Minimal cluster computation for four planar regions with the same area

TL;DR: In this paper, the authors describe the computation used to check that the symmetric cluster with the given topology is indeed the unique minimal cluster, where the topology of a minimal cluster of four planar regions with equal areas and smallest possible perimeter was found.

Abstract: Abstract The topology of a minimal cluster of four planar regions with equal areas and smallest possible perimeter was found in [9]. Here we describe the computation used to check that the symmetric cluster with the given topology is indeed the unique minimal cluster.

Summary

Introduction

• The authors consider the problem of enclosing and separating N regions of R with prescribed area and with the minimal possible interface length.
• The case N = corresponds to the celebrated isoperimetric problem whose solution, the circle, was known since antiquity.
• Wichiramala [14] proved that for N = in R the three regions of any minimizer are delimited by six circular arcs joining in four points.
• In this paper the authors describe the computations backing such a conjecture.

general results

• The perimeter of the cluster represents the total length of the interfaces between the regions.
• Almost every point in the boundary of the cluster belongs to two adjacent regions, hence the factor in the previous de nition.
• The authors are interested in the problem of nding the clusters with minimal perimeter among all clusters with prescribed areas.
• The authors recall the regularity result for the planar case.

Theorem 2.2 (regularity of planar clusters).

• Each arc separates two components of di erent regions.
• The arcs meet in triples at their end points (which the authors call vertices) with equal angles of degrees.
• One can check (see [13] , [5] , [15] , [11] ) that stationarity is also preserved by circle inversion: if three circular arcs meet in a point with equal angles of 120 degrees of course their circular inversion are also arcs (or straight line segments) which also meet in a point with the angles preserved.
• In particular it happens (see [5] ) that all double bubbles (i.e. the minimal clusters with two regions, see [2] ) can be obtained, up to rescaling, by projecting the geodesic network on the sphere given by three meridians joining in two antipodal points with equal angles of 120 degrees.
• To further investigate the geometry of minimal clusters the authors point out a general result which can be stated for a triangular region of any stationary cluster (see [15] ).

Theorem 2.3 (removal of triangular regions).

• The sum of the signed curvatures of these three arcs is zero.
• Hence, if the authors remove the triangle T and prolong the three arcs, they obtain a new stationary cluster.
• Let z and w be any two points on the complex plane and consider the circular arc going from z to w which has direction β in the point z (i.e. the half line starting from z tangent to the arc towards w identi es and angle β with respect to the axis of positive real numbers).
• The authors need to compute the oriented area, so they consider the area positive if the path going from z to w along the arc and then following the trapezoid is counter-clockwise oriented.
• Notice that when α is negative the area computed above is also negative.

Minimal planar clusters with four regions of equal area

• From now on the authors will consider the particular case of four planar regions with equal area.
• In the recent paper [9] the authors have obtained the following result.
• Let us identify the ambient plane with the set of complex numbers C. Up to translating, rotating and rescaling the authors might suppose that the two vertices of the resulting double bubble are the two complex numbers and .
• Suppose also that E is the upper quadrangular region, E is the lower quadrangular, E is the right hand side triangular and E is the left hand side triangular region .
• The Releaux triangle can be uniquely determined by means of the distance of its vertices from the center: let call ρ such a distance.

Figures

Figure 3: Plot of |E 1 | − |E 2 | with respect to parameter θ.

Figure 2: Cluster construction by means of circular inversion.

Figure 1: The symmetric cluster with sandwich topology.

Full text

Open Access. © 2018 E. Paolini and A. Tamagnini, published by De Gruyter. This work is licensed under the Creative Commons Attribution-
Non-Commercial-NoDerivs 4.0 License.
Geometric Flows 2018; 3:90–96
Research Article Open Access
E. Paolini* and A. Tamagnini
Minimal cluster computation for four planar
regions with the same area
https://doi.org/10.1515/geofl-2018-0007
Received September 15, 2017; accepted November 7, 2017
Abstract: The topology of a minimal cluster of four planar regions with equal areas and smallest possible
perimeter was found in [9]. Here we describe the computation used to check that the symmetric cluster with
the given topology is indeed the unique minimal cluster.
MSC: 49K10, 65K10, 51M25
1 Introduction
We consider the problem of enclosing and separating N regions of R
2
with prescribed area and with the
minimal possible interface length.
The case N = 1 corresponds to the celebrated isoperimetric problem whose solution, the circle, was
known since antiquity.
For N ≥ 1 rst existence and partial regularity in R
n
was given by Almgren [1] while Taylor [12] describes
the singularities for minimizers in R
3
. Existence and regularity of minimizers in R
2
was proved by Morgan [6]
(see also [4]): the regions of a minimizer in R
2
are delimited by a nite number of circular arcs which meet in
triples at their end-points.
Foisy et al. [2] proved that for N = 2 in R
2
the two regions of any minimizer are delimited by three circular
arcs joining in two points (standard double bubble) and are uniquely determined by their enclosed areas.
Wichiramala [14] proved that for N = 3 in R
2
the three regions of any minimizer are delimited by six circular
arcs joining in four points. Such conguration (standard triple bubble) is uniquely determined by the given
enclosed areas, as shown by Montesinos [5].
Recently, in [9], the case of four regions with equal areas has been considered and in this case the topology
of minimal cluster has been determined: the cluster is composed by four connected regions, two of them E
1
,
E
2
are quadrangular and are adjacent to each other, two of them E
3
, E
4
are triangular and are adjacent to the
two quadrangular regions. Such conguration is called sandwich and is depicted in Figure 1.
It is conjectured that in this topology there is a unique minimizer which is the cluster with two axes of
symmetry: regions E
1
and E
2
are congruent to each other and the same is true for regions E
3
and E
4
. In this
paper we describe the computations backing such a conjecture. The actual code to produce the computation
is available in [10].
*Corresponding Author: E. Paolini: Dipartimento di Matematica, Largo Bruno Pontecorvo 5, Università di Pisa, Italy
E-mail: emanuele.paolini@unipi.it
A. Tamagnini: Dipartimento di Matematica e Informatica “U. Dini”, Viale Morgagni 67/a, Università degli Studi di Firenze, Italy,
E-mail: andrea.tamagnini@uni.it

Minimal cluster computation for four planar regions with the same area | 91
E
1
E
2
E
3
E
4
Figure 1: The symmetric cluster with sandwich topology.
2 general results
Let us denote with E = (E
1
, . . . , E
N
) an N-uple of disjoint measurable subsets of R
2
. We will say that E is a
cluster and that E
1
, . . . , E
N
are its regions. We dene the external region E
0
as
E
0
= R
2
\
N
[
i=1
E
i
.
The sets E
0
, E
1
, . . . , E
N
are hence a partition of the whole plane R
2
.
We denote with m(E) = (| E
1
| , . . . , |E
N
|) the N-uple of the measures of the regions of the cluster. Also we
dene the perimeter of the cluster as
P(E) :=
1
2
N
X
i=0
P(E
i
)
where P(E
i
) is the perimeter (the length of the boundary) of the measurable set E
i
. The perimeter of the cluster
represents the total length of the interfaces between the regions. Almost every point in the boundary of the
cluster belongs to two adjacent regions, hence the factor
1
2
in the previous denition.
We are interested in the problem of nding the clusters with minimal perimeter among all clusters with
prescribed areas. Such clusters will be called minimal cluster.
The following result gives the existence of minimal clusters (see [1], [6] and [4]).
Theorem 2.1 (existence of minimal clusters). Given a ∈ R
N
+
there exists a cluster E in R
n
such that m(E) = a
and such that
P(E) ≤ P(F)
for all F such that m(F) = a.
Minimal clusters have also very good regularity properties. In particular the structure of minimal cluster has
been widely studied when the ambient space is R
2
(see [6]) or R
3
(see [12]). We recall the regularity result for
the planar case.
Theorem 2.2 (regularity of planar clusters). Let E be a minimal cluster in R
2
. Then each region E
k
of E is
composed by a nite number of connected components. Each connected component is delimited by a nite
number of circular arcs or straight line segments. Each arc separates two components of dierent regions. The
arcs meet in triples at their end points (which we call vertices) with equal angles of 120 degrees. The sum of the
signed curvatures of the three arcs joining in a vertex is zero.
We will say that a cluster E satisfying the regularity properties stated in the previous theorem is stationary.
Of course stationarity is preserved under isometries and rescalings of the plane. One can check (see [13],
[5], [15], [11]) that stationarity is also preserved by circle inversion: if three circular arcs meet in a point with
equal angles of 120 degrees of course their circular inversion are also arcs (or straight line segments) which
also meet in a point with the angles preserved. But not only that: if the sum of the signed curvatures is zero,

92 | E. Paolini and A. Tamagnini
this condition is also preserved under circular inversion (while the single curvatures might change). This
means that stationarity of clusters is preserved by all Moebius transformations of the plane. So, if a stationary
cluster is obtained as the stereographic projection of a partition on the sphere, we can rotate the sphere as
we like, project back the partition to the plane and obtain another stationary cluster.
In particular it happens (see [5]) that all double bubbles (i.e. the minimal clusters with two regions, see [2])
can be obtained, up to rescaling, by projecting the geodesic network on the sphere given by three meridians
joining in two antipodal points with equal angles of 120 degrees. All triple bubbles (i.e. the minimal clusters
with three regions, see [15]) are obtained, up to rescaling, by projecting the geodesic network of a regular
tetrahedron on the sphere.
To further investigate the geometry of minimal clusters we point out a general result which can be stated
for a triangular region of any stationary cluster (see [15]).
Theorem 2.3 (removal of triangular regions). Let T be a triangular region of a stationary cluster E. Consider
the three arcs which meet the three vertices of T and are not edges of T. The sum of the signed curvatures of
these three arcs is zero. Moreover, if prolonged inside T, these arcs meet in a point P inside T with equal angles
of 120 degrees. Hence, if we remove the triangle T and prolong the three arcs, we obtain a new stationary cluster.
To explicitly compute the area of a region bounded by circular arcs we develop some formulas. We are going
to use complex numbers to represent the points of R
2
.
Let z and w be any two points on the complex plane and consider the circular arc going from z to w which
has direction β in the point z (i.e. the half line starting from z tangent to the arc towards w identies and angle
β with respect to the axis of positive real numbers).
We are going to separately compute the area of the trapezoid below the straight segment zw and the area
of the circle segment bounded by the arc and the line segment. We need to compute the oriented area, so
we consider the area positive if the path going from z to w along the arc and then following the trapezoid is
counter-clockwise oriented.
To compute the area of the circle segment we need to nd the angle α between the arc and the chord
going from z to w. The angle of the arc is θ and the angle of the chord is¹ arg(w − z) hence
α = θ − arg(w − z).
Since arg(w − z) is dened up to multiples of 2π we require that α ∈ [−π, π] by possibly adding an appropriate
multiple of 2π. Notice that the arc spans an angle 2α on the circle containing it. So we can compute the radius
of the circle
R =
| w − z|
2 · sin(α)
and the area of the circular sector is αR
2
while the area of the circular segment is obtained removing the area
of the triangle with angle 2α between two equal sides of length R:
αR
2
− R
2
sin α cos α = R
2
(α − sin α cos α).
Notice that when α is negative the area computed above is also negative. This is correct because the region
below the arc (imagine that z is on the right with respect to w) is concave and should hence be removed from
the area of the trapezoid.
The area of the trapezoid below the straight segment zw can be easily computed as
1
2
Re(z − w) Im(z + w) =
1
8
(z
2
−
¯
z
2
)(w
2
−
¯
w
2
).
We have hence found the signed area below the arc zw as
A(z, w, θ) =
1
8
(z
2
−
¯
z
2
)(w
2
−
¯
w
2
) + R
2
(α − sin α cos α) (1)
1 Remember arg z is such an angle that z = |z| · exp(i · arg(z)).

Minimal cluster computation for four planar regions with the same area | 93
where R and α are obtained by z, w and θ as described above.
Another computation we will need in the following, is a formula to nd the direction of an arc after a
circular inversion is performed.
In general suppose to have an arc Γ starting at point z in direction α. The angle α is the angle between
Γ and the ray H starting in z and with the same direction as the positive real numbers (horizontal ray from
z). Let z
0
= 1/
¯
z be the circular inversion of the point z. Now consider the ray R starting from 0 and passing
trought z. This ray has direction β = arg z. The angle α between Γ and H is the sum α = γ + β where γ is the
(oriented) angle between R and Γ. We want to compute the angle α
0
in z
0
between the inversion Γ
0
of Γ and
the inversion H
0
of H. We know that R
0
= R but the direction of R
0
in z
0
is opposite to the direction of R in z: it
is π + β. From this direction we subtract γ to nd the direction α
0
of Γ
0
. But since γ = α − β we nally obtain
α
0
= π + β − γ = π + 2β − α = π + 2 arg(z) − α. (2)
3 Minimal planar clusters with four regions of equal area
From now on we will consider the particular case of four planar regions with equal area. In the recent paper
[9] we have obtained the following result.
Theorem 3.1. Let E = (E
1
, E
2
, E
3
, E
4
) be a cluster with N = 4 regions in the plane which is minimal with
prescribed equal areas a = (a, a, a, a). Then all the four regions are connected. Moreover two of them (say E
1
and E
2
) have four edges and two of them (say E
3
and E
4
) have three edges.
Consider now the minimal cluster E with four planar regions E
1
, E
2
, E
3
and E
4
of equal area. Suppose E
1
and
E
2
are the quadrangular regions while E
3
and E
4
are triangular. If we remove both triangular regions E
3
and
E
4
prolonging the three remaining arcs we obtain a double bubble delimited by three arcs and two vertices.
Let us identify the ambient plane with the set of complex numbers C. Up to translating, rotating and
rescaling we might suppose that the two vertices of the resulting double bubble are the two complex numbers
0 and 1. Suppose also that E
1
is the upper quadrangular region, E
2
is the lower quadrangular, E
3
is the right
hand side triangular and E
4
is the left hand side triangular region (see Figure 2).
If we now perform the circular inversion z 7→ 1/
¯
z the vertex in 1 remains xed, the vertex in 0 goes to
∞ and the three arcs joining the two vertices become three straight half lines going from 1 to ∞ with equal
angles of 120 degrees. Let us call θ the direction (angle with respect to the positive real axis) of the half line
which is obtained as the inversion of the arc which is the interface between the two regions of the double
bubble. The triangular component E
3
around the vertex at 1 becomes a Releaux triangle centered in 1 i.e. a
region bounded by three isometric circular arcs with vertices on the straight lines, such that the arc between
two vertices has center on the third vertex. The Releaux triangle can be uniquely determined by means of the
distance of its vertices from the center: let call ρ such a distance.
We will now show that the whole minimal cluster E is determined by the two parameters θ and ρ. Let us
explicitly compute the coordinates of the vertices of E.
We start by identifying the vertices of the Releaux triangle, they are:
z
k
= 1 + ρ · exp

iθ +
2
3
iπk

, k = 0, 1, 2.
The corresponding vertices of E
3
are the circle inversion of the vertices of the Releaux triangle:
w
k
=
1
¯
z
k
.
More precisely w
0
is the right hand side internal vertex of the cluster, and then w
1
and w
2
are the other two
vertices of the triangular component in clockwise order.
Notice that the Releaux triangle with vertices z
0
, z
1
, z
2
is increasing (in the sense of set inclusion) with
respect to the parameter ρ. Hence also the triangular region E
3
(which is the inversion of the Releaux triangle)

94 | E. Paolini and A. Tamagnini
ρ
E
1
E
2
E
3
E
4
θθ
z
0
z
1
z
2
0
1
w
0
w
1
w
2
w
3
w
4
w
5
Figure 2: Cluster construction by means of circular inversion.
is increasing in ρ. This means that the area of such region is also increasing in ρ. So, if the area of the region
is xed, then ρ is uniquely determined. As a consequence the two regions E
1
, E
2
and their areas |E
1
| , |E
2
| are
decreasing in ρ.
Also, if we repeat the construction with the triangle centered in 0, if we impose that the two triangles have
the same area we nd that they correspond to the same parameter ρ because the double bubble is symmetric
with respect to the axis x = 1/2. This means that the minimal cluster E is itself symmetric with respect to
the axis x = 1/2 and the left hand side triangular region E
4
can be obtained by symmetry once E
3
has been
determined. More precisely the three vertices of E
4
are given by
w
k+3
= σ(w
k
), k = 0, 1, 2
where σ is the symmetry σ(x + iy) = (1 − x) + iy. The point w
3
is the internal vertex of E on the left hand side
and w
4
, w
5
are the other two vertices of E
4
in counter-clockwise order.
All the coordinates of the vertices of E have been found as a function of the two parameters θ and ρ.
We are now going to compute the areas of the regions. To compute the areas of the regions of the cluster E
using formula (1) we now need to nd the direction of the arcs bounding each region. We know which are the
directions of the arcs bounding the Releaux triangle, so we are going to use the formula (2) to compute the
direction of the circular inversion of an arc.
This formula allows us to compute the direction of the three arcs which start from the three vertices
w
0
, w
1
, w
2
and end on the three vertices of the left hand side triangular region w
3
, w
4
, w
5
. These directions
are
β
k
= π + 2 arg(z
k
) −

θ +
2
3
kπ

, k = 0, 1, 2.
The direction of the arcs bounding the triangular region can be computed accordingly, since the three arcs
joining in a vertex form equal angles of 120 degrees.
So, the area of the right hand side triangular region E
3
(which is equal to the area of the other triangular
region E
4
) is given as a function of θ and ρ:
| E
3
| = |E
4
| = A(w
0
, w
1
, β
0
+ 2π/3) + A(w
1
, w
2
, β
1
+ 2π/3) + A(w
2
, w
0
, β
2
+ 2π/3).

Frequently asked questions

Q1. What are the contributions in this paper?

In this paper, the authors considered the problem of enclosing and separating N regions of R2 with prescribed area and with the minimal possible interface length. 

Q2. What is the simplest way to obtain a stationary cluster?

if a stationarycluster is obtained as the stereographic projection of a partition on the sphere, the authors can rotate the sphere aswe like, project back the partition to the plane and obtain another stationary cluster. 

Q3. How do the authors get the triple bubbles?

All triple bubbles (i.e. the minimal clusterswith three regions, see [15]) are obtained, up to rescaling, by projecting the geodesic network of a regulartetrahedron on the sphere. 

Q4. How many arcs are there in R2?

Wichiramala [14] proved that for N = 3 in R2 the three regions of any minimizer are delimited by six circular arcs joining in four points. 

Q5. What is the simplest way to identify the ambient plane?

Let us identify the ambient plane with the set of complex numbers C. Up to translating, rotating and rescalingwemight suppose that the two vertices of the resulting double bubble are the two complex numbers 0 and 1. 

Q6. What is the topology of the cluster?

It is conjectured that in this topology there is a unique minimizer which is the cluster with two axes of symmetry: regions E 1 and E 2 are congruent to each other and the same is true for regions E 3 and E 4 . 

Q7. What is the simplest way to check stationarity?

One can check (see [13],[5], [15], [11]) that stationarity is also preserved by circle inversion: if three circular arcs meet in a point withequal angles of 120 degrees of course their circular inversion are also arcs (or straight line segments) whichalso meet in a point with the angles preserved. 

Q8. What is the significance of the rescaling of the plane?

In particular it happens (see [5]) that all double bubbles (i.e. theminimal clusterswith two regions, see [2])can be obtained, up to rescaling, by projecting the geodesic network on the sphere given by three meridiansjoining in two antipodal points with equal angles of 120 degrees. 

Q9. What is the vertices of the Releaux triangle?

Notice that the Releaux triangle with vertices z 0 , z 1 , z 2 is increasing (in the sense of set inclusion) with respect to the parameter ρ. 

Q10. What is the way to determine the existence of a minimizer in R2?

Existence and regularity of minimizers inR2 was proved byMorgan [6] (see also [4]): the regions of a minimizer inR2 are delimited by a nite number of circular arcs which meet in triples at their end-points. 

Q11. what is the area of the upper region of the cluster?

The area of the upper region E 1 is given by|E 1 | = 2A(w 0 , w 2 , β 0 − 2π/3)A(w 2 , w 5 , β 1 ) − A(w 0 , w 3 , β 0 )and the area of the lower region E 2 is given by|E 2 | = −2A(w 0 , w 1 , β 0 + 2π/3) − A(w 1 , w 4 , β 1 ) + A(w 0 , w 3 , β 0 ). 

Q12. What is the symmetry of the vertices of the Releaux triangle?

This means that the minimal cluster E is itself symmetric with respect to the axis x = 1/2 and the left hand side triangular region E 4 can be obtained by symmetry once E 3 has been determined. 

Q13. What is the formula for calculating the area of the cluster E?

This allows us to nd ρ = ρ(θ) such that |E 3 | = |E 2 | using a simple bisection method since |E 3 | − |E 2 | is decreasing in ρ. 

Q14. What is the simplest way to determine the area of the cluster?

Theorem 3.1. Let E = (E 1 , E 2 , E 3 , E 4 ) be a cluster with N = 4 regions in the plane which is minimal with prescribed equal areas a = (a, a, a, a).