Abstract
Abstract. We consider Lotka-Volterra systems in three dimensions depending on three real parameters. By using elementary algebraic methods we classify the Darboux polynomials (also known as second integrals) for such systems for various values of the parameters, and give the explicit form of the corresponding cofactors. More precisely, we show that a Darboux polynomial of degree greater than one is reducible. In fact, it is a product of linear Darboux polynomials and first integrals.
Introduction
The Lotka-Volterra model is a basic model of predator-prey interactions. The model was developed independently by Alfred Lotka (1925) , and Vito Volterra (1926) . It forms the basis for many models used today in the analysis of population dynamics. It has other applications in Physics, e.g. laser Physics, plasma Physics (as an approximation to the Vlasov-Poisson equation), and neural networks. In three dimensions it describes the dynamics of a biological system where three species interact.
Introduction
The most general form of Lotka-Volterra equations iṡ
Introduction
a ij x i x j , i = 1, 2, . . . , n.
Introduction
We consider Lotka-Volterra equations without linear terms (ε i = 0), and where the matrix of interaction coefficients A = (a ij ) is skew-symmetric. This is a natural assumption related to the principle that crowding inhibits growth.
Introduction
The most famous special case of Lotka-Volterra system is the KM system (also known as the Volterra system) defined by (1)ẋ i = x i (x i+1 − x i−1 ) i = 1, 2, . . . , n, where x 0 = x n+1 = 0. It was first solved by Kac and van-Moerbeke in [14] , using a discrete version of inverse scattering due to Flaschka [11] . In [17] Moser gave a solution of the system using the method of continued fractions, and in the process he constructed actionangle coordinates. Equations (1) In fact, a transformation of Hénon connects the two systems:
Introduction
The systems which we consider are all integrable in the sense of Liouville. In other words, there are enough integrals in involution to ensure the complete integrability of the system.
Introduction
Any constant value of a first integral defines a submanifold which is invariant under the flow of the Hamiltonian vector field. A second integral is a function which is constant on a specific level set. While a first integral satisfiesḟ = 0, a second integral is characterized by the propertyḟ = λf , for some function λ which is called the cofactor of f . Second integrals are also called special functions, stationary solutions, and in the case of polynomials, eigenpolynomials, or, more frequently, Darboux polynomials. In systems which have a Lie theoretic origin (e.g. the full Kostant Toda lattice), they arise from semi-invariants of group actions. The importance of Darboux polynomials lies in the following simple fact. If f and g are relatively prime Darboux polynomials, with the same cofactor, then their quotient is a first integral. We propose to understand the behavior of a system based on the algebraic properties of its Darboux polynomials.
Introduction
As a starting point we consider the system (2)ẋ
Introduction
where r, s, t ∈ R. Our main result is the following:
Theorem 1. An arbitrary Darboux polynomial of the system (2) is reducible. In fact, it is a product of linear Darboux polynomials.
The method of proof that we use follows the approach of Labrunie in [15] for the so called ABC system.
Theorem 1. An arbitrary Darboux polynomial of the system (2) is reducible. In fact, it is a product of linear Darboux polynomials.
The system (2) is Hamiltonian. We define the following quadratic Poisson bracket in R by the formula
Theorem 1. An arbitrary Darboux polynomial of the system (2) is reducible. In fact, it is a product of linear Darboux polynomials.
Generically, the rank of this Poisson bracket is 2 and it possesses a Casimir given by
Theorem 1. An arbitrary Darboux polynomial of the system (2) is reducible. In fact, it is a product of linear Darboux polynomials.
The function H = x 1 + x 2 + x 3 is always a constant of motion. In fact, taking H as the Hamiltonian and using the Poisson bracket (3) we obtain equations (2).
Theorem 1. An arbitrary Darboux polynomial of the system (2) is reducible. In fact, it is a product of linear Darboux polynomials.
Lotka-Volterra systems have been studied extensively, see e.g. [4] , [13] , [20] . The Darboux method of finding integrals of finite dimensional vector fields and especially for various types of Lotka-Volterra systems has been used by several authors, e.g. [6] , [7] , [15] [16] , [18] , [19] .
Theorem 1. An arbitrary Darboux polynomial of the system (2) is reducible. In fact, it is a product of linear Darboux polynomials.
The paper is organized as follows. In Section 2, we recall a few basic facts about Darboux polynomials. In Section 3 we prove Theorem 1 under general conditions for r, s, t, and we also give the explicit form of the cofactors. Section 4 deals with the case s = t. We did not examine other such cases since the method of proof is identical with these two cases.
Theorem 1. An arbitrary Darboux polynomial of the system (2) is reducible. In fact, it is a product of linear Darboux polynomials.
Finally in Section 5 we present in detail three examples which include the open and periodic KM-system in three dimensions.
Darboux polynomial preliminaries
Consider a system of ordinary differential equations
Darboux polynomial preliminaries
where the functions v i are smooth on a domain U ⊂ K n . Here K = R or K = C, and we denote by K[x], x = (x 1 , ..., x n ), the ring of polynomials in n variables over K. Let φ : I → U be a solution of (4) defined on an open non-empty interval I of the real axis. A continuous function F : U → R is called a first integral of system (4) if it is constant along its solution, i.e. if the function F • φ is constant on its domain of definition for arbitrary solution φ of (4). When F is differentiable, it is a first integral of system (4) if
Darboux polynomial preliminaries
where L v is the Lie derivative along the vector field v = (v 1 , ..., v n ). If A is any function of x, then the Lie derivative of A is the time derivative of A, i.e.Ȧ = dA dt = L v (A). The vector field generates a flow φ t that maps a subset U of K n to K n in such a way that a point in U follows the solution of the differential equation. That is,φ(x)(t) = v(φ(x)) ∀ x ∈ U. The time derivative is also called the derivative along the flow since it describes the variation of a function of x with respect to t as x evolves according to the differential system.
Darboux polynomial preliminaries
Many first integral search techniques, such as the Prelle-Singer procedure, are based on the Darboux polynomials Thus, the search for Darboux polynomials can be restricted to irreducible polynomials, and, if the system is homogeneous, to homogeneous polynomials. Since the dynamical system (2) is homogeneous of degree 2, the cofactor of any Darboux polynomial of the system will be a linear combination of the variables x 1 , x 2 , x 3 . It follows that any Darboux polynomial f of system (2) will satisfy
Darboux polynomial preliminaries
where α, β, γ are constants. If it is not clear from the context, we shall denote these constants by α(f ), β(f ), γ(f ) respectively.
Darboux polynomials of the Lotka-Volterra system
We carry out our analysis aiming at maximum generality, that is, imposing as few conditions on the parameters r, s, t as possible. In this section we make such assumptions in propositions 9, 10 and Theorem 2, however, as we note in remark 3, one can obtain the results by making assumptions about the cofactor of the Darboux polynomial instead of the parameters. An important role in this work plays the homogeneity property, as can be seen in the following two propositions.
Darboux polynomials of the Lotka-Volterra system
Proof. Since the polynomial f is homogeneous, we use Euler's identity
Darboux polynomials of the Lotka-Volterra system
Using equation (8) we substitute for x 1 ∂f ∂x 1 in equation (7) to obtain
Darboux polynomials of the Lotka-Volterra system
Setting x 1 = 0, x 2 = 0, and letting F (x 3 ) = f (0, 0, x 3 ) we have
Darboux polynomials of the Lotka-Volterra system
If s = 0, γ = 0, equation (9) implies that F = 0. Otherwise, if s = 0 we have
Darboux polynomials of the Lotka-Volterra system
, for some constant κ. Since f is homogeneous of degree m, the only term containing only x 3 is necessarily x m 3 . Thus, if γ = 0 we must have F = 0 also in this case, and the proposition is proved.
Darboux polynomials of the Lotka-Volterra system
We shall use the following notation:
Darboux polynomials of the Lotka-Volterra system
, and for any number r, N m r = {nr : n ∈ N m }.
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
We also have the following statements for α and β:
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
Proof. We prove the statements for γ. The proof of the statements for α and β is similar. Assume that γ = 0. Then, it follows from Proposition 3 that f = x 1 φ 1 + x 2 ψ 1 , where
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
are either homogeneous polynomials of degree m − 1, or zero (but they are not both zero). Setting this in equation (7) yields
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
Setting x 2 = 0 in equation (10) we have
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
The operator φ 1 →φ 1 commutes with the derivations with respect to x 1 and x 3 , and therefore we obtain
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
If s = 0 thenφ 1 = 0, which implies that φ 1 is divisible by x 2 and that f = x 1 φ 1 + x 2 ψ 1 is divisible by x 2 . Suppose now that s = 0, degφ 1 = deg φ 1 = m − 1, and that γ = ns, n ∈ N m . The r.h.s. of (11) is divisible by x 1 , and since γ − s = 0, it follows that
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
where φ 2 is a homogeneous polynomial of degree m − 2. Then, we have
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
and from (11) we obtain
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
Since γ − 2s = 0, φ 2 is divisible by x 1 . Continuing in the same way we obtain
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
where deg φ m−1 = 1, and x 1 |φ m−1 . Thus, φ m−1 = const. x 1 , and from the above equation we have
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
. By equating coefficients we obtain γ = ms, which is a contradiction. Therefore, we must haveφ 1 = 0, which implies that f is divisible by x 2 . Setting x 1 = 0 in (10) and using (7) we obtain
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
If t = 0 then ψ 1 = 0, hence ψ 1 is divisible by x 1 and so f is divisible by x 1 . Suppose that t = 0, deg ψ 1 = deg ψ 1 = m − 1, and γ = nt, n ∈ N m . Then it can be shown in a similar way as above that ψ 1 is divisible by x 1 , which implies that f is divisible by x 1 , and the proposition is proved.
Proposition 4. Let f be a homogeneous Darboux polynomial of degree
This leads to the characterization of the cofactors of Darboux polynomials of system (2), as follows.
Proposition 5. Let f be a homogeneous Darboux polynomial of degree m. We have either
Suppose that γ(f 1 ) = 0, i.e. γ = t, and that γ(f 1 ) = ns, n ∈ N m−1 , that is γ = ns + t, n ∈ N m−1 . Then, again by proposition 4 it follows that f 1 is divisible by x 2 , and writing
Proposition 5. Let f be a homogeneous Darboux polynomial of degree m. We have either
If γ = 2t, and γ = ns + 2t, n ∈ N m−2 , then f 2 is divisible by x 2 . Continuing in the same way, after m − 1 steps we obtain
Proposition 5. Let f be a homogeneous Darboux polynomial of degree m. We have either
where deg f m−1 = 1. If γ = (m − 1)t and γ = s + (m − 1)t, then x 2 |f m−1 , and thus f m−1 = const. x 2 . From equation (12) we then have
Proposition 5. Let f be a homogeneous Darboux polynomial of degree m. We have either
, and by equating coefficients we obtain γ = mt. We therefore conclude that we have either γ = 0, or γ = ns, or γ = nt, n ∈ N m , or γ = γ 1 s + γ 2 t, γ 2 = 1, 2, ..., m − 1, γ 1 ∈ N m−γ 2 , and the proposition is proved.
Remark 1.
We note that in the proof of proposition 5 we can make the successive assump- 
Remark 1.
Proof. The proof is similar to the proof of proposition 5.
Remark 1.
The following propositions give further analysis on the cofactors, and their relation with the parameters and the form of the Darboux polynomials. Proposition 7. Let r, s, t be non-zero, r\s = q 1 , r\t = q 2 , and s\t = q 3 . Let f be a homogeneous Darboux polynomial of degree m, and α 1 , α 2 , β 1 , β 2 , γ 1 , γ 2 the integers which appear in propositions 5 and 6.
Remark 1.
Proof. We prove statement (a). The proof of statements (b) and (c) is similar. If α 2 or β 2 is non-zero, then by hypothesis we have α(f ) = −(α 1 + α 2 1 q 1 )r = 0 and α(f ) = −nr, n ∈ N m , or β(f ) = (β 1 − β 2 1 q 2 )r = 0 and β(f ) = nr, n ∈ N m , respectively. In either case, it follows
Remark 1.
By the same argument as above, if we do not have α 2 (f 1 ) = β 2 (f 1 ) = 0, i.e. if we do not have α 2 = β 2 = 1, then we have either α(f 1 ) = −(α 1 + (α 2 − 1)
Remark 1.
)r = 0 and β(f 1 ) = nr, n ∈ N m−1 , and f 1 is divisible by ∈ N m . The other case is similar. Then we have α(f ) = −(α 1 q 1 + α 2 )s = 0 and α(f ) = −ns, n ∈ N m . From proposition 4 it follows that f is divisible by x 2 , so that f = x 2 f 1 for some homogeneous polynomial f 1 of degree m − 1, and we have (14) L(f 1 ) = ((α + r)x 1 + βx 2 − tx 3 )f 1 .
Remark 1.
Equation (14) shows that f 1 is a Darboux polynomial with γ(f 1 ) = −t. However, from proposition 5 we have γ(f 1 ) = γ 1 s + γ 2 t for non-negative integers γ 1 , γ 2 ∈ {0, 1, 2, ..., m − 1}. Therefore, γ 1 s + γ 2 t = −t, which is possible only if γ 1 = 0, in which case s = − (1+γ 2 ) γ 1 t, and the statement is proved with p = These results allow us to characterize the Darboux polynomials of system (2).
Proposition 9. Let f be a Darboux polynomial of system
2 f 2 , where f 2 is a Darboux polynomial with γ(f 2 ) = 0. Here, the non-negative integers γ 1 , γ 2 are such that γ(f ) = γ 1 s + γ 2 t.
Proposition 9. Let f be a Darboux polynomial of system
Proof. If γ(f ) = 0, then the result in each case follows by setting γ 1 = γ 2 = 0, f 1 = f 2 = f . Suppose that γ(f ) = 0 and s = 0. Then, by proposition 4 f is divisible by x 2 , and writing
Proposition 9. Let f be a Darboux polynomial of system
. Let this procedure be repeated as many times as it can, and let γ 2 be the number of times that it can. We have f = x γ 2 2 f 1 , where f 1 is a Darboux polynomial with γ(f 1 ) = γ − γ 2 t = 0 since we had to stop the division procedure by x 2 , and equation (15) 
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
have the same sign, they are both rational, and s/t < 1/m or s/t > m, (e) s, t have the same sign, they are both irrational, their ratio is rational, and s/t < 1/m or s/t > m.
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
Remark 3. In proposition 9, instead of the condition N m s ∩ N m t = ∅, we can make an alternative assumption as follows. First let s\t = q 3 , and let f k , k = 0, 1, 2, ..., f = f 0 , be a sequence of Darboux polynomials as we describe below. We denote γ(
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
In particular, if γ 1 (f k ) = 0 then we require condition (i) to hold, whereas if γ 2 (f k ) = 0 then we require condition (ii) to hold (if γ 1 (f k ) = 0 and γ 2 (f k ) = 0 then we can have either condition
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
If condition (ii) holds, then γ(f k ) = 0 and γ(f k ) = nt, n ∈ N m−k , which implies that x 1 |f k . In this case we have
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
Following this procedure, after γ 1 + γ 2 steps we obtain f = x
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
The following proposition states similar results in terms of the constants α and β. The proof is similar to the proof of proposition 9. 
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
Proof. Consider the case s = 0 and N m r ∩ (−N m t) = ∅. The other case is similar. We use equation (15) of proposition 9 and the equations in statements (i) and (iii) of proposition 10 in the following algorithm.
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
(1) Set n = 0 and f n = f . (2) Applying proposition 10 for α (statement (i)) yields
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
If f n+1 is a first integral, go to the final step, else increment n by one. (3) Applying proposition 10 for β (statement (iii)) yields
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
If f n+1 is a first integral, go to the final step, else increment n by one. (4)Applying proposition 9 for γ (eq. (15)) yields
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
If f n+1 is a first integral, go to the final step, else increment n by one and return to step 2. (5) (Final step) Set I = f n+1 and using the sequence of equations linking f l to f l+1 , l = 1, ..., n given by the algorithm determine the exponents i, j, k in eq. (18) . At every step one has deg f l+1 ≤ deg f l ; when three consecutive terms of the sequence are of the same degree, they are equal and α(f l ) = β(f l ) = γ(f l ) = 0, so f l is a first integral. Thus the algorithm converges in a finite number of steps. Equation (19) follows from simple properties of Darboux polynomials.
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
If condition (ii) holds, then the proof is the same but now in steps 2 and 4 of the algorithm we use the equation in statement (ii) of proposition 10, and equation (16) 
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
l , where i, j, k, l are non-negative integers, are Darboux polynomials. We show that a Darboux polynomial will have this form with l > 0, provided its cofactor satisfies some conditions which depend on the ratio r\s.
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
Proposition 11. Suppose that r, s, t are non-zero, s = t, and let r\s = q 1 . Let f be a homogeneous Darboux polynomial of degree m which does not have the form (18), and let α 1 ,  α 2 , β 1 , β 2 , γ 1 , γ 2 be the integers which appear in propositions 5 and 6. For j = 0, 1, 2, . .., m− 1, suppose that α 1 + (α 2 − j)
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
Proof. Relation (i) is statement (a) of proposition 7. We prove the inequality (ii). Suppose on the contrary that α 1 + β 1 > γ 1 + γ 2 . By arguments that we have used repeatedly in this paper (for example see proposition 8), f is divisible α 1 times by x 2 and β 1 times by x 1 . Thus we have f = x
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
By proposition 5 there exist non-negative integers γ
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
If
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
f ′ , and our assumptions imply that f ′ is divisible α 2 times by x 3 (proposition 8). So we have
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
3 I, and therefore f = x
t. This condition is satisfied in each of the following cases: (a) one of s, t is positive and the other is negative, (b) s, t have the same sign but one is rational and the other irrational, (c) s, t have the same sign, they are both irrational, and their ratio is irrational, (d) s, t
3 I, where I is a first integral. Since we assume that f does not have the form (18) we may exclude this possibility, and the proof is completed.
Proposition 12.
Suppose that r, s, t are non-zero, s = t, and let r\s = q 1 . Let f be a homogeneous Darboux polynomial of degree m which does not have the form (18) . With the same assumptions as in proposition 11 we have f = (x 1 + x 2 )f 1 , for some polynomial f 1 .
Proposition 12.
Proof. From proposition 8 it follows that we may assume γ(f ) = 0. By proposition 3 f does not have an x m 3 term and we can write f = x 1 φ 1 + x 2 ψ 1 , for some polynomials φ 1 , ψ 1 . For a polynomial f = f (x 1 , x 2 , x 3 ) we denote by f the polynomial obtained from f by setting
Proposition 12.
, and letting h 1 = φ 1 − ψ 1 we have f = x 1 h 1 . Setting s = t and x 2 = −x 1 in equation (10) we obtain
Proposition 12.
Setting s = t and x 2 = −x 1 in equation (7) we obtain
Proposition 12.
Combining equations (20) and (21), and noting that α 2 = β 2 (proposition 11), we obtain
Proposition 12.
From proposition 11 we also have α 1 + β 1 < γ 1 + γ 2 , which implies that the term −(α 1 + β 1 − 1)rx 1 + (γ 1 + γ 2 − 1)sx 3 is not a constant multiple of (rx 1 − sx 3 ). Since (rx 1 − sx 3 ) divides the right-hand side of equation (22), it divides h 1 . Therefore we have
Proposition 12.
Substituting for
Proposition 12.
from equations (23), (24) respectively in equation (22) we obtain
Proposition 12.
and simplifying further we have
Proposition 12.
The term −(α 1 + β 1 − 2)rx 1 + (γ 1 + γ 2 − 1)sx 3 is not a constant multiple of (rx 1 − sx 3 ), and so (rx 1 − sx 3 )| h 2 . Continuing in the same way we find that f is divisible by an infinity of powers of (rx 1 − sx 3 ), which is a contradiction. Therefore we must have f = 0. This implies that f = (x 1 + x 2 )f 1 , for some polynomial f 1 , and the proof of the proposition is completed. 
Proposition 12.
where I is a first integral and i, j, k, l are non-negative integers.
Proposition 12.
Proof. Note first that if γ(f ) = 0 then by proposition 8 it follows that we must have α 1 = β 1 = 0 and f = x 1 +x 3 ) k I.
Examples

5.1.
Example: The KM-system. We give a complete description of Darboux polynomials for the case of the KM system (r = 1, s = 0, t = 1).
5.1.
(27)ẋ
5.1.
The Hamiltonian description of system (1) can be found in [10] and [8] . We will follow [8] and use the Lax pair of that reference. The Lax pair in the case n = 3 is given bẏ
5.1.
This is an example of an isospectral deformation; the entries of L vary over time but the eigenvalues remain constant. It follows that the functions
5.1.
Tr L i are constants of motion. We note that
5.1.
corresponds to the total momentum and
5.1.
Using (3) we define the following quadratic Poisson bracket, {x i , x i+1 } = x i x i+1 , i = 1, 2, and {x 1 , x 3 } = 0. For this bracket detL = x 2 1 x 2 3 is a Casimir and the eigenvalues of L are in involution. Taking the function H 1 = x 1 + x 2 + x 3 as the Hamiltonian we obtain equations (27). Therefore the system has a Casimir given by F = x 1 x 3 and a constant of motion x 1 + x 2 + x 3 corresponding to the Hamiltonian. Note that H 2 = H 2 1 − 2F . In the following table we present all Darboux polynomials of degree ≤ 3 and the corresponding cofactors.
5.1.
Darboux polynomial cofactor 1 Table 3 . Cubic Darboux polynomials and corresponding cofactors; c 1 , ..., c 6 are constants 5.2. Periodic KM-system. The periodic system (r = 1, s = −1, t = 1) The periodic KM-system is given with the same equations (1) plus a periodicity condition x i = x i+n . In the case n = 3 we obtain:ẋ
5.1.
We give a different type of Lax pair for this system from [1] .
It follows that the functions H
Tr L i are constants of motion. We note that H 1 = 0 , H 2 = x 1 + x 2 + x 3 and H 3 = 1 + x 1 x 2 x 3 . As expected the function H 2 = x 1 + x 2 + x 3 plays the role of the Hamiltonian with respect to the poisson bracket (3) while F = x 1 x 2 x 3 is a Casimir.
It follows that the functions H
In the following 
It follows that the functions H
We list below all linear, quadratic, and cubic Darboux polynomials of the above system which do not have the form (18) , and their corresponding cofactors. Table 9 . Cubic Darboux polynomials and corresponding cofactors
Acknowledgement
We thank the Cyprus Research Promotion Foundation for support through the grant CRPF0506/03.

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