TITLE: If $a_n =\sum_{k=1}^m c_k a_{n-k} $, find the generating function $\sum_{n=0}^{\infty} a_n x^n $
QUESTION [2 upvotes]: (This is undoubtedly a duplicate,
but I just wanted to work this out
from scratch.
The algebra and rearrangements of summations
is complicated enough
so that,
as usual,
I am hoping that
someone might have
a simpler derivation.)
Given the linear recurrence
$a_n
=\sum_{k=1}^m c_k a_{n-k}
$
for $n \ge m$.
Find the generating function
$A(x)
=\sum_{n=0}^{\infty} a_n x^n
$
in terms of
$a_{0..m-1}$
and
$c_{1..m}$.
My answer is
$A(x)
=\dfrac{\sum_{r=0}^{m-1}  x^r(a_r-\sum_{k=1}^{r} c_{k} a_{r-k})}{1-\sum_{k=1}^m c_k x^k}
$.
My derivation.
Let
$A_j(x)
=\sum_{n=0}^{j} a_n x^n
$
for
$j = 0$ to $m-1$.
$\begin{array}\\
A(x)
&=\sum_{n=0}^{\infty} a_n x^n\\
&=\sum_{n=0}^{m-1} a_n x^n+\sum_{n=m}^{\infty} a_n x^n\\
&=A_{m-1}(x)+\sum_{n=m}^{\infty}  x^n\sum_{k=1}^m c_k a_{n-k}\\
&=A_{m-1}(x)+\sum_{k=1}^m c_k\sum_{n=m}^{\infty}  x^n  a_{n-k}\\
&=A_{m-1}(x)+\sum_{k=1}^m c_k x^k\sum_{n=m}^{\infty}  x^{n-k}  a_{n-k}\\
&=A_{m-1}(x)+\sum_{k=1}^m c_k x^k\sum_{n=m-k}^{\infty}  x^{n}  a_{n}\\
&=A_{m-1}(x)+\sum_{k=1}^m c_k x^k(\sum_{n=0}^{\infty}  x^{n}  a_{n}-\sum_{n=0}^{m-k-1}  x^{n}  a_{n})\\
&=A_{m-1}(x)+\sum_{k=1}^m c_k x^k(A(x)-A_{m-k-1}(x))\\
&=A_{m-1}(x)+\sum_{k=1}^m c_k x^kA(x)-\sum_{k=1}^m c_k x^kA_{m-k-1}(x)\\
\end{array}
$
so
$\begin{array}\\
A(x)(1-\sum_{k=1}^m c_k x^k)
&=A_{m-1}(x)-\sum_{k=1}^m c_k x^kA_{m-k-1}(x)\\
&=A_{m-1}(x)-\sum_{k=0}^{m-1} c_{m-k} x^{m-k}A_{k-1}(x)\\
&=A_{m-1}(x)-\sum_{k=0}^{m-1} c_{m-k} x^{m-k}\sum_{n=0}^{k-1} a_n x^n\\
&=A_{m-1}(x)-\sum_{n=0}^{m-2}\sum_{k=n+1}^{m-1} c_{m-k} x^{m-k} a_n x^n\\
&=A_{m-1}(x)-\sum_{n=0}^{m-2}\sum_{k=n+1}^{m-1} c_{m-k} a_n x^{n+m-k}\\
&=A_{m-1}(x)-\sum_{n=0}^{m-2}\sum_{r=n+m-m+1}^{n+m-n-1} c_{m-n-m+r} a_n x^{r}
\qquad r = n+m-k, k = n+m-r\\
&=A_{m-1}(x)-\sum_{n=0}^{m-2}\sum_{r=n+1}^{m-1} c_{r-n} a_n x^{r}\\
&=A_{m-1}(x)-\sum_{r=1}^{m-1}\sum_{n=0}^{r-1} c_{r-n} a_n x^{r}\\
&=\sum_{r=0}^{m-1} a_r x^r-\sum_{r=1}^{m-1}x^r\sum_{n=0}^{r-1} c_{r-n} a_n\\
&=a_0+\sum_{r=1}^{m-1}  x^r(a_r-\sum_{n=0}^{r-1} c_{r-n} a_n)\\
&=\sum_{r=0}^{m-1}  x^r(a_r-\sum_{n=0}^{r-1} c_{r-n} a_n)\\
&=\sum_{r=0}^{m-1}  x^r(a_r-\sum_{n=1}^{r} c_{n} a_{r-n})\\
&=\sum_{r=0}^{m-1}  x^r(a_r-\sum_{k=1}^{r} c_{k} a_{r-k})\\
\end{array}
$
Therefore
$A(x)
=\dfrac{\sum_{r=0}^{m-1}  x^r(a_r-\sum_{k=1}^{r} c_{k} a_{r-k})}{1-\sum_{k=1}^m c_k x^k}
$.

REPLY [0 votes]: We can turn the recurrence into a proper Cauchy product by defining $c_k = 0$ for $k=0$ and for $k>m$. Take $C(x) \equiv \sum_{k\geq 0} c_k x^k = \sum_{k=1}^mc_kx^k$ and consider
$$A(x) C(x) = \sum_{k\geq 0} d_k x^k\tag{1}$$
where $d_n = \sum_{k=0}^n c_k a_{n-k} = \sum_{k=1}^n c_k a_{n-k}$ which reduces to $d_n = a_n$ for $n\geq m$. Now it's just a matter of splitting the sum in $(1)$ into $k<m$ and $k\geq m$ and rearranging to get the result
$$A(x) = \frac{\sum_{k=0}^{m-1}[a_k - d_k]x^k}{1-C(x)} = \frac{\sum_{k=0}^{m-1}[a_k - \sum_{i=1}^k c_ia_{k-i}]x^k}{1-C(x)}$$


TITLE: Why measuring a liquid volume from the bottom of the meniscus isn't considered an underestimation?
QUESTION [0 upvotes]: I know from my OL chemistry and physics classes that the reading of a liquid volume should always be taken from the bottom of the meniscus, but isn't this an underestimation of the real volume. I mean there's some water sticking upward to the surface of the container and, say, if these water fell and made a flat surface they would add to the volume of water.

I realize that if we try to take the reading from the highest point the liquid reached this would be an overestimation so I was wandering if measuring from a point that's between the meniscus and the highest point the liquid reached would be more accurate.

REPLY [1 votes]: The markings on volumetric glassware are already calibrated to account for the volume above the meniscus*. The location of the mark on the glass is where the bottom of the meniscus should be in order to have the "true" volume be what's marked. So by measuring from the bottom of the meniscus, you're synchronizing your measurement procedure with the procedure of the people who originally calibrated the markings on the glass.
*For volumetric flasks where this kind of precision is very important, they are usually calibrated for a specific solution with a specific meniscus height. Otherwise, the general assumption, as far as I can tell, is that your solution will have a vaguely similar surface tension to water, with a vaguely similar meniscus height. For example, calibration of general graduated cylinders is often done with distilled water. Typically the error introduced by the difference in meniscus height is not the dominant source of measurement error, so this is often ignored. In cases where you're using something with a very different surface tension than water, and precision to that level actually matters, explicit calibration is usually recommended.

As an aside, while researching this answer, I came across the following entertaining scenario on the official USGS Water Science School site:

In your high-school chemistry final exam you mistakenly read a meniscus as 72 milliliters (ml) instead of the correct 66 ml (in this picture), and thus you get an 89 on the test instead of a 90. Your GPA falls from 4.00 to 3.99 and you don't get into that engineering college program you wanted. Consequently, you don't get that prestigious engineering job, where, 20 years later, you would have invented a new water-based chemical to allow rubber to grip better. Sadly, 10 years later, a mother and her adorable 4-year old daughter are leaving the ice cream store and the little girl, whose shoes don't have your un-invented coating, slips on a napkin and drops her ice cream cone. She cries at her loss ... because you misread the meniscus in the 12th grade.

The moral of this fictional tale is that it is important to read the measurement correctly, and yes, in the picture (top right) the true volume in the graduated cylinder is at the bottom of the water level—21.7 milliliters, not 21.9.


TITLE: Homeomorphism between $S^1$ and $K$ (square in the taxicab metric)
QUESTION [0 upvotes]: Let us consider $\mathbb{R}^2$ in the euclidean metric, and let $S^1 = \left\{(x,\,y) \in \mathbb{R}^2 \colon x^2 + y^2 = 1\right\}$ and $K \doteq \left\{(x,\,y) \in \mathbb{R}^2 \colon \lvert x \rvert + \lvert y \rvert = 1\right\}$ as subspaces. I have to prove that $S^1$ and $K$ are homeomorphic.My obvious guess was to define $f \colon S^1 \longrightarrow K$ as
$$f(x,\,y) \doteq \left(\frac{x^2}{x^2 + y^2},\,\frac{y^2}{x^2 + y^2}\right).$$ It is continuous, since both $u(x,\,y) \doteq \frac{x^2}{x^2\,+\,y^2}$ and $v(x,\,y) \doteq \frac{y^2}{x^2\,+\,y^2}$ are continuous in $S^1$, and $f(x,\,y) \in K$, for all $(x,\,y) \in S^1$, but I'm not sure how to prove it's bijective and it's inverse $g$ such that $g(u,\,v) = (x,\,y)$ is continuous as well.  My guess is that it fails to be injective, since $f(a,\,b) = f(x,\,y)$ would imply $ay = bx$, and we can't guarantee the uniqueness of the solution. Any ideas will be appreciated, thanks in advance.

REPLY [2 votes]: Use $f(x,y) = \frac{1}{|x|+|y|}(x,y)$, from $S^1$ to $K$ with inverse
$g(x,y) = \frac{1}{\sqrt{x^+y^2}}(x,y)$...


TITLE: How to do a group StructureDescription in Sage?
QUESTION [2 upvotes]: In Sage I executed the following commands:
 F.<x, y> = FreeGroup()
 G=F / [x^2*y^-1, x^3*y^2, x*y*x^-1*y^-1]

Now G is isomorphic to C7. In GAP I could check this with the StructureDescription command, how can I check this in Sage?

REPLY [2 votes]: This works:
sage: G.as_permutation_group().structure_description()
'C7'

The function should probably be exposed in G directly, however.
Nathann


TITLE: The smallest positive number $x_0$ exists such that $f(x_0)=0$
QUESTION [1 upvotes]: Assumption:
$$f\ is\ continuous\ function\ on\ \mathbb{R^1}$$
$$\exists x>0\ \ s.t\ \ f(x)=0$$
$$f(0)\neq 0$$
and we know
$$The\ set\ of\ zeros\ of\ a\ continuous\ function\ is\ closed$$
for $x\in(0,\infty]$, prove existence of smallest positive number $x_0$ such that $f(x_0)=0$

REPLY [2 votes]: Let $C=\{ t | t\in [0,x], f(t) = 0 \}$. The set $C$ is compact since $f$
is continuous.
Let $x_0 = \inf C$, then $x_0 \in C$ and since we are given that $0 \notin C$, we see that $x_0 > 0$.


TITLE: If $(a_n)$ is Cauchy it has a subsequence $(a_{n_i})$ such that $d(a_{n_{i+1}},a_{n_i})<2^{-i}$ for all $i$.
QUESTION [2 upvotes]: Let $(X,d)$ be a metric space and $(a_n)$ a Cauchy sequence in $X$. How to show that there exists $n_1<n_2<\cdots$ such that
$$d(a_{n_{i+1}},a_{n_i})<2^{-i},$$
for all $i\in\mathbb{N}$?
I tried to use induction, but it doesn't work. The base case is OK, but if we have
$$d(a_{n_2},a_{n_1})<2^{-1},\ldots,d(a_{n_k},a_{n_{k-1}})<2^{-(k-1)},$$
I don't see any reason for to exist $n_{k+1}>n_k$ such that $d(a_{n_{k+1}},a_{n_k})<2^{-k}$.

REPLY [2 votes]: What does it mean for a sequence to be Cauchy?  It means for each $\epsilon > 0$, there is some $N \in \mathbb{N}$ such that $n, m \geq N$ implies $d(x_{n}, x_{m}) < \epsilon$.
Since this is true for every $\epsilon > 0$, it is true if $\epsilon = \frac{1}{2}$, for example.  That is, there is some $N_{1} \in \mathbb{N}$ such that $n, m \geq N_{1}$ implies $d(x_{n}, x_{m}) < \frac{1}{2}$.  Let $n_{1} = N_{1}$.
Similarly, we know if $\epsilon= \frac{1}{2^{2}}$, there is some $N_{2} \in \mathbb{N}$ such that $n, m \geq N_{2}$ implies $d(x_{n}, x_{m}) < \frac{1}{2^{2}}$.  But if this is true for all $n, m \geq N_{2}$, then it is true for all $n, m \geq N_{2} + 1$.  And for all $n, m \geq N_{2} + 2$.  And for all $n, m \geq p > N_{2}$ where $p$ is any integer greater than $N_{2}$.  In particular, this is true for $\max \{n_{1}, N_{2} \} + 1$.  So we can let $n_{2} = \max \{n_{1}, N_{2} \} + 1$, and clearly, $n_{2} > n_{1}$, and if $n, m \geq n_{2}$, $d(x_{n}, x_{m}) < \frac{1}{2^{2}}$.
Similarly, we know if $\epsilon = \frac{1}{2^{3}}$, there is some $N_{3} \in \mathbb{N}$ such that $n, m \geq N_{3}$ implies $d(x_{n}, x_{m}) < \frac{1}{2^{3}}$.  If we let $n_{3} = \max \{ n_{2}, N_{3} \} + 1$, then clearly $n_{3} > n_{2}$ and if $n, m \geq n_{3}$, then $d(x_{n}, x_{m}) < \frac{1}{2^{3}}$.
Note that because $n_{1} < n_{2} < n_{3} < \dots$ by our construction, and since we picked $n_{i}$ such that $d(x_{n}, x_{m}) < \frac{1}{2^{i}}$ if $n, m \geq n_{i}$, AND $n_{i + 1}, n_{i} \geq n_{i}$, it follows that $d(x_{n_{i + 1}}, x_{n_{i}}) < \frac{1}{2^{i}}$.


text:
\begin{document}

\allowdisplaybreaks

\renewcommand{\PaperNumber}{051}

\FirstPageHeading

\ShortArticleName{Linearization of Second-Order ODEs by Generalized Sundman Transformations}

\ArticleName{Linearization of Second-Order Ordinary Dif\/ferential\\
Equations by Generalized Sundman Transformations}

\Author{Warisa NAKPIM and Sergey V. MELESHKO}

\AuthorNameForHeading{W.~Nakpim and S.V.~Meleshko}
\Address{School of Mathematics, Institute of Science, Suranaree University of
Technology,
\\ Nakhon Ratchasima, 30000, Thailand}
\Email{\href{mailto:warisa_2006@hotmail.com}{warisa{\_}2006@hotmail.com}, \href{mailto:sergey@math.sut.ac.th}{sergey@math.sut.ac.th}}

\ArticleDates{Received January 18, 2010, in f\/inal form June 03, 2010;  Published online June 15, 2010}

\Abstract{The linearization problem of a second-order ordinary dif\/ferential
equation  by the generalized Sundman transformation was considered
earlier by Duarte, Moreira and Santos  using the Laguerre form.
The results obtained in the present paper demonstrate that their
solution  of the linearization problem for a second-order ordinary
dif\/ferential equation via the generalized Sundman transformation
is not complete. We also give examples which show that the
Laguerre form is not suf\/f\/icient for the linearization problem via
the generalized Sundman transformation.}

\Keywords{linearization problem; generalized Sundman transformations; nonlinear second-order ordinary dif\/ferential equations}

\Classification{34A05; 34A25}

\section{Introduction}
The basic problem in the modeling of physical and other phenomena is to f\/ind
 solutions of dif\/ferential equations.
Many methods of solving dif\/ferential equations use a change of
variables that transforms a given dif\/ferential equation into another
equation with known properties. Since the class of linear equations
is considered to be the simplest class of equations, there arises the
problem of transforming a given dif\/ferential equation into a linear
equation. This problem is called a linearization
problem\footnote{The linearization problem has been studied in many
publications. A short review can be found in \cite{Meleshko,Ibragimov}.}.

The linearization problem  of a second-order ordinary dif\/ferential
equation via point transformations was solved by Sophus Lie~\cite{Lie}. He also noted that all second-order ordinary dif\/ferential equations
can be mapped into each other by means of contact transformations.
Hence, the solution of the linearization problem via contact transformations is trivial.

Comparing with the set of contact transformations the set of generalized Sundman transformations is
weaker: not any second-order ordinary dif\/ferential equation can be transformed to a~linear equation.
Hence, it is interesting to study an application of the set of genera\-li\-zed Sundman transformations to the
linearization problem of second-order ordinary dif\/ferential equations.

 The linearization problem via a generalized
Sundman transformation for second-order ordinary dif\/ferential
equations was investigated in \cite{DuarteMoreiraSantos}. The
authors of \cite{DuarteMoreiraSantos} obtained that any
second-order linearizable ordinary dif\/ferential equation which can
be mapped into the equation $u''=0$ via a generalized Sundman
transformation has to be of the form
\begin{gather}
\label{equa:MMM}
 y^{\prime \prime }+\lambda _2(x,y)y'^2+\lambda
_1(x,y)y^{\prime }+\lambda _0(x,y)=0.
\end{gather}
Using the functions
\[
\lambda_3=\lambda_{1y}-2\lambda_{2x},\qquad
\lambda_4=2\lambda_{0yy}-2\lambda_{1xy}+2\lambda_0\lambda_{2y}-\lambda_{1y}\lambda_1+2\lambda_{0y}\lambda_2+2\lambda_{2xx},
\]
they showed that equation \eqref{equa:MMM}
can be mapped into
the equation $u''=0$ via a generalized Sundman transformation if the
coef\/f\/icients $\lambda_i(x,y)$ $(i=0,1,2)$ satisfy the conditions:

(a) if $\lambda_3=0$, then $\lambda_4=0$;

(b) if $\lambda_3\neq0$, then $\lambda_4\neq0$ and the following
 equations have to be satisf\/ied
\begin{gather*}
\lambda_4^2+2\lambda_{3x}\lambda_4-2\lambda_3^2\lambda_{1x}+4\lambda_3^2\lambda_{0y}
+4\lambda_3^2\lambda_0\lambda_2-2\lambda_3\lambda_{4x}-\lambda_3^2\lambda_1^2=0,
\\
\lambda_{3y}\lambda_4+\lambda_3^2\lambda_{1y}-2\lambda_3^2\lambda_{2x}-\lambda_3\lambda_{4y}=0.
\end{gather*}

The generalized Sundman transformation was also applied
\cite{EulerWolfLeachEuler,NakpimMeleshko} for obtaining necessary and
suf\/f\/icient conditions for a third-order ordinary dif\/ferential
equation to be equivalent to a linear equation in the Laguerre form.
Some applications of the generalized Sundman transformation to
ordinary dif\/ferential equations were considered in~\cite{Berkovich2001} and earlier papers, which are summarized
in the book~\cite{Berkovich2002}.

According to the Laguerre theorem in any linear ordinary dif\/ferential equation the two terms of order
below next to highest can be simultaneously removed by a point transformation. For example, the Laguerre
form of a second-order ordinary dif\/ferential equation is the
linear equation $u''=0$.
For obtaining
this form, several point transformations are applied
consecutively. Since the composition of point transformations is a
point transformation, the f\/inal transformation is again a point
transformation. This is not the case for generalized Sundman
transformations: the composition of a point transformation and a
generalized Sundman transformation is not necessarily a
generalized Sundman transformation. Hence, for the linearization
problem via generalized Sundman transformations it is not
suf\/f\/icient to use the Laguerre form.

In this paper, we demonstrate that the solution of the
linearization problem via the gene\-ra\-li\-zed Sundman transformation
of second-order ordinary dif\/ferential equations given in
\cite{DuarteMoreiraSantos} only gives particular criteria
for linearizable equations.
Complete analysis of the compatibility of arising equations is given for the case $F_x=0$.

\section{Generalized Sundman transformations}

A generalized Sundman transformation is a non-point
transformation def\/ined by the formulae
\begin{gather}
\label{equa:01}
u(t) = F(x,y),\qquad dt = G(x,y)dx,\qquad F_yG\ne 0.
\end{gather}
Let us explain how the generalized Sundman transformation maps one
function into another.

Assume that $y_0(x)$ is a given function. Integrating
the second equation of (\ref{equa:01}),
one obtains $t=Q(x)$, where
\[
Q(x)= t_0+\int_{x_0}^{x}  G(s,y_0(s))\,ds
\]
with some initial conditions $t_0$ and $x_0$.
Using the inverse function
theorem, one f\/inds $x=Q^{-1}(t)$. Substituting $x$ into the
function $F(x,y_0(x)) $, one gets the transformed function
\[
u_0(t) = F\left(Q^{ - 1} (t),y_0 (Q^{ - 1} (t))\right).
\]

Conversely, let $u_0(t)$ be a given function of $t$. Using the
inverse function theorem, one solves the equation
\[
u_0 (t) = F(x,y)
\]
with respect to $y$: $y=\phi \left( {x,t}\right)$. Solving the
ordinary dif\/ferential equation
\[
\frac{dt} {dx} = G(x,\phi (x,t)),
\]
one f\/inds $t=H(x)$. The function $H(x)$ can be written as an action of a functional $H={\cal L}(u_0)$.
Substituting $t=H(x)$ into the function $\phi(x,t)$, the transformed function $y_0(x)=\phi
(x,H(x))$ is obtained.

Notice that for the case
$G_y=0$ the action of the functional ${\cal L}$ does not depend on the function $u_0(t)$.
In this case the generalized Sundman transformation becomes a point transformation.
Conversely, since for a point transformation the value $dt$ in the generalized Sundman transformation
is the total dif\/ferential of $t$, then the compatibility condition for $dt$ to be a~total dif\/ferential
leads to the equation $G_y=0$. Hence, the generalized Sundman transformation is a~point transformation if and only if $G_y=0$.

Formulae (\ref{equa:01}) also allow  us to obtain the derivatives
of $u_0(t)$ through the derivatives of the function $y_0(x)$, and
vice versa.

Hence, using transformation (\ref{equa:01}), one can relate the
solutions of two dif\/ferential equations
$Q(x,y,y',\ldots,y^{(n)})=0$ and $P(t,u,u',\ldots,u^{(n)})=0$.
Therefore the knowledge of the general solution of one of them
gives the general solution of the other equation, up to solving
one ordinary dif\/ferential equation of f\/irst-order and f\/inding two
inverse functions.

\section{Necessary conditions}

We start with obtaining necessary conditions for the linearization
problem.

First, one f\/inds the general form of a second-order ordinary
dif\/ferential equation
\[
y''  = H\big( x,y,y' \big),
\]
which can be mapped via a generalized Sundman transformation into
the linear equation
\begin{gather}
\label{equa:KL}
u''+\beta u'+\alpha u=\gamma,
\end{gather}
where $\alpha(t)$, $\beta(t)$ and $\gamma(t)$ are some functions.
Notice that the Laguerre form of a linear second-order ordinary
dif\/ferential equation corresponds to $\alpha=0$, $\beta=0$ and
$\gamma=0$.

The function $u$ and its derivatives $u'$ and $u''$ are def\/ined by the f\/irst formula (\ref{equa:01})
and its derivatives with respect to $x$:
\begin{gather}
u^{\prime }G=F_x+F_yy^{\prime },\nonumber
\\
u^{\prime \prime }G^2+u^{\prime }(G_x+G_yy^{\prime })=F_yy^{\prime
\prime }+2F_{xy}y^{\prime }+F_{yy}y^{\prime 2}+F_{xx}.\label{equa:02}
\end{gather}
The independent variable $t$ is def\/ined by the functional  ${\cal L}(u)$.
As noted above, if $G_y\neq 0$, then the action of the functional ${\cal L}$ depends on the function $u$.
Hence, if one of the coef\/f\/icients (\ref{equa:KL})
is not constant and $G_y\neq 0$, then the substitution of $t$ into equation
(\ref{equa:KL})
gives a functional equation. Since the case $G_y=0$ reduces the generalized Sundman
transformation to a point
transformation\footnote{Later it will be shown that in our study for $F_x=0$
this case is automatically excluded from
consideration because in the process of studying
the compatibility this case leads to the conditions $\lambda_3=0$ and $\lambda_4=0$,
which were considered in~\cite{DuarteMoreiraSantos}.},
the irreducible generalized Sundman transformation maps equation~(\ref{equa:KL})
into a~dif\/ferential equation only for constant coef\/f\/icients
$\alpha$, $\beta$ and $\gamma$.
Thus, f\/inding the derivatives~$u^{\prime }$,~$u^{\prime \prime }$ from
(\ref{equa:02}), and substituting them into (\ref{equa:KL}) with constant coef\/f\/icients, one
has the following equation
\begin{gather}
\label{equa:03}
y^{\prime \prime  }+\lambda _2(x,y)y'^2+\lambda _1(x,y)y^{\prime
}+\lambda _0(x,y)=0,
\end{gather}
where the coef\/f\/icients  $\lambda _i(x,y)$ $(i=0,1,2)$ are related to the functions
$F$ and $G$:
\begin{gather}
 \label{equa:04} \lambda _2 = (F_{yy}G - F_yG_y)/K, \\
   \label{equa:05}\lambda _1= (2F_{xy}G- F_xG_y -F_yG_x + F_y\beta G^2)/K, \\
  \label{equa:06} \lambda_0=(F_{xx}G - F_xG_x + F_x\beta G^2 + \alpha FG^3- G^3\gamma)/K,
\end{gather}
where $K=GF_y\neq0$.

Equation (\ref{equa:03}) presents the necessary form of a
second-order ordinary dif\/ferential equation which can be mapped
into a linear equation (\ref{equa:KL}) via a generalized Sundman
transformation.

\section{Suf\/f\/icient conditions}

 For obtaining suf\/f\/icient conditions, one has to solve the
compatibility problem (\ref{equa:04})--(\ref{equa:06}), con\-si\-de\-ring
(\ref{equa:04})--(\ref{equa:06}) as an overdetermined system of
partial dif\/ferential equations  for the functions $F$ and $G$ with
the given coef\/f\/icients $\lambda _i(x,y)$ $(i=0,1,2)$. Notice that
the compatibility conditions (\ref{equa:04})--(\ref{equa:06}) for
the particular case $\alpha=0$, $\beta=0$ and $\gamma=0$ were
obtained in~\cite{DuarteMoreiraSantos}. This case corresponds to the Laguerre form of a
linear second-order ordinary dif\/ferential equation. It is shown
here that for the linearization problem via generalized Sundman
transformations it is not suf\/f\/icient to use the Laguerre form.

The compatibility analysis depends on the value of $F_x$. A
complete study of all cases is cumbersome. Here a
complete solution is given for the case where $F_x=0$.

Solving  equations (\ref{equa:04})--(\ref{equa:06}) with respect to
$F_{yy}$, $\beta$ and $\gamma$, one f\/inds
\begin{gather}
\label{equa:07}
F_{yy}= (G_yF_y +F_yG\lambda_2)/G,
\\
\label{equa:08}
\beta= (  G_x  + G\lambda_1)/ G^2,
\\
\label{equa:09}
\gamma=(- F_y\lambda_0 + \alpha FG^2)/G^2 .
\end{gather}
Since $F_x=0$, then dif\/ferentiating $F_{yy}$ with respect to $x$, one obtains
\begin{gather}
\label{equa:100}
GG_{xy}-G_xG_y + \lambda_{2x}G^2=0.
\end{gather}
Dif\/ferentiating (\ref{equa:08}) and (\ref{equa:09}) with respect
to $x$ and $y$, one obtains the following equations
\begin{gather}
\label{equa:097}
 G_{xx}= (2G_x^2 + G_xG\lambda_1 -
\lambda_{1x}G^2)/G,
\\
\label{equa:096}
 G_{xy}= G\lambda_3- G_y\lambda_1 ,
\\
\label{equa:00} 2G_x\lambda_0 - \lambda_{0x}G=0,
\\
\label{equa:MN} \alpha= ( - G_y\lambda_0 + G(\lambda_{0y} +
\lambda_0\lambda_2))/G^3,
\end{gather}
where
\[
\lambda_3=\lambda_{1y} - 2\lambda_{2x}.
\]
Substituting (\ref{equa:096}) into (\ref{equa:100}), this becomes
\begin{gather}
\label{equa:098}   G_xG_y + G_yG\lambda_1 - G^2(\lambda_{2x} +
\lambda_3)=0.
\end{gather}
Comparing the mixed derivatives $(G_{xy})_x=(G_{xx})_y$, one
obtains the equation
\begin{gather}
\label{equa:000}
G_x\lambda_3 - G(\lambda_{2xx} + \lambda_{2x}\lambda_1 +
\lambda_{3x})=0.
\end{gather}
Dif\/ferentiating $\alpha$ with respect to $x$ and $y$, one has
\begin{gather}
\label{equa:001}
 2G_x(\lambda_{0y} + \lambda_0\lambda_2) +G_y( \lambda_{0x} +
2\lambda_0\lambda_1) - G(\lambda_{0xy} +\lambda_{0x}\lambda_2

+4\lambda_{2x}\lambda_0 + 2\lambda_0\lambda_3)=0 ,
\\
\label{equa:002}
2GG_{yy}\lambda_0-6G_y^2\lambda_0 + 2G_yG( 3\lambda_{0y} +
2\lambda_0\lambda_2) - G^2(\lambda_4

+ 2\lambda_5-\lambda_1\lambda_3)=0,
\end{gather}
where
\begin{gather*}
\lambda_4=2\lambda_{0yy}-2\lambda_{1xy}+2\lambda_0\lambda_{2y}-\lambda_{1y}\lambda_1+2\lambda_{0y}\lambda_2+2\lambda_{2xx},
\\
\lambda_5=\lambda_{2xx} + \lambda_{2x}\lambda_1 + \lambda_{3x} +
\lambda_1\lambda_3.
\end{gather*}

Further analysis of the compatibility depends on $\lambda_3$.

\subsection[Case lambda_3 not= 0]{Case $\boldsymbol{\lambda_3 \neq 0}$}

From equations (\ref{equa:000}), one f\/inds
\begin{gather}
\label{equa:AAA}
G_x= G(\lambda_{2xx} + \lambda_{2x}\lambda_1 +
\lambda_{3x})/\lambda_3.
\end{gather}
Substituting $G_x$ into equations (\ref{equa:00}),
(\ref{equa:098}),  (\ref{equa:097}) and (\ref{equa:096}), one
obtains the equations
\begin{gather}
\label{equa:1b}
 \lambda_{0x}= 2\lambda_0( - \lambda_1\lambda_3 +
\lambda_5)/\lambda_3,
\\
\label{equa:1a}
\lambda_{2xxy}= - \lambda_{2xy}\lambda_1- \lambda_{3xy} -
2\lambda_{2x}^2 - 2\lambda_{2x}\lambda_3
 - \lambda_{3y}\lambda_1 + (\lambda_{3y}\lambda_5)\lambda_3^{-1},
\\
\label{equa:1d}
\lambda_{2xxx}=- \lambda_{3xx}-
\lambda_{1x}\lambda_{2x} - \lambda_{1x}\lambda_3 +
\lambda_{2x}\lambda_1^2 + \lambda_1^2\lambda_3

-2\lambda_1\lambda_5+\lambda_3^{-1}\lambda_5(\lambda_{3x}+\lambda_5) ,
\\
\label{equa:1c}
 G_y\lambda_5-
G\lambda_3(\lambda_{2x}+\lambda_3)=0.
\end{gather}

\subsubsection[Case lambda_5 not= 0]{Case $\boldsymbol{\lambda_5 \neq 0}$}

Equation (\ref{equa:1c}) gives
\begin{gather}
\label{equa:ZZZ} G_y=
G\lambda_3(\lambda_{2x}+\lambda_3)/\lambda_5.
\end{gather}
Substituting $G_y$ into equations (\ref{equa:096}),
(\ref{equa:001}) and (\ref{equa:002}) and comparing the mixed
derivatives $(G_{x})_y=(G_{y})_{x}$, one gets
\begin{gather}
\lambda_3\lambda_5(6\lambda_{0y}\lambda_{2x}+2\lambda_{2xy}\lambda_0
+4\lambda_{2x}\lambda_0\lambda_2+2\lambda_{3y} \lambda_0+4\lambda_0\lambda_2\lambda_3
+\lambda_1\lambda_5)\nonumber\\
\qquad{}
-\lambda_3^2( 6\lambda_{2x}^2\lambda_0+
12\lambda_{2x}\lambda_0\lambda_3- 6\lambda_{0y}\lambda_{5}+
6\lambda_0\lambda_3^2)- \lambda_4\lambda_5^2 - 2\lambda_5^3=0.\label{equa:1g}
\end{gather}

\subsubsection[Case lambda_5=0]{Case $\boldsymbol{\lambda_5 = 0}$}

Equations (\ref{equa:1b}), (\ref{equa:1c}), (\ref{equa:1a}),
(\ref{equa:1d}), (\ref{equa:001}) and (\ref{equa:002}) become
\begin{gather}
\label{equa:1h} \lambda_{0x}=  - 2\lambda_0\lambda_1,
\\
\label{equa:1hhh} \lambda_{2x}=-\lambda_3,
\\
\label{equa:1i}
 2GG_{yy}\lambda_0-6G_y^2\lambda_0 + 2G_yG( 3\lambda_{0y} +
2\lambda_0\lambda_2) - G^2(\lambda_4 - \lambda_1\lambda_3)=0.
\end{gather}

If $\lambda_0 \neq0$, then equation (\ref{equa:1i}) def\/ines
\begin{gather}
\label{equa:ABC} G_{yy}=(6G_y^2\lambda_0 - 2G_yG(  3\lambda_{0y}
+2\lambda_0\lambda_2 ) + G^2( \lambda_4 - \lambda_1\lambda_3
))/(2G\lambda_0).
\end{gather}
In this case, $(G_{yy})_x=(G_{xy})_y$ and
$(G_{x})_{yy}=(G_{yy})_x$ are satisf\/ied. Hence, there are no other
compatibility conditions. Thus, if $\lambda_3\neq 0$ , $\lambda _5=0$ and
$\lambda _0\neq0$, then conditions (\ref{equa:1h}) and
(\ref{equa:1hhh}) are suf\/f\/icient for equation (\ref{equa:03}) to
be linearizable by a generalized Sundman transformation.

If $\lambda_0 =0$, there is no other conditions.

\begin{remark}
If  $\lambda_5 = 0$, equations (\ref{equa:1b}),
 (\ref{equa:1a}),
(\ref{equa:1d}), (\ref{equa:1c})
and (\ref{equa:1g}) become conditions (\ref{equa:1h}) and
(\ref{equa:1hhh}) respectively.
\end{remark}

Thus, suf\/f\/icient conditions for equation (\ref{equa:03}) in the case $\lambda_3\neq 0$
to be linearizable by  generalized Sundman transformations are (\ref{equa:1b}), (\ref{equa:1a}), (\ref{equa:1d})
and (\ref{equa:1g}).

\subsection[Case lambda_3=0]{Case $\boldsymbol{\lambda_3 = 0}$}

Notice that the particular case $\lambda_3=0$ and $\lambda_4=0$
was studied in \cite{DuarteMoreiraSantos}. Here the case $\lambda_3=0$ and
$\lambda_4\neq0$ is considered.

Equation (\ref{equa:002}) for $\lambda_3=0$ becomes
\begin{gather}
\label{equa:2d}
2GG_{yy}\lambda_0-6G_y^2\lambda_0 + 2G_yG( 3\lambda_{0y} +
2\lambda_0\lambda_2) - G^2\lambda_4 =0.
\end{gather}

The assumption $\lambda_0=0$ leads to the contradiction that
$\lambda_4=0$. Hence, one has to assume that $\lambda_0\neq0$.

Equations (\ref{equa:000}), (\ref{equa:00}) and (\ref{equa:001})
become
\begin{gather}
\label{equa:2a}
 \lambda_{2xx}= -\lambda_{2x}\lambda_1,
\\
\label{equa:2b} G_x= (G\lambda_{0x})/(2\lambda_0),
\\
\label{equa:2c} G_y\lambda_0\lambda_6 - G(\lambda_{6y}\lambda_0-
\lambda_{0y}\lambda_6 )=0,
\end{gather}
where
\[
\lambda_6=\lambda_{0x} + 2\lambda_0\lambda_1.
\]
Substituting $G_x$ into equations (\ref{equa:096})  and
(\ref{equa:097}), one gets
\begin{gather}
\label{equa:2e}
 \lambda_{6y}= (\lambda_{0y}\lambda_6 +
2\lambda_{2x}\lambda_0^2)/\lambda_0,
\\
\label{equa:2f}\lambda_{6x}=(3\lambda_6( \lambda_6-
2\lambda_0\lambda_1 ))/(2\lambda_0).
\end{gather}

\subsubsection[Case lambda_6 not= 0]{Case $\boldsymbol{\lambda_6 \neq0}$}

From equations (\ref{equa:2c}), one f\/inds
\[
G_y=G( - \lambda_{0y}\lambda_6 +
\lambda_{6y}\lambda_0)/(\lambda_0\lambda_6).
\]
Substituting $G_y$ into equations  (\ref{equa:096}) and
(\ref{equa:2d}),  and comparing the mixed derivatives
$(G_{x})_y=(G_{y})_{x}$, one obtains
\begin{gather}
\label{equa:2g}
 \lambda_{4x}= ( - 24\lambda_{2x}^2\lambda_0^3 -
4\lambda_0\lambda_1\lambda_4\lambda_6 +
\lambda_4\lambda_6^2)/(2\lambda_0\lambda_6).
\end{gather}

\subsubsection[Case lambda_6 =0]{Case $\boldsymbol{\lambda_6 =0}$}

In this case equation (\ref{equa:2c}) is satisf\/ied. One needs to
check the only condition $(G_{yy})_x=(G_{x})_{yy}$, which is
\begin{gather}
\label{equa:2hh} \lambda_{4x}=  - 2\lambda_1\lambda_4.
\end{gather}
Equation (\ref{equa:2e}) becomes
\begin{gather}
\label{equa:2h} \lambda_{2x}=0.
\end{gather}

\begin{remark}
If  $\lambda_6 = 0$, equation (\ref{equa:2e}) becomes a condition (\ref{equa:2h}).
\end{remark}

All obtained results can be summarized in the theorem.

\begin{theorem}\label{theorem}
Sufficient conditions for equation \eqref{equa:03}
to be linearizable via a generalized Sundman transformation
with $F_x=0$ are as follows.
\begin{enumerate}\itemsep=0pt
\item[$(a)$] If $\lambda_3\neq 0$, then the
conditions are \eqref{equa:1b}, \eqref{equa:1a}, \eqref{equa:1d}
and \eqref{equa:1g}.

\item[$(b)$] If $\lambda_3= 0$, $\lambda_6\neq0$, then the conditions are
\eqref{equa:2a}, \eqref{equa:2e}, \eqref{equa:2f} and
\eqref{equa:2g}.

\item[$(c)$]  If $\lambda_3= 0$, $\lambda_6=0$, then the conditions are
\eqref{equa:2a}, \eqref{equa:2e}, \eqref{equa:2f} and
\eqref{equa:2hh}.
\end{enumerate}
\end{theorem}

\begin{remark}
These conditions  extend the criteria obtained in
\cite{DuarteMoreiraSantos} to the case $\alpha, \beta,\gamma\neq 0$ in (3), for restricted ($F_x=0$) generalized
Sundman transformations.
\end{remark}

\begin{remark}
Notice that a discussion of the case
$\lambda_i=\lambda_i(y)$ $(i=0,1,2)$ and $G_x=0$, is also given in~\cite{Berkovich2001}.
\end{remark}

\begin{remark}
Recall S.~Lie's results \cite{Lie} on linearization of a second-order
ordinary dif\/ferential equation via a change of the independent and
dependent variables (point transformations).
The necessary form of a linearizable equation $y''=f(x,y,y')$ has to be the following
 form
\begin{gather}
\label{eq:Oct1604.4}
y^{\prime \prime }+a(x,y)y^{\prime }{}^3+b(x,y)y^{\prime
}{}^2+c(x,y)y^{\prime }+d(x,y)=0.
\end{gather}
Equation (\ref{eq:Oct1604.4}) is linearizable if and only if its coef\/f\/icients
satisfy the conditions
\begin{gather}
3a_{xx}-2b_{xy}+c_{yy}-3a_xc+3a_yd+2b_xb-3c_xa-c_yb+6d_ya=0, \nonumber\\
b_{xx}-2c_{xy}+3d_{yy}-6a_xd+b_xc+3b_yd-2c_yc-3d_xa+3d_yb=0.\label{eq:Oct1604.11}
\end{gather}

Despite that the form (\ref{equa:03}) is a particular case of (\ref{eq:Oct1604.4}),
suf\/f\/icient conditions of linearization via point transformations (\ref{eq:Oct1604.11}) and the generalized Sundman transformation dif\/fer.
Hence, the second class of equations is not contained in the f\/irst class due
to dif\/ferences in conditions on arbitrary elements of the classes. At
the same time, these classes have a nonempty intersection.
\end{remark}

\section{Examples}

\begin{example}
Consider the nonlinear ordinary dif\/ferential
equation
 \begin{gather}
 \label{ex:01}
y''+(1/y)y'^2 + yy'+1/2=0.
 \end{gather}
Since this equation does not satisfy Lie criteria for linearization \cite{Lie}
it is not linearizable by point transformations.
Equation (\ref{ex:01}) is of the form (\ref{equa:03}) with
coef\/f\/icients
\begin{gather}
 \lambda_2 = 1/y, \qquad  \lambda_1 = y,  \qquad \lambda_0 =1/2.
 \label{ex:BB}
\end{gather}
One can check that the coef\/f\/icients (\ref{ex:BB}) obey the
conditions (\ref{equa:1b}), (\ref{equa:1a}), (\ref{equa:1d}) and
(\ref{equa:1g}). Thus, equation (\ref{ex:01}) is linearizable via
generalized Sundman transformation.

For f\/inding the functions $F$ and $G$ one has to solve equations
(\ref{equa:07}), (\ref{equa:AAA})  and (\ref{equa:ZZZ}), which
become
\begin{gather*}
   F_{x} =0, \qquad F_{yy}=(2F_y)/y,  \qquad G_x =0, \qquad G_y=
G/y.
\end{gather*}

We take the simplest solution,  $F=y^3$ and $G=y$, which satisf\/ies
(\ref{equa:07}), (\ref{equa:AAA})  and (\ref{equa:ZZZ}). One
obtains the transformation
\begin{gather}
\label{ex:BBN} u = y^3,\qquad dt = ydx.
\end{gather}
Equations (\ref{equa:08}), (\ref{equa:09}) and (\ref{equa:MN})
give
\begin{gather*}
 \beta =1, \qquad \gamma=-3/2, \qquad \alpha=0.
\end{gather*}
Hence equation (\ref{ex:01}) is mapped by the transformation
(\ref{ex:BBN}) into the linear equation
 \begin{gather}
 \label{ex:16}
u''+u'+3/2=0.
 \end{gather}
The general solution of equation (\ref{ex:16}) is
\[
u=c_1+c_2e^{-t}-3t/2,
\]
where $c_1$, $c_2$  are arbitrary constants. Applying the generalized Sundman
transformation~(\ref{ex:BBN}) to equation~(\ref{ex:01}) one
obtains that the general solution of equation (\ref{ex:01}) is
\[
y(x)=(c_1+c_2e^{-\phi(x)} -3\phi(x)/2)^{1/3},
\]
where the function $t=\phi(x)$ is a solution of the equation
\[
 \frac{dt}{dx} =\big(c_1+c_2e^{-t} -3t/2\big)^{1/3}.
\]
For example, if $c_1=c_2=0$, then one obtains the solution of
equation (\ref{ex:01}):
\[
y=(-x)^{1/2}.
\]
\end{example}

\begin{example}
Consider the nonlinear ordinary dif\/ferential
equation
  \begin{gather}
 \label{ex:02}
y''+xy'^2 + yy'+1/{e^{2xy}}=0.
 \end{gather}
Equation (\ref{ex:02}) is of the form~(\ref{equa:03}) with the
coef\/f\/icients
\begin{gather}
 \lambda_2 = x, \qquad  \lambda_1 = y, \qquad \lambda_0 = 1/{e^{2xy}}.
 \label{ex:CC}
\end{gather}
One can check that the coef\/f\/icients (\ref{ex:CC}) do not satisfy
the conditions of linearizability by point transformations, but
they obey the conditions (\ref{equa:1h}) and
(\ref{equa:1hhh}). Thus, equation~(\ref{ex:02}) is linearizable
via a generalized Sundman transformation.

For f\/inding the functions $F$ and $G$ one has to solve equations
(\ref{equa:07}), (\ref{equa:AAA})  and (\ref{equa:ABC}), which
become
\begin{gather*}
    F_{x} =0, \qquad F_{yy}=(G_yF_y + F_yGx)/G, \\
     G_x =-yG, \qquad G_{yy}= (3G_y^2 + 4G_yGx + 2G^2x^2)/G.
\end{gather*}

We take the simplest solution,  $F=y$ and $G=e^{-xy}$, which
satisf\/ies (\ref{equa:07}), (\ref{equa:AAA})  and (\ref{equa:ABC}).
The linearizing generalized Sundman transformation is
\begin{gather}
\label{ex:15} u = y,\qquad dt = e^{-xy}dx.
\end{gather}
Equations (\ref{equa:08}), (\ref{equa:09}) and (\ref{equa:MN})
give
\begin{gather*}
   \beta =0, \qquad \gamma=-1, \qquad \alpha=0.
\end{gather*}
Hence equation (\ref{ex:02}) is mapped by the transformation
(\ref{ex:15}) into the linear equation
 \begin{gather}
 \label{ex:BBM}
u''+1 =0.
 \end{gather}
The general solution of equation (\ref{ex:BBM}) is
\[
u=-t^2/2+c_1t+c_2,
\]
where $c_1$, $c_2$  are arbitrary constants. Applying the generalized Sundman
transformation (\ref{ex:15}) to equation~(\ref{ex:02}) one obtains
that the general solution of equation~(\ref{ex:02}) is
\[
y(x)=-\phi(x)^2/2+c_1\phi(x)+c_2,
\]
where the function $t=\phi(x)$ is a solution of the equation
\[
 \frac{dt}{dx} =e^{-x(-t^2/2+c_1t+c_2)}.
\]
\end{example}

\begin{example}
Consider the nonlinear second-order ordinary dif\/ferential
equation
\begin{gather}
\label{ex:00}y^{\prime \prime }+\mu _3y^{k_3}y'^2+\mu _2y^{k_2}y^{\prime }+\mu
_1y^{k_1}=0,
\end{gather}
where $k_1$, $k_2$, $k_3$, $\mu _1$, $\mu _2$ and $\mu_3\neq 0$ are arbitrary constants. The
Lie criteria \cite{Lie} show that the nonlinear equation
(\ref {ex:00}) is linearizable by a point transformation if and
only if $\mu_1=0$ and $\mu _2=0$.

From equation (\ref {ex:00}), the coef\/f\/icients are
\begin{gather*}
\lambda _0=\mu _1y^{k_1},\qquad \lambda _1=\mu_2y^{k_2},\qquad \lambda
_2=\mu_3y^{k_3},\qquad \lambda
_3=\mu_2k_2y^{k_2}/y, \nonumber\\
\lambda _4=2\mu _1y^{(k_1+k_3)+1}(k_1\mu_3+k_3\mu_1)+2\mu_1y^{k_1}(k_1^2-k_1)-k_2\mu_2^2y^{2k_2+1}/y^2,\nonumber\\ \lambda _5=k_2\mu_2^2y^{2k_2}/y.
\end{gather*}

If $\mu_2\neq0$ and $\mu_1=0$, then $\lambda _3\neq0$ and
$\lambda _5\neq0$. One can check that the coef\/f\/icients obey the
conditions (\ref{equa:1b}), (\ref{equa:1a}), (\ref{equa:1d}) and
(\ref{equa:1g}). Thus, equation
\begin{gather}
\label{ex:VV}y^{\prime \prime }+\mu _3y^{k_3}y'^2+\mu _2y^{k_2}y^{\prime }=0
\end{gather}
is linearizable by
a generalized Sundman transformation.

For f\/inding the functions $F$ and $G$ one has to solve equations
(\ref{equa:07}), (\ref{equa:AAA})  and (\ref{equa:ZZZ}), which
become
\begin{gather*}
   F_{x} =0, \qquad F_{yy}=F_y(\mu_3y^{k_3+1}+k_2)/y, ~ G_x =0, \qquad G_y=
Gk_2/y.
\end{gather*}

For example, if $k_2=k_3$, one takes the simplest solution,
$F=\frac{1}{\mu_3}e^\frac{\mu_3y^{k_2+1}}{k_2+1}$ and $G=y^{k_2}$, and
the generalized Sundman transformation becomes
\begin{gather}
\label{ex:ZA} u =\frac{1}{\mu_3}e^\frac{\mu_3y^{k_2+1}}{k_2+1} ,\qquad dt = y^{k_2}dx.
\end{gather}
Equations (\ref{equa:08}), (\ref{equa:09}) and (\ref{equa:MN})
give
\begin{gather*}
    \beta =\mu_2,\qquad  \gamma=0,\qquad  \alpha=0.
\end{gather*}
Hence equation (\ref{ex:VV}) is mapped by the transformation
(\ref{ex:ZA}) into the linear equation
 \begin{gather*}

u''+\mu_2u'=0.
 \end{gather*}

If $\mu_3=0$, then equation (\ref{ex:00}) is
\begin{gather}
\label{ex:000}
y^{\prime \prime }+\mu _2y^{k_2}y^{\prime }+\mu _1y^{k_1}=0,
\end{gather}
where $\mu _2\neq 0$. The
Lie criteria \cite{Lie} show that the nonlinear equation (\ref{ex:000}) is linearizable by a~point transformation if and only if $k_1=3$, $
k_2=1$ and $\mu _1=(\mu _2/3)^2$. In the particular case, $k_1=3$, $k_2=1$, $\mu
_1=1$ and $\mu _2=3$, one has the equation
\begin{gather}
\label{ex:aa}y^{\prime \prime }+3yy^{\prime }+y^3=0.
\end{gather}
Equation (\ref{ex:aa}) arises in many areas. Some of these are the analysis
of the fusion of pellets, the theory of univalent functions, the stability
of gaseous spheres, operator Yang--Baxter equations, motion of a free
particle in a space of constant curvature, the stationary reduction of the
second member of the Burgers hierarchy \cite{KarasuLeach}.
\begin{remark}
Equation (\ref{ex:aa}) is linearizable by a point
transformation and by a generalized Sundman transformation into
the equation $u^{\prime \prime }=0$ and $u^{\prime \prime
}+3u^{\prime }+2u=0$, respectively.
\end{remark}

Without loss of the generality\footnote{For example, scaling of the independent variable: $\overline{x}=\mu_2x$.},
one can assume that $\mu _2=1$. Hence, equation (\ref{ex:000}) becomes
\begin{gather}
\label{ex:bb}
y^{\prime \prime }+y^{k_2}y^{\prime }+\mu _1y^{k_1}=0.
\end{gather}
For this equation the coef\/f\/icients are
\begin{gather*}
\lambda _0=\mu _1y^{k_1},\qquad \lambda _1=y^{k_2},\qquad \lambda _2=0,\qquad \lambda
_3=k_2y^{k_2-1}, \\
\lambda _4=\mu _1k_1(k_1-1)y^{k_1-2}-k_2y^{2k_2-1},\qquad \lambda
_5=k_2y^{2k_2+1}.
\end{gather*}

If $k_2=0$, then $\lambda _5=0$ and equation (\ref{ex:bb}) is linearizable
by a generalized Sundman transformation.

If $k_2\neq 0$, then $\lambda _5\neq 0$ and conditions (\ref{equa:1b}), (\ref{equa:1a}), (\ref{equa:1d}), (\ref{equa:1g}) are reduced to
\begin{gather}
\label{eq:aug0509.1}\mu _1(2k_2+1-k_1)(k_2-k_1)=0.
\end{gather}
If conditions (\ref{eq:aug0509.1}) are satisf\/ied, then equation (\ref{ex:bb}) is
linearizable by a generalized Sundman transformation. Notice that in
the case $\mu _1(k_2-k_1)=0$, equation (\ref{ex:bb}) is trivially integrated
by using the substitution $y^{\prime }=H(y)$. A nontrivial case is $k_1=2k_2+1$.
In this case the functions~$F$ and~$G$ are solutions of the compatible overdetermined
system of equations
\begin{gather}
\label{eq:aug0509.2}F_x=0,\qquad F_{yy}=k_2F_y/y,\qquad G_x=0,\qquad G_y=k_2G/y.
\end{gather}
The general solution of equations (\ref{eq:aug0509.2}) depends on the value of the
constant $k_2$. For example, if $k_2\neq -1$, then a particular solution of system
(\ref{eq:aug0509.2}) is
\[
F=y^{k_2+1},\qquad G=y^{k_2}.
\]
Thus, the generalized Sundman transformation reduces equation (\ref
{ex:bb}) into the linear equation
\begin{gather*}

u^{\prime \prime }+u^{\prime }+(\mu _1(k_2+1))u=0.
\end{gather*}
\end{example}

\begin{remark}
Since equations (\ref{ex:01}), (\ref{ex:00}) and (\ref{ex:000})
are autonomous, their order can be reduced by the substitution $y'=f(y)$. It is worth to note that for
equations (\ref{ex:00}) and (\ref{ex:000})
the dif\/f\/iculties in using the generalized Sundman transformation are similar to solving the original equation by this reduction.
 \end{remark}

\section{Conclusion}

Application of the generalized Sundman transformation for the
linearization problem was analyzed in the paper. Since the method is well-known,
the ef\/f\/iciency of the method is not discussed in the paper.
The paper just warns that a researcher has
to be careful when using the well-known method for the linearization problem.
In particular, our examples show that, in
contrast to point transformations (S.~Lie results), for a
linearization problem via the generalized Sundman transformation
one needs to use the general form of a linear second-order
ordinary dif\/ferential equation instead of the Laguerre form.

\subsection*{Acknowledgements}
\label{Acknowledgements}

This research was supported by the Royal Golden Jubilee Ph.D.
Program of Thailand (TRF).

\pdfbookmark[1]{References}{ref}

config: arxiv


TITLE: Correlation between numbers and percentages
QUESTION [1 upvotes]: I am trying to understand the correlation, given by my data, between numbers and percentages.
I have four numbers : 1, 2, 3 and 4. And there percentages are respectively: 0, 0.3333, 0.6667 and 1.
By using a rule of 3, i'm getting : 0.25, 0.5, 0.75 and 1.
What i know from the start is that my lowest number's percentage is 0 and my highest number's percentage is 1.
If i add a number 5, i'll get the following percentages : 0, 0.25, 0.5, 0.75 and 1.
What can replace a rule of 3 to get the right percentages ? Thank you

REPLY [0 votes]: The correlation is $1$: all the points lie on an upward sloping straight line as shown here.

In statistics you could use simple linear regression to find the line and the correlation, but this example is easier than that since the observations are colinear.  Each step of $1$ to the right takes you up by $\frac13$ so the slope is $\frac13$; a step back from $(1,0)$ then takes you to $(0,-\frac13)$ so the vertical intercept is $-\frac13$.  That gives the equation of the line as $$y=\frac13 x - \frac13$$ which you can also write as $y=\frac{x-1}{3}$.
If you knew that the line passes through $(a,0)$ and $(b,1)$ then the equation would be $y=\frac{x-a}{b-a}$, i.e. $y=\frac{1}{b-a}x -\frac{a}{b-a}$, with $a=1$ and $b=4$ in your example.


TITLE: Non trivial counterexample: $H_1\cap H_2$ is normal subgroup $\not\Rightarrow H_1$ and $ H_2$ are normal.
QUESTION [1 upvotes]: I know that if $H_1, H_2$  are normal subgroups then $H_1 \cap H_2$ is normal.
The converse is not true, e.g. $H_1 = \langle(1, 2)\rangle, H_2 = \langle(3, 4)\rangle $ (permutations) are not normal, but $H_1 \cap H_2 = \{ e \}$ is normal.
Are they any non trivial ($|H_1\cap H_2| > 1$) counterexamples?

REPLY [2 votes]: Take $H_1, H_2 \subset S_4$ as in your example. Now, define $G := S_4 \times H$, where $H$ is your favorite non-trivial group.
Set now $H_1' := H_1 \times H$, $H_2' := H_2 \times H$. Then $H_1' \cap H_2' = \{e\} \times H$ is normal in $G$, but $H_1'$ and $H_2'$ are not normal in $G$, since $H_1, H_2$ are not normal in $S_4$.


TITLE: Which areas of arithmetic algebraic geometry can be learned as "black boxes" and are there any references where they are treated as such?
QUESTION [12 upvotes]: In Matthew Emerton's comment on Terry Tao's blog, he speaks about learning etale cohomology or the theory of Neron models as "black boxes". By this he means that you can learn what the theory is about and how to use it, without going into the detailed proofs of why they can be used.
Which theories (e.g. etale cohomology) can be learned as black boxes?
And where would one go (e.g. find lecture notes) to learn something like that?
Notes on something like this would ideally give you an idea of what is going on, give examples, and most importantly illustrate how they would be used to solve problems. I am mainly interested in arithmetic algebraic geometry and algebraic number theory, so I would especially like to know about "black boxes" in this direction, though "black boxes" in other areas might also be worth knowing about.

REPLY [8 votes]: I find Hodge theory pretty scary stuff with its compact inclusions of Sobolev spaces, pseudodifferential operators and parametrixes for elliptic differential operators. However it is very easy to use the results of Hodge theory as emanating from a black box. I remember how exhilarated I was by the argument that a Hopf surface, homeomorphic to $S^1 \times S^3$, could not be Kähler, and much less projective, just because its first Betti number is $b_1=1$. Whereas by Hodge theory a compact Kähler manifold $X$  has   betti numbers $b_q(X)$  which are even whenever  $q$ is odd.


TITLE: On Interior and closure under Topology
QUESTION [1 upvotes]: Supposing $A$ and $B$ are subsets of the topological space $X$, $C$ is a closed set in $X$, and $A$ belongs to $C$, then prove that $\text{cl}(A)$ belongs to $C$.

REPLY [0 votes]: Let's take the following definition of the closure : the closure of $D$, noted $\overline{D}$, is the smallest (for the inclusion) closed set that contains $D$.
Here, you have $A\subset C$, so that $C$ is a closed set that contains $A$. Therefore, $\overline{A}\subset C$.


TITLE: How does one derive the Lamb shift for the Hydrogen atom?
QUESTION [2 upvotes]: I've been perusing my copies of Srednicki and Peskin & Schroeder, and I can't seem to find an explanation of how one derives the Lamb shift that I can follow.
How does one derive the Lamb shift? What order in perturbation theory do you have to go up to? Does one use a path integral or canonical quantization approach?

REPLY [3 votes]: A derivation is here:

http://en.wikipedia.org/wiki/Lamb_shift#Derivation

or in Landau-Lifshitz. Bethe's original derivation is found e.g. in Matt Schwartz's Harvard lecture here

http://isites.harvard.edu/fs/docs/icb.topic792163.files/20-LambShift.pdf

The leading contribution to the Lamb shift is the one-loop level (the first non-classical correction) but of course, the effect receives corrections at every higher order, too.
One may derive it in the operator approach or path integral approach, much like pretty much everything in physics. These are just equivalent languages to do physics.
The Lamb shift has to deal with an atom which is not quite an elementary particle. So the usual perturbative rules of QED have to be "generalized" to deal with the composite object. However, otherwise it's about a virtual photon emitted and reabsorbed by the atom. If the atom were elementary, it would be a simple "photon loop" correction to the atom's propagator. A divergent term has to be removed – equivalently, one has to find out sensible limits of the integral – and what is left is some "truncated logarithmic divergence" that produces those 1,000 MHz for the relevant levels.


TITLE: Proving that if $G$ has a Hamiltonian circuit, then the line graph $L(G)$ also has a Hamiltonian Circuit
QUESTION [6 upvotes]: I have a graph G = (V, E). I know that the line graph L(G) of G is defined as follows: each vertex in L(G) corresponds to an edge in G, and two vertices are connected by an edge in L(G) if and only if the corresponding edges in G are adjacent.
Now I am trying to (a) prove that if G has a Hamiltonian circuit, then L(G) also has a Hamiltonian circuit, and (b) give an example of a graph G such that G does not have a Hamiltonian circuit but L(G) does.
However, I'm extremely stuck on how to do this, and any help would be greatly appreciated. Thanks so much!

REPLY [7 votes]: For (a), we're essentially after a sequence of the edges such that consecutive edges share an endpoint (and the first and last edges in the sequence also share an endpoint).
We start off with the sequence of edges defined by the Hamiltonian circuit:
$$(1,2),(2,3),\ldots,(n-1,n),(n,1)$$
after relabeling the vertices such that $(1,2,\ldots,n)$ is the Hamiltonian circuit. Then we add edges outside of the Hamiltonian circuit to the sequence. Here's an illustration as to how we would do it:
The line graph of

has the Hamiltonian circuit:
$$(1,2),\color{orange}{(2,4)},\color{orange}{(2,5)},(2,3),(3,4),\color{blue}{(4,6)},(4,5),(5,6),(6,1).$$
For (b), there's some 4-vertex example. Graphs with cut vertices don't have Hamiltonian circuits.


text:
\subsection{Crank Shaft}

To show the overall performance of the implementation described, a crank shaft is analysed. The geometry is depicted in \figref{fig:crankshaft}.
The material parameters are ${E=\SI{210}{\giga\pascal}}$ and ${\nu=0.25}$. An arbitrary constant load is applied to the crank pins. The flywheel as well as the axle on the other end of the crank are fixed. Because of the constant loading, Neumann boundary conditions on the crank pins are described by means of a coarse discretisation and no refinement is required. All other surfaces are allocated with homogeneous boundary conditions. The calculation is performed with errors ${\err_{\h}=\{10^{-3},10^{-5},10^{-7}\}}$ for the matrix approximation. The minimal leafsize and the admissibility factor are set to $n_{min}=30$ and $\eta=1$.

\tikzfig{tikz/crankshaft}{1.2}{Crank shaft geometry}{crankshaft}{figure24}

\Figref{fig:crankshaft_compression} shows the total compression rates $c_{tot}$ for the left hand side $\mat{L}$ and the right hand side $\mat{R}$ of the block system \eqref{eq:BIEdis} with respect to the total number of degrees of freedom $n$. For the matrix $\mat{L}$, which is assigned to the unknown boundary data, the compression is defined by $c_{tot}=c_{\h}$. For the system matrix $\mat{R}$, which is assigned to the unknowns, the total compression rate is calculated by $c_{tot}=c_{\h}\: c_{sub}$. Homogeneous boundary conditions are not considered for the right hand side and the corresponding matrix coefficients are not calculated as for the cantilever in \secref{sec:cantilever}. It is envisaged in \figref{fig:crankshaft_compression}, that for $\mat{L}$ the gradient of the compression rate
\revdel{becomes almost linear}{}
\revadd{is of $\order( n \log n )$}{}
with increasing degrees of freedom $n$. Since no further refinement for the description of the known constant tractions is necessary, the number of rows in $\mat{R}$ is determined by the number of collocation points while the number of columns stays constant for all discretisations. The additional compression by means of \hmatrices{} is thus negligible and the slope of $c_{tot}$ is at least linear with respect to $n$.

\tikzfig{plots/crankshaft_compression}{1.0}{
  \revcomment{The gradient of the figure on the left has been updated to demonstrate the difference compression rates of L and R. }
  Total compression rates $c_{tot}$ for the system matrices of the crankshaft example with respect to the total degrees of freedom $n$.
  \revcomment{Added to emphasise the difference between the two compressions.}
  The matrix $\mat{L}$ is solely compressed by the approximation of \hmatrices{} ($c_{tot}=c_{\h}$), whereas $\mat{R}$ is also compressed due to the application of subparametric patches ($c_{tot}=c_{\h}\: c_{sub}$).  }{crankshaft_compression}{figure25}

The resulting absolute displacement for the third level of refinement is illustrated in \figref{fig:crankshaft_displacement}.
\tikzfig{tikz/crankshaft_displ}{1.2}{The absolute displacement $|u|$ of the crank shaft for the third level of refinement.}{crankshaft_displacement}{figure26}


TITLE: Is this a subspace? Simple example
QUESTION [0 upvotes]: Question:
$S := \{x = (x_1, x_2, x_3, x_4, x_5) | x_1^2 + x_2^2 = x_3^2 + x_4^2 + x_5^2\}$
Is subset $S$ a valid subspace of $\mathbb{R}^5$?
I am not sure how to prove or disprove it if it's closed under addition.
Because from:
$$(x_1+y_1)^2 + (x_2+y_2)^2 = (x_3+y_3)^2 + (x_4+y_4)^2 + (x_5+y_5)^2$$
I get:
$$x_1y_1+x_2y_2 = x_3y_3 + x_4y_4 + x_5y_5$$
and I don't know how to continue.

REPLY [0 votes]: To prove it's not closed under addition, you need just one counterexample.
Here's one for you to check:  $(1,1,0,1,1),(1,1,1,0,1)\in S$ but is their sum?


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\listfiles


TITLE: how to simplify scalar product of 2 scalar triple products.
QUESTION [0 upvotes]: Can anyone explain to me how to simplify following expression (a scalar product of two scalar triple products),
$$(a\cdot(b \times c))( p\cdot(q\times r))
\quad\quad ( \text{ or }\;\; [a,b,c][p,q,r]\;),
$$
where $\times$ is cross product, $\cdot$ is dot product and $a,b,c,p,q,r$ are vectors.

REPLY [0 votes]: Maybe it's too long for a comment/ not informative enough for an answer, but I think it may help:
Let $\vec{a}=u_a\hat{i}+v_a\hat{j}+w_a\hat{k}$, $\vec{b}=u_b\hat{i}+v_b\hat{j}+w_b\hat{k}$, etc.
Then
\begin{align}
(a\cdot(b \times c))( p\cdot(q\times r))&=
        \begin{vmatrix}
        u_a & v_a & w_a \\
        u_b & v_b & w_b \\
        u_c & v_c & w_c \\
        \end{vmatrix}\cdot \begin{vmatrix}
        u_p & v_p & w_p \\
        u_q & v_q & w_q \\
        u_r & v_r & w_r \\
        \end{vmatrix}
\end{align}


TITLE: Can an element in a Noetherian domain have arbitrarily long factorizations?
QUESTION [3 upvotes]: I tried to answer this question two days ago. Unfortunately, I said ring, rather than domain, which is what I wanted. So I try again.

Let $R$ be a Noetherian commutative domain and let $r\in R$.  Is it possible that $r$ have arbitrarily long factorizations? That is, is it possible that for all $n>0$, there are elements $a_{1n}, a_{2n},\ldots,a_{nn}$, all non-units, such that $r=a_{1n}a_{2n}\cdots a_{nn}$?

REPLY [3 votes]: No, it's not possible.

If $R$ is a noetherian domain then there are no non-zero elements having arbitrary long factorizations.

Let $r\in R$ be a non-zero non-unit, and $\mathfrak p_1,\dots,\mathfrak p_m$ the height one primes of $R$ containing $r$. Suppose that some non-unit $r'\in R$ divides $r$. Then there is $j\in\{1,\dots,m\}$ such that $r'\in\mathfrak p_j$. If $r=r_1\cdots r_n$ with $r_i$ non-units and $n$ arbitrarily large, then there is $j\in\{1,\dots,m\}$ such that $r\in\bigcap_{k\ge1}\mathfrak p_j^k=(0)$, a contradiction.


TITLE: Letter permutations with restriction.
QUESTION [1 upvotes]: I need help with this problem.
How many ways con you organize  P A P E L O N, so that the three vowels are not consecutive.
What I did was calculate the total number without restrictions first which is 7!. Then I tried to calculate for those which do have the 3 vowels together in order to subtract them.
To do this I realized there was 5 ways for each 3 vowel order, and each of those 5 had 4!. So 5(4!)= 5!. And then since each 3 vowel order can be arranged in 3! ways, 5!(3!).
Total would then be 7! -5!3!.
Could you tell me if its okay? Thank you very much!

REPLY [3 votes]: You're close, but there is one problem. There are two instances of the letter P. By the generalized permutation formula, the total number of permutations of PAPELON is actually $$\frac{7!}{2}$$
And when you're counting the number of permutations where all three vowels are adjacent, the factor of $4!$ counts the number of ways to order the consonants. So once again, you must divide by 2 to account for the duplicate P:
$$\frac{5!\cdot 3!}2$$
And the difference is your final answer: $$\frac{7!-3!\cdot 5!}{2}=2160$$


TITLE: How to observe Floquet state?
QUESTION [4 upvotes]: The Schrodinger equation is
$$i\hbar\partial_t\psi(t)=H(t)\psi(t).$$
Now, given that the situation that the Hamiltonian is periodically driven, i.e., $H(t+T)=H(t)$, then the equation can be solved by the Floquet ansatz,
$$\psi(t)=e^{-i\varepsilon_n t/\hbar}u_n(t),$$
where $\varepsilon_n$ is called the quasienergy, and the Floquet state $u_n(t)$ satisfies the periodic condition $u_n(t+T)=u_n(t)$.
This can be written in a compact form by defining the Floquet Hamiltonian as
\begin{align}
H_F(t) & =H(t)-i\partial_t\\
H_F(t)u_n(t) & =\varepsilon_n u_n(t).
\end{align}
Then if we treat time $t$ as a new degree of freedom, the $H_F$ is then defined in the composite Hilbert space $\mathcal{R}\otimes\mathcal{T}$. Formally, $H_F$ is analogous to the Hamiltonian for stationary states of the "time"-independent form and all mathematical conclusion of the stationary theory can be obtained.
My question is, although we can formally define the Floquet state $u_n(t)$ and the quasienergy $\varepsilon_n$, how can we detect them? I think that only the real wave function $\psi$ describes the real physics, so how can we draw the information of the Floquet state $u_n(t)$ from $\psi$?

REPLY [2 votes]: They have been observed in the following experiment:

Observation of Floquet-Bloch States on the Surface of a Topological Insulator.
  YH Wang, H Steinberg, P Jarillo-Herrero and N. Gedik. Science 342, 453 (2013), arXiv:1310.7563.

In this paper, there is a periodic laser pulse hitting the sample which gives a time-periodic Hamiltonian. They then used ARPES (angle-resolved photoemission) to see the band structure repeated in energy steps, $\epsilon_n=\hbar \omega_n$. Without the time-dependent laser pulse, however, the "replica bands" are not visible. Of course the experiment is not perfect, which manifests itself in the fact that there are not infinitely many "replica bands" but at least this is strong evidence that under perfect experimental conditions, a perfect Floquet state could exist.


TITLE: What is the time complexity of conjugate gradient method?
QUESTION [11 upvotes]: I have been trying to figure out the time complexity of the conjugate gradient method.
I have to solve a system of linear equations given by
$$Ax=b$$
where $A$ is a sparse, symmetric, positive definite matrix.
What would be the time complexity of the conjugate gradient method?

REPLY [15 votes]: $O(m\sqrt{k})$, where $m$ is the number of nonzero entries in $A$ and $k$ is its condition number.
See Chapter 10 in this excellent tutorial.


TITLE: Intermediate Value Theorem in Munkres : Topology A First Course
QUESTION [0 upvotes]: This is the Intermediate Value Theorem as stated and proved in Munkres: Topology A First Course

I understand everything except the claim that $(-\infty, r) \subset Y$ and $(r, +\infty) \subset Y$, I don't see why $-\infty, \infty \in Y$, since all we know about $Y$ is that it is an ordered set, we don't know anything about the structure or the elements of $Y$.

REPLY [2 votes]: In an ordered space $(S,<)$, the order topology is generated by the subbase consisting of sets of the form $\{x \in S: x < r\}, r \in S$, and of the form $\{x \in S: x >r\}, r \in S$ which are sometimes denoted $(-\infty,r)$, or $(\leftarrow, r)$ and $(r, +\infty)$ or $(r, \rightarrow)$, respectively. So the sets denoted by Munkres are just subbasic open sets in the alternative notation. They are always well-defined in any ordered set, regardless of further structure.


text:
\section{Preliminaries}
\label{sec:preliminaries}

\subsection{Initial data sets and MOTS}
\label{sec:initial_data_sets_and_MOTS} In this section we introduce
the notation, sign conventions, and terminology used in this
survey. Classical references for this material are \cite{HawEll} and
\cite{Wald}.

An {\it initial data set} is a triple $(M, g, k)$ where $M$
is a complete $3$-dimensional manifold, possibly with boundary,
together with a positive definite metric $g$ and a symmetric $(0, 2)$
tensor $k$. In the context of general relativity, such triples arise
as embedded spacelike hypersurfaces of time-orientable Lorentzian
manifolds $(\bar{M}, \bar g)$, referred to as the spacetime, with
induced metric $g$ and (future directed) second fundamental form
$k$. Hence, if $\eta$ is a future directed normal vector field of $M
\subset \bar{M}$ such that $\bar g(\eta, \eta) \equiv -1$, and if
$\xi, \zeta \in T_p M \subset T_p \bar M$ where $p \in M$, then
$k(\xi, \zeta) = \bar{g} (\bar D_\xi \eta, \zeta)$. Here, $\bar D$ is
the Levi-Civita connection of the spacetime.

Now let $\Sigma \subset M$ be an embedded two-sided
$2$-surface and let $\nu$ be a unit normal vector field of $\Sigma
\subset M$. We write $h$ for the second fundamental form of $\Sigma$
with respect to $\nu$ so that $h(\xi, \zeta) = g(D_\xi \nu, \zeta)$
for tangent vectors $\xi, \zeta \in T_p\Sigma$ for any $p \in \Sigma$,
where $D$ is the Levi-Civita connection of $(M, g)$. We may think of
$\nu$ as a vector field along $\Sigma \subset \bar{M}$ so that $l =\nu
+ \eta$ is a future directed null vector field of $\Sigma$ when viewed
as a spacelike $2$-surface of the spacetime. Note that then $\chi
(\xi, \zeta) := (h + k)(\xi, \zeta) = \bar g (\bar D_\xi l, \zeta)$
for tangent vectors $\xi, \zeta \in T_p \Sigma$. This symmetric $(0,
2)$-tensor $\chi$ is the null second fundamental form of $\Sigma
\subset M$ with respect to $l$. Its trace $\theta_\Sigma = \tr_\Sigma
\chi$ is called the expansion of $\Sigma$ with respect to the
null-vector field $l$. Note that $\theta_\Sigma = \mc_\Sigma +
\tr_\Sigma k$ where $\mc_\Sigma$ is the mean curvature of $\Sigma
\subset M$ with respect to the unit normal $\nu$.

In many places in this survey, $\Sigma \subset M$ has a clearly
designated {\it outward} unit normal. If we compute the null-second
fundamental form and expansion of such a surface $\Sigma$ with respect
to the corresponding future-directed \emph{outward} null-normal, we
will write $\chi^+_\Sigma$ and $\theta^+_\Sigma$ for emphasis. If
$\Sigma$ is such a $2$-surface, with $\theta_\Sigma^+ \equiv 0$ on
$\Sigma$, then we say that $\Sigma$ is a {\it marginally outer trapped
  surface} or {\it MOTS} for short.

If we use the future-directed \emph{inward} unit normal to compute the
expansion we write $\chi^-_\Sigma$ and $\theta^-_\Sigma$.  If
$\theta^-_\Sigma \equiv 0$ we say that $\Sigma$ is a {\it marginally
  inner trapped surface} or {\it MITS}. For ease of exposition, we say
that a two-sided surface $\Sigma \subset M$ is an \emph{apparent horizon} if
it is either a MOTS or a MITS, i.e., if $\mc_\Sigma + \tr_\Sigma(k)
\equiv 0$ holds for one of the two possible consistent choices to
compute the mean curvature scalar.

The following spacetime analogue of the Bonnet-Myers theorem, whose
proof uses the Raychaudhuri rather than the Riccati equation, lies at
the heart of the Penrose-Hawking singularity theorems of general
relativity (cf. \cite[Proposition 4.4.3]{HawEll}): if the \emph{null
  energy condition} holds in $\bar{M}$, i.e. if for the spacetime
Ricci tensor $\overline \Ric(\upsilon, \upsilon) \geq 0$ holds for all
points $q \in \bar{M}$ and null vectors $\upsilon \in T_q \bar{M}$,
and if the expansion $\theta_\Sigma$ of $\Sigma \subset \bar{M}$ with
respect to the null vector field $l$ along $\Sigma$ is negative at $p
\in \Sigma$, then a null geodesic emanating from $p$ in direction
$l(p)$ has conjugate points within a finite affine
distance. Physically, this means that the surface area of a shell of
light emanating from $\Sigma$ near $p$ will go to zero before it has a
chance to `escape to infinity'.

Frequently, additional assumptions are included in the
definition of an initial data set in the literature. In this survey we
will always state such extra hypotheses explicitly when needed. Two
such extra assumptions will be particularly relevant for us. First,
recall \cite[p. 219]{Wald} that a spacetime $(\bar{M}, \bar g)$ is
said to satisfy the {\it dominant energy condition} if its
stress-energy tensor $\T := \overline{\Ric} - \frac{1}{2}\bar \R \bar
g$ has the property that for every $p \in \bar{M}$ the vector dual to
the one form $ - \T(\eta, \cdot)$ with respect to $\bar {g}$ is a
future directed causal vector in $T_p \bar{M}$ for every future
directed causal vector $\eta \in T_p \bar{M}$. Here $\overline \Ric$
and $\overline \R$ respectively denote the spacetime Ricci tensor and
spacetime scalar curvature. Note that this dominant energy condition
implies the null energy condition used above
in connection with the formation of caustics along light-like
geodesics in the spacetime. If $\eta$ arises as above as the future
directed normal vector field of a spacelike hypersurface $M \subset
\bar{M}$, then one can use the Gauss and the Codazzi equations to
express the normal-normal component (abbreviated by $\mu$) and the
normal-tangential part of $\T$ (written as a one form ${J})$ along $M$
entirely in terms of the initial data $(M, g, k)$. Explicitly,
\begin{align}
  \label{eqn:mu}
  \frac{1}{2} \left( \R_M - |k|_M^2 + (\tr_M(k))^2 \right) &=: \mu
  \\
  \div_M \left( k - \tr_M(k)g \right) &=: J
  \label{eqn:J}
\end{align}
where $\R_M$ is the scalar curvature of $(M, g)$, $D$ is its
Levi-Civita connection, and where $|k|_M^2$ and $\tr_M(k)$ are the
square length and trace of $k$ with respect to $g$. Physically, $\mu$
and $ J$ are, respectively, the energy density
and the current
density of an observer traveling with $4$-velocity $\eta$. A
consequence of the dominant energy condition for the spacetime
$(\bar{M}, \bar g)$ is that $\mu \geq | J|$ must hold on $M$. By
slight abuse of language one says that an initial data $(M, g, p)$
satisfies the {\it dominant energy condition} if $\mu \geq | J|$
holds. An important special case of the dominant energy condition is
when $M$ is {\it a maximal slice} of the spacetime, i.e. when
$\tr_M(k) \equiv 0$, so that the dominant energy condition implies
that the scalar curvature $\R_M \geq 0$ is non-negative. In the case
where $k \equiv 0$, $M \subset \bar M$ is called a {\it time symmetric
  slice} or maybe more precisely a {\it totally geodesic slice}. In
the case of time-symmetric data $(M, g, k \equiv 0)$, the dominant
energy condition is equivalent to $R_M \geq 0$, and apparent horizons
$\Sigma \subset M$ are precisely minimal surfaces.

For our discussion of the positive mass and positive energy theorem in
subsection \ref{sec:posit-mass-theor} we will also need the following
definition (stated here as in \cite{Schoen-Yau:1981:PMTII}): An
initial data set $(M, g, k)$ is said to be \emph{asymptotically flat}
if there is a compact set $K \subset M$ so that $M \setminus K$ is
diffeomorphic to a finite number of copies of $\mathbb{R}^3 \setminus
\bar B(0, 1)$ (each corresponding to an \emph{end}), and such that
under these diffeomorphisms
\begin{align*}
  &|g_{ij} - \delta_{ij}|
  + |x| |\partial_p g_{ij}|
  + |x|^2 | \partial^2_{pq} g_{ij}|
  = O(|x|^{-1})
  \text { and }
  |\R_M| + |\partial_p \R_M|
  =
  |x|^{-4}
  \intertext { as well as }
  &|k_{ij}|
  + |x| |\partial_p k_{ij}|
  + |x|^2 |\partial^2_{pq} k_{ij}|
  = O (|x|^{-2})
  \text { and }
  \left|\sum_{i=1}^3 k_{ii}\right|
  = O(|x|^{-3})
\end{align*}
as $|x| := \sqrt{(x^1)^2 + (x^2)^2 + (x^3)^2} \to \infty$ on
each end. For the rigidity part of the positive mass theorem it will
also be necessary to assume that the third and forth order derivatives
of the metric are $O(|x|^{-4})$.
These conditions guarantee that the ADM energy
\begin{align*}
  E_\text{ADM}
  &=
  \frac{1}{16 \pi} \lim_{r \to \infty } \sum_{i, j=1}^3
  \int_{|x| = r} \left( \partial_i g_{ij} - \partial_j g_{ii}\right) \frac{x^j}{|x|} d \mathcal{H}^2
  \intertext{and the ADM linear momentum}
  P_{\text{ADM}}^l
  &=
  \frac{1}{ 8 \pi} \lim_{r \to \infty} \sum_{j=1}^3
  \int_{|x| = r} |x|^{-1}\left( x^j k^l_j - x^l k^j_{j} \right) d\mathcal{H}^2
\end{align*}
are well-defined, see \cite{ADM:1961,
Bartnik:1986,omurchadha:1986}.
The spacetime coordinate transformations which leave the asympototic
conditions invariant are asymptotically Lorentz transformations and the ADM
4-momentum vector $P_\text{ADM}^\mu = (E_\text{ADM},
P_\text{ADM}^l)$ is Lorentz covariant under such transformations.
In particular, the ADM mass $m_\text{ADM} = \sqrt {-
  P_\text{ADM}^\mu (P_\text{ADM})_\mu}$ is a coordinate independent
quantity.
Note that we will be sloppy
in the sequel and refer to the positive energy theorem, i.e. the
question whether $E_\text{ADM}\geq 0$, by the term \emph{positive mass
  theorem}, which would rather be appropriate for the statement
$m_\text{ADM}\geq 0$.

\subsection{Linearization of the expansion}
\label{sec:second_variation_formulae}
Marginally outer trapped surfaces $\Sigma \subset M \subset \bar{M}$
in general initial data sets are not known to arise as critical points
(or indeed to occur as minimizers) of a standard variational problem
described in terms of the data $(M, g, k)$. (However, note that {by
definition} a MOTS $\Sigma \subset M$ is a critical point for the area
functional inside the future-directed null-cone of $\Sigma \subset
\bar{M}$.) In recent years, properties of MOTS akin to those of
minimal surfaces have been introduced and investigated. In this
subsection we discuss the linearization of the expansion
$\theta_\Sigma$ with respect to variations in $M$. This sets the
ground for a discussion of the natural notion of stability of MOTS in
subsection \ref{sec:stability_of_MOTS}.

Let $\Sigma \subset M$ be a two-sided hypersurface and let $\varphi_s$
be a smooth family of diffeomorphisms of $M$ parametrized by $s \in
(-\delta, \delta)$ so that $\varphi_0$ is the identity. Assume that
$\frac{d}{ds}|_{s =0} \varphi_s = \sigma + f \nu$ on $\Sigma$, where
$\sigma \in \Gamma (T \Sigma)$ is a tangential vector field, where $f$
is a smooth scalar function on $\Sigma$, and where $\nu$ is a smooth unit
normal vector field of $\Sigma$. The mean curvature scalar and
tangential trace of $k$ of $\Sigma_s := \varphi_s (\Sigma)$ can be
viewed as functions on $\Sigma$ via pullback by $\varphi_s$, and they
are smooth functions of $s$. Here we agree that the mean curvature
scalar of $\Sigma$ is computed as the tangential divergence
$\div_\Sigma (\nu)$ of $\nu$. It follows that
\begin{equation}
  \label{eqn:variation_mean_curvature}
  \begin{split}
    - \la D_\Sigma \mc_\Sigma, \sigma \ra + \frac{d}{ds}|_{s = 0}
    \varphi_s^* \mc_{\Sigma_s}
    &=
    - \Delta_\Sigma f - \left(|h|_\Sigma^2 + \Ric (\nu, \nu) \right)
    f,
    \text{ and}
    \\
    - \la D_\Sigma \tr_\Sigma(k), \sigma \ra + \frac{d}{ds}|_{s=0}
    \varphi_s^* \tr_{\Sigma_s} (k)
    &=
    2 k (\nu, D_{\Sigma } f) + D_\nu (\tr_M (k)) f - (D_\nu k)(\nu, \nu) f.
  \end{split}
\end{equation}
Here, $\Delta_\Sigma$ and $D_\Sigma$ denote respectively the
non-positive Laplacian and the gradient operator of $\Sigma$ with
respect to the induced metric, $|h|_\Sigma$ is the length of the
second fundamental form of $\Sigma$, and $\Ric$ and $D$ are the
ambient Ricci curvature tensor and covariant derivative operator. The
literature often refers to the first identity as the Riccati or Jacobi
equation, in particular when the tangential part $\sigma$  of the variation
 vanishes identically. If $X \in \Gamma (T \Sigma)$ denotes the
tangential part of the vector field dual to $k(\nu, \cdot)$ one can
compute that
\begin{equation*}
  (D_\nu k)(\nu, \nu)
  =
  - \mc_\Sigma k(\nu, \nu) + \la h, k \ra_\Sigma + (\div_M k) (\nu) - \div_\Sigma X.
\end{equation*}
Using the Gauss equation and the expressions for the
 mass density $\mu$ in (\ref{eqn:mu}) to substitute for
$\Ric(\nu, \nu)$, and the  current density $ J$ in (\ref{eqn:J})
to re-write $\div_M k$, it follows that
\begin{equation}
  \begin{split}
    &- \la D_\Sigma \theta_\Sigma, \sigma \ra
    + \frac{d}{ds}|_{s=0} \varphi_s^* \theta_{\Sigma_s}
    \\
    &\quad=
    - \Delta_\Sigma f + 2 \la X, D_\Sigma f \ra
    \\
    &\qquad + \left( \tfrac{1}{2} \R_\Sigma - \tfrac{1}{2} |h + k|_\Sigma^2 - J (\nu)
      - \mu + \div_\Sigma X - |X|^2 + \tfrac{1}{2} \theta_\Sigma(2 \tr_M (k)
      - \theta_\Sigma) \right) f
    \\
    &\quad =: \L_\Sigma f
  \end{split}
\end{equation}
where $\theta_{\Sigma_s} := \mc_{\Sigma_s} + \tr_{\Sigma_s} (k)$ is
the expansion of $\Sigma_s$. See \cite[(1)]{Andersson-Mars-Simon:2005}
and \cite[(2.3)]{Galloway-Schoen:2006}, and also
\cite[(2.25)]{Schoen-Yau:1981:PMTII} for an important special case of
this formula that we will come back to in subsection \ref{sec:stability_of_MOTS}. Following \cite{Galloway-Schoen:2006} we point
out that when $\Sigma$ is a MOTS
and $f>0$, one can rearrange and express $\frac{1}{f} \L_\Sigma f$
more compactly as
\begin{equation}
  \label{eqn:Galloway-Schoen_rearrangement}
  \div_\Sigma (X - D_\Sigma \log f) - |X - D_\Sigma \log f|_\Sigma^2 +
  \frac{1}{2} \R_\Sigma - |h + k|_\Sigma^2 - {J}(\nu) - \mu.
\end{equation}
Note that if the dominant energy condition $\mu \geq | J|$ is assumed,
then the last three terms make a non-positive contribution to this
expression.


TITLE: Do square integrable function and its derivative imply the function tends to $0$ for $|x|$ large?
QUESTION [2 upvotes]: I know a square integrable function does not necessarily tend to $0$ for $|x|$ large. Now suppose $f\colon\mathbb{R}\to\mathbb{R}$ is a smooth function satisfying
\begin{equation}
f, f' \in L^2(\mathbb{R}).
\end{equation}
Then is it true that
\begin{equation}
\lim_{x\to\pm\infty} f(x) = 0 \quad \text{ and } \quad \lim_{x\to\pm\infty} f'(x) = 0 \, ?
\end{equation}
Is it also true for any derivative of $f$?

REPLY [4 votes]: It is true that $\lim_{x\to\infty} f(x) = 0$.  To see this, let $\chi_R(x)$ be a smooth function with $\chi_R(x) = 0$ when $|x|<R$ and $\chi_R(x) = 1$ for $|x|>2R$.  Given an $\epsilon>0$ we can find large enough $R$ so that
$$
\|\chi_R f\|_{L^2} + \|(\chi_R f)'\|_{L^2} < \epsilon.
$$
Together these imply that $\|\chi_R f\|_{L^\infty} < C \epsilon$ (this is a Sobolev embedding, but a simpler case).  This implies the first limit is true (at least once you modify $f$ on a set of measure zero).
Let me explain that point a bit more in detail.  Let $g \in L^2(\mathbb{R})$ be a square integrable function with $g'\in L^2(\mathbb{R})$.  We want a bound on
$|g(x)|$.  Momentarily assume that $g\in C_c^\infty(\mathbb{R})$ is smooth and compactly supported.  Then by the fundamental theorem of calculus, and the Cauchy-Schwarz inequality,
\begin{align*}
|g(x)^2| &\leq \int_{-\infty}^x |(g(x)^2)'|\,dx \\
&= \int_{-\infty}^x 2|g(x) g'(x)|\,dx  \\
&\leq 2 \left(\int_{-\infty}^x |g(x)|^2\right)^{1/2}\left(\int_{-\infty}^x |g'(x)|^2\,dx\right)^{1/2}.
\end{align*}
Taking square roots we obtain the expression
$$
|g(x)| \leq \sqrt{2} \|g\|_{L^2} \|g'\|_{L^2}.
$$
Now that we know this is true for smooth compactly supported functions, it remains true for functions with $f,f'\in L^2(\mathbb{R})$ by a limiting argument.
The second limit, $\lim_{x\to\infty} f'(x)$ does not need to exist or tend to $0$ because all we know about $f'$ is that $f'\in L^2$.  We construct an example here for completeness.  Let $h(x)$ be a triangular "hat" function centered at $0$, with support in $[-1,1]$, and with slope $\pm 1$ on either side of the origin.  Observe that $2^{-n} h(2^nx-4^{n})$ is centered at $2^n$, has width $2^{1-n}$, and slope $\pm 1$ on either side of its peak at $2^n$.  Thus
$$
f(x) = \sum_{n=0}^\infty 2^{-n} h(2^nx - 4^{n})
$$
is square integrable and has square integrable derivative, but the derivative $f'(x)$ does not tend to $0$ as $x\to\infty$.


TITLE: Showing $\bigg( \frac{x_1}{1-x_{n+1}},..., \frac{x_n}{1-x_{n+1}} \bigg)=y$ as onto?
QUESTION [0 upvotes]: Am I allowed to use an index of $y \in \mathbb{R}^n$ as constant, when showing onto ($\exists$ s.t. $f(x)=y$)?
Particularly,
I'm trying to show
$\bigg( \frac{x_1}{1-x_{n+1}},..., \frac{x_n}{1-x_{n+1}} \bigg)=y$
And I find an expression
$$x_i=y_i(1-x_{n+1})=y_i-y_ix_{n+1}$$
$\iff$
$$x_i+y_i x_{n+1}=y_i$$
But by previous examples on surjective functions I think I'd need to have all $x_i$ on the other side, while all $y_i$ on the other side.

There are some topological shortcuts for this, but I was just wondering, whether this algebraic way is feasible.

REPLY [0 votes]: If $f(\vec x)=\bigg( \frac{x_1}{1-x_{n+1}},..., \frac{x_n}{1-x_{n+1}} \bigg),\ $ you can check directly that
$g(\vec y)=\left ( \frac{2y_1}{1+\|y\|},\cdots \frac{2y_n}{1+\|y\|},1-\frac{2}{1+\|y\|} \right )$ is inverse to $f$.
In fact, $f$ is a homeomorphism: $S^n\setminus \left \{ (0,0,\cdots, 1) \right \}\to \mathbb R^n.$