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\headline={\ifnum\pageno=1 \hfill\else%
{\tenrm\ifodd\pageno\rightheadline \else
\leftheadline\fi}\fi}
\def\rightheadline{EJDE--1997/02\hfil Qualitative behavior of
axial-symmetric solutions\hfil\folio }
\def\leftheadline{\folio\hfil Andrew F. Acker \& Kirk E. Lancaster
\hfil EJDE--1997/02}
\voffset=2\baselineskip
\vbox {\eightrm\noindent\baselineskip 9pt %
Electronic Journal of Differential Equations,
Vol.\ {\eightbf 1997}(1997) No.\ 02, pp. 1--24.\hfill\break
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\hfil\break ftp (login: ftp) 147.26.103.110 or 129.120.3.113 }
\footnote{}{\vbox{\hsize=10cm\eightrm\noindent\baselineskip 9pt %
1991 {\eighti Subject Classification:} Primary 35J65, 35R35; Secondary 35B99
\hfil\break
{\eighti Key words and phrases:} Free boundary problem, curves of constant
gradient direction.
\hfil\break
\copyright 1997 Southwest Texas State University and
University of North Texas.\hfil\break
Submitted September 22, 1996. Published January 8, 1997.} }
\bigskip\bigskip
\centerline{QUALITATIVE BEHAVIOR OF AXIAL-SYMMETRIC SOLUTIONS}
\centerline{OF ELLIPTIC FREE BOUNDARY PROBLEMS}
\medskip
\centerline{Andrew F. Acker \& Kirk E. Lancaster}
\bigskip\bigskip
{\eightrm\baselineskip=10pt \narrower
\centerline{\eightbf Abstract}
A general free boundary problem in $\Bbb R^3$ is investigated for
axial-symmetric solutions and qualitative geometric properties of the
free boundary are compared to those of the fixed boundary for the axial
and radial directions. Counterexamples obtained previously by the first
author show that our results cannot hold in the same generality as those
for similar free boundary problems in $\Bbb R^2$.
\bigskip}
\bigbreak
\centerline{\bf \S 0. INTRODUCTION} \medskip\nobreak\noindent
Let $G$ be the quasilinear, elliptic, second-order
partial differential operator on $\Bbb R\sp N$ given by
$$
GU=\sum_{i,j=1}^N A_{ij}(X,DU(X))D_i D_j U
+ B(X,DU(X)), \quad X\in{\cal{O}},
\eqno{(1)}
$$
for $U\in C^2({\cal{O}})$, where ${\cal{O}}$ is any open set
in $\Bbb R^N$,
$A_{ij}\in C\sp{1,\delta}(\Bbb R^N\times\Bbb R^N)$,
$i,j=1,\dots,N$, satisfies
$\sum_{i,j=1}\sp{N} A_{ij}(X,P)\xi_i\xi_j > 0$
for $X,P\in\Bbb R^N$ and $\Xi=(\xi_1,\xi_2,\dots,\xi_N)
\in\Bbb R^N\backslash \{0\}$, and
$B\in C\sp{2,\delta}(\Bbb R^N\times\Bbb R^N)$ for some $\delta\in(0,1)$.
For ${\cal{S}}^*$ a closed hypersurface in $\Bbb R^N$
and ${\cal{S}}$ a closed hypersurface in $\Bbb R^N$ which
surrounds ${\cal{S}}^*$, we denote by ${\cal{O}}({\cal{S}}^*,
{\cal{S}})$ the open region between ${\cal{S}}^*$ and ${\cal{S}}$.
The purpose of this paper is to study the qualitative geometric properties
of axial-symmetric solutions of the following ``Bernouli'' free boundary
problem when $N=3$.
\medskip
\noindent {\bf N-dimensional free boundary problem.}
Given a closed hypersurface
${\cal{S}}^*\subset\Bbb R^N$ and a positive constant $\lambda$, find a
closed $C^1$ hypersurface ${\cal{S}}={\cal{S}}\sb\lambda\subset\Bbb R^N$
which surrounds ${\cal{S}}^*$ and
$U=U_\lambda\in C^2({\cal{O}})\cap C^0({\overline{\cal{O}}})\cap
C^1({\cal{O}}\cup{\cal{S}})$ such that
$$
GU=0\quad\rm{in}\ {\cal{O}},
\eqno{(2a)}
$$
$$
\ U=1\quad\rm{on}\ {\cal{S}}^*,
\eqno{(2b)}
$$
$$
\ U=0\quad\rm{on}\ {\cal{S}},
\eqno{(2c)}
$$
$$
\vert DU\vert=\lambda\quad\rm{on}\ {\cal{S}},
\eqno{(2d)}
$$
where ${\cal{O}}={\cal{O}}({\cal{S}}^*,{\cal{S}})$.
We will call ${\cal{S}}^*$ the {\it fixed boundary} and ${\cal{S}}$ the
{\it free boundary} of this problem.
General existence results related to this free boundary problem were
obtained by Alt, Caffarelli, and Friedman ([10]) using the method of
variational inequalities which is discussed in greater generality in
books by Friedman ([12]) and Kinderlehrer ([14]). However, their solutions
might not be classical solutions and need not be doubly connected.
When $G$ is the Laplace operator and ${\cal{S}}^*$ is starlike
relative to all points in a sufficiently small ball, the free boundary
problem has a unique, starlike, classical solution
$({\cal{S}}\sb\lambda,U\sb\lambda)$ such that
${\cal{S}}={\cal{S}}\sb\lambda$
is symmetric with respect to some line whenever
${\cal{S}}^*$ is symmetric with respect to that line ([9]).
When $G$ is the p-Laplace operator with $1
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% end of figure 1
\noindent Clearly, this problem is actually
two-dimensional, since the surfaces ${\cal{S}}$ and ${\cal{S}}^*$
are generated by the corresponding arcs
$$
\eqalignno{\Gamma&=\{(x,y):(x,y,0)\in{\cal{S}},y>0\}&(3a)\cr
\Gamma^*&=\{(x,y):(x,y,0)\in{\cal{S}}^*,y>0\}.&(3b)\cr}
$$ % figure 2
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\noindent
Notice that the endpoints of $\Gamma$ and $\Gamma^*$ are the
intersections of ${\cal{S}}$ and ${\cal{S}}^*$ respectively with the
$x_1-$axis. Our conclusions regarding the qualitative geometric
properties of ${\cal{S}}$ and ${\cal{S}}^*$ can then be expressed
entirely in terms of the qualitative geometric properties of $\Gamma$ and
$\Gamma^*$.
In order to discuss our results, let us adopt the notation of [2] and [4].
Thus $\Gamma$ and $\Gamma^*$ are oriented curves with initial and
terminal points lying on the $x_1-$axis such that
the $x_1-$coordinate of each initial point is smaller than the
$x_1-$coordinate of the corresponding terminal point.
We define $\vec n(x,y)$ to be the unit normal vector to $\Gamma\cup
\Gamma^*$ at $(x,y)\in\Gamma\cup\Gamma^*$ which points to the
right of the curve (with respect to the direction of the curve).
Further, we have the following:
\smallskip
\noindent {\bf Definition.}
Given a unit vector $\vec\nu$, we call $(x_0,y_0)\in\Gamma$
a $\vec\nu$-minimum ($\vec\nu$-maximum)
of $\Gamma$ if $\vec n(x_0,y_0)=\vec\nu$ and
$(x_0,y_0)$ is a strict local minimum (maximum)
relative to $\Gamma$ of $f(x,y)=\vec\nu\cdot (x,y)$
(see, for example, Figures 2 and 3 in [4]).
\smallskip
\noindent {\bf Definition.}
Given a unit vector $\vec\nu$, we call $(x_0,y_0)\in\Gamma^*$
a $\vec\nu$-minimum ($\vec\nu$-maximum)
of $\Gamma^*$ if $\vec n(x_0,y_0)=\vec\nu$ and either
$(x_0,y_0)$ is a strict local minimum (maximum)
relative to $\Gamma$ of $f(x,y)=\vec\nu\cdot (x,y)$
or there is a closed line segment $\gamma^*\subset\Gamma^*$
such that $(x_0,y_0)\in\gamma^*$ and
$\vec\nu\cdot(x,y)>\ (<)\ \vec\nu\cdot(x_0,y_0)$ for
$(x,y)\in\Gamma^*\backslash\gamma^*$ near $\gamma^*$.
Here $\gamma^*$ is considered
as a single local extremum.
\smallskip
\noindent We may define $\vec\nu-$inflection points of $\Gamma$ and
$\Gamma^*$ similarly (see [2], [16]). Notice that Lemma 2(b.) implies
the definitions of $\vec\nu-$extrema are equivalent.
The following figures illustrate the definition of $\vec\nu-$extrema of
$\Gamma;$ the letters $a,A,b,B,c,C$ represent points at which $\Gamma$
has a $-\vec j-$minimum, $-\vec j-$maximum, $-\vec i-$minimum,
$-\vec i-$maximum, $\vec i-$minimum, and $\vec i-$maximum respectively.
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Let us assume that $\Gamma^*$ contains a finite number
of maximal line segments (including isolated points) on which
$\vec n(x,y)=\pm \vec i$ or $\vec n(x,y)=-\vec j$.
Let equation (2a) be either Laplace's equation (i.e. (15) ) or the
minimal surface equation (i.e. (18) )
in $\Bbb R^3$, ${\cal{S}}^*$ be a closed surface in
$\Bbb R^3$, and $({\cal{S}},U)$ be a solution of the free boundary
problem for some $\lambda>0$. Suppose ${\cal{O}}={\cal{O}}({\cal{S}}^*,
{\cal{S}})$ is rotationally symmetric with respect to the $x_1-$axis
and set
$$
W=\{(x,y)\in\Bbb R^2:(x,y,0)\in{\cal{O}}\}.
\eqno{(4)}
$$
Let $\partial_iW$ and $\partial_oW$ denote the inner boundary
and outer boundary of $W$ respectively and let $\Gamma$ and
$\Gamma^*$ be given by (3).
Then our main results, which are given in \S 1, include the
following as a special case:
\proclaim Theorem 1.
Suppose there exists $u\in C^2(W\cup\partial_o W\cup\Gamma^*)
\cap C^1({\overline{W}})$ such that
$$
U(x_1,y\cos(\theta),y\sin(\theta))=u(x_1,y)
\eqno{(5)}
$$
for $(x_1,y)\in W$ and $\theta\in\Bbb R$. Suppose also that $\Gamma^*$
and $\partial_oW$ are $C^2$ curves and $W$ satisfies an interior
sphere condition at each point of $\partial_iW$. Then
$\Gamma$ has no more $\vec\nu-$minima (maxima) than does $\Gamma^*$
and each $\vec\nu-$minimum (maximum) of $\Gamma$ can be joined to a
(distinct) $\vec\nu-$minimum
(maximum) of $\Gamma^*$ by a curve along which $\nabla u$ has a
constant direction, for each $\vec\nu=-\vec i,\vec i,-\vec j$.
In particular, if $\Gamma^*$ is a graph over the $x-$axis, then
$\Gamma$ is also a graph over the $x-$axis.
\centerline{\epsffile{fig6.ps}}
The study of the relationship between $\vec\nu-$extrema of the free and
fixed boundaries of solutions of the $N-$dimensional
free boundary problem has previously been restricted to the
case $N=2$. In this case, we have a quasilinear elliptic partial
differential operator $Q$ on $\Bbb R^2$, a constant $\lambda>0$, and
a Jordan curve $\Gamma^*$ in $\Bbb R^2$ and the free boundary problem
consists of finding a Jordan curve $\Gamma$ in $\Bbb R^2$ which surrounds
$\Gamma^*$ and a function $u\in C^2(\Omega)\cap
C^1(\Omega\cup\Gamma) \cap C^0(\overline{\Omega})$
such that
$$
Qu=0 \quad \rm{in} \ \ \Omega,
\eqno{(6a)}
$$
$$
u=1 \quad \rm{on} \ \ \Gamma^*,
\eqno{(6b)}
$$
$$
u=0,\ \vert\nabla u\vert = \lambda \quad \rm{on}\ \ \Gamma,
\eqno{(6c)}
$$
where $\Omega={\cal{O}}(\Gamma^*,\Gamma)$.
The ``geometric study'' of this two-dimensional free boundary
problem began with the consideration of the case in which $Q$ is the
Laplace operator. In this case, the principal model for later work
was established by the first author in [1], [2], and [4], where a
method of curves of constant gradient direction was developed and
applied in an analysis of the number and ordering of the directional
extrema and inflection points of the free boundary.
At approximately the same time, curves of constant gradient direction were
independently used to study ideal fluid flows by Friedman and Vogel ([11]).
The use of curves of constant gradient direction was extended
to solutions of the two-dimensional free boundary problem by Vogel ([18])
and the first author ([3]) when (6a) is Poisson's equation,
by the authors when (6a) is the minimal surface equation ([7])
or the heat equation ([8]), and by the second author ([16]) when $Q$ is
any elliptic partial differential operator of the form
$$
Qu\equiv au_{xx}+2bu_{xy}+cu_{yy},
\eqno{(7)}
$$
where $a,b,c$ depend on $x,y,u_x$, and $u_y$. The conclusion
obtained (in the elliptic cases) is that if $\Gamma$ and $u$
constitute a solution of the
free boundary problem, $\Omega$ is a $C^2$ domain, and
$u\in C^2(\overline{\Omega})$, then each $\vec\nu-$extremum
of the free boundary can be joined to a corresponding (distinct)
$\vec\nu-$extremum of the fixed boundary by a curve $(\gamma)$ along which
$\nabla u$ remained parallel to $\vec\nu$
(i.e. $\nabla u(x,y)=\vert\nabla u(x,y)\vert\ \vec\nu$ for each
$(x,y)$ on $\gamma)$ and, in particular, $\Gamma$ has no more
$\vec\nu-$minima $(\vec\nu-$maxima) than does $\Gamma^*$, for each
$\vec\nu$. In addition, the number of $\vec\nu-$inflection points of
$\Gamma$ cannot exceed the number of $\vec\nu-$inflection points of
$\Gamma^*$.
When the three-dimensional free boundary problem is symmetric with respect
to the $x_1-$axis and $({\cal{S}},U)$ is an axial-symmetric solution,
the function $u(x,y)=U(x,y,0)$ is the solution of a related two-dimensional
free boundary problem. In fact, we obtain immediately the following
\proclaim Proposition.
Suppose $({\cal{S}},U)$ is a solution of the three-dimensional free
boundary problem, $U\in C^1(\overline{{\cal{O}}})$, and there exists
$u\in C^2(W)\cap C^1(\overline{W})$ which satisfies (5)
for $(x_1,y)\in W$ and $\theta\in\Bbb R$, where
$W=\{(x,y): (x,y,0)\in\cal{O}\}$. Let $x=x_1$ and
define $\Omega=\{(x,y)\in W: y>0\}$ and $Q$ to be the quasilinear, elliptic
operator given by
$$
Qu(x,y)=a(x,y,\nabla u)u_{xx}+2b(x,y,\nabla u)u_{xy}
+c(x,y,\nabla u)u_{yy}+d(x,y,\nabla u)
\eqno{(8)}
$$
for $u\in C^2(\Omega)$ and $(x,y)\in\Omega$,
where $a(x,y,p,q)=A_{11}(x,y,0,p,q,0)$,
$b(x,y,p,q)=A_{12}(x,y,0,p,q,0)$,
$c(x,y,p,q)=A_{22}(x,y,0,p,q,0)$, and
$d(x,y,p,q)=B(x,y,0,p,q,0)+{q\over{y}}A_{33}(x,y,0,p,q,0)$.
Then $u$ is a solution of free boundary problem (6) when $Q$ is given by
(8).\par
It is natural to conjecture that the results obtained for the two-dimensional
free boundary problem (6) with $Q$ given by (7) apply to solutions of the
$N-$dimensional free boundary problem for arbitrary $N\ge 3$.
Such a generalization, if true, would be esthetically more satisfactory
than the (3-dimensional) axial-symmetric results we obtain. However,
this conjecture is incorrect, as the first author ([6]) has shown
by means of a counterexample in which $N=3$, $G=\triangle$ is the Laplace
operator, $\lambda>0$, the fixed boundary ${\cal{S}}^*$
has precisely one $\vec\nu-$minimum, and the free boundary
${\cal{S}}={\cal{S}}\sb\lambda$ has two distinct
$\vec\nu-$minima, for some direction $\vec\nu$.
The study of qualitative properties of axial-symmetric solutions in
$\Bbb R^3$ is suggested by the facts that the properties in question seem
to correspond to two-dimensional problems and axial-symmetric
solutions of three-dimensional free boundary problems are of physical
interest (e.g. [15]).
The results in Theorems 1 and 3 about the directional extrema of $\Gamma$
would be more appealing if they applied to arbitrary directions in
$\Bbb R^2$. However, when such a problem is reduced to the two-dimensional
free boundary problem (6), the differential operator (8) may contain a
lower order term (i.e. $d$) which complicates the situation. The
conjecture that the solution of (6) has the same qualitative
properties with regard to arbitrary directions is false.
The first author ([6]) has obtained a counterexample
when $N=3$, $G=\triangle$, and $\lambda>0$
in which the generator $\Gamma^*$ has only one $\vec\nu-$minimum while
the free boundary $\Gamma=\Gamma\sb\lambda$ has two
$\vec\nu-$minima, for some direction $\vec\nu$ (which is not an axial or
radial direction). Thus, while our results seem somewhat restricted, the
most natural and appealing generalizations are false.
The paper is organized as follows. In \S 1, we state our main results.
In \S 2, we present some examples of free boundary problems in
$\Bbb R^3$ to which our results apply. The statements of our
preliminary results, which consist of nine lemmas, are given in \S 3
and these lemmas are proven in \S 4; the statements are separated from
their proofs in the hope of making the paper more readable.
Our main results are proven in \S 5 and we include some concluding
remarks in \S 6.
\bigbreak
\centerline{\bf \S 1. MAIN RESULTS} \medskip\nobreak\noindent
Suppose $({\cal{S}},U)$ is a solution of the free boundary problem,
$U\in C^1(\overline{{\cal{O}}})$, ${\cal{O}}$ is
axial-symmetric, $W$ is given by (4),
and there exists $u\in C^2(W)\cap C^1(\overline{W})$
which satisfies (5). Let us write $x=x_1$. We set
$\Gamma^*=\{(x,y):(x,y,0)\in {\cal{S}}^*,y>0\}$,
$\Gamma=\{(x,y):(x,y,0)\in {\cal{S}},y>0\}$, and
$\Omega=\{(x,y)\in W: y>0\}$.
Define $Q$ to be the quasilinear, elliptic operator given by (8).
Then $u$ is a solution of the free boundary problem (6).
We will assume that linear functions of the form $U(x,y,z)=\alpha x+\beta$
are solutions of (2a); this is equivalent to assuming
$$
B(x,y,0,p,0,0)=0.
\eqno{(9)}
$$
Let us define the ratio of the coefficient $a$ of $u_{xx}$ in $Q$ to
the lower order term $d$ in $Q$ to be
$$
g(x,y,p,q)={d(x,y,p,q)\over{a(x,y,p,q)}}\equiv
{{qA_{33}(x,y,0,p,q,0)+yB(x,y,0,p,q,0)}
\over{yA_{11}(x,y,0,p,q,0)}}.
$$
Notice that $g(x,y,p,0)=0$, and so
${\partial g\over{\partial y}}(x,y,p,0)=0$, for all $x\in\Bbb R,y>0$.
\vskip .2 true in
\proclaim Theorem 2.
Let us assume the three-dimensional free boundary problem
(2) has a solution $({\cal{S}},U)$, $U$ is in
$C^2({\cal{O}}\cup{\cal{S}})\cap C^1({\overline{\cal{O}}})$,
the solution $({\cal{S}},U)$ is axial-symmetric, and
condition (9) holds. Let $\partial_iW$ be the inner portion
of the boundary of $W$ and assume $W$ satisfies an interior
sphere condition at each point of $\partial_iW$.
If we define $\Gamma$ and $\Gamma^*$ as
above and if $\Gamma^*$ is the graph of a $C^1$ function,
then $\Gamma$ is the graph of a $C^2$ function. \par
If we are willing to assume that additional conditions are satisfied,
we can obtain a result which is stronger than that of Theorem 2.
Let us define the function
$$
h(x,y,p,q)={d(x,y,p,q)\over{c(x,y,p,q)}}\equiv
{{qA_{33}(x,y,0,p,q,0)+yB(x,y,0,p,q,0)}
\over{yA_{22}(x,y,0,p,q,0)}}.
\eqno{(10a)}
$$
Let us assume that
$$
{\partial h\over{\partial x}}(x,y,0,q)=0
\eqno{(10b)}
$$
and there is a $C^1$ function $\Phi(y,q)$
satisfying
$$
\Phi(y,q)<0,
\eqno{(10c)}
$$
$$
{\partial\Phi\over{\partial y}}(y,q)<0,
\eqno{(10d)}
$$
$$
{\partial\Phi\over{\partial q}}(y,q)>0,
\eqno{(10e)}
$$
$$
{\partial\Phi\over{\partial y}}(y,q)=h(x,y,0,q)
{\partial\Phi\over{\partial q}}(y,q),
\eqno{(10f)}
$$
$$
{q_1\over{\Phi(y,q_1)}}\le{q_2\over{\Phi(y,q_2)}}
\quad{\rm{when}}\ \ q_20$, and $q<0$.
If there exist $C^0$ functions
$k:(-\infty,0)\to(-\infty,0)$ and $l:(0,\infty)\to(0,\infty)$
such that
$$
k(q)\ge {1\over{q}}
\eqno{(11a)}
$$
and
$$
{d(x,y,0,q)\over{c(x,y,0,q)}}={l(y)\over{k(q)}},
\eqno{(11b)}
$$
for $x\in\Bbb R,y>0,q<0$, then $\Phi(y,q)=K(q)L(y)$ satisfies the
conditions (10c)-(10g) above, where
$$
K(q)=-\exp\bigg(\int_{-1}^qk(t)dt\bigg)
\eqno{(11c)}
$$
and
$$
L(y)=\exp\bigg(\int_1^yl(t)dt\bigg).
\eqno{(11d)}
$$
Recall that we have oriented $\Gamma$ $(\Gamma^*)$ so that $\Omega$
lies locally to the right of $\Gamma$ (left of $\Gamma^*)$.
Notice that the definition of the unit normal $\vec n$ on $\partial W$
implies
$$
\nabla u(x,y)=\vert\nabla u(x,y)\vert\ \vec n(x,y),\qquad
(x,y)\in\Gamma\cup\Gamma^*.
\eqno{(12)}
$$
We will assume that $\Gamma^*$ contains a finite number of
maximal line segments (including isolated points) on which
$\vec n(x,y)=\pm \vec i$ or $\vec n(x,y)=-\vec j$.
\vskip .3 true in
\proclaim Theorem 3.
Suppose $W$ is an open, doubly connected region in the plane which
is symmetric with respect to the $x-$axis and conditions (9) and
(10) hold. Let $\partial_i W$ be the inner portion of the
boundary of $W$ and $\partial_o W$ be the outer portion. Let
$\Gamma^*=\{(x,y)\in\partial_i W :y>0\}$ and
$\Gamma=\{(x,y)\in\partial_o W :y>0\}$. Assume
$\partial_o W$ and $\Gamma^*$ are $C^2$ curves
and that $W$ satisfies an interior sphere condition at
each point of $\partial_i W$.
Let $\lambda$ be a positive constant.
\hfil\break\noindent
Suppose there exists
$u\in C^2(W\cup\partial_o W\cup\Gamma^*)
\cap C^1({\overline{W}})$ such that
$$\eqalign{
Qu&=0 \qquad\qquad \rm{in}\ \ W, \cr
u&=1 \qquad\qquad \rm{on}\ \ \partial_i W, \cr
u&=0 \qquad\qquad \rm{on}\ \ \partial_o W, \cr
\vert\nabla u\vert&=\lambda\qquad\qquad\rm{on}\ \ \partial_o W, \cr}
\eqno{(13)}
$$
and $u(x,-y)=u(x,y)$ for $(x,y)\in W$.
Let $E_1$ be the set of $\pm\vec i-$extrema of $\Gamma$,
$E_2$ be the set of $-\vec j-$extrema of $\Gamma$, and
$E=E_1\cup E_2$. Also let $I_1$ be the set of
$\pm\vec i-$inflection points, $I_2$ be the set of
$-\vec j-$inflection points, and $I=I_1\cup I_2$.
\hfil\break\noindent
Then every point $p\in E$ can be joined to a
point $p^*\in\Gamma^*$ by a directed
simple arc $\gamma_p\subset\overline{\Omega}$
(with $\gamma_p\cap\Omega$ piecewise $C^1)$
and every point $q\in I$ can be joined to two distinct points
$q^*$ and $q\sp{**}$ by directed simple
arcs $\sigma_q$ and $\sigma_{qq}$ in $\overline{\Omega}$
(with $\sigma_q\cap\Omega,\sigma_{qq}\cap\Omega$
piecewise $C^1)$ such that:
\item{(i)} If $p,q\in E$ and $p\neq q$, then $p^*\neq
q^*$ and $\gamma_p\cap\gamma_q=\emptyset$.
If $p\in E$ and $q\in I$, then $p^*,q^*,q\sp{**}$ are
distinct and the curves $\gamma_p,\sigma_q,\sigma_{qq}$
are disjoint.
If $p,q\in I$, then $p^*,p\sp{**},q^*,q\sp{**}$ are
distinct and the curves $\gamma_p,\gamma_{pp},\sigma_q,
\sigma_{qq}$ are disjoint.
\item{(ii)} If $p=(x,y)\in E$, $p^*=(x^*,y^*)$,
and $(s,t)\in\gamma_p$, then
$\nabla u(s,t)$ is parallel to $\vec n(x,y)$ and so
$\vec n(x^*,y^*)=\vec n(x,y)$.
\item{(iii)} If $q=(x,y)\in I$ and $\vec\nu=\vec n(x,y)$,
then $\nabla u(s,t)$ is parallel to $\vec\nu$ for
$(s,t)\in\sigma_q\cup\sigma_{qq}$ and $\Gamma^*$ has a
$\vec\nu-$minimum at $q^*$ and a $\vec\nu-$maximum at $q\sp{**}$.
\item{(iv)} If $(x,y)\in E$, $\vec\nu=\vec n(x,y)$, and
$(x,y)$ is a $\vec\nu-$minimum $(\vec\nu-$maximum) of $\Gamma$,
then $(x^*,y^*)$ is a $\vec\nu-$minimum $(\vec\nu-$maximum)
of $\Gamma^*$.
\item{(v)} Suppose $p=(x,y)\in E_1$ and $\vec\nu=\vec n(x,y)$.
If $(x,y)$ is a $\vec\nu-$minimum of $\Gamma$, then
$u_x^2$ is strictly increasing on $\gamma_p$,
$(q-p)\cdot\vec\nu>0$ for each point $q\in\gamma_p$ with
$q\neq p$, $\vert\nabla u(p^*)\vert>\lambda$, and
$0<(p^*-p)\cdot\vec\nu<{1\over{\lambda}}$.
\item{(vi)}
Suppose $p=(x,y)\in E_1$ and $\vec\nu=\vec n(x,y)$.
If $(x,y)$ is a $\vec\nu-$maximum of $\Gamma$, then
$u_x^2$ is strictly decreasing on $\gamma_p$,
$(p^*-q)\cdot\vec\nu>0$ for each point $q\in\gamma_p$ with
$q\neq p^*$, $\vert\nabla u(p^*)\vert<\lambda$, and
$(p^*-p)\cdot\vec\nu>{1\over{\lambda}}$.
\item{(vii)} If $p=(x,y)$ is a $-\vec j-$minimum of $\Gamma$ and
$p^*=(x^*,y^*)$, then $\Phi(y,u_y)$ is strictly
decreasing on $\gamma_p$,
$\Phi(y,-\lambda)\big(v^p(x_1^*,y_1^*)
-v^p(x,y)\big)<1$, and
$y>t$ for all points $q=(s,t)\in\gamma_p$ with $q\neq p$,
where $(x_1^*,y_1^*)$
is the first point of $\gamma_p$ at which $y_1^*=y^*$.
\item{(viii)} If $p=(x,y)$ is a $-\vec j-$maximum of $\Gamma$ and
$p^*=(x^*,y^*)$, then $\Phi(y,u_y)$ is strictly
increasing on $\gamma_p$,
$\Phi(y,-\lambda)\big(v^p(x^*,y^*)-v^p(x,y)\big)>1$,
and $t>y^*$
for all points $q=(s,t)\in\gamma_p$ with $q\notin\Gamma^*$.
Here
$$
v^p(s,t)=\int_{\gamma(s,t)} {u_y\over{\Phi(y,u_y)}}dy,
\quad (s,t)\in\gamma_p,
\eqno{(14)}
$$
and $\gamma(s,t)$ is the portion of $\gamma_p$ between
$(x_0,y_0)$ and $(s,t)$.\par
\proclaim Corollary.
Let $\Gamma^*,\lambda,\Gamma,\Omega$, and $u$ be as in
Theorem 3. Suppose $\Gamma^*$ is the graph of a $C^2$
function $g^*(x)$, $\Gamma^*=\{(x,g^*(x))\}$.
Then $\Gamma$ is the graph of a $C^2$ function $g(x)$
and each point $(x,g(x))$ at which $g$ has a relative
maximum (minimum) corresponds to a distinct point
$(x^*,g^*(x^*))$ at which $g^*$
has a relative maximum (minimum).\par
The proof of Theorem 2 will follow from Lemma 9, which does not depend
on assumption (10). The proof of Theorem 3 will make use of nine preliminary
lemmas, which constitute the bulk of the paper.
Specifically, Lemmas 1, 4, 8, and 9 consider properties of the
set $\{(x,y)\in{\overline{\Omega}}:u_y(x,y)=0\}$,
Lemmas 2, 3, 6, and 7 consider properties of the set
$\{(x,y)\in{\overline{\Omega}}:u_x(x,y)=0\}$, and Lemma 5
shows that the gradient of $u$ does not vanish on ${\overline{\Omega}}$.
Theorem 1 is a special case of Theorem 3.
\bigbreak
\centerline{\bf \S 2. EXAMPLES} \medskip\nobreak\noindent
\noindent{\bf Laplace's Equation.}
Suppose $G$ is the Laplacian, so that equation (2a) is
$$
U_{x_1x_1}+U_{x_2x_2}+U_{x_3x_3}=0;
\eqno{(15)}
$$
then equation (6a) becomes
$$
u_{xx}+u_{yy}+{1\over{y}}u_y=0
\eqno{(16)}
$$
and we observe that
$$
k(q)={1\over{q}},\quad l(y)={1\over{y}},\quad \Phi(y,q)=yq.
\eqno{(17)}
$$
Also $v(x,y)=\ln(y)$, the conclusions of Theorem 3 apply to solutions
of (6), and the condition
$\Phi(y_0,-\lambda)(v(x,y)-v(x_0,y_0))<(>) 1$ becomes
$y_0\exp(-(\lambda y_0)\sp{-1})<(>)y$.
\medskip
\noindent{\bf Minimal Surface Equation.}
Suppose $G$ is the minimal surface operator on $\Bbb R^3$, so that
equation (2a) becomes
$$
\big(1+\vert DU\vert^2\big)\sp{{3\over{2}}}
{\rm{div}}\bigg({DU\over{\sqrt{1+\vert DU\vert^2}}}\bigg)=0.
\eqno{(18)}
$$
The conclusions of Theorem 3 apply to solutions of (6),
since (6a) is
$$
(1+u_y^2)u_{xx}-2u_xu_yu_{xy}
+(1+u_x^2)u_{yy}+{1\over{y}}(1+u_x^2+u_y^2)u_y=0,
\eqno{(19)}
$$
and
$$
k(q)={1\over{q(1+q^2)}},\quad l(y)={1\over{y}},\quad
\Phi(y,q)={yq\over{\sqrt{1+q^2}}}.
\eqno{(20)}
$$
\medskip
\noindent{\bf A Contrived Equation.}
Suppose (2a) is
$$
U_{x_1x_1}+U_{x_2x_2}+U_{x_ 3x_3}
-{x_2\over{x_2^2+x_3^2}}U_{x_2}
-{x_3\over{x_2^2+x_3^2}}U_{x_3}=0.
\eqno{(21)}
$$
Then (6a) becomes
$$
u_{xx}+u_{yy}=0
\eqno{(22)}
$$
and the results of [2] imply that the geometry of $\Gamma$ is simpler
than that of $\Gamma^*$ with respect to all $\vec\nu-$extrema
of $\Gamma$.
\medskip
\noindent{\bf p-Laplace Equation.}
Suppose (2a) is
$$
{\rm{div}}\big(\vert DU\vert\sp{{\rm{p}}-2}DU\big)=0
\eqno{(23)}
$$
for p $>1$. Then (6a) becomes
$$
{\rm{div}}\big(\vert\nabla u\vert\sp{{\rm{p}}-2}\nabla u\big)
+{1\over{y}}\vert\nabla u\vert\sp{{\rm{p}}-2}u_y=0
\eqno{(24)}
$$
and
$$
k(q)={1\over{q}},\quad l(y)={1\over{({\rm{p}}-1)y}},\quad
\Phi(y,q)=y\sp{{\rm{p}}-1}q.
\eqno{(25)}
$$
If $U:{\overline{{\cal{O}}}}\to\Bbb R$ is a $C^2$ solution of (2)
with $\vert DU\vert>0$ on ${\overline{{\cal{O}}}}$, then the
conclusions of Theorem 3 apply to this solution.
\medskip
\noindent{\bf A Class of Operators.}
Suppose (2a) has the form
$GU=\sum_{i,j=1}^3 A_{ij}(X,DU(X))D_i D_j U$,
where $G$ is elliptic; hence $B\equiv 0$.
Let $({\cal{S}},U)$ be a solution of the Dirichlet problem (2) with
$U\in C^2({\cal{O}}\cup{\cal{S}})\cap C^1(\overline{{\cal{O}}})$.
If $U$ should happen to be axial-symmetric (with respect to the $x_1-$axis),
the conclusions of Theorem 2 would apply to this solution.
While our operator $G$ above appears to be quite general, the
assumption that $U$ is axial-symmetric may impose some symmetry condition
on $G$.
\bigbreak
\centerline{\bf \S 3. PRELIMINARY RESULTS} \medskip\nobreak\noindent
In \S 3 and \S 4, we will suppose the assumptions given at the beginning
of \S 1 hold. In particular, we assume $u$ is given by (5) and conditions
(9) and (10) hold. Notice, however, that Lemmas 1, 2, 4(a), 4(c), 5, 8,
and 9 do not depend on condition (10).
\vskip .2 true in
\proclaim Lemma 1.
Suppose $u\in C^2(\Omega)$ satisfies $Qu=0$ in $\Omega$.
Define $T_0=\{(x,y)\in\Omega:u_y(x,y)=0\}$.
Suppose $(x_0,y_0)\in T_0$, $\vert\nabla u(x_0,y_0)\vert
\neq 0$, and $D^2u(x_0,y_0)\neq \vec 0$.
Then locally near $(x_0,y_0)$, the set $T_0$ is a
simple, $C^1$ curve $\sigma$ which divides its complement into
two connected components on which $u_y$ has opposite signs.
Further, $u_x^2$ is strictly increasing on $\sigma$ if we
choose the forward direction on $\sigma$ such that $u_x u_y>0$
locally to the right of $\sigma$ (or $u_x u_y<0$ locally to the
left of $\sigma)$.\par
\proclaim Lemma 2. \hfil\break
(a.) Let $\gamma=\{(x,y)\in{\overline{\Omega}}:
u_x(x,y)=0\}$ and $\Sigma=\{(x,y)\in\Omega\cup\Gamma:
u_{xx}(x,y)=u_{xy}(x,y)=0\}$.
Then $\gamma\backslash\Sigma$ is dense in $\gamma$.
\item{(b.)} $\Gamma$ does not contain any line segments.\par
Let us define
$$
\phi(x,y)=\Phi(y,u_y(x,y))
\eqno{(26a)}
$$
and
$$
\psi(x,y)=u_x(x,y).
\eqno{(26b)}
$$
\proclaim Lemma 3.
Suppose $u\in C^2(\Omega)$ satisfies $Qu=0$ in $\Omega$.
Define $\Lambda_0=\{(x,y)\in\Omega:u_x(x,y)=0\}$.
Suppose $(x_0,y_0)\in\Lambda_0$,
$\vert\nabla u(x_0,y_0)\vert
\neq 0$, and $\vert\nabla u_x(x_0,y_0)\vert\neq 0$.
Then locally near $(x_0,y_0)$, the set $\Lambda_0$ is a
simple, $C^1$ curve $\gamma$ which divides its complement into
two connected components on which $u_x$ has opposite signs.
Further, $\phi$ is strictly decreasing on $\gamma$ if we
choose the forward direction on $\gamma$ such that $u_x>0$
locally to the right of $\gamma$ (or $u_x<0$ locally to the
left of $\gamma)$.\par
\proclaim Lemma 4. \hfil\break
(a.) Let $(x_0,y_0)\in W$ and suppose $u_y(x_0,y_0)=0$.
Define
$$
v(x,y)=u(x_0,y_0)+u_x(x_0,y_0)(x-x_0).
\eqno{(27)}
$$
Then there is an integer $n\ge 2$
such that the zeros of $u-v$ in a neighborhood of
$(x_0,y_0)$ lie on $n$ $C^1$ curves $\delta_1,\dots,
\delta_n$ which divide a neighborhood of $(x_0,y_0)$
into $2n$ disjoint open sectors such that $u-v$ has opposite
signs on adjacent sectors and $\vert\nabla(u-v)\vert\neq 0$
in a deleted neighborhood of $(x_0,y_0)$.
\hfil\break
(b.) Let $(x_1,y_1)\in W$. Suppose $u_x(x_1,y_1)=0$
and $\vert\nabla u_x(x_1,y_1)\vert=0$.
Then there is an integer $m\ge 2$
such that the zeros of $u_x$ in a neighborhood of
$(x_1,y_1)$ lie on $m$ $C^1$ curves $\gamma_1,\dots,
\gamma_m$ which divide a neighborhood of $(x_1,y_1)$
into $2m$ disjoint open sectors such that $u_x$ has opposite
signs on adjacent sectors and $\vert\nabla u_x\vert\neq 0$
in a deleted neighborhood of $(x_1,y_1)$.
\hfil\break
(c.) Let $(x_2,y_2)\in W$. Suppose $u_y(x_2,y_2)=0$
and $\vert\nabla u_y(x_2,y_2)\vert=0$.
Then there is an integer $m\ge 2$
such that the zeros of $u_y$ in a neighborhood of
$(x_2,y_2)$ lie on $m$ $C^1$ curves $\sigma_1,\dots,
\sigma_m$ which divide a neighborhood of $(x_2,y_2)$
into $2m$ disjoint open sectors such that $u_y$ has opposite
signs on adjacent sectors and $\vert\nabla u_y\vert\neq 0$
in a deleted neighborhood of $(x_1,y_1)$. \par
\proclaim Lemma 5.
$\vert\nabla u\vert>0$ on $\overline{\Omega}$.
\proclaim Lemma 6\hfil\break
(a.) Suppose $\Gamma$ has a $-\vec j-$minimum at $(x_0,y_0)\in\Gamma$,
$\gamma$ is a directed curve in $\overline{\Omega}$ starting at
$(x_0,y_0)$ along which $u_x=0$ and $\phi$ is strictly
decreasing. Then $yy^*$ for each point $(x,y)$ of $\gamma$
with $(x,y)\notin\Gamma^*$. \par
\proclaim Lemma 7.
Suppose $\Gamma$ has a $-\vec j-$maximum ($-\vec j-$minimum)
at $(x_0,y_0)\in\Gamma$.
Let $\Lambda=\{(x,y)\in{\overline{\Omega}}:u_x(x,y)=0\}$.
Then there exists a directed curve $\gamma$ in $\Lambda$
(with $\gamma\cap\Omega$ piecewise $C^1)$
starting at $(x_0,y_0)$ along
which $\phi$ is strictly increasing (decreasing) and which
is maximal in the sense that $\gamma=\sigma$ whenever $\sigma$
is a directed curve in $\Lambda$ starting at
$(x_0,y_0)$ along which $\phi$ is strictly increasing
(decreasing) and $\gamma\subset\sigma$.
Further, if $\gamma$ is any such curve, then:
\item{(a.)} $\gamma$ does not intersect itself and has no
terminal or accumulation points in $\Omega$.
\item{(b.)} $\gamma$ does not intersect the $x-$axis and intersects
$\Gamma^*$ only at points of $\Lambda$.
\item{(c.)} If $\Gamma$ has $-\vec j-$minimum at $(x_0,y_0)$,
then $\gamma$ does not return to $\Gamma$ after leaving
$(x_0,y_0)$ and terminates at a point
$(x_1,y_1)\in\Gamma^*$ at which $\Gamma^*$ has a
$-\vec j-$minimum. Further, $y_11$,
where $v(x,y)$ is defined as in (c.). \par
\proclaim Lemma 8.
\item{(a.)}
Suppose $\sigma$ is a directed curve in $\overline{\Omega}$
starting at $(x_0,y_0)\in\Gamma$ along which $u_y=0$,
$u_x^2$ is strictly increasing, and $u_x>0$ $(u_x<0)$.
For each point $(x,y)$ of $\sigma$ with $(x,y)\neq(x_0,y_0)$,
we have $x_0x)$.
\item{(b.)}
Suppose $\sigma$ is a directed curve in $\overline{\Omega}$
along which $u_y=0$, $u_x^2$ is strictly decreasing,
and $u_x>0$ $(u_x<0)$. Suppose $\sigma$ terminates
at a point $(x^*,y^*)\in\Gamma^*$.
For each point $(x,y)$ of $\sigma$ with $(x,y)\neq(x_0,y_0)$,
we have $x^*x)$. \par
\proclaim Lemma 9.
Suppose $T$ has a $\pm\vec i-$maximum ($\pm\vec i-$minimum)
at $(x_0,y_0)\in\Gamma$.
Let $\Sigma =\{(x,y)\in{\overline{\Omega}}:u_y(x,y)=0\}$.
Then there exists a directed curve $\sigma$ in $T$
(with $\sigma\cap\Omega$ piecewise $C^1)$
starting at $(x_0,y_0)$ along
which $u_x^2$ is strictly decreasing (increasing) and which
is maximal in the sense that $\sigma=\sigma_0$ whenever $\sigma_0$
is a directed curve in $\Sigma $ starting at
$(x_0,y_0)$ along which $u_x^2$ is strictly decreasing
(increasing) and $\sigma\subset\sigma_0$.
Further, if $\sigma$ is any such curve, then:
\item{(a.)} $\sigma$ does not intersect itself and has no
terminal or accumulation points in $\Omega$.
\item{(b.)} $\sigma$ does not intersect the $x-$axis and intersects
$\Gamma^*$ only at points of $T$.
\item{(c.)} Let $\vec\nu=\pm\vec i$.
If $\Gamma$ has $\vec\nu-$minimum at $p_0=(x_0,y_0)$,
then $\sigma$ does not return to $\Gamma$ after leaving
$(x_0,y_0)$ and terminates at a point $p^*=
(x^*,y^*)\in\Gamma^*$ at which $\Gamma^*$ has a
$\vec\nu-$minimum.
Also, $(p^*-p_0)\cdot\vec\nu<{1\over{\lambda}}$ and
$(q-p_0)\cdot\vec\nu>0$ for each $q=(x,y)\in\sigma$ with
$q\neq p_0$.
\item{(d.)} Let $\vec\nu=\pm\vec i$.
If $\Gamma$ has $\vec\nu-$maximum at $p_0=(x_0,y_0)$,
then $\sigma$ does not return to $\Gamma$ after leaving
$(x_0,y_0)$ and terminates at a point $p^*=
(x^*,y^*)\in\Gamma^*$ at which $\Gamma^*$ has a
$\vec\nu-$maximum.
Also, $(p^*-p_0)\cdot\vec\nu>{1\over{\lambda}}$ and
$(p^*-q)\cdot\vec\nu>0$ for each $q=(x,y)\in\sigma$ with
$q\neq p^*$. \par
\bigbreak
\centerline{\bf \S 4. PROOFS OF LEMMAS} \medskip\nobreak\noindent
\noindent{\bf Proof of Lemma 1.}
If we set $r=u_{xx}, s=u_{xy}, t=u_{yy}$, we see that
$0=r(ar+2bs+ct+d)$ and so
$rt-s^2=-{dr\over{c}}-{1\over{c}}(ar^2+2brs+cs^2)$.
Similarly, we see that
$rt-s^2=-{dt\over{a}}-{1\over{a}}(as^2+2bst+ct^2)$.
Recall that $d(x,y,p,q)=B(x,y,0,p,q,0)+{q\over{y}}A_{33}(x,y,0,p,q,0)$
and $B(x,y,0,p,0,0)=0$, so $d=0$ on $T_0$.
Since $Q$ is elliptic, $a\xi_1^2 + 2b\xi_1\xi_2
+ c\xi_2^2>0$ if and only if $\vec\xi=(\xi_1,\xi_2)\neq 0$.
Thus
$$
u_{xx}u_{yy}-u_{xy}^2\le 0 \quad{\rm{on}}\ \ T_0.
\eqno{(29)}
$$
Also, on $T_0$, $rt-s^2=0$ iff $r=s=t=0$.
Since $D^2u(x_0,y_0)\neq 0$, $rt-s^2<0$ near
$(x_0,y_0)$ on $T_0$.
Then $\vert\nabla u_y(x_0,y_0)\vert >0$ and so the
first part follows from the implicit function theorem. The monotonicity
of $u_x^2$ follows from Lemma 1 of [16] with the choice
$\alpha=0$ or $\alpha=\pi$. \hfill Q.E.D.
\bigbreak
\noindent {\bf Proof of Lemma 2.}
(a.) Notice that $\Sigma\neq{\overline{\Omega}}$ since $u$
cannot be a linear function.
Let int$(\gamma)$ be the interior of $\gamma$ in $\Bbb R^2$.
If int$(\gamma)\neq\emptyset$, then the proof of Theorem 8.19
of [13]\ implies $u_x\equiv 0$, which is a
contradiction. Suppose $\gamma\backslash\Sigma$ is not
dense in $\gamma$. Then there exists a connected set $K\subset
\gamma$ which is relatively open in $\gamma$ such that
$\overline{K} \subset \gamma\cap\Sigma$.
Choose a point $(x_1,y_1)\in \Omega\backslash\gamma$
such that dist$((x_1,y_1),K)<$dist$((x_1,y_1),
\partial\Omega\cup\gamma\backslash K)$,
which is possible since $\gamma$ is a closed set.
Let $r=$dist$((x_1,y_1),K)>0$ and let $B=
B((x_1,y_1),r)$. Then $\partial B\cap K
\neq\emptyset$ and $B\cap\gamma=\emptyset$.
Let $(x_2,y_2)\in\partial B\cap K$.
Then $u_x>0$ or $u_x<0$ in $B$ and
$u_x(x_2,y_2)=0;$ the Hopf boundary point lemma ([13])
implies ${\partial\over{\partial\vec\eta}}(u_x)\neq 0$ at
$(x_2,y_2)$, where
$\vec\eta$ is a unit normal direction to $\partial B$
at $(x_2,y_2)$. This contradicts the fact that
$u_{xx}=u_{xy}=0$ on $K$.
(b.) Suppose first that $\gamma$ is a line segment
parallel to the $x-$axis (i.e. a horizontal line segment).
Then $u_x=u_{xx}=0$ on $\lambda$. Also $u_y$ is constant
$(=\pm\lambda)$ on $\gamma$ and so $u_{xy}=0$ on $\gamma$, in
contradiction to (a.). If $\Gamma$ contains a line segment $\sigma$, we
may rotate $\Omega$ so that $\sigma$ is horizontal, thereby possibly
changing $Q$, and apply the argument above to obtain a contradiction.
\hfill Q.E.D.
\bigbreak
\noindent {\bf Proof of Lemma 3.}
The first part follows from the implicit function theorem.
Notice that $\nabla\psi=(u_{xx},u_{xy})$ and
$$
\nabla\phi=\bigg({\partial\Phi(y,u_y)\over{\partial q}}u_{xy},
{\partial\Phi(y,u_y)\over{\partial q}}u_{yy}+
{\partial\Phi(y,u_y)\over{\partial y}}\bigg).
\eqno{(30)}
$$
From the proof of Lemma 1, we see that
$$
u_{xx}u_{yy}-u_{xy}^2\le{-d\over{c}}u_{xx}
\eqno{(31)}
$$
with equality only when $u_{xx}=u_{xy}=0$.
Now $(\nabla\psi)\sp\bot\equiv (-u_{xy},u_{xx})$
is a forward tangent vector to $\gamma$ and
$$\eqalign{
\nabla\phi\cdot(\nabla\psi)\sp\bot
&={\partial\Phi\over{\partial q}}
(u_{xx}u_{yy}-u_{xy}^2) +
{\partial\Phi\over{\partial y}}u_{xx} \cr
&<\big({\partial\Phi\over{\partial y}}
-{d\over{c}} {\partial\Phi\over{\partial q}}\big)u_{xx} \cr
&=0 \cr}
\eqno{(32)}
$$
since ${\partial\Phi\over{\partial q}}>0$ and
$c(x,y,0,q)\Phi_y(y,q)=d(x,y,0,q)\Phi_q(y,q)(y,q)$.
\hfill Q.E.D.
\bigbreak
\noindent {\bf Proof of Lemma 4.}
Let us define the operator $R$ by
$$
Rw=a^0w_{xx}+2b^0w_{xy}+c^0w_{yy}+d,
\eqno{(33)}
$$
where $a^0(x,y)=a(x,y,u_x(x,y),u_y(x,y))$,
$b^0(x,y)=b(x,y,u_x(x,y),u_y(x,y))$, and
$c^0(x,y)=c(x,y,u_x(x,y),u_y(x,y))$.
Then $R(u-v)=0$ and $\vert\nabla(u-v)\vert=0$ at $(x_0,y_0)$.
Since $d(x,y,0,0)=0$, (a.) follows from the Proposition in [17].
Since $h_x(x,y,0,q)=0$ and $h\in C\sp{1,\delta}(\Bbb R^4)$,
$h_x(x,y,p,q)=ph_1(x,y,p,q)$ for some function
$h_1\in C\sp{0,\delta}(\Bbb R^4)$. Also, since
$u_x(x_1,y_1)=0$ and $u\in C\sp{2,\delta}(\Omega)$,
we see that
$$
u_x(x,y)=\mu(x,y)u_{xx}(x,y)+\chi(x,y)u_{xy}(x,y)
\eqno{(34)}
$$
for some functions $\mu,\chi\in C\sp{0,\delta}$
in some neighborhood of $(x_1,y_1)$ with
$\mu(x_1,y_1)=\chi(x_1,y_1)=0$. Now define
$$\eqalign{
e^0(x,y,p,q)&=c^0(x,y)\bigg(\bigg({a^0\over{c^0}}\bigg)_x
+h_p(x,y,u_x,u_y)
+\mu(x,y)h_1(x,y,u_x,u_y)\bigg)p \cr
&+c^0(x,y)\bigg(\bigg({2b^0\over{c^0}}\bigg)_x
+h_q(x,y,u_x,u_y)
+\chi(x,y)h_1(x,y,u_x,u_y)\bigg)q. \cr}
\eqno{(35)}
$$
Notice that $e^0(x,y,0,0)=0$.
Let $L$ be the linear operator given by
$$
Lw=a^0w_{xx}+2b^0w_{xy}+c^0w_{yy}+e^0,
\eqno{(36)}
$$
where $e^0=e^0(x,y,w_x,w_y)$.
As in [13], 13.2, we see that $L(u_x)=0$ and part (b.) follows
from the Proposition in [17].
The proof of (c.) follows in a manner similar to that of (b.).
\hfill Q.E.D.
\bigbreak
\noindent {\bf Proof of Lemma 5.}
Let us first observe that $00$ on $\Gamma^*$. To see
this, let $(x_0,y_0)\in\Gamma^*$ and define $w=1-u$. Then
$Rw=0$ as above, $w\ge 0$ on $\overline{\Omega}$, and
$w(x_0,y_0)=0$. The Hopf boundary point lemma then implies
$\vert\nabla u(x_0,y_0)\vert>0$ and so our observation holds.
Thus $\vert\nabla u\vert>0$ on $\partial W$.
Suppose $\vert\nabla u(x_0,y_0)\vert=0$ for some
$(x_0,y_0)\in W$ and set $z_0=u(x_0,y_0)$.
Then $0z_0$ on $\omega_2\cup\omega_4\cup
\dots\cup\omega_{2m}$. The Jordan curve theorem and the fact that
$W$ is an annular domain implies that there is a component $\omega$ of
$\{(x,y)\in W:u(x,y)\neq z_0\}$ whose closure does not intersect
$\partial W$ and the maximum principle implies $u\equiv z_0$
in $\omega$, in contradiction to the fact that $u\neq z_0$ in
$\omega$. Thus $\vert\nabla u\vert>0$ in $W$. \hfill Q.E.D.
\bigbreak
\noindent {\bf Proof of Lemma 6.} (a.)
Let $\gamma$ be directed curve in $\overline{\Omega}$ starting
at $(x_0,y_0)\in\Gamma$ along which $u_x=0$ and
$\phi$ is strictly decreasing.
From Lemma 2 (b.) and the fact that $\Gamma$ has a $-\vec j-$minimum
at $(x_0,y_0)$, we see that $\Gamma$ is ``strictly concave
down'' near $(x_0,y_0)$.
Assume the claim is false and $\gamma$ ends at a point
$(x_1,y_1)\in{\overline{\Omega}}$
with $y_1=y_0$ such that
$yFrom Lemma 4, we see that
$F=\{(x,y)\in\gamma: u_{xx}(x,y)=u_{xy}(x,y)=0\}$ is finite
in every compact subset of $\Omega$.
Since $\phi$ is strictly monotonic, $\gamma$ cannot intersect itself,
and so we see that $\gamma\cap\Omega$ is a piecewise $C^1$ curve(s).
We may write $\gamma=\{(x(t),y(t)):0\le t\le 1\}$, with
$(x(0),y(0))=(x_0,y_0)$ and $(x(1),y(1))=(x_1,y_1)$,
such that $x(\cdot),y(\cdot)\in C^0([0,1])\cap
C^1([0,1]\backslash D_1)$ and
$x'(t)=y'(t)=0$ if $t\in D_2$, where
$D_1=\{t\in[0,1]:(x(t),y(t))\in\partial\Omega\cap F\}$ and
$D_2=\{t\in[0,1]:(x(t),y(t))\in\Omega\cap F\};$
notice that $D_1$ is a discrete subset of $[0,1]\backslash D_1$.
Then $u_x(x(t),y(t))=0$ for $0\le t\le 1$ and $y(t)-\infty$, and
$C_2=\sup\sb\gamma {u\over{\phi^2}}<\infty$.
Suppose $(s,t)\subset [0,1]\backslash D_1$. Integration by parts
yields
$$
\int_s^t {(u(\vec\gamma(\tau)))'\over{\phi(\vec\gamma(\tau))}}
d\tau= {u(\vec\gamma(\tau))\over{\phi(\vec\gamma(\tau))}}\vert_s^t
+\int_s^t{u(\vec\gamma(\tau))\over{\phi^2(\vec\gamma(\tau))}}
\big(\phi(\vec\gamma(\tau))\big)'d\tau.
\eqno{(39)}
$$
Since $(\phi(\vec\gamma(\tau)))'\le 0$ and $u\ge 0$, we observe
that the second integral exists and equals $-\infty$ or a
finite nonpositive number. Using the fact that
$0\le {u\over{\phi^2}}\le C_2$ on $\gamma$, we obtain
$$
{u(\vec\gamma(\tau))\over{\phi(\vec\gamma(\tau))}}\vert_s^t
\ge \int_s^t{\big(u(\vec\gamma(\tau))\big)'\over
{\phi(\vec\gamma(\tau))}}d\tau
\ge {u(\vec\gamma(\tau))\over{\phi(\vec\gamma(\tau))}}\vert_s^t
+C_2\phi(\vec\gamma(\tau))\vert_s^t
\ge C_1+C_2\phi(\vec\gamma(\tau))\vert_0^1
\eqno{(40)}
$$
and therefore $\int_s^t {\big(u(\vec\gamma(\tau))\big)'\over
{\phi(\vec\gamma(\tau))}}d\tau$ is a well-defined, finite number.
Suppose $D_1=\{t_1,t_2,\dots,t_n\}$ with
$0\le t_1t$ and $y(s)=y(t)$, the facts that
$\phi(\vec\gamma(t))$ is decreasing in $t$ and
${\partial\Phi\over{\partial q}}(y,q)>0$ imply that
$u_y(\vec\gamma(t))>u_y(\vec\gamma(s))$.
Suppose first that $0\le t_10$ on $(t_3,t_4)\backslash D_1$,
$y(t_1)=y(t_4)$, and $y(t_2)=y(t_3)$.
For each $t\in (t_1,t_2)$, there is exactly one solution
in $(t_3,t_4)$ to the equation $y(\cdot)=y(t);$
let us denote this value by $s(t)$. Set $s(t_1)=t_4$ and
$s(t_2)=t_3$. Then $y(t)=y(s(t))$ for $t\in [t_1,t_2]$.
Notice that $y'(t)=y'(s(t))s'(t)$ and $s(t)>t$ for each
$t\in (t_1,t_2)\backslash D_1$. Then
$$\eqalign{
\int_{t_1}\sp{t_2} v'(t) dt
&=\int_{t_1}\sp{t_2} {u_y(\vec\gamma(t))y'(t)
\over{\phi(\vec\gamma(t))}} dt \cr
&\le \int_{t_1}\sp{t_2}
{u_y(\vec\gamma(s(t)))y'(s(t))s'(t)\over{\phi(\vec\gamma(s(t)))}}
dt \qquad\rm{by}\ \ (10g) \cr
&=\int_{t_4}\sp{t_3} {u_y(\vec\gamma(s))y'(s)
\over{\phi(\vec\gamma(s))}} ds \cr
&=-\int_{t_3}\sp{t_4} v'(s) ds \cr }
\eqno{(42)}
$$
and so $v(t_4)-v(t_1)+v(t_2)-v(t_1)
=\int_{t_1}\sp{t_2}v'(t) dt +\int_{t_3}\sp{t_4}v'(t) dt
\ge 0$.
Recall that $y(0)=y(1)$, $y(t)0$ on $(c_n,d_n)\backslash D_1$.
Since
$$
v(d_n)-v(c_n)+v(b_n)-v(a_n)\ge 0,
\eqno{(44)}
$$
we see that
$$
v(1)-v(0)=\sum_n \big(v(d_n)-v(c_n)+v(b_n)-v(a_n)\big)
\ge 0.
\eqno{(45)}
$$
(If $D_1\neq\emptyset$, then we do not know if $v\in L^1([0,1])$
and so we may need to modify our argument slightly. Lemma 5 implies that
for some $\epsilon_0>0$, $\{(x,y)\in\overline{\Omega}:{\rm{dist}}
((x,y),\gamma)\le\epsilon_0\}\cap\{(x,y)\in\overline{\Omega}:y=0
\ \ {\rm{or}}\ \ u_y(x,y)=0\}=\emptyset$. If $I$ is a compact
subset of $[0,1]\backslash D_1$, then $v\in L^1(I)$.
For $\epsilon\in (0,\epsilon_0)$, let $I\sb\epsilon=\cup_{j=1}
\sp{n+1}[t_{j-1}^+,t_j^-]$ be a compact subset of
$[0,1]\backslash D_1$ such that
$\sum_{j=1}\sp{n+1}(t_j^- -t_{j-1}^+)>1-\epsilon$,
$t_j^-0$
and $l(t)=0$ if $t\notin I\sb\epsilon$ with $y'\sb\epsilon(t)\le 0$,
where $C_3=\sup\{{u_y(x,y)\over{\phi(x,y)}}:
\rm{dist}((x,y),\gamma)\le\epsilon\}$.
Using an argument similar to that for (45), we see that
$$
\int_0^1 l(t)y'\sb\epsilon(t)\ dt\ge 0,
\eqno{(46)}
$$
since $l\ y\sb\epsilon'\in L^1([0,1])$, and so
$$
\int_{I\sb\epsilon}{u_y(\vec\gamma(t))\over{\phi(\vec\gamma(t))}}
y'(t)\ dt\ge -(n+1)C_3\epsilon.
\eqno{(47)}
$$
Letting $\epsilon\to 0$, we obtain $v(1)-v(0)\ge 0$.\ )
Now suppose $t\in(0,1);$ then $y(t)0$
when $y'(s)>0$. Now setting
$t_3=\sup\{s\in [t_2,1]:
y(\tau)\le y(t_2)$ for all $\tau\in[t_2,s]\}$,
obtaining $v(t_2)\le v(t_3)$, and continuing to argue in this
manner, we see that $v(t)t$ and $y(s)=y(t)$, the facts that
$\phi(\vec\gamma(t))$ is increasing in $t$ and
${\partial\Phi\over{\partial q}}(y,q)>0$ imply that
$u_y(\vec\gamma(t))y^*$ for $0v(1)$ for $00$ on the other side.
Setting $\Pi=\{(x,y)\in{\overline{\Omega}}:u_x(x,y)>0\}$ and
letting $\gamma_0$ be a component of $\Omega\cap\partial M$
with $(x_0,y_0)\in\gamma$, where $\gamma$ is the closure
of $\gamma_0$,
we see that there is at least one directed curve $\gamma$
from $(x_0,y_0)$ into $\Omega$ along which $u_x=0$.
Suppose $(x_0,y_0)$ is a $-\vec j-$minimum of $\Gamma;$
then $u_x>0$ locally to the left of $(x_0,y_0)$
on $\Gamma$ (i.e. preceeding $(x_0,y_0)$ on $\Gamma)$
and $u_x<0$ locally to the right (i.e. following
$(x_0,y_0)$ on $\Gamma)$. Let $F$ be as in the proof
of Lemma 6 (i.e. $F=\{u_{xx}=u_{xy}=0\}\cap\Lambda)$
and let $\gamma_1$ be the component of
${\overline{\partial\Pi\cap\Omega\backslash F}}$ which
contains $(x_0,y_0)$, where $\Pi$ is the component
of $\{(x,y)\in\Omega:u_x(x,y)>0\}$ whose closure contains
a portion of $\Gamma$ immediately preceeding $(x_0,y_0)$.
>From Lemma 3, we see that $\phi$ is strictly decreasing on
$\gamma_1$ and so $\gamma_1$ cannot return to
$(x_0,y_0)$. Now define $\gamma$ to contain
$\gamma_1$ and to be maximal with respect to forward
continuation in $\Lambda\backslash\gamma_1$ under the
condition that $\phi$ remain strictly decreasing.
>From Lemmas 3 and 4(b.), we see that $\gamma\cap\Omega$ is piecewise
$C^1$. If $(x_0,y_0)$ is a $-\vec j-$maximum,
then the existence of a curve $\gamma$ with the required
properties follows in a similar manner.
Let $\gamma$ be any maximal curve along which $u_x=0$ and
$\phi$ is strictly increasing (decreasing).
>From the monotonicity of $\phi$, we observe that $\gamma$
does not intersect itself. Suppose $\gamma$ terminates at a point
$(x_2,y_2)\in\Omega$. If $\vert\nabla u_x\vert\neq 0$
at $(x_2,y_2)$, then Lemma 3 implies that
$\gamma$ can be extended beyond $(x_2,y_2)$ with
$u_x<0$ $(u_x>0)$ locally to the right of $\gamma$
and $\phi$ is strictly increasing (decreasing) along $\gamma$,
in violation of the maximal property of $\gamma$.
Suppose $\vert\nabla u_x\vert=0$ at $(x_2,y_2)$.
For some $\epsilon>0$ and integer $m\ge 2$, Lemma 4 implies
$N=\{(x,y):\vert(x-x_2,y-y_2)\vert<\epsilon\}$
is contained in $\Omega$ and the set $\{(x,y):u_x(x,y)<0\}\cap N$
$(\ \{(x,y):u_x(x,y)>0\}\cap N\ )$
contains $m$ components; denote by $V$ the component whose
closure contains an interval of $\gamma$. Then we may extend
$\gamma$ by adding the component of $\partial V\cap N$
which includes $(x_2,y_2)$. Lemma 3 implies $\phi$
is strictly increasing (decreasing) on the extension of $\gamma$,
in contradiction to the maximal property of $\gamma$.
Thus $\gamma$ cannot terminate at a point of $\Omega$.
The fact that $\gamma$ has no accumulation points follows
from the characterization of the
points of $\gamma$ given by Lemmas 3 and 4; thus (a.) holds.
The monotonicity of $\phi$ implies $\gamma$ does not intersect
itself. Also, since $u\in C^1({\overline{\Omega}})$,
$\gamma\cap\Gamma^*\subset\Lambda$. Therefore (b.) holds.
Let $(x_0,y_0)$ be a $-\vec j-$minimum of $\Gamma$.
Suppose $\gamma$ intersects $\Gamma$ at a point $(x_2,y_2)
\neq (x_0,y_0)$.
>From Lemma 6, we see that $y_2\le y_0$. Since $u_y<0$
on $\gamma$, $u_y(x_0,y_0)=u_y(x_2,y_2)=-\lambda$.
Now $\Phi(y_0,u_y(x_0,y_0))>
\Phi(y_2,u_y(x_2,y_2))$ and so $y_00$,
$y>y_2$ when $(x,y)\in\Gamma^*$ satisfies
$\alpha-\epsilon0\}$ and
$L=\{(x,y)\in\Omega:u_x(x,y)<0\}$. Then $\partial P$
contains a neighborhood of $\Gamma^*$ immediately following
$(\beta,y_2)$ and $\partial L$ contains a neighborhood
of $\Gamma^*$ immediately preceeding $(\alpha,y_2)$.
Since $u_x>0$ locally to the right of $\gamma$ and $u_x<0$
locally to the left of $\gamma$ at all but a finite number of
points of $\gamma$, $\gamma$ may be extended beyond $(x_2,y_2)$
so that $u_x=0$ and $\phi$ is strictly decreasing on $\gamma;$
for example, if $P_1$ is the component of $P$ which lies to
the immediate right of $\gamma$ near $(x_2,y_2)$
and $\partial_1 P$ is the component of $\partial P_1$ which
contains $(x_2,y_2)$, then
$\gamma\cup\partial_1 P$ is one of the two possible extensions.
If $\Gamma^*$ has a $-\vec j-$inflection point at $(x_2,y_2)$,
then there is a (possibly degenerate) line segment $\sigma$ as above
and either $u_x<0$ locally preceeding $(\alpha,y_2)$ or
$u_x>0$ locally following $(\beta,y_2)$ on $\Gamma^*$.
Then $\gamma$ may be extended past $(x_2,y_2)$ either as
illustrated above (if $u_x<0$ locally preceeding $(\alpha,y_2)\ )$
or by replacing $\gamma$ by $\gamma\cup \partial_1 L$, where
$L_1$ is the component of $L$ which lies immediately to the
left of $\gamma$ near $(x_2,y_2)$ and $\partial_1 L$ is
the component of $\partial L_1$ which contains $(x_2,y_2)$.
Since $\gamma$ extends beyond $-\vec j-$maximum and
$-\vec j-$inflection
points of $\Gamma^*$, it must terminate at a $-\vec j-$minimum
of $\Gamma^*$. Further, if $\Gamma$ has a $-\vec j-$maximum at
$(x_0,y_0)$ and $\gamma$ does not return to $\Gamma$, then
a similar argument shows $\gamma$ must terminate at a point
$(x_1,y_1)\in\Gamma^*$ at which $\Gamma^*$ has a
$-\vec j-$maximum.
Let us consider the last part of (c.). Let $\gamma_0=
\{(x(t),y(t)):0\le t\le 1\}$ be the portion of $\gamma$ from
$\vec\gamma(0)=(x_0,y_0)$ to
$\vec\gamma(1)=(x_1^*,y_1^*)$. Notice that
$v(t)=v(\vec\gamma(t))$ is given by (36) and $v(t)>v(1)$ for
$0\le t<1$. Then
$$\eqalign{
1&\ge u(x_1^*,y_1^*)-u(x_0,y_0) \cr
&=v(t)\phi(\vec\gamma(t))\vert_0^1
-\int_0^1 v(t)\big(\phi(\vec\gamma(t)))' dt \cr
&>\big(v(t)\phi(\vec\gamma(t))-v(1)\phi(\vec\gamma(t))\big)
\vert_0^1 \cr
&=\phi(x_0,y_0)(v(1)-v(0)). \cr}
\eqno{(49)}
$$
Together with Lemma 6(a.), this implies (c.). To see that the last
part of (d.) is valid, notice that $v(t)x_0$ $(x(t)0$ $(u_x<0)$ on $\sigma$. If we set
$\vec\gamma(t)=(x(t),y(t))$, we see
$$\eqalign{
0&\le u(p_1)-u(p_0) \cr
&=x(t)u_x(\vec\gamma(t))\vert_0^1
-\int_0^1 x(t)d\big(u_x(\vec\gamma(t))\big) \cr
&<\big(x(t)u_x(\vec\gamma(t))
-x(0)u_x(\vec\gamma(t))\big)\vert_0^1 \cr
&=u_x(p_1)(x_1-x_0)=0. \cr}
\eqno{(51)}
$$
This contradiction implies $x>x_0$ $(xx^*$ $(x(t)0$ $(u_x<0)$ on $\sigma$. Then
$u(p^*)=1$ and so
$$\eqalign{
0&\le u(p^*)-u(p_1) \cr
&<\big(x(t)u_x(\vec\sigma(t))
-x^* u_x(\vec\sigma(t))\big)\vert_0^1 \cr
&=u_x(p_1)(x^*-x_1)=0. \cr}
\eqno{(52)}
$$
This contradiction implies (b.) holds.
\hfill Q.E.D.
\bigbreak
\noindent {\bf Proof of Lemma 9.}
Using Lemmas 1 and 4 in a manner
similar to the proof of Lemma 7, we see that there is a maximal
directed curve $\sigma$ along which
$u_y=0$ and $u_x^2$ is strictly monotonic as claimed.
By Lemma 1, $\sigma$ does not intersect itself.
Let us suppose for a moment that $\sigma$ does not
intersect the $x-$axis. The monotonicity of $u_x^2$
implies $\sigma$ does not return to $\Gamma$.
The fact that $\sigma$ has no terminal or accumulation
points in $\Omega$ follows as in Lemma 6 from Lemmas 1 and 4.
The fact that $\sigma$ terminates at a point $p^*=(x^*,y^*)$
at which $\Gamma^*$ has a $\pm\vec i-$extrema
of the same type as that of $\Gamma$ at $(x_0,y_0)$
follows as in the proof of Lemma 7.
Let us consider the claim that $\sigma$ does not intersect
the $x-$axis. Let $V=\{(x,0)\in{\overline{\Omega}}\}$ and
$p\sp\pm =(x\sp\pm,0)\in\partial_o W$ with $x^+0$ on
$\Gamma^+$. From Lemma 7, we see that there is a curve
$\gamma^0$ starting at $p^0$ and ending at a point $p^1$
of $\Gamma^*$ such that $u_x=0$ on $\gamma^0$ and
$\Phi(y,u_y)$ is strictly decreasing along $\gamma^0$.
>From Lemma 5, we see that any curve in $\overline{\Omega}$
along which $u_y=0$ cannot intersect $\gamma^0$.
Suppose $\sigma$ intersects $V^+$. Then
$p_0=(x_0,y_0)\in\Gamma^+$, since $\sigma$
cannot cross $\gamma^0$, and so $p_0$ is a $\vec i-$extrema
of $\Gamma$. Let $p_1=(x_1,0)$ be the first point at which
$\sigma$ intersects $V^+$ and denote by $\sigma_1$
the portion of $\sigma$ between $p_0$ and $p_1$.
Let $W=\{(x,y)\in\Omega:u_y(x,y)\neq 0\}$ and let
$W_0$ be any component of $W$ contained in the open
set bounded by $\sigma_1$, the portion of $\Gamma$ between
$p^+$ and $p_0$, and the portion of $V^+$ between
$p^+$ and $p_1$. Let us suppose that $u_y>0$ in
$W_0$. If we orient $\partial W_0$ so $W_0$ lies
to the right, then $u_x^2$ is strictly increasing on
$\partial W_0\backslash\Gamma$. Since $u_x=\lambda$
at $p^+$ and at each point of $\Gamma^+$ at which $u_y=0$,
we have a contradiction. Therefore $\sigma$ does not intersect
$V^+;$ similar reasoning implies $\sigma$ does not intersect
$V^-$.
The proof of the lemma will be complete once we have shown that
$(p^*-p_0)\cdot\vec\nu<1$ if $p_0$ is a $\vec\nu-$minimum
of $\Gamma$ and $(p^*-p_0)\cdot\vec\nu>1$
if $p_0$ is a $\vec\nu-$maximum of $\Gamma$, since
Lemma 8 implies $(q-p_0)\cdot\vec\nu>0$ for $q\in\sigma$,
$q\neq p_0$ when $p_0$ is a $\vec\nu-$minimum and
$(p^*-q)\cdot\vec\nu>0$ for $q\in\sigma$,
$q\neq p^*$ when $p_0$ is a $\vec\nu-$maximum.
Suppose first that $p_0$ is a $\vec\nu-$minimum and let
$p_1=(x_1,y_1)$ be the first point on $\sigma$ at which
$x_1=x^*$. Let $\sigma_1$ be the portion of $\sigma$
between $p_0$ and $p_1$ and write
$\sigma_1=\{(x(t),y(t)):0\le t\le 1\}$. Then $x(t)>x_0$
if $\vec\nu=\vec i$ and $x(t)0$. If we write $\vec\sigma(t)=(x(t),y(t))$, we have
$$\eqalign{
1&= u(p_1)-u(p_0)
=\int_0^1u_x(\vec\sigma(t))x'(t)\ dt \cr
&=x(t)u_x(\vec\sigma(t))\vert_0^1
-\int_0^1 x(t)\ d\big(u_x(\vec\sigma(t))\big) \cr
&>x(t)u_x(\vec\sigma(t))
-x^*\big(u_x(p_1)-u_x(p_0)\big)\vert_0^1\cr
&=u_x(p_0)(x^*-x_0)=\lambda(p^*-p_0)\cdot\vec\nu. \cr}
\eqno{(53)}
$$
Suppose next that $p_0$ is a $\vec\nu-$maximum. Let us write
$\sigma=\{(x(t),y(t)):0\le t\le 1\}$. Then $x(t)x^*$ if $\vec\nu=-\vec i$
for each $t>0$. We have
$$\eqalign{
1&=u(p_1)-u(p_0)
=x(t)u_x(\vec\sigma(t))\vert_0^1
-\int_0^1 x(t)\ d\big(u_x(\vec\sigma(t))\big) \cr
&y$ and $y_2>y$.
Suppose $\gamma=\gamma_p$ returns to $\Gamma$ at a point
$q=(s,t)\in\Gamma$. Since $\phi(q)>\phi(p)$, $u_y(q)=u_y(p)=
-\lambda$, and ${\partial\Phi\over{\partial y}}<0$, we see that
$\Phi(t,-\lambda)>\Phi(y,-\lambda)$ and so $t\phi(q)>\phi(p)$ and
$u_y(p_1)=u_y(p)=-\lambda$, we see that
$\Phi(y_1,-\lambda)>\Phi(y,-\lambda)$ and so $y_1y$ and implies $\gamma$
and $\gamma_1$ do not intersect. Therefore $\gamma$ ends at
a $-\vec j-$maximum of $\Gamma^*$. The fact that $\gamma$
lies between $\gamma_1$ and $\gamma_2$ and does not
intersect either of them implies that $\gamma_1$ and
$\gamma_2$ cannot intersect. Similarly, if $p_3$ is a
$-\vec j-$maximum of $\Gamma$ and $p_3\neq p$, then
$\gamma_{p_3}$ and $\gamma$ cannot intersect.
Using the monotonicity of $u_x^2$ or $\phi$ on $\gamma_p$
and Lemma 5 as above, we see that if $p,q\in E$
with $p\neq q$, then $\gamma_p\cap\gamma_q=\emptyset$.
Using Lemmas 7 and 9, we see that the claims involving points
$p\in E$ follow.
We will next consider those claims which refer only to points $q\in I;$
these include (iii) and a portion of (i).
Let $q\in I_1;$ then $u_y(q)=0$ and $u_y(p)\ge (\le) 0$
for $p\in\Gamma$ near $q$. Suppose, for a moment, that $u_y>(<) 0$
in $\Omega\cap V$, where $V$ is a deleted neighborhood of $q$.
Since $\vert\nabla u\vert=\lambda$, Lemma 2(b.) implies $u_y>(<) 0$
on $\Gamma\cap V$. Thus $u_y$ has a local minimum (maximum) at $q$.
As in the proof of Lemma 4, we see that $u_y$ is the solution of the
linear equation
$$
Mw=a^0w_{xx}+b^0w_{xy}+c^0w_{yy}+f^0,
\eqno{(55)}
$$
where $a^0,b^0,c^0$ are as in the proof of Lemma 4,
$f^0=f^0(x,y,w_x,w_y)$, and $f^0$ is defined in a
similar manner to the definition of $e^0$ in (35).
Since $u\in C^2({\overline{\Omega\cap V}})$, $M$ is uniformly
elliptic near $q$. The Hopf boundary point lemma
then implies $u_{xy}(q)=(u_y)_x(q)\neq 0$. Now $u_x^2$
equals $\lambda^2$ at $q$, $u_x^2<\lambda^2$ on
$\Gamma\cap V$, and $\vec j$ is a tangent vector to $\Gamma$ at
$q$, so $u_{xy}(q)=(u_x)_y(q)=0$, a contradiction.
Therefore $u_y$ changes signs in $\Omega\cap V$ for every
neighborhood $V$ of $q$.
Let us assume $q$ is a $-\vec i-$inflection point (so $u_x(q)<0)$
and $u_y\le 0$ on $\Gamma$ near $q$.
Let $\Pi$ be a component of $\{(x,y)\in\Omega:u_y(x,y)>0\}$
whose closure contains $q$ and let $\sigma_1$ and $\sigma_2$
be distinct directed curves which are the closures of components of
$\partial\Pi\cap\Omega\backslash F$ and each begin at $q$. In fact,
we may assume that, in some neighborhood of $q$,
$u_y<0$ between $\sigma_1$ and $\Gamma_q^+$ and
$u_y>0$ between $\sigma_1$ and $\sigma_2$, where
$\Gamma_q^+$ is the portion of $\Gamma$ following $q$.
Thus $\sigma_1$ is the curve in $T_0$ adjacent to
$\Gamma_q^+$ and $\sigma_2$ is the curve adjacent to
$\sigma_1$. Now let $\sigma_1$ $(\sigma_2)$ represent
a maximal extension of $\sigma_1$ $(\sigma_2)$ with respect
to forward continuation under the conditions that $u_y=0$ and
$u_x^2$ be strictly decreasing (increasing) on $\sigma_1$
$(\sigma_2)$. As in the proof of Lemma 9, we see that
$\sigma_1$ terminates at a point $q^*\in\Gamma^*$ at
which $\Gamma^*$ has a $-\vec i-$maximum, $\sigma_2$
terminates at a point $q\sp{**}\in\Gamma^*$ at which $\Gamma^*$
has a $-\vec i-$minimum, and $q^*$ follows $q\sp{**}$ on
$\Gamma^*$.
Suppose now that $q$ is any element of $I_1$ and $\vec\nu=
\vec n(q)$. In a similar manner to that above, we see there are
two directed simple curves $\sigma_1$
and $\sigma_2$ and two points $q^*$ and $q\sp{**}$ on
$\Gamma^*$ such that $\Gamma^*$ has a $\vec\nu-$minimum
at one of these points and has a $\vec\nu-$maximum at the other.
Using the monotonicity of $u_x^2$, it is easy to see that
the points $p_1^*,p_2^*,\dots,p_k^*$, $q_1^*,
q_1\sp{**},\dots,q_l^*,q_l\sp{**}$ are all distinct
$\pm\vec i-$extrema of $\Gamma^*$ if
$p_1,p_2,\dots,p_k\in E_1$ and
$q_1,\dots,q_l\in I_1$.
Suppose $q=(x_0,y_0)\in I_2$. As above, the Hopf boundary
point lemma implies there are two curves $\sigma_1$ and
$\sigma_2$ starting at $q$ along which $u_x=0$ such that
$\phi$ is strictly increasing on $\sigma_1$ and strictly
decreasing on $\sigma_2$. Let $\sigma_1$ and $\sigma_2$
denote maximal extensions (with respect to forward continuation
under the conditions $u_x=0$ and $\phi$ be strictly monotonic)
of $\sigma_1$ and $\sigma_2$. Once we know that $\sigma_1$
and $\sigma_2$ do not return to $\Gamma$ after leaving $q$,
the remainder of the proof follows as for the case of
$\pm\vec i-$inflection points. Suppose $u_x(q)=0$ and
$u_x\le 0$ on $\Gamma$ near $q$. We may assume that, in
a neighborhood of $q$, $u_x<0$ between $\sigma_1$ and
$\Gamma_q^+$ and $u_x>0$ between $\sigma_1$ and
$\sigma_2$. Then $\phi$ is decreasing on $\sigma_1$ and
increasing on $\sigma_2;$ hence $\sigma_1$ can only return
to $\Gamma$ at a point above the line $y=y_0$ and $\sigma_2$
can return to $\Gamma$ at a point below $y=y_0$. Since
$\sigma_1$ lies to the right of $\sigma_2$, $\sigma_1$
and $\sigma_2$ cannot intersect any of the curves $\gamma_p$
for $p\in E$, $\Gamma_q^+\cap V$ lies below the line $y=y_0$,
and $\Gamma_q^-\cap V$ lies above $y=y_0$, where
$\Gamma_q^+$ $(\Gamma_q^-)$ is the portion of $\Gamma$
following (preceeding) $q$ and $V$ is some neighborhood of $q$,
we see that $\sigma_2$ and
$\sigma_2$ cannot return to $\Gamma$. The case when
$u_x(q)=0$ and $u_x\ge 0$ on $\Gamma$ near $q$ is similar.
Finally, we will consider the proof of the remaining portion of (i)
concerning points $p\in E$ and $q\in I$.
We claim that if $p,q\in E\cup I$ and $p\neq q$, then $\gamma_p$
(or $\sigma_p)$ and $\gamma_q$ (or $\sigma_q)$ are disjoint.
If $p$ is a $\vec\nu$ extreme or inflection point, $q$ is a
$\vec\mu$ extreme or inflection point, and the appropriate curves
starting at $p$ and $q$ intersect, then Lemma 5 implies
$\vec\nu=\vec\mu$. To illustrate that two such curves cannot meet,
suppose $p$ is a $\vec i-$minimum, $q$ is the next point (i.e.
following on $\Gamma)$, and $q$ is a $\vec i-$inflection point.
Then $u_y<0$ between $p$ and $q$ on $\Gamma$. If $\sigma_q$
is the first curve (i.e. adjacent to $\Gamma_q^-)$ leaving
$q$ along which $u_y=0$, then $u_x^2$ is strictly
increasing along $\gamma_p$ and strictly decreasing along
$\sigma_q$. Since $u_x(p)=u_x(q)=\lambda$, we see
that $\gamma_p\cap\sigma_q=\emptyset$. The general case
is similar and so the theorem follows.
\hfill Q.E.D.
\vskip .2 true in
\noindent
{\bf{Proof of Theorem 2:}}
If $\vec\nu=\pm\vec i$, $q\in\Gamma$, and $\Gamma$ has a
$\vec\nu-$extrema or $\vec\nu-$inflection point at $q$, then,
as in Lemma 9, there is a curve $\gamma$ in ${\overline{\Omega}}$
starting at $q$ along which $u_y=0$ and ending at a point
$q^*\in\Gamma^*$. Since there are no points on $\Gamma^*$
at which $u_y=0$, we see that $\Gamma$ has no $\vec\nu$
extreme or inflection points; the theorem then follows.
\hfill Q.E.D.
\vskip .2 true in
In fact, if we do not assume conditions (10b)-(10g) hold but otherwise
assume the hypotheses of Theorem 3, then
we claim that the conclusions of
the theorem which concern $\pm\vec i$ extreme and inflection points
of $\Gamma$ continue to hold. To see this, recall
that Lemmas 1,2,4(a.),4(c.),5,8, and 9 do not depend on
condition (10). The proof of our claim will follow from
the proof of the theorem once we know that none of the curves
$\gamma_p$ or $\sigma_q$ can intersect the $x-$axis.
If $p$ is a $\vec i-$extrema or $\vec i-$inflection point of $\Gamma$,
then $\gamma_p$ cannot intersect
$V^+$ as in the proof of Lemma 9 and cannot intersect $V^-$
since $u_x>0$ on $\gamma_p$ and, from Lemma 5, $u_x<0$
on $V^-$. Similarly, if $p$ is a $-\vec i$ extreme or inflection
point, then $\gamma_p$ cannot intersect the $x-$axis.
\bigbreak
\centerline{\bf \S 6. CONCLUDING REMARKS} \medskip\nobreak\noindent
The results of this paper were obtained
by the first author ([5]) in the case where $G$ is the Laplacian and
${\cal{O}}$ has analytic boundary and were extended to the general case of
smooth domains and equations satisfying conditions (9) and (10)
by the second author. We regard this work as one extension
to three dimensions of the qualitative theory for two-dimensional
free boundary problems given in [1]-[4], [7], [8], [11], [16], and [18].
Other authors might consider additional free boundary problems in
$\Bbb R^N$ $(N\ge 3)$
which reduce to two-dimensional free boundary problems,
determine conditions on the partial differential operator and/or the
boundary conditions which allow one to compare the $\vec\nu-$extrema
of the free boundary with the $\vec\nu-$extrema of the fixed boundary,
and so obtain other extensions of the two-dimensional qualitative theory.
It would be interesting to determine genuine three-dimensional (or
$N-$dimensional) generalizations of the qualitative geometric theory.
While it seems unlikely that a relationship between $\vec\nu-$extrema
of the fixed and free boundaries exists in some generic sense (e.g. for
almost all operators $G$ and almost all fixed boundaries ${\cal{S}}^*)$,
perhaps other kinds of geometric information, such as sectional or mean
curvature, of the free and fixed boundaries can be compared. Unfortunately,
we have no idea at the moment of an appropriate genuine higher dimensional
generalization of this work.
\bigbreak
\centerline{\bf REFERENCES} \medskip\nobreak\noindent
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\bigskip \noindent
Andrew Acker \& Kirk Lancaster \hfil\break
Department of Mathematics and Statistics\hfil\break
Wichita State University\hfil\break
Wichita, Kansas 67260-0033\hfil\break
E-mail address: acker@twsuvm.uc.twsu.edu\hfil\break
E-mail address: lancaste@twsuvm.uc.twsu.edu
\vfill
\end