This is a preprint of: “A dynamic Parrondo’s paradox for continuous seasonal systems”, Anna Cima, Armengol Gasull, Vı́ctor Mañosa, Nonlinear Dynam., 2020. DOI: [10.1007/s11071-020-05656-w] A dynamic Parrondo’s paradox for continuous seasonal systems Anna Cima(1) , Armengol Gasull(1) and Vı́ctor Mañosa(2) (1) Departament de Matemàtiques, Facultat de Ciències, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Spain cima@mat.uab.cat, gasull@mat.uab.cat (2) Departament de Matemàtiques, Universitat Politècnica de Catalunya Colom 11, 08222 Terrassa, Spain victor.manosa@upc.edu Abstract 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. As a byproduct of our approach we also prove that there are locally invertible orientation preserving planar maps that cannot be the time-1 flow map of any smooth planar vector field. Mathematics Subject Classification 2010: Primary: 37C75; 34D20. Secondary: 37C25. Keywords: Continuous dynamical systems with seasonality, non-hyperbolic critical points, local asymptotic stability, Parrondo’s dynamic paradox. 1 Introduction and Main results For dynamical systems given by differential equations, alternating systems take the form    ẋ(t) = X1 (x(t)) for t such that t (mod T ) ∈ [0, T1 ),      ẋ(t) = X2 (x(t)) for t such that t (mod T ) ∈ [T1 , T1 + T2 ), .. (1)   .      ẋ(t) = Xn (x(t)) for t such that t (mod T ) ∈ [T1 + · · · + Tn−1 , T1 + · · · + Tn ), 1 Pn where T = j=1 Tj , with Tj > 0 for j = 1, 2, . . . , n, and Xj being class C 1 vector fields. They can be used to model continuous seasonal systems with n seasons of durations T1 , T2 , . . . , Tn . 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 har- vesting, see [12, 19, 20] and references therein (and [5, 9, 14, 15] for discrete examples); or also in epidemiological models with periodic contact rate, see [3] and the references therein. The existence of chaotic behaviors in this kind of systems have been reported, see for instance the proof of the existence of topological horseshoes in the Poincaré maps associated to the flow of a 2-seasonal Lotka–Volterra system of type (1) in [18]. The study of chaotic dynamics in continuous seasonal systems is a challenge that will continue to require a lot of attention. In this work we expose a collateral aspect that appears, however, in regular regimes: one type of dynamic Parrondo’s paradox. 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 [11, 16]. 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 f1 and f2 to give rise to chaos or order. The existence of discrete systems that exhibit (numerically) chaotic dynamics by alternating regular, or more pre- cisely, 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, F1 and F2 , with the stability of this point for F2 ◦ F1 . We prove that when the fixed points have complex conjugated eigenvalues of modulus 1 (i.e. elliptic fixed points), a common attracting character of the common fixed point of F1 and F2 , can be reversed for F2 ◦ F1 . 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. Remember that a singular point p is said to be stable if for every neighborhood U of p there is a neighborhood V ⊂ U such that every solution ϕ(t; p0 ) with p0 ∈ V is defined and lies in U for all t > 0. If, in addition, V is such that limt→+∞ ϕ(t; p0 ) = p then the point is (locally) asymptotically stable (LAS from now on). The point is a repeller if it is not stable and limt→−∞ ϕ(t; p0 ) = p, [10]. 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 2 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 LAS 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 system ẋ = Xi (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 T1 = T2 = 1. Hence systems of the form   ẋ(t) = X1 (x(t)) for t ∈ [2k, 2k + 1), (2)  ẋ(t) = X (x(t)) for t ∈ [2k + 1, 2k + 2), k ∈ N ∪ {0}, 2 with x(t) ∈ R2 . Our main result is: Theorem 1. There exist planar polynomial vector fields X1 and X2 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, [13]. 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 real planar vector fields U (x, y)∂/∂x + V (x, y)∂/∂y with (x, y) ∈ R2 , we will consider the same vector fields but in complex notation X(z, z̄)∂/∂z + X̄(z, z̄)∂/∂ z̄ where z = x + iy ∈ C , with associated differential equation ż = X(z, z̄), where, of course z̄˙ = X̄(z, z̄). One of the key ingredients in our approach will be to know whether for a given local polynomial diffeomorphism of the form n X iα F (z, z̄) = e z + fj,k z j z̄ k , α ∈ (0, 2π), (3) j+k=2 of degree at most n, that has a non-hyperbolic elliptic fixed point at the origin, there exists 3 of a polynomial vector field with n X X(z, z̄) = iαz + aj,k z j z̄ k (4) j+k=2 and such that its associated flow ϕ(t; z, z̄) satisfies ϕ(1; z, z̄) = F (z, z̄) + O(n + 1), (5) for every (z, z̄), for z in a small enough neighborhood of z = 0, and where f (z, z̄) = O(k) if there exist two constants δ > 0 and C such that |f (z, z̄)| ≤ C|z|k for all z with |z| < δ. 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 12. 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 diffeomorphism of the form (3), where eiα 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 X whose flow ϕ(t; z, z̄) satisfies Equation (5). The above result implies the existence of planar local diffeomorphisms, preserving ori- entation, 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). We notice that it is known that the “embedding problem” (that is the answer to the question: can an orientation preserving diffeomorphism be embedded as the time-one map of a flow of an autonomous vector field? ) has negative answer in the general setting, see [17] for instance. It is also known that for 4 any k ∈ N ∪ {∞} the subset of local C k diffeomorphisms having an embedding vector field with some smoothness is dense in the set of all local C k diffeomorphisms with the coefficient topology, see [21]. Also as a consequence of [22, Theorem 1.3 (a)] all planar C ∞ diffeo- morphisms with a hyperbolic fixed point that have a C ∞ embedding flow are characterized. Our result provide a simple proof and a very simple example in the case of non-hyperbolic fixed points, and gives some insight on the open problem about the embedding problem for planar diffeomorphisms with non-hyperbolic fixed points stated in this last reference. 2 Vector fields with prescribed maps as time-1 map 2.1 A recurrent procedure We start establishing the structure equation that must satisfy the first terms of a flow map associated with a vector field. First we prove that if a flow map satisfies Equations (3)–(5), then the vector field must have the form X(z, z̄) = iαz + O(2), so we will work with vector fields with this fixed linear part. Lemma 3. A planar C 2 vector field X(z, z̄) such that its associate flow satisfies ϕ(1; z, z̄) = eiαt z + O(2) has the form X(z, z̄) = iαz + O(2). Proof. We set X(z, z̄) = a1,0 z + a0,1 z̄ + O(2). By plugging the the associate flow ϕ(t) = ϕ1,0 (t)z + ϕ0,1 (t)z̄ + O(2) into the differential equation ż = X(z, z̄) (using that ϕ̇(t) = a1,0 ϕ(t) + a0,1 ϕ(t) + O(2)), and by a power comparison argument we have ϕ̇1,0 (t) = a1,0 ϕ1,0 (t) + a0,1 ϕ̄0,1 (t), ϕ̇0,1 (t) = a0,1 ϕ̄1,0 (t) + a1,0 ϕ0,1 (t). We solve the linear initial value problem given by the above equations and their conjugates with the initial conditions ϕ1,0 (0) = 1 and ϕ0,1 (0) = 0 (since ϕ(0; z, z̄) = z), and we obtain the solutions ϕ1,0 and ϕ0,1 , that we omit for the sake of shortness. At this stage, by imposing that ϕ1,0 (1) = eiα and ϕ0,1 (1) = 0, we get that a1,0 = iα and a0,1 = 0, which proves the result. Hence, form now on, if we impose that X is polynomial of degree n we can write P X(z, z̄) = iαz + nj+k=2 aj,k z j z̄ k . When we only assume that it is of class C n+1 , near the ori- P gin we can write it as X(z, z̄) = iαz+ nj+k=2 aj,k z j z̄ k +O(n+1). In any case, by plugging the Taylor expansion of ϕ(t; z, z̄) in the expression of the differential system ż = X(z, z̄), that is P by imposing dϕ(t; z, z̄)/dt = X(ϕ(t), ϕ(t)) = iαϕ(t; z, z̄) + nj+k=2 aj,k ϕj (t, z, z̄)ϕ̄k (t, z, z̄) + O(n + 1), and from a power comparison argument we get the following result: 5 Pn j z̄ k Lemma 4. Let X(z, z̄) = iαz + j+k=2 aj,k z + O(n + 1) with α ∈ (0, 2π) be a planar C n+1 vector field. Then, in a neighborhood of the origin, its flow is given by ϕ(t; z, z̄) = P eiαt z + nj+k=2 ϕj,k (t)z j z̄ k + O(n + 1), where for each j, k ∈ N such that 2 ≤ j + k ≤ n the functions ϕj,k (t) satisfy the linear differential equation ϕ̇j,k (t) = iα ϕj,k (t) + aj,k ei(j−k)αt + bj,k (t) with ϕj,k (0) = 0, (6) P where bj,k (t) = γ∈Sj,k Pγ (t)eγit and Sj,k ⊂ Z is a finite set, Pγ depends on the values on the coefficients a`,m and the functions ϕ`,m (t) with 2 ≤ ` + m < j + k. By using the above result, given a map as in (3), we want either to obtain a planar polynomial vector field X(z, z̄) such that in a neighborhood of the origin its flow satisfies (5) or to prove that there is no C n+1 vector field whose 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 aj,k with j + k = κ, we solve the initial value problem (6) and impose Equation (5). If j − k − 1 = 0, then  Z 1  iα −iατ ϕj,k (1) = e aj,k + bj,k (τ )e dτ = fj,k . 0 In this case we can isolate the coefficient aj,k , thus contributing to determinate the expres- sion of the vector field. If j − k − 1 6= 0, then we have    Z 1  iα aj,k i(j−k−1)α −iατ ϕj,k (1) = e e −1 + bj,k (τ )e dτ = fj,k . (7) i(j − k − 1)α 0 From the above equation we always can isolate the coefficient aj,k except in the case that ei(j−k−1)α − 1 = 0, or, in other words if eiα is a |j − k − 1|-root of the unity. In this case we say that there appears a resonance associated with the coefficient aj,k , and the equation (7) is satisfied for every value of aj,k (thus leading to a parametric family of vector fields) if and only if it is satisfied the compatibility equation corresponding to the coefficient aj,k : Z 1 eiα bj,k (τ )e−iατ dτ = fj,k . (8) 0 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). 6 In fact, observe that if for any couple ` and m with 2 ≤ ` + m < j + k there is not a P resonance, then the function bj,k (t) = γ∈Sj,k Pγ (t)eγit introduced in Lemma 4, depends on the values of the previous coefficients a`,m , thus on the previous coefficients f`,m . On the contrary, if there exists a couple of values ` and m with 2 ≤ ` + m < j + k giving rise to a resonance (that is, eiα is a |` − m − 1|-root of the unity) and the compatibility condition associated with a`,m is satisfied, then the function bj,k (t) also depends on the parameter a`,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. Remark 5. Fixing an order n, if we consider the pairs (j, k) with j + k = n we get that |j−k−1| ∈ {0, 2, 4, . . . , n+1} if n is odd and |j−k−1| ∈ {1, 3, 5, . . . , n+1} if n is even. Hence, if a resonance appears at order n and it has not appeared at order k < n, then eiα 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 6. Consider a polynomial map F of degree n of the form (3). (i) If eiα 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, z̄) = F (z, z̄) + O(n + 1). (ii) If eiα 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 aj,k is not satisfied, then there is no C n+1 vector field such that its associated flow satisfies ϕ(1; z, z̄) = F (z, z̄) + O(n + 1). (iii) If there are ` couples j, k with j + k ∈ {0, 1, . . . , n} such that eiα is a |j − k − 1|-root of the unity and the compatibility equations (8) corresponding to the coefficients aj,k are satisfied, then there exists an `-parametric family of polynomial vector fields of degree at most n satisfying ϕ(1; z, z̄) = F (z, z̄) + 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. 7 2.2 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 = eiα 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. P Proposition 7. Set F (z, z̄) = ωz + j+k=2 fj,k z j z̄ k , where ω = eiα with α ∈ (0, 2π). Then (a) If ω is not a cubic root of unity, then there exists a unique quadratic vector field satis- P fying (5) with n = 2, given by X(z, z̄) = iαz + j+k=2 aj,k z j z̄ k where iα f2,0 iα f1,1 3iα ω 2 f0,2 a2,0 = , a1,1 = , a0,2 = . (9) ω (ω − 1) ω−1 ω3 − 1 (b) Assume that ω is a cubic root of unity. If f0,2 = 0, then there exists an one-parameter family of quadratic vector fields satisfying (5) with n = 2. In this case the coefficients a2,0 and a1,1 of such a vector field are the ones given in Equation (9) and a0,2 is the free parameter. If f0,2 6= 0 then there is no C 3 vector field satisfying (5) with n = 2. Proof. Consider a quadratic map F (z, z̄) = eiα z + f2,0 z 2 + f1,1 z z̄ + f0,2 z̄ 2 and a vector field of the form X(z, z̄) = iαz + a2,0 z 2 + a1,1 z z̄ + a0,2 z̄ 2 . If we search for its associated flow ϕ(t; z, z̄) = eiαt z + ϕ2,0 (t)z 2 + ϕ1,1 (t)z z̄ + ϕ0,2 (t)z̄ 2 + O(3), by plugging this expression in the differential equation ż = X(z, z̄), we get that ż = iαeiαt z + ϕ̇2,0 (t)z 2 + ϕ̇1,1 (t)z z̄ + ϕ̇0,2 (t)z̄ 2 + O(3) = X(z, z̄) = = iαeiαt z + (iαϕ2,0 (t) + a2,0 e2iαt )z 2 + (iαϕ1,1 (t) + a1,1 )z z̄ + (iαϕ0,2 (t) + a0,2 e−2iαt )z̄ 2 + O(3), so we obtain the differential equations: ϕ̇2,0 (t) = iαϕ2,0 (t) + a2,0 e2iαt , ϕ̇1,1 (t) = iαϕ1,1 (t) + a1,1 , ϕ̇0,2 (t) = iαϕ0,2 (t) + a0,2 e−2iαt . with the conditions ϕ2,0 (0) = 0, ϕ1,1 (0) = 0 and ϕ0,2 (0) = 0 (remember that ϕ(0; z, z̄) = z). By integrating them, evaluating their solutions at time t = 1 and imposing ϕ(1; z, z̄) = F (z, z̄) + O(3) we get the corresponding equations (6): i  ϕ2,0 (1) = a2,0 1 − eiα eiα = f2,0 , α i  ϕ1,1 (1) = a1,1 1 − eiα = f1,1 , (10) α i  ϕ0,2 (1) = − a0,2 1 − e−3iα eiα = f0,2 . 3α 8 Since α ∈ (0, 2π), the first two equations can always be solved giving the values for a2,0 and a1,1 in Equation (9). The third equation fixes a value of a0,2 unless ω is a third root of unity, obtaining the expressions in (9), thus proving (a). If ω is a cubic root of the unity, then the compatibility condition (8) associated with the coefficient a0,2 is f0,2 = 0, and the result in statement (b) follows. 2.3 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 eiα 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 5. According to Theorem 6, if eiα is not such a root of the unity there exists a unique polynomial vector field satisfying (5). Also according to Theorem 6, if eiα is a third root of the unity and the compatibility condition associated to a0,2 is satisfied, then there exists an one-parameter family of vector fields satisfying (5). If eiα is a square root of the unity (hence it also is a fourth-root of unity) and the compatibility condition associated with a3,0 , a1,2 and a0,3 are fulfilled, then there exists a three-parametric family of such vector fields. And finally, if eiα is a primitive quartic root of the unity and the compatibility condition associated with a0,3 holds, then there exists an one-parameter family of such vector fields. All these four cases are studied in the next four propositions: P3 j z̄ k , Proposition 8. Set F (z, z̄) = ωz + j+k=2 fj,k z where ω = eiα with α ∈ (0, 2π). If ω is not a quadratic, cubic or fourth root of unity, then there exists a unique cubic vector field P satisfying (5) with n = 3, X(z, z̄) = iαz + 3j+k=2 aj,k z j z̄ k , where the coefficients a2,0 , a1,1 and a0,2 are the ones given in (9), and −iα P3,0 a3,0 = , ω 2 (ω 3 − 1) (ω + 1) with    P3,0 = f0,2 f1,1 − 2 f3,0 ω 3 + 2 f0,2 f1,1 + f2,0 2 − f3,0 ω 2 + 2 f2,0 2 2 − f3,0 ω + 2 f2,0 ; −i P2,1 a2,1 = , ω 2 (ω 3 − 1)2 9 with     P2,1 = (i + α) |f1,1 |2 + if2,1 ω 7 + (i + 2 α) |f1,1 |2 − 2 if1,1 f2,0 ω 6 +   (i + 3 α) |f1,1 |2 + (2 i + 6 α) |f0,2 |2 − f1,1 f2,0 (i + α) ω 5 +     (−i + 2 α) |f1,1 |2 − (i + 2 α) f2,0 f1,1 − 2 if2,1 ω 4 + 3 f1,1 f2,0 − |f1,1 |2 ×   (i − α) ω 3 + −i |f1,1 |2 − 2 i |f0,2 |2 + (i − 2 α) f1,1 f2,0 ω 2 + (f2,0 f1,1 (i − α) + if2,1 ) ω − if1,1 f2,0 ; −iα P1,2 a1,2 = , (ω 3 − 1) (ω + 1) with  P1,2 =f1,1 f2,0 ω 4 + 2 f1,1 f0,2 + f2,0 f1,1 − 2 f1,2 ω 3 + 4 f1,1 f0,2 + f2,0 f1,1 + 4 f0,2 f2,0 + 2   f1,1 − 2 f1,2 ω 2 + 2 f0,2 f2,0 + f1,1 2 2 − 2 f1,2 ω + f1,1 ; and −iα ω 2 P0,3 a0,3 = , (ω 2 + 1) (ω 3 − 1) (ω + 1) with   P0,3 =2f0,2 f2,0 ω 4 + 4 f0,2 f2,0 − f0,3 ω 3 + 6 f0,2 f2,0 + 3 f0,2 f1,1 − 4 f0,3 ω 2 + 2 (f0,2 f1,1 − 2 f0,3 ) ω + f0,2 f1,1 . Proof. Consider the cubic map F (z, z̄) and a cubic vector field X(z, z̄). We follow the recursive procedure based on Lemma 4. Observe that the coefficients a3,0 , a2,1 , a1,2 and a0,3 will depend on the expressions of the quadratic coefficients a2,0 , a1,1, and a0,2 that must be P computed firstly. We plug the expression ϕ(t; z, z̄) = eiα z + 3j+k=2 ϕj,k (t)z j z̄ k + O(4) into the differential equation ż = X(z, z̄), getting the equation 3 X 3 X iαt j k iαt iαe z+ ϕ̇j,k (t)z z̄ + O(4) = iαe z+ Rj,k (t)z j z̄ k + O(4), (11) j+k=2 j+k=2 where each term Rj,k (t) has the form of the right-hand side of equation (6). By a simple power comparison argument we arrive to each equation (6) for j + k = 2, 3. The equations corresponding to the quadratic terms are obtained by repeating the steps proof of Proposition 7. After this steps we obtain the quadratic coefficients given in (9) and also the functions ϕ2,0 , ϕ1,1 and ϕ0,2 , in terms of the quadratic coefficients of the map f2,0 , f1,1 and f0,2 . 10 For the cubic terms we follow the same procedure. To obtain the coefficient a3,0 we equal the terms with z 3 in both sides of equation (11), and we get iα Q3,0 (t) ϕ̇3,0 (t) = iα ϕ3,0 (t) + a3,0 e3iα t + , ω 2 (ω − 1)2 (ω 2 + ω + 1) where  2 2 iα t  2  3 iα t Q3,0 (t) = −2 ω 2 + ω + 1 f2,0 e + f0,2 f1,1 ω 3 + 2 ω 2 + ω + 1 f2,0 e − f0,2 f1,1 ω 3 . By integrating this differential equation and imposing Equation (5), we obtain the corre- sponding Equation (7):   2 −i ω 2 − 1 ω a3,0 f0,2 f1,1 ω 3 + 2 f0,2 f1,1 + 2 f2,0 2 2 ω + 2 f2 ω + 2 f2,0 2,0 + = f3,0 , 2α 2ω (ω 2 + ω + 1) thus we get the expression of the coefficient a3,0 in the statement. The expressions of the rest of cubic coefficients are obtained in an analogous way. P3 j z̄ k , Proposition 9. Set F (z, z̄) = ωz + j+k=2 fj,k z where ω = eiα is a primitive third root of unity. Then there exists a cubic vector field X satisfying (5) for n = 3 if and only if f0,2 = 0. In this case, there is an one-parameter family of cubic vector fields X satisfying (5) for n = 3, whose coefficients a2,0 and a1,1 are the ones given in Equation (9), a0,2 is a free parameter, and a0,2 f1,1 (ω − 1)2 − 6 iαf3,0 ω + 6 iαf2,0 2 i P2,1 a3,0 = , a2,1 = , 3 ω (ω − 1) 3α(ω − 1)2 with   P2,1 = 9 iα f1,1 f2,0 − 3 f2,0 f1,1 α2 + 3 iα f2,1 − 2 (|a0,2 |)2 ω 2   + 3 i (|f1,1 |)2 α + 3 (|f1,1 |)2 α2 − 3 iα f1,1 f2,0 + 3 iα f2,1 + 4 (|a0,2 |)2 ω − 3 i (|f1,1 |)2 α − 6 iα f1,1 f2,0 − 6 iα f2,1 − 2 (|a0,2 |)2 ; i P1,2 a1,2 = , 3 (1 − ω) with   P1,2 = −6 if1,2 α + 2 f1,1 a0,2 + 4 f2,0 a0,2 ω 2 + i 3 iα f1,1 2 − 4 f1,1 a0,2 − 2 f2,0 a0,2 ω  + i 3 if2,0 α f1,1 + 2 f1,1 a0,2 − 2 f2,0 a0,2 ; and P0,3 a0,3 = , 3 (1 − ω) with   P0,3 = − 6 a0,2 f2,0 + f1,1 a0,2 ω 2 + 2 a0,2 f2,0 − f1,1 a0,2 ω − 12 if0,3 α + 4 a0,2 f2,0 + 2 f1,1 a0,2 . 11 Proof. We follow the same steps as in the proof of Proposition 8, but we take into account that as mentioned before, and as can be seen in the proof of Proposition 7, when eiα is a third root of the unity, the only compatibility condition that appears is the one associated with the coefficient a0,2 , and it is f0,2 = 0. Assuming now this condition, we obtain a new Equation (11). Setting a0,2 as a free parameter and fixing the values of the coefficients a2,0 and a1,1 as the ones in Equation (9), we proceed to compute the coefficients of the cubic term. To obtain a3,0 , we equal the terms with z 3 in both sides of the corresponding equation (11), and we get e 3,0 (t) Q ϕ̇3,0 (t) = iα ϕ3,0 (t) + a3,0 e3iα t + , 3 ω 2 (ω − 1)2 where   e 3,0 (t) = −6 iαf 2 e2 iα t + −a0,2 f1,1 ω 2 + 6 iα f 2 + a0,2 f1,1 e3 iα t + a0,2 f1,1 ω 2 − 1 . Q 2,0 2,0 By integrating this equation, imposing Equation (5) and taking into account that ω 3 = 1, we get that the corresponding Equation (7) is:   2 2 i (ω − 1) a3,0 i a f 0,2 1,1 (ω − 1) + 6 iαf2,0 − = f3,0 . 2α 6ω α Thus we get the expression of the coefficient a3,0 in the statement. The expressions of the other coefficients are obtained similarly. If ω = eiα is a squared root of the unity, then α = π (since α 6= 0). In this case the compatibility conditions (8) are the ones associated with the coefficients a3,0 and a1,2 but also a0,3 , because ω 2 = 1 implies ω 4 = 1. Proceeding as in the previous results, we obtain: P Proposition 10. Set F (z, z̄) = −z + 3j+k=2 fj,k z j z̄ k . Then there exists a cubic vector field P X(z, z̄) = iπz + 3j+k=2 aj,k z j z̄ k satisfying (5) for n = 3, if and only if 1 2 f3,0 = − f1,1 f0,2 − f2,0 , 2 1 2 1 f1,2 = − f1,1 − f1,1 f0,2 − f2,0 f0,2 − f0,2 f1,1 , 2 2 1  f0,3 = − f0,2 2 f2,0 + f1,1 . 2 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 πi πi 3πi a2,0 = f2,0 , a1,1 = − a0,2 = − f1,1 , f0,2 , 2 2 2   1 1 3 πi a2,1 = (iπ − 2) |f1,1 |2 + (3 iπ − 2) |f0,2 |2 − + f2,0 f1,1 − f2,1 , 4 2 2 4 and a3,0 , a1,2 and a0,3 are free parameters. 12 The resonant case that appears when ω = eiα is a primitive fourth root of the unity is studied in the following result: P3 j z̄ k . Proposition 11. Set F (z, z̄) = ωz + j+k=2 fj,k z Then, (a) If ω = i (that is α = π2 ), then there exists a cubic vector field X(z, z̄) = i π2 z + P3 j k j+k=2 aj,k z z̄ satisfying (5) for n = 3, if and only if 1  f0,3 = f0,2 (2 + 2 i) f2,0 + (1 − i) f1,1 . (12) 2 If this equation is fulfilled, then there is an one-parameter family of vector fields satis- fying (5) for n = 3, given by π π 3π a2,0 = − (1 + i) f2,0 , a1,1 = (1 − i) f1,1 , a0,2 = (1 + i) f0,2 4 4 4 and     π i 2 a3,0 =− − 1+ f1,1 f0,2 + if2,0 + f3,0 , 2 2 1 1 a2,1 = (−2 + (π − 2) i) |f1,1 |2 + (−2 + (3π + 2) i) |f0,2 |2 4 2 1 + (6 + (π − 2) i) f1,1 f2,0 − if2,1 , 4   π i i 2 a1,2 = − (2 + i) f0,2 f1,1 − f1,1 f2,0 − (2 − i) f2,0 f0,2 + f1,1 + f1,2 , 2 2 2 being a0,3 the free parameter. (b) If ω = −i (that is α = 3π 3π 2 ), then there exists a cubic vector field X(z, z̄) = i 2 z + P3 j k j+k=2 aj,k z z̄ satisfying (5) for n = 3, if and only if 1  f0,3 = f0,2 (1 + i) f1,1 + (2 − 2 i) f2,0 . 2 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 3π 3π 9π a2,0 = (1 − i) f2,0 , a1,1 = − (1 + i) f1,1 , a0,2 = (−1 + i) f0,2 4 4 4 and    3π i 2 a3,0 =− 1− f1,1 f0,2 + if2,0 − f3,0 , 2 2 1 1 a2,1 = (−2 + (3π + 2) i) |f1,1 |2 + (−2 + (9π − 2) i) |f0,2 |2 4    2 3 1 3 + +i + π f2,0 f1,1 + if2,1 , 2 2 4   3π i i 2 a1,2 = (2 − i) f0,2 f1,1 − f2,0 f1,1 + (2 + i) f2,0 f0,2 + f1,1 − f1,2 , 2 2 2 being a0,3 the free parameter. 13 3 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 12. The two polynomial maps 1 √  F1 (z, z̄) = iz + (1 − 3i)z 2 + z z̄ and F2 (z, z̄) = 1 + i 3 z − z 2 z̄, 2 have the origin as a LAS fixed point for both of them, while the composition map F2 ◦ F1 has the origin as a repeller fixed point. We notice that the maps F1 and F2 in the above result are the complex representation of the real maps  f1 (x, y) = −y + 2x2 + 6xy, x − 3x2 + 2xy + 3y 2 , √ √ ! x 3 2 2 3 1 2 2 f2 (x, y) = − y − x (x + y ), x + y − y (x + y ) , 2 2 2 2 introduced in [8, Example 7]. Proof of Proposition 12. Let U be a small enough neighborhood of the origin. Recall that for a suitable m ∈ N∪{0}, a C 2m+2 map F in U with an elliptic fixed point whose eigenvalues λ, λ̄ = 1/λ, are not roots of unity of order ` for 0 < ` ≤ 2m + 1, is locally conjugate to its Birkhoff normal form:  m X  FB (z, z̄) = λz 1 + Bj (z z̄)j + O(2m + 2), j=1 see [1]. The first non-vanishing number Bj is called the jth Birkhoff constant. If Vj = Re(Bj ) < 0 (resp. Vj > 0), then the point p is LAS (resp. repeller), see [8, Lem. 4.1] for instance. The quantity Vj is called the jth Birkhoff stability constant. This is so, because the fact that Vj 6= 0 implies that the function zz = |z|2 is a strict Lyapunov function at the origin for the normal form map FB of F. In [8], both the Birkhoff and the Birkhoff stability constants of F1 and F2 are computed √ 3 obtaining that B1 (F1 ) = − 12 − 11 2 i and B1 (F2 ) = − 12 + 2 i. So V1 (Fj ) = − 12 < 0 for j = 1, 2, and the origin is LAS for both maps F1 and F2 . Also in this reference it is proved √  that V1 (F2 ◦ F1 ) = 12 3 3 − 5 > 0, so that the origin is a repeller fixed point for the composition map. 14 Proposition 13. Consider the planar polynomial vector fields     iπ i 1 i X1 (z, z̄, µ) = z − 1 − πz 2 + − πz z̄ − (3 − 4i) πz 3 2 2 4 4    3π 1 π 11 3π 2 + − +i − z 2 z̄ + z z̄ + µ z̄ 3 , 4 2 2 2 4 where µ is a free parameter, and √ ! iπ 1 i 3 X2 (z, z̄) = z + − + z 2 z̄. 3 2 2 Let ϕj (t; z, z̄), j = 1, 2 be their respective associated flows. Then, for z in a small enough neighborhood of the origin ϕj (1; z, z̄) = Fj (z, z̄) + O(4), j = 1, 2, where the maps Fj are given in Proposition 12. Proof. Observe that F1 has the form (3) with α = π/2, so that eiα is a primitive fourth root of the unity. Since the compatibility condition (12) is satisfied, by using the expression in Proposition 11(a) we can find an one-parameter family of vector fields X1 (z, z̄, µ) satisfying ϕ1 (1; z, z̄) = F1 (z, z̄) + O(4). This is the family of vector fields X1 given in the statement, where µ is the free parameter a0,3 . Also observe that F2 has also the form (3) with α = π/3, so that eiα is a primitive sixth root of the unity. By using Proposition 8 we can find a unique vector field X2 satisfying ϕ2 (z, z̄) = F2 (z, z̄) + O(4). This X2 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 Propo- sition 13 provide the desired example with X1 and X2 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 −X1 and −X2 . The key point is to realize that if ϕ(t; z, z̄) denotes the flow of (2) it holds that  ϕ(2; z, z̄) = ϕ2 (1; ϕ1 (1; z, z̄), ϕ1 (1; z, z̄)) = F2 F1 (z, z̄ + O(4)) + O(4) = F2 ◦ F1 (z, z̄) + O(4). Now, a crucial step is that the first Birkhoff stability constant V1 (F ) only depends on the third order jet of F at the fixed point, see [8, Equation (3)]. Hence V1 (ϕj (1; z, z̄)) = V1 (Fj ), j = 1, 2 and V1 (ϕ(2; z, z̄)) = V1 (F2 ◦ F1 ). It is clear that for the vector fields X1 , X2 and the one in (2) the stability of the origin coincides with the one of the corresponding flows ϕ1 (1; ·, ·), ϕ2 (1; ·, ·) and ϕ(2; ·, ·) 15 respectively. Equivalently, these stabilities coincide with the ones of the origin for the maps F1 , F2 and F2 ◦ F1 . Since, by Proposition 12, these maps provide a discrete dynamic Parrondo’s paradox, we have that both X1 and X2 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. 4 Proof of Theorem 2 (i) This is a corollary of statement (i) of Theorem 6. (ii) We will use item (ii) of Theorem 6. For each n ≥ 2 we will prove that the polynomial map 2π F (z, z̄) = eiα z + z̄ n , with α = , (13) n+1 satisfies the statement of the theorem. The result for n = 2 is a direct consequence of Proposition 7. When n = 3, the result follows by item (a) of Proposition 11 because the compatibility condition (12) does not hold. Now suppose that n ≥ 4. We claim that for each 2 ≤ m ≤ n − 1, if eiα is a primitive (n + 1)-root of unity and Xm is a vector field with associated flow of the form ϕm (t; z, z̄) = eiαt z + O(m + 1), then it satisfies Xm (z, z̄) = 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 7 the result is true for m = 2. Assume that the result is true for m < n−1. As a consequence, for any vector field of the form X Xm+1 (z, z̄) = iαz + aj,k z j z̄ k + O(m + 2), j+k=m+1 its associated flow has the form X ϕm+1 (t; z, z̄) = eiαt z + ϕj,k (t)z j z̄ k + O(m + 2). j+k=m+1 By plugging the above expression into the differential system ż = Xm+1 (z, z̄), we get that for j + k = m + 1: ϕ0j,k (t) = iαϕj,k (t) + aj,k e(j−k)iαt , (14) and since ϕj,k (0) = 0 we obtain that   aj,k   eiαt e(j−k−1)iαt − 1 , j = 6 k + 1, ϕj,k (t) = (j − k − 1)iα (15)  aj,k t eiαt , j = k + 1. 16 Since eiα is a primitive (n + 1)-root of unity, e(j−k−1)iα 6= 1 for j + k = m + 1 < n (see Remark 5). Now, if we assume that ϕm+1 (1; z, z̄) = eiα z + O(m + 1) then ϕj,k (1) = 0 for j + k = m + 1 and from (15) aj,k = 0 and ϕj,k (t) ≡ 0. So, the claim is proved. Now we proceed by contradiction. We consider the map (13), and suppose that there exists a vector field X whose flow satisfies ϕ(1; z, z̄) = F (z, z̄) + O(n + 1) = eiα z + z̄ n + O(n + 1). (16) P j z̄ k From the claim, X must have the form X(z, z̄) = iαz + j+k=n aj,k z + O(n + 1). For these kind of vector fields the associated flow must have the form ϕ(t; z, z̄) = eiαt + P j k j+k=n ϕj,k (t)z z̄ + O(n + 1). For j + k = n, the functions ϕj,k (t) also satisfy (14) and hence (15). In particular, 1 2πi  ϕ0,n (t) = − a0,n e n+1 t e−2πit − 1 , 2πi and therefore ϕ0,n (1) = 0. But this is in contradiction with Equation (16), which implies that ϕ0,n (1) = 1. Funding The authors are supported by Ministry of Economy, Industry and Competitiveness–State Research Agency of the Spanish Government through grants MTM2016-77278-P (MINE- CO/AEI/FEDER, UE, first and second authors) and DPI2016-77407-P (MINECO/AEI/FE- DER, UE, third author). The first and second authors are also supported by the grant 2017-SGR-1617 from AGAUR, Generalitat de Catalunya. The third author acknowledges the group’s research recognition 2017-SGR-388 from AGAUR, Generalitat de Catalunya. 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