Non-Integrability of a Hamiltonian System and Legendre Functions

1. Introduction

We consider Hamiltonian systems of type

$$H=\sum _i{\displaystyle \frac{1}{2}}{p}_i^2+V\left(q\right),$$

(1)

where $q=(q_1,\dots ,q_n)$ and $p=(p_1,\dots ,p_n)$ are canonical coordinates in a symplectic linear space ${\mathbb{C}}^{2n}$ and $V\left(q\right)$ is a homogeneous polynomial function. If $F=F(q,p)$ and $G=G(q,p)$ are two functions, then their Poisson bracket is defined by $\{F,G\}={\sum }_i({\displaystyle \frac{\partial F}{\partial q_i}}{\displaystyle \frac{\partial G}{\partial p_i}}-{\displaystyle \frac{\partial F}{\partial p_i}}{\displaystyle \frac{\partial G}{\partial q_i}})$. The functions F and G are in involution if $\{F,G\}=0$ is satisfied. Non-constant function F is the first integral for the Hamiltonian system with Hamiltonian H if $\{F,H\}=0$. Since the Poisson bracket is anti-symmetric, it is clear that H is always the first integral. A Hamiltonian system is called integrable if a set of n independent first integrals in involution exists for it (see Appendix A for further details).

The study of Hamiltonian systems with a homogeneous polynomial potential is an important problem in modern mechanics and physics. In our opinion, the reason for this research interest is the quite important physical problems can be considered as Hamiltonian systems with a homogeneous potentials. The integrability of Hamiltonian systems of type (1) has been studied for a long time, but the results of this work are limited to a rather small class of solved problems. The approaches to the study of this problem are: direct method [12,13] and Painleve test [3,4,5,6,7,8]. The most remarkable and perhaps most general result was obtained in [9]. In 1982 in [14] and 1983 in [15], Ziglin reduced the study of the integrability of Hamiltonian systems to the investigation of the monodromy group of the linear equations obtained by the variation of the system in a neighborhood of its non-equilibrium solution. The Ziglin’s theory gives the opportunity for the study complex Hamiltonian systems for integrability. Using this theory, Yoshida [10] was able to obtain a criterion for the integrability of two-dimensional systems with a homogeneous polynomial potential. Later in [11] Yoshida generalized his result to n degrees of freedom. In the late 20-th century, the Ziglin’s theory was linked to the differential Galois theory in the works of Churchill, Baider, Rod, Singer in [16,17], Moralez-Ruiz, Ramis, and Simo in [18,19,20]. The idea here is to move from studying the discrete monodromy group to the continuous algebraic differential Galois group for linear variational equations. The closure of the monodromy group in the Zariski topology gives us the differential Galois group, as noted in the Picard–Vessiot theory (see Appendix A.)

The use of differential Galois theory to study the integrability of Hamiltonian systems with homogeneous potentials in [18]. In the papers [21,22], the cases of homogeneous polynomial potentials of degrees 3 and 4 are scrutinized. Homogeneous potentials with a negative degree of homogeneity are considered in [1,2].

The approach implemented to study the non-integrability of $V_6$ in this paper differs from [18,21,22]. Here, the search for Darboux points is skipped. With a change of variables $V{E}_1$, which is a Fuchsian differential equation with 5 regular singularities, is reduced to a Legendre equation with 3 singularities.

We scrutinize systems with Hamiltonian $H={\sum }_i{\displaystyle \frac{1}{2}}{p}_i^2+V\left(q\right)$, with the potential $V\left(q\right)$ shown in the form $V\left(q\right)={\sum }_l{V}_l\left(q\right)$, here $V_l\left(q\right)$ are homogeneous polynomials of degree $l\in \mathbb{N}$, $l\ge 2$.

Every homogeneous polynomial potential $V\left(q\right)$ is represented as a sum of homogeneous polynomials of not greater than $V\left(q\right)=V_{min}\left(q\right)+\dots +V_{max}\left(q\right)$ degrees. We call the potential $V\left(q\right)$ integrable if the Hamiltonian system $H={\sum }_i{\displaystyle \frac{1}{2}}{p}_i^2+V\left(q\right)$ is integrable. If a homogeneous potential $V\left(q\right)=V_{min}\left(q\right)+\dots +V_{max}\left(q\right)$ is integrable then $V_{min}\left(q\right)$ and $V_{max}\left(q\right)$ are also integrable. This remarkable property can be seen in [13].

In this paper, our study will be limited to assessing the non-integrability of such problem. We assume that ${\mathbb{C}}^4$ is a symplectic linear space with canonical variables $q=(q_1,q_2)$ and $p=(p_1,p_2)$. We are studying the Hamiltonian systems defined by the Hamiltonian $H={\displaystyle \frac{1}{2}}{\sum }_1^2{p}_i^2+V_l\left(q\right)$, where $V_l\left(q\right)$ is a homogeneous function of degree l. The equations of motion of the above system are

$$\begin{array}{ccc}\hfill {\dot{q}}_i& =& p_i,\hfill \\ \hfill {\dot{p}}_i& =& -{\displaystyle \frac{\partial V(q_1,q_2)}{\partial q_i}},i=1,2.\hfill \end{array}$$

We say that a Hamiltonian system with 2 degrees of freedom is integrable in the sense of Liouville if there exists a second (other than H) first integral F for which the gradients $dH$ and $dF$ are linearly independent (see [1] for details).

In this paper we are studying a two dimensional model with sixth-order homogeneous potential

$$H={\displaystyle \frac{1}{2}}(p_r^2+p_z^2)+r^6+A{r}^2{z}^4+D{r}^3{z}^3+B{r}^4{z}^2+E{r}^{5}z+C{z}^6,$$

(2)

where A, B, C, D and E are appropriate real constants for existing additional meromorphic integral of motion.

Let us justify the absence of the coefficient in front of $r^{5}z$: If we assume that this coefficient in $V_6$ is $\alpha$, i.e. we have

$$V_6=r^6+\alpha r^{5}z+A{r}^2{z}^4+D{r}^3{z}^3+B{r}^4{z}^2+E{r}^{5}z+C{z}^6.$$

Now let us change the variables $r\to {\displaystyle \frac{6r-\alpha z}{36}}$ and $z\to {\displaystyle \frac{z}{6}}$ (a linear change, we can also define a canonical one of the Hamiltonian system). This is enough to change the coefficient $\alpha$ to 0 (see [12]).

The Hamiltonian equations are:

$$\begin{array}{cc}\hfill \dot{r}& =p_r,{\dot{p}}_r=-(2Ar{z}^4+4B{r}^3{z}^2+3D{r}^2{z}^3+5E{r}^{4}z+6r^5),\hfill \\ \hfill \dot{z}& =p_z,{\dot{p}}_z=-(4A{r}^2{z}^3+2B{r}^{4}z+6C{z}^5+3D{r}^3{z}^2+E{r}^5),\hfill \end{array}$$

(3)

(here as usual $\dot{\phantom{y}}={\displaystyle \frac{d}{dt}}$).

The organization of the text is as follows: In Section 2 we are investigating the solvability of Legendre’s equation. In Section 3 we are trying to find a suitable partial solution, we so, we have identified the variational equations $V{E}_1$, and have related them to Legendre’s equation. In Section 4 we are going through cases that have not been achieved using the results of Section 3. In this section the main theorem of our paper is proved. In the Appendix A we have given some facts from the Differential Galois theory necessary for our text.

2. The Associated Legendre Equation

In this section, we are studying for solvability of the associated Legendre equation

$$(1-z^2){\displaystyle \frac{d^{2}w}{d{z}^2}}-2z{\displaystyle \frac{dw}{dz}}+\left(p(p+1)-{\displaystyle \frac{q^2}{1-z^2}}\right)w=0,p,q\in \mathbb{R},p+q\ne -1,-2,-3\dots ,$$

(4)

and we have applied the result to study the potential

$$V_6(r,z)=r^6+A{r}^2{z}^4+D{r}^3{z}^3+B{r}^4{z}^2+E{r}^{5}z+C{z}^6,$$

($A,B,C,D,E\in \mathbb{R}$) for integrability within the scope of Liouville.

Let us write down the known facts about the solutions of (4) and outline some properties required for solvability. (We follow [23].) With

$$\begin{array}{ccc}\hfill P_p^q\left(z\right)& =& {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{q/2}{\phantom{y}}_2{F}_1(p+1,-p;1-q,{\displaystyle \frac{1}{2}}-{\displaystyle \frac{z}{2}}),\hfill \\ \hfill Q_p^q\left(z\right)& =& {\displaystyle \frac{\pi }{2sin\left(q\pi \right)}}(cos\left(q\pi \right){\left({\displaystyle \frac{1+z}{1-z}}\right)}^{q/2}{\phantom{y}}_2{F}_1(p+1,-p;1-q,{\displaystyle \frac{1}{2}}-{\displaystyle \frac{z}{2}})\hfill \\ & -& {\left({\displaystyle \frac{1-z}{1+z}}\right)}^{q/2}{\displaystyle \frac{\mathrm{\Gamma }(p+q+1)}{\mathrm{\Gamma }(p-q+1)}}{\phantom{y}}_2{F}_1(p+1,-p;1+q,{\displaystyle \frac{1}{2}}-{\displaystyle \frac{z}{2}}))\hfill \end{array}$$

(5)

we can show the solutions of (4) (P and Q Legendre functions), shown via the use of the hypergeometric function ${\phantom{y}}_2{F}_1(a,b;c,z)$. The singularities of equation (4) are the points $z=-1$, $z=1$ and $z=\infty$, which are regular. The following equalities are valid

$$\begin{array}{ccc}\hfill P_p^{-q}\left(z{e}^{si\pi }\right)& =& e^{spi\pi }{P}_p^{-q}\left(z\right)+{\displaystyle \frac{2isin(p+1/2)s\pi }{cosp\pi \mathrm{\Gamma }(q-p)}}{Q}_p^q\left(z\right)\hfill \\ \hfill Q_p^q\left(z{e}^{si\pi }\right)& =& {(-1)}^s{e}^{-spi\pi }{Q}_p^q\left(z\right).\hfill \end{array}$$

(6)

The indicative equations for the singular points $\pm 1$ are ${\displaystyle {\rho }^2-\rho +\frac{1-q^2}{4}=0}$ (see [26]) with a roots (exponents) ${\displaystyle {\rho }_{1,2}=\frac{1\pm q}{2}}$, and for ∞ it is ${\lambda }^2+\lambda -p^2-p=0$ with an exponents ${\lambda }_1=-p-1$, ${\lambda }_2=p$.

Therefore, we can specify the generators of the local monodromy for the points $\pm 1$. These are the matrices $\left(\begin{array}{cc}{e}^{(1+q)i\pi }& {\alpha }_1\\ 0& e^{-(1+q)i\pi }\end{array}\right)$, and $\left(\begin{array}{cc}{e}^{(1-q)i\pi }& {\alpha }_2\\ 0& e^{-(1-q)i\pi }\end{array}\right)$.

For the point ∞, they are $\left(\begin{array}{cc}{e}^{-2(1+p)i\pi }& {\alpha }_3\\ 0& e^{2(1+p)i\pi }\end{array}\right)$, and $\left(\begin{array}{cc}{e}^{2pi\pi }& {\alpha }_4\\ 0& e^{-2pi\pi }\end{array}\right)$ (${\alpha }_j\in \mathbb{C}$).

It is known from theory (see Appendix A) that a necessary condition for the solvability of a linear differential equation is that the monodromy group should be commutative. With the generators found, we can conclude that our equation does not have Liouville solutions (is not solvable) if $porq\notin \mathbb{Q}$.

Proposition 1.

The equation (4) is non solvable if at least one of p, $q\notin \mathbb{Q}$.

In the next of this paper, we assume that $p,q\in \mathbb{Q}$. We now have an expressions for the solution of (5) utilizing the P and Q Legendre functions $P_p^q\left(z\right)$ and $Q_p^q\left(z\right)$, which otherwise we can express using the Hypergeometric function ${\phantom{y}}_2{F}_1(p+1,-p;1-q,{\displaystyle \frac{1}{2}}-{\displaystyle \frac{z}{2}})$. Here we can apply Kimura’s conditions [24,25]. Thus we get the following

Theorem 1.

Let p, $q\in \mathbb{Q}$, then the equation (4) is non solvable if:

(i) all of numbers $2p+1$, $2(q-p)+1$ and $2(p+q)+1$ are not odd integer;

(ii) at least one of $p\ne 1/2(-1\pm (1/2+m\left)\right)$ or $q\ne \pm (1/2+l)$, for $l,m\in \mathbb{Z}$;

(iii) at least one of $p\ne 1/2(-1\pm (1/3+m\left)\right)$ or $q\ne \pm (2/3+l)$, for $l,m\in \mathbb{Z}$, m is odd;

(iv) at least one of $p\ne 1/2(-1\pm (2/5+m\left)\right)$ or $q\ne \pm (2/5+l)$, for $l,m\in \mathbb{Z}$, m is even;

(v) at least one of $p\ne 1/2(-1\pm (1/5+m\left)\right)$ or $q\ne \pm (4/5+l)$, for $l,m\in \mathbb{Z}$, m is odd.

It is not difficult to conclude that if $q=0$, then Legendre’s the equation 4 is not solvable for $p\notin \mathbb{Z}$.

3. Sixth-Order Homogeneous Potential

First, we are finding a non equilibrium partial solution for (3). Let we put $r=p_r=0$ in (3) and we have

$$\ddot{z}=-6C{z}^5,$$

multiplied by $\dot{z}$ and integrated by the time t we get

$${\dot{z}}^2=-2(C{z}^6+C{h}^3),$$

(7)

where h is a real constant.To this end, it we need to find a non branching solution, so, we get $w=z^2$ (finite ramified covering of the curve $y^2=-2(C{z}^6+h)$). We obtain ${\displaystyle \dot{w}=2z\dot{z}=\frac{d{z}^2}{dt}}$, then we have

$$\begin{array}{ccc}\hfill {\dot{w}}^2& =& -8(C{w}^4+C{h}^{3}w),\hfill \\ \hfill \ddot{w}& =& -4C(4w^3+h^3).\hfill \end{array}$$

(8)

Further, we follow the procedures for Ziglin-Morales-Ramis theory and we detect an invariant manifold here w is the solution of (8). According to theory, the solution of (8) must be a rational function of Weierstrass ℘-function. It is convenient to choose for field of constants $K=\mathbb{C}\left(w\right)$ - rational functions over a complex variable w. After having found the Variation Equations (VE) we have ${\xi }_{11}=dr$, ${\eta }_{11}=d{p}_r$, ${\xi }_{12}=dz$, ${\eta }_{12}=d{p}_z$, and we achieve:

$$\begin{array}{cc}\hfill {\ddot{\xi }}_{11}& =-2A{z}^4{\xi }_{11},\hfill \\ \hfill {\ddot{\xi }}_{12}& =-30C{z}^4{\xi }_{12},\hfill \end{array}$$

(9)

and after changing $w=z^2$ we have

$$\begin{array}{cc}\hfill {\ddot{\xi }}_{11}& =-2A{w}^2{\xi }_{11},\hfill \\ \hfill {\ddot{\xi }}_{12}& =-30C{w}^2{\xi }_{12},\hfill \end{array}$$

(10)

Now let us change the variables to (10) $t\to w\left(t\right)$, and we obtain $V{E}_1$-equations. Let us write denote with ${\phantom{y}}^{\prime }={\displaystyle \frac{d}{dw}}$ then we get for the $V{E}_1$ two Fuchsian linear differential equations:

$$\begin{array}{c}\hfill {\xi }_{11}^{″}+{\displaystyle \frac{4w^3+h^3}{2w(w^3+h^3)}}{\xi }_{11}^{\prime }-{\displaystyle \frac{A}{4C}}{\displaystyle \frac{w}{(w^3+h^3)}}{\xi }_{11}=0,\\ \hfill {\xi }_{12}^{″}+{\displaystyle \frac{4w^3+h^3}{2w(w^3+h^3)}}{\xi }_{12}^{\prime }-{\displaystyle \frac{15}{4}}{\displaystyle \frac{w}{(w^3+h^3)}}{\xi }_{12}=0.\end{array}$$

(11)

The equations ( 11) are a Fuchsian and have five regular singularities 0, $-h$, ${\displaystyle \frac{h}{2}}(1\pm \sqrt{3}i)$ and ∞. There are three possible ways to investigate (11) the solvability: The first is to apply the Kovacic’s algorithm [27] to the first equation (it is not really simple, but it is still an option). The second – is studying the existence of a hyperexponential solution [28,29]. The third one - is to change of variables

$$z^2:=(1+{\displaystyle \frac{w^3}{h^3}}),{\tilde{\xi }}_{11}:={\displaystyle \frac{{\xi }_{11}}{{\left(h^3(z^2-1)\right)}^{1/12}}},$$

(12)

we get an easy to investigate equation (for simplicity, we have omitted the tilde).

$${\displaystyle \frac{d^2{\xi }_{11}}{d{z}^2}}-{\displaystyle \frac{2z}{1-z^2}}{\displaystyle \frac{d{\xi }_{11}}{dz}}+\left({\displaystyle \frac{2A-5C}{36C}}-{\displaystyle \frac{1/6}{1-z^2}}\right){\xi }_{11}=0.$$

(13)

This change is a finite branched covering map ${\mathbb{CP}}^1\to {\mathbb{CP}}^1$. In general, the differential Galois group is changed under such transformation, but the identity component remains unchanged (see [18], p. 28).

The equation (13) is associated Legendre equation with ${\displaystyle p=-\frac{1}{2}\pm \frac{1}{2}\sqrt{\frac{4C+2A}{9C}}}$ and ${\displaystyle q=\frac{1}{6}}$. We note $\tau =\pm \sqrt{{\displaystyle \frac{2A+4C}{9C}}}$ ($C\ne 0$) and apply the results of Proposition 1 and Theorem 1:

Proposition 2.

The system (3) is non-integrable for ${\displaystyle p=-\frac{1}{2}+\frac{1}{2}\tau \notin \mathbb{Q}}$.

Proposition 3.

For $p\in \mathbb{Q}$, the system (3) is non-integrable for ${\displaystyle p=-\frac{1}{2}+\frac{1}{2}\tau \ne \pm (k-\frac{1}{6})}$, $k\in \mathbb{Z}$.

Now we need to go through the case ${\displaystyle p=-\frac{1}{2}+\frac{1}{2}\tau =\pm (k-\frac{1}{6})}$, $k\in \mathbb{Z}$, (which is equivalent to ${\displaystyle \tau =-2k+\frac{4}{3}}$, and ${\displaystyle \tau =2k+\frac{2}{3}}$ for $k\in \mathbb{Z}$) to obtain additional non-integrability conditions.

4. Additional Cases

In this section we are considering cases $\tau =-2k+{\displaystyle \frac{4}{3}}$, and $\tau =2k+{\displaystyle \frac{2}{3}}$ for $k\in \mathbb{Z}$ that complete the research from the previous section.

Let us find the second variations of the Hamiltonian system with Hamiltonian (2). We denote it with

$$\begin{array}{ccc}\hfill r& =& \epsilon {\xi }_{11}+{\epsilon }^2{\xi }_{21}+\dots ,\hfill \\ \hfill z& =& z\left(t\right)+\epsilon {\xi }_{12}+{\epsilon }^2{\xi }_{22}+\dots ,\hfill \\ \hfill p_r& =& \epsilon {\eta }_{11}+{\epsilon }^2{\eta }_{21}+\dots ,\hfill \\ \hfill p_z& =& \dot{z}\left(t\right)+\epsilon {\eta }_{12}+{\epsilon }^2{\eta }_{22}+\dots ,\hfill \end{array}$$

here $(p_r,r,p_z,z)=(0,0,\dot{z}\left(t\right),z\left(t\right))$ is an invariant manifold of the system (3).

We do the replacement in the system (3) and we compare the coefficients at ${\epsilon }^2$. After changing the variables $t\to z\left(t\right)\to w\left(t\right)$ we obtain

$$\begin{array}{ccc}\hfill {\xi }_{21}^{\prime \prime }& +& {\displaystyle \frac{4w^3+h^3}{2w(w^3+h^3)}}{\xi }_{21}^{\prime }-\left({\displaystyle \frac{9{\tau }^2-4}{8}}\right){\displaystyle \frac{w}{(w^3+h^3)}}{\xi }_{21}=K_2^{\left(1\right)}\hfill \\ \hfill {\xi }_{22}^{\prime \prime }& +& {\displaystyle \frac{4w^3+h^3}{2w(w^3+h^3)}}{\xi }_{22}^{\prime }-{\displaystyle \frac{15}{4}}{\displaystyle \frac{w}{(w^3+h^3)}}{\xi }_{22}=K_2^{\left(2\right)},\hfill \end{array}$$

(14)

and we have

$$\begin{array}{ccc}\hfill K_2^{\left(1\right)}& =& \left({\displaystyle \frac{9{\tau }^2-4}{2}}\right){\displaystyle \frac{w^{1/2}{\xi }_{11}{\xi }_{12}}{(w^3+h^3)}}-\left({\displaystyle \frac{3D}{8C}}\right){\displaystyle \frac{w^{1/2}{\left({\xi }_{11}\right)}^2}{(w^3+h^3)}},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill K_2^{\left(2\right)}& =& \left(\left({\displaystyle \frac{9{\tau }^2-4}{4}}\right){\displaystyle \frac{w^{1/2}{\left({\xi }_{11}\right)}^2}{(w^3+h^3)}}+{\displaystyle \frac{15}{2}}{\displaystyle \frac{w^{1/2}{\left({\xi }_{12}\right)}^2}{(w^3+h^3)}}\right).\hfill \end{array}$$

We will focus on the singular point ∞. To this end, we change variables $w={\displaystyle \frac{1}{x}}$ into (11) and (14) (here we assume ${\phantom{y}}^{\prime }={\displaystyle \frac{d}{dx}}$). We have

$$\begin{array}{ccc}\hfill {\xi }_{11}^{\prime \prime }& +& {\displaystyle \frac{3h^3{x}^2}{2(h^3{x}^3+1)}}{\xi }_{11}^{\prime }-\left({\displaystyle \frac{9{\tau }^2-4}{16}}\right){\displaystyle \frac{{\xi }_{11}}{x^2(h^3{x}^3+1)}}=0,\hfill \\ \hfill {\xi }_{12}^{\prime \prime }& +& {\displaystyle \frac{3h^3{x}^2}{2(h^3{x}^3+1)}}{\xi }_{12}^{\prime }-{\displaystyle \frac{15}{4}}{\displaystyle \frac{{\xi }_{12}}{x^2(h^3{x}^3+1)}}=0,\hfill \end{array}$$

(15)

$$\begin{array}{ccc}\hfill {\xi }_{21}^{\prime \prime }& +& {\displaystyle \frac{3h^3{x}^2}{2(h^3{x}^3+1)}}{\xi }_{21}^{\prime }-\left({\displaystyle \frac{9{\tau }^2-4}{16}}\right){\displaystyle \frac{{\xi }_{21}}{x^2(h^3{x}^3+1)}}\hfill \\ & =& \left({\displaystyle \frac{9{\tau }^2-4}{2}}\right){\displaystyle \frac{{\xi }_{11}{\xi }_{12}}{x^{3/2}(h^3{x}^3+1)}}+\left({\displaystyle \frac{3D}{8C}}\right){\displaystyle \frac{{\left({\xi }_{11}\right)}^2}{x^{3/2}(h^3{x}^3+1)}},\hfill \\ \hfill {\xi }_{22}^{\prime \prime }& +& {\displaystyle \frac{3h^3{x}^2}{2(h^3{x}^3+1)}}{\xi }_{22}^{\prime }+{\displaystyle \frac{15}{4}}{\displaystyle \frac{{\xi }_{22}}{x^2(h^3{x}^3+1)}}\hfill \\ & =& \left({\displaystyle \frac{9{\tau }^2-4}{4}}\right){\displaystyle \frac{{\left({\xi }_{11}\right)}^2}{x^{3/2}(h^3{x}^3+1)}}+{\displaystyle \frac{15}{2}}{\displaystyle \frac{{\left({\xi }_{12}\right)}^2}{x^{3/2}(h^3{x}^3+1)}}.\hfill \end{array}$$

(16)

We would like to underline some known facts we will need for our further research. Solutions should be provided for the equations (15) or (16) where Wronsky determinant is a constant. This happens exactly when the coefficient in front of the first derivative in the differential equation disappears. This can be achieved with the following standard manipulation. We have

$$\xi {\left(x\right)}^{\prime \prime }+a\left(x\right)\xi {\left(x\right)}^{\prime }+b\left(x\right)\xi \left(x\right)=K_2\left(x\right),$$

and let $\xi \left(x\right)=\zeta \left(x\right)e^{-{\displaystyle \frac{1}{2}}\int a\left(x\right)dx}$, then we obtain equation in normal form

$$\zeta {\left(x\right)}^{\prime \prime }-r\left(x\right)\zeta =-K_2\left(x\right)e^{{\displaystyle \frac{1}{2}}\int a\left(x\right)dx},$$

where $r\left(x\right)={\displaystyle \frac{1}{2}}a{\left(x\right)}^{\prime }+{\displaystyle \frac{1}{4}}{\left(a\left(x\right)\right)}^2-b\left(x\right)$.

Now we change the form of (11) and (14). Let us replace $\xi$ with $\zeta$ by

$$\xi =\zeta e^{-{\displaystyle \frac{1}{2}}\int {\displaystyle \frac{3h^3{x}^2}{2(h^3{x}^3+1)}}dx}=\zeta .{(h^3{x}^3+1)}^{-{\displaystyle \frac{1}{4}}},$$

and we obtain

$$\begin{array}{c}\hfill {\zeta }_{11}^{″}-r_1\left(x\right){\zeta }_{11}=0,\end{array}$$

$$\begin{array}{c}\hfill {\zeta }_{12}^{″}-r_2\left(x\right){\zeta }_{12}=0,\\ \hfill {\zeta }_{21}^{″}-r_1\left(x\right){\zeta }_{21}=\widetilde{K_2^{\left(1\right)}},\end{array}$$

(17)

$$\begin{array}{c}\hfill {\zeta }_{22}^{″}-r_2\left(x\right){\zeta }_{22}=\widetilde{K_2^{\left(2\right)}},\end{array}$$

(18)

here

$$\begin{array}{ccc}\hfill r_1\left(x\right)& =& {\displaystyle \frac{-h^6{x}^6+h^3(9{\tau }^2+20)x^3+9{\tau }^2-4}{16x^2{(h^3{x}^3+1)}^2}},\hfill \\ \hfill r_2\left(x\right)& =& {\displaystyle \frac{-h^6{x}^6+84h^3{x}^3+60}{16x^2{(h^3{x}^3+1)}^2}}.\hfill \end{array}$$

We also need to give the solutions of (17) in series near 0.

$$\begin{array}{c}\hfill {\zeta }_{11}^{\left(1\right)}\left(x\right)=x^{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{3\tau }{4}}}\left(1-{\displaystyle \frac{(9{\tau }^2-28)h^3}{144+72\tau }}{x}^3+\dots \right),\end{array}$$

$$\begin{array}{c}\hfill {\zeta }_{11}^{\left(2\right)}\left(x\right)=x^{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{3\tau }{4}}}\left(1+{\displaystyle \frac{(9{\tau }^2-28)h^3}{-144+72\tau }}{x}^3+\dots \right),\\ \hfill {\zeta }_{12}^{\left(1\right)}\left(x\right)=x^{{\displaystyle \frac{5}{2}}}\left(1-{\displaystyle \frac{3h^3}{28}}{x}^3+\dots \right),\end{array}$$

(19)

$$\begin{array}{c}\hfill {\zeta }_{12}^{\left(2\right)}\left(x\right)=x^{-{\displaystyle \frac{3}{2}}}\left(-144-108h^3{x}^3+\dots \right).\end{array}$$

(20)

Now we have

$$\begin{array}{ccc}\hfill {\widetilde{K_2}}^{\left(1\right)}({\zeta }_{11},{\zeta }_{12})& =& -K_2^{\left(1\right)}(h^3{x}^3+1{)}^{1/4}\hfill \\ & =& \left(-{\displaystyle \frac{\left(9{\tau }^2-4\right)}{2}}{\displaystyle \frac{{\zeta }_{11}{\zeta }_{12}}{x^{3/2}{(h^3{x}^3+1)}^{3/4}}}-\left({\displaystyle \frac{3D}{8C}}\right){\displaystyle \frac{{\left({\zeta }_{11}\right)}^2}{x^{3/2}{(h^3{x}^3+1)}^{3/4}}}\right)\hfill \\ \hfill {\widetilde{K_2}}^{\left(2\right)}({\zeta }_{11},{\zeta }_{12})& =& -K_2^{\left(2\right)}(h^3{x}^3+1{)}^{1/4}\hfill \\ & =& \left(-{\displaystyle \frac{\left(9{\tau }^2-4\right)}{2}}{\displaystyle \frac{{\left({\zeta }_{11}\right)}^2}{x^{3/2}{(h^3{x}^3+1)}^{3/4}}}-{\displaystyle \frac{15}{2}}{\displaystyle \frac{{\left({\zeta }_{12}\right)}^2}{x^{3/2}{(h^3{x}^3+1)}^{3/4}}}\right).\hfill \end{array}$$

Without loosing a community, we can assume that

${\zeta }_{11}^{\left(1\right)}{\left({\zeta }_{11}^{\left(2\right)}\right)}^{\prime }-{\zeta }_{11}^{\left(2\right)}{\left({\zeta }_{11}^{\left(1\right)}\right)}^{\prime }=1$ and ${\zeta }_{12}^{\left(1\right)}{\left({\zeta }_{12}^{\left(2\right)}\right)}^{\prime }-{\zeta }_{12}^{\left(2\right)}{\left({\zeta }_{12}^{\left(1\right)}\right)}^{\prime }=1$ . Then the fundamental matrix of (17) and its inverse are

$$X\left(z\right)=\left(\begin{array}{cccc}{\zeta }_{11}^{\left(1\right)}& {\zeta }_{11}^{\left(2\right)}& 0& 0\\ {\left({\zeta }_{11}^{\left(1\right)}\right)}^{\prime }& {\left({\zeta }_{11}^{\left(2\right)}\right)}^{\prime }& 0& 0\\ 0& 0& {\zeta }_{12}^{\left(1\right)}& {\zeta }_{12}^{\left(2\right)}\\ 0& 0& {\left({\zeta }_{12}^{\left(1\right)}\right)}^{\prime }& {\left({\zeta }_{12}^{\left(2\right)}\right)}^{\prime }\end{array}\right),$$

(21)

$$X^{-1}\left(z\right)=\left(\begin{array}{cccc}{\left({\zeta }_{11}^{\left(2\right)}\right)}^{\prime }& -{\zeta }_{11}^{\left(2\right)}& 0& 0\\ -{\left({\zeta }_{11}^{\left(1\right)}\right)}^{\prime }& {\zeta }_{11}^{\left(1\right)}& 0& 0\\ 0& 0& {\left({\zeta }_{12}^{\left(2\right)}\right)}^{\prime }& -{\zeta }_{12}^{\left(2\right)}\\ 0& 0& -{\left({\zeta }_{12}^{\left(1\right)}\right)}^{\prime }& {\zeta }_{12}^{\left(2\right)}\end{array}\right).$$

(22)

We show that a logarithmic term appears in local solution of (${VE}_2$). For this purpose, it is sufficient to show that at least one component of $X^{-1}{f}_2$ has a nonzero residue at 0. We are calculating of $X^{-1}{f}_2$, which looks like

$${(-{\zeta }_{11}^{\left(2\right)}\widetilde{K_2^{\left(1\right)}},{\zeta }_{11}^{\left(1\right)}\widetilde{K_2^{\left(1\right)}},-{\zeta }_{12}^{\left(2\right)}\widetilde{K_2^{\left(2\right)}},-{\zeta }_{12}^{\left(1\right)}\widetilde{K_2^{\left(2\right)}})}^T.$$

We have $\tau \in \{-2k+4/3,2k+2/3\}$ for $k\in \mathbb{Z}$. The presence of a non-zero residue in the above expression is possible in the cases;

1) ${\zeta }_{11}^{\left(2\right)}\widetilde{K_2^{\left(1\right)}}({\zeta }_{11}^{\left(1\right)},{\zeta }_{12}^{\left(1\right)})$, ${\zeta }_{11}^{\left(2\right)}\widetilde{K_2^{\left(1\right)}}({\zeta }_{11}^{\left(1\right)},{\zeta }_{12}^{\left(2\right)})$ and ${\zeta }_{11}^{\left(1\right)}\widetilde{K_2^{\left(1\right)}}({\zeta }_{11}^{\left(2\right)},{\zeta }_{12}^{\left(1\right)})$, then we obtain conditions for non-integrability k is odd and $D\ne 0$;

2) ${\zeta }_{11}^{\left(1\right)}\widetilde{K_2^{\left(1\right)}}({\zeta }_{11}^{\left(1\right)},{\zeta }_{12}^{\left(2\right)})$, then we have k is even and $k\ne 0$;

3) ${\zeta }_{11}^{\left(2\right)}\widetilde{K_2^{\left(1\right)}}({\zeta }_{11}^{\left(2\right)},{\zeta }_{12}^{\left(2\right)})$, then $k\in \mathbb{Z}\setminus \left\{1\right\}$.

For sub-cases $k=0$ we obtain non zero residue in the case ${\zeta }_{11}^{\left(1\right)}\widetilde{K_2^{\left(2\right)}}({\zeta }_{11}^{\left(2\right)},{\zeta }_{12}^{\left(1\right)})$, and for $k=1$ we have the condition $D\ne 0$.

We proved the following

Theorem 2.

Let $\tau =\pm \sqrt{{\displaystyle \frac{2A+4C}{9C}}}$ ($C\ne 0$), then the system (3) is non integrable if at least one of the conditions is available:

(i) $\tau \notin \mathbb{Q}$;

(ii) $\tau \in \mathbb{Q}\setminus \{-2k+4/3,2k+2/3\}$, for $k\in \mathbb{Z}$;

(iii) $\tau \in \{-2k+4/3,2k+2/3\}$, for $k\in \mathbb{Z}\setminus \left\{1\right\}$;

(iiii) for $\tau =-2/3$ ($k=1$) and $D\ne 0$.

If we use the homogeneity of the potential $V_6$, and we perform the transformation $z\leftrightarrow r$, we may get a similar result. Since the coefficient in front of $r^6$ is previously assumed to be 1, the non-integrability condition reduces to $D\ne 0$. Thus we proved the main theorem in this work:

Theorem 3.

Let the condition $D\ne 0$ is held for the potential $V_6$, then the Hamiltonian system (3) is non-integrable.

Examples of integrable homogeneous potentials of degree 6 can be found in [13]. There, the potential $V_6=r^6+80r^2{z}^4+24r^4{z}^2+64z^6$ has been taken into account. From the presentation in [2], we can come to the conclusion that $V_6=r^6+3r^2{z}^4+3r^4{z}^2+z^6$ is integrable. In both examples, $D=0$ is satisfied.