Journal ArticleDOI

Minimal cluster computation for four planar regions with the same area

01 Apr 2018-Vol. 3, Iss: 1, pp 90-96

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.
Topics: Cluster (physics) (56%)

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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]
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
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).

Citations
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Journal ArticleDOI
Emanuele Paolini1, V. M. Tortorelli1Institutions (1)
Abstract: In this paper we make the final step in finding the optimal way to enclose and separate four planar regions with equal area. In Paolini and Tamagnini (ESAIM COCV 24(3):1303–1331, 2018) the graph-topology of the optimal cluster was found reducing the set of candidates to a one-parameter family of different clusters. With a simple argument we show that the minimal set has a further symmetry and hence is uniquely determined up to isometries.

1 citations

Posted Content
Abstract: In this paper we discuss the Steiner property for minimal clusters in the plane with an anisotropic double density. This means that we consider the classical isoperimetric problem for clusters, but volume and perimeter are defined by using two densities. In particular, the perimeter density may also depend on the direction of the normal vector. The classical "Steiner property" for the Euclidean case (which corresponds to both densities being equal to $1$) says that minimal clusters are made by finitely many ${\rm C}^{1,\gamma}$ arcs, meeting in finitely many "triple points". We can show that this property holds under very weak assumptions on the densities. In the parallel paper "On the Steiner property for planar minimizing clusters. The isotropic case" we consider the isotropic case, i.e., when the perimeter density does not depend on the direction, which makes most of the construction much simpler. In particular, in the present case the three arcs at triple points do not necessarily meet with three angles of $120^\circ$, which is instead what happens in the isotropic case.

References
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Book
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408 citations

Journal ArticleDOI
Abstract: In this paper we provide a complete classification of the local structure of singularities in a wide class of two-dimensional surfaces in R3 collected under the adjective (M, i, a) minimal by Almgren [A3] (see I(8)). The results, Theorems II. 4, IV. 5, IV. 8, are that the singular set of an (M, i, a) minimal set consists of H6lder continuously differentiable curves along which three sheets of the surface meet (Holder continuously) at equal (120?) angles, together with isolated points at which four such curves meet bringing together six sheets of the surface (H6lder continuously) at equal anglesin fact, in a neighborhood of each singular point, the surface is H6lder continuously diffeomorphic to either the surface Y of Figure 1 or the surface T of Figure 2 (both of which are defined in I(11)). The results apply to (idealizations of) many actual surfaces which are governed by surface tension, such as soap films as in Figure 4 and compound soap bubbles as in Figure 3 (and therefore to aggregates of some kinds of biological and metallurgical cells) (Corollary IV. 9 (i), (ii)), and thus are a proof of a result deduced experimentally by Plateau over 100 years ago [P]. They also apply to surfaces which minimize integrals which equal the area integral times some Holder continuous function on R3. A necessary first step in classifying singularities is to determine all possible area minimizing cones (Proposition II. 3). (In 1864 Lamarle claimed to make such a determination but his analysis of the technically most difficult case Figure 12 (p. 503)-was wrong.) Also included is a proof that the surface T of Figure 2 is in fact area minimizing (Theorem IV. 6); it seems to require the full force of Theorem IV.5 and I have never seen it proved elsewhere. The methods of this paper are

407 citations

"Minimal cluster computation for fou..." refers background or methods in this paper

• ...For N ≥ 1 first existence and partial regularity in Rn was given by Almgren [1] while Taylor [12] describes the singularities for minimizers in R3....

[...]

• ...For N ≥ 1 rst existence and partial regularity in Rn was given by Almgren [1] while Taylor [12] describes the singularities for minimizers inR(3)....

[...]

• ...In particular the structure of minimal cluster has been widely studied when the ambient space isR(2) (see [6]) orR(3) (see [12])....

[...]

Journal ArticleDOI
Abstract: This is a research announcement of results [Al] the full details and proofs of which have been submitted for publication elsewhere. We study the structure of m dimensional subsets of R which are well behaved with respect to deformations of R" and also show the existence of such sets as solutions to geometric variational problems satisfying various constraints. Suppose, for example, one is given several positive numbers av a2, • • *, aN and is asked to find disjointed regions Av A2, • • • , AN in R n such that At has volume at for each i and the n 1 dimensional area of S = (J{Boundary(At): i = 1, • • • , N} is as small as possible. For n = 3 this is a common formulation of a variational problem associated with compound soap bubbles. As a variant of this problem one could set A0 = R n ~ (J i Closure^ z.) and attempt to minimize the sum of the weighted areas of the various interfaces {Boundary^ t) n Boundary^ )}z., or perhaps the weighted integrals over these interfaces of various geometric integrands. For n = 2, 3 such minimal partitioning hypersurfaces have been the subject of numerous papers in mathematics, physics, ajid especially biology for the past several centuries (see, for example, [TD, Chapter 4, pp. 88—125] for a discussion and references). Among other things we give the first mathematical proof of the general existence of such surfaces. The methods are representative of those required to show the existence and regularity of solutions to a variety of geometric variational problems with constraints; e.g. capillarity problems, minimal surfaces avoiding obstacles, variational problems with partially free boundaries, etc.

343 citations

"Minimal cluster computation for fou..." refers background or methods in this paper

• ...For N ≥ 1 first existence and partial regularity in Rn was given by Almgren [1] while Taylor [12] describes the singularities for minimizers in R3....

[...]

• ...The following result gives the existence of minimal clusters (see [1], [6] and [4])....

[...]

• ...For N ≥ 1 rst existence and partial regularity in Rn was given by Almgren [1] while Taylor [12] describes the singularities for minimizers inR(3)....

[...]

Book ChapterDOI
01 Jan 2012-

202 citations

"Minimal cluster computation for fou..." refers background in this paper

• ...The following result gives the existence of minimal clusters (see [1], [6] and [4])....

[...]

• ...Existence and regularity of minimizers inR(2) was proved byMorgan [6] (see also [4]): the regions of a minimizer inR(2) are delimited by a nite number of circular arcs which meet in triples at their end-points....

[...]

Journal ArticleDOI
Abstract: Of course the circle is the least-perimeter way to enclose a region of prescribed area in the plane. This paper proves that a certain standard «double bubble» is the least-perimeter way to enclose and separate two regions of prescribed areas. The solution for three regions remains conjectural

101 citations

"Minimal cluster computation for fou..." refers background in this paper

• ...[2] proved that for N = 2 inR(2) the two regions of anyminimizer are delimited by three circular arcs joining in two points (standard double bubble) and are uniquely determined by their enclosed areas....

[...]

• ...theminimal clusterswith 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....

[...]

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