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2021

Non-integrability of a three dimensional generalized

Henon-Heiles system

Ognyan ChristovFaculty of Mathematics and Informatics, Sofia University,

5 J. Bouchier blvd., 1164 Sofia, Bulgaria

Abstract

In recent paper Fakkousy et al. show that the 3D Henon-Heiles system with HamiltonianH = 1

2(p21 + p22 + p23) +

12(Aq

21 +Cq22 +Bq23) + (αq21 + γq22)q3 +

β3 q

33 is integrable in sense

of Liouville when α = γ, αβ

= 1, A = B = C; or α = γ, αβ

= 16 , A = C, B-arbitrary;

or α = γ, αβ

= 116 , A = C, A

B= 1

16 (and of course, when α = γ = 0, in which casethe Hamiltonian is separable). It is known that the second case remains integrable forA,C,B arbitrary. Using Morales-Ramis theory, we prove that there are no other casesof integrability for this system.

Keywords: three dimensional Henon-Heiles system, Morales-Ramis theory, Lame typeequation, higher order variational equations

1 Introduction

The generalized two-degrees of freedom Henon-Heiles system (H-H2) is defined by the Hamil-tonian

H =1

2(p21 + p22) +

1

2(Aq21 +Bq22) + αq21q2 +

β

3q32. (1.1)

The original Henon-Heiles Hamiltonian [1] agrees with (1.1) when A = B = α = 1, β = −1.This famous system appears as an important model in a large set of problems in Celestial,Statistical and Quantum Mechanics, see for example the references in [2, 3, 4, 5, 6, 7, 8].

It is well known that the integrable cases of the above Hamiltonian are:

(0) α = 0 – the Hamiltonian is separable;(1) β = α and A = B;(2) β = 6α and A,B - arbitrary;(3) β = 16α and B = 16A.

It seems that Ito [9] was the first who rigorously studied the non-integrability of (1.1) usingthe Ziglin’s monodromy approach (see Sect. 2). Assuming that A = B, he established that

1

there exists an additional independent integral only in the above cases. Later, Morales-Ruiz[10] studied the non-integrability using Differential Galois approach. Again assuming A = B,he showed that (1.1) is integrable in Liouvillian sense only in the above cases. Finally, WenleiLi et al [2] proved using Differential Galois approach without assuming priory A = B that(1.1) is Liouville integrable only in the above cases.

Let us recall some generalizations of (H-H2). In fact, Henon and Heiles have studied in [1]an axisymmetric three dimensional natural Hamiltonian system describing galactic motion.By means of the angular momentum integral they have reduced the considered system toa two-degrees of freedom one and have carried out numerical experiments in searching foradditional integral on the zero level of the angular momentum integral. Later, Ferrer et al[3, 4], again for three dimensional axisymmetric Henon-Heiles systems have studied normalforms, relative equilibria, bifurcations and existence of periodic orbits.

Several multi-dimensional integrable generalizations of the case (2) above are known. InEilbeck et al [5] a Lax pair for such a generalization is given, and hence, the commutingintegrals are obtained. Kostov et al [6] have shown that the complete integrability is preserved

in [5] even considering the additional terms of the kindC2

j

q2j

with arbitrary constants Cj. Zeng

[7] has constructed a hierarchy of multidimensional Henon-Heiles systems with the help ofthe x- and tn- constrained flows of the KdV hierarchy.

In recent paper Fakkousy et al. [8] have studied for integrability the following generaliza-tion of the Henon-Heiles system (H-H3)

H =1

2(p21 + p22 + p23) +

1

2(Aq21 + Cq22 +Bq23) + (αq21 + γq22)q3 +

β

3q33, (1.2)

assuming further that A = C and α = γ. Due to this symmetry there exists an additionalintegral, this is clear after introducing spherical coordinates. On the zero level of that integralthe Hamiltonian (1.2) coincides with the (H-H2) Hamiltonian (1.1). Then, starting fromthe integrable cases listed above, the authors have succeeded in finding a complete set ofcommuting first integrals for (H-H3) in the following cases:

(i) α = γ, αβ= 1, A = B = C;

(ii) α = γ, αβ= 1

6, A = C, B-arbitrary;

(iii) α = γ, αβ= 1

16, A = C, A

B= 1

16.

Again we have to add to the above list the case α = γ = 0 in which the Hamiltonian (1.2)is separable. As we point out earlier, the integrability in the second case (ii) can be extendedfor A,B,C arbitrary.

Numerical study of (H-H3), carried out in [8] near the cases of integrability reveals chaoticbehavior which prevents integrability. But there is still a possibility for existence of otherintegrable cases far from those already found. The motivation for our study is to establishwhether this is true. Without assuming a priori that A = C and α = γ we establish thefollowing result.

Theorem 1. The three dimensional Henon-Heiles system corresponding to (1.2) is non-integrable by means of meromorphic first integrals except for the cases given above.

2

The paper is organized as follows. In Section 2 we recall some notions and results fromwhat is now called Ziglin-Morales-Ramis theory, including its generalization with higher ordervariational equations. For the reader’s convenience, we also add two appendices in which wesummarize known results concerning necessary conditions for integrability of Hamiltoniansystems with homogeneous potentials and of Hamiltonian systems with variational equationsof Lame type. In Section 3, the proof of Theorem 1 is carried out in several steps.

First, we study the necessary conditions for integrability of the homogeneous Hamiltonianrelated to (1.2). Making use of the results in Appendix A1, we narrow the range of the valuesof the parameters, for which an additional integral would exist. In Step 2, already for theinitial Hamiltonian (1.2), we study the necessary conditions for integrability of the variationalequations along a particular solution, which are of Lame type. In this way, we recover theintegrability case (iii). Still, there are values of the parameters for which we can not able toconclude non-integrability. To resolve these cases, in Step 3 we derive the higher variationalequations up to order 3. Then, we use these equations in Steps 4 and 5, and along the way werecover the integrable case (i). Finally, in Step 6 we present the additional to H commutingintegrals in the integrable case (ii) for A,B,C arbitrary.

Thus, we get a complete answer to the question about integrability and non-integrabilityof the considered system.

2 Theory

In this section, we summarize some notions and results related to Ziglin-Morales-Ramis the-ory. We refer the reader to [10, 11, 12]) for a more complete exposition on Differential Galoisapproach to integrability of Hamiltonian systems. A very detailed presentation about Differ-ential Galois Theory may be found in the book of van der Put and Singer [13]).

Given an analytic Hamiltonian H , defined on a complex 2n-dimensional manifold Mdetermining the system

x = XH(x), t ∈ C, x ∈ M (2.1)

Recall that such Hamiltonian system is completely integrable (or integrable in Liouville sense)if it admits n independent first integrals F1 = H,F2, . . . , Fn in involution.

Assume that the system (2.1) has a non-equilibrium solution ϕ(t), t ∈ I ⊂ C. The imageof I by ϕ is a Riemann surface Γ. We can write the equation in variation (VE) along thissolution

ξ = DXH(ϕ(t))ξ, ξ ∈ TΓM. (2.2)

Further, using the integral dH we can reduce the variational equation. Consider the normalbundle of Γ, F := TΓM/TM and let π : TΓM → F be the natural projection. The equation(2.2) induces an equation on F

η = π∗(DXH(ϕ(t))(π−1η), η ∈ F . (2.3)

which is called the normal variational equation (NVE). Each meromorphic first integral ofthe Hamiltonian system (2.1) in a vicinity of Γ gives rise to a meromorphic first integral of

3

(NVE) [14, 15]. Therefore, the problem of complete integrability of (2.1) reduces to the studyof integrability of the linear systems (2.2) or (2.3).

Consider such a linear system

x = A(t)x, x ∈ Cn, t ∈ Γ. (2.4)

The continuation of the solutions along nontrivial loops on Γ defines a linear automorphismsof the space of solutions, called monodromy transformations. More precisely, let Y (t) bea fundamental solution of (2.4), analytic in some neighborhood of any non-singular pointt0. The linear automorphism ∆γ , associated with the loop γ ∈ π1(Γ, t0) corresponds to amultiplication of Y (t) from the right by a constant matrix Mγ - monodromy matrix.

∆γY (t) = Y (t)Mγ .

The set of these matrices constitute the monodromy group.In 1982, Ziglin [14, 15] used the relation between the monodromy group of (VE) or (NVE)

and branching of solutions to obtain necessary conditions for integrability of complex Hamil-tonian systems. Let us mention one more application of Ziglin’s theory which is related toour study: in 1987, Yoshida [16] found a criterion for non-existence of an additional integralin two degrees of freedom Hamiltonian systems with a homogeneous potential.

Morales-Ruiz and Ramis extended the Ziglin’s approach in studying the integrability ofHamiltonian systems by investigating the structure of the differential Galois group of (NVE)( or (VE)) along certain non-equilibrium solution. We briefly recall some notions and facts.

Denote the coefficient field in (2.4) by K. A differential field K is a field with derivation∂ =′, i.e. an additive mapping satisfying Leibnitz rule. A differential automorphism of K isan automorphism commuting with the derivation. Let yij be the elements of the fundamentalmatrix Y (t). Let L(yij) be the extension ofK generated byK and yij – a differential field. Thisextension is called Picard-Vessiot extension. Similarly to classical Galois Theory we define theGalois group G := GalK(L) = Gal(L/K) to be the group of all differential automorphisms ofL leaving the elements of K fixed. The Galois group is, in fact, an algebraic group. It has aunique connected component G0 which contains the identity and which is a normal subgroupof finite index. The Galois group G can be represented as an algebraic linear subgroup ofGL(n,C) by

σ(Y (t)) = Y (t)Rσ,

where σ ∈ G and Rσ ∈ GL(n,C) (see e.g. [13]).The solutions of (2.2) define an extension L1 of the coefficient field K of (VE). This

naturally defines a differential Galois groupG = Gal(L1/K). Then, the following fundamentalresult has been established

Theorem 2. (Morales-Ruiz-Ramis [10]) Suppose that a Hamiltonian system has n mero-morphic first integrals in involution. Then the identity component G0 of the Galois groupG = Gal(L1/K) is abelian.

If it turns out that G0 is not abelian, the studied Hamiltonian system is non-integrable inthe Liouville sense. But since the above result is only a necessary condition if G0 is abelianthat doesn’t imply integrability.

4

To obtain other obstacles to the integrability, a method based on the higher variationalequations has been introduced in [10] and the previous Theorem has been extended in [11].Before formulating this result let us give an idea of higher variational equations. For thesystem (2.2) with a particular solution ϕ(t) we put

x = ϕ(t) + εξ(1) + ε2ξ(2) + . . .+ εkξ(k) + . . . , (2.5)

where ε is a formal small parameter. Substituting the above expression into (2.1) and compar-ing terms with the same order in ε we obtain the following chain of linear non-homogeneousequations

ξ(k) = A(t)ξ(k) + rk(ξ(1), . . . , ξ(k−1)), k = 1, 2, . . . , (2.6)

where A(t) = DXH(ϕ(t)) and r1 ≡ 0. The equation (2.6) is called k-th variational equation(VEk). Let Y (t) be the fundamental matrix of (VE1)

Y = A(t)Y.

Then the solutions of (VEk), k > 1 can be found by

ξ(k) = Y (t)c(t), (2.7)

where c(t) is a solution ofc = Y −1(t)rk. (2.8)

As we mention above, the higher variational equations (VEk) are not homogeneous equations,but they can be made such, and therefore, one can define successive extensions K ⊂ L1 ⊂L2 ⊂ . . . ⊂ Lk, where Lk is the extension obtained by adjoining the solutions of (VEk).Then, naturally one can define the Galois groups Gal(L1/K), . . . , Gal(Lk/K). The followingtheorem is proven in [11].

Theorem 3. If the Hamiltonian system (2.1) is integrable in Liouville sense then the identitycomponent of every Galois group Gal(Lk/K) is abelian.

Notice that we apply Theorem 3 in the situation when the identity component of theGalois group Gal(L1/K) is abelian. This means that the first variational equation is solvable.Once we have the solution of (VE1), then the solutions of (VEk) can be found successivelyby the method of variations of constants in the way explained above. In turn, this impliesthat the Galois groups Gal(Lk/K) are solvable. One possible way to show that some of themis not abelian is to find a logarithmic term in the corresponding local solution (see detaileddescriptions and explanations in [10, 11, 12]).

3 Proof of Theorem 1

For the proof we use the strategy from [2] applied in studying the non-integrability of the twodimensional Henon-Heiles system.

5

Any natural Hamiltonian with a polynomial potential can be represented in the form

H =1

2

∑

p2j + V (q) =1

2

∑

p2j + Vmin(q) + . . .+ Vmax(q), (3.1)

where Vmin(q)(Vmax(q)) is the lowest (highest) order term of V (q). It is established by Hi-etarinta [17] and Maciejevski, Przybylska [18] that if the Hamiltonian system defined by H isintegrable, then integrable are its subsystems

Hmin =1

2

∑

p2j + Vmin(q), Hmax =1

2

∑

p2j + Vmax(q).

On the other hand, if it turns out that some of the above subsystems is non-integrable, thenthe original Hamiltonian system is non-integrable.

In our case Hmin is trivially integrable, so we consider

H = T + Vmax =1

2(p21 + p22 + p23) + (αq21 + γq22)q3 +

β

3q33, (3.2)

where α, β, γ are assumed nonzero. After rescaling t we get

H =1

2(p2x + p2y + p2z) + (α1x

2 + γ1y2)z +

1

3z3 (3.3)

with α1 = α/β and γ1 = γ/β.First, we study the integrability of the Hamiltonian system corresponding to the Hamil-

tonian (3.3).

Step 1. Necessary conditions for the integrability of (3.3)

The equations of motion for the Hamiltonian (3.3) are

q1 = −2α1q1q3, q2 = −2γ1q2q3, q3 = −α1q21 − γ1q

22 − q23. (3.4)

We are looking for a particular solution of the form

q1 = c1u(t), q2 = c2u(t), q3 = c3u(t), (3.5)

where

u = −u2, u2 =2

3(1− u3). (3.6)

Then, c = (c1, c2, c3)T satisfies the following system

c1 = 2α1c1c3, c2 = 2γ1c2c3, c3 = α1c21 + γ1c

22 + c23. (3.7)

There are several possibilities for solutions of (3.7) which we are going to use:

1) c1 = 0, c2 =1

2γ1

√

2− 1γ1, c3 =

12γ1

.

6

The variational equations (VE) along the found particular solution read

ξ = −u(t)V ′′(c)ξ, (3.8)

where ξ = (ξ1, ξ2, ξ3)T and

V ′′(c) =

2α1c3 0 2α1c10 2γ1c3 2γ1c2

2α1c1 2γ1c2 2c3

=

α1

γ10 0

0 1√

2− 1γ1

0√

2− 1γ1

1γ1

. (3.9)

The eigenvalues of V ′′(c) are

λ1 =α1

γ1, λ2 =

1

γ1− 1, λ3 = 2. (3.10)

2) c2 = 0, c1 =1

2α1

√

2− 1α1

, c3 =1

2α1

.

The eigenvalues of V ′′(c) in this case are

λ1 =γ1α1

, λ2 =1

α1− 1, λ3 = 2. (3.11)

3) c1 = 0, c2 = 0, c3 = 1.The eigenvalues of V ′′(c) in this case are

λ1 = 2α1, λ2 = 2γ1, λ3 = 2. (3.12)

Now, if the Hamiltonian system is integrable, the pairs (k = 3, λj) have to belong to oneof the following cases (1), (11), (12), (13), (14), (18) of Theorem 4 (see Appendix A1).

Denote

g1(s) = s+ s(s− 1)3/2, g2(s) = − 1

24+

1

24(2 + 6s)2, g3(s) = − 1

24+

1

24

(

3

2+ 6s

)2

,

g4(s) = − 1

24+

1

24

(

6

5+ 6s

)2

, g5(s) = − 1

24+

1

24

(

12

5+ 6s

)2

, g6(s) =1

3+

3

2s(s+ 1)

and observe that gj(s) > 0 for s ∈ Z. Then, if λ2 = gj(s) for some integer s in (3.10), (3.11)we get that

α1, γ1 ∈ (0, 1], α1, γ1 ∈ Q. (3.13)

Now, to obtain necessary conditions for existence of additional first integral we incorporatethe condition (A.8) from Theorem 5 (see Appendix A1) for λ1, λ2 in any of above possibilities.We have

1

6

√

1 + 24α1

γ1=

1

6

√

1 + 24

(

1

γ1− 1

)

+ l1, (3.14)

7

1

6

√

1 + 24γ1α1

=1

6

√

1 + 24

(

1

α1− 1

)

+ l2, (3.15)

1

6

√1 + 48α1 =

1

6

√

1 + 48γ1 + l3, (3.16)

for some l1, l2, l3 ∈ Z. It is known from [19] that all radicals in these expressions are in factrational numbers.

Consider the relation (3.16). Due to (3.13)

1 <√1 + 48α1,

1

6

√

1 + 48γ1 ≤ 7.

Hence, (3.16) is valid only for l3 = 0, but this gives

γ1 = α1.

Alternatively, the equality γ1 = α1 can be obtained also by exploring the necessaryconditions from Theorem 4. Suppose λ1 = α1

γ1:= w from (3.10) takes values in some

gi(s), i = 1, . . . , 6, s ∈ Z, i.e., w = gi(s). Similarly, for integrability λ1 = γ1α1

:= 1w

from

(3.11) also has to take values in some gj(q), j = 1, . . . , 6, q ∈ Z, that is, 1w= gj(q). Then we

must have

gj(q) =1

gi(s)

for some i, j = 1, . . . , 6 and some q, s ∈ Z. But this can be achieved only for i = j = 1 andq = s = 1, which implies that w = 1, or γ1 = α1.

This result has two major consequences: firstly, the Hamiltonian system (3.4) admits anadditional integral F = q1p2 − q2p1, and secondly, we get a representation from (3.15) for α1

for which we may have complete integrability, namely

5

6=

1

6

√

1 + 24

(

1

α1− 1

)

+ l, l ∈ Z, (3.17)

which is equivalent to

1

α1− 1 = − 1

24+

1

24(5− 6l)2 or α1 =

24

23 + (5− 6l)2. (3.18)

We will use the above representation to obtain those α1 for which the Hamiltonian system(3.4) is necessarily integrable by investigating whether there are some s, l ∈ Z, such that λ2

from (3.11) and (3.12) takes values in the six families gj(s), j = 1, 2, . . . , 6. Starting with λ2

from (3.11) we have

1

α1− 1 = − 1

24+

1

24(5− 6l)2 = gj(s), j = 1, . . . , 6. (3.19)

Trivial computations show that there are no such integers s, l for j = 2, . . . , 6. However,

− 1

24+

1

24(5− 6l)2 = g1(s) =

3

2s2 − 1

2s

is reduced to l + s = 1, which in turn gives

8

• s = 0, l = 1, and therefore, α1 = 1;

• s = 1, l = 0, and therefore, α1 =12.

Further, we turn to λ2 from (3.12). Using that α1 = γ1 and (3.18), we have to find s, l ∈ Z

for which the following relation

2α1 =48

23 + (5− 6l)2= gj(s), for some s, l ∈ Z. (3.20)

is fulfilled. Since 0 < 4823+(5−6l)2

≤ 1, we should have 0 < gj(s) ≤ 1 and this gives a very fewpossibilities for s. Indeed,

48

23 + (5− 6l)2= g1(s) =

3

2s2 − 1

2s

is fulfilled for s = 0 and l = 1, which amounts to α1 = 1/2, but we know that.Next,

48

23 + (5− 6l)2= g2(s) = − 1

24+

1

24(6s+ 2)2

is fulfilled for s = 0 and l = 4, which gives α1 =116. There are no s, l ∈ Z which satisfy

48

23 + (5− 6l)2= gj(s), j = 3, 4, 5.

Finally,48

23 + (5− 6l)2= g6(s) =

1

2

[2

3+ 3s(s+ 1)

]

is fulfilled for s = 0 and l = −1, and hence, α1 =16.

Summarizing, the necessary conditions for the system (3.4) to be integrable are α1 = γ1and α1 = 1, 1

2, 16, 116. Therefore, we have the following

Proposition 1. The Hamiltonian system (3.4) corresponding to the Hamiltonian with thehomogeneous potential (3.3) is non-integrable if

(i) γ1 6= α1, or(ii) γ1 = α1, α1 ∈ C \

{

1, 12, 16, 116

}

�

Now, consider the original Hamiltonian (1.2) assuming that A,B and C are nonzero.Again after rescaling t we get

H =1

2(p21 + p22 + p23) +

1

2(A1q

21 + C1q

22 +B1q

23) + (α1q

21 + γ1q

22)q3 +

1

3q33, (3.21)

where A1 = A/β,B1 = B/β, C1 = C/β and α1 = γ1 are as before. The Hamiltonian systemcorresponding to (3.21) reads

q1 = −A1q1 − 2α1q1q3, q2 = −C1q2 − 2α1q2q3, q3 = −B1q3 − α1(q21 + q22)− q23 . (3.22)

9

The above system admits the following particular solution

Γh : q1 = q2 = p1 = p2 = 0, q3 = q03 = −1

2B1 + ℘(τ, g2, g3), p3 = q03, (3.23)

where τ = i√6t, i2 = −1, g2 = 3B2

1 , g3 = 12h− B31 .

The variational equation (VE) along this solution, written with respect to the independentvariable τ , (′ = d/dτ) are

ξ′′1 − [12α1℘(τ) + 6(A1 − B1α1)]ξ1 = 0, (3.24)

ξ′′2 − [12α1℘(τ) + 6(C1 −B1α1)]ξ2 = 0, (3.25)

ξ′′3 − 12℘(τ)ξ3 = 0. (3.26)

Clearly, the equation (3.26) forms the tangential part of (VE), while the equations (3.24) and(3.25) form (NVE). Notice that (NVE) are given by two independent second order equationsof Lame type.

Step 2. Analysis of (NVE)

In what follows, we apply the results described in the Appendix A.2. Consider first (3.24)written as

ξ′′1 − f1(τ, h)ξ1 = 0

with f1(τ, h) = 12α1℘(τ) + 6(A1 − B1α1). After some calculations we get

f ′21 = P (f1, h) =

1

3α1

f 31 +

6

α1

(A1 − B1α1)f21 + [−36α1B

21 +

36

α1

(A1 − B1α1)2]f1 (3.27)

− 216α1B21(A1 − B1α1) +

72

α1

(A1 −B1α1)3 + (12)2α2

1B31 − h(12)3α2

1.

We write down the coefficients of the polynomial P (f1, h) in a way we need them

a1 =1

3α1

, a2 = 0,

b1 =6

α1

(A1 +B1α1), b2 = 0,

c1 = −36α1B21 +

36

α1

(A1 − B1α1)2, c2 = 0,

d1 = −216α1B21(A1 −B1α1) +

72

α1

(A1 − B1α1)3 + (12)2α2

1B31 , d2 = −(12)3α2

1.

Now, we explore the necessary conditions for integrability as they are given in the Appen-dix A2. The first necessary condition from Theorem 5 – I. a1 = 4

n(n+1)for some n ∈ N gives

that α1 =n(n+1)

12for some n ∈ N. Recall that we have to deal only with the cases

n = 1

(

α1 =1

6

)

, n = 2

(

α1 =1

2

)

, and n = 3 (α1 = 1) (3.28)

10

because of Proposition 1.Next, we proceed with the cases of condition II.The case II1. m = 1 gives that α1 =

116, while the condition b1 = 0 amounts to

A1

B1

= α1 =1

16

(

A

B=

1

16

)

. (3.29)

The case II2. m = 2 gives α1 = 5/16, but for values of α1 different from {1, 1/2, 1/6, 1/16} wehave non-integrability again due to the results from Step 1.

The case II3. m = 3 does not occur here since c2 = 0 and 16a1d2+3b1c2 = 0 yield a1d2 = 0,which is not possible.

The case IIm. m > 3 is also not possible due to d2 6= 0 and the assumptions A1 6= 0 andB1 6= 0.

Finally, the case III does not occur, because after expressing all necessary conditions weget B1 = 0 which is a contradiction to our assumption.

Similarly, (3.25) can be written as

ξ′′2 − f2(τ, h)ξ1 = 0

with f2(τ, h) = 12α1℘(τ) + 6(C1 − B1α1). The polynomial P (f2, h) is exactly as (3.27) withA1 replaced by C1. The necessary conditions I, II2, II3, IIm and III give nothing new. In thecase II1 where m = 1 and α1 = 1/16 we obtain from the condition b1 = 0 that

C1

B1= α1 =

1

16

(

C

B=

1

16

)

. (3.30)

This condition, together with (3.29) results in

A = C,A

B=

1

16, α = γ,

α

β=

1

16, (3.31)

but we know that in this case the Hamiltonian (1.2) is integrable.Summarizing, we get a more definitive answer for the case α1 = 1/16: together with

already established condition α1 = γ1(α = γ) the Hamiltonian (1.2) is necessarily integrable ifA1 = C1(A = C), but this gives the already known integrable case (iii) from the Introduction.

Step 3. Higher variational equations

Further, it remains to deal with the cases (3.28). In any of these cases the Galois groupof variational equation is abelian (see Apppendix A.2). Hence, we need to study the Galoisgroups of higher variational equations. As it was explained in Section 2, to write them weput

q1 = 0 + εξ(1)1 + ε2ξ

(2)1 + ε3ξ

(3)1 + . . . , (3.32)

q2 = 0 + εξ(1)2 + ε2ξ

(2)2 + ε3ξ

(3)2 + . . . ,

q3 = q03 + εξ(1)3 + ε2ξ

(2)3 + ε3ξ

(3)3 + . . .

11

and substitute these expressions into (3.22). Comparing the terms with the same order in ε,we obtain the variational equations up to order three.

We are going to write these higher variational equations as second-order linear equationswith respect to the independent variable τ . The first variational equation (VE1) is nothingbut (3.24) – (3.26)

(ξ(1)1 )′′ − [12α1℘(τ) + 6(A1 − B1α1)]ξ

(1)1 = 0, (3.33)

(ξ(1)2 )′′ − [12α1℘(τ) + 6(C1 − B1α1)]ξ

(1)2 = 0, (3.34)

(ξ(1)3 )′′ − 12℘(τ)ξ

(1)3 = 0. (3.35)

For the second variational equation (VE2) we get

(ξ(2)1 )′′ − [12α1℘(τ) + 6(A1 −B1α1)]ξ

(2)1 = W

(2)1 , (3.36)

(ξ(2)2 )′′ − [12α1℘(τ) + 6(C1 − B1α1)]ξ

(2)2 = W

(2)2 , (3.37)

(ξ(2)3 )′′ − 12℘(τ)ξ

(2)3 = W

(2)3 , (3.38)

whereW

(2)j = 12α1ξ

(1)j ξ

(1)3 , j = 1, 2, W

(2)3 = 6α1[(ξ

(1)1 )2 + (ξ

(1)2 )2] + 6(ξ

(1)3 )2. (3.39)

The third variational equation (VE3) is

(ξ(3)1 )′′ − [12α1℘(τ) + 6(A1 −B1α1)]ξ

(3)1 = W

(3)1 , (3.40)

(ξ(3)2 )′′ − [12α1℘(τ) + 6(C1 − B1α1)]ξ

(3)2 = W

(3)2 , (3.41)

(ξ(3)3 )′′ − 12℘(τ)ξ

(3)3 = W

(3)3 . (3.42)

HereW

(3)j = 12α1(ξ

(1)j ξ

(2)3 + ξ

(2)j ξ

(1)3 ), j = 1, 2. (3.43)

We do not need W(3)3 , but it is calculated to be W

(3)3 = 12α1(ξ

(1)1 ξ

(2)1 + ξ

(1)2 ξ

(2)2 ) + 12ξ

(1)3 ξ

(2)3 .

Keeping our notation from Section 2 we have

r2 = (0,W(2)1 , 0,W

(2)2 , 0,W

(2)3 ), r3 = (0,W

(3)1 , 0,W

(3)2 , 0,W

(3)3 ). (3.44)

To prove non-integrability of (3.21) we have to show that the identity component of theGalois group of (VEk, k = 2, 3) is not abelian. To get this let us note that the variationalequations have a singular point at τ = 0. Then it is enough to obtain a logarithm around thesingular point, or equivalently, a residue different from zero in some integrand (see [10]). Nowwe are going to calculate the local solutions around τ = 0 and to show that a logarithmicterm appears in the solutions of (VE3). Notice that all expansions below are convergent insome vicinity of τ = 0 [20].

Let ξ(1)j,1 , ξ

(1)j,1 , j = 1, 2, 3 be two linearly independent solutions of (3.33)-(3.35) with unit

Wronskian. Then, the fundamental matrix Y (τ) of (VE1) and its inverse have the block-diagonal form

Y (τ) =

Y1 0 00 Y2 00 0 Y3

, Y −1(τ) =

Y −11 0 00 Y −1

2 00 0 Y −1

3

, (3.45)

12

where

Yj =

(

ξ(1)j,1 ξ

(1)j,2

(ξ(1)j,1 )

′ (ξ(1)j,2 )

′

)

, Y −1j =

(

(ξ(1)j,2 )

′ −ξ(1)j,2

−(ξ(1)j,1 )

′ ξ(1)j,1

)

. (3.46)

Now, we return to the integrability analysis of the cases (3.28).

Step 4. Non-integrability of the case n = 3, α1 = 1.

One can find the following expansions of the solutions of (VE1) (3.33), (3.34) and (3.35)with α1 = 1.

ξ(1)1,1 =

1

τ 3

[

1− 3

5(A1 − B1) τ

2 +3

10

(

A21 − 2A1B1 +

B21

2

)

τ 4 + . . .]

,

ξ(1)1,2 =

τ 4

7

[

1 +1

3(A1 − B1)τ

2 + . . .]

. (3.47)

Similarly, we have

ξ(1)2,1 =

1

τ 3

[

1− 3

5(C1 − B1) τ

2 +3

10

(

C21 − 2C1B1 +

B21

2

)

τ 4 + . . .]

,

ξ(1)2,2 =

τ 4

7

[

1 +1

3(C1 − B1)τ

2 + . . .]

(3.48)

and also,

ξ(1)3,1 =

1

τ 3− g2

20τ − g3

14τ 3 +

g22400

τ 5 + . . . , ξ(1)3,2 =

τ 4

7

[

1 +3

220g2τ

4 + . . .]

. (3.49)

There are no logarithmic terms in the expansions around τ = 0 of the local solutions of(VE2) and in what follows we write down the expansions of their local solutions only in thecases we will use further.

For the first equation of (VE2) with a specific choice of the right-hand terms

(ξ(2)1 )′′ − [12α1℘(τ) + 6(A1 − B1)]ξ

(2)1 = 12ξ

(1)1,2ξ

(1)3,1 (3.50)

we find the following expansions around τ = 0

ξ(2)1,j = ξ

(1)1,j + τ 3

[

− 2

7− 1

7(A1 − B1)τ

2 + . . .]

, j = 1, 2. (3.51)

Similarly, for the second equation

(ξ(2)2 )′′ − [12α1℘(τ) + 6(C1 − B1)]ξ

(2)2 = 12ξ

(1)2,2ξ

(1)3,1 (3.52)

we obtain

ξ(2)2,j = ξ

(1)2,j + τ 3

[

− 2

7− 1

7(C1 −B1)τ

2 + . . .]

, j = 1, 2. (3.53)

13

Finally, for the third equation in (VE2) with the following choice of the right-hand terms

(ξ(2)3 )′′ − 12℘(τ)ξ

(2)3 = 6

[

(ξ(1)1,1)

2 + (ξ(1)2,1)

2 + (ξ(1)3,1)

2]

(3.54)

we get the expansions

ξ(2)3,j = ξ

(1)3,j +

1

τ 4

[9

4+

6

5(C1 + A1 − 2B1)τ

2 +K4τ4 + . . .

]

, j = 1, 2, (3.55)

where K4 is

K4 := −339

400B2

1 −12

25C2

1 +24

25A1B1 +

24

25C1B1 −

12

25A2

1.

Now we will study the local solutions of (VE3). Consider the first equation with thefollowing specific representatives on the right-hand side.

(ξ(3)1 )′′ − [12α1℘(τ) + 6(A1 −B1)]ξ

(3)1 = 12

(

ξ(1)1,1ξ

(2)3,2 + ξ

(2)1,2ξ

(1)3,2

)

= W(3)1 . (3.56)

Denote

(ν(3)1,1 , ν

(3)1,2) = Y −1

1

(

0

W(3)1

)

.

Then, for ν(3)1,1 = −12ξ

(1)1,2

[

ξ(1)1,1ξ

(2)3,2 + ξ

(2)1,2ξ

(1)3,2

]

we obtain

Resτ=0 ν(3)1,1 = −36

35(A1 + 2C1 − B1). (3.57)

Similar computations for (3.41)

(ξ(3)2 )′′ − [12α1℘(τ) + 6(C1 − B1)]ξ

(3)2 = 12

(

ξ(1)2,1ξ

(2)3,2 + ξ

(2)12,2ξ

(1)3,2

)

= W(3)2 (3.58)

give

(ν(3)2,1 , ν

(3)2,2) = Y −1

2

(

0

W(3)2

)

and

Resτ=0 ν(3)2,1 = −36

35(C1 + 2A1 − B1). (3.59)

Suppose for a moment that the both residues (3.57) and (3.59) are zero, that is,

A1 + 2C1 − 3B1 = 0, C1 + 2A1 − 3B1 = 0.

The only solution of this system is A1 = B1 = C1(A = B = C), but for these values of theparameters we recover the already known integrable case (i) from the Introduction.

On the other hand, if A1 6= B1 6= C1, then at least one of the above residues is non-trivial. Thus, we have obtained a nonzero residue at τ = 0, which implies a logarithm in thesolutions of (VE3). Then, the Galois group of (VE3) is solvable, but not abelian. Hence, thenon-integrability of the Hamiltonian system (3.22) in this case follows from Theorem 3.

14

Step 5. Non-integrability of the case n = 2, α1 =12.

Here we will show that for α1 = 1/2 the Hamiltonian system (3.22) is not integrable,whatever the other parameters are. The same line of considerations as in the previous Stepis taken.

One can find the following expansions of the solutions of (VE1) (3.33), (3.34) and (3.35)with α1 =

12.

ξ(1)1,1 =

1

τ 2

[

1−(

A1 −B1

2

)

τ 2 +3

2

(

A21 −A1B1 +

B21

10

)

τ 4 + . . .]

,

ξ(1)1,2 =

τ 3

5

[

1 +3

7

(

A1 −B1

2

)

τ 2 + . . .]

. (3.60)

To get the expansions of the solutions of (3.34) we only need to replace A1 with C1.

ξ(1)2,1 =

1

τ 2

[

1−(

C1 −B1

2

)

τ 2 +3

2

(

C21 − C1B1 +

B21

10

)

τ 4 + . . .]

,

ξ(1)2,2 =

τ 3

5

[

1 +3

7

(

C1 −B1

2

)

τ 2 + . . .]

. (3.61)

Notice that we already have the expansion for ξ(1)3 in (3.49).

There are no logarithmic terms in the expansions around τ = 0 of the local solutions of(VE2). We calculate the expansions of the solutions of (VE2) only in the cases we need in thesequel.

For the first equation of (VE2) with a specific choice of the right-hand terms

(ξ(2)1 )′′ −

[

6α1℘(τ) + 6

(

A1 −B1

2

)

]

ξ(2)1 = 6ξ

(1)1,1ξ

(1)3,2 (3.62)

we find the following expansions

ξ(2)1,j = ξ

(1)1,j + τ 4

[1

7+

1

14(A1 − B1/2)τ

2 + . . .]

, j = 1, 2. (3.63)

Similarly, for the second equation

(ξ(2)2 )′′ −

[

6α1℘(τ) + 6

(

C1 −B1

2

)

]

ξ(2)2 = 6ξ

(1)2,1ξ

(1)3,2 (3.64)

we obtain

ξ(2)2,j = ξ

(1)1,j + τ 4

[1

7+

1

14(C1 − B1/2)τ

2 + . . .]

, j = 1, 2. (3.65)

Finally, for the third equation in (VE2) with the following choice of the right-hand terms

(ξ(2)3 )′′ − 12℘(τ)ξ

(2)3 = 3

[

(ξ(1)1,1)

2 + (ξ(1)2,1)

2]

+ 6(ξ(1)3,2)

2 (3.66)

15

we get the expansions

ξ(2)3,j = ξ

(1)3,j +

1

τ 2

[

− 1 +1

2(A1 −B1 + C1)τ

2 + . . .]

, j = 1, 2. (3.67)

It is enough for our purposes to consider only the first equation of (VE3) with the followingchoice of the right-hand terms.

(ξ(3)1 )′′ −

[

6α1℘(τ) + 6

(

A1 −B1

2

)

]

ξ(3)1 = 6

[

ξ(1)1,2ξ

(2)3,2 + ξ

(2)1,2ξ

(1)3,1

]

= W(3)1 . (3.68)

Denoting again

(ν(3)1,1 , ν

(3)1,2) = Y −1

1

(

0

W(3)1

)

we have, this time for ν(3)1,2 = 6ξ

(1)1,1

[

ξ(1)1,2ξ

(2)3,2 + ξ

(2)1,2ξ

(1)3,1

]

Resτ=0 ν(3)1,2 = −12

356= 0. (3.69)

Thus, we have a nonzero residue at τ = 0, which implies a logarithm in the solutions of (VE3).Then, the Galois group of (VE3) is solvable, but not abelian. Again, the non-integrability ofthe Hamiltonian system (3.22) in this case follows from Theorem 3.

Step 6. Integrability of the case n = 1, α1 =16.

As we point out in the Introduction this integrable case has a multi-dimensional genera-lization. Here we merely present the commuting integrals in our three-dimensional case forA,B,C arbitrary.

With γ = α, β = 6α and after rescaling t the Hamiltonian (1.2) reads

H =1

2(p21 + p22 + p23) +

1

2(Aq21 + Cq22 + Bq23) + (q21 + q22)q3 + 2q33, (3.70)

where A = A/α, B = B/α, C = C/α. Then the corresponding first integrals are (see e.g.[5, 6])

G1 =(q2p1 − q1p2)

2

A− C− q41 − q21q

22 − 4p3q1p1 + (B − 4A+ 4q3)(p

21 + Aq21), (3.71)

G2 = −(q2p1 − q1p2)2

A− C− q42 − q21q

22 − 4p3q2p2 + (B − 4C + 4q3)(p

22 + Cq22).

This completes the proof of Theorem 1.�

Acknowledgements.This work is partially supported by grant DN 02-5 of Bulgarian Fund ”Scientific Research”.

16

A Some applications of the Morales-Ramis theory

A.1 Homogeneous potentials

In this part we follow mainly the treatment in [10], which is a generalization of a non-integrability theorem by Yoshida [16] for the case n = 2 based on Ziglin’s theorem.

Consider a Hamiltonian system with n - degrees of freedom (n ≥ 2) governed by a naturalHamiltonian

H = T + V =1

2(y21 + . . .+ y2n) + V (x1, x2, . . . , xn), (A.1)

where V is a homogeneous function of degree k 6= 0.Taking advantage of homogeneity, it is possible to obtain a particular solution of the form

Γ : xj = u(t)cj, yj = u(t)cj , j = 1, . . . , n, (A.2)

where u(t) is a solution of the hyperelliptic equation u2 = 2k(1− uk) and c = (c1, . . . , cn) is a

solution of the nonlinear systemc = V ′(c). (A.3)

Such solutions are called Darboux points. Then the (VE) along Γ are given by

ξ = −u(t)k−2V ′′(c)ξ. (A.4)

Since V ′′(c) is diagonalizable, we can write (A.4) as a direct sum of second order equations

ξj = −u(t)k−2λjξj, j = 1, 2, . . . , n, (A.5)

where λj are the eigenvalues of V ′′(c) (usually called Yoshida coefficients).In order to get (NVE) we rule out the equation corresponding to λn = k − 1

η = −u(t)k−2diag(λ1, . . . , λn−1)η, (A.6)

with η = (ξ1, . . . , ξn−1).Further, we change the independent variable by x := (u(t))k and obtain a system of

hypergeometric differential equations

x(1 − x)d2ξjdx2

+

(

k − 1

k− 3k − 2

2kx

)

dξjdx

+λj

2kξj = 0, j = 1, . . . , n− 1. (A.7)

It is clear that the identity component of the Galois group of (A.6) is solvable (abelian) if,and only if, each one of the identity components of the Galois groups of the hypergeometricequations (A.7) is solvable (abelian).

Finally, exploring the Galois groups of the hypergeometric equations (A.7) the followingresult is obtained.

17

Theorem 4. (Theorem 5.1 [10]) If the Hamiltonian system with the Hamiltonian (A.1) iscompletely integrable with holomorphic or meromorphic first integrals, then each pair (k, λj)belongs to one of the following lists

(1) (k, s + s(s-1)k/2) (2) (2,C)(3) (−2,C) (4) (−5, 49

40− 1

40(103+ 10s)2)

(5) (−5, 4940

− 140(4 + 10s)2) (6) (−4, 9

8− 1

8(43+ 4s)2)

(7) (−3, 2524

− 124(2 + 6s)2) (8) (−3, 25

24− 1

24(32+ 6s)2)

(9) (−3, 2524

− 124(65+ 6s)2) (10) (−3, 25

24− 1

24(125+ 6s)2)

(11) (3,− 124

+ 124(2 + 6s)2) (12) (3,− 1

24+ 1

24(32+ 6s)2)

(13 (3,− 124

+ 124(65+ 6s)2) (14) (3,− 1

24+ 1

24(125+ 6s)2)

(15) (4,−18+ 1

8(43+ 4s)2) (16) (5,− 9

40+ 1

40(103+ 10s)2)

(17) (5,− 940

+ 140(4 + 10s)2) (18) (k, 1

2(k−1

k+ s(s+ 1)k))

where s is an arbitrary integer and C stands for an arbitrary complex number.

The above result is extended by Maciejewski, Przybilska and Yoshida in several ways. Wewill use the following

Theorem 5. (Theorem 1.3 [19]) Assume that the Hamiltonian system defined by Hamiltonian(A.1) with a homogeneous potential V of degree k ∈ Z \ {0} satisfies the following conditions:

(1) there exists a non-zero c ∈ Cn such that c = V ′(c), and

(2) matrix V ′′(c) is diagonalizable with eigenvalues λ1, . . . , λn−1, λn = k − 1;

(3) the system admits an additional first integral F , which is meromorphic in a connectedneighborhood U of Γ.

Then either

(i) there exist 1 ≤ r < n such that the pair (k, λr) belongs to the table in Theorem 4, or

(ii) there exist 1 ≤ i < j < n such that

1

2k

√

(k − 2)2 + 8kλi =1

2k

√

(k − 2)2 + 8kλj + l, (A.8)

for some l ∈ Z.

A.2 Necessary conditions for integrability of Hamiltonian systems

which have (NVE) of Lame type

Here we recall some facts concerning the integrability of Hamiltonian systems with two degreesof freedom, an invariant plane and which (NVE) are of Lame type. More details can be foundin [21, 10]. In our case the (NVE) splits into two equations of Lame type, and therefore, thesearguments can be applied.

18

Classically the Lame equation is written in the form

ξ − [n(n + 1)℘(t) +B]ξ = 0, (A.9)

where ℘(t) is the Weierstrass function with invariants g2 and g3, satisfying v2 = 4v3−g2v−g3with ∆ = g32 − 27g23 6= 0.

The known (mutually exclusive) cases of closed form solutions of (A.9) are:(i) The Lame and Hermite solutions. In this case n ∈ Z and g2, g3, B are arbitrary

parameters;(ii) The Brioschi-Halphen-Crowford solutions. Here m := n+1/2 ∈ N and the parameters

g2, g3, B must satisfy an algebraic equation.(iii) The Baldassarri solutions. Now n+1/2 ∈ 1

3Z∪ 1

4Z∪ 1

5Z \Z with additional algebraic

relations between the other parameters.Note that in the case (i) the identity component of the Galois group G0 is of the form

(

1 0ν 1

)

and in the cases (ii) and (iii) G0 = id (G is finite). And these are the all cases when

the Lame equation is integrable.Now consider a natural two degrees of freedom Hamiltonian

H =1

2(p21 + p22) + V (q1, q2), (A.10)

qj(t) ∈ C, pj(t) = qj , j = 1, 2. We assume that there exists a family of solutions of the form

Γh : q2 = p2 = 0, q1 = q1(t, h), p1(t, h) = q1(t, h)

and q1(t, h) is a solution of1

2q21 + ϕ(q1) = h, h ∈ R.

The (NVE) along Γh isξ − f(t, h)ξ = 0, (A.11)

where f(t, h) = f(q1(t, h)) is such that (A.11) is of type (A.9).In [21, 10] the type of the potentials V with this property are obtained as well as the

necessary conditions for the integrability of the Hamiltonian systems with the Hamiltonian(A.10). In order to formulate the result we need certain additional quantities.

Since f(t, h) depends linearly on ℘(t), then f 2 is a cubic polynomial in f , depending alsoin h, namely

f 2 := P (f, h) = P1(f) + hP2(f). (A.12)

The following coefficients are introduced

P (f, h) = (a1 + ha2)f3 + (b1 + hb2)f

2 + (c1 + hc2)f + (d1 + hd2). (A.13)

Now we are ready to give the corresponding result. Note that the following Theorem givesnecessary conditions only from the analysis of the first variational equation.

19

Theorem 6. (Theorem 6.2 [10]). Assume that a natural Hamiltonian system has (NVE) ofLame type, associated to the family of solutions Γh, lying on the plane q2 = 0 and param-eterized by the energy h. Then, a necessary conditions for integrability is that the relatedpolynomials P1 and P2 satisfy a2 = 0, and one of the following conditions holds:

I. a1 =4

n(n+1)for some n ∈ N;

II. a1 =16

4m2−1for some m ∈ N. Then, assuming the conjecture above is true, one should have

b2 = 0 and we should be in one of the following cases:

II1 m = 1 and b1 = 0,

II2 m = 2 and c2 = 0, 16a1c1 + 3b21 = 0,

II3 m = 3 and 16a1d2 + 11b1c2 = 0, 1024a21d1 + 704a1b1c1 + 45b31 = 0,

IIm m > 3. Then, we should have b1 = 0 and, furthermore, either c1 = c2 = 0 if m iscongruent with 1, 2, 4 or 5 modulo 6, or d1 = d2 = 0 if m is odd;

III. a1 =4

n(n+1)with n+ 1/2 ∈ 1

3Z ∪ 1

4Z ∪ 1

5Z \ Z, b2 = 0 and either c2 = 0, b21 − 3a1c1 = 0 or

c2b1 − 3a1d2 = 0, 2b31 − 9a1b1c1 + 27a21d1 = 0.

It is clear that the condition I. in the above Theorem gives the Lame and Hermite solutions(i), the condition II.– the Brioschi-Halphen-Crowford solutions (ii), and the condition III. –the Baldassarri solutions (iii).

References

[1] M. Henon, C. Heiles, The Applicability of the Third Integral of Motion: Some NumericalExperiments, The Astron. J. 69 7-79 (1964)

[2] Wenlei Li, Shaoyun Shi, Non-integrability of Henon-Heiles system, Celest. Mech DynAstr, 2010, DOI 10.1007/s10569-010-9315-1

[3] S. Ferrer, J. Palacian, J.F. San Juan, A. Viartola, P. Yanguas, On the Henon and Heilessystem in three dimensions: The role of the axial symmetry, Phys. lett. A 228 255-260(1997)

[4] S. Ferrer, M. Lara, J. Palacian, J.F. San Juan, A. Viartola, P. Yanguas, The Henonand Heiles problem in three dimensions. I. Periodic orbits near the origin, Int. J. ofBifurcation and Chaos, 8, No. 6, 1199-1213 (1998)

[5] J.C. Eilbeck, V. Z. Enol’skii, V.B. Kuznetsov C and D.V. Leykin, Linear r-matrix algebrafor systems separable in parabolic coordinates, Physics Letters A 180 (1993) 208—214

[6] N.A. Kostov, V.S. Gerdjikov, V. Miok, Exact solutions for a class of integrable Henon-Heiles-type system, J. Math. Phys. 51 022702 (2010), doi.org/10.1063/1.3300310

20

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