text:
\label{appendix:generalization_error}

In this section, we present the computation of the asymptotic generalization error
\begin{align}
e_{\rm g}(\alpha) \equiv \lim_{\ndim \to \infty} \EE_{y, \vec{x}}  \id\[ y \ne \hat{y}\(\hat{\vec{w}}(\alpha); \vec{x} \) \]\,,
\end{align}
 leading to expressions in Proposition.~\ref{main:thm:generalization_errors} and Thm.~\ref{main:thm:fixed_point_equations_bayes}. The computation at finite dimension is similar if we do not consider the limit $d\to \infty$.

\subsection{General case}
\label{appendix:sec:generalization_error:general}

The generalization error $e_{\rm g}$ is the prediction error of the estimator $\hat{\vec{w}}$ on new samples $\{\vec{y}, \mat{X}\}$, where $\mat{X}$ is an \iid Gaussian matrix and $\vec{y}$ are $\pm 1$ labels generated according to \eqref{appendix:teacher_model}:
\begin{align}
	\vec{y} = \varphi_{\out^\star}\(\vec{z}\) \hhspace \textrm{ with } \hhspace \vec{z} = \frac{1}{\sqrt{\ndim}} \mat{X} \vec{w}^\star	\,.
\end{align}
 As the model fitted by ERM may not lead to binary outputs, we may add a non-linearity $\varphi:\bbR \mapsto \{\pm 1\}$ (for example a sign) on top of it to insure to obtain binary outputs $\hat{\vec{y}} = \pm 1$ according to
 \begin{align}
 		\hat{\vec{y}} = \varphi\(\hat{\vec{z}}\) \hhspace \textrm{ with } \hhspace \hat{\vec{z}} = \frac{1}{\sqrt{\ndim}} \mat{X} \hat{\vec{w}} \,.
 \end{align}
The classification generalization error is given by the probability that the predicted labels $\hat{y}$ and the true labels $y$ do not match. To compute it, first note that the vectors $(\vec{z},\hat{\vec{z}})$
averaged over all possible ground truth vectors $\vec{w}^\star$ (or equivalently labels $y$) and input matrix $\mat{X}$
follow in the large size limit a joint Gaussian distribution with zero mean and covariance matrix
\begin{align}
	\sigma = \lim_{\ndim \to \infty} \EE_{\vec{w}^\star, \mat{X}} \frac{1}{\ndim}
\begin{bmatrix}
	\vec{w}^{\star \intercal} \vec{w}^\star & \vec{w}^{\star \intercal} \hat{\vec{w}} \\
	\vec{w}^{\star \intercal} \hat{\vec{w}} & \hat{\vec{w}}^{\intercal} \hat{\vec{w}}
\end{bmatrix}
\equiv
\begin{bmatrix}
	\sigma_{\w^\star} & \sigma_{\w^\star \hat{\w}} \\
	\sigma_{\w^\star \hat{\w}} & \sigma_{\hat{\w}}
\end{bmatrix}\,.
\end{align}
The asymptotic generalization error depends only on the covariance matrix $\sigma$ and as the samples are \iid it reads
\begin{align}
\begin{aligned}
	e_{\rm g}(\alpha) &=\lim_{\ndim \to \infty} \EE_{y, \vec{x}}  \id\[ y \ne \hat{y}\(\hat{\vec{w}}(\alpha); \vec{x} \) \] = 1 - \bbP[y = \hat{y}\(\hat{\vec{w}}(\alpha); \vec{x} \)] =  1 - 2  \int_{\(\bbR^+\)^{2}} d\vec{x} \mN_{\vec{x}} \( \vec{0}, \sigma\)  \\
	&=  1 - \( \frac{1}{2}  + \frac{1}{\pi}  \atan\(\sqrt{\frac{\sigma_{\w^\star\hat{\w}}^2}{\sigma_{\w^\star} \sigma_{\hat{\w}} - \sigma_{\w^\star \hat{\w}}^2}}\) \)  = \frac{1}{\pi} \acos\(\frac{\sigma_{\w^\star\hat{\w}}}{\sqrt{\sigma_{\w^\star} \sigma_{\hat{\w}}}}\) \,,

\end{aligned}
\end{align}
where we used the fact that $\atan(x) = \frac{\pi}{2} - \frac{1}{2}\acos\(\frac{x^2-1}{1+x^2}\)$ and $\frac{1}{2} \acos(2x^2-1) = \acos(x)$. Finally
\begin{align}

	e_{\rm g}(\alpha) &\equiv \lim_{\ndim \to \infty} \EE_{y, \vec{x}}  \id\[ y \ne \hat{y}\(\hat{\vec{w}}(\alpha); \vec{x} \) \] =\frac{1}{\pi} \acos\(\frac{\sigma_{\w^\star\hat{\w}}}{\sqrt{\rho_{\w^\star} \sigma_{\hat{\w}}}}\)\,,
	\label{appendix:generalization_error:general}

\end{align}
with
\begin{align*}
 	\sigma_{\w^\star\hat{\w}} &\equiv \lim_{\ndim \to \infty} \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \hat{\vec{w}}^\intercal \vec{w}^\star \,,
 	&& \rho_{\w^\star} \equiv \lim_{\ndim \to \infty}  \EE_{\vec{w}^\star} \frac{1}{\ndim}  \|\vec{w}^\star\|_2^2 \,,
 	&& \sigma_{\hat{\w}} \equiv \lim_{\ndim \to \infty}  \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \|\hat{\vec{w}}\|_2^2\,.
\end{align*}

\subsection{Bayes-optimal generalization error}
The Bayes-optimal generalization error for classification is equal to eq.~\eqref{appendix:generalization_error:general} where the Bayes estimator $\hat{\vec{w}}$ is the average over the posterior distribution eq.~\eqref{appendix:posterior_distribution} denoted $\langle . \rangle$, knowing the teacher prior $P_{\w^\star}$ and channel $P_{\out^\star}$ distributions: $\hat{\vec{w}}=\langle \vec{w} \rangle_{\vec{w}}$. Hence the parameters $\sigma_{\hat{\w}}$ and $\sigma_{\w^\star\hat{\w}}$ read in the Bayes-optimal case
\begin{align*}
	\sigma_{\hat{\w}} &\equiv \lim_{\ndim \to \infty}  \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \|\hat{\vec{w}}\|_2^2 = \lim_{\ndim \to \infty}  \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \|\langle \vec{w} \rangle_{\vec{w}}\|_2^2  \equiv q_{\rm b}\,,\\
	\sigma_{\w^\star\hat{\w}} &\equiv \lim_{\ndim \to \infty} \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \hat{\vec{w}}^\intercal \vec{w}^\star = \lim_{\ndim \to \infty} \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \langle \vec{w} \rangle_{\vec{w}}^\intercal \vec{w}^\star \equiv m_{\rm b} \,.
\end{align*}
Using Nishimori identity \cite{Nishimori_1980}, we easily obtain $m_{\rm b}=q_{\rm b}$ which is solution of eq.~\eqref{main:fixed_point_equations_bayes}. Therefore the generalization error simplifies
\begin{align}

	e_{\rm g}^{\rm bayes}(\alpha) = \frac{1}{\pi} \acos\( \sqrt{\eta}_\bayes\)\,, &\text{~~with~~~} \eta_\bayes = \frac{q_\bayes}{\rho_{\w^\star}} \,.
	\label{appendix:generalization_error:bayes}

\end{align}

\subsection{ERM generalization error}
The generalization error of the ERM estimator is given again by eq.~\eqref{appendix:generalization_error:general} with parameters
\begin{align*}
	\sigma_{\hat{\w}} &\equiv \lim_{\ndim \to \infty}  \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \|\hat{\vec{w}}\|_2^2 = \lim_{\ndim \to \infty}  \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \| \hat{\vec{w}}^{\rm erm}\|_2^2  \equiv q\,,\\
	\sigma_{\w^\star\hat{\w}} &\equiv \lim_{\ndim \to \infty} \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \hat{\vec{w}}^\intercal \vec{w}^\star =  \lim_{\ndim \to \infty} \EE_{\vec{w}^\star,\mat{X}} \frac{1}{\ndim} \(\hat{\vec{w}}^{\rm erm}\)^\intercal \vec{w}^\star \equiv m \,.
\end{align*}
where the parameters $m, q$ are the asymptotic ERM overlaps solutions of eq.~\eqref{main:fixed_point_equations_replicas} and that finally lead to the ERM generalization error for classification:

\begin{align}
	e_{\rm g}^{\rm erm}(\alpha) &= \frac{1}{\pi} \textrm{acos}\( \sqrt{\eta} \)\,,
	&& \text{with~~} \eta \equiv \frac{m^2}{\rho_{\w^\star}q}\,.
	\label{appendix:generalization_error:erm}

\end{align}


TITLE: Expected value of steps required to reach 50 m
QUESTION [1 upvotes]: A monkey starts at the origin. Every time we pat on his back he takes any number of steps, from $0$ up to $50$ steps. What is the expected number of pats required to make the monkey reach from origin to $50$m? Assume selecting the step has a uniform distribution.

REPLY [1 votes]: I know this is not at all the answer you want, but I got around 2.69 pats running a computer simulation with 100 000 monkeys
Python code if someone wants to check :
import random

def sim():#With one monkey
   nbpats=0
   position=0
   while position<50 :
       nbpats =nbpats+1
       position =position+random.randint(0,50)
   return nbpats

#main code here
i=0
s=0
while i<100000:
   s=s+sim()
   i=i+1
print (s/i)


text:
\begin{document}
\title[]{Relaxed Elastic Line on an Oriented Surface in the Galilean Space}
\author{Tevfik \c{S}ahin}
\address{\textit{Amasya University,} \textit{Faculty of Sciences and Arts,}
\textit{Department of Mathematics,} \textit{Amasya-Turkey}}
\email{tevfiksah@gmail.com}
\subjclass{53A35, 53A55, 49Q20}
\keywords{Galilean space, relaxed elastic line of second kind, Variational
method, Geodesic}

\begin{abstract}
In this paper, we consider the classical variational problem in the Galilean space. we develop the Euler-Lagrange equations for a elastic line on an oriented surface in the Galilean 3-dimensional space $G_{3}.$ Using the variation method, we will try to give some characterization for the solution curve (the elastic line) of energy equation described by the total squared curvature function of a curve on an oriented surface in $G_3$. Finally, we will investigate whether or not the relaxed elastic curves are on a geodesic.

\end{abstract}

\maketitle

\section{ Introduction}

There has been in recent years a wide variety of applications of variational methods to various fields of mechanics and technology. Calculus of variation is an important theory which has common applications in geometry, analysis, physics, chemistry and engineering. Variation method is used to find the maxima or the minima of expressions involving unknown functions called functionals. Therefore, the main problem in the calculus of variations is to minimize (or maximize) not only functions but also functionals. For instance; a typical variational problem is to find the fixed length $l$ of a curve joining two points on a surface. The Euler-Lagrange equations is used for the solution of the problem.

In this paper, we will first define the total squared curvature function for a curve in a Galilean space. Using the variation method, we will try to give some characterization for the solution curve (the relaxed elastic line) of energy equation described by the total squared curvature function of a curve on an oriented surface in Galilean space. Finally, we will investigate whether or not the relaxed elastic curves are on a geodesic.

 There has been in recent years a wide variety of applications of variational methods to various geometries. For instance, several geometers were
interested in studying of calculus of variations in Euclidean Space
\cite{manning,nickerson,unan,yilmaz}. Nevertheless, similar applications of this theory in
Minkowski space can be found in \cite{tutar}. The main point of the studying the
variation calculus
 is defining the Euler-Lagrange equations for a relaxed elastic
line on an oriented surface.

The natural variational integrals in geometry are the common integrals on
space curves $\alpha \left( s\right) $. These include the length $L\left(
\alpha \right) =\dint ds$, total squared curvature $K\left( \alpha \right)
=\dint \kappa ^{2}ds$ used in \cite{manning, nickerson}, total squared torsion $T\left( \alpha
\right) =\dint \tau ^{2}ds$ used in \cite{tutar,unan} and the integral $H\left( \alpha
\right) =\dint \kappa ^{2}\tau ds$ used in \cite{sahin}. An elastic line of length $l$
is defined as a curve with associated energy equation $K=\dint
\limits_{0}^{l}\kappa ^{2}\left( s\right) ds$ by Nickerson and Manning in
\cite{nickerson}, where $s$ is the arc-length along the curve, $\kappa ^{2}\left(
s\right) $ is the square curvature in there. The integral $K$ is called the
total square curvature. If no boundary conditions are imposed at $s=l$, and
if no external forces act at any $s,$ the elastic line is \textit{relaxed,}
[manning,nickerson].
Hilbert and Cohn-Vossen stated in \cite{hilbert} that a relaxed elastic line with
specified position and tangent at $s=0$ always has the trajectory of a
geodesic. However, Manning have proved in \cite{manning} that the
conslusion of Hilbert and Cohn-Vossen is incorrect. Manning in
\cite{manning} has obtained Euler-Lagrange equations of the relaxed elastic line on an
oriented surface as a model of DNA molecule.

Besides Euclidean Geometry, a range of new types of geometries have been
invented and developed in the last two centuries. They can be introduced in
a variety of ways. One possible way is through projective manner, where one
can express metric properties through projective relations. Among these
geometries, there is also Galilean geometry which is our matter in this
paper.

In literature, there is very ? study on relaxed elastic line on an oriented surface in Galilean space $
G_{3}$. Gokce and Sahin in \cite{gokce,sahin} have obtained intrinsic equations of the relaxed elastic line on an oriented surface in $G_3$ for different energy functions. However,  Gokce and Sahin in \cite{gokce,sahin} have not use variation method. Hence, in
literature, there is no any study that using of variation method on the relaxed elastic line on an oriented surface in $G_{3}$.

Therefore, firstly, we will derive the Euler-Lagrange
equations for a relaxed elastic line on an oriented surface in $G_{3}$.

Hence, we have proved in this paper that the conslusion of Hilbert and Cohn-Vossen is incorrect in $G_3$.
The main result of this paper is Theorem $2$. We describe the geometric
meaning of Theorem $2$ in section 4.

\section{Preliminaries on Galilean Geometry}

\qquad \textquotedblright All geometry is projective
geometry.\textquotedblright\ (A. Cayley). From A. Cayley point of view, $G_{3
\text{ }}$ is a real 3-dimensional projective space $P^{3}(\mathbb{R}) $, is the set of equivalence classes of $\sim $ on $\mathbb{R}^{4}-\left\{ 0\right\} $ by equivalence relation $x\sim y$ iff
$x\lambda y$
for some $\lambda \in \mathbb{R}\backslash \{ 0\} $. Thus, $P^{3}(\mathbb{R}) $ obtained as a factor space on $\mathbb{R}^{4}\backslash \{ 0\} $ by $\sim $, i.e. $P^{3}(\mathbb{R}) \widetilde{=}( \mathbb{R}^{4}-\{ 0\} ) /\sim $ \cite{cox}. We can think of $P^{3}(\mathbb{R}) $ more geometrically as a set of lines through the origin in
$\mathbb{R}^{4}$. $G_{3}$ is a real Cayley-Klein space equipped with the projective
metric of signature $( 0,0,+,+) $, as showed in \cite{molnar}. The
absolute of the Galilean geometry is an ordered triple $\{w,f,I\}$, where $w$
is the ideal (absolute) plane, $f$ \ is the line (absolute line) in $w$ and $
I$ is the fixed elliptic involution of points of $f$. \ The points, the
lines and the planes of $P^{3}( \mathbb{R}) $ are the one-dimensional, two-dimensional and three-dimensional
subspaces of $\mathbb{R}^{4}$, respectively \cite{casse}. Therefore, $G_{3}$ contains $\mathbb{R}^{3}$ as a proper subset and the complement in $G_{3}$ to $w$ is
diffeomorphic to $\mathbb{R}^{3}$.

Let $P$ be any point of $\mathbb{R}^{3}$ with affine coordinates $( x,y,z) $. Write $(
x,y,z) $ as $( \frac{X_{1}}{X_{0}},\frac{X_{2}}{X_{0}},\frac{X_{3}
}{X_{0}}) $, where $X_{0}$ is some common deminator. Call $(
X_{0},X_{1},X_{2},X_{3}) $ the \textit{homogeneous coordinates of }$P$
. Thus, the homogeneous coordinates $( X_{0}:X_{1}:X_{2}:X_{3}) $
and $\rho ( X_{0}:X_{1}:X_{2}:X_{3}) $ refer to the same point
for all $\rho \in \mathbb{R}-\left\{ 0\right\} ,$ \cite{casse}. Now,we can introduce homogeneous coordinates in
$G_{3}$ in such a way that the absolute plane $w$ is given by $X_{0}=0$, the
absolute line $f$ by $X_{0}=X_{1}=0$ and the elliptic involution $I$ by
\begin{equation*}
( 0:0:X_{2}:X_{3}) \rightarrow ( 0:0:X_{3}:-X_{2}) .
\end{equation*}

In the nonhomogeneous coordinates the isometries group $B_{6}$ has the form
\begin{eqnarray}
\overline{x} &=&a+x  \notag \\
\overline{y} &=&b+cx+y\cos \varphi +z\sin \varphi \\
\overline{z} &=&d+ex-y\sin \varphi +z\cos \varphi  \notag
\end{eqnarray}
where $a,b,c,d,e$ and $\varphi $ are real numbers. The group of motions of $
G_{3}$ is a six-parameter group \cite{pavkovic}.\newline
The Galilean norm of the vector $\textbf{v}=(x,y,z)$ defined by
\begin{equation}
\left\Vert \mathbf{v}\right\Vert _{G}=
\begin{Bmatrix}
x\ \ ,\  & if\ \ \ x\neq 0 \\
\sqrt{ y^{2}+z^{2} }, & if\ \ \ x=0\
\end{Bmatrix}
.
\end{equation}
A vector $\mathbf{v}=( x,y,z) $ is said to be non-isotropic if $x\neq 0$.
All unit non-isotropic vectors are of the form $\left( 1,y,z\right) $.
For a curve $\alpha :I\rightarrow G_{3}$ , \ $I\subset \mathbb{R}$ parametrized by the arc-length parameter $s=x$, given in the coordinate form
\begin{equation}
\alpha \left( x\right) =\left( x,y\left( x\right) ,z\left( x\right) \right) ,
\end{equation}
the curvature $\kappa \left( x\right) $ and the torsion $\tau \left(
x\right) $ are defined by
\begin{eqnarray}
\kappa \left( x\right) &=&\Vert \alpha''(x) \Vert_{G}=\sqrt{y^{\prime \prime }\left( x\right)
^{2}+z^{\prime \prime }\left( x\right) ^{2}}, \\
\tau \left( x\right) &=&\frac{\det \left( \alpha ^{\prime }\left( x\right)
,\alpha ^{\prime \prime }\left( x\right) ,\alpha ^{\prime \prime \prime
}\left( x\right) \right) }{\kappa ^{2}\left( x\right) }
\end{eqnarray}
and the associated moving trihedron is given by
\begin{eqnarray}
T\left( x\right) &=&\alpha ^{\prime }\left( x\right) ,  \notag \\
N\left( x\right) &=&\frac{\alpha''(x)}{\kappa \left( x\right) } , \\
B\left( x\right) &=&\frac{1}{\kappa \left( x\right) }\left( 0,-z^{\prime
\prime }\left( x\right) ,y^{\prime \prime }\left( x\right) \right) .  \notag
\end{eqnarray}
The vectors $T\left( x\right) ,N\left( x\right) $ and $B\left( x\right) $
are called the vectors of the tangent, principal normal and the binormal
line, respectively \cite{pavkovic,sahinac}. Therefore, the Frenet-Serret formulas can be
written in matrix notation as
\begin{equation}
\begin{bmatrix}
T \\
N \\
B
\end{bmatrix}
^{\prime }=
\begin{bmatrix}
0 & \kappa & 0 \\
0 & 0 & \tau \\
0 & -\tau & 0
\end{bmatrix}
\begin{bmatrix}
T \\
N \\
B
\end{bmatrix}.
\end{equation}
Let $\mathbf{a=}( a_{1},a_{2},a_{3}) $, $\mathbf{b=}(b_{1},b_{2},b_{3}) $ be two vectors in $G_{3}.$

For any regular curve $\alpha $ the following formulas hold:
\begin{equation}
\kappa =\frac{\left\Vert \gamma ^{\prime }\times _{G}\gamma ^{\prime \prime
}\right\Vert_{G} }{\left\Vert \gamma ^{\prime }\right\Vert ^{3}_{G}},\text{ \ }\tau =
\frac{\det \left( \gamma ^{\prime },\gamma ^{\prime \prime },\gamma ^{\prime
\prime \prime }\right) }{\left\Vert \gamma ^{\prime }\times _{G}\gamma
^{\prime \prime }\right\Vert ^{2}_{G}}.
\end{equation}
Here,the Galilean cross product $\times _{G}$ is defined by
\begin{equation}
\mathbf{a}\times _{G}\mathbf{b=}
\begin{vmatrix}
0 & e_{2} & e_{3} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{vmatrix}
\end{equation}
as in \cite{sipus}.

The Galilean Sphere $S_{G}^{2}$ is defined by $S_{G}^{2}=\{ (
x,y,z) \in G_{3}~\ :~\vert x-x_{0}\vert =r\} .$
\newline
For more facts about the Galilean geometry, we refer the reader to
\cite{gokce, pavkovic, sahinuk, sahin, roschel, yaglom, sipus} and references therein.

\begin{theorem}
Let $S$ be a surface in $G_{3}$ and $\alpha $ be a curve on $S$. At a point
$\alpha (x)$ of $\alpha$, let $T$ denote
the unit tangent vector of $\alpha$ at $\alpha(x)$, $n$ the unit normal to $S$ and $
n\times _{G}T=Q$ the tangential-normal. Then $\{ T,Q,n\} $ is an
orthonormal basis at $\alpha(x)$ in $S$. This frame is called Galilean Darboux frame or tangent-normal frame. The Frenet-Serret formulas the for the Galilean Darboux frame can be written in matrix notation as
\begin{equation}
\left[
\begin{array}{c}
T \\
Q \\
n
\end{array}
\right] ^{\prime }=\left[
\begin{array}{ccc}
0 & \kappa _{g} & \kappa _{n} \\
0 & 0 & \tau _{g} \\
0 & -\tau _{g} & 0
\end{array}
\right] \left[
\begin{array}{c}
T \\
Q \\
n
\end{array}
\right]
\end{equation}
where $\kappa _{g}$,$\kappa _{n},\tau _{g}$ are geodesic curvature, normal
curvature and geodesic torsion, respectively.
\end{theorem}

Eq.(2.10) implies the important relation $\kappa ^{2}=T^{\prime }._{G}$ $
T^{\prime }=\kappa _{g}^{2}+\kappa _{n}^{2}$ where $\kappa ^{2}$ is the
square curvature of $\alpha $.\newline

\begin{proof} It is clear?
$( 2.7) $ and $(2.10)$ implies the important relation.
\end{proof}

We shall treat a $C^{3}$- surface, $r\geq 1$, as a subset $S\subset G_{3}$
for which there exists an open subset $D$ of $\mathbb{R}^{2}$ and a $C^{3}$- mapping $\psi :D\rightarrow G_{3}$ satisfying $S=\psi\left( D\right) $. A $C^{3}$- surface $S\subset G_{3}$ is called regular if $\psi $ is an immersion, and simple if $\psi $ is an embedding.
Therefore, the oriented surface $S$ represented by $\psi (u_{1},u_{2}) =( \mathbf{X}( u_{1},u_{2}) ,\mathbf{Y}( u_{1},u_{2}) ,\mathbf{Z}( u_{1},u_{2}) ) $.
Here $\mathbf{X,Y}$ and $\mathbf{Z}$ are the coordinate functions of $\psi$
. A unit normal field of surface at a point $p$ is defined by
\begin{equation}
n=\frac{\psi _{1}\times _{G}\psi _{2}}{\left\Vert \psi _{1}\times _{G}\psi
_{2}\right\Vert }=\frac{1}{W}\left( 0,\mathbf{X}_{1}\mathbf{Z}_{2}-\mathbf{X}
_{2}\mathbf{Z}_{1},\mathbf{X}_{2}\mathbf{Y}_{1}-\mathbf{X}_{1}\mathbf{Y}
_{2}\right) ,
\end{equation}
\newline
where $\psi _{i}=\frac{\partial \psi }{\partial u_{i}}, W=\Vert \psi
_{u_{1}}\times _{G}\psi _{u_{2}}\Vert, \, \mathbf{X}_{i}=\frac{\partial \mathbf{X}}{
\partial u_{i}}, \mathbf{Y}_{i}=\frac{\partial \mathbf{Y}}{\partial u_{i}}$ and $\mathbf{Z}_{i}=\frac{
\partial \mathbf{Z}}{\partial u_{i}}$, $i=1,2.$ In a tangent plane of the surface at the
point $p$, there is a unique isotropic direction defined by the condition $
\mathbf{X}_{1}du_{1}+\mathbf{X}_{2}du_{2}=0$. A side tangential vector $Q=
\frac{1}{W}( \mathbf{X}_{2}\psi _{1}-\mathbf{X}_{1}\psi _{2}) $ is a unit isotropic
vector in a tangent plane. The curve $\alpha $ on the oriented surface $S$
is specified by $\psi(x) =\psi( u_{1}(x), u_{2}(x)).$ Then the unit tangent vector along $\alpha $
is
\begin{equation}
T=\frac{d\psi }{dx}=\psi _{1}\overset{\centerdot }{u_{1}}+\psi _{2}\overset{
\centerdot }{u_{2}},
\end{equation}
where $\overset{\centerdot }{u_{1}}=\frac{du_{1}}{dx}$ and $\overset{
\centerdot }{u_{2}}=\frac{du_{2}}{dx}$. \newline
The first fundamental form of a surface is introduced in the following way
\begin{equation}
dx^{2}=\left( \mathbf{X}_{1}du_{1}+\mathbf{X}_{2}du_{2}\right)
^{2}+\varepsilon \left\{ \left( \mathbf{Y}_{1}du_{1}+\mathbf{Y}
_{2}du_{2}\right) ^{2}+\left( \mathbf{Z}_{1}du_{1}+\mathbf{Z}
_{2}du_{2}\right) ^{2}\right\} ,
\end{equation}
\newline
where
\begin{equation}
\varepsilon =\left\{
\begin{array}{c}
0,\text{ \ \ }dx\text{\ is non-isotropic,} \\
1,\text{ \ }dx\text{ \ \ \ \ \ is isotropic }
\end{array}
\right\} .
\end{equation}
\newline
Since $T.T=1,$ we have a constraining relation between any pair of functions
$\left( u_{1}\left( x\right) ,u_{2}\left( x\right) \right) $ that define the
curve $\alpha $ on the surface $S:$
\begin{equation}
g\left( u_{1},u_{2},\overset{\centerdot }{u_{1}},\overset{\centerdot }{u_{2}}
\right) =1,
\end{equation}
\newline
where
\begin{equation}
g=\tsum\limits_{i,j=1}^{2}g_{ij}\dot{u}_{i}\dot{u}_{j},
\end{equation}
\newline
and $g_{1}=\mathbf{X}_{1},$ $g_{2}=\mathbf{X}_{2}$ and $g_{ij}=g_{i}.g_{j},$
$i,j=1,2,$ where $g_{ij},$ $i,j=1,2,$ are the coefficients of the first
fundamental form. The normal curvature can be expressed in terms of the
coordinates $\left( u_{1}\left( x\right) ,u_{2}\left( x\right) \right) $
along the curve $\alpha $ as
\begin{equation}
\kappa _{n}=\tsum\limits_{i,j=1}^{2}L_{ij}u_{i}^{\prime }u_{j}^{\prime },
\end{equation}
where $L_{11},L_{12}$ and $L_{22}$ are the coefficients of the second
fundamental form $II$ and the normal components of $\psi _{11},\psi _{12}$
and $\psi _{22}$, respectively. It holds
\begin{equation}
L_{ij}=\frac{\mathbf{X}_{2}\psi _{ij}-\mathbf{X}_{ij}\psi _{2}}{\mathbf{X}
_{2}}\cdot n\text{ or }L_{ij}=\frac{\mathbf{X}_{1}\psi _{ij}-\mathbf{X}
_{ij}\psi _{1}}{\mathbf{X}_{1}}\cdot n.
\end{equation}
\newline
Similarly, for the geodesic torsion $\tau _{g}$ we have
\begin{equation}
\tau _{g}=\tsum\limits_{i,j=1}^{2}g^{i}L_{ij}u_{i}^{\prime },
\end{equation}
\newline
where $g^{1}=\frac{\mathbf{X}_{2}}{W}$, $g^{2}=-\frac{\mathbf{X}_{1}}{W}$
and $g^{ij}=g^{i}.g^{j}.$ On the other hand, the square geodesic curvature can
be obtained as
\begin{equation}
\kappa _{g}^{2}=\tsum\limits_{i,j=1}^{2}g_{ij}\gamma _{i}\gamma _{j},
\end{equation}
\newline
where
\begin{equation}
\gamma _{i}=\overset{\cdot \cdot }{u}_{i}+\tsum\limits_{k,l=1}^{2}\Gamma
_{kl}^{i}\dot{u}_{k}\dot{u}_{l},i=1,2
\end{equation}
\newline
and the quantities $\Gamma _{kl}^{i}$ are the Christoffel symbols of the
second kind; available formulas are expressed as functions $g_{ij}$, and
their first partial derivatives with respect to $u_{i}$. Furthermore, the
Christoffel symbols can be written as follows:
\begin{equation}
\Gamma _{kl}^{1}=\frac{\mathbf{X}_{2}\psi _{ij}-\mathbf{X}_{ij}\psi _{2}}{W}
\cdot Q,\text{ }\Gamma _{kl}^{2}=\frac{\mathbf{X}_{1}\psi _{ij}-\mathbf{X}
_{ij}\psi _{1}}{W}\cdot Q,\text{ }k,l=1,2.
\end{equation}
\newline
Hence, the equations of a geodesic curve, which is characterized by
identically vanishing $\kappa _{g}$, must be given by $\gamma _{i}=0,$ and
indeed they are \cite{carmo}.

\section{The incomplete and complete variational problems}

Now we can give the definition of elastic line on an oriented surface $S$ in $G_{3}.$\newline
\subsection{The incomplete variational problem}
\begin{definition}
Let $\alpha $ be a $C^{2}$-curve with parametrized by arc-length $x,$ $0\leq
x\leq \ell ,$ on an oriented surface $S$ in $G_{3}.$ An elastic line of length $\ell $ is defined as a curve with associated energy
\begin{equation}
K=\dint\limits_{0}^{\ell }\kappa ^{2}dx,
\end{equation}
where $\kappa ^{2}$ is the square curvature of the curve $\alpha$ \cite{manning}.
The integral $K$ is called the \textit{total square curvature}. Therefore, the arc $\alpha$ is called an elastic line if it is an extremal for the variational problem of minimizing the value of $K$ with in the family of all arcs of length $\ell $ on $S$, having the same initial point and initial direction as $\alpha .$
\end{definition}

The pertinent problem is to identify that curve in a family of curves of
fixed length $\ell $ which has the least total square curvature. In general the minimizing curve is not a geodesic line. It is a geodesic line, however, in the circumstances described below.\newline
Let
\begin{equation}
K_{n}=\dint\limits_{0}^{\ell } \kappa^2 _{n}(x)dx.
\end{equation}
\newline
and let $\alpha $ be the curve that minimizes $K_{n}$ among all curves of
length $\ell $ on a surface $S$. \newline
For any curve $\alpha $ that minimizes $K_n$ on $S$, if $\alpha $ is geodesic line, then the
geodesic curvature $\kappa _{g}$ identically vanishes. Hence $K=K_{n}$. We call this problem \textit{incomplete}
because it seeks to minimize $K_{n}$, not the total square curvature $K$. If a curve that minimises $K_n$ is a geodesic, then the curve is minimise $K$. Therefore, the relaxed elastic curve lie on a geodesic trajectory.
\newline
Through $\kappa_n^2$ has a dependence
on $u_{1},u_{2},\dot{u}_{1}$ and $\dot{u}_{2}$. Therefore, the Euler-Lagrange equations for the
incomplete variational problem are
\begin{eqnarray}
H_{u_{1}}-( H_{\dot{u}_{1}}) ^{\cdot } &=&0, \\
H_{u_{2}}-( H_{\dot{u}_{2}}) ^{\cdot } &=&0,
\end{eqnarray}
\newline
where
\begin{equation}
H=\kappa _{n}^{2}+\lambda ( g-1)
\end{equation}
\newline
and $\lambda =\lambda \left( x\right) $ is a Lagrange multiplier function.
Equations $(3.3) ,(3.4) $ and $(3.5) $ are
a system of three equations to determine the three functions $u_{1}(x)$, $u_{2}(x)$ and $\lambda(x)$.
Equations $(3.3, 3.4) $ and $(3.5) $ are second order in $u_{1},$ $u_{2}$ and first order in $\lambda$. Therefore, the system is fifth order. To get the system in normal form, Eq. $(2,15)$
must be differentiated one times with respect to $x$. Reintegration of $g'=0$ gives that the constant of integration equal unity. The other four constants of integrations in the general solution are determined by the boundary conditions.\newline
We turn now to the boundary conditions at $x=0$ and $x=\ell$ that determine
the other  four constants of integration of the equations.
This conditions are
\begin{eqnarray}
H_{\dot{u}_1}=H_{\dot{u}_2} =0, \, (x=0, \ell)
\end{eqnarray}
If the initial point of the elastic line is fixed say,
\begin{equation}
u_1(0)=u_{1_0}, \, u_2(0)=u_{2_0}
\end{equation}
but the end of the elastic line is free, then
\begin{equation}
H_{\dot{u}_1}=H_{\dot{u}_2} =0, \, (x=\ell).
\end{equation}
Therefore, the four constants of integration are now determined by above two equations.

\subsection{The complete variational problem}
If a curve that minimises $K_n$ is not a geodesic, then the curve is or not minimise $K$. Therefore, the incomplete variational problem is not anything say about the relaxed elastic curve that minimises K. In this case, we have no choice but is to find the curve that minimises the integral
$$K=\dint\limits_{0}^{\ell }\kappa ^{2}dx=\dint\limits_{0}^{\ell }(\kappa_n ^{2}+\kappa_g^2)dx$$
The relaxed elastic curve that is not a geodesic curve is one of solutions of the complete variational problem.
Through $\kappa_n^2+\kappa_g^2$ has a dependence
on $u_{1},u_{2},\dot{u}_{1}, \dot{u}_{2}, \ddot{u}_{1}$ and $\ddot{u}_{2}$. Therefore, the Euler-Lagrange equations for the complete variational problem are
\begin{eqnarray}
H_{u_{1}}-( H_{\dot{u}_{1}})^{\cdot}+ ( H_{\ddot{u}_{1}})^{\cdot\cdot}&=&0, \\
H_{u_{2}}-( H_{\dot{u}_{2}}) ^{\cdot }+( H_{\ddot{u}_{2}})^{\cdot\cdot} &=&0,
\end{eqnarray}
Now, let
\begin{equation}
H=\kappa^{2}+\lambda \left( g-1\right)
\end{equation}
with the constrain $g=1$ given by Eqs. $(2.15) $, $(
2.16) $ and $\lambda =\lambda(x)$ again a Lagrange
multiplier function to be determined in the course of solving the system of
differential equations, then we call this variation problem \textit{complete}
because it seeks to minimize the total square curvature.
Equations $(3.9), (3.10) $ and $(3.11)$ are
a system of three equations to determine the three functions $u_{1}(x) $, $u_{2}(x)$ and $\lambda(x)$.
Equations $(3.9)$ and $(3.10)$ are fourth  order in $u_{1},$ $u_{2}$ and first order in $\lambda $. Therefore, the system is of order nine. To get the system in normal form, Eq. $(2.15)$
must be differentiated three times with respect to $x$. Reintegration of $g'''=0$ gives the respective fixed values $0, 0,$ and $1$ to the resulting three constants of integration. The other six constants of integrations in the general solution are determined by the boundary conditions.
Because the side condition Eq. (2.15) does not involve $\ddot u_1$ and $\ddot u_2$, the boundary terms of the variation are not determined in the usual straightforward way. We nevertheless find the following six boundary terms if Eq. (2.15) can be solved for $\dot u_2$ as a function of $u_1, u_2,$ and $\dot u_1$ to give $\dot u_2= U_2(u_1,u_2,\dot u_1)$:
\begin{eqnarray}
\big\{H_{\ddot u_1}+{H_{\ddot u_2}}{(U_2)}_{\dot u_1}\big\} \delta{\dot u_1}    &\hskip 1cm (s=0, \ell)& \\
\big\{H_{\dot{u}_{1}}- ( H_{\ddot{u}_{1}})^{\cdot}+{H_{\ddot u_2}}{(U_2)}_{u_1}\big\} \delta{u_1} &\hskip 1cm (s=0, \ell)& \\
\big\{H_{\dot{u}_{2}}- ( H_{\ddot{u}_{2}})^{\cdot}+{H_{\ddot u_2}}{(U_2)}_{u_2}\big\} \delta{u_2} &\hskip 1cm (s=0, \ell)&
\end{eqnarray}
On the other hand, if ${U_2}_{\dot u_1}$ is singular, then we must use a different set of six boundary terms, after solving Eq. (2.15) for $\dot u_1= U_1(u_1,u_2,\dot u_2)$:
\begin{eqnarray}
\big\{H_{\ddot u_2}+{H_{\ddot u_1}}{(U_1)}_{\dot u_2}\big\}\delta{\dot u_2}    &\hskip 1cm (s=0, \ell)& \\
\big\{H_{\dot{u}_{2}}- ( H_{\ddot{u}_{2}})^{\cdot}+{H_{\ddot u_1}}{(U_1)}_{u_2}\big\} \delta{u_2} &\hskip 1cm (s=0, \ell)& \\
\big\{H_{\dot{u}_{1}}- ( H_{\ddot{u}_{1}})^{\cdot}+{H_{\ddot u_1}}{(U_1)}_{u_1}\big\} \delta{u_1} &\hskip 1cm (s=0, \ell)&
\end{eqnarray}
Suppose that we place no restrictions at all on the elastic line other than confinement to the surface. Then six integration constants are determined by setting equal to zero each of the six factors multiplying $\delta u_1, \delta u_2,$ and $\delta \dot{u_1}$ in the "$U_1$ boundary terms," or those multiplying $\delta u_1, \delta u_2,$ and $\delta \dot{u_2}$ in the "$U_2$ boundary terms." These boundary conditions are \textit{completely natural (free)} \cite{manning}.

If natural conditions are allowed to prevail at $x=\ell$, but the initial position and direction of the elastic line are specified, then we get a different set of equations to determine the six integration constants. We illustrate with the $U_2$ boundary terms. Three boundary conditions are obtained by setting equal to zero each of the three factors multiplying $\delta u_1, \delta u_2,$ and $\delta \dot{u_1}$, with the factors evaluated at $x=\ell$. Two more boundary conditions are specified by the values of $u_1$ and $u_2$ at $x=0$: ${u_1}(0)={u_1}_0, \, {u_2}(0)={u_2}_0$. The unit tangent vector $T(x)$ at any point $x$ of the curve is given by $\phi_{u_1}(u_1,u_2){\dot {u_1}(x)} +\phi_{u_2}(u_1,u_2){\dot {u_2}(x)}$, and so the initial tangent $T(0)=\phi_{u_1}({u_1}_0,{u_2}_0){\dot {u_1}(0)} +\phi_{u_2}({u_1}_0,{u_2}_0){\dot {u_2}(0)}$. Specifications of $\dot{u_1}(0)=\dot{u_1}_0$ therefore completely determines the initial direction, since $\dot{u_2}(0)$ is then fixed at the value $U_2({u_1}_0,{u_2}_0, \dot{u_1}_0)$. Thus, the required sixth boundary condition is that $\dot{u_1}(0)$ have a specified value.

\section{Elastic Lines On Some Surfaces in Galilean Space}

\subsection{Elastic lines on Galilean plane}

Let $S$ be a Galilean plane. Then, for the coefficients of the first and
second fundamental forms we have
\begin{eqnarray*}
g_{11} &=&1,g_{12}=0,g_{22}=1 \\
L_{11} &=&L_{12}=L_{22}=0.
\end{eqnarray*}
Thus, we obtain $\Gamma _{kl}^{j}=0,1\leq i,k,l\leq 2.$ Therefore, we can
say that the normal curvature $\kappa _{n}$ and the geodesic torsion $\tau
_{g}$ of any curve on $S$ vanish in all points of this curve from Eq.(2.17). In particular, any geodesic line on the plane minimizes $K_n$ (giving to it the value zero), Hence, the curve minimizes K. Therefore, the relaxed elastic lines on Galilean plane lies along a geodesic line.

It follows from Eq.(2.21) that the differential equations of geodesic curves on a plane become $u_1''=u_2''=0$. The solutions of differential equations are straight lines $u_1=ax+b, u_2=cx+d, a^2+c^2=1$.
The relaxed elastic line on a plane assumes the form of a straight line.

\subsection{Elastic lines on Galilean sphere}

Non-isotropic Galilean unit sphere is defined a set of non-isotropic unit
vectors in Galilean space. Therefore, Non-isotropic Galilean unit sphere is
described as set of points $\left( \pm 1,y,z\right) ,$ i.e. is two parallel
plane given by $x=\pm 1.$ The non-isotropic Galilean unit sphere with
parametrization is obtained by $\varphi (u_1,u_2)=( \pm 1,u_1,u_2).$ Its
first fundamental form is given by
\begin{equation*}
ds^{2}=du_1^{2}+du_2^{2}.
\end{equation*}
Then, the coefficients of the first and second fundamental forms are
\begin{eqnarray*}
g_{11} &=&1,g_{12}=0,g_{22}=1, \\
L_{11} &=&L_{12}=L_{22}=0.
\end{eqnarray*}
Thus, we obtain $\Gamma _{kl}^{j}=0,1\leq i,k,l\leq 2.$

Therefore, we can say that the relaxed elastic line on the Galilean sphere assumes the form of a straight line.

\subsection{Elastic lines on a cylinder}
Let $M$ be the cylinder over a curve $C: f(y,z)=c$ in the $yz$ plane. If $\alpha=(0, \alpha_1,\alpha_2)$ is a parametrization of $C$, we assert that $$\chi(u,v)=(u, \alpha_1(v),\alpha_2(v))$$
is a parametrization of $M$. If $\alpha=(0, R\cos(v/R),R\sin(v/R))$ is a parametrization of $C$, we have  $$\chi(u,v)=(u, R\cos(v/R),R\sin(v/R))$$

From Eq.(2.11-2.22), we obtain the following coefficients
\begin{eqnarray*}
g_{11} &=&1,g_{12}=g_{22}=0, \\
L_{11} &=&L_{12}=0, L_{22}=R^{-1}, \\
\Gamma _{kl}^{j}&=&0.
\end{eqnarray*}
where $1\leq i,k,l\leq 2.$  From Eq.(2.20), (2.21) and (2.22) the geodesic curvature along any curve $[u(x),v(x)]$ on cylinder is given by $\kappa_g=\vert  \overset{..}{u}\vert$. It is a characteristic of geodesic curves that the geodesic curvature $\kappa_g$ vanishes identically along them. Therefore, the geodesic on a cylinder have $u(x)=ax+b, \, a^2=1$. Consequently, any curve $[u(x),v(x)]$ on cylinder is geodesic if and only if $u(x)=\pm x+b$.

From Eq.(2.17) the square normal curvature along any curve $[u(x),v(x)]$ on the cylinder is given by
\begin{equation}
\kappa_N^2(x)=R^{-1}\overset{.}{v}^2
\end{equation}
We obtain following $H$  for the incomplete variational problem from Eq.(2.16) and (3.5)
$$H=R^{-2}\overset{.}{v}^4+\lambda(\overset{.}{u}^2-1).$$
Therefore, we get from the Euler equations (3.3) and (3.4), and also (3.6)
$$H_{\dot{u}}= H_{\dot{v}}=0.$$
Therefore, we have the system of three differential equations for $u(x), v(x),$ and $\lambda(x)$ is
\begin{eqnarray}
2\lambda{\dot{u}}&=&0, \\
4R^{-2}{\dot{v}^3}&=&0, \\
\dot{u}^2&=&1,
\end{eqnarray}
where Eq.(4.4) is side condition Eq.(2.15) for any curve on a cylinder.
Hence, any solution of this system automatically satisfies the boundary conditions.
From Eq.(4.4), we have $u(x)=\pm x+b$. We must choose $\lambda=0$ and $v=const.$ to satisfy  Eq.(4.2) and (4.3). Hence, all solutions of this system are any vertical generator $u(x)=\pm x+b$, $v=const.$ together with $\lambda=0$. From Eq.(4.1) the total normal curvature is zero for a vertical generator, and this solution provides a minimum for $K_N.$ In the same time, the vertical generator is a geodesic on cylinder. Consequently, we can say that minimum-energy trajectory on a cylinder adopts the form of a vertical generator. That is,  a relaxed elastic line on cylinder is lying on a geodesic.

Now, we examine the complete variational problem. In this case, we have following $H$  for the complete variational problem from Eq.(3.11)
$$H=R^{-2}\overset{.}{v}^4+ \overset{..}{u}^2+\lambda(\overset{.}{u}^2-1).$$
Therefore, we get from the Euler equations (3.9) and (3.10)
\begin{eqnarray*}
H_{\dot{u}}- (H_{\ddot{u}})'&=&const, \\
H_{\dot{v}}- (H_{\ddot{v}})'&=&const.
\end{eqnarray*}

Therefore, we have the system of three differential equations for $u(x), v(x),$ and $\lambda(x)$ is
\begin{eqnarray}
\lambda{\dot{u}}-\dddot{u}&=&0, \\
4R^{-2}{\dot{v}^3}&=&0, \\
\dot{u}^2&=&1,
\end{eqnarray}

\subsection{Elastic lines on helical surfaces}

Let $\alpha $ be a plane curve and $d$ a given line. A helical surface (or
generalized helicoid) $H$ is a surface obtained when $\alpha $ is displaced
in a rigid screw motion about $d.$ The normal form of the 1-parameter group
of motions called the screw motions in the Galilean space is given by
\begin{eqnarray*}
\overline{x} &=&pt+x, \\
\overline{y} &=&y\cos t+z\sin t, \\
\overline{z} &=&-y\sin t+z\cos t,
\end{eqnarray*}
where $p\in IR.$ By substituting $p=0$, the Euclidean rotations about the $x-
$axis are described, and obtained surface is a surface of revolution.\newline
Since a plane curve $\alpha $ can lie in a Euclidean or in a isotropic
plane, we treat these two cases separately.\newline
Let a plane curve $\alpha $, given by $\gamma (v) =(0,f(v) ,g(v)) ,$ where $f,g\in C^{2}$, be an
admissible curve in a Euclidean plane. If $\alpha $ undergoes the screw
motion, the helical surface $H_{p}$ with parametrization is obtained by
\begin{equation*}
\varphi (u,v) =(pu,f(v) \cos u+g(v) \sin u,-f(v) \sin u+g(v) \cos u) .
\end{equation*}
Its first fundamental form is given by
\begin{equation*}
dx^{2}=( pdu) ^{2}+\epsilon [( f^{2}+g^{2})
du^{2}+2( gf^{\prime }-fg^{\prime }) dudv+( f^{\prime
2}+g^{\prime 2}) dv^{2}] .
\end{equation*}
It is convenient to assume that the curve $\alpha $ is parametrized by the
Euclidean arc length, that is, $f^{\prime 2}+g^{\prime 2}=1.$ Then, the
coefficients of the first and second fundamental forms are
\begin{eqnarray*}
g_{11} &=&p^{2}, \, g_{12}=0, \, g_{22}=0, \\
L_{11} &=&fg^{\prime }-f^{\prime }g, \, L_{12}=-sgn(p), \, L_{22}=\frac{
g^{\prime \prime }}{f^{\prime }}, \\
\Gamma _{11}^{2} &=&p^{2}\left( -ff^{\prime }-gg^{\prime }\right)
\end{eqnarray*}
and other the Christoffel symbols are zero. Therefore, we have
\begin{eqnarray*}
\kappa _{n} &=&(fg^{\prime }-f^{\prime }g)( \overset{.}{u}) ^{2}-2sgn(p) \overset{.}{u}\overset{.}{v}+\frac{g^{\prime \prime }}{f^{\prime }}( \overset{.}{v}) ^{2}, \\
\tau _{g} &=&p\overset{.}{u}-sgn(p) p\frac{g^{\prime \prime }}
{f^{\prime }}\overset{.}{v}, \\
\kappa _{g} &=&\vert p\vert \overset{..}{u}.
\end{eqnarray*}
\newline
Now, let a plane curve $\alpha $ be an admissible curve in an isotropic
plane. We can treat two cases: the first one when the curve $\alpha $ is
parametrized by $\gamma \left( v\right) =\left( g\left( v\right) ,f\left(
v\right) ,0\right) $ and the second when the curve $\alpha $ is parametrized
by $\gamma \left( v\right) =\left( g\left( v\right) ,0,f\left( v\right)
\right) .$ Then the helical surface $H_{i}$ is given by
\begin{equation*}
\varphi \left( u,v\right) =\left( pu+g\left( v\right) ,f\left( v\right) \sin
u,f\left( v\right) \cos u\right)
\end{equation*}
and
\begin{equation*}
\varphi \left( u,v\right) =\left( pu+g\left( v\right) ,f\left( v\right) \cos
u,f\left( v\right) \sin u\right)
\end{equation*}
respectively.\newline
The first fundamental form of a helical surface $H_{i}$ obtained in this way
is given by
\begin{equation*}
dx^{2}=\left( pdu+g^{\prime }\left( v\right) dv\right) ^{2}+\epsilon \left(
f^{2}\left( v\right) du^{2}+f^{\prime 2}\left( v\right) dv^{2}\right) .
\end{equation*}
If the curve $\alpha $ is parametrized by the isotropic arc length, that is
if $g\left( v\right) =v$, then the above form can be simplified. Therefore,
the coefficients of the first and second fundamental forms are
\begin{eqnarray*}
g_{11} &=&p^{2},g_{12}=p,g_{22}=1, \\
L_{11} &=&\frac{f^{2}}{\sqrt{f^{2}+p^{2}f^{\prime 2}}},L_{12}=\frac{
-pf^{\prime 2}}{\sqrt{f^{2}+p^{2}f^{\prime 2}}},L_{22}=\frac{-ff^{\prime
\prime }}{\sqrt{f^{2}+p^{2}f^{\prime 2}}}, \\
\Gamma _{11}^{1} &=&\frac{pff^{\prime }}{f^{2}+p^{2}f^{\prime 2}},\Gamma
_{12}^{1}=\frac{ff^{\prime }}{f^{2}+p^{2}f^{\prime 2}},\Gamma _{22}^{1}=
\frac{-pf^{\prime }f^{\prime \prime }}{f^{2}+p^{2}f^{\prime 2}}, \\
\Gamma _{11}^{2} &=&\frac{-p^{2}ff^{\prime }}{f^{2}+p^{2}f^{\prime 2}}
,\Gamma _{12}^{2}=\frac{-pff^{\prime }}{f^{2}+p^{2}f^{\prime 2}},\Gamma
_{22}^{2}=\frac{p^{2}f^{\prime }f^{\prime \prime }}{f^{2}+p^{2}f^{\prime 2}}.
\end{eqnarray*}
Therefore, we have
\begin{eqnarray*}
\kappa _{n} &=&\left( \frac{f^{2}}{\sqrt{f^{2}+p^{2}f^{\prime 2}}}\right)
\left( \overset{.}{u}\right) ^{2}+2\frac{pf^{\prime 2}}{\sqrt{
f^{2}+p^{2}f^{\prime 2}}}\overset{.}{u}\overset{.}{v}+\left( \frac{
-ff^{\prime \prime }}{\sqrt{f^{2}+p^{2}f^{\prime 2}}}\right) \left( \overset{
.}{v}\right) ^{2}, \\
\tau _{g} &=&\overset{.}{\left( \frac{f^{2}+p^{2}f^{\prime 2}}{
f^{2}+p^{2}f^{\prime 2}}\right) \overset{.}{u}+}\left( \frac{pff^{\prime
\prime }-pf^{\prime 2}}{f^{2}+p^{2}f^{\prime 2}}\right) \overset{.}{v}, \\
\kappa _{g} &=&\left\vert p\overset{..}{u}+\overset{..}{v}\right\vert .
\end{eqnarray*}

config: arxiv


text:
\section{Square-exponential kernel cannot yield Euler's method}
\label{sec:se-kernel}

We show that the square-exponetial (SE, aka. RBF, Gaussian) kernel
cannot yield Euler's method for finite length-scales.

\newcommand{\tsq}{\theta^2}
\newcommand{\lsq}{\lambda^2}

The SE kernel and its derivatives are
\begin{align}
k(t,t') &= \tsq \exp(\nicefrac{-\norm{t-t'}^2}{2\lsq})\\
\kd(t,t') &= \frac{(t-t')}{\lsq}k(t,t')\\
\dkd(t,t') &= \left[\frac{1}{\lsq}
            - \left(\frac{(t-t')}{\lsq}\right)^2\right]k(t,t')
\end{align}

To show that this choice does not yield Euler's method, we proceed as in the
case for the GMRK methods. The predictive mean after observing $x_0$ and $y_1$
is given by
\begin{align}
\mu_{\vert x_0, y_1} (t_0 + s) &=
  \begin{bmatrix}k(t_0+s,t_0) &\kd(t_0+s,t_0)\end{bmatrix}
  \underbrace{\begin{bmatrix}k(t_0,t_0) & \kd(t_0,t_0)\\
                             \dk(t_0,t_0)&\dkd(t_0,t_0)\end{bmatrix}^{-1}}_{\ec K}
  \begin{pmatrix} x_0\\y_1\end{pmatrix}\notag\\
&= \begin{bmatrix} \tsq \exp(\nicefrac{-\norm{s}^2}{2\lsq}) &
                   s \frac{\tsq \exp(\nicefrac{-\norm{s}^2}{2\lsq})}{\lsq}
   \end{bmatrix}
  \begin{bmatrix} \tsq & 0\\ 0 & \nicefrac{\tsq}{\lsq}\end{bmatrix}^{-1}
  \begin{pmatrix} x_0\\y_1\end{pmatrix}\notag\\
&=\begin{bmatrix} \exp(\nicefrac{-\norm{s}^2}{2\lsq}) &
                s \exp(\nicefrac{-\norm{s}^2}{2\lsq})\end{bmatrix}
 \begin{pmatrix} x_0\\y_1\end{pmatrix}\notag\\
&=\exp(\nicefrac{-\norm{s}^2}{2\lsq})x_0 + s \exp(\nicefrac{-\norm{s}^2}{2\lsq})y_1\label{eq:se-posterior}
\intertext{evaluated at $h$ yields}
&=\exp(\nicefrac{-\norm{h}^2}{2\lsq})x_0 + h \exp(\nicefrac{-\norm{h}^2}{2\lsq})y_1\label{eq:se-post-at-h}
\end{align}
An interesting observation, left out in the main paper to avoid
confusion, is that Eq.~\eqref{eq:se-post-at-h} \emph{does} indeed
produce the weights for Euler's method for the limit $\lim_{\lambda
  \to \infty}$. In fact, it can even be used to derive second and
third order Runge-Kutta means, too. Future work will provide more
insight into this property. However, this limit in the length-scale
yields a Gaussian process posterior that has little use as a
probabilistic numerical method, because its posterior covariance
vanishes everywhere. This is in contrast to the integrated Wiener
processes discussed in the paper, which yield proper finite,
interpretable posterior variances, even after in the limit in
$\tau$. Finally, SE kernel-GPs are not Markov. Inference in these
models has cost cubic in the number of observations, reducing their
utility as numerical methods.


text:
\subsection{Generalized duality theorem}
\label{sec:dual}
We now present the main theorem of this paper, the generalized duality theorem of \kdro~\eqref{eq:kdro}.

\begin{theorem}[\textbf{Generalized Duality}]
\label{thm:dro_main_thm}

Under Assumption~\ref{asm:compact}, \eqref{eq:kdro} is equivalent

to

\begin{equation}

	\begin{aligned}
		(D):=&\min_{ \var,f_0\in\mathbb R, f\in\rkhs}&&{\  f_0 + \suppc (f)  }\\
		&\sjt& &\semiinfcons
	\end{aligned}
	\label{eq:kdro_rc}

\end{equation}
where $\suppc (f)  :=\sup_{\mu\in\mathcal C} \hip{f}{\mu} $ is the support function of $\mathcal C$, i.e., $(P)=(D)$, \emph{strong duality} holds for the inner moment problem for any \var point-wise.

\end{theorem}
The theorem holds regardless of the dependency of $l$ on \var, e.g., non-convexity.
If $l$ is convex in \var, then \eqref{eq:kdro_rc} is a \emph{convex program}.
Formulation \eqref{eq:kdro_rc} has a clear geometric interpretation: we find a function $f_0+f$ that \emph{majorizes} $l(\var,\cdot)$ and subsequently minimize a surrogate loss involving $f_0$ and $f$. This is illustrated in Figure~\ref{fig:3in1_kdro}. Note the term duality here refers to the inner moment problem.

The statement can be further simplified by replacing $f_0+f$ with $f$. However, we choose the current notation for the sake of its explicit connection to RO.

\paragraph*{Proof sketch.}
Our weak duality proof follows standard paradigms of Lagrangian relaxation by introduing dual variables.
Notably, we associate the functional constraint $\int{\phi }{\ dP} = \mu$ with a dual function $f\in \rkhs$, which is the decision variable in the dual problem~\eqref{eq:kdro_rc}.
Using the reproducing property of RKHSs and conic duality, we arrive at~\eqref{eq:kdro_rc} with weak duality.
Our strong duality proof is an extension of the conic strong duality in Eulidean spaces.
We rely on the existance of separating hyperplnes between convex sets in locally convex function spaces, e.g., \rkhs. See the illustration in Figure~\ref{fig:separate_main}. In our generalized duality theorem, this separating hyperplane is determined by the witness function $f^*$, which is the optimal dual variable in \eqref{eq:kdro_rc}. See the appendix for the full proof.

\begin{figure}[h!]
	\vspace{-0.2cm}
	\centering
	\includegraphics[width=0.6\columnwidth]{draft/fig/separate.pdf}

	\vspace{-0.5cm}
	\caption{Illustration of a separating hyperplane in \rkhs}
	\label{fig:separate_main}
	\vspace{-0.5cm}
  \end{figure}

Theorem~\ref{thm:dro_main_thm} generalizes the classical bounds in generalized moment problems \cite{isiiSharpnessTchebychefftypeInequalities1962,lasserreBoundsMeasuresSatisfying2002,
shapiroDualityTheoryConic2001,
bertsimasOptimalInequalitiesProbability2005,popescuSemidefiniteProgrammingApproach2005,vandenbergheGeneralizedChebyshevBounds2007,
vanparysGeneralizedGaussInequalities2016} to infinitely many moments using RKHSs.
A distinction between Theorem~\ref{thm:dro_main_thm} and other DRO approaches is that it uses the density of universal RKHSs to find a surrogate which can sharply bound the worst-case risk. This means that we do not require the loss $l(\var,\cdot)$ to be affine, quadratic, or living in a known RKHS, nor do we require the knowledge of Lipschitz constant or RKHS norm of $l(\var,\cdot)$.
To our knowledge, existing works

typically require one of such assumptions.

Moreover, Theorem~\ref{thm:dro_main_thm} generalizes existing RO and DRO in the sense that it gives us a unifying tool to work with various ambiguity and ambiguity sets, which may be customized for specific applications.

We outline a few closed-form expressions of the support function $\suppc (f)$ in Table~\ref{tbl:part}, and more in Table~\ref{tbl:sets}.

We now return IPM-DRO with a duality result.
\begin{corol}[IPM-DRO duality]
	\label{thm:ipm_dual}
Given the integral probability metric $\ipm (P,\hat{P}):=\sup_{f\in \mathcal F}\int f d (P-\hat{P})$, a dual program to \eqref{eq:ipm_primal} is given by
	\begin{equation}
		\label{eq:ipm0}
		\begin{opt1}

			\MINST{\var, \lambda\geq 0, f_0\in\mathbb R, f\in \mathcal F}{\           {f_0+\frac{1}N\sum_{i=1}^N \lambda f(\xi_i)+\lambda\epsilon }
			}{l(\var, \xi)\leq  f_0 + \lambda f(\xi)   ,\ \forall \xi \in\mathcal X.}
		\end{opt1}
	\end{equation}
\end{corol}
The reduction to \eqref{eq:kdro_rc} as a special case can be seen by replacing $\lambda f$ with $f$ and choosing $\mathcal F=\rkhs$.

We now establish an explicit connection between DRO and stochastic optimization with expectation constraint, whose solution methods using stochastic approximation are an topic of active research~\cite{lanAlgorithmsStochasticOptimization2020a,xuPrimalDualStochasticGradient2020}.
\begin{corol}(Stochastic optimization with expectation constraint)
	\label{thm:expect_cons}
	Under the Assumption~\ref{asm:compact}, the optimal value of \eqref{eq:kdro} coincides with that of
	\begin{equation}
		\begin{opt1}
		\label{eq:sa_exp_cons}
		\MINST{ \var,f_0\in\mathbb R, f\in\rkhs}{\  f_0 + \suppc (f)  }
		{\mathbb{E}h\left(l\left(\var, \zeta\right) -   f_0 - \lambda f\left(\zeta\right) \right) \leq 0}
		\end{opt1}
	  \end{equation}
	  for some function $h$ that satisfies $h(t) = 0 \text{ if } t\leq 0, h(t)>0 \text{ if } t>0$, and random variable $\zeta\sim\mu$ whose probability measure places positive mass on any non-empty open subset of \domain, i.e.,
	  $
	  \mu (B)>0,\ \forall B\subseteq \domain, B\neq\emptyset, B\text{ is open}
	  $.
\end{corol}
A choice for $h$ is $h(\cdot) = [\cdot]_+$, which is used in the conditional value-at-risk~\cite{rockafellarOptimizationConditionalValueatrisk2000a}.
We will see the computational implication of Corollary~\ref{thm:expect_cons}

in Section~\ref{sec:practical}.


text:
\begin{definition}[Definition:Linear Ring Action]
Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $M$ be an [[Definition:Abelian Group|abelian group]].
=== [[Definition:Linear Ring Action/Left|Left Ring Action]] ===
{{:Definition:Linear Ring Action/Left}}
=== [[Definition:Linear Ring Action/Right|Right Ring Action]] ===
{{:Definition:Linear Ring Action/Right}}
\end{definition}

config: wiki


TITLE: Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt79+\sqrt81}$
QUESTION [3 upvotes]: Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}$
I've thought about multiplying every fraction by 1, but like this $\frac{1} {\sqrt1+\sqrt3} *\frac{\sqrt1+\sqrt3}{\sqrt1+\sqrt3}$, $\frac{1} {\sqrt3+\sqrt5} *\frac{\sqrt3+\sqrt5}{\sqrt3+\sqrt5}...$
After multiplying I end up with $\frac{\sqrt1+\sqrt3} {(\sqrt1+\sqrt3)^2} + \frac{\sqrt3+\sqrt5} {(\sqrt3+\sqrt5)^2}  ...$ but I don't have any more ideas than this, am I on the right path?
I apologize for the question's simplicity, but I'm a younger math enthusiast so questions like this are harder for me.

REPLY [0 votes]: $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}=\frac{1}{\sqrt3+\sqrt1} *\frac{\sqrt3-\sqrt1}{\sqrt3-\sqrt1}+ \frac{1}{\sqrt5+\sqrt3}*\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3} + \frac{1}{\sqrt7+\sqrt5}*\frac{\sqrt7-\sqrt5}{\sqrt7-\sqrt5} + ... \frac{1}{\sqrt{81}+\sqrt{79}} *\frac{\sqrt81-\sqrt79}{\sqrt81-\sqrt79}=\frac{\sqrt3-\sqrt1}{2}+\frac{\sqrt5-\sqrt3}{2}+\frac{\sqrt7-\sqrt5}{2}+...+\frac{\sqrt81-\sqrt79}{2} =(\frac{1}{2})((\sqrt3-\sqrt1)+(\sqrt5-\sqrt3)+(\sqrt7-\sqrt5)+...(\sqrt81-\sqrt79))=(\frac{1}{2})(\sqrt81-\sqrt1)=4$
So the problem is done.


TITLE: Convergence in $\ell^p$
QUESTION [1 upvotes]: Let $0\neq x=(x_1,x_2,\cdots,x_n,\cdots) \in\ell^4$. For which of the following  values of $p$, the series $$\sum_{i=1}^\infty x_iy_i$$ converges for every $y=(y_1,y_2,\cdots,y_n,\cdots) \in\ell^p $ ?
(a) $p=1$ $\hspace{2cm}$ (b) $p=2$$\hspace{2cm}$ (c) $p=3$ $\hspace{2cm}$ (d) $p=4$

I know this result: For any $a=(a_i) \in \ell^q$, the series $\sum_{i=1}^\infty a_iy_i$ converges absolutely, for all $y=(y_i) \in \ell^p$ where $q \in (1,\infty]$ and $p \in [1,\infty)$ with $\frac{1}{p}+\frac{1}{q}=1$
So by above result, $\sum_{i=1}^\infty x_iy_i$ converges absolutely, for all $y=(y_i) \in \large\ell^\frac{4}{3}$. But how about these four options ?
Here $$ \sum_{i=1}^\infty \vert x_iy_i \vert \leq \vert\vert x \vert \vert_{\large\ell^4}\vert\vert y \vert \vert_{\large\ell^p} $$ How to choose suitable $p$ to make the RHS to be small?  Any Guidance please?

REPLY [2 votes]: The correct answer is 1). Use the fact that $p \leq \frac 4 3 $ and $(y_n) \in \ell^{p}$ implies $(y_n) \in \ell^{4/3}$


TITLE: How to prove that this function is unimodal?
QUESTION [0 upvotes]: I have a function f(x) = $\sum_{i=1}^n c_i|x-h_i|$ where both $c_i$ and $h_i$ are positive integers. Now I wanted to prove that this function is unimodal but I am unable to do so.
Actually, I was learning Ternary Search so I wanted to apply that technique to a problem and this was the equation that I came up with for solving the particular problem. The answer to the problem is the minimum value of this function. Though I could solve the problem using Ternary Search, I couldn't figure out that why this function is unimodal. How can one prove whether a function is unimodal or not (especially this one)?

REPLY [1 votes]: The function $f(x)=\sum_{i=1}^nc_i|x-h_i|$ is piecewise linear. Without loss of generality, let's assume that $h_1\lt h_2\lt\cdots\lt h_n$ (otherwise one can first permute the index notation). Then on each interval $(h_i,h_{i+1})$, the slope of the linear function is $m_i=(c_1+\cdot+c_i)-(c_{i+1}+\cdots+c_n)$, with $m_0=-(c_1+\cdots+c_n)$ for the interval $(-\infty,h_1)$ and $m_n=c_1+\cdots+c_n$ for the interval $(h_n,\infty)$. Since all the $c_i$'s are positive (by assumption), the slopes are strictly increasing (i.e., $m_i-m_{i-1}=2c_i\gt0$), so there cannot be more than a single local minimum. Finally, since $m_0\lt0$ and $m_n\gt0$, there is a minimum, and thus the function is unimodal.
Remark: The only crucial assumption here is that the $c_i$'s are all positive; they do not need to be integers, nor is any assumption on the $h_i$'s needed at all.


TITLE: Permutations with a restriction
QUESTION [2 upvotes]: Let $n\geq 0$ be an integer and let $0 \leq j \leq n$, $0 \leq t \leq n$.
Let $S_n(j, t)$ be the set of all permutations $\sigma$ of $\{1, \dots, n\}$ with the property that
$$\sigma(i) \leq j \quad \text{for} \quad i \leq t.$$
Is there a simple formula for the cardinality of $S_n(j, t)$?
Example If $t>j$ then $S_n(j,t)=0$ by the pigeonhole principle ($\{1, \dots, t\}$ cannot map injectively into $\{1, \dots, j\}$ ).
Example If $t=j$ then the datum of an element of $S_n(j, t)$ is equivalent to the datum of a permutation of $\{1, \dots, j\}$ and a permutation of its complement in $\{1, \dots, n\}$. Hence $\# S_n(j,j) = j!(n-j)!$.
What about the general case $t\leq j$?
This question arose in a multilinear albgera calculation.

REPLY [2 votes]: Normally the number of permutations of $\left\{1,2,\ldots,n\right\}$ is $n! = \prod_{k=1}^n k$ by stating the first element has $n$ choices, the next $n-1$, etc.
Here we see that the first element has $j$ choices, the next $j-1$, etc, until $j + 1 - t$ after which the next element has no further restrictions so it has $n - t$ choices, then $n - t - 1$, etc, so we get $\prod_{k=1}^t (j + 1 - k) \times \prod_{k=t+1}^n (n + 1-k)$ as our answer.


TITLE: Suppose $F: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a continuous functions. If $f$ and $g$ are measurable, then
QUESTION [1 upvotes]: Suppose $F: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a continuous functions. If $f$ and $g$ are measurable, then $h(x) = F( f(x), g(x) ) $ is also measurable.
Proof:
For all $a$, $ \{ x : h(x) > a \} = \{ x : (f(x),g(x)) \in G_a \} $ where $G_a := \{ (u,v) : F(u,v) > a \} = F^{-1}((a, \infty))$. Since $F$ is continuous, then given any $(a, \infty)$, we must have $F^{-1}((a, \infty))$ open set. Since open sets are union of open rectangles on $\mathbb{R}^2$, then it must be the case that
$$ G_a = \bigcup_{n=1}^{\infty} (a_n,b_n) \times (c_n,d_n) $$ Therefore, Can we say the next step follows from the above? Im having trouble proving this holds. Can someone help me?
$$ \{ x : h(x) > a \} = \bigcup_{n=1}^{\infty} \{ x : f(x) \in (a_n,b_n) \} \cap \{ x : g(x) \in (c_n, d_n) \} $$
And the above is measurable since it is union of measurable sets, thence $F$ must be measurable.

REPLY [1 votes]: \begin{align*}
h(x) > a &\iff
(f(x),g(x)) \in G_a
\\
&\iff
(f(x),g(x)) \in (a_n,b_n) \times (c_n, d_n) \text{ for some } n
\\
&\iff f(x) \in (a_n,b_n) \text{ and } g(x) \in (c_n, d_n) \text{ for some } n
\\
\end{align*}
So
\begin{align*}
\{x : h(x) > a\} = \bigcup_n \{x : f(x) \in (a_n,b_n)\} \cap \{x : g(x) \in (c_n, d_n)\}
\end{align*}


TITLE: I Need help creating an intuitive answer to the sum of $1(1!) + 2(2!) + 3(3!) +\cdots+ n(n!)$
QUESTION [9 upvotes]: Given the sum $1+\sum_{i=1}^n i(i)! = (n+1)!$, is there an intuitive way to think about this sum? I understand the algebraic manipulation to get to that answer, and also how to use induction to prove the answer, but say could we use a combinatorial proof to get to the same answer? I cant seem to think of a way to think of anything that could intuitively explain this. I tried thinking about it as the number of ways to create license plates with restrictions on the number of different characters that are allowed to be used at any given spot in the license plate, whereas the right-hand side counts them all at once, but that didn't get me to an answer. Any help is greatly appreciated.

REPLY [7 votes]: If we want to arrange $n+1$ distinct objects in a line ($(n+1)!$ ways in total), we can take the last object and either

place it in any one of the first $n$ positions, after which we have $n!$ ways to permute the remaining objects ($n(n!)$ ways)
place it in the last position

In the second case, we repeat the whole process with the first $n$ objects and free positions. If we place the second-last object in the second-last position, we continue with the first $n-1$ objects and free positions, etc. This recursion stops when we have only one object left to place in one position. The total number of arrangements that can be generated by this process is
$$n(n!)+(n-1)(n-1)!+\cdots+1(1!)+1$$
Every permutation can be achieved this way, so the sum is equal to $(n+1)!$.


TITLE: If $f \neq 0$ there exist $x_0$ with $|f(x_0)| < |f'(x_0)|$
QUESTION [1 upvotes]: I am facing the following problem.

Let $f$ be a continuous and derivable function from $[0, 1]$ to $R$,
  such that $f(0) = 0$. Show that if $f \neq 0$ then there exist $x_0$
  such that $|f'(x_0)| > |f(x_0)|$,

I have been thinking for a while on this problem but I have no idea how to demonstrate it. Can someone provide a hint?

REPLY [1 votes]: Since $|f|$ is continuous on the closed interval $[0,1]$, $|f|$ assumes a maximum value, $b$ say.

Since $f\ne 0$, it follows that $b > 0$.

Let $S=\{x\in [0,1]{\;:\;}|f(x)|=b\}$.

Since $|f|$ is continuous, $S$ is closed.

Since $S\subseteq [0,1]$, $S$ is compact, hence has a least element, $a$ say.

Since $0\not\in S$, we have $a>0$.

Since $f$ is differentiable, applying the Mean Value Theorem, there exists $c\in (0,a)$ such that
$$f'(c)=\frac{f(a)-f(0)}{a-0}=\frac{f(a)}{a}$$
hence
$$|f'(c)|=\left|\frac{f(a)}{a}\right|=\frac{|f(a)|}{|a|}=\frac{b}{|a|}\ge \frac{b}{1}=b > |f(c)|$$


TITLE: The existence of a 1-1 continuous map between two topological spaces.
QUESTION [3 upvotes]: Show that there is no one-to-one continuous map $f$ from $\mathbb{R}^n$ to $\mathbb{R}^2$ for $n\gt 2$ with $f(0)=0$.  I tried using the hint:  consider $f:\mathbb{R}^n-\{0\}\rightarrow \mathbb{R}^2-\{0\}$ and the induced map: $f_*:\pi_1 (\mathbb{R}^n-\{0\}) \rightarrow \pi_1(\mathbb{R^2}-\{0\})$.

REPLY [3 votes]: On Georges' request, here is a detailed proof expanding Michael's incomplete argument.
Lemma 1. Let $(K_i)$ be an exhaustion of $R^n$ by a sequnce of compact subsets:
$$
K_1\subset K_2\subset ...
$$
Then $\exists i$ such that $K_i$ has nonempty interior.
Proof. This is an immediate corollary of Baire's theorem ($R^n$ cannot be a countable union of nowhere dense subsets). qed
Lemma 2. $X=R^n \setminus \{0\}$ is simply-connected if $n\ge 3$.
 Proof.  $X$ is homeomorphic to $S^{n-1}\times R$, hence, homotopy-equivalent to $S^{n-1}$. To prove simple connectivity of the latter, note that every loop in it is homotopic to a piecewise-circular loop. The latter has nowehere dense image (as $n-1\ge 2$), hence, can be regarded as a loop in $R^{n-1}$. The latter is contractible. Hence, the original loop is null-homotopic. qed
(If you know Seifert - van Kampen theorem, it gives an alternative proof, but what I wrote is much easier than S-vK Theorem.)
Lemma 3. Let $f: R^n\to R^2$ be a  continuous bijection. Then there exists a round circle $C\subset R^2$ such  that the restriction $f^{-1}|_C$ is continuous.
Proof. Let $(B_i)$ denote an exhaustion of $R^n$ by closed round balls. Their images $K_i=f(B_i)$ are compact and, obviously, form an exhaustion of $R^2$. By Lemma 1, there exists $K_i$ with nonempty interior. Take a round circle $C$ contained in $K_i$. The map $f: B_i\to K_i$ is a continuous bijection; $B_i$ is compact and  $K_i$ is Hausdorff. Hence, $f^{-1}|_{K_i}$ is continuous. Lemma follows. qed
Lemma 4. There does not  exists a continuous bijection $f: R^n\to R^2$, $n\ge 3$.
Proof. Suppose that $f$ exists. Let $C$ be a circle as in Lemma 3. Let $p$ be the center of this circle and $r$ its radius. The map $g=f^{-1}|_C$ is continuous (Lemma 3). Then $g: C\to X=R^n \setminus \{q\}$ is also continuous, where $q=f^{-1}(p)$. Since $X$ is simply-connected (Lemma 2), the map $g$ extends to a continuous map $h$ of the disk $B(p,r)$ (centered and $p$ and of radius $r$). Composing $j=f\circ h$, we get a continuous map $j: B(p,r)\to R^2\setminus \{p\}$ which is the identity on $C$.  Contradiction. (OK: here you need to know something about nontriviality of the 1st homology or fundamental group of the circle, but I assume that you already know this.) qed


TITLE: Interpretation of dispersion relation
QUESTION [2 upvotes]: I'm going through Terry Tao's book on nonlinear dispersive PDE, and am a little confused by some of the things going on. The setup is that we are working with a PDE given by
$$
\partial_t u = Lu.
$$
In particular, on page 56, he defines the "frequency operator" $D = \frac{1}{i}\nabla$, and says that we can write a general partial differential operator $L$ as $L = ih(D)$ where $h$ is some polynomial. While I do not doubt this, I just don't understand the usefulness of doing any of it. I am completely new to dispersive PDE, so this might be the issue.
He also calls $h$ the "dispersion relation" of the dispersive PDE $\partial_t u = Lu$, and I don't understand a) what a dispersion relation is, or b) why this polynomial should be called it.
If anyone could point me in the right direction, it would be much appreciated!

REPLY [2 votes]: In order to give a complete notion of what a dispersive partial differential equation is, consider the one-dimensional frame. We look for plane wave solutions of the form $$
u(x,t)=Ae^{i(kx-\omega t)},
$$
where $A$, $k$ and $\omega$ are constants representing the amplitude, the wavenumber, and the frequency, respectively. Hence $u$ will be a solution of $$\partial_t u=Lu, \qquad Lu:=\sum_{\vert \alpha \vert\leq k}c_\alpha \partial^\alpha_xu(x),$$ if and only if $$
\omega+\sum_{\alpha\leq k}c_\alpha i^{\alpha-1}k^\alpha=0.
$$
This equation is called the dispersion relation. A commonly used defining criteria for dispersive equations is that $\omega(k)$ is a real valued function of $k$ such that $\tfrac{d^2\omega}{dk^2}\neq 0$. In the physical context this means that different frequencies in this equation will tend to propagate at different velocities, thus when time evolves, the different waves disperse in the medium, with the result that a single hump breaks into wave-trains.
The importance of these definitions comes from the fact that the relation between $\omega$ and $k$ characterizes the plane wave motion. Consider for instance the linear Schrödinger equation $$
iu_t+u_{xx}=0,
$$
and the plane wave $u(t,x)=e^{i(k x-\omega t)}$. Then, $u(t,x)$ is a solution of the equation if and only if the dispersion relation $\omega=k^2$ holds. Note that in this case $\omega$ is a real valued function of the frequency. Now, consider the quantity $$
\nu_p(k):=\dfrac{\omega}{k},
$$
which is commonly call the phase velocity of the wave. With this definition one can re-write the solutions of the linear Schrödinger equation as: $$
u(t,x)=e^{ik(x-\nu_p(k)t)}=u(0,x-\nu_p(k)t),
$$
and thus conclude that the wave travels with velocity $\nu_p(k)$. In particular, large frequency data travel faster than smaller ones.


TITLE: Functions satisfying a given condition
QUESTION [3 upvotes]: Let $m,n \geq 2$ be integers. Find all functions $f:[0,\infty) \to \mathbb{R}$ continuous at at least one point in $[0,\infty)$ and such that $$f\biggl(\frac{1}{n} \sum\limits_{i=1}^{n} x_{i}^{m} \biggr)=\frac{1}{n} \sum\limits_{i=1}^{n}(f(x_i))^{m} \quad \text{for} \ x_{i} \geq 0, i=1,2,...,n$$
I am not getting any idea to proceed.

REPLY [4 votes]: Let $x_1=x_2=\ldots=x_n=x$, then $f(x^m)=f(x)^m$, and $f(0)=f(0)^m$, so $f(0)$ is either 0,1 or -1.
Now let $y_i=x_i^m$, then we have $f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$ for any $y_i\in \mathbb{R}$.
If $f(0)=0$, then $y_1=y$, $y_2=\ldots=y_n=0$ implies that $f(\frac{1}{n} y)=\frac{1}{n}f(y)$.
This implies $\frac{1}{n}f(\sum_{i=1}^n y_i)=f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$.
Letting $y_3=\ldots=y_n=0$, we get that $f(y_1+y_2)=f(y_1)+f(y_2)$, which is Cauchy's Functional Equation. Since $f$ is continuous at one point, it's linear, so $f(x)=ax$.
We get $a=a^m$, so the solutions in this case are $f(x)=0$, $f(x)=x$ and $f(x)=-x$ (if $m$ is odd).
If $f(0)\neq 0$, WLOG let $f(0)=1$ (or else replace $f$ by $-f$ if $m$ is odd).
Analogously $f(\frac{1}{n}y)=\frac{f(y)+n-1}{n}$, and $\frac{f(\sum_{i=1}^n y_i)+n-1}{n}=f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$, so that $f(\sum_{i=1}^n y_i)+n-1=\sum_{i=1}^n f(y_i)\Rightarrow g(\sum_{i=1}^n y_i)=\sum_{i=1}^n g(y_i)$ where $g(y)=f(y)-1$.
From this we have that $g$ satisfies Cauchy's functional equation, so that $f(y)=ay+1$. But $ay^m+1=f(y^m)=f(y)^m=(ay+1)^m$, which implies $a=0$ by comparing coefficients. So in this case the solutions are $f(x)=1$ and $f(x)=-1$ (if $m$ is odd).
We conclude that the solutions are $f(x)=0$, $f(x)=x$, $f(x)=1$ and if $m$ is odd $f(x)=-x$ and $f(x)=-1$.


TITLE: Understanding induction proof with inequalities
QUESTION [1 upvotes]: I'm having a hard time proving inequalities with induction proofs. Is there a pattern involved in proving inequalities when it comes to induction? For example:
Prove ( for any integer $n>4$ ):
$$2^n > n^2 \\ $$
Well, the skipping ahead to the iduction portion, here's what I've got so far:
let $n=k$, then $2^{k+1} > (k+1)^2 $.
Starting with the LHS
$$\begin{align}     \\  2^{k+1}=2\cdot2^k  \\  > 2\cdot k^2  \end{align}$$
And that 's where it ends for me. What would be the next step, and what suggestions might you give for future proofs like this? Do we always need to prove that the left hand side is equal to the other?

REPLY [2 votes]: Working off Eric's answer/approach:
Let P(n) be the statement that $2^n \gt n^2$.
Basis step: (n=5) $2^5\gt 5^2 $ which is true.
Inductive step: We assume the inductive hypothesis that $P(k)$ is true for an arbitrary integer $k\ge5$.
$$ 2^{k} \gt k^2 \text{ (IH)}$$
Our goal is to show that P(k+1) is true. Let's multiply each side by two:
$$ 2*2^{k}>2(k)^2 $$
$$ 2^{k+1}>2k^2 $$
If this statement is true, then $2k^2$ $\ge$ $(k+1)^2$ must also be true when $k\ge5$:
$$ 2k^{2}\ge(k+1)^2 $$
$$ 2k^{2} \ge k^2+2k+1$$
$$k^2\ge2k+1$$
And it is. Therefore $\forall{_{k\ge5}}P(k) \implies P(k+1)$ and $\forall{_{n\ge5}}P(n)$ follows by induction.


text:
\begin{document}

\title{Nonautonomous Food-Limited Fishery Model With Adaptive Harvesting}

\author{
   L. V. Idels
   \thanks{Research supported by a grant from Vancouver Island University (VIU)}.
      \\
   Department of Mathematics  \\
   Vancouver Island University  \\
   900 Fifth St. Nanaimo BC, V9S5S5  \\
   email:  {\tt{lev.idels@viu.ca}} \\\\
   A. Khokhlov\\
 Department of Mathematics \\   Chelyabinsk South Ural State University\\
   80a Lenin Ave, Chelyabinsk  454080, \\Russia \\ email:
   {\tt{Artur.Khokhlov@rocketsoftware.com}}\\\\
   Corresponding Author:  Lev V. Idels {\vspace{-1mm}}}

\maketitle \vfill{}
\newpage
\begin{abstract}
We will introduce the biological motivation of the $\gamma$-
food-limited model with variable parameters. New criteria are
established for the existence and global stability of positive
periodic solutions. To prove the existence of steady-state
solutions, we used the upper-lower solution method where the
existence of at least one positive periodic solution is obtained by
constructing a pair of upper and lower solutions and application of
the Friedreichs Theorem. Numerical simulations illustrate effects of periodic variation in
the values of the basic biological and environmental parameters and how the adaptive harvesting strategies
affect fishing stocks.
\end{abstract}
{\bf Keywords}-Fishery models, Nonautonomous Differential Equations,
Harvesting Strategies, Food-Limited Model, Adaptive Harvesting,
Periodic Solutions, Stability.\\
Math Subject Classifications: 34C25, 34D23, 92B05

\section{Time-Varying Fishery Models: Biological motivation}

Most populations experience regular or recurring fluctuations in
biological and environmental factors which affect demographic
parameters \cite{Bo1}--\cite{Bo3}, \cite{Cla}, \cite{Cush1},
\cite{Cush2}, \cite{Hs},  and mathematical models cannot ignore
for example, year-to-year changes in weather, the global climate
variability, habitat destruction and exploitation, the expanding
food surplus, and other factors that affect the population growth
\cite{Bo1}--\cite{Bra2}, \cite{Cush2}, \cite{Hut},
\cite{La}--\cite{Ro} and \cite{Vlad}. In most models of population dynamics, increases in population due to birth are assumed to be time-independent,
but many species reproduce only during a single period of the year. There are several biological parameters that can vary
seasonally, including some cyclical changes in control parameters.
For example, in  temperature or polar zones growth frequently slows down, or even
ceases in winter.
Careful analysis in \cite{Me} shows that there might be a relationship between
asymptotic recruitment and bottom temperature, i.e., stocks located in warmer waters had lower asymptotic recruitment.

Consider the following autonomous model for the harvested population with size
$N(t)$ at time $t$ :

\begin{equation}\label{Eq0}
\frac{1}{N} \frac{dN}{dt} = G(N)-E
\end{equation}
where
\begin{equation}\label{pol}
G(0) = r \,, \qquad G(K) = 0 .\nonumber
\end{equation}
We assume that $G(N)$ is strictly
decreasing, and define the intrinsic growth rate $r>0$ for $N \approx 0$, the
carrying capacity $K>0$, and the effort
function $E$.\\
The linearity assumption in the logistic model is
violated for nearly all populations, e.g. for a food-limited population Smith \cite{SM} (see \cite{Tso})
reported a snag in the classical logistic model, i.e. it did not fit
experimental data, and suggested a modification of the logistic
equation
\begin{equation}\label{Sm}
\frac{1}{N}\frac{dN}{dt}=G(N|r,K,\beta)-E,
\end{equation}
where $G(N|r,K,\beta)=r(1-N/K)(1+\beta N/K)^{-1}$, and Smith
\cite{SM} called the coefficient $(1+\beta N/K)^{-1}$ the delaying
factor.\\
Let $x(t)=\frac{N(t)}{K}$, then equation \eqref{Sm} has the form
\begin{equation}\label{Sm0}
\frac{dx}{dt}=rx\frac{1-x}{1+\beta x}-Ex \nonumber
\end{equation}
To take into consideration a crowding factor , we
introduce a new function
\begin{equation}\label{Eq1}
G(x|r,\beta,\gamma) = r(1-x^{\gamma})(1+\beta x^{\gamma})^{-1},
\end{equation}
where $\gamma > 0$.
Then equation \eqref{Eq0} takes a form
\begin{equation}\label{Smg}
\frac{dx}{dt}=rx\frac{1-x^\gamma}{1+\beta x^\gamma}-Ex.\nonumber
\end{equation}
It is clear, if $\beta=0$, the function $G(x)$ in \eqref{Eq1} is
concave and $G(x) \rightarrow -\infty$ as $x \rightarrow \infty$. If
$ \beta>0$, then $G(x)$ is sigmoidal with $- \frac{r}{\beta} < G(x)
\leq r$.\\

{\bf {Remark.}} Let $G_{1}(x)=x(1-x^{\gamma})(1+\beta
x^{\gamma})^{-1}$. Clearly (see for example Fig 1 below) function $G_{1}(x)$ has a unique maximum
on the interval $x \in (0,1)$,
\begin{figure}[ht]
    \begin{center}
    \includegraphics[width=2.5in, viewport=83 425 533 731,clip=true]{graph1il}
    \label{fig1}
    \caption{Different forms of  function $G_1(x)$  (L1: $\beta=0, \gamma=1$;
    L2: $\beta=0.2, \gamma=5$; L3: $\beta=4, \gamma=0.5$) }
    \end{center}
  \end{figure}
and the maximum value and the
position of a critical point depend on some combinations of the
parameters $\gamma$ and $\beta$. If $\beta=0$ and $\gamma=1$ then
$G_{1}(x)$ is a classical symmetrical logistic function.\\
We construct a time-varying (nonautonomous) model based on equation \eqref{Sm} by allowing $r(t)$ and $K(t)$ to vary, while maintaining
constant parameters $ (\beta,\gamma)$. Similar model for unharvested population was studied in
\cite{Cush2} under the assumption that $K$ and $r$ oscillate with
small amplitudes. For an exploited marine population we introduce a varying effort function
$E=E(t)$.
\begin{equation}\label{vary}
\frac{1}{N}\frac{dN}{dt}=r(t)\left[ 1-\left( \frac{N}{K(t)}\right)
^{\gamma }\right] \left[ 1+\beta \left( \frac{N}{K(t)}\right)
^{\gamma }\right] ^{-1}-E(t),
\end{equation}
where $N|_{t=0} = N_0 >0 .$ \\In population dynamics the term
$\left(\frac{N}{K}\right)^{\gamma}$ is refereed to as the
Richards' nonlinearity \cite{Tso}. \\
 Denote
$$\overline{N}=\int_{0}^{T}N(s)ds$$
and
$$\overline{K}=\int_{0}^{T}K(s)ds ,$$
where $T > 0$ is the period of the system. Illustrative periodic functions $K(t)$  and $r(t)$ can be of
the forms:

\begin{equation}\label{rok}
r(t) = r_{0} \left[ 1 + \alpha_r \sin \frac{2\pi (t-\phi_r)}{T_r}
\right]
\end{equation}

\begin{equation}\label{K}
K(t) = K_{0} \left[ 1 + \alpha_K \sin \frac{2\pi (t-\phi_K)}{T_K}
\right]
\end{equation}
Parameters $r_{0}$, $K_{0}$, $\alpha $, and $T $  can be different for $r$ and $K$. We assume, however, that periods have a lowest common multiple,
which defines the period for the system. \\
For a canonical logistic
equation it was proven that $\overline N$ decreased as the magnitude
of variation in $K$ increases, and that $\overline N \leq \overline
K$ irrespective to $r$. However, it was shown in \cite{Cush2} that
the effect of the environmental cycles is "a very
model-dependent phenomenon".

Harvesting population models with a periodic function $E(t)$ have been studied extensively in recent
years \cite{Bra2}--\cite{Co}, \cite{Ha},
\cite{Id}-- \cite{La}. However, function $E(t)$ is not a periodic function, in fact, it is a function of many (continuous) variables which fishery managers can manipulate and this function can be defined by the fishery strategies. For example, adaptive harvesting strategies in fisheries \cite{F} rests on a combination of three elements:
 (a) deduction from prior knowledge of the ecosystem's components, (b) experience with similar ecosystems elsewhere, coupled with (c) mathematical modeling. Adaptive fishery strategies fit better with new multi-species management, with more emphasis on an "ecosystem approach" to sustainable fishery management.
The study of the dynamical behaviors of the Fox harvesting
models
\begin{equation}\label{Fox}
\frac{1}{N} \frac{dN}{dt}=r(t)\ln^{\gamma}\frac{K(t)}{N}-E(t).
\end{equation}
 in periodical environments were introduced in \cite{Bra2} and \cite{Id}.\\

The paper is organized as follows. In the next section we study qualitative
behavior of the solutions of a
harvesting model in constant environments and obtain the explicit conditions for the
existence of a unique positive solution of equation. In Section 3
for equation \eqref{vary} we will prove that it possess positive,
bounded, asymptotically stable periodic solutions. In Section 4 we
will investigate numerically the effects of periodic variation in
the values of the basic population parameters $r(t)$
and $K(t)$ and discuss adaptive  vs static harvesting strategies. \\
\section{Autonomous Model with Proportional Harvesting}
Consider equation \eqref{Smg} with proportional harvesting

\begin{equation}\label{base}
\frac{dx}{dt}=rx[ 1-x ^{\gamma }] [ 1+\beta  x ^{\gamma }] ^{-1}-Ex.
\end{equation}
If $\beta =0$ then \eqref{base} is an alternative to the logistic
fishing model with the Richards' nonlinearity
\begin{equation}\label{y}
\frac{dx}{dt}=rx( 1-x ^{\gamma }) -Ex.
\end{equation}
Equation \eqref{base} has a nontrivial  equilibrium point
$$x_{Ge}=\sqrt[\gamma]{\frac{1-E^{*}}{1+\beta E^{*}}},$$  and
equation \eqref{y} has a nonzero equilibrium point

$$x _{Le}=\sqrt[\gamma]{1-E^{*}} ,$$ where $E ^{*}=E/r $ and $0<E<r$.

For the corresponding annual equilibrium harvests
$Y_{Ge}(E)=Ex_{Ge}$ and $ Y_{Le}(E)=Ex_{Le}$, \\
 $x_{Ge}< x_{Le}$ and $Y_{Ge}\ <Y_{Le} $. Note that the maximum sustainable yield
($\max_{E}Y(E)$) exists for $E \in (0,r/q)$. \\

Let $u=\ln x$ then equations \eqref{base} has the following form $$
\frac{du}{dt}=\Phi (u)-E ,$$ where
$$
\Phi (u)=r \frac{1-e^{u \gamma}}{1+\beta e^{u \gamma}}.$$ Note
$$
\frac{d\Phi }{du}=\Phi ^{\prime }(u)=-\gamma re^{\gamma u}\frac{
1+\beta }{\left( 1+\beta e^{\gamma u}\right) ^{2}}<0
$$
and positive equilibrium
$$x^*=\sqrt[\gamma]{\frac{1-E^{*}}{1+\beta
E^{*}}}$$ of \eqref{base} is locally asymptotically stable.

\begin{Th} If in equation \eqref{base} we assume that $r,K,\beta, E $ and $ \gamma $
are all positive constants, then

$a1)$ If  $r>E$ then for every solution $x(t)$
$$
\lim_{t\rightarrow \infty }x(t)= x^{*}$$ exists.

$a2)$ If $E\equiv 0$ then $\lim_{t\rightarrow \infty }x(t)=1.$
\end{Th}
{\bf {Proof.}} The solution to equation \eqref{base} has an implicit
form
$$
E^{*}-1+(\beta E^{*} +1)x^{\gamma}=C x^{\alpha \gamma }\exp (-r\alpha
\gamma (1-E^{*})t)\
$$
 with $$\alpha =\frac{
1+E^{*}\beta  }{1+\beta },$$ $C$ is an arbitrary constant. As $
t\rightarrow \infty $
$$
x(t)\rightarrow \sqrt[\gamma ]{\frac{1-E^{*}}{1+E^{*}\beta
}}=x^{*}\text{ }.
$$

To prove the second part of the theorem, we note that if $E\equiv 0$
then the solution of \eqref{base} takes a form

$$
-1+x^{\gamma} =Cx^{\alpha\gamma}\exp (-\gamma \alpha rt)\
$$
thus $$\lim_{t\rightarrow \infty}x(t)=1 .$$

\section{Nonautonomous Model with Seasonal Harvesting}
Let us assume that all parameters in \eqref{vary} are continuous
functions and for all $t\geq 0$ \begin{equation} \label{con}
\frac{1}{N}\frac{dN}{dt}=r(t) \frac{K^{\gamma}(t)-N^{\gamma
}}{K^{\gamma}(t)+\beta N^{\gamma }}-E(t) ,
\end{equation} where $\gamma
>0$ and $\beta
>0.$

\begin{Def}
We say that a positive solution $N^{\ast}(t)$ of equation
(\ref{con}) is a global attractor or globally asymptotically stable
(GAS) if for any positive solution $N(t)$
$$
\lim_{t\rightarrow\infty}|N(t)-N^{\ast}(t)|=0.
$$
\end{Def}
Usually $N^{\ast}(t)$ is a positive equilibrium  or a positive
periodic solution of  equation (\ref{con}) if it exists. In general,
we will use the following definition.
\begin{Def}
We say that equation  (\ref{con}) is GAS, if for every two positive
solutions $N_1(t)$ and $N_2(t)$ of equation (\ref{con}) we have
$$
\lim_{t\rightarrow\infty}|N_1(t)-N_2(t)|=0.
$$
\end{Def}

Note that canonical logistic equation with variable parameters has
been well-studied, and the questions of the existence and stability
in this case are easily handled since the equation is solvable as a
Riccati equation in a closed form \cite{Bo1}--\cite{Bo3},
\cite{Cush1}, \cite{La} and \cite{Nis}.

\begin{Th}\label{teo2}
Assume that
\begin{equation} \label {eigen}
E_{1}(t)=E(t)+\frac{1}{K}\frac{dK}{dt}>0
\end{equation} and
$r(t),K(t)$ and $E(t)$ are positive functions. If $0<N(0)<K(0)$,
then every solution of equation \eqref{con} satisfies the following
inequality $0<N(t)<K(t)$ for all $t\geq 0.$
\end{Th}
{\bf {Proof.}} Let $v(t)=\ln\frac{N(t)}{K(t)},$ then equation (11)
has a form
\begin{equation} \label{mi}
\frac{dv}{dt}=r(t)\frac{1-e^{v\gamma }}{1+\beta e^{v\gamma
}}-E_{1}(t),
\end{equation}
where $v(0)<0.$ Let us prove that $v(t)<0$ for all $t\geq 0.$
Suppose there exists $t_{1}>0$ such that $v(t_{1})=0.$ Then
$$
\frac{dv}{dt} \mid_{t_{1}}=-E_{1}(t_{1})<0.
$$
Therefore in some interval ($t_{1}-a, t_{1})$ the function $v(t)$ is
decreasing and $v(t_{1})=0.$ But this is impossible, therefore
$v(t)<0$ follows by $N(t)<K(t).$ \\
To prove our next theorems for the periodic models we will use
\cite{Hart} (see also \cite{Bra2}).

\begin{Th}(Friedrichs Theorem)\label{Fr}. Suppose that $G(t,N)$ is a smooth
function with period $T$ in $t$ for every $N$. Suppose also that
there exist constants $a,b$ with $a<b$ such that $G(t,b)<0<G(t,a)$
for every $t$. Then there is a periodic solution $N_{0}(t)$ of the
differential equation $dN/dt=G(t,N)$ with period $T$ and $N(0)=c$
for some $c\in(a,b).$
\end{Th}

\begin{Th}\label{teo}
Consider equation \eqref{vary}. For all $\gamma >0,\beta \geq 0$ and
$t\geq 0$ we assume that $E(t),r(t),K(t)$ are all positive
$T-$periodic functions, and $r(t)-E_{1}(t)>0$
for all $t\geq 0.$ Then there exists a positive nonconstant periodic
solution $N_{0}(t)$ such that
$$
K(0)e^{b_{0}}<N_{0}(0)<K(0),
$$
where
$$
b_{0}<\frac{1}{\gamma }\min\lim_{t\in \lbrack 0,T]}\ln
\frac{r(t)-E_{1}(t) }{r(t)+\beta E_{1}(t)}<0.
$$\end{Th}

{\bf {Proof.}} Let $v=\ln\frac{N(t)}{K(t)},$ then equation
\eqref{vary} has a form \begin{equation}\label{op}
\frac{dv}{dt}=r(t)\frac{1-e^{v\gamma }}{1+\beta e^{v\gamma
}}-E_{1}(t)=\Phi (t,v) .\end{equation}
 Clearly, $\Phi
(t,0)=-E_{1}(t)$ $<0$ and

$$
\Phi (t,b_{0})=r(t)\frac{1-e^{b_{0}\gamma }}{1+\beta e^{b_{0}\gamma
}} -E_{1}(t)\ >0. $$ The last inequality is equivalent to the
inequality

$$
\text{\ }\frac{1}{\gamma }\ln \frac{r(t)-E_{1}(t)}{r(t)+\beta
E_{1}(t)}
>b_{0}.
$$
Therefore based on Theorem \eqref{Fr}, there exists periodic
solution $v_{0}(t)$ of equation \eqref{op}, such that
$b_{0}<v_{0}(t)<0.$ That yields the existence of the periodic
solution $N_{0}(t)$ of equation \eqref{vary} such that
$$
K(0)e^{b_{0}}<N_{0}(0)<K(0).\text{\ }
$$
\begin{Th}\label{Base} Suppose all conditions of Theorem \eqref{teo} hold.
Then there exists a unique periodic solution $N_{0}(t)>0$ of
equation \eqref{vary} such that $ 0<N_{0}(0)<K(0)$ \ and
$$
\lim_{t\rightarrow \infty }[N(t)-N_{0}(t)]=0 .$$
\end{Th}

{\bf {Proof.}} According to Theorem \eqref{teo2} the
solution  $v(t)$ of equation \eqref{mi} is negative for all $ t\geq 0.$
Firstly, let us prove that

\begin{equation}\label{ika}
\lim_{t\rightarrow \infty }\inf v(t)>-\infty .\end{equation} Assume
that

$$
\lim_{t\rightarrow \infty }v(t)=-\infty.
$$
Then
$$\lim_{t\rightarrow \infty }\inf
\frac{dv}{dt}=\lim_{t\rightarrow \infty }\inf [r(t)-E_{1}(t)]>0, $$

and $$\lim_{t\rightarrow \infty }v(t)=\lim_{t\rightarrow \infty
}\left[v(0)+\int_{0}^{t}\frac{dv}{ds}\right]=\infty .$$ We have a
contradiction. If $\lim_{t\rightarrow \infty }v(t)$ does not
exist, then there exists a sequence $\{t_{n}\}$ such that
$v(t_{n})\rightarrow -\infty $ and
\begin{equation}\label{mom}\frac{dv}{ dt}|_{t_{n}}=0.
\end{equation}
 Then equality \eqref{mom} yields
$$r(t_{n})-E_{1}(t_{n})\rightarrow 0.$$ That contradiction proves
statement \eqref{ika}. Suppose $N_{1}(t)$ and $N_{2}(t)$ are two
positive solutions of equation \eqref{mi} with $0<N_{i}(t)<K(0)$
$i=1,2$. Then $$v_{1}(t)=\ln \frac{N_{1}(t)}{K(t)}$$ and
$$v_{2}(t)=\ln \frac{N_{2}(t)}{K(t)}$$ are two solutions of equation
\eqref{mi}. Assume that
$v_{1}(t)<v_{2}(t).$ Let
$$v(t)=v_{2}(t)-v_{1}(t).$$ Then equation \eqref{mi} takes the form

\begin{equation} \label{oka}
\frac{dv}{dt}=r(t)\left[\frac{1-e^{v_{2}\gamma }}{1+\beta
e^{v_{2}\gamma }}-\frac{ 1-e^{v_{1}\gamma }}{1+\beta e^{v_{1}\gamma
}}\right].\end{equation} Application of the Mean-Value Theorem
transforms equation \eqref{oka} to \begin{equation}\label{mom}
\frac{dv}{dt}=-a(t)v(t)\end{equation} with
$$
a(t)=r(t)\left\{\frac{(1+\beta )\gamma e^{c(t)\gamma }}{[1+\beta
e^{c(t)\gamma }]^{2}}\right \}, $$ where
$$\int_{0}^{\infty}a(s)ds=\infty$$ and $$v_{1}(t)<c(t)<v_{2}(t).$$ Based on the last inequality
$$\liminf_{t\rightarrow \infty }v(t)>-\infty , $$
 $$\liminf_{t\rightarrow \infty }c(t)>-\infty .$$
Hence
$\liminf_{t\rightarrow \infty }a(t)>0,$ therefore for every
solution of \eqref{mom} we have $\lim_{t\rightarrow \infty
}v(t)=0,$ or
$ \lim_{t\rightarrow \infty }[v_{1}(t)-v_{2}(t)]=0.$
Similarly, if $v_{1}(t)> v_{2}(t)$, then $ \lim_{t\rightarrow \infty }[v_{2}(t)-v_{1}(t)]=0.$
Summing up, we conclude
$\lim_{t\rightarrow \infty }[N_{1}(t)-N_{2}(t)]=0. $

{\bf {Remark.}} If $N_{1}(t)$ and $N_{2}(t)$ are any two solutions
of equation \eqref{vary}, then statement of Theorem 3.4 is true without
the assumption that all functions $r(t)$, $K(t)$ and $E(t)$ are
$T$-periodic.\\
\newpage
\section{Numerical Experiments}
Fig 2 illustrates dynamics of the population  for a different set of parameters $\alpha$ and $\beta$.
\begin{figure}[ht]
    \begin{center}
    \includegraphics[width=2.5in, viewport=67 309 560 717,clip=true]{graph2il}
    \label{fig2}
    \caption{Size of the population for different models: N1: $\beta=0, \gamma=1$ logistic;
    N2: $\beta=0.2, \gamma=5$; N3: $\beta=4, \gamma=0.5$ }
    \end{center}
  \end{figure}

 Fig 3  supports a well-known feature of the population models:
increase in the amplitude of the carrying capacity $K(t)$, defined by equation \eqref{K}, yields
decrease in the average size of the population, whereas,  \begin{figure}
    \begin{center}
    \includegraphics[width=2.5in, viewport=57 368 544 700,clip=true]{graph3il}
    \label{fig3a}
  \caption{Change of environment. Population size for different values of $\alpha_{K}$
   (N1: $\alpha_{K}=0.1$, N2: $\alpha_{K}=0.5$ and N3: $\alpha_{K}=0.7$).}
    \end{center}
  \end{figure}

  Fig 4 proves that the qualitative behavior of the system is unchanged  \cite{Nis} by oscillations
in $r(t)$, defined by equation \eqref{rok}, alone.\\
{\bf {Remark.}} In all numerical experiments below we used parameters $E(t)$ and $K(t)$ that satisfy the condition \eqref{eigen}.
  \newpage
  The qualitative behavior of the system depends critically on the phase difference of the oscillations in $r(t)$ and $K(t)$.
  For example, if functions $r(t)$ and $K(t)$ have a $180^{\circ}$ phase shift, then
  the relationship between the average population size and environmental and demographic variations (see Fig 5) is
  very different from the corresponding Fig 3.
  \begin{figure}
    \begin{center}
    \includegraphics[width=2.5in, viewport=69 397 524 725,clip=true]{graph4il}
    \label{fig4a}
  \caption{Change of demography. Population size for different values of $\alpha_{r}$
   (N1: $\alpha_{r}=0.1$, N2: $\alpha_{r}=0.5$ and N3: $\alpha_{r}=0.9$).}
    \end{center}
  \end{figure}
  \begin{figure}
    \begin{center}
    \includegraphics[width=2.5in, viewport=57 368 544 700,clip=true]{graph5il}
    \label{fig3b}
  \caption{Change of phase shift. Population size for different values of $\alpha_{K}$
   (N1: $\alpha_{K}=0.1$, N2: $\alpha_{K}=0.5$ and N3: $\alpha_{K}=0.7$).}
    \end{center}
  \end{figure} Often debated questions are the choice of the harvesting strategies and timing of harvesting.
For simplicity, we consider static vs adaptive fishing strategies.\\

{\bf Static Fishing Strategy.}
Consider a fishery manager who has no access to previous fishery data. He starts fishing
(on Fig 6 curve $N1$) all year with an annual quota of 12 tons and realized  that in one-year period  a fishstock is decreased significantly. Then the manager  decides to shorten a fishing season, starts it in June and fishes for six months with a monthly quota of 2 tons (curve $N2$), and thereafter, he starts in September with the quota of 4 tons per month and fishes three months ( curve $N3$ ),
\begin{figure}[ht]
    \begin{center}
    \includegraphics[width=2.5in, viewport=76 303 584 719 clip=true]{graph6il}
    \label{fig6}
     \caption{Static Fishing Strategy}
    \end{center}
  \end{figure}
  \\
 \newpage
    {\bf Adaptive Fishing Strategy.}
Let a fishery manager have access (Fig 7 first graph) to the fishery data. He decides to fish in March  because at that time the population attains its maximum. Curve $N2$ represents a six-month fishing, starting in March
with 2 tons per month, whereas on curve $N3$, fishing takes place with the quota of 4 tons per month but in a three-month period. Clearly, the latter strategy is more efficient than the static strategy (curve $N1$ on Fig 7) which
represents a greater risk of depletion of the fishstock.
\begin{figure}[ht]
    \begin{center}
    \includegraphics[width=2.5in, viewport=74 270 530 750,clip=true]{graph7il}
    \label{fig6}
   \caption{ Adaptive Fishing Strategy}
    \end{center}
  \end{figure}

\section{Discussion}
In this paper we illustrate various effects of the varying
environmental carrying capacity and intrinsic rates on dynamics of
marine populations. Application of the food-limited model with $\gamma$-nonlinearity is consistent with
fishery data, and supports a well-known feature of the population models: average population size decreases
with an increase of amplitude of variation in the carrying capacity $K(t)$, but $\overline{N}\leq \overline{K}$ and the dynamics of the population must ride on this highly variable resource. On other hand,
even large oscillations in $r(t)$ alone leave the system's behavior practically unchanged. Note that similar results were obtained for the Fox fishery model \eqref{Fox} in \cite{Id}. However, the qualitative behavior of the system depends critically on the phase difference between the oscillations in $r(t)$ and $K(t)$. For example, if functions $r(t)$ and $K(t)$ have a $180^{\circ}$ phase shift then
the relationship between the average population size and environmental and demographic variations is
very different. \\
 With no access to fisheries data a static fishing strategy represents a greater risk of extinction of marine populations under severe harvesting, compared with adaptive strategies that save fishstock in a cost effective manner.\\

{\bf Acknowledgements}\\
We wish to express thanks to Dr. D. Barker (Fisheries and Aquaculture department at Vancouver Island University) whose comments significantly improved the text.