TITLE: Prove that $n!>n^2$ for all integers $n \geq 4$.
QUESTION [5 upvotes]: I am working on induction problems to prep for Real Analysis for the fall semester. I wanted proof verification and editing suggestions for part (a), and assistance understanding part (b). For part (b), the portion that has the additional indentation is where I am unclear.

The principle of mathematical induction can be extended as follows. A list $P_m, >P_{m+1}, \cdots$ of propositions is true provided (i) $P_m$ is true, (ii) >$P_{n+1}$ is true whenever $P_n$ is true and $n \geq m$.
(a) Prove $n^2 > n + 1$ for all integers $n \geq 2$.
Assume for $P_n$: $n^2 > n + 1$, for all integers $n \geq 2$. Observe for $P _2$:
$P_2: 2^2=4 > 2+1 = 3$,
thus the basis step holds. Now, let $n=k$ such that $k^2 > k + 1$, and assume this also holds. We now consider the case $P_{k+1} : (k+1)^2 > (k+1) + 1$.
Observe:
$(k+1)^2 = k^2 + 2k + 1$
= $k(k+2) + 1$
Clearly, $k(k+2)+1$ must be greater than $(k+1)+1$. Thus, by the principle of mathematical induction, the case holds for all $n \geq 2$.
*I am aware that part (a) does not require induction to prove, but the book problem suggests doing so...

(b) Prove $n! > n^2$ for all integers $n \geq 4$.
Assume for $P_n$: $n!>n^2$ for all integers $n \geq 4$. Observe for $P_4$:
$P_4: 24 = 4! > 16 = 4^2$,
thus the basis step holds. Let $n=k$ such that $k! > k^2 $, and assume this also holds. We now consider the case $P_{k+1} : (k+1)! > (k+1)^2$. Observe:

$(k+1)! = (k+1)k!$
$> (k+1)k^2$
$= k^3 + k^2$
$> k^2 + 2k + 1$
$= (k + 1)^2$

I was able to write some of this on my own, and I used my book + internet to help me figure out how to write this out. First, I am not clear why we are adding $(k+1)$ to the right hand side (the $k^2$) side of the equation (or what that's allowed, really). Also, I'm not clear on the jump from $k^3+k^2>k^2 + 2k + 1$.
Additionally, while looking around MSE, I have noticed many people talk about induction with LHS and RHS notation. I have not seen this in any books--would someone be able to explain using that method as a form of bookkeeping, or be able to suggest a site or stack that could do that?

REPLY [5 votes]: Note: Your question should really be two questions since they're completely distinct and separate. You are simply trying to cover too much for one single question on this site, but I'll help with what I can.
Before addressing both parts of your question, I would encourage you to read the following three posts because I think they would go a long way in helping you with induction proofs in general.

How to write a clear induction proof.
The difference between weak and strong induction.
Use of LHS/RHS notation or terminology in an induction answer.

Now on to your question(s).
Part (a): Your write-up here is really struggling in a number of ways. I'm going to be straight with you--your statement of strong induction is sloppy (you don't actually need strong induction even if you do use an inductive argument which is why I included (2) above) and your inductive "proof" never even uses the inductive hypothesis. See (1) above for how to actually write a clear induction proof; this will more or less force you to understand how the proof actually works, where you use the inductive hypothesis, etc. Also, as Hirshy notes, you actually assume what you are trying to prove (you do this again in part (b)). Regardless, the following is one way you could structure the main part of your induction proof for part (a):
\begin{align}
(k+1)^2&= \color{blue}{k^2}+2k+1\tag{expand}\\[0.5em]
&\color{blue}{> (k+1)}+2k+1\tag{by inductive hypothesis}\\[0.5em]
&= 3k+2\tag{simplify}\\[0.5em]
&> k+2\tag{since $k\geq 2$}\\[0.5em]
&= (k+1)+1.\tag{simplify}
\end{align}
Did you see how that worked? Can you see how the inductive hypothesis is used in the part highlighted with $\color{blue}{\mathrm{blue}}$?
Part (b): The part you have indented is actually not too bad at all when you actually write out what is happening and why at each step; nonetheless, you still need to clean up the beginning, as you are again assuming what you are trying to prove. The goal here is to move from the left-hand side (LHS) of the statement $P_{n+1}$ to the right-hand side (RHS) of $P_{n+1}$. To that end, note the following (the indented steps have explanatory notes in the margin now):
\begin{align}
(k+1)! &= (k+1)k!\tag{by definition}\\[0.5em]
       &> (k+1)k^2\tag{by induction hypothesis}\\[0.5em]
       &= k^3+k^2\tag{expand}\\[0.5em]
       &> k^2+2k+1\tag{since $k\geq 4$}\\[0.5em]
       &= (k+1)^2.\tag{factor}
\end{align}
Can you see how all of that worked?


TITLE: Prove equilateral triangle if vertices $a,b,c$ satisfy $|a|=|b|=|c|$ and $\sum_{cyc}\frac1{8-a/b-b/a-a/c-c/a}=0.3$
QUESTION [1 upvotes]: If $a,b,c\in \mathbb{C}$ are different complex numbers in pairs and $|a|=|b|=|c|=1$ and $P(a),Q(b),R(c)$ are the vertices of a triangle and $\displaystyle \sum_{\bf{cyc}}\frac{1}{8-\frac{a}{b}-\frac{b}{a}-\frac{a}{c}-\frac{c}{a}}=0.3.$ Then show that $PQ=QR=RP.$

What I tried
Let $a=e^{i\alpha}=\cos \alpha+i\sin \alpha$
and $b=e^{i\beta}=\cos \beta+i\sin \beta$
and $c=e^{i\gamma}=\cos \gamma+i\sin \gamma$
Now $\displaystyle \frac{a}{b}+\frac{b}{a}=e^{i(\alpha-\beta)}+e^{-i(\alpha-\beta)}=2\cos (\alpha-\beta)$
and $\displaystyle \frac{a}{c}+\frac{c}{a}=e^{-i(\gamma-\alpha)}+e^{i(\gamma-\alpha)}=2\cos(\gamma-\alpha)$
$\displaystyle \Longrightarrow \sum_{\bf{cyc}}\frac{1}{8-2\cos(\alpha-\beta)-2\cos(\gamma-\alpha)}=0.3$
$\displaystyle \Longrightarrow \sum_{\bf{cyc}}\frac{1}{4-\cos(\alpha-\beta)-\cos(\gamma-\alpha)}=0.6$
Let $\alpha-\beta=A,\beta-\gamma=B,\gamma-\alpha=C$
$\displaystyle \Longrightarrow \sum_{\bf{cyc}}\frac{1}{4-\cos A-\cos C}=0.6$
How do  I solve it after that
Please help.

REPLY [2 votes]: Apply the Jensen's inequality
$$f(x_1) + f(x_2) + f(x_3) \ge 3f\left(\frac{x_1+x_2+x_3}3\right)$$
for the convex function $f(x)=\frac1x$ to get
$$0.3=\sum_{{cyc}}\frac{1}{8-\frac{a}{b}-\frac{b}{a}-\frac{a}{c}-\frac{c}{a}}
\ge \frac{1}{24-2\left(\frac{a}{b}+\frac{b}{a}+\frac bc+\frac cb+\frac{c}{a}+\frac{a}{c}\right)}
$$
or,
$$\frac{a}{b}+\frac{b}{a}+\frac bc+\frac cb+\frac{c}{a}+\frac{a}{c}\le -3\tag 1$$
Given that $|a|=|b|=|c|$, assume, without loss of generality $\alpha, \ \beta, \ \gamma >0$,
$$\frac ba =e^{i\alpha},\>\>\>\>\>\frac cb =e^{i\beta},\>\>\>\>\>\frac ac=e^{i\gamma}
\>\>\>\>\>\alpha + \beta + \gamma = 2\pi$$
Then, the inequality (1) becomes,
$$\cos\alpha + \cos\beta + \cos \gamma \le -\frac32\tag 2$$
Note that,
$$\cos\alpha + \cos\beta + \cos \gamma
= 2\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2 + 2\cos^2\frac\gamma2-1$$
$$= 2\cos^2\frac\gamma2 - 2\cos\frac{\gamma}2\cos\frac{\alpha-\beta}2 -1
\ge 2\cos^2\frac\gamma2 - 2\cos\frac{\gamma}2 -1 $$
$$= 2\left(\cos\frac\gamma2 - \frac12\right)^2 -\frac32 \ge-\frac32\tag 3$$
From (2) and (3), we have
$$\cos\alpha + \cos\beta + \cos \gamma=-\frac32$$
and, according to (3), the equality occurs at $\cos\frac\gamma2 - \frac12=0$ and $\cos\frac{\alpha-\beta}2 =1$, which leads to $\alpha=\beta =\gamma = 120^\circ$. Thus, the neighboring vertexes of equal module all have the same argument angle between them, hence, forming an equilateral triangle.


TITLE: How to evaluate $ \int_{0}^{{\frac{\pi}{2}}} \cos(ax) \cos(a\cos(bx)) \, dx$
QUESTION [1 upvotes]: Evaluate
$$ \int_{0}^{{\frac{\pi}{2}}} \cos(ax) \cos(a\cos(bx)) \, dx \ ; \ a,b \in \mathbb{Z^+}$$

I tried taylor series and finding a reduction formula, but couldn't evaluate the integral. Please avoid contour integration as I don't know that yet.
Any help will be appreciated.
Thanks.

REPLY [2 votes]: I doubt that there's a "closed-form" formula.  In the case $b=1$ with $a$ an odd positive integer, it seems to be:
$$ (-1)^{(a-1)/2} \dfrac{1 - \text{LommelS1}(1,a,a)}{a} $$
EDIT: if $a$ is an even integer and $b$ is an integer, by symmetry your integral is
$$ \dfrac{1}{4} \int_0^{2\pi} \cos(ax) \cos(a \cos(bx))\; dx $$
and we can express the second cosine in a power series:
$$ \dfrac{1}{4} \sum_{n=0}^\infty  \dfrac{a^n}{n!} \int_0^{2\pi} \cos(ax) \cos(bx)^n\; dx$$
and $\cos(bx)^n$ can be expressed as a combination of $\cos(jbx)$ for $0 \le j \le n$.  If $a$ is not a multiple of $b$, then by orthogonality the result will be $0$.


TITLE: How to solve this inequality $3> 2^x+2^{1-x}$
QUESTION [1 upvotes]: By hit and trial I get the answer as $x\in\mathbb(0,1)$ but I don't think about the actual process by which this inequality can solve. Plz tell any process by which I solve this ? I take log both sides , but this can't make any sense.

REPLY [0 votes]: Another way is to note that $2^t$ is convex and we have $x\in (0,1)\implies (1,0)\succ (x,1-x)$ while $x\not \in (0,1)\implies (x,1-x)\succ (1,0)$. Then Karamata’s inequality ( https://en.m.wikipedia.org/wiki/Karamata%27s_inequality) directly gives the solution set is $x\in (0,1)$


TITLE: Let $f_n: D \rightarrow \mathbb{R}: f_n(x) = g(x)^n, n≥1$. Necessary and sufficient conditions such that $f_n$ converges?
QUESTION [2 upvotes]: The Assignment:

Let $D := [a,b]$ with $a<b$ and $g: D \rightarrow \mathbb{R}$ be continuous. Observe the sequence of functions $f_n: D \rightarrow \mathbb{R}: f_n(x) = g(x)^n, n≥1$. List and explain necessary and sufficient conditions such that $f_n$ converges (a) pointwise and (b) uniformly.

What I've gathered so far is that, $|g(x)|<1$ is necessary, $ g(x) = 1$ would be sufficient for both pointwise and uniform convergence. I also don't know where the continuity of $g$ actually matters.
Any help would be appreciated

REPLY [2 votes]: Your conditions on $g$ are also necessary, since of $|g(x)|>1$ for some $x\in[a,b]$ then the sequence $(f_n(x))$ does not converge. You may also let $g(x)=1$,for some (or all) $x$ for guarantee pointwise convergence
For uniform convergence, if $M$ is the maximum of $|g|$ on the interval (which exists because it is compact), you may require that $M<1$: in this case, $f_n\to 0$ as $n\to \infty$ so, given $\varepsilon>0$,
$$
|f_n(x)|=|g(x)|^n\leq M^n<\varepsilon
$$
If $n$ is large enough. Note that this is valid for every $x$ in the domain, so the convergence is uniform.
If $M=1$, I'm going to show that the convergence is not uniform. Let $c\in[a,b]$ be such that $|g(c)|=M$. You can take a sequence $(x_n)$ of points such that $x_n\to c$. By continuity, the sequence $(|g(x_n)|)$ will converge to $1$.
If $m\in\mathbb{N}$, then there's a $x_{n_m}$ in the sequence such that $|g(x_{n_m})|>\tfrac{1}{2^{1/m}}$
 This follows from the fact that $\lim\limits_{n\to\infty}|g(x_n)|=1$ so we can get as closer as we want to 1 and since $\tfrac{1}{2^{1/m}}<1$, the sequence will eventually be greater than $\tfrac{1}{2^{1/m}}$ and we can let $x_{n_m}$ be one of those numbers that satisfies this. By this way we can construct a subsequence $(x_{n_m})$ such that $|f_m(x_{n_m})|=|g(x_{n_m})|^m\geq 1/2$. Note that this imply that the convergence is not uniform, because if we take $0< \varepsilon<1/2$ then you can not find an $N\in\mathbb{N}$ such that $n\geq N$ imply $|f_n(x)|<\varepsilon$ for all $x\in[a,b]$ since there's always and $m>N$ such that $|f_m(x_{n_m})|\geq 1/2>\varepsilon$.
Briefly: if the maximum of $g$ is less than 1, there's uniform convergence, if not, there's no uniform convergence (and maybe not even pointwise convergence).


TITLE: how we can prove or disprove this mutual information equation?
QUESTION [1 upvotes]: $$I(X_{1:n};Y_{1:n})=\sum_{i=1}^n [H(Y_i|Y_{1:i-1})-H(Y_i,X_{i:n}|Y_{1:i-1},X_{1:i-1})+H(X_{i:n}|Y_{1:i-1},X_{1:i-1})]$$

Edited:
I know a little how to go further
$$I(X_{1:n};Y_{1:n})=H(Y_{1:n})+H(Y_{1:n}|X_{1:n})$$
The first part of this, match the fist part of the result so we should prove this instead :
$$H(Y_{1:n}|X_{1:n})=\sum_{i=1}^n [H(Y_i,X_{i:n}|Y_{1:i-1},X_{1:i-1})-H(X_{i:n}|Y_{1:i-1},X_{1:i-1})] $$

REPLY [1 votes]: Starting with $$I(X_{1:n};Y_{1:n})=H(Y_{1:n})+H(Y_{1:n}\mid X_{1:n}) \tag{1}$$
For the first term we have (chain rule) :
$$H(Y_{1:n})=H(Y_n\mid Y_1 \cdots Y_{n-1})+H(Y_{n-1}\mid Y_1 \cdots Y_{n-2}) +\cdots =\sum_{i=1}^n H(Y_i\mid Y_{1,i-i}) \tag{2}$$
The second is equivalent, only that conditioned on $X_{1:n}$, then we can write
$$H(Y_{1:n}\mid X_{1:n}) =\sum_{i=1}^n H(Y_i\mid Y_{1,i-i},X_{1:n}) \tag{3}$$
Now, because we want the get "up" the $X_{i:n}$ part, we split the multidimensional $X$ variable in two parts:
$$ H(Y_i\mid Y_{1,i-i},X_{1:n})=H(Y_i\mid Y_{1,i-i},X_{1:i-1},X_{i:n}) \tag{4}$$
Recalling $H(A \mid  B) = H(A,B) - H(B)$, and $H(A \mid  B,C) = H(A,B\mid C) - H(B\mid C)$ we get (taking $A=Y_i$, $B=X_{i:n}$, $C=(Y_{1,i-i},X_{1:i-1})$ ) :
$$ H(Y_i\mid Y_{1,i-i},X_{1:i-1},X_{i:n})= H(Y_i,X_{i:n}\mid Y_{1,i-i},X_{1:i-1})-
H(X_{i:n}\mid Y_{1,i-i},X_{1:i-1}) \tag{5}$$
I leave to you the task of putting all together.


TITLE: A rigorous definition of the intuitive concept of the PEMDAS language.
QUESTION [0 upvotes]: I apologize if this question is similar to a previous question, but I didn't get an answer and it is still bothering me. Suppose our alphabet is a countably infinite set of variables $VAR$, well-ordered by the natural numbers, along with the symbols $\{+,*,-,div,(,)\}$. $($ and $)$ are parentheses, and $+$, $*$, $-$, $div$ represent addition, multiplication, subtraction, and division, respectively. I want to know how to rigorously define the PEMDAS language, where parenthesis is done first, exponentiation is done second, multiplication and division have equal precedence and are done from left to right, and addition and subtraction have equal precedence and are done from left to right, and also where superfluous parentheses are eliminated. So, for instance, $(x_0*x_1)+x_2$ would not be allowed. Also, since this is a purely formal language, we allow terms like $x_0 div (x_0 - x_0)$. In other words, terms that denote division by zero are allowed. Is there some book where this language is explained?

REPLY [1 votes]: If I understand you correctly, you want to define a language where parentheses that could be omitted without changing the abstract parse tree are not allowed.
You could do something like this:
$$\begin{align}
   S \to{} & E_3 \mid E_2 \mid E_1 \\
   E_3 \to{} & L_3 + R_3 \mid L_3 - R_3 \\
   L_3 \to{} & E_3 \mid E_2 \mid E_1 \\
   R_3 \to{} & (\;E_3\;) \mid E_2 \mid E_1 \\
   E_2 \to{} & L_2 * R_2 \mid L_2 \mathbin{\tt div} R_2 \\
   L_2 \to{} & (\;E_3\;) \mid E_2 \mid E_1 \\
   R_2 \to{} & (\;E_3\;) \mid (\;E_2\;) \mid E_1 \\
   E_1 \to{} & \mathtt{x}_0 \mid \mathtt{x}_1 \mid \mathtt{x}_2 \mid \cdots
 \end{align} $$
Some optimizations are obviously possible here, but the principle should be clear.


TITLE: Determining if a function satisfies a Lipschitz condition
QUESTION [3 upvotes]: I have $f(t, y) = e^{t - y}$. I want to see if $f$ satisfies a Lipschitz condition on $D = \{(t, y): 0 \le t \le 1, \; - \infty < y < \infty   \}$.
I am using this definition:

A function $f(t, y)$ satisfies a LC in the variable $y$ on the set $D \subset \mathbb R^2$ if there is a constant $L > 0$ such that $|f(t, y_1) - f(t, y_2)| \le L |y_1 - y_2|$ with $(t, y_1)$ and $(t, y_2)$ in $D$.

I tried:
$$ |e^{t - y_1} - e^{t - y_2} |$$
$$= |e^t(e^{-y_1} - e^{-y_2})|$$
$$= |e^t||e^{-y_1} - e^{-y_2}|$$
$$\le e^1 |y_1 - y_2| $$
At this step, can I say that this does not satisfy a LC, since if $y_1 = -1$ and $y_2 = -2$, the inequality would not be true?

REPLY [2 votes]: Suppose $y <0$, then $|f(0,y)-f(0,0)| = |e^{-y}-1| = \sum_{k=1}^\infty \frac{1}{n!} |y|^n\geq \frac{1}{2} y^2$, so $f$ cannot be uniformly Lipschitz on the domain specified (for any $L>0$, you can choose $|y|>L$ to violate the Lispschitz bound).
If $f$ satisfies a Lipschitz bound in the second variable, then you would have $|f(0,y)-f(0,0)| \leq L |y-0| = L|y|$, for some $L>0$. However, we have shown that $|f(0,y)-f(0,0)|\geq \frac{1}{2} y^2$, so if both bounds hold, then you would have $\frac{1}{2} y^2 \leq L|y|$ for $y \in \mathbb{R}$. Choosing $y\neq 0$ and dividing across by $|y|$ gives $\frac{1}{2}|y| \leq L$, which clearly cannot hold for all $y \in \mathbb{R}$.


text:
\begin{document}
\baselineskip = 24pt

\begin{titlepage}
\nonumber
\vspace{1cm}
\begin{center}{\large {\bf Phase space geometry for
constrained lagrangian systems}}\\
\vspace{1.5cm}
V.P. Pavlov \\
\vspace{0.5 cm}
{\it Steklov Mathematical Institute,} \\
{\it Vavilov 42, GSP-1, 117966, Moscow, Russia}\\
\vspace{1.5 cm}
 A.O.Starinets \footnote[1]{E-mail: aos2839@scires.nyu.edu}\\
\vspace{0.5 cm}
{\it Department of Physics, New York University,} \\
{\it 4 Washington Pl., New York, NY, 10003-6621} \\
\vspace{2.5cm}

\begin{abstract}
We study geometry of the phase space for finite dimensional dynamical systems with degenerate Lagrangians. The Lagrangian and Hamiltonian constraint formalisms are treated as the different local-coordinate pictures of the same invariant procedure. The invariant description is given in terms of geometrical objects associated with the structure of foliation on the phase space.
\end{abstract}

\end{center}
\end{titlepage}

\section*{I. INTRODUCTION} Since the appearance of the fundamental work of
Dirac \cite{Dirac}, a certain progress has been made in understanding
geometry which underlies  constrained dynamics \cite{Faddeev} --
\cite{T}. For a system with regular (nondegenerate) Lagrangian an
appropriate phase space in the  case of Lagrangian (Hamiltonian) formalism
is the tangent bundle $TQ$ (cotangent bundle $T^*Q$) of the
configurational space $Q$. In consideration of degenerate Lagrangians it
is traditionally assumed that the phase space of the system is a subspace
of $TQ$ (Lagrangian formalism) or $T^*Q$ (Hamiltonian formalism). Tangent
and cotangent bundles are treated separatively within the framework of the
constrained dynamics, being related by the standard Legendre transformation. In this paper another point of view is advocated.

Given a Lagrangian $l\in C^{\infty}(TQ)$, the action $S$ can be written as
a functional \begin{equation} S[s] = \int_X s^*\varphi \label{1}
\end{equation} acting on the sections $s$ of the bundle $TQ\bigotimes R$
$\rightarrow$ $R$, where $X\in R$ is a compact set, $\varphi =
-Edt+\theta_l$, $\theta_l =\frac{\partial l}{\partial v}dq$ is Cartan -
Liouville one-form and $E$ is the energy of the system. The Legendre
transformation $L$: $TQ\rightarrow T^*Q$ is a diffeomorphism for
nondegenerate $l$, so the transition to the Hamiltonian formalism can be
viewed as a change of variables in (1), $s$ being the section of $T^*Q\bigotimes R$ $\rightarrow$ $R$, \ \ $\varphi = -hdt+\theta$, where
$\theta = pdq$, $\theta_l = L^*\theta$ and $h$ is the Hamiltonian,
$E=L^*h$.

For degenerate Lagrangians the map $L$ is in general neither surjection
nor injection. In this case the pleasant symmetry between the Lagrangian and
Hamiltonian formalisms is broken, since the image of $TQ$ by $L$ now is
only a subspace of the cotangent bundle. If one wishes to proceed in the
usual way, the machinery of Dirac - Bergmann algorithm \cite{Dirac,Bergmann} is applicable either in local-coordinate approach or
by using the invariant (coordinate-free) description in the spirit of
Gotay - Nester's work
 \cite{Gotay1, Gotay2, Gotay3}.

The invariant description of the constrained Lagrangian dynamics begins
with the basic notion of foliation on the tangent bundle and should in
principle end up with the well-defined procedure of the reduction of the
system to the submanifold of the lower dimension (which itself must admit
a tangent bundle structure) with no second class constraints left in the
theory. The next step is to compare the result with that one of the
reduction procedure in the Hamiltonian formalism and check that the
reduced submanifolds are connected by the nondegenerate Legendre
transformation. For the specific choice of the Lagrangian function the
reduction procedure of that kind has been described in \cite{Car2} by
making use of the tangent bundle geometry formalism developed earlier in
\cite{Crampin}.

The Hamiltonian reduction is controlled by the standard symplectic
structure on $T^*Q$ given by the closed nondegenerate two-form $\omega
=dp\wedge dq$, whereas the analog of $\omega$ in the Lagrangian formalism,
$\omega_l = L^*\omega$, is degenerate for  degenerate $\ell$.

The usual treatment of Lagrangian dynamics starts from the invariant form
of the Euler - Lagrange equations on $TQ$, \begin{equation} \dot{f} =
X_{E}f,\label{2} \end{equation} $f\in C^{\infty}(TQ)$, where for any $g\in
C^{\infty}(TQ)$ $X_{g}$ is defined by \begin{equation}
i_{X_{g}}\omega_{\ell} = -dg.\label{3} \end{equation} One may note that
the relation (3) also plays the key role in geometric quantization
\cite{Kirillov, W}.

In this paper we pay more attention to the action (1) itself. In the case of degenerate Lagrangians the bundle $TQ\bigotimes R\rightarrow R$ appears not to be an appropriate phase space and should be replaced by the bundle $W\bigotimes R\rightarrow R$, where $W=TQ\bigoplus T^*Q$ is a Whitney sum
of $TQ$ and $T^*Q$\footnote{After this work was completed, we learned that similar approach was previously considered (in local coordinates) by
W. Kundt \cite{kundt} and by R. Skinner and R. Rusk (using coordinate-free formulation)\cite{rusk}} . The equations which come from the variational principle can still be written in the form of (2) but the relation (3) must be deformed taking into account the structure of the foliation on the original phase space $W$. This modification allows us to include into the scheme those functions (such as the second class constraints) which are not constant along the leaves of foliation. Then the equivalence of the Lagrangian and Hamiltonian formalisms becomes apparent and the constraints in both pictures are in one-to-one correspondence with each other. For nondegenerate $\ell$ the original phase space $W$ decouples trivially into $TQ$ and $T^*Q$ at the first step of the constraint algorithm procedure.

 It follows that the most general situation one encounters in the invariant
description of the classical dynamics is that of the presymplectic
manifold rather than the symplectic one.

The organization of the paper is as follows. In Sec. II a brief review of
the invariant formulation of the Lagrangian and Hamiltonian dynamics is
given for the systems with nondegenerate Lagrangians. The degenerate case
is treated in local coordinates in Sec.III and the invariant formulation
is provided in Sec. IV. Several illustrative examples are given in Sec.V.
The Appendix contains definitions of some geometric objects used.

\section*{II. NONDEGENERATE LAGRANGIANS}
\subsection*{A. LAGRANGIAN FORMALISM}

For a given Lagrangian $l(q,\dot{q})$ the action
\begin{equation}
S[q(t)] = \int \ell (q,\dot{q})dt \label{4}
\end{equation}
with the appropriate boundary conditions can be deliberately changed to
\begin{equation}
S[q, v] = \int dt \left( \ell (q, v) -
v^i\frac{\partial\ell}{\partial v^i}+\dot{q}^i\frac{\partial\ell}{\partial v^i}\right)dt\label{5}
\end{equation}
to be considered as a functional on the tangent bundle $TQ$ of the configurational space (smooth manifold) $Q$. We denote the independent local coordinates on $TQ$ as $(q^i , v^i)$, $i=1,...N$, $N=$dim $Q$, $l(q,v)\in C^{\infty}(TQ)$.

The Lagrangian $l$ is called nondegenerate (regular) if the matrix $\Gamma_{ij} = \partial^2 l / \partial v_i \partial v_j$ has maximal rank $R=N$. Otherwise the Lagrangian is called degenerate (singular). Note that $R$ does not depend on the particular choice of the local coordinates.

The variational principle applied to (5) gives \begin{equation}
\frac{\delta S}{\delta q^i} = \frac{\partial\ell}{\partial q^i} -
\frac{\partial^2 \ell}{\partial v^k\partial q^i}v^k+\frac{\partial^2
\ell}{\partial v^k\partial q^i}\dot{q}^k - \frac{d}{dt}\left(
\frac{\partial\ell}{\partial v^i}\right)=0, \label{6} \end{equation}
\begin{equation} \frac{\delta S}{\delta v^i} = \Gamma_{ij}(\dot{q}^j -
v^j) = 0.\label{7} \end{equation}
 For nondegenerate $l$ the last equation
implies $\dot{q}^i = v^i$. Then (7) gives the Euler - Lagrange equations.
One can define a special vector field $X_E \in \cal{X}$ $(TQ)$
corresponding to (6), \begin{equation} X_E = v^i \frac{\partial}{\partial
q^i}+\Gamma^{ij}\alpha_j \frac{\partial}{\partial v^i},\label{8}
\end{equation} where \begin{equation} \alpha_j =
\frac{\partial\ell}{\partial q^i} - \frac{\partial^2 \ell}{\partial
v^j\partial q^k}v^k \end{equation} and
$\Gamma^{ik}\Gamma_{kj}=\delta_{ij}$ such that for any $f\in
C^{\infty}(TQ)$ its time derivative is given by \begin{equation} \dot{f} =
X_E f. \end{equation} In particular, for energy $E =
\frac{\partial\ell}{\partial v^i}v^i - \ell$ we have \begin{equation}
\dot{E} = X_E E =0. \end{equation} The invariant description can be given
as follows. Let $Q$ be a finite-dimensional smooth manifold, $(TQ, \pi ,
Q)$ -- its tangent bundle, $\ell\in C^{\infty}(TQ)$. The Cartan -
Liouville one-form $\theta_L\in \Lambda^1 (TM)$ is defined by
\begin{equation} \theta_L = d\ell\circ\tau , \end{equation} where $\tau$
is a canonical type $(1,1)$ tensor field on $TQ$ \cite{Vershik, Morandi}.
Then the closed nondegenerate (for nondegenerate $l$) two-form $\omega_L
=d\theta_L$ provides $TQ$ with the symplectic structure. In local
coordinates \begin{equation}\theta_L = \frac{\partial\ell}{\partial
v^i}dq^i ,\end{equation} \begin{equation} \omega_L = \Gamma_{ij}dv^i\wedge
dq^j + \frac{1}{2}M_{ij}dq^i\wedge dq^j, \end{equation} where $$ M_{ij} =
\frac{\partial^2 \ell}{\partial q^i\partial v^j}- \frac{\partial^2
\ell}{\partial v^i\partial q^j}. $$ For $f, g\in C^{\infty}(TQ)$ one can
define the Lagrangian bracket by \begin{equation} \{ f,g\}_L =
\omega_L\left( X_f,X_g \right) , \end{equation} where the vector field
$X_f\in \cal{X}$ $(TM)$is defined by \begin{equation} i_{X_f}\omega_L =
-df.\label{g} \end{equation} Since $\omega_L$ is nondegenerate, the map
$\Omega^1(TQ)\rightarrow\cal{X}$ $(TQ)$ (16) is an isomorphism. In local
coordinates we have \begin{equation} \{ f,g\}_L = \Gamma^{ij}\left(
\frac{\partial f}{\partial v^i}\frac{\partial g}{\partial q^j} -
\frac{\partial f}{\partial q^i}\frac{\partial g}{\partial v^j}\right)-
M^{ij}\frac{\partial f}{\partial v^i}\frac{\partial g}{\partial v^j},
\end{equation} where $\Gamma^{ik}\Gamma_{kj}=\delta^{i}_{j}$, $M^{ik} =
\Gamma^{ik}M_{kl}\Gamma^{lj}$. This bracket is antysimmetric. It obeys
Jacobi identity since $d\omega_L =0$. The energy $E$ can be introduced as
a smooth function on $TQ$ by \begin{equation} E = i_Z\theta_L -\ell,
\end{equation} where $Z=\left\{ X\in \cal{X}(TQ):\tau X =\triangle
\right\}$, $\triangle$ is the Liouville vector field.

Putting all this together, we can write the action (5) in the invariant form (1). For the energy (18) one can define the corresponding vector field using (16). A simple local-coordinate calculation shows that this vector field is equivalent to the field $X_E$ (8) which originates from the variational principle. Therefore for
any $ f\in C^{\infty}(TQ)$ we can write \begin{equation} \dot{f} = X_E f =
df(X_E) = \omega_L\left( X_E,X_f\right) = \{ E, f\}_L . \end{equation} In
particular, \begin{equation} \dot{E} = dE(X_E) = 0. \end{equation}

\subsubsection*{B. Hamiltonian formalism}
The transition to the Hamiltonian form of dynamics is provided by the Legendre transformation
 $L: T_q Q\rightarrow T_{q}^*Q$
\begin{equation}
(q,v)\rightarrow (q,p),\hspace{0.5 cm} p_i = \frac{\partial\ell}{\partial
v^i},
\end{equation}
$(q,p)\in T^*Q$ which is a diffeomorphism for nondegenerate $l$. In the new variables the action (1) becomes
$$
S[q,v] \rightarrow S[q,p],
$$
\begin{equation}
S = \int_X s^*\varphi ,
\end{equation}
where $\varphi = -hdt +\theta$,
$\varphi \in\Omega^1 (T^*Q\bigotimes R)$, $X\subset R^1$,
$h(q,p)=(L^*)^{-1}(E)$ is the Hamiltonian, $\theta =p_idq^i$ is the canonical one-form on $T^*Q$, $\theta_L = L^*(\theta )$.

The variational principle applied to (22) gives the standard Hamilton's equations which can be associated with the vector field
$X_h \in \cal{X}$ $(T^*Q)$(the counterpart of $X_E$ (8)),
\begin{equation}
X_h = \frac{\partial h}{\partial
p^i}\frac{\partial}{\partial q^i} - \frac{\partial
h}{\partial q^i}\frac{\partial}{\partial p^i}.
\end{equation}
The field $X_h$ can be defined invariantly by
\begin{equation}
i_{X_h}\omega = -dh,\label{h}
\end{equation}
where $\omega =d\theta$ is the nondegenerate two-form which provides $T^*Q$ with the standard symplectic structure. Then  $\forall f\in C^{\infty}(T^*M)$ we have
\begin{equation}
\dot{f} = X_H f = df(X_H) = \omega\left( X_H,
X_f\right) = \{ H, f\},
\end{equation}
where  $\{ f,g\}$ is the Poisson bracket. In particular,
\begin{equation}
\dot{h} = dh(X_h) = 0.
\end{equation}

\section*{III. DEGENERATE LAGRANGIANS}

For degenerate Lagrangians the Legendre transformation (21) is not a bijection. This means in particular that $R$= rank
$\Gamma_{ij}$ $\leq$ $N$ so the condition $\dot{x}^i =v^i$
is valid now only for a part of coordinates, i.e. for $\dot{x}^a =v^a$, a=1,...$R$. The transition to the Hamiltonian formulation can no longer be considered as a change of variables in the action (1). To cure this situation one should extend the phase space by adding $N-R$ new independent variables  $p_{\mu}$($\mu$=1,...$N-R$) and write the action $S$ as follows \cite{Pavlov}
\begin{equation}
S[q,v,p_{\mu}] = \int dt \left( \ell (q,v) - v^a
\frac{\partial\ell}{\partial v^a} + \dot{q}^a
\frac{\partial\ell}{\partial v^a} +p_{\mu}(\dot{q}^{\mu}-
v^{\mu})\right)\label{ol}
\end{equation}
which can also be presented in the form
\begin{equation}
S = \int \varphi_1 ,\hspace{1cm}
\end{equation}
with $\varphi_1 = -E_1dt +
\theta_1$,
where
\begin{equation}
E_1 = v^a \frac{\partial\ell}{\partial v^a} +
p_{\mu}v^{\mu} -\ell ,
\end{equation}
\begin{equation}
 \theta_1 =
\frac{\partial\ell}{\partial v^a}dq^a + p_{\mu}dq^{\mu
}.\end{equation}
The variation of $S$ gives the equations
\begin{equation}
\dot{v}^a = \Gamma^{ab} \frac{\partial\ell}{\partial q^b}-
\Gamma^{ab} \frac{\partial^2 \ell}{\partial v^b\partial
q^i}v^i - \Gamma^{ab}\Gamma_{b\mu}\dot{v}^{\mu},
\end{equation}
\begin{equation}
\dot{p}_{\mu} = \frac{\partial\ell}{\partial q^{\mu}},
\end{equation}
\begin{equation}
\dot{q}^i - v^i=0,
\end{equation}
\begin{equation}
p_{\mu} - \frac{\partial\ell}{\partial v^{\mu}} = 0.
\end{equation}
The last equation is a constraint in the space of variables $q^i , v^i , p_{\mu}$. The time derivative of a smooth function $f(q^i , v^i , p_{\mu})$ can be written now as
\begin{equation}
	\dot{f} = Y^{(1)}f + \dot{v}^{\mu}Y_{\mu}^{(1)}f, \end{equation}
\begin{equation}
	\phi_{\mu} = p_{\mu }-\frac{\partial l}{\partial v^{\mu}}=0,
\end{equation}
where
\begin{equation}
Y^{(1)} =v^i \frac{\partial}{\partial q^i} +\Gamma^{ab}
\alpha_b\frac{\partial }{\partial v^a} +
\frac{\partial\ell}{\partial q^{\mu}}\frac{\partial
}{\partial p^{\mu}},
\end{equation}
\begin{equation}
\alpha_b = \frac{\partial\ell}{\partial q^b} -
\frac{\partial^2 \ell}{\partial v^b\partial q^i}v^i ,
\end{equation}
\begin{equation}
Y_{\mu}^{(1)} = b^i_{\mu} \frac{\partial}{\partial v^i},
\end{equation}
and $b^{i}_{\mu}$ are the null-vectors of $\Gamma_{ij}$.

The action (28), however, does not have an invariant form. For the sake of invariance one should write $S$ as
\begin{equation}
S[q,v,p] = \int dt\left( l-p_iv^i + p_i\dot{q}^i \right).
\end{equation}
Application of the variational principle gives dynamical equations
\begin{equation}
\dot{p}_i = \frac{\partial l}{\partial v^i},
\end{equation}
\begin{equation}
\dot{q}^i = v^i
\end{equation}
as well as the {\it primary} constraints
\begin{equation}
\phi_i = p_i -\frac{\partial l}{\partial v^i} =0.
\end{equation}
The action (40) can be considered as the basic one for both the Lagrangian and the Hamiltonian formulations for either degenerate or nondegenerate Lagrangians.

Equations (41)-(42) imply that the evolution of any smooth function $f(q,v,p)$ is given by
\begin{equation}
	\dot{f} = Yf + \dot{v}^{i}Y_{i}f,
\end{equation}
\begin{equation}
	\phi_{i} = p_{i}-\frac{\partial l}{\partial v^{i}}=0,
\end{equation}
where
\begin{equation}
Y =v^i \frac{\partial}{\partial q^i} +
\frac{\partial\ell}{\partial q^{i}}\frac{\partial
}{\partial p^{i}},
\end{equation}
\begin{equation}
Y_{i} = \frac{\partial}{\partial v^i}.
\end{equation}
Since $Y_i\phi_j$=-$\Gamma_{ij}$, the Dirac's compatibility condition $\dot{\phi}_i=0$ leads to determination of $R$ accelerations
\begin{equation}
\dot{v}^a = (Y\phi_b )\Gamma^{ba} - \dot{v}^{\mu}\Gamma_{\mu b}\Gamma^{ba}
\end{equation}
as well as to the new constraints
\begin{equation}
\phi^{(1)}_{\mu }= b^{i}_{\mu }\left( Y\phi_i \right) =0.
\end{equation}
At the same time a part of constraints in (45), namely those for which
\begin{equation}
\mbox{det}(Y_a\phi_b )\neq 0,
\end{equation}
$Y_a = \frac{\partial}{\partial v^a}$, can be resolved. Because of the condition (50), we are free to choose either $v^a$ or $p_a$ as the local coordinates on the resulting surface $M_1 = \left\{ q^i, p_i , v^i : \phi_a = 0, a=1,...R\right\}$. As a result of this choice one gets either Lagrangian or Hamiltonian formulation, correspondingly.
\subsection*{A. LAGRANGIAN FORMALISM}
In the Lagrangian scheme the system (44)-(45) on $M_1$ becomes
	\begin{equation}
	\dot{f} = Y^{(1)}f + \dot{v}^{\mu}Y_{\mu}^{(1)}f,
\end{equation}
\begin{equation}
	\phi_{\mu}\mid_{M_1} = 0,
\end{equation}
\begin{equation}
        \phi_{\mu}^{(1)}\mid_{M_1} = 0,
\end{equation}
where $Y^{(1)}$ and $Y^{(1)}_{\mu}$ have been defined in (37)-(39). The action $S$ (40) confined to $M_1$ is given by (27). We will call the primary constraints for which (50) (and consequently (48)) holds {\it the primary constraints of the second class}. The rest of them are called the primary constraints of the first class.

Next the condition $\dot{\phi}_{\mu}^{(1)}=0$ is to be verified. This gives $R_1$ determined accelerations,
\begin{equation}
\dot{v}^{a_1} = - \left( Y^{(1)}\phi_{b_1}^{(1)}\right)\stackrel{(1)}{\gamma}^{b_1a_1} - \dot{v}^{\mu_1}\left( Y^{(1)}_{\mu_1}\phi_{b_1}^{(1)}\right)\stackrel{(1)}{\gamma}^{b_1 a_1},
\end{equation}
and a new set of constraints,
\begin{equation}
\phi^{(2)}_{\mu_1}= b^{(2)\mu}_{\mu_1}\left(
Y^{(1)}\phi^{(1)}_{\mu}\right)=0. \end{equation}
Here $\stackrel{(1)}{\gamma}_{\mu\nu}$ = $Y^{(1)}_{\mu}\phi_{\nu}^{(1)}$,
 $R_1$=rank $\stackrel{(1)}{\gamma}_{\mu\nu}$, $a_1$=1,...$R_1$,
 $\mu_1$=1,...$N-R-R_1$ and $b^{(2)\mu}_{\mu_1}=\delta^{\mu}_{\mu_1} -
 \delta^{\mu}_{a_1}\gamma^{a_1b_1}\gamma_{b_1\mu_1}$ are the null vectors of $ \stackrel{(1)}{\gamma}_{\mu\nu}$. Again,
 among the secondary constraints (53) there are $R_1$ constraints of the second class (det$\gamma_{ab}\neq 0$) as well as $N-R-R_1$ those of the first class. The system therefore can be reduced to the surface $M_2 \subset M_1$ of the second class secondary constraints $\phi^{(1)}_{a_1}=0$ (the corresponding number of the primary constraints ($R_1$) must also be resolved, since $\dot{\phi}_{a_1}\sim \phi_{a_1}^{(1)}$). The evolution then is given by
\begin{equation}
	\dot{f} = Y^{(2)}f + \dot{v}^{\mu_1}Y_{\mu_1}^{(2)}f,
\end{equation}
\begin{equation}
	\phi_{\mu_1}\mid_{M_2} = 0,
\end{equation}
\begin{equation}
        \phi_{\mu_1}^{(1)}\mid_{M_2} = 0,
\end{equation}
\begin{equation}
     \phi_{\mu_1}^{(2)}\mid_{M_2} = 0,
\end{equation}
where
\begin{equation}
Y^{(2)}= \left( Y^{(1)} -\left( Y^{(1)}\phi^{(1)}_{b_1}\right)\stackrel{(1)}{\gamma}^{b_1 a_1}Y^{(1)}_{a_1}\right)\mid_{M_2},
\end{equation}
\begin{equation}
Y^{(2)}_{\mu_1}= b^{(2)\mu}_{\mu_1}Y^{(1)}_{\mu}\mid_{M_2}.
\end{equation}
The following iterations are straightforward. After the k-th step the system becomes
\begin{equation}
	\dot{f} = Y^{(k)}f + \dot{v}^{\mu_{k-1}}Y_{\mu_{k-1}}^{(k)}f,
\end{equation}
plus the set of constraints
\begin{equation}
     	\phi_{\mu_{k-1}}^{\alpha}\mid_{M_k} = 0,
\end{equation}
where $\alpha = 0,1,...k$, $\phi_{\mu_{k-1}}^{(0)} =\phi_{\mu_{k-1}}$,
\begin{equation}
Y^{(k)}= \left( Y^{(k-1)} -\left( Y^{(k-1)}\phi^{(k-1)}_{b_{k-1}}\right)\stackrel{(k-1)}{\gamma}^{b_{k-1} a_{k-1}}Y^{(k-1)}_{a_{k-1}}\right)\mid_{M_k},
\end{equation}
$M_k =\left\{ x\in M_{k-1}: \phi^{(\alpha)}_{a_{k-1}}(x)=0, \alpha
 = 0,...k \right\}$, $M_k\subset M_{k-1}\subset M_{k-1} \subset \cdots
\subset W$.

The iteration process terminates, if:

1) for certain $k$ we have
$$
R_k = N - R_1 - \cdots - R_{k-1},
$$
where $R_k$= rank $\stackrel{(k)}{\gamma}_{b_{k-1} a_{k-1}}$. In this case all the accelerations are determined and the dynamics confined to the final constraint surface $M_k$ is totally fixed.

2) all the constraints of generation $k+1$ are reducible to those of the previous generations or are identically zero. In this case $N-R_1-\cdots - R_k$ accelerations remain undetermined and the system posesses certain ``gauge freedom''.

\subsection*{B. HAMILTONIAN FORMALISM}
Let us rewrite the system (44) - (45) in a slightly different form
\begin{equation}
	\dot{f} = \left\{ E, f\right\} +\dot{v}^i Y_i f,
\end{equation}
\begin{equation}
	\phi_{i} = 0,
\end{equation}
where $E(p,q,v) = p_iv^i -l$ and $\left\{ f,g \right\}$ is the usual Poisson bracket generated by the two-form $dp\wedge dq$ (which is of course degenerate in the space of variables $p, q, v$). As it was in (54) - (55), we determine $R$ accelerations,
\begin{equation}
\dot{v}^a = \left\{ E, \phi_b \right\}\Gamma^{ba} - \dot{v}^{\mu}\Gamma_{\mu b}\Gamma^{ba},
\end{equation}
and a set of new constraints,
\begin{equation}
\phi^{(1)}_{\mu}=b^{i}_{\mu}\left\{ E, \phi_i \right\} = 0.
\end{equation}
The condition (50) allows us to resolve a certain part of
 constraints in (66). To produce the Hamiltonian scheme, one should
 choose $p_a$ rather than $v^a$ as a set of local coordinates on the surface
 $M_1$ of the resolved constraints. Due to the identity
 $\Gamma_{\mu\nu}=\Gamma_{\mu a}\Gamma^{ab}\Gamma_{b\nu}$ the Routh's function
\begin{equation}
R(q^i, p_a, v^{\mu}) = \left( l - p_av^a \right)\mid_{M_1}
\end{equation}
is linear in $v^{\mu}$:
\begin{equation}
R = -h(q^i, p_a) +v^{\mu}\psi_{\mu}(q^i, p_a)
\end{equation}
(here $\psi_{\mu}(q^i, p_a) =\partial l /\partial v^{\mu}\mid_{M_1}$), so
the action (40) being confined to $M_1$ in coordinates $p_a, q^i, v^{\mu}$ reads
\begin{equation}
S[q^i, p_a, v^{\mu}]=\int dt \left( -h_T + p_i\dot{q}^i\right),
\end{equation}
where
\begin{equation}
h_T=h(q^i, p_a) +v^{\mu}\phi_{\mu},
\end{equation}
$$
\phi_{\mu}=p_{\mu} - \psi_{\mu}(q^i, p_a).
$$
Correspondingly, the system (65) - (66) becomes
\begin{equation}
	\dot{f} = \left\{ h, f\right\} + v^{\mu}\left\{\phi_{\mu}, f\right\} +\dot{v}^{\mu} Y_{\mu}f,
\end{equation}
\begin{equation}
	\phi_{\mu}\mid_{M_1} = 0,
\end{equation}
\begin{equation}
\phi_{\mu}^{(1)}\mid_{M_1} = 0,
\end{equation}
Note that (28) and (71) are equivalent. The transition between the Lagrangian and the Hamiltonian formulations is a smooth change of variables in the action, as it was for nondegenerate Lagrangians. There is no need to add voluntarily the primary constraints $\phi_{\mu}$ to the function $h(q^i, p_a)$, since the old Dirac's construction $h_T$ naturally appears in (71) and in (73).

The condition $\dot{\phi}^{(1)}_{\mu}$=$0$ leads to the determination of some accelerations,
\begin{equation}
\dot{v}^{a_1} =- \left\{ h_T, \phi^{(1)}_{b_1}\right\}\gamma^{b_1 a_1} -
 \dot{v}^{\mu_1}\gamma_{\mu_1 b_1}\gamma^{b_1  a_1},
\end{equation}
and to the new constraints,
\begin{equation}
\phi^{(2)}_{\mu_1} = b^{(2)\mu}_{\mu_1} \left\{ h_T, \phi^{(1)}_{\mu_1}\right\}=0.
\end{equation}
Here $\gamma_{\mu\nu}$=$\partial (\phi^{(1)}_{\nu}\mid_{M_1})/\partial v^{\mu}$. Now we can reduce the system to the surface $M_2\subset  M_1$ of the second class secondary constraints $\phi^{(1)}_{a_1}=0$. In the Hamiltonian scheme we can do this explicitly by expressing $v^{a_1}$:
\begin{equation}
v^{a_1} =- \left\{ h, \phi_{b_1}\right\}\gamma^{b_1 a_1} -
 v^{\mu_1}\left\{ \phi_{\mu_1},\phi_{b_1}\right\}\gamma^{b_1  a_1}.
\end{equation}
Then the system (73)-(75) on $M_2$ can be written as
\begin{equation}
	\dot{f} = \left\{ h, f\right\}^{*}_1  + v^{\mu}\left\{\phi_{\mu_1}, f\right\}^{*}_1 +\dot{v}^{\mu_1} Y_{\mu_1}f,
\end{equation}
\begin{equation}
	\phi_{\mu_1}\mid_{M_2} = 0,
\end{equation}
\begin{equation}
\phi_{\mu_1}^{(1)}\mid_{M_2} = 0,
\end{equation}
\begin{equation}
 \phi_{\mu_1}^{(2)}\mid_{M_2} = 0
	\end{equation}
where
\begin{equation}
\left\{ f,g\right\}^*_1=\{ f,g\}-\{ f,\phi_{b_1}\}\gamma^{b_1
a_1}\{ \phi_{a_1} ,g\}
\end{equation}
is the Dirac bracket in respect to the second class primary constraints. Next iteration gives us another set of the determined accelerations,

\begin{equation}
\dot{v}^{a_2} =- \left\{ h, \phi^{(2)}_{b_2}\right\}^*_1\stackrel{(2)}{\gamma}^{b_2 a_2} - v^{\nu_1}\left\{ \phi_{\nu_1}, \phi^{(2)}_{b_2}\right\}^*_1\stackrel{(2)}{\gamma}^{b_2 a_2}-
 \dot{v}^{\mu_2}\stackrel{(2)}{\gamma}_{\mu_2 b_2}\gamma^{b_2  a_2},
\end{equation}
velocities,
\begin{equation}
v^{a_2} =- \left\{ h, \phi^{(1)}_{b_2}\right\}^*_1\stackrel{(2)}{\gamma}^{b_2 a_2} -
 v^{\mu_2}\left\{ \phi_{\mu_2},\phi_{b_2}^{(1)}\right\}^*_1\stackrel{(2)}{\gamma}^{b_2  a_2},
\end{equation}
where
\begin{equation}
\stackrel{(2)}{\gamma}_{\mu_1\nu_1}=\partial \phi_{\nu_1}^{(2)}/\partial v^{\mu_1} = \left\{ \phi_{\mu_1},\phi_{\nu_1}^{(1)}\right\}^*_1,
\end{equation}
and constraints
\begin{equation}
\phi^{(3)}_{\mu_2} = b^{(3)\mu_1}_{\mu_2}\left( \left\{ h, \phi^{(2)}_{\mu_1}\right\}^*_1 + v^{\nu_1} \left\{ \phi_{\nu_1}, \phi^{(2)}_{\mu_1}\right\}^*_1\right) =0.
\end{equation}
The system (79)-(82) then becomes
\begin{equation}
	\dot{f} = \left\{ h, f\right\}^{*}_2  + v^{\mu_2}\left\{\phi_{\mu_2}, f\right\}^{*}_2 +\dot{v}^{\mu_2} Y_{\mu_2}f,
\end{equation}
\begin{equation}
	\phi_{\mu_2}\mid_{M_3} = 0,
\end{equation}
\begin{equation}
\phi_{\mu_2}^{\alpha}\mid_{M_3} = 0, \alpha =1,2,3,
\end{equation}
where
\begin{equation}
\left\{ f,g\right\}^*_2=\left\{ f,g\right\}^*_1-\left\{ f,\phi_{b_2}^{(1)}\right\}^*_1\stackrel{(2)}{\gamma}^{b_2
a_2}\left\{ \phi_{a_2} ,g\right\}^*_1 +
\left\{ g,\phi_{b_2}^{(1)}\right\}^*_1\stackrel{(2)}{\gamma}^{b_2
a_2}\left\{ \phi_{a_2} ,f\right\}^*_1 .
\end{equation}
The iterative procedure continues until either all acceleration ( and velocities) are determined (and the final phase space is a submanifold of $T^*Q$) or the constraints of the k-th generation are reducible to the previous ones (or are identically zero).

\subsection*{IV. INVARIANT FORMULATION}

Let $M$ be a real smooth $N$-dimensional manifold, $(TM, \pi_1 , M)$
 its tangent bundle, $(T^*M, \pi_2 , M)$ its cotangent bundle. Let $W$=$
 TM\bigoplus T^*M$ denotes a Whitney sum of these two bundles (Fig.1).

 \begin{picture}(100,100)
\put (50,80){W} \put (50,20){M}
\put (20,50){$TM$} \put (70,50){$T^*M$}

\put (30,65){$\rho_1 \swarrow$}
 \put (60,65){$\searrow\rho_2$}
\put (30,35){$\pi_1\searrow$} \put (60,35){$\swarrow\pi_2$}
\put (45,5){Fig.1}
 \end{picture}

  Consider  $\rho^*_2\theta\in
\Omega^1(W)$, where $\theta$ is the canonical one-form on $T^*M$. To simplify
 notations we denote $\rho^*_2\theta$ by $\theta$ again. The two-form
 $\omega =d\theta $ is closed and degenerate on $W$, Ker $\omega =
 {\cal X}^V$(TM)$\subset {\cal X}$(W). The elements of Ker $\omega$ form
 an integrable distribution ${\cal D}$ on $W$ thus defining a foliation
 $F$ of $W$, the leaves of $F$ being the maximal integral manifolds of
 ${\cal D}$. The foliation $F$ has codimension $q=3N-$rank${\cal D}$ = dim
\ Coker $\omega$=$2N$. We assume that the space of leaves ${\cal L}$ has
 a structure of manifold, so $F$ is a regular foliation \cite{Marmo} of
 codimension $q$. Now consider the de Rham complex of $F$. We have $\Omega^1 W/{\cal  L}$=$T^*W/{\cal L}$, $\Omega^i W/{\cal L}$=$\Lambda^i\Omega^1 W/{\cal L}$,
 $\Omega  W/{\cal L}$=$\bigoplus_{i\geq 0} \Omega^i W/{\cal L}$. The
 first order differential operator $d_q$:
$\Omega W/{\cal L}$ $\rightarrow$ $\Omega W/{\cal L}$ has the following properties \cite{Manin}:

i) $d_q$ : $C^{\infty}(W)\rightarrow \Omega^1 W/{\cal L}$ is a composition
 $C^{\infty}(W)\stackrel{d}{\rightarrow}\Omega^1 W\rightarrow T^*W/{\cal L}$,

ii) $d_q$($\omega^p\wedge\omega^s$)= $d_q\omega^p\wedge \omega^s +(-1)^p
\omega^p\wedge d_q\omega^s$, $\omega^p\in \Omega^pW/{\cal L}$,

iii) $d_{q}^2$ =0.

Note also that $d_0=d$ : $\Omega W\rightarrow\Omega W$. In any Frobenius
neighborhood $U\subset W$ one can choose the local coordinate system
 $(x^1,...x^d, y^{d+1}...y^{d+q})$ such that ${\cal D}_U$ is generated by
 the set $\{ \partial /\partial x^{\mu} \}$, $\mu$=1,...$d$=rank ${\cal D}$. Then the action of $d_q$ on any $\alpha\in \Omega^p W/{\cal L}$,
$$
\alpha = a_{\mu_1 ...\mu_p}(x,y)dx^1\wedge \cdots \wedge dx^{\mu_p},
$$
can be described as follows
$$
d_q \alpha = \frac{\partial a_{\mu_1 ...\mu_p}(x,y)}{\partial x^{\nu}}dx^{\nu}
\wedge dx^1\wedge \cdots \wedge dx^{\mu_p},
$$
$\mu ,\nu$=$1,...d$. In other words, provided $\left\{ K_i\right\}^{d}_{i=1}$ is a basis in ${\cal D}$ and $\left\{ \theta^i\right\}^{d}_{i=1}$
is a dual basis in ${\cal D}^*$, the action of $d_q$ is given by
$$
d_q\alpha =\left( K_i a_{i_1\cdots i_p}\right)\theta^i\wedge
\theta^1\wedge\cdots\wedge\theta^{i_p}. $$
Let us denote $\omega^{q}_{f}=d_qf$, $f\in C^{\infty}(W)$. Now we are ready to proceed with the invariant formulation of dynamics.

Let $Z=\{ X\in {\cal X}(W): \tau (\rho_{1*}X)=\triangle \}$, where $\tau$ is the vertical endomorphism, $\triangle$ is the Liouville field. Then for any Lagrangian $l\in C^{\infty}(TM)$ the energy $E\in C^{\infty}(W)$ is defined as follows
\begin{equation}
E=i_Z\theta -\rho^{*}_1 l .
\end{equation}
The action is a functional on the smooth sections of $W\bigotimes R$$\rightarrow R$ ,
\begin{equation}
S[s] = \int_X s^*\varphi ,
\end{equation}
where $X\in R$ is a compact set, $\varphi = -Edt +\theta$, $\varphi
 \in \Omega^1 (W\bigotimes R)$. Since $W$ is only a presymplectic manifold,
 the relation (16) between one-forms and vector fields on $W$ now makes sense
 only for functions constant along the leaves of foliation $F$.
 In particular, (16) cannot be used for the second class constraints of
 any generation. However, in the local coordinate approach to the constrained
 dynamics we do not encounter such restriction. To cure this, we deform
 the correspondence (16) and put it to be
\begin{equation}
i_{X_f}\omega = -df +\omega^q_f,\label{kmkk}
\end{equation}
where $q$ is the codimension of the foliation generated by Ker $\omega$. Then
the system of equations (44) - (45) which follows from the variation of $S$ can be written in the form
\begin{equation}
\dot{f}=df(X_E),
\end{equation}
\begin{equation}
\omega^{q}_E=0,
\end{equation}
where $X_E$ is defined by (94). Using (94) we can rewrite (95) as
\begin{equation}
\dot{f}= \omega \left( X_E , X_f\right) +\omega^{q}_f \left( X_E \right).
\end{equation}
Note that (94) -(95) give $\dot{E}=0$.
One can decompose $X_E$ as
\begin{equation}
X_E = Y +K,
\end{equation}
where $K\in {\cal D}$ is an arbitrary vector. Locally, $K=\alpha^iK_i$,
 where $\{ K_i \}^{d}_{i=1}$ is a basis in ${\cal D}$. Since ${\cal D}\in
{\cal X}^V (TW)$, the yet undetermined multipliers $\alpha^i$ are interpreted
 as accelerations $\dot{v}^i$ in any given chart. The equation (96) then reads
\begin{equation}
\omega^{q}_E = \left( K_i E\right)\theta^i = 0,
\end{equation}
with $\theta^i\left( K_i\right) =\delta^{i}_j$. This defines a subspace
 $S\subset W$ by
\begin{equation}
\phi_i = dE\left( K_i\right) =0.
\end{equation}
For compatibility of (95) - (96) one must have therefore $X_E\in TS$, i.e.
\begin{equation}
d\phi_i \left( X_E \right) =0.
\end{equation}
If $S$ is transversal to the leaves of $F$ (in this case ${\cal D}\cap
 TS=\emptyset$ and rank $\mid d\phi_i (K_j )\mid$=$N$), the vector
 field $X_E$ is completely fixed by the condition (101) in the sense that
 all coefficients $\alpha^i$ are determined. Then we can confine the system to $S$ and choose either p's or v's as the set of local coordinates on $S$ thus generating either Hamiltonian ($S=T^*Q$) or Lagrangian ($S=TQ$) form of dynamics.

In the generic case, however, ${\cal D}\cap TS$=${\cal D}_1$
 $\neq \emptyset$, rank $\mid d\phi_i (K_j )\mid$= $R_1$ $\leq N$,
 so only $R_1$ out of $N$ accelerations are determined by (101). Instead
of $S$ we can choose now an intermediate transverse subspace $M_1$
($S\subset M_1\subset W$) such that $d_1 =$dim (${\cal
D}\cup TM_1$) = $N-R_1$. The two-form $\omega_1 =\omega\mid_{M_1}$
is degenerate, dim Ker $\omega_1$ = $d_1$. Now the scheme repeats itself. The system (95) -- (96) becomes
\begin{equation}
\dot{f}=df\left(X_{E_1}\right),
\end{equation}
\begin{equation}
\omega^{q}_E\mid_{M_1}=0,
\end{equation}
\begin{equation}
\omega^{q_1}_{E_1}\mid_{M_1}=0,
\end{equation}
where $E_1 =E\mid_{M_1}$, $q_1 = 2N$ is a codimension of the foliation of
 $M_1$ generated by $\omega_1$. This system can be expressed equivalently in
 terms of either Lagrangian ( (51) -- (53) ) or Hamiltonian ( (73) -- (75) )
 local coordinates. A new subspace $S_1 \subset M_1$ defined by (103)
arises here. The requirement
$X_{E_1}$ $\subset TS_1$ gives $R_2\leq N-R_1$ determined accelerations.
 We choose $M_2$ such that $d_2 =$ dim (${\cal D}\cap TM_2$) = $N-R_1-R_2$.
 Now $\omega_2 =\omega_1 \mid_{M_2}$ is degenerate and dim Ker $\omega_2$ = $d_2$. The dynamics is governed by the system
\begin{equation}
\dot{f}=df\left(X_{E_2}\right) = \omega_2 \left( X_{E_2}, X_f\right) + \omega^{q_2}_f\left( X_{E_2}\right),
\end{equation}
\begin{equation}
\omega^{q}_E\mid_{M_2}=0,
\end{equation}
\begin{equation}
\omega^{q_1}_{E_1}\mid_{M_2}=0,
\end{equation}
\begin{equation}
\omega^{q_2}_{E_2}\mid_{M_2}=0,
\end{equation}
The sequence stops when for certain $n$ $S_n$ is not foliated, i.e. when ${\cal D}_{n-1}\cap TM_{n-1}$ = $\emptyset$ and $\omega_n$ is nondegenerate. In this case the dynamics on $S_n = M_n\subset W$ is fixed and one has
\begin{equation}
\dot{f}= \omega_n \left( X_{E_n}, X_f\right)
\end{equation}
for any $f\in C^{\infty}(S_n)$. As an alternative, one may discover,
 however, that for some $n$ the condition $X_{E_{n-1}}\in TS_{n-1}$ is
 automatically satisfied while $S_{n-1}$ is still foliated.
 This means that $N-R_1-\dots -R_{n-1}$ accelerations remain undetermined
 and the evolution is described by the equation
\begin{equation}
\dot{f}= \omega_n \left( X_{E_n}, X_f\right) + \omega^{q_n}_f\left( X_{E_n}\right),
\end{equation}
together with the final set of the first-class constraints
\begin{equation}
\omega^{q}_{E}\mid_{M_n}=0,\cdots \omega^{q_n}_{E_n}\mid_{M_n}=0.
\end{equation}
The invariant geometrical treatment of the first-class constraints case including gauge-fixing procedure was given by L.D. Faddeev \cite{Faddeev}.

\subsection*{V. EXAMPLES}

{\it Example 1}:
Let us consider the simplest nondegenerate Lagrangian,
\begin{equation}
l= \frac{v^2}{2}-U\left( q\right).
\end{equation}
We start with the phase space $W=TR^1\bigoplus T^*R^1$. The canonical two-form $\omega =dp\wedge dq$ is degenerate on $W$, Ker $\omega$= span$\{ \partial /\partial v\}$. The energy $E\in C^{\infty}(W)$ (92) is given by
\begin{equation}
E= pv-\frac{v^2}{2}+U\left( q\right).
\end{equation}
The primary constraint one-form $\omega_E$ can be written as
\begin{equation}
\omega_E= d_q E = \frac{\partial E}{\partial v}dv = \phi dv =0,
\end{equation}
where $\phi = p-v$. The relation (94) gives the evolution vector field $X_E$ on $W$,
\begin{equation}
X_E= v\frac{\partial}{\partial q} - \frac{\partial U}{\partial q}\frac{\partial }{\partial p} + \dot{v} \frac{\partial}{\partial v}.
\end{equation}
Thus the dynamics on $W$ is given by $\{ \dot{f}=X_E f, \phi =0 \}$.
 The compatibility condition (101) allows to determine the acceleration
 $\dot{v}=\partial U/\partial q$. This means that $\phi$ is the
 primary constraint of the second class. Therefore, we reduce the system to $S_1$ = $\{ x\in W:\phi =0 \}$. One can choose either $(q,v)$ or $(q,p)$ as the local coordinates on $S_1$. In the first case one gets the Lagrangian scheme,
\begin{equation}
E_1= E\mid_{S_1} = \frac{v^2}{2}+U\left( q\right),
\end{equation}
\begin{equation}
X_{E_1}= v\frac{\partial}{\partial q} - \frac{\partial U}{\partial q}\frac{\partial }{\partial v},
\end{equation}
in the second case - the Hamiltonian description,
\begin{equation}
h= E\mid_{S_1} = \frac{p^2}{2}+U\left( q\right),
\end{equation}
\begin{equation}
X_{h}= p\frac{\partial}{\partial q} - \frac{\partial U}{\partial q}\frac{\partial }{\partial p}.
\end{equation}
In this case $W$ trivially decouples into $TQ$ and $T^*Q$.

{\it Example 2}: Consider the Lagrangian (Ref. \cite{Nesterenko})
\begin{equation}
l= \frac{v_{1}^2}{2}- v_2q_3 .
\end{equation}
The initial phase space is $W$=$TR^3\bigoplus T^*R^3$ with local coordinates $q_i, p_i, v_i$, $i=1,2,3$. The fundamental two-form $\omega = dp_i\wedge dq_i$ is degenerate on $W$, Ker $\omega$ = span $\{ \partial /\partial v_1,
\partial /\partial v_2, \partial / \partial v_3 \}$. The energy $E\in C^{\infty}(W)$,
\begin{equation}
E=p_1v_1 + p_2v_2 + p_3v_3 - \frac{1}{2}v_{1}^2 + v_2q_3,
\end{equation}
defines (by (94)) the evolution vector field $X_E =Y+K$, where
\begin{equation}
Y= v_i\frac{\partial}{\partial q_i} - v_2\frac{\partial }{\partial p_3},
\end{equation}
\begin{equation}
K=\dot{v}_i\frac{\partial}{\partial v_i}
\end{equation}
with yet undetermined multipliers $\dot{v}_i$. The one-form (96) of the primary constraints is given by
\begin{equation}
\omega_E = d_q E =\frac{\partial E}{\partial v_i}dv_i =\phi_i dv_i =0,
\end{equation}
where $\phi_1 = p_1 -v_1$, $\phi_2 = p_2 +q_3$, $\phi_3 =p_3$. The compatibility condition (101) determines one of the accelerations, $\dot{v}_1 =0$, and produces two new constraints,
\begin{equation}
\phi_{2}^{(1)} =v_3 =0, \hspace{1 cm}
\phi_{3}^{(1)} =v_2 =0 .
\end{equation}
Thus the primary constraint $\phi_1$ is of the second class and can be resolved. On the intermediate transverse subspace $M_1 = \{ x\in W:\phi_1 =0\}$ the evolution is given by
\begin{equation}
\dot{f}= v_i\frac{\partial f}{\partial q_i} - v_2\frac{\partial f}{\partial p_3}+ \dot{v}_2\frac{\partial f}{\partial v_2} + \dot{v}_3\frac{\partial f}{\partial v_3} ,
\end{equation}
together with the set of constraints
\begin{equation}
M_2 = \left\{ \phi_2 =0, \phi_3 =0,
 \phi^{(1)}_2 =0, \phi^{(1)}_3 =0 \right\}.
\end{equation}
 Since $\dot{\phi}^{(1)}_2 =
\dot{v}_3 =0$, $\dot{\phi}^{(1)}_3 =\dot{v}_2 =0$, all the secondary
 constraints are of the second class and the final constrained submanifold
 $S_2 =M_2$. For any $f\in C^{\infty}(S_2)$ one has then $\dot{f} = v_1\frac{\partial f}{\partial q_1}$. The dynamics is totally fixed.

To consider the Hamiltonian form of dynamics, one should choose $p_1$ instead of $v_1$ as the local coordinate on $M_1$. Note that
\begin{equation}
E\mid_{M_1} = \frac{1}{2}p_{1}^2 + v_2q_3 +p_2v_2 +p_3v_3 = h+v_2\phi_2 +v_3\phi_3.
\end{equation}
The equation (126) can be written in the form
  \begin{equation}
\dot{f}= \left\{ h, f\right\} +v_2\left\{ \phi_2 , f\right\} +v_3\left\{ \phi_3 , f\right\} +  \dot{v}_2\frac{\partial f}{\partial v_2} + \dot{v}_3\frac{\partial f}{\partial v_3} ,
\end{equation}
together with the set $M_2$ (127). The compatibility condition again gives $\dot{v}_2 =0$, $\dot{v}_3 =0$. One can make the canonocal transformation,
$P_1 = p_1$, $Q_1 = q_1$, $P_2 =p_2$, $Q_2 = q_2 +p_3$, $P_3 = \phi_3 = p_3$,
 $Q_3 = q_3 +p_2 =\phi_2$. In terms of these new variables $S_2 = \{ P_2 =0,
 Q_2 =0, P_3 =0, Q_3 =0\}$. One has $h\mid_{S_2} = P_{1}^2$ and $\dot{f}
 = \{ h, f\}$ for any $f\in C^{\infty}(S_2)$.

{\it Example 3}: Let
$
l = \frac{1}{2} \left( q_1v_2 -q_2v_1 -q_{1}^2 - q_{2}^2\right)
$ (Ref. \cite{olga}). The two-form $\omega = dp_i\wedge dq_i$, $i=1,2$, is degenerate on $W= TR^2\bigoplus T^*R^2$, Ker $\omega$ = span $\{ \partial /\partial v_1, \partial / \partial v_2 \}$. The energy is
\begin{equation}
E=p_1v_1 +p_2v_2 -l
\end{equation}
and the evolution vector field $X_E$ is given by
\begin{equation}
X_E= v_1\frac{\partial }{\partial q_1} +v_2\frac{\partial }{\partial q_2} +
\left( \frac{v_2}{2} -q_1\right)\frac{\partial }{\partial p_1}-
\left( \frac{v_1}{2} -q_2\right)\frac{\partial }{\partial p_2}
+ \dot{v}_1\frac{\partial }{\partial v_1} + \dot{v}_2\frac{\partial }{\partial v_2} ,
\end{equation}
The set of primary constraints is defined by
\begin{equation}
\omega_E = d_q E=\phi_i dv_i =0,
\end{equation}
where $\phi_1 = p_1 +q_2/2$, $\phi_2= p_2 -q_1/2$. The condition (101) gives
 only the new constraints, $\phi^{(1)}_1 = v_2 -q_1 =0$, $\phi^{(1)}_2 = v_1
 +q_2 =0$. The next iteration $\left( \dot{\phi}^{(1)}_{1}=0,
 \dot{\phi}^{(1)}_{2}=0 \right)$, however, allows to determine $\dot{v}_1
 =-v_2$ and $\dot{v}_2 =v_1$. Therefore, the final constrained submanifold
 is $$S=\{ \phi_1 =0, \phi_2 =0, \phi^{(1)}_{1} =0, \phi^{(1)}_{2}=0\}.$$
The dynamics on $S$ is completely fixed and $\forall f\in C^{\infty}(S)$ we have  $\dot{f} = v_1\frac{\partial f}{\partial q_1} - q_1\frac{\partial f}
{\partial v_1}.
$

The energy (130) has the form
\begin{equation}
E = \frac{1}{2}\left( q_1^2 +q_2^2\right) +v_1\phi_1 +v_2\phi_2 .
\end{equation}
Instead of (131), (132) we can write
 \begin{equation}
\dot{f}= \left\{ E, f\right\} +  \dot{v}_1\frac{\partial f}{\partial v_1} + \dot{v}_2\frac{\partial f}{\partial v_2} ,
\end{equation}
\begin{equation}
\phi_1 =0,
\end{equation}
\begin{equation}
\phi_2 =0,
\end{equation}
$\forall f\in C^{\infty}(W)$.
The reduction to $S$ must be ``canonical'' in Hamiltonian formalism. The
 canonical transformation $P_1 = p_1 -q_2/2$, $Q_1 = q_1 /2 +p_2$, $P_2
 =p_2 - q_1 /2 =\phi_2$, $Q_2 = q_2 /2  +p_1 =\phi_1$ gives $S=\{ P_2 =0,
 Q_2 =0\}$ and $h=E\mid_{S} = \frac{1}{2}\left( P_1^2 +Q_1^2\right)$. The evolution
equation (134) on $S$ becomes $\dot{f} =\{ h, f\}$, $f\in C^{\infty}(S)$.

{\it Example 4}: Now consider the Lagrangian (Ref.\cite{H})
$$
l = \frac{1}{2}e^{q_2}v_{1}^{2} .
$$
As it was in the previous example, $W=TR^2\bigoplus T^*R^2$, Ker $\omega$ =
 span $\{ \partial /\partial v_1 , \partial /\partial v_2 \}$. The energy
\begin{equation}
E=p_1v_1 +p_2v_2 - \frac{1}{2}e^{q_2}v_{1}^{2}
\end{equation}
produces (by (94)) the evolution vector field
\begin{equation}
X_E= v_1\frac{\partial }{\partial q_1} +v_2\frac{\partial }{\partial q_2} +
      \frac{1}{2}e^{q_2}v_{1}^{2}\frac{\partial }{\partial p_2}
+ \dot{v}_1\frac{\partial }{\partial v_1} + \dot{v}_2\frac{\partial }
{\partial v_2}
\end{equation}
and the primary constrained form
\begin{equation}
\omega_E = \phi_1 dv_1 + \phi_2 dv_2 =0,
\end{equation}
where $\phi_1 = p_1 - e^{q_2}v_1$, $\phi_2 =p_2$. The condition (101) gives $\dot{v}_1 =-v_1v_2$ as well as the new constraint $\phi^{(1)}_2 =v_1 =0$. Thus the primary constraint $\phi_1$ is of the second class. On $M_1 = \{ x\in W:\phi_1 =0\}$ we have
\begin{equation}
\dot{f} =  v_1\frac{\partial f}{\partial q_1} +v_2\frac{\partial f}{\partial q_2} +\frac{1}{2}e^{q_2}v_{1}^{2}\frac{\partial }{\partial p_2}
-v_1v_2\frac{\partial f}{\partial v_1} + \dot{v}_2\frac{\partial f}{\partial v_2} ,
\end{equation}
\begin{equation}
\phi_2 =p_2 =0,
\end{equation}
\begin{equation}
\phi_2^{(1)} =v_1 =0.
\end{equation}
Since $\dot{\phi}_{2}^{(1)} \sim \phi^{(1)}_2$, the acceleration $\dot{v}_2$ remains undetermined. The constraints $\phi_2$ and $\phi_{2}^{(1)}$ are of the first class. We have also
\begin{equation}
E\mid_{M_1} = \frac{1}{2}p_1^2e^{-q_2} + p_2v_2 = h +v_2\phi_2,
\end{equation}
and the system (140) -(142) in Hamiltonian form reads
\begin{equation}
\dot{f}= \left\{ h, f\right\} + v_2\left\{ \phi_2 , f\right\}+
 \dot{v}_2\frac{\partial f}{\partial v_2} ,
\end{equation}
\begin{equation}
\phi_2 = p_2 =0,
\end{equation}
\begin{equation}
\phi_2^{(1)}=p_1 =0 .
\end{equation}
Once again, the acceleration $\dot{v}_2$ remains arbitrary.

{\it Example 5}: Finally, consider the Lagrangian (Ref. \cite{sundermeyer})
$$
l= \frac{1}{2}v_1^2 +q_2v_1 +(1-\alpha )q_1v_2 +\frac{\beta}{2}(q_1 -q_2)^2
$$
The form $\omega = dp_i\wedge dq_i$ is degenerate on  $W=TR^2\bigoplus T^*R^2$,
 Ker $\omega$ = span $\{ \partial /\partial v_1 , \partial /\partial v_2 \}$. The energy
\begin{equation}
E=p_1v_1 +p_2v_2 - l
\end{equation}
gives  the evolution vector field
\begin{equation}
X_E= v_1\frac{\partial }{\partial q_1} +v_2\frac{\partial }{\partial q_2} +
 \left( v_2(1 -\alpha )+\beta (q_1 -q_2)\right)\frac{\partial }{\partial p_1}
+\left( v_1 -\beta (q_1 -q_2)\right)\frac{\partial }{\partial p_2}
+ \dot{v}_1\frac{\partial }{\partial v_1} + \dot{v}_2\frac{\partial }{\partial v_2}
\end{equation}
as well as  the primary constraint form
\begin{equation}
\omega_E = \phi_1 dv_1 + \phi_2 dv_2,
\end{equation}
where $\phi_1 = p_1 -q_2 -v_1$, $\phi_2 = p_2 -(1-\alpha)q_1$. For simplicity we consider here only some particular cases.

Case A: $\alpha =0, \beta =0$.

The condition (101) gives $\dot{v}_1 =0$, so $\phi_1$ is the primary constraint of the second class. On $M_1 = \{ x\in W: \phi_1 =0\}$ we get
\begin{equation}
X_{E_1}= v_1\frac{\partial }{\partial q_1} +v_2\frac{\partial }{\partial q_2} +
 v_1\frac{\partial }{\partial p_2}
 + \dot{v}_2\frac{\partial }{\partial v_2}.
\end{equation}
The acceleration $\dot{v}_2$ cannot be determined. The dynamics is controlled by
\begin{equation}
\dot{f} =X_{E_1}f,
\end{equation}
\begin{equation}
\phi_2 = p_2 -q_1 =0 .
\end{equation}
Since
\begin{equation}
E\mid_{M_1} = \left( v_1 +q_2\right) v_1 +p_2v_2 -l = \frac{1}{2}\left( p_1 -q_2\right)^2 +v_2\phi_2 = h +v_2\phi_2,
\end{equation}
the Hamiltonian analog of (151) -(152) is
\begin{equation}
\dot{f}= \left\{ h, f\right\} + v_2\left\{ \phi_2 , f\right\}+
 \dot{v}_2\frac{\partial f}{\partial v_2} ,
\end{equation}
\begin{equation}
\phi_2 = p_2 - q_1 =0 .
\end{equation}

Case B: $\alpha =0, \beta \neq 0$.

In this case the compatibility condition allows to determine $\dot{v}_1 =\beta (q_1 - q_2)$
 and gives the secondary constraint $\phi^{(1)}_{2}= q_2 -q_1$.
 The primary constraint $\phi_1$ is of the second class. On $M_1 = \{ x\in W:
 \phi_1 =0 \}$ the evolution is given by
\begin{equation}
\dot{f}= v_1\frac{\partial f}{\partial q_1} +v_2\frac{\partial f}{\partial q_2} +
 \left( v_1 -\beta(q_1 -q_2) \right)\frac{\partial f}{\partial p_2}
+ \beta (q_1 -q_2 )\frac{\partial f}{\partial v_1} + \dot{v}_2\frac{\partial f}{\partial v_2},
\end{equation}
\begin{equation}
\phi_2 = p_2 - q_1 =0 ,
\end{equation}
\begin{equation}
\phi^{(1)}_2 = q_2 - q_1 =0 .
\end{equation}
The next iteration produces $\phi^{(2)}_{2} =v_2 -v_1$ and the next one gives $\dot{v}_2 =\beta (q_1 -q_2)$.
 As a result, on $S_3\subset M_1$, $S_3 = \{ x\in M_1 : \phi_2 =0,
 \phi^{(1)}_{2} =0,  \phi^{(2)}_{2} =0\}$, we have $\dot{f} =
 v_1\frac{\partial f}{\partial q_1}$ and $E\mid_{S_3} = \frac{1}{2}v_1^2$.
To work out the Hamiltonian scheme, note that
\begin{equation}
E\mid_{M_1} =\frac{1}{2}\left( p_1 -q_2\right)^2 - \frac{\beta}{2}\left(
q_1 - q_2\right)^2 +v_2\phi_2 = h+v_2\phi_2 .
\end{equation}
The equation (145) then reads
\begin{equation}
\dot{f} = \left\{ h, f\right\} +v_2\left\{ \phi_2 , f\right\} +
\dot{v}_2\frac{\partial f}{\partial v_2}.
\end{equation}
Again, $\dot{\phi}^{(1)}_{2} = v_2 -v_1 = \phi^{(2)}_{2}$ and
$\dot{\phi}^{(2)}_{2} =\dot{v}_2 =0$. Since $v_2 =-\{h,\phi^{(1)}_{2}\}$,
we can write (149) as
\begin{equation}
\dot{f} = \left\{ h, f\right\} - \left\{ h, \phi^{(1)}_{2}\right\}\left\{
\phi_2, f\right\} + \left\{ h, \phi_2\right\}\left\{ \phi^{(1)}_{2},
f\right\}
\end{equation}
(we have added the last term "by hand" to antisymmetrize the bracket,
since $\{ h, \phi_2 \}$ $\sim$ $\phi^{(1)}_{2}=0$ on the surface of the
second class constraints. The canonical transformation of the form $P_1 =
p_1+p_2-q_1-q_2$, $Q_1 =q_1$, $P_2 =p_2 -q_1 =\phi_2$, $Q_2 =q_2 -q_1
=\phi^{(1)}_{2}$ allows to represent the dynamics on $S_3 = \{ P_2 =0,
Q_2 =0, \phi^{(2)}_{2} =0 \}$ in a simple way, $\dot{f} =\{ h, f\}_{P_1,
Q_1}$, where $h=P_1^2/2$.

\subsection*{ACKNOWLEDGMENTS}
A.S. gratefully acknowledges support from Russian Foundation of Fundamental Research.

\subsection*{APPENDIX: THE GLOSSARY}

For convenience of the reader we give some notations and definitions of
geometric objects used throughout the paper. More detailed information
can be found in Ref. \cite{Marmo, Morandi}.

 \begin{enumerate}
  \item $M$ (or $Q$) denotes finite-dimensional real smooth manifold,
$(TM, \pi_1 , M)$ --  its tangent bundle with local coordinates $\{ q^i,
v^i\}$,  $(T^{*}M, \pi_2 , M)$ -- its cotangent bundle with
local coordinates $\{ q^i, p_i\}$, $i=1,\dots N$ =dim $M$.
\item ${\cal X}$(M) denotes the Lie algebra of smooth vector fields on
$M$. Locally  $X\in {\cal X}(M)$  is $X=X^i(q)\frac{\partial}{\partial
q^i}$, $X^i(q) \in C^{\infty}(M)$.
 \item Correspondingly, $Y\in  {\cal X}(TM)$ in local coordinates
reads $Y=Y^i(q,v)\frac{\partial}{\partial
q^i}+V^i(q,v)\frac{\partial}{\partial v^i}$, where $Y^i$ and $V^i$ are
smooth functions on $TM$.
\item ${\cal X}^V(TM)$=$\{ X\in {\cal
X}(TM):\pi_* X = 0 \}$ is the algebra of vertical vector fields on $TM$.
Any $X\in {\cal X}^V(TM)$ can be locally represented as
$X=X^i(q,v)\frac{\partial }{\partial v^i}$.
 \item  Lift $\gamma$: ${\cal X}(M)\rightarrow {\cal X}^V(TM)$
maps      $X=X^i(q)\frac{\partial}{\partial q^i}$ $\in {\cal X}(M)$ to
$X^V = X^i(q)\frac{\partial}{\partial
v^i}\in {\cal X}^V(TM)$, where $X^V$ is the generator of
the one-parameter group of the fiber diffeomorphisms  $t\rightarrow ( q^i,
v^i+X^i(q)t)$.
 \item The canonical (1,1) tensor $\tau$ on $TM$ is defined as
the composition  $\tau = \gamma\circ
\pi_*$. In coordinates we have $\tau =
dq^i\otimes\frac{\partial}{\partial v^i}$.
\item Cartan -- Liouville one-form is defined for  $f\in C^{\infty}(TM)$
by
 $\theta_f = df\circ\gamma$. If $\Gamma_{ij}=\frac{\partial^2 f}{\partial
v^i\partial v^j}$ has  maximal rank, $\omega_f =
d\theta_f$  defines the symplectic structure on $TM$.
\item    Liouville vector field $\triangle \in {\cal X}^V(TM)$ is the
generator of the one-parameter group of dilations $(q,v) \rightarrow (q,
e^tv)$. Locally,  $\triangle = v^i \frac{\partial}{\partial
v^i}$.
 \item   Whitney sum  $E_1\bigoplus E_2$ of two bundles $E_1$ and $E_2$
with the same base $B$ is the bundle with the base $B$ and the fiber
 $E_{1x}\bigoplus E_{2x}$   $\forall x \in B$.

\end{enumerate}

config: arxiv


TITLE: Does reheating occur when inflation ends or when the field reaches minimum?
QUESTION [0 upvotes]: Link for the picture

The inflaton field can be represented as a ball rolling down a hill. During inflation, the energy density is approximately constant, driving the tremendous expansion of the universe. When the ball starts to oscillate around the bottom of the hill, inflation ends and the inflaton energy decays into particles. In certain cases, the coherent oscillations of the inflaton could generate a resonant production of particles which soon thermalize, reheating the universe.

When does reheating occur?
a) after inflation ends ; before the field reaches minimum
b) after the field reaches minimum of the potential
c) when inflaton oscillates around the minimum of the potential

REPLY [1 votes]: This link may be useful for you because it shows how model dependent the term reheating and its behavior is.

who suggest that the decay can undergo broad parametric resonance, with extremely efficient transfer of energy from the coherent oscillations of the inflaton field. This initial transfer has been dubbed preheating. With such an efficient start to the reheating process, it now appears possible that the reheating epoch may be very short indeed and hence that most of the energy density in the inflaton field at the end of inflation may be available for conversion into thermalized form.

Reheating happens at the end of inflation, as far as this discussion goes, it signal's its end.
Thus the simplistic plot you show is an analogue, not the true story, which needs a deeper understanding of the particular models, referenced in the link. Also this link goes into preheating and reheating.


TITLE: Number of ordered choices of $n$ balls, using exponential generating functions
QUESTION [0 upvotes]: I want to understand how I would go about solving counting problems like this

Use exponential generating functions to determine the number $a_n$ of ordered choices of $n$ balls such that there are 2 or 4 red balls, an even number of green balls, and an arbitrary number of blue balls

I understand how I would start solving this type of problem, getting a multiplication that starts like:
$(\dfrac{x^2}{2!} + \dfrac{x^4}{4!})(\dfrac{1}{2}(e^x + e^{-x}))(e^x)$
corresponding to the red, green, blue balls respectively.
I am aware of an existing  question asking precisely this. However the answer there does not make sense to me question is here. Precisely speaking, I cannot understand how the multiplication of the generating functions is expanded in the first expansion in the answer there. Also please note that I cannot comment in the answer to the original question there, since I am a beginner and my reputation doesn't allow me to do that :)

REPLY [0 votes]: If I understand your question correctly, you want to see the intermediate steps to show that if
$$f(x)={\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}{\frac12\left(e^x+e^{-x}\right)}{e^x}$$
then
$$f(x)=\frac12\cdot\frac{x^2}{2!} + \frac12\cdot\frac{x^4}{4!} + \sum_{n=2}^\infty \frac1{16}2^nn(n-1)\frac{x^n}{n!} + \\\sum_{n=4}^\infty\frac1{768}2^nn(n-1)(n-2)(n-3)\frac{x^n}{n!}. \tag{1}$$
OK, here goes:
$$\begin{align}
f(x)&={\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}{\frac12\left(e^x+e^{-x}\right)}{e^x} \\
&={\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)} \frac{1}{2}(e^{2x}+1) \\
&=\frac{1}{2}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)} + \frac{1}{2}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}e^{2x} \\
&=\frac{1}{2}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)} + \frac{1}{2}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)} \sum_{n=0}^{\infty} \frac{1}{n!} 2^n x^n \tag{2} \\
&=\frac{1}{2}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)} + \frac{1}{2} \cdot \frac{x^2}{2!} \sum_{n=0}^{\infty} \frac{1}{n!} 2^n x^n + \frac{1}{2} \cdot \frac{x^4}{4!} \sum_{n=0}^{\infty} \frac{1}{n!} 2^n x^n \\
&=\frac{1}{2} \cdot \frac{x^2}{2!}+\frac{1}{2} \cdot \frac{x^4}{4!} + g(x) + h(x) \tag{3}
\end{align}$$
where for the sake of keeping our equations short we define
$$g(x) = \frac{1}{2} \cdot \frac{x^2}{2!} \sum_{n=0}^{\infty} \frac{1}{n!} 2^n x^n$$
and
$$h(x) = \frac{1}{2} \cdot \frac{x^4}{4!} \sum_{n=0}^{\infty} \frac{1}{n!} 2^n x^n$$
Then
$$\begin{align}
g(x) &= \frac{1}{2} \cdot \frac{x^2}{2!} \sum_{n=0}^{\infty} \frac{1}{n!} 2^n x^n \\
&= \frac{1}{4} \sum_{n=0}^{\infty} \frac{1}{n!} 2^n x^{n+2}\\
&= \frac{1}{4} \sum_{i=2}^{\infty} \frac{1}{(i-2)!} 2^{i-2} x^i \tag{4}\\
&= \frac{1}{4} \sum_{i=2}^{\infty} \frac{1}{(i-2)!} \cdot \frac{i(i-1)}{i(i-1)} \frac{2^i}{4} x^i\\
&= \frac{1}{16} \sum_{i=2}^{\infty} \frac{i(i-1)}{i!} 2^i x^i \tag{5}
\end{align}$$
and
$$\begin{align}
h(x) &= \frac{1}{2} \cdot \frac{x^4}{4!} \sum_{n=0}^{\infty} \frac{1}{n!} 2^n x^n\\
&= \frac{1}{48} \sum_{n=0}^{\infty} \frac{1}{n!} 2^n x^{n+4}\\
&= \frac{1}{48} \sum_{j=4}^{\infty} \frac{1}{(j-4)!} 2^{j-4} x^j \tag{6}\\
&= \frac{1}{48} \sum_{j=4}^{\infty} \frac{1}{(j-4)!} \cdot \frac{j(j-1)(j-2)(j-3)}{j(j-1)(j-2)(j-3)} \frac{2^j}{16} x^j\\
&= \frac{1}{768} \sum_{j=4}^{\infty} \frac{j(j-1)(j-2)(j-3)}{j!} 2^j x^j \tag{7}
\end{align}$$
Now substitute $(5)$ and $(7)$ into equation $(3)$, change $i$ and $j$ to $n$, and we have the desired result $(1)$.
Notes:

$(2)$ by use of the infinite series for $e^x$
$(4)$ shifting the index in the summation by letting $i = n+2$
$(6)$ shifting the index in the summation by letting $j = n+4$


TITLE: Complex scalar theory: annihilation and creation operators give wrong commutators with Hamiltonian
QUESTION [2 upvotes]: The theory of a real (hermitian) scalar field can be found in many books and everywhere online. On the other hand, if we take the field non-hermitian, then I can only find notes on path integrals. I can't find anything about canonical commutations, so I tried to derive it myself. The thing is, I find a wrong commutator with the Hamiltonian, and I can't spot my mistake. I'd find it really helpful if somebody has anything to say about this.
I will probably miss some numerical factors and I might misplace some daggers/signs along this post, but these issues are not important right now. I apologise if there are some minor mistakes, please don't mind them. On the other hand, if there is any major mistake, please mind it and tell me :)
First step: the fields are $\phi(x)$, $\phi^\dagger(x)$, $\pi(x)$ and $\pi^\dagger(x)$. These commute as follow
$$
[\phi,\phi^\dagger]=[\pi,\pi^\dagger]=[\phi,\pi^\dagger]=0\\
[\phi,\pi]=[\phi^\dagger,\pi^\dagger]=i\delta(\boldsymbol x-\boldsymbol y)
$$
(we obviously take the fields at equal times)
Second step: from KG equation, we solve for $\phi$
$$
\phi(x)=\int\frac{\mathrm d\boldsymbol k}{2\omega(\boldsymbol k)}\ \mathrm e^{-ikx}a(\boldsymbol k)+\mathrm e^{+ikx}b^\dagger(\boldsymbol k)
$$
and similar equations for the other fields. Inverting these, we find
$$
a(\boldsymbol k)=\int \mathrm d\boldsymbol x\ \mathrm e^{ikx} \left[\omega(\boldsymbol k)\phi(x)+i\pi^\dagger(x)\right]
$$
and similar equations for $b(\boldsymbol k)$.
Third step: write the hamiltonian as
$$
H=\int\frac{\mathrm d \boldsymbol k}{2\omega(\boldsymbol k)} \frac{1}{2}\omega(\boldsymbol k)\left[ a^\dagger(\boldsymbol k) a(\boldsymbol k)+b^\dagger(\boldsymbol k) b(\boldsymbol k)\right]
$$
The problem comes with the next (and final) step: if I work out the commutator of $a$ and $b$ with $H$, I get an unexpected result. For example,
$$
[H,a(\boldsymbol k)]=-\frac{1}{2}\omega(\boldsymbol k)\left[a(\boldsymbol k)+b(\boldsymbol k)\right]
$$
(I was expecting something like $[H,a]=-\omega a$; note that this would we had if $a=b$, i.e., the field was hermitian)
From this, we can see that $a$ can not be used as an annihilation operator, because if $|E\rangle$ is an state with energy $E$, then $a(\boldsymbol k)|E\rangle$ will not be another eigenstate with energy $E-\omega(\boldsymbol k)$; to see this, note that
$$
Ha(\boldsymbol k)|E\rangle=\big(a(\boldsymbol k)H+[H,a(\boldsymbol k)]\big)|E\rangle=\big(a(\boldsymbol k)E-\frac{1}{2}\omega(\boldsymbol k)\left[a(\boldsymbol k)+b(\boldsymbol k)\right]\big)|E\rangle
$$
If the field $\phi$ was hermitian, then $a=b$ so that the last equality would read $Ha|E\rangle=(E-\omega)a|E\rangle$, so $a$ would be an annihilation operator. On the other hand, if $a\neq b$, then nor $a$ nor $b$ decrese the energy of eigenstates. Where did I go wrong?

REPLY [1 votes]: The correct commutation relations for creation/annihilation operators are the following:
$$ [a(k), a^{\dagger} (k')] = (2\pi)^3 \delta^{(3)} (k - k') $$
$$ [b(k), b^{\dagger} (k')] = (2\pi)^3 \delta^{(3)} (k - k') $$
Everything else commutes, including $[a, b^{\dagger}]$.
From this, it is clear that $[H, a]$ can not depend on $b$.


TITLE: Is it safe to say that $Z_p$ is always a finite group? And that the order of $Z_p$ is p?
QUESTION [2 upvotes]: I know that seems a dumb question, but I want to make sure.

REPLY [1 votes]: I am writing here maybe it completes Tarnation's answer.

Definition: Let $n$ is a fixed positive integer number and $h, k$ are two arbitrary integers so the number $r$: $$h+k=nq+r, 0\leq r\leq n$$ is called the sum of $h$ and $h$ modulo $n$.
  Now consider the set: $$\{1,2,3,...,n-1\}$$ then:
Theorem: the above set forms a cyclic group, called $\mathbb Z_n$, under addition modulo $n$.

Of course if $h+k=r$ happens in group $\mathbb Z_n$ then for common addition in $\mathbb Z$ we see $h+k\equiv r$ (mod $n$). The order of set $\mathbb Z_n$ is finite so it is a finite group of order $n$. In you problem if $n=p$ a prime number, as Tarnatian's second comment, all elements of $\mathbb Z_p$ could generate the whole group.


TITLE: Inequality involving perimeter and area
QUESTION [3 upvotes]: I am studying an article: The parametric problem of capillarity: the case of two and three fluids, by U. Massari. In one of his proofs, he uses an inequality I can't manage to prove. It is like this:

Let $\Omega \subset \Bbb{R}^n$ be an open, bounded set, with Lipschitz boundary with constant $L$. The following inequality holds:
  $$(1) \ \ \int_{\partial \Omega} \chi_E d\mathcal{H}^{n-1} \leq \sqrt{1+L^2}\int_{\Omega_\varepsilon}| \nabla \chi_E|+c \int_{\Omega_\varepsilon} \chi_E dV $$
  where $E$ is a measurable set of finite perimeter (i.e. $\chi_E \in BV(\Omega)$), and the integral on $\partial \Omega$ is in fact the integral of the trace of $\chi_E$ on the boundary of $\Omega$.

where $\Omega_\varepsilon = \lbrace x \in \Omega : d(x,\partial \Omega) <\varepsilon \rbrace$
The previous inequality is stated without proof or reference in the article, but there is another similar inequality, with a reference to a proof:

Let $\Omega \subset \Bbb{R}^n$ be an open, bounded set, with the property that there exists $\rho>0$ such that for every $x \in \Omega$ there is a ball $B_\rho$ of radius $\rho$ (not necessarily centered in $x$) with $x \in B_\rho \subset \Omega$. The following inequality holds:
  $$ (2) \ \ \int_{\partial \Omega} \chi_E d\mathcal{H}^{n-1} \leq \int_{\Omega_\varepsilon}| \nabla \chi_E|+c \int_{\Omega_\varepsilon} \chi_E dV $$
  where $E$ is a measurable set of finite perimeter (i.e. $\chi_E \in BV(\Omega)$), and the integral on $\partial \Omega$ is in fact the integral of the trace of $\chi_E$ on the boundary of $\Omega$. The constant $c$ depends on $\varepsilon, \rho, \Omega$ and $n$.

This inequality is proved in I. Tamanini: Il problema della capillarita su domini non regolari. The hypothesys with the interior spheres of radius $\rho$ is used heavily in the proof. First $\Omega$ is written as a countable union of balls of radius $\rho$.
The question is: do you know an article which proves the inequality $(1)$? If not, it is possible to deduce $(1)$ using $(2)$? Thank you.

REPLY [4 votes]: I have found an article which deals with this kind of inequalities. It is available in the following link:  Funzioni BV e Tracce


TITLE: Proof of the Markov Property when Taking "Two steps at a time".
QUESTION [1 upvotes]: I have seen the solution, but I would like to understand how it actually works.
A new variable is defined, $Z_n = (X_n, X_{n-1})$, this is kind of like the join probability of 2 consecutive steps; the smallest value of n in this case is 1. The proof proceeds as follows:

$P(Z_n = z_n|Z_{n-1}, Z_{n-2}...Z_1 = z_1)$
The above is equal to : $P(X_n = x_n, X_{n-1} = x_{n-1}|X_{n-1} = x_{n-1}, X_{n-2} = x_{n-2}...X_1 = x_1)$
Which is equal to: $P(X_n = x_n|X_{n-1} = x_{n-1})$
Which finally, is equal to : $P(Z_n=z_n|Z_{n-1} = z_{n-1})$

I'm struggling to understand how the RHS of the 2nd step came about, and also why the 3rd step automatically implies the last step is true.
Your help would be appreciated greatly, thanks in advance !

REPLY [1 votes]: You can use the rule $P(AB|C) = P(A|BC)P(B|C)$ on step 2, where $A$ is the event $X_n=x_n$, $B$ is the event $X_{n-1}=x_{n-1}$, and $C$ is the event $X_{n-1}=x_{n-1}, \; ... \; X_1=x_1$. By the Markov property, you can see that $P(A|BC)$ will simply reduce to $P(A|B)$. Given the definition of $B$ and $C$, it is trivial to see why $P(B|C) = 1$. This will lead you from step 2 to step 3.
Let's expand step 4:
$P(Z_n=z_n|Z_{n-1}=z_{n-1})$
$=P(X_n=x_n, X_{n-1}=x_{n-1}|X_{n-1}=x_{n-1},X_{n-2}=x_{n-2})$
$=P(X_n=x_n|X_{n-1}=x_{n-1}, X_{n-1}=x_{n-1},X_{n-2}=x_{n-2})$ (using $P(AB|C) = P(A|BC)P(B|C)$)
$=P(X_n=x_n|X_{n-1}=x_{n-1},X_{n-2}=x_{n-2})$ (removing redundant term)
$=P(X_n=x_n|X_{n-1}=x_{n-1})$ (Markov property)


TITLE: How to find the inverse system of a given one
QUESTION [0 upvotes]: what is the inverse formula of y[n]=x[n]*x[n+1] ?
And how can I find the inverse formula/system of a given one in general? I'm having some troubles with this when it comes to some formulas.

REPLY [0 votes]: What does it mean to find an inverse? Here we have a function, $y$, which maps from $n \rightarrow x[n]x[n+1]$. We would like to find $y^{-1}$, such that $y^{-1}[y[n]] = n$.
However, without knowing the sequence of $x$ values, there's no telling that this inverse exists. Suppose $x[i] = 0$ for all $i$. then $y[i] = 0$ for all $i$.
We would say that $y$ is not one-to-one. Meaning multiple inputs go to the same output. This means $y$ is not invertible. If we found an inverse, there would be no way to tell whether $0 = y[1]$ or $0 = y[2]$ or $0 = y[14]$, and so on. This is a more general form of the vertical line test from high-school.
In general, you are looking for a function which maps back from the outputs of your function to the inputs. Here, this is not possible for all sequences $x[1], x[2], ...$.
Hope that helps!


TITLE: Cardinality of collection of subfields of $\mathbb C$
QUESTION [1 upvotes]: The question is just curiosity on my part. The title says it all. I can see that the cardinality is at least $\aleph_1$ (take simple extensions by an uncountable family of transcendental numbers). But is the cardinality the same as that of all subsets of $\mathbb C$? As a proof strategy, I'd wonder about finding an uncountable set such that any subset is "mutually transcendental", but this sounds like a lot to ask.

REPLY [3 votes]: Obviously, the set $\{ \mbox{subfields of } \Bbb{C} \}$ has cardinality almost  $2^{\mathfrak{c}}$.
If you fix a transcendence basis $B$ of the extension $\Bbb{C} / \Bbb{Q}$, necessarily $|B| = \mathfrak{c}$.
Now, $$\{ \Bbb{Q}(S): S \in \mathcal{P}(B)\}$$
is a set of distinct subfields of $\Bbb{C}$ whose cardinality is exactly $2^{\mathfrak{c}}$. It follows that the cardinality of the set $\{ \mbox{subfields of } \Bbb{C} \}$ is exactly $2^{\mathfrak{c}}$.


TITLE: Second order differential equation check. Find general solution
QUESTION [1 upvotes]: I could use a process check. In general since the solution is $$y = y_c + y_p$$ where $y_c$ is the complementary solution and $y_p$ is the particular, I can break it down.
The equation in question is this:
$$y" - 3y' = sin2x$$
Let's find the complementary solution first:
$$y_c = c_1e^{r_1x} + c_2e^{r_2x}$$
so let's solve the auxiliary equation:
$$r^2 - 3r = 0$$
$$r(r - 3) = 0$$
so the roots are $r = {0, 3}$
so the complementary equation is:
$$y = c_1 + c_2e^{3x}$$
Let's find the particular solution second:
so the form of the particular since it's a trig function on the right side is:
$$y_p = A\cos{kx} + B\sin{kx}$$
where k is the coefficient for the trig function radian measure.
$$y'p = -2A\sin{2x} + 2B\cos{2x}$$
$$y''p = -4A\cos{2x} - 4B\sin{2x}$$
so substituting:
$$(-4A\cos2x - 4B\sin2x) - 3(-2A\sin2x + 2B\cos2x) = \sin{2x}$$
$$(-4B + 6A)sin2x + (-4A - 6B)cos2x = sin2x$$
so matching coefficients
$(-4B + 6A) = 1$ and $-6B - 4A = 0$ and $-6B = 4A$ so $B = \frac{-2}{3}A$
solving further:
$\frac{8}{3}A + 6A = 1$ so $\frac{26}{3}A = 1$ so $A = \frac{3}{26}$ and so $B = \frac{-12}{216}$
so then the general solution is:
$$y = c_1 + c_2e^{3x} + \frac{3}{26}cos2x - \frac{12}{216}sin2x$$
Is my process right? Did I make an error?

REPLY [1 votes]: Your process is correct. Your  answer is wrong because you have a mistake in finding your constants $A$ and $B$.
The correct values for $A$ and $B$ are $A=\frac {3}{26}$ and $B=\frac {-1}{13}$
Fix this minor miscalculation and your solution is as good as gold.