On the Period Function $T_1$

Abstract

Associated with every effective potential, $U_{\mathrm{eff}}(r, \ell )=U_Q(r)+\frac{\ell ^2}{2r^2}$ is a period function $T_1(\cdot , \ell )$ that is defined for certain energies $e\in [e_{\mathrm{min}}(\ell ), e_0]$, for which periodic solutions of $\ddot{r}=-U'_{\mathrm{eff}}(r, \ell )$ do exist; see Appendix I, Section A.1, for more information.

Associated with every effective potential $U_{\mathrm{eff}}(r, \ell )=U_Q(r)+\frac{\ell ^2}{2r^2}$ is a period function $T_1(\cdot , \ell )$ that is defined for certain energies $e\in [e_{\mathrm{min}}(\ell ), e_0]$, for which periodic solutions of $\ddot{r}=-U'_{\mathrm{eff}}(r, \ell )$ do exist; see Appendix I, Sect. A.1, for more information. According to (A.20), this period function is given by

$$ T_1(e, \ell )=2\int _{r_-(e, \ell )}^{r_+(e, \ell )} \frac{dr}{\sqrt{2(e-U_{\mathrm{eff}}(r, \ell ))}}, $$

where $r_\pm =r_\pm (e, \ell )$ are the zeros of $0=2(e-U_{\mathrm{eff}}(r, \ell ))$ and satisfy $0<r_-(e, \ell )<r_+(e, \ell )$. In addition, for every $\ell >0$, the potential minimum $\inf \,\{U_{\mathrm{eff}}(r, \ell ): r\ge 0\}$ is attained at some unique $r_0(\ell )\in ]r_-(e, \ell ), r_+(e, \ell )[$. The corresponding frequency function is $\omega _1(e, \ell )=\frac{2\pi }{T_1(e, \ell )}$.

3.1 Upper Boundedness of $T_1$

Recall that

$$\begin{aligned} D=\{(e, \beta ): \beta \in [0, \beta _*], e\in [e_{\mathrm{min}}(\beta ), e_0]\}, \end{aligned}$$

(3.1)

and

$$\begin{aligned} \mathring{D}=\{(e, \beta ): \beta \in ]0, \beta _*[, e\in ]e_{\mathrm{min}}(\beta ), e_0[\} \end{aligned}$$

is its interior. We are going to show that $T_1$ is bounded from above (or equivalently, $\omega _1$ is bounded from below), uniformly in $\mathring{D}$, which is the set of relevant $(e, \ell )$, where $T_1$ is defined. As $T_1$ will be shown to be continuous in D (see Theorem 3.13 below), this is of course for free, but since the direct argument in Theorem 3.2 could be of general interest, we include it anyhow; the same remark applies to Theorem 3.5 on the lower boundedness of $T_1$.

We start with an auxiliary lemma that will be useful for the proof of Theorem 3.2 and beyond.

Lemma 3.1

The following assertions are verified.

• (a) If $r>s>0$, then

  $$\begin{aligned} \frac{2\pi }{3}\rho _Q(r)(r^2-s^2) \le U_Q(r)-U_Q(s)\le \frac{2\pi }{3}\rho _Q(0)(r^2-s^2). \end{aligned}$$

  (3.2)

  Moreover, for $r_Q\ge r>s>0$,

  $$\begin{aligned} U_Q(r)-U_Q(s)\ge \frac{\pi }{12}\,\rho _Q\Big (\frac{r_Q}{2}\Big )(r^2-s^2). \end{aligned}$$

  (3.3)

• (b) One has

  $$ \frac{\pi }{6}\,\rho _Q\Big (\frac{r_Q}{2}\Big )\,r_-^2 r_+^2 \le \ell ^2\le \frac{4\pi }{3}\,\rho _Q(0)\,r_-^2 r_+^2. $$

• (c) One has

  $$\begin{aligned} r_0\le \bigg (\frac{6}{\pi \rho _Q(\frac{r_Q}{2})}\bigg )^{1/4}\,\sqrt{\ell }. \end{aligned}$$

Proof

• (a) According to (A.2), we have by changing variables $s=r\tau$, $ds=r d\tau$,

  $$\begin{aligned} U'_Q(r)=\frac{4\pi }{r^2}\int _0^r s^2\rho _Q(s)\,ds=4\pi r\int _0^1\tau ^2\rho _Q(r\tau )\,d\tau . \end{aligned}$$

  (3.4)

In particular, $U'_Q(r)\ge 0$. Furthermore, for $r>s>0$ and putting $t=\sigma /r$, $dt=d\sigma /r$,

$$\begin{aligned} U_Q(r)-U_Q(s)= & {} \int _s^r U'_Q(\sigma )\,d\sigma =4\pi \int _s^r d\sigma \sigma \int _0^1 d\tau \tau ^2\rho _Q(\sigma \tau ) \nonumber \\= & {} 4\pi \int _0^1 d\tau \tau ^2\int _s^r d\sigma \sigma \rho _Q(\sigma \tau ) \end{aligned}$$

(3.5)

$$\begin{aligned}= & {} 4\pi r^2\int _0^1 d\tau \tau ^2\int _{\frac{s}{r}}^1 dt\,t\,\rho _Q(r\tau t). \end{aligned}$$

(3.6)

Due to (A.32), we have that $\rho '_Q(r)\le 0$, i.e., $\rho _Q$ is radially decreasing. Thus, if $\tau \in [0, 1]$ and $\sigma \in [s, r]$, then $\rho _Q(r)\le \rho _Q(\sigma \tau )\le \rho _Q(0)$ and (3.2) follows from (3.5). To establish (3.3), we use (3.6). To begin with, since $\rho _Q\ge 0$,

$$ U_Q(r){-}U_Q(s)\ge 4\pi r^2\int _0^{\frac{1}{2}} d\tau \tau ^2\int _{\frac{s}{r}}^1 dt\,t\,\rho _Q(r\tau t). $$

Owing to $r\le r_Q$, $\tau \in [0, \frac{1}{2}]$, and $t\le 1$, we have $r\tau t\le \frac{r_Q}{2}$, so that $\rho _Q(r\tau t)\ge \rho _Q(\frac{r_Q}{2})$. It follows that

$$ U_Q(r)-U_Q(s){\ge } 4\pi r^2\rho _Q\Big (\!\frac{r_Q}{2}\!\Big )\int _0^{\frac{1}{2}} d \tau \tau ^2\int _{\frac{s}{r}}^1 dt\,t {=}4\pi r^2\rho _Q\Big (\!\frac{r_Q}{2}\Big )\frac{1}{48}\Big (1-\Big (\frac{s}{r}\Big )^2\Big ), $$

which is (3.3). (b) The condition $U_{\mathrm{eff}}(r_\pm , \ell )=e$ means that $U_Q(r_\pm )+\frac{\ell ^2}{2r_\pm ^2}=e$, and hence

$$\begin{aligned} 2r_\pm ^2 U_Q(r_\pm )+\ell ^2=2r_\pm ^2 e. \end{aligned}$$

(3.7)

Therefore,

$$\begin{aligned} 2(r_+^2-r_-^2)e= & {} 2(r_+^2 U_Q(r_+)-r_-^2 U_Q(r_-))\\= & {} 2(r_+^2-r_-^2)U_Q(r_+)+2r_-^2(U_Q(r_+)-U_Q(r_-)), \end{aligned}$$

so that

$$ (r_+^2-r_-^2)\frac{\ell ^2}{r_+^2}=2(r_+^2-r_-^2)(e-U_Q(r_+)) =2r_-^2(U_Q(r_+)-U_Q(r_-)). $$

It remains to use (3.3) and the upper bound from (3.2). (c) First note that $\rho _Q(\frac{r_Q}{2})>0$, as otherwise $\mathrm{supp}\,\rho _Q\subset [0, \frac{r_Q}{2}]$. By Lemma A.7(a), (3.4) and since $\rho _Q$ is non-negative and radially decreasing,

$$\begin{aligned} \ell ^2= & {} r_0^3\,U'_Q(r_0)=4\pi r_0^4\int _0^1\tau ^2\rho _Q(r_0\tau )\,d\tau \ge 4\pi r_0^4\int _0^{\frac{1}{2}}\tau ^2\rho _Q(r_0\tau )\,d\tau \\\ge & {} 4\pi r_0^4\int _0^{\frac{1}{2}}\tau ^2\rho _Q\Big (\frac{r_Q}{2}\Big )\,d\tau =\frac{\pi }{6}\,\rho _Q(\frac{r_Q}{2})\,r_0^4. \end{aligned}$$

We will derive a more precise asymptotics of $r_0$ as $\ell \rightarrow 0^+$ below in (A.34).    $\square$

Now, we are in a position to derive a uniform lower bound on $\omega _1$ or equivalently a uniform upper bound on $T_1$.

Theorem 3.2

We have

$$\begin{aligned} \delta _1=\inf \,\{\omega _1(e, \ell ): (e, \ell )\in \mathring{D}\}>0. \end{aligned}$$

Proof

Put $a_Q=\rho _Q(\frac{r_Q}{2})>0$. Then in particular $a_Q\le \rho _Q(0)$, so that

$$\begin{aligned} \delta _Q=1-\sqrt{\frac{a_Q}{16\rho _Q(0)}}\in \Big [\frac{1}{2}, 1\Big [. \end{aligned}$$

Let $r_{\pm }=r_{\pm }(e, \ell )$ and $r_0=r_0(\ell )$ be as before. From Lemma 3.1(c), we recall that

$$\begin{aligned} r_0\le \bigg (\frac{6}{\pi a_Q}\bigg )^{1/4}\,\sqrt{\ell }. \end{aligned}$$

(3.8)

Case 1: $r_0\ge (1-\delta _Q)r_+$. Then Lemma A.10(b) in conjunction with (3.8) implies that

$$\begin{aligned} T_1(e, \ell )\le \pi \,\frac{\sqrt{r_- r_+}}{\ell }\,(r_- + r_+) \le 2\pi \,\frac{r_+^2}{\ell }{\le }\frac{2\pi }{(1-\delta _Q)^2}\,\frac{r_0^2}{\ell } \le \frac{2\pi }{(1-\delta _Q)^2}\,\bigg (\frac{6}{\pi a_Q}\bigg )^{1/2}. \end{aligned}$$

(3.9)

Case 2: $r_0\le (1-\delta _Q)r_+$. This is the nontrivial part of the argument. Here, we split up the integral as

$$\begin{aligned} T_1(e, \ell )= & {} 2\int _{r_-}^{r_+}\frac{dr}{\sqrt{2(e-U_{\mathrm{eff}}(r, \ell ))}} \\= & {} 2\int _{r_-}^{r_0}\frac{dr}{\sqrt{2(e-U_{\mathrm{eff}}(r, \ell ))}} +2\int _{r_0}^{r_+}\frac{dr}{\sqrt{2(e-U_{\mathrm{eff}}(r, \ell ))}} \\=: & {} T_1^-(e, \ell )+T_1^+(e, \ell ). \end{aligned}$$

Using Lemma A.10(a), we can bound $T_1^-$ as

$$\begin{aligned} T_1^-(e, \ell )\le & {} 2\,\frac{\sqrt{r_- r_+}}{\ell } \int _{r_-}^{r_0}\frac{r\,dr}{\sqrt{(r-r_-)(r_+ -r)}} \nonumber \\\le & {} 2\,\frac{\sqrt{r_- r_+}}{\ell }\,\frac{r_0}{\sqrt{r_+ -r_0}}\int _{r_-}^{r_0}\frac{dr}{\sqrt{r-r_-}} \nonumber \\= & {} 4\,\frac{\sqrt{r_- r_+}}{\ell }\,\frac{r_0}{\sqrt{r_+ -r_0}}\,\sqrt{r_0-r_-}. \end{aligned}$$

(3.10)

It follows from $r_0\le (1-\delta _Q)r_+$ that $\sqrt{r_+}\le \delta _Q^{-1/2}\sqrt{r_+ -r_0}$. Thus, by (3.8) and since $\delta _Q\ge 1/2$,

$$\begin{aligned} T_1^-(e, \ell )\le 4\delta _Q^{-1/2}\,\frac{\sqrt{r_-}}{\ell }\,r_0\,\sqrt{r_0-r_-} \le 4\delta _Q^{-1/2}\,\frac{r_0^2}{\ell }\le 4\sqrt{2}\,\bigg (\frac{6}{\pi a_Q}\bigg )^{1/2}. \end{aligned}$$

(3.11)

Regarding $T_1^+$, we can invoke Lemma A.7(a) to get for $r\in [r_0, r_+]$ by also using Lemma A.6(a),

$$\begin{aligned} \frac{\ell ^2}{2r_+^2 r^2}\le & {} \frac{\ell ^2}{2r_+^2 r_0^2}=\frac{r_0\,U'_Q(r_0)}{2r_+^2} \le \frac{r_0\,U'_Q(r_0)}{2r_0^2}\,(1-\delta _Q)^2=\frac{1}{2}\,(1-\delta _Q)^2\,A(r_0) \nonumber \\\le & {} \frac{1}{2}\,(1-\delta _Q)^2 A(0)=\frac{2\pi }{3}\,(1-\delta _Q)^2\,\rho _Q(0). \end{aligned}$$

(3.12)

We then deduce from (3.3) in Lemma 3.1(a) and (3.12) that for $r\in [r_0, r_+]$,

$$\begin{aligned} e-U_{\mathrm{eff}}(r, \ell )= & {} U_{\mathrm{eff}}(r_+, \ell )-U_{\mathrm{eff}}(r, \ell ) =U_Q(r_+)+\frac{\ell ^2}{2r_+^2}-U_Q(r)-\frac{\ell ^2}{2r^2} \\= & {} U_Q(r_+)-U_Q(r)-\frac{\ell ^2}{2r_+^2 r^2}\,(r_+^2-r^2) \\\ge & {} \bigg [\frac{\pi }{12}\,a_Q-\frac{2\pi }{3}\,(1-\delta _Q)^2\,\rho _Q(0)\bigg ](r_+^2-r^2) \\= & {} \frac{\pi }{24}\,a_Q (r_+^2-r^2), \end{aligned}$$

the latter owing to the choice of $\delta _Q$. This in turn yields

$$\begin{aligned} T_1^+(e, \ell )= & {} 2\int _{r_0}^{r_+}\frac{dr}{\sqrt{2(e-U_{\mathrm{eff}}(r, \ell ))}} \le \sqrt{2}\,\sqrt{\frac{24}{\pi a_Q}}\int _{r_0}^{r_+}\frac{dr}{\sqrt{r_+^2-r^2}} \\\le & {} \frac{4\sqrt{3}}{\sqrt{\pi a_Q}}\,\frac{1}{\sqrt{r_+}}\int _{r_0}^{r_+}\frac{dr}{\sqrt{r_+-r}} =\frac{8\sqrt{3}}{\sqrt{\pi a_Q}}\,\frac{1}{\sqrt{r_+}}\,\sqrt{r_+ -r_0} \le \frac{8\sqrt{3}}{\sqrt{\pi a_Q}}. \end{aligned}$$

Adding this to (3.11), we have shown that

$$\begin{aligned} T_1(e, \ell )\le 4\sqrt{2}\,\bigg (\frac{6}{\pi a_Q}\bigg )^{1/2} +\frac{8\sqrt{3}}{\sqrt{\pi a_Q}}=\frac{16\sqrt{3}}{\sqrt{\pi a_Q}}. \end{aligned}$$

(3.13)

Hence, the boundedness of $T_1$ from above is a consequence of (3.9) and (3.13).    $\square$

Observe that in the proof of Theorem 3.2 actually no properties of the sets $\mathring{D}$ or D from (3.1) have been used, apart from the fact that $T_1(e, \ell )$ is defined for $(e, \ell )\in \mathring{D}$.

3.2 Lower Boundedness of $T_1$

It is the purpose of this section to verify that $T_1$ is bounded from below (or equivalently, $\omega _1$ is bounded from above), uniformly in $\mathring{D}$.

In some cases, it will be convenient to be able to re-express the period function

$$\begin{aligned} T_1(e, \beta )=2\int _{r_-(e, \beta )}^{r_+(e, \beta )}\frac{dr}{\sqrt{2(e-U_{\mathrm{eff}}(r, \beta ))}} \end{aligned}$$

(3.14)

from (A.20), written in terms of $\beta =\ell ^2$, by means of an integral with fixed limits of integration; this is more or less taken from [11, Section 2].

Lemma 3.3

We have

$$ T_1(e, \beta )=\sqrt{2}\int _{-\pi /2}^{\pi /2} \,\frac{d\theta }{\frac{\partial h}{\partial s}(s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta ), \beta )}, $$

where

$$\begin{aligned} h(s, \beta )=s\bigg (\frac{V(s, \beta )}{s^2}\bigg )^{1/2},\quad h(0, \beta )=0, \end{aligned}$$

for

$$\begin{aligned} V(s, \beta )=U_{\mathrm{eff}}(r_0(\beta )+s, \beta )-e_{\mathrm{min}}(\beta ). \end{aligned}$$

Also, $\hat{e}(\beta )=e-e_{\mathrm{min}}(\beta )$, and $R\mapsto s(R, \beta )=s$ denotes the inverse mapping to $s\mapsto h(s, \beta )=R$. Explicitly,

$$\begin{aligned} \frac{\partial h}{\partial s}(s, \beta ) =\frac{\mathrm{sgn}(s)}{2}\,\frac{U'_{\mathrm{eff}}(r_0(\beta )+s, \beta )}{\sqrt{U_{\mathrm{eff}}(r_0(\beta )+s, \beta )-e_{\mathrm{min}}(\beta )}}\ge 0, \end{aligned}$$

(3.15)

so that also

$$\begin{aligned} T_1(e, \beta )=2\,\sqrt{2}\int _{-\pi /2}^{\pi /2} d\theta \, \frac{[\int _0^1 (1-\rho )\,U''_{\mathrm{eff}}(r_0(\beta )+\rho s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta ), \beta ) \,d\rho ]^{1/2}}{\int _0^1 U''_{\mathrm{eff}}(r_0(\beta )+\rho s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta ), \beta )\,d\rho }. \end{aligned}$$

(3.16)

Proof

Let $s_\pm (e, \beta )=r_\pm (e, \beta )-r_0(\beta )$. Setting $s=r-r_0(\beta )$, $ds=dr$, we obtain

$$\begin{aligned} T_1(e, \beta )= & {} 2\int _{s_-(e, \beta )}^{s_+(e, \beta )}\frac{ds}{\sqrt{2(e-e_{\mathrm{min}}(\beta ) -[U_{\mathrm{eff}}(r_0(\beta )+s, \beta )-e_{\mathrm{min}}(\beta )])}} \nonumber \\= & {} 2\int _{s_-(e, \beta )}^{s_+(e, \beta )}\frac{ds}{\sqrt{2(\hat{e}(\beta )-V(s, \beta ))}}. \end{aligned}$$

(3.17)

Note that $V(\cdot , \beta )$ is increasing in $[0, s_+(e, \beta )]$, decreasing in $[s_-(e, \beta ), 0]$ and such that

$$\begin{aligned} V(s_\pm (e, \beta ), \beta )=e-e_{\mathrm{min}}(\beta )=\hat{e}(\beta ). \end{aligned}$$

Furthermore, $V(0, \beta )=U_{\mathrm{eff}}(r_0(\beta ), \beta )-e_{\mathrm{min}}(\beta )=0$ by definition and $\frac{\partial V}{\partial s}(0, \beta )=U'_{\mathrm{eff}}(r_0(\beta ), \beta )$ by (A.35), i.e., $V(\cdot , \beta )$ is at least quadratic about $s=0$. The next change of variables to be applied is

$$\begin{aligned} s\mapsto R=h(s, \beta ),\quad dR=\frac{\partial h}{\partial s}\,ds,\quad R^2=V(s, \beta ). \end{aligned}$$

Then (3.17) transforms into

$$ T_1(e, \beta )=2\int _{-\sqrt{\hat{e}(\beta )}}^{\sqrt{\hat{e}(\beta )}} \,\frac{dR}{\frac{\partial h}{\partial s}(s(R, \beta ), \beta )\sqrt{2(\hat{e}(\beta )-R^2)}}. $$

Finally, put $R=\sqrt{\hat{e}(\beta )}\sin \theta$, $dR=\sqrt{\hat{e}(\beta )}\cos \theta \,d\theta$. This yields

$$ T_1(e, \beta )=\sqrt{2}\int _{-\pi /2}^{\pi /2} \,\frac{d\theta }{\frac{\partial h}{\partial s}(s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta ), \beta )}, $$

and thus the claimed formula for $T_1$. The relation (3.15) is straightforward, whereas (3.16) follows from Lemma A.9.    $\square$

Corollary 3.4

If $s\in [r_-(e, \beta )-r_0(\beta ), 0]$, then

$$ 0\le \frac{\partial h}{\partial s}(s, \beta )\le \frac{1}{\sqrt{2B(r_Q)}} \,\bigg (3\beta \int _0^1\frac{d\rho }{(r_0(\beta )+\rho s)^4} +\frac{28\pi }{3}\,\rho _Q(0)\bigg ). $$

Proof

Let $s_-=s_-(e, \beta )-r_0(\beta )$. If $s\in [s_-, 0]$, then

$$ 0\le \frac{\partial h}{\partial s}(s, \beta ) =\Big |\frac{\partial h}{\partial s}(s, \beta )\Big | =\frac{1}{2}\,\frac{|U'_{\mathrm{eff}}(r_0(\beta )+s, \beta )|}{\sqrt{U_{\mathrm{eff}}(r_0(\beta )+s, \beta )-e_{\mathrm{min}}(\beta )}} $$

by (3.15) in Lemma 3.3. Thus, it remains to use (A.37) and (A.38) from Lemma A.9.

   $\square$

Theorem 3.5

We have

$$\begin{aligned} \Delta _1=\sup \,\{\omega _1(e, \ell ): (e, \ell )\in \mathring{D}\}<\infty . \end{aligned}$$

Proof

As above, we write $r_{\pm }=r_{\pm }(e, \beta )$ and $r_0=r_0(\beta )$. If $r\in [r_-, r_+]$, then by Lemma 3.1(a),

$$\begin{aligned} e-U_{\mathrm{eff}}(r, \beta )= & {} U_Q(r_+)-U_Q(r)-\frac{\beta }{2r_+^2 r^2}\,(r_+^2-r^2) \nonumber \\\le & {} U_Q(r_+)-U_Q(r)\le \frac{2\pi }{3}\rho _Q(0)(r_+^2-r^2) \le \frac{4\pi }{3}\rho _Q(0)r_+(r_+-r).\nonumber \\ \end{aligned}$$

(3.18)

Case 1: $r_+/2\ge r_0$. Here (3.18) implies that

$$\begin{aligned} T_1(e, \beta )= & {} 2\int _{r_-}^{r_+}\frac{dr}{\sqrt{2(e-U_{\mathrm{eff}}(r, \beta ))}} \ge \sqrt{2}\int _{r_0}^{r_+}\frac{dr}{\sqrt{e-U_{\mathrm{eff}}(r, \beta )}} \\\ge & {} \sqrt{\frac{3}{2\pi \rho _Q(0)}}\,\frac{1}{\sqrt{r_+}} \int _{r_0}^{r_+}\frac{dr}{\sqrt{r_+ -r}} =2\,\sqrt{\frac{3}{2\pi \rho _Q(0)}}\,\sqrt{\frac{r_+-r_0}{r_+}} \ge \sqrt{\frac{3}{\pi \rho _Q(0)}}. \end{aligned}$$

Case 2: $r_-\le r_+/2\le r_0$. Similarl to the first case, we obtain

$$\begin{aligned} T_1(e, \beta )= & {} 2\int _{r_-}^{r_+}\frac{dr}{\sqrt{2(e-U_{\mathrm{eff}}(r, \beta ))}} \ge \sqrt{2}\int _{r_+/2}^{r_+}\frac{dr}{\sqrt{e-U_{\mathrm{eff}}(r, \beta )}} \\\ge & {} \sqrt{\frac{3}{2\pi \rho _Q(0)}}\,\frac{1}{\sqrt{r_+}} \int _{r_+/2}^{r_+}\frac{dr}{\sqrt{r_+ -r}} =\sqrt{\frac{6}{\pi \rho _Q(0)}}. \end{aligned}$$

Case 3: $0\le r_+/2\le r_-$. Then $r_+/2\le r_-\le r_+$ and also $r_-\le r_0\le r_+\le 2r_-$ as well as $r_0\le r_+\le 2r_-\le 2r_0$, so all of $r_-$, $r_0$ and $r_+$ are of comparable size. In particular, if $r\in [r_-, r_+]$, then $r_0/2\le r\le 2r_0$. In the following, we are going to use the notation from the proof of Lemma 3.3. Let $R=\sqrt{\hat{e}(\beta )}\sin \theta$. If $\theta \in [-\pi /2, 0]$, then $R\in [-\sqrt{\hat{e}(\beta )}, 0]$ and hence $s(R, \beta )\in [s_-, 0]$. Thus, if furthermore $\rho \in [0, 1]$, then $r_0+\rho s(R, \beta ) \in r_0+[s_-, 0]=[r_-, r_0]$, so that

$$\begin{aligned} \frac{1}{2}\,r_0\le r_0+\rho s(R, \beta )\le 2r_0. \end{aligned}$$

(3.19)

Since $s(R, \beta )\in [s_-, 0]$, Corollary 3.4 and (3.19) imply that

$$\begin{aligned} 0\le & {} \frac{\partial h}{\partial s}(s(R, \beta ), \beta )\le \frac{1}{\sqrt{2B(r_Q)}} \,\bigg (3\beta \int _0^1\frac{d\rho }{(r_0+\rho s(R, \beta ))^4} +\frac{28\pi }{3}\,\rho _Q(0)\bigg ) \nonumber \\\le & {} \frac{1}{\sqrt{2B(r_Q)}} \,\bigg (\frac{48\beta }{r_0^4}+\frac{28\pi }{3}\,\rho _Q(0)\bigg ) \end{aligned}$$

(3.20)

for $\theta \in [-\pi /2, 0]$. By (A.34) from Lemma A.7, we have

$$ r_0^4=\frac{1}{A(0)}\,\beta +\mathcal{O}(\beta ^{5/4}) =\beta \Big (\frac{1}{A(0)}+\mathcal{O}(\beta ^{1/4})\Big ) $$

as $\beta \rightarrow 0^+$. Hence, there is $\beta _0\in ]0, \beta _*[$ such that

$$ \frac{\beta }{2A(0)}\le r_0^4\le \frac{2\beta }{A(0)}, \quad \beta \in ]0, \beta _0]. $$

Accordingly, owing to Lemma A.7(a), we can find a constant $c_0>0$ so that $r_0\ge c_0$ for $\beta \in [\beta _0, \beta _*]$. If we now distinguish the cases $\beta \in ]0, \beta _0]$ and $\beta \in [\beta _0, \beta _*]$, by using the foregoing estimates, we deduce that in any case

$$\begin{aligned} \frac{\beta }{r_0^4}\le \max \Big \{2A(0), \frac{\beta _*}{c_0^4}\Big \}. \end{aligned}$$

Upon going back to (3.20), it follows that

$$ 0\le \frac{\partial h}{\partial s}(s(R, \beta ), \beta ) \le \frac{1}{\sqrt{2B(r_Q)}} \,\bigg (48 \max \Big \{2A(0), \frac{\beta _*}{c_0^4}\Big \} +\frac{28\pi }{3}\,\rho _Q(0)\bigg )=:C_1 $$

for $\theta \in [-\pi /2, 0]$. Since generally $\frac{\partial h}{\partial s}\ge 0$, we finally get from Lemma 3.3

$$ T_1(e, \beta )=\sqrt{2}\int _{-\pi /2}^{\pi /2} \,\frac{d\theta }{\frac{\partial h}{\partial s}(s(R, \beta ), \beta )} \ge \sqrt{2}\int _{-\pi /2}^0 \,\frac{d\theta }{\frac{\partial h}{\partial s}(s(R, \beta ), \beta )} \ge \frac{\pi }{\sqrt{2}\,C_1}, $$

which completes the proof, as we have found a positive lower bound on $T_1$ in all three cases.    $\square$

3.3 Further Properties of $T_1$

First, we discuss some regularity properties of $T_1$.

Theorem 3.6

We have $T_1\in C^1(\mathring{D})$.

Proof

The continuity of $T_1$ may be shown directly from (3.14), as we already know that $r_\pm \in C^2(\mathring{D})$ by Remark A.3; we omit the details. To prove the differentiability, we use a method that is known and that we learned from R. Ortega. It is considerably less painful than differentiating an explicit relation for $T_1$ like (3.14). For $(e, \beta )\in \mathring{D}$, we consider

$$ \ddot{r}=-U'_{\mathrm{eff}}(r, \beta ),\quad r(0)=r_-(e, \beta ), \quad \dot{r}(0)=0, $$

where $r(t)=r(t, e, \beta )$. Defining

$$\begin{aligned} F: \mathbb R\times \mathring{D}\rightarrow \mathbb R,\quad F(t, e, \beta )=\dot{r}(t, e, \beta ), \end{aligned}$$

we have $F\in C^1(\mathbb R\times \mathring{D})$ by Lemma A.11(a). Next observe that $F(t, e, \beta )=0$ exactly for

$$ t=0,\,\,t=\pm \frac{1}{2}\,T_1(e, \beta ),\,\,t=\pm T_1(e, \beta ), \,\,t=\pm \frac{3}{2}\,T_1(e, \beta ),\,\,\ldots . $$

Fix $(\tilde{e}, \tilde{\beta })\in \mathring{D}$ and define $\tilde{t}=T_1(\tilde{e}, \tilde{\beta })$. Then $F(\tilde{t}, \tilde{e}, \tilde{\beta })=0$ by the above. Furthermore,

$$ \frac{\partial F}{\partial t}(t, e, \beta )=\ddot{r}(t, e, \beta ) =-U'_{\mathrm{eff}}(r(t, e, \beta ), \beta ) $$

and $r(\tilde{t}, \tilde{e}, \tilde{\beta })=r(T_1(\tilde{e}, \tilde{\beta }), \tilde{e}, \tilde{\beta }) =r(0, \tilde{e}, \tilde{\beta })=r_-(\tilde{e}, \tilde{\beta })$ in conjunction with Lemma A.4 imply that

$$ \frac{\partial F}{\partial t}(\tilde{t}, \tilde{e}, \tilde{\beta }) =-U'_{\mathrm{eff}}(r_-(\tilde{e}, \tilde{\beta }), \tilde{\beta })>0. $$

Hence, the implicit function theorem yields the existence of a $C^1$-function $t=t(e, \beta )$ that is defined for $(e, \beta )$ in a neighborhood $U\subset \mathring{D}$ of $(\tilde{e}, \tilde{\beta })$, such that

$$ F(t(e, \beta ), e, \beta )=0\,\,\text{ for }\,\,(e, \beta )\in U \quad \text{ and }\quad t(\tilde{e}, \tilde{\beta })=\tilde{t}=T_1(\tilde{e}, \tilde{\beta }). $$

According to our previous remarks, for every $(e, \beta )\in U$, we must have

$$\begin{aligned} t(e, \beta )=k(e, \beta )\,\frac{1}{2}\,T_1(e, \beta ) \end{aligned}$$

for some $k(e, \beta )\in \mathbb Z$. Then k is continuous in U and such that $k(\tilde{e}, \tilde{\beta })=2$, which means that $k=2$ throughout U. Thus, $T_1=t$ in U shows that $T_1\in C^1(U)$.    $\square$

Remark 3.7

If $\rho _Q\in C^k$, then $U_Q\in C^{k+2}$. As a consequence, $r_-\in C^{k+2}(\mathring{D})$ by the argument from Remark A.3. Comparing to Lemma A.11(a), this entails $F=\dot{r}\in C^{k+1}(\mathbb R\times \mathring{D})$, so that $t=t(e, \beta )\in C^{k+1}(U)$ for the solution function in the proof of Theorem 3.6. Hence, we get $T_1\in C^{k+1}(\mathring{D})$ in this case. $\diamondsuit$

Now, we are going to show that $T_1$ can be extended continuously from $\mathring{D}$ to D. We start with the continuous extension to $\{(e, \beta ): \beta \in ]0, \beta _*], e=e_{\mathrm{min}}(\beta )\}$.

Lemma 3.8

Let $\hat{\beta }\in ]0, \beta _*]$. Then

$$\begin{aligned} T_1(e, \beta )\rightarrow \frac{2\pi }{\sqrt{B(r_0(\hat{\beta }))}} \quad \text{ as }\quad \mathring{D}\ni (e, \beta )\rightarrow (e_{\mathrm{min}}(\hat{\beta }), \hat{\beta }). \end{aligned}$$

(3.21)

Proof

This relies on the representation (3.16) of $T_1(e, \beta )$, which we recall as

$$\begin{aligned} T_1(e, \beta )=2\,\sqrt{2}\int _{-\pi /2}^{\pi /2} d\theta \, \frac{[\int _0^1 (1-\rho )\,U''_{\mathrm{eff}}(r_0(\beta )+\rho s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta ), \beta ) \,d\rho ]^{1/2}}{\int _0^1 U''_{\mathrm{eff}}(r_0(\beta )+\rho s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta ), \beta )\,d\rho }. \end{aligned}$$

(3.22)

Here, $h(s, \beta )=s(\frac{V(s, \beta )}{s^2})^{1/2}$ and $h(0, \beta )=0$ for $V(s, \beta )=U_{\mathrm{eff}}(r_0(\beta )+s, \beta )-e_{\mathrm{min}}(\beta )$. Furthermore, $\hat{e}(\beta )=e-e_{\mathrm{min}}(\beta )$ and $R\mapsto s(R, \beta )=s$ denotes the inverse mapping to $s\mapsto h(s, \beta )=R$. Due to $\beta \rightarrow \hat{\beta }>0$, we can assume that $\beta \ge \hat{\beta }/2$ throughout the argument. If $r\in [r_-, r_+]$ and $\beta \in ]0, \beta _*]$, then Lemma A.6(c) and (A.28) yields

$$\begin{aligned} U'''_{\mathrm{eff}}(r, \beta )= & {} -\frac{12\beta }{r^5}+B'(r)-3A'(r) =-\frac{12\beta }{r^5}+4\pi \rho '_Q(r)-2A'(r) \\= & {} -\frac{12\beta }{r^5}+4\pi \rho '_Q(r) -\frac{8\pi }{r^4}\int _0^r s^3\rho '_Q(s)\,ds. \end{aligned}$$

Therefore, (Q4) gives the bound

$$\begin{aligned} |U'''_{\mathrm{eff}}(r, \beta )|\le C\Big (1+\frac{1}{r_-^5}\Big ), \quad r\in [r_-, r_+],\quad \beta \in ]0, \beta _*], \quad e\in [e_{\mathrm{min}}(\beta ), e_0]. \end{aligned}$$

(3.23)

By definition, we have $U_Q(r_-)+\frac{\beta }{2r_-}=e$. Hence, $U'_Q(r)\ge 0$ leads to

$$\begin{aligned} \frac{\beta }{2r_-^2}\le U_Q(r_-)-U_Q(0)+\frac{\beta }{2r_-^2}=e-U_Q(0), \end{aligned}$$

and thus

$$\begin{aligned} r_-\ge \sqrt{\frac{\beta }{2(e-U_Q(0))}}\ge \sqrt{\frac{\hat{\beta }}{4(e-U_Q(0))}} \end{aligned}$$

for $\beta \in [\hat{\beta }/2, \beta _*]$ and $e\in [e_{\mathrm{min}}(\beta ), e_0]$; note that we will have $e\rightarrow e_{\mathrm{min}}(\hat{\beta })>U_Q(0)$. Going back to (3.23), we obtain

$$\begin{aligned} |U'''_{\mathrm{eff}}(r, \beta )|\le C, \quad r\in [r_-, r_+],\quad \beta \in [\hat{\beta }/2, \beta _*], \quad e\in [e_{\mathrm{min}}(\beta ), e_0]. \end{aligned}$$

(3.24)

Next, we assert that

$$\begin{aligned} \lim _{\beta \rightarrow \hat{\beta },\,e\rightarrow e_{\mathrm{min}}(\hat{\beta })} \,\sup _{\theta \in [-\pi /2, \pi /2]} \,|s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta )|=0. \end{aligned}$$

(3.25)

Otherwise, there would be $\varepsilon _0>0$ and sequences $(\beta _j)$, $(e_j)$ and $(\theta _j)$ such that $\beta _j\rightarrow \hat{\beta }$, $\theta _j\rightarrow \hat{\theta }\in [-\pi /2, \pi /2]$, $\hat{e}(\beta _j)=e_j-e_{\mathrm{min}}(\beta _j)\rightarrow e_{\mathrm{min}}(\hat{\beta }) -e_{\mathrm{min}}(\hat{\beta })=0$ as well as $|s(\sqrt{\hat{e}(\beta _j)}\sin \theta _j, \beta _j)|\ge \varepsilon _0$ for all $j\in \mathbb N$; here it was used that $e_{\mathrm{min}}(\beta )=U_{\mathrm{eff}}(r_0(\beta ), \beta )$ is continuous in $\beta \in ]0, \beta _*[$, cf. Remark A.3. Thus, $\sqrt{\hat{e}(\beta _j)}\sin \theta _j\rightarrow 0$ and $s(\sqrt{\hat{e}(\beta _j)}\sin \theta _j, \beta _j) \rightarrow s(0, \hat{\beta })=0$, which is a contradiction. For the latter convergence, note that $s\mapsto h(s, \beta )$ for $s\in [s_-, s_+]$ is an increasing function that connects $-\sqrt{\hat{e}(\beta )}$ to $\sqrt{\hat{e}(\beta )}$. Since $\hat{e}(\beta _j)\rightarrow 0$, we must also have $s_\pm (e_j, \beta _j)\rightarrow 0$: for instance, if we had $s_+(e_j, \beta _j)\rightarrow \hat{s}_+>0$ (along a subsequence), then $h(s, \hat{\beta })=0$ for $s\in [0, \hat{s}_+]$, which is impossible. Thus, $s_\pm (e_j, \beta _j)\rightarrow 0$, and due to $|s(R, \beta )|\le \max \{|s_-(e, \beta )|, s_+(e, \beta )\}$, we obtain $s(\sqrt{\hat{e}(\beta _j)}\sin \theta _j, \beta _j)\rightarrow 0$ as claimed.

Coming back to (3.22) and using Lemma A.7(d) and (3.24), we estimate

$$\begin{aligned}&{\bigg |\int _0^1 U''_{\mathrm{eff}} (r_0(\beta )+\rho s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta ), \beta )\,d\rho -B(r_0(\beta ))\bigg |} \nonumber \\&= \bigg |\int _0^1 [U''_{\mathrm{eff}}(r_0(\beta )+\rho s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta ), \beta ) -U''_{\mathrm{eff}}(r_0(\beta ), \beta )]\,d\rho \,\bigg | \nonumber \\&\le C\int _0^1 |s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta )|\,d\rho \le CS(e, \beta ), \nonumber \\&\quad \quad S(e, \beta )=\sup _{\theta \in [-\pi /2, \pi /2]} \,|s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta )|. \end{aligned}$$

(3.26)

Similarly,

$$\begin{aligned} \bigg |\int _0^1 (1-\rho )\,U''_{\mathrm{eff}}(r_0(\beta ) +\rho s(\sqrt{\hat{e}(\beta )}\sin \theta , \beta ), \beta )\,d\rho -\frac{1}{2}\,B(r_0(\beta ))\bigg | \le CS(e, \beta ). \end{aligned}$$

(3.27)

From (3.26), (3.27) and (3.25), in conjunction with Lebesgue’s dominated convergence theorem and $B(r_0(\hat{\beta }))>0$, we deduce (3.21).    $\square$

Remark 3.9

Note that $T_1(e, \beta )$ is defined for $e=e_0$ and $\beta \in ]0, \beta _*]$; it is the period of the orbit of $\ddot{r}=-U''_{\mathrm{eff}}(r, \beta )$ that has the largest energy $e=e_0$. Therefore, it is straightforward that

$$ T_1(e_0, \beta )=2\int _{r_-(e_0, \beta )}^{r_+(e_0, \beta )} \frac{dr}{\sqrt{2(e_0-U_{\mathrm{eff}}(r, \beta ))}} $$

extends $T_1$ continuously to $\{(e, \beta ): e=e_0, \beta \in ]0, \beta _*]\}$. $\diamondsuit$

There is yet another way to represent $T_1$; see [24, Exercise 1, p. 40].

Lemma 3.10

Define

$$\begin{aligned}&\chi (r, e, \beta )=\int _0^1 d\tau \,(1-\tau )\int _0^1 d\sigma \,U''_{\mathrm{eff}}(\tau r_+(e, \beta )+\sigma (1-\tau )r\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad +(1-\sigma )(1-\tau )r_-(e, \beta ), \beta ). \end{aligned}$$

Then

$$\begin{aligned} T_1(e, \beta )=\sqrt{2}\int _{r_-(e, \beta )}^{r_+(e, \beta )} \frac{dr}{\sqrt{(r_+(e, \beta )-r)(r-r_-(e, \beta ))\,\chi (r, e, \beta )}}. \end{aligned}$$

(3.28)

Proof

If $r>s$, then

$$ U_{\mathrm{eff}}(r, \beta )-U_{\mathrm{eff}}(s, \beta ) =(r-s)\int _0^1 U'_{\mathrm{eff}}(\tau r+(1-\tau )s, \beta )\,d\tau , $$

and in particular $U_{\mathrm{eff}}(r_\pm , \beta )=e$ yields $\int _0^1 U'_{\mathrm{eff}}(\tau r_+ +(1-\tau )r_-, \beta )\,d\tau =0$. Therefore, we can write

$$\begin{aligned} {e-U_{\mathrm{eff}}(r, \beta )}= & {} U_{\mathrm{eff}}(r_+, \beta )-U_{\mathrm{eff}}(r, \beta ) \\= & {} (r_+ -r)\int _0^1 U'_{\mathrm{eff}}(\tau r_+ +(1-\tau )r, \beta )\,d\tau \\= & {} (r_+ -r)\int _0^1 [U'_{\mathrm{eff}}(\tau r_+ \!+\!(1-\tau )r, \beta ) \!-\!U'_{\mathrm{eff}}(\tau r_+ \!+\!(1\!-\!\tau )r_-, \beta )]\,d\tau \\= & {} (r_+ -r)(r-r_-)\int _0^1 d\tau \,(1-\tau )\int _0^1 d\sigma \,U''_{\mathrm{eff}}(\tau r_+ + \sigma (1-\tau )r\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad +\,(1-\sigma )(1-\tau )r_-, \beta ), \end{aligned}$$

which leads to (3.28).    $\square$

Lemma 3.11

We have

$$\begin{aligned} T_1(e, \beta )\rightarrow \frac{2\pi }{\sqrt{B(0)}} \quad \text{ as }\quad \mathring{D}\ni (e, \beta )\rightarrow (U_Q(0), 0). \end{aligned}$$

(3.29)

Proof

First, we note that, although it won’t be used, $e-e_{\mathrm{min}}(\beta )\ge 0$ together with Lemma A.7(f) implies $e-U_Q(0)\ge e_{\mathrm{min}}(\beta )-U_Q(0) \sim \sqrt{U''_Q(0)}\sqrt{\beta }$ as $\beta \rightarrow 0$, which means that as $e\rightarrow U_Q(0)$, the quantity $e-U_Q(0)$ can’t be too small in terms of $\beta \rightarrow 0$; due to $U''_Q(r)+\frac{2}{r}\,U'_Q(r)=4\pi \rho _Q(r)$, we have $U''_Q(0)=\frac{4\pi }{3}\,\rho _Q(0)>0$.

To actually verify (3.29), we are going to write

$$\begin{aligned} T_1(e, \beta )=\sqrt{2}\int _{r_-}^{r_+}\frac{dr}{\sqrt{(r_+-r)(r-r_-)\,\chi (r)}} \end{aligned}$$

(3.30)

as in (3.28) from Lemma 3.10, where

$$\begin{aligned}&\chi (r)=\chi (r, e, \beta )=\int _0^1 d\tau \,(1-\tau )\int _0^1 d\sigma \,U''_{\mathrm{eff}}(\tau r_+ +\sigma (1-\tau )r\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \!\!+(1-\sigma )(1-\tau )r_-, \beta ). \end{aligned}$$

Due to Lemma A.6(c), we have $U''_{\mathrm{eff}}(r, \beta )=\frac{3\beta }{r^4}+B(r)-3A(r)$. By explicit integration,

$$\begin{aligned}&{3\beta \int _0^1 d\sigma \int _0^1 d\tau \,(1-\tau ) \,\frac{1}{(\tau r_+ +\sigma (1-\tau )r+(1-\sigma )(1-\tau )r_-)^4}} \\= & {} \frac{\beta }{2r_+^2}\int _0^1 d\sigma \, \frac{2r_+ +\sigma r+(1-\sigma )r_-}{(\sigma r+(1-\sigma )r_-)^3} \\= & {} \frac{\beta }{r_+}\int _0^1 d\sigma \,\frac{1}{(\sigma r+(1-\sigma )r_-)^3} +\frac{\beta }{2r_+^2}\int _0^1 d\sigma \,\frac{1}{(\sigma r+(1-\sigma )r_-)^2} \\= & {} \frac{\beta }{r_+}\,\frac{r+r_-}{2r^2 r_-^2} +\frac{\beta }{2r_+^2}\,\frac{1}{rr_-} \\= & {} \frac{\beta }{2}\,\frac{r(r_- +r_+) +r_- r_+}{r^2 r_-^2 r_+^2} \\= & {} \frac{\beta }{2}\,\frac{(r_+ +r)(r_- +r)}{r^2 r_-^2 r_+^2} -\frac{\beta }{2}\,\frac{1}{r_-^2 r_+^2}. \end{aligned}$$

Hence, we obtain

$$\begin{aligned} \chi (r)&= \frac{\beta }{2}\,\frac{(r_+ +r)(r_- +r)}{r^2 r_-^2 r_+^2}+\chi _2(r), \\ \chi _2(r)&= \int _0^1 d\tau \,(1-\tau )\int _0^1 d\sigma \,(B-3A)(\tau r_+ +\sigma (1-\tau )r \nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \!\!+ (1-\sigma )(1-\tau )r_-) -\frac{\beta }{2}\,\frac{1}{r_-^2 r_+^2}, \nonumber \end{aligned}$$

(3.31)

and $\chi _2(r)=\chi _2(r, e, \beta )$. From Lemma A.6(a) and (b), we get $(B-3A)(0)=\frac{16\pi }{3}\,\rho _Q(0)-\frac{12\pi }{3}\,\rho _Q(0)=\frac{4\pi }{3}\,\rho _Q(0)=U''_Q(0)$. Since $\tau r_+ +\sigma (1-\tau )r+(1-\sigma )(1-\tau )r_-\in [r_-, r_+]\subset [0, r_+]$ for $\tau , \sigma \in [0, 1]$ and $r\in [r_-, r_+]$, it follows from

$$\begin{aligned} \chi _2(r)= & {} \int _0^1 d\tau \,(1-\tau )\int _0^1 d\sigma \,[(B-3A)(\tau r_+ +\sigma (1-\tau )r\\&\qquad \qquad \qquad \qquad \qquad \qquad \quad +\,(1-\sigma )(1-\tau )r_-)-(B-3A)(0)] \\&+\frac{1}{2}\,U''_Q(0)-\frac{\beta }{2}\,\frac{1}{r_-^2 r_+^2} \end{aligned}$$

and (A.26) in Lemma A.5 that

$$\begin{aligned} \sup _{r\in [r_-, r_+]} |\chi _2(r, e, \beta )|\le & {} \frac{1}{2}\sup _{s\in [0, r_+(e, \beta )]}\,|(B-3A)(s)-(B-3A)(0)| \nonumber \\&+\frac{1}{2}\sup _{s\in [0, r_+(e, \beta )]}\,|U''_Q(s)-U''_Q(0)| \end{aligned}$$

(3.32)

for $(e, \beta )\in \mathring{D}$ and $r_\pm =r_\pm (e, \beta )$.

Next, we assert that

$$\begin{aligned} r_+(e, \beta )\rightarrow 0 \quad \text{ as }\quad \mathring{D}\ni (e, \beta )\rightarrow (U_Q(0), 0). \end{aligned}$$

(3.33)

To establish this claim, we will use the relation

$$ U_Q(r_+)-U_Q(r_0) =e-\frac{\beta }{2r_+^2}-e_{\mathrm{min}}(\beta )+\frac{\beta }{2r_0^2} =e-e_{\mathrm{min}}(\beta )+\frac{\beta }{2r_0^2 r_+^2}(r_+^2-r_0^2). $$

Hence, (3.3) from Lemma 3.1 yields

$$\begin{aligned} \frac{\pi }{12}\,\rho _Q\Big (\frac{r_Q}{2}\Big )(r_+^2-r_0^2) \le e-U_Q(0)+U_Q(0)-e_{\mathrm{min}}(\beta )+\frac{\beta }{2r_0^2 r_+^2}(r_+^2-r_0^2). \end{aligned}$$

(3.34)

Due to Lemma A.7(f), we have $|e_{\mathrm{min}}(\beta )-U_Q(0)|=\mathcal{O}(\beta ^{1/2})$ and $r_0=\mathcal{O}(\beta ^{1/4})$ as $\beta \rightarrow 0$. Thus, if $r_+(e, \beta )\rightarrow \hat{r}_+>0$ as $(e, \beta )\rightarrow (U_Q(0), 0)$, and along some subsequence, then (3.34) would imply that $\frac{\pi }{12}\,\rho _Q(\frac{r_Q}{2})\,\hat{r}_+^2\le 0$, which is a contradiction and confirms (3.33). Since both $(B-3A)(s)$ and $U''_Q(s)$ are continuous at $s=0$, (3.32) in turn shows that

$$\begin{aligned} \lim _{e\rightarrow U_Q(0),\,\beta \rightarrow 0}\,\sup _{r\in [r_-, r_+]} |\chi _2(r, e, \beta )|=0. \end{aligned}$$

(3.35)

A further preparatory step is to rewrite (3.31) as

$$\begin{aligned} \chi (r)= & {} \frac{\beta }{2}\,\frac{(r_+ +r)(r_- +r)}{r^2 r_-^2 r_+^2}\,(1+\chi _3(r)), \\ \chi _3(r)= & {} \frac{2}{\beta }\,\frac{r^2 r_-^2 r_+^2}{(r_+ +r)(r_- +r)}\,\chi _2(r), \nonumber \end{aligned}$$

(3.36)

for $\chi _3(r)=\chi _3(r, e, \beta )$. Owing to Lemma 3.1(b), we have

$$\begin{aligned} r_-^2 r_+^2\le C\beta . \end{aligned}$$

(3.37)

Since also $\frac{r^2}{(r_+ +r)(r_- +r)}\le 1$, it follows from (3.35) that

$$\begin{aligned} \lim _{e\rightarrow U_Q(0),\,\beta \rightarrow 0}\,\sup _{r\in [r_-, r_+]} |\chi _3(r, e, \beta )|=0. \end{aligned}$$

(3.38)

Coming back to (3.29), consider sequences $e_j\rightarrow U_Q(0)$ and $\beta _j\rightarrow 0$. Let $\varepsilon >0$ be given. According to (3.38), there is $j_0\in \mathbb N$ such that $\sup _{r\in [r_-(e_j, \beta _j), r_+(e_j, \beta _j)]} |\chi _3$ $(r, e_j, \beta _j)|\le \varepsilon$ for $j\ge j_0$. Due to (3.36), this yields for $j\ge j_0$ and $r\in [r_-(e_j, \beta _j),$ $r_+(e_j, \beta _j)]$

$$ \frac{\beta _j}{2}\,\frac{(r_{+,\,j}+r)(r_{-,\,j}+r)}{r^2\,r_{-,\,j}^2\,r_{+,\,j}^2}\,(1-\varepsilon ) \le \chi (r, e_j, \beta _j) \le \frac{\beta _j}{2}\,\frac{(r_{+,\,j}+r)(r_{-,\,j}+r)}{r^2\,r_{-,\,j}^2\,r_{+,\,j}^2}\,(1+\varepsilon ), $$

where $r_{\pm ,\,j}=r_\pm (e_j, \beta _j)$. Therefore, (3.30) leads to

$$ \frac{r_{-,\,j}\,r_{+,\,j}}{\sqrt{1+\varepsilon }}\,\frac{2}{\beta _j^{1/2}}\,I_j \le T_1(e_j, \beta _j)\le \frac{r_{-,\,j}\,r_{+,\,j}}{\sqrt{1-\varepsilon }}\,\frac{2}{\beta _j^{1/2}}\,I_j $$

for $j\ge j_0$, where

$$\begin{aligned} I_j=\int _{r_{-,\,j}}^{r_{+,\,j}}\frac{r}{(r_{+,\,j}^2-r^2)^{1/2}(r^2-r_{-,\,j}^2)^{1/2}}\,dr. \end{aligned}$$

Setting $s=r^2$, $ds=2r\,dr$, this integral may be evaluated as $I_j=\pi /2$. Thus, we obtain

$$\begin{aligned} \frac{1-\varepsilon }{\pi ^2}\,\frac{\beta _j}{r_{-,\,j}^2\,r_{+,\,j}^2} \le \frac{1}{T_1(e_j, \beta _j)^2}\le \frac{1+\varepsilon }{\pi ^2}\,\frac{\beta _j}{r_{-,\,j}^2\,r_{+,\,j}^2} \end{aligned}$$

(3.39)

for $j\ge j_0$. From (A.26) in Lemma A.5, we know that

$$ \bigg |\frac{\beta _j}{r_{-,\,j}^2\,r_{+,\,j}^2}-U''_Q(0)\bigg | \le \sup _{r\in [0, r_{+,\,j}]}\,|U''_Q(r)-U''_Q(0)|. $$

As $r_{+,\,j}\rightarrow 0$ by (3.33), we may assume that $j_0$ is already taken so large that

$$\begin{aligned} U''_Q(0)-\varepsilon \le \frac{\beta _j}{r_{-,\,j}^2\,r_{+,\,j}^2}\le U''_Q(0)+\varepsilon \end{aligned}$$

for $j\ge j_0$. Therefore, (3.39) implies that

$$ \frac{1-\varepsilon }{\pi ^2}\,(U''_Q(0)-\varepsilon ) \le \frac{1}{T_1(e_j, \beta _j)^2}\le \frac{1+\varepsilon }{\pi ^2}\,(U''_Q(0)+\varepsilon ) $$

for $j\ge j_0$. Altogether, this shows that $\lim _{j\rightarrow \infty } T_1(e_j, \beta _j)=\frac{\pi }{\sqrt{U''_Q(0)}}$, and it remains to recall that $B(0)=\frac{16\pi }{3}\,\rho _Q(0)=4U''_Q(0)$, cf. Lemma A.6(a), (b).    $\square$

Lemma 3.12

Let $\hat{e}\in ]U_Q(0), e_0]$. Then

$$\begin{aligned} T_1(e, \beta )\rightarrow 2\int _0^{\hat{r}(\hat{e})}\frac{dr}{\sqrt{2(\hat{e}-U_Q(r))}} \quad \text{ as }\quad \mathring{D}\ni (e, \beta )\rightarrow (\hat{e}, 0), \end{aligned}$$

(3.40)

where $\hat{r}(e)\in [0, r_Q]$ is the unique solution to $U_Q(\hat{r}(e))=e$.

Proof

First, we are going to show that $r_+$ stays away from zero in the limiting case that we are considering here. For this, we may assume that $r_+\le r_Q/2$. Due to (3.6), we have

$$\begin{aligned} r_+^2\varphi (r_+)+\frac{\beta }{2r_+^2} =U_Q(r_+)-U_Q(0)+\frac{\beta }{2r_+^2}=e-U_Q(0) \end{aligned}$$

(3.41)

for

$$\begin{aligned} \varphi (r_+)=4\pi \int _0^1 d\tau \tau ^2\int _0^1 dt\,t\,\rho _Q(\tau t r_+). \end{aligned}$$

Since $\rho _Q$ is radially decreasing and $0\le \tau tr_+\le r_Q/2$, it follows that

$$\begin{aligned} 0<c_1=\frac{2\pi }{3}\rho _Q\Big (\frac{r_Q}{2}\Big )\le \varphi (r_+) \le \frac{2\pi }{3}\rho _Q(0)=C_1. \end{aligned}$$

(3.42)

In (3.41), solving the resulting quadratic equation for $r_+^2$, we obtain

$$\begin{aligned} r_+^2=\frac{e-U_Q(0)\pm \sqrt{(e-U_Q(0))^2-2\varphi (r_+)\beta }}{2\varphi (r_+)}. \end{aligned}$$

(3.43)

Let us suppose that the sign were ‘−’, along (a subsequence) of $e\rightarrow \hat{e}$ and $\beta \rightarrow 0$.

Then

$$\begin{aligned} r_+^2=\frac{\beta }{e-U_Q(0)+\sqrt{(e-U_Q(0))^2-2\varphi (r_+)\beta }} \end{aligned}$$

together with $\hat{e}-U_Q(0)>0$ and (3.42) would yield $c_2\beta \le r_+^2\le C_2\beta$ for suitable constants $C_2>c_2>0$. By Lemma 3.1(b), we have the general estimate

$$\begin{aligned} c\beta \le r_-^2 r_+^2. \end{aligned}$$

As a consequence,

$$\begin{aligned} \frac{c}{C_2}\le r_-^2. \end{aligned}$$

However, $r_-^2\le r_0^2=\mathcal{O}(\beta ^{1/2})$ as $\beta \rightarrow 0$ by Lemma A.7(f), which gives a contradiction. To summarize, we may suppose that the sign is ‘$+$’ in (3.43). Hence,

$$ r_+^2=\frac{e-U_Q(0)+\sqrt{(e-U_Q(0))^2-2\varphi (r_+)\beta }}{2\varphi (r_+)} \ge \frac{1}{2C_1} (e-U_Q(0)) $$

for $\beta \le \frac{1}{2C_1}(e-U_Q(0))^2$ yields the desired lower bound for $r_+$. Thus, in what follows, we can assume that $r_+(e, \beta )\ge \eta _0>0$ for an appropriate constant $\eta _0$ and $(e, \beta )\rightarrow (\hat{e}, 0)$.

Next, we are going to show that

$$\begin{aligned} T_1^-(e, \beta )=2\int _{r_-}^{r_0}\frac{dr}{\sqrt{2(e-U_{\mathrm{eff}}(r, \beta ))}} \rightarrow 0\quad \text{ as }\quad \mathring{D}\ni (e, \beta )\rightarrow (\hat{e}, 0). \end{aligned}$$

(3.44)

For, owing to (3.10), we get

$$ T_1^-(e, \beta )\le 4\,\frac{\sqrt{r_- r_+}}{\sqrt{\beta }} \,\frac{r_0}{\sqrt{r_+ -r_0}}\,\sqrt{r_0-r_-} \le 4\,\frac{\sqrt{r_- r_+}}{\sqrt{\beta }} \,\frac{r_0^{3/2}}{\sqrt{r_+ -r_0}}. $$

Since $r_-^2 r_+^2\le C\beta$ by (3.37) and $r_0=\mathcal{O}(\beta ^{1/4})$ by Lemma A.7(f), $r_+\ge \eta _0$ yields

$$\begin{aligned} T_1^-(e, \beta )\le C\beta ^{3/8} \end{aligned}$$

and completes the argument for (3.44).

Thus, in order to establish (3.40), we need to prove that

$$\begin{aligned} \int _{r_0}^{r_+}\frac{dr}{\sqrt{e-U_{\mathrm{eff}}(r, \beta )}} \rightarrow \int _0^{\hat{r}(\hat{e})}\frac{dr}{\sqrt{\hat{e}-U_Q(r)}} \quad \text{ as }\quad \mathring{D}\ni (e, \beta )\rightarrow (\hat{e}, 0); \end{aligned}$$

(3.45)

note that $U_Q(r_+)\le U_Q(r_+)+\frac{\beta }{2r_+^2}=e=U_Q(\hat{r}(e))$ implies $r_+\le \hat{r}(e)$. In addition, using (3.3), we obtain

$$\begin{aligned} \frac{\pi }{12}\,\rho _Q\Big (\frac{r_Q}{2}\Big )\eta _0\,(\hat{r}(e)-r_+)\le & {} \frac{\pi }{12}\,\rho _Q\Big (\frac{r_Q}{2}\Big )(\hat{r}(e)^2-r_+^2) \nonumber \\\le & {} U_Q(\hat{r}(e))-U_Q(r_+)=\frac{\beta }{2r_+^2}\le \frac{1}{2\eta _0^2}\,\beta . \end{aligned}$$

Similarly, by (3.2),

$$\begin{aligned} \frac{\beta }{2r_Q^2}\le & {} \frac{\beta }{2r_+^2}=U_Q(\hat{r}(e))-U_Q(r_+)\\\le & {} \frac{2\pi }{3}\rho _Q(0)(\hat{r}(e)^2-r_+^2)\\\le & {} \frac{4\pi }{3}\rho _Q(0)\,r_Q\,(\hat{r}(e)-r_+), \end{aligned}$$

so that $c_3\beta \le \hat{r}(e)-r_+\le C_3\beta$. To validate (3.45), we are going to show

$$\begin{aligned}&\bigg |\int _{r_0}^{r_+}\frac{dr}{\sqrt{e-U_{\mathrm{eff}}(r, \beta )}} -\int _0^{\hat{r}(e)}\frac{dr}{\sqrt{e-U_Q(r)}}\bigg |\rightarrow 0, \end{aligned}$$

(3.46)

$$\begin{aligned}&\bigg |\int _0^{\hat{r}(e)}\frac{dr}{\sqrt{e-U_Q(r)}} -\int _0^{\hat{r}(\hat{e})}\frac{dr}{\sqrt{\hat{e}-U_Q(r)}}\bigg |\rightarrow 0, \end{aligned}$$

(3.47)

both as $\mathring{D}\ni (e, \beta )\rightarrow (\hat{e}, 0)$; the second relation is independent of $\beta$.

To begin with,

$$\begin{aligned} \int _0^{r_0}\frac{dr}{\sqrt{e-U_Q(r)}}\rightarrow 0 \quad \text{ and }\quad \int _{r_+}^{\hat{r}(e)}\frac{dr}{\sqrt{e-U_Q(r)}}\rightarrow 0. \end{aligned}$$

(3.48)

For the first claim, if $0\le r\le r_0=\mathcal{O}(\beta ^{1/4})$ and $e\rightarrow \hat{e}>U_Q(0)$, we may suppose that $e-U_Q(r)\ge \eta _1>0$ for the e and r in question; therefore, the first claim in (3.48) follows. Regarding the second assertion, we write

$$\begin{aligned} e-U_Q(r)=U_Q(\hat{r}(e))-U_Q(r)=(\hat{r}(e)-r)\int _0^1 U'_Q(\tau \hat{r}(e)+(1-\tau )r)\,d\tau \end{aligned}$$

for $r\in [\frac{r_+}{2}, \hat{r}(e)]$. If $s\in [\frac{r_+}{2}, \hat{r}(e)]$, then the fact that $\rho _Q$ is radially decreasing yields

$$\begin{aligned} U'_Q(s)=\frac{4\pi }{s^2}\int _0^s\sigma ^2\rho _Q(\sigma )\,d\sigma \ge \frac{4\pi }{r_Q^2}\int _0^{r_+/2}\sigma ^2\rho _Q(\sigma )\,d\sigma \ge \frac{\pi r_+^3}{6r_Q^2}\,\rho _Q\Big (\frac{r_+}{2}\Big )\ge \eta _2>0. \end{aligned}$$

(3.49)

Hence,

$$\begin{aligned} e-U_Q(r)\ge \eta _2(\hat{r}(e)-r),\quad r\in \Big [\frac{r_+}{2}, \hat{r}(e)\Big ], \end{aligned}$$

(3.50)

and accordingly,

$$ \int _{r_+}^{\hat{r}(e)}\frac{dr}{\sqrt{e-U_Q(r)}} \le \eta _2^{-1/2}\int _{r_+}^{\hat{r}(e)}\frac{dr}{\sqrt{\hat{r}(e)-r}} =2\eta _2^{-1/2}\sqrt{\hat{r}(e)-r_+}\le C\beta ^{1/2}\rightarrow 0. $$

Thus, both relations in (3.48) hold, and therefore (3.46) comes down to proving that

$$ \bigg |\int _{r_0}^{r_+}\frac{dr}{\sqrt{e-U_{\mathrm{eff}}(r, \beta )}} -\int _{r_0}^{r_+}\frac{dr}{\sqrt{e-U_Q(r)}}\bigg |\rightarrow 0 \quad \text{ as }\quad \mathring{D}\ni (e, \beta )\rightarrow (\hat{e}, 0). $$

If $r\in [r_0, (1-\beta ^{1/4})r_+]$, then $\frac{\beta }{2r^2}\le \frac{\beta }{2r_0^2}\le C\beta ^{1/2}$, as $r_0=\mathcal{O}(\beta ^{1/4})$. Therefore, (3.3) yields

$$\begin{aligned} e-U_Q(r)\ge & {} e-U_{\mathrm{eff}}(r, \beta )=e-U_Q(r)-\frac{\beta }{2r^2}\\\ge & {} U_Q(\hat{r}(e))-U_Q((1-\beta ^{1/4})r_+)-C\beta ^{1/2}\\\ge & {} \frac{\pi }{12}\,\rho _Q\Big (\frac{r_Q}{2}\Big )\, (\hat{r}(e)^2-(1-\beta ^{1/4})^2 r_+^2)-C\beta ^{1/2}\\\ge & {} \frac{\pi }{12}\,\rho _Q\Big (\frac{r_Q}{2}\Big )\,\eta _0\, (\hat{r}(e)-r_+ +\beta ^{1/4}r_+)-C\beta ^{1/2}\\\ge & {} c_4\beta +c_5\beta ^{1/4}-C\beta ^{1/2}\\\ge & {} c_6\beta ^{1/4}. \end{aligned}$$

From the estimate $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\le \frac{b-a}{a\sqrt{b}}$ for $b\ge a>0$, we hence infer

$$\begin{aligned}&{\bigg |\int _{r_0}^{(1-\beta ^{1/4})r_+}\frac{dr}{\sqrt{e-U_{\mathrm{eff}}(r, \beta )}} -\int _{r_0}^{(1-\beta ^{1/4})r_+}\frac{dr}{\sqrt{e-U_Q(r)}}\bigg |} \nonumber \\\le & {} \frac{\beta }{2}\int _{r_0}^{(1-\beta ^{1/4})r_+} \frac{1}{(e-U_{\mathrm{eff}}(r, \beta ))\sqrt{e-U_Q(r)}}\,\frac{dr}{r^2} \nonumber \\\le & {} \frac{\beta }{2r_0^2}\int _{r_0}^{(1-\beta ^{1/4})r_+} \frac{1}{c_6^{3/2}\beta ^{3/8}}\,dr\le C\beta ^{1/8}\rightarrow 0. \end{aligned}$$

(3.51)

For the remaining part, $r\in [(1-\beta ^{1/4})r_+, r_+]$, we note that for such r, by (3.50),

$$\begin{aligned} e-U_Q(r)\ge \eta _2(\hat{r}(e)-r)\ge \eta _2(\hat{r}(e)-r_+)\ge \eta _2 c_3\,\beta . \end{aligned}$$

In addition,

$$\begin{aligned} e-U_{\mathrm{eff}}(r, \beta )= & {} U_{\mathrm{eff}}(r_+, \beta )-U_{\mathrm{eff}}(r, \beta ) \nonumber \\= & {} (r_+ -r)\int _0^1 U'_{\mathrm{eff}}(\tau r_+ +(1-\tau )r, \beta )\,d\tau . \end{aligned}$$

(3.52)

If $r{\in } [(1{-}\beta ^{1/4})r_+, r_+]$, then $s{=}\tau r_+ +(1-\tau )r\in [(1-\beta ^{1/4})r_+, r_+] \subset [\frac{r_+}{2}, \hat{r}(e)]$ for instance, and

$$ U'_{\mathrm{eff}}(s, \beta )=U'_Q(s)-\frac{\beta }{s^3} \ge \eta _2-\frac{8\beta }{r_+^3}\ge \eta _2-\frac{8\beta }{\eta _0^3}\ge \frac{1}{2}\,\eta _2 $$

by (3.49), if $\beta$ is small enough. Thus, (3.52) leads to

$$ e-U_{\mathrm{eff}}(r, \beta )\ge \frac{1}{2}\,\eta _2\,(r_+ -r), \quad r\in [(1-\beta ^{1/4})r_+, r_+]. $$

If we now use that $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\le \frac{b-a}{\sqrt{a}b}$ for $b\ge a>0$, we obtain the bound

$$\begin{aligned}&{\bigg |\int _{(1-\beta ^{1/4})r_+}^{r_+}\frac{dr}{\sqrt{e-U_{\mathrm{eff}}(r, \beta )}} -\int _{(1-\beta ^{1/4})r_+}^{r_+}\frac{dr}{\sqrt{e-U_Q(r)}}\bigg |} \nonumber \\\le & {} \frac{\beta }{2}\int _{(1-\beta ^{1/4})r_+}^{r_+} \frac{1}{\sqrt{e-U_{\mathrm{eff}}(r, \beta )}(e-U_Q(r))}\,\frac{dr}{r^2} \nonumber \\\le & {} \frac{\beta }{2}\,\frac{4}{\eta _0^2}\,\frac{1}{\eta _2 c_3\,\beta }\,\sqrt{\frac{2}{\eta _2}} \int _{(1-\beta ^{1/4})r_+}^{r_+}\frac{1}{\sqrt{r_+ -r}}\,dr\le C\beta ^{1/8}\rightarrow 0. \end{aligned}$$

(3.53)

By (3.51) and (3.53), the proof of (3.46) is complete.

Therefore, it remains to check that (3.47) is satisfied. This is not worked out, since it is just the continuity of the standard period function in the potential $V(x)=U_Q(x)-U_Q(0)$ for $x\ge 0$ and $V(x)=U_Q(-x)-U_Q(0)$ for $x\le 0$, for energies $\hat{e}=e-U_Q(0)\in ]0, e_0-U_Q(0)]$.    $\square$

If we now summarize Lemma 3.8, Remark 3.9 and Lemmas 3.11 and 3.12, then we have shown the following result (note that $e_{\mathrm{min}}(0)=U_Q(0)$ and $r_0(0)=0$).

Theorem 3.13

We have $T_1\in C(D)$. The extensions to $\partial D$ are given by

where $\hat{r}(e)\in [0, r_Q]$ is the unique solution to $U_Q(\hat{r}(e))=e$.

In the remaining part of this section, we will discuss some monotonicity properties of $T_1$.

Lemma 3.14

The function $[0, \beta _*]\ni \beta \mapsto T_1(e_{\mathrm{min}}(\beta ), \beta )$ is strictly increasing.

Proof

We know from Lemma A.7(e) that $\beta \mapsto r_0(\beta )$ is strictly increasing, and furthermore $r\mapsto B(r)$ is strictly decreasing by Lemma A.6(b). Hence, the claim follows from $T_1(e_{\mathrm{min}}(\beta ), \beta )=\frac{2\pi }{\sqrt{B(r_0(\beta ))}}$.    $\square$

Lemma 3.15

The function $[U_Q(0), e_0]\ni e\mapsto T_1(e, 0)$ is strictly increasing.

Proof

The argument is analogous to the fact that for a one degree of freedom oscillator $\ddot{x}=-V'(x)$ about a stable center, where $V(0)=V'(0)=0$ and $V(-x)=V(x)$ for simplicity, the condition $V'(x)>0$ and $V''(x)>V'(x)/x$ for $x>0$ guarantees that the period function of the periodic orbits about $x=0$ is decreasing in the energy $e=\frac{1}{2}\,\dot{x}^2+V(x)$. The first reference to point this out seems to be [64] (which we basically follow); related papers are [11, 78, 79]. To see the connection, first observe that, by (1.13), Remark A.1 and (A.32),

$$ U''_Q(r)-\frac{U'_Q(r)}{r}=4\pi \rho _Q(r)-\frac{3}{r}\,U'_Q(r) =\frac{4\pi }{r^3}\int _0^r s^3\rho '_Q(s)\,ds<0,\quad r>0. $$

Thus, $(U'_Q(r)/r)'=(rU''_Q(r)-U'_Q(r))/r^2<0$ for $r>0$, and it follows that

$$\begin{aligned} \frac{1}{p}\,U'_Q(pr)<U'_Q(r),\quad p>1,\quad r>0. \end{aligned}$$

(3.54)

The function $\hat{r}$ is strictly increasing, due to $1=U'_Q(\hat{r}(e))\,\hat{r}'(e)$ and $U'_Q(r)>0$ for $r>0$. Therefore, its inverse $[0, r_Q]\ni \hat{r}\mapsto e(\hat{r})\in [U_Q(0), e_0]$ is well-defined and strictly increasing too; note that $\hat{r}(U_Q(0))=0$ and $\hat{r}(e_0)=r_Q$. Let

$$\begin{aligned} \hat{T}(\hat{r})=2\int _0^{\hat{r}}\frac{dr}{\sqrt{2(U_Q(\hat{r})-U_Q(r))}}. \end{aligned}$$

Then

$$ T_1(e, 0)=2\int _0^{\hat{r}(e)}\frac{dr}{\sqrt{2(U_Q(\hat{r}(e))-U_Q(r)}} =\hat{T}(\hat{r}(e)), $$

which implies that $e\mapsto T_1(e, 0)$ is increasing if and only if $\hat{r}\mapsto \hat{T}(\hat{r})$ is increasing. If $p>1$ and $s\in [0, \hat{r}]$, then by (3.54), one has

$$ U_Q(p\hat{r})-U_Q(ps){=}p\int _s^{\hat{r}} U'_Q(p\tau )\,d\tau {\le } p^2\int _s^{\hat{r}} U'_Q(\tau )\,d\tau {=}p^2(U_Q(\hat{r})-U_Q(s)). $$

As a consequence,

$$\begin{aligned} \hat{T}(p\hat{r})= & {} 2\int _0^{p\hat{r}}\frac{dr}{\sqrt{2(U_Q(p\hat{r})-U_Q(r))}} =2p\int _0^{\hat{r}}\frac{ds}{\sqrt{2(U_Q(p\hat{r})-U_Q(ps))}} \\\ge & {} 2\int _0^{\hat{r}}\frac{ds}{\sqrt{2(U_Q(\hat{r})-U_Q(s))}} =\hat{T}(\hat{r}), \end{aligned}$$

which completes the proof.    $\square$

Corollary 3.16

Suppose that $\delta _1=\inf _{\mathring{D}}\omega _1=\min _D\omega _1$ is attained at some point $(\hat{e}, \hat{\beta })\in \partial D$. Then $(\hat{e}, \hat{\beta })$ lies on the ‘upper line’ $\{(e, \beta ): e=e_0, \beta \in [0, \beta _*]\}$ of the boundary.

Proof

This follows from $\omega _1=\frac{2\pi }{T_1}$ together with Lemmas 3.14 and 3.15.    $\square$

Remark 3.17

It will certainly be important to gain a better understanding of the monotonicity properties of $\omega _1$ (or, equivalently, $T_1$) in D. In particular, we expect that it should be significant to locate those points in D, where $\omega _1$ attains its minimum $\delta _1$. Some relations for $\frac{\partial T_1}{\partial e}$ and $\frac{\partial T_1}{\partial \beta }$ are stated in Lemma A.12(b), (c). For instance, we have

$$\begin{aligned} \frac{\partial T_1}{\partial \beta }(e, \beta ) =-\frac{1}{2}\,\frac{\partial }{\partial e}\int _0^{T_1(e, \beta )}\frac{ds}{r(s)^2} =-\frac{\partial }{\partial e}\int _{r_-(e, \beta )}^{r_+(e, \beta )}\frac{dr}{r^2 p_r}, \end{aligned}$$

(3.55)

which could provide a way to approach the monotonicity of $T_1$ in $\beta$. To see this, we apply the transformation $\rho =\sqrt{\beta }\,r^{-1}$, $d\rho =-\sqrt{\beta }\,r^{-2}\,dr$, like for the ‘apsidal angle’ [77]. Defining

$$ \tilde{U}(\rho , \beta )=\frac{1}{2}\,\rho ^2+U_Q\Big (\frac{\sqrt{\beta }}{\rho }\Big ), \quad \rho _{\mp }(e, \beta )=\frac{\sqrt{\beta }}{r_\pm (e, \beta )}, $$

and recalling that $p_r=\sqrt{2(e-U_Q(r)-\frac{\beta }{2r^2})}$, we get

$$ \frac{\partial T_1}{\partial \beta }(e, \beta ) =-\frac{1}{\sqrt{\beta }}\,\frac{\partial }{\partial e}\int _{\rho _-(e, \beta )}^{\rho _+(e, \beta )} \frac{d\rho }{\sqrt{2(e-\tilde{U}(\rho , \beta ))}}. $$

At fixed $\beta$, this has turned the integral on the right-hand side of (3.55) into the period function

$$ \tilde{T}(e)=\int _{\rho _-(e, \beta )}^{\rho _+(e, \beta )} \frac{d\rho }{\sqrt{2(e-\tilde{U}(\rho , \beta ))}} $$

for the transformed potential $\tilde{U}$; note that $0<\rho _-<\rho _+$ and $\tilde{U}(\rho _\pm , \beta )=e$. One could study the monotonicity of $\tilde{T}(e)$ in the energy e by checking the criteria that have been listed in the papers we mentioned in the proof of Lemma 3.15 or which can be found in similar works. Let us state a remarkable relation that could be useful in this respect. Writing $\tilde{U}(\rho )=\tilde{U}(\rho , \beta )$, it is calculated that

$$ \tilde{U}'(\rho )=-\frac{\sqrt{\beta }}{\rho ^2}\,U'_Q\Big (\frac{\sqrt{\beta }}{\rho }\Big )+\rho , \quad \tilde{U}''(\rho )=\frac{\beta }{\rho ^4}\,U''_Q\Big (\frac{\sqrt{\beta }}{\rho }\Big ) +\frac{2\sqrt{\beta }}{\rho ^3}\,U'_Q\Big (\frac{\sqrt{\beta }}{\rho }\Big )+1, $$

and using (1.13) this yields

$$\begin{aligned} \tilde{U}''(\rho )-\frac{\tilde{U}'(\rho )}{\rho }= & {} \frac{\beta }{\rho ^4}\,U''_Q\Big (\frac{\sqrt{\beta }}{\rho }\Big ) +\frac{3\sqrt{\beta }}{\rho ^3}\,U'_Q\Big (\frac{\sqrt{\beta }}{\rho }\Big ) \\= & {} \frac{\beta }{\rho ^4}\,\bigg [4\pi \rho _Q\Big (\frac{\sqrt{\beta }}{\rho }\Big ) -\frac{2\rho }{\sqrt{\beta }}\,U'_Q\Big (\frac{\sqrt{\beta }}{\rho }\Big )\bigg ] +\frac{3\sqrt{\beta }}{\rho ^3}\,U'_Q\Big (\frac{\sqrt{\beta }}{\rho }\Big ) \\= & {} \frac{\beta }{\rho ^4}\,\bigg [4\pi \rho _Q\Big (\frac{\sqrt{\beta }}{\rho }\Big ) +\frac{\rho }{\sqrt{\beta }}\,U'_Q\Big (\frac{\sqrt{\beta }}{\rho }\Big )\bigg ] \\= & {} \frac{\beta }{\rho ^4}\,B\Big (\frac{\sqrt{\beta }}{\rho }\Big ). \end{aligned}$$

In other words,

$$\begin{aligned} \bigg (\frac{\tilde{U}'(\rho )}{\rho }\bigg )'=\frac{\beta }{\rho ^5}\,B\Big (\frac{\sqrt{\beta }}{\rho }\Big ), \end{aligned}$$

and the function B is strictly positive. Comparing to the reasoning in Lemma 3.15, this looks promising for proving that $\tilde{T}(e)$ is increasing in e, i.e., that $\frac{\partial T_1}{\partial \beta }<0$. However, the argument does not seem to work properly, since the integral defining $\tilde{T}(e)$ is on $[\rho _-, \rho _+]$, instead of it beginning at zero, as is the case in Lemma 3.15. $\diamondsuit$

3.4 $\lambda _*\le \delta _1^2$

From (1.20), recall the definition of $\lambda _*$.

Lemma 3.18

We have $\lambda _*\le \delta _1^2$.

Proof

From (1.18), cf. Corollary B.19 and Lemma B.8(c), we deduce that, for $u\in X^2_{\mathrm{odd}}$,

$$\begin{aligned} {(Lu, u)}_Q= & {} \int _{\mathbb R^3}\int _{\mathbb R^3}\frac{dx\,dv}{|Q'(e_Q)|}\,|\mathcal{T}u|^2 -\frac{1}{4\pi }\int _{\mathbb R^3} |\nabla _x U_{\mathcal{T}u}|^2\,dx \\\le & {} \int _{\mathbb R^3}\int _{\mathbb R^3}\frac{dx\,dv}{|Q'(e_Q)|}\,|\mathcal{T}u|^2 ={\Vert \mathcal{T}u\Vert }^2_{X^0}=16\pi ^3\sum _{k\ne 0} k^2 {\Vert \omega _1 u_k\Vert }^2_{L^2_{\frac{1}{|Q'|}}(D)}. \end{aligned}$$

Since $u_{-k}=-u_k$ by Lemma B.3(b), this yields

$$\begin{aligned} \lambda _*\le 32\pi ^3\sum _{k=1}^\infty k^2 {\Vert \omega _1 u_k\Vert }^2_{L^2_{\frac{1}{|Q'|}}(D)} \end{aligned}$$

(3.56)

for all $u\in X^2_{\mathrm{odd}}$ such that ${\Vert u\Vert }_{X^0}={\Vert u\Vert }_Q=1$. Now we specialize (3.56) to $u\cong (\ldots , 0, u_{-1}, 0, u_1, 0, \ldots ) =(\ldots , 0, -u_1, 0, u_1, 0, \ldots )$ to find that

$$\begin{aligned} \lambda _*\le & {} 32\pi ^3 {\Vert \omega _1 u_1\Vert }^2_{L^2_{\frac{1}{|Q'|}}(D)} =32\pi ^3\iint \limits _D dI\,d\ell \,\ell \,\frac{1}{|Q'(e)|}\,\omega _1^2(I, \ell )\,|u_1(I, \ell )|^2 \nonumber \\= & {} 32\pi ^3\iint \limits _D de\,d\ell \,\ell \,\frac{1}{|Q'(e)|}\,\omega _1(e, \ell )\,|u_1(e, \ell )|^2 \end{aligned}$$

(3.57)

for all $u_1=u_1(I, \ell )=u_1(e, \ell )\in {L^2_{\frac{1}{|Q'|}}(D)}$ satisfying

$$\begin{aligned} 1= & {} 32\pi ^3\iint \limits _D dI\,d\ell \,\ell \,\frac{1}{|Q'(e)|}\,|u_1(I, \ell )|^2\\= & {} 32\pi ^3\iint \limits _D de\,d\ell \,\ell \,\frac{1}{|Q'(e)|}\,\frac{1}{\omega _1(e, \ell )}\,|u_1(e, \ell )|^2; \end{aligned}$$

see Definition B.1 and cf. (A.18). Let $\varepsilon >0$. Since $\delta _1=\inf _{\mathring{D}}\omega _1$, there is $(\hat{e}, \hat{l})\in \mathring{D}$ such that $\omega _1(\hat{e}, \hat{l})<\delta _1+\varepsilon /2$. As $\omega _1$ is continuous in $\mathring{D}$ by Theorem 3.6, there is an open neighborhood $U\subset \mathring{D}$ of $(\hat{e}, \hat{l})$ with the property that $\omega _1(e, l)<\delta _1+\varepsilon$ for $(e, l)\in U$; then $\iint \limits _U de\,d\ell \,\ell >0$. Define

and $u_1(e, \ell )=a\,|Q'(e)|^{1/2}\omega _1(e, \ell )^{1/2}\,\chi (e, \ell )$ for $a=(32\pi ^3\iint \limits _U de\,d\ell \,\ell )^{-1/2}$. It follows that

$$ 32\pi ^3\iint \limits _D de\,d\ell \,\ell \,\frac{1}{|Q'(e)|\omega _1}\,|u_1|^2 =32\pi ^3 a^2\iint \limits _U de\,d\ell \,\ell =1. $$

Thus, by (3.57),

$$\begin{aligned} \lambda _*\le & {} 32\pi ^3\iint \limits _D de\,d\ell \,\ell \,\frac{\omega _1}{|Q'|}\,|u_1|^2 =32\pi ^3 a^2\iint \limits _U de\,d\ell \,\ell \,\omega _1^2 \\\le & {} 32\pi ^3 a^2\,(\delta _1+\varepsilon )^2\iint \limits _U de\,d\ell \,\ell =(\delta _1+\varepsilon )^2. \end{aligned}$$

As $\varepsilon \rightarrow 0^+$, we get $\lambda _*\le \delta _1^2$.    $\square$