# Hyperbolic and Bi-hyperbolic solutions in the planar restricted $(N+1)$-body
problem

Guowei Yu  Chern Institute of Mathematics and LPMC, Nankai University,
Tianjin, China<EMAIL_ADDRESS>
###### Abstract.

Consider the planar restricted $(N+1)$-body problem with trajectories of the
$N(\geq 2)$ primaries forming a collision-free periodic solution of the
$N$-body problem, for any positive energy $h$ and directions
$\theta_{\pm}\in[0,2\pi)$, we prove that starting from any initial position
$x$ at any initial time $t_{x}$, there are hyperbolic solutions
$\gamma^{\pm}|_{[t_{x},\pm\infty)}$ satisfying $\gamma^{\pm}(t_{x})=x$ and

$\lim_{t\to\pm\infty}\gamma^{\pm}(t)/|\gamma^{\pm}(t)|=e^{i\theta_{\pm}(\text{mod
}2\pi)},\;\;\lim_{t\to\pm\infty}\dot{\gamma}^{\pm}(t)=\pm\sqrt{2h}e^{i\theta_{\pm}(\text{mod
}2\pi)}.$

Moreover we also prove the existence of a bi-hyperbolic solution
$\gamma|_{\mathbb{R}}$ satisfying

$\lim_{t\to\pm\infty}\gamma(t)/|\gamma(t)|=e^{i\theta_{\pm}(\text{mod
}2\pi)},\;\;\lim_{t\to\pm\infty}\dot{\gamma}(t)=\pm\sqrt{2h}e^{i\theta_{\pm}(\text{mod
}2\pi)}.$

This work is supported by the National Key R&D Program of China
(2020YFA0713303), NSFC (No. 12171253), Nankai Zhide Fundation and the
Fundamental Research Funds for the Central Universities.

##  1\. Introduction

The $N$-body problem studies motion of $N$ point masses, $m_{i}>0$,
$i\in\mathbf{N}:=\\{1,\dots,N\\}$, under Newton’s universal gravitational law.
The trajectories of the masses $q(t)=(q_{i}(t))_{i\in\mathbf{N}}$ satisfy

(1)
$\ddot{q}_{i}(t)=-\sum_{j\in\mathbf{N}\setminus\\{i\\}}\frac{m_{j}(q_{i}(t)-q_{j}(t))}{|q_{i}(t)-q_{j}(t)|^{3}},\;\;\forall
i\in\mathbf{N}.$

One of the most important aspects of the problem is to understand final
motions of the masses as time goes to infinity. When $N=3$, a complete
classification of the final behavior was given by Chazy [8] (also see [2], a
similar classification can be obtained for $N>3$). Although a solution of (1)
may end at a finite time due to singularity, by results from [31], [32] and
[17], such solutions should be rare in the measure sense.

By introducing an additional massless body into the system, we get the so-
called _restricted $(N+1)$-body problem_, where the motion of the massless
body is governed by the gravitational forces of the $N$ positive masses (will
be referred as primaries), while the massless body does not produce any
gravitational force, so the motion of the primaries are not affected by its
presence. As a result, the trajectory $z(t)$ of the massless body satisfies

(2)
$\ddot{z}(t)=\partial_{z}U(z(t),t)=-\sum_{i\in\mathbf{N}}\frac{m_{i}(z(t)-q_{i}(t))}{|z(t)-q_{i}(t)|^{3}},$

where $U(z,t)$ is a time-dependent potential function

(3) $U(z,t)=\sum_{i\in\mathbf{N}}\frac{m_{i}}{|z-q_{i}(t)|}.$

The restricted $(N+1)$-body problem can be seen as certain limit of the
$(N+1)$-body problem, when one of the masses is much smaller than the others.
Hence it is a good model in describing the motion of comets and satellites in
gravitational systems, and has been studies by many authors including Poincaré
and Birkhoff.

Solutions of (2) may also experience singularities at finite time. However if
we assume there is no collision between the primaries, then all these
singularities must be binary collisions between the massless body and one of
the primaries. As is well-known such singularities can be regularized (see
[23], [29] and [35]), so solutions can be extended through these binary
collisions as elastic bounce. Hence the maximum domain of any solution of (2)
will be the whole time interval $\mathbb{R}$. Then following Chazy, a complete
classification of all the solutions of (2) according to the final behaviors of
the massless body can be given as below.

  * •

hyperbolic solution: $|z(t)|\to\infty$ and $|\dot{z}(t)|\to c>0$, as
$t\to\pm\infty$;

  * •

parabolic solution: $|z(t)|\to\infty$ and $|\dot{z}(t)|\to 0$, as
$t\to\pm\infty$;

  * •

bounded solution: $\limsup_{t\to\pm\infty}|z(t)|<\infty$;

  * •

oscillatory solution: $\limsup_{t\to\pm\infty}|z(t)|<\infty$ and
$\liminf_{t\to\pm\infty}|z(t)|<\infty$.

######  Remark 1.1.

$z|_{\mathbb{R}}$ will be called bi-hyperbolic, if both $z|_{[0,\infty)}$ and
$z|_{(-\infty,0]}$ are hyperbolic solutions.

In this paper, we will study (bi)-hyperbolic solutions of the planar
restricted $(N+1)$-body problem, and we set $\mathbb{R}^{2}=\mathbb{C}$ for
the rest of the paper.

######  Theorem 1.1.

Assume $q(t)=(q_{i}(t))_{i\in\mathbf{N}}$ is a planar collision-free periodic
solution of (1), i.e.,

(*) $\begin{cases}&q(t)=q(t+T)\in\mathbb{C}^{N},\;\forall
t\in\mathbb{R},\;\text{ for some }T>0;\\\
&\rho_{0}=\inf\\{|q_{i}(t)-q_{j}(t)|:t\in\mathbb{R},\;i\neq
j\in\mathbf{N}\\}>0.\par\end{cases}$

Given arbitrarily an energy constant $h>0$ and asymptotic directions
$\theta_{\pm}\in[0,2\pi)$, for any $x\in\mathbb{C}$ and $t_{x}\in\mathbb{R}$,
there are hyperbolic solutions $\gamma^{\pm}:[t_{x},\pm\infty)\to\mathbb{C}$
of (2) satisfies $\gamma^{\pm}(t_{x})=x$ and

$\lim_{t\to\pm\infty}\gamma^{\pm}(t)/|\gamma^{\pm}(t)|=e^{i\theta_{\pm}(\text{mod
}2\pi)},\;\;\lim_{t\to\pm\infty}\dot{\gamma}^{\pm}(t)=\pm\sqrt{2h}e^{i\theta_{\pm}(\text{mod
}2\pi)}.$

Moreover $\gamma^{\pm}(t)$ must be collision-free, whenever $t\neq t_{x}$.

######  Remark 1.2.

For any $t\in\mathbb{R}$, we say $\gamma(t)\in\mathbb{C}$ is collision-free,
if $\gamma(t)\neq q_{i}(t)$, for all $i\in\mathbf{N}$.

Next we will compare the above result with some previous results of the
$N$-body problem.

######  Definition 1.1.

Let $q|_{[t_{0},\pm\infty)}$ be a solution of (1). We say it is a hyperbolic
solution, if

$|q_{i}(t)-q_{j}(t)|\simeq O(|t|),\text{ as }t\to\pm\infty,\;\forall\\{i\neq
j\\}\subset\mathbf{N},$

it is a parabolic solution, if

$|q_{i}(t)-q_{j}(t)|\simeq O(|t|^{\frac{2}{3}}),\text{ as
}t\to\pm\infty,\;\forall\\{i\neq j\\}\subset\mathbf{N},$

and it is a hyperbolic-elliptic solution, if there is a
$\mathbf{K}\subset\mathbf{N}$ with $2\leq|\mathbf{K}|<N$, such that

$|q_{i}(t)-q_{j}(t)|\leq C_{1}<\infty,\;\forall|t|\geq t_{0}\text{ and
}\forall\\{i\neq j\\}\subset\mathbf{K};$ $|q_{i}(t)-q_{j}(t)|\simeq
O(|t|),\text{ as
}t\to\pm\infty,\;\forall(i,j)\in\mathbf{N}\times(\mathbf{N}\setminus\mathbf{K})\text{
with }i\neq j.$

######  Remark 1.3.

Recall that the energy of a solution of the $N$-body is a constant. In
particular the energy of a hyperbolic solution must be positive, and of a
parabolic solution must be zero. In contrary, the energy of a hyperbolic-
elliptic solution can be any real number.

In [26] Maderna and Venturelli showed that for any $h>0$, starting from any
initial configuration, there is a $h$-energy hyperbolic solution with any
prescribed non-collision limiting shape at infinity (a different proof of this
result was given in [24] recently). A similar result for parabolic solutions
was obtained by Maderna and Venturelli in [25], when the energy is zero and
the limiting shapes at infinity are minimal central configurations. While
hyperbolic and parabolic solutions only affect dynamics on positive and zero
energy surfaces, hyperbolic-elliptic solutions will affect dynamics at any
energy surface as there is no restriction on energies of these solutions.
However so far it is unclear whether similar results hold for hyperbolic-
elliptic solutions.

Since hyperbolic solutions of the restricted $(N+1)$-body problem could be
seen as certain limit of hyperbolic-elliptic solutions of the $(N+1)$-body
problem with a single escaping mass, Theorem 1.1 may also help us understand
hyperbolic-elliptic solutions of the unrestricted problem.

Under different coordinates, solutions mentioned above form stable manifolds
(when time goes to positive infinity) or unstable manifolds (when time goes to
negative infinity) of different invariant subsets at infinity parameterized by
the limiting energy and shape or direction (see [27], [30] and [15]). The
geometric and topological properties of these manifolds have great influences
on the global dynamics. A basic question is: _with given limiting energies and
shapes or directions at infinity corresponding to negative and positive
infinite time, does the corresponding stable and unstable manifolds
intersect?_ The intersection of these stable and unstable manifolds is
equivalent to the existence of solutions with the given limiting energies and
shapes as time goes to negative and positive infinity.

For the $N$-body problem, this is a difficult problem with very few results.
Up to our knowledge there is only some partial result in [15] for hyperbolic
solutions. Here we obtain the following result for the restricted $(N+1)$-body
problem.

######  Theorem 1.2.

Under the assumption (* ‣ 1.1), given arbitrarily an energy constant $h>0$ and
a pair of asymptotic directions $\theta_{\pm}\in[0,2\pi)$. There is a bi-
hyperbolic solution $\gamma:\mathbb{R}\to\mathbb{C}$ of (2) satisfying

$\lim_{t\to\pm\infty}\gamma(t)/|\gamma(t)|=e^{i\theta_{\pm}(\text{mod
}2\pi)},\;\;\lim_{t\to\pm\infty}\dot{\gamma}(t)=\pm\sqrt{2h}e^{i\theta_{\pm}(\text{mod
}2\pi)}.$

Moreover $\gamma|_{\mathbb{R}}$ has at most one collision.

For the $N$-center problem, which is a simplified model of the restricted
$(N+1)$-problem with the primaries fixed, there are much more results. In [21]
and [22], (bi)-hyperbolic solutions of the planar and spatial $N$-center
problem were studied correspondingly. More recently using variational methods,
[6] and [3] studied bi-parabolic solutions and bi-hyperbolic solutions for the
spatial $N$-center problem correspondingly, and [4] studied bi-parabolic
solutions in the planar $N$-center problem.

Our proofs of Theorem 1.1 and Theorem 1.2 are variational in nature. Since the
work of Chenciner and Montgomery [14], variational methods have had a lot of
success in the $N$-body problem, first in proving periodic solutions (see
[16], [9], [39] and the references within), and recently in proving parabolic
and hyperbolic solutions (see [25] and [26] given above). The success is
similar in the $N$-center problem. Besides references mentioned above,
periodic solutions were studied in [33], [38], [7] and [12], while homoclinic
and heteroclinic solutions were studied in [13].

Compare to the $N$-body problem and the $N$-center problem, very few results
were obtained using variational methods in the restricted $(N+1)$-body
problem. Up to our knowledge, so far there are only a few results for periodic
solutions, see [1], [20] and [11], and no result seems to be available for
parabolic or hyperbolic solutions.

The main difficulties in proving (bi)-hyperbolic solutions of the restricted
$(N+1)$-body problems using variational methods are as below. First, energy is
not conserved in this case as the system is time periodic, and the energy
conservation plays an important role in controlling the asymptotic behaviors
at infinity. Second, to obtain a bi-hyperbolic solution as a minimizer, we
need to find some proper topological constraint. Without it the minimizer is
likely to be a trivial solution staying at infinity all the time. On the other
hand, if the topological constraint was not chosen properly then the
corresponding minimizer may be a collision-ejection solution that is not
regularizable, as indicated by Gordon’s result [19] for the Kepler problem.

Our paper is organized as following: Section 2 contains results that are
essential for applying action minimization methods to the restricted
$(N+1)$-body problem; Section 3 and Section 4 are devoted to the proofs of
Theorem 1.1 and Theorem 1.2 correspondingly; Appendix A shows how to lower the
action of a collision-ejection solution near an isolated collision by local
deformation; Appendix B contains proofs of two propositions essential to
control the asymptotic behavior of the massless body at infinity.

Notations. The following notations will be used throughout the paper.

  1. (1)

for any $t\in\mathbb{R}$, $\\{t\\}\in[0,1)$ represents the fractional part of
$t$.

  2. (2)

$C_{i}$, $i=1,2,\cdots$, are finite positive constants that vary in different
properties and proofs.

  3. (3)

For any $x\in\mathbb{C}$ and $\delta>0$,
$B_{\delta}(x)=\\{y\in\mathbb{C}:|y-x|\leq\delta\\}$ and
$B_{\delta}=\\{y\in\mathbb{C}:|y|\leq\delta\\}.$

  4. (4)

$H^{1}([t_{1},t_{2}],\mathbb{C})$ denotes the set of all Sobolev paths defined
on $[t_{1},t_{2}]$.

  5. (5)

For any $x,y\in\mathbb{C}$ and $t_{1}<t_{2}\in\mathbb{R}$,

$H^{1}_{t_{1},t_{2}}(x,y)=\\{\gamma\in
H^{1}([t_{1},t_{2}],\mathbb{C}):\;\gamma(t_{1})=x,\gamma(t_{2})=y\\}.$

  6. (6)

For any $\gamma\in H^{1}_{t_{1},t_{2}}(x,y)$,
$\Delta(\gamma)=\\{t\in[t_{1},t_{2}]:\gamma(t)=q_{i}(t),\text{ for some
}i\in\mathbf{N}\\}.$

  7. (7)

For any $\gamma\in H^{1}([t_{1},t_{2}],\mathbb{C})$ and $\xi\in
H^{1}([\tau_{1},\tau_{2}],\mathbb{C})$ with $\gamma(t_{2})=\xi(\tau_{1})$ and
$\\{t_{2}\\}=\\{\tau_{1}\\}$,
$(\gamma*\xi)|_{[t_{1},t_{2}+\tau_{2}-\tau_{1}]}=\gamma|_{[t_{1},t_{2}]}*\xi|_{[\tau_{1},\tau_{2}]}$
represents the concatenation of $\gamma$ and $\xi$ as

$(\gamma*\xi)(t)=\begin{cases}\gamma(t),\;&\text{ when }t\in[t_{1},t_{2}];\\\
\xi(t-t_{2}+\tau_{1}),\;&\text{ when
}t\in[t_{2},t_{2}+\tau_{2}-\tau_{1}].\end{cases}$

##  2\. Properties of action minimizers

In this section we prove some basic results that will be needed when applying
action minimization methods to the restricted $(N+1)$-body problem. Some ideas
are similarly to those used in the $N$-body problem.

For any constant $h$, a collision-free critical point of the action functional

(4)
$\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\int_{t_{1}}^{t_{2}}L(\gamma(t),\dot{\gamma}(t),t)+h\,dt;\;\;L(\gamma,\dot{\gamma},t)=\frac{1}{2}|\dot{\gamma}(t)|^{2}+U(\gamma(t),t),$

is a solution of the restricted problem (2).

######  Definition 2.1.

We say $\gamma\in H^{1}_{t_{1},t_{2}}(x,y)$ is _a fixed-end local minimizer_
of $\mathcal{A}_{h}$, if $\exists\varepsilon>0$, such that

$\mathcal{A}_{h}(\gamma;t_{1},t_{2})\leq\mathcal{A}_{h}(\xi;t_{1},t_{2}),\;\;\forall\xi\in
H^{1}_{t_{1},t_{2}}(x,y)\text{ satisfying
}\|\xi-\gamma\|_{H^{1}}\leq\varepsilon.$

We say $\gamma\in H^{1}_{t_{1},t_{2}}(x,y)$ _a fixed-end minimizer_ of
$\mathcal{A}_{h}$, if

$\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\psi_{h}(x,y;t_{1},t_{2}):=\inf\\{\mathcal{A}_{h}(\gamma):\;\gamma\in
H^{1}_{t_{1},t_{2}}(x,y)\\}.$

For simplicity, a fixed-end (local) minimizer will be called a (local)
minimizer in the following.

######  Proposition 2.1.

If $\gamma\in H^{1}([t_{1},t_{2}],\mathbb{C})$ is a local minimizer of
$\mathcal{A}_{h}$ with finite action value, then its set of collision moments
$\Delta(\gamma)$ is an isolated subset in $[t_{1},t_{2}]$.

###### Proof.

As for any $t\in\Delta(\gamma)$, $U(\gamma(t),t)=\infty$, the Lebesgues
measure of $\Delta(\gamma)$ must be zero. Therefore
$[t_{1},t_{2}]\setminus\Delta(\gamma)$ is the union of some (at most
countable) open intervals.

Choose a $t_{0}\in\Delta(\gamma)\cap(t_{1},t_{2})$ (if $t_{1}$ or
$t_{2}\in\Delta(\gamma)$, the proof is similar and will be omitted). Then
$\gamma(t_{0})=q_{i_{0}}(t_{0})$, for some $i_{0}\in\mathbf{N}$. By (* ‣ 1.1),
we can find a $\delta>0$ small enough, such that

(5) $\forall
t\in[t_{0}-\delta,t_{0}+\delta],\;\begin{cases}|\gamma(t)-q_{i}(t)|\leq\rho_{0}/2,\;&\text{if
}i=i_{0};\\\ |\gamma(t)-q_{i}(t)|\geq\rho_{0}/2,\;&\text{if
}i\in\mathbf{N}\setminus\\{i_{0}\\}.\end{cases}$

Notice that for any
$[\tau_{1},\tau_{2}]\subset[t_{0}-\delta,t_{0}+\delta]\setminus\Delta(\gamma)$,
$\gamma|_{[\tau_{1},\tau_{2}]}$ is a smooth solution of (2), as it is a
collision-free local minimizer of $\mathcal{A}_{h}$. Then $E_{i_{0}}\in
C^{1}([t_{0}-\delta,t_{0}+\delta]\setminus\Delta(\gamma),\mathbb{R})$, where

(6)
$E_{i_{0}}(t)=\frac{1}{2}|\dot{\gamma}(t)-\dot{q}_{i_{0}}(t)|^{2}-\frac{m_{i_{0}}}{|\gamma(t)-q_{i_{0}}(t)|}.$

Moreover we have the following lemma, whose proof will be postponed.

######  Lemma 2.1.

$E_{i_{0}}(t)$ can be extended continuously to $[t_{0}-\delta,t_{0}+\delta]$.

By a contradiction argument, let’s assume $t_{0}$ is not isolated in
$\Delta(\gamma)$. Set $I(t)=|\gamma(t)-q_{i_{0}}(t)|^{2}$, for any $t$, the we
can find a sequence of times $\\{\tau_{n}(\neq t_{0})\\}_{n\in\mathbb{N}}$
satisfying

$\lim_{n\to\infty}\tau_{n}=t_{0}\;\text{ and }\;I(\tau_{n})=0,\;\forall
n\in\mathbb{N}.$

We may further assume $\\{\tau_{n}\\}_{n\in\mathbb{N}}$ is monotone and
$(\tau_{2k-1},\tau_{2k})\cap\Delta(\gamma)=\emptyset$, $\forall
k\in\mathbb{N}$. Then for each $k\in\mathbb{N}$, there is a
$s_{k}\in(\tau_{2k-1},\tau_{2k})$, such that

(7) $\dot{I}(s_{k})=0,\;\forall k\in\mathbb{N}.$

By (5) and Lemma 2.1, for any
$t\in[t_{0}-\delta,t_{0}+\delta]\setminus\Delta(\gamma)$, there is a $C_{1}$,
such that

(8) $\displaystyle\ddot{I}(t)$
$\displaystyle=2\frac{m_{i_{0}}}{|\gamma(t)-q_{i_{0}}(t)|}+4E_{i_{0}}(t)-2\langle\gamma(t)-q_{i_{0}}(t),\sum_{i\neq
i_{0}}\frac{m_{i}(\gamma(t)-q_{i}(t))}{|\gamma(t)-q_{i}(t)|^{3}}+\ddot{q}_{i_{0}}(t)\rangle$
$\displaystyle\geq 2m_{i_{0}}I^{-\frac{1}{2}}(t)-C_{1}.$

Since $I(t_{0})=0$, we can find a $\delta_{0}\in(0,\delta)$, such that

(9) $\ddot{I}(t)>0,\;\forall
t\in[t_{0}-\delta_{0},t_{0}+\delta_{0}]\setminus\Delta(\gamma)$

As $\lim_{k\to\infty}s_{k}=t_{0}$, we have $\dot{I}(s_{k})>0$, for $k$ large
enough, which is a contradiction to (7). ∎

###### Proof of Lemma 2.1.

Since
$\max\\{|\ddot{q}_{i_{0}}(t)|,|\dot{q}_{i_{0}}(t)|:\;t\in[t_{1},t_{2}]\\}<C_{2}$,
for some constant $C_{2}$,

$\displaystyle\int_{t_{0}-\delta}^{t_{0}+\delta}|\dot{\gamma}-\dot{q}_{i_{0}}|^{2}\,dt$
$\displaystyle\leq\int_{t_{0}-\delta}^{t_{0}+\delta}|\dot{\gamma}|^{2}\,dt+2C_{2}\int_{t_{0}-\delta}^{t_{0}+\delta}|\dot{\gamma}|\,dt+2\delta
C_{2}^{2}$
$\displaystyle\leq\int_{t_{0}-\delta}^{t_{0}+\delta}|\dot{\gamma}|^{2}\,dt+2C_{2}\left(2\delta\int_{t_{0}-\delta}^{t_{0}+\delta}|\dot{\gamma}|^{2}\,dt\right)^{\frac{1}{2}}+2\delta
C_{2}^{2}$ $\displaystyle\leq
2\mathcal{A}(\gamma;t_{1},t_{2})+4C_{2}\sqrt{\delta\mathcal{A}(\gamma;t_{1},t_{2})}+2\delta
C_{2}^{2}<\infty.$

By the definition of $E_{i_{0}}(t)$,

$\int_{t_{0}-\delta}^{t_{0}+\delta}|E_{i_{0}}(t)|\,dt\leq\frac{1}{2}\int_{t_{0}-\delta}^{t_{0}+\delta}|\dot{\gamma}-\dot{q}_{i_{0}}|^{2}\,dt+\mathcal{A}(\gamma;t_{1},t_{2})<\infty.$

Meanwhile for any $t\in[t_{1},t_{2}]\setminus\Delta(\gamma)$,

$\dot{E}_{i_{0}}=\langle\ddot{\gamma}-\ddot{q}_{i_{0}},\dot{\gamma}-\dot{q}_{i_{0}}\rangle+\langle\frac{m_{i_{0}}(\gamma-
q_{i_{0}})}{|\gamma-
q_{i_{0}}|^{3}},\dot{\gamma}-\dot{q}_{i_{0}}\rangle=-\langle\ddot{q}_{i_{0}}+\sum_{i\neq
i_{0}}\frac{m_{i}(\gamma-q_{i})}{|\gamma-
q_{i}|^{3}},\dot{\gamma}-\dot{q}_{i_{0}}\rangle.$

By (5), $|\dot{E}_{i_{0}}|\leq C_{3}|\dot{\gamma}-\dot{q}_{i_{0}}|$, for some
finite $C_{3}$. Then the following implies $\dot{E}_{i_{0}}\in
L^{1}([t_{0}-\delta,t_{0}+\delta],\mathbb{R})$

$\int_{t_{0}-\delta}^{t_{0}+\delta}|\dot{\gamma}-\dot{q}_{i_{0}}|\,dt\leq\left(2\delta\int_{t_{0}-\delta}^{t_{0}+\delta}|\dot{\gamma}-\dot{q}_{i_{0}}|^{2}\,dt\right)^{\frac{1}{2}}$

As a result, $E_{i_{0}}\in
W^{1,1}([t_{0}-\delta,t_{0}+\delta],\mathbb{R})\subset
C([t_{0}-\delta,t_{0}+\delta],\mathbb{R}).$ ∎

By the above result, let’s assume $\gamma\in
H^{1}([t_{0}-\delta,t_{0}+\delta],\mathbb{R})$ is a collision-ejection
solution of (2), i.e., $\gamma(t)$ is collision-free and satisfies (2),
$\forall t\neq t_{0}$, and $\gamma(t_{0})=q_{i_{0}}(t_{0})$, for some
$i_{0}\in\mathbf{N}$. Then we have the following two propositions, whose
proofs can be found in Appendix A.

######  Proposition 2.2.

There exist $\sigma_{\pm}\in\mathbb{C}$ and $f^{\pm}\in
C^{0}([t_{0}-\delta,t_{0}+\delta],\mathbb{C})\cap
C^{1}([t_{0}-\delta,t_{0}+\delta]\setminus\\{t_{0}\\},\mathbb{C})$ with
$f^{\pm}(t)\simeq o(|t-t_{0}|^{2/3})$ and $\dot{f}^{\pm}(t)\simeq
o(|t-t_{0}|^{-1/3})$, when $t\to t_{0}$, such that the following hold.

  1. (a).

$\sigma_{\pm}=\lim_{t\to
t_{0}^{\pm}}\frac{\gamma(t)-q_{i_{0}}(t)}{|\gamma(t)-q_{i_{0}}(t)|}.$

  2. (b).

When $t\in(t_{0},t_{0}+\delta]$,

$\displaystyle\gamma(t)-q_{i_{0}}(t)$
$\displaystyle=(\frac{9}{2}m_{i_{0}})^{\frac{1}{3}}(t-t_{0})^{\frac{2}{3}}\sigma_{+}+f^{+}(t);$
$\displaystyle\dot{\gamma}(t)-\dot{q}_{i_{0}}(t)$
$\displaystyle=\frac{2}{3}(\frac{9}{2}m_{i_{0}})^{\frac{1}{3}}(t-t_{0})^{-\frac{1}{3}}\sigma_{+}+\dot{f}^{+}(t).$

  3. (c).

When $t\in[t_{0}-\delta,t_{0})$,

$\displaystyle\gamma(t)-q_{i_{0}}(t)$
$\displaystyle=(\frac{9}{2}m_{i_{0}})^{\frac{1}{3}}(t-t_{0})^{\frac{2}{3}}\sigma_{-}+f^{-}(t);$
$\displaystyle\dot{\gamma}(t)-\dot{q}_{i_{0}}(t)$
$\displaystyle=\frac{2}{3}(\frac{9}{2}m_{i_{0}})^{\frac{1}{3}}(t-t_{0})^{-\frac{1}{3}}\sigma_{-}+\dot{f}^{-}(t).$

######  Proposition 2.3.

When $\sigma_{-}\neq\sigma_{+}$, for any $\varepsilon>0$ small enough, there
are two collision-free paths $\eta^{\pm}\in
H^{1}_{t_{0}-\delta,t_{0}+\delta}(\gamma(t_{0}-\delta),\gamma(t_{0}+\delta))$
satisfying
$\mathcal{A}(\eta^{\pm};t_{0}-\delta,t_{0}+\delta)<\mathcal{A}(\gamma;t_{0}-\delta,t_{0}+\delta)$
and

(10) $|\eta^{\pm}(t)-\gamma(t)|\leq\varepsilon,\;\;\forall
t\in[t_{0}-\delta,t_{0}+\delta].$

Moreover for $\delta$ small enough, we have

(11)
$\displaystyle\text{Arg}(\eta^{+}(t_{0}+\delta)-q_{i_{0}}(t_{0}+\delta))-\text{Arg}(\eta^{+}(t_{0}-\delta)-q_{i_{0}}(t_{0}-\delta))$
$\displaystyle\in(0,2\pi);$
$\displaystyle\text{Arg}(\eta^{-}(t_{0}+\delta)-q_{i_{0}}(t_{0}+\delta))-\text{Arg}(\eta^{-}(t_{0}-\delta)-q_{i_{0}}(t_{0}-\delta))$
$\displaystyle\in(-2\pi,0).$

######  Remark 2.1.

A similar result as the above proposition was obtained in [5] for the
perturbed Kepler problem. However it does not imply our result, as the
trajectories of the primaries were given a priori, so can not be deformed.

######  Proposition 2.4.

If $\gamma\in H^{1}([t_{1},t_{2}],\mathbb{C})$ is a local minimizer of
$\mathcal{A}_{h}$, then $\gamma|_{(t_{1},t_{2})}$ is a collision-free solution
of (2).

###### Proof.

This follows immediately from Proposition 2.1 and 2.3. ∎

######  Proposition 2.5.

For any $x,y\in\mathbb{C}$ and $t_{1}<t_{2}$, there is a $\gamma\in
H^{1}_{t_{1},t_{2}}(x,y)$ satisfying

$\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\psi_{h}(x,y;t_{1},t_{2}).$

###### Proof.

As $\mathcal{A}_{h}$ is weakly lower semi-continuous in
$H^{1}_{t_{1},t_{2}}(x,y)$, by a standard argument from the direct method of
calculus of variation, the existence of such a $\gamma$ follows once we can
show $\mathcal{A}_{h}$ is coercive in $H^{1}_{t_{1},t_{2}}(x,y)$, i.e., if a
sequence $\\{\gamma_{n}\in H^{1}_{t_{1},t_{2}}(x,y)\\}$ satisfies
$\|\gamma_{n}\|_{H^{1}}\to\infty$, as $n\to\infty$, then
$\mathcal{A}_{h}(\gamma_{n})\to\infty$ as well.

By a contradiction argument, let’s assume $\|\gamma_{n}\|_{H^{1}}\to\infty$,
as $n\to\infty$ and there is a finite $C_{1}$, such that
$\mathcal{A}_{h}(\gamma_{n})\leq C_{1}$, $\forall n$. Then the Cauchy-Schwartz
inequality implies

$\int_{t_{1}}^{t_{2}}|\dot{\gamma}_{n}|\,dt\leq(t_{2}-t_{1})^{\frac{1}{2}}\left(\int_{t_{1}}^{t_{2}}|\dot{\gamma}_{n}(t)|^{2}\,dt\right)^{\frac{1}{2}}\leq\sqrt{2(t_{2}-t_{1})C_{1}}.$

As a result,

$|\gamma_{n}(t)|\leq|x|+\int_{t_{1}}^{t}|\dot{\gamma}_{n}(\tau)|\,d\tau\leq|x|+\sqrt{2(t_{2}-t_{1})C_{1}},\;\forall
t\in[t_{1},t_{2}].$

Then the following inequality holds for all $n$, which is a contradiction.

$\|\gamma_{n}\|^{2}_{H^{1}}=\int_{t_{1}}^{t_{2}}|\gamma(t)|^{2}+|\dot{\gamma}(t)|^{2}\,dt\leq(|x|+\sqrt{2(t_{2}-t_{1})C_{1}})^{2}(t_{2}-t_{1})+2C_{1}.$

∎

By (* ‣ 1.1) or the periodicity of $q(t)$, we can find an $R_{0}>0$, such that

(12) $\sup\\{|q_{i}(t)|:t\in\mathbb{R},\;i\in\mathbf{N}\\}=R_{0}-1.$

######  Lemma 2.2.

Given any $(x,y)\in\mathbb{C}\times(\mathbb{C}\setminus B_{R_{0}}^{o})$ and
$t_{1}<t_{2}$, there is a $C_{1}$ with

$\sup\\{\psi_{h}(x,z;t_{1},t_{2}):z\in B_{1/2}(y)\\}\leq C_{1}.$

###### Proof.

By Proposition 2.5, there is a $\gamma\in H^{1}_{t_{1},t_{2}}(x,y)$ with
$\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\psi_{h}(x,y;t_{1},t_{2})$. Then

$\tau_{0}=\min\\{t\in[t_{1},t_{2}]:\gamma(t)\in B_{1/2}(y)\\}<t_{2}.$

Notice that for any $z\in B_{1/2}(y)$, $|z|\geq|y|-|y-z|\geq R_{0}-1/2$, then
there is a $C_{2}>0$, such that

(13) $U(z,t)\leq C_{2},\;\forall z\in B_{1/2}(y)\text{ and }t\in\mathbb{R}.$

Now for any $z\in B_{1/2}(y)$, we define a new path

$\xi(t)=\begin{cases}\gamma(t),&\text{ if }t\in[t_{1},\tau_{0}],\\\
\gamma(\tau_{0})+\frac{t-\tau_{0}}{t_{2}-\tau_{0}}(z-\gamma(\tau_{0})),&\text{
if }t\in[\tau_{0},t_{2}].\end{cases}$

Since $\xi(t)\in B_{1/2}(y)$, $\forall t\in[\tau_{0},t_{0}]$, by (13),

$\displaystyle\psi_{h}(x,z;t_{1},t_{2})-\psi_{h}(x,y;t_{1},t_{2})$
$\displaystyle\leq\int_{t_{1}}^{t_{2}}L(\xi,\dot{\xi},t)-L(\gamma,\dot{\gamma},t)\,dt\leq\int_{\tau_{0}}^{t_{2}}L(\xi,\dot{\xi},t)\,dt$
$\displaystyle=\int_{\tau_{0}}^{t_{2}}\frac{1}{2}|\xi(t)|^{2}+U(\xi(t),t)\,dt\leq\frac{|z-\gamma(\tau_{0})|^{2}}{2(t_{2}-\tau_{0})}+C_{2}(t_{2}-\tau_{0})$
$\displaystyle\leq\frac{1}{2(t_{2}-\tau_{0})}+C_{2}(t_{2}-\tau_{0}).$

As a result,
$C_{1}=\psi_{h}(x,y;t_{1},t_{2})+C_{2}(t_{2}-\tau_{0})+2^{-1}(t_{2}-\tau_{0})^{-1}$.
∎

######  Proposition 2.6.

Fix arbitrarily an $x\in\mathbb{C}$ and a $t_{1}\in\mathbb{R}$,
$(y,t_{2})\mapsto\psi_{h}(x,y;t_{1},t_{2})$ is locally Lipschitz continuous in
$(\mathbb{C}\setminus
B_{R_{0}}^{o})\times\\{t_{2}\in\mathbb{R}:t_{2}>t_{1}\\}$.

###### Proof.

Since $h$ is a constant, it is enough to prove the result for $h=0$.

Let’s fix an arbitrary $t_{2}>t_{1}$, and prove the local Lipschitz continuity
of $\phi_{0}$ with respect to $y$ first. Choose a $y_{0}\in\mathbb{C}\setminus
B^{o}_{R_{0}}$ and a $\delta\in(0,1/4]$ small enough. For any $y\in
B_{\delta}(y_{0})$, by Proposition 2.5, there is a $\gamma_{y}\in
H^{1}_{t_{1},t_{2}}(x,y)$ satisfying
$\mathcal{A}_{0}(\gamma_{y};t_{1},t_{2})=\psi_{0}(x,y;t_{1},t_{2})$. Let

$\tau_{y}=\min\\{t_{0}\in[t_{1},t_{2}]:\gamma_{y}(t)\in
B_{1/2}(y_{0}),\;\forall t\in[t_{0},t_{2}]\\}.$

We claim

(14) $\tau=\sup\\{\tau_{y}:y\in B_{\delta}(y_{0})\text{ with
}\tau_{y}>t_{1}\\}<t_{2}.$

Notice that we only need to consider the cases, when $\tau_{y}>t_{1}$, and in
these case, we must have $|\gamma_{y}(\tau_{y})-y_{0}|=1/2$. Then by Lemma
2.2, there is a $C_{1}>0$, such that

(15) $\displaystyle C_{1}\geq\mathcal{A}_{0}(\gamma_{y};t_{1},t_{2})$
$\displaystyle\geq\frac{1}{2}\int_{\tau_{y}}^{t_{2}}|\dot{\gamma}_{y_{n}}|^{2}\,dt\geq\frac{1}{2(t_{2}-\tau_{y})}\left(\int_{\tau_{y}}^{t_{2}}|\dot{\gamma}_{y}|\,dt\right)^{2}$
$\displaystyle\geq\frac{|y-\gamma_{y}(\tau_{y})|^{2}}{2(t_{2}-\tau_{y})}\geq\frac{(|\gamma_{y}(\tau_{y})-y_{0}|-|y_{0}-y|)^{2}}{2(t_{2}-\tau_{y})}\geq\frac{(\frac{1}{2}-\delta)^{2}}{2(t_{2}-\tau_{y})}.$

As a result $t_{2}-\tau_{y}\geq(1-2\delta)^{2}/(8C_{1})>0$, and this implies
(14).

Next we show that there must be an $s_{y}\in[\tau,t_{2}]$ with

$|\dot{\gamma}_{y}(s_{y})|\leq\sqrt{2C_{1}(t_{2}-t_{1})}/(t_{2}-\tau).$

Otherwise
$\int_{\tau}^{t_{2}}|\dot{\gamma}_{y}|\,dt>\sqrt{2C_{1}(t_{2}-t_{1})}$, which
is a contradiction to

$\int_{\tau}^{t_{2}}|\dot{\gamma}_{y}|\,dt\leq\int_{t_{1}}^{t_{2}}|\dot{\gamma}_{y}|\,dt\leq\sqrt{2(t_{2}-t_{1})\mathcal{A}_{0}(\gamma_{y};t_{1},t_{2})}\leq\sqrt{2C_{1}(t_{2}-t_{1})}.$

Meanwhile since $\forall t\in[\tau,t_{2})$, $\gamma_{y}(t)$ satisfies (2), and
$\gamma_{y}(t)\in B_{1/2}(y_{0})\subset(\mathbb{C}\setminus
B^{o}_{R_{0}-1/2})$, there is a $C_{2}>0$, such that

$|\ddot{\gamma}_{y}(t)|\leq|\nabla U(\gamma_{y}(t),t)|\leq C_{2}.$

Therefore for any $t\in[\tau,t_{2}]$ and $y\in B_{\delta}(y_{0})$,

(16)
$|\dot{\gamma}_{y}(t)|\leq|\dot{\gamma}_{y}(s_{y})|+\int_{\tau}^{t_{2}}|\ddot{\gamma}_{y}(s)|\,ds\leq\frac{\sqrt{2C_{1}(t_{2}-t_{1})}}{t_{2}-\tau}+C_{2}(t_{2}-\tau)=C_{3}.$

Let $\delta_{0}=\min\\{\delta,t_{2}-\tau\\}$. For any $y\in
B_{\delta_{0}}(y_{0})$, define a new path $\xi\in H^{1}_{t_{2}-\Delta
y,t_{2}}(\gamma_{y_{0}}(t_{2}-\Delta y),y_{0})$ as below, where $\Delta
y=|y-y_{0}|$,

$\xi_{y}(t)=\gamma_{y_{0}}(t_{2}-\Delta y)+\frac{t+\Delta y-t_{2}}{\Delta
y}\big{(}y-\gamma_{y_{0}}(t_{2}-\Delta y)\big{)},\;t\in[t_{2}-\Delta
y,t_{2}].$

By the triangle inequality and (16),

$\displaystyle\int_{t_{2}-\Delta y}^{t_{2}}|\dot{\xi}_{y}|^{2}\,dt$
$\displaystyle\leq\frac{1}{\Delta y}\left(\int_{t_{2}-\Delta
y}^{t_{2}}|\dot{\xi}_{y}|\,dt\right)^{2}=\frac{|y-\gamma_{y_{0}}(t_{2}-\Delta
y)|^{2}}{\Delta y}$
$\displaystyle\leq\frac{(|y-y_{0}|+|y_{0}-\gamma_{y_{0}}(t_{2}-\Delta
y)|)^{2}}{\Delta y}\leq\frac{\left(\Delta y+\int_{t_{2}-\Delta
y}^{t_{2}}|\dot{\gamma}_{y_{0}}|\,dt\right)^{2}}{\Delta
y}\leq(C_{3}+1)^{2}\Delta y.$

As there is a $C_{4}>0$, such that $U(z,t)\leq C_{4}$, for any
$z\in\mathbb{C}\setminus B^{0}_{R_{0}-1/2}$ and $t\in\mathbb{R}$,

$\psi_{0}(x,y;t_{1},t_{2})-\psi_{0}(x,y_{0};t_{1},t_{2})\leq\int_{t_{2}-\Delta
y}^{t_{2}}\frac{1}{2}|\dot{\xi}_{y}|^{2}+U(\xi_{y}(t),t)\,dt\leq\frac{1}{2}(C_{3}+1)^{2}\Delta
y+C_{4}\Delta y.$

Meanwhile we can also define a path $\eta_{y}\in H^{1}_{t_{2}-\Delta
y,t_{2}}(\gamma_{y}(t_{2}-\Delta y),y)$ as

$\eta_{y}(t)=\gamma_{y}(t_{2}-\Delta y)+\frac{t+\Delta y-t_{2}}{\Delta
y}\big{(}y-\gamma_{y}(t_{2}-\Delta y)\big{)},\;t\in[t_{2}-\Delta y,t_{2}].$

Then a similar argument as above shows

$\psi_{0}(x,y_{0};t_{1},t_{2})-\psi_{0}(x,y;t_{1},t_{2})\leq\int_{t_{2}-\Delta
y}^{t_{2}}\frac{1}{2}|\dot{\eta}_{y}|^{2}+U(\eta_{y}(t),t)\,dt\leq\frac{1}{2}(C_{3}+1)^{2}\Delta
y+C_{4}\Delta y.$

These results imply that for any $y\in B_{\delta}(y_{0})$,

$|\psi_{0}(x,y;t_{1},t_{2})-\psi_{0}(x,y_{0};t_{1},t_{2})|\leq\frac{(C_{3}+1)^{2}+2C_{4}}{2}|y-y_{0}|=C_{5}|y-y_{0}|.$

This finishes our proof of the local Lipschitz continuity of $\phi_{0}$ with
respect to $y$.

To show the local Lipschitz continuity of $\phi_{0}$ with respect to $t_{2}$.
Let $\gamma_{y}\in H^{1}_{t_{1},t_{2}}(x,y)$ be a minimizer of
$\mathcal{A}_{0}$. Now choose a $\Delta t>0$ small enough, then similar
arguments as above show

(17) $|\dot{\gamma}_{y}(t)|\leq C_{3}\text{ and }U(\gamma_{y}(t),\tau)\leq
C_{4},\;\forall t\in[t_{2}-\Delta t,t_{2}]\text{ and }\tau\in\mathbb{R}.$

This implies

$|y-\gamma_{y}(t_{2}-\Delta t)|\leq\int_{t_{2}-\Delta
t}^{t_{2}}|\dot{\gamma}_{y}(t)|\,dt\leq C_{3}\Delta t;$
$\mathcal{A}_{0}(\gamma_{y};t_{2}-\Delta t,t_{2})=\int_{t_{2}-\Delta
t}^{t_{2}}\frac{1}{2}|\dot{\gamma}_{y}(t)|^{2}+U(\gamma_{y}(t),t)\,dt\leq(\frac{1}{2}C_{3}^{2}+C_{4})\Delta
t.$

Combining these inequalities with the Lipschitz continuity with respect to
$y$, we get

$\displaystyle|\psi_{0}(x,y;t_{1},t_{2})$
$\displaystyle-\psi_{0}(x,y;t_{1},t_{2}-\Delta t)|$
$\displaystyle\leq\mathcal{A}_{0}(\gamma_{y};t_{2}-\Delta
t,t_{2})+|\psi_{0}(x,\gamma_{y}(t_{2}-\Delta t);t_{1},t_{2}-\Delta
t)-\psi_{0}(x,y;t_{1},t_{2}-\Delta t)|$
$\displaystyle\leq(\frac{1}{2}C_{3}^{2}+C_{4})\Delta
t+C_{5}|y-\gamma_{y}(t_{2}-\Delta
t)|\leq(\frac{1}{2}C_{3}^{2}+C_{4}+C_{3}C_{5})\Delta t.$

On the other hand,

$\displaystyle|\psi_{0}(x,y;$ $\displaystyle t_{1},t_{2}+\Delta
t)-\psi_{0}(x,y;t_{1},t_{2})|$
$\displaystyle\leq|\psi_{0}(\gamma_{y}(t_{2}-\Delta t),y;t_{2},t_{2}+\Delta
t)|+|\psi_{0}(x,\gamma_{y}(t_{2}-\Delta
t);t_{1},t_{2})-\psi_{0}(x,y;t_{1},t_{2})|$
$\displaystyle\leq|\psi_{0}(\gamma_{y}(t_{2}-\Delta t),y;t_{2},t_{2}+\Delta
t)|+C_{6}|y-\gamma_{y}(t_{2}-\Delta t)|$
$\displaystyle\leq|\psi_{0}(\gamma_{y}(t_{2}-\Delta t),y;t_{2},t_{2}+\Delta
t)|+C_{3}C_{6}\Delta t.$

Now define a new path $\tilde{\gamma}_{y}\in H^{1}_{t_{2},t_{2}+\Delta
t}(\gamma_{y}(t_{2}-\Delta t),y)$ as
$\tilde{\gamma}_{y}(s)=\gamma_{y}(s-\Delta t)$, $\forall
s\in[t_{2},t_{2}+\Delta t]$. With (17), we get

$|\psi_{0}(\gamma_{y}(t_{2}-\Delta t),y;t_{2},t_{2}+\Delta
t)|\leq\mathcal{A}_{0}(\tilde{\gamma}_{y};t_{2},t_{2}+\Delta
t)\leq(\frac{1}{2}C_{3}^{2}+C_{4})\Delta t.$

As a result,

$|\psi_{0}(x,y;t_{1},t_{2}+\Delta
t)-\psi_{0}(x,y;t_{1},t_{2})|\leq(\frac{1}{2}C_{3}^{2}+C_{4}+C_{3}C_{6})\Delta
t.$

This proves the local Lipschitz continuity of $\phi_{0}$ with respect to
$t_{2}$. ∎

To control the asyptotic energy of the massless body as it goes to infinity,
instead of minimizer with the same starting and ending time, we need to
consider minimizers with varying starting and ending time. However as the
system is $T$-periodic the time difference must belong to $T\mathbb{Z}$. For
simplicity, we will only state and prove these results for $T=1$ (it is easy
to generalize the following results to arbitrary $T>0$). Notice that

In the rest of the section, we assume $q(t)$ is $1$-periodic, i.e., $T=1$ in
(* ‣ 1.1).

######  Definition 2.2.

For any $x,y\in\mathbb{C}$ and $s_{1},s_{2}\in[0,1)$, we say $\gamma\in
H^{1}_{t_{1},t_{2}}(x,y)$ is _a free-time minimizer of $\mathcal{A}_{h}$_, if
$\\{t_{i}\\}=s_{i}$, $i=1,2$, and
$\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\phi_{h}(x,y;s_{1},s_{2})$, where

$\phi_{h}(x,y;s_{1},s_{2}):=\inf\\{\mathcal{A}_{h}(\gamma):\gamma\in
H^{1}_{\tau_{1},\tau_{2}}(x,y)\text{ with
}\tau_{1}<\tau_{2},\;\\{\tau_{i}\\}=s_{i},\;i=1,2\\};$

######  Remark 2.2.

Since the Lagrangian $L$ is $1$-periodic, if $\gamma|_{[t_{1},t_{2}]}$ is a
free-time minimizer of $\mathcal{A}_{h}$, then so is
$\gamma|_{[\tau_{1},\tau_{2}]}$, for any $t_{1}\leq\tau_{1}<\tau_{2}\leq
t_{2}$.

######  Proposition 2.7.

When $h>0$, for any $x\neq y\in\mathbb{C}$ and $s_{1},s_{2}\in[0,1)$, there
exist $t_{1}<t_{2}$ with $\\{t_{i}\\}=s_{i}$, $i=1,2$ and a $\gamma\in
H^{1}_{t_{1},t_{2}}(x,y)$ satisfying
$\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\phi_{h}(x,y;s_{1},s_{2})$. Moreover
$\gamma|_{(t_{1},t_{2})}$ is a collision-free solution of (2).

###### Proof.

First let’s assume $s_{1}<s_{2}$, then

$\phi_{h}(x,y;s_{1},s_{2})=\inf_{n\in\mathbb{Z}^{+}\cup\\{0\\}}\psi_{h}(x,y;s_{1},s_{2}+n).$

Since $h>0$, $\psi_{h}(x,y;s_{1},s_{2}+n)\geq h(s_{2}-s_{1}+n)\to\infty$, when
$n\to\infty$. As a result,

$\phi_{h}(x,y;s_{1},s_{2})=\psi_{h}(x,y;s_{1},s_{2}+n_{0}),\;\text{ for some
}n_{0}\in\mathbb{Z}^{+}\cup\\{0\\}.$

By Proposition 2.5, there is a $\gamma\in H^{1}_{s_{1},s_{2}+n_{0}}(x,y)$
satisfying

$\mathcal{A}_{h}(\gamma_{n};s_{1},s_{2}+n)=\psi_{h}(x,y;s_{1},s_{2}+n_{0})=\phi_{h}(x,y;s_{1},s_{2}),$

and $\gamma|_{(s_{1},s_{2}+n_{0})}$ is a collision-free solution of (2).

When $s_{1}\geq s_{2}$,
$\phi_{h}(x,y;s_{1},s_{2})=\inf_{n\in\mathbb{Z}^{+}}\psi_{h}(x,y;s_{1},s_{2}+n)$,
and the rest of the proof is the exactly the same as above. ∎

######  Proposition 2.8.

When $h>0$, for any $x\neq y$ and $s\in[0,1)$, there is a $s^{*}\in[0,1)$,
such that

$\phi_{h}(x,y;s,s^{*})=\inf\\{\phi_{h}(x,y;s,t):t\in[0,1)\\}.$

###### Proof.

As the infimum in the above equation is finite, there is a sequence
$\\{s_{n}\in[0,1)\\}_{n\in\mathbb{Z}^{+}}$ with

(18)
$\lim_{n\to\infty}\phi_{h}(x,y;s,s_{n})=\inf\\{\phi_{h}(x,y;s,t):t\in[0,1)\\}\leq
C_{1}.$

By Proposition 2.7, for each $n\in\mathbb{Z}^{+}$, there is a
$k_{n}\in\mathbb{Z}^{+}\cup\\{0\\}$ and a $\gamma_{n}\in
H^{1}_{s,s_{n}+k_{n}}(x,y)$ satisfying

$\mathcal{A}_{h}(\gamma_{n};s,s_{n}+k_{n})=\phi_{h}(x,y;s,s_{n}).$

We claim $\\{k_{n}\\}_{n\in\mathbb{Z}^{+}}$ must have a finite upper bound.
Otherwise

$\phi_{h}(x,y;s,s_{n})=\mathcal{A}_{h}(\gamma_{n};s,s_{n}+k_{n})\geq
hk_{n}\to\infty,\text{ as }n\to\infty,$

which is a contradiction to (18).

This implies the existence of a finite $t^{*}\in\mathbb{R}$ with (after
passing to a subsequence) $s_{n}+k_{n}\to t^{*}$, as $n\to\infty$. Since
$\gamma_{n}(s)=x$, $\forall n$ and

$\int_{s}^{s_{n}+k_{n}}|\dot{\gamma}_{n}|^{2}\,dt\leq
2\mathcal{A}_{h}(\gamma_{n};s,s_{n}+k_{n})=2\phi_{h}(x,y;s,s_{n})\leq 2C_{1}.$

By a similar argument as in the proof of Proposition 2.5, we can show
$\\{\gamma_{n}|_{[s,s_{n}+k_{n}]}\\}_{n\in\mathbb{Z}^{+}}$is a bounded
sequence under the $H^{1}$-norm. Then there is a $\gamma\in
H^{1}_{s,t^{*}}(x,y)$, such that (after passing to a proper subsequence)
$\gamma_{n}$ will converge weakly to $\gamma$ with respect to the
$H^{1}$-norm, as $n\to\infty$. The weakly lower semi-continuity of
$\mathcal{A}_{h}$ then implies

$\phi_{h}(x,y;s,s^{*})\leq\mathcal{A}_{h}(\gamma;s,t^{*})\leq\liminf_{n\to\infty}\mathcal{A}_{h}(\gamma_{n};s,s_{n}+k_{n})=\inf\\{\phi_{h}(x,y;s,t):t\in[0,1)\\},$

where $s^{*}=\\{t^{*}\\}$. The desired result then follows from the fact that

$\inf\\{\phi_{h}(x,y;s,t):t\in[0,1)\\}\leq\phi_{h}(x,y;s,s^{*}).$

∎

##  3\. Existence of hyperbolic solutions

In this section, we assume $q(t)$ is a $1$-periodic solution of (1), and give
a detailed proof of Theorem 1.1, until the last proof, where the cases with
$T\neq 1$ will be treated by a blow-up argument. Moreover only the proof of
$\gamma^{+}|_{[t_{x},\infty)}$ will be given, as it is the same for
$\gamma^{-}|_{(-\infty,t_{x}]}$.

In the following it is more convenient to rewrite the potential function $U$
given in (3) as

(19) $U(z,t)=\frac{m}{|z|}+W(z,t),\text{ where }\;m=\sum_{i=1}^{N}m_{i}.$

A direct computation shows

(20) $W(z,t)\simeq O(m|z|^{-2}),\;\text{ as }|z|\to\infty.$

Therefore we can find a constant $R_{1}\geq R_{0}$, and positive constants
$\alpha_{1}$, $\alpha_{2}$, such that

(21)
$|W(z,t)|\leq\frac{\alpha_{1}}{|z|^{2}}\leq\frac{m}{|z|},\;\;|\partial_{z}W(z,t)|\leq\frac{\alpha_{2}}{|z|^{3}}\leq\frac{m}{|z|^{2}},\;\;\forall|z|\geq
R_{1}.$

Let $\gamma(t)=r(t)e^{i\theta(t)}\in C^{2}([t_{1},t_{2}],\mathbb{C})$ be a
collision-free solution of (2), we have the following results, which are
useful in controlling the asymptotic behavior of the massless body near
infinity.

######  Proposition 3.1.

If $r(t_{1})\geq R_{1}$ and $\dot{r}(t_{1})\geq\sqrt{6m/r(t_{1})}$, then
$\dot{r}(t)>\frac{1}{2}\dot{r}(t_{1})$, $\forall t\in[t_{1},t_{2}]$.

######  Proposition 3.2.

If $r(t_{1})\geq R_{1}$ and $\dot{r}(t)\geq v_{0}$, $\forall t\geq t_{1}$, for
some constant $v_{0}>0$, then

  1. (a).

$\sup\\{|\omega(t)|:\;t\in[t_{1},t_{2}]\\}\leq|\omega(t_{1})|+\frac{\alpha_{2}}{v_{0}r(t_{1})},$,
where $\omega(t)=\gamma(t)\wedge\dot{\gamma}(t)$;

  2. (b).

when $t_{2}=\infty$, $\lim_{t\to\infty}\theta(t)=\theta_{0}$, for some
$\theta_{0}\in\mathbb{R}$.

Proofs of the above two propositions will be given in Appendix B.

######  Lemma 3.1.

For any $x,y\in\mathbb{C}$ satisfying $\sqrt{2}R_{1}\leq|x|,|y|\leq R$, the
following results hold.

  1. (a).

$\psi_{0}(x,y;t_{1},t_{2})\leq
16\frac{R^{2}}{t_{2}-t_{1}}+12m\frac{t_{2}-t_{1}}{R}$, $\forall t_{1}<t_{2}$.

  2. (b).

For any $h>0$, there exist positive constants $\beta_{1},\beta_{2}$
independent of $R$, such that

$\sup\\{\phi_{h}(x,y;s_{1},s_{2}):s_{1},s_{2}\in[0,1)\\}\leq\beta_{1}R+\beta_{2}.$

###### Proof.

(a). Choose a point $z\in\mathbb{C}$ with $|z|=R$. Let
$\overrightarrow{ox},\overrightarrow{oy}$ and $\overrightarrow{oz}$ denote the
three rays starting from the origin and passing the points $x,y$ and $z$
correspondingly. Moreover we assume $\overrightarrow{oz}$ divide the angle
between $\overrightarrow{ox}$ and $\overrightarrow{oy}$ into two equal angles
with each angle less than or equal to $\pi/2$.

Define a straight line segment $\xi(t)=(1-\lambda(t))x+\lambda(t)z$,
$t\in[t_{1},\frac{t_{1}+t_{2}}{2}]$ with

$\lambda(t)=\begin{cases}\mu\left[1-\left(\frac{\tau-t}{\tau-
t_{1}}\right)^{\frac{2}{3}}\right],&\text{ if }t\in[t_{1},\tau];\\\
\mu+(1-\mu)\left(\frac{t-\tau}{\frac{t_{1}+t_{2}}{2}-\tau}\right)^{\frac{2}{3}},&\text{
if }t\in[\tau,\frac{t_{1}+t_{2}}{2}],\end{cases}$

satisfying $\lambda(t_{1})=0$ and $\lambda(\frac{t_{1}+t_{2}}{2})=1$. For
this, we need to let

$\mu=\frac{|x|}{|x|+|z|}\in(0,1),\;\;\tau=t_{1}+\frac{\mu^{3/2}}{\mu^{3/2}+(1-\mu)^{3/2}}\frac{t_{2}-t_{1}}{2}\in(t_{1},\frac{1}{2}(t_{1}+t_{2})).$

By the triangle inequality, for any $t\in[t_{1},\frac{t_{1}+t_{2}}{2}]$,

$|\xi(t)|=|(1-\lambda(t))x+\lambda(t)z|\geq\big{|}(1-\lambda(t))|x|-\lambda(t)|z|\big{|}=|\mu-\lambda(t)|\big{(}|x|+|z|\big{)}.$

Since $|x|,|z|\geq\sqrt{2}R_{1}$, $|\xi(t)|\geq R_{1}$, $\forall
t\in[t_{1},\frac{t_{1}+t_{2}}{2}]$. Then (21) and the above inequality imply

$U(\xi(t),t)\leq\frac{2m}{|\xi(t)|}\leq\frac{2m}{(|x|+|z|)|\mu-\lambda(t)|}\leq\frac{2m}{|z|\cdot|\mu-\lambda(t)|}.$

Using this, we get

$\displaystyle\int_{t_{1}}^{\tau}L(\xi,\dot{\xi},t)\,dt$
$\displaystyle=\int_{t_{1}}^{\tau}\frac{1}{2}|\dot{\xi}|^{2}+U(\xi,t)\,dt\leq\frac{|z-x|^{2}}{2}\int_{t_{1}}^{\tau}|\dot{\lambda}(t)|^{2}\,dt+\frac{2m}{|z|}\int_{t_{1}}^{\tau}\frac{1}{|\mu-\lambda(t)|}\,dt$
$\displaystyle=\frac{2|z-x|^{2}}{9}\frac{\mu^{2}}{(\tau-
t_{1})^{2}}\int_{t_{1}}^{\tau}\left(\frac{\tau-t}{\tau-
t_{1}}\right)^{-\frac{2}{3}}\,dt+\frac{2m}{|z|}\frac{1}{\mu}\int_{t_{1}}^{\tau}\left(\frac{\tau-t}{\tau-
t_{1}}\right)^{-\frac{2}{3}}\,dt$
$\displaystyle=\frac{2}{3}|z-x|^{2}\frac{\mu^{2}}{\tau-
t_{1}}+\frac{6m}{|z|}\frac{\tau-t_{1}}{\mu}$
$\displaystyle=\frac{4}{3}\mu^{\frac{1}{2}}\left(\mu^{\frac{3}{2}}+(1-\mu)^{\frac{3}{2}}\right)\frac{|z-x|^{2}}{t_{2}-t_{1}}+3m\frac{\mu^{\frac{1}{2}}}{\mu^{\frac{3}{2}}+(1-\mu)^{\frac{3}{2}}}\frac{t_{2}-t_{1}}{|z|};$
$\displaystyle\int_{\tau}^{\frac{t_{1}+t_{2}}{2}}L(\xi,\dot{\xi},t)\,dt$
$\displaystyle=\int_{\tau}^{\frac{t_{1}+t_{2}}{2}}\frac{1}{2}|\dot{\xi}(t)|^{2}+U(\xi(t),t)\,dt$
$\displaystyle\leq\frac{|z-x|^{2}}{2}\int_{\tau}^{\frac{t_{1}+t_{2}}{2}}|\dot{\lambda}(t)|^{2}\,dt+\frac{2m}{|z|}\int_{\tau}^{\frac{t_{1}+t_{2}}{2}}\frac{1}{|\mu-\lambda(t)|}\,dt$
$\displaystyle=\left(\frac{2|z-x|^{2}}{9}\frac{(1-\mu)^{2}}{(\frac{t_{1}+t_{2}}{2}-\tau)^{2}}+\frac{2m}{|z|}\frac{1}{1-\mu}\right)\int_{\tau}^{\frac{t_{1}+t_{2}}{2}}\left(\frac{t-\tau}{\frac{t_{1}+t_{2}}{2}-\tau}\right)^{-\frac{2}{3}}\,dt$
$\displaystyle=\frac{2}{3}|z-x|^{2}\frac{(1-\mu)^{2}}{\frac{t_{1}+t_{2}}{2}-\tau}+\frac{6m}{|z|}\frac{\frac{t_{1}+t_{2}}{2}-\tau}{1-\mu}$
$\displaystyle=\frac{4}{3}(1-\mu)^{\frac{1}{2}}\left(\mu^{\frac{3}{2}}+(1-\mu)^{\frac{3}{2}}\right)\frac{|z-x|^{2}}{t_{2}-t_{1}}+3m\frac{(1-\mu)^{\frac{1}{2}}}{\mu^{\frac{3}{2}}+(1-\mu)^{\frac{3}{2}}}\frac{t_{2}-t_{1}}{|z|}.$

These two estimates imply

(22)
$\int_{t_{1}}^{\frac{t_{1}+t_{2}}{2}}L(\xi,\dot{\xi},t)\,dt\leq\frac{4}{3}f_{1}(\mu)f_{2}(\mu)\frac{|z-x|^{2}}{t_{2}-t_{1}}+3m\frac{f_{1}(\mu)}{f_{2}(\mu)}\frac{t_{2}-t_{1}}{|z|},$

where

$f_{1}(\mu)=\mu^{\frac{1}{2}}+(1-\mu)^{\frac{1}{2}},\;f_{2}(\mu)=\mu^{\frac{3}{2}}+(1-\mu)^{\frac{3}{2}}.$

Since $\mu\in(0,1)$,

$f_{1}(\mu)=2\left(\frac{1}{2}\mu^{\frac{1}{2}}+\frac{1}{2}(1-\mu)^{\frac{1}{2}}\right)\leq
2\left(\frac{1}{2}\mu+\frac{1}{2}(1-\mu)\right)^{\frac{1}{2}}=\sqrt{2};$
$1\geq
f_{2}(\mu)=2\left(\frac{1}{2}\mu^{\frac{3}{2}}+\frac{1}{2}(1-\mu)^{\frac{3}{2}}\right)\geq
2\left(\frac{1}{2}\mu+\frac{1}{2}(1-\mu)\right)^{\frac{3}{2}}\geq\frac{\sqrt{2}}{2}.$

Plug these into (22), we have

$\int_{t_{1}}^{\frac{t_{1}+t_{2}}{2}}L(\xi,\dot{\xi})\,dt\leq\frac{4\sqrt{2}}{3}\frac{|z-x|^{2}}{t_{2}-t_{1}}+6m\frac{t_{2}-t_{1}}{|z|}\leq\frac{16\sqrt{2}}{3}\frac{R^{2}}{t_{2}-t_{1}}+6m\frac{t_{2}-t_{1}}{R}.$

A similarly straight line segment $\zeta|_{[\frac{t_{1}+t_{2}}{2},t_{2}]}$
going from $z$ to $y$ can be found, which satisfies

$\int_{\frac{t_{1}+t_{2}}{2}}^{t_{2}}L(\zeta,\dot{\zeta},t)\,dt\leq\frac{16\sqrt{2}}{3}\frac{R^{2}}{t_{2}-t_{1}}+6m\frac{t_{2}-t_{1}}{R}.$

As a result,

$\psi_{0}(x,y;t_{1},t_{2})\leq\int_{t_{1}}^{\frac{t_{1}+t_{2}}{2}}L(\xi,\dot{\xi},t)\,dt+\int_{\frac{t_{1}+t_{2}}{2}}^{t_{2}}L(\zeta,\dot{\zeta},t)\,dt\leq
16\frac{R^{2}}{t_{2}-t_{1}}+12m\frac{t_{2}-t_{1}}{R}.$

(b). Let $t_{1}=s_{1}$, we can always find a $t_{2}\in[s_{1}+R,s_{1}+R+1]$
satisfying $\\{t_{2}\\}=s_{2}.$ Then $R\leq t_{2}-t_{1}=t_{2}-s_{1}\leq R+1$.
Hence

$\displaystyle\phi_{h}(x,y;s_{1},s_{2})$
$\displaystyle\leq\psi_{h}(x,y;t_{1},t_{2})\leq
16\frac{R^{2}}{t_{2}-t_{1}}+6m\frac{t_{2}-t_{1}}{R}+h(t_{2}-t_{1})$
$\displaystyle\leq 16R+6m(1+R^{-1})+hR+h$
$\displaystyle\leq(16+h)R+6m+\frac{3\sqrt{2}m}{R_{1}}+h.$

∎

For the rest of the section, we fix arbitrarily a positive constant $h$, a
position $x\in\mathbb{C}$, a moment $t_{x}$ (without loss of generality we
further assume $t_{x}\in[0,1)$) and an angle $\theta_{+}\in[0,2\pi)$. Set

(23) $\Lambda_{\theta_{+}}(R):=\\{re^{i\theta}\in\mathbb{C}:r\geq R\text{ and
}\theta=\theta_{+}(\text{mod }2\pi)\\},\;\text{ for }R>0.$

By Proposition 2.7 and 2.8, for any $y\in\Lambda_{\theta_{+}}(|x|+1)$, there
is a $\gamma_{y}\in H^{1}_{t_{x},t_{y}}(x,y)$ satisfying

(24)
$\mathcal{A}_{h}(\gamma_{y};t_{x},t_{y})=\phi_{h}(x,y;t_{x},\\{t_{y}\\})=\inf_{s\in[0,1)}\phi_{h}(x,y;t_{x},s).$

######  Lemma 3.2.

For any $R_{2}\geq\max\\{2\sqrt{2}R_{1},|x|+1\\}$ and
$y\in\Lambda_{\theta_{+}}(R_{2})$, let

(25) $\tau_{y}:=\min\\{t\in[t_{x},t_{y}]:|\gamma_{y}(t)|=R_{2}\\}.$

The following results hold.

  1. (a).

$t_{x}<\inf\\{\tau_{y}:\;y\in\Lambda_{\theta_{+}}(R_{2})\\}\leq\sup\\{\tau_{y}:\;y\in\Lambda_{\theta_{+}}(R_{2})\\}<\infty$.

  2. (b).

When $R_{2}$ is large enough,
$\dot{r}_{y}(t)\geq\sqrt{3m/2}R_{2}^{-\frac{1}{2}}$, $\forall
t\in[\tau_{y},t_{y}]$, where $r_{y}(t)=|\gamma_{y}(t)|$.

###### Proof.

(a). The second inequality in the statement is trivial. Let’s give a proof of
the third inequality first. Fix an $n_{0}\in\mathbb{Z}^{+}$, by Proposition
2.6,

$\sup\\{\psi_{h}(x,z;t_{x},t):\;z\in\partial B_{R_{2}}\text{ and
}t\in[n_{0},n_{0}+1]\\}\leq C_{1}.$

Since $\gamma_{y}|_{[t_{x},t_{y}]}$ is a free-time minimizer, so is
$\gamma_{y}|_{[t_{x},\tau_{y}]}$. Then

(26)
$h(\tau_{y}-t_{x})\leq\mathcal{A}_{h}(\gamma_{y};t_{x},\tau_{y})=\phi_{h}(x,\gamma_{y}(\tau_{y});t_{x},\\{\tau_{y}\\})\leq\psi_{h}(x,\gamma_{y}(\tau_{y});t_{x},n_{0}+\\{\tau_{y}\\})\leq
C_{1}.$

This implies

$\sup\\{\tau_{y}:\;y\in\Lambda_{\theta_{+}}(R_{2})\\}\leq
t_{x}+C_{1}/h<\infty.$

Meanwhile by the Cauchy-Schwartz inequality,

$(\tau_{y}-t_{x})\int_{t_{x}}^{\tau_{y}}|\dot{\gamma}_{y}|^{2}\,dt\geq\left(\int_{t_{x}}^{\tau_{y}}|\dot{\gamma}_{y}|\,dt\right)^{2}\geq|\gamma(\tau_{y})-x|^{2}\geq(R_{2}-|x|)^{2}.$

Combining this with (26), we get

$C_{1}\geq\mathcal{A}_{h}(\gamma_{y};t_{x},\tau_{y})\geq\frac{1}{2}\int_{t_{x}}^{\tau_{y}}|\dot{\gamma}_{y}|^{2}\,dt\geq\frac{(R_{2}-|x|)^{2}}{2(\tau_{y}-t_{x})}.$

Assuming the first inequality in the statement does not hold, then we can find
a sequence $\\{y_{n}\in\Lambda_{\theta_{+}}(R_{2})\\}_{n\in\mathbb{Z}^{+}}$
with the corresponding $\\{\tau_{y_{n}}\\}_{n\in\mathbb{Z}^{+}}$ satisfying
$\lim_{n\to\infty}\tau_{y_{n}}=t_{x}$. Then

$C_{1}\geq\mathcal{A}_{h}(\gamma_{y_{n}};t_{x},\tau_{y_{n}})\geq\frac{(R_{2}-|x|)^{2}}{2(\tau_{y_{n}}-t_{x})}\to\infty,\;\text{
as }n\to\infty,$

which is absurd.

(b). For each $y\in\Lambda_{\theta_{+}}(R_{2})$, let
$a_{y}=\max\\{t\in[t_{x},\tau_{y}]:r_{y}(t)=R_{2}/2\\}$. Then

$r_{y}(t)\geq R_{2}/2\geq R_{1},\;\;\forall t\in[a_{y},\tau_{y}].$

Now it is enough to show $\dot{r}_{y}(t_{0})\geq\sqrt{6m/r_{y}(t_{0})}$, for
some $t_{0}\in[a_{y},\tau_{y}]$, as Proposition 3.1 implies

$\dot{r}_{y}(t)\geq\frac{1}{2}\dot{r}(t_{0})\geq\frac{1}{2}\sqrt{6m/r_{y}(t_{0})}\geq\sqrt{3m/2}R^{-\frac{1}{2}}_{2},\;\;\forall
t\in[t_{0},t_{y}],$

where the last inequality follows from $r_{y}(t_{0})\leq
r_{y}(\tau_{y})=R_{2}$.

By a contradiction argument, let’s assume $\dot{r}_{y}(t)<\sqrt{6m/r_{y}(t)}$,
$\forall t\in[a_{y},\tau_{y}].$ Then

$\frac{R_{2}}{2}=r_{y}(\tau_{y})-r_{y}(a_{y})=\int^{\tau_{y}}_{a_{y}}\dot{r}_{y}(t)\,dt<\int^{\tau_{y}}_{a_{y}}\sqrt{6m/r_{y}(t)}\,dt\leq\sqrt{\frac{12m}{R_{2}}}(\tau_{y}-a_{y}).$

Notice that $r_{y}(t)\geq R_{2}/2$, $\forall t\in[a_{y},\tau_{y}]$. As a
result,

$\mathcal{A}_{h}(\gamma_{y};a_{y},\tau_{y})\geq
h(\tau_{y}-a_{y})>\frac{h}{4\sqrt{3m}}R_{2}^{\frac{3}{2}}.$

However with Lemma 3.1, they imply

$\frac{h}{4\sqrt{3m}}R_{2}^{\frac{3}{2}}<\mathcal{A}_{h}(\gamma_{y};a_{y},\tau_{y})=\phi_{h}(\gamma_{y}(a_{y}),\gamma_{y}(\tau_{y});\\{a_{y}\\},\\{\tau_{y}\\})\leq\beta_{1}R_{2}+\beta_{2},$

which is absurd, when $R_{2}$ is large enough. ∎

In the rest of the section, we fix an
$R_{2}\geq\max\\{2\sqrt{2}R_{1},|x|+1\\}$ large enough, such that the above
lemma always holds.

######  Lemma 3.3.

For any $\tau>\sup\\{\tau_{y}:\;y\in\Lambda_{\theta_{+}}(R_{2})\\}$, the
following results hold.

  1. (a).

There is a constant $R=R(\tau,R_{2})>0$ (independent of $y$), such that for
any $y\in\Lambda_{\theta_{+}}(R_{2})$ with $t_{y}\geq\tau$, we have
$|\gamma_{y}(t)|\leq R$, $\forall t\in[t_{x},\tau]$.

  2. (b).

$\\{\gamma_{y}|_{[t_{x},\tau]}:y\in\Lambda_{\theta_{+}}(R_{2})\text{ with
}t_{y}\geq\tau\\}$ is a bound subset in $H^{1}([t_{x},\tau],\mathbb{C})$.

###### Proof.

(a). By (25), $|\gamma_{y}(t)|\leq R_{2}$, when $t\in[t_{x},\tau_{y}]$.
Meanwhile Lemma 3.2 implies $|\gamma_{y}(t)|$ is strictly increasing, when
$t\in[\tau_{y},\tau]$. Hence it is enough to show $|\gamma_{y}(\tau)|\leq R$,
for some finite constant $R$ independent of $y$. By the Cauchy-Schwartz
inequality,

$|\gamma_{y}(\tau)|\leq|\gamma_{y}(\tau_{y})|+|\gamma_{y}(\tau)-\gamma_{y}(\tau_{y})|\leq
R_{2}+\int_{\tau_{y}}^{\tau}|\dot{\gamma}_{y}|\,dt\leq
R_{2}+\left((\tau-\tau_{y})\int_{\tau_{y}}^{\tau}|\dot{\gamma}_{y}|^{2}\,dt\right)^{\frac{1}{2}}.$

At the same time, Lemma 3.1 implies

$\int_{\tau_{y}}^{\tau}|\dot{\gamma}_{y}|^{2}\,dt\leq
2\mathcal{A}_{h}(\gamma_{y};\tau_{y},\tau)=2\phi_{h}(\gamma_{y}(\tau_{y}),\gamma_{y}(\tau);\\{\tau_{y}\\},\\{\tau\\})\leq
2\beta_{1}|\gamma_{y}(\tau)|+2\beta_{2}.$

These two estimates implies $|\gamma_{y}(\tau)|\leq
R_{2}+\sqrt{2\tau}(\beta_{1}|\gamma_{y}(\tau)|+\beta_{2})^{\frac{1}{2}}$,
which is equivalent to

$|\gamma_{y}(\tau)|^{2}-2(R_{2}+\beta_{1}\tau)|\gamma_{y}(\tau)|+R_{2}^{2}-2\beta_{2}\tau\leq
0.$

Solving this inequality, we get

$|\gamma_{y}(\tau)|\leq
R_{2}+\beta_{1}\tau+(\beta_{1}^{2}\tau^{2}+2\beta_{1}\tau
R_{2}+2\beta_{2}\tau)^{\frac{1}{2}}:=R.$

(b). Let $a=\inf\\{\tau_{y}:y\in\Lambda_{\theta_{+}}(R_{2})\text{ and
}\tau_{y}\geq\tau\\}$,
$b=\sup\\{\tau_{y}:y\in\Lambda_{\theta_{+}}(R_{2})\text{ and
}\tau_{y}\geq\tau\\}.$ By Lemma 3.2, $t_{x}<a<b<\infty$. As
$|\gamma_{y}(\tau_{y})|=R_{2}$, $\forall y\in\Lambda_{\theta_{+}}(R_{2})$ with
$t_{y}\geq\tau$. By Proposition 2.6

$\mathcal{A}_{h}(\gamma_{y};t_{x},\tau_{y})=\psi_{h}(x,\gamma_{y}(\tau_{y});t_{x},\tau_{y})\leq\sup_{t\in[a,b],z\in\partial
B_{R_{2}}}\psi_{h}(x,z;t_{x},t)\leq C_{2}.$

Meanwhile as $\sqrt{2}R_{1}\leq
R_{2}=|\gamma_{y}(\tau_{y})|\leq|\gamma_{y}(\tau)|\leq R$, Lemma 3.1 implies

$\mathcal{A}_{h}(\gamma_{y};\tau_{y},\tau)=\phi_{h}(\gamma_{y}(\tau_{y}),\gamma_{y}(\tau);\\{\tau_{y}\\},\\{\tau\\})\leq\beta_{1}R+\beta_{2}.$

Combining these two estimates, we get

$\int_{t_{x}}^{\tau}|\dot{\gamma}_{y}|^{2}\,dt\leq
2\mathcal{A}_{h}(\gamma_{y};t_{x},\tau)\leq 2(\beta_{1}R+\beta_{2}+C_{2}).$

On the other hand, property (a) implies

$\int_{t_{x}}^{\tau}|\gamma_{y}(t)|^{2}\,dt\leq(\tau-t_{x})R^{2}.$

As a result, for any $y\in\Lambda_{\theta_{+}}(R_{2})$ with $t_{y}\geq\tau$,

$\|\gamma_{y}|_{[t_{x},\tau]}\|_{H^{1}}\leq\int_{t_{x}}^{\tau}|\gamma_{y}|^{2}+|\dot{\gamma}_{y}|^{2}\,dt\leq(\tau-
t_{x})R^{2}+2(\beta_{1}R+\beta_{2}+C_{2}).$

∎

The next result is a simple exercise from real analysis, which will be needed.

######  Proposition 3.3.

Given a sequence $\\{f_{n}\\}_{n\in\mathbb{N}}\subset
C^{2}([T_{1},T_{2}],\mathbb{C})$, whose $C^{2}$-norm is bounded. If there is a
$f\in C^{2}([T_{1},T_{2}],\mathbb{C})$ with $\|f_{n}-f\|_{L^{\infty}}\to 0$,
as $n\to\infty$, then $\|f_{n}-f\|_{C^{1}}\to 0$, as $n\to\infty$.

Now let’s choose a sequence of points
$\\{y_{n}\in\Lambda_{\theta_{+}}(R_{2})\\}_{n\in\mathbb{N}}$ with
$|y_{n}|\to\infty$, as $n\to\infty$. For each $n$, let $\gamma_{y_{n}}\in
H^{1}_{t_{x},t_{y_{n}}}(x,y_{n})$ be a free-time minimizer satisfying

$\mathcal{A}_{h}(\gamma_{y_{n}};t_{x},t_{y_{n}})=\phi_{h}(x,y_{n};t_{x},\\{t_{y_{n}}\\})=\inf_{s\in[0,1)}\phi_{h}(x,y_{n};t_{x},s).$

######  Proposition 3.4.

Under the above notations, the following results hold.

  1. (a).

$t_{y_{n}}\to\infty$, as $n\to\infty$.

  2. (b).

There exists a $\gamma\in H^{1}([t_{x},\infty),\mathbb{C})$, such that (after
passing to a proper subsequence) $\gamma_{y_{n}}\to\gamma$ strongly in
$L^{\infty}_{\text{loc}}$-norm and weakly in $H^{1}_{\text{loc}}$-norm, as
$n\to\infty$,

  3. (c).

$\forall t>t_{x}$,
$\mathcal{A}_{h}(\gamma;t_{x},t)=\phi_{h}(x,\gamma(\tau);t_{x},\\{t\\})$, and
$\gamma|_{(t_{x},\infty)}$ is a collision-free solution of (2).

  4. (d).

$\gamma_{y_{n}}\to\gamma$ in $C^{1}_{\text{loc}}$-norm, as $n\to\infty$, i.e.,
for any compact subset $I\subset(t_{x},\infty)$,
$\|\gamma_{y_{n}}|_{I}-\gamma|_{I}\|_{C^{1}}$, as $n\to\infty$.

######  Remark 3.1.

For the rest of the section, we set $\gamma_{n}=\gamma_{y_{n}}$,
$t_{n}=t_{y_{n}}$ and $\tau_{n}=\tau_{y_{n}}$(given as in (25)).

###### Proof.

(a). Since for $n$ large enough, $|\gamma_{n}(\tau_{n})|$ and
$|\gamma_{n}(t_{n})|=|y_{n}|\geq\sqrt{R_{1}}$, by Lemma 3.1,

$\displaystyle\beta_{1}|y_{n}|+\beta_{2}$
$\displaystyle\geq\phi_{h}(\gamma_{n}(\tau_{n}),\gamma_{n}(t_{n});\\{\tau_{n}\\},\\{t_{n}\\})\geq\frac{1}{2}\int_{\tau_{n}}^{t_{n}}|\dot{\gamma}_{n}|^{2}\,dt\geq\frac{1}{2(t_{n}-\tau_{n})}\left(\int_{\tau_{n}}^{t_{n}}|\dot{\gamma}_{n}|\,dt\right)^{2}$
$\displaystyle\geq\frac{|y_{n}-\gamma_{n}(\tau_{n})|^{2}}{2(t_{n}-\tau_{n})}\geq\frac{(|y_{n}|-|\gamma_{n}(\tau_{n})|)^{2}}{2t_{n}}\geq\frac{|y_{n}|^{2}-2R_{2}|y_{n}|+R_{2}^{2}}{2t_{n}}.$

This then implies

$t_{n}\geq\frac{|y_{n}|^{2}-2R_{2}|y_{n}|+R_{2}^{2}}{2(\beta_{1}|y_{n}|+\beta_{2})}\geq\frac{|y_{n}|-2R_{2}+R^{2}_{2}|y_{n}|^{-1}}{2\beta_{1}+2\beta_{2}|y_{n}|^{-1}}\to\infty,\;\text{as
}|y_{n}|\to\infty.$

(b). For any $\tau>t_{x}$ large enough, by Lemma 3.3,
$\\{\gamma_{n}|_{[t_{x},\tau]}\\}_{n\in\mathbb{N}}$ is bounded under the
$H^{1}$-norm, so (after passing to a proper subsequence) it converges to a
$\xi\in H^{1}([t_{x},\tau],\mathbb{C})$ weakly in $H^{1}$-norm and strongly in
$L^{\infty}$-norm.

Since $t_{n}\to\infty$, when $n\to\infty$, by a diagonal argument, we can find
a $\gamma\in H^{1}([t_{x},\infty),\mathbb{C})$, such that (after passing to a
proper subsequence) $\gamma_{n}\to\gamma$ strongly in $L^{\infty}_{loc}$-norm
and weak in $H^{1}_{loc}$-norm. Moreover the weakly lower semi-continuity of
$\mathcal{A}_{h}$ implies

(27)
$\mathcal{A}_{h}(\gamma;t_{x},\tau)\leq\liminf_{n\to\infty}\mathcal{A}_{h}(\gamma_{n};t_{x},\tau),\;\;\forall\tau>t_{x}.$

(c). Notice that is enough to prove the result for all $t$ large enough. By a
contradiction argument, let’s assume there is a $t>t_{x}$ large, such that

$\mathcal{A}_{h}(\gamma;t_{x},t)>\phi_{h}(x,\gamma(\tau);t_{x},\\{t\\}).$

Then by Proposition 2.7, there is a $\xi\in H^{1}_{t_{x},\tau}(x,\gamma(t))$
with $\\{\tau\\}=\\{t\\}$(see Figure 1), such that

$\mathcal{A}_{h}(\xi;t_{x},\tau)=\phi_{h}(x,\gamma(t);t_{x},\\{t\\})\leq\mathcal{A}_{h}(\gamma;t_{x},t)-3\varepsilon,$

for some $\varepsilon>0$ small enough. Then by (27), when $n$ is large enough,

(28)
$\mathcal{A}_{h}(\gamma_{n};t_{x},t)\geq\mathcal{A}_{h}(\gamma;t_{x},t)-\varepsilon\geq\mathcal{A}_{h}(\xi;t_{x},\tau)+2\varepsilon.$

Since $\xi(\tau)=\gamma(t)$ must be collision-free, for $\tau$ large enough,
so is $\xi|_{[\tau^{\prime},\tau]}$ for $\tau^{\prime}$ close enough to
$\tau$. Then $\mathcal{A}_{h}$ is continuous in a small neighborhood of
$\xi|_{[\tau^{\prime},\tau]}$ in $H^{1}([\tau^{\prime},\tau],\mathbb{C})$. As
$\gamma_{n}(t)\to\gamma(t)=\xi(\tau)$, when $n\to\infty$. For each $n$ large
enough, we can find a $\eta_{n}\in
H^{1}_{\tau^{\prime},\tau}(\xi(\tau),\gamma_{n}(t))$ with

$|\mathcal{A}_{h}(\xi;\tau^{\prime},\tau)-\mathcal{A}_{h}(\eta_{n};\tau^{\prime},\tau)|\leq\varepsilon.$

Then
$\zeta_{n}|_{[t_{x},\tau]}=\xi|_{[t_{x},\tau^{\prime}]}*\eta_{n}|_{[\tau^{\prime},\tau]}\in
H^{1}_{t_{x},\tau}(x,\gamma_{n}(t))$ and

$\mathcal{A}_{h}(\xi;t_{x},\tau)-\mathcal{A}_{h}(\zeta_{n};t_{x},\tau)=\mathcal{A}_{h}(\xi;\tau^{\prime},\tau)-\mathcal{A}_{h}(\eta_{n};\tau^{\prime},\tau)\geq-\varepsilon.$

Combining this with (28), we get the following inequality, which is absurd.

$\phi_{h}(x,\gamma_{n}(t);t_{x},\\{t\\})=\mathcal{A}_{h}(\gamma_{n};t_{x},t)\geq\mathcal{A}_{h}(\xi;t_{x},\tau)+2\varepsilon\geq\mathcal{A}_{h}(\zeta_{n};t_{x},\tau)+\varepsilon.$

Figure 1.

(d). Given an arbitrary compact subset $I\subset(t_{x},\infty)$, by property
(c), $\gamma|_{I}$ is a collision-free solution of (2). As a result,

$|\ddot{\gamma}(t)|=\left|\sum_{i\in\mathbf{N}}\frac{m_{i}(\gamma(t)-q_{i}(t))}{|\gamma(t)-q_{i}(t)|^{3}}\right|\leq\sum_{i\in\mathbf{N}}\frac{m_{i}}{|\gamma(t)-q_{i}(t)|}\leq
C_{1},\;\;\forall t\in I.$

Since each $\gamma_{n}|_{I}$ is also a free-time minimizer, $\gamma_{n}|_{I}$
is a collision-free solution of (2) as well. By property (a),
$\|\gamma_{n}|_{I}-\gamma|_{I}\|_{L^{\infty}}\to 0$, as $n\to\infty$. Then for
$n$ large enough,

$|\ddot{\gamma}_{n}(t)|=\left|\sum_{i\in\mathbf{N}}\frac{m_{i}(\gamma_{n}(t)-q_{i}(t))}{|\gamma_{n}(t)-q_{i}(t)|^{3}}\right|\leq\sum_{i\in\mathbf{N}}\frac{m_{i}}{|\gamma_{n}(t)-q_{i}(t)|}\leq
2C_{1},\;\;\forall t\in I.$

The desired result now follows directly from Proposition 3.3. ∎

For $\gamma$ and $\gamma_{n}$ from above, in polar coordinate, we let
$\gamma(t)=r(t)e^{i\theta(t)}$ and $\gamma_{n}(t)=r_{n}(t)e^{i\theta_{n}(t)}$.

######  Proposition 3.5.

$\lim_{t\to\infty}\theta(t)=\theta_{+}(\text{mod }2\pi).$

###### Proof.

Notice that Proposition 3.4 implies

(29)
$\lim_{n\to\infty}|\gamma_{n}(t)-\gamma(t)|=0,\;\;\lim_{n\to\infty}|\dot{\gamma}_{n}(t)-\dot{\gamma}(t)|=0,\;\;\forall
t>t_{x}.$

Meanwhile by a direct computation,

$\displaystyle|\dot{r}_{n}-\dot{r}|$
$\displaystyle=\left|\frac{\langle\gamma_{n},\dot{\gamma}_{n}\rangle}{|\gamma_{n}|}-\frac{\langle\gamma,\dot{\gamma}\rangle}{|\gamma|}\right|=\left|\frac{\langle|\gamma|\gamma_{n},\dot{\gamma}_{n}\rangle-\langle|\gamma_{n}|\gamma,\dot{\gamma}\rangle}{|\gamma_{n}||\gamma|}\right|$
$\displaystyle=\left|\frac{\langle|\gamma|\gamma_{n},\dot{\gamma}_{n}\rangle-\langle|\gamma|\gamma_{n},\dot{\gamma}\rangle+\langle|\gamma|\gamma_{n},\dot{\gamma}\rangle-\langle|\gamma|\gamma,\dot{\gamma}\rangle+\langle|\gamma|\gamma,\dot{\gamma}\rangle-\langle|\gamma_{n}|\gamma,\dot{\gamma}\rangle}{|\gamma_{n}||\gamma|}\right|$
$\displaystyle\leq\frac{|\langle|\gamma|\gamma_{n},\dot{\gamma}_{n}-\dot{\gamma}\rangle|+|\langle|\gamma|(\gamma_{n}-\gamma),\dot{\gamma}\rangle|+|\langle(|\gamma|-|\gamma_{n}|)\gamma,\dot{\gamma}\rangle|}{|\gamma_{n}||\gamma|}$
$\displaystyle\leq\frac{|\gamma|\cdot|\gamma_{n}|\cdot|\dot{\gamma}_{n}-\dot{\gamma}|+|\gamma|\cdot|\gamma_{n}-\gamma|\cdot|\dot{\gamma}|+|\gamma-\gamma_{n}|\cdot|\gamma|\cdot|\dot{\gamma}|}{|\gamma|\cdot|\gamma_{n}|}$

Then by (29),

(30) $\lim_{n\to\infty}|\dot{r}_{n}(t)-\dot{r}(t)|=0,\;\forall t>t_{x}.$

Recall
$\tau_{n}=\tau_{y_{n}}=\min\\{t\in[t_{x},t_{y_{n}}]:|\gamma_{y_{n}}(t)|=R_{2}\\}$.
By Lemma 3.2, $\tau=\sup\\{\tau_{n}:\;n\geq 1\\}<\infty$. From now on we
always assume $t_{n}=t_{y_{n}}>\tau$. Then again by Lemma 3.2,

(31) $\dot{r}_{n}(t)\geq
v_{0}=(3m/2)^{\frac{1}{2}}R_{2}^{-\frac{1}{2}},\;\;\forall t\geq\tau\text{ and
}\forall n.$

Then (30) implies

(32) $\dot{r}(t)\geq v_{0},\;\;\forall t\geq\tau.$

As $|\gamma(\tau)|=\lim_{n\to\infty}|\gamma_{n}(\tau)|\geq R_{2}$, Proposition
3.2 implies,

$\lim_{t\to\infty}\theta(t)=\theta_{0},\;\text{ for some
}\theta_{0}\in\mathbb{R}.$

What’s left now is to show $\theta_{0}=\theta_{+}(\text{mod }2\pi)$. For this,
we need an uniform upper bound of the angular momenta $\omega(t)$ and
$\omega_{n}(t)$. First by Proposition 3.2, for all $t\geq\tau$,

(33) $\displaystyle|\omega(t)|$
$\displaystyle\leq|\omega(\tau)|+\frac{\alpha_{2}}{v_{0}r(\tau)}\leq|\omega(\tau)|+\frac{\alpha_{2}}{v_{0}R_{2}};$
$\displaystyle|\omega_{n}(t)|$
$\displaystyle\leq|\omega_{n}(\tau)|+\frac{\alpha_{2}}{v_{0}r_{n}(\tau)}\leq|\omega_{n}(\tau)|+\frac{\alpha_{2}}{v_{0}R_{2}}.$

Meanwhile notice that

$|\omega-\omega_{n}|=|\gamma\wedge\dot{\gamma}-\gamma\wedge\dot{\gamma}_{n}+\gamma\wedge\dot{\gamma}_{n}-\gamma_{n}\wedge\dot{\gamma}_{n}|\leq|\gamma|\cdot|\dot{\gamma}-\dot{\gamma}_{n}|+|\dot{\gamma}_{n}|\cdot|\gamma-\gamma_{n}|.$

Then (29) implies, $\lim_{n\to\infty}|\omega(\tau)-\omega_{n}(\tau)|=0$.
Combining this with (33), we get

(34) $|\omega(t)|,|\omega_{n}(t)|\leq C_{1},\;\;\forall t\geq\tau\text{ and
}\forall n.$

For any $s_{2}>s_{1}\geq\tau$, using (32) and (34), we get

(35) $\displaystyle|\theta(s_{2})-\theta(s_{1})|$
$\displaystyle\leq\int_{s_{1}}^{s_{2}}|\dot{\theta}(t)|\,dt\leq\int_{s_{1}}^{s_{2}}\frac{|\omega(t)|}{r^{2}(t)}\,dt\leq\frac{C_{1}}{v_{0}^{2}}\int_{s_{1}}^{s_{2}}\left(t-s_{1}+\frac{r(s_{1})}{v_{0}}\right)^{-2}\,dt.$
$\displaystyle\leq\frac{C_{1}}{v_{0}^{2}}\left(\frac{v_{0}}{r(s_{1})}-\frac{v_{0}}{v_{0}(s_{2}-s_{1})+r(\tau)}\right)\leq\frac{v_{0}^{-1}C_{1}}{r(s_{1})},$

By (31) and (34), a similar computation as above shows that for any $n$,

(36)
$|\theta_{n}(s_{2})-\theta_{n}(s_{1})|\leq\frac{v_{0}^{-1}C_{1}}{r_{n}(s_{1})}.$

By a contradiction argument, let’s assume $\theta_{0}\neq\theta_{+}(\text{mod
}2\pi)$, then there is a $\varepsilon>0$, such that

(37) $|\theta_{0}-\theta_{+}-2k\pi|\geq 6\varepsilon,\;\;\forall
k\in\mathbb{Z}.$

As $r(t)\to\infty$, when $t\to\infty$, there must be a $s_{0}\geq\tau$ with
$r(s_{0})\geq C_{1}/(v_{0}\varepsilon)$. Then (35) implies

(38) $|\theta(t)-\theta(s_{0})|\leq\varepsilon,\;\;\forall t\geq s_{0}.$

Since $\gamma_{n}(s_{0})\to\gamma(s_{0})$, when $n\to\infty$. For each $n$
large enough, there is a $j_{n}\in\mathbb{Z}$, such that

(39) $|\theta(s_{0})-\theta_{n}(s_{0})-2j_{n}\pi|\leq\varepsilon.$

Notice that
$r_{n}(s_{0})\geq\frac{r(s_{0})}{2}\geq\frac{C_{1}}{2v_{0}\varepsilon}$, for
$n$ large enough. Then (36) implies

(40) $|\theta_{n}(t)-\theta_{n}(s_{0})|\leq 2\varepsilon,\;\;\forall t\geq
s_{0}.$

Since $y_{n}=\gamma_{n}(t_{n})\in\Lambda_{\theta_{+}}(R_{2})$, for each $n$,
there is a $k_{n}\in\mathbb{Z}$, such that

(41) $\theta_{n}(t_{n})-\theta_{+}=2k_{n}\pi.$

Meanwhile as $t_{n}\to\infty$, when $n\to\infty$, we have

(42) $|\theta_{0}-\theta(t_{n})|\leq\varepsilon,\;\text{ for $n$ large
enough}.$

Then by (38), (39), (40), (41) and (42), for $n$ large enough, we have

$\displaystyle|\theta_{0}-\theta_{+}$
$\displaystyle-2k_{n}\pi-2j_{n}\pi|=|\theta_{0}-\theta_{n}(t_{n})-2j_{n}\pi|$
$\displaystyle=|\theta_{0}-\theta(t_{n})+\theta(t_{n})-\theta(s_{0})+\theta(s_{0})-\theta_{n}(s_{0})-2j_{n}\pi+\theta_{n}(s_{0})-\theta_{n}(t_{n})|\leq
5\varepsilon,$

which is a contradiction to (37). ∎

######  Proposition 3.6.

$\lim_{t\to\infty}\dot{\gamma}(t)=\sqrt{2h}e^{i\theta_{+}}$.

###### Proof.

First we claim $\lim_{t\to\infty}\dot{r}(t)=v_{1}$, for some $v_{1}>0$.

Since $r(t)\geq R_{2}\geq R_{1}$, $\forall t\geq\tau$, (34) and the first
equation in (93) imply

$|\ddot{r}(t)|\leq\frac{2m}{r^{2}(t)}+\frac{|\omega^{2}(t)|}{r^{3}(t)}\leq\frac{2m}{r^{2}(t)}+\frac{C_{1}^{2}}{r^{3}(t)}.$

Meanwhile by (32), $r(t)\geq v_{0}(t-\tau)+r(\tau)$, $\forall t\geq\tau$. As a
result, $\int_{\tau}^{\infty}|\ddot{r}(t)|\,dt<\infty$. This implies
$\lim_{t\to\infty}\dot{r}(t)$ exists and is finite. Moreover (32) shows it is
also positive. This proves the claim.

Since $|\omega(t)|\leq C_{1}$, $\forall t\geq\tau$ and $r(t)\to\infty$, as
$t\to\infty$,

$\lim_{t\to\infty}r(t)\dot{\theta}(t)=\lim_{t\to\infty}(\omega(t)/r(t))=0.$

Then Proposition 3.5 and the above claim imply

$\lim_{t\to\infty}\dot{\gamma}(t)=\lim_{t\to\infty}\big{(}\dot{r}(t)+ir(t)\dot{\theta}(t)\big{)}e^{i\theta(t)}=v_{1}e^{i\theta_{+}(\text{mod
}2\pi)}.$

Now we only need to show $v_{1}=\sqrt{2h}$. By a contradiction argument,
assume $v_{1}\neq\sqrt{2h}$. Notice that $\sqrt{2h}$ is the unique global
minimum of the following continuous function

$f:(0,\infty)\to\mathbb{R};\;v\mapsto v/2+h/v.$

Then $\delta=(f(v_{1})-\sqrt{2h})/5>0$, and for $\varepsilon\in(0,\delta)$
small enough, $f(v_{1}+\varepsilon)-\sqrt{2h}\geq 4\delta$.

As $\lim_{t\to\infty}|\dot{\gamma}(t)|=v_{1}$, we can find an $s_{1}>\tau$
large enough, such that

$v_{1}-\varepsilon\leq|\dot{\gamma}(t)|\leq v_{1}+\varepsilon,\;\forall t\geq
s_{1}.$

Then for any $s_{2}>s_{1}$,
$|\gamma(s_{2})-\gamma(s_{1})|\leq\int_{s_{1}}^{s_{2}}|\dot{\gamma}(t)|\,dt\leq(v_{1}+\varepsilon)(s_{2}-s_{1})$,
which implies

$s_{2}-s_{1}\geq|\gamma(s_{2})-\gamma(s_{1})|/(v_{1}+\varepsilon).$

Using the above estimates, we get

(43) $\displaystyle\mathcal{A}_{h}(\gamma;s_{1},s_{2})$
$\displaystyle\geq\int_{s_{1}}^{s_{2}}\frac{1}{2}|\dot{\gamma}|^{2}+h\,dt\geq\left(\frac{1}{2}(v_{1}-\varepsilon)^{2}+h\right)\frac{|\gamma(s_{2})-\gamma(s_{1})|}{v_{1}+\varepsilon}$
$\displaystyle\geq\left(\frac{1}{2}(v_{1}+\varepsilon)+\frac{h}{v_{1}+\varepsilon}-2\varepsilon\right)|\gamma(s_{2})-\gamma(s_{1})|$
$\displaystyle=(f(v_{1}+\varepsilon)-2\varepsilon)|\gamma(s_{2})-\gamma(s_{1})|\geq(\sqrt{2h}+4\delta-2\varepsilon)|\gamma(s_{2})-\gamma(s_{1})|$
$\displaystyle\geq(\sqrt{2h}+2\delta)|\gamma(s_{2})-\gamma(s_{1})|.$

For any $s_{2}>s_{1}$ as above, we can always find an $s\in[0,1)$, such that

$s^{*}_{2}=s_{1}+(2h)^{-\frac{1}{2}}|\gamma(s_{2})-\gamma(s_{1})|+s\;\text{
satisfying }\;\\{s^{*}_{2}\\}=\\{s_{2}\\}.$

We will show $\xi\in H^{1}_{s_{1},s^{*}_{2}}(\gamma(s_{1}),\gamma(s_{2}))$
given as below satisfying
$\mathcal{A}_{h}(\xi;s_{1},s^{*}_{2})<\mathcal{A}_{h}(\gamma;s_{1},s_{2})$.

$\xi(t)=\gamma(s_{1})+\frac{t-s_{1}}{s^{*}_{2}-s_{1}}(\gamma(s_{2})-\gamma(s_{1})),\;\forall
t\in[s_{1},s_{2}^{*}].$

However this is a contradiction to
$\mathcal{A}_{h}(\gamma;s_{1},s_{2})=\phi_{h}(\gamma(s_{1}),\gamma(s_{2});\\{s_{1}\\},\\{s_{2}\\})$.

To estimate $\int|\dot{\xi}|^{2}\,dt$, consider the following strictly
increasing continuous function,

$g:[0,\infty)\to\mathbb{R};\;v\mapsto(1+v)+(1+v)^{-1}.$

As $|\gamma(s_{2})-\gamma(s_{1})|\to\infty$, when $s_{2}\to\infty$,

$g\left(\frac{\sqrt{2h}s}{|\gamma(s_{2})-\gamma(s_{1})|}\right)\leq
2+\sqrt{2/h}\delta,\;\text{ for }s_{2}\text{ large enough}.$

As a result, for $s_{2}$ large enough,

(44) $\displaystyle\int_{s_{1}}^{s^{*}_{2}}\frac{1}{2}|\dot{\xi}|^{2}$
$\displaystyle+h\,dt=\left(\frac{1}{2}\frac{|\gamma(s_{2})-\gamma(s_{1})|^{2}}{(s^{*}_{2}-s_{1})^{2}}+h\right)\left(\frac{|\gamma(s_{2})-\gamma(s_{1})|}{\sqrt{2h}}+s\right)$
$\displaystyle=\left[\left(1+\frac{\sqrt{2h}s}{|\gamma(s_{2})-\gamma(s_{1})|}\right)+\left(1+\frac{\sqrt{2h}s}{|\gamma(s_{2})-\gamma(s_{1})|}\right)^{-1}\right]\sqrt{\frac{h}{2}}|\gamma(s_{2})-\gamma(s_{1})|$
$\displaystyle\leq(\sqrt{2h}+\delta)|\gamma(s_{2})-\gamma(s_{1})|.$

To estimate $\int U(\xi(t),t)\,dt$, we set $\theta_{+}=0$ for simplicity. For
any $s_{2}>s_{1}$ with $s_{1}$ large enough,

(45) $|\theta(s_{2})|,|\theta(s_{1})|\leq\pi/12.$

By further assuming $s_{2}-s_{1}$ is large enough, we can get

(46) $|\gamma(s_{2})-\gamma(s_{1})|>|\gamma(s_{1})|.$

Figure 2. the two triangles.

Consider the triangle with $\gamma(s_{1}),\gamma(s_{2})$ and the origin as the
three vertices, and $\sigma_{i}$, $i=1,2,3$, as the three angles (see Figure
2). By (45), $0\leq\sigma_{1}\leq\pi/6$. Then (46) implies
$0\leq\sigma_{3}\leq\sigma_{1}\leq\pi/6$. As a result,
$2\pi/3\leq\sigma_{2}\leq\pi$.

Meanwhile for any $t\in[s_{1},s_{2}^{*}]$, there is a triangle with
$\gamma(s_{1}),\xi(t)$ and the origin as the three vertices, and $\sigma_{i}$,
$i=2,4,5$ as the three angles (see Figure 2). Then
$2\pi/3\leq\sigma_{2}\leq\pi$ implies $0\leq\sigma_{4},\sigma_{5}\leq\pi/3$
and as a result, for any $t\in[s_{1},s_{2}^{*}]$,

$|\xi(t)|=|\xi(t)-\gamma(s_{1})|\cos\sigma_{4}+|\gamma(s_{1})|\cos\sigma_{5}\geq\frac{1}{2}\left(\frac{t-s_{1}}{s^{*}_{2}-s_{1}}|\gamma(s_{2})-\gamma(s_{1})|+|\gamma(s_{1})|\right)\geq
R_{1}.$

Using this, we get

(47) $\displaystyle\int_{s_{1}}^{s^{*}_{2}}|U(\xi(t),t)|\,dt$
$\displaystyle\leq\int_{s_{1}}^{s_{2}^{*}}\frac{2m}{|\xi(t)|}\,dt\leq\int_{s_{1}}^{s^{*}_{2}}\frac{4m(s^{*}_{2}-s_{1})}{|\gamma(s_{2})-\gamma(s_{1})|}\left(t-s_{1}+\frac{(s^{*}_{2}-s_{1})|\gamma(s_{1})|}{|\gamma(s_{2})-\gamma(s_{1})|}\right)^{-1}\,dt$
$\displaystyle=\int_{s_{1}}^{s^{*}_{2}}\frac{\frac{4m}{\sqrt{2h}}\left(1+\frac{\sqrt{2h}s}{|\gamma(s_{2})-\gamma(s_{1})|}\right)}{t-s_{1}+\frac{|\gamma(s_{1})|}{\sqrt{2h}}\left(1+\frac{\sqrt{2h}s}{|\gamma(s_{2})-\gamma(s_{1})|}\right)}\,dt$
$\displaystyle=\frac{4m}{\sqrt{2h}}\left(1+\frac{\sqrt{2h}s}{|\gamma(s_{2})-\gamma(s_{1})|}\right)\log\left(1+\frac{|\gamma(s_{2})-\gamma(s_{1})|}{|\gamma(s_{1})|}\right)$

Combining the above inequality with (43) and (44), we get

$\displaystyle\mathcal{A}_{h}$
$\displaystyle(\gamma;s_{1},s_{2})-\mathcal{A}_{h}(\xi;s_{1},s^{*}_{2})$
$\displaystyle\geq\delta|\gamma(s_{2})-\gamma(s_{1})|-\frac{4m}{\sqrt{2h}}\left(1+\frac{\sqrt{2h}s}{|\gamma(s_{2})-\gamma(s_{1})|}\right)\log\left(1+\frac{|\gamma(s_{2})-\gamma(s_{1})|}{|\gamma(s_{1})|}\right)>0,$

for $s_{2}$ large enough, which is absurd. ∎

Up to now, we have proved the existence of the desired hyperbolic solution
$\gamma^{+}|_{[t_{x},\infty)}$, when $T=1$. By a similar argument, we can
obtain the corresponding $\gamma^{-}|_{(-\infty,t_{x}]}$, when $T=1$. To
generalize the result to $T\neq 1$, we use a blow-up argument similar to the
one that have been used quite a lot in the variational study of the $N$-body
problem (see [37] and [16]).

###### Proof of Theorem 1.1.

Recall that $q(t)=(q_{i}(t))_{i\in\mathbf{N}}$ is a collision-free
$T$-periodic solution of the $N$-body problem (1). Define
$q^{\lambda}(t)=(q_{i}^{\lambda}(t))_{i\in\mathbf{N}}$ as

$q_{i}^{\lambda}(s)=\lambda^{\frac{2}{3}}q_{i}(t)=\lambda^{\frac{2}{3}}q_{i}(s/\lambda),\;i\in\mathbf{N},\;\text{
where }\lambda=T^{-1}.$

Then $q^{\lambda}(t)$ is a collision-free $1$-periodic solution of (1), as a
direct computation shows

$\ddot{q}^{\lambda}_{i}(s)=-\sum_{j\in\mathbf{N}\setminus\\{i\\}}\frac{m_{j}(q^{\lambda}_{i}(s)-q^{\lambda}_{j}(s))}{|q^{\lambda}_{i}(s)-q^{\lambda}_{j}(s)|^{3}},\;\forall
i\in\mathbf{N}.$

By previous results of this section, for $h^{\lambda}=\lambda^{-\frac{2}{3}}h$
and $\theta_{+}\in[0,2\pi)$, there is a $\gamma^{\lambda}(s)$, $s\in[\lambda
t_{x},\infty)$, with $\gamma^{\lambda}(\lambda t_{x})=\lambda^{\frac{2}{3}}x$,
which is a solution of

$\ddot{\gamma}^{\lambda}(s)=-\sum_{i\in\mathbf{N}}\frac{m_{i}(\gamma(s)-q_{i}^{\lambda}(s))}{|\gamma(s)-q_{i}^{\lambda}(s)|^{3}},$

and satisfies

$\lim_{s\to\infty}\gamma^{\lambda}(s)/|\gamma^{\lambda}(s)|=e^{i\theta_{+}(\text{mod
}2\pi)},\;\lim_{s\to\infty}\dot{\gamma}^{\lambda}(s)=\sqrt{2h^{\lambda}}e^{i\theta_{+}(\text{mod
}2\pi)}.$

Then $\gamma^{+}(t)=\lambda^{-\frac{2}{3}}\gamma^{\lambda}(\lambda
t)=\lambda^{-\frac{2}{3}}\gamma^{\lambda}(s)$, $t\in[t_{x},\infty)$, is a
solution of (2) satisfying $\gamma^{+}(t_{x})=x$ and

$\lim_{t\to\infty}\gamma^{+}(t)/|\gamma^{+}(t)|=\lim_{s\to\infty}\gamma^{\lambda}(s)/|\gamma^{\lambda}(s)|=e^{i\theta_{+}(\text{mod
}2\pi)}$

and

$\lim_{t\to\infty}\dot{\gamma}^{+}(t)=\lim_{t\to\infty}\lambda^{\frac{1}{3}}\dot{\gamma}^{\lambda}(s)=\lambda^{\frac{1}{3}}\sqrt{2h^{\lambda}}e^{i\theta_{\pm}(\text{mod
}2\pi)}=\sqrt{2h}e^{i\theta_{\pm}(\text{mod }2\pi)}.$

The proof of $\gamma^{-}|_{(-\infty,t_{x}]}$ is exactly the same and will be
omitted. ∎

##  4\. Existence of bi-hyperbolic solutions

Like in the previous section we assume $T=1$ throughout this section except
the last proof.

######  Definition 4.1.

For any $x,y\in\mathbb{C}\setminus B^{o}_{R_{0}}$ and $\gamma\in
H^{1}_{t_{1},t_{2}}(x,y)$ with $\gamma([t_{1},t_{2}])\cap
B_{R_{0}}\neq\emptyset$, let

(48) $\displaystyle s_{y}$
$\displaystyle=s_{y}(\gamma)=\max\\{t\in[t_{1},t_{2}]:|\gamma(t)|=R_{0}\\};$
$\displaystyle s_{x}$
$\displaystyle=s_{x}(\gamma)=\min\\{t\in[t_{1},t_{2}]:|\gamma(t)|=R_{0}\\}.$

We say such a $\gamma$ is _tied with $m_{i_{0}}$ and $m_{i_{1}}$_
($\\{i_{0}\neq i_{1}\\}\subset\mathbf{N}$), if for any $\xi\in
H^{1}_{s_{x},s_{y}}(\gamma(s_{x}),\gamma(s_{y}))$ with
$\xi([s_{x},s_{y}])\subset\mathbb{C}\setminus B^{o}_{R_{0}}$, there is no
fixed-end homotopy between $\gamma|_{[s_{x},s_{y}]}$ and
$\xi|_{[s_{x},s_{y}]}$ avoiding the trajectories of $m_{i_{0}}$ and
$m_{i_{1}}$, i.e., there is no continuous paths $P(t,\tau)$,
$(t,\tau)\in[s_{x},s_{y}]\times[0,1]$, in $H^{1}([s_{x},s_{y}],\mathbb{C})$
satisfying $P(t,0)=\gamma(t),\;P(t,1)=\xi(t),\;\forall t\in[s_{x},s_{y}]$, and

$P(t,\tau)\in\mathbb{C}\setminus\\{q_{i_{0}}(t),q_{i_{1}}(t)\\},\;\forall(t,\tau)\in[s_{x},s_{y}]\times[0,1].$

######  Remark 4.1.

The named _tied_ was used as in the 3-dim space $\mathbb{C}\times\mathbb{R}$,
the trajectories $q_{i_{0}}|_{[s_{x},s_{y}]},q_{i_{1}}|_{[s_{x},s_{y}]}$ and
$\gamma|_{[s_{x},s_{y}]}$ can be seen as three braids. It has been used by
Gordon [19] and Montgomery [28].

For the rest of the section, we shall always assume $x,y\in\mathbb{C}\setminus
B^{o}_{R_{0}}$ unless otherwise stated.

######  Definition 4.2.

For any $t_{1}<t_{2}\in\mathbb{R}$ and $s_{1},s_{2}\in[0,1)$, we define

$\hat{\Gamma}^{i_{0},i_{1}}_{t_{1},t_{2}}(x,y)=\\{\gamma\in
H^{1}_{t_{1},t_{2}}(x,y):\gamma\text{ is tied with }m_{i_{0}}\text{ and
}m_{i_{1}}\\};$ $\Gamma^{i_{0},i_{1}}_{t_{1},t_{2}}(x,y)=\text{ the weak
closure of }\hat{\Gamma}^{i_{0},i_{1}}_{t_{1},t_{2}}(x,y)\text{ in
}H^{1}_{t_{1},t_{2}}(x,y);$
$\hat{\Xi}_{s_{1},s_{2}}^{i_{0},i_{1}}(x,y)=\cup_{\\{t_{i}\\}=s_{i},i=1,2}\hat{\Gamma}^{i_{0},i_{1}}_{\tau_{1},\tau_{2}}(x,y);\;\;\Xi_{s_{1},s_{2}}^{i_{0},i_{1}}(x,y)=\cup_{\\{t_{i}\\}=s_{i},i=1,2}\Gamma^{i_{0},i_{1}}_{\tau_{1},\tau_{2}}(x,y).$

######  Remark 4.2.

An arbitrary pair of indices $\\{i_{0}\neq i_{1}\\}\subset\mathbf{N}$ will be
fixed for the rest of the section, and for simplicity, they will be omitted
from the notations introduced in the above definition.

The next result is a simple corollary of the above definition, but will be
useful in our proof.

######  Lemma 4.1.

For any $\gamma\in H^{1}_{t_{1},t_{2}}(x,y)$, if there is a
$t_{0}\in(t_{1},t_{2})$, such that $\gamma(t_{0})=q_{i_{0}}(t_{0})$ or
$q_{i_{1}}(t_{0})$, then $\gamma\in\Gamma_{t_{1},t_{2}}(x,y)$.

We further introduce the following two functions:

(49)
$\Psi_{h}(x,y;t_{1},t_{2})=\inf\\{\mathcal{A}_{h}(\gamma;t_{1},t_{2}):\gamma\in\Gamma_{t_{1},t_{2}}(x,y)\\},$
(50)
$\Phi_{h}(x,y;s_{1},s_{2})=\inf\\{\mathcal{A}_{h}(\gamma):\gamma\in\Xi_{s_{1},s_{2}}(x,y)\\}.$

In the previous section, we assume $x\neq y$ in several lemmas and
propositions to give a positive lower bound of
$\int_{t_{1}}^{t_{2}}|\dot{\gamma}|\,dt$, $\forall\gamma\in
H^{1}_{t_{1},t_{2}}(x,y)$. Here it is not needed, because of the following
result.

######  Lemma 4.2.

For any $x,y\in\mathbb{C}\setminus B^{o}_{R_{0}}$ and
$\gamma\in\Gamma_{t_{1},t_{2}}(x,y)$,
$\int_{t_{1}}^{t_{2}}|\dot{\gamma}(t)|\,dt\geq 2$.

###### Proof.

By the definition $\Gamma_{t_{1},t_{2}}(x,y)$, there must be a
$t_{0}\in(t_{1},t_{2})$ with$|\gamma(t_{0})|\leq R_{0}-1$, then

$\int_{t_{1}}^{t_{2}}|\dot{\gamma}(t)|\,dt\geq|\gamma(t_{1})-\gamma(t_{0})|+|\gamma(t_{0})-\gamma(t_{2})|\geq
2.$

∎

######  Proposition 4.1.

There is a $\gamma\in\Gamma_{t_{1},t_{2}}(x,y)$ satisfying
$\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\Psi_{h}(x,y;t_{1},t_{2})$. Moreover

  1. (a).

for any $\tau_{0}\in(t_{1},t_{2})\setminus\Delta(\gamma)$, there is a
$\delta>0$ small enough, such that
$\gamma|_{[\tau_{0}-\delta,\tau_{0}+\delta]}$ is a collision-free local
minimizer and a solution of (2);

  2. (b).

$\Delta(\gamma)=\emptyset$ or $\\{t_{0}\\}\subset(t_{1},t_{2})$ and in the
latter case, $\gamma(t_{0})=q_{j_{0}}(t_{0})$ with $j_{0}\in\\{i_{0},i_{1}\\}$
and the following limits exist

(51) $\lim_{t\to
t_{0}}\frac{1}{2}|\dot{\gamma}(t)-\dot{q}_{j_{0}}(t)|^{2}-\frac{m_{i}}{|\gamma(t)-q_{j_{0}}(t)|},\;\;\lim_{t\to
t_{0}}\frac{\gamma(t)-q_{j_{0}}(t)}{|\gamma(t)-q_{j_{0}}(t)|}.$

  3. (c).

$\gamma|_{[t_{1},t_{2}]}$ is a solution of (2) with at most one collision, and
if there is a collision, it is regularizable.

###### Proof.

By Lemma 4.2, the existence of such a $\gamma\in\Gamma_{t_{1},t_{2}}(x,y)$
follows from the same argument given in the proof of property (a) in
Proposition 2.5. In particular $\mathcal{A}_{h}(\gamma;t_{1},t_{2})$ must be
finite.

(a). Since $\mathcal{A}_{h}(\gamma;t_{1},t_{2})$ is finite, the Lebesgue
measure of $\Delta(\gamma)$ must be zero with
$(t_{1},t_{2})\setminus\Delta(\gamma)$ being the union of at most countably
many open intervals. Then
$\forall\tau_{0}\in(t_{1},t_{2})\setminus\Delta(\gamma)$, there is a
$\delta>0$ small enough, such that
$[\tau_{0}-\delta,\tau_{0}+\delta]\subset(t_{1},t_{2})\setminus\Delta(\gamma)$
with $\gamma|_{[\tau_{0}-\delta,\tau_{0}+\delta]}$ being collision-free.

As a result, we can find a $\varepsilon>0$ small enough, such that
$\forall\xi\in
H^{1}_{\tau_{0}-\delta,\tau_{0}+\delta}(\gamma(\tau_{0}-\delta),\gamma(\tau_{0}+\delta))$
satisfying $\|\xi-\gamma\|_{H^{1}}\leq\varepsilon$, the following new path
still belongs to $\Gamma_{t_{1},t_{2}}(x,y)$,

(52)
$\tilde{\gamma}|_{[t_{1},t_{2}]}=\gamma|_{[t_{1},\tau_{0}-\delta]}*\xi|_{[\tau_{0}-\delta,\tau_{0}+\delta]}*\gamma|_{[\tau_{0}+\delta,t_{2}]}\in\Gamma_{t_{1},t_{2}}(x,y).$

This means $\gamma|_{[\tau_{0}-\delta,\tau_{0}+\delta]}$ is a collision-free
local minimizer and a solution of (2), as otherwise

(53)
$\mathcal{A}_{h}(\tilde{\gamma};t_{1},t_{2})<\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\Psi_{h}(x,y;t_{1},t_{2}),$

which is absurd.

(b). By the above property (a), $\gamma(t)$,
$t\in[t_{1},t_{2}]\setminus\Delta(\gamma)$, satisfies (2). Then similar
arguments as in the proof of Proposition 2.1 show that $\Delta(\gamma)$ must
be isolated in $[t_{1},t_{2}]$.

Choose an arbitrary $\tau_{0}\in\Delta(\gamma)$. Then
$\gamma(\tau_{0})=q_{j_{0}}(\tau_{0})$ for some $j_{0}\in\mathbf{N}$. First we
will show $j_{0}\in\\{i_{0},i_{1}\\}$. Otherwise we can find
$\delta,\varepsilon>0$ small enough, such that $\forall\xi\in
H^{1}_{\tau_{0}-\delta,\tau_{0}+\delta}(\gamma(\tau_{0}-\delta),\gamma(\tau_{0}+\delta))$
satisfying
$\|\gamma|_{[\tau_{0}-\delta,\tau_{0}+\delta]}-\xi|_{[\tau_{0}-\delta,\tau_{0}+\delta]}\|_{H^{1}}\leq\varepsilon$,
the new path $\tilde{\gamma}$ given as in (52) is still contained in
$\Gamma_{t_{1},t_{2}}(x,y)$. Meanwhile by Proposition 2.3, one of such a $\xi$
satisfies

(54)
$\mathcal{A}_{h}(\xi;\tau_{0}-\delta,\tau_{0}+\delta)<\mathcal{A}_{h}(\gamma;\tau_{0}-\delta,\tau_{0}+\delta).$

This then leads to (53), which is absurd.

Next we will show $|\Delta(\gamma)|\leq 1$. By a contradiction argument,
assume there are two different collision moments $\tau_{0}$ and $\tau_{1}$.
Then $\gamma(\tau_{k})=q_{j_{k}}(\tau_{k})$ with $j_{k}\in\\{i_{0},i_{1}\\}$,
for $k=1,2$. By Proposition 2.3, there is a $\xi\in
H^{1}_{\tau_{0}-\delta,\tau_{0}+\delta}(\gamma(\tau_{0}-\delta),\gamma(\tau_{0}+\delta))$
satisfying
$\|\gamma|_{[\tau_{0}-\delta,\tau_{0}+\delta]}-\xi|_{[\tau_{0}-\delta,\tau_{0}+\delta]}\|_{H^{1}}\leq\varepsilon$
and (54), for $\delta,\varepsilon>0$ small enough. Let $\tilde{\gamma}$ be a
path given as in (52), then (53) holds. However by Lemma 4.1,
$\tilde{\gamma}\in\Gamma_{t_{1},t_{2}}(x,y)$, which is absurd.

What’s left now is to prove (51). Without loss of generality, let’s assume
$\gamma(\tau_{0})=q_{i_{0}}(\tau_{0})$. Then the first limit in (51) follows
directly from Lemma 2.1. For the second limit, by Proposition 2.2, the
following two one-sided limits exist

$\sigma_{\pm}=\lim_{t\to\tau_{0}^{\pm}}\frac{\gamma(t)-q_{i}(t)}{|\gamma(t)-q_{i}(t)|}.$

Then it is enough show $\sigma_{-}=\sigma_{+}$. Assume this is not true. By
Proposition 2.3, for $\delta,\varepsilon>0$ small enough, there are two paths
$\eta^{\pm}\in
H^{1}_{\tau_{0}-\delta,\tau_{0}+\delta}(\gamma(\tau_{0}-\delta),\gamma(\tau_{0}+\delta)$
satisfying (10), (11) and

$\mathcal{A}_{h}(\eta^{\pm};\tau_{0}-\delta,\tau_{0}+\delta)<\mathcal{A}_{h}(\gamma;\tau_{0}-\delta,\tau_{0}+\delta).$

Let
$\gamma^{\pm}|_{[t_{1},t_{2}]}=\gamma|_{[t_{1},\tau_{0}-\delta]}*\xi^{\pm}|_{[\tau_{0}-\delta,\tau_{0}+\delta]}*\gamma|_{[\tau_{0}+\delta,t_{1}]}$.
Then
$\mathcal{A}_{h}(\gamma^{\pm};t_{1},t_{2})<\mathcal{A}_{h}(\gamma;t_{1},t_{2}).$
Since one of $\gamma^{\pm}$ must belong to $\Gamma_{t_{1},t_{2}}(x,y)$, we get
a contradiction.

(c). By property (c), $\gamma|_{[t_{1},t_{2}]}$ contains at most one collision
moment. If $t_{0}$ is such a collision moment, then $t_{0}\in(t_{1},t_{2})$
and the two limits in (51) hold. This implies $\gamma|_{[t_{1},t_{2}]}$ is
regularizable, where the details can be found in [40]. ∎

######  Lemma 4.3.

For any $t_{1}<t_{2}$ and $x,y\in\mathbb{C}\setminus B^{o}_{R_{0}}$, there is
a $C_{1}=C_{1}(x,y,t_{1},t_{2},h)$, such that

$\sup\\{\Psi_{h}(x^{\prime},y^{\prime};t_{1},t_{2}):(x^{\prime},y^{\prime})\in
B_{1/2}(x)\times B_{1/2}(y)\text{ with }|x^{\prime}|,|y^{\prime}|\geq
R_{0}\\}\leq C_{1}.$

###### Proof.

By Proposition 4.1, there is a $\gamma\in\Gamma_{t_{1},t_{2}}(x,y)$ with
$\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\Psi_{h}(x,y;t_{1},t_{2})$. Then

$\tau_{1}=\max\\{\tau\in[t_{1},t_{2}]:\gamma(t)\in B_{1/2}(x),\;\forall
t\in[t_{1},\tau]\\};$ $\tau_{2}=\min\\{\tau\in[t_{1},t_{2}]:\gamma(t)\in
B_{1/2}(y),\;\forall t\in[\tau,t_{2}]\\},$

are well-defined. Moreover $t_{1}<\tau_{1}<\tau_{2}<t_{2}$ and
$|\gamma(\tau_{1})-x|=|\gamma(\tau_{2})-y|=1/2$.

For any $x^{\prime}\in B_{1/2}(x)\cap(\mathbb{C}\setminus B^{o}_{R_{0}})$ and
$y^{\prime}\in B_{1/2}(y)\cap(\mathbb{C}\setminus B^{o}_{R_{0}})$, we can
define a new path

$\xi(t)=\begin{cases}\gamma(\tau_{2})+\frac{t-\tau_{2}}{t_{2}-\tau_{2}}(y^{\prime}-\gamma(t_{2})),&\text{
if }t\in[\tau_{2},t_{2}],\\\ \gamma(t),&\text{ if }t\in[\tau_{1},\tau_{2}],\\\
x^{\prime}+\frac{t-t_{1}}{\tau_{1}-t_{1}}(\gamma(\tau_{1})-x^{\prime}),&\text{
if }t\in[t_{1},\tau_{1}].\end{cases}$

Notice that $\xi\in\Gamma_{t_{1},t_{2}}(x^{\prime},y^{\prime})$. Moreover
$\xi(t)\in B_{1/2}(x)$, $\forall t\in[t_{1},\tau_{1}]$ and $\xi(t)\in
B_{1/2}(y)$, $\forall t\in[\tau_{2},t_{2}]$. By (12), there is a
$C_{2}=C_{2}(x,y)$ with $U(\xi(t),t)\leq C_{2}$, for all
$t\in[t_{1},\tau_{1}]\cup[\tau_{2},t_{2}]$. As a result,

$\displaystyle\Psi_{h}(x^{\prime},y^{\prime};t_{1},t_{2})$
$\displaystyle\leq\mathcal{A}_{h}(\xi;t_{1},t_{2})-\mathcal{A}_{h}(\gamma;t_{1},t_{2})+\Psi_{h}(x,y;t_{1},t_{2})$
$\displaystyle\leq\int_{t_{1}}^{\tau_{1}}L(\xi,\dot{\xi},t)\,dt+\int_{\tau_{2}}^{t_{2}}L(\xi,\dot{\xi},t)\,dt+\Psi_{h}(x,y;t_{1},t_{2})$
$\displaystyle\leq\frac{|\gamma(\tau_{1})-x^{\prime}|^{2}}{2(\tau_{1}-t_{1})}+\frac{|\gamma(\tau_{2})-y^{\prime}|^{2}}{2(t_{2}-\tau_{2})}+C_{2}(t_{2}-t_{1})+\Psi_{h}(x,y;t_{1},t_{2})$
$\displaystyle\leq\frac{1}{2(\tau_{1}-t_{1})}+\frac{1}{2(t_{2}-\tau_{2})}+C_{2}(t_{2}-t_{1})+\Psi_{h}(x,y;t_{1},t_{2}):=C_{1}.$

∎

The next three propositions are similarly to Proposition 2.6, 2.7 and 2.8, so
are their proofs.

######  Proposition 4.2.

$\Psi_{h}$ is locally Lipschitz continuous in
$\\{(x,y,t_{1},t_{2}):x,y\in\mathbb{C}\setminus
B^{o}_{R_{0}},\;t_{2}>t_{1}\\}\subset\mathbb{C}^{2}\times\mathbb{R}^{2}$.

###### Proof.

With Lemma 4.3, this can be proven by a similar argument as Proposition 2.6. ∎

######  Proposition 4.3.

When $h>0$, for any $s_{1},s_{2}\in[0,1)$ and $x,y\in\mathbb{C}\setminus
B^{o}_{R_{0}}$. There exist $t_{1}<t_{2}\in\mathbb{R}$ satisfying
$\\{t_{i}\\}=s_{i}$, $i=1,2$, and a $\gamma\in\Gamma_{t_{1},t_{2}}(x,y)$, such
that

$\mathcal{A}_{h}(\gamma;t_{1},t_{2})=\Psi_{h}(x,y;t_{1},t_{2})=\Phi_{h}(x,y;s_{1},s_{2}).$

###### Proof.

With Proposition 4.1, this can be proven by the same arguments as Proposition
2.7. ∎

For any $x,y\in\mathbb{C}\setminus B^{o}_{R_{0}}$ and $t_{1}<t_{2}$, we set

(55) $\Omega_{t_{1},t_{2}}(x,y)=\\{\xi\in
H^{1}_{t_{1},t_{2}}(x,y):|\xi(t)|\geq R_{0},\;\forall t\in[t_{1},t_{2}]\\}.$

######  Proposition 4.4.

For any $h>0$ and $x,y\in\mathbb{C}\setminus B^{o}_{R_{0}}$, there is a
$\gamma_{xy}\in\Gamma_{t_{x},t_{y}}(x,y)$ satisfying:

  1. (a).

$\mathcal{A}_{h}(\gamma_{xy};t_{x},t_{y})=\Phi_{h}(x,y;\\{t_{x}\\},\\{t_{y}\\})=\inf\\{\Phi_{h}(x,y;s_{1},s_{2}):s_{1},s_{2}\in[0,1)\\};$

  2. (b).

for any $\tau_{1}\in[t_{x},s_{x}]$ and $\tau_{2}\in[s_{y},t_{y}]$ (with
$s_{x}$ and $s_{y}$ given as in (48)),

$\mathcal{A}_{h}(\gamma_{xy};\tau_{1},\tau_{2})=\Phi_{h}(\gamma_{xy}(\tau_{1}),\gamma_{xy}(\tau_{2});\\{\tau_{1}\\},\\{\tau_{2}\\}).$

  3. (c).

for any $[\tau_{1},\tau_{2}]\subset[t_{x},s_{x}]$ or $[s_{y},t_{y}]$,

$\mathcal{A}_{h}(\gamma_{xy};\tau_{1},\tau_{2})=\inf\\{\mathcal{A}_{h}(\xi;\tau_{1},\tau_{2}):\;\xi\in\Omega_{\tau_{1},\tau_{2}}(\gamma_{xy}(\tau_{1}),\gamma_{xy}(\tau_{2})\\}.$

###### Proof.

(a). By Proposition 4.3, we can find a sequence of paths
$\\{\gamma_{n}\in\Gamma_{t_{n},\tau_{n}}(x,y)\\}_{n\in\mathbb{N}}$ satisfying

$\lim_{n\to\infty}\mathcal{A}_{h}(\gamma_{n};t_{n},\tau_{n})=\Phi_{h}(x,y;\\{t_{n}\\},\\{\tau_{n}\\})=\inf\\{\Phi_{h}(x,y;s_{1},s_{2});s_{1},s_{2}\in[0,1)\\}.$

Since the above infimum must be finite,
$\\{\tau_{n}-t_{n}\\}_{n\in\mathbb{N}}$ is bounded. Otherwise the following
holds, which is absurd.

$\mathcal{A}_{h}(\gamma_{n};t_{n},\tau_{n})\geq
h(\tau_{n}-t_{n})\to\infty,\;\text{ as }n\to\infty.$

As we may assume $t_{n}\in[0,1]$, $\forall n$, the rest of property (a) can be
proven just like Proposition 2.8.

(b). For simplicity, set $\gamma=\gamma_{xy}$. By a contradiction argument,
let’s assume the desired result does not hold. Then there exist $t_{1}<t_{2}$
with $\\{t_{i}\\}=\\{\tau_{i}\\}$, $i=1,2$, and a
$\xi\in\Gamma_{t_{1},t_{2}}(\gamma(\tau_{1}),\gamma(\tau_{2}))$ satisfying
$\mathcal{A}_{h}(\xi;t_{1},t_{2})<\Phi_{h}(\gamma(\tau_{1}),\gamma(\tau_{2});\\{\tau_{1}\\},\\{\tau_{2}\\}).$
Then

$\tilde{\gamma}|_{[t_{x},\tau_{1}+t_{2}-t_{1}+t_{y}-\tau_{2}]}=\gamma|_{[t_{x},\tau_{1}]}*\xi|_{[t_{1},t_{2}]}*\gamma|_{[\tau_{2},t_{y}]}\in\Xi_{\\{t_{x}\\},\\{t_{y}\\}}(\gamma(t_{x}),\gamma(t_{y}))$

and
$\mathcal{A}_{h}(\tilde{\gamma};t_{1},\tau_{1}+t_{2}-t_{1}+t_{y}-\tau_{2})<\mathcal{A}_{h}(\gamma;t_{x},t_{y})=\Phi_{h}(x,y;\\{t_{n}\\},\\{\tau_{n}\\})$,
which is absurd.

(c). This follows directly from Definition 4.1 and (55). ∎

A lemma similar to Lemma 3.1 will be needed. However because of the
topological constraint imposed in Definition 4.1, we only consider paths from
$\Omega$ as defined in (55).

######  Lemma 4.4.

For any $x,y\in\mathbb{C}$ satisfying $\sqrt{2}R_{1}\leq|x|,|y|\leq R$, the
following results hold.

  1. (a).

For any $t_{1}<t_{2}$,
$\inf\\{\mathcal{A}(\gamma;t_{1},t_{2}):\gamma\in\Omega_{t_{1},t_{2}}(x,y)\\}\leq
16\frac{R^{2}}{t_{2}-t_{1}}+12m\frac{t_{2}-t_{1}}{R}.$

  2. (b).

For any $h>0$, there exist positive constant $\beta_{1},\beta_{2}$ independent
of $R$, such that

$\inf\\{\mathcal{A}_{h}(\gamma;\tau_{1},\tau_{2}):\gamma\in\cup_{\tau_{1}<\tau_{2}}\Omega_{\tau_{1},\tau_{2}}(x,y)\\}\leq\beta_{1}R+\beta_{2}.$

###### Proof.

This can be proven exactly the same as Lemma 3.1. ∎

For the rest of the section, let’s fix arbitrarily an $h>0$, a pair of angles
$\theta_{\pm}\in[0,2\pi)$ and set

$\Lambda_{\theta_{-}}^{\theta_{+}}(R)=\\{(r_{1}e^{i\theta_{1}},r_{2}e^{i\theta_{2}}):r_{1},r_{2}\geq
R,\;\theta_{1}=\theta_{-}(\text{mod }2\pi)\text{ and
}\theta_{2}=\theta_{+}(\text{mod }2\pi)\\}.$

For any $(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{0})$, let
$\gamma_{xy}\in\Gamma_{t_{x},t_{y}}(x,y)$ be the minimizer given by
Proposition 4.4.

######  Lemma 4.5.

For any $(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})$ with $R_{2}\geq
2\sqrt{2}R_{1}$, let

(56)
$\tau_{y}=\min\\{t\in[s_{y},t_{y}]:|\gamma_{xy}(t)|=R_{2}\\},\;\;\tau_{x}=\max\\{t\in[t_{x},s_{x}]:|\gamma_{xy}(t)|=R_{2}\\},$

with $s_{x},s_{y}$ defined as in (48). The following results hold.

  1. (a).

$0<\inf\\{s_{y}-s_{x}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}\leq\sup\\{s_{y}-s_{x}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}<\infty.$

  2. (b).

$0<\inf\\{\tau_{y}-s_{y}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}\leq\sup\\{\tau_{y}-s_{y}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}<\infty.$

  3. (c).

$0<\inf\\{s_{x}-\tau_{x}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}\leq\sup\\{s_{x}-\tau_{x}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}<\infty.$

  4. (d).

Set $r_{xy}(t)=|\gamma_{xy}(t)|$, when $R_{2}$ is large enough,

(57) $\dot{r}_{xy}(t)\geq\sqrt{3m/2}R_{2}^{-\frac{1}{2}},\;\forall
t\in[\tau_{y},t_{y}];\quad\dot{r}_{xy}(t)\leq-\sqrt{3m/2}R_{2}^{-\frac{1}{2}},\;\forall
t\in[t_{x},\tau_{x}].$

Figure 3.

###### Proof.

(a). Fixed an $n_{0}\geq 2$, by Proposition 4.2,

$\sup\\{\Psi_{h}(x_{0},y_{0};s_{1},s_{2}+n_{0}):x_{0}\in\partial
B_{R_{2}},\;y_{0}\in\partial B_{R_{2}},s_{1},s_{2}\in[0,1]\\}\leq C_{1}.$

Then By Proposition 4.4, for any
$[\tau_{1},\tau_{2}]\subset[\tau_{x},\tau_{y}]$ (notice that
$[s_{x},s_{y}]\subset[\tau_{x},\tau_{y}]$, see Figure 3),

(58) $\displaystyle\mathcal{A}_{h}(\gamma_{xy};\tau_{1},\tau_{2})$
$\displaystyle\leq\mathcal{A}_{h}(\gamma_{xy};\tau_{x},\tau_{y})=\Phi_{h}(\gamma_{xy}(\tau_{x}),\gamma_{xy}(\tau_{y});\\{\tau_{x}\\},\\{\tau_{y}\\})$
$\displaystyle\leq\Psi_{h}(x,y;\\{\tau_{x}\\},\\{\tau_{y}\\}+n_{0})\leq
C_{1}.$

This gives us the following estimates, which imply the third inequality in
(a),

$s_{y}-s_{x}\leq\mathcal{A}_{h}(\gamma_{xy};s_{x},s_{y})/h\leq\mathcal{A}_{h}(\gamma_{xy};\tau_{x},\tau_{y})/h\leq
C_{1}/h.$

Meanwhile by Lemma 4.2 and the Cauchy-Schwartz inequality,

$(s_{y}-s_{x})\int_{s_{x}}^{s_{y}}|\dot{\gamma}_{xy}|^{2}\,dt\geq\left(\int_{s_{x}}^{s_{y}}|\dot{\gamma}_{xy}|\,dt\right)^{2}\geq
4.$

Therefore

$\frac{2}{s_{y}-s_{x}}\leq\frac{1}{2}\int_{s_{x}}^{s_{y}}|\dot{\gamma}_{xy}|^{2}\,dt\leq\mathcal{A}_{h}(\gamma_{xy};s_{x},s_{y})\leq
C_{1}.$

This implies the first inequality in (a), while the second inequality in (a)
is trivial.

(b) & (c). Since $[\tau_{x},s_{x}],[s_{y},\tau_{y}]\subset[\tau_{x},\tau_{y}]$
and

$\int_{s_{y}}^{\tau_{y}}|\dot{\gamma}_{xy}|\,dt,\int_{\tau_{x}}^{s_{x}}|\dot{\gamma}_{xy}|\,dt\geq
R_{2}-R_{0}>0,$

the proofs are similar to those given in part (a) and will be omitted.

(d). With Lemma 4.4, this can be proven by a similar argument as the proof of
property (b) in Lemma 3.2. ∎

Like in the previous section, we fix an $R_{2}\geq 2\sqrt{2}R_{1}$ for the
rest of the section, such that the above lemma always holds. Moreover since
the system is $1$-periodic, we may further assume

(59)
$s_{x}\in[0,1),\;\forall(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2}).$

Under the above assumption, Lemma 4.5 implies

(60)
$\sup\\{|\tau_{x}|,|\tau_{y}|:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}<\infty.$

######  Lemma 4.6.

Given an arbitrary
$\tau\geq\sup\\{|\tau_{x}|,|\tau_{y}|:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}$,
the following results hold.

  1. (a).

For any $(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})$ with
$[-\tau,\tau]\subset[t_{x},t_{y}]$, there is a finite $R=R(\tau,R_{2})$
independent of $(x,y)$, such that $|\gamma_{xy}(t)|\leq R$, $\forall
t\in[-\tau,\tau].$

  2. (b).

$\\{\gamma_{xy}|_{[-\tau,\tau]}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\text{
with }[-\tau,\tau]\subset[t_{x},t_{y}]\\}$ is a bound subset of
$H^{1}([-\tau,\tau],\mathbb{C})$.

###### Proof.

Throughout this proof, we only consider
$(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})$ with
$[-\tau,\tau]\subset[t_{x},t_{y}]$.

(a). Since $|\gamma_{xy}(t)|\leq R_{2}$, $\forall t\in[\tau_{x},\tau_{y}]$. By
property (d) in Lemma 4.5, it is enough to show $|\gamma_{xy}(\pm\tau)|\leq
R$, for some $R$ independent of $(x,y)$. Meanwhile by property (c) in
Proposition 4.4 and property (b) in Lemma 4.4,

(61) $\int_{\tau_{y}}^{\tau}|\dot{\gamma}_{xy}|^{2}\,dt\leq
2\mathcal{A}_{h}(\gamma_{xy};\tau_{y},\tau)\leq 2\beta_{1}R+2\beta_{2};$ (62)
$\int_{-\tau}^{\tau_{x}}|\dot{\gamma}_{xy}|^{2}\,dt\leq
2\mathcal{A}_{h}(\gamma_{xy};-\tau,\tau_{x})\leq 2\beta_{1}R+2\beta_{2}.$

The desired result then follows from the same argument as in property (a) of
Lemma 3.3.

(b). It is enough to show $\int_{-\tau}^{\tau}|\dot{\gamma}_{xy}|^{2}\,dt\leq
C_{1}$ for some $C_{1}$ independent of $\gamma_{xy}$. By Lemma 4.5,

$0<a^{-}=\inf\\{\tau_{x}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}<a^{+}=\sup\\{\tau_{x}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}<\infty;$
$0<b^{-}=\inf\\{\tau_{y}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}<b^{+}=\sup\\{\tau_{y}:(x,y)\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}<\infty.$

Then Proposition 4.2 implies

$\sup\\{\Psi_{h}(z_{1},z_{2};t_{1},t_{2}):z_{1},z_{2}\in\partial
B_{R_{2}},t_{1}\in[a^{-},a^{+}],\;t_{2}\in[b^{-},b^{+}]\\}\leq C_{2}.$

By Proposition 4.4,

$\mathcal{A}_{h}(\gamma_{xy};\tau_{x},\tau_{y})=\Phi_{h}(\gamma_{xy}(\tau_{x}),\gamma_{xy}(\tau_{y});\\{\tau_{x}\\},\\{\tau_{y}\\})\leq\Psi_{h}(\gamma_{xy}(\tau_{x}),\gamma_{xy}(\tau_{y});\\{\tau_{x}\\},\\{\tau_{y}\\}).$

Since $|\gamma_{xy}(\tau_{x})|=|\gamma_{xy}(\tau_{y})|=R_{2}$, the above two
estimates imply

$\int_{\tau_{x}}^{\tau_{y}}|\dot{\gamma}_{xy}|^{2}\,dt\leq
2\mathcal{A}_{h}(\gamma_{xy};\tau_{x},\tau_{y})\leq
2\Psi_{h}(\gamma_{xy}(\tau_{x}),\gamma_{xy}(\tau_{y});\\{\tau_{x}\\},\\{\tau_{y}\\})\leq
2C_{2}.$

Combining this with (61) and (62), we get the desired result. ∎

For the rest of the section, let’s choose a sequence
$\\{(x_{n},y_{n})\in\Lambda_{\theta_{-}}^{\theta_{+}}(R_{2})\\}_{n\in\mathbb{N}}$
satisfying

$\lim_{n\to\infty}|x_{n}|=\lim_{n\to\infty}|y_{n}|=\infty.$

For each $n$, by Proposition 4.4, there is a
$\gamma_{n}\in\Gamma_{t_{x_{n}},t_{y_{n}}}(x_{n},y_{n})$, such that

$\mathcal{A}_{h}(\gamma_{n};t_{x_{n}},t_{y_{n}})=\Phi_{h}(x_{n},y_{n};\\{t_{x_{n}}\\},\\{t_{y_{n}}\\})=\inf\\{\Phi_{h}(x_{n},y_{n};s_{1},s_{2}):s_{1},s_{2}\in[0,1)\\}.$

######  Proposition 4.5.

Under the above notations.

  1. (a).

$t_{x_{n}}\to-\infty$ and $t_{y_{n}}\to\infty$, as $n\to\infty$.

  2. (b).

There is a $\gamma\in H^{1}(\mathbb{R},\mathbb{C})$, such that (after passing
to a proper subsequence) $\gamma_{n}\to\gamma$ strongly in
$L^{\infty}_{\text{loc}}$-norm and weakly in $H^{1}_{\text{loc}}$-norm, as
$n\to\infty$.

  3. (c).

For any $t_{0}>0$ large enough,
$\gamma|_{[-t_{0},t_{0}]}\in\Gamma_{-t_{0},t_{0}}(\gamma(-t_{0}),\gamma(t_{0}))$
and

$\mathcal{A}_{h}(\gamma;-t_{0},t_{0})=\Phi_{h}(\gamma(-t_{0}),\gamma(t_{0});\\{-t_{0}\\},\\{t_{0}\\}).$

  4. (d).

$\gamma(t)$ is a solution of (2) with at most one regularizable collision, and
there is a $\tau_{0}>0$ large enough, such that for any compact interval
$I\subset(-\infty,\tau_{0})$ or $(\tau_{0},\infty)$,
$\|\gamma_{n}|_{I}-\gamma|_{I}\|_{C^{1}}\to 0$, as $n\to\infty$.

###### Proof.

(a). We give the details for $t_{y_{n}}\to\infty$, while $t_{x_{n}}\to\infty$
can be proven similarly.

By Proposition 4.4, Lemma 4.4 and the Cauchy-Schwartz inequality,

$\displaystyle\beta_{1}|y_{n}|+\beta_{2}$
$\displaystyle\geq\mathcal{A}_{h}(\gamma_{n};\tau_{y_{n}},t_{y_{n}})\geq\frac{1}{2}\int_{\tau_{y_{n}}}^{t_{y_{n}}}|\dot{\gamma}_{n}|^{2}\,dt\geq\frac{1}{2(t_{y_{n}}-\tau_{y_{n}})}\left(\int_{\tau_{y_{n}}}^{t_{y_{n}}}|\dot{\gamma}_{n}|\,dt\right)^{2}$
$\displaystyle\geq\frac{|y_{n}-\gamma_{n}(\tau_{y_{n}})|^{2}}{2(t_{y_{n}}-\tau_{y_{n}})}\geq\frac{|y_{n}|^{2}-2R_{2}|y_{n}|+R_{2}^{2}}{2t_{y_{n}}}.$

This implies

$t_{y_{n}}\geq\frac{|y_{n}|^{2}-2R_{2}|y_{n}|+R_{2}^{2}}{2(\beta_{1}|y_{n}|+\beta_{2})}\geq\frac{|y_{n}|-2R_{2}+R_{2}^{2}|y_{n}|^{-1}}{2\beta_{1}+2\beta_{2}|y_{n}|^{-1}}\to\infty,\text{
and }|y_{n}|\to\infty.$

(b). With property (a), it follows from a similarly argument as in the proof
of property (b) in Lemma 3.4.

(c). Choose an arbitrary
$t_{0}>\sup\\{|\tau_{x_{n}}|,|\tau_{y_{n}}|:n\in\mathbb{N}\\}$. By (60), the
supremum must be finite. Then for $n$ large enough,
$\gamma_{n}|_{[-t_{0},t_{0}]}\in\Gamma_{-t_{0},t_{0}}(\gamma_{n}(-t_{0}),\gamma_{n}(t_{0}))$.
Then property (b) implies
$\gamma|_{[-t_{0},t_{0}]}\in\Gamma_{-t_{0},t_{0}}(\gamma(-t_{0}),\gamma(t_{0}))$.
To obtain the rest of the property, by a contradiction argument, let’s assume
there is a $t_{0}$ large enough, such that

$\mathcal{A}_{h}(\gamma;-t_{0},t_{0})<\Phi_{h}(\gamma(-t_{0}),\gamma(t_{0});\\{t_{0}\\},\\{t_{0}\\}).$

By Proposition 4.4, there is a $\xi\in
H^{1}_{t_{1},t_{2}}(\gamma(-t_{0}),\gamma(t_{0}))$ with
$\\{t_{1}\\}=\\{-t_{0}\\}$ and $\\{t_{2}\\}=\\{t_{0}\\}$(see Figure 4), and an
$\varepsilon>0$ small enough, such that

(63)
$\mathcal{A}_{h}(\xi;t_{1},t_{2})=\Phi_{h}(\gamma(-t_{0}),\gamma(t_{0});\\{-t_{0}\\},\\{t_{0}\\})\leq\mathcal{A}_{h}(\gamma;-t_{0},t_{0})-4\varepsilon.$

By property (b) above and the lower semi-continuity of $\mathcal{A}_{h}$, for
$n$ large enough,

(64)
$\mathcal{A}_{h}(\gamma_{n};-t_{0},t_{0})\geq\mathcal{A}_{h}(\gamma;-t_{0},t_{0})-\varepsilon\geq\mathcal{A}_{h}(\xi;t_{1},t_{2})+3\varepsilon.$

Notice that $\xi(t_{1})=\gamma(-t_{0})$ and $\xi(t_{2})=\gamma(t_{0})$ are
collision-free, for $t_{0}$ large enoguh. Then so is
$\gamma|_{[t_{1},\tau_{1}]}$ and $\gamma|_{[\tau_{2},t_{2}]}$, for $\tau_{i}$
close enough to $t_{i}$, $i=1,2$. As a result, $\mathcal{A}_{h}$ is continuous
in a small neighborhood of $\xi|_{[t_{1},\tau_{1}]}$ in
$H^{1}_{t_{1},\tau_{1}}(\xi(t_{1}),\xi(\tau_{1}))$, and of
$\xi|_{[\tau_{2},t_{2}]}$ in
$H^{1}_{\tau_{2},t_{2}}(\xi(\tau_{2}),\xi(t_{2}))$. Since

$\gamma_{n}(-t_{0})\to\gamma(-t_{0})=\xi(t_{1}),\;\gamma_{n}(t_{0})\to\gamma(t_{0})=\xi(t_{2}),\;\text{
when }n\to\infty.$

For each $n$ large enough, there is a $\eta_{n}\in
H^{1}_{t_{1},\tau_{1}}(\gamma_{n}(-t_{0}),\xi(\tau_{1}))$ and
$\tilde{\eta}_{n}\in H^{1}_{\tau_{2},t_{2}}(\xi(\tau_{2}),\gamma_{n}(t_{0}))$
with

$|\mathcal{A}_{h}(\xi;t_{1},\tau_{1})-\mathcal{A}_{h}(\eta_{n};t_{1},\tau_{1})|,|\mathcal{A}_{h}(\xi;\tau_{2},t_{2})-\mathcal{A}_{h}(\tilde{\eta}_{n};\tau_{2},t_{2})\leq\varepsilon.$

Let
$\zeta_{n}|_{[t_{1},t_{2}]}=\eta_{n}|_{[-t_{0},\tau_{1}]}*\xi|_{[\tau_{1},\tau_{2}]}*\tilde{\eta}_{n}|_{[\tau_{2},t_{2}]}$.
Then

$\mathcal{A}_{h}(\xi;t_{1},t_{2})-\mathcal{A}_{h}(\zeta_{n};t_{1},t_{2})\geq-2\varepsilon.$

Combining this with (64), we get

$\mathcal{A}_{h}(\gamma_{n};-t_{0},t_{0})-\mathcal{A}_{h}(\zeta_{n};t_{1},t_{2})\geq\varepsilon.$

This is absurd, as
$\zeta_{n}|_{[t_{1},t_{2}]}\in\Xi_{\\{-t_{0}\\},\\{t_{0}\\}}(\gamma_{n}(-t_{0}),\gamma_{n}(t_{0}))$
and Proposition 4.4 implies

$\mathcal{A}_{h}(\gamma_{n};-t_{0},t_{0})=\Phi_{h}(\gamma_{n}(-t_{0}),\gamma_{n}(t_{0});\\{-t_{0}\\},\\{t_{0}\\}).$

Figure 4.

(d). With property (c), Proposition 4.1 shows $\gamma|_{\mathbb{R}}$ is a
solution of (2) with at most one regularizable collision. Then we can find a
$\tau>0$ large enough, such that $\gamma(t)$ is collision-free, for all
$t\in(-\infty,\tau_{0}]\cup[\tau_{0},\infty)$. The result then follows from
similar arguments as in the proof of property (d) in Proposition 3.4. ∎

With the above results, similar arguments as in the proofs of Proposition 3.5
and 3.6 can show

######  Proposition 4.6.

$\lim_{t\to\pm\infty}\frac{\gamma(t)}{|\gamma(t)|}=e^{i\theta_{\pm}(\text{mod
}2\pi)}$ and
$\lim_{t\to\pm\infty}\dot{\gamma}(t)=\pm\sqrt{2h}e^{i\theta_{\pm}(\text{mod
}2\pi)}.$

###### Proof of Theorem 1.2.

When $T=1$, the desired result follows from previous results obtained in this
section. When $T\neq 1$, it follows from a blow-up argument as in the proof of
Theorem 1.1 given in the end of Section 3. ∎

##  5\. Appendix A: asymptotic analysis near an isolated collision

We give proofs of Proposition 2.2 and 2.3 in this appendix. For this it is
more convenient to consider the relative position between the massless body
and the primary $m_{i_{0}}$, which is

(65) $\zeta(t)=\gamma(t)-q_{i_{0}}(t),\;\forall
t\in[t_{0}-\delta,t_{0}+\delta].$

Then the Lagrangian $L$ becomes

(66)
$L(\zeta,\dot{\zeta},t)=\frac{1}{2}|\dot{\zeta}(t)+\dot{q}_{i_{0}}(t)|^{2}+\frac{m_{i_{0}}}{|\zeta(t)|}+\sum_{i\neq
i_{0}}\frac{m_{i}}{|\zeta(t)+q_{i_{0}}(t)-q_{i}(t)|}\,dt.$

###### Proof of Proposition 2.2.

Since $\gamma(t)$ is a collision-ejection solution of (2), $\zeta(t)$
satisfies

(67)
$\ddot{\zeta}(t)=-\frac{m_{i_{0}}\zeta(t)}{|\zeta(t)|^{3}}-\ddot{q}_{i_{0}}(t)-\sum_{i\neq
i_{0}}\frac{m_{i}(\zeta(t)+q_{i_{0}}(t)-q_{i}(t))}{|\zeta(t)+q_{i_{0}}(t)-q_{i}(t)|^{3}},\;\;\forall
t\neq t_{0}.$

Notice that there is a $C_{1}>0$, such that for any
$t\in[t_{0}-\delta,t_{0}+\delta]$,

$\Big{|}\ddot{q}_{i_{0}}(t)+\sum_{i\neq
i_{0}}\frac{m_{i}(\zeta(t)+q_{i_{0}}(t)-q_{i}(t))}{|\zeta(t)+q_{i_{0}}(t)-q_{i}(t)|^{3}}\Big{|}\leq
C_{1}.$

This means $\zeta(t)$ is a collision-ejection solution of the Kepler problem
with bounded perturbation. Then the desired results follow from Sperling [34].
∎

For the proof of Proposition 2.3, we need some results of the Kepler problem

(68) $\ddot{\xi}(s)=-m_{i_{0}}\frac{\xi(s)}{|\xi(s)|^{3}},$

which is the Euler-Lagrangian equation of the following Lagrangian action
functional

(69)
$\bar{\mathcal{A}}(\xi;s_{1},s_{2})=\int_{s_{1}}^{s_{2}}\bar{L}(\xi,\dot{\xi})\,ds,\;\text{
where
}\bar{L}(\xi,\dot{\xi})=\frac{1}{2}|\dot{\xi}|^{2}+\frac{m_{i_{0}}}{|\xi|}.$

Recall that for $\sigma_{\pm}\in\mathbb{C}$ with $|\sigma_{\pm}|=1$, the
following $\bar{\zeta}(s)$ is _a parabolic homothetic collision-ejection
solution_ of the Kepler problem (68) with an isolated collision at the moment
$t_{0}$,

(70)
$\bar{\zeta}(s)=\begin{cases}(\frac{9}{2}m_{i_{0}})^{\frac{1}{3}}|s|^{\frac{2}{3}}\sigma_{-},&\text{
if }s\leq t_{0};\\\
(\frac{9}{2}m_{i_{0}})^{\frac{1}{3}}|s|^{\frac{2}{3}}\sigma_{+},&\text{ if
}s\geq t_{0}.\end{cases}$

######  Proposition 5.1.

When $\sigma_{-}\neq\sigma_{+}$, for any $T>0$, there are two collision-free
solutions of (68), $\xi^{\pm}\in
C^{\infty}([-T,T],\mathbb{C}\setminus\\{0\\})$ satisfying
$\bar{\mathcal{A}}(\xi^{\pm};-T,T)<\bar{\mathcal{A}}(\bar{\zeta};-T,T)$ and

  1. (a).

$\xi^{+}(\pm T)=\bar{\zeta}(\pm T)$ and
$\text{Arg}(\xi^{+}(T))-\text{Arg}(\xi^{+}(-T))\in(0,2\pi)$;

  2. (b).

$\xi^{-}(\pm T)=\bar{\zeta}(\pm T)$ and
$\text{Arg}(\xi^{-}(T))-\text{Arg}(\xi^{-}(-T))\in(-2\pi,0).$

######  Remark 5.1.

As the angular momentum of a collision-free solution of the Kepler problem is
a non-zero constant, $\text{Arg}(\xi^{+}(t))$ is strictly increasing and
$\text{Arg}(\xi^{-}(t))$ is strictly decreasing, as $t$ goes from $-T$ to $T$.

###### Proof.

This is a well-known result with several different proofs, see [36], [18],
[38] or [10]. ∎

A blow-up argument will be used in order to apply this result.

######  Definition 5.1.

For any $\lambda>0$, define the $\lambda$-blow-up of
$\zeta|_{[t_{0}-\delta,t_{0}+\delta]}$ and
$q_{i}|_{[t_{0}-\delta,t_{0}+\delta]}$, $i\in\mathbf{N}$, as

(71)
$\zeta^{\lambda}(s)=\lambda^{\frac{2}{3}}\zeta(t_{0}+\lambda^{-1}s),\;\;q^{\lambda}_{i}(s)=\lambda^{\frac{2}{3}}q_{i}(t_{0}+\lambda^{-1}s),\;\;\forall
s\in[-\lambda\delta,\lambda\delta].$

The time is shifted only to simplify notations. As a result,
$\zeta^{\lambda}(s)$, $s\in[-\lambda\delta,\lambda\delta]\setminus\\{0\\}$,
satisfies

(72)
$\ddot{\zeta}^{\lambda}(s)+\ddot{q}_{i}^{\lambda}(s)=-m_{i_{0}}\frac{\zeta^{\lambda}(s)}{|\zeta^{\lambda}(s)|^{3}}-\sum_{i\neq
i_{0}}\frac{m_{i}(\zeta^{\lambda}(s)+q^{\lambda}_{i_{0}}(s)-q^{\lambda}_{i}(s))}{|\zeta^{\lambda}(s)+q^{\lambda}_{i_{0}}(s)-q^{\lambda}_{i}(s)|^{3}},$

which is the Euler-Lagrangian equation of the action functional

(73)
$\mathcal{A}^{\lambda}(\zeta^{\lambda};s_{1},s_{2})=\int_{s_{1}}^{s_{2}}L^{\lambda}(\zeta^{\lambda}(s),\dot{\zeta}^{\lambda}(s),s)\,ds,$

where

(74)
$L^{\lambda}(\zeta^{\lambda}(s),\dot{\zeta}^{\lambda}(s),s)=\frac{1}{2}|\dot{\zeta}^{\lambda}(s)+\dot{q}^{\lambda}_{i_{0}}(s)|^{2}+\frac{m_{i_{0}}}{|\zeta^{\lambda}(s)|}+\sum_{i\neq
i_{0}}\frac{m_{i}}{|\zeta^{\lambda}(s)+q^{\lambda}_{i_{0}}(s)-q^{\lambda}_{i}(s)|}.$

######  Lemma 5.1.

For any $\delta_{1}\in(0,\delta]$,
$\mathcal{A}^{\lambda}(\zeta^{\lambda};-\lambda\delta_{1},\lambda\delta_{1})=\lambda^{\frac{1}{3}}\mathcal{A}(\zeta;t_{0}-\delta_{1},t_{0}+\delta_{1}).$

###### Proof.

By a direct computation,

$L^{\lambda}(\zeta^{\lambda}(s),\dot{\zeta}^{\lambda}(s),s)=\lambda^{-\frac{2}{3}}L(\zeta(t),\dot{\zeta}(t),t).$

The result then follows from the fact that $t=\lambda^{-1}s+t_{0}$, as it
implies $ds=\lambda dt$. ∎

######  Lemma 5.2.

For any $T>0$, as $\lambda$ goes to infinity,

  1. (a).

$\zeta^{\lambda}(s)$ converges uniformly to $\bar{\zeta}(s)$ on $[-T,T]$;

  2. (b).

$\dot{\zeta}^{\lambda}(s)$ converges uniformly to $\dot{\bar{\zeta}}(s)$ on
any compact subset of $[-T,T]\setminus\\{0\\}$;

###### Proof.

It follows from Proposition 2.2 and direct computations. ∎

Now we are ready to give a proof of Proposition 2.3.

###### Proof of Proposition 2.3.

Fix a $T>0$ for the rest of the proof. Let $\xi$ be either $\xi^{+}$ or
$\xi^{-}$ obtained in Proposition 5.1. Then there is an $\varepsilon_{0}>0$,
such that

(75)
$\bar{\mathcal{A}}(\bar{\zeta};-T,T)-\bar{\mathcal{A}}(\xi;-T,T)=6\varepsilon_{0}.$

As $\sigma_{\pm}=\frac{\bar{\zeta}(\pm T)}{|\bar{\zeta}(\pm T)}=\frac{\xi(\pm
T)}{|\xi(\pm T)|}$, we may further assume
$\text{Arg}(\sigma_{\pm})=\text{Arg}(\xi(\pm T)).$ By Proposition 5.1,

(76)
$\text{Arg}(\sigma_{+})-\text{Arg}(\sigma_{-})\in\begin{cases}(0,2\pi),\;&\text{
if }\xi=\xi^{-};\\\ (-2\pi,0),\;&\text{ if }\xi=\xi^{+}.\end{cases}$

By Lemma 5.2, $\\{\zeta^{\lambda}|_{[-2T,2T]}:\lambda>2T/\delta\\}$ is bound
in $H^{1}([-2T,2T],\mathbb{C})$, so we can find a sequence of positive numbers
$\\{\lambda_{n}\nearrow\infty\\}_{n=1}^{\infty}$ with
$\zeta^{\lambda_{n}}|_{[-2T,2T]}$ converging weakly to $\zeta|_{[-2T,2T]}$, as
$n$ goes to infinity. The weakly lower semi-continuity of $\bar{\mathcal{A}}$
then implies

$\lim_{n\to\infty}\bar{\mathcal{A}}(\zeta^{\lambda_{n}};-2T,2T)\geq\bar{\mathcal{A}}(\bar{\zeta};-2T,2T).$

As a result, for $n$ large enough,

(77)
$\bar{\mathcal{A}}(\zeta^{\lambda_{n}};-2T,2T)\geq\bar{\mathcal{A}}(\bar{\zeta};-2T,2T)-\varepsilon_{0}.$

By Lemma 5.2, we can also find a sequence of positive numbers
$\\{\delta_{n}\searrow 0\\}_{n=1}^{\infty}$, such that

(78)
$|\zeta^{\lambda_{n}}(s)-\bar{\zeta}(s)|\leq\delta_{n},\;\;|\dot{\zeta}^{\lambda_{n}}(s)-\dot{\bar{\zeta}}(s)|\leq\delta_{n},\;\;\forall
s\in[-2T,2T]\setminus[-T,T].$

For each $n$, we define a path $\xi^{\lambda_{n}}\in
H^{1}([-\delta/\lambda_{n},\delta/\lambda_{n}],\mathbb{C})$ as

(79) $\xi^{\lambda_{n}}(s)=\begin{cases}\zeta^{\lambda_{n}}(s),\;&\text{ when
}s\in[2T,\lambda_{n}\delta],\\\
\frac{2T-s}{\delta_{n}}(\bar{\zeta}(s)-\zeta^{\lambda_{n}}(s))+\zeta^{\lambda_{n}}(s),\;&\text{
when }s\in[2T-\delta_{n},2T],\\\ \bar{\zeta}(s),\;&\text{ when
}s\in[T,2T-\delta_{n}],\\\ \xi(s),\;&\text{ when }s\in[-T,T],\\\
\bar{\zeta}(s),\;&\text{ when }s\in[-2T+\delta_{n},-T],\\\
\frac{s+2T-\delta_{n}}{\delta_{n}}(\bar{\zeta}(s)-\zeta^{\lambda_{n}}(s))+\bar{\zeta}(s),\;&\text{
when }s\in[-2T,-2T+\delta_{n}],\\\ \zeta^{\lambda_{n}}(s),\;&\text{ when
}s\in[-\lambda_{n}\delta,-2T],\end{cases}$

and a corresponding path $\eta^{n}\in
H^{1}([t_{0}-\delta,t_{0}+\delta],\mathbb{C})$ as

$\eta^{\lambda_{n}}(t)=\lambda_{n}^{-\frac{2}{3}}\xi^{\lambda_{n}}(\lambda_{n}(t-t_{0})),\;\forall
t\in[t_{0}-\delta,t_{0}+\delta].$

We will show that for large $n$,
$(\eta^{\lambda_{n}}+q_{i_{0}})|_{[t_{0}-\delta,t_{0}+\delta]}$ are the
desired paths we are looking for.

Notice that $\xi^{\lambda_{n}}(s)$ and $\zeta^{\lambda_{n}}(s)$ are
correspondingly the $\lambda_{n}$-blow-up of $\eta^{\lambda_{n}}(t)$ and
$\zeta(t)$. By (79), for any $n$ large enough, we have the following
inequality, which verifies (10).

(80)
$|\eta^{\lambda_{n}}(t)-\zeta(t)|=|\eta^{\lambda_{n}}(t)+q_{i_{0}}(t)-\gamma(t)|\leq\varepsilon,\;\;\forall
t\in[t_{0}-\delta,t_{0}+\delta].$

In particular this implies that, for $\varepsilon$ small enough,

$\eta^{\lambda_{n}}(t)+q_{i_{0}}(t)\neq q_{i}(t),\;\forall
t\in[t_{0}-\delta,t_{0}+\delta].$

At the same time Proposition 5.1 and (79) also implies
$|\xi^{\lambda_{n}}(s)|\neq 0$, $\forall
s\in[-\lambda_{n}\delta,\lambda_{n}\delta]$. Then

$\eta^{\lambda_{n}}(t)\neq 0,\;\forall t\in[t_{0}-\delta,t_{0}+\delta].$

As a result, for $n$ large enough,
$(\eta^{\lambda_{n}}+q_{i_{0}})|_{[t_{0}-\delta,t_{0}+\delta]}$ must be
collision-free.

Meanwhile by Proposition 2.2, $\lim_{\delta\to
0}\zeta(t_{0}\pm\delta)/|\zeta(t_{0}\pm\delta)|=\sigma_{\pm}$. Therefore we
may assume

(81) $\lim_{\delta\to
0}\text{Arg}(\zeta(t_{0}\pm\delta))=\text{Arg}(\sigma_{\pm}).$

Then by (76), for $\delta$ small enough,

(82)
$\text{Arg}(\zeta(t_{0}+\delta))-\text{Arg}(\zeta(t_{0}-\delta))\in\begin{cases}(0,2\pi),&\text{
if }\xi=\xi^{+};\\\ (-2\pi,0),&\text{ if }\xi=\xi^{-}.\end{cases}$

Since
$\xi^{\lambda_{n}}(\pm\lambda_{n}\delta)=\zeta^{\lambda_{n}}(\pm\lambda_{n}\delta)$,
we may assume

$\text{Arg}(\xi^{\lambda_{n}})(\pm\lambda_{n}\delta)=\text{Arg}(\zeta^{\lambda_{n}})(\pm\lambda_{n}\delta).$

This then implies

$\text{Arg}(\eta^{\lambda_{n}})(t_{0}\pm\delta)=\text{Arg}(\zeta)(t_{0}\pm\delta).$

Combining this with (82), we get

$\text{Arg}(\eta^{\lambda_{n}}(t_{0}+\delta))-\text{Arg}(\eta^{\lambda_{n}}(t_{0}-\delta))\in\begin{cases}(0,2\pi),&\text{
if }\xi=\xi^{+};\\\ (-2\pi,0),&\text{ if }\xi=\xi^{-}.\end{cases}$

This implies
$\eta^{\lambda_{n}}|_{[t_{0}-\delta,t_{0}+\delta]}+q_{i_{0}}|_{[t_{0}-\delta,t_{0}+\delta]}$
satisfies (11), for $n$ large enough.

We claim that for $n$ large enough,

(83)
$\mathcal{A}^{\lambda_{n}}(\xi^{\lambda_{n}};-\lambda_{n}\delta,\lambda_{n}\delta)<\mathcal{A}^{\lambda_{n}}(\zeta^{\lambda_{n}};-\lambda_{n}\delta,\lambda_{n}\delta).$

Then (83) and Lemma 5.1 imply

$\mathcal{A}(\eta^{\lambda_{n}};t_{0}-\delta,t_{0}+\delta)<\mathcal{A}(\zeta;t_{0}-\delta,t_{0}+\delta).$

As a result, our proof is finished once (83) is established. To show this,
notice that

(84)
$\displaystyle\mathcal{A}^{\lambda_{n}}(\zeta^{\lambda_{n}};-\lambda_{n}\delta,\lambda_{n}\delta)$
$\displaystyle-\mathcal{A}^{\lambda_{n}}(\xi^{\lambda_{n}};-\lambda_{n}\delta,\lambda_{n}\delta)$
$\displaystyle=\mathcal{A}^{\lambda_{n}}(\zeta^{\lambda_{n}};-2T,2T)-\mathcal{A}^{\lambda_{n}}(\xi^{\lambda_{n}};-2T,2T)$
$\displaystyle\geq\bar{\mathcal{A}}(\zeta^{\lambda_{n}};-2T,2T)-\bar{\mathcal{A}}(\xi^{\lambda_{n}};-2T,2T)+\int_{-2T}^{2T}\langle\dot{\zeta}^{\lambda_{n}}-\dot{\xi}^{\lambda_{n}},\dot{q}_{i_{0}}^{\lambda_{n}}\rangle\,ds$
$\displaystyle-\int_{-2T}^{2T}\sum_{i\neq
i_{0}}\frac{m_{i}}{|\xi^{\lambda_{n}}+q^{\lambda_{n}}_{i_{0}}-q^{\lambda_{n}}_{i}|}\,ds.$

Meanwhile by (75) and (79),

(85)
$\bar{\mathcal{A}}(\bar{\zeta};-2T,2T)-\bar{\mathcal{A}}(\xi^{\lambda_{n}};-2T,2T)\geq
6\varepsilon_{0}-\bar{\mathcal{A}}(\xi^{\lambda_{n}};-2T,-2T+\delta_{n})-\bar{\mathcal{A}}(\xi^{\lambda_{n}};2T-\delta_{n},2T).$

As (78) and (79) imply

$|\dot{\xi}^{\lambda_{n}}(s)|\leq\begin{cases}1+|\dot{\zeta}^{\lambda_{n}}(s)|,&\text{
if }s\in[2T-\delta_{n},2T];\\\ 1+|\dot{\bar{\zeta}}(s)|,&\text{ if
}s\in[-2T,-2T+\delta_{n}],\end{cases}$

By Lemma 5.2, for $n$ large enough, there exist positive $C_{1}$ and $C_{2}$,
such that

(86) $\displaystyle|\dot{\xi}^{\lambda_{n}}(s)|\leq C_{1},$
$\displaystyle\forall s\in[-2T,-2T+\delta_{n}]\cup[2T-\delta_{n},2T];$
$\displaystyle|\xi^{\lambda_{n}}(s)|\leq C_{2}^{-1},$ $\displaystyle\forall
s\in[-2T,-2T+\delta_{n}]\cup[2T-\delta_{n},2T].$

As a result, for $n$ large enough,

$\displaystyle\bar{\mathcal{A}}(\xi^{\lambda_{n}};$
$\displaystyle-2T,-2T+\delta_{n})+\bar{\mathcal{A}}(\xi^{\lambda_{n}};2T-\delta_{n},2T)$
$\displaystyle\leq\int_{-2T}^{-2T+\delta_{n}}+\int_{2T-\delta_{n}}^{2T}\frac{1}{2}|\dot{\xi}^{\lambda_{n}}(s)|^{2}+\frac{m_{i_{0}}}{|\xi^{\lambda_{n}}(s)|}\,ds\leq
2\left(\frac{1}{2}C_{1}^{2}+m_{i_{0}}C_{2}\right)\delta_{n}\leq\varepsilon_{0}.$

Plug the above inequality into (85), we get

$\bar{\mathcal{A}}(\bar{\zeta};-2T,2T)-\bar{\mathcal{A}}(\xi^{\lambda_{n}};-2T,2T)\geq
5\varepsilon_{0}.$

Then (77) implies

(87)
$\bar{\mathcal{A}}(\zeta^{\lambda_{n}};-2T,2T)-\bar{\mathcal{A}}(\xi^{\lambda_{n}};-2T,2T)\geq
4\varepsilon_{0}.$

Since $\rho_{1}=\min\\{|\gamma(t)-q_{i}(t)|:\forall
t\in[t_{0}-\delta,t_{0}+\delta]\text{ and }i\neq i_{0}\\}>0$, $\forall i\neq
i_{0}$, (80) implies,

$|\eta^{\lambda_{n}}(t)+q_{i_{0}}(t)-q_{i}(t)|\geq|\zeta(t)+q_{i_{0}}(t)-q_{i}(t)|-|\eta^{\lambda_{n}}(t)-\zeta(t)|\geq\rho_{1}-\varepsilon,\;\forall
t\in[t_{0}-\delta,t_{0}+\delta].$

As a result, when $0<\varepsilon\leq\rho_{1}/2$, for $n$ large enough, we have

(88) $\displaystyle\int_{-2T}^{2T}$ $\displaystyle\sum_{i\neq
i_{0}}\frac{m_{i}}{|\xi^{\lambda_{n}}(s)+q^{\lambda_{n}}_{i_{0}}(s)-q^{\lambda_{n}}_{i}(s)|}\,ds=\lambda_{n}^{\frac{1}{3}}\int_{t_{0}-\frac{2T}{\lambda_{n}}}^{t_{0}+\frac{2T}{\lambda_{n}}}\sum_{i\neq
i_{0}}\frac{m_{i}}{|\eta^{\lambda_{n}}(t)+q_{i_{0}}(t)-q_{i}(t)|}\,dt$
$\displaystyle\leq\lambda_{n}^{\frac{1}{3}}\int_{t_{0}-\frac{2T}{\lambda_{n}}}^{t_{0}+\frac{2T}{\lambda_{n}}}\frac{\sum_{i\neq
i_{0}}m_{i}}{\rho_{1}-\varepsilon}\,dt\leq\frac{8T(\sum_{i\neq
i_{0}}m_{i})}{\rho_{1}}\lambda_{n}^{-\frac{2}{3}}\leq\varepsilon_{0}.$

Plug (87) and (88) into (84), we get that when $n$ is large enough,

$\mathcal{A}^{\lambda_{n}}(\zeta^{\lambda_{n}};-\delta/\lambda_{n},\delta/\lambda_{n})-\mathcal{A}^{\lambda_{n}}(\xi^{\lambda_{n}};-\delta/\lambda_{n},\delta/\lambda_{n})\geq
3\varepsilon_{0}+\int_{-2T}^{2T}\langle\dot{\zeta}^{\lambda_{n}}(s)-\dot{\xi}^{\lambda_{n}}(s),\dot{q}_{i}^{\lambda_{n}}(s)\rangle\,ds.$

As a result, it is enough to show that when $n$ is large enough,

$\int_{-2T}^{2T}|\dot{\zeta}^{\lambda_{n}}(s)|\cdot|\dot{q}_{i_{0}}^{\lambda_{n}}(s)|\,ds+\int_{-2T}^{2T}|\dot{\xi}^{\lambda_{n}}(s)|\cdot|\dot{q}_{i_{0}}^{\lambda_{n}}(s)|\,ds\leq
2\varepsilon_{0}.$

By the definition of $\lambda_{n}$-blow up,

(89)
$\int_{-2T}^{2T}|\dot{\zeta}^{\lambda_{n}}(s)|\cdot|\dot{q}_{i_{0}}^{\lambda_{n}}(s)|\,ds=\lambda_{n}^{\frac{1}{3}}\int_{t_{0}-\frac{2T}{\lambda_{n}}}^{t_{0}+\frac{2T}{\lambda_{n}}}|\dot{\zeta}(t)|\cdot|\dot{q}_{i_{0}}(t)|\,dt$

Since $q|_{[t_{0}-\delta,t_{0}+\delta]}$, is a collision-free solution of (2),

(90)
$C_{3}=\sup\\{|\dot{q}_{i_{0}}(t)|:t\in[t_{0}-\delta,t_{0}+\delta]\\}<\infty.$

Meanwhile by Proposition 2.2, there exist two positive constants
$C_{4},C_{5}$,

(91) $|\dot{\zeta}(t)|\leq C_{4}|t-t_{0}|^{-\frac{1}{3}}+C_{5},\;\forall t\neq
t_{0}\text{ small enough}.$

As a result, for $n$ large enough,

$\displaystyle\int_{-2T}^{2T}|\dot{\zeta}^{\lambda_{n}}(s)|\cdot|\dot{q}_{i_{0}}^{\lambda_{n}}(s)|\,ds$
$\displaystyle\leq\lambda_{n}^{\frac{1}{3}}C_{3}\int_{t_{0}-\frac{2T}{\lambda_{n}}}^{t_{0}+\frac{2T}{\lambda_{n}}}C_{4}|t-t_{0}|^{-\frac{1}{3}}+C_{5}\,dt$
$\displaystyle=C_{3}(3C_{4}(2T)^{\frac{2}{3}}\lambda_{n}^{-\frac{1}{3}}+4C_{5}T\lambda_{n}^{-\frac{2}{3}})\leq\varepsilon_{0}.$

By (79) and the first inequality in (86), we can find a $C_{6}$, such that for
$n$ large enough,

$\int_{-2T}^{2T}|\dot{\xi}^{\lambda_{n}}(s)|\,ds\leq C_{6}.$

Since
$|\dot{q}^{\lambda_{n}}_{i_{0}}(s)|=\lambda_{n}^{-\frac{1}{3}}|\dot{q}_{i_{0}}(t_{0}+\lambda_{n}^{-1}s)|\leq\lambda_{n}^{-\frac{1}{3}}C_{3}$,
$\forall s\in[-2T,2T]$, for $n$ large enough, we have

$\int_{-2T}^{2T}|\dot{\xi}^{\lambda_{n}}(s)|\cdot|\dot{q}_{i_{0}}^{\lambda_{n}}(s)|\,ds\leq\lambda_{n}^{-\frac{1}{3}}C_{3}C_{6}\leq\varepsilon_{0}.$

This finishes our proof.

∎

##  6\. Appendix B: The Perturbed Kepler problem

In this appendix, instead of the restricted $(N+1)$-body problem, we will
prove two propositions of the perturbed Kepler problem (92), that imply
Proposition 3.1 and 3.2, as we believe these results will be useful in more
general setting.

(92) $\ddot{z}=-\frac{mz}{|z|^{3}}+F(z,t),\;\;F\in
C^{0}(\mathbb{C}\times\mathbb{R},\mathbb{R}).$

Let $z|_{[t_{1},t_{2}]}$, is a solution of (92) for the rest of the appendix.
Rewrite it in polar coordinates with $z(t)=r(t)e^{i\theta(t)}$, we get

(93) $\begin{cases}&\ddot{r}-r\dot{\theta}^{2}=-m/r^{2}+F_{1}(r,\theta,t);\\\
&r\ddot{\theta}+2\dot{r}\dot{\theta}=F_{2}(r,\theta,t),\end{cases}$

where $F_{1}(r,\theta,t)=\text{Re}(F(re^{\theta},t)e^{-i\theta}),$ and
$F_{2}(r,\theta,t)=\text{Im}(F(re^{\theta},t)e^{-i\theta})$.

######  Proposition 6.1.

Assume there are two positive constants $R$ and $C$, such that

(94) $|F(x,t)|\leq C/|x|^{2},\;\;\forall x\in\mathbb{C}\setminus B_{R}^{o}.$

If $r(t_{1})\geq R$ and $\dot{r}(t_{1})\geq\sqrt{\frac{3(m+C)}{r(t_{1})}}>0$,
then $\dot{r}(t)>\frac{1}{2}\dot{r}(t_{1}),\;\forall t\in[t_{1},t_{2}]$.

###### Proof.

Assume the desired result does not hold, then there is a
$\tau_{1}\in(t_{1},t_{2}]$, such that

$\dot{r}(\tau_{1})=\dot{r}(t_{1})/2\text{ and
}\dot{r}(t)>\dot{r}(t_{1})/2,\;\forall t\in[t_{1},\tau_{1}).$

By (94), for $i=1,2$,

(95) $|F_{i}(r,\theta,t)|<C/r^{2},\;\;\forall r\geq R.$

Let $g(t)=\frac{1}{2}\dot{r}(t)^{2}-\frac{m+C}{r(t)}$. Then

$\dot{g}=\dot{r}(\ddot{r}+\frac{m+C}{r^{2}}).$

Combining this with (93) and (95), we get

$\ddot{r}=r\dot{\theta}^{2}-\frac{m}{r^{2}}+F_{1}\geq-\frac{m+C}{r^{2}},\;\;\forall
t\in[t_{1},\tau_{1}].$

As a result, $\dot{g}(t)\geq 0$, $\forall t\in[t_{1},\tau_{1}]$. This implies
$g(\tau_{1})\geq g(t_{1})$ and

$\frac{\dot{r}^{2}(t_{1})}{4}=\dot{r}^{2}(\tau_{1})>2g(\tau_{1})\geq
2g(t_{1})\geq\dot{r}^{2}(t_{1})-\frac{2(m+C)}{r(t_{1})},$

which is a contradiction to the condition that
$\dot{r}(t_{1})\geq\sqrt{3(m+C)/r(t_{1})}$.

∎

######  Proposition 6.2.

Assume there are two positive constants $R$ and $C$, such that

(96) $|F(x,t)|\leq C/|x|^{3},\;\;\forall x\in\mathbb{C}\setminus B^{o}_{R}.$

If $r(t_{1})\geq R$ and $\dot{r}(t)\geq v_{0}$, $\forall t\geq t_{1}$, for
some constant $v_{0}>0$, then

  1. (a).

$\sup\\{|\omega(t)|:\;t\in[t_{1},t_{2}]\\}\leq|\omega(t_{1})|+\frac{C}{v_{0}r(t_{1})}$,
where $\omega(t)=z(t)\wedge\dot{z}(t)$;

  2. (b).

when $t_{2}=\infty$, $\lim_{t\to\infty}\theta(t)=\theta_{0}$, for some
$\theta_{0}\in\mathbb{R}$.

###### Proof.

(a). By (92) and (96),

$|\dot{\omega}(t)|=|z(t)\wedge\ddot{z}(t)|=|z(t)\wedge
F(z,t)|\leq\frac{C}{r^{2}(t)}.$

For any $t\in[t_{1},t_{2}]$, since $r(t)\geq v_{0}(t-t_{1})+r(t_{1})$,

$\displaystyle|\omega(t)|$
$\displaystyle\leq|\omega(t_{1})|+\int_{t_{1}}^{t}\frac{C}{r^{2}(\tau)}d\tau\leq|\omega(t_{1})|+\int_{t_{1}}^{t}\frac{C/v_{0}^{2}}{(\tau-\tau_{0}+\frac{r(t_{1})}{v_{0}})^{2}}d\tau$
$\displaystyle=|\omega(t_{1})|+\frac{C}{v_{0}^{2}}\left(\frac{v_{0}}{r(t_{1})}-\frac{v_{0}}{v_{0}(t-t_{1})+r(t_{1})}\right)$
$\displaystyle\leq|\omega(t_{1})|+\frac{C}{v_{0}r(t_{1})}.$

(b). Notice that in polar coordinates $\omega(t)=r^{2}(t)\dot{\theta}(t)$.
Hence

$\dot{\theta}(t)=\frac{\omega(t)}{r^{2}(t)}\leq\frac{v_{0}r(t_{1})|\omega(t_{1})|+C}{v_{0}^{3}r(t_{1})}\frac{1}{(t-t_{1}+r(t_{1})/v_{0})^{2}}.$

Therefore for any $t>t_{1}$,

(97)
$|\theta(t)-\theta(t_{1})|\leq\int_{t_{1}}^{t}|\dot{\theta}(\tau)|d\tau\leq\frac{|\omega(t_{1})|}{v_{0}r(t_{1})}+\frac{C}{v_{0}^{2}r^{2}(t_{1})}.$

Since $\dot{r}(t)\geq v_{0}>0$, $\forall t\geq t_{0}$, $r(t)\to\infty$ as
$t\to\infty$. As a result, the upper bound given in the above inequality goes
to zero, as $t_{1}$ goes to infinity. By Cauchy’s criterion for convergence,
$\theta(t)$ converges to a finite $\theta_{0}$, as $t$ goes to infinity. ∎

By (20), Proposotion 3.1 and 3.2 follows directly from the above two results
correspondingly.

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