Closed 1/2-Elasticae in the 2-Sphere

Abstract

We study critical trajectories in the sphere for the 1/2-Bernoulli’s bending functional with length constraint. For every Lagrange multiplier encoding the conservation of the length during the variation, we show the existence of infinitely many closed trajectories which depend on a pair of relatively prime natural numbers. A geometric description of these numbers and the relation with the shape of the corresponding critical trajectories is also given.

Appendices

Appendix A. The Curvature of the Extrema and the Complete Elliptic Integral $\Psi _\lambda$

This appendix has two parts. In the first one we will show how to build the $\mu$-invariant from incomplete elliptic integrals of the third kind and how to compute its least period in terms of complete elliptic integrals. In the second part we will decompose the integral $\Psi _\lambda$ and compute its limits as $e_1$ approaches the boundaries of its domain.

1.1 Part I: The Curvature of the Extrema

Let $K(\phi ,\delta )$ and $\Pi (\zeta ,\phi ,\delta )$ be the Legendre’s incomplete elliptic integrals of the first and third kind, defined as

$$\begin{aligned}{} & {} K(\phi ,\delta )=\int _0^{\phi } \frac{1}{\sqrt{1-\delta \sin ^2 (\theta )}}\,\textrm{d}\theta ,\\{} & {} \Pi (\zeta ,\phi ,\delta )= \int _0^{\phi } \frac{1}{(1-\zeta \sin ^2(\theta ))\sqrt{1-\delta \sin ^2(\theta )}}\,\textrm{d}\theta , \end{aligned}$$

and $K(\delta )=K(\pi /2,\delta )$, $\Pi (\zeta ,\delta )=\Pi (\zeta , \pi /2, \delta )$ be the corresponding complete elliptic integrals. Let $\textrm{am}(u,\delta )$ be the Jacobi’s amplitude with parameter $\delta$ and $\textrm{sn}(u,\delta ) =\sin (\textrm{am}(u,\delta ))$ the associated Jacobi’s elliptic function. For simplicity, we denote by

$$\begin{aligned}{} & {} \alpha \equiv \alpha (\lambda ,e_1)=\frac{e_2-e_1}{e_2-e_4},\quad \beta \equiv \beta (\lambda ,e_1)=\frac{2}{\sqrt{(e_1-e_3)(e_2-e_4)}},\\{} & {} \qquad \qquad \qquad \delta \equiv \delta (\lambda ,e_1)=\frac{(e_1-e_2)(e_3-e_4)}{(e_1-e_3)(e_2-e_4)}, \end{aligned}$$

and

$$\begin{aligned} \zeta \equiv \zeta (\lambda ,e_1)=\frac{e_4(e_2-e_1)}{e_1(e_2-e_4)}, \end{aligned}$$

where $e_1>e_2>0$ and $e_3,e_4$ are the roots of the polynomial Q. Recall that $e_2, e_3$ and $e_4$ are functions of the fundamental parameters $\lambda$ and $e_1$. Let $\mu$ be the solution of (4) with $\mu (0)=e_2$ and $\omega >0$ be its least period. By construction, $\mu$ is strictly increasing on $[0,\omega /2]$ and $\mu (\omega /2)=e_1$. Let $h:[e_2,e_1]\rightarrow [0,\omega /2]$ be defined by

$$\begin{aligned} h(y)= \int _{e_2}^{y}\frac{1}{x\sqrt{-Q(x)}}\,\textrm{d}x = \frac{\omega }{2}-\int _y^{e_1}\frac{1}{x\sqrt{-Q(x)}}\,\textrm{d}x. \end{aligned}$$

Then, from (4) it follows that $\mu |_{[0,\omega /2]}=h^{-1}$. Since $\mu$ is even, this is enough to reconstruct $\mu$ on the whole real axis. Using 257.12 and 340.04 of Byrd and Friedman (1954), we obtain

$$\begin{aligned} \int _y^{e_1}\frac{1}{x\sqrt{-Q(x)}}\,\textrm{d}x=\frac{\beta }{e_1}\left( \frac{\alpha }{\zeta }u(y)-\frac{\alpha -\zeta }{\zeta }\Pi (\zeta ,\textrm{am}(u(y),\delta ), \delta )\right) , \end{aligned}$$

where

$$\begin{aligned} u(y)=\textrm{sn}^{-1}\left( \sqrt{\frac{(e_2-e_4)(e_1-y)}{(e_1-e_2)(y-e_4)}},\delta \right) . \end{aligned}$$

Putting $y=e_2$, we see that the least period of $\mu$ is

$$\begin{aligned} \omega = 2\frac{\beta }{e_1}\left( \frac{\alpha }{\zeta }K(\delta )-\frac{\alpha -\zeta }{\zeta }\Pi (\zeta ,\delta ) \right) . \end{aligned}$$

Figure 14 reproduces the graphs of the h-function and of the $\mu$-function on the intervals $[e_2,e_1]$ and $[0,\omega ]$, when $e_1=2$, $e_2=1$, $e_3=(-3+i\sqrt{23})/8$, $e_4=\overline{e_3}$. The h-function is evaluated via the Mathematica library of elliptic functions while the $\mu$ function is evaluated solving numerically (2) with initial conditions $\mu (0)=e_2$ and ${\dot{\mu }}(0)=0$. The black-dashed portion of the graph of $\mu$ on $[0,\omega /2]$ is obtained by symmetrizing the graph of h with respect to the bisector of the first quadrant, showing that the two methods are in agreement with each other.

Fig. 14

On the left: the graph of the h-function for $e_1=2$ and $e_2=1$. On the right: the graph of the $\mu$ function for the same values of $e_1$ and $e_2$

The 1/2-Bernoulli’s bending energy of a B-string can also be evaluated in terms of the wave number and complete elliptic integrals of the first kind as

$$\begin{aligned} {{\mathcal {B}}}_{\lambda }(\gamma _{m, n})=2n\int _{e_2}^{e_1}\frac{1}{\sqrt{-Q(y)}}\,\textrm{d}y+n\lambda \omega =n\left( 2\beta K(\delta )+\lambda \omega \right) . \end{aligned}$$

1.2 Part II: The Complete Elliptic Integral $\Psi _\lambda$

Using (4) to make a change of variable in the definition of $\Psi _\lambda$, (14), we have

$$\begin{aligned} \Psi _\lambda (e_1)= & {} 2\xi \int _0^\omega \frac{\mu ^2\left( \mu +2\lambda \right) }{1-4\xi ^2\mu ^2}\,\textrm{d}s\\= & {} 4\xi \int _{e_2}^{e_1}\frac{\mu \left( \mu +2\lambda \right) }{(1-4\xi ^2\mu ^2)\sqrt{-(\mu -e_1)(\mu -e_2)(\mu -e_3)(\mu -e_4)}}\,\textrm{d}\mu . \end{aligned}$$

This integral can be solved in terms of complete elliptic integrals of the first and third kind.

For simplicity, we denote by

$$\begin{aligned}{} & {} \zeta _+\equiv \zeta _+(\lambda ,e_1)=-\frac{(e_1-e_2)(-1+2e_4\xi )}{(e_2-e_4)(-1+2e_1\xi )}\,,\\{} & {} \zeta _-\equiv \zeta _-(\lambda ,e_1)=-\frac{(e_1-e_2)(1+2e_4\xi )}{(e_2-e_4)(1+2e_1\xi )}\,. \end{aligned}$$

We then have

$$\begin{aligned} \Psi _\lambda =2\pi \left( I+[1-\chi ] II+III\right) , \end{aligned}$$

where $\chi$ is the indicator function of the exceptional locus ${\mathcal {P}}_*$ and,

$$\begin{aligned} I= & {} \frac{\beta }{4\pi \xi }\left( -2+\alpha \left[ \frac{-1-4\lambda \xi }{\zeta _+(-1+2e_1\xi )}+\frac{1-4\lambda \xi }{\zeta _-(1+2e_1\xi )}\right] \right) K(\delta )\,,\end{aligned}$$

(19)

$$\begin{aligned} II= & {} \frac{\beta (\alpha -\zeta _+)(1+4\lambda \xi )}{4\pi \zeta _+\xi (-1+2e_1\xi )}\,\Pi (\zeta _+,\delta )\,,\end{aligned}$$

(20)

$$\begin{aligned} III= & {} \frac{\beta (\alpha -\zeta _-)(-1+4\lambda \xi )}{4\pi \zeta _-\xi (1+2e_1\xi })\,\Pi (\zeta _-,\delta )\,. \end{aligned}$$

(21)

These formulas follow from three standard elliptic integrals. The first one (cf. 340.01 and 341.03 of Byrd and Friedman 1954) is

$$\begin{aligned} \int _0^{K(\delta )}\frac{1-a\, \textrm{sn}^2(u,\delta )}{1-b\, \mathrm{sn^2}(u,\delta )}\,\textrm{d}u=\frac{a}{b}K(\delta )-\frac{a-b}{b}\Pi (b,\delta ). \end{aligned}$$

The second elliptic integral (cf. 257 and 259 of Byrd and Friedman (1954)) is

$$\begin{aligned} \int _{e_2}^{e_1}\frac{\textrm{d}\mu }{\sqrt{-(\mu -e_1)(\mu -e_2)(\mu -e_3)(\mu -e_4)}}=\beta K(\delta )\,, \end{aligned}$$

where $\beta$ and $\delta$ are as above. The third relevant elliptic integral (cf. 257.39 and 259.04 of Byrd and Friedman 1954) is

$$\begin{aligned}{} & {} \int _{e_2}^{e_1}\frac{\textrm{d}\mu }{(p-\mu )\sqrt{-(\mu -e_1)(\mu -e_2)(\mu -e_3)(\mu -e_4)}}\\{} & {} \quad =\frac{\beta }{p-e_1}\int _0^{K(\delta )}\frac{1-\alpha \, \textrm{sn}^2(u,\delta )}{1-{\widetilde{\zeta }}\,\mathrm{sn^2}(u,\delta )}\,\textrm{d}u\\{} & {} \quad =\frac{\beta }{p-e_1}\left( \frac{\alpha }{{\widetilde{\zeta }}}K(\delta )-\frac{\alpha -{\widetilde{\zeta }}}{{\widetilde{\zeta }}}\Pi ({\widetilde{\zeta }},\delta )\right) , \end{aligned}$$

where $p\ne e_1$, $\alpha$, $\beta$ and $\delta$ are as above, and

$$\begin{aligned} {\widetilde{\zeta }}=\frac{(p-e_4)(e_1-e_2)}{(e_1-p)(e_2-e_4)}. \end{aligned}$$

We begin proving that $\Psi _\lambda \rightarrow 2\pi p(\lambda )$ when $e_1$ approaches $\eta _\lambda$ from the right. By construction we have that $e_2\rightarrow \eta _\lambda$ too, and, hence, it follows that

$$\begin{aligned} \left\{ \begin{aligned}&\lim _{e_1\rightarrow \eta _\lambda ^+} e_3(\lambda ,e_1)=\frac{-1+\sqrt{1-\eta _\lambda ^4}}{\eta _\lambda ^3} \\&\lim _{e_1\rightarrow \eta _\lambda ^+} e_4(\lambda ,e_1)=\frac{-1-\sqrt{1-\eta _\lambda ^4}}{\eta _\lambda ^3} \end{aligned}\right. , \end{aligned}$$

and

$$\begin{aligned} \lambda =\frac{1-\eta _\lambda ^4}{2\eta _\lambda ^3}\,. \end{aligned}$$

We now see that the coefficients in (19)–(21) tend to, respectively,

$$\begin{aligned} \lim _{e_1\rightarrow \eta _\lambda ^+} \alpha (\lambda ,e_1)=0\,,\quad \quad \quad \lim _{e_1\rightarrow \eta _\lambda ^+} \beta (\lambda ,e_1)=\frac{2\eta _\lambda }{\sqrt{3+\eta _\lambda ^4}}\,,\quad \quad \quad \lim _{e_1\rightarrow \eta _\lambda ^+} \delta (\lambda ,e_1)=0\,, \end{aligned}$$

while

$$\begin{aligned} \lim _{e_1\rightarrow \eta _\lambda ^+} \xi (\lambda ,e_1)=\frac{\sqrt{1+\eta _\lambda ^4}}{2\eta _\lambda ^3}\,,\quad \quad \quad \lim _{e_1\rightarrow \eta _\lambda ^+} \zeta _+(\lambda ,e_1)=0\,,\quad \quad \quad \lim _{e_1\rightarrow \eta _\lambda ^+} \zeta _-(\lambda ,e_1)=0\,, \end{aligned}$$

and

$$\begin{aligned}{} & {} \lim _{e_1\rightarrow \eta _\lambda ^+} \frac{\alpha }{\zeta _+}=\frac{\eta _\lambda ^4\left( \eta _\lambda ^2-\sqrt{1+\eta _\lambda ^4}\right) }{\sqrt{1+\eta _\lambda ^4}+\sqrt{1-\eta _\lambda ^8}+\eta _\lambda ^6},\,\,\\{} & {} \lim _{e_1\rightarrow \eta _\lambda ^+} \frac{\alpha }{\zeta _-}=-\frac{\eta _\lambda ^4\left( \eta _\lambda ^2+\sqrt{1+\eta _\lambda ^4}\right) }{\sqrt{1+\eta _\lambda ^4}+\sqrt{1-\eta _\lambda ^8}-\eta _\lambda ^6}\,. \end{aligned}$$

Finally, recalling that $K(0)=\Pi (0,0)=\pi /2$, using the above limits and (15) we conclude that

$$\begin{aligned} \lim _{e_1\rightarrow \eta _\lambda ^+} \Psi _\lambda =-2\pi \sqrt{\frac{1+\eta _\lambda ^4}{3+\eta _\lambda ^4}}=2\pi p(\lambda ). \end{aligned}$$

(22)

In what follows, we prove the limit when $e_1\rightarrow \infty$. This limit will depend on the sign of $\lambda$. More precisely, we will see that

$$\begin{aligned} \lim _{e_1\rightarrow \infty } I(\lambda ,e_1)=\lim _{e_1\rightarrow \infty } III(\lambda ,e_1)=0\,, \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{aligned}&\lim _{e_1\rightarrow \infty } II(\lambda ,e_1)=-\frac{1}{2},&\quad \text {if}\,\lambda \ge 0\\&\lim _{e_1\rightarrow \infty } II(\lambda ,e_1)=\frac{1}{2},&\quad \text {if}\,\lambda <0 \end{aligned}\right. \,. \end{aligned}$$

In order to prove these limits we observe that, as $e_1\rightarrow \infty$, the following asymptotic estimates hold true:

$$\begin{aligned}&e_2\sim 1/e_1\,,\quad e_3\sim -1/e_1\,,\quad e_4\sim -e_1\,,\quad \xi \sim e_1/2\,,\quad \beta \sim 2/e_1\,,\\&\delta \sim 1-4/e_1^2\,,\quad \alpha \sim (1-e_1^2)/(1+e_1^2)\,,\quad \zeta _-\sim 1-4/e_1^2\,, \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{aligned}&\zeta _+\sim 1-\frac{4\lambda ^2}{e_1^4},&\quad \text {if}\,\lambda \ne 0\\&\zeta _+\sim 1-\frac{1}{e_1^6},&\quad \text {if}\,\lambda =0 \end{aligned}\right. . \end{aligned}$$

(23)

Moreover, recall that $K(\delta )\sim -\log (1-\delta )/2$ as $\delta \rightarrow 1^-$ and so, in our case, we have

$$\begin{aligned} K\left( \delta (\lambda ,e_1)\right) \sim \frac{-1}{2}\log \frac{4}{e_1^2}, \end{aligned}$$

as $e_1\rightarrow \infty$. Combining this and the above estimates we conclude that

$$\begin{aligned} I(\lambda ,e_1)\sim \frac{1}{\pi e_1^2}\log \frac{4}{e_1^2}, \end{aligned}$$

as $e_1\rightarrow \infty$. This proves the first limit.

For the other limits we need some basic properties of the complete elliptic integral of the third kind $\Pi (\zeta ,\delta )$. Let $\Lambda =\{(\zeta ,\delta )\in [0,1)\times [0,1)\,|\,\zeta \ge \delta \}$ and consider the function

$$\begin{aligned} f:(\zeta ,\delta )\in \Lambda \longmapsto \frac{2}{\pi }\sqrt{1-\zeta }\sqrt{1-\delta }\,\Pi (\zeta ,\delta )\in {\mathbb {R}}\,. \end{aligned}$$

This function f is bounded below by $2/\pi$ and above by 1. In addition, $f(\zeta ,0)=1$ for every $\zeta \in [0,1)$ and $f(\zeta ,\delta )\rightarrow 1$ when $\zeta \rightarrow 1^-$, for every value of $\delta \in [0,1)$. Moreover, for every $\zeta \in [0,1)$,

$$\begin{aligned} f(\zeta ,\zeta )=\frac{2}{\pi }\sqrt{1-\zeta }\,E\left( \frac{\zeta }{\zeta -1}\right) \end{aligned}$$

where E is the complete elliptic integral of the second kind. From this we infer that

$$\begin{aligned} \lim _{\zeta \rightarrow 1^-} f(\zeta ,\zeta )=\frac{2}{\pi }\,. \end{aligned}$$

From these properties we deduce the following facts:

• 1. If $\gamma :(a,\infty )\longrightarrow \Lambda$ is a smooth curve such that $\gamma (t)\rightarrow (1,1)$ when $t\rightarrow \infty$ and $\zeta =\delta$ is an asymptote of $\gamma$ as $t\rightarrow \infty$, then

  $$\begin{aligned} \lim _{t\rightarrow \infty }f(\gamma (t))=\frac{2}{\pi }\,. \end{aligned}$$

• 2. If ${\widetilde{\gamma }}:(a,\infty )\longrightarrow \Lambda$ is a smooth curve such that ${\widetilde{\gamma }}(t)\rightarrow (1,1)$ when $t\rightarrow \infty$ and $\zeta =1$ is an asymptote of ${\widetilde{\gamma }}$ as $t\rightarrow \infty$, then

  $$\begin{aligned} \lim _{t\rightarrow \infty }f({\widetilde{\gamma }}(t))=1\,. \end{aligned}$$

Combining both things, it follows that as $t\rightarrow \infty$,

$$\begin{aligned} \left\{ \begin{aligned}&\Pi (\gamma (t))\sim \frac{1}{\sqrt{1-\gamma _1(t)}\sqrt{1-\gamma _2(t)}}\\&\Pi ({\widetilde{\gamma }}(t))\sim \frac{\pi }{2\sqrt{1-{\widetilde{\gamma }}_1(t)}\sqrt{1-{\widetilde{\gamma }}_2(t)}} \end{aligned}\right. \,. \end{aligned}$$

In view of these properties, fix $\tau _\lambda$ sufficiently large and consider the curves

$$\begin{aligned} \left\{ \begin{aligned}&\gamma _\lambda :e_1\in (\tau _\lambda ,\infty )\longmapsto (\zeta _-(\lambda ,e_1),\delta (\lambda ,e_1))\in \Lambda \\&{\widetilde{\gamma }}_\lambda :e_1\in (\tau _\lambda ,\infty )\longmapsto (\zeta _+(\lambda ,e_1),\delta (\lambda ,e_1))\in \Lambda \end{aligned}\right. \,. \end{aligned}$$

From above estimates when $e_1\rightarrow \infty$, we have the following asymptotic behavior for $\gamma _\lambda$,

$$\begin{aligned} \gamma _\lambda \sim \left( 1-\frac{4}{e_1^2},1-\frac{4}{e_1^2}\right) , \end{aligned}$$

while for ${\widetilde{\gamma }}_\lambda$, it depends on the value of $\lambda$,

$$\begin{aligned} \left\{ \begin{aligned}&{\widetilde{\gamma }}_\lambda \sim \left( 1-\frac{4\lambda ^2}{e_1^4},1-\frac{4}{e_1^2}\right) ,&\quad \text {if}\,\lambda \ne 0\\&{\widetilde{\gamma }}_0\sim \left( 1-\frac{1}{e_1^6},1-\frac{4}{e_1^2}\right) ,&\quad \text {if}\,\lambda =0 \end{aligned}\right. \,. \end{aligned}$$

Thus, $\gamma _\lambda \rightarrow (1,1)$ and $\zeta =\delta$ is an asymptote of $\gamma _\lambda$ as $e_1\rightarrow \infty$. Similarly, ${\widetilde{\gamma }}_\lambda \rightarrow (1,1)$ as $e_1\rightarrow \infty$. Hence, it follows from above facts that

$$\begin{aligned} \left\{ \begin{aligned}&\Pi \left( \zeta _-(\lambda ,e_1),\delta (\lambda ,e_1)\right) \sim \frac{e_1^2}{e_1^2-4}\\&\Pi \left( \zeta _+(\lambda \ne 0,e_1),\delta (\lambda \ne 0,e_1)\right) \sim \frac{\pi e_1^3}{8|\lambda |}\\&\Pi \left( \zeta _+(\lambda =0,e_1),\delta (\lambda =0,e_1)\right) \sim \frac{\pi e_1^4}{4}\\ \end{aligned}\right. \,, \end{aligned}$$

when $e_1\rightarrow \infty$.

It is then clear, combining this and above estimates, that

$$\begin{aligned} \left\{ \begin{aligned}&\lim _{e_1\rightarrow \infty }III(\lambda ,e_1)=0\\&\lim _{e_1\rightarrow \infty }II(\lambda \ge 0,e_1)=-\frac{1}{2}\\&\lim _{e_1\rightarrow \infty }II(\lambda <0,e_1)=\frac{1}{2}\\ \end{aligned}\right. \,. \end{aligned}$$

This completes the proof about the claimed limits for $\Psi _\lambda$ when $e_1\rightarrow \infty$.

Appendix B. Closed 1/2-Elasticae in the Plane

We briefly comment about 1/2-elasticae in ${{\mathbb {R}}}^2$ in order to clarify some assertions made in the Introduction. We begin with the non-existence of closed convex 1/2-elasticae other than circles. In ${{\mathbb {R}}}^2$ the phase curves for convex 1/2-elasticae are the singular rational curves (see the picture on the left of Fig. 16)

$$\begin{aligned} {{\mathcal {C}}}_{e_1,e_2}\,:\, y^2+x^4\left( x^2-[e_1+e_2]x + e_1e_2\right) =0\,, \end{aligned}$$

where $e_1>e_2>0$. Then, following the general argument of the Introduction, $\mu$ is a solution of ${\dot{\mu }}^2+\mu ^4(\mu ^2-[e_1+e_2]\mu +e_1e_2)=0$ and $(\mu ,{\dot{\mu }})$ is a periodic, regular parameterization of the smooth component of the phase curve lying in the positive half-plane ${{\mathbb {H}}}^2=\{(x,y)\,|\, x>0\}$.

Integrating by quadratures, the arc-length parameterization of a critical curve, up to rigid motions, is

$$\begin{aligned} \gamma _{\lambda ,d}(s)=\sqrt{\frac{1}{d}}\left( \lambda s +\frac{1}{2}\int \mu (s)\textrm{d}s,-\frac{1}{2\mu (s)}\right) . \end{aligned}$$

Let $\omega$ be the least period of $\mu$. Then,

$$\begin{aligned} \gamma (\omega )-\gamma (0)=\sqrt{\frac{1}{d}}\left( \lambda \omega +\frac{1}{2}\int _0^{\omega }\mu (s)\textrm{d}s,0\right) . \end{aligned}$$

On the other hand,

$$\begin{aligned} \omega = 2\int _{e_2}^{e_1}\frac{\textrm{d}\mu }{\mu ^2\sqrt{-(\mu -e_1)(\mu -e_2)}}=\pi \frac{e_1+e_2}{(e_1e_2)^{3/2}}\,, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{2}\int _0^{\omega }\mu (s)\textrm{d}s = \int _{e_2}^{e_1}\frac{\textrm{d}\mu }{\mu \sqrt{-(\mu -e_1)(\mu -e_2)}}=\frac{\pi }{\sqrt{e_1e_2}}\,. \end{aligned}$$

Hence

$$\begin{aligned} \gamma (\omega )-\gamma (0)=-\frac{\pi (e_1-e_2)}{(e_1e_2)^{3/2}}(1,0). \end{aligned}$$

This implies that the trajectory of $\gamma _{\lambda ,d}$ is invariant by the subgroup generated by a non-trivial translation along the Ox-axis. In particular, it is unbounded.

Next we focus on the non-existence of non-convex critical curves with periodic curvature. In this case $e_1>0>e_2$. By contradiction, suppose that $\kappa$ is periodic and non-constant. Without loss of generality $\kappa (0)=\textrm{max}(\kappa )>0$. Let $J=(a,b)$, $a<0<b$ be the connected component of $\{s\in {{\mathbb {R}}}\,|\, \kappa (s)>0\}$ containing the origin. Since $\kappa$ is not strictly positive, at least one among a or b is finite. Put $\mu =\sqrt{\kappa |_{J}}$. Then, $m:s\in J\rightarrow (\mu ,\mu ')\in {{\mathbb {H}}}^2$ is a parameterization of an open arc contained in ${{\mathcal {C}}}^+_{e_1,e_2}:= {{\mathcal {C}}}_{e_1,e_2}\cap {{\mathbb {H}}}^2$ (see the picture on the right of Fig. 15; ${{\mathcal {C}}}^+_{e_1,e_2}$ is represented in the black part). By construction, $\mu (0)=e_1$ and, in addition, there exist $\epsilon >0$ such that $\mu '>0$ on $(-\epsilon ,0)$ and $\mu '<0$ on $(0,\epsilon )$. Taking into account that $\mu$ is a solution of the Euler–Lagrange equation, m is an integral curve of the vector field

$$\begin{aligned} \textbf{X}|_{(x,y)}=y\,\partial _{y}+\frac{1}{x}\left( 2y^2+\frac{1}{2}[e_1+e_2]x^5-x^6\right) \partial _{y}\in {{\mathfrak {X}}}({{\mathbb {H}}}^2)\,. \end{aligned}$$

Fig. 15

Phase curves of convex planar 1/2-elasticae. On the left, $e_1=2$ and $e_2=1$; while, on the right, $e_1=3$ and $e_2=-2$ (Color Figure Online)

Fig. 16

On the left, the functions $h_+$ (black) and $h_-$ (red). On the right, the graph of the $\mu$-invariant of a planar 1/2-elastic curve with $e_1=3$ and $e_2=-2$ (Color Figure Online)

Thus, m(s) cannot invert his motion along ${{\mathcal {C}}}^+_{e_1,e_2}$. Since ${{\mathcal {C}}}^+_{e_1,e_2}$ is homeomorphic to ${{\mathbb {R}}}$, this implies that m is a homeomorphism onto its image. Therefore, m((a, b)) intersects the Ox-axis at one point, namely at $m(0)=(e_1,0)$. Hence, $\mu '>0$ on (a, 0) and $\mu '<0$ on (0, b). Let $h_{+}$ and $h_{-}$ be the inverses of $\mu |_{(a,0)}$ and $\mu |_{(0,b)}$ respectively (see the picture on the left of Fig. 16; the graph of $h_+$ is shown in black while the graph of $h_-$ in red). Then,

$$\begin{aligned} \begin{aligned} h_{\pm }(m)&=\int _{e1}^{m}\frac{1}{\mu ^2\sqrt{-\mu ^2+(e_1+e_2)\mu -e_1e_2}}\textrm{d}\mu \\&\quad = \mp \frac{1}{(e_1^2e_2^2)^{3/4}m}\\&\qquad \times \, \left( \sqrt{e_1|e_2|(e_1-m)(m-e_2)} -(e_1+e_2)m\,\textrm{arctanh} \sqrt{\frac{|e_2|(e_1-m)}{e_1(m-e_2)}} \right) . \end{aligned}\end{aligned}$$

Consequently, $\mu$ can be extended to a function ${\hat{\mu }}:{{\mathbb {R}}}\rightarrow (0,e_1]$ attaining its maximum at $s=0$, strictly increasing on $(-\infty ,0)$ and strictly decreasing on $(0,+\infty )$ (see the picture on the right of Fig. 16) such that

$$\begin{aligned} \lim _{s \rightarrow \infty }{\hat{\mu }}(s)= \lim _{s \rightarrow +\infty }{\hat{\mu }}(s)=0\,. \end{aligned}$$

This implies $a=-\infty$ and $b=+\infty$, which is a contradiction.