# Equilibria and energy minimizers for an interaction model on the hyperbolic
space

Razvan C. Fetecau Department of Mathematics, Simon Fraser University, 8888
University Dr., Burnaby, BC V5A 1S6, Canada<EMAIL_ADDRESS>Hansol Park
Department of Mathematics, Simon Fraser University, 8888 University Dr.,
Burnaby, BC V5A 1S6, Canada<EMAIL_ADDRESS>
###### Abstract

We study an intrinsic model for collective behaviour on the hyperbolic space
$\mathbb{H}^{n}$. We investigate the equilibria of the aggregation equation
(or equivalently, the critical points of the associated interaction energy)
for interaction potentials that include Newtonian repulsion. By using the
method of moving planes, we establish the radial symmetry and the monotonicity
of equilibria supported on geodesic balls of $\mathbb{H}^{n}$. We find several
explicit forms of equilibria and show that one such equilibrium is a global
energy minimizer. We also consider more general potentials and utilize a
technique used for $\mathbb{R}^{n}$ to establish the existence of compactly
supported global minimizers. Numerical simulations are presented, suggesting
that some of the equilibria studied here are global attractors. The key tool
in our investigations is a family of isometries of $\mathbb{H}^{n}$ that we
have developed for this purpose.

Keywords: swarming on manifolds, hyperbolic space, energy minimizers,
Newtonian potential

##  1 Introduction

In recent years there has been extensive work on investigating minimizers of
nonlocal interaction energies, as motivated by a wide range of applications in
a variety of disciplines, such as biology, physics, economics and social
sciences [7, 16, 39, 2, 10, 24]. With very few exceptions, such works have
studied the Euclidean case, where individual particles in $\mathbb{R}^{n}$
interact via an interaction potential that depends on the Euclidean distance
between them. Since our goal in this paper is to consider nonlocal interaction
energies set up on Riemannian manifolds (in particular, on the hyperbolic
space), we will present the general setup from the start.

Consider an $n$-dimensional Riemannian manifold $M$, and denote by
$\mathcal{P}(M)$ the set of probability measures on $M$. Define the
interaction energy $E:\mathcal{P}(M)\to\mathbb{R}\cup\\{\infty\\}$ by

$E[\rho]=\frac{1}{2}\iint_{M\times
M}K(x,y)\mathrm{d}\rho(x)\mathrm{d}\rho(y),$ (1)

where $K:M\times M\to\mathbb{R}$ is an interaction potential that models
attractive and repulsive interactions. More precisely, the negative manifold
gradient $-\nabla_{M}K(x,y)$ (the gradient with respect to $x$) of $K$
provides the interaction force that a point mass located at $x$ feels by
interacting with a point mass located at $y$ [7, 23].

The interaction energy (1) is closely related to the following evolution
equation for a population density $\rho$:

$\partial_{t}\rho-\nabla_{M}\cdot(\rho\nabla_{M}K\ast\rho)=0,$ (2)

where $\nabla_{M}\cdot$ denotes the manifold divergence on $M$. Here, for a
time-dependent measure $\rho_{t}$ on $M$, the convolution $K\ast\rho_{t}$ is
defined by

$K*\rho_{t}(x)=\int_{M}K(x,y)\mathrm{d}\rho_{t}(y).$ (3)

Equation (2) is in the form of an active transport equation for the density
$\rho$, with a nonlocal velocity field given by

$\mathbf{v}[\rho](x,t)=-\nabla_{M}K*\rho_{t}(x).$ (4)

Note that $\displaystyle{\mathbf{v}[\rho]=-\nabla_{M}\frac{\delta
E[\rho]}{\delta\rho}}$, which sets (at least formally) equation (2) as the
gradient flow of the energy (1). One can check by a direct calculation that
indeed, formally, the energy $E$ decays in time along solutions of equation
(2) [23]. Equilibria (steady states) of the aggregation model (2) are the
critical points of the interaction energy, and from a dynamical point of view,
we expect that solutions to model (2) approach asymptotically as $t\to\infty$,
local minimizers of the energy.

There has been extensive literature in recent years on interaction energies of
type (1), with the vast majority of the works focusing on the case
$M=\mathbb{R}^{n}$. One approach is to use the techniques developed in [1] and
formulate the gradient flow of the energy $E$ on the space of probability
measures with finite second (or other order) moments, equipped with the
$2$-Wasserstein metric [13, 9, 41]. Another approach is to use direct methods
of the calculus of variations and establish the existence of global minimizers
(ground states) of the interaction energy [16, 39, 24]. There are also various
studies of quantitative and qualitative properties of minimizers, such as
their regularity [8] and compactness and dimensionality of their support [7,
2, 10].

Model (2) can exhibit a very diverse range of swarming or self-organized
behaviour. We refer to [33, 40, 21, 3, 12] for explicit calculations and
numerical illustrations of equilibria for the model in $\mathbb{R}^{n}$, which
include aggregations on disks, annuli, rings, and soccer balls. The model on
manifolds is far less investigated, but nevertheless, a similarly diverse set
of equilibria was demonstrated in [23] for the hyperbolic plane and also,
interesting aggregation patterns were shown in [19] for the model set up on
the rotation group $SO(3)$. Applications of model (2) are numerous, e.g., to
flocking and swarming of biological organisms [11], material science and
granular media [13], self-assembly of nanoparticles [31], robotics [25, 32]
and opinion formation [35].

In our study we will follow the intrinsic approach pursued in various recent
papers [23, 19, 22] and assume that $K(x,y)$ only depends on the geodesic
distance $d(x,y)$ between the two points. By an abuse of notation we write
$K(x,y)=K(d(x,y))$. In contrast, when $M$ comes with a canonical embedding in
a larger Euclidean space (e.g., a surface in $\mathbb{R}^{3}$), $K(x,y)$ can
be assumed to depend on the Euclidean distance $|x-y|$ in the ambient space
between $x$ and $y$; we refer to this case as the extrinsic approach. The two
approaches yield very different models (unless $M=\mathbb{R}^{n}$, in which
case the two models are the same). For extrinsic models we refer to [41, 36]
for studies on well-posedness by gradient flow techniques, and to various
works on the emergent behaviours of the extrinsic models on manifolds, e.g.,
on sphere [15], unitary matrices [34, 30, 29], hyperbolic space [27], and
Stiefel manifolds [28]. Relevant to the present paper, we note the work in
[27], where the authors study synchronization behaviour on the hyperbolic
space with an extrinsic quadratic potential (i.e., $K(x,y)=|x-y|^{2}$) for the
discrete analogue of (2). A related study is done in [37]. The emergence of
self-synchronization has been widely studied in the literature, due to its
numerous occurrences in biological, physical and chemical systems (e.g.,
flashing of fireflies, neuronal synchronization in the brain, quantum
synchronization) – see [34, 27] and references therein.

For intrinsic models, the sign of $K^{\prime}$ determines whether the
interactions are repulsive or attractive in nature. Indeed, provided $x$ and
$y$ are not in the cut locus of each other, by chain rule one gets

$-\nabla_{M}K(d(x,y))=K^{\prime}(d(x,y))\frac{\log_{x}y}{d(x,y)},$

where $\log_{x}y$ denotes the Riemannian logarithm map (i.e., the inverse of
the Riemannian exponential map) on $M$. Therefore, by interacting with the
point mass at $y$, the point mass at $x$ is driven by a force of magnitude
proportional to $|K^{\prime}(d(x,y))|$, to move either towards $y$ (if
$K^{\prime}(d(x,y))>0$) or away from $y$ (if $K^{\prime}(d(x,y))<0$). To
obtain non-trivial swarming behaviours, repulsion and attraction must balance
each other. Typically, such interaction potentials have short-range repulsion
and long-range attraction, i.e., $K^{\prime}(r)<0$ for small $r$ and
$K^{\prime}(r)>0$ for large $r$ (see Figure 2).

The interaction energy for intrinsic models is invariant under isometries.
Indeed, if $f:M\to M$ is an isometry of $M$ and $f_{\\#}\rho$ is the push-
forward by $f$ of the measure $\rho$, then it is immediate to show that

$E[f_{\\#}\rho]=E[\rho].$ (5)

Consequently, the energy minimizers can only be expected to be unique up to
isometries.

In this paper we are exclusively concerned with studying the nonlocal
aggregation model on the hyperbolic space $\mathbb{H}^{n}$, represented as a
one-sheeted hyperboloid in $\mathbb{R}^{n+1}$ endowed with the Lorentzian
inner product. Based on the Lorentz transform, we construct a family of
isometries of $\mathbb{H}^{n}$ and a binary operation on
$\mathbb{H}^{n}\times\mathbb{H}^{n}$, along with a concept of coordinates on
$\mathbb{H}^{n}$, that are the key tools in the results we establish in the
present work.

In Section 3 we consider attractive-repulsive interaction potentials that
include Newtonian repulsion, i.e., potentials for which the repulsion
component is given by the Green’s function of the negative Laplacian on
$\mathbb{H}^{n}$. We note here that the Green’s function on $\mathbb{H}^{n}$
has an explicit expression [17] – see (28). For the model on $\mathbb{R}^{n}$,
potentials in this form have been given extensive attention in recent years
[6, 21, 20, 12, 38], and a potential with Newtonian repulsion was also
considered on the hyperbolic plane [23]. In this section we only consider
equilibria that are absolutely continuous with respect to the canonical volume
measure on $\mathbb{H}^{n}$. This choice is motivated by results in the
Euclidean space $\mathbb{R}^{n}$ which have established that for interaction
potentials with Newtonian repulsion, the local minimizers of the interaction
energy are absolutely continuous with respect to the Lebegue measure [12].

We study equilibria supported in geodesic balls of $\mathbb{H}^{n}$ and show
that such equilibria must be radially symmetric and monotone with respect to
the centre of the ball. To show this result (Theorem 3.3) we use the method of
moving planes, a well-known method for studying qualitative properties of
positive solutions of elliptic equations [26]. More recently, the method found
applications in integral equations [14, 20]. The challenge here is to adapt
the method to the hyperboloid; to the best of our knowledge this the first
application of the method of moving planes to integral equations on manifolds.
We also find several explicit forms of radially symmetric equilibria,
corresponding to particular choices of the attractive part of the potential.
For one such attractive potential, we show that the equilibrium we computed is
in fact the global energy minimizer. We also present various numerical
simulations in Section 5.

In Section 4 we investigate the existence of compactly supported global
minimizers of the energy (1) for $M=\mathbb{H}^{n}$. As we apply and follow
closely the technique used by Cañizo et al. [7] for the Euclidean case, we
only highlight the main differences from their work. In other words, we are
not aiming for a self-contained presentation in Section 4, as some of results
from [7] would transfer with no modifications to $\mathbb{H}^{n}$.
Nevertheless, we show how the family of isometries and the concept of
coordinates on $\mathbb{H}^{n}$ that we developed, enable a similar approach
as for the Euclidean case. In particular, minimizers cannot have too large
gaps, as otherwise, an isometric displacement of one of the components would
decrease the energy (Lemma 4.3).

Finally, we provide some numerical results in Section 5. Using various
interaction potentials, we illustrate the dynamical evolution to steady states
of radially symmetric solutions to (2). The equilibria we present are
qualitatively different in terms in their monotonicity, in consistency with
the analytical findings. Numerics suggest these equilibria are global
attractors for the dynamics (at least for radial initial densities).

The summary of the paper is the following. Section 2 provides key background
needed for the paper, on the variational formalism and on the geometrical
properties of $\mathbb{H}^{n}$. Section 3 contains analytical results for
interaction potentials that contain Newtonian repulsion (characterization of
monotonicity, explicit forms for equilibria, convexity of energy). In Section
4 we establish the existence of compactly supported global minimizers,
following the approach for $\mathbb{R}^{n}$ from [7]. Proof of various lemmas
are deferred to the Appendix.

##  2 Preliminaries

In this section we provide some background on the variational formalism used
in our paper, as well as on the various geometrical properties of
$\mathbb{H}^{n}$, in particular on a family of isometries of the hyperbolic
space that plays a key role in our studies.

###  2.1 Variational approach: Euler-Lagrange equations

For the interaction energy set up on the Euclidean space $\mathbb{R}^{n}$,
there exists a well-established approach to characterize its local minimizers
(by Euler-Lagrange equations) in the topology of transport distances [2]. As
this approach extends immediately to arbitrary manifolds, we will simply list
below the main notations and results.

Take $p$ to be a fixed point in $M$, and denote by $\mathcal{P}_{2}(M)$ the
set of probability measures that have finite moments of order $2$ (with
respect to $p$), i.e.,

$\mathcal{P}_{2}(M)=\left\\{\mu\in\mathcal{P}(M):\int_{M}d^{2}(x,p)\mathrm{d}\mu(x)<\infty\right\\}.$

The $2$-Wasserstein distance between two measures
$\mu,\sigma\in\mathcal{P}_{2}(M)$ is given by:

$d_{2}^{2}(\mu,\sigma)=\inf_{\pi\in\Pi(\mu,\sigma)}\iint_{M\times
M}d^{2}(x,y)\mathrm{d}\pi(x,y),$

where $\Pi(\mu,\sigma)\subset\mathcal{P}(M\times M)$ is the set of transport
plans between $\mu$ and $\sigma$, i.e., the set of probability measures on the
product space $M\times M$ with first and second marginals $\mu$ and $\sigma$,
respectively.

A probability measure $\mu\in\mathcal{P}_{2}(M)$ is a local minimizer of $E$
with respect to the $d_{2}$-distance provided there exists $\epsilon>0$ such
that $E[\sigma]\geq E[\mu]$, for all $\sigma\in B(\mu,\epsilon)$, where
$B(\mu,\epsilon)$ denotes the open ball (in $d_{2}$) centred at $\mu$ and of
radius $\epsilon$.

We will pay particular attention in this paper to probability measures that
are absolutely continuous with respect to the canonical volume measure
$\mathrm{d}x$; we denote this space by $\mathcal{P}^{ac}(M)$. Throughout the
paper, we will refer to an absolutely continuous measure directly by its
density $\rho$, and by abuse of notation write $\rho\in\mathcal{P}^{ac}(M)$ to
mean $\mathrm{d}\rho(x)=\rho(x)\mathrm{d}x\in\mathcal{P}^{ac}(M)$. Finally, we
denote by $\mathcal{P}^{ac}_{c}(M)$ the space of absolutely continuous
probability measures with compact support in $M$.

The following result, stated originally for $M=\mathbb{R}^{n}$, extends
trivially to arbitrary Riemannian manifolds.

######  Proposition 2.1 ([2], Theorem 4 and Remark 4).

Assume the interaction potential $K$ is bounded from below and lower
semicontinuous. Let $\rho\in\mathcal{P}^{ac}(M)\cap\mathcal{P}_{2}(M)$ be a
local minimizer of $E$ with respect to $d_{2}$. Then,

$\displaystyle\Lambda(x)$ $\displaystyle=\lambda,\qquad\text{ for a.e.
}x\in\mathrm{supp}(\rho),$ (6a) $\displaystyle\Lambda(x)$
$\displaystyle\geq\lambda,\qquad\text{ for a.e. on
}M\setminus\mathrm{supp}(\rho),$ (6b)

where

$\displaystyle\Lambda(x):=\int_{M}K(x,y)\rho(y)\mathrm{d}y,$ (7)

and $\lambda=2E[\rho]$ is a constant.

Condition (6a) is shown by taking perturbations of $\rho$ supported in
$\mathrm{supp}(\rho)$. The condition is equivalent to $\rho$ being a critical
point of $E$, at which the first variation of the energy vanishes. The
constant $\lambda$, which plays the role of a Lagrange multiplier in the
derivation of (6a), has a physical interpretation [4]: it represents the
energy per unit mass felt by a point mass at position $x$ due to interaction
with all points in $\mathrm{supp}(\rho)$.

We also note that critical points of the interaction energy represent
equilibrium solutions (steady states) of the aggregation model (2). This can
be inferred from the fact that the velocity
$\mathbf{v}(x)=-\nabla_{M}K\ast\rho(x)=-\nabla_{M}\Lambda(x)$ vanishes for
$x\in\mathrm{supp}(\rho)$, given that $\Lambda$ is constant in
$\mathrm{supp}(\rho)$. We will use these concepts (critical point of the
energy, equilibrium solution, steady state) interchangeably in the sequel.

The necessary condition (6b) is found by taking perturbations of $\rho$ that
can be supported anywhere in $M$, in particular outside $\mathrm{supp}(\rho)$.
The interpretation of (6b) is that transporting mass from the support of
$\rho$ into its complement increases the total energy (hence $\rho$ is a local
minimizer, being in a favourably energetic state).

###  2.2 Geometric properties of the hyperbolic space $\mathbb{H}^{n}$

In this part, we present some geometric properties of the hyperbolic space
$\mathbb{H}^{n}$ which will be used in the paper.

####  2.2.1 Hyperboloid model, polar coordinates and notations

We consider the space $\mathbb{R}^{n+1}$ endowed with the (negative)
Lorentzian inner product:

$\langle x,y\rangle=x_{0}y_{0}-x_{1}y_{1}-\cdots-x_{n}y_{n},$ (8)

where $x=(x_{0},x_{1},\dots,x_{n})$,
$y=(y_{0},y_{1},\dots,y_{n})\in\mathbb{R}^{n+1}$. The inner product (8)
induces the (complex-valued) norm defined by

$\|x\|=\langle x,x\rangle^{\frac{1}{2}},\qquad\text{ for
}x\in\mathbb{R}^{n+1}.$ (9)

We use the hyperboloid model of the hyperbolic space and take:

$\mathbb{H}^{n}:=\\{x\in\mathbb{R}^{n+1}:-x_{0}^{2}+x_{1}^{2}+\dots+x_{n}^{2}=-1,x_{0}>0\\}\subset\mathbb{R}^{n+1},$

or, written differently,

$\mathbb{H}^{n}=\\{x\in\mathbb{R}^{n+1}:x^{\top}\eta x=-1,x_{0}>0\\},$

where $\eta=\mathrm{diag}(-1,\underbrace{1,1,\cdots,1}_{n-\text{times}})$ is a
matrix of size $(n+1)\times(n+1)$.

The geodesic distance $d(x,y)$ on $\mathbb{H}^{n}$ between two points $x$ and
$y$, is given by:

$d(x,y)=\cosh^{-1}(\langle
x,y\rangle)=\cosh^{-1}(x_{0}y_{0}-x_{1}y_{1}-\cdots-x_{n}y_{n}).$ (10)

For $x\in\mathbb{H}^{n}$ and $R>0$, we denote by $B_{R}(x)$ the open geodesic
ball in $\mathbb{H}^{n}$ centred at $x$ of radius $R$, i.e.,

$B_{R}(x)=\\{y\in\mathbb{H}^{n}:d(x,y)<R\\}.$

We use the hyperbolic polar coordinates on $\mathbb{H}^{n}$, given by:

$\displaystyle(\theta,\xi)\in[0,\infty)\times\mathbb{S}^{n-1}\mapsto(\cosh\theta,\sinh\theta\,\xi)\in\mathbb{H}^{n}\subset\mathbb{R}^{n+1},$
(11)

where $\mathbb{S}^{n-1}$ denotes the $(n-1)$-dimensional unit sphere embedded
in $\mathbb{R}^{n}$. We denote by $v$ the vertex of $\mathbb{H}^{n}$,
corresponding to $\theta=0$, i.e., $v=(1,0,\dots,0)\in\mathbb{R}^{n+1}$. The
vertex $v$ plays the role of the origin in the Euclidean space.

In polar coordinates, the Laplace-Beltrami operator $\Delta_{\mathbb{H}^{n}}$
can be written as:

$\displaystyle\begin{aligned}
\Delta_{\mathbb{H}^{n}}f(\theta,\xi)&=\sinh^{1-n}\theta\,\partial_{\theta}\left(\sinh^{n-1}\theta\partial_{\theta}f\right)+\sinh^{-2}\theta\Delta_{\xi}f\\\
&=\partial_{\theta\theta}f+(n-1)\coth\theta\,\partial_{\theta}f+\sinh^{-2}\theta\Delta_{\xi}f,\end{aligned}$
(12)

where $\Delta_{\xi}$ is the Laplace–Beltrami operator on $\mathbb{S}^{n-1}$,
with respect to the variable $\xi$. Also, the canonical volume form on
$\mathbb{H}^{n}$ in these coordinates is given by

$\mathrm{d}x=\sinh^{n-1}\theta\,\mathrm{d}\theta\mathrm{d}\sigma_{\xi}.$ (13)

We denote by $\alpha(n)$ the volume of the $n$-dimensional unit ball in
$\mathbb{R}^{n}$; consequently, $n\alpha(n)$ is the surface area of
$\mathbb{S}^{n-1}$ in $\mathbb{R}^{n}$. Finally, we denote the standard
Euclidean norm in $\mathbb{R}^{n}$ by $|\cdot|$, i.e.,

$|w|=\left(w_{1}^{2}+\cdots+w_{n}^{2}\right)^{\frac{1}{2}},\qquad\text{ for
}w=(w_{1},\dots,w_{n})\in\mathbb{R}^{n}.$

####  2.2.2 Lorentz transform

For a fixed vector $w=(w_{1},\dots,w_{n})\in\mathbb{R}^{n}$ with $|w|<1$,
consider the following matrix $B(w)$ of size $(n+1)\times(n+1)$ with indices
$0,1,\dots,n$:

$\displaystyle B(w)_{00}=\gamma,\quad B(w)_{i0}=B(w)_{0i}=-\gamma w_{i},\quad
B(w)_{ij}=\delta_{ij}+(\gamma-1)\frac{w_{i}w_{j}}{|w|^{2}},\quad\forall
i,j=1,2,\cdots,n,$ (14)

where $\gamma=(1-|w|^{2})^{-1/2}$. Equivalently, the matrix expression of
$B(w)$ is

$B(w)=\begin{bmatrix}\gamma&-\gamma w^{\top}\\\ -\gamma
w&I_{n}+(\gamma-1)\frac{ww^{\top}}{|w|^{2}}\end{bmatrix}.$

We denote this matrix by $B$, since the corresponding concept is the Lorentz
boost.

######  Lemma 2.2.

For a fixed $w\in\mathbb{R}^{n}$ with $|w|<1$, $B(w)$ maps $\mathbb{H}^{n}$
into $\mathbb{H}^{n}$, i.e.,

$y=B(w)x\in\mathbb{H}^{n},\qquad\text{ for all
}x\in\mathbb{H}^{n}\subset\mathbb{R}^{n+1}.$

Here, $y$ is the matrix product of $B(w)\in\mathbb{R}^{(n+1)\times(n+1)}$ and
$x\in\mathbb{R}^{n+1}$.

###### Proof.

By direct calculations (see Appendix A for details) one can show:

$y_{0}^{2}-y_{1}^{2}-\cdots-y_{n}^{2}=x_{0}^{2}-x_{1}^{2}-\cdots-x_{n}^{2},$

which implies the desired result. ∎

From the proof of Lemma 2.2, for any fixed $w\in\mathbb{R}^{n}$ with $|w|<1$,
it holds that:

$x^{\top}\eta x=(B(w)x)^{\top}\eta B(w)x=x^{\top}(B(w)^{\top}\eta
B(w))x,\qquad\forall~{}x\in\mathbb{R}^{n+1}.$

This yields

$x^{\top}(\eta-B(w)^{\top}\eta B(w))x=0,\qquad\forall~{}x\in\mathbb{R}^{n+1},$

and hence,

$\displaystyle B(w)^{\top}\eta B(w)=\eta.$ (15)

Identity (15) holds for all $w\in\mathbb{R}^{n}$ with $|w|<1$.

######  Lemma 2.3.

For a fixed $w\in\mathbb{R}^{n}$ with $|w|<1$, the map
$F_{w}:\mathbb{H}^{n}\to\mathbb{H}^{n}$ defined by

$F_{w}(x):=B(w)x,$ (16)

is an isometry. In particular, $F_{0}$ is the identity map on
$\mathbb{H}^{n}$.

###### Proof.

The fact that $F_{0}$ is the identity map follows immediately from
$B(0)=\text{Id}$. The general result follows from a simple calculation that
uses (15); see Appendix A for details. ∎

######  Lemma 2.4.

Let $x\in\mathbb{H}^{n}$ and $w\in\mathbb{R}^{n}$ with $|w|<1$. Then, the
following are equivalent:

  1. 1.

$F_{w}(1,0,\cdots,0)=(x_{0},-x_{1},\cdots,-x_{n})$,

  2. 2.

$F_{w}(x)=(1,0,\cdots,0)$,

  3. 3.

$w_{i}=\frac{x_{i}}{x_{0}}$, for all $1\leq i\leq n$.

###### Proof.

The proof is provided in Appendix A. ∎

####  2.2.3 Translation on $\mathbb{H}^{n}$

In this part, we define the translation on the hyperbolic space using the
isometry $F_{w}$ defined in (16). For any
$x=(x_{0},x_{1},\cdots,x_{n})\in\mathbb{H}^{n}$, we define a $n$-dimensional
vector $\hat{x}$ by

$\hat{x}:=\left(\frac{x_{1}}{x_{0}},\cdots,\frac{x_{n}}{x_{0}}\right).$

From the definition of $\hat{x}$ and Lemma 2.4, we have that

$F_{\hat{x}}(x)=v\qquad\text{ and }\qquad
F_{\hat{x}}(v)=(x_{0},-x_{1},\cdots,-x_{n}).$

######  Definition 2.1.

We define a binary operation
$+^{\prime}:\mathbb{H}^{n}\times\mathbb{H}^{n}\to\mathbb{H}^{n}$ as follows:

$x+^{\prime}y:=F_{-\hat{y}}(x)\in\mathbb{H}^{n},\qquad\forall~{}x,y\in\mathbb{H}^{n}.$

Since $F_{w}$ is an isometry for all $w\in\mathbb{R}^{n}$ with $|w|<1$, we
know that the map $x\mapsto x+^{\prime}y$ is also an isometry for all
$y\in\mathbb{H}^{n}$. The following lemma provides the coordinate expression
of $x+^{\prime}y$.

######  Lemma 2.5.

Let $x,y\in\mathbb{H}^{n}$, then we have

$\displaystyle\begin{aligned}
&(x+^{\prime}y)_{0}=x_{0}y_{0}+x_{1}y_{1}+\cdots+x_{n}y_{n},\\\
&(x+^{\prime}y)_{j}=x_{0}y_{j}+x_{j}+\frac{y_{j}}{y_{0}+1}(x_{1}y_{1}+\cdots+x_{n}y_{n}),\qquad\forall~{}1\leq
j\leq n.\end{aligned}$ (17)

###### Proof.

The proof follows from direct calculations, see Appendix A. ∎

######  Remark 2.6.

We can also express $(x+^{\prime}y)_{j}$ for any $1\leq j\leq n$, as

$\displaystyle(x+^{\prime}y)_{j}=x_{0}y_{j}+x_{j}+\frac{y_{j}}{y_{0}+1}((x+^{\prime}y)_{0}-x_{0}y_{0})=x_{j}+\frac{y_{j}}{y_{0}+1}((x+^{\prime}y)_{0}+x_{0}).$
(18)

We investigate now the algebraic structure of the operator $+^{\prime}$.

(1) Identity of $+^{\prime}$. Recall that
$v=(1,\underbrace{0,\cdots,0}_{n-\text{times}})$ denotes the vertex of
$\mathbb{H}^{n}$. Since $\hat{v}$ is the $n$-dimensional zero vector, we can
easily find that

$x+^{\prime}v=F_{0}(x)=x,$

since $F_{0}$ is the identity map on $\mathbb{H}^{n}$. Now, we calculate
$v+^{\prime}x$. From the definition of $+^{\prime}$, we have

$(v+^{\prime}x)_{0}=v_{0}x_{0}+v_{1}x_{1}+\cdots+v_{n}x_{n}=x_{0}$

and

$(v+^{\prime}x)_{j}=v_{0}x_{j}+v_{j}+\frac{x_{j}}{x_{0}+1}(v_{1}x_{1}+\cdots+v_{n}x_{n})=x_{j},\qquad\text{
for }j=1,\dots,n.$

This implies that

$v+^{\prime}x=x.$

We conclude that $v$ is an identity element for $+^{\prime}$. The uniqueness
of identities can be shown easily, so $v$ is the unique identity element of
$+^{\prime}$.

(2) Inverse of $+^{\prime}$. We want to find the inverse $-x$ of $x$, which
satisfies

$x+^{\prime}(-x)=(-x)+^{\prime}x=v.$

For $x=(x_{0},x_{1},\cdots,x_{n})$, set:

$-x:=(x_{0},-x_{1},-x_{2},\cdots,-x_{n}).$ (19)

It is easy to check that $-x$ defined above is indeed the inverse of $x$. In
polar coordinates, if $(\theta,\xi)\in[0,\infty)\times\mathbb{S}^{n-1}$ are
the coordinates of $x$, then $-x$ has coordinates $(\theta,-\xi)$. So one can
see that the inverse operator is quite natural. We will prove the uniqueness
of the inverse later (see Corollary 2.9).

Using the definition (19) of the inverse, we define the operation $-^{\prime}$
by

$x-^{\prime}y:=x+^{\prime}(-y),\qquad\forall~{}x,y\in\mathbb{H}^{n}.$ (20)

We have the following simple lemma which will be used in further calculations.

######  Lemma 2.7.

Let $x$ and $y$ be two points on $\mathbb{H}^{n}$. Then we have the following
relation:

$-(x+^{\prime}y)=(-x)+^{\prime}(-y).$

###### Proof.

The proof is provided in Appendix A. ∎

(3) Non-commutativity of $+^{\prime}$. Using (LABEL:a-a-10), we can show
easily that $+^{\prime}$ is not commutative.

(4) Non-associativity of $+^{\prime}$. By direct (but tedious) calculations,
one can check that in general,
$((x+^{\prime}y)+^{\prime}z)_{0}\neq(x+^{\prime}(y+^{\prime}z))_{0}$. Hence,
the operation $+^{\prime}$ is not associative.

To prove the uniqueness of the inverse, we provide the following lemma.

######  Lemma 2.8.

For any $x,y\in\mathbb{H}^{n}$, we have

$x=(x+^{\prime}y)-^{\prime}y.$

###### Proof.

The proof is provided in Appendix A. ∎

From the previous lemma, we have the uniqueness of the inverse for
$+^{\prime}$.

######  Corollary 2.9.

Let $x,y,z\in\mathbb{H}^{n}$ that satisfy

$x+^{\prime}y=v,\qquad\text{and}\qquad x+^{\prime}z=v.$

Then, we have $y=z$.

###### Proof.

From Lemma 2.8, we have

$(x+^{\prime}y)-^{\prime}y=x,\qquad(x+^{\prime}z)-^{\prime}z=x.$

If we use $x+^{\prime}y=x+^{\prime}z=v$, we get

$v-^{\prime}y=x,\qquad v-^{\prime}z=x.$

This relation directly yields $y=z$. ∎

Note that one can express the geodesic distance on $\mathbb{H}^{n}$ using
$-^{\prime}$. Let $x,y\in\mathbb{H}^{n}$, $x=(x_{0},x_{1},\cdots,x_{n})$ and
$y=(y_{0},y_{1},\cdots,y_{n})$. Then we have

$(x-^{\prime}y)_{0}=x_{0}y_{0}-x_{1}y_{1}-\cdots-x_{n}y_{n},$ (21)

and by (10),

$(x-^{\prime}y)_{0}=\cosh\Big{(}d(x,y)\Big{)}.$ (22)

####  2.2.4 Coordinate grid on $\mathbb{H}^{n}$

In this part, we define a coordinate grid on $\mathbb{H}^{n}$. In the
Euclidean space, the coordinate grid is set by the hyperplanes
$x_{i}=\text{const.}$, for all coordinates $x_{i}$. However, we should be
careful when we define this concept on $\mathbb{H}^{n}$.

For a fixed $k\in\\{1,2,\cdots,n\\}$, we define the hypersurface:

$P_{k}(0):=\\{x\in\mathbb{H}^{n}:x_{k}=0\\}.$

For $a\in\mathbb{R}$, define

$u_{k}(a):=\left(\cosh a,0,\cdots,0,\sinh
a,0,\cdots,0\right)\in\mathbb{H}^{n},$ (23)

where $\cosh a$ is the $0^{th}$ index and $\sinh a$ is the $k^{th}$ index.
Then consider translates of $P_{k}(0)$ by $u_{k}(a)$:

$P_{k}(a):=P_{k}(0)+^{\prime}u_{k}(a),\qquad a\in\mathbb{R}.$ (24)

We refer to Figure 1 for an illustration of $P_{k}(0)$ and some of its
translates for the $2$-dimensional hyperbolic space.

|
---|---
(a) | (b)
Figure 1: Illustration of the translation (24) on the $2$-dimensional
hyperbolic space $\mathbb{H}^{2}$. (a) The curve $P_{1}(0)$ and its translates
$P_{1}(a)$ for $a=0.5$ and $a=1$. The points $u_{1}(a)$ are also indicated on
the plot. (b) Projections of $P_{1}(a)$ on the $x_{1}x_{2}$-plane for various
values of $a$.

######  Lemma 2.10.

Let $x\in P_{k}(0)$ and $a,b\in\mathbb{R}$, then we have

$(x+^{\prime}u_{k}(a))+^{\prime}u_{k}(b)=x+^{\prime}u_{k}(a+b).$

###### Proof.

The proof is provided in Appendix A. ∎

######  Remark 2.11.

(1) If we put $x=v$, we have

$\displaystyle
u_{k}(a)+^{\prime}u_{k}(b)=u_{k}(a+b),\qquad\forall~{}a,b\in\mathbb{R}.$

(2) Since $u_{k}(0)=v$, we also have

$u_{k}(a)=-u_{k}(-a),\qquad\forall~{}a\in\mathbb{R}.$

The following lemma is key for the concept of coordinate grid on
$\mathbb{H}^{n}$. In particular, it shows that the family (24) of translates
of $P_{k}(0)$ covers the entire $\mathbb{H}^{n}$; this is similar to how
translating a hyperplane $x_{i}=0$ along the $x_{i}$-axis covers the Euclidean
space.

######  Lemma 2.12.

For any $x\in\mathbb{H}^{n}$ there exists a unique pair
$(a,y)\in\mathbb{R}\times P_{k}(0)$ such that

$\displaystyle x=y+^{\prime}u_{k}(a).$ (25)

Furthermore, $a$ and $y$ can be expressed as

$a=\tanh^{-1}\left(\frac{x_{k}}{x_{0}}\right),\qquad y=x-^{\prime}u_{k}(a).$
(26)

###### Proof.

The proof is provided in Appendix A. ∎

To illustrate the decomposition (25), we show in Figure 1(a) a point $x\in
P_{1}(1)\subset\mathbb{H}^{2}$ and its corresponding $y\in P_{1}(0)$ such that
$x=y+^{\prime}u_{1}(1)$. In summary, any point $x\in\mathbb{H}^{n}$ lies on
the hypersurface $P_{k}(a)$ with $a$ given by (26), where $P_{k}(a)$ is the
translation of $P_{k}(0)$ by $u_{k}(a)$. One can think of $u_{k}(a)$ as the
$k$-th coordinate of $x$. Given the uniqueness of the decomposition (25)-(26),
we will be using in the sequel the map $\pi_{k}:\mathbb{H}^{n}\to\mathbb{R}$
defined by

$\pi_{k}(x)=\tanh^{-1}\left(\frac{x_{k}}{x_{0}}\right).$ (27)

######  Corollary 2.13.

Let $x\in\mathbb{H}^{n}$ with $x\in\pi_{k}^{-1}(a)$ for some $a\in\mathbb{R}$.
Then we have

$x+^{\prime}u_{k}(b)\in\pi_{k}^{-1}(a+b),\qquad\text{or
equivalently,}\quad\pi_{k}(x+^{\prime}u_{k}(b))=a+b.$

###### Proof.

From Lemma 2.12, there exists $y\in P_{k}(0)$ such that

$x=y+^{\prime}u_{k}(a).$

Then, by Lemma 2.10, we get

$x+^{\prime}u_{k}(b)=(y+^{\prime}u_{k}(a))+^{\prime}u_{k}(b)=y+^{\prime}u_{k}(a+b).$

Therefore, $x+^{\prime}u_{k}(b)\in\pi_{k}^{-1}(a+b)$. ∎

We conclude this section with a simple inequality which will be used later in
the paper.

######  Lemma 2.14.

Let $x$ and $y$ be two points in $\mathbb{H}^{n}$. Then we have

$d(x,y)\geq|\pi_{k}(x)-\pi_{k}(y)|,\qquad\forall~{}k=1,2,\cdots,n.$

###### Proof.

The proof is provided in Appendix A. ∎

##  3 Interaction potentials with Newtonian repulsion

In this section, we consider interaction potentials that include Newtonian
repulsion, i.e., potentials for which the repulsion component is given by the
Green’s function of the negative Laplacian on $\mathbb{H}^{n}$ [23].

Define

$\displaystyle\Phi(\theta):=\begin{cases}\displaystyle-\frac{1}{2\pi}\log\left(\tanh\frac{\theta}{2}\right)&\text{
for }n=2,\\\\[10.0pt]
\displaystyle\frac{1}{n\alpha(n)}\int^{\infty}_{\theta}\frac{1}{\sinh^{n-1}\zeta}\mathrm{d}\zeta\qquad&\text{
for }n\geq 3.\end{cases}$ (28)

Note that the case $n=2$ can in fact be recovered from the expression for
general $n$, by calculating explicitly the integral. It was shown in [17] that
the Green’s function for $-\Delta_{\mathbb{H}^{n}}$ is given by

$G(x,y):=\Phi(d(x,y)),$ (29)

i.e., $G$ satisfies

$\Delta_{x}G(x,y)=-\delta_{y}.$ (30)

In this section we consider interaction potentials in the form

$K(x,y)=G(x,y)+A(d(x,y)),$ (31)

where $G$ is given by (29) and $A:[0,\infty)\to\mathbb{R}$. As discussed in
[23], the Green’s function component of $K$ (referred to in the sequel as
Newtonian potential) models repulsive interactions. To balance the repulsion,
we take the $A$ component to model attractive interactions. We refer to Figure
2 for an illustration of several interaction potentials in the form (31); the
attraction components are specified later – see (56), (58) and (89).

Figure 2: Interaction potentials in the form (31), with attraction given by
(56), (58) and (89), respectively.

Our main goals here are to investigate the qualitative properties of critical
points of the interaction energy (1) with $K$ in the form (31), as well as to
study explicit forms of minimizers for certain specific examples of attraction
potentials $A$. All the equilibria investigated in this section are absolutely
continuous probability measures $\rho(x)\mathrm{d}x$, with $\rho\in
L^{1}(\mathrm{d}x)$.

######  Remark 3.1.

Similar to the Euclidean case, the energy (1) corresponding to the Newtonian
potential is the square of the $H^{-1}$ norm. Indeed, for $\rho\in
H^{-1}(\mathbb{H}^{n})$ with compact support, by definition of the $H^{-1}$
norm and the divergence theorem, we have:

$\displaystyle\left\lVert\rho\right\rVert_{H^{-1}(\mathbb{H}^{n})}^{2}$
$\displaystyle=\int_{\mathbb{H}^{n}}\|\nabla
u(x)\|_{\mathbb{H}^{n}}^{2}\mathrm{d}x=-\int_{\mathbb{H}^{n}}(\Delta_{\mathbb{H}^{n}}u)(x)u(x)\mathrm{d}x,$

where $u$ satisfies (in the weak sense):
$\displaystyle{-\Delta_{\mathbb{H}^{n}}u=\rho}$. Further, using the
fundamental solution, write
$\displaystyle{u(x)=\int_{\mathbb{H}^{n}}G(x,y)\rho(y)\mathrm{d}y}$, to reach

$\left\lVert\rho\right\rVert_{H^{-1}(\mathbb{H}^{n})}^{2}=\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}G(x,y)\rho(x)\rho(y)\mathrm{d}x\mathrm{d}y.$
(32)

Regarding the Euler-Lagrange equations, for an interaction potential in the
form (31), $\Lambda$ defined in (7) reads:

$\Lambda(x)=\int_{\mathbb{H}^{n}}\left(G(x,y)+A(d(x,y)\right)\rho(y)\mathrm{d}y.$
(33)

Then, by (30), one finds:

$\Delta\Lambda(x)=\begin{cases}\begin{aligned}
&-\rho(x)+\int_{\mathbb{H}^{n}}\Delta_{x}A(d(x,y))\rho(y)\mathrm{d}y\qquad\text{
for }x\in\mathrm{supp}(\rho),\\\
&\int_{\mathbb{H}^{n}}\Delta_{x}A(d(x,y))\rho(y)\mathrm{d}y\hskip
66.86414pt\text{ for }x\notin\mathrm{supp}(\rho).\end{aligned}\end{cases}$
(34)

For critical points of the energy (or equivalently, equilibrium solutions of
model (2)), $\Lambda(x)$ is identically equal to a constant in the support of
$\rho$ (see (6a)). Hence, $\Delta\Lambda=0$ in $\mathrm{supp}(\rho)$, and the
first equation in (34) leads to the following integral equation for
equilibrium densities $\rho$:

$\rho(x)=\int_{\mathrm{supp}(\rho)}\Delta_{x}A(d(x,y))\rho(y)\mathrm{d}y,\qquad\text{
for }x\in\mathrm{supp}(\rho).$ (35)

###  3.1 Equilibria supported on geodesic balls

We consider equilibria of the aggregation model (2) that are supported on
geodesic balls centred at the vertex $v$, and study their qualitative
properties. To prove the main result (Theorem 3.3) on the radial symmetry and
the monotonicity of such states, we employ the method of moving planes by
adapting it to the hyperbolic space.

Before we present the main result, we introduce the reflection on the
hyperbolic space $\mathbb{H}^{n}$. In the Euclidean space $\mathbb{R}^{n}$,
the reflection by a hyperplane $x_{1}=a$ is given by the map:

$\displaystyle(x_{1},x_{2},\cdots,x_{n})\mapsto(2a-x_{1},x_{2},\cdots,x_{n}).$
(36)

The reflection map (36) can be divided into three maps as follows:

$\displaystyle(x_{1},x_{2},\cdots,x_{n})\mapsto(x_{1}-a,x_{2},\cdots,x_{n})\mapsto(a-x_{1},x_{2},\cdots,x_{n})\mapsto(2a-x_{1},x_{2},\cdots,x_{n}).$
(37)

The first map is the translation in the $x_{1}$ coordinate by $-a$, the second
map is the reflection by the hyperplane $x_{1}=0$, and the third map is the
translation in the $x_{1}$ coordinate by $a$.

We apply now a similar argument to define the reflection on $\mathbb{H}^{n}$.
We present the reflection corresponding to coordinate $x_{1}$, the reflections
corresponding to coordinates $x_{2},\dots,x_{n}$ are similar. First, we define
the reference hypersurface for the reflection. For the Euclidean space, this
is a regular flat hyperplane, however, for the hyperbolic space we choose the
hypersurface to be $P_{1}(a)$ defined in (24) (see also Figure 1). For the
translation on $\mathbb{H}^{n}$ we use the $+^{\prime}$ operation (see
Definition 2.1), and take the translation by $\pm a$ in the $x_{1}$
coordinate, to be $x\mapsto x\pm^{\prime}u_{1}(a)$ (see (24)). Finally,
analogous to the reflection by the hyperplane $x_{1}=0$ in the Euclidean
space, on $\mathbb{H}^{n}$ we have reflection by the hypersurface
$P_{1}(0)=\\{x\in\mathbb{H}^{n}:x_{1}=0\\}$, which we define it to be:

$R_{1}:(x_{0},x_{1},x_{2},\cdots,x_{n})\in\mathbb{H}^{n}\shortmid\\!\longrightarrow(x_{0},-x_{1},x_{2},\cdots,x_{n})\in\mathbb{H}^{n}.$

Following the map sequence in (37) (translation by $-a$, reflection,
translation by $a$), we express the reflection across an arbitrary
hypersurface $P_{1}(a)$ on $\mathbb{H}^{n}$ as:

$x\shortmid\\!\longrightarrow x-^{\prime}u_{1}(a)\shortmid\\!\longrightarrow
R_{1}(x-^{\prime}u_{1}(a))\shortmid\\!\longrightarrow
R_{1}(x-^{\prime}u_{1}(a))+^{\prime}u_{1}(a).$ (38)

This sequence of three maps is illustrated schematically in Figure 3. Written
compactly, the reflection map above reads:

$x\shortmid\\!\longrightarrow
x^{a}:=R_{1}(x-^{\prime}u_{1}(a))+^{\prime}u_{1}(a).$ (39)

For a better understanding of the reflection map, let us express the reflected
point $x^{a}$ in coordinates. From the definition of $-^{\prime}$ and
$u_{1}(a)$ (see (20) and (23)), we have

$\begin{cases}(x-^{\prime}u_{1}(a))_{0}=x_{0}\cosh a-x_{1}\sinh a,\\\
(x-^{\prime}u_{1}(a))_{1}=x_{1}\cosh a-x_{0}\sinh a,\\\
(x-^{\prime}u_{1}(a))_{j}=x_{j},\qquad\forall~{}2\leq j\leq n.\end{cases}$

Then, by the definition of the reflection $R_{1}$, we have

$\begin{cases}(R_{1}(x-^{\prime}u_{1}(a)))_{0}=x_{0}\cosh a-x_{1}\sinh a,\\\
(R_{1}(x-^{\prime}u_{1}(a)))_{1}=-x_{1}\cosh a+x_{0}\sinh a,\\\
(R_{1}(x-^{\prime}u_{1}(a)))_{j}=x_{j},\quad\forall~{}2\leq j\leq
n.\end{cases}$

Finally, we use the definition of $+^{\prime}$ and $u_{1}(a)$ to get

$\begin{cases}(R_{1}(x-^{\prime}u_{1}(a))+^{\prime}u_{1}(a))_{0}&=(x_{0}\cosh
a-x_{1}\sinh a)\cosh a+(-x_{1}\cosh a+x_{0}\sinh a)\sinh a\\\ &=x_{0}\cosh
2a-x_{1}\sinh 2a,\\\\[2.0pt]
(R_{1}(x-^{\prime}u_{1}(a))+^{\prime}u_{1}(a))_{1}&=(-x_{1}\cosh a+x_{0}\sinh
a)\cosh a-(x_{0}\cosh a-x_{1}\sinh a)\sinh a\\\ &=-x_{1}\cosh 2a+x_{0}\sinh
2a,\\\\[2.0pt]
(R_{1}(x-^{\prime}u_{1}(a))+^{\prime}u_{1}(a))_{j}&=x_{j},\quad\forall~{}2\leq
j\leq n.\end{cases}$

Therefore, we can express the reflection (39) in coordinates, by:

$x=(x_{0},x_{1},\cdots,x_{n})\shortmid\\!\longrightarrow x^{a}=(x_{0}\cosh
2a-x_{1}\sinh 2a,-x_{1}\cosh 2a+x_{0}\sinh 2a,x_{2},\cdots,x_{n}).$

$x_{1}$$x_{2}$$P_{1}(a)$$A$$B$$C$$D$$x\mapsto
x-^{\prime}u_{1}(a)$$(x_{0},x_{1},x_{2})\mapsto(x_{0},-x_{1},x_{2})$$x\mapsto
x+^{\prime}u_{1}(a)$ Figure 3: Schematic illustration (in the
$x_{1}x_{2}$-plane) of the reflection on $\mathbb{H}^{2}$ across the
hypersurface $P_{1}(a)$ – see (39). Point $A$ reflects into point $D$, as the
result of three maps $A\mapsto B$ (translation by $-u_{1}(a)$), $B\mapsto C$
(reflection across $P_{1}(0)=x_{0}x_{2}-$plane), and $C\mapsto D$ (translation
by $u_{1}(a)$.

######  Remark 3.2.

The following properties can be checked easily:

  1. 1.

$x=(x^{a})^{a}$, for all $x\in\mathbb{H}^{n}$.

  2. 2.

The set of fixed points for the reflection (39) is $P_{1}(a)$.

  3. 3.

The reflection map (39) is an isometry on $\mathbb{H}^{n}$.

We present now the main result of this subsection.

######  Theorem 3.3.

Consider the aggregation model (2) with an interaction potential in the form
(31). Let $\rho$ be a bounded equilibrium solution of model (2) that is
supported in the geodesic ball $B_{R}(v)$, for some $R>0$. Let
$F(d(x,y)):=\Delta_{x}A(d(x,y))$ and assume $F(d(x,\cdot))$ is locally
integrable for all $x$. Then, provided $F$ is a monotone function, $\rho$ is
radially symmetric and monotone about the vertex $v$. Specifically, (i) if $F$
is decreasing, then $\rho$ is radially symmetric and monotonically decreasing
about the vertex $v$, and (ii) if $F$ is increasing, then $\rho$ is radially
symmetric and monotonically increasing about $v$.

###### Proof.

Since $\rho$ is an equilibrium solution, we use (35) to write:

$\rho(x)=\begin{cases}\displaystyle\int_{B_{R}(v)}F(d(x,y))\rho(y)\mathrm{d}y\qquad&\text{
for }x\in B_{R}(v),\\\\[10.0pt] 0&\text{otherwise}.\end{cases}$ (40)

We define the reflection $\rho_{a}$ of $\rho$ across the plane $P_{1}(a)$ by:

$\rho_{a}(x)=\rho(x^{a}).$

From (40) we have

$\rho_{a}(x)=\begin{cases}\displaystyle\int_{B_{R}(v)}F(d(x^{a},y))\rho(y)\mathrm{d}y\qquad&x\in
B_{R}(v^{a}),\\\\[10.0pt] 0&\text{otherwise}.\end{cases}$

Also, we set

$\Sigma_{a}=\\{x\in B_{R}(v):\pi_{1}(x)\geq a\\}.$

We apply the method of moving planes and compare $\rho(x)$ and $\rho_{a}(x)$
on $\Sigma_{a}$. If $x\not\in B_{R}(v^{a})$, then we know that
$\rho_{a}(x)=0$. Since $\rho(x)\geq 0$, we can conclude that
$\rho(x)-\rho_{a}(x)\geq 0$ for all $x\in\Sigma_{a}\setminus B_{R}(v^{a})$.

Consider the case $x\in\Sigma_{a}\cap B_{R}(v^{a})$. By (40) we write
$\rho(x)$ as

$\rho(x)=\int_{B_{R}(v)\cap\Sigma_{a}}F(d(x,y))\rho(y)\mathrm{d}y+\int_{B_{R}(v)\setminus\Sigma_{a}}F(d(x,y))\rho(y)\mathrm{d}y.$
(41)

We use that the reflection map (39) is an isometry to rewrite the second
integral above in reflected form (reflect the domain of integration, as well
as $x$ and $y$). The image of $B_{R}(v)\setminus\Sigma_{a}$ under the
reflection map (39) is $B_{R}(v^{a})\cap\Sigma_{a}$. Also use
$\rho(y)=\rho_{a}(y^{a})$ to get from (41):

$\displaystyle\rho(x)$
$\displaystyle=\int_{B_{R}(v)\cap\Sigma_{a}}F(d(x,y))\rho(y)\mathrm{d}y+\int_{B_{R}(v^{a})\cap\Sigma_{a}}F(d(x^{a},y^{a}))\rho_{a}(y^{a})\mathrm{d}y^{a}$
$\displaystyle=\int_{B_{R}(v)\cap\Sigma_{a}}F(d(x,y))\rho(y)\mathrm{d}y+\int_{B_{R}(v^{a})\cap\Sigma_{a}}F(d(x^{a},y))\rho_{a}(y)\mathrm{d}y,$

where for the second equal sign we simply renamed the variable $y^{a}$ to $y$.

Since $\rho$ and $\rho_{a}$ are zero outside $B_{R}(v)$ and $B_{R}(v^{a})$,
respectively, we can also write the expression above as:

$\rho(x)=\int_{\Sigma_{a}}F(d(x,y))\rho(y)\mathrm{d}y+\int_{\Sigma_{a}}F(d(x^{a},y))\rho_{a}(y)\mathrm{d}y.$
(42)

By replacing $x$ with $x^{a}$ in (42), we can obtain the following expression
for $\rho_{a}$:

$\rho_{a}(x)=\int_{\Sigma_{a}}F(d(x^{a},y))\rho(y)dy+\int_{\Sigma_{a}}F(d(x,y))\rho_{a}(y)\mathrm{d}y.$
(43)

From (42) and (43) we calculate the difference $\rho(x)-\rho_{a}(x)$ for
$x\in\Sigma_{a}\cap B_{R}(v^{a})$ as:

$\rho(x)-\rho_{a}(x)=\int_{\Sigma_{a}}(F(d(x,y))-F(d(x^{a},y)))(\rho(y)-\rho_{a}(y))\mathrm{d}y.$
(44)

We can show that

$d(x,y)\leq d(x^{a},y),\qquad\text{ for all }x,y\in\Sigma_{a}.$ (45)

Indeed, without loss of generality, it is enough to proof the case $a=0$. Let
$x=(x_{0},x_{1},\cdots,x_{n})$ and $y=(y_{0},y_{1},\cdots,y_{n})$. Since
$x,y\in\Sigma_{0}$, we have $x_{1},y_{1}\geq 0$. This yields that

$\cosh\left(d(x,y)\right)=x_{0}y_{0}-x_{1}y_{1}-\cdots-x_{n}y_{n}\leq
x_{0}y_{0}+x_{1}y_{1}-\cdots-x_{n}y_{n}=\cosh\left(d(x^{0},y)\right),$

which implies the desired result.

Consider now the following two cases.

Case (i): $F$ is decreasing. Since $F$ is decreasing and (45) holds, we have

$F(d(x,y))-F(d(x^{a},y))\geq 0,\qquad\text{ for all }x\in\Sigma_{a}\cap
B_{R}(v^{a})\text{ and }y\in\Sigma_{a}.$ (46)

We will show that

$\rho_{a}\leq\rho,\qquad\text{ on }\Sigma_{a},$

for all $a\in(-R,0)$, which implies that $\rho$ is monotonically decreasing
about the vertex, in the $x_{1}$ direction.

Assume first that there exists $a_{0}\in(-R,0)$ such that

$\rho_{a_{0}}(x)\leq\rho(x),\qquad\text{ for all }x\in\Sigma_{a_{0}},$ (47)

and show that $a_{0}$ can be extended to the origin. Indeed, suppose by
contradiction that (47) holds, but $a_{0}$ cannot be extended, and take
$a\in(a_{0},a_{0}+\epsilon)$, with $\epsilon>0$. Define

$\Sigma_{a}^{-}=\\{x\in\Sigma_{a}:\rho(x)<\rho_{a}(x)\\}.$

Then, from (44) and (46), for any $x\in\Sigma_{a}^{-}$, we have:

$0<\rho_{a}(x)-\rho(x)\leq\int_{\Sigma_{a}^{-}}(F(d(x,y))-F(d(x^{a},y)))(\rho_{a}(y)-\rho(y))\mathrm{d}y.$

Therefore,

$\|\rho_{a}-\rho\|_{L^{\infty}(\Sigma_{a}^{-})}\leq\|\rho_{a}-\rho\|_{L^{\infty}(\Sigma_{a}^{-})}\sup_{x\in\Sigma_{a}^{-}}\int_{\Sigma_{a}^{-}}(F(d(x,y))-F(d(x^{a},y)))\mathrm{d}y.$
(48)

The function under the integral on the right-hand-side is locally integrable.
Also, by (44), it holds that $\rho(x)>\rho_{a_{0}}(x)$ in the interior of
$\Sigma_{a_{0}}$, implying that the closure of $\Sigma_{a_{0}}^{-}$ has
measure $0$. As $\lim_{a\to
a_{0}}\Sigma_{a}^{-}\subset\overline{\Sigma_{a_{0}}^{-}}$, we infer that the
measure of $\Sigma_{a}^{-}$ approaches $0$ as $a\to a_{0}$. Therefore, we can
choose $\epsilon$ small enough such that

$\sup_{x\in\Sigma_{a}^{-}}\int_{\Sigma_{a}^{-}}(F(d(x,y))-F(d(x^{a},y)))\mathrm{d}y\leq\frac{1}{2}.$

From (48) we then infer that
$\|\rho_{a}-\rho\|_{{L^{\infty}}(\Sigma_{a}^{-})}$ = 0, which implies that
$\Sigma_{a}^{-}$ is empty. Consequently, $a_{0}$ is not maximal and we reached
a contradiction.

To show that there exists $a_{0}\in(-R,0)$ with the property (47), note that
(47) holds for $a_{0}=-R$. Then, by an argument entirely similar to the one
made in the previous step, one can infer that the plane $a_{0}=-R$ can be
moved further to the right, while (47) still holds.

A similar argument can be made using hypersufaces with $a>0$. Finally, since
the direction $x_{1}$ can be chosen arbitrarily, we conclude that $\rho$ is
radially symmetric and decreasing about the vertex $v$. This concludes the
proof for case (i).

Case (ii): $F$ is increasing. From a similar argument as for case (i), one can
show that $\rho$ is radially symmetric and increasing about the vertex $v$. ∎

We now turn the attention on radially symmetric equilibria that are local
minimizers of the energy (see Lemma 2.1).

######  Proposition 3.4.

Let $K$ be an interaction potential in the form (31), and let $\rho$ be a
bounded, radially symmetric equilibrium solution of model (2) that is
supported in the geodesic ball $B_{R}(v)$, for some $R>0$. Assume that
$A:[0,\infty)\to\mathbb{R}$ satisfies $\Delta_{x}A(d(x,y))\geq 0$ for all
$x,y\in\mathbb{H}^{n}$. Then $\rho$ satisfies the necessary condition (6) for
being a local energy minimizer.

###### Proof.

Since $\rho$ is an equilibrium solution, it satisfies the first equation of
(6). It remains to show (6b). By using (34), we write:

$\displaystyle\Delta\Lambda(x)=\begin{cases}0\qquad&\text{if }x\in
B_{R}(v),\\\\[3.0pt]
\displaystyle\int_{\mathbb{H}^{n}}\Delta_{x}A(d(x,y))\rho(y)\mathrm{d}y,\qquad&\text{if
}x\not\in B_{R}(v).\end{cases}$ (49)

As $K$ and $\rho$ are radially symmetric, $\Lambda$ has radial symmetry as
well. Hence, we can define $\tilde{\Lambda}:[0,\infty)\to R$ by:

$\tilde{\Lambda}(r):=\Lambda(x),$ (50)

for some $x\in\partial B_{r}(v)$. In particular, by (6a),
$\tilde{\Lambda}(r)=\lambda$ for $0\leq r\leq R$.

For $r>R$, use the divergence theorem in $B_{r}(v)$ to obtain that

$\int_{B_{r}(v)}\Delta\Lambda(x)dx=\int_{\partial
B_{r}(v)}\nabla\Lambda(x)\cdot\mathrm{d}\vec{n}=|\partial
B_{r}(v)|\tilde{\Lambda}^{\prime}(r).$ (51)

From $\Delta_{x}A(d(x,y))\geq 0$ and (49), it follows that
$\Delta\Lambda(x)\geq 0$ for all $x$. Consequently, using the equation above,
we infer that $\tilde{\Lambda}$ is a non-decreasing function, and conclude
that

$\begin{cases}\tilde{\Lambda}(r)=\lambda\qquad\text{if }0\leq r\leq
R,\\\\[3.0pt] \tilde{\Lambda}(r)\geq\lambda\qquad\text{if }r>R.\end{cases}$

This shows that the necessary condition (6b) for a local minimizer is also
satisfied. ∎

The following proposition shows that for potentials with $\Delta A>0$,
radially symmetric local minimizers must have a geodesic ball as their
support.

######  Proposition 3.5.

Let $K$ be an interaction potential in the form (31), where
$A:[0,\infty)\to\mathbb{R}$ satisfies $\Delta_{x}A(d(x,y))>0$ for all
$x,y\in\mathbb{H}^{n}$. If $\rho$ is a bounded, radially symmetric local
minimizer of the interaction energy, then there exists $R>0$ such that
$\mathrm{supp}(\rho)=B_{R}(v)$.

###### Proof.

As $\rho$ is bounded and radially symmetric, its support can be expressed as

$\mathrm{supp}(\rho)=\bigcup_{r\in C}\partial B_{r}(v),$

where $C\subset[0,\infty)$ is a compact set.

First note that $\Lambda$ is radially symmetric (see (50)) and
$\tilde{\Lambda}(r)=\lambda$ for $r\in C$. Also, from

$\Delta\Lambda(x)=\begin{cases}0\qquad&\text{if
}x\in\mathrm{supp}(\rho),\\\\[3.0pt]
\displaystyle\int_{\mathbb{H}^{n}}\Delta_{x}A(d(x,y))\rho(y)\mathrm{d}y>0\qquad&\text{if
}x\not\in\mathrm{supp}(\rho),\end{cases}$

and (51), we find that:

$\displaystyle\tilde{\Lambda}^{\prime}(r)>0\qquad\text{if }r\not\in C.$ (52)

Since $C$ is a compact set in $[0,\infty)$, we can express $C$ as a union of
disjoint closed intervals as follows:

$C=[a_{1},b_{1}]\cup[a_{2},b_{2}]\cup\cdots\cup[a_{k},b_{k}],$

where $0\leq a_{1}<b_{1}<a_{2}<b_{2}<\cdots<a_{k}<b_{k}$. If $k\geq 2$, then

$\tilde{\Lambda}(b_{1})=\tilde{\Lambda}(a_{2})\;(=\lambda).$

However, this contradicts (52), so we conclude that $k=1$. Now assume that
$0<a_{1}$. From (52), we have

$\tilde{\Lambda}(0)<\tilde{\Lambda}(a_{1}),$

but this cannot occur for a local minimizer (see (6b)).

We conclude that $a_{1}=0$, and $C=[0,b_{1}]$. By setting $R=b_{1}$, we then
have

$\mathrm{supp}(\rho)=\bigcup_{r\in[0,R]}\partial B_{r}(v)=B_{R}(v).$

∎

###  3.2 Explicit forms of radially symmetric equilibria

We first find some explicit expressions of equilibria for certain interaction
potentials in the form (31). These equilibria are radially symmetric and with
$\mathrm{supp}(\rho)=D_{R}(v)$, $R>0$.

Denote

$\tilde{\rho}(d(v,x)):=\rho(x).$ (53)

Then, the integral equation (35) reads:

$\tilde{\rho}(\theta_{x})=\int_{0}^{R}\int_{\mathbb{S}^{n}}\left(A^{\prime\prime}(\theta_{xy})+(n-1)\coth(\theta_{xy})A^{\prime}(\theta_{xy})\right)\tilde{\rho}(\theta_{y})\sinh^{n-1}\theta_{y}\,\mathrm{d}\sigma_{y}\mathrm{d}\theta_{y},$
(54)

that holds for all $x\in D_{R}(v)$, where $\theta_{x}:=d(v,x)$ and
$\theta_{xy}:=d(x,y)$. Note that we used (12) to write:

$\displaystyle\Delta_{x}A(d(x,y))=A^{\prime\prime}(d(x,y))+(n-1)\coth(d(x,y))A^{\prime}(d(x,y)).$
(55)

The two main specific examples of attractive potentials investigated here are:

(i) Constant Laplacian attraction:

$\displaystyle
A_{c}(\theta)=\int_{0}^{\theta}\frac{\int_{0}^{\theta_{*}}\sinh^{n-1}\xi\mathrm{d}\xi}{\sinh^{n-1}\theta_{*}}\,\mathrm{d}\theta_{*}.$
(56)

This is a very important example for applications; $A_{c}$ has the property
that

$\Delta_{x}A_{c}(d(x,y))=1,$ (57)

and leads to equilibria that are constant within their support.

(ii) Hyperbolic-cosine attraction:

$\displaystyle A_{h}(\theta)=\frac{\cosh\theta-1}{n}.$ (58)

This potential satisfies (use (55)):

$\Delta_{x}A_{h}(d(x,y))=\cosh(d(x,y)).$

We will show that the energy functional in this case is strictly convex, and
we will find explicitly the global minimizer.

##### Constant Laplacian attraction.

Consider the interaction potential (31) with attraction given by (56). Since
$A_{c}$ satisfies (57), it is immediate to check that the following density
distribution is a solution of (54):

$\displaystyle\rho_{c}(x)=\begin{cases}1\qquad\text{ if }d(v,x)\leq
R_{c},\\\\[2.0pt] 0\qquad\text{ if }d(v,x)>R_{c},\end{cases}$ (59)

where $R_{c}$ satisfies

$\displaystyle\mathrm{Vol}(D_{R_{c}}(v))=n\alpha(n)\int_{0}^{R_{c}}\sinh^{n-1}\theta\mathrm{d}\theta=1.$
(60)

Hence, $\rho_{c}$ is an equilibrium solution, and by Proposition 3.4, it
satisfies the necessary conditions for a local minimizer. This configuration
was studied in [23] as an attractor for the aggregation equation (2).

##### Hyperbolic-cosine attraction.

We consider now the interaction potential (31) with attraction given by (58).
We will show that the following density distribution is a global minimizer of
the interaction energy:

$\displaystyle\rho_{h}(x)=\begin{cases}\displaystyle\frac{\cosh(d(v,x))}{\alpha(n)\sinh^{n}R_{h}}&\qquad\text{
if }d(v,x)\leq R_{h},\\\\[10.0pt] 0&\qquad\text{ if }d(v,x)>R_{h},\end{cases}$
(61)

where $R_{h}>0$ satisfies

$\displaystyle
n\alpha(n)\int_{0}^{R_{h}}\sinh^{n-1}\theta\cosh^{2}\theta\,\mathrm{d}\theta=1.$
(62)

We will show that (61) is a solution of (54). By rotational symmetry, use
notation:

$\tilde{\rho}_{h}(d(v,x)):=\rho_{h}(x).$

The hyperbolic law of cosines on $\mathbb{H}^{n}$ states:

$\displaystyle\cosh(d(x,y))=\cosh(d(x,v))\cosh(d(y,v))-\sinh(d(x,v))\sinh(d(y,v))\cos\angle(xvy).$
(63)

From (63) we compute, using polar coordinates for $y$ and the rotational
symmetry of $\rho_{h}$:

$\displaystyle\int_{\mathbb{H}^{n}}\cosh(d(x,y))\rho_{h}(y)\mathrm{d}y$
$\displaystyle\quad=\int_{0}^{R_{h}}\int_{\mathbb{S}^{n-1}}\left(\cosh(d(x,v))\cosh\theta-\sinh(d(x,v))\sinh\theta\cos\angle(xvy)\right)\tilde{\rho}_{h}(\theta)\sinh^{n-1}\theta\,\mathrm{d}\sigma_{y}\mathrm{d}\theta,$

where we also used that $\mathrm{supp}(\rho_{h})=D_{R_{h}}(v)$. Then, using

$\displaystyle\int_{\mathbb{S}^{n-1}}\cos\angle(xvy)\mathrm{d}\sigma_{y}$
$\displaystyle=\frac{1}{2}\int_{\mathbb{S}^{n}}(\cos\angle(xvy)+\cos\angle(xv(-y))\mathrm{d}\sigma_{y}$
$\displaystyle=\frac{1}{2}\int_{\mathbb{S}^{n}}(\cos\angle(xvy)+\cos(\pi+\angle(xvy))\mathrm{d}\sigma_{y}$
$\displaystyle=0,$

in the equation above, we have

$\int_{\mathbb{H}^{n}}\cosh(d(x,y))\rho_{h}(y)\mathrm{d}y=n\alpha(n)\cosh(d(v,x))\int_{0}^{R_{h}}\cosh\theta\tilde{\rho}_{h}(\theta)\sinh^{n-1}\theta\,\mathrm{d}\theta.$
(64)

Consequently, to show that $\rho_{h}$ is a solution of (54), we need to show:

$\tilde{\rho}_{h}(d(v,x))=n\alpha(n)\cosh(d(v,x))\int_{0}^{R_{h}}\cosh\theta\tilde{\rho}_{h}(\theta)\sinh^{n-1}\theta\,\mathrm{d}\theta.$

Indeed, by substituting (61) into the right-hand-side above, one computes:

$\displaystyle
n\alpha(n)\cosh(d(v,x))\int_{0}^{R_{h}}\frac{\cosh^{2}\theta\sinh^{n-1}\theta}{\alpha(n)\sinh^{n}R_{h}}d\theta$
$\displaystyle=\frac{\cosh(d(v,x))}{\alpha(n)\sinh^{n}R_{h}}\,\underbrace{n\alpha(n)\int_{0}^{R_{h}}\cosh^{2}\theta\sinh^{n-1}\theta
d\theta}_{=1\text{ by \eqref{d-8-1}}}$
$\displaystyle=\tilde{\rho}_{h}(d(v,x)).$

This shows that $\rho_{h}$ is an equilibrium solution of model (2). By
Proposition 3.4, it also satisfies the necessary conditions to be a local
energy minimizer. In Section 3.3 we will prove that the energy functional
corresponding to this potential is strictly convex and hence, $\rho_{h}$ is a
global energy minimizer.

###  3.3 Hyperbolic cosine attraction: convexity of the energy

In this subsection we consider the potential (31) with hyperbolic cosine
attraction given by (58), and show that the equilibrium solution $\rho_{h}$
from (61) is a global energy minimizer in
$\mathcal{A}:=\mathcal{P}^{ac}_{c}(\mathbb{H}^{n})\cap
H^{-1}(\mathbb{H}^{n})$.

Consider the energy (1) defined on $\mathcal{A}$. Then use Remark 3.1 (see
(32)) to write the energy as

$\displaystyle E[\rho]$
$\displaystyle=\frac{1}{2}\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}(G(x,y)+A_{h}(d(x,y)))\rho(x)\rho(y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=\frac{1}{2}\|\rho\|^{2}_{H^{-1}(\mathbb{H}^{n})}+\frac{1}{2n}\mathcal{K}[\rho]-\frac{1}{2n},$
(65)

where the functional $\mathcal{K}[\rho]$ is defined by

$\mathcal{K}[\rho]=\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}\cosh(d(x,y))\rho(x)\rho(y)\mathrm{d}x\mathrm{d}y.$
(66)

First note that the repulsion component of the energy is given in terms of the
square of the $H^{-1}$ norm, which is strictly convex. Since the last term in
(65) is simply a constant, it remains to investigate the convexity of the
functional $\mathcal{K}$.

By (22), we write

$\displaystyle\mathcal{K}[\rho]=$
$\displaystyle=\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}(x_{0}y_{0}-x_{1}y_{1}-\cdots-
x_{n}y_{n})\rho(x)\rho(y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=\left\langle\int_{\mathbb{H}^{n}}(x_{0},x_{1},\cdots,x_{n})\rho(x)\mathrm{d}x,\int_{\mathbb{H}^{n}}(y_{0},y_{1},\cdots,y_{n})\rho(y)\mathrm{d}y\right\rangle$
$\displaystyle=\left\|\int_{\mathbb{H}^{n}}(x_{0},x_{1},\cdots,x_{n})\rho(x)\mathrm{d}x\right\|^{2},$

where we also used (8) and (9).

Denote

$c_{\rho}:=\int_{\mathbb{H}^{n}}(x_{0},x_{1},\cdots,x_{n})\rho(x)\mathrm{d}x,$
(67)

and hence, write

$\mathcal{K}[\rho]=\left\|c_{\rho}\right\|^{2}.$ (68)

By a direct calculation, we then have

$\displaystyle
t\mathcal{K}[\rho_{1}]+(1-t)\mathcal{K}[\rho_{2}]-\mathcal{K}[t\rho_{1}+(1-t)\rho_{2}]=t\|c_{\rho_{1}}\|^{2}+(1-t)\|c_{\rho_{2}}\|^{2}-\|c_{t\rho_{1}+(1-t)\rho_{2}}\|^{2}$
$\displaystyle\hskip
56.9055pt=t\|c_{\rho_{1}}\|^{2}+(1-t)\|c_{\rho_{2}}\|^{2}-\|tc_{\rho_{1}}+(1-t)c_{\rho_{2}}\|^{2}$
$\displaystyle\hskip
56.9055pt=t(1-t)\|c_{\rho_{1}}\|^{2}+t(1-t)\|c_{\rho_{2}}\|^{2}-2t(1-t)\left\langle
c_{\rho_{1}},c_{\rho_{2}}\right\rangle$ $\displaystyle\hskip
56.9055pt=t(1-t)\|c_{\rho_{1}}-c_{\rho_{2}}\|^{2},$ (69)

for any density functions $\rho_{1}$ and $\rho_{2}$. Note that the right-hand-
side above is not necessarily non-negative, as the inner product (8) is not
positive definite. So the positivity of this term needs to be shown, which is
what we will focus on next.

Consider the following set:

$\mathcal{P}^{ac}_{0}(\mathbb{H}^{n}):=\left\\{\rho\in\mathcal{P}^{ac}(\mathbb{H}^{n}):\int_{\mathbb{H}^{n}}x_{j}\rho(x)\mathrm{d}x=0,\text{
for all }1\leq j\leq n\right\\}.$

######  Lemma 3.6.

For any $\rho\in\mathcal{P}^{ac}(\mathbb{H}^{n})$, define $\hat{c}_{\rho}$ to
be the normalized $c_{\rho}$, i.e.,

$\hat{c}_{\rho}=\frac{c_{\rho}}{\|c_{\rho}\|}.$

Then,

$f_{\\#}\rho\in\mathcal{P}^{ac}_{0}(\mathbb{H}^{n}),\qquad\text{ where
}f(x)=x-^{\prime}\hat{c}_{\rho}.$

###### Proof.

Write $c_{\rho}=(c_{0},c_{1},\cdots,c_{n})$ using coordinates, i.e.,

$c_{k}=\int_{\mathbb{H}^{n}}x_{k}\,\rho(x)\mathrm{d}x,\qquad\text{ for all
}0\leq k\leq n.$ (70)

First note that $\langle c_{\rho},c_{\rho}\rangle>0$, so $\|c_{\rho}\|$ is
real-valued. Indeed, it is easy to show that
$c_{\rho}\in\mathrm{conv}(\mathbb{H}^{n})$, where
$\mathrm{conv}(\mathbb{H}^{n})$ denotes the convex hull of $\mathbb{H}^{n}$,
defined as:

$\mathrm{conv}(\mathbb{H}^{n}):=\\{(x_{0},x_{1},\cdots,x_{n})\in\mathbb{R}^{n+1}:x_{0}^{2}-x_{1}^{2}-\cdots-
x_{n}^{2}\geq 1\\}.$

Consequently,

$\langle c_{\rho},c_{\rho}\rangle=c_{0}^{2}-c_{1}^{2}-\cdots-c_{n}^{2}\geq
1>0.$

By a change of coordinates, one gets:

$\int_{\mathbb{H}^{n}}x_{j}f_{\\#}\rho(x)\mathrm{d}x=\int_{\mathbb{H}^{n}}(x-^{\prime}\hat{c}_{\rho})_{j}\rho(x)\mathrm{d}x,\qquad\text{
for all }1\leq j\leq n.$ (71)

To have $f\\#\rho\in\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})$, the expression
above needs to be $0$ for all $1\leq j\leq n$.

By (20) and (LABEL:a-a-10), we compute:

$\int_{\mathbb{H}^{n}}(x-^{\prime}\hat{c}_{\rho})_{j}\rho(x)dx=\int_{\mathbb{H}^{n}}\left(x_{j}-x_{0}(\hat{c}_{\rho})_{j}+\frac{(\hat{c}_{\rho})_{j}}{(\hat{c}_{\rho})_{0}+1}(x_{1}(\hat{c}_{\rho})_{1}+\cdots+x_{n}(\hat{c}_{\rho})_{n})\right)\rho(x)dx.$

From here, by the definition of $\hat{c}_{\rho}$ and (70), we get:

$\displaystyle\int_{\mathbb{H}^{n}}(x-^{\prime}\hat{c}_{\rho})_{j}\rho(x)dx$
$\displaystyle=c_{j}-c_{0}(\hat{c}_{\rho})_{j}+\frac{(\hat{c}_{\rho})_{j}}{(\hat{c}_{\rho})_{0}+1}(c_{1}(\hat{c}_{\rho})_{1}+\cdots+c_{n}(\hat{c}_{\rho})_{n})$
$\displaystyle=\|c_{\rho}\|(\hat{c}_{\rho})_{j}-\|c_{\rho}\|(\hat{c}_{\rho})_{0}(\hat{c}_{\rho})_{j}+\frac{\|c_{\rho}\|(\hat{c}_{\rho})_{j}}{(\hat{c}_{\rho})_{0}+1}((\hat{c}_{\rho})_{1}^{2}+\cdots+(\hat{c}_{\rho})_{n}^{2})$
$\displaystyle=\|c_{\rho}\|\Big{(}(\hat{c}_{\rho})_{j}-(\hat{c}_{\rho})_{0}(\hat{c}_{\rho})_{j}+(\hat{c}_{\rho})_{j}((\hat{c}_{\rho})_{0}-1)\Big{)}$
$\displaystyle=0,$ (72)

where for the third equal sign we used that $\hat{c}_{\rho}$ has norm $1$,
which implies
$(\hat{c}_{\rho})_{1}^{2}+\cdots+(\hat{c}_{\rho})_{n}^{2}=(\hat{c}_{\rho})_{0}^{2}-1$.

The conclusion now follows from (71) and (72). ∎

From (66) and the fact that the operation $-^{\prime}$ is an isometry, we get:

$\displaystyle\mathcal{K}[f_{\\#}\rho]$
$\displaystyle=\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}\cosh(d(x,y))f_{\\#}\rho(x)f_{\\#}\rho(y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}\cosh(d(x-^{\prime}\hat{c}_{\rho},y-^{\prime}\hat{c}_{\rho}))\rho(x)\rho(y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}\cosh(d(x,y))\rho(x)\rho(y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=\mathcal{K}[\rho].$

Since $\mathcal{K}[\rho]=\mathcal{K}[f_{\\#}\rho]$, we can restrict
$\mathcal{K}$ to
$\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})\subset\mathcal{P}^{ac}(\mathbb{H}^{n})$
for the purpose of investigating its minimizers.

######  Lemma 3.7.

The functional $\mathcal{K}$ defined in (66) is convex on
$\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})$.

###### Proof.

Take $\rho_{1},\rho_{2}\in\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})$; also note
here that $\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})$ is a convex set. Then,
$c_{\rho_{1}}=\alpha_{1}(1,0,\cdots,0)$ and
$c_{\rho_{2}}=\alpha_{2}(1,0,\cdots,0)$, for some positive constants
$\alpha_{1},\alpha_{2}>0$. Therefore,

$\|c_{\rho_{1}}-c_{\rho_{2}}\|^{2}=(\alpha_{1}-\alpha_{2})^{2}\geq 0,$

and by using (69), we find that:

$t\mathcal{K}[\rho_{1}]+(1-t)\mathcal{K}[\rho_{2}]\geq\mathcal{K}[t\rho_{1}+(1-t)\rho_{2}],\qquad\text{
for all }0<t<1.$

This shows the claim. ∎

Next, we state and prove the main result of this subsection.

######  Theorem 3.8.

Consider the interaction potential (31) with hyperbolic cosine attraction
given by (58). Then, the equilibrium solution $\rho_{h}$ from (61) is a global
minimizer of the interaction energy on $\mathcal{A}$.

###### Proof.

As noted above, the square of the $H^{-1}$ norm, representing the repulsion
component of the energy, is strictly convex on $\mathcal{A}$. Since the
attraction component $\mathcal{K}$ is convex on
$\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})$, the sum of the two functionals (and
hence $E$) is strictly convex on
$\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})\cap\mathcal{A}$. We conclude from here
that the critical point $\rho_{h}$ given by (61) is the unique global
minimizer of $E$ in $\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})\cap\mathcal{A}$.

By Lemma 3.6, any density can be translated by an isometry into
$\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})$. Since the interaction energy is
invariant under isometries (see (5)), we infer that the global minimizer
$\rho_{h}$ of $E[\cdot]$ on
$\mathcal{P}^{ac}_{0}(\mathbb{H}^{n})\cap\mathcal{A}$ is also a global
minimizer of $E[\cdot]$ on $\mathcal{A}$ itself. ∎

######  Remark 3.9.

A similar result as in Theorem 3.8 holds in the standard Euclidean space
$\mathbb{R}^{n}$– see Theorem 2.4 in [16]. There, it is shown that for an
interaction potential that consists of Newtonian repulsion and quadratic
attraction (i.e., $A(x)=\frac{1}{2}|x|^{2}$), the interaction energy is
strictly convex and the global energy minimizer consists in a uniform density
supported on a ball. However, for the model on $\mathbb{H}^{n}$ we could not
show the convexity of the energy for the attractive potential $A_{c}$ (that
leads to constant density equilibria), but we showed it instead for $A_{h}$.

##  4 Existence of compactly supported minimizers

In this section we study the existence of compactly supported global
minimizers of the energy (1). The arguments we use here follow the layout and
the general technique used by Cañizo et al. [7] for the Euclidean case. For
this reason, we use similar notations, and we only highlight the main
differences.

Consider the interaction energy (1) defined on $\mathcal{P}(\mathbb{H}^{n})$.
Retaining the numbering from [7], we make the following assumptions on the
interaction potential $K$:

$(\mathcal{H}1)$: $K$ is bounded from below by a finite constant $K_{min}<0$.

$(\mathcal{H}2)$: $K$ is locally integrable, i.e.,
$\displaystyle{\int_{d(x,v)<R}|K(d(x,v))|dx<\infty}$, for any $R>0$.

$(\mathcal{H}3)$: Interactions are symmetric, i.e.,
$\displaystyle{K(x,y)=K(d(x,y))}$ (by an abuse of notation), for all
$x,y\in\mathbb{H}^{n}$.

$(\mathcal{H}4)$: The limit
$\displaystyle{K_{\infty}=\lim_{\theta\to\infty}K(\theta)}$ exists and $K$ is
unstable. Here, $K$ is unstable if there exists
$\rho\in\mathcal{P}(\mathbb{H}^{n})$ such that
$E[\rho]<\frac{1}{2}K_{\infty}$.

$(\mathcal{H}5)$: $K$ is lower semi-continuous.

$(\mathcal{H}6)$: There exists $R_{6}>0$ such that $K$ is strictly increasing
on $[R_{6},\infty)$.

Note that hypothesis $(\mathcal{H}3)$ has been used throughout the entire
paper, but we list it here separately for consistency with the numbering in
[7]. In the Euclidean case, this hypothesis shows in a weaker form in fact,
specifically $K(x)=K(-x)$ for all $x\in\mathbb{R}^{n}$.

Assumptions $(\mathcal{H}1)-(\mathcal{H}6)$ were considered in [7] for the
Euclidean space $\mathbb{R}^{n}$. Under such assumptions, the main result in
[7] (Theorem 1.4) states that there exists a global minimizer of the energy in
$\mathcal{P}(\mathbb{R}^{n})$. Moreover, there exists a uniform bound for the
supports of all minimizers of $E$. A very similar result holds for
$\mathbb{H}^{n}$. We state the result below and then we sketch its proof.

######  Theorem 4.1.

Assume that $K:[0,\infty)\to\mathbb{R}\cup\\{\infty\\}$ satisfies
$(\mathcal{H}1)-(\mathcal{H}6)$. Then, there exists a global minimizer
$\rho\in\mathcal{P}(\mathbb{H}^{n})$ of the energy $E[\rho]$ defined in (1).
In addition, there exists $D>0$ (depending only on $K$ and the dimension $n$)
such that every minimizer of $E$ has compact support with diameter at most
$D$.

The layout of the proof of the analogous theorem for $\mathbb{R}^{n}$ from [7]
is the following (we write it for $\mathbb{H}^{n}$ instead):

Part 1. Fix $R\geq 0$ and set
$\mathcal{P}_{R}(\mathbb{H}^{n})=\\{\rho\in\mathcal{P}(\mathbb{H}^{n}):\mathrm{supp}(\rho)\subset\overline{B_{R}(v))}\\}$.
Show the existence of global minimizers on $\mathcal{P}_{R}(\mathbb{H}^{n})$.
This can be done as follows [7, Lemma 2.1]. Consider a minimizing sequence
$(\rho_{k})_{k\geq 1}$ of $E$ restricted to $\mathcal{P}_{R}(\mathbb{H}^{n})$.
As $\mathcal{P}_{R}(\mathbb{H}^{n})$ is tight and weakly-$*$ closed in
$\mathcal{P}(\mathbb{H}^{n})$, by Prokhorov’s Theorem, there exists a
subsequence $(\rho_{k_{l}})_{l\geq 1}$ that converges weakly-$*$ to
$\rho_{R}\in\mathcal{P}_{R}(\mathbb{H}^{n})$. By the weakly-$*$ lower
semicontinuity of $E$, $\rho_{R}$ is a global minimizer of $E$ on
$\mathcal{P}_{R}(\mathbb{H}^{n})$.

Part 2. Show that there exists $D>0$ (depending only on $K$ and $n$) such that
for all $R\geq 0$ and any global minimizer $\rho_{R}$ of $E$ on
$\mathcal{P}_{R}(\mathbb{H}^{n})$, the diameter of the support of $\rho_{R}$
is bounded by $D$. This is the core of the proof and follows from several key
lemmas which we will discuss after presenting the layout.

Part 3. Consider $\rho^{\prime}$ a global minimizer on
$\mathcal{P}_{D}(\mathbb{H}^{n})$, where $D$ is the upper bound on the
diameters of minimizers from Part 2. Show that $\rho^{\prime}$ is in fact a
global minimizer on $\mathcal{P}(\mathbb{H}^{n})$. This can be argued as
follows [7, Lemma 2.10]. First note that for any
$\rho\in\mathcal{P}(\mathbb{H}^{n})$ with compact support, $E[\rho]\geq
E[\rho^{\prime}]$. Indeed, for $R\geq 0$ fixed, such that
$\rho\in\mathcal{P}_{R}(\mathbb{H}^{n})$, take $\rho_{R}$ a global minimizer
of $E$ on $\mathcal{P}_{R}(\mathbb{H}^{n})$. Then, $E[\rho]\geq
E[\rho_{R}]\geq E[\rho^{\prime}]$, where the second inequality holds by the
translation invariance of $E$ (as $\rho_{R}$ can be translated to a measure in
$\mathcal{P}_{D}(\mathbb{H}^{n}))$. Second, for any
$\rho\in\mathcal{P}(\mathbb{H}^{n})$, approximate $\rho$ by compactly
supported measures $\rho_{k}$ with $E[\rho_{k}]\to E[\rho]$ as $k\to\infty$.
Since $E[\rho_{k}]\geq E[\rho^{\prime}]$, for all $k$, infer $E[\rho]\geq
E[\rho^{\prime}]$ and conclude $\rho^{\prime}$ global energy minimizer on
$\mathcal{P}(\mathbb{H}^{n})$.

As noted above, Part 2 is the core of the proof. It is based on two
fundamental lemmas. The first lemma states that for any point in the support
of a minimizer $\rho_{R}$ on $\mathcal{P}_{R}(\mathbb{H}^{n})$, there has to
be some mass not far from it. Specifically, the following result holds.

######  Lemma 4.2.

[7, Lemma 2.6] Suppose the potential $K$ satisfies hypotheses
$(\mathcal{H}1)-(\mathcal{H}5)$. Then there exist constants $r,m>0$ (depending
only on $K$) such that for all sufficiently large $R$, and all global
minimizers $\rho_{R}$ of the energy on $\mathcal{P}_{R}(\mathbb{H}^{n})$ we
have

$\int_{B_{r}(x_{0})}\mathrm{d}\rho_{R}(x)\geq m,\qquad\text{ for all
}x_{0}\in\mathrm{supp}(\rho_{R}).$

Note that the constants $m$ and $r$ are universal (they do not depend on $R$).
The proof of the lemma can be immediately adapted from that of [7, Lemma 2.6].
We only point out here that for its proof, Hypotheses $(\mathcal{H}1)$ and
$(\mathcal{H}4)$ are used in a fundamental way, to bound the interaction
potential $K$ at short and long distances, respectively.

The second fundamental lemma states that minimizers $\rho_{R}$ cannot have
large gaps. The general idea of its proof is the same as for [7, Lemma 2.7],
but the details are more involved. In place of the projection on the $k$-th
coordinate (as used in $\mathbb{R}^{n}$), we will use the map $\pi_{k}$
defined in (27). This lemma is the only place where hypothesis
$(\mathcal{H}6)$ is being used.

######  Lemma 4.3.

(Generalization of [7, Lemma 2.7]) Assume that the potential $K$ satisfies
$(\mathcal{H}1)-(\mathcal{H}6)$. Let $R\geq 0$ and suppose that $\rho_{R}$ is
a global minimizer of the energy (1) on $\mathcal{P}_{R}(\mathbb{H}^{n})$.
Then the support of $\rho_{R}$ cannot have “gaps” larger than $2R_{6}$ in each
coordinate ($R_{6}$ is the constant from $(\mathcal{H}6)$): if
$k\in\\{1,\dots,n\\}$ and $a_{k}\in\mathbb{R}$ is such that
$\displaystyle{\pi_{k}^{-1}([a_{k}-R_{6},a_{k}+R_{6}])\subseteq\mathbb{H}^{n}\backslash\mathrm{supp}(\rho_{R})}$,
then either
$\displaystyle{\pi_{k}^{-1}((-\infty,a_{k}-R_{6}])\subseteq\mathbb{H}^{n}\backslash\mathrm{supp}(\rho_{R})}$
or
$\displaystyle{\pi_{k}^{-1}([a_{k}+R_{6},\infty))\subseteq\mathbb{H}^{n}\backslash\mathrm{supp}(\rho_{R})}$.

###### Proof.

Suppose by contradiction that there exists an index $k\in\\{1,2,\cdots,n\\}$
and $a_{k}\in\mathbb{R}$ with
$\displaystyle{\pi_{k}^{-1}([a_{k}-R_{6},a_{k}+R_{6}])\subseteq\mathbb{H}^{n}\backslash\mathrm{supp}(\rho_{R})}$,
such that the support of $\rho_{R}$ has nontrivial intersections with both the
“left” and “right” parts, defined as

$H_{\mathrm{L}}:=\pi_{k}^{-1}((-\infty,a_{k}-R_{6}])\quad\text{and}\quad
H_{\mathrm{R}}:=\pi_{k}^{-1}([a_{k}+R_{6},\infty)),$

respectively. Also, fix $0<\epsilon<R_{6}$ and consider

$\bar{\rho}_{R}:=\rho_{R}|_{{}_{H_{\mathrm{L}}}}+{T_{\epsilon}}_{\\#}\bigl{(}\rho_{R}|_{{}_{H_{\mathrm{R}}}}\bigr{)},$

where $T_{\epsilon}$ is the isometry defined as
$T_{\epsilon}(x)=x-^{\prime}u_{k}(\epsilon)$ (recall (23)).

It follows immediately that

$\displaystyle\mathrm{supp}(\rho_{R}|_{{}_{H_{\mathrm{L}}}})\subset
H_{\mathrm{L}}\quad\textrm{ and
}\quad\mathrm{supp}\bigl{(}{T_{\epsilon}}_{\\#}\bigl{(}\rho_{R}|_{{}_{H_{\mathrm{R}}}}\bigr{)}\bigr{)}\subset
H_{\mathrm{R}}-^{\prime}u_{k}(\epsilon).$

From Corollary 2.13 we have

$H_{\mathrm{R}}-^{\prime}u_{k}(\epsilon)=\pi_{k}^{-1}([a_{k}+R_{6},\infty))-^{\prime}u_{k}(\epsilon)=\pi_{k}^{-1}([a_{k}+R_{6}-\epsilon,\infty)).$

Hence, we get

$\mathrm{supp}(\rho_{R}|_{{}_{H_{\mathrm{L}}}})\subset\pi_{k}^{-1}((-\infty,a_{k}-R_{6}])\quad\textrm{
and
}\quad\mathrm{supp}\bigl{(}{T_{\epsilon}}_{\\#}\bigl{(}\rho_{R}|_{{}_{H_{\mathrm{R}}}}\bigr{)}\bigr{)}\subset\pi_{k}^{-1}([a_{k}+R_{6}-\epsilon,\infty)).$

From this fact and $0<\epsilon<R_{6}$, we infer that
$\mathrm{supp}(\rho_{R}|_{{}_{H_{\mathrm{L}}}})$ and
$\mathrm{supp}\bigl{(}{T_{\epsilon}}_{\\#}\bigl{(}\rho_{R}|_{{}_{H_{\mathrm{R}}}}\bigr{)}\Bigr{)}$
are disjoint. This yields the following calculation:

$\displaystyle E(\bar{\rho}_{R})$
$\displaystyle=E(\rho_{R}|_{{}_{H_{\mathrm{L}}}})+E({T_{\epsilon}}_{\\#}\bigl{(}\rho_{R}|_{{}_{H_{\mathrm{R}}}}\bigr{)})+\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}K(d(x,y))\,\mathrm{d}\rho_{R}|_{{}_{H_{\mathrm{L}}}}(x)\,\mathrm{d}{T_{\epsilon}}_{\\#}\bigl{(}\rho_{R}|_{{}_{H_{\mathrm{R}}}}\bigr{)}(y)$
$\displaystyle=E(\rho_{R}|_{{}_{H_{\mathrm{L}}}})+E(\rho_{R}|_{{}_{H_{\mathrm{R}}}})+\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}K(d(x,y-^{\prime}u_{k}(\epsilon)))\,\mathrm{d}\rho_{R}|_{{}_{H_{\mathrm{L}}}}(x)\,\mathrm{d}\rho_{R}|_{{}_{H_{\mathrm{R}}}}(y),$
(73)

where for the second equal sign we used the invariance under isometries of the
energy (see (5)) and the change of variable $z=y+^{\prime}u_{k}(\epsilon)$ in
the integral.

Now take $x\in H_{\mathrm{L}}$ and $y\in H_{\mathrm{R}}$, and recall (22), by
which we write

$\cosh(d(x,y-^{\prime}u_{k}(\epsilon)))=(x-^{\prime}(y-^{\prime}u_{k}(\epsilon)))_{0}.$

Next, we will compare $(x-^{\prime}(y-^{\prime}u_{k}(\epsilon)))_{0}$ with
$(x-^{\prime}y)_{0}$.

Using (LABEL:a-a-10), (19) and (20), we compute

$\displaystyle(x-^{\prime}(y-^{\prime}u_{k}(\epsilon)))_{0}$
$\displaystyle=x_{0}(y-^{\prime}u_{k}(\epsilon))_{0}-\sum_{l=1}^{n}x_{l}(y-^{\prime}u_{k}(\epsilon))_{l}$
$\displaystyle=x_{0}(y_{0}\cosh\epsilon-
y_{k}\sinh\epsilon)-\sum_{l=1}^{n}x_{l}(-y_{0}\delta_{kl}\sinh\epsilon+y_{l}+y_{l}\delta_{kl}(\cosh\epsilon-1))$
$\displaystyle=x_{0}(y_{0}\cosh\epsilon-
y_{k}\sinh\epsilon)+x_{k}y_{0}\sinh\epsilon-\sum_{l=1}^{n}x_{l}y_{l}+(1-\cosh\epsilon)x_{k}y_{k}.$

Furthermore, add and subtract $x_{0}y_{0}$ to the right-hand-side above and
use (21) to get

$\displaystyle(x-^{\prime}(y-^{\prime}u_{k}(\epsilon)))_{0}-(x-^{\prime}y)_{0}$
$\displaystyle=$
$\displaystyle(\cosh\epsilon-1)(x_{0}y_{0}-x_{k}y_{k})+\sinh\epsilon(x_{k}y_{0}-x_{0}y_{k})$
(74) $\displaystyle=$ $\displaystyle
2\sinh^{2}\left(\frac{\epsilon}{2}\right)x_{0}y_{0}\left(1-\frac{x_{k}y_{k}}{x_{0}y_{0}}\right)+2\sinh\left(\frac{\epsilon}{2}\right)\cosh\left(\frac{\epsilon}{2}\right)x_{0}y_{0}\left(\frac{x_{k}}{x_{0}}-\frac{y_{k}}{y_{0}}\right)$
$\displaystyle=$ $\displaystyle
2\sinh\left(\frac{\epsilon}{2}\right)x_{0}y_{0}\left(\sinh\left(\frac{\epsilon}{2}\right)\left(1-\tanh(\pi_{k}(x))\tanh(\pi_{k}(y))\right)+\cosh\left(\frac{\epsilon}{2}\right)(\tanh(\pi_{k}(x))-\tanh(\pi_{k}(y)))\right)$
$\displaystyle=$ $\displaystyle
2\sinh\left(\frac{\epsilon}{2}\right)x_{0}y_{0}(1-\tanh(\pi_{k}(x))\tanh(\pi_{k}(y)))\left(\sinh\left(\frac{\epsilon}{2}\right)+\cosh\left(\frac{\epsilon}{2}\right)\tanh(\pi_{k}(x)-\pi_{k}(y))\right)$
$\displaystyle=$
$\displaystyle\left(\frac{2\sinh\left(\frac{\epsilon}{2}\right)x_{0}y_{0}(1-\tanh\pi_{k}(x)\tanh\pi_{k}(y))}{\cosh(\pi_{k}(x)-\pi_{k}(y))}\right)\sinh\left(\pi_{k}(x)-\pi_{k}(y)+\frac{\epsilon}{2}\right).$
(75)

Since

$\frac{2\sinh\left(\frac{\epsilon}{2}\right)x_{0}y_{0}(1-\tanh\pi_{k}(x)\tanh\pi_{k}(y))}{\cosh(\pi_{k}(x)-\pi_{k}(y))}$

is always positive for any $\epsilon>0$, we only have to study a sign of
$\sinh\left(\pi_{k}(x)-\pi_{k}(y)+\frac{\epsilon}{2}\right)$. As $x\in H_{L}$,
$y\in H_{R}$, and $\epsilon<R_{6}$, we have

$\pi_{k}(x)-\pi_{k}(y)+\frac{\epsilon}{2}<(a_{k}-R_{6})-(a_{k}+R_{6})+\frac{R_{6}}{2}=-\frac{3}{2}R_{6}<0,$

which by (75) it implies that

$(x-^{\prime}(y-^{\prime}u_{k}(\epsilon)))_{0}<(x-^{\prime}y)_{0}\quad\Longleftrightarrow\quad
d(x,y-^{\prime}u_{k}(\epsilon))<d(x,y).$ (76)

Now apply Lemma 2.14 for $x$ and $y-^{\prime}u_{k}(x)$, and use
$\pi_{k}(x)<a_{k}-R_{6}$ and
$\pi_{k}(y-^{\prime}u_{k}(\epsilon))>a_{k}+R_{6}-\epsilon$, to get

$d(x,y-^{\prime}u_{k}(\epsilon))>|(a_{k}+R_{6}-\epsilon)-(a_{k}-R_{6})|=2R_{6}-\epsilon>R_{6}.$
(77)

By combining (76) and (77) we now get

$R_{6}<d(x,y-^{\prime}u_{k}(\epsilon))<d(x,y),$

and furthermore,

$K(d(x,y-^{\prime}u_{k}(\epsilon)))<K(d(x,y)),$

since $K$ is strictly increasing on $[R_{6},\infty)$ by Hypothesis
$(\mathcal{H}6)$.

Finally, we substitute this result into (73) to get

$\displaystyle E(\bar{\rho}_{R})$
$\displaystyle=E(\rho_{R}|_{{}_{H_{\mathrm{L}}}})+E(\rho_{R}|_{{}_{H_{\mathrm{R}}}})+\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}K(d(x,y-^{\prime}u_{k}(\epsilon)))\,\mathrm{d}\rho_{R}|_{{}_{H_{\mathrm{L}}}}(x)\,\mathrm{d}\rho_{R}|_{{}_{H_{\mathrm{R}}}}(y)$
$\displaystyle<E(\rho_{R}|_{{}_{H_{\mathrm{L}}}})+E(\rho_{R}|_{{}_{H_{\mathrm{R}}}})+\iint_{\mathbb{H}^{n}\times\mathbb{H}^{n}}K(d(x,y))\,\mathrm{d}\rho_{R}|_{{}_{H_{\mathrm{L}}}}(x)\,\mathrm{d}\rho_{R}|_{{}_{H_{\mathrm{R}}}}(y)$
$\displaystyle=E(\rho_{R}).$

However, this is a contradiction to the fact that $\rho_{R}$ is a global
minimizer on $\mathcal{P}_{R}(\mathbb{H}^{n})$. ∎

Lemmas 4.2 and 4.3 lead to the following last result, that concludes Part 2 of
the proof of Theorem 4.1. The statement and its proof are identical to those
of [7, Lemma 2.9].

######  Lemma 4.4.

[7, Lemma 2.9] Assume $K$ satisfies $(\mathcal{H}1)-(\mathcal{H}6)$. Then
there exists $D>0$ (depending only on $K$ and $n$) such that for all $R\geq 0$
and any global minimizer $\rho_{R}$ of the energy on
$\mathcal{P}_{R}(\mathbb{H}^{n})$, the diameter of the support of $\rho_{R}$
is bounded by $D$.

###### Proof.

The proof of the lemma follows that of [7, Lemma 2.9] by replacing the $k$-th
coordinate of a point $x$ by $\pi_{k}(x)$ as defined in (27). We sketch it
here. First, given that there is at least an amount of mass $m$ in any
geodesic ball $B_{r}(x)$ with $x\in\mathrm{supp}(\rho_{R})$ (by Lemma 4.2),
the support of $\rho_{R}$ can be covered by a finite number of geodesic balls
$B_{2r}(x_{i})$, $i=0,1,\dots,N$, with $N+1\leq[1/m]$ ($[\cdot]$ denotes the
ceiling of a number). Note that the points $x_{i}$ are chosen so that
$x_{i}\notin B_{2r}(x_{j})$, for any $i\neq j$, so the balls $B_{r}(x_{i})$
are disjoint.

With $k$ fixed ($1\leq k\leq n$), order the points so that
$\pi_{k}(x_{0})<\dots<\pi_{k}(x_{N})$. Then,

$\pi_{k}(x_{i+1})-\pi_{k}(x_{i})\leq 4r+2R_{6},\qquad i=0,1,\dots N-1,$

as otherwise this would contradict Lemma 4.3 (gaps larger than $2R_{6}$). From
here, get that

$\pi_{k}(x_{N})-\pi_{k}(x_{0})\leq N(4r+2R_{6})\leq([1/m]-1)(4r+2R_{6}).$ (78)

Now set $x_{l}$ and $x_{m}$ such that

$d(x_{l},x_{m})=\max_{0\leq i,j\leq N}d(x_{i},x_{j}).$

Consider the geodesic curve between $x_{l}$ and $x_{m}$, and the midpoint
between the two points on this geodesic. Take the isometry $T$ that maps the
midpoint into the vertex $v$, and $x_{l}\mapsto z$, $x_{m}\mapsto-z$ (see
(19)), where $z=(\cos\alpha,\sin\alpha,0,\dots,0)$. Then,

$d(x_{l},x_{m})=d(z,-z)=2\alpha,$

where the second equal sign comes from the fact that (see (10))

$d(-z,z)=2d(v,z)=2\cosh^{-1}(\cosh\alpha)=2\alpha.$

On the other hand, by (27), we get

$2\alpha=\pi_{1}(z)-\pi_{1}(-z).$

Finally, by (78) (used for the image under the isometry $T$ of
$\mathrm{supp}(\rho_{R})$), one has

$\pi_{1}(z)-\pi_{1}(-z)\leq([1/m]-1)(4r+2R_{6}).$

Combining these considerations, we find

$\text{diam}(\mathrm{supp}(\rho_{R}))\leq([1/m]-1)(4r+2R_{6})+4r=:D.$

∎

##  5 Numerical simulations

In this section, we provide numerical simulations of the model (2) for some
specific interaction potentials in the form (31). For this purpose, we write
model (2) in characteristic form; this form of the equation is widely used for
both analysis and numerical simulations [5, 18, 20, 23].

##### Model (2) in characteristic form.

Consider the flow generated by the velocity field $\mathbf{v}[\rho]$ given by
(4). The characteristic paths $x(t)$ then satisfy

$\frac{\mathrm{d}x}{\mathrm{d}t}=\mathbf{v}[\rho](x(t),t).$ (79)

To reach the characteristic form of model (2), expand the divergence

$\nabla\cdot(\rho\mathbf{v}[\rho])=\nabla_{\mathbf{v}[\rho]}\rho+\rho\nabla\cdot\mathbf{v}[\rho],$

where $\nabla_{\mathbf{v}[\rho]}\rho$ denotes the covariant derivative of
$\rho$ along $\mathbf{v}[\rho]$. Then write (2) as

$\displaystyle\partial_{t}\rho+\nabla_{\mathbf{v}[\rho]}\rho$
$\displaystyle=-\rho\nabla\cdot\mathbf{v}[\rho]$
$\displaystyle=\rho(t,x)\int_{\mathbb{H}^{n}}\Delta_{x}K(x,y)\rho(t,y)\mathrm{d}y,$
(80)

where for the second equal sign we used (4). Note that the left-hand-side
above represents the material derivative (along characteristic paths) of
$\rho$.

For numerical simulations we consider only radially symmetric solutions. In
such case, the characteristic paths $x(t)$ are geodesic curves passing through
the vertex $v$. Denote the radial coordinate

$\theta_{x}:=d(x,v).$

By writing the characteristic equations (79) in radial coordinates, we then
find

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\theta_{x(t)}=-\hat{n}_{x}\cdot\nabla_{x}\int_{\mathbb{H}^{n}}K(x,y)\rho(t,y)\mathrm{d}y,$
(81)

where $\hat{n}_{x}:=\frac{x}{\theta_{x}}$.

The interaction potentials in the numerical simulations are in the form (31).
Using (31), the radial symmetry of $\rho$, the divergence theorem and (30), we
compute starting from (81):

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\theta_{x(t)}=-\hat{n}_{x}\cdot\nabla_{x}\int_{\mathbb{H}^{n}}(G(x,y)+A(d(x,y))\rho(t,y)\mathrm{d}y$
$\displaystyle\quad=-\frac{1}{|\partial B_{v}(\theta_{x})|}\int_{\partial
B_{v}(\theta_{x})}\hat{n}_{x}\cdot\nabla_{x}\int_{\mathbb{H}^{n}}G(x,y)\rho(t,y)\mathrm{d}y\mathrm{d}\sigma_{x}-\hat{n}_{x}\cdot\nabla_{x}\int_{\mathbb{H}^{n}}A(d(x,y))\rho(t,y)\mathrm{d}y$
$\displaystyle\quad=-\frac{1}{|\partial
B_{v}(\theta_{x})|}\int_{B_{v}(\theta_{x})}\Delta_{x}\int_{\mathbb{H}^{n}}G(x,y)\rho(t,y)\mathrm{d}y\mathrm{d}\sigma_{x}-\hat{n}_{x}\cdot\nabla_{x}\int_{\mathbb{H}^{n}}A(d(x,y))\rho(t,y)\mathrm{d}y$
$\displaystyle\quad=-\frac{1}{n\alpha(n)\sinh^{n-1}\theta_{x}}\int_{B_{v}(\theta_{x})}\left(-\rho(t,x)\right)\mathrm{d}x-\hat{n}_{x}\cdot\nabla_{x}\int_{\mathbb{H}^{n}}A(d(x,y))\rho(t,y)dy.$

Furthermore, using the notation $\tilde{\rho}(t,\theta_{x})=\rho(t,x)$ and the
hyperbolic law of cosines, we write

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\theta_{x(t)}=\frac{1}{\sinh^{n-1}\theta_{x}}\int_{0}^{\theta_{x}}\tilde{\rho}(t,\lambda)\sinh^{n-1}\lambda\,\mathrm{d}\lambda$
$\displaystyle\quad-\int_{\mathbb{H}^{n}}\partial_{\theta_{x}}A\left(\cosh^{-1}\left(\cosh\theta_{x}\cosh\theta_{y}-\sinh\theta_{x}\sinh\theta_{y}\cos\angle
xvy\right)\right)\tilde{\rho}(t,\theta_{y})\mathrm{d}y$
$\displaystyle=\frac{1}{\sinh^{n-1}\theta_{x}}\int_{0}^{\theta_{x}}\tilde{\rho}(t,\lambda)\sinh^{n-1}\lambda\,\mathrm{d}\lambda-(n-1)\alpha(n-1)$
$\displaystyle\quad\times\int_{0}^{\pi}\int_{0}^{\infty}\partial_{\theta_{x}}A\left(\cosh^{-1}\left(\cosh\theta_{x}\cosh\theta_{y}-\sinh\theta_{x}\sinh\theta_{y}\cos\beta\right)\right)\tilde{\rho}(t,\theta_{y})\sinh^{n-1}\theta_{y}\sin^{n-2}\beta\,\mathrm{d}\theta_{y}\mathrm{d}\beta.$
(82)

For the evolution of the density along characteristic paths, we calculate from
(80):

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\rho(t,x(t))=\rho(t,x)\int_{\mathbb{H}^{n}}\Delta_{x}K(x,y)\rho(t,y)\mathrm{d}y$
$\displaystyle=\rho(t,x)\left(-\rho(t,x)+\int_{\mathbb{H}^{n}}\Delta_{x}A(d(x,y))\rho(t,y)\mathrm{d}y\right)$
$\displaystyle=-\rho(t,x)^{2}+\rho(t,x)\int_{\mathbb{H}^{n}}\Delta
A\left(\cosh^{-1}\left(\cosh\theta_{x}\cosh\theta_{y}-\sinh\theta_{x}\sinh\theta_{y}\cos\angle
xvy\right)\right)\tilde{\rho}(t,\theta_{y})\mathrm{d}y$
$\displaystyle=-\rho(t,x)^{2}+(n-1)\alpha(n-1)\rho(t,x)$
$\displaystyle\quad\times\int_{0}^{\pi}\int_{0}^{\infty}\Delta
A\left(\cosh^{-1}\left(\cosh\theta_{x}\cosh\theta_{y}-\sinh\theta_{x}\sinh\theta_{y}\cos\beta\right)\right)\tilde{\rho}(t,\theta_{y})\sinh^{n-1}\theta_{y}\sin^{n-2}\beta\,\mathrm{d}\theta_{y}\mathrm{d}\beta.$
(83)

Due to the radial symmetry assumption, the evolution equations (82) and (83)
are one-dimensional. For simplicity, denote $\theta(t)=\theta_{x(t)}$ and use
the notation $\tilde{\rho}$, to write (82) and (83) as a system of two
integro-differential equations for $\theta(t)$ and
$\tilde{\rho}(t,\theta(t))$:

$\displaystyle\begin{cases}&\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\theta(t)=\frac{1}{\sinh^{n-1}\theta}\int_{0}^{\theta}\tilde{\rho}(t,\lambda)\sinh^{n-1}\lambda\,\mathrm{d}\lambda-(n-1)\alpha(n-1)\\\\[12.0pt]
&\quad\displaystyle\times\int_{0}^{\pi}\int_{0}^{\infty}\partial_{\theta}A\left(\cosh^{-1}\left(\cosh\theta\cosh\theta_{y}-\sinh\theta\sinh\theta_{y}\cos\beta\right)\right)\tilde{\rho}(t,\theta_{y})\sinh^{n-1}\theta_{y}\sin^{n-2}\beta\,\mathrm{d}\theta_{y}\mathrm{d}\beta\\\\[15.0pt]
&\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\tilde{\rho}(t,\theta(t))=-\tilde{\rho}(t,\theta)^{2}+(n-1)\alpha(n-1)\tilde{\rho}(t,\theta)\\\\[12.0pt]
&\quad\displaystyle\times\int_{0}^{\pi}\int_{0}^{\infty}\Delta
A\left(\cosh^{-1}\left(\cosh\theta\cosh\theta_{y}-\sinh\theta\sinh\theta_{y}\cos\beta\right)\right)\tilde{\rho}(t,\theta_{y})\sinh^{n-1}\theta_{y}\sin^{n-2}\beta\,\mathrm{d}\theta_{y}\mathrm{d}\beta,\end{cases}$
(84)

where $\Delta
A(\theta)=A^{\prime\prime}(\theta)+(n-1)A^{\prime}(\theta)\coth\theta$.

##### Numerical results.

All numerical results we present are for the hyperbolic plane ($n=2$), for
which (84) simplifies to

$\displaystyle\begin{cases}&\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\theta(t)=\frac{1}{\sinh\theta}\int_{0}^{\theta}\tilde{\rho}(t,\lambda)\sinh\lambda\,\mathrm{d}\lambda\\\\[12.0pt]
&\qquad\displaystyle-2\int_{0}^{\pi}\int_{0}^{\infty}\partial_{\theta}A\left(\cosh^{-1}\left(\cosh\theta\cosh\theta_{y}-\sinh\theta\sinh\theta_{y}\cos\beta\right)\right)\tilde{\rho}(t,\theta_{y})\sinh\theta_{y}\,\mathrm{d}\theta_{y}\mathrm{d}\beta\\\\[15.0pt]
&\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\tilde{\rho}(t,\theta)=-\tilde{\rho}(t,\theta)^{2}\\\\[12.0pt]
&\qquad\displaystyle+2\tilde{\rho}(t,\theta)\int_{0}^{\pi}\int_{0}^{\infty}\Delta
A\left(\cosh^{-1}\left(\cosh\theta\cosh\theta_{y}-\sinh\theta\sinh\theta_{y}\cos\beta\right)\right)\tilde{\rho}(t,\theta_{y})\sinh\theta_{y}\,\mathrm{d}\theta_{y}\mathrm{d}\beta.\end{cases}$
(85)

For all simulations, the (radially symmetric) initial density $\tilde{\rho}$
is taken to be

$\displaystyle\tilde{\rho}_{0}(\theta)=e^{-5\theta}(0.01+\theta-\theta^{2})/C,$
(86)

where $C$ is a normalization constant which satisfies
$\int_{0}^{\infty}\tilde{\rho}_{0}(\theta)\sinh\theta\mathrm{d}\theta=C$. We
use the classical fourth order Runge-Kutta method with $\Delta t=0.02$ to
evolve (85) in time. The initial particle positions are set at
$\theta_{i}(0)=(i-1)\Delta x$ with $\Delta x=0.02$, $i=1,\dots,N$, with
$N=51$.

We consider the interaction potentials (in the form (31)) from Section 3.2,
for which we have explicit expressions of the equilibria. First, take
attraction given by (56) and (58), for which we computed the explicit
solutions (59) and (61). In $n=2$ dimensions, these solutions (written in the
radial notation, using the tilde symbol) read

$\displaystyle\begin{aligned}
&\tilde{\rho}_{c}(\theta)=\begin{cases}\displaystyle 1\qquad\text{if
}0\leq\theta<\cosh^{-1}\Bigl{(}1+\frac{1}{2\pi}\Bigr{)},\\\\[5.0pt]
0\qquad\text{otherwise},\end{cases}\end{aligned}$ (87)

and

$\displaystyle\begin{aligned}
&\tilde{\rho}_{h}(\theta)=\begin{cases}\displaystyle\frac{\cosh\theta}{\pi\left(\left(1+\frac{3}{2\pi}\right)^{2/3}-1\right)}&\displaystyle\qquad\text{if
}0\leq\theta<\cosh^{-1}\biggl{(}\Bigl{(}1+\frac{3}{2\pi}\Bigr{)}^{1/3}\biggr{)},\\\\[10.0pt]
0&\qquad\text{otherwise}.\end{cases}\end{aligned}$ (88)

Note the monotonicity of these equilibria, which is consistent with Theorem
3.3.

The time evolution of the model with Newtonian repulsion and attraction (56)
and (58), is illustrated in Figure 4. In both cases, the solution approaches
the equilibria in (87) and (88), respectively. The simulation confirms the
theoretical findings in Section 3.3 (see also Theorem 3.8): the energy
functional is strictly convex and solutions approach asymptotically the global
energy minimizer. On the other hand, numerical simulations with a variety of
other initial conditions suggest that the equilibrium (87) is also a global
attractor; this was first conjectured in [23].

|
---|---
(a) | (b)
Figure 4: Time evolution of a radially symmetric solution to the aggregation
model (2) on $\mathbb{H}^{2}$, for $K$ in the form (31), with attraction given
by (a) $A=A_{c}$ and (b) $A=A_{h}$ – see (56) and (58). The initial density
(blue dashed-line marked with circles) is given in (86). The solutions
approach asymptotically the equilibria (87) and (88), respectively, indicated
by red solid lines.

For numerical simulations, we also considered linear combinations of the
potentials $A_{c}$ and $A_{h}$, i.e., potentials in the form

$\displaystyle A_{m}:=b_{0}A_{c}+b_{1}A_{h},$ (89)

with $b_{0},b_{1}$ constants. Note that $\Delta
A_{m}(\theta)=b_{0}+b_{1}\cosh\theta$.

The following proposition provides an explicit expression for the equilibria
corresponding to mixed attraction $A_{m}$.

######  Proposition 5.1.

Consider the aggregation model (2) on $\mathbb{H}^{2}$, with
$K(x,y)=G(x,y)+A_{m}(d(x,y))$, and $b_{0},b_{1}\geq 0$ such that
$(b_{0},b_{1})\neq(0,0)$. Then, there exists a unique radially symmetric
equilibrium in the form

$\tilde{\rho}_{m}(\theta)=\begin{cases}\displaystyle b_{0}+\left(\frac{1-2\pi
b_{0}\alpha_{b_{0}b_{1}}}{\pi\alpha_{b_{0}b_{1}}(\alpha_{b_{0}b_{1}}+2)}\right)\cosh\theta\qquad&\text{
if }0\leq\theta\leq\cosh^{-1}(\alpha_{b_{0}b_{1}}+1),\\\\[10.0pt]
0\qquad&\text{ if }\theta>\cosh^{-1}(\alpha_{b_{0}b_{1}}+1),\end{cases}$

where $\alpha_{b_{0}b_{1}}$ is a solution of the polynomial

$1-2\pi(b_{0}+b_{1})x-2\pi b_{1}x^{2}-\frac{2\pi
b_{1}}{3}x^{3}+\frac{\pi^{2}b_{0}b_{1}}{3}x^{4}=0.$

###### Proof.

See Appendix B. ∎

######  Remark 5.2.

While we cannot prove Proposition 5.1 for general
$(b_{0},b_{1})\in\mathbb{R}^{2}\backslash\\{(0,0)\\}$, similar arguments can
be used for some specific cases. For example, the result in Proposition 5.1
holds for $(b_{0},b_{1})=(2,-1)$, although $b_{1}<0$. We will present below
numerical simulations for $(b_{0},b_{1})=(2,-1)$, as in this case the
equilibrium solution $\tilde{\rho}_{m}$ is qualitatively different from
$\tilde{\rho}_{c}$ and $\tilde{\rho}_{h}$. Indeed, $\tilde{\rho}_{m}$ is
monotonically decreasing, following the monotonicity of $\Delta A_{m}$ (see
Theorem 3.3).

For $(b_{0},b_{1})=(2,-1)$, the equilibrium in Proposition 5.1 can be
calculated as

$\displaystyle\begin{aligned}
&\tilde{\rho}_{m}(\theta)\simeq\begin{cases}2-1.0956\cosh\theta&\qquad\text{
if }0\leq\theta<0.6227,\\\ 0&\qquad\text{ otherwise}.\end{cases}\end{aligned}$
(90)

Figure 5(a) shows the time evolution of the initial density (86), for a
potential with mixed attraction (89), and $(b_{0},b_{1})=(2,-1)$. The solution
approach asymptotically the steady state $\tilde{\rho}_{m}$. Simulations with
a variety of other initial densities (including non-symmetric ones) suggest
that $\tilde{\rho}_{m}$ is in fact a global attractor. We clarify here that
for non-symmetric densities we use a particle method and evolve in time the
discrete analogue of (2) – see [23] for the particle method formulation of the
model (2) on $\mathbb{H}^{n}$. Indeed, starting from random distributions,
particles approach a symmetric steady state supported on a geodesic ball (we
do not present these simulations here).

Finally, we provide a simulation with an interaction potential in the form
(31) for which $\Delta A$ is not monotone (so Theorem 3.3 does not hold).
Specifically, we take

$A(\theta)=\frac{1}{6}\sinh^{2}\theta-\frac{3}{2}\cosh\theta+6\ln\left(\cosh\frac{\theta}{2}\right),$
(91)

for which $\Delta A(\theta)=\cosh^{2}\theta-3\cosh\theta+8/3$. Figure 5(b)
shows this simulation, starting from the same initial density (86). Note that
in this case, the steady state is not monotone, as in the previous examples.
Also, we do not have any explicit form for the steady state, or any result
about its uniqueness; this is simply a numerically obtained equilibrium.
Nevertheless, simulations with various other symmetric and non-symmetric
initial data suggest a large basin of attraction for this equilibrium,
possibly being a global attractor as well.

|
---|---
(a) | (b)
Figure 5: Time evolution of a radially symmetric solution to the aggregation
model (2) on $\mathbb{H}^{2}$, for $K$ in the form (31), with attraction given
by (a) $A=A_{m}$ with $(b_{0},b_{1})=(2,-1)$ – see (89), and (b) $A$ given by
(91). The initial data is given by (86). For simulation (a), the solution
approaches asymptotically the equilibrium (90); note that the steady state is
monotonically decreasing about the vertex, consistent with the result in
Theorem 3.3. The steady state for the simulation (b) is not monotone; this
case is not covered by Theorem 3.3.

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##  Appendix A Proofs of several lemmas from Section 2.2

###### Proof of Lemma 2.2.

From the definition of the matrix product, we have

$y_{\alpha}=\sum_{\beta=0}^{n}B(w)_{\alpha\beta}x_{\beta},\qquad\forall~{}\alpha=0,1,\cdots,n.$

We can simplify it as follows:

$y_{0}=\sum_{\beta=0}^{n}B(w)_{0\beta}x_{\beta}=B(w)_{00}x_{0}+\sum_{i=1}^{n}B(w)_{0i}x_{i}=\gamma
x_{0}-\gamma\sum_{i=1}^{n}w_{i}x_{i},$

and for $j=1,\dots,n$,

$\displaystyle y_{j}$
$\displaystyle=\sum_{\beta=0}^{n}B(w)_{j\beta}x_{\beta}=B(w)_{j0}x_{0}+\sum_{i=1}^{n}B(w)_{ji}x_{i}$
$\displaystyle=-\gamma
w_{j}x_{0}+\sum_{i=1}^{n}\left(\delta_{ij}+(\gamma-1)\frac{w_{i}w_{j}}{|w|^{2}}\right)x_{i}$
$\displaystyle=-\gamma
w_{j}x_{0}+x_{j}+(\gamma-1)\frac{w_{j}}{|w|^{2}}\sum_{i=1}^{n}x_{i}w_{i}.$

This yields

$\displaystyle y_{0}^{2}-y_{1}^{2}-\cdots-y_{n}^{2}$
$\displaystyle=\gamma^{2}\left(x_{0}-\sum_{i=1}^{n}w_{i}x_{i}\right)^{2}-\sum_{j=1}^{n}\left(-\gamma
w_{j}x_{0}+x_{j}+(\gamma-1)\frac{w_{j}}{|w|^{2}}\sum_{i=1}^{n}x_{i}w_{i}\right)^{2}$
$\displaystyle=\gamma^{2}x_{0}^{2}-2\gamma^{2}x_{0}\sum_{i=1}^{n}w_{i}x_{i}+\gamma^{2}\left(\sum_{i=1}^{n}w_{i}x_{i}\right)^{2}-\sum_{j=1}^{n}\left(\gamma^{2}w_{j}^{2}x_{0}^{2}+x_{j}^{2}+(\gamma-1)^{2}\frac{w_{j}^{2}}{|w|^{4}}\left(\sum_{i=1}^{n}x_{i}w_{i}\right)^{2}\right)$
$\displaystyle\quad-\sum_{j=1}^{n}\left(-2\gamma
w_{j}x_{j}x_{0}-2\gamma(\gamma-1)\frac{w_{j}^{2}x_{0}}{|w|^{2}}\sum_{i=1}^{n}x_{i}w_{i}+2(\gamma-1)\frac{x_{j}w_{j}}{|w|^{2}}\sum_{i=1}^{n}x_{i}w_{i}\right)$
$\displaystyle=\gamma^{2}x_{0}^{2}-2\gamma^{2}x_{0}\sum_{i=1}^{n}w_{i}x_{i}+\gamma^{2}\left(\sum_{i=1}^{n}w_{i}x_{i}\right)^{2}-\gamma^{2}x_{0}^{2}|w|^{2}-\sum_{j=1}^{n}x_{j}^{2}-(\gamma-1)^{2}\frac{1}{|w|^{2}}\left(\sum_{i=1}^{n}x_{i}w_{i}\right)^{2}$
$\displaystyle\quad+2x_{0}\gamma\sum_{i=1}^{n}x_{i}w_{i}+2\gamma(\gamma-1)x_{0}\sum_{i=1}^{n}x_{i}w_{i}-2(\gamma-1)\frac{1}{|w|^{2}}\left(\sum_{i=1}^{n}x_{i}w_{i}\right)^{2}$
$\displaystyle=\gamma^{2}x_{0}^{2}+\gamma^{2}\left(\sum_{i=1}^{n}w_{i}x_{i}\right)^{2}-\gamma^{2}x_{0}^{2}|w|^{2}-\sum_{j=1}^{n}x_{j}^{2}-(\gamma^{2}-1)\frac{1}{|w|^{2}}\left(\sum_{i=1}^{n}x_{i}w_{i}\right)^{2}$
$\displaystyle=\gamma^{2}(1-|w|^{2})x_{0}^{2}-\sum_{j=1}^{n}x_{j}^{2}=x_{0}^{2}-x_{1}^{2}-\cdots-
x_{n}^{2}.$

If we use $x\in\mathbb{H}^{n}$, then we have $x_{0}^{2}-x_{1}^{2}-\cdots-
x_{n}^{2}=1$. This implies that

$y_{0}^{2}-y_{1}^{2}-\cdots-y_{n}^{2}=1,$

which is the desired result. ∎

###### Proof of Lemma 2.3.

Let $a,b$ be tangent vectors at $x\in\mathbb{H}^{n}$. Since $\mathbb{H}^{n}$
is embedded in $\mathbb{R}^{n+1}$, tangent vectors $a$ and $b$ can also be
considered as vectors in $\mathbb{R}^{n+1}$ which satisfy

$a_{0}x_{0}-a_{1}x_{1}-\cdots-a_{n}x_{n}=b_{0}x_{0}-b_{1}x_{1}-\cdots-
b_{n}x_{n}=0.$

The induced metric defines the inner product between two tangent vectors as

$a\cdot b=a_{0}b_{0}-a_{1}b_{1}-\cdots-a_{n}b_{n}=-a^{\top}\eta b.$

From the definition of the push forward of a tangent vector, we have

$dF_{w}(a)=B(w)a,\quad dF_{w}(b)=B(w)b.$

This yields

$dF_{w}(a)\cdot
dF_{w}(b)=-(dF_{w}(a))^{\top}\eta\,dF_{w}(b)=-a^{\top}B(w)^{\top}\eta
B(w)b=-a^{\top}\eta b=a\cdot b,$

where we used (15) for the third equal sign. Finally, this implies that
$F_{w}$ is an isometry on $\mathbb{H}^{n}$ for all $w$ such that $|w|<1$. ∎

###### Proof of Lemma 2.4.

By a simple calculation, for any $w\in\mathbb{R}^{n}$ with $|w|<1$, we know
the vertex of $\mathbb{H}^{n}$ maps to

$F_{w}(1,0,\cdots,0)=(\gamma,-\gamma w_{1},\cdots,-\gamma w_{n}).$

The equation above shows that $1\iff 3$.

Now, we investigate the relationship between $x\in\mathbb{H}^{n}$ and
$w\in\mathbb{R}^{n}$ so that:

$F_{w}(x)=(1,0,\cdots,0).$

From the definition of $F_{w}$, we have

$\displaystyle 1=\gamma x_{0}-\gamma(w_{1}x_{1}+\cdots+w_{n}x_{n}),$ (92)

and

$\displaystyle 0=-\gamma
w_{i}x_{0}+x_{i}+(\gamma-1)\frac{w_{i}}{|w|^{2}}\sum_{j=1}^{n}w_{j}x_{j},\quad\forall~{}i=1,2,\cdots,n.$
(93)

Equation (93) yields

$\displaystyle w_{i}=\alpha x_{i},\qquad\forall~{}i=1,2,\cdots,n,$ (94)

for some $\alpha\in\mathbb{R}$. We substitute (94) into (92) to get

$1=\gamma x_{0}-\gamma\alpha(x_{1}^{2}+\cdots+x_{n}^{2})=\gamma
x_{0}-\gamma\alpha(x_{0}^{2}-1),$

which gives

$\displaystyle\gamma=\frac{1}{x_{0}-\alpha(x_{0}^{2}-1)}.$ (95)

By (94), we also have:

$|w|^{2}=w_{1}^{2}+w_{2}^{2}+\cdots+w_{n}^{2}=\alpha^{2}(x_{1}^{2}+\cdots+x_{n}^{2})=\alpha^{2}(x_{0}^{2}-1),$

where for the last equality we used that $x\in\mathbb{H}^{n}$. From the
definition of $\gamma$ and the above relation, we get

$\displaystyle\gamma^{2}=\frac{1}{1-|w|^{2}}=\frac{1}{1-\alpha^{2}(x_{0}^{2}-1)}.$
(96)

Finally, we compare (95) and (96) to get

$(x_{0}-\alpha(x_{0}^{2}-1))^{2}=1-\alpha^{2}(x_{0}^{2}-1),$

which is equivalent to

$(x_{0}^{2}-1)(1-\alpha x_{0})^{2}=0.$ (97)

If $w=0$, then $F_{w}$ is the identity map. So, in this case, we have
$x=(1,0,\cdots,0)$. Now assume that $|w|>0$. Then, (94) implies $x_{i}\neq 0$
for some $1\leq i\leq n$, which yields

$x_{0}^{2}=1+x_{1}^{2}+\cdots+x_{n}^{2}\geq 1+x_{i}^{2}>1.$

Hence, from (97), we get the unique solution

$\alpha=\frac{1}{x_{0}},$

and this gives the relation

$w=\left(\frac{x_{1}}{x_{0}},\cdots,\frac{x_{n}}{x_{0}}\right).$

Note that this relation also holds for $w=0$. We have thus shown that $2\iff
3$, which concludes the proof. ∎

###### Proof of Lemma 2.5.

First, we have

$\displaystyle\gamma(-\hat{y})$
$\displaystyle=(1-\|\hat{y}\|^{2})^{-1/2}=\left(1-\frac{y_{1}^{2}}{y_{0}^{2}}-\cdots-\frac{y_{n}^{2}}{y_{0}^{2}}\right)^{-1/2}$
$\displaystyle=\frac{y_{0}}{y_{0}^{2}-y_{1}^{2}-\cdots-y_{n}^{2}}=y_{0}.$

From direct calculations, we have

$\displaystyle(x+^{\prime}y)_{0}$
$\displaystyle=(B(-\hat{y})x)_{0}=\sum_{\alpha=0}^{n}[B(-\hat{y})]_{0\alpha}x_{\alpha}=y_{0}x_{0}+\sum_{j=1}^{n}(-\gamma(-\hat{y}))(-\hat{y}_{j})x_{j}$
$\displaystyle=x_{0}y_{0}+x_{1}y_{1}+\cdots+x_{n}y_{n}.$

This is the first desired result. By a similar calculation, we get for
$j=1,\dots,n$,

$\displaystyle(x+^{\prime}y)_{j}$
$\displaystyle=\sum_{\alpha=0}^{n}[B(-\hat{y})]_{j\alpha}x_{\alpha}=[B(-\hat{y})]_{j0}x_{0}+\sum_{k=1}^{n}[B(-\hat{y})]_{jk}x_{k}$
$\displaystyle=\gamma(-\hat{y})\hat{y}_{j}x_{0}+\sum_{k=1}^{n}\left(\delta_{jk}+(\gamma(-\hat{y})-1)\frac{\hat{y}_{j}\hat{y}_{k}}{\|\hat{y}\|^{2}}\right)x_{k}=x_{0}y_{j}+x_{j}+(y_{0}-1)\sum_{k=1}^{n}\frac{y_{j}y_{k}x_{k}}{y_{0}^{2}\|\hat{y}\|^{2}}$
$\displaystyle=x_{0}y_{j}+x_{j}+(y_{0}-1)\sum_{k=1}^{n}\frac{y_{j}y_{k}x_{k}}{y_{0}^{2}-1}$
$\displaystyle=x_{0}y_{j}+x_{j}+\frac{y_{j}}{y_{0}+1}(x_{1}y_{1}+\cdots+x_{n}y_{n}).$

This is the second part of the desired result. ∎

###### Proof of Lemma 2.7.

From direct calculations, we have

$\displaystyle(-(x+^{\prime}y))_{0}=(x+^{\prime}y)_{0}=x_{0}y_{0}+x_{1}y_{1}+\cdots+x_{n}y_{n},$

and

$\displaystyle(-(x+^{\prime}y))_{j}=-(x+^{\prime}y)_{j}=-x_{0}y_{j}-x_{j}-\frac{y_{j}}{y_{0}+1}(x_{1}y_{1}+\cdots+x_{n}y_{n}).$

Also, we have

$((-x)+^{\prime}(-y)^{\prime})_{0}=x_{0}y_{0}+x_{1}y_{1}+\cdots+x_{n}y_{n},$

and

$((-x)+^{\prime}(-y)^{\prime})_{j}=-x_{0}y_{j}-x_{j}-\frac{y_{j}}{y_{0}+1}(x_{1}y_{1}+\cdots+x_{n}y_{n}).$

Since we have $(-(x+^{\prime}y))_{0}=((-x)+^{\prime}(-y))_{0}$ and
$(-(x+^{\prime}y)_{j})=((-x)+^{\prime}(-y))_{j}$ for all $1\leq j\leq n$, we
obtain the desired result. ∎

###### Proof of Lemma 2.8.

First, we have the following calculation:

$\displaystyle\begin{aligned}
((x+^{\prime}y)+^{\prime}z)_{0}&=(x+^{\prime}y)_{0}z_{0}+\sum_{j=1}^{n}(x+^{\prime}y)_{j}z_{j}\\\
&=x_{0}y_{0}z_{0}+z_{0}\sum_{j=1}^{n}x_{j}y_{j}+\sum_{j=1}^{n}z_{j}\left(x_{0}y_{j}+x_{j}+\frac{y_{j}}{y_{0}+1}\sum_{k=1}^{n}x_{k}y_{k}\right)\\\
&=x_{0}y_{0}z_{0}+z_{0}\sum_{j=1}^{n}x_{j}y_{j}+x_{0}\sum_{j=1}^{n}y_{j}z_{j}+\sum_{j=1}^{n}x_{j}z_{j}+\frac{1}{y_{0}+1}\sum_{j=1}^{n}y_{j}z_{j}\sum_{k=1}^{n}x_{k}y_{k}.\end{aligned}$
(98)

We substitute $z=-y$ into (98) to get

$((x+^{\prime}y)-^{\prime}y)_{0}=x_{0}y_{0}^{2}+y_{0}\sum_{j=1}^{n}x_{j}y_{j}-x_{0}\sum_{j=1}^{n}y_{j}^{2}-\sum_{j=1}^{n}x_{j}y_{j}+\frac{1}{y_{0}+1}\sum_{j=1}^{n}(-y_{j}^{2})\sum_{k=1}^{n}x_{k}y_{k}.$

If we use $y_{0}^{2}-y_{1}^{2}-\cdots-y_{n}^{2}=1$, then we can simplify the
above relation to get

$((x+^{\prime}y)-^{\prime}y)_{0}=x_{0}.$

From the above result and (18), we find for $j=1,\dots,n$,

$\displaystyle((x+^{\prime}y)-^{\prime}y)_{j}$
$\displaystyle=(x+^{\prime}y)_{j}-\frac{y_{j}}{y_{0}+1}(((x+^{\prime}y)-^{\prime}y)_{0}+(x+^{\prime}y)_{0})$
$\displaystyle=x_{j}+\frac{y_{j}}{y_{0}+1}((x+^{\prime}y)_{0}+x_{0})-\frac{y_{j}}{y_{0}+1}(x_{0}+(x+^{\prime}y)_{0})$
$\displaystyle=x_{j}.$

We infer that $(x+^{\prime}y)-^{\prime}y=x$. ∎

###### Proof of Lemma 2.10.

Set

$x=(x_{0},x_{1},\cdots,x_{n})\in\mathbb{H}^{n},\quad x_{k}=0.$

By a simple calculation, we have

$\displaystyle\begin{aligned} (x+^{\prime}u_{k}(a))_{0}&=x_{0}\cosh
a+x_{k}\sinh a,\\\ (x+^{\prime}u_{k}(a))_{j}&=x_{0}\sinh
a\,\delta_{jk}+x_{j}+\frac{\sinh a\,\delta_{jk}}{\cosh a+1}(\sinh a\,x_{k})\\\
&=x_{j}(1+\delta_{jk}(\cosh a-1))+x_{0}\delta_{jk}\sinh a,\qquad\forall 1\leq
j\leq n,\end{aligned}$ (99)

where $\delta$ is the Kronecker delta symbol. We substitute $a+b$ instead of
$a$ in (99) to get

$\displaystyle\begin{aligned}
(x+^{\prime}u_{k}(a+b))_{0}&=x_{0}\cosh(a+b)+x_{k}\sinh(a+b),\\\
(x+^{\prime}u_{k}(a+b))_{j}&=x_{j}(1+\delta_{jk}(\cosh(a+b)-1))+x_{0}\delta_{jk}\sinh(a+b).\end{aligned}$
(100)

We substitute $x+^{\prime}u_{k}(a)$ instead of $x$ and $b$ instead of $a$ into
(99) to get

$\displaystyle\begin{aligned}
((x+^{\prime}u_{k}(a))+^{\prime}u_{k}(b))_{0}&=(x+^{\prime}u_{k}(a))_{0}\cosh
b+(x+^{\prime}u_{k}(a))_{k}\sinh b,\\\
((x+^{\prime}u_{k}(a))+^{\prime}u_{k}(b))_{j}&=(x+^{\prime}u_{k}(a))_{j}(1+\delta_{jk}(\cosh
b-1))+(x+^{\prime}u_{k}(a))_{0}\delta_{jk}\sinh b.\end{aligned}$ (101)

From the first equality of (101) and (99), we have

$\displaystyle((x+^{\prime}u_{k}(a))+^{\prime}u_{k}(b))_{0}$
$\displaystyle=(x_{0}\cosh a+x_{k}\sinh a)\cosh b+(x_{k}\cosh a+x_{0}\sinh
a)\sinh b$ $\displaystyle=x_{0}\cosh(a+b)+x_{k}\sinh(a+b).$ (102)

Also, from the second equality of (101) and (99), we have

$\displaystyle((x+^{\prime}u_{k}(a))+^{\prime}u_{k}(b))_{j}$
$\displaystyle=\big{(}x_{j}(1+\delta_{jk}(\cosh a-1))+x_{0}\delta_{jk}\sinh
a\big{)}(1+\delta_{jk}(\cosh b-1))$ $\displaystyle\quad+(x_{0}\cosh
a+x_{k}\sinh a)\delta_{jk}\sinh b$
$\displaystyle=x_{j}\big{(}(1+\delta_{jk}(\cosh a-1))(1+\delta_{jk}(\cosh
b-1))+\sinh a\delta_{jk}\sinh b\big{)}$
$\displaystyle\quad+x_{0}\big{(}\delta_{jk}\sinh a(1+\delta_{jk}(\cosh
b-1))+\cosh a\delta_{jk}\sinh b\big{)}$
$\displaystyle=x_{j}(1+\delta_{jk}(\cosh(a+b)-1)+x_{0}\delta_{jk}\sinh(a+b).$
(103)

Finally, we combine (100), (102), and (103) to conclude that

$(x+^{\prime}u_{k}(a))+^{\prime}u_{k}(b)=x+^{\prime}u_{k}(a+b).$

∎

###### Proof of Lemma 2.12.

$\diamond$ (Existence of the decomposition) First, we will show the existence
of the decomposition (25). Set $x=(x_{0},x_{1},\cdots,x_{n})$, and define

$a=\tanh^{-1}\left(\frac{x_{k}}{x_{0}}\right).$

Then we can check that

$\displaystyle(x-^{\prime}u_{k}(a))_{k}$
$\displaystyle=((x_{0},x_{1},\cdots,x_{n})-^{\prime}(\cosh a,0,\cdots,0,\sinh
a,0,\cdots,0))_{k}$ $\displaystyle=((x_{0},x_{1},\cdots,x_{n})+^{\prime}(\cosh
a,0,\cdots,0,-\sinh a,0,\cdots,0))_{k}$ $\displaystyle=-x_{0}\sinh
a+x_{k}+\frac{-\sinh a}{\cosh a+1}(-x_{k}\sinh a)$ $\displaystyle=x_{k}\cosh
a-x_{0}\sinh a=0.$

Therefore,

$x-^{\prime}u_{k}(a)\in P_{k}(0).$

If we set $y=x-^{\prime}u_{k}(a)$, then we have

$y+^{\prime}u_{k}(a)=(x-^{\prime}u_{k}(a))+^{\prime}u_{k}(a)=x,$

by using Lemma 2.8.

$\diamond$ (Uniqueness of the decomposition) Now, we show the uniqueness of a
pair $(a,y)$ introduced in the existence part. If there exists another pair
$(b,z)\in\mathbb{R}\times P_{k}(0)$ which satisfies

$x=z+^{\prime}u_{k}(b),$

then we have

$y+^{\prime}u_{k}(a)=z+^{\prime}u_{k}(b).$

From Lemma 2.10, we get

$(y+^{\prime}u_{k}(a))-^{\prime}u_{k}(a)=(z+^{\prime}u_{k}(b))-^{\prime}u_{k}(a)\quad\Rightarrow\quad
y=z+^{\prime}u_{k}(b-a).$

This implies that

$\displaystyle y_{k}=z_{0}\sinh(b-a)+z_{k}\cosh(b-a).$

Since $y_{k}=z_{k}=0$ and $z_{0}\geq 1$, we have $a=b$. This yields $y=z$,
showing the uniqueness of the decomposition (25). ∎

###### Proof of Lemma 2.14.

Without loss of generality, we assume that $\pi_{k}(x)\leq\pi_{k}(y)$. Then,
there exists $R$ such that

$\pi_{k}(x)=a_{k}-R,\qquad\text{and}\qquad\pi_{k}(y)=a_{k}+R.$

If we define

$\tilde{x}=x-^{\prime}u_{k}(a_{k}),\quad\tilde{y}=y-^{\prime}u_{k}(a_{k}),$

we have

$\displaystyle\begin{cases}\pi_{k}(\tilde{x})=\pi_{k}(x-^{\prime}u_{k}(a_{k}))=\pi_{k}(x)-a_{k}=-R,\\\
\pi_{k}(\tilde{y})=\pi_{k}(y-^{\prime}u_{k}(a_{k}))=\pi_{k}(y)-a_{k}=R,\end{cases}$

by Corollary 2.13. On the other hand,

$\pi_{k}(\tilde{x})=\tanh^{-1}\left(\frac{\tilde{x}_{k}}{\tilde{x}_{0}}\right)\quad\text{and}\quad\pi_{k}(\tilde{y})=\tanh^{-1}\left(\frac{\tilde{y}_{k}}{\tilde{y}_{0}}\right)$

which implies

$\frac{\tilde{x}_{k}}{\tilde{x}_{0}}=-\tanh
R\quad\text{and}\quad\frac{\tilde{y}_{k}}{\tilde{y}_{0}}=\tanh R.$

We use the above relation and the following facts:

$\tilde{x}_{0}\geq\sqrt{1+\tilde{x}_{k}^{2}}\quad\text{and}\quad\tilde{y}_{0}\geq\sqrt{1+\tilde{y}_{k}^{2}},$

to get

$\tanh(\sinh^{-1}\tilde{x}_{k})=\frac{\tilde{x}_{k}}{\sqrt{1+\tilde{x}_{k}^{2}}}\leq-\tanh
R\quad\text{and}\quad\tanh(\sinh^{-1}\tilde{y}_{k})=\frac{\tilde{y}_{k}}{\sqrt{1+\tilde{y}_{k}^{2}}}\geq\tanh
R.$

Since $\tanh$ and $\sinh$ are increasing function, we get

$\tilde{x}_{k}\leq-\sinh R\quad\text{and}\quad\tilde{y}_{k}\geq\sinh R.$

We substitute the above result into the following calculation:

$\displaystyle(\tilde{x}-^{\prime}\tilde{y})_{0}=$
$\displaystyle=\tilde{x}_{0}\tilde{y}_{0}-\tilde{x}_{1}\tilde{y}_{1}-\cdots-\tilde{x}_{n}\tilde{y}_{n}$
$\displaystyle=\sqrt{(1+\tilde{x}_{1}^{2}+\cdots+\tilde{x}_{n}^{2})(1+\tilde{y}_{1}^{2}+\cdots+\tilde{y}_{n}^{2})}-\tilde{x}_{1}\tilde{y}_{1}-\cdots-\tilde{x}_{n}\tilde{y}_{n}$
$\displaystyle\geq
1+\sum_{l=1}^{n}|\tilde{x}_{l}\tilde{y}_{l}|-\sum_{l=1}^{n}\tilde{x}_{l}\tilde{y}_{l}\geq
1-2\tilde{x}_{k}\tilde{y}_{k},$

to get

$(\tilde{x}-^{\prime}\tilde{y})_{0}\geq 1+2\sinh^{2}R=\cosh(2R).$

Since $z\mapsto z-^{\prime}u_{k}(a_{k})$ is an isometry on $\mathbb{H}^{n}$,
we finally get

$\cosh(d(x,y))=(x-^{\prime}y)_{0}=(\tilde{x}-^{\prime}\tilde{y})_{0}\geq\cosh(2R),$

which yields

$d(x,y)\geq 2R=|\pi_{k}(x)-\pi_{k}(y)|.$

∎

##  Appendix B Proof of Proposition 5.1

We look for a solution of (see (35)):

$\tilde{\rho}(\theta_{x})=\int_{B_{v}(R)}\Delta_{x}A(\theta_{xy})\tilde{\rho}(\theta_{y})\mathrm{d}y,\qquad\text{
for }0\leq\theta_{x}\leq R,$ (104)

where $\theta_{x}=\mathrm{dist}(x,v)$ and $\theta_{xy}=\mathrm{dist}(x,y)$.
Note that the radius $R$ is also unknown.

We first show that if $\Delta A(\theta)$ is in the form of a sum of powers of
$\cosh\theta$, then necessarily, $\tilde{\rho}(\theta)$ is in the same form.
Therefore, assume

$\Delta A(\theta):=\sum_{m=0}^{N}b_{m}\cosh^{m}\theta.$

Recall the hyperbolic cosine law:

$\cosh\theta_{xy}=\cosh\theta_{x}\cosh\theta_{y}-\sinh\theta_{x}\sinh\theta_{y}\cos\angle(xvy).$

This yields that

$\displaystyle\begin{aligned}
\cosh^{m}\theta_{xy}&=\left(\cosh\theta_{x}\cosh\theta_{y}-\sinh\theta_{x}\sinh\theta_{y}\cos\angle(xvy)\right)^{m}\\\
&=\sum_{k=0}^{m}(-1)^{k}{m\choose
k}\cosh^{m-k}\theta_{x}\cosh^{m-k}\theta_{y}\sinh^{k}\theta_{x}\sinh^{k}\theta_{y}\cos^{k}\angle(xvy).\end{aligned}$
(105)

From a simple observation,
$\cos\angle(xvy)=\cos(\pi-\angle(xv(-y)))=-\cos\angle(xv(-y))$. Hence, if $k$
is an odd number, then
$\displaystyle{\int_{\mathbb{S}^{n-1}}\cos^{k}\angle(xvy)\mathrm{d}\sigma_{y}=0}$,
and by integrating (105) over $\mathbb{S}^{n-1}$, we get

$\int_{\mathbb{S}^{n-1}}\cosh^{m}\theta_{xy}\mathrm{d}\sigma_{y}=\sum_{l=0}^{[m/2]}{m\choose
2l}\mathcal{A}_{l}\cosh^{m-2l}\theta_{x}\sinh^{2l}\theta_{x}\cosh^{m-2l}\theta_{y}\sinh^{2l}\theta_{y},$

where

$\mathcal{A}_{l}:=\int_{\mathbb{S}^{n-1}}\cos^{2l}\angle(xvy)\mathrm{d}\sigma_{y}.$

Then, we have from (104):

$\displaystyle\tilde{\rho}(\theta_{x})=\int_{B_{v}(R)}\sum_{m=0}^{N}b_{m}\cosh^{m}\theta_{xy}\tilde{\rho}(\theta_{y})\mathrm{d}y$
$\displaystyle=\sum_{m=0}^{N}b_{m}\int_{0}^{R}\left(\int_{\mathbb{S}^{n-1}}\cosh^{m}\theta_{xy}\mathrm{d}\sigma_{y}\right)\sinh^{n-1}\theta_{y}\tilde{\rho}(\theta_{y})\mathrm{d}\theta_{y}$
$\displaystyle=\sum_{m=0}^{N}b_{m}\int_{0}^{R}\left(\sum_{l=0}^{[m/2]}{m\choose
2l}\mathcal{A}_{l}\cosh^{m-2l}\theta_{x}\sinh^{2l}\theta_{x}\cosh^{m-2l}\theta_{y}\sinh^{2l}\theta_{y}\right)\sinh^{n-1}\theta_{y}\tilde{\rho}(\theta_{y})\mathrm{d}\theta_{y}$
$\displaystyle=\sum_{m=0}^{N}\sum_{l=0}^{[m/2]}b_{m}{m\choose
2l}\mathcal{A}_{l}\cosh^{m-2l}\theta_{x}\sinh^{2l}\theta_{x}\int_{0}^{R}\cosh^{m-2l}\theta_{y}\sinh^{2l+n-1}\theta_{y}\tilde{\rho}(\theta_{y})\mathrm{d}\theta_{y}$
$\displaystyle=\sum_{m=0}^{N}\sum_{l=0}^{[m/2]}b_{m}{m\choose
2l}\mathcal{A}_{l}\left(\int_{0}^{R}\cosh^{m-2l}\theta_{y}\sinh^{2l+n-1}\theta_{y}\tilde{\rho}(\theta_{y})\mathrm{d}\theta_{y}\right)\cosh^{m-2l}\theta_{x}(\cosh^{2}\theta_{x}-1)^{l}.$

The calculation above implies that $\tilde{\rho}$ can be expressed as

$\tilde{\rho}(\theta)=\sum_{m=0}^{N}a_{m}\cosh^{m}\theta,\qquad\text{ for
}0\leq\theta<R.$

Furthermore, this yields that

$\displaystyle\sum_{m=0}^{N}a_{m}\cosh^{m}\theta_{x}=\sum_{k,m=0}^{N}\sum_{l=0}^{[m/2]}a_{k}b_{m}{m\choose
2l}\mathcal{A}_{l}\left(\int_{0}^{R}\cosh^{m-2l+k}\theta_{y}\sinh^{2l+n-1}\theta_{y}\mathrm{d}\theta_{y}\right)\cosh^{m-2l}\theta_{x}(\cosh^{2}\theta_{x}-1)^{l}.$
(106)

For the purpose of proving Proposition 5.1, we only need the case $N=1$;
nevertheless, we presented above the case of general $N$ for its own interest.
Hence, from now on assume $b_{m}=0$, $m\geq 2$.

Consider three cases:

Case 1: $b_{0}>0$ and $b_{1}=0$. In this case, (106) reduces to

$\sum_{m=0}^{0}a_{m}\cosh^{m}\theta_{x}=\sum_{k=0}^{0}a_{k}b_{0}\mathcal{A}_{0}\left(\int_{0}^{R}\cosh^{k}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right),$

which implies
$1=b_{0}\mathcal{A}_{0}\int_{0}^{R}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}$.
Since $\mathcal{A}_{0}=|\mathbb{S}^{n-1}|$, we can determine $R$ uniquely.
Also, we can determine $a_{0}=1$. We thus reached the equilibrium $\rho_{c}$
from (59).

Case 2: $b_{0}=0$ and $b_{1}>0$. In this case, (106) reduces to

$\sum_{m=0}^{1}a_{m}\cosh^{m}\theta_{x}=\sum_{k=0}^{1}a_{k}b_{1}\mathcal{A}_{0}\left(\int_{0}^{R}\cosh^{k+1}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)\cosh\theta_{x},$

which implies
$1=\mathcal{A}_{0}b_{1}\int_{0}^{R}\cosh^{2}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}$.
One can then argue as in Case 1 to obtain the uniqueness of $R$, $a_{0}$ and
$a_{1}$. This is the equilibrium solution from (61).

Case 3: $b_{0}>0$ and $b_{1}>0$. In this case, (106) reads

$\displaystyle\sum_{m=0}^{1}a_{m}\cosh^{m}\theta_{x}$
$\displaystyle=\sum_{k=0}^{1}a_{k}b_{0}\mathcal{A}_{0}\left(\int_{0}^{R}\cosh^{k}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)$
$\displaystyle\quad+\sum_{k=0}^{1}a_{k}b_{1}\mathcal{A}_{0}\left(\int_{0}^{R}\cosh^{k+1}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)\cosh\theta_{x}.$

This yields

$\displaystyle a_{0}$
$\displaystyle=\sum_{k=0}^{1}a_{k}b_{0}\mathcal{A}_{0}\left(\int_{0}^{R}\cosh^{k}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)$
$\displaystyle=b_{0}\mathcal{A}_{0}\left(a_{0}\int_{0}^{R}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}+a_{1}\int_{0}^{R}\cosh\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)$

and

$\displaystyle a_{1}$
$\displaystyle=\sum_{k=0}^{1}a_{k}b_{1}\mathcal{A}_{0}\left(\int_{0}^{R}\cosh^{k+1}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)$
$\displaystyle=b_{1}\mathcal{A}_{0}\left(a_{0}\int_{0}^{R}\cosh\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}+a_{1}\int_{0}^{R}\cosh^{2}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right).$

From the two equations above we find

$\displaystyle\begin{aligned}
a_{0}:a_{1}&=\left(b_{0}\mathcal{A}_{0}\int_{0}^{R}\cosh\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right):\left(1-b_{0}\mathcal{A}_{0}\int_{0}^{R}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)\\\\[2.0pt]
&=\left(1-b_{1}\mathcal{A}_{0}\int_{0}^{R}\cosh^{2}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right):\left(b_{1}\mathcal{A}_{0}\int_{0}^{R}\cosh\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right).\end{aligned}$
(107)

Therefore, $R$ satisfies

$\displaystyle\left(b_{0}\mathcal{A}_{0}\int_{0}^{R}\cosh\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)\left(b_{1}\mathcal{A}_{0}\int_{0}^{R}\cosh\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)$
$\displaystyle=\left(1-b_{0}\mathcal{A}_{0}\int_{0}^{R}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right)\left(1-b_{1}\mathcal{A}_{0}\int_{0}^{R}\cosh^{2}\theta_{y}\sinh^{n-1}\theta_{y}\mathrm{d}\theta_{y}\right),$

which yields

$\displaystyle
b_{0}b_{1}\mathcal{A}_{0}^{2}\left(\int_{0}^{R}\cosh\theta\sinh^{n-1}\theta\mathrm{d}\theta\right)^{2}$
$\displaystyle=1-b_{0}\mathcal{A}_{0}\int_{0}^{R}\sinh^{n-1}\theta\mathrm{d}\theta-
b_{1}\mathcal{A}_{0}\int_{0}^{R}\cosh^{2}\theta\sinh^{n-1}\theta\mathrm{d}\theta$
$\displaystyle\quad+b_{0}b_{1}\mathcal{A}_{0}^{2}\left(\int_{0}^{R}\sinh^{n-1}\theta\mathrm{d}\theta\right)\left(\int_{0}^{R}\cosh^{2}\theta_{y}\sinh^{n-1}\theta\mathrm{d}\theta\right).$

From now on, we only consider the case $n=2$. From the definition of
$\mathcal{A}_{l}$, we get $\mathcal{A}_{0}=2\pi$. Then, from the equation
above we get

$\displaystyle
4\pi^{2}b_{0}b_{1}\left(\int_{0}^{R}\cosh\theta\sinh\theta\mathrm{d}\theta\right)^{2}$
$\displaystyle=1-2\pi b_{0}\int_{0}^{R}\sinh\theta\mathrm{d}\theta-2\pi
b_{1}\int_{0}^{R}\cosh^{2}\theta\sinh\theta d\theta$
$\displaystyle\quad+4\pi^{2}b_{0}b_{1}\left(\int_{0}^{R}\sinh\theta
d\theta\right)\left(\int_{0}^{R}\cosh^{2}\theta\sinh\theta\mathrm{d}\theta\right),$

which can be calculated as

$\pi^{2}b_{0}b_{1}(\cosh^{2}R-1)^{2}=1-2\pi b_{0}(\cosh R-1)-\frac{2\pi
b_{1}}{3}(\cosh^{3}R-1)+\frac{4\pi^{2}b_{0}b_{1}}{3}(\cosh
R-1)(\cosh^{3}R-1).$

By some simplifications, we finally reach

$\displaystyle 0=1-2\pi b_{0}(\cosh R-1)-\frac{2\pi
b_{1}}{3}(\cosh^{3}R-1)+\frac{\pi^{2}b_{0}b_{1}}{3}(\cosh R-1)^{4}.$ (108)

We will prove that there exist unique $R$, $a_{0}$, and $a_{1}$, provided
$b_{0}$ and $b_{1}$ are positive real numbers. Substitute $x=\cosh R-1$ into
(108) to get

$0=1-2\pi(b_{0}+b_{1})x-2\pi b_{1}x^{2}-\frac{2\pi
b_{1}}{3}x^{3}+\frac{\pi^{2}b_{0}b_{1}}{3}x^{4}=:f(x).$

We want to find non-negative solutions of the above equation. We know that
$f(0)=1$, $f^{\prime}(0)=-2\pi(b_{0}+b_{1})<0$, and

$f^{\prime}(x)=-2\pi(b_{0}+b_{1})-4\pi b_{1}x-2\pi
b_{1}x^{2}+\frac{4\pi^{2}b_{0}b_{1}}{3}x^{3}.$

As $f^{\prime}(0)<0$, and $\lim_{x\to\infty}f^{\prime}(x)=\infty$, by the
intermediate value theorem, there exists at least one positive solution
$x_{+}$ such that $f^{\prime}(x_{+})=0$. Now, we will show that the positive
solution of $f^{\prime}(x)=0$ is unique. Since the product of the three roots
of $f^{\prime}(x)=0$ is positive, if there exist more than two positive
solutions then the other solution must also be positive. If
$x_{1},x_{2},x_{3}$ are three positive solutions of $f^{\prime}(x)=0$, then
$x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}=-\frac{4\pi
b_{1}}{4\pi^{2}b_{0}b_{1}/3}=-\frac{3}{\pi b_{0}}<0$, which leads to a
contradiction. We conclude that $f^{\prime}(x)=0$ has a unique positive
solution, which we denote by $x_{+}$.

From the above, we infer that $f(x)$ is decreasing on $x\in[0,x_{+}]$ and
increasing on $x\in[x_{+},\infty)$. Finally, we use the intermediate value
theorem to conclude that if $f(x_{+})<0$ then there exist two distinct
positive solutions of $f(x)=0$. We will now show that indeed, $f(x_{+})<0$.
From a simple calculation, we have

$\displaystyle f\left(\frac{1}{2\pi(b_{0}+b_{1})}\right)$
$\displaystyle=1-1-\frac{2\pi b_{1}}{(2\pi(b_{0}+b_{1}))^{2}}-\frac{2\pi
b_{1}}{3(2\pi(b_{0}+b_{1}))^{3}}+\frac{\pi^{2}b_{0}b_{1}}{3(2\pi(b_{0}+b_{1}))^{4}}$
$\displaystyle=-\frac{b_{1}}{2\pi(b_{0}+b_{1})^{2}}-\frac{b_{1}}{12\pi^{2}(b_{0}+b_{1})^{3}}+\frac{b_{0}b_{1}}{48\pi^{2}(b_{0}+b_{1})^{4}}$
$\displaystyle<0-\frac{b_{1}}{12\pi^{2}(b_{0}+b_{1})^{3}}+\frac{b_{1}}{48\pi^{2}(b_{0}+b_{1})^{3}}<0.$

So we can conclude that $\displaystyle{f(x_{+})\leq
f\left(\frac{1}{2\pi(b_{0}+b_{1})}\right)<0}$.

Denote the two distinct roots of $f(x)=0$ by
$\alpha_{b_{0}b_{1}}<\beta_{b_{0}b_{1}}$. There are two choices for the radius
$R$: $\cosh R=\alpha_{b_{0}b_{1}}+1$ and $\cosh R=\beta_{b_{0}b_{1}}+1$. We
will show that the latter cannot be possible.

As needed below, we calculate $f\left(\frac{1+\sqrt{1+4\pi b_{0}}}{2\pi
b_{0}}\right)$. Set $\lambda:=\frac{1+\sqrt{1+4\pi b_{0}}}{2\pi b_{0}}$; then
$\pi b_{0}\lambda^{2}=\lambda+1$. From this fact, we have

$\displaystyle f\left(\lambda\right)$
$\displaystyle=1-2\pi(b_{0}+b_{1})\lambda-2\pi b_{1}\lambda^{2}-\frac{2\pi
b_{1}}{3}\lambda^{3}+\frac{\pi^{2}b_{0}b_{1}}{3}\lambda^{4}$
$\displaystyle=1-2\pi(b_{0}+b_{1})\lambda-2\pi b_{1}\lambda^{2}-\frac{2\pi
b_{1}}{3}\lambda^{3}+\frac{\pi b_{1}\lambda^{2}}{3}(\pi b_{0}\lambda^{2})$
$\displaystyle=1-2\pi(b_{0}+b_{1})\lambda-2\pi b_{1}\lambda^{2}-\frac{2\pi
b_{1}}{3}\lambda^{3}+\frac{\pi b_{1}\lambda^{2}}{3}(\lambda+1)$
$\displaystyle=1-2\pi(b_{0}+b_{1})\lambda-\frac{5\pi}{3}b_{1}\lambda^{2}-\frac{\pi
b_{1}}{3}\lambda^{3}$ $\displaystyle<1-2\pi b_{0}\lambda<0.$

Note that

$\alpha_{b_{0}b_{1}}<\frac{1}{2\pi(b_{0}+b_{1})}<\lambda=\frac{1+\sqrt{1+4\pi
b_{0}}}{2\pi b_{0}}<\beta_{b_{0}b_{1}}.$

By (107), we have

$a_{0}:a_{1}=\left(\pi b_{0}(\cosh^{2}R-1)\right):\left(1-2\pi b_{0}(\cosh
R-1)\right),$ (109)

which implies

$\tilde{\rho}(\theta)=C\left(\pi b_{0}(\cosh^{2}R-1)+\left(1-2\pi b_{0}(\cosh
R-1)\right)\cosh\theta\right),\qquad\text{ for }0\leq\theta<R.$ (110)

for some constant $C>0$.

Now, assume that $\cosh R=\beta_{b_{0}b_{1}}+1$. Then, we have

$\tilde{\rho}(0)=C\left(\pi
b_{0}\beta_{b_{0}b_{1}}(\beta_{b_{0}b_{1}}+2)+(1-2\pi
b_{0}\beta_{b_{0}b_{1}})\right)=C(\pi b_{0}\beta_{b_{0}b_{1}}^{2}+1),$

and

$\displaystyle\tilde{\rho}(R)$ $\displaystyle=C\left(\pi
b_{0}(\cosh^{2}R-1)+(1-2\pi b_{0}(\cosh R-1))\cosh R\right)$
$\displaystyle=C\left(\pi
b_{0}\beta_{b_{0}b_{1}}(\beta_{b_{0}b_{1}}+2)+(1-2\pi
b_{0}\beta_{b_{0}b_{1}})(\beta_{b_{0}b_{1}}+1)\right)$ $\displaystyle=C(-\pi
b_{0}\beta_{b_{0}b_{1}}^{2}+\beta_{b_{0}b_{1}}+1).$

Since $\frac{1\pm\sqrt{1+4\pi b_{0}}}{2\pi b_{0}}$ are the two solutions of
$-\pi b_{0}x^{2}+x+1=0$ and $\beta_{b_{0}b_{1}}>\frac{1+\sqrt{1+4\pi
b_{0}}}{2\pi b_{0}}$, then $-\pi
b_{0}\beta_{b_{0}b_{1}}^{2}+\beta_{b_{0}b_{1}}+1<0$. However, this implies
$\tilde{\rho}(R)<0$, which is not possible. We conclude that $\cosh R$ cannot
be $\beta_{b_{0}b_{1}}+1$, but instead, $\cosh R=\alpha_{b_{0}b_{1}}+1$ ($R$
is unique).

In this case, by (109), we know that $a_{0}$ and $a_{1}$ have the same sign
since $1-2\pi b_{0}(\cosh R-1)=1-2\pi
b_{0}\alpha_{b_{0}b_{1}}>1-\frac{b_{0}}{b_{0}+b_{1}}>0$. From (110) we can
express the equilibrium as

$\tilde{\rho}(\theta)=\begin{cases}\displaystyle C\left(\pi
b_{0}\alpha_{b_{0}b_{1}}(\alpha_{b_{0}b_{1}}+2)+(1-2\pi
b_{0}\alpha_{b_{0}b_{1}})\cosh\theta\right),\qquad&\text{if }0\leq\theta\leq
R,\\\ 0,\qquad&\text{if }\theta>R.\end{cases}$

Here, $C$ should satisfy
$1=2\pi\int_{0}^{\cosh^{-1}(\alpha_{b_{0}b_{1}}+1)}\tilde{\rho}(\theta)\sinh\theta
d\theta$ (unit mass condition). So we have

$\displaystyle\frac{1}{2\pi}$
$\displaystyle=\int_{0}^{\cosh^{-1}(\alpha_{b_{0}b_{1}}+1)}\tilde{\rho}(\theta)\sinh\theta\mathrm{d}\theta$
$\displaystyle=C\int_{0}^{\cosh^{-1}(\alpha_{b_{0}b_{1}}+1)}\left(\pi
b_{0}\alpha_{b_{0}b_{1}}(\alpha_{b_{0}b_{1}}+2)\sinh\theta+(1-2\pi
b_{0}\alpha_{b_{0}b_{1}})\cosh\theta\sinh\theta\right)\mathrm{d}\theta$
$\displaystyle=C\left(\pi
b_{0}\alpha_{b_{0}b_{1}}^{2}(\alpha_{b_{0}b_{1}}+2)+(1-2\pi
b_{0}\alpha_{b_{0}b_{1}})\frac{1}{2}\alpha_{b_{0}b_{1}}(\alpha_{b_{0}b_{1}}+2)\right)$
$\displaystyle=\frac{1}{2}C\alpha_{b_{0}b_{1}}(\alpha_{b_{0}b_{1}}+2).$

This implies that $C=\frac{1}{\pi\alpha_{b_{0}b_{1}}(\alpha_{b_{0}b_{1}}+2)}$.

Finally, derive

$\tilde{\rho}(\theta)=\begin{cases}\displaystyle b_{0}+\left(\frac{1-2\pi
b_{0}\alpha_{b_{0}b_{1}}}{\pi\alpha_{b_{0}b_{1}}(\alpha_{b_{0}b_{1}}+2)}\right)\cosh\theta,\qquad&\text{if
}0\leq\theta\leq\cosh^{-1}(\alpha_{b_{0}b_{1}}+1),\\\ 0,\qquad&\text{if
}\theta>\cosh^{-1}(\alpha_{b_{0}b_{1}}+1).\end{cases}$

∎