A dynamic Parrondo's paradox for continuous seasonal systems

We show that planar continuous alternating systems, which can be used to model systems with seasonality, can exhibit a type of Parrondo's dynamic paradox, in which the stability of an equilibrium, common to all seasons is reversed for the global seasonal system.

It is not necessary to recall the importance of these kind of systems in mathematical biology, for instance in population models for which the seasonality has an effect in the reproduction and mortality rates due to environmental circumstances or to human intervention like harvesting, see [11,16,17] and references therein (and [5,9,13,14] for discrete examples); or also in epidemiological models with periodic contact rate, see [3] and the references therein.
The so called Parrondo's paradox is a paradox in game theory, that in a few words says that a combination of losing strategies can become a winning strategy, see [10,15].
Several dynamical versions of related paradoxes are presented in [4,6,7,8] for discrete non-autonomous dynamical systems. In the first paper the authors combine periodically one-dimensional maps f 1 and f 2 to give rise to chaos or order. The existence of discrete systems that exhibit (numerically) chaotic dynamics by alternating regular, or more precisely, integrable systems, has been referred in [6] and [7]. In this last reference also are shown alternating systems with regular (integrable) dynamics obtained by alternating an integrable map and a numerically chaotic one. In [8] we study a local problem, but in any dimension.
In particular, we relate the stability of a common fixed point of two planar maps, F 1 and F 2 , with the stability of this point for F 2 • F 1 . We prove that in the non-hyperbolic case, with complex conjugated eigenvalues (elliptic fixed points), a common attracting character of the common fixed point of F 1 and F 2 , can be reversed for F 2 • F 1 . This phenomenon is the one that we named Parrondo's dynamic type paradox for 2-periodic discrete dynamical systems. In this work we will show that a similar dynamical paradox appears for continuous seasonal systems.
As noted in [3], the asymptotic stability of the equilibria of a seasonal system, for instance the disease-free equilibrium of an epidemiological model, is a more complex issue than in the autonomous case. In this note we evidence that a seasonal system of type (1) can exhibit a dynamic-type Parrondo's paradox, in which the stability of an equilibrium common to all stations (either locally asymptotically stable, LAS from now on, or a repeller), is reversed for the seasonal system (1). That is, we show that there exist systems (1) with a common singular point which is LAS (resp. repeller) for each season systemẋ = X i (x) for i = 1, . . . , n and such it is a repeller (resp. LAS) for the global seasonal system.
To simplify the problem, we prove the existence of the Parrondo's-type paradox for planar differential with two seasons, both with duration T 1 = T 2 = 1. Hence systems of the with x(t) ∈ R 2 . Our main result is: There exist planar polynomial vector fields X 1 and X 2 sharing a common singular point which is LAS (resp. repeller) for both of their associated differential systems, and such that it is a repeller (resp. LAS) for the 2-seasonal differential system (2).
Notice that this theoretical result opens a practical interesting situation. Let us consider a system where the state variables represent the density of individuals of an age-structured population of a species that can be potentially dangerous to humans, like for instance mosquitoes, [12]. Let us assume that for two different environmental situations (the two seasons) the zero solution is a repeller. Of course, this corresponds to unwanted scenarios since, in each season, for an arbitrary small initial density of individuals the amount of them increases over time. Then, it might happen that alternating both situations we get a system with the origin as a LAS critical point, implying the population decline (and long term extinction) of the dangerous species.
In the following, we will use complex notation in order to simplify the expressions.
Hence instead of taking planar vector fields U (x, y)∂/∂x + V (x, y)∂/∂y with (x, y) ∈ R 2 , we will consider the same vector fields but in complex notation X(z,z) = F (z,z)∂/∂z where z = x + iy ∈ C , with associated differential equationż = X(z,z).
One of the key ingredients in our approach will be to know whether for a given local polynomial diffeomorphism of the form of degree at most n, that has a non-hyperbolic elliptic fixed point at the origin, there exists of a polynomial vector field of type X(z,z) = iαz + n j+k=2 a j,k z jzk (4) and such that its associated flow ϕ(t; z,z) satisfies for every (z,z), for z in a small enough neighborhood of z = 0. As we will see, for our purposes we only will need to consider the cases n = 2 or n = 3. This question is solved in next section.
We also would like to comment that very few planar polynomial maps are exactly a flow at a fixed time, i.e. the remainder term O(n + 1) in (5) is identically zero. They are the so-called polynomial flows, and the normal forms of their corresponding vector fields are given in [2,Thm. 4.3].
In fact, ultimately, the proof of Theorem 1 relies on the fact that, near a critical point, the flow of some suitable vector fields are such, up to certain fixed order on the initial conditions, their associated time-1 maps are the ones given in Example 7 of [8]. We recall them in Proposition 11. These maps display the features of the Parrondo's dynamic paradox for the dynamics induced by iterating maps and this fact translates to alternating systems of differential equations. This proof is given in Section 3.
As a byproduct of our study we obtain the following result that we believe is interesting by itself. Its proof is given in Section 4.
Theorem 2. It holds: (i) Consider a local diffemorphism of the form (3), where e iα is not a root of the unity.
Then, for any n ≥ 2 there is a unique polynomial vector field of the form (4) and degree at most n such that its flow satisfies Equation (5).
(ii) For any n ≥ 2, there exists a map F of the form (3) with α = 2 π/(n + 1) for which there is no C n+1 vector field whose flow satisfies Equation (5).
Notice that the above result implies the existence of planar polynomial local diffeomorphisms, preserving orientation, that can not be given as the flow at a fixed time of smooth planar vector fields. Particular examples of such maps are given in (13).
2 Vector fields with prescribed maps as time-1 map

A recurrent procedure
First we establish the structure equation that must satisfy the first terms of a flow map associated with a vector field. It is easy to prove that if a flow map satisfies Equations (3)-(5), then the vector field must have the form X(z,z) = iαz + O(2), so we must work with vector fields with this fixed linear part. If we impose that X is polynomial of degree n we can write X(z,z) = iαz + n j+k=2 a j,k z jzk . When we only assume that it is of class C n+1 , near the origin we can write it as X(z,z) = iαz + n j+k=2 a j,k z jzk + O(n + 1). In any case, by plugging the Taylor expansion of ϕ(t; z,z) in the expression of the differential systemż = X(z,z), that is by imposing dϕ(t; z,z)/dt = X(ϕ(t), ϕ(t)) = iαϕ(t; z,z) + n j+k=2 a j,k ϕ j (t, z)φ k (t, z) + O(n + 1), and from a power comparison argument we get the following result: be a planar C n+1 vector field. Then, in a neighborhood of the origin, its flow is given by ϕ(t; z,z) = where b j,k (t) = γ∈S j,k P γ (t)e γit and S j,k ⊂ Z is a finite set, P γ depends on the values on the coefficients a ℓ,m and the functions ϕ ℓ,m (t) with 2 ≤ ℓ + m < j + k.
By using the above result, given a map (3), we want either to obtain a planar polynomial vector field X(z,z) such that in a neighborhood of the origin its flow satisfies (5) or to prove that there is no C n+1 vector field which flow satisfies (5). We do it by a recursive procedure.
Indeed, suppose that we have computed the coefficients of X up to order κ − 1 for 2 < κ ≤ n. To compute any coefficient a j,k with j + k = κ, we solve the initial value problem (6) and impose Equation (5). If j − k − 1 = 0, then In this case we can isolate the coefficient a j,k , thus contributing to determinate the expression of the vector field.
From the above equation we always can isolate the coefficient a j,k except in the case that or, in other words if e iα is a |j − k − 1|-root of the unity. In this case we say that there appears a resonance associated with the coefficient a j,k , and the equation (7) is satisfied for every value of a j,k (thus leading to a parametric family of vector fields) if and only if it is satisfied the compatibility equation corresponding to the coefficient a j,k : Otherwise, we get an obstruction for F to be the time-1 map of a polynomial (or C n+1 ) vector field, see the proof of Theorem 2 for examples of polynomial maps for which there is no vector field whose flow satisfies (5).
In fact, observe that if for any couple ℓ and m with 2 ≤ ℓ + m < j + k there is not a resonance, then the function b j,k (t) = γ∈S j,k P γ (t)e γit introduced in Lemma 3, depends on the values of the previous coefficients a ℓ,m , thus on the previous coefficients f ℓ,m . On the contrary, if there exists a couple of values ℓ and m with 2 ≤ ℓ + m < j + k giving rise to a resonance (that is, e iα is a |ℓ − m − 1|-root of the unity) and the compatibility condition associated with a ℓ,m is satisfied, then the function b j,k (t) also depends on the parameter a ℓ,m .
Also observe that a resonance may appear at different order levels, so that in order to obtain the associated vector field, we must identify the first order in which a resonance appears and verify that each compatibility equation is fulfilled. In that case, we can proceed by solving the different equations (6) for higher orders, by carrying the expressions of the indeterminate terms, and verifying that the next different compatibility equations are also satisfied. if a resonance appears at order n and it has not appeared at order k < n, then e iα is an m-root of unity with m ∈ {2, 4, . . . , n + 1} if n is odd and m ∈ {3, 5, . . . , n + 1} if n is even.
Summarizing the above recursive procedure we obtain the following result: Theorem 5. Consider a polynomial map F of degree n of the form (3).
(i) If e iα is not a |j −k −1|-root of the unity for all couple j, k with j +k ∈ {0, 1, . . . , n} then there exists a unique polynomial vector field of degree at most n such that its associated flow satisfies ϕ(1; z,z) = F (z,z) + O(n + 1).
(i) If e iα is a |j − k − 1|-root of the unity for certain j, k with j + k ∈ {0, 1, . . . , n} and the compatibility equation (8) corresponding to the coefficient a j,k is not satisfied, then there is no C n+1 vector fiel such that its associated flow satisfies ϕ(1; z,z) = F (z,z)+O(n+1). In the next sections we present the explicit expressions for the vector fields associated with quadratic and cubic maps of the form (3), satisfying Equation (5) for n = 2 and n = 3 respectively.

Vector fields for quadratic maps
The whole scene in the quadratic case is described in the next proposition. Observe that in the above scheme, at order two a resonance can only occur if ω = e iα is a cubic root of unity. This can be seen by taking the function r(j, k) = |j − k − 1| and observing that it takes the values r(2, 0) = r(1, 1) = 1 and r(0, 2) = 3.
Proposition 6. Set F (z,z) = ωz + j+k=2 f j,k z jzk , where ω = e iα with α ∈ (0, 2π). Then (a) If ω is not a cubic root of unity, then there exists a unique quadratic vector field satisfying (5) with n = 2, given by X(z,z) = iαz + j+k=2 a j,k z jzk where (b) Assume that ω is a cubic root of unity. If f 0,2 = 0, then there exists an one-parameter family of quadratic vector fields satisfying (5) with n = 2. In this case the coefficients a 2,0 and a 1,1 of such a vector field are the ones given in Equation (9) and a 0,2 is the free parameter. If f 0,2 = 0 then there is no C 3 vector field satisfying (5) with n = 2.
If ω is a cubic root of the unity, then the compatibility condition (8) associated with the coefficient a 0,2 is f 0,2 = 0, and the result in statement (b) follows.

Vector fields for cubic maps
Given a cubic map, to search a cubic vector field satisfying (5) with n = 3, first we notice that the resonances only occur when e iα is a third root of the unity, when is a square root of the unity, or when is a primitive fourth root of the unity, see Remark 4. According to Theorem 5, if e iα is not such a root of the unity there exists a unique polynomial vector field satisfying (5).
Also according to Theorem 5, if e iα is a third root of the unity and the compatibility condition associated to a 0,2 is satisfied, then there exists an one-parameter family of vector fields satisfying (5). If e iα is a square root of the unity (hence it also is a fourth-root of unity) and the compatibility condition associated with a 3,0 , a 1,2 and a 0,3 are fulfilled, then there exists a three-parametric family of such vector fields. And finally, if e iα is a primitive quartic root of the unity and the compatibility condition associated with a 0,3 holds, then there exists an one-parameter family of such vector fields. All these four cases are studied in the next four propositions: If ω is not a quadratic, cubic or fourth root of unity, then there exists a unique cubic vector field satisfying (5) with n = 3, X(z,z) = iαz + 3 j+k=2 a j,k z jzk , where the coefficients a 2,0 , a 1,1 and a 0,2 are the ones given in (9), and and a 0,3 = −iα ω 2 P 0,3 (ω 2 + 1) (ω 3 − 1) (ω + 1) , Proof. Consider the cubic map F (z,z) and a cubic vector field X(z,z). To search for the flow ϕ(t; z,z) = e iα z + 3 j+k=2 ϕ j,k (t)z jzk + O(4) associated with X(z,z), we plug this expression in the differential equationż = X(z,z), and we get the corresponding equations (6). The equations corresponding to the quadratic terms are the ones obtained in the proof of Proposition 6, that is Equations (10)-(11), thus we obtain the same terms for the quadratic terms of the vector field. To obtain the cubic terms we follow the same procedure. For reasons of space we omit the steps to obtain the expression of all the four equations (6) and its corresponding solutions. We only show the case of the equation corresponding to the coefficient a 3,0 . Indeed, we get: ϕ 3,0 (t) = iα ϕ 3,0 (t) + a 3,0 e 3iα t + iα Q 3,0 (t) By integrating this differential equation and imposing Equation (5), we obtain the corresponding Equation (7): thus we get the expression of the coefficient a 3,0 in the statement. The other expressions are obtained in a similar way.
Proof. As mentioned before, and as can be seen in the proof of Proposition 6, when e iα is a third root of the unity, the only compatibility condition that appears is the one associated with the coefficient a 0,2 , and it is f 0,2 = 0. Assuming now this condition, setting a 0,2 as a free parameter and fixing the values of the coefficients a 2,0 and a 1,1 as the ones in Equation (9), we proceed to compute the coefficients of the cubic term. As in the proof of Proposition 7, we only show how to obtain the the coefficient a 3,0 . Indeed, we get: where Q 3,0 (t) = −6 iαf 2 2,0 e 2 iα t + −a 0,2 f 1,1 ω 2 + 6 iα f 2 2,0 + a 0,2 f 1,1 e 3 iα t + a 0,2 f 1,1 ω 2 − 1 .
By integrating this equation, imposing Equation (5) and taking into account that ω 3 = 1, we get that the corresponding Equation (7) is: Thus we get the expression of the coefficient a 3,0 in the statement. The other expressions are obtained similarly.
If ω = e iα is a squared root of the unity, then α = π (since α = 0). In this case the compatibility conditions (8) are the ones associated with the coefficients a 3,0 and a 1,2 but also a 0,3 , because ω 2 = 1 implies ω 4 = 1. Proceeding as in the previous results, we obtain: Then there exists a cubic vector field satisfying (5) for n = 3, if and only if If these equations are fulfilled, then there is a three-parameter family of cubic vector fields satisfying (5) for n = 3, and it is given by and a 3,0 , a 1,2 and a 0,3 are free parameters.
The resonant case that appears when ω = e iα is a primitive fourth root of the unity is studied in the following result: , then there exists a cubic vector field satisfying (5) for n = 3, if and only if If this equation is fulfilled, then there is an one-parameter family of vector fields satisfying (5) for n = 3, given by being a 0,3 the free parameter.
(b) If ω = −i (that is α = 3π 2 ), then there exists a cubic vector field satisfying (5) for n = 3, if and only if If this equation is fulfilled, then the there is an one-parameter family of cubic vector fields satisfying (5) for n = 3, and it is given by being a 0,3 the free parameter.

Proof of Theorem 1
Theorem 1 is a consequence of the following two results. The first one is proved in [8] but for completeness we include a sketch of its proof. The second one is a consequence of the results in the previous section.
Proposition 11. The two polynomial maps have the origin as a LAS fixed point for both of them, while the composition map F 2 • F 1 has the origin as a repeller fixed point.
Proof. Let U be a small enough neighborhood of the origin. A C 2m+2 map F in U with an elliptic fixed point whose eigenvalues λ,λ = 1/λ, are not roots of unity of order ℓ for 0 < ℓ ≤ 2m + 1, is locally conjugate to its Birkhoff normal form: see [1]. The first non-vanishing number B j is called the jth Birkhoff constant. If V j = Re(B j ) < 0 (resp. V j > 0), then the point p is LAS (resp. repeller), see [8,Lem. 4.1] for instance. The quantity V j is called the jth Birkhoff stability constant. This is so, because the fact that V j = 0 implies that the function zz = |z| 2 is a strict Lyapunov function at the origin for the normal form map F B of F.
In [8], both the Birkhoff and the Birkhoff stability constants of F 1 and F 2 are computed obtaining that B 1 (F 1 ) = − 1 2 − 11 2 i and B 1 (F 2 ) = − 1 2 + Let ϕ j (t; z,z), j = 1, 2 be their respective associated flows. Then, for z in a small enough neighborhood of the origin where the maps F j are given in Proposition 11.
Proof. Observe that F 1 has the form (3) with α = π/2, so that e iα is a primitive fourth root of the unity. Since the compatibility condition (12) is satisfied, by using the expression in Proposition 10(a) we can find an one-parameter family of vector fields X 1 (z,z, µ) satisfying . This is the family of vector fields X 1 given in the statement, where µ is the free parameter a 0,3 . Also observe that F 2 has also the form (3) with α = π/3, so that e iα is a primitive sixth root of the unity. By using Proposition 7 we can find a unique vector field X 2 satisfying ϕ 2 (z,z) = F 2 (z,z) + O(4). This X 2 is the second vector field given in the statement.
Proof of Theorem 1. We will prove that the vector fields given in the statement of Proposition 12 provide the desired example with X 1 and X 2 having the origin as a singular LAS point and with the origin being a repeller for the 2-seasonal differential system (2). Then, the converse situation will hold simply by considering the vector fields −X 1 and −X 2 .
It is clear that for the vector fields X 1 , X 2 and the one in (2) the stability of the origin coincides with the one of the corresponding flows ϕ 1 (1; ·, ·), ϕ 2 (1; ·, ·) and ϕ(2; ·, ·) respectively. Equivalently, these stabilities coincide with the ones of the origin for the maps F 1 , F 2 and F 2 • F 1 . Since, by Proposition 11, these maps provide a discrete dynamic Parrondo's paradox, we have that both X 1 and X 2 have a LAS singular point at the origin, and the corresponding 2-seasonal system (2) has a repeller point at the origin, as we wanted to prove.

Proof of Theorem 2
(i) This is a corollary of statement (i) of Theorem 5.
(ii) We will use item (ii) of Theorem 5. For each n ≥ 2 we will prove that the polynomial map F (z,z) = e iα z +z n , with α = 2 π n + 1 , satisfies the statement of the theorem.
The result for n = 2 is a direct consequence of Proposition 6. When n = 3, the result follows by item (a) of Proposition 10 because the compatibility condition (12) does not hold.
Now suppose that n ≥ 4. We claim that for each 2 ≤ m ≤ n − 1, if e iα is a primitive (n + 1)-root of unity and X m is a vector field with associated flow of the form ϕ m (t; z,z) = e iαt z + O(m + 1), then it satisfies X m (z,z) = iαz + O(m + 1). We will prove the claim by induction on m, by using the same method and notations introduced in Section 2.1.
By Proposition 6 the result is true for m = 2. Assume that the result is true for m < n−1.
From the claim, X must have the form X(z,z) = iαz + j+k=n a j,k z jzk + O(n + 1).