Introduce a Cartesian frame *Oxyz* such that the axis *Z* is directed down along the cone axis and the origin *O* is cone’s vertex. Introduce also a cylindrical coordinate frame such that
$$\begin{array}{}{\displaystyle (z,r,\psi ),\phantom{\rule{1em}{0ex}}x=r\mathrm{cos}\psi ,\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}\psi .}\end{array}$$

The cone is given by the equation
$$\begin{array}{}{\displaystyle z=ar,\phantom{\rule{1em}{0ex}}a=\mathrm{cot}\alpha >0.}\end{array}$$

Let
$$\begin{array}{}{\displaystyle r=r(\psi ),\phantom{\rule{1em}{0ex}}r(\psi +2\pi )=r(\psi )}\end{array}$$

be the equation of the curve described by the chain. Infinitesimal arclength element of such a curve is as follows
$$\begin{array}{c}{\displaystyle ds=\sqrt{d{x}^{2}+d{y}^{2}+d{z}^{2}}=\sqrt{(1+{a}^{2})({r}^{\prime}{)}^{2}+{r}^{2}}d\psi ,}\\ {\displaystyle {r}^{\prime}=\frac{d}{d\psi}r(\psi ).}\end{array}$$

Then *Z*–coordinate of the centre of mass of the chain is expressed by the formula.
$$\begin{array}{}{\displaystyle Z[r(\cdot )]=\frac{a}{l}\underset{0}{\overset{2\pi}{\int}}r(\psi )ds.}\end{array}$$

So we are looking for conditional extremals of the functional *r*(⋅)↦ *Z*[*r*(⋅)] in class of 2*π*–periodic functions *r*(ψ) under the condition
$$\begin{array}{}{\displaystyle \underset{0}{\overset{2\pi}{\int}}ds=l.}\end{array}$$(2.1)

The corresponding Lagrange function is
$$\begin{array}{}{\displaystyle L(r,{r}^{\prime})=(\frac{a}{l}r+\lambda )\sqrt{(1+{a}^{2})({r}^{\prime}{)}^{2}+{r}^{2}},}\end{array}$$
with a Lagrange multiplier *λ* [1].

Since the Lagrangian *L* does not depend on *ψ* we have the following integral of the Lagrange equations
$$\begin{array}{}{\displaystyle h={r}^{\prime}\frac{\mathrm{\partial}L}{\mathrm{\partial}{r}^{\prime}}-L,}\end{array}$$

or
$$\begin{array}{}{\displaystyle -(\frac{a}{l}r+\lambda ){r}^{2}=h\sqrt{(1+{a}^{2})({r}^{\prime}{)}^{2}+{r}^{2}}.}\end{array}$$(2.2)

So that the constant *h* and the expression $\begin{array}{}{\displaystyle -(\frac{a}{l}r+\lambda )}\end{array}$ must have the same sign for all *ψ* ∈ ℝ i.e.
$$\begin{array}{}{\displaystyle \mathrm{s}\mathrm{g}\mathrm{n}\phantom{\rule{thinmathspace}{0ex}}h=-\mathrm{s}\mathrm{g}\mathrm{n}\phantom{\rule{thinmathspace}{0ex}}(\frac{a}{l}r+\lambda ).}\end{array}$$(2.3)

Under this assumption (we justify this assumption in the sequel) take square from both sides of (2.2)
$$\begin{array}{}{\displaystyle (1+{a}^{2})({r}^{\prime}{)}^{2}+{r}^{2}-\frac{1}{{h}^{2}}(\frac{a}{l}r+\lambda {)}^{2}{r}^{4}=0,}\end{array}$$

and perform a change of variables *r* ↦ *ρ* by the formula
$$\begin{array}{}{\displaystyle r=\frac{\lambda l}{a}\rho ,\phantom{\rule{1em}{0ex}}\rho =\rho (\psi )}\end{array}$$(2.4)

to obtain
$$\begin{array}{}{\displaystyle (1+{a}^{2})({\rho}^{\prime}{)}^{2}+{\rho}^{2}-{u}^{2}(\rho +1{)}^{2}{\rho}^{4}=0,}\end{array}$$(2.5)

where
$$\begin{array}{}{\displaystyle {u}^{2}=\frac{{\lambda}^{4}{l}^{2}}{{h}^{2}{a}^{2}}.}\end{array}$$(2.6)

The condition (2.1) takes the form
$$\begin{array}{}{\displaystyle \underset{0}{\overset{2\pi}{\int}}\sqrt{(1+{a}^{2})({\rho}^{\prime}{)}^{2}+{\rho}^{2}}d\psi =\frac{a}{|\lambda |}.}\end{array}$$(2.7)

Our plan is as follows. Choosing a constant *u*^{2} we will find 2*π*–solution *ρ*(*ψ*) of equation (2.5) and then take |*λ*| to satisfy (2.7). By reason that will be clear in the sequel we assume that
$$\begin{array}{}{\displaystyle \lambda <0.}\end{array}$$(2.8)

Then already known *u*^{2} and |*λ*| we will substitute to (2.6) and find a constant *h*^{2}.

The solution *ρ* must also be such that the expression
$$\begin{array}{}{\displaystyle (\frac{a}{l}r+\lambda )=\lambda (\rho +1)}\end{array}$$

is sign-definite for all *ψ*. If the solution *ρ*(*ψ*) provides this condition then (2.3) is satisfied by choosing the sign of *h*.

Introducing new variable $\begin{array}{}{\displaystyle t=\psi /\sqrt{{a}^{2}+1},}\end{array}$ rewrite equation (2.5) as follows
$$\begin{array}{}{\displaystyle \frac{1}{(\rho +1{)}^{2}{\rho}^{4}}{\dot{\rho}}^{2}+\frac{1}{{\rho}^{2}(\rho +1{)}^{2}}={u}^{2},\phantom{\rule{1em}{0ex}}\dot{\rho}=\frac{d\rho}{dt}.}\end{array}$$(2.9)

Now we are looking for the solution *ρ*(*t*) of equation (2.9) that has period
$$\begin{array}{}{\displaystyle \frac{2\pi}{\sqrt{{a}^{2}+1}}.}\end{array}$$

Equation (2.9) has the form of energy integral of a classical mechanical system with kinetic energy
$$\begin{array}{}{\displaystyle E=\frac{1}{(\rho +1{)}^{2}{\rho}^{4}}{\dot{\rho}}^{2}}\end{array}$$

and a potential energy
$$\begin{array}{}{\displaystyle V(\rho )=\frac{1}{{\rho}^{2}(\rho +1{)}^{2}}.}\end{array}$$
Use this observation to analyze system (2.9).

Having a graph of the function *V* we see that the point *C* = (−1/2,0) on the phase plane (*ρ*,$\begin{array}{}{\displaystyle \dot{\rho}}\end{array}$) is an equilibrium of the type ”centre”. This equilibrium is surrounded with closed curves that are squeezed between vertical lines *ρ* = −1 and *ρ* = 0. These trajectories are marked with parameter *u*^{2}. There are no periodic orbits in other domains of the phase space.

Therefore all the periodic solutions satisfy the condition
$$\begin{array}{}{\displaystyle \rho (t)\in (-1,0),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}}\end{array}$$

consequently one has *ρ*(*t*)+1 > 0. By virtue of formulas (2.8) and (2.4) it follows that *r* > 0.

The equilibrium *C* corresponds to the value *u*^{2} = 16 and gives the trivial equilibrium of the chain.

Separating variables in (2.9), find period of the trajectory *ρ*(*t*) by the formula:
$$\begin{array}{}{\displaystyle T(u)=-2\underset{{\rho}_{-}}{\overset{{\rho}_{+}}{\int}}\frac{d\rho}{\rho \sqrt{{u}^{2}(\rho +1{)}^{2}{\rho}^{2}-1}},\phantom{\rule{1em}{0ex}}{\rho}_{\pm}=\frac{-1\pm \sqrt{1-4/u}}{2}.}\end{array}$$

The function *T*(*u*) is defined for *u* > 4. It is not hard to show that the function *T* is continuous and decreased.

Thus each root *u* of the equation
$$\begin{array}{}{\displaystyle T(u)=\frac{2\pi}{\sqrt{1+{a}^{2}}}}\end{array}$$

corresponds to the solution of (2.9) with desired period. This solution gives the oblique equilibrium.

The proof of the theorem is concluded by a lemma.

#### Lemma 1

*The following formulas hold*

$$\begin{array}{}{\displaystyle \underset{u\to 4+}{lim}T(u)=\sqrt{2}\pi ,\phantom{\rule{1em}{0ex}}\underset{u\to \mathrm{\infty}}{lim}T(u)=\pi .}\end{array}$$(2.10)

Indeed, the oblique equilibrium shows up iff
$$\begin{array}{}{\displaystyle \pi <\frac{2\pi}{\sqrt{1+{a}^{2}}}<\sqrt{2}\pi ,}\end{array}$$

or
$$\begin{array}{}{\displaystyle \pi /6<\alpha <\pi /4.}\end{array}$$

The theorem is proved.

## 2.1 Proof of the Lemma

The first one of the formulas (2.10) is the most simple; we prove it by using a little bit of informal mechanical argument. Rigorous proof of this formula follows by the same manner as we employ below to prove the second formula of (2.10).

Up to third order terms in the neighbourhood of the point *C* equation (2.9) has the form
$$\begin{array}{}{\displaystyle {\dot{\xi}}^{2}+2{\xi}^{2}=const,\phantom{\rule{1em}{0ex}}\rho =-\frac{1}{2}+\xi .}\end{array}$$

So that the period of small oscillations is equal to $\begin{array}{}{\displaystyle 2\pi /\sqrt{2}}\end{array}$ or
$$\begin{array}{}{\displaystyle \underset{u\to 4+}{lim}T(u)=\sqrt{2}\pi .}\end{array}$$

Let us check the second formula. Observe that
$$\begin{array}{}{\displaystyle T(u)=-\frac{2}{u}\underset{{\rho}_{-}}{\overset{{\rho}_{+}}{\int}}\frac{d\rho}{\rho \sqrt{(\rho -{\rho}_{-})(\rho -{\rho}_{+})(\rho -{\stackrel{~}{\rho}}_{-})(\rho -{\stackrel{~}{\rho}}_{+})}},}\end{array}$$

where
$$\begin{array}{}{\displaystyle {\stackrel{~}{\rho}}_{\pm}=\frac{-1\pm \sqrt{1+4/u}}{2}.}\end{array}$$

Introducing a small parameter *ε* = 1/*u* → 0 as *u* → ∞, we get
$$\begin{array}{}{\displaystyle {\rho}_{+}=-\u03f5+O({\u03f5}^{2}),\phantom{\rule{2em}{0ex}}{\rho}_{-}=-1+\u03f5+O({\u03f5}^{2}),}\\ {\displaystyle {\stackrel{~}{\rho}}_{+}=\u03f5+O({\u03f5}^{2}),\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{\rho}}_{-}=-1-\u03f5+O({\u03f5}^{2}).}\end{array}$$

For small *ε* the following inequalities hold
$$\begin{array}{}{\displaystyle {\rho}_{+}<-\u03f5,\phantom{\rule{1em}{0ex}}{\rho}_{-}>-1+\u03f5.}\end{array}$$

It is easy to see that
$$\begin{array}{}{\displaystyle T(u)=-2\u03f5\underset{-\frac{1}{2}}{\overset{{\rho}_{+}}{\int}}\frac{d\rho}{\rho \sqrt{(\rho -{\rho}_{-})(\rho -{\rho}_{+})(\rho -{\stackrel{~}{\rho}}_{-})(\rho -{\stackrel{~}{\rho}}_{+})}}+O(\sqrt{\u03f5}).}\end{array}$$

Transform this integral as follows:
$$\begin{array}{}{\displaystyle \underset{-\frac{1}{2}}{\overset{{\rho}_{+}}{\int}}\frac{d\rho}{\rho \sqrt{(\rho -{\rho}_{-})(\rho -{\rho}_{+})(\rho -{\stackrel{~}{\rho}}_{-})(\rho -{\stackrel{~}{\rho}}_{+})}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\underset{-\frac{1}{2}}{\overset{{\rho}_{+}}{\int}}\frac{(1+O(\u03f5))\phantom{\rule{thinmathspace}{0ex}}d\rho}{\rho (\rho +1)\sqrt{(\rho -{\rho}_{+})(\rho +{\rho}_{+}-{\rho}_{+}-{\stackrel{~}{\rho}}_{+})}}.}\end{array}$$

Sinse $\begin{array}{}{\displaystyle -{\rho}_{+}-{\stackrel{~}{\rho}}_{+}=O({\u03f5}^{2})}\end{array}$ the last integral equals
$$\begin{array}{}{\displaystyle \underset{-\frac{1}{2}}{\overset{{\rho}_{+}}{\int}}\frac{(1+O(\u03f5))\phantom{\rule{thinmathspace}{0ex}}d\rho}{\rho (\rho +1)\sqrt{{\rho}^{2}-{\rho}_{+}^{2}}}.}\end{array}$$

The integral
$$\begin{array}{}{\displaystyle \underset{-\frac{1}{2}}{\overset{{\rho}_{+}}{\int}}\frac{\phantom{\rule{thinmathspace}{0ex}}d\rho}{\rho (\rho +1)\sqrt{{\rho}^{2}-{\rho}_{+}^{2}}}}\end{array}$$

is computed explicitly, nevertheless the formula is very large and we do not bring it; write down the asymptotic
$$\begin{array}{}{\displaystyle \underset{-\frac{1}{2}}{\overset{{\rho}_{+}}{\int}}\frac{\phantom{\rule{thinmathspace}{0ex}}d\rho}{\rho (\rho +1)\sqrt{{\rho}^{2}-{\rho}_{+}^{2}}}=-\frac{\pi}{2\u03f5}+O(\mathrm{ln}\frac{1}{\u03f5}).}\end{array}$$

Gathering all these formulas we yield $\begin{array}{}{\displaystyle T(u)=\pi +O(\sqrt{\u03f5}).}\end{array}$

The Lemma is proved.

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