text:
\begin{document}

\title[Singular differential equations]{On solutions of singular differential equations of the second order}

\author[A.O. Remizov]{A.O. Remizov}
\address{
Moscow Institute of Physics and Technology,
Inststitutskii per. 9, 141700 Dolgoprudny, Russia}
\email{alexey-remizov@yandex.ru}

\subjclass[2010]{Primary 34A12; Secondary 34C05}

\keywords{vector fields, singular point, normal form, resonance, oscillation}

\begin{abstract}
We study the behaviour of solutions of second-order differential equations at their singular points,
where the coefficient of the second-order derivative vanishes. In particular, we consider solutions issuing from
a singular point without definite tangential direction. Great attention is paid to second-order differential equations, whose
right-hand sides are cubic polynomials by the first-order derivative.
\end{abstract}

\maketitle

\begin{flushright}
\it {Dedicated to the memory of A.\,F.~Filippov}
\end{flushright}

\section{Introduction}

A.\,F.~Filippov showed in \cite{Fil} that a system of ordinary differential equations
\begin{equation}
F(x,y,p) = 0, \ \ p=dy/dx,
\label{1}
\end{equation}
where $x \in \bR^1$, $y \in \bR^n$ and $F: \bR^{2n+1} \to \bR^n$ is a smooth\footnote{
By smooth we mean $C^\infty$, if the otherwise is not stated.
}
vector function, may have solutions without definite tangential direction at a certain point.
This motivates the following terminology.

A solution $y(x)$ of system \eqref{1} is called {\it oscillating} at a point $(x_0,y_0)$, if $y(x)$ is a vector function
differentiable on the interval $(x_0, x_0+\delta)$ or $(x_0-\delta,x_0)$, $\delta>0$, such that
$y(x) \to y_0$ but $y'(x)$ has neither finite not infinite limit as $x \to x_0$.
A solution $y(x)$ is called {\it proper} at a point $(x_0,y_0)$, if $y'(x)$ has limit (finite or infinite) as $x \to x_0$.

Let $J^1$ be the 1-jet space of smooth functions $y(x)$, with the coordinates $(x,y,p)$.
Filippov showed that if for a point $T_0 \in J^1$ there exists $T_0' \in J^1$ such that $x_0=x_0'$, $y_0=y_0'$, $p_0 \neq p_0'$,
and the matrix $F_p$ degenerates at $T_0'$, then besides a unique proper solution passing through $(x_0,y_0)$ with the tangential direction $p_0$,
system \eqref{1} may have oscillating solutions, which pass through $(x_0,y_0)$ and even have definite tangential direction $p_0$ at $(x_0,y_0)$.

The following example is taken from~\cite{Fil}. Consider solutions of the system
\begin{equation}
\left\{
\begin{aligned}
& p_1(1-p_1^2-p_2^2) + 8xy_1 +4y_2 = 0, \\
& p_2(1-p_1^2-p_2^2) + 8xy_2 -4y_1 = 0,
\end{aligned}
\right.
\label{2}
\end{equation}
that pass through the origin with the tangential direction $p_i=0$, that is, satisfy the conditions $y_i(0)=y'_i(0)=0$, $i=1,2$.
The matrix $F_p$ is non-degenerate at the point $T_0=0$, and system \eqref{2} has a unique proper solution $y_i(x) \equiv 0$, which
obviously satisfies the required conditions. However, system \eqref{2} has an infinite number of oscillating solutions given by the formula
$$
y_1(x) = x^2 \cos (x^{-1}+c), \ \
y_2(x) = x^2 \sin (x^{-1}+c), \ \
c=\const,
$$
for $x \neq 0$, and $y_i(0)=0$.
The both derivatives $y_i'(x)$ are zero at $x=0$, but the limits $y_i'(x)$ as $x \to 0$ do not exist.

On the other hand, Filippov showed that systems with oscillating solutions are a sort of exception.
Given a point $q_{0}=(x_{0},y_{0}) \in \bR^{n+1}$, we define the set
$$
Q(q_{0}) = \{ T=(q_0,p) \in J^1 : F(T)=0, \ \det (F_p(T))=0 \}.
$$

\begin{theorem}[\cite{Fil}]
\label{T0}
Assume that at least one of the following conditions holds true:

1. $p_0 \notin {\rm co}\,Q(q_{0})$, where ${\rm co}$ denotes the convex hull,

2. the set $Q(q_{0})$ is at most countable (i.e., countable, or finite, or empty).

Then system \eqref{1} has no solution oscillating at $q_{0}$ with tangential direction $p_0$ at $q_{0}$.
Generically, the condition~2 holds true at all points, and system \eqref{1} has no oscillating solutions.
\end{theorem}

Although Theorem~\ref{T0} has a very general character, it gives only a trivial result for systems
\begin{equation}
A(x,y)\,p=b(x,y), \ \ p=dy/dx,
\label{3}
\end{equation}
where $A$ is a $n \times n$ matrix depending on $(x,y)$, $b$ is a vector function.
Theorem~\ref{T0} guarantees the absence of oscillating solution at those points where $\det A \neq 0$,
which obviously follows from the fact that then system \eqref{3} is locally equivalent to $p = A^{-1}(x,y)b(x,y)$.

System \eqref{3} are of great interest themselves (see, e.g., \cite{ZhS} and the references therein), but in this paper
we restrict ourselves with a partial case of \eqref{3}. Namely, we shall consider the second-order equations
\begin{equation}
\Delta(x,y) \frac{dp}{dx} = M(x,y,p),  \ \ p=dy/dx,
\label{4}
\end{equation}
which are equivalent to systems \eqref{3} of a special type.

A great attention is paid to the case that the right-hand side of equation \eqref{4}
is a cubic polynomial in $p$, that is,
\begin{equation}
M(x,y,p) = \sum_{i=0}^3 \mu_i(x,y)p^i.
\label{5}
\end{equation}
This attention to equations \eqref{4}, \eqref{5} is motivated by their role in the description of various geometric structures,
for instance, the affine connection and the projective connection.
Equations of this class were studies by Sophus Lie, A.~Tresse, J.~Liouville, E.~Cartan, to name a few.
(See, e.g., \cite{AA, Cartan, Tresse, Yu} and the references therein.)
For instance, geodesics in the Levi--Civita connection generated by the metric tensor
\begin{equation}
ds^2 = a(x,y)\,dx^2 + 2b(x,y)\,dxdy + c(x,y)\,dy^2
\label{6}
\end{equation}
are solutions of the equation in the form \eqref{4}, \eqref{5}, where
\begin{equation}
\begin{aligned}
\Delta &= ac-b^2,  \\
\mu_0 &= a(a_y-2b_x) +a_xb, \\
\mu_1 &= b(3a_y-2b_x) + a_xc - 2ac_x, \\
\mu_2 &= b(2b_y-3c_x) + 2a_yc - ac_y, \\
\mu_3 &= c(2b_y-c_x) - bc_y.
\end{aligned}
\label{7}
\end{equation}

Here the coefficient $\Delta$ vanishes at point that the quadratic form \eqref{6} degenerates.
In Riemannian geometry, such is not the case, since the inequality $\Delta>0$ holds true everywhere.

However, degenerate quadratic forms generically appear on surfaces embedded into pseudo-Euclidean spaces:
the metric induced on a surface into pseudo-Euclidean space degenerates at those points where the surface tangents
the light cone of the ambient space. Singularities of the geodesic equation appearing at degenerate points of the metric \eqref{6} are
studied in a recent series of papers \cite{PR2019, R2009, RT2016}. See also the survey \cite{PR2018}.

It should be remarked that in all papers mentioned above (as well as in other works known to us)
only proper geodesics were considered, while the possibility of oscillating geodesics is not studied yet.
In the present paper, we shall fill this gap.

The paper is organized as follows.
In the next section, we establish a sufficient condition for equation \eqref{4}, \eqref{5} to have no oscillating solutions at a given point.
This implies that a generic equation of this type has no oscillating solutions at all.
The closing section of the paper is devoted to the behaviour of proper solutions of equations \eqref{4}, \eqref{5}
at its generic singular points.

\section{Oscillating solutions}

Consider the differential equation
\begin{equation}
\Delta(x,y) \frac{dp}{dx} = M(x,y,p),  \ \ p=dy/dx,
\label{8}
\end{equation}
where $\Delta(x,y)$, $M(x,y,p)$ are smooth functions.

A point $q_0=(x_0,y_0)$ is called a {\it singular point} of equation \eqref{8}, if $\Delta(x_0,y_0)=0$.
Denote the locus of singular points by $\Gamma$.
Generically, $\Gamma$ is a curve on the $(x,y)$-plane, but we do not assume it in what follows.

When dealing with singular points, it makes sense to refine the notion of solutions.

Let $q_{0} = (x_{0},y_{0}) \in \Gamma$.
A solution of equation \eqref{8} {\it issuing} from the point $q_{0}$ is a function $y(x)$ such that

1) \ $y(x)$ is continuous on the segment $I_{\varepsilon}$ with endpoints $x_{0}$, $x_{0}+\varepsilon$, $\varepsilon>0$ or $\varepsilon<0$,

2) \ $y(x)$ is differentiable and it satisfies \eqref{8} at all inner points of $I_{\varepsilon}$,

3) \ $y(x_{0})=y_{0}$ and the graph of $y(x)$ has no common points with $\Gamma$ except for $q_0$.

If in addition to the above conditions, the derivative $y'(x)$ has a (finite or infinite) limit as $x \to x_{0}$,
the solution $y(x)$ is called {\it proper}. Otherwise the solution $y(x)$ is called {\it oscillating} at~$q_{0}$.
See Fig.~\ref{pict1}.

Finally, we shall give one more definition.
A solution of equation \eqref{8} {\it passing through} the point $q_{0}$ is a function $y(x)$ differentiable and satisfying \eqref{8} at all points
of an open interval $I$ that contains $x_0$ such that $y(x_{0})=y_{0}$ the graph of $y(x)$, $x \in I$, does not intersects $\Gamma$ except for the point $q_0$. Obviously, any solutions passing through $q_{0}$ is the union of two solutions issuing from $q_{0}$, but the inverse is not true.

The advantage of the given definitions becomes clear from the following simple example.

\begin{example}
\label{Ex1}
{\rm
The parabolas $y = \alpha x^2$, $\alpha =\const$, are integral curves of the second-order equation $2y {dp}/{dx} = p^2$
passing through the origin.
However, the graphs of functions defined by the formula $y = \alpha_1 x^2$ for $x < 0$ and
$y = \alpha_2 x^2$ for $x \ge 0$ are also integral curves, and all such functions are solution of the given equation
passing through the origin.
To avoid such ambiguity, it is sufficient to use the notion of a solution issuing from a point defined above.
Then solutions issuing from the origin are the branches of the parabolas $y = \alpha x^2$, $\alpha =\const$, and only them.
}
\end{example}

In the previous example, all solutions are proper.
Now we give two examples of equations \eqref{8} that possess oscillating solutions.

\begin{example}
\label{Ex2}
{\rm
The equation $x^4 {dp}/{dx} = 2x^3p - (2x^2+1)y$ has the family of solutions defined by the formula
\begin{equation}
y = x^2  (\alpha \cos x^{-1} + \beta \sin x^{-1}) \ \ \textrm{for} \ \ x \neq 0,  \ \ \alpha, \beta = \const,
\label{X9}
\end{equation}
and zero for $x=0$.
These functions are differentiable on the whole real axis, but their derivatives are not continuous at the origin except for $\alpha =\beta = 0$.
According to the given definitions, formula \eqref{X9} gives a family of solutions passing through the origin with the tangential direction $p=0$.
For $\alpha =\beta = 0$, we have the regular solution $y=0$, while for all others the solutions \eqref{X9} are oscillating at the origin.
}
\end{example}

\begin{example}
\label{Ex3}
{\rm
The equation $x^2 {dp}/{dx} = xp-2y$  has the family of solutions
\begin{equation}
y= x  (\alpha \cos \ln |x| + \beta \sin \ln |x|), \ \ \alpha, \beta = \const,
\label{X10}
\end{equation}
issuing from the origin.
Except for $\alpha =\beta = 0$, all these solutions are oscillating at the origin and (unlike the previous example)
they  have no definite tangential directions at the origin.

Moreover, using formula \eqref{X10}, one can construct an equation of the form \eqref{8} with oscillating solution,
whose right-hand side is a polynomial of arbitrary degree in $p$.
For instance, one can see that the function \eqref{X10} with $\alpha=\beta=1$
is a solution of the first-order equations $f_i(x,y,p)=0$, where
\begin{equation*}
\begin{aligned}
f_2 &= (xp)^2 - 2xyp + 2(y^2-x^2), \\
f_3 &= (xp)^3 + y(xp)^2 - 2xp (x^2+2y^2) + 6y(y^2-x^2).
\end{aligned}
\end{equation*}
Therefore, $y= x(\cos \ln |x| + \sin \ln |x|)$ is an oscillating solution of the second-order equations
$$
x^2 \frac{dp}{dx} = xp-2y + f_i(x,y,p), \ \ p=dy/dx,
$$
where $f_i$ is a polynomial in $p$ of the degree~$i=2,3$.
}
\end{example}

We presented several examples demonstrating the existence of oscillating solutions.
The following theorem shows that oscillating solutions do not appear generically.

\begin{theorem}
\label{T1}
Let $q_{0} \in \Gamma$ and $M(q_{0},p)$ is an analytic function not identically zero.
Then equation \eqref{8} has no oscillating solutions issuing from~$q_{0}$.
\end{theorem}

{\bf Proof.}
Assume that the above conditions hold true, but equation \eqref{8} has an oscillating solution $y(x)$ that issues from $q_0$
without definite tangential direction. Without loss of generality, assume that $y(x)$ is defined on the segment
$I_{\varepsilon} = [x_0, x_0+\varepsilon]$ and $\Delta(x,y(x))>0$ at all its inner points.

From the absence of the limit $y'(x)$ as $x \to x_{0}$ it follows that there exist two sequences
$x_n' \to x_{0}+0$ and $x_n'' \to x_{0}+0$ such that $p(x_n') \to p'$  and $p(x_n'') \to p''$, $p' \neq p''$.
For definiteness, assume that $p' < p''$. Since the function $y'(x)$ is continuous on the interval $x_{0} <x< x_{0}+\varepsilon$,
for every $p_* \in (p', p'')$ there exist two sequences $\xi_n \to x_{0}+0$ and $\xi_n' \to x_{0}+0$ such that
\begin{equation}
\lim_{n \to \infty} p(\xi_n) = \lim_{n \to \infty} p(\xi_n') = p_*, \ \
\frac{dp}{dx}(\xi_n)>0, \ \ \frac{dp}{dx}(\xi_n')<0.
\label{X11}
\end{equation}
See Fig.~\ref{pict1} (right).
Substituting the solution $y(x)$  into \eqref{8}, at the points $x=\xi_n$ and $x=\xi_n'$ we have the equalities
\begin{equation}
\Delta(\xi_n,y(\xi_n)) \frac{dp}{dx}(\xi_n) = M(\xi_n,y(\xi_n),p(\xi_n)),
\label{X12-1}
\end{equation}
\begin{equation}
\Delta(\xi_n',y(\xi_n')) \frac{dp}{dx}(\xi_n') = M(\xi_n',y(\xi_n'),p(\xi_n')).
\label{X12-2}
\end{equation}

The right-hand sides of both equalities \eqref{X12-1}, \eqref{X12-2} have the finite limit $M(x_{0},y_{0},p_*)$, whence their
left-hand sides have the same limit $M(x_{0},y_{0},p_*)$. On the other hand,
$\Delta(\xi_n,y(\xi_n))$ and $\Delta(\xi_n',y(\xi_n'))$ are positive for every $n$, and
from \eqref{X11} it follows that the left-hand sides of equalities \eqref{X12-1}, \eqref{X12-2} have different signs.
Therefore, their limit $M(x_{0},y_{0},p_*)=0$.

Thus, we proved that $M(q_{0},p_*)=0$ for every $p_* \in (p', p'')$.
Since the function $M(q_{0},p)$ is analytic in $p$, this implies $M(q_{0},p)=0$ for every~$p$.
This completes the proof.

\medskip

\begin{figure}[!htp]
\centering
\includegraphics[scale=0.85]{pict1.jpg}
\caption{An oscillating solution issuing from the origin (on the left) and its derivative (on the right).
}
\label{pict1}
\end{figure}

\begin{corollary}
\label{T2}
Let $q_0$ be a singular point of equation \eqref{4}, \eqref{5}.
If the coefficients $\mu_0, \ldots, \mu_3$ of the cubic monomial $M$ do not vanish at $q_{0}$ simultaneously,
there are no oscillating solutions issuing from $q_{0}$.
Moreover, in the case of the geodesic equation,
$d\Delta(q_{0}) = 0$ is a necessary condition for the existence of an oscillating solution issuing from $q_{0}$.
\end{corollary}

{\bf Proof.}
The first statement is obvious, let us prove the second one.

The geodesic equation in metric \eqref{6} has the form \eqref{4}, \eqref{5}, where $\Delta$ and $\mu_0, \ldots, \mu_3$ are expressed through $a,b,c$
by formula~\eqref{7}. By Theorem~\ref{T1}, for the non-existence of solutions oscillating at $q_{0} \in \Gamma$, it is sufficient to prove
that if all coefficients $\mu_0, \ldots, \mu_3$ simultaneously vanish at $q_{0}$, then $d\Delta (q_{0}) = 0$.

Let us assume on the contrary that $\mu_i(q_{0})=0$ for all $i$ and $d\Delta(q_{0})\neq 0$.
To simplify the calculations, we choose local coordinates centered at $q_{0} \in \Gamma$ such that $b(q_{0})=0$,
which is always possible to attain by an appropriate linear transformation.
Then from the equalities $\Delta(q_{0})= 0$, $d\Delta(q_{0})\neq 0$ it follows that $|a(q_{0})| + |c(q_{0})| \neq 0$.
Without loss of generality one can assume that $a(q_{0}) \neq 0$ and $c(q_{0}) = 0$.
Then from the equalities $\mu_i(q_{0})=0$ and \eqref{7} it follows $c_x(q_{0}) = c_y(q_{0})=0$.
Finally, from the equalities
$$
b(q_{0}) = c(q_{0}) = c_x(q_{0}) = c_y(q_{0})=0
$$
it follows $d\Delta (q_{0})=0$.
The obtained contradiction completes the proof.

\section{Singularities of second-order equations cubic in the first-order derivative}

In this section, we shall examine generic singularities of equation \eqref{4}, whose right-hand side has the form \eqref{5}.
Here we assume that $\Delta(x,y)$ and $\mu_i(x,y)$ are smooth functions not connected with each other, in contrast with the geodesic equation, whose
coefficients are expressed through three functions $a,b,c$ (formula~\eqref{7}), and therefore, they are connected with some
functional relations. According to the genericity assumption, we shall assume that the coefficients $\mu_0, \ldots, \mu_3$ of the cubic monomial \eqref{5}
do not vanish simultaneously and the set of singular points $\Gamma$ is a regular curve, which locally separates the $(x,y)$-plane
into two domains.

Under the above conditions, equation \eqref{4}, \eqref{5} has no oscillating solutions (Theorem~\ref{T1}).
Our purpose is to study its proper solutions issuing from generic singular points.
Further, we shall omit the adjective {\it proper}.
We start with a brief survey of the basic facts established in the papers \cite{PR2018} -- \cite{RT2016}
for geodesic equations that are valid for all equations of the form \eqref{4}, \eqref{5}, not necessarily geodesic.

Denote by $\pi$ the standard projection acting from the space $J^1$ to the $(x,y)$-plane along to the $p$-direction, which we call {\it vertical},
that is, $\pi: (x,y,p) \mapsto (x,y)$. Equation \eqref{4}, \eqref{5} is obviously connected with the vector field
\begin{equation}
\dot x = \Delta(x,y), \ \ \dot y = p\Delta(x,y), \ \ \dot p = M(x,y,p).
\label{X13}
\end{equation}
The $\pi$-projection of any integral curve of the field \eqref{X13} different from a straight vertical line, is a solution of equation \eqref{4}.
Conversely, the Legendrian lift of every solution of \eqref{4} is an integral curve of the field \eqref{X13}.

\begin{lemma}
\label{L0}
For any point $q_{0} \in \Gamma$, the vertical line $\pi^{-1}(q_{0})$ is an integral curve of the field \eqref{X13}.
Moreover, if $p_{0}$ satisfies the condition $M(q_{0},p_{0}) \neq 0$, then $\pi^{-1}(q_{0})$
is the only integral curve of the field \eqref{X13} that passes through the point $(q_{0},p_{0})$.
\end{lemma}

The proof is trivial.

From Lemma \ref{L0} it follows that solutions of equation \eqref{4}, \eqref{5} can issue from a singular point $q_{0} \in \Gamma$
only in so-called {\it admissible} tangential directions $p$, which correspond to real roots of the cubic polynomial $M(q_{0},p)$.
Here we shall understand the term {\it cubic polynomial} in the broad sense: the higher coefficients $\mu_3$ can vanish at some points
or even identically. Under such a convention, the class of equations \eqref{4}, \eqref{5} is invariant with respect to changes
of the variables $x,y$. It includes, in particular, the Bessel equation and the Gaussian hypergeometric equation.

The tangential direction $p=\infty$ is admissible at $q_{0}$, if and only if $\mu_3(q_{0})=0$.
Indeed, interchanging the variables $x,y$, we obtain the equation
$$
\Delta(y,x) \frac{dp}{dx} = -M^*(y,x,p) = - \sum_{i=0}^{3} \mu_i(x,y) p^{3-i},
$$
where $M^*$ is the reciprocal polynomial, and the direction $p=\infty$ becomes $p=0$.
The equality $\mu_3(q_{0})=0$ is equivalent to the condition that $p=0$ is a root of $M^*(q_{0},p)$.
Therefore, we shall consider $p=\infty$ as a root of the cubic polynomial $M(q_{0},p)$ of multiplicity $k$ iff
$p=0$ as a root of the cubic polynomial $M^*(q_{0},p)$ of multiplicity~$k$.

\begin{definition}
We shall call a singular point of equation \eqref{4}, \eqref{5} {\it generic}, if
$d\Delta(q_0) \neq 0$,
all roots of the cubic polynomial $M(q_{0},p)$ are prime and the corresponding admissible directions are
transversal to the curve $\Gamma$.
\end{definition}

In a neighborhood of every generic singular point $q_{0}$, there exist local coordinates such that $\mu_3(q_{0}) \neq 0$,
and consequently, all admissible directions are finite. In what follows, we shall assume that this condition is satisfied.
Then the polynomial $M(q_{0},p)$ has either a unique real $p_0$ or three different real roots $p_0, p_1, p_2$.

By Lemma \ref{L0}, all solutions of equation \eqref{4}, \eqref{5} that issue from a generic point $q_{0} \in \Gamma$
are $\pi$-projections of integral curves of the field \eqref{X13} that tend to its singular point $(q_{0},p)$ as
time tends to infinity (positive or negative). Singular points of the field \eqref{X13} are given by the equations
\begin{equation}
\Delta(x,y) = 0, \ \  M(x,y,p)=0,
\label{X14}
\end{equation}
which define a single curve or three non-intersecting curves, each of them corresponds to one of the roots $p_0, p_1, p_2$.
The spectrum $\Sigma$ of the linear part of \eqref{X13} at its singular point $(q_{0},p)$ has the form
\begin{equation}
\Sigma = (0, \lambda_1, \lambda_2), \ \ \lambda_1=\Delta_x + p\Delta_y, \ \ \lambda_2=M_p.
\label{X15}
\end{equation}

From the condition of genericity (see the above definition) it follows that $\lambda_{1,2} \neq 0$.
It is worth observing that $\partial_p$ is the eigenvector with the eigenvalue $\lambda_2=M_p$, while
the eigenvector with the eigenvalue $\lambda_1$ is not vertical.
Denote
\begin{equation}
\lambda = \lambda_2/\lambda_1.
\label{X16}
\end{equation}

Recall several basic definitions concerning local equivalence of vector fields (see, e.g., \cite{AI}).

Let $V_1$ and $V_2$ be the germs of two smooth vector fields in $\bR^n$.
The germs $V_1$ and $V_2$ are called $C^k$-smoothly (resp. topologically) equivalent, if there exists a $C^k$-diffeomorphism (resp.  homeomorphism)
$h \colon \bR^n \to \bR^n$ that conjugates their phase flows $g^t_1$ and $g^t_2$, i.e., $h \circ g^t_1 \equiv g^t_2 \circ h$.
Here $k \ge 1$ is an integer number (finite-smooth equivalence) or $\infty$ (infinite-smooth equivalence).

The germs $V_1$ and $V_2$ are called {\it orbitally} $C^k$-smoothly (resp. topologically) equivalent,
if there exists a $C^k$-diffeomorphism (resp.  homeomorphism) $h \colon \bR^n \to \bR^n$
that conjugates their integral curves (orbits of their phase flows).
In other words, $V_1$ and $V_2$ are {orbitally} $C^k$-smoothly (resp. topologically) equivalent,
if there exists a non-vanishing scalar function $\varphi$ such that $V_1$ and $\varphi V_2$ are $C^k$-smoothly (resp. topologically) equivalent.

The second definition (orbital equivalence) slightly differs from the generally accepted definition of the orbital equivalence,
where coincidence of the orientation of integral curves is also required.
In fact, our definition is naturally related to directions fields, whose integral curves
do not have an orientation a priori.

Consider the germ of a smooth vector field $V$ in $\bR^3$, whose component belong to the ideal $I$
(in the ring of smooth functions) generated by two of them.
In appropriate local coordinates, such a field has the form
\begin{equation}
\dot \xi = v, \ \
\dot \eta = w, \ \
\dot \zeta = \alpha v + \beta w,
\label{X17}
\end{equation}
where $v,w$ and $\alpha, \beta$ are arbitrary smooth functions on the variables $\xi, \eta, \zeta$.
The set of singular points of \eqref{X17} is given by two equations $v=w=0$ and the spectrum of the linear part of \eqref{X17} at every singular point
has the form $\Sigma = (\lambda_1, \lambda_2, 0)$. For instance, vector field \eqref{X13} has the form \eqref{X17}, in this case
the ideal $I$ is generated by the components $\Delta$ and $M$.

Let $T_0$ be a singular point of \eqref{X17} such that the eigenvalues $\lambda_{1,2}$ are real and non-zero.
Then in a neighborhood of $T_0$, the set of singular points is the center manifold of \eqref{X17}, and the reduction principle
(see \cite{AI, HPS}) leads to the following result:

\begin{lemma}
\label{L1}
The germ of vector field \eqref{X17} at $T_0$ is orbitally topologically equi\-va\-lent to
\begin{equation}
\dot \xi = \xi, \ \ \dot \eta = \pm \eta, \ \ \dot \zeta = 0.
\label{X18}
\end{equation}
For vector field \eqref{X13}, the sign $\pm$ coincides with the sign of $\lambda$ defined in \eqref{X16}.
\end{lemma}

\medskip

Lemma~\ref{L1} shows that vector field \eqref{X17} has a local invariant foliation that is topologically equivalent to $\zeta = \const$.
Under certain additional conditions, this invariant foliation can be brought to the form $\zeta = \const$ by means of a local diffeomorphism,
finitely or infinitely smooth. The main obstacle for the existence of such a diffeomorphism is the resonance relations
\begin{equation}
p \lambda_1 + q\lambda_2 = 0, \ \ p,q \in \bZ_+, \ \  p+q \ge 1,
\label{X19}
\end{equation}
the integer number $p+q$ is called the order of resonance \eqref{X19}.
For instance, using the results of \cite{Sam1}, it is not hard to prove the following statement.

Following \cite{Sam1}, for every integer $k \ge 1$ we define the number $N(k) = 2 \left[ (2k+1) m_1/m_2 \right] + 2$, where
\begin{equation*}
m_1 = \max |\lambda_{1,2}|, \ \ \, m_2 = \min |\lambda_{1,2}|,

\end{equation*}
and the square brackets denote the integer part.
If $\lambda_{1,2}$ have no resonances \eqref{X19} up to the order $N(k)$,
then there exists a local $C^k$-smooth diffeomorphism that brings the germ of \eqref{X17} to the form
\begin{equation}
\dot \xi = X(\xi, \eta, \zeta), \ \ \, \dot \eta = Y(\xi, \eta, \zeta), \ \ \, \dot \zeta = 0,
\label{X21}
\end{equation}
where $X,Y$ are smooth functions that the ideal $I = \<X,Y\> = \<\xi,\eta\>$.

There are examples showing that if $\lambda_{1,2}$ have resonances \eqref{X19}, such diffeomorphism does not exist or at least it has a restricted smoothness. For instance, for a special class of vector fields \eqref{X17}, whose non-zero eigenvalues have resonance \eqref{X19}
with relatively prime $p \ge q$, it is shown that the reduction to \eqref{X21} holds true by mean of $C^{p-1}$-smooth diffeomorphism,
but not $C^{p}$-smooth. See Theorem~5 in \cite{PavRem2020}.

Further we shall deal with the case that $\lambda_{1,2}$ have no resonances \eqref{X19} of all orders.
There are two different cases: $\lambda$ defined in \eqref{X16} is either positive or negative irrational.
In the both cases, for every integer $k \ge 1$ there exists a $C^k$-smooth diffeomorphism
\begin{equation}
f_k: \ U_k \to f_k (U_k),
\label{X22}
\end{equation}
where $U_k$ is a neighborhood of the point $T_0$, that brings \eqref{X17} in $U_k$ to the form \eqref{X21}.
If $\lambda$ is positive, there exists a neighborhood $U$ and a $C^\infty$-smooth diffeomorphism
$f: U \to f(U)$ that brings \eqref{X17} in $U$ to the form \eqref{X21}.
This statement is proved in \cite{Rouss}, a~simpler proof can be obtained by the homotopy method; see \cite{IYa} (Sections 1.2\,--\,1.5).

If $\lambda$ is negative irrational, a $C^\infty$-smooth diffeomorphism $f: U \to f(U)$
that brings \eqref{X17} to the form \eqref{X21} exists only in the exceptionally rare case:
if the value $\lambda$ is the same at all singular points in a neighborhood of~$T_0$.
See, e.g. \cite{Rouss, PavRem2020}. Otherwise, every neighborhood of the point $T_0$ contains a singular point
of the field \eqref{X17} with resonance \eqref{X19}, and there required diffeomorphism $f$ does not exist.

Taking into account the previous reasonings and using results from \cite{IYa, Sam1, Sam2}, one can get the following result.
As before, we put $\lambda = \lambda_2/\lambda_1$. Define the set $\bN^{-1} = \{1/n, \ n \in \bN\}$.

\begin{lemma}
\label{L2}
Assume that $T_0$ is a singular point of vector field \eqref{X17} such that $\lambda$ is either positive or negative irrational.
Then for any integer $k \ge 1$ the germ of vector field \eqref{X17} at the point $T_0$ is $C^k$-smoothly equivalent
to one of the germs at the origin:
\begin{eqnarray}
&&
\dot \xi = a_1(\zeta)\xi, \ \
\dot \eta = a_2(\zeta)\eta, \ \ \phantom{111111111}
\dot \zeta = 0,
\quad \textrm{if} \ \ \lambda \notin \bN \cup \bN^{-1},
\label{X23}\\
&&
\dot \xi = a_1(\zeta)\xi, \ \
\dot \eta = a_2(\zeta)\eta + b_2(\zeta)\xi^n, \ \
\dot \zeta = 0,
\quad \textrm{if} \ \ \lambda_2 = n\lambda_1, \ n>1,
\label{X24} \\
&&
\dot \xi = a_1(\zeta)\xi + b_1(\zeta)\eta^n, \ \
\dot \eta = a_2(\zeta)\eta, \ \
\dot \zeta = 0,
\quad \textrm{if} \ \ \lambda_1 = n\lambda_2, \ n > 1,
\label{X25} \\

&&
\dot \xi = a_1(\zeta)\xi + b_1(\zeta)\eta, \ \
\dot \eta = a_2(\zeta)\eta + b_2(\zeta)\xi, \ \
\dot \zeta = 0,
\quad \textrm{if} \ \ \lambda_1 = \lambda_2,
\label{X255}
\end{eqnarray}
where $n$ is integer, $a_i, b_i$ are smooth functions, $a_i(0)=\lambda_i$.
Moreover, if $\lambda>0$, then the same statement holds true with $k=\infty$.
\end{lemma}

Lemma \ref{L2} states that there exists a $C^k$-smooth diffeomorphism \eqref{X22} bringing the field \eqref{X17} to one of the normal forms
\eqref{X23} -- \eqref{X255}, where $U_k$ is a neighborhood of the point $T_0$ and $f_k(U_k)$ is a neighborhood of the origin,
$f_k(T_0)=0$. All vector fields \eqref{X23} -- \eqref{X255} have the invariant foliation
$$
S_c = \{\zeta = c\}, \ \ c= \const.
$$
This yields the corresponding invariant foliation of the field \eqref{X13} with the leaves $f_k^{-1}(S_c)$.
The restriction of \eqref{X13} to the leaf $f_k^{-1}(S_c)$ is a saddle if $\lambda<0$ or a node if $\lambda>0$.

It is worth observing that \eqref{X23} -- \eqref{X255} are a natural generalization of the Poincar\'e--Dulac normal form
for vector fields \eqref{X17}. The monomial $b_2(\zeta)\xi^n$ in \eqref{X24} and $b_1(\zeta)\eta^n$ in \eqref{X25} are
corollaries of the resonances $\lambda_2 = n\lambda_1$, $\lambda_1 = n\lambda_2$, respectively.
In the case $\lambda_1 = \lambda_2$, the both resonances hold true, whence the normal form \eqref{X255}
contains the both monomials $b_1(\zeta)\eta$, $b_2(\zeta)\xi$.
Generically, none of them can be simplified identically to zero. See, e.g., \cite{Arn71}.

\begin{lemma}
\label{L3}
Consider the germ of vector field \eqref{X13} at its singular point $T_0=(q_0,p_i)$ with non-zero eigenvalues $\lambda_{1,2}$.
If $\lambda_1 = n\lambda_2$ with some integer $n \ge 1$, then the coefficient $b_1(\zeta)$ in normal forms \eqref{X25}, \eqref{X255}
vanishes at $\zeta=0$.
\end{lemma}

{\sc Proof}.
Lemma \ref{L2} states that there exists a $C^\infty$-smooth diffeomorphism $f: U \to f(U)$ bringing the field \eqref{X13} to the normal form
\eqref{X25}, where $U$ is a neighborhood of the point $T_0$ and $f(U)$ is a neighborhood of the origin, $f(T_0)=0$.
Consider the restriction of \eqref{X13} to the leaf $f^{-1}(S_0)$ passing through the point $T_0$.
It is a resonant node that has a single integral curve
with the tangential direction corresponding to the eigenvector with $\lambda_1$
and an infinite family of integral curves with the tangential direction $\partial_p$, the eigendirection with $\lambda_2$.
This family contains the vertical straight line $\pi^{-1}(q_0)$. See Fig.~\ref{pict2}~(a).

Consider the case $\lambda_1 = n\lambda_2$ with integer $n > 1$.
Integrating the restriction of the field \eqref{X25} to the invariant plane $S_0$,
one can get the exact formula for the images of integral curves of this family under the diffeomorphism $f$:

\begin{equation}
\xi = \eta^n(c+\varepsilon \ln|\eta|), \ \ \zeta = 0, \ \ c=\const,
\label{X26}
\end{equation}
where $\varepsilon = {b_1(0)}/{a_2(0)}$.
If $\varepsilon=0$, then all curves of the family \eqref{X26} are $C^\infty$-smooth.
If $\varepsilon \neq 0$, they are only $C^{n-1}$-smooth but not $C^{n}$ at the origin.
This contradicts the fact that one of the curves \eqref{X26} is the image of the vertical straight line $\pi^{-1}(q_0)$
under the $C^\infty$-smooth diffeomorphism $f$. Thus, we proved that $\varepsilon=0$, and consequently, $b_1(0)=0$.

The above reasoning is not applicable if $\lambda_1 = \lambda_2$, but in this case the situation is even simpler. The restriction of \eqref{X255} to the invariant plane $S_0$ is a linear vector field, and its matrix can be brought to the Jordan normal form by an appropriate linear transformation.

\begin{figure}[!htp]
\centering
\includegraphics[width=360pt,height=160pt]{pict20.jpg}
\caption{
Integral curves of the field \eqref{X13} and their $\pi$-projections passing through a fixed point $q \in \Gamma$.
From left to right: $\lambda<0$ (a), $0<\lambda<1$ (b), $\lambda>1$ (c).
The curve $\Gamma$ is depicted as a dotted line.
}
\label{pict2}
\end{figure}

\medskip

Let $q_{0} \in \Gamma$ be a generic singular point of equation \eqref{4}, \eqref{5}
and $p_i$ be an admissible direction at $q_{0}$, that is, $p_i$ is a real root of the cubic polynomial $M(q_{0},p)$ in $p$.
Then $T_0=(q_0,p_i)$ is the corresponding singular point of vector field \eqref{X13}
with non-zero eigenvalues $\lambda_{1,2}$ defined in \eqref{X15}.
As before, we use the notation $\lambda = \lambda_2/\lambda_1$.

Before formulating this theorem, we remark that from the condition $d\Delta(q_0) \neq 0$ it follows that in a neighborhood of $q_{0}$,
$\Gamma$ is a regular curve that locally splits the $(x,y)$-plane into two open domains.
Denote them by $\Gamma_+$ and $\Gamma_-$.
Then solutions of equation \eqref{4}, \eqref{5} issuing from a generic singular point $q_{0} \in \Gamma$
with tangential direction $p_i$ are described by the following theorem.

\begin{theorem}
\label{T3}
Assume that $\lambda$ is either positive or negative irrational.
Then the following statements hold true:

1. If $\lambda<0$, then there exists a $C^\infty$-smooth solution passing through $q_{0}$
and there are no other solutions issuing from $q_{0}$ with tangential direction $p_i$.

2.
If $\lambda>0$, then there exist an infinite number of solutions issuing from $q_{0}$ with tangential direction $p_i$
in $\Gamma_+$ and an infinite number of solutions issuing from $q_{0}$ with tangential direction $p_i$ in $\Gamma_-$.
There exist local coordinates centered at $q_{0}$ such that the curve $\Gamma$ coincides with the axis $x=0$ and the infinite
family of solutions mentioned above has the form
\begin{eqnarray}
&&
y = F(x, c|x|^{\lambda}),
\quad \textrm{if} \ \ \lambda \notin \bN,

\label{X27}\\
&&
y = F(x, x^n(c+\varepsilon \ln|x|)),
\quad \textrm{if} \ \ \lambda = n \in \bN,

\label{X28}
\end{eqnarray}
where $\varepsilon$ is a real number, $F$ is a smooth function on two variables, $c=\const$.
\end{theorem}

{\bf Proof.}
Recall that solutions of equation \eqref{4}, \eqref{5} issuing from $q_{0}$ with tangential direction $p_i$
are $\pi$-projections of integral curves of the vector field \eqref{X13} that enter its singular point $T_0=(q_0,p_i)$
as time tends to infinity. From Lemmas \ref{L1}\,--\,\ref{L3} it follows that such integral curves of the field \eqref{X13}
and their $\pi$-projections are located as it presented in Fig.~\ref{pict2}
(from the left to right: the cases $\lambda<0$, $0<\lambda<1$, $\lambda>1$).

If $\lambda<0$, then the field \eqref{X13} has two integral curves that enter $T_0$~--
the stable and unstable manifolds of the field, both one-dimensional and $C^\infty$-smooth.
One of them coincides with the vertical straight line $\pi^{-1}(q_{0})$,
and its $\pi$-projection is the point $q_0$.
Another one has a non-vertical tangential direction, and its $\pi$-projection is a unique solution passing
through $q_{0}$ with the tangential direction $p_i$. See Fig.~\ref{pict2}~(a).

If $\lambda>0$, then the field \eqref{X13} has a one-parameter family of integral curves that enter the point $T_0$
as time tends to plus or minus infinity. Taking into account Lemmas \ref{L2}, \ref{L3} and
exploiting the notations used in the proofs of these lemmas,
one can see that this family belongs to the invariant surface $f^{-1}(S_0)$,
where $f: U \to f(U)$ is a diffeomorphism bringing \eqref{X13}
to one of the normal forms \eqref{X23} -- \eqref{X255}.
The restriction of these normal forms to the invariant plane $S_0$ is
\begin{eqnarray}
&&
\dot \xi = \xi, \ \  \dot \eta = \lambda \eta,

\quad \textrm{if} \ \ \lambda \notin \bN,
\label{X29}\\
&&
\dot \xi = \xi, \ \  \dot \eta = \lambda \eta + \varepsilon \xi^{n},

\quad \textrm{if} \ \ \lambda = n \in \bN.
\label{X30}
\end{eqnarray}
Integrating, we get the families of integral curves
\begin{eqnarray}
&&
\{\eta = c |\xi|^{\lambda}, \ \ c=\const\}, \ \ \{\xi = 0\},

\quad \textrm{if} \ \ \lambda \notin \bN,
\label{X31}\\
&&
\{\eta = \xi^n(c+\varepsilon \ln|\xi|), \ \ c=\const\}, \ \ \{\xi = 0\},

\quad \textrm{if} \ \ \lambda = n \in \bN.
\label{X32}
\end{eqnarray}
The $\pi$-projections of the images of the curves \eqref{X31}, \eqref{X32} under the diffeomorphism $f^{-1}$
are infinite families of solutions mentioned in the second statement of the theorem.
See Fig.~\ref{pict2}~(b,\,c). It is important to remark that the family of solutions obtained from the curves \eqref{X31}
contains at least one $C^\infty$-smooth solution, which corresponds either to $\eta=0$ or to $\xi=0$.
On the contrary, if $\varepsilon \neq 0$, then the family of solutions obtained from the curves \eqref{X32}
contains no $C^\infty$-smooth solutions, since the only smooth curve $\xi=0$ corresponds to the vertical straight line $\pi^{-1}(q_0)$.
This fact will be used below.

It remains to prove the existence of local coordinates such that the families of solution obtained from the curves
\eqref{X31}, \eqref{X32}  have the forms \eqref{X27}, \eqref{X28}, respectively.
Let us choose smooth local coordinates on the $(x,y)$-plane centered in $q_{0}$ such that the curve $\Gamma$
coincides with the axis $x=0$ and the admissible direction $p_i=0$.
Moreover, in the case \eqref{X31} one can choose local coordinates such that one of solutions issuing from $q_0=0$
lies on the axis $y=0$.

The invariant surface $f^{-1}(S_0)$ is locally presented as the graph of a smooth function
\begin{equation}
y = F_0(x,p).

\label{X33}
\end{equation}
After multiplication by an appropriate scalar factor, the restriction of the field \eqref{X13} to the invariant surface
$f^{-1}(S_0)$ has the form
\begin{equation}
\dot x = x, \ \ \dot p = p A(x,p) + xB(x),
\label{X34}
\end{equation}
where $A,B$ are smooth functions, $A(0)=\lambda$.
By $f_0$ denote the restriction of the local diffeomorphism $f$ to the surface $f^{-1}(S_0)$, it brings \eqref{X34} to the form \eqref{X29} if
$\lambda \notin \bN$ or to the form \eqref{X30} if $\lambda \in \bN$.
Comparing the vector field \eqref{X34} with \eqref{X29}, \eqref{X30}, one can see that $f_0$ can be chosen in the form
\begin{equation}
x = \xi, \ \ p = \eta \varphi(\xi,\eta) + \xi \psi(\xi,\eta),
\label{X35}
\end{equation}
where $\varphi, \psi$ are smooth functions, $\varphi(0)=1$.
For according to the choice of local coordinates, in the case $\lambda \notin \bN$
the axis $p=0$ is an integral curve of \eqref{X34}, whence $B \equiv 0$ in \eqref{X34} and $\psi \equiv 0$ in~\eqref{X35}.
Thus, if $\lambda \notin \bN$, then from \eqref{X31} and \eqref{X35} with $\psi \equiv 0$
we have the following expression for the derivative of solutions: $p=\eta \varphi(x,\eta)$, where $\eta = c|x|^{\lambda}$.
Substituting it into \eqref{X33}, we have
$$
y = F_0(x,\eta \varphi(x,\eta)) = F(x, \eta) = F(x, c|x|^{\lambda}), \ \ c=\const,
$$
with a smooth function $F$. The remaining integral curve $\{\xi = 0\}$ corresponds to the axis $x=0$ on the $(x,y)$-plane,
and therefore, it does not give a solution.

Similarly, if $\lambda = n \in \bN$, then from formulas \eqref{X32} and \eqref{X35} we obtain the expression
$$
p=\eta \eta \varphi(x,\eta) + x \psi(x,\eta), \ \  \eta = x^n(c+\varepsilon \ln|x|).
$$
Substituting it into \eqref{X33}, we obtain
$$
y = F_0(x,\eta \eta \varphi(x,\eta) + x \psi(x,\eta)) = F(x, \eta) = F(x, x^n(c+\varepsilon \ln|x|)), \ \ c=\const,
$$
with a smooth function $F$.
The proof of theorem is completed.

\begin{remark}
\label{Rem1}
{\rm
Theorem~\ref{T3} is valid

for a single admissible direction $p_i$ that required the only condition that the corresponding value $\lambda$
is either positive or negative irrational, even if other admissible directions at the given singular point do not satisfy this condition.
}
\end{remark}

\begin{example}
\label{Ex4}
{\rm
Integrating the differential equation
\begin{equation}
x \frac{dp}{dx} = \alpha p(p^2-1), \ \ \alpha \neq 0,
\label{X36}
\end{equation}
we get two families
\begin{equation}
p= \pm  \frac{1}{\sqrt{1+c|x|^{2\alpha}}},  \ \ c={\rm const},
\label{X37}
\end{equation}
and the single solution $p=0$, which can be considered as the limit of \eqref{X37} as $c = +\infty$.

The polynomial $M(p)=\alpha p(p^2-1)$ has three roots $p_0=0$, $p_{1,2}=\pm 1$, which correspond to singular points of the vector field
\begin{equation}
\dot x = x, \ \ \dot y = px, \ \ \dot p = \alpha p(p^2-1).
\label{X38}
\end{equation}
At every singular point $x=0$, $p=0$, the value $\lambda = -\alpha$, while
at every singular point $x=0$, $p=\pm 1$, the value $\lambda = 2\alpha$.
This explains the difference between the cases $\alpha > 0$ and $\alpha < 0$.

If $\alpha > 0$, then for every point $q_0 \in \Gamma = \{x=0\}$ there exist
two infinite families of solutions issuing from $q_0$ with tangential directions $p_{1,2}= \pm 1$.
Solutions of the both families are obtained from \eqref{X37} when the constant $c$ runs through the real numbers.
Moreover, there exists the single solution $y=y_0$ passing through the point $q_0$ with the direction $p_0=0$.

If $\alpha < 0$, then there exist two solutions passing through the point $q_0$ with the admissible directions $p_{1,2}$:
$y=y_0 \pm x$, which can be obtained from \eqref{X37} with $c=0$.
Moreover, there exists a one-parameter family of solutions issuing from $q_0$ with the admissible direction $p_0=0$,
which can be obtained from formula \eqref{X37} with all possible $c>0$, including the value $c=+\infty$
(the solution $y=y_0$).

Everything that was said above is in accord with Theorem~\ref{T3}.
All one-parameter families of solutions of equation \eqref{X36} have the form \eqref{X27},
while the form \eqref{X28} with $\varepsilon \neq 0$ does not appear.
The latter property is also in accord with the general theory.
Indeed, the field \eqref{X38} has the form \eqref{X34} with $B \equiv 0$,
which yield $\varepsilon = 0$ in the final form~\eqref{X28}.

In the next example, the logarithm appears.
}
\end{example}

\begin{example}
\label{Ex5}
{\rm
Consider the equation
\begin{equation}
x \frac{dp}{dx} = \alpha p + f(x), \ \ f(x) = \sum_{i\ge 1} f_i x^i, \ \ \alpha \neq 0,
\label{X39}
\end{equation}
where $f(x)$ can be an analytic function or a polynomial of any degree.
From the general viewpoint, here we have the cubic polynomial
$M(p) = \mu_3 p^3 + \mu_2 p^2 + \alpha p + f(x)$, $\mu_2=\mu_3=0$.
Since $f(0)=0$, this yields the admissible direction $p=0$ and the double admissible direction $p=\infty$.
Theorem~\ref{T3} is applicable for studying solutions that issue from singular points of equation \eqref{X39}
with the tangential direction $p=0$ (see Remark~\ref{Rem1}).
It claims that in the case $\alpha<0$ there exists a unique solution of such type, while
in the case $\alpha>0$ there exists an infinite family of such solutions.

Linear equation \eqref{X39} can be explicitly solved, for instance, using the method of variation of parameters.
For simplicity (avoiding modules), we consider it for $x \ge 0$. Then the general solution has the form
$p = A(x) x^{\alpha}$, where $A'(x) x^{\alpha+1} = f(x)$, i.e., $A'(x) = \sum_{}f_i x^{i-\alpha-1}$.
Assume that $\alpha$ in not integer. Then, after integration the latter equality, we get
\begin{equation*}
p = x^{\alpha} A(x) = \sum_{i \ge 1}g_i x^{i} + cx^{\alpha}, \ \ \ g_i = \frac{f_i}{i-\alpha}, \ \ c=\const.
\end{equation*}
Integrating by $x$, we obtain the expression \eqref{X27} for all solutions
that issue from singular points of equation \eqref{X39} with the tangential direction $p=0$.

Now consider the case that $\alpha$ in integer, let us put $\alpha=n$.
Then, after integration the equality $A'(x) = \sum_{}f_i x^{i-\alpha-1}$, we get
\begin{equation*}
p = x^{\alpha} \Bigl( \sum_{i \neq n}g_i x^{i-n} + f_{n} \ln x + c \Bigr) =
\sum_{i \neq n}g_i x^{i} + x^n (c + f_{n} \ln x),
\end{equation*}
where $g_i$, $i \neq n$, are the same as above and $c=\const$.
Integrating by $x$, we obtain the expression \eqref{X28} for all solutions
that issue from singular points with the tangential direction $p=0$.
}
\end{example}

\textbf{Acknowledgements}.

This work is supported
by the Russian Foundation for Basic Research (projects 19-51-50005 and 20-01-00610).
and by the Ministry of Science and Higher Education of the Russian Federation (Goszadaniye N.~075-00337-20-03, project 0714-2020-0005).

\small


TITLE: Simplify arbitrary union expression in set theory
QUESTION [1 upvotes]: Let $A_k$ be the the set  $A_k=\left\{x\in\left.\mathbb{R}\right|x^2-\left(4k+1\right)x+4k^2+2k<0\right\}$
I need to find a simplified way to rewrite the following group : $\mathbb{R}\setminus\left(\bigcup_{k\in z} A_k\right)$
By simplified I mean - not using arbitrary union or arbitrary intersection
The only equality I could think of is the following:$$\mathbb{R}\setminus\left(\bigcup_{k\in z} A_k\right) = \bigcup_{k\in z}\left[2k+1,2k+2\right]$$
But I still had to use arbitrary union, which I can't.
I would appriciate your assitance with finding the answer and prove the  equality.

REPLY [1 votes]: If you solve the inequality in $A_k$ you get $A_k = (2k,2k+1)$.
Then $\mathbb{R} \setminus \cup_k A_k = \cup_k[2k+1,2k+2]$.
You could write this as $[1,2]+2 \mathbb{Z}$.


TITLE: In theory, could we heat up trash to a plasma for recycling purposes?
QUESTION [0 upvotes]: To my knowledge, any substance when heated up enough to the point of a plasma should completely break any molecular bonds, since the electrons are no longer bound to the atoms. So a molecule like a hydrocarbon should, at plasma temperatures, completely break apart into carbon nuclei, hydrogen nuclei, and a sea of electrons, since the concept of molecules just doesn't exist at those temperatures. We can then use magnetic fields in order to attract the plasma towards a horizontal plate, then apply a strong magnetic field in order to separate everything out. Since we know due to the atomic mass exactly how much a carbon nucleus would bend it's path, exactly how much hydrogen would bend it's path (Relatively a lot more than carbon due to it's much lesser mass), and also the same for electrons, which should bend quite significantly, if in the other direction since it's reversed charge.
So, could we not take a handful of unwanted trash, heat it up to the point of everything becoming a plasma (Hydrocarbons, radioactive materials, toxic materials, etc all turned to this plasma), and then take the massive mixture of elements and feed it through a magnetic seperator in order to get out clumps of pure carbon (Either in the form of coal, diamonds, or some other form), oxygen (Which should form either ozone or oxygen gas), nitrogen gas, phosperous, sulphur, etc once it all cools down?
I don't see why such a thing couldn't work (Albeit taking a metric ton of energy in order to do so). It could be a completely overkill way to recycle any form of trash at all, and recycle it directly back into it's atomic components. Isotopes would obviously be a pain to work with since many isotopes have very similiar masses and charges (Causing the bending through the magnetic plates to have almost the exact same effect to the exact same intensity), but I see this as a completely overkill "Recycle anything" solution.
So would such an idea work, or am I missing something obvious?

REPLY [2 votes]: You appear to be describing a mass spectrometer, a device that heats materials to a plasma and separates them into parts of different charge/mass ratio to identify them. Mass spectrometers are normally designed to process only very tiny quantities, but I understand the same principle has been proposed for large-scale separation of Uranium isotopes for bomb making, a device called a calutron.
Such a device would be physically possible, but ridiculously uneconomic for the purpose of trash recycling.


TITLE: Need help understanding a relation between the fundamental forms
QUESTION [6 upvotes]: The book I am reading briefly mentions this relation between the fundamental forms but gives no explanation of how they got it.
Take the following as the Weingarten Map/Shape Operator where $\nu$ is the gauss map and f is a surface element.
$$L =  - D\nu  \circ {(Df)^{ - 1}}$$
Then it defines the fundamental forms as...
$$I(X,Y) = \left\langle {X,Y} \right\rangle $$
$$II(X,Y) = I(LX,Y)$$
$$III(X,Y) = I(LX,LY)$$
It goes through all of this and explains it but then it gives me the following relation with no support...
$$III - tr(L)II + \det (L)I = 0$$
Can someone show me how we would go about verifying it?

REPLY [4 votes]: Notice that this equation is a $2$-dimensional phenomenon: it only holds on surfaces. In this case we can use the identities $\mathrm{tr} L = 2 H$ and $\det L = K$, where $H$ and $K$ are the mean curvature and the Gaussian curvature, respectively.
A straightforward verification can be done by computing the left hand side in a suitable basis and appealing to the tensoriality of all the terms. Clearly, at any point of the surface we can take a principal basis $\{e_1,e_2\}$ and compute $III - 2 H\, II + K\, I$ on all combinations $\{e_i,e_j\}$ for $i,j=1,2$ to see that this expression indeed vanishes (here I am following A.Gray et al., Modern Differential Geometry of curves and surfaces with Mathematica. Third Ed., p.403, see e.g.here).
Pretty similarly, W.Klingenberg in "A course in differential geometry", p.48, takes the quantity $IV:=(D \nu + \kappa_1 D f)(D \nu + \kappa_2 D f)$ and observes that $IV(e_1,X)=IV(X,e_2)=0$ for an arbitrary $X$ tangent to the surface. Expanding the definitions, one sees that $IV = III - 2 H\, II + K\, I$.
Another way is to observe that $III - \mathrm{tr}(L)II + \det (L)I$ is in fact the Cayley-Hamilton equation for the Weingarten operator $L$.

As an aside, I want to take the chance and continue the propaganda of the abstract index notation. I will derive this equation along with a particular case of the Cayley-Hamilton theorem, following M.Eastwood, "Invariant Theory in Differential Geometry", Proceedings of the Centre for Mathematics and its Applciations, 1994, pp.71-78, see here.
Since I have spent a bit of time explaining the abstract index notation on this site, see e.g. my answers here and here, I will use the conventions form those answers extensively. In particular, the tangent space $T_p S$ to the surface $S$ at a point $p$ will be denoted by $E^a$. The vector bundles are denoted by the same letters as the fibers, so, e.g., $E_a$ will mean both the cotangent space at some point, and the cotangent bundle.
Let us write the Weingarten operator as $L_a{}^b$, and notice that the third fundamental form can be presented as $III_{a b} = g_{c d} L_a{}^c L_b{}^d$, where $g_{a b}$ is the inner product in the tangent space of the chosen point. Also, we write $\mathrm{tr}L = L_a{}^a$. In fact, we use $g_{a b}$ and its inverse $g^{a b}$ tacitly to raise and lower indices ("musical isomorphisms"). Furthermore, $III_a{}^b = (L^2)_a{}^b$. The identity operator  is presented by (the abstract-index version of) the Kronecker symbol $\mathrm{id}_a{}^b \equiv \delta_a{}^b$.
The Cayley-Hamilton theorem is equivalent to an obvious fact that on $n$-dimensional vector space all exterior forms of degree $> n$ vanish identically (the latter fact is collectively called the Cayley-Hamilton identity). Moreover, this holds for exterior forms with values in any vector space!
Consider $L_{[a}{}^d L_{b}{}^e \delta_{c]}{}^f \in E_{[abc]}\otimes E^{d e f}$, a $3$-form on $T_p S$ with values in $\otimes^3 T_pS$. From the Cayley-Hamilton identity, this form is identically zero if $\dim T_p S = 2$. In particular, all the traces of this quantity are zeroes!
Hopefully, the above paragraphs justify enough the following calculation.
$$
\begin{align}
0_a{}^d & = 6 \, L_{[a}{}^b L_b{}^c \delta_{c]}{}^d\\
& = L_a{}^b L_b{}^c \delta_c{}^d + L_b{}^b L_c{}^c \delta_a{}^d + L_c{}^b L_a{}^c \delta_b{}^d - L_b{}^b L_a{}^c \delta_c{}^d - L_a{}^b L_c{}^c \delta_b{}^d - L_c{}^b L_b{}^c \delta_a{}^d \\
& = (L^2)_a{}^d + (\mathrm{tr}L)^2 \delta_a{}^d + (L^2)_a{}^d - (\mathrm{tr} L) L_a{}^d - (\mathrm{tr} L) L_a{}^d + (\mathrm{tr}L^2) \delta_a{}^d \\
& = 2 (L^2)_a{}^d - 2(\mathrm{tr}L)L_a{}^d + [(\mathrm{tr}L)^2 - \mathrm{tr}(L^2)]\delta_a{}^d
\end{align}
$$
Exercise. Show that $(\mathrm{tr}L)^2 - \mathrm{tr}(L^2) = 2\,\det L$ for any $2\times 2$-matrix $L$.


TITLE: How many minimal spanning trees has this particular graph?
QUESTION [1 upvotes]: I'm counting minimal spanning trees and my suggestion was the following, when I have a graph with 10 vertices and 13 edges:
For a minimal spanning tree I need 1 edge less then vertices, so (10,9). I can choose now 4 edges to be deleted to get a minimal spanning tree.
However I have to take care to not disconnect the graph. But I'm missing something. Do I have to take care of 4 or 5 circles in the following example? And how can I compute then the amount of spanning trees? Is there a closed formula (not Prims algorithm)?
It should be mentioned, that the normal edges are weighted twice as the bolded edges.

REPLY [0 votes]: Problems like this typically require a bit of case-work.
Let's view the graph as two triangles connected by three vertical paths.
We need to delete four edges. Since we want minimal spanning trees, we should delete as many high-weight (thin) edges as possible.
However, we have to delete at least one edge from the top triangle and one edge from the bottom triangle. (There are 9 ways to do this.)
The remaining two deletions can be thin edges: We just need to make sure we cut two of the three vertical paths. Depending on which two paths we cut, there are either 2, 4, or 2 ways to do this.
So in total there are $9 \times (2+4+2) = 72$ minimal spanning trees.


TITLE: How to prove that $\mathbb{Z}[X]/g\mathbb{Z}[X]$ is not a field?
QUESTION [5 upvotes]: How to prove that $\mathbb{Z}[X]/g\mathbb{Z}[X]$ is not a field, where $g$ is a non constant polynomial in $\mathbb{Z}[X]$?

(The key is to show that $f : \mathbb{Z}[X]/g\mathbb{Z}[X] \to \mathbb{Z}/p\mathbb{Z}$ is not injective.)
edit 1:
Now I'm trying to find the characteristic of the given ring.
Consider the map: $g: \mathbb{Z} \to \mathbb{Z}[X]/g\mathbb{Z}[X]$, where $m \mapsto 1 + 1 + \cdots + 1 $ (m times). Since the $\ker f$ is an ideal of $\mathbb{Z}$, it has to be of the form $n\mathbb{Z}$, where $n$ is the characteristic. So I guess the characteristic is $0$...no?
edit 2: Thank you for all the replies below! I looked through the answer but it's a little advanced for me... I figured it out eventually using some more basic ideas and I'll put it here when I get some time...Thanks again!
edit 3:
My solution:
Step 1: Prove there is a $a \in \mathbb{Z}$ such that $g(a) \neq 0, \pm 1$ and let $p$ be a prime number that divides $g(a)$.
Suppose $g(X) = \sum_{i = 1}^na_iX^i$, then $g(X) = 0$ has at most $n$ integer solutions, $p(X) = g(X) -1 = 0$ has at most $n$ integer solutions, and so is $q(X) = g(X)+1 = 0$. There are at most $3n$ integers that satisfy $g(a) = 0,\pm1$, but $\mathbb{Z}$ is infinite, so there has to be a $a \in \mathbb{Z}$ such that $g(a) \neq 0, \pm 1$.
Step 2: Prove that there is a unique well-defined surjective morphism of rings $f: \mathbb{Z}[X] \to \mathbb{Z}/p\mathbb{Z}$ sending $X$ to $a$ mod $p$ and that it yields a homomorphism of rings $\phi: \mathbb{Z}[X]/g\mathbb{Z}[X] \to \mathbb{Z}/p\mathbb{Z}$.
Observe that $g\mathbb{Z}[X]$ is in the kernel, so the map factor through $\mathbb{Z}[X]/g\mathbb{Z}[X]$.
Step 3: Show that the resulting map is not injective.
Note that the map $\phi$ is injective if and only if $\ker f = g\mathbb{Z}[X]$, but actually $g\mathbb{Z}[X] \subset \ker f$. Alternatively, according to my prof, if the map was injective, $\mathbb{Z}[X]/g\mathbb{Z}[X]$ would identify with a subring of $\mathbb{Z}/p\mathbb{Z}$ so it would have characteristic $p$, contradiction!
Step 4: We use that if $g: K \to A$ where $K$ is a field, then $g$ is either injective or the zero map.
Since the above map is not a zero map and we proved that it's not injective, $\mathbb{Z}[X]/g\mathbb{Z}[X]$ can't be a field.

REPLY [6 votes]: My answer expands on JonHales comment. Suppose your polynomial is $g(x)=a_nx^n + \cdots + a_0$, with $a_i\in\Bbb{Z}$, $a_n\ne 0$, and $n > 0$. Choose a prime $p$ with $p\nmid a_n$. Then mod $p$, $g(x)$ still has degree $n$.
Thus $\Bbb{Z}[x]/(g,p)\simeq \Bbb{F}_p[x]/(g)\ne 0$, so $(g,p)$ is a proper ideal of $\Bbb{Z}[x]$. Moreover, it properly contains $(g)$, since $(g)\cap \Bbb{Z}=(0)$. Thus $(g)$ is not maximal, and $\Bbb{Z}[x]/(g)$ is not a field.
Edit
In response to the OP's edit, saying that they found this solution a bit advanced, I thought I'd translate the solution into more basic ideas for future readers.
Again, if our polynomial is $g(x)=a_nx^n+\cdots + a_0$, then if $p$ is a prime integer with $p\nmid a_n$, $p$ is nonzero, but not a unit in $\Bbb{Z}[x]/(g)$. Therefore $\Bbb{Z}[x]/(g)$ is not a field.
Proof.
By contradiction. Assume $p \cdot c(x) \equiv 1 \pmod{g(x)}$ for some polynomial $c(x)$. Then $pc(x)-1 = g(x)h(x)$ for some polynomial $h(x)$, so $pc(x) = g(x)h(x)-1$. Write $h(x)=pq(x)+r(x)$, where the coefficients of $r$ are either between $0$ and $p-1$ (inclusive).
Then $$pc(x)=g(x)(pq(x)+r(x))-1=pg(x)q(x) + g(x)r(x)-1,$$
and
$$p(c(x)-g(x)q(x))=g(x)r(x)-1.$$
Now if $r(x)=0$, we have that the left hand side is divisible by $p$, and the right hand side is $-1$, so this is impossible.
On the other hand, if $r(x)\ne 0$, $g(x)r(x)$ has positive degree, and it's leading term is $a_nb_m$, where $b_m$ is the leading term of $r(x)$. However $p$ must divide $g(x)r(x)-1$, so it must divide the leading term, so $p\mid a_nb_m$. However $p\nmid a_n$, and $1\le b_m\le p-1$, so $p\nmid b_m$ either. Thus $p\nmid a_nb_m$. Contradiction.


TITLE: A clear reference on regular conditional distributions?
QUESTION [3 upvotes]: I've been trying to learn about regular conditional distributions from Klenke's book on probability theory, but I'm incredibly confused. I looked at Durrett's book, but his chapter on regular conditional distributions is short and lacks examples. Does anyone know of another reference that is gentle and gives many examples? It would be great if it also included a treatment of the Borel-Kolmogorov paradox, but perhaps that is too much to ask.
I am not familiar with the probability literature, so perhaps this is an easy question and another standard textbook gives a good treatment. If so, I welcome that as an answer!

REPLY [1 votes]: David Williams, Probability with Martingales. About 250 pages in A5 format (and not in small print), exercises included, extremely congenial.


TITLE: $\infty$-categorical understanding of Bridgeland stability?
QUESTION [8 upvotes]: On triangulated categories we have a notion of Bridgeland stability conditions.
Is there any known notion of "derived stability conditions" on a stable $\infty$-category $C$ such that they become Bridgeland stability conditions after passing to the homotopy category $hC$?

REPLY [6 votes]: Yes, this is chapter 7 of Fosco Loregian's thesis, linked from his webpage. The paper Simone Virili linked to is one of 3 papers making up the thesis. Specifically, Section 7.2.1 discusses the topology, after Bridgeland, and proves the comparison you asked for. The setting is a slight generalization of that of Bridgeland's original papers, as Loregian makes clear.


TITLE: Equating coefficients of binomial expansion modulo p
QUESTION [0 upvotes]: In this answer: https://math.stackexchange.com/a/652909
Ted equates mod $p$ the coefficients of $$\sum_{n=0}^{pa} \binom{pa}{n} x^n$$ and $$\sum_{i=0}^{a} \binom{a}{i} x^{pi}$$ to get that $$\binom{pa}{pb} \equiv \binom{a}{b} \mod p$$
But one of the summations has $pa$ terms, and the other summation has $a$ terms, so what does it mean to equate the coefficients mod $p$? Shouldn't it be that several different elements of $\sum \binom{pa}{n} x^n$ will collapse into one element of $\sum \binom{a}{i} x^{pi}$ since $x^{pb} \equiv x^{p^kb} \mod p$ so that after modding by $p$, the coefficients of each $x^{p^kb}$ will contribute to the coefficient of $x^{pb}$ i.e. $$\binom{pa}{b} + \binom{pa}{pb} + \binom{pa}{p^2b} \dots \equiv \binom{a}{b} \mod p$$
Is Ted's proof complete? And if not, what would a complete proof look like?

REPLY [1 votes]: In the original post, this was not my proof, but clark's.  I merely edited some typos in it.  But I can address your concerns.
First, you cannot write something like $x^{pb} \equiv x^{p^k b} \mod{p}$.  Not sure how you got that, so I'll just say this: When working with polynomials mod $p$, the exponents do not reduce modulo $p$. The exponents are ordinary (nonnegative) integers.  You can only equate powers that are exactly equal as integers (you cannot reduce them modulo $p$).  Only the coefficients can be reduced modulo $p$.
So what do we get when we equate coefficients of equal powers of $x$?  On the right side (the second sum in your question above), the term $i=b$ gives $\binom{a}{b}$ as the coefficient of $x^{pb}$.  This corresponds to the term $n=pb$ in the first sum, which has coefficient $\binom{pa}{pb}$.  These can be equated modulo $p$, giving $\binom{pa}{pb} \equiv \binom{a}{b} \mod p$.
To address your concern that the first sum has $pa$ terms and the second sum only $a$ terms, what's happening is that most of the terms in the first sum are 0.  If $n$ is not divisible by $p$, then $\binom{pa}{n}$ is divisible by $p$ and therefore becomes 0 when working modulo $p$.  The argument in the previous paragraph is a proof of this (since the second sum obviously has no terms that are not multiples of $p$), but you can also prove it directly.


TITLE: How to compute the $ \lim_{x \to 0+} x \int_x^1\frac{\cos t}{t^2}\, dt$?
QUESTION [2 upvotes]: I am working out the proof that
$$ \lim_{→0+} x \int_x^1\frac{\cos t}{t^2}\, dt =1 $$
My attempt is firstly handle the integral parts:
$$ \int_x^1\frac{\cos t}{t^2}\, dt = \left(-\left(-\text{Si}(x)-\frac{\cos (x)}{x}\right)-\text{Si}(1)-\frac{\cos (1)}{1}\right. $$
where
$$ \text{Si}(z)=\int _0^z\frac{\sin(t)}{t}\, dt $$
Then
$$x \int_x^1\frac{\cos t}{t^2} = xSi(x) + \cos(x)-xSi(1)-x\cos(1)$$
so,
$$\lim_{→0+} x \int_x^1\frac{\cos t}{t^2} = \lim_{→0+} (xSi(x) + \cos(x)-xSi(1)-x\cos(1))$$
And also,
$$ \lim_{→0+} xSi(1) = 0$$
$$\lim_{→0+}x\cos(1)=0 $$
$$\lim_{→0+}\cos(x)=1 $$
$$\lim_{→0+} xSi(x)=0 ??$$
$$$$
then $$\lim_{→0+} x \int_x^1\frac{\cos t}{t^2} = 1$$
I am uncertain about my proof. Is it correct? Thanks in advance.

REPLY [3 votes]: A simpler way is to apply the Hopital's rule to and then the Fundamental theorem of calculus:
$$\lim_{x→0+} \frac{\int_x^1\frac{\cos(t)}{t^2}\, dt}{1/x}=
\lim_{x→0+} \frac{-\frac{\cos(x)}{x^2}}{-1/x^2}=\lim_{x→0+}\cos(x)=1.$$


TITLE: Probability (Family of 3 children)
QUESTION [2 upvotes]: Suppose that from an arbitrary family of $3$ children we choose one of them at random. If the chosen one is a girl, what is the probability that she has a sister and a brother?
(Note: suppose that the probability of being a boy or girl is the same and that the events are independent. i.e the sex of one child doesn't affect the probability of the sex of the other ones.)

What I have tried:
So this is probably wrong, but my approach was saying this is the same as asking what's the probability of two of them being girls and the other one a boy. What is really bugging me is that BGG, GGB, and GBG appear to be the same, so I don't really know what the sample space is...
So my next approach was saying, this can be rephrased as saying:
$$P(2\text{ girls, }1\text{ boy)}=P(2\text{ girls})-P(3\text{ girls})$$
So $P(2\text{ girls})$ must be equal to $P(2\text{ boys})$ since any family has at least 2 boys or 2 girls and neither can happen in the same family, meaning $P(2\text{ girls}) + P(2\text{ boys})=1$, then $P(2\text{ girls})=\frac{1}{2}$. $P(3\text{ girls})$ must be $\frac{1}{8}$ and so $P(2\text{ girls, }1\text{ boy})= \frac{1}{2} - \frac{1}{8} = \frac{3}{8}$
Is this correct or am I misunderstanding the problem? Or could it be $P(1\text{ boy, }2\text{ girls}|1\text{ girl})$ ?  Thanks in advance.
Oh, and the problem says you can use Bayes to solve this but I dont see how to be honest...

REPLY [0 votes]: There are $2^3$ outcomes in the preliminary sample space $\{B,G\}^3$, and these may be assumed to be reasonably equally probable. That is eight.  We then select the first, second, or third child without bias, and it so happens that we do select a girl.
Intuitively, take eight boxes and place three blue-or-green marbles in each as: one box with three green marbles, three boxes with two green marbles, three boxes with one green marble, and the last box has no green marbles.
There are $3+6+3+0$ green marbles, from which we may select without bias, of which $6$ are in boxes with two green and one blue marble.

Mathematically, let $G$ be the count of girls, and $S$ the event of selecting a girl.   Then by conditional probability and the law of total probability:
$$\begin{align}\mathsf P(G=2\mid S)&=\dfrac{\mathsf P(S\cap G=2)}{\mathsf P(S)}\\[1ex]&=\dfrac{\mathsf P(S\cap G=2)}{\mathsf P(S\cap G=1)+\mathsf P(S\cap G=2)+\mathsf P(S\cap G=3)}\\[1ex]&=\dfrac{\tfrac 23\binom 32\tfrac 18}{\tfrac 13\binom 31\tfrac 18+\tfrac 23\binom 32\tfrac 18+\tfrac 33\binom 33\tfrac 18}\\[1ex]&=\dfrac{6}{3+6+3}\\[1ex]&=\dfrac{1}{2}\end{align}$$


TITLE: Point-Set Topology Proof Verification
QUESTION [0 upvotes]: I am self-studying Baby Rudin (after some idleness) and his second chapter is on basic point-set topology.
Now, he asks the following:
Is every point of every open set $E\subset\mathbb{R}^2$ a limit point of $E$. Answer the same question for closed sets.
My answer was:
For closed sets, this is obviously false as seen through the case of a singleton set.
For open sets: (When I write a neighborhood of $x$, I mean the deleted neighborhood rather than the one with $x$)
Let $x$ be a non-limit point of $E$. Thus, $\exists N_1(x,r_1)\subset E^c$.
Assume $x\in E$ then $\exists N_2(x,r_2)$ such that $N_2(x,r_2)\subset E$.
There arises two cases:
a) $N_2\subset N_1\Rightarrow N_2\subset E^c \Rightarrow N_2=\phi$ since $N_2\subset E$ as well. Thus, $E=\{x\}$ which contradicts the fact that $E$ is open.
b) $N_1\subset N_2\Rightarrow N_1\subset E \Rightarrow N_1=\phi$ since $N_1\subset E^c$ as well.
However, in $\mathbb{R}^2$, no neighborhood is empty. (I believe this can be trivially proved for any $\mathbb{R}^n$ by connectedness of $\mathbb{R}$).
Thus, we have a contradiction and $x$ doesn't belong to $E$.
By law of contraposition, we proved that every point of $E$ is a limit point of $E$.
My doubt was the fact that I barely used the fact that $E\subset R^2$. I believe this is the fact that differentiates $E$ from another open set in some metric space where a neighborhood is allowed to be empty.
I hope someone would verify my proof. Thanks in advance.

REPLY [1 votes]: Your proof looks correct, but I think your statements about "no neighborhood being empty" are missing the main idea here. In fact, $\emptyset$ is open in every topology, so it's a neighborhood in $\mathbb R^2$. It's just that since it doesn't contain any points, it's not a neighborhood of any point.
What you're meaning to say is that the punctured neighborhoods $N_1 \setminus \{x\}$ and $N_2 \setminus \{x\}$ cannot be empty. This is equivalent to the statement that no singleton set in $\mathbb R^2$ is open. This really is something which is special to $\mathbb R^2$ (and many other topological spaces) but doesn't hold even for general metric spaces. For instance, $\mathbb Z \subset \mathbb R$ is a metric space, but it has the discrete topology, meaning that every subset of $\mathbb Z$ is open in its topology. So indeed, not every point of an open subset of $\mathbb Z$ is a limit point of that set. In fact, no point is a limit point for any subset of $\mathbb Z$!


TITLE: Calculus: Integration problem.
QUESTION [0 upvotes]: Does there exist a function $f(x) \neq \dfrac{1}{x}$  such that $\int_1^{\infty} f(x) \mathrm d{x} = \log t$ for all $t>0$ ?
Note: the exception $f(x) = \dfrac{1}{x}$ was treated as the limit of $\int_1^{t} f(x) \mathrm d{x}$ as $t$ tends to infinity.

REPLY [0 votes]: This is just a minor modification of Simon's answer:
Suppose $\int_1^{\infty} f(x) \mathrm {d}x = \log t$ for all $t > 0$
This be written in the form $\lim_ {t \to \infty} \int_1^{t} f(x) \mathrm {d}x = \log t$
From here we can then apply the Fundamental Theorem of Calculus and differentiate both sides and conclude that the only possibility is $f(x) = x^{-1}$.


TITLE: Understanding a proof from the APMO 1998 on inequalities.
QUESTION [4 upvotes]: I was having trouble with proving the following inequality.The question was from the book Secrets to Inequalities by Pham Kim Hung.
$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{xyz} }$
I know that the AM-GM inequality must be applied but I am having a hard time proving the the above inequality. In the book the author was able to show that by applying the trick shown below and then applying AM-GM we can prove the inequality holds.
$3\left (\frac{x}{y}+\frac{y}{z}+\frac{z}{x}  \right ) = \left ( \frac{2x}{y} +\frac{y}{z}\right ) + \left ( \frac{2y}{z}+\frac{z}{x} \right ) + \left ( \frac{2z}{x}+\frac{x}{y} \right ) \geq \frac{3x}{\sqrt[3]{xyz}} + \frac{3y}{\sqrt[3]{xyz}} + \frac{3z}{\sqrt[3]{xyz}}$
The problem I had was after applying the AM-GM to the LHS I could not prove the RHS.

REPLY [1 votes]: You can get the Pham Kim Hung's proof by the following way.
We'll try to find non-negatives $a$, $b$ and $c$ such that $a+b+c=1$ and
$$a\cdot\frac{x}{y}+b\cdot\frac{y}{z}+c\cdot\frac{z}{x}\geq\frac{x}{\sqrt[3]{xyz}}$$
Now, by AM-GM
$$a\cdot\frac{x}{y}+b\cdot\frac{y}{z}+c\cdot\frac{z}{x}\geq\left(\frac{x}{y}\right)^a\left(\frac{y}{z}\right)^b\left(\frac{z}{x}\right)^c=x^{a-c}y^{b-a}z^{c-b}.$$
Thus, we need $$x^{a-c}y^{b-a}z^{c-b}=\frac{x}{\sqrt[3]{xyz}}$$ or
$$x^{a-c}y^{b-a}z^{c-b}=x^{\frac{2}{3}}y^{-\frac{1}{3}}z^{-\frac{1}{3}},$$ which gives the following system.
$$a-c=\frac{2}{3},$$
$$b-a=-\frac{1}{3}$$ and $$a+b+c=1.$$
After solving of this system we obtain:
$$(a,b,c)=\left(\frac{2}{3},\frac{1}{3},0\right),$$ which gives that by AM-GM
$$\frac{2x}{3y}+\frac{y}{3z}\geq\frac{x}{\sqrt[3]{xyz}}$$ and since
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\sum_{cyc}\frac{x}{y}=\sum_{cyc}\left(\frac{2x}{3y}+\frac{y}{3z}\right)\geq\sum_{cyc}\frac{x}{\sqrt[3]{xyz}}=\frac{x+y+z}{\sqrt[3]{xyz}},$$
we are done!


text:
\begin{document}

\title{3-Manifolds Admitting Toric Integrable Geodesic Flows}
\author{Christopher R. Lee}
\address{Department of Mathematics \\ University of Illinois \\ Urbana, IL 61801}
\email{crlee@math.uiuc.edu}
\date{\today}
\subjclass[2000]{Primary 53D25; Secondary 53D10}

\begin{abstract}
The geodesic flow of a Riemannian metric on a compact manifold $Q$ is said to be toric integrable if it is completely integrable and the first integrals of motion generate a homogeneous torus action on the punctured cotangent bundle $T^*Q\setminus{Q}$. If the geodesic flow is toric integrable, the cosphere bundle admits the structure of a contact toric manifold. By comparing the Betti numbers of contact toric manifolds and cosphere bundles, we are able to provide necessary conditions for the geodesic flow on a compact, connected 3-dimensional Riemannian manifold to be toric integrable.
\end{abstract}

\maketitle

\section{Introduction}
Let $(Q,g)$ be an $n$-dimensional Riemannian manifold. Consider, for each $t \in \R$, the diffeomorphism
\begin{eqnarray*}
\phi_{t}:TQ &\to& TQ \\
\phi_{t}(q,v) &=& (\gamma(t),\dot\gamma(t))
\end{eqnarray*}
where $\gamma$ is the unique geodesic with initial conditions $\gamma(0) = q$ and $\dot\gamma(0)=v$. Unique dependence of geodesics on initial conditions provides $\phi_{t} \circ \phi_{s} = \phi_{t+s}$, i.e. the family $\{\phi_{t}\}$ is actually a flow, the {\bf geodesic flow} of the metric $g$ on the manifold $Q$.

Equivalently, we may describe the geodesic flow in terms of symplectic geometry. Recall that a {\bf symplectic manifold} is a manifold equipped with a closed, nondegenerate two-form $\omega$. The punctured cotangent bundle $T^*Q\setminus{Q}$ is a symplectic manifold; its symplectic form is $\omega = d\alpha$, where $\alpha$ is the one-form defined by $\alpha(v) = \eta(d\pi(v))$. Here, $\pi:T^*Q \to Q$ is the standard bundle projection, $\eta \in T^*Q$ and $v \in T_{\eta}(T^*Q)$. The form $\alpha$ is called the {\bf Liouville one-form}.

Define a smooth function $h:T^*Q\setminus{Q} \to \R$ by
\begin{displaymath}
h(q,p) = \sqrt{g^*_{q}(p,p)}
\end{displaymath}
where $g^*$ is the metric on $T^*Q$ induced by $g$. The {\bf Hamiltonian vector field} $X_{h}$ of $h$ is then the unique solution to the equation
\begin{displaymath}
\iota_{X_{h}}\omega = dh.
\end{displaymath}
The geodesic flow of the metric $g$ on $TQ$ can then be identified with the flow of the Hamiltonian vector field $X_{h}$ on $T^*Q$.

We say that the geodesic flow is {\bf toric integrable} if there exist $n$ smooth functions $h=f_1,\ldots,f_n$ on $T^*Q\setminus{Q}$ such that
\begin{enumerate}
\item $X_{f_i}(f_j)=0$ for all $i,j=1,\ldots,n$,
\item the set $\{df_i\}_{i=1}^{n}$ is linearly independent on an open, dense subset of $T^*Q\setminus{Q}$,
\item each $f_i$ is homogeneous of degree 1: if $\rho_{t}(p,q) = (p,e^tq)$ is the $\R$-action given by dilations, then $\rho_{t}^*f_i = e^tf_i$ for each $i$, and
\item the $\R^n$-action defined by the flows of the vector fields $X_{f_{i}}$ descends to a well-defined action of $\T^n = \R^n/\Z^n$.
\end{enumerate}
In the case that the geodesic flow is toric integrable, there is an effective action of the $n$-torus $\T^n$ on $T^*Q\setminus{Q}$ that preserves both the Liouville one-form $\alpha$ and the function $h$, and which also commutes with dilations.

Since the $\T^n$-action preserves $h$, it preserves to the {\bf cosphere bundle} $S(T^*Q)$ defined as the set of unit vectors in  $T^*Q\setminus{Q}$ with respect to the metric $g^*$. Recall that $S(T^*Q)$ is a {\bf contact manifold}, that is, the Liouville one-form $\alpha$ restricted to $S(T^*Q)$ satisfies $\alpha \wedge (d\alpha)^{n-1} \neq 0$. The $\T^n$-action generated by the integrals of the geodesic flow preserves $\alpha$, and so gives $S(T^*Q)$ the structure of a {\bf contact toric} manifold. The moral is thus:

\begin{center}
{\em If a Riemannian manifold $Q$ admits a toric integrable geodesic flow, then the cosphere bundle $S(T^*Q)$ of $Q$ is a contact toric manifold.}
\end{center}

The notion of toric integrability of the geodesic flow was important to Toth and Zelditch who discovered, while examining the spectra of Laplacians on compact Riemannian manifolds in \cite{tz}, that this condition provided substatntial simplification. However, toric integrability of the geodesic flow seems to be a rather restrictive geometric condition. It is known (see \cite{ls}) that the only metric which support a toric integrable geodesic flow on a torus is the flat metric. On the spheres $S^2$ and $S^3$, the geodesic flow of the round metrics are also toric integrable. Y. Colin de Verdiere in \cite{CdV} has constructed other metrics on $S^2$ which are toric integrable, but a complete classification of such metrics on low dimensional spheres is incomplete.

We instead tackle a less formidable problem: for a given manifold $Q$, are there topological obstructions to toric integrability of the geodesic flow? In light of the preceeding discussion, this amounts to finding topological obstructions to cosphere bundles being contact toric manifolds. The main result of this paper is

\begin{theorem}
Suppose $Q$ is a $3$-dimensional compact, connected Riemannian manifold admitting a toric integrable geodesic flow. Then, the cosphere bundle $S(T^*Q)$ is diffeomorphic to either $S^3 \times S^2$ if $Q$ is simply connected or $\T^3 \times S^2$ if the fundamental group of $Q$ is infinite. If the fundamental group of $Q$ is finite and nontrivial, then $S(T^*Q)$ is homotopy equivalent to $(S^3/\Gamma) \times S^2$ where $\Gamma$ is a finite cyclic group.
\end{theorem}

\section*{Acknowledgements} I would like to thank Eugene Lerman for his invaluable guidance and inspiration as well as Donald Yau for assistance with a calculation and beneficial conversations. In addition, Ian Hambleton provided some references and remarks concerning group actions on spheres for which I am grateful.

\section{Contact Toric Manifolds}
We begin by reviewing some basic facts about contact manifolds. Recall that a $(2n-1)$-dimensional contact manifold $M$ is {\bf cooriented} if there exists a globally defined contact form $\alpha$ on $M$. In this case, the {\bf contact structure} $\xi \subset TM$ is defined as the kernel of the contact form $\alpha$. We make the banket assumption in this paper that all contact manifolds in question are cooriented. If there is an effective action of $\T^n$ on $M$ preserving $\xi$, we say that $M$ is a {\bf contact toric} manifold.

Recall that given a Hamiltonian action of a compact Lie group $G$ on a symplectic manifold $X$, we may associate to this action a (symplectic) moment map $\Phi:X \to \mathfrak{g}^{*}$ (here, $\mathfrak{g}^*$ is the dual of the Lie algebra of $G$). If $M$ is a contact toric manifold, consider the annihilator $\xi^{0} \subset T^*M$ of $\xi \subset TM$. The $\T^n$-action on $M$ lifts to a Hamiltonian $\T^n$-action on a component $\xi_{+}^{0}$. The {\bf contact moment map} is then defined to be the moment map $\Psi:\xi_{+}^{0} \to \mathfrak{t}^{*}$ of this action. Additionally, the image $\Psi(\xi_{+}^{0}) \cup \{0\}$ of the contact moment map together with the origin in $\mathfrak{t}^{*}$ is a rational polyhedral cone, called the {\bf moment cone} of the contact toric manifold $M$.

The classification of compact contact toric manifolds was initiated by Banyaga and Molino in \cite{bm} and completed by Lerman in \cite{l2}. Lerman's result makes use of the moment cone of a contact toric manifold by singling out certain types of polyhedral cones in $\R^n$ he calls ``good cones'' and showing that they correspond to unique contact toric manifolds. We sum his results up in dimensions greater than 3:

\begin{theorem}{\cite{l2}}
Suppose $M^{2n-1}$ is a contact toric manifold with $n>2$. Then,
\begin{enumerate}
\item if the $\T^n$-action is free, $M$ can be given the structure of a principal $\T^n$-bundle over $S^{n-1}$. Moreover, each principal $\T^n$-bundle over $S^{n-1}$ corresponds to a unique compact contact toric manifold.
\item If the $\T^n$ action is not free, the moment cone of $M$ is good. Conversely, every good cone corresponds to a unique compact contact toric manifold.
\end{enumerate}
\end{theorem}

We can say a bit more about the second part of the above classification theorem. If the moment cone of a contact toric manifold contains a subspace of positive dimension, then the contact toric manifold is diffeomorphic to $\T^{k} \times S^{2n-k-1}$ for some $k>0$. If the moment cone contains no lines, the contact toric manifold is said to be of {\bf Reeb type}.

Contact toric manifolds of Reeb type are not, in general, easily identifiable. To every contact manifold $(M,\alpha)$ we may associated a distinguished vector field, the {\bf Reeb vector field} $R_{\alpha}$ which is uniquely determined by the equations
\begin{eqnarray*}
\alpha(R_{\alpha}) &=& 1, \\
\iota_{R_{\alpha}}d\alpha&=&0.
\end{eqnarray*}
The Reeb vector field provides a splitting $TM = \xi \oplus \R{R_{\alpha}}$. Since the vector bundle $\xi$ is symplectic, there exists an almost complex structure $J$ which is compatible with the form $d\alpha|_{\xi}$ so that $\tilde{g} = d\alpha|_{\xi}(\cdot,J\cdot)$ is a metric on $\xi$. Since $TM$ splits, we can extend $\tilde{g}$ (by zero) to a metric on $TQ$. The Riemannian metric $g = \tilde{g} \oplus \alpha \otimes \alpha$ on $M$ is such that $\R{R_{\alpha}}$ and $\xi$ are orthogonal. If the Reeb vector field is Killing with respect to $g$, i.e. the Lie derivative of $g$ in the direction of $R_{\alpha}$ vanishes, we say that $M$ is a {\bf K-contact} manifold. What's important to us, as proved in Theorem 3.1 in \cite{l1}, is that a contact manifold is K-contact if and only if it is a contact toric manifold of Reeb type.

Since we endeavor to study the geodesic flow of a compact, connected 3-dimensional Riemannian manifold, we should be interested in finding all compact contact toric manifolds in dimension 5. So far, our list includes
\begin{itemize}
\item principal $\T^3$-bundles over $S^2$,
\item $S^1 \times S^4$, $S^2 \times S^3$, and
\item K-contact manifolds.
\end{itemize}
By the classification theorem, this list is exhaustive in dimension $5$. Luckily, thanks to Yamazaki, we also have at our disposal a complete list of all simply connected $5$-dimensional K-contact manifolds:

\begin{theorem}{\cite{yama}}
Suppose $M$ is a closed, simply connected 5-manifold. Denote by $w_{2}(M)$ the second Stiefel-Whitney class of the bundle $TM \to M$. Then, $M$ admits an effective $\T^3$ action and a $\T^3$-invariant $K$-contact structure with $r$ closed orbits of the Reeb flow if and only if $M$ is diffeomorphic to
\begin{enumerate}
\item $S^5$, if $r=3$,
\item $\sharp^{r-3}(S^2 \times S^3)$, if $w_{2}(M)=0$,
\item $(S^2 \tilde\times S^3) \sharp^{r-4} (S^2 \times S^3)$, if $w_2(M) \neq 0$.
\end{enumerate}
where $S^2 \tilde\times S^3$ is the oriented, nontrivial $S^3$ bundle over $S^2$ and $\sharp$ denotes the connected sum operation.
\end{theorem}

It is important to note here that if $M$ does indeed admit the structure of a contact toric manifold
with $r$ closed orbits of the Reeb flow, then $r \geq 3$. In particular, $r \geq 4$ if $w_{2}(M) \neq 0$.

\section{Cosphere Bundles and Simply Connected Contact Toric Manifolds}
In this section, we commence the proof of Theorem 1.1. We split the statement into cases, the first being the assumption that our Riemannian manifold $Q$ is simply connected. Of course, if $Q$ is simply connected, so is $S(T^*Q)$. This is easily seen by considering the long exact sequence of homotopy groups associated to the fibration $S^2 \to S(T^*Q) \to Q$. In particular, we prove
\begin{prop}
Suppose $(Q,g)$ is a $3$-dimensional compact, simply connected Riemannian manifold admitting a toric integrable geodesic flow. Then, the cosphere bundle of $Q$ is diffeomorphic to $S^3 \times S^2$.
\end{prop}

The crux of our computations involve the following theorem, due to McCord, Meyer and Offin (\cite{mmo}) which place necessary conditions on a manifold to be a cosphere bundle. We adopt the following notation: $\beta_{i}(M)$ is the $i$th Betti number of $M$ and $T_{i}(M)$ is the torsion subgroup of the homology group $H_{i}(M,\Z)$.
\begin{theorem}\cite{mmo}
Suppose $M$ is a connected, $(2n-1)$-dimensional manifold. Then, if $M$ arises as the cosphere bundle of some $n$-manifold $Q$,
\begin{itemize}
\item[(i)] $\beta_{2n-1}(M)=1$ and $T_{n-1}(M)$ is cyclic or trivial if $Q$ is compact and orientable,
\item[(ii)] $\beta_{2n-1}(M)=1$ and $T_{n-1}(M)$ has order $4$ if $Q$ is compact and nonorientable.
\end{itemize}
In addition, if $T_{n-1}(M)$ is trivial and $M$ is the cosphere bundle of some compact, orientable manifold $Q$, we have
\begin{eqnarray}
0 \leq \beta_{n-1}(M)-\beta_{2n-2}(M) &=& \beta_{n}(M)-\beta_{1}(M) \leq 1 \\
\left|\sum_{i=0}^{n-2} (-1)^{i}\beta_{i}(M)+(-1)^{n}(1-\beta_{2n-2})\right| &=& 1+\beta_{1}(M)-\beta_{n}(M).
\end{eqnarray}
\end{theorem}

From the results of the previous section, namely Theorem 2.2, our candidates for simply connected, contact toric cosphere bundles are $S^5$and connected sums of $S^2 \times S^3$ and $S^3 \tilde\times S^2$. Our first application of Theorem 3.2 will be to show that $S^5$ is not the cosphere bundle of any manifold $Q$. We then explicitly compute the Betti numbers of the remaining simply connected contact toric manifolds on our list.
\begin{lemma}
The contact toric manifold $S^5$ does not arise as the cosphere bundle of any compact, simply connected 3-manifold.
\end{lemma}
\begin{proof}
Suppose that $S^5$ arose as the cosphere bundle of a compact manifold. Then, by Theorem 3.2, we would have
\begin{displaymath}
|\beta_{0}(S^5)-\beta_{1}(S^5)-1+\beta_{4}(S^5)| = 1+\beta_{1}(S^5)-\beta_{3}(S^5).
\end{displaymath}
But, $\beta_{k}(S^5)$ is trivial for $0<k<5$, and $\beta_{0}(S^5)=1$. Hence, we would then have $0=1$, a contradiction.
\end{proof}

\begin{lemma}
For $k>0$, we have
\item
\begin{displaymath}
\beta_{i}(\sharp^{k}(S^2 \times S^3))= \left\{ \begin{matrix}  1 & \textrm{if } i=0,5 \\ k & \textrm{if } i=2,3 \\ 0 & \textrm {if } i= 1,4 \textrm{ or } i >5  \end{matrix} \right.
\end{displaymath}
\end{lemma}
\begin{proof}
The case $k=1$ follows from applying the K\"unneth formula to $S^2 \times S^3$ to get Betti numbers
\begin{displaymath}
\beta_{i}(S^2 \times S^3) = \left\{ \begin{matrix} 1 & \textrm{if } i=0,2,3,5 \\ 0 & \textrm {if } i= 1,4 \textrm{ or } i >5 \end{matrix} \right.
\end{displaymath}

For $0<i<m$ and connected, orientable $m$-manifolds $M,N$, we have $H_{i}(M \sharp N) \cong H_{i}(M) \oplus H_{i}(N)$. Also, $\dim_{\Z}{H_0(M \sharp N)} = 1$ by connectivity and $\dim_{\Z}{H_m(M \sharp N)} = 1$ by orientability. Our result then follows by induction.
\end{proof}

\begin{lemma}
For $k \geq 0$,
$$
\beta_{i}((S^2 \tilde\times S^3) \sharp (\sharp^{k}(S^2 \times S^3)))= \left\{ \begin{matrix}  1 & \textrm{if } i=0,5 \\ k+1 & \textrm{if } i=2,3 \\ 0 & \textrm {if } i= 1,4 \textrm{ or } i >5  \end{matrix} \right.
$$
\end{lemma}
\begin{proof}
When $k=0$, we are making a statement about the Betti numbers of $S^2 \tilde\times S^3$. Now, there is an exact sequence in homology, called the Wang sequence, which is useful for fibrations $F \hookrightarrow E \to S$ whose base is a sphere:
$$
\cdots\to H_{m}(F) \to H_{m}(E)\to H_{m-n}(F) \to H_{m-1}(F) \to \cdots.
$$
The Wang sequence for the fiber bundle $S^3 \hookrightarrow S^2 \tilde\times S^3 \to S^2$ thus yields
$$
\beta_{i}(S^2 \tilde\times S^3) = \left\{ \begin{matrix} 1 & \textrm{if } i=0,2,3,5 \\ 0 & \textrm {if } i= 1,4 \textrm{ or } i >5 \end{matrix} \right.
$$
Induction and Lemma 3.4 now combine to complete the argument.
\end{proof}

The previous results narrow down possible candidates for contact toric cosphere bundles:

\begin{cor}
The only possible simply connected, compact contact toric cosphere bundles are $S^2 \times S^3$ and $S^2 \tilde\times S^3$.
\end{cor}

\begin{proof}
When $w_2(M) =0$ and $r \geq 3$, equation $(2)$ of Theorem 3.2 applied to $\sharp^{r-3} (S^2 \times S^3)$ gives $0=4-r$.
By equation $(1)$ of the same theorem, the only possibilities for $r$ are $r=3$ or $r=4$. But $0 \neq 4-3= 1$, and
when $r=4$ we have $S^2 \times S^3$.

Similarly, when $w_2(M) \neq 0$, we have $r \geq 4$. Equation $(1)$ again provides $0=4-r$. This
is only possible when $r=4$ and hence $S^2 \tilde\times S^3$ is a possibility.
\end{proof}

We are now in a position to prove Proposition 3.1. Before we do this, however, note that the Lie group $S^3$ is parallelizable. Indeed the cosphere bundle is diffeomorphic to $S^3 \times S^2$. To complete the proof of Proposition 3.1, then, it suffices to show that $S^3 \tilde\times S^2$ cannot arise as the cosphere bundle of any manifold.

\begin{proof}{(of Proposition 3.1)}
Suppose $S^3 \tilde\times S^2$ is the cosphere bundle of some compact Riemannian manifold $Q$. Note that, by Theorem 3.2(i), $Q$ must be orientable since $T_2(S^2 \tilde\times S^3)$ is trivial. So, $Q$ is compact and orientable and by a theorem of Stiefel (see \cite{parker} for a proof using obstruction theory) it is parallelizable. Hence, $TQ = Q \times \R^3$. So, $S(TQ) = Q \times S^2$.  Again, since $Q$ is parallelizable, $w_2(Q \times S^2) = w_2(S^2)$. But, $w_2(S^2)=0$ by naturality of the Stiefel-Whitney classes (see \cite{milnor} for details). Since $w_{2}(S^2 \tilde\times S^3) \neq 0$, it cannot occur as the cosphere bundle of $Q$.
\end{proof}

\section{Non-Simply Connected Contact Toric Cosphere Bundles}
In the previous section, we found that the only compact, simply connected contact toric manifold which is also a cosphere bundle is $S^3 \times S^2$. We now turn our attention to the non-simply connected case.

Recall that $\T^n$ with the flat metric always has a toric integrable geodesic flow. As it is a Lie group, $\T^n$ is parallelizable and its cosphere bundle is $\T^n \times S^{n-1}$. From Theorem 2.1, we see that $\T^n \times S^{n-1}$ is a contact toric manifold. The main result of this section is

\begin{prop}
Suppose $Q\neq \T^3$ is a compact, connected Riemannian manifold with nontrivial, finite fundamental group. Then, if $Q$ admits a toric integrable geodesic flow, the cosphere bundle of $Q$ is homotopy equivalent to the cosphere bundle of $S^3/\Gamma$, where $\Gamma$ is a finite cyclic group.
\end{prop}

The proof of Proposition 4.1 will proceed as follows: We first show that if $Q \neq \T^3$ and the cosphere bundle $S(T^*Q)$ is contact toric, then the fundamental group of $S(T^*Q)$ is finite. We then show that if $M$ is a compact, connected contact toric manifold which arises as a cosphere bundle, the universal cover $\widetilde{M}$ is also a compact contact toric manifold which is a cosphere bundle. Since universal covers are simply connected, we apply the results of the previous section.

\begin{lemma} Suppose that $M$ is a non-simply connected 5-manifold, $M = S(T^*Q)$ for some compact, connected
3-manifold $Q \neq \T^3$. Then, the fundamental group of $M$ is finite.
\end{lemma}

\begin{proof}
In Theorem 2.2, we see that the only $5$-dimensional contact toric manifolds which are not simply connected and have infinite fundamental group are diffeomorphic to principal $\T^3$ bundles over $S^2$, $\T^2 \times S^3$ or $S^1 \times S^4$. Now, applying the K\"unneth formula to the latter two spaces, we see that $\beta_{2}(\T^2 \times S^3)-\beta_{4}(\T^2 \times S^3) = 1-2 <0$ and $\beta_{2}(S^1 \times S^4)-\beta_{4}(S^1 \times S^4) = 0-1 <0$. So, by Theorem 3.2, neither of these spaces arise as the cosphere bundle of any manifold.

It remains to show that a (nontrivial) principal $\T^3$-bundle $P$ over $S^2$ cannot arise as the cosphere bundle over any compact 3-manifold $Q$. In order to calculate the Betti numbers of such bundles, we employ the cohomology Serre spectral sequence with rational coefficients. Denote by $E_{n}^{p,q}$ the $(p,q)$th entry on the $n$th page of the spectral sequence. Denote by $a,b,c$ the generators of $H^1(\T^3;\Q)$ and by $x$ the generator of $H^2(S^2;\Q)$. Then, using the multiplicative structure on the spectral sequence we have
$$
\begin{array}{llll}
E_{2}^{0,0} = \Q & E_{2}^{0,1} = \Q{a}\oplus\Q{b}\oplus\Q{c} & E_{2}^{0,2} = \Q{ab}\oplus\Q{ac}\oplus\Q{bc} & E_{2}^{0,3} = \Q{abc} \\
E_{2}^{2,0} = \Q{x} & E_{2}^{2,1} = \Q{ax}\oplus\Q{bx}\oplus\Q{cx} & E_{2}^{2,2} = \Q{abx}\oplus\Q{acx}\oplus\Q{bcx} & E_{2}^{2,3} = \Q{abcx}
\end{array}
$$
and all other entries of the sequence are 0. Thus, the only nonzero differentials are
\begin{eqnarray*}
d_{2}^{1}:E_{2}^{0,1} \to E_{2}^{2,0} \\
d_{2}^{2}:E_{2}^{0,2} \to E_{2}^{2,1} \\
d_{2}^{3}:E_{2}^{0,3} \to E_{2}^{2,2}
\end{eqnarray*}

For the generators $a,b,c$ of $H^1(\T^3;\Q)$, we write $d_{2}^{1}(a) = a'x$, $d_{2}^{1}(b) = b'x$, $d_{2}^{1}(c) = c'x$ where
$a',b',c' \in \Q$. We then have
\begin{eqnarray*}
d_{2}^{2}(ab) &=& d_{2}^{2}(a)b-ad_{2}^{2}(b) \\
&=& a'bx-b'ax
\end{eqnarray*}
and similarly, $d_{2}^{2}(ac) = a'cx-c'ax$, $d_{2}^{2}(bc) = b'cx-c'bx$. In addition,
\begin{eqnarray*}
d_{2}^{3}(abc) &=& d_{2}^{3}(a)bc-ad_{2}^{3}(bc) \\
&=& a'bcx-b'acx+c'abx
\end{eqnarray*}

Now, we may break our computations down into essentially 4 cases: all of $a',b',c'$ are zero, none are zero, at most one is zero and at most two are zero. These cases cover all possible situations as the above expressions represent a symmetric situation. Now, when $a'=b'=c'=0$, the principal bundle is trivial so we may assume at least one is nonzero.

For illustration, we now compute the Betti numbers of $P$ when none of $a',b',c'$ are zero. In this
case, $E_{\infty}^{p,q} = E_{3}^{p,q}$, $E_{3}^{0,i} = \ker{d_{2}^{i}}$ and $E_{3}^{2,i} = \coker{d_{2}^{i}}$.
Since we are only looking for Betti numbers, we need only find the sums of the ranks of kernels and
cokernels along the ``diagonals''. Now, $\ker{d_{2}^{1}}$ contains terms like $b'c-c'b$ and is hence free
on 3 generators. So, $\beta_{1}(P) = 3$ since $E_{3}^{1,0} =0$. Similarly, we can see that $\beta_2(P) = 1$,
$\beta_{3}(P) = 2$, $\beta_{4}(P) = 2$ and $\beta_{5}(P) = 1$. By Theorem 3.1 (2), we see that in this case
$P$ cannot arise as the cosphere bundle of a compact, connected manifold $Q$.

In a similar fashion, we may compute the Betti numbers of $P$ in each of the above cases. As it turns out,
when exactly one or two of $a',b',c'$ is zero we have
$$
\beta_{i}(P) = \left\{ \begin{array}{cc} 2 & \textrm{if } i=1,4\\  1 & \textrm{if } i=0,2,3,5\end{array}\right.
$$
which again violates conditions imposed in Theorem 3.2. It follows that none of the possible total
spaces $P$ under consideration arise as cosphere bundles.
\end{proof}

\begin{lemma}
Suppose $M = S(T^*Q)$ for some manifold $Q$. Then, if $\widetilde{M}$ is the universal cover of $M$, $\widetilde{M} = S(T^*\widetilde{Q})$
where $\widetilde{Q}$ is the universal cover of $Q$.
\end{lemma}
\begin{proof}
The covering map $p:\widetilde{Q} \to Q$ is a local diffeomorphism. Hence, it induces an isomorphism on fibers $S(T^*_{q}Q) \to S(T^*_{p(q)}\widetilde{Q})$. Since $\widetilde{Q}$ is simply connected, so is $S(T^*\widetilde{Q})$. It follows that $\widetilde{M} = S(T^*\widetilde{Q})$.
\end{proof}

\begin{lemma}
Suppose $M$ is a contact toric manifold with finite fundamental group. Denote by $\widetilde{M}$ the universal cover of $M$. Then, $\widetilde{M}$ is a compact contact toric manifold.
\end{lemma}
\begin{proof}
We have to verify both that $\widetilde{M}$ admits a contact structure and that the torus action on $M$ lifts to a torus
action on $\widetilde{M}$.
First note that being a contact manifold is a local condition. Since a covering map is a local diffeomorphism,
ths contact structure on $M$ lifts to a contact structure on $\widetilde{M}$.

We now lift the torus action from $M$ to $\widetilde{M}$. We claim it is enough to lift an $S^1$ action
since a product of groups acts on a manifold if and only if each component does and these actions commute.

So, suppose $S^1$ acts smoothly on $M$. This action generates a vector field $X$ on $M$. Denote by $\phi_{t}$ the
flow of $X$. If we view $S^1$ as $\R/\Z$, we see that $\phi_{t} = \phi_{t+1}$ for any $t$. Since
the covering map $\pi: \widetilde{M} \to M$ is a local diffeomorphism, we may lift $X$ to a vector field $\widetilde{X}$
on $\widetilde{M}$ with corresponding flow $\tilde{\phi}$. We claim that for some positive integer
$k$, $\phi_{t} = \phi_{t+k}$ for all $t$. Hence, the $S^1$ action will lift to $\widetilde{M}$ (with perhaps
a different parameterization of $S^1$). Now, the fundamental group of $M$ is finite, so for any point $x \in M$, the fiber
$\pi^{-1}(x)$ is finite, say cardinality $m$. Since $M$ is assumed connected, the number of elements
in the fiber is constant over each point in $M$ so we need only argue pointwise. Now, for any $\tilde{x} \in \pi^{-1}(x)$,
note that $\tilde{\phi}(\tilde{x}) \in \pi^{-1}(x)$ as well since $\tilde{\phi}$ was gotten by lifting
$\phi$ locally. In other words, $\pi(\tilde{\phi}(\tilde{x})) = \phi(x)$. Since the fibers have only finitely
many elements, and the flow of $\tilde{X}$ always stays within fibers, there must be some $k$
with $\tilde{\phi}_{t}(\tilde{x}) = \tilde{\phi}_{t+k}(\tilde{x})$ for all $t$.
\end{proof}

We now provide a proof of Proposition 4.1. which, in turn, completes the proof of Theorem 1.1.
\begin{proof}{(Proposition 4.1)}
Let $M = S(T^*Q)$. Then, by Lemmas 4.3 and 4.4, the universal cover $\widetilde{M}$ is a contact toric manifold and the cosphere bundle of $\widetilde{Q}$. Since $\widetilde{M}$ is simply connected, Proposition 3.1 implies that $\widetilde{M}$ is diffeomorphic to $S^3 \times S^2$. Since $\widetilde{Q}$ is a closed $3$-manifold that is simply connected, Whitehead's theorem (along with the Hurewicz theorem) implies that $\widetilde{Q}$ is homotopy equivalent to $S^3$. Therefore, $Q$ is homotopy equivalent to $S^3/\pi_1(Q)$. Now, consider the bundle $S^2 \hookrightarrow M \to Q$. Since $S^2$ is simply connected, by the long exact sequence in homotopy we can see that $\pi_1(M) \cong \pi_1(Q)$. Hence, $Q$ is homotopy equivalent to $S^3/\pi_1(M)$. By passing to a double cover, if necessary, we may assume without loss of generality that $Q$ is orientable. Hence, it is parallelizable and the cosphere bundle of $Q$ is homotopy equivalent to $S^3/\Gamma \times S^2$.

In \cite{l1} it is shown that the fundamental group of a contact toric $\T^n$-manifold of Reeb type is a quotient of $\Z^n$ by a syblattice. So,  $\pi_1(M)$ is isomorphic to a product of at most 3 finite cyclic groups.

We now show that $\Gamma = \pi_1(M)$ is indeed a cyclic group. Write $\Gamma=\Z_{m} \times \Z_{l} \times \Z_{k}$. If the greatest common divisor $(m,l,k) :=\textrm{gcd}(m,l,k)=1$, then $\Gamma$ is isomorphic to the finite cyclic group $\Z_{(m,l,k)}$. So, suppose not. Then, $\textrm{gcd}(m,l,k)=s \neq 1$ and there is some prime $p$ dividing $s$ and hence each of $m,l,k$. So, $\Gamma$ contains $\Z_p \times \Z_p \times \{0\}$ as a noncyclic subgroup of order $p^2$. A theorem of Madsen,Thomas and Wall \cite{mtw} says that a finite group cannot act freely on $S^3$ if it contains a noncyclic subgroup of order $p^2$. Hence,  $\Gamma$ cannot act freely on $S^3$ and cannot be an isomorphic copy of $\pi_1{M}$.  A similar argument applies if $\Gamma$ is assumed to be  a product of two finite cyclics, and so the only way $\Gamma$ can act freely on $S^3$ is if it is a finite cyclic group.
\end{proof}

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