Entire parabolic trajectories as minimal phase transitions

Abstract

For the class of anisotropic Kepler problems in $\mathbb{R }^d\setminus \{0\}$ with homogeneous potentials, we seek parabolic trajectories having prescribed asymptotic directions at infinity and which, in addition, are Morse minimizing geodesics for the Jacobi metric. Such trajectories correspond to saddle heteroclinics on the collision manifold, are structurally unstable and appear only for a codimension-one submanifold of such potentials. We give them a variational characterization in terms of the behavior of the parameter-free minimizers of an associated obstacle problem. We then give a full characterization of such a codimension-one manifold of potentials and we show how to parameterize it with respect to the degree of homogeneity.

Appendices

Appendices

For the reader’s convenience, we collect here the proofs of some slight modifications of rather standard arguments.

1.1 The Maupertuis’ principle

Let us consider the (sufficiently smooth) maps

• $K:\mathbb{R }^{2d}\ni (\dot{x},x)\mapsto K(\dot{x},x)\in [0,+\infty ),$ with $K(\lambda \dot{x},x)= \lambda ^2K(\dot{x},x),$ for every $\lambda ,$ and

• $P:\mathbb{R }^{d}\ni x \mapsto P(x)\in (0,+\infty ].$

 

Lemma 7.1

Let $x\in H^1\left( (a,b); \mathbb{R }^d\right)$ be a fixed path, such that

$$\begin{aligned} A([a,b];x) := \int \limits _a^b \left[ K(\dot{x}(t),x(t))+P(x(t))\right] {\,\mathrm dt }\in (0,+\infty ) \end{aligned}$$

and

$$\begin{aligned} K(\dot{x}(t),x(t))\ge \delta >0 \text{ on} (a,b). \end{aligned}$$

Let us consider the set of all re-parameterizations of $x,$ namely

$$\begin{aligned} \Gamma _{x}:=\left\{ \begin{array}{ll} ((0,T),f) :&\,f:(0,T)\rightarrow (a,b),\,\text{ Lipschitz} \text{ continuous} \text{ and}\\&\text{ increasing,} \text{ such} \text{ that} x\circ f\in H^1\left( (0,T); \mathbb{R }^d\right) \end{array}\right\} . \end{aligned}$$

Finally, let us define

$$\begin{aligned} \vartheta (t):=\int \limits _a^t\sqrt{\frac{K(\dot{x}(t),x(t))}{P(x(t))}}{\,\mathrm dt },\quad \hat{T}:=\int \limits _a^b\sqrt{\frac{K(\dot{x}(t),x(t))}{P(x(t))}}{\,\mathrm dt }, \end{aligned}$$

$$\begin{aligned} \hat{f} := \vartheta ^{-1} :[0,\hat{T}]\rightarrow [a,b]. \end{aligned}$$

Then

$$\begin{aligned} \min _{\Gamma _x} A([0,T],x\circ f) \text{ is} \text{ achieved} \text{ by} \left(\left(0,\hat{T}\right),\hat{f}\right). \end{aligned}$$

Moreover, writing $\hat{x} := x\circ \hat{f},$ we have that, for (almost) every $\vartheta ,$

$$\begin{aligned} K({\hat{x}}^{\prime }(\vartheta ),{\hat{x}}(\vartheta ))=P({\hat{x}}(\vartheta )) \quad \left( ^{\prime }:= \frac{\text{ d}}{\text{ d}\vartheta } \right). \end{aligned}$$

Proof

Let us observe that, for any $f \in \Gamma _x,$

$$\begin{aligned} A([0,T],x\circ f)&= \int \limits _0^T \left[K(\dot{x}(f(\vartheta ))f^{\prime }(\vartheta ),x(f(\vartheta )))+P(x(f(\vartheta )))\right] \text{ d}\vartheta \\&= \int \limits _0^T \left[f^{\prime }(\vartheta )^2 K(\dot{x}(f(\vartheta )),x(f(\vartheta )))+P(x(f(\vartheta )))\right] \text{ d}\vartheta \\&\ge 2\left( \int \limits _0^T f^{\prime }(\vartheta )^2 K(\dot{x}(f(\vartheta )),x(f(\vartheta ))) \text{ d}\vartheta \cdot \int \limits _0^T P(x(f(\vartheta ))) \text{ d}\vartheta \right)^{\!\!1/2}\\&\ge 2\int \limits _0^T \sqrt{K(\dot{x}(f(\vartheta )),x(f(\vartheta ))) \cdot P(x(f(\vartheta )))} f^{\prime }(\vartheta )\text{ d}\vartheta \\&= 2\int \limits _a^b \sqrt{K(\dot{x}(t),x(t)) \cdot P(x(t))} {\,\mathrm dt }, \end{aligned}$$

and equality holds if and only if

$$\begin{aligned} f^{\prime }(\vartheta )^2 K(\dot{x}(f(\vartheta )),x(f(\vartheta ))) = P(x(f(\vartheta ))) \end{aligned}$$

almost everywhere. Since the last term in the previous inequality does not depend on $f,$ we have that $f$ minimizes $A([0,T],x\circ f)$ if and only if the last equality holds. This is equivalent to satisfy

$$\begin{aligned} f^{\prime }(\vartheta ) = \sqrt{\frac{P(x(f(\vartheta )))}{K(\dot{x}(f(\vartheta )),x(f(\vartheta )))}}. \end{aligned}$$

Since $f$ is strictly increasing, we can use its inverse in order to write $\vartheta = \vartheta (t),$ and the lemma follows. $\square$

Corollary 7.2

Let $\Gamma$ be a set of paths closed under re-parametrization and let $\bar{x} \in \Gamma$ be such that $A(\bar{x}) =\min _{x \in \Gamma }A(x).$ Then, for (almost) every $t,$

$$\begin{aligned} K(\dot{\bar{x}}(t),{\bar{x}}(t))=P({\bar{x}}(t)). \end{aligned}$$

1.2 A stability theorem

Theorem 7.3

Let us assume that, for $n \in \mathbb{N },$

• 1. $\alpha _n \rightarrow \alpha \in (0,2), \liminf _n T_{1,n} = T_1, \limsup _n T_{2,n} = T_2$;

• 2. $V_n \in \mathcal P$ is $\alpha _n$-homogeneous, $V \in \mathcal P$ is $\alpha$-homogeneous, and $V_n|_\mathbb{S ^{d-1}} \rightarrow V|_\mathbb{S ^{d-1}}$ in $\mathcal C ^1(\mathbb S ^{d-1})$;

• 3. $z_n \in \mathcal C ^2(T_{1,n},T_{2,n};\mathbb{R }^d \setminus B_{\varepsilon }(0))$ satisfies $\ddot{z}_n = \nabla V_n(z_n)$;

• 4. $($up to time translations$)$ $T_{1,n}< \bar{t}_n <T_{2,n}$ and $|z_n(\bar{t}_n)| + |\dot{z}_n(\bar{t}_n)| \le C,$ for some $\bar{t}_n \rightarrow \bar{t},$ and $C>0.$

Then there exists a subsequence $\left( {z_n}_{k}\right)_{k} \subset \left( {z}_{n}\right)_{n}$ and a function $\bar{z} \in \mathcal C ^2(T_1,T_2)$ such that $z_{n_k}|_I$ converges in $\mathcal C ^2$ to $\bar{z}|_I,$ for every $I \subset (T_1,T_2)$ compact.

Proof

First of all let us observe that there exists a constant $C>0$ such that, for every $n,$

$$\begin{aligned} \left| \nabla V_n(x) \right| \le C, \quad \forall x \in \mathbb{R }^d \setminus B_{\varepsilon }(0). \end{aligned}$$

Let $k \in \mathbb{N }$ and $I_k = [\bar{t}-k,\bar{t}+k] \cap [T_1,T_2].$ We infer that, up to a subsequence, $\ddot{z}_n$ is (defined and) bounded on $I_k.$ Integrating assumption 4. we obtain that $|z_n| + |\dot{z}_n| \le C$ on $I_k.$ Ascoli’s Theorem guarantees that, again up to a subsequence, there exists $\bar{z}$ such that $z_n \rightarrow \bar{z}$ in $\mathcal C ^1(I_k).$ Passing to the limit in the equations we have that the convergence is indeed $\mathcal C ^2,$ and that $\bar{z}$ satisfies the limiting equation. By a diagonal procedure we easily conclude. $\square$

1.3 Properties of zero-energy trajectories

The aim of this appendix is to sum up some well known results about the behavior of zero energy trajectories. The first ones concern homothetic trajectories, which are motions with constant angular part.

Lemma 7.4

Let us fix $\gamma >0$ and consider the functional

$$\begin{aligned} \mathcal{A }_\mathrm{rad ,\gamma } \left([a,b],r\right) := \int \limits _{a}^{b}\left[ \frac{1}{2} \dot{r}^2(t) + \frac{\gamma }{r^\alpha (t)} \right] {\,\mathrm dt }\end{aligned}$$

defined on $H^1\left((a,b);[0,+\infty )\right).$ Then, for any $r_+ \ge r_- \ge 0,$

$$\begin{aligned} \mathrm hom (r_-,r_+,\gamma )&:= \inf \left\{ \mathcal{A }_\mathrm{rad ,\gamma } \left([-T,T],r\right) :\, \begin{array}{l} T\ge 0, r \in H^1(-T,T),\\ r(\pm T)=r_{\pm } \end{array} \right\} \\&= \frac{\sqrt{2\gamma }}{\alpha _*}\left( r_+^{\alpha _*} - r_-^{\alpha _*}\right). \end{aligned}$$

Proof

If $r_-=r_+$ then the result is trivial. Otherwise, arguing as in Sect. 3 we deduce the existence of a monotone increasing minimizer $\bar{r}.$ From Corollary 7.2 we deduce that

$$\begin{aligned} \frac{1}{2} \dot{\bar{r}}^2(t) = \frac{\gamma }{\bar{r}^\alpha (t)} \implies \dot{\bar{r}}(t) = \sqrt{2\gamma }\bar{r}^{-\alpha /2} (t). \end{aligned}$$

Integrating the last equation and imposing the boundary conditions we obtain the explicit expression of $\bar{r}(t)$

$$\begin{aligned} \bar{r}(t) = \left[\frac{\alpha +2}{2}\sqrt{2\gamma }\,t + \frac{1}{2}\left(r_+^{(2+\alpha )/{2}} + r_-^{(2+\alpha )/{2}}\right)\right]^{{2}/({2+\alpha })} \end{aligned}$$

which is defined on $[-\bar{T},\bar{T}],$ where $\bar{T} = \left[(\alpha +2)\sqrt{2\gamma }\right]^{-1} \left(r_+^{(2+\alpha )/{2}} - r_-^{(2+\alpha )/{2}}\right). \square$

Lemma 7.5

Let $\xi \in \mathbb S ^{d-1}$ and $r_+ \ge r_- \ge 0.$ Then

$$\begin{aligned} \mathrm hom (r_-,r_+,V_{\min })&\le \inf \left\{ \mathcal{A }([-T,T];x):\, \begin{array}{l} T>0, \, x \in H^1(-T,T),\\ x({\pm }T)= r_{\pm } \xi \end{array} \right\} \\&\le \mathrm hom (r_-,r_+,V(\xi )). \end{aligned}$$

In particular, if $V(\xi )=V_{\min },$ then equality holds $($and the infimum is achieved by a path with constant direction $\xi ).$

Proof

The estimate from below follows straightforwardly from the previous lemma, once one notices that, for any $x$ satisfying the constraint,

$$\begin{aligned} \mathcal{A }(x) = \int \limits _{a}^b \left[ \frac{1}{2} \dot{r}^2 + \frac{1}{2} r^2 |\dot{s}|^2 +\frac{V(s)}{r^\alpha }\right] \ge \int \limits _{a}^b \left[ \frac{1}{2} \dot{r}^2 + \frac{V_{\min }}{r^\alpha }\right]. \end{aligned}$$

On the other hand,

$$\begin{aligned} x(t) = \bar{r} (t)\xi , \quad t \in [-\bar{T},\bar{T}], \end{aligned}$$

where $\bar{r} (t)$ and $\bar{T}$ have been defined in the proof of the previous lemma, satisfies the constraint, providing the estimate from above. $\square$

Lemma 7.6

Let us suppose that $x=rs$ satisfies both (1.1) and (1.2) on $(t_0,+\infty ).$ Then

• 1. $r(t) \rightarrow +\infty$ and $\dot{r}(t)>0,$ as $t \rightarrow +\infty$;

• 2. $\dot{r}(t) \rightarrow 0$ and $|\dot{s}(t)| \rightarrow 0,$ as $t \rightarrow +\infty .$

 

Proof

The first part follows from the (strict) convexity of $r^2(t),$ which is implied by the Lagrange-Jacobi identity (3.2) (see also Corollary 3.4). On the other hand, from (1.3), we immediately deduce that both $\dot{r}^2(t)$ and $r^2(t)\dot{s}^2(t)$ tend to 0 as $t \rightarrow +\infty .$ Since $r(t),$ by assumption, diverges the second assertion follows. $\square$

In the next theorem we prove the asymptotic estimates for parabolic solutions as time diverges (see for instance [4–6, 13, 24]). The proof we propose (that can be extended to non necessarily homogeneous potentials) is different from the classical ones, and it is similar to the one that the first two authors exploited for the asymptotic behavior near collisions for $N$-body type systems in [1].

Theorem 7.7

Let us suppose that $x=rs$ satisfies both (1.1) and (1.2) on $(t_0,+\infty ).$ Then there exists $\gamma >0$ such that:

• (a) $\displaystyle r(t) \sim (Kt)^{{2}/(2+\alpha )},$ as $t \rightarrow +\infty ,$ where $K := \frac{\alpha +2}{2}\sqrt{2\gamma }$;

• (b) $\displaystyle \dot{r}(t) \sim \sqrt{2\gamma }(Kt)^{-\alpha /(2+\alpha )},$ as $t \rightarrow +\infty$;

• (c) $\displaystyle \lim _{t \rightarrow +\infty }V(s(t)) = \gamma$;

• (d) $\displaystyle \lim _{t \rightarrow +\infty } \nabla _T V(s(t)) = 0$;

• (e) $\displaystyle \lim _{t \rightarrow +\infty } \mathrm{dist}\left(C^{\gamma },s(t)\right) =0,$ where $C^{\gamma } = \left\{ s : V(s) =\gamma , \nabla _T V(s)=0\right\} .$

 

Proof

By the previous lemma, we can assume that $\dot{r}(t)>0$ on $(t_0,+\infty ).$

In order to prove (a) we define the function

$$\begin{aligned} \Gamma (t) := \frac{1}{2} r^{\alpha + 2}(t)|\dot{s}(t)|^2 - V(s(t)) = -\frac{1}{2} r^{\alpha }(t)\dot{r}^2(t), \quad t \in (t_0,+\infty ) \end{aligned}$$

(the last equality follows from the conservation of energy). Since $r(t)>0$ and $\dot{r}(t)>0, \Gamma (t)$ is a strictly negative and bounded quantity, indeed

$$\begin{aligned} -V_{\max }\le - V(s(t)) \le \Gamma (t) < 0. \end{aligned}$$

Multiplying the Euler–Lagrange equation in $s$ (see (1.3)) by $\dot{s}$ we obtain

$$\begin{aligned} r^{2+\alpha } \ddot{s} \dot{s} + 2r^{1+\alpha }\dot{r}|\dot{s}|^2 - \nabla _T V(s)\dot{s} = 0; \end{aligned}$$

hence the derivative of the function $\Gamma$ satisfies, for large $t,$

$$\begin{aligned} \dot{\Gamma }(t) = -\frac{2-\alpha }{2} r^{1+\alpha }(t)\dot{r}(t)|\dot{s}(t)|^2 < 0. \end{aligned}$$

$\Gamma (t)$ is then bounded and (strictly) decreasing; hence there exists $\gamma >0$ such that

$$\begin{aligned} \lim _{t \rightarrow +\infty } \Gamma (t) = -\gamma \quad \text{ and} \quad \lim _{t \rightarrow +\infty } r^{{\alpha }/{2}}(t)\dot{r}(t) = \sqrt{2\gamma }. \end{aligned}$$

Therefore, using de l’Hopital rule we have

$$\begin{aligned} \lim _{t \rightarrow +\infty }\frac{r^{{\alpha }/{2}+1}(t)}{\sqrt{2\gamma }\left(\frac{\alpha }{2}+1 \right)t} = \lim _{t \rightarrow +\infty } \frac{r^{{\alpha }/{2}}(t)\dot{r}(t)}{\sqrt{2\gamma }} = 1 \end{aligned}$$

and we deduce the asymptotic behavior of $r(t)$ as $t \rightarrow +\infty .$ Straightforwardly we now prove (b), indeed, we define $K := \frac{\alpha +2}{2}\sqrt{2\gamma }$ and we obtain

$$\begin{aligned} \lim _{t \rightarrow +\infty } \frac{\dot{r}(t)}{\sqrt{2\gamma } (Kt)^{-{\alpha }/(2+\alpha )} } = \lim _{t \rightarrow +\infty } \frac{r^{{\alpha }/{2}}(t)\dot{r}(t)}{\sqrt{2\gamma }} \left[\frac{(Kt)^{{2}/(2+\alpha )}}{r(t)}\right]^{{\alpha }/{2}} = 1. \end{aligned}$$

In order to claim (c) we remark that $\Gamma$ is bounded on $(t_0,+\infty ),$ hence its derivative has a finite integral on the same interval, that is

$$\begin{aligned} \int \limits _{t_0}^{+\infty }\frac{\dot{r}(t)}{r(t)} r^{2+\alpha }(t)|\dot{s}(t)|^2{\,\mathrm dt }< +\infty . \end{aligned}$$

Since ${\dot{r}(t)}/{r(t)} \sim {2}/(2+\alpha )\,t^{-1}$ as $t \rightarrow +\infty ,$ then necessarily

$$\begin{aligned} \liminf _{t \rightarrow +\infty } r^{2+\alpha }(t)|\dot{s}(t)|^2 = 0, \end{aligned}$$

or, equivalently,

$$\begin{aligned} \liminf _{t \rightarrow +\infty } V(s(t)) = \gamma . \end{aligned}$$

In order to conclude we need to show that also the superior limit of $V(s),$ as $t \rightarrow +\infty ,$ is $\gamma .$ By the sake of contradiction let us assume that for some $C>0$

$$\begin{aligned} \limsup _{t \rightarrow +\infty } V(s(t)) = \gamma + C, \quad \text{ that} \text{ is} \quad \limsup _{t \rightarrow +\infty } r^{2+\alpha }(t)|\dot{s}(t)|^2 = 2C. \end{aligned}$$

Then there exists a sequence $\left( {t}_{k}\right)_{k}$ such that

$$\begin{aligned}&t_k \rightarrow +\infty , \text{ as} k \rightarrow +\infty , \\&r^{2+\alpha }(t_{2k})|\dot{s}(t_{2k})|^2 = \frac{2C}{3}, \forall \, k,\quad \text{ and} V(s(t_{2k})) \rightarrow \frac{C}{3} + \gamma , \text{ as} k \rightarrow +\infty , \\&r^{2+\alpha }(t_{2k+1})|\dot{s}(t_{2k+1})|^2 = \frac{4C}{3}, \forall \, k,\quad \text{ and} V(s(t_{2k+1})) \rightarrow \frac{2C}{3} + \gamma , \text{ as} k \rightarrow +\infty , \\&r^{2+\alpha }(t)|\dot{s}(t)|^2 \in \left(\frac{2C}{3},\frac{4C}{3}\right)\quad \text{ for} \text{ every} t \in \bigcup _k(t_{2k},t_{2k+1}). \end{aligned}$$

Using the monotone convergence of $\Gamma (t)$ to $\gamma$ we deduce that there exists $\bar{t}_0 \ge t_0$ such that, for every $t \ge \bar{t}_0$

$$\begin{aligned} \sqrt{\gamma } \le r^{{\alpha }/{2}}(t)\dot{r}(t) \le \sqrt{2\gamma }, \end{aligned}$$

and, integrating on $[\bar{t}_0,t],$

$$\begin{aligned} \frac{2+\alpha }{2}\sqrt{\gamma }\,t + C_{\sqrt{\gamma }} \le r^{(2+\alpha )/2}(t) \le \frac{2+\alpha }{2}\sqrt{2\gamma }\,t + C_{\sqrt{2\gamma }}, \end{aligned}$$

where $C_{\eta }:= r^{(2+\alpha )/2}(\bar{t}_0)-\bar{t}_0\frac{2+\alpha }{2}\eta , \eta \in \{\sqrt{\gamma },\sqrt{2\gamma } \}.$ We will obtain a contradiction using the properties of the sequence $\left( {t}_{k}\right)_{k}$ and the previous two estimates. Indeed, on one hand we have

$$\begin{aligned} +\infty&> \int \limits _{t_0}^{+\infty }\frac{\dot{r}(t)}{r(t)} r^{2+\alpha }(t)|\dot{s}(t)|^2{\,\mathrm dt }\ge \sum _k \int \limits _{t_{2k}}^{t_{2k+1}} \frac{\dot{r}(t)}{r(t)} r^{2+\alpha }(t)|\dot{s}(t)|^2{\,\mathrm dt }\\&\ge \frac{2C}{3} \sum _k \log \frac{r(t_{2k+1})}{r(t_{2k})} \ge \frac{4C}{3(2+\alpha )} \sum _k \log \frac{ (2+\alpha )\sqrt{\gamma }\,t_{2k+1} + 2C_{\sqrt{\gamma }}}{ (2+\alpha )\sqrt{2\gamma }\,t_{2k} + 2C_{\sqrt{2\gamma }}}. \end{aligned}$$

On the other hand, since the continuity of $V(s(\cdot ))$ implies the existence of $M >0$ such that $|s(t_{2k})-s(t_{2k+1})| > M$ for every $k,$ we have

$$\begin{aligned} M^2&< |s(t_{2k})-s(t_{2k+1})|^2 \le \left(\int \limits _{t_{2k}}^{t_{2k+1}} |\dot{s}(t)| {\,\mathrm dt }\right)^ {\!\! 2}\\&\le \int \limits _{t_{2k}}^{t_{2k+1}} \frac{\dot{r}(t)}{r(t)} r^{2+\alpha }(t)|\dot{s}(t)|^2{\,\mathrm dt }\int \limits _{t_{2k}}^{t_{2k+1}} \frac{{\,\mathrm dt }}{r^{1+\alpha }(t)\dot{r}(t)}. \end{aligned}$$

Since

$$\begin{aligned} r^{1+\alpha }(t)\dot{r}(t) = r^{{\alpha }/{2}}(t)\dot{r}(t)\cdot r^{(2+\alpha )/2}(t) \ge \frac{\sqrt{\gamma }}{2}\left((2+\alpha ) \sqrt{\gamma }\,t + 2C_{\sqrt{\gamma }}\right), \end{aligned}$$

then

$$\begin{aligned} \int \limits _{t_{2k}}^{t_{2k+1}} \frac{{\,\mathrm dt }}{r^{1+\alpha }(t)\dot{r}(t)}&\le \frac{2}{\sqrt{\gamma }}\int \limits _{t_{2k}}^{t_{2k+1}} \frac{{\,\mathrm dt }}{(2+\alpha )\sqrt{\gamma }\,t + 2C_{\sqrt{\gamma }}}\\&= \frac{2}{(2+\alpha )\gamma } \log \frac{(2+\alpha ) \sqrt{\gamma }\,t_{2k+1} + 2 C_{\sqrt{\gamma }}}{(2+\alpha ) \sqrt{\gamma }\,t_{2k} + 2 C_{\sqrt{\gamma }}} \end{aligned}$$

and

$$\begin{aligned} +\infty&> \int \limits _{t_0}^{+\infty }\frac{\dot{r}(t)}{r(t)} r^{2+\alpha }(t)|\dot{s}(t)|^2{\,\mathrm dt }\\&\ge M^2 \frac{(2+\alpha )\gamma }{2} \sum _k \left[ \log \frac{(2+\alpha ) \sqrt{\gamma }\,t_{2k+1} + 2C_{\sqrt{\gamma }}}{(2+\alpha ) \sqrt{\gamma }\,t_{2k} + 2C_{\sqrt{\gamma }}} \right]^{-1} \end{aligned}$$

Hence,

$$\begin{aligned} +\infty&> \int \limits _{t_0}^{+\infty }\frac{\dot{r}(t)}{r(t)} r^{2+\alpha }(t)|\dot{s}(t)|^2{\,\mathrm dt }\ge \frac{2C}{3(2+\alpha )} \sum _k \log \frac{ (2+\alpha )\sqrt{2\gamma }\,t_{2k+1} + 2C_{\sqrt{2\gamma }}}{ (2+\alpha )\sqrt{\gamma }\,t_{2k} + 2C_{\sqrt{\gamma }}} \\&+ M^2 \frac{(2+\alpha )\gamma }{4} \sum _k \left[ \log \frac{(2+\alpha ) \sqrt{\gamma }\,t_{2k+1} + 2C_{\sqrt{\gamma }}}{(2+\alpha ) \sqrt{\gamma }\,t_{2k} + 2C_{\sqrt{\gamma }}} \right]^{-1}. \end{aligned}$$

We obtain a contradiction when we impose the convergence of both series (with positive terms) at the right hand side.

We now turn to assertion (d), that is equivalent to

$$\begin{aligned} \lim _{t \rightarrow +\infty } r^{2+\alpha }(t) \ddot{s}(t)=0, \end{aligned}$$

indeed in the Euler–Lagrange equation in the variable $s$ (see (1.3)) both terms $-2r^{1+\alpha }\dot{r}\dot{s} = -2r^{(2+\alpha )/2}\dot{s}\,r^{{\alpha }/{2}}\dot{r}$ and $-r^{2+\alpha }|\dot{s}|^2 s$ are infinitesimal as $t \rightarrow +\infty$ (indeed $r^{2+\alpha }|\dot{s}|^2$ is infinitesimal and $r^{{\alpha }/{2}}\dot{r}$ remains bounded). We proceed by contradiction: suppose there exists $\left( {t}_{k}\right)_{k}$ such that, as $k \rightarrow +\infty , t_k \rightarrow +\infty$ and $\nabla _TV(s(t_k)) \not \rightarrow 0.$ Up to subsequences we deduce the existence of $\bar{s} \in \mathbb S ^{d-1}$ and $\sigma \ne 0$ such that

$$\begin{aligned} s(t_k) \rightarrow \bar{s}, \text{ as} k \rightarrow +\infty \text{ and} \nabla _TV(\bar{s}) = \sigma . \end{aligned}$$

For any $h>0$ and $\varepsilon >0$ we have that, since, for $t$ large, $|\dot{s}(t)|<\varepsilon ,$

$$\begin{aligned} |s(t)-s(t_k)|<\varepsilon h, \quad \forall t \in \bigcup _k [t_k,t_k+h] \end{aligned}$$

and

$$\begin{aligned} |s(t)-\bar{s}| \le |s(t)-s(t_k)|+|s(t_k)-\bar{s}|<\varepsilon (h+1), \quad \forall t \in \bigcup _k [t_k,t_k+h]. \end{aligned}$$

The convergence of $s(t)$ to $\bar{s}$ is then uniform on $\bigcup _k [t_k,t_k+h],$ for any $h>0,$ hence, by continuity,

$$\begin{aligned} \sup _{t \in [t_k,t_k+h]} |\nabla _TV(s(t))-\sigma | \rightarrow 0 \quad \text{ and} \quad \sup _{t \in [t_k,t_k+h]} |r^{2+\alpha }(t) \ddot{s}(t)-\sigma | \rightarrow 0, \end{aligned}$$

as $k \rightarrow +\infty .$ In order to obtain a contradiction we perform the time scaling (see [19])

$$\begin{aligned} \tau = \int \limits _{t_0}^{+\infty } \frac{{\,\mathrm dt }}{r^{(2+\alpha )/2}(t)} \end{aligned}$$

which maps $[t_0,+\infty )$ into $[0,+\infty )$ (we have used here the asymptotic for $r(t)).$ Letting $^\prime {}:= {{\,\mathrm dm }}/{{\,\mathrm dm }\tau }$ we have

$$\begin{aligned} \lim _{\tau \rightarrow +\infty } |s^{\prime }(\tau )|^2 = \lim _{t \rightarrow +\infty } r^{2+\alpha }(t)|\dot{s}(\tau )|^2 = 0 \end{aligned}$$

while, by contradiction, for any fixed $h>0$

$$\begin{aligned} \sup _{t \in [t_k,t_k+h]} |r^{2+\alpha }(t) \ddot{s}(t)-\sigma | = \sup _{\tau \in [\tau _k,\tau _k+\tilde{h}]} |s^{\prime \prime }(\tau )-\sigma | \rightarrow 0, \quad \text{ as} k \rightarrow +\infty . \end{aligned}$$

We obtain the contradiction

$$\begin{aligned} 0 = \lim _{k \rightarrow +\infty } |s^{\prime }(\tau _k +h) - s^{\prime }(\tau _k)| = \lim _{k \rightarrow +\infty } \left| \int \limits _{\tau _k}^{\tau _k+h}s^{\prime \prime }(\tau ) {\,\mathrm d{\tau } }\right| = h |\sigma | \ne 0. \end{aligned}$$

Finally, assertion (e) follows directly from (c) and (d). $\square$

Corollary 7.8

Let us suppose that $x=rs$ satisfies both (1.1) and (1.2) on $(-\infty ,t_*)\cup (t_{**},+\infty ).$ Then there exist constants $\Upsilon _\pm >0$ such that

$$\begin{aligned} \lim _{t \rightarrow \pm \infty } \frac{r^{(2+\alpha )/2}(t)}{t} = \Upsilon _\pm >0 \end{aligned}$$