text:
\begin{document}
\maketitle

\begin{abstract}
Common meadows are fields expanded with a total inverse function.
Division by zero produces an
additional value denoted with $\bot$ that propagates through all operations
of the meadow signature (this additional value can be interpreted as an error element).
We provide a basis theorem for so-called common cancellation meadows of characteristic zero,
that is, common meadows of characteristic zero that admit a certain cancellation law.
\\[5mm]
\emph{Keywords and phrases:}
Meadow, common meadow, division by zero, additional value, abstract datatype.
\end{abstract}

{\small\tableofcontents}

\section{Introduction}
\label{sec:Intro}
Elementary mathematics is uniformly taught around the world with a focus on
natural numbers, integers, fractions, and fraction calculation.
The mathematical basis of that part of mathematics seems to reside in the field of
rational numbers.
In elementary teaching material the incorporation of rational numbers in a field is
usually not made explicit.
This leaves open the possibility that some other abstract datatype or some
alternative abstract datatype specification improves upon fields in providing a
setting in which such parts of elementary mathematics can be formalized.

In this paper we will propose the signature for | and model class of |
\emph{common meadows}
and we will provide a \emph{loose} algebraic specification of common meadows
by way of a set of equations.
In the terminology of Broy and Wirsing~\cite{BW81,Wir90}, the semantics of a loose
algebraic specification $S$ is given by the class of all models of $S$, that is, the
semantic approach is not restricted to the isomorphism class of initial algebras.
For a loose specification it is expected that its initial algebra is an important
member of its model class, worth of independent investigation.
In the case of common meadows this aspect is discussed in the last remark
of Section~\ref{sec:4} (Concluding remarks).

A common meadow (using inversive notation) is an extension of a field equipped with a
multiplicative inverse function $(...)^{\inverse 1}$ and an additional element
$\bot$ that serves as the inverse of zero and propagates through all operations.
It should be noticed that the use of the constant $\bot$ is a matter of convenience
only because it merely constitutes a derived constant with defining equation
$\bot = 0^{-1}$.
This implies that all uses of $\bot$ can be removed from the story of common meadows
(a further comment on this can be found in Section~\ref{sec:4}).

The inverse function of a common meadow is not an involution because
$(0^{\inverse 1})^{\inverse 1}=\bot$.
We will refer to meadows with zero-totalized inverse, that is, $0^{\inverse 1}=0$,
as \emph{involutive meadows} because inverse becomes an involution.
By default a ``meadow'' is assumed to be an involutive meadow.

The key distinction between meadows and fields, which we consider to be so
important that it justifies a different name, is the presence of an operator symbol
for inverse in the signature (inversive notation, see~\cite{BM2011a}) or for
division (divisive notation, see~\cite{BM2011a}),
where divisive notation $x/y$ is defined as $x\cdot y^{\inverse 1}$.
A major consequence is that fractions can be viewed as terms over the
signature of (common) meadows.
Another distinction between meadows and fields is that we do not require a meadow
to satisfy the separation axiom $0\ne 1$.

The paper is structured as follows: below we conclude this section
with a brief introduction to some aspects of involutive meadows that will play
a role later on, and a discussion on why common meadows can be preferred over
involutive meadows. In Section~\ref{sec:2} we formally define common meadows and
present some elementary results. In Section~\ref{sec:3} we define ``common
cancellation meadows'' and provide a basis theorem for common cancellation meadows
of characteristic zero, which we consider our main result. Section~\ref{sec:4}
contains some concluding remarks.

\subsection{Common Meadows versus Involutive Meadows}
\label{subsec:versus}
Involutive meadows, where instead of choosing $1/0 = \bot$, one calculates with
$1/0 = 0$,
constitute a different solution to the question how to deal with the value of
$1/0$  once the design decision has been made to work with the signature of meadows,
that is to include a function name for inverse or for division (or both) in
an extension of the syntax of fields.
Involutive meadows feature a definite advantage over
common meadows in that, by avoiding an extension of the domain with an additional
value, theoretical work is very close to classical algebra of fields.
This conservation property, conserving the domain, of involutive meadows
has proven helpful for the development of theory about
involutive meadows in~\cite{BBP2013a,BBP2013,BP2013,BM2011a,BR10,BT07}.
Earlier and comparable work on the equational
theory of fields was done by Komori~\cite{K75} and Ono~\cite{Ono83}: in 1975,
Komori introduced the name \emph{desirable pseudo-field} for what was introduced
as a ``meadow'' in~\cite{BT07}.\footnote{\cite{BT07} was published in 2007; the
  finding of~\cite{K75,Ono83} is mentioned
  in~\cite{BM2011a} (2011) and was found via Ono's 1983-paper~\cite{Ono83}.}

An equational axiomatization $\Md$ of involutive meadows is given in
Table~\ref{tab:Md}, where $^{\inverse 1}$ binds stronger than $\cdot$, which in
turn binds stronger than $+$.
From the axioms in $\Md$ the following identities are derivable:
\begin{align*}
0\cdot x  &= 0,
&0^{-1}  &= 0,\\
x\cdot (-y) &= -(x\cdot y),
&(-x)^{-1} &= -(x^{-1}),\\
-(-x) &= x,&(x \cdot  y)^{-1}
&= x^{-1} \cdot  y^{-1}.
\end{align*}

\begin{table}[t]
\centering
\hrule
\begin{align*}
(x+y)+z &= x + (y + z)
&x \cdot  y &= y \cdot  x\\
x+y &= y+x
&1\cdot x &= x\\
x+0 &= x
&x \cdot  (y + z) &= x \cdot  y + x \cdot  z\\
x+(-x) &= 0
&(x^{-1})^{-1} &= x \\
(x \cdot y) \cdot  z &= x \cdot  (y \cdot  z)
&x \cdot (x \cdot x^{-1}) &= x
\end{align*}
\hrule
\caption{The set $\Md$ of axioms for (involutive) meadows}
\label{tab:Md}
\end{table}

Involutive \emph{cancellation meadows} are involutive
meadows in which the following cancellation law holds:
\begin{equation}
\label{eq:CL}
\tag{$\CL$}
(x\ne 0 \wedge x\cdot y =x\cdot z) \rightarrow y=z.
\end{equation}
Involutive cancellation meadows form an important subclass of involutive meadows:
in~\cite[Thm.3.1]{BBP2013} it is shown that the axioms in
Table~\ref{tab:Md} constitute a complete axiomatization of the equational theory
of involutive cancellation meadows.
We will use a consequence of this result in Section~\ref{sec:3}.

A definite disadvantage of involutive meadows against common meadows is that
$1/0 = 0$ is quite remote from common intuitions regarding the partiality of
division.

\subsection{Motivating a Preference for Common Meadows}
Whether common meadows are to be preferred over involutive meadows
depends on the applications one may have in mind.
We envisage as an application area the development of alternative foundations of
elementary mathematics from a perspective of abstract datatypes, term rewriting,
and mathematical logic.
For that objective we consider common meadows to be the preferred option over
involutive meadows.
At the same time it can be acknowledged
that a systematic investigation of involutive meadows constitutes a necessary
stage in the development of a theory of common meadows by facilitating in a
simplified setting the determination of results which might be obtained about
common meadows. Indeed each result about involutive meadows seems to suggest a
(properly adapted) counterpart in the setting of common meadows, while proving or
disproving such counterparts is not an obvious matter.

\section{Common Meadows}
\label{sec:2}
In this section we formally define ``common meadows'' by fixing their signature
and providing an equational axiomatization. Then, we consider some  conditional
equations that follow from this axiomatization.
Finally, we discuss some conditional laws that can be used to define
an important subclass of common meadows.

\subsection{Meadow Signatures}
The signature $\Sigma_{f}^S$ of fields (and rings) contains a sort (domain) $S$,
two constants 0, and 1, two two-place functions $+$ (addition) and $\cdot$
(multiplication) and the one-place function $-$  (minus) for the inverse of addition.

We write $\Sigma_{\md}^S$ for the signature of meadows in inversive notation:
\[\Sigma_{\md}^S = \Sigma_f^S \cup \{\_^{-1}\colon S \rightarrow S\},\]
and we write $\Sigma_{\md,\bot}^S$ for the signature of meadows in inversive notation
with an $\bot$-totalized inverse operator:
\[\Sigma_{\md,\bot}^S = \Sigma_{\md}^S \cup \{\bot \colon S\}.\]
The interpretation of $\bot$ is called the additional value and we write $\bota$
for this value.
Application of any function to the additional value returns that same value.

When the name of the carrier is fixed it need not be mentioned explicitly in a
signature.
Thus, with this convention in mind, $\Sigma_{\md}$ represents $\Sigma_{\md}^S$
and so on.
If we want to make explicit that we consider terms over some signature
$\Sigma$ with variables in set $X$, we write $\Sigma(X)$.

Given a field several meadow signatures and meadows can be connected with it.
This will now be exemplified with the field $\Rat$ of rational numbers.
The following meadows are distinguished in this case:
\begin{description}
\item[$\Rat_0,$] the meadow of rational numbers with zero-totalized inverse:
$\Sigma(\Rat_0) = \Sigma_{\md}^\Rat$.

\item[$\Rat_\bot,$] the meadow of rational numbers with $\bot$-totalized inverse:
$\Sigma(\Rat_\bot) = \Sigma_{\md,\bot}^{\Rat_a}$.
The additional value $\bota$ interpreting $\bot$ has been taken outside $|\Rat|$
so that $|\Rat_{\bota}| = |\Rat| \cup \{\bota\}$.
\end{description}

\subsection{Axioms for Common Meadows}

\begin{table}[t]
\centering
\hrule
\begin{align}
\label{eq:1}
(x+y)+z &= x + (y + z)\\
\label{eq:2}
x+y     &= y+x\\
\label{eq:3}
x+0     &= x\\
\label{eq:4}
x+(-x)  &= 0 \cdot x \\
\label{eq:5}
(x \cdot y) \cdot  z &= x \cdot  (y \cdot  z)\\
\label{eq:6}
x \cdot  y &= y \cdot  x\\
\label{eq:7}
1\cdot x &= x \\
\label{eq:8}
x \cdot  (y + z) &= x \cdot  y + x \cdot  z\\[0mm]
\label{eq:9}
-(-x) &= x\\
\label{eq:10}
0 \cdot (x \cdot x) &= 0 \cdot x \\[0mm]
\label{eq:11}
(x^{\inverse 1})^{\inverse 1} &= x + 0 \cdot x^{\inverse 1} \\
\label{eq:12}
x \cdot x^{\inverse 1} &= 1  + 0 \cdot x^{\inverse 1}\\
\label{eq:13}
(x \cdot y)^{\inverse 1} &= x^{\inverse  1} \cdot y^{\inverse 1}\\
\label{eq:14}
1^{\inverse 1}&=1\\[0mm]
\label{eq:15}
0^{\inverse 1} &= \bot \\
\label{eq:16}
x + \bot &= \bot\\
\label{eq:17}
x \cdot \bot &= \bot
\end{align}
\hrule
\caption{$\Md_\bot$, a set of axioms for common meadows}
\label{Mda}
\end{table}

The axioms in Table~\ref{Mda} define the class (variety) of \emph{common meadows},
where we adopt the convention that $\_^{\inverse 1}$ binds stronger than $\cdot$,
which in turn binds stronger than $+$.
Some comments: Axioms~$\eqref{eq:15}-\eqref{eq:17}$ take care of $\bot$'s propagation
through all operations, and for the same reason, axioms~\eqref{eq:11} and
\eqref{eq:12} have their particular form. Axiom~\eqref{eq:4} is a variant of the
common axiom on additional inverse, which also serves $\bot$'s propagation.
Axioms~\eqref{eq:13} and \eqref{eq:14} are further
identities needed for manipulation of $(...)^{-1}$-expressions.
Finally, axiom~\eqref{eq:10} is needed to reason with expressions of the form
$0\cdot t$.

The following proposition provides some typical identities for common meadows.

\begin{proposition}
\label{prop:1}
Equations that follow from $\Md_{\bot}$ (see Table~\ref{Mda}):
\begin{align}
\label{e1}\tag{e1}
0\cdot 0&=0,
\\[.4mm]
\label{e2}\tag{e2}
-0&=0,
\\[.4mm]
\label{e3}\tag{e3}
0\cdot x&=0\cdot (-x),
\\[.4mm]
\label{e4}\tag{e4}
0\cdot (x\cdot y)&=0\cdot (x+y),
\\[.4mm]
\label{e5}\tag{e5}
-(x\cdot y)&=x\cdot (-y),
\\[.4mm]
\label{e6}\tag{e6}
(-1)\cdot x&=-x,
\\[.4mm]
\label{e7}\tag{e7}
(-x)^{\inverse 1}&=-(x^{\inverse 1}),
\\[.4mm]
\label{e8}\tag{e8}
(x\cdot x^{\inverse 1})\cdot x^{\inverse 1}&=x^{\inverse 1}
\\[.4mm]
\label{e9}\tag{e9}
-\bot&=\bot,
\\[.4mm]
\label{e10}\tag{e10}
\bot^{\inverse 1}&=\bot.
\end{align}
\end{proposition}

\begin{proof}
Most derivations are trivial.
\begin{description}
\item[$\eqref{e1}.$]
By axioms~\eqref{eq:3}, \eqref{eq:7}, \eqref{eq:8}, \eqref{eq:2} we find
$x=(1+0)\cdot x=x+0\cdot x=0\cdot x + x$, hence
$0=0\cdot 0+0$, so by axiom~\eqref{eq:3}, $0=0\cdot 0$.

\item[$\eqref{e2}.$]
By axioms~\eqref{eq:3}, \eqref{eq:2}, \eqref{eq:4} and \eqref{e1} we find
$-0=(-0)+0=0+(-0)=0\cdot 0=0$.

\item[$\eqref{e3}.$]
By axioms~\eqref{eq:2}, \eqref{eq:4}, \eqref{eq:9} we find
$0\cdot x=x+(-x)=(-x)+-(-x)=0\cdot (-x)$.

\item[$\eqref{e4}.$]
First note $0\cdot x+0\cdot x=(0+0)\cdot x = 0\cdot x$.
By axioms~$\eqref{eq:2}-\eqref{eq:4}$, \eqref{eq:6}, \eqref{eq:8}, \eqref{eq:10}
we find
$0\cdot (x+y)=0\cdot ((x+y)\cdot(x+y))
=(0\cdot x+0\cdot(x\cdot y))+(0\cdot y+0\cdot(x\cdot y))
=(0+ 0\cdot y)\cdot x+(0+0\cdot x)\cdot y=0\cdot(x\cdot y)+0\cdot(x\cdot y)=
0\cdot(x\cdot y)$.

\item[$\eqref{e5}.$] We give a detailed derivation:
\begin{align*}
-(x\cdot y)&=-(x\cdot y)+0\cdot -(x\cdot y)
&&\text{by $x=x+0\cdot x$}\\
&=-(x\cdot y)+0\cdot(x\cdot y)
&&\text{by \eqref{e3}}\\
&=-(x\cdot y)+x\cdot(0\cdot y)
&&\text{by axioms \eqref{eq:5} and \eqref{eq:6}}\\
&=-(x\cdot y)+x\cdot(y+(-y))
&&\text{by axiom \eqref{eq:4}}\\
&=-(x\cdot y)+(x\cdot y+x\cdot(-y))
&&\text{by axiom \eqref{eq:8}}\\
&=(-(x\cdot y)+x\cdot y)+x\cdot(-y)
&&\text{by axiom \eqref{eq:1}}\\
&=0\cdot(x\cdot y)+x\cdot(-y)
&&\text{by axioms \eqref{eq:2} and \eqref{eq:4}}\\
&=0\cdot(x\cdot -y)+x\cdot(-y)
&&\text{by axioms \eqref{eq:6} and \eqref{eq:5}, and \eqref{e3}}\\
&=x\cdot(-y).
&&\text{by $x=0\cdot x+x$}
\end{align*}
Thus, with axiom~\eqref{eq:9} it follows that $(-x)\cdot(-y)=x\cdot y$.

\item[$\eqref{e6}.$]
From \eqref{e5} with $y=1$ we find $-x=-(x\cdot 1)=x\cdot (-1)=(-1)\cdot x$.

\item[$\eqref{e7}.$]
By axiom~\eqref{eq:12}, $(-1)\cdot (-1)^{\inverse 1}=1+0\cdot(-1)^{\inverse 1}$,
hence $(-1)^{\inverse 1}=(-1)+0\cdot (-1)\cdot(-1)^{\inverse 1}
=(-1)+0\cdot (-1)^{\inverse 1}$.
Now derive $1=((-1)\cdot (-1))^{\inverse 1}=(-1)^{\inverse 1}\cdot (-1)^{\inverse 1}
=(-1)^{\inverse 1}\cdot((-1)+0\cdot (-1)^{\inverse 1})
=(-1)\cdot(-1)^{\inverse 1}+0\cdot(-1)^{\inverse 1}\cdot (-1)^{\inverse 1}
=(-1)\cdot(-1)^{\inverse 1}+0\cdot(1)^{\inverse 1}=(-1)\cdot(-1)^{\inverse 1}$, thus
$(-1)^{\inverse 1}=-1$.
Hence, $(-x)^{\inverse 1}=(-1\cdot x)^{\inverse 1}
=(-1)^{\inverse 1}\cdot x^{\inverse 1}=(-1)\cdot x^{\inverse 1}=-(x^{\inverse 1})$.

\item[$\eqref{e8}.$]
By axioms~\eqref{eq:12} and~\eqref{eq:10}, $(x\cdot x^{\inverse 1})\cdot x^{\inverse 1}=
(1+0\cdot x^{\inverse 1})\cdot x^{\inverse 1}=x^{\inverse 1}+0\cdot x^{\inverse 1}
=x^{\inverse 1}$.

\item[$\eqref{e9}.$]
By~\eqref{e5} and axioms~\eqref{eq:6} and \eqref{eq:17}, $-\bot=-(\bot\cdot 1)
=\bot\cdot(-1)=\bot$.

\item[$\eqref{e10}.$]
By axioms~\eqref{eq:11} and $\eqref{eq:15} - \eqref{eq:17}$,
$\bot^{\inverse 1}=(0^{\inverse 1})^{\inverse 1}=0+0\cdot \bot=\bot$.
\end{description}
\end{proof}

The next proposition establishes a generalization of a familiar identity concerning
the addition of fractions.

\begin{proposition}
\label{prop:2a}
$\Mda\vdash \displaystyle x \cdot y^{\inverse  1}  + u \cdot v^{\inverse  1}
= (x \cdot v + u \cdot y) \cdot (y \cdot v)^{\inverse  1}$.
\end{proposition}

\begin{proof} We first derive
\begin{align}
\nonumber
x \cdot y \cdot y^{\inverse  1}
&=x\cdot(1+0\cdot y^{\inverse  1})
&&\text{by axiom~\eqref{eq:12}}\\
\nonumber
&=x+0\cdot x\cdot y^{\inverse  1}\\
\nonumber
&=x+0\cdot x+0\cdot y^{\inverse  1}
&&\text{by~\eqref{e4}}\\
\label{eq:cv}
&=x+0\cdot y^{\inverse  1}.
\end{align}
Hence,
\begin{align*}
(x \cdot v + u \cdot y) \cdot (y \cdot v)^{\inverse  1}&=
x \cdot y^{\inverse  1} \cdot v \cdot v^{\inverse  1}+
u \cdot v^{\inverse  1} \cdot y \cdot y^{\inverse  1}\\
&=(x \cdot y^{\inverse  1}+0\cdot v^{\inverse  1})+
(u \cdot v^{\inverse  1} +0 \cdot y^{\inverse  1})
&&\text{by \eqref{eq:cv}}\\
&=(x \cdot y^{\inverse  1}+0\cdot y^{\inverse  1})+
(u \cdot v^{\inverse  1} +0 \cdot v^{\inverse  1})\\
&=x \cdot y^{\inverse  1}+u \cdot v^{\inverse  1}.
\end{align*}
\end{proof}

We end this section with two more propositions that characterize typical
properties of common meadows and that are used in the proof of
Theorem~\ref{thm:basis}. The first of these establishes that each (possibly open)
term over $\Sigma_{\md,\bot}$ has a simple representation in the syntax of meadows.

\begin{proposition}
\label{prop:3}
For each term $t$ over $\Sigma_{\md,\bot}(X)$ with variables in $X$
there exist terms $r_1,r_2$ over $\Sigma_f(X)$ such that
$\Md_\bot\vdash t=r_1\cdot r_2^{-1}$ and $\VAR(t)=\VAR(r_1)\cup\VAR(r_2)$.
\end{proposition}

\begin{proof}
By induction on the structure of $t$, where the $\VAR(t)$-property follows
easily in each case.

\begin{description}
\item[\normalfont If $t\in\{0,1,x,\bot\}$,]this follows trivially
(for the first three cases we need $1^{-1}=1$).

\item[\normalfont Case $t\equiv t_1+t_2$.]
By Proposition~\ref{prop:2a}.

\item[\normalfont Case $t\equiv t_1\cdot t_2$.] Trivial.

\item[\normalfont Case $t\equiv-t_1$.] By Proposition~\ref{prop:1}  (\ref{e5}).

\item[\normalfont Case $t\equiv t_1^{\inverse 1}$.] By induction there exist
$r_i\in \Sigma_f(X)$ such that $\Md_\bot\vdash t_1=r_1\cdot r_2^{\inverse 1}$.
Now derive
$t_1^{\inverse 1}=r_1^{\inverse 1}\cdot(r_2^{\inverse 1})^{\inverse 1}
=r_1^{\inverse 1}\cdot(r_2+0\cdot r_2^{\inverse 1})
=r_2\cdot r_1^{\inverse 1}+0\cdot r_1^{\inverse 1}+ 0\cdot r_2^{\inverse 1}
=r_2\cdot r_1^{\inverse 1}+ 0\cdot r_2^{\inverse 1}
$ and apply Proposition~\ref{prop:2a}.
\end{description}
\end{proof}

The next proposition shows how a term of the form $0\cdot t$ with $t$ a
(possibly open) term over $\Sigma_f(X)$ can  be simplified (note that
$0\cdot x = 0$ is not valid, since $0\cdot\bot=\bot$).
\begin{proposition}
\label{prop:3a}
For each term $t$ over $\Sigma_f(X)$,
$\Md_\bot\vdash 0\cdot t=0\cdot \sum_{x\in\VAR(t)}x$, where
$\sum_{x\in\emptyset}x=0$.
\end{proposition}

\begin{proof}
By induction on the structure of $t$, where identity~\eqref{e4}
(Proposition~\ref{prop:1}) covers the multiplicative case.
\end{proof}

\subsection{Conditional Equations}
We discuss a number of conditional equations that will turn
out useful,
and we start off with a few that follow directly from $\Mda$.

\begin{proposition}
\label{prop:2}
Conditional equations that follow from $\Md_{\bot}$ (see Table~\ref{Mda}):
\begin{align}
\label{ce1}\tag{ce1}
x \cdot y = 1 &\rightarrow 0 \cdot y = 0,
\\[.4mm]
\label{ce2}\tag{ce2}
x \cdot y = 1 &\rightarrow x^{\inverse  1} = y,
\\[.4mm]
\label{ce25}\tag{ce3}
0\cdot x=0\cdot y&\rightarrow 0 \cdot (x\cdot y) = 0\cdot x,
\\[.4mm]
\label{ce3}\tag{ce4}
0 \cdot x \cdot y  = 0 &\rightarrow 0 \cdot x = 0,
\\[.4mm]
\label{ce4}\tag{ce5}
0 \cdot (x + y ) = 0 &\rightarrow 0 \cdot x = 0,
\\[.4mm]
\label{ce5}\tag{ce6}
0 \cdot x^{\inverse 1} =0 &\rightarrow 0 \cdot x = 0,
\\[.4mm]
\label{ce7}\tag{ce7}
0 \cdot x = \bot &\rightarrow x = \bot.
\end{align}
\end{proposition}

\begin{proof}
Most derivations are trivial.
\begin{description}

\item[$\eqref{ce1}.$]
By axiom~\eqref{eq:10},
$0\cdot x\cdot y=
0\cdot x\cdot y\cdot y
=0\cdot x\cdot y+0\cdot y\cdot y
=(0\cdot x+0\cdot y)\cdot y$,
and hence by assumption,
$0 = 0 \cdot 1 = 0 \cdot x \cdot y = (0 \cdot x + 0 \cdot y) \cdot y =
 0 \cdot x  \cdot y + 0 \cdot y \cdot y = 0 + 0 \cdot y = 0 \cdot y$.

\item[$\eqref{ce2}.$]
By assumption and axioms~\eqref{eq:13} and \eqref{eq:14},
$x^{\inverse 1}\cdot y^{\inverse 1}=1$, and thus by \eqref{ce1},
$0\cdot x^{\inverse 1}=0$, so by axiom~\eqref{eq:12},
$y=(1+0\cdot x^{\inverse 1})\cdot y=(x\cdot x^{\inverse 1})\cdot y=
(x\cdot y)\cdot x^{\inverse 1}=x^{\inverse 1}$.

\item[$\eqref{ce25}.$]
By assumption, identity~\eqref{e4}, and axiom~\eqref{eq:8},
$0\cdot(x\cdot y)=0\cdot x+0\cdot y=0\cdot x + 0\cdot x=0\cdot x$.

\item[$\eqref{ce3}.$]
By assumption,
$0\cdot x=0\cdot x+0\cdot x\cdot y=x\cdot(0+0\cdot y)=0\cdot(x\cdot y)=0$.

\item[$\eqref{ce4}.$]
Apply identity~\eqref{e4} to \eqref{ce3}.

\item[$\eqref{ce5}.$]
By axiom~\eqref{eq:12} and assumption,
$x\cdot x^{\inverse 1}=1+0\cdot x^{\inverse 1}=1$,
so by~\eqref{ce1}, $0\cdot x=0$.

\item[$\eqref{ce7}.$]
By $x=x+0\cdot x$ and assumption, $x = x + \bot = \bot$.
\end{description}
\end{proof}

Note that~\eqref{ce1} and \eqref{ce2} immediately imply
\begin{equation*}
x \cdot y = 1 \rightarrow 0 \cdot x^{\inverse  1} = 0.
\end{equation*}

In Table~\ref{tab:laws} we define various conditional laws that we will use
to single out certain classes of common meadows in Section~\ref{sec:3}:
the Normal Value Law ($\NVL$), the Additional Value Law
($\AVL$), and  the Common Inverse Law ($\CIL$).
Here we use the adjective ``normal'' to express that values different from $\bot$
(more precisely, the interpretation of $\bot$) are at stake.
We conclude this section by interrelating these laws.

\begin{table}[t]
\centering
\hrule
\begin{align*}
\tag{$\NVL$}
x \neq \bot& ~\rightarrow~ 0\cdot x =0
&&\text{Normal Value Law}\\
\tag{$\AVL$}
x^{\inverse 1}= \bot& ~\rightarrow~ 0\cdot x =x
&&\text{Additional Value Law}\\
\tag{$\CIL$}
x \neq 0 \wedge x \neq \bot& ~\rightarrow~ x\cdot x^{-1} =1
&&\text{Common Inverse Law}
\end{align*}
\hrule
\caption{Some conditional laws for common meadows}
\label{tab:laws}
\end{table}

\begin{proposition}
\label{prop:4}~
\begin{enumerate}
\item[$1.$]
\label{4.1}
$\Md_\bot+\NVL\vdash (x \cdot y = \bot \wedge x \neq \bot) \rightarrow   y = \bot$,

\item[$2.$]
\label{4.2}
$\Md_\bot+\NVL\vdash x^{\inverse 1}  \neq \bot \rightarrow 0 \cdot x = 0$,

\item[$3.$]
\label{4.3}
$\Md_\bot+\NVL+\AVL\vdash \CIL$,

\item[$4.$]$\Md_\bot+\CIL\vdash\NVL$,

\item[$5.$]
$\Mda+\CIL\vdash \AVL$.
\end{enumerate}
\end{proposition}

\begin{proof}~
\begin{enumerate}
\item
By $\NVL$,
$x\ne \bot\rightarrow 0\cdot x=0$, so $0\cdot y=(0\cdot x)\cdot y=
0\cdot (x\cdot y)=0\cdot\bot=\bot$
and hence $y=(1+0)\cdot y=y+0\cdot y=y+\bot=\bot$.

\item
By $\NVL$, $0\cdot x^{\inverse 1}=0$ and hence by axiom~\eqref{eq:12},
$x\cdot x^{\inverse 1}=1$ and by~\eqref{ce1}, $0 \cdot x= 0$.

\item
From $x \neq \bot$ we find $0 \cdot x = 0$.
There are two cases: $ x^{\inverse 1} = \bot$ which implies by $\AVL$
that $x= 0$  contradicting the assumptions of $\CIL$, and
$ x^{\inverse 1} \neq \bot$ which implies by $\NVL$ that $0 \cdot x^{\inverse 1}=0$,
and this implies  $x \cdot x^{\inverse 1} =1$ by axiom~\eqref{eq:12}.

\item
Assume that $x \neq \bot$. If $x = 0$ then also $0 \cdot x=0$.
If $x \neq 0$ then by $\CIL$,
$ 0 = 0 \cdot 1 = 0 \cdot x \cdot x^{\inverse 1}$, so
$0 \cdot x= 0$ by~\eqref{ce1}.

\item
We distinguish three cases: $x=0$, $x=\bot$, and $x \neq 0 \wedge x \neq \bot$.
In the first two cases it immediately follows that $0\cdot x=x$.
In the last case it follows by
$\CIL$ that $x\cdot x\cdot x^{\inverse 1}=x$, so $x^{\inverse 1}=\bot$ implies
$x=\bot$, and thus $x=0\cdot x$.

\end{enumerate}
\end{proof}

\section{Models and Model Classes}
\label{sec:3}
In this section we define ``common cancellation meadows'' as common meadows that
satisfy the so-called ``inverse cancellation law'', a law that is equivalent with
the Common Inverse Law $\CIL$.
Then, we provide a basis theorem for common cancellation meadows of
characteristic zero.

\subsection{Common Cancellation Meadows}
In~\cite[Thm.3.1]{BBP2013} we prove a generic basis theorem that implies that the
axioms in Table~\ref{tab:Md} constitute a complete axiomatization of the equational
theory of the involutive cancellation meadows (over signature $\Sigma_{\md}$).
The cancellation law used in that result (that is, \ref{eq:CL} in
Section~\ref{subsec:versus}) has various
equivalent versions, and a particular one is $x\ne 0 \rightarrow x\cdot x^{-1}=1$,
a version that is close to $\CIL$.

Below we define common cancellation meadows, using a cancellation law that is
equivalent with $\CIL$, but first we establish a correspondence between models of
$\Md_\bot+\NVL+\AVL$ and involutive cancellation meadows.

\begin{proposition}
\label{prop:7}~~
\begin{itemize}
\item[$1.$]
\label{7.1}
Every field can be extended with an additional value $\bota$ and subsequently it
can be expanded with a constant $\bot$ and an inverse function in such a way that
the equations of common meadows as well as $\NVL$
and $\AVL$ are satisfied, where the interpretation of $\bot$ is $\bota$.

\item[$2.$]
\label{7.2}
A model of $\Md_\bot+\NVL+\AVL$ extends a field with an additional value $\bota$ (the
interpretation of $\bot$) and expands it with the $\bot$-totalized inverse.
\end{itemize}
\end{proposition}

\begin{proof}
Statement 1 follows immediately. To prove 2,
consider the substructure of elements $b$ of the domain that satisfy $0 \cdot b = 0$.
Only $\bota$ is outside this
subset. For $b$ with $0 \cdot b = 0$ we must check that $0 \cdot b^{\inverse 1} = 0$
unless $b=0$.
To see this distinguish two cases: $b^{\inverse 1} = \bot$
(which implies $b=0$ with help of $\AVL$),
and $b^{\inverse 1} \neq \bot$ which implies $0 \cdot b^{\inverse 1} = 0$ by $\NVL$.
\end{proof}

As a consequence, we find the following result.
\begin{theorem}
\label{thm:one-one}
The models of $\Md_\bot+\NVL+\AVL$ that satisfy $0\ne 1$ are in one-to-one
correspondence with the involutive cancellation meadows
satisfying  $\Md$ (see Table~\ref{tab:Md}).
\end{theorem}

\begin{proof}
An involutive cancellation meadow can be expanded to a model of $\Md_\bot+\NVL+\AVL$
by extending its domain with a constant $\bota$ in such a way that the equations of
common meadows as well as $\NVL$ and $\AVL$ are satisfied, where the interpretation
of $\bot$ is $\bota$ (cf.~Proposition~\ref{prop:7}.1).

Conversely, given a model $\M$ of $\Md_\bot+\NVL+\AVL$, we  construct a cancellation
meadow $\M'$ as follows:
$|\M'|=|\M|\setminus\{\bota\}$ with $\bota$ the interpretation of $\bot$, and
$0^{-1}=0$ (by $0\ne 1$, $|\M'|$ is non-empty).
We find by $\NVL$ that $0\cdot x=0$ and by
$\CIL$ (thus by $\NVL+\AVL$, cf.~Proposition~\ref{prop:4}.3) that
$x\ne 0\rightarrow x\cdot x^{-1}=1$, which shows that $\M'$ is a cancellation meadow.
\end{proof}

We define a \emph{common cancellation meadow} as a common meadow that satisfies
the following \emph{inverse cancellation law} \eqref{CL}:
\begin{equation}
\label{CL}
\tag{$\iCAP$}
(x \neq 0 \wedge x \neq \bot \wedge x^{\inverse 1}\cdot y = x^{\inverse 1}\cdot z)
\rightarrow y=z.
\end{equation}
The class $\CCM$ of common cancellation meadows is axiomatized by $\Mda+\CIL$
in Table~\ref{Mda} and Table~\ref{tab:laws}, respectively.
In combination with $\Mda$, the laws $\ICL$ and $\CIL$ are equivalent:
first, $\Mda + \ICL \vdash \CIL$ because
\[
(x \neq 0 \wedge x \neq \bot)\stackrel{\eqref{e8}}\rightarrow
(x \neq 0 \wedge x \neq \bot\wedge x^{\inverse 1}\cdot x \cdot x^{\inverse 1}
= x^{\inverse 1} \cdot 1)\stackrel{\ICL}\rightarrow x\cdot x^{\inverse 1}=1.
\]
Conversely,
$\Mda+\CIL\vdash\ICL$:
\[
(x \neq 0 \wedge x \neq \bot \wedge x^{\inverse 1}\cdot y
= x^{\inverse 1}\cdot z)\rightarrow x\cdot x^{\inverse 1}\cdot y=
x\cdot x^{\inverse 1} \cdot z\stackrel{\CIL}\rightarrow y=z.
\]

\subsection{A Basis Theorem for Common Cancellation Meadows of Characteristic Zero}

\begin{table}[t]
\centering
\hrule
\begin{align*}
\underline{n+1}\cdot (\underline{n+1})^{-1} &=1&&\text{($n\in\Nat$)}
\tag{$\CMCZ$}\\[2mm]
\underline{0}&=0&&\text{(axioms for }\\
\underline{1}&=1&&\text{~numerals, }\\
\underline{n+1}&=\underline{n} + 1&&\text{~$n\in\Nat$ and $n\geq 1$})
\end{align*}
\hrule
\caption{$\CMCZ$, the set of axioms for meadows of characteristic zero and numerals}
\label{CMCZ}
\end{table}

As in our paper~\cite{BBP2013a}, we use numerals $\underline{n}$
and the axiom scheme $\CMCZ$ defined in Table~\ref{CMCZ} to single out common
cancellation meadows of characteristic zero. In this section we prove that
$\Md_\bot+\CMCZ$ constitutes an axiomatization for common cancellation meadows of
characteristic zero. In~\cite[Cor.2.7]{BBP2013a} we prove that $\Md+\CMCZ$
(for $\Md$ see Table~\ref{tab:Md}) constitutes
an axiomatization for involutive cancellation meadows of characteristic zero.
We define $\CCM_0$ as the class of common cancellation meadows of characteristic
zero.

\bigskip

We further write
$~\dfrac{t}{r}~$ (and sometimes
$t/r$ in plain text) ~for $~t\cdot r^{\inverse 1}$.

\begin{theorem}
\label{thm:basis}
$\Md_\bot+\CMCZ$ is a basis for the equational theory of $\CCM_0$.
\end{theorem}

\begin{proof}
Soundness holds by definition of $\CCM_0$.

Assume $\CCM_0\models t=r$ and $\CCM_0\models t=\bot$.
Then, by axioms~$\eqref{eq:15}-\eqref{eq:17}$ and
identities~$\eqref{e9}-\eqref{e10}$, $t$ and $r$
are provably equal to $\bot$, that is, $\Mda\vdash t=r$.

Assume $\CCM_0\models t=r$ and $\CCM_0\not\models t=\bot$.
By Proposition~\ref{prop:3}
we can bring $t$ in the form $t_1/t_2$
and $r$ in the form $r_1/r_2$ with $t_i,r_i$ terms over $\Sigma_f(X)$,
thus
\begin{equation}
\label{v0}
\CCM_0\models \frac{t_1}{t_2}=\frac{r_1}{r_2}.
\end{equation}
We will first argue that \eqref{v0} implies that the following three equations
are valid in $\CCM_0$:
\begin{align}
\label{p1}
0\cdot t_2^{\inverse 1}&= 0\cdot r_2^{\inverse 1},\\
\label{p2}
0\cdot t_1+0\cdot t_2&=0\cdot r_1+0\cdot r_2,\\
\label{p3}
t_2\cdot r_2\cdot(t_1\cdot r_2+ (-r_1)\cdot t_2)+0\cdot t_2^{\inverse 1}+0\cdot r_2^{\inverse 1}
&=0\cdot t_1+0\cdot t_2^{\inverse 1}+0\cdot r_1+0\cdot r_2^{\inverse 1}.
\end{align}
\underline{Ad~\eqref{p1}}.
Assume this is not the case, then there exists a common cancellation meadow
$\M\in\CCM_0$ and an interpretation of the variables in $t_2$ and $r_2$ such that
one of $\smash{t_2^{\inverse 1}}$ and $\smash{r_2^{\inverse 1}}$ is interpreted as
$\bota$ (the interpretation of $\bot$), and the other is not.
This contradicts~\eqref{v0}.
\\[2mm]
\underline{Ad~\eqref{p2}}.
This equation characterizes that $t_1/t_2$ and $r_1/r_2$ contain the same
variables, and is related to Proposition~\ref{prop:3a}.
Assume this is not the case, say $t_1$ and/or $t_2$ contains a variable $x$ that
does not occur in $r_1$ and $r_2$. Since $\CCM_0\not\models r_1/r_2=\bot$, there is
an instance of $r_i$'s variables, say $\overline{r_i}$ such that
$\CCM_0\models \overline{r_1}/\overline{r_2}\ne\bot$.
But then $x$ can be instiantiated with $\bot$, which contradicts~\eqref{v0}.
\\[2mm]
\underline{Ad~\eqref{p3}}. It follows from~\eqref{v0} that in~\eqref{p3}
both the lefthand-side and the right\-hand-side equal zero in all involutive
cancellation meadows.
By Theorem~\ref{thm:one-one} we find $\CCM\models \eqref{p3}$, and hence
$\CCM_0\models \eqref{p3}$.

\bigskip

We now argue that~$\eqref{p1} - \eqref{p3}$
are derivable from $\Mda + \CMCZ$, and that from those~\eqref{v0} is derivable
from $\Mda + \CMCZ$.
\\[2mm]
\underline{Ad~\eqref{p1}}. The statement
$\CCM_0\models 0\cdot t_2^{\inverse 1}= 0\cdot r_2^{\inverse 1}$
implies that $t_2$ and $r_2$ have the same
zeros in the algebraic closure $\smash{\overline{\Rat}}$ of $\Rat$
(if this were not the case, then
$\overline{\Rat}_\bot\not\models0\cdot t_2^{\inverse 1}= 0\cdot r_2^{\inverse 1}$,
but $\overline{\Rat}_\bot\in \CCM_0$).
We may assume that the gcd of $t_2$'s coefficients is 1, and similar for $r_2$:
if not, then $t_2=k\cdot t'$ with $t'$ a polynomial with that property, and since
$k$ is a fixed numeral, we find $0\cdot k=0$ (also in fields with a characteristic
that is a factor of $k$), and hence $0\cdot t_2=0\cdot t'$.
We can apply~\cite[Cor.2.4 (Ch.IV)]{Lang}: because $t_2$ and $r_2$ are
polynomials in $\Sigma_f(\VAR(t_2,r_2))$ with the property that they have the
same zeros and that the gcd of their coefficients is 1, they have equal
factorization in primitive polynomials.
So, in common cancellation meadows of characteristic zero (thus, models in $\CCM_0$),
each such factor of $t_2$ is one of $r_2$, and vice versa.
Application of axiom~\eqref{eq:10} (that is, $0\cdot (x\cdot x)=0\cdot x$) then yields
\begin{equation}
\label{v3}
\Mda+ \CMCZ\vdash 0\cdot t_2^{\inverse 1}= 0\cdot r_2^{\inverse 1}.
\end{equation}

\noindent
\underline{Ad~\eqref{p2}}.
From Proposition~\ref{prop:3a} and validity of~\eqref{p2}
it follows that
\begin{equation}\textstyle
\label{v4}
\Mda\vdash 0\cdot t_1+0\cdot t_2=0\cdot \sum_{x\in\VAR(t_1/t_2)}x
=0\cdot \sum_{x\in\VAR(r_1/r_2)}x=0\cdot r_1+0\cdot r_2.
\end{equation}

\noindent
\underline{Ad~\eqref{p3}}. We first derive
\begin{align*}
\nonumber
\Mda\vdash 0\cdot t_1+0\cdot t_2^{\inverse 1}
&=0\cdot t_1+0\cdot(1+0\cdot t_2^{\inverse 1})\\
\nonumber
&=0\cdot t_1+0\cdot t_2\cdot t_2^{\inverse 1}
&&\text{with axiom~\eqref{eq:12}}\\
\nonumber
&=0\cdot t_1+0\cdot t_2+0\cdot t_2^{\inverse 1},
\end{align*}
and in a similar way one derives $\Mda\vdash 0\cdot r_1+0\cdot r_2^{\inverse 1}=
0\cdot r_1+0\cdot r_2+0\cdot r_2^{\inverse 1}$.
Hence, we find with~\eqref{v3} and \eqref{v4}
that
\begin{align}
\label{v5}
\Mda+ \CMCZ\vdash
0\cdot t_1+0\cdot t_2^{\inverse 1}&=(0\cdot t_1+0\cdot t_2^{\inverse 1})+
(0\cdot r_1+0\cdot r_2^{\inverse 1})\\
\label{v6}
&=0\cdot r_1+0\cdot r_2^{\inverse 1}.
\end{align}

From $\CCM_0\models\eqref{p3}$ it follows from the completeness result on the
class of involutive meadows of characteristic zero (see~\cite[Cor.2.7]{BBP2013a})
that $\Md+\CMCZ\vdash \eqref{p3}$, and hence $\Mda+\CMCZ\vdash \eqref{p3}$.

We now show the derivability of $t_1/t_2=r_1/r_2$.
Multiplying both sides of~\eqref{p3} with
\\
$(t_2\cdot r_2)^{\inverse 1}\cdot (t_2\cdot r_2)^{\inverse 1}$ implies by~\eqref{e8},
$0\cdot x+0\cdot x=0\cdot x$, and axiom~\eqref{eq:10} that
\begin{align*}
\Mda+\CMCZ\vdash
(t_2\cdot r_2)^{\inverse 1}\cdot(t_1\cdot r_2+ (-r_1)\cdot t_2)
+0\cdot t_2^{\inverse 1}+0\cdot r_2^{\inverse 1}
=\hspace{16mm}\\
0\cdot t_1+0\cdot t_2^{\inverse 1}+0\cdot r_1+0\cdot r_2^{\inverse 1},
\end{align*}
which implies by Proposition~\ref{prop:2a} that
\begin{align*}
\Mda+\CMCZ\vdash\frac{t_1}{t_2}+\frac{-r_1}{r_2}+0\cdot t_2^{\inverse 1}+0\cdot r_2^{\inverse 1}
&=0\cdot t_1+0\cdot t_2^{\inverse 1}
+0\cdot r_1+0\cdot r_2^{\inverse 1},
\end{align*}
and thus
\begin{align}
\nonumber
\Mda+\CMCZ\vdash\frac{t_1}{t_2}+\frac{-r_1}{r_2}+0\cdot t_1+0\cdot t_2^{\inverse 1}
+0\cdot r_1+0\cdot r_2^{\inverse 1}=\hspace{19mm}\\
\label{e:A}
0\cdot t_1+0\cdot t_2^{\inverse 1}
+0\cdot r_1+0\cdot r_2^{\inverse 1},
\end{align}
and hence
\begin{align*}
\Mda+ \CMCZ\vdash\frac{t_1}{t_2}
&=\frac{t_1}{t_2}+0\cdot t_1+0\cdot t_2^{\inverse 1}\\
&=\frac{t_1}{t_2}+0\cdot t_1 +0\cdot t_2^{\inverse 1}+0\cdot r_1+0\cdot r_2^{\inverse 1}
&&\text{by \eqref{v5}}\\
&=\frac{t_1}{t_2}+(\frac{r_1}{r_2}+\frac{-r_1}{r_2})
+0\cdot t_1 +0\cdot t_2^{\inverse 1}+0\cdot r_1+0\cdot r_2^{\inverse 1}
&&\text{}\\
&=(\frac{t_1}{t_2}+\frac{-r_1}{r_2})+\frac{r_1}{r_2}
+0\cdot t_1 +0\cdot t_2^{\inverse 1}+0\cdot r_1+0\cdot r_2^{\inverse 1}
\\
&=\frac{r_1}{r_2}
+0\cdot t_1 +0\cdot t_2^{\inverse 1}+0\cdot r_1+0\cdot r_2^{\inverse 1}
&&\text{by \eqref{e:A}}\\
&=\frac{r_1}{r_2}+0\cdot r_1+0\cdot r_2^{\inverse 1}
&&\text{by \eqref{v6}}\\
&=\frac{r_1}{r_2}.
\end{align*}
\end{proof}

\section{Concluding Remarks}
\label{sec:4}

\paragraph{\bf Open Question.}
~It is an open question whether there exists a basis result for the equational
theory of $\CCM$.
We notice that in~\cite{BM14} a basis result for one-totalized non-involutive
cancellation meadows is provided,
where the multiplicative inverse of 0 is 1 and cancellation is defined as usual
(that is, by the cancellation law~\ref{eq:CL} in Section~\ref{subsec:versus}).

\paragraph{\bf Common Intuitions and Related Work.}
~Common meadows are motivated as being the most intuitive modelling of a
totalized inverse function to the best of our knowledge.
As stated in Section~\ref{sec:Intro} (Introduction), the use of the constant
$\bot$ is a matter of convenience only because it merely
constitutes a derived constant with defining equation $\bot = 0^{-1}$, which
implies that all uses of $\bot$ can be removed.\footnote{
  We notice that $0 = 1 + (-1)$, from which it follows that $0$ can also be
  considered a derived constant over a reduced signature. Nevertheless, the
  removal of $0$ from the signature of fields is usually not considered helpful.}
We notice that considering $\bot = 0^{-1}$ as an error-value supports the intuition
for the equations of $\Mda$.

As a variant of involutive and common meadows, \emph{partial meadows} are
defined in~\cite{BM2011a}. The specification method used in this paper
is based on meadows and therefore it is more simple, but
less general than the construction of Broy and Wirsing~\cite{BW81} for the specification of
partial datatypes.

The construction of common meadows is related to the construction of \emph{wheels}
by Carlstr\"om~\cite{Carl}.
However, we have not yet found a structural
connection between both constructions which differ in quite important details.
For instance, wheels are involutive whereas common meadows are non-involutive.

\paragraph{\bf Quasi-Cancellation Meadows of Characteristic Zero.}
~Following Theorem~\ref{thm:basis},
a common meadow of characteristic zero can alternatively be defined as a structure
that satisfies all equations true of all common cancellation meadows of
characteristic zero.
We write $\CM_0$ for the class of all common meadows of characteristic zero.

With this alternative definition in mind, we define a
\emph{common quasi-can\-cellation meadow of characteristic zero}
as a structure that satisfies all conditional equations which are true of all
common cancellation meadows of characteristic zero.
We write $\CQCM_0$ for the class of all common quasi-cancellation meadows of
characteristic zero.

It is easy to show that $\CQCM_0$ is strictly larger than $\CCM_0$.
To see this one extends the signature of common meadows
with a new constant $c$. Let $L_{ccm,0}$ be the set of conditional equations
true of all structures in $\CCM_0$.
We consider the initial algebra of $L_{ccm,0}$ in the signature extended with $c$.
Now neither $L_{ccm,0} \vdash c= \bot$ can hold
(because $c$ might be interpreted as say 1), nor  $L_{ccm,0} \vdash 0 \cdot c = 0$
can hold (otherwise $L_{ccm,0} \vdash 0 = 0\cdot \bot = \bot$ would hold).
For that reason in the initial algebra of $L_{ccm,0}$ in the extended signature
interprets $c$ as an entity $e$ in such a way  that neither
$c = \bot$  nor $0 \cdot c = 0$ is satisfied. For that reason
$c$ will be interpreted by a new entity that refutes $\CIL$.

$\CM_0$ is strictly larger than $\CQCM_0$.
To see this let $E_{ccm,0}$ denote the set of equations valid in all common
cancellation meadows of characteristic zero.
Again we add an extra constant $b$ to the signature of common meadows.
Consider the initial algebra $I$ of
$E_{ccm,0}+ (b^{-1} =\bot)$ in the extended signature.
In $I$ the interpretation of $b$ is a new object because it cannot be
proven equal to 0 and not to $\bot$ and not to any other closed term over the
signature of common meadows.
Now we transform $E_{ccm,0} + (b^{-1} =\bot)$ into its set of closed consequences
$E_{ccm,0}^{cl,b}$ over the extended signature.
We claim that $b = 0 \cdot b$ cannot be proven from $E_{ccm,0} + (b^{-1} =\bot$).
If that were the case at some stage in the derivation an $\bot$ must appear from
which it follows that $b=\bot$ is provable as well, because $\bot$ is propagated by
all operations.
But that cannot be the case as we have already concluded that $b$ differs from
$\bot$ in the initial algebra $I_0$ of $E_{ccm,0}^{cl,b}$.
Thus, $b\ne\bot\rightarrow  0 \cdot b=0$ (an instance of $\NVL$) is not valid in
$I_0$.

However, at this stage we do not know the answers to the following two questions:
\begin{itemize}
\item Is there a finite equational basis for the class $\CM_0$ of common meadows
of characteristic zero?
\item Is there a finite conditional equational basis for the class $\CQCM_0$ of
common quasi-cancellation meadows of characteristic zero?
\end{itemize}

\paragraph{\bf The Initial Common Meadow.}
~In~\cite{BP14} we introduce \emph{fracpairs} with a definition that
is very close to that of the field of fractions of an integral
domain. Fracpairs are defined over a commutative ring $R$
that is \emph{reduced}, i.e., $R$ has no nonzero nilpotent
elements. A fracpair over $R$ is an expression $\dfrac pq$ with $p,q\in R$
(so $q=0$ is allowed) modulo the equivalence generated by
\begin{align*}
	 \frac{x \cdot z }{y \cdot (z \cdot z) } &= \frac{x}{y \cdot z}.
\end{align*}
This rather simple equivalence appears to be a congruence with respect to the
common meadow signature $\Sigma_{\md,\bot}$
when adopting natural definitions:
\begin{align*}
&0=\frac 01, \quad 1=\frac 11, \quad \bot=\frac 10,
\quad
\lh\frac pq\rh+\lh\frac rs\rh=
\frac{p\cdot s+r\cdot q}{q\cdot s},
\\
&\lh\frac pq\rh\cdot\lh\frac rs\rh=\frac{p\cdot r}{q\cdot s},
\quad
-\lh\frac pq\rh=\frac{-p}q,
\quad\text{and}\quad
\lh\frac pq\rh^{-1}=\frac{q\cdot q}{p\cdot q}.
\end{align*}

In~\cite{BP14} we prove that the initial common meadow is isomorphic to
the initial algebra of fracpairs over the integers $\Int$.
Moreover, we prove that the initial algebra of fracpairs over $\Int$ constitutes
a homomorphic pre-image of the common meadow $\Rat_\bot$,
and we define ``rational fracpairs'' over $\Int$ that constitute
an initial algebra that is isomorphic to $\Rat_\bot$. Finally, we consider
some term rewriting issues for meadows.

These results reinforce our idea that common meadows can be used
in the development of alternative foundations of elementary (educational)
mathematics from a perspective of abstract datatypes, term rewriting and
mathematical logic.

\subsection*{Acknowledgement}
We thank Bas Edixhoven (Leiden University) for helpful comments concerning the
proof of Theorem~\ref{thm:basis},
including his suggestion to use~\cite{Lang} as a reference for this proof.
Furthermore, we thank two anonymous reviewers for valuable comments.

\addcontentsline{toc}{section}{References}


text:
\begin{document}
\title[Global solution to the
Cauchy problem on UFM] {Global solution to the Cauchy problem on a
universe fireworks model}

\author{Zhenglu Jiang}

\address{Department of Mathematics, Zhongshan University,
Guangzhou 510275, China} \email{mcsjzl@mail.sysu.edu.cn}

\thanks{This work  was supported by NSFC 10271121 and
the Scientific Research Foundation for the Returned Overseas Chinese
Scholars, the Ministry of Education of China, and sponsored by joint
grants of NSFC 10511120278/10611120371 and RFBR 04-02-39026.}

\author{Hongjiong Tian}
\address{Department of Mathematics,  Shanghai Teachers' University,
Shanghai 200234,  China} \email{hongjiongtian@263.net}

\subjclass[2000]{76P05; 35Q75}

\date{\today.}

\keywords{Cauchy problem; global solution; universe fireworks}

\begin{abstract}
We prove  existence and uniqueness of the global solution to the
Cauchy problem on a universe fireworks model with finite total mass
at the initial state when the ratio of the mass surviving the
explosion, the probability of the explosion of fragments and the
probability function of the velocity change of a surviving particle
satisfy the corresponding physical conditions. Although the
nonrelativistic Boltzmann-like equation modeling the universe
fireworks is mathematically easy, this paper leads rather
theoretically to an understanding of how to construct contractive
mappings in a Banach space for the proof of the existence and
uniqueness by means of methods taken from the famous work by DiPerna
\& Lions about the Boltzmann equation. We also show both the
regularity and the time-asymptotic behavior of solution to the
Cauchy problem.
\end{abstract}

\maketitle

\section{Introduction}
\label{intro} We are concerned with existence and uniqueness of the
global solution to the Cauchy problem on a universe fireworks model
(hereafter, UFM) without gravity described by the following
nonrelativistic
 Boltzmann-like equation \cite{wl}
\begin{equation}
\frac{\partial f}{\partial t}+\xi \frac{\partial f}{\partial x}=
\eta(t, x, \xi)\int_{{\bf R}^3}\Gamma(t, x, \xi_1)P(-\xi\cdot\xi_1)f_1d\xi_1
-\Gamma(t, x, \xi)f
\label{(2.1.1)}
\end{equation}
where $f\equiv f(t, x, \xi)$ is the total
mass distribution with respect to time
$t$,  space $x$ and velocity $\xi$ during the explosion of a huge cloud
of matter; $f_1\equiv f(t, x, \xi_1)$; $\eta\equiv
\eta(t, x, \xi)$ is the mass ratio which is a measure of the mass
surviving the explosion; $\Gamma\equiv
\Gamma(t, x, \xi)$ is the explosion probability per unit time for a fragment;
$P\equiv P(-\xi\cdot\xi_1)$
 is the probability density so that $P(-\xi\cdot\xi_1)d\xi_1$
 is the probability that a surviving particle will change its
 velocity  $\xi$ to $\xi_1\in d\xi_1$ during explosion.

Assume that $\Gamma,  \eta$ and $P$ satisfy the following physical
conditions
\begin{equation}
0\leq\eta\leq 1,  0\leq\Gamma\leq 1, 0{\leq} P,
\label{(2.1.2)}
\end{equation}
\begin{equation}
\int_{{\bf R}^3} P(-\xi\cdot\xi_1)d\xi_1=1 \hbox{  for  }
\forall \xi\in {{\bf R}^3},
\label{(2.1.3)}
\end{equation}
and that the initial total mass distribution $f_0\equiv f_0(x, \xi)$ satisfies
\begin{equation}
 0\leq f_0(x, \xi)\in L^1({{\bf R}^3}
\times{{\bf R}^3}),
\label{(2.1.4)}
\end{equation}
i.e., having finite total mass at the initial state.

In this paper we shall prove existence and uniqueness of the global
solution to the Cauchy problem on the nonrelativistic Boltzmann-like
equation~(\ref{(2.1.1)}) of the distribution $f$ with finite total
mass at the initial state when the ratio $\eta,$ the probability
$\Gamma $  and the probability density $P$ satisfy the corresponding
physical conditions (\ref{(2.1.2)}) and (\ref{(2.1.3)}). We shall
also show both the regularity and the time-asymptotic behavior of
solution to this problem.

The nonrelativistic Boltzmann-like equation is mathematically almost
trivial since it is a linear transport equation with an integrable
kernel and bounded multiplicative coefficients. However, this paper
leads rather theoretically to an understanding of how to construct
contractive mappings in a Banach space by means of methods taken
from the famous work by DiPerna \& Lions \cite{dl} about the
Boltzmann equation.

The fireworks model is one of the metagalaxy models which are used
to give a physical explanation of the expansion. Alfv\'{e}n
\cite{ah} suggested the first fireworks model for the evolution of
the universe
 as an alternative cosmology to big bang and described
a possible mechanism for this model in an original work in 1983, but
many details remain unexplained because of the extremely complicated
physics involved \cite{wl}. Some assumptions on the mechanism of the
explosion, such as  that matter and antimatter are mixed, are
invalid in a metagalaxy model given by Alfv\'{e}n \& Klein \cite{ak}
for the highest redshift observed. Also, the highest redshift
observed cannot be accounted for even if it is assumed that there is
an upper limit to the velocities that could be reached by one
explosion of about $0.7c$ ($c$ is the light velocity) \cite{lbc}. A
special relativistic model was investigated by Laurent \& Carlqvist
\cite{lc} but the evolution of the distributions in configuration
space was not extended to general relativity by use of their
approach. Due to these, Widlund \cite{wl} gave a fireworks model
(defined by equation~(\ref{(2.1.1)}))  without prescribing the
details of the explosion mechanism and developed the model using a
fully general relativistic treatment of the fireworks. The
nonrelativistic fireworks model, which is given by
equation~(\ref{(2.1.1)}), is a special case of the nonrelativistic
limit to the general relativistic model considered by Widlund
\cite{wl}. Widlund has explained the observed redshifts only from
the start of the study of an approximate equation of
equation~(\ref{(2.1.1)}) because it is a differential equation and
has an exact solution for
 many weak explosions. Our results mentioned above are not only a kind of
 reasoning to adopt the research method of replacing equation~(\ref{(2.1.1)})
 by an approximate equation  as Widlund has done, but additionally
 make it possible to do some
  research work about the redshifts directly from the start of
   the solution to equation~(\ref{(2.1.1)}).

\section{Existence and Uniqueness}
One of our results mentioned in the previous section can be
described as follows.
\begin{theorem}
Equation~(\ref{(2.1.1)}) with physical conditions (\ref{(2.1.2)})
and (\ref{(2.1.3)}) has a unique nonnegative global distributional
solution $f(t,x, \xi)$ through a total mass distribution $f_0(x,
\xi)$ satisfying a finite total mass condition (\ref{(2.1.4)}) at
the initial state, and this solution $f(t,x, \xi)$ belongs to the
Banach space $ L^\infty((0,+\infty);L^1({\bf R}^3\times{\bf R}^3)).$
\label{th01}
\end{theorem}

Theorem~\ref{th01} shows existence and uniqueness of solution to the
Cauchy problem on UFM without gravity no matter how long the
fireworks era lasts.

The proof of Theorem~\ref{th01} is as follows. We first define a
mild solution to equation~(\ref{(2.1.1)}) as DiPerna \& Lions did in
\cite{dl}. Then,  we establish a Banach space and a mapping of this
space into itself which uniformly decreases distances,  and shows
that this mapping has a unique fixed point which is a unique global
mild solution to equation~(\ref{(2.1.1)}). This is an application of
 the well-known Banach fixed point theorem. Finally, according to
 the relations between mild solution
 and distributional solution (e.g., \cite{dl}, \cite{j98a}),
 we know that this mild solution is also
 a global distributional solution.

\begin{definition}
Let $f=f(t, x, \xi)$ be a nonnegative function which belongs to
$L_{loc}^1((0, T)\times{{\bf R}^3}\times{{\bf R}^3}), $ and assume
that for almost all $(x, \xi)\in{{\bf R}^3}\times{{\bf R}^3}, $
\begin{equation}
Q^\#(t, x, \xi)\in L^1(0, T_1)\hbox{  } (\forall T_1\in(0, T))
\label{(2.2.0)}
\end{equation}
and
\begin{equation}
f^\#(t, x, \xi)-f^\#(s, x, \xi)=\int^t_sQ^\#(\sigma, x, \xi)d\sigma\hbox{  }
(\forall 0{\leq}s<t{\leq}T),
\label{(2.2.1)}
\end{equation}
where $h^\#$ denotes, for any measurable function $h$ on $(0, +\infty)
\times{{\bf R}^3}\times{{\bf R}^3}, $
the following restriction to characteristics:
$$h^\#(t, x, \xi)=h(t, x+t\xi, \xi).$$
If $0<T<+\infty, $ $f$ is called a local-in-time mild solution to
the equation
\begin{equation}
 \frac{\partial f}{\partial t}+\xi \frac{\partial f}{\partial x}=Q
\label{(2.2.2)}
\end{equation}
in the time interval $[0, T].$ If $T=+\infty, $ $f$ is called a global
 mild solution to equation~(\ref{(2.2.2)}).
\label{def01}
\end{definition}

By Definition~\ref{def01},  we have the following result:

\begin{lemma}
Suppose that $f, h\in L_{loc}^1({\bf R}\times{{\bf R}^3}\times{{\bf
R}^3}), $ and that for almost all $x, \xi\in{{\bf R}^3}, $ $f^\#$ is
absolutely continuous with respect to $t, $ $h^\#\in L_{loc}^1({\bf
R}).$ Then $f$ is a distributional solution to
equation~(\ref{(2.2.2)}) if and only if $f$ is a mild solution to
equation~(\ref{(2.2.2)}). \label{lemma1}
\end{lemma}

This result is similar to that given by DiPerna \& Lions \cite{dl}
for the Boltzmann equation. To show Theorem~\ref{th01}, we also have
to define a mapping as follows.

\begin{definition}
A mapping $J$ is defined as follows: for any $
f\in {\cal F}$ and $t\in [0, T], $
$$
J(f)(t, x, \xi)=f_0(x-t\xi, \xi)-\int^t_0\Gamma(\sigma, x+(\sigma-t)\xi, \xi)
f(\sigma, x+(\sigma-t)\xi, \xi)d\sigma $$
\begin{equation}
+\int^t_0\int_{{\bf R}^3}\eta(\sigma, x+(\sigma-t)\xi, \xi)\Gamma(\sigma, x+(\sigma
-t)\xi, \xi_1)P(-\xi\cdot\xi_1)f(\sigma, x+(\sigma-t)\xi, \xi_1)d\xi_1d\sigma
  \label{(2.2.3)}
\end{equation}
 where ${\cal F}=\left\{f\equiv f(t, x, \xi):
f\in L^\infty((0, T);L^1({{\bf R}^3}\times{{\bf R}^3}))
 \hbox{ for all  }  T\in [0, +\infty)\right\}.$
\label{def02}
\end{definition}

Let the Lebesgue measure be used for all the integrals in this paper.
Since the Lebesgue measure defined in ${\bf R}\times{\bf R}^3\times{\bf R}^3$
 is $\sigma$-finite,  by use of the Fubini-Tonelli Theorem
and the physical condition (\ref{(2.1.3)}),
we know
that $f, f^\#,  \int_{{\bf R}^3}P(-\xi\cdot\xi_1)f_1d\xi_1,
\left[\int_{{\bf R}^3}P(-\xi\cdot\xi_1)f_1d\xi_1\right]^\#
\in L_{loc}^1(0, +\infty)$ for almost every $(x, \xi)\in
{\bf R}^3\times{\bf R}^3, $ i.e.,  the product $P(-\xi\cdot\xi_1)f(\cdots)$
 in equation (\ref{(2.2.3)}) is integrable with respect to $\xi_1.$
 Thus Definition~\ref{def02} is valid.

Take $T_0$ as a definite time the fireworks era lasts. Then, if $0\leq T_0
<\frac{1}{2},$   equation~(\ref{(2.1.1)}) has a local-in-time mild solution in the time interval
$[0, T_0], $ or say,  the total mass distribution of the universe fireworks is
uniquely determined by the initial total mass data during the  fireworks
 era. Indeed,  by Definition~\ref{def02} and with the help of the physical conditions
(\ref{(2.1.2)}) and (\ref{(2.1.3)}),  we can
 easily prove that $J(f)\in {\cal F}$ for all $ f\in {\cal F}, $
 and that the following
 inequality holds:
\begin{equation}
\mathop{\max}\limits_{0\leq t \leq T_0}\|J(f)-J(h)\|_{L^1({{\bf R}^3}\times{{\bf R}^3})}
 \leq 2T_0\mathop{\max}\limits_{0\leq t \leq T_0}\|f-h\|_{L^1(
 {{\bf R}^3}\times{{\bf R}^3})}
\label{(2.2.4)}
\end{equation}
 for all $ f, h \in {\cal F}.$ The inequality (\ref{(2.2.4)}) shows that $J$ is a contractive mapping from
 ${\cal F}$ into itself with the same norm as in
 $C((0, T_0);L^1({\bf R}^3\times{\bf R}^3)).$
  Therefore there exists a unique
  element $h_1\in {\cal F}$ such that $J(h_1)=h_1$ for a.e. $(t, x, \xi)\in
  (0, T_0) \times{{\bf R}^3}\times{{\bf R}^3}.$
  In order to prove that $h_1$ is a local-in-time mild
  solution to equation~(\ref{(2.1.1)}),  it is enough to show that $h_1$ is nonnegative for almost
  every $(t, x, \xi)\in (0, T_0)\times{{\bf R}^3}\times{{\bf R}^3}.$

 Let us take another mapping $J_1$ as follows: $J_1(f)=\max(0, J(f))$ for all
$ f \in {\cal F}^+=\{f:f\in {\cal F} \hbox{ and }
f\mathop{\geq}\limits^{a.e.}0\}.$ Obviously,
 ${\cal F}^+$ is a subset of ${\cal F}.$ Similarly, we can easily show that the
 mapping $J_1$ maps ${\cal F}^+$ into itself and is uniformly contractive
  with the same norm as mentioned above.
  Then there exists a unique element $f_1\in {\cal F}^+$
  such that $J_1(f_1)=f_1$ for almost every $(t, x, \xi)\in[0, T_0]\times
  {{\bf R}^3}\times{{\bf R}^3}.$ Thus,  if $f_1=J(f_1)$ for almost every $(t, x, \xi)\in [0, T_0]
  \times{{\bf R}^3}\times{{\bf R}^3}, $ $f_1$ is a local-in-time mild solution to equation~(\ref{(2.1.1)}) through
  $f_0$ in the time interval $[0, T_0], $ and $f_1\mathop{=}\limits^{a.e.}h_1.$
   We will below show that $J(f_1)\mathop{=}\limits^{a.e.}f_1, $ or equivalently,
   $J(f_1)^\#\mathop{=}\limits^{a.e.}f_1^\#.$  In fact,  by equation (\ref{(2.2.3)}),  we know that
   $J(f_1)^\#$ is absolutely continuous with respect to $t\in [0, T_0]$ for
   almost every $(x, \xi)\in {{\bf R}^3}\times{{\bf R}^3}.$ We may assume without loss of
    generality  that $J(f_1)^\#(t, x, \xi)$ is continuous for all $(t, x, \xi).$
To prove that $J(f_1)^\#=f_1^\#, $ it suffices to prove that $J(f_1)^\#
\geq 0.$ If $J(f_1)^\#(t_0, x_0, \xi_0)<0$ for some point $(t_0, x_0, \xi_0)
\in [0, T_0]\times{{\bf R}^3}\times{{\bf R}^3}, $ we can find some value $t_1 \in [0, T_0]$
such that $J(f_1)^\#(t, x_0, \xi_0)<0$ for all $t\in[t_1, t_0], $ so that $
f_1^\#(t, x_0, \xi_0)=0$ for all $t\in [t_1, t_0].$ By equation (\ref{(2.2.3)}),  we can know that
$$0>J(f_1)^\#(t, x_0, \xi_0)\geq J(f_1)^\#(t_1, x_0, \xi_0)~~(t_0>t>t_1).$$
Repeating the above analysis,  we can conclude that
$J(f_1)^\#(0, x_0, \xi_0)<0, $ i.e.,  $f_0(x_0, \xi_0)<0, $ which is a
contradiction. Therefore $h_1=f_1=J(f_1)\geq 0$ for almost every $
(t, x, \xi)\in [0, T_0]\times{{\bf R}^3}\times{{\bf R}^3}.$

How about $T_0\geq\frac{1}{2}?$ It suffices to consider the
$T_0=+\infty$ case,  i.e.,  to prove Theorem~\ref{th01}. In order to
do this,  we have to establish
 a subset ${\cal B}$ of ${\cal F}, $ such that $J$ is a contractive
 mapping from ${\cal B}$
 into itself with some norm in the time interval
 $[0, +\infty).$ Now,  let us take ${\cal F}_a$ as follows:
 $${\cal F}_a=\left\{ f: \|f\|_a\equiv \mathop{\sup}\limits_{0\leq t<+\infty}
 \left\{e^{-at}\int\int_{{{\bf R}^3}\times{{\bf R}^3}}|f(t, x, \xi)|dxd\xi
 \right\}\leq A,
 f\in {\cal F}\right\}$$
 where $a>0, A=
 \frac{a}{2}\|f_0\|_{L^1({{\bf R}^3}\times{{\bf R}^3})}.$
 Then $({\cal F}_a, \|\cdot\|_a)$ is a complete Banach space. By Definition~\ref{def02} and
 with the help of the physical conditions (\ref{(2.1.2)}) and (\ref{(2.1.3)}),  it is also easy to see that
 \begin{equation}
 \|J(f)\|_a\leq A,
\label{(2.2.5)}
\end{equation}
\begin{equation}
\|J(f)-J(h)\|_a\leq \frac{2}{a}\|f-h\|_a.
\label{(2.2.6)}
\end{equation}
 for all $f, h \in {\cal F}_a.$
 We know from the two inequalities (\ref{(2.2.5)}) and (\ref{(2.2.6)})
that $J$ is a contractive mapping of
  ${\cal F}_a$ into itself with the norm $\|\cdot\|_a$ in
  the time interval $[0, +\infty)$
  for any given $a>2.$ Thus,  choosing a real $a_0$ such that $a_0>2$ and
  taking ${\cal B}={\cal F}_{a_0}, $ we can have a unique element $h_0\in {\cal B}$ such that
  $J(h_0)=h_0$ in $[0, +\infty)\times{\bf R}^3\times{\bf R}^3.$
  As stated above about the fixed point nonnegativity in the
 $0\leq T_0<\frac{1}{2}$ case,  we can easily show that $h_0$ is nonnegative for
 almost every $(t, x, \xi)\in [0, +\infty)\times{{\bf R}^3}\times{{\bf R}^3}.$
Therefore equation (\ref{(2.1.1)}) with the physical conditions
(\ref{(2.1.2)}) and (\ref{(2.1.3)}) has a global mild solution
through the total mass distribution $f_0$ satisfying the finite
total mass condition (\ref{(2.1.4)}) at the initial state. Finally,
by Lemma~\ref{lemma1},  Theorem~\ref{th01} holds true.

We can also construct another contractive mapping to prove
Theorem~\ref{th01}. Let us write
$F^\#(t)=\int_0^t\Gamma^\#(\sigma,x,\xi)d\sigma. $ Then, using the
same device as given by DiPerna \& Lions \cite{dl} for the Boltzmann
equation, we have
\begin{lemma}
 $f=f(t,x,\xi)$ is a
mild solution to equation~(\ref{(2.1.1)}) if and only if
$f=f(t,x,\xi)$ satisfies
$$f^\#(t,x,\xi)-f^\#(s,x,\xi)e^{-[F^\#(t)-F^\#(s)]} $$ $$
=\int_s^t \int_{{\bf R}^3}\eta^\#(\sigma, x, \xi)\Gamma^\#(\sigma, x, \xi_1)P(-\xi\cdot\xi_1)f^\#(\sigma,x, \xi_1)
e^{-[F^\#(t)-F^\#(\sigma)]} d\xi_1 d\sigma $$
for almost all $(x,\xi) \in {\bf R}^3\times{\bf R}^3$ and all $0<s<t<+\infty.$
\label{lemma2}
\end{lemma}
\begin{definition}
A mapping $J^+$ is defined as follows:  $\forall
f\in {\cal F}^+$ and $\forall  t\in [0, T], $
$$ J^+(f)(t, x, \xi)=e^{-\int_0^t\Gamma(\sigma,x+(\sigma-t)\xi,\xi)d\sigma}[f_0(x-t\xi, \xi) $$
\begin{equation}
+\int^t_0Q^+(\sigma,x+(\sigma-t)\xi, \xi_1)
e^{-\int_0^\sigma\Gamma(\tau,x+(\tau-t)\xi,\xi)d\tau}d\sigma],
  \label{(2.2.7)}
\end{equation}
 where $${\cal F}^+=\left\{f\equiv f(t, x, \xi): f\geq 0 \hbox{ and }
f\in L^\infty((0, T);L^1({{\bf R}^3}\times{{\bf R}^3}))
 \hbox{ for all }  T\in [0, +\infty) \right\},$$
$$ Q^+(t,x, \xi)
=\int_{{\bf R}^3}\eta(t, x, \xi)\Gamma(t, x, \xi_1)P(-\xi\cdot\xi_1)f(t,x, \xi_1)d\xi_1.$$
\label{def03}
\end{definition}

Using Lemma \ref{lemma2} and Definition \ref{def03}, we can give a
different and brief proof of Theorem~\ref{th01}. Let us take ${\cal
F}_a^+$ as follows:
 $${\cal F}_a^+=\left\{ f: \|f\|_a\equiv \mathop{\sup}\limits_{0\leq t<+\infty}
 \left\{e^{-at}\int\int_{{{\bf R}^3}\times{{\bf R}^3}}|f(t, x, \xi)|dxd\xi
 \right\}\leq A,
 f\in {\cal F}^+\right\}$$
 where $a>0, A=
 a\|f_0\|_{L^1({{\bf R}^3}\times{{\bf R}^3})}.$
 Then $({\cal F}_a^+, \|\cdot\|_a)$ is a complete metric space
with the distance $dist(\cdot,\cdot)=\|\cdot-\cdot\|_a.$
 Under the physical conditions (\ref{(2.1.2)}) and (\ref{(2.1.3)}),
it can be easily seen from Definition~\ref{def03} that
 \begin{equation}
 \|J^+(f)\|_a\leq A,
\label{(2.2.8)}
\end{equation}
\begin{equation}
\|J^+(f)-J^+(h)\|_a\leq \frac{1}{a}\|f-h\|_a.
\label{(2.2.9)}
\end{equation}
 for all $f, h \in {\cal F}_a^+.$ By (\ref{(2.2.8)}) and (\ref{(2.2.9)}),
it follows that $J^+$ is a contractive mapping from
  ${\cal F}_a^+$ into itself with the distance $\|\cdot-\cdot\|_a$ in
  the time interval $[0, +\infty)$
  for any given $a>1.$ This implies that if we put $a>1,$ then
there exists a unique element $g_0\in {\cal F}_a^+$ such that
$J^+(g_0)=g_0.$  Hence, by Lemma~\ref{lemma2}, Theorem~\ref{th01}
holds true.

Similarly,
 we can further show the regularity of solution to the Cauchy
problem, that is,
\begin{theorem}
Let $k$ be any fixed natural and $f$ a unique nonnegative global
distributional solution under assumptions in Theorem~\ref{th01}.
Support that $\eta$ and $\Gamma$ are in $C^k([0,+\infty)\times{\bf
R}^3\times{\bf R}^3)$ and that $P\in C^k({\bf R}).$ If $f_0\in
C^k({\bf R}^3\times{\bf R}^3),$ then $f\in C^k([0,+\infty)\times{\bf
R}^3\times{\bf R}^3).$ \label{th02}
\end{theorem}

In fact, under all the assumptions in Theorem~\ref{th02}, we can
find that $J^+$ defined by (\ref{(2.2.7)}) is still a contractive
mapping from ${\cal F}_a^+\cap C^k([0,+\infty)\times{\bf
R}^3\times{\bf R}^3)$ into itself for any fixed $a>1,$ and then
easily know that Theorem~\ref{th02} holds.

\begin{remark} In the relativistic case,  the total rest mass
distribution of UFM without gravity is uniquely determined by the
initial total rest mass data as in the nonrelativistic case. Since
the velocity $\xi$ in the relativistic case is not greater than the
light velocity $c$ (for convenience,  write $c=1$),  we
 have to choose a transformation: $\xi=\frac{p}{\sqrt{1+p^2}}$ where $p$ is
a variable of momentum for the relativistic UFM, and regard the
 total rest mass distribution as that of the three variables of time $t,$ space $x$ and momentum $p.$
Thus we can easily show the results mentioned above as done in the
nonrelativistic case,
  although the form of the transport operator $\frac{\partial}{\partial t}
  +\frac{p}{\sqrt{1+p^2}}\frac{\partial}{\partial x}$ of the relativistic
  Boltzmann-like equation \cite{wl} is different from that of
the nonrelativistic Boltzmann-like equation (\ref{(2.1.1)}). It is
here worthwhile to mention that many results about the Cauchy
problems relative to the relativistic Boltzmann or Enskog equation
have been given in a number of papers (e.g., \cite{g06}, \cite{j99},
\cite{j07a}, \cite{j07b}).
\end{remark}

\section{Asymptotic Behavior}
In this section we shall study the time-asymptotic behavior of the
solution to the Cauchy problem on UFM. We can find that this
solution is time-asymptotically convergent to the free motion in
$L^1$-norm under some suitable assumptions.

To do so, we first have to show the following lemma:
\begin{lemma}
Let $\delta$ be a positive constant less than $1.$  In addition to
all the assumptions in Theorem~\ref{th01}, if $\eta$ and $P$ satisfy
the following inequality:
\begin{equation}
\int_{{\bf R}^3}\eta(t,x,\xi)P(-\xi\xi_1)d\xi\leq\delta
 \label{con04}
\end{equation}
for all the three variables $(t,x,\xi_1),$ then the nonnegative
solution $f$ to the Boltzmann-like equation (\ref{(2.1.1)})
satisfies
\begin{equation}
\frac{d}{dt}\iint_{{\bf R}^3\times{\bf R}^3}f(t,x,\xi)dxd\xi
+(1-\delta)\iint_{{\bf R}^3\times{\bf
R}^3}\Gamma(t,x,\xi)f(t,x,\xi)dxd\xi\leq 0.
 \label{ineqfun01}
\end{equation}
Furthermore, we have
\begin{eqnarray}
\iint_{{\bf R}^3\times{\bf R}^3}f(s,x,\xi)dxd\xi \hspace{8cm}\nonumber \\
\leq(\delta-1)\int_0^s\iint_{{\bf R}^3\times{\bf
R}^3}\Gamma(t,x,\xi)f(t,x,\xi)dtdxd\xi+ \iint_{{\bf R}^3\times{\bf
R}^3}f_0(x,\xi)dxd\xi
 \label{ineqfun02}
\end{eqnarray}
for any $s\in(0,+\infty).$
 \label{lem03}
\end{lemma}

\begin{proof}
Note that $f^\#(t,x,\xi)$ is absolutely continuous with respect to
the time $t$ and that the solution $f$ satisfies
\begin{eqnarray}
\frac{\partial}{\partial t}f^\#(t,x,\xi)=
-\Gamma^\#(t,x,\xi)f^\#(t,x,\xi) \hspace*{3.5cm}\nonumber \\
+\eta^\#(t, x, \xi)\int_{{\bf R}^3}\Gamma(t, x+t\xi,
\xi_1)P(-\xi\cdot\xi_1)f(t, x+t\xi, \xi_1)d\xi_1 .
 \label{(3.1.1)}
\end{eqnarray}
Integrating equation (\ref{(3.1.1)}) over the two variables
$(x,\xi)$ and using assumptions (\ref{(2.1.2)}) and (\ref{con04}),
we can know that (\ref{ineqfun01}) holds true. Furthermore, with the
help of (\ref{(2.1.4)}), integrating (\ref{ineqfun01}) over the time
$t$ from zero to $s$ gives (\ref{ineqfun02}).
\end{proof}

\begin{remark}
The inequality (\ref{ineqfun01}) implies that the total mass in UFM
does not increase as time increases.
\end{remark}

Then, using Lemma~\ref{lem03}, we can show that
\begin{theorem}
Under all the assumptions in Lemma~\ref{lem03}, the solution
$f(t,x,\xi)$ to the Boltzmann-like equation (\ref{(2.1.1)})
converges in $L^1({\bf R}^3\times{\bf R}^3)$ to
$f_{\infty}(t,x,\xi)$ as $t\to+\infty,$ where
\begin{eqnarray}
f_{\infty}(t,x,\xi)=f_0(x-t\xi,\xi)-\int_0^{+\infty}\Gamma(\sigma,x+(\sigma-t)\xi,\xi)f(\sigma,x+(\sigma-t)\xi,\xi)d\sigma
\hspace*{0.5cm}\nonumber \\
+\int_0^{+\infty}\int_{{\bf
R}^3}\eta(\sigma,x+(\sigma-t)\xi,\xi)P(-\xi\xi_1)\Gamma(\sigma,x+(\sigma-t)\xi,\xi_1)f(\sigma,x+(\sigma-t)\xi,\xi_1)d\sigma
d\xi_1~~~~~~ \label{(3.1.2)}
\end{eqnarray}
for any point $(t,x,\xi).$
 \label{th03}
\end{theorem}

\begin{proof}
Note that $f$ is a nonnegative solution to the Boltzmann-like
equation (\ref{(2.1.1)}). Then, by (\ref{(3.1.2)}), we can easily
know that
\begin{eqnarray}
\iint_{{\bf R}^3\times{\bf
R}^3}|f(t,x,\xi)-f_{\infty}(t,x,\xi)|dxd\xi\leq\int_t^{+\infty}
\iint_{{\bf R}^3\times{\bf
R}^3}\Gamma(\sigma,x,\xi)f(\sigma,x,\xi)dxd\xi d\sigma
\hspace*{0.5cm}\nonumber \\
+\int_t^{+\infty}\iint_{{\bf R}^3\times{\bf R}^3}\left[\int_{{\bf
R}^3}\eta(\sigma,x,\xi)P(-\xi\xi_1)d\xi_1\right]\Gamma(\sigma,x,\xi_1)f(\sigma,x,\xi_1)dxd\xi
 d\sigma . \label{(3.1.3)}
\end{eqnarray}
By (\ref{con04}), it then follows that
\begin{eqnarray}
\iint_{{\bf R}^3\times{\bf
R}^3}|f(t,x,\xi)-f_{\infty}(t,x,\xi)|dxd\xi\leq
(1+\delta)\int_t^{+\infty}
\iint_{{\bf R}^3\times{\bf
R}^3}\Gamma(\sigma,x,\xi)f(\sigma,x,\xi)dxd\xi d\sigma.
\label{(3.1.4)}
\end{eqnarray}
Since the inequality (\ref{ineqfun02}) in Lemma~\ref{lem03} implies
that the integral on the right side of (\ref{(3.1.4)}) is convergent
to zero as $t\to +\infty,$ the left side of (\ref{(3.1.4)})
converges to zero as time goes to infinity. This completes our proof
of this theorem.
\end{proof}

\begin{remark}
Since $f_{\infty}(t,x,\xi)$ defined by (\ref{(3.1.2)}) describes  a
free motion in UFM, Theorem~\ref{th03} shows that the solution to
the Cauchy problem on UFM is time-asymptotically convergent to the
free motion in $L^1$-norm under some assumptions. It is worth
mentioning that this time-asymptotic convergence of solution still
holds true for the Boltzmann or Enskog equation under some suitable
assumptions (e.g., \cite{p89}).
\end{remark}

\vspace*{0.5cm}\noindent {\bf Acknowledgement.} The authors would
like to thank the referees of this paper for their valuable comments
and suggestions.

config: arxiv


TITLE: How can I determine whether an arc is clockwise or anti-clockwise?
QUESTION [3 upvotes]: I have 3 points on an Arc - Start, MidPoint & EndPoint.
I also know the Radius and the Center.
How can I determine whether an arc is clockwise or anti-clockwise relative to it's Center?

REPLY [4 votes]: Let ${\bf s}=(s_1,s_2)$, ${\bf m}=(m_1,m_2)$, ${\bf e}=(e_1,e_2)$ be the three points. We have to check whether the triangle they form is positively oriented. Therefore compute the quantity
$$\sigma:=({\bf m}-{\bf s})\wedge({\bf e}-{\bf s})=(m_1-s_1)(e_2-s_2)-(m_2-s_2)(e_1-s_1)\ .$$
If $\sigma>0$ the arc is counterclockwise, if $\sigma<0$ the arc is clockwise.


TITLE: For what conditions will two paths not induce the same isomorphism?
QUESTION [2 upvotes]: A problem in my book states:

Under what conditions will two path classes, $\gamma$ and $\gamma'$, from $x$ to $y$ give rise to the same isomorphism of $\pi(X,x)$ onto $\pi(X,y)$?

For the two isomorphisms to be the same, then we need:
$$\forall\beta \in \pi(X,x) : \gamma^{-1}\beta\gamma = (\gamma')^{-1}\beta(\gamma')$$
$$\forall \beta \in \pi(X,x) : \gamma' \gamma^{-1} \beta = \beta \gamma' \gamma^{-1}$$
Hence $\gamma'\gamma^{-1}$ commutes with every element $\beta$. This doesn't seem very satisfactory though, is there a geometrical interpretation of this condition? What does it mean for this path to commute?

REPLY [0 votes]: The algebraic model of this situation that one needs for algebraic topology is that of groupoids, which can be found developed in these downloadable books: Categories and Groupoids, Topology and Groupoids, so I'll assume you know the basic definitions, and relation with the funamental groups and  groupoid.
Here is something which is not in these books!   Let $G$ be a groupoid. There is a groupoid $AUT(G)$ whose objects are the same as those of $G$ and whose arrows $x \to y$ are the isomorphisms of object groups $G(x) \to G(y)$.There is a morphism of groupoids $\chi: G \to AUT(G)$ which sends $a: x \to y$ in $G$ to the conjugation isomorphism $G(x) \to G(y) $ given by $u \mapsto a^{-1}ua$. (There is a question   of the order in which one writes this and the conventions in these two books differ, but I'll leave you to draw the pictures and sort it out. The point is:  does one write the composite (product) in $G$ of $a: x\to y$ and $b: y \to z$ as $ab$ or $ba$?)
So you can now work out that: if $a,b: x \to y$ in $G$ then $\chi(a)= \chi(b)$ if and only if  $ba^{-1} u= u ba^{-1}$ for all $u \in G(x)$.


TITLE: ratio of alcohol to water
QUESTION [0 upvotes]: A cask initially contains pure alcohol up to the brim. The cask can be emptied by removing exactly 5 liters at a time . Each time this is done, the cask must be filled back to the brim with water. The capacity of the cask is 15 liters. When the cask is completely emptied and filled back to the brim two times, what is the ratio of alcohol to water in the cask?
now  i did not understand problem,if  it says that cask can be emptied by  removing exactly $5$ liter it means that,right now there is $5$  liter alcohol but  when it is removed and  water is  poured,then we have not  alcohol and ratio of alcohol to water  how could i calculate?i did not understand condition of problem,please could help me

REPLY [3 votes]: Since at each step you remove one third of the mixture and don't fill back any alcohol, the absolute amount of alcohol gets multiplied with $\frac23$ at each turn, so after $n$ turns, the absolute amount of alcohol is $\left(\frac23\right)^n\cdot 15$ litres and the rest is water.


TITLE: Is string theory over a time varying background a conformal field theory to all orders in perturbation theory?
QUESTION [9 upvotes]: When computing the first order perturbative corrections to string theory over a curved background, we find the background has to be Ricci-flat if the dilaton is constant and we have no fluxes. Such is the case for Calabi-Yau compactifications. However, to fourth order in perturbation theory, we find nonzero contributions to the beta function. But this can be resolved by perturbative modifications to the background metric which cancels the beta function order by order in perturbation theory.
Does this procedure work for a generic time-varying background which is Ricci-flat to first order in perturbation theory? If not, does that tell us we can't apply first quantized string theory to such backgrounds?

REPLY [2 votes]: Note: This is a repost of a former upvoted answer which was plagiarized from a MathOverflow answer by Jeff Harvey. It is reproduced below as a community wiki, since visitors of this question apparently found it helpful nevertheless.

One must distinguish between quantum/classical on the string world-sheet and in spacetime.
  Both of your statements are basically correct, but should read something like  "CFT theory is the space of classical solutions to the spacetime equations of string theory" and "Quantization of the
  the world-sheet sigma model of a string theory gives rise to a CFT."
In a little more detail,
  the sigma-model describing string theory propagation on some manifold M is a 2-dimensional
  quantum field theory which in order to describe a consistent string theory must be a conformal
  field theory. The "classical limit" of this 2-dimensional field theory is a limit in which some
  measure of the curvature of M is small in units of the string tension. To construct a CFT one
  must solve the sigma-model exactly, including world-sheet quantum effects.
The coupling constants of the sigma-model are fields in spacetime such as the metric $g_{\mu \nu}(X(\sigma))$ on $M$ where $X: \Sigma \rightarrow M$ define the embedding of the string world-sheet  $\Sigma$ into $M$. Now there
  is also a spacetime theory of these fields. You can think of it as a ``string field theory".  At low-energies it can sometimes be usefully approximated by a theory of gravity coupled to some finite number
  of quantum fields, but in full generality it is a theory of an infinite number of quantum fields. Roughly speaking, each operator in the CFT gives rise to a field in spacetime.  The spacetime string field theory lives
  in 10 dimensions for the superstring or 26 dimensions for the bosonic string and it also has a classical limit. The classical limit is $g_s \rightarrow 0$ where $g_s$ is
  a dimensionless coupling constant. It appears in perturbative string theory as a factor which
  weights the  contribution of a Riemann surface by the Euler number of the surface. It can also be
  thought of as the constant (in spacetime) mode of a scalar spacetime field known as the dilaton.
The main point is that there are two notions of classical/quantum in string theory, one involving
  the world-sheet theory, the other the spacetime theory. In order to avoid confusion one must be clear which is being discussed. Unfortunately string theorists often assume it is clear from the context.
In response to the further question about the space of string fields, I would suggest that you have a look at the introductory material in http://arXiv.org/pdf/hep-th/9305026. You may also find http://arXiv.org/pdf/hep-th/0509129 useful. I should add that while string field theory has had some success recently in the description of D-brane states, it is not widely thought to be a completely satisfactory definition of non-perturbative string theory.


TITLE: Twin paradox with one twin in orbit, one in radial free-fall
QUESTION [6 upvotes]: Many questions on Physics SE relate to the twin paradox, but I did not find any that ask this specific question. Suppose that object A is in a circular orbit around a spherically symmetric, non-rotating planet (Schwarzschild metric). Suppose that object B is in radial free-fall, with a worldline that meets object A on the way up (passing arbitrarily closely without colliding) and again on the way down, as depicted here:

Object A executes exactly one period of its orbit between the two meetings, and both objects remain in free-fall between the two meetings.
The question: Which object ages less between the two meetings?
Here's what I already know, expressed using coordinates $t,r,\phi$ where the equation for an object's proper time $\tau$ is
$$
    d\tau^2=\left(1-\frac{R}{r}\right)\,dt^2-\frac{dr^2}{1-\frac{R}{r}}
  - r^2d\phi^2,
\tag{1}
$$
where $R$ is the Schwarzschild radius. (This is in the plane that contains both objects' worldlines, so only $3$ coordinates are needed.) I already know that object A ages less when the coordinate-radius $r$ of object A's orbit is $r=3R/2$, because then A's worldline is lightlike and B's worldline is still timelike.
But does A age less than B for all radii $r>3R/2$? If not, then at what value(s) of $r$ do the two objects age the same amount between meetings?

REPLY [1 votes]: Approach
My question asks whether or not the elapsed proper time between meetings is always less for A than it is for B, for every possible radius of A's orbit.
One way to answer this would be to calculate the ratio of the two elapsed proper times as in reference [1] cited by user m4r35n357, but my question is less demanding: I only want to know which of the two elapsed proper times is smaller. This less-demanding question can be answered using a less-demanding approach.
Start with the proper-time equation shown in the question, hereafter called equation (1), and consider any worldline of the form
$$
 r(t) = r_A
\hskip2cm
 \phi(t) = \omega t
\tag{2}
$$
where $r_A$ and $\omega$ are constants, meaning that they are independent of the coordinate $t$. Object A's circular orbit has the form (2) for a special value of $\omega$ that we won't need here except for the fact that it's not zero. Now, consider one cycle of A's orbit. The second of equations (2) says that the elapsed coordinate time during one cycle is
$$
 \Delta t=\frac{2\pi}{\omega}
\tag{3}
$$
According to equation (1) in the question, the corresponding elapsed proper time is
$$
 \Delta \tau_A = \left(1-\frac{R}{r_A}-r_A^2\omega^2\right)^{1/2}
  \,\Delta t.
\tag{4}
$$
Now let $\Delta\tau_0$ denote the elapsed proper time for an object that hovers at $r=r_A$ and $\phi=0$. This object meets A once per orbit. Equation (4) gives the elapsed proper time between meetings according to A. The elapsed proper time between meetings according to the hovering object is
$$
 \Delta \tau_0 = \left(1-\frac{R}{r_A}\right)^{1/2}\,\Delta t
\tag{5}
$$
with the same value (3) of $\Delta t$. (The elapsed coordinate time depends only on the two meeting-events, not on the worldline that connects them.) Equations (4)-(5) give
$$
 \Delta\tau_A < \Delta\tau_0.
\tag{6}
$$
This doesn't answer the question yet, because the hovering object is not the one we want. We want an object B that is in radial free-fall, meeting A twice with coordinate time interval (3) between the two meetings.
Let $\Delta\tau_B$ elapsed proper time between meetings according to the radially-falling object B. The goal is to prove that
$$
 \Delta\tau_0 < \Delta\tau_B.
\tag{7}
$$
If we can prove (7), then combining the inequalities (6)-(7) will answer the question: the orbiting object A ages less than B between meetings, regardless of the orbital radius $r_A$.
 How to prove (7)
Object B is in free-fall (as is A). In other words, the worldline of object B is a geodesic. To prove (7), I will use this fact:

Among worldlines with the given endpoints, a geodesic gives a local maximum of the elapsed proper time.

A geodesic doesn't necessarily give a global maximum (that's the point of the question), but it does give a local maximum. To exploit this, we can continuously deform the hovering worldline into B's worldline in such a way that the elapsed proper time continuously increases by virtue of this local-maximum principle.
Start with the hovering worldline. The coordinate-times of the two meetings with object A are  $t=0$ and $t=\Delta t$, where $\Delta t$ is given by equation (3). Write $[a,b]$ for the coordinate-time interval from $a$ to $b$. Choose an intermediate time $T$, with $T\in[0,\Delta t]$, and consider the worldline $W_T$ defined like this:

During the interval $t\in[0,T]$, the worldline $W_T$ is a radial geodesic that starts and ends at $r=r_A$ (at times $0$ and $T$, respectively).
During the interval $t\in[T,\Delta t]$, the worldline $W_T$ is that of an object hovering at $r=r_A$.

The angular coordinate is $\phi=0$ during both segments, so the motion (if any) is purely radial. For any value of $T$, the worldline $W_T$ meets A twice, namely at times $t=0$ and $t=\Delta t$. By varying $T$ continuously from $0$ to $\Delta t$, the worldline $W_T$ varies continuously from that of an always-hovering object to that of an object that is in radial free fall during the whole interval between meetings. The latter case, namely $W_{T=\Delta t}$, corresponds to object B in the question.
From here, the proof is easy. The quantity $\Delta\tau_0$ in the inequality (6) corresponds to the worldline $W_{T=0}$. For any $T$, the worldline $W_{T}$ is a geodesic at times $t\in[0, T]$. For infinitesimal $\delta T$, the worldline $W_{T-\delta T}$ is not a geodesic during part of the interval $[0,T]$, and it differs only infinitesimally from $W_T$, which is a geodesic during all of that same interval. Therefore, the elapsed proper time for $W_T$ in the interval $[0,T]$ is greater than the elapsed proper time for $W_{T-\delta T}$ in the same interval $[0,T]$, because a geodesic locally maximizes the proper time. Since the two worldlines are identical to each other during the remaining interval $[T,\Delta t]$, we infer that the elapsed proper time for $W_T$ in the full interval $[0,\Delta t]$ is greater than the elapsed proper time for $W_{T-\delta T}$ in $[0,\Delta t]$. This is true for all $T$, so the elapsed proper time for B's worldline $W_{\Delta t}$ is greater than the elapsed proper time for the hovering worldline $W_0$. This gives the inequality (7), and combining this with (6) gives $\Delta \tau_A<\Delta \tau_B$, for arbitrary $r_A$. This answers the question.
Answer: Between meetings, the orbiting object A ages less than the radially-falling object B, regardless of the orbital radius $r_A$. This conclusion is consistent with the result of the more laborious calculations in [1], where the ratio of the elapsed proper times is calculated quantitatively.

 Comments

This approach does not assume that A's worldline is a geodesic, only that A's worldline follows a circular path. In other words, the approach works even if A is moving faster or slower than orbital speed. However, the question is most interesting when A's worldline is a geodesic, because then it illustrates the fact that geodesics don't necessarily maximize the proper time globally, even though they do locally.
This approach was inspired by the worldline-deformation arguments used in Witten's excellent review [2], which also mentions a variant of the scenario I described here.

References:
[1] "Proper Time for Intersecting Orbits", https://www.mathpages.com/rr/s6-05/6-05.htm
[2] Witten (2019), "Light rays, singularities, and all that," https://arxiv.org/abs/1901.03928


TITLE: As a child i used to do multiplication through this method. Now i want to understand the logic behind its working. Help! please
QUESTION [2 upvotes]: As a child i used to do multiplication through this method. Now i want to understand the logic behind its working. Help! please
As you can see from the image, i lined numbers and subtract them from their nearest 10th number and multiply the difference and add the original numbers put them together and cancel the first digit. it worked always but how?
!http://i.imgur.com/ct1kami.jpg?1

REPLY [2 votes]: It is quite simple.
Let we have two numbers. Say, $a$ and $b$.
Now, comsider the product -
 $(100-a)(100-b)$ ...[We are considering this product because firstly you have subtracted given two numbers from 100.]
Now, $(100-a)(100-b) = 10000-100b-100a+ab$
In the above expression, we only want $ab$[We want to find product of numbers $a$ and $b$]
So, we have to subtract $10000-100b-100a$ from above expression to get final result $ab$.
Now, subtracting $10000-100b-100a$ means adding $-10000+100b+100a=-10000+100(a+b)$ in the expression . Note that you put $a+b$ in the hundread's place which means to add $100(a+b) $ in the expression,  and removed $1$ from last, which means to subtract $10000$ from the expression.


TITLE: High-dimensional polytopes
QUESTION [1 upvotes]: I have two questions regarding polytopes in high dimensions, $\mathbb{R}^d$ where $d > 3$, that I could not find resources for on the web for. Suppose I have a polytope that is non-convex:

How can I tell if a point (in the same dimension $d$) is within the polytope, and also the distance to the closest "edge"/constraint? If it was convex and I could use the H-representation $\{x: Ax+b \geq 0\}$ to check if all constraints are positive. But since it is non-convex, what representation is suitable?

Another way to solve this is to do a convex decomposition of the complex shape to simpler and convex shapes, and then I could use the H-representation. I saw algorithms like HACD but I could not find convex decomposition for dimensions higher than $d=3$. Does something like this exist?

EDIT:
I left the representation of the non-convex polytope vague on purpose because I will conform myself to what is best suited here. I am starting from defining a physical object in $d=2$ or $3$, so I imagine that I have a bunch of vertices and edges, and from that I concatenate them (lets say 2 together) so they are in a higher dimension. The two concatenated objects are independent (if it helps). So for example, lets say I have an L-shape in the x-y plane, $\{(0,0), (1,0), (1,0.5), (0.5,0.5), (0.5, 1), (0,1), (0,0)\}$ and I concatenate two of these objects. How do I know, if the point $p=(0.1,0.8,0.0,-0.1)$ is in this polyhedron or not?
You could tell me to check each object by itself, but I might also add dimensions that are not coming from the physical space. So even one object could have $d>3$ dimensions to it.
Thank you.

REPLY [1 votes]: First of all you could consider the facet hyperplanes.
Then you might check whether those run through the origin or not.
If none does you might decompose your polytope into the various facet-centri-pyramids.
If the facets themselves aren't convex you might iterate that same process again.
Thus you finally get left with convex polytopes only and you could do your original quest with those (in the known manner).
(Sure, that does not help for the most general cases with holes running through, but it well solves for the prominent class of starshaped polytopes.)
--- rk


TITLE: Converge uniformly on close interval implies on open interval
QUESTION [0 upvotes]: Suppose $f_n(x)$ converges uniformly to $f(x)$ on $(a,b)$. Does this imply that $f_n(x)$ converges uniformly to $f(x)$ on $[a,b]$?
I have a theorem in my book that states the argument is correct if  $f_n(x)$ are all continuous on $[a,b]$. But is it true for other cases?

REPLY [1 votes]: Correct me if wrong:
$f_n(x) = 0$ for $x \in (a,b).$
$f_n(x)=1$ for $x=a$, or $x=b.$
Theorem:
$ f_n$, continuos, converges uniformly on [a,b] to f(x) implies $f(x)$ continuos on $[a,b]$
Hence choose a sequence $f_n(x)$ that is not continuous at $x=a$ or $x=b.$


TITLE: Determine divergence of $\int_0^\infty\frac{ e^x}{x}$ with limit comparison test.
QUESTION [3 upvotes]: Today I was asked if you can determine the divergence of $$\int_0^\infty \frac{e^x}{x}dx$$ using the limit comparison test.
I've tried things like $e^x$, $\frac{1}{x}$, I even tried changing bounds by picking $x=\ln u$, then $dx=\frac{1}{u}du$. Then the integral, with bounds changed becomes $\int_1^\infty \frac{1}{\ln u}du$ This didn't help either.
This problem intrigued me, so any helpful pointers would be greatly appreciated.

REPLY [0 votes]: ∫  ε to 1  e^x / x dx > ∫  ε to 1 1/x dx,  → ∞  as ε → 0; and
∫  1 to N  e^x / x dx > ∫  1 to N  1/x dx,  → ∞  as  N → ∞;
and so we are done.


text:
\section{Introduction}
Moving horizon estimation (MHE) is an optimization-based state estimation approach that computes an estimate of the state of a dynamical system by using a finite number of the most recent measurements. More specifically, a suitable optimization problem is solved to compute the optimal estimate at each time instant and the horizon of measurements is shifted forward in time whenever a new measurement becomes available. Various MHE formulations have been proposed and investigated for stability and are by now well-established in state estimation area~\cite{rao2001constrained,bookDiehl,alessandri2003receding,sui2014linear,rao2003constrained,haseltine2005critical,rawlings2012optimization}. Practical issues related to the online solution of MHE has drawn special attention since the underlying optimization problem has to be solved online at each time instant. In order to overcome this computational burden, fast optimization strategies based on interior-point methods~\cite{jorgensen2004numerical,haverbeke2009structure} are proposed, however, with no theoretical guarantees. In~\cite{alessandri2008moving},  approximation schemes are considered, in which suboptimal solutions for minimizing quadratic cost functions with a given accuracy  are allowed and upper bounds on the estimation errors are derived under observability assumptions. However, no  optimization algorithm is specified.
 A similar convergence analysis is carried in \cite{zavala2010stability} for the MHE algorithm presented in~\cite{zavala2008fast}, where a nominal background problem is solved based on predicted future measurements and when the true measurement arrives, the actual state is computed using a fast online correction step. To show that the generated  estimation errors remain bounded, the associated approximate cost  and resulting suboptimality are taken into account in the analysis.  Particularly interesting are works which explicitly consider the dynamics of the optimization algorithm in the convergence analysis \cite{alessandri2017fast,alessandri2020fast}.
 In~\cite{alessandri2017fast}, a fast MHE implementation is achieved by performing single or multiple iterations of gradient or Newton methods to minimize least-squares cost functions.  For linear systems, global exponential stability of the estimation errors is shown based on an explicit representation of the error dynamics. However, the required  observability assumption restricts the choice of the horizon length  and implies that it has to be greater than the state dimension.  Moreover, variants of the so-called real-time iteration scheme~\cite{kuhl2011real} which performs a single  Gauss-Newton iteration per time instant are proposed.
 The local convergence results derived in~\cite{wynn2014convergence}  are established for the unconstrained case, i.e. no inequality constraints are considered, and hold under the assumptions of observability and a sufficiently small initial estimation  error.
 Real-time implementations of MHE are also successfully carried out in real-world applications, such as structural vibration applications~\cite{abdollahpouri2017real}, induction machines~\cite{favato2019fast} and industrial separation processes~\cite{kupper2009efficient}.  However, theoretical studies that consider both stability as well as performance
of MHE schemes under rather mild assumptions are to the best of our knowledge rarely addressed in the literature.

\textbf{Statement of contributions.} In this work, we present a novel MHE iteration scheme for constrained linear discrete-time systems, which is based on
the idea of proximity MHE (pMHE), recently introduced in~\cite{gharbi2018proximity,gharbi2020anIteration}.
The pMHE framework exploits the advantages of a stabilizing a priori estimate from which stability can provably be inherited for any horizon length while allowing for a flexible design via rather general convex stage costs. \rev{In \cite{gharbi2018proximity}, stability is investigated under the assumption that a solution of the optimization problem is available at each time instant. In \cite{gharbi2020anIteration}, an unconstrained MHE problem in which the inequality constraints are incorporated into the cost function by means of so-called relaxed barrier functions is considered. Although only a limited number of iterations are executed at each time instant, the derived stability conditions tailored to this relaxed barrier function based formulation are rather conservative.}  The contribution  of this paper is fourfold. First, we present a pMHE iteration scheme where at each time instant, a limited number of optimization iterations are carried out and a state estimate is delivered in real-time. The underlying optimization algorithm consists of a proximal point algorithm~\cite{beck2003mirror} and is warm-started by a stabilizing a priori estimate constructed based on the Luenberger observer. Second, we establish global exponential stability  of the underlying estimation errors under minimal assumptions and by means of a Lyapunov analysis. In particular, the iteration scheme can be considered as an \textit{anytime} algorithm in which stability  is guaranteed   after any number of optimization algorithm iterations, including the case of a single iteration per time instant. \rev{Third, in contrast to the pMHE scheme in \cite{gharbi2018proximity}, the a priori estimate is only used to warm-start the proposed algorithm. Nevertheless, stability is inherited from it, despite its stabilizing effect is fading away with each iteration.} Forth, we study the performance of the pMHE iteration scheme by using the notion of regret, which is widely used in the field of online convex optimization to characterize performance~\cite{hazan2008adaptive,hazan2016introduction,hall2013dynamical}, and adapting it to our setting. More specifically, we define the regret as the difference of the accumulated costs generated by the iteration scheme relative to a comparator sequence and show that this regret can be upper bounded. Furthermore, we prove that, for any given comparator sequence, this bound can be rendered smaller by increasing the number of optimization iterations,  and that a constant regret bound can be derived for the special case of exponentially stable comparator sequences.  Overall, we present a novel anytime pMHE iteration scheme that is designed based on rather general convex stage cost functions, ensures stability after each iteration as well as for any horizon length,  and for which performance guarantees are provided and characterized in terms of rigorously derived  regret bounds.

\textbf{Organization.} The paper is  organized as follows. The constrained MHE problem for discrete-time linear systems is stated in Section \ref{sec: setup}. The proposed pMHE iteration scheme is described in details in Section \ref{sec: iteration scheme} and its stability properties are established in Section \ref{sec: stability}. In Section \ref{sec: regret}, the focus is on the performance properties of the iteration scheme which are reflected by the derived regret upper bounds. A simulation example that illustrates both the stability and performance properties is presented in Section \ref{sec: sim}. Finally, conclusions are drawn in Section \ref{sec: conclusion}.
\\
~\\
  \textit{Notation:} Let $\mathbb{N}_{+}$ denote the set of positive natural numbers, $\mathbb{R}_{+}$ and $\mathbb{R}_{+\!\!+}$  the sets of nonnegative real and positive real numbers, respectively, and $\mathbb{S}_{+}^n$ and $\mathbb{S}_{+\!\!+}^n$ the sets of symmetric positive semi-definite and positive-definite matrices of dimension $n \in \mathbb{N}_{+}$, respectively. For a vector $v\in \mathbb{R}^n$, let  $\left\Vert v \right\Vert_P \coloneqq \sqrt{v^{\top}P\,v}$  for any $P \in \mathbb{S}_{+}^n $.  Moreover, let $\mathbf{0} \coloneqq \begin{bmatrix} 0 & \cdots & 0 \end{bmatrix}^\top$.


text:
{\bf Problem.} If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1,$ find $b.$
{\bf Level.} Level 3
{\bf Type.} Intermediate Algebra
{\bf Solution.} If $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1,$ then the other factor must be linear, where the coefficient of $x$ is $a,$ and the constant coefficient is $-1.$  Thus
\[(x^2 - x - 1)(ax - 1) = ax^3 + bx^2 + 1.\]Expanding, we get
\[ax^3 - (a + 1) x^2 + (1 - a) x + 1 = ax^3 + bx^2 + 1.\]Matching coefficients, we get
\begin{align*}
-(a + 1) &= b, \\
1 - a &= 0.
\end{align*}Hence, $a = 1.$  Then $b = -(a + 1) = \boxed{-2}.$

set_name: MATH


TITLE: Notation to express $(n^1+n^2+...+n^k)$.
QUESTION [0 upvotes]: What are some mathematical conventions for expressing $(n^1+n^2+...+n^k)$ in a simpler format?

REPLY [2 votes]: This is a geometric series.  You could write it as $\sum \limits_{j=1}^k n^j$
If $S = n^1+n^2+n^3+n^4 +\cdots +n^k$
then  $nS=n^2+n^3+n^4 +\cdots +n^k+n^{k+1}$
so by subtraction $(n-1)S= n^{k+1} -n^1$
and thus $S = \dfrac{n}{n-1}(n^k-1)$

REPLY [1 votes]: There is a sum-notation:
$$\sum_{i=1}^k n^i=n+n^2+\dotso +n^{k-1}+n^k$$
Also this is a geometric sum so there is an explicit formula for it given by:
$$\frac{n(n^k-1)}{n-1}$$


TITLE: Computation of 2nd homology using Hopf's formula
QUESTION [5 upvotes]: Let $G$ be a group and $G'$ be a group obtained from $G$ by adding a one generator $x$ and relations $gx=xg$.
That is, $G'=G\times\mathbb{Z}$ and $H_2(G')=H_2(G)$.
Problem. Prove $H_2(G')=H_2(G)$ using Hopf's formula.
Suppose that we fix a presentation of $G$, say $G=F/R$.
Then, $G'=F'/R'$, where $F'=F\ast \mathbb{Z}\langle x\rangle$ and $R'$ is a subgroup generated by $R$ and $[y,x]$ for all $y\in F$.
Hopf's formula shows $H_2(G)=\frac{R\cap [F,F]}{[F,R]}$ and $H_2(G')=\frac{R'\cap [F',F']}{[F',R']}$.
Since $[y,x]\in [F',F']$ for all $y\in F$, $R'\cap [F',F']$ contains $R\cap [F,F]$ and all words of the form $[y,x]$. However, I could not see that why $[y,x]$'s lie in $[F',R']$.
Any help?

REPLY [1 votes]: It follows from the Kunneth Theorem that $$H_2(G\times {\mathbb Z})=(H_2(G)\otimes H_0({\mathbb Z}))\oplus (H_1(G)\otimes H_1({\mathbb Z}))$$
$$=H_2(G)\oplus H_1(G)$$
So your statement is only true if $H_1(G)=0$.


TITLE: Measures for homogeneity of a finite sequence - approximating an image with lines.
QUESTION [1 upvotes]: I am trying to approximate a grayscale image with lines, using a greedy algorithm. For our purposes, the line is a finite sequence, with each number being between $0$ and $1$, where $0$ is pure white and $1$ pure black. I evaluate the fitness of a line as follows:
Treat a line as a set of coordinates (pixels) $L \subset \mathbb{N_0^2}$. I use the Bresenham Line Algorithm to decide what pixels belong on the line.
The function $f(p_{i,j}) \in [0, 1]$ takes the color value of a pixel at position $i, j$ and outputs the color value that that pixel will take, after the line goes over it. In the simplest case, where the lines are always black, it just outputs $1$, but it could also output $p_{i,j} + \frac{1 - p_{i,j}}{2}$ for 50% transparent lines etc.
The function $g(i, j) \in [0, 1]$ gives the color value of the pixel at $i, j$ in the target image.
Then the fitness functions is given as the sum of differences between the pixels on the line in the target image and the current image after is drawn: $$F(L) = \frac{1}{|L|} \cdot \sum_{(i, j) \in L} |f(p_{i,j}) - g(i,j)|$$
The best line is the one that has lowest value of $F(L)$.
The problem that I've encountered is that an image like this:
is being approximated like that:  because vertical lines'fit' the image the best. However, this destroys the details that humans, who would look at the picture, care about.
To counter that I thought of factoring in the 'homogeneity' of a line, by treating the line as a sequence instead of a set, and adding a penalty to the fitness score, if a line goes over a section that contains a lot of changes.
Question: What suitable measures of 'homogeneity' in this context are there?

For reference, I would like the result to look something like that:  here I changed the color of the background from white to gray to achieve the result, but it only works because this image is so simple, the same approach wouldn't work with more complex images.

REPLY [1 votes]: Measure the 'Mean Number of Pixels Since Last Change'?
The model/intuition/analogy here is from reliability theory, specifically system that alternates between Working vs Broken. Here Working = Draw As Black vs Broken = Draw As White.
One of the common measures in such systems is the Mean Time To Failure. By extrapolation, it seems you want lines with the equivalent of a long Mean Time To Failure rather than a short one.