TITLE: How to find a constraint equation to relate the two tension forces in this arrangement?
QUESTION [0 upvotes]: Consider this arrangement of pulley:

The problem is to find the acceleration of $m_1$.
This is my solution:

The problem explicitly wants $\ddot y_1$.
So starting by applying Newton's second law we get:
$$T_1-m_1g=m_1\ddot y_1$$
$$T_2-m_2g=m_2\ddot y_2$$
Where $T$'s are tension forces acting on the masses with the same subscript.
We need two more equations. First is obtained by geometry of the system:
$$l_2=y_3+(y_3-y_2)+\pi R=2y_3-y_2+\pi R\implies 2\ddot y_3=\ddot y_2$$
$$l_1=h-y_1+h-y_3+\pi R=2h-y_1-y_3+\pi R\implies \ddot y_1=-\ddot y_3$$
$$\ddot y_2=-2\ddot y_3$$
But I still need another constraint to relate $T_1$ to $T_2$.
How to write that?

REPLY [3 votes]: Since the pulley is weightless, according to Newton's second law the net force acting on it should be zero. so: $$T_1=2T_2$$


TITLE: Show that $R\cong Re\times R(1-e)$
QUESTION [0 upvotes]: An element $e\in R$  is called idempotent if $e^2=e$. Assume that $e$ is idempotent and $er=re$ $\forall r\in R$. Prove that $Re$ and $R(1-e)$   are two sided ideals of $R$ and $R\cong  Re\times R(1-e)$.
I  have successfully proved the other results and stuck on the last one. Applying Chinese Remainder Theorem we have that $R/(Re\cap R(1-e))\cong R/Re\times R/R(1-e)$ .
Also $Re+R(1-e)=R$ and $Re\cap R(1-e)=\{0\}$. Thus $R\cong  R/Re\times R/R(1-e)$.
where am I going wrong?

REPLY [2 votes]: Let $f=1-e$, so that $ef=fe$ and $f^2=f,e^2=e$. We have a morphism of rings (prove it is one!) $R\to Re\times Rf$ that sends $a=ae+af$ to $ (af,ae)$. The rings $Re$ and $Rf$ have units $e,f$ respectively. This is an epimorphism since $ae+bf$ maps to $(ae,bf)$ (for $ef=fe$ and $f,e$ are idempotent), and a monomorphism since $af=ae=0$ gives $af+ae=a=0$. The point of this is you don't need to meddle with CRT and all those isomorphism theorems: central idempotents disconnect your ring "nicely" and you can make concrete computations with them.


TITLE: chi-square test for uniform distribution
QUESTION [5 upvotes]: So, i have a hash function which maps a set of possible inputs to a defined range of outputs. I want to test if the mapped outputs are uniformly distributed over the defined range. Wikipedia seems to suggest that the chi-square test can be used to measure the uniformity of the result - how would i go about doing this?

REPLY [3 votes]: Set up "bins" such that the expected number of data points in each bin is greater than 5 (this recommendation is due to the fact that the Chi-square test relies on the central limit theorem, so that results form an approximate multivariate normal distribution).
Then, subtract the expected number of points in each bin from the actual number of points in each bin. Use these differences to form your chi-square statistic (per the wiki page) and then apply the test.
Since you are testing for uniformity, the expected number of points in each bin bill be proportional to either the length or the number of outputs in each bin (depending on if your hash function is continuous or discrete valued, respectively.)


TITLE: Show that a complex number is on a line.
QUESTION [0 upvotes]: Let $v$ and $w$ be  two distinct complex numbers such that $v + t (w-v)$ is a line $l$, where $t \in \mathbb{R}$. Prove that:
If $\frac{z - w}{z-v}$ is a real number, for instance $t$, then $z$ is on the line $l$.
I have already tried using the hint as follows:
\begin{equation}
\frac{z - w}{z-v}=t
\end{equation}
\begin{equation}
z - w=t(z-v)
\end{equation}
\begin{equation}
z = \frac{-v t +w}{1-t}
\end{equation}
This doesn't really get me anywhere, at least I think it doesn't because I don't recognise the form of my desired line in here. I also tried using the following algorithm:
\begin{equation}
\frac{z - w}{z-v}(w-v)=t(w-v)
\end{equation}
\begin{equation}
\frac{z - w}{z-v}(w-v)+v=v+ t(w-v)
\end{equation}
Which can be rewritten as:
\begin{equation}
\frac{zw-w^2 +vw-v^2}{z-v}=v+ t(w-v)
\end{equation}
I would have hoped it to simplify to $z$. Do you people have any pointers or tips?

REPLY [1 votes]: Just one more step.
$$z=\frac{-vt+w}{1-t}=v+\frac{w-v}{1-t}=v+t'(w-v)$$
with $t'=\frac{1}{1-t}$.


text:
\begin{document}

 \title{Self-adjoint extensions of Coulomb systems in 1,2 and 3 dimensions}
\author{C\'esar R. de Oliveira {\small and} Alessandra A. Verri\\
\vspace{-0.6cm}
\small
\it Departamento de Matem\'{a}tica -- UFSCar, \small \it S\~{a}o Carlos, SP,
13560-970
Brazil\\ \\}
\date{\today}

\maketitle

\begin{abstract} We study the nonrelativistic quantum Coulomb hamiltonian (i.e., inverse of distance
potential) in
$\R^n$,
$n=1,2,3$. We characterize their self-adjoint extensions and, in the unidimensional case, present a discussion of
controversies in the literature,  particularly the question of the permeability of the origin. Potentials given by
fundamental solutions of Laplace equation are also briefly considered.
\end{abstract}

\section{Introduction}

In principle the unidimensional ({\sc1D}) hydrogen atom is a simplification of the three-dimensional
({\sc3D}) model which has been invoked in theoretical and numerical studies \cite{JSS,DKS,LCO3}; note that
Cole and Cohen
\cite{CC} and Wong et al.\ \cite{Wong} have
reported some experimental evidence for the {\sc1D} hydrogen atom.  In a particular situation the
{\sc1D} eigenvalues coincide with the well-known
eigenvalues of the {\sc3D} hydrogen model, as discussed ahead.

Apparently the  {\sc1D} hydrogen atom was first considered in 1928  by Vrkljan \cite{VRs}. However, it was a work of
Loudon \cite{Lou} published in 1959,  whose potential model is
\[
V_C(x)= -\frac{\kappa}{|x|},\quad \kappa>0,
\] that increased attention to the subject which has become
interesting and quite controversial. We refer to $V_C$ as {\it the Coulomb potential}.

Loudon stated that the {\sc1D} hydrogen atom was twofold degenerate, having
even and odd eigenfunctions for each
eigenvalue, except for the (even) ground state having infinite binding
energy. Typically {\sc1D} systems have no
degenerate eigenvalues, and Loudon justified the double degeneracy as a
consequence of the singular atomic
potential. Andrews \cite{And1} questioned the existence of a ground
state with infinite binding energy. Ten
years later Haines and Roberts \cite{HR} revised Loudon's work and
obtained that their even wave functions,
with continuous eigenvalues, were complementary to odd functions, but
such results were criticized by
Andrews \cite{And2}, who did not accept the continuous eigenvalues.
Gomes and Zimerman \cite{GZ} argued
that the even states with finite energy should be excluded. Spector and
Lee \cite{SL} presented a
relativistic treatment that removed the problem of infinite binding
energy of the ground state. Several other
works \cite{DPST,BKB,NVS,LC,OL,FLM,XDD,LL,LCO6} (see also references therein)
have discussed this apparently
simple problem.

In this work we advocate that the roots of such controversies is a lack of sufficient mathematical care in some
papers: in {\sc 1D} the Coulomb singularity is so severe that it is not a trivial problem to assign boundary
conditions at the origin. The main question is how to properly define the self-adjoint realization(s) of
\[
\dot H=-\frac{\hbar^2}{2m}\Delta + V_C(x), \quad \dom \dot H = C_0^\infty(\R\setminus\{0\}).
\]The domain choice is  because functions $\psi$ in $C_0^\infty(\R\setminus\{0\})$ are kept far enough from the origin
(i.e., zero does not belong to their support), and so $\dot H\psi$ is well defined. $\dot H$ is hermitian but not
self-adjoint, and it turns out that it has deficiency index
$n_+=2=n_-$ and so an  infinite family of self-adjoint extensions (see Section~\ref{SAEsection1D}). Although such
extensions appear in
\cite{FLM}, details of how they were obtained are missing; in Section~\ref{SAEsection1D} we find such extensions by
another approach, that is, we use a (modified) boundary form as discussed in \cite{CRdO}. These extensions are the
candidates for the energy operator of the {\sc1D} hydrogen atom.

With these extensions at hand, we discuss the question of permeability of the origin, that is, whether in {\sc1D} the
Coulomb singularity acts as barrier that allows the electron to pass through it or not. This is one of the important
questions considered in the literature. It is found that the permeability depends on the self-adjoint extension and we
present explicit examples of both behaviors.

In Sections~\ref{SAEsection3D} and \ref{SAEsection2D} we present the self-adjoint extensions of $\dot H$ in 3 and 2
dimensions, respectively. It is well-known that in 3D the operator $H$, with the same action as $\dot H$ but domain
$C_0^\infty(\R^3)$, is essentially self-adjoint, that is, it has just one self-adjoint extension, whose domain is the
Sobolev space
$\mathcal{H}^2(\R^3)$ (this is known as Kato-Rellich Theorem; see ahead). However, if the origin is  removed and the
initial  domain
$C_0^\infty(\R^3\setminus\{0\})$ is considered, then also in {\sc3D} there are infinitely many self-adjoint extensions.
Note that in both {\sc1D} and {\sc2D} the origin must be removed in order to get well-defined initial operators $\dot
H$.

It  is sometimes assumed that the right potential describing the coulombian interaction is given by the fundamental
solutions of Laplace equation, that is,
\[
V_1(x) = \kappa |x|,\quad V_2(x)= \kappa \ln |x|,\quad V_3(x)=-\frac{\kappa}{|x|},
\]in {\sc 1D}, {\sc 2D}  and {\sc 3D}, respectively. For example, in the statistical mechanics of the
Coulomb gas in {\sc 2D} the potential $V_2$ is often considered,  instead of $V_3$, and the so-called Kosterlitz-Thouless
transition is  obtained. So  in Section~\ref{PotLaplacesection}  we consider the Schr\"odinger
operator with such potentials and argue  that they are always essentially self-adjoint (in suitable domains),
independently of dimension. Finally, the conclusions are reported in Section~\ref{ConclusionsSection}.

A
notational detail: the dot in $\dot H$ means that the origin has ben removed from the domain of the initial hermitian
operator; e.g., in {\sc 1D} $\dom \dot H$ is $C_0^\infty(\R\setminus\{0\})$ and so on.

\section{Self-adjoint extensions: {\sc 3D}}\label{SAEsection3D}
The initial hermitian
operator modelling the nonrelativistic quantum {\sc 3D} hydrogen atom is (write $r=|x|,\theta,\varphi$ for the
spherical coordinates)
\[
H=-\frac{\hbar^2}{2m}\Delta + V_C(x), \quad \dom H = C_0^\infty(\R^3)\subset \LL^2(\R^3),
\]which is well defined since for $\psi\in\dom H$
\[
\|V_C\psi\|^2 = \kappa^2\int_{\R^3} \frac{|\psi(x)|^2}{|x|^2} dx =\kappa^2 \int_0^\infty dr \int_{0}^{\pi} d\theta
\sin\theta
\int_0^{2\pi}d\varphi |\psi(r,\theta,\varphi)|^2 <\infty.
\]

The Kato-Rellich Theorem \cite{RS2,deOIST} applies to this case, since $V_C\in \LL^2(\R^3)+\LL^\infty(\R^3)$, and $H$
has just one self-adjoint extension  whose domain is the Sobolev space $\mathcal H^2(\R^3)$. Thus, the Schr\"{o}dinger
operator is well established in this case, so the quantum dynamics, and this is the standard operator discussed in
textbooks on quantum mechanics (usually with less mathematical details).

It is worth mentioning that in $\R^n$,
$n\ge4$, Kato-Rellich Theorem implies unique self-adjointness for potentials $V\in \LL^p(\R^n)+\LL^\infty(\R^n)$ with
$p>n/2$; so in dimensions $n\ge4$ the Schr\"odinger operators $H$ with potential $V_C$ and domain $C_0^\infty(\R^n)$
are always essentially self-adjoint.

However, in {\sc 1D} and {\sc 2D} the condition $\|V_C\psi\|^2<\infty$ for all $\psi\in C_0^\infty(\R^3)$
requires
$\psi(0)=0$ for $x$ in a neighbourhood of the origin, that is,
$\psi\in C_0^\infty(\R^n\setminus\{0\})$, $n=1,2$. The self-adjoint extensions in such cases will be discussed in other
sections; for question of comparison with other dimensions, now we consider what happens if the origin is removed also
in
$\R^3$, that is, if the initial hermitian operator is
\[
\dot H=-\frac{\hbar^2}{2m}\Delta + V_C(x), \quad \dom \dot H = C_0^\infty(\R^3\setminus\{0\}).
\]

Write $\xi=(\theta,\varphi)$ for the angular variables in the unit sphere $S^2$ and $d\xi=\sin\theta \,d\theta
d\varphi$. Let
$\mathcal D$ denote the set of linear combinations of products $f(r)w(\xi)$ with $f\in C_0^\infty(0,\infty)\subset
\LL^2((0,\infty),r^2dr)$ and
$w\in \LL^2(S^2,d\xi)$. Due to the decomposition in spherical coordinates
\[
\LL^2(\R^3) = \LL^2((0,\infty),r^2dr) \otimes \LL^2(S^2,d\xi),
\] $\mathcal D$ is a dense set in $\LL^2(\R^3)$. For functions $\phi(r,\xi)=f(r)w(\xi)\in\mathcal D$ the
operator
$\dot H$ takes the form
\[
\dot H f(r)w(\xi) = \left[-\frac{\hbar^2}{2m} \left(\partial_r^2 + \frac2r \partial_r  \right)f(r)-\frac\kappa r f(r)
\right]w(\xi) + \frac{\hbar^2}{2m}\frac{f(r)}{r^2}\mathcal Bw(\xi),
\]where $\mathcal B$ is the Laplace-Beltrami operator \cite{Msh}
\[
(\mathcal Bw)(\xi) = -\frac1{\sin\theta}\left[ \partial_\theta \left(\sin\theta \partial_\theta w  \right)
+\frac1{\sin\theta} \partial^2_\varphi w\right]
\]acting in $\LL(S^2,d\xi).$ $\mathcal B$ with domain $C_0^\infty(S^2)$ is essentially self-adjoint and its
eigenfunctions are the spherical harmonics $Y_{l,m}(\xi)$, which constitute an orthonormal basis of $\LL^2(S^2,d\xi)$;
recall that
\[
(\mathcal BY_{l,m})(\xi) = l(l+1) Y_{l,m}(\xi),\quad l\in\N,\, -l\le m\le l.
\]

Denote by $\mathcal J_l$ the subspace spanned by $\{Y_{l,m}:  -l\le m\le l \}$, that is, the subspace corresponding
to the eigenvalue $l(l+1)$ and $\LL_l :=\LL^2((0,\infty),r^2dr) \otimes \mathcal J_l$; thus
\[
\LL^2(\R^3) = \LL^2((0,\infty),r^2dr) \otimes \LL^2(S^2,d\xi) = \bigoplus_{l=0}^\infty \LL_l.
\]
If $I_l$ is the identity operator on $\mathcal J_l$, the restriction of $\dot H$ to $\mathcal D_l = \mathcal D\cap
\LL_l$ is given by $\left. \dot H\right|_{\mathcal D_l} = \dot H_l \otimes I_l,$ with
\[
\dot H_l = -\frac{\hbar^2}{2m} \left(\frac{d^2}{dr^2} + \frac2r \frac{d}{dr}
-\frac{l(l+1)}{r^2}\right)-\frac\kappa r,
\]and our task is reduced to finding the self-adjoint extensions of $\dot H_l$ with domain $C_0^\infty(0,\infty)$. It
is convenient to introduce the unitary transformation $U:\LL^2((0,\infty),r^2\,dr)\to \LL^2(0,\infty)$,
$(U\phi)(r)=r\phi(r)$, which maps $C_0^\infty(0,\infty)$ to itself and
\[
h_l:= U\dot H_l U^{-1} = -\frac{\hbar^2}{2m} \left(\frac{d^2}{dr^2} -\frac{l(l+1)}{r^2}\right)-\frac\kappa r,
\] with domain $\dom H_l = C_0^\infty(0,\infty)$. Standard arguments gives that the adjoint $h_l^*$ has the same
action as $h_l$ but with domain
\[
\dom h_l^* = \left\{\phi\in \LL^2(0,\infty): \phi,\phi'\in \mathrm{AC}(0,\infty), h_l^*\phi\in \LL^2(0,\infty)
\right\}.
\] If $\Omega$ is an open subset of $\R$, $\mathrm{AC}(\Omega)$ indicates the set of absolutely continuous functions
in every bounded and closed subinterval of $\Omega$.

By adapting the analysis  of the free hamiltonian in
$\R^3$ with the origin removed, which was performed in
\cite{AJS}, one proves the following result:

\begin{teor2}\label{teorR3partes}
$h_0$ has deficiency indices equal to 1, while $h_l$, $l\ne0$, is essentially self-adjoint.
\end{teor2}

Hence, for $l\ne0$ the unique self-adjoint extension of $h_l$ is $h_l^*$, while $h_0$ has infinitely many self-adjoint
extensions. In order to find such extensions in case $l=0$, we will make use of the following lemma
\cite{Kuras,Kuras2,Moshinsky}, whose proof we adapt and reproduce.

\begin{lema2}\label{lemah0} If $\phi\in\dom h_0^*$, then the lateral limits $\phi(0^+):=
\lim_{r\to 0^+}\phi(r)$ and
\[\tilde\phi(0^+)  := \lim_{r\to0^+} \left( \phi'(r) + \frac{2m\kappa}{\hbar^2} \phi(r) \ln (\kappa r) \right)
\]  exist  (and are finite).
\end{lema2}
\proof For $\phi\in\dom h_0^*$ one has
\[  - h_0^*\phi = \frac{\hbar^2}{2m} \frac{d^2\phi}{dr^2} +\frac\kappa r\phi:=u \in\LL^2(0,\infty),
\]and one can write $\phi = \phi_1+\phi_2$ with  $\frac{\hbar^2}{2m}\phi_1''=u$, $\phi_1(0^+)=0$ and
$\frac{\hbar^2}{2m}\phi_2''+\kappa/r \phi=0.$ Since $\phi_j\in {\mathcal H}^2(\varepsilon,\infty)$,
$j=1,2$, for all $\veps>0$, and $u \in\LL^2$, it follows that these functions are of class $C^1(0,\infty)$.

Consider an interval $[r,c]$,  $0<r<c<\infty$. Since
\[
\phi_1'(r)-\phi_1'(c) = \frac{2m}{\hbar^2}\int_r^c u(s)\,ds,
\]
$\phi_1'(r)$ has a lateral limit
\[
\phi_1'(0^+)= \phi'(c)+ \frac{2m}{\hbar^2}\int_0^c u(s)\,ds.
\] On integrating successively twice over the interval $[r,c]$ one gets
\[
\phi_2'(c) -\phi_2'(r) = -\frac{2m\kappa}{\hbar^2}\int_r^c \frac{\phi(s)}{s}\,ds,
\]and then
\[
\phi_2(r) = \phi_2(c) - (c-r)\phi_2'(c) -\frac{2m\kappa}{\hbar^2}  \int_r^c dv \int_v^c
ds\frac{\phi(s)}{s}
\]
\[
= \phi_2(c) - (c-r)\phi_2'(c) -\frac{2m\kappa}{\hbar^2}  \int_r^c ds\phi(s)\,\frac{s-r}{s},
\]and since $0\le(s-r)/s<1$, by dominate convergence the last integral converges to $\int_0^1 \phi(s)$ as $r\to
0^+$. Therefore
$\phi_2(0^+)$ exists and
\[
\phi_2(0^+)  =\phi_2(c) - c\phi_2'(c) - \frac{2m\kappa}{\hbar^2} \int_0^c\phi(s)\,ds.
\]
Now,\[
\left| \phi_2(r)-\phi_2(0^+) \right| \le r|\phi_2'(c)| + \frac{2m\kappa}{\hbar^2} \int_0^r |\phi(s)|\,ds +
\frac{2m\kappa}{\hbar^2} r \int_r^c ds\, \frac{|\phi(s)|}{s}.
\]Taking into account that $\phi$ is bounded, say $|\phi(r)|\le C$, $\forall r$, Cauchy-Schwarz in $\LL^2$
implies
\[
\int_0^r\,|\phi(s)|ds = \int_0^r\,1\,|\phi(s)|ds \le C \sqrt{r},
\] and so, for $r$ small enough and fixing $c=1$,
\[
\int_r^c ds \frac{\phi(s)}{s} \le C \left(c|\ln c|+r|\ln  r|\right)\le \tilde C \sqrt r,
\] for some constant $\tilde C$. Such inequalities imply $\phi(r) = \phi(0^+) +
O(\sqrt r)$, and on substituting this into
\[
\phi'(r) =\phi'(1) + \frac{2m\kappa}{\hbar^2}\int_r^1 \frac{\phi(s)}{s}\,ds
\] (recall that $\phi_1'(0^+)$ is finite) it is found that there is $b$ so that, as $r\to 0^+$,
\[
\phi'(r) = \phi'(1) -
\frac{2m\kappa}{\hbar^2}\phi(0^+) \ln(\kappa r) + b + o(1);
\]thus, the derivative $\phi'$ has a a logarithmic divergence as $r\to0$ and the statement in the lemma also
follows.
\endproof

 For $\phi,\psi\in \dom h_0^*$ integration by parts gives
\[
\la h_0^*\psi,\phi\ra - \la \psi,h_0^*\phi\ra =\Gamma(\psi,\phi),
\]where
\[
\Gamma(\psi,\phi):=-\frac{\hbar^2}{2m}\;\lim_{r\to 0^+}\left(\psi(r)\overline{\phi'(r)}
-\psi'(r)\overline{\phi(r)}
\right)
\] is called a {\it boundary form} for $h_0$ \cite{CRdO}. Note that although $\Gamma(\psi,\phi)$ is finite, the lateral
limit $\phi'(0^+)$ can diverge; however, by Lemma~\ref{lemah0} it is readily verified that
\[
\Gamma(\psi,\phi)=-\frac{\hbar^2}{2m}\left(\psi(0^+)\overline{\tilde\phi(0^+)}
-\tilde\psi(0^+)\overline{\phi(0)}
\right)
\]and now all lateral limits are finite. The self-adjoint extensions of $h_0$ are restrictions of $h_0^*$ to suitable
subspaces $D$ so that $\dom h_0\subset D\subset \dom h_0^*$ and $\Gamma|_D=0$, that is, $\Gamma(\psi,\phi)=0$ for all
$\phi,\psi\in D$.

We introduce the unidimensional vector spaces $X=\{\rho_1(\psi):=\psi(0^+)+i\tilde\psi(0^+): \psi\in\dom h_0^*  \}$ and
$Y=\{\rho_2(\psi):=\psi(0^+)-i\tilde\psi(0^+): \psi\in\dom h_0^*  \}$ and note that
\[
\frac{4mi}{\hbar^2} \Gamma(\psi,\phi) = \la \rho_1(\psi),\rho_1(\phi)\ra_X - \la \rho_2(\psi),\rho_2(\phi)\ra_Y ,
\]with inner products in $X,Y$, as indicated. Hence, the subspaces $D$ for which $\Gamma$ vanishes are related to maps
that preserve inner products, that is, unitary maps from $X$ to $Y$ (see details in \cite{CRdO}), and since these
vector spaces are unidimensional such maps are multiplication by the complex numbers
$e^{i\theta}$, $0\le\theta<2\pi$. Therefore, for each $\theta$ a self-adjoint extension of $h_0$ is characterized by
the functions $\psi\in\dom h_0^*$ so that $\rho_2(\psi)=e^{i\theta}\rho_1(\psi)$, that is,
\[
(1-e^{i\theta})\psi(0^+) = i (1+e^{i\theta})\tilde\psi(0^+).
\]If $\theta\ne0$ this condition reduces to
\[
\psi(0^+) = \lambda\tilde\psi(0^+),\quad \lambda = i\frac{1+e^{i\theta}}{1-e^{i\theta}}\in \R,
\]and writing $\lambda=\infty$ in case $\theta=0$, the desired self-adjoint extensions $h_0^\lambda$ are
described by
\[
\dom h_0^\lambda = \left\{\psi\in\dom h_0^*:  \psi(0^+) = \lambda\tilde\psi(0^+) \right\},\quad \lambda\in \R\cup
\{\infty\},
\]and $h_0^\lambda\psi = h_0^*\psi.$ The Dirichlet boundary condition corresponds to $\lambda=0$. With such results at
hand, we have

\begin{teor2}\label{teorExtAA3D}
The self-adjoint extensions of $\dot H$ in {\sc 3D} are
\[
H^\lambda = \left( U^{-1} h_0^\lambda U \otimes I_0 \right) \bigoplus_{l=1}^\infty \left( U^{-1} h_l^* U \otimes
I_l \right), \quad \lambda\in \R\cup \{\infty\}.
\]
\end{teor2}

This should be compared with the case without removing the origin, for which there is just one self-adjoint extension.
The eigenvalue equation for the Dirichlet case $\lambda=0$ can be exactly solved in terms of Whittaker functions, and
the negative eigenvalues are
\[
E_n^0 = -\frac{\kappa^2 m}{2\hbar^2}\frac{1}{n^2},\quad n=1,2,3,\cdots,
\]each one with multiplicity $n^2$. For $\lambda\ne0$ the manipulations become more involved and numerical procedures
must be employed to find roots of implicit functions, and so the eigenvalues.

\section{Self-adjoint extensions: {\sc 2D}}\label{SAEsection2D}
Although our main interest is in the {\sc 1D} and {\sc 3D} cases, we say something about the Coulomb system in {\sc
2D}. As already mentioned,  the origin must be excluded from the domain of the Coulomb potential in {\sc 2D} and
the initial hermitian operator is
\[
\dot H=-\frac{\hbar^2}{2m}\Delta + V_C(x), \quad \dom \dot H = C_0^\infty(\R^2\setminus\{0\}).
\] To find its self-adjoint extensions, introduce polar coordinates $(r,\varphi)$ so that
\[
\LL^2(\R^2) = \LL^2((0,\infty),rdr) \otimes \LL^2(S^1,d\varphi),
\] ($S^1$ is the usual unit circle in $\R^2$) and the set ${\mathcal D}$ of linear combinations of the products
$f(r)g(\varphi)$,
$f\in C_0^\infty(0,\infty)$ and $g\in C_0^\infty(S^1)$, is dense in $\LL^2(\R^2)$. Now
\[
\dot H f(r)g(\varphi) = \left[-\frac{\hbar^2}{2m} \left(\partial_r^2 + \frac1r \partial_r  \right)f(r)-\frac\kappa r
f(r)
\right]g(\varphi) - \frac{\hbar^2}{2m}\frac{f(r)}{r^2}\mathcal Bg(\varphi),
\]
where $\mathcal B=\partial_\varphi^2$ is the Laplace-Beltrami operator acting in $\LL^2(S^1,d\varphi)$. This operator
with domain $C_0^\infty(S^1)$ is essentially self-adjoint, its eigenvectors
$g_l(\varphi)=e^{il\varphi}/\sqrt{2\pi}$ constitute an orthonormal basis of $\LL^2(S^1,d\varphi)$ and
\[
(\mathcal Bg_l)(\varphi) = -l^2 g_l(\varphi),\quad l\in\Z.
\]

Let $[g_l]$ denote the subspace spanned by $g_l$ and $\LL_l=\LL^2((0,\infty),rdr)\otimes[g_l]$; thus
\[
\LL^2(\R^2) = \bigoplus_{l\in\Z} \LL_l,
\]and if $I_l$ is the identity operator on $[g_l]$, the restriction
of $\dot H$ to $\mathcal D_l = \mathcal D\cap
\LL_l$ is given by $\left. \dot H\right|_{\mathcal D_l} = \dot H_l \otimes I_l,$ with
\[
 \dot  H_l = -\frac{\hbar^2}{2m} \left(\partial_r^2 + \frac1r \partial_r -\frac{l^2}{r^2}
\right)-\frac\kappa r,
\] with domain $C_0^\infty(0,\infty)$, and the question is to find the self-adjoint extensions of such restrictions.
By using the unitary operator $U: \LL^2((0,\infty),rdr)\to  \LL^2(0,\infty) $, $(U\phi)(r) = r^{1/2}\phi(r)$, one has
\[
h_l:= U\dot H_lU^{-1}= -\frac{\hbar^2}{2m} \left(\partial_r^2 +  \left( \frac14-l^2
\right)\frac{1}{r^2}
\right)-\frac\kappa r
\]with $\dom h_l= C_0^\infty(0,\infty)$ (since this set is invariant under $U$). By standard results it follows
that the adjoint
$h_l^*$ has the same action as $h_l$ but with domain
\[
\dom h_l^* = \left\{\phi\in \LL^2(0,\infty): \phi,\phi'\in \mathrm{AC}(0,\infty), h_l^*\phi\in\LL^2(0,\infty)
\right\}.
\]

\begin{teor2}\label{teor2D}
The operators $h_l$ are essentially self-adjoint if, and only if, $l\ne0$, whereas $h_0$ has deficiency indices equal
to one.
\end{teor2}
\proof Weyl's limit point-limit circle criterion will be used \cite{RS2}. Thus we consider the solutions of $h_l^* \phi
= i
\phi$, that is,
\[
-\frac{\hbar^2}{2m} \phi'' -
\left[ \frac{\hbar^2}{2m} \left( \frac{1}{4} - l^2 \right)  \frac{1}{r^2}
+ \frac{\kappa}{r} + i\right] \phi = 0,\quad \phi\in \dom h_l^*.
\]
By writing $p =   \frac{2m\kappa}{\hbar^2}$,
$q =   \frac{2mi}{\hbar^2}$ and performing the change of variable
$y = (-4 q)^{1/2} r$, one gets
\[
\phi''+ \left[ \left( \frac{1}{4} - l^2 \right) \frac{1}{y^2} +
\frac{\tau}{y}-\frac{1}{4} \right] \phi = 0,
\]
with $\tau = p /(- 4 q)^{1/2}$.
This equation has two linearly independent solutions given by the Whittaker functions
\cite{Abro,Grad}
\[\phi_1 (y) = {\cal M}_{\tau, |l|}(y) \quad  \hbox{e} \quad
\phi_2 (y) = {\cal W}_{\tau, |l|}(y),
\]whose asymptotic behaviors as $|y| \to \infty$ are
\[\phi_1 (y)  \sim e^{y/2} (-y)^{-\tau}
\quad  \hbox{e} \quad
\phi_2(y) \sim  e^{-y/2} y^{\tau}.
\]

Since there is no $c \in \mathbb R$ so that $\phi_2 \in \LL^2(c, \infty)$,
it follows that $h_l$ is limit point at $\infty$. Note that the above asymptotic behaviors as $|y| \to \infty$
do not depend on  $l$; however, at the origin we need to separate the cases $l=0$ and $l \neq 0$.

For $l= 0$ \cite{Abro,Grad},
\[
\phi_1(0^+) = 0 \quad  \hbox{and} \quad  \phi_2(0^+)=0.
\]
In this case there is $c>0$ so that $\phi_1,\phi_2 \in \LL^2(0, c)$ and $h_0$ is limit circle at 0. Therefore,
$h_0$ is not essentially self-adjoint but has deficiency indices equal to 1.

For $l \neq 0$,
$\phi_1 (0^+) = 0$ while
$\phi_2 (y)$ diverges as
\[
 \sum_{k=0}^{2 |l|-1} \frac{\Gamma(2 |l| - k)}{k!}\Gamma(k - |l|- \tau + 1/2) (-y)^{-2|l|+k},
\]
for $y \to 0^+$ (here $\Gamma$ denotes the well-known Gamma function).
Therefore, $h_l$ ($l \neq 0$)  is limit point at $0$. By Weyl criterion, $h_l$ is essentially self-adjoint if $l
\neq 0$.
\endproof

Since $h_l$, $l\ne0$, is essentially self-adjoint, its unique self-adjoint extension is exactly $h_l^*$.  According to
the proof of Theorem~\ref{teor2D}, the deficiency subspace
$K_-(h_0)$ is spanned by
$\phi_-(r)={\cal M}_{\tau, 0}((-4 q)^{1/2}r)$ and the deficiency subspace $K_+(h_0)$ spanned by
$\phi_+(r)=\overline{\phi_-(r)}$. The von Neumann theory of self-adjoint extensions \cite{RS1,deOIST} characterizes
them by unitary maps from, say, $K_-$ to $K_+$, and since such subspaces are unidimensional these unitary maps are just
multiplication by $e^{i\theta}$, $0\le
\theta<2\pi$. Thus there is a self-adjoint extension $h_0^\theta$ of $h_0$, for each $\theta$, and if $\overline{h_0}$
denotes the closure of $h_0$, it is given by
\[
\dom h_0^\theta = \left\{ \psi+c \left(\phi_+ - e^{i\theta}\phi_-  \right): \psi\in\dom\overline{h_0}, \, c\in\C
\right\},
\]
\[
h_0^\theta \psi = h_0^*\psi,\quad \psi\in\dom h_0^\theta.
\] In summary:

\begin{teor2}\label{teorExtAA2D}
The self-adjoint extensions of $\dot H$ in {\sc 2D} are
\[
H^\theta = \left( U^{-1} h_0^\theta U \otimes I_0 \right) \bigoplus_{l\in\Z,l\ne0}^\infty \left( U^{-1} h_l^* U \otimes
I_l \right), \quad \theta\in [0,2\pi).
\]
\end{teor2}

\section{Hydrogen atom: {\sc 1D}}\label{SAEsection1D}
In the unidimensional case the initial hermitian operator is
\[
\dot H = -\frac{\hbar^2}{2m} \frac {d^2}{dx^2} - \frac{\kappa}{|x|},\quad \dom \dot H = C_0^\infty(\R\setminus\{0\}),
\]and the origin naturally decomposes the space
\[
\LL^2(\R\setminus\{0\}) = \LL^2(-\infty,0) \oplus \LL^2(0,\infty),
\]
and also the domain of $\dot H$ into $C_0^\infty(-\infty,0)$ and
$C_0^\infty(0,\infty)$; let $\dot H_+$ and $\dot H_-$ denote the restriction of $\dot H$ to these subspaces,
respectively. Thus, we have
\[
\dot H = \dot H_- \oplus \dot H_+.
\]From the physical point of view, an important question is about the behavior of the system at the origin, e.g., is
it impermeable, so that the system actually decomposes into a right one and a left one, or is it permeable? In the
latter possibility, what do happen with wavefunctions at the transition point (the origin)? As discussed ahead, there
are  plenty of possibilities, due to infinitely many self-adjoint extensions.

\subsection{Self-adjoint extensions}\label{subsectSae}
The adjoint operator $\dot H_+^*$ has the same action as $\dot H$ but with domain
\[
\dom \dot H_+^* = \left\{\phi\in \LL^2(0,\infty): \phi,\phi'\in \mathrm{AC}(0,\infty), \,H_+^*\phi\in  \LL^2(0,\infty)
\right\},
\]and an analogous expression for $\dot H_-^*$ and its domain.

To find the deficiency subspace
$K _+(\dot H_+)$ we look for solutions of
\[
- \frac{\hbar^2}{2m}  \phi '' + \left(- \frac{\kappa}{x} + i \right) \phi = 0
\]
that belong to $\dom \dot H_+^*$.
Write  $p =   \frac{2m \kappa}{\hbar^2}$, $q = -   \frac{2 m i}{\hbar^2}$ and perform the change of variable
$y = (-4 q)^{1/2} x$, so that this equation takes the form
\[ \phi''(y) + \left( \frac{\tau}{y} - \frac{1}{4} \right) \phi(y) = 0,
\]
with $\tau = p/(-4 q)^{1/2}$, which has exactly two linearly independent solutions \cite{Abro,Grad}
\[\phi_{1+} (y) = {\cal W}_{\tau, 1/2} (y) ,\quad
\phi_{2+} (y) = {\cal M}_{\tau, 1/2} (y).
\]
These solutions have finite limites as $y \to 0$. The asymptotic behaviors for $|y| \rightarrow \infty$ are
\[\phi_{1+} (y) \sim y^{\tau} e^{-y/2} \quad \mathrm{and}\quad
\phi_{2+} (y) \sim (-y)^{-\tau} e^{y/2}.
\]

Hence, in the original variable the deficiency subspace $K_+(\dot H_+)$
is unidimensional and spanned by $\phi_{1+}(x)= {\cal W}_{\tau, 1/2}((-4q)^{1/2}x)$. A similar analysis implies that
$K_- (\dot H_+)$ is also unidimensional and spanned by $ \overline{\phi_{1+}(x)}$. Therefore $\dot H_+$ has
both deficiency indices equal to 1.

Similarly one finds that the deficiency subspaces $K_+(\dot H_-)$ and $K_-(\dot H_-)$ of  $\dot H_-$ are spanned,
respectively, by
$\phi_{1-}(x) = {\cal W}_{\tau, 1/2} ((-4q)^{1/2}|x|)$ and
$ \overline{\phi_{1-}(x)}$.
Therefore $\dot H_-$ also has both deficiency indices equal to 1.

Now, due to the above results, the adjoint operator $\dot H^*$ has the same action as $\dot H$ but domain (write
$\mathcal H=\LL^2 ( \R\setminus \{0\})$)
\[\dom H^*  =
\left\{ \phi \in \mathcal H  : \phi, \phi' \in \mathrm{AC} (\R\setminus \{0\} ),
- \frac{\hbar^2}{2m}  \phi'' - \frac{\kappa}{|x|} \phi  \in \mathcal H \right\},
\]
and both deficiency subspaces $K_\pm (\dot H)$ have dimension $2$; $K_+ (\dot H)$  is spanned by
\[
\psi_1 (x)  = \left \{
\begin{array}{rll}
\phi_{1+}(x)   &  \mathrm{if}  &  x > 0 \\
        0                  &  \mathrm{if}  &  x < 0
\end{array} \right. \quad \mathrm{and}\quad
\psi_2 (x)  =  \left \{
\begin{array}{rll}
        0                   &  \mathrm{if}  &  x > 0 \\
\phi_{1-}(x)  &  \mathrm{if}  &  x < 0
\end{array} \right.,
\]and $K_- (\dot H)$ by $\overline{\psi_1}$ and $\overline{\psi_2}$.

Therefore,
$\dot H$ has both deficiency indices equal to $2$ and it has infinitely many self-adjoint extensions.
A boundary form will be used in order to get such extensions. The following lemma will be needed, whose proof is
similar to the proof of Lemma~\ref{lemah0} (note
the difference of signs in the definitions of $ \tilde{\phi}(0^+)$ and $ \tilde{\phi}(0^-)$ below).

\begin{lema2}\label{lila}
If $\phi \in \dom  \dot H^*$, then the lateral limits $\phi(0^\pm):=
\lim_{x\to 0^\pm}\phi(x)$ and
\[
  \tilde{\phi} (0^{\pm}) :=   \lim_{x \rightarrow 0^{\pm}}
\left( \phi'(x) \pm \frac{2m \kappa}{\hbar^2} \phi(x) \ln( \pm \kappa x) \right)
\]
exist and are finite.
\end{lema2}

\

For $\psi,\phi\in\dom \dot H^*$ one has, upon integrating by parts,
\[
\la \dot H^*\psi,\phi\ra - \la \psi,\dot H^*\phi\ra =\Gamma(\psi,\phi),
\]where
\[
-\frac{2m}{\hbar^2}\;\Gamma(\psi,\phi)=
\]
\[=\left[\lim_{x\to 0^+}\left(
\psi(x) \overline{\phi'(x)} - \psi'(x) \overline{\phi(x)}\right) + \lim_{x\to 0^-}\left(
- \psi(x) \overline{\phi'(x)} + \psi'(x) \overline{\phi(x)}\right)
\right]
\]
and, using  Lemma~\ref{lila}, straightforward computation gives
\[\Gamma (\psi, \phi)   =
\frac{-\hbar^2}{2m}\left(
\psi(0^+) \overline{\tilde{\phi}(0^+)} - \tilde{\psi}(0^+) \overline{\phi(0^+)} -
\psi(0^-) \overline{\tilde{\phi}(0^-)} + \tilde{\psi}(0^-) \overline{\phi(0^-)}
\right),\]
and now each lateral limit is finite.

Introduce two linear maps
$\rho_1, \rho_2: \dom \dot H^* \to \C^2$:
\[ \rho_1(\psi) = \left(\begin{array}{ll}
\tilde{\psi}(0^+) + i \psi (0^+)\\
\tilde{\psi}(0^-) - i \psi (0^-)
          \end{array}\right)
\quad\mathrm{and}\quad
\rho_2(\psi) = \left(\begin{array}{ll}
\tilde{\psi}(0^+) - i \psi (0^+)\\
\tilde{\psi}(0^-) + i \psi (0^-)
          \end{array}\right),
\]
so that
\[
\langle \rho_1(\psi), \rho_1(\phi) \rangle_{\C^2} -
\langle \rho_2(\psi), \rho_2(\phi) \rangle_{\C^2} =
-\frac{4m}{\hbar^2} i \,\Gamma (\psi, \phi),
\quad \forall \psi, \phi \in \dom \dot H^*.
\]
As in Section~\ref{SAEsection3D}, the self-adjoint extensions of $\dot H$ are restrictions of $\dot H^*$ to
suitable subspaces $D$ so that $\dom \dot H\subset D\subset \dom \dot H^*$ and $\Gamma|_D=0$, that is,
$\Gamma(\psi,\phi)=0$ for all
$\phi,\psi\in D$ \cite{CRdO}.

Vanishing of $\Gamma$ on domains $D$ is equivalent to the preservation of the inner products in $\C^2$,
and so it corresponds to unitary  $2 \times 2$ matrices $\hat U$, and each of such matrices characterizes a
self-adjoint extension $\dot H_{\hat U}$ of $\dot H$, so that $\dom \dot H_{\hat U}$ is composed of
$\psi \in \dom H^*$
so that $\rho_2(\psi) = \hat U \rho_1(\psi)$; also $\dot H_{\hat U}\psi = \dot H^*\psi$ for $\psi\in\dom H_{\hat U}$.

The condition $\rho_2(\psi) = \hat U \rho_1(\psi)$ is then written
\[
(I - \hat U) \left(\begin{array}{cc}
\tilde{\psi}(0^+) \\
\tilde{\psi}(0^-)
          \end{array}\right) =
- i (I + \hat U) \left(\begin{array}{cc}
-\psi(0^+) \\
\psi(0^-)
          \end{array}\right),
\]and we have explicitly got the boundary conditions characterizing the desired self-adjoint extensions.

In case $(I-\hat U)$ is invertible (similarly if $(I+\hat U)$ is invertible) it is possible to write the above boundary
conditions in the form
\[
\left(\begin{array}{cc}
\tilde{\psi}(0^+) \\
\tilde{\psi}(0^-)
          \end{array}\right)
= A
\left(\begin{array}{cc}
-\psi(0^+) \\
\psi(0^-)
          \end{array}\right),
\]
with $A = -i (I-\hat U)^{-1} (I+\hat U)$  being a $2 \times 2$ self-adjoint matrix.

In \cite{FLM} the authors have got this form for the self-adjoint extensions, and it was claimed that by allowing the
entries of $A$ taking infinity all self-adjoint extensions are found; we think it is a hard task to cover all
possibilities above (i.e., via $\hat U$) with this representation via self-adjoint matrices $A$.

Particular choices of the matrix $\hat U$ ($I$ is the identity matrix)
\[
\mathrm{a)}\;I,\qquad \mathrm{b)}\;-I,\qquad\mathrm{c)}\;\left(\begin{array}{cc}
0   &   1 \\
1   &     0
\end{array}\right),\qquad \mathrm{d)} \;\left(\begin{array}{cc}
0   &   -1 \\
-1   &     0
\end{array}\right),
\]impose, respectively, the boundary conditions: a) $\psi(0^-)=0=\psi(0^+)$ (Dirichlet); b)
$\tilde\psi(0^-)=0=\tilde\psi(0^+)$ (``Neumann''); c) $\psi(0^-)=\psi(0^+)$ and
$\tilde\psi(0^-)=\tilde\psi(0^+)$ (periodic); d)
$\psi(0^-)=-\psi(0^+)$ and $\tilde\psi(0^-)=-\tilde\psi(0^+)$ (antiperiodic).

Recall that the general form of a $2\times 2$ unitary matrix is
\[
\hat U = e^{i \theta}\left(
\begin{array}{cc}
a   &   - \overline{b} \\
b   &     \overline{a}
\end{array}\right),
\quad \theta \in [0, 2 \pi), \quad  a,b \in \mathbb C,
|a|^2+ |b|^2 = 1.
\]This form will be used ahead.

\subsection{Negative eigenvalues}\label{subsectNE1} In this subsection we discuss the negative eigenvalues of some
self-adjoint extensions $\dot H_{\hat U}$. The main goal is to remark that the eigenvalues,  their multiplicities and
parity of eigenfunctions depend on the boundary conditions. This becomes important since in the past
some authors have assumed particular hypotheses on the eigenfunctions (see references in the Introduction), but
without specifying the self-adjoint extension they were working with; this was the main source of controversies in the
studies of the unidimensional hydrogen atom.

Our first analysis is for $\hat U=I$, i.e.,   Dirichlet boundary condition,
and denote by $H_D$ the operator
\[
H_D =   - \frac{\hbar^2}{2m}  \frac{d^2}{dx^2} - \frac{\kappa}{|x|}, \quad
\dom H_D = \{\phi \in \dom \dot H^*:
\phi(0^+) = 0 = \phi(0^-) \}.
\]
As in \cite{FLM} we consider the Green function of $(H_D - E)^{-1}$, denoted by
$G(x,y)$, that is
\[
(H_D - E)^{-1} u(x)   =   \Theta(x) \int_0^x
G(x,y)  u(y) dy +  \Theta(-x) \int_x^0
G(x,y)  u(y) dy.
\] Here $\Theta(x)=1$ if $x>0$ and vanishes if $x<0$.

If $u \in\mathrm{rng} (H_D - E)$,
we search for solutions $\phi$ of
\begin{equation}\label{equackernel1d}
(H_D - E) \phi = u,
\end{equation}
 and the discussion is for $(-\infty, 0)$ and $(0, \infty)$ separately.

For $x \in (0, \infty)$, by the method of variation of parameters, one finds the solution
\[
\phi(x) =  \phi_1(x) \int_{0}^{x}  - \frac{\phi_2(y) u(y)}{W_x(\phi_1, \phi_2)} dy  +
\phi_2(x) \int_{0}^{x}  \frac{\phi_1(y) u(y)}{W_x(\phi_1, \phi_2)}  dy,
\]
where $\phi_1$ and $\phi_2$ are independent solutions of the homogeneous equation
$(H_D - E) \phi = 0$, that is,
\begin{equation}\label{autovalorE}
- \frac{h^2}{2m}  \phi '' - \left(\frac{\kappa}{x} + E \right) \phi = 0,
\end{equation}
and $W_x(\phi_1, \phi_2)$ the corresponding Wronskian.

Writing $p =  \frac{2m \kappa}{h^2}$, $q =  \frac{2 m E}{h^2}$, $\tau =  p/ (-4 q)^{1/2}$ and $z = (-4 q)^{1/2} x$,
the last equation takes the form
\[
 \phi '' + \left(\frac{\tau}{z} - \frac{1}{4} \right) \phi = 0,
\]
whose independent solutions are
\[
\phi_1 (z) = {\cal W}_{\tau, 1/2} (z) \quad  \mathrm{and} \quad
\phi_2 (z) = {\cal M}_{\tau, 1/2} (z).
\]

Recall that ${\cal W}_{\tau, 1/2} (z) \sim e^{-z/2} z^\tau$ and
${\cal M}_{\tau, 1/2} (z) \sim e^{z/2} (-z)^{-\tau}$, as $z \to \infty$.
In the original variable\[
W_x(\phi_1, \phi_2) =
- \frac{(-4 q)^{1/2}}{\Gamma(1-\tau)},
\]
 and the unique solution satisfying $\phi(0^+)=0$ is
\begin{eqnarray*}
\phi(x) =
 \int_{0}^{x} \frac{\Gamma (1-\tau)}{(-4q)^{1/2}}
&  \left({\cal W}_{\tau, 1/2}( (-4 q)^{1/2} x) \right. &
{\cal M}_{\tau, 1/2}( (-4 q)^{1/2} y) - \\
&  {\cal M}_{\tau, 1/2}( (-4 q)^{1/2} x) &  \left.{\cal W}_{\tau, 1/2}( (-4 q)^{1/2} y)
\right) u(y) dy.
\end{eqnarray*}

Similarly, for $x \in (-\infty, 0)$, the unique solution satisfying $\phi(0^-)=0$ is
\begin{eqnarray*}
\phi(x) = \int_{x}^{0} \frac{\Gamma (1-\tau)}{(-4q)^{1/2}}
&\left({\cal W}_{\tau, 1/2}( (-4 q)^{1/2} |x|) \right.  &
{\cal M}_{\tau, 1/2}( (-4 q)^{1/2} |y|) - \\
&{\cal M}_{\tau, 1/2}( (-4 q)^{1/2} |x|) & \left. {\cal W}_{\tau, 1/2}((-4 q)^{1/2}|y|)
\right) u(y) dy.
\end{eqnarray*}

Summing up, the Green function of the resolvent operator  ($H_D - E$)$^{-1}$  is given by
\[
G(x,y)  =
\Theta (xy) \frac{\Gamma (1-\tau)}{(-4q)^{1/2}}
\]
\[
\times\left[
\Theta ( |x| - |y| ) {\cal W}_{\tau, 1/2}( (-4 q)^{1/2} |x|)
{\cal M}_{\tau, 1/2}( (-4 q)^{1/2} |y|)  -
( x \leftrightarrow y) \right].
\]

The values $E$ for which ($H_D - E$)$^{-1}$  does not exist constitute the eigenvalues of
$H_D$, and they are obtained from the points for which the Gamma function $\Gamma(1 - \tau)$ is not defined, that is,
$1 -\tau$ is a negative integer number. By recalling the expressions of $p$, $q$ and
$\tau$, the condition
$1 - \tau = - n$, $n =  0,1, 2,3, \cdots$, gives
\[
E_n = - \frac{\kappa^2m}{2 \hbar^2}\, \frac{1}{n^2} \quad
n = 1, 2, 3,\cdots,
\]which coincide with the eigenvalues of the usual {\sc 3D} hydrogen atom model.

These eigenvalues are twofold degenerated and a basis $\{ \phi_{n,1}, \phi_{n,2} \}$ of the subsequent eigenspace
is
\[
\phi_{n,k}(x) = \Theta((-1)^k x) \; {\cal W}_{\tau, 1/2} ((-4q)^{1/2}|x|),
\quad  k=1,2.
\]

The negative eigenvalues of other extensions $\dot H_{\hat U}$ are harder to get and numerical computation should be
employed. In the following particular cases of interest are selected.
Let $\Psi(x) := \frac{d}{dx}(\ln{\Gamma(x)})$ and define
\[
\omega(E) := \frac{2m\kappa}{\hbar^2}\left[\ln\left(\frac{\hbar^2}{2m}\tau\right)
+ 2 \Psi(1) - \Psi(1-\tau)-1\right]-\frac{(-2Em)^{1/2}}{\hbar}
\] with  $[\Gamma(1-\tau)]^{-1}$ and $\pm \omega(E) [\Gamma(1-\tau)]^{-1}$ denoting,
respectively, the values of the lateral limits
$ \lim_{x \to 0^\pm} {\cal W}_{\tau, 1/2} ((-4q)^{1/2}|x|)$ and
\[
\lim_{x \rightarrow 0^\pm}
\left(\frac{d}{dx}{\cal W}_{\tau, 1/2}((-4q)^{1/2}|x|) \pm
\frac{2m\kappa}{\hbar^2} {\cal W}_{\tau, 1/2} ((-4q)^{1/2} |x|)
\ln( \pm \kappa x)\right).
\]
Given a unitary matrix $\hat U$, the candidates for eigenfunctions of $\dot H_{\hat U}$ must satisfy the corresponding
boundary conditions.

\begin{exem2}\label{exemU1} Let's take  $\theta = \frac{\pi}{2}$, $a=1$ and $b=0$ so that
\[
\hat U = i
\left(\begin{array}{ll}
1  &  0 \\
0  &  1
\end{array} \right),
\]
and the values of $E$ for which
$\omega (E) = -1$ are the eigenvalues of
$\dot H_{\hat U}$ and with multiplicity two; the corresponding eigenspace is spanned by
\[
\phi_k(x) = \Theta((-1)^k x) {\cal W}_{\tau, 1/2} ((-4q)^{1/2}|x|),
\quad k=1,2.
\]
\end{exem2}

\begin{exem2}\label{exemU2}
Consider another case: $\theta = \frac{\pi}{2}$, $a=i$ and $b=0$ so that
\[
\hat U = i
\left(\begin{array}{ll}
i  &  0 \\
0  &  -i
\end{array} \right)
\]
and the values of $E$ for which $\omega (E) = 0$ are found to be nondegenerate eigenvalues of
$\dot H_{\hat U}$, and for each eigenvalue the corresponding eigenspace is spanned by
\[
\phi(x) = \Theta(x) {\cal W}_{\tau, 1/2} ((-4q)^{1/2}|x|).
\]
\end{exem2}

These two cases illustrate that the behavior of eigenfunctions
are not related only to the parity of the potential $V_C(x)$, since  there are
cases for which the eigenfunctions do not have a definite parity and cases with eigenvalues simple as well as with
multiplicity two. These different possibilities are directly related to the singularity of the potential and depend
on the selected self-adjoint extension.

\subsection{Permeability of the origin}
Another question that has been discussed in the literature is about the permeability of the origin in the
unidimensional hydrogen atom; see, for instance \cite{And1,Lou,Kuras, Kuras2,Moshinsky,FLM2,RNewton}. Some authors
consider the origin an impermeable barrier, while others assume it is permeable. Again the answer strongly depends on
the self-adjoint extension considered (see also \cite{FLM2}), as illustrated ahead. Here the definition of permeability
is through the probability current density; for simplicity we assume $\hbar=1$ and $m=1$.

Recall that the probability current density $j(x)$ in {\sc 1D} is given by
\[
j(x) = \frac{i}{2} \left(\phi(x) \overline{\phi'(x)} - \phi'(x) \overline{\phi(x)} \right),
\quad \phi \in \dom \dot H_{\hat U},
\]and it satisfies
the continuity equation
\[
 \frac{\partial}{\partial t} | \phi(t,x)|^2 +
\frac{\partial}{\partial x} j(t,x) =0.
\] Our previous results in Subsection~\ref{subsectSae} show that $\lim_{x\to0^\pm}j(x)$ do exist (see ahead).

For each  $\phi \in \dom \dot H_{\hat U}$, on integrating by parts we get
\[
0=\langle \dot H_{\hat U} \phi, \phi \rangle -
\langle  \phi, \dot H_{\hat U} \phi \rangle =
i \lim_{\varepsilon \to 0}
[j(\varepsilon)- j(-\varepsilon)].
\]
Hence, the function $j(x)$ can be continuously defined at the origin $j(0)$ via lateral limits.
Physically this relation means that the current density is isotropic at the origin, in spite of the strong
singularity there. A simple observation shows that
\[
j(x) = \im (\phi'(x) \overline{\phi(x)}),
\quad \phi \in \dom \dot H_{\hat U},
\] where $\im$ indicates imaginary part, so that
\[
j(0) =   \lim_{x \to 0^+} \im (\phi'(x) \overline{\phi(x)})
       =   \lim_{x \to 0^-} \im (\phi'(x)\overline{\phi(x)}).
\]

Since $\phi'(0^+)$ and $\phi'(0^-)$ can be divergent, we use using Lemma~\ref{lila} to obtain
\[
j(0)  =  \lim_{x \to 0^+} \im (\tilde{\phi}(x) \overline{\phi(x)} )
        =  \lim_{x \to 0^-} \im (\tilde{\phi}(x) \overline{\phi(x)}).
\]
Note that it is exactly this relation that guaranties that $j(0)$ is well defined and finite.

We are now in position of giving a rigorous definition of permeability: If  $j(0)=0$, $\forall \phi \in \dom \dot
H_{\hat U}$, the electron is completely reflected when approaching the origin, and so we say the origin is not
permeable (or is impermeable), so that the regions $x<0$ and
$x>0$ are kept separated by the singularity. If
$j(0)\ne0$ we say the origin is permeable.

Next we study the current density related to $\dot H_{\hat U}$ in three cases.

\

\noindent
{\bf Case 1.}
$(I-\hat U)$ is invertible.  Since $A = -i(I-\hat U)^{-1}(I+\hat
U)$ is a self-adjoint matrix, the boundary conditions of $\dot H_{\hat U}$ become
\[
\left(\begin{array}{cc}
\tilde{\phi}(0^+) \\
\tilde{\phi}(0^-)
          \end{array}\right)
=
\left(\begin{array}{cc}
u   &     z  \\
\overline{z}   &    v
\end{array}\right)
\left(\begin{array}{cc}
-\phi(0^+) \\
\phi(0^-)
          \end{array}\right),
\quad  u, v \in \R, z \in \C,
\]
and $u$, $z$ and $v$ are functions of the entries of  $\hat U$.
The boundary conditions become
\begin{eqnarray}\label{denscorrecapar}
\lim_{x \to 0^+} \tilde{\phi}(x) & = & - u \lim_{x \to 0^+} \phi(x) +
z \lim_{x \to 0^-} \phi(x) \\
\lim_{x \to 0^-} \tilde{\phi}(x) & = & - \overline{z} \lim_{x \to 0^+} \phi(x) +
v \lim_{x \to 0^-} \phi(x)
\end{eqnarray}

Multiply the first equation (before taking limits) by
$ \overline{\phi(x)}$ to get
\[
\lim_{x \to 0^+} \tilde{\phi}(x) \overline{\phi(x)}  =
-u \lim_{x \to 0^+} | \phi(x) |^2
+ z \lim_{x \to 0^+} \phi(-x) \overline{\phi(x)},
\]
and so
\[
j(0) = \lim_{x \to 0^+} {\rm Im}( \tilde{\phi}(x) \overline{\phi(x)}) =
\lim_{x \to 0^+} {\rm Im} (z  \phi(-x) \overline{\phi(x)}).
\]

Therefore, if $z = 0$ then $j(0)=0$,
$\forall \phi \in \dom \dot H_{\hat U}$, and we have found a family of self-adjoint extensions for which the origin is
not permeable. Example~\ref{exemU1} above corresponds to the self-adjoint matrix
\[
-i(I-\hat U)^{-1}(I+\hat U) =
\left(\begin{array}{cc}
1   &     0  \\
0   &     1
\end{array}\right),
\]
and the origin is not permeable in this case.

\

\noindent
{\bf Case 2.} $(I+\hat U)$ is invertible. The matrix
$A = i (I+\hat U)^{-1}(I-\hat U)$ is also self-adjoint and the boundary conditions of $\dot H_{\hat U}$ take the form
\[
\left(\begin{array}{cc}
u   &     z  \\
\overline{z}   &    v
\end{array}\right)
\left(\begin{array}{cc}
\tilde{\phi}(0^+) \\
\tilde{\phi}(0^-)
          \end{array}\right)
=
\left(\begin{array}{cc}
-\phi(0^+) \\
\phi(0^-)
          \end{array}\right),
\quad  u,v \in \mathbb R, z \in \mathbb C,
\]
and $u$, $z$ and $v$ are functions of the entries of $\hat U$. Table~1 shows the current density at the origin for
various values of $u$, $z$ and $v$.

\begin{center}
\def\tablename{Table}\label{tabelinha}
\begin{table}[htb!]
\caption{Current density for invertible $(I+\hat U)$.} \vspace*{0.1cm}
\centering
\begin{tabular}{|c|c|c|} \hline
$u$, $z$, $v$ $\neq 0$        &
$j(0) =  {\rm Im} \left(- \frac{z}{u}
\lim_{x \to 0^-}  \tilde{\phi}(x) \overline{\phi(-x)}  \right)$
\\ \hline
$z = 0$               &
j(0) = 0
\\ \hline
$z \neq 0$ and $u = 0$               &
$j(0) =  {\rm Im} \left(- \frac{1}{z}
\lim_{x \to 0^+}  \phi(x) \overline{\phi(-x)} \right)$
\\ \hline
$z \neq 0$ and $v = 0$               &
$j(0) = {\rm Im} \left( \frac{1}{\overline{z}}
\lim_{x \to 0^-}  \phi(x) \overline{\phi(-x)} \right)$
\\ \hline \hline
\end{tabular}
\end{table}
\end{center}

Note that Dirichlet boundary condition (so $\hat U=I$) is a particular case with $z=0$ (the matrix $A=0$), and since
this case can not be an extension of another self-adjoint extension of $\dot H$,  we conclude that the
current density vanishes for all
$\phi\in \dom \dot H_{\hat U}$ if, and only if, $z = 0$. In other words, if $(I+\hat U)$ is invertible, the origin is
impermeable precisely if $z=0$. As expected, Dirichlet boundary condition implies the origin is impermeable.

\

\noindent
{\bf Case 3.} Both $(I+\hat U)$ and $(I-\hat U)$ are not invertible. This case amounts to
\[
\det(I+\hat U)=0=\det (I-\hat U),
\]
which turns out to be equivalent to the following matrix representation
\[
\hat U = \left(
\begin{array}{cc}
-u &  v \\
\overline{v} &  u
\end{array}\right), \quad
u \in \mathbb R, v \in \mathbb C, \quad
|u|^2+|v|^2 = 1.
\]
The current density always vanishes at the  origin if, and only if, $v=0$, that is, the matrix $\hat U$ equals
\[
\left(
\begin{array}{cc}
-1 &  0 \\
0 &  1
\end{array}\right) \quad  \hbox{or}  \quad
\left(
\begin{array}{cc}
1 &  0 \\
0 &  -1
\end{array}\right).
\]

If $v \neq 0$, we have
\[
j(0) = {\rm Im}  \left(\frac{v}{1+u}\lim_{x \to 0^-}  \tilde{\phi}(x) \overline{\phi(-x)}  -
\frac{i v}{(1+u)} \lim_{x \to 0^-}  \phi(x) \overline{\phi(-x)} +
i \frac{1-u}{1+u} \lim_{x \to 0^+} | \phi(x)|^2 \right).
\]

\

\begin{exem2}\label{exemiiej}{\rm
Consider the self-adjoint extension $\dot H_{\hat U}$ which corresponds to the unitary matrix
\[
\hat U = i
\left(
\begin{array}{ccc}
\sqrt2 /2  & -\sqrt2 /2  \\
\sqrt2 /2  & \sqrt2 /2
\end{array}\right).
\]
The domain of $\dot H_{\hat U}$ constitutes of the  $\psi \in \dom \dot H^*$ so that
\[
\left(\begin{array}{cc}
\tilde{\psi}(0^+) \\
\tilde{\psi}(0^-)
          \end{array}\right)
=
\left(\begin{array}{cc}
\sqrt 2  &     -i  \\
i        &   \sqrt 2
\end{array}\right)
\left(\begin{array}{cc}
-\psi(0^+) \\
\psi(0^-)
          \end{array}\right),
\]
and such conditions imply
\[
j(0) =   \lim_{x \to 0^+} \im (-i \psi(-x) \overline{\psi(x)}).
\]
The values of $E$ for which
$\omega(E)= - \sqrt 2$ are eigenvalues of $\dot H_{\hat U}$ of multiplicity two, and the corresponding eigenfunctions
are
\[
\phi_k(x) = \Theta((-1)^k x) {\cal W}_{\tau, 1/2} ((-4q)^{1/2}|x|),
\quad k=1,2.
\]
By taking the linear combination $\psi(x) = \phi_1(x) + \phi_2(x)$, and the asymptotic behavior of such
eigenfunctions near zero, discussed in Subsection~\ref{subsectNE1},
we obtain
\[
j(0) =   -\Gamma(1 - \tau)^{-2} \neq 0,
\]
that is,  if the electron is in this eigenstate it is transmitted through the origin.
}
\end{exem2}

We conclude that there are extensions for which the origin is permeable and for others it is impermeable. Andrews
\cite{And1} defines
$j(x) = i[\overline{F'(x)}F(x) - \overline{F(x)} F'(x)]$ but computes $j(0)$ only for eigenfunctions; it is clear that
$j$ vanishes if the eigenvalue is nondegenerate, since the corresponding eigenfunction can be taken real. Andrews
mentioned the possibility of zero current in case of degenerated eigenvalues. In our Example~\ref{exemU1} the
eigenvalues have multiplicity two and the current density is zero, whereas for the operator in Example~\ref{exemiiej}
the eigenvalues have multiplicity two and the origin is permeable; therefore both possibilities are allowed in case of
multiple eigenvalues.

We note that the  analysis of Moshinsky
\cite{Moshinsky}, although interesting, considers the ``eigenfunctions''  ${\cal W}_{\lambda}(-z)$ and ${\cal
W}_{\lambda}(z)$  that do no belong to $L^2(\mathbb R \setminus\{0\})$.

\section{Potentials Via Laplace Equation}\label{PotLaplacesection}
This brief section deals with  Schr\"odinger operators with potentials $V$ along the fundamental solutions of Laplace
equation
\[
\Delta V =0.
\]

As mentioned in the Introduction, in physics sometimes one assumes that in each dimension the potential describing the
Coulomb interaction is the fundamental solutions of Laplace equation \cite{Barton,Vladimirov};
recall that these solutions are (take $\kappa>0$)
\[
V_1(x) = \kappa |x|,\quad V_2(x)= \kappa \ln |x|,\quad V_3(x)=-\frac{\kappa}{|x|},
\]in {\sc 1D}, {\sc 2D}  and {\sc 3D}, respectively.

The case of $V_3$ in {\sc 3D} is standard and was recalled in Section~\ref{SAEsection3D}; we underline that the
operator
\[
{H}  =  - \frac{\hbar^2}{2m}  \Delta  + V_3(x), \quad \dom {H} = C_0^\infty(\mathbb R^3),
\]
is essentially self-adjoint and its unique self-adjoint extension has the same action but domain $\mathcal
H^2(\R^3)$; this extension has nonempty discrete and essential spectra.

The  case of potential $V_2$ was analyzed by Gesztesy and Pittner \cite{GesztesyPittner} and they state the following
result:

\begin{teor2}
The operator
\[
H = -\frac{\hbar^2}{2m} \Delta + \kappa \ln|x|, \quad
\dom H = C_0^\infty(\mathbb R^2),
\]
is essentially self-adjoint, and its unique self-adjoint extension has empty essential spectrum.
\end{teor2}

Now we consider the unidimensional Schr\"odinger operator
\[
H  = - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \kappa |x|, \quad
\dom H = C_0^\infty (\mathbb R),
\]
whose adjoint $H^*$ has the same action as $H$ but domain
\[
\dom H^* = \left\{ \phi \in \LL^2(\R): \phi, \phi' \in \mathrm{AC}(\mathbb R), H^* \phi \in \LL^2(\mathbb R) \right\}.
\]Since $V_1$ is a bounded from below and continuous potential, with
\[
\lim_{|x|\to\infty} V_1(x)=\infty,
\] the
following theorem follows from general results \cite{RS2,deOIST}.

\begin{teor2}\label{liinderedsi}
The above unidimensional operator $H$ is essentially self-adjoint, its unique self-adjoint extension $H^*$ is bounded
from below and has empty essential spectrum.
\end{teor2}

By solving the eigenvalue equation
\[
-\frac{\hbar^2}{2m} \phi'' + \left( \kappa |x| - E \right) \phi = 0
\]in terms of Airy functions we have found the eigenvalues $E_n$ are simple and with asymptotic behavior
\[
E_n\sim \frac{\hbar^2}{2m}
\left[ \frac{m\kappa}{\hbar^2} \frac{3\pi}{4} (4n-3) \right]^{2/3},
\quad n \to \infty.
\]

Hence, for  Schr\"odinger operators $H$ with potentials along the fundamental solutions of the Laplace equation, we
have:

\begin{enumerate}
\item $H$ is essentially self-adjoint in  $C_0^\infty(\mathbb R^3)$ and its self-adjoint extension
has both nonempty discrete and essential spectra.

\item
 $H$ is essentially self-adjoint in $C_0^\infty(\mathbb R^n)$ and its self-adjoint extension has purely discrete
spectrum for $n=1,2$.
\end{enumerate}

However, for Schr\"odinger operators whose potential is the Coulomb one, i.e., $V_C(x)$,
we have:

\begin{enumerate}
\item
 The deficiency indices are equal to 0 in $C_ 0^\infty(\mathbb R^3)$.

\item
 The deficiency indices are equal to 1, 1 and 2 in $C_ 0^\infty(\mathbb R^3 \setminus \{0\})$, $C_ 0^\infty(\mathbb R^2
\setminus \{0\})$ and $C_ 0^\infty(\mathbb R \setminus \{0\})$, respectively.
\end{enumerate}

\section{Conclusions}\label{ConclusionsSection}

Although the {\sc3D} usual model hamiltonian $H$ with Coulomb potential $V_C$, $\dom H=C_0^\infty(\R^3)$, is essentially
self-adjoint, in $\R^n$ the $1/|x|$ singularity imposes the initial domain must be $C_0^\infty(\R^n\setminus\{0\})$,
$n=1,2$, and the corresponding operators $\dot H$ have deficiency indices equal to $2$ an $1$, respectively; hence with
infinitely many self-adjoint extensions. For the sake of comparison, we have also considered the origin removed in
$\R^3$, that is,
$\dot H$ with domain $C_0^\infty(\R^3\setminus\{0\})$: the deficiency indices are equal to 1 in this case. In each
case, all self-adjoint extensions have been found.

In {\sc1D} the question of permeability of the origin was analyzed and the answer depends strongly on the
self-adjoint extension considered. Due to particular examples discussed, we conclude that the multiplicity two of the
eigenvalues does not determine the permeability.

We have paid particular attention to the {\sc 1D} case, since there are many papers in the literature about this
model and occasionally with conflicting conclusions. We have found that these conflicting positions have been
originated from boundary conditions imposed mainly on ``physical basis'' that can fail for strong singularities, as is
the case of $V_C$ in one-dimension. We expect to have clarified the situation, and the next step could be
presenting  arguments to select the extension(s) to be considered natural, with the consequent implications as, for
instance, the  permeability of the origin.

Finally, we have found remarkable that, for potentials in $\R^n$, $n=1,2,3,$ given by
fundamental solutions of Laplace equation, the corresponding initial hermitian operators with domain
$C_0^\infty(\R^n)$ are always essentially self-adjoint.

 \subsubsection*{Acknowledgments} {\small AAV was  supported by CAPES (Brazil). CRdO acknowledges partial
support from CNPq (Brazil).}

\

config: arxiv


TITLE: Vectors transforming under change of coordinates
QUESTION [2 upvotes]: I was watching a lecture on tensors and the professor said that a defining feature of a vector $v$ is that it transforms under a coordinate transformation $x^{\mu} \rightarrow x^{\mu'}$ as
$$v^{\mu'}(x^{\mu'}) \equiv \frac{\partial x^{\mu'} }{\partial x^\mu} v^{\mu}(x^{\mu}(x^{\mu'})) $$
Basically the last last term on the right hand side is the $x^{\mu}$ coordinates expressed in terms of $x^{\mu'}$ coordinates.
I am trying to understand this in $\mathbb R^2$ for the Cartesian and Polar coordinates. If the $x^{\mu'} \equiv (x,y)$ and  $x^{\mu} \equiv (r,\theta)$, then I get
$$\bigg(\begin{matrix}x\\y \end{matrix}\bigg) \equiv
\bigg[
\begin{matrix}
\cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta
\end{matrix}\bigg] \bigg(\begin{matrix}r\\\theta \end{matrix}\bigg)$$
which doesn't work out to be true. Can you please tell me what I'm missing here?

REPLY [2 votes]: This transformation equation applies to a vector $v$ that "lives at" at the point $x$ (i.e., is in the tangent space at $x$). The point $x$ is described by the coordinates $(x, y)$ or $(r, \theta)$, but these coordinate tuples are not vectors at $x$. Examples of vectors at $x$ are $(dx, dy)$ and $(dr, d\theta)$, and your matrix equation applies to them:
$$\bigg(\begin{matrix}dx\\dy \end{matrix}\bigg) =
\bigg[
\begin{matrix}
\cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta
\end{matrix}\bigg] \bigg(\begin{matrix}dr\\d\theta \end{matrix}\bigg)$$


TITLE: Infinite Real Symmetric Toeplitz Matrix Reference
QUESTION [1 upvotes]: I am looking for a good starting point (book or articles) for studying Toeplitz matrices.  Specifically as mentioned in the title, I am most interested in the case where they are of the form
$$A = \{\phi(i-j)\}_{i,j\in\mathbb{Z}}$$
where $\phi:\mathbb{R}\to\mathbb{R}$ and $\phi(i-j)=\phi(j-i)$.
I so far have looked at "Analysis of Toeplitz Operators" by Bottcher and Silbermann, but wonder if there might be some more references that address my specific interest, and if anything can be said about the explicit form of the inverse.
I apologize if this has been answered before, but I did not find much in searching old posts.

REPLY [0 votes]: I strongly recommend the following book
Spectra and Pseudospectra: The Behavior of Nonnormal Matrices and Operators
Lloyd N. Trefethen & Mark Embree
The first chapter is partly devoted to Toeplitz matrices, although their interest is focused on the non-selfadjoint case. Anyhow, the book is pleasant to read and very informative.


TITLE: Prove area under curve is between two values
QUESTION [2 upvotes]: The question reads:
Prove that $$0 \leq \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{\sin(x)}{x} \, dx \leq \frac{\sqrt2}{2} \ .$$
I know that I should apply $$ m(b-a) \leq \int_a^b f(x) \, dx \leq M(b-a)$$
where $ m \leq f(x) \leq M $ on $[a, b]$.
If this where simply $\sin(x)$ it would easier to evaluate, but dividing by the x term is throwing me off. Any suggestions on where to start?

REPLY [1 votes]: Hint 1:
$$f(x)=\dfrac{\sin{x}}{x}\le f(\dfrac{\pi}{4})$$
because $f'(x)<0,x\in (\pi/4,\pi/2)$
Hint 2
Use Cauchy-Schwarz inequality
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\dfrac{\sin{x}}{x}dx\le\sqrt{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin^2{x}dx\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{x^2}dx}=\sqrt{\dfrac{\pi+2}{4\pi}}<\dfrac{\sqrt{2}}{2}$$


TITLE: Proving continuity on spaces of distributions?
QUESTION [6 upvotes]: Let $\mathcal{D}'(\Omega)$ be the space of distributions on an open set $\Omega$, and $\mathcal{E}'(\Omega)$ the compactly supported ones.
When you have a linear operator $T:\mathcal{D}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ or $T:\mathcal{E}'(\Omega)\rightarrow\mathcal{E}'(\Omega)$, I often find it easy to prove that convergent sequences get mapped to convergent sequences (in the weak topology), but proving continuity with respect to the weak topology is much harder. Since these spaces are not first-countable (I think), one must work with a local basis around 0.
For example, if you have a pseudodifferential operator $P:\mathcal{E}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ which is properly supported, then it continuously extends to a (necessarily unique) operator $P:\mathcal{D}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ and maps $\mathcal{E}'(\Omega)$ continuously into $\mathcal{E}'(\Omega)$. I was able to prove the first statement, but I was able to prove the second statement only in terms of sequential continuity. Note that the weak topology of $\mathcal{E}'(\Omega)$ is not the subspace topology that it inherits from the weak topology of $\mathcal{D}'(\Omega)$ (I think), so this is not trivial.
In many textbooks involving distributions, authors don't seem to be careful about this. They give an argument proving sequential continuity and claim that it is continuous in the more general sense. Is there a general scheme for converting sequential continuity arguments to actual continuity arguments for these spaces? Or can you point me to a detailed proof of the aforementioned fact about proper PDOs?

REPLY [4 votes]: Every linear mapping $T:\mathcal D'\to \mathcal D'$ which is $\sigma(\mathcal D',\mathcal D)$-sequentially continuous is $\sigma(\mathcal D',\mathcal D)$-continuous.
Perhaps, this is the same as Peter Michor's answer, but let me try to explain that the major point is the bornologicity of $\mathcal D'$ endowed with the strong topology
$\beta (\mathcal D',\mathcal D)$ (uniform convergence on the bounded subsets of $\mathcal D$) which follows e.g. from a beautiful theorem of Laurent Schwartz (the strong dual a every complete Schwartz space is bornological).
Moreover, since $\mathcal D$ is barrelled, the strong and the weak-$*$ dual have the same bounded sets.
If now $T: (\mathcal D',\sigma(\mathcal D',\mathcal D)) \to (\mathcal D',\sigma(\mathcal D',\mathcal D))$ is sequentially continuous then every bounded sequence of $\mathcal D'$ is mapped to a bounded sequence and hence $T$ is bounded on bounded sets
of $\mathcal D'$. As $(\mathcal D',\beta(\mathcal D',\mathcal D))$ is bornological
$T:(\mathcal D',\beta(\mathcal D',\mathcal D)) \to (\mathcal D',\beta(\mathcal D',\mathcal D))$ is continuous and therefore
$T:(\mathcal D',\sigma(\mathcal D',\mathcal D''))\to (\mathcal D',\sigma(\mathcal D',\mathcal D''))$ is continuous. By reflexivity, you get the desired weak-$*$ continuity.
In general, you are perfectly right with your skepticism against careless arguments
involving sequential continuity. For example, it may very well be that for a linear partial differential operator with constant coefficients $T=P(D): \mathcal D'(\Omega) \to \mathcal D'(\Omega)$ the inverse of $P(-D): \mathcal D(\Omega) \to \mathrm{Im} (P(-D))$ is sequentially continuous without being continuous (the former condition is equivalent to $P$-convexity, the latter to strong $P$-convexity)


TITLE: A generalization of integral Poincaré duality
QUESTION [5 upvotes]: In this paper, Felix, Halperin and Thomas define the notion of a Gorenstein space over a field $\mathbb{k}$:

An augmented differential graded algebra $R$ over $\mathbb{k}$ is Gorenstein if $\text{Ext}_R(\mathbb{k},R)$ is concentrated in a single degree and has $\mathbb{k}$-dimension one.
$X$ is Gorenstein over $\mathbb{k}$ if the cochain algebra
$C^*(X,\mathbb{k})$ is Gorenstein.

This definition is motivated by their subsequent results on this being a generalization of the notion of a Poincaré duality space.
Does there exist a parallel notion of a Gorenstein space over $\mathbb{Z}$ which similarly generalizes Poincaré duality over $\mathbb{Z}$?
EDIT: Alternatively, is it thought that no such generalization is available, so that one needs to use the machinery of  symmetric spectra to get such a generalization over $\mathbb{Z}$?

REPLY [3 votes]: Prior to Dwyer-Greenlees-Iyengar, Dwyer and myself (independently) defined Gorenstein conditions for group rings over the sphere $S[G]$, i.e., the suspension spectrum of a topological group.
The definition easily extends to the case of a morphism $R\to k$.
In the orientable case the definition goes like this:
$R\to k$ is said to be Gorenstein of dimension $d$ if:
1) $k$ is finitely dominated as an $R$-module, i.e., $k$ is a retract up to homotopy of a finite $R$-module (this finiteness condition shouldn't be ignored!), and
2) the derived mapping spectrum $\hom_R(k,R)$ is weakly equivalent as an $R$-module to $k[-d] := \Sigma^{-d}k$.
If one wishes to have an unoriented Gorenstein condition, then it seems to me one needs to replace $k[-d]$ by a twisted version of it. In the special case when $R$ is an augmented $k$-algebra spectrum, we can simply require that the $k$-module $\hom_R(k,R)$ is equivalent to $k[-d]$ as a $k$-module (but not necessarily as an $R$-module).
If $R$ is isn't a $k$-algebra, we can fix another $R$-module structure on $k$, call it $k^\xi$, and require that $\hom_R(k,R)$ is
equivalent to $k^\xi[-d]$ as an $R$-module.
Returning the the group ring case $S[G]$, we have:
Theorem. Assume in addition $\pi_0(G)$ is finitely presented.  Then
following are equivalent:
1) $S[G] \to S$ is Gorenstein in dimension $d$.
2) $BG$ is a (finitely dominated) Poincaré duality space in dimenson $d$.
And yes, there is a parallel story over $\Bbb Z$, but it is enough to work with dgas instead of ring spectra (for example, the derived $\hom_{R}(\Bbb Z,R)$ is a differential graded module for a discrete ring homomorphism $R\to \Bbb Z$).


TITLE: Different definitions of conditional expectation
QUESTION [4 upvotes]: As I'm trying to learn more about conditional expectation, I noticed that there are different definitions for that, depending on the book. For the following, let $(\Omega,\mathcal{F},P)$ be a measure space.
Definition 1:

Let $(X,Y)$ be a two-dimensional random variable on $\Omega$ with a joint distribution function $f_{X,Y}(x,y)$. The $(X=x)$-conditional expectation of $Y$ is given by
$$E(Y\mid X=x):=\int_{-\infty}^{\infty} yf_{Y\mid X=x}(y)dy$$
with
$$f_{Y\mid X=x}(y):=\frac{f_{X,Y}(x,y)}{f_X(x)}.$$
($f_{Y\mid X=x}(y)$ is also called conditional density function)

Definition 2:

Let $X$ be a random variable on $\Omega$ and $\mathcal{G}$ a sub $\sigma$-field of $\mathcal{F}$. A $\mathcal{G}$-measurable random variable $U$ with
$$\int_G U \ dP = \int_G X \ dP$$
for all $G \in \mathcal{G}$ is called $\mathcal{G}$-conditional expectation of $X$.

Intuitively, both definitions make sense to me. But how does one show that the definitions are equivalent?
Thank you in advance.

REPLY [3 votes]: Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $(X,Y)$ be random vector with probability density function $g_{(X,Y)}$. Finally, let $f$ be any borel function, such that $\mathbb E[|f(X,Y)|] < \infty$. Then it holds: $\mathbb E[f(X,Y)|Y] = h(Y)$, where:
$$ h(y) = \frac{\int_{\mathbb R} f(x,y)g_{(X,Y)}(x,y)dx}{\int_{\mathbb R} g_{(X,Y)}(x,y)dx} $$when $\int_{\mathbb R} g_{(X,Y)}(x,y)dx \neq 0$, and $h(y) = 0$ otherwise.
Firstly, we can put $0$ in the second case, because the set $S=\{ \omega \in \Omega : \int_{\mathbb R} g_{(X,Y)}(x,Y(\omega)) dx = 0 \}$ has measure $0$. Clearly $\mathbb P(S) = \mathbb P(Y \in S_Y)$, where $S_Y = \{ y \in \mathbb R: \int_{\mathbb R} g_{(X,Y)}(x,y)dx = 0 \}$. Then $\mathbb P(Y \in S_Y) = \int_{S_Y} g_Y(y) dy $, where $g_Y$ is marginal density (It exists due to Fubini + existence of joint density of rv $(X,Y)$ ). But note $g_Y(y) = \int_{\mathbb R} g_{(X,Y)}(x,y)dx $, so we're just integrating $0$ function (cause we on $S_Y$ where it's $0$), so $\mathbb P(S) = 0$. This + the fact that Conditional Expectation is up to the set of measure $0$ allows us to forget about the case when $g_Y(y) = 0$.
So, we have to prove $2$ things:
1) $h(Y)$ is $\sigma(Y)$ measurable. Clearly both $\int_{\mathbb R} g_{(X,Y)}(x,Y) dx$ and $\int_{\mathbb R} g_{(X,Y)}(x,Y) f(x,Y) dx$ are $\sigma(Y)$ measurable due to Fubinii theorem (integrals of $\sigma(Y) -$ measurable functions are $\sigma(Y)$ measurable (We here used the fact that $g_{(X,Y)}$ is bounded and $\mathbb E[f(X,Y)]$ is finite to be able to apply Fubini's theorem.
2) For any $A \in \sigma(Y)$ we have to show $\int_A f(X,Y) d\mathbb P = \int_A h(Y) d\mathbb P$. Note that $A$ is of the form $Y^{-1}(B)$ where $B \in \mathcal B(\mathbb R)$ (borel set).
Note that $\int_A f(X,Y) d\mathbb P = \mathbb E[ f(X,Y) \cdot \chi_{_{Y \in B}} ]$ and $\int_A h(Y) d\mathbb P = \mathbb E[ h(Y) \cdot \chi_{_{Y \in B}} ]$
We'll use the fact, that if random variable/vector (in $\mathbb R^n$) $V$ has density function $g_V$, then for any borel function $\phi: \mathbb R^n \to \mathbb R^n$, we have $\mathbb E[\phi(V)] = \int_{\mathbb R^n} \phi(v) g_V(v) d\lambda_n(v)$.
Then: $$\mathbb E[ f(X,Y) \cdot \chi_{_{Y \in B}} ] = \int_{\mathbb R^2} f(x,y)\chi_{_{B}} g_{(X,Y)}(x,y) d\lambda_2(x,y) = \int_{B} \int_{\mathbb R} f(x,y)g_{(X,Y)}(x,y)dxdy $$
That last split of integrals due to fubinii (function is integrable due to our assumption with $f$ ).
And now similarly at the beggining:
$$ \mathbb E[ h(Y) \cdot \chi_{_{Y \in B}} ] = \int_{B} h(y) (\int_{\mathbb R} g_{(X,Y)}(x,y)dx)dy$$
Now due to our assumption of $h$ (that is getting rid of that case when denominator is $0$ due to its being $0$-measurable set). We have:
$$ \int_{B} (h(y)) (\int_{\mathbb R} g_{(X,Y)}(x,y)dx) dy = \int_{B} (\frac{\int_{\mathbb R} g_{(X,Y)}(x,y)f(x,y)dx}{\int_{\mathbb R} g_{(X,Y)}(x,y)dx}) (\int_{\mathbb R} g_{(X,Y)}(x,y)dx )dy$$
After simplification we get $\mathbb E[ h(Y) \cdot \chi_{_{Y \in B}} ] = \int_{B} \int_{\mathbb R} f(x,y)g_{(X,Y)}(x,y)dxdy = \mathbb E[ f(X,Y) \cdot \chi_{_{Y \in B}} ]$, what we wanted to prove.
Now your "definition $1$" follows when you take $f(x,y) = x$. Then $ h(y) = \mathbb E[X|Y=y] $


TITLE: Finding the "square root" of a permutation
QUESTION [0 upvotes]: Suppose $r$ is odd, and ${\rm ord}(\alpha)=r$. ($\alpha,\beta$ are cycles.)
Now, $\alpha=(a_1\cdots a_n)$.
I need to find $\beta$ that will make $$\alpha = \beta^2$$
How can I show it?

What I tried?
${\rm ord}(\alpha)=r$ so $a^{r+1}=\alpha$.
Let say that $\beta=\alpha^{\frac{r+1}{2}}$, $\frac{r+1}{2}$ is an integer, how do I prove that $\alpha^\frac{r+1}{2}$ is a cycle?
Because then $\alpha=\beta^2$...
Thank you!

REPLY [2 votes]: Let $\alpha$ be the cycle given by:
$$(a_1~a_2~\ldots~a_n)$$
We are furthermore given that $r$ is the smallest natural number such that $\alpha^r = ()$.
Using the cycle notation, we easily determine that $\alpha^k(a_1) = a_{k+1}$ for any $k$ (where $a_{n+1} = a_1$ and so on) -- similarly for the other entries. It follows that $r = n$.

Now to find out why $\beta = \alpha^{\frac{r+1}2}$ is also a cycle. Let us contemplate what it means for $\beta$ to consist of shorter cycles.
Recall or observe that:

The order of a cycle is the length of that cycle;
Disjoint cycles commute.

Therefore, $\beta$ will consist of one $r$-cycle if and only if there are no $a_i$ and $k < r$ such that $\beta^k(a_i) = a_i$. (For, otherwise, $(a_i~\beta(a_i)~\beta^2(a_i)~\beta^{k-1}(a_i))$ would be a subcycle of the permutation $\beta$. Conversely, any subcycle is of that form.)

Aiming for a contradiction, suppose $\beta^k(a_i) = a_i$ for some $0 < k < r$ and $i$.
By definition of $\beta$ and $r = {\rm ord}(\alpha)$, we know that$\beta^k = \alpha^{k\frac{r+1}2}$ leaves some $a_i$ fixed if and only if $r \mid k \frac{r+1}2$ (since $\alpha$ is a cycle).
But $r$ and $\frac{r+1}2$ are coprime. For, suppose that, for some prime $p$, $p \mid r$ and $p \mid \frac{r+1}2$. Then $p \mid 2(\frac{r+1}2)-r = 1$, a contradiction.
It therefore follows that $r \mid k$. But we assumed $0<k < r$, yielding yet another contradiction. Therefore, ${\rm ord}(\beta) = r$, and, being a power of $\alpha$, it must be an $r$-cycle, as desired.


TITLE: Is $g(x)= \frac{x}{1-\left | x \right |}$ uniformly continuous?
QUESTION [2 upvotes]: I was asked to prove that the function:
$$
g: B \subset \mathbb{R}^n \rightarrow \mathbb{R}^n
$$
Is uniformly continuous on $B$, where $B$ is the open ball $B(0,1) \subset \mathbb{R}^n$ y $g(x)= \frac{x}{1-\left | x \right |}$
My attempt
I tried to applied the negation of the definition of uniform continuity
$$
\exists \space \epsilon_0 > 0 \space | \space \forall \delta > 0, \exists \space x_0, y_0 \in B
$$
with
$$
\left | x_0 - y_0 \right | < \delta
$$
such as
$$
\left | f(x_0) - f(y_0)  \right | \geq \epsilon_0
$$
So I took $\epsilon_0 = 1/2$ and for any $\delta > 0$ I took $x_0=0$ and $y_0 = \frac{\delta}{1+\delta}e_1$, where $e_1 = (1,0,\dots,0)$ so $x_0, y_0 \in B$ and $\left |x_0 - y_0 \right | < \delta$.
But I just obtain: $\left | f(x_0)-f(y_0) \right | = \delta$ which is not always greater than $1/2$.
Any help would be really appreciated.

REPLY [1 votes]: Let us first work with the case $n=1$. There is a straightforward way to reduce the general case to this one. Since uniform continuity is a property that a function has to have on its entire domain of definition, if we find that it is not uniformly continuous on a subset of its domain, it is not uniformly continuous, period. So, let us focus on $f(x) = \frac{x}{1-x}$ for $x > 0$.
The idea now to find an appropriate $\epsilon_0$ is to start with one we suspect might work, and typically with functions with vertical asymptotes like this one, any old value for $\epsilon_0$ will work. We want to show that we can make the difference $\Delta f$ very large even though the difference $\Delta x$ may be small. The tricks I will use below aren't strictly speaking necessary, but they make the work simple at each step, even if they may seem like they come out of thin air if you are new to proofs like this. The spirit is always the same: make $\Delta f$ large for arbitrarily small $\Delta x$.
Let's start with the observation that $1 = \frac{1}{1-x} - f(x)$, so $f(x) = \frac{1}{1-x} - 1$ for $x > 0$. Now, a function $g(x)$ is uniformly continuous if and only if the vertical translations $g(x) - a$ with $a\in \Bbb R$ are uniformly continuous, as you can verify with the definition of uniform continuity. So, we can shift our focus to $F(x) = \frac{1}{1-x}$ for $0 < x < 1$. Applying the symmetry $x\mapsto 1-x$ (or changing variables $u = 1-x$ if you like), we can switch our attention again to the function $G(x) = \frac{1}{x}$ with $0 < x < 1$.
Let $\delta>0$ be arbitrary. Then by the mean value theorem, $\big|G\big(\frac{3\delta}{2}\big)-G(\delta)\big| = |G'(c)|\cdot\frac{\delta}{2}$ for some $\delta < c < \frac{3\delta}{2}$. Let's impose $\delta < 1$ since the length of the domain of our function is $1$, so the condition for uniform continuity is trivially not satisfied for $\delta\ge 1$ because the function $G$ is unbounded. Then since $|G'(c)| = \frac{1}{c^2}$, we have $\frac{1}{\delta^2} > |G'(c)| > \frac{4}{9\delta^2}$. Hence $\big|G\big(\frac{3\delta}{2}\big)-G(\delta)\big| > \frac{4}{9\delta^2}\cdot\frac{\delta}{2} > \frac{2}{9}$. Setting $\epsilon_0 = \frac{2}{9}$ gives us the desired result by proving that $G$ is not uniformly continuous for $0 < x < 1$, hence neither is $f$.


TITLE: ODE Solution for Canonical State Space form
QUESTION [0 upvotes]: Suppose I have a first order ODE
$$
X'(t) = A(t)X(t), \quad \quad X(t_0) = X_0
$$
where $X\in R^n$, and time-dependent $A(t)$ has a canonical state-space form
$$
\begin{bmatrix}
0 & 1 & & &\\
0 & 0 & 1 & &\\
 \vdots & \cdots  &  & &\\
a_1(t) & a_2(t) & a_3(t) & \cdots & a_n(t)
\end{bmatrix}
$$
I have two questions:

Could I express the solution as

$$
X(t) = X_0\,e^{\int^t_{t_0}A(s)ds}
$$
in this particular case? Does $A(t)$ commutes with its integral?

If no, how about if I say the coefficients $a_1(t),a_2(t)\cdots$ are selected such that $A(t)$ has $n$ identical real negative eigenvalue $r(t)$? If again no, what could be a condition of $A(t)$ such that it could?

REPLY [2 votes]: For general $A(t)$ the solution is given by the Magnus Series.
Important special cases are when $A(t), A(t')$ commute or when $A(t)$ is constant. Then the solutions are given by the Matrix Exponential
\begin{align}
A(t)A(t') &= A(t')A(t)
&&\implies
&&X(t_1) = \exp \left(\int_{t_0}^{t_1} A(t') \, dt' \right) X_0
\\
A(t) &= A_0 = \text{const}
&&\implies
&&X(t_1) = \exp \left( A_0 (t_1-t_0) \right) X_0\,.
\end{align}

For your $A(t)$: in general it does not commute. Thus, you have to use the magnus series which will probably not help a lot.


TITLE: 2-D heat equation being parabolic?
QUESTION [0 upvotes]: This is a mighty dumb question but I can not seem to figure out how a 2-D heat equation is parabolic.
  $$ \frac{\partial u}{\partial t}=\Delta u $$
I know that the discriminant : $b^2-ac=0$ for a parabolic PDE.
Here, $b=0,a=c=-1$ . Then how is it parabolic ?
I think I am missing out something.

REPLY [1 votes]: $u$ is a function of $x$ and $t$. In the discriminant, $c$ is the co-efficient of $\frac{\partial^2 u}{\partial t^2}$, which is $0$, so the discriminant is indeed $0$.


TITLE: Finding the point on a graph where two lines intersect?
QUESTION [2 upvotes]: I have the following problem: "An ant has traveled from $(4,8)$ to $(4,4)$ in $30~\text{seconds}$. A mouse located at $(-4,2)$ traveling at $10~\text{units/minute}$ wants to intercept the ant. At what point on the line $x = 4$ will the mouse intercept the ant?".
So far I have calculated what the speed of the ant is by using the distance formula
\begin{align*}
D & = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\
  & = \sqrt{(4-4)^2+(4-8)^2}\\
  & = \sqrt{0+16}\\
  & = \sqrt{16}\\
  & = 4
\end{align*}
Knowing that the distance is $4$ and the time is $30 \text{ seconds}$, i then plugged these numbers in to solve for speed.
\begin{align*}
S & = \frac{D}{T}\\
  & = \frac{4}{30}
\end{align*}
seconds.
(then i converted seconds into minutes)
$$4/30 \text{ seconds} \times 60 \text{ seconds/minute}=8 \text{ units/minute}$$
so the speed of the ant is $8 \text{ units/minute}$.
However, I am unsure of where to go from here and I don't even fully know if i am even on the right track. I would greatly appreciate any help I can get.

REPLY [1 votes]: If the first ant can continue down past (4,4)...
Since the first ant is travelling straight down, let's say they meet at point $(4,Y)$
Let t=time in minutes.
The first ant's y position can be defined as $8-8t$.  From there we can create an equation with the distance ant2 travels using $D=RT$ & the distance formula for time t.
$$10t=\sqrt{(4--4)^2+(y-2)^2}$$
$$10t=\sqrt{(8)^2+(8-8t-2)^2}$$
$$10t=\sqrt{64+(6-8t)^2}$$
$$t=.801$$
So they will meet at point $(4,8-8\times(0.801))$ or $(4,1.592)$


TITLE: Colouring $2n$ Points in $\mathbb{R}^2 $
QUESTION [6 upvotes]: Definition: A strip is the area in the plane enclosed between some two parallel lines. An axis-
parallel strip is a strip whose bounding lines are parallel to one of the axes.
Let $P$
be a set of $2n$ points in the plane (for some integer $n$). Prove that the points of $P$
can be colored with 'red' or `blue' such that in any axis-parallel strip the difference
between the number of red points and the number of blue points is at most 2.
The direction I was thinking about is definig monotone non-decreasing seq of the points, relatively to the $x$-axis, and another non-decreasing monotone sequence relatively to y-axis, from the given points. Then I would like taking a graph, such that the vertices are the points and the edges connect consecutive terms of both sequences. Im pretty stuck right now.

REPLY [0 votes]: The next answer is not mine' but of my proffesor.
As been offered in the answer above me , we counstruct a graph which it's vertices are our points, and the edges constructed by pairs of points- by a non dec seq in x- axis and y-axis.
Now, since our edges are realy a $ matching$ , every cycle of it has even length.

Our graph is , by the bipartite graph characterization, bipartite, i.e $V= X \cup Y $ where X and Y are disjoint sets. Color X with blue and Y with red.
For given parellel strip, we can see that between every pair of one color points, there is a point colored by the opposite color. Hence, exept the "end pints" of the strip, we have split our strip into pairs of different colors. There are at most 2 end pints to a given strip- so we are done :)

My proffesor mentioned that it's just an example to much deeper idea in math, called Discrepancy.


text:
\section{Benchmark description}\label{chap:benchmark}
In this section, we describe the geometry and mathematical model of the FSI benchmark from \cite{turek2006proposal}.
It studies the flow of an incompressible Newtonian fluid through a 2D channel as the fluid interacts with a submerged structure. The channel is a rectangle $[0,2.2]\times[0,0.41]$. The structure consists of a rigid disk $B_{0.05}(0.2,0.2)$ and a flexible beam $[0.2,0.6]\times[0.19,0.21]\backslash B_{0.05}(0.2,0.2)$ which is attached at its left end to the boundary of the disk. Figure \ref{fig:geometry} illustrates the setting; note that the geometry is intentionally non-symmetric.

\begin{figure*}
	\centering
	\includegraphics[height=3cm]{images/fig1a.pdf}\\
	\includegraphics[height=2cm]{images/fig1b.pdf}
	\caption{Top: the flow channel and the submerged structure in the initial configuration. Bottom: close-up on the structure.}
	\label{fig:geometry}
\end{figure*}

The top and bottom walls of the channel are impermeable to the fluid. The fluid enters the channel through the left wall with a prescribed velocity and exits through the right wall freely. The presence of the submerged structure changes the flow of the fluid and, in response, the fluid exerts a certain force on the structure. Thus, this is an FSI problem. Depending on the prescribed material parameters of the fluid and beam as well as on the inflow fluid velocity, this fluid force can result in a noticeable deformation of the beam. The beam deformation alters the shape of the flow channel, the flow itself and, as a consequence, the force exerted on the structure by the fluid. Such systems with a two-way interaction between components are called coupled systems.

In what follows, when we use the word \say{structure}, we refer to the flexible beam only. If the rigid disk is also considered, we explicitly mention it. Throughout this work, we use subscripts $s$ and $f$ to distinguish between  objects related to the structure and fluid respectively. Moreover, we use subscript $a$ when dealing with ALE mappings and related objects. Let $\Omega_f(t) \subset \mathbb{R}^2$ and $\Omega_s(t) \subset \mathbb{R}^2$ denote domains occupied by the fluid and structure respectively at time $t\in[0,T]$. Additionally, we use $\Omega^0_f=\Omega_f(0)$ and $\Omega^0_s=\Omega_s(0)$ to denote the fluid and structure domains at time $t=0$. Furthermore, we use $\Gamma(t)$ to denote the FSI interface $\partial\Omega_f(t)\cap\partial\Omega_s(t)$ where the interaction between the fluid and structure takes place. Correspondingly, $\Gamma^0=\Gamma(0)$ denotes the FSI interface at time $t=0$. In the rest of this section, we briefly formulate the equations that we use to describe the motion of the beam and the fluid as well as their interaction. The main goal here is to fix the notation necessary for the ensuing description of mesh deformation techniques (MDTs). We encourage the reader to consult the following references for more information on: finite element discretization for the Navier-Stokes equations for incompressible flows \cite{john2016finite}; finite element discretization for nonlinear elasticity problems \cite{wriggers2008nonlinear,bernal2013isogeometric}; ALE formulation of FSI problems \cite{richter2010numerical,bazilevs2013}; partitioned solution approach to FSI problems \cite{kuttler2008fixed}.

\subsection{Structure motion}
We assume that the structure behavior can be characterized as elastic and compressible. Let the structure in its initial (undeformed) configuration occupy the domain $\Omega_s^0$. We can describe the structure motion in terms of a displacement field $\f{u}_s: \Omega^0_s\times[0,T] \to \mathbb{R}^2$. Note that displacement $\f{u}_s$ can describe a rigid body motion with no deformation involved. Information on whether actual deformation of $\Omega_s^0$ takes place is contained in the deformation gradient $\f{F}_s = \f{I} + \nabla\f{u}_s$. Two important objects derived from the deformation gradient are the Green-Lagrange strain tensor $\f{E}_s = (\f{F}_s^T\f{F}_s-\f{I})/2$ and the Jacobian determinant $J_s=\det\f{F}_s$.

For the material behavior, we use the St.~Venant-Kirchhoff constitutive law which links the second Piola-Kirchhoff stress tensor $\f{S}_s$ to the Green-Lagrange strain tensor $\f{E}_s$:
\begin{equation}\label{eq:StVK}
\f{S}_s = \lambda_s\tr(\f{E}_s)\f{I} + 2\mu_s\f{E}_s.
\end{equation}
The material law (\ref{eq:StVK}) includes the Lam\'e parameters $\lambda_s$ and $\mu_s$, which are constitutive parameters describing physical properties of the material. They can be computed from Young's modulus $E_s$ and Poisson's ration $\nu_s$ as
\begin{equation}
\lambda_s = \frac{\nu_s E_s}{(1+\nu_s)(1-2\nu_s)}\sg\text{and}\sg\mu_s = \frac{E_s}{2(1+\nu_s)}.
\end{equation}
Second Piola-Kirchhoff stress tensor $\f{S}_s$ measures forces appearing in the deformed structure with respect to its initial configuration $\Omega_s^0$. In FSI applications, it is important to have an ability to express these forces with respect to the deformed configuration $\Omega_s(t)$. This can be achieved by means of the Cauchy stress tensor $\pmb{\sigma}_s$, which is related to $\f{S}_s$ by
\begin{equation}
\pmb{\sigma}_s = \frac{1}{J_s}\f{F}_s\f{S}_s\f{F}_s^T.
\end{equation}

In the presence a given external acceleration $\f{g}:\Omega^0_s\times[0,T]\to\mathbb{R}^2$, the displacement $\f{u}_s$ should satisfy the local conservation equations of linear momentum
\begin{equation}\label{eq:beamBalance}
\rho_s\ddot{\f{u}}_s = \dv\f{P}_s + \rho_s\f{g} \text{ in }\Omega_s^0.
\end{equation}
Here, $\rho_s$ is the structure density which we assume to be constant in $\Omega_s^0$, and $\f{P}_s = \f{F}_s\f{S}_s$ is the first Piola-Kirchhoff stress tensor. Note that equations (\ref{eq:beamBalance}) are formulated in the stationary domain $\Omega_s^0$.

\subsection{Fluid motion}
To describe the fluid motion, we use the Navier-Stokes equations for incompressible flows. If a computational domain $\Omega_f$ does not change with time, the Navier-Stokes equations have the following form:
\begin{align}
\rho_f\dot{\f{v}}_f + \rho_f\nabla\f{v}_f\cdot\f{v}_f&=\dv\pmb{\sigma}_f + \rho_f\f{g},\\
\dv\f{v}_f &= \f{0} \text{ in }\Omega_f.
\end{align}
Here, $\f{v}_f:\Omega_f\times[0,T]\to\mathbb{R}^2$ is a vector field describing the fluid velocity at a given point in $\Omega$, $\rho_f$ is a constant fluid density, and $\pmb{\sigma}_f$ denotes a Cauchy stress tensor. Behavior of an incompressible Newtonian fluid is characterized by the following constitutive law:
\begin{equation}
\pmb{\sigma}_f = -p_f\f{I} + \rho_f\nu_f(\nabla\f{v}_f + \nabla\f{v}_f^T),
\end{equation}
where $p_f:\Omega_f\times[0,T]\to\mathbb{R}$ is a pressure field, and $\nu_f$ denotes the kinematic viscosity of the fluid.

In FSI applications, one has to account for the motion of the fluid domain. In this work, we consider a common strategy based on the arbitrary Lagrangian-Eulerian (ALE) mappings. An ALE mapping describes the motion of the fluid domain in terms of an auxiliary displacement field $\f{u}_a:\Omega_f^0\times[0,T]\to\mathbb{R}^2$ such that $\Omega_f(t) = \Omega^0_f + \f{u}_a(\cdot,t)$. With an ALE mapping, the Navier-Stokes equations can be formulated in the moving domain $\Omega_f(t)$:
\begin{align}
\label{eq:NS_ALE1}\rho_f\dot{\f{v}}_f + \rho_f\nabla\f{v}_f\cdot(\f{v}_f-\dot{\f{u}}_a)&=\dv\pmb{\sigma}_f+ \rho_f\f{g},\\
\label{eq:NS_ALE2}\dv\f{v}_f &= \f{0} \text{ in }\Omega_f(t).
\end{align}

Since the fluid domain motion is driven by the structure deformation, the ALE displacement $\f{u}_a$ has to comply with the structure displacement $\f{u}_s$ on the FSI interface:
\begin{equation}
\f{u}_a = \f{u}_s\text{ on }\Gamma(t).
\end{equation}
Inside the fluid domain, $\f{u}_a$ can be chosen arbitrary, hence the term ALE. The only condition is that the ALE displacement should define an invertible deformation of the fluid domain. This means that the following condition has to hold:
\begin{equation}\label{eq:bijectivity}
J_a = \det\f{F}_a = \det(\f{I}+\nabla\f{u}_a) > 0.
\end{equation}
The scope of this work is to compare different options for defining the ALE displacement in the fluid domain provided the structure displacement on the FSI interface.

\subsection{Interaction conditions}
Physical interaction between the fluid and structure takes place on the FSI interface $\Gamma(t)$. It is characterized by the following two coupling conditions: the kinematic continuity
\begin{equation}
\f{v}_f = \dot{\f{u}}_s \text{ on }\Gamma(t),
\end{equation}
which assures that the fluid stays attached to the structure; and the dynamic continuity
\begin{align}
\nonumber&\pmb{\sigma}_s\cdot\f{n} = \pmb{\sigma}_f\cdot\f{n} \text { on }\Gamma(t) \sg\Leftrightarrow\\
&\f{P}_s\cdot\f{n}= J_a\pmb{\sigma}_f\f{F}_a^{-T}\cdot\f{n} \text{ on }\Gamma^0,
\end{align}
which maintains the balance of forces on the FSI interface. If a partitioned approach to FSI is used, one has to enforce the coupling conditions by exchanging information between the fluid and structure solvers.

\subsection{Initial and boundary conditions}
To complete the definition of the benchmark, we provide suitable initial and boundary conditions. The system is initialized with zero initial conditions for the fluid velocity, structure displacement and structure velocity:
\begin{align}
\nonumber&\f{v}_f(\cdot,0) = \f{0} \text{ in }\Omega^0_f,\\
&\f{u}_s(\cdot,0)=\f{0}\text{ in }\Omega_s^0,
\bg\dot{\f{u}}_s(\cdot,0)=\f{0}\text{ in }\Omega_s^0.
\end{align}
The main energy source of the system is an inflow boundary condition on the fluid velocity prescribed on the left end of the channel $\partial\Omega^\text{in}_f$. The condition prescribes a parabolic velocity profile
\begin{equation}
\f{v}_\text{par} = \left(\begin{array}{c}
v_{\max}\big(\frac{2}{0.41}\big)^2y(0.41-y)\\0\end{array}\right)
\end{equation}
with the maximum inflow velocity $v_\text{max}$ serving as an adjustable parameter. In order to comply with the initial conditions, $\f{v}_\text{par}$ is made time-dependent by scaling it with time:
\begin{equation}
\f{v}_\text{in}(t) = \left\{\begin{array}{c}
\f{v}_\text{par}\frac{1-\cos(\pi t/2)}{2}\;\sg\text{if }t<2\text{s},\\
\f{v}_\text{par}\sg\bg\bg\bg\sg\text{if }t\geqslant 2\text{s}.\end{array}\right.
\end{equation}
The resulting time-dependent inflow boundary condition $\f{v}_f = \f{v}_\text{in}$ provides a smooth warm-up phase for the simulation. On the right wall of the channel $\partial\Omega^\text{out}_f$, a do-nothing condition $\pmb{\sigma}_f\cdot\f{n} = \f{0}$ is prescribed. Additionally, a no-slip boundary condition $\f{v}_f = \f{0}$ is set on the rest of the channel wall $\partial\Omega^\text{ns}_f=\partial\Omega_f^0\setminus\partial\Omega^\text{in}_f\setminus\partial\Omega^\text{out}_f\setminus\Gamma^0$. Finally, the structure is fixed ($\f{u}_s=\f{0}$) on its left end $\partial\Omega_s^0\setminus\Gamma^0$.

\subsection{Geometry parametrization}
The research we present in this work has been conducted in the framework of isogeometric analysis \cite{Hughes2005,Cottrell.2009}.  In the spirit of IGA, we model computational domains as collections of tensor-product NURBS patches \cite{Piegl1997}. For the structure, we use a single quadratic NURBS patch, whereas the fluid domain is modeled with seven patches. Figure \ref{fig:flowMesh} illustrates the isoparametric mesh in the fluid domain after several applications of uniform $h$-refinement. For simplicity, we use matching parametrizations for neighboring fluid patches as well as for the structure patch and the surrounding fluid patches. This choice does not restrict the applicability of our results; however, it significantly simplifies the exchange of coupling information between the fluid and structure solvers.

\begin{figure*}
	\centering
	\includegraphics[width=0.75\textwidth]{images/fig2.png}\\
	\caption{Computational mesh in the fluid domain. Three patches adjacent to the beam are chosen for ALE deformation.}
	\label{fig:flowMesh}
\end{figure*}

Let us consider one of the NURBS patches forming the initial configuration of the fluid domain. Its parametrization $\f{G}_f^0:[0,1]^2\to\Omega_f^0$ can be written in terms of control points $\f{c}_k\in\mathbb{R}^2$ and tensor-product NURBS basis functions $N_k:[0,1]^2\to\mathbb{R}$ as
\begin{equation}
\f{G}_f^0(\xi,\eta) = \sum_{k=1}^n\f{c}_k N_k(\xi,\eta).
\end{equation}
In IGA, we the same NURBS basis functions to approximate solutions to PDEs. In our case, we use $N_k$ to discretize the ALE displacement $\f{u}_a$ in space. Assuming that we use time moments $t_0=0,\dots,t_i,\dots,t_N = T$ for the time discretization, we denote the ALE displacement at time $t_i$ by $\f{u}_a^i$. Its NURBS representation can be written as
\begin{equation}
\f{u}_a^i(\xi,\eta) = \sum_{k=1}^n\f{d}_k^i N_k(\xi,\eta),
\end{equation}
where $\f{d}_k^i$ are the corresponding control points. Given $\f{u}_a^i$, we can easily construct a NURBS parametrization $\f{G}_f^i$ of the deformed fluid domain $\Omega_f^i=\Omega_f(t_i) = \Omega_f^0 + \f{u}_a^i$ as
\begin{equation}\label{eq:deformDomain}
\f{G}_f^i(\xi,\eta) =  \sum_{k=1}^n(\f{c}_k+\f{d}_k^i)N_k(\xi,\eta).
\end{equation}
Now $\f{G}_f^i$ can be used to discretize the Navier-Stokes equations (\ref{eq:NS_ALE1}--\ref{eq:NS_ALE2}) in the deformed configuration of the fluid domain.

As a final comment, let us mention that it is possible to deform only a portion of the fluid domain while keeping the rest fixed. An obvious advantage of this approach is a considerable reduction in computational cost associated with construction of ALE mappings. In this work, we choose to deform only three patches of the fluid domain that are directly adjacent to the FSI interface. In Figure \ref{fig:flowMesh}, they are highlighted in red. In our experience, these three patches are more than capable of absorbing deformation of the structure that appears in this benchmark.


TITLE: Can an uncountable group have a countable number of subgroups?
QUESTION [16 upvotes]: Can an uncountable group have only a countable number of subgroups?
Please give examples if any exist!

Edit: I want a group having uncountable cardinality but having a countable number of subgroups.
By countable number of subgroups, I mean the collection of all subgroups of a group is countable.

REPLY [33 votes]: No. Suppose $G$ is an uncountable group. Every element $g$ of $G$ belongs to a countable subgroup of $G$, namely the cyclic subgroup $\langle g\rangle$. Thus $G$ is the union of all of its countable subgroups. Since a countable union of countable sets is countable, $G$ must have uncountably many countable subgroups.

REPLY [11 votes]: EDIT: bof's answer is the right one, but the construction below - while completely pointless overkill - is still an example of a useful technique, so I'm leaving this answer up.

No, this cannot happen.
Suppose $G$ is a group and $A$ is a countable subset of $G$. Then the closure of $A$ under the group operations ($*$ and $^{-1}$) of $G$, $\langle A\rangle$, is again countable - this is a good exercise (HINT: the set of finite strings from a countable set is countable).
With this in hand, if $G$ is an uncountable group we can build an uncountable chain of subgroups of $G$, as follows:

We will define a countable subgroup $A_\delta$ for every countable ordinal $\delta$. There are uncountably many of these, so if we can do this we'll be done.
We let $A_0$ be the trivial subgroup.
Having defined $A_\eta$ for every $\eta<\delta$, we let $a$ be some element of $G$ not in $\bigcup_{\eta<\delta}A_\eta$ - which exists, since this is a countable union of countable subgroups, and $G$ is uncountable - and let $A_\delta=\langle (\bigcup_{\eta<\delta}A_\eta)\cup\{a\}\rangle$.
It's easy to prove by transfinite induction that $(A_\delta)_{\delta<\omega_1}$ is a strictly increasing chain of countable subgroups of $G$, so we're done.


text:
\section{Conclusion and open questions}
In this paper we have exhibited the usefulness of entropic regularization with debiasing for the estimation of the squared Wasserstein distance: it may increase both accuracy and efficiency when the problem has a smooth nature. Numerical experiments suggest that the theory could be extended in several directions. First, the Sinkhorn divergence estimator appears at least as statistical efficient as the plug-in estimator, while our bound is slightly weaker. Also, the estimation of Kantorovich potentials seems to enjoy similar guaranties, but this is not covered by our theory.


TITLE: $M\cong N$ iff $[M:N]_R$ is a principal fractional ideal
QUESTION [16 upvotes]: Let $R$ be a Dedekind ring, $K$ its field of fractions, $U$ a finite vector space over $K$, and $M,N$ finitely generated $R$-modules that span $U$, i.e. contain a basis of $U$.
For every $\mathfrak p \subset R$, define $[M:N]_{\mathfrak p} = (\det \phi_{\mathfrak p})$, where $\phi_{\mathfrak p}$ is a linear transformation that takes $M_{\mathfrak p}$ onto $N_{\mathfrak p}$. Recall here $R_{\mathfrak p}$ is a principal ideal domain $\Rightarrow$ $M_{\mathfrak p},N_{\mathfrak p}$ are free and one can indeed find such a transformation.
Now define $[M:N]_R := \prod_i \mathfrak p_i^{v_{\mathfrak p_i}([M:N]_{\mathfrak p_i})}$. According to Froehlich, one can show that $M\cong N$ iff $[M:N]_R$ is a principal fractional ideal.
One direction is easy. If we have a general linear transformation $\Psi$ of $U$ such that $\Psi (M)=N$, then $(\det \phi_{\mathfrak p_i}) = (\det \Psi_{\mathfrak p_i})$ for every $i$, so that $[M:N]_R =\prod_i \mathfrak p_i^{v_{\mathfrak p_i}([M:N]_{\mathfrak p_i})} = (\det \Psi)$, which is a principal fractional ideal.
But how do I get the other direction?

REPLY [0 votes]: First consider the case $\dim U=1$. Let $[M:N]_R=\alpha R$. By localizing is then easy to see that $x\mapsto \alpha x$ is an isomorphism $M\cong N$.
Now for the general case: Let $n=\dim U$. An isomorphism $\phi_{\frak p}:M_{\frak p} \to N_{\frak p}$ induces an isomorphism $\wedge^nM_{\frak p}\to \wedge^n N_{\frak p}\subset \wedge^nU$ with the same determinant, so $[\wedge^nM:\wedge^nN]_R=[M:N]_R$. By the dimension $1$ case we conclude $\wedge^nM\cong\wedge^nN$. From this it follows that $M\cong N$. E.g. we can write $M\cong R^{n-1}\oplus I$, $N\cong R^{n-1}\oplus J$ for ideals $I,J$ and then $I\cong \wedge^nM\cong\wedge^nN\cong J$, so $M\cong N$.