Integrability of the Rucklidge system

We study the Darboux and the analytic integrability of the Rucklidge system.

is a famous model (see for instance [9]), where x, y, z ∈ R 3 are the state variables, (a, b) are real parameters and the dot denotes the derivative with respect to the time t. This model considers the problem of two-dimensional convection in a horizontal layer of Boussinesq fluid with lateral constants. It provides an accurate description of convection in the parameter regime where the chaotic solutions appear. Despite its simplicity it has a reach local dynamical behavior as chaotic attractors for some values of the parameters a and b (for example when a = 2, b = 6.7 or a = −0.1 and b = −1) and has been widely analyzed (see for instance [10,11] and the references therein). We note that system (1) is a family of quadratic systems in a three-dimensional space. Quadratic systems in R 3 are the simplest systems after the linear ones. Examples of such systems are the well-known Lorenz system, Rössler system, Rikitake system, among others. These have been investigated in the last decades from different dynamical points of view. Despite their simplicity, quadratic systems are not completely understood from the view point of the integrability, see for instance [8].
The aim of this paper is to study the existence of Darboux and analytic first integrals of system (1). We recall that a first integral of Darboux type is a first integral H which is a function of Darboux type (see below (2) for a precise definition). The study of the integrability is a classical problem in the theory of differential equations.
The vector field associated to system (1) is Let U ⊂ C 3 be an open set. We say that the nonconstant function H : U → C is a first integral of the polynomial vector field X on U if H (x(t), y(t), z(t)) is constant for all values of t for which the solution (x(t), y(t), z(t)) of X is defined on U . Clearly H is a first integral of X on U if and only if In this paper, we want to study the so-called Darboux first integrals of the polynomial differential systems (1), using the Darboux theory of integrability (originated in the papers [2]). For a present state of this theory see the Chap. 8 of [3], the paper [5], and the references quoted in them. Moreover, we also study the analytic integrability of system (1), i.e., the existence of a global analytic first integral H : R 3 → R.
We emphasize that the study of the existence of first integrals is a classical problem in the theory of differential systems, because the knowledge of first integrals of a differential system can be very useful in order to understand and simplify the topological structure of their orbits. Thus, their existence or not can also be viewed as a measure of the complexity of a differential system.
We recall that a first integral is of Darboux type if it is of the form where f 1 , . . . , f p are Darboux polynomials (see Sect. 2 for a definition), F 1 , . . . , F q are exponential factors (see Sect. 2 for a definition), and λ j , μ k ∈ C for all j and k. The functions of the form (2) are called Darboux functions, and they are the base of the Darboux theory of integrability, which looks when these functions are first integrals or integrating factors. In this last case, the first integrals associated to integrating factors given by Darboux functions are the Liouvillian first integrals, see for more details [3,5].
The Darboux theory of integrability is essentially an algebraic theory of integrability based in the invariant algebraic hypersurfaces that a polynomial differential system has. In fact to every Darboux polynomial there is associated some invariant algebraic hypersurface (see again Sect. 2), and the exponential factors appear when an invariant algebraic surface has multiplicity larger than 1, for more details see [1,3,5]. As far as we know is the unique theory of integrability which is developed for studying the first integrals of polynomial differential systems. In general the other theories of integrability do not need that the differential system be polynomial.
The main results related to the integrability problem are summarized in the next theorems.
As a straightforward consequence of this result we have the following result.

Corollary 2 System (1) has no polynomial first integrals or invariant algebraic surfaces.
We prove Theorem 1 and Corollary 2 in Sect. 3. The next two results states the no existence of Darboux first integrals and analytic first integrals (for a set of the values of the parameters (a, b)) for system (1).

Theorem 3 System (1) has no Darboux first integrals.
Theorem 4 System (1) has no global analytic first integral except perhaps in a set S of zero Lebesgue measure in the plane of parameters (a, b). The set S is contained in a countable set of segments.
Remark 5 A more precise statement of Theorem 4 is given in Theorem 13. We prove Theorem 3 in Sect. 3 and Theorem 13 in Sect. 4. Since the Darboux theory of integrability of a polynomial differential system is based on the existence of Darboux polynomials and their multiplicity, the study of the existence or not of Darboux first integrals needs to look for the Darboux polynomials. So the main steps for proving Theorem 3 are Theorem 10 and 1.

Basic results
Let h = h(x, y, z) ∈ C[x, y, z]\C. As usual C[x, y, z] denotes the ring of all complex polynomials in the variables x, y, z. We say that h is a Darboux polynomial of system (1) if it satisfies Xh = K h, the polynomial K = K (x, y, z) ∈ C[x, y, z] is called the cofactor of h and has degree at most one. Every Darboux polynomial h defines an invariant algebraic hypersurface h = 0, i.e., if a trajectory of system (1) has a point in h = 0, then the whole trajectory is contained in h = 0, see for more details [3]. When K = 0 the Darboux polynomial h is a polynomial first integral.
We recall the following auxiliary result that was proved in [1]. An exponential factor E of system (1) is a function of the form

Lemma 6 Let f be a polynomial and f =
for some polynomial L = L(x, y, z) of degree at most one, called the cofactor of E.
A geometrical meaning of the notion of exponential factor is given by the next result.

Proposition 7
If E = exp(g/ h) is an exponential factor for the polynomial differential system (1) and h is not a constant polynomial, then h = 0 is an invariant algebraic hypersurface, and eventually e g can be exponential factors, coming from the multiplicity of the infinite invariant hyperplane.
The proof of Proposition 7 can be found in [1,6]. We explain a little the last part of the statement of Proposition 7. If we extend to the projective space PR 3 the polynomial differential system (1) defined in the affine space R 3 , then the hyperplane at infinity always is invariant by the flow of the extended differential system. Moreover, if this invariant hyperplane has multiplicity higher than 1, then it creates exponential factors of the form e g , see for more details [6].

Theorem 8
Suppose that the polynomial vector field X of degree m defined in C 4 admits p invariant algebraic hypersurfaces f i = 0 with cofactors K i , for i = 1, . . . , p and q exponential factors E j = exp(g j / h j ) with cofactors L j , for j = 1, . . . , q. Then there exists λ i , μ j ∈ C not all zero such that if and only if the function of Darboux type Theorem 8 is proved in [3]. The following result is well-known.

Lemma 9
Assume that exp(g 1 / h 1 ), . . . , exp(g r / h r ) are exponential factors of some polynomial differential system with P, Q, R ∈ C[x, y, z] with cofactors L j for j = 1, . . . , r . Then is also an exponential factor of system (3) with cofactor L = r j=1 L j .

Proof of Theorems 1 and 3.
We consider the automorphism τ : The following result characterizes the polynomial first integrals of system (1), i.e., it characterizes the Darboux polynomial with zero cofactor that are invariant by τ .

Theorem 10
There are no polynomial first integrals of system (1) that are invariant by τ .
Proof Let h = h(x, y, z) be a polynomial first integral of system (1) that is invariant by τ . We write it as a sum of homogeneous polynomials as where each h j is a homogeneous polynomial of degree j. Without loss of generality we can assume that h 0 = 0, n > 0 and h n = 0. Note that if h is a polynomial first integral then Now we define the following linear partial differ- To solve system (5) is equivalent to look for a polynomial first integral of systeṁ We introduce the change of variables Then we can rewrite system (6) aṡ which implies that system (5) becomes Since h n is invariant by τ we must have τ (h n ) = h n that is Hence, a l,k = 0 for l odd. This yields that l must be even and thus Now we will show by induction that for j = 1, . . . , m, where and O(X −1 ) denotes polynomials in the variable X −1 (with coefficients depending in the variables Y, Z ). We compute the terms of order n = 2m in (4) we get which is equivalent with the change of variables in (7) as Note that in the right hand of equation (11) the parts that will provide the highest degree in X are so we only need to consider them. Hence, using (9) we obtain Integrating it with respect to Z we obtaiñ Sinceh n−1 must be invariant byτ ( p(X, Y, Z )) = p(X, −Y, Z ) and must have dimension 2m − 1 we get thatg n−1 (X, Y ) = 2l+2k=2m−1 c l Y 2l X k which is not possible and thusg n−1 = 0. Hence, Sinceh n−1 is a polynomial we must have b l = 0 for l = 0, 1 and thus which satisfies (10) for j = 1 with c l,0 = b l . Now assume (10) is true for j = 1, . . . , k and we will prove it for j = k + 1. Hence, if we compute the terms of order n − k in (4) we get Proceeding as in (11) using the new variables (7) we only need to consider the terms in (12) for each one of the sums when we let r = 0, . . . , [k/2]. Hence, using equation (10) with j = k we get and so, Integrating (13) with respect to Z we obtaiñ If k is even then this is not possible, and thusg n−k−1 = 0. Note that in this case Sinceh n−k−1 is a polynomial in Y we must have that it is zero when l = 2(k −2r ) and when l = 2(k −2r )+1. We emphasize that A k+1 = 0 for l = 2(k − 2r ) or l = 2(k − 2r ) + 1 and thus c l,r = 0 for l = 2(k − 2r ) and l = 2(k−2r )+1. This implies that l ≥ 2(k+1−2r ) which proves (10) for j = k + 1 and k even. If k is odd then we get that

Note that in this casẽ
where for r = (k + 1)/2 we get that d s = c l,r . As before, sinceh n−k−1 is a polynomial in Y we must have that it is zero when l = 2(k − 2r ) and when l = 2(k − 2r ) + 1. We emphasize that A k+1 = 0 for l = 2(k − 2r ) or l = 2(k − 2r ) + 1 and thus c l,r = 0 for l = 2(k − 2r ) and l = 2(k − 2r ) + 1. This implies that l ≥ 2(k + 1 − 2r ) which proves (10) for j = k + 1 with k odd. It follows from (10) with r = 0 and j = 0, . . . , m that b l = 0 for l = 0, . . . , m. This implies that h m is a zero which is not possible since h is a first integral. This completes the proof of the theorem. Now we compute the Darboux polynomials of system (1) that are invariant by τ .

Proposition 11
If g is an irreducible Darboux polynomial for system (1) with cofactor then f = g · τ g is a Darboux polynomial invariant by τ with a cofactor of the form Proof Since system (1) is invariant under τ , then τ g is a Darboux polynomial of system (1) and with cofactor τ (K ). Moreover, by Lemma 6, g · τ g is also a Darboux polynomial of system (1) with cofactor K + τ (K ). Therefore again by Lemma 6, the cofactor of f is K τ = K + τ (K ) which is given in (15).
Proof Let g = g(x, y, z) be a Darboux polynomial of system (1) with nonzero cofactor K τ . Then it satisfies We write g in its homogeneous parts as where each g j is a homogeneous polynomial of degree j. Computing the terms of degree n + 1 in (16), we get L (g n ) = 2α 3 zg n .
Solving it we get g n = G n y, x y + z 2 2 e where G n is any function in the variables y and x y+ z 2 2 . Since g n must be a polynomial we must have α 3 = 0. Then we get that g n can be written as in (8). Hence, proceeding exactly as we did in the proof of Theorem 10 taking into account that in the computation of g n−1 we have that 2α 0 g n is in the O(X −1 ) we can prove in the same manner that (10) holds. Then, g n = K n a constant and n = 0. Thus, it follows from (16) that 0 = 2α 0 g 0 that is α 0 = 0. This concludes the proof.
Proof of Theorem 1 Let g be an irreducible Darboux polynomial of system (1) of degree n and with cofactor K of the form given in (14). Then, from Proposition 11, we can assume that f = g · τ g is a Darboux polynomial of system (1) and invariant by τ , with degree 2n and non-zero cofactor K τ of the form in (15). From Proposition 12, we get that K τ = 0. Hence, f must be a polynomial first integral of system (1) that is invariant by τ . By Theorem 10 this is not possible. Note that this proof states for Darboux polynomials with zero and nonzero cofactor. Now we proceed with the proof of Theorem 3.

Proof of Theorem 3
It follows from Theorems 8 and 1 and Proposition 7 that in order to have a first integral of Darboux type we must have q exponential factors E j = exp(g j ) with cofactors L j such that q j=1 μ j L j = 0. Let G = q j=1 μ j g j , then E = exp(G) is an exponential factor of system (1) with cofactor L = q j=1 μ j L j = 0 (see Lemma 9). So G is a polynomial first integral of system (1). By Theorem 1 this is not possible.

Proof of Theorem 13
In this section we prove the next result Theorem 13. A corollary of Theorem 13 is Theorem 4 Theorem 13 System (1) has no analytic first integral for the values of the parameters a, b ∈ R satisfying: all (a, b) such that ab = 0. Before proving Theorem 13 we need an auxiliary lemma that characterizes all the first integrals of the linear part of system (1).

Lemma 14
The linear part of system (1) has the two independent first integrals: Note that the homogeneous terms of degree 1 in (17) satisfy the differential equation: L 1 (h 1 ) = 0 where L 1 is the linear partial differential operator given by

is a first integral
of the linear vector field X (x, y, z) = (by, x, −z).

1−4b
and h 2n+1 = 2nα 2n (−x 2 +by 2 ) n−1 (x 2 +xy+by 2 )z . For the homogeneous terms of degree 2n + 2 of (17) we have that has polynomial solution just when α 2n = 0 and the homogeneous solution of degree 2n + 2 is given by h 2n+2 = α 2n+1 (−x 2 + by 2 ) n+1 . So we have by induction under the degree of homogeneity that h k ≡ 0 for all k ≥ 1 and system (1) Note that the homogeneous terms of degree 1 in (19) satisfy the differential equation: L 2 (h 1 ) = 0 where L 2 is the linear partial differential operator given by i.e. h 1 is a first integral of the linear vector field X (x, y, z) = y 4 , x, −z . As h 1 is polynomial it follows from Lemma 14 that h 1 (x, y, z) = α 1 (4x 2 − y 2 ) l . But, as h 1 is of degree 1 we must have α 1 = 0 and so h 1 ≡ 0. Now the homogeneous terms of degree 2 in (19) satisfy L 2 (h 2 ) = yz ∂h 1 ∂ x = 0. So, we get h 2 = α 2 4x 2 − y 2 l 2 for some l 2 ∈ N. Using the fact that h 2 is homogeneous of degree 2 we obtain h 2 = α 2 4x 2 − y 2 . Concerning the terms of degree 3 we have This equation has polynomial solution only when α 2 = 0 and, in this case, h 3 = α 3 (4x 2 −y 2 ) l 3 for some l 3 ∈ N. But as h 3 is homogeneous of degree 3 we get α 3 = 0 and so h 3 ≡ 0. Now using this fact the homogeneous terms of degree 4 in (19) satisfy that have homogeneous polynomial solution of degree 4 of the form h 4 = α 4 (4x 2 − y 2 ) 2 with α 4 ∈ R.
Suppose that we have h 1 , h 2 , · · · , h 2n−1 ≡ 0 and h 2n = α 2n (−x 2 + by 2 ) n with α 2n ∈ R. So, the terms of degree 2n + 1 satisfy Then the homogeneous terms of degree 2n + 1 satisfy that has polynomial solution only if α 2n = 0 and, in this case, the homogeneous polynomial solution of this system is given by h 2n+1 = α 2n+1 (4x 2 − y 2 ) n+1 . So we have by induction under the degree of homogeneity that h k ≡ 0 for all k 1 and system (1) restricted to case (i.1.1) has no analytic first integral.
Suppose that a = −1/n with n ∈ N. So β 2 must be zero and the homogeneous terms of degree 3 in (20) satisfy ∂z that has polynomial solution only when α 1 = 0 and, in this case, In a similar way the homogeneous terms of degree 4 are solutions of the partial differential equation that are polynomial. The solution exists only for α 2 = 0 and is given by

Now suppose by induction that
with n ≥ 4. Then the homogeneous terms of degree n + 1 satisfy L 3 (h n+1 ) = y ∂h n ∂ x − y 2 ∂h n ∂z z that just admit polynomial solutions if α n−1 = 0 and, in this case, we have So, h k ≡ 0 for all k ≥ 1 and system (1) The homogeneous terms of degree 1 satisfy L 4 (h 1 ) The characteristic equations associated to the partial differential equation They have the general solutions x − y = c 1 and zx = c 2 , where c 1 and c 2 are constants of integration. So doing the change of coordinates the equation L 4 (h 1 ) = 0 is written as This implies thath 1 (u, v, w) = g 1 (u, v) = g 1 (x − y, xz), and as h 1 has degree 1 we obtain h 1 (x, y, z) = α 1 (x − y). The homogeneous terms of degree 2 in (21) satisfy In the new coordinates u, v, w this last equation becomes Therefore,h 2 (u, v, w) = α 1 v u u+w + log(u + w) + f 2 (u, v). Going back to the x, y, z coordinates we have As h 2 is a homogeneous polynomial of degree 2 we have α 1 = 0 and h 2 (x, y, z) Performing with the homogeneous terms of degree 3 in the same way than before we obtain This has polynomial solution only when α 2,0 = 0, and considering that h 3 is homogeneous of degree 3 we have Solving (21) with respect to the homogeneous terms of degree 4 we must solve This has a polynomial solution only if α 2,1 = α 3,0 = α 3,1 = 0 and, in this case, the homogeneous part of degree 4 is h 4 = 2 i=0 α 4,i (y − x) 4−2i (xz) i . Taking into account the expression of h 4 and solving the homogeneous part of (21) of degree 5 we obtain We consider, by induction that for fixed n ≥ 1 we have h 2n−1 = − (n − 1)α 2(n−1),n−1 3 (xz) n−2 So the homogeneous terms of degree 2n + 1 in (21) satisfy This has a polynomial solution only when α 2n, j = 0 for j = 0, · · · n − 1 and, in this case, we get The homogeneous terms of degree 2n + 2 satisfy This has a polynomial solution only when α 2n,n = α 2n+1, j = 0 for j = 0, · · · n, and, we have All together we can conclude that H is constant and so system (1)  The terms of degree 1 in (23) satisfy L 5 (h 1 ) = 0, where L 5 is the linear partial differential operator given by The characteristic equations associated to the partial differential equation They have the general solution y − nx = c 1 and x n z = c 2 , where c 1 and c 2 are constants of integration. So doing the change of coordinates whose inverse change is its solution ish 1 (u, v, w) = g 1 (u, v) = g 1 (y − nx, x n z). As h 1 has degree 1 we obtain h 1 (x, y, z) = α 1 (y − nx). Now the homogeneous terms of degree 2 in (23) satisfy In the new coordinates u, v, w it becomes dh 2 dw = − n n+1 α 1 vw (−u + w) n−1 .
The homogeneous terms of degree 3 satisfy that in the u, v, w coordinates is written in the form It has polynomial solutions only for α 1 = 0 and, in this case, the homogeneous solution of degree 3 written in coordinates x, y, z is of the form h 3 = 2nα 2 z(y − nx)(nx + (n − 1)y) n − 1 + α 3 (nx − y) 3 .
We can do this for all m < n + 1 (this is due to the degree of the polynomial x n z), and we conclude that h k ≡ 0 for all k < n. Now, for the homogeneous part of h of degree k ≥ n we have h n = (n − 1)nα n−1 z(y − nx) n−2 (nx + (n − 1)y) n − 1 +α n (y − nx) n .
Solving (23) with respect to the homogeneous terms of degree n + 1 we obtain polynomial solutions only if α n−1 = 0, and this is h n+1 = n 2 α n z(y − nx) n−1 (nx + (n − 1)y) n − 1 In order to find the homogeneous part of degree n+2 satisfying (23) we have to solve Note that when we consider this partial differential equation in the u, v, w coordinates as before the left hand part of the resultant ordinary differential equation is (−u+w) n dh n+2 dw , and the right hand part is composed of two parts. The first part is the term n 2 α n z(y − nx) n−1 (nx + (n − 1)y) n − 1 in the new coordinates, and the second part is 1 i=0 α n+1,i (y − nx) (n+1)(1−i) (x n z) i also in the new coordinates. Moreover, by the linearity of the integral which provides the solution of this ordinary differential equation, we can study these two parts separately.