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%\textup{2000} Mathematics Subject Classification}

\dateposted{July 17, 2000}
\PII{S 1079-6762(00)00079-2}
\copyrightinfo{2000}{Michael Hutchings, Frank Morgan, Manuel Ritor\'{e}, and
Antonio Ros}






\title{Proof of the double bubble conjecture}

\author{Michael Hutchings}
\address{Department of Mathematics, Stanford University, Stanford, CA 94305}

\author{Frank Morgan}
\address{Department of Mathematics, Williams College, Williamstown, MA

\author{Manuel Ritor\'{e}}
\address{Departamento de Geometr\'{i}a y Topolog\'{i}a, Universidad de
Granada, E-18071 Granada, Espa\~{n}a}

\author{Antonio Ros}
\address{Departamento de Geometr\'{i}a y Topolog\'{i}a, Universidad de
Granada, E-18071 Granada, Espa\~{n}a}

\commby{Richard Schoen}
\date{March 3, 2000}

\subjclass[2000]{Primary 53A10; Secondary 53C42}

\keywords{Double bubble, soap bubbles, isoperimetric problems, stability}

We prove that the standard double bubble provides the least-area way
to enclose and separate two regions of prescribed volume in ${\mathbb R}^3$.



Archimedes and Zenodorus (see \cite[p. 273]{K}) claimed and Schwarz
\cite{S} proved that the round sphere is the least-perimeter way to
enclose a given volume in ${\mathbb R}^3$.  The Double Bubble
Conjecture, long assumed true (\cite[pp. 300--301]{P}, \cite[p.
120]{B}) but only recently stated as a conjecture \cite[Section 3]{F1},
says that the familiar double soap bubble in
Figure~\ref{fig:standard}, consisting of two spherical caps separated
by a spherical cap (or flat disc), meeting at 120-degree angles,
provides the least-perimeter way to enclose and separate two given
volumes.  The analogous result in ${\mathbb R}^2$ was proved by the
1990 Williams College ``SMALL'' undergraduate research Geometry Group
\cite{F2}.  In 1995, Hass, Hutchings, and Schlafly [HHS] announced a
computer-assisted proof for the case of equal volumes in ${\mathbb
R}^3$.  (See \cite{M1}, \cite{HS1}, \cite{HS2}, \cite{Hu}, \cite[Chapter
13]{M2}.)  Here we announce a proof \cite{HMRR} of the general Double
Bubble Conjecture, using stability arguments.

In ${\mathbb R}^3$, the unique perimeter-minimizing double bubble
enclosing and separating regions $R_1$ and $R_2$ of prescribed volumes
$v_1$ and $v_2$ is a standard double bubble as in
Figure~\ref{fig:standard}, consisting of three spherical caps meeting
along a common circle at $120$-degree angles.  (For equal volumes, the
middle cap is a flat disc.)

Reichardt et al. \cite{RHLS} have generalized our results to ${\mathbb
R}^4$ and certain higher dimensional cases (when at least one region
is known to be connected).  The 2000 edition of \cite{M2} treats
bubble clusters through these current results.

\section{Previous results}

(See \cite[Chapters 13 and 14]{M2}.)  F. Almgren \cite[Thm.\ VI.2]{A}
proved the existence and almost everywhere regularity of
perimeter-minimizing bubble clusters enclosing $k$ prescribed volumes
in ${\mathbb R}^n$, using geometric measure theory. Taylor \cite{T}
proved that minimizers in ${\mathbb R}^3$ consist of smooth
constant-mean-curvature surfaces meeting in threes at 120-degree
angles along curves, which in turn meet in fours at isolated points.
An idea suggested by White, written up by Foisy \cite[Thm.\ 3.4]{F1}
and Hutchings \cite[Thm.\ 2.6]{Hu}, shows that any
perimeter-minimizing double bubble in ${\mathbb R}^n$ has rotational
symmetry about some line. A major complication is that the regions are
not {\em a priori} known to be connected.  (If one tries to require
the regions to be connected, they might in principle disconnect in the
minimizing limit as thin connecting tubes shrink away.)  Hutchings
\cite{Hu} developed new concavity and decomposition arguments to rule
out ``empty chambers'' (bounded components of the exterior) and to
bound the number of connected components of the two regions of a
minimizer. In particular, a nonstandard minimizer in ${\mathbb R}^n$
consists of a central bubble with nested toroidal bands. For equal
volumes in ${\mathbb R}^2$, there is only one band and at most a
two-parameter family of possibilities. Hass and Schlafly \cite{HS2}
carried out a rigorous computer search of this family to eliminate
these possibilities and prove the Double Bubble Conjecture for equal
volumes. Earlier computer experiments of Hutchings and Sullivan had
suggested that no such double bubbles were stable.

%[Fig. 1a is the right half of http://www.math.uiuc.edu/~jms/Images/double/]
%[Fig. 1b is attached]
\caption{The standard double bubble provides the least-perime- ter
way to enclose and separate two prescribed volumes.  [Computer
graphics copyright John M. Sullivan, University of Illinois, 
{\em http://www.math.uiuc.edu/$\sim$jms/Images}.]} \label{fig:standard}

\section{The proof}

As indicated above, a perimeter-minimizing double bubble is known to
exist, to have rotational symmetry, and to consist of
constant-mean-curvature surfaces of revolution (``surfaces of
Delaunay''), meeting in threes at 120-degree angles along circles of
revolution.  In ${\mathbb R}^3$ the convexity and decomposition
arguments of Hutchings \cite[Thm.\ 4.2]{Hu} imply easily that the
larger region is connected and with careful computation that the
smaller region has at most two components, as in
Figure~\ref{fig:nonstandard}.  Actually we give a less computational,
stability argument to show that the smaller region has at most two

%[Fig 2a is the left half of http://www.math.uiuc.edu/~jms/Images/double/]
%[Fig 2b is attached]
\caption{In this nonstandard double bubble, the smaller region has two
components---a central bubble and a thin toroidal bubble around the
outside---while the larger region is another toroidal bubble in
between.  Note that although this example consists of constant mean
curvature surfaces meeting at $120$-degree angles, it is not in
equilibrium, because the two components of the disconnected region
have unequal pressures.  [Computer graphics copyright John M. Sullivan,
University of Illinois, {\em
http://www.math.uiuc.edu/ $\sim$jms/Images}.]}
%[Fig 3 is attached]
\caption{Case by case, one finds an ``axis of instability'' $A$
perpendicular to the $x$-axis of symmetry. The places `` $|$ ''
where the rotational vector field is tangent to the surface divide
the bubble into four pieces. [Drawing by James F. Bredt, copyright
2000 Frank Morgan.]} \label{fig:cases}
To rule out nonstandard minimizers, we use the following stability
argument.  Suppose that there is a nonstandard minimizer, and
consider rotations about an axis $A$ orthogonal to the axis of
symmetry as in Figure~\ref{fig:cases}. This axis can be chosen so
that the places where the rotation vector field $v$ is tangent to
the double bubble separate the bubble into (at least) four
pieces. Some linear combinations of the restrictions of $v$ to
the four pieces vanish on one piece and respect the two volume
constraints. By regularity for eigenfunctions, $v$ must vanish on
certain associated parts of the bubble, which therefore must be
spherical or flat. When three surfaces meet and two are spherical
or flat, so is the third. Lots of spherical or flat surfaces lead
easily to a contradiction.

This argument is inspired by Courant's Nodal Domain Theorem
\cite[p.\ 452]{CH}, which says for example that the first
eigenfunction is nonvanishing. Other applications of this
principle to isoperimetric problems and to the study of
volume-preserving stability have been given by Ritor\'{e} and Ros
\cite{RR}, by Ros and Vergasta \cite{RV}, by Ros and Souam
\cite{RS}, and by Pedrosa and Ritor\'{e} \cite{PR}. Perhaps the
simplest such phenomenon without constraints is that a circle of
longitude on the unit sphere is unstable because the rotation
vector field vanishes at the poles (conjugate points): rotating
just half of it can be smoothed to reduce length.

Finding the required axis $A$ requires consideration of a number of
cases, as shown in Figure~\ref{fig:cases}.


Much of this work was carried out while Morgan was visiting
the University of Granada in the spring of 1999. Morgan has partial
support from a National Science Foundation grant.  Ritor\'{e} and Ros
have partial support from DGICYT research group no. PB97-0785.

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