A double bubble is pair of bubbles which intersect and are separate...
Here's an interesting way to prove the circle is the shape of maxim...
The proof for ℝ³ was actually achieved 2 years later by [Joel Hass,...
To calculate R, A and L let's first draw the entire circumference.
...
If the cluster is not connected as can be seen in the image on the ...
Here's a plot of A''(x)
...
Here's an illustration of the calculation of the total perimeter fo...
PACIFIC JOURNAL OF MATHEMATICS
Vol. 159, No. 1, 1993
THE STANDARD DOUBLE SOAP BUBBLE IN R
2
UNIQUELY MINIMIZES PERIMETER
JOEL
FOISY,
MANUAL
ALFARO,
JEFFREY
BROCK,
NICKELOUS
HODGES
AND
JASON
ZIMBA
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" i s the least-perimeter way to enclose and separate two
regions of prescribed areas. The solution for three regions remains
conjectural.
1.
Introduction. Soap bubbles naturally tend to minimize surface
area for given volumes. This paper considers the two-dimensional
analog of soap bubbles, seeking the way to fence in prescribed ar-
eas using the least amount of perimeter. For one prescribed quantity
of area, the answer is of course a circle. This paper shows that for
two prescribed quantities of area, the unique answer is the "standard
double bubble" of Figure 1.0.1, and not the non-standard competitors
admitted by the general existence theory
([Al],
[M]) with disconnected
bubbles or exterior (see Figure
1.0.2).
FIGURE
1.0.1. The standard double-bubble is the
unique least-perimeter way to enclose and separate
two prescribed areas.
A B C
FIGURE
1.0.2.
Some non-standard double-bubbles.
(A) has connected bubbles but the exterior is dis-
connected. (B) has a connected exterior, but its B\
bubble is disconnected. (C) has both disconnected
bubbles and a disconnected exterior.
47
48
J. FOISY, M. ALFARO, J. BROCK, N. HODGES AND J. ZIMBA
FIGURE 1.0.3. It is an open question whether the
standard triple bubble is the least-perimeter way to
enclose and separate three prescribed areas.
FIGURE
1.0.4.
It is an open question whether the
standard double bubble is the least-area way to en-
close and separate two given volumes in R
3
.
For three or more areas, the question remains open (see Figure
1.0.3).
In R
3
it is also an open question whether the standard double
bubble is the least-area way to enclose and separate two prescribed
volumes (see Figure
1.0.4).
Proofs 1.1. The proofs first treat the case when there are no bounded
components of the exterior trapped inside the configuration (Lemma
2.4).
In this case, the bubble must be a combinatorial tree. In a non-
standard double bubble, part of an extreme bubble can be reflected to
contradict known regularity (cf. Figure
1.1.1).
The general case involves filling presumptive bounded components
of the exterior to obtain a contradiction of certain monotonicity prop-
erties of the least-perimeter function.
Existence 1.2. F. J. Almgren [Al, Theorem VI.2] and [M] have estab-
lished the existence of perimeter-minimizing bubble clusters in general
DOUBLE
BUBBLES
49
FIGURE 1.1.1. In the
proof,
presumptive extreme
extra components can be reflected to contradict
regularity. In the pictured reflection, there is a for-
bidden meeting of four arcs at the point t.
dimensions. Their results admit the possibility of disconnected bub-
bles and exterior. In the category of clusters with connected bubbles
and connected exterior, existence remains open, although it is an easy
mistake to think it obvious (cf. [B]).
The following regularity theorem, proved for R
3
by Jean Taylor
([Ta],
[ATa]), appears in [M].
REGULARITY THEOREM 1.3. A perimeter-minimizing bubble cluster
in R
2
consists of
arcs
of
circles
(or line segments) meeting in threes at
angles of 120°.
DEFINITIONS 1.4. A cluster of bubbles (or bubble cluster) in R
n
is a
collection of finitely many pairwise disjoint open sets, B\, 2?2
?
^3
?
?
B
k
. Each open set B\, , B
k
is called a bubble. We do not require
bubbles to be connected.
Call a cluster of exactly two bubbles a double bubble and a cluster
of exactly three bubbles a triple bubble.
In R
2
, a standard double bubble has three circular arcs (in this
paper, a line segment is a circular arc) meeting at two vertices at angles
of 120°.
The perimeter of a cluster is given by the one-dimensional Hausdorff
measure of the topological boundaries of the bubble:
Haus
1
A cluster is perimeter-minimizing if no other cluster enclosing the same
areas has less perimeter.
50 J.
FOISY,
M.
ALFARO,
J.
BROCK,
N. HODGES AND J. ZIMBA
Ac kno wle dgme nt s .
This paper consists of some main results of the
Geometry
Group [A] of the Williams College SMALL Undergradu
ate
Research Project, summer 1990, a National Science Foundation
site for Research Experiences for Undergraduates. Professor Frank
Morgan
advised the group. The work was further revised and devel
oped in Joel Foisy's senior thesis [F] u n d e r advisor Dave Witte and by
the
SMALL 1991 Geometry
Group,
consisting of Thomas Colthurst,
Chris
Cox, Joel Foisy, Hugh Howards, Kathryn Kollett, Holly Lowy,
and
Steve Root. A
very
special thanks to Professors Morgan and Witte
for all their help.
2.
The
double
bubble
in R
2
. Theorem 2.3 establishes the existence
and
uniqueness of the standard double bubble enclosing any two pre
scribed quantities of area. Lemma 2.4 shows that, if the exterior
is connected, a perimeter minimizing double bubble has connected
bubbles. Proposition 2.8 shows that a perimeter minimizing double
bubble must have a connected exterior. Finally, the Main Theorem
2.9 establishes the standard double bubble as the unique perimeter
minimizing enclosure for two prescribed quantities of area.
We begin with some basic formulas and omit th e easy computations.
PROPOSITION
2.1. Let S be an
edge
of a
perimeter minimizing
bub
ble
cluster
in R
2
.
Define
C to be the
distance
between
the
endpoints
of S,
with
θ the
angle
between
S and the
line
segment
connecting
its
endpoints
(see
Figure
2.1.1). Then the
radius
of
curvature
R of S, the
area
A of the
region
between
S and the
line
segment
connecting
its
endpoints,
and the
length
L of S are
given
by
c
AM r\ C
2
(θ sin(θ)cos(θ))
A{0C) = , ana
,
2
4sur(0)
1
c '
FIGURE
2.1.1. The circular arc S has radius of
curvature R, area A and length L.
DOUBLE BUBBLES
51
PROPOSITION
2.2. I f a
perimeter minimizing
double
bubble
in R
2
has
connected
bubbles
and a
connected
exterior,
then
it is a
standard
double
bubble.
Proof.
The
wh o le cluster must
be
connected. Otherwise, sliding
components
together until their boundaries
are
t an gen t yields
a
con
tradiction
of
the Regularity Theorem
1.3.
In
addition, Euler's formula concerning
the
vertices, edges,
and
faces
of a
planar graph
gives
V E + F = 1.
Since each vertex
has degree
3, 2E = 3V. We
also know that
F = 2.
Solving these
equations
yields that
V = 2 and E = 3 . By t h e
R e g u l a r i t y Theorem
1.3, the cluster indeed consists
of
t h r ee circular arcs
all
m e e t i n g
in two
points
at
an gles
of 120°.
THEOREM
2.3.
For a n y t w o
p r e s c r i b e d
qu a nt ities
of
area,
there
exists
a unique
standard
double
bubble.
REMARK.
The
proof shows that given
a
family
of
standard double
bubbles with vertices
a
fixed
distance apart, increasing
the
cu r vat ure
on
the separator arc decreases the area enclosed
by
t h e smaller bubble
and
increases
the
a r e a enclosed
by t h e
l a r g e r bubble.
Proof.
We
will
show that given
any λ
with
0 < λ < 1,
there exists,
up
to
c o n gr u e n c e,
a
un ique standard double bubble
B\, Bι
such that
the
ratio (area(J?i))/area(2?2))
of
the
two
a r e a s
it
encloses
is λ.
Consider
a
standard double bubble with
the
d ist a n ce between
its
two vertices
fixed
to b e 1.
Note
that
its
edges
are
c ir c u lar arcs that
meet
at
these
two
vertices.
For t h e
area underneath
an arc o f o u r
enclosure, Proposition
2.1
yields
A straightforward calculation shows that
„
=
20(2
+
cos(2fl)) 3(sin(2fl))
1
]
4sin
4
0
It
can
be
ro u t in ely shown that
A"(θ) > 0 on (0, π) .
For
any θ e [0,
π / 3 ) ,
an
angle formed
by a
circular
arc a n d the
line segment
of
distance
1
t h a t joins
its
en dpoin ts,
one c a n
construct
a
standard double bubble
(see
F i g u r e 2.3.1).
52
J.
FOISY, M. ALFARO, J. BROCK, N .
H OD G ES
AND J.
ZIMBA
FIGURE
2.3.1. For any θ, one can construct a stan
dard double bubble.
The
enclosed areas
satisfy
(1)
area B
x
= A((2π/ 3) θ) + A(θ),
area B
2
= A((2π/ 3) + θ) A(θ).
For
θ e[0, π/3), any ratio of A\ to A
2
will
be uniquely represented.
Indeed,
let F{θ) = (area(5!))/(area(5
2
)). Since A"(θ) > 0 for θ
in
(0, π), in (0, π/3), area ^i = A((2π/ 3) θ) + A(θ) is strictly
decreasing and area B
2
= A((2π/3) + θ) A(θ) is strictly increasing.
In
general, increasing θ
will
decrease the area enclosed by the smaller
FIGURE
2.3.2. As θ varies, a unique standard
double bubble represents
every
possible ratio of
area(2?i) to area(i?
2
).
DOUBLE BUBBLES
53
bubble and increase the area enclosed by the larger bubble. Thus F(θ)
is strictly decreasing on the interval [0, π/ 3). In addition, F(0) = 1
and
F(θ) > 0 as θ > π/ 3. Thus F: [0, π/3) + ( 0, 1] is bijective
(see Figure 2.3.2). Since F is bijective, a standard double bubble
enclosing any two prescribed quantities of area uniquely
exists
for
every value of θ e [0, π / 3).
We now have to show that any double bubble that contains bo u n d ed
components
of the exterior or disconnected bubbles is not perimeter
minimizing.
LEMMA 2.4. A
perimeter minimizing
double
bubble
whose
exterior
is
connected
must be
standard.
Proof.
Let U be a perimeter minimizing double bubble with con
nected
exterior. If U is not standard, by Proposition 2.2, U has a dis
connected
bubble. We
will
show that U is not perimeter minimizing.
Consider
a graph formed by placing a vertex inside each bubble
component
of U, with an edge between vertices of adjacent compo
nents.
For any U with a connected exterior, the corresponding graph
has no cycles. Thus there
will
be a component of U that lies at an
endpoint
of the corresponding graph. It must have exactly two edges
and
exactly two vertices (see Figure 2.4.1).
FIGURE
2.4.1. Since the exterior is assumed to be
connected
as in A, the associated graph has an
endpoint
in a component with two edges and two
vertices. If the exterior were disconnected, then a
cycle as in B could result.
54
J.
FOISY,
M.
ALFARO,
J.
BROCK,
N.
HODGES
AND J .
ZIMBA
\
FIGURE
2.4.2.
An
extreme component
F
bounded
by only
two
edges.
Let
F be a
component
of U
that
has
exactly
two
edges
and
exactly
two vertices,
r and q. Let t be a
vertex
of U
that
is
adjacent
to
r
but is not a
vertex
of F (see
Figure 2.4.2).
Let S be the
edge
connecting
r and t.
Let
p = r and
define
a new
bubble cluster,
U
p
, by
replacing
the
component
F by its
reflection across
the
perpendicular bisector
of
the line segment
qp (= qr). If we let the
point
p
move continuously
along
arc S
from point
r to
point
t and
reflect component
F and
arc
rp
across
the
perpendicular bisector
of
line
qp, the
bubble cluster
changes,
but
initially
the
perimeter
and
enclosed quantities
of
area
remain constant.
FIGURE
2.4.3.
The
reflected component
F may
touch another component, contradicting regular-
ity.
DOUBLE BUBBLES 55
FIGURE
2.4.4.
Otherwise the reflected component
F eventually touches t, also contradicting regular
ity.
As p
varies
continuously from r to t, two things could happen:
either there
will
be a point p for which the reflection
will
result in
the
touching of another bubble component and the creation of a new
vertex with four edges leading to it (see Figure 2.4.3), or p
will
even
tually coincide with point t , and there
will
be an instance of four
edges meeting at a vertex (see Figure 2.4.4).
This operation creates a new bubble cluster of the same perime
ter, enclosing the same prescribed quantities of area, that contradicts
regularity. Therefore, the original bubble
itself
must not be perimeter
minimizing.
We
will
soon show that a perimeter minimizing double bubble must
have a connected exterior.
LEMMA
2.5.
Increasing
the
larger
o f the two
prescribed
a r e a s
enclosed
by a
standard
double
bubble
will
increase
total
perimeter.
Proof.
F r o m Proposition 2.1, the length function for a circular arc
with endpoints distance 1 apart and with θ, the angle between the
segment connecting the endpoints and the arc, is
The
perimeter of the standard double bubble with angle θ between
the
line segment connecting its vertices (distance 1 apart) and the arc
separating its two bubbles is
given
by
perim (θ) = L (θ) + L{(2π/ 3) + θ) + L ( ( 2π / 3) θ).
56
J.
F OI SY,
M.
ALF AR O,
J.
BR OC K,
N.
H OD G E S
AND
J .
Z I M B A
A
routine c a lcu la t io n sh ows
that
,
sin0 0cos0
L
W
=
sm
z
0
sin
2
0 + 2 0
c o s
2
0 sin 2(9
ΓTί
sπ r
0
It
can
easily
be
shown that
L'(θ) > 0 on (0,
π ) ,
and
t h u s
L(θ) is
increasing
on [0,
π / 3 ) .
In
ad d ition , Z/'(0)
> 0 on (0,
π ) ; thus
o n
[0,
π/3), L((2π/3)
+ 0 )
4 L ( ( 2 π / 3 )
0) is
increasing. This implies
that
perim(0)
is
increasing
on [0,
π / 3 ) .
Thus
increasing
0
will
increase
the
t o t a l perimeter.
In
a d d i t i o n ,
it
follows
from Theorem
2.3 ( c f .
R e m a r k ) that increasin g
0
will
decrease
the
area enclosed
by
t h e smaller bubble and increase the area enclosed
by the larger bubble.
By
sc a l i n g
up
t h e double bubble until the smaller
bubble contains
its
original area, only
the
a r e a enclosed
by the
la r ge r
bubble
will
increase.
In t h e
p r o c e s s , total perimeter also increases.
DEFINITION
2.6. For two prescribed quantities
of
area,
A\ and A
2 ?
let
P(Aχ, A
2
) be t h e
p e r i m e t e r
of
the least perimeter double bubble
enclosing areas
A\ and A
2
. Let
P§(A\,
A
2
) be t h e
p e r i m e t e r
of
the
standard
double bubble enclosing areas
of
size
A\ and A
2
.
LEMMA
2.7. For a n y
fixed
A\ , A
2
> 0,
the
function
P(A, A
2
) has
a minimum
for A e [A\, oo).
Proof.
By the
isoperimetric inequality, P(A
y
A
2
)
>
perim(Z)),
where
D is a
disk
of
area
A.
Hence
P(A, A
2
)
—•
oo as A
—•
oo.
Since
P is
continuous,
P has a
minimum
for A e [Aι, oo).
PROPOSITION
2.8. The
exterior
of a
perimeter minimizing
double
bubble
must
be
connected.
Proof.
G iven
two
quan tities
of
area,
A\ and A
2
,
without
loss
of
generality assume
A\ > A
2
.
Suppose that
the
e x t e r i o r
of a
p er im e t er
minimizing double bubble
B\, B
2
enclosing
A\ and A
2
is
discon
nected.
By
Le m m a
2.7, w e c an
c h o o s e some
A\ e [A\, oo)
that min
imizes
P(A\, A
2
). In
p a r t i c u l a r ,
P(A\, A
2
) < P(A
{
, A
2
).
We now show that
if B[, B
2
is
t h e perimeter minimizing enclosure
of the quantities
of
area
A[ and A
2
,
respectively, then
the
ext erio r
of
B[, B
2
is
connected. Suppose,
to
o b t a in
a
contradiction, t h at
t h e
exterior
is
disconnected.
By
incorporating
the
b o u n d e d components
DOUBLE BUBBLES
5 7
of the exterior into the bubble B[ and then removing the edges that
were separating the bounded components of the exterior from B[,
a
new double bubble, B", B
2
, with
less
perimeter
than
B[, B
2
is
formed. This contradicts the choice of A\.
Thus B[ Φ B
x
and hence A
λ
< A\. By Lemma 2.5, P
0
(Aι, A
2
) <
P
0
(A[,
A
2
). By Lemma 2.4,
P
0
(A[,
A
2
) = P(A\, A
2
). By definition,
P(A\, A
2
) < PQ(A\ , A
2
). In summary:
P{A
X
, ^
2
) < Po(A
x
, ^
2
) < P
o
( 4 , A
2
) = P ( 4 , ^
2
) < ^ ( ^i, ^2)
This is a contradiction. We conclude that the exterior must be con
nected.
MAIN THEOREM 2.9. For any two
prescribed
quantities
of
area,
the
standard
double
bubble
is the
unique
perimeter minimizing
enclosure
of the
prescribed
quantities
of
area,
{See
Figure
2.9.1.)
FIGURE
2.9.1. Theorem 2.9 shows that a perimeter
minimizing double bubble must look
like
the
first
one,
not the second two, which have disconnected
bubbles or exteriors.
Proof.
By Theorem 1.6, we know that a perimeter minimizing dou
ble bubble
exists.
By Proposition 2.8, the exterior of this double bub
ble must be connected. Then, by Lemma 2.4, the bubble cluster must
be standard. Therefore, only a standard double bubble is perimeter
minimizing. By Theorem 2.3, it
exists
uniquely for any two prescribed
quantities of area.
Lemma 2.5 showed that increasing the larger quantity of area en
closed by a standard double bubble increases perimeter. It
follows
from our main theorem that increasing either quantity of area en
closed by a standard double bubble increases perimeter.
COROLLARY 2.10.
Increasing
either
given
area
A\, A
2
increases
the
perimeter
of the
perimeter minimizing
double
bubble.
58
J.
FOISY,
M.
ALFARO,
J.
BROCK,
N.
HODGES
AND J.
ZIMBA
FIGURE
2.10.1.
After
we
shorten
an
outside
arc,
the resulting double bubble
has
less perimeter,
and
2?i
has
less area.
Proof. If not, for
some
A\, A2
some slight increase
in, say, A\ de-
creases
the
least perimeter
of the
minimizing double bubble.
Now A\
can
be
decreased back
to its
original value
by
shortening
an
outside
arc,
as in
Figure
2.10.1,
and
further decreasing perimeter, contradict-
ing
the
minimizing property
of the
original double bubble.
Added
in proof.
There
has
been recent progress
on the
existence
of
connected bubbles
[M] and on the
triple bubble:
Chris
Cox,
Lisa Harrison, Michael Hutchings, Susan
Kim,
Janette
Light, Andrew Mauer,
Meg
T i l t o n ,
The
shortest enclosure
of
three
con-
nected areas
in R
2
, NSF
SMALL undergraduate research Geometry
Group, Williams College,
1992.
Also
see the
survey:
Frank Morgan, Mathematicians, including undergraduates, look
at
soap bubbles, Amer. Math. Monthly, (1993),
in
press.
REFERENCES
[A] Manuel Alfaro, Jeffrey Brock, Joel Foisy, Nickelous Hodges,
and
Jason Zimba,
Compound soap bubbles
in the
plane, SMALL undergraduate research project,
Williams College,
1990.
[Al]
F. J.
Almgren,
Jr.,
Existence
and
regularity almost everywhere
of
solutions
to elliptic variational problems with constraints,
Mem.
Amer. Math.
Soc, 35
(1976).
[ATa]
F. J.
Almgren,
Jr. and J. E.
Taylor,
The
geometry
of
soap films
and
soap
bubbles, Scientific American, July
1976,
82-9 3 .
[B] Michael
H.
Bleicher, Isoperimetric division into
a
finite number
of
cells
in the
plane, Studia
Sci.
Math. Hung.,
22
(1987), 123-137.
[F] Joel Foisy, Soap Bubble Clusters
in R
2
and R
3
,
Senior Honors Thesis,
Williams College,
1991.
DOUBLE BUBBLES 59
[M] Frank Morgan, Soap bubbles in R
2
and in surfaces, preprint (1992).
[Ta] Jean Taylor, The structure of singularities in soap-bubble-like and soap-film-like
minimal surfaces, Annals of Math., 103 (1976), 489-539.
Received August 12, 1991. Thi s research was partially supported by the National Sci-
ence Foundation Research Experiences for Undergraduates Program, the Ford Foun-
dation, the New England Consortium for Undergraduate Science Education, the Shell
Foundation, G.T.E., and the Bronfman Science Center at Williams College.
c/o FRANK MORGAN
WILLIAMS COLLEGE
WILLIAMSTOWN, MA 01267
E-mail address: Frank.Morgan@williams.edu
The proof for ℝ³ was actually achieved 2 years later by [Joel Hass, Michael Hutchings and Roger Schlafly](https://www.emis.de/journals/ERA-AMS/1995-03-001/1995-03-001.html)
Here's a plot of A''(x)
As can be seen A''(x)>0 on (0,$\pi$).
Here's an illustration of the calculation of the total perimeter for the standard double bubble.
Here's an interesting way to prove the circle is the shape of maximal area in fixed perimeter (which is equivalent of saying that the circle minimizes the perimeter for a certain area) using some intuition about physics.
Imagine a rope surrounding a 2D gas, with vacuum outside the rope. The gas will expand, pushing the rope to enclose a maximal area at equilibrium.
When the system reaches equilibrium the net force applied on the rope is going to be zero in every direction, otherwise it would keep expanding. Thus, the pressure the gas exerts must be cancelled by the tension of the rope in every point. Drawing the force diagram for an infinitesimal piece of rope we have
The force due to the gas is just the pressure times the length of the segment: $F_{GAS}=PL$. The force due to the tension is the projection along the direction of the gas expansion times 2 since the rope is under tension on both sides: $2T \sin(\frac{\theta}{2}) = 2T \sin (\frac{L}{2R})$.
Since the pressure is the same everywhere, and the force from pressure must be canceled by the force from tension, the net tension force must be the same for any rope segment of the same length. That means the radius is the same everywhere, so the rope must be a circle.
To calculate R, A and L let's first draw the entire circumference.
It's not difficult to see that $\cos (90-\theta)=\frac{C/2}{R}$ and so $R=\frac{C}{2 \sin (\theta)}$.
The length of the arc (L) is just the angle $2\theta$ times the radius of curvature R: $L=2\theta R$. Using the expression for R we get $L=\frac{C\theta }{\sin(\theta)}$
Finally to calculate the area A we just need to compute the area of the semi-circle and subtract the sections A1 and A2.
$$
A = A_{semi-circle} - 2A1-2A2 \\
A = \theta R^2 - \frac{C^2 \cos (\theta)}{4 \sin (\theta)}\\
A = \frac{\theta C^2 - \cos (\theta)\sin (\theta) C^2}{4 \sin^2(\theta)}
$$
A double bubble is pair of bubbles which intersect and are separated by a membrane bounded by the intersection.
The Double Bubble Conjecture states that the “standard” familiar double soap bubble provides the least-perimeter way to enclose and separate two given spaces. It was first conjectured in the 19th century by the Belgian physicist Joseph Plateau while observing the structure of soap films.
However it was only in 1993 that a group of undergraduate students from Williams College proved the conjecture was true for ℝ² and in 1995, Hass, Hutchings, and Schlafly announced a computer-assisted
proof for the case of equal volumes in ℝ³.
If the cluster is not connected as can be seen in the image on the left it is possible to slide the interior component so that the boundaries are tangent and so instead of having a group of 3 arcs meeting in point P we have 4 which contradicts Theorem 1.3!
