text:
\begin{document}
\maketitle
\begin{abstract}
We prove that the interval $(5/6,\, 1)$ contains no $3$-dimensional canonical
thresholds.
\end{abstract}

\section{Introduction}

Let $(X\ni P)$ be a three-dimensional
canonical singularity
and let $S\subset X$ be a $\QQ$-Cartier divisor.
The \emph{canonical threshold} of the pair $(X,S)$ is
\[
\ct(X,S):=\sup \{ c \mid \text{the pair $(X,cS)$ is canonical}\}.
\]
It is easy to see that $\ct(X,S)$ is rational and non-negative.
Moreover, if $S$ is effective and integral, then $\ct(X,S)\in [0,\, 1]$.
Define the subset $\Tc_n\subset [0,\, 1]$ as follows
\[
\Tc_n:=\{\ct(X,S) \mid \text{$\dim X=n$, $S$ is
integral and effective}
\}.
\]

The following conjecture is an analog of
corresponding conjectures for log canonical
thresholds and minimal discrepancies, see \cite{Shokurov1988p},
\cite{Utah}, \cite{Kollar-1995-pairs}, \cite{Prokhorov-McKernan-2004}, \cite{kollar-2008}.

\begin{conjecture}
\label{conj-acc}
The set $\Tc_n$ satisfies the ascending chain condition.
\end{conjecture}
The conjecture is interesting for applications to
birational geometry, see, e.g., \cite{Corti1995}.
It was shown in \cite{Birkar-Shokurov} that much more general form of \ref{conj-acc}
follows from ACC for minimal log discrepancies
and weak Borisov-Alexeev conjecture.
The important particular case of \ref{conj-acc} is the following
\begin{conjecture}[cf. {\cite{kollar-2008}}]
\label{conj-gap}
$\epsilon_n^{\mathrm{can}}:=1-\sup (\Tc_n\setminus \{1\})>0$.
\end{conjecture}
\noindent
The aim of this note is to prove Conjecture \ref{conj-gap} for $n=3$ in a
precise form:

\begin{theorem}
\label{main-1}
$\epsilon^{\mathrm{can}}_3=1/6$.
\end{theorem}
An analog of this theorem for log canonical thresholds was
proved by J. Koll\'ar \cite{Kollar1994}: $\epsilon^{\mathrm{lc}}_3=1/42$.

Note that replacing $(X\ni P)$ with its terminal $\QQ$-factorial
modification we may assume that $(X\ni P)$ is terminal.
Thus the following is a stronger form of Theorem \ref{main-1}:

\begin{theorem}
\label{main-2}
Let $(X\ni P)$ be a three-dimensional terminal singularity
and let $S\subset X$ be an \textup(integral\textup)
effective Weil $\QQ$-Cartier divisor
such that the pair $(X,S)$ is not canonical.
Then $\ct(X,S)\le 5/6$.
Moreover, if $(X\ni P)$ is singular, then $\ct(X,S)\le 4/5$.
\end{theorem}
The proof is rather standard.
We use the classification of terminal singularities and
weighted blowups techniques, cf. \cite{Kawamata-1992-discr}, \cite{Kollar1994},
\cite{Markushevich-1996-discr}.

\section{Preliminaries}

\begin{case} \textbf{Notation.}
For a polynomial $\phi$,
$\ord_0 \phi$ denotes the order of vanishing of
$\phi$ at $0$ and
$\phi_d$ is the homogeneous
component of degree $d$.

Throughout this paper we let $(X\ni P)$ be the germ of
a three-dimensional terminal singularity and let
$S\subset X$ be an
effective Weil $\QQ$-Cartier divisor
such that the pair $(X,S)$ is not canonical.
Put $c:=\ct(X,S) >0$. Since $(X,S)$ is not canonical, $c<1$.

We work over the complex number field $\CC$.
\end{case}

\begin{lemma}
In the above notation the singularity $(S\ni P)$ is not Du Val.
\end{lemma}

\begin{proof}
This is well-known, see e.g. \cite[Th. 2.6]{Reid-1980can}.
\end{proof}

\begin{case}
We use the techniques of weighted blowups.
For definitions and basic properties we refer, for example,
to \cite{Markushevich-1996-discr}, \cite{Reid-YPG1987}.
By fixing coordinates $x_1,\dots,x_n$
we regard the affine space $\CC^n$ as a toric variety.
Let $\alphaa=(\alpha_1,\dots,\alpha_n)$ be a weight
(a primitive lattice vector in the positive octant)
and let $\sigma_{\alpha}\colon \CC^n_{\alphaa}\to \CC^n$
be the weighted blowup with weight $\alphaa$ ($\alphaa$-blowup).
The exceptional divisor $E_{\alphaa}$ is irreducible and
determines a discrete valuation $v_{\alphaa}$ of the function field
$\CC(\CC^n)$ such that $v_{\alphaa}(x_i)=\alpha_i$.
\end{case}

\begin{case}
Now let $X\subset \CC^n$ be a hypersurface given by
the equation $\phi=0$ and let $X_{\alphaa}\subset \CC^n_{\alphaa}$
be its proper transform.
Fix an irreducible component $G$ of $E_{\alphaa}\cap X_{\alphaa}$
such that $X_{\alphaa}$ is smooth at the generic point of $G$.
Let $v_G$ be the corresponding discrete valuation of $\CC(X)$.
Write
\[
E_{\alphaa}\mid_{ X_{\alphaa}}=m_G G+(\text{other components}).
\]
Assume that $d_G=1$ and $G$ is not a toric subvariety
in $\CC^n_{\alphaa}$. Then the discrepancy of $G$ with respect to $K_X$
is computed by the formula
\[
a(G,K_X)=|\alphaa|-1-v_{\alphaa}(\phi), \quad |\alphaa|=\sum \alpha_i,
\]
see \cite{Markushevich-1996-discr}. Let $S\subset X$ be a Cartier divisor and let $\psi$
be a local defining equation of $S$ in $\OOO_{0,X}$.
Then $v_G(\psi)=v_{\alphaa}(\psi)$ and
the discrepancy of $G$ with respect to $K_X+cS$
is computed by the formula
\[
a(G,K_X+cS)=a(G,K_X)-cv_{G}(\psi)=|\alphaa|-1-v_{\alphaa}(\phi)-cv_{\alphaa}(\psi).
\]
Therefore,
\[
c \le a(G,K_X)/v_{\alphaa}(\psi)= (|\alphaa|-1-v_{\alphaa}(\phi))/v_{\alphaa}(\psi).
\]
\end{case}

\begin{definition}[cf. {\cite{Markushevich-1996-discr}}]
A weight $\alphaa$ is said to be \textit{admissible} if
$E_{\alphaa}\cap X_{\alphaa}$ contains at least one
reduced non-toric component.
\end{definition}

\section{Gorenstein case}
In this section we consider the case where $(X\ni P)$ is either smooth or
an index one singularity.
\begin{lemma}
\label{l-G-1}
If $(X\ni P)$ is smooth, then $c\le 5/6$.
\end{lemma}
\begin{proof}
Let $c>5/6$.
We may assume that $X=\CC^3$.
Let $\psi(x,y,z)=0$ be an equation of $S$.
Consider a weighted blowup
$\sigma_{\alphaa}\colon \CC^3_{\alphaa}\to \CC^3$ with a suitable
weight $\alphaa$. Let $E_\alphaa$ be the exceptional divisor.
Recall that $(S\ni P)$ is not Du Val.
Up to analytic coordinate change there are the following cases
(cf. \cite[4.25]{Kollar-Mori-19988}):

\begin{case}
\textbf{Case $\ord_0 \psi\ge 3$.}
Take $\alphaa=(1,1,1)$ (usual blowup of $0$).
Then $a(E_{\alphaa}, K_X) =2$, $v_{\alphaa}(\psi)=\ord_0 \psi\ge 3$. Hence
$c\le a(E_{\alphaa}, K_X)/v_{\alphaa}(\psi) \le 2/3$, a contradiction.
\end{case}

\begin{case} \textbf{Case $\psi=x^2+\eta(y,z)$, where $\ord_0 \eta\ge 4$.}
Take $\alphaa=(2,1,1)$.
Then $a(E_{\alphaa}, K_X) =3$, $v_{\alphaa}(\psi)=4$. Hence
$c\le a(E_{\alphaa}, K_X) /v_{\alphaa}(\psi)\le 3/4$, a contradiction.
\end{case}

\begin{case} \textbf{Case $\psi=x^2+y^3+\eta(y,z)$, where $\ord_0 \eta\ge 4$.}
Here $\eta$ contains no terms $yz^l$, $l\le 3$ and $z^l$, $l\le 5$
(see, e.g., \cite[4.25]{Kollar-Mori-19988}).
Take $\alphaa=(3,2,1)$.
Then $a(E_{\alphaa}, K_X) =5$, $v_{\alphaa}(\psi)=6$. Hence
$c\le a(E_{\alphaa}, K_X) /v_{\alphaa}(\psi)= 5/6$, a contradiction.
\end{case}
\end{proof}

\begin{lemma}
\label{l-G-2}
Assume that $(X\ni P)$ is a Gorenstein terminal singularity
and $(X\ni P)$ is not smooth. Then $c\le 4/5$.
\end{lemma}

\begin{proof}
Let $c>4/5$.
We may assume that $X$ is a hypersurface in $\CC^4$
(it is an isolated cDV-singularity \cite{Reid-1980can}).
Let $\phi(x,y,z,t)=0$ be the equation of $X$.
Since $(X\ni P)$ is a cDV-singularity,
$\ord_0 \phi=2$. According to \cite{Markushevich-1996-discr},
in a suitable coordinate system $(x,y,z,t)$, there is an admissible weighted blowup
$\sigma_{\alphaa} \colon \CC^4_{\alphaa}\to \CC^4$ such that at least
for one component $G$ of $E_{\alphaa}\cap X_{\alphaa}$
we have $a(G,K_X)=1$. Then $c\le 1/v_{\alphaa}(\psi)$, so $v_{\alphaa}(\psi)=1$.
This means, in particular, that $\ord_0 \psi=1$.
Up to coordinate change we may assume that
$\psi=t$. Write
\[
\phi=\eta(x,y,z)+t \zeta (x,y,z,t).
\]
Then $S$ is a hypersurface in $\CC^3_{x,y,z}$
given by $\eta(x,y,z)=0$.
As in the proof of Lemma \ref{l-G-1},
using Morse Lemma we get the following cases:

\begin{case} \textbf{Case $\ord_0 \eta\ge 3$.}
Take $\alphaa=(1,1,1,2)$.
By the terminality condition \cite[Th. 4.6]{Reid-YPG1987},
we have $4=v_{\alphaa}(xyzt)-1>v_{\alphaa}(\phi)$.
Hence, $v_{\alphaa}(\eta)\le 3$ and $\eta_3\neq 0$.
We claim that $\alphaa$ is
admissible whenever $\eta_3$ is not a cube of a linear form.
Indeed, in the affine chart $U_x:=\{x\neq 0\}$
the map $\sigma_{\alphaa}^{-1}$ is given by
\begin{equation}
\label{eq-odd3}
x \mapsto x',\quad y \mapsto y'x',\quad z\mapsto z'x',\quad
t \mapsto t'x'^2.
\end{equation}
First we assume that $\zeta$ contains the term $x$.
After the coordinate change $x\mapsfrom \zeta (x,y,z,t)$ we obtain
\[
\phi=\eta(x,y,z)+tx.
\]
Using \eqref{eq-odd3} we see that
$E_{\alphaa}\cap X_{\alphaa}$ is given in $\sigma_{\alphaa}^{-1}(U_x)\simeq
\CC^4$ by
\[
x'=\eta_3(1,y',z')+t'=0.
\]
Hence $\alphaa$ is admissible, i.e.,
$E_{\alphaa}\cap X_{\alphaa}$ has
a reduced non-toric component $G$. Then
$a(G, K_X) =1$,
$v_{G}(\psi)=2$ and $c\le a(G, K_X) /v_{G}(\psi)=1/2$, a contradiction.

Thus by symmetry we may assume that $\zeta$ contains no terms $x$, $y$, $z$.
Since $\ord_0 \phi=2$, $\zeta$ contains $t$. So,
\[
\phi=\eta(x,y,z)+t^2+ t\xi (x,y,z,t),\quad \ord_0 \xi\ge 2.
\]
As above, $E_{\alphaa}\cap X_{\alphaa}$ is given in $\CC^4$ by
$x'=\eta_3(1,y',z')=0$.
If $\eta_3$ is not a cube of a linear
form, then $E_{\alphaa}\cap X_{\alphaa}$ has a reduced non-toric component $G$.
Then, as above,
$c\le 1/2$, a contradiction.

Consider the case where $\eta_3$ is a cube of a linear form.
Then we may assume that
$\eta_3(x,y,z)=y^3$, so
\[
\phi=y^3+\eta^{\bullet}(x,y,z)+t^2+ t\xi (x,y,z,t),\quad \ord_0 \xi\ge 2, \quad
\ord_0\eta^{\bullet} \ge 4.
\]
Put $\alphaa'=(2,2,2,3)$. Again,
in the affine chart $U_x:=\{x\neq 0\}$
the map $\sigma_{\alphaa'}^{-1}$ is given by
$x \mapsto x'^2$, $y \mapsto y'x'^2$, $z\mapsto z'x'^2$,
$t \mapsto t'x'^3$,
where $\sigma_{\alphaa'}^{-1}(U_x)\simeq \CC^4/\muu_2(1,0,0,1)$ and
\[
E_{\alphaa'}\cap X_{\alphaa'}\cap \sigma_{\alphaa'}^{-1}(U_x)=
\{
x'=0,\ y'^3+t'^2 =0\}.
\]
Thus $\alphaa'$ is admissible and for some component $G'$ of
$X_{\alphaa'}\cap E_{\alphaa'}$ we have
$a(G', K_X)=2$, $v_{G'}(\psi)=3$, $c\le 2/3$, a contradiction.
\end{case}

\begin{case} \textbf{Case $\eta=x^2+\xi(y,z)$, where $\ord_0 \xi\ge 4$.}
By Morse Lemma we may assume that $\zeta$ does not depend on $x$.
Write $\zeta_1=\delta_1y+\delta_2z+\delta_3t$, $\delta_i\in \CC$.
Take $\alphaa=(2,1,1,3)$.
In the affine chart $U_y:=\{y\neq 0\}$
the map $\sigma_{\alphaa}^{-1}$ is given by
$x \mapsto x'y'^2$, $y \mapsto y'$, $z\mapsto z'y'$,
$t \mapsto t'y'^3$ and
\[
E_{\alphaa}\cap X_{\alphaa}\cap \sigma_{\alphaa}^{-1}(U_y)=
\{ y'=0,\ x'^2+\xi_4(1,z')+\delta_1t' +\delta_2t'z'=0\}.
\]
If either $\delta_1\neq 0$ or $\delta_2\neq 0$ or $\xi_4\neq 0$, then
$E_{\alphaa}\cap X_{\alphaa}$ is reduced (at least over $U_y$).
Hence,
$\alphaa$ is
admissible and for some component $G$ of $E_{\alphaa}\cap X_{\alphaa}$
we have
$c\le a(G, K_X) /v_{G}(\psi)= 2/3$, a contradiction.
Thus $\delta_1=\delta_2= 0$ and $\xi_4=0$.
Then we can write
\[
\phi=x^2+\xi(y,z)+\delta_3t^2+t\zeta^{\bullet}(y,z,t), \quad \ord_0\xi\ge 5,
\quad \ord_0\zeta^{\bullet}\ge 2.
\]
Take $\alphaa'=(2,1,1,2)$.
In the affine chart $U_y:=\{y\neq 0\}$
the map $\sigma_{\alphaa'}^{-1}$ is given by
$x \mapsto x'y'^2$, $y \mapsto y'$, $z\mapsto z'y'$,
$t \mapsto t'y'^2$ and
\[
E_{\alphaa'}\cap X_{\alphaa'}\cap \sigma_{\alphaa'}^{-1}(U_y)=
\{ y'=0,\ x'^2+\delta_3t'^2+t\lambda(1,z')=0\},
\]
where $\lambda$ is the degree $2$
homogeneous part of $\zeta(y,z,0)$.
If $\delta_3\neq 0$ or $\lambda\neq 0$,
as above, $\alphaa'$ is
admissible and $c\le 1/2$, a contradiction.
Thus $\delta_3=0$, $\lambda= 0$, and
\[
\phi=x^2+\xi(y,z)+\delta t^3+t\zeta^{\circ}(y,z,t), \quad \delta\in \CC, \quad\ord_0\xi\ge 5,
\quad \ord_0\zeta^{\circ}\ge 3.
\]
Applying the terminality condition \cite[Th. 4.6]{Reid-YPG1987}
with weight $(2,1,1,1)$ we get that $\delta\neq 0$.

Take $\alphaa''= (3,1,1,2)$. As above we get that $\alphaa''$
is admissible and then
$c\le 1/2$, a contradiction.
\end{case}

\begin{case} \textbf{Case $\eta=x^2+y^3+\xi(y,z)$, where $\ord_0 \xi\ge 4$.}
Here $\xi$ contains no terms $yz^l$, $l\le 3$ and $z^l$, $l\le 5$
(see, e.g., \cite[4.25]{Kollar-Mori-19988}).
Write $\zeta_1=cz+\ell(x,y,t)$ and
$\xi=\xi_{(6)}+\xi_{(7)}+\cdots$, where $\xi_{(d)}$ is
the degree $d$
weighted homogeneous part of $\xi$ with respect to $\wt(y,z)=(2,1)$.
Here $\xi_{(6)}$ is a linear combination of $z^6$, $yz^4$,  $y^2z^2$.
Take $\alphaa=(3,2,1,5)$.
In the affine chart $U_z:=\{z\neq 0\}$
the map $\sigma_{\alphaa}^{-1}$ is given by
$x \mapsto x'z'^3$, $y \mapsto y'z'^2$, $z\mapsto z'$,
$t \mapsto t'z'^5$ and
\[
E_{\alphaa}\cap X_{\alphaa}\cap \sigma_{\alphaa}^{-1}(U_z)=
\{z'=0,\ x'^2+y'^3+\xi_{(6)}(y',1)+\delta t'=0\},
\]
where $\delta$ is a constant and $\xi_{(6)}(y',1)$ contains no $y'^3$.
Hence $\alphaa$ is admissible, i.e.,
$E_{\alphaa}\cap X_{\alphaa}$ has
a reduced non-toric component $G$. Then
$a(G, K_X) =4$, $v_{G}(\psi)=5$, and
$c\le a(G, K_X) /v_{G}(\psi)\le 4/5$, a contradiction.
\end{case}
\end{proof}

The following examples show that
bounds $\ct(X,S)\le 5/6$ and $\le 4/5$ in
Theorem \ref{main-2} are sharp.

\begin{example}
Let $X=\CC^3$ and let $S=S^d$ is given by $x^2+y^3+z^d$, $d\ge 6$.
Then $\ct(\CC^3,S^d)=5/6$. We prove this by descending induction on $\down{d/6}$.
Take $\alphaa=(3,2,1)$ and consider the $\alphaa$-blowup
$\sigma_{\alphaa}\colon \CC^3_{\alphaa}\to \CC^3$.
Let $S_{\alphaa}\subset X_{\alphaa}$ be the proper transform of $S$.
We have $a(E_{\alphaa}, K_X) =5$ and $v_{\alphaa}(\psi)=6$. Hence, $\ct(\CC^3,S^d)\le 5/6$.
Further,
\[
\textstyle
K_{\CC^3_{\alphaa}}+\frac56S_{\alphaa}=\sigma_{\alphaa}^*(K_{\CC^3}+\frac56S).
\]
Thus it is sufficient to show that $\ct(X_{\alphaa},\frac 56 S_{\alphaa})$ is canonical.
We have three affine charts:
\begin{itemize}
 \item
$U_x:=\{x\neq 0\}$. Here $\sigma_{\alphaa}^{-1}\colon$
$x \mapsto x'^3$, $y \mapsto y'x'^2$, $z\mapsto z'x'$,
$S_{\alphaa}$ is given in $\sigma_{\alphaa}^{-1}(U_x)\simeq\CC^3/\muu_3(-1,2,1)$ by the equation
$1+y'^3+z'^dx'^{d-6}=0$. Hence, in this chart,
$S_{\alphaa}$ is smooth and does not pass through a (unique) singular point
of $\sigma_{\alphaa}^{-1}(U_x)$.
 \item
$U_y:=\{y\neq 0\}$. Here $\sigma_{\alphaa}^{-1}\colon$
$x \mapsto x'y'^3$, $y \mapsto y'^2$, $z\mapsto z'y'$,
$S_{\alphaa}$ is given in $\sigma_{\alphaa}^{-1}(U_y)\simeq \CC^3/\muu_2(3,-1,1)$ by the equation
$x'^2+1+z'^dy'^{d-6}=0$. Again, in this chart,
$S_{\alphaa}$ is smooth and does not pass through a (unique) singular point
of $\sigma_{\alphaa}^{-1}(U_y)$.

 \item
$U_z:=\{z\neq 0\}$. Here $\sigma_{\alphaa}^{-1}\colon$
$x \mapsto x'z'^3$, $y \mapsto y'z'^2$, $z\mapsto z'$,
$S_{\alphaa}$ is given in $\sigma_{\alphaa}^{-1}(U_z)\simeq \CC^3$ by the equation
$x'^2+y'^3+z'^{d-6}=0$. In this chart, $(X_{\alphaa},S_{\alphaa})\simeq (\CC^3, S^{d-6})$.
\end{itemize}
Thus $X_{\alphaa}$ has only terminal singularities,
$S_{\alphaa}$ does not pass through any singular point of
$X_{\alphaa}$,
and the pair $(X_{\alphaa}, S_{\alphaa})$ is terminal in
charts $U_x$ and $U_y$. In the chart $U_z$ the pair
by induction $(X_{\alphaa},\frac 56 S_{\alphaa})$ is canonical
(moreover, $(X_{\alphaa}, S_{\alphaa})$ is canonical if $d\le 11$).
Therefore, $\ct(X,S)=5/6$.
\end{example}

\begin{example}
Let $X\subset \CC^4$ is given by
$x^2+y^3+z^d+tz=0$, $d\ge 7$ and let $S$ cut out by $t=0$.
Take $\alphaa=(3,2,1,5)$ and
consider the $\alphaa$-blowup $\sigma_{\alphaa} \colon X_{\alphaa}\to X$.
Let $S_{\alphaa}\subset X_{\alphaa}$ be the proper transform of $S$.
We see below that $\alphaa$ is admissible.
Moreover, the exceptional divisor $G:=E_{\alphaa}\cap X_{\alphaa}$
is reduced and irreducible.
We have four charts:
\begin{itemize}
 \item
$U_x:=\{x\neq 0\}$. Here $\sigma_{\alphaa}^{-1}\colon
x \mapsto x^3$, $y \mapsto yx^2$, $z\mapsto zx$, $t\mapsto tx^5$,
$X_{\alphaa}$ is given in $\sigma_{\alphaa}^{-1}(U_x)\simeq\CC^4/\muu_3(-1,2,1,5)$ by the equation
$1+y^3+z^dx^{d-6}+tz=0$ and $S_{\alphaa}$ by two equations $x=1+y^3+tz=0$.
Hence, in this chart, both $X_{\alphaa}$ and
$S_{\alphaa}$ are smooth.
 \item
$U_y:=\{y\neq 0\}$. Here
$\sigma_{\alphaa}^{-1}\colon
x \mapsto xy^3$, $y \mapsto y^2$, $z\mapsto zy$, $t\mapsto ty^5$,
$\sigma_{\alphaa}^{-1}(U_y)\simeq \CC^4/\muu_2(3,-1,1,5)$,
$X_{\alphaa}=\{x^2+1+z^dy^{d-6}+tz=0\}$, and $S_{\alphaa}=\{y=x^2+1+tz=0\}$.
As above, both $X_{\alphaa}$ and
$S_{\alphaa}$ are smooth in this chart.
 \item
$U_z:=\{z\neq 0\}$. Here $\sigma_{\alphaa}^{-1}\colon
x \mapsto xz^3$, $y \mapsto yz^2$, $z\mapsto z$, $t\mapsto tz^5$,
$\sigma_{\alphaa}^{-1}(U_z)\simeq \CC^4$,
$X_{\alphaa}=\{x^2+y^3+z^{d-6}+t=0\}$, and $S_{\alphaa}=\{z=x^2+y^3+t=0\}$.
As above, both $X_{\alphaa}$ and
$S_{\alphaa}$ are smooth in this chart.

 \item
$U_t:=\{t\neq 0\}$. Here $\sigma_{\alphaa}^{-1}\colon
x \mapsto xt^3$, $y \mapsto yt^2$, $z\mapsto zt$, $t\mapsto t^5$,
$\sigma_{\alphaa}^{-1}(U_t)\simeq \CC^4/\muu_5(3,2,1,-1)$,
$X_{\alphaa}=\{x^2+y^3+z^dt^{d-6}+z=0\}$, and $S_{\alphaa}=\{t=x^2+y^3+z=0\}$.
The variety
$X_{\alphaa}$ has a unique singular point $Q$ at the origin and this point is terminal of type
$\frac15(3,2,-1)$ the surface
$S_{\alphaa}$ is smooth and does not pass through $Q$.
\end{itemize}
Thus we have $a(G, K_X) =4$, $v_{\alphaa}(\psi)=5$, and $a(G, K_X+ \frac45 S) =0$.
Therefore,
\[
\textstyle
K_{X_{\alphaa}}+\frac45S_{\alphaa}=\sigma_{\alphaa}^*(K_{X}+\frac45S).
\]
Since the pair $K_{X_{\alphaa}}+\frac45S_{\alphaa}$ is canonical, $\ct(X,S)=4/5$.
\end{example}

\section{Non-Gorenstein case}
Now we assume that $(X\ni P)$ is a (terminal) point of index $r>1$.
Let $\pi \colon (X^\sharp\ni P^\sharp)\to (X\ni P)$ be the index-one
cover and let $S^\sharp:=\pi^{-1}(S)$.

\begin{lemma}
\label{l-nG-1}
If $(X\ni P)$ is a cyclic quotient singularity, then $\ct(X,S)\le 1/2$.
\end{lemma}

\begin{proof}
By our assumption we have
$X\simeq \CC^3/\muu_r(a,-a,1)$ for some $r\ge2$, $1\le a <r$, $\gcd(a,r)=1$.
Assume that $c=\ct(X,S)> 1/2$.
Let
$\psi=0$ be a defining equation of $S^\sharp$.
Consider the weighted blowup $\sigma_{\alphaa}\colon X_{\alphaa}\to X$
with weights $\alphaa =\frac1r (a,r-a,1)$.
Then $a(E_{\alphaa}, K_X)=1/r$.
Since $a(E_{\alphaa}, K_X)-c v_\alphaa (\psi)\ge 0$, we have
$v_\alphaa (\psi)\le a(E_{\alphaa}, K_X)/c< 2a(E_{\alphaa}, K_X)= 2/r$
and so $v_\alphaa (\psi)=1/r$.
Thus we may assume that $\psi$ contains $x_3$
(if $a\equiv \pm 1$ we possibly have to permute coordinates).
Then $S^\sharp\simeq \CC^2$ is smooth and $S\simeq \CC^2/\muu_r(a,-a)$, i.e.,
$S$ is Du Val of type $A_{r-1}$.
\end{proof}

\begin{lemma}
\label{l-nG-2}
If $(X\ni P)$ is a terminal singularity of index $r>1$
and $\ct(X,S)>1/2$, then $K_X+S\sim 0$.
\end{lemma}

\begin{proof}
By Lemma \ref{l-nG-1} $(X\ni P)$ is not a cyclic quotient singularity.
There is an analytic $\muu_r$-equivariant
embedding $(X^\sharp, P^\sharp) \subset (\CC^4,0)$.
Let $(x_1,x_2,x_3,x_4)$ be coordinates in $\CC^4$, let $\phi=0$ be an
equation of $X^\sharp$, and let $\psi=0$ be an equation of $S^{\sharp}$.
We can take $(x_1,x_2,x_3,x_4)$ and $\phi$
to be semi-invariants
such that one of the following holds \cite{Reid-YPG1987}:
\begin{enumerate}
\item[-]
\textbf{Main series.}
$\wt (x_1,x_2,x_3,x_4; \phi)\equiv (a,-a, 1, 0;0)\mod r$, where $\gcd (a,r)=1$.
\item[-]
\textbf{Case $cAx/4$.} $r=4$,
$\wt (x_1,x_2,x_3,x_4; \phi)\equiv (1,3, 1, 2;2)\mod 4$.
\end{enumerate}
In both cases $\wt (x_1x_2x_3x_4)- \wt \phi\equiv \wt x_3\mod r$.
According to \cite{Kawamata-1992-discr} there is a weight
$\alphaa$
such that for the corresponding $\alphaa$-blowup
$\sigma_{\alphaa}\colon X_{\alphaa}\subset W\to X\subset \CC^4/\muu_r$
the exceptional divisor $E_{\alphaa}\cap X_{\alphaa}$
has a reduced component $G$ of
discrepancy $a(G, K_X)=1/r$.
Moreover, $r\alpha_i\equiv \wt x_i \mod r$, $i=1,2,3,4$.
Since $c>1/2$, we have $1/r-cv_\alphaa (\psi)\ge 0$, i.e.,
$rv_\alphaa (\psi)< 2$, so $rv_\alphaa (\psi)=1$.
In particular, $\wt \psi\equiv 1 \mod r$.

Let $\omega$
be a section of $\OOO_X(-K_X)$. Then $\omega$ can be written as
\[
\omega=\lambda (\partial \phi/\partial x_4)(dx_1 \wedge dx_2\wedge dx_3)^{-1},
\]
where
$\lambda$ is a semi-invariant function with
\[
\wt \lambda-\wt (x_1x_2x_3x_4)+\wt \phi\equiv \wt\omega \equiv 0\mod r.
\]
Thus, $\wt \psi \equiv \wt \lambda\mod r$. Hence, $S\sim -K_X$.
\end{proof}

\begin{lemma}
\label{l-nG-3}
If $(X\ni P)$ is a terminal singularity of index $r>1$,
then $c\le 4/5$.
\end{lemma}

\begin{proof}
Since $\pi$ is \'etale in codimension one,
we have $K_{X^\sharp }+cS^\sharp=\pi^*(K_X+cS)$. Hence
the pair $(X^\sharp,\, cS^\sharp)$ is canonical (see, e.g.,
\cite[3.16.1]{Kollar-1995-pairs}). Assume that $c> 4/5$.
By Lemma \ref{l-nG-1} the point $(X^\sharp\ni P^\sharp)$ is singular.
Then by Lemma \ref{l-G-2} the pair
$(X^\sharp,\, S^\sharp)$ is canonical.
Therefore, $(S^\sharp\ni P^\sharp)$ is a Du Val singularity.
Then the singularity $(S\ni P)=(S^\sharp\ni P^\sharp)/\muu_r$
is log terminal. On the other hand, by Lemma \ref{l-nG-2}
the divisor $K_S$ is Cartier. Hence, $(S\ni P)$ is Du Val, a contradiction.
\end{proof}


TITLE: The growth rate (order) of a particular function
QUESTION [0 upvotes]: I have a function $$g(m)=m-m^{(1-1/\ln^2(m))}$$ I want to describe the limit behavior of $g(m)$ when $m$ approaches to infinity. More specifically, I need $O(g(m))$, where $O(.)$ denotes Big O notation. From my first view, $g(m)$ should be equal to $O(m)$.
At least I want to prove that $$O(g(m))>O(\sqrt{m}) \text{  as } m\to \infty$$ which means that the grow rates of $g(m)$ is larger than $\sqrt{m}$ when $m$ is large enough.
I have conducted some simulations with Matlab, and the result is right. However, is there any efficient method to identify or compare the order of a function?

REPLY [1 votes]: $$m-m^{1-1/\ln(m)^2}=m\left(1-\mathrm e^{-\ln(m)/\ln(m)^2}\right)
=m\left(1-\mathrm e^{-1/\ln(m)}\right)=m\left(1-1+\frac{1}{\ln(m)}+O\left(\frac{1}{\ln(m)^2}\right)\right)=\frac{m}{\ln m}+O\left(\frac{m}{\ln(m)^2}\right)$$


TITLE: How does the Earth remain in its shape?
QUESTION [0 upvotes]: I know only stars do nuclear fusion to keeps its shape from gravity. Then what about the Earth? Earth doesn't do nuclear fusion. How does the Earth keep its shape even though the gravity keeps pushing Earth inward?

REPLY [2 votes]: Stars are (mostly) in hydrostatic equilibrium: at each depth in the star the pressure gradient balances the gravity, preventing it from compressing the star. This happens because a combination of the equation of state (relating temperature, pressure and volume; see also polytropes) and that the fusion processes heat up as the star gets compressed. The polytrope for stars is typically of the form $\rho=K P^{n/(n+1)}$, where $\rho$ is the density and $P$ the pressure, with $n\approx 3$ for main sequence stars.
Planets are in hydrostatic equilibrium because their equation of state typically behaves like $\rho = c + K P^n$ with $n\approx 0.5$. The reason is that they are solid or liquid rather than gaseous (gas giants are obviously slightly different). Inter-molecular forces increase a lot when one tries to compress such matter, and hence the density does not increase that much: the matter is stiff. For heavy planets degeneracy pressure starts to matter: when squeezing matter together electrons resist it fiercely since they would be forced into the same quantum states.


TITLE: Where I'm wrong? Why my answer is different from the book?
QUESTION [4 upvotes]: The question is to integrate
$$\int x\cos^{-1}x dx$$
Answer in my book
$$(2x^2-1)\frac{\cos^{-1}x}{4}-\frac{x}{4}\sqrt{1-x^2}+C$$
I'm learning single variable calculus right now and at current about integration with part. I'm confused in a problem from sometime. I don't know where I am wrong. Please have a look at the images.

Solution. $\newcommand{\dd}{\; \mathrm{d}}$
  $$
\begin{align}
&\cos^{-1} x \int x \dd x - \int \left[\frac{-1}{\sqrt{1-x^2}}\cdot\frac{x^2}2\right] \dd x=\\
=&\cos^{-1}x \cdot \frac{x^2}2 + \int \frac{x^2}{2\sqrt{1-x^2}} \dd x2=\\
=&\frac{x^2}2 \cdot \cos^{-1}x + \frac12 \int \frac{x}{\sqrt{1-x^2}} \dd x =
\begin{vmatrix}x=\sin t\\ \dd x=\cos t\dd t\end{vmatrix} = \\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{\sin^2t \cos t}{\sqrt{1-\sin^2t}} \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{\sin^2t \cos t}{\cos t} \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \sin^2t \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{1-\cos2t}2 \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac12 \dd t -\frac 14 \int \cos2t \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 t - \frac18 \sin 2t + C =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin2\sin^{-1}x + C =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin(2\sin^{-1}x) + C
\end{align}
$$

Please help. Thankyou in advance.

REPLY [0 votes]: With the help of Bernard's answer you can further transform your term
\begin{align}
\int x \cos^{-1}(x)dx&=\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin(2\sin^{-1}x)\\
&=\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac 14 x\sqrt{1-x^2}+c \ \text{ but this does not equal}\\
&\neq (2x^2-1)\frac{\cos^{-1}x}{4}-\frac{x}{4}\sqrt{1-x^2}+c \\&=\frac{x^2\cos^{-1}x}2 - \frac14 \cos^{-1}x - \frac 14 x\sqrt{1-x^2}+c
\end{align}
so additionally to the fact that you can further simplify your answer the given answer in the book is wrong.


TITLE: Why does thermal conductivity of an alloy becomes nearly flat from $0.2<x<0.8$ composition?
QUESTION [1 upvotes]: I am trying to understand the effect of alloying on the thermal conductivity of an crystalline alloy. I have found a great many papers where I see thermal conductivity sharply decreases from 0 to 0.2 X of the alloy composition, then becomes nearly flat, then they start increasing again after 0.8X composition. But not one paper properly explains what thermal conductivity becomes flat from 0.2<x<0.8. Can anyone please help me with this?

REPLY [0 votes]: bufferlab gives a good answer, but I'll dumb it down a bit. Since you tag phonons (and not electrons), I'm assuming that most of your thermal transport is due to phonons.
When $x$ is near 0 or 1, then you basically have a crystal with some impurities. E.g. if you're dealing with $Si_xGe_{1-x}$, then if $x$ is close to 0, then you more or less have Ge, and if $x$ is close to 1, then you more or less have Si.
However, if $x$ is in the middle, then you just have a disordered mess where your phonons are scattered all over the place. In fact, it's a little difficult to even define phonons at this point, because phonons as we normally think of them exist only in regular lattice --- not alloys. Arguably, $x=0.4$ is "more" disordered than $x=0.2$, but in either case, it's so disordered that it doesn't make a huge difference. Hence, thermal conductivity is basically the same for a fairly wide range of $x$.


TITLE: continuity after taking partial derivative
QUESTION [1 upvotes]: Suppose $f:(x,y)\in K\times \mathbb{R}\to \mathbb{R}$ be a continuous function (in $x$ and $y$) and $K$ is a compact set. If $f$ is continuously differentiable in $y$. Is $f_y$ still continuous in $x$?

I am not sure whether it is correct or under what condition it is correct. Briefly speaking, if $f$ is continuously differentiable in $y$, then for any $x\in K$, I have
$$
f(x,y+h)=f(x,y)+f_y(x,y)h+o(h),
$$
as $h\to0$.
I would like $f_y(x,y)$ to be bounded for all $x\in K$ and $|y|\le C$, and a sufficient condition is to show $f_y$ is still continuous in $x$.

REPLY [1 votes]: Try $f(x,y) = \begin{cases} \sqrt{|x|} \arctan {y \over x},& x \neq 0 \\
0, & \text{otherwise}\end{cases}$.
Then ${\partial f(x,y) \over \partial y} = \begin{cases} {\sqrt{|x|} \over x^2+y^2},& x \neq 0 \\
0, & \text{otherwise}\end{cases}$.
Then, for a fixed $x$, the function $y \mapsto {\partial f(x,y) \over \partial y}$ is continuous,
but $x \mapsto {\partial f(x,0) \over \partial y}$ is not continuous at $x=0$.


TITLE: Show that non entire function is constant
QUESTION [0 upvotes]: Let $f : \Bbb C^* \to \Bbb C$ be a holomorphic function such that $\lvert f(z) \rvert \leq \sqrt{\lvert z \rvert} + \frac{1}{\sqrt{\lvert z \rvert}}$. Show that $f$ is constant.
I cannot use Liouville theorem because the function is not entire and we do not know what happens near $0$.
I noticed that $\lvert f(\frac1z)\rvert \leq \sqrt{\lvert z \rvert} + \frac{1}{\sqrt{\lvert z \rvert}}$. I also noticed that for $z$ such that $\lvert z \rvert \leq 1$, we have that $\lvert f(z) \rvert \leq 2$. I tried to look at what happens for $1/f$ or $e^f$ but no interesting result. How can I do it ?
I wrote $\lvert f(z) \rvert \leq \sqrt{\lvert z \rvert} + \frac{1}{\sqrt{\lvert z \rvert}}=\frac{\lvert z \rvert +1}{\sqrt{\lvert z \rvert} }$. For $\lvert z \rvert \geq 1,$ $\lvert f(z) \rvert \leq \frac{2 \lvert z \rvert}{\sqrt{\lvert z \rvert}}=2\sqrt{\lvert z \rvert}$ but it does not help a lot.

REPLY [2 votes]: Note that $|zf(z)| \to 0$ as $z\to 0$, so $f$ has a removable singularity at $z = 0$. Let $F$ be a holomorphic extension of $f$ to an entire function. Use Cauchy’s estimates to obtain $|F^{(n)}(0)| \le n!\, (r^{1/2-n}+r^{-n-1/2})$ for all $r > 0$ and $n\ge 1$. Let $r\to \infty$ to deduce $F$ is constant. Then $f$ is constant.


TITLE: $\partial A = \emptyset \iff$ there is a map $f: X \to \Bbb R$ for which $\overline{f[A]} \cap \overline{f[X \setminus A]} = \emptyset$
QUESTION [2 upvotes]: Let $A \subset X$. Show that the following statements are equivalent.
\begin{align}(1) & \ \partial A = \emptyset \\ (2) & \text{ There exists a continuous map $f: X \to \Bbb R$ for which $\overline{f[A]} \cap \overline{f[X \setminus A]} = \emptyset$}\end{align}

I’ve managed to prove $(2) \to (1)$ as follows.
Since $f[\overline{A}] \subset \overline{f[A]}$ and $f[\overline{X \setminus A}] \subset  \overline{f[X \setminus A]}$ we have that $$f[\overline{A}] \cap f[\overline{X \setminus A}]  = \emptyset.$$ Due to continuity this means that $\overline{A} \cap \overline{X \setminus A} = \emptyset$, but $\overline{A} \cap \overline{X \setminus A}  = \partial A$, thus $\partial A = \emptyset$.
However I cannot get from $(1) \to (2)$. The fact that $\partial A = \emptyset \implies A$ is closed and open. How can I use this fact to show $\overline{f[A]} \cap \overline{f[X \setminus A]} = \emptyset$?

REPLY [1 votes]: For $(2) \to (1)$, the observation that $$f[\overline{A}] \cap f[\overline{X\setminus A}] = \emptyset\tag{a}$$ is correct (and uses the continuity of $f$).
Then the last observation you have to make (nothing to do with continuity): if $\partial A \neq \emptyset$ then any $p \in \partial A$ is both in $\overline{A}$ and $\overline{X\setminus A}$ and so $f(p)$ would contradict $(a)$. So $\partial A = \emptyset$ must hold.
For the other direction, use the fact that if $\partial A = \emptyset$ the map $f(x)=0$ if $x \in A$ and $f(x)=1$ if $x \notin A$ is continuous, because $X$ is the union of the two disjoint closed sets $\overline{A}$ and $\overline{X\setminus A}$ (the intersection is $\partial A$ which is empty) and $f$ is continuous (even constant) on both parts.
It clearly fulfills the disjoint closure condition from $(2)$.


TITLE: What are CL contours?
QUESTION [0 upvotes]: I am reading about the precise measurement of top quark mass from: http://arxiv.org/abs/1510.04483
In the second page there is this statement

"The 68% and 95% CL contours for the indirect determination of mW and mtop from global SM red, dashed curves, respectively. For negative values of λ, the lifetime of the SM fits to electroweak precision data"

I am a noob in analysis and did not understand what CL contours mean.

REPLY [1 votes]: Stands for "Confidence level" contours.


TITLE: Positivity of a matrix
QUESTION [0 upvotes]: Let $A$ be a $3\times 3$ matrix defined in the following way:
$$A=\begin{bmatrix} a & c & 0\\c & b &-c\\0 & -c & 1-a-b\end{bmatrix}$$
I wish to show that $A=BB^t$ for some matrix $B$. Further suppose that $a,b,c\ge0$ and $abc\ne0$.
Since there is a characterization of positive elements in matrices which says that $A$ is positive iff $A$ can be written as $BB^t$ for some matrix $B$.
Towards that end, for $u=[x,y,z] \in \mathbb{R}^3$, I looked at the inner product $\langle Au,u \rangle$ and found out that $$\langle Au,u \rangle=ax^2+by^2+(1-a-b)z^2+2cxy-2cyz$$
By Hilbert's seventeenth problem, a polynomial of degree $n$ is positive iff it can be written as a sum of $2^n$ positive polynomials. How do I write this polynomial as a sum of such polynomials?
Thanks for the help!!

REPLY [1 votes]: In order for your matrix to be positive definite, all its leading principal minors must be positive.
So in addition to $a > 0$ you need $ab - c^2 > 0$ and $ab + b c^2 - a^2 b - a b^2 - c^2 > 0$.
EDIT: For your matrix to be positive semidefinite, you need
$  ab + b c^2 - a^2 b - a b^2 - c^2 \ge 0$ and $a+b-a^2-ab-b^2-2c^2\ge 0$.


TITLE: Proving that $\lceil x-\lceil y \rceil \rceil =\lfloor \lceil x \rceil - y \rfloor$
QUESTION [1 upvotes]: Having trouble proving this equality for all real numbers. I understand that the left side simplifies to $\lceil x \rceil$ - $\lceil y \rceil$, however I cannot for the life of me get the right side simplified to equal the left side.
$\lceil x-\lceil y \rceil \rceil =\lfloor \lceil x \rceil - y \rfloor$

REPLY [0 votes]: For any real numbers, $x, y$ there are integers $n_x$ and $n_y$ such that
$n_x \le x < n_x + 1$ and $n_y \le y < n_y + 1$.
Case 1:  Neither $x$ nor $y$ are integers.
$\lceil x-\lceil y \rceil \rceil = \lceil x-(n_y + 1) \rceil= \lceil x\rceil-(n_y + 1)= n_x + 1 - n_y - 1 = n_x - n_y$
While $\lfloor \lceil x \rceil - y \rfloor = \lfloor n_x + 1 - n_y - (y - n_y) \rfloor = n_x - n_y + \lfloor 1 - (y - n_y) \rfloor = n_x - n_y + 0 = n_x - n_y$
Case 2: $x = n_x$ is an integer.  $y$ is not.
$\lceil x-\lceil y \rceil \rceil =   x-(n_y + 1)= n_x  - n_y - 1 $
While $\lfloor \lceil x \rceil - y \rfloor = \lfloor n_x  - n_y - (y - n_y) \rfloor = n_x - n_y + \lfloor  - (y - n_y) \rfloor = n_x - n_y -1$
Case 3: $x$ is not and integer. $y = n_y$ is.
$\lceil x-\lceil y \rceil \rceil = \lceil x-n_y \rceil= \lceil x\rceil-n_y= n_x + 1 - n_y $
While $\lfloor \lceil x \rceil - y \rfloor = \lfloor n_x + 1 - n_y  \rfloor = n_x +1 - n_y$
Case 4: $x = n_x$ and $y = n_y$ are both integers.
$\lceil x-\lceil y \rceil \rceil =\lceil x_n-y_n \rceil = x_n - y_n= \lfloor x_n - y_n \rfloor = \lfloor \lceil x \rceil - y \rfloor$


TITLE: Is $\mathbf F_3(t,t^{1/3})/\mathbf F_3(t)$ a normal extension? Is it separable?
QUESTION [2 upvotes]: I'm used to checking this sort of thing for Archimedian fields, but I'm not sure what to do in this case.
I know that $[\mathbf F_3(t,t^{1/3}):\mathbf F_3(t)]=3$ since $x^3-t$ is irreducible over $\mathbf F_3(t)$ by Eisenstein with prime $t$. I believe that this extension is not separable, since $x^3-t=(x-t^{1/3})^3$ so we have repeated factors. How do I go about checking whether or not this extension is normal?

Edit: To clarify, the definition of normal I am using is the following: $L/K$ is normal if for all $\alpha\in L$, the minimal polynomial $m_{\alpha,K}$ has all its roots in $L$.

REPLY [1 votes]: I showed it this way: Let $K=\mathbf F(t)$, and $L=K(t^{1/3})$, so $\{1,t^{1/3},t^{2/3}\}$ is a $K$-basis for $L$. Consider a generic element $\alpha=a+bt^{1/3}+ct^{2/3}\in L$. Then the minimal polynomial $m_{\alpha,K}$ of $\alpha$ divides $$f:=x^3-\alpha^3=x^3-a^3-bt-ct^2 \in K[x].$$
But $f=(x-\alpha)^3$, so clearly $f$ splits into linear factors over $L$ (and therefore so does $m_{\alpha,K}$). Thus $L/K$ is normal.


TITLE: External force and centre of mass
QUESTION [0 upvotes]: I’ve read that linear momentum of a system will be constant when there is no external force on the system. Thus, the velocity of the centre of mass will be the same. Thus, this means that the position on COM of the system is not constant, but instead moves with a constant velocity.
Then, in the link given below, gravity, an external force is acting on the system in the y-direction. Thus, there is no force on the x-direction and thus, the velocity of the x-com is constant. Now, as the initial velocity of the x-com was zero, the final velocity of the x-com will also be zero and thus the position of x-coordinate of the com does not change. This realisation is being exploited to solve the question.
Is my understanding correct?
The question being referenced

REPLY [1 votes]: Thus, this means that the position on COM of the system is not constant, but instead moves with a constant velocity.

That depends on whether there was an initial momentum, an initial speed. As you say yourself in the next paragraph, if the initial speed (of the centre-of-mass) is zero, then it remains zero if there are no external forces acting. Your understanding seems correct.


TITLE: Obtaining the discriminant of the characteristic polynomial directly from the matrix
QUESTION [6 upvotes]: Let $M \in \mathbb{Z}_{n \times n}$ be a square matrix with integer coefficients. Let $P(x)$ be its characteristic polynomial
$$
    P(x) = \det\left(x \cdot \mathbb{I}_{n \times n}- M\right)
$$
I would like to compute the discriminant of $P(x)$, and I am wondering if it can be obtained from $M$ directly.
The intent is to determine whether $M$ has distinct eigenvalues.
I am looking for references, ideas, algorithms. Thank you.

REPLY [1 votes]: Every polynomial $p \in \mathbb{Z}[x]$ has a corresponding companion matrix, whose characteristic polynomial is $p(x)$. A companion matrix of $M$ is naturally similar to $M$ itself.
The companion matrix of $M$ can be found by bringing $M$ to its Frobenius normal form, which can be done in the field $\mathbb{Q}$.  Characteristic polynomial of $M$ is readily obtained by the Frobenius normal form (which is block-diagonal matrix of companion matrices), and its discriminant is then computed as $\gcd(p(x),p^\prime(x))$.
See "Faster Algorithms for characteristic polynomial" by A. Storjohann and C. Pernet.


TITLE: Definition of an E-infinity algebra
QUESTION [42 upvotes]: Can anyone give me a plain-and-simple
definition of an E-infinity algebra without using
the words "operad," "ring spectrum," or
"stable homotopy"?
Sorry, but I honestly couldn't find it using
all on-line resources at my disposal.
Thanks!

REPLY [8 votes]: A completely explicit definition that works over any ring is given as Proposition 18 in the following preprint of Malte Dehling and Bruno Vallette that was posted on the arXiv today. http://arxiv.org/abs/1503.02701


text:
\begin{document}
\date{}
\maketitle

\begin{abstract}
We consider two simple models for the formation of polymers where at the initial time, each monomer has a certain number of potential links (called arms in the text)
that are consumed when aggregations occur.  Loosely speaking, this imposes restrictions on the number of aggregations. The dynamics of concentrations are governed by modifications of Smoluchowski's coagulation equations.
Applying classical techniques based on generating functions, resolution of quasi-linear PDE's, and Lagrange inversion formula, we obtain explicit solutions to these non-linear systems of ODE's. We also discuss the asymptotic behavior of the solutions and point at some connexions with certain known solutions to Smoluchowski's coagulation equations with
additive or multiplicative kernels.
\vskip 2mm

 \noindent {\bf Keywords : } Coagulation equations, generating function, quasi-linear PDE, Lagrange inversion formula, gelation.

 \noindent{\bf A.M.S. Classification } Primary : 34A34 ; Secondary : 82C23, 82D60.

 \end{abstract}

\begin{section}{Introduction}

The coagulation equations of Smoluchowski \cite{Smoluch}  describe the evolution of concentrations of particles in a medium in which particles coalesce pairwise as time passes. The surveys by Aldous \cite{Aldous} and Lauren\c{c}ot and Mischler \cite{LM} provide stimulating introductions to the subject.
We also refer to Drake \cite{Drake} and Dubovski \cite{Dubov} for detailed accounts, and e.g. to \cite{EM, FL, MenPeg} and references therein  for some more recent works in this area.

In the discrete version of  this model, particles are characterized by an integer $m\geq 1$ which should be though of as a size or a mass, in the sense that the  result  of  the coagulation of a pair of particles, say $\{m,m'\}$,  is $m+m'$. The dynamics are determined in terms  of some symmetric kernel $\kappa: \N^*\times \N^*\to \R_+$ such that, loosely speaking, every  pair of particles $\{m,m'\}$ coalesces at
rate  $\kappa(m,m')$. This means that if
 $c_t(m)$ denotes  the concentration of particles with size $m$ at time $t$, then
the evolution of concentrations is  governed by the infinite system
\begin{equation}\label{EQ1}
\d c_t(m)=\frac{1}{2}\sum_{m'=1}^{m-1}
c_t(m')c_t(m-m')\kappa(m',m-m')
-c_t(m)\sum_{m'=1}^{\infty}c_t(m')\kappa(m,m')\,,
\end{equation}
where the first term in the right-hand side accounts for the creation of particles of size $m$ by the coagulation of a pair of particles $\{m',m-m'\}$, the factor $\frac{1}{2}$ stems from an obvious symmetry, and the second term accounts for the disappearance of particles with size $m$ as the result of a coagulation with other particles.

In general \eqref{EQ1} cannot be solved explicitly; however there are a few important exceptions.  The fundamental  examples of solvable Smoluchowski's equations occur when the coagulation  kernel $\kappa$ is constant, additive, or multiplicative.  For instance, for a monodisperse initial condition (this means that at time $t=0$, all particles have unit size),
it is known since the original work of Smoluchowski \cite{Smoluch} that
\begin{equation}\label{EQ2}
c_t(m) = \left(1+\frac{t}{2}\right)^{-2}\left(\frac{t}{2+t}\right)^{m-1}\,,\qquad 0\leq t < \infty
\end{equation}
solves \eqref{EQ1} for $\kappa(m,m')=1$.
Further, if
$$B(\lambda,m)= (\lambda m)^{m-1}\e^{-\lambda m}/m!\,,\qquad m\in\N^*$$
denotes the Borel probability function with parameter $\lambda\in[0,1]$, then, according to Golovin \cite{Golov},
\begin{equation}\label{EQ3}
c_t(m)=\e^{-t}B(1-\e^{-t},m)\,,\qquad 0\leq t < \infty
\end{equation}
is a solution to \eqref{EQ1} for $\kappa(m,m')=m+m'$. Finally,  it has been shown by McLeod \cite{McLeod} that
\begin{equation}\label{EQ4}
c_t(m)=m^{-1}B(t,m)\,,\qquad 0\leq t < 1
\end{equation}
solves \eqref{EQ1} for $\kappa(m,m')=mm'$.
More generally, explicit though complicated expressions can be obtained for solutions to \eqref{EQ1} when the coagulation kernel $\kappa$ is a linear combination of these three basic kernels; see \cite{Trub, Spouge, vDE1}.

The key reason why Smoluchowski's equations can be solved explicitly in these cases  is that considering generating functions transforms \eqref{EQ1} into a solvable  PDE; see e.g. Deaconu and Tanr\'e \cite{DT}.
The purpose of the present work is to point out that the same techniques can be applied successfully to investigate a related but more complex model for coagulation.
Roughly speaking, the dynamics described by
Smoluchowski's equations depend only on an additive quantity, namely the size of particles. Norris  \cite{Norris}  developed the much more sophisticated setting of cluster coagulation models in which the rates of transition may depend on
further parameters, such as typically
the shapes of the clusters of particles. In this direction,  we will deal here with  a toy model for the formation of polymers in which
coagulation rates are functions of a non-additive quantity that depends on the history of polymers. More precisely,  the medium consists initially in
monomers (elementary particles with unit size), such that each monomer has a certain number of potential links which we call  {\it arms}. Specifically, we suppose that each particle is determined by a pair of integers $(a,m)$, where $a\geq 0$ is the number of  arms of the particle and $m\geq 1$ represents its size. The arms serve to perform aggregations, i.e. to connect pairs of particles, and are consumed each time an aggregation occurs. We will consider two different models.

 The first model will be studied in Section 2. It is {\it oriented}, in the sense that a coagulation occurs when a particle uses one of its arms to grab another particle. Only one arm is consumed for each coagulation event; specifically,
when a particle $(a,m)$ with $a\geq 1$ grabs a particle $(a',m')$, these two particles merge into a single one $(a+a'-1, m+m')$. We assume further that each arm grabs other particles at uniform rate; the precise formulation will be given in Section 2.1. In Section 2.2 we use generating functions to connect these dynamics  to a quasi-linear PDE that is then solved by the method of characteristics. Section 2.3 presents a version of Lagrange inversion formula which will be needed to invert the generating functions. The main results on the oriented model are given in Section 2.4, and finally some illustrative examples are discussed in Section 2.5. In particular, we point at  some resemblances between on the one hand,
the oriented model when the number of arms of monomers is distributed according to the standard Poisson law, and on the other hand, Smoluchowski's coagulation equation for the additive kernel.

The second model will be studied in Section 3. It is {\it symmetric}, in the sense that  each coagulation consumes one arm for each of the two particles involved into a coagulation event~:
when a pair of particles $\{(a,m),(a',m')\}$ with $aa'\geq 1$ coalesces, the resulting particle is \linebreak  $(a+a'-2, m+m')$.  Furthermore, we assume that each pair of arms is activated uniformly. Section 3.1 introduces the setting rigorously, and points at a critical time $\Gamma_{\infty}$ in the evolution of the system which resembles the gelation time (i.e. the instant when particles with infinite size appear in the medium)
 in Smoluchowski's coagulation equations. We also use generating functions to make the connexion with a quasi-linear PDE closely related to the one that arises for the oriented model. The main results and some  examples are given in Section 3.2.  For instance, we determine the critical time $\Gamma_{\infty}$ in terms of the initial data (interestingly,  $\Gamma_{\infty}$ may be finite or infinite) and note some similarities between the symmetric model started from monomers with arms distributed according to the standard Poisson law and McLeod's solution to Smoluchowski's coagulation equation for the multiplicative kernel at gelation time.

We  conclude this Introduction by explaning the term {\it limited} in the title. Roughly speaking, the effect
of introducing arms in a coagulation model is that it imposes some restrictions to the number of aggregations. Indeed, in the oriented model, we shall show that if at the initial time the mean number of arms is less than the total concentration of particles, then the system converges as time tends to infinity to some limiting concentrations (and more precisely the system does not create particles with infinite size); see Corollary \ref{C1}. A similar phenomenon is observed for the symmetric model; see Corollary \ref{C2} and the remark thereafter. Perhaps this is more surprising for the symmetric model, as the latter bears some similarities with the classical Smoluchowski's coagulation for the multiplicative kernel,
and it is well-known that gelation  then always occurs in that case.

\end{section}

\begin{section}{The oriented model}
\subsection{Setting}
We now describe precisely the dynamics for oriented coagulation with arms. For every $t\geq 0$,
$a\in\N:=\{0,1,2,\ldots\}$ and $m\in\N^*:=\{1,2,\ldots\}$, let $c_t(a,m)$ denote the density of particles $(a,m)$ at time $t$. Loosely speaking, we suppose that at any time, each arm may grab each particle at the same unit rate. Rigorously, this means in terms of concentrations that the transition
\begin{equation}\label{EQ4'}
\{(a,m),(a',m')\}\longrightarrow (a+a'-1,m+m')
\end{equation}
occurs at time $t$ with intensity
$$ac_t(a,m)c_t(a',m') + a'c_t(a',m')c_t(a,m)=(a+a')c_t(a,m)c_t(a',m').$$
The evolution of the concentration functions is thus governed by an infinite non-linear system of ODE's
\begin{eqnarray}\label{EQ5}
\d c_t(a,m)
 \ =&&\frac{1}{2}\sum_{a'=0}^{a+1}\sum_{m'=1}^{m-1} (a+1)
c_t(a',m')c_t(a-a'+1,m-m')\nonumber \\
&&-c_t(a,m)\sum_{a'=0}^{\infty}\sum_{m'=1}^{\infty}(a+a')c_t(a',m')\,.
\end{eqnarray}

We point out that
$$
 \frac{1}{2}\sum_{a'=0}^{a+1} \sum_{m'=1}^{m-1}(a+1)
c_t(a',m')c_t(a-a'+1,m-m')
= \sum_{a'=0}^{a+1} \sum_{m'=1}^{m-1} a'
c_t(a',m')c_t(a-a'+1,m-m')\,,$$
and since in the sum in the right hand side, the term corresponding to $a'=0$ is null,
we can reformulate \eqref{EQ5} as
$$\d c_t(a,m)=\sum_{a'=1}^{a+1}\sum_{m'=1}^{m-1} a'
c_t(a',m')c_t(a-a'+1,m-m')
-c_t(a,m)\sum_{a'=0}^{\infty}\sum_{m'=1}^{\infty}(a+a')c_t(a',m')\,.
$$
 The latter expression may be closer to the asymmetric description of the dynamics
of the system.

We also introduce the notation
$$C_t:=\sum_{a=0}^{\infty}\sum_{m=1}^{\infty}c_t(a,m)$$
for the total concentration of particles at time $t$, and
$$A_t:=\sum_{a=0}^{\infty}\sum_{m=1}^{\infty}ac_t(a,m)$$
for the mean number of arms, and stress that \eqref{EQ5} only makes sense as long as
$A_t$ and $C_t$ are finite. But it is easy to check  (see Lemma \ref{L0} below) that it suffices to assume that at the initial time $A_0$ and $C_0$ are finite.

It will be convenient to re-express the system \eqref{EQ5} in a slightly different form by introducing the notation
$$\langle c_t,f\rangle := \sum_{a=0}^{\infty}\sum_{m=1}^{\infty} f(a,m)c_t(a,m)\,,$$
where $f: \N\times \N^* \to \R_+$ stands for a generic  nonnegative function.

\begin{lemma}\label{L0} For any solution to \eqref{EQ5}, the function
$t\to A_t+C_t$ is non-increasing; as a consequence $A_t+C_t\leq A_0+C_0$ for all $t\geq 0$. Further we have
\begin{equation}\label{EQ6}
\d \langle c_t,f\rangle=\frac{1}{2}
\sum_{a,a'=0}^{\infty}\sum_{m,m'=1}^{\infty} (f(a+a'-1,m+m')- f(a,m)-f(a',m'))(a+a')
c_t(a,m)c_t(a',m')
\end{equation}
whenever $f: \N\times \N^* \to \R$ is bounded.
\end{lemma}

\proof First observe that
 \eqref{EQ5} is precisely \eqref{EQ6}  when $f$ is a Dirac function. By linearity, \eqref{EQ6} thus holds when $f$ has finite support.

Next fix $n\in\N^*$ and define
 $$\varphi_n(a):=\left\{ \begin{matrix}
1+a &\hbox{ if }1+a\leq n, \\
2n-1-a &\hbox{ if }n\leq 1+a\leq 2n, \\
0 &\hbox{ otherwise.}\\
\end{matrix}\right.
 $$
Then consider the function with finite support $f_n(a,m):={\bf 1}_{\{m\leq n\}}\varphi_n(a)$. It is readily checked that
$$f_n(a+a'-1,m+m')- f_n(a,m)-f_n(a',m')\leq 0$$
for all $ a,a'\in\N $ and $
m,m'\in\N^*$. We deduce from \eqref{EQ6} that $t\to  \langle c_t,f_n\rangle$ is non-increasing. Further, $f_n(a,m)$ increases to $1+a$ as $n\uparrow\infty$, so by
 monotone convergence we have that $A_t+C_t=\lim_{n\uparrow \infty} \langle c_t,f_n\rangle$, and we conclude that $t\to A_t+C_t$ is non-increasing.

As we already know that \eqref{EQ6} holds when $f$ has finite support, a standard approximation procedure yields the extension when
$f$ is only bounded by dominated convergence. \QED

We now start the analysis by pointing at an elementary property which should be intuitively obvious, as at each coagulation event, the number of particles and the total number of arms both decrease by one unit. The rigorous proof requires however some care.

\begin{lemma}\label{L1} For any solution to \eqref{EQ5}, the difference $D:=C_t-A_t$ remains constant as time passes.
More precisely :

\noindent{\rm (i)}  If $C_0=A_0$, then
$$C_t=A_t=C_0/(1+tC_0)\,.$$

\noindent{\rm (ii)}  If $C_0-A_0={D} \neq0$, then
$$C_t=A_t+{D} = {D} \frac{C_0 \e^{{D} t}}{C_0
(\e^{{D} t}-1)+{D}}\,.$$
\end{lemma}

\proof We get from \eqref{EQ6} for $f\equiv 1$ that
$\d C_t=-A_tC_t$, and in particular the total concentration decreases with $t$.
Similarly, fix $\alpha>0$ and apply \eqref{EQ6} for $f^{(\alpha)}(a,m)=\alpha\wedge a$
to see that $A^{(\alpha)}_t:=\langle c_t,f^{(\alpha)}\rangle$ fulfills $\d A^{(\alpha)}_t\leq -A_tC_t $. As $A_t=\lim_{\alpha\uparrow \infty}\uparrow A^{(\alpha)}_t$, this entails that
$A_t$ also decreases with $t$.

Then observe that
$$|f^{(\alpha)}(a+a'-1,m+m')-f^{(\alpha)}(a,m)-f^{(\alpha)}(a',m')|\leq (a\wedge a')+1\,,$$
and that therefore
\begin{equation}\label{EQ7}
|f^{(\alpha)}(a+a'-1,m+m')-f^{(\alpha)}(a,m)-f^{(\alpha)}(a',m')|(a+a')\leq 2 aa'+a+a'\,.
\end{equation}
Since
$$\lim_{\alpha\to\infty}(f^{(\alpha)}(a+a'-1,m+m')-f^{(\alpha)}(a,m)-f^{(\alpha)}(a',m'))
=-1\,,$$
it is easy to get from \eqref{EQ6} by dominated convergence that $\d A_t=-A_tC_t$.
We conclude that the difference $C_t-A_t$ is constant and
the explicit expressions in the statement follow readily. \QED

\subsection{A quasi-linear PDE}
 The system of evolution equations \eqref{EQ6} resembles Smoluchowski's coagulation equation for the additive kernel. In the latter case, it is well-known that considering exponential functions yields a quasi-linear PDE related to the transport equation which can be solved explicitly, see e.g. \cite{DT}.
This invites us to introduce the generating functions
$$g_t(x,y):=\langle c_t,f_{x,y}\rangle \hbox{ with } f_{x,y}: (a,m)\to x^a y^m
\hbox{ and }x,y\in[0,1]\,.$$

For the sake of simplicity, we will focus on the situation where the total concentration of particles equals $1$  at the initial time, i.e. $C_0=1$. The general case can easily be reduced to that one by a linear time-substitution.

\begin{lemma}\label{L2} {\rm (i)} Consider initial concentrations $(c_0(a,m): a\in\N\hbox{ and }m\in\N^*)$ such that  $C_0=A_0=1$. The system \eqref{EQ5} is then equivalent to
the quasi-linear PDE
$$\d g_t(x,y)=\left(g_t(x,y)-\frac{x}{1+t}\right)\p g_t(x,y)-\frac{1}{1+t}g_t(x,y)$$
 for the generating functions, where $t\geq 0$ and $x,y\in[0,1]$. The latter possesses a unique solution which is given in terms of  its value at boundary $t=0$ by
 $$g_t(x,y) = (1+t)^{-1}g_0(h_t(x,y),y)= t^{-1} h_t(x,y)-\frac{x}{t^2+t}\,,$$
where $h_t(\cdot,y): [0,1]\to [0,1]$ is the (unique) inverse of the function $x\to (1+t)x-tg_0(x,y)$.

\noindent {\rm (ii)} Consider initial concentrations $(c_0(a,m): a\in\N\hbox{ and }m\in\N^*)$ such that $C_0=1$ and $C_0-A_0={D}$ for some ${D}\in
(-\infty,1)\backslash\{0\}$. The system \eqref{EQ5} is then equivalent to
the quasi-linear PDE
$$\d g_t(x,y)=\left(g_t(x,y)-x\frac{ {D} \e^{{D} t}}{\e^{{D} t}-1+{D}}\right)
\p g_t(x,y)+\left({D}- \frac{ {D} \e^{{D} t}}{\e^{{D} t}-1+{D}}
\right)g_t(x,y) $$
 for the generating functions, where $t\geq 0$ and $x,y\in[0,1]$. The latter possesses a unique solution which is given in terms of  its value at boundary $t=0$ by
  \begin{eqnarray*}
g_t(x,y) &=& \frac{ {D} \e^{{D} t}}{\e^{{D} t}-1+{D}}
g_0(h_t(x,y),y) \\
&=&  \frac{ {D} \e^{{D} t}}{\e^{{D} t}-1} h_t(x,y)- \frac{ {D}^2 \e^{{D} t}}
{(\e^{{D} t}-1+{D})(\e^{{D} t}-1)}\, x\,,
\end{eqnarray*}
where $h_t(\cdot,y): [0,1]\to [0,1]$ is the (unique) inverse of the function
$$x\to {D}^{-1}\left(\left(\e^{{D} t}-1+{D}\right)x-\left(\e^{{D} t}-1\right)g_0(x,y)\right)\,.$$
\end{lemma}

\proof (i) From \eqref{EQ6} for $f=f_{x,y}$, we get
$$\d g_t(x,y)= (\alpha-\beta-\gamma)/2$$
where $\alpha$ is given by
\begin{eqnarray*}
& &
\sum_{a,a'=0}^{\infty}\sum_{m,m'=1}^{\infty} ax^{a-1}y^m
c_t(a,m)x^{a'} y^{m'}c_t(a',m')
+ \sum_{a,a'=0}^{\infty}\sum_{m,m'=1}^{\infty} a'x^{a'-1}y^{m'}
c_t(a',m') x^{a} y^{m} c_t(a,m)\\
&=& 2 g_t(x,y) \p g_t(x,y)\,,
\end{eqnarray*}
and $\beta=\gamma$ by
\begin{eqnarray*}
& &x
\sum_{a,a'=0}^{\infty}\sum_{m,m'=1}^{\infty} ax^{a-1}y^m
c_t(a,m)c_t(a',m')
+ \sum_{a,a'=0}^{\infty}\sum_{m,m'=1}^{\infty} a'x^{a}y^{m}
 c_t(a,m) c_t(a',m')\\
&=& x C_t\p g_t(x,y) + g_t(x,y) A_t\,.
\end{eqnarray*}
Recall from Lemma \ref{L1} that
$C_t=A_t=1/(1+t)$ to get the PDE in the statement.

Note that this PDE does not involve partial derivatives with respect to the variable $y$, and is quasi-linear. It can be solved using the methods of characteristics.
Specifically we consider functions $t=t(r,s)$, $x=x(r,s)$ and $u=u(r,s)$ for $0\leq r \leq   1$ and $s\geq 0$, with the boundary conditions $t(r,0)=0$, $x(r,0)=r$, $u(r,0)=g_0(r,y)$ and
such that
$$\frac{{\rm d}t}{{\rm d}s}=1\quad,\quad \frac{{\rm d}x}{{\rm d}s}= \frac{x}{1+t}-u\quad,\quad
\frac{{\rm d}u}{{\rm d}s}=-\frac{u}{1+t}\,.$$
We deduce
 $$t(r,s)=s\quad,\quad u(r,s)=g_0(r,y)/(1+s)\quad,\quad x(r,s)=r(1+s)-sg_0(r,y)\,;$$
and then the comparison with the quasi-linear PDE gives $g_t(x,y)=u(r,s)$.

The requirements $C_0=g_0(1,1)=1$ and $A_0=\p g_0(1,1)=1$ ensure that for any $s\geq 0$ and $y\in[0,1]$, the
function $r\to r(1+s)-sg_0(r,y)$ has derivative $1+s(1-\p g_0(r,y))\geq 1$, and since
$-sg_0(0,y)<0<1\leq (1+s)-sg_0(1,y)$, there is a unique solution $h_s(\cdot,y):[0,1]\to[0,1]$ to the equation
$h_s(x,y)=h_s(x,y)(1+s)-sg_0(h_s(x,y),y)$. This provides the solution given in the statement.

 Conversely, if $(g_t: t\geq0)$ is a family of  generating functions solving this PDE with a boundary condition $g_0(x,y)=\langle c_0,f_{x,y}\rangle$ and such that
$g_0(1,1)=C_0=1$ and $\p g_0(1,1)=A_0=1$, then we know from above that
 $$g_t(x,y) = (1+t)^{-1}g_0(h_t(x,y),y)= t^{-1} h_t(x,y)-\frac{x}{t^2+t}\,,$$
with $h_t(\cdot,y): [0,1]\to [0,1]$ the (unique) inverse of  $x\to (1+t)x-tg_0(x,y)$.
Note that $h_t(1,1)=1$ for all $t$, which gives $g_t(1,1)=1/(1+t)$. Further,
taking a (left) derivative with respect to the variable $x$ at $x=1$ in the identity
$$x= (1+t)h_t(x,1)-tg_0(h_t(x,1),1)$$
entails $\p h_t(1,1)=1$ and then
$$\p g_t(1,1)=t^{-1}\p h_t(1,1)-1/(t^2+t)=1/(1+t)\,.$$
Thus the family $(g_t: t\geq 0)$ also solves the equation
$$\d g_t(x,y)=\left(g_t(x,y)-x g_t(1,1)\right)\p g_t(x,y)-g_t(x,y)\p g_t(1,1)\,.$$
Since $g_t(1,1)=\langle c_t,1\rangle=C_t$ and $\p g_t(1,1)=\langle c_t,a\rangle=A_t$ , we recover \eqref{EQ5} by inverting the generating functions.

(ii) The derivation of the PDE is obtained as in the proof of (i), applying
the second part of Lemma \ref{L1} in place of the first.
It is solved again by the methods of characteristics; using the same notation as in (i), we are led to consider the system of ODE's
$$\frac{{\rm d}t}{{\rm d}s}=1\quad,\quad \frac{{\rm d}x}{{\rm d}s}= x\,\frac{ {D} \e^{{D} t}}{\e^{{D} t}-1+{D}}-u\quad,\quad
\frac{{\rm d}u}{{\rm d}s}=D\left(1-\frac{  \e^{{D} t}}{\e^{{D} t}-1+{D}}\right) u$$
 with the boundary conditions $t(r,0)=0$, $x(r,0)=r$, $u(r,0)=g_0(r,y)$.
We obtain
 $$t(r,s)=s\quad,\quad u(r,s)=g_0(r,y) \frac{ {D} \e^{{D} s}}{\e^{{D} s}-1+{D}}$$
 and
 $$ x(r,s)=D^{-1}\left(r\left(\e^{{D} s}-1+{D}\right)
 -g_0(r,y)\left( \e^{{D} s}-1\right)\right)\,.$$
The proof can then be completed just as in the case (i). \QED

\subsection{A version of Lagrange inversion formula}
The final step of our analysis consists in checking that the function $(x,y)\to h_t(x,y)$ that appears in Lemma \ref{L2}
is the generating function of some finite measure on $\N\times \N^*$
that can be inverted explicitly, at least under some natural hypotheses. It is an easy application of Lagrange inversion formula (see, e.g. Section 5.1 in \cite{Wilf}).

\begin{lemma}\label{L3} Let $\mu=(\mu(a))_{a\in\N}$ be a finite measure on $\N$, $\mu\not\equiv 0$. We write
$$g(x)=\sum_{a=0}^{\infty} x^a\mu(a)\,,\qquad x\in[0,1]$$
for its generating function. For every $p,q>0$ and for every $x,y>0$ sufficiently small,
the equation
$$h(x,y)=yg(px +qh(x,y))$$
has a unique solution which  is analytic in $x$ and $y$ and given by
$$h(x,y)=\sum_{a=0}^{\infty}\sum_{m=1}^{\infty}x^{a}y^m\frac{1}{m} \left(  \begin{matrix} m+a-1\\ a\\
\end{matrix}\right)q^{m-1}p^{a}\mu^{*m}(m+a-1)\,,$$
where  $\mu^{*m}$ stands for the $m$-th convolution power of $\mu$.
\end{lemma}

\proof  We fix $p,q>0$ and $x$ sufficiently small so that $px<1$. Then we define $\tilde g(y):=g(px+qy)$
 for $y\geq 0$ with $px+qy\leq 1$, which is the generating function of the sigma-finite measure
$$\tilde \mu(k):=\sum_{n=k}^{\infty} \left(  \begin{matrix} n\\ k\\
\end{matrix}\right) (px)^{n-k} q^k\mu(n)\,,\qquad k\in\N\,. $$
Observe also that $\tilde g(0)=g(px)>0$.
According to the Lagrange inversion formula, the equation $h(x,y)=y\tilde g(h(x,y))$
has a unique solution for $y>0$ sufficiently small, which is analytic in $y$ and can be expressed as
$$h(x,y)=\sum_{n=1}^{\infty}y^n n^{-1}\tilde \mu^{*n}(n-1)\,.$$
Then observe that the generating function of the  measure $\tilde \mu^{*n}$
is $\tilde g^n(y)=g^n(px+qy)$, so
$$\tilde \mu^{*n}(k)=\sum_{j=k}^{\infty} \left(  \begin{matrix} j\\ k\\
\end{matrix}\right) (px)^{j-k} q^k\mu^{*n}(j)\,. $$
We deduce that
$$
h(x,y)=\sum_{n=1}^{\infty}y^n n^{-1} \sum_{j=n-1}^{\infty} \left(  \begin{matrix} j\\ n-1\\
\end{matrix}\right) (px)^{j-n+1} q^{n-1}\mu^{*n}(j)\,,
$$
and the change of variables $j=a+n-1$ completes the proof. \QED

\subsection{Explicit solutions}
We are now able to solve \eqref{EQ5} explicitly in the situation when the initial concentrations $c_0(a,m)$ are carried by particles having unit size; recall also
the assumption that the initial total concentration is $1$. In this direction, it may be interesting to point out that even though the number of arms of a particle is not
an additive quantity, the simple change of variables $\alpha=a-1$ yields a new parametrization of particles that is additive on coagulation, in the sense that
\eqref{EQ4'} then reads
\begin{equation}\label{EQ4"}
\{(\alpha,m),(\alpha',m)\}\longrightarrow (\alpha+\alpha', m+m')\,.
\end{equation}
This observation makes the solvability of \eqref{EQ5} by techniques based on generating functions easier to understand. Note however that the parameter $\alpha$ may take the
value $-1$ and that the rate of the coagulation \eqref{EQ4"} is now
$a+a'=\alpha+\alpha'+2$ and hence not additive in the alternative parameter $\alpha$.

More precisely,
we consider $\mu=(\mu(a))_{a\in\N}$ a probability measure on $\N$, and
denote for every $m\in\N^*$ by $\mu^{*m}$ the $m$-th convolution power of $\mu$.

\begin{theorem}\label{T1} The system \eqref{EQ5} has a unique solution
$(c_t(a,m): a\in\N, m\in\N^* \hbox{ and } t\geq0)$  started from
$$c_0(a,m)={\bf 1}_{\{m=1\}}\mu(a)\,,\qquad a\in\N \hbox{ and } m\in\N^*$$
which is given as follows :

\noindent {\rm (i)} If $\sum_{a\in\N}a\mu(a)=1$,
then
$$c_t(a,m)= m^{-1}t^{m-1}(1+t)^{-(a+m)} \left(  \begin{matrix} a+m-1\\ a\\
\end{matrix}\right) \mu^{*m}(a+m-1)\,.$$
\noindent {\rm (ii)} If $\sum_{a\in\N}a\mu(a)=1-D$
for some $D\in(-\infty,1)\backslash\{0\}$,
then
$$c_t(a,m)= \e^{Dt}m^{-1} D^{a+1}\left(\e^{Dt}-1\right)^{m-1}
\left(\e^{Dt}-1+D\right)^{-(a+m)} \left(  \begin{matrix} a+m-1\\ a\\
\end{matrix}\right) \mu^{*m}(a+m-1)\,.$$
\end{theorem}

\proof We shall focus on (i), the argument for (ii) being similar
(note also that taking the limit as $D\to0$
in part (ii) yields the solution in the case (i), as it should be expected).

According to Lemma \ref{L2} and the present hypotheses,
we have to consider the initial generating function
$$g_0(x,y)=\langle c_0, f_{x,y}\rangle =\sum_{a=0}^{\infty} x^a y \mu(a)=yg(x)$$
where $g$ is as in Lemma \ref{L3}, and then the solution $h_t(x,y)$ to
$$(1+t)h_t(x,y)-tyg(h_t(x,y))=x\,.$$
Lemma \ref{L3} invites us to introduce
$$h(x,y):=t^{-1}((1+t)h_t(x,y)-x)\,,$$
so that
$$h(x,y)=yg\left(\frac{x}{1+t} +\frac{t}{1+t}h(x,y)\right)\,.$$
According to Lemma \ref{L3} (with $p=1/(1+t)$ and $q=t/(1+t)$), the solution to this equation can be expressed in the form
$$h(x,y)=\sum_{a=0}^{\infty}\sum_{m=1}^{\infty}x^{a}y^m\frac{1}{m} \left(  \begin{matrix} m+a-1\\ a\\
\end{matrix}\right)\left(\frac{t}{1+t}\right)^{m-1}(1+t)^{-a}\mu^{*m}(m+a-1)\,,$$
whenever $x,y$ are sufficiently small.
Recall that
$$h_t(x,y)=(1+t)^{-1}(th(x,y)+x)\,,$$
so we get
$$h_t(x,y)=\frac{x}{1+t}+\sum_{a=0}^{\infty}\sum_{m=1}^{\infty}x^{a}y^m\frac{1}{m} \left(  \begin{matrix} m+a-1\\ a\\
\end{matrix}\right)t^m(1+t)^{-(a+m)}\mu^{*m}(m+a-1)\,.$$

We then known that the unique solution to the quasi-linear PDE in Lemma \ref{L2}(i)  is\begin{eqnarray*}
g_t(x,y)&= &t^{-1} h_t(x,y)-\frac{x}{t^2+t}\\
&=&\sum_{a=0}^{\infty}\sum_{m=1}^{\infty}x^{a}y^m\frac{1}{m} \left(  \begin{matrix} m+a-1\\ a\\
\end{matrix}\right)t^{m-1}(1+t)^{-(a+m)}\mu^{*m}(m+a-1)\,.
\end{eqnarray*}
Hence  $g_t$ coincides with the generating function of the concentrations that appear in the statement, and the proof can be completed by an appeal to Lemma \ref{L2}(i).
\QED

We now conclude this section describing the limiting behavior of concentrations as time tends to infinity, starting with the case
$D>0$ (i.e. at the initial time, the mean number of arms is less than the total concentration).

\begin{corollary}\label{C1} Let $\mu=(\mu(a))_{a\in\N}$ be a probability measure on $\N$
 with mean
$$\sum_{a\in\N}a\mu(a)=1-D<1\,.$$
Then the solution $(c_t(a,m): a\in\N, m\in\N^*
\hbox{ and } t\geq0)$ to the system \eqref{EQ5}  started from
$$c_0(a,m)={\bf 1}_{\{m=1\}}\mu(a)\,,\qquad a\in\N \hbox{ and } m\in\N^*$$
has a limit as $t\to\infty$ in $\ell^1(\N\times \N^*)$ given by
$$c_{\infty}(a,m)= {\bf 1}_{\{a=0\}}\frac{D}{m}\mu^{*m}(m-1)\,.$$
 \end{corollary}

\proof The pointwise convergence (i.e. with $a$ and $m$ fixed) should be plain from the expression given in Theorem \ref{T1}(ii). Recall from Lemma \ref{L1}
that the total concentration $C_t$ tends to $D$ as $t\to \infty$.
On the other hand, as $\mu$ is a probability measure on $\N$ with mean less than $1$, it is easily seen (see e.g. the remark after this proof) that $m^{-1}\mu^{*m}(m-1)$ also defines a probability measure on $\N^*$.
So
$$ \sum_{a=0}^{\infty}\sum_{m=1}^{\infty}c_{\infty}(a,m)=D\,,$$
and we can complete the proof invoking Scheff\'e's lemma (see for example \cite{Billing}). \QED

 It is well-known and easy to check  that if $\mu$ is a probability measure on $\N$ with mean $1-D<1$, then  $\breve{\mu}:=(m^{-1}\mu^{*m}(m-1))_{m\in\N^*}$ is a probability measure on $\N^*$ with mean
$$\sum_{m=1}^{\infty}\mu^{*m}(m-1)=D^{-1}\,.$$
Indeed, we know from the Lagrange inversion formula (see for instance Theorem 5.1.1 in \cite{Wilf}) that the generating functions $g$ of $\mu$ and $h$ of $\breve{\mu}$ are related by the equation $h(x)=xg(h(x))$. The condition on
the mean of $\mu$ reads $g'(1)=1-D<1$; this readily entails that $h(1)=1$, so that $\breve{\mu}$ has total mass $1$. Then taking the (left) derivative at $x=1$, we get
$h'(1)=g(h(1))+ h'(1)g'(h(1))$, which yields $h'(1)=1/D$.
Thus Corollary \ref{C1} implies that
$$\sum_{m=1}^{\infty}mc_{\infty}(0,m)=1\,.$$
This property can be interpreted as follows. Suppose that at the initial time, we tag
a monomer uniformly at random. Then the distribution of the polymer
at time $t$ that contains this tagged particle
has the distribution
$$\sum_{a=0}^{\infty}\sum_{m=1}^{\infty}m c_t(a,m)\delta_{(a,m)}\,,$$
and as time passes, this family of measures remains tight. Physically, this means that
{\it if at the initial time  the mean number of arms is less than the total concentration of monomers, then the oriented model does not produce particles with infinite size as time tends to infinity.}

\noindent {\bf Remark. } The recent paper \cite{BSV} deals with a system of randomly interacting particles which is closely related to the present deterministic model, and sheds a probabilistic light on Corollary \ref{C1}. More precisely, in \cite{BSV}, time is discrete and at the initial time
there are $n$ particles with arms such that the sequence $\xi_1, \ldots, \xi_n$ of the number of arms of particles is i.i.d. with a fixed distribution $\mu$ on $\N$.
Arms are enumerated uniformly at random, which specifies the order of activation. When an arm is activated, it grabs uniformly at random one of the particles in the system which had not been grabbed previously and which does not belong to its own cluster either. The polymerization procedure terminates when all arms have been activated, and the terminal configuration is given by a forest of  trees.
Roughly speaking, the main result in \cite{BSV} is that if $\mu$ is subcritical (i.e. its first moment is $1-D<1$), then as $n\to\infty$, the distribution
of a tree picked uniformly at random in the terminal configuration converges to that of a Galton-Watson tree with reproduction law $\mu$. According to Dwass \cite{Dwass}, the probability that the size of a Galton-Watson tree with reproduction law $\mu$ is $m$
equals $m^{-1}\mu^{*m}(m-1)$. Further, by the law of large numbers, the number of trees in the terminal forest is approximately $Dn$, so that the density of trees with size $m$ at the terminal time is $Dm^{-1}\mu^{*m}(m-1)$. This corroborates  Corollary \ref{C1}, proving another example of the deep connexions between coagulation models and branching processes (see, e.g. \cite{DT, Spouge'}).

\subsection{Examples}

Let us now discuss some explicit examples.

Consider first the degenerate case when $\mu = \delta_1$ is the Dirac mass at $1$, so at the initial time, there is a unit concentration of  particles with unit size having exactly one arm, and all the other concentrations are $0$. It is clear that coagulations then always produce polymers with exactly one arm, and therefore in this specific situation,  the notion of arms plays no role in the evolution of the system. In particular it is not surprising that a known solution should emerge, however it may be interesting to discuss this case as a verification of our general formulas. Specifically we have $ \mu^{*m}=\delta_m$
so we find
$$c_t(1,m)= t^{m-1}(1+t)^{-(1+m)}\,,$$
and $c_t(a,m)=0$ for $a\neq 1$.
As a check, observe that summing these quantities for $m\in\N^*$ is in
agreement with Lemma \ref{L2}.
It is also interesting to compare with Smoluchowski's solution \eqref{EQ2}. Specifically
we see that  Smoluchowski's coagulation equation for the kernel $\kappa(m,m')=2$
and monodisperse  initial condition can be viewed as the present oriented coagulation with arms when at the initial time all particles have unit size and exactly one arm.
This observation can be established directly by an elementary analysis of the transition rates in both models.

 More generally, consider the case when $\mu$ is the binomial law with
 parameter $(n,1/n)$, where $n\geq 2$ is some integer, i.e.
 $$\mu(a)= \left(  \begin{matrix} n\\ a\\
\end{matrix}\right)n^{-n}(n-1)^{n-a}\qquad \hbox{for }a=0, \ldots, n\,.$$
 Then $\mu^{*m}$ is the binomial distribution with parameter $(mn,1/n)$, and we get
$$c_t(a,m)=
\frac{(mn)!}{(mn+1-a-m)!m! a!} t^{m-1}(1+t)^{-(a+m)}
n^{-mn}(n-1)^{mn-a-m+1}
$$
when $a+m-1\leq mn$, and $c_t(a,m)=0$ otherwise.

Then we let $n$ tend to $\infty$ and thus consider the case when $\mu$ is the standard Poisson law, i.e. $\mu(a)=1/( a! \e)$ for $a\in\N$. Then
$\mu^{*m}$ is the Poisson distribution with parameter $m$ and we get
$$c_t(a,m)= \e^{-m}
m^{a+m-1}t^{m-1}(1+t)^{-(a+m)}  \frac{1}{a! m!}\,,\qquad m\in \N\,.$$
It is interesting to point out that summing this quantity over $a\in\N$ yields the total concentration of particles with size $m$,
$$C_t(m):=\sum_{a\in\N}c_t(a,m)=\frac{1}{1+t}\, \frac{(tm/(1+t))^{m-1}}{m!}\e^{-mt/(1+t)}
= \frac{1}{1+t} B(t/(1+t),m)\,. $$
The comparison with Golovin's solution \eqref{EQ3} suggests that, loosely speaking,
 Smoluchowski's coagulation equation for the additive kernel and monodisperse  initial condition coincides after the logarithmic time change $t\to \log(1+t)$, with the present oriented coagulation with arms when at the initial time, particles have unit size and the number of arms is distributed according to the Poisson law.

  We leave to the interested reader the task of developing similar calculations for binomial or Poisson laws with mean $\neq 1$ to illustrate Theorem \ref{T1}(ii).
Here is a final example in this vein. We assume that $\mu$ is the Negative Binomial distribution with parameters $r>0$ and $p\in(0,1)$, viz.
$$\mu(a)=\frac{\Gamma(r+a)}{a!\Gamma(r)} p^r (1-p)^a\,,\qquad a\in\N\,.$$
Recall that
$$\sum_{a=0}^{\infty}a\mu(a)=r(1-p)/p:=1-D\,,$$
and that $\mu^{*m}$ is Negative Binomial distribution with parameters $mr$ and $p$.
Assuming that $D\neq 0$, we then find that the concentration $c_t(a,m)$ is given by
 $$\e^{Dt}m^{-1} D^{a+1}\left(\e^{Dt}-1\right)^{m-1}
\left(\e^{Dt}-1+D\right)^{-(a+m)}\frac{\Gamma(mr+a+m-1)}{a!(m-1)!\Gamma(rm)} p^{mr} (1-p)^{a+m-1}
\,.$$
\end{section}

\begin{section}{The symmetric model}

\subsection{Setting and relation to a quasi-linear PDE}
We next turn our attention to the symmetric model of coagulation with arms, keeping the notation for the oriented one.
This means that now each aggregation event consumes two arms, one for each particle involved, and that any pair of arms is activated at the same unit rate.
So the transition
\begin{equation}\label{EQ8'}
\{(a,m),(a',m')\}\longrightarrow (a+a'-2,m+m')
\end{equation}
occurs at time $t$ with intensity
$$aa'c_t(a,m)c_t(a',m')\,,$$
and the evolution of the concentration functions is specified by the infinite non-linear system of ODE's
\begin{eqnarray}\label{EQ8}
\d c_t(a,m)  =& &\frac{1}{2}\sum_{a'=1}^{a+1}\sum_{m'=1}^{m-1} a'(a-a'+2)
c_t(a',m')c_t(a-a'+2,m-m')\nonumber \\
&-&c_t(a,m)\sum_{a'=1}^{\infty}\sum_{m'=1}^{\infty}aa'c_t(a',m').
\end{eqnarray}
Again \eqref{EQ8} only makes sense as long $A_t<\infty$, but it is readily seen that $t\to A_t$ decreases as $t$ grows, so it suffices to require that $A_0<\infty$.

Before starting the analysis, let us point at the special role of particles with no arms. Indeed, particles with no arms are inactive in the symmetric model, in the sense that they cannot coagulate with other particles (this was not the case in the oriented model as a particle with no arm could  still be grabbed by some other particle).
Analytically, this is seen from the fact that the sub-system \eqref{EQ8} for $(a,m)\in\N^*\times \N^*$ is autonomous.
We also stress that  particles with no arms are produced by the coagulation of two particles both with a single arm, and more precisely,  specializing  \eqref{EQ8}
yields the simple identity
\begin{equation}\label{EQ9}
\d c_t(0,m)  =\frac{1}{2}\sum_{m'=1}^{m-1}
c_t(1,m')c_t(1,m-m')\,.
\end{equation}

Just as for the oriented model, it is convenient to re-express the system \eqref{EQ8}
 as
\begin{equation}\label{EQ10}
\d \langle c_t,f\rangle=\frac{1}{2}
\sum_{a,a'=1}^{\infty}\sum_{m,m'=1}^{\infty} (f(a+a'-2,m+m')- f(a,m)-f(a',m')) aa'
c_t(a,m)c_t(a',m')\,,
\end{equation}
where $f:\N\times \N^*\to \R_+$ is a generic  nonnegative and bounded function.

The equations \eqref{EQ8} resembles Smoluchowski's coagulation equation \eqref{EQ1} for the multiplicative kernel. In the latter case, it is well-known that a phenomenon of gelation occurs, in the sense that the total mass is not a preserved quantity for all times
as one might expect naively. Informally, this is due to the formation of particles of infinite size  in finite time; see e.g. \cite{Aldous, Jeon, vDE2}.
A similiar phenomenon may (or may not) happen in the present case, and we shall study \eqref{EQ8} and \eqref{EQ10} before that critical time. Specifically, we
introduce for any $r>0$
$$\Gamma_r:=\inf\{t\geq 0 : \langle c_t, a^2\rangle \geq r\}\,,$$
where by a slight abuse in notation, we write $a^2$ for the function $(a,m)\to a^2$,
and then
$$\Gamma_{\infty}:=\sup\{\Gamma_r: r\geq 0\}\,.$$
Recall also that $A_t:=\langle c_t,a\rangle$ is the mean number of arms at time $t$.

\begin{lemma}\label{L4} For the symmetric model, we have
$$A_t=\frac{A_0}{1+tA_0}\qquad \hbox{for all }t<\Gamma_{\infty}\,.$$
\end{lemma}

\proof The argument is similar to that in Lemma \ref{L1}. Using the same notation as there, we first specify \eqref{EQ10}  to the function
$f^{(\alpha)}$ and let $\alpha\to \infty$. The bound
$$|f^{(\alpha)}(a+a'-2,m+m')-f^{(\alpha)}(a,m)-f^{(\alpha)}(a',m')|\leq (a\wedge a')+2\,,$$
enables us to apply the theorem of dominated convergence provided that $\sup_{0\leq s\leq t}\langle c_s, a^2 \rangle <\infty$ and we get the equation
$ \d A_t=-A_t^2$. \QED

We stress that the formula in Lemma \ref{L4} may fail when $t$ is too large. Indeed,
applying \eqref{EQ10} to $f(a,m)\equiv 2$, we also get $ \d (2C_t)=-A_t^2$
without  requiring that $t<\Gamma_{\infty}$. Thus if Lemma \ref{L4} was always valid,
then the difference $2C_t-A_t$ would remain constant. But this is absurd when
$2C_0<A_0$ since then one would have $2C_t=2C_0+A_t-A_0\to 2C_0-A_0<0$ as $t\to \infty$.

For the sake of simplicity, we shall focus in the rest of this section on the case when
$A_0:=\langle c_0,a\rangle=1$, which induces no significant loss of generality as the general case
can be reduced to that one by a linear time-change (provided that of course $A_0<\infty$). Recall the notation $g_t(x,y)$ for the generating function of $c_t$ and introduce
$$k_t(x,y):=\p g_t(x,y)\,,\qquad x,y\in[0,1]$$
which should be viewed as the generating function :
$$k_t(x,y)=\sum_{a=0}^{\infty}\sum_{m=1}^{\infty} x^ay^m (a+1) c_t(a+1,m)\,. $$
The following statement is a partial counter-part of Lemma \ref{L2} for the symmetric model.

\begin{lemma}\label{L5} Assume that $A_0=1$. Then for any solution to the system \eqref{EQ8} we have the equation
$$\d k_t(x,y)=\left(k_t(x,y)-x A_t\right)\p k_t(x,y)-A_t k_t(x,y)\,,$$
and for  $t< \Gamma_{\infty}$, the latter can be rewritten as the quasi-linear PDE
$$\d k_t(x,y)=\left(k_t(x,y)-\frac{x}{1+t}\right)\p k_t(x,y)-\frac{1}{1+t}k_t(x,y)\,.$$
\end{lemma}

\proof
Just as in the proof of Lemma \ref{L2}, the equation \eqref{EQ10} specialized to
$f(a,m)=f_{x,y}(a,m)=x^a y^m$ yields
$$\d g_t(x,y)=\frac{1}{2}
\left(\p g_t(x,y)\right)^2 -x A_t \p g_t(x,y)\,. $$
Then taking the partial derivative with respect to $x$, we obtain the first equation in the statement.
The second follows from an application of Lemma \ref{L4} and the assumption $A_0=1$.
\QED

We observe that the second PDE in Lemma \ref{L5}  is the
same as in Lemma \ref{L2}(i) for the oriented model
with $k_t(x,y)$ replacing $g_t(x,y)$. In this direction, we recall that a similar relation between solutions of Smoluchowski's coagulation equations for the additive and the multiplicative kernels holds, see e.g. Theorem 3.9 in  \cite{DT}.

\subsection{Explicit solutions and examples}
We are now able to solve \eqref{EQ8} up-to the critical time $\Gamma_{\infty}$ when at the initial time, all particles are monomers, i.e. each particle has unit size and its number of arms is arbitrary. Just as for the oriented model, it may be interesting to observe that  the re-parameterization $\alpha=a-2$ is additive on coagulation, in the sense that
\eqref{EQ8'} then reads
\begin{equation}\label{EQ8"}
\{(\alpha,m),(\alpha',m')\}\longrightarrow (\alpha+\alpha',m+m')
\end{equation}
This also makes the solvability of \eqref{EQ5} easier to understand; note however that the parameter $\alpha$ may take negative values and that the rate of the coagulation \eqref{EQ8"} is now
$aa'=(\alpha+2)(\alpha'+2)$ and hence not multiplicative in the alternative parameter $\alpha$.

We consider a measure
$\mu=(\mu(a))_{a\in\N}$ on $\N$ with unit mean
and finite second moment (we stress that we do not require $\mu$ to be a probability measure), and introduce the probability measure $\nu=(\nu(a))_{a\in\N}$ given by
$$\nu(a)=(a+1)\mu(a+1)\,,\qquad a\in\N\,.$$
We denote  the first moment of $\nu$ by
$$M:=\sum_{a=0}^{\infty} a \nu(a)=\sum_{a=1}^{\infty}a(a-1)\mu(a)\,,$$
and then define
$$T =\left\{  \begin{matrix} \infty &\hbox{ if }M\leq 1\\
1/(M-1) &\hbox{ otherwise.} \\
\end{matrix}\right. $$
Recall also that we write $\nu^{*m}$ for the $m$-th convolution product of $\nu$.

\begin{theorem}\label{T2}
The system \eqref{EQ8} has a unique solution
$(c_t(a,m): a, m\in\N^* \hbox{ and } t<T)$  started from
$$c_0(a,m)={\bf 1}_{\{m=1\}}\mu(a)\,,\qquad a,m\in\N^*$$
which is given for $a,m\geq 1$ by
$$c_t(a,m)=\frac{(a+m-2)!}{a! m!}  t^{m-1}(1+t)^{-(a+m-1)}\nu^{*m}(a+m-2)\,.$$
Further, $T$ coincides with the critical time $\Gamma_{\infty}$.
\end{theorem}

\proof
The initial conditions invite us to introduce the generating function of $\mu$,
$$g(x):=\sum_{a=0}^{\infty} x^a\mu(a)\,,\qquad x\in[0,1]\,,$$
and to set for $y\in[0,1]$
$$g_0(x,y)=yg(x)\quad \hbox{and} \quad
k_0(x,y):=\p g_0(x,y)=y\sum_{a=0}^{\infty} x^a\nu(a).$$

Consider the second PDE of Lemma \ref{L5}, viz.
\begin{equation}\label{EQ11}
\d k_t(x,y)=\left(k_t(x,y)-\frac{x}{1+t}\right)\p k_t(x,y)-\frac{1}{1+t}k_t(x,y)\,,
\end{equation}
with boundary value $k_0(x,y)$ defined above. If we replace $k_t$ by $g_t$, this
is precisely  the PDE that has been solved by the method of characteristics  in Lemma \ref{L2}(i). There is however a difference that requires some attention : here we assume that
$\p k_0(1,1)=M<\infty$ whereas we had the stronger hypothesis $\p g_0(1,1)=1$ in Lemma \ref{L2}(i). Nonetheless, the condition  $t<T$ implies $1+t-t M>0$, and thus
the derivative of the function $x\to (1+t)x-tk_0(x,y)$ is strictly positive for $x\in[0,1]$. As
$-tk_0(0,y)<0<1\leq 1+t-tk_0(1,y)$, this ensures the existence of a unique inverse function
$\ell_t(\cdot,y):[0,1]\to[0,1]$ to $x\to (1+t)x-tk_0(x,y)$.
The argument in the proof of Lemma \ref{L2}(i) is thus still valid, and we conclude that
for every $t<T$  and $x,y\in[0,1]$
$$k_t(x,y) = (1+t)^{-1}k_0(\ell_t(x,y),y)= t^{-1} \ell_t(x,y)-\frac{x}{t^2+t}\,.$$

The comparison with Theorem \ref{T1}(i) now yields that for $t<T$ and $x,y\in[0,1]$
$$k_t(x,y)=\sum_{a=0}^{\infty}\sum_{m=1}^{\infty}x^ay^m
 m^{-1}t^{m-1}(1+t)^{-(a+m)} \left(  \begin{matrix} a+m-1\\ a\\
\end{matrix}\right) \nu^{*m}(a+m-1)\,.$$
We thus see that if we define $c_t(a,m)$ for $a,m\in\N^*$ and $0\leq t< T$ as in the statement, and let $k_t$ be the generating function of $((a+1)c_t(a+1,m): a\in\N \hbox{ and }m\in\N^*)$, then
$k_t$ solves \eqref{EQ11} for $0\leq t<T$.

We next observe that $\ell_t(1,1)=1$ since $k_0(1,1)=1$, and thus
$$k_t(1,1)=\langle c_t, a\rangle =1/(1+t)\,.$$
Hence $k_t$ also solves
$$\d k_t(x,y)=\left(k_t(x,y)-x\langle c_t, a\rangle\right)\p k_t(x,y)-\langle c_t, a\rangle k_t(x,y)\,,$$
which is the first PDE in Lemma \ref{L5}.
Inverting the generating functions, we conclude that $(c_t(a,m): a,m\in\N^*\hbox{ and }
0\leq t < T)$ is a solution to \eqref{EQ8}.

We then check that the critical time
$\Gamma_{\infty}$ of this solution coincides with $T$.
In this direction, we first recall that $\p k_0(1,1)=M$, $\ell_t(1,1)=1$ and, by definition, that
$$(1+t)\ell_t(x,1)=tk_0(\ell_t(x,1),1)+x\,.$$
We take the (left) derivative with respect to the variable $x$ at $x=1$
and obtain
$$(1+t)\p \ell_t(1,1)=tM\p \ell_t(1,1) + 1\,,$$
so
$$\p \ell_t(1,1)=\frac{1}{1+t(1-M)}\,,\qquad t<T\,.$$
It follows that
$$\p k_t(1,1) = t^{-1}\p \ell_t(1,1)-(t+t^2)^{-1}=\frac{M}{(1+t)(1+t(1-M))}$$
remains bounded on compact intervals in $[0,T[$, and further explodes as $t \uparrow T$ when $M>1$ (i.e. $T<\infty$). As $\p k_t(1,1)=\langle c_t, a^2-a\rangle$,
we conclude that $T=\Gamma_{\infty}$. Finally, the uniqueness of the solution to \eqref{EQ8} up to the critical time should be plain from a perusal of the preceding arguments.
 \QED

Theorem \ref{T2} does not provide an expression for concentrations of particles with no arms. However, these can be recovered from the concentrations of particles with exactly one arm via the equation \eqref{EQ9} (note that the concentration of monomers with no arm does not evolve, so we may focus on particles with size $m\geq 2$ in the following statement).

\begin{corollary}\label{C2} Under the same assumptions as in Theorem \ref{T2}, and assuming also that $\nu(0)>0$,
we have for every $m\geq 2$ and $t<T$
$$c_t(0,m)=\frac{1}{m(m-1)}(1+1/t)^{1-m}\nu^{*m}(m-2)\qquad \hbox{ for }m\geq 2\,.$$

As a consequence, in the case when $M\leq 1$ (that is when $T=\infty$), the concentration of particles $(a,m)$ at time $t$ has a limit $c_{\infty}(a,m)$
as $t\to \infty$ which is given by
$$c_{\infty}(a,m)= {\bf 1}_{\{a=0\}}\frac{1}{m(m-1)}\nu^{*m}(m-2)\qquad \hbox{ for }
a\in\N \hbox{ and }m\geq 2\,.$$
\end{corollary}

\proof We know from Theorem \ref{T2} that for every $t<T$ and $m\geq 1$
$$c_t(1,m)=t^{m-1}(1+t)^{-m}m^{-1}\nu^{*m}(m-1)\,.$$
According to the classical Lagrange inversion formula (cf. Theorem 5.1.1 in \cite{Wilf}),
$m^{-1}\nu^{*m}(m-1)$ appears as the $m$-th coefficient in the analytic expansion
of the entire function $u$ which solves $u(x)=x\phi(u(x))$, where $\phi$ is the generating function of the probability measure $\nu$ (the assumption $\nu(0)>0$ ensures that
$\phi(0)>0$).
Another application of the Lagrange inversion formula now shows that
 the $m$-th coefficient in the analytic expansion of the entire function $u^2(x)$
 is $2m^{-1}\nu^{*m}(m-2)$. In terms of convolution product, this reads
$$c_t(1,\cdot)^{*2}(m): =\sum_{m'=1}^{m-1}
c_t(1,m')c_t(1,m-m')=
 t^{m-2}(1+t)^{-m} \frac{2}{m}\nu^{*m}(m-2)\,.$$
Since $c_0(1,m)=0$ for $m\geq 2$ by assumption and
$$\int_0^t  s^{m-2}(1+s)^{-m}{\rm d} s
=(m-1)^{-1}(1+1/t)^{1-m}\,,$$
our first claim follows from \eqref{EQ9}. The second follows immediately from the first for $a=0$, and from Theorem \ref{T2} for $a\geq 1$. \QED

 We stress that when $M\leq 1$ and $\nu\neq \delta_1$, then
\begin{equation}\label{EQ12}
\sum_{a=0}^{\infty}\sum_{m=2}^{\infty}m c_{\infty}(a,m)=\sum_{m=2}^{\infty}\frac{1}{(m-1)}\nu^{*m}(m-2)
=\sum_{a=0}^{\infty}\frac{1}{a+1}\nu(a)=\mu(\N^*)\,.
\end{equation}
Just as for the oriented model, this identity can be interpreted as a tightness property
  as time passes
for the distribution of the polymer
at time $t$ that contains a randomly tagged monomer. Physically, this means that the symmetric coagulation model does not produce particles with infinite size as time tends to $\infty$.

 Indeed, the identity \eqref{EQ12} is easily checked as follows :
\begin{eqnarray*}
\sum_{m=2}^{\infty}\frac{1}{(m-1)}\nu^{*m}(m-2)&=&\sum_{a=0}^{\infty}\nu(a)\sum_{m=2}^{\infty}\frac{1}{(m-1)}\nu^{*(m-1)}(m-2-a)\\
&=&\sum_{a=0}^{\infty}\nu(a)\sum_{n=1}^{\infty}n^{-1}\nu^{*n}(n-a-1)\,.\\
\end{eqnarray*}
As $\nu\neq \delta_{1}$ is a probability measure with mean $M\leq 1$, it follows easily from the Lagrange inversion formula (see the argument in the remark after the proof of Corollary \ref{C1}) that for any $a\geq 0$,
$$\sum_{n=1}^{\infty}\frac{a+1}{n}\nu^{*n}(n-a-1)=1\,.$$
We stress that this is where the assumption  $\nu\neq \delta_{1}$ is needed as otherwise the generating function of $\nu$ would vanish at $0$, impeding the application of Lagrange inversion formula. We thus have checked that
$$\sum_{m=2}^{\infty}\frac{1}{(m-1)}\nu^{*m}(m-2)=\sum_{a=0}^{\infty}(a+1)^{-1}\nu(a)\,,$$
the other equalities in  \eqref{EQ12} are obvious.

To summarize this observation, if we exclude the case when $\nu=\delta_1$
(or equivalently, $\mu=\frac{1}{2}\delta_2$) that will discussed further below, we have shown that there either gelation occurs at a finite time (by  Theorem \ref{T2}, this happens if and only if $M> 1$), or the mass distribution of particles remains tight as time tends to infinity.

\noindent{\bf Remark. } The limiting concentrations in Corollary \ref{C2}  bear a striking
resemblance with the distribution of the total population generated by a (sub)-critical Galton-Watson branching process with reproduction law $\nu$  and started from two ancestors. We refer the interested reader to
the recent work \cite{BS} for a probabilistic interpretation that relies on the study of the random configuration model. The latter can be used to create typical graphs with specified degree sequence, and is constructed via a stochastic algorithm which
can be thought of as the probabilistic counterpart of the present symmetric model of coagulations with limited aggregations.

We now conclude this work by illustrating Theorem \ref{T2} with some examples.
The simplest is when $\mu=\delta_1$, that is, initially, there is a unit concentration of monomers with a single arms. So $\nu=\delta_0$ and one finds
$$c_t(1,1)=\frac{1}{1+t}\quad,\quad c_t(0,2)=\frac{t}{2t+2}\,,\qquad 0\leq t < T=\infty\,,$$
and all the other concentrations are zero.

Next, suppose that at the initial time, there is a concentration $1/2$ of monomers with two arms, i.e. $\mu=\frac{1}{2}\delta_2$. Then $\nu=\delta_1$ and we get
$$c_t(2,m)=\frac{1}{2} t^{m-1}(1+t)^{-(m+1)}\,,\qquad m\in\N^*,0\leq t < T=\infty\,,$$
and all the other concentrations are zero. Note the similarity with the oriented model started from a unit density of monomers with a single arm. Of course, this property could also be observed by a direct argument.

We then turn our attention to the simplest example with a finite critical time, namely the situation where at the initial time, there is a concentration $1/3$ of monomers with three arms, i.e. $\mu=\frac{1}{3}\delta_3$. Then $\nu=\delta_2$ and $M=2$, so the critical time is $T=1$.
One gets
$$c_t(m+2,m)=\frac{(2m)!}{(m+2)! m!}  t^{m-1}(1+t)^{-(2m+1)}\,,\qquad m\in\N^*\hbox{ and }t<1\,,$$
and all the other concentrations are $0$. It is easily checked  that
$$\lim_{t\to 1-}\sum_{m=1}^{\infty}(m+2)^2c_t(m+2,m)=\infty\,,$$
in agreement with Lemma \ref{L4}.

Finally, consider the case when $\mu$ is the standard Poisson law, then $\nu$ is also the standard Poisson law and in particular $T=\infty$. As $\nu^{*m}$ is the Poisson law with parameter $m$, we get from Theorem \ref{T2} that
$$c_t(a,m)=\frac{m^{a+m-2}}{a! m!}  t^{m-1}(1+t)^{-(a+m-1)}\e^{-m}\,,\quad a,m\in\N^*, t\geq 0\,,$$
and from Corollary \ref{C2} that
$$
c_t(0,1)=\e^{-1} \quad \hbox{and}\quad
c_t(0,m) = \e^{-m}\frac{m^{m-2}}{m!}(1+1/t)^{1-m} \hbox{ for }m\geq 2\,.
$$
It is interesting to point out that
$$\lim_{t\to\infty}c_t(0,m)=\e^{-m}\frac{m^{m-2}}{m!}=m^{-1}B(1,m)\,,\qquad m\in\N^*$$
and to compare with McLeod's solution \eqref{EQ4} to Smoluchowski's coagulation equation for the multiplicative kernel and monodisperse initial condition.
We see that the terminal state for the symmetric model of coagulation started from monomers with arms distributed according to the standard Poisson law is the same as
the state at gelation time in Smoluchowski's model for the multiplicative kernel and monodisperse initial condition.

\end{section}

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