Nonexistence results for some nonlinear elliptic and parabolic inequalities with functional parameters

We obtain results on nonexistence of nontrivial nonnegative solutions for some elliptic and parabolic inequalities with functional parameters involving the p(x)Laplacian operator. The proof is based on the test function method.


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@ujat.mxwhere 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 h(x) represents the growth rate of the prey population in absence of the predators, and the functions f (x) and g(y) are the functional responses for the mesopredator and the superpredator, respectively.The parameters c 1 , c 2 , c 3 and d 2 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 h(x) is a logistic map and the functional responses f and g are Holling type II.Using the averaging theory, they proved that the system (1.1) 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 (1.1) has been analyzed in Ref. [2], when h(x) 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 h(x) 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 (1.1) 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 (1.1) when the functional responses f (x) and g(y) 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 h(x) = ρx, and the logistic case taking h(x) = ρx(1 − x R ).The functions f and g will be where a 1 , b 1 , a 2 , b 2 are positive constants.Explicitly, we will study the differential system 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 2 and 3.

Linear case
In this section we consider the differential system (1.2) with a linear growth for the prey, this means that the function h(x) = ρx and then the differential system becomes (2.1) 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 2.1.The differential system (2.1) has only one equilibrium point p 0 = (x 0 , y 0 , Moreover, if ones of above condition does not hold, then the differential system (2.1) does not have any equilibrium point in Ω.
Proof.The equilibrium points of the differential system (2.1) are the solutions of By multiplying the above equations by the term b 1 + x 2 (b 2 + y), (which is always positive in Ω), we obtain that an equilibrium point in Ω must satisfy the system From the third equation in system (2.2), y 0 = d 2 b 2 a 2 c 3 −d 2 and it is positive by hypothesis (a).Substituting y = y 0 in the first equation of (2.2), 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 (2.2), we have that the unique solution z = z 0 of this equation is positive, if and only if, c 2 b 1 + x 2 0 − a 1 c 1 x 0 < 0, but, from the first equation in system (2.2), we have that Clearly, if ones of the conditions a 2 c 3 − d 2 > 0, a 1 y 0 − ρb 1 > 0 or c 2 y 0 − c 1 x 0 ρ < 0 does not hold then the differential system (2.1) 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 2.2.If the parameters of the system (2.1) satisfy the conditions (a), (b) and (c) in Lemma 2.1, then there exist k 1 > 0, k 2 > 0 and k 3 > 0, such that the parameters a 1 , a 2 and b 2 involved in the differential system (2.1) can be written as and the unique equilibrium point of the system (2.1) in Ω, is Proof.The solutions of system (2.2) are ∈ Ω, by Lemma 2.1 p 0 ∈ Ω, and ρ(a 2 c 3 − d 2 ) > 0, then there exists k 1 > 0 such that , and Lemma 2.3.Under the hypothesis of Lemma 2.2 and considering that the parameters a 1 , a 2 and b 1 satisfy the conditions (2.3) and then the equilibrium point p 0 is given by and the eigenvalues of the linear approximation of system (2.1) at p 0 are α = 64ρ 39 and ± iω, where Proof.Taking into account the assignations of the parameters a 1 , a 2 and b 1 given by (2.3), the characteristic polynomial of the linear approximation M p 0 of differential system (2.1) at the equilibrium point p 0 is If we consider the assignments for k 2 , k 3 and d 2 given by (2.4), then A 1 , A 2 and A 3 reduce to The characteristic polynomial P(λ) = λ 3 + A 1 λ 2 + A 2 λ + A 3 has a pair of purely imaginary roots ±iω and a real root α if and only if Thus comparing coefficients, P(λ) has a pair of purely imaginary roots ±iω and a real root α if and only if A 2 > 0 and where 1521 , solving equation (2.5) for the parameter c 2 , we have that if Thus, we conclude that the characteristic polynomial P(λ) has a pair of purely imaginary roots ±iω and a real root α, where α = 64ρ 39 and ω = k 1 2 .The equilibrium point p 0 becomes In order to compute the Lyapunov coefficients and a regularity condition, from now in this section b 1 = 1, Remark 2.4.If the assumptions of Lemma 2.3 are satisfied, then the linear approximation of the differential system (2.1) at p 0 has the eigenvalues α = 64ρ 39 and ± i 2 , when c 2 = c 20 (ρ), hence, by continuity on the eigenvalues, the linear approximation of the differential system at p 0 has a pair of complex eigenvalues, when c 2 is in a neighborhood of c 20 (ρ).
In order to compute the first Lyapunov coefficient 1 , we apply the Kuznetsov formula, (see Ref. [7]).Taking into account the assumptions of Lemma 2.3 and using the Mathematica software, we obtain the first Lyapunov coefficient of the differential system (2.1) at the equilibrium point p 0 .Lemma 2.5.If the hypotheses of Lemma 2.3 hold, then the first Lyapunov coefficient of the differential system (2.1) at the equilibrium point p 0 is . Corollary 2.6.There exists a unique real number ρ 0 > 0 such that 1 (p 0 , c 20 (ρ 0 ), ρ 0 ) = 0.
We know proceed to show the regularity condition in order to obtain a Bautin bifurcation.
Proof.By hypothesis, the linear approximation of the differential system (2.1) at p 0 depends only on the parameters c 2 and ρ, let M p 0 (c 2 , ρ) be this linear approximation.By Lemma 2.5, the complex numbers ± i 2 are eigenvalues of M p 0 (c 0 , ρ 0 ), hence, the real part of the complex eigenvalues of M p 0 (c 0 , ρ 0 ), are ξ(p 0 , c 0 , ρ 0 ) = 0, (2.7) let p and q be eigenvectors of M p 0 (c 0 , ρ 0 ) for the corresponding eigenvalues − i 2 and i 2 , respectively, such that q tr • q = 1 and p tr • q = 1.
Proof.From Lemma 2.3, the linearization M p 0 (c 2 , ρ) of differential system (2.1) at p 0 has a positive real eigenvalue and a conjugate pair of pure imaginary eigenvalues if c 2 = c 20 (ρ).
From Lemma 2.8, the derivative of the real part of the complex eigenvalues is which is negative for ρ = 0, and hence the transversality condition holds.The Lemma 2.5 and Corollary 2.6 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][8][9]) hold and we conclude the proof.
Lemma 2.10 (Second Lyapunov coefficient).If we have the assumptions given in Lemma 2.3, then the second Lyapunov coefficient of the differential system (2.1) at the equilibrium point p 0 is given by where Moreover, if ρ = ρ 0 , then 2 (p 0 , c 20 (ρ), ρ) = 0, where ρ 0 is given in the Corollary 2.6.
Proof.In order to compute the second Lyapunov coefficient 2 , 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 2 (p 0 , c 20 (ρ), ρ), of the differential system (2.1) at the equilibrium point p 0 is given by (2.14) and 2 (p 0 , c 20 (ρ 0 ), ρ 0 ) ≈ 7.40065.
Corollary 2.6, Lemma 2.8 and Lemma 2.10 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 2.11 (Bautin bifurcation in linear growth).If the parameters a 1 , a 2 , b 2 , k 2 , k 3 and d 2 satisfy the hypothesis given in Lemma 2.3, then the differential system (2.1) exhibits a Bautin bifurcation at p 0 , with respect to the parameters c 2 and ρ and its critical bifurcation value is (c 20 (ρ 0 ), ρ 0 ).

Logistic case
In this section we consider the differential system (1.2) with a logistic growth for the prey, this means that the function h(x) = ρx 1 − x R and we will analyze the differential system ( In order to make ecological sense we assume that all parameters of the system (3.1) are positive.
Lemma 3.1.If the parameters a 1 , a 2 , b 1 , c 1 and R, satisfy then the unique equilibrium points of the differential system (3.1) in the region Ω are Proof.The equilibrium points of the differential system (3.1) are the solutions of the system, Multiplying the above equations by b 1 + x 2 (b 2 + y), (which is always positive in the region Ω), we obtain that the equilibrium point must satisfy (3.4).Correspondingly each solution of (3.4) must be an equilibrium point of the differential system (3.1).
A point (x 0 , y 0 , z 0 ) ∈ Ω is an equilibrium point of the differential system (3.1) if We suppose x 0 > 0, y 0 > 0 and z 0 > 0. Note that the first equation of the system (3.5) is a linear equation in terms of a 1 , and it has the unique solution, Since a 1 > 0, R − x 0 must be positive, so there exists k 2 > 0 such that R = x 0 + k 2 .A similar argument using the third equation of system (3.5),we obtain that: Using the values of a 1 , a 2 and R, and solving the second equation of system (3.5) for z 0 , we have that Then Therefore, if a 1 , a 2 , R and c 1 satisfy (3.2), then (x 0 , y 0 , z 0 ) is a solution of system (3.4) in Ω, where . Moreover, the system (3.4) takes the form . Solving the third equation of system (3.6) for y, we have that y = y 0 .Moreover, the first equation of system (3.6)reduce to: Hence, the solutions of this equation are Thus, a necessary condition to have at least two solutions of system (3.6) in Substituting b 1 , k 4 , k 5 , k 2 , y = y 0 and x = x 1 in the second equation of system (3.6) and solving this equation for z, we have that In the same way, replacing y = y 0 and x = x 2 in (3.6), we obtain x 0 , and z 1 , z 2 becomes Therefore, the unique equilibrium points of the differential system (3.1) in the region Ω, are: which completes the proof.

Remark 3.2.
Choosing the values k 7 , k 8 adequately, we obtain one, two or three equillibria.

One equilibrium point of the differential system
In this subsection, we assume that the parameters a 1 , a Proof.Under the hypothesis of this subsection, the characteristic polynomial of the linear approximation M p 1 of differential system (3.1) at p 1 is The eigenvalues of M p 1 are Since λ 1 = 0, the equilibrium point p 1 of differential system (3.1) is not hyperbolic.Moreover, if λ 2 and λ 3 are complex then It is not difficult to see that if λ 2 and λ 3 are real, then λ 2 > 0 and λ 3 > 0. Therefore the equilibrium point p 1 of differential system (3.1) has a local unstable manifold of dimension 2.
Proposition 3.5.The equilibrium point p 1 is not hyperbolic and it has a local unstable manifold of dimension 2.
Proof.Considering the assignations for the parameters a 1 , a 2 , b 1 , c 1 , R, k 2 , k 4 , k 5 , k 6 and k 7 given in this subsection, the characteristic polynomial of the linear approximation M p 1 of differential system (3.1) at p 1 is and the eigenvalues of M p 1 are Then the equilibrium point p 1 of differential system (3.1) is not hyperbolic.Moreover, if λ 2 and λ 3 are complex then And it can be verify that if λ 2 and λ 3 are real then λ 2 > 0 and λ 3 > 0. Therefore the equilibrium point p 1 of differential system (3.1) is not hyperbolic and has a local unstable manifold of dimension 2.
Corollary 3.6.The differential system (3.1)does not exhibit an Andronov-Hopf bifurcation at the equilibrium point p 1 = x 0 , y 0 , 2k 3 d 2 k 8 +6d 2 x 0 .Whereas the equilibrium point p 1 does not have an Andronov-Hopf bifurcation, we will show that the equilibrium point p 2 can have a pair of purely imaginary eigenvalues and consequently it can exhibit an Andronov-Hopf bifurcation.
Proof.Let M p 2 be the Jacobian matrix of the differential system (3.1)evaluated at the equilibrium point p 2 , then the characteristic polynomial where , Taking k 8 , b 2 and k 3 satisfying (3.7), then A 1 , A 2 and A 3 are reduced to Following the proof of Lemma 2.3, the characteristic polynomial P(λ) has a pair of purely imaginary roots ±iω and a real root α if and only if A 2 > 0 and where ω = √ A 2 and α = −A 1 . Since , solving equation (3.8) for the parameter d 2 , we have that Taking into account this assignment for d 2 , the coefficient Therefore, the characteristic polynomial P(λ) has a pair of purely imaginary roots ±i and a real root α = − Hence, by continuity on the eigenvalues, the linear approximation of the differential system at p 2 has a pair of complex eigenvalues, In order to compute the first Lyapunov coefficient 1 , we make the following assignations.
Lemma 3.9.If we have the assumptions given in Lemma 3.7, then the eigenvalues of the linear approximation of system (3.1) at the equilibrium point p 2 are α = −  Let p and q be eigenvectors of M p 2 (d 0 , ρ 0 ) for the corresponding eigenvalues −i and i, respectively, such that q tr • q = 1 and p tr • q = 1.Taking into account the values of q, p and , and using the Mathematica software, we obtain the partial derivative of the real part of the eigenvalues ξ(d 2 , ρ) .12) Applying Lemma 2.7 one more time, it follows from the values of q, p and .13) Using the Wolfram Mathematica software, we have that Re
Corollary 3.10, Lemma 3.11 and Lemma 3.13 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 3.14.If the parameters k 8 , b 2 , k 3 , c 2 and ρ satisfy the relations (3.7) of Lemma 3.7, then the differential system (3.1)exhibits a Bautin bifurcation at p 2 , with respect to the parameters d 2 and ρ and its critical bifurcation value is (d 20 (ρ 0 ), ρ 0 ).

Three equilibrium points of the differential system
From now on in this subsection, we assume that the parameters a 1 , a 2 , b 1 , c 1 , R, k 2 , k 4 , k 5 and k 6 satisfy the conditions (3.2) and (3.3) of Lemma 3.1, k 7 > 0, k 8 > 0 and k 7 = k 8 .
Then by Remark 3.2, are the unique three equilibrium points of differential system (3.1) in Ω.

Local dynamics and bifurcation at
then the equilibrium point and the eigenvalues of the linear approximation at p 1 are α = − 13832ρ 3 2448ρ 2 + 32234193 and ± i.
Proof.The characteristic polynomial of the linear approximation M p 1 of differential system (3.1), at the equilibrium point p 1 is , By hypothesis k 7 , k 8 , b 2 and k 3 satisfy (3.14), then A 1 , A 2 and A 3 reduce to As in the proof of Lemma 2.3, the characteristic polynomial P(λ) has a pair of purely imaginary roots ±iω and a real root α if and only if A 2 > 0 and where ω = √ A 2 and α = −A 1 .In this case, Solving the equation (3.15) for the parameter d 2 , we have that , we have A 2 = 1.Since c 2 > 0, then the parameter ρ must satisfy Therefore, the characteristic polynomial P(λ) has a pair of purely imaginary roots ±i and a real root α = − hence, by continuity on the eigenvalues, the linear approximation of the differential system at p 1 has a pair of complex eigenvalues, when d 2 is in a neighborhood of d 20 (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.
Lemma 3.17.If the assumptions given in Lemma 3.15 hold, then the eigenvalues of the linear approximation of the differential system (3.1) at the equilibrium point p 1 are α = − 13832ρ 3 153(16ρ 2 +210681) and ±i.The first Lyapunov coefficient is Let p and q be eigenvectors of M p 1 (d 0 , ρ 0 ) for the corresponding eigenvalues −i and i, respectively, such that q tr • q = 1 and p tr • q = 1.
By Lemma 2.7 and taking into account the values of q, p, which is positive if ρ ∈ (0, 459/ √ 10358), and hence the transversality condition holds.Lemma 3.17 and Corollary 3.18 imply that the first Lyapunov coefficient is negative if ρ > ρ 0 , and is positive if ρ < ρ 0 , (see Figure 3.1).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 2 .Applying the Kuznetsov formula, and using the Mathematica software, we obtain that the second Lyapunov coefficient 2 (p 1 , d 20 (ρ), ρ), of the differential system (3.1) at the equilibrium point p 1 takes the value 2 (p 1 , d 20 (ρ 0 ), ρ 0 ) ≈ −26718.1.We summarize the results in the following theorem.
Theorem 3.29.The differential system (3.1)exhibits a Bautin bifurcation at p 2 , with respect to the parameters d 2 and ρ and its critical bifurcation value is (d 21 (ρ 1 ), ρ 1 ).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 2.3, the characteristic polynomial P(λ) = λ 3 + A 1 λ 2 + A 2 λ + A 3 has a pair of purely imaginary roots ±iω and a real root α if and only if A 2 > 0 and where ω = √ A 2 and α = −A 1 .By hypothesis, if M p 3 is the linear approximation of differential system (3.1) at p 3 then where, , we have that Therefore, the differential system (3.1)does not exhibit an Andronov-Hopf bifurcation at the equilibrium point p Proof.From Theorem 3.30, the characteristic polynomial P(λ) 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.Since ρ * < ρ 0 < ρ 1 , by Theorems 3.20 and 3.27, this simultaneous Andronov-Hopf bifurcation is subcritical at p 1 and p 2 (the Figure 3.1 (b) shows the first Lyapunov coefficient corresponding to p 1 or p 2 ).In this case, the limit cycle bifurcating from p 1 and the limit cycle bifurcating from p 2 are unstable.By Theorems 3.22 and 3.29 the differential systems exhibits a Bautin bifurcation, then there are two limit cycles bifurcating from p 1 or p 2 , where one is stable and the other is unstable.In this case, we have two stable limit cycle each one bifurcating from the equilibrium points p 1 and p 2 .In the Figure 3.2 we show two trajectories whose ω−limit are the stable periodic orbits, where d 2 = d 20 + 1/100.

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.

p 1 p 3 p 2 Figure 3 . 2 :
Figure 3.2: Phase space of differential system (3.1) with three equilibria and two limit cycles.
If the parameters a 1 , a 2 , b 2 , k 2 , k 3 and d 2 satisfy the hypothesis given in Lemma 2.3, then the differential system (2.1) exhibits an Andronov-Hopf bifurcation at p 0 = Theorem 2.9.
2 , b 1 , c 1 , R, k 2 , k 4 , k 5 and k 6 satisfy the conditions (3.2) and (3.3) of Lemma 3.1, and k 7 = k 8 = 0. Then according to Remark 3.2,p 1 = x 0 , y 0 , k 33d 2 x 0 is the unique equilibrium point of the differential system (3.1) in Ω.The equilibrium point p 1 is not hyperbolic and it has a local unstable manifold of dimension 2.

3.3.3 Local dynamics at p 3 Theorem 3.30.
If the parameters k 3 , k 7 , k 8 , b 2 , c 2 and ρ satisfy the relations (3.14) of Lemma 3.15, then, the differential system (3.1)does not exhibit a Hopf bifurcation at the equilibrium point