Compact normalisations in the elliptic restricted three body problem

Abstract

This paper considers the spatial elliptic restricted three body problem in the case that the particle with negligible mass is revolving around one of the primaries. The system is modelled in an inertial frame through a Hamiltonian function representing a non-autonomous dynamical system with three degrees of freedom that depends periodically on time. Three Lie transformations are applied at first order to eliminate successively the mean anomaly of the infinitesimal particle’s motion, the time dependence of the system and the argument of the node of the particle with negligible mass. All the transformations are implemented in a compact way, that is, carrying out the computations by means of infinite series. This approach can be very useful to deal with certain artificial satellite problems or, in general, with systems such that the ratio between the distance of the infinitesimal particle to the body around it orbits and the distance between the two primaries is smaller than one but close to it. In this case the Legendre expansion of the potential converges slowly and many terms of the series must be taken into consideration.

Appendix: Integrals of powers of sines and cosines

In this Appendix we describe how to obtain the integrals appearing in Sect. 3 to get the averaged Hamiltonians and the associated generating functions. We discuss three cases.

1. Case: $\int (\cos x)^{2n+1}(\sin x)^{k}\,\mathrm{d}x$

We make the change $u=\sin x$, so $\mathrm{d}u=\cos x\,\mathrm{d}x$, arriving at

$$\begin{aligned} & \int (\cos x)^{2n+1}(\sin x)^{k}\,\mathrm{d}x \\ &\quad= \int (\cos x)^{2n}u^{k} \,\mathrm{d}u \\ &\quad= \int \bigl(1-u^{2}\bigr)^{n} u^{k}\, \mathrm{d}u \\ &\quad= \int \sum_{s\ge 0}^{n} \binom{n}{s}(-1)^{s}u^{2s+k}\,\mathrm{d}u \\ &\quad=\sum_{s\ge 0}^{n} \binom{n}{s}\frac{(-1)^{s}}{2s+k+1} u^{2s+k+1} \\ &\quad= \sum_{s\ge 0}^{n} \binom{n}{s}\frac{(-1)^{s}}{2s+k+1} (\sin x)^{2s+k+1}. \end{aligned}$$

Then, it is readily concluded that

$$\begin{aligned} \frac{1}{2\pi }\int_{0}^{2\pi }(\cos x)^{2n+1}(\sin x)^{k}\,\mathrm{d}x=0 \end{aligned}$$

(55)

and

$$\begin{aligned} &\int_{\text{pp}}(\cos x)^{2n+1}(\sin x)^{k}\,\mathrm{d}x \\ &\quad=\sum_{s\ge 0} ^{n}\binom{n}{s}\frac{(-1)^{s}}{2s+k+1}(\sin x)^{2s+k+1}. \end{aligned}$$

(56)

The terms $(\sin x)^{2s+k+1}$ may be expressed as combinations of sines and cosines of multiples of $x$ by using the following identities, see Wolfram (2017):

$$\begin{aligned} &(\sin x)^{p} = \frac{2}{2^{p}}\sum_{u\ge 0}^{\frac{p-1}{2}} (-1)^{ ( \frac{p-1}{2}-u ) } \binom{p}{u}\sin \bigl((p-2u)x\bigr) \\ &\quad \mbox{if } p \mbox{ is odd}, \end{aligned}$$

(57)

$$\begin{aligned} &\!\begin{aligned} (\sin x)^{p} &= \frac{1}{2^{p}}\binom{p}{p/2} \\ &\quad{}+\frac{2}{2^{p}} \sum_{u\ge 0}^{ \frac{p}{2}-1} (-1)^{ ( \frac{p}{2}-u ) } \binom{p}{u}\cos \bigl((p-2u)x\bigr) \end{aligned} \\ &\quad \mbox{if } p \mbox{ is even.} \end{aligned}$$

(58)

If $p$ is even we can remove the term $\frac{1}{2^{p}}\binom{p}{p/2}$ of the indefinite integral $\int_{\text{pp}}$, as the constant terms will not affect the generating function.

2. Case: $\int (\cos x)^{n}(\sin x)^{2k+1}\,\mathrm{d}x$

We make the change of variable $u=\cos x$. Thence, $\mathrm{d}u=-\sin x \,\mathrm{d}x$ and therefore,

$$\begin{aligned} &\int (\cos x)^{n}(\sin x)^{2k+1}\,\mathrm{d}x \\ &\quad=-\int u^{n}(\sin x)^{2k} \,\mathrm{d}u \\ &\quad=-\int u^{n}(1-u^{2})^{k}\,\mathrm{d}u \\ &\quad=-\int \sum_{s\ge 0}^{k} \binom{k}{s}(-1)^{s}u^{2s+n}\,\mathrm{d}u \\ &\quad= -\sum_{s\ge 0}^{k} \binom{k}{s}(-1)^{s}\int u^{2s+n}\,\mathrm{d}u \\ &\quad= \sum_{s\ge 0}^{k} \binom{k}{s}\frac{(-1)^{s+1}}{2s+n+1} u^{2s+n+1} \\ &\quad= \sum_{s\ge 0}^{k} \binom{k}{s} \frac{(-1)^{s+1}}{2s+n+1} (\cos x)^{2s+n+1}. \end{aligned}$$

Thus, it follows that

$$\begin{aligned} \frac{1}{2\pi }\int_{0}^{2\pi }(\cos x)^{n}(\sin x)^{2k+1}\,\mathrm{d}x=0, \end{aligned}$$

(59)

together with

$$\begin{aligned} &\int_{\text{pp}}(\cos x)^{n}(\sin x)^{2k+1}\,\mathrm{d}x \\ &\quad =\sum_{s\ge 0} ^{k}\binom{k}{s}\frac{(-1)^{s+1}}{2s+n+1}(\cos x)^{2s+n+1}. \end{aligned}$$

(60)

We may express $(\cos x)^{2s+n+1}$ as combinations of sines and cosines of multiples of $x$, see for instance Wolfram (2017):

$$\begin{aligned} &(\cos x)^{p}=\frac{2}{2^{p}}\sum_{u\ge 0}^{ \frac{p-1}{2}} \binom{p}{u}\cos \bigl((p-2u)x\bigr) \quad \mbox{if } p \mbox{ is odd}, \end{aligned}$$

(61)

$$\begin{aligned} &(\cos x)^{p}=\frac{1}{2^{p}}\binom{p}{p/2}+\frac{2}{2^{p}}\sum _{u \ge 0}^{\frac{p}{2}-1}\binom{p}{u}\cos \bigl((p-2u)x \bigr) \\ &\quad \mbox{if } p \mbox{ is even.} \end{aligned}$$

(62)

If $p$ is even we can remove the term $\frac{1}{2^{p}}\binom{p}{p/2}$ from the indefinite integral $\int_{\text{pp}}$, as the constant terms do not affect the form of the generating function.

3. Case: $\int (\cos x)^{2n}(\sin x)^{2k}\,\mathrm{d}x$

In this case we do not need to perform any change of variable, since it is enough to put the powers of $\sin x$ in terms of powers of $\cos x$, or vice versa. Concretely we make

$$\begin{aligned} &\int (\cos x)^{2n}(\sin x)^{2k}\,\mathrm{d}x \\ &\quad=\int (\cos x)^{2n}(1-\cos ^{2}x)^{k}\,\mathrm{d}x \\ &\quad= \int (\cos x)^{2n}\sum_{s\ge 0}^{k} \binom{k}{s}(-1)^{s}(\cos^{2}x)^{s}\,\mathrm{d}x \\ &\quad= \sum_{s\ge 0}^{k}\binom{k}{s}(-1)^{s}\int (\cos x)^{2n+2s}\,\mathrm{d}x. \end{aligned}$$

(63)

We express $(\cos x)^{2n+2s}$ as combinations of sines and cosines of multiples of $x$ using (62), arriving at

$$\begin{aligned} &(\cos x)^{2n+2s} \\ &\quad= \frac{1}{2^{2n+2s}}\binom{2n+2s}{n+s} \\ &\qquad{}+ \frac{2}{2^{2n+2s}}\sum_{u \ge 0}^{n+s-1} \binom{2n+2s}{u} \cos \bigl(2(n+s-u)x\bigr). \end{aligned}$$

(64)

Therefore, (63) is transformed into

$$\begin{aligned} & \int (\cos x)^{2n}(\sin x)^{2k}\,\mathrm{d}x \\ &\quad= \sum_{s\ge 0}^{k} \binom{k}{s}(-1)^{s} \int \frac{1}{2^{2n+2s}}\binom{2n+2s}{n+s}\,\mathrm{d}x \\ &\qquad{} +\sum_{s\ge 0}^{k} \binom{k}{s}(-1)^{s}\ \frac{2}{2^{2n+2s}}\sum _{u\ge 0}^{n+s-1}\binom{2n+2s}{u} \\ &\qquad{}\times \int \cos \bigl(2(n+s-u)x\bigr)\,\mathrm{d}x \end{aligned}$$

from where we get

$$\begin{aligned} &\frac{1}{2\pi }\int_{0}^{2\pi }(\cos x)^{2n}(\sin x)^{2k}\,\mathrm{d}x \\ &\quad= \sum_{s\ge 0}^{k}\binom{k}{s} \binom{2n+2s}{n+s} \frac{(-1)^{s}}{2^{2n+2s}}, \end{aligned}$$

(65)

and

$$\begin{aligned} &\int_{\text{pp}} (\cos x)^{2n}(\sin x)^{2k}\,\mathrm{d}x \\ &\quad= \sum_{s \ge 0}^{k}\binom{k}{s}(-1)^{s} \frac{2}{2^{2n+2s}}\sum_{u\ge 0}^{n+s-1} \binom{2n+2s}{u} \\ &\qquad{} \times \frac{\sin (2(n+s-u)x)}{2n+2s-2u} \\ &\quad= \sum_{s\ge 0}^{k} \sum_{u\ge 0}^{n+s-1}\binom{k}{s} \binom{2n+2s}{u}\frac{(-1)^{s}}{2^{2n+2s}} \\ &\qquad{} \times\frac{\sin (2(n+s-u)x)}{n+s-u}. \end{aligned}$$

(66)