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International Journal of Differential Equations
Volume 2010, Article ID 287969, 41 pages
Research Article

A Predator-Prey Model in the Chemostat with Time Delay

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.


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)=0𝐷𝑠(𝑡)0𝑥(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)=0𝐷𝑠(𝑡)0𝑥(𝑡)𝑓(𝑠(𝑡))𝜂,𝑦̇𝑥(𝑡)=𝑥(𝑡)(𝐷+𝑓(𝑠(𝑡)))(𝑡)𝑔(𝑥(𝑡))𝜉,̇𝑦(𝑡)=Δ𝑦(𝑡)+𝑒Δ𝜏𝑦(𝑡𝜏)𝑔(𝑥(𝑡𝜏)).(1.2) For 𝑡[𝜏,0],

𝑠(0)=𝑠0int+[],(𝑥(𝑡),𝑦(𝑡))=(𝜙,𝜓)𝜏,0,int2+.(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 𝐷𝐷0, and Δ𝐷0. The additional parameter 𝜏 is a nonnegative constant modelling the time required for the conversion process. Hence, 𝑒Δ𝜏𝑦(𝑡𝜏) represents the concentration of the predator population in the growth chamber at time 𝑡 that were available at time 𝑡𝜏 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)=0𝐷𝑠(𝑡)0𝛼𝑥(𝑡)𝑠(𝑡)𝜂,̇𝑥(𝑡)=𝑥(𝑡)(𝐷+𝛼𝑠(𝑡))𝑘𝑥(𝑡)𝑦(𝑡)𝜉,̇𝑦(𝑡)=Δ𝑦(𝑡)+𝑘𝑒Δ𝜏𝑦(𝑡𝜏)𝑥(𝑡𝜏).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)=1d𝑡𝑠0𝐷0𝑠0𝐷𝑠(𝑡)0𝛼𝑥(𝑡)𝑠(𝑡)𝜂=1𝑠(𝑡)𝑠0𝛼𝑠0𝐷0𝑥(𝑡)𝑠0𝜂𝑠(𝑡)𝑠0̆𝑡̆𝑡̆𝑡,̆𝑡=1̆𝑠̆𝛼̆𝑥̆𝑠d̆𝑥d̆𝑡=1𝑠0𝜂d𝑥(𝑡)d𝑡d𝑡d̆𝑡=1𝑠0𝜂𝐷0d𝑥(𝑡)=1d𝑡𝑠0𝜂𝐷0𝑥(𝑡)(𝐷+𝛼𝑠(𝑡))𝑘𝑥(𝑡)𝑦(𝑡)𝜉=𝑥(𝑡)𝑠0𝜂𝐷𝐷0+𝛼𝑠0𝐷0𝑠(𝑡)𝑠0𝑘𝑠0𝜂𝐷0𝑥(𝑡)𝑠0𝜂𝑦(𝑡)𝑠0̆𝑡̆̆𝑡̆̆𝑡̆𝑡,̆𝑡𝜂𝜉=̆𝑥𝐷+̆𝛼̆𝑠𝑘̆𝑥̆𝑦d̆𝑦d̆𝑡=1𝑠0𝜂𝜉d𝑦(𝑡)d𝑡d𝑡d̆𝑡=1𝑠0𝜂𝜉𝐷0d𝑦(𝑡)=1d𝑡𝑠0𝜂𝜉𝐷0Δ𝑦(𝑡)+𝑘𝑒Δ𝜏=𝑦(𝑡𝜏)𝑥(𝑡𝜏)Δ𝑦(𝑡)𝑠0𝜂𝜉𝐷0+𝑘𝑒Δ𝜏𝑠0𝜂𝜉𝐷0=𝑦(𝑡𝜏)𝑥(𝑡𝜏)Δ𝐷0𝑦(𝑡)𝑠0+𝜂𝜉𝑘𝑠0𝜂𝐷0𝑒(Δ/𝐷0)𝐷0𝜏𝑦(𝑡𝜏)𝑠0𝜂𝜉𝑥(𝑡𝜏)𝑠0𝜂̆̆𝑡+̆=Δ̆𝑦𝑘𝑒̆Δ̆𝜏̆̆.̆𝑦𝑡̆𝜏̆𝑥𝑡̆𝜏(2.2) With this change of variables, omitting the ̆’s for convenience, system (2.1) becomes

̇𝑠(𝑡)=1𝑠(𝑡)𝛼𝑥(𝑡)𝑠(𝑡),̇𝑥(𝑡)=𝑥(𝑡)(𝐷+𝛼𝑠(𝑡))𝑘𝑦(𝑡)𝑥(𝑡),̇𝑦(𝑡)=Δ𝑦(𝑡)+𝑘𝑒Δ𝜏𝑦(𝑡𝜏)𝑥(𝑡𝜏),(2.3) where Δ1 and 𝐷1, 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 1𝐷/𝛼Δ𝐷/𝑘<0, the model has a second equilibrium point (𝐷/𝛼,(𝛼𝐷)/𝛼𝐷,0) and it is globally asymptotically stable. When 1𝐷/𝛼Δ𝐷/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))int+×([𝜏,0],int2+), 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 𝑡𝜏[𝜏,0],𝑥(𝑡𝜏)=𝜙(𝑡𝜏), and 𝑦(𝑡𝜏)=𝜓(𝑡𝜏). System (2.3) becomes ̇𝑠(𝑡)=1𝑠(𝑡)𝛼𝑥(𝑡)𝑠(𝑡),̇𝑥(𝑡)=𝑥(𝑡)(𝐷+𝛼𝑠(𝑡))𝑘𝑦(𝑡)𝑥(𝑡),̇𝑦(𝑡)=Δ𝑦(𝑡)+𝑘𝑒Δ𝜏𝜙(𝑡𝜏)𝜓(𝑡𝜏),(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 𝑡0.
Now we prove 𝑠(𝑡)>0 for all 𝑡>0. From the first equation of (2.3),
̇𝑠(𝑡)=1𝑠(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) 𝑡̇𝑠𝑡=1𝑠𝑡𝛼𝑥𝑠𝑡=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
𝑡̇𝑦𝑡=Δ𝑦+𝑘𝑒Δ𝜏𝑦𝑡𝑥𝑡𝜏𝜏=𝑘𝑒Δ𝜏𝑦𝑡𝑥𝑡𝜏𝜏>0,(2.8) a contradiction.
To prove the boundedness of solutions, define
𝜔(𝑡)=𝑠(𝑡)+𝑥(𝑡)+𝑒Δ𝜏𝑦(𝑡+𝜏)1,for𝑡0.(2.9) It follows that ̇𝜔(𝑡)=1𝑠(𝑡)𝐷𝑥(𝑡)Δ𝑒Δ𝜏𝑦(𝑡+𝜏)1𝑠(𝑡)𝑥(𝑡)𝑒Δ𝜏𝑦(𝑡+𝜏)𝜔(𝑡),(2.10) where the first inequality holds since 𝐷1, Δ1, 𝑥(𝑡)>0 and 𝑦(𝑡+𝜏)>0. It follows that 𝑠(𝑡)+𝑥(𝑡)+𝑒Δ𝜏𝑠𝑦(𝑡+𝜏)1+0+𝑥(0)+𝑒Δ𝜏𝑒𝑦(𝜏)1𝑡1as𝑡.(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,(𝛼𝐷)/𝛼𝐷,0), and

𝐸+=𝑠+(𝜏),𝑥+(𝜏),𝑦+=1(𝜏)1+(𝛼Δ/𝑘)𝑒Δ𝜏,Δ𝑘𝑒Δ𝜏,𝛼𝑘+𝛼Δ𝑒Δ𝜏𝐷𝑘.(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=1Δ𝑘lnΔ1𝐷1𝛼.(3.2) Note that if (𝑘/Δ)(1/𝐷1/𝛼)1, the equilibrium 𝐸+ does not exist for any 𝜏 (0), and if (𝑘/Δ)(1/𝐷1/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)1𝛼𝑥𝛼𝑠0𝛼𝑥𝐷+𝛼𝑠𝑘𝑦𝑘𝑥00Δ1𝑧(𝑡)2𝑧(𝑡)3+(𝑡)0000000𝑘𝑒Δ𝜏𝑦𝑘𝑒Δ𝜏𝑥𝑧1𝑧(𝑡𝜏)2𝑧(𝑡𝜏)3.(𝑡𝜏)(3.3) The associated characteristic equation is given by

det1𝛼𝑥𝜆𝛼𝑠0𝛼𝑥𝐷+𝛼𝑠𝑘𝑦𝜆𝑘𝑥0𝑘𝑒Δ𝜏𝜆𝜏𝑦Δ+𝑘𝑒Δ𝜏𝜆𝜏𝑥𝜆=0.(3.4) Direct calculation of the left-hand side of (3.4) gives

Δ+𝑘𝑒(Δ+𝜆)𝜏𝑥𝜆(1𝛼𝑥𝜆)(𝐷+𝛼𝑠𝑘𝑦𝜆)+𝛼2𝑠𝑥+𝑘𝑥𝑘𝑒(Δ+𝜆)𝜏𝑦((1𝛼𝑥𝜆)=(Δ𝜆)1+𝛼𝑥+𝜆)(𝐷𝛼𝑠+𝑘𝑦+𝜆)+𝛼2𝑠𝑥+𝑒(Δ+𝜆)𝜏×𝑘𝑥𝑘𝑦(1𝛼𝑥𝜆)+(1+𝛼𝑥+𝜆)(𝐷𝛼𝑠+𝑘𝑦+𝜆)+𝛼2=𝑠𝑥(Δ𝜆)(1+𝛼𝑥+𝜆)(𝐷𝛼𝑠+𝑘𝑦+𝜆)+𝛼2𝑠𝑥+𝑒(Δ+𝜆)𝜏𝑘𝑥(1+𝛼𝑥+𝜆)(𝐷𝛼𝑠+𝜆)+𝛼2=𝑠𝑥(Δ𝜆){(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)=(Δ𝜆){(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+𝐷𝛼)=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=𝑘𝐷1𝛼.(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 (𝛼𝐷)/𝛼𝐷=(1𝑠)/𝛼𝑠 at 𝐸2, ||𝑃(𝜆)𝐸2×={(𝜆+1)(𝜆+𝐷)+𝛼𝑥(𝜆+𝐷)𝛼𝑠(𝜆+1)}𝜆Δ+𝑒(Δ+𝜆)𝜏=𝑘𝑥(𝜆+1)(𝜆+𝐷)+1𝑠𝑠×(𝜆+𝐷)𝐷(𝜆+1)𝜆Δ+𝑒(Δ+𝜆)𝜏𝑘𝛼𝐷=𝛼𝐷𝜆(𝜆+1)(𝜆+𝐷)+𝜆+𝐷𝑠𝜆Δ+𝑘𝛼𝐷𝑒𝛼𝐷(Δ+𝜆)𝜏𝜆=2+𝛼𝐷𝜆+𝛼𝐷𝜆+Δ𝑘𝛼𝐷𝑒𝛼𝐷(Δ+𝜆)𝜏=𝑒(Δ+𝜆)𝜏𝜆𝜆1𝜆𝜆2(𝜆+Δ)𝑒(Δ+𝜆)𝜏1𝑘𝐷1𝛼=0,(3.9) where 𝜆1+𝜆2=𝛼/𝐷 and 𝜆1𝜆2=𝛼𝐷>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=𝑘𝐷1𝛼.(3.10) Solving for 𝜏 gives 1𝜏=Δ𝑘lnΔ1𝐷1𝛼=𝜏𝑐.(3.11)

Theorem 3.4. Assume that 𝐷1, Δ1, 𝑘>0, 𝛼>0, and (𝑘/Δ)(1/𝐷1/𝛼)1 so that 𝜏𝑐0. 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, Δ1, and (𝑘/Δ)(1/𝐷1/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 𝜆Δ. 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/𝐷1/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))=𝑘𝐷1𝛼𝑒𝛾𝜏.(3.12) Equating the real parts and imaginary parts of the equation, we have 1𝛾cos(𝛽𝜏)𝛽sin(𝛽𝜏)=𝑘𝐷1𝛼𝑒𝛾𝜏𝛾sin(𝛽𝜏)+𝛽cos(𝛽𝜏)=0.(3.13) Squaring both equations, adding, and taking the square root on both sides give 𝛾2+𝛽2𝑒𝛾𝜏1=𝑘𝐷1𝛼.(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}=𝛾Δ<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 𝑘𝑒Δ𝜏<Δ. In this case, choose 𝜖0>0 small enough such that 𝑘𝑒Δ𝜏(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, 𝑘𝑒Δ𝜏𝑥(𝑡𝜏)<Δ. In Example 5.1 of Kuang ([10, page 32]), choose 𝜌(𝑡)=𝜏, 𝑎(𝑡)=Δ, 𝑏(𝑡)=𝑘𝑒Δ𝜏𝑥(𝑡𝜏), and 𝛼=Δ/2. We obtain (𝑘𝑒Δ𝜏(1+𝜖0))2<Δ2=4(Δ𝛼)𝛼. Therefore 𝑦(𝑡)0 as 𝑡. Let 𝑧(𝑡)=𝑠(𝑡)+𝑥(𝑡). Noting 𝐷=1, from (2.3), we have ̇𝑧(𝑡)=1𝑧(𝑡)𝑘𝑥(𝑡)𝑦(𝑡). Multiply by the integrating factor 𝑒𝑡, (𝑧(𝑡)𝑒𝑡)=𝑒𝑡(1𝑘𝑥(𝑡)𝑦(𝑡)). Integrating both sides from 0 to 𝑡 gives
𝑧(𝑡)=𝑒𝑡𝑧(0)+𝑒𝑡𝑒𝑡1𝑒𝑡𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠=1+𝑒𝑡(𝑧(0)1)𝑒𝑡𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠.(3.15) If lim𝑡𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠<, then lim𝑡𝑒𝑡𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠=0. Therefore lim𝑡𝑧(𝑡)=1. If lim𝑡𝑡0𝑒𝑠𝑘𝑥(𝑠)𝑦(𝑠)d𝑠=, by L'Hôspital's rule, lim𝑡+𝑒𝑡𝑡0𝑒𝑠𝑘𝑥(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 lim𝑡𝑦(𝑡)=0. It again follows that limt𝑧(𝑡)=1. Hence lim𝑡𝑠(𝑡)+𝑥(𝑡)=1.(3.17)
We show that lim𝑡𝑠(𝑡)=1/𝛼 and lim𝑡𝑥(𝑡)=(𝛼1)/𝛼. First assume that the limits exist, that is, lim𝑡𝑠(𝑡)=𝑠 and lim𝑡𝑥(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 lim𝑡̇𝑠(𝑡)=0 and lim𝑡̇𝑥(𝑡)=0. Note that lim𝑡𝑦(𝑡)=0. Letting 𝑡 in (2.3) gives
1𝑠𝛼𝑥𝑠=0,𝑥1+𝛼𝑠=0.(3.18) Either (𝑠,𝑥)=(1,0) or (𝑠,𝑥)=(1/𝛼,(𝛼1)/𝛼). Assume that (𝑠,𝑥)=(1,0), that is, lim𝑡𝑠(𝑡)=1 and lim𝑡𝑥(𝑡)=0. Note that 𝛼>𝐷. There exists 𝜖>0 such that 𝛼𝐷(𝛼+𝑘)𝜖>0. For such 𝜖, there exists a sufficiently large 𝑡 so that 𝑠(𝑡)>1𝜖 and 0<𝑦(𝑡)<𝜖. Recalling that 𝑥(𝑡)>0, by (2.3) ̇𝑥(𝑡)>𝑥(𝑡)(𝐷+𝛼(1𝜖)𝑘𝜖)=𝑥(𝑡)(𝛼𝐷𝛼𝜖𝑘𝜖)>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,(𝛼1)/𝛼).
Now suppose that the limits do not exist. In particular if 𝑥(𝑡) does not converge, then let 𝑥=limsup𝑡𝑥(𝑡) and 𝑥=liminf𝑡𝑥(𝑡). By Lemma A.2 in the appendix, there exists {𝑡𝑚} and {𝑠𝑚} such that
lim𝑚𝑥𝑡𝑚=𝑥,lim𝑚𝑡̇𝑥𝑚=0,lim𝑚𝑥𝑠𝑚=𝑥lim𝑚𝑠̇𝑥𝑚=0.(3.20) From (2.3), 𝑥𝑡𝑚𝑡𝐷+𝛼𝑠𝑚𝑡+𝑘𝑦𝑚=0.(3.21) Noting that 𝑥(𝑡𝑚)>0, we have 𝑠(𝑡𝑚)=(1𝑘𝑦(𝑡𝑚))/𝛼. Since lim𝑡𝑦(𝑡)=0, lim𝑡𝑠(𝑡𝑚)=1/𝛼. By (3.17), lim𝑡𝑥(𝑡𝑚)=lim𝑡(𝑥(𝑡𝑚)+𝑠(𝑡𝑚))𝑠(𝑡𝑚)=11/𝛼=(𝛼1)/𝛼. Therefore 𝑥=(𝛼1)/𝛼. Similarly we can show that 𝑥=(𝛼1)/𝛼. This implies that lim𝑡𝑥(𝑡)=(𝛼1)/𝛼, a contradiction.
Since 𝑠(𝑡)+𝑥(𝑡) converges and 𝑥(𝑡) converges, then 𝑠(𝑡) must also converge. Hence lim𝑡𝑠(𝑡)=1/𝛼 and lim𝑡𝑥(𝑡)=(𝛼1)/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+=(Δ𝜆)1+𝛼𝑥+(𝜏)+𝜆𝐷𝛼𝑠+(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)=(Δ𝜆)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)=(Δ𝜆)𝑠+(𝜏)+𝜆𝜆+𝛼1𝑠+(𝜏)+𝑒𝜆𝜏Δ(𝜆+1)(𝜆+𝐷)+1𝑠+(𝜏)𝑠+(𝜏)(𝜆+𝐷)𝛼𝑠+𝜆(𝜏)(𝜆+1)=(Δ𝜆)2+𝜆𝑠+(𝜏)+𝛼1𝑠+(𝜏)+Δ𝑒𝜆𝜏1𝜆+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)+𝛼1𝑠+(𝜏)+𝑒𝜆𝜏𝜆+𝛼𝑠+1(𝜏)𝑠+𝜆(𝜏)=(𝜆+1)2+𝑝(𝜏)𝜆+𝛽(𝜏)+𝑒𝜆𝜏(𝑞𝜆+𝑐(𝜏))=0,(4.2) where

1𝑝(𝜏)=𝑠+(𝜏),𝛽(𝜏)=𝛼1𝑠+(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


Lemma 4.1. Assuming 𝑘>0, 𝛼>0, and 𝑘(11/𝛼)1 so that 𝜏𝑐=ln(𝑘(11/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)=𝛼𝑠+(𝜏)=0.(4.5) Noting that 𝜏𝑐>0 if and only if 𝑘(11/𝛼)>1, for any 0<𝜏<𝜏𝑐, 1𝛼𝑠+𝛼(𝜏)=𝛼1𝑘𝑒𝜏1>𝛼1𝛼1𝛼=0,(4.6) a contradiction.

Lemma 4.2. Assume 𝑘>0, 𝛼>0, 𝑘(11/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+𝛼𝑠+(0).(4.7) Both coefficients are positive, since 1𝑠+𝛼(0)1=𝑘1>0,𝛼𝑠+𝛼(0)=𝛼1𝑘1=𝛼1𝛼1𝑘>0,(4.8) and 𝑘(11/𝛼)>1 implies 11/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))𝑒𝜆𝜏+𝛽(𝜏) in Kuang [10, Theorem 1.4, page 66] gives limsupRe𝜆>0,|𝜆|||𝜆2𝑔||(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 𝜏0, assuming 𝜆=𝑖𝜔 (𝜔>0) is a root of 𝑃(𝜆)|𝐸+=0,
𝜔2+𝑖𝑝(𝜏)𝜔+𝛽(𝜏)+𝑒𝑖𝜔𝜏(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)=𝜔2𝛽(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+𝑞𝜔2𝛽(𝜏)𝑐(𝜏)2+𝑞2𝜔2,𝜔cos(𝜔𝜏)=𝑐(𝜏)2𝛽(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)𝑞2𝜔2𝛽(𝜏)2+𝛽2(𝜏)𝑐2(𝜏)=0.(4.14) Solving for 𝜔, we obtain two roots 𝜔1(𝜏) and 𝜔2(𝜏): 𝜔11(𝜏)=2𝑞2𝑝2(𝜏)+2𝛽(𝜏)+𝑞2𝑝2(𝜏)+2𝛽(𝜏)24(𝛽2(𝜏)𝑐2(𝜏))1/2=1𝑠+(𝜏)21𝑠+(𝜏)2𝛼𝑠2+(𝜏)𝑠+(+𝜏)1𝑠2+(𝜏)12+4𝛼𝑠2+𝑠(𝜏)2+(𝜏)11𝑠+(𝜏)+4𝑠2+(𝜏)𝛼𝑠2+(𝜏)121/2𝜔21(𝜏)=2𝑞2𝑝2(𝜏)+2𝛽(𝜏)𝑞2𝑝2(𝜏)+2𝛽(𝜏)2𝛽42(𝜏)𝑐2(𝜏)1/2=1𝑠+(𝜏)21𝑠+(𝜏)2𝛼𝑠2+(𝜏)𝑠+(𝜏)1𝑠2+(𝜏)12+4𝛼𝑠2+𝑠(𝜏)2+(𝜏)11𝑠+(𝜏)+4𝑠2+(𝜏)𝛼𝑠2+(𝜏)121/2.(4.15)
Define conditions (𝐻1) and (𝐻2) as follows:

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 𝜏𝐼 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 𝜏𝐼. If no interval exists where either (𝐻1) or (𝐻2) holds, then there are no positive real roots of (4.14).

Define the interval

𝑘𝐽=ln𝛼11/4+𝑘1/16+1/(2𝛼)1,ln4𝛼1𝛼.(4.18) When the end points of 𝐽 are real and 𝐽, define

𝐼1=0,𝜏𝑐𝐽.(4.19) We prove that (𝐻1) holds for any 𝜏𝐼1.

From 𝐷=Δ=1,

𝜏𝑐=1Δ𝑘lnΔ1𝐷1𝛼=ln𝑘(𝛼1)𝛼.(4.20) If 𝛼>1, then 𝛼>4𝛼. It follows that

𝜏𝑐𝑘>ln4𝛼1𝛼.(4.21) Therefore,


Theorem 4.5. Assume 𝛼>(7+35)/2 and 𝑘>𝛼/(4𝛼1), then 𝐼1 is not empty, and for any 𝜏𝐼1, but 𝜏ln((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 11/4𝛼>0, and therefore 14𝛼+1252<147+35+1/2252=0.(4.23) Hence, 14𝛼1421+1162𝛼2=1𝛼124𝛼1=2𝛼124𝛼14𝛼32+14𝛼=124𝛼114𝛼14𝛼2+14𝛼=1124𝛼114𝛼14𝛼+12254=124𝛼114𝛼14𝛼+125214𝛼+12+52<0.(4.24) Therefore, 1/4𝛼1/4<1/16+1/(2𝛼). Since 1/4+1/16+1/(2𝛼)<1/4+1/16+1/(7+35)<1, it follows that 14𝛼<14+1+1162𝛼<1.(4.25) Hence, 𝑘ln𝛼11/4+𝑘1/16+1/(2𝛼)1<ln4𝛼1𝛼.(4.26) From 𝑘>𝛼/(4𝛼1), we have ln(𝑘(4𝛼1)/𝛼)>0. Therefore, 𝑘max0,ln𝛼11/4+𝑘1/16+1/(2𝛼)1<ln4𝛼1𝛼,(4.27) and so 𝐼1 is not empty. Noting 𝑠+(𝜏)=1/(1+(𝛼Δ/𝑘)𝑒Δ𝜏) and Δ=1, for any 𝜏𝐼1, but 𝜏ln((𝑘/𝛼)(1/(1/4+1/16+1/(2𝛼))1)), we have 𝑠+(𝜏)[1/4𝛼,1/4+1/16+1/(2𝛼)).
In what follows, we intend to show that for any such 𝜏, condition (𝐻1) holds. From (4.3),
𝑞2𝑝2(𝜏)+2𝛽(𝜏)=(1)21𝑠2+(𝜏)+2𝛼1𝑠+=(𝜏)1𝑠+(𝜏)2𝛼𝑠2+𝑠(𝜏)+(𝜏)2𝑠+(𝜏)12𝛼=2𝛼1𝑠+(𝜏)2𝛼𝑠2+𝑠(𝜏)+1(𝜏)4𝛼2116𝛼21.2𝛼(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/𝛼1/(4𝛼)=(1/𝛼)(11/(4𝛼))>0, and 116𝛼2+12𝛼21𝛼14𝛼2=116𝛼2+112𝛼𝛼+12𝛼𝛼116𝛼2=112𝛼𝛼1<0.(4.29) Since 1/𝛼<1/4𝛼 for 𝛼>(7+35)/2, 1+4𝛼116𝛼2+1<12𝛼𝛼<14𝛼.(4.30) For any 𝑠+(𝜏)>1/4𝛼, 𝑠+1(𝜏)14𝛼4𝛼1>4𝛼116𝛼2+1.2𝛼(4.31) Hence, 𝑠+1(𝜏)4𝛼2116𝛼2+12𝛼.(4.32) Next consider the second inequality in (𝐻1). For 𝛼>(7+35)/2, since 1/4𝛼>1/𝛼, 𝑠+(𝜏)1/4𝛼>1/𝛼. Therefore, 𝛼𝑠+(𝜏)>1. For 𝑠+(𝜏)[1/4𝛼,1/4+1/16+1/(2𝛼))𝛽2(𝜏)𝑐2=(𝜏)=(𝛽(𝜏)𝑐(𝜏))(𝛽(𝜏)+𝑐(𝜏))𝛼2𝛼𝑠+1(𝜏)+𝑠+(1𝜏)𝛼𝑠+(𝜏)=2𝛼𝑠2+𝑠(𝜏)2+𝑠(𝜏)+(𝜏)212𝛼𝛼𝑠+(𝜏)1=2𝛼𝑠2+𝑠(𝜏)+1(𝜏)4211162𝛼𝛼𝑠+(𝜏)1>0.(4.33) Finally, 𝑞2𝑝2(𝜏)+2𝛽(𝜏)2𝛽42(𝜏)𝑐2=𝑞(𝜏)2𝑝2𝑞(𝜏)2𝑝2(𝜏)+4𝛽(𝜏)+4𝑐2=1(𝜏)1𝑠2+(1𝜏)1𝑠2+(𝜏)+4𝛼1𝑠+(𝜏)+4𝛼𝑠+1(𝜏)𝑠+(𝜏)2=11𝑠2+(𝜏)21+4𝛼1𝑠2+(𝜏)1𝑠+(𝜏)+4𝑠+(𝜏)2𝛼248𝛼+𝑠2+(𝜏)=4𝑠+(𝜏)2𝛼21+4𝛼1𝑠2+(𝜏)1𝑠+1(𝜏)8𝛼+1𝑠2+(𝜏)2+4𝑠2+(𝜏)=4𝑠+(𝜏)2𝛼21+4𝛼1𝑠2+(𝜏)1𝑠++1(𝜏)21+𝑠2+(𝜏)2=𝛼𝛼1𝛼𝛼2,(4.34) where 𝛼1=211/𝑠2+(𝜏)1𝑠++(𝜏)1/𝑠2+(𝜏)+2𝑠+(𝜏)+11/𝑠+(𝜏)122𝑠2+,𝛼(𝜏)2=211/𝑠2+(𝜏)1𝑠+(𝜏)1/𝑠2+(𝜏)+2𝑠+(𝜏)+11/𝑠+(𝜏)122𝑠2+.(𝜏)(4.35) Since 𝑠+(𝜏)<1, 121𝑠2+(𝜏)1𝑠+(𝜏)=𝑠+(𝜏)+1+1𝑠+(𝜏)𝑠2+1(𝜏)>0,21𝑠2+(𝜏)1𝑠+(𝜏)2>121𝑠2+(𝜏)1𝑠+(𝜏)2𝑠2+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<211/𝑠2+(𝜏)1𝑠++(𝜏)1/𝑠2+(𝜏)+2/𝑠+(𝜏)+11/𝑠+(𝜏)122𝑠2+=(𝜏)211/𝑠2+(𝜏)1𝑠++(𝜏)1/𝑠+(𝜏)+11/𝑠+(𝜏)12𝑠2+=(𝜏)21𝑠+(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)>1/4𝛼, we have 𝛼>1/𝑠4+(𝜏)>𝛼1>𝛼2. This leads to 𝑞2𝑝2(𝜏)+2𝛽(𝜏)2𝛽42(𝜏)𝑐2=(𝜏)𝛼𝛼1𝛼𝛼2>0.(4.38) Therefore, (𝐻1) holds for any 𝜏𝐼1. 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
𝑞2𝑝2(𝜏)+2𝛽(𝜏)>0,𝛽2(𝜏)𝑐2𝑞(𝜏)=0,2𝑝2(𝜏)+2𝛽(𝜏)2𝛽42(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,𝜏𝑐𝑘,ln𝛼11/4+1/16+1/(2𝛼)1.(4.40) In the following theorem, we consider the case that parameters are chosen so that


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 𝜏𝐼2, (𝐻2) holds and hence 𝜔1(𝜏)>0.

Proof. Assume 𝛼>1. Letting 1𝐺(𝛼)=4+1+11612𝛼𝛼,(4.42) then d1d𝛼𝐺(𝛼)=4𝛼2+11/16+1/(2𝛼)𝛼2=1𝛼211+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 1𝛼1>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(𝛼1)𝛼>𝑘𝛼11/4+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(𝛼1)𝛼