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Liouvillian Integrability of a Modified Michaelis-Menten Equation

In this work we consider the modified Michaelis-Menten equation in biochemistry It models the enzyme kinetics. We contribute to the understanding of its global dynamics, or more precisely, to the topological structure of its orbits by studying the integrability problem. We prove that a = 0, or r = 0, or E = 0 are the unique values of the parameters for which the system is integrable, and in this case we provide an explicit expression for its first integrals.


Introduction
In biochemistry, the Michaelis-Menten equation models the enzime kinetics, that is, the chemical reactions that are catalysed by enzymes. In particular, it describes the rates of irreversible enzymatic reactions by relating reaction rates to the concentrations of the substrate. It was given by Michaelis and Menton in [7].
In this work we consider the Michaelis-Menten equation in the following form (see [7]): x = −a(E − y)x + by,ẏ = a(E − y)x − (b + r)y,ż = ry. (1.1) where x, y, z are the thickness of matters, a, b, r are rates of response, which are nonnegative real numbers and E is the quantity of enzyme, which is a constant. A number of dynamic aspects of system (1.1) have been analysed. In [3] it was discussed the existence of periodic solutions and in [4], the authors give an equation, called the Goldbeter-Koshland equation that gives the steady state solution for a chemical equilibrium modeled by the Michaelis-Menton kinetics.
Here we further contribute to the understanding of the complexity, or more precisely of the topological structure of the dynamics of system (1.1) by studying its integrability.
For the three dimensional system of differential equations the existence of one first integral reduces the complexity of its dynamics and the existence of two first integrals that are functionally independent solves completely the problem (at least theoretically) of determining its phase portraits. In general for a given differential system it is a difficult problem to determine the existence or nonexistence of first integrals. Thus, for proving our main results we shall use the information about invariant algebraic surfaces of this system. This is the basis of the so called Darboux theory of integrability, for more details see Section 2.
We first consider the case in which either a = 0, or r = 0, or E = 0. We start with the following result.
(a) if a = 0, it is integrable with the polynomial first integrals is a first integral of system (1.1). Therefore, setting H = h, h ∈ R we can rewrite system (1.1) as the following system in the plane (z = H − x − y): From now on instead of working with system (1.1) we will work with system (1.2). In particular, we introduce the variables X = x + y,Y = y. We have thaṫ (1. 3) The following theorem is the main result of this paper. In particular, it states the non-existence of Liouvillian first integrals for system (1.1) when arE ∈ R \ {0}. As far as we know the Liouvillian first integral of some multi-parameter family of planar polynomial differential systems has been classified for very few families of systems (see for instance [1,6,8]).
The paper is organised as follows. In Section 2 we introduce some basic definitions and results related to the Darboux theory of integrability that we shall need in order to prove one of our main results. In Section 3 we prove Theorem 1.2.

Preliminary results
During recent years the interest in the study of integrability of differential equations has attracted much attention. Darboux theory of integrability plays a central role in the integrability of the polynomial differential models. It gives a sufficient condition for the integrability inside the family of Darboux functions. More precisely, the significance of this method is that we can compute Darboux first integrals by knowing a sufficient number of algebraic invariant surfaces (the so-called Darboux polynomials) and of the so-called exponential factors. We would like to highlight that it works for real or complex polynomial ordinary differential equations. The study of complex invariant algebraic curves is necessary for obtaining all the real first integrals of a real polynomial differential equation, for more details see [5]. We associate to system (1.1) the following vector field , called the cofactor of h. Note that K has degree at most 1. The polynomial h is called a Darboux polynomial, and we also say that K is the cofactor of the Darboux polynomial h. We note that a Darboux polynomial with a zero cofactor is a polynomial first integral.
The following proposition provides a basic property of the Darboux polynomials and it states that to study Darboux polynomials it is sufficient to study the irreducible ones. Let g, h ∈ C[X ,Y ] be coprime. We say that a non-constant function E = e h/g is an exponential factor of the vector field X given in (2.1) if it satisfies XE = LE, for some polynomial L = L(X ,Y ) ∈ C[X ,Y ], called the cofactor of E and having degree at most 1. Note that this relation is equivalent to For a geometrical and algebraic meaning of the exponential factors see [2].
A first integral G of system (1.1) is called of Darboux type if it is of the form where f 1 ,... , f p are Darboux polynomials, E 1 ,... ,E q are exponential factors and λ j , μ k ∈ C for j = 1,... , p, k = 1,... ,q. For more information on the Darboux theory of integrability see, for instance, [5] and the references therein.
For a proof of the next proposition see [2]. A Liouvillian first integral is a first integral H which is a Liouvillian function, that is, roughly speaking which can be obtained "by quadratures" of elementary functions. For a precise definition see [9]. The study of the Liouvillian first integrals is a classical problem of the integrability theory of the differential equations which goes back to Liouville, see for details again [9].
A non-constant complex function R : C 2 → C is an integrating factor of the polynomial vector field X in (2.1) on U , if one of the following three equivalent conditions holds on U and where P = −rY and Q = −(b + r + aE)Y + aEX − aY X + aY 2 . As usual the divergence of the vector field X is given by To prove the results related with Liouvillian first integrals we use the following result proved in [9].

3) has a Liouvillian first integral if and only if it has an integrating factor of Darboux type.
We also need the following well-known statement.
if and only if the function of Darboux type (2.2) is an integrating factor of the vector field X.

Proof of Theorem 1.2
We separate the proof of Theorem 1.2 into different propositions.
Without loss of generality we can write where each h j = h j (X ,Y ) is a homogeneous polynomial of degree j and we assume that h n = 0.
Computing the terms of degree n + 1 in (3.1) we get Solving this differential equation we obtain h n = h n [X ]. Since h n = 0 is a homogeneous polynomial of degree n ≥ 1, we conclude that h n = α n X n , with α n ∈ C.
Computing the terms of degree n in (3.1) we get Evaluating it on X = Y it yields −rnα n Y n = 0, which implies α n = 0, and this is not possible. This concludes the proof of the proposition. Proof. Let h be an irreducible Darboux polynomial of system (1.1) with nonzero cofactor K, where K = α 0 + α 1 X + α 2 Y , with α i ∈ C for i = 0, 1, 2 not all zero.
It is easy to see by direct computations that if h has degree one then h is different from zero if and only if b = 0 and in this case h = Y − X . Now we assume that h is irreducible and has degree greater or equal to two. Thus we decompose h as a sum of homogeneous polynomials similarly as in (3.2), where n ≥ 2 and h n = 0.
Computing the terms of degree n + 1 in (3.5) we get Solving this linear differential equation we obtain where K n is a function in the variable X . Since h n is a homogeneous polynomial of degree n we must have α 1 = 0 and α 2 = am with m ∈ N. Hence where c n ∈ C \ {0} (since h n = 0). Computing the terms of degree n in (3.5) we get Substituting (3.6) into equation (3.7) and solving it with respect to h n−1 we obtain Since arE = 0, c n = 0 and h n−1 is a homogeneous polynomial of degree n − 1, we get the conditions m = n and α 0 = −aEn. Hence Now we consider two cases.
Substituting (3.8) into equation (3.9) and solving it with respect to h n−2 we obtain Since h n−2 is a homogeneous polynomial of degree n− 2 and aEb = 0 we get a contradiction. Hence this case is not possible.
Case 2: b = 0. In this case h n = c n (Y − X ) n and h n−1 = 0. We considerh to be the restriction of h to Y = X . If we writeh in sum of its homogeneous parts we get thath = ∑ n j=0h j where each h j =h j (X ) is a homogeneous polynomial of degree j andh j is the restriction of h j to Y = X . Since h is irreducible we must haveh = 0. However we have thath n = 0 and thenh = 0, which is not possible. This completes the proof of the proposition. Proof. We consider two different cases.
In this case it follows from Proposition 2.2 that we can write E = e g and g satisfies where α i ∈ C, for i = 0, 1, 2 are not all zero.
We first prove that g is a polynomial of degree two. We proceed by contradiction. Assume that g is polynomial of degree n ≥ 3. We write it as a sum of its homogeneous parts as in equation (3.2) with h j replaced by g j . Since the right-hand side of equation (3.10) has degree at most one, computing the terms of degree n + 1 in equation (3.10) we get Since G is a first integral it must satisfy XG = 0, that is, Hence, since r = 0 we get that if b = 0 then μ 1 = 0 and thus G is a constant. Moreover, since aE = 0 if b = 0 we must have λ 1 = 0 and then also μ 1 = 0 and thus G is also a constant. In both cases we get that G is a constant, in contradiction with the fact that G is a first integral. This completes the proof of statement (d).
Now we prove statement (e). By Theorem 2.1 in order that system (1.3) has a Liouvillian first integral it must have an integrating factor of Darboux type (see (2.2)). From Theorem 2.
with λ 1 , μ 1 ∈ C. Since a = 0 the equality in (3.11) is not possible because the right-hand side is independent of X . This ends the proof of the theorem.