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International Journal of Differential Equations
Volume 2010 (2010), Article ID 287969, 41 pages
doi:10.1155/2010/287969
Research Article

A Predator-Prey Model in the Chemostat with Time Delay

Guihong Fan1,2 and Gail S. K. Wolkowicz1

1Department of Mathematics and Statistics, McMaster University, Hamilton, ON, L8S 4K1, Canada
2Department of Mathematics and Statistics, York University, 4700 Keele Street, Toronto, ON, M3J 1P3, Canada

Received 1 November 2009; Accepted 11 January 2010

Academic Editor: Yuri V. Rogovchenko

Copyright © 2010 Guihong Fan and Gail S. K. Wolkowicz. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The aim of this paper is to study the dynamics of predator-prey interaction in a chemostat to determine whether including a discrete delay to model the time between the capture of the prey and its conversion to viable biomass can introduce oscillatory dynamics even though there is a globally asymptotically stable equilibrium when the delay is ignored. Hence, Holling type I response functions are chosen so that no oscillatory behavior is possible when there is no delay. It is proven that unlike the analogous model for competition, as the parameter modeling the delay is increased, Hopf bifurcations can occur.

1. Introduction

The chemostat, also known as a continuous stir tank reactor (CSTR) in the engineering literature, is a basic piece of laboratory apparatus used for the continuous culture of microorganisms. It has potential applications for such processes as wastewater decomposition and water purification. Some ecologists consider it a lake in a laboratory. It can be thought of as three vessels, the feed bottle that contains fresh medium with all the necessary nutrients, the growth chamber where the microorganisms interact, and the collection vessel. The fresh medium from the feed bottle is continuously added to the growth chamber. The growth chamber is well stirred and its contents are then removed to the collection vessel at a rate that maintains constant volume. For a detailed description of the importance of the chemostat and its application in biology and ecology, one can refer to [1, 2].

The following system describes a food chain in the chemostat where a predator population feeds on a prey population of microorganisms that in turn consumes a nonreproducing nutrient that is assumed to be growth limiting at low concentrations

𝑠 ̇ 𝑠 ( 𝑡 ) = 0 𝐷 − 𝑠 ( 𝑡 ) 0 − 𝑥 ( 𝑡 ) 𝑓 ( 𝑠 ( 𝑡 ) ) 𝜂 , 𝑦 ̇ 𝑥 ( 𝑡 ) = 𝑥 ( 𝑡 ) ( − 𝐷 + 𝑓 ( 𝑠 ( 𝑡 ) ) ) − ( 𝑡 ) 𝑔 ( 𝑥 ( 𝑡 ) ) 𝜉 , ̇ 𝑦 ( 𝑡 ) = − Δ 𝑦 ( 𝑡 ) + 𝑦 ( 𝑡 ) 𝑔 ( 𝑥 ( 𝑡 ) ) . ( 1 . 1 ) Here 𝑠 ( 𝑡 ) represents the concentration of the growth limiting nutrient, 𝑥 ( 𝑡 ) the density of the prey population, and 𝑦 ( 𝑡 ) the density of the predator population. Parameter 𝑠 0 denotes the concentration of the growth limiting nutrient in the feed vessel, 𝐷 0 the dilution rate, 𝜂 ( 𝜉 ) the growth yield constant, 𝐷 ( Δ ) the sum of the dilution rate 𝐷 0 and the natural species specific death rate of the prey (predator) population, respectively. Here 𝑓 ( 𝑠 ) denotes the functional response of the prey population on the nutrient and 𝑔 ( 𝑥 ) denotes the functional response of the predator on the prey.

Butler et al. [3] considered the coexistence of two competing predators feeding on a single prey population growing in the chemostat. As a subsystem of their model, they studied the global stability of system (1.1) with both 𝑓 ( 𝑠 ) and 𝑔 ( 𝑥 ) taking the form of Holling type II. They proved that under certain conditions the interior equilibrium is globally asymptotically stable with respect to the interior of the positive cone. However, they also proved that for certain ranges of the parameters there is at least one nontrivial limit cycle and conjectured that the limit cycle is unique and would be a global attractor with respect to the noncritical orbits in the open positive octant. This conjecture was partially solved by Kuang [4]. He showed that there is a range of parameters for which a unique periodic orbit exists and roughly located the position of the limit cycle.

Bulter and Wolkowicz [5] studied predator mediated coexistence in the chemostat assuming 𝐷 0 = 𝐷 = Δ . Model (1.1) was studied as a submodel. For general monotone response functions, Bulter and Wolkowicz showed that (1.1) is uniformly persistent if the sum of the break even concentrations of substrate and prey is less than the input rate of the nutrient 𝑠 0 . However they showed that it is necessary to specify the form of the response functions in order to discuss the global dynamics of the model. If 𝑓 ( 𝑠 ) is modelled by Holling type I or II and 𝑔 ( 𝑥 ) by Holling type I, Bulter and Wolkowicz proved that (1.1) could have up to three equilibrium points and that there is a transfer of global stability from one equilibrium point to another as different parameters are varied making conditions favorable enough for a new population to survive. In this case, there are no periodic solutions. However, even if 𝑓 ( 𝑠 ) is given by Holling type I, if 𝑔 ( 𝑥 ) is given by Holling type II, they showed that a Hopf bifurcation can occur in (1.1), and numerical simulations indicated that the bifurcating periodic solution was asymptotically stable.

We include a time delay in (1.1) to model the time between the capture of the prey and its conversion to viable biomass. Our aim is to show that such a delay can induce nontrivial periodic solutions in a model where there is always a globally asymptotically stable equilibrium when delay is ignored, and hence no such periodic solutions are possible otherwise. For this reason we select the response functions of the simplest form; that is, we choose the Holling type I form for both 𝑓 ( 𝑠 ) and 𝑔 ( 𝑥 ) , so that (1.2) always has a globally asymptotically stable equilibrium when the conversion process is assumed to occur instantaneously. It is interesting to note that in the analogous model of competition between two species in the chemostat, delay cannot induce oscillatory behavior for any reasonable monotone response functions (see Wolkowicz and Xia [6]).

With delay modelling the time required for the predator to process the prey after it has been captured, the model is given by

𝑠 ̇ 𝑠 ( 𝑡 ) = 0 𝐷 − 𝑠 ( 𝑡 ) 0 − 𝑥 ( 𝑡 ) 𝑓 ( 𝑠 ( 𝑡 ) ) 𝜂 , 𝑦 ̇ 𝑥 ( 𝑡 ) = 𝑥 ( 𝑡 ) ( − 𝐷 + 𝑓 ( 𝑠 ( 𝑡 ) ) ) − ( 𝑡 ) 𝑔 ( 𝑥 ( 𝑡 ) ) 𝜉 , ̇ 𝑦 ( 𝑡 ) = − Δ 𝑦 ( 𝑡 ) + 𝑒 − Δ 𝜏 𝑦 ( 𝑡 − 𝜏 ) 𝑔 ( 𝑥 ( 𝑡 − 𝜏 ) ) . ( 1 . 2 ) For 𝑡 ∈ [ − 𝜏 , 0 ] ,

𝑠 ( 0 ) = 𝑠 0 ∈ i n t ℝ + [ ] , ( 𝑥 ( 𝑡 ) , 𝑦 ( 𝑡 ) ) = ( 𝜙 , 𝜓 ) ∈ ℂ − 𝜏 , 0 , i n t ℝ 2 + . ( 1 . 3 ) Here variables 𝑠 ( 𝑡 ) , 𝑥 ( 𝑡 ) , 𝑦 ( 𝑡 ) , and parameters 𝑠 0 , 𝐷 0 , 𝜂 , 𝜉 , 𝐷 0 , 𝐷 , and Δ have the same interpretation as for model (1.1). Note therefore that 𝐷 ⩾ 𝐷 0 , and Δ ⩾ 𝐷 0 . The additional parameter 𝜏 is a nonnegative constant modelling the time required for the conversion process. Hence, 𝑒 − Δ 𝜏 𝑦 ( 𝑡 − 𝜏 ) represents the concentration of the predator population in the growth chamber at time 𝑡 that were available at time 𝑡 − 𝜏 to capture prey and were able to avoid death and washout during the 𝜏 units of time required to process the captured prey.

We analyze the stability of each equilibrium and prove that the coexistence equilibrium can undergo Hopf bifurcations. Numerical simulations appear to show that (1.2) can have a stable periodic solution bifurcating from the coexistence equilibrium as the delay parameter increases from zero. This periodic orbit can then disappear through a secondary Hopf bifurcation as the delay parameter increases further.

2. Scaling of the Model and Existence of Solutions

Suppose that functions 𝑓 ( 𝑠 ) and 𝑔 ( 𝑠 ) are of Holling type I form, that is, 𝑓 ( 𝑠 ) = 𝛼 𝑠 ( 𝛼 > 0 ) and 𝑔 ( 𝑥 ) = 𝑘 𝑥 ( 𝑘 > 0 ). System (1.2) reduces to

𝑠 ̇ 𝑠 ( 𝑡 ) = 0 𝐷 − 𝑠 ( 𝑡 ) 0 − 𝛼 𝑥 ( 𝑡 ) 𝑠 ( 𝑡 ) 𝜂 , ̇ 𝑥 ( 𝑡 ) = 𝑥 ( 𝑡 ) ( − 𝐷 + 𝛼 𝑠 ( 𝑡 ) ) − 𝑘 𝑥 ( 𝑡 ) 𝑦 ( 𝑡 ) 𝜉 , ̇ 𝑦 ( 𝑡 ) = − Δ 𝑦 ( 𝑡 ) + 𝑘 𝑒 − Δ 𝜏 𝑦 ( 𝑡 − 𝜏 ) 𝑥 ( 𝑡 − 𝜏 ) . 𝑡 > 0 , ( 2 . 1 ) Introducing the following change of variables gives:

̆ 𝑡 = 𝐷 0 ̆ 𝑡 = 𝑡 , ̆ 𝑠 𝑠 ( 𝑡 ) 𝑠 0 ̆ 𝑡 = , ̆ 𝑥 𝑥 ( 𝑡 ) 𝑠 0 𝜂 ̆ 𝑡 = , ̆ 𝑦 𝑦 ( 𝑡 ) 𝜉 𝑠 0 𝜂 , ̆ 𝜏 = 𝐷 0 ̆ 𝐷 𝜏 , 𝐷 = 𝐷 0 , ̆ Δ Δ = 𝐷 0 , ̆ 𝑘 = 𝑘 𝑠 0 𝜂 𝐷 0 , ̆ 𝛼 = 𝛼 𝑠 0 𝐷 0 , ̆ 𝑡 d ̆ 𝑠 d ̆ 𝑡 = 1 𝑠 0 d 𝑠 ( 𝑡 ) d 𝑡 d 𝑡 d ̆ 𝑡 = 1 𝑠 0 𝐷 0 d 𝑠 ( 𝑡 ) = 1 d 𝑡 𝑠 0 𝐷 0 𝑠 0 𝐷 − 𝑠 ( 𝑡 ) 0 − 𝛼 𝑥 ( 𝑡 ) 𝑠 ( 𝑡 ) 𝜂 = 1 − 𝑠 ( 𝑡 ) 𝑠 0 − 𝛼 𝑠 0 𝐷 0 𝑥 ( 𝑡 ) 𝑠 0 𝜂 𝑠 ( 𝑡 ) 𝑠 0 ̆ 𝑡 ̆ 𝑡 ̆ 𝑡 , ̆ 𝑡 = 1 − ̆ 𝑠 − ̆ 𝛼 ̆ 𝑥 ̆ 𝑠 d ̆ 𝑥 d ̆ 𝑡 = 1 𝑠 0 𝜂 d 𝑥 ( 𝑡 ) d 𝑡 d 𝑡 d ̆ 𝑡 = 1 𝑠 0 𝜂 𝐷 0 d 𝑥 ( 𝑡 ) = 1 d 𝑡 𝑠 0 𝜂 𝐷 0 𝑥 ( 𝑡 ) ( − 𝐷 + 𝛼 𝑠 ( 𝑡 ) ) − 𝑘 𝑥 ( 𝑡 ) 𝑦 ( 𝑡 ) 𝜉 = 𝑥 ( 𝑡 ) 𝑠 0 𝜂 − 𝐷 𝐷 0 + 𝛼 𝑠 0 𝐷 0 𝑠 ( 𝑡 ) 𝑠 0 − 𝑘 𝑠 0 𝜂 𝐷 0 𝑥 ( 𝑡 ) 𝑠 0 𝜂 𝑦 ( 𝑡 ) 𝑠 0 ̆ 𝑡 − ̆ ̆ 𝑡 − ̆ ̆ 𝑡 ̆ 𝑡 , ̆ 𝑡 𝜂 𝜉 = ̆ 𝑥 𝐷 + ̆ 𝛼 ̆ 𝑠 𝑘 ̆ 𝑥 ̆ 𝑦 d ̆ 𝑦 d ̆ 𝑡 = 1 𝑠 0 𝜂 𝜉 d 𝑦 ( 𝑡 ) d 𝑡 d 𝑡 d ̆ 𝑡 = 1 𝑠 0 𝜂 𝜉 𝐷 0 d 𝑦 ( 𝑡 ) = 1 d 𝑡 𝑠 0 𝜂 𝜉 𝐷 0 − Δ 𝑦 ( 𝑡 ) + 𝑘 𝑒 − Δ 𝜏 = 𝑦 ( 𝑡 − 𝜏 ) 𝑥 ( 𝑡 − 𝜏 ) − Δ 𝑦 ( 𝑡 ) 𝑠 0 𝜂 𝜉 𝐷 0 + 𝑘 𝑒 − Δ 𝜏 𝑠 0 𝜂 𝜉 𝐷 0 = 𝑦 ( 𝑡 − 𝜏 ) 𝑥 ( 𝑡 − 𝜏 ) − Δ 𝐷 0 𝑦 ( 𝑡 ) 𝑠 0 + 𝜂 𝜉 𝑘 𝑠 0 𝜂 𝐷 0 𝑒 − ( Δ / 𝐷 0 ) 𝐷 0 𝜏 𝑦 ( 𝑡 − 𝜏 ) 𝑠 0 𝜂 𝜉 𝑥 ( 𝑡 − 𝜏 ) 𝑠 0 𝜂 ̆ ̆ 𝑡 + ̆ = − Δ ̆ 𝑦 𝑘 𝑒 − ̆ Δ ̆ 𝜏 ̆ ̆ . ̆ 𝑦 𝑡 − ̆ 𝜏 ̆ 𝑥 𝑡 − ̆ 𝜏 ( 2 . 2 ) With this change of variables, omitting the ̆ ‌ ’s for convenience, system (2.1) becomes

̇ 𝑠 ( 𝑡 ) = 1 − 𝑠 ( 𝑡 ) − 𝛼 𝑥 ( 𝑡 ) 𝑠 ( 𝑡 ) , ̇ 𝑥 ( 𝑡 ) = 𝑥 ( 𝑡 ) ( − 𝐷 + 𝛼 𝑠 ( 𝑡 ) ) − 𝑘 𝑦 ( 𝑡 ) 𝑥 ( 𝑡 ) , ̇ 𝑦 ( 𝑡 ) = − Δ 𝑦 ( 𝑡 ) + 𝑘 𝑒 − Δ 𝜏 𝑦 ( 𝑡 − 𝜏 ) 𝑥 ( 𝑡 − 𝜏 ) , ( 2 . 3 ) where Δ ⩾ 1 and 𝐷 ⩾ 1 , with initial data given by (1.3). For biological significance, a point is assumed to be an equilibrium point of (2.3) only if all of its components are nonnegative.

Let 𝜏 = 0 . Model (2.3) reduces to a special case of the model considered in [7]. If 𝐷 > 𝛼 , the model has only one equilibrium point ( 1 , 0 , 0 ) and it is globally asymptotically stable. If 𝐷 < 𝛼 and 1 − 𝐷 / 𝛼 − Δ 𝐷 / 𝑘 < 0 , the model has a second equilibrium point ( 𝐷 / 𝛼 , ( 𝛼 − 𝐷 ) / 𝛼 𝐷 , 0 ) and it is globally asymptotically stable. When 1 − 𝐷 / 𝛼 − Δ 𝐷 / 𝑘 > 0 , the model has a third equilibrium point ( 𝑘 / ( 𝑘 + 𝛼 Δ ) , Δ / 𝑘 , 𝛼 / ( 𝑘 + 𝛼 𝐷 ) − 𝐷 / 𝑘 ) and it is the global attractor. Therefore, model (2.3) has no periodic solutions when the time delay is ignored. If 𝑔 ( 𝑥 ) is of Holling type II form, Butler and Wolkowicz [5] proved that a Hopf bifurcation is possible resulting in a periodic solution for a certain range of parameter values. We emphasize again here, that it is for this reason that in this paper we restrict our attention to the simplest case for both response functions, that is, Holling type I, in order to see whether delay can be responsible for periodic solutions in (1.2).

Theorem 2.1. Assuming ( 𝑠 0 , 𝜙 ( 𝜃 ) , 𝜓 ( 𝜃 ) ) ∈ i n t ℝ + × ℂ ( [ − 𝜏 , 0 ] , i n t ℝ 2 + ) , then there exists a unique solution ( 𝑠 ( 𝑡 ) , 𝑥 ( 𝑡 ) , 𝑦 ( 𝑡 ) ) of (2.3) passing through ( 𝑠 0 , 𝜙 ( 𝜃 ) , 𝜓 ( 𝜃 ) ) with 𝑠 ( 𝑡 ) > 0 , 𝑥 ( 𝑡 ) > 0 and 𝑦 ( 𝑡 ) > 0 for 𝑡 ∈ [ 0 , ∞ ) . The solution is bounded. In particular, given any 𝜖 0 > 0 , 𝑥 ( 𝑡 ) < 1 + 𝜖 0 for all sufficiently large 𝑡 .

Proof. For 𝑡 ∈ [ 0 , 𝜏 ] , one has 𝑡 − 𝜏 ∈ [ − 𝜏 , 0 ] , 𝑥 ( 𝑡 − 𝜏 ) = 𝜙 ( 𝑡 − 𝜏 ) , and 𝑦 ( 𝑡 − 𝜏 ) = 𝜓 ( 𝑡 − 𝜏 ) . System (2.3) becomes ̇ 𝑠 ( 𝑡 ) = 1 − 𝑠 ( 𝑡 ) − 𝛼 𝑥 ( 𝑡 ) 𝑠 ( 𝑡 ) , ̇ 𝑥 ( 𝑡 ) = 𝑥 ( 𝑡 ) ( − 𝐷 + 𝛼 𝑠 ( 𝑡 ) ) − 𝑘 𝑦 ( 𝑡 ) 𝑥 ( 𝑡 ) , ̇ 𝑦 ( 𝑡 ) = − Δ 𝑦 ( 𝑡 ) + 𝑘 𝑒 − Δ 𝜏 𝜙 ( 𝑡 − 𝜏 ) 𝜓 ( 𝑡 − 𝜏 ) , ( 2 . 4 ) a system of nonautonomous ordinary differential equations with initial conditions 𝑠 ( 0 ) = 𝑠 0 , 𝑥 ( 0 ) = 𝜙 ( 0 ) , and 𝑦 ( 0 ) = 𝜓 ( 0 ) . Since the right-hand side of (2.4) is differentiable in both 𝑥 and 𝑦 , by Theorems 2 . 3 , 3 . 1 , and Corollary 4 . 3 in Miller and Michel [8], there exists a unique solution defined on [ 0 , 𝜏 ] satisfying (2.4). By using the method of steps in Bellman and Cooke [9], it can be shown that the solution through ( 𝑠 0 , 𝜙 ( 𝜃 ) , 𝜓 ( 𝜃 ) ) is defined for all 𝑡 ⩾ 0 .
Now we prove 𝑠 ( 𝑡 ) > 0 for all 𝑡 > 0 . From the first equation of (2.3),
̇ 𝑠 ( 𝑡 ) = 1 − 𝑠 ( 𝑡 ) − 𝛼 𝑥 ( 𝑡 ) 𝑠 ( 𝑡 ) . ( 2 . 5 ) Proceed using the method of contradiction. Suppose that there exists a first 𝑡 ⋆ such that 𝑠 ( 𝑡 ⋆ ) = 0 and 𝑠 ( 𝑡 ) > 0 for 𝑡 ∈ [ 0 , 𝑡 ∗ ) . Then ̇ 𝑠 ( 𝑡 ⋆ ) ⩽ 0 . But from the first equation of (2.3) 𝑡 ̇ 𝑠 ⋆ 𝑡 = 1 − 𝑠 ⋆ 𝑡 − 𝛼 𝑥 ⋆ 𝑠 𝑡 ⋆ = 1 > 0 , ( 2 . 6 ) a contradiction.
To prove 𝑥 ( 𝑡 ) > 0 for 𝑡 ∈ [ 0 , ∞ ) , assume there is a first 𝑡 > 0 such that 𝑥 ( 𝑡 ) = 0 , and 𝑥 ( 𝑡 ) > 0 for 𝑡 ∈ [ 0 , 𝑡 ) . Divide both sides of the second equation of (2.3) by 𝑥 ( 𝑡 ) and integrate from 0 to 𝑡 , to obtain
𝑥 𝑡 = 𝜙 ( 0 ) e x p 𝑡 0 ( − 𝐷 + 𝛼 𝑠 ( 𝑡 ) − 𝑘 𝑦 ( 𝑡 ) ) d 𝑡 > 0 , ( 2 . 7 ) contradicting 𝑥 ( 𝑡 ) = 0 .
To show that 𝑦 ( 𝑡 ) is positive on [ 0 , ∞ ) , suppose that there exists 𝑡 ⋆ > 0 such that 𝑦 ( 𝑡 ⋆ ) = 0 , and 𝑦 ( 𝑡 ) > 0 for 𝑡 ∈ [ 0 , 𝑡 ∗ ) . Then ̇ 𝑦 ( 𝑡 ⋆ ) ⩽ 0 . From the third equation of (2.3), we have
𝑡 ̇ 𝑦 ⋆ 𝑡 = − Δ 𝑦 ⋆ + 𝑘 𝑒 − Δ 𝜏 𝑦 𝑡 ⋆ 𝑥 𝑡 − 𝜏 ⋆ − 𝜏 = 𝑘 𝑒 − Δ 𝜏 𝑦 𝑡 ⋆ 𝑥 𝑡 − 𝜏 ⋆ − 𝜏 > 0 , ( 2 . 8 ) a contradiction.
To prove the boundedness of solutions, define
𝜔 ( 𝑡 ) = 𝑠 ( 𝑡 ) + 𝑥 ( 𝑡 ) + 𝑒 Δ 𝜏 𝑦 ( 𝑡 + 𝜏 ) − 1 , f o r 𝑡 ⩾ 0 . ( 2 . 9 ) It follows that ̇ 𝜔 ( 𝑡 ) = 1 − 𝑠 ( 𝑡 ) − 𝐷 𝑥 ( 𝑡 ) − Δ 𝑒 Δ 𝜏 𝑦 ( 𝑡 + 𝜏 ) ⩽ 1 − 𝑠 ( 𝑡 ) − 𝑥 ( 𝑡 ) − 𝑒 Δ 𝜏 𝑦 ( 𝑡 + 𝜏 ) ⩽ − 𝜔 ( 𝑡 ) , ( 2 . 1 0 ) where the first inequality holds since 𝐷 ⩾ 1 , Δ ⩾ 1 , 𝑥 ( 𝑡 ) > 0 and 𝑦 ( 𝑡 + 𝜏 ) > 0 . It follows that 𝑠 ( 𝑡 ) + 𝑥 ( 𝑡 ) + 𝑒 Δ 𝜏 𝑠 𝑦 ( 𝑡 + 𝜏 ) ⩽ 1 + 0 + 𝑥 ( 0 ) + 𝑒 Δ 𝜏 𝑒 𝑦 ( 𝜏 ) − 1 − 𝑡 ⟶ 1 a s 𝑡 ⟶ ∞ . ( 2 . 1 1 ) Therefore, the solution ( 𝑠 ( 𝑡 ) , 𝑥 ( 𝑡 ) , 𝑦 ( 𝑡 ) ) is bounded, and given any 𝜖 0 > 0 , 𝑥 ( 𝑡 ) < 1 + 𝜖 0 for all sufficiently large 𝑡 .

3. Equilibria and Stability

Model (2.3) has three equilibrium points: 𝐸 1 = ( 1 , 0 , 0 ) , 𝐸 2 = ( 𝐷 / 𝛼 , ( 𝛼 − 𝐷 ) / 𝛼 𝐷 , 0 ) , and

𝐸 + = 𝑠 + ( 𝜏 ) , 𝑥 + ( 𝜏 ) , 𝑦 + = 1 ( 𝜏 ) 1 + ( 𝛼 Δ / 𝑘 ) 𝑒 Δ 𝜏 , Δ 𝑘 𝑒 Δ 𝜏 , 𝛼 𝑘 + 𝛼 Δ 𝑒 Δ 𝜏 − 𝐷 𝑘 . ( 3 . 1 ) We call 𝐸 1 the washout equilibrium, 𝐸 2 the single species equilibrium, and 𝐸 + the coexistence equilibrium. For the sake of biological significance, 𝐸 + exists (distinct from 𝐸 2 ) if and only if its third coordinate 𝑦 + ( 𝜏 ) = ( 𝛼 𝑠 + ( 𝜏 ) − 𝐷 ) / 𝑘 > 0 , that is, 𝑠 + ( 𝜏 ) > 𝐷 / 𝛼 , or equivalently, 𝜏 lies between 0 and 𝜏 𝑐 , where

𝜏 𝑐 = 1 Δ 𝑘 l n Δ 1 𝐷 − 1 𝛼 . ( 3 . 2 ) Note that if ( 𝑘 / Δ ) ( 1 / 𝐷 − 1 / 𝛼 ) ⩽ 1 , the equilibrium 𝐸 + does not exist for any 𝜏 ( ⩾ 0), and if ( 𝑘 / Δ ) ( 1 / 𝐷 − 1 / 𝛼 ) = 1 , then 𝐸 + = 𝐸 2 .

The linearization of (2.3) about an equilibrium ( 𝑠 , 𝑥 , 𝑦 ) is given by

⎡ ⎢ ⎢ ⎢ ⎣ ̇ 𝑧 1 ( 𝑡 ) ̇ 𝑧 2 ( 𝑡 ) ̇ 𝑧 3 ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎣ 𝑧 ( 𝑡 ) − 1 − 𝛼 𝑥 − 𝛼 𝑠 0 𝛼 𝑥 − 𝐷 + 𝛼 𝑠 − 𝑘 𝑦 − 𝑘 𝑥 0 0 − Δ 1 𝑧 ( 𝑡 ) 2 𝑧 ( 𝑡 ) 3 ⎤ ⎥ ⎥ ⎥ ⎦ + ⎡ ⎢ ⎢ ⎢ ⎣ ( 𝑡 ) 0 0 0 0 0 0 0 𝑘 𝑒 Δ 𝜏 𝑦 𝑘 𝑒 Δ 𝜏 𝑥 ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎣ 𝑧 1 𝑧 ( 𝑡 − 𝜏 ) 2 𝑧 ( 𝑡 − 𝜏 ) 3 ⎤ ⎥ ⎥ ⎥ ⎦ . ( 𝑡 − 𝜏 ) ( 3 . 3 ) The associated characteristic equation is given by

⎡ ⎢ ⎢ ⎢ ⎣ d e t − 1 − 𝛼 𝑥 − 𝜆 − 𝛼 𝑠 0 𝛼 𝑥 − 𝐷 + 𝛼 𝑠 − 𝑘 𝑦 − 𝜆 − 𝑘 𝑥 0 𝑘 𝑒 − Δ 𝜏 − 𝜆 𝜏 𝑦 − Δ + 𝑘 𝑒 − Δ 𝜏 − 𝜆 𝜏 ⎤ ⎥ ⎥ ⎥ ⎦ 𝑥 − 𝜆 = 0 . ( 3 . 4 ) Direct calculation of the left-hand side of (3.4) gives

− Δ + 𝑘 𝑒 − ( Δ + 𝜆 ) 𝜏 𝑥 − 𝜆 ( − 1 − 𝛼 𝑥 − 𝜆 ) ( − 𝐷 + 𝛼 𝑠 − 𝑘 𝑦 − 𝜆 ) + 𝛼 2 𝑠 𝑥 + 𝑘 𝑥 𝑘 𝑒 − ( Δ + 𝜆 ) 𝜏 𝑦 ( ( − 1 − 𝛼 𝑥 − 𝜆 ) = ( − Δ − 𝜆 ) 1 + 𝛼 𝑥 + 𝜆 ) ( 𝐷 − 𝛼 𝑠 + 𝑘 𝑦 + 𝜆 ) + 𝛼 2 𝑠 𝑥 + 𝑒 − ( Δ + 𝜆 ) 𝜏 × 𝑘 𝑥 𝑘 𝑦 ( − 1 − 𝛼 𝑥 − 𝜆 ) + ( 1 + 𝛼 𝑥 + 𝜆 ) ( 𝐷 − 𝛼 𝑠 + 𝑘 𝑦 + 𝜆 ) + 𝛼 2 = 𝑠 𝑥 ( − Δ − 𝜆 ) ( 1 + 𝛼 𝑥 + 𝜆 ) ( 𝐷 − 𝛼 𝑠 + 𝑘 𝑦 + 𝜆 ) + 𝛼 2 𝑠 𝑥 + 𝑒 − ( Δ + 𝜆 ) 𝜏 𝑘 𝑥 ( 1 + 𝛼 𝑥 + 𝜆 ) ( 𝐷 − 𝛼 𝑠 + 𝜆 ) + 𝛼 2 = 𝑠 𝑥 ( − Δ − 𝜆 ) { ( 𝜆 + 1 ) ( 𝜆 + 𝐷 + 𝑘 𝑦 ) + 𝛼 𝑥 ( 𝜆 + 𝐷 + 𝑘 𝑦 ) − 𝛼 𝑠 ( 𝜆 + 1 ) } + 𝑒 − ( Δ + 𝜆 ) 𝜏 𝑘 𝑥 { ( 𝜆 + 1 ) ( 𝜆 + 𝐷 ) + 𝛼 𝑥 ( 𝜆 + 𝐷 ) − 𝛼 𝑠 ( 𝜆 + 1 ) } . ( 3 . 5 ) For convenience, define 𝑃 ( 𝜆 ) ∶ = ( − Δ − 𝜆 ) { ( 𝜆 + 1 ) ( 𝜆 + 𝐷 + 𝑘 𝑦 ) + 𝛼 𝑥 ( 𝜆 + 𝐷 + 𝑘 𝑦 ) − 𝛼 𝑠 ( 𝜆 + 1 ) } + 𝑒 − ( Δ + 𝜆 ) 𝜏 𝑘 𝑥 { ( 𝜆 + 1 ) ( 𝜆 + 𝐷 ) + 𝛼 𝑥 ( 𝜆 + 𝐷 ) − 𝛼 𝑠 ( 𝜆 + 1 ) } . ( 3 . 6 )

Theorem 3.1. Equilibrium 𝐸 1 is stable if 𝛼 < 𝐷 and unstable if 𝛼 > 𝐷 .

Proof. Evaluating the characteristic equation at 𝐸 1 gives | | 𝑃 ( 𝜆 ) 𝐸 1 = − ( Δ + 𝜆 ) ( 𝜆 + 1 ) ( 𝜆 + 𝐷 − 𝛼 ) = 0 . ( 3 . 7 ) The eigenvalues − 1 and − Δ are both negative. The third eigenvalue is − 𝐷 + 𝛼 . Therefore the equilibrium 𝐸 1 is stable if 𝛼 < 𝐷 and unstable if 𝛼 > 𝐷 .

Remark 3.2. If 𝛼 < 𝐷 , then there is only one equilibrium, 𝐸 1 . If 𝛼 > 𝐷 , equilibrium 𝐸 2 also exists.

Lemma 3.3. Assume 𝛼 > 𝐷 . The characteristic equation evaluated at 𝐸 2 has two negative eigenvalues, and the remaining eigenvalues are solutions of ( 𝜆 + Δ ) 𝑒 ( 𝜆 + Δ ) 𝜏 1 = 𝑘 𝐷 − 1 𝛼 . ( 3 . 8 ) In addition, the characteristic equation evaluated at 𝐸 2 has zero as an eigenvalue if and only if 𝜏 = 𝜏 𝑐 .

Proof. Assume 𝛼 > 𝐷 . Equilibrium 𝐸 2 exists. Consider the characteristic equation at 𝐸 2 . Since ( 𝛼 − 𝐷 ) / 𝛼 𝐷 = ( 1 − 𝑠 ) / 𝛼 𝑠 at 𝐸 2 , | | 𝑃 ( 𝜆 ) 𝐸 2 × = { ( 𝜆 + 1 ) ( 𝜆 + 𝐷 ) + 𝛼 𝑥 ( 𝜆 + 𝐷 ) − 𝛼 𝑠 ( 𝜆 + 1 ) } − 𝜆 − Δ + 𝑒 − ( Δ + 𝜆 ) 𝜏 = 𝑘 𝑥 ( 𝜆 + 1 ) ( 𝜆 + 𝐷 ) + 1 − 𝑠 𝑠 × ( 𝜆 + 𝐷 ) − 𝐷 ( 𝜆 + 1 ) − 𝜆 − Δ + 𝑒 − ( Δ + 𝜆 ) 𝜏 𝑘 𝛼 − 𝐷 = 𝛼 𝐷 𝜆 ( 𝜆 + 1 ) − ( 𝜆 + 𝐷 ) + 𝜆 + 𝐷 𝑠 − 𝜆 − Δ + 𝑘 𝛼 − 𝐷 𝑒 𝛼 𝐷 − ( Δ + 𝜆 ) 𝜏 𝜆 = − 2 + 𝛼 𝐷 𝜆 + 𝛼 − 𝐷 𝜆 + Δ − 𝑘 𝛼 − 𝐷 𝑒 𝛼 𝐷 − ( Δ + 𝜆 ) 𝜏 = − 𝑒 − ( Δ + 𝜆 ) 𝜏 𝜆 − 𝜆 1 𝜆 − 𝜆 2 ( 𝜆 + Δ ) 𝑒 ( Δ + 𝜆 ) 𝜏 1 − 𝑘 𝐷 − 1 𝛼 = 0 , ( 3 . 9 ) where 𝜆 1 + 𝜆 2 = − 𝛼 / 𝐷 and 𝜆 1 𝜆 2 = 𝛼 − 𝐷 > 0 . Therefore, 𝜆 1 and 𝜆 2 have negative real parts. The rest of the eigenvalues are roots of (3.8).
Assuming that 𝜆 = 0 is a root of (3.8), we have
Δ 𝑒 Δ 𝜏 1 = 𝑘 𝐷 − 1 𝛼 . ( 3 . 1 0 ) Solving for 𝜏 gives 1 𝜏 = Δ 𝑘 l n Δ 1 𝐷 − 1 𝛼 = 𝜏 𝑐 . ( 3 . 1 1 )

Theorem 3.4. Assume that 𝐷 ⩾ 1 , Δ ⩾ 1 , 𝑘 > 0 , 𝛼 > 0 , and ( 𝑘 / Δ ) ( 1 / 𝐷 − 1 / 𝛼 ) ⩾ 1 so that 𝜏 𝑐 ⩾ 0 . Equilibrium 𝐸 2 is locally asymptotically stable if 𝜏 > 𝜏 𝑐 and unstable if 𝜏 < 𝜏 𝑐 . If 𝐷 = 1 , then equilibrium 𝐸 2 is globally asymptotically stable for 𝜏 > ( 1 / Δ ) l n ( 𝑘 / Δ ) .

Proof. Assume that 𝜏 > 𝜏 𝑐 . Assumptions 𝑘 > 0 , Δ ⩾ 1 , and ( 𝑘 / Δ ) ( 1 / 𝐷 − 1 / 𝛼 ) ⩾ 1 imply 1 / 𝐷 > 1 / 𝛼 , or equivalently 𝛼 > 𝐷 . By Lemma 3.3, to prove that equilibrium 𝐸 2 is locally asymptotically stable, one only needs to show that (3.8) admits no root with nonnegative real part.
Consider the real roots of (3.8) first. Note that 1 / 𝐷 > 1 / 𝛼 . Equation (3.8) has no solution for 𝜆 ⩽ − Δ . Otherwise the left-hand side would be less than zero, but the right-hand side would be greater than zero. Assume 𝜆 > − Δ . The left-hand side of (3.8) is a monotone increasing function in both 𝜆 and 𝜏 , takes value 0 at 𝜆 = − Δ , and goes to positive infinity as 𝜆 → + ∞ or 𝜏 → + ∞ . By Lemma 3.3, when 𝜏 = 𝜏 𝑐 , then 𝜆 = 0 is a solution of (3.8). Thus for 𝜏 > 𝜏 𝑐 , any real root 𝜆 of (3.8) must satisfy − Δ < 𝜆 < 0 .
For any 𝜏 = ̃ 𝜏 < 𝜏 𝑐 , we have ( 𝜆 + Δ ) 𝑒 ( 𝜆 + Δ ) 𝜏 | 𝜏 = ̃ 𝜏 , 𝜆 = 0 < 𝑘 ( 1 / 𝐷 − 1 / 𝛼 ) and l i m 𝜆 → + ∞ ( 𝜆 + Δ ) 𝑒 ( 𝜆 + Δ ) ̃ 𝜏 = + ∞ . Therefore there exists at least one ̃ 𝜆 = 𝜆 > 0 such that ̃ ( ̃ 𝜏 , 𝜆 ) is a solution of (3.8). Equilibrium 𝐸 2 is unstable if 𝜏 < 𝜏 𝑐 .
In what follows, we prove that if 𝜏 > 𝜏 𝑐 all complex eigenvalues of (3.8) have negative real parts. Suppose that 𝜆 + Δ = 𝛾 + 𝑖 𝛽 ( 𝛽 > 0 ) is a solution of (3.8). Using the Euler formula, we have
1 𝛾 c o s ( 𝛽 𝜏 ) − 𝛽 s i n ( 𝛽 𝜏 ) + 𝑖 ( 𝛾 s i n ( 𝛽 𝜏 ) + 𝛽 c o s ( 𝛽 𝜏 ) ) = 𝑘 𝐷 − 1 𝛼 𝑒 − 𝛾 𝜏 . ( 3 . 1 2 ) Equating the real parts and imaginary parts of the equation, we have 1 𝛾 c o s ( 𝛽 𝜏 ) − 𝛽 s i n ( 𝛽 𝜏 ) = 𝑘 𝐷 − 1 𝛼 𝑒 − 𝛾 𝜏 𝛾 s i n ( 𝛽 𝜏 ) + 𝛽 c o s ( 𝛽 𝜏 ) = 0 . ( 3 . 1 3 ) Squaring both equations, adding, and taking the square root on both sides give 𝛾 2 + 𝛽 2 𝑒 𝛾 𝜏 1 = 𝑘 𝐷 − 1 𝛼 . ( 3 . 1 4 ) The left-hand side of (3.14) is monotonically increasing in 𝛾 , 𝛽 , and 𝜏 provided that 𝛾 > 0 . Since (3.14) has solution 𝛾 = Δ , 𝛽 = 0 at 𝜏 = 𝜏 𝑐 , any roots of (3.14) must satisfy 𝛾 < Δ since 𝜏 > 𝜏 𝑐 . Hence R e { 𝜆 } = 𝛾 − Δ < 0 . Therefore (3.8) has no complex eigenvalue with nonnegative real part and so 𝐸 2 is locally asymptotically stable for 𝜏 > 𝜏 𝑐 .
Assume that 𝐷 = 1 . Now we prove that 𝐸 2 is globally asymptotically stable when 𝜏 > ( 1 / Δ ) l n ( 𝑘 / Δ ) , or equivalently 𝑘 𝑒 − Δ 𝜏 < Δ . In this case, choose 𝜖 0 > 0 small enough such that 𝑘 𝑒 − Δ 𝜏 ( 1 + 𝜖 0 ) < Δ . By Theorem 2.1, for such 𝜖 0 , there exists a 𝑇 > 0 so that 0 < 𝑥 ( 𝑡 ) < 1 + 𝜖 0 for 𝑡 > 𝑇 . Hence, for 𝑡 > 𝑇 + 𝜏 , 𝑘 𝑒 − Δ 𝜏 𝑥 ( 𝑡 − 𝜏 ) < Δ . In Example 5 . 1 of Kuang ([10, page 32]), choose 𝜌 ( 𝑡 ) = 𝜏 , 𝑎 ( 𝑡 ) = Δ , 𝑏 ( 𝑡 ) = 𝑘 𝑒 − Δ 𝜏 𝑥 ( 𝑡 − 𝜏 ) , and 𝛼 = Δ / 2 . We obtain ( 𝑘 𝑒 − Δ 𝜏 ( 1 + 𝜖 0 ) ) 2 < Δ 2 = 4 ( Δ − 𝛼 ) 𝛼 . Therefore 𝑦 ( 𝑡 ) → 0 as 𝑡 → ∞ . Let 𝑧 ( 𝑡 ) = 𝑠 ( 𝑡 ) + 𝑥 ( 𝑡 ) . Noting 𝐷 = 1 , from (2.3), we have ̇ 𝑧 ( 𝑡 ) = 1 − 𝑧 ( 𝑡 ) − 𝑘 𝑥 ( 𝑡 ) 𝑦 ( 𝑡 ) . Multiply by the integrating factor 𝑒 𝑡 , ( 𝑧 ( 𝑡 ) 𝑒 𝑡 ) = 𝑒 𝑡 ( 1 − 𝑘 𝑥 ( 𝑡 ) 𝑦 ( 𝑡 ) ) . Integrating both sides from 0 to 𝑡 gives
𝑧 ( 𝑡 ) = 𝑒 − 𝑡 𝑧 ( 0 ) + 𝑒 − 𝑡 𝑒 𝑡 − 1 − 𝑒 − 𝑡 𝑡 0 𝑒 𝑠 𝑘 𝑥 ( 𝑠 ) 𝑦 ( 𝑠 ) d 𝑠 = 1 + 𝑒 − 𝑡 ( 𝑧 ( 0 ) − 1 ) − 𝑒 − 𝑡 𝑡 0 𝑒 𝑠 𝑘 𝑥 ( 𝑠 ) 𝑦 ( 𝑠 ) d 𝑠 . ( 3 . 1 5 ) If l i m 𝑡 → ∞ ∫ 𝑡 0 𝑒 𝑠 𝑘 𝑥 ( 𝑠 ) 𝑦 ( 𝑠 ) d 𝑠 < ∞ , then l i m 𝑡 → ∞ 𝑒 − 𝑡 ∫ 𝑡 0 𝑒 𝑠 𝑘 𝑥 ( 𝑠 ) 𝑦 ( 𝑠 ) d 𝑠 = 0 . Therefore l i m 𝑡 → ∞ 𝑧 ( 𝑡 ) = 1 . If l i m 𝑡 → ∞ ∫ 𝑡 0 𝑒 𝑠 𝑘 𝑥 ( 𝑠 ) 𝑦 ( 𝑠 ) d 𝑠 = ∞ , by L'Hôspital's rule, l i m 𝑡 → + ∞ 𝑒 − 𝑡 𝑡 0 𝑒 𝑠 𝑘 𝑥 ( 𝑠 ) 𝑦 ( 𝑠 ) d 𝑠 = l i m 𝑡 → + ∞ ∫ 𝑡 0 𝑒 𝑠 𝑘 𝑥 ( 𝑠 ) 𝑦 ( 𝑠 ) d 𝑠 𝑒 𝑡 = l i m 𝑡 → + ∞ 𝑒 𝑡 𝑘 𝑥 ( 𝑡 ) 𝑦 ( 𝑡 ) 𝑒 𝑡 = l i m 𝑡 → + ∞ 𝑘 𝑥 ( 𝑡 ) 𝑦 ( 𝑡 ) = 0 , ( 3 . 1 6 ) since 𝑥 ( 𝑡 ) is bounded and l i m 𝑡 → ∞ 𝑦 ( 𝑡 ) = 0 . It again follows that l i m t → ∞ 𝑧 ( 𝑡 ) = 1 . Hence l i m 𝑡 → ∞ 𝑠 ( 𝑡 ) + 𝑥 ( 𝑡 ) = 1 . ( 3 . 1 7 )
We show that l i m 𝑡 → ∞ 𝑠 ( 𝑡 ) = 1 / 𝛼 and l i m 𝑡 → ∞ 𝑥 ( 𝑡 ) = ( 𝛼 − 1 ) / 𝛼 . First assume that the limits exist, that is, l i m 𝑡 → ∞ 𝑠 ( 𝑡 ) = 𝑠 and l i m 𝑡 → ∞ 𝑥 ( 𝑡 ) = 𝑥 . From (2.3), we know that ̇ 𝑠 ( 𝑡 ) and ̇ 𝑥 ( 𝑡 ) are uniformly continuous since 𝑠 ( 𝑡 ) , 𝑥 ( 𝑡 ) , and 𝑦 ( 𝑡 ) are bounded. By Theorem A.3, it follows that l i m 𝑡 → ∞ ̇ 𝑠 ( 𝑡 ) = 0 and l i m 𝑡 → ∞ ̇ 𝑥 ( 𝑡 ) = 0 . Note that l i m 𝑡 → ∞ 𝑦 ( 𝑡 ) = 0 . Letting 𝑡 → ∞ in (2.3) gives
1 − 𝑠 − 𝛼 𝑥 𝑠 = 0 , 𝑥 − 1 + 𝛼 𝑠 = 0 . ( 3 . 1 8 ) Either ( 𝑠 , 𝑥 ) = ( 1 , 0 ) or ( 𝑠 , 𝑥 ) = ( 1 / 𝛼 , ( 𝛼 − 1 ) / 𝛼 ) . Assume that ( 𝑠 , 𝑥 ) = ( 1 , 0 ) , that is, l i m 𝑡 → ∞ 𝑠 ( 𝑡 ) = 1 and l i m 𝑡 → ∞ 𝑥 ( 𝑡 ) = 0 . Note that 𝛼 > 𝐷 . There exists 𝜖 > 0 such that 𝛼 − 𝐷 − ( 𝛼 + 𝑘 ) 𝜖 > 0 . For such 𝜖 , there exists a sufficiently large 𝑡 so that 𝑠 ( 𝑡 ) > 1 − 𝜖 and 0 < 𝑦 ( 𝑡 ) < 𝜖 . Recalling that 𝑥 ( 𝑡 ) > 0 , by (2.3) ̇ 𝑥 ( 𝑡 ) > 𝑥 ( 𝑡 ) ( − 𝐷 + 𝛼 ( 1 − 𝜖 ) − 𝑘 𝜖 ) = 𝑥 ( 𝑡 ) ( 𝛼 − 𝐷 − 𝛼 𝜖 − 𝑘 𝜖 ) > 0 , ( 3 . 1 9 ) for all sufficiently large 𝑡 . Therefore it is impossible for 𝑥 ( 𝑡 ) to approach 0 from above giving a contradiction. Therefore, we must have ( 𝑠 , 𝑥 ) = ( 1 / 𝛼 , ( 𝛼 − 1 ) / 𝛼 ) .
Now suppose that the limits do not exist. In particular if 𝑥 ( 𝑡 ) does not converge, then let 𝑥 = l i m s u p 𝑡 → ∞ 𝑥 ( 𝑡 ) and 𝑥 = l i m i n f 𝑡 → ∞ 𝑥 ( 𝑡 ) . By Lemma A.2 in the appendix, there exists { 𝑡 𝑚 } ↑ ∞ and { 𝑠 𝑚 } ↑ ∞ such that
l i m 𝑚 → ∞ 𝑥 𝑡 𝑚 = 𝑥 , l i m 𝑚 → ∞ 𝑡 ̇ 𝑥 𝑚 = 0 , l i m 𝑚 → ∞ 𝑥 𝑠 𝑚 = 𝑥 l i m 𝑚 → ∞ 𝑠 ̇ 𝑥 𝑚 = 0 . ( 3 . 2 0 ) From (2.3), 𝑥 𝑡 𝑚 𝑡 − 𝐷 + 𝛼 𝑠 𝑚 𝑡 + 𝑘 𝑦 𝑚 = 0 . ( 3 . 2 1 ) Noting that 𝑥 ( 𝑡 𝑚 ) > 0 , we have 𝑠 ( 𝑡 𝑚 ) = ( 1 − 𝑘 𝑦 ( 𝑡 𝑚 ) ) / 𝛼 . Since l i m 𝑡 → ∞ 𝑦 ( 𝑡 ) = 0 , l i m 𝑡 → ∞ 𝑠 ( 𝑡 𝑚 ) = 1 / 𝛼 . By (3.17), l i m 𝑡 → ∞ 𝑥 ( 𝑡 𝑚 ) = l i m 𝑡 → ∞ ( 𝑥 ( 𝑡 𝑚 ) + 𝑠 ( 𝑡 𝑚 ) ) − 𝑠 ( 𝑡 𝑚 ) = 1 − 1 / 𝛼 = ( 𝛼 − 1 ) / 𝛼 . Therefore 𝑥 = ( 𝛼 − 1 ) / 𝛼 . Similarly we can show that 𝑥 = ( 𝛼 − 1 ) / 𝛼 . This implies that l i m 𝑡 → ∞ 𝑥 ( 𝑡 ) = ( 𝛼 − 1 ) / 𝛼 , a contradiction.
Since 𝑠 ( 𝑡 ) + 𝑥 ( 𝑡 ) converges and 𝑥 ( 𝑡 ) converges, then 𝑠 ( 𝑡 ) must also converge. Hence l i m 𝑡 → ∞ 𝑠 ( 𝑡 ) = 1 / 𝛼 and l i m 𝑡 → ∞ 𝑥 ( 𝑡 ) = ( 𝛼 − 1 ) / 𝛼 . It follows that 𝐸 2 is globally asymptotically stable.

4. Hopf Bifurcations at 𝐸 + Assuming 𝐷 = Δ = 1

Now consider the stability of 𝐸 + . The characteristic equation at 𝐸 + is

| | 𝑃 ( 𝜆 ) 𝐸 + = ( − Δ − 𝜆 ) 1 + 𝛼 𝑥 + ( 𝜏 ) + 𝜆 𝐷 − 𝛼 𝑠 + ( 𝜏 ) + 𝑘 𝑦 + ( 𝜏 ) + 𝜆 + 𝛼 2 𝑠 + ( 𝜏 ) 𝑥 + ( 𝜏 ) + 𝑒 − ( Δ + 𝜆 ) 𝜏 𝑘 𝑥 + ( 𝜏 ) ( 𝜆 + 1 ) ( 𝜆 + 𝐷 ) + 𝛼 𝑥 + ( 𝜏 ) ( 𝜆 + 𝐷 ) − 𝛼 𝑠 + ( 𝜏 ) ( 𝜆 + 1 ) = ( − Δ − 𝜆 ) 1 + 𝛼 𝑥 + ( 𝜏 ) + 𝜆 𝜆 + 𝛼 2 𝑠 + ( 𝜏 ) 𝑥 + ( 𝜏 ) + 𝑒 − ( Δ + 𝜆 ) 𝜏 𝑘 𝑥 + ( 𝜏 ) ( 𝜆 + 1 ) ( 𝜆 + 𝐷 ) + 𝛼 𝑥 + ( 𝜏 ) ( 𝜆 + 𝐷 ) − 𝛼 𝑠 + 1 ( 𝜏 ) ( 𝜆 + 1 ) = ( − Δ − 𝜆 ) 𝑠 + ( 𝜏 ) + 𝜆 𝜆 + 𝛼 1 − 𝑠 + ( 𝜏 ) + 𝑒 − 𝜆 𝜏 Δ ( 𝜆 + 1 ) ( 𝜆 + 𝐷 ) + 1 − 𝑠 + ( 𝜏 ) 𝑠 + ( 𝜏 ) ( 𝜆 + 𝐷 ) − 𝛼 𝑠 + 𝜆 ( 𝜏 ) ( 𝜆 + 1 ) = ( − Δ − 𝜆 ) 2 + 𝜆 𝑠 + ( 𝜏 ) + 𝛼 1 − 𝑠 + ( 𝜏 ) + Δ 𝑒 − 𝜆 𝜏 1 𝜆 + 𝑠 + ( 𝜏 ) ( 𝜆 + 𝐷 ) − 𝛼 𝑠 + ( 𝜏 ) ( 𝜆 + 1 ) = 0 . ( 4 . 1 ) By assumption Δ = 𝐷 = 1 , and so

| | 𝑃 ( 𝜆 ) 𝐸 + 𝜆 = − ( 𝜆 + 1 ) 2 + 𝜆 𝑠 + ( 𝜏 ) + 𝛼 1 − 𝑠 + ( 𝜏 ) + 𝑒 − 𝜆 𝜏 − 𝜆 + 𝛼 𝑠 + 1 ( 𝜏 ) − 𝑠 + 𝜆 ( 𝜏 ) = − ( 𝜆 + 1 ) 2 + 𝑝 ( 𝜏 ) 𝜆 + 𝛽 ( 𝜏 ) + 𝑒 − 𝜆 𝜏 ( 𝑞 𝜆 + 𝑐 ( 𝜏 ) ) = 0 , ( 4 . 2 ) where

1 𝑝 ( 𝜏 ) = 𝑠 + ( 𝜏 ) , 𝛽 ( 𝜏 ) = 𝛼 1 − 𝑠 + ( 𝜏 ) , 𝑞 = − 1 , 𝑐 ( 𝜏 ) = 𝛼 𝑠 + 1 ( 𝜏 ) − 𝑠 + . ( 𝜏 ) ( 4 . 3 ) The characteristic equation at 𝐸 + has one eigenvalue equal to − 1 and the others are given by solutions of the equation

𝜆 2 + 𝑝 ( 𝜏 ) 𝜆 + 𝛽 ( 𝜏 ) + 𝑒 − 𝜆 𝜏 ( 𝑞 𝜆 + 𝑐 ( 𝜏 ) ) = 0 . ( 4 . 4 )

Lemma 4.1. Assuming 𝑘 > 0 , 𝛼 > 0 , and 𝑘 ( 1 − 1 / 𝛼 ) ⩾ 1 so that 𝜏 𝑐 = l n ( 𝑘 ( 1 − 1 / 𝛼 ) ) ⩾ 0 , then 𝐸 + has no zero eigenvalue for 𝜏 ∈ ( 0 , 𝜏 𝑐 ) .

Proof. Assume that 𝜏 ∈ ( 0 , 𝜏 𝑐 ) . By the method of contradiction, suppose that there exists a zero root of (4.4). Therefore 1 𝛽 ( 𝜏 ) + 𝑐 ( 𝜏 ) = 𝛼 − 𝑠 + ( 𝜏 ) = 0 . ( 4 . 5 ) Noting that 𝜏 𝑐 > 0 if and only if 𝑘 ( 1 − 1 / 𝛼 ) > 1 , for any 0 < 𝜏 < 𝜏 𝑐 , 1 𝛼 − 𝑠 + 𝛼 ( 𝜏 ) = 𝛼 − 1 − 𝑘 𝑒 𝜏 1 > 𝛼 − 1 − 𝛼 1 − 𝛼 = 0 , ( 4 . 6 ) a contradiction.

Lemma 4.2. Assume 𝑘 > 0 , 𝛼 > 0 , 𝑘 ( 1 − 1 / 𝛼 ) > 1 . Equilibrium 𝐸 + is asymptotically stable when 𝜏 = 0 .

Proof. For 𝜏 = 0 , (4.4) reduces to 𝜆 2 + 𝑝 ( 0 ) 𝜆 + 𝛽 ( 0 ) + ( 𝑞 𝜆 + 𝑐 ( 0 ) ) = 𝜆 2 + 1 𝑠 + 1 ( 0 ) − 1 𝜆 + 𝛼 − 𝑠 + ( 0 ) . ( 4 . 7 ) Both coefficients are positive, since 1 𝑠 + 𝛼 ( 0 ) − 1 = 𝑘 1 > 0 , 𝛼 − 𝑠 + 𝛼 ( 0 ) = 𝛼 − 1 − 𝑘 1 = 𝛼 1 − 𝛼 − 1 𝑘 > 0 , ( 4 . 8 ) and 𝑘 ( 1 − 1 / 𝛼 ) > 1 implies 1 − 1 / 𝛼 > 1 / 𝑘 . Therefore, all the roots of the characteristic equation have negative real parts.

Lemma 4.3. As 𝜏 is increased from 0 , a root of (4.4) with positive real part can only appear if a root with negative real part crosses the imaginary axis.

Proof. Taking 𝑛 = 2 and 𝑔 ( 𝜆 , 𝜏 ) = 𝑝 ( 𝜏 ) 𝜆 + ( 𝑞 𝜆 + 𝑐 ( 𝜏 ) ) 𝑒 − 𝜆 𝜏 + 𝛽 ( 𝜏 ) in Kuang [10, Theorem 1 . 4 , page 66] gives l i m s u p R e 𝜆 > 0 , | 𝜆 | → ∞ | | 𝜆 − 2 𝑔 | | ( 𝜆 , 𝜏 ) = 0 < 1 . ( 4 . 9 ) Therefore, no root of (4.4) with positive real part can enter from infinity as 𝜏 increases from 0 . Hence roots with positive real part can only appear by crossing the imaginary axis.
For 𝜏 ≠ 0 , assuming 𝜆 = 𝑖 𝜔 ( 𝜔 > 0 ) is a root of 𝑃 ( 𝜆 ) | 𝐸 + = 0 ,
− 𝜔 2 + 𝑖 𝑝 ( 𝜏 ) 𝜔 + 𝛽 ( 𝜏 ) + 𝑒 − 𝑖 𝜔 𝜏 ( 𝑖 𝑞 𝜔 + 𝑐 ( 𝜏 ) ) = 0 . ( 4 . 1 0 ) Substituting 𝑒 𝑖 𝜃 = c o s 𝜃 + 𝑖 s i n 𝜃 into (4.10) gives − 𝜔 2 + 𝛽 ( 𝜏 ) + 𝑞 𝜔 s i n ( 𝜔 𝜏 ) + 𝑐 ( 𝜏 ) c o s ( 𝜔 𝜏 ) + 𝑖 ( 𝑝 ( 𝜏 ) 𝜔 + 𝑞 𝜔 c o s ( 𝜔 𝜏 ) − 𝑐 ( 𝜏 ) s i n ( 𝜔 𝜏 ) ) = 0 . ( 4 . 1 1 ) Separating the real and imaginary parts, we obtain 𝑐 ( 𝜏 ) c o s ( 𝜔 𝜏 ) + 𝑞 𝜔 s i n ( 𝜔 𝜏 ) = 𝜔 2 − 𝛽 ( 𝜏 ) , 𝑐 ( 𝜏 ) s i n ( 𝜔 𝜏 ) − 𝑞 𝜔 c o s ( 𝜔 𝜏 ) = 𝑝 ( 𝜏 ) 𝜔 . ( 4 . 1 2 ) Solving for c o s ( 𝜔 𝜏 ) and s i n ( 𝜔 𝜏 ) gives 𝜔 s i n ( 𝜔 𝜏 ) = 𝑐 ( 𝜏 ) 𝑝 ( 𝜏 ) 𝜔 + 𝑞 𝜔 2 − 𝛽 ( 𝜏 ) 𝑐 ( 𝜏 ) 2 + 𝑞 2 𝜔 2 , 𝜔 c o s ( 𝜔 𝜏 ) = 𝑐 ( 𝜏 ) 2 − 𝛽 ( 𝜏 ) − 𝑞 𝑝 ( 𝜏 ) 𝜔 2 𝑐 ( 𝜏 ) 2 + 𝑞 2 𝜔 2 . ( 4 . 1 3 ) Noting s i n 2 ( 𝜔 𝜏 ) + c o s 2 ( 𝜔 𝜏 ) = 1 , squaring both sides of equations (4.13), adding, and rearranging gives 𝜔 4 + 𝑝 2 ( 𝜏 ) − 𝑞 2 𝜔 − 2 𝛽 ( 𝜏 ) 2 + 𝛽 2 ( 𝜏 ) − 𝑐 2 ( 𝜏 ) = 0 . ( 4 . 1 4 ) Solving for 𝜔 , we obtain two roots 𝜔 1 ( 𝜏 ) and 𝜔 2 ( 𝜏 ) : 𝜔 1 1 ( 𝜏 ) = √ 2 𝑞 2 − 𝑝 2 ( 𝜏 ) + 2 𝛽 ( 𝜏 ) + 𝑞 2 − 𝑝 2 ( 𝜏 ) + 2 𝛽 ( 𝜏 ) 2 − 4 ( 𝛽 2 ( 𝜏 ) − 𝑐 2 ( 𝜏 ) ) 1 / 2 = 1 𝑠 + √ ( 𝜏 ) 2 1 − 𝑠 + ( 𝜏 ) 2 𝛼 𝑠 2 + ( 𝜏 ) − 𝑠 + ( + 𝜏 ) − 1 𝑠 2 + ( 𝜏 ) − 1 2 + 4 𝛼 𝑠 2 + 𝑠 ( 𝜏 ) 2 + ( 𝜏 ) − 1 1 − 𝑠 + ( 𝜏 ) + 4 𝑠 2 + ( 𝜏 ) 𝛼 𝑠 2 + ( 𝜏 ) − 1 2 1 / 2 𝜔 2 1 ( 𝜏 ) = √ 2 𝑞 2 − 𝑝 2 ( 𝜏 ) + 2 𝛽 ( 𝜏 ) − 𝑞 2 − 𝑝 2 ( 𝜏 ) + 2 𝛽 ( 𝜏 ) 2 𝛽 − 4 2 ( 𝜏 ) − 𝑐 2 ( 𝜏 ) 1 / 2 = 1 𝑠 + √ ( 𝜏 ) 2 1 − 𝑠 + ( 𝜏 ) 2 𝛼 𝑠 2 + ( 𝜏 ) − 𝑠 + − ( 𝜏 ) − 1 𝑠 2 + ( 𝜏 ) − 1 2 + 4 𝛼 𝑠 2 + 𝑠 ( 𝜏 ) 2 + ( 𝜏 ) − 1 1 − 𝑠 + ( 𝜏 ) + 4 𝑠 2 + ( 𝜏 ) 𝛼 𝑠 2 + ( 𝜏 ) − 1 2 1 / 2 . ( 4 . 1 5 )
Define conditions ( 𝐻 1 ) and ( 𝐻 2 ) as follows:
𝐻 1 𝑞 2 − 𝑝 2 ( 𝜏 ) + 2 𝛽 ( 𝜏 ) > 0 , 𝛽 2 ( 𝜏 ) − 𝑐 2 𝑞 ( 𝜏 ) > 0 , 2 − 𝑝 2 ( 𝜏 ) + 2 𝛽 ( 𝜏 ) 2 𝛽 − 4 2 ( 𝜏 ) − 𝑐 2 ( 𝜏 ) ⩾ 0 , ( 4 . 1 6 ) 𝐻 2 𝛽 2 ( 𝜏 ) − 𝑐 2 ( 𝜏 ) < 0 , o r 𝛽 2 ( 𝜏 ) − 𝑐 2 ( 𝜏 ) = 0 a n d 𝑞 2 − 𝑝 2 ( 𝜏 ) + 2 𝛽 ( 𝜏 ) > 0 . ( 4 . 1 7 )

Lemma 4.4. If ( 𝐻 1 ) holds for all 𝜏 in some interval 𝐼 , then (4.14) has two positive roots 𝜔 1 ( 𝜏 ) ⩾ 𝜔 2 ( 𝜏 ) for all 𝜏 ∈ 𝐼 with 𝜔 1 ( 𝜏 ) > 𝜔 2 ( 𝜏 ) when all the inequalities in ( 𝐻 1 ) are strict. If ( 𝐻 2 ) holds for all 𝜏 in some interval 𝐼 , then (4.14) has only one positive root, 𝜔 1 ( 𝜏 ) for all 𝜏 ∈ 𝐼 . If no interval exists where either ( 𝐻 1 ) or ( 𝐻 2 ) holds, then there are no positive real roots of (4.14).

Define the interval

⎡ ⎢ ⎢ ⎣ 𝑘 𝐽 = l n 𝛼 1 √ 1 / 4 + ⎛ ⎜ ⎜ ⎝ 𝑘 1 / 1 6 + 1 / ( 2 𝛼 ) − 1 , l n 4 √ 𝛼 − 1 𝛼 ⎞ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎦ . ( 4 . 1 8 ) When the end points of 𝐽 are real and 𝐽 ≠ ∅ , define

𝐼 1 = 0 , 𝜏 𝑐 ∩ 𝐽 . ( 4 . 1 9 ) We prove that ( 𝐻 1 ) holds for any 𝜏 ∈ 𝐼 1 .

From 𝐷 = Δ = 1 ,

𝜏 𝑐 = 1 Δ 𝑘 l n Δ 1 𝐷 − 1 𝛼 = l n 𝑘 ( 𝛼 − 1 ) 𝛼 . ( 4 . 2 0 ) If 𝛼 > 1 , then 𝛼 > 4 √ 𝛼 . It follows that

𝜏 𝑐 ⎛ ⎜ ⎜ ⎝ 𝑘 > l n 4 √ 𝛼 − 1 𝛼 ⎞ ⎟ ⎟ ⎠ . ( 4 . 2 1 ) Therefore,

𝐼 1 = ⎡ ⎢ ⎢ ⎣ 𝑘 m a x 0 , l n 𝛼 1 √ 1 / 4 + ⎛ ⎜ ⎜ ⎝ 𝑘 1 / 1 6 + 1 / ( 2 𝛼 ) − 1 , l n 4 √ 𝛼 − 1 𝛼 ⎞ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎦ . ( 4 . 2 2 )

Theorem 4.5. Assume √ 𝛼 > ( 7 + 3 5 ) / 2 and 𝑘 > 𝛼 / ( 4 √ 𝛼 − 1 ) , then 𝐼 1 is not empty, and for any 𝜏 ∈ 𝐼 1 , but √ 𝜏 ≠ l n ( ( 𝑘 / 𝛼 ) ( 1 / ( 1 / 4 + 1 / 1 6 + 1 / ( 2 𝛼 ) ) − 1 ) ) , condition ( 𝐻 1 ) holds and 𝜔 1 ( 𝜏 ) > 𝜔 2 ( 𝜏 ) > 0 . If √ 𝜏 = l n ( ( 𝑘 / 𝛼 ) ( 1 / ( 1 / 4 + 1 / 1 6 + 1 / ( 2 𝛼 ) ) − 1 ) ) ∈ 𝐼 1 , then 𝜔 1 ( 𝜏 ) > 𝜔 2 ( 𝜏 ) = 0 .

Proof. For any √ 𝛼 > ( 7 + 3 5 ) / 2 , we have 1 − 1 / 4 √ 𝛼 > 0 , and therefore 1 4 √ 𝛼 + 1 2 − √ 5 2 < 1 4 √ 7 + 3 5 + 1 / 2 2 − √ 5 2 = 0 . ( 4 . 2 3 ) Hence, 1 4 √ 𝛼 − 1 4 2 − 1 + 1 1 6 2 𝛼 2 = 1 √ 𝛼 − 1 2 4 √ 𝛼 − 1 = 2 𝛼 − 1 2 4 √ 𝛼 ⎛ ⎜ ⎜ ⎝ 1 4 √ 𝛼 3 2 + 1 − 4 √ 𝛼 ⎞ ⎟ ⎟ ⎠ = 1 2 4 √ 𝛼 1 1 − 4 √ 𝛼 ⎛ ⎜ ⎜ ⎝ 1 4 √ 𝛼 2 + 1 4 √ 𝛼 ⎞ ⎟ ⎟ ⎠ = 1 − 1 2 4 √ 𝛼 1 1 − 4 √ 𝛼 ⎛ ⎜ ⎜ ⎝ 1 4 √ 𝛼 + 1 2 2 − 5 4 ⎞ ⎟ ⎟ ⎠ = 1 2 4 √ 𝛼 1 1 − 4 √ 𝛼 1 4 √ 𝛼 + 1 2 − √ 5 2 1 4 √ 𝛼 + 1 2 + √ 5 2 < 0 . ( 4 . 2 4 ) Therefore, 1 / 4 √ √ 𝛼 − 1 / 4 < 1 / 1 6 + 1 / ( 2 𝛼 ) . Since √ 1 / 4 + 1 / 1 6 + 1 / ( 2 𝛼 ) < 1 / 4 + √ 1 / 1 6 + 1 / ( 7 + 3 5 ) < 1 , it follows that 1 4 √ 𝛼 < 1 4 + 1 + 1 1 6 2 𝛼 < 1 . ( 4 . 2 5 ) Hence, 𝑘 l n 𝛼 1 √ 1 / 4 + ⎛ ⎜ ⎜ ⎝ 𝑘 1 / 1 6 + 1 / ( 2 𝛼 ) − 1 < l n 4 √ 𝛼 − 1 𝛼 ⎞ ⎟ ⎟ ⎠ . ( 4 . 2 6 ) From 𝑘 > 𝛼 / ( 4 √ 𝛼 − 1 ) , we have l n ( 𝑘 ( 4 √ 𝛼 − 1 ) / 𝛼 ) > 0 . Therefore, 𝑘 m a x 0 , l n 𝛼 1 √ 1 / 4 + ⎛ ⎜ ⎜ ⎝ 𝑘 1 / 1 6 + 1 / ( 2 𝛼 ) − 1 < l n 4 √ 𝛼 − 1 𝛼 ⎞ ⎟ ⎟ ⎠ , ( 4 . 2 7 ) and so 𝐼 1 is not empty. Noting 𝑠 + ( 𝜏 ) = 1 / ( 1 + ( 𝛼 Δ / 𝑘 ) 𝑒 Δ 𝜏 ) and Δ = 1 , for any 𝜏 ∈ 𝐼 1 , but √ 𝜏 ≠ l n ( ( 𝑘 / 𝛼 ) ( 1 / ( 1 / 4 + 1 / 1 6 + 1 / ( 2 𝛼 ) ) − 1 ) ) , we have 𝑠 + ( 𝜏 ) ∈ [ 1 / 4 √ √ 𝛼 , 1 / 4 + 1 / 1 6 + 1 / ( 2 𝛼 ) ) .
In what follows, we intend to show that for any such 𝜏 , condition ( 𝐻 1 ) holds. From (4.3),
𝑞 2 − 𝑝 2 ( 𝜏 ) + 2 𝛽 ( 𝜏 ) = ( − 1 ) 2 − 1 𝑠 2 + ( 𝜏 ) + 2 𝛼 1 − 𝑠 + = ( 𝜏 ) 1 − 𝑠 + ( 𝜏 ) 2 𝛼 𝑠 2 + 𝑠 ( 𝜏 ) + ( 𝜏 ) 2 − 𝑠