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International Journal of Differential Equations
Volume 2010, Article ID 287969, 41 pages
http://dx.doi.org/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, Canada L8S 4K1
2Department of Mathematics and Statistics, York University, 4700 Keele Street, Toronto, ON, Canada M3J 1P3

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

𝑠̇𝑠(𝑡)=0𝐷−𝑠(𝑡)0−𝑥(𝑡)𝑓(𝑠(𝑡))𝜂,𝑦̇𝑥(𝑡)=𝑥(𝑡)(−𝐷+𝑓(𝑠(𝑡)))−(𝑡)𝑔(𝑥(𝑡))𝜉,̇𝑦(𝑡)=−Δ𝑦(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 𝑠0 denotes the concentration of the growth limiting nutrient in the feed vessel, 𝐷0 the dilution rate, 𝜂(𝜉) the growth yield constant, 𝐷(Δ) the sum of the dilution rate 𝐷0 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 𝐷0=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 𝑠0. 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

𝑠̇𝑠(𝑡)=0𝐷−𝑠(𝑡)0−𝑥(𝑡)𝑓(𝑠(𝑡))𝜂,𝑦̇𝑥(𝑡)=𝑥(𝑡)(−𝐷+𝑓(𝑠(𝑡)))−(𝑡)𝑔(𝑥(𝑡))𝜉,̇𝑦(𝑡)=−Δ𝑦(𝑡)+𝑒−Δ𝜏𝑦(𝑡−𝜏)𝑔(𝑥(𝑡−𝜏)).(1.2) For 𝑡∈[−𝜏,0],

𝑠(0)=𝑠0∈intℝ+[],(𝑥(𝑡),𝑦(𝑡))=(𝜙,𝜓)∈ℂ−𝜏,0,intℝ2+.(1.3) Here variables 𝑠(𝑡), 𝑥(𝑡), 𝑦(𝑡), and parameters 𝑠0, 𝐷0, 𝜂, 𝜉, 𝐷0, 𝐷, and Δ have the same interpretation as for model (1.1). Note therefore that 𝐷⩾𝐷0, and Δ⩾𝐷0. The additional parameter 𝜏 is a nonnegative constant modelling the time required for the conversion process. Hence, 𝑒−Δ𝜏𝑦(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)=𝛼𝑠 (𝛼>0) and 𝑔(𝑥)=𝑘𝑥 (𝑘>0). System (1.2) reduces to

𝑠̇𝑠(𝑡)=0𝐷−𝑠(𝑡)0−𝛼𝑥(𝑡)𝑠(𝑡)𝜂,̇𝑥(𝑡)=𝑥(𝑡)(−𝐷+𝛼𝑠(𝑡))−𝑘𝑥(𝑡)𝑦(𝑡)𝜉,̇𝑦(𝑡)=−Δ𝑦(𝑡)+𝑘𝑒−Δ𝜏𝑦(𝑡−𝜏)𝑥(𝑡−𝜏).𝑡>0,(2.1) Introducing the following change of variables gives:

̆𝑡=𝐷0̆𝑡=𝑡,̆𝑠𝑠(𝑡)𝑠0̆𝑡=,̆𝑥𝑥(𝑡)𝑠0𝜂̆𝑡=,̆𝑦𝑦(𝑡)𝜉𝑠0𝜂,̆𝜏=𝐷0̆𝐷𝜏,𝐷=𝐷0,̆ΔΔ=𝐷0,̆𝑘=𝑘𝑠0𝜂𝐷0,̆𝛼=𝛼𝑠0𝐷0,̆𝑡d̆𝑠d̆𝑡=1𝑠0d𝑠(𝑡)d𝑡d𝑡d̆𝑡=1𝑠0𝐷0d𝑠(𝑡)=1d𝑡𝑠0𝐷0𝑠0𝐷−𝑠(𝑡)0−𝛼𝑥(𝑡)𝑠(𝑡)𝜂=1−𝑠(𝑡)𝑠0−𝛼𝑠0𝐷0𝑥(𝑡)𝑠0𝜂𝑠(𝑡)𝑠0̆𝑡̆𝑡̆𝑡,̆𝑡=1−̆𝑠−̆𝛼̆𝑥̆𝑠d̆𝑥d̆𝑡=1𝑠0𝜂d𝑥(𝑡)d𝑡d𝑡d̆𝑡=1𝑠0𝜂𝐷0d𝑥(𝑡)=1d𝑡𝑠0𝜂𝐷0𝑥(𝑡)(−𝐷+𝛼𝑠(𝑡))−𝑘𝑥(𝑡)𝑦(𝑡)𝜉=𝑥(𝑡)𝑠0𝜂−𝐷𝐷0+𝛼𝑠0𝐷0𝑠(𝑡)𝑠0−𝑘𝑠0𝜂𝐷0𝑥(𝑡)𝑠0𝜂𝑦(𝑡)𝑠0̆𝑡−̆̆𝑡−̆̆𝑡̆𝑡,̆𝑡𝜂𝜉=̆𝑥𝐷+̆𝛼̆𝑠𝑘̆𝑥̆𝑦d̆𝑦d̆𝑡=1𝑠0𝜂𝜉d𝑦(𝑡)d𝑡d𝑡d̆𝑡=1𝑠0𝜂𝜉𝐷0d𝑦(𝑡)=1d𝑡𝑠0𝜂𝜉𝐷0−Δ𝑦(𝑡)+𝑘𝑒−Δ𝜏=𝑦(𝑡−𝜏)𝑥(𝑡−𝜏)−Δ𝑦(𝑡)𝑠0𝜂𝜉𝐷0+𝑘𝑒−Δ𝜏𝑠0𝜂𝜉𝐷0=𝑦(𝑡−𝜏)𝑥(𝑡−𝜏)−Δ𝐷0𝑦(𝑡)𝑠0+𝜂𝜉𝑘𝑠0𝜂𝐷0𝑒−(Δ/𝐷0)𝐷0𝜏𝑦(𝑡−𝜏)𝑠0𝜂𝜉𝑥(𝑡−𝜏)𝑠0𝜂̆̆𝑡+̆=−Δ̆𝑦𝑘𝑒−̆Δ̆𝜏̆̆.̆𝑦𝑡−̆𝜏̆𝑥𝑡−̆𝜏(2.2) With this change of variables, omitting the ̆‌’s for convenience, system (2.1) becomes

̇𝑠(𝑡)=1−𝑠(𝑡)−𝛼𝑥(𝑡)𝑠(𝑡),̇𝑥(𝑡)=𝑥(𝑡)(−𝐷+𝛼𝑠(𝑡))−𝑘𝑦(𝑡)𝑥(𝑡),̇𝑦(𝑡)=−Δ𝑦(𝑡)+𝑘𝑒−Δ𝜏𝑦(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−Δ𝐷/𝑘<0, the model has a second equilibrium point (𝐷/𝛼,(𝛼−𝐷)/𝛼𝐷,0) and it is globally asymptotically stable. When 1−𝐷/𝛼−Δ𝐷/𝑘>0, the model has a third equilibrium point (𝑘/(𝑘+𝛼Δ),Δ/𝑘,𝛼/(𝑘+𝛼𝐷)−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 (𝑠0,𝜙(𝜃),𝜓(𝜃))∈intℝ+×ℂ([−𝜏,0],intℝ2+), then there exists a unique solution (𝑠(𝑡),𝑥(𝑡),𝑦(𝑡)) of (2.3) passing through (𝑠0,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 𝜖0>0, 𝑥(𝑡)<1+𝜖0 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)−𝛼𝑥(𝑡)𝑠(𝑡),̇𝑥(𝑡)=𝑥(𝑡)(−𝐷+𝛼𝑠(𝑡))−𝑘𝑦(𝑡)𝑥(𝑡),̇𝑦(𝑡)=−Δ𝑦(𝑡)+𝑘𝑒−Δ𝜏𝜙(𝑡−𝜏)𝜓(𝑡−𝜏),(2.4) a system of nonautonomous ordinary differential equations with initial conditions 𝑠(0)=𝑠0, 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 (𝑠0,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)−𝛼𝑥(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) 𝑡̇𝑠⋆𝑡=1−𝑠⋆𝑡−𝛼𝑥⋆𝑠𝑡⋆=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
𝑥𝑡=𝜙(0)exp𝑡0(−𝐷+𝛼𝑠(𝑡)−𝑘𝑦(𝑡))d𝑡>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
𝑡̇𝑦⋆𝑡=−Δ𝑦⋆+𝑘𝑒−Δ𝜏𝑦𝑡⋆𝑥𝑡−𝜏⋆−𝜏=𝑘𝑒−Δ𝜏𝑦𝑡⋆𝑥𝑡−𝜏⋆−𝜏>0,(2.8) a contradiction.
To prove the boundedness of solutions, define
𝜔(𝑡)=𝑠(𝑡)+𝑥(𝑡)+𝑒Δ𝜏𝑦(𝑡+𝜏)−1,for𝑡⩾0.(2.9) It follows that ̇𝜔(𝑡)=1−𝑠(𝑡)−𝐷𝑥(𝑡)−Δ𝑒Δ𝜏𝑦(𝑡+𝜏)⩽1−𝑠(𝑡)−𝑥(𝑡)−𝑒Δ𝜏𝑦(𝑡+𝜏)⩽−𝜔(𝑡),(2.10) 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)+𝑒Δ𝜏𝑠𝑦(𝑡+𝜏)⩽1+0+𝑥(0)+𝑒Δ𝜏𝑒𝑦(𝜏)−1−𝑡⟶1as𝑡⟶∞.(2.11) Therefore, the solution (𝑠(𝑡),𝑥(𝑡),𝑦(𝑡)) is bounded, and given any 𝜖0>0, 𝑥(𝑡)<1+𝜖0 for all sufficiently large 𝑡.

3. Equilibria and Stability

Model (2.3) has three equilibrium points: 𝐸1=(1,0,0),𝐸2=(𝐷/𝛼,(𝛼−𝐷)/𝛼𝐷,0), and

𝐸+=𝑠+(𝜏),𝑥+(𝜏),𝑦+=1(𝜏)1+(𝛼Δ/𝑘)𝑒Δ𝜏,Δ𝑘𝑒Δ𝜏,𝛼𝑘+𝛼Δ𝑒Δ𝜏−𝐷𝑘.(3.1) We call 𝐸1 the washout equilibrium, 𝐸2 the single species equilibrium, and 𝐸+ the coexistence equilibrium. For the sake of biological significance, 𝐸+ exists (distinct from 𝐸2) if and only if its third coordinate 𝑦+(𝜏)=(𝛼𝑠+(𝜏)−𝐷)/𝑘>0, that is, 𝑠+(𝜏)>𝐷/𝛼, or equivalently, 𝜏 lies between 0 and 𝜏𝑐, where

𝜏𝑐=1Δ𝑘lnΔ1𝐷−1𝛼.(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+=𝐸2.

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

⎡⎢⎢⎢⎣̇𝑧1(𝑡)̇𝑧2(𝑡)̇𝑧3⎤⎥⎥⎥⎦=⎡⎢⎢⎢⎣⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣𝑧(𝑡)−1−𝛼𝑥−𝛼𝑠0𝛼𝑥−𝐷+𝛼𝑠−𝑘𝑦−𝑘𝑥00−Δ1𝑧(𝑡)2𝑧(𝑡)3⎤⎥⎥⎥⎦+⎡⎢⎢⎢⎣(𝑡)0000000𝑘𝑒Δ𝜏𝑦𝑘𝑒Δ𝜏𝑥⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣𝑧1𝑧(𝑡−𝜏)2𝑧(𝑡−𝜏)3⎤⎥⎥⎥⎦.(𝑡−𝜏)(3.3) The associated characteristic equation is given by

⎡⎢⎢⎢⎣det−1−𝛼𝑥−𝜆−𝛼𝑠0𝛼𝑥−𝐷+𝛼𝑠−𝑘𝑦−𝜆−𝑘𝑥0𝑘𝑒−Δ𝜏−𝜆𝜏𝑦−Δ+𝑘𝑒−Δ𝜏−𝜆𝜏⎤⎥⎥⎥⎦𝑥−𝜆=0.(3.4) Direct calculation of the left-hand side of (3.4) gives

−Δ+𝑘𝑒−(Δ+𝜆)𝜏𝑥−𝜆(−1−𝛼𝑥−𝜆)(−𝐷+𝛼𝑠−𝑘𝑦−𝜆)+𝛼2𝑠𝑥+𝑘𝑥𝑘𝑒−(Δ+𝜆)𝜏𝑦((−1−𝛼𝑥−𝜆)=(−Δ−𝜆)1+𝛼𝑥+𝜆)(𝐷−𝛼𝑠+𝑘𝑦+𝜆)+𝛼2𝑠𝑥+𝑒−(Δ+𝜆)𝜏×𝑘𝑥𝑘𝑦(−1−𝛼𝑥−𝜆)+(1+𝛼𝑥+𝜆)(𝐷−𝛼𝑠+𝑘𝑦+𝜆)+𝛼2=𝑠𝑥(−Δ−𝜆)(1+𝛼𝑥+𝜆)(𝐷−𝛼𝑠+𝑘𝑦+𝜆)+𝛼2𝑠𝑥+𝑒−(Δ+𝜆)𝜏𝑘𝑥(1+𝛼𝑥+𝜆)(𝐷−𝛼𝑠+𝜆)+𝛼2=𝑠𝑥(−Δ−𝜆){(𝜆+1)(𝜆+𝐷+𝑘𝑦)+𝛼𝑥(𝜆+𝐷+𝑘𝑦)−𝛼𝑠(𝜆+1)}+𝑒−(Δ+𝜆)𝜏𝑘𝑥{(𝜆+1)(𝜆+𝐷)+𝛼𝑥(𝜆+𝐷)−𝛼𝑠(𝜆+1)}.(3.5) For convenience, define 𝑃(𝜆)∶=(−Δ−𝜆){(𝜆+1)(𝜆+𝐷+𝑘𝑦)+𝛼𝑥(𝜆+𝐷+𝑘𝑦)−𝛼𝑠(𝜆+1)}+𝑒−(Δ+𝜆)𝜏𝑘𝑥{(𝜆+1)(𝜆+𝐷)+𝛼𝑥(𝜆+𝐷)−𝛼𝑠(𝜆+1)}.(3.6)

Theorem 3.1. Equilibrium 𝐸1 is stable if 𝛼<𝐷 and unstable if 𝛼>𝐷.

Proof. Evaluating the characteristic equation at 𝐸1 gives ||𝑃(𝜆)𝐸1=−(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 𝐸1 is stable if 𝛼<𝐷 and unstable if 𝛼>𝐷.

Remark 3.2. If 𝛼<𝐷, then there is only one equilibrium, 𝐸1. If 𝛼>𝐷, equilibrium 𝐸2 also exists.

Lemma 3.3. Assume 𝛼>𝐷. The characteristic equation evaluated at 𝐸2 has two negative eigenvalues, and the remaining eigenvalues are solutions of (𝜆+Δ)𝑒(𝜆+Δ)𝜏1=𝑘𝐷−1𝛼.(3.8) In addition, the characteristic equation evaluated at 𝐸2 has zero as an eigenvalue if and only if 𝜏=𝜏𝑐.

Proof. Assume 𝛼>𝐷. Equilibrium 𝐸2 exists. Consider the characteristic equation at 𝐸2. Since (𝛼−𝐷)/𝛼𝐷=(1−𝑠)/𝛼𝑠 at 𝐸2, ||𝑃(𝜆)𝐸2×={(𝜆+1)(𝜆+𝐷)+𝛼𝑥(𝜆+𝐷)−𝛼𝑠(𝜆+1)}−𝜆−Δ+𝑒−(Δ+𝜆)𝜏=𝑘𝑥(𝜆+1)(𝜆+𝐷)+1−𝑠𝑠×(𝜆+𝐷)−𝐷(𝜆+1)−𝜆−Δ+𝑒−(Δ+𝜆)𝜏𝑘𝛼−𝐷=𝛼𝐷𝜆(𝜆+1)−(𝜆+𝐷)+𝜆+𝐷𝑠−𝜆−Δ+𝑘𝛼−𝐷𝑒𝛼𝐷−(Δ+𝜆)𝜏𝜆=−2+𝛼𝐷𝜆+𝛼−𝐷𝜆+Δ−𝑘𝛼−𝐷𝑒𝛼𝐷−(Δ+𝜆)𝜏=−𝑒−(Δ+𝜆)𝜏𝜆−𝜆1𝜆−𝜆2(𝜆+Δ)𝑒(Δ+𝜆)𝜏1−𝑘𝐷−1𝛼=0,(3.9) where 𝜆1+𝜆2=−𝛼/𝐷 and 𝜆1𝜆2=𝛼−𝐷>0. Therefore, 𝜆1 and 𝜆2 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
Δ𝑒Δ𝜏1=𝑘𝐷−1𝛼.(3.10) Solving for 𝜏 gives 1𝜏=Δ𝑘lnΔ1𝐷−1𝛼=𝜏𝑐.(3.11)

Theorem 3.4. Assume that 𝐷⩾1, Δ⩾1, 𝑘>0, 𝛼>0, and (𝑘/Δ)(1/𝐷−1/𝛼)⩾1 so that 𝜏𝑐⩾0. Equilibrium 𝐸2 is locally asymptotically stable if 𝜏>𝜏𝑐 and unstable if 𝜏<𝜏𝑐. If 𝐷=1, then equilibrium 𝐸2 is globally asymptotically stable for 𝜏>(1/Δ)ln(𝑘/Δ).

Proof. Assume that 𝜏>𝜏𝑐. 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 𝐸2 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=𝜏𝑐, then 𝜆=0 is a solution of (3.8). Thus for 𝜏>𝜏𝑐, any real root 𝜆 of (3.8) must satisfy −Δ<𝜆<0.
For any 𝜏=̃𝜏<𝜏𝑐, we have (𝜆+Δ)𝑒(𝜆+Δ)𝜏|𝜏=̃𝜏,𝜆=0<𝑘(1/𝐷−1/𝛼) and lim𝜆→+∞(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 𝐸2 is unstable if 𝜏<𝜏𝑐.
In what follows, we prove that if 𝜏>𝜏𝑐 all complex eigenvalues of (3.8) have negative real parts. Suppose that 𝜆+Δ=𝛾+𝑖𝛽(𝛽>0) is a solution of (3.8). Using the Euler formula, we have
1𝛾cos(𝛽𝜏)−𝛽sin(𝛽𝜏)+𝑖(𝛾sin(𝛽𝜏)+𝛽cos(𝛽𝜏))=𝑘𝐷−1𝛼𝑒−𝛾𝜏.(3.12) Equating the real parts and imaginary parts of the equation, we have 1𝛾cos(𝛽𝜏)−𝛽sin(𝛽𝜏)=𝑘𝐷−1𝛼𝑒−𝛾𝜏𝛾sin(𝛽𝜏)+𝛽cos(𝛽𝜏)=0.(3.13) Squaring both equations, adding, and taking the square root on both sides give 𝛾2+𝛽2𝑒𝛾𝜏1=𝑘𝐷−1𝛼.(3.14) 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=𝜏𝑐, any roots of (3.14) must satisfy 𝛾<Δ since 𝜏>𝜏𝑐. Hence Re{𝜆}=𝛾−Δ<0. Therefore (3.8) has no complex eigenvalue with nonnegative real part and so 𝐸2 is locally asymptotically stable for 𝜏>𝜏𝑐.
Assume that 𝐷=1. Now we prove that 𝐸2 is globally asymptotically stable when 𝜏>(1/Δ)ln(𝑘/Δ), or equivalently 𝑘𝑒−Δ𝜏<Δ. In this case, choose 𝜖0>0 small enough such that 𝑘𝑒−Δ𝜏(1+𝜖0)<Δ. By Theorem 2.1, for such 𝜖0, there exists a 𝑇>0 so that 0<𝑥(𝑡)<1+𝜖0 for 𝑡>𝑇. Hence, for 𝑡>𝑇+𝜏, 𝑘𝑒−Δ𝜏𝑥(𝑡−𝜏)<Δ. In Example 5.1 of Kuang ([10, page 32]), choose 𝜌(𝑡)=𝜏, 𝑎(𝑡)=Δ, 𝑏(𝑡)=𝑘𝑒−Δ𝜏𝑥(𝑡−𝜏), and 𝛼=Δ/2. We obtain (𝑘𝑒−Δ𝜏(1+𝜖0))2<Δ2=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)−𝑘𝑥(𝑡)𝑦(𝑡). Multiply by the integrating factor 𝑒𝑡, (𝑧(𝑡)𝑒𝑡)=𝑒𝑡(1−𝑘𝑥(𝑡)𝑦(𝑡)). Integrating both sides from 0 to 𝑡 gives
𝑧(𝑡)=𝑒−𝑡𝑧(0)+𝑒−𝑡𝑒𝑡−1−𝑒−𝑡𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠=1+𝑒−𝑡(𝑧(0)−1)−𝑒−𝑡𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠.(3.15) If lim𝑡→∞∫𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠<∞, then lim𝑡→∞𝑒−𝑡∫𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠=0. Therefore lim𝑡→∞𝑧(𝑡)=1. If lim𝑡→∞∫𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠=∞, by L'Hôspital's rule, lim𝑡→+∞𝑒−𝑡𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠=lim𝑡→+∞∫𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠𝑒𝑡=lim𝑡→+∞𝑒𝑡𝑘𝑥(𝑡)𝑦(𝑡)𝑒𝑡=lim𝑡→+∞𝑘𝑥(𝑡)𝑦(𝑡)=0,(3.16) since 𝑥(𝑡) is bounded and lim𝑡→∞𝑦(𝑡)=0. It again follows that limt→∞𝑧(𝑡)=1. Hence lim𝑡→∞𝑠(𝑡)+𝑥(𝑡)=1.(3.17)
We show that lim𝑡→∞𝑠(𝑡)=1/𝛼 and lim𝑡→∞𝑥(𝑡)=(𝛼−1)/𝛼. First assume that the limits exist, that is, lim𝑡→∞𝑠(𝑡)=𝑠 and lim𝑡→∞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 lim𝑡→∞̇𝑠(𝑡)=0 and lim𝑡→∞̇𝑥(𝑡)=0. Note that lim𝑡→∞𝑦(𝑡)=0. Letting 𝑡→∞ in (2.3) gives
1−𝑠−𝛼𝑥𝑠=0,𝑥−1+𝛼𝑠=0.(3.18) Either (𝑠,𝑥)=(1,0) or (𝑠,𝑥)=(1/𝛼,(𝛼−1)/𝛼). Assume that (𝑠,𝑥)=(1,0), that is, lim𝑡→∞𝑠(𝑡)=1 and lim𝑡→∞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)−𝑘𝜖)=𝑥(𝑡)(𝛼−𝐷−𝛼𝜖−𝑘𝜖)>0,(3.19) 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=limsup𝑡→∞𝑥(𝑡) and 𝑥=liminf𝑡→∞𝑥(𝑡). By Lemma A.2 in the appendix, there exists {𝑡𝑚}↑∞ and {𝑠𝑚}↑∞ such that
lim𝑚→∞𝑥𝑡𝑚=𝑥,lim𝑚→∞𝑡̇𝑥𝑚=0,lim𝑚→∞𝑥𝑠𝑚=𝑥lim𝑚→∞𝑠̇𝑥𝑚=0.(3.20) From (2.3), 𝑥𝑡𝑚𝑡−𝐷+𝛼𝑠𝑚𝑡+𝑘𝑦𝑚=0.(3.21) Noting that 𝑥(𝑡𝑚)>0, we have 𝑠(𝑡𝑚)=(1−𝑘𝑦(𝑡𝑚))/𝛼. Since lim𝑡→∞𝑦(𝑡)=0, lim𝑡→∞𝑠(𝑡𝑚)=1/𝛼. By (3.17), lim𝑡→∞𝑥(𝑡𝑚)=lim𝑡→∞(𝑥(𝑡𝑚)+𝑠(𝑡𝑚))−𝑠(𝑡𝑚)=1−1/𝛼=(𝛼−1)/𝛼. Therefore 𝑥=(𝛼−1)/𝛼. Similarly we can show that 𝑥=(𝛼−1)/𝛼. This implies that lim𝑡→∞𝑥(𝑡)=(𝛼−1)/𝛼, a contradiction.
Since 𝑠(𝑡)+𝑥(𝑡) converges and 𝑥(𝑡) converges, then 𝑠(𝑡) must also converge. Hence lim𝑡→∞𝑠(𝑡)=1/𝛼 and lim𝑡→∞𝑥(𝑡)=(𝛼−1)/𝛼. It follows that 𝐸2 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+𝛼𝑥+(𝜏)+𝜆𝐷−𝛼𝑠+(𝜏)+𝑘𝑦+(𝜏)+𝜆+𝛼2𝑠+(𝜏)𝑥+(𝜏)+𝑒−(Δ+𝜆)𝜏𝑘𝑥+(𝜏)(𝜆+1)(𝜆+𝐷)+𝛼𝑥+(𝜏)(𝜆+𝐷)−𝛼𝑠+(𝜏)(𝜆+1)=(−Δ−𝜆)1+𝛼𝑥+(𝜏)+𝜆𝜆+𝛼2𝑠+(𝜏)𝑥+(𝜏)+𝑒−(Δ+𝜆)𝜏𝑘𝑥+(𝜏)(𝜆+1)(𝜆+𝐷)+𝛼𝑥+(𝜏)(𝜆+𝐷)−𝛼𝑠+1(𝜏)(𝜆+1)=(−Δ−𝜆)𝑠+(𝜏)+𝜆𝜆+𝛼1−𝑠+(𝜏)+𝑒−𝜆𝜏Δ(𝜆+1)(𝜆+𝐷)+1−𝑠+(𝜏)𝑠+(𝜏)(𝜆+𝐷)−𝛼𝑠+𝜆(𝜏)(𝜆+1)=(−Δ−𝜆)2+𝜆𝑠+(𝜏)+𝛼1−𝑠+(𝜏)+Δ𝑒−𝜆𝜏1𝜆+𝑠+(𝜏)(𝜆+𝐷)−𝛼𝑠+(𝜏)(𝜆+1)=0.(4.1) By assumption Δ=𝐷=1, and so

||𝑃(𝜆)𝐸+𝜆=−(𝜆+1)2+𝜆𝑠+(𝜏)+𝛼1−𝑠+(𝜏)+𝑒−𝜆𝜏−𝜆+𝛼𝑠+1(𝜏)−𝑠+𝜆(𝜏)=−(𝜆+1)2+𝑝(𝜏)𝜆+𝛽(𝜏)+𝑒−𝜆𝜏(𝑞𝜆+𝑐(𝜏))=0,(4.2) where

1𝑝(𝜏)=𝑠+(𝜏),𝛽(𝜏)=𝛼1−𝑠+(𝜏),𝑞=−1,𝑐(𝜏)=𝛼𝑠+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

𝜆2+𝑝(𝜏)𝜆+𝛽(𝜏)+𝑒−𝜆𝜏(𝑞𝜆+𝑐(𝜏))=0.(4.4)

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

Proof. Assume that 𝜏∈(0,𝜏𝑐). By the method of contradiction, suppose that there exists a zero root of (4.4). Therefore 1𝛽(𝜏)+𝑐(𝜏)=𝛼−𝑠+(𝜏)=0.(4.5) Noting that 𝜏𝑐>0 if and only if 𝑘(1−1/𝛼)>1, for any 0<𝜏<𝜏𝑐, 1𝛼−𝑠+𝛼(𝜏)=𝛼−1−𝑘𝑒𝜏1>𝛼−1−𝛼1−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 𝜆2+𝑝(0)𝜆+𝛽(0)+(𝑞𝜆+𝑐(0))=𝜆2+1𝑠+1(0)−1𝜆+𝛼−𝑠+(0).(4.7) Both coefficients are positive, since 1𝑠+𝛼(0)−1=𝑘1>0,𝛼−𝑠+𝛼(0)=𝛼−1−𝑘1=𝛼1−𝛼−1𝑘>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+(𝑞𝜆+𝑐(𝜏))𝑒−𝜆𝜏+𝛽(𝜏) in Kuang [10, Theorem 1.4, page 66] gives limsupRe𝜆>0,|𝜆|→∞||𝜆−2𝑔||(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=𝑖𝜔 (𝜔>0) is a root of 𝑃(𝜆)|𝐸+=0,
−𝜔2+𝑖𝑝(𝜏)𝜔+𝛽(𝜏)+𝑒−𝑖𝜔𝜏(𝑖𝑞𝜔+𝑐(𝜏))=0.(4.10) Substituting 𝑒𝑖𝜃=cos𝜃+𝑖sin𝜃 into (4.10) gives −𝜔2+𝛽(𝜏)+𝑞𝜔sin(𝜔𝜏)+𝑐(𝜏)cos(𝜔𝜏)+𝑖(𝑝(𝜏)𝜔+𝑞𝜔cos(𝜔𝜏)−𝑐(𝜏)sin(𝜔𝜏))=0.(4.11) Separating the real and imaginary parts, we obtain 𝑐(𝜏)cos(𝜔𝜏)+𝑞𝜔sin(𝜔𝜏)=𝜔2−𝛽(𝜏),𝑐(𝜏)sin(𝜔𝜏)−𝑞𝜔cos(𝜔𝜏)=𝑝(𝜏)𝜔.(4.12) Solving for cos(𝜔𝜏) and sin(𝜔𝜏) gives 𝜔sin(𝜔𝜏)=𝑐(𝜏)𝑝(𝜏)𝜔+𝑞𝜔2−𝛽(𝜏)𝑐(𝜏)2+𝑞2𝜔2,𝜔cos(𝜔𝜏)=𝑐(𝜏)2−𝛽(𝜏)−𝑞𝑝(𝜏)𝜔2𝑐(𝜏)2+𝑞2𝜔2.(4.13) Noting sin2(𝜔𝜏)+cos2(𝜔𝜏)=1, squaring both sides of equations (4.13), adding, and rearranging gives 𝜔4+𝑝2(𝜏)−𝑞2𝜔−2𝛽(𝜏)2+𝛽2(𝜏)−𝑐2(𝜏)=0.(4.14) Solving for 𝜔, we obtain two roots 𝜔1(𝜏) and 𝜔2(𝜏): 𝜔11(𝜏)=√2𝑞2−𝑝2(𝜏)+2𝛽(𝜏)+𝑞2−𝑝2(𝜏)+2𝛽(𝜏)2−4(𝛽2(𝜏)−𝑐2(𝜏))1/2=1𝑠+√(𝜏)21−𝑠+(𝜏)2𝛼𝑠2+(𝜏)−𝑠+(+𝜏)−1𝑠2+(𝜏)−12+4𝛼𝑠2+𝑠(𝜏)2+(𝜏)−11−𝑠+(𝜏)+4𝑠2+(𝜏)𝛼𝑠2+(𝜏)−121/2𝜔21(𝜏)=√2𝑞2−𝑝2(𝜏)+2𝛽(𝜏)−𝑞2−𝑝2(𝜏)+2𝛽(𝜏)2𝛽−42(𝜏)−𝑐2(𝜏)1/2=1𝑠+√(𝜏)21−𝑠+(𝜏)2𝛼𝑠2+(𝜏)−𝑠+−(𝜏)−1𝑠2+(𝜏)−12+4𝛼𝑠2+𝑠(𝜏)2+(𝜏)−11−𝑠+(𝜏)+4𝑠2+(𝜏)𝛼𝑠2+(𝜏)−121/2.(4.15)
Define conditions (𝐻1) and (𝐻2) as follows:
𝐻1𝑞2−𝑝2(𝜏)+2𝛽(𝜏)>0,𝛽2(𝜏)−𝑐2𝑞(𝜏)>0,2−𝑝2(𝜏)+2𝛽(𝜏)2𝛽−42(𝜏)−𝑐2(𝜏)⩾0,(4.16)𝐻2𝛽2(𝜏)−𝑐2(𝜏)<0,or𝛽2(𝜏)−𝑐2(𝜏)=0and𝑞2−𝑝2(𝜏)+2𝛽(𝜏)>0.(4.17)

Lemma 4.4. If (𝐻1) holds for all 𝜏 in some interval 𝐼, then (4.14) has two positive roots 𝜔1(𝜏)⩾𝜔2(𝜏) for all 𝜏∈𝐼 with 𝜔1(𝜏)>𝜔2(𝜏) when all the inequalities in (𝐻1) are strict. If (𝐻2) holds for all 𝜏 in some interval 𝐼, then (4.14) has only one positive root, 𝜔1(𝜏) for all 𝜏∈𝐼. If no interval exists where either (𝐻1) or (𝐻2) holds, then there are no positive real roots of (4.14).

Define the interval

⎡⎢⎢⎣𝑘𝐽=ln𝛼1√1/4+⎛⎜⎜⎝𝑘1/16+1/(2𝛼)−1,ln4√𝛼−1𝛼⎞⎟⎟⎠⎤⎥⎥⎦.(4.18) When the end points of 𝐽 are real and 𝐽≠∅, define

𝐼1=0,𝜏𝑐∩𝐽.(4.19) We prove that (𝐻1) holds for any 𝜏∈𝐼1.

From 𝐷=Δ=1,

𝜏𝑐=1Δ𝑘lnΔ1𝐷−1𝛼=ln𝑘(𝛼−1)𝛼.(4.20) If 𝛼>1, then 𝛼>4√𝛼. It follows that

𝜏𝑐⎛⎜⎜⎝𝑘>ln4√𝛼−1𝛼⎞⎟⎟⎠.(4.21) Therefore,

𝐼1=⎡⎢⎢⎣𝑘max0,ln𝛼1√1/4+⎛⎜⎜⎝𝑘1/16+1/(2𝛼)−1,ln4√𝛼−1𝛼⎞⎟⎟⎠⎤⎥⎥⎦.(4.22)

Theorem 4.5. Assume √𝛼>(7+35)/2 and 𝑘>𝛼/(4√𝛼−1), then 𝐼1 is not empty, and for any 𝜏∈𝐼1, but √𝜏≠ln((𝑘/𝛼)(1/(1/4+1/16+1/(2𝛼))−1)), condition (𝐻1) holds and 𝜔1(𝜏)>𝜔2(𝜏)>0. If √𝜏=ln((𝑘/𝛼)(1/(1/4+1/16+1/(2𝛼))−1))∈𝐼1, then 𝜔1(𝜏)>𝜔2(𝜏)=0.

Proof. For any √𝛼>(7+35)/2, we have 1−1/4√𝛼>0, and therefore 14√𝛼+12−√52<14√7+35+1/22−√52=0.(4.23) Hence, 14√𝛼−142−1+1162𝛼2=1√𝛼−124√𝛼−1=2𝛼−124√𝛼⎛⎜⎜⎝14√𝛼32+1−4√𝛼⎞⎟⎟⎠=124√𝛼11−4√𝛼⎛⎜⎜⎝14√𝛼2+14√𝛼⎞⎟⎟⎠=1−124√𝛼11−4√𝛼⎛⎜⎜⎝14√𝛼+122−54⎞⎟⎟⎠=124√𝛼11−4√𝛼14√𝛼+12−√5214√𝛼+12+√52<0.(4.24) Therefore, 1/4√√𝛼−1/4<1/16+1/(2𝛼). Since √1/4+1/16+1/(2𝛼)<1/4+√1/16+1/(7+35)<1, it follows that 14√𝛼<14+1+1162𝛼<1.(4.25) Hence, 𝑘ln𝛼1√1/4+⎛⎜⎜⎝𝑘1/16+1/(2𝛼)−1<ln4√𝛼−1𝛼⎞⎟⎟⎠.(4.26) From 𝑘>𝛼/(4√𝛼−1), we have ln(𝑘(4√𝛼−1)/𝛼)>0. Therefore, 𝑘max0,ln𝛼1√1/4+⎛⎜⎜⎝𝑘1/16+1/(2𝛼)−1<ln4√𝛼−1𝛼⎞⎟⎟⎠,(4.27) and so 𝐼1 is not empty. Noting 𝑠+(𝜏)=1/(1+(𝛼Δ/𝑘)𝑒Δ𝜏) and Δ=1, for any 𝜏∈𝐼1, but √𝜏≠ln((𝑘/𝛼)(1/(1/4+1/16+1/(2𝛼))−1)), we have 𝑠+(𝜏)∈[1/4√√𝛼,1/4+1/16+1/(2𝛼)).
In what follows, we intend to show that for any such 𝜏, condition (𝐻1) holds. From (4.3),
𝑞2−𝑝2(𝜏)+2𝛽(𝜏)=(−1)2−1𝑠2+(𝜏)+2𝛼1−𝑠+=(𝜏)1−𝑠+(𝜏)2𝛼𝑠2+𝑠(𝜏)+(𝜏)2−𝑠+(𝜏)−12𝛼=2𝛼1−𝑠+(𝜏)2𝛼𝑠2+𝑠(𝜏)+1(𝜏)−4𝛼2−116𝛼2−1.2𝛼(4.28) Since 𝑠+(𝜏)<1, to show that the first inequality in (𝐻1) holds, it suffices to show that the factor on the right-hand side of the above expression is positive. Since √𝛼>(7+35)/2, √1/√𝛼−1/(4𝛼)=(1/√𝛼)(1−1/(4𝛼))>0, and 116𝛼2+12𝛼2−1√𝛼−14𝛼2=116𝛼2+1−12𝛼𝛼+1√2𝛼𝛼−116𝛼2=112𝛼√𝛼−1<0.(4.29) Since √1/𝛼<1/4√𝛼 for √𝛼>(7+35)/2, 1+4𝛼116𝛼2+1<12𝛼√𝛼<14√𝛼.(4.30) For any 𝑠+(𝜏)>1/4√𝛼, 𝑠+1(𝜏)−⩾14𝛼4√𝛼−1>4𝛼116𝛼2+1.2𝛼(4.31) Hence, 𝑠+1(𝜏)−4𝛼2⩾116𝛼2+12𝛼.(4.32) Next consider the second inequality in (𝐻1). For √𝛼>(7+35)/2, since 1/4√𝛼>1/𝛼, 𝑠+(𝜏)⩾1/4√𝛼>1/𝛼. Therefore, 𝛼𝑠+(𝜏)>1. For 𝑠+(𝜏)∈[1/4√√𝛼,1/4+1/16+1/(2𝛼))𝛽2(𝜏)−𝑐2=(𝜏)=(𝛽(𝜏)−𝑐(𝜏))(𝛽(𝜏)+𝑐(𝜏))𝛼−2𝛼𝑠+1(𝜏)+𝑠+(1𝜏)𝛼−𝑠+(𝜏)=−2𝛼𝑠2+𝑠(𝜏)2+𝑠(𝜏)−+(𝜏)2−12𝛼𝛼𝑠+(𝜏)−1=−2𝛼𝑠2+𝑠(𝜏)+1(𝜏)−42−1−1162𝛼𝛼𝑠+(𝜏)−1>0.(4.33) Finally, 𝑞2−𝑝2(𝜏)+2𝛽(𝜏)2𝛽−42(𝜏)−𝑐2=𝑞(𝜏)2−𝑝2𝑞(𝜏)2−𝑝2(𝜏)+4𝛽(𝜏)+4𝑐2=1(𝜏)1−𝑠2+(1𝜏)1−𝑠2+(𝜏)+4𝛼1−𝑠+(𝜏)+4𝛼𝑠+1(𝜏)−𝑠+(𝜏)2=11−𝑠2+(𝜏)21+4𝛼1−𝑠2+(𝜏)1−𝑠+(𝜏)+4𝑠+(𝜏)2𝛼24−8𝛼+𝑠2+(𝜏)=4𝑠+(𝜏)2𝛼21+4𝛼1−𝑠2+(𝜏)1−𝑠+1(𝜏)−8𝛼+1−𝑠2+(𝜏)2+4𝑠2+(𝜏)=4𝑠+(𝜏)2𝛼21+4𝛼1−𝑠2+(𝜏)1−𝑠++1(𝜏)−21+𝑠2+(𝜏)2=𝛼−𝛼1𝛼−𝛼2,(4.34) where 𝛼1=2−1−1/𝑠2+(𝜏)1−𝑠++(𝜏)1/𝑠2+(𝜏)+2𝑠+(𝜏)+11/𝑠+(𝜏)−122𝑠2+,𝛼(𝜏)2=2−1−1/𝑠2+(𝜏)1−𝑠+−(𝜏)1/𝑠2+(𝜏)+2𝑠+(𝜏)+11/𝑠+(𝜏)−122𝑠2+.(𝜏)(4.35) Since 𝑠+(𝜏)<1, 12−1−𝑠2+(𝜏)1−𝑠+(𝜏)=𝑠+(𝜏)+1+1−𝑠+(𝜏)𝑠2+1(𝜏)>0,2−1−𝑠2+(𝜏)1−𝑠+(𝜏)2>12−1−𝑠2+(𝜏)1−𝑠+(𝜏)2−𝑠2+1(𝜏)1+𝑠2+(𝜏)2=1𝑠2+(𝜏)+2𝑠+1(𝜏)+1𝑠+(𝜏)−12>0.(4.36) It follows that 0<𝛼2<𝛼1. Again noting that 𝑠+(𝜏)<1, 𝛼1<2−1−1/𝑠2+(𝜏)1−𝑠++(𝜏)1/𝑠2+(𝜏)+2/𝑠+(𝜏)+11/𝑠+(𝜏)−122𝑠2+=(𝜏)2−1−1/𝑠2+(𝜏)1−𝑠++(𝜏)1/𝑠+(𝜏)+11/𝑠+(𝜏)−12𝑠2+=(𝜏)2−1−𝑠+(𝜏)−1/𝑠2+(𝜏)+1/𝑠+(𝜏)+1/𝑠2+(𝜏)−12𝑠2+=𝑠(𝜏)+(𝜏)+2/𝑠2+(𝜏)−1/𝑠+(𝜏)2𝑠2+=1(𝜏)22𝑠4+−1(𝜏)𝑠3+−1(𝜏)𝑠+<1(𝜏)22𝑠4+=1(𝜏)𝑠4+.(𝜏)(4.37) Hence, for any 𝑠+(𝜏)>1/4√𝛼, we have 𝛼>1/𝑠4+(𝜏)>𝛼1>𝛼2. This leads to 𝑞2−𝑝2(𝜏)+2𝛽(𝜏)2𝛽−42(𝜏)−𝑐2=(𝜏)𝛼−𝛼1𝛼−𝛼2>0.(4.38) Therefore, (𝐻1) holds for any 𝜏∈𝐼1. By Lemma 4.4, both 𝜔1(𝜏)>0 and 𝜔2(𝜏)>0.
If √𝜏=ln((𝑘/𝛼)(1/(1/4+1/16+1/(2𝛼))−1))∈𝐼1, we have 𝑠+√(𝜏)=1/4+1/16+1/(2𝛼). Noting (4.25), we obtain
𝑞2−𝑝2(𝜏)+2𝛽(𝜏)>0,𝛽2(𝜏)−𝑐2𝑞(𝜏)=0,2−𝑝2(𝜏)+2𝛽(𝜏)2𝛽−42(𝜏)−𝑐2(𝜏)>0.(4.39) By (4.15), it follows that 𝜔1(𝜏)>0 and 𝜔2(𝜏)=0.

Now we define interval 𝐼2 and prove that (𝐻2) holds on 𝐼2

𝐼2∶=0,𝜏𝑐∩𝑘−∞,ln𝛼1√1/4+1/16+1/(2𝛼)−1.(4.40) In the following theorem, we consider the case that parameters are chosen so that

𝐼2=𝑘0,ln𝛼1√1/4+1/16+1/(2𝛼)−1.(4.41)

Theorem 4.6. Assume 𝛼>1 and √𝑘>𝛼(1/(1/4+1/16+1/(2𝛼))−1)−1. Interval 𝐼2 given by (4.41) is not empty. For any 𝜏∈𝐼2, (𝐻2) holds and hence 𝜔1(𝜏)>0.

Proof. Assume 𝛼>1. Letting 1𝐺(𝛼)=4+1+116−12𝛼𝛼,(4.42) then d1d𝛼𝐺(𝛼)=−4𝛼2√+11/16+1/(2𝛼)𝛼2=1𝛼2−1√1+8/𝛼+1>0.(4.43)𝐺(𝛼) is an increasing function of 𝛼 and 𝐺(1)=0. 𝐺(𝛼)>𝐺(1) implies that √1/4+1/16+1/(2𝛼)−1/𝛼>0. Therefore 11>4+1+116>12𝛼𝛼.(4.44) This gives 1𝛼−1>√1/4+1/16+1/(2𝛼)−1>0.(4.45) By assumption √𝑘>𝛼(1/(1/4+1/16+1/(2𝛼))−1)−1, we obtain 𝑘(𝛼−1)𝛼>𝑘𝛼1√1/4+1/16+1/(2𝛼)−1>1.(4.46) Noting that 𝐷=Δ=1 and recalling the definition of 𝜏𝑐 given in (3.2), 𝜏𝑐=ln𝑘(𝛼−1)𝛼𝑘>