Sudan's central bank is increasing dollar supplies to local banks to stop a slide of the Sudanese pound after it generated more than $300 million from gold sales, a central bank official said on Sunday.
The African country is undergoing a severe economic crisis after losing three-quarters of its oil production when South Sudan became independent in July under a 2005 peace deal. Oil is the lifeline of both economies.
Sudan has avoided an Arab spring like Egypt or Tunisia but the economic crisis has sparked small protests against high food price inflation.
The pound hit a historic low on the black market last week after losing more than 70 per cent of its value since July as the loss of oil has cut off the main source of dollars.
"The central bank has sold 6.5 tonnes of gold since January which has generated more than $300 million," central bank spokesman Hazem Abdul-Qadir told Reuters.
"Therefore we are increasing dollar supplies to banks and foreign exchange offices," he said.
He gave no figures but said dollar supplies would top the $35 million banks currently get per week from the central bank.
He said Sudan had managed to get some foreign aid but declined to elaborate. Sudanese media have said that Qatar will grant a substantial loan but nothing has been confirmed so far.
Sudan is locked in a row with South Sudan over oil payments as the landlocked new nation needs to export its crude through northern facilities.
Both sides have failed to agree on a transit fee. South Sudan shut down its entire output last month after Sudan seized southern oil as compensation for what it calls unpaid fees.
Sudan hopes rising gold exports will replace oil revenues but experts are sceptical since much of the gold comes from small-scale miners whose output is hard to verify.


text:
\section{Introduction}

A surface $S$ in the $3$-dimensional Euclidean space
is called a Weingarten surface if there exists some relation
\begin{equation}\label{relW}
	W(\lambda_1,\lambda_2)=0,
\end{equation}
among its principal curvatures $\lambda_1$ and $\lambda_2$. Since the principal
curvatures of a surface can always be determined from its mean curvature $H$
and its Gaussian curvature $K$, and vice-versa, the relation \eqref{relW} can
always be rewritten as a relation $U(H,K)=0$.

\bigskip

Weingarten surfaces is a classical topic in Differential Geometry that began
with the works of Weingarten in the middle of the 19th century
\cite{Wei1,Wei2} and that has been a subject of interest for many authors
since then (see Chern \cite{Che1}, Hartman and
Winter \cite{HW}, Hopf \cite{Hop1}, Voss \cite{Vos}, Rosenberg and S\'{a}
Earp \cite{RSE}, K\"{u}hnel and Steller \cite{KS}, L\'{o}pez
\cite{Lop1,Lop3,Lop,Lop4}, to name just a few).
\bigskip

Minimal surfaces, surfaces with constant mean curvature and surfaces with
constant Gaussian curvature are classical examples of Weingarten surfaces.
Another well known class (generalizing the previous ones) is that of the
linear
Weingarten surfaces, \ie Weingarten surfaces verifying either the relation
\begin{equation}\label{relLWI}
	W(\lambda_1,\lambda_2)= a\lambda_{1}+b\lambda_{2}= c
\end{equation}
\noindent or the relation
\begin{equation}\label{relLWII}
	U(H,K)=aH+bK=c,
\end{equation}
where $a,b,c\in\mathbb{R}$ are constants such that $a$ and $b$ do not vanish simultaneously.

\bigskip

The complete classification of Weingarten surfaces is far from being
achieved. The
existent results deal mostly with the linear case,
sometimes making use of additional topological/geometric hypothesis and/or
working with important subclasses of surfaces such as revolution surfaces \cite{Hop1,Lop,KS,RSE}, tubes along curves and cyclic surfaces \cite{Lop3,Lop4}, ruled surfaces
and helicoidal surfaces \cite{Kuh}, translation surfaces \cite{DGW,LM}, etc. In general, the approaches used to treat the linear case do not apply to the non-linear case. Therefore, results concerning non-linear Weingarten surfaces are more rare \cite{RSE,KS}.

\bigskip

In this paper we study rotational surfaces  in the $3$-dimensional Euclidean
space whose second fundamental forms have constant length (recall that the
squared length
$|A|^2$ of the second fundamental form of a surface in $\mathbb R^3$ is
defined as the trace of $A^2$, where $A$ is its shape operator). In other
words, we study rotational Weingarten surfaces
that satisfy the non-linear relation
\begin{equation}\label{relacao segforma cte}%
	W(\lambda_1,\lambda_2)=\left(  \lambda_{1}\right)  ^{2}+\left(
	\lambda_{2}\right)^{2}=c,
\end{equation}
or equivalently
\begin{equation}\label{relacao segforma cte2}%
U(H,K)=4H^{2}-2K=c,
\end{equation}
for some $c>0$.

\bigskip

In this case we prove the following result (notice that since the property of
having constant $|A|$ is invariant by homotheties in $\mathbb R^3$,
we
can assume without loss of generality that  $c=1$):

\begin{theorem}\label{Family}
There are two infinite families $\mathcal F_1$ and $\mathcal F_2$ of complete
non embedded rotational surfaces in $\mathbb R^3$ with $|A|=1$. The family
$\mathcal F_1$ is one parameter and its members are periodic $C^{\infty}$
surfaces, while the members of $\mathcal F_2$ are $C^3$ surfaces. Moreover, any
complete rotational
$C^2$ surface with $|A|=1$ is either a round sphere of radius $\sqrt{2}$, a
circular cylinder of radius 1 or, up to a rigid motion in $\mathbb R^3$, a
member of one of the two families.
\end{theorem}

\begin{cor}\label{Embedded}
	The only complete embedded rotational $C^2$-surfaces in $\mathbb R^3$ with
	second
	fundamental form of constant length are the
	round sphere and the circular cylinder.
\end{cor}

Non trivial examples of compact surfaces (embedded or immer\-sed) with second
fundamental form of constant length are unknown by the authors.
In view of Theorem 1.1 and Corollary 1.2, one can then formulate the following question:

\begin{question}
- Is there a compact surface, other than the round sphere, embedded/immersed in
the Euclidean 3-space whose second fundamental form has constant length?
\end{question}

It is worth to point out that the above question has a negative answer in the class of the
compact surfaces with positive Gaussian curvature \cite[Theorem 5 on p. 347 and
Section 4]{Ale}  (see also \cite{Kor} or \cite[Theorem 2.3]{FS}).

\bigskip

The importance of the class of hypersurfaces whose second fundamental forms
have constant length goes beyond the context of Weingarten  surfaces. We
mention \cite{DX} (see also \cite{Gua}), where it is proved that the
generalized cylinders are the only complete embedded self-shrinkers in $\R^3$
with polynomial volume growth whose second fundamental forms have constant
length.

\bigskip

The paper is organized as follows. In Section \ref{convexprofcurve}, we prove
that the profile curve of any rotational surface in $\mathbb R^3$ with
$|A|\equiv 1$ is convex, \ie its
signed curvature does not change signal. This fact enable us to reduce the
study of rotational surfaces with $|A|\equiv 1$ to the study of the
trajectories of a certain vector field in the plane.
The study of this vector field is made in Section \ref{phasefundvectfield}.
Finally, we prove Theorem \ref{Family} in Section \ref{proofmaintheorem}.

\subsection*{Acknowledgments} The authors would like to thank Thiago de Melo
(IGCE-UNESP) for helpful conversations during the preparation of this work and
for his help with the figures.

\section{Convexity of the profile curves}\label{convexprofcurve}

Our goal in this section is to prove that the profile curve $\mathcal{C}$ of
any rotational $C^2$-surface $M\subset\mathbb R^3$, whose shape operator
$A$ has length $|A|\equiv 1$, is convex. By applying a rigid motion of
$\mathbb R^3$ if necessary, we can assume that $\mathcal{C}$ is contained in
the $xz$-plane and that the axis of revolution is the $x$-axis.

Let $\alpha(t)=(x(t),0,z(t)),\;t\in\mathbb (a,b)$, be a
parametrization of $\mathcal{C}$ such that $||\alpha'(t)||=1$ and $z(t)>0$
for all $t$, and let $\theta:(a,b)\to\mathbb R$ be a continuous
(and, hence, of class $C^1$) function satisfying
\begin{equation}\label{Rot1}
\alpha'(t)=(x'(t),0,z'(t))=(\cos\theta(t),0,\sin\theta(t)),\;\;\;t\in (a,b).
\end{equation}
It is easy to see that the function $\theta$ satisfying
\Eqref{Rot1} is unique up to an integer multiple of $2\pi$. The
principal curvatures of $M$ are given by (see \eg \cite{Lop})
\begin{equation}\label{Rot2}
\lambda_1(t)=\theta'(t),\;\;\;\;\lambda_2(t)=-\frac{\cos\theta(t)}{z(t)}.
\end{equation}
Since $|A|^2=\lambda_1^2+\lambda_2^2\equiv 1$ by hypothesis, one
then has
\begin{equation}\label{Rot3}
\theta'(t)^2+\frac{\cos^2\theta(t)}{z^2(t)}=1,\;\;\;t\in
(a,b).
\end{equation}

As we observed in the introduction, the proof of Theorem
\ref{Family} will be based on a careful study of the trajectories
of a suitable vector field in the plane. The fundamental property
of the profile curves that makes this approach possible is
provided by the following proposition (recall that the signed curvature
of $\alpha$ is $\theta'$):

\begin{prop}\label{monotona}
	The function $\theta:(a,b)\to\mathbb R$ is monotone.
\end{prop}

Before proceeding to the proof of Proposition \ref{monotona}, let us
explain how to relate profile curves with the trajectories of a specific
vector field.

Let $\alpha$ and $\theta$ be as above. Assuming that
$\theta$ is monotone,
re\-pa\-ra\-me\-tri\-zing $\alpha$ we can assume that $\theta'\geq 0$. Then,
by \Eqref{Rot1} and \eqref{Rot3},
\begin{equation}\label{sistema}
\begin{cases}
\theta'(t)=\sqrt{1-\frac{\cos^2\theta(t)}{z^2(t)}},\\
z'(t)=\sin\theta(t).
\end{cases}
\end{equation}
Let
$X:\Omega\to\mathbb R^2$ be the (smooth) vector field defined by
\begin{equation}\label{VectorField}
X(\theta,z)=\left(\sqrt{1-\frac{\cos^2\theta}{z^2}},\sin\theta\right),
\end{equation}
where $\Omega=\{(\theta,z)\in\mathbb R^2:z>|\cos\theta|\}$.
As long as $z(t)>|\cos\theta(t)|$, the system in \Eqref{sistema} can
be rewritten as
\begin{equation}\label{Trajetoria}
(\theta'(t),z'(t))=X(\theta(t),z(t)),
\end{equation}
and so the curve $t\mapsto(\theta(t),z(t))$ is a trajectory of
$X$. A representation of $\Omega$ and the vector field $X$ can be seen in
Figure \ref{Figura1}.
\begin{figure}[htp!]
	\begin{tikzpicture}%
	[line cap=round,>=latex,x=1cm,y=1cm]
	%
	\clip(-.5,-1.) rectangle (7,4.6)
	%
	\begin{scope}[domain=0:2*pi
	\draw[dashed,black!50,thick] plot[smooth] (\x,{abs(cos(\x r))});
	\end{scope}
	%
	\begin{axis}
	anchor=origin,
	x=1cm,y=1cm,
	domain=0.05:2*pi,
	axis lines=none,
	axis line style={},
	xtick=\empty,
	ytick=\empty,
	]
	\addplot[%
	blue,
	quiver={u=0,v=0.3 * sin (\x r)}
	-{Stealth[length=3.2pt,width=3.2pt,inset=2pt]},
	line width=.4pt,
	samples=18,
	] {abs(cos(\x r))};
	\end{axis}
	%
	\begin{scope}[line width=.6pt,-
		Stealth[length=3.2pt,width=3.2pt,inset=2pt]%

	}]
	\clip(0,0) rectangle (6.6,4.6)
	\input{vetores2}
	\end{scope}

	\draw[black!15] (1.57075,0) -- (1.57075,4) (3.1415,0) -- (3.1415,4)
	(4.7125,0) -- (4.7125,4) (6.283,0) -- (6.283,4);
	%

	\draw[fill] (0,1) circle (1.2pt) (3.1415,1) circle (1.2pt) (6.2831,1)
	circle (1.2pt) ;
	%
	%
	\begin{scope}[line width=.5pt]
	\draw[->] (-.5,0) -- (6.8,0)node[below]{\small$\theta$}
	\draw[->] (0,-.5) -- (0,4.5)node[left]{\small$z$}

	\foreach \x/\l in
{1.57075/{\frac{\pi}{2}},3.1415/{\pi},4.71225/{\frac{3\pi}{2}},6.283/{2\pi}}{%
		\draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt);
		\draw[shift={(\x,0)}] (0,-8pt)node[fill=white,inner sep=0pt]
		{\small$\l$};
	}

	\foreach \y in {1}{
		\draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {\small
			$\y$};
	}

	\draw (0,0) node[below left] {\small$0$};
	\end{scope}
	\end{tikzpicture}
	\caption{Graphic representation of $\Omega$ and the vector field
	$X$ for $\theta\in[0,2\pi]$.}\label{Figura1}
\end{figure}

Conversely, given a trajectory
$\varphi(t)=(\theta(t),z(t)),\;t\in(a,b),$ of $X$ and
$t_0\in(a,b)$, consider the curve
$\alpha(t)=(x(t),0,z(t)),\;t\in(a,b),$ where
$$
x(t)=\int_{t_0}^t\cos\theta(s)ds.
$$
Using \Eqref{Rot2}, \eqref{VectorField} and \eqref{Trajetoria} one
easily proves that the surface in $\mathbb R^3$ obtained by the
rotation of the image of $\alpha$ around the $x$-axis satisfies
$|A|\equiv 1$.

In the proof of Proposition \ref{monotona}, as well as in the proofs of
later results, we will use the following technical lemma. In its statement,
$\alpha(t)=(x(t),0,z(t))$ and $\theta(t)$ are as in the beginning
of this section.

\begin{lemma}\label{ThetaLinhaZero}
	For any $t\in(a,b)$, the following assertions hold:

	\begin{enumerate}
		\item[(i)] $\theta'(t)=0$ if, and only if, $z(t)=1$ and
		$\sin\theta(t)=0$.

		\item[(ii)] If $z(t)<1$ then $|z'(t)|\geq|\theta'(t)|$.

		\item[(iii)] If $\theta'(t)=0$, then there exists $\delta>0$
		such that $z(s)\geq 1$, $s\in(t-\delta,t+\delta)$.
	\end{enumerate}
\end{lemma}

\begin{proof} (i) If $\theta'(t)=0$ then, by \Eqref{Rot3}, the function
	$\cos^2\theta/z^2$ attains a maximum at $t$. Hence,
	\begin{eqnarray}
	0&=&\left[\frac{\cos^2\theta}{z^2}\right]'(t)
	=\left[\frac{-2\theta'\cos\theta\sin\theta z^2-2\cos^2\theta
		zz'}{z^4}\right](t)\nonumber\\
	&=&\frac{-2\cos^2\theta(t)z'(t)}{z^3(t)}.\nonumber
	\end{eqnarray}
	Since $\cos\theta(t)\neq 0$, one obtains from the above equality and
	\Eqref{Rot1} that
	$$
	\sin\theta(t)=z'(t)=0.
	$$
	Using this information in \Eqref{Rot3}, one concludes that
	$z(t)=1$. The converse is an immediate consequence of
	\Eqref{Rot3}.

	\vskip5pt

	\noindent(ii) From \Eqref{Rot3} and $z(t)<1$ one obtains
	$$
	1=\theta'(t)^2+\frac{\cos^2\theta(t)}{z^2(t)}\geq\theta'(t)^2+\cos^2\theta(t),
	$$
	and so
	$$
	\theta'(t)^2\leq 1-\cos^2\theta(t)=\sin^2\theta(t)=z'(t)^2.
	$$
	The conclusion now follows by taking square roots in the above inequality.

	\vskip5pt

	\noindent(iii) Supposing, by contradiction, that the conclusion
	does not hold, we have $z(t_n)<1$ for some sequence $(t_n)$ that
	converges to $t$. Since, by (i), $t_n\neq t$ for all $n$,
	passing to a subsequence and reparametrizing $\alpha$ if necessary, one can assume that $t_n<t$, for all
	$n$.

	We claim that
	\begin{equation}\label{Rot3bb}
	z(s)<1,\;\;\;s\in(a,t).
	\end{equation}
	Indeed, if $z(c)\geq 1$ for some $c\in(a,t)$ then, since $t_n\to
	t$ and $z(t_n)<1$ for all $n$, there is $d\in(c,t)$ such that
	\begin{equation}\label{Rot3bc}
	z(d)=\inf\{z(s):c\leq s\leq t\}<1.
	\end{equation}
	Hence, $z'(d)=0$. On the other hand, from (i), (ii) and \Eqref{Rot3bc}
	one
	obtains $|z'(d)|\geq|\theta'(d)|>0$.
	This contradiction proves \Eqref{Rot3bb}.

	By (i), (ii) and \Eqref{Rot3bb},
	\begin{equation}\label{Rot3bbb}
	\theta'(s)\neq 0\;\;\text{and}\;\;z'(s)\neq 0,\;\;\;s\in(a,t).
	\end{equation}
	Then, by \Eqref{Rot3bb} and $z(t)=1$,
	\begin{equation}\label{Rot3c}
	\sin\theta(s)=z'(s)>0,\;\;\;s\in(a,t).
	\end{equation}
	Since $\sin\theta(t)=0$ by (i), we have two possibilities:

	\vskip5pt

	\noindent{a)} $\theta(t)=2k\pi$, for some $k\in\mathbb Z$.

	\vskip5pt

	\noindent{b)} $\theta(t)=(2k+1)\pi$, for some $k\in\mathbb Z$.

	\vskip5pt

	Assuming a), from \Eqref{Rot3c} one obtains
	$$
	\theta(t)+\pi>\theta(s)>\theta(t),\;\;\;s\in(a,t).
	$$
	Then, by the first inequality of \Eqref{Rot3bbb},
	\begin{equation}\label{Rot3d}
	\theta'(s)<0,\;\;\;s\in(a,t).
	\end{equation}
	From \Eqref{Rot3bb}, \eqref{Rot3c}, \eqref{Rot3d} and (ii), we obtain
	$$
	z'(s)\geq -\theta'(s),\;\;\;s\in(a,t).
	$$
	Hence, for fixed $s_1\in(a,t)$, we have
	$$
	1=z(t)=z(s_1)+\int_{s_1}^{t}z'(s)ds\geq
	z(s_1)-\int_{s_1}^{t}\theta'(s)ds=z(s_1)+\theta(s_1)-\theta(t),
	$$
	and so
	$$
	1-z(s_1)\geq\theta(s_1)-\theta(t).
	$$
	It now follows from \Eqref{Rot3d} and the fact that the cosine function
	is decreasing on $[0,\pi]$, that
	\begin{equation}\label{Rot3f}
	\begin{aligned}
	\cos(1-z(s_1)) \leq& \ \cos(\theta(s_1)-\theta(t))\\
	=& \ \cos\theta(s_1)\cos\theta(t)+\sin\theta(s_1)\sin\theta(t)\\
	=& \ \cos\theta(s_1).
	\end{aligned}
	\end{equation}
	Since, by \Eqref{Rot3} and \eqref{Rot3d},
	$z(s_1)>|\cos\theta(s_1)|$, inequalities \Eqref{Rot3bb} and
	\eqref{Rot3f} imply
	\begin{equation}\label{Rot3g}
	\cos(1-z(s_1))<z(s_1)<1,
	\end{equation}
	contradicting the fact, easily verified, that $\cos(1-x)>x$,
	for all $x\in[0,1)$.

	A reasoning entirely similar to the above shows that b) cannot
	occur either. Hence, $z(s)\geq 1$ on a
	neighbourhood of $t$.
\end{proof}

\noindent{\bf Proof of Proposition \ref{monotona}:} Suppose, by
contradiction, that $\theta$ is not monotone. Then there exists
$t_1<t_2<t_3$ in $(a,b)$ such that either i) or ii) below holds:

\vskip5pt

\noindent i) $\theta(t_1)<\theta(t_2)$ and
$\theta(t_2)>\theta(t_3)$.

\vskip5pt

\noindent ii) $\theta(t_1)>\theta(t_2)$ and
$\theta(t_2)<\theta(t_3)$.

\vskip5pt

Assuming i), we have
$$
\lambda:=\sup\{\theta(t):t\in[t_1,t_3]\}>\max\{\theta(t_1),\theta(t_3)\}.
$$
Define
$$
\xi:=\inf\{t\geq t_1:\theta(t)=\lambda\},\;\;\;\eta:=\sup\{t\leq
t_3:\theta(t)=\lambda\}.
$$
Since $\theta$ attains a local maximum at $\xi$ and at $\eta$, we
have $\theta'(\xi)=\theta'(\eta)=0$. Then, by Lemma
\ref{ThetaLinhaZero} (i), $z(\xi)=z(\eta)=1$ and
$\sin\theta(\xi)=\sin\theta(\eta)=0$. The latter implies that
either $\theta(\xi)=\theta(\eta)=2k\pi$ or
$\theta(\xi)=\theta(\eta)=(2k+1)\pi$, for some $k\in\mathbb Z$. If
$\theta(\xi)=\theta(\eta)=2k\pi$, one has  $\mu\in (\eta,t_3]$, such that
$$
z'(t)=\sin\theta(t)<0,\;\;\;t\in(\eta,\mu).
$$
Then, $z(t)<z(\eta)=1$, for all $t\in(\eta,\mu)$, contradicting
Lemma \ref{ThetaLinhaZero} (iii).

If $\theta(\xi)=\theta(\eta)=(2k+1)\pi$, there exists $\nu\in [t_1,\xi)$
such that
$$
z'(t)=\sin\theta(t)>0,\;\;\;t\in(\nu,\xi).
$$
Then, $z(t)<z(\xi)=1$, for all  $t\in(\nu,\xi)$, which also
contradicts Lemma \ref{ThetaLinhaZero} (iii).

A reasoning entirely similar to the above shows that ii) can not
occur either. Hence, the function $\theta$ is monotone.\qed

\section{Phase portrait of the fundamental vector field}\label{phasefundvectfield}

With the aim to prove Theorem \ref{Family}, we study in this section the trajectories of the
vector field $X$ defined by \Eqref{VectorField}. This study will
be carried out through a series of technical lemmas.

Since the trajectories
of $X$ are invariant by horizontal translations by multiples of
$2\pi$ (that is, if $\varphi(t)=(\theta(t),z(t))$ is a trajectory
of $X$ then so is the curve $\psi(t)=(\theta(t)+2n\pi,z(t))$ for
any $n\in\mathbb Z$), it is sufficient to consider the
trajectories that pass through some point
$(\theta_0,z_0)\in\Omega$ such that $0\leq\theta_0\leq 2\pi$.

\begin{lemma}\label{SuperaBarreira}
	Let $(\theta_0,z_0)$ be a point in $\Omega$ such that
	$0<\theta_0<\pi$ and $z_0\leq 1$. If
	$\varphi(t)=(\theta(t),z(t)),\;t\in(a,b),$ is the maximal integral
	curve of $X$ satisfying $\varphi(0)=(\theta_0,z_0)$, then there
	exists $c\in(0,b)$ such that $z(c)>1$.
\end{lemma}

\begin{proof}
	We can assume that $z_0<1$, for otherwise the conclusion follows
	immediately from $z'(0)=\sin\theta(0)>0$. Suppose, by
	contradiction, that the conclusion does not hold. Then, by the
	definition of $\Omega$,
	\begin{equation}\label{Rot12}
	\theta_0<\theta(t)<\pi,\;\;\;t\in(0,b),
	\end{equation}
	and so
	\begin{equation}\label{Rot13}
	z'(t)=\sin\theta(t)>0\;\;\text{and}\;\;z(t)<1,\;\;\;t\in(0,b).
	\end{equation}
	Let
	\begin{equation}\label{Rot15}
	\theta_+=\lim_{t\to
		b}\theta(t)\;\;\;\text{and}\;\;\;z_+=\lim_{t\to b}z(t).
	\end{equation}
	By \Eqref{Rot12} and \eqref{Rot13}, $\frac{\pi}{2}<\theta_+\leq\pi$ and $z_+\leq
	1$. From the maximality of $\varphi$ and the fact that $X$ has no
	singularities in $\Omega$, one obtains
	$(\theta_+,z_+)\in\partial\Omega$, and so
	\begin{equation}\label{Rot16}
	z_+=|\cos\theta_+|=-\cos\theta_+.
	\end{equation}

	We have two cases to consider:

	\vskip5pt

	\noindent i) $\theta_+<\pi$ (and so $z_+<1$).

	\vskip5pt

	\noindent ii) $\theta_+=\pi$ (and so $z_+=1$).

	\vskip5pt

	Since the vectors of $X$ on the boundary of  $\Omega$ points inward, we
	can use transversality to conclude that case i) can not occur.  However,
	we will discard this case by a direct argument.
	 Consider the (positive) function
	$\xi:(0,b)\to\mathbb R$ defined by $\xi(t)=z(t)+\cos\theta(t)$. By
	\Eqref{Rot15} and \eqref{Rot16},
	\begin{equation}\label{Rot17}
	\lim_{t\to b}\xi(t)=z_++\cos\theta_+=0.
	\end{equation}
	Using now \Eqref{sistema}, \eqref{Rot15} and \eqref{Rot16}, one
	obtains
	\begin{equation*}\label{Rot18}
	\begin{aligned}
	\lim_{t\to b}\xi'(t)=&\ \lim_{t\to
		b} \ \Bl \sin\theta(t)-\theta'(t)\sin\theta(t) \Br\\
	=&\ \sin\theta_+-\sin\theta_+\sqrt{1-\frac{\cos^2\theta_+}{z_+^2}}\\
	=&\ \sin\theta_+>0.
	\end{aligned}
	\end{equation*}
	Then, there is $t_0\in(0,b)$
	such that $\xi'(t)>\sin\theta_+/2,\;t\in[t_0,b),$ and so
	\begin{equation*}\label{Rot20}
	\xi(t)-\xi(t_0)=\int_{t_0}^t\xi'(s)ds>\frac{\sin\theta_+}{2}(t-t_0),\;\;\;
	t>t_0.
	\end{equation*}
	Letting $t\to b$ in the above inequality, and using \Eqref{Rot17},
	one obtains
	$$
	-\xi(t_0)\geq\frac{\sin\theta_+}{2}(b-t_0)>0,
	$$
	contradicting the fact that $\xi(t)>0$, for all $t$.

	\vskip5pt

	Suppose now ii). From \Eqref{Rot13} and Lemma \ref{ThetaLinhaZero}
	(ii), we obtain $z'(t)\geq\theta'(t)$ for all $t\in(0,b)$, and so
	\begin{equation*}\label{Rot21}
	1>z(t)=z(0)+\int_0^tz'(s)ds \geq
	z(0)+\int_0^t\theta'(s)ds=z_0+\theta(t)-\theta_0,
	\end{equation*}
	for every $t\in(0,b)$. Taking the limit when $t\to b$ in the above
	inequality, and using \Eqref{Rot15} and b), we obtain
	\begin{equation}\label{Rot21a}
	1\geq z_0+\theta_+-\theta_0=z_0+\pi-\theta_0>\pi-\theta_0,
	\end{equation}
	and thus $0<\pi-\theta_0<1$. Choosing $k\in\mathbb N$ such that
	\begin{equation}\label{Rot23}
	\frac{1}{k+1}\leq\pi-\theta_0<\frac{1}{k},
	\end{equation}
	one has, since the cosine function is decreasing on $(0,\pi)$,
	\[
	0<\cos\left(\frac{1}{k}\right)<\cos(\pi-\theta_0)=-\cos\theta_0\leq|\cos\theta_0|.
	\]
	Then, by \Eqref{Rot3} and the above inequality,
	\begin{equation*}
	\frac{\cos^2(1/k)}{z_0^2}<\frac{\cos^2\theta(0)}{z^2(0)}\leq 1,
	\end{equation*}
	and so $z_0>\cos(1/k)$. Using now that $\cos x>1/(1+x)$ for every
	$x\in(0,1]$, one concludes that $z_0>k/(k+1)$. Hence, by
	\Eqref{Rot21a} and \eqref{Rot23},
	\begin{equation*}
	1\geq z_0+\pi-\theta_0>\frac{k}{k+1}+\frac{1}{k+1}=1,
	\end{equation*}
	which is obviously false. This contradiction finishes the proof of
	the lem\-ma.
\end{proof}

\vskip10pt

\begin{lemma}\label{Alcance}
	Let $(\theta_0,z_0)\in\Omega$ such that $0\leq\theta_0<\pi$. If
	$\varphi(t)=(\theta(t),z(t)),\;t\in(a,b),$ is the maximal integral
	curve of $X$ satisfying $\varphi(0)=(\theta_0,z_0)$, then there
	exists $t_0\in(0,b)$ such that $\theta(t_0)=\pi$.
\end{lemma}

\begin{proof}
	Assuming, by contradiction, that the conclusion
	does not hold, one has $0<\theta(t)<\pi,\;t\in(0,b)$, and so
	\begin{equation}\label{Rot29}
	z'(t)=\sin\theta(t)>0,\;\;t\in(0,b).
	\end{equation}
	Let $c\in[0,b)$ such that $z(c)>1$ (such a number $c$ exists by
	Lemma \ref{SuperaBarreira}). Since $\theta$ is bounded above and, by \Eqref{sistema} and
	\eqref{Rot29},
	\begin{equation*}
	\theta'(t)\geq\sqrt{1-\frac{1}{z^2(t)}}>\sqrt{1-\frac{1}{z^2(c)}}>0,\;\;\;t\in(c,b),
	\end{equation*}
	one concludes that $b<\infty$. Then, since $z'(t)\leq 1$, one also has that $z$ is bounded.
	Hence, $\varphi(t)=(\theta(t),z(t))$ converges to a point in $\Omega$ when $t\to b$, but this can not occur because $b<\infty$ (see, for instance, \cite[p.
	91]{Per}).
\end{proof}

\begin{lemma}\label{Simetrico} Given $z_0>1$ and $n\in\mathbb Z$,
	let $\varphi(t)=(\theta(t),z(t)),\;t\in(a,b),$ be the maximal
	integral curve of $X$ satisfying $\varphi(0)=(n\pi,z_0)$. Then,
	$a=-b$ and $\varphi(-t)=R(\varphi(t))$ for every $t\in(-b,b)$,
	where $R$ denotes the reflection in $\mathbb R^2$ with respect to
	the line $\theta=n\pi$. In short, $\varphi$ is symmetric with
	respect to the line $\theta=n\pi$.
\end{lemma}

\begin{proof}
	Consider the curve $\sigma:(-b,-a)\to\Omega$ defined by
	\begin{equation}\label{Rot24}
	\sigma(t)=R(\varphi(-t))=(2n\pi-\theta(-t),z(-t)).
	\end{equation}
	It is easy to see that $\sigma$ is an integral curve of $X$. Since
	$\sigma(0)=(n\pi,z_0)=\varphi(0)$, it follows from the maximality
	of $\varphi$ that $a=-b$ and
	$$
	\varphi(t)=\sigma(t)=R(\varphi(-t)),\;\;\;t\in(-b,b).
	$$
\end{proof}

Lemmas \ref{Alcance} and \ref{Simetrico} tell us that to
obtain a picture of the phase portrait of $X$ it is sufficient to
consider the family of trajectories
$\{\varphi_{\lambda}\}_{\lambda>1}$, where
$\varphi_{\lambda}:(-b_{\lambda},b_{\lambda})\to\Omega$ is the
maximal integral curve of $X$ such that $\varphi_
{\lambda}(0)=(\pi,\lambda)$.

From Lemma \ref{Alcance} and the fact that $z'(t)= \sin\theta(t)$ is positive
on $\theta^{-1}((0,\pi)) $, one concludes, for each $\lambda>1$, that the
trajectory
$\varphi_{\lambda}:(-b_{\lambda},b_{\lambda})\to\Omega$ either
crosses the ray $\{(\theta,z)\in\mathbb
R^2:\theta=0\;\text{and}\;z>1\}$ or converges to a point
$p_{\lambda}\in\partial\Omega$ when $t\to -b_{\lambda}$. Moreover,
each point $(\theta,z)\in\partial\Omega$ such that
$0\leq\theta<\pi$ is the limit point $p_{\lambda}$ of some
$\varphi_{\lambda}$. The later is clear when $\theta\not = 0$ and $\theta \not = \frac{\pi}{2}$,
as the vector field $X$ can be continuously extended, without singularities,
to a neighbourhood of $(\theta,z)$, and follows easily for the other two
values of $\theta$ by a continuity argument.

The following lemma shows that two distinct
trajectories of the family $\{\varphi_{\lambda}\}_{\lambda>1}$ can
not converge to the same point in $\partial\Omega$. Note that this
fact does not follow from the standard theory of ordinary
differential equations, because the vector field $X$ does not admit a
differentiable extension to a neighbourhood of any given point in
$\partial\Omega$.

\begin{lemma}\label{UnicLimPoint}
	With the same notation as above, assume for some
	$(\theta_0,z_0)\in\partial\Omega$ that
	\begin{equation}\label{Rot33A}
	\lim_{t\to -b_{\lambda_1}}\varphi_{\lambda_1}(t)=(\theta_0,z_0)=\lim_{t\to
		-b_{\lambda_2}}\varphi_{\lambda_2}(t).
	\end{equation}
	Then $\lambda_1=\lambda_2$ (and hence
	$\varphi_{\lambda_1}=\varphi_{\lambda_2}$).
\end{lemma}

\begin{proof}
	Assume, by contradiction, that $\lambda_1\neq\lambda_2$, say
	$\lambda_1<\lambda_2$. Setting
	$\varphi_{\lambda_1}=(\theta_1,z_1)$ and
	$\varphi_{\lambda_2}=(\theta_2,z_2)$, from \Eqref{Rot33A} one
	obtains that $\theta_1=\theta_1|_{(-b_{\lambda_1},0]}$ (respectively,
	$\theta_2=\theta_2|_{(-b_{\lambda_2},0]}$) is a diffeomorphism
	from $(-b_{\lambda_1},0]$ (respectively, $(-b_{\lambda_2},0]$) to
	$(\theta_0,\pi]$. Let
	$\psi=\theta_2^{-1}\circ\theta_1:(-b_{\lambda_1},0]\to(-b_{\lambda_2},0]$.
	By the Chain Rule and the Inverse Function Theorem,
	\begin{equation}\label{Rot33B}
	\psi'(t)=(\theta_2^{-1})'(\theta_1(t))\theta_1'(t)=
	\frac{\theta_1'(t)}{\theta_2'(\psi(t))}>0,\;\;\;t\in(-b_{\lambda_1},0].
	\end{equation}
	Since $\lambda_1<\lambda_2$ and $\theta_2(\psi(t))=\theta_1(t)$
	for $t\in(-b_{\lambda_1},0]$, we have $z_2(\psi(t)) > z_1(t)$ and so
	\[
	\theta_2'(\psi(t))=\sqrt{1-\frac{\cos^2\theta_2(\psi(t))}{z_2^2(\psi(t))}}
	\geq \sqrt{1-\frac{\cos^2\theta_1(t)}{z_1^2(t)}}=\theta_1'(t),
	\]
	for all $t\in (-b_{\lambda_1},0]$. Using this inequality in
	\Eqref{Rot33B}, we obtain
	\begin{equation}\label{Rot33C}
	\psi'(t)\leq 1,\;\;\;t\in(-b_{\lambda_1},0].
	\end{equation}
	Using again the equality $\theta_2(\psi(t))=\theta_1(t)$, it follows from
	the
	Chan\-ge of
	Variables Formula that
	\begin{equation*}\label{Rot33D}
	\begin{aligned}
	z_2(0)-z_2(\psi(t))=& \ \int_{\psi(t)}^0z_2'(s)ds=\int_t^0z_2'(\psi(u))
	\psi'(u)du\\
	=& \ \int_t^0\sin\theta_2(\psi(u))\psi'(u)du
	=\int_t^0\sin\theta_1(u)\psi'(u)du,
	\end{aligned}
	\end{equation*}
	for every $t\in (-b_{\lambda},0]$.
	Hence, by \Eqref{Rot33C},
	\begin{equation*}\label{Rot33E}
	\begin{aligned}
	z_2(0)-z_2(\psi(t))=& \ \int_t^0z_1'(u)\psi'(u)du\\
	\leq& \ \int_t^0z_1'(u)du =z_1(0)-z_1(t), \;\;\;
	t\in(-b_{\lambda_1},0].
	\end{aligned}
	\end{equation*}
	Taking the limit when $t\to
	-b_{\lambda_1}$, and using \Eqref{Rot33A}, one obtains
	$\lambda_2=z_2(0)\leq z_1(0)=\lambda_1$, contradicting our
	assumption $\lambda_1<\lambda_2$. Hence $\lambda_1=\lambda_2$.
\end{proof}

\begin{figure}[htp!]
	\begin{tikzpicture}%
	[line cap=round,>=latex,x=1.5cm,y=1.5cm]
	%
	\clip(-.5,-1.) rectangle (7,4.6)
	%

	%
	\begin{scope}[line width=.7pt,cyan!80
	\clip(0,0) rectangle (2 * pi,4.6)
	\input{curva1}
	\end{scope}
	%
	\begin{scope}[line width=.7pt,magenta!80
	\clip(0,0) rectangle (2 * pi,4.6)
	\input{curva3}
	\input{curva4}
	\end{scope}
	\begin{scope}[line width=.7pt,green!80
	\clip(0,0) rectangle (2 * pi,4.6);
	\input{curva5}
	\end{scope}

	%
	\begin{scope}[line width=1.5pt,red!80
	\clip(0,0) rectangle (2 * pi,4.6)
	\input{curva2}
	\end{scope}
	%
	\begin{scope}[line width=1.5pt,blue!80
	\clip(0,0) rectangle (2 * pi,4.6)
	\input{sphere}
	\end{scope}
	%
	\begin{scope}[domain=0:2*pi
	\draw[dashed,black!50,thick] plot[smooth] (\x,{abs(cos(\x r))});
	\end{scope}
	%

	\draw (3,3.25) node[fill=white,above,inner sep=0pt] {\footnotesize
		$\lambda_0$};
	\draw (3.3,1.45) node[fill=white,above,inner sep=0pt] {\footnotesize
		$\sqrt{2}$};

	\draw[dashed,black!50] (3.1415,0) -- (3.1415,4) (6.283,0) -- (6.283,4);
	%

	\draw[fill] (0,1) circle (1.2pt) (3.1415,1) circle (1.2pt) (6.2831,1)
	circle (1.2pt)  (3.1415,3.22) circle (1.2pt) (3.1415,1.4142) circle
	(1.2pt);
	%
	%
	\begin{scope}[line width=.5pt]
	\draw[->] (-.5,0) -- (6.8,0)node[below]{\small$\theta$}
	\draw[->] (0,-.5) -- (0,4.5)node[left]{\small$z$}

	\foreach \x/\l in
	{1.57075/{\pi/2},3.1415/{\pi},4.71225/{3\pi/2},6.283/{2\pi}}{%
		\draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt);
		\draw[shift={(\x,0)}] (0,-8pt)node[fill=white,inner sep=0pt]
		{\small$\l$};
	}

	\foreach \y in {1}{
		\draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {\small
			$\y$};
	}

	\draw (0,0) node[below left] {\small$0$};
	\end{scope}
	\end{tikzpicture}
	\caption{Graphic representation of the phase portrait of the vector field
	$X$ for
	$\theta\in[0,2\pi]$.}\label{Figure2}
\end{figure}

\section{Proof of Theorem \ref{Family}}\label{proofmaintheorem}

As before, for each $\lambda>1$ denote by
$\varphi_{\lambda}(t)=(\theta(t),z(t)),\;t\in(-b_{\lambda},b_{\lambda}),$
the maximal integral curve of $X$ such that
$\varphi_{\lambda}(0)=(\pi,\lambda)$. From Lemma
\ref{UnicLimPoint} and the discussion that precedes its statement
one concludes that there is a unique $\lambda_0>1$ such that
$\varphi_{\lambda_0}(t)\to (0,1)$ when $t\to -b_{\lambda_0}$.
Moreover, when $\lambda\neq\lambda_0$, the trajectory $\varphi_{\lambda}$ either crosses the ray
$\{(\theta,z)\in\mathbb R^2:\theta=0\;\text{and}\;z>1\}$ or converges to a
point in $\partial\Omega$ depending on whether $\lambda>\lambda_0$ or
$1<\lambda<\lambda_0$ (see Figure \ref{Figure2}).

For each $\lambda>1$, consider the curve
$\alpha_{\lambda}:(-b_{\lambda},b_{\lambda})\to\mathbb R^3$ defined by
$\alpha_{\lambda}(t)=(x(t),0,z(t))$, where
\begin{equation}\label{Rot59b}
x(t)=\int_0^t\cos\theta(s)ds,
\end{equation}
and the surface $M_{\lambda}$ of $\mathbb R^3$ obtained
by the rotation of the image of $\alpha_{\lambda}$ around the
$x$-axis. As we have seen in Section \ref{convexprofcurve}, the length of the
shape
operator of $M_{\lambda}$ equals 1 at every point. The detailed classification
of the surfaces $M_\lambda$ reads:

\begin{theorem}\label{surfaces}
	Let $M_{\lambda}$ be as above.

	\vskip5pt

	\noindent(i) If $\lambda>\lambda_0$ then $M_{\lambda}$ is a
	complete $C^{\infty}$-surface. Moreover, $M_{\lambda}$ is periodic
	and has self-intersections.

	\vskip5pt

	\noindent(ii) $M_{\lambda_0}$ is incomplete, but it can be extended in infinite many
	ways to a complete $C^3$-surface satisfying $|A|\equiv 1$. Any
	such extension has self-intersections.

	\vskip5pt

	\noindent(iii) $M_{\sqrt{2}}$ is the sphere with center at
	$(-\sqrt{2},0,0)$ and radius $\sqrt{2}$ (minus two points).

	\vskip5pt

	\noindent(iv) If $\sqrt{2}<\lambda<\lambda_0$ or
	$1<\lambda<\sqrt{2}$ then $M_{\lambda}$ is incomplete and cannot
	be extended to a surface with $|A|\equiv 1$.
\end{theorem}

Concerning the Gaussian curvature of the surfaces obtained in the above theorem, we observe that
the only surfaces with positive Gaussian curvature are the surfaces $M_{\lambda}$
with $1<\lambda\leq \sqrt{2}$. For all the others, the Gaussian curvature
changes the signal.

	\vskip10pt

Theorem \ref{surfaces} is a direct consequence of the following result:
\begin{theorem}\label{curves}
	Let $\alpha_{\lambda}:(-b_{\lambda},b_{\lambda})\to\mathbb R^3$ be as
	above.

	\vskip5pt

	\noindent(i) If $\lambda>\lambda_0$ then $b_{\lambda}=+\infty$ and
	$\alpha_{\lambda}$ is of class $C^{\infty}$. Moreover,
	$\alpha_{\lambda}$ is periodic and has self-intersections.

	\vskip5pt

	\noindent(ii) $b_{\lambda_0}<+\infty$ and $\alpha_{\lambda_0}$ can
	be extended in infinite many ways to a profile curve of class
	$C^3$ defined on $\mathbb R$. Any such extension has
	self-intersections.

	\vskip5pt

	\noindent(iii) $\alpha_{\sqrt{2}}$ is a parametrization by arc
	length of the semicircle in the $xz$-plane with center at
	$(-\sqrt{2},0,0)$
	and
	radius $\sqrt{2}$.

	\vskip5pt

	\noindent(iv) If $\sqrt{2}<\lambda<\lambda_0$ or
	$1<\lambda<\sqrt{2}$ then $b_{\lambda}<+\infty$ and
	$\alpha_{\lambda}$ cannot be extended to a profile curve defined
	on an open interval containing $(-b_{\lambda},b_{\lambda})$ properly.
\end{theorem}

\begin{figure}
	\begin{subfigure}[b]{0.4\linewidth}
		\centering
		\begin{tikzpicture}
		\coordinate (0) at (0,0);
		\coordinate (x) at (4,0);
		\coordinate (y) at (0,3);
		\draw[->] (0)--(x)node[below]{$x$};
		\draw[->] (0)--(y)node[left]{$z$};
		\node[anchor=south west] at (-.15,1.5)
		{\includegraphics[scale=0.8]{drawing.pdf}};
		\draw[dashed] (0,1)node[left]{$1$}--(4,1);
		\node[left] at (0,0) {\phantom{$\sqrt{2}$}};
		\end{tikzpicture}
		\caption{$\alpha_{\lambda}$ with $\lambda>\lambda_0$ }
		\vspace{4ex}
	\end{subfigure}

	\begin{subfigure}[b]{0.4\linewidth}
		\centering
		\begin{tikzpicture}
		\coordinate (0) at (0,0);
		\coordinate (x) at (4,0);
		\coordinate (y) at (0,3);
		\draw[->] (0)--(x)node[below]{$x$};
		\draw[->] (0)--(y)node[left]{$z$};
		\node[anchor=south west] at (-.15,.85)
		{\includegraphics[scale=0.8]{drawing-1.pdf}};
		\draw[dashed] (0,1)node[left]{$1$}--(4,1);
		\node[left] at (0,0) {\phantom{$\sqrt{2}$}};
		\end{tikzpicture}
		\caption{$\alpha_{\lambda}$ with $\lambda=\lambda_0$}
		\label{Figb}
		\vspace{4ex}
	\end{subfigure}
	\begin{subfigure}[b]{0.4\linewidth}
		\centering
		\begin{tikzpicture}
		\coordinate (0) at (0,0);
		\coordinate (x) at (4,0);
		\coordinate (y) at (0,3);
		\draw[->] (0)--(x)node[below]{$x$};
		\draw[->] (0)--(y)node[left]{$z$};
		\node[anchor=south west] at (-.15,.85)
		{\includegraphics[scale=0.8]{drawing-6.pdf}};
		\draw[dashed] (0,1)node[left]{$1$}--(4,1);
		\node[left] at (0,0) {\phantom{$\sqrt{2}$}};
		\end{tikzpicture}
		\caption{$\alpha_{\lambda}$ with $\lambda=\lambda_0$}
		\vspace{4ex}
	\end{subfigure}
	\begin{subfigure}[b]{0.4\linewidth}
		\centering
		\begin{tikzpicture}
		\coordinate (0) at (0,0);
		\coordinate (x) at (4,0);
		\coordinate (y) at (0,3);
		\draw[->] (0)--(x)node[below]{$x$};
		\draw[->] (0)--(y)node[left]{$z$};
		\node[anchor=south west] at (-.15,.85)
		{\includegraphics[scale=0.8]{drawing-4.pdf}};
		\draw[dashed] (0,1)node[left]{$1$}--(4,1);
		\node[left] at (0,0) {\phantom{$\sqrt{2}$}};
		\end{tikzpicture}
		\caption{$\alpha_{\lambda}$ with $\lambda=\lambda_0$}
		\vspace{4ex}
	\end{subfigure}
	\begin{subfigure}[b]{0.4\linewidth}
		\centering
		\begin{tikzpicture}
		\coordinate (0) at (0,0);
		\coordinate (x) at (4,0);
		\coordinate (y) at (0,3);
		\draw[->] (0)--(x)node[below]{$x$};
		\draw[->] (0)--(y)node[left]{$z$};
		\node[anchor=center] at (2,1.1)
		{\includegraphics[scale=1]{drawing-5.pdf}};
		\draw[dashed] (0,1)node[left]{$1$}--(4,1);
		\draw[dashed]
		(0,1.4142)node[left]{$\sqrt{2}$}--(4,1.4142);
		\end{tikzpicture}
		\caption{$\alpha_\lambda$ with
			$\sqrt{2}<\lambda<\lambda_0$}
		\vspace{4ex}
	\end{subfigure}
	\begin{subfigure}[b]{0.4\linewidth}
		\centering
		\begin{tikzpicture}
		\coordinate (0) at (0,0);
		\coordinate (x) at (4,0);
		\coordinate (y) at (0,3);
		\draw[->] (0)--(x)node[below]{$x$};
		\draw[->] (0)--(y)node[left]{$z$};
		\node[anchor=center] at (2,1.2)
		{\includegraphics[scale=1]{drawing-2.pdf}};
		\draw[dashed] (0,1)node[left]{$1$}--(4,1);
		\draw[dashed]
		(0,1.4142)node[left]{$\sqrt{2}$}--(4,1.4142);
		\end{tikzpicture}
		\caption{$\alpha_\lambda$ with
			$\sqrt{2}<\lambda<\lambda_0$}
		\vspace{4ex}
	\end{subfigure}
	\begin{subfigure}[b]{0.4\linewidth}
		\centering
		\begin{tikzpicture}
		\coordinate (0) at (0,0);
		\coordinate (x) at (4,0);
		\coordinate (y) at (0,3);
		\draw[->] (0)--(x)node[below]{$x$};
		\draw[->] (0)--(y)node[left]{$z$};
		\node[anchor=north] at (2,1.22)
		{\includegraphics[scale=01]{drawing-3.pdf}};
		\draw[dashed] (0,1)node[left]{$1$}--(4,1);
		\draw[dashed]
		(0,1.4142)node[left]{$\sqrt{2}$}--(4,1.4142);
		\end{tikzpicture}
		\caption{$\alpha_\lambda$ with $1<\lambda<\sqrt{2}$}
		\vspace{4ex}
	\end{subfigure}
	\caption{Graphic representation of the curves $a_\lambda$}\label{Figure3}
\end{figure}

\newpage

\begin{proof}
(i) From
$\lambda>\lambda_0$ and the discussion in the beginning of this
section one infers that there exists $t_0>0$ such that
$\theta(-t_0)=0$. Applying Lemma \ref{Simetrico} with $n=0$ and
$n=1$ one concludes that $b_{\lambda}=+\infty$. Being the
trajectory of a vector field of class $C^{\infty}$,
$\varphi_{\lambda}$, and hence $\alpha_{\lambda}$, is of class
$C^{\infty}$.

We will now prove that $\alpha_{\lambda}$ is periodic. Since, by
\Eqref{VectorField} and Lemma \ref{Simetrico}, the maps
$t\in\mathbb R\mapsto\varphi_{\lambda}(t+2t_0)$ and $t\in\mathbb
R\mapsto(\theta(t)+2\pi,z(t))$ are both trajectories of $X$
passing through $(3\pi,\lambda)$, one has
\begin{equation}\label{Rot30}
\varphi_{\lambda}(t+2t_0)=\big(\theta(t)+2\pi,z(t)\big),\;\;\;t\in\mathbb
R.
\end{equation}
On the other hand, by \Eqref{Rot59b} one has
\begin{equation}\label{Rot48b}
\begin{aligned}
x(t+2t_0)=&\ \int_{0}^{t+2t_0}\cos\theta(s)ds\\
=&\int_{0}^{t}\cos\theta(s)ds
+\int_{t}^{t+2t_0}\cos\theta(s)ds\\
=&\ x(t)+\int_{t}^{2t_0}\cos\theta(s)ds+\int_{2t_0}^{2t_0+t}\cos\theta(s)ds\\
=&\ x(t)+\int_{t}^{2t_0}\cos\theta(s)ds+\int_{0}^{t}\cos\theta(s+2t_0)ds.
\end{aligned}
\end{equation}
Since $\theta(s+2t_0)=\theta(s)+2\pi$ by \Eqref{Rot30}, it follows
that
\begin{equation}\label{Rot48c}
\begin{aligned}
x(t+2t_0)=&\ x(t)+\int_{t}^{2t_0}\cos\theta(s)ds
+\int_{0}^{t}\cos\theta(s)ds\\
=&\ x(t)+\int_{0}^{2t_0}\cos\theta(s)ds=\ x(t)+x(2t_0),\;\;\;t\in\mathbb R.
\end{aligned}
\end{equation}
Since $z(t+2t_0)=z(t)$ for all $t\in\mathbb R$ by \Eqref{Rot30},
the curve $\alpha_{\lambda}$ is periodic.

To complete the proof of (i), it remains to show that
$\alpha_{\lambda}$ is non-embedded. In fact, we will show that the
restriction of $\alpha_{\lambda}$ to the interval $(-t_0,t_0)$ has
already self-intersections. For that observe first that, since
$\theta(-t_0)=0$, $\theta(0)=\pi$ and $\theta'>0$, the function
$\theta=\theta|_{[-t_0,0]}$ is a diffeomorphism from $[-t_0,0]$ to
$[0,\pi]$. In particular, there exists a unique $t_1\in(0,t_0)$
such that $\theta(-t_1)=\pi/2$. Let $\xi:[-t_0,-t_1]\to[-t_1,0]$ be
defined by
\begin{equation}\label{Rot35}
\xi(t)=\theta^{-1}(\pi-\theta(t)).
\end{equation}
Clearly, $\xi$ is a diffeomorphism, $\xi(-t_0)=0$ and
$\xi(-t_1)=-t_1$. Moreover,
\begin{equation}\label{Rot36}
\xi'(t)=-\frac{\theta'(t)}{\theta'(\xi(t))}\,,\;\;\;t\in[-t_0,-t_1],
\end{equation}
and
\begin{equation}\label{Rot36a}
\cos\theta(\xi(t))=\cos(\pi-\theta(t))=-\cos\theta(t),\;\;\;t\in[-t_0,-t_1].
\end{equation}
By \Eqref{sistema} and \eqref{Rot36a},
\begin{equation}\label{Rot37}
\theta'(\xi(t))=\sqrt{1-\frac{\cos^2\theta(\xi(t))}{z^2(\xi(t))}}
=\sqrt{1-\frac{\cos^2\theta(t)}{z^2(\xi(t))}}.
\end{equation}
Since, by \Eqref{sistema},
\begin{equation*}
z(\xi(t))-z(t)=\int_t^{\xi(t)}z'(s)ds
=\int_t^{\xi(t)}\sin\theta(s)ds>0,\;\;\;t\in[-t_0,-t_1),
\end{equation*}
one then has
\begin{equation}\label{Rot39}
\theta'(\xi(t))>\sqrt{1-\frac{\cos^2\theta(t)}{z^2(t)}}
=\theta'(t),\;\;\;t\in[-t_0,-t_1).
\end{equation}

Using the informations collected above, we will now compare the
values of $x(t)$ for $t=-t_0$, $t=-t_1$ and $t=0$. Since
$\pi/2\leq\theta(t)\leq\pi$ for $t\in[-t_1,0]$, from \Eqref{Rot59b}
we obtain
\begin{equation}\label{Rot40a}
x(-t_1)=\int_0^{-t_1}\cos\theta(s)ds=-\int_{-t_1}^{0}\cos\theta(s)ds>0=x(0).
\end{equation}
On the other hand, by \Eqref{Rot36}, \eqref{Rot36a} and \eqref{Rot39}, one has
\begin{equation}\label{Rot41}
\begin{aligned}
\int_{-t_1}^0\cos\theta(t)dt =& \ \int_{-t_1}^{-t_0}\cos\theta(\xi(s))\xi'(s)ds
=\int_{-t_0}^{-t_1}\frac{\cos\theta(\xi(s))\theta'(s)}{\theta'(\xi(s))}ds\\
>&\ \int_{-t_0}^{-t_1}\cos\theta(\xi(s))ds=-\int_{-t_0}^{-t_1}\cos\theta(s)ds.
\end{aligned}
\end{equation}
Hence, by \Eqref{Rot59b} and inequality above,
\begin{equation}\label{Rot42}
x(-t_0)=-\int_{-t_0}^0\cos\theta(t)dt
=-\int_{-t_0}^{-t_1}\cos\theta(t)dt-\int_{-t_1}^0\cos\theta(t)dt<0.
\end{equation}

The curve $\alpha_{\lambda}$ is symmetric with respect to the line
$x=x(0)=0$. Indeed, by Lemma \ref{Simetrico} one has
\begin{equation}\label{Rot43}
\theta(t)=2\pi-\theta(-t),\;\;z(t)=z(-t),\;\;\;t>0,
\end{equation}
and so
\begin{equation}\label{Rot44}
\begin{aligned}
x(t)=&\ \int_{0}^{t}\cos\theta(s)ds=\int_{0}^{t}\cos\theta(-s)ds\\
=&\ -\int_{0}^{-t}\cos\theta(s)ds=-x(-t),
\end{aligned}
\end{equation}
for every $t>0$.

Since $x'>0$ on $(-t_0,-t_1)$ and, by \Eqref{Rot40a} and
\eqref{Rot42}, $x(-t_0)<0<x(-t_1)$, there exists a unique
$t_2\in(t_1,t_0)$ such that $x(-t_2)=0=x(0)$. Then, by \Eqref{Rot44},
\begin{equation}\label{Rot48}
x(t_2)=-x(-t_2)=0=x(-t_2).
\end{equation}
Since $z(t_2)=z(-t_2)$ by \Eqref{Rot43}, it follows that
$\alpha_{\lambda}(t_2)=\alpha_{\lambda}(-t_2)$. Hence, the
restriction of $\alpha_{\lambda}$ to the interval $(-t_0,t_0)$ has
a self-intersec-tion.

\vskip10pt

\noindent(ii) We begin by showing that $b_{\lambda_0}<+\infty$.
Let $t_1\in(-b_{\lambda_0},0)$ such that $\theta(t_1)=\pi/2$. For
every $t\in(-b_{\lambda_0},t_1]$ we have
\begin{equation}\label{Finito7}
\begin{aligned}
t_1-t=& \ \int_t^{t_1}1ds=\int_{\theta(t)}^{\pi/2}(\theta^{-1})'(u)du
=\int_{\theta(t)}^{\pi/2}\frac{1}{\theta'(\theta^{-1}(u))}du\\
=& \
\int_{\theta(t)}^{\pi/2}\frac{z}{\sqrt{z^2-\cos^2\theta}}
(\theta^{-1}(u))du\\
\leq&
\lambda_0\int_{\theta(t)}^{\pi/2}\frac{1}{\sqrt{z-\cos\theta}}
(\theta^{-1}(u))\,du,
\end{aligned}
\end{equation}
where in the last inequality we used the fact that
$1<z\circ\theta^{-1}<\lambda_0$ and the cosine function is nonnegative on $(0,\pi/2]$.

\begin{claim}
There is $C_1>0$ such that
\begin{equation}\label{Finito6}
\frac{1}{\sqrt{z(t)-\cos\theta(t)}}\leq\frac{C_1}{\sin^{\frac{3}{4}}\theta(t)},\;\;\;\;t\in(-b_{\lambda_0},t_1].
\end{equation}
\end{claim}

Indeed, since
$\theta(t)\to 0$ and $z(t)\to 1$ when $t\to -b_{\lambda_0}$, from
\Eqref{sistema} we obtain

\begin{equation}\label{ThetaLinhavaiparazero}
    \lim\limits_{t\to -b_{\lambda_0}}\theta'(t)=0.
\end{equation}
Then, again by \Eqref{sistema},
\begin{equation}\label{Finito1}
\begin{aligned}
\lim_{t\to
	-b_{\lambda_0}}\frac{\sin^2\theta(t)}{\theta'(t)^2}=& \ \lim_{t\to
	-b_{\lambda_0}}\frac{z^2(t)}{z(t)+\cos\theta(t)}\,
\frac{\sin^2\theta(t)}{z(t)-\cos\theta(t)}\\
=&\ \frac{1}{2}\lim_{t\to
	-b_{\lambda_0}}\frac{\sin^2\theta(t)}{z(t)-\cos\theta(t)}\\
=& \ \frac{1}{2}\lim_{t\to-b_{\lambda_0}}
\frac{2\theta'(t)\sin\theta(t)\cos\theta(t)}{\sin\theta(t)(1+\theta'(t))}\\
=&\ \frac{1}{2}\lim_{t\to
	-b_{\lambda_0}}\frac{2\theta'(t)\cos\theta(t)}{1+\theta'(t)}=0,
\end{aligned}
\end{equation}
and so
\begin{equation}\label{Finito2}
\begin{aligned}
\lim_{t\to -b_{\lambda_0}}\frac{z(t)-\cos\theta(t)}{\sin\theta(t)}
=& \ \lim_{t\to -b_{\lambda_0}}
\frac{\sin\theta(t)(1+\theta'(t))}{\theta'(t)\cos\theta(t)}\\
=& \ \lim_{t\to
	-b_{\lambda_0}}\frac{\sin\theta(t)}{\theta'(t)}=0.
\end{aligned}
\end{equation}
From \Eqref{sistema} and the above equality one obtains
\begin{equation}\label{Finito4}
\lim_{t\to
	-b_{\lambda_0}}\frac{\theta'(t)^2}{\sin\theta(t)}=\lim_{t\to
	-b_{\lambda_0}}\frac{z(t)+\cos\theta(t)}{z^2(t)}\,\frac{z(t)-\cos\theta(t)}{\sin\theta(t)}=0.\nonumber
\end{equation}
Therefore,
\begin{equation}\label{Finito3}
\begin{aligned}
\lim_{t\to
	-b_{\lambda_0}}\frac{\sin^{\frac{3}{2}}\theta(t)}{z(t)-\cos\theta(t)}
=& \ \lim_{t\to-b_{\lambda_0}}\frac{\frac{3}{2}\theta'(t)
	\sin^{\frac{1}{2}}\theta(t)\cos\theta(t)}{\sin\theta(t)(1+\theta'(t))}\\
=& \ \frac{3}{2}\lim_{t\to-b_{\lambda_0}}\frac{\theta'(t)}{\sin^{\frac{1}{2}}
	\theta(t)}=0,
\end{aligned}
\end{equation}
and the claim follows.

From \Eqref{Finito7} and \eqref{Finito6}, we obtain
\begin{equation}\label{Finito8}
t_1-t\leq\lambda_0C_1\int_{\theta(t)}^{\pi/2}\sin^{-\frac{3}{4}}udu,\;\;\;
-b_{\lambda_0}<t<t_1.
\end{equation}
Setting $C_2=\inf\left\{\sin u/u:u\in(0,\frac{\pi}{2}]\right\}$, one has
\begin{equation*}\label{Finito9}
\sin^{-\frac{3}{4}}u\leq
C_2^{-\frac{3}{4}}u^{-\frac{3}{4}},\;\;\;\;u\in(0,\frac{\pi}{2}].
\end{equation*}
Using this information in \Eqref{Finito8}, we obtain
\begin{equation}\label{Finito10}
\begin{aligned}
t_1-t\leq\lambda_0C_1C_2^{-\frac{3}{4}}\int_{\theta(t)}^{\pi/2}
u^{-\frac{3}{4}}du
=& \ 4\lambda_0C_1C_2^{-\frac{3}{4}}u^{\frac{1}{4}}\Big|_{\theta(t)}^{\pi/2}\\
=&\ 4 \lambda_0C_1C_2^{-\frac{3}{4}}\left[\left(\pi/2\right)^{\frac{1}{4}}
-\theta(t)^{\frac{1}{4}}\right]\\
<& \ 4\lambda_0C_1C_2^{-\frac{3}{4}}(\pi/2)^{\frac{1}{4}},
\end{aligned}
\end{equation}
for every $t\in(-b_{\lambda_0},t_1)$. Therefore,
$b_{\lambda_0}<+\infty$.

In order to prove that $\alpha_{\lambda_0}$ can be extended to a
profile curve of class $C^3$ defined on $\mathbb R$, we need to
evaluate the limits of $\theta''$ and $\theta'''$ when
$t\to -b_{\lambda_0}$.
From \Eqref{sistema} one obtains, after some work,
\begin{equation}\label{Rot52}
\theta''(t)=\frac{\sin\theta(t)}{z(t)\theta'(t)}
+\frac{\sin\theta(t)\cos\theta(t)}{z^2(t)}
-\frac{\theta'(t)\sin\theta(t)}{z(t)}.
\end{equation}
Since $\theta(t)\to 0$, $z(t)\to 1$ and
$\theta'(t)\to 0$ when $t\to -b_{\lambda_0}$, it follows from \Eqref{Finito2} and \eqref{Rot52} that
\begin{equation}\label{Rot53}
\lim_{t\to -b_{\lambda_0}}\theta''(t)=\lim_{t\to
	-b_{\lambda_0}}\frac{\sin\theta(t)}{\theta'(t)}=0.
\end{equation}
As to $\lim\limits_{t\to -b_{\lambda_0}}\theta'''(t)$,
observe that, since $z(t)\to 1$, $\theta(t)\to 0$, $\theta'(t)\to
0$ and $\theta''(t)\to 0$, the derivatives of either of the last
two terms on the right hand side of \Eqref{Rot52} goes to zero
when $t\to -b_{\lambda_0}$. Therefore,
\begin{equation}\label{Rot55}
\begin{aligned}
\lim_{t\to -b_{\lambda_0}}\theta'''(t)=&\ \lim_{t\to
	-b_{\lambda_0}}\left(\frac{\sin\theta}{z\theta'}\right)'(t)\\
=&\ \lim_{t\to
	-b_{\lambda_0}}\left\{-\frac{\sin\theta}{z^2}\frac{\sin\theta}{\theta'}
+\frac{1}{z}\left(\frac{\sin\theta}{\theta'}\right)'\right\}(t),\\
=&\ \lim_{t\to
	-b_{\lambda_0}}\left(\frac{\sin\theta}{\theta'}\right)'(t),
\end{aligned}
\end{equation}
where in the last equality we used \Eqref{Finito1}. Since, by
\Eqref{sistema},
\begin{equation}\label{Rot55a}
\frac{\sin\theta}{\theta'}=\frac{z}{\sqrt{z+\cos\theta}}\,
\frac{\sin\theta}{\sqrt{z-\cos\theta}}\,,
\end{equation}
one has
\begin{equation}\label{Rot57}
\begin{aligned}
\left(\frac{\sin\theta}{\theta'}\right)'=&\ \left(
\frac{\sin\theta}{\sqrt{z+\cos\theta}}
-\frac{z\sin\theta(1-\theta')}{2(z+\cos\theta)^{3/2}}\right)
\frac{\sin\theta}{\sqrt{z-\cos\theta}}\\
&\ +\cos\theta-\frac{z(1+\theta')}{2\sqrt{z+\cos\theta}}\,
\frac{\sin^2\theta}{(z-\cos\theta)^{3/2}}.
\end{aligned}
\end{equation}
The first term on the right hand side goes to zero by
\Eqref{Finito1} and \eqref{Rot55a}. Hence,
\begin{equation}\label{Rot58}
\lim_{t\to
	-b_{\lambda_0}}\left(\frac{\sin\theta}{\theta'}\right)'=1-\frac{1}{2\sqrt{2}}\lim_{t\to
	-b_{\lambda_0}}\frac{\sin^2\theta}{(z-\cos\theta)^{3/2}}.
\end{equation}
Using \Eqref{sistema} again, one obtains
$$
\lim_{t\to
	-b_{\lambda_0}}\frac{\sin^2\theta}{(z-\cos\theta)^{3/2}}=\lim_{t\to
	-b_{\lambda_0}}\frac{4\cos\theta\sqrt{z+\cos\theta}}{3z(1+\theta')}=\frac{4\sqrt{2}}{3}.
$$
It now follows from \Eqref{Rot55}, \Eqref{Rot58} and the above
equality that
\begin{equation}\label{Rot59}
\lim_{t\to
	-b_{\lambda_0}}\theta'''(t)=1-\frac{1}{2\sqrt{2}}\frac{4\sqrt{2}}{3}=\frac{1}{3}.
\end{equation}

It is possible to extend $\alpha_{\lambda_0}$ gluing together copies of
$\alpha_{\lambda_0}$. By \Eqref {ThetaLinhavaiparazero}, \eqref{Rot53} and
\eqref{Rot59}, this extension is (at least) $C^4$ (recall that if a profile
curve is of class $C^s$ then its corresponding angle function is of class
$C^{s-1}$). Another way to extend $\alpha_{\lambda_0}$ is gluing together
copies of $\alpha_{\lambda_0}$ and horizontal segments with any length and
with height equal to 1. By the same equations, these extensions are $C^3$ but
not $C^4$ (see Figure \ref{Figure3}, items (B), (C) and (D), for a sample of
these
extensions).

To complete the proof of (ii), it remains to show that any extension of
$\alpha_{\lambda_0}$ is non-embedded. But clearly this follows from the fact
that $\alpha_{\lambda_0}$ has a self-intersection, which in turn can be
proved as in (i) (with $b_{\lambda_0}$ playing the role of $t_0$).

\vskip10pt

\noindent(iii) As can be easily seen, the curve
\begin{equation*}\label{Rot70}
\psi(t)=(\pi+t/\sqrt{2},\sqrt{2}\cos(t/\sqrt{2})),\;\;\;
t\in(-\sqrt{2}\pi/2,\sqrt{2}\pi/2),
\end{equation*}
is a trajectory of $X$. Since $\psi(0)=(\pi,\sqrt{2})$, one has
$\varphi_{\sqrt{2}}=\psi$.
Then, by \Eqref{Rot59b},
$$
\alpha_{\sqrt{2}}(t)=(-\sqrt{2}\sin(t/\sqrt{2})-\sqrt{2},
\sqrt{2}\cos(t/\sqrt{2})),\;\;\;t\in(-\frac{\sqrt{2}\pi}{2},
\frac{\sqrt{2}\pi}{2}),
$$
which is a parametrization by arc length of the portion of the circle in the
$xz$-plane with
center $(-\sqrt{2},0,0)$ and radius $\sqrt{2}$ that is above the $x$-axis.

\vskip10pt

\noindent(iv) Let $(\theta_0,z_0)=\lim_{t\to
	-b_{\lambda}}\varphi_{\lambda}(t)$. Since $\lambda\neq\sqrt{2}$
and, by item (iii), $\varphi_{\sqrt{2}}(t)\to(\pi/2,0)$ when $t\to
-b_{\sqrt{2}}=-\sqrt{2}\pi/2$, it follows from Lemma
\ref{UnicLimPoint} that $0<z_0<1$ and either $0<\theta_0<\pi/2$ or
$\pi/2<\theta_0<\pi$.

Suppose, by contradiction, that $b_{\lambda}=+\infty$. Since
$\theta(t)\to\theta_0$ when $t\to -b_{\lambda}$ and either
$0<\theta_0<\pi/2$ or $\pi/2<\theta_0<\pi$, there exist
$\varepsilon>0$ and $t_1\in\mathbb R$ such that
$$
\sin\theta(t)>\varepsilon,\;\;\;t\leq t_1.
$$
Then, by \Eqref{Rot1},
$$
z(t_1)-z_0>z(t_1)-z(t)=\int_t^{t_1}\sin\theta(s)ds>\varepsilon(t_1-t),
$$
for every $t<t_1$, a contradiction. Hence, $b_{\lambda}<\infty$.

Suppose, by contradiction, that $\alpha_{\lambda}$ can be extended
to a profile curve $\widetilde\alpha_{\lambda}(t)=(\widetilde
x(t),0,\widetilde z(t)),\;t\in(a,b),$ where $a<-b_{\lambda}$, say.
Let $\widetilde\theta:(a,b)\to\mathbb R$ be a function satisfying
$$
\widetilde\alpha_{\lambda}'(t)=(\cos\widetilde\theta(t),0,\sin\widetilde\theta
(t)),\;t\in(a,b).
$$
Since $\widetilde\alpha_{\lambda}'(t)=\alpha_{\lambda}'(t)$ for
all $t\in(-b_{\lambda},b_{\lambda})$, and, by \Eqref{Rot59b},
$$
\alpha_{\lambda}'(t)=(x'(t),0,z'(t))=(\cos\theta(t),0,\sin\theta(t)),
\;\;\;t\in(-b_{\lambda},b_{\lambda}),
$$
one concludes that $\theta$ and the restriction of
$\widetilde\theta$ to the interval $(-b_{\lambda},b_{\lambda})$
differ by an integer multiple of $2\pi$. Assuming without loss of
generality that
$\theta=\widetilde\theta|_{(-b_{\lambda},b_{\lambda})}$, one has
$$
\widetilde\theta'(-b_{\lambda})=\lim_{t\to
	-b_{\lambda}}\widetilde\theta'(t)=\lim_{t\to
	-b_{\lambda}}\theta'(t)=0
$$
and
$$
\widetilde z(-b_{\lambda})=\lim_{t\to -b_{\lambda}}\widetilde
z(t)=\lim_{t\to -b_{\lambda}}z(t)=z_0,
$$
which contradicts Lemma \ref{ThetaLinhaZero} (i) as $0<z_0<1$.
\end{proof}

Now we finish the prove of our main theorem.

\vskip5pt

\noindent{\bf Completion of proof of Theorem \ref{Family}:} Let $\alpha_{\lambda}$ and $M_{\lambda}$, $\lambda>1$, be as in the beginning of this section, and let $\mathcal F_1=\{M_{\lambda}:\lambda>\lambda_0\}$. Let $\mathcal A$ be the set of extensions of $\alpha_{\lambda_0}$ constructed in the proof of Theorem \ref{curves} (ii), and $\mathcal F_2$ the (infinite) family of surfaces obtained by the rotation of the images of the curves in $\mathcal A$ around the $x$-axis. By Theorem \ref{surfaces} (i), the surfaces in $\mathcal F_1$ have the properties  stipulated in the statement of the theorem. That the surfaces in $\mathcal F_2$ also meet the stipulated conditions is a consequence of the proof of Theorem \ref{curves} (ii). This establishes the first part of the theorem.

Let $M$ be a complete rotational surface of class $C^2$ satisfying $|A|\equiv
1$. Applying a rigid motion if necessary, we can assume that its axis of revolution is
the $x$-axis and that its profile curve $\mathcal C$ is contained in the $xz$-plane. Let  $\alpha(t)=(x(t),0,z(t))$, $t\in(a,b)$, be a
parametrization of $\mathcal C$ such that $z(t)>0$ and
$||\alpha'(t)||=1$ for all $t\in(a,b)$. Let $\theta:(a,b)\to\mathbb R$ be a
$C^1$-function such that
$$
(x'(t),0,z'(t))=(\cos\theta(t),0,\sin\theta(t)),\;\;\;t\in(a,b).
$$
As we have seen in Section \ref{convexprofcurve}, the function $\theta$
satisfies
$$
\theta'^2(t)+\frac{\cos^2\theta(t)}{z^2(t)}=1,\;\;\;t\in(a,b).
$$

If $\theta$ is constant, by the above equality one has that $z$ is constant.
Then $\sin\theta=z'=0$ and so $z=|\cos\theta|\equiv 1$. Being complete, $M$
is then a right circular cylinder of radius 1.

Suppose now that $\theta$ is not constant. Since $\theta$ is monotone by Proposition \ref{monotona}, reparametrizing $\alpha$ if necessary we can assume that $\theta'(t) \geq 0$ for all $t\in(a,b)$. Let
$t_0\in(a,b)$ such that $\theta'(t_0)>0$. Without loss of generality, we can
assume that $0\leq\theta(t_0)<2\pi$. Let $(a_1,b_1)\subset(a,b)$ be the
maximal interval containing $t_0$ on which $\theta'>0$. As we have seen in
Section \ref{convexprofcurve}, the map
$t\in(a_1,b_1)\mapsto\varphi(t):=(\theta(t),z(t))$ is an integral curve of the
vector field $X$ defined by \Eqref{VectorField}. By the results in Section
\ref{phasefundvectfield}, there is a unique trajectory
$\varphi_{\lambda}:(-b_{\lambda},b_{\lambda})\to\Omega$ in the family $\{\varphi_{\lambda}\}_{\lambda>1}$ that passes
through $(\theta(t_0),z(t_0))$. Let $s_0\in(-b_{\lambda},b_{\lambda})$
such that $\varphi_{\lambda}(s_0)=(\theta(t_0),z(t_0))=\varphi(t_0)$.
Assuming without loss of generality that $t_0=s_0$, it follows from the
maximality of $\varphi_{\lambda}$ that
$\varphi=\varphi_{\lambda}|_{(a_1,b_1)}$. Write
$\varphi_{\lambda}=(\theta_{\lambda},z_{\lambda})$ and consider its associated
profile curve
$t\in(-b_{\lambda},b_{\lambda})\mapsto\alpha_{\lambda}(t)=(x_{\lambda}(t),0,z_{\lambda}(t))$
 (\cf the beginning of this section). Since
$$
x_{\lambda}'(t)=\cos\theta_{\lambda}(t)=\cos\theta(t)=x'(t),\;\;\;t\in(a_1,b_1),
$$
it holds that
\begin{equation}\label{new}
\alpha(t)=\alpha_{\lambda}(t)+(d,0,0),\;\;\;t\in(a_1,b_1),
\end{equation}
for some $d\in\mathbb R$.

\vskip5pt

\noindent{\bf Claim.} $a\leq -b_{\lambda}$ and $b\geq b_{\lambda}$.

\vskip5pt

Assuming, by contradiction, that $-b_{\lambda}<a$, one has $-b_{\lambda}<a_1$. Then, since $\theta_{\lambda}'>0$ on $(-b_{\lambda},b_{\lambda})$
and $\theta|_{(a_1,b_1)}=\theta_{\lambda}|_{(a_1,b_1)}$,
$$
\lim_{t\to a_1}\theta'(t)=\lim_{t\to a_1}\theta_{\lambda}'(t)=\theta_{\lambda}'(a_1)>0.
$$
From the above inequality and definition of $(a_1,b_1)$ one obtains $a=a_1$. It now follows from \Eqref{new} that $\alpha$ can be extended to an interval containing $(a,b)$ properly, contradicting the completeness of $M$. This contradiction proves that $a\leq -b_{\lambda}$. In the same way, one proves that $b\geq b_{\lambda}$.

It follows from the Claim that $a_1=-b_{\lambda}$. Indeed, if we had $-b_{\lambda}<a_1$, reasoning as above one would obtain $\lim_{t\to a_1}\theta'(t)>0$. On the other hand, from $a<a_1$ one would obtain  $\lim_{t\to a_1}\theta'(t)=0$, a contradiction. In the same manner, one proves that $b_1=b_{\lambda}$. Hence $(a_1,b_1)=(-b_{\lambda},b_{\lambda})$ and, by \Eqref{new},
\begin{equation}\label{new3}
\alpha(t)=\alpha_{\lambda}(t)+(d,0,0),\;\;\;t\in(-b_{\lambda},b_{\lambda}),
\end{equation}
for some $d\in\mathbb R$.

From \Eqref{new3} one obtains that either $\lambda\geq\lambda_0$ or $\lambda=\sqrt{2}$. In fact, if we had $1<\lambda<\sqrt{2}$ or $\sqrt{2}<\lambda<\lambda_0$, from Theorem \ref{curves} (iv) we would obtain that $M$ is a translation of $M_{\lambda}$, and so $M$ would be incomplete, contradicting the hypothesis.

In the case $\lambda>\lambda_0$, it follows from \Eqref{new3} and Theorem \ref{curves} (i) that $\alpha(t)=\alpha_{\lambda}(t)+(d,0,0)$ for all $t\in\mathbb R$, and therefore $M=M_{\lambda}$ (up to translation).

In the case $\lambda=\sqrt{2}$, it follows from \Eqref{new3} and Theorem \ref{curves} (iii) that $\alpha(t)=\alpha_{\sqrt{2}}(t)+(d,0,0),\;t\in(-b_{\sqrt{2}},b_{\sqrt{2}})$. Since $M$ is complete, one concludes that $M$ is, up to translation, the sphere with center at $(-\sqrt{2},0,0)$ and radius $\sqrt{2}$.

Finally, consider the case $\lambda=\lambda_0$. By \Eqref{new3}, $M$ is an extension of $M_{\lambda_0}$. Since $M$ is complete and, by Theorem \ref{surfaces} (ii), $M_{\lambda_0}$ is incomplete, one has $a<-b_{\lambda_0}$ and $b>b_{\lambda_0}$. We will conclude that, up to congruence, $\alpha$ belongs to the family $\mathcal A$ and so $M\in\mathcal F_2$. For that we can assume that $\theta'$ is not identically zero on $(a,-b_{\lambda_0})\cup(b_{\lambda_0},b)$, for otherwise $z=1$ outside $(-b_{\lambda_0},b_{\lambda_0})$ and the conclusion holds trivially. We claim that for any $s\in(a,-b_{\lambda_0})\cup(b_{\lambda_0},b)$ at which $\theta'(s)>0$, there is an open interval $I$ of length $2b_{\lambda_0}$ containing $s$ such that $\alpha(I)$ differs from the image of $\alpha_{\lambda_0}$ by a horizontal vector. We will prove the claim in the case $b_{\lambda_0}<s<b$ (the proof in the case $a<s<-b_{\lambda_0}$ is analogous). Denote by $(a_2,b_2)\subset(a,b)$ the maximal interval
 containing $s$ on which $\theta'>0$. By what we have already proved (\cf
 \Eqref{new3}), $\alpha((a_2,b_2))$ coincides with a horizontal translation of
 the image of $\alpha_{\lambda}$, for some $\lambda>1$. Since $a_2>-\infty$ and
 $z(a_2)=1$ (by Lemma \ref{ThetaLinhaZero}(i), since $\theta'(a_2)=0$), we have
 $\lambda=\lambda_0$ and the claim is proved. Let $I,J$ be subintervals of
 $(a,b)$ such that $\alpha(I)$ and $\alpha(J)$ are both horizontal translations
 of the image of $\alpha_{\lambda_0}$. If the distance between $I$ and $J$ is
 positive but less than $2b_{\lambda_0}$, then, by the previous claim, one has
 $\theta'=0$, and hence $z=1$, in the interval between $I$ and $J$. It is now
 clear that the image of $\alpha$ is made up of curves congruent to
 $\alpha_{\lambda_0}$ and eventually of horizontal segments of height equal to
 1. Therefore, $\alpha\in\mathcal A$ and so $M\in\mathcal F_2$.\qed

redpajama_set_name: RedPajamaArXiv


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The following is a list of pipeline accidents in the United States in 2015. It is one of several lists of U.S. pipeline accidents. See also list of natural gas and oil production accidents in the United States.

Incidents

This is not a complete list of all pipeline accidents. For natural gas alone, the Pipeline and Hazardous Materials Safety Administration (PHMSA), a United States Department of Transportation agency, has collected data on more than 3,200 accidents deemed serious or significant since 1987.

A "significant incident" results in any of the following consequences:
 fatality or injury requiring in-patient hospitalization
 $50,000 or more in total costs, measured in 1984 dollars
 liquid releases of five or more barrels (42 US gal/barrel)
 releases resulting in an unintentional fire or explosion

PHMSA and the National Transportation Safety Board (NTSB) post incident data and results of investigations into accidents involving pipelines that carry a variety of products, including natural gas, oil, diesel fuel, gasoline, kerosene, jet fuel, carbon dioxide, and other substances. Occasionally pipelines are repurposed to carry different products.

 On January 14, a fire broke out at an oil pipeline pump station, in Texas City, Texas. Texas City fire officials said that company officials reported that there had been issues with the pump station over the weekend. There were no injuries.
 On January 14, during work to free a trapped inline inspection unit, a leak was discovered on the Evangeline Pipeline, near Cameron Parish, Louisiana. This pipeline had been given a Corrective Action Order in October 2014, due to a number of leaks.
 Also on January 14, a gas pipeline exploded near the Ross Barnett Reservoir in Brandon, Mississippi, creating a sizable crater in the ground and burning 6 acres of vegetation before the fire was extinguished. No injuries were reported. The failure was due to a "hard spot" from manufacturing, that already had a repair sleeve on it. There are 788 sleeves on the Index 129 pipeline from Edna, Texas, to Sterlington, Louisiana; and, 726 sleeves on the Index 130 pipeline from Marchand Junction, Louisiana to Kosciusko, Mississippi. Both were built from pipe made in 1952.
 On January 16, a transmission pipeline operated by Kinder Morgan subsidiary Southern Natural Gas had an equipment malfunction in Walthall County, Mississippi. The cause was cold weather; ice build-up caused a regulator to malfunction.
 On January 17, oil from a broken pipeline seeped into the Yellowstone River, and contaminated the water supply 10 miles south of Glendive, Montana. The release was from Bridger Pipeline LLC's 12-inch Poplar line, which can carry 42,000 barrels a day of crude from the Bakken Formation and runs from Canada south to Baker, Montana. Bridger Pipeline is a subsidiary of True Cos., a privately held Wyoming-based company. The company said in a statement that the pipeline was shut down within an hour of the leak. About 30,000 gallons of crude were spilled, with about 28,000 gallons of crude being lost.
 On January 21, a petroleum products pipeline in Honolulu, Hawaii ruptured, due to external corrosion, spilling about 42,000 gallons of petroleum product, of which about 22,000 gallons was lost.
 On January 21, a crude oil pipeline pump station caught fire northwest of Texas City, Texas. Texas City fire officials said that company officials reported that there had been issues with the pump station over the weekend.
 On January 26, a 20-inch Enterprise Products ATEX pipeline carrying ethane exploded and burned, in Brooke County, West Virginia. Despite snow in the area, five acres of woodlands burned, and 1,283,000 gallons of ethane were consumed, or lost. The fireball melted siding on nearby homes and damaged power lines; it is believed that day's snowy weather lessened the damage. Reports suspect a girth weld failure from ductile tensile overload, with the pipeline being less than two years old. Survey showed that the pipeline had slumped three feet since being built. A geotechnical survey conducted by Pennsylvania Soil and Rock determined that the failed pipe was installed across a transition area or "head wall" of an old underground mine and surface strip mine. There were no injuries.
 On January 29, near Bowling Green, Missouri, a rupture in a Rockies Express 42-inch-diameter natural gas pipeline blew a 20 by 20-foot crater and forced a six-hour evacuation of 50 families. The rupture occurred in a vacant field a few yards east of Pike County Road 43. Strong winds dissipated gas until a temporary cap was put in place. This explosion did $2,672,345 in property damage and was caused by a faulty weld; the pipe was manufactured in 2008.
 On February 10 in Hopkinton, Massachusetts, low temperatures caused freezing rain to do "natural force damage" to a Kinder Morgan Tennessee Gas Pipeline, causing $55,150 worth of property damage.
 On February 17, a suspected electrical arc made a hole in a Marathon Petroleum pipeline in Shively, Kentucky, spilling about 6,700 gallons of jet fuel. More than 2,500 tons of soil were removed to clean up the area.
 On February 25, a 26-inch crude oil pipeline in Navarro County near the Town of Dawson, Texas, failed, spilling about 50 barrels of crude oil. Near the failure, investigation showed that the pipe had lost about 80% of its thickness due to external corrosion. This anomaly was not seen in a 2011 test of this pipeline.
 On March 2, a Kinder Morgan Tennessee Gas Pipeline leaked due to equipment failure, causing $281,890 of property damage in Marshall, Mississippi. The pipe was manufactured in 1944.
 On March 13 a pipeline Patrol pilot identified an oil sheen on a pond near Tehuacana Creek, Texas which was then linked to a leaking 10-inch petroleum products pipeline. About 50 barrels of diesel fuel were spilled.
 On March 20, a pipe owned by Kinder Morgan subsidiary Southern Natural Gas failed in Rolling Fork, Mississippi; the pipe was manufactured in 1930.
 On March 23, another Southern Natural Gas pipeline failed due to a cracked weld at Augusta, Georgia, causing $311,785 in property damage.
 On April 9, 2 Williams Companies pipelines broke within hours of each other in Marshall County, West Virginia. A 4-inch condensate pipeline broke at 8 pm local time, spilling about 132 barrels of condensate into a creek. Around 10:50 pm local time, a 12-inch gas pipeline ruptured. There was no fire or injuries. Heavy rains were said to be the cause of the failures.
 On April 10, a landowner, in Glen Mills, Pennsylvania, reported a petroleum odor to the Sunoco Control Center. The source of the odor was traced to the Point Breeze to Montello 12" refined products pipeline system. About 1,600 gallons of mixed refined products was leaked. The cause was external corrosion.
 On April 13, a Kinder Morgan / Natural Gas Pipeline Company of America pipeline exploded and burned near Borger, Texas. One home was evacuated, but there were no injuries. The explosion, caused by equipment failure due to environmental cracking, caused $455,000 in property damage. A 60-foot section and a 30-foot section of pipe were ejected and landed about 60 feet and 100 feet away from the failure site, respectively. Metallurgical analysis revealed the cause of the leak was stress corrosion cracking. The pipe was manufactured in 1967.
 On April 17, a 12-inch natural gas pipeline near Fresno, California operated by Pacific Gas and Electric Company was ruptured by a backhoe. The resulting explosion killed 1 person and injured 12 others.
 On May 15, Kinder Morgan's Tennessee Gas Pipeline leaked in Powell County, Kentucky causing $23,400 in property damage.
 On May 19, a Plains All American Pipeline oil pipeline ruptured near Refugio State Beach, also near Goleta, California, spilling about 124,000 gallons of crude oil. It is referred to as the Refugio Oil Spill.
 On May 31, a 24-inch diameter Texas Eastern natural gas pipeline under the flooding Arkansas River ruptured in Little Rock, Arkansas, damaging a tugboat which "sustained damage on the port side, the smokestack and main deck." The pipeline exploded with enough force to raise geysers of water in the river and also threw cement chunks onto the tugboat's deck. The pipeline is owned by Houston-based Spectra Energy. Analysis concluded that high water flow in the river eroded the cover over the pipeline and scoured away the soil support under the pipeline. Once the pipeline was exposed as a free span, rushing water caused vertical oscillation of the pipe, resulting in metal fatigue, cracking the pipe adjacent to a girth weld. The pipe was manufactured in 1952.
 On June 9 in Moorehouse Parish, Louisiana, Kinder Morgan's Tennessee Gas Pipeline equipment failed due to environmental cracking, and leaked, causing $73,395 in property damage. The pipe was manufactured in 1948.
 On June 9, a 24-inch natural gas pipeline ruptured in Lycoming County, Pennsylvania. About 130 individuals were evacuated from their homes. No injuries or damage reported. There was no fire. The cause was stress corrosion cracking. The pipe was installed in 1963. When it failed, it was owned by Transcontinental Gas Pipe Line Company.
 On June 10, Kinder Morgan's El Paso Natural Gas control/relief equipment failed and leaked in Gray County, Texas. Technicians believed debris or trash may have caused the valve to malfunction.
 On June 13, an Energy Transfer Partners 42-inch gas gathering pipeline exploded and burned near Cuero, Texas. Seven homes were evacuated, but there were no injuries. The heat from the flames melted electric lines, cutting off power to 130 homes. Sixteen people were evacuated from homes near the explosion, which caused $500,000 in damage. Energy Transfer Company reported to the Texas Commission on Environmental Quality that as much as 165,732 pounds of volatile organic compounds burned. The Railroad Commission of Texas, in its Pipeline Safety Evaluation, and Inspection Package number 111385, published November 2, 2015, concluded that material failure, as a result of bending stress overload that caused a weld to fail, led to the rupture, explosion, and fire. The pipeline was constructed and installed in 2012.
 On June 15, Kinder Morgan's Natural Gas Pipeline Co. of America equipment failed adjacent to a weld seam for unknown causes, with $260,555 of property damage in Marshall, Texas (that area's third documented Kinder Morgan leak). The pipe was manufactured in 1967.
 On June 18, in Victoria Texas, Kinder Morgan's Tennessee Gas Pipeline pipe failed due to external corrosion and caused $159,346 in property damage. The pipe was manufactured in 1944.
 On June 22, a truck driver was killed when his rig veered off a highway and broke above ground facilities for a propylene pipeline in Houston, Texas. The highway was closed for several hours while the gas dissipated.
 Four workers were hurt on June 25, when a 4-inch gas pipeline exploded at a gas pipeline facility, near White Deer, Texas. 2 of the workers were critically injured. The cause of the explosion was not immediately known.
 On July 10, a fitting on a 20-inch Plains All American Pipeline crude oil pipeline broke, spilling 4200 gallons of crude oil near Grantfork, Illinois. Much of the crude reached a nearby creek. There were no injuries.
 On July 15, two workers were hurt by an explosion, when a bulldozer hit a 4-inch gas pipeline, at an EQT gas compressor station in Worthington, Pennsylvania. One of the workers later on died from the injuries. While the contractor did use the 811 One call system, the pipeline hit was not a participant of 811. Later this incident was a factor in restricting 811 exemptions.
 On August 3, a Kinder Morgan Tennessee Gas Pipeline 16-inch diameter pipe exploded; the blowout created a 70-foot by 30-ft crater 350 feet from a house and did $191,498 in property damage. Investigation found a longitudinal split in the pipe about 55 feet long along an ERW seam, but the failure was from stress corrosion cracking. The pipe was manufactured in 1947.
 On August 7, a natural gas liquids pipeline in Weld County, Colorado burned after being struck by a third party.
 On August 13, crew working for Colonial Pipeline damaged one of Colonial's lines in Kannapolis, North Carolina, spilling about 6,000 gallons of petroleum product. About 1,000 gallons of product was lost.
 On August 13, a natural gas pipeline in Cypress, Texas, ruptured and leaked while a contract crew worked in the area. The pipeline was owned by Gulfsouth Pipeline. There were no injuries or immediate damage; residents were evacuated.
 On August 17, a 4-inch gas pipeline was hit and ruptured by an excavator, in Winfield, West Virginia. During the shut down of the pipeline, the gas ignited, burning a repair crew worker.
 On August 26, two maintenance divers were injured while working on a pipeline owned by Boardwalk/Gulf South Pipeline Co. 25 miles offshore of Louisiana when the pipeline ruptured and the gas ignited.
On September 5, 2015, Great Lakes Gas Transmission Company had an explosion at a compressor station in Kittson County, Minnesota, about a mile south of the Canadian border. The cause was mechanical failure caused by stress corrosion cracking. The pipe was installed in 1972.
 On September 16, a Shell Oil Company crude oil pipeline leaked near Tracy, California, spilling about 37,800 gallons of crude. An in line inspection report of that pipeline two weeks before had detected no anomalies.
 On September 21, a Colonial Pipeline 32-inch main line leaked in Centreville, Virginia, spilling about 4,000 gallons of gasoline and forcing several nearby businesses to close.
 On October 8, natural gas ignited and exploded at a Williams Companies pipeline facility in Gibson, Louisiana. Four employees were killed and one injured. Cost of the accident was $7,545,044. The cause of the explosion was from procedure not being followed during welding work.
 On November 3, 2015, a leak occurred due to a crack in a girth weld on a Texas Eastern Transmission 24-inch diameter pipeline near Onalaska, Texas. The pipe was manufactured in 1954.
 On November 13, at Bakersfield, California, a heavy equipment operator doing agricultural work struck a 30-inch-diameter PG&E natural gas transmission line, causing a release of gas. The gas ignited and in the resulting explosion one person died and three others were sent to the hospital with second- and third-degree burns. A nearby home and barn were damaged by the fire.
 On November 15, work was being performed on a flow control valve, on a Sunoco 10-inch crude oil pipeline, in Wortham, Texas, when the valve failed, injuring five workers, and spilling some crude oil. It was later determined that the valve was under 400 psi of nitrogen pressure when it was being worked on.
On November 21, Paiute Pipeline personnel were running an inline inspection tool through the 8-inch-diameter Carson lateral pipeline. The leading eye of the tool struck the inside of a long radius 45-degree elbow, puncturing the pipe and resulting in a rupture. Public and private property damage included six vehicles and a landscape retaining wall as a result of the blowing dirt and rocks.
 On November 24, Transco reported a leak on a "dead-leg" section of piping, located on a 20-inch header/regen pipe, within a station in Austin, Pennsylvania. There was no fire or injuries. The facility was shut down for 14 days for repairs and the cost of the damage was approximately $224,528. The cause was internal microbial corrosion.
 On November 30, about 11,000 gallons of gasoline, butane and propane leaked from a pipeline, in eastern Summit County, Utah.
On December 1, personnel from Enterprise Crude Pipeline, LLC (Enterprise Products), discovered a spill at their West Cushing Tank Farm in Cushing, Oklahoma. Approximately 42,000 gallons of crude oil were released within the terminal. A tank line had failed from internal corrosion.
 On December 8, a contractor drilled into an , that transports oil from a holding station in Ventura, to a Wilmington refinery, near Long Beach, while setting new poles for Southern California Edison along State Route 118 near Somis that spilled about .

References

Lists of pipeline accidents in the United States

redpajama_set_name: RedPajamaWikipedia


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Q: Convert Reports values into numbers from strings in a script activity I want to perform a script activity on an Account Send Summary report I dropped into a Data Extension. The current numbers inside of the system cannot be logically filtered because they contain "," and "%". Is there a way to remove them and convert the string into numbers using ssjs in a script activity for automation?
I would prefer using ampscript but I can't use it inside of the script activity. If you know another way of me converting the numbers and filtering only the rows meeting my requirements, that would be helpful.
My requirements is to show results in an email where the campaign had less than 95% deliverable rate.

A: I would highly recommend that you take care of this via SQL instead of AMPscript or SSJS as it would be much quicker and more efficient.
Below are 2 sample snippets to help you to remove the commas and percentages and turn them into numbers.
Comma:
CONVERT(NUMERIC(10,0), REPLACE(r.Bounces, ',','') )

Input: 2,945
Output: 2945

Percent:
CONVERT(NUMERIC(10,2), REPLACE(r.BounceRate, '%','') )

Input:  24.95%
Output: 24.95

Please note with the percent that it will be changed to a decimal and will not include the percent sign. I would not use any formatting to turn this back into a percent, as it would take 24.95 and make it 2495% on output.  I would recommend doing a concat and adding a % at the end, whether via SQL, AMPscript or SSJS when sending to output in the email/CP.

redpajama_set_name: RedPajamaStackExchange


Maybe the house's main room? Many practical tasks require a high standard of quality and design – the kitchen shall invite the efficiency and well-being. Based on this we choose our kitchen items. We also put importance to assortment structure across the item groups. Porcelain/ stoneware, cutlery and glass to the covered tables are produced mainly in China, peculiar developed for the Norwegian market in close cooperation with suppliers / factories. Kitchen items and series of steel, wood, glass, plastics / acrylics and more are produced mainly in the Far East.
Our own brand in the kitchen is Modern House.
We are now ready to present Modern House Kitchen at our showroom in Bryne. The Modern House brand has now established a great selection of all product groups within this product range. Have a look through our catalogue here and sample some of our products. You are also welcome to visit our showroom in Bryne.


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East Lansing, Haslett, Okemos, Webberville & Williamston
Quick links... Home Neighborhoods Charlotte & Eaton Rapids Dansville, Leslie, Mason & Stockbridge Delhi Township, Holt & Southside Lansing Delta Township & Grand Ledge Dewitt & St. Johns Downtown, Old Town & REO Town East Lansing, Haslett, Okemos, Webberville & Williamston Jackson & Hillsdale Counties MSU Campus State Capitol
NeighborhoodsCharlotte - Eaton Rapids
Charlotte will welcome home Medal of Honor recipient Francis Charles Flaherty
Courtesy: Navy Personnel Command Public Affairs
Francis Charles Flaherty
By: Cali Montana
CHARLOTTE, Mich. — Charlotte native Francis Charles Flaherty died aboard the USS Oklahoma during the attack on Pearl Harbor on Dec. 7, 1941. This week, his body will be returned home to be buried.
Flaherty was a posthumous Medal of Honor recipient. During the attack by Japanese forces, as his ship was capsizing, he stayed in a gun turret and held a flashlight, helping others to escape at the cost of his own life.
Flaherty's nephew John Flaherty was a month old when the attack on Pearl Harbor occurred, so he never met his uncle. He said the day before the attack on Pearl Harbor, his uncle mailed the family Christmas presents and his gift was three straw and wooden dolls. He still has them.
Courtesy of John Flaherty
The dolls Francis Flaherty sent to his nephew for Christmas.
He's deeply saddened about his uncle's loss but proud of his heroism.
"Out of all the wars that we have been involved in, there have been 4,000 Medal of Honor recipients, and my uncle is one of them. So yeah, I'm proud," John Flaherty said.
Francis Flaherty's awards and decorations include the Purple Heart Medal, Combat Action Ribbon, American Defense Service Medal, and the World War II Victory Medal. The escort ship USS Flaherty was named in his honor.
Francis Flaherty memorial in Charlotte
Flaherty's remains were identified in 2019 using DNA reference samples from family members. According to the Navy Personnel Command Public Affairs Office, scientists at the Defense POW/MIA Accounting Agency work to identify sailors who died in past conflicts.
"What an honor it is to finally be able to relocate his remains through the efforts of the POW/MIA project, which identified his remains and return him home to Charlotte, where he grew up," said John Flaherty.
John Flaherty said it gives him chills knowing his uncle will be back home, and he and several family members will be at the burial on Saturday.
World War II Pilot missing for nearly 80 years receives special burial
Remains of a soldier killed during Korean War returned after 71 years
Want to see more local news? Visit the FOX47News Website.
Sign up for newsletters emailed to your inbox.
Select from these options: Breaking News, Severe Weather, School Closings, Daily Headlines, and Daily Forecasts.
Cali Montana
Neighborhood Reporter

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This high capacity double-decker hot towel cabinet and ultraviolet sterilizer performs double duty for your salon or spa. Each super-high capacity cabinet serves dual purpose as an Ultraviolet Sterilizer and a Hot Towel Warmer. With separate function switches, you can run the Sterilizer or the Heat function individually - or both at the same time! Keep your towels at a toasty 175 degrees Fahrenheit and use powerful ultraviolet rays to sterilize your tools and implements.
140+ disposable facial towel capacity!
Ultraviolet rays provide sterilization of bacteria and microbes on glasses, brushes, electrodes, or any kind of manicure/pedicure instruments you have. You can rest assured that your implements are safe, clean and free of bacteria for use on your next client. Use this as a sterile overnight storage cabinet.
You get an extra Ultraviolet bulb for FREE with purchase!
Our ultra-soft gentle exfoliating microfiber towels are especially beneficial for maintaining personal health & beauty. Super thin fibers allow for more thorough cleansing to remove more dirt, bacteria, oils and residual cosmetics than non-microfiber towels. People who suffer from acne or allergies will enjoy the powerful yet gentle cleansing and exfoliation they can achieve with our microfiber towels.
Our plush, ultra-soft, flexible towels are finer than cotton and soft as silk. Use them to Remove makeup with just water and reduce allergies caused by chemicals. Our towels contain no chemical additives and dry quickly between use.
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Dotted with dainty nonpareils, these molasses-spiked morsels are stealth health sweets — no one will guess they're secretly made with whole wheat flour.
In medium bowl, whisk both flours, ginger, baking soda, cinnamon, and salt until blended.
In large bowl, with mixer on low speed, beat sugar and oil spread until blended. On high speed, beat until light and creamy, scraping bowl with rubber spatula. Beat in eggs and molasses. On low speed, blend in flour mixture. Cover dough and refrigerate until easier to handle (dough will be sticky), 1 hour.
Preheat oven to 350 degrees F. With greased hands, shape dough into 1-inch balls. Dip tops of balls in nonpareils; place, 2 1/2 inches apart, on ungreased cookie sheets.
Bake cookies until tops are slightly cracked, 9 to 11 minutes. (Cookies will be very soft.) Cool cookies on cookie sheets 1 minute. With metal spatula, transfer cookies to wire rack to cool. Repeat with remaining dough. Store in airtight container at room temperature up to 3 days or in freezer up to 1 month.


HealthyStuff.org, a website run by the Ecology Center, a Michigan-based non-profit, released a report that several national vendors sell low-cost jewelry that contain toxic chemicals known to cause cancer and allergies. The products, which included inexpensive jewelry for adults and children, were tested for chemicals including lead, cadmium, arsenic, mercury, bromine, and chlorine (PVC).
More than half of the tested items resulted in a "high" level of concern because one or more of the dangerous chemicals were detected at high levels. The presence of hazardous substances were found by using an X-ray fluorescence detector. Retailers such as Target, Walmart, Claire's, and H&M were among several popular national chains represented. To find the items tested by brand name and vendor, click here and see the individual contaminants in each product.
A total of 27 percent of the 99 tested items contained more than 300 ppm lead in one or more components, which is higher than the legal limit for children's products. Additionally, ten percent contained cadmium, 93 percent chromium, 30 percent nickel, seven percent brominated flame retardants, and 12 percent PVC. Click here to view a summary of the results as well as other chemicals found.
Cadmium and chromium are known carcinogens. Nickel is carcinogenic in addition to causing skin and lung problems. Brominated flame retardants contain more than 1,000 ppm bromine, which causes nervous system damage and disrupts genetic materials. Chlorine, which is contained in PVC at 25,000 ppm, does not have consistently documented long-term effects, but is know to irritate the skin, lungs, and eyes.
However, the CPSC has failed to regulate cadmium in children's products in general, instead opting for a voluntary standard developed by the industry. Six states, including California, Connecticut, Illinois, Maryland, Minnesota, and Washington, have moved to regulate the harmful chemical through state law. Also, the Toxics Substance Control Act is in the process of being reformed through a bill, the Safe Chemicals Act (S.847), introduced by Senator Lautenburg in 2011 with 15 co-sponsors.
Click here to watch a video and contact your senators about the Safe Chemical Act.
Is Mysterious Illness Fault of an Old Chemical Spill?
West Nile Aerial Spraying: Is it Safe?


This post is part of a series of tutorials on customizing Divi with the Engined child theme.
Customizing Divi social icons is probably one of points that have raised more questions within the Divi users community, so we will begin this Engined tutorial series addressing it.
First of all, you should have your CSS output set to "Development Mode" in the "Engined Output Mode" section of the Customizer, so you can preview your changes. If you have never set it yet to "Production Mode" then you´re ok, because "Development Mode" is the default mode in Engined.
The first social icons option you will find in Engined is a dual button that lets you choose between the default Divi social icons settings (found in the Divi ET epanel) and the Engined customization settings.
Even if you choose to customize your social icons instead of using the settings from the epanel, Engined will automatically import any setting values you might have already entered via your epanel.
You´ll notice that a couple of new options appear underneath the button as soon as you hit it: one to open the social icon links in a new browser tab and the other to add nofollow attribute to links. You´ll no doubt find the first option very useful because you´ve probably noticed that in the default Divi condition the links open in the same browser tab, and there was no way to change this without adding some code to your site.
You will also see a whole sub-section appearing below those options. It is titled Individual social media settings, and here is where you can type or paste your social media urls and set which social icons you want to display on your site. Here is also where the data imported from your epanel settings is used to fill the appropriate default settings.
As you can see in the image above, Engined comes with settings for a few other social media networks that have been added to the default Divi options, since only Facebook, Twitter, Google+ and RSS are included in Divi.
With Engined you can also set your social icons for Pinterest, Linkedin, Instagram, Skype, Flickr, Youtube and Vimeo.
Once your social icons have been set, you´re ready to start customizing the way they are displayed on your Divi site. We´ll begin having fun with the footer social icons.
The Divi footer is divided in two parts: the Main Footer where the footer widgets are displayed and the so called Footer Bottom or the bottom band where you can see your site credits. In Divi out of the box, the footer Social Icons are displayed on the right side of the Footer Bottom.
Engined includes two separate Customizer sections for the Main Footer and the Footer Bottom (besides yet another one for the Footer Credits, but we will take care of that one on another tutorial). So go ahead and open the Footer Bottom section you will find inside the Engined General Settings panel.
Next, if you scroll down the Engined Footer Bottom section of the Customizer, you´ll find the Social Icons sub-section where you have several settings.
The first option lets you hide or display your social icons in the footer. The Divi "Show social icons in footer" option is always forced to on by the Engined child theme, so here is where you can disable the display of footer social icons if you don´t want to show them.
Regarding the social icons colors, you have the option to display all the icons with the same color (Divi default) or to use the color taken from the original color used by each media network (so that Engined uses Facebook blue, Google+ red, and so on).
If you select "Use Divi Default", you can choose the color and the hover color you want to display for all the social icons, whereas if you select "Use Original Icon Colors" each icon will be displayed with its own original color, the hover color being a darker shade of it.
Social icons size: This slider control lets you change the size of your icons if you like it to be different from the Divi default 24px size.
Social icons separation: The spacing between social icons can be set via a slider control that lets you modify the left margin for each icon.
The other place where social icons are shown in Divi is on the Top Header. By default, Divi displays the phone and email info at the left side of the header and the social icons at the right side, but Engined lets you reverse this layout in the Top Header section of the Customizer, also inside the Engined General Settings panel.
Finally, remember that once you´re done with all your customizations, you can switch the Engined Output Mode back to Production Mode in order to make your site work faster.
And that´s it. I hope you´ve found this tutorial useful. See you in the next one!
Brilliant, thank you Luis. I will definitely be using Engined on my next website, need to read up a bit more about it to see if I could and should be using it existing sites which were created with the DIVI Children plugin and Child Theme.
Just one thing: I LOVE Engined, it makes my life so much easier!! I have one problem though… when I try to set the social icons like you explained, in the Customizer, it works in the pre-mode, but when I save and publish, it still shows the default icons from the epanel. Have you any idea what is happening here? Thank you so much!!
I´ve checked somebody else´s site last week with a similar issue, and it was a problem related to the initialization of setting values imported from the epanel. The way to solve it was really simple: I checked the one icon she didn´t want, saved the changes, unchecked the unwanted icon again, saved once more… and it worked. Can you please check this out?
Sorry about the inconvenience. I still have to find the right way to fix this, but I´m afraid I´m going to be too busy with all the changes coming with the Divi 2.4 update to have this ready soon.
Hi Luis! Yesssss it worked!! Thanks a lot, you just made my day!
Best of luck with the changes for the update and keep on rockin'!!
First of all, I would like to thank you for all the efforts you've made for the great improvement / developer's ease of use experience in using DIVI. I am aware about what happened in the other fence.
I am a newbie and your product Engined will great help me in developing. How or what will I do to install Engined flawlessly…?
Can you check whether you have a standard installation of Divi like yoursite.com/wp-content/themes/Divi/ or not?
I see, that´s why you´re having problems: the parent theme folder should be named just 'Divi', not 'Divi2.1.2', and it should be within the '/themes/' folder.
Just wanted to let you know that this helped me a lot today! Could not have figured it out alone.
Is it ok if I rename it using Filezilla? Btw, there are Engined folder there already. There are 6 Engined folders coz I tried 6x. Should I delete the Engined folders and I'l do an Add New theme within WP?
Is that the right thing to do in my scenario?
Yes you can delete all those repeated Engined folders and install Engined from scratch once you have renamed the parent theme folder.
Great Luis, it is now working.
I can now start customizing my page.
Is it possible to change the screenshot of the engined theme?
Not in an automated way as it´s possible with Divi Children, but you can always replace the screenshot file manually.
its a fresh installation of WP & DIVI, then i want the Engined to be there and i am getting the above error. what shall i do??
luis, prob solved. i just activated the divi theme and engined is working now.


This boxy-cut OSKA pullover has a pleasantly loose-fitting width. It has soft and round falling shoulders for a flattering silhouette and a wonderfully casual flair. Great combination partners are the new flared skirts and all shapes of trousers – from narrow to wide, straight-cut to round.