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\newtheorem{claim}{Claim}[section]
\newtheorem{corollary}{Corollary}[section]
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\title{Limit Surfaces of Riemann Examples}
\author{David Hoffman, Wayne Rossman}

\date{}

\maketitle

\section{Introduction}

The only connected minimal surfaces foliated by circles and lines are
domains on one
of the following surfaces: the helicoid, the catenoid, the plane, and the
examples of Riemann (\cite{Ri} p329-33, \cite{En} p403-6,
\cite{Ni} p85-6).  All these surfaces are complete and embedded.
Topologically they are planar domains: the helicoid is simply-connected,
the catenoid is an annulus (conformally a twice-punctured sphere), and each
Riemann example (see Figure 1) is conformal to the plane minus the points
$\{ (n,0), \, (\frac{1}{n},0) \, | \, n \in \bfZ \}$ \cite{HKR}.
In this section we will
show that the plane, helicoid, and catenoid arise naturally as limits of
well-chosen and properly normalized sequences of Riemann examples.
The local behavior of domains on Riemann examples accounts for
the existence of these limits, thus allowing a change of topology in
the limit surfaces.

\begin{theorem}
	Any sequence of Riemann examples that converges to a surface in
$\bfR^3$ converges either to a Riemann example, a helicoid, a catenoid,
an infinite set of equally spaced parallel planes, or a single plane.
Furthermore, there exist sequences of Riemann examples that converge
to each of these possibilities.
\end{theorem}

\begin{figure}[t]
\hspace{1.3in}

\epsfysize = 2.5in	\epsffile{fig1.eps}
\hfill \break

\caption{The Riemann Example ${\cal R}_\lambda, \; \lambda = 1$}
\end{figure}

In order to be precise, we state what we mean by convergence of
surfaces in $\bfR^3$ in this context.

\begin{definition}
	A sequence of surfaces $\{{\cal S}_i\}_{i=1}^\infty$
{\em converges} as $i \rightarrow \infty$ to a surface $\cal S$ in
$\bfR^3$ if,
for any compact region $B \subseteq \bfR^3$, there exists an
integer $N_B$ such that for $i > N_B$, ${\cal S}_i \cap B$
is a normal graph over
$\cal S$, and $\{ {\cal S}_i \cap B \}_{i=N_B}^\infty$
converges to $\cal S$ in the $C^\infty$-topology.
\end{definition}

One can define analogously the convergence of a one-parameter family
${\cal S}_t, t \in \bfR$, as
$t \rightarrow t_0$.  Since this convergence is in the
$C^\infty$-topology, it requires that derivatives of all orders
of the coordinate functions on
the graphs converge to the corresponding derivatives of the coordinate
functions on the limit-surface.

We observe that catenoids, helicoids, and planes are
limit-surfaces of sequences of Riemann examples.  Therefore the
collection of connected embedded minimal surfaces foliated
by circles and lines is itself connected in the topology associated with
this convergence.

We wish to thank Rob Kusner, Pascal Romon, Ed Thayer, Johannes Nitsche
and Harold Rosenberg for helpful
conversations.

\section{The Family of Riemann Examples}

\subsection{Definition of the Riemann Examples}

	We now describe properties of the Riemann examples, and we give a
Weierstrass representation for these surfaces.  Proofs of the
statements in this section may be found in \cite{HKR}.

	For any Riemann example $\cal R$, there exists a translation $T$ of
$\bfR^3$ so that $T$: leaves $\cal R$ invariant; is orientation-preserving on
$\cal R$; generates the full cyclic orientation-preserving
translation symmetry group of
$\cal R$.  The quotient-surface ${\cal R}/T$ is a twice-punctured
rectangular torus.  Thus,
Riemann examples can be parametrized via a Weierstrass
representation on such a torus.
Fix $\lambda > 0$.  We first define $\bar{M_\lambda}$:
\begin{equation}
\bar{M_\lambda} = \{ \; (z,w) \in (\bfC \cup \{\infty\})^2 :
w^2=z(z-\lambda)(z+\frac{1}{\lambda}) \; \} \; .
\end{equation}
Let $M_\lambda$ be the twice-punctured torus
\begin{equation}
M_\lambda = \bar{M_\lambda} \setminus \{ (0,0), \, (\infty,\infty) \} \; .
\end{equation}
With the Weierstrass data
\begin{equation}
g(z,w) = z \; , \; \eta(z,w) = \frac{dz}{zw} \; ,
\end{equation}
the Weierstrass representation for the Riemann examples
${\cal R}_\lambda$ is
\begin{equation}
R_\lambda(p) := Re\int_{p_0}^{p} \Phi_{(g,\eta)} \, dz \; , \; \; \;
\Phi_{(g,\eta)} := \left( \begin{array}{c}
	\frac{(1-z^2)dz}{zw} \\
	\frac{i(1+z^2)dz}{zw} \\
	\frac{2dz}{w}
	\end{array}
\right) \; , \; \; \; p \in M_\lambda \; .
\end{equation}
The family ${\cal R}_\lambda$ of Riemann examples is parametrized by
$\lambda \in (0,\infty)$.

	The surface $M_\lambda$ can be considered to be a double covering
of the punctured $z$-plane, branched at $\lambda$ and $\frac{-1}{\lambda}$.
Consider a circle $\alpha$ in the $z$-plane with center
$\frac{-1}{2\lambda}$ and radius $\frac{1}{2}(\lambda + \frac{1}{\lambda})$,
and lift $\alpha$ to a closed curve $\hat{\alpha}$ in $M_\lambda$.
The Weierstrass integral about $\hat{\alpha}$ results in a
nonzero period-vector, and this is the translation $T$.

	Letting $\acute{z} = -z$ and $\acute{w} = iw$, we have
\[ \acute{w}^2 = \acute{z}(\acute{z} - \frac{1}{\lambda})
(\acute{z} + \lambda) \; . \]
Thus we may consider the map $(z,w) \rightarrow (\acute{z},
\acute{w})$ to be a conformal diffeomorphism between $M_\lambda$ and
$M_{\frac{1}{\lambda}}$.

	Because the metric on ${\cal R}_\lambda$ (\cite{Os} p65,
\cite{HKR}) is
\[ ds^2 = \frac{(1 + |z|^2)^2}{4|z|^2|w|^2} |dz|^2 \; , \]
the maps $(z,w) \rightarrow (\bar{z},\bar{w})$ and
$(z,w) \rightarrow (\bar{z},-\bar{w})$ both restrict to isometries of
$M_\lambda$ in the induced metric.  The set
${\cal A} = \{ (t,w) \in M_\lambda \, | \, t \in \bfR \}$, which consists of
the union of their fixed-point sets, is mapped by the Weierstrass integral to
geodesics in ${\cal R}_\lambda$.  It is clear from the
Weierstrass integral that the intervals in $\cal A$ of the form
$t \leq \frac{-1}{\lambda}$ or $0 < t \leq \lambda$ represent lines in
${\cal R}_\lambda$,
while intervals in $\cal A$ of the form $\frac{-1}{\lambda}
\leq t < 0$ or $t \geq \lambda$ represent planar geodesics in
${\cal R}_\lambda$.  Planar geodesics must be principal curves.

	The surfaces ${\cal R}_\lambda$
have the following additional properties:
\newcounter{num}
\begin{list}
{\arabic{num})}{\usecounter{num}\setlength{\rightmargin}{\leftmargin}}

\item ${\cal R}_\lambda$ is
foliated by circles and lines in horizontal planes.
All the lines in ${\cal R}_\lambda$ are the image
under the Weierstrass representation of points in $\cal A$ such
that $t \leq \frac{-1}{\lambda}$ or $0 < t \leq \lambda$, which is the
fixed-point set of the map $(z,w(z)) \rightarrow (\bar{z},-\bar{w}(z))$.

\item ${\cal R}_\lambda$ has an infinite number of equally spaced horizontal
flat ends.  These ends correspond to the punctures of $\bar{M_\lambda}$ at
$z = 0, \infty$.

\item ${\cal R}_\lambda$ is invariant under
reflection in a plane parallel to the
$(x_1,x_3)$-plane.  The intersection of this plane with ${\cal R}_\lambda$
consists of the image
under the Weierstrass representation of points in $\cal A$ such
that $\frac{-1}{\lambda} \leq z < 0$ or $z \geq \lambda$, which is the
fixed-point set of the map $(z,w(z)) \rightarrow (\bar{z},\bar{w}(z))$.

\item ${\cal R}_\lambda$ is
invariant under rotation about horizontal lines that are
perpendicular to the $(x_1,x_3)$-plane, and meet the surface orthogonally at
the points where $g(z) = \pm i$.
The map $(z,w) \rightarrow (\frac{-1}{z},\frac{w}{z^2})$ restricted to
$M_\lambda$ represents this rotation.

\end{list}

We will consider a Riemann example to be any surface that is the image of one
of these surfaces ${\cal R}_\lambda$
under an isometry or homothety of $\bfR^3$.

\subsection{Conjugate Pairs of Riemann Examples}

\begin{proposition}
	For any positive $\lambda \neq 1$, the Riemann examples
${\cal R}_\lambda$ and
${\cal R}_{\frac{1}{\lambda}}$ are conjugate but not congruent.
${\cal R}_1$ is self-conjugate.
\end{proposition}

\begin{proof}
	Choose $\lambda > 1$.  The conjugate surface of ${\cal R}_\lambda$ has
Weierstrass data
\[g = z \; , \] \[i\eta = \frac{idz}{zw} = \frac{dz}{(-z)(iw)} \; . \]
The Weierstrass data for the conjugate surface,
expressed in terms of $(\acute{z},\acute{w}) = (-z,iw)$ on
$M_{\frac{1}{\lambda}}$, is \[\tilde{g}(\acute{z},\acute{w}) = -\acute{z}
\; , \] \[\tilde{\eta}(\acute{z},\acute{w}) =
\frac{-d\acute{z}}{\acute{z}\acute{w}} \; . \]
Because
\[ \Phi_{(\tilde{g},\tilde{\eta})} = \left( \begin{array}{rrr}
	-1 & 0 & 0 \\
	0 & -1 & 0 \\
	0 & 0 & 1
	\end{array}
\right) \Phi_{(g,\eta)} \; , \]
it follows from equation (2.4) that
the minimal surface in $\bfR^3$ with this Weierstrass data is the image
of ${\cal R}_{\frac{1}{\lambda}}$ under a rotation by $\pi$ about a
vertical axis.  Hence
${\cal R}_\lambda$ and ${\cal R}_{\frac{1}{\lambda}}$ are conjugate surfaces.

We use the Weierstrass data to show that ${\cal R}_\lambda$ and
${\cal R}_{\frac{1}{\lambda}}$ cannot be congruent for $\lambda \neq 1$.
Consider the points on
${\cal R}_\lambda$ where $g = \pm 1$.  As shown in the previous
Section~2.1 (Property~1),
when $\lambda > 1$, these points lie
on the lines in ${\cal R}_\lambda$.  By contrast, the points of
${\cal R}_{\frac{1}{\lambda}}$ where $g(z) = \pm 1$ lie on the planar
geodesics in the vertical plane of reflective symmetry of the surface.
Suppose there exists a symmetry of $\bfR^3$ taking ${\cal R}_\lambda$ to
${\cal R}_{\frac{1}{\lambda}}$,
then this symmetry must take lines to lines, ends to ends, and planar
geodesics to planar geodesics.
Since the lines in these two surfaces are all parallel to the
$x_2$-axis and the ends are horizontal, the symmetry of $\bfR^3$ must
be of the form
\[ \left( \begin{array}{rrr}
\pm 1 & 0 & 0 \\
0 & \pm 1 & 0 \\
0 & 0 & \pm 1
\end{array} \right) \; . \]
Therefore points of ${\cal R}_\lambda$ where
$g = \pm 1$ must be mapped to points of ${\cal R}_{\frac{1}{\lambda}}$ where
$g = \pm 1$.  This implies that the points of ${\cal R}_\lambda$ and
${\cal R}_{\frac{1}{\lambda}}$ where $g = \pm 1$ are the
points where the lines
and planar geodesics in the surface intersect i.e.\ where $ g=\pm
\lambda ^{\pm 1} $.  But we are assuming $ \lambda \neq 1 $.
Thus ${\cal R}_\lambda$ and
${\cal R}_{\frac{1}{\lambda}}$ are not congruent.
\end{proof}

\begin{remark}
One checks that the mapping $ (z,w)\rightarrow (z',w') $ from
$ M_\lambda $ to $ M_{\frac{1}{\lambda}} $ has the property that
on fundamental cycles it preserves the periods (one of which is
always zero) of (2.4).  Hence this mapping induces an isometry
of $ {\cal R}_\lambda $ and $ {\cal R}_{\frac{1}{\lambda}} $.
\end{remark}

Upon considering the behavior of ${\cal R}_\lambda$ and
${\cal R}_{\frac{1}{\lambda}}$ as $\lambda \rightarrow 0$,
one finds that ${\cal R}_\lambda$ and ${\cal R}_{\frac{1}{\lambda}}$ do
not have limits that are regular surfaces.  However, if these surfaces are
normalized
appropriately they converge to a catenoid and its conjugate surface, the
helicoid, which is consistent with Proposition~2.1.
We shall prove this in the next section.

\section{Proof of the Main Result}

To find a sequence of Riemann examples that converges to a catenoid or
helicoid, we renormalize the surfaces ${\cal R}_\lambda$ as follows.
Begin the integration in the Weierstrass representation
at the point $z_0
 = 1$.  If $\lambda > 1$, rescale by
$\sqrt{\lambda}$; if $\lambda < 1$, rescale
by $\frac{1}{\sqrt{\lambda}}$.  We name
these normalized Riemann examples $\acute{\cal R}_\lambda$.  Note that the
rescaling is accomplished by multiplying $\eta$ in the Weierstrass
representation by $\sqrt{\lambda}$ and $\frac{1}{\sqrt{\lambda}}$,
respectively.  Thus, the
Weierstrass representation for $\acute{\cal R}_\lambda$ is
\[ g = z \; , \; \eta = \frac{\sqrt{\lambda}dz}{zw} \; , \; \lambda \geq 1 \;
, \]
\[ g = z \; , \; \eta = \frac{dz}{\sqrt{\lambda}zw} \; , \; \lambda \leq 1 \;
. \]  One can check as in Proposition~2.1 that
$\acute{\cal R}_\lambda$ and
$\acute{\cal R}_{\frac{1}{\lambda}}$ are conjugate.

Let $ K $ denote Gauss curvature.

\begin{lemma}
	There exists a universal bound $c$ such that $|K| \leq c$ on
$\acute{\cal R}_\lambda$ for all $\lambda \in (0,\infty)$.
\end{lemma}

\begin{proof}
With Weierstrass data given locally as $g$ and $\eta = fdz$,
the Gauss curvature of a minimal surface is (\cite{Os}, p76)
\begin{equation} K = -(\frac{4|g^\prime|}{|f|(1+|g|^2)^2})^2 \; .
\end{equation}
In the case of $\acute{\cal R}_\lambda$ a computation using equations (2.1)
and (3.5) yields
\[|K| = 16\lambda^{\alpha}\frac{|z-\lambda||z+{\lambda^{-1}}|}{
|z|(|z|+|z|^{-1})^4} \; . \]
where $\alpha = 1$ if $\lambda \leq 1$, and $\alpha = -1$ if $\lambda
\geq 1$.  If $\lambda \leq 1$, then
\[|K| \leq 16\frac{(|z|+1)\lambda(|z|+{\lambda^{-1}})}{
|z|(|z|+|z|^{-1})^4} \; \leq \;
16\frac{(|z|+1)^2}{|z|(|z|+|z|^{-1})^4} \; \leq 4 \; . \]
Similarly, if $\lambda \geq 1$, then
\[|K| \leq 16\frac{{\lambda^{-1}}(|z|+\lambda)(|z|+1)}{
|z|(|z|+|z|^{-1})^4} \; \leq \;
16\frac{(|z|+1)^2}{|z|(|z|+|z|^{-1})^4} \; \leq 4 \; . \]
Thus, $|K|$ has a universal upper bound.
\end{proof}

\begin{remark}
More calculation shows that $|K|$ on ${\cal R}_\lambda$
is maximized at a fixed point of one of the
symmetries of the surface.  Thus $K$ is maximized in one of the
following three places: somewhere along a straight line in
${\cal R}_\lambda$; somewhere along a planar geodesic in
the plane of reflective symmetry parallel to $\{x_2 = 0\}$; or at the
points where $g(z) = \pm i$, which are the fixed points of the normal
rotation.  At the points where $z = g(z) = \pm i$, the value of
$|K|$ on ${\cal R}_\lambda$ is $\lambda + \frac{1}{\lambda}$.  Hence
on $ \acute{\cal R}_\lambda $ the value is $ 1+\frac{1}{\lambda^{2}}
(\mbox{resp. }1+\lambda^2) $ when $ \lambda > 1 (\mbox{resp.\ when }
\lambda < 1) $.

The following conjecture has been verified numerically:
The absolute value of the Gaussian curvature on ${\cal R}_\lambda$
is maximized at the points where $z = g(z) = \pm i$.  This implies
that the optimal value in Lemma~3.1 is $ c=2 $.
\end{remark}

\begin{figure}[p]
   \hspace{2in}
   \epsfxsize = 1.7in	\epsffile{fig2a.eps}
\hfill \break
   \caption{Portion of the Riemann Example
	$\acute{\cal R}_\lambda, \; \lambda = 0.1$}
\end{figure}

\begin{figure}[p]
   \hspace{2in}
   \epsfxsize = 1.7in	\epsffile{fig2b.eps}
\hfill \break
   \caption{Portion of the Riemann Example
	$\acute{\cal R}_\lambda, \; \lambda = 0.5$}
\end{figure}

\begin{figure}[p]
   \hspace{2in}
   \epsfxsize = 1.7in	\epsffile{fig3a.eps}
\hfill \break
   \caption{Portion of the Riemann Example
	$\acute{\cal R}_\lambda, \; \lambda = 1.0$}
\end{figure}

\begin{figure}[p]
   \hspace{2in}
   \epsfxsize = 1.7in	\epsffile{fig3b.eps}
\hfill \break
   \caption{Portion of the Riemann Example
	$\acute{\cal R}_\lambda, \; \lambda = 3.0$}
\end{figure}

\begin{figure}[p]
   \hspace{2in}
   \epsfxsize = 1.7in	\epsffile{fig4a.eps}
\hfill \break
   \caption{Portion of the Riemann Example
	$\acute{\cal R}_\lambda, \; \lambda = 5.0$}
\end{figure}

\begin{figure}[p]
   \hspace{2in}
   \epsfxsize = 1.7in	\epsffile{fig4b.eps}
\hfill \break
   \caption{Portion of the Riemann Example
	$\acute{\cal R}_\lambda, \; \lambda = 10.0$}
\end{figure}

\begin{proposition}
	There exist sequences of Riemann examples that converge to a
helicoid and sequences that converge to a catenoid.
\end{proposition}

\begin{proof}
	We will prove this proposition by showing that the
limit, as $\lambda \rightarrow 0$, of $\acute{\cal R}_\lambda$ is a
catenoid, and the limit, as $\lambda \rightarrow \infty$, of
$\acute{\cal R}_\lambda$ is a helicoid (see Figures 2 through 4).  We
consider first the case $\lambda \rightarrow 0$.

	For a large positive number
$L$, let $A_L$ be the annular ring
$\{z \in \bfC : \frac{1}{L} < |z| < L\}$.  One can think of $z$ in
(2.1) as a map
from $M_\lambda$ to $\bfC \setminus \{0\}$.  Then
for $\lambda$ sufficiently
small, $z^{-1}(A_L)$ consists of two disjoint sets
in $M_\lambda$.  Choose either one and name it $\acute{A}_{L,\lambda}$.
Since $z$ parametrizes $\acute{A}_{L,\lambda}$, we shall refer to points
in $\acute{A}_{L,\lambda}$ by their $z$ coordinates.

Let $B_r$ be a ball centered at the origin in $\bfR^3$ with radius $r$.
Let $S_{r}$ be the horizontal slab
$\{(x_1,x_2,x_3) \in \bfR^3 \, | \, -r \leq x_3 \leq r\}$,
with boundary planes $x_3 = \pm r$.

Let $\cal C$ be the catenoid obtained by using the Weierstrass data
\[ g(z) = z \; , \; \eta(z) = \frac{dz}{z^2} \; , \]
and integrating over $\bfC \setminus \{0\}$ with base point $z_0 = 1$.
The Weierstrass representation for the catenoid $\cal C$ is
\[C(p) := Re\int_{1}^{p} \left( \begin{array}{c}
	\frac{(1-z^2)dz}{z^2} \\
	\frac{i(1+z^2)dz}{z^2} \\
	\frac{2dz}{z}
	\end{array}
\right) \; , \; p \in \bfC \setminus \{0\} \; .\]
Let $\hat{\cal C}_L$ be the image of the restriction of $C(p)$
to $A_L$.  By examination of the
third coordinate of the Weierstrass representation, we see that
$\hat{\cal C}_L = {\cal C} \cap S_{2 \ln L}$.
Choose a large value for $r$, and
choose $L > e^{r}$  so that ${\cal C} \cap S_{2r} \subseteq \hat{\cal C}_L$.

	For $\lambda < 1$, the Weierstrass representation
for $\acute{\cal R}_\lambda$ is
\begin{equation}
\acute{R}_\lambda(p) := Re\int_{1}^{p} \left( \begin{array}{c}
	\frac{(1-z^2)dz}{\sqrt{\lambda}zw} \\
	\frac{i(1+z^2)dz}{\sqrt{\lambda}zw} \\
	\frac{2dz}{\sqrt{\lambda}w}
	\end{array}
\right) = C(p) +
Re\int_{1}^{p} \left( \begin{array}{c}
	f_{0,\lambda}(z)\frac{(1-z^2)dz}{z^2} \\
	f_{0,\lambda}(z)\frac{i(1+z^2)dz}{z^2} \\
	f_{0,\lambda}(z)\frac{2dz}{z}
	\end{array}
\right) \; ,
\end{equation}
where
\[ f_{0,\lambda}(z) = \frac{\sqrt{z}}{\sqrt{(z-\lambda)(\lambda z+1)}}
 - 1 \; . \]

	If $\lambda$ is close to zero, then the Weierstrass integral
$\acute{R}_\lambda$ along any closed curve in
$\acute{A}_{L,\lambda}$ is zero.  Therefore the Weierstrass
integral depends only on the endpoints of a path
in $\acute{A}_{L,\lambda}$.  All paths from $z_0 = 1$ to $p \in A_L$ may be
chosen to be of length less than $cL$, where $c$ is some fixed constant.

	Note that the functions
\[ \frac{1-z^2}{z^2} \; , \; \frac{1+z^2}{z^2} \; , \;
\frac{2}{z} \] are bounded on $A_L$.
Also, for all $z \in A_L$,
\[\lim_{\lambda \rightarrow 0}|f_{0,\lambda}(z)| = 0 \; . \]
It follows that the last integral in
equation~(3.6) becomes arbitrarily small,
uniformly on $\acute{A}_{L,\lambda}$, as $\lambda \rightarrow 0$.
In particular, for all $p \in A_L$, $\lim_{\lambda \rightarrow 0}
\acute{R}_\lambda(p) = C(p)$.

Let $\hat{\cal R}_\lambda$ be the image of $\acute{A}_{L,\lambda}$
under the map $\acute{R}_\lambda(p)$.  It follows that
$\hat{\cal R}_\lambda$ is a small perturbation of $\hat{\cal C}_L$
for $\lambda$ close to zero.  We conclude that
$\acute{\cal R}_\lambda \cap B_r \subseteq \hat{\cal R}_\lambda$.
Thus to examine the behavior of $\acute{\cal R}_\lambda$,
as $\lambda \rightarrow 0$, in arbitrary compact
regions of $\bfR^3$, it is enough to examine
$\hat{\cal R}_\lambda$ in $B_r$ for arbitrarily large fixed $r$.

We claim that the surface $\acute{\cal R}_\lambda \cap B_r$ is a
graph over a subset of the catenoid $\cal C$, for $\lambda$ close to $0$.
By Lemma~3.1, the Gaussian curvature $K$ is uniformly bounded for
all $\acute{\cal R}_\lambda$.  This implies that the normal curvatures are
also uniformly founded.

This in turn implies $\acute{\cal R}_\lambda \cap B_r$ is a
union of graphs over a subset of the catenoid $\cal C$.
Since the surface $\acute{\cal R}_\lambda$ is foliated by
lines and circles in horizontal planes, there is a single graph.

We now show that $\acute{\cal R}_\lambda \cap B_r$ converges to
${\cal C} \cap B_r$, in the $C^\infty$-topology, as
$\lambda \rightarrow 0$.  Because the curvature of $\acute{\cal R}_\lambda $
is bounded uniformly in $ \lambda $, it follows from standard
elliptic theory that the convergence is $ C^\infty $.  (Curvature
bounds give $ C^1 $ estimates.)  However, we can give in our
specific case a direct proof.  Let $p \in {\cal C} \cap B_r$.
Let $z_p \in A_L$ be the point such that $C(z_p) = p$, and let
$B_h(z_p) \subseteq A_L$ be a ball of radius $h$ about $z_p$.  Note that
$\acute{R}_\lambda(B_h(z_p))$
consists of disks in $\bfR^3$ converging in the $C^0$-norm,
as $\lambda \rightarrow 0$,
to a disk containing $p$ and lying on the catenoid $\cal C$.  Let
$x_{i,\lambda}(z)$ be the $i$'th coordinate function of
$\acute{\cal R}_\lambda$, let $x_i(z)$ be the $i$'th coordinate
function on $\cal C$,
and let $f_\lambda(z) = x_{i,\lambda}(z) - x_i(z)$.  Note that
\[ \max_{w \in B_h(z_p)} \left| f_\lambda(w) \right| \rightarrow 0 \, ,
\mbox{ as } \lambda \rightarrow 0 \; . \]
Since $f_\lambda(z)$ is harmonic, it is the real part of a holomorphic
function $F_\lambda(z)$, and $F_\lambda(z)$ can be chosen so that
\[ \max_{w \in B_h(z_p)} \left| F_\lambda(w) \right| \rightarrow 0 \, ,
\mbox{ as } \lambda \rightarrow 0 \; . \]
Assuming $z \in B_{\frac{h}{2}}(z_p)$, and using the Cauchy integral formula,
we have
\[ \left| f^{(k)}_\lambda(z) \right| \leq \left| F^{(k)}_\lambda(z) \right|
= \left| \frac{k!}{2\pi i} \int_{\partial B_h(z_p)} \frac{F(\xi)}
{(\xi - z)^{k+1}} d\xi \right| \]
\[ \leq \frac{k!}{2\pi} \left(\frac{2}{h}\right)^{k+1} 2\pi h
\max_{w \in B_h(z_p)} \left| F_\lambda(w) \right| \; , \]
and so we have that $|f^{(k)}_\lambda(z)| \rightarrow 0 \; \forall k, \;
\forall z \in B_{\frac{h}{2}}(z_p)$,
as $\lambda \rightarrow 0$.  Therefore $|f^{(k)}_\lambda(z)|
\rightarrow 0 \; \forall k, \; \forall z \in A_L$,
and the convergence is in the $C^\infty$-topology in $B_r$.

Since $r$ is arbitrary, we have completed the proof that
$\acute{\cal R}_\lambda$
converges to the catenoid $\cal C$ as $\lambda \rightarrow 0$.

	It is now intuitively clear that the surfaces
$\acute{\cal R}_\lambda$ converge to
a helicoid for the following reason: as $\lambda \rightarrow \infty$,
their conjugate surfaces $\acute{\cal R}_{\frac{1}{\lambda}}$
converge to a catenoid.  One can give an explicit proof of this,
similar to the proof just given for the case $\lambda \rightarrow
0$.  However, there are some differences in the proof of the
case $\lambda \rightarrow \infty$.
For $\lambda$ close to $\infty$, the representation for $\acute{\cal R}_
\lambda$ is
\[ \acute{R}_\lambda(p) := Re\int_{1}^{p} \left( \begin{array}{c}
	\frac{\sqrt{\lambda}(1-z^2)dz}{zw} \\
	\frac{i\sqrt{\lambda}(1+z^2)dz}{zw} \\
	\frac{2\sqrt{\lambda}dz}{w}
	\end{array}
\right) = \]
\begin{equation}
-Re\int_{1}^{p} \left( \begin{array}{c}
	\frac{i(1-z^2)dz}{z^2} \\
	\frac{-(1+z^2)dz}{z^2} \\
	\frac{2idz}{z}
	\end{array}
\right) +
 Re\int_{1}^{p} \left( \begin{array}{c}
	f_{\infty,\lambda}(z)\frac{i(1-z^2)dz}{z^2} \\
	f_{\infty,\lambda}(z)\frac{-(1+z^2)dz}{z^2} \\
	f_{\infty,\lambda}(z)\frac{2idz}{z}
	\end{array}
\right) \; ,
\end{equation}
where
\[ f_{\infty,\lambda}(z) = 1 -
\frac{\sqrt{z}}{\sqrt{(1-\frac{z}{\lambda})(z+\frac{1}{\lambda})}}
 \; .  \]

Note that the first integral in the above sum in
equation~(3.7) is the Weierstrass integral for a helicoid,
since Weierstrass data for a helicoid is
\[ g(z) = z \; , \; \eta(z) = \frac{idz}{z^2} \; . \]

If $\lambda$ is
large, then the vertical distance between adjacent planar ends of
$\acute{\cal R}_\lambda$ is approximately $2\pi$.  This can be verified by
integrating the third function in the integrand
of the Weierstrass representation from $+1$ to
$-1$ along the half of the unit circle lying
in the upper half of the complex plane.
This curve is $z = e^{it}, 0 \leq t \leq \pi$.  The distance between adjacent
planar ends for large $\lambda$ is
\[Re\int_0^\pi\frac{2\sqrt{\lambda}ie^{it}dt}
{\sqrt{e^{it}(e^{it}-\lambda)(e^{it}+\frac{1}{\lambda})}} \approx
Re\int_0^\pi\frac{2\sqrt{\lambda}idt}{\sqrt{-\lambda}} = 2\pi \; . \]

The major difference from the proof in the case $\lambda
\rightarrow 0$ is this: When $\lambda$ is close to
$\infty$, then a homologically nontrivial loop in $\acute{A}_
{L,\lambda}$ has a nonzero real period with respect to the Weierstrass
integral.  Now one may only assume that a path from $z=1$ to
$z=p$ in $A_L$ has length less than
$c n L$, where $n$ is the number of
times that the path wraps about the origin, and $c$ is some fixed constant.
Since the distance between adjacent ends is approximately $2\pi$,
it follows that if the wrapping number $n$ about the origin
of a path from 1 to $p$ is sufficiently large, then $z=p$ will be mapped by
the Weierstrass integral to a point outside $B_r$.
Thus, we may assume there is an upper bound $N$ for $|n|$, depending only
on $r$.  Hence, for large $\lambda$, we may assume that any path from 1
to $p$ has length less than $\acute{c} L$, where $\acute{c} = c N$.

The other parts of the proof of the case $\lambda \rightarrow
\infty$ transfer directly from the proof of the case $\lambda \rightarrow 0$.
\end{proof}

\begin{corollary}
	As $\lambda \rightarrow \infty$, the radius of any
level-circle of $\acute{\cal R}_\lambda$ diverges to infinity.
\end{corollary}

\begin{corollary}
	The planar
curve described by the set of centers of the horizontal
circles that foliate $\acute{\cal R}_\lambda$ has the property that
its maximum curvature
approaches zero as $\lambda \rightarrow \infty$.
\end{corollary}

	It remains to determine what other surfaces are limits
of Riemann examples.  Three possibilities are clear:
a single flat plane; an infinite number of equally-spaced flat planes; and
another Riemann example.  The following proposition states that these are
the only other possibilities, completing the proof of Theorem~1.1.

\begin{proposition}
Any convergent sequence of Riemann examples converges to one of
the following surfaces:

	a Riemann example;

	a helicoid;

	a catenoid;

	a single flat plane;

	an infinite number of equally-spaced flat planes.
\end{proposition}

\begin{proof}
	Let $\{{\cal R}_j\}_{j=1}^\infty$ be a sequence of Riemann examples
converging to a surface $\cal S$.  From the definition of convergence,
$\cal S$ is clearly minimal, properly embedded, and complete.
Without loss of generality we may assume that the

limiting normals $\vec{v_j}$ at the
ends of the surfaces ${\cal R}_j$ converge in $\bf S^2$ to
a vertical vector.  With this normalization, the surfaces ${\cal R}_j$ are
foliated by circles and lines lying in planes that are becoming
horizontal as $j \rightarrow \infty$.  These almost-horizontal planes
intersect ${\cal R}_j$
in connected sets, and these connected sets each consist of a
single circle or a single line.  Thus every curve ${\cal S} \cap \{x_3 = c\}$
is the limit set of a sequence of
circles and/or lines.  It follows that $ {\cal S} \cap \{x_{3}=c\} $ is
either a circle, a line, the empty set or the plane $ \{x_{3}=c\} $.

Recall from the introduction that this implies that each component of
$ \cal S $ is either a helicoid, a catenoid, a Riemann example or a
plane.  Moreover: the helicoid and catenoid must have vertical
axes; the plane and the ends of the Riemann example must be horizontal.
If $ \cal S $ has more than one component, they must all be horizontal
planes (because $ \cal S $ is fibred by circles or lines, one in each
horizontal plane).  Suppose $ \cal S $ consists of two or more horizontal
planes.
Since the points on the surfaces ${\cal R}_j$ where
the Gauss map is vertical must
approach the ends of ${\cal R}_j$ as $j \rightarrow \infty$,
it is the equally spaced-ends of the surfaces ${\cal R}_j$ that
converge to $\cal S$.
Therefore $\cal S$ is an {\em infinite} collection of equally-spaced planes.
\end{proof}

config: arxiv


TITLE: Closest point to a surface using Lagrange multipliers
QUESTION [0 upvotes]: I know how to solve constrained problems if I am given the surface and the restriction. But I'm a little lost on the application of this for finding the closest point.
Using Lagrange multipliers, how could I find the closest points of $z = x^2-y^2+1$ to the origin? What would be the reasoning for it?
What about if the it was the closest from another set, let's say, the closest points from $y = x$ to $y = x^2+2$. What is the reasoning behind solving these kinds of problems using Lagrange's Multipliers?

REPLY [1 votes]: You want to minimize the distance from a point $p$ that lies on the surface to the origin.
First, you have three variables $x$, $y$, $z$, i.e. the three coordinates of the point $p$.
Second, the objective function is
$$
f(x,y,z) = (x-0)^2+ (y-0)^2 + (z-0)^2 = x^2+y^2+z^2.
$$
Note that we use the distance squared. This is because the distance squared is minimal if and only if the distance is minimal. But the distance squared function makes the calculations easier; there is no square root, so the derivative becomes easier. This is a standard trick.
And finally, the constraint is given by $g(x,y,z) = x^2 - y^2 + 1 - z = 0$, which is basically the equation for the surface.
Now you can write down the Lagrangian function $L(x,y,z,\lambda) = f(x,y,z) - \lambda g(x,y,z)$ and proceed to solve the problem.
Your second problem goes as follows: We have two points $p$ and $q$, each lying on a surface. What is their minimal distance? This problem is basically the same, but you have more variables and constraints.
First, you have 6 variables, namely 3 for the point $p=(x_1,y_1,z_1)$ and 3 for the point $q=(x_2,y_2,z_2)$.
The objective function now becomes $f(x_1,\ldots,z_2)=(x_1-x_2)^2+(y_1-y_2)^2 + (z_1-z_2)^2$. And finally, there are two constraints, namely $g(x_1,y_1,z_1)=x_1-y_1 = 0$ and $h(x_2,y_2,z_2)=x_2+2 - y_2 = 0$.
Now the Lagrangian becomes
$$ L(x_1,\ldots,z_2,\lambda,\mu) = f(x_1,\ldots,z_2) - \lambda g(x_1,y_1,z_1) - \mu h(x_2,y_2,z_2).$$


TITLE: How to differentiate $\left(\frac{d}{dt}\left(x\right)\right)^2$?
QUESTION [1 upvotes]: How does
$\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{d}{dt}\left(x\right)\right)^2+\frac{1}{2}kx^2\right)$ = 0
equals to
m$\frac{d^2}{dt^2}\left(x\right)$ + kx = 0

REPLY [1 votes]: \begin{align}\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{d}{dt}\left(x\right)\right)^2+\frac{1}{2}kx^2\right) &= 0 \\ \Longrightarrow \frac{1}{2} m \frac{d}{dt}\big(\frac{d}{dt}x\big)^2+\frac{1}{2}k \frac{d}{dt}x^2 &= 0 \\ \Longrightarrow  m \frac{d}{dt} x\frac{d}{dt}\frac{d}{dt}x+kx \frac{d}{dt}x &= 0 ~\text{ (chain rule)}\\ \Longrightarrow m \frac{d}{dt}\frac{d}{dt}x+k x &= 0 ~\text{ (assuming that $\dot x \neq 0$)} \\ \Longrightarrow m\frac{d^2}{dt^2}x + kx &= 0\end{align}

REPLY [0 votes]: Hint: $$1/2m(2\dot{x})\ddot{x}+1/2k(2x)\dot{x}=0$$


TITLE: Sonic Doppler shift from a distant source
QUESTION [0 upvotes]: Probably most of us have heard the pitch of an ambulance change as it’s passed us, but that’s from a nearby sound source. If a sound source is far away, does the Doppler shift change? I find it hard to believe that the sound wave compression would be the same over a mile as it is over a couple of feet. At the same time, astronomy uses Doppler shift over distances of many light years, but that’s also through space, where we don’t have air molecules bumping into each other and losing energy.
(I heard a Doppler shift from an ambulance that passed me while I was driving yesterday. When I heard another ambulance late at night that I suspect was about a mile away, I started wondering if the pitch shift would be the same.)

REPLY [0 votes]: Ideally the doppler shift does not depends on the distance of sound source from the observer, so the doppler shift should not change. But there is a phenomenon where higher frequency sound wave scatter more than a lower frequency sound wave (for electromagnetic wave it is called Rayleigh Scattering). So if a sound source is far away, the high frequency wave got scatter mostly and one can only hear to the lower frequency waves.


TITLE: Rate of convergence of an irrational rotation
QUESTION [4 upvotes]: Let $\alpha, \beta \in \mathbb{R}$. Let $\{x\}$ denote the fractional part of $x$ and let $\|x\| = \min(\{x\}, 1-\{x\})$.
If we assume that $\alpha$ is irrational, then there exists an increasing sequence of integers $(n_k)_{k \in \mathbb{N}}$ such that $\|n_k \alpha - \beta\| \to 0$. Is it possible that the rate of convergence is exponential, that is, that there exists some $\eta > 1$ such that $\eta^{n_k} \|n_k \alpha - \beta\| \to 0$?
I believe that the fact that the sequence $\{n \alpha\}_{n \in \mathbb{N}}$ is equidistributed implies that if this is true for some $\eta > 1$, then it must be true for every $\eta > 1$, so this seems to be a rather strong condition.
If the rate may be exponential for general $\alpha$ and $\beta$, what about the case when $e^{2\pi i \alpha}$ and $e^{2\pi i \beta}$ are algebraic?

REPLY [7 votes]: For the algebraic case a lot is known. There is a deep result of Gelfond (see A. O. Gelfond, Transcendental and Algebraic Numbers, Dover, New York, 1960) that if $\lambda$ is an algebraic number of absolute value 1, then for every $\epsilon>0$ there is a constant $C>0$ such that $$|\lambda^n-1|>Ce^{-n\epsilon}$$ for all $n\ge1$. More recent work of Feldman (see Chapter 9 in  Stolarsky's book Algebraic Numbers and Diophantine Approximation, Dekker, 1974) actually shows that there are effectively computable numbers $C=C(\lambda)>0$ and $N=N(\lambda)\ge1$ such that
$$|\lambda^n-1|>Cn^{-N}$$
for all $n\ge1$.
These diophantine results are very useful while investigating the dynamical properties of maps defined algebraically, e.g. automorphisms of tori. See for example my paper "Dynamical Properties of Quasihyperbolic Toral Automorphisms", Ergod. Th. Dynam. Sys. 2 (1982), 49--68.


TITLE: Direct sum of dual space
QUESTION [0 upvotes]: Let $E$ be a finite vector space and $L_1$, $L_2$ are subspaces.
I want to show that :
if $E=L_1\oplus L_2$ than $E^*=L_1^*\oplus L_2^*$.
I attempted:
$\forall x\in E, x=x_1+x_2$ where $x_i\in L_i$
now if $\phi\in E^*$, we have $\phi(x)= \phi(x_1)+\phi(x_2) $.
here I want to consider a forme $\phi_1$ that take $\phi (x_1)$ as image, and the same for $\phi_2$ but I had some strugle.
Furthermore, I search here for a response but I found just this post, so I' wondering : Do I need other conditions to show this?

REPLY [1 votes]: Note that with finite-dimensional spaces, there is an isomorphism $E \cong E^*$. Suppose it is over a field $\mathbb{F}$, so $L_1 \cong \mathbb{F}^m$, $L_2 \cong \mathbb{F^n}$ and $E \cong \mathbb{F}^{n+m}$.
Then, $L_1^* \cong L_1 \cong \mathbb{F}^n$ and similar for $L_2$.
Hence, $E^* \cong \mathbb{F}^{n+m} \cong \mathbb{F}^n \oplus \mathbb{F}^m \cong L_1^* \oplus L_2 ^*$


TITLE: Simplifying the summation $ (3^i\sqrt{n/3^i})/\log_2(n/3^i)$ from $i=0 \text{ to }k-1$
QUESTION [3 upvotes]: I have been trying to simplify the following summation with the intention of breaking it into less complex summations, but I keep getting stuck no matter what I try:
$$\sum_{i=0}^{k-1} 3^{i} \cdot \frac{\sqrt{\frac{n}{3^i}}}{\log_{2}\frac{n}{3^i}}$$
Among the things I tried was raising to the power of $1/2$* the upper part of the fraction to cancel out the $3^i$, but after that I'm left with $\sqrt{n\,3^i}$ anyways without any clear steps to simplify further. Also tried putting the lower part of the fraction as a difference of logarithms, but got stuck in the same way. I even tried to use the change of base identity to put the lower part of the fraction in terms of $\log_3(2)$ and see if I could have canceled something but no luck.
Just simplifying the sum would be good enough since I can try to take it from there by substituting $k\; (k = \log_3(n)$ in case that is useful).

REPLY [1 votes]: The issue expression can be presented in the form of
$$S={\large\sum\limits_{i=0}^{k-1}}3^i\dfrac{\sqrt{\dfrac n{3^i}}}{\log_2\dfrac{n}{3^i}}
= \sqrt n \log_32{\large\sum\limits_{i=0}^{k-1}}\dfrac{(\sqrt3)^i}{\log_3{n}-i},$$
so
$$S=\sqrt n \log_32\left(3^{k/2}\Phi(\sqrt3,1,k-\log_3n)-\Phi(\sqrt3,1,-\log_3n)\right)$$
(see also Wolfram Alpha), where

$$\Phi(z,s,a) = {\large\sum\limits_{m=0}^\infty\,\dfrac
{z^m}{(m+a)^s}}$$
  is Lerch transcendent.

Assuming $k=\lceil\log_3n -1\rceil,$ can be built the plots for the issue sums

and via Lerch transcendent

Easy to see that summation via Lerch transcendent is correct.