Non-integrability of the Truncated Zonal Satellite Problem

In this section we present some results about this problem recently obtained by the authors [2], although the proof has been considerably shortened.

By using cylindrical coordinates $(\rho ,\varphi ,z; p_\rho ,p_\varphi , p_z)$, the homogeneous Hamiltonian for this problem can be written as

$$H = \displaystyle\frac{1}{2}\left( p_\rho ^2 + p_z^2\right)......} +\sum_{k=2}^{n} \displaystyle\frac{V_k}{r^{k+1}} +p_0\, ,$$ (1)

where $\, r=\sqrt{\rho ^2 +z^2}\,$, $\,V_k =\varepsilon _k P_k(z/r)\,$, $\varepsilon _k$ are constant and P_k(x) is the Legendre polynomial of order k.

From the fact that $p_\varphi$ is a first integral of (1) it follows that, if this Hamiltonian admits a meromorphic integral independent from the other two and in involution with them, such an integral is analytic with respect to $p_\varphi$. Consequently, the integrability in the Liouville sense of (1) implies that of the Hamiltonian

$$H = \displaystyle\frac{1}{2}\left( p_\rho ^2 + p_z^2\right)......} +\sum_{k=2}^{n} \displaystyle\frac{V_k}{r^{k+1}} +p_0\, ,$$ (2)

which is obtained by making $p_\varphi =0$ in (1). It is immediate that (2) admits a particular solution with $\rho =p_\rho =0$. However, since an explicit expression for the associate monodromy matrices which are required by Ziglin's method is not known, it is convenient to introduce a suitable limit problem.

The change of scale defined by

$$\begin{array}{lll}t=\beta \bar{t}\, , &\qquad \rho =\bet......qquad p_z=\beta ^{-\frac{n+1}{n+3}}\;\bar{p}_z\, ,\end{array}$$

gives a new homogeneous Hamiltonian, which after multiplying by $\beta ^{\frac{2(n+1)}{n+3}}$ and taking limits as $\beta\rightarrow 0$, reduces to

$$K = \displaystyle\frac{1}{2}\left( p_\rho ^2 + p_z^2\right)......arepsilon _n}{r^{n+1}}P_n\left(\displaystyle\frac{z}{r}\right)$$ (3)

and the bar in the new variables has been removed to simplify the notation.

The Hamiltonian (3) is of the form required by Yoshida's theorem [11], since the potential V is homogeneous of degree k=-n-1. As is well known, the Hessian of V computed on the vector ${\bf c}$ quoted in section 2 always has the eigenvalue k-1, which in our case turns out to be -n-2. It is easy to establish that this is precisely the value of $\partial ^2V/\partial z^2$, so that the integrability coefficient reduces to $\partial ^2V/\partial\rho^2$. Moreover, due to the symmetry with respect to the OZ axis, it is obvious that

$$\displaystyle\frac{\partial ^2V}{\partial\rho ^2}=\displays......{\partial y^2} \quad\mbox{in}\quad\rho =x=y=0\, ,\; z =c\, .$$

Since V verifies Laplace equation, we get

$$\left.\displaystyle\frac{\partial ^2V}{\partial\rho ^2}\rig......\ [-.2cm]z = c\end{array}} = 1+\displaystyle\frac{n}{2}\, ,$$

a value which lies in the non-integrability region S_-n-1 and coincides with that previously obtained by the authors in [2] but with rather more tedious calculations.

Once we have established that the limit problem (3) does not have a complete system of meromorphic integrals, it follows that neither does the problem (2) in a neighbourhood of $\beta =0$.

Now, it is enough to apply the usual continuity argument - see [12] or [3] -, since the non-integrability regions correspond to those for which the trace of the monodromy matrices is >2.

However, notice that such an argument is more complicated than simply checking the aforementioned inequality. In fact, it is necessary to verify that the Riemann surface, the poles and the loops which are used to calculate the monodromy matrices do not suffer significant qualitative changes for any problem close to the limit problem, i.e., in a neighbourhood of $\beta =0$ after performing the change of scale. Nevertheless, this can be verified in an immediate way.

To this end it is enough to realize that the basic solution can be considered of the form $\rho =0$, $z=\psi (t)$, where $\psi$ is a solution of the differential equation

$$\ddot\psi + \displaystyle\frac{\partial V}{\partial z}(0,\psi ) = 0\, ,$$

which admits as a first integral

$$\displaystyle\frac{1}{2}(\dot\psi )^2 + V(0,\psi ) = h\, ,$$

and that the Riemann surface involved in Ziglin's theorem is defined by the integral

$$t = \int\displaystyle\frac{dw}{\sqrt{p(w)}}\, ,\quad \mbox{with}\;\;p(w) = 2(h - V(0,w))\, .$$

The monodromy group of the normal variation equation is then built by taking closed circuits around the branch points corresponding to the roots of p(w)=0.

In the limit problem, we can take $h=\varepsilon _n\,P_n(\psi /\sqrt{\psi ^2})$ and n+1 simple poles, roots of wⁿ⁺¹=1, appear. If the potential is a finite sum of zonal harmonics it is obvious that the solution $\rho =0$ still exists and the equation wⁿ⁺¹=1 would be replaced by another one with the same leading and independent terms and with lower powers of w multiplied by factors which are positive powers of $\beta ^{\frac{2}{n+3}}$ and hence it goes on having n+1 simple roots provided $\beta$ is small enough.

Notice that this result includes as a particular case the non-integrability through meromorphic integrals of the truncated two fixed centres problem in the asymmetric case. The non-integrability of such a problem in the symmetric case was first established by Irigoyen [4] by following a scheme similar to that used for the J₂-problem in [3]. We would like to point out that the starting solution used in Irigoyen's work is not suitable for the study of odd harmonics.

On the other hand, although in (1) only the harmonics with negative powers have been included, it is easy to realize that if we add up a finite number of zonal harmonics of the form r^kV_k to (1), the conclusion still holds. Such terms could be envisaged as coming from the truncation of the development of a perturbation due to a third body by simply assuming some suitable conditions (e.g. the Keplerian orbit is circular and equatorial).