TITLE: Show that $\langle y_i, y_j\rangle = 0 \forall i \neq j.$
QUESTION [1 upvotes]: Let $y_1, y_2, . . .$ be a sequence in a Hilbert space.
Let $V_n$ be the linear span
of $\{y_1, y_2, . . . , y_n\}$.
Assume that for $n \ge 1, ∥y_{n+1}∥ \le ∥y − y_{n+1}∥$ for each
$y \in V_n$. Show that $\langle y_i, y_j\rangle  = 0 \forall i \neq j.$oblem
Please give some hints on how to start the problem.I feel totally clueless

REPLY [1 votes]: W.l.o.g I will assume $\|y_i\|=1$ for any $i$.
By induction, we only have to show that $\langle y_{n+1}, y_i \rangle = 0$ for any $i = 1, \dotsc, n$.
Consider $y = \sum_{i=1}^n \langle y_{n+1},y_i\rangle y_i$ and $z = y_{n+1}-y$. By assumption we have $\| y_{n+1} \|^2 \leq \|z\|^2$.
Computing $\|z\|^2=\langle z,z \rangle$ we get
$$\langle z,z \rangle = \|y_{n+1}\|^2-\sum_{i=1}^n \langle y_{n+1},y_i \rangle^2,$$
hence the inequality can only hold if $\langle y_{n+1},y_i \rangle=0$ for any $1 \leq i \leq n$.
This has the following geometric interpretation:
$\inf_{y \in V_n }\|y-y_{n+1}\|$ is the smallest distance from $y_{n+1}$ to $V_n$. The infimum is well known to be taken on, when $y$ is the orthogonal projection of $v_{n+1}$ on $V_n$, i.e. the assumption of your exercise states that the orthogonal projection of $y_{n+1}$ onto $V_n$ is zero, which means that $v_{n+1}$ is orthogonal to $V_n$.


TITLE: Roots of trigonometric function.
QUESTION [2 upvotes]: I have a function:
$ f(x) = (ax+b) \cdot \sin(x) + (cx+d) \cdot \cos(x) + e$
for which I want to determine the roots.
I know that for $ax \cdot \sin(x) + cx \cdot \cos(x)$
the roots are $2(\tan^{-1}(\frac{a \pm \sqrt{a^2+c^2}}{c})+\pi n)$
and for $b\cdot \sin(x) + d \cdot \cos(x)$
the roots are $2(\tan^{-1}(\frac{b\pm\sqrt{b^2+d^2}}{d})+\pi n)$
but I have some trouble in finding the roots of the sum of both. Can anybody explain how to find it?

REPLY [0 votes]: As already answered by Julián Aguirre, equations which contains polynomial and trigonometric functions do not show explicit solutions (this is already the case for $x=\cos(x)$) and only numerical methods could find the roots (one at the time because there is an infinite number of solutions).
If you located one root, probably the simplest would be Newton method which, starting from a "reasonable" guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
Let us consider the case of the equation you gave and use $a=1,b=2,c=3,d=4,e=5$ and we look for the root located close to $x_0=15$. The iterates will be $$x_1=14.5438284668645$$ $$x_2=14.5706131998069$$ $$x_3=14.5705976124983$$ $$x_4=14.5705976124952$$ which is the solution for fifteen significant figures.


TITLE: Show $S_N = \sum\limits_{n=1}^{N} \text{sign}(Y-X_n)$ is Markov, $(X_n),Y $ iid Uniform(0,1)
QUESTION [3 upvotes]: Let $(X_n)$ and Y be i.i.d. Uniform$(0,1)$ random variables and let
$$S_N = \sum\limits_{n=1}^{N} \text{sign}(Y-X_n)$$
Show that $S_n$ is a Markov Chain and find its transition probabilities.
Any help with this one?
Its clear that $\mathbb{P}(S_N = x\,\,\vert\,\, S_{n-1},...S_0) = \mathbb{P}([(S_N = x)\cap S_{N-1} = x-1]\cup[(S_N = x)\cap S_{N-1} = x+1]\,\,\vert\,\, S_{n-1},...S_0) = \mathbb{P}([(S_N = x)\cap S_{N-1} = x-1]\cup[(S_N = x)\cap S_{N-1} = x+1]\,\,\vert\,\, S_{N-1})$
How would you find the transitional probabilities?

REPLY [0 votes]: Notice that $u=(S_n+n)/2$ is $Beta-Binomial(n,1,1)=U[0,n]$. Hence $$Y|S_n\sim Beta(1+\frac{n+S_n}2,1+\frac{n-S_n}2)=Beta(1+u,1+d)$$ with average $\hat y_n=\frac {1+u}{2+u+d}=\frac{1+u}{n+2}$ and hence $$S_{n+1}-S_n|S_n\sim B(P((Y|S_n)>X_{n+1}))= B(\hat y_n)$$
where $B(p)$ is Bernoulli on $\{-1,1\}$ with $P(B(p)=1)=p$.


TITLE: Why is $ \|T(x) \|\le \| T\| \|x \|$?
QUESTION [3 upvotes]: Let $X$ and $Y$ be normed linear spaces and let $T$ be a bounded linear operator from $X$ to $Y$. The norm of $T$ is defined as
$$\|T \|=\sup\{\|T(x) \|\;:\;\|x \|\le 1\}. $$
From the definition of the norm, we can say that $\|T(x) \|\le \|T \|\;\forall \;x,$ and that $\|T \|\|x \|\le \|T \|$ as $\|x\|\le 1$.
However I don't understand why we write $\|T(x)\|\le \|T\|\|x\|$?

REPLY [7 votes]: Clearly $\Vert T(x) \Vert \le \Vert T \Vert$ if $\Vert x \Vert = 1$.
Now, just remember that we're dealing with linear operators.
So if $\Vert x \Vert = c$, then $\Vert \frac{x}{c} \Vert = 1$.
So $$\left\Vert T(x) \right\Vert = c \left\Vert T\left(\frac{x}{c} \right) \right\Vert \le c \Vert T \Vert$$


TITLE: find the limit of $\lim_{x \to 0} \frac{1}{8}\sin(x)^{1/x^2}$
QUESTION [2 upvotes]: $$\lim_{x \to 0} \frac{\sin(x)}{8}^{\frac{1}{x^2}}$$

$y=\lim_{x \to 0}\frac{1}{x^2}\cdot \lim_{x \to 0} \ln(\frac{\sin(x)}{8})=-\infty$
So we have $e^{-\infty}=0$?

REPLY [0 votes]: The function is not defined for $x<0$ because of the problem of raising negative numbers to arbitrary powers. For $x>0,$ we have $0<\sin x < 1/2$ for $0< x< \pi/6.$ So for small $x>0,$
$$0<(\sin x)^{1/x^2} < (1/2)^{1/x^2}.$$
Because $0<1/2<1$ and $1/x^2 \to \infty,$ $ (1/2)^{1/x^2} \to 0$ as $x\to 0^+.$ Thus our limit (modified to require $x\to 0^+$) is $0$ by the squeeze theorem.


text:
\section{Gibbard's Collapse Result}\label{sec:Gibbard}

The Law of Import-Export is an important bridge between different types of conditionals: it permits to transform right-nested conditionals into simple ones. Import-Export is of specific interest in  suppositional accounts of indicative conditionals that assess the assertability of a conditional by the corresponding conditional probability \citep[as per Adams' thesis, viz.][]{Adams1965}. Import-Export is then an indispensable tool for providing a probabilistic analysis of embedded conditionals.
However, when Adams' Thesis, originally limited to conditionals with Boolean antecedent and consequent, is extended to nested conditionals, Import-Export creates unexpected problems.\footnote{The unrestricted version of Adams's equation is often called Stalnaker's Thesis \citep[going back to][]{Stalnaker1970} or simply ``The Equation'', with the latter name being prevalent in the psychological literature. Adams defends it in his \citeyear{Adams1975} monograph, too.} For example, a famous result by David \citet{lewis1976probabilities} shows that combining this latter equation with the usual laws of probability and an unrestricted application of Import-Export trivializes the probability of the indicative conditional.\footnote{On the reasons to defend Import-Export in relation to probabilities of conditionals, see \citet{mcgee1989conditional} and \cite{arlo2001bayesian}. A discussion of the links between Gibbardian collapse and Lewisian triviality lies beyond the scope of this paper, but we refer to \cite{lassiter2019trivalent} for a survey of Lewisian triviality results and their treatment in a trivalent framework.} Gibbard establishes a second difficulty with Import-Export, namely that any conditional satisfying Import-Export in combination with other intuitive principles collapses to the material conditional.

Gibbard's original proof \citep[][234--235]{gibbard1980two}---in reality more of an outline---was based on semantic considerations and left various assumptions implicit. Here we provide a more formal derivation. In particular, Gibbard only stressed conditions (i)--(iii) below, but implicitly assumed two further conditions, here highlighted as (iv) and (v), as well as structural constraints on the underlying consequence relation. In what follows we use $\models_{\sf CL}$ for classical consequence, and $\equiv_L$ for the conjunction of $\models_L$ and its converse. Under (v) we mean that $\supset$ obeys a classical law whenever it obeys a classical inference or a classical metainference.\footnote{An inference is a relation between (sets of) formulae: for instance the relation between $(A\supset B)\wedge A$ and $A\wedge B$ ; a metainference is a relation between inferences, for example the relation between $A\models B$ and $\models A\supset B$.}

\begin{thm}[Gibbard] Suppose $L$ is a logic whose consequence relation $\models_{L}$ is at least transitive, with $\supset$ and $\rightarrow$ two binary operators, obeying principles (i)-(v) for every formulae $A, B, C$. Then $\rightarrow$ and $\supset$ are provably equivalent in $L$.\\

\begin{tabular}{lll}

(i) & $A \rightarrow (B \rightarrow C)\equiv_{L} (A \wedge B) \rightarrow C$ & Import-Export \\

(ii) & $A \rightarrow B\models_{L} (A \supset B)$ & Stronger-than-Material\\

(iii) & If $A\models_{\sf CL} B$, then $\models_{L} A \rightarrow B$ & Supraclassicality\\

(iv) & If $A\equiv_{L} A'$ then $A\rightarrow B\equiv_{L} A'\rightarrow B$ & Left Logical Equivalence \\

(v) & $\supset$ obeys classical laws in $L$ & Classicality of $\supset$
\end{tabular}\label{T:Gibbard}
\end{thm}

\begin{proof}\

\begin{tabular}{lll}

1. & $(A\supset B) \rightarrow (A \rightarrow
B)\equiv_{L} ((A\supset B) \wedge A)\rightarrow B$ & by (i)\\

2. &  $((A\supset B) \wedge A)\equiv_{L} (A \wedge B)$ & by (v) (classical inferences)\\

3. & $(A\supset B) \wedge A)\rightarrow B\equiv_{L} (A\wedge B)
\rightarrow B$ & by 2 and (iv)\\

4. & $ (A\wedge B) \rightarrow B \equiv_{L} (A\supset B) \rightarrow (A \rightarrow
B)$ & 1, 3 and the transitivity of $\models_L$ \\

5 & $A \wedge B\models_{\sf CL} B$ &  Conjunction Elimination \\

6. & $\models_L (A\wedge B) \rightarrow B$ &  5 and (iii) \\

7. & $\models_L (A\supset B) \rightarrow (A \rightarrow B)$ & 4, 6, and the transitivity of $\models_L$ \\

8. & $(A\supset B) \rightarrow (A \rightarrow B)\models_L
(A\supset B) \supset (A \rightarrow B)$ & by (ii)  \\

9. & $\models_L (A\supset B) \supset (A \rightarrow B)$ & 7, 8 and the transitivity of $\models_L$\\

10. & $A\supset B \models_L  A \rightarrow B$ & by 9 and (v) (classical metainference)

\end{tabular}
\end{proof}
This is not the only proof of Gibbard's result. In particular \citet{Fitelson2013} and \citet{khoo2019triviality} give more parsimonious derivations. But it closely matches the structure of his original argument: first Gibbard shows that $(A\supset B) \rightarrow (A \rightarrow B)$ is a theorem of $L$ (step 1--7), from that he derives $\models_L (A\supset B) \supset (A \rightarrow B)$ (step 8--9) and finally, he infers $A\supset B \models_L  A \rightarrow B$ (step 10).

With Gibbard we can grant that the assumptions (ii) and (iii) introduced alongside Import-Export are fairly weak. Stronger-than-Material is shared by all theories that classify an indicative conditional with true antecedent and false consequent as false.\footnote{The name MP is sometimes used for this principle, see \citet{unterhuber2014completeness}, or \citet{khoo2019triviality} who call it Modus Ponens.

We find more appropriate to use `Stronger-than-Material' since Modus Ponens is strictly speaking a two-premise argument form. The two principles are not necessarily equivalent: in the system {\sf DF/TT} for instance, Stronger-than-Material holds but not Modus Ponens (in the form $A\to B, A\models B$).
} Supraclassicality, a restricted version of the principle of Conditional Introduction, means that deductive relations are supported by the corresponding conditional. Even that could be weakened by just assuming the conditional to support conjunction elimination as in step 6. In section \ref{sec:FitRes} we discuss more general conditions for Gibbardian collapse proposed by Branden \citet{Fitelson2013}.

Assumptions (iv) and (v), on the other hand, are stronger than meets the eye. While the substitution rule LLE was taken for granted by Gibbard, likely on grounds of compositionality, it raises issues in relation to counterpossibles and other forms of hyperintensionality \citep[see][]{nute1980topics,fine2012counterfactuals}. However, even if one is inclined to give up principle (iv), one may not find fault with applying it in this particular case. Similarly, (v) implies that the material conditional supports classical absorption laws (step 2 of the proof) and (meta-inferential) Modus Ponens (step 10) in $L$ --- two properties not necessarily retained in non-classical logics.

Gibbard's result also leaves a number of questions unanswered. One of them concerns the implication of the mutual entailment between $\rightarrow$ and $\supset$. Does the collapse imply that the two conditionals can be replaced by one another in all contexts, for example? The answer to this question is in fact negative, as we proceed to show using trivalent logic in the next section.

config: arxiv


TITLE: Interval of convergence question
QUESTION [1 upvotes]: $$f\left(x\right) = \frac{3}{2x-1},\; c = 2$$
The power series is centered at 2, determine the interval of convergence.
I get (same with answer sheet) this geometric series: $$f\left(x\right) = \sum \limits_{n=0}^{\infty} \left[ \frac{-2}{3}\left(x-2\right)\right]^n$$
I have $r = \frac{-2}{3}\left(x-2\right)$ but the answer sheet has $r = x -2$.
Did I miss something?
I get (1/2, 7/2) for the interval.

REPLY [1 votes]: I think I've at last understood your approach.
Now you have the expression of $r=-\frac{2}{3}(x-2)$. You will use the series $\frac{1}{1-r}=\sum_{n=0}^\infty r^n$, so you must look for $x$ values that make $|r|<1$, so that this series converges.
$$|-\frac{2}{3}(x-2)|<1$$
$$\frac{2}{3}|x-2|<1$$
$$|x-2|<\frac{3}{2}$$
$$-\frac{3}{2}<x-2<\frac{3}{2}$$
$$2-\frac{3}{2}<x<\frac{3}{2}+2$$
$$\frac{1}{2}<x<\frac{7}{2}$$
So your series converges for $x \in ]\frac{1}{2},\frac{7}{2}[$. Is that the kind of thing you are looking for?


TITLE: How do I find $\cot[\sin^{-1}(-\sqrt2/3)]$?
QUESTION [1 upvotes]: My first problem is finding $\sin^{-1}(-\sqrt2/3)$, since $-\sqrt2/3$ isn't in the unit circle. Hence, I'm not sure how to find the angle.
Then once I get the angle, how do I find $\cot(x)$?

REPLY [4 votes]: Hint: $$\cot(\sin^{-1} (x)) = \frac{\cos(\sin^{-1} (x))}{\sin(\sin^{-1} (x))}=\frac{\sqrt{1-x^2}}{x}$$


TITLE: Separation of Variables for PDE - wave equation
QUESTION [0 upvotes]: I've been posed the following question:

I'm trying to solve for the solution to the partial differential equation but cannot seem to get it right.
the question specifies using the "separation of variables method" to solve it
My current attempt includes:
$\theta (x,t) = \sum_{n=1}^{\infty} u_n (x,t)$
$ = \sum_{n=1}^{\infty} (B_n cos \lambda _n t + C_n sin \lambda t) sin (\frac {n \pi x}{2 \pi})$
I'm not quite sure if this is the correct approach or how to solve for the coefficients.
Any help with this question will be greatly appreciated!

REPLY [1 votes]: This equation can be solved analytically using for instance, the Laplace transform. You can use it as a means to certify your variable separation method.
In large steps, applying the Laplace transform
$$
s^2\Theta(x,s)-\theta(x,0)-s\theta_t(x,0) = \Theta_{xx}(x,s)
$$
or
$$
s^2\Theta(x,s)-6\cos(3x)-s(5+4\cos(4x)) = \Theta_{xx}(x,s)
$$
Solving for $x$
$$
\Theta(x,t) = c_1(s) e^{\frac{s x}{3}}+c_2(s) e^{-\frac{s x}{3}}+\frac{4 s^4 \cos (4 x)+5 s^4+6 s^3 \cos (3 x)+324 s^2 \cos (4 x)+1125
   s^2+864 s \cos (3 x)+58320}{s \left(s^4+225 s^2+11664\right)}
$$
and now we have
$$
\Theta_x(0,s) = \Theta_x(2\pi,s)
$$
giving
$$
c_1(s)-c_2(s)=e^{\frac{2 \pi  s}{3}} c_1(s)-e^{-\frac{2 \pi  s}{3}}
   c_2(s)
$$
or
$$
c_1(s) = -e^{-\frac{2 \pi  s}{3}} c_2(s)
$$
and finally
$$
\Theta(x,s) = c_2(s) \left(e^{-\frac{s x}{3}}-e^{\frac{1}{3} s (x-2 \pi )}\right)+\frac{6 \cos (3 x)}{s^2+81}+\frac{4 s \cos (4
   x)}{s^2+144}+\frac{5}{s}
$$
Here $c_2(s)$ represents the Laplace transform of any $f(t)$. Assuming now $\theta(0,t) = \sin(t)$ we have $c_2(s) = \frac{1}{s^2+1}$ and then
$$
\Theta(x,s) = \frac{e^{-\frac{s x}{3}}-e^{\frac{1}{3} s (x-2 \pi )}}{s^2+1}+\frac{6 \cos (3 x)}{s^2+81}+\frac{4 s \cos (4
   x)}{s^2+144}+\frac{5}{s}
$$
and anti-transforming
$$
\theta(x,t) = u \left(t-\frac{x}{3}\right) \sin \left(t-\frac{x}{3}\right)-u \left(t+\frac{1}{3} (x-2 \pi )\right) \sin
   \left(t+\frac{1}{3} (x-2 \pi )\right)+4 \cos (12 t) \cos (4 x)+\frac{2}{3} \sin (9 t) \cos (3 x)+5
$$
where $u(\cdot)$ is the Heaviside step function.
For $f(t) = 0$ as requested, then
$$
\theta(x,t) = 4 \cos (12 t) \cos (4 x)+\frac{2}{3} \sin (9 t) \cos (3 x)+5
$$
NOTE
For the problem
$$
\theta_{tt}=c^2\theta_{xx},\ \ \theta(0,t) = \theta(L,t) = 0, \ \ \theta(x,0) = \phi(x), \theta_t(x,0) = \psi(x)
$$
the variables separation method says that
$$
\theta(x,t) = \sum_{n}^{\infty}\left(a_n\cos\left(\frac{n\pi c t}{L}\right)+b_n\sin\left(\frac{n\pi c t}{L}\right)\right)\sin\left(\frac{n\pi x}{L}\right)
$$
with
$$
a_n = \frac 2L\int_0^L\phi(x)\sin\left(\frac{n\pi x}{L}\right)dx
$$
and
$$
b_n = \frac {2}{n\pi c}\int_0^L\psi(x)\sin\left(\frac{n\pi x}{L}\right)dx
$$
Here $L = 2\pi, \ \ c = 3, \ \ \phi(x) = 6\cos(3x),\ \ \psi(x) = 5+4\cos(4x)$
giving
$$
a_n = \frac{12 n (1-\cos (\pi  n))}{\pi  \left(n^2-36\right)}\\
b_n = \frac{12 \left(n^2-20\right) (1-\cos (\pi  n))}{\pi  n^2 \left(n^2-36\right)}
$$


text:
\begin{document}

\author{Yulia Kempner $^{1}$\\$^{1}$ Holon Institute of Technology, ISRAEL\\yuliak@hit.ac.il
\and Vadim E. Levit $^{1,2}$\\$^{2}$ Ariel University Center of Samaria, ISRAEL\\levitv@ariel.ac.il}
\title{Geometry of antimatroidal point sets}
\date{}
\maketitle

\begin{abstract}
The notion of "antimatroid with repetition" was conceived by Bjorner, Lovasz
and Shor in 1991 as a multiset extension of the notion of antimatroid
\cite{ChipF}. When the underlying set consists of only two elements, such
two-dimensional antimatroids correspond to point sets in the plane. In this
research we concentrate on geometrical properties of antimatroidal point sets
in the plane and prove that these sets are exactly parallelogram polyominoes.
Our results imply that two-dimensional antimatroids have convex dimension $2$.
The second part of the research is devoted to geometrical properties of
three-dimensional antimatroids closed under intersection.

\bigskip\textbf{Keywords: } antimatroid, convex dimension, lattice animal, polyomino.

\end{abstract}

\section{Preliminaries}

An antimatroid is an accessible set system closed under union \cite{BZ}. An
algorithmic characterization of antimatroids based on the language definition
was introduced in \cite{BF}. Another algorithmic characterization of
antimatroids that depicted them as set systems was developed in \cite{KL}.
While classical examples of antimatroids connect them with posets, chordal
graphs, convex geometries, etc., game theory gives a framework in which
antimatroids are interpreted as permission structures for coalitions
\cite{BJ}. There are also rich connections between antimatroids and cluster
analysis \cite{Malta}. In mathematical psychology, antimatroids are used to
describe feasible states of knowledge of a human learner \cite{Eppstein}.

In this paper we investigate the correspondence between antimatroids and
polyominoes. In the digital plane ${\mathbb{Z}}^{2}$, \emph{a polyomino}
\cite{Golomb} is a finite connected union of unit squares without cut points.
If we replace each unit square of a polyomino by a vertex at its center, we
obtain an equivalent object named \emph{a lattice animal} \cite{LA}. Further,
we use the name polyomino for the two equivalent objects.

A polyomino is called column-convex (resp. row-convex) if all its columns
(resp. rows) are connected (in other words, each column/row has no holes). A
convex polyomino is both row- and column-convex. The parallelogram polyominoes
\cite{DeLungo,DV,DDD}, sometimes known as staircase polygons \cite{BM,NB}, are
a particular case of this family. They are defined by a pair of monotone paths
made only with north and east steps, such that the paths are disjoint, except
at their common ending points.

In this paper we prove that antimatroidal point sets in the plane and
parallelogram polyominoes are equivalent.

Let $E$ be a finite set. A \textit{set system }over $E$ is a pair
$(E,\mathcal{F})$, where $\mathcal{F}$ is a family of sets over $E$, called
\textit{feasible }sets. We will use $X\cup x$ for $X\cup\{x\}$, and $X-x$ for
$X-\{x\}$.

\begin{definition}
\cite{Greedoids}A finite non-empty set system $(E,\mathcal{F})$ is an
\textit{antimatroid }if

$(A1)$ for each non-empty $X\in\mathcal{F}$, there exists $x\in X$ such that
$X-x\in\mathcal{F}$

$(A2)$ for all $X,Y\in\mathcal{F}$, and $X\nsubseteq Y$, there exists $x\in
X-Y$ such that $Y\cup x\in\mathcal{F}$.
\end{definition}

\smallskip Any set system satisfying $(A1)$ is called \textit{accessible.}

In addition, we use the following characterization of antimatroids.

\begin{proposition}
\cite{Greedoids} For an accessible set system $(E,\mathcal{F})$ the following
statements are equivalent:

$(i)$ $(E,\mathcal{F})$ is an antimatroid

$(ii)$ $\mathcal{F}$ is closed under union ($X,Y\in\mathcal{F}\Rightarrow
X\cup Y\in\mathcal{F}$)
\end{proposition}

Consider another property of antimatroids.

\begin{definition}
A set system $(E,\mathcal{F})$ satisfies the \textit{chain property }if for
all $X,Y\in\mathcal{F}$, and $X\subset Y$, there exists a chain $X=X_{0}
\subset X_{1}\subset...\subset X_{k}=Y$ such that $X_{i}=X_{i-1}\cup
x_{i\text{ }}$and $X_{i}\in\mathcal{F}$ for $0\leq i\leq k$.
\end{definition}

It is easy to see that the chain property follows from $(A2)$, but these
properties are not equivalent.

A poly-antimatroid \cite{NM} is a generalization of the notion of the
antimatroid to multisets. \emph{A poly-antimatroid} is a finite non-empty
multiset system $(E,S)$ that satisfies the antimatroid properties $(A1)$ and
$(A2)$.

Let $E=\{x,y\}$. In this case each point $A=(x_{A},y_{A})$ in the digital
plane ${\mathbb{Z}}^{2}$ may be considered as a multiset $A$ over $E$, where
$x_{A}$ is a number of repetitions of an element $x$, and $y_{A}$ is a number
of repetitions of an element $y$ in multiset $A$. Consider a set of points in
the digital plane ${\mathbb{Z}}^{2}$ that satisfies the properties of an
antimatroid. That is a two-dimensional poly-antimatroid.

\begin{definition}
\label{Def} A set of points $S$ in the digital plane ${\mathbb{Z}}^{2}$ is an
antimatroidal point set if

$(A1)$ for every point $(x_{A},y_{A})\in S$, such that $(x_{A},y_{A}
)\neq(0,0)$,

either $(x_{A}-1,y_{A})\in S$ or $(x_{A},y_{A}-1)\in S$

$(A2)$ for all $A\nsubseteq B\in S$,

if $x_{A}\geq x_{B}$ and $y_{A}\geq y_{B}$ then either $(x_{B}+1,y_{B})\in S$
or $(x_{B},y_{B}+1)\in S$

if $x_{A}\leq x_{B}$ and $y_{A}\geq y_{B}$ then $(x_{B},y_{B}+1)\in S$

if $x_{A}\geq x_{B}$ and $y_{A}\leq y_{B}$ then $(x_{B}+1,y_{B})\in S$
\end{definition}

Accessibility implies that $\emptyset\in S$.

For example, see an antimatroidal point set in Figure \ref{int_fig1}.

\begin{figure}[ptbh]
\setlength{\unitlength}{1cm}
\par
\begin{picture}(8,4.5)(-3,-0.5)\thicklines
\put(0,0){\vector(1,0){7.5}} \put(0,0){\vector(0,1){3.5}}
\multiput(0,0)(0.5,0){5}{\circle*{0.15}}
\multiput(0.5,0.5)(0.5,0){4}{\circle*{0.15}}
\multiput(1,1)(0.5,0){5}{\circle*{0.15}}
\multiput(1,1.5)(0.5,0){5}{\circle*{0.15}}
\multiput(3,2)(0.5,0){6}{\circle*{0.15}}
\multiput(4,2.5)(0.5,0){5}{\circle*{0.15}}
\multiput(4.5,3)(0.5,0){4}{\circle*{0.15}}
\put(7.3,-0.3){\makebox(0,0){x}} \put(-0.2,3.3){\makebox(0,0){y}}
\put(-0.1,-0.3){\makebox(0,0){0}} \put(0.5,-0.3){\makebox(0,0){1}}
\put(1,-0.3){\makebox(0,0){2}} \put(-0.3,0.5){\makebox(0,0){1}}
\put(3.75,2.7){\makebox(0,0){$A$}}
\put(5.75,1.9){\makebox(0,0){$B$}}
\end{picture}
\caption{An antimatroidal point set.}
\label{int_fig1}
\end{figure}

A three-dimensional point set is defined similarly to a two-dimensional set.

\section{Two-dimensional antimatroidal point sets and polyominoes}

In this section we consider a geometric characterization of two-dimensional
antimatroidal point sets. The following notation \cite{Digital} is used. If
$A=(x,y)$ is a point in a digital plane, the \textit{4-neighborhood}
$N_{4}(x,y)$ is the set of points
\[
N_{4}(x,y)=\{(x-1,y),(x,y-1),(x+1,y),(x,y+1)\}
\]
and \textit{8-neighborhood} $N_{8}(x,y)$ is the set of points
\[
N_{8}(x,y)=\{(x-1,y),(x,y-1),(x+1,y),(x,y+1),(x-1,y-1),
\]
\[
(x-1,y+1),(x+1,y-1),(x+1,y+1)\}.
\]
Let $m$ be any of the numbers $4$ or $8$. A sequence $A_{0},A_{1},...,A_{n}$
is called an $N_{m}$-\textit{path }if $A_{i}\in N_{m}(A_{i-1})$ for each
$i=1,2,...n$. Any two points $A,B\in S$ are said to be $N_{m}$-connected in
$S$ if there exists an $N_{m}$-path $A=A_{0},A_{1},...,A_{n}=B$ from $A$ to
$B$ such that $A_{i}\in S$ for each $i=1,2,...n$.\ A digital set $S$ is an
$N_{m}$-\textit{connected} set if any two points $P$,$Q$ from $S$ are $N_{m}
$-connected in $S$. An $N_{m}$-\textit{connected component} of a set $S$ is a
maximal subset of $S$, which is $N_{m}$-connected.

An $N_{m}$-path $A=A_{0},A_{1},...,A_{n}=B$ from $A$ to $B$ is called
\textit{a monotone increasing} $N_{m}$-path if $A_{i}\subset A_{i+1}$ for all
$0\leq i<n$, i.e.,
\[
(x_{A_{i}}<x_{A_{i+1}})\wedge(y_{A_{i}}\leq y_{_{A_{i+1}}})\text{ or
}(x_{_{A_{i}}}\leq x_{_{A_{i+1}}})\wedge(y_{_{A_{i}}}<y_{_{A_{i+1}}}).
\]

The chain property and the fact that the family of feasible sets
of an antimatroid is closed under union mean that for each two
points $A,B$: if $B\subset A$, then there is a monotone increasing
$N_{4}$-path from $B$ to $A$, and if $A$ is non-comparable with
$B$, then there is a monotone increasing $N_{4}$-path from both
$A$ and $B$ to $A\cup B=(\max(x_{A},x_{B}),\max(y_{A},y_{B}))$. In
particular, for each $A\in S$ there is a monotone decreasing
$N_{4}$-path from $A$ to $0$. So, we can conclude that an
antimatroidal point set is an $N_{4}$-connected component in the
digital plane ${\mathbb{Z}}^{2}$.

\begin{definition}
A point set $S\subseteq{\mathbb{Z}}^{2}$ is defined to be orthogonally convex
if, for every line L that is parallel to the x-axis ($y=y^{\ast}$) or to the
y-axis ($x=x^{\ast}$), the intersection of $S$ with $L$ is empty, a point, or
a single interval ($[(x_{1},y^{\ast}),(x_{2},y^{\ast})]=\{(x_{1},y^{\ast
}),(x_{1}+1,y^{\ast}),...,(x_{2},y^{\ast})\}$).
\end{definition}

It follows immediately from the chain property that any antimatroidal point
set $S$ is an orthogonally convex connected component\textit{.}

Thus, antimatroidal point sets are convex polyominoes.

In the following we prove that antimatroidal point sets in the plane closed
not only under union, but under intersection as well.

\begin{lemma}
\label{Square_P} An antimatroidal point set in the plane is closed under
intersection, i.e., if two points $A=(x_{A},y_{A})$ and $B=(x_{B},y_{B})$
belong to an antimatroidal point set $S$, then the point $A\cap B=(min(x_{A}
,x_{B}),min(y_{A},y_{B}))\in S$.
\end{lemma}

\begin{proof}
The proposition is evident for two comparable points. Consider two
non-comparable points $A$ and $B$, and assume without loss of generality that
$x_{A}<x_{B}$ and $y_{A}>y_{B}$. Then there is a monotone decreasing $N_{4}
$-path from $A$ to $0$, and so there is a point $C=(x_{C},y_{B})\in S$ on this
path with $x_{C}\leq x_{A}$. Hence, the point $A\cap B$ belongs to $S$, since
it is located on the monotone increasing $N_{4}$-path from $C$ to $B$.
\end{proof}

Lemma \ref{Square_P} implies that every antimatroidal point set is a union of
rectangles built on each pair of non-comparable points. The following theorem
shows that antimatroidal point sets in the plane and parallelogram polyominoes
are equivalent.

\begin{theorem}
\label{Antimatroid_Set Theorem} A set of points $S$ in the digital plane
${\mathbb{Z}}^{2}$ is an antimatroidal point set if and only if it is an
orthogonally convex $N_{4}$-connected set that is bounded by two monotone
increasing $N_{4}$-paths between $(0,0)$ and $(x_{max},y_{max})$.
\end{theorem}

To prove the "if" part of Theorem \ref{Antimatroid_Set Theorem} it remains to
demonstrate that every antimatroidal point set $S$ is bounded by two monotone
increasing $N_{4}$-paths between $(0,0)$ and $(x_{max},y_{max})$. To give a
definition of the boundary we begin with the following notions.

A point $A$ in set $S$ is called an\textit{\ interior point} in $S$ if
$N_{8}(A)\in S$. A point in $S$ which is not an interior point is called a
\textit{boundary point}. All boundary points of $S$ constitute the boundary of
$S$. We can see an antimatroidal point set with its boundary in Figure
\ref{int_fig2}.

\begin{figure}[ptbh]
\setlength{\unitlength}{1cm}
\par
\begin{picture}(8,4.5)(-3,-0.5)\thicklines
\put(0,0){\vector(1,0){8}} \put(0,0){\vector(0,1){3.5}}
\multiput(0,0)(0.5,0){5}{\circle*{0.15}}
\multiput(0.5,0.5)(0.5,0){4}{\circle*{0.15}}
\multiput(1,1)(0.5,0){5}{\circle*{0.15}}
\multiput(1,1.5)(0.5,0){5}{\circle*{0.15}}
\multiput(3,2)(0.5,0){6}{\circle*{0.15}}
\multiput(4,2.5)(0.5,0){5}{\circle*{0.15}}
\multiput(4.5,3)(0.5,0){4}{\circle*{0.15}}
\put(3,2){\line(1,0){2.5}} \put(0.5,0.5){\line(1,0){0.5}}
\put(1,1.5){\line(1,0){2}} \put(4,2.5){\line(1,0){0.5}}
\put(4.5,3){\line(1,0){1.5}} \put(2,1){\line(1,0){1}}
\put(5.5,2.5){\line(1,0){0.5}} \put(0.5,0){\line(0,1){0.5}}
\put(1,0.5){\line(0,1){1}} \put(2,0){\line(0,1){1}}
\put(3,1){\line(0,1){1}} \put(4,2){\line(0,1){0.5}}
\put(5.5,2){\line(0,1){0.5}} \put(4.5,2.5){\line(0,1){0.5}}
\put(6,2.5){\line(0,1){0.5}} \put(7.8,-0.3){\makebox(0,0){x}}
\put(-0.2,3.3){\makebox(0,0){y}}
\put(-0.1,-0.3){\makebox(0,0){$B_{0}$}}
\put(0.5,-0.3){\makebox(0,0){$B_{1}$}}
\put(2,-0.3){\makebox(0,0){$B_{4}$}}
\put(3,0.7){\makebox(0,0){$B_{8}$}}
\put(8,3){\makebox(0,0){$B_{18}=(x_{max},y_{max})$}}
\put(2.5,2){\makebox(0,0){$B_{10}$}}
\end{picture}
\caption{A boundary of an antimatroidal point set.}
\label{int_fig2}
\end{figure}

Since antimatroidal sets in the plane are closed under union and under
intersection, there are six types of boundary points that we divide into two
sets -- lower and upper boundary:
\[
\mathcal{B}_{lower}=\{(x,y)\in S:(x+1,y)\notin S\vee(x,y-1)\notin
S\vee(x+1,y-1)\notin S\}
\]
\[
\mathcal{B}_{upper}=\{(x,y)\in S:(x-1,y)\notin S\vee(x,y+1)\notin
S\vee(x-1,y+1)\notin S\}
\]

It is possible that $\mathcal{B}_{lower}\cap\mathcal{B}_{upper}\neq\emptyset$.
For example, the point $B_{10}$ in Figure \ref{int_fig2} belongs to both the
lower and upper boundaries.

\begin{lemma}
\label{Increase}The lower boundary is a monotone increasing path between
$(0,0)$ and $(x_{max},y_{max})$.
\end{lemma}

\begin{proof}
Let $B_{0}=(0,0),B_{1},...,B_{k}=(x_{max},y_{max})$ be a lexicographical order
of $\mathcal{B}_{lower}$. We will prove that it is a monotone increasing path.
Suppose the opposite. Then there is a pair of non-comparable points $B_{i}$
and $B_{j}$, such that $i<j$, $x_{B_{i}}<x_{B_{j}}$ and $y_{B_{i}}>y_{B_{j}}$.
Then Lemma \ref{Square_P} implies that there is a rectangle built on the pair
of non-comparable points. Hence, the point $B_{i}$ does not belong to the
lower boundary, which is a contradiction.
\end{proof}

\begin{lemma}
\label{Inside}For each $B_{k}=(x,y)\in\mathcal{B}_{lower}$ holds:

(i) if $A=(x,z)\in S$ and $A\notin\mathcal{B}_{lower}$ then $z>y$;

(ii) if $A=(z,y)\in S$ and $A\notin\mathcal{B}_{lower}$ then $z<x$.
\end{lemma}

\begin{proof}
Suppose the opposite, i.e., there is $A=(x,z)\in S$ such that $A\notin
\mathcal{B}_{lower}$ and $z<y$. Then the point $(x+1,z)\in S$, so there is the
last right point $(x^{\ast},z)\in S$, such that $x^{\ast}>x$. Hence, the point
$(x^{\ast},z)$ belongs to $\mathcal{B}_{lower}$, contradicting Lemma
\ref{Increase}. The Property (\textit{ii}) is proved similarly.
\end{proof}

\begin{lemma}
The lower boundary is an $N_{4}$-path between $(0,0)$ and $(x_{max},y_{max})$.
\end{lemma}

\begin{proof}
Prove that $|B_{i+1}-B_{i}|=1$ for every $1\leq i<k$. If $x_{B_{i}}
=x_{B_{i+1}}$. then from the chain property and Lemma \ref{Inside} it
immediately follows that $y_{B_{i+1}}=y_{B_{i}}+1$.

If $x_{B_{i}}<x_{B_{i+1}}$, then the chain property implies that either
$(x_{B_{i}}+1,y_{B_{i}})\in S$ or $(x_{B_{i}},y_{B_{i}}+1)\in S$. Since the
point $(x_{B_{i}},y_{B_{i}}+1)\notin\mathcal{B}_{lower}$, we can conclude that
$(x_{B_{i}}+1,y_{B_{i}})\in S$ in any case. Now Lemma \ref{Increase} implies
that this point belongs to $\mathcal{B}_{lower}$, i.e., $B_{i+1}=(x_{B_{i}
}+1,y_{B_{i}}).$
\end{proof}

The upper boundary case is validated in the same way.

Thus an antimatroidal point set is an orthogonally convex $N_{4}$-connected
set bounded by two monotone increasing $N_{4}$-paths.

The following lemma is the "only-if" part of Theorem
\ref{Antimatroid_Set Theorem}.

\begin{lemma}
An orthogonally convex $N_{4}$-connected set $S$ that is bounded by two
monotone increasing $N_{4}$-paths between $(0,0)$ and $(x_{max},y_{max})$ is
an antimatroidal point set.
\end{lemma}

\begin{proof}
By Definition \ref{Def} we have to check the two properties (A1) and (A2):

(A1) Let $A=(x,y)\in S$. If $A$ is an interior point in $S$ then $(x-1,y)\in
S$ and $(x,y-1)\in S$. If $A$ is a boundary point, then the previous point on
the boundary ($(x,y-1)$ or $(x-1,y)$) belongs to $S$.

(A2) Let $A\nsubseteq B\in S$. Consider two cases:

(i) $x_{a}\geq x_{b}$ and $y_{a}\geq y_{b}$. If $B$ is an interior point in
$S$ then $(x_{b}+1,y_{b})\in S$ and $(x_{b},y_{b}+1)\in S$. If $B$ is a
boundary point, then the next point on the boundary ($(x_{b},y_{b}+1)$ or
$(x_{b}+1,y_{b})$) belongs to $S$.

(ii)$x_{a}\leq x_{b}$ and $y_{a}\geq y_{b}$. We have to prove that
$(x_{b},y_{b}+1)\in S$. Suppose the opposite. Then the point $B$ is an upper
boundary point. Since $y_{a}\geq y_{b}$ there exists an upper boundary point
$(x_{a},y)$ with $y\geq y_{b}$ that contradicts the monotonicity of the boundary.
\end{proof}

\begin{corollary}
Any antimatroidal point set $S$ may be represented by its boundary in the
following form:
\[
S=\mathcal{B}_{lower}\vee\mathcal{B}_{upper}=\{X\cup Y:X\in\mathcal{B}
_{lower},Y\in\mathcal{B}_{upper}\}
\]

\end{corollary}

This result shows that the \textit{convex dimension} of a two-dimensional
poly-antimatroid is two.

\begin{definition}
Convex dimension \cite{Greedoids} $c\dim(S)$ of any antimatroid $S$ is the
minimum number of maximal chains
\[
\emptyset=X_{0}\subset X_{1}\subset...\subset X_{k}=X_{\max}\text{ with }
X_{i}=X_{i-1}\cup x_{i\text{ }}
\]
whose union gives the antimatroid $S$.
\end{definition}

The boundary of any poly-antimatroid may be found by using the following
algorithm. Starting with the maximum point $(x_{\max},y_{\max})$ it follows
the upper boundary down to point $(0,0)$ in the first pass, and it follows the
lower boundary in the second pass.

\begin{algorithm}
\label{Alg}\textbf{Upper} \textbf{boundary tracing algorithm}

1. $i:=0$; $x:=x_{\max}$; $y:=y_{\max}$;

2. $B_{i}:=(x,y)$

3. do

\ \ \ \ \ \ \ \ \ \ \ 3.1\ if $(x-1,y)\in S$ then $x:=x-1$, else $y:=y-1$;

\ \ \ \ \ \ \ \ \ \ \ 3.2 $\ i:=i+1$;

\ \ \ \ \ \ \ \ \ \ \ 3.3 \ $B_{i}:=(x,y)$;

\ \ \ \ until $B_{i}=(0,0)$

4. Return the sequence $\mathcal{B}=B_{i},B_{i-1},...,B_{0}$
\end{algorithm}

It is easy to check that Algorithm \ref{Alg} returns a monotone increasing
$N_{4}$-path that only passes over the upper boundary points from
$\mathcal{B}_{upper}$ (for each $i$ point $B_{i}=(x^{i},y^{i})$ has the
maximum $y-$coordinate, i.e., $y^{i}=\underset{y}{\arg\max}\{(x^{i},y)\in
S\}$). Hence, the Upper boundary tracing algorithm returns the upper boundary
of poly-antimatroids.

The Lower boundary tracing algorithm differs from Algorithm \ref{Alg} on
search order only: $(y,x)$ instead of $(x,y)$. Step 3.1 of Algorithm \ref{Alg}
will be as follows:
\[
3.1\text{ if }(x,y-1)\in S\text{ then }y:=y-1\text{,else }x:=x-1\text{;}
\]

Correspondingly, the Lower boundary tracing algorithm returns the lower
boundary of poly-antimatroids.

It turned out that a two-dimensional poly-antimatroid known in this section as
an antimatroidal point set is equivalent to special cases of polyominoes or
staircase polygons and is defined by a pair of monotone paths made only with
north and east steps. In the next section we research the path structure of
three-dimensional antimatroidal point sets.

\section{Three-dimensional antimatroidal point set}

In this section we consider a particular case of three-dimensional
antimatroidal point sets.

\begin{definition}
A\ finite non-empty set $C$ of points in digital $3D$ space is a digital
cuboid if
\[
C=\{(x,y,z)\in{\mathbb{Z}}^{3}:x_{\min}\leq x\leq x_{\max}\wedge y_{\min}\leq
y\leq y_{\max}\wedge z_{\min}\leq z\leq z_{\max}\}\text{ and }|C|>1
\]

\end{definition}

A digital cuboid is specified by the coordinates of opposite corners.

Consider the following construction of $n$ cuboids:

\begin{definition}
\label{staircase}The sequence of $n$ cuboids $C_{1},C_{2},...,C_{n}$ is called
regular if \

(a) $x_{\min}^{0}=y_{\min}^{0}=z_{\min}^{0}=0$

(b) $x_{\min}^{i}\leq x_{\min}^{i+1}\wedge y_{\min}^{i}\leq y_{\min}
^{i+1}\wedge z_{\min}^{i}\leq z_{\min}^{i+1}$

\ \ \ \ and at least one of the inequality is strong \ for each $1\leq i\leq
n-1$

(c) $x_{\min}^{i+1}\leq x_{\max}^{i}\wedge y_{\min}^{i+1}\leq y_{\max}
^{i}\wedge z_{\min}^{i+1}\leq z_{\max}^{i}$ for each $1\leq i\leq n-1$

(d) $x_{\max}^{i}\leq x_{\max}^{i+1}\wedge y_{\max}^{i}\leq y_{\max}
^{i+1}\wedge z_{\max}^{i}\leq z_{\max}^{i+1}$ \

\ \ \ \ and at least one of the inequality is strong \ for each $1\leq i\leq
n-1$
\end{definition}

\begin{definition}
The union of elements of regular sequence $\mathcal{C}=C_{1}\cup C_{2}
\cup...\cup C_{n}$  is called an $n$-step staircase.
\end{definition}

An example of a $2$-step staircase is depicted in Figure $\ref{int_fig3}$.

\begin{figure}[ptbh]
\setlength{\unitlength}{1cm}
\par
\begin{picture}(8,4.5)(-4.5,-0.5)\thicklines
\multiput(0,0.5)(1.0,0){2}{\circle*{0.25}}
\multiput(0,1.5)(1.0,0){2}{\circle*{0.25}}
\multiput(0,2.5)(1.0,0){2}{\circle*{0.25}}
\put(0.7,3.2){\circle*{0.25}}
\multiput(1.4,3.9)(1.0,0){4}{\circle*{0.25}}
\multiput(1.3,0)(1.0,0){3}{\circle*{0.25}}
\multiput(1.3,1)(2,0){2}{\circle*{0.25}}
\multiput(1.3,2)(1.0,0){3}{\circle*{0.25}}
\multiput(2.0,2.7)(2,0){2}{\circle*{0.25}}
\multiput(2.7,3.4)(1.0,0){3}{\circle*{0.25}}
\put(4.0,0.7){\circle*{0.25}}
\multiput(4.7,1.4)(0,1.0){2}{\circle*{0.25}}
\put(0,0.5){\line(1,0){1.3}} \put(0,1.5){\line(1,0){1.3}}
\put(0,0.5){\line(0,1){2.0}} \put(1.3,0){\line(0,1){2.0}}
\put(1.3,0){\line(1,0){2.0}}\put(1.3,2){\line(1,0){2.0}}
\put(3.3,0){\line(0,1){2.0}}\put(4.7,1.4){\line(0,1){2.0}}
\put(2.7,3.4){\line(1,0){2.0}} \put(1.4,3.9){\line(1,0){3.0}}
\put(0,2.5){\line(1,1){1.4}}\put(1.3,2.0){\line(1,1){1.4}}
\put(3.3,2.0){\line(1,1){1.4}}\put(3.3,0.0){\line(1,1){1.4}}
\put(0,2.5){\line(1,0){1.8}} \put(4.4,3.4){\line(0,1){0.5}}
\put(3.9,3.4){\line(1,1){0.5}}
\end{picture}
\caption{$2$-step staircase $C$.}
\label{int_fig3}
\end{figure}

Each point $A=(x_{A},y_{A},z_{A})$ in the digital space ${\mathbb{Z}}^{3}$ may
be considered as a multiset $A$ over $\{x,y,z\}$, where $x_{A}$ is the number
of repetitions of an element $x$, and $y_{A}$ is the number of repetitions of
an element $y$, and $z_{A}$ is the number of repetitions of an element $z$ in
multiset $A$. Then a $1$-step staircase (cuboid) is a poly-antimatroid, since
the family of points is accessible and closed under union ($(x^{1},y^{1}
,z^{1}),(x^{2},y^{2},z^{2})\in C\Rightarrow(x^{1},y^{1},z^{1})\cup(x^{2}
,y^{2},z^{2})=(\max(x^{1},x^{2}),\max(y^{1},y^{2}),\max(z^{1},z^{2}))\in C$).

\begin{lemma}
An $n$-step staircase $\mathcal{C}$ is a poly-antimatroid.
\end{lemma}

\begin{proof}
Consider some point $A\neq(0,0,0)\in\mathcal{C}$. Then there exists $i$ such
that $A\in C_{i}$. If $A\neq(x_{\min}^{i},y_{\min}^{i},z_{\min}^{i})$, then,
without lost of generality, $x_{\min}^{i}<x_{A}$, and so $(x_{A}-1,y_{A}
,z_{A})\in C_{i}\subseteq\mathcal{C}$. If $A=(x_{\min}^{i},y_{\min}
^{i},z_{\min}^{i})$, then, from Definition \ref{staircase} (b,c) it follows
that $A\in C_{i-1}$and $A\neq(x_{\min}^{i-1},y_{\min}^{i-1},z_{\min}^{i-1})$.
Hence, $(x_{A}-1,y_{A},z_{A})\in C_{i-1}\subseteq\mathcal{C}$ or $(x_{A}
,y_{A}-1,z_{A})\in C_{i-1}\subseteq\mathcal{C}$ or $(x_{A},y_{A},z_{A}-1)\in
C_{i-1}\subseteq\mathcal{C}$. Thus, an $n$-step staircase $\mathcal{C}$ is accessible.

To prove that $\mathcal{C}$ is closed under union consider two points
$A,B\in\mathcal{C}$. Let $A\in C_{i}$, $B\in C_{j}$, and $i\leq j$. From
Definition \ref{staircase} (b) it follows that $x_{\min}^{j}\leq\max
(x_{A},x_{B})$. From Definition \ref{staircase} (d), $\max(x_{A},x_{B})\leq
x_{\max}^{j}$. So, $A\cup B=(\max(x_{A},x_{B}),\max(y_{A},y_{B}))\in C_{j}$,
i.e., $A\cup B\in\mathcal{C}.$
\end{proof}

Our goal is to find the minimum number of maximal chains which describe an
$n$-step staircase.

First, consider a cuboid $C$ given by two points $(x_{\min},y_{\min},z_{\min
})$ and $(x_{\max},y_{\max},z_{\max})$. We will denote by $P_{X}$ the maximal
chain connecting the points $(x_{\min},y_{\min},z_{\min})$ and $(x_{\max
},y_{\min},z_{\min})$, i.e.,
\[
P_{X}=(x_{\min},y_{\min},z_{\min}),(x_{\min}+1,y_{\min},z_{\min}
),...,(x_{\max},y_{\min},z_{\min})\text{.}
\]
$P_{Y}$ and $P_{Z}$ are defined correspondingly. It is easy to see that
\[
C=P_{X}\vee P_{Y}\vee P_{Z}=\{A_{1}\cup A_{2}\cup A_{3}:A_{1}\in P_{X}
,A_{2}\in P_{Y},A_{3}\in P_{Z}\}
\]

So, if we have three maximal chains $B_{1}$,$B_{2}$,$B_{3}$ from the cuboid
$C$, which pass over all the points of $P_{X}$, $P_{Y}$ and $P_{Z}$, then
$C\subseteq B_{1}\vee B_{2}\vee B_{3}$. On the other hand, since $C$ is closed
under union, we have $B_{1}\vee B_{2}\vee B_{3}\subseteq C$. Thus, $B_{1}\vee
B_{2}\vee B_{3}=C$.

Now, consider the three-dimensional version of Algorithm \ref{Alg}
for search order $(x,y,z)$. Let $(x_{\max},y_{\max},z_{\max})$ be
the maximum point of $n$-step staircase $\mathcal{C}$. The
algorithm builds the chain connecting the maximum point and the
point $(0,0,0)$.

\begin{algorithm}
\label{Alg3}XYZ-Boundary tracing algorithm

1. $i:=0$; $x:=x_{\max}$; $y:=y_{\max}$; $z:=z_{\max}$;

2. $B_{i}:=(x,y,z)$

3. do

\ \ \ \ \ \ \ \ \ \ \ 3.1\ if $(x-1,y,z)\in S$ then $x:=x-1$, else if
$(x,y-1,z)\in S$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ then $y:=y-1$, else $z:=z-1$;

\ \ \ \ \ \ \ \ \ \ \ 3.2 $\ i:=i+1$;

\ \ \ \ \ \ \ \ \ \ \ 3.3 \ $B_{i}:=(x,y,z)$;

\ \ \ \ until $B_{i}=(0,0,0)$

4. Return the sequence $\mathcal{B}=B_{i},B_{i-1},...,B_{0}$
\end{algorithm}

Repeat Algorithm \ref{Alg3} for search order $(y,z,x)$ and for search order
$(z,x,y)$. As a result, we obtain three monotone increasing $N_{6}$-paths
\cite{Digital} from $(0,0,0)$ to $(x_{\max},y_{\max},z_{\max})$, denoted by
$\mathcal{B}_{Z}$, $\mathcal{B}_{X}$ and $\mathcal{B}_{Y}$, respectively. The
length of any of these chains is equal to $(x_{\max}+y_{\max}+z_{\max})$.

\begin{theorem}
\label{3convex}Any $n$-step staircase $\mathcal{C}=\mathcal{B}_{X}
\vee\mathcal{B}_{Y}\vee\mathcal{B}_{Z}$.
\end{theorem}

To show this we prove a stronger statement.

Change Algorithm \ref{Alg3} in the following way. Let the XYZ-Boundary tracing
algorithm begin from some point $(x,y,z_{\max})\in\mathcal{C}$ and return
chain $\mathcal{H}_{Z}$; the YZX-Boundary tracing algorithm begins from some
point $(x_{\max},y,z)\in\mathcal{C}$ and returns chain $\mathcal{H}_{X}$; and
the ZXY-Boundary tracing algorithm begins from some point $(x,y_{\max}
,z)\in\mathcal{C}$ and returns chain $\mathcal{H}_{Y}$.

\begin{lemma}
Any $n$-step staircase $\mathcal{C}=\mathcal{H}_{X}\vee\mathcal{H}_{Y}
\vee\mathcal{H}_{Z}$.
\end{lemma}

\begin{proof}
Let us proceed by induction on $n$.

For $n=1$ the XYZ-Boundary tracing algorithm beginning from the point
$(x,y,z_{\max})\in\mathcal{C}$ returns chain $\mathcal{H}_{Z}$ that begins
from the point $(0,0,0)$, moves to $(x_{\min},y_{\min},z_{\max})$, then to
$(x_{\min},y,z_{\max})$, and finishes at the point $(x,y,z_{\max})$. So, this
chain passes over all the points of $P_{Z}$. In the same way, the YZX-Boundary
tracing algorithm returns chain $\mathcal{H}_{X}$ that passes over all the
points of $P_{X}$, and the ZXY-Boundary tracing algorithm returns chain
$\mathcal{H}_{Y}$ that passes over all the points of $P_{Y}$. Thus,
$\mathcal{C}=\mathcal{H}_{X}\vee\mathcal{H}_{Y}\vee\mathcal{H}_{Z}$.

Assume that the proposition is correct for all $k<n$ and prove it for $n$. It
is easy to see that the XYZ-Boundary tracing algorithm begins from some point
$(x,y,z_{\max})\in C_{n}$, reaches the point $(x_{\min}^{n},y_{\min}
^{n},z_{\max}^{n})$, moves down to the point $(x_{\min}^{n},y_{\min}
^{n},z_{\max}^{n-1})\in C_{n-1}$, and then continues to build chain
$\mathcal{H}_{Z}$. In the same way, the YZX-Boundary tracing algorithm moves
through the point $(x_{\max}^{n-1},y_{\min}^{n},z_{\min}^{n})\in C_{n-1}$, and
the ZXY-Boundary tracing algorithm moves through the point $(x_{\min}
^{n},y_{\max}^{n-1},z_{\min}^{n})\in C_{n-1}$. The induction assumption
implies $C_{1}\cup C_{2}\cup...\cup C_{n-1}\subseteq\mathcal{H}_{X}
\vee\mathcal{H}_{Y}\vee\mathcal{H}_{Z}$.

It remains to show that $C_{n}\subseteq\mathcal{H}_{X}\vee\mathcal{H}_{Y}
\vee\mathcal{H}_{Z}$. To this end we prove that $P_{X}^{n}\subseteq
\mathcal{H}_{X}\vee\mathcal{H}_{Y}\vee\mathcal{H}_{Z}$, $P_{Y}^{n}
\subseteq\mathcal{H}_{X}\vee\mathcal{H}_{Y}\vee\mathcal{H}_{Z}$, and
$P_{Z}^{n}\subseteq\mathcal{H}_{X}\vee\mathcal{H}_{Y}\vee\mathcal{H}_{Z}$. We
have already seen that the XYZ-Boundary tracing algorithm returns chain
$\mathcal{H}_{Z}$ that covers the part $[(x_{\min}^{n},y_{\min}^{n},z_{\max
}^{n}),(x_{\min}^{n},y_{\min}^{n},z_{\max}^{n-1})]$ from $P_{Z}^{n}$, and the
remaining part $[(x_{\min}^{n},y_{\min}^{n},z_{\max}^{n-1}),(x_{\min}
^{n},y_{\min}^{n},z_{\min}^{n})]\subseteq C_{1}\cup C_{2}\cup...\cup C_{n-1}$.
So $P_{Z}^{n}\subseteq\mathcal{H}_{X}\vee\mathcal{H}_{Y}\vee\mathcal{H}_{Z}$.
It is easy to verify two other cases.

Finally, $\mathcal{C}=C_{1}\cup C_{2}\cup...\cup C_{n}\subseteq(\mathcal{H}
_{X}\vee\mathcal{H}_{Y}\vee\mathcal{H}_{Z})\vee(\mathcal{H}_{X}\vee
\mathcal{H}_{Y}\vee\mathcal{H}_{Z})\vee(\mathcal{H}_{X}\vee\mathcal{H}_{Y}
\vee\mathcal{H}_{Z})=\mathcal{H}_{X}\vee\mathcal{H}_{Y}\vee\mathcal{H}_{Z}$,
since each $\mathcal{H}$ is a chain.

On the other hand, $\mathcal{H}_{X}\vee\mathcal{H}_{Y}\vee\mathcal{H}
_{Z}\subseteq\mathcal{C}$, and so $\mathcal{C}=\mathcal{H}_{X}\vee
\mathcal{H}_{Y}\vee\mathcal{H}_{Z}$.
\end{proof}

Theorem \ref{3convex} shows that it is enough to know only three maximal
chains $\mathcal{B}_{X}$,$\mathcal{B}_{Y}$, and $\mathcal{B}_{Z}$ to describe
an $n$-step staircase.

\begin{corollary}
An n-step staircase has convex dimension at most $3$.
\end{corollary}

Note that the three points $(0,0,1)$, $(0,1,0)$ and $(1,0,0)$ cannot be
covered by two chains only, since each such point (multiset) cannot be formed
as a union of smaller multisets. So, there exist $n$-step staircases of convex
dimension $3$.

However, the convex dimension of an arbitrary three-dimensional
poly-antimatroid may be arbitrarily large \cite{Eppstein}. Let $S$ be a set of
points:
\[
S=\{(x,y,z):(0\leq x,y\leq N)\wedge(0\leq z\leq1)\wedge(z=1\Rightarrow x+y\geq
N)\}
\]

It is easy to check that $S$ is a three-dimensional poly-antimatroid. Consider
$N+1$ points $(x,y,1)$ with $x+y=N$. Since each of these points cannot be
represented as a union of any points from $S$ with smaller coordinates, the
convex dimension of $S$ is at least $N+1$ \cite{Eppstein}.

In order to characterize the family of three-dimensional poly-antimatroids of
convex dimension $3$, consider a particular case of antimatroids called
\textit{poset antimatroids} \cite{Greedoids}. A poset antimatroid has as its
feasible sets the lower sets of a poset (partially ordered set). The poset
antimatroids can be characterized as the unique antimatroids which are closed
under intersection \cite{Greedoids}.\ We extend this definition to poly-antimatroids.

\begin{definition}
A poly-antimatroid is called a poset poly-antimatroid if it is closed under intersection.
\end{definition}

Now, note that $n$-step staircases are closed under intersection too. Indeed,
consider two points $A,B\in\mathcal{C}$. Let $A\in C_{i}$, $B\in C_{j}$, and
$i\leq j$. From Definition \ref{staircase} (b) it follows that $\min
(x_{A},x_{B})\geq x_{\min}^{i}$. On the other hand, (Definition
\ref{staircase} (d)), $\min(x_{A},x_{B})\leq x_{\max}^{i}$. So, $A\cap
B=(\min(x_{A},x_{B}),\min(y_{A},y_{B}))\in C_{i}$, i.e., $A\cap B\in
\mathcal{C}.$

So, our conjectures are as follows.

\begin{conjecture}
A three-dimensional poset poly-antimatroid is a step staircase.
\end{conjecture}

\begin{corollary}
The convex dimension of a three-dimensional poset poly-antimatroid is at most
$3$.
\end{corollary}

Moreover,

\begin{conjecture}
The convex dimension of an $n$-dimensional poset poly-antimatroid is at most
$n$.
\end{conjecture}

\begin{acknowledgement}
The authors are grateful to the referees for their constructive criticism and
correction of style.
\end{acknowledgement}


text:
\begin{document}
\maketitle

\begin{abstract}
A thin two-layered waveguide is considered.
The governing equations
for this waveguide is a  matrix Klein--Gordon equation of dimension~2.
A formal solution of this system
in the form of a double integral
can be obtained by using Fourier transformation.
Then, the double integral can be reduced to a single integral with the help
of residue integration with respect to the time frequency.
However, such an integral can be difficult to estimate since it involves branching and oscillating functions. This integral is studied asymptotically.
A zone diagram technique is proposed to represent the set of possible
asymptotic formulae.
The zone diagram generalizes the
concept of far-field and near-field zones.
\end{abstract}

\section{Problem formulation}
Consider a waveguide composed of two one-dimensional layers weakly coupled with each other.
The subsystems (the layers) are described by unknown functions
$u_1(t, x)$ and $u_2(t, x)$ (here $t$ is time, and $x$ is longitudinal coordinate).
Assume that the waveguide is described by a {\em matrix Klein-Gordon equation\/}
of dimension~2
(another naming for this equation is WFEM or WaveFEM):
\begin{equation}
\label{MKGE}
\left[ \begin{pmatrix} c_1^2 \quad 0 \\ 0 \quad c_2^2 \end{pmatrix} \partial_x^2 + \begin{pmatrix} -\Omega_1^2 \qquad \mu \\ \mu \qquad -\Omega_2^2
\end{pmatrix} -
\begin{pmatrix} 1 \quad 0 \\ 0 \quad 1 \end{pmatrix} \partial_t^2 \right] \begin{pmatrix} u_1(t,x) \\ u_2(t,x) \end{pmatrix} =
\begin{pmatrix} f_1 \\ f_2 \end{pmatrix} \delta(t) \delta(x)
\end{equation}
$\begin{pmatrix} f_1 , f_2 \end{pmatrix}^T$ is a constant vector describing the excitation. The
excitation is a short pulse localized at~$x = 0$. Indeed, we are looking for a causal
solution equal to zero for~$t < 0$.
For simplicity, we take
\[
\left( \begin{array}{cc} f_1  \\ f_2 \end{array} \right) =
\left( \begin{array}{cc} 1  \\ 0 \end{array} \right).
\]

Parameter
$\mu$ is a small value describing a coupling between the subsystems.
A system with $\mu=0$ is {\em unperturbed}. Matrix coefficients
of the unperturbed system are diagonal,
and the waves in the subsystems
propagate
independently of each other.
Each subsystem is described by a scalar Klein--Gordon equation
\[
( c_j^2 \ptl_x^2 - \ptl_t^2 - \Omega_j^2) u_j (t, x) = f_j \, \delta(t) \, \delta(x).
\]
The parameters $\Omega_j$ are
cut-off frequencies of the unperturbed subsystems.

Our aim is to obtain
a formal solution of \eqref{MKGE} (see Section~\ref{FSsection})
and to get
asymptotic estimations of the wave field for different values of $t$ and~$x$.
A classical way to do this is to apply a
{\em  saddle point method\/} (see Section~\ref{SPsection}). This method allows one to describe the wave field for $x \rightarrow \infty$ and $t \rightarrow \infty$. However, for values of $x$ and $t$ that are moderately large,
one needs some more
sophisticated asymptotics. Section~\ref{ZDsection} is devoted to them.

\section{Formal solution of the matrix Klein--Gordon equation}
\label{FSsection}
\subsection{Double integral solution}

Apply Fourier transform with respect to $x$ and Laplace transform with respect to $t$ to
\eqref{MKGE}. As a result, obtain a double integral representation of the waveguide field:
\begin{equation}
\label{Double_int_repr}
\begin{pmatrix} u_1(t,x)\\u_2(t,x)\end{pmatrix} =
\int\limits_{-\infty + i\varepsilon}^{\infty + i \varepsilon}\int\limits_{-\infty}^{\infty}
\frac{{\rm A}(\omega,k)}{D(\omega,k)} e^{-i\omega t + i k x}  dk \, d\omega  ,
\end{equation}
where
\begin{equation}
\label{A_det}
{\rm A}(\omega,k)= \begin{pmatrix} -k^2 c_2^2-\Omega_2^2 + \omega^2 \\ -\mu \end{pmatrix},
\end{equation}
\begin{equation}
\label{D_det}
D(\omega,k) = (-k^2c_1^2 + \omega^2 -\Omega_1^2)(-k^2c_2^2 + \omega^2 -\Omega_2^2)-\mu^2 ,
\end{equation}
$\varepsilon$ is an arbitrary  positive real number (it provides causality of the solution).
We assume that $\epsilon$ is small.

\subsection{Series--integral solution}

Let be $x > 0$.
\eqref{Double_int_repr} can be transformed into a sum of single integrals.
For that purpose, one has to close the contour of integration with respect to $k$ in the upper half plane of $k$ (it is possible due to Jordan's lemma) and apply the residue method. As a result, obtain the following representation of the solution
represented by \eqref{Double_int_repr}:
\begin{equation}
\label{Ser_int_repr}
\begin{pmatrix} u_1(t,x)\\u_2(t,x)\end{pmatrix} = 2\pi i
\sum _{m=1}^{2}\int\limits_{-\infty + i\varepsilon}^{\infty + i \varepsilon}
\frac{{\rm A}(\omega,k_m(\omega))}{
\ptl_k D(\omega,k_m(\omega))} e^{i k_m(\omega) x - i\omega t } d\omega  ,
\end{equation}
where $\ptl_k D$ is a partial derivative of function $D$ with respect to the second argument.
Functions $k_m(\omega)$, $m = 1,2$ are two of four roots of the {\em dispersion equation}
\begin{equation}
\label{DE}
D(\omega,k)=0.
\end{equation}
solved for a fixed $\omega$.
Among four roots of this equation, we select two roots with ${\rm Im}[k_m (\omega)] > 0$
as follows.
Four roots of \eqref{DE} can be split into pairs: $\pm k_m(\omega)$, $m = 1,2$.
We denote
by $k_1(\omega)$ and $k_2(\omega)$ the solutions having positive group velocities
\[
v_{\rm gr}(\omega) \equiv \left( \partial_{\omega} k \right) ^{-1}.
\]
Solutions $k_1(\omega)$ and $k_2(\omega)$ have ${\rm Im}[k]>0$ for ${\rm Im}[\omega] = \varepsilon > 0$. This can be shown by the Taylor's series:
\begin{equation}
\label{eq0}
{\rm Im} [k(\omega' + i\varepsilon)] \approx {\rm Im}[k(\omega') + \frac{dk(\omega')}{d\omega} \cdot i\varepsilon] = \varepsilon \frac{dk(\omega')}{d\omega}, \quad \omega' = {\rm Re}[\omega].
\end{equation}

\subsection{Dispersion diagram}

A {\em dispersion diagram\/} is a graphical representation of the roots of \eqref{DE}.
We will use a real dispersion diagram, which is the set of all real points $(\omega,k)$
obeying \eqref{DE}, and a complex dispersion diagram, which is the set of complex
$(\omega,k)$ obeying \eqref{DE}.

In the real dispersion diagram,
we assign indices 1 or 2 to the branches,
depending on their asymptotics as $\omega \rightarrow \infty$.
Namely, we assume that  $k_1(\omega) \rightarrow \omega/c_1$, $k_2(\omega) \rightarrow \omega/c_2$.

\begin{figure}[!h]
\begin{center}
\includegraphics[scale=0.8]{DD_Sh.pdf}
\includegraphics[scale=0.8]{vgr_full2.pdf}
\end{center}
	\caption{
    Real dispersion diagram (left) and group velocities (right)
    for the system described by \eqref{MKGE}}
\label{DD}
\end{figure}

For a working example throughout this paper,
let us choose some values of the waveguide parameters in such a way that an {\em exchange pulse} is observed \cite{Shanin_2018}. In our case $c_1 > c_2$, $\Omega_1 <\Omega_2$. Dispersion diagram of such a system looks like in \figurename~\ref{DD}~left, black.

Dispersion diagram of the corresponding unperturbed system is
presented by
two intersecting red lines in \figurename~\ref{DD}~left. The point of intersection
$(\omega_{\rm sh},k_{\rm sh})$ is a referred to as a
Shestopalov's singular point \cite{Shestopalov1996}:
\begin{equation}
\label{Shest_Point}

\omega_{\rm sh} = \sqrt{(c_1^2 \Omega_2^2 - c_2^2 \Omega_1^2)/(c_1^2 - c_2^2)},
\qquad

k_{\rm sh} = \sqrt{(\Omega_2^2 - \Omega_1^2)/(c_1^2 - c_2^2)}.
\end{equation}
One can see that the perturbed real dispersion diagram displays an {\em avoiding crossing\/}
near the Shestopalov's point.

In \figurename~\ref{DD}~right,
the group velocities  for the system with $\mu \neq 0$ (black) and with $\mu = 0$ (red)
are displayed.
Let
$v_1'$ and $v_2'$ be a local maximum and a local minimum
of the group velocity. The corresponding frequencies are $\omega_1^e$ and $\omega_2^e$, respectively. Obviously, these points are inflection points of the real dispersion diagram.

Let $v_1$ and $v_2$ be
the values of the group velocity of the unperturbed system taken at $\omega_{\rm sh}$.
They can be easily computed:
\begin{equation}
\label{Vgr_Shest_form}
v_1 = c_1 \sqrt{\omega_{\rm sh}^2 - \Omega_{1}^2} \, / \omega_{\rm sh}, \qquad
v_2 = c_2 \sqrt{\omega_{\rm sh}^2 - \Omega_{2}^2} \, / \omega_{\rm sh}.
\end{equation}
For the systems, which we chose, $v_1 > v_2$. Obviously, $v_1 < c_1$ and $v_2 < c_2$. It can be $v_1 > c_2$ or $v_1 < c_2$ or even $v_1 = c_2$. We take $v_1 < c_2$ for definiteness.

\begin{figure}[!h]
\centering
\includegraphics[scale = 0.9]{surface_AV-eps-converted-to}
\caption{Scheme of the Riemann surface of $k(\omega)$.
Blue lines are cuts. Equal numbers denote the sides that should be attached to each other. Integration contours $\gamma_1$ and $\gamma_2$ are the red and green lines, respectively}
\label{RS}
\end{figure}

Instead of four functions $\pm k_1(\omega)$, $\pm k_2(\omega)$, we consider a multivalued function $k(\omega)$ on the complex plane $\omega$. A scheme of Riemann surface of
the function $k(\omega)$ is shown in \figurename~\ref{RS}. This
Riemann surface has four sheets. Denote the sheet corresponding to $k_1(\omega)$ by ``Sheet~1'', the sheet for $k_2(\omega)$ by ``Sheet~2'', the sheet for $-k_1(\omega)$ by ``Sheet~3'', and the sheet for $-k_2(\omega)$ by ``Sheet~4''.

The Riemann surface has branch points of two types:
{\em cut-off branch points\/} and {\em exchange branch points}.
The cut-off branch points, are solutions of the equation

$
D(\omega,k=0) = 0.
$

These branch points
connect sheets corresponding to the solutions of the dispersion equation with opposite signs
(i.~e.\ $k_1(\omega)$ with $-k_1(\omega)$, and $k_2(\omega)$ with $-k_2(\omega)$).
The cut-off frequencies belong to the real axis of~$\omega$.

The rest of the branch points are the exchange branch points.
These points exist due to interaction between the subsystems. They connect
the sheets corresponding to different modes. In \figurename~\ref{RS} the
exchange branch points correspond to the values of $\omega$ that are not real.

One can consider \eqref{Ser_int_repr} as a contour integral of a multivalued function,
whose Riemann surface is the same of the Riemann surface of $k(\omega)$. The contour of integration is $\gamma_1 + \gamma_2$, where $\gamma_1$
and $\gamma_2$ are shown in \figurename~\ref{RS}. These contours are preimages of
the real axis of $\omega$ on Sheet~1 and Sheet~2,
slightly shifted into the upper half plane.

\section{Analysis of solutions by saddle point method (far zone)} \label{SPsection}

\subsection{Saddle point approach}

According to the Cauchy's theorem, the contour of integration
in \eqref{Ser_int_repr}
can be homotopically deformed in the area of analyticity of the integrand.
Using this fact, one can obtain some asymptotical assessments of  \eqref{Ser_int_repr}. Particularly, the saddle point method allows to estimate
the wave field in a far zone.

Introduce a {\em  formal velocity\/}  $V$ by
\[
V = x/t.
\]
Let $V$ be fixed, and let be $x \to \infty$. Indeed, this means $t \to \infty $ as well.

As it was stated above, \eqref{Ser_int_repr} can be rewritten as follows:
\begin{equation}
\label{SIR2}
u_j(t,x) = \int \limits_{\gamma_1 + \gamma_2} h_j(\omega,k(\omega)) \exp\{ ixg(\omega) \}
\, d\omega,
\quad
h_j(\omega,k) = \frac{A_j(\omega,k)}{\ptl_k D(\omega,k)},
\quad j = 1,2,
\end{equation}
where $A_j$ are components of the vector ${\rm A}$, and
\begin{equation}
\label{g_det}
g(\omega) \equiv k(\omega) - \omega / V.
\end{equation}

Since the exponential factor $\exp\{ixg\}$ contains the large parameter $x$,
while $h_j$ has not, the exponential factor mainly
determines the magnitude of the integrand. Namely,  the
integrand is {\em exponentially large\/} if ${\rm Im} [g] < 0$, and is
{\em exponentially small\/} if ${\rm Im}[g] > 0$.

Such integrals can be asymptotically estimated using the saddle point method \cite{Borovikov_1994}. Since the integration is held on the Riemann surface of a multivalued function, one has to use a multi-contour version of the saddle point method
\cite{Shanin_2018,Shanin_2017}.

According to the saddle point method,
the contour $\gamma_1 + \gamma_2$ can be deformed into a sum of several {\em saddle point contours} $\gamma_{\ast m}$:
\begin{equation}
\label{SPsum}
u_j(t,x) = \sum_m I_{j,\ast m},
\quad
I_{j,\ast m} = \int \limits_{\gamma_{\ast m}} h_j(\omega,k(\omega)) \exp\{ ixg(\omega) \}d\omega.
\end{equation}
Each saddle point contour $\gamma_{j,\ast m}$ passes through {\em a saddle point\/}
$\omega_{\ast m}$.
A saddle point $\omega_{\ast m}$ is a root of the first derivative of $g$
with an additional condition:
\begin{equation}
\label{SPdet}
\partial_{\omega} g(\omega_{\ast m}) = 0, \quad \partial_{\omega}^2 g(\omega_{\ast m}) \neq 0.
\end{equation}
After substitution of \eqref{g_det} into \eqref{SPdet} one obtains
\begin{equation}
\label{SPeq}
\frac{dk(\omega_{\ast m})}{d \omega} = V^{-1}
\quad \mbox{or} \quad
v_{\rm gr} (\omega_{\ast m}) = V.
\end{equation}

The contours $\gamma_{\ast m}$ are the contours of steepest growth of ${\rm Im}[g]$.
Consequently, the magnitude of the integrand is significant only in some vicinity of
$\omega_{\ast m}$.

Note that not all the saddle points satisfying the \eqref{SPeq} are passed by saddle point contours obtained by an appropriate deformation of $\gamma_1 + \gamma_2$.

\subsection{Wave component due to a saddle point}

Since the saddle point integration is held over some vicinity of $\omega_{\ast m}$,
$h_j(\omega,k(\omega))$ and $g(\omega)$
can be approximated by a constant $h_j(\omega_{\ast m}, k(\omega_{\ast m}))$
and by a quadratic function
\begin{equation}
\label{gTaylor}
g(\omega) \approx g(\omega_{\ast m}) + \frac{1}{2} \frac{d^2 g (\omega_{\ast m}) }{d \omega^2} (\omega - \omega_{\ast m})^2.
\end{equation}
Using \eqref{g_det} and \eqref{gTaylor},
one can obtain an approximation of the integral $I_{j,\ast m}$:
\begin{equation}
\label{SPint1}
I_{j,\ast m} = h_j(\omega_{\ast m}, k(\omega_{\ast m}))
\sqrt{\frac{2 \pi}{|\alpha| x}}
\exp\{ ik(\omega_{\ast m})x - i \omega_{\ast m} t + \mbox{sign}(\alpha) \cdot i\pi/4\}, \quad \alpha = \partial_{\omega}^2k(\omega_{\ast m}).
\end{equation}

\subsection{Merging of two saddle points near an inflection point.  Airy wave}

Consider an inflection point of the real dispersion diagram, for example
the local minimum of the group velocity (a local maximum is described in
a similar way). Such a point is shown in
\figurename~\ref{DD},~right, as the point $\omega_2^e$.
If $V$ is slightly higher than $v_2'$, there should be two real saddle points
near~$\omega_2^e$. If $V = v_2'$ these two saddle points merge with each other.
An asymptotics of wave field with $V \approx \omega_2^e$ is described by an Airy function.

Approximate function $k(\omega)$ as follows:
\begin{equation}
\label{gTaylor2}
k(\omega) \approx k(\omega^e) + \frac{\omega - \omega^e}{v'_2} - \frac{1}{3}
\alpha (\omega - \omega^e_2)^3,
\quad
\alpha = - \frac{1}{2}\partial_{\omega}^3 k(\omega_2^e),
\quad
\alpha > 0.
\end{equation}

The integral over some vicinity of $\omega^e_2$ can be approximated by an Airy function \cite{Borovikov_1994}, \cite{Brekhovskih_1980}:
\begin{equation}
\label{AiryInt1}
I_j(V, x) \approx h_j(\omega_2^e, k (\omega_2^e))\, \frac{2 \pi}{(x \alpha)^{1/3}} \,
\mbox{Ai}\left( \frac{x^{2/3}}{\alpha^{2/3}} \left(V^{-1} - (v'_2)^{-1}\right) \right) .
\end{equation}

\subsection{An overview of the saddle points  evolution}

Let us list the {\em real\/} roots of \eqref{SPeq} for different $V$:
\begin{itemize}

\item
$V>c_1$: there are no roots. This case is shown by an orange line in \figurename~\ref{SPnot}.
Moreover, for $V>c_1 > c_2$ the wave field is equal to zero identically.
This can be shown by closing the integration contours in the upper half-plane.

\item
$c_1 > V > c_2$: there is a single real root $\omega_{\ast1}$.
The case is shown by red lines in \figurename~\ref{SPnot}.

\item
$c_2 > V > v_1'$: there are two real roots $\omega_{\ast1}$ and $\omega_{\ast4}$
shown by blue lines  in \figurename~\ref{SPnot}.

\item
$v_1' > V > v_2'$: there are four
real roots $\omega_{\ast1}$, $\omega_{\ast2}$, $\omega_{\ast3}$, $\omega_{\ast4}$.
They are shown by magenta lines in \figurename~\ref{SPnot}.

\item
$v_2' > V > 0$: there are two real roots $\omega_{\ast1}$ and $\omega_{\ast2}$. They are  shown
by  green lines in \figurename~\ref{SPnot}.
\end{itemize}

\begin{figure}[!h]
\centering
\includegraphics[scale = 0.8]{SP.pdf}
\caption{Solving \eqref{SPeq} graphically for real $\omega$}
\label{SPnot}
\end{figure}

After deformation of the contour $\gamma_1 + \gamma_2$,
some
steepest descend contours pass through each of the listed saddle points.
There also exists a set of negative saddle points denoted as follows:
\[
\omega_{\ast -n}, \quad n=1,\dots,6, \quad \omega_{\ast -n} = - \omega_{\ast n}.
\]
They should be taken into account in just a similar way.

\begin{figure}[!h]
	\centering
	\begin{subfigure}{0.4\textwidth}
		\includegraphics[width=\textwidth]{w2w3_2.pdf}
		\caption{Vicinity of $V = v_1'$}
	\end{subfigure}
	\vspace{0.1cm}
    \hspace{8ex}
	\begin{subfigure}{0.4\textwidth}
		\includegraphics[width=\textwidth]{w3w4_2.pdf}
		\caption{Vicinity of $V = v_2'$}
	\end{subfigure}
	\caption{Evolution of the saddle points in vicinities of local extrema of the group velocity. Black circles: points of an extremum, green circles: saddle points passed by steepest descend contours, red circles: saddle points not passed by steepest descend contours. The arrows
	correspond to growth of~$V$}
\label{VgrExtr}
\end{figure}

The analysis made above, together with \eqref{SPint1} and
\eqref{AiryInt1}, displays the basic structure of the field for very large~$x$.
The list contains the description of the zones, in which there exists a different number
of asymptotic terms. The formulae for the terms are given by \eqref{SPint1}.
There are also two intermediate zones ($V \approx v_1'$ and $V \approx v_2'$)
described by the Airy asymptotics of the form of \eqref{AiryInt1}.

All the wave components described above do not decay exponentially as $x$ grows.
There exist also saddle points with complex $\omega_{\ast m}$ possessing an exponential
growth. Sometimes, corresponding terms cannot be ignored (if $x$ is not large, decay is
not fast, or the decaying wave is the only wave observed for corresponding $x$ and~$t$).
Let us consider evolution of saddle points position for $V$ changing in some vicinity of $v_1'$. As $V$ increases from $v_1' - \nu$ to $v_1' + \nu$, where $\nu$ is a small real positive number, saddle points $\omega_{\ast 2}$ and $\omega_{\ast 3}$ evolve into two complex saddle points (\figurename~\ref{VgrExtr}~(a)). Among the saddle points for $V > v_1'$, only saddle point
$\omega_{\ast 5}$ is passed by deformed a contour because the initial contour of integration is shifted in the direction of growth of ${\rm Im}[g]$.

The case when $V$ grows near $v_2'$ is analogous. However, for $V < v_2'$ the deformed contour passes through the saddle point $\omega_{\ast 6}$ with negative imaginary part (\figurename~\ref{VgrExtr}~(b)).

Thus, for $v_1'<V$ and for $V<v_2'$ there may exist a complex saddle point, which gives
a decaying saddle point asymptotics.

\section{Zone diagrams} \label{ZDsection}
\subsection{The concept of zone diagram}

The saddle point method does not provide a correct asymptotics of the wave field
if $x$ is not large enough. Each saddle point has a domain of influence (DOI)
introduced in \cite{Borovikov_1994},
which is a spot on the saddle point contour on which the
integrand is not negligibly small. Typically, the size of such a spot in the
$\omega$-plane has size $\sim x^{-1/2}$. if $x$ is not very large, it may happen
that the DOI of two or more saddle points overlap. In this case the saddle point
approximation is not adequate.

An example of such a configuration is the Airy wave mentioned above, which is
a result of merging of two saddle points.
Depending on $V$ and $x$, there can be more sophisticated asymptotics.

The concept of merging of DOI of saddle points is formalized as follows.
 Let $(\omega_{\ast},k_{\ast})$ be a saddle point, and let $\gamma_{\ast}$ be the corresponding steepest descend contour. Introduce the length coordinate $l$ directed along $\gamma_{\ast}$. Let $l = 0$ correspond to the saddle point, and the coordinate direction coincide with the contour direction. The integral along $\gamma_{\ast}$ can be represented as
\begin{equation}
\label{eq1}
I_{\ast}(x) = \int \limits _{-\infty}^{\infty} h^*(l) \exp \{ ix g^*(l) \} \, dl
\end{equation}
where $h^*$ and $g^*$ are restrictions of functions
$h \, d\omega / dl$ and $g$ onto the contour.

The imaginary part of $g^*(l)$ has minimum at $l = 0$, and grows monotonically as $|l|$ grows. The DOI is a segment $[a_1,a_2]$ on the $l$-axis, such that $a_1 < 0 < a_2$, and
\begin{equation}
\label{InflDom}
x({\rm Im}[g^*(a_1)]-{\rm Im}[g^*(0)]) = x({\rm Im}[g^*(a_2)]-{\rm Im}[g^*(0)]) = {S}.
\end{equation}
Here and below,
${S}$ is some arbitrary positive moderately large real number of order 1
(a good choice is ${S} = 3$).
We find convenient to write $a > S$ instead of $a \gg 1$.

The segment $[a_1,a_2]$ is, indeed, just an estimation of the DOI.
One can see that
the size of the DOI depends on $x$, and the domain becomes smaller as $x$ grows.

One needs an Airy representation (\eqref{AiryInt1})
when the DOIs of the saddle points $\omega_{\ast 2}$ and
$\omega_{\ast 3}$  (or $\omega_{\ast 3}$ and $\omega_{\ast 4}$) overlap. Note that the Airy function ${\rm Ai}(z)$ has two asymptotics: for $z \gg 1$ and for $-z \gg 1$. One of these asymptotics correspond to two real saddle points, while another corresponds to a single
complex saddle point.

Evaluate the phase difference for any pair of
{\em neighbouring} saddle points $\omega_{\ast m}$ and $\omega_{\ast n}$:
\begin{equation}
\label{Phase_deff}
\Delta_{m,n} \equiv |(k(\omega_{\ast m})x - \omega_{\ast m} t) -
(k(\omega_{\ast n})x - \omega_{\ast n} t) |
=
x\, |\, g^*(\omega_{\ast m}) -  g^*(\omega_{\ast n}) \,|.
\end{equation}
If $\Delta_{m,n}$ is bigger than ${S}$ then
$\omega_{\ast m}$ and $\omega_{\ast n}$ can be considered as separate saddle points.
If  $\Delta_{m,n} < {S}$
then one should introduce some more sophisticated asymptotics.

Let us say when we consider saddle points $\omega_{\ast m}$ and $\omega_{\ast n}$ being neighboring. This is when there exists a path between $\omega_{\ast m}$ and $\omega_{\ast n}$ along the Riemann surface of $k(\omega)$, on which the phase $xg(\omega)$
does not change strongly (for more than ${S}$).

The asymptotics of a waveguide can be organized in {\em a zone diagram}.
We plot the zone diagram in the coordinates $(t,V) = (t , x/t)$,
The diagram is branched, i.~e.\ over each point $(t, V)$ there may exist several sheets
of the zone diagram. Each sheet corresponds to a separate asymptotic term.

There are parent / child relations between the asymptotics on the diagram. The
``children'' are obtained by taking some asymptotics of the ``parents''.

\subsection{A simple example: a zone diagram for a scalar Klein--Gordon equation}

Consider a scalar inhomogeneous Klein--Gordon equation
\begin{equation}
(c^2 \ptl_x^2 - \ptl_t^2 - \Omega^2) u(t, x) = \delta(t) \, \delta(x).
\label{SKGE}
\end{equation}
A causal solution of this equation can be easily obtained:
\begin{equation}
u(t, x) = \left\{ \begin{array}{ll}
-(2c)^{-1} J_0 (\Omega \sqrt{t^2 - x^2 / c^2}) & t > x /c \\
0                                              & 0 < t < x /c
\end{array} \right.
\label{SKGEsol}
\end{equation}
where $J_0$ is the Bessel function.

One can see that there are two asymptotics of the solution:
if the argument of the Bessel function is much greater than~1 (we prefer to say that it is greater than
${S}$), then one can represent the Bessel function in terms of two exponential functions:
\begin{equation}
u(t, x) \approx -\frac{1}{2c \sqrt{2 \pi z}}
(\exp \{i z - i \pi / 4 \} + \exp \{-i z + i \pi / 4 \}) ,
\quad
z = \Omega \sqrt{t^2 - x^2 / c^2},
\quad
z > {S}.
\label{SKGEas1}
\end{equation}
This is the ``far field'' for the waveguide.

If the argument is small, there is another asymptotics:
\begin{equation}
u(t, x) \approx - (2c)^{-1},
\quad
z < {S}^{-1}.
\label{SKGEas2}
\end{equation}
This is the ``near field''.

A zone diagram for the scalar Klein--Gordon equation is plot in
\figurename~\ref{SKGEdiagram}. The left part of the figure is the diagram projected onto
the $(t, V)$-plane, and the left part shows the branching of the diagram.
One can see that the zone of the diagram labeled as ``far field'' is represented by two sheets.
These sheets denote two terms of \eqref{SKGEas1}. Indeed, these terms can be treated as
the saddle point contributions of a well-known integral representation
for the Bessel function.

\begin{figure}[!h]
	\centering
	\begin{subfigure}{0.4\textwidth}
		\includegraphics[width=0.8 \textwidth]{figB5201-eps-converted-to}
		\caption{Zone diagram in the $(t, V)$-plane}
	\end{subfigure}
	\vspace{0.1cm}
    \hspace{8ex}
	\begin{subfigure}{0.4\textwidth}
		\includegraphics[width=0.8 \textwidth]{figB5202-eps-converted-to}
		\caption{Branching of the diagram}
	\end{subfigure}
	\caption{Zone diagram for the scalar Klein--Gordon equation}
\label{SKGEdiagram}
\end{figure}

The central part of the diagram
denoted by $J_0$ is the domain where one cannot use any asymptotics, and should use the
initial expression given by \eqref{SKGEsol}.
In the ``near field'' zone the field is the constant given by \eqref{SKGEas2}.

The green arrows show the parent / child relations between the zones.

\subsection{Zone diagram for 2D matrix Klein--Gordon equation}

Building of a zone diagram for the matrix Klein--Gordon equation (\eqref{MKGE}) is a  sophisticated task. The reason is that there are four saddle points
($\omega_{\ast 1} , \dots , \omega_{\ast 4}$) that can merge.
As a result, the zone diagram becomes complicated. Here we do not build the
complete zone diagram, namely, we don't show the near field zones.

A sketch of the boundaries between the zones is shown in \figurename~\ref{ZD1}.
Corresponding DOIs overlap if the point $(t , V)$ is located to the left of a boundary.

\begin{figure}[!h]
\centering
\includegraphics[scale = 1.1]{ZD1-eps-converted-to}
\caption{Boundaries between zones for \eqref{MKGE}}
\label{ZD1}
\end{figure}

The central part of the zone diagram is shown in more details
in \figurename~\ref{ZD2}.
The letter notations describe the asymptotic terms that exist
in corresponding zones.

Note that for the sake of compact notations, only the terms produced by
saddle points located in the right half-plane of $\omega$ are indicated.
In a complete description of terms, our notations should be doubled.

Let us list the types of the asymptotics.

\begin{figure}[!h]
\centering
\includegraphics[scale = 1]{ZD2-eps-converted-to}
\caption{Central part of the zone diagram}
\label{ZD2}
\end{figure}

{\bf SP}: A term due to a real saddle point (see \eqref{SPint1}).

{\bf SPe}: A term due to a saddle point with complex $\omega_{\ast m}$. Such a term has
a structure close to  \eqref{SPint1}, but the exponential factors decays as $x$ grows.

{\bf Ai}: A term having Airy-type asymptotics of the form (\eqref{AiryInt1}).
The diagram indicates that the ``Ai'' term degenerates into
two ``SP'' terms or into a single ``SPe'' term.

{\bf J}: Such a term can be observed in the zone where the DOIs of the saddle points
$\omega_{\ast 1}$ and $\omega_{\ast 3}$ merge, while the saddle points
$\omega_{\ast 2}$ and $\omega_{\ast 4}$ are isolated.
\figurename~\ref{cont_J}
shows the integration contour $\gamma_1 + \gamma_2$
after an appropriate deformation
corresponding to this point. Bold parts of the contours show the points of the complex plane
$\omega$, where the magnitude of the integrand is not exponentially small. The points
$\omega_{\ast 2}$ and $\omega_{\ast 4}$ produce standard saddle point terms.
Another significant part of the integral is computed  along the contours encircling the cuts. This term is the J asymptotics.
We should note that this term
describes the exchange pulse \cite{Shanin_2018}.

\begin{figure}[!h]
\centering
\includegraphics[scale = 1]{cont_J-eps-converted-to}
\caption{The deformed contour of integration for the points
$(t, V)$, for which the asymptotics J exists}
\label{cont_J}
\end{figure}

Let us compute the J-type asymptotics of the field.
Introduce new coordinates
\begin{equation}
\label{new_var}
\omega' = \omega - \omega_{\rm sh}, \quad k' = k - k_{\rm sh}.
\end{equation}
We assume that
\[
\frac{k'}{k_{\rm sh}} \sim \frac{\omega'}{\omega_{\rm sh}} << 1.
\]
Applying the perturbation method one can obtain the following representation of the dispersion function $D(\omega,k)$ (\eqref{D_det}):
\begin{equation}
\label{D_as}
D \approx 4c_1^2 c_2^2 k_{\rm sh}^2 \left( k' - \frac{\omega'}{v_1} \right)\left( k' - \frac{\omega'}{v_2} \right) - \mu^2.
\end{equation}
Substituting \eqref{D_as} into \eqref{DE} one can obtain the following representation of the function $k(\omega)$:
\begin{equation}
\label{k_asympt}
k(\omega) \approx k_{\rm sh} + \frac{v_1^{-1} + v_2^{-1}}{2}\omega' \pm \sqrt{\frac{(v_2^{-1} - v_1^{-1})^2 (\omega')^2}{4} + \frac{\mu^2}{4c_1^2 c_2^2 k_{\rm sh}^2}}.
\end{equation}
Introduce a stretched variable
\begin{equation}
\label{coord}
\tau = \xi \omega' ,
\end{equation}
where
\[
\xi = \frac{c_1 c_2 k_{\rm sh} (v_2^{-1} - v_1^{-1})}{\mu}.
\]
The
J-asymptotics can be expressed as an integral
\begin{equation}
\label{J_int}
I_j \approx -\frac{i \pi}{2 c_1^2 c_2^2 k_{\rm sh}^2 (v_2^{-1} - v_1^{-1})}
A_j(\omega_{\rm sh},k_{\rm sh})
\,
\exp\{ ik_{\rm sh}x -i \omega_{\rm sh} t  \}
\,\times
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
\end{equation}
\[
\qquad \qquad \qquad \qquad \qquad
\int \limits_{\sigma}
\exp\left\{
-i\frac{\tau}{\xi} \left( t - x\frac{ (v_1^{-1} + v_2^{-1})}{2} \right) +
i  \frac{x\, \mu \sqrt{1+\tau^2}}{2 c_1 c_2 k_{\rm sh}}
\right\}
\frac{d\tau}{\sqrt{1 + \tau^2}}
,
\]
where $\sigma$ is the contour encircling the cut $[-i,i]$ in the positive direction.
This integral can be expressed in terms of Bessel function $J_0$:
\begin{equation}
\label{Jasympt}
I_j = -\frac{\pi^2}{c_1^2c_2^2 k_{\rm sh}^2
(v_2^{-1} - v_1^{-1}) }
A_j(\omega_{\rm sh},k_{\rm sh}) \exp\{ik_{\rm sh}x -i \omega_{\rm sh} t \}
J_0(b),
\end{equation}
\[
b = \frac{ \mu \sqrt{(t - x / v_1)(x / v_2 - t)}}{c_1 c_2 k_{\rm sh} (v_2^{-1} - v_1^{-1})}
\]

{\bf Q}: This asymptotics is valid when the DOIs of the saddle points
$\omega_{\ast 1}$, $\omega_{\ast 2}$, and $\omega_{\ast 3}$
overlap (or when the DOIs
of $\omega_{\ast 1}$, $\omega_{\ast 3}$, and $\omega_{\ast 4}$ overlap). Although we don't present an explicit expression for the Q-asymptotics here, we describe a way to obtain it.

In \figurename~\ref{cont_Q} we plot deformed contours of integration for
the lower domain with the Q-asymptotics
when the DOIs of $\omega_{\ast 1}$, $\omega_{\ast 3}$, and $\omega_{\ast 4}$ overlap,
and $\omega_{\rm ast 2}$ is isolated.
The bold green contour on Sheet~2 going along the cut corresponds to the Q-asymptotics.

\begin{figure}[!h]
\centering
\includegraphics[scale = 1]{cont_Q-eps-converted-to}
\caption{The deformed contour of integration for the points
$(t, V)$ with the asymptotics Q}
\label{cont_Q}
\end{figure}

Some detailed consideration shows that
an
 approximation of the field corresponding to the Q-asymptotics
  can be expressed through a new special function
\begin{equation}
\label{asymptQ}
Q(\beta,z) =  \int \limits_{\Gamma} \frac{1}{\sqrt{1 + \tau^2}}
\exp\left\{i (\sqrt{1 + \tau^2} + z\tau + \beta \tau^2)\right\} d \tau,
\end{equation}
where $\beta$, and $z$ are some constants, which depend on the waveguide parameters,
contour $\Gamma$ is shown in \figurename~\ref{cont_gamma_Q}.
\begin{figure}[!h]
\centering
\includegraphics[scale = 0.9]{cont_gamma_Q-eps-converted-to}
\caption{Contour $\Gamma$ for defenition of $Q(\beta,z)$}
\label{cont_gamma_Q}
\end{figure}

{\bf B}: In this case the DOIs of all saddle points overlap. This zone is the parent for all, and
no simplification of the integral
\eqref{Ser_int_repr} can be made.

\section{Conclusion}

The saddle point method allows one to obtain asymptotic
estimations of the field in the far zone.
This zone is the set of points $(t, x)$,
such  that the
domains of influence of different saddle points do not overlap.

If the domains of influences of different saddle points overlap, one has to build new asymptotics. A parent / child relation between the new and old asymptotics
can be established.

To visualize all the obtained asymptotics, we propose a zone diagram.
This is a branched diagram in the $(t, V)$-plane.

In this work we built such set of asymptotics for a system described by \eqref{MKGE}.
A formula for the J-asymptotics is derived.
A sketch of obtaining a Q-asymptotics is given.

\section{Acknowledgements}
The work is supported by the RFBR grant 19-29-06048.

\bibliography{biblio}
\bibliographystyle{ieeetr}

\end{document}


TITLE: LIGO: Importance of photons
QUESTION [1 upvotes]: In LIGO, a photon beam travels in two perpendicular direction and time taken by each beam is noted. Non zero time difference is a signature of GWs here. What if I use electron beam travelling at velocity (say, 0.8c) instead of photons? Will I be able to detect GWs?
I think this will not work since electrons dont travel along null geodesics. But I am not sure.

REPLY [2 votes]: In LIGO, a photon beam travels in two perpendicular direction and time taken by each beam is noted. Non zero time difference is a signature of GWs here.

Not correct, see this description  and this video, at about 27', of the web cast . It is the interference pattern, the phase difference, that generates the signal.

Simplified operation of a gravitational wave observatory
Figure 1: A beamsplitter (green line) splits coherent light (from the white box) into two beams which reflect off the mirrors (cyan oblongs); only one outgoing and reflected beam in each arm is shown, and separated for clarity. The reflected beams recombine and an interference pattern is detected (purple circle).
Figure 2: A gravitational wave passing over the left arm (yellow) changes its length and thus the interference pattern.

The LIGO set up is made so that phases at the light detector cancel out (see video at 27'), no light arrives,  unless the relative distances of the two arms change, then an interference pattern  will appear in the detector.
Electrons do not build up a space wave the same way photons build an electromagnetic wave, due to the charges introducing repulsive forces and magnetic fields attractive, the probability wave describing an electron beam would need a density matrix, not a coherent sinusoidal beam with phases. Thus this experiment could not be done with electrons with any velocity.


TITLE: Geometric intersection with incompressible surfaces
QUESTION [13 upvotes]: Let $M$ be a oriented compact $3$-manifold, closed or with boundary.
For any incompressible surface $F$, define a function $i_F$ on the set of homotopy classes of closed curves in $M$ by $$i_F (\alpha) = \alpha \cap F $$ the geometric intersection number of $\alpha$ with $F$.
Is it true that two incompressible, $\partial$-incompressible surfaces $F$ and $F'$ are isotopic if $i_F =i_{F'}$?

REPLY [11 votes]: Yes, this is true (with an appropriate definition of $i_F(\alpha)$).
An incompressible surface gives rise to an action of $\pi_1(M)$ on a tree (see for example Chapter 1 of Shalen's notes).
For a homotopy class of loops $\alpha$, define $i_F(\alpha)=\|\alpha\|$ to be the translation length of $\alpha$ acting on this tree (well-defined up to conjugacy). In fact, $\alpha$ can be homotoped so that the geometric intersection number is $\|\alpha\|$. Then this defines a length function on conjugacy classes in $\pi_1(M)$, satisfying the Culler-Morgan axioms (1.11):

Culler and Morgan show that their axioms uniquely characterize a minimal action of a group on a tree, if the action is semi-simple. It's not hard to see that if the surface is not a fiber or semi-fiber, that the action is semi-simple, essentially from the cocompactness of the action. In the fiber or semi-fiber case, the intersection number determines a homomorphism to $\mathbb{Z}$ or $D_\infty$, whose kernel is finitely generated and determines the surface group. In the semifibered case, I don't think $I_F(\alpha)$ distinguishes the two non-orientable surfaces. However, your assumption of incompressible surface usually means the surface is orientable, and thus this case won't appear.
Thus, suppose two surfaces induce the same length function. Then they induce the same action on a tree, and thus the edge stabilizer will be the same, implying that the surfaces are homotopic. But homotopic incompressible surfaces are isotopic, by a result of Waldhausen (Corollary 5.5; Waldhausen assumed that the manifold is irreducible, but I believe this was only to avoid the pitfall of a potential counterexample to the Poincaré conjecture as a connect summand).


TITLE: sum of two random variables
QUESTION [1 upvotes]: Can any of you help me?
I have some problem with this exercise of "Probability and Statistics" :

Calculate the probability density function (PDF) of $Z=X+Y$
where $Y$ is discrete random variable which is be equal to $-1,1$ with equal probability;
$X$ is standard Gaussian random variable independent from $Y$.

I know that the PDF of sum of two continuous independent variables is given by the convolution of the marginal PDF
$f_z(z)=f_x*f_y$
but if one on the two variable is discrete what should I do?
thanks all!

REPLY [2 votes]: If one variable is discrete, you can use the Dirac delta function to preform the integration:
$$P(X=x) = \frac{1}{2}\delta(x-1) + \frac{1}{2}\delta(x+1)$$
Down-votes from purist mathematicians in 3.. 2.. 1..


TITLE: Circle $\Omega$ is tangent to circles $\omega_1$ and $\omega_2$ at points $A_1$ and $A_2$, respectively. Let $B$ be a point on $\Omega$
QUESTION [1 upvotes]: Circle $\Omega$ is tangent to circles $\omega_1$ and $\omega_2$ at points $A_1$ and $A_2$, respectively.  Let $B$ be a point on $\Omega$, let $K_1$ and $K_2$  be the other intersection points of lines $A_1B$ and $A_2B$ with $\omega_1$ and $\omega_2$, respectively. Prove that if $K_1K_2$ is tangent to $\omega_1$, then it is also tangent to $\omega_2$.
Please help with this problem, I tried using directed angles but made no progress at all

REPLY [0 votes]: Suppose $\omega_1$ touches $K_1K_2$.
Homothety at $A_1$ which takes $\omega _1$ to $\Omega$ takes $K_1$ to $B$ and $K_1K_2$ to (parallel) tangent $t$ to $\Omega$ at $B$. Homothety at $A_2$ which takes $\Omega $ to $\omega _2$ takes $B$ to $K_2$ and $t$ to (parallel) tangent $\ell$ to $\omega _2$ at $K_2$. Since $\ell|| K_1K_2$ we have $\ell = K_1K_2$. Q.E.D.


TITLE: Plane passing through $(0,0,0)$
QUESTION [0 upvotes]: Let $P$ be a plane passing through $(0,0,0)$. Show that there is an invertible matrix $M$ such that
$P=\left\{M\begin{bmatrix} x \\ y \\ 0 \end{bmatrix} \mid x,y \in  \mathbb{R} \right\}$.
My attempt: The equation of the plane $P$ is $ax+by+cz+d=0$. Since $P$ is passing through $(0,0,0)$, we have $0a+0b+0c+d=0$, then $d=0$. Thus $\dots$ [incorrect here]

REPLY [3 votes]: $P$ is spanned by two linearly independent vectors $u_1, u_2 \in \mathbb{R}^3$. We can extend this to base $[u_1, u_2, b]$ of $\mathbb{R}^3$. $M$ is the matrix with those vectors as columns.