On the Liouville integrability of Edelstein’s reaction system in R 3

We consider Edelstein’s dynamical system of three reversible reactions in R 3 and show that it is not Liouville (hence also not Darboux) integrable. To do so, we characterize its polynomial ﬁrst integrals, Darboux polynomials and exponential factors.

constraints imposed by the network) and when it has multiple such states.
In renowned work by Horn, Jackson and Feinberg, conditions for a unique asymptotically stable steady state have been given in terms of the structure of the network (so-called deficiency theorems) [1,2].More recently, injectivity theorems have flourished, giving sufficient conditions for a network to have a single steady state for all parameter values [3,4,5], or for each set of parameter values individually [6].
Reaction networks might, however, exhibit very complex behaviour such as oscillations and limit cycles [7,8,9], and even chaos [10].Chaos is essentially a dynamical concept, in which two initially close trajectories may be divergent after a finite time [11].Contrary to intuition, chaotic behaviour is not an exclusive property of dissipative or random systems but of conservative and deterministic systems as well.Focusing on the last ones, for non-integrable systems, the irregular behaviour that a system may exhibit is associated with the appearance of chaos, because it seems impossible to predict the long-term evolution of a non-integrable system -at least with the most commonly used perturbation techniques.To the contrary, if the system is integrable then the existence of constants of motion (also called integrals and hence the name) are responsible for the regular evolution of the phase-space trajectories of the system in well-defined regions of the phase space.
We consider Edelstein's system of three reversible chemical reactions [12]: where α 1 , α 2 , β 1 , β 2 , γ 1 , γ 2 are positive reaction rate constants.Under mass-action kinetics the evolution of the species concentrations is described by the following ODE system of degree 2, where x, y, z denote the (non-negative) concentrations of the species A, B, C, respectively.Edelstein designed the system as an example of a system with three steady states, two stable and one unstable, for some choices of reaction rate constants.For other choices there is a single stable steady state.He characterized the region of multiple steady states by computational means.Furthermore, Edelstein suggested that the system's "analytical simplicity" could serve as a potential model system, although the scheme had not been chemically demonstrated [12].Despite its apparent simplicity, many aspects of the system are mathematically hard to analyse.In this paper, we study the Liouville (including Darboux) integrability of the system by characterising its Darboux polynomials and exponential factors.
The Edelstein system is an example of a deficiency one reaction network [1,2].This implies that it has a unique positive and asymptotically stable steady state for each positive value of the conserved quantity H = y + z, for any parameter values fulfilling the condition ( * ) α 1 β 1 γ 1 = α 2 β 2 γ 2 , a 5dimensional set with co-dimension one (the deficiency).Furthermore, using a structural criterion in [13], one can show that the system is persistent (any trajectory starting in the interior of R 3 is bounded away from the boundary, uniformly in times) for all parameter values, which in turn implies that the asymptotically stable steady state is globally stable, assuming condition ( * ) [14].Finally, using [15, Proposition 1], that the trajectories of the system are attracted towards a compact set, again for all parameter values.The use of linear first integrals in reaction network theory is common.Edelstein's system has one such first integral, namely H = y + z, which restricts two coordinates of the trajectories to a compact set given by H.With some exceptions the non-linear first integrals have rarely been considered, see [16,17,18,19] for some general considerations and [20,21,22,7] for specific examples.Non-linear first integrals have been useful for studying the dynamics of a reaction network, see for example [7,18].However, they are generally hard to find, even for polynomial dynamical systems [23,24,25].
Here we show that Edelstein's system in R 3 has no polynomial nor rational first integrals, except for H (and transformations thereof).Additionally, we show that it is neither Darboux nor Liouville integrable.

Main theorems
System (2) has a single linear first integral for any choice of positive rate constants, which is a consequence of the graphical structure of the reaction network (1) [1].Using the fact that the set {(x, y, z) ∈ R 3 | y + z = w} is invariant under the flow generated by (2) for any w ∈ R ≥0 , we can reduce the state space by one dimension.Hence, letting w = y + z, system (2) is transformed into the system where the six rate constants c 1 , . . ., c 6 > 0 are defined as Note that by definition the constants fulfil c 6 > c 5 > c 4 , but this is not important for the arguments that follow.The system (3) cannot be interpreted as a reaction network with mass-action kinetics, because of the term −c 4 y in the equation for ẋ.The other terms, like c 4 w, can be interpreted in terms of reactions, for example, D → D + A in the case of c 4 w, where D is a species with concentration w.
Since w is a constant, we set w = 0 and consider the planar differential system with only five parameters, Note that the condition w = 0 is biochemically uninteresting as x, y, z are non-negative concentrations in this context.
We will first prove a theorem for system (4) and then use this result to prove a theorem for the original system.Let X be the vector field associated system (4), Then the following holds.
Theorem 1. Suppose that c 2 /c 3 ∈ Q + , where Q + denotes the positive rational numbers.Then the following statements hold for system (4).As a consequence of Theorem 1 we can state a theorem for system (3).
Theorem 2. Suppose that c 2 /c 3 ∈ Q + .Then the following statements hold for system (3).(e) It is not Darboux integrable.
(f ) It is not Liouville integrable.
Likewise Theorem 2(a) is true for all parameter values and not only for It is possible to show that the results hold for some rational choices of c 2 /c 3 .However, we do not have a proof in general.

Conclusion
We have shown that the Edelstein system has a polynomial first integral H = y + z, and that, restricted to the surface H = 0, it has no other Liouvillian first integrals.In particular, it is not completely integrable (in the sense of Liouville) and hence, it is neither possible to express the solution as a sequence of integrals, nor is it possible to describe completely the phase portrait (in the sense of the existence of regular foliations by submanifolds within the phase space that are invariant under the flow).
However, more can be said.Since the restricted system has no invariant algebraic curves (that is, Darboux polynomials) we cannot describe in a clear way the behaviour of the flow of the system at infinity, the so-called α-limit and ω-limit sets.

Preliminary results
Consider an n-dimensional polynomial differential system of degree d ∈ N where and the dot denotes derivative with respect to the independent variable t.
A function H(x) is a first integral of system (5) if it is continuous and defined on a full Lebesgue measure subset Ω ⊆ R n , is not locally constant on any positive Lebesgue measure subset of Ω and moreover is constant along each orbit in Ω of system (5).If X is the vector field associated with the system ( 5) and H is C 1 , then we have For an n-dimensional system of differential equations the existence of some first integrals reduces the complexity of its dynamics and the existence of n−1 functionally independent first integrals solves completely the problem (at least theoretically) of determining its phase portrait.In general for a given differential system it is a difficult problem to determine the existence or non-existence of first integrals.
During recent years the interest in the study of the integrability of differential equations has attracted much attention from the mathematical community.Darboux theory of integrability plays a central role in the integrability of the polynomial differential systems since it gives a sufficient condition for the integrability inside a wide family of functions.We highlight that it works for real or complex polynomial differential systems and that the study of complex algebraic solutions is necessary for obtaining all real first integrals of a real polynomial differential system.
A Darboux polynomial of ( 5) is a polynomial which is called the cofactor of f , has degree at most d − 1, where d = max{deg P 1 , . . ., deg P n } is the degree of system ( 5).An invariant algebraic surface is a surface given by f = 0. Note that it is invariant by the dynamics in the sense that if a trajectory starts on the surface it does not leave it.
An exponential factor of ( 5) is a function such that It is widely known that in this case f is a Darboux polynomial of (5) and that X (g) = kg + Lf , where k is the cofactor of f .
An inverse integrating factor of ( 5) is a function V such that We note that, in case V is a polynomial, then it is a Darboux polynomial whose cofactor is the divergence of the system.
The following results, proved in [26], explain how to find Darboux and Liouville first integrals.
Theorem 3. Assume that a polynomial differential system X of degree m defined in C 2 admits p Darboux polynomials f i with cofactors k i , i = 1, . . ., p, and q exponential factors F j = exp(g j /h j ) with cofactors L j , j = 1, . . ., q.
Then, the following statements hold: (a) There exist λ i , µ j ∈ C not all zero such that p i=1 if and only if the function is a first integral of X .Such a function is called a Darboux function.
(b) There exist λ i , µ j ∈ C not all zero such that p i=1 if and only if the function of Darboux type (6) is an inverse integrating factor of X .Here div (X ) stands for the divergence of the system.
To prove the results related with the Liouville first integrals we used the following result proved in [27].
Theorem 4. The polynomial differential system (4) has a Liouville first integral if and only if it has an integrating factor which is a Darboux function.
The previous theorem says that the method of Darboux finds all Liouville first integrals.

Statement (a)
Let H(x, y) be a polynomial first integral of degree m ≥ 1 of system (4).
We write H = m i=0 H i (x, y), with H i being homogeneous polynomials for all i, and split the PDE X (H) = 0 into a system of m + 2 homogeneous ODEs.
The equation of degree m + 1 is using ( 4).The solution when c 2 = c 3 is where f is here an arbitrary function.Again this cannot be a polynomial.
Therefore H m is not a polynomial and hence no such H can exist.Statement (a) follows.
Remark 5. We note that, in the proof of statement (a), we do not need the It holds for all positive values of the constants.
To show that there is a most one other Darboux polynomial we follow the techniques of [28] and [29].To system (4), we first apply the change of variables (x, y) = (1, u)/v that sets the line at infinity in the horizontal axis.
We obtain the cubic system Let Y be the associated vector field.We note that u and v are Darboux polynomial of (8) since u| u and v| v.Note also that we have written the expressions of u and v as polynomials in v.
Let f (x, y) be an irreducible Darboux polynomial of degree m ≥ 1 of system (4) with cofactor k(x, y) (8), where The degree of K is 2. It follows that This PDE can be transformed into an ODE system by writing it as a polynomial equation in the variable v with coefficients depending on u [28].The coefficients of the PDE give rise to the equations of the ODE system and these equations can be solved recursively to obtain the polynomials g i , i = 0, . . ., m, that form g [28].For the monomial v i , i = 0, . . ., m, we extract an equation where i = 0, . . ., m, and g −1 ≡ 0. The key point is that all these ODEs depend only on the variable u.
From equation ( 10) with i = 0 we have up to a non-zero constant which we might take as one because v g (as g is irreducible).We notice that the quotient of the eigenvalues of the Jacobian of ( 8) at the singular point (0, 0) is (c 2 − c 3 )/c 2 , which by assumption is not rational.Hence Theorem 8 of [29] assures that only two analytic curves pass through this point.Since u = 0 and v = 0 are Darboux polynomials of the system (8), g cannot pass through (0, 0) and therefore the exponent of u in g 0 must be equal to zero.That is, As the degree of g is m, we have 0 ≤ n ≤ m.
From equation (10) with i = 1 we obtain where (j + 1)c 2 − jc 3 = 0, because c 2 /c 3 ∈ Q + , and Since g 1 is a polynomial and c 2 /c 3 is not rational, we must have ∆ = 0 to cancel the infinite sum.Hence from this equation we obtain an expression for k 0 .The expression of g 1 becomes From equation (10) with i = 2 we obtain where p 2 is a polynomial of degree 2, Λ is a constant depending on m, n and the coefficients of the system, and 2 F 1 is the hypergeometric function.
Since this function is not a polynomial, we must have Λ = 0.This equation provides a well-defined expression for m: Of course this expression must be a natural number, but this is not important at this moment for the argument.
From equation (10) with i = 3 we obtain three different situations in order to obtain a polynomial expression for g 3 : In the first two cases we have n = m from ( 11) by direct computations.
Moreover further direct computations show that there is a linear solution It remains to deal with the third case.We aim to show that this case does not lead to more Darboux polynomials.From equation (10) with i = 4 we obtain two different situations that provide a polynomial expression for We do not provide the solutions g 3 and g 4 because they are very long and not relevant for the proofs.
Concerning the cases (iv) and (v), we discard the first case because by insertion of (iv) into (iii), we get c 1 < 0, which is not possible.We study the second case, for which we have m = 2n by insertion of (iii) and (v) into (11).
We claim that under the stated hypotheses system (8) has no Darboux polynomials but the axes.If we prove the claim then statement (b) of the theorem follows.
Assume (iii) and (v) hold.In order to prove the claim we first show that deg g i (u) = n for all 0 ≤ i ≤ m.Using the expressions obtained for k 0 , k 1 , k 2 that hold generally, and (iii) and (v), the solution of equation ( 10) can be directly computed; where is a polynomial (this is consistent with the expressions obtained for g 0 , g 1 , g 2 ).Now write C i (u) as a power series in u.Plugging this expression into (10) and reducing, then the first i+1 coefficients (corresponding to the monomials u j , j = 0, . . ., i) of the power series can be obtained from a determined linear system of equations.The rest of the coefficients of the power series is solved from an infinite homogeneous determined linear system of equations, hence they are all zero.This last step assumes c 2 /c 3 ∈ Q + , which holds by assumption.Thus C i is a polynomial of degree i and therefore g i has degree n.
In particular, this implies that g(u, v) = n i=0 g i (u)v i , that is g i ≡ 0 for i = n + 1, . . ., m.This further implies that (10) where m, k 0 , c 1 , c 4 must be substituted by their respective values.From this equation we get where C n is a constant.Since this is a polynomial and c 2 /c 3 ∈ Q + , we must have C n = 0, and hence g n ≡ 0. However, this contradicts the fact that g has degree m.Therefore no new Darboux polynomials are obtained in this case and hence statement (b) of the theorem follows.

Statement (d)
We divide the proof of statement (d) into different partial results.Lemma 6. System (4) has no exponential factors of the form exp(g), with Proof.Suppose that exp(g) is an exponential factor of system (4) with cofactor L and deg g = m ∈ N. It is clear that g satisfies the equation X (g) = L.We can write this equation as a system of homogeneous ODE, g(x, y) = m i=0 g i (x, y), where g i is a homogenous polynomial of degree i in x, y.The equation of degree m + 1 is According to the proof of statement (a) in Section 4.1.1,such polynomial g cannot exist and the lemma follows.Proof.Suppose that exp(g/y n ) is an exponential factor of system (4) with cofactor L. Then Let g = g| y=0 ≡ 0 (by assumption), which is a polynomial in x.Evaluating the above equation on y = 0 we get This equation has solution where C = 0 is a constant.The function g is not a polynomial since it has degree nc 3 /c 2 ∈ Q + by assumption.Hence the lemma follows.An exponential factor with f = y n for some n ∈ N is excluded by Lemma 7.
Assume that (c 2 − c 3 )c 4 + c 3 (c 1 + c 6 ) = 0.In this case, system (4) has exactly two Darboux polynomials with non-zero cofactor, namely F 1 = y and F 2 .Consequently, if it has an exponential factor, then it must be of the form , where y, F 2 g and with n 1 ∈ N ∪ {0} and n 2 ∈ N. Proceeding as in the proof of Lemma 7, we have n 1 = 0.So the exponential factor must be of the form e g/F n 2 , with n ∈ N. Let L be its cofactor.Since F n 2 has cofactor of nK 2 , we have where K 2 = (c 1 − c 2 x) follows from statement (b).We take F 2 = 0, that is, x/c 4 and let g = g| F 2 =0 ≡ 0. Then Solving this equation we obtain where C = 0 is a constant.This expression is not a polynomial since the exponent of x is negative, because n = 0, or since it has degree nc 2 /c 3 ∈ Q.
Hence we get a contradiction and statement (d) follows in this case.
Consequently, if it has an exponential factor, then it must be of the form e g/(y n 1 F n 2 3 ) , where y, F 3 g and with n 1 ∈ N ∪ {0} and n 2 ∈ N. Proceeding as in the proof of Lemma 7, we have n 1 = 0.So the exponential factor has the form e g/F n 3 , with n ∈ N. Let m = deg g ∈ N. We can assume that m < n.
Indeed, if m ≥ n, then there exist polynomials q and r such that g = qF n 3 +r, with deg r < n.Hence e g/F n 3 = e q e r/F n 3 and therefore e q is an exponential factor, in contradiction with Lemma 6.Thus m < n.
From statement (b) we have that nK 3 = −c 2 x is the cofactor of F n 3 .Proceeding as in the previous case, this leads to the equation where L = 0 + 1 x + 2 y is the cofactor of the exponential factor.Let g = m i=0 g i (x, y), where g i is a homogeneous polynomial of degree i, for all not possible to construct rational first integrals nor Darboux first integrals without Darboux polynomials.Statement (f) follows as in the proof of Theorem 1.Hence, we need only prove the statements (a), (b) and (d).We prove them separately.Instead of system (2) we shall consider its equivalent system (3).

Statement (a)
Let H(x, y, w) be an irreducible polynomial first integral of degree m ∈ N of system (3).Since w is also a polynomial first integral of ( 3) and H is irreducible, we can assume that w H. Let H 0 = H| w=0 ≡ 0. Clearly, H 0 (x, y) is a polynomial first integral of system (4).Thus by Theorem 1(a) we have H 0 ≡ 0, which is a contradiction.Therefore statement (a) follows.
Remark 8. We note that, in the proof of statement (a), we do not need the restriction c 2 /c 3 ∈ Q + as the proof only depends on Theorem 1(a).

Statement (b)
Let f (x, y, w) be an irreducible Darboux polynomial of degree m ∈ N We first assume that (c 2 −c we can take c 0 = 1 as it cannot be zero because f is irreducible (and in particular w f ).Indeed, we must have n = m, otherwise there is a factor w in the highest degree terms of f , and this is not possible because w = 0 is not invariant at infinity (we only have the directions xy((c 2 − c 3 )x + c 3 y) = 0, which are fulfilled for singular points).Since f is invariant under the flow of system (3), we have where the terms of system (3) have be reordered and we have used f 0 = y m .
The equation of degree i in w obtained from ( 13) is Equation ( 14) for i = 0 is trivial.Equation ( 14) for i = 1 writes as We can repeat this argument to show that y m−i |f i , for all i < m.That is, f i = y m−i f i , with y f i .Now equation ( 14) for i = m writes as Let g = g| w=0 ≡ 0. If n > 0 then the above equation on w = 0 writes as X ( g) = 0, and hence g is a polynomial first integral of system (4), which is not possible by Theorem 1.
Hence, any exponential factor of system (3) must be of the form exp(g), with g ∈ C[x, y, w].Now let exp(g) be an exponential factor of system (3) with cofactor L and deg g = m ∈ N, that is, X (g) = L.We can write this equation as a system of homogeneous ODEs.Let g m be the homogeneous polynomial of degree m of g, and let g m = g m | w=0 .The equation of degree This equation is identical to equation ( 7) of the proof of Theorem 1(a).
Hence the same conclusions apply.In particular we conclude that it has no polynomial solutions of positive degree.Hence g m ≡ 0, which yields g = w j ḡ with j ∈ N and ḡ ∈ C[x, y, w], w ḡ.Moreover since deg L ≤ 1 because the system is quadratic we must have L = αw, where α ∈ C \ {0}, and then j = 1.We note that L has no constant term because j > 0.
We end the proof by showing that indeed α = 0.After simplifying by w, we have that ḡ satisfies the equation X (ḡ) = α.The arguments in the proof of Theorem 1(a) show that ḡ does not depend on x, y, that is ḡ = ḡ(w) (because ẇ = 0), so X (ḡ) = 0, which means that α = 0, a contradiction.
of this work was done while CW was visiting Universitat Politècnica de Catalunya in Spring 2017.

( a )
It has no polynomial first integrals.(b) It has at most two irreducible Darboux polynomials, all of them of degree one.Indeed, F 1 = y is a Darboux polynomial with cofactor K 1 = −c 6 − c 3 x and either: (b.1) F 2 = (c 1 + c 6 )x − c 4 y is another Darboux polynomial with cofactor

c 3 c 6 2 − c 3 )c 4 + c 3 (c 1 + c 6 ) = c 2 c 4 + c 1 c 3 + c 3 (c 6 − c 4 )(c 2 − c 3 )c 4 + c 3 c 6 = c 2 c 4 + c 3 (c 6 − c 4 )
= 0; or (b.3) there are no more Darboux polynomials, otherwise.(c) It has no rational first integrals.(d) It has no exponential factors.(e) It is not Darboux integrable.(f ) It is not Liouville integrable.We remark that Theorem 1(a) and (b.1) are true for all values of the parameters, and not only for c 2 /c 3 ∈ Q + .The two conditions in (b.1) and (b.2) cannot be fulfilled at the same time.With the added biochemical constraints on the parameters we observe that the quantity in (b.1) (c > 0 and the quantity in (b.2) > 0 are always positive.Hence with these restrictions there is a unique irreducible Darboux polynomial F 1 = y for all parameter values, provided that c 2 /c 3 ∈ Q + .

( a )
The unique irreducible polynomial first integral is H = y +z.Any other polynomial first integral is a polynomial function of H. (b) It has no Darboux polynomials with non-zero cofactor.(c) It has no rational first integrals except rational functions of H.(d) The unique exponential factors are F = e p(H) , where p ∈ C[H].
the first case and g(u, v) = (c 3 (c 4 − c 6 )(c 6 − c 4 u) − c 1 c 4 c 6 v) m in the second case.By transforming back to the coordinates (x, y), we obtain the Darboux polynomials F 2 and F 3 , respectively, as given in the theorem.No more Darboux polynomials are obtained in these two cases.

Lemma 7 .
System (4) has no exponential factors of the form exp(g/y n ), with g ∈ C[x, y], y g and n ∈ N.

and some f 1
such that y f 1 .Plugging the expression of f 1 into the previous equation and simplifying we obtain an ODE with unknown f 1 .Direct computations show that this equation has no polynomial solutions unless k = m − 1.Thus f 1 = y m−1 f 1 .

c 5 f
m−1 + m(c 6 + c 3 x)f m − k 3 f m−1 y = 0, where both fm−1 and f m = 0 are constant.Since c 3 > 0 we must have m = 0. Therefore statement (b) follows in this case.Now assume that (c 2 − c 3 )c 4 + c 3 (c 1 + c 6 ) = 0.In this case y and F 2 are Darboux polynomials of system (4).It follows from Theorem 1(b) thatf 0 (x, y) = y n 1 F n 2 2 with n 1 , n 2 ∈ N ∪ {0}.It can be proved in a similar way as in the previous case that if n 1 > 0 then we have y|f .Thus we have n 1 = 0.Hence we can write f 0 (x, y) = F n 2 with n ∈ N∪{0}.Now the same arguments explained above but restricting to F 2 = 0 instead of restricting to y = 0 lead to F 2 |f .Hence no irreducible Darboux polynomial f can exist and statement (b) follows in this case.The case (c 2 −c 3 )c 4 +c 3 c 6 = 0 follows using the same arguments, replacing F 2 by F 3 .Hence all cases have been considered and statement (b) is proved.4.3.Proof of statement (d)It follows from statements (a) and (b) that if system (3) has an exponential factor with cofactor L, then it must be of the form exp(g/w n ), with g ∈ C[x, y, w], w g and n ∈ N ∪ {0}.Moreover X (g) = Lw n .