# Linear Stability of Periodic Trajectories in Inverse Magnetic Billiards

Sean Gasiorek<EMAIL_ADDRESS>School of Mathematics and
Statistics, Carslaw Building F07, University of Sydney, NSW 2006, Australia

###### Abstract.

We study the stability of periodic trajectories of planar inverse magnetic
billiards, a dynamical system whose trajectories are straight lines inside a
connected planar domain $\Omega$ and circular arcs outside $\Omega$. Explicit
examples are calculated in circles, ellipses, and the one parameter family of
curves $x^{2k}+y^{2k}=1$. Comparisons are made to the linear stability of
periodic billiard and magnetic billiard trajectories.

###### Key words and phrases:

Inverse magnetic billiards, stability, linear stability, periodic orbits,
periodic trajectories

###### 2020 Mathematics Subject Classification:

37J25, 37J46, 70H12, 70H14, 78A35

##  1\. Introduction

Mathematical billiards serves as a fundamental example of a dynamical system
and has been studied extensively over the last century. Connecting geometry
and dynamics, mathematical billiards concerns the motion of a free particle
(the “billiard ball”) under inertia in a domain (the “billiard table”) which
moves at constant speed and undergoes perfectly elastic collisions with the
boundary of the table. The collisions with the boundary follow the billiard
reflection law “angle of incidence equals angle of reflection”, where the
tangential component of the velocity is conserved while the normal component
instantly changes sign. See e.g. [4, 15, 20] for a survey.

While mathematical billiards serves as a model of certain physical phenomena,
such as wave fronts and geometric optics, magnetic variants of billiards,
where the billiard ball is interpreted as a charged particle moving under the
influence of a magnetic field $\boldsymbol{B}$ which satisfies the reflection
law at the boundary of the billiard table, provides an extension of these same
ideas to various geometric settings (e.g. [3, 12, 21]) and to problems in
dynamics and mathematical physics (e.g. [2, 9, 17, 18, 19]).

The definition of the magnetic field $\boldsymbol{B}$ greatly affects the
dynamics but are also informed by the problem which is to be solved. The study
of charged-particle dynamics in piecewise-constant magnetic fields appears in
a variety of settings, such as nano- and condensed-matter physics,
semiconductor design, and quantum mechanics [6, 13, 14, 22]. Classical,
semiclassical, and quantum approaches to this system are each addressed to a
degree in compact or unbounded domains depending upon the applications of
interest.

The magnetic billiard of interest is that of _inverse magnetic billiards_ ,
following the naming by [22], which has only been studied in detail recently
[10, 11]. Given a connected domain $\Omega\subset\mathbb{R}^{2}$, define a
constant magnetic field $\boldsymbol{B}$ orthogonal to the plane which has
strength 0 on $\Omega$ and strength $B\neq 0$ on its complement. The classical
motion of a charged particle of charge $e$ and mass $m$ with constant speed
$|v|$ throughout $\Omega$ and its complement are continuous curves which are
circular arcs outside $\Omega$ and straight chords inside $\Omega$. The
charged particle is subject to the Lorentz force outside $\Omega$, and the
resulting motion is circular arcs of fixed Larmor radius $\mu=m|v|/|eB|$. We
take $eB<0$ so the Larmor arcs are traversed in the anticlockwise direction.

This paper is organized as follows. Section 2 constructs the inverse magnetic
billiard map, its derivative, and relevant properties. Section 3 establishes a
linear stability criteria for 2-periodic trajectories in inverse magnetic
billiards, provides explicit examples, and contrasts the stability criteria
with the existing linear stability criteria for standard and magnetic
billiards. In Section 4, we give examples of the linear stability of 3- and
4-periodic trajectories with symmetries in various domains.

##  2\. Properties of the Inverse Magnetic Billiard Map

We give a brief review of inverse magnetic billiards in a convex set
$\Omega\subset\mathbb{R}^{2}$ and note that additional details can be found in
[10, 11].

Suppose $\Omega\subset\mathbb{R}^{2}$ is strictly convex and parametrize the
boundary $\partial\Omega=\Gamma(s)$ by arc length, $s$ in the anticlockwise
direction and let $L=|\partial\Omega|$. Provided $\Gamma(s)$ is sufficiently
smooth, the convexity of $\Omega$ implies the radius of curvature
$\rho(s)=1/\kappa(s)$ of $\Gamma(s)$ satisfies
$0<\rho_{min}\leq\rho(s)\leq\rho_{max}<\infty$. In the study of magnetic
billiards, Robnik and Berry [18, 19] classified the dynamics based upon three
_curvature regimes_ , depending upon the relative sizes of $\mu$,
$\rho_{min}$, and $\rho_{max}$:

$\mu<\rho_{min},\qquad\rho_{min}<\mu<\rho_{max},\qquad\mu>\rho_{max}.$

The motion of the charged particle (the “billiard”) in this setting can be
described in terms of its two geometric components: its straight-line standard
billiard component and its magnetic Larmor arc. Suppose a trajectory starts at
a point $P_{0}:=\Gamma(s_{0})$ with an initial velocity vector $v_{0}$
pointing to the interior of $\Omega$. This initial velocity vector makes an
angle $\theta_{0}\in(0,\pi)$ with the positively-oriented tangent vector to
$\Gamma(s)$ at $P_{0}$, and as the billiard travels in a straight line inside
$\Omega$, it will eventually meet $\Gamma(s)$ at a new point
$P_{1}:=\Gamma(s_{1})$. Let $\ell_{1}=|P_{0}P_{1}|$ be the chord length and
$\theta_{1}$ be the angle between the tangent vector to $\Gamma(s)$ at $P_{1}$
and the velocity vector $\overrightarrow{P_{0}P_{1}}$. This completes the
billiard-like component of motion.

Next, because $\partial\Omega$ acts as a permeable boundary between magnetic
and non-magnetic regions, the motion starting at $P_{1}$ in the direction
$v_{0}$ moves along a circular Larmor arc $\gamma$ of radius $\mu$ until
intersecting $\Gamma(s)$ at a point $P_{2}:=\Gamma(s_{2})$ where the billiard
re-enters $\Omega$. The velocity vector $v_{2}$ of the billiard at $P_{2}$ is
the tangent vector to the Larmor circle at $P_{2}$ and makes an angle
$\theta_{2}$ with the tangent vector to $\Gamma(s)$ at $P_{2}$. The re-entry
point $P_{2}$ is well-defined, as the Larmor circle is tangent to the ray
$\overrightarrow{P_{0}P_{1}}$ at $P_{1}$. Let $\ell_{2}=|P_{1}P_{2}|$ be the
chord length connecting the exit and re-entry points $P_{1}$ and $P_{2}$ of
$\Omega$; let $\chi_{0,2}$ be the angle between $v_{0}$ and the ray
$\overrightarrow{P_{1}P_{2}}$; let $\mathcal{S}_{0,2}$ be the area inside the
Larmor arc $\gamma_{0,2}$ but outside $\Omega$; let
$\kappa_{i}:=\kappa(s_{i})$ be the curvature of $\Gamma$ at $s_{i}$; and let
$\mathcal{A}_{0,2}$ denote the area of the “cap” between the chord $\ell_{2}$
and $\Gamma(s)$ that is inside the Larmor circle. See figure 1.

Figure 1. A labeled diagram of both components of motion of the inverse
magnetic billiard trajectory.

######  Remark 2.1.

In general, we will keep the same numbering convention described above, so
that the quantities $\theta_{2i}$, $\theta_{2i+1}$, $\theta_{2i+2}$,
$\chi_{2i,2i+2}$, $\ell_{2i+1}$, $\ell_{2i+2}$, etc. are all associated with
the trajectory from $P_{2i}\to P_{2i+2}$.

From the above definitions, we can encode the dynamics through the inverse
magnetic billiard map $T$ on the open annulus
$\mathbb{R}/L\mathbb{Z}\times(0,\pi)\approx[0,L)\times(0,\pi)$, which sends
each successive reentry point and angle of the inward pointing velocity vector
to that of the next reentry point and angle:

$T:[0,L)\times(0,\pi)\to[0,L)\times(0,\pi),\qquad(s_{i},\theta_{i})\mapsto(s_{i+2},\theta_{i+2}).$

By assuming that motion stops when $\theta=0$ or $\pi$, we can extend the
domain of this map to the closed annulus $\mathbb{A}:=[0,L)\times[0,\pi]$;
that is, by continuity, $T(s,0)=(s,0)$ and $T(s,\pi)=(s,\pi)$.

The map $T$ preserves the symplectic area form $\sin(\theta)ds\wedge d\theta$
on the annulus $\mathbb{A}$. This informs the definition of area-preserving
coordinates, $u_{i}:=-\cos(\theta_{i})$.

######  Proposition 2.2 ([10, 11]).

Suppose $\partial\Omega$ is of class $C^{k}$. Then the inverse magnetic
billiard map $T$ is $C^{k-1}$ and $DT=\left(\begin{array}[]{cc}\frac{\partial
s_{2}}{\partial s_{0}}&\frac{\partial s_{2}}{\partial u_{0}}\\\ \frac{\partial
u_{2}}{\partial s_{0}}&\frac{\partial u_{2}}{\partial
u_{0}}\end{array}\right)$ with

$\displaystyle\frac{\partial s_{2}}{\partial s_{0}}$
$\displaystyle=\frac{\kappa_{0}\ell_{1}\sin(2\chi_{0,2}-\theta_{1})-\sin(\theta_{0})\sin(2\chi_{0,2}-\theta_{1})-\kappa_{0}\ell_{2}\cos(\chi_{0,2})\sin(\theta_{1})}{\sin(\theta_{1})\sin(\theta_{2})}$
$\displaystyle\frac{\partial s_{2}}{\partial u_{0}}$
$\displaystyle=\frac{\ell_{1}\sin(2\chi_{0,2}-\theta_{1})-\ell_{2}\cos(\chi_{0,2})\sin(\theta_{1})}{\sin(\theta_{0})\sin(\theta_{1})\sin(\theta_{2})}$
$\displaystyle\frac{\partial u_{2}}{\partial s_{0}}$
$\displaystyle=\frac{\kappa_{2}\sin(\theta_{0})\sin(2\chi_{0,2}-\theta_{1})}{\sin(\theta_{1})}+\frac{2\sin(\chi_{0,2})\sin(2\chi_{0,2}-\theta_{1}-\theta_{2})(\kappa_{0}\ell_{1}-\sin(\theta_{0}))}{\ell_{2}\sin(\theta_{1})}$
$\displaystyle\;\;\;\;\;-\kappa_{0}\left(\sin(2\chi_{0,2}-\theta_{2})+\frac{\kappa_{2}\ell_{1}\sin(2\chi_{0,2}-\theta_{1})}{\sin(\theta_{1})}-\kappa_{2}\ell_{2}\cos(\chi_{0,2})\right)$
$\displaystyle\frac{\partial u_{2}}{\partial u_{0}}$
$\displaystyle=\frac{2\ell_{1}\sin(\chi_{0,2})\sin(2\chi_{0,2}-\theta_{1}-\theta_{2})-\kappa_{2}\ell_{1}\ell_{2}\sin(2\chi_{0,2}-\theta_{1})}{\ell_{2}\sin(\theta_{0})\sin(\theta_{1})}$
$\displaystyle\;\;\;\;\;+\frac{\kappa_{2}\ell_{2}\cos(\chi_{0,2})-\sin(2\chi_{0,2}-\theta_{2})}{\sin(\theta_{0})}.$

Furthermore, $\det(DT)=1$.

######  Remark 2.3.

The preceding proposition shows that the entries of the derivative matrix $DT$
do not directly depend upon the magnetic field or Larmor radius $\mu$.
However, simple geometry yields the equation
$\ell_{2i+2}=2\mu\sin(\chi_{2i,2i+2})$, so the quantities above can be stated
using $\mu$. Our analysis in the next sections makes use of this equation.

The derivative matrix $DT$ from Prop. 2.2, interpreted as a linearization of
$T$, can be used to determine the stability of periodic trajectories. If we
let $z=(s,u)$, the _stability matrix_ is

$S_{n}(z)=DT(T^{n-1}(z))\cdots DT(T(z))DT(z).$ (2.1)

If $z$ is part of an orbit of period $n$, the orbit is _hyperbolic_ if
$|\operatorname{Tr}S_{n}(z)|>2$, _parabolic_ if
$|\operatorname{Tr}S_{n}(z)|=2$, and _elliptic_ if
$|\operatorname{Tr}S_{n}(z)|<2$. To simplify notation, we will drop the
dependence upon $z$ and largely refer to $\operatorname{Tr}S_{n}$.

##  3\. Linear Stability of 2-Periodic Trajectories

The linear stability of standard billiards has been studied recently [15, 16],
though the formulation was known earlier in the 20th century [1, 23]. In
particular, the linear stability of 2-periodic billiard trajectories can be
simply stated in terms of the length of the trajectory and the radii of
curvature of the boundary at the two points of impact, $\rho_{1}$ and
$\rho_{2}$.

######  Proposition 3.1 ([15]).

Suppose $\rho_{1}<\rho_{2}$ are the radii of curvature of the billiard table
at the impact points of a 2-periodic billiard trajectory and let $l$ denote
the distance between the two impact points. If either of the inequalities
$\rho_{1}<l<\rho_{2}$ or $l>\rho_{1}+\rho_{2}$ are satisfied, the 2-periodic
billiard trajectory will be hyperbolic, while in the case $0<l<\rho_{1}$ or
$\rho_{2}<l<\rho_{1}+\rho_{2}$ the trajectory is elliptic. If $l=\rho_{1}$,
$\rho_{2}$, or $\rho_{1}+\rho_{2}$, then the trajectory is parabolic.

The proof of this proposition follows from a generating function argument [15]
or by a calculation and analysis of $\operatorname{Tr}S_{2}$. The argument
does not depend upon the convexity of $\Omega$ and still holds if one or both
of $\rho_{1}$, $\rho_{2}$ are negative. With this stability criteria for
2-periodic billiard trajectories in mind, we investigate the linear stability
of 2-periodic trajectories in the setting of inverse-magnetic billiards.

###  3.1. 2-Periodic Trajectories in a Strictly Convex Set

Suppose $\Omega$ is strictly convex and consider a 2-periodic inverse magnetic
billiard trajectory. In particular, the periodicity means
$\kappa_{4}=\kappa_{0}$ and $\theta_{4}=\theta_{0}$. Geometrically, 2-periodic
trajectories are in the shape of stadia – a rectangle capped by two
semicircles on opposite sides – and was stated as an offhand remark in [11]
(and is clear if one draws a picture). The following proposition adds more
structure to this statement.

######  Proposition 3.2.

An inverse magnetic billiard trajectory in a strictly convex set is 2-periodic
if and only if the angles $\chi_{0,2}$, $\chi_{2,4}$ in successive iterations
of $T$ are both equal to $\pi/2$.

###### Proof.

Suppose a trajectory is 2-periodic and $\chi_{0,2},\chi_{2,4}\neq\pi/2$.
Without loss of generality, suppose $\chi_{0,2}>\pi/2$. Then the tangent lines
to the Larmor arc $\gamma_{0,2}$ at $P_{1}$, $P_{2}$ must intersect at some
point $Q$. Then the second Larmor arc $\gamma_{2,4}$ must also be tangent to
these two lines at the points $P_{3}$ and $P_{4}=P_{0}$ and lie within the
compact set bounded by $Q$, the tangent lines, and $\gamma_{0,2}$. However,
since the two Larmor arcs $\gamma_{0,2}$, $\gamma_{2,4}$ have equal radii, the
only way this can happen is if $\gamma_{0,2}\cup\gamma_{2,4}$ is the complete
Larmor circle of radius $\mu$, making the trajectory not 2-periodic.

Suppose $\chi_{0,2}=\chi_{2,4}=\pi/2$. Then $\ell_{1}$ and $\ell_{3}$ are
parallel and the Larmor arcs $\gamma_{0,2}$, $\gamma_{2,4}$ are semicircles.
If $\ell_{1}>\ell_{3}$, then $\partial\Omega$ cannot be convex, as $P_{0}$,
$P_{1}$, and $P_{4}$ are collinear. A similar argument holds for
$\ell_{1}<\ell_{3}$. Thus $\ell_{1}=\ell_{3}$ and the trajectory is a stadium.
∎

######  Remark 3.3.

As 2-periodic trajectories are stadia, we also mention the strict convexity of
$\Omega$ necessarily implies that each $\theta_{2i},\theta_{2i+1}$ must have
angle measure strictly less than $\pi/2$. Further, it is also the case that
$\ell_{2}=\ell_{4}=2\mu$.

With the preceding proposition, we can simplify $DT$ from Prop. 2.2 into

$DT=\left(\begin{array}[]{cc}\frac{\kappa_{0}\ell_{1}-\sin(\theta_{0})}{\sin(\theta_{2})}&\frac{\ell_{1}}{\sin(\theta_{0})\sin(\theta_{2})}\\\
\frac{(\kappa_{0}\ell_{1}-\sin(\theta_{0}))(\sin(\theta_{1}+\theta_{2})-\kappa_{2}\mu\sin(\theta_{1}))}{\mu\sin(\theta_{1})}-\kappa_{0}\sin(\theta_{2})&\frac{\ell_{1}(\sin(\theta_{1}+\theta_{2})-\kappa_{2}\mu\sin(\theta_{1}))}{\mu\sin(\theta_{0})\sin(\theta_{1})}-\frac{\sin(\theta_{2})}{\sin(\theta_{0})}\end{array}\right)$
(3.1)

which in turn allows us to compute the stability matrix $S_{2}(s_{0},u_{0})$
and its trace

$\operatorname{Tr}S_{2}=2-2\alpha(\beta+\delta)+\alpha^{2}\beta\delta\\\ $
(3.2)

with

$\alpha=\ell_{1}/\mu,\;\;\;\beta=\cot(\theta_{0})+\cot(\theta_{3}),\;\;\;\delta=\cot(\theta_{1})+\cot(\theta_{2}).$
(3.3)

By Remark 3.3, $\beta$ and $\delta$ are positive, and it follows that $\alpha$
is positive as it is the ratio of two positive quantities. A direct
calculation proves the following theorem.

######  Theorem 3.4.

Let $\alpha$, $\beta$, $\delta$ be defined as above and suppose $\Omega$ is
strictly convex. Let $m=\min\\{2/\beta,2/\delta\\}$ and
$M=\max\\{2/\beta,2/\delta\\}$. If $\beta\neq\delta$, then the 2-periodic
trajectory is

  * •

parabolic if and only if $\alpha=m$, $M$ or $m+M$;

  * •

elliptic if and only if $\alpha\in(0,m)\cup(M,m+M)$;

  * •

hyperbolic if and only if $\alpha\in(m,M)\cup(m+M,\infty)$.

If $\beta=\delta$, then $m=M$ and the 2-periodic trajectory is

  * •

parabolic if and only if $\alpha=m$ or $2m$;

  * •

elliptic if and only if $\alpha\in(0,m)\cup(m,2m)$;

  * •

hyperbolic if and only if $\alpha\in(2m,\infty)$.

Figure 2 presents a graphical representation of the above statement.

As stated in Proposition 3.1, the linear stability of a 2-periodic billiard
trajectory is dependent upon the relative sizes of the length of the chord and
the curvature of the boundary at the points of impact. In magnetic billiards,
the linear stability is dependent upon the curvature at the points of impact
and the associated diametric chord length ([17, 18]). The linear stability
criteria is precisely the same criteria as standard billiards, implying the
linear stability of magnetic billiards is independent of the magnetic field.
However, the linear stability of periodic trajectories in the inverse magnetic
billiard setting is indirectly dependent upon the magnetic field through the
term $\alpha$. And in contrast to the standard billiard or magnetic billiard
settings, the linear stability of inverse magnetic billiards is not dependent
upon the curvature of the boundary at the exit and re-entry points.

0$E$$m$$P$$H$$M$$P$$E$$m+M$$P$$H$$\alpha$0$E$$m$$P$$E$$2m$$P$$H$$\alpha$
Figure 2. A stability diagram illustrating Theorem 3.4 when $\beta\neq\delta$
(above) and $\beta=\delta$ (below).

######  Example 3.5.

Consider the case where $\partial\Omega$ is the circle of radius $R$. Then
$\theta_{0}=\theta_{1}=\theta_{2}=\theta_{3}$ and hence $\beta=\delta$. It is
then necessarily true that $\mu<R$, as $\mu\geq R$ makes the stadia larger
than $\Omega$ itself. A quick calculation shows that $\alpha=4/\beta=2m$ and
hence all 2-periodic trajectories in the circle are parabolic.

######  Example 3.6.

Consider the case where $\partial\Omega$ is the ellipse
$\mathcal{E}:x^{2}/a^{2}+y^{2}/b^{2}=1$ with $a>b>0$. In standard billiards,
the only 2-periodic trajectories in the ellipse are the major and minor axes
whose stability are hyperbolic and elliptic, respectively. In inverse magnetic
billiards, the 2-periodic trajectories are aligned along and symmetric about
the major and minor axes of the ellipse; that is, the centers of the Larmor
circles lie on the coordinate axes of the ellipse. See figure 3. By symmetry,
the angles $\theta_{0}=\theta_{1}=\theta_{3}=\theta_{4}$ and we are again in
the case $\beta=\delta$.

Consider the 2-periodic trajectory aligned along the major axis of the
ellipse. This constrains the Larmor radius to $0<\mu<b$. Two direct
calculations show that

$\alpha=\frac{2a\sqrt{b^{2}-\mu^{2}}}{b\mu}\qquad\text{and}\qquad\cos(\theta_{0})=\frac{a\mu}{\sqrt{b^{4}+\mu^{2}(a^{2}-b^{2})}}.$
(3.4)

As $\beta=2\cot(\theta_{0})$, this gives

$\beta=\frac{2a\mu}{b\sqrt{b^{2}-\mu^{2}}}.$ (3.5)

Combining these two equations gives us $\alpha$ in terms of $\beta$,

$\alpha=\frac{a^{2}}{b^{2}}\frac{4}{\beta}.$ (3.6)

In terms of Theorem 3.4 and figure 2, this 2-periodic trajectory has
$\alpha=\frac{a^{2}}{b^{2}}\cdot 2m$, and hence the trajectory is hyperbolic.

In the case of the 2-periodic trajectory aligned along the minor axis of the
ellipse, we now have $0<\mu<a$. Repeating the above calculations yields

$\alpha=\frac{b^{2}}{a^{2}}\frac{4}{\beta}.$ (3.7)

In terms of Theorem 3.4 and figure 2, this 2-periodic trajectory has
$\alpha=\frac{b^{2}}{a^{2}}\cdot 2m$, and hence the trajectory is elliptic
except for when $2b^{2}=a^{2}$, in which case the trajectory is parabolic. It
is noteworthy that this same obstruction to ellipticity occurs for standard
billiards in the ellipse [16]. We discuss this further in section 3.3.

Figure 3. 2-periodic trajectories in the ellipse along the major axis (dashed)
and minor axis (dotted).

An interesting feature of the two families of 2-periodic trajectories is that
the stability is only dependent upon the lengths of the semi-major and semi-
minor axes of the ellipse. Moreover, the imposed restrictions on $\mu$ in both
cases apply to the regime $\mu<\rho_{min}$ and to a subinterval of the
$\rho_{min}<\mu<\rho_{max}$ regime. This adds to the limited knowledge of
inverse magnetic billiards in the intermediate curvature regime.

For comparison, consider standard billiards in the ellipse $\Omega$. The only
2-periodic billiard trajectories are the major and minor axes. A quick
calculation shows that the major axis of the ellipse is hyperbolic while the
minor axis is elliptic except when $a^{2}=2b^{2}$, in which case the
trajectory is parabolic.

######  Example 3.7.

Consider the case when $\partial\Omega$ is the one-parameter family of curves
$x^{2k}+y^{2k}=1$, $k\in\mathbb{N}$. As the case $k=1$ corresponds to the unit
circle, we also assume $k>1$. By symmetry, we again have
$\theta_{0}=\theta_{1}=\theta_{2}=\theta_{3}$ and are again in the case
$\beta=\delta$.

The axes of symmetry of $\partial\Omega$ gives rise to two families of
2-periodic trajectories: the stadia are oriented along the horizontal or
vertical axes of symmetry, and the stadia which are oriented along the
diagonal axes of symmetry, $y=\pm x$.

Consider first the trajectories that are oriented about the horizontal (and
equivalently, vertical) axis. See figure 4a. Geometrically, 2-periodic
trajectories have Larmor radius $\mu$ limited to $0<\mu<1$. Two calculations
show that

$\alpha=\frac{2(1-\mu^{2k})^{\frac{1}{2k}}}{\mu}\qquad\text{and}\qquad\cos(\theta_{0})=\left((\mu^{-2k}-1)^{\frac{2k-1}{k}}+1\right)^{-1/2}.$
(3.8)

Again, $\beta=2\cot(\theta_{0})$, and hence

$\beta=2(\mu^{-2k}-1)^{\frac{1-2k}{2k}}.$ (3.9)

Combining these to get $\alpha$ in terms of $\beta$, we get

$\alpha=\beta(\mu^{-2k}-1).$ (3.10)

In terms of Theorem 3.4 and figure 2, we now must find the relative size of
$\alpha$ with $m=2/\beta$ and $2m$. Through some algebra, we see that
$\alpha=m$ and $\alpha=2m$ at the values

$\mu^{*}=\left(2^{\frac{k}{k-1}}+1\right)^{-\frac{1}{2k}}\qquad\text{and}\qquad\mu^{**}=2^{-\frac{1}{2k}},$
(3.11)

respectively, and the stability can be summarized in the following way:

  * •

Parabolic if and only if $\mu=\mu^{*}$ or $\mu^{**}$;

  * •

Elliptic if and only if $\mu\in(0,\mu^{*})\cup(\mu^{*},\mu^{**})$;

  * •

Hyperbolic if and only if $\mu\in(\mu^{**},1)$.

(a)  | (b)
---|---
Figure 4. 2-periodic trajectories in the curve $x^{2k}+y^{2k}=1$ along the
horizontal axis (a) and diagonal axis (b).

Next, consider the case where the trajectory is oriented along the diagonal
$y=x$, as in figure 4b. The diagonal of $\partial\Omega$ is of length $2\cdot
2^{\frac{k-1}{2k}}$, so any Larmor circle must have diameter smaller than this
diameter, meaning $0<\mu<2^{\frac{k-1}{2k}}$.

We take a slightly different approach than previously. Suppose
$P_{0}=(x_{0},y_{0})=(x_{0},(1-x^{2k})^{\frac{1}{2k}})$ is the point in figure
4b with the largest value of $y_{0}$ (i.e. the point that is “on top”). This
means $x_{0}\in(-2^{-\frac{1}{2k}},2^{-\frac{1}{2k}})$ and $y_{0}>x_{0}$. In
this construction,

$\ell_{1}=(x_{0}+y_{0})\sqrt{2},\qquad\mu=\frac{y_{0}-x_{0}}{\sqrt{2}}\qquad\implies\qquad\alpha=2\frac{y_{0}+x_{0}}{y_{0}-x_{0}}.$
(3.12)

Using implicit differentiation, we find the angle $\theta_{0}$ can be written
simply in terms of $x_{0}$, and $y_{0}$:

$\tan(\theta_{0})=\frac{y_{0}^{2k-1}+x_{0}^{2k-1}}{y_{0}^{2k-1}-x_{0}^{2k-1}}.$
(3.13)

Just as in the previous examples, $\beta=2\cot(\theta_{0})$. Because
$m=2/\beta=\tan(\theta_{0})$, the stability depends upon the relative values
of $\alpha$ and $m=\tan(\theta_{0})$ and $2m=2\tan(\theta_{0})$. Through
factoring the above equation, we get

$\alpha=2mf_{k}(x_{0},y_{0})$ (3.14)

with

$f_{k}(x_{0},y_{0})=\frac{y_{0}^{2k-2}+y_{0}^{2k-3}x_{0}+\cdots+y_{0}x_{0}^{2k-3}+x_{0}^{2k-2}}{y_{0}^{2k-2}-y_{0}^{2k-3}x_{0}+\cdots-
y_{0}x_{0}^{2k-3}+x_{0}^{2k-2}}.$ (3.15)

Thus determining the stability of the diagonal 2-periodic orbit reduces to
determining the values of $f_{k}(x_{0},y_{0})$. Luckily $f_{k}(x_{0},y_{0})$
is a continuous, monotone increasing function on
$(-2^{-\frac{1}{2k}},2^{-\frac{1}{2k}})$, has limiting values $1/(2k-1)$ and
$2k-1$ as $x_{0}\to-2^{-\frac{1}{2k}}$ and $x_{0}\to 2^{-\frac{1}{2k}}$,
respectively, and satisfies $f_{k}(0,1)=1$. By the intermediate value theorem,
there exists an $\tilde{x_{0}}\in(-2^{-\frac{1}{2k}},0)$ with corresponding
$\tilde{y_{0}}$ such that $f_{k}(\tilde{x_{0}},\tilde{y_{0}})=1/2$, which in
turn means $\alpha=m$. Moreover, $f_{k}(0,1)=1$ corresponds to $\mu=2^{-1/2}$
and $f_{k}(\tilde{x_{0}},\tilde{y_{0}})=1/2$ corresponds to a value
$\tilde{\mu}=(\tilde{y_{0}}-\tilde{x_{0}})/\sqrt{2}$ given by the equations
above.

Ultimately, the stability can be summarized as follows:

  * •

Parabolic if and only if $\mu=2^{-1/2}$ or $\tilde{\mu}$ if and only if
$x_{0}=0$ or $\tilde{x_{0}}$;

  * •

Elliptic if and only if $\mu\in(0,2^{-1/2})$ if and only if
$x_{0}\in(0,2^{-\frac{1}{2k}})$;

  * •

Hyperbolic if and only if $\mu\in(2^{-1/2},2^{\frac{k-1}{2k}})$ if and only if
$x_{0}\in(-2^{-\frac{1}{2k}},0)\setminus\\{\tilde{x_{0}}\\}$.

For comparison, we can apply Proposition 3.1 to this family of curves to study
the linear stability of 2-periodic billiard trajectories. Such 2-periodic
billiard trajectories are the horizontal and vertical axes of symmetry, and
the diagonals, which are the widths and diameters of $\Omega$, respectively.

Consider first a width of $\Omega$ as the horizontal segment connecting the
points $(-1,0)$ and $(0,1)$. As the curvature of $\partial\Omega$ vanishes at
these points, the radii of curvature are infinite. An analysis of the billiard
stability matrix yields

$\operatorname{Tr}S_{2}=2-4l\left(\frac{1}{\rho_{1}}+\frac{1}{\rho_{2}}\right)+\frac{4l^{2}}{\rho_{1}\rho_{2}},$
(3.16)

and hence this trajectory is parabolic (that is, Proposition 3.1 does not
directly apply at points of infinite radii of curvature).

Consider next a diameter of $\Omega$ as the segment connecting the points
$(-2^{-\frac{1}{2k}},-2^{-\frac{1}{2k}})$ and
$(2^{-\frac{1}{2k}},2^{-\frac{1}{2k}})$. At these points the radii of
curvature of $\partial\Omega$ are at their minimum,
$\rho_{1}=\rho_{2}=2^{\frac{1-k}{2k}}/(2k-1).$ Because $l=2\cdot
2^{\frac{k-1}{2k}}=2^{\frac{3k-1}{2k}}$, it is quick to show that
$l>2\rho_{1}$ for all $k>1$ and hence the diagonal 2-periodic billiard is
hyperbolic.

###  3.2. 2-Periodic Trajectories in a Non-Strictly Convex or Concave Set

We again consider the case when $n=2$ and we suppose $\Omega$ is either not
strictly convex (i.e. contains points of vanishing curvature) or concave (i.e.
contains points of negative curvature). We address the same propositions as
the convex case.

######  Proposition 3.8.

Let $\Omega$ be convex or concave. If an inverse magnetic billiard trajectory
is 2-periodic, then the angles $\chi_{0,2}=\chi_{2,4}=\pi/2$.

The proof of this is identical to the first part of the proof of Proposition
3.2. With the assumption that $\Omega$ can be convex or concave, this now
allows $\theta_{i}$ to be $\geq\pi/2$. The equations for $DT$,
$\operatorname{Tr}S_{2}$, $\alpha$, $\beta$, and $\delta$ are unchanged but
now we have the possibility that $\beta$ and $\delta$ could be nonpositive.
Again, a direct calculation proves the following.

######  Theorem 3.9.

Let $\alpha$, $\beta$, $\delta$ be defined as in equation 3.3 and suppose
$\Omega$ is not strictly convex or is concave.

  1. (i)

If $\beta=\delta=0$, then the trajectory is parabolic for all
$\alpha\in(0,\infty)$;

  2. (ii)

If $\beta\leq 0$ and $\delta\leq 0$ and $\beta$ and $\delta$ are not both
zero, then the trajectory is hyperbolic for all $\alpha\in(0,\infty)$;

  3. (iii)

If either

    1. a)

$\beta>0$ and $\delta=0$; or

    2. b)

$\beta>0$ and $\delta<0$, and
$\displaystyle\frac{2}{\beta}+\frac{2}{\delta}\leq 0$,

then the trajectory is

     * •

parabolic if and only if $\alpha=\dfrac{2}{\beta}$;

     * •

elliptic if and only if $\alpha\in\left(0,\dfrac{2}{\beta}\right)$;

     * •

hyperbolic if and only if $\alpha\in\left(\dfrac{2}{\beta},\infty\right)$;

  4. (iv)

If $\beta>0$ and $\delta<0$ and $\dfrac{2}{\beta}+\dfrac{2}{\delta}>0$, then
the trajectory is

     * •

parabolic if and only if $\alpha=\dfrac{2}{\beta}$ or
$\alpha=\dfrac{2}{\beta}+\dfrac{2}{\delta}$;

     * •

elliptic if and only if
$\alpha\in\left(\dfrac{2}{\beta}+\dfrac{2}{\delta},\dfrac{2}{\beta}\right)$;

     * •

hyperbolic if and only if
$\alpha\in\left(0,\dfrac{2}{\beta}+\dfrac{2}{\delta}\right)\cup\left(\dfrac{2}{\beta},\infty\right)$;

  5. (v)

If $\beta>0$ and $\delta>0$, then the stability is determined by the statement
of Theorem 3.4.

Identical statements to (iii) and (iv) hold when $\beta$ and $\delta$ are
switched.

As noted in the literature on the stability of 2-periodic billiard
trajectories (e.g. [15, 16, 23]), the stability criteria therein allow for the
cases where the boundary is convex or concave at the point of reflection. As
in the strictly convex case, the stability criterion for inverse magnetic
billiards is only dependent upon the angles $\theta_{i}$, and the ratio
$\ell_{1}/\mu$. These criteria are again not dependent upon the curvature of
the boundary at the exit or reentry points nor is it directly dependent on the
magnetic field.

######  Example 3.10.

Suppose $\Omega$ is the stadium: a rectangle with side lengths $L$ and $2R$
capped by semicircles of radius $R$ on opposite sides (see figure 5a). If the
points $P_{0},P_{1},P_{2},P_{3}$ of a 2-periodic trajectory are all on the
straight sides of $\partial\Omega$ with $P_{0},P_{3}$ on one side and
$P_{1},P_{2}$ on the opposite side, then the angles $\theta_{i}=\pi/2$ for
$i=0,1,2,3$ and necessarily $2\mu<L$. Then $\beta=\delta=0$ and the trajectory
is parabolic.

If instead the trajectory has the points $P_{0},$ $P_{3}$ on one semicircular
cap and the points $P_{1},P_{2}$ on the other semicircular cap of the stadium,
then necessarily $\mu<R$ and $\beta=\delta>0$ as all angles $\theta_{i}$ are
equal. We use the second part of Theorem 3.4 to analyze the stability. A
direct calculation yields

$\ell_{1}=L+2\sqrt{R^{2}-\mu^{2}},\qquad\cos(\theta_{i})=\frac{\mu}{R},\qquad
m=\frac{\sqrt{R^{2}-\mu^{2}}}{\mu}.$ (3.17)

It follows that $\alpha=L/\mu+2m>2m$, and hence the trajectory is hyperbolic
for all $\mu<R$.

######  Example 3.11.

Suppose $\Omega$ is the “telephone curve” with the 2-periodic trajectory
pictured in figure 5b. Suppose $P_{0}$ is the lower right point on
$\partial\Omega$. Then $\theta_{1}=\theta_{2}>\pi/2$ while
$\theta_{0}=\theta_{3}<\pi/2$, and so $\delta<0$ and $\beta>0$. Thus we are in
either case (iii) or (iv) of Theorem 3.9, depending upon the sign of
$\frac{2}{\beta}+\frac{2}{\delta}$, which in turn equals
$\tan(\theta_{0})+\tan(\theta_{1})$ in this setting. Knowing or calculating
the values of $\ell_{1}$, $\mu$, $\theta_{0},$ and $\theta_{1}$ will then
determine the stability of the pictured trajectory.

(a)  | (b)
---|---
Figure 5. (a) A parabolic (dotted) and hyperbolic (dashed) 2-periodic
trajectory in the stadium ; (b) A 2-periodic trajectory in the concave
“telephone curve”.

###  3.3. A Digression on the Ellipse and Rotation Numbers

Example 3.6 illustrates that the linear stability criterion for inverse
magnetic billiards in an ellipse is the same as standard billiards in an
ellipse: the major axis-aligned 2-periodic inverse magnetic billiard is
hyperbolic and so is the standard billiard trajectory along the major axis;
the minor axis-aligned 2-periodic inverse magnetic billiard trajectory is
elliptic, as is the standard billiard trajectory along the minor axis, and
both fail to be elliptic when $a^{2}=2b^{2}$. We investigate this exception
further in the context of standard billiards.

The standard billiard in an ellipse is a well-known example of an integrable
system and is conjectured to be the only integrable planar billiard [4].
Billiard trajectories in the ellipse have the following caustic property: if
one segment of a trajectory is tangent to an ellipse confocal to the billiard
table, then after every reflection the trajectory will remain tangent to the
same confocal ellipse; if one segment (or its extension) of a trajectory is
tangent to a hyperbola which is confocal to the billiard table, then every
segment or its extension will be tangent to the same confocal hyperbola; and
if a segment of the billiard trajectory passes through one focus, then after
reflection the trajectory will pass through the other focus. The confocal
ellipse or hyperbola with the aforementioned tangency property are called
_caustics_ of the billiard trajectory. If we write the ellipse
$\mathcal{E}:x^{2}/a^{2}+y^{2}/b^{2}=1$ for $a>b>0$, then the confocal family
of $\mathcal{E}$ is given by

$\mathcal{E}_{\lambda}:\frac{x^{2}}{a^{2}-\lambda}+\frac{y^{2}}{b^{2}-\lambda}=1$
(3.18)

for $\lambda\in\mathbb{R}$. The caustics are members of the confocal family
which are ellipses for $\lambda\in(0,b^{2})$ and hyperbolas for
$\lambda\in(b^{2},a^{2})$. The caustics are degenerate and are contained in
the major and minor axes for $\lambda=b^{2}$ and $a^{2}$, respectively. If
$\lambda<0$ or $\lambda>a^{2}$, then the curves $\mathcal{E}_{\lambda}$ are
outside $\mathcal{E}$ or imaginary, respectively.

Given a periodic billiard trajectory in the ellipse $\mathcal{E}$, define the
following _rotation function_ as the ratio of elliptic integrals

$\text{rot}:\left(0,b^{2}\right)\cup\left(b^{2},a^{2}\right)\to\mathbb{R},\qquad\text{rot}(\lambda)=\frac{\displaystyle\int_{0}^{\min(b^{2},\lambda)}\frac{dt}{\sqrt{(\lambda-t)(b^{2}-\lambda)(a^{2}-\lambda)}}}{\displaystyle
2\int_{\max(b^{2},\lambda)}^{a^{2}}\frac{dt}{\sqrt{(\lambda-t)(b^{2}-\lambda)(a^{2}-\lambda)}}}.$
(3.19)

See e.g. [5, 7] for a modern treatment, though these elliptic integrals were
known to Jacobi and other mathematicians of the 19th century.

A periodic trajectory will have $\text{rot}(\lambda)=m/n\in\mathbb{Q}$ for
coprime integers $m\geq n$. When $\mathcal{E}_{\lambda}$ is an ellipse, $m$ is
the winding number and $n$ is the minimal period of the trajectory. In
particular, $\text{rot}(\lambda)$ increases monotonically from 0 to 1 as
$\lambda$ increases monotonically from 0 to $b^{2}$. If
$\mathcal{E}_{\lambda}$ is a hyperbola, then $m$ is the number of times the
trajectory crosses the $y$-axis and $n$ is the minimal period. This
necessarily implies the trajectory must have even period when the caustic is a
hyperbola. As $\lambda$ increases monotonically from $b^{2}$ to $a^{2}$,
$\text{rot}(\lambda)$ decreases monotonically from 1 to $r>0$, where $r$ can
be calculated explicitly, as we will see below. Proving the (decreasing)
monotonicity of $\text{rot}(\lambda)$ when the caustic is a hyperbola is
nontrivial and was proved in [8]. A simpler proof was given recently [7] using
a different technique.

To determine the limiting value of $\text{rot}(\lambda)$ for a hyperbolic
caustic, $r>0$, consider the approach of [8], where the confocal family has
fixed foci at $(\pm 1,0)$ and can be written in the form:

$\mathcal{C}_{\nu}:\frac{x^{2}}{\nu}+\frac{y^{2}}{\nu-1}=1.$ (3.20)

The confocal curves are hyperbolas for $\nu\in(0,1)$ and ellipses for
$\nu\in(1,\infty)$. Any ellipse of the form
$\mathcal{E}:x^{2}/a^{2}+y^{2}/b^{2}=1$ can be homothetically scaled to a
member of the confocal family $\mathcal{C}_{\nu_{0}}$ for some $\nu_{0}>1$.
Suppose the billiard boundary is the member of this confocal family
$\mathcal{C}_{\nu}$ corresponding to $\nu=\nu_{0}>1$. Then the limiting
rotation number for a hyperbolic caustic is only dependent upon $\nu_{0}$ and
is given by

$r=\displaystyle\lim_{\lambda\nearrow
a^{2}}\text{rot}(\lambda)=\frac{1}{\pi}\arccos\left(1-\frac{2}{\nu_{0}}\right).$
(3.21)

See Section 11.2 and specifically subsection 11.2.3.5 of [8] for details.

In the case of the isolated parabolic billiard trajectory along the minor axis
of the ellipse, the ellipse $\mathcal{E}$ has parameters $a$ and $b$
satisfying $a^{2}=2b^{2}$. The confocal family $\mathcal{C}_{\nu}$ has
$a^{2}=\nu_{0}$ and $b^{2}=\nu_{0}-1$, so that together these three equations
imply that the billiard table $\mathcal{C}_{\nu_{0}}$ has $\nu_{0}=2$.
Moreover, this implies that $r=1/2$. The conclusion we draw from this is the
isolated parabolic billiard trajectory in the minor axis of the ellipse has
rotation number $1/2$, and that as $\lambda$ increases from $b^{2}$ to
$a^{2}$, the rotation function $\text{rot}(\lambda)$ decreases monotonically
from $1$ to $1/2$. That is, the isolated parabolic billiard trajectory along
the minor axis of the ellipse is characterized by being the limiting
trajectory of the unique elliptical billiard table whose trajectories with
hyperbolic caustics have limiting rotation number $1/2$.

##  4\. Linear Stability of 3- and 4-Periodic Trajectories

###  4.1. 3-Periodic Trajectories

The study of 3-periodic trajectories can be performed in a similar fashion.
However, we now have two distinct types of 3-periodic trajectories: those with
rotation number $1/3$ and those with rotation number $2/3$. For an arbitrary
3-periodic trajectory, $\operatorname{Tr}S_{3}$ is cubic in $1/\mu$ with
coefficients that are expressions in terms of $\ell_{1}$, $\ell_{3}$,
$\ell_{5}$, and sines and cosines of combinations of $\chi_{0,2}$,
$\chi_{2,4}$, $\chi_{4,6}$, and $\theta_{i}$ for $i=0,\ldots,5$. As in the
2-periodic case, no curvature terms $\kappa_{i}$ appear.

Figure 6. The shape of 3-periodic trajectories with dihedral symmetry with
rotation number $1/3$ (left) and $2/3$ (right).

To make the 3-periodic trajectories simpler to study, consider 3-periodic
trajectories with dihedral symmetry. See figure 6. This assumption means
$\ell_{1}=\ell_{3}=\ell_{5}=:\ell$ and
$\chi_{0,2}=\chi_{2,4}=\chi_{4,6}=:\chi$ are both constant with the latter
equal to $\pi/3$ or $2\pi/3$ when the trajectory has rotation number $1/3$ or
$2/3$, respectively. Then $\operatorname{Tr}S_{3}$ is cubic in
$\alpha:=\ell/\mu$ and has coefficients which can be written in terms of sums,
products, and differences of cotangents of the $\theta_{i}$. For example, the
constant term of $\operatorname{Tr}S_{3}$ can be written as

$\displaystyle
2-\frac{3}{4}C_{23}C_{45}-\frac{3}{8}C_{01}(2C_{2345}\mp\sqrt{3}C_{23}C_{45})$
(4.1)

where $C_{A}=\sum_{i\in A}\cot(\theta_{i})$, the minus sign corresponds to the
trajectory with rotation number $1/3$, and the plus sign corresponds to the
trajectory with rotation number $2/3$. The cubic coefficient can be written as

$\displaystyle-\mp\frac{\cos(\frac{\pi}{6}\mp(\theta_{1}+\theta_{2}))\cos(\frac{\pi}{6}\mp(\theta_{3}+\theta_{4}))\cos(\frac{\pi}{6}\mp(\theta_{0}+\theta_{5}))}{\sin(\theta_{0})\sin(\theta_{1})\sin(\theta_{2})\sin(\theta_{3})\sin(\theta_{4})\sin(\theta_{5})}$
(4.2)

with the same sign convention stated above applying to all $\mp$ terms.

Ultimately, the linear stability of 3-periodic trajectories is difficult to
analyze in general.

######  Example 4.1.

Consider the case when all $\theta_{i}=:\theta$ are equal. Then

$\displaystyle\operatorname{Tr}S_{3}(s,\theta)$
$\displaystyle=2-9\cot^{2}(\theta)-3\sqrt{3}\cot^{3}(\theta)$
$\displaystyle\;\;\;\;\;+\alpha\left[\frac{3\cos(\theta)}{4\sin^{4}(\theta)}\left(5\sqrt{3}\cos(\theta)+\sqrt{3}\cos(3\theta)-3\sin(\theta)+9\sin(3\theta)\right)\right]$
$\displaystyle\;\;\;\;\;+\alpha^{2}\left[-\frac{3\sin(\frac{\pi}{3}+2\theta)(\cos(\theta)+\sin(\frac{\pi}{6}+3\theta))}{\sin^{5}(\theta)}\right]$
$\displaystyle\;\;\;\;\;+\alpha^{3}\left[\frac{\cos^{3}(\frac{\pi}{6}-2\theta)}{\sin^{6}(\theta)}\right]$

and

$\displaystyle\operatorname{Tr}S_{3}(s,\theta)$
$\displaystyle=2-9\cot^{2}(\theta)+3\sqrt{3}\cot^{3}(\theta)$
$\displaystyle\;\;\;\;\;+\alpha\left[\frac{3\cos(\theta)}{4\sin^{4}(\theta)}\left(-5\sqrt{3}\cos(\theta)-\sqrt{3}\cos(3\theta)-3\sin(\theta)+9\sin(3\theta)\right)\right]$
$\displaystyle\;\;\;\;\;+\alpha^{2}\left[\frac{3\sin(\frac{\pi}{3}-2\theta)(\cos(\theta)+\sin(\frac{\pi}{6}-3\theta))}{\sin^{5}(\theta)}\right]$
$\displaystyle\;\;\;\;\;+\alpha^{3}\left[-\frac{\cos^{3}(\frac{\pi}{6}+2\theta)}{\sin^{6}(\theta)}\right]$

when the trajectory has rotation number $1/3$ and $2/3$, respectively. Once
$\theta$ and $\ell$ are known, the value of $|\operatorname{Tr}S_{3}|$ can be
computed to determine the stability.

######  Example 4.2.

Consider the case when $\partial\Omega$ is the circle of radius $R$. By
symmetry, $\theta_{i}$ is a constant of motion and we are in the same
situation as the previous example. First, suppose the trajectory has a
rotation number $1/3$ and hence $\chi=\pi/3$. Then for each $\mu<R$, there is
a unique $\theta\in(0,\pi/3)$ such that $(s,\theta)$ is the initial condition
for the 3-periodic trajectory. In fact, we can say that this $\theta$
satisfies the equation

$\sin\left(\frac{\pi}{3}-\theta\right)=\frac{\mu\sin(\theta)}{\sqrt{R^{2}+\mu^{2}-2R\mu\cos(\theta)}}$
(4.3)

(see [11]) which has functional solution

$\cos(\theta)=\frac{3\mu+\sqrt{4R^{2}-3\mu^{2}}}{4R}.$ (4.4)

In the case of the circle, $\ell=2R\sin(\theta)$, which turns equation for
$\operatorname{Tr}S_{3}$ into an equation in terms of $R$ and $\mu$. In
particular, the equation $|\operatorname{Tr}S_{3}|=2$ is satisfied for all
$\mu<R$, so the trajectory is parabolic.

Next, consider the case when the rotation number is $2/3$ and so
$\chi=2\pi/3$. Again, for each $\mu<R$ there exists a unique
$\theta\in(\pi/3,2\pi/3)$ such that $(s,\theta)$ is the initial condition for
a 3-periodic trajectory. Just as before, this value of $\theta$ satisfies

$\sin\left(\frac{\pi}{3}-\theta\right)=\frac{\mu\sin(\theta)}{\sqrt{R^{2}+\mu^{2}-2R\mu\cos(\theta)}}$
(4.5)

and has functional solution

$\cos(\theta)=\frac{3\mu-\sqrt{4R^{2}-3\mu^{2}}}{4R}.$ (4.6)

Evaluating the expression for $\operatorname{Tr}S_{3}$ for this value of
$\theta$ yields $|\operatorname{Tr}S_{3}|=2$ for all values of $\mu<R$.
Therefore both types of 3-periodic trajectories in the circle are parabolic.

###  4.2. 4-Periodic Trajectories

The study of 4-periodic trajectories can be performed in a similar fashion.
Again, we have two distinct types of 4-periodic trajectories: those with
rotation number $1/4$ and those with rotation number $3/4$. For an arbitrary
4-periodic trajectory, $\operatorname{Tr}S_{4}$ is quartic in $1/\mu$ with
coefficients that are expressions in terms of $\ell_{i}$ for
$i\in\\{1,3,5,7\\}$ and sines and cosines of combinations of $\chi_{0,2}$,
$\chi_{2,4}$, $\chi_{4,6}$, $\chi_{6,8}$, and $\theta_{i}$ for $i=0,\ldots,7$.
As in the 2-periodic case, no curvature terms $\kappa_{i}$ appear.

Again, we examine examples which incorporate symmetry to simplify the quartic
expression for $\operatorname{Tr}S_{4}$.

######  Example 4.3.

Consider the case when $\partial\Omega$ is the circle of radius $R$. By
symmetry, $\theta_{i}$ is a constant of motion and
$\ell_{1}=\ell_{3}=\ell_{5}=\ell_{7}=:\ell$ and
$\chi_{0,2}=\chi_{2,4}=\chi_{4,6}=\chi_{6,8}=:\chi$. First, suppose the
trajectory has a rotation number $1/4$ and hence $\chi=\pi/4$. Then for each
$\mu<R$, there is a unique $\theta\in(0,\pi/4)$ such that $(s,\theta)$ is the
initial condition for the 4-periodic trajectory. In fact, we can say that this
$\theta$ satisfies the equation

$\sin\left(\frac{\pi}{4}-\theta\right)=\frac{\mu\sin(\theta)}{\sqrt{R^{2}+\mu^{2}-2R\mu\cos(\theta)}}$
(4.7)

(see [11]) which has functional solution

$\cos(\theta)=\frac{R^{2}+\mu^{2}-\mu^{2}\csc^{2}(\frac{\pi}{4}-\theta)\sin^{2}(\theta)}{2R\mu}.$
(4.8)

In the case of the circle, $\ell=2R\sin(\theta)$, which turns equation for
$\operatorname{Tr}S_{4}$ into an equation in terms of $R$ and $\mu$. Letting
$\alpha=R/\mu$, the equation becomes

$\begin{split}\operatorname{Tr}S_{4}&=\frac{2(1-10\cos^{2}(\theta)+17\cos^{4}(\theta))}{\sin^{4}(\theta)}+\alpha\left[\frac{32\cos(2\theta)(3\cos^{2}(\theta)-1)}{\sin^{4}(\theta)}\right]\\\
&\;\;\;\;\;+\alpha^{2}\left[\frac{8\cos^{2}(2\theta)(5+7\cos(2\theta))}{\sin^{4}(\theta)}\right]-\alpha^{3}\left[\frac{64\cos^{3}(2\theta)\cos(\theta)}{\sin^{4}(\theta)}\right]\\\
&\;\;\;\;\;+\alpha^{4}\left[\frac{16\cos^{4}(2\theta)}{\sin^{4}(\theta)}\right].\end{split}$
(4.9)

For the specific value of $\theta$ which produces the 4-periodic trajectory
with rotation number $1/4$, this expression for $\operatorname{Tr}S_{4}$
simplifies to $2$, and hence the trajectory is parabolic.

An analogous calculation can be done to show the 4-periodic trajectory with
rotation number $3/4$ is parabolic.

######  Example 4.4.

Consider the case when $\partial\Omega$ is the ellipse
$x^{2}/a^{2}+y^{2}/b^{2}=1$ with $a>b>0$. Further, suppose the 4-periodic
trajectories share the same horizontal and vertical axes of symmetry as the
ellipse (see figure 7). In particular, this symmetry assumption means that
$\theta_{0}=\theta_{1}=\theta_{4}=\theta_{5}$,
$\theta_{2}=\theta_{3}=\theta_{6}=\theta_{7}$, $\ell_{1}=\ell_{5}$,
$\ell_{3}=\ell_{7}$, and $\chi_{0,2}=\chi_{2,4}=\chi_{4,6}=\chi_{6,8}=:\chi$,
which in turn implies $\ell_{2}=\ell_{4}=\ell_{6}=\ell_{8}$. Suppose the
Cartesian coordinates $(x_{0},-y_{0})$ are the initial point $P_{0}$ in the
fourth quadrant of a 4-periodic trajectory.

Figure 7. Symmetric 4-periodic trajectories in the ellipse with rotation
number $1/4$ (dashed) and $3/4$ (dotted), each with different Larmor radii.

Consider first the case when the trajectory has rotation number $1/4$, and
hence $\chi=\pi/4$. We aim to express the stability of the trajectory in terms
of the Cartesian coordinates of $P_{0}$. Clearly $P_{1}=(x_{0},y_{0})$, and
the point $P_{2}$ will be the intersection of the ellipse with the line
$y=-x+x_{0}+y_{0}$ other than the point $P_{1}$, which is the same as the
second intersection point of the Larmor circle. Further, this means the only
possible initial $x$-coordinates are
$x_{0}\in(a(a^{2}-b^{2})/(a^{2}+b^{2}),a)$ – otherwise the quarter Larmor
circle cannot intersect the ellipse to create the symmetric 4-periodic
trajectory. Since $\chi=\pi/4$, we can write
$P_{2}=(x_{2},y_{2})=(x_{0}-\mu,y_{0}+\mu)$. A few calculations yield the
following:

$\begin{split}\ell_{1}&=2y_{0}\\\
x_{2}&=\frac{a^{2}(x_{0}+y_{0})-ab\sqrt{a^{2}+b^{2}-(x_{0}+y_{0})^{2}}}{a^{2}+b^{2}}\\\
y_{2}&=\frac{b^{2}(x_{0}+y_{0})+ab\sqrt{a^{2}+b^{2}-(x_{0}+y_{0})^{2}}}{a^{2}+b^{2}}\\\
\mu&=\frac{2ab\sqrt{a^{2}+b^{2}-(x_{0}+y_{0})^{2}}}{a^{2}+b^{2}}=\pm\frac{2(b^{2}x_{0}-a^{2}y_{0})}{a^{2}+b^{2}}\\\
\ell_{3}&=2x_{2}\\\
\cos(\theta_{0})&=\frac{b^{2}x_{0}}{\sqrt{a^{4}y_{0}^{2}+b^{4}x_{0}^{2}}}\\\
\cos(\theta_{2})&=\frac{a^{2}y_{2}}{\sqrt{a^{4}y_{2}^{2}+b^{4}x_{2}^{2}}}=\frac{a^{2}(y_{0}+\mu)}{\sqrt{a^{4}(y_{0}+\mu)^{2}+b^{4}(x_{0}-\mu)^{2}}}\\\
\end{split}$ (4.10)

where the second equation for $\mu$ uses plus if $\chi=\pi/4$ and minus if
$\chi=3\pi/4$.

Plugging in each of the above equations into $\operatorname{Tr}S_{4}$ yields a
rational function in $x_{0}$ and $y_{0}$ with coefficients in $a$ and $b$:

$\begin{split}\operatorname{Tr}S_{4}&=\left[\right.16b^{4}\left(a^{2}-b^{2}\right)^{4}x_{0}^{4}-16b^{2}\left(a^{2}-b^{2}\right)^{3}\left(2a^{4}-3a^{2}b^{2}+2b^{4}\right)x_{0}^{3}y_{0}\\\
&\;\;\;\;\;2\left(a^{2}-b^{2}\right)^{2}\left(8a^{8}-40a^{6}b^{2}+49a^{4}b^{4}-40a^{2}b^{6}+8b^{8}\right)x_{0}^{2}y_{0}^{2}\\\
&\;\;\;\;\;8a^{2}\left(a^{2}-b^{2}\right)\left(4a^{8}-10a^{6}b^{2}+13a^{4}b^{4}-10a^{2}b^{6}+4b^{8}\right)x_{0}y_{0}^{3}\\\
&\;\;\;\;\;8a^{4}\left(2a^{8}-4a^{6}b^{2}+5a^{4}b^{4}-4a^{2}b^{6}+2b^{8}\right)y_{0}^{4}\left.\right]/\left[a^{4}b^{4}y_{0}^{2}\left(b^{2}x_{0}-a^{2}\left(x_{0}+2y_{0}\right)\right)^{2}\right].\end{split}$
(4.11)

As $y_{0}$ can be written as a function of $x_{0}$, we can then express the
stability of the 4-periodic trajectory in terms of the coordinate $x_{0}$.

Repeating the above calculations in the case with rotation number $3/4$ and
$\chi=3\pi/4$ yields nearly the same equations as in (4.10) and ultimately
yields the same equation for $\operatorname{Tr}S_{4}$ given above in (4.11).
See Remark 4.5 below for additional commentary on this case.

To further illustrate the stability analysis, consider the specific case when
$a=3$ and $b=2$. Then $|\operatorname{Tr}S_{4}|=2$ at the values

$x_{0}^{*}=\frac{291}{9\sqrt{13}},\qquad
x_{0}^{**}=\sqrt{\frac{88731+1575\sqrt{217}}{14534}},\qquad
x_{0}^{***}=\frac{291}{13\sqrt{61}}.$ (4.12)

Analyzing $|\operatorname{Tr}S_{4}|$ yields that the 4-periodic trajectory is

  * •

elliptic for $x_{0}\in(x_{0}^{*},x_{0}^{**})\cup(x_{0}^{**},x_{0}^{***})$;

  * •

parabolic for $x_{0}\in\\{x_{0}^{*},x_{0}^{**},x_{0}^{***}\\}$; and

  * •

hyperbolic for $x_{0}\in(15/13,x_{0}^{*})\cup(x_{0}^{***},3).$

######  Remark 4.5.

The symmetric 4-periodic trajectories in the ellipse with rotation number
$1/4$ and $3/4$ are related to one another in the following way. Consider the
eight ordered points $P_{0},P_{1},\ldots,P_{7}$ on $\partial\Omega$ which
determine a trajectory with rotation number $1/4$. Then a trajectory in the
order $P_{0},P_{5},P_{6},P_{3},P_{4},P_{1},P_{2},P_{7}$ is a symmetric
4-periodic trajectory with rotation number $3/4$. See figure 8. This is partly
due to the fact that for each segment $\ell_{2j}$, there are two supplementary
values of $\chi$ which realize this segment. This in turn creates two Larmor
arcs whose union (after a suitable reflection across $\ell_{2j}$) is a
complete Larmor circle. As such, these two 4-periodic trajectories are
complementary to one another.

Figure 8. Symmetric 4-periodic trajectories in the ellipse with rotation
number $1/4$ (dashed) and $3/4$ (dotted), each with the same Larmor radii and
same points $P_{i}$.

######  Remark 4.6.

There is a second family of symmetric 4-periodic trajectories in the ellipse:
those whose Larmor circles are symmetric about the axes of symmetry of the
ellipse. The stability of these trajectories can be analyzed in a similar
fashion to those above, though we do not address this example here.

######  Example 4.7.

Consider the case when $\partial\Omega$ is the one-parameter family of curves
$x^{2k}+y^{2k}=1$, $k\in\mathbb{N}$. As the case $k=1$ corresponds to the unit
circle, we also assume $k>1$. By the dihedral symmetry of $\partial\Omega$,
the 4-periodic trajectories have constant $\theta_{i}:=\theta$, $\ell_{2j}$,
$\ell_{2j+1}$, and $\chi_{2l,2l+2}=:\chi$. In addition to the two families of
trajectories determined by the rotation number, we can also consider two
further families based upon whether the centers of the Larmor circles are on
the diagonal axes of symmetry of $\partial\Omega$ or on the horizontal and
vertical axes of symmetry of $\partial\Omega$.

First, consider the case when the rotation number is $1/4$ and the centers of
the Larmor circles are on the diagonal axes of symmetry of $\partial\Omega$.
Suppose the point $P_{0}=(x_{0},-y_{0})\in\partial\Omega$ is in the fourth
quadrant and the trajectory has initial velocity in the direction $(0,1)$ so
that $P_{1}=(x_{0},y_{0})\in\partial\Omega$ is in the first quadrant. To
produce the 4-periodic trajectory, it must be the case that
$x_{0}\in(2^{-\frac{1}{2k}},1)$ and $P_{2}=(y_{0},x_{0})$. Further, we know
$\chi=\pi/4$, which implies the center of the first Larmor arc is at
$(y_{0},y_{0})$ and hence $\mu=x_{0}-y_{0}$.

Due to the curvature of $\partial\Omega$, it is possible for the Larmor arc of
this first segment of the trajectory to intersect $\partial\Omega$ in up to
four different locations. To ensure the point $P_{2}$ is the proper point of
intersection of the Larmor circle with $\partial\Omega$, there must be exactly
two intersection points between the Larmor circle and $\partial\Omega$ so that
the billiard can “round the corner” of $\partial\Omega$. That is, there exists
$\hat{x}\in(2^{-\frac{1}{2k}},1)$ such that

$(x-y_{0})^{2}+(y-y_{0})^{2}=\mu^{2}\qquad\text{and}\qquad x^{2k}+y^{2k}=1$
(4.13)

has exactly three real roots in $x$: $x=\hat{x}$, $x=2^{-\frac{1}{2k}}$, and
$x=y_{0}$, with $y_{0}<2^{-\frac{1}{2k}}<\hat{x}$. Alternately, we can say
that $\hat{x}$ is the value of $x_{0}$ such that the distance from the center
of the Larmor circle, $(y_{0},y_{0})$ and
$(2^{-\frac{1}{2k}},2^{-\frac{1}{2k}})$ is exactly $\mu$. For this value
$\hat{x}$, these two equations (4.13) have exactly two real roots for
$x_{0}\in(2^{-\frac{1}{2k}},\hat{x})$, three real roots when $x_{0}=\hat{x}$,
and four real roots when $x_{0}\in(\hat{x},1)$. This behavior is analogous to
a circle whose center lies on the diagonal of a square and can have multiple
intersections with the sides and corners of the square. We now assume
$x_{0}\in(2^{-\frac{1}{2k}},\hat{x})$.

We can now compute several of the relevant quantities that appear in
$\operatorname{Tr}S_{4}$ :

$\begin{split}\ell_{1}&=2y_{0}\\\ \mu&=x_{0}-y_{0}\\\ \ell_{2}&=2\mu\\\
\tan(\theta)&=\frac{y_{0}^{2k-1}}{x_{0}^{2k-1}}.\\\ \end{split}$ (4.14)

These quantities produce the following expression for $\operatorname{Tr}S_{4}$
in terms of $x_{0}$ and $y_{0}$:

$\displaystyle\begin{split}\operatorname{Tr}S_{4}&=2+\left[16x_{0}^{2}\left(x_{0}^{2k-2}-y_{0}^{2k-2}\right)\left(x_{0}^{4k-2}-y_{0}^{4k-2}\right)\left(x_{0}^{2k-2}-y_{0}^{2k-2}+2x_{0}^{-2}y_{0}^{2k-1}\right)\right.\\\
&\;\;\;\;\;\cdot\left.\left(y_{0}^{4k-2}-x_{0}^{4k-2}+(x_{0}-y_{0})x_{0}^{2k-1}y_{0}^{2k-2}\right)^{2}\right]/\left[y_{0}^{16k-12}(x_{0}-y_{0})^{4}\right]\end{split}$
(4.15)

By assumption, $0<y_{0}<x_{0}<1$, and each term in the above rational function
will be strictly positive for $x_{0}\in(2^{-\frac{1}{2k}},\hat{x})$. Therefore
this 4-periodic trajectory will be hyperbolic for all
$x_{0}\in(2^{-\frac{1}{2k}},\hat{x})$ and integer $k\geq 2$.

Repeating the above calculations in the case when the rotation number is $3/4$
and $x_{0}\in(-1,2^{-\frac{1}{2k}})$ yields the same equation for
$\operatorname{Tr}S_{4}$ as (4.15). Similar to before, this expression will be
larger than 2 for
$x_{0}\in(-1,2^{-\frac{1}{2k}})\setminus\\{-2^{-\frac{1}{2k}}\\}$ and be equal
to 2 for $x_{0}=-2^{-\frac{1}{2k}}$. Therefore the 4-periodic trajectory with
rotation number $3/4$ is hyperbolic for all
$x_{0}\in(-1,2^{-\frac{1}{2k}})\setminus\\{-2^{-\frac{1}{2k}}\\}$ and
parabolic for $x_{0}=-2^{-\frac{1}{2k}}$. There is also the same duality
between this trajectory and the trajectory with rotation number $1/4$ as is
discussed in Remark 4.5.

(a)  | (b)
---|---
Figure 9. Symmetric 4-periodic trajectories in the curve $x^{2k}+y^{2k}=1$
with rotation number $1/4$ (dashed) and $3/4$ (dotted), with different Larmor
radii, and with Larmor centers on (a) the lines $y=\pm x$, and (b) on the
coordinate axes.

Next, consider the case when the rotation number is $1/4$ and the Larmor
centers are on the coordinate axes. Then, as before, all $\theta_{i}$,
$\ell_{2i}$, $\ell_{2i+1}$, $\chi_{2i,2i+2}$ are constant. If the point
$P_{0}=(x_{0},y_{0})$ is in the first quadrant, then
$x_{0}\in(2^{-\frac{1}{2k}},1)$, $P_{1}=(y_{0},x_{0})$, and
$P_{2}=(-y_{0},x_{0})$. Simple geometry yields

$\begin{split}\ell_{1}&=\sqrt{2}(x_{0}-y_{0})\\\ \mu&=\sqrt{2}y_{0}\\\
\ell_{2}&=2y_{0}\\\
\tan(\theta)&=\frac{x_{0}^{2k}y_{0}-x_{0}y_{0}^{2k}}{x_{0}^{2k}y_{0}+x_{0}y_{0}^{2k}}\end{split}$
(4.16)

and the Larmor centers are at the points $(0,\pm(x_{0}-y_{0}))$ and
$(\pm(x_{0}-y_{0}),0)$.

These quantities produce the following expression for $\operatorname{Tr}S_{4}$
in terms of $x_{0}$ and $y_{0}$:

$\displaystyle\begin{split}\operatorname{Tr}S_{4}&=2-\left[64x_{0}^{2k+1}y_{0}^{2k}\left(x_{0}^{2k}y_{0}^{2}-x_{0}^{2}y_{0}^{2k}\right)\left(x_{0}^{2k}(x_{0}-2y_{0})+2x_{0}y_{0}^{2k}\right)\right.\\\
&\;\;\;\;\;\cdot\left.\left(x_{0}^{4k}y_{0}^{2}-x_{0}^{2}y_{0}^{4k}-2(x_{0}-y_{0})x_{0}^{2k+1}y_{0}^{2k}\right)^{2}\right]/\left[x_{0}^{2k}y_{0}-x_{0}y_{0}^{2k}\right]^{8}\end{split}$
(4.17)

The above rational function equals zero at exactly one point,
$x_{0}=\check{x}$ in the interval $(2^{-\frac{1}{2k}},1)$. A quick analysis of
this rational function and the above expression tells us the trajectory will
be hyperbolic for $x_{0}\in(2^{-\frac{1}{2k}},\check{x})$, parabolic for
$x_{0}=\check{x}$, and elliptic for $x_{0}\in(\check{x},1)$.

We can repeat the above calculations in the case when the rotation number is
$3/4$. In particular, this case is valid for $x_{0}\in(-2^{-\frac{1}{2k}},1)$.
Suppose $P_{0}=(x_{0},y_{0})$ is in the first quadrant and
$P_{1}=(-y_{0},x_{0})$ is in the third quadrant. Then $P_{2}=(y_{0},-x_{0})$
is in the fourth quadrant and the point $P_{0}$ and $P_{1}$ determine the
start of a 4-periodic trajectory. In this case we have

$\begin{split}\ell_{1}&=\sqrt{2}(x_{0}+y_{0})\\\ \mu&=\sqrt{2}y_{0}\\\
\ell_{2}&=2y_{0}\\\
\tan(\theta)&=\frac{x_{0}^{2k}y_{0}+x_{0}y_{0}^{2k}}{x_{0}y_{0}^{2k}-x_{0}^{2k}y_{0}}\end{split}$
(4.18)

and the Larmor centers are at the points $(0,\pm(x_{0}+y_{0}))$ and
$(\pm(x_{0}+y_{0}),0)$.

These quantities produce a similar expression for $\operatorname{Tr}S_{4}$ in
terms of $x_{0}$ and $y_{0}$ as the previous case:

$\displaystyle\begin{split}\operatorname{Tr}S_{4}&=2-\left[64x_{0}^{2k}y_{0}^{2k-2}\left(x_{0}^{2k-2}-y_{0}^{2k-2}\right)\left(x_{0}^{2k-1}(x_{0}+2y_{0})+y_{0}^{2k}\right)\right.\\\
&\;\;\;\;\;\cdot\left.\left(x_{0}^{4k-2}-y_{0}^{4k-2}-2(x_{0}+y_{0})x_{0}^{2k-1}y_{0}^{2k-2}\right)^{2}\right]/\left[x_{0}^{2k-1}+y_{0}^{2k-1}\right]^{8}.\end{split}$
(4.19)

An analysis of this equation yields five values of
$x_{0}\in(-2^{-\frac{1}{2k}},1)$ for which $\operatorname{Tr}S_{4}$ is
parabolic, call them $\tilde{x}_{j}$, $j\in\\{1,\ldots,5\\}$ with
$x_{i}<x_{j}$ for $i<j$. In particular $\tilde{x}_{1}=0$. Further analysis
yields that the trajectory is

  * •

hyperbolic for
$x_{0}\in(-2^{-\frac{1}{2k}},\tilde{x}_{1})\cup(\tilde{x}_{1},\tilde{x}_{2})$

  * •

parabolic for
$x_{0}\in\\{\tilde{x}_{1},\tilde{x}_{2},\tilde{x}_{3},\tilde{x}_{4},\tilde{x}_{5}\\}$

  * •

elliptic for
$x_{0}\in(\tilde{x}_{2},\tilde{x}_{3})\cup(\tilde{x}_{3},\tilde{x}_{4})\cup(\tilde{x}_{4},\tilde{x}_{5})\cup(\tilde{x}_{5},1)$.

##  5\. Conclusions

We have established a linear stability criterion for 2-periodic inverse
magnetic billiard trajectories in convex and concave domains and have applied
this to examples such as the ellipse and one-parameter family of curves given
by $x^{2k}+y^{2k}=1$ for $k\in\mathbb{N}$. Further, comparisons have been made
to the linear stability criteria for standard and magnetic billiards, noting
the similarities (e.g. analogous stability results for billiards and inverse
magnetic billiards in the ellipse) and differences (e.g. the linear stability
of inverse magnetic billiards does not depend upon the curvature of the
boundary at the exit and reentry points compared to the linear stability of
standard billiards which does depend upon the curvature of the boundary at the
reflection points). We also established the linear stability of symmetric 3-
and 4-periodic inverse magnetic billiards trajectories in the same domains.
Further, we made a geometric characterization of an isolated parabolic
2-periodic billiard trajectory in an ellipse in terms of rotation numbers and
caustics.

A further direction of research in this area is reconciling the following: the
approach of [15] establishes linear stability criterion for billiards using
the generating function $\mathcal{L}$ of the billiard map and its Hessian
determinant evaluated at nondegenerate critical points. This criterion is
identical to the criterion constructed via the stability matrix $S_{n}$.
However, this generating function approach to linear stability in the setting
of inverse magnetic billiards does not produce viable results in the
2-periodic case: the generating function $G$ for inverse magnetic billiards is
not sufficiently differentiable when $\chi=\pi/2$. An equivalent statement in
the magnetic billiards setting is made in [2] in the context of asymptotic
expansions of the magnetic billiard map. The technique of [15] establishes a
method to transform a degenerate critical point to a nondegenerate critical
point, and finding magnetic and inverse magnetic analogues to this method
constitute one path to overcoming this obstruction.

## Acknowledgements

This research is supported by the Discovery Project No. DP190101838 _Billiards
within confocal quadrics and beyond_ from the Australian Research Council. A
portion of this research is based upon work supported by the National Science
Foundation under Grant No. DMS-1440140 while the author was in residence at
the Mathematical Sciences Research Institute in Berkeley, California, during
the Fall 2018 semester.

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