On the Dynamics of a Model with Coexistence of Three Attractors: A Point, a Periodic Orbit and a Strange Attractor

For a dynamical system described by a set of autonomous differential equations, an attractor can be either a point, or a periodic orbit, or even a strange attractor. Recently a new chaotic system with only one parameter has been presented where besides a point attractor and a chaotic attractor, it also has a coexisting attractor limit cycle which makes evident the complexity of such a system. We study using analytic tools the dynamics of such system. We describe its global dynamics near the infinity, and prove that it has no Darboux first integrals.


Introduction and Statement of the Main Results
In this paper we study the differential systeṁ x = a + yz,  1 where a ∈ R is a parameter, x, y, z ∈ R and the independent variable is the time t. This is probably the easiest known polynomial differential systems for which coexists at the same time for a = 1/100 three attractors given by an equilibrium point, a periodic orbit and a strange attractor (see [20] for more details). Many chaotic differential systems in R 3 have been studied after the seminal work made by Lorenz [12], see for instance [2, 5-7, 9, 13-19, 21-25, 27].
In general, to describe the global dynamics of a nonlinear differential system in R 3 is a hard problem, usually unsolved. The goal of this paper is to analyze what can be said using analytical tools about the dynamics of system (1). We shall describe the global dynamics near the infinity and study the Darboux integrability of system (1).
In order to detect if a polynomial differential system is integrable, in the sense that it has some first integral, usually one starts looking for a Darboux first integral, because some of the easiest integrable polynomial differential systems are the ones having a Darboux first integral, due to the fact that these first integrals can be detected using the invariant algebraic surfaces or the so-called Darboux polynomials of the system. For this reason trying to know if a polynomial system is integrable or not, which usually is not an easy problem, a first step in this direction is to know if it has or not a Darboux first integral. A polynomial differential system in R 3 can be extended in a unique way to an analytic differential system on the Poincaré ball, where the boundary of this ball, a 2-dimensional sphere S 2 , corresponds to the infinity of R 3 (each point of S 2 provides a different direction for going to infinity), and its interior corresponds to R 3 , see the Poincaré compactification in Section 2. A such compactification cannot be done in general for analytic or smooth differential systems. This compactification of the polynomial differential systems allows to describe the dynamics of the system near the infinity in a relative easy way, such a dynamics near the infinity is in general very difficult to describe for analytic and smooth differential systems.
The following result shows that the dynamics in a neighborhood of the infinity for system (1) does not depend on the parameter a.

Proposition 1
For all values of the parameter a the phase portrait of system (1) on the infinity of the Poincaré ball is topologically equivalent to the one of Figure 1. There are two pairs of singular points at infinity. The pair which is at the endpoints of the y-axis are nodes, unstable the endpoint of the positive half-axis and stable the endpoint of the negative half-axis. The pair which is at the endpoints of the z-axis are nilpotent cusps.
Proposition 1 is proved in Section 2. Let C[x, y, z] be the ring of the real polynomials in the variables x, y and z. We where k = k(x, yz) is a real polynomial of degree at most 1 called the cofactor of F (x, yz), and ∇F is the gradient of F . If the cofactor is zero, then F (x, y, z) is a polynomial first integral of system (1). If F (x, y, z) is a real Darboux polynomial with nonzero cofactor, then the surface F (x, y, z) = 0 is an invariant algebraic surface, that is, if an orbit of system (1) has a point on this surface, then the whole orbit is contained in it.
The vector field associated to system (1) is Let U be an open subset of R 3 such that R 3 \ U has zero Lebesgue measure. We say that a non-constant real function H = H (x, y, z) : U → R is a first integral if H (x(t), y(t), z(t)) is constant on all solutions (x(t), y(t), z(t)) of X contained in U , i.e. X H | U = 0. The existence of a first integral for a differential system in R 3 allows to reduce the study of the differential system in one dimension. This is the main reason to look for first integrals. An exponential factor F (x, y, z) of the vector field X is an exponential function of the form exp(g/ h) with g and h coprime polynomials in C[x, y, z], satisfying X F = LF for some L ∈ C[x, y, z] with degree one. The exponential factors appear when some Darboux polynomial has multiplicity larger than one, for more details see [3,10].
A first integral of system (1) is called of Darboux type if its is a first integral of the form where f 1 , · · · , f p are Darboux polynomials and F 1 , · · · , F q are exponential factors.

Theorem 2 System
(1) has no first integrals of Darboux type for all a ∈ R.
The proof of Theorem 2 is given in Section 3.
In [26] the authors have studied the Hopf bifurcations of system (1).

The Poincaré Compactification
In what follows we present a summary on the Poicaré compactification of a polynomial vector field in R 3 , for more details see [4]. We consider the polynomial differential systeṁ in R 3 , or equivalently its associated polynomial vector field X = (P 1 , P 2 , P 3 ). The degree n of X is defined as n = max{deg(P i ) : i = 1, 2, 3}. Let S 3 = {y = (y 1 , y 2 , y 3 , y 4 ) ∈ R 4 : y = 1} be the unit sphere in R 4 , and S + = {y ∈ S 3 : y 4 > 0} and S − = {y ∈ S 3 : y 4 < 0} be the northern and the southern hemispheres, respectively. We denote by T y S 3 the tangent space to S 3 at the point y. We identify R 3 with the tangent hyperplane Doing central projections of the hyperplane T (0,0,0,1) S 3 on the sphere S 3 we get two copies of our vector field X on S 3 , one in the open nothern hemisphere S + and the other in the open southern hemisphere S − . Now the equator S 2 = S 3 ∩ {y 4 = 0} plays the role of the infinity of R 3 . There is a unique extension of the two copies of the polynomial vector field X on S + ∪ S − to an analytic vector field p(X) on S 3 . This vector field p(X) on S 3 is called the Poincaré compactification of X.
Note that the projection of S + ∪ S 2 on the hyperplane y 4 = 0 through (y 1 , y 2 , y 3 , y 4 ) → (y 1 , y 2 , y 3 ) is the unit closed ball centered at the origin of R 3 = {(y 1 , y 2 , y 3 )}. The interior of this ball is diffeomorphic to R 3 and its boundary S 2 corresponds to the infinity of R 3 . This ball is called the Poincaré ball.
We consider the following eight local charts on S 3 : for i = 1, 2, 3, 4. Then the analytical field p(X) in the local chart U 1 becomes where In a similar way the expression of p(X) in U 2 is where In U 4 we have z n+1 3 (P 1 , P 2 , P 3 ) in the expression for p(X) where P i = P i (z 1 , z 2 , z 3 ). The expression for p(X) in the local chart V i is the same as in U i multiplied by (−1) n−1 .
When we work with the expression of the compactified vector field p(X) in the local charts we shall omit the common factor 1/( z) n−1 . We can do that through a rescaling of the time.
We remark that all the points on the sphere at infinity in the coordinates of any local chart have z 3 = 0.
In this section we study the behavior of the differential system (1) near the infinity using the Poincaré compactification.
Proof of Proposition 1 From (3) the Poincaré compactification of system (1) in the local chart U 1 isż . We look for the equilibria (z 1 , z 2 , z 3 ) with z 3 = 0, which are the ones that are at infinity and we do not find any because the system in the local chart U 1 restricted to the infinity z 3 = 0 becomeṡ So, the dynamics on U 1 is trivially conjugated to straight lines.
From (4) the Poincaré compactification of system (1) in the local chart U 2 iṡ . This system with z 3 = 0 has a unique equilibrium point, the origin, i.e., the endpoints of the positive y-axis. The system in the local chart U 2 restricted to z 3 = 0 writesż 1 = z 2 − z 3 1 ,ż 2 = −z 2 1 z 2 . This system has the first integral Doing blow-ups we can study the local phase portrait at the origin U 2 , obtaining a stable node, for more details on the blow-up change of variables see Chapters 2 and 3 of [8]. Finally, from (5) the Poincaré compactification of system (1) in the local chart U 3 isż Again the origin of this system with z 3 = 0 is the unique equilibrium point, i.e., the end-points of the positive z-axis. The system in the local chart U 3 restricted to z 3 = 0 writesż 1 = z 2 ,ż 2 = z 2 1 .
This system has the first integral The origin of U 3 is a nilpotent singular point. Using Theorem 3.15 of [8] its local phase portrait is a cusp.

Darboux Integrability: Proof of Theorem 2
To prove Theorem 2 first we state and show some auxiliary results.

Proposition 3 System (1) has no polynomial first integrals.
Proof Let f be a polynomial first integral and let n ≥ 1 be its degree. Then the homogeneous part of degree n of f denoted by f n = f n (x, y, z) satisfies the quasilinear partial differential equation Its general solution is F (z, 3y 2 z − 2x 3 ) being F an arbitrary C 1 function. Since f n must be a homogeneous polynomial of degree n we must have a j z n−3j (y 2 z − 2x 3 ) j , a j ∈ C.
Note that the homogeneous part of degree n−1 of f denoted by f n−1 = f n−1 (x, y, z) satisfies We do the change of variables In these variables (7) becomes wheref n−1 is f n−1 written in the variables u, v, w. Integrating this equation we getf where F ( |m) and E( |m) are the elliptic integral of the first and second kind, respectively, for more details see [1]; and Now introducing the inverse change of (8), i.e.
we obtain Since f n−1 must be a homogeneous polynomial of degree n − 1, we must have and Clearly, from (10) we get that n j =0 ja j u n−3j v j −1 = 0, which implies that a j = 0 for j = 1, · · · , n. So, equation (9) becomes −14na 0 z n−1 = 0, which yields a 0 = 0. So, a j = 0 for j = 0, · · · , n and thus f n = 0 in contradiction with the fact that f n is a polynomial of degree n ≥ 1. This concludes the proof of the proposition.

Proposition 4 System (1) has no Darboux polynomials with non-zero cofactor.
Proof Let f (x, y, z) be a Darboux polynomial of system (1) with nonzero cofactor k = k 0 + k 1 x + k 2 y + k 3 z. Clearly, if follows from (2) that f satisfies We expand f as a power series in the variable z and let m be the degree of the polynomial in the variable z, so The terms of degree m + 1 in z in (11) satisfy where K m (y) is a C 1 -function in the variable y. Since f m must be a polynomial we get that either k 3 = 0 and K m (y) is a polynomial in the variable y, or if k 3 = 0 then f m = 0. In this last case f = f (x, y) and satisfies Since f does not depend on z, in particular (12) must hold for z = 0 and so a ∂f ∂x Solving (13) we get that if a = 0 then and if a = 0, where K 0 is a C 1 -function in its variables. When a = 0, f (x, y) is never a polynomial unless k 0 = k 1 = k 2 = 0 and K 0 is constant, but then f (x, y) is constant, in contradiction with the fact that f is a Darboux polynomial.
When a = 0 in order that f is a polynomial we must have k 1 = k 2 = 0 and then f (x, y) = (y − x 2 ) −k 0 p(x) where k 0 ∈ N − and p(x) ∈ C [x]. Imposing that f must satisfy (12) we get that where c 0 is a constant. Since k 3 = 0 we get a contradiction with the fact that p(x) must be a polynomial, so this case is also not possible. In short, k 3 = 0. Now we expand f as a power series in the variable y and let be the degree of the polynomial in the variable y, so The terms of degree + 1 in y in (11) satisfy where K (z) is a C 1 -function in the variable z. Since f must be a polynomial we get that either k 2 = 0 and K (z) is a polynomial in the variable z, or if k 2 = 0 then f = 0. In this last case f = f (x, z) and satisfies Solving (14) we get where K 0 is a C 1 function in its variable. Since f must be a polynomial we get that k 0 = k 1 = k 2 = 0 in contradiction with the fact that k 2 = 0. So, k 2 = 0 and k = k 0 + k 1 x. Now let n be the degree of the polynomial f . Then the homogeneous part of degree n of f denoted by f n = f n (x, y, z) satisfies yz ∂f n ∂x + x 2 ∂f n ∂y = k 1 xf n Solving this partial differential equation we obtain where K n (z, 3y 2 z − 2x 3 ) is a C 1 function in the variables z and 3y 2 z − 2x 3 , and 2 F 1 is the hypergeometric function, see [1]. Since f n must be a homogeneous polynomial of degree n ≥ 1 (otherwise f would not be a Darboux polynomial) we must have k 1 = 0, then k = k 0 and also a j z n−3j (y 2 z − 2x 3 ) j , a j ∈ C. Now proceeding as in the proof of Proposition 3 we get that in order that f be a Darboux polynomial k 0 must be 0. But this is not possible because k = k 0 = 0. This concludes the proof.
Before proving Theorem 2 we also recall the following two well-known results of the Darboux theory of integrability. For the first one see for instance Chapter 8 of [8], and for the second one see [10,11].
Theorem 5 (Darboux theory of integrability) Suppose that a polynomial vector field X defined in R n of degree m admits p Darboux polynomials f i with cofactors K i for i = 1, · · · , p, and q exponential factors F j = exp(g j /h j ) with cofactors L j for j = 1, · · · , q. If there exist λ i , μ j ∈ C not all zero such that μ j L j = 0, then the following real (multivalued) function of Darboux type f λ 1 1 · · · f λ p p F μ 1 1 · · · F μ q q substituting f λ i i by |f i | λ i if λ i ∈ R, is a first integral of the vector field X .

Proposition 6
The following statements hold.
(1) If e g/ h is an exponential factor for the polynomial differential system (1) and h is a non-constant real polynomial, then h is a Darboux polynomial of system (1). (2) Eventually e g can be an exponential factor, coming from the multiplicity of the infinite invariant plane.
Proof of Theorem 2 It follows from Propositions 6, 3 and 4 that if system (1) has an exponential factor then it must be of the form e g with g ∈ C[x, y, z]. Then g must satisfy (a + yz) ∂g ∂x where L = l 0 + l 1 x + l 2 y + l 3 z. Let n be the degree of g. If n ≥ 2 then if we write g in its homogeneous parts as g = n j =0 g j (x, y, z) where g j = g j (x, y, z) is a homogeneous polynomial in the variables x, y, z we have that g n satisfies (6). Proceeding as in the proof of Proposition 3 we obtain that g n = 0. Therefore, g has degree one, that is, g = g 0 + g 1 x + g 2 y + g 3 z. Introducing g in (15) we get that g = c 0 + c 1 z with L = c 1 (1 − 4x), c 0 , c 1 ∈ C. Now Theorem 5 implies that system (1) has no first integrals of Darboux type.