bush and Cook [1]. They have investigated a model of growth of one organism in the chemostat with a delay in the intrinsic growth rate of the microorganism but with no delay in the substrate equation. (As a consequence there is a delay in the growth equation but no delay in the consumption term in the nutrient equation.) This approach was extended to the competitive situation in [5]. Moreover, in [4], Freedman et al. proposed and studied a model of growth of one organism in the chemostat with a time delay that defines as the time that elapses between the uptake of nutrient by cells and the incorporation of this nutrient as biomass. Ellermeyer [8] extended the results of [4] to the model for competition between two microorganisms in the chemostat. Under the assumption that uptake functions are monotone increasing, the sufficient conditions one established in [8] under which at least one species is persistent. In [7], El-Owaidy and Ismail, improved on and extended the results of [8,9] by introducing three auxiliary functions (which improve the proofs of the main results of [9]) and the uptake functions was relaxed to nonmonotone functions (the necessity for this can be found in [3]). They showed that at most one competitor survives, and that the substrate and the surviving competitor (if two exist) approach limiting values, provided that some parameters are distinct. The purpose of this paper is to extend the number of competing species to be an n microorganism. We introduce three extended auxiliary functions to extend the proofs of the main results of [7]. This paper is organized as follow. In Section 2, we present some preliminary results about the model (2.1). Our main results are stated and proven in Section 3. Finally, we conclude with a brief discussion in Section 4.
2. The model We consider, as a model of an n microbial populations competing for a single nutrient in a chemostat with delays in uptake conversion, the following system of retarded functional differential equations: 9 8 n X dsðtÞ > > 0 > > ¼ Dðs sðtÞÞ Pi ðsðtÞÞxi ðtÞ; = < dt i¼1 ð2:1Þ > > > ; : dxi ¼ Dxi ðtÞ þ eDsi Pi ðsðt si ÞÞxi ðt si Þ: > dt In these equations, xi ðtÞ, sðtÞ and s0 denote the concentration of the ith population of microorganisms at time t, the concentration of substrate at time t
H.M. El-Owaidy, A.A. Moniem / Appl. Math. Comput. 147 (2004) 147–161
149
and the concentration of substrate in the feed bottle, respectively. Also, D denotes the input rate from the feed bottle containing the substrate and the washout rate of substrate, microorganisms and byproducts from culture vessel; Pi ðsÞ is the function that represents the rate of conversion of nutrient to biomass (i.e. the density-dependent uptakes of nutrient by population xi ). There is a delay of time si in converting nutrient into xi . Through our analysis of the system (2.1), we assume that si are positive constants with s ¼ maxfs1 ; s2 ; . . . ; sn g, i ¼ 1; 2; . . . ; n. Also, we assume that D and s0 are positive constants. In [3], we assume that the Pi satisfy the following: ii(i) i(ii) (iii) (iv)
Pi : R þ ! R þ , Pi is continuously differential, Pi ð0Þ ¼ 0, There exist unique, positive, extended real numbers ki and li with ki < li such that Pi ðsÞ < D
if s 62 ½ki ; li and Pi ðsÞ > D
if s 2 ðki ; li Þ
ð2:2Þ
Here ki and li represent the break-even concentration of the substrate for the ith competitor. The interpretation of (iv) is that species xi increases when ki < s < li and decreases if 0 6 s 6 ki or if li < s < 1 (if li 6¼ 1). In particular, if 0 < ki < li < 1 then xi grows at moderate nutrient concentration (ki < s < li ) and declines at low concentration (s < ki ) and at high concentration (li < s). Remark. A further assumption of a generic character is that all the finite ki and li is distinct from each other and from s0 (which is biologically reasonable). nþ1 Let Rþ ¼ fðs; x1 ; x2 ; . . . ; xn Þ : s P 0; xi P 0; i ¼ 1; 2; . . . ; ng and let C þ ð½s; nþ1 nþ1 0 ; Rþ Þ denote the space of continuous functions from ½s; 0 into Rþ . Also, we assume that the initial conditions for the system (2.1) take the following form:
8 9 < sðhÞ ¼ U0 ðhÞ; xi ðhÞ ¼ Ui ðhÞ; i ¼ 1; 2; . . . ; n; = nþ1 ðU ; U ; . . . ; Un Þ 2 C þ ð½s; 0 ; Rþ Þ; : 0 1 ; Ui ð0Þ > 0; i ¼ 1; 2; . . . ; n:
ð2:3Þ
One can easily show that solutions of the system (2.1) corresponding to the inequality (2.2) are defined for all t P 0 and remain positive for all t P 0. Such solutions will be called positive solutions. The following lemma shows that positive solutions of the system (2.1) are bounded for all positive time.
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Lemma 2.1. If ðsðtÞ; x1 ðtÞ; x2 ðtÞ; . . . ; xn ðtÞÞ is a positive solution of system (2.1), then sðtÞ þ
n X
eDsi xi ðt þ si Þ ¼ s0 þ eðtÞ for t P 0;
i¼1
where eðtÞ ! 0 exponentially as t ! þ1. Pn Proof. Let ZðtÞ ¼ sðtÞ þ i¼1 eDsi xi ðt þ si Þ. From (2.1) we obtain Z 0 ðtÞ ¼ Dðs0 ZðtÞÞ, from which we obtain ZðtÞ ¼ s0 þ ðZ0 s0 ÞeDt , where Z0 ¼ Zð0Þ. This completes the proof. We shall need the following two lemmas: Lemma 2.2 [2]. Let f ðtÞ 2 C 2 ½t0 ; 1Þ, f ðtÞ P 0 and L > 0. i(i) If f 0 ðtÞ P 0, f ðtÞ is bounded and f 00 ðtÞ < L for all t P t0 , then f 0 ðtÞ ! 0 as t ! þ1. (ii) If f 0 ðtÞ 6 0 and f 00 ðtÞ P L > 1 for all t P t0 , then f 0 ðtÞ ! 0 as t ! þ1. Lemma 2.3 [9]. Let f ðtÞ 2 C 3 ½t0 ; 1Þ and 0 6 f ðtÞ 6 L for all t P t0 . Suppose that f1 ¼ lim inf t!þ1 f ðtÞ and f 1 ¼ lim supt!þ1 f ðtÞ. If f ðtÞ is not eventually monotone, then there exist sequences ftn g and ftn g, tn ! þ1, tn ! þ1 as n ! þ1, such that f ðtn Þ ! f 1 and f ðtn Þ ! f1 as n ! þ1 and f 0 ðtn Þ ¼ f 0 ðtn Þ ¼ 0 for all n.
3. Main results The principal result of this paper will be stated and proved in this current section. Theorem 3.1. If the break-even concentration of the substrate for the ith competitor is greater than or equal to the concentration of substrate in the feed bottle; then limt!þ1 xi ðtÞ ¼ 0 for all positive solutions of the system (2.1). Proof. Let ðsðtÞ; x1 ; x2 ; . . . ; xn Þ be a positive solution of system (2.1) and let 1 x1 i ¼ lim supt!þ1 xi ðtÞ. We consider that xi > 0 to derive a concentration by considering two cases. Case 1. Suppose that xi ðtÞ is eventually monotone. Then xi ðtÞ ! x1 as i t ! þ1. By Lemma 2.2, we see that x0i ðtÞ ! 0 as t ! þ1. For fixed i, Lemma 2.1 implies that
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x0i ðtÞ ¼ Dxi ðtÞ þe
Dsi
" 0
Dsi
Pi s þ eðt si Þ e xi ðtÞ
n X
151
# e
Dsj
xj ðt þ sj si Þ xi ðt si Þ:
j¼1 j6¼i
From this we have " 0
Dsi
Pi s þ eðt si Þ e xi ðtÞ
n X
# e
Dsj
xj ðt þ sj si Þ ¼
j¼1 j6¼i
½x0i ðtÞ þ Dxi ðtÞ eDsi : xi ðt si Þ
It follows that when t ! þ1, we obtain: " 0
Dsi
Pi s þ eðt si Þ e xi ðtÞ
n X
# e
Dsj
xj ðt þ sj si Þ ! DeDsi :
j¼1 j6¼i
Note that Pi ðsÞ ! DeDsi if s ! ki or li . It follows from the boundedness of sðtÞ and continuity of Pi ðsÞ that: " 0
lim
t!þ1
Dsi
s þ eðt si Þ e xi ðtÞ
n X
# e
Dsj
xj ðt þ sj si Þ ¼ ki or li :
j¼1 j6¼i
Assume that the left hand side of the above equation equal to ki . Since xj ðtÞ are positive for all j 6¼ i, j ¼ 1; 2; . . . ; n, we have s0 eDsi x1 i P ki , which yields Dsi 0 1 x1 6 e ðs k Þ 6 0, which contradict x > 0. If we replace li by ki , we find i i i that similar argument leads to a contradiction. Case 2. Suppose that xi ðtÞ is not eventually monotone. Then, Lemma 2.3 implies that there exists a sequence ftk g, tk ! þ1 as k ! þ1, such that x0i ðtk Þ ¼ 0, k ¼ 1; 2; . . . ; xi ðtk Þ ! x1 i as k ! þ1. From second equation of the system (2.1) and by Lemma 2.1, we get " 0
Dsi
Pi s þ eðtk si Þ e xi ðtk Þ
n X
# e
Dsj
xj ðtk þ sj si Þ ¼
j¼1 j6¼1
Dxi ðtk ÞeDsi : xi ðtk si Þ
Recall that lim supt!þ1 xi ðtÞ ¼ x1 i . For any g > 0 there is an N ðgÞ > 0 such that xi ðtk si Þ < x1 þ g for k > N . As consequence, i " 0
Dsi
Pi s þ eðtk si Þ e xi ðtk Þ
n X j¼1 j6¼i
# e
Dsj
xj ðtk þ sj si Þ P
Dxi ðtk ÞeDsi x1 i þg
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for k > N . It follows that: 2 6 lim inf Pi 4s0 þ eðtk si Þ eDsi xi ðtk Þ
n X
tk !þ1
Since g > 0 is arbitrarily fixed. We have 2 6 lim inf Pi 4s0 þ eðtk si Þ eDsi xi ðtk Þ
j¼1 j6¼i
n X
tk !þ1
j¼1 j6¼i
3 Dx1 eDsi 7 : eDsj xj ðtk þ sj si Þ5 P 1i xi þ g
3 7 eDsj xj ðtk þ sj si Þ5 P DeDsi :
Note that Pi ðsÞ < DeDsi when s 2 ð0; ki Þ. We have 2 3 n X 6 7 lim inf 4s0 þ eðtk si Þ eDsi xi ðtk Þ eDsj xj ðtk þ sj si Þ5 P ki ; tk !þ1
j¼1 j6¼i
1 Dsi 0 from which we obtain s0 eDsi x1 ðs ki Þ 6 0. We i P ki , which yields xi 6 e are led to a contradiction, and therefore, the proof of Theorem 3.1 is completed.
Due to Theorem 3.1, we shall assume that the populations are labeled so that ki < kj < s0 for i < j. Set bi ¼ eDsi ðs0 ki Þ, i ¼ 1; 2; . . . ; n. By the same type of arguments as those in the proof of Theorem 3.1, one can obtain the following lemma. Lemma 3.1. Let ki < kj < s0 for i < j and i; j ¼ 1; 2; . . . ; n. Set bi ¼ eDsi ðs0 ki Þ, for fixed i; then any positive solution ðsðtÞ; x1 ; x2 ; . . . ; xn Þ of the system (2.1) satisfies lim supt!þ1 xi ðtÞ 6 bi , i ¼ 1; 2; . . . ; n. For mathematical reasons, we shall assume that s0 < l in the following theorem. Theorem 3.2. If ki < kj < s0 < l for i < j; then any positive solution of the system (2.1) satisfies limt!þ1 sðtÞ ¼ ki , limt!þ1 xi ðtÞ ¼ bi and limt!þ1 xj ðtÞ ¼ 0, i; j ¼ 1; 2; . . . ; n. To prove this theorem, we need the following lemma. Lemma 3.2. If ki < kj < s0 < l for i < j; then lim inf t!þ1 xi ðtÞ P 0 for all positive solutions of system (2.1), i; j ¼ 1; 2; . . . ; n.
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153
Proof. Fix n > 0 small enough such that kj 2n > ki . By Lemmas 2.1 and 3.1, we see that there is a t0 > 0 such that 2 3 n X 60 7 eDsj xj ðt þ sj si Þ5 P kj n for t P t0 : ð3:1Þ 4s þ eðt si Þ j¼1 j6¼i
Also, let d > 0 be small enough such that n > deDsi :
ð3:2Þ
Suppose lim inf t!þ1 xi ðtÞ ¼ 0. Then there is a sufficiently large t1 such that xi ðt1 Þ < d. Set m ¼ mint2½t1 si ;t1 xi ðtÞ. Since t1 is large enough and xi ðtÞ is positive when t is large, we must have m > 0. Set t2 ¼ supft : t P t1 ; xi ðuÞ P m for all u 2 ½t1 si ; t g. Since xi ðuÞ P m for all u 2 ½t1 si ; t1 , the set ft : t P t1 ; xi ðuÞ P m for all u 2 ½t1 si ; t g is empty. Now we have two cases for t2 . Case 1. If t2 is infinite, then x0i ðtÞ P m > 0 for all t P t1 si . This would contradict lim inf t!þ1 xi ðtÞ ¼ 0. Case 2. If t2 is finite, it is clear that x0i 6 ðt2 Þ 6 0. However, x0i ðt2 Þ ¼ Dxi ðt2 Þ 2 6 þ eDsi Pi 4s0 þ eðt2 si Þ eDsi xi ðt2 Þ x0i ðt2 Þ P xi ðt2 Þ 2
n X j¼1 j6¼i
3 7 eDsj xj ðt2 þ sj si Þ5xi ðt2 si Þ;
2
6 6 4 D þ eDsi Pi 4s0 þ eðt2 si Þ eDsi xi ðt2 Þ
By (3.1) and (3.2), we have 2 60 4s þ eðt2 si Þ eDsi d
n X j¼1 j6¼i
n X j¼1 j6¼i
33 77 eDsj xj ðt2 þ sj si Þ55:
3 7 eDsj xj ðt2 þ sj si Þ5 > ki :
It follows from s0 < l that: 2 2 6 6 4 D þ eDsi Pi 4s0 þ eðt2 si Þ eDsi xi ðt2 Þ
n X j¼1 j6¼i
33 77 eDsj xj ðt2 þ sj si Þ55 > 0:
Consequently, x0i ðt2 Þ > 0. We are led to a contradiction. The proof of Lemma 3.2 is completed.
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Proof of Theorem 3.2. Let ðsðtÞ; x1 ; x2 ; . . . ; xn Þ be a positive solution of the system (2.1). We need to prove that limt!þ1 sðtÞ ¼ ki , limt!þ1 xi ðtÞ ¼ bi and limt!þ1 xj ðtÞ ¼ 0 for i < j and i; j ¼ 1; 2; . . . ; n. In view of Lemma 2.1, it suffices to show that limt!þ1 xi ðtÞ ¼ bi and limt!þ1 xj ðtÞ ¼ 0 for i < j. We get x0i ðtÞ ¼ Dxi ðtÞ
2
6 þ eDsi Pi 4s0 þ eðt si Þ eDsi xi ðtÞ
n X j¼1 j6¼i
3 7 eDsj xj ðt þ sj si Þ5xi ðt si Þ: ð3:3Þ
Let we define three auxiliary functions fi ðxj Þ, gi ðxk Þ and hi ðxL Þ with k < L < j, n Y eDsm ; fi ðxj Þ ¼ s0 ki xj eDsj
i; j ¼ 1; 2; . . . ; n;
m¼1 m6¼j
n Y gi ðxk Þ ¼ s0 ki xk eDsk eDsm ;
i; k ¼ 1; 2; . . . ; n;
m¼1 m6¼k
n Y hi ðxL Þ ¼ s0 ki xL eDsL eDsm ;
i; L ¼ 1; 2; . . . ; n:
m¼1 m6¼L
By the definition of fi ðxj Þ, gi ðxk Þ and hi ðxL Þ; we can see that if common sm ¼ 0 between gi ðxk Þ and hi ðxL Þ except sk ; sL 6¼ 0 in hi ðxL Þ and in gi ðxk Þ, respectively, then i(i) xL ¼ gi ðxk Þ are inverse functions of xk ¼ hi ðxL Þ for each corresponding i, (ii) gL ð0Þ ¼ bL . Also, if common sm ¼ 0 between fi ðxj Þ and hi ðxL Þ except sj 6¼ 0 in hi ðxL Þ and sL 6¼ 0 in fi ðxj Þ, we obtain i(i) xj ¼ hi ðxL Þ are inverse functions of xL ¼ fi ðxj Þ for each corresponding i, (ii) hj ð0Þ ¼ bj . Finally, for common sm ¼ 0 between fi ðxj Þ and gi ðxk Þ except sj 6¼ 0 in gi ðxk Þ and sk 6¼ 0 in fi ðxj Þ, we get i(i) xj ¼ gi ðxk Þ are inverse functions of xk ¼ fi ðxj Þ for each corresponding i, (ii) fk ð0Þ ¼ bk .
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Let ek ¼ fk ðbk Þ, b0 ¼ gk ð0Þ. Since kk < kL < kj < s0 , we have bj > 0, bL > 0, bk > 0 and b0 > bL . Notice that fi ðxj Þ is strictly decreasing, we see that 0 < ek < bk . Set eL ¼ fL ð0Þ. It easy to see that eL < bk . Let we consider that lim inf t!þ1 xk ðtÞ ¼ p. Then p > 0 by Lemma 3.2. At this point we need the following. Claim. If p P eL , then bk lim xm ðtÞ ¼ 0 t!þ1
if m ¼ k ; if m ¼ 6 k
k; m ¼ 1; 2; . . . ; n:
Proof. Set lim supt!þ1 xL ðtÞ ¼ qL > 0 and lim supt!þ1 xi ðtÞ ¼ r ¼ 0, L; i ¼ 1; 2; . . . ; n. Case 1. Assume that xL ðtÞ and xi ðtÞ are eventually monotone and converge to qL and ri , respectively, as t ! þ1. Then by Lemmas 2.1 and 2.2, x0i ðtÞ ! 0 as t ! þ1 and x0L ðtÞ ! 0 as t ! þ1. From the equation of system (3.3), replace i by L, one obtains 2 3 n X eDsi xi ðt þ si sL Þ5 PL 4s0 þ eðt sL Þ eDsL xL ðtÞ eDsk xk ðt þ sk sL Þ i¼1 i6¼k6¼L
¼
½x0L ðtÞ þ DxL ðtÞ eDsL ; xL ðt sL Þ
which leads to 2 lim inf PL 4s0 þ eðt sL Þ eDsL xL ðtÞ eDsk xk ðt þ sk sL Þ t!þ1
n X
3 eDsi xi ðt þ si sL Þ5 P DeDsL ;
i¼1 i6¼k6¼L
consequently, 2 lim inf 4s0 þ eðt sL Þ eDsL xL ðtÞ eDsk xk ðt þ sk sL Þ t!þ1
n X i¼1 i6¼k6¼L
3 eDsi xi ðt þ si sL Þ5 P kL :
156
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Thus, we have 2
3 n X eDsi ri 5 P kL : lim inf 4s0 eDsL qL eDsk p t!þ1
i¼1 i6¼k6¼L
As a consequence, 2
3 n X eDsi ri 5 qL 6 eDsL 4s0 kL eDsk p i¼1 i6¼k6¼L
6 eDsL ½s0 kL eDsk eL ¼ gL ðeL Þ ¼ 0; which contradicts qL > 0. Case 2. Assume that xL ðtÞ and xi ðtÞ are not eventually monotone. Then by Lemma 2.3, there exists a sequence ftn g, tn ! þ1 as n ! þ1, such that x0L ðtn Þ ¼ x0i ðtn Þ ¼ 0 and 8 9 < qL > 0 if m ¼ L = xi ðtn Þ ¼ ri ¼ 0 if i ¼ m ; : ; p > 0 if m ¼ k
i 6¼ k 6¼ L; i; k; L ¼ 1; 2; . . . ; n:
Thus, from the equation of system (3.3) and i is replaced by L, we have 2 PL 4s0 þ eðtn sL Þ eDsL xL ðtn Þ eDsk xk ðtn þ sk sL Þ
n X
3 eDsi xi ðtn þ si sL Þ5 ¼
i¼1 i6¼k6¼L
DxL ðtn ÞeDsL : xL ðtn sL Þ
For any g > 0, there is an N ðgÞ > 0 such that xL ðtn sL Þ < qL þ g when N ðgÞ < n and lim supt!þ1 xL ðtn Þ ¼ qL . As a consequence, if N ðgÞ < n and as tn ! þ1, we get 2 lim inf PL 4s0 þ eðtn sL Þ eDsL xL ðtn Þ eDsk xk ðtn þ sk sL Þ tn !þ1
n X i¼1 i6¼k6¼L
3 eDsi xi ðtn þ si sL Þ5 P
DqL eDsL : qL þ g
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157
Since g > 0 is arbitrarily fixed, we have 2 lim inf PL 4s0 þ eðtn sL Þ eDsL xL ðtn Þ eDsk xk ðtn þ sk sL Þ tn !þ1
n X
3 eDsi xi ðtn þ si sL Þ5 P DeDsL :
i¼1 i6¼k6¼L
Furthermore, we can note that PL ðsÞ < DeDsL for 0 6 s 6 kL . We conclude that, for any f > 0, there is an N ðfÞ > 0 such that 2 lim inf 4s0 þ eðtn sL Þ eDsL xL ðtn Þ eDsk xk ðtn þ sk sL Þ tn !þ1
n X
3 eDsi xi ðtn þ si sL Þ5 P kL f;
i¼1 i6¼k6¼L
for n > N ðfÞ. One obtains 2 4s0 qL eDsL peDsk
n X
3 ri eDsi 5 P kL f;
i¼1 i6¼k6¼L
from which, we get 2 qL 6 eDsL 4s0 kL þ f peDsk
n X
3 ri eDsi 5:
i¼1 i6¼k6¼L
Since ri ¼ 0 and f > 0 is arbitrarily fixed, we have qL 6 eDsL ½s0 kL eL eDsk ¼ gL ðeL Þ ¼ 0, which contradicts qL > 0. Thus, in all cases, we must have qL ¼ 0, and therefore, p if m ¼ k lim xm ðtÞ ¼ ; m; k ¼ 1; 2; . . . ; n: 0 if m 6¼ k t!þ1 Now, we want to show that limt!þ1 xk ðtÞ ¼ p ¼ bk . Since lim supt!þ1 xk ðtÞbk by Lemma 3.1. Therefore, we must show that lim inf t!þ1 xk ðtÞ ¼ p ¼ bk . Suppose that eL 6 p < bk is in the following two possible cases. Case 1. Assume that xk ðtÞ is eventually monotone and converge to p as t ! þ1. Then, by Lemmas 2.1 and 2.2, x0k ðtÞ ! 0 as t ! þ1. From the equation of system (3.3), replace i by k, one obtains
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2 Pk 4s0 þ eðt sk Þ eDsk xk ðtÞ
n X
3 eDsi xi ðt þ si sk Þ5
i¼1 i6¼k
¼
½x0k ðtÞ þ Dxk ðtÞ eDsk : xk ðt sk Þ
From the result of the claim and limt!þ1 x0k ðtÞ ¼ 0, one obtains Pk ½s0 peDsk ¼ DeDsk , which, together with s0 < l, leads to ½s0 peDsk ¼ kk . Consequently, p ¼ ½s0 kk eDsk ¼ fk ð0Þ ¼ bk , contrary to p < bk . Case 2. Assume that xk ðtÞ is not eventually monotone. Then, by Lemma 2.3, there exists a sequence ftn g, tn ! þ1 as n ! þ1, such that x0k ðtn Þ ¼ 0 and xk ðtn Þ ! p as n ! þ1. Thus, from the equation of system (3.3) and i is replaced by k, we have 2 3 n X Dxk ðtn ÞeDsk : Pk 4s0 þ eðtn sk Þ eDsk xk ðtn Þ eDsi xi ðtn þ si sk Þ5 ¼ xk ðtn sk Þ i¼1 i6¼k
Since lim inf t!þ1 xk ðtÞ ¼ p, it follows that: 2 lim sup Pk 4s0 þ eðtn sk Þ eDsk xk ðtn Þ tn !þ1
n X
3 eDsi xi ðtn þ si sk Þ5 6 DeDsk ;
i¼1 i6¼k
which yields Pk ½s0 peDsk 6 DeDsk . Since s0 < l, we have ½s0 peDsk 6 kk . Hence, p P ½s0 kk eDsk ¼ fk ð0Þ ¼ bk , which contradicts p < bk . Summarizing the discussions of the last two cases, we conclude that p ¼ bk , and hence limt!þ1 xk ðtÞ ¼ bk . This completes the proof of the claim. To complete the proof of Theorem 3.2, it suffices to show that lim inf t!þ1 xk ðtÞ ¼ p ¼ bk > eL , k < L. First we verify the following assertion. Assertion 1. If ek ¼ fk ðbk Þ, then lim inf t!þ1 xk ðtÞ ¼ p ¼ bk > ek . Proof. Suppose that p < ek is in the following two possible cases. Case 1. Assume that xk ðtÞ is eventually monotone and converge to p as t ! þ1. Then, x0k ðtÞ ! 0 as t ! þ1 From the equation of system (3.3), replace i by k and k < j, one has 2 Pk 4s0 þ eðt sk Þ eDsk xk ðtÞ eDsj xj ðt þ sj sk Þ
n X i¼1 i6¼k6¼j
3 eDsi xi ðt þ si sk Þ5 ¼
½x0k ðtÞ þ Dxk ðtÞ eDsk ; xk ðt sk Þ
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159
from which, we get 2 lim 4s0 þ eðt sk Þ eDsk xk ðtÞ eDsj xj ðt þ sj sk Þ
t!þ1
n X
3 eDsi xi ðt þ si sk Þ5 ¼ kk ;
i¼1 i6¼k6¼j
which leads to 2 4s0 þ eðt sk Þ eDsk lim xk ðtÞ eDsj lim xj ðtÞ t!þ1
t!þ1
n X i¼1 i6¼k6¼j
3 eDsi lim xi ðtÞ5 ¼ kk : t!þ1
By using Lemma 3.1, we have 2 4s0 bj eDsj
n X
3 bi eDsi kk 5eDsk 6 p:
i¼1 i6¼k6¼j
Setting bi ¼ 0 for i 6¼ j, this reduces to p P ½s0 kk bj eDsj eDsk :
ð3:4Þ
Since kk < kj , we have p P ½s0 kk bk eDsj eDsk ¼ fk ðbk Þ ¼ ek , contrary to p < ek . Case 2. Assume that xk ðtÞ is not eventually monotone. Then, there exists a sequence ftn g, tn ! þ1 as n ! þ1, such that x0k ðtn Þ ¼ 0 and xk ðtn Þ ! p as n ! þ1. Thus, from the equation of system (3.3) and i is replaced by k, k < j, we have 2 Pk 4s0 þ eðtn sk Þ eDsk xk ðtn Þ eDsj xj ðt þ sj sk Þ
n X i¼1 i6¼k6¼j
3 eDsi xi ðtn þ si sk Þ5 ¼
Dxk ðtn ÞeDsk ; xk ðtn sk Þ
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from which, we get 2 lim sup Pk 4s0 þ eðtn sk Þ eDsk xk ðtn Þ eDsj xj ðt þ sj sk Þ tn !þ1
n X
3 eDsi xi ðtn þ si sk Þ5 6 DeDsk :
i¼1 i6¼k6¼j
Thus, for any g > 0, there is an N ðgÞ > 0 such that 2 lim sup4s0 þ eðtn sk Þ eDsk xk ðtn Þ eDsj xj ðt þ sj sk Þ tn !þ1
n X
3 eDsi xi ðtn þ si sk Þ5 6 kk þ g;
i¼1 i6¼k6¼j
for n > N ðgÞ. By using Lemma 3.1 and since g > 0 is arbitrarily fixed, we have 2 3 n X 0 Ds Ds 4s kk bj e j bi e i 5eDsk 6 p: i¼1 i6¼k6¼j
Setting bi ¼ 0, it leads to p P ½s0 kk bj eDsj eDsk . As a consequence of the Eq. (3.4) in the first case, we obtain p > ek , contrary to p < ek . Thus, we proved that in all cases p > ek . Now, we must show that eL < lim inf t!þ1 xk ðtÞ ¼ p ¼ bk . Suppose that p < eL . Thus, there are two cases to be considered. Case 1. From the Eq. (3.4), we have p P eDsk ½s0 kk bj eDsj . Putting bj ¼ 0, say by the claim, and for all k < L, we have p P eDsk ½s0 kk > eDsk ½s0 kL ¼ fL ð0Þ ¼ eL , which contradicts p < eL . Case 2. Furthermore, for all kL > kk , from the Eq. (3.4) and the claim, we have p > eDsk ½s0 kL ¼ fL ð0Þ ¼ eL , which contradicts p < eL . Thus in all cases, p > eL . Therefore, the proof of Theorem 3.2 is completed. Remark. Theorem 3.2 implies that limt!þ1 sðtÞ ¼ kk , limt!þ1 xk ðtÞ ¼ eDsk ½s0 kk and limt!þ1 xm ðtÞ ¼ 0, 8m 6¼ k. This indicates that Theorem 3.2 is generalization of the results of [7] where three populations are considered. Assertion 2. If kL > kk , then eL > ek . Proof. It easy to show that bk ¼ ek ½1 eDðsk sj Þ 1 ¼ eL þ ðkL kk ÞeDsk ;
H.M. El-Owaidy, A.A. Moniem / Appl. Math. Comput. 147 (2004) 147–161
161
which leads to eL þ ðkL kk ÞeDsk ¼ ek þ bk eDsj : 0
Since s ¼ kL þ bL e therefore we have
DsL
¼ kk þ bk e
ðkL kk ÞeDsk < bk eDsj :
Dsk
ð3:5Þ
, from which, one obtains kL kk 6 bk eDsk , ð3:6Þ
By (3.5) and (3.6), we have ek þ bk eDsj < eL þ bk eDsj , which implies that ek < eL . 4. Discussion In this paper, a chemostat model with delayed response in growth is considered. The model incorporates time delays that allow the competing organisms to be stored the nutrient (so that the free nutrient concentration does not reflect the nutrient available for growth) and allow growth response functions to be nonmonotone. The main results of El-Owaidy and Ismail are generalized. Our results show that at most one competitor survives, and the substrate and the surviving competitor (if two exist) approaches limiting values, provided that some parameters are distinct. Thus the principal of competitive exclusion remains valid for that model. This indicates that this kind of time delay does not influence the competitive outcome of the organisms. References [1] A.W. Bush, A.E. Cook, The effect of time delay and growth rate inhibition in the bacterial treatment of wastewater, J. Theor. Biol. 63 (1975) 385–395. [2] G.B. Hsu, S. Hubbell, P. Waltman, A mathematical theory for single-nutrient competition in continuous cultures of microorganisms, SIAM J. Appl. Math. 32 (1977) 366–383. [3] G.J. Butler, G.S. Wolkowicz, A mathematical model of the chemostat with a general class of functions describing nutrient uptake, SIAM J. Appl. Math. 45 (1985) 138–151. [4] H.I. Freedman, J.W.H. So, P. Waltman, Chemostat competition with time delays, in: R. Vichnevetsky, P. Borne, J. Vignes (Eds.), Proceeding of IMACS (1988): 12th World Congress on Scientific Computation: Modeling and Simulation of System Paris, Gerfidn Cite Scientifique, France, 1988, pp. 102–104. [5] H.I. Freedman, J.W.H. So, P. Waltman, Coexistence in a model of competition in the chemostat incorporating discrete delays, SIAM J. Appl. Math. 49 (1989) 859–870. [6] H.L. Smith, P. Waltman, The Theory of Chemostat, Cambridge University Press, Cambridge, UK, 1995. [7] H.M. El-Owaidy, M. Ismail, Asymptotic behavior of the chemostat model with delayed response in growth, Chaos, Solitons Fract. 13 (2002) 787–795. [8] S.F. Ellermeyer, Competition in the chemostat: global asymptotic behavior of a model with delayed response in growth, SIAM J. Appl. Math. 54 (1994) 456–465. [9] W. Wendi, M. Zhien, Converges in the chemostat model with delayed response in growth, J. Syst. Sci. Math. Sci. 12 (1999) 23–32.