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\chapter{Variation of the domain}
$ M\grave{\eta} ~\epsilon\hat{\iota}\nu\alpha\iota~
\beta\alpha\sigma\iota\lambda\iota\kappa\grave{\eta}\nu~
\dot{\alpha}\tau\rho\alpha\pi\grave{o}\nu~
\grave{\epsilon}\pi\grave{\iota}~
\gamma\epsilon\omega\mu\epsilon\tau\rho\grave{\iota}\alpha\nu~~
 [E\grave{\upsilon}\kappa\lambda\epsilon\iota\delta\eta\nu]$

{\it There is no royal road to geometry [Euclid]}
\section{Variation of the domains: Setting }



Here, we derive the necessary conditions of optimality for the
domain $\GO$ where the Lagrangian in defined. The one-dimensional
analog of this problem is the variation of the interval that leads
to the transversality conditions, see Section \eq{transv}. First, we derive optimality condition  for the isoperimetric and related problem, where the Lagrangian is degenerative: it depends solemnly on the domain (like volume, inertia moment, or perimeter). Then, we consider a general problem for Lagrangian depending on a minimizer inside the domain.


%\subsection{Setting}

Consider a region $\CO$  ($\CO$ can coincide with $R_2$) and a  set of admissible domains $\GO$ with the  twice differentiable boundary$\Gg$ that belongs to the interior of $\CO$. Consider also a differentiable minimizer $u(x)$, $x\in \CO$, and a twice differentiable Lagrangian $ F(x, u, \Grad u) $. Consider the variational problem (Mayer-Bolza problem).%, see \eq{?} {\bf do we need the name?})
 \beq 
 J(\GO) =\min_{u(x)} \left\{
 \int_\GO F(x, u, \Grad u) dx + \int_\Gg f(x,
u) d s \right\}, \quad x\in \GO
 \eeq{261.problem1}
that contains the bulk and the boundary integrals. The objective
functional $J(\GO)$ depends on the domain $\GO$. 

Consider domain optimization:  Find $\GO$ that minimizes $J(\GO)$
We come to a  variational problem for the variable domain
 \beq I = \min_{\GO\subset \CO}  J(\GO).
  \eeq{261.problem12}
Here, we study stationarity condition for optimal $\GO$. In order to define variation of the domain, 
assume that functions $F(x, u,\Grad u) $ and $f(x, u)$ are defined everywhere in a larger region
$\CO$. For simplicity, we first discuss the two-dimensional problem.

\paragraph{Variation of the boundary}

We describe the boundary variation  $\Gd \Gg$ and related variation $
\delta \GO $.  Consider an admissible domain $\GO \cup \delta \GO $
that differs from  $\GO$ by an infinitely thin domain $\delta \GO
$ with a twice differentiable boundary $\Gg \cup \GG$, as on
Figure \ref{f261.1}. Denote the added boundary of this domain by
$\GG(s)$. In other words, we consider the added/subtracted domain
$\delta \GO$ as an infinitesimally thin strip of the width $\Gd
\Gn$  and the boundary $\Gg \cup \GG$. Asusual, we compute the variation of
the objective
$$
\Gd I =J(\GO  \cup \delta \GO)-J(\GO)
$$
and analyze the stationary condition $\Gd I =0$ which provide an
additional boundary condition at the unknown optimal boundary $\Gg$. This way,
we formulate the {\em free boundary problem}: a boundary value
problem in an unknown domain but with an additional condition for
this domain.

\begin{figure} \label{f261.1} 
  % Requires \usepackage{graphicx}
 % \includegraphics[width=]{}\\
  \caption{Infinitesimal variation of the boundary of the domain}
  \vspace {1.5in}
\end{figure}

\begin{remark}
The considered variation does not change the topology
of the domain: No new components of the boundary were added. %The variations that change the topology of the domain (inclusions) are discussed below, in Section \ref{wei2}
\end{remark}



We start with establishing the coordinate system in proximity of $\Md \GO$,
particularly, the correspondence between points of the stationary
boundary $\Gg$ and the varied close-by boundary $\GG$. Define normal $n=n(s)$ to each point $\Gg$;  the normal of a twice-di?erentiable curve is a continuous function of $s$. Introduce the following coordinates in an infinitisimal proximity of $\Gg$: a distance $s$ along the curve and the (infinitesimal) distance $\eta$ along the normal $n(s)$. Particularly, each point of $\Gg$ is represented by $(s, 0)$.  The admissible close-by curve $\GG$ is represented by infinitesimal distance $\Gd \eta$, $ \GG: ~\{ s, \Gd \eta(s) \} $. Notice that $\Gd \eta(s) $ may have any sign, the variation can go inside or outside of $\GO$. 


A point in an infinitesimal strip $\Gd \GO$ can be defined by two
coordinates: The distance $s$ along the boundary line $\Gg$ and
the distance $z \Gd \Gn$ from this line computed along the normal, where $z\in[0, 1]$. Particularly, points at $\Gd$ correspond to $z=0$ and points at $\GG$ -- to $z=1$. 

We consider "thin and small" variations assuming that 
 \beq
\Gd \Gn = 0 ~ \If s \not \in (s_0, \,  s_0+\Ge), \quad \Gd \eta \ll \Ge 
 \eeq{261.1.10}

If the curvature $k $ of $\Gg$ is not zero, the length of the arch varies with the normal. The infinitesimal length $dS$ along $\GG$ is related to the  infinitesimal length $ds$ along $\Gg$ as
 \beq
dS= ds (1+k \Gd  \Gn ) + o(\Gd  \Gn)
 \eeq{261.arch} This formula is known in differential geometry as ???? , see \cite{}
(see for example \ref{diffgeo})  Figure
\ref{f261.2} illustrates it:
\begin{figure}
  % Requires \usepackage{graphicx}
%  \includegraphics[width=.5]{}\\
  \caption{The variation of the differential of a curve due to its curvature}\label{f261.2}
\end{figure}
\hspace{1.5in}


\begin{remark}
Here, we assume that $\GO$ belongs to an open set $\CO$. If this
set is closed, the consideration must be supplemented by
variational inequalities (see Section \ref{varinq}).
  Additional
constraints on $\GO$ such as the prescription of its volume or
perimeter can be conventionally accounted by the Lagrange
multipliers.
\end{remark}
\begin{remark}
The variation of singular points of the domain where the normal is
not defined must be considered separately. Such variations will be
shown on the examples below.
\end{remark}


\section{Geometric Lagrangian}

\subsection{Stationarity conditions}
First we show how the bulk and boundary integrals in
\eq{261.problem1} vary due to the variation of the boundary in the
case when the integrands are independent of $u$ and are
continuous bounded functions of $x$ only, $ F=F(x) $ and $f=f(x)$.
In other words, consider the variational problem
 \beq I =
\min_{\GO\subset \CO}  J(\GO), \quad  J(\GO) =
 \int_\GO F(x) dx + \int_\Gg f(x) d s , \quad x\in \GO.
 \eeq{261.1.1}
that  depends only on the shape of the region $\GO$.

\paragraph{Variation of the integral over $\Omega$}
 The integral of a continuous function $F(x)$ over an infinitely thin domain is
estimated through the boundary integral
 \beq
 I_{\Gd \GO}=\int_{ \delta \GO}F(x) dx =  \int_{ \Gg}F(s) \Gd  \Gn(s) d s + o\|\Gd  \Gn(s)\|.
 \eeq{261.1.2}
To obtain this formula, we compute the integral $ I_{\Gd
\GO}$ as the repeated integral
 \beq
 I_{\Gd \GO}=\int_{\Gg} \Gd \Gn \left( \int_0^1 F(s, \Gd \Gn\,z) dz \right) d s
 \eeq{261.1.21}
 integrating first along the normal $n$ and then over the arch
 $\Gg$.
The inner integral is estimated using the continuity of $F$ and
the smallness of $\Gd \Gn$, as
$$
\int_0^1 F(s, \Gd \Gn z)  dz = \Gd \Gn \,F(s, 0) + o(\Gd \Gn)
$$
and the formula \eq{261.1.2} is obtained by computing the first-order term. Here, we denote $ F(s, 0)$ as $F(s)$.

\paragraph{Variation of the boundary integral}

The value of the differentiable function $f(x)$ at the point $S\in
\GG$ is expressed as
 \beq
 f(S)=f(s)+ \Gd \Gn \pd{f}{n} + o(|\Gd  \Gn|)
 \eeq{261.1.66}
where $\pd{f}{n}= n^T \Grad f $ is the normal derivative of $f$.

The variation of an integral of a function $f(x)$ can be computed
using \eq{261.arch} and \eq{261.1.66} and rounding to $|\Gd  \Gn|$
 \beqa
  \int_{ \GG}f(S) d S- \int_{ \Gg}f(s) d s &=&
\int_{ \Gg}\left[\left( f(s)+ + \Gd \Gn \pd{f}{n} \right)(1+k \Gd
\Gn) - f(s))\right] ds \nonum &=& \int_{ \Gg}\left( k f(s)+
\pd{f(s)}{n} \right) \Gd \Gn \, ds  .
 \eeqa{261.1.3}


 \paragraph{Stationarity conditions}
Adding together the two increments  \eq{261.1.2} and \eq{261.1.3},
we find that the increment $\Gd I$ in \eq{261.1.1} is equal
to
$$
\Gd I=\int_{ \Gg}\left( F(s) + k f(s)+ \pd{f(s)}{n} \right) \Gd
\Gn \, ds.
$$
As usual, we use the arbitrariness of $ \Gd \Gn $ and  obtain the
stationary condition
 \beq
 F(s) + \left( k + \pd{}{n}\right) f(s) =0 \quad \forall s\in \Gg
 \eeq{261.1.4}
for this simplest case \eq{261.1.1} where the Lagrangian is a fixed function of $x$.

\begin{example}[Isoperimetric problem]

 Consider the isoperimetric problem: Maximize the area of $\GO$
keeping its perimeter equal to one:
    \beq \max_\GO \int_\GO dx \quad
    \mbox{subject to }\int_\Gg ds=1
    \eeq{14.2.1}
or
$$\max_\GO \left\{\int_\GO dx  + \GL \int_{\Md \GO} ds \right\}
$$
where $\GL$ is the Lagrange multiplier by the isoperimetric
constraint. Here we put
$$ F=1, \quad f=\Lambda$$

The necessary condition  \eq{261.1.4} gives $1+ k \Lambda=0$
    \beq k=
    -\f{1}{\GL}=\mbox{constant}
    \eeq{14.2.4}
 at the optimal boundary. The unknown boundary is a circle (or a
part of a circle).
\end{example}


\begin{remark}[Comparison with symmetrization technique]

The isoperimetric problem can be solved by the symmetrization
method, see Section{symm}. Contrary to the symmetrization method,
the variational technique is applicable to a wider range of
problems. Here, for instance, the necessary conditions are
applicable for the case if a part of the boundary of $\GO$ is
fixed. The unfixed boundary components are the arcs of constant
curvature, joined by the fixed boundary components. The curvature
of the unfixed components is constant everywhere and can be found
from the isoperimetric condition.

On the other hand, the obtained condition do not prove that the
shape with constant curvature deliver the global minimum of the
isoperimetric problem because only close-by trajectories were
compared, while the symmetrization method guarantees the global
minimum but cannot withstand these additional constraints.

\end{remark}

\begin{example}
[A domain with an extremal moment and fixed perimeter] Consider
the problem of the symmetric domain with maximal moment of inertia
$M$
\beq
M=\int_{\GO} x^2 dx
\eeq{261.1.ex1}
that passes through two given points $(-a, 0)$ and $(a, 0)$ and
has a fixed perimeter.
 We set $F= x^2$, $f=\Gl$, where $\Gl$ is the Lagrange multiplier.
The optimality condition is $ \Gl k=x^2 $. In a Cartesian
coordinates, the boundary of the domain consists of a pair of  curves $\pm
y(x)$, the curvature is given by the known expression $  k=
\f{y''}{1+ y'^2}^{\f{3}{2}}$, and the optimality condition leads to the
equation
$$  y''=\f{1}{\Gl} x^2 (1+ y'^2)^{\f{3}{2}}
$$

Separating the variables and using symmetry $(y'(0)=0)$, we obtain
$$  y'(x)=\f{x^3 \lambda}{\sqrt{9-x^6\lambda^2}} $$
Next , we find $y(x)$ and calculate $\Gl$ using condition \eq{261.1.ex1}. 
The graph of the optimal curve is shown at Figure \ref{261.4}.

Notice that the curve is smooth everywhere except the points $(-a,
0)$ and $(a, 0)$ where its values are prescribed.

\end{example}

\subsection{Geometric applications}

\subsection{Cluster of domains with minimal boundary length}

\paragraph{Boundary components}
Assume that two  finite domains $\GO_1$ and $\GO_2$  of given
areas $A_1$ and $A_2$, respectively, have a common component $\Gg_{12}$ of
the boundary. Consider the problem of the shapes of these domain
that minimizes the total length of the boundary.


 The domain
$\GO=\GO_1 \cup \ldots \GO_N$ has outer and inner boundaries. The
optimality conditions for variation of the outer boundary is
similar to the condition \eq{14.2.4} in the isoperimetric problem:
$$ \Gd  \Gn_i:  \quad k_i + \GL_i=0, \quad i=1, 2.
$$
The conditions indicate that the outer bound is composed of two
circular arcs with radii $R_1= 1/k_1$ and $ R_2=1/k_2$,
respectively.

The variation of an inner component
$\Gg_{12}$ of the boundary 
$\Gg_{12} $  increases the volume of one of the neighbor and
decreases the volume of the second one by the same
amount. The resulting condition is:
$$ \Gd  \Gn_{12}:  \quad k_{12} + \GL_1- \GL_2=0
$$
This implies that the dividing lines are also circular, with the
radius (a reciprocal to the curvature) $R_{ij}$ equal to
    \beq
    { 1\over R_{12}}={ 1\over R_{1}}-{ 1\over R_{2}}
    \eeq{14.2.10}
In particular, the boundary between two domains of equal areas is
straight. Smaller domain remains convex when it is joined with the
larger one, and the larger is of crescent shape.


\paragraph{Singular points}

To complete the consideration, we determine the angle at
the point $s_0$ where three components of the boundary meet.
This problem requires special consideration because the boundary curve is
not smooth, the normal at the point $s_0$ is not defined, so the
standard variations are not possible. 





\begin{figure}
\caption{Optimal angle between normals in the singular point
\newline Left: The angle between any two normal is less than $180^\circ$
\newline Right: The angle between any two normal equals
$120^\circ$ } 
\vspace{1.5in}
\end{figure}

First, it is easy to show that the components stay continuously differentiable until they meet. 
Assume that a boundary component has an angular point $s_0$ and the normal is discontinuous there, denote the angle $\Gf$. 
Consider two points infinitely close points $s_1$ and $s_2$ at the boundary at different sides of $s_0$ such that $|s_1-s_0|=|s_2-s_0|=\Ge$ and a triangle $s_0s_1s_2$. Assume that the varied boundary replaces curve $s_2s_0s_1$ with the straight line $s_1s_2$ of the length $|s_1 s_2|=  2 \Ge \cos \Gf$. The variation $\Gd l$ of boundary lengths is negative $ |Gd l=2 $ (which is evident from triangle inequality) and is of the order $\Ge$, the variation of the area  $ \Gd A=\Ge^2 (\cos \Gf -1)$ is neglected in linear part of increment. We conclude that there cannot be kinks at an opptimal boundary.

Consider now a meeting point $s_0$ of three boundary components. Again, choose three points $s_1$, $s_2$ and $s_3$ at the corresponding branches of the boundaries in $\Ge$-neighborhood of $s_0$, so that $s_i\neq s_0$. $s_0$ is optimal, if
$$
L(s_0)=\sum_{i=1}^3 |s_i - s_0|
$$
is minimal. Differentiation with respect to $s_0$, we obtain equation for an optimal $s_0$,
$$
\od{L(s_0)}{s_0}=\sum_{i=1}^3 \f{s_i - s_0}{|s_i - s_0|}=0.
$$
If the sum of three unit vectors is zero, these vectors are directed at $120^\circ$ to each other. 
\begin{remark}
The minimization problem also has another solution, $s_0=s_2$ that is optimal when the angle between 
$(s_1-s_2)$ and  $(s_2-s_3)$ is less than a threshold (see Problem \ref{?}). Here, however, we assumed that the branch point is not equal to any of  $s_1$, $s_2$ and $s_3$, for optimal contours.
\end{remark}


Again, the increment of the variation of lengths is of the order of $\Ge$ and the increment of areas is of the order of $\Ge^2$ and is neglected in linear term.
We conclude that 
\begin{theorem}
Optimal boundary components meet at the angle $120^\circ$ to each other.
\end{theorem}




\paragraph{Definition of parameters}
We found that the boundaries of two-domain configuration are
consist of three arches of circles which centers lie on a line and
radii are connected by the condition \eq{14.2.10}; that intersect
in two symmetric points, the angle between the arches equal to
$120^\circ$. To draw the configuration, it is enough to find
radii $R_1$, $R_2$ and $R_{12}$, and angles $\Gt_1$, $\Gt_2$ of the arches. To find these five parameters we
have two equations that express the conditions of intersections
$$
R_1 \sin \Gt_1=R_2 \sin \Gt_2=R_{12} \sin \Gt_{12}
$$
 equation  \eq{14.2.10}, and two equations that fix the areas of the
domains
$$
A_1= R_1 (\pi -\Gt_1)+ \hf R_1^2 \sin\Gt_1 \cos \Gt_1 + R_{12}
\Gt_{12}- \hf R_{12}^2 \sin\Gt_{12} \cos \Gt_{12}
$$
and
$$
A_2= R_2 (\pi -\Gt_2)+ \hf R_2^2 \sin\Gt_2 \cos \Gt_2 - R_{12}
\Gt_{12}+ \hf R_{12}^2 \sin\Gt_{12} \cos \Gt_{12}
$$
and the condition
$$
\Gt_1-\Gt_{12}=\Gt_2-\Gt_{12}=2\pi/3
$$
{\bf too many conditions!}

Summing up, the formulate the theorem:
\begin{theorem}
The optimal boundary consists of three circular arches that meet
at two symmetric points at $120^\circ$; the curvatures of the
arches are related as $ k_{12}= k_1 - k_2$.
\end{theorem}




For instance, if the areas are very distinct, $A_1 \ll A_2$, the
smaller area is a lens made of the circles of the close-by radii
circle, and the larger area is a circle without a lens-shaped
area.

\paragraph{Multiple domains}

A natural generalization of the previous problem is the problem of
$N$ separated domains $\GO_i$, $i=1, \ldots, N$ of given areas
$A_i$ with the minimal length of separating boundary. The domain
$\GO=\GO_1 \cup \ldots \GO_N$  has outer and inner boundaries. The
optimality conditions for variation of the outer boundary are
similar to the condition \eq{} in the isoperimetric problem:
$$ \Gd  \Gn_i:  \quad k_i + \GL_i=0
$$
Let us find the conditions of optimality of an inner component of
the boundary.  Suppose that two domains $\GO_i$ and $\GO_j$ are
neighbors. The variation of their common boundary $\Md_{ij} $
results in the condition
$$ \Gd  \Gn_{ij}:  \quad k_{ij} + \GL_i- \GL_j=0
$$
This implies that the dividing lines are also circular, with the
radius (a reciprocal to the curvature) $R_{ij}$ equal to
$$
{ 1\over R_{ij}}={ 1\over R_{i}}-{ 1\over R_{j}}
$$
%In particular, the boundary between two domains of equal area is straight. 
%Smaller domain remains convex when it is joined with the
%larger one, and the larger is of crescent type.

 If the number of domains with different areas is larger of equal to
four, the question arise what domains should be placed inside the
configuration and do not have an outer boundary. More generally,
the question is which domains should be neighbors. This problem
requires combinatoric methods since it is needed to compare
several not close-by configurations. We do not discuss the problem
in this text but encourage the reader to try to solve it.

\begin{example}
An infinite system of domains of equal areas forms a honeycomb
structure. Bees know variational calculus! Indeed, all the inner
boundaries are straight due to symmetry, and all angles where the
boundaries meet are equal to $120^\circ$.

% If one cell is smaller (larger) that the other, the system is
% disturbed as follows: the boundaries around the distinguished cell
% are curved, while the other boundaries are straight. The cells are
% hexagonal, but they are not perfect hexagons.

\end{example}

\paragraph{Optimal shapes in a bounded domain}
When a shape touches the boundary of the domain, it meets the this boundary at a right
angle.    Indeed, the angle which correspond to minimal length of $\Gg$ in the  to the boundary is 
a right angle.  The rest of the consideration is as before.

\paragraph{Natural cracks in Badlands}

Photo! Observe the cracks meeting either at $120^o$ or at $90^o$
Discuss the history of development of the cracks.

\subsection{Three-dimensional problems: Minimal surface and shape of bubbles}

\paragraph{Three-dimensional problem} The consideration of the
three-dimensional case is quit the same, but the formula for the
variation of the boundary arch is replaced by the formula for the
variation of the boundary surface element:
 \beq
 dS=d s(1+k_1 \Gd  \Gn )(1+k_2\Gd  \Gn)  =d s(1+
(k_1+k_2)\Gd  \Gn ) =o(\Gd \Gn)
 \eeq{261.3d1}
  where $k_1$ and $k_2$ are the main
curvatures on the boundary surface. The calculation of the
stationary condition is performed as before. The necessary
condition:
 \beq
 F(x) + \left(k_1+k_2 + \pd{}{n}\right) f  = 0
 \eeq{261.1.3d2}
differs from the two-dimensional case by replacing the curvature
$k$ of the boundary line with the mean curvature $k_1+k_2$ of the
boundary surface.

The simplest problem is the minimal surface problem: Find a
surface of minimal area that is attached to a given contour. Here,
$$F=0 \quad f=1
$$
\begin{remark} The surface may be not closed, but this is of no
  importance since we consider only local variation of the
  ``boundary'' surface.
\end{remark}
The stationarity condition
$$
 k_1+k_2=0
$$
shows that the minimal surface has zero mean curvature.

 \paragraph{Bubble}  The
bubble problem is the three-dimensional analog of the
isoperimetric problem: Find a domain of maximal volume enclosed in
the surface of fixed area. Here
$$
F=1 \quad f=\GL
$$
and the optimality condition
$$
 k_1+k_2=\GL
$$
stays that the mean curvature is constant. A sphere is an obvious
solution. Beside it there are many other shapes: a circular
cylinder, for example. One may experiment with an air-balloon to
find these shapes. The problem is formulated as
a partial differential equation that expressed the constancy of
the mean curvature of the surface with the boundary conditions
that correspond to the requirements that the surface passes
through a given contour. For references, see \cite{}.


\paragraph{Cluster of bubbles}
Similarly to the two-dimensional case, a variational problem of
minimization the total area around several spacial domains with
given volumes is considered. Physically, this is the problem of a
cluster of bubbles.

The cases of two, three, and four equal bubbles can be handled out
analytically using the obvious symmetries. In the general case,
the outer and interior surfaces of the cluster satisfy the
following conditions:
\begin{theorem}[Bubble theorem] ~ The following statements
describe the cluster of the bubbles:

(1) The outer boundary of the cluster consists of the surface with
piece-wise constant mean curvature.

(2) Any two volumes (bubbles) are divided by a surface of constant
mean curvature.

(3) Any three bubbles are divided by a curves formed by
intersection of three mean curvature surfaces with normals that
meet at the angles $2\pi/3$ independently of the  volumes of
bubbles.

(4) Any four bubbles meet at the points where four  mean curvature
surfaces meet. The normals to these surfaces at the meeting point
meet at the angles  $2\pi/3$ as the planes meet in the center of a
symmetric tetragon that pass through the sides of this tetragon.
The angles between the meeting planes are fixed independently of
the  volumes of bubbles.


{\bf Find a reference}

 \bigskip {\it Hints for the proof}

(1) Apply the necessary conditions on a free boundary

(2) Apply the necessary conditions on a dividing boundary

  (3) Consider an infinitesimally thin cylinder abound the line of intersection.

(4) Consider an infinitesimal sphere or tetragon around the point
of intersection.

 \end{theorem}


\section{General type Lagrangian}
\subsection{Formulas for the increment}


Here we consider the general form of the functional
\eq{261.problem12}, assuming that that the integrants depend on a
minimizer $u(x)$, $x\in \GO$ and the domain $\GO$ is varied. We derive an additional boundary condition at the optiaml domain which serves for finding this domain.

The variation of the functional in \eq{261.problem12} is computed as
the difference
\begin{eqnarray}
\delta J_n  &= & \delta I_1 + \delta  I_2 \nonum \Gd I_1 &=&
\int_{\GO \cup \delta \GO} F(x, u+ \Gd u,\Grad (u + \Gd u)) dx -
\int_{\GO} F(x,u,\Grad u) dx
 \nonum
  \Gd I_2&=& \int_{\GG} f(x, u+
\Gd u) d S   - \int_\Gg f(x, u) d s
\end{eqnarray}
between the cost of the Lagrangians of an admissible solution in
an admissible domain and of the extremal solution in the optimal
domain. 

\paragraph{Variation of the bulk integral}
First, we work out the increment $\Gd I_1$ of bulk integrals. We
 rewrite it as
\beqa
\Gd I_1 = \int_{\GO \cup \delta \GO}F_u \Gd u \, dx
+ \int_{\delta \GO} F(x, u, \Grad u) dx \nonum +
\int_\Gamma n \cdot \pd{F}{\Grad u} \Gd u(\GG)\, dS + o(\|\delta u\| , \|\Gd  \Gn \|)
\eeqa{261.2.2}
where {\color{red} refer~to~formula instead}
$$
F_u= -\Div \pd{F}{\Grad u} + \pd{F}{u} 
 $$
 To derive this formula, we add and subtract the integral over $\delta \GO $ of $F(x, u, \Grad u)$ and compute the fist variation  $\Gd I_1$ with respect to $\Gd u$ in $\GO \cup \delta \GO$.
 
 
The variation  $\delta u$ at $\GG=\Gg+ \Gd \Gn$ that is a sum of
the variation of minimizer $\delta u(\Gg)$ at stationary
boundary $\Gg$ and shift $\pd{F}{u} \Gd \Gn$ of the minimizer due to
variation of the boundary,
$$
\delta u(\GG)= \delta u(\Gg)+ \pd{u}{n}\Gd \Gn
$$
In other words, variation $\delta u(\Gg)$ is expressed as the linear
combination of two free variations $\delta u(\GG)$ and $ \Gd \Gn
$:
$$
\delta u(\Gg)= \delta u(\GG)- \pd{u}{n}\Gd \Gn
$$
Integral over $\Md \GO$ in the right-hand side of \eq{261.2.2} is estimated as
$$
\int_{\GO \cup \delta \GO} F(x, u, \Grad u) dx= \int_{\Gg} F(x, u, \Grad u) \Gd \Gn dx+ o(\|\Gd\Gn\|)
$$


It remains to substitute these expressions into \eq{261.2.2} and group the terms:
\beqa
\Gd I_1= \int_{\GO } F_u(x, u, \Grad u) \delta u \, dx +
 \int_{\Gg} A_u \delta u\,ds+  \int_{\Gg} A_\Gn \delta
 \Gn\,ds
\nonum
A_u=n \cdot \pd{F}{\Grad u} \quad A_\Gn=F- \left(n \cdot \pd{F}{\Grad u} \right) \pd{u}{n}
\eeqa{261.1.6}
The first two terms are standart conditions of stationarity of $u$, the last term expressed the stationarity of $\GO$. 
Increment $A_\Gn$ is analogous to the transversality
condition \eq{} in the one-dimensional problem. It is the sum of 
the accumulated value of the Lagrangian over the added domain $\Gd
\GO$ and (term $-n^T \pd{F}{\Grad u} \Gd \Gn $)  a shift of natural boundary conditions
from the admissible to the stationary boundary.  $A_\Gn$  is an analog of
the term $-u'\pd{F}{u'}\Gd x$ in the one-dimensional
transversality condition.

If the boundary integral is zero, the additional
(transversality) boundary condition
$ A_\Gn=0$
holds at the unknown boundary $\Gg$ and serves to find this
boundary. 

\paragraph{First variation of the boundary integral}
To compute the curve integral $I_2$, we expand $f$ as follows:
$$
f(X, u(X))|_{X\in \GG}=f(x)|_{x\in \Gg}+ \left(\pd{f}{x}\pd{x}{n}
+ \pd{f}{u}\pd{u}{n}\right) \Gd  \Gn + \pd{f}{u} \Gd u + o(\| \Gn\|)
$$
and take into account the variation of the boundary arch
\eq{261.arch} $( dS= ds (1+ k \Gd  \Gn))$, obtaining:
$$
\Gd I_2 = \int_{\Gg} \left[ f(x, u) + \pd{f}{u} \Gd u  + \left(\pd{f}{x}\pd{x}{n}+
\pd{f}{u}\pd{u}{n}\right) \Gd  \Gn \right](1+ k \Gd  \Gn)ds-  \int_{\Gg}
f(x, u) ds
$$
Rounding to $\Gd  \Gn$, we have
$$ \Gd  I_2=\int_{\GG}\left[\left(k\, f +\pd{f}{n}+
\pd{f}{u}\pd{u}{n}\right)\Gd  \Gn\, + \pd{f}{u} \Gd u\right]ds
$$
The first terms in parenthesis accounts to increase of the length of varied curve $\GG$, the next two terms show the shift of  the $f(s)$ due to variation of $s$, and the last term is the variation due to variation of $u$. 

\paragraph{Stationary conditions}



Adding $\Gd I_1$ and $\Gd I_2$, we finally obtain the stationarity conditions:


The variations with respect to $u$ returns the familiar
expressions: $F_u$ is the left-hand-side of  the Euler equation
$$   \Gd u(x): \quad F_u=\Div \pd{F}{\Grad u} -\pd{F}{u} = 0 \quad \In \GO,  $$
natural boundary condition,
    \beq   \Gd u(s): \quad  \GF_u \Gd u(s)=0, \quad  \GF_u= \pd{F}{\Grad u}\Bn -
    \pd{f}{u} =  0\quad \On \Gg,     \eeq{261.1.15}
and the stationarity of the domain itself:
$$
  \Gd \Gn: \quad  F(x, u, \Grad u) + \left[-\pd{F}{\Grad u} \cdot
n + \pd{f}{u} \right]\pd{u}{n} + \pd{f}{n} + k\, f(x,
u)=0.
$$
%{\color{red} chain rule}
 
 If no boundary condition is prescribed (free boundary, or variational boundary conditions), then $\GF_u=0$  due to \eq{261.1.15}, and the last condition is simplified to 
         \beq
R_F=   F(x, u, \Grad u)  + \pd{f}{n} + k\, f(x,
u)=0.
    \eeq{261.2.bound} 
   It serves to determine an unknown boundary with variational boundary conditions. 
\begin{remark}
The last condition involves the
dependence of $\Grad u$ which we represent in local coordinates as
$\Grad u= \pd{u}{n} \Bn
  +\pd{u}{s} \Bt $; generally, it leads to a differential equation
$$ \Psi \left( \pd{u(s)}{s}, u(s), \pd{u(s)}{n} \right)=0
$$
along the boundary. In this equation, $u(s)$ and $\pd{u(s)}{n}$
are independent functions. To define both of them, we consider
also the natural boundary condition \eq{261.1.15} that expresses
another relation $\psi\left(u(s),\pd{u(s)}{n} \right)=0$ between
these functions.
\end{remark}

\begin{example}[Vibrating membrane of minimal frequency]

Consider the problem of the vibration membrane of a fixed area with minimal first frequency
$F$ in the form $ F=\hf \Grad^2 u - \hf c^2  u^2 + \GL$ 
where $\GL$ is Lagrange multiplier by the condition fixing the area, $u$ is the deflection, and $\Go$ i the frequency. 
on the boundary $\Gg$, the membrane is fixed, $u=0$. In this problem, $f=0$, and $\pd{u}{s}=0$ on $\Gg$. 

It is convenient to compute  $R_F$
in local coordinates
 -- the tangent $t$  and the normal $ n$ to the boundary. In this
coordinates, we compute
$$
F=\hf\left( \pd{u}{t}\right)^2 + \left( \pd{u}{n}\right)^2- \hf
c^2 u^2+ \GL= \hf \left(\pd{u}{t}\right)^2+\GL .$$
We have
$$
R_F=- \hf \left( \pd{u}{n}\right)^2+ \GL=0
\quad \Or ~
\pd{u}{n} =\Cnst \quad \On \Gg
 $$
and it serves to find  the unknown boundary $\Gg$. One easily
guesses that the circular membrane has an optimal shape.
\end{example}


\paragraph{Three-dimensional problem} In  the three-dimensional case,
 the formula for the variation  becomes:
$$
F(x, u, \nabla u)  +  \pd{f}{n}+(k_1+k_2) f=0
$$
Here, we use formula \eq{} for the variation of boundary element.


\paragraph{Conditions on the free boundary}

Assume that the natural boundary conditions \eq{} are imposed on
the unknown component of the boundary. In many physical
applications,  $F=F( u, \nabla u)$ means the bulk energy, and
$f=f(u)$ is the surface energy. We also assume that both forms of
energy do not explicitly depend on the position $x$,
$$
F=F( u, \nabla u), \quad f=f(u)
$$
Due to the listed conditions, the optimality condition on the
unknown moving boundary simplifies and becomes:
 \beq
 F( u, \nabla u)  -(k_1+k_2) f(u)=0
 \eeq{261.2.2}
We see that the mean curvature of the optimal boundary surface is
the proportionality coefficient between the bulk and surface
energies.

In particular, the constant value of $f$  corresponds to
the prescription of the surface area. In this case, the bulk
energy density on the boundary is proportional to its mean
curvature.



\paragraph{Inner boundary}
Similar consideration is applicable to the problem of optimal
position of a variable boundary $\Gg$ dividing two volumes $\GO_1$
and $\GO_2$. The problem is to minimize the energy
 \beq
 \int_{\GO_1} F_1(u, \Grad u) dx +  \int_{\GO_2} F_2(u, \Grad u)
 dx+
  \int_{\Gg} f(u) dx
  \eeq{}
  The necessary conditions for the boundary $\Gg$ are
  $$
  n^T \pd{}{\Grad u}[F( u, \nabla u)]^+_-+ \pd{f(u)}{u}=0 \quad \On \Gg
  $$
  (the natural boundary condition on any boundary between the
  volumes, obtained by the variation of $u$), and
  $$
 [F( u, \nabla u)]^+_-  + (k_1+k_2) f(u)=0 \quad \On \Gg
  $$
Here, $[F]^+_-= F_2-F_1$ denotes the jump: The difference in the
values of the function $F$ at both sides of $\Gg$. Notice that the
solution $u$ is continuous at the boundary, and so is its
tangential derivative $\pd{u}{t}$ but the normal derivative
$\pd{u}{n}$ is, generally, discontinuous.

%\paragraph{Phase equilibrium}

\subsection{Application: Optimal Shape of Conducting Domains}

 Next problems ask
for the shape of the domain of extremal resistivity. Consider a
domain $\GO$ with the boundary $\Gg= \Gg_0\cup \Gg_1\cup \Gg_i $
divided into three components. Assume that the domain is filled
with an isotropic material with unit conductivity.  Assume that
the potentials $u=0$ and $u=1$ are applied to the two components
$\Gg_0$ and $\Gg_1$ of the boundary, respectively. The
supplementary component $\Gg_i$ of the boundary is insolated,
 \beqa
 u=0 \quad \On \Gg_0, \quad  u=1 \quad \On \Gg_1,\quad
 \pd{u}{n}=0 \quad \On \Gg_i,
 \eeqa{262.3.1}
  The resistivity is defined as the total conducting
energy of the domain. Indeed, the total normal current $ j=
\pd{u}{n} $ through the $\Gg_1$ component of the domain is equal
to the total energy of it:
$$
\int_{\Gg_1} \pd{u}{n} ds=\int_{\Gg} u\pd{u}{n} ds=\min_u \hf
\int_{\GO} (\Grad u)^2 \,dx
$$
The first equality follows from the boundary conditions
\eq{262.3.1} and the second -- from Green's formula combined with
the stationary condition $\nabla^2 u=0$ in $\GO$.

Let us formulate an optimization problem: Maximize the total
current through the $\Gg_1$ or, equivalently, minimize the total
conductivity of the domain by varying its boundary. Additional
geometrical restriction can be  assigned.


 Assume in addition that the area $s$ of the plain domain $\GO$ is fixed.
 Consider the conductivity problem in a domain of the fixed
volume and a partly known boundary
$$
\min_{\Gg_i} \left(\min_{u \mbox{ as in }\eq{262.3.1}} \hf
\int_\GO (\Grad w)^2 dx  + \Gl \int_\GO dx \right)
$$
where $\Gl$ is the Lagrange multiplier. The potential $u$ on the
known components is prescribed and natural boundary conditions are
satisfied on $\Gg_i$.

The augmented Lagrangian is
$$ F=(\Grad u)^2+\Gl, \quad f=0 $$
and  $\Gl$ is the Lagrange multiplier.
 The Euler-Lagrange
equation and natural boundary condition are
$$
\Grad^2 u =0 \quad \In \GO, \quad \pd{u}{n}=0 \quad \On \Gg_i
$$

One more condition is needed to define the shape of unknown
boundary component $\Gg$. This condition on the unknown boundary
is
 \beq
n^T \pd{F}{\Grad u}- F=  \left(\pd{u}{n}\right)^2-
\left(\pd{u}{t}\right)^2 -\Gl  =0
 \eeq{261.3.8}
 where $t$ is the tangent to the boundary.

 \begin{example}[Conditions on optimal insolated boundary component]

Assume that the shape of boundary components $\Gg_0$ and $\Gg_1$
is known, but the insolated component $\Gg_i$ is movable.
Combining the optimality condition  \eq{261.3.8} with the
stationary condition $ \pd{u}{n}=0 \On \Gg_i$ we obtain the
condition for the unknown boundary:
$$w_t= \sqrt{-\Gl}=\mbox{constant} \quad  \On \Gg
$$
This condition tells that the optimal isolated boundary is also
the current line: the current density along it is constant.

For example, a rectangular domain is optimal if the potentials on
two opposite sides of it are constants.
\end{example}

\begin{example}[Optimal boundary component with the given
potential]
 Suppose now that the boundary component $\Gg_0$ where the main boundary
condition $w=C$ is imposed should be found from the
optimality requirements. Again, we use the condition \eq{261.3.8}
combining it with the prescribed boundary conditions in the form
$$
\pd{u}{t}=0 \quad \On \Gg_0, \And \Gg_1
$$
The optimality of this component is expressed as
$$w_n =\sqrt{\lambda},
$$
This condition implies the constancy of the normal current on the
optimal boundary where the potential is constant.
\end{example}


\begin{example}[Optimal boundary of the domain with given perimeter]
 Consider the previous conductivity
 problem in a domain with the boundary
of the fixed perimeter but of arbitrary volume; this time
$$ F=(\Grad w)^2,\quad f=\GL $$
where $\GL$ is the Lagrange multiplier.
The condition on the unknown boundary is
$$F+\Lambda k =0 \quad\mbox{or } w_n^2 + w_t^2+\Lambda k=0
$$
Combining it with the stationary condition
$$ w_n=0  \On \Gg
$$ with respect to
$w$,
we obtain the condition
$$w_t^2 = \GL k.
$$
which says that the square of the flux density along the boundary is
proportional to its curvature.

\end{example}


\begin{example}[Optimal shape of the membrane with minimal eigenfrequency]
 Consider the problem of the shape of
a membrane of minimal eigenfrequency and given volume. Here
$$
F= (\Grad w)^2+ c^2 w^2 + \GL, \quad f=0
$$
Here $\GL $ is the Lagrange multiplier accounting for the volume
constraint, $w$ is deflection of the membrane which is zero on its
boundary, $$ w =0 \quad \On \Gg.$$ The tangent derivative among
the boundary $\pd{w}{t}$ is zero as well. The optimality condition
becomes
$$
\pd{w}{n}=\mbox{constant},
$$
it shows that the normal derivative of the deflection is constant
along the boundary. One easily guesses that the optimal shape is
the circle.
\end{example}

 % \subsection{Optimal shape of the hole in an elastic plane}

%\input hole.tex
\end{document}