First integrals for the Kepler problem with linear drag

Abstract

In this work we consider the Kepler problem with linear drag, and prove the existence of a continuous vector-valued first integral, obtained taking the limit as $t\rightarrow +\infty$ of the Runge–Lenz vector. The norm of this first integral can be interpreted as an asymptotic eccentricity $e_{\infty }$ with $0\le e_{\infty } \le 1$. The orbits satisfying $e_{\infty } <1$ approach the singularity by an elliptic spiral and the corresponding solutions $x(t)=r(t)e^{i\theta (t)}$ have a norm r(t) that goes to zero like a negative exponential and an argument $\theta (t)$ that goes to infinity like a positive exponential. In particular, the difference between consecutive times of passage through the pericenter, say $T_{n+1} -T_n$, goes to zero as $\frac{1}{n}$.

Appendix

Proof of Lemma 2.2

To understand the need for all the subtleties involved in the proof it is convenient to start with a special case. By now we impose the additional assumption $K\cap M_0 =\emptyset$ with $M_0 =\{ (x,v)\in \Omega : x\wedge v=0\}$. In this case we know that the energy $E(t;x_0,v_0)$ tends decreasingly to $-\infty$ as $t\rightarrow +\infty$ for each $(x_0,v_0)\in K$. By continuous dependence we can find an instant $t_* =t_* (x_0,v_0)>0$ and an open neighbourhood $\mathcal{U} =\mathcal{U} (x_0,v_0)$ such that

$$\begin{aligned} E(t_*;x_0,v_0)<-\frac{1}{2}\quad \mathrm{if}\quad (x_0,v_0)\in \mathcal{U}. \end{aligned}$$

(53)

Next we use the compactness of K to find a finite covering of the type $\left\{ \mathcal{U} (x_0^{(i)},v_0^{(i)}) \right\} _{1\le i\le n}$. If we define

$$\begin{aligned} \tau =\max \{ t_* (x_0^{(i)},v_0^{(i)}) : 1\le i\le n\}, \end{aligned}$$

we obtain

$$\begin{aligned} E(\tau ;x_0,v_0)<-\frac{1}{2}\quad \mathrm{for \; each}\quad (x_0,v_0)\in K. \end{aligned}$$

(54)

Then

$$\begin{aligned} |x(t; x_0,v_0)|<\frac{1}{|E(t ;x_0,v_0)|} <2 \quad \mathrm{if}\quad t\ge \tau . \end{aligned}$$

(55)

By continuous dependence we know that the number

$$\begin{aligned} \hat{m}_K =\max \{ |x(t; x_0,v_0)|: t\in [0,\tau ], (x_0,v_0)\in K\} \end{aligned}$$

(56)

is finite. The claim holds with $m_K =\max \{ 2,\hat{m}_K \}$.

The previous argument cannot be employed when $K\cap M_0 \ne \emptyset$. In this case some solutions are only defined in the future on an interval $[0,\omega [$ with $\omega <\infty$ and the limit of the energy as $t\rightarrow \omega$ can be any number (see again Margheri et al. 2014). However, if we replace classical solutions on $M_0$ by bouncing solutions (obtained by gluing collision-ejection solutions in a sequence, adjacent ejection and collision occurring with the same energy) then they will be defined for all future times and the energy will eventually become negative. This observation will be made rigorous via the Levi–Civita regularization that was presented in Margheri et al. (2014). We recall that, after the natural identification of $x=(x_1,x_2)$ with the complex number $x_1+ix_2$, the Levi–Civita regularization is defined by the change of variables

$$\begin{aligned} x=w^2,\quad ds=\frac{dt}{|x|}. \end{aligned}$$

(57)

Using this regularization Eq. (1) is transformed into the polynomial system of ODEs in the new time s

$$\begin{aligned} w'=v,\quad v'=\frac{Ew}{2} -\epsilon |w|^2 v, \quad E'=-2\epsilon (E|w|^2 +1). \end{aligned}$$

(58)

This system has to be considered on the invariant manifold

$$\begin{aligned} \mathcal{M}=\{ (w,v,E)\in \mathbb {C}^2 \times \mathbb {R} :\, E|w|^2 +1-2|v|^2 =0\}, \end{aligned}$$

(59)

which contains all the physically meaningful solutions. Let (w(s), v(s), E(s)) be a solution lying on $\mathcal{M}$. We claim that this solution is defined on $[0,\infty [$ and that the energy E(s) eventually becomes negative. Assume that $[0,\sigma [$ is the maximal interval to the right. From the equation defining the manifold $\mathcal{M}$ we deduce that

$$\begin{aligned} |w'(s)|= \sqrt{\frac{1+E(s)|w(s)|^2}{2}}. \end{aligned}$$

This leads to the differential inequality for |w(s)|,

$$\begin{aligned} \frac{d}{ds} |w(s)|\le |w'(s)|\le \sqrt{\frac{1+E(0)|w(s)|^2}{2}}, \end{aligned}$$

(60)

and a standard continuation argument shows that $\sigma =\infty$. To prove that the energy is eventually negative we reason by contradiction and assume that E(s) has a limit $E_{\infty } \ge 0$ as $s\rightarrow +\infty$. Then $E(s)>0$ for each $s>0$ and

$$\begin{aligned} E(s)=E(0)-2\epsilon \int _0^s (E(\xi )|w(\xi )|^2 +1)d\xi \le E(0)-2\epsilon s \rightarrow -\infty , \end{aligned}$$

(61)

a contradiction with $E_{\infty } \ge 0$. We are ready to present the complete proof of the lemma. Given $(x_0,v_0)\in K$ we can produce two initial conditions $(w_0,\hat{v}_0, E_0)$ for the regularized system (58)–(59). After identifying $x_0$ and $v_0$ with complex numbers, $w_0$ will be a square root of $x_0$, $\hat{v}_0 =\frac{|x_0 |v_0}{2w_0}$ and $E_0 =\frac{1}{2}|v_0 |^2 -\frac{1}{|x_0 |}$. We observe that the triplets $(w_0 ,\hat{v}_0,E_0)$ lie on a compact subset $\mathcal {K}$ of $\mathcal {M}$. The same type of compactness argument as in the previous case shows the existence of a number $s_*$ depending only on $\mathcal {K}$ (and hence on K) such that

$$\begin{aligned} E(s;w_0,\hat{v}_0,E_0)<-\frac{1}{2} \quad \mathrm{if} \quad s\ge s_* \quad \mathrm{and}\quad (w_0,\hat{v}_0,E_0)\in \mathcal {K}. \end{aligned}$$

(62)

Define

$$\begin{aligned} \hat{m}_\mathcal{{K}} =\max \{ |w(s;w_0,\hat{v}_0,E_0)|^2 : s\in [0,s_* ], (w_0,\hat{v}_0,E_0)\in \mathcal {K} \} <\infty . \end{aligned}$$

(63)

Going back to the original variables and time we deduce that

$$\begin{aligned} |x(t; x_0,v_0)|\le \hat{m}_\mathcal{{K}}\quad \mathrm{if}\quad 0\le t\le t_* \quad \mathrm{and}\quad |x(t;x_0,v_0)|\le 2\quad \mathrm{if}\quad t\ge t_*, \end{aligned}$$

(64)

where

$$\begin{aligned} t_* = \int _0^{s_*} |w(s;w_0,\hat{v}_0,E_0)|^2 ds \quad \mathrm{for\; each} \quad (w_0,\hat{v}_0,E_0)\in \mathcal {K}. \end{aligned}$$

(65)

The instant $t_*$ will be different for each initial condition but this creates no trouble because the number $m_K =\max \{ 2, \hat{m}_\mathcal{{K}}\}$ only depends on K. $\square$