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(𝑑)=0ξ€Έπ·βˆ’π‘ (𝑑)0βˆ’π‘₯(𝑑)𝑓(𝑠(𝑑))πœ‚,𝑦̇π‘₯(𝑑)=π‘₯(𝑑)(βˆ’π·+𝑓(𝑠(𝑑)))βˆ’(𝑑)𝑔(π‘₯(𝑑))πœ‰,̇𝑦(𝑑)=βˆ’Ξ”π‘¦(𝑑)+𝑦(𝑑)𝑔(π‘₯(𝑑)).(1.1) Here 𝑠(𝑑) represents the concentration of the growth limiting nutrient, π‘₯(𝑑) the density of the prey population, and 𝑦(𝑑) the density of the predator population. Parameter 𝑠0 denotes the concentration of the growth limiting nutrient in the feed vessel, 𝐷0 the dilution rate, πœ‚(πœ‰) the growth yield constant, 𝐷(Ξ”) 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(π‘₯) 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(π‘₯) 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=Ξ”. 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(π‘₯) 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(π‘₯) 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(π‘₯), 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(𝑑)=0ξ€Έπ·βˆ’π‘ (𝑑)0βˆ’π‘₯(𝑑)𝑓(𝑠(𝑑))πœ‚,𝑦̇π‘₯(𝑑)=π‘₯(𝑑)(βˆ’π·+𝑓(𝑠(𝑑)))βˆ’(𝑑)𝑔(π‘₯(𝑑))πœ‰,̇𝑦(𝑑)=βˆ’Ξ”π‘¦(𝑑)+π‘’βˆ’Ξ”πœπ‘¦(π‘‘βˆ’πœ)𝑔(π‘₯(π‘‘βˆ’πœ)).(1.2) For π‘‘βˆˆ[βˆ’πœ,0],

𝑠(0)=𝑠0∈intℝ+ξ€·[],(π‘₯(𝑑),𝑦(𝑑))=(πœ™,πœ“)βˆˆβ„‚βˆ’πœ,0,intℝ2+ξ€Έ.(1.3) Here variables 𝑠(𝑑), π‘₯(𝑑), 𝑦(𝑑), and parameters 𝑠0, 𝐷0, πœ‚, πœ‰, 𝐷0, 𝐷, and Ξ” 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, π‘’βˆ’Ξ”πœπ‘¦(π‘‘βˆ’πœ) represents the concentration of the predator population in the growth chamber at time 𝑑 that were available at time π‘‘βˆ’πœ 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(π‘₯)=π‘˜π‘₯ (π‘˜>0). System (1.2) reduces to

𝑠̇𝑠(𝑑)=0ξ€Έπ·βˆ’π‘ (𝑑)0βˆ’π›Όπ‘₯(𝑑)𝑠(𝑑)πœ‚,Μ‡π‘₯(𝑑)=π‘₯(𝑑)(βˆ’π·+𝛼𝑠(𝑑))βˆ’π‘˜π‘₯(𝑑)𝑦(𝑑)πœ‰,̇𝑦(𝑑)=βˆ’Ξ”π‘¦(𝑑)+π‘˜π‘’βˆ’Ξ”πœπ‘¦(π‘‘βˆ’πœ)π‘₯(π‘‘βˆ’πœ).𝑑>0,(2.1) Introducing the following change of variables gives:

̆𝑑=𝐷0̆𝑑=𝑑,̆𝑠𝑠(𝑑)𝑠0̆𝑑=,Μ†π‘₯π‘₯(𝑑)𝑠0πœ‚ξ€·Μ†π‘‘ξ€Έ=,̆𝑦𝑦(𝑑)πœ‰π‘ 0πœ‚,Μ†πœ=𝐷0Μ†π·πœ,𝐷=𝐷0,̆ΔΔ=𝐷0,Μ†π‘˜=π‘˜π‘ 0πœ‚π·0,̆𝛼=𝛼𝑠0𝐷0,̆𝑑d̆𝑠d̆𝑑=1𝑠0d𝑠(𝑑)d𝑑d𝑑d̆𝑑=1𝑠0𝐷0d𝑠(𝑑)=1d𝑑𝑠0𝐷0𝑠0ξ€Έπ·βˆ’π‘ (𝑑)0βˆ’π›Όπ‘₯(𝑑)𝑠(𝑑)πœ‚ξ‚Ά=1βˆ’π‘ (𝑑)𝑠0βˆ’π›Όπ‘ 0𝐷0π‘₯(𝑑)𝑠0πœ‚π‘ (𝑑)𝑠0̆𝑑̆𝑑̆𝑑,̆𝑑=1βˆ’Μ†π‘ βˆ’Μ†π›ΌΜ†π‘₯̆𝑠dΜ†π‘₯d̆𝑑=1𝑠0πœ‚dπ‘₯(𝑑)d𝑑d𝑑d̆𝑑=1𝑠0πœ‚π·0dπ‘₯(𝑑)=1d𝑑𝑠0πœ‚π·0ξ‚΅π‘₯(𝑑)(βˆ’π·+𝛼𝑠(𝑑))βˆ’π‘˜π‘₯(𝑑)𝑦(𝑑)πœ‰ξ‚Ά=π‘₯(𝑑)𝑠0πœ‚ξ‚΅βˆ’π·π·0+𝛼𝑠0𝐷0𝑠(𝑑)𝑠0ξ‚Άβˆ’π‘˜π‘ 0πœ‚π·0π‘₯(𝑑)𝑠0πœ‚π‘¦(𝑑)𝑠0ξ€·Μ†π‘‘βˆ’Μ†ξ€·Μ†π‘‘βˆ’Μ†ξ€·Μ†π‘‘ξ€Έξ€·Μ†π‘‘ξ€Έ,ξ€·Μ†π‘‘ξ€Έπœ‚πœ‰=Μ†π‘₯𝐷+Μ†π›ΌΜ†π‘ ξ€Έξ€Έπ‘˜Μ†π‘₯̆𝑦d̆𝑦d̆𝑑=1𝑠0πœ‚πœ‰d𝑦(𝑑)d𝑑d𝑑d̆𝑑=1𝑠0πœ‚πœ‰π·0d𝑦(𝑑)=1d𝑑𝑠0πœ‚πœ‰π·0ξ€·βˆ’Ξ”π‘¦(𝑑)+π‘˜π‘’βˆ’Ξ”πœξ€Έ=𝑦(π‘‘βˆ’πœ)π‘₯(π‘‘βˆ’πœ)βˆ’Ξ”π‘¦(𝑑)𝑠0πœ‚πœ‰π·0+π‘˜π‘’βˆ’Ξ”πœπ‘ 0πœ‚πœ‰π·0=𝑦(π‘‘βˆ’πœ)π‘₯(π‘‘βˆ’πœ)βˆ’Ξ”π·0𝑦(𝑑)𝑠0+πœ‚πœ‰π‘˜π‘ 0πœ‚π·0π‘’βˆ’(Ξ”/𝐷0)𝐷0πœπ‘¦(π‘‘βˆ’πœ)𝑠0πœ‚πœ‰π‘₯(π‘‘βˆ’πœ)𝑠0πœ‚Μ†ξ€·Μ†π‘‘ξ€Έ+Μ†=βˆ’Ξ”Μ†π‘¦π‘˜π‘’βˆ’Μ†Ξ”Μ†πœξ€·Μ†ξ€Έξ€·Μ†ξ€Έ.Μ†π‘¦π‘‘βˆ’Μ†πœΜ†π‘₯π‘‘βˆ’Μ†πœ(2.2) With this change of variables, omitting the Μ†β€Œβ€™s for convenience, system (2.1) becomes

̇𝑠(𝑑)=1βˆ’π‘ (𝑑)βˆ’π›Όπ‘₯(𝑑)𝑠(𝑑),Μ‡π‘₯(𝑑)=π‘₯(𝑑)(βˆ’π·+𝛼𝑠(𝑑))βˆ’π‘˜π‘¦(𝑑)π‘₯(𝑑),̇𝑦(𝑑)=βˆ’Ξ”π‘¦(𝑑)+π‘˜π‘’βˆ’Ξ”πœπ‘¦(π‘‘βˆ’πœ)π‘₯(π‘‘βˆ’πœ),(2.3) where Ξ”β©Ύ1 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 1βˆ’π·/π›Όβˆ’Ξ”π·/π‘˜<0, the model has a second equilibrium point (𝐷/𝛼,(π›Όβˆ’π·)/𝛼𝐷,0) and it is globally asymptotically stable. When 1βˆ’π·/π›Όβˆ’Ξ”π·/π‘˜>0, the model has a third equilibrium point (π‘˜/(π‘˜+𝛼Δ),Ξ”/π‘˜,𝛼/(π‘˜+𝛼𝐷)βˆ’π·/π‘˜) and it is the global attractor. Therefore, model (2.3) has no periodic solutions when the time delay is ignored. If 𝑔(π‘₯) 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,πœ™(πœƒ),πœ“(πœƒ))∈intℝ+Γ—β„‚([βˆ’πœ,0],intℝ2+), then there exists a unique solution (𝑠(𝑑),π‘₯(𝑑),𝑦(𝑑)) of (2.3) passing through (𝑠0,πœ™(πœƒ),πœ“(πœƒ)) with 𝑠(𝑑)>0, π‘₯(𝑑)>0 and 𝑦(𝑑)>0 for π‘‘βˆˆ[0,∞). The solution is bounded. In particular, given any πœ–0>0, π‘₯(𝑑)<1+πœ–0 for all sufficiently large 𝑑.

Proof. For π‘‘βˆˆ[0,𝜏], one has π‘‘βˆ’πœβˆˆ[βˆ’πœ,0],π‘₯(π‘‘βˆ’πœ)=πœ™(π‘‘βˆ’πœ), and 𝑦(π‘‘βˆ’πœ)=πœ“(π‘‘βˆ’πœ). System (2.3) becomes ̇𝑠(𝑑)=1βˆ’π‘ (𝑑)βˆ’π›Όπ‘₯(𝑑)𝑠(𝑑),Μ‡π‘₯(𝑑)=π‘₯(𝑑)(βˆ’π·+𝛼𝑠(𝑑))βˆ’π‘˜π‘¦(𝑑)π‘₯(𝑑),̇𝑦(𝑑)=βˆ’Ξ”π‘¦(𝑑)+π‘˜π‘’βˆ’Ξ”πœπœ™(π‘‘βˆ’πœ)πœ“(π‘‘βˆ’πœ),(2.4) a system of nonautonomous ordinary differential equations with initial conditions 𝑠(0)=𝑠0, π‘₯(0)=πœ™(0), and 𝑦(0)=πœ“(0). Since the right-hand side of (2.4) is differentiable in both π‘₯ 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,πœ™(πœƒ),πœ“(πœƒ)) is defined for all 𝑑⩾0.
Now we prove 𝑠(𝑑)>0 for all 𝑑>0. From the first equation of (2.3),
̇𝑠(𝑑)=1βˆ’π‘ (𝑑)βˆ’π›Όπ‘₯(𝑑)𝑠(𝑑).(2.5) Proceed using the method of contradiction. Suppose that there exists a first 𝑑⋆ such that 𝑠(𝑑⋆)=0 and 𝑠(𝑑)>0 for π‘‘βˆˆ[0,π‘‘βˆ—). Then ̇𝑠(𝑑⋆)β©½0. But from the first equation of (2.3) 𝑑̇𝑠⋆𝑑=1βˆ’π‘ β‹†ξ€Έξ€·π‘‘βˆ’π›Όπ‘₯⋆𝑠𝑑⋆=1>0,(2.6) a contradiction.
To prove π‘₯(𝑑)>0 for π‘‘βˆˆ[0,∞), assume there is a first 𝑑>0 such that π‘₯(𝑑)=0, and π‘₯(𝑑)>0 for π‘‘βˆˆ[0,𝑑). Divide both sides of the second equation of (2.3) by π‘₯(𝑑) and integrate from 0 to 𝑑, to obtain
π‘₯ξ€·π‘‘ξ€Έξƒ©ξ€œ=πœ™(0)exp𝑑0ξƒͺ(βˆ’π·+𝛼𝑠(𝑑)βˆ’π‘˜π‘¦(𝑑))d𝑑>0,(2.7) contradicting π‘₯(𝑑)=0.
To show that 𝑦(𝑑) is positive on [0,∞), suppose that there exists 𝑑⋆>0 such that 𝑦(𝑑⋆)=0, and 𝑦(𝑑)>0 for π‘‘βˆˆ[0,π‘‘βˆ—). Then ̇𝑦(𝑑⋆)β©½0. From the third equation of (2.3), we have
𝑑̇𝑦⋆𝑑=βˆ’Ξ”π‘¦β‹†ξ€Έ+π‘˜π‘’βˆ’Ξ”πœπ‘¦ξ€·π‘‘β‹†ξ€Έπ‘₯ξ€·π‘‘βˆ’πœβ‹†ξ€Έβˆ’πœ=π‘˜π‘’βˆ’Ξ”πœπ‘¦ξ€·π‘‘β‹†ξ€Έπ‘₯ξ€·π‘‘βˆ’πœβ‹†ξ€Έβˆ’πœ>0,(2.8) a contradiction.
To prove the boundedness of solutions, define
πœ”(𝑑)=𝑠(𝑑)+π‘₯(𝑑)+π‘’Ξ”πœπ‘¦(𝑑+𝜏)βˆ’1,for𝑑⩾0.(2.9) It follows that Μ‡πœ”(𝑑)=1βˆ’π‘ (𝑑)βˆ’π·π‘₯(𝑑)βˆ’Ξ”π‘’Ξ”πœπ‘¦(𝑑+𝜏)β©½1βˆ’π‘ (𝑑)βˆ’π‘₯(𝑑)βˆ’π‘’Ξ”πœπ‘¦(𝑑+𝜏)β©½βˆ’πœ”(𝑑),(2.10) where the first inequality holds since 𝐷⩾1, Ξ”β©Ύ1, π‘₯(𝑑)>0 and 𝑦(𝑑+𝜏)>0. It follows that 𝑠(𝑑)+π‘₯(𝑑)+π‘’Ξ”πœξ€·π‘ π‘¦(𝑑+𝜏)β©½1+0+π‘₯(0)+π‘’Ξ”πœξ€Έπ‘’π‘¦(𝜏)βˆ’1βˆ’π‘‘βŸΆ1asπ‘‘βŸΆβˆž.(2.11) Therefore, the solution (𝑠(𝑑),π‘₯(𝑑),𝑦(𝑑)) is bounded, and given any πœ–0>0, π‘₯(𝑑)<1+πœ–0 for all sufficiently large 𝑑.

3. Equilibria and Stability

Model (2.3) has three equilibrium points: 𝐸1=(1,0,0),𝐸2=(𝐷/𝛼,(π›Όβˆ’π·)/𝛼𝐷,0), and

𝐸+=𝑠+(𝜏),π‘₯+(𝜏),𝑦+ξ€Έ=ξ‚΅1(𝜏)1+(𝛼Δ/π‘˜)π‘’Ξ”πœ,Ξ”π‘˜π‘’Ξ”πœ,π›Όπ‘˜+π›ΌΞ”π‘’Ξ”πœβˆ’π·π‘˜ξ‚Ά.(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)βˆ’π·)/π‘˜>0, that is, 𝑠+(𝜏)>𝐷/𝛼, or equivalently, 𝜏 lies between 0 and πœπ‘, where

πœπ‘=1Ξ”ξ‚€π‘˜lnΞ”ξ‚€1π·βˆ’1𝛼.(3.2) Note that if (π‘˜/Ξ”)(1/π·βˆ’1/𝛼)β©½1, the equilibrium 𝐸+ does not exist for any 𝜏 (β©Ύ0), and if (π‘˜/Ξ”)(1/π·βˆ’1/𝛼)=1, then 𝐸+=𝐸2.

The linearization of (2.3) about an equilibrium (𝑠,π‘₯,𝑦) is given by

βŽ‘βŽ’βŽ’βŽ’βŽ£Μ‡π‘§1(𝑑)̇𝑧2(𝑑)̇𝑧3⎀βŽ₯βŽ₯βŽ₯⎦=⎑⎒⎒⎒⎣⎀βŽ₯βŽ₯βŽ₯βŽ¦βŽ‘βŽ’βŽ’βŽ’βŽ£π‘§(𝑑)βˆ’1βˆ’π›Όπ‘₯βˆ’π›Όπ‘ 0𝛼π‘₯βˆ’π·+π›Όπ‘ βˆ’π‘˜π‘¦βˆ’π‘˜π‘₯00βˆ’Ξ”1𝑧(𝑑)2𝑧(𝑑)3⎀βŽ₯βŽ₯βŽ₯⎦+⎑⎒⎒⎒⎣(𝑑)0000000π‘˜π‘’Ξ”πœπ‘¦π‘˜π‘’Ξ”πœπ‘₯⎀βŽ₯βŽ₯βŽ₯βŽ¦βŽ‘βŽ’βŽ’βŽ’βŽ£π‘§1𝑧(π‘‘βˆ’πœ)2𝑧(π‘‘βˆ’πœ)3⎀βŽ₯βŽ₯βŽ₯⎦.(π‘‘βˆ’πœ)(3.3) The associated characteristic equation is given by

⎑⎒⎒⎒⎣detβˆ’1βˆ’π›Όπ‘₯βˆ’πœ†βˆ’π›Όπ‘ 0𝛼π‘₯βˆ’π·+π›Όπ‘ βˆ’π‘˜π‘¦βˆ’πœ†βˆ’π‘˜π‘₯0π‘˜π‘’βˆ’Ξ”πœβˆ’πœ†πœπ‘¦βˆ’Ξ”+π‘˜π‘’βˆ’Ξ”πœβˆ’πœ†πœβŽ€βŽ₯βŽ₯βŽ₯⎦π‘₯βˆ’πœ†=0.(3.4) Direct calculation of the left-hand side of (3.4) gives

ξ€·βˆ’Ξ”+π‘˜π‘’βˆ’(Ξ”+πœ†)𝜏π‘₯βˆ’πœ†ξ€Έξ€½(βˆ’1βˆ’π›Όπ‘₯βˆ’πœ†)(βˆ’π·+π›Όπ‘ βˆ’π‘˜π‘¦βˆ’πœ†)+𝛼2𝑠π‘₯+π‘˜π‘₯π‘˜π‘’βˆ’(Ξ”+πœ†)πœπ‘¦ξ€½((βˆ’1βˆ’π›Όπ‘₯βˆ’πœ†)=(βˆ’Ξ”βˆ’πœ†)1+𝛼π‘₯+πœ†)(π·βˆ’π›Όπ‘ +π‘˜π‘¦+πœ†)+𝛼2𝑠π‘₯+π‘’βˆ’(Ξ”+πœ†)πœΓ—ξ€½π‘˜π‘₯π‘˜π‘¦(βˆ’1βˆ’π›Όπ‘₯βˆ’πœ†)+(1+𝛼π‘₯+πœ†)(π·βˆ’π›Όπ‘ +π‘˜π‘¦+πœ†)+𝛼2ξ€Ύ=𝑠π‘₯(βˆ’Ξ”βˆ’πœ†)(1+𝛼π‘₯+πœ†)(π·βˆ’π›Όπ‘ +π‘˜π‘¦+πœ†)+𝛼2𝑠π‘₯+π‘’βˆ’(Ξ”+πœ†)πœξ€½π‘˜π‘₯(1+𝛼π‘₯+πœ†)(π·βˆ’π›Όπ‘ +πœ†)+𝛼2ξ€Ύ=𝑠π‘₯(βˆ’Ξ”βˆ’πœ†){(πœ†+1)(πœ†+𝐷+π‘˜π‘¦)+𝛼π‘₯(πœ†+𝐷+π‘˜π‘¦)βˆ’π›Όπ‘ (πœ†+1)}+π‘’βˆ’(Ξ”+πœ†)πœπ‘˜π‘₯{(πœ†+1)(πœ†+𝐷)+𝛼π‘₯(πœ†+𝐷)βˆ’π›Όπ‘ (πœ†+1)}.(3.5) For convenience, define 𝑃(πœ†)∢=(βˆ’Ξ”βˆ’πœ†){(πœ†+1)(πœ†+𝐷+π‘˜π‘¦)+𝛼π‘₯(πœ†+𝐷+π‘˜π‘¦)βˆ’π›Όπ‘ (πœ†+1)}+π‘’βˆ’(Ξ”+πœ†)πœπ‘˜π‘₯{(πœ†+1)(πœ†+𝐷)+𝛼π‘₯(πœ†+𝐷)βˆ’π›Όπ‘ (πœ†+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(πœ†)𝐸1=βˆ’(Ξ”+πœ†)(πœ†+1)(πœ†+π·βˆ’π›Ό)=0.(3.7) The eigenvalues βˆ’1 and βˆ’Ξ” are both negative. The third eigenvalue is βˆ’π·+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 (πœ†+Ξ”)𝑒(πœ†+Ξ”)πœξ‚€1=π‘˜π·βˆ’1𝛼.(3.8) In addition, the characteristic equation evaluated at 𝐸2 has zero as an eigenvalue if and only if 𝜏=πœπ‘.

Proof. Assume 𝛼>𝐷. Equilibrium 𝐸2 exists. Consider the characteristic equation at 𝐸2. Since (π›Όβˆ’π·)/𝛼𝐷=(1βˆ’π‘ )/𝛼𝑠 at 𝐸2, ||𝑃(πœ†)𝐸2Γ—ξ€·={(πœ†+1)(πœ†+𝐷)+𝛼π‘₯(πœ†+𝐷)βˆ’π›Όπ‘ (πœ†+1)}βˆ’πœ†βˆ’Ξ”+π‘’βˆ’(Ξ”+πœ†)πœξ€Έ=ξ‚†π‘˜π‘₯(πœ†+1)(πœ†+𝐷)+1βˆ’π‘ π‘ ξ‚‡Γ—ξ‚€(πœ†+𝐷)βˆ’π·(πœ†+1)βˆ’πœ†βˆ’Ξ”+π‘’βˆ’(Ξ”+πœ†)πœπ‘˜π›Όβˆ’π·ξ‚=ξ‚†π›Όπ·πœ†(πœ†+1)βˆ’(πœ†+𝐷)+πœ†+π·π‘ ξ‚‡ξ‚€βˆ’πœ†βˆ’Ξ”+π‘˜π›Όβˆ’π·π‘’π›Όπ·βˆ’(Ξ”+πœ†)πœξ‚ξ‚€πœ†=βˆ’2+π›Όπ·πœ†+π›Όβˆ’π·ξ‚ξ‚€πœ†+Ξ”βˆ’π‘˜π›Όβˆ’π·π‘’π›Όπ·βˆ’(Ξ”+πœ†)πœξ‚=βˆ’π‘’βˆ’(Ξ”+πœ†)πœξ€·πœ†βˆ’πœ†1ξ€Έξ€·πœ†βˆ’πœ†2ξ€Έξ‚€(πœ†+Ξ”)𝑒(Ξ”+πœ†)πœξ‚€1βˆ’π‘˜π·βˆ’1𝛼=0,(3.9) where πœ†1+πœ†2=βˆ’π›Ό/𝐷 and πœ†1πœ†2=π›Όβˆ’π·>0. Therefore, πœ†1 and πœ†2 have negative real parts. The rest of the eigenvalues are roots of (3.8).
Assuming that πœ†=0 is a root of (3.8), we have
Ξ”π‘’Ξ”πœξ‚€1=π‘˜π·βˆ’1𝛼.(3.10) Solving for 𝜏 gives 1𝜏=Ξ”ξ‚€π‘˜lnΞ”ξ‚€1π·βˆ’1𝛼=πœπ‘.(3.11)

Theorem 3.4. Assume that 𝐷⩾1, Ξ”β©Ύ1, π‘˜>0, 𝛼>0, and (π‘˜/Ξ”)(1/π·βˆ’1/𝛼)β©Ύ1 so that πœπ‘β©Ύ0. Equilibrium 𝐸2 is locally asymptotically stable if 𝜏>πœπ‘ and unstable if 𝜏<πœπ‘. If 𝐷=1, then equilibrium 𝐸2 is globally asymptotically stable for 𝜏>(1/Ξ”)ln(π‘˜/Ξ”).

Proof. Assume that 𝜏>πœπ‘. Assumptions π‘˜>0, Ξ”β©Ύ1, and (π‘˜/Ξ”)(1/π·βˆ’1/𝛼)β©Ύ1 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 πœ†β©½βˆ’Ξ”. Otherwise the left-hand side would be less than zero, but the right-hand side would be greater than zero. Assume πœ†>βˆ’Ξ”. The left-hand side of (3.8) is a monotone increasing function in both πœ† and 𝜏, takes value 0 at πœ†=βˆ’Ξ”, and goes to positive infinity as πœ†β†’+∞ or πœβ†’+∞. By Lemma 3.3, when 𝜏=πœπ‘, then πœ†=0 is a solution of (3.8). Thus for 𝜏>πœπ‘, any real root πœ† of (3.8) must satisfy βˆ’Ξ”<πœ†<0.
For any 𝜏=Μƒπœ<πœπ‘, we have (πœ†+Ξ”)𝑒(πœ†+Ξ”)𝜏|𝜏=Μƒπœ,πœ†=0<π‘˜(1/π·βˆ’1/𝛼) and limπœ†β†’+∞(πœ†+Ξ”)𝑒(πœ†+Ξ”)Μƒπœ=+∞. Therefore there exists at least one Μƒπœ†=πœ†>0 such that Μƒ(Μƒπœ,πœ†) is a solution of (3.8). Equilibrium 𝐸2 is unstable if 𝜏<πœπ‘.
In what follows, we prove that if 𝜏>πœπ‘ all complex eigenvalues of (3.8) have negative real parts. Suppose that πœ†+Ξ”=𝛾+𝑖𝛽(𝛽>0) is a solution of (3.8). Using the Euler formula, we have
ξ‚€1𝛾cos(π›½πœ)βˆ’π›½sin(π›½πœ)+𝑖(𝛾sin(π›½πœ)+𝛽cos(π›½πœ))=π‘˜π·βˆ’1π›Όξ‚π‘’βˆ’π›Ύπœ.(3.12) Equating the real parts and imaginary parts of the equation, we have ξ‚€1𝛾cos(π›½πœ)βˆ’π›½sin(π›½πœ)=π‘˜π·βˆ’1π›Όξ‚π‘’βˆ’π›Ύπœπ›Ύsin(π›½πœ)+𝛽cos(π›½πœ)=0.(3.13) Squaring both equations, adding, and taking the square root on both sides give 𝛾2+𝛽2π‘’π›Ύπœξ‚€1=π‘˜π·βˆ’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=Ξ”, 𝛽=0 at 𝜏=πœπ‘, any roots of (3.14) must satisfy 𝛾<Ξ” since 𝜏>πœπ‘. Hence Re{πœ†}=π›Ύβˆ’Ξ”<0. Therefore (3.8) has no complex eigenvalue with nonnegative real part and so 𝐸2 is locally asymptotically stable for 𝜏>πœπ‘.
Assume that 𝐷=1. Now we prove that 𝐸2 is globally asymptotically stable when 𝜏>(1/Ξ”)ln(π‘˜/Ξ”), or equivalently π‘˜π‘’βˆ’Ξ”πœ<Ξ”. In this case, choose πœ–0>0 small enough such that π‘˜π‘’βˆ’Ξ”πœ(1+πœ–0)<Ξ”. By Theorem 2.1, for such πœ–0, there exists a 𝑇>0 so that 0<π‘₯(𝑑)<1+πœ–0 for 𝑑>𝑇. Hence, for 𝑑>𝑇+𝜏, π‘˜π‘’βˆ’Ξ”πœπ‘₯(π‘‘βˆ’πœ)<Ξ”. In Example 5.1 of Kuang ([10, page 32]), choose 𝜌(𝑑)=𝜏, π‘Ž(𝑑)=Ξ”, 𝑏(𝑑)=π‘˜π‘’βˆ’Ξ”πœπ‘₯(π‘‘βˆ’πœ), and 𝛼=Ξ”/2. We obtain (π‘˜π‘’βˆ’Ξ”πœ(1+πœ–0))2<Ξ”2=4(Ξ”βˆ’π›Ό)𝛼. Therefore 𝑦(𝑑)β†’0 as π‘‘β†’βˆž. Let 𝑧(𝑑)=𝑠(𝑑)+π‘₯(𝑑). Noting 𝐷=1, from (2.3), we have ̇𝑧(𝑑)=1βˆ’π‘§(𝑑)βˆ’π‘˜π‘₯(𝑑)𝑦(𝑑). Multiply by the integrating factor 𝑒𝑑, (𝑧(𝑑)𝑒𝑑)ξ…ž=𝑒𝑑(1βˆ’π‘˜π‘₯(𝑑)𝑦(𝑑)). Integrating both sides from 0 to 𝑑 gives
𝑧(𝑑)=π‘’βˆ’π‘‘π‘§(0)+π‘’βˆ’π‘‘ξ€·π‘’π‘‘ξ€Έβˆ’1βˆ’π‘’βˆ’π‘‘ξ€œπ‘‘0π‘’π‘ π‘˜π‘₯(𝑠)𝑦(𝑠)d𝑠=1+π‘’βˆ’π‘‘(𝑧(0)βˆ’1)βˆ’π‘’βˆ’π‘‘ξ€œπ‘‘0π‘’π‘ π‘˜π‘₯(𝑠)𝑦(𝑠)d𝑠.(3.15) If limπ‘‘β†’βˆžβˆ«π‘‘0π‘’π‘ π‘˜π‘₯(𝑠)𝑦(𝑠)d𝑠<∞, then limπ‘‘β†’βˆžπ‘’βˆ’π‘‘βˆ«π‘‘0π‘’π‘ π‘˜π‘₯(𝑠)𝑦(𝑠)d𝑠=0. Therefore limπ‘‘β†’βˆžπ‘§(𝑑)=1. If limπ‘‘β†’βˆžβˆ«π‘‘0π‘’π‘ π‘˜π‘₯(𝑠)𝑦(𝑠)d𝑠=∞, by L'HΓ΄spital's rule, lim𝑑→+βˆžπ‘’βˆ’π‘‘ξ€œπ‘‘0π‘’π‘ π‘˜π‘₯(𝑠)𝑦(𝑠)d𝑠=lim𝑑→+βˆžβˆ«π‘‘0π‘’π‘ π‘˜π‘₯(𝑠)𝑦(𝑠)d𝑠𝑒𝑑=lim𝑑→+βˆžπ‘’π‘‘π‘˜π‘₯(𝑑)𝑦(𝑑)𝑒𝑑=lim𝑑→+βˆžπ‘˜π‘₯(𝑑)𝑦(𝑑)=0,(3.16) since π‘₯(𝑑) is bounded and limπ‘‘β†’βˆžπ‘¦(𝑑)=0. It again follows that limtβ†’βˆžπ‘§(𝑑)=1. Hence limπ‘‘β†’βˆžπ‘ (𝑑)+π‘₯(𝑑)=1.(3.17)
We show that limπ‘‘β†’βˆžπ‘ (𝑑)=1/𝛼 and limπ‘‘β†’βˆžπ‘₯(𝑑)=(π›Όβˆ’1)/𝛼. First assume that the limits exist, that is, limπ‘‘β†’βˆžπ‘ (𝑑)=𝑠 and limπ‘‘β†’βˆžπ‘₯(𝑑)=π‘₯. From (2.3), we know that ̇𝑠(𝑑) and Μ‡π‘₯(𝑑) are uniformly continuous since 𝑠(𝑑), π‘₯(𝑑), and 𝑦(𝑑) are bounded. By Theorem A.3, it follows that limπ‘‘β†’βˆžΜ‡π‘ (𝑑)=0 and limπ‘‘β†’βˆžΜ‡π‘₯(𝑑)=0. Note that limπ‘‘β†’βˆžπ‘¦(𝑑)=0. Letting π‘‘β†’βˆž in (2.3) gives
ξ€·1βˆ’π‘ ξ€Έβˆ’π›Όπ‘₯𝑠=0,π‘₯ξ€·βˆ’1+𝛼𝑠=0.(3.18) Either (𝑠,π‘₯)=(1,0) or (𝑠,π‘₯)=(1/𝛼,(π›Όβˆ’1)/𝛼). Assume that (𝑠,π‘₯)=(1,0), that is, limπ‘‘β†’βˆžπ‘ (𝑑)=1 and limπ‘‘β†’βˆžπ‘₯(𝑑)=0. Note that 𝛼>𝐷. There exists πœ–>0 such that π›Όβˆ’π·βˆ’(𝛼+π‘˜)πœ–>0. For such πœ–, there exists a sufficiently large 𝑑 so that 𝑠(𝑑)>1βˆ’πœ– and 0<𝑦(𝑑)<πœ–. Recalling that π‘₯(𝑑)>0, by (2.3) Μ‡π‘₯(𝑑)>π‘₯(𝑑)(βˆ’π·+𝛼(1βˆ’πœ–)βˆ’π‘˜πœ–)=π‘₯(𝑑)(π›Όβˆ’π·βˆ’π›Όπœ–βˆ’π‘˜πœ–)>0,(3.19) for all sufficiently large 𝑑. Therefore it is impossible for π‘₯(𝑑) to approach 0 from above giving a contradiction. Therefore, we must have (𝑠,π‘₯)=(1/𝛼,(π›Όβˆ’1)/𝛼).
Now suppose that the limits do not exist. In particular if π‘₯(𝑑) does not converge, then let π‘₯=limsupπ‘‘β†’βˆžπ‘₯(𝑑) and π‘₯=liminfπ‘‘β†’βˆžπ‘₯(𝑑). By Lemma A.2 in the appendix, there exists {π‘‘π‘š}β†‘βˆž and {π‘ π‘š}β†‘βˆž such that
limπ‘šβ†’βˆžπ‘₯ξ€·π‘‘π‘šξ€Έ=π‘₯,limπ‘šβ†’βˆžξ€·π‘‘Μ‡π‘₯π‘šξ€Έ=0,limπ‘šβ†’βˆžπ‘₯ξ€·π‘ π‘šξ€Έ=π‘₯limπ‘šβ†’βˆžξ€·π‘ Μ‡π‘₯π‘šξ€Έ=0.(3.20) From (2.3), π‘₯ξ€·π‘‘π‘šξ€·π‘‘ξ€Έξ€·βˆ’π·+π›Όπ‘ π‘šξ€Έξ€·π‘‘+π‘˜π‘¦π‘šξ€Έξ€Έ=0.(3.21) Noting that π‘₯(π‘‘π‘š)>0, we have 𝑠(π‘‘π‘š)=(1βˆ’π‘˜π‘¦(π‘‘π‘š))/𝛼. Since limπ‘‘β†’βˆžπ‘¦(𝑑)=0, limπ‘‘β†’βˆžπ‘ (π‘‘π‘š)=1/𝛼. By (3.17), limπ‘‘β†’βˆžπ‘₯(π‘‘π‘š)=limπ‘‘β†’βˆž(π‘₯(π‘‘π‘š)+𝑠(π‘‘π‘š))βˆ’π‘ (π‘‘π‘š)=1βˆ’1/𝛼=(π›Όβˆ’1)/𝛼. Therefore π‘₯=(π›Όβˆ’1)/𝛼. Similarly we can show that π‘₯=(π›Όβˆ’1)/𝛼. This implies that limπ‘‘β†’βˆžπ‘₯(𝑑)=(π›Όβˆ’1)/𝛼, a contradiction.
Since 𝑠(𝑑)+π‘₯(𝑑) converges and π‘₯(𝑑) converges, then 𝑠(𝑑) must also converge. Hence limπ‘‘β†’βˆžπ‘ (𝑑)=1/𝛼 and limπ‘‘β†’βˆžπ‘₯(𝑑)=(π›Όβˆ’1)/𝛼. It follows that 𝐸2 is globally asymptotically stable.

4. Hopf Bifurcations at 𝐸+ Assuming 𝐷=Ξ”=1

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

||𝑃(πœ†)𝐸+=(βˆ’Ξ”βˆ’πœ†)ξ€·ξ€·1+𝛼π‘₯+(𝜏)+πœ†ξ€Έξ€·π·βˆ’π›Όπ‘ +(𝜏)+π‘˜π‘¦+ξ€Έ(𝜏)+πœ†+𝛼2𝑠+(𝜏)π‘₯+ξ€Έ(𝜏)+π‘’βˆ’(Ξ”+πœ†)πœπ‘˜π‘₯+ξ€·(𝜏)(πœ†+1)(πœ†+𝐷)+𝛼π‘₯+(𝜏)(πœ†+𝐷)βˆ’π›Όπ‘ +ξ€Έ(𝜏)(πœ†+1)=(βˆ’Ξ”βˆ’πœ†)ξ€·ξ€·1+𝛼π‘₯+ξ€Έ(𝜏)+πœ†πœ†+𝛼2𝑠+(𝜏)π‘₯+ξ€Έ(𝜏)+π‘’βˆ’(Ξ”+πœ†)πœπ‘˜π‘₯+ξ€·(𝜏)(πœ†+1)(πœ†+𝐷)+𝛼π‘₯+(𝜏)(πœ†+𝐷)βˆ’π›Όπ‘ +ξ€Έ1(𝜏)(πœ†+1)=(βˆ’Ξ”βˆ’πœ†)𝑠+ξ‚Άξ€·(𝜏)+πœ†πœ†+𝛼1βˆ’π‘ +ξ€Έξ‚Ά(𝜏)+π‘’βˆ’πœ†πœΞ”ξ‚΅(πœ†+1)(πœ†+𝐷)+1βˆ’π‘ +(𝜏)𝑠+(𝜏)(πœ†+𝐷)βˆ’π›Όπ‘ +ξ‚Άξ‚΅πœ†(𝜏)(πœ†+1)=(βˆ’Ξ”βˆ’πœ†)2+πœ†π‘ +ξ€·(𝜏)+𝛼1βˆ’π‘ +(ξ€Έξ‚Άπœ)+Ξ”π‘’βˆ’πœ†πœ1ξ‚΅ξ‚΅πœ†+𝑠+ξ‚Ά(𝜏)(πœ†+𝐷)βˆ’π›Όπ‘ +ξ‚Ά(𝜏)(πœ†+1)=0.(4.1) By assumption Ξ”=𝐷=1, and so

||𝑃(πœ†)𝐸+ξ‚΅πœ†=βˆ’(πœ†+1)2+πœ†π‘ +ξ€·(𝜏)+𝛼1βˆ’π‘ +ξ€Έ(𝜏)+π‘’βˆ’πœ†πœξ‚΅βˆ’πœ†+𝛼𝑠+1(𝜏)βˆ’π‘ +ξ€·πœ†(𝜏)ξ‚Άξ‚Ά=βˆ’(πœ†+1)2+𝑝(𝜏)πœ†+𝛽(𝜏)+π‘’βˆ’πœ†πœ(ξ€Έπ‘žπœ†+𝑐(𝜏))=0,(4.2) where

1𝑝(𝜏)=𝑠+ξ€·(𝜏),𝛽(𝜏)=𝛼1βˆ’π‘ +ξ€Έ(𝜏),π‘ž=βˆ’1,𝑐(𝜏)=𝛼𝑠+1(𝜏)βˆ’π‘ +.(𝜏)(4.3) The characteristic equation at 𝐸+ has one eigenvalue equal to βˆ’1 and the others are given by solutions of the equation


Lemma 4.1. Assuming π‘˜>0, 𝛼>0, and π‘˜(1βˆ’1/𝛼)β©Ύ1 so that πœπ‘=ln(π‘˜(1βˆ’1/𝛼))β©Ύ0, then 𝐸+ has no zero eigenvalue for 𝜏∈(0,πœπ‘).

Proof. Assume that 𝜏∈(0,πœπ‘). By the method of contradiction, suppose that there exists a zero root of (4.4). Therefore 1𝛽(𝜏)+𝑐(𝜏)=π›Όβˆ’π‘ +(𝜏)=0.(4.5) Noting that πœπ‘>0 if and only if π‘˜(1βˆ’1/𝛼)>1, for any 0<𝜏<πœπ‘, 1π›Όβˆ’π‘ +𝛼(𝜏)=π›Όβˆ’1βˆ’π‘˜π‘’πœξ‚€1>π›Όβˆ’1βˆ’π›Ό1βˆ’π›Όξ‚=0,(4.6) a contradiction.

Lemma 4.2. Assume π‘˜>0, 𝛼>0, π‘˜(1βˆ’1/𝛼)>1. Equilibrium 𝐸+ is asymptotically stable when 𝜏=0.

Proof. For 𝜏=0, (4.4) reduces to πœ†2+𝑝(0)πœ†+𝛽(0)+(π‘žπœ†+𝑐(0))=πœ†2+ξ‚΅1𝑠+ξ‚Ά1(0)βˆ’1πœ†+π›Όβˆ’π‘ +(0).(4.7) Both coefficients are positive, since 1𝑠+𝛼(0)βˆ’1=π‘˜1>0,π›Όβˆ’π‘ +𝛼(0)=π›Όβˆ’1βˆ’π‘˜ξ‚€1=𝛼1βˆ’π›Όβˆ’1π‘˜ξ‚>0,(4.8) and π‘˜(1βˆ’1/𝛼)>1 implies 1βˆ’1/𝛼>1/π‘˜. 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(πœ†,𝜏)=𝑝(𝜏)πœ†+(π‘žπœ†+𝑐(𝜏))π‘’βˆ’πœ†πœ+𝛽(𝜏) in Kuang [10, Theorem 1.4, page 66] gives limsupReπœ†>0,|πœ†|β†’βˆž||πœ†βˆ’2𝑔||(πœ†,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 πœβ‰ 0, assuming πœ†=π‘–πœ” (πœ”>0) is a root of 𝑃(πœ†)|𝐸+=0,
βˆ’πœ”2+𝑖𝑝(𝜏)πœ”+𝛽(𝜏)+π‘’βˆ’π‘–πœ”πœ(π‘–π‘žπœ”+𝑐(𝜏))=0.(4.10) Substituting π‘’π‘–πœƒ=cosπœƒ+𝑖sinπœƒ into (4.10) gives βˆ’πœ”2+𝛽(𝜏)+π‘žπœ”sin(πœ”πœ)+𝑐(𝜏)cos(πœ”πœ)+𝑖(𝑝(𝜏)πœ”+π‘žπœ”cos(πœ”πœ)βˆ’π‘(𝜏)sin(πœ”πœ))=0.(4.11) Separating the real and imaginary parts, we obtain 𝑐(𝜏)cos(πœ”πœ)+π‘žπœ”sin(πœ”πœ)=πœ”2βˆ’π›½(𝜏),𝑐(𝜏)sin(πœ”πœ)βˆ’π‘žπœ”cos(πœ”πœ)=𝑝(𝜏)πœ”.(4.12) Solving for cos(πœ”πœ) and sin(πœ”πœ) gives ξ€·πœ”sin(πœ”πœ)=𝑐(𝜏)𝑝(𝜏)πœ”+π‘žπœ”2ξ€Έβˆ’π›½(𝜏)𝑐(𝜏)2+π‘ž2πœ”2,ξ€·πœ”cos(πœ”πœ)=𝑐(𝜏)2ξ€Έβˆ’π›½(𝜏)βˆ’π‘žπ‘(𝜏)πœ”2𝑐(𝜏)2+π‘ž2πœ”2.(4.13) Noting sin2(πœ”πœ)+cos2(πœ”πœ)=1, squaring both sides of equations (4.13), adding, and rearranging gives πœ”4+𝑝2(𝜏)βˆ’π‘ž2ξ€Έπœ”βˆ’2𝛽(𝜏)2+𝛽2(𝜏)βˆ’π‘2(𝜏)=0.(4.14) Solving for πœ”, we obtain two roots πœ”1(𝜏) and πœ”2(𝜏): πœ”11(𝜏)=√2ξ‚΅π‘ž2βˆ’π‘2(𝜏)+2𝛽(𝜏)+ξ€·π‘ž2βˆ’π‘2ξ€Έ(𝜏)+2𝛽(𝜏)2βˆ’4(𝛽2(𝜏)βˆ’π‘2ξ‚Ά(𝜏))1/2=1𝑠+√(𝜏)2ξ€·ξ€·1βˆ’π‘ +(𝜏)ξ€Έξ€·2𝛼𝑠2+(𝜏)βˆ’π‘ +(ξ€Έ+ξ”πœ)βˆ’1𝑠2+ξ€Έ(𝜏)βˆ’12+4𝛼𝑠2+𝑠(𝜏)2+(𝜏)βˆ’1ξ€Έξ€·1βˆ’π‘ +ξ€Έ(𝜏)+4𝑠2+ξ€·(𝜏)𝛼𝑠2+ξ€Έ(𝜏)βˆ’12ξ‚Ά1/2πœ”21(𝜏)=√2ξ‚΅π‘ž2βˆ’π‘2(𝜏)+2𝛽(𝜏)βˆ’ξ€·π‘ž2βˆ’π‘2ξ€Έ(𝜏)+2𝛽(𝜏)2ξ€·π›½βˆ’42(𝜏)βˆ’π‘2ξ€Έξ‚Ά(𝜏)1/2=1𝑠+√(𝜏)2ξ€·ξ€·1βˆ’π‘ +(𝜏)ξ€Έξ€·2𝛼𝑠2+(𝜏)βˆ’π‘ +ξ€Έβˆ’ξ”(𝜏)βˆ’1𝑠2+ξ€Έ(𝜏)βˆ’12+4𝛼𝑠2+𝑠(𝜏)2+(𝜏)βˆ’1ξ€Έξ€·1βˆ’π‘ +ξ€Έ(𝜏)+4𝑠2+ξ€·(𝜏)𝛼𝑠2+ξ€Έ(𝜏)βˆ’12ξ‚Ά1/2.(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 πœ”1(𝜏)β©Ύπœ”2(𝜏) for all 𝜏∈𝐼 with πœ”1(𝜏)>πœ”2(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, πœ”1(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

βŽ‘βŽ’βŽ’βŽ£ξƒ©π‘˜π½=ln𝛼1√1/4+βŽ›βŽœβŽœβŽπ‘˜ξ‚€1/16+1/(2𝛼)βˆ’1ξƒͺξƒͺ,ln4βˆšξ‚π›Όβˆ’1π›ΌβŽžβŽŸβŽŸβŽ βŽ€βŽ₯βŽ₯⎦.(4.18) When the end points of 𝐽 are real and π½β‰ βˆ…, define

𝐼1=ξ€Ί0,πœπ‘ξ€Έβˆ©π½.(4.19) We prove that (𝐻1) holds for any 𝜏∈𝐼1.

From 𝐷=Ξ”=1,

πœπ‘=1Ξ”ξ‚€π‘˜lnΞ”ξ‚€1π·βˆ’1𝛼=lnπ‘˜(π›Όβˆ’1)𝛼.(4.20) If 𝛼>1, then 𝛼>4βˆšπ›Ό. It follows that

πœπ‘βŽ›βŽœβŽœβŽπ‘˜ξ‚€>ln4βˆšξ‚π›Όβˆ’1π›ΌβŽžβŽŸβŽŸβŽ .(4.21) Therefore,

𝐼1=βŽ‘βŽ’βŽ’βŽ£ξƒ―ξƒ©π‘˜max0,ln𝛼1√1/4+βŽ›βŽœβŽœβŽπ‘˜ξ‚€1/16+1/(2𝛼)βˆ’1ξƒͺξƒͺξƒ°,ln4βˆšξ‚π›Όβˆ’1π›ΌβŽžβŽŸβŽŸβŽ βŽ€βŽ₯βŽ₯⎦.(4.22)

Theorem 4.5. Assume βˆšπ›Ό>(7+35)/2 and π‘˜>𝛼/(4βˆšπ›Όβˆ’1), then 𝐼1 is not empty, and for any 𝜏∈𝐼1, but βˆšπœβ‰ ln((π‘˜/𝛼)(1/(1/4+1/16+1/(2𝛼))βˆ’1)), condition (𝐻1) holds and πœ”1(𝜏)>πœ”2(𝜏)>0. If √𝜏=ln((π‘˜/𝛼)(1/(1/4+1/16+1/(2𝛼))βˆ’1))∈𝐼1, then πœ”1(𝜏)>πœ”2(𝜏)=0.

Proof. For any βˆšπ›Ό>(7+35)/2, we have 1βˆ’1/4βˆšπ›Ό>0, and therefore 14βˆšπ›Ό+12βˆ’βˆš52<14ξ‚™ξ‚€βˆš7+35+1/22βˆ’βˆš52=0.(4.23) Hence, 14βˆšπ›Όβˆ’14ξƒͺ2βˆ’ξƒ©ξ‚™1+116ξƒͺ2𝛼2=1βˆšπ›Όβˆ’124βˆšπ›Όβˆ’1=2π›Όβˆ’124βˆšπ›ΌβŽ›βŽœβŽœβŽξƒ©14βˆšπ›Όξƒͺ32+1βˆ’4βˆšπ›ΌβŽžβŽŸβŽŸβŽ =124βˆšπ›Όξƒ©11βˆ’4βˆšπ›ΌξƒͺβŽ›βŽœβŽœβŽξƒ©14βˆšπ›Όξƒͺ2+14βˆšπ›ΌβŽžβŽŸβŽŸβŽ =1βˆ’124βˆšπ›Όξƒ©11βˆ’4βˆšπ›ΌξƒͺβŽ›βŽœβŽœβŽξƒ©14βˆšπ›Ό+12ξƒͺ2βˆ’54⎞⎟⎟⎠=124βˆšπ›Όξƒ©11βˆ’4βˆšπ›Ό1ξƒͺ4βˆšπ›Ό+12βˆ’βˆš521ξƒͺ4βˆšπ›Ό+12+√52ξƒͺ<0.(4.24) Therefore, 1/4βˆšβˆšπ›Όβˆ’1/4<1/16+1/(2𝛼). Since √1/4+1/16+1/(2𝛼)<1/4+√1/16+1/(7+35)<1, it follows that 14βˆšπ›Ό<14+ξ‚™1+1162𝛼<1.(4.25) Hence, ξƒ©π‘˜ln𝛼1√1/4+βŽ›βŽœβŽœβŽπ‘˜ξ‚€1/16+1/(2𝛼)βˆ’1ξƒͺξƒͺ<ln4βˆšξ‚π›Όβˆ’1π›ΌβŽžβŽŸβŽŸβŽ .(4.26) From π‘˜>𝛼/(4βˆšπ›Όβˆ’1), we have ln(π‘˜(4βˆšπ›Όβˆ’1)/𝛼)>0. Therefore, ξƒ―ξƒ©π‘˜max0,ln𝛼1√1/4+βŽ›βŽœβŽœβŽπ‘˜ξ‚€1/16+1/(2𝛼)βˆ’1ξƒͺξƒͺξƒ°<ln4βˆšξ‚π›Όβˆ’1π›ΌβŽžβŽŸβŽŸβŽ ,(4.27) and so 𝐼1 is not empty. Noting 𝑠+(𝜏)=1/(1+(𝛼Δ/π‘˜)π‘’Ξ”πœ) and Ξ”=1, for any 𝜏∈𝐼1, but βˆšπœβ‰ ln((π‘˜/𝛼)(1/(1/4+1/16+1/(2𝛼))βˆ’1)), we have 𝑠+(𝜏)∈[1/4βˆšβˆšπ›Ό,1/4+1/16+1/(2𝛼)).
In what follows, we intend to show that for any such 𝜏, condition (𝐻1) holds. From (4.3),
π‘ž2βˆ’π‘2(𝜏)+2𝛽(𝜏)=(βˆ’1)2βˆ’1𝑠2+ξ€·(𝜏)+2𝛼1βˆ’π‘ +ξ€Έ=ξ€·(𝜏)1βˆ’π‘ +ξ€Έ(𝜏)2𝛼𝑠2+𝑠(𝜏)+(𝜏)2βˆ’π‘ +(𝜏)βˆ’12𝛼=ξ€·2𝛼1βˆ’π‘ +ξ€Έ(𝜏)2𝛼𝑠2+𝑠(𝜏)+1(𝜏)βˆ’ξ‚4𝛼2βˆ’116𝛼2βˆ’1ξ‚Ά.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 βˆšπ›Ό>(7+35)/2, √1/βˆšπ›Όβˆ’1/(4𝛼)=(1/βˆšπ›Ό)(1βˆ’1/(4𝛼))>0, and 116𝛼2+1ξƒͺ2𝛼2βˆ’ξƒ©1βˆšπ›Όβˆ’1ξƒͺ4𝛼2=116𝛼2+1βˆ’12𝛼𝛼+1√2π›Όπ›Όβˆ’116𝛼2=112π›Όβˆšπ›Όξƒͺβˆ’1<0.(4.29) Since √1/𝛼<1/4βˆšπ›Ό for βˆšπ›Ό>(7+35)/2, 1+ξ‚™4𝛼116𝛼2+1<12π›Όβˆšπ›Ό<14βˆšπ›Ό.(4.30) For any 𝑠+(𝜏)>1/4βˆšπ›Ό, 𝑠+1(𝜏)βˆ’β©Ύ14𝛼4βˆšπ›Όβˆ’1>ξ‚™4𝛼116𝛼2+1.2𝛼(4.31) Hence, 𝑠+1(𝜏)βˆ’ξ‚4𝛼2β©Ύ116𝛼2+12𝛼.(4.32) Next consider the second inequality in (𝐻1). For βˆšπ›Ό>(7+35)/2, since 1/4βˆšπ›Ό>1/𝛼, 𝑠+(𝜏)β©Ύ1/4βˆšπ›Ό>1/𝛼. Therefore, 𝛼𝑠+(𝜏)>1. For 𝑠+(𝜏)∈[1/4βˆšβˆšπ›Ό,1/4+1/16+1/(2𝛼))𝛽2(𝜏)βˆ’π‘2=ξ‚΅(𝜏)=(𝛽(𝜏)βˆ’π‘(𝜏))(𝛽(𝜏)+𝑐(𝜏))π›Όβˆ’2𝛼𝑠+1(𝜏)+𝑠+(1𝜏)ξ‚Άξ‚΅π›Όβˆ’π‘ +(ξ‚Άπœ)=βˆ’2𝛼𝑠2+𝑠(𝜏)2+𝑠(𝜏)βˆ’+(𝜏)2βˆ’1ξ‚Άξ€·2𝛼𝛼𝑠+ξ€Έ(𝜏)βˆ’1=βˆ’2𝛼𝑠2+𝑠(𝜏)+1(𝜏)βˆ’42βˆ’1βˆ’116ξ‚Άξ€·2𝛼𝛼𝑠+ξ€Έ(𝜏)βˆ’1>0.(4.33) Finally, ξ€·π‘ž2βˆ’π‘2ξ€Έ(𝜏)+2𝛽(𝜏)2ξ€·π›½βˆ’42(𝜏)βˆ’π‘2ξ€Έ=ξ€·π‘ž(𝜏)2βˆ’π‘2π‘ž(𝜏)ξ€Έξ€·2βˆ’π‘2ξ€Έ(𝜏)+4𝛽(𝜏)+4𝑐2=1(𝜏)1βˆ’π‘ 2+(1𝜏)ξƒͺ1βˆ’π‘ 2+(ξ€·πœ)+4𝛼1βˆ’π‘ +(ξ€Έξƒͺξ‚΅πœ)+4𝛼𝑠+1(𝜏)βˆ’π‘ +ξ‚Ά(𝜏)2=11βˆ’π‘ 2+ξƒͺ(𝜏)21+4𝛼1βˆ’π‘ 2+ξƒͺξ€·(𝜏)1βˆ’π‘ +ξ€Έ(𝜏)+4𝑠+(𝜏)2𝛼24βˆ’8𝛼+𝑠2+(𝜏)=4𝑠+(𝜏)2𝛼21+4𝛼1βˆ’π‘ 2+ξƒͺξ€·(𝜏)1βˆ’π‘ +ξ€Έξƒͺ1(𝜏)βˆ’8𝛼+1βˆ’π‘ 2+ξƒͺ(𝜏)2+4𝑠2+(𝜏)=4𝑠+(𝜏)2𝛼21+4𝛼1βˆ’π‘ 2+ξƒͺξ€·(𝜏)1βˆ’π‘ +ξ€Έξƒͺ+1(𝜏)βˆ’21+𝑠2+ξƒͺ(𝜏)2=ξ€·π›Όβˆ’π›Ό1ξ€Έξ€·π›Όβˆ’π›Ό2ξ€Έ,(4.34) where 𝛼1=ξ€·2βˆ’1βˆ’1/𝑠2+(𝜏)ξ€Έξ€·1βˆ’π‘ +ξ€Έ+(𝜏)ξ€·1/𝑠2+(𝜏)+2𝑠+(𝜏)+1ξ€Έξ€·1/𝑠+ξ€Έ(𝜏)βˆ’122𝑠2+,𝛼(𝜏)2=ξ€·2βˆ’1βˆ’1/𝑠2+(𝜏)ξ€Έξ€·1βˆ’π‘ +ξ€Έβˆ’ξ”(𝜏)ξ€·1/𝑠2+(𝜏)+2𝑠+(𝜏)+1ξ€Έξ€·1/𝑠+ξ€Έ(𝜏)βˆ’122𝑠2+.(𝜏)(4.35) Since 𝑠+(𝜏)<1, 12βˆ’1βˆ’π‘ 2+ξƒͺξ€·(𝜏)1βˆ’π‘ +ξ€Έ(𝜏)=𝑠+(𝜏)+1+1βˆ’π‘ +(𝜏)𝑠2+1(𝜏)>0,2βˆ’1βˆ’π‘ 2+ξƒͺξ€·(𝜏)1βˆ’π‘ +ξ€Έξƒͺ(𝜏)2>12βˆ’1βˆ’π‘ 2+ξƒͺξ€·(𝜏)1βˆ’π‘ +ξ€Έξƒͺ(𝜏)2βˆ’π‘ 2+1(𝜏)1+𝑠2+ξƒͺ(𝜏)2=1𝑠2+(𝜏)+2𝑠+ξƒͺξ‚΅1(𝜏)+1𝑠+ξ‚Ά(𝜏)βˆ’12>0.(4.36) It follows that 0<𝛼2<𝛼1. Again noting that 𝑠+(𝜏)<1, 𝛼1<ξ€·2βˆ’1βˆ’1/𝑠2+(𝜏)ξ€Έξ€·1βˆ’π‘ +ξ€Έ+(𝜏)ξ€·1/𝑠2+(𝜏)+2/𝑠+(𝜏)+1ξ€Έξ€·1/𝑠+ξ€Έ(𝜏)βˆ’122𝑠2+=ξ€·(𝜏)2βˆ’1βˆ’1/𝑠2+(𝜏)ξ€Έξ€·1βˆ’π‘ +ξ€Έ+ξ€·(𝜏)1/𝑠+(𝜏)+1ξ€Έξ€·1/𝑠+ξ€Έ(𝜏)βˆ’12𝑠2+=ξ€·(𝜏)2βˆ’1βˆ’π‘ +(𝜏)βˆ’1/𝑠2+(𝜏)+1/𝑠+ξ€Έ(𝜏)+1/𝑠2+(𝜏)βˆ’12𝑠2+=𝑠(𝜏)+(𝜏)+2/𝑠2+(𝜏)βˆ’1/𝑠+(𝜏)2𝑠2+=1(𝜏)22𝑠4+βˆ’ξƒ©1(𝜏)𝑠3+βˆ’1(𝜏)𝑠+<1(𝜏)ξƒͺξƒͺ22𝑠4+=1(𝜏)𝑠4+.(𝜏)(4.37) Hence, for any 𝑠+(𝜏)>1/4βˆšπ›Ό, we have 𝛼>1/𝑠4+(𝜏)>𝛼1>𝛼2. This leads to ξ€·π‘ž2βˆ’π‘2ξ€Έ(𝜏)+2𝛽(𝜏)2ξ€·π›½βˆ’42(𝜏)βˆ’π‘2ξ€Έ=ξ€·(𝜏)π›Όβˆ’π›Ό1ξ€Έξ€·π›Όβˆ’π›Ό2ξ€Έ>0.(4.38) Therefore, (𝐻1) holds for any 𝜏∈𝐼1. By Lemma 4.4, both πœ”1(𝜏)>0 and πœ”2(𝜏)>0.
If √𝜏=ln((π‘˜/𝛼)(1/(1/4+1/16+1/(2𝛼))βˆ’1))∈𝐼1, we have 𝑠+√(𝜏)=1/4+1/16+1/(2𝛼). Noting (4.25), we obtain
π‘ž2βˆ’π‘2(𝜏)+2𝛽(𝜏)>0,𝛽2(𝜏)βˆ’π‘2ξ€·π‘ž(𝜏)=0,2βˆ’π‘2ξ€Έ(𝜏)+2𝛽(𝜏)2ξ€·π›½βˆ’42(𝜏)βˆ’π‘2ξ€Έ(𝜏)>0.(4.39) By (4.15), it follows that πœ”1(𝜏)>0 and πœ”2(𝜏)=0.

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

𝐼2ξ€ΊβˆΆ=0,πœπ‘ξ€Έβˆ©ξƒ©ξƒ©π‘˜βˆ’βˆž,ln𝛼1√1/4+1/16+1/(2𝛼)βˆ’1ξƒͺξƒͺξƒ­.(4.40) In the following theorem, we consider the case that parameters are chosen so that


Theorem 4.6. Assume 𝛼>1 and βˆšπ‘˜>𝛼(1/(1/4+1/16+1/(2𝛼))βˆ’1)βˆ’1. Interval 𝐼2 given by (4.41) is not empty. For any 𝜏∈𝐼2, (𝐻2) holds and hence πœ”1(𝜏)>0.

Proof. Assume 𝛼>1. Letting 1𝐺(𝛼)=4+ξ‚™1+116βˆ’12𝛼𝛼,(4.42) then d1d𝛼𝐺(𝛼)=βˆ’4𝛼2√+11/16+1/(2𝛼)𝛼2=1𝛼2ξƒ©βˆ’1√ξƒͺ1+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)βˆ’1/𝛼>0. Therefore 11>4+ξ‚™1+116>12𝛼𝛼.(4.44) This gives 1π›Όβˆ’1>√1/4+1/16+1/(2𝛼)βˆ’1>0.(4.45) By assumption βˆšπ‘˜>𝛼(1/(1/4+1/16+1/(2𝛼))βˆ’1)βˆ’1, we obtain π‘˜(π›Όβˆ’1)𝛼>π‘˜π›Όξƒ©1√1/4+ξƒͺ1/16+1/(2𝛼)βˆ’1>1.(4.46