domain: info math

text:
Delphine Boucher
email: delphine.boucher@univ-rennes1.fr

Jacques-Arthur Weil
email: jacques-arthur.weil@unilim.fr

Linear Differential Equations, Differential Galois Groups, First Integrals of Differential Systems

C

and give a criterion to decide between either case.

Introduction

Linear differential equations are ubiquitous in mathematics and their applications. The Galois theory of linear differential equations aims at describing properties of solutions of linear differential equations by studying the algebraic relations between solutions (and their derivatives). This algebraic object also has some analytic interpretations; in a sense, one may think of the differential Galois group as mesuring "what algebra sees of the dynamics" (Malgrange). Thanks to its algebraic nature, the differential Galois group is an object many of whose properties can be established/detected by symbolic computation.

Among the many applications of the algebraic theory of linear differential equations, we will highlight two families.

A first one stems from the fact that solutions of linear differential operators ("D-finite functions") have good "closure" properties: their sums, products, derivatives, etc are also D-finite. An extensive illustration of this aspect will be found in the lectures of J.-M. Maillard where applications to statistical mechanics are described.

A second family of applications is the study of non-linear differential systems. We are given a (non-linear) differential system and we assume that we know a solution curve; linearizing along this solution curve yields a first order linear differential system, the "variational system". Several properties of the original system, in particular existence of first integrals in some "good" class of functions, induce similar properties on the variational system; the algebraic theory of linear differential equations then yields methods and algorithms to detect such properties This talk is split into two lectures. In the first part, we will focus on differential Galois groups and describe the basic foundations of the algorithms; this is adapted from notes for lectures by the second author in a CIMPA school in 2001. The second part will be devoted to detection of first integrals of non-linear differential systems by the variational approach; we will particularly emphasize on hamiltonian systems. These notes are introductory material. Although proofs will be supplied, I'll refer to the original papers for technical aspects or up-to-date algorithms not covered here.

Part I

Galois Theory of Linear Differential Equations 1 Differential Galois Theory

Differential Galois theory is, as in the clasical case, a tool to study the algebraic relations among solutions of linear differential equations; as we will see in the next lecture, it can also be used to study dynamical properties on solutions of some non-linear systems. A general reference for differential Galois theory is the fundamental book [START_REF] Van Der Put | Galois theory of linear differential equations[END_REF].

Differential Equations and Differential Systems

We consider a differential field k with a derivation that we note or ∂ if the context is not clear. For the rest of these notes, we assume that the field of constants C of k is algebraically closed (i.e any polynomial over C has all its roots in C) and of characteristic zero. Typical examples are Q or C.

Differential Operators

Let k denote a field (here k = C(x)) equipped with a derivation ∂ (naturally ∂ = d dx ). One says that k is a differential field. We define the ring of differential operators as the set k[∂] with the (non-commutative) multiplication defined by ∂.a = a.∂ + ∂(a) for a ∈ k. This multiplication corresponds to the composition of differential operators (compute the derivative of ay for a in k and an unknown function y). Let L be a linear differential operator :

L = ∂ n + • • • + a 1 ∂ + a 0 , a i ∈ k.
K is a Picard-Vessiot extension associated to the homogeneous linear differential equation L(y) = 0 if K is generated over k by y 1 , . . . , y n and their derivatives where y 1 , . . . , y n are linearly independant solutions of L(y) = 0 and if the constant fields of k and K are equal.

Equivalent differential systems and differential modules

A linear differential equation is an equation of the form L(y) = y (n) + a n-1 y (n-1) + . . . + a 0 y = 0 with the a i being functions in k (not constants). Solving this equation is equivalent to solving the companion system

      y 1 y 2 • • • y n-1 y n       =         0 1 0 • • • 0 0 0 1 0 . . . . . . . . . . . . 0 0 • • • • • • 0 1 a 0 a 1 • • • a n-2 a n-1               y 1 y 2 • • • y n-1 y n      
Conversely, suppose that (y 1 , y 2 , • • • , , y n ) satisfy a homogeneous first order linear differential system Y = AY of size n. To find an equation associated with Y = AY, we would like to find a system in companion form whose solutions are equivalent (or "isomorphic") to the ones of the original system (precisely: obtained with a change of variables Y = PZ with P an invertible matrix with coefficients in k). This is done by the following cyclic vector process (see e.g [ [START_REF] Barkatou | An algorithm for computing a companion block diagonal form for a system of linear differential equations[END_REF][START_REF] Van Der Put | Galois theory of linear differential equations[END_REF] for references and other methods). Consider Λ ∈ k n and let z 1 = ΛY = λ 1 y 1 + • • • + λ n y n . We compute z 2 = z 1 , . . . , z n+1 = z (n)

1 by using the relation Y = AY. We obtain n + 1 linear expressions in the n variables y i and so they are linearly dependant: this provides a linear differential equation L(z 1 ) = 0 for z 1 . Letting Z = (z 1 , • • • , z n ) t , we now have a relation Z = PY and Z = BY; If the matrix P is invertible, Λ is called a cyclic vector for the system and Z = (BP -1 )Z is a companion system, equivalent to the first one. It can be shown that the cyclic vectors form a Zariski open set, and almost all choices of Λ will fit [START_REF] Van Der Put | Galois theory of linear differential equations[END_REF].

Thus, in the following, everything that is stated for first order systems is valid for n-th order equations and vice-versa.

A useful tool to describe an equivalence class of differential systems (intrinsically) is the notion of differential module: Definition 1.1. A differential module over (k, ∂) is finite-dimensional k-vector space M k n with an operator D satisfying a Leibniz rule:

∀ f ∈ k, v ∈ M : D( f .v) = ∂( f ).v + f .D(v)
Differential modules are sometimes also called "modules with a connection" in the litterature. To a differential system Y = AY, one may cannonically associate a differential module with a basis e = (e 1 , . . . , e n ) and the operator D acting by D(e i ) = ∑ n j=1 -a j,i e j . This way, a solution Y = y 1 e 1 + . . . + y n e n of Y = AY is caracterized by the relation D(Y) = 0 (check this) so that we may write D = -A. It can be checked that two differential systems are equivalent if and only if they are associated to the same differential module (in fact, the change of variable that sends a system to the other can be viewed as a change of basis in the differential module). The cyclic vector method above amounts to finding an element v such that v, D(v), . . . , D ( n -1)(v) are a basis of the differential module M (hence the term "cyclic").

First integrals of linear differential systems

Let U denote a fundamental matrix of solutions of Y = AY (i.e U is invertible and its collumns are solutions of Y = AY). A linear first integral of the system is a linear function R such that R(Y) is a constant whenever Y satisfies Y = AY. Let Y i denote the columns of U and R i denote the rows of U -1 . As U -1 U = 1, we see that R i .Y j = δ i, j (where δ i, j is the kronecker symbol, δ i,i = 1 and δ i, j = 0 when i = j). It follows that the rows R i define linear first integrals of the system. We see that the columns of t U -1 are coefficients of linear first integrals; they are also solutions of the adjoint system Z = -t AZ. We see that solutions of the adjoint (or dual) system yield first integrals of the system. Similarly, one can show [START_REF] Weil | First integrals and Darboux polynomials of homogeneous linear differential systems, Applied algebra, algebraic algorithms and error-correcting codes[END_REF][START_REF] Juan | Meromorphic nonintegrability of Hamiltonian systems[END_REF] that polynomial first integrals will be homogeneous, and obtained from rational solutions of symmetric powers of the dual system. This simple fact is useful in the Morales-Ramis theory of non-integrability of Hamiltonian systems.

The Differential Galois Group

We will proceed as in classical Galois theory: first, we construct a field generated by all the solutions (and their derivatives)

Definition 1.2. A differential field extension K ⊃ k is said to be a Picard-Vessiot extension of k (for L(y) = 0) if 1. K = k(y 1 , y 1 , . . . , y ( j) i , . . . , y (n-1) n ),
where the y i are a basis of solutions of L(y) = 0 (i.e K is the differential field generated 1 by the solutions of L.

K and k have the same field of constants.

As the constant field of k is algebraically closed of characteristic 0, one can show that Picard-Vessiot extensions exist and are unique up to differential isomorphism [START_REF] Van Der Put | Galois theory of linear differential equations[END_REF], the original proof goes back to Kolchin in 1948 [START_REF] Kolchin | Selected works of Ellis Kolchin with commentary[END_REF]). In the sequel, the term "solution" will always denote a solution in the Picard-Vessiot extension K.

Example 1.3. Let k = C(x) and consider the equation L(y) = y -1 3x y. We know that the solution is x 1 3 but let's construct the Picard-Vessiot extension (hence the solution) like we did in the the preliminary lecture on Galois groups. We consider the ring k[y] with the derivation D = d dx + 1 3x d dy , where y is an indeterminate. In this construction, y satisfies D(y) -1 3x y = 0. However, it is easily checked that D( y 3

x ) = 0 so this ring contains a new constant. Take the ideal I = (y 3x); it is prime (in fact, it is maximal) and stable under the derivation. We now let K = k[y]/I. This is now a differential field and it now has no new constant (the "new constant" is included in the relation defined by I). This is now a Picard-Vessiot extension. Note the similarity with the second construction of the splitting field in the preliminary Galois lecture.

In fact, the existence of Picard-Vessiot extension can be achieved through this construction (see [START_REF] Van Der Put | Galois theory of linear differential equations[END_REF][START_REF] Magid | Lectures on differential Galois theory[END_REF]). Consider the ring R := k[X 1,1 , . . . , X n,n , W] where the X i, j are indeterminates and

W. det    X 1,1 . . . X 1,n . . . . . . X n,1 . . . X n,n    = 1.
Extend the derivation on k to a derivation D on R by letting D(X i, j ) = X i, j+1 for 1 ≤ j < n and D(X i,n ) = -sum n-1 l=0 a l X i,l+1 . This way, we have formally realized that L(X i,1 ) = 0. Let J denote a differential ideal in R (i.e D(J) = J) which is maximal among differential ideals. It can be shown that then J is a prime ideal. So R/J has no zero-divisor and we may let K := Frac(R/J). Now K has no new constant: if P Q was a new constant, then (P -Q) would be a differential ideal in R/J, contradicting the maximality of J. It follows that this construction yields a Picard-Vessiot extension K. The ideal J is called the ideal of relations among solutions. Definition 1.4. We call a differential k-automorphism of K an automorphism g of K which leaves k fixed et which commutes with the dérivation,i.e:

2. ∀y ∈ k, g(y) = y The differential Galois group G = Gal ∂ (K/k) of a differential extension K ⊃ k is the group of differential k-automorphisms of K.
The differential Galois group G of L(y) = 0 is defined as the differential Galois group of K/k, where K is a Picard-Vessiot extension of k for L.

Consider the n-dimensional C-vector space V of solutions of L(y) = 0 in K, generated by the y i over C. Let g ∈ G and let y denote a solution of L(y) = 0. Then, g(y) is also a solution. Indeed, as a i ∈ k, we have g(a i ) = a i and :

L(g(y)) = g(y) (n) + a n-1 g(y) (n-1) + . . . + a 0 g(y) = g(y (n) ) + g(a n-1 )g(y (n-1) ) + g(a 0 )g(y) = g(L(y)) = g(0) = 0.
As any solution is a linear combination of the y i , we deduce that there exists constants c i, j such that g(y j ) = ∑ i c i, j y i . As an automorphism is fully determined by its action on the generators y i of K, this gives us a faithful representation of G as a subgroup of the group GL(n, C) of invertible n × n matrices. In fact, one can show that G is a linear algebraic group, i.e the entries c i, j of the matrices are defined as solutions of a set of algebraic equations; the reason for this is the fact (intuitively clear, but still deep) that G can also be viewed as the set of such automorphisms that preserve all the relations among the solutions. Recall the construction of the Picard-Vessiot extension; G can be viewed as the set of matrices with a right action on the X i, j that preserve the ideal of relations J, and this is what makes it a linear algebraic group. In what follows, we will identify G to this representation as a group of matrices acting on solutions.

A little bit on linear algebraic groups Before proceeding with Galois theory, a quick incursion into linear algebraic groups. Recall that an affine algebraic variety V over C is defined as the set of solution of some polynomial equations (e.g straight lines, a circle, conics,etc..). The ideal I(V) associated to the variety is the set of polynomials that vanish at every point of the variety. Conversely, to any ideal we associate the variety of points that annul all polynomials of the ideal. The variety is called irreducible if I(V) is prime. As every ideal is the intersection of a finite number of prime ideals, we see that any affine variety is the union of a finite number of irreducible varieties.

A linear algebraic group G is a group of n × n matrices whose entries form an algebraic variety of C n 2 +1 . The reason for adding one dimension is to include the condition that the determinant is not zero: if we add one variable, the condition det(g) = 0 is rephrased as u. det(g) -1 = 0 which is now a polynomial equation. This way, we see that GL(n, C) is a linear algebraic group.

If G is a linear algebraic group and G is irreducible as a variety, we will say that it is connected (for the Zariski topology). If G is not connected, then G is the finite union of irreducible ("connected") varieties. Among those, the one that contains the identity element is called the connected component of the identity in G, and denoted by G

• . One can show that G • is a normal subgroup of finite index in G.
A linear algebraic group is said to virtually have a property if G • has that property; for example, G is called virtually abelian if G • is abelian (as a group). The dimension of G is defined as the transcendence degree of G • over C. 1. GL(n, C) and SL(n, C) (defined by det(g) = 1).

2. The group of upper triangular matrices T (defined by T i, j = 0 for j < i).

3. Let I n denote the identity matrix of size n and the standard symplectic matrix

J = 0 I n -I n 0
The set of matrices M that satisfy t M.J.M = J (this relation induces a finite set of polynomial relations on the entries of M) is called the Symplectic group Sp(2n, C) and will be central in the applications to symplectic mechanics.

Any finite group of matrices (check this!)

To measure properties of the connected component of the identity G • , one uses its Lie algebra. The Lie algebra of a linear algebraic group is the tangent space at the identity (this makes sense: the group is an algebraic variety, so there is a natural tangent space). This is simply computed with the epsilon-trick (see [START_REF]Galois theory of differential equations, algebraic groups and Lie algebras[END_REF][START_REF] Van Der Put | Galois theory of linear differential equations[END_REF]) as follows. Let denote an object satisfying 2 = 0 (think of as being ideally small). A matrix M is in the tangent space at the identity if and only if Id + M satisfies the equations of the group (or "is a C[ ]-point of the variety"). This defines the Lie algebra Lie(G). The dimension of G as a variety equals the dimension of Lie(G) as a vector space. If you compute the Lie algebra for the groups in the above examples (easy computation), you will find that sl(n, C) := Lie(SL(n, C)) is the set of matrices with zero trace, that the Lie algebra of T is T itself, that sp(2n, C) := Lie(Sp(2n, C)) is the set of matrices M satisfying t MJ + J M = 0 (or M = JS, with S any symmetrical matrix), and that the Lie algebra of a finite group is equal to 0.

Some Essential Properties

As in classical Galois theory, there is a Galois correspondence between algebraic subgroups of G and differential subfields of K. We will admit here a weak version of this correspondence which will be enough for our purposes (see e.g [START_REF] Van Der Put | Galois theory of linear differential equations[END_REF] for a full proof).

Theorem 1.6 (Galois normality). Let K denote a Picard-Vessiot extension of k, let G be its differential Galois group, and let z ∈ K. Then:

z ∈ k ⇐⇒ ∀g ∈ G, g(z) = z.
A first application of this concerns unimodularity of the Galois group: recall that a matrix is called unimodular if its determinant is equal to 1, and that the group of unimodular matrices is denoted by SL(n, C). Define the Wronskian matrix W = (y ( j-1) i ) i, j=1..n and the Wronskian determinant w = det(W). Exercise 2.

1. Show that y 1 , . . . , y n are linearly independent over C if and only if the Wronskian determinant w is not equal to zero.

2. Show that w = a 1 w 3. Show that, ∀g ∈ G, g(w) = w. det(g) (Hint: show that g acts on W by multiplication on the right).

unimodular Lemma 1.7. There exists f ∈ k such that a n-1 = f f if and only if G ⊂ SL(n, C).

Proof. Assume that G ⊂ SL(n, C). By the above exercise, we have g(w) = w for all g in G hence the normality theorem shows that w ∈ k and it is thus a solution in k of w = a n-1 w.

Conversely, if there exists f in k such that a n-1 = f f , then we must have w = c. f with c ∈ C (because K contains no new constants) so w ∈ k and thus the relation g(w) = w. det(g) implies that det(g) = 1.

Note that we can always arrange that the Galois group be unimodular by letting y = z.e a n-1 n , we see that z satisfies a linear differential over k where we have no term in z (n-1) any more, hence the wronskian is a constant (necessarily in k) and the Galois group is unimodular.

If z ∈ K, the orbit of z under G, noted Orb G (z) is the set of elements of the form g(z) for some g in G.

algebrique Proposition 1.8. Let z ∈ K. Then z is algebraic of degree m over k if and only if

Orb G (z) has exactly m elements. All solutions of L(y) = 0 are algebraic if and only if G is a finite group.

Proof. Assume that z is algebraic; let Q be its minimum polynomial. Let

P = ∏ g∈G (Y -g(z)) ∈ K[Y].
Of course, P(z) = 0 and we now show that the coefficients are in k. Let g 0 ∈ G. As left multiplication by g 0 is a bijection of G, we have

∏ g∈G (Y -g 0 .g(y)) = ∏ g∈G (Y -g(y)) .
So the coefficients of P are fixed by the group and hence, by Galois normality, they are all in k. The roots of P are exactly Orb G (z) by construction.

As Q is the minimum polynomial of z, Q is a divisor of P so the roots of Q are in Orb G (z). Now, the image of a root of Q (here: z) by any element of G is again a root of Q. We conclude that there are as many roots of

Q as elements in Orb G (z), hence deg(Q) = card (Orb G (z)).
Conversely, if Orb G (z) has exactly m elements, call g 1 , . . . , g m elements of G such that Orb G (z) = {g 1 (z), . . . , g m (z)}. As above, we form P = ∏ i=1..m (Yg i (z)). Let g ∈ G; as it is an automorphisms, g(g i (z)) = g(g j (z)) when i = j so g acts as a transitive permutation on the g i (z). It follows that P has coefficients in k. Now, if it was reducible, z would be a zero of a factor and its orbit would hence have less than m element. We conclude that z is algebraic of degree m. Now, if G is a finite group, the above shows that any element in the Picard-Vessiot extension is algebraic; conversely, if all solutions are algebraic, then the y i have only a finite number of possible images under G so, as an automorphism of K is defined by its action on the generators of K, the group G must be finite.

Exercise 3. Let z ∈ K be algebraic of degree m. Let Stab G (z) := {g ∈ G, g(z) = z} denote the stabilizer of z in G. Show that m is the index in G of Stab G (z) (i.
e the cardinal of the quotient).

exponentiel Proposition 1.9. An non-zero element z of K is exponential over k if and only if, for all g ∈ G, there exists a constant c g ∈ C such that g(y) = c g .y.

Proof. Let g ∈ G.

g(z) z = g(z ) z - z z 2 g(z) = g(z) z g( z z ) - z z .
So, g(z)/z is a constant for all g ∈ G if and only if z /z is left fixed by all g ∈ G. By Galois normality, this is indeed equivalent with the fact that z /z ∈ k.

Note that in fact, this means that an element z of K is exponential if and only if the straight-line z.k is globally invariant under G.

Exercise 4.[Put99c]

We consider the equation y = ay with a ∈ C(x).

Show that any proper algebraic subgroup of C

* is finite and cyclic.

2. By considering the possible Galois groups, show that an algebraic solution y = ay must satisfy y m = f where f ∈ C(x) (i.e y is radical over C(x)). Show that the equation y = ay has an algebraic solution if and only if there exists a positive integer m such that the equation f = ma f has a solution f ∈ C(x).

3. For m ∈ N, show that the equation f = ma f has a rational solution if and only if a = ∑ i n i m(x-x i ) with n i ∈ Z having their gcd prime to m. What is the Galois group in this case. 4. Deduce from this a method which decides if the equation y = ay has an algebric solution, computes it, and gives the differential Galois group.

5. Application: compute the Galois group for y = y, for y = 1 4x y, and for y = α x y.

Exercise 5.

[Logarithms]

1. Consider the field K = C(x, log(x)). Show that K is a Picard-Vessiot extension of C(x) corresponding to a homogeneous linear differential equation of order 2 and that its Galois group is conjugate to

G a = { 1 c 0 1 , c ∈ C}.
2. Show that G a is abelian, isomorphic to the additive group (C, +), and that its only algebraic subgroups are itself and {Id}.

3. Show that the reasonning of (i) applies for any element f such that f = a ∈ C(x). Deduce from this that either f is transcendental or f is in C(x) (theorem of Liouville).

These two exercises show that when one adjoins to k an exponential or an integral, the Galois group of the extension is abelian. Recall that the Liouvillian functions are the elements of fields obtained by adjoining successively exponentials, integrals, or algebraic elements to C(x). Exercise 6. We say that a field K is a purely Liouvillian extension of C(x) if it is constructed via a tower of fields C(x) = K 0 ⊂ K 1 ⊂ . . . ⊂ K N = K where K i+1 is obtained from K i by adjoining either an exponential or an integral. Show that the differential Galois group of a purely Liouvillian extension K is solvable.

If we allow arbitrary algebraic extensions, their Galois group G is finite and needs not be abelian any more (in general, it is not!). However, G • is then reduced to the identity. This gives us the first step to the following theorem of Kolchin (which we will admit) Theorem 1.10. A linear differential has a basis of liouvillian solutions if and only if its differential Galois group G is virtually solvable.

Kolchin actually proved the Lie-Kolchin theorem: G • is solvable if and only if its matrices can be put simultaneously in triangular form. The above theorem then follows without too many difficulties. To get convinced, you may study an exemple of this situation where a linear differential equation has a solvable differential Galois group.

Exercise 7. Consider the differential equation exp-red

L(y) = y + (-g(x) -2) y + (g(x) + 1) y = 0 with g a non-zero rational function in C(x).
1. Show that it admits y 1 = e x as a solution.

2. Using variation of constants, compute a second solution from y 1 and integrals.

3. Show that the differential Galois group of L is solvable.

4. Show that the differential Galois group of L is either

c 0 0 c , c ∈ C

Second Order Differential Equations

We will now show how to use the differential Galois group for solving linear differential equations. For simplicity, we focus on second order equations. The idea is to first classify the possible Galois groups and then, on the basis of this classification, to study the corresponding properties of the solutions. We consider the differential equation L(y) = ya 1 ya 0 y = 0.

Subgroups of SL(2, C)

Recall from lemma unimodular unimodular 1.7 that we can assume that the differential Galois group is unimodular, i.e a subgroup of SL(2, C). The subgroups of SL(2, C) are classified and we now go through this construction.

Case I: Reducible Case

Definition 2.1. Let G be a linear group acting on a vector space. We say that (the action of) G is reducible if there exists a non-trivial subspace W ⊂ V such that G(W) ⊂ W.

In our case, dim(V) = 2 so W has to be of dimension 1 and the matrices in the Galois group are triangular (or diagonal). By proposition exponentiel exponentiel 1.9 (and the remark right after it), this is equivalent with the fact that the differential equation has an exponential solution. We also see that the group is diagonal if there are (at least) two exponential solutions (this case is handled further in exercise compred compred 9). In this case, W admits a complement subspace that is also stable under G and the group is then said to be completely reducible.

To the equation L(y) = 0, we associate the Riccati equation, i.e the equation satisfied by u = y y . Simple computation shows that u = a 0 + a 1 uu 2 . We see that the equation is reducible if and only if the Riccati equation has a rational solution.

If G is reducible, then L itself is reducible, in the sense that it can be written as a composition of two operators (check this in exercise exp-red exp-red 7); you may check that L has a right factor ∂u if and only if u is a solution of the Riccati equation.

If G does not act reducibly, it is called irreducible.

Case II: Imprimitive Case

Definition 2.2. Let G be an irreducible group acting on a vector space V. We say that G is imprimitive if there exist subspaces V i such that V = V 1 ⊕ . . . ⊕ V r and G permutes transitively the V i :

∀i = 1, . . . , r ∀g ∈ G, ∃ j ∈ {1, . . . , r} : g(V i ) = V j .
We then say that V 1 , . . . , V r form a system of imprimitivity for G.

In our case, we must have r = 2 and dim(

V 1 ) = dim(V 2 ) = 1. The matrices have the form a 0 0 a -1 or 0 b -b -1 0 with a, b ∈ C * .
Lemma 2.3. Assume that G is irreducible. Then the Riccati equation has an algebraic solution of degree 2 if and only if G is imprimitive.

Proof. Let P denote the minimum polynomial of an algebraic Riccati solution of degree 2. Let u 1 , u 2 be the roots of P. As u 1 , u 2 satisfy the Riccati equation, there exists solutions y i of L(y) = 0 such that y i /y i = u i . As G permutes the u i , it permutes the lines V i = C. < y i > and the V i form a system of imprimitivity.

Reciprocally, let V i = C. < y i > be a system of imprimitivity. If we let u i = y i /y i , then G permutes the u i and proposition algebrique algebrique

1.8 shows that they are algebraic of degree 2 and conjugate (you may also check that the symmetric functions in the u i are left fixed by G).

Case III: Primitive case

Definition 2.4. If G is irreducible and not imprimitive, we say that it is primitive.

One can show ([SU93a]

) that an equation whose Galois group is an infinite primitive subgroup of SL(n, C) does not have Liouvillian solutions; And if the group is finite, all solutions are algebraic (proposition algebrique algebrique 1.8) and hence Liouvillian.

In the case of SL(2, C), there are three primitive groups (e.g [Kov86, SU93b, SU93a] The tetraedral group (A SL 2 4 ) of order 24, the octaedral group (S SL 2 4 ) of order 48, and the icosaedral group (A SL 2 5 ) of order 120. We will now review them The tetraedral group A SL 2 4 of order 24 is generated by matrices

M 1 = ξ 0 0 ξ -1 and M 2 = 1 3 (2ξ -1) 1 1 2 -1
where ξ denotes a primitive sixth root of unity, i.e ξ 2ξ + 1 = 0. The subgroup < M 1 > generated by M 1 has order 6. Assume the second order operator L has A SL 2 4 as its Galois group. Let y 1 , y 2 denote a basis of the solution space on which these matrices act and have this form. We see that the line generated by y 1 is left fixed by < M 1 >; this imposes that u 1 :=

y 1 y 1 is left fixed by < M 1 > (recall proposition exponentiel exponentiel
1.9) so its orbit under G has length at most 4 (in fact, direct computation shows that it has length exactly 4). By the Galois correspondenc (proposition algebrique algebrique 1.8), it follows that u 1 is algebraic of degree 4. We conclude that when the Galois group is A SL 2 4 , the Riccati equation has a solution which is algebraic of degree 4. Another way to see this result is the following. Computing the conjugates

of y 1 under G shows that the polynomial Y 1 (2 Y 2 + Y 1 ) 4 Y 2 2 -2 Y 1 Y 2 + Y 1 2
is a semi-invariant of the Galois group (i.e it is sent to a multiple of itself by the group acting by linear substitution, right action). It can also be checked that its cube is an invariant. We will use this below to compute the minimum polynomial of u 1 .

The octaedral group S SL 2 4 of order 48, is generated by matrices

M 1 = ξ 0 0 ξ -1 and 1/2 ξ ξ 2 + 1 1 1 1 -1
where ξ denotes a primitive eighth root of unity, i.e ξ 4 + 1 = 0. The subgroup < M 1 > generated by M 1 has order 8 and reasoning as above shows that the riccati solution u 1 is algebraic of degree 6 = 48 8 . The group admits the semi-invariant

Y 5 1 Y 2 -Y 1 Y 5 2 , whose square is an in- variant.
The icosaedral group A SL 2 5 of order 120. is generated by matrices

M 1 = ξ 0 0 ξ -1 and φ ψ ψ -φ
where ξ denotes a primitive tenth root of unity, i.e ξ 4ξ 3 + ξ 2ξ + 1 = 0, and φ = 1 5 (ξ 3ξ 2 + 4ξ -2) and ψ = 1 5 (ξ 3 + 3ξ 2 -2ξ + 1). The subgroup < M 1 > generated by M 1 has order 10 and reasoning as above shows that the riccati solution u 1 is algebraic of degree 12 = 120 10 . The group admits the invariant

Y 11 1 Y 2 -11Y 6 1 Y 6 2 -Y 1 Y 11 2 .
In the cases of these three groups, all solutions are algebraic and hence all solutions of the Riccati equation are of course also algebraic. By developping on propositions algebrique algebrique 1.8 et exponentiel exponentiel 1.9, the study of those groups (see [?, ?] or push further the above calculations) shows that :

• For A SL 2 4 :
The Riccati has algebraic solutions of degrees 4,6 ou 12.

• For S SL 2 4 : The Riccati has algebraic solutions of degrees 6,8,12, ou 24.

• For A SL 2 5 : The Riccati has algebraic solutions of degrees 12,20,30, ou 60.

Lastly, if G = SL(2, C), the differential equation does NOT have liouvillian solutions. In view of the applications to hamiltonian mechanics ([?]), we see that the only subgroup(s) of SL(2, C) that will yield obstructions to integrability are . . . only SL(2, C) itself.

This classification work is summarized in Kovacic's theorem kovacic Theorem 2.5 [START_REF] Kovacic | An algorithm for solving second order linear homogeneous differential equations[END_REF]). Let L(y) = 0 be a linear differential equation with coefficients in k It has Liouvillian solutions if and only if it has solutions of the form y = e u where u is an algebraic solution of degree 1 (reducible case I), degree 2 (Imprimitive case II), or degree in {4, 6, 12} (Primitive case III) of the associated Riccati equation.

The Kovacic algorithm

We now give a (simplified) version of the algorithm of Kovacic for solving second order linear differential equations.

Symmetric powers

Let P = U m + b m-1 U m-1 + . . . b 0 be the minimum polynomial of an algebraic solution of the Riccati equation. Let u 1 , . . . , u m be the roots of P, and y i the solutions of L(y) = 0 of which they are logarithmic derivatives. We then have

b m-1 = -(u 1 + • • • + u m ) = -( y 1 y 1 + . . . y m y m ) = - (∏ y i ) ∏ y i .
The coefficient -b m-1 is thus the logarithmique derivative of a product of m solutions of L(y) = 0.

sympow Lemma 2.6. Let y 1 , y 2 be a basis of solutions of L(y) = 0. There exists a linear differential equation L s m whose solution space is the set of homogeneous polynomials of degree m in y 1 , y 2 with coefficients in C.

Proof. Let y be a generic solution of L(y) = 0. Let z = y m . Compute z , z ,. . . ,z m+1 by always replacing y by its expression given by L(y) = 0. The z (i) are linear combinations of monomials of degree m in y, y . These monomials form a k-vector space of dimension m + 1 ; if we have m + 2 elements of such a space, they are linearly dependent, so z,z , z ,. . . ,z m+1 satisfy a linear dependence relation over k that we note L s m (z) = 0 : what we know is that it has order at most m + 1.

Let A be the differential ring K[X 1 , X 2 ] where the derivation is given by

X 1 = X 2 = 0. By construction, we have L s m ((X 1 y 1 + X 2 y 2 ) m ) = 0.
We easily infer that any monomial of degree m en y 1 , y 2 is a solution of L s m (z) = 0. But, if these monomials were linearly dependent, then y 1 , y 2 would be linearly dependent (any homogeneous polynomial in two variables factors as a product of linear factors over C) : So they form a vector space of dimension m + 1, then L s m (z) is order m + 1, and its solution space is precisely that vector space.

We can calculate the linear dependence between z and z using standard linear algebra, but a faster method is the following: Exercise 8.[[BMW97]] Let L(y) = y + ay + by = 0. we define recursively a sequence of operators L i by :

   L 0 (y) = y, L 1 (y) = y , L i+1 (y) = L i (y) + iaL i (y) + i(m -i + 1)bL i-1 (y).
1. Let y be a solution of L(y) = 0. Show by induction that

L i (y m ) = m(m -1) • • • (m -i + 1)y m-i (y ) i . 2. Deduce that L m+1 = L s m .

Algebraic Solutions of the Riccati Equation

Theorem 2.7. The Riccati equation has a solution algebraic of degree at most m if and only if the symmetric power L s m (z) = 0 has an exponential solution.

Proof. If the Riccati equation has an algebraic solution of degree m, we have seen that the coefficient b m-1 of its minimum polynomial is the logarithmic derivative of an exponential solution of L s m (z) = 0. Conversely, let z be an exponential solutions of L s m (z) = 0. Lemma sympow sympow

2.6

shows that there exists a polynomial Q(y 1 , y 2 ) homogeneous of degree m such that z = Q(y 1 , y 2 ). Let v be the logarithmic derivative of z. As Q(y 1 , y 2 ) factors as a product of linear factors over C, let u 1 , . . . , u m be the logarithmic derivatives of these factors. A linear combination of solutions is a solution so the u i are Riccati solutions. For any g ∈ G, as g(v) = v, g(u i ) must be one of the u j : by proposition algebrique algebrique 1.8, it follows that the u i are algebraic of degree at most m.

If the u i do not have degree m, you may check that the product P of their minimum polynomials will be of degree m and its coefficient b m-1 will be given by b m-1 = -v /v.

In [START_REF] Ulmer | Note on Kovacic's algorithm[END_REF], it is shown how, in many cases, one can also ask for rational solutions of L s m (y), which simplifies the algorithm.

Recall that the Riccati equation associated with y + a 1 y + a 0 y is u = -a 0a 1 uu 2 . Differentiating the identity P(u) = 0 and replacing u by its expression, we obtain a polynomial relation of degree m + 1 for u. The remainder of the Euclidean division of this polynomial of degree m + 1 by P must be zero which gives us the following recursion to obtain all coefficients b i once b m-1 is known.

( ) m :        b m = 1 b i-1 = -b i + b m-1 b i + a 1 (i -m)b i + a 0 (i + 1)b i+1 m -i + 1 , m -1 ≥ i ≥ 0 b -1 = 0
Finally, we obtain the following algorithm to calculate algebraic solutions of Riccati equations: For m ∈ {1, 2, 4, 6, 12} :

• Compute L s m (y)
• Seek exponential solutions f

• If there are some: let b m-1 = -f f , compute the other coefficients b i of P by using ( ) m , and return P else, proceed with next m.

If no solution is found this way, there are no Liouvillian solutions.

Exercise 9. Consider the differential equation L(y) = yry = 0. compred 1. Write the Riccati equation R(u) = 0 satisfied by u = y /y.

2. Show that L s 2 (y) = y -4ry -2r y .

3. We assume that R has an algebraic solution of degree 2.

(a) Show that its minimum polynomial has the form

P = u 2 - f f u + f 2 f -r where L s 2 ( f ) = 0. (b) Show that f disc(P) = c

Local and Global Differential Galois Theory

A powerful method to obtain information on differential Galois groups is to first compute some local information at the singularities, and then try to patch (or glue) together these local informations. This is how one will compute rational solutions, exponential solutions, factorizations, etc. We will now sketch some (easiest) algebraic aspects of this local approach, and how to realize actual computations with it.

Local Solutions

Power Series Solutions series

We make the coefficients of our differential equation polynomials: L(y) = a n y (n) + a n-1 y (n-1) + . . . + a n y = 0 where the a i are now polynomials (in fact, analytic would be enough). If a n (x 0 ) = 0, then Cauchy's theorem shows that the equation has a basis of analytic solutions around zero.

Computing these power series is achieved the following way.

Let T = x -x0 if x 0 ∈ C of T = 1 x if x 0 = ∞.
Perform the change of variables x → T in the equation and plug ∑ i c i T i into the equation: you will obtain a recurrence relation for the c i and they will be uniquely determined by their first n terms (exercise: prove this cleanly). To fix notations, let's make this recursion explicit. The operator L can be viewed as an endomorphism of the infinite dimensional vector space C[[x]]. Assume that a i = ∑ m i i=1 a i, j x j . We write the action of L on a basis of C[[x]]:

we get L(x N ) = ∑ n i=0 ∑ m i j=0 N(N -1) . . . (N -i + 1)a i, j x N+ j-i . Now, L( ∞ ∑ N=0 c N x N ) = ∞ ∑ N=0 n ∑ i=0 m i ∑ j=0 N(N -1) . . . (N -i + 1)c N a i, j x N+ j-i . (3.1) pre-rec
So, grouping powers of x, we obtain a recurrence relation

(R N ) : m ∑ l=0 E l (N)c N-l = 0 (3.2) recurrence
with E l being a polynomial. In particular, if ν is the valuation of the power series (the smallest integer such that c ν = 0), we must have E 0 (ν) = 0 (all the c ν-l are zero). If a n (x 0 ) = 0, then direct computation shows that E 0 (N) = N(N -1) . . . (N -(n -1)).

Exercise 10. Use this to show the formal part of Cauchy's theorem i.e the existence of a basis of solutions in power series.

Exponents and Quasi-Series exponents

We now turn to the singular case. For notational convenience, we assume that the considered singular point is zero (i.e a n (0) = 0). We say that a solution y is a quasi-series if it is of the form y = x α φ with φ analytic and α ∈ C. If α ∈ Z, this is a Laurent series; if α ∈ Q, this is a Puiseux series. If further φ has valuation zero (i.e its constant term is not zero), then α is well defined and we call it the exponent of the quasi-series y.

Example 3.1. The Euler homogeneous equation ( [START_REF] Ince | Ordinary Differential Equations[END_REF]). Consider the equation L(y) = x 2 y + c 1 xy + c 2 y = 0 with c 1 , c 2 constants (this equation is often called, as many other equations, an Euler equation). Computation shows that L(x α ) = E 0 (α)x α with E 0 (α) = α(α -1) + c 1 α + c 0 . We see that there is a solution of the form x α if and only if E 0 (α) = 0.

If E 0 has two distinct roots α 1 , α 2 , then we have distinct solutions x α 1 and x α 2 (check that they are linearly independent over C). If E 0 has a double root α 1 , we have E 0 (α 1 ) = E 0 (α 1 ) = 0. We differentiate the relation L(x α ) = E 0 (α)x α with respect to α (note that ∂ ∂α (x α ) = x α log(x)): we obtain L(x α log(x)) = (E 0 (α) + log(x)E 0 (α))x α , from which it follows that L(x α 1 log(x)) = 0 so a second solution is x α 1 log(x). Exercise 11. Show that the equation L(y) := 2xy + 3y + 2y = 0 has a basis of quasi-series solutions at x = 0. Hint: Compute L(x α+N ) for an arbitrary integer N and show that L(∑ ∞ N=0 c N x α+N ) = 0 if and only if we both have that E 0 (α) = 0 for some polynomial E 0 and the c N are solutions of a recursion relation (compute it).

To make this general, look at equation pre-rec pre-rec

3.1 and replace N by α + N in there. Relation

recurrence recurrence 3.2 then becomes (R N ) : m ∑ l=0 E l (α + N)c N-l = 0 (3.3) recurrence2
A necessary condition for the existence of a quasi-series solution with exponent α is that E 0 (α) = 0 (from the case N = 0 in recursion recurrence2 recurrence2

3.3).

Definition 3.2. . The polynomial E 0 is called the indicial polynomial of L at zero. The roots of E 0 are called the exponents of L at zero. If E 0 has degree exactly n, then zero is called a regular singularity; otherwise, E 0 has degree strictly less than n and zero is an irregular singularity.

Exercise 12. Let α be an exponent at zero such that, for all i ∈ N, α + i is Not an exponent. Prove that L admits a quasi-series solution of exponent α.

More generally, if the singularity is regular singular, then (see [?] or [START_REF] Ince | Ordinary Differential Equations[END_REF]) either there is a basis of quasi-series solutions, or there is a basis formed of quasi-series and of solutions of the form x α 1 φ 1 , x α 1 -n 1 (φ 2 + x n 1 φ 1 log(x)), . . . (this may happen in the case when two exponents differ by an integer). Moreover, the power series φ i are analytic. So in this case, the formal theory and the analytical theory coincide.

Generalised Exponents local

If the singularity zero is not regular, then quasi-series and logarithms are clearly not enough to define solutions. For example, consider the equation x 2 y + y = 0: the solution e 1

x can not be written as a quasi-series at zero, so we need to add exponentials to our formal local objects.

Definition 3.3. An element e i ∈ C[x -1
r ] is called a generalized exponent if there is a formal solution of the form e e i x φ i where

φ i ∈ C[[x]
][e i , log(x)] and the valuation (with respect to x, not counting the log) of φ i is equal to zero. If r > 1, then r is called the ramification index of the generalized exponent.

Note that exponents themselves are generalized exponents: indeed

x α = e α x .
To compute generalized exponents, one looks for formal Puiseux series solutions for the Riccati equation associated with L (i.e solutions in C((x 1 r )) for some r ∈ N) and keeps only the parts of such solutions whose valuation is less or equal to -1; the degrees of the generalized exponents can be measured from the Newton polygon of L at zero.

One can show ([vdPS03, ?]) that one can compute a basis of formal solutions of the form e e i x φ i where

φ i ∈ C[[x]][e i , log(x)].

The Formal Local Galois group formal-group

We still assume that we work at zero (otherwise, take a local parameter t = xx 0 at x 0 ∈ C or t = 1

x at infinity and work with t). We consider the field C((x)) as our base field. The formal local Galois group Ĝ0 at zero is defined as the differential Galois group of a Picard-Vessiot extension of C((x)) for L.

Because we know the structure of the formal solutions, we may describe the structure of the formal local Galois group: for each i, we may write e e i x = x α i e P i (with P i of negative degree in x). The formal local monodromy is defined as the Galois group over C((x)) of 1. either C((x))(x α 1 , . . . , x α n ) if there are no logarithms in the solutions (in which case it is a torus) 2. or C((x))(x α 1 , . . . , x α n , log(x)) if there are logarithms in the solutions (in this case, it contains a unipotent element)

The exponential Torus is defined as the Galois group of C((x))(e P 1 , . . . , e P n ).

One readily sees that these two groups generate the formal local Galois group; moreover, they can be easily computed from the given of local solutions.

Lemma 3.4. The formal local Galois group can be embedded into a subgroup of the differential Galois group of L over C(x).

Proof. We use the fact that C(x) can be embedded in C((x)) so we view it as a subfield. Consider the following Kaplansky diagram: K Ĝ0 q q q q q q q q q q q q I I I

I I I I I I I I I C((x)) L L L L L L L L L L K H v v v v v v v v v v G C((x)) ∩ K C(x)
where H is the differential Galois group of K. Galois theory shows that H is a subgroup of K. Then, the above diagram shows ([?]) that the Galois groups H and Ĝ0 are isomorphic. Thus, Ĝ0 can be viewed as a subgroup of G.

If we now take for our base field the field C({x}) of convergent power series, we define the local Galois group G 0 as the differential Galois group of a Picard-Vessiot extension of C({x}). A Kaplansky diagram again shows that Ĝ0 ⊂ G 0 ⊂ G.

At a regular singularities, we have Ĝ0 = G 0 and the Schlesinger density theorem ([?]) shows that the global Galois group is generated by its local Galois groups.

At irregular singularities, though, then new phenomena may occur (Stokes phenomenon) and in this case Ĝ0 G 0 .

A very simple illustration of a link between local and global information is given in the following finite Proposition 3.5. Assume that the Global galois group is finite. Then the exponents at all singularities are rational.

Proof. The local Galois group is embedded in the global Galois group and hence finite. The exponential torus is infinite so there cannot be irregular points and all points must be regular. Now, if there are logarithms, the monodromy contains an additive subgroup and is infinite (alternatively: a logarithm is transcendental hence not algebraic, contradicting the fact that the group is finite). So the monodromy must be diagonal. But, because it is finite, it is cyclic and hence the exponents must be rational.

We note that this result is proved more naturally using Puiseux expansions of algebraic functions, but this proof gives light on the power of the Galois theoretic tools.

Local and Global Algorithms

Rational Solutions ratsols

Let S denote the set of singular points (i.e the zeroes of a n and possibly infinity). We search for a method to check if our differential equation has a rational solution. Let y be a rational function. Then y can be written as y = ∏ x i ∈S (xx i ) α i .(p m x m + p m-1 x m-1 + . . . + p 0 ). So to compute y, we need to find the α i , the degree m, and the coefficients p i . Expansion in Laurent series (or partial fraction decomposition) shows that the α i must be exponents of L at x i . Now, expansion at infinity (expand in powers of 1 x ) shows that there must exist an integer exponent α ∞ at infinity such that m = -α ∞ -∑ x i ∈S α i . We thus obtain the following algorithm, whose solutions are a basis of rational solutions (if any) of L(y) = 0:

1. Select the minimal integer exponents α i at all singularities, including ∞. If one singularity does not have integer exponents, then STOP.

Let m

:= -α ∞ -∑ x i ∈S α i . if m is not positive, then STOP. Plug y = ∏ x i ∈S (x -x i ) α i .(p m x m + p m-1 x m-1 + . . . + p 0 ) into the equa- tion
3. solve the resulting linear system in the p i .

Radical and Global Solutions

The same reasoning applies to radical solutions, i.e the exponents may be used also to compute solutions having some power which is rational: one can similarly prove (see e.g [?, ?]) that there is a radical solution only if there are rational exponents e i at all singular points x i ∈ S such that m := -e ∞ -∑ x i ∈S e i is a positive integer. The solution would again be y = P. ∏ x i ∈S (xx i ) e i with P of degree m. Pluging this expression L(y) = 0 with indeterminate coefficients of P gives a linear system for the coefficients of P, any (non-zero) solution of this system leading to a solution of L; Note that unlike the case of rational solutions, there may be different combinations of the e i to be tested. Also, note that if some factors of a n are irreducible polynomials, then we may have to compute with a splitting field of those to check for combinations, and this can make the algorithm more costly.

For the more general case of exponential solutions, the process is similar, though a little bit more technical, see [?] or [START_REF] Van Der Put | Galois theory of linear differential equations[END_REF]. Now we would like to see how to use these tools in the Kovacic algorithm. We introduce the following useful trick from [?]: Note that if we have two formal solutions e e i x φ i , then their product is e e 1 +e 2

x φ 1 φ 2 , hence e 1 + e 2 is a generalized exponent for the symmetric square of L. In general, it is easy to verify that the expressions ie 1 + (mi)e 2 form the generalized exponents of L s m . So we can check necessary conditions on rational (or radical) solutions of L s m without having to compute this equation.

Reducibility and Factorization

• Let G ⊂ GL(n, C) be a linear algebraic group. The representation in GL(n, C) of the group G is said to be reducible if there exists a nontrivial subspace W which is left globally invariant by G ; it is called completely reducible if any subspace W of C n which is left globally invariant by G admits a complement V in C n which is invariant under G. In our case (characteristic zero), this is equivalent with being reductive (i.e all linear representations are completely reducible).

• A differential operator is called reducible if it can be written as a product L = L 1 .L 2 with L i ∈ k[∂] but not in k. Right-hand factors correspond to subspaces of the solution space that are stable under the differential Galois group. The ring of differential operators is seen being left and right euclidean ring, one can define (effectively) notions of greatest common right divisors (gcrd) and least common left multiples (lclm). A differential operator is called completely reducible if it is the least common (left) multiple of irreducible operators (equivalently: if its Galois group is reductive).

In terms of linear differential system, here is another possible formulation.

• A system [A] Y = AY with A ∈ M n (k)
is said to be reducible if there exists a matrix P ∈ Gl n (k) such that the change of variable Y = PZ transforms the system [A] into a system of the form

Z = × 0 × F Z
where F is a square matrix of dimension k (it corresponds to a factor).

Otherwise the system is said to be irreducible.

The system is said to be completely reducible if it can be transformed into a block diagonal system

Z =        F 1 F 2 . . . F s        Z
where each system ν = F i ν is irreducible.

The Art of Computing Galois Groups

We now have tools to smoothly use and apply the Kovacic algorithm (and generalisations like [SU97, vHRUW99]). We will show on examples how to combine all these tools together to obtain differential Galois groups.

Example 3.6. Consider the Airy equation L(y) = yxy = 0. The only singularity is infinity. The local (generalised) exponents are -1 4 ± 2 √ x 3 at infinity.

Because of the ramification at infinity, we see that the equation cannot have exponential solutions. Now we look at the second symmetric power.

Looking at sums of the exponents, we see that the only rational exponent of the second symmetric power L s 2 will be ( 1 2 ) at infinity. This cannot be the degree of a polynomial. so there cannot exists a radical solution the the second symmetric power. Now, as the equation is irregular at infinity, the group cannot be finite by proposition finite finite 3.5: this excludes case 3 of the Kovacic algorithm. Finally, the only possibility is that the Galois group is SL(2, C).

Example 3.7. Consider the equation

L(y) = y + 1/4 (x -1) -1 + 5/4 (x -1) -2 + 3/16 x -2 y = 0
The local (generalised) exponents are ( 1 4 , 3 4 ) at zero, the roots of X 2 -2X + 5 at 1 and -1 4 ± 2 √ -x at infinity. Because of the ramification at infinity, we see that the equation cannot have exponential solutions. Now we look at the second symmetric power. Looking at sums of the exponents, we see that the only rational exponents of the second symmetric power L s 2 will be ( 1 2 , 1, 3 2 ) at zero, (1) at 1 and (-1 2 ) at infinity. Taking the lowest e 0 , e 1 , e ∞ possible, we have -e ∞e 1e 0 = -1 < 0 and the latter cannot be the degree of a polynomial so there cannot exists a radical solution the the second symmetric power. Now, as the equation is irregular at infinity (and does not have rational exponents at 1), the group cannot be finite by proposition finite finite 3.5: this excludes case 3 of the Kovacic algorithm. Finally, the only possibility is that the Galois group is SL(2, C).

Exercise 13. Consider the equation L(y) = x 3 y + (x 2 + x)yy = 0 1. Show that the exponents at ∞ are (0, 0) and that the generalized exponents at 0 are (1, -1 x ).

Show that e 1

x is an exponential solution, the only one (up to scalar multiplication). Show that there is a unique power series solution at zero, and that it is divergent (and Gevrey). Compute the formal local Galois group at 0.

3. Show that the formal solutions at infinity are of the form ŷ1,∞ and ŷ1,∞ log(x) + ŷ2,∞ where the ŷi,∞ are power series in 1

x . Compute the formal local Galois group at infinity. 1. Show (using the above examples) that for generic values of λ the Galois group is SL(2, C).

2. Show that the exponents at zero are rational if and only if λ = 12 n(n-1)

with n ∈ Q.

3. In this case, the equation is y -1 4 -n(n-1) x 2 . Show that the exponents at zero are (n, 1n) and ± x 2 at infinity. Prove that the Galois group is SL(2, C) unless n is an integer. 4. Perform the change of variables y(x) = e x 2 f (x). Search for f as a power series: its coefficients u N satisfy the recursion Nu N -(n + N) * (n -1 -N)u(N + 1) = 0 and u(0) = 0. Conclude that when n is an integer, f is a polynomial and hence L has one (in fact, two) exponential solutions.

In this exercise, we see that these tools give strong necessary conditions. However, question (4) shows that when there are parameters and the necessary conditions are satisfied, then it is not that easy to decide if there actually exists a solution. In this case, it was feasible; in general, it is not (see [Bou99]) and even sometimes undecidable. Still, in the applications to Hamiltonian mechanics, we encounter many systems where "mysteriously" reasonings like the above (and many other tricks) allow one to say a lot about non-integrability of entire families of equations. This will be the topic of the next lecture.

Part II

Integrability of Hamiltonian Differential Systems

The Hamiltonian systems are differential systems which describe the equations of motion of mechanical systems whose mechanical energy (the Hamiltonian) is conserved. One says that the mechanical energy is a first integral for the Hamiltonian system. The notion of complete integrability of a Hamiltonian system refers to the given of sufficiently many independent first integrals (see definitions below). In recent decades, many significant improvements have been made in the research regarding complete (meromorphic) integrability of Hamiltonian systems after the pioneering papers of Ziglin ([Zig83a, Zig83b], [START_REF] Ziglin | Branching of solutions and the non-existence of first integrals in hamiltonian mechanics ii[END_REF]Zig83b], 1982) relating integrability to properties of the monodromy group of a variational equation (or variational system) along a known solution. Churchill Let (S) be a Hamiltonian system, x 0 (t) be a particular solution of (S) and G be the differential Galois group of the normal variational equation of (S) computed along the solution x 0 (t). If the system (S) is completely integrable, then the connected component of the identity in the group G, denoted G 0 , is an abelian group.

This theorem was generalized to 'higher variational equations' in [START_REF] Juan | Integrability of hamiltonian systems and differential galois groups of higher variational equations[END_REF]. At first sight, it may not be obvious that the conclusion of the statement (i.e. G 0 abelian) is any easier to test than its premices. For second order equations, many methods are known after the work of Kovacic [START_REF] Kovacic | An algorithm for solving second order linear homogeneous differential equations[END_REF], see also [START_REF] Ulmer | Note on Kovacic's algorithm[END_REF], [DLR92], . . . ). They enabled many results of non integrability, let us cite only some of them : [Tsy01b, Tsy01a, MSS01, Mac02, Aud03, MP03, MPW04, MP04, MP05, Aud01, Aud02, Aud03] The goal of these notes is to provide tools that seem to be practically efficient in handling families of Hamiltonian systems. We would like to promote the following idea. The variational system has an infinitesimally symplectic matrix. This fact induces a number of structural properties on the system that can be used to simplify computations. So our approach is to try to exploit and preserve as much as possible this symplectic structure. Another point is that integrability questions are asked on families of Hamiltonians depending on parameters. This induces families of variational equations which also depend on parameters and this is an obstruction to algorithms that test Morales-Ramis theorem and the criterium we deduced from it. However, we will show below that the symplectic structure allows one to confront a part of this obstruction. We will emphasize on systems of order greater than 2.

The topic of this lecture is at the crossing point of two theories for which we assume the reader has a working knowledge (or is ready to accept some statements and references ot litterature): Integrability of Hamiltonian systems (see [START_REF] Audin | Les systèmes hamiltoniens et leur intégrabilité[END_REF][START_REF]Intégrabilité et non-intégrabilité de systèmes hamiltoniens[END_REF] [MRR01a], [START_REF] Juan | Meromorphic nonintegrability of Hamiltonian systems[END_REF][START_REF]A note on the non-integrability of some Hamiltonian systems with a homogeneous potential[END_REF], [START_REF] Churchill | Differential algebraic techniques in Hamiltonian dynamics, Differential algebra and related topics[END_REF], [START_REF] Juan | Integrability of hamiltonian systems and differential galois groups of higher variational equations[END_REF] . . . ) and Differential Galois theory (see the previous lecture, and the remarkable reference book [START_REF] Van Der Put | Galois theory of linear differential equations[END_REF], or also [Sin90b], [START_REF] Magid | Lectures on differential Galois theory[END_REF]. . . ).

Basic Facts on First Integrals of (Hamiltonian) Differential Systems section2

In this first section we recall basic facts on Hamiltonian systems, differential galois theory and the Ziglin and Morales-Ramis theorems ([MR01], [START_REF]Galoisian obstructions to integrability of Hamiltonian systems. I, II[END_REF]).

Variational equations and First Integrals

Definition 4.1. Let K be a field of characteristic zero. Let us consider F = ( f 1 , . . . f n ) where f i is of class C k on an open set U of K n . A dynamical system is a system of differential equations

dx i dt = f i (x 1 (t), . . . , x n (t)), i = 1, . . . , n
which can be written, in vectorial notation as

dx dt = F(x)
We will focus on conserved quantities for such systems:

Definition 4.2. A function G : U → R is a first integral of the dynamical system ẋ = F(x) if, for all solution x(t) of the system, d dt G(x(t)) = 0.
The first integrals give the geometry of the solution curve. Indeed the solutions of the system lie on the hypersurfaces G = cte.

To answer questions regarding the existence of analytic first integrals, Poincaré studied the solutions which are infinitesimaly close to a given particular solution of the differential system. The behavior of these solutions is given by a homogeneous linear differential system, the variational system, which is obtained by linearization of the differential system along the particular solution.

Definition 4.3. The variational system along a solution x 0 (t) of a differential system is the linear differential system:

h (t) = Jac(F)(x 0 (t)) h(t)
where Jac(F) denotes the Jacobian matrix of F at x 0 (t). [START_REF] Audin | Les systèmes hamiltoniens et leur intégrabilité[END_REF] for a nicer definition) weak-ziglin Lemma 4.4 (Ziglin, weak form). If (S) admits m algebraically independent first integrals f 1 , . . . , f m , then it admits m algebraically independent first integrals g 1 , . . . , g m such that their junior parts g • 1 , . . . , g • m are algebraically independent. Furthermore they are solutions to the variational system.

Let G : U → R and let k in N such that (dG) x = • • • = (d k-1 G) x = 0. The junior part G • of G is d k G (see

Hamiltonian Systems and Complete Integrability

Definition 4.5. Let n ∈ N * , x = (x 1 , . . . , x 2n ) = (q 1 , . . . , q n , p 1 , . . . , p n ) ∈ R 2n . A Hamiltonian system on a non empty domain U of R 2n is a system of differential equations of the form:

                   dq i dt = ∂H ∂p i (q, p) dp i dt = - ∂H ∂q i (q, p) i = 1, . . . , n
where H : U → R is the Hamiltonian function. The variables p i and q i are conjugate variables. The positive integer n is called the number of degrees of freedom.

Remark 15 : A Hamiltonian system can also be written in the following form:

x (t) = J ∇H(x(t))

where J = 0 I -I 0 , I is the order n identity matrix and ∇H(x) is the gradient of H at x. The Hamiltonian H is conserved, it is a first integral for the hamiltonian system. Indeed, for each solution x of the system, dH(x(t)) = 0. A first integral G of the Hamiltonian system is also characterized by the equality {G(x), H(x)} = 0

(or {G(x), H(x)} = {H(x), G(x)}) where {G 1 , G 2 } = n ∑ i=1 ∂G 1 ∂p i ∂G 2 ∂q i - ∂G 1 ∂q i ∂G 2 ∂p i =< ∇G 1 (x), J ∇G 2 (x) > is the Poisson bracket of G 1 and G 2 .
Definition 4.6. A Hamiltonian system with number of degrees of freedom equal to n is called completely integrable if it has n first integrals G 1 , . . . , G n such that:

• G 1 , . . . , G n are functionally independant (∇G 1 , . . . , ∇G n are linearly independant)

• G 1 , . . . , G n are in involution: for all solution x of the Hamiltonian system,

{G i (x), G j (x)} = {G j (x), G i (x)} (G 1 , . . . , G n commute for the Poisson bracket).
It is necessary to give a more precise sense to this notion of complete integrability by asking which class of functions we want the first integrals to belong to (analytic, algebraic, meromorphic functions, . . . ).

To answer the question of the analytic complete integrability, Poincaré studied the solutions which are infinitesimaly close to a particular solution of the Hamiltonian system. The behavior of these solutions is given by a homogeneous linear differential system, the variational system, which is obtained by linearization of the Hamiltonian system along the particular solution.

Definition 4.7. The variational system along a solution x 0 (t) of a Hamiltonian system is the linear differential system:

h (t) = J H(H, x 0 (t)) h(t)
where H(H, x 0 (t)) is the hessian of H at x 0 (t).

Remark 16 : The matrix A = J H(H, x 0 (t)) belongs to the simple Lie algebra sp(2n, C) = {M ∈ Gl(2n, C)/ t M J + J M = 0} which is the Lie Algebra of the symplectic group Sp(2n, C) = {g ∈ Gl(2n, C) / t g J g = J}. One says that A is an infinitesimally symplectic matrix.

As x 0 (t) is a particular solution of the variational system along the solution x 0 (t), one can reduce it and get a normal variational system of order 2n -2. During the last twenty years many results on the meromorphic non integrability of the Hamiltonian systems were deduced from the study of this (normal) variational system. Exercise 17. We consider here the example of the system of H Őnon-Heiles from [START_REF] Juan | Integrability of hamiltonian systems and differential galois groups of higher variational equations[END_REF]. The initial Hamiltonian is

H = 1 2 (y 2 1 + y 2 2 ) + 1 2 x 2 1 + 1 2 x 2 2 + 1 3 x 3 1 + 1 2 x 1 x 2 2
A particular solution of it is ([MRRS05]) :

x 0 (t) = t 1 2 - 3 2 tanh t 2 , 0, - 3 2 tanh t 2 1 -tanh 2 t 2 , 0
The variational system along this particular solution is :

h (t) =     0 0 1 0 0 0 0 1 -2 + 3 tanh 2 ( t 2 ) 0 0 0 0 -3 2 + 3 2 tanh 2 ( t 2 ) 0 0     h(t)
If one wants rational coefficients one makes the change x = tanh(t/2) and one gets

h (x) =      0 0 -2 x 2 -1 0 0 0 0 -2 x 2 -1 -2(3x 2 -2) x 2 -1 0 0 0 0 -3 0 0      h(x)
To reduce it to a normal variational system one makes the symplectic change of variable (see [START_REF]Galoisian obstructions to integrability of Hamiltonian systems. I, II[END_REF] or annex of [START_REF]Sur les équations différentielles linéaires dépendant de paramètres ; application aux systèmes hamiltoniens[END_REF], [Bou00a, BW03] ) h = PY where the first column of the matrix P is x 0 (t) and P satisfies t P J P -J = 0. One gets an equivalent linear differential sytem from which one extracts the rows and columns 2 and 4. One gets the normal variational system

Y (x) = 0 3 2 x 2 -1 0 Y(x)
In 1982, Ziglin ([Zig83a, Zig83b], [Zig83a, Zig83b]) established necessary conditions of meromorphic complete integrability on the monodromy group of the variational system. In 1995, Baider, R.C. Churchill, Rod and M. Singer ( [CRS95, BCRS96]) and in 1998, J.J. Morales and J.P. Ramis ([MRR01a]) established necessary conditions of meromorphic complete integrability on the differential Galois group of the variational system.

Ziglin, Morales-Ramis

We may now restate lemma weak-ziglin weak-ziglin 4.4 in differential Galois terms (this is not Ziglin's original formulation, but rather the one from [CRS95, BCRS96] or [START_REF] Audin | Les systèmes hamiltoniens et leur intégrabilité[END_REF]).

Theorem 4.8 (Ziglin, strong form). Consider a differential system (S) : x = F(x) and the variational system [A] : Y = AY along a solution x 0 (t) of (S). If (S) admits m algebraically independent meromorphic first integrals, then Gal([A]) admits m algebraically independent rational invariants.

For completely integrable hamiltonian systems, the additional involution property translates into the celebrated result of Morales and Ramis:

theoRM Theorem 4.9 (J.J. Morales, J.P. Ramis). Consider a hamiltonian system (S) and the variational system [A] : Y = AY along a solution x 0 (t) of (S). If the system (S) is completely integrable, then Gal([A]) is virtually abelian.

In next section we present a criterion deduced from Morales-Ramis theorem to get sufficient conditions of non complete integrability of a Hamiltonian system (see [START_REF]Sur les équations différentielles linéaires dépendant de paramètres ; application aux systèmes hamiltoniens[END_REF]Bou00a,[START_REF] Boucher | theorem to test the non-complete integrability of the planar three-body problem, From combinatorics to dynamical systems[END_REF] for a preliminary version of it).

A criterion of non complete integrability of Hamiltonian systems int-criteria

The analysis of the local solutions around singularities can give information on the integrability of nonlinear systems of ordinary differential equations such as Hamiltonian systems (Kovalevskaya and Painlevé test) However, the criterion we deduce from Morales Ramis theorem requires not only a local study of the normal variational equation but also a global study of it. In next subsection we give introducory examples illustrating that the study of the local solutions for the variational system will not be sufficient in general to give an obstruction to the virtual abelianity of the corresponding Galois group. In the second subsection, we give a non integrability criterion based on local and global considerations.

Examples

Before stating the criterion, let us focus on two introducing examples.

• Let us consider the first order linear differential symplectic system Y = 0 1 x 0 0 Y which can be also written

xY (x) = 0 1 0 0 Y(x)
This system has formal solutions at zero with logarithmic terms as the local monodromy is given by the matrix 0 1 0 0 (whose single eigenvalue is 0 and who is not diagonalizable). However its differential Galois group is additive, so abelian. One can notice that this system is equivalent to the linear differential equation xy (x) + y (x) = 0 (using the cyclic vector (0, 1)). Its space of solutions is generated by 1 and ln(x). The presence of logarithmic terms in the formal solutions does not suffice to imply the non virtual abelianity of the group G. Furthermore, one notices here that the equation was reducible without being completely reducible. Indeed,

x∂ 2 + ∂ = (x∂ + 1)∂ = ∂(x∂ + 1).
This is the crucial point that is at the origin of the proposition prop2 prop2

5.2 and theorem new new 5.3 (or corollary new2 new2 5.5).

• Let us consider the normal variational system of the example 1

Y (x) =   0 3 2 x 2 -1 0   Y(x)
The system can be written

(x -1)Y (x) =    0 3(x -1) 2 x + 1 0    Y(x) = 0 0 1 0 + 0 3 -1 2 0 (x -1) + • • • Y(x)
So the monodromy matrix at 1 is given by the matrix 0 0 1 0 and there are logs at 1.

As the exponents at 1 and -1 are both equal to 0, the exponential solutions of the system correspond to the polynomial solutions. The exponents at infinity are 2 and -3 so one searchs a polynomial solution of degree 3. One finds (using package ISOLDE in Maple) t (z(z 2 -1), z 2 -1 3 ) so the system is reducible. One can also see the normal variational system as an equivalent linear differential equation

y (x) - 6 x 2 -1 y(x) = 0
The exponents at the singularities are 0 and 1 at the points -1 and 1; -3 and 2 at the point infinity, so an exponential solution is of the form p(x) where p is a polynomial of degree 3. One finds one single solution z 3z so the system is reducible without being completely reducible.

Again, the differential Galois group of this normal variational system is abelian, so the presence of logarithmic terms does not suffice to conclude to the non virtual abeliannity and Morales-Ramis theorem does not enable to conclude here. Furthermore, this special example was given a complete answer in [START_REF] Juan | Integrability of hamiltonian systems and differential galois groups of higher variational equations[END_REF] where the use of higher variational equations enabled to conclude to the non complete integrability of the system.

A non-integrability criterion

We propose a criterion where added assumptions about the factorization of the equation suffice to make the group non virtually abelian. We derive this criterion using both the presentation of linear differential equations and the presentation of linear differential systems. First we need a lemma (established independently in [Bou00b, ?]).

lem1 Lemma 5.1 ([Bou00b, BW03, ?]). Let G ⊂ GL(n, C) be a completely reducible linear algebraic group acting on V = C n . The following assertions are equivalent: (a) G 0 is solvable, (b) G 0 is diagonalizable, (c) G 0 is abelian.

Proof. The implications (b) ⇒ (c) and (c) ⇒ (a) are immediate. Now assume that (a) holds. First we assume that V is irreducible under G. Because G • is solvable, it is triangularizable. In particular, all its elements have a common eigenvector v 1 . Let g ∈ G and h ∈ G

• . Because G • is normal in G, we have h(g(v 1 )) = g( h(v 1 )) with h ∈ G • . But, as v 1 is a common eigenvector for G • , we have h(v 1 ) = χ h.v 1 with χ h ∈ C. But now h(g(v 1 )) = g(χ h.v 1 ) = χ h.g(v 1
) so all g(v 1 ) are eigenvectors of G • . The linear space spanned by the g(v 1 ) (for g ∈ G) is a subspace of V invariant under G. By irreducibility, it is equal to V. Because it it is generated by the g(v 1 ), G • acts diagonally on it. Now, if V is reducible, it is a direct sum of subspaces which are irreducible under G, and we apply inductively the above reasonning to these irreducible summands.

If G is the differential Galois group of a linear differential equation L(y) = 0, then it is easier to find sufficient conditions of non-diagonality of G • using the local information that we can read in the formal solutions prop2 Proposition 5.2. Let L(y) = 0 be a homogeneous linear differential equation with Galois group G. Assume that the equation L(y) = 0 is completely reducible.

1. If it has formal solutions around some point which contain logarithmic terms, then the connected component of the identity in the group G is not an abelian group.

2. If it has a non-trivial Stokes multiplier around some point (an irregular singularity), then the connected component of the identity in the group G is not an abelian group.

Proof. Let L be a differential operator of degree n with coefficients in a field k with differential Galois group G.

Let us assume that L is irreducible. If the group G 0 is abelian, then according to Lemma lem1 lem1

5.1, it is diagonalizable. However, if a logarithmic term appears locally, then the corresponding local group has a non trivial unipotent subgroup ([?]). So the group G 0 contains a non trivial unipotent subgroup, which contradicts the diagonality of G 0 . If L is completely reducible one concludes in the same way as in the proof of Lemma We extend now the proposition prop2 prop2 5.2 to a more general one: new Theorem 5.3. Let L be a linear differential operator such that L = L 1 L 2 . . . L s .

If there exists i in {1, . . . , s} such that L i is completely reducible and the linear differential equation L i (y) = 0 has formal solutions with logarithmic terms, then the differential Galois group G of the equation L(y) = 0 is not virtually abelian.

Proof. If s = 1 one concludes using proposition prop2 prop2

5.2. All we now need is to prove that if there exists i in {1 . . . s} such that the differential Galois group of L i (y) = 0 is not virtually abelian then the differential Galois group of L(y) = 0 is also not virtually abelian. It is enough to prove this in the case of two factors (s = 2), the result then being an easy induction. So we assume that L = L 1 .L 2 .

Lemma 5.4. Let K denote a Picard-Vessiot of k associated to L. If F is an intermediate differential Picard-Vessiot extension (k ⊂ F ⊂ K), then virtual abelianity of Gal d (K/k) implies that of Gal d (F/k).

Proof. From [?] (lemma 5.10 page 38), we know that there is a natural surjection from Gal d (K/k) to Gal d (F/k). In particular, the abelianity of Gal d (K/k) • implies that of Gal d (F/k) • .

We now finish the proof of the theorem. As the solutions of L 2 are solutions of L, K contains a Picard-Vessiot extension for L 2 and the result follows for L 2 from the lemma. Now, L 2 can be viewed as a morphism from Sol(L) to Sol(L 1 ). Hence, K contains a full set of solutions of L 1 and we may apply the lemma again to conclude for L 1 . We now conclude applying proposition prop2 prop2 5.2 to the L i .

Using the point of view of the systems, we get : new2 Corollary 5.5. Let Y = AY be a linear differential system such that it is equivalent to a block-triangular system Z =

     B 1 0 • • • 0 × . . . 0 × × B s-1 0 × × × B s     
Z If there exists i in {1, . . . , s} such that the system ν i = B i ν i is completely reducible and has formal solutions with logarithmic terms then the differential Galois group G of the system Y = AY is not virtually abelian.

Remark 18 : As we already had noted in Proposition prop2 prop2

5.2 the conclusion of the above theorem still holds if we replace the presence of logarithms by the presence of a non trivial Stokes multiplier (or any recognizable nondiagonalizable element) at an irregular singularity. The reason is that these Stokes matrices (see [?] for example) induce a unipotent element in the differential Galois group and hence yield an obstruction to diagonalizability.

An example: the Friedman-Robertson-Walker model

Let us consider the Hamiltonian of the system FRW (Friedmann-Robertson-Walker) cosmological model

H = 1 2 ȧ2 + φ2 + 1 2 a 2 + 1 2 φ 2 - 1 2 m 2 a 2 φ 2 + 1 4 λφ 4 + 1 4 Λa 4
where a is the scale factor of the universe; Λ is the cosmological constant; φ is the scalar field with self-coupling constant λ and with mass m (which we will assume to be non zero).

The approach via constructive Morales-Ramis criteria

• We first select a particular solution X 0 to the hamiltonian system associated to H: X 0 = (0, φ 0 , 0, φ0 )

where φ 2 0 = - 2 λ ℘ (
(E) 4 27 (3x -2) (3x + 1) 2 y (x) + 2 3 (3x -1) (3x + 1) y (x)+ 2m 2 λ x + 1 + 2 3 m 2 λ y(x) = 0
• There are logarithmic terms in some formal solution of (E) at the singular point -1/3. Namely, the formal solutions at that point are

s 1 = √ 3x + 1 (1 + • • • ) and - m 2 2λ
s 1 ln(3x + 1) + s 2 where s 2 is a formal series.

• The equation (E) is irreducible if λ = - 2m 2 (n + 2)(n + 1)
with n ∈ N. Indeed, the equation (E) is irreducible if, and only if, it possesses no exponential solution y(x) = (3x + 1) -1/2 (3x -2) e p(x) where p(x) is a polynomial of degree d; e = 0 or

= - 2m 2 (n + 2)(n + 1)
with n ∈ N then the FRW system is not completely integrable.

Higher variational approaches

If λ = - 2m 2 (n + 2)(n + 1)
with n ∈ N then the variational equation is a direct sum of two Lamé-Hermite type equations. One then needs to go to higher variational equations. The variational equation is an approximation at order 1 for the linearization. Approximation at order 2, 3, etc, yields successive linear differential systems (with growing size). Their Galois groups G k form a tower of groups; the result of Morales-Ramis-Simo is that each of these G k must be virtually abelian. Now, as the variational equation is a direct sum of two Lamé-Hermite type equations, one may reduce the Morales-Ramis-Simo criterion to a local study of the differential at zero. This local study of the third variational equations enables to conclude that the system is not completely integrable for n ∈ {2, . . . , 10} (an additional global study is not required in this special case). There are two particular cases : Λ = λ when n = 0 and n = 1. For both these special cases an additional polynomial first integral I n is found:

• I 0 = φ ȧ -a φ • I 1 = ȧ φ + aφ - m 2 3 (aφ 3 + a 3 φ)
If n ≥ 11, we conjecture that he FRW system is not completely integrable.

7 Some properties to factorize symplectic linear differential systems fact The theorem new new 5.3 can be applied very easily when the coefficients of the normal variational system (or equation) are in Q(x) as there exist efficient algorithms to compute formal solutions of linear differential equations (systems) and to factorize linear differential operators (systems). However the equations we deal with here are parameterized equations and the existing algorithms are not adapted to this situation (problems of indecidability). In the following we see how the symplectic structure of the normal variational system can help the factorization of the normal variational system and hence the application of the criterion. This symplectic structure can be read very simply on the matrix of the normal variational system this is why we choose to keep the system instead of transforming it to a linear differential equation (using a cyclic vector computation).

Factorization in practice

The factorization of linear differential systems was studied and implemented in [START_REF] Yu | Complexity of factoring and calculating the GCD of linear ordinary differential operators[END_REF][START_REF] Bronstein | On the factorisation of linear ordinary differential operators[END_REF]. Another family of algorithms uses the notion of eigenring [Sin96], [START_REF] Barkatou | On the equivalence problem of linear differential systems and its application for factoring completely reducible systems[END_REF] In this section we first give a property of the eigenring in the symplectic case, which enables a simpler computation of it. Then we see that despite the presence of parameters one can look easily for some elements of the eigenring. Let k denote a differential field (e.g k = C(x)) and let A be infinitesimally symplectic.

Properties of the eigenring

The differential Galois group G of the differential system Y = AY is a subgroup of the symplectic group ([Aud01] page 65). Let V denote the solution space of Y = AY. If U is a fundamental solution matrix, then J.U is a fundamental solution matrix for the dual system Y = -t AY. Hence, J is the matrix of the musical isomorphism between V and its dual V * , i.e these are isomorphic G-modules.

The Eigenring of A is the set E (A) := {P ∈ GL(2n, k)|P = AP -PA}. It describes ([Sin96]) the ring End G (V) of G-endomorphisms of V; in particular, the presence of a non-trivial element in the Eigenring provides a factorization of the differential system. Now, as End(V) = V ⊗ V * and V G V * , we see that End(V) G V ⊗ V = Sym 2 (V) ⊕ Ext 2 (V). The image of the identity in this morphism is the symplectic form (hence an element of Ext 2 (V)) that is preserved by the symplectic group. In fact, this isomorphism can be made even more explicit: eigenring Proposition 7.1. Let A be an infinitesimally symplectic matrix ( t A J + J A = 0)

and let E (A) be its eigenring. Then

E (A) = E + (A) ⊕ E -(A)
where E + (A) = {T ∈ E (A), t T J + J T = 0} E -(A) = {T ∈ E (A), t T J -J T = 0}.

Proof. Let A be an infinitesimally symplectic matrix ( t A J + J A = 0) and let T be in E (A). We first need the following lemma:

Lemma 7.2. Each matrix of M of M n (C) can be uniquely written as the sum of a matrix M 1 such that M 1 J -J t M 1 = 0 and a matrix M 2 such that M 2 J + J t M 2 = 0 (M 2 is an infinitesimally symplectic matrix).

Proof. Let M be in M n (C) and let us consider the two matrices M 1 = -1 2 (M J + J t M) J and M 2 = -1 2 (M J -J t M) J. One checks: M = M 1 + M 2 , M 1 J -J t M 1 = 0, M 2 J + J t M 2 = 0 (M 2 is an infinitesimally symplectic matrix).

According to the lemma, the matrix T can be written as a unique sum of two matrices T 1 and T 2 such that T 1 J -J t T 1 = T 2 J + J t T 2 = 0.

It suffices to prove now that both T 1 and T 2 are in E (A). As T is in E (A), it satisfies

T = A T -T A so T 1 + T 2 = (A T 1 -T 1 A) + (A T 2 -T 2 A).
(7.4) eqT1T2

A quick computation enables to see

T 1 J -J t T 1 = (A T 1 -T 1 A) J -J t (A T 1 -T 1 A) = 0 and T 2 J + J t T 2 = (A T 2 -T 2 A) J + J t (A T 2 -T 2 A) = 0.
According to the previous lemma, the equality ( eqT1T2 eqT1T2

7.4) enables to conclude:

T 1 = A T 1 -T 1 A and T 2 = A T 2 -T 2 A
so T 1 and T 2 are in E (A) and one has written the matrix T as the direct sum of two elements of E -(A) and E + (A).

Remark 19 : If we define the Eigenring not over k but over a Picard-Vessiot extension K then the elements of Sym 2 (V) correspond to E + (A) and those of Ext 2 (A) correspond to E -(A). We recover, in particular, the well-known fact that the Lie algebra of the symplectic group is isomorphic to the symmetric square of its faithful representation of order 2n.

The elements of the eigenring of A correspond to rational solutions of the tensor system Z = (A ⊗ I -I ⊗ t A) Z

(see [?] or [Sin96]).

If A = (a i, j ) 1≤i, j≤n (with n = 2m even), then the matrix A ⊗ I -I ⊗ t A is the n 2 × n 2 matrix (a i, j I)diag( t A). At first sight, one needs to compute a rational vector solution with n 2 components. However, thanks to proposition eigenring eigenring

7.1, one can reduce the number of unknown components. Indeed elements of E + (A) (resp. E -(A)) have m 2 + m(m + 1) (resp. m 2 + m(m -1)) unknows instead of 4m 2 . So one can reduce the number of unknows while searching rational solutions of the tensor system.

Properties of the exponents at the singularities

To compute rational solutions of the tensor system Z = (A ⊗ I -I ⊗ t A) Z one needs to compute the exponents at the singular points of this system. When the matrix A depends on parameters then these exponents may also depend on the parameters, which makes the search of rational solutions difficult (in particular there are problems of indecidability). However if the matrix A is infinitesimally symplectic then some of the exponents of the tensor system will not depend on the parameters. In this case the search of the rational solutions of the tensor system (i.e. the elements of the eigenring of A) can be effectively achieved.

Proposition 7.3. Let us consider a symplectic linear differential system. Let us assume that it is regular singular at the point 0 and that it can be written:

xY (x) = A(x) Y(x)
where the matrix A(x) is an infinitesimally symplectic matrix defined at the point 0.

Then the exponents at the point 0 can be gathered in the following way : (e 1 , -e 1 , e 2 , -e 2 ).

Proof. The exponents at 0 are the eigenvalues of the matrix A(0). But this matrix is infinitesimally symplectic ( t A(0) J + J A(0) = 0) so its eigenvalues can be gathered in the following way : (e 1 , -e 1 , e 2 , -e 2 ).

So in this particular case the structure of the matrix of the system enables to conclude that two sums of two exponents are equal to 0 and so are independant of the parameters. As the exponents at the singularities of the tensor system are sums of exponents of the initial system, some of the exponents for the tensor system are independant of the parameters (even if all the exponents of the initial system depend on the parameters). This proposition can be also seen with the point of view of the linear differential equations. exponents Proposition 7.4. Let L(y) = 0 be a linear differential homogeneous equation and let G be its differential Galois group. Assume that the group G is symplectic. Let p be a singular point, and let r denote the ramification index at this point (note: if the point is regular singular, then r = 1).

The generalized exponents at p are pairewise gathered in the following way :

(e, me) where m is a rational in 1 r Z . Proof. Let us assume that we work at the point 0 and let y(x) = x e Σ(x) be a formal solution at the point 0 with Σ(x) ∈ Q[ln(x)]((x 1 r )). Let g be in the formal local Galois group G 0 at the point 0; then there exists c such that g(x e ) = cx e and c is an eigenvalue for g. But the group G is symplectic so 1 c is also an eigenvalue and there exists ẽ such that g(x ẽ) = 1 c x ẽ.

Then g(x e-ẽ) = x e-ẽ and x e-ẽ ∈ Q((x 1 r )). So the difference between the two exponents, eẽ, belongs to Q. Furthermore if there is no ramification (r = 1), eẽ is an integer.

Application to the factorization of linear differential systems

If constructions like exterior powers, exponents in local solutions are obtained from the original local solutions by sums (and adding integers).

For example, in the second exterior powers, the exponents will be (modulo an integer) sums pairwise of exponents of the original solutions; symplecticity garantees that some of these sums will not depend on parameters, thus making computations easier. Similar observations hold for other constructions (on the dual, because symplecticity makes it isomorphic to the original system, on the Endomorphism ring of the differential module, etc).

This makes factorization of symplectic parametrized differential systems much easier because some factors will be constructively computable. This turns out to be powerful in applications -although the success is still not fully explained.

Remark 20 : If the original Hamiltonian system admits a first integral, independent from the Hamiltonian whose initial form is linear (resp. quadratic), then the differential Galois group admits an invariant of degree 1 (resp. 2). In both case, this introduces an invariant element Sym 2 (V) and, by the above result, in the eigenring. So in such a case, the system will be reducible. Similar observations can be found in the book of M. Audin.

Example 1. 5 .

 5 Here are classical examples of linear algebraic groups.

  where c is a constant. (c) We assume that L s 2 ( f ) = 0 has a solution f ∈ C(x) and that disc(P) = 0. Show that the Riccati equation has one or two rational solutions; Compute them, and deduce that L admits the liouvillian solutions y = f e ± √ c 2 f . (d) Conversely, show that if L(y) = 0 has two linearly independent exponential solutions, then the matrices of the Galois group are diagonal matrices and L s 2 (y) = 0 has a rational solution (recall that the Galois group is unimodular) (e) Application : solve y -c 16x 2 y = 0 où c ∈ C.

4.

  Compute the global Galois group, and compare it with the formal local Galois groups. Exercise 14. The Whittaker equation L(y) = y -( 1 4 + 12 λ )y = 0. The exponents at zero are the roots of λX 2 -λX -12 and at infinity ± x 2 .

  , Rod and Singer ( [CRS95, BCRS96], 1996) and Morales and Ramis ([MRR01a], 1998; [MRRS05], 2005) then recast and extended these ideas in terms of the differential Galois group of this variational equation. To quickly summarize, this culminated in the following criterion of Morales and Ramis (from [MRR01a]) :

  ) is proved similarly because the presence of a Stokes multiplier induces a unipotent element in the differential Galois group (see[?] or [?] for definition and properties of Stokes multipliers).

  = ℘(t, g 2 , g 3 ) is the Weierstrass function with parameters The variational equation can then be seen as the direct sum of two Lamé equations (the tangential part and the normal part of the variational equation). After the change of variable x = ℘(t), we get the following normal variational equation:

			t) -	2 3λ
	and ℘(t) g 2 =	4 3	+ 4 h λ and g 3 =	8 27	+	4 3	h λ
	•

Note that as L(y i ) = 0, we have y (n) i and the higher derivatives in K, which really makes it a differential field

AN EXAMPLE: THE FRIEDMAN-ROBERTSON-WALKER MODEL 39

∀y ∈ K, g(y) = g(y


est destinée au dépôt

Implementation of the Magnetic Anisotropy in 2D Finite Element Method Using the Theory of Orientation Distribution Functions

Guilherme Costa Ayres Tolentino, Jean Vianei Leite, Guillaume Parent, N. J.

Batistela

To cite this version: Abstract-In this work the magnetic anisotropy phenomenon in ferromagnetic materials was modeled using orientation distribution function (ODF). The ODF is a methodology that can be used to model the first magnetization curve of electrical steel sheet along a several different angles. The model is inserted in a Finite Element (FE) field calculation software (2D) with potential magnetic vector formulation.

Index Terms-Magnetic anisotropy, Grain oriented steel (GO).

I. ORIENTATION DISTRIBUTION FUNCTION (ODF)

The ODF theory can be employed to model any physical property of a polycrystalline material [START_REF] Bunge | Texture and Magnetic Properties[END_REF]. In particular, Bunge [START_REF] Bunge | Texture and Magnetic Properties[END_REF] states that the ODF theory can be used to express the magnetic induction density inside an electrical steel sheet as a series of i th order :

B(H, θ) = n i=1 A i (H)cos[2(i -1)θ] (1)
where θ is the angle between the magnetic field H and some reference and A i is a coefficient depending on the considered material. In [START_REF] De Campos | Anisotropy of steel sheets and consequence for Epstein test: I theory[END_REF], the authors used (1) at an order equal to 3 to express the magnetic flux density in Grain Oriented (GO) steel sheets. In [START_REF] Jiang | Anisotropy model for modern grain oriented electrical steel based on orientation distribution function[END_REF], the authors showed that a 3 order is not enough to correctly represent the behavior of modern GO steel sheets. They improved the method by expressing coefficients A i as cosine series as well and by determining them by using FFT from experimental data.

In [START_REF] De Campos | Anisotropy of steel sheets and consequence for Epstein test: I theory[END_REF] and [START_REF] Jiang | Anisotropy model for modern grain oriented electrical steel based on orientation distribution function[END_REF], the authors focused on the representation of the behavior of the electrical steel sheets. This communication present the insertion of ODF into a field calculation tool.

II. FINITE ELEMENT COUPLING

In several works the coupling between the material modeling and the FE computation was accomplished through a 2D permeability tensor [START_REF] Bastos | Electromagnetic modeling by finite element methods[END_REF][START_REF] Liu | Anisotropic magnetic material modeling using finite element method[END_REF]. In the present work the ODF model coupling with FE was performed assuming that there is no interdependence between the magnetic permeability for different directions of excitation, which means that outside of the main diagonal of the tensor, the terms of the permeability tensor are null. This simplification must be done, since the ODF model in its origin requires BH curves obtained with alternating fields (usually by using an Epstein's frame). Such a relationship is valid as it considers that the induction and field are collinear. This approach leads to the permeability representation by a scalar, which is dependent of the angle of field with respect to a given origin.

When carrying out FE calculation, for each element on mesh at each time step, the magnetic induction vector inside a finite element is obtained. With the induction module and the angle θ, the ODF coefficients are obtained by B(θ) = µ(θ)H(θ).

With the calculated coefficients it is possible to obtain the corresponding BH curve for any desired angle θ. It is possible to obtain from the generated curve BH, the field component for the respective induction value in the element. With the H tracked on the BH curve generated by the ODF model, one can then find the permeability µ(θ) for the considered θ angle through the defined permeability.

III. RESULTS

For illustration of the methodology here proposed it was implemented an ODF model on MEF, using 10 order approach. Fig. 1a show results for a Rotation Single Sheet Tester (RSST) with a magnetic flux pulsation at 45 • with respect to the rolling direction. The BH curve obtained at the middle of the sample is showed in Fig. 1b. The results are quite promising. More details about the numerical implementation and its challenges will be given in the full paper.

Guilherme

  Costa Ayres Tolentino, Jean Vianei Leite, Guillaume Parent, N. J. Batistela. Implementation of the Magnetic Anisotropy in 2D Finite Element Method Using the Theory of Orientation Distribution Functions. 19th Biennial IEEE Conference on Electromagnetic Field Computation (CEFC 2020), Nov 2020, Pise, Italy. pp.398. hal-04114193

Fig. 1 .

 1 Fig. 1. Results of the simulation the ODF function on FEM

  Implementation of the Magnetic Anisotropy in 2D Finite Element Method Using the Theory of Orientation Distribution Functions G. C. A. Tolentino 1 , J. V. Leite 2 , G. Parent 1 , N. J. Batistela 2

	1 Univ. Artois, EA 4025, Laboratoire Systèmes Électrotechniques et Environnement (LSEE), F-62400 Béthune, France
	2 Universidade Federal de Santa Catarina, GRUCAD, Florianópolis, Santa Catarina, CEP: 88040-970, Brasil
	guilherme costaayrestolentino@ens.univ-artois.fr