Andronov Hopf and Bautin bifurcation in a tritrophic food chain model with Holling functional response types IV and II

Electronic Journal of Qualitative Theory of Differential Equations 018 No 78 1 7; https://doiorg/10143/ejqtde018178 wwwmathu-szegedhu/ejqtde/ Andronov Hopf and Bautin bifurcation in a tritrophic food chain model with Holling functional response types IV and II Gamaliel Blé 1 Víctor Castellanos 1 and Iván Loreto-Hernández 1 División Académica de Ciencias Básicas UJAT Km 1 Carretera Cunduacán Jalpa de Méndez Cunduacán Tabasco cp 86690 México División Académica de Ciencias Básicas CONACyT-UJAT Km 1 Carretera Cunduacán Jalpa de Méndez Cunduacán Tabasco cp 86690 México Received 1 March 018 appeared 19 September 018 Communicated by Hans-Otto Walther Abstract The existence of an Andronov Hopf and Bautin bifurcation of a given system of differential equations is shown The system corresponds to a tritrophic food chain model with Holling functional responses type IV and II for the predator and superpredator respectively The linear and logistic growth is considered for the prey In the linear case the existence of an equilibrium point in the positive octant is shown and this equilibrium exhibits a limit cycle For the logistic case the existence of three equilibrium points in the positive octant is proved and two of them exhibit a simultaneous Hopf bifurcation Moreover the Bautin bifurcation on these points are shown Keywords: model Andronov Hopf bifurcation Bautin bifurcation limit cycle food chain 010 Mathematics Subject Classification: 37G15 34C3 37C75 34D45 9D40 9D5 1 Introduction In the task of understanding the complexity presented by the interactions among the different populations living in a habitat the mathematical modeling has been a very important rule in ecology in the last decades Some of the models which have been studied are the tritrophic systems see Ref [3] and references therein In particular in this work we analyzed a tritrophic model given by the following differential equation system Corresponding author Email: vicas@ujatmx dx = hx f xy dt dy dt = c 1y f x gyz c y dz dt = c 3gyz d z 11

G Blé V Castellanos and I Loreto-Hernández where x represents the density of a prey that gets eaten by a predator of density y mesopredator and the species y feeds the top predator z superpredator The function hx represents the growth rate of the prey population in absence of the predators and the functions f x and gy are the functional responses for the mesopredator and the superpredator respectively The parameters c 1 c c 3 and d are positive and we are interested to find the stable solutions in the positive octant Ω = {x > 0 y > 0 z > 0} There are different proposals of functional responses in literature among which are the Holling type see Refs [5 10 11] In Ref [3] is considered the case when hx is a logistic map and the functional responses f and g are Holling type II Using the averaging theory they proved that the system 11 has an equilibrium point which exhibits a triple Andronov Hopf bifurcation It implies the existence of a stable periodic orbit contained in the domain of interest The local dynamics of the differential system 11 has been analyzed in Ref [] when hx is a linear map and the functional responses f and g are Holling type III They proved the existence of two equilibrium points which exhibit simultaneously a zero-hopf bifurcation in Ω In Ref [1] the authors analyzed the case when hx is a linear map and the functional responses f and g are Holling type III and Holling type II respectively They proved that there is a domain in the parameter space where the system 11 has a stable periodic orbit which results from an Andronov Hopf bifurcation In this paper we are interested in analyzed the dynamics of the differential system 11 when the functional responses f x and gy are Holling type IV and II respectively In particular we are interested in stable equilibrium points or stable limit cycles inside the positive octant Ω We consider two cases the linear case taking hx = ρx and the logistic case taking hx = ρx1 x R The functions f and g will be f x = a 1x b 1 + x gy = a y b + y where a 1 b 1 a b are positive constants Explicitly we will study the differential system dx dt = hx a 1xy x + b 1 dy dt = c 1a 1 yx x a yz + b 1 b + y c y dz dt = c 3a yz b + y d z Along this manuscript the terms linear or logistic case will be used to refer cases when the prey has either linear or logistic growth rate respectively The main results in this paper are contained in Sections and 3 1 Linear case In this section we consider the differential system 1 with a linear growth for the prey this means that the function hx = ρx and then the differential system becomes ẋ = a 1xy b 1 + x + ρx ẏ = c y + a 1c 1 xy b 1 + x a yz b + y 1 ż = d + a c 3 y b + y z

Andronov Hopf and Bautin bifurcation 3 In next lemma we show the existence of an equilibrium point in the positive octant Ω under certain conditions on the parameters involved in the system of differential equations Lemma 1 The differential system 1 has only one equilibrium point p 0 = x 0 y 0 z 0 Ω if a a c 3 d > 0 b a 1 y 0 ρb 1 > 0 c c y 0 c 1 x 0 ρ < 0 Moreover if ones of above condition does not hold then the differential system 1 does not have any equilibrium point in Ω Proof The equilibrium points of the differential system 1 are the solutions of a 1xy + ρx = 0 b 1 + x c y + a 1c 1 xy b 1 + x a yz b + y = 0 d + a c 3 y b + y z = 0 By multiplying the above equations by the term b 1 + x b + y which is always positive in Ω we obtain that an equilibrium point in Ω must satisfy the system ρ b 1 + x a 1 y = 0 b + y c b1 + x a 1 c 1 x + a z b 1 + x = 0 d b + y a c 3 y = 0 From the third equation in system y 0 = d b a c 3 d and it is positive by hypothesis a Substituting y = y 0 in the first equation of we obtain a unique positive solution x = x 0 by hypothesis b Now substituting x = x 0 and y = y 0 in the second equation of system we have that the unique solution z = z 0 of this equation is positive if and only if c b1 + x0 a1 c 1 x 0 < 0 but from the first equation in system we have that b 1 + x0 = a 1y 0 /ρ then c b1 + x0 a1 c 1 x 0 = a 1 ρ c y 0 c 1 x 0 ρ and z 0 > 0 by hypothesis c Clearly if ones of the conditions a c 3 d > 0 a 1 y 0 ρb 1 > 0 or c y 0 c 1 x 0 ρ < 0 does not hold then the differential system 1 has no equilibrium points in Ω In order to simplify the expression of the equilibrium point p 0 we introduce a new parameters given by the next result Lemma If the parameters of the system 1 satisfy the conditions a b and c in Lemma 1 then there exist k 1 > 0 k > 0 and k 3 > 0 such that the parameters a 1 a and b involved in the differential system 1 can be written as a = d ρ + k 1 b = b 1k 1 + k a 1 = b 1c k 1 + c k + k 3 3 c 3 ρ a 1 d c 1 k 1 k and the unique equilibrium point of the system 1 in Ω is p 0 = k c 1 k ρ b 1 k 1 + k c 1 c 3 k k 3 ρ k 1 k 1 b 1 c k 1 + c k + k 3 b 1 c d k 3 1 + c d k 1 k + d k 1 k 3

4 G Blé V Castellanos and I Loreto-Hernández Proof The solutions of system are a1 b d + b 1 ρd a c 3 b p 0 = d c 1c 3 ρa c 3 d 1 b c c 3 d ρa c 3 d a c 3 d d a c 3 d a1 b d + b 1 ρd a c 3 b p 1 = d c 3 c 1 ρa c 3 d 1 + b c d ρa c 3 d a c 3 d d a c 3 d 1 = a 1 b d + b 1 ρd a c 3 Since p 1 / Ω by Lemma 1 p 0 Ω and ρa c 3 d > 0 then there exists k 1 > 0 such that a = d ρ+k 1 c 3 ρ Hence a p 0 = 1 b d b 1 k 1 b c 3 ρ d ρ k 1 k 1 c 1 k 1 a 1 b d b 1 k 1 b c d d k 1 Moreover a 1 b d b 1 k 1 > 0 then there exists k > 0 such that b = b 1k 1 +k a 1 d then k a 1 c 1 k 1 c k b 1 c k 1 p 0 = k k 1 ρ b 1 k 1 + k c 3 ρ a 1 k 1 a 1 d k 1 Since k a 1 c 1 k 1 c k b 1 c k 1 > 0 then there exists k 3 > 0 such that a 1 = b 1c k 1 +c k +k 3 and p 0 = k c 1 k ρ b 1 k 1 + k c 1 c 3 k k 3 ρ k 1 k 1 b 1 c k 1 + c k + k 3 b 1 c d k 3 1 + c d k 1 k + d k 1 k 3 c 1 k 1 k Lemma 3 Under the hypothesis of Lemma and considering that the parameters a 1 a and b 1 satisfy the conditions 3 and k = b 1 k 1 d = 1b 4 1k 1 k 3 = 3 5k 3 b 1k 1 ρ and c = c 0 ρ := 9k 1 + 38ρ 4 5ρ then the equilibrium point p 0 is given by p 0 = b 1 5 b 1 c 1 ρ 9k 1 + 64ρ 65 b1 c 1 c 3 ρ 4 18k 4 1 + 18k 1 ρ and the eigenvalues of the linear approximation of system 1 at p 0 are where α = 64ρ 39 and ω = k 1 4 > 0 ± iω

Andronov Hopf and Bautin bifurcation 5 Proof Taking into account the assignations of the parameters a 1 a and b 1 given by 3 the characteristic polynomial of the linear approximation M p0 of differential system 1 at the equilibrium point p 0 is Pλ = detλi M p0 = λ 3 + A 1 λ + A λ + A 3 where k A 1 = ρ d ρ + k 1 + d k 3 b 1 k 1 + k d ρ + k 1 A = d ρ b 1 k 1 + k h 1 + k 1 ρ b 1 k 1 k h + d k 1 k 3 b 1 k 1 + k b 1 k 1 + k d ρ + k 1 d k A 3 = 1 k k 3 ρ b 1 k 1 + k d ρ + k 1 h 1 = b 1 c k 1 c k + k 3 h = b 1 c k 1 + c k + k 3 If we consider the assignments for k k 3 and d given by 4 then A 1 A and A 3 reduce to A 1 = 64ρ 39 A = 1 ρ 6c + 19ρ + 4k 1 and A 3 = 16k 1 ρ 78 39 The characteristic polynomial Pλ = λ 3 + A 1 λ + A λ + A 3 has a pair of purely imaginary roots ±iω and a real root α if and only if Pλ = λ αλ + ω = λ 3 αλ + ω λ αω Thus comparing coefficients Pλ has a pair of purely imaginary roots ±iω and a real root α if and only if A > 0 and A 1 A A 3 = 0 5 where ω = A and α = A 1 Since A 1 A A 3 = 16ρ 5c ρ+9k 1 +38ρ 151 solving equation 5 for the parameter c we have that if c = 9k 1 + 38ρ 5ρ then A = k 1 4 > 0 Thus we conclude that the characteristic polynomial Pλ has a pair of purely imaginary roots ±iω and a real root α where α = 64ρ 39 and ω = k 1 The equilibrium point p 0 becomes p 0 = b 1 5 b 1 c 1 ρ 9k 1 + 64ρ 65 b1 c 1 c 3 ρ 4 18k 4 1 + 18k 1 ρ In order to compute the Lyapunov coefficients and a regularity condition from now in this section b 1 = 1 k 1 = 1 c 1 = 1 and c 3 = 1 Remark 4 If the assumptions of Lemma 3 are satisfied then the linear approximation of the differential system 1 at p 0 has the eigenvalues α = 64ρ 39 and ± i when c = c 0 ρ

6 G Blé V Castellanos and I Loreto-Hernández hence by continuity on the eigenvalues the linear approximation of the differential system at p 0 has a pair of complex eigenvalues when c is in a neighborhood of c 0 ρ λp 0 c ρ = ξp 0 c ρ ± iωp 0 c ρ In order to compute the first Lyapunov coefficient l 1 we apply the Kuznetsov formula see Ref [7] Taking into account the assumptions of Lemma 3 and using the Mathematica software we obtain the first Lyapunov coefficient of the differential system 1 at the equilibrium point p 0 Lemma 5 If the hypotheses of Lemma 3 hold then the first Lyapunov coefficient of the differential system 1 at the equilibrium point p 0 is l 1 p 0 c 0 ρ ρ = 64ρ +930074175488ρ 8 +986601040ρ 6 50409194ρ 4 1131103197ρ 80677701 169ρ 3 4096ρ +15116384ρ +151100ρ 4 +15ρ +81 Corollary 6 There exists a unique real number ρ 0 > 0 such that l 1 p 0 c 0 ρ 0 ρ 0 = 0 Proof By Lemma 5 l 1 p 0 c 0 ρ ρ = 0 if and only if 30074175488ρ 8 + 986601040ρ 6 50409194ρ 4 1131103197ρ 80677701 = 0 6 According to the Descartes rule there is a unique real number ρ 0 > 0 such that l 1 p 0 c 0 ρ 0 ρ 0 = 0 Indeed solving numerically equation 6 for the parameter ρ we have that ρ 0 05771 Since l 1 p 0 c 0 ρ ρ takes positive and negative values we will verify the transversality conditions to have Andronov Hopf or Bautin bifurcation At first we state the following result proposed as an exercise in Ref [6] whose proof is straight forward and we omit the details Lemma 7 Let Mτ be a parameter-dependent real n n-matrix which has a simple pair of complex eigenvalues ξτ ± iωτ such that ξτ 0 = 0 and ωτ 0 := ω 0 > 0 Then the derivative of the real part of the complex eigenvalues at τ 0 is given by dξ dτ τ 0 = Re p tr dm dτ τ 0 q where p q C n are eigenvectors satisfying the normalization conditions Mτ 0 q = iω 0 M tr τ 0 p = iω 0 q tr q = 1 and p tr q = 1 We know proceed to show the regularity condition in order to obtain a Bautin bifurcation Lemma 8 Bautin regularity condition If the parameters a 1 a b k k 3 and d satisfy the hypothesis of Lemma 3 then the map c ρ ξp 0 c ρ l 1 p 0 c ρ is regular at c 0 ρ 0 where ξp 0 c ρ is given in Remark 4 and c 0 := c 0 ρ 0 Proof By hypothesis the linear approximation of the differential system 1 at p 0 depends only on the parameters c and ρ let M p0 c ρ be this linear approximation By Lemma 5 the complex numbers ± i are eigenvalues of M p 0 c 0 ρ 0 hence the real part of the complex eigenvalues of M p0 c 0 ρ 0 are ξp 0 c 0 ρ 0 = 0 7

Andronov Hopf and Bautin bifurcation 7 let p and q be eigenvectors of M p0 c 0 ρ 0 for the corresponding eigenvalues i and i respectively such that q tr q = 1 and p tr q = 1 8 By 7 and 8 we can apply Lemma 7 to the linear approximation M p0 c ρ then taking into account the values of p q and M p 0 c ρ c which we compute with the Mathematica software we have that the partial derivative of the real part of the eigenvalues ξc ρ ± iωc ρ of M p0 c ρ with respect to the parameter c at the point c 0 ρ 0 takes the value c 0 ρ 0 =p tr Mp0 c 0 ρ 0 1664ρ 0 q = c c 16384ρ 9 0 + 151 Applying Lemma 7 one more time and taking into account the values of p q and Mc ρ ρ it follows that ρ c 0 ρ 0 = p tr Mp0 c 0 ρ 0 q = 3 38ρ 0 9 ρ 16384ρ 10 0 + 151 From the Kuznetsov formula see Ref [4] the first Lyapunov coefficient at the equilibrium point p 0 is given by l 1 p 0 c ρ = Re C 1c ρ ωc ρ ξc ρ Im C 1c ρ ω c ρ 11 where C 1 c ρ is a function that takes complex values as a differentiable function in the variables c ρ Notice that from Corollary 6 7 11 and since ωc 0 ρ 0 = 1/ Re C 1 c 0 ρ 0 = 0 1 Hence from 11 7 and 1 the partial derivative of l 1 c ρ with respect to c at the point c 0 ρ 0 is given by l 1 1 C1 c 0 ρ 0 = c ω ωc 0 ρ 0 Re c 0 ρ 0 Im C c 0 ρ 0 1 c 0 ρ 0 c 0 ρ 0 c c and the partial derivative of l 1 c ρ with respect to ρ at the point c 0 ρ 0 is given by l 1 ρ c 1 C1 0 ρ 0 = ω ωc 0 ρ 0 Re c 0 ρ 0 ρ c 0 ρ 0 Im C 1 c 0 ρ 0 ρ c 0 ρ 0 thus the determinant of interest reduces to c det c 0 ρ 0 ρ c 0 ρ 0 l 1 l c c 0 ρ 0 1 ρ c 0 ρ 0 c c 0 ρ 0 Re C1 ρ = c 0 ρ 0 ρ c 0 ρ 0 Re C1 c c 0 ρ 0 13 ωc 0 ρ 0 Numerically one has that Re C 1 c c 0 ρ 0 09053 and Re C 1 ρ c 0 ρ 0 4835 and by Corollary 6 ρ 0 05771 Then by 9 10 and 13 c det c 0 ρ 0 ρ c 0 ρ 0 l 1 l c c 0 ρ 0 1 ρ c 01805 0 ρ 0 Hence the map c ρ ξp 0 c ρ l 1 p 0 c ρ is regular at c 0 ρ 0

8 G Blé V Castellanos and I Loreto-Hernández Theorem 9 If the parameters a 1 a b k k 3 and d satisfy the hypothesis given in Lemma 3 then the differential system 1 exhibits an Andronov Hopf bifurcation at p 0 = 5 ρ 64ρ +9 65ρ 4 18ρ +18 with respect to the parameter c and its bifurcation value is c 0 ρ where ρ > 0 and ρ = ρ 0 Moreover if ρ > ρ 0 the bifurcation is subcritical and if ρ < ρ 0 the bifurcation is supercritical Proof From Lemma 3 the linearization M p0 c ρ of differential system 1 at p 0 has a positive real eigenvalue and a conjugate pair of pure imaginary eigenvalues if c = c 0 ρ From Lemma 8 the derivative of the real part of the complex eigenvalues is 1664ρ c 0 ρ ρ = c 16384ρ + 151 which is negative for ρ = 0 and hence the transversality condition holds The Lemma 5 and Corollary 6 give a negative first Lyapunov coefficient if ρ < ρ 0 and a positive first Lyapunov coefficient if ρ > ρ 0 Then the hypotheses of Andronov Hopf bifurcation Theorem see Refs [7 9] hold and we conclude the proof Lemma 10 Second Lyapunov coefficient If we have the assumptions given in Lemma 3 then the second Lyapunov coefficient of the differential system 1 at the equilibrium point p 0 is given by l p 0 c 0 ρ ρ 64ρ + 9 s1 ρ = 65804544ρ 9 4096ρ + 151 3 16384ρ + 151 3 16384ρ + 13689 s ρ 14 where s 1 ρ = 168408831837157778187004408361897984ρ 6 11593545316357771314958979006464ρ 4 + 844510001357516308069633148790890496ρ + 88370770375111636106636084589363ρ 0 109618776714701834747940641433301549056ρ 18 19437015837810733849076799850637950ρ 16 113113899478591363401500081175360ρ 14 331896113104957376711495416864784816ρ 1 51371580818961494819571679640576ρ 10 30599875788551890754596438876606303ρ 8 + 3044031039595973584678047367564897ρ 6 + 703136698566108044013136776088ρ 4 + 499370844954383765695584ρ + 1343578315861950477388915456 s ρ = 100ρ 4 + 15ρ + 81 Moreover if ρ = ρ 0 then l p 0 c 0 ρ ρ = 0 where ρ 0 is given in the Corollary 6

Andronov Hopf and Bautin bifurcation 9 Proof In order to compute the second Lyapunov coefficient l we apply the Kuznetsov formula see Ref [4] Taking into account the assumptions of this Lemma and using the Mathematica software we obtain that the second Lyapunov coefficient l p 0 c 0 ρ ρ of the differential system 1 at the equilibrium point p 0 is given by 14 and l p 0 c 0 ρ 0 ρ 0 740065 Corollary 6 Lemma 8 and Lemma 10 provide the validity of the necessary and sufficient conditions to apply the Bautin bifurcation theorem see Ref [4] In summary we have the following result Theorem 11 Bautin bifurcation in linear growth If the parameters a 1 a b k k 3 and d satisfy the hypothesis given in Lemma 3 then the differential system 1 exhibits a Bautin bifurcation at p 0 with respect to the parameters c and ρ and its critical bifurcation value is c 0 ρ 0 ρ 0 3 Logistic case In this section we consider the differential system 1 with a logistic growth for the prey this means that the function hx = ρx 1 x R and we will analyze the differential system ẋ = ρx 1 x a 1xy R b 1 + x ẏ = a 1c 1 xy b 1 + x a yz b + y c y a c 3 y ż = z b + y d In order to make ecological sense we assume that all parameters of the system 31 are positive Lemma 31 If the parameters a 1 a b 1 c 1 and R satisfy a 1 = ρ b 1 + x 0 R x 0 Ry 0 a = d b + y 0 c 3 y 0 R = k + x 0 31 c 1 = c c 3 y 0 k + x 0 + k 3 c 3 k ρx 0 b 1 = k x 0 + k 4 3 then the unique equilibrium points of the differential system 31 in the region Ω are p 1 = p = p 3 = k 3 x 0 y 0 d k 7 + k 8 + 6x 0 k7 + x 0 y 0 c c 3 k 7 y 0 k 8 + x 0 k 7 + k 8 + 6x 0 + k 3 k 7 + x 0 k 8 + 4x 0 d x 0 k 7 + k 8 + 4x 0 k 7 + k 8 + 6x 0 k8 + x 0 y 0 c c 3 k 8 y 0 k 7 + x 0 k 7 + k 8 + 6x 0 + k 3 k 7 + 4x 0 k 8 + x 0 d x 0 k 7 + k 8 + 4x 0 k 7 + k 8 + 6x 0 Where x 0 > 0 y 0 > 0 k 3 > 0 k 7 0 k 8 0 and k = 4x 0 + k 7 + k 8 k 4 = 1 4 k 5k 6 k 5 = x 0 + k 7 k 6 = x 0 + k 8 33

10 G Blé V Castellanos and I Loreto-Hernández Proof The equilibrium points of the differential system 31 are the solutions of the system ρx 1 x a 1xy R b 1 + x = 0 a 1 c 1 xy b 1 + x a yz b + y c y = 0 a c 3 y z b + y d = 0 Multiplying the above equations by b 1 + x b + y which is always positive in the region Ω we obtain that the equilibrium point must satisfy 34 Correspondingly each solution of 34 must be an equilibrium point of the differential system 31 a 1 Ry ρ b 1 + x R x = 0 b + y c b1 + x a 1 c 1 x + a z b 1 + x = 0 34 d b + y a c 3 y = 0 A point x 0 y 0 z 0 Ω is an equilibrium point of the differential system 31 if a 1 Ry 0 ρ b 1 + x 0 R x0 = 0 b + y 0 c b1 + x 0 a1 c 1 x 0 + a z 0 b1 + x 0 = 0 d b + y 0 a c 3 y 0 = 0 We suppose x 0 > 0 y 0 > 0 and z 0 > 0 Note that the first equation of the system 35 is a linear equation in terms of a 1 and it has the unique solution a 1 = ρ b 1 + x 0 R x 0 Ry 0 Since a 1 > 0 R x 0 must be positive so there exists k > 0 such that R = x 0 + k A similar argument using the third equation of system 35 we obtain that: a = d b + y 0 c 3 y 0 Using the values of a 1 a and R and solving the second equation of system 35 for z 0 we have that z 0 = c 1c 3 k ρx 0 c c 3 y 0 k + x 0 d k + x 0 Since z 0 > 0 there must exists k 3 > 0 such that c 1 c 3 k ρx 0 c c 3 y 0 k + x 0 = k 3 Then c 1 = c c 3 y 0 k + x 0 + k 3 c 3 k ρx 0 Therefore if a 1 a R and c 1 satisfy 3 then x 0 y 0 z 0 is a solution of system 34 in Ω k where z 0 = 3 Moreover the system 34 takes the form d k +x 0 k ρy b 1 + x 0 ρ b 1 + x k x + x 0 = 0 y 0 35 b + y c b1 + x Q + d z b 1 + x b + y 0 c 3 y 0 = 0 36 b d y 0 y y 0 = 0

Andronov Hopf and Bautin bifurcation 11 where Q = xb 1+x 0 c c 3 y 0 k +x 0 +k 3 Solving the third equation of system 36 for y we have c 3 x 0 y 0 k +x 0 that y = y 0 Moreover the first equation of system 36 reduce to: Hence the solutions of this equation are x 0 x 1 := 1 ρx x 0 x k x + b 1 k x 0 = 0 k k k + 4x 0 4b 1 x := 1 k + k k + 4x 0 4b 1 Thus a necessary condition to have at least two solutions of system 36 in Ω is that k k + 4x 0 4b 1 0 On the other hand x 1 > 0 if and only if 0 < k k k + 4x 0 4b 1 = 4b 1 k x 0 then x 1 > 0 if and only if there exists k 4 > 0 such that b 1 = k x 0 + k 4 which is a hypothesis in 3 Let k 5 = k k 4k 4 > 0 then k 4 = 1 4 k 5k 6 where k 6 = k k 5 > 0 Hence x 1 = k 5 and x = k 6 Substituting b 1 k 4 k 5 k y = y 0 and x = x 1 in the second equation of system 36 and solving this equation for z we have that z 1 = c c 3 k 6 y 0 k 5 x 0 k 5 + k 6 + x 0 + k 3 k 5 k 6 + x 0 d x 0 k 5 + k 6 k 5 + k 6 + x 0 Moreover if k 5 x 0 0 then z 1 > 0 In the same way replacing y = y 0 and x = x in 36 we obtain z = c c 3 k 5 y 0 k 6 x 0 k 5 + k 6 + x 0 + k 3 k 6 k 5 + x 0 d x 0 k 5 + k 6 k 5 + k 6 + x 0 Also if k 6 x 0 0 then z > 0 Let k 7 = k 5 x 0 and k 8 = k 6 x 0 then x 1 = k 7 + x 0 x = k 8 + x 0 and z 1 z becomes z 1 = c c 3 k 7 y 0 k 8 + x 0 k 7 + k 8 + 6x 0 + k 3 k 7 + x 0 k 8 + 4x 0 d x 0 k 7 + k 8 + 4x 0 k 7 + k 8 + 6x 0 z = c c 3 k 8 y 0 k 7 + x 0 k 7 + k 8 + 6x 0 + k 3 k 7 + 4x 0 k 8 + x 0 d x 0 k 7 + k 8 + 4x 0 k 7 + k 8 + 6x 0 Therefore the unique equilibrium points of the differential system 31 in the region Ω are: which completes the proof p 1 = x 0 y 0 z 0 p = x 1 y 0 z 1 and p 3 = x y 0 z Remark 3 Choosing the values k 7 k 8 adequately we obtain one two or three equillibria 1 If k 7 = k 8 = 0 then p 1 = p = p 3 = k x 0 y 0 3 3d x 0 is the unique equilibrium point of the differential system 31 in Ω If k 7 = 0 and k 8 > 0 then p 1 = p = x 0 y 0 p 3 = k 3 d k 8 +6d x 0 and k8 + x 0 y 0 c c 3 k 8 y 0 k 8 + 6x 0 + 4k 3 k 8 + x 0 d k 8 + 4x 0 k 8 + 6x 0 hence there are two equilibrium points of the differential system 31 in Ω

1 G Blé V Castellanos and I Loreto-Hernández 3 If k 7 > 0 k 8 > 0 and k 7 = k 8 then k 3 p 1 = x 0 y 0 d k 7 + k 8 + 6x 0 p = p 3 = k7 + x 0 y 0 c c 3 k 7 y 0 k 8 + x 0 k 7 + k 8 + 6x 0 + k 3 k 7 + x 0 k 8 + 4x 0 d x 0 k 7 + k 8 + 4x 0 k 7 + k 8 + 6x 0 k8 + x 0 y 0 c c 3 k 8 y 0 k 7 + x 0 k 7 + k 8 + 6x 0 + k 3 k 7 + 4x 0 k 8 + x 0 d x 0 k 7 + k 8 + 4x 0 k 7 + k 8 + 6x 0 are three different equilibrium points of system 31 in Ω 31 One equilibrium point of the differential system In this subsection we assume that the parameters a 1 a b 1 c 1 R k k 4 k 5 and k 6 satisfy the conditions 3 and 33 of Lemma 31 and k 7 = k 8 = 0 Then according to Remark 3 p 1 = k x 0 y 0 3 3d x 0 is the unique equilibrium point of the differential system 31 in Ω Proposition 33 The equilibrium point p 1 is not hyperbolic and it has a local unstable manifold of dimension Proof Under the hypothesis of this subsection the characteristic polynomial of the linear approximation M p1 of differential system 31 at p 1 is Pλ = λ 3 k 3 + λ ρb + y 0 3c c 3 x 0 y 0 + k 3 + 3b d k 3 λ 3b c 3 x 0 + 3c 3 x 0 y 0 9c 3 x 0 y 0 b + y 0 The eigenvalues of M p1 are and λ 1 = 0 k 3 y 0 y 0 4c 3 x 0 b + y 0 ρb + y 0 3c c 3 x 0 y 0 + k 3 3b d k 3 + k 3 y 0 λ = 6c 3 x 0 y 0 b + y 0 λ 3 = k 3 y 0 + y 0 4c 3 x 0 b + y 0 ρb + y 0 3c c 3 x 0 y 0 + k 3 3b d k 3 + k 3 y 0 6c 3 x 0 y 0 b + y 0 Since λ 1 = 0 the equilibrium point p 1 of differential system 31 is not hyperbolic Moreover if λ and λ 3 are complex then Reλ = Reλ 3 = k 3 y 0 6c 3 x 0 y 0 b + y 0 > 0 It is not difficult to see that if λ and λ 3 are real then λ > 0 and λ 3 > 0 Therefore the equilibrium point p 1 of differential system 31 has a local unstable manifold of dimension Corollary 34 The differential system 31 does not exhibit an Andronov Hopf bifurcation at the equilibrium point p 1 = k x 0 y 0 3 3d x 0

Andronov Hopf and Bautin bifurcation 13 3 Two equilibrium points of the differential system From now on this subsection we assume that the parameters a 1 a b 1 c 1 R k k 4 k 5 and k 6 satisfy the conditions 3 and 33 of Lemma 31 k 7 = 0 and k 8 > 0 By Remark 3 p 1 = k x 0 y 0 3 d k 8 +6d x 0 and p = k 8 + x 0 y 0 c c 3 k 8 y 0 k 8 +6x 0 +4k 3 k 8 +x 0 d k 8 +4x 0 k 8 +6x 0 are the unique two equilibrium points of differential system 31 in Ω Proposition 35 The equilibrium point p 1 is not hyperbolic and it has a local unstable manifold of dimension Proof Considering the assignations for the parameters a 1 a b 1 c 1 R k k 4 k 5 k 6 and k 7 given in this subsection the characteristic polynomial of the linear approximation M p1 of differential system 31 at p 1 is Pλ = λ 3 + and the eigenvalues of M p1 are λ 1 = 0 λ = λ 3 = k 3 c 3 b + y 0 k 8 + 6x 0 λ ρb + y 0 k 8 + x 0 c c 3 y 0 k 8 + 6x 0 + k 3 + b d k 3 k 8 + 6x 0 c 3 y 0 b + y 0 k 8 + 6x 0 λ k 3 c 3 b + y 0 k 8 + 6x 0 k 3 c 3 b + y 0 k 8 + 6x 0 + Q1 c 3 y 0 b + y 0 k 8 + 6x 0 Q1 c 3 y 0 b + y 0 k 8 + 6x 0 Q 1 = b c 3 d k 3 k 8 + 6x 0 c 3 ρb + y 0 k 8 + x 0 c c 3 y 0 k 8 + 6x 0 + k 3 + k 3 y 0 k 3 b c 3 d k 8 + 6x 0 Then the equilibrium point p 1 of differential system 31 is not hyperbolic Moreover if λ and λ 3 are complex then Reλ = Reλ 3 = k 3 c 3 b + y 0 k 8 + 6x 0 > 0 And it can be verify that if λ and λ 3 are real then λ > 0 and λ 3 > 0 Therefore the equilibrium point p 1 of differential system 31 is not hyperbolic and has a local unstable manifold of dimension Corollary 36 The differential system 31 does not exhibit an Andronov Hopf bifurcation at the equilibrium point p 1 = k x 0 y 0 3 d k 8 +6d x 0 Whereas the equilibrium point p 1 does not have an Andronov Hopf bifurcation we will show that the equilibrium point p can have a pair of purely imaginary eigenvalues and consequently it can exhibit an Andronov Hopf bifurcation Lemma 37 If the parameters k 8 b k 3 c ρ and d satisfy the conditions k 8 = x 0 b = c 3x 0 y 0 35c + ρ + 60k 3 k 3 = c c 3 x 0 y 0 c = 3c 3 ρx 0 581875 ρ < 5877 d = d 0 ρ := 30060160ρ 3 116375 873ρ 581875 5877ρ 581875 5877ρ 143640ρ 37

14 G Blé V Castellanos and I Loreto-Hernández then the equilibrium point p of differential system 31 is given by p = 3x 0 y 0 c 3 581875 5877ρ 116375 873ρ y 0 84687918336000ρ 4 and the eigenvalues of the linear approximation of system 31 at p are 59ρ α = 3 475 9ρ + 3065 and ± i Proof Let M p be the Jacobian matrix of the differential system 31 evaluated at the equilibrium point p then the characteristic polynomial Pλ = detλi M p = λ 3 + A 1 λ + A λ + A 3 where A 1 = λ k 8 +4x 0 c c 3 k 8 y 0 k 8 +6x 0 +4k 3 k 8 +x 0 c 3 k 8 ρb +y 0 k 8 +x 0 c 3 b +y 0 k 8 +4x 0 k 8 +6x 0 A = b d k 8 +6x 0 k 8 +4x 0 c c 3 k 8 y 0 k 8 +6x 0 +4k 3 k 8 +x 0 B 1 c 3 y 0 b +y 0 k 8 +4x 0 3 k 8 +6x 0 ρy 0k 8 +x 0 k 8 +4x 0 c c 3 y 0 k 8 +6x 0 k 8 +4k 8 x 0 16x 0 +4k 3 k 8 +4x 0 k 8 x 0 c 3 y 0 b +y 0 k 8 +4x 0 3 k 8 +6x 0 A 3 = b d k 8 ρk 8 +x 0 c c 3 k 8 y 0 k 8 +6x 0 +4k 3 k 8 +x 0 B 1 = 8b ρx 0 k 8 + x 0 c 3 y 0 b +y 0 k 8 +4x 0 3 k 8 +6x 0 8x 0 k 8 c c 3 y 0 k 8 + 6x 0 + k 3 Taking k 8 b and k 3 satisfying 37 then A 1 A and A 3 are reduced to 1ρ A 1 = 17595c + 4ρ A = c 665c 475d + 16ρ + ρ335d + 5877ρ 61595c + 4ρ A 3 = 57c d ρ95c + ρ 61595c + 4ρ Following the proof of Lemma 3 the characteristic polynomial Pλ has a pair of purely imaginary roots ±iω and a real root α if and only if A > 0 and where ω = A and α = A 1 A 1 A A 3 = 0 38 Since A 1 A A 3 = 3c ρ3000815c d +665c ρ475d 864ρ 3508ρ 3 107187595c +4ρ solving equation 38 for the parameter d we have that d = 36ρ 15960c + 653ρ 315875c 95c + ρ Taking into account this assignment for d the coefficient A = 9ρ15960c +653ρ 581875 > 0 which is equal to 1 when c = 581875 5877ρ 143640ρ Moreover c > 0 if ρ < 581875 5877

Andronov Hopf and Bautin bifurcation 15 Therefore the characteristic polynomial Pλ has a pair of purely imaginary roots ±i and a 59ρ 3 real root α = 4759ρ +3065 The equilibrium point p of system 31 becomes 3x 0 p = y 0 c 3 581875 5877ρ 116375 873ρ y 0 84687918336000ρ 4 Remark 38 If the assumptions of Lemma 37 are satisfied then the linear approximation of the differential system 31 at p has two eigenvalues purely imaginary when d = d 0 ρ Hence by continuity on the eigenvalues the linear approximation of the differential system at p has a pair of complex eigenvalues when d is in a neighborhood of d 0 ρ λp d ρ = ξp d ρ ± iωp d ρ In order to compute the first Lyapunov coefficient l 1 we make the following assignations c 3 = 1 y 0 = 1 and x 0 = 1 Applying the Kuznetsov formula see Ref [7] and using the Mathematica software we obtain the first Lyapunov coefficient l 1 p d 0 ρ ρ of the differential system 31 at the equilibrium point p Lemma 39 If we have the assumptions given in Lemma 37 then the eigenvalues of the linear approximation of system 31 at the equilibrium point p are α = Lyapunov coefficient 59ρ 3 4759ρ +3065 and ±i and the first where l 1 p d 0 ρ ρ = 37791360ρ5 9ρ + 3065 s 3 ρ 49 5877ρ 581875 s 4 ρs 6 ρs 5 ρ s 3 ρ = 667911733488169984885774464ρ 14 + 6174013587685160995638930875ρ 1 44174467587991308958764393437500ρ 10 + 338793464197339538653798974609375ρ 8 + 17775906705145400843981933593750000ρ 6 39 841493581845049559575439006408984375ρ 4 + 1078380414077910199417846085034179687500ρ 433196073499560640944003313779830936171875 and s 4 ρ = 11611576565 + 9ρ 1381953150 + 03065ρ + 18664ρ 4 s 5 ρ = 11611576565 + 9ρ 1381953150 + 03065ρ + 746496ρ 4 s 6 ρ = 458541644971313476565 + 9ρ 8741951684675781500 + 7ρ 877811316768150 + 79497391678300ρ + 108361587ρ 4

16 G Blé V Castellanos and I Loreto-Hernández Corollary 310 If we have the assumptions given in Lemma 37 then there exists a unique real number 581875 0 < ρ 0 < 5877 such that the first Lyapunov coefficient l 1p d 0 ρ 0 ρ 0 = 0 Proof By Lemma 39 and equation 39 the first Lyapunov coefficient l 1 p d 0 ρ ρ = 0 if and only if s 3 ρ = 0 By Descartes rule of signs there are 1 3 or 5 positive real numbers ρ such that s 3 ρ = 0 Indeed numerically this equation has three positive solutions but only 581875 ρ 0 976907 is less than 5877 Lemma 311 Bautin regularity condition If the parameters k 8 b k 3 c and ρ satisfy the relations 37 of Lemma 37 then the map d ρ ξp d ρ l 1 p d ρ ρ is regular at d 0 ρ 0 where ξp d ρ is given in Remark 38 and d 0 := d 0 ρ 0 Proof By hypothesis the linear approximation of the differential system 31 at p depends only on the parameters d and ρ let M p d ρ be this linear approximation By Lemma 37 the real part of the complex eigenvalues of M p d 0 ρ 0 are ξp d 0 ρ 0 = 0 310 Let p and q be eigenvectors of M p d 0 ρ 0 for the corresponding eigenvalues i and i respectively such that q tr q = 1 and p tr q = 1 311 By 310 and 311 we can apply Lemma 7 to the linear approximation M p d ρ Taking into account the values of q p and M p 0 d ρ d and using the Mathematica software we obtain the partial derivative of the real part of the eigenvalues ξd ρ ± iωd ρ of M p d ρ d 0 ρ 0 = 5 581875 5877ρ 0 116375 873ρ 0 31 d Q Q = 4939 9 746496ρ 4 0 + 03065ρ 0 + 1381953150 ρ 0 + 11611576565 Applying Lemma 7 one more time it follows from the values of q p and M p d ρ ρ that ρ d 0 ρ 0 = 9700ρ 0 171007ρ 4 0 + 39730450ρ 0 6771570315 313 Q 3 Q 3 = 873ρ 0 116375 9 746496ρ 4 0 + 03065ρ 0 + 1381953150 ρ 0 + 11611576565 Using the Wolfram Mathematica software we have that Re C 1 d d 0 ρ 0 0637 and Re C 1 ρ d 0 ρ 0 15886065 where C 1 d ρ is the function given in the proof of Lemma 8 and by Corollary 310 ρ 0 976907 Then by 31 313 and the analogous of formula 13 given in the proof of Lemma 8 we have that det d d 0 ρ 0 l 1 d d 0 ρ 0 ρ d 0 ρ 0 l 1 ρ d 00091456 0 ρ 0 Hence the map d ρ ξp d ρ l 1 p d ρ is regular at d 0 ρ 0

Andronov Hopf and Bautin bifurcation 17 Theorem 31 If the parameters k 8 b k 3 c and ρ satisfy the relations 37 of Lemma 37 then the differential system 31 exhibits an Andronov Hopf bifurcation at p = 3 1 581875 5877ρ 116375 873ρ 84687918336000ρ 4 581875 with respect to the parameter d and its critical bifurcation value is d 0 ρ where ρ 0 5877 and ρ = ρ 0 Moreover if ρ < ρ 0 the bifurcation is subcritical and if ρ > ρ 0 the bifurcation is supercritical Proof From Lemma 37 the linear approximation M p d ρ of differential system 31 at p has a negative real eigenvalue and a pair of purely imaginary eigenvalues if d = d 0 ρ From Lemma 311 the derivative of the real part of the complex eigenvalues d 0 ρ ρ = 5 581875 5877ρ 116375 873ρ d Q 4 Q 4 = 4939 9 746496ρ 4 + 03065ρ + 1381953150 ρ + 11611576565 which is positive if ρ 0 581875 5877 and hence the transversality condition holds By Corollary 310 the first Lyapunov coefficient is negative if ρ > ρ 0 and is positive if ρ < ρ 0 Then the hypotheses of Andronov Hopf bifurcation Theorem hold and we conclude the proof see Refs [7 9] Lemma 313 Second Lyapunov coefficient If we have the assumptions given in Lemma 37 and ρ = ρ 0 then the second Lyapunov coefficient of differential system 31 at the equilibrium point p l p d 0 ρ 0 ρ 0 = 0 Proof In order to compute the second Lyapunov coefficient l we apply the Kuznetsov formula see Ref [4] Taking into account the assumptions of this Lemma and using the Mathematica software we obtain that the second Lyapunov coefficient l p d 0 ρ ρ of the differential system 31 at the equilibrium point p takes the value l p d 0 ρ 0 ρ 0 889415 if ρ = ρ 0 Corollary 310 Lemma 311 and Lemma 313 provide the validity of the necessary and sufficient conditions to apply the Bautin bifurcation theorem see Ref [4] Then we have obtained the following Theorem 314 If the parameters k 8 b k 3 c and ρ satisfy the relations 37 of Lemma 37 then the differential system 31 exhibits a Bautin bifurcation at p with respect to the parameters d and ρ and its critical bifurcation value is d 0 ρ 0 ρ 0 33 Three equilibrium points of the differential system From now on in this subsection we assume that the parameters a 1 a b 1 c 1 R k k 4 k 5 and k 6 satisfy the conditions 3 and 33 of Lemma 31 k 7 > 0 k 8 > 0 and k 7 = k 8

18 G Blé V Castellanos and I Loreto-Hernández Then by Remark 3 p 1 = p = p 3 = k 3 x 0 y 0 d k 7 + k 8 + 6x 0 k7 + x 0 y 0 c c 3 k 7 y 0 k 8 + x 0 k 7 + k 8 + 6x 0 + k 3 k 7 + x 0 k 8 + 4x 0 d x 0 k 7 + k 8 + 4x 0 k 7 + k 8 + 6x 0 k8 + x 0 y 0 c c 3 k 8 y 0 k 7 + x 0 k 7 + k 8 + 6x 0 + k 3 k 7 + 4x 0 k 8 + x 0 d x 0 k 7 + k 8 + 4x 0 k 7 + k 8 + 6x 0 are the unique three equilibrium points of differential system 31 in Ω 331 Local dynamics and bifurcation at p 1 Lemma 315 If the parameters k 7 k 8 b k 3 c and ρ satisfy the conditions k 7 = x 0 k 8 = x 0 / b = 7k 3 + y 0 k 3 = c c 3 x 0 y 0 c = 459 10358ρ c 3 ρx 0 140049ρ ρ < 459 1349380771ρ d = d 0 ρ := 3 314 10358 459 5171ρ 459 10358ρ then the equilibrium point p 1 = x 0 y 0 8c 3 459 10358ρ 459 5171ρ y 0 9401813691643ρ 4 and the eigenvalues of the linear approximation at p 1 are 1383ρ α = 3 448ρ + 334193 and ± i Proof The characteristic polynomial of the linear approximation M p1 of differential system 31 at the equilibrium point p 1 is Pλ = detλi M p1 = λ 3 + A 1 λ + A λ + A 3 where A 1 = k 3 c 3 b +y 0 k 7 k 8 ρx 0 k 7 +4x 0 k 8 +4x 0 B 3 A = ρy 0 c c 3 y 0 B 3 B + k 3 k 7 + 4x 0 k 8 + 4x 0 k 7 + k 8 + x 0 c 3 y 0 b + y 0 k 7 + 4x 0 k 8 + 4x 0 B 3 A 3 = + b ρb c c 3 y 0 B 3 + k 3 + b d k 3 k 7 + 4x 0 k 8 + 4x 0 B 3 c 3 y 0 b + y 0 k 7 + 4x 0 k 8 + 4x 0 B 3 4b d k 3 k 7 k 8 ρx 0 c 3 y 0 b + y 0 k 7 + 4x 0 k 8 + 4x 0 B 3 B = k 7 + k 8 + 4x 0 4x 0 k 7 + k 8 + k 7 k 8 + 8x 0 B 3 = k 7 + k 8 + 6x 0

Andronov Hopf and Bautin bifurcation 19 By hypothesis k 7 k 8 b and k 3 satisfy 314 then A 1 A and A 3 reduce to 8ρ A 1 = 4597c + ρ A = c 81c 61d + 179ρ + ρ918d + 5179ρ 78037c + ρ A 3 = 16c d ρ7c + ρ 78037c + ρ As in the proof of Lemma 3 the characteristic polynomial Pλ has a pair of purely imaginary roots ±iω and a real root α if and only if A > 0 and where ω = A and α = A 1 In this case A 1 A A 3 = 0 315 A 1 A A 3 = 8c ρ 669c d + 81c ρ179ρ 306d + 10358ρ 3 35815777c + ρ Solving the equation 315 for the parameter d we have that d = ρ 140049c + 10358ρ 4786c 7c + ρ Taking c = 459 10358ρ 140049ρ we have A = 1 Since c > 0 then the parameter ρ must satisfy ρ < 459 10358 Therefore the characteristic polynomial Pλ has a pair of purely imaginary roots ±i and a 1383ρ 3 real root α = 448ρ +334193 If k 7 k 8 b k 3 c ρ and d satisfy the relations given by 37 then p 1 = x 0 y 0 8c 3 459 10358ρ 459 5171ρ y 0 9401813691643ρ 4 Remark 316 If the assumptions of Lemma 315 are satisfied then the linear approximation of the differential system 31 at p 1 has two purely imaginary eigenvalues when d = d 0 ρ hence by continuity on the eigenvalues the linear approximation of the differential system at p 1 has a pair of complex eigenvalues when d is in a neighborhood of d 0 ρ λp 1 d ρ = ξp 1 d ρ ± iωp 1 d ρ In order to compute the first and second Lyapunov coefficients we make the following assignations c 3 = 1 y 0 = 1 and x 0 = 1 Applying the Kuznetsov formula and using the Mathematica software we obtain the next result

0 G Blé V Castellanos and I Loreto-Hernández Lemma 317 If the assumptions given in Lemma 315 hold then the eigenvalues of the linear approximation of the differential system 31 at the equilibrium point p 1 are α = The first Lyapunov coefficient is where l 1 p 1 d 0 ρ ρ = s 3 ρ = 170748805865381069386373596ρ 14 3194771ρ5 16ρ + 459 s 3 ρ 4 10358ρ 459 s 4 ρs 5 ρs 6 ρ + 49697966755843616165963008ρ 1 10816145463670610005586850486688ρ 10 11670758978607800175094619917568716ρ 8 948599985830719899667794878711506ρ 6 + 8965980591097747554775891941073773130ρ 4 + 14441398419369459748638034904616060180ρ 1840686148630673837681774164640096430491 s 4 ρ = 4590086517896384ρ 8 + 556754388989669584ρ 6 1383ρ 3 15316ρ +10681 + 147414771406943567049ρ 4 + 4695185844785496768881ρ + 3155590506486781936 s 5 ρ = 16 989441ρ 4 + 374544ρ + 9863663058 ρ + 103904319836149 s 6 ρ = 3 597888ρ 4 + 1877ρ + 493183159 ρ + 103904319836149 and ±i Corollary 318 If the assumptions given in Lemma 315 hold then there exists a unique real number 0 < ρ 0 < 459 such that the first Lyapunov coefficient 10358 l 1 p 1 d 0 ρ 0 ρ 0 = 0 Proof By Lemma 317 the first Lyapunov coefficient l 1 p 1 d 0 ρ ρ = 0 if and only if s 3 ρ = 0 According to Descartes rule of signs there are 1 or 3 positive real numbers ρ such that s 3 ρ = 0 Indeed numerically this equation has three positive solutions but only ρ 0 371999 is less than 459 10358 Lemma 319 Bautin regularity condition If the parameters k 3 k 7 k 8 b c and ρ satisfy the relations 314 of Lemma 315 then the map d ρ ξp 1 d ρ l 1 p 1 d ρ ρ is regular at d 0 ρ 0 where ξp 1 d ρ is given in Remark 316 and d 0 := d 0 ρ 0 Proof By hypothesis the linear approximation of the differential system 31 at p 1 depends only on the parameters d and ρ Let M p1 d ρ be this linear approximation By Lemma 315 the complex numbers ±i are eigenvalues of M p1 d 0 ρ 0 Hence the real part of the complex eigenvalues of M p1 d 0 ρ 0 is ξp 1 d 0 ρ 0 = 0 Let p and q be eigenvectors of M p1 d 0 ρ 0 for the corresponding eigenvalues i and i respectively such that q tr q = 1 and p tr q = 1

Andronov Hopf and Bautin bifurcation 1 By Lemma 7 and taking into account the values of q p M p 1 d ρ d and M p 1 d ρ ρ we obtain d 0 ρ 0 = 8 459 10358ρ 0 459 5171ρ 0 316 d 46683Q 5 Q 5 = 3 597888ρ 4 0 + 1877ρ 0 + 493183159 ρ 0 + 103904319836149 ρ d 0 ρ 0 = 1058148ρ 0 Q 7 = 3 5356118ρ 4 0 + 37166549ρ 0 13315945183 5171ρ 0 10681 Q 7 317 597888ρ 4 0 + 1877ρ 0 + 493183159 ρ 0 + 103904319836149 Numerically taking ρ 0 976907 we have that Re C 1 d d 0 ρ 0 06034 and Re C 1 ρ d 0 ρ 0 147856 where C 1 d ρ is as in the proof of Lemma 8 Then by 316 317 and the analogous formula of 13 given in Lemma 8 det d d 0 ρ 0 l 1 d d 0 ρ 0 ρ d 0 ρ 0 l 1 ρ d 00091 0 ρ 0 Hence the map d ρ ξp 1 d ρ l 1 p 1 d ρ is regular at d 0 ρ 0 Theorem 30 If the parameters k 3 k 7 k 8 b c and ρ satisfy the relations 314 of Lemma 315 then the differential system 31 exhibits an Andronov Hopf bifurcation at p 1 = 1 1 8 459 10358ρ 459 5171ρ 9401813691643ρ 4 with respect to the parameter d and its critical bifurcation value is d 0 ρ where ρ 0 459/ 10358 and ρ = ρ 0 Moreover if ρ < ρ 0 the bifurcation is subcritical and if ρ > ρ 0 the bifurcation is supercritical Proof From Lemma 315 the linearization M p1 d ρ of differential system 31 at p 1 has a negative real eigenvalue and a pair of purely imaginary eigenvalues if d = d 0 ρ From Lemma 319 the derivative of the real part of the complex eigenvalues d 0 ρ ρ = 8 459 10358ρ 459 5171ρ d 46683Q 8 Q 8 = 3 597888ρ 4 + 1877ρ + 493183159 ρ + 103904319836149 which is positive if ρ 0 459/ 10358 and hence the transversality condition holds Lemma 317 and Corollary 318 imply that the first Lyapunov coefficient is negative if ρ > ρ 0 and is positive if ρ < ρ 0 see Figure 31 Then the hypotheses of Andronov Hopf bifurcation theorem hold and we conclude the proof In order to show the Bautin bifurcation we compute the second Lyapunov coefficient l Applying the Kuznetsov formula and using the Mathematica software we obtain that the second Lyapunov coefficient l p 1 d 0 ρ ρ of the differential system 31 at the equilibrium point p 1 takes the value l p 1 d 0 ρ 0 ρ 0 67181

G Blé V Castellanos and I Loreto-Hernández Lemma 31 Second Lyapunov coefficient If we have the assumptions given in Lemma 315 then the second Lyapunov coefficient of differential system 31 at the equilibrium point p 1 l p 1 d 0 ρ 0 ρ 0 = 0 From Corollary 318 Lemma 319 and Lemma 31 we have the necessary and sufficient conditions to apply the Bautin bifurcation theorem Therefore we obtain the following Theorem 3 If the parameters k 3 k 7 k 8 b c and ρ satisfy the relations 314 of Lemma 315 then the differential system 31 exhibits a Bautin bifurcation at p 1 with respect to the parameters d and ρ and its critical bifurcation value is d 0 ρ 0 ρ 0 From Theorems 30 and 3 we have shown the existence of limit cycles in Ω to differential system 31 near to p 1 Now we will analyze the local dynamics at equilibrium point p 33 Local dynamics and bifurcation at p In this subsection we assume that the parameters k 3 k 7 k 8 b c and ρ satisfy the relations 314 of Lemma 315 and x 0 = y 0 = c 3 = 1 In the same way of the previous subsection we obtain the next results relative to p Lemma 33 If d = d 1 ρ := 4ρ 39818709 1509ρ 16966ρ + 1430179 544749 459 5171ρ 459 10358ρ then the equilibrium point p of the differential system 31 is given by 3869893 459 10358ρ 459 5171ρ p = 1 588058ρ 39818709 1509ρ 16966ρ + 1430179 and the eigenvalues of the linear approximation of system 31 at p are α = ρ 16966ρ + 1430179 5967 16ρ and ± 4 39818709 1509ρ + 10681 1 i 38779 Lemma 34 If we have the assumptions given in Lemma 33 the first Lyapunov coefficient at p is l 1 p d 1 ρ ρ = 1587777577 47ρ 5 16ρ + 10681 1509ρ 39818709 σ 3 ρ 314 651537313 3878444ρ 10358ρ 10681 σ 4 ρσ 5 ρσ 6 ρ where σ 3 ρ = 6749435441467886954839873977006431186560ρ 14 + 8051848708654778807704908091467190331979785960ρ 1 833510517876964764784536789656777949050385850640ρ 10 39908334741334897543661396558797698093938169581979104ρ 8 + 189394169707008875741540495071978539336677498753530ρ 6 103337845134816394773870756117998401045950936989864334ρ 4 + 045655110149866337445394616316667453076170351157348636599ρ 13861057196949501517943399705176765759635757614394568670153

Andronov Hopf and Bautin bifurcation 3 σ 4 ρ = 10987383148633964ρ 6 + 1537746581714574ρ 4 9645018613960399565ρ + 846039668614879118144 σ 5 ρ = 10988745581469548ρ 6 + 15595061076430844ρ 4 6961875797976491437ρ + 061509917153119779536 and σ 6 ρ = 8930934748868470660ρ 8 93757116749758615866756ρ 6 + 7535731011874816587361581ρ 4 + 1115868587553717803656518ρ + 764308163618984385308053 Corollary 35 There exists a unique real number 0 < ρ 1 < 459 10358 such that the first Lyapunov coefficient l 1 p d 1 ρ 1 ρ 1 = 0 Indeed ρ 1 436757 Lemma 36 Bautin regularity condition The map d ρ ξp d ρ l 1 p d ρ is regular at d 1 ρ 1 where ξp d ρ is as in Lemma 319 and d 1 := d 1 ρ 1 Theorem 37 The differential system 31 exhibits an Andronov Hopf bifurcation at p with respect to the parameter d and its critical bifurcation value is d 1 ρ where ρ 0 459/ 10358 and ρ = ρ 1 Moreover if ρ < ρ 1 the bifurcation is subcritical and if ρ > ρ 1 the bifurcation is supercritical In order to show a Bautin bifurcation we compute the second Lyapunov coefficient Lemma 38 Second Lyapunov coefficient If we have the assumptions given in Lemma 33 then the second Lyapunov coefficient of differential system 31 l p d 1 ρ 1 ρ 1 is negative Indeed l p d 1 ρ 1 ρ 1 16116 We summarize the results in the following theorem Theorem 39 The differential system 31 exhibits a Bautin bifurcation at p with respect to the parameters d and ρ and its critical bifurcation value is d 1 ρ 1 ρ 1 333 Local dynamics at p 3 Theorem 330 If the parameters k 3 k 7 k 8 b c and ρ satisfy the relations 314 of Lemma 315 then the differential system 31 does not exhibit a Hopf bifurcation at the equilibrium point 5 94 10681 10358ρ p 3 = 1 4 3095089d ρ Proof A necessary condition for a differential system to exhibit an Andronov Hopf bifurcation at an equilibrium point is that the characteristic polynomial of its linear approximation has a pair of purely imaginary roots According to the proof of Lemma 3 the characteristic polynomial Pλ = λ 3 + A 1 λ + A λ + A 3 has a pair of purely imaginary roots ±iω and a real root α if and only if A > 0 and A 1 A A 3 = 0 where ω = A and α = A 1 By hypothesis if M p3 is the linear approximation of differential system 31 at p 3 then Pλ = detλi M p3 = λ 3 + A 1 λ + A λ + A 3

4 G Blé V Castellanos and I Loreto-Hernández where A 1 = ρ417c + 10ρ 6637c + ρ A = c 81c 10387d + 14070ρ + ρ31161d + 84655ρ 146537c + ρ A 3 = 470c d ρ7c + ρ 146537c + ρ Since c = 459 10358ρ 140049ρ > 0 when 0 < ρ < 459 10358 we have that c ρq 9 A 1 A A 3 = 971447497c + ρ < 0 Q 9 = 3759854c d + 4754390c ρ + 878740c d ρ + 46697835c ρ + 846550ρ 3 Therefore the differential system 31 does not exhibit an Andronov Hopf bifurcation at the equilibrium point p 3 = 5 4 1 9410681 10358ρ 3095089d ρ Theorem 331 If the parameters k 3 k 7 k 8 b c and ρ satisfy the relations 314 of Lemma 315 then the equilibrium point p 3 = 5 4 1 9410681 10358ρ 3095089d ρ of differential system 31 is locally unstable Proof From Theorem 330 the characteristic polynomial Pλ has three sign changes in its coefficients By the Descartes rule of signs we have that there exists at least one positive eigenvalue for the linearization at p 3 Then p 3 is unstable 334 Simultaneous periodic orbits at p 1 and p If the parameters k 3 k 7 k 8 b c and ρ satisfy the relations 314 of Lemma 315 then according to Theorem 30 the differential system 31 exhibits an Andronov Hopf bifurcation at p 1 with respect to the parameter d with critical bifurcation value d = d 0 ρ By Theorem 37 the differential system 31 exhibits an Andronov Hopf bifurcation at p with respect to the parameter d with critical bifurcation value d = d 1 ρ In order to find a parameter value where the differential system exhibits a simultaneous Andronov Hopf bifurcation we solve the equation d 1 ρ d 0 ρ = 0 The unique solution in the interval 0 459 10358 is 478 ρ := 86 38779 10580386691137 + 16081609691 08189 The Figure 31 a shows the graph of critical bifurcation value in terms of ρ for each equilibrium point and its intersection at ρ d 0 ρ Therefore the differential system 31 exhibits a simultaneous Andronov Hopf bifurcation at p 1 and p with respect to the parameter d with critical bifurcation value d 0 ρ = d 1 ρ 159399484 478 38779 10580386691137 + 16081609691 = 76178503957 00803 10580386691137 + 4807846464819189

Andronov Hopf and Bautin bifurcation 5 Since ρ < ρ 0 < ρ 1 by Theorems 30 and 37 this simultaneous Andronov Hopf bifurcation is subcritical at p 1 and p the Figure 31 b shows the first Lyapunov coefficient corresponding to p 1 or p In this case the limit cycle bifurcating from p 1 and the limit cycle bifurcating from p are unstable By Theorems 3 and 39 the differential systems exhibits a Bautin bifurcation then there are two limit cycles bifurcating from p 1 or p where one is stable and the other is unstable 10 08 06 04 0 00 d a d 0 ρ d 1 ρ ρ ρ 05 10 15 0 06 04 0 00-0 -04 l 1 b l 1 p ρ l 1 p 1 ρ 1 3 4 ρ ρ 0 ρ ρ 1 Figure 31: a Parameter bifurcation value d b First Lyapunov coefficient at p 1 and p Example 33 Taking a 1 = 7118 a = 98473 b 1 = 575 b = 110108 c 1 = 00060473 c = 00164717 c 3 = 1 and ρ = 44 > ρ 1 > ρ 0 then d 0 = 468666 d 1 = 4901 and the parameters involved in the differential system 31 satisfy the hypothesis established in the Subsection 33 Hence we have three equilibrium points p 1 p and p 3 The first Lyapunov coefficient at p 1 and p are l 1 p 1 d 0 = 1443 l 1 p d 1 = 00135894 In this case we have two stable limit cycle each one bifurcating from the equilibrium points p 1 and p In the Figure 3 we show two trajectories whose ω limit are the stable periodic orbits where d = d 0 + 1/100 4 Conclusion When the prey has a linear growth the differential system has only one equilibrium point in the positive octant Ω and around this point appear an stable periodic orbit generated by an Andronov Hopf bifurcation or a Bautin bifurcation On the other hand if the growth of the prey is logistic the differential system can have even three equilibrium points in Ω In particular when there is only one equilibrium point in Ω it is not hyperbolic When there are two equilibria one is not hyperbolic and the other exhibits an limit cycle generated by an Andronov Hopf bifurcation or Bautin bifurcation In the case when there are three equilibrium points two of them can present Andronov Hopf and Bautin bifurcation in fact they can appear simultaneously Thus the differential system exhibits bi-stability The other equilibrium point is always unstable This analysis shows that the condition to have coexistence of the three populations is better in the logistic growth

6 G Blé V Castellanos and I Loreto-Hernández p p 3 p 1 Figure 3: Phase space of differential system 31 with three equilibria and two limit cycles References [1] G Blé V Castellanos J Llibre Existence of limit cycles in a tritrophic food chain model with Holling functional responses of type II and III Math Methods Appl Sci 391996 No 14 3996 4006 https://doiorg/10100/mma384; MR3536517; Zbl 1345911 [] V Castellanos J Llibre I Quilantan Simultaneous periodic orbits bifurcating from two zero-hopf equilibria in a tritrophic food chain model Journal of Applied Mathematics and Physics 1013 31 38 https://doiorg/10436/jamp01317005; [3] J P Françoise J Llibre Analytical study of a triple Hopf bifurcation in a tritrophic food chain model Appl Math Comput 17011 No 17 7146 7154 https://doiorg/ 101016/jamc01101109; MR781107; Zbl 1119057 [4] J Guckenheimer Y A Kuznetsov Bautin bifurcation Scholarpedia 007 No 5 1853 https://doiorg/10449/scholarpedia1853 [5] C S Holling Some characteristics of simple types of predation and parasitism Can Entomol 911959 No 7 385 398 https://doiorg/104039/ent91385-7; [6] Y A Kuznetsov Elements of applied bifurcation theory Applied Mathematical Sciences Vol 11 Springer-Verlag New York 3rd ed 004 https://doiorg/101007/ 978-1-4757-41-9; MR071006; [7] Y A Kuznetsov Andronov Hopf bifurcation Scholarpedia 1006 No 10 1858 https: //doiorg/10449/scholarpedia1858; [8] J E Marsden M McCracken The Hopf bifurcation and its applications Applied Mathematical Sciences Vol 1 Springer-Verlag New York 1976 https://doiorg/101007/ 978-1-461-6374-6; MR494309;