Computing necessary integrability conditionsfor planar parametrized homogeneous potentials\ast

Computing necessary integrability conditions
for planar parametrized homogeneous potentials


Let be a rationally parametrized planar homogeneous potential of homogeneity degree . We design an algorithm that computes polynomial necessary conditions on the parameters such that the dynamical system associated to the potential is integrable. These conditions originate from those of the Morales-Ramis-Simó integrability criterion near all Darboux points. The implementation of the algorithm allows to treat applications that were out of reach before, for instance concerning the non-integrability of polynomial potentials up to degree . Another striking application is the first complete proof of the non-integrability of the collinear three body problem.

Computing necessary integrability conditions

for planar parametrized homogeneous potentials

Alin Bostan
INRIA (France)
Thierry Combot
Institut de Mathématiques de Bourgogne UMR CNRS 5584
Univ. de Bourgogne (France)
Mohab Safey El Din
Sorbonne Universities
Univ. Pierre et Marie Curie (Paris 06)
INRIA Paris Rocquencourt, POLSYS Project
Institut Universitaire de France




Categories and Subject Descriptors:
I.1.2 [Computing Methodologies]: Symbolic and Algebraic Manipulations — Algebraic Algorithms

General Terms: Algorithms, Theory.

Keywords: Integrability, potentials, algorithms.

Let us consider the Hamiltonian system


with , called the potential. System (Computing necessary integrability conditions for planar parametrized homogeneous potentials) describes the motion of a particle in the plane submitted to the force field . It always admits the so-called Hamiltonian as a rational first integral. The potential  is called (rationally) integrable if system (Computing necessary integrability conditions for planar parametrized homogeneous potentials) admits another rational first integral , functionally independent on . Intuitively, the integrability of  is equivalent to the fact that (Computing necessary integrability conditions for planar parametrized homogeneous potentials) can be solved in explicit terms.

Integrability is a rare phenomenon and it is in general a difficult task to determine whether a given potential is integrable or not. For homogeneous potentials in , necessary conditions for integrability were given by Morales-Ramis [?] and by Morales-Ramis-Simó [?]. Building on these works, we design in this article an algorithm which takes as input a family of rational homogeneous potentials depending on parameters and which computes a set of constraints on the parameter values that are necessary for the integrability of . These constraints turn out to be of polynomial nature in .

There are several difficulties in this parameterized setting. The first one is that the integrability constraints provided by the Morales-Ramis theory —on which our whole approach relies—, are expressed in terms of quantities (eigenvalues of Hessian matrices at Darboux points, see Section Computing necessary integrability conditions for planar parametrized homogeneous potentials) which are not easily accessible. We circumvent this basic difficulty by using an equation that relates the eigenvalues, but this brings a new technical complication since the equation is of Diophantine type. A third difficulty is that the number of Darboux points itself may depend on the parameters, leading to singular cases.

We follow a classical approach, inspired mostly by ideas in [?]. Our contribution to the topic is effective and algorithmic, as we provide a complete, proven and implemented algorithm for the problem of computing necessary integrability conditions for planar parametrized homogeneous potentials, with precise output specifications. Our algorithm uses classical tools in computer algebra, such as polynomial ideal elimination based on Gröbner bases techniques. An important feature is the use of (complex) polar coordinates to represent homogeneous potentials by univariate rational functions with parameters . This change of representation considerably simplifies the computations and the proofs. For instance, in polar representation, singular cases are those with non-generic multiplicity of the roots/poles of . They are treated by our algorithm, which builds a tree containing each possible singular case. This approach is related with comprehensive Gröbner bases [?], which are avoided here thanks to some a priori knowledge about singular cases.

In summary, our strategy for computing necessary integrability conditions for  consists in 4 steps: (i) rewrite in polar coordinates; (ii) set up a Diophantine equation whose solutions belong to the so-called Morales-Ramis table (that contains all possible eigenvalues of the Hessian of at Darboux points of ); (iii) solve this Diophantine equation; (iv) rewrite the condition of having prescribed eigenvalues at Darboux points as polynomial conditions on .

Some prior works used a similar strategy, but it was unclear which cases were possible to tackle, in particular for singular ones. The approach was not fully automatized and this explains that results were only available for special families of potentials, for instance polynomials of small degree (3 or 4) [?, ?, ?, ?], as the number of singular cases grows very fast (already for polynomials of degree ). By contrast, our treatment is unified and fully automated, and it allows not only to retrieve (and sometimes correct) known results, but more importantly, to treat potentials of degrees previously unreached (up to 9). By applying our algorithm to polynomial potentials, we found three new cases admissible for integrability at degree  (but still not proved to be integrable), and various new families for higher degrees. An even more striking application of our algorithm is the first complete proof of the non-integrability of the collinear three body problem, on which only partial results were known [?, ?, ?]. The direct approach that consists in searching first integrals [?, ?] is complementary to our (non-)integrability analysis, as our algorithm helps either proving that the lists in [?, ?] are complete, or finding new unknown cases.

Warning: We will assume throughout the article that the homogeneity degree is different from and . (This is because the Morales-Ramis theory is much less powerful when .)

Convention of notation: to avoid confusion, we will use bold letters for variables/parameters, and italic letters for parameter values.

There exist strong integrability constraints (see Theorem 1 below). They require to deal with Darboux points, whose definition we now recall.

Definition 1

Let be a homogeneous rational function of homogeneity degree . A point is called a (proper) Darboux point of if it satisfies the equations


Note that, by homogeneity, we could have chosen an arbitrary normalization non-zero constant on the right-hand side of (1). In the literature, this normalization constant is frequently chosen equal to  [?]. However, our choice is deliberate, see the remark after Theorem 1.

The following result (which is an application of a more general criterion due to Morales and Ramis [?]) provides necessary conditions for integrability under the form of constraints on eigenvalues of Hessian matrices at each Darboux point. It is the basic ingredient for numerous non-integrability proofs [?, ?, ?, ?, ?, ?, ?, ?, ?]. Roughly, its main idea is as follows. A Darboux point leads to a straight line orbit of the dynamical system (Computing necessary integrability conditions for planar parametrized homogeneous potentials) associated to , around which the system (Computing necessary integrability conditions for planar parametrized homogeneous potentials) can be linearized. If the whole system is integrable, then the linearized system, which in our case corresponds to a hypergeometric equation, is also integrable. Thus the integrability table of Theorem 1 below is reminiscent of Kimura’s classification [?] of solvable hypergeometric equations.

Theorem 1

(Morales-Ramis [?]) Let be a homogeneous rational function of homogeneity degree , and let be a Darboux point of . If the potential is integrable, then for any eigenvalue  of the Hessian matrix of at , the pair belongs to the following table, for some .

This table will be called throughout the article the Morales-Ramis table. For a fixed homogeneity degree , we will denote by the infinite set of allowed eigenvalues in the table, corresponding to .

Note two differences with the classical statement of the Morales-Ramis theorem. First, due to our choice of the normalization constant in Definition 1 ( instead of ), the eigenvalues displayed in the previous table are times larger than those of [?, Theorem 3]. Our choice is motivated by the fact that it simplifies the computations, and it has the nice and useful property that the eigenvalue sets in the table are lower bounded. Second, both the original proof [?] and the statement of the Morales-Ramis theorem [?, Theorem 1.2], require the additional assumption that the Hessian matrix of at is diagonalizable; but in fact, [?, Theorem 1.3(1)] shows that this hypothesis is not necessary.

We now illustrate the basic notion of Darboux points and the use of Theorem 1 on a toy parametrized example. The example is simple enough so that the eigenvalues are accessible by a direct computation.

  • Consider the homogeneous potential

    The homogeneity degree is and the Darboux point equation missing(1) is

    For parameter values such that , its solutions read

    The Hessian matrices at these points are

    The eigenvalues of the first matrix are . The (a priori unexpected) fact that none of them depend on the parameter values comes from a relation on eigenvalues at Darboux points that will be proved later (Theorem 3). Now, Theorem 1 tells us that , the set of allowed eigenvalues for homogeneity degree , is the set of the rational numbers of the form

    where . The eigenvalue is allowed (by choosing in the first sequence), but the eigenvalue is not. This can be seen by simply solving for integers six quadratic equations. Thus, by Theorem 1, the potential is not integrable when , and a necessary condition for integrability is .

We will use complex polar coordinates in order to represent a given rational homogeneous potential in a simpler way, by a pair , where is a univariate rational function, and is an integer. In this new representation, various quantities attached to , such as Darboux points and eigenvalues of the Hessian of , are much easier to express, including a useful relation (11) on these eigenvalues. This representation has already been used for non-integrability proofs [?, ?, ?]. This section provides an overview on some results on polar coordinates with useful properties needed to prove our algorithm (see Theorem 2 below).

In the rest of the article, we will use the notation and for the following subdomains of :

and for the map defined by .

Proposition 1

The map is differentiable on , and its image is a double covering of (i.e., each fiber contains exactly two points).


The functions and are differentiable on , and thus is differentiable on . The relation implies that the image of is contained in . Let us compute the inverse of . If and with , then and The first relation determines up to a sign; since there are always exactly two possible choices . After this sign choice, is uniquely determined, thus is determined up to translation by . Since , then is uniquely determined. ∎

Proposition 2

Any homogeneous potential can be written in complex polar coordinates


where is the homogeneity degree of , and is a rational function in having the same parity as .

Moreover, if , then belongs to .


Let be the rational function in defined by

Then is equal to , and homogeneity of allows to conclude that for and we have

Using again that is -homogeneous, we obtain that

and thus has the same parity as . The last assertion is obvious by definition of . ∎

Proposition 2 shows that the homogeneous rational potential is represented in polar coordinates by a pair , where is a univariate rational function, and is an integer. We now write the equation of a Darboux point  of a potential and the eigenvalues of the corresponding Hessian matrix in polar coordinates.

Proposition 3

Let be a homogeneous potential with polar representation , and let be a Darboux point for . Then for we have


Moreover, and are eigenvectors of the Hessian matrix , with respective eigenvalues


We start from the relations

From there we deduce, by differentiation, the equalities

These equalities imply that the derivatives of write


Combining these last two equations yields


(The first one is Euler’s relation for -homogeneous functions.) Evaluating equalities (8) at the Darboux point , and using the Darboux point equation (1), yields the proof of (4).

Let us now prove the last assertion of the proposition. By differentiating the first equality in (8) with respect to  and , by evaluating at , and by using (4), we obtain . Thus is an eigenvector of , with corresponding eigenvalue .

Similarly, differentiating the second equality in (8) and specializing the result at yields , where denotes the vector . This concludes the proof. ∎

Proposition 3 motivates the following definition of Darboux points in polar representation, and of associated eigenvalues.

Definition 2

Let be the polar representation of a homogeneous potential A complex number is called a Darboux point of if and . A Darboux point is said to be multiple if is a multiple root of ; else it is said to be simple.

If is a Darboux point for , we will call associated eigenvalues the values and .

The map naturally sends Darboux points in polar representation to Darboux points in Cartesian coordinates in , also carrying the definition of associated eigenvalues.

We now prove the main result of this subsection; it gives a necessary condition for integrability of a homogeneous potential in terms of its polar representation. We recall that is the set of allowed values in the Morales-Ramis table for degree .

Theorem 2

Let be a homogeneous potential with polar representation and let be the following set


Let denote the union of the sets taken over all Darboux points of . Then


Moreover, if is integrable, then .


We first prove equality (10). Proposition 3 readily yields the inclusion Indeed, if is in , then there exists a Darboux point of such that . Then letting , Proposition 3 implies that satisfies , , and . Therefore, is equal to , and thus it belongs to .

Conversely, let be in . There exists a such that , and . Write this as for some with , and write as with . Then, Equations (6) and (7) imply that is a Darboux point of in . By Proposition 3, belongs to . Equality (10) is now proven.

To prove the last assertion, assume that is integrable. Then Theorem 1 shows that . Since the eigenvalue belongs to the Morales-Ramis table (first sequence with ), we also have . The desired inclusion is then a consequence of equality (10). ∎

There are infinitely many allowed eigenvalues for integrability in the Morales-Ramis table. We now prove an interesting equation that relates the eigenvalues of the Hessian matrix of at Darboux points. This will allow us to bound the possible eigenvalues allowed for integrability.

Definition 3

For a rational function , we denote by and two integers such that and , with . These integers will be called asymptotic exponents of (at zero and infinity).

Theorem 3

Let be a homogeneous potential with polar representation , let be the set defined in (9) (Theorem 2), counting multiplicities, and be the asymptotic exponents of . If and if  has only simple Darboux points, then:


This result is a generalization of [?, Theorem 2.3] for polynomial homogeneous potentials, and was already proved under an equivalent form in [?, Theorem 1.7]. Still, in [?], the polar representation is not used, leading to a more complicated description of the set , a less readable relation (11) (but not harder to compute in practice) and a more complicated proof. This is why we display here a simple self-contained proof of Theorem 3.


Consider the rational function . We study its poles. Since with , the origin is a pole of . Moreover, the power series expansion of at gives which shows that is a pole of order of , with residue . Similarly, is a pole of order  of with residue .

Let now be a finite pole of . It is either a pole of , or a root of . Any nonzero pole of of order is a pole of of order , and thus it is not a pole of . Thus, is necessarily a root of . By the assumption that all Darboux points of are simple, is not a root of . Therefore, is a pole of of order , and the series expansion shows that the residue of at is . Recognizing this expression as for some , and using Cauchy’s residue formula, proves Equation (11). ∎

Theorem 3 contains two assumptions, that and that Darboux points of are simple. The first assumption does not always hold, and then the possible eigenvalues could be unbounded, as proven for instance by a family of potentials in [?], for which stronger integrability conditions were needed (there is a similar difficulty in [?]). The second hypothesis (simple Darboux points) is not always satisfied, but [?, Theorem 1] provides a classification of integrable potentials with a multiple Darboux point: they are invariant by rotation, i.e. with constant. This motivates the following definition.

Definition 4

We say that a homogeneous potential with polar representation has property if it satisfies one of the following conditions:

  1. All Darboux points of are simple and the associated eigenvalues belong to the Morales-Ramis table
    (By Theorem 2, this condition is equivalent to .)

  2. is finite and nonzero either at the origin, or at infinity (i.e., ).

Remark that condition (1) includes the case (since then ), and that condition (2) includes the case constant nonzero.

In the case of odd homogeneity degree , the function is odd due to Proposition 2. Thus the asymptotic exponents are odd, and so condition (2) of cannot occur. Therefore, for odd homogeneity degrees, is equivalent to condition (1), which matches exactly the integrability conditions given by [?, Theorem 1] and Theorem 2.

Equation (11) in Theorem 3 provides a constraint on the possible eigenvalues for a homogeneous potential . We are thus naturally led to study the equation


where is a rational number, the homogeneity degree of , and an integer related to the number of Darboux points of .

Assume that is integrable and the assumption of Theorem 3 are satisfied. With our notation, Theorem 2 states that , where is the set of allowed values in the Morales-Ramis table for degree . Since the relation (11) holds, the aim is to solve equation (12) for unknowns in . We will prove that there are only finitely many solutions of this type.

Proposition 4

For any solution of the equation (12) the following holds


In the case , at least one term in the sum (12) should be negative, and thus . Let us now look at the case . Assume that we have for all . Then and thus , which is a contradiction with (12). This proves the proposition. ∎

Let us now remark that all entries of the Morales-Ramis table are bounded below by (and this minimum is reached for or ). Starting from this observation, we design a recursive algorithm that finds all the solutions of equation (12) that belong to the Morales-Ramis table.

Input: The parameters of the equation (12).
Output: The set of all solutions in of (12), up to permutation.

  1. If and , then return . If and , then return if it belongs to , else . If , beginning by , compute the elements of , up to the bound of Proposition 4. This yields a set .

  2. For each entry of , recursively run the algorithm on the input , with output .

  3. Return the set of solutions .

Due to Proposition 4, the equation (12) has finitely many solutions in (this was already proved in [?, Lemma B.1]), and algorithm MoralesRamisDiophantineSolve always terminates. In practice, this algorithm is very costly. The case with the constraint leads to the equation analysed in [?], for which an optimal bound on is found. This bound is doubly exponential in  (which in our problem is the number of Darboux points). It is natural to conjecture that a similar doubly exponential bound holds in our case.

Let be parameters and be a parametrized homogeneous potential. In the sequel, we assume that is given in canonical form, i.e. the coefficients of its numerator and denominator lie in .

Our goal is to compute a subset in the set of parameter values  such that is a necessary condition for the integrability of .

In Section Computing necessary integrability conditions for planar parametrized homogeneous potentials, we have defined the polar representation of a homogeneous potential with coefficients in . We can do the same in the context of parametrized homogeneous potential by defining the function as in the proof of Proposition 2. With this definition, the following lemma is an immediate consequence of Proposition 2.

Lemma 1

Let be the complementary of the common solutions of the coefficients of the denominator of . For all , is the polar representation of .

This allows us to define the following set. We let be the set of values such that and has property (Definition 4).

Corollary 1

Let be a parametrized homogeneous potential, and let be its polar representation, . If is integrable, then .


Assume that is integrable. Then thanks to [?, Theorem 1], if has a multiple Darboux point, then is constant and thus . If has only simple Darboux points, Theorem 2 implies that eigenvalues at Darboux points are all in . Thus condition (1) of is satisfied, and thus . ∎

Our main algorithm IntegrabilityConditions in Section Computing necessary integrability conditions for planar parametrized homogeneous potentials will take as input a parametrized homogeneous potential and will compute polynomial constraints in that define the Zariski closure of .

It uses a subroutine that takes as input special parametrized polar representations and computes polynomial constraints that define the intersection of with the subset of the parameter space over which the valuation and number of roots/poles of counted with multiplicities is constant.

Definition 5

We say that is a model function if either identically, or there exist with finitely many non zero ’s, such that


with . We then denote this function by ; we will write for the number of parameters of .

In the following, when there will be no ambiguity on (which will be mostly the case), they will be omitted in the subscripts.

Definition 6

Assume . We let be the subset of the parameter space defined by where

where denotes the resultant of two polynomials with respect to .

Remark that for a , the roots of the ’s are all simple and non zero. Moreover, the ’s do not have any common root.

Definition 7

Assume . Let be a finite subset of . We define the polynomials in


where denotes the numerator of .

The rest of this section is devoted to the design of an algorithm called IntegrabilityConditionsModelFamily, that takes as input a model family and an integer and returns a set of polynomial equations and inequalities in that define the intersection of and .

We are now ready to describe our algorithm.

Input: A model family and an integer .
Output: A pair such that is a list of lists of polynomials and is defined by the union of the zero-sets of the polynomials in and at which for .

  1. If then return .

  2. Compute the polynomial .

  3. If then return .

  4. Else

    1. Compute the coefficients of the relation with and .

    2. Solve this equation using algorithm DiophantineSolve; let be its output.

    3. For each solution in , build the polynomial .

    4. Compute the remainder for the Euclidean division of by in and let be the sequence of polynomials for .

    5. Let be the concatenation of all for ; return .

Before proving the correctness of our algorithm, we will first prove the following two lemmas.

Lemma 2

Assume that , and let . The set of Darboux points of is equal to the set of roots of . Moreover, if is a simple Darboux point of , then is a simple root of .


Let us first prove that any root of is a Darboux point of . Let be a root of . We need to prove the following

Consider the logarithmic derivative of

Taking the numerator of this expression, we obtain


Evaluating this expression at gives

This quantity is non zero as (due to Definition 6). Thus . Let us now prove that any non zero root and pole of is not a root of . Let be a root or a pole of . Then cancels one and only one of the factors (let’s say ) of (because for , the ’s have no common root due to Definition 6). Now evaluating the expression (15) at , we obtain

This quantity is non zero because (in Definition 6, the ’s have only simple roots). Thus is not a root of , and therefore is not a root nor a pole of . To conclude, we have


The function is well defined at (i.e. has a finite value), the ’s do not vanish at , and thus .

Let us now prove the reverse. If is a Darboux point of , then is finite and non zero, and . Using equality (16) at , we obtain .

Finally, let us look at multiplicity. If is a simple Darboux point, then . So we differentiate relation (16) in and evaluate it at . On the right-hand side, all terms vanish except . As is a Darboux point, it is neither a pole of , nor a root of any , and thus . Therefore, is a simple root of . ∎

Lemma 3

Assume that , and . Let be a finite set with and let . The polynomial divides if and only if and all Darboux points of are simple.


Let us first assume that divides . Thus all roots of are roots of . By Lemma 2, the set of roots of is the set of Darboux points of . Let be a Darboux point of . As , we have and thus at least one factor of the product defining (Eq. 14) is zero. So there exists such that