text:
\begin{document}

\newcommand{\D}{\mathbb{D}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}

\def\diam{\operatorname{diam}}
\def\dist{\operatorname{dist}}
\def\diver{\operatorname{div}}
\def\ess{\operatorname{ess}}
\def\inner{\operatorname{int}}
\def\osc{\operatorname{osc}}
\def\sign{\operatorname{sign}}
\def\supp{\operatorname{supp}}

\newcommand{\BMO}{BMO(\Omega)}
\newcommand{\LOne}{L^{1}(\Omega)}
\newcommand{\LOnen}{(L^{1}(\Omega))^d}
\newcommand{\LTwo}{L^{2}(\Omega)}
\newcommand{\Lq}{L^{q}(\Omega)}
\newcommand{\Lp}{L^{2}(\Omega)}
\newcommand{\Lpn}{(L^{2}(\Omega))^d}
\newcommand{\LInf}{L^{\infty}(\Omega)}
\newcommand{\HOneO}{H^{1,0}(\Omega)}
\newcommand{\HTwoO}{H^{2,0}(\Omega)}
\newcommand{\HOne}{H^{1}(\Omega)}
\newcommand{\HTwo}{H^{2}(\Omega)}
\newcommand{\HmOne}{H^{-1}(\Omega)}
\newcommand{\HmTwo}{H^{-2}(\Omega)}

\newcommand{\LlogL}{L\log L(\Omega)}

\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits}

\numberwithin{equation}{section}
\title{Generalised global supersolutions with  mass control for systems with taxis}
\author{Anna~Zhigun}
\renewcommand\Affilfont{\itshape\small}
\affil{ Queen's University Belfast, School of Mathematics and Physics\\
University Road, Belfast BT7 1NN, Northern Ireland, UK\\
  e-mail: {A.Zhigun@qub.ac.uk}}
\date{}
\maketitle
\begin{abstract}
 The existence of  generalised global supersolutions with a control upon the total mass is established for a wide family of  parabolic-parabolic chemotaxis systems and general integrable initial data in any space dimension. It is verified that as long as a  supersolution  of this sort remains smooth, it coincides with the  classical solution. At the same time, the proposed construction provides solvability beyond a blow-up time.
 The considered class of systems includes the basic form of the Keller-Segel model as well as the case of a chemorepellent.

\\\\
{\bf Keywords}: chemotaxis;  generalised supersolution; global existence; measure-valued solutions.
\\
MSC 2010:
35B45,
92C17,
35D30,
35D99,
35K55.
\end{abstract}

\section{Introduction}
Ever since it was introduced \cite{KS70,KS71} in the 1970's, the Keller-Segel model for chemotaxis and its modifications have been extensively studied, compare, e.g., reviews \cite{Horstmann,BBTW} on the available results. The  best-known model representative has the form
\begin{subequations}\label{KScl}
 \begin{alignat}{3}
  &\partial_tu=\nabla\cdot\left(\nabla u-\chi u\nabla v\right)\quad&&\text{ in }\R^+\times\Omega,\\
  &\partial_t v=\Delta v-v+u\quad&&\text{ in }\R^+\times\Omega,\\
  &\partial_{\nu} u=\partial_{\nu} v=0\quad&&\text{ in }\R^+\times\partial\Omega,\\
  &u(0,\cdot)=u_0,\ v(0,\cdot)=v_0\quad&&\text{ in }\Omega,
 \end{alignat}
\end{subequations}
where $\Omega$ is a smooth bounded domain in $\R^N$, $N\in\N$, with the corresponding outer normal unit vector $\nu$ on $\partial\Omega$. The constant $\chi>0$ is the so-called chemotactic sensitivity coefficient.

It has long been suspected that in higher dimensions  \cref{KScl} possesses solutions which blow-up in finite time. The later means that there are solutions which start as classical at $t=0$ but exist only up to some finite time  $T_{max}\in\R^+$ when the aggregation and collapse of the $u$-component occur, i.e., \begin{align*}
\underset{t\uparrow T_{max}}{\lim\sup}\, \|u(t,\cdot)\|_{L^{\infty}(\Omega)}=\infty.                                                                                \end{align*}

This hypothesis  was finally confirmed in  {\cite{Winkler2013}} for the case when $\Omega$ is a ball.

One of the  questions naturally arising in connection with  the blow-up phenomena is the following: how can a solution be reasonably extended  beyond its blow-up time? It is clear that the only nonlinear term in the system, $\chi u\nabla v$, creates difficulties. Indeed, it is not obvious as to how the product between $u$ and $\nabla v$ should be interpreted if, for instance, already
\begin{align*}
 \underset{t\uparrow T_{max}}{\lim\sup}\, \|uv(t,\cdot)\|_{L^1(\Omega)}=\infty
\end{align*}
can occur in the blow-up case {\cite{Winkler2013}}.
This issue has been addressed with success for some  parabolic-elliptic modifications of \cref{KScl}  in dimension $N=2$. Usually one introduces  a suitable family of regular problems parametrised by a small parameter $\varepsilon>0$ in such a way that by taking $\varepsilon=0$  the original system is formally recovered.
The main challenges are then to find the correct limit problem for $\varepsilon\rightarrow 0$ and  to rigorously prove convergence. Formal asymptotics undertaken in   \cite{Velazquez2004_1,Velazquez2004_2}  for one class of such regularisations revealed the structure of the blow-up solutions and indicated what the corresponding limit problem should be. The local in time well-posedness of the later was also verified \cite{Velazquez2004_3}.
Several subsequent works  deal with the construction of  generalised weak solutions in $\R^2$ \cite{DolbSchm2009} or a bounded domain \cite{LuckSugVel} and establish convergence to it. An important mark of the resulting limit problems  is that they involve measure-valued terms, which, moreover, turn out to be dependent upon the choice of a specific regularisation. Measures necessarily appear in those parts of the solution formulation where integrability can be lost due to a potential blow-up. The  solution concepts in both \cite{DolbSchm2009} and \cite{LuckSugVel}  rely heavily upon the properties of the Green function of the Laplace operator in planar domains, as well as  a certain symmetry of the model: both the coefficient before $\nabla v$ in the convective flux in the equation for $u$ and the production term in the $v$-equation are linear in $u$. It has remained open as to how these constructions could be extended to the cases in which the space dimensions $N$ is larger than two, and/or the system is fully parabolic,  and/or $\chi$ depends upon $u$. The present study aims to fill this gap.

System \cref{KScl} can also be brought into  comparison with its modification involving a  logarithmic chemotactic sensitivity function: in this case the term $\chi u\nabla v$ is replaced by $\chi u\nabla \ln(v)=\chi \frac{u}{v}\nabla v$. Owing to the damping effect of $\frac{1}{v}$
at large values of $v$, the corresponding solutions are less prone to formation of strong singularities. Very recently a new, generalised, solution concept has been introduced \cite{LankWink2017} and further developed \cite{ZhigunKSLog} which makes an extensive use of this property. The underlying idea was to replace the equation for $u$ by a variational inequality which describes the evolution of a certain coupling of both $u$ and $v$. There results a supersolution concept with a mass control from above, and it has been shown that such a supersolution, should it be regular, actually coincides with the  classical solution. By switching to an inequality, one avoids the necessity to include a non-negative measure-valued term, of which, other than in the cases studied in \cite{DolbSchm2009,LuckSugVel}, not much is presently known. As to the employed  coupled quantities, they enjoy better regularity than $u$ generally does. It ultimately allows to  sustain some sort of  solvability even if a blow-up does take place in some parts of the system. This approach has previously been used in the context of degenerate haptotaxis systems \cite{ZSH,ZSU}.

In this paper we elaborate a global solvability concept  for  the following generalisation of \cref{KScl}:

\begin{subequations}\label{KS}
 \begin{alignat}{3}
  &\partial_tu=\nabla\cdot\left(\nabla u-c(u,v)\nabla v\right)\quad&&\text{ in }\R^+\times\Omega,\label{Equ}\\
  &\partial_t v=\Delta v-v+u\quad&&\text{ in }\R^+\times\Omega,\label{Eqv}\\
  &\partial_{\nu} u=\partial_{\nu} v=0\quad&&\text{ in }\R^+\times\partial\Omega,\label{bc}\\
  &u(0,\cdot)=u_0,\ v(0,\cdot)=v_0\quad&&\text{ in }\Omega,
 \end{alignat}
\end{subequations}
where
\begin{align*}
 &c(u,v):=\chi(u,v)u
\end{align*}
for an arbitrary
\begin{align}
 \chi\in C(\R_0^+\times \R_0^+).\nonumber
\end{align}
It is important to note that no further restrictions upon $\chi$ are required. In fact, just as it is the case for the solvability in the classical sense, the upper-triangular structure of system \cref{KS} turns out to be a decisive factor for the generalised solvability as well. More precisely, the outcome of this work can be summarised as follows: the classical theory  for upper-triangular systems (see, e.g., \cite{Amann1}) implies that \cref{KS} possesses a  solution which is classical as long as it doesn't blow-up, and if it does, then it continuous to exist as generalised in terms of {\cref{defsol}} given below.

Our construction is based on the ideas developed in   \cite{LankWink2017,ZhigunKSLog} for the case of a logarithmic sensitivity (which is not covered by this work since we assume $\chi$ to be continuous). The main difference lies in the choice of suitable couplings of $u$ and $v$: we use cut-offs in order to overcome possible unboundedness of any of the two components.
As in \cite{LankWink2017,ZhigunKSLog}, the  generalised solution is obtained by means of a limit procedure for an   approximation family which in the present case involves a regularisation

in the chemotactic coefficient (see \cref{KSe} below). However, other  regularisations are also possible. It is to expect that just as in \cite{DolbSchm2009,LuckSugVel} the resulting solution depends upon the choice of a particular approximation.

The rest of the paper is organised in the following way. We begin with a preliminary {\cref{SecPrelim}} where we fix some notations which will be used throughout and also recall some results involving measures.
In {\cref{SecSol}} we introduce our supersolution concept for \cref{KS} (cl. {\cref{defsol}}), and   two main results: {\cref{TheoEx}} deals with existence of such supersolutions, while {\cref{classol}} establishes a link to the classical solvability. In {\cref{secreg}} we prepare some  ingredients which are necessary to our proof of  {\cref{TheoEx}} in {\cref{SecEx}}. {\cref{classol}} is proved in the closing {\cref{SecCl}}.

\section{Preliminaries}\label{SecPrelim}
A number of notational conventions are used in this paper for the sake of conciseness:
\begin{Notation}\label{Notation}~
\begin{enumerate}[(1)]
 \item For any index $i$, a quantity $C_i$ denotes a positive constant or function.
 \item

Dependence upon such parameters as: the space { dimension $N$}, domain $\Omega$, function $c$,  the $L^1$-norms of the initial conditions $u_0$ and $v_0$, as well as constants $a$ and $b$ (are introduced below) is mostly
{\bf not} indicated in an explicit way.
\item\label{cut}  Given $n,l\in\N$, we introduce the cut-offs
 \begin{align}
 &\overline{u}:=(n-u)_+,\qquad \overline{v}:=(l-v)_+. \nonumber
\end{align}
Terms $\overline{u_k}, \overline{u_{k_m}},\overline{v_k}$, $\overline{v_{k_m}}$  are to be understood in the same fashion.

\end{enumerate}
\end{Notation}
\subsection{Functional spaces}\label{SecFSp}
We assume the reader to be familiar with the standard spaces of continuous and continuously differentiable functions, as well as the Lebesgue, Sobolev,    and Bochner spaces and standard results concerning them.

As usual, ${\cal M}(\overline{\Omega})$ denotes the Banach space of signed Radon measures in $\Omega$, while   ${\cal M}_+(\overline{\Omega})$ stands for the set of all positive Radon measures.

We also make use of the Banach space
\begin{align*}
L^{\infty}_{w-*}(\R^+;{\cal M}(\overline{\Omega})):=&\left\{\mu:\R^+\rightarrow {\cal M}(\overline{\Omega})\quad\text{is weak}-*\text{ measurable and}\right.\\
&\ \left. \left\|\mu\right\|_{L^{\infty}_{w-*}(\R^+;{\cal M}(\overline{\Omega}))}:=\left\|\|\mu(\cdot)\|_{{\cal M}(\overline{\Omega})}\right\|_{L^{\infty}(\R^+)}<\infty\right\}
\end{align*}
and its closed subset
\begin{align*}
&L^{\infty}_{w-*}(\R^+;{\cal M}_+(\overline{\Omega})):=\left\{\mu:\R^+\rightarrow {\cal M}_+(\overline{\Omega})|\ \mu\in L^{\infty}_{w-*}(\R^+;{\cal M}(\overline{\Omega}))\right\}.
\end{align*}
 We identify functions which coincide a.e. in $\R^+$. It is known \cite[sections 8.18.1-8.18.2]{Edwards} that
$L^{\infty}_{w-*}(\R^+;{\cal M}(\overline{\Omega}))$ is isometrically isomorphic to the continuous dual of the Bochner space $L^1(\R^+{;}C(\overline{\Omega}))$ via the duality paring
\begin{align*}
 \left<\mu,\varphi\right>=\int_{\R^+}\int_{\overline{\Omega}}\varphi(t)(x)\, (d\mu(t))(x)dt.
\end{align*}
Since $L^1(\R^+{;}C(\overline{\Omega}))$ is separable, the Banach-Alaoglu theorem implies that { closed} balls in $L^{\infty}_{w-*}(\R^+;{\cal M}(\overline{\Omega}))$ are weak-$*$ sequentially compact.

\subsection{Divergence-measure fields and their normal traces}\label{AppA}
In this sequel some facts concerning the  divergence-measure fields are  { presented in the same way as  in \cite[Appendix A]{ZhigunKSLog}}. We use them in order to give weak reformulations of the Neumann boundary conditions (see {\cref{defsol}} below).

  Recall the definition of the Banach space of divergence-measure fields \cite{ChenFrid2001} and its norm:
\begin{align}
&{\cal DM}^p(\Omega):=\{F\in (L^p(\Omega))^N|\ \ \nabla\cdot F\in {\cal M}(\overline{\Omega})\},\nonumber\\
 &\|F\|_{{\cal DM}^p(\Omega)}:=\|F\|_{(L^p(\Omega))^N}+\|\nabla\cdot F\|_{{\cal M}(\overline{\Omega})}.\nonumber
\end{align}
Here we assume that $p\in\left(1,\frac{N}{N-1}\right)$, which is sufficient for our needs.

Following \cite{ChenFrid2001} we introduce a generalisation of the normal trace over the boundary of $\partial\Omega$ which automatically satisfies a  Gauss-Green formula:
\begin{align}
 \left<F\cdot\nu|_{\partial\Omega},\varphi\right>:=\int_{\Omega}F\cdot\nabla ({\cal E}\varphi)\,dx+\int_{\overline{\Omega}} ({\cal E}\varphi)\, d(\nabla\cdot F)\qquad\text{for all }\varphi\in  W^{\frac{1}{p},\frac{p}{p-1}}(\partial\Omega).\label{GG}
\end{align}
Here ${\cal E}: W^{\frac{1}{p},\frac{p}{p-1}}(\partial\Omega)\rightarrow W^{1,\frac{p}{p-1}}(\Omega)$ is a usual   extension operator, i.e., a continuous right inverse of the  corresponding trace operator.
  It is known (see \cite[Theorem 2.1]{ChenFrid2001}) that  $F\cdot\nu|_{\partial\Omega}\in W^{-\frac{1}{p},p}(\partial\Omega)$ and  doesn't depend upon the particular choice of ${\cal E}$. Formula \cref{GG} ensures the following implication:
  \begin{align}
   \begin{rcases}F_n\underset{m\rightarrow\infty}{{\rightharpoonup}}F&\text{ in }(L^p(\Omega))^N,\\ \nabla \cdot F_n\underset{m\rightarrow\infty}{\overset{*}{\rightharpoonup}}\nabla\cdot F&\text{ in }{\cal M}(\overline{\Omega})
   \end{rcases}\quad\Rightarrow\quad  F_n\cdot\nu|_{\partial\Omega}\underset{m\rightarrow\infty}{\overset{*}{\rightharpoonup}}F\cdot\nu|_{\partial\Omega}\quad\text{in }W^{-\frac{1}{p},p}(\partial\Omega).\label{contitr}
  \end{align}

\section{Generalised supersolutions to \texorpdfstring{\cref{KS}}{}}\label{SecSol}
We make the following general assumptions on the data involved:
\begin{align}
&{ c(u,v)=\chi(u,v)u,\qquad u,v\in \R_0^+,}\label{cchi}\\
&\chi\in C(\R_0^+\times \R_0^+),\label{chiC}\\
 &u_0,v_0\in L^1(\Omega),\label{uv0L1}\\
 &0\leq u_0,v_0\not\equiv0.\label{uvp}
\end{align}
Motivated by an idea from  \cite{LankWink2017} and our previous work \cite{ZhigunKSLog}, we introduce a  concept of a generalised supersolution with a mass control.

\begin{Definition}[Generalised supersolution]\label{defsol} { Let $c$ and $(u_0,v_0)$ satisfy  \cref{cchi}-\cref{uvp}}.  We call a pair of measurable functions $(u,v):\R_0^+\times\overline{\Omega}\rightarrow {\R_0^+}\times {\R_0^+}$ a \underline{generalised supersolution} to system \cref{KS} if there exist some:
\begin{enumerate}[(1)]
\item element $\mu\in L^{\infty}_{w-*}(\R^+;{\cal M}_+(\overline{\Omega}))$,
 \item numbers $a,b>2$,
 \item function $M:\N\times\N\rightarrow \R^+_0$,  $M=M(n,l)$,
\end{enumerate}
such that { for all $n,l\in\N$}:
\begin{enumerate}[(i)]
  \item\label{defi} $u\in L^{\infty}(\R^+;L^1(\Omega))$,\  $v\in L^1_{loc}(\R_0^+;W^{1,1}(\Omega))$,\ $v^{-1}\in L_{loc}^{\infty}(\R^+\times\overline{\Omega})$;
  \item $\overline{u}^{a}\overline{v}^{b},\ \overline{v}^{b}\in L^2_{loc}(\R^+_0;H^1(\Omega))$;
  \item \label{inDMp} $\int_0^{\infty}\psi\left(\left(\nabla\left( \overline{u}^{a}\overline{v}^{b}\right)-\overline{u}^{a}\nabla \overline{v}^{b}\right)+\frac{1}{b}\left(b\left({ M} +\overline{u}^{a}\right)-ac(u,v)\overline{u}^{a-1}\overline{v}\right)\nabla \overline{v}^{b}\right)\,ds$,\\   $\int_0^{\infty}\psi \nabla v\,ds\in {\cal DM}^{p}(\Omega)$   for all $\psi\in C^1_0(\R_0^+)$ for some $p\in\left(1,\frac{N}{N-1}\right)$;
\item for all $0\leq\varphi\in C^1(\overline{\Omega})$ and $0\leq\psi\in C^1_0(\R_0^+)$ it holds that
\begin{align}
&-\int_0^{\infty}\partial_t\psi\int_{\Omega}\left({ M} +\overline{u}^{a}\right)\overline{v}^{b}\varphi\,dxdt-\psi(0)\int_{\Omega}\left({ M} +\overline{u_0}^{a}\right)\overline{v_0}^{b}\,dx\nonumber\\
 \leq&\int_0^{\infty}\psi\int_{\Omega}-4\left(\frac{a-1}{a}\left|\nabla \left(\overline{u}^{\frac{a}{2}}\overline{v}^{\frac{b}{2}}\right)-\overline{u}^{\frac{a}{2}}\nabla \overline{v}^{\frac{b}{2}}\right|^2\right.\nonumber\\
 &\phantom{aaaaaa}\left.+\frac{1}{b}\overline{u}^{\frac{a}{2}-1}\left(2b\overline{u}-(a-1)c(u,v)\overline{v}\right)\left(\nabla \left(\overline{u}^{\frac{a}{2}}\overline{v}^{\frac{b}{2}}\right)-\overline{u}^{\frac{a}{2}}\nabla \overline{v}^{\frac{b}{2}}\right)\cdot\nabla \overline{v}^{\frac{b}{2}}\right.\nonumber\\
 &\phantom{aaaaaa}\left.+\frac{1}{b}\left((b-1)\left({ M}+\overline{u}^{a}\right)-ac(u,v)\overline{u}^{a-1}\overline{v}\right)\left|\nabla \overline{v}^{\frac{b}{2}}\right|^2\right)\varphi\nonumber\\
 &\phantom{aaaaaa}-\left(\left(\nabla\left( \overline{u}^{a}\overline{v}^{b}\right)-\overline{u}^{a}\nabla \overline{v}^{b}\right)+\frac{1}{b}\left(b\left({ M} +\overline{u}^{a}\right)-ac(u,v)\overline{u}^{a-1}\overline{v}\right)\nabla \overline{v}^{b}\right)\cdot\nabla\varphi\nonumber\\
 &\phantom{aaaaaa}-b\left({ M} +\overline{u}^{a}\right)\overline{v}^{b-1}\left(-v+u\right)\varphi\,dxdt;\label{superu}
 \end{align}
\item for all { $\varphi\in C^1(\overline{\Omega})$ and $\psi\in C^1_0(\R_0^+)$} it holds that
\begin{align}
 -\int_0^{\infty}\partial_t\psi\int_{\Omega}v\varphi\,dxds-\psi(0)\int_{\Omega}v_0\varphi\,dx=&\int_0^{\infty}\psi\int_{\Omega}-\nabla v\cdot\nabla\varphi+(-v+u)\varphi\,dx\nonumber\\
 &+\int_{\overline{\Omega}}\varphi \,(d\mu(s))(x)ds,\label{superv}
\end{align}
and
\begin{align}
 \|u(t,\cdot)\|_{L^1(\Omega)}+\int_{\overline{\Omega}} \,(d\mu(t))(x)= \|u_{0}\|_{L^1(\Omega)}\qquad\text{for a.a. }t>0;\label{massu0}
\end{align}

\item
for all $\psi\in C^1_0(\R_0^+)$ it holds in $W^{-\frac{1}{p},p}(\partial\Omega)$ that
\begin{align}
&\int_0^{\infty}\psi\left(\left(\nabla\left( \overline{u}^{a}\overline{v}^{b}\right)-\overline{u}^{a}\nabla \overline{v}^{b}\right)+\frac{1}{b}\left(b\left({ M} +\overline{u}^{a}\right)-ac(u,v)\overline{u}^{a-1}\overline{v}\right)\nabla \overline{v}^{b}\right)\,ds\cdot\nu|_{\partial\Omega}\nonumber\\
=&0,\label{prweakbc}\\
 &\int_0^{\infty}\psi\nabla v\,ds\cdot\nu|_{\partial\Omega}=0.\label{vweakbc}
\end{align}
\end{enumerate}
 Here $\bar u$ and $\bar v$ are as defined in {\cref{Notation}\eqref{cut}}, i.e., they implicitly depend upon $n$ and $l$.
\end{Definition}

\begin{Remark}[Boundary conditions] The variational reformulations \cref{prweakbc}-\cref{vweakbc} of the boundary conditions \cref{bc} are consistent with the regularity assumptions in \eqref{inDMp}, cl. \cite[Theorem 2.1]{ChenFrid2001}.
\end{Remark}

Our result on existence now reads:
\begin{Theorem}[Existence of generalised supersolutions]\label{TheoEx}
   Let $c$ and $(u_0,v_0)$ satisfy  \cref{cchi}-\cref{uvp}. Then there exists a generalised supersolution $(u,v)$  to system \cref{KS} in terms of {\cref{defsol}}.
\end{Theorem}
The proof of this theorem is based on a suitable regularisation and a series of priori estimates in {\cref{secreg}} leading into  a limit procedure in {\cref{SecEx}}.
\begin{Remark}[Choice of $M$]
As we see later on in the proof, a solution in terms of {\cref{defsol}} can be constructed using any function $M$ which satisfies
\begin{align}
 M{>} M_{*}{ [c]},
\end{align}
where, { for any function $\sigma\in C(\R_0^+\times \R_0^+)$,}
\begin{align}
 M_{*}{[\sigma]}=&M_{*}{[\sigma]}(n,l)\nonumber\\
 :=&{\frac{1}{4}}\frac{ab}{(a-1)(b-1)}\underset{(u,v)\in[0,n]\times[0,l]}{\max}\left\{\frac{(a-1)^2}{b^2}{\sigma}^2(u,v)\overline{u}^{a-2}\overline{v}^2+4\frac{a+b-1}{ab}\overline{u}^{a}\right\}.\label{condM}
\end{align}

\end{Remark}

Our interest in the introduced concept of generalised supersolutions is supported by the following result:
\begin{Theorem}[Classical solutions]\label{classol}
 Let
\begin{align}
\chi\in C^1(\R_0^+\times \R_0^+).\label{chi1}\end{align}
Let a pair $(u,v)$ be a supersolution in terms of {\cref{defsol}}  such that for some $T\in{(0,\infty]}$
 \begin{align}
 u,v\in C([0,T)\times\overline{\Omega})\cap C^{1,2}((0,T)\times\overline{\Omega}).\label{uvsmooth}
\end{align}
 Then $(u,v)$ is a classical solution to   \cref{KS} in ${ [0,T)}\times \overline{\Omega}$.
\end{Theorem}
The proof of {\cref{classol}} is given in {\cref{SecCl}}.

\section{Smooth regularisations for  \texorpdfstring{\cref{KS}}{}}\label{secreg}

{ Consider a sequence of functions \begin{align*}
\chi_k\in C_0^2(\R_0^+\times \R_0^+),\qquad k\in\N,                                              \end{align*}

which approximates $\chi$ on compact sets:
\begin{align*}
 \chi_k\underset{k\rightarrow\infty}{\rightarrow}\chi\qquad \text{in }C([0,n]\times[0,l])\qquad\text{for all }n,l\in\N.
\end{align*}
Set
\begin{align*}
 c_k(u,v):=\chi_k(u,v)u\qquad\text{for all }n,l\in\N.
\end{align*}
The assumptions on $\chi_k$ readily imply that
\begin{align*}
  &c_k\in C_0^2(\R_0^+\times \R_0^+), \\
  & c_k\underset{k\rightarrow\infty}{\rightarrow}c\qquad \text{in }C([0,n]\times[0,l])\qquad\text{for all }n,l\in\N,
\end{align*}
and
\begin{align}
 M_{*}[c_k](n,l)\underset{k\rightarrow\infty}{\rightarrow}M_{*}[c](n,l)\qquad\text{for all }n,l\in\N,\label{convM*}
\end{align}
with $M_*$ as defined in \cref{condM}.
Now we are ready to introduce} a family of regularisations of system \cref{KS}:
\begin{subequations}\label{KSe}
 \begin{alignat}{3}
  &\partial_t u_k=\nabla\cdot\left(\nabla u_k-{ c_k}(u_k,v_k)\nabla v_k\right)\quad&&\text{ in }(0,T)\times\Omega,\label{ue}\\
  &\partial_t v_k=\Delta v_k-v_k+{ u_k}\quad&&\text{ in }(0,T)\times\Omega,\label{ve}\\
  &\partial_{\nu} u_k=\partial_{\nu} v_k=0\quad&&\text{ in }(0,T)\times\partial\Omega,\label{uvebc}\\
  &u_k(0,\cdot)=u_{k0},\ v_k(0,\cdot)=v_{k0}\quad&&\text{ in }\Omega.
 \end{alignat}
\end{subequations}
We choose the regularised initial data  $u_{k0}$ and $v_{k0}$ so as to satisfy
\begin{align}
&0< u_{k0},v_{k0}\in W^{1,\infty}(\Omega),\label{RegIni}\\
&\|v_{k0}\|_{L^1(\Omega)}\geq \|v_0\|_{L^1(\Omega)},\label{aproxminv0}
\end{align}
and
\begin{alignat}{3}
&u_{k0}\underset{k\rightarrow\infty}{\rightarrow}u_0&&\qquad\text{in }L^1(\Omega)\text{ and a.e. in }\Omega,\label{aproxiniu_}\\
&v_{k0}\underset{k\rightarrow\infty}{\rightarrow}v_0&&\qquad\text{in }L^1(\Omega)\text{ and a.e. in }\Omega.\label{aproxiniv_}
\end{alignat}
Classical theory  for upper-triangular systems (see, e.g., \cite{Amann1}) implies that \cref{KSe} possesses a unique global  bounded classical solution  $(u_k,v_k)$  provided that both solution components are a priori bounded. The latter is the case here. Indeed, we observe first that  due to the maximum principle  both solution components are strictly positive in $\R^+_0\times \overline{\Omega}$.  Further, since $c_k$ is compactly supported, the maximum
 principle implies that $u_k$ is bounded above, which, together with the comparison principle, leads to boundedness  of  $v_k$.  Solutions to \cref{KSe} are further studied in  \cref{basic,VarF,keyest}.

 Throughout the rest of this section we  assume that
 \begin{align*}
  { n,l,k\in\N}
 \end{align*}
are arbitrary but fixed.
\begin{Notation}\label{Notation2}~
\begin{enumerate}[(1)]
 \item\label{Notitem} In addition to our previous conventions (see {\cref{Notation}})  we do {\bf not} indicate in an explicit way the  dependence of quantities $B_i,C_i,D,F$  (are introduced in what follows) and that of $M_*$  (as defined by \cref{condM}) upon $n$, $l$, and $M_0$ (yet another parameter which is used below).
\item   On the other hand, we stress that all estimates which we derive in this section are uniform w.r.t. $k$. In particular, all $C_i$'s are  independent of this number.

\end{enumerate}
\end{Notation}
\subsection{Basic properties of    \texorpdfstring{\cref{KSe}}{}}\label{basic}
Integrating equations \cref{ue} and \cref{ve} over $\Omega$ and using the boundary conditions and partial integration we obtain the following information about the total masses: for all $t\geq0$
\begin{align}
 &\|u_k(t,\cdot)\|_{L^1(\Omega)}= \|u_{k 0}\|_{L^1(\Omega)},\label{massue}\\
 &\|v_k(t,\cdot)\|_{L^1(\Omega)}\leq\left(1-e^{-t}\right)\|u_{k 0}\|_{L^1(\Omega)}+e^{-t}\|v_{k 0}\|_{L^1(\Omega)}.\label{massve}
\end{align}
Due to \cref{massue}-\cref{massve} and a classical result based on duality (see, e.g., the proof of Lemma 5 in \cite[Appendix A]{BOTHE2010120}) we have for all \begin{align}
(r,s)\in\left[1,\frac{N+2}{N}\right)\times\left[1,\frac{N+2}{N+1}\right)\nonumber                                                                                \end{align}
that

\begin{align}
 \left\{\left(v_k,\nabla v_k\right)\right\}_{k\in(0,1]} \quad \text{is precompact in }L^r_{loc}(\R_0^+\times\Omega)\times L^s_{loc}(\R_0^+\times\Omega).\label{vcomp}
\end{align}
Further, using the maximum principle and the strict positivity of the Neumann heat kernel, we conclude from   \cref{aproxminv0} that $v_{k}$ can be controlled  from below in the following way:
\begin{align}
 \inf_{(\tau,T)\times \Omega}v_{k}\geq &\inf_{(\tau,T)\times \Omega}e^{-t}e^{t\Delta}v_{k 0}\nonumber\\
 \geq &\C{\left(\tau,T\right)}\|v_{k 0}\|_{L^1(\Omega)}\nonumber\\
 \geq &\Cl{vmin}{\left(\tau,T\right)}>0\qquad\text{for all }0<\tau<T<\infty.\label{estvmin}
\end{align}
\subsection{A variational formulation for  \texorpdfstring{\cref{KSe}}{}}\label{VarF}
Let
\begin{align*}
 a,b>2,\qquad M_0\geq0
\end{align*}
be some numbers and consider the function
\begin{align}
 &F(u,v):=\left(M_0 +\overline{u}^{a}\right)\overline{v}^{b}.\label{F}
\end{align}
We have that
\begin{align*}
 F\in C^2_b(\R_0^+\times \R_0^+),
\end{align*}
and the  derivatives of $F$ up to order two are:
\begin{subequations}\label{FDir}
\begin{align}
&\partial_uF(u,v)=-a\overline{u}^{a-1}\overline{v}^{b},\label{Fu}\\
&\partial_vF(u,v)=-b\left(M_0 +\overline{u}^{a}\right)\overline{v}^{b-1},\label{Fv}\\
 &\partial_{uu}F(u,v)=a(a-1)\overline{u}^{a-2}\overline{v}^{b},\label{Fuu}\\
 &\partial_{uv}F(u,v)=ab\overline{u}^{a-1}\overline{v}^{b-1},\label{Fuv}\\
 &\partial_{vv}F(u,v)=b(b-1)\left(M_0+\overline{u}^{a}\right)\overline{v}^{b-2}.\label{Fvv}
\end{align}
\end{subequations}
Multiplying  \cref{ue} and \cref{ve} by $\partial_{u}F(u_k,v_k)$ and $\partial_{v}F(u_k,v_k)$, respectively, adding the results together, and using the chain rule and expressions \cref{FDir} where necessary, we compute  that
\begin{align}
 &\partial_t\left(\left(M_0 +\overline{u_k}^{a}\right)\overline{v_k}^{b}\right) \nonumber\\
 =&-\left(\nabla u_k-{ c_k}(u_k,v_k)\nabla v_k\right)\cdot\left(a(a-1)\overline{u_k}^{a-2}\overline{v_k}^{b}\nabla u_k+ab\overline{u_k}^{a-1}\overline{v_k}^{b-1}\nabla v_k\right)\nonumber\\
 &+\nabla\cdot\left(-a\overline{u_k}^{a-1}\overline{v_k}^{b}\left(\nabla u_k-{ c_k}(u_k,v_k)\nabla v_k\right)\right)\nonumber\\
 &-\nabla v_k \cdot\left(ab\overline{u_k}^{a-1}\overline{v_k}^{b-1}\nabla u_k+b(b-1)\left(M_0+\overline{u_k}^{a}\right)\overline{v_k}^{b-2}\nabla v_k\right)\nonumber\\
 &+\nabla\cdot\left(-b\left(M_0 +\overline{u_k}^{a}\right)\overline{v_k}^{b-1}\nabla v_k\right)\nonumber\\
 &-b\left(M_0 +\overline{u_k}^{a}\right)\overline{v_k}^{b-1}\left(-v_k+{ u_k}\right)
 \nonumber\\
 =&-\left(a(a-1)\overline{u_k}^{a-2}\overline{v_k}^{b}\left|\nabla u_k\right|^2\right.\nonumber\\
 &\phantom{aaa}\left.+\left(2ab\overline{u_k}^{a-1}\overline{v_k}^{b-1}-a(a-1){ c_k}(u_k,v_k)\overline{u_k}^{a-2}\overline{v_k}^{b}\right)\nabla u_k\cdot\nabla v_k\right.\nonumber\\
 &\phantom{aaa}\left.+\left(b(b-1)\left(M_0+\overline{u_k}^{a}\right)\overline{v_k}^{b-2}-ab{ c_k}(u_k,v_k)\overline{u_k}^{a-1}\overline{v_k}^{b-1}\right)\left|\nabla v_k\right|^2\right)\nonumber\\
 &+\nabla\cdot\left(-a\overline{u_k}^{a-1}\overline{v_k}^{b}\nabla u_k+\left(-b\left(M_0 +\overline{u_k}^{a}\right)\overline{v_k}^{b-1}+a{ c_k}(u_k,v_k)\overline{u_k}^{a-1}\overline{v_k}^{b}\right)\nabla v_k\right)\nonumber\\
 &-b\left(M_0 +\overline{u_k}^{a}\right)\overline{v_k}^{b-1}\left(-v_k+{ u_k}\right)\nonumber\\
 =&-4\left(\frac{a-1}{a}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)-\overline{u_k}^{\frac{a}{2}}\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\right.\nonumber\\
 &\phantom{aaaa}\left.+\frac{1}{b}\overline{u_k}^{\frac{a}{2}-1}\left(2b\overline{u_k}-(a-1){ c_k}(u_k,v_k)\overline{v_k}\right)\left(\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)-\overline{u_k}^{\frac{a}{2}}\nabla \overline{v_k}^{\frac{b}{2}}\right)\cdot\nabla \overline{v_k}^{\frac{b}{2}}\right.\nonumber\\
 &\phantom{aaaa}\left.+\frac{1}{b}\left((b-1)\left(M_0+\overline{u_k}^{a}\right)-a{ c_k}(u_k,v_k)\overline{u_k}^{a-1}\overline{v_k}\right)\left|\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\right)\nonumber\\
 &+\nabla\cdot\left(\left(\nabla\left( \overline{u_k}^{a}\overline{v_k}^{b}\right)-\overline{u_k}^{a}\nabla \overline{v_k}^{b}\right)+\frac{1}{b}\left(b\left(M_0 +\overline{u_k}^{a}\right)-a{ c_k}(u_k,v_k)\overline{u_k}^{a-1}\overline{v_k}\right)\nabla \overline{v_k}^{b}\right)\nonumber\\
 &-b\left(M_0 +\overline{u_k}^{a}\right)\overline{v_k}^{b-1}\left(-v_k+{ u_k}\right).
\label{KEYab_}
\end{align}
The first big bracket on the  right-hand side of  \cref{KEYab_} is a quadratic form with respect to
\begin{align*}
 \nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)-\overline{u_k}^{\frac{a}{2}}\nabla \overline{v_k}^{\frac{b}{2}}\qquad\text{and}\qquad \nabla \overline{v_k}^{\frac{b}{2}}.
\end{align*}
Its discriminant $D{ [c_k]}=D{ [c_k]}(u,v)$ satisfies
\begin{align}
 &D{[\sigma]}(u,v)\nonumber\\
 =&\frac{1}{b^2}\overline{u}^{a-2}\left(2b\overline{u}-(a-1){\sigma}(u,v)\overline{v}\right)^2-4\frac{a-1}{ab}\left((b-1)\left(M_0+\overline{u}^{a}\right)-a{\sigma}(u,v)\overline{u}^{a-1}\overline{v}\right)\nonumber\\
 =&\frac{(a-1)^2}{b^2}{\sigma}^2(u,v)\overline{u}^{a-2}\overline{v}^2+4\frac{a+b-1}{ab}\overline{u}^{a}-{ 4}\frac{(a-1)(b-1)}{ab}M_0\nonumber\\
 \leq&{ 4}\frac{(a-1)(b-1)}{ab}(M_{*}{ [\sigma]}-M_0),\nonumber
\end{align}
with $M_{*}$ is as defined in \cref{condM}. Thus, \cref{KEYab_} takes the form
\begin{align}
 &\partial_tF(u_k,v_k)\nonumber\\
 =&-4\frac{a-1}{a}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)+\Cl[B]{B1}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\nonumber\\
 &-\left(4\frac{b-1}{b}(M_0-M_{*}{ [c_k]})+\Cl[B]{B2}{[c_k]}(u_k,v_k)\right)\left|\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\nonumber\\
 &+\nabla\cdot\left(\nabla\left(\overline{u_k}^{a}\overline{v_k}^{b} \right)+\Cl[B]{B4}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{b}\right)+{\Cl[B]{Bnew}}(u_k,v_k)-{\Cl[B]{B5}(u_k,v_k)}{ u_k},
\label{KEYab}
\end{align}
where, in order to simplify the exposition, we have introduced
\begin{align*}
 &\Cr{B1}{ [\sigma]}(u,v):=-\overline{u}^{\frac{a}{2}}+\frac{a}{2(a-1)}\frac{1}{b}\overline{u}^{\frac{a}{2}-1}\left(2b\overline{u}-(a-1){ \sigma}(u,v)\overline{v}\right),\\
 &\Cr{B2}{ [\sigma]}(u,v):=-\frac{a}{a-1}D{ [\sigma]}(u,v)-4\frac{b-1}{b}(M_0-M_{*}{ [\sigma]}),\\
 &\Cr{B4}{ [\sigma]}(u,v):=-\overline{u}^{a}+\frac{1}{b}\left(b\left(M_0 +\overline{u}^{a}\right)-a{ \sigma}(u,v)\overline{u}^{a-1}\overline{v}\right),\\
 &{\Cr{Bnew}(u,v):=b\left(M_0 +\overline{u}^{a}\right)\overline{v}^{b-1}v,}\\
 &\Cr{B5}(u,v):=b\left(M_0 +\overline{u}^{a}\right)\overline{v}^{b-1}.
\end{align*}
Observe that
\begin{alignat}{5}
 &{ B_i[c_k]\in C_b(\R_0^+\times\R_0^+),\qquad }&&{\|B_i[c_k]\|_{C_b(\R_0^+\times\R_0^+)}\leq{\Cl{CB}}\qquad}&&{\text{for }i\in\{ 1,2,3\},}\label{Bbound1}\\
 &B_i\in C_b(\R_0^+\times\R_0^+),\qquad &&\|B_i\|_{C_b(\R_0^+\times\R_0^+)}\leq{\Cr{CB}}\qquad&&\text{for }i\in\{{ 4,5}\},\label{Bbound2}\\
 &\Cr{B2}{[c_k]},\Cr{B5}\geq0,&&&&\label{Bpos}
\end{alignat}
and
\begin{align}
 &{ B_i[c_k]\underset{k\rightarrow\infty}{\rightarrow}B_i[c]\qquad\text{in }C_b(\R_0^+\times\R_0^+)\qquad\text{for }i\in\{ 1,2,3\}.}\label{Bconv}
\end{align}
Multiplying \cref{KEYab} by an arbitrary function $\psi\in C^1_0(\R_0^+)$ and integrating by parts w.r.t. $t$ yields for all $x\in\Omega$ that
\begin{align}
 &-\int_0^{\infty}F(u_k,v_k)\partial_t\psi\,dt-F(u_{k0},v_{k0})\psi(0)\nonumber\\
 =&-\int_0^{\infty}4\frac{a-1}{a}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)+\Cr{B1}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\psi\,dt\nonumber\\
 &-\int_0^{\infty}\left(4\frac{b-1}{b}(M_0-M_{*}{ [c_k]})+\Cr{B2}{[c_k]}(u_k,v_k)\right)\left|\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\psi\,dt\nonumber\\
 &+\nabla\cdot\int_0^{\infty}\psi\left(\nabla\left(\overline{u_k}^{a}\overline{v_k}^{b} \right)+\Cr{B4}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{b}\right)\,dt\nonumber\\
 &+\int_0^{\infty}{\Cr{Bnew}}(u_k,v_k)\psi-{\Cr{B5}}(u_k,v_k){ u_k}\psi\,dt.
\label{KEYabt}
\end{align}
On the other hand, multiplying \cref{KEYab} by an arbitrary function $\varphi\in H^1(\Omega)$ and integrating by parts w.r.t. $x$ and using the boundary conditions where necessary yields that
\begin{align}
 \int_{\Omega}\partial_tF(u_k,v_k)\varphi\,dx
 =&-\int_{\Omega}4\frac{a-1}{a}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)+\Cr{B1}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{\frac{b}{2}}\right|^2{\varphi}\,dx\nonumber\\
 &-\int_{\Omega}\left(4\frac{b-1}{b}(M_0-M_{*}{ [c_k]})+\Cr{B2}{[c_k]}(u_k,v_k)\right)\left|\nabla \overline{v_k}^{\frac{b}{2}}\right|^2{\varphi}\,dx\nonumber\\
 &-\int_{\Omega}\left(\nabla\left(\overline{u_k}^{a}\overline{v_k}^{b} \right)+\Cr{B4}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{b}\right)\cdot\nabla\varphi\,dx\nonumber\\
 &+\int_{\Omega}{\Cr{Bnew}}(u_k,v_k)\varphi-{\Cr{B5}}(u_k,v_k){ u_k}\varphi\,dx.
\label{KEYabx}
\end{align}
Finally, if we multiply \cref{KEYab} by the product $\psi\varphi$ and integrate by parts w.r.t. $t$ and $x$, then we arrive at the following variational reformulation:
\begin{align}
 &-\int_0^{\infty}\partial_t\psi\int_{\Omega}F(u_k,v_k)\varphi\,dxdt-\psi(0)\int_{\Omega}F(u_{k0},v_{k0}){\varphi}\,dx\nonumber\\
 =&-\int_0^{\infty}\psi\int_{\Omega}4\frac{a-1}{a}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)+\Cr{B1}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{\frac{b}{2}}\right|^2{\varphi}\,dxds\nonumber\\
 &-\int_0^{\infty}\psi\int_{\Omega}\left(4\frac{b-1}{b}(M_0-M_{*}{ [c_k]})+\Cr{B2}{[c_k]}(u_k,v_k)\right)\left|\nabla \overline{v_k}^{\frac{b}{2}}\right|^2{\varphi}\,dxds\nonumber\\
 &-\int_0^{\infty}\psi\int_{\Omega}\left(\nabla\left(\overline{u_k}^{a}\overline{v_k}^{b} \right)+\Cr{B4}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{b}\right)\cdot\nabla\varphi\,dxds\nonumber\\
 &+\int_0^{\infty}\psi\int_{\Omega}{\Cr{Bnew}}(u_k,v_k)\varphi-{\Cr{B5}}(u_k,v_k){ u_k}\varphi\,dxds.
\label{superue}
\end{align}
For equation \cref{ve} a standard procedure yields the following reformulations:
for all ${\varphi\in H^1(\Omega)}$ and ${\psi\in C^1_0(\R_0^+)}$ it holds that
\begin{align}
 -\int_0^{\infty}v_k\partial_t\psi \,ds-\psi(0)v_{k0}=\nabla\cdot\int_0^{\infty}\psi\nabla v_k\,dt+\int_0^{\infty}\left(-v_k+{ u_k}\right)\psi\,dt.\label{supervte}
\end{align}
and
\begin{align}
 -\int_0^{\infty}\partial_t\psi\int_{\Omega}v_k\varphi\,dxds-\psi(0)\int_{\Omega}v_{k0}\varphi\,dx
 =&\int_0^{\infty}\psi\int_{\Omega}-\nabla v_k\cdot\nabla\varphi+\left(-v_k+{ u_k}\right)\varphi\,dxds.\label{superve}
\end{align}

\subsection{Further uniform \texorpdfstring{{ (w.r.t. $k$)}}{} estimates for \texorpdfstring{\cref{KSe}}{}}\label{keyest}
Choosing $\varphi\equiv 1$ and
\begin{align*}
M_0={ M_{*}[c_k]+1}
\end{align*}
in \cref{KEYabx}  and using \cref{Bbound1}-\cref{Bpos} yields

that
\begin{align}
 &\frac{d}{dt}\int_{\Omega}{ \left(M_{*}[c_k]+1 +\overline{u}_k^{a}\right)\overline{v}_k^{b}}\,dx\nonumber\\
 =&-\int_{\Omega}4\frac{a-1}{a}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)+\Cr{B1}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\,dx\nonumber\\
 &-\int_{\Omega}\left({ 4\frac{b-1}{b}}+\Cr{B2}{[c_k]}(u_k,v_k)\right)\left|\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\,dx\nonumber\\
 &+\int_{\Omega}{\Cr{Bnew}}(u_k,v_k)-{\Cr{B5}}(u_k,v_k){ u_k}\,dx\nonumber\\
 \leq&-\int_{\Omega}4\frac{a-1}{a}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)+\Cr{B1}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{\frac{b}{2}}\right|^2{+4\frac{b-1}{b}}\left|\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\,dx+\Cr{CB}|\Omega|.
\label{KeyApr}
\end{align}
 Integrating the differential inequality \cref{KeyApr}  over $(0,T)$ for any $T>0$ we conclude that
\begin{align}
 &\int_{\Omega}{ \left(M_{*}[c_k]+1 +\overline{u}_k^{a}\right)\overline{v}_k^{b}}(T)\,dx\nonumber\\
 &+ \int_0^T\int_{\Omega}4\frac{a-1}{a}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)+\Cr{B1}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{\frac{b}{2}}\right|^2{+4\frac{b-1}{b}}\left|\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\,dxds\nonumber\\
 \leq& \int_{\Omega}{ \left(M_{*}[c_k]+1 +\overline{u}_{k0}^{a}\right)\overline{v}_{k0}^{b}}\,dx+\C(T).\label{est1}
\end{align}
{ Since $M_{*}[c_k]$ is nonnegative and uniformly bounded w.r.t. $k$ (recall \cref{convM*}),}
the integral inequality \cref{est1} directly implies
\begin{align}
 &\int_0^T\int_{\Omega}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)+\Cr{B1}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\,dxds\leq \Cl{C1new}(T),\label{estnew}\\
 &{\int_0^T}\int_{\Omega}\left|\nabla \overline{v_k}^{\frac{b}{2}}\right|^2\,dxds\leq\Cr{C1new}(T).\label{est12new}
\end{align}
Combining \cref{estnew}-\cref{est12new} and using the fact that { $\Cr{B1}{[c_k]}$ is uniformly bounded w.r.t. $k$}, we also get
\begin{align}
 &{\int_0^T}\int_{\Omega}\left|\nabla \left(\overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right)\right|^2\,dxds\leq\Cl{C2new}(T),\label{est11new}
\end{align}
which due to the boundedness of the cut-offs immediately implies that
\begin{align}
 &{\int_0^T}\left\| \overline{u_k}^{\frac{a}{2}}\overline{v_k}^{\frac{b}{2}}\right\|_{H^1(\Omega)}^2\,ds\leq\Cl{C2new1}(T),\label{est11new1}
\end{align}
Taking $M_0=0$ in \cref{KEYabx} we obtain { thanks to  \cref{estnew}-\cref{est11new}} and the boundedness of $F$ and $B_i$'s that
\begin{align}

 &\int_0^T\left\|\partial_t\left(\overline{u_k}^{a}\overline{v_k}^{b}\right)\right\|_{(H^1(\Omega))^*}\leq \Cl{C3}({ T}).\label{estprt}
\end{align}
Further, using \cref{estnew}-\cref{est11new}  and the boundedness of $F$ and $B_i$'s, we conclude from  \cref{KEYabt}
that
\begin{align}
 \left\|\nabla\cdot\int_0^{\infty}\psi\left(\nabla\left(\overline{u_k}^{a}\overline{v_k}^{b} \right)+\Cr{B4}{[c_k]}(u_k,v_k)\nabla \overline{v_k}^{b}\right)\,dt\right\|_{L^1(\Omega)}\leq {\C(\psi)}.\label{estintnabla_}
\end{align}
Similarly, using \cref{aproxiniv_} and \cref{massue}-\cref{massve} we deduce from \cref{supervte} that
\begin{align}
 \left\|\nabla\cdot\int_0^{\infty}\psi\nabla v_{k}\,dt\right\|_{L^1(\Omega)}\leq \C(\psi).\label{estintDelta_}
\end{align}
\section{Construction of a generalised supersolution to \texorpdfstring{\cref{KS}}{}: proof of {\texorpdfstring{\cref{TheoEx}}{}}}\label{SecEx}
\begin{Notation}
 Below $k_m$ always denotes a suitably chosen subsequence, i.e.,   such that the stated convergence holds. It may thus become 'thinner' from one instance to another.

\end{Notation}

\subsection{Convergence in the \texorpdfstring{$v$-}{}component}\label{SubSecv}

To begin with, we use  properties established in {\cref{basic}} and  conclude that there exists a  measurable function $v:[0,\infty)\times \overline{\Omega}\rightarrow\R^+$ such that:\\
due to \cref{vcomp},
\begin{alignat}{3}
 &v_{k_m}\underset{m\rightarrow\infty}{\rightarrow}v&&\qquad\text{in }L^r_{loc}(\R_0^+\times\overline{\Omega})\text{ and a.e. in }\R^+\times\Omega,\label{convv}\\
 &\nabla v_{k_m}\underset{m\rightarrow\infty}{\rightarrow}\nabla v\in[0,\infty)&&\qquad\text{in }L^s_{loc}(\R_0^+\times\overline{\Omega})\text{ and a.e. in }\R^+\times\Omega;\label{convnv}
\end{alignat}
due to \cref{estvmin} and \cref{convv}, for all $0<\tau<T<\infty$
\begin{align}
 \underset{(\tau,T)\times \Omega}{\ess\inf}v\geq \Cr{vmin}(\tau,T)>0,\label{vmin}
\end{align}
so that $v^{-1}\in L_{loc}^{\infty}(\R^+\times\overline{\Omega})$.
\subsection{Almost everywhere convergence in the \texorpdfstring{$u$-}{}component}\label{SubSecue}
Next, we address the a.e. convergence of the $u$-component. We deduce that:\\
 due to \cref{est11new1} for $a=b=6$ and \cref{estprt} for $a=b=3$, the Lions-Aubin lemma \cite[Corollary 4]{Simon},  and the continuity of the root function,
\begin{align}
{\text{for each }n,l\in\N\text{ a subsequence of }}\left(n-u_{ k}\right)_+\left(l-v_{ k}\right)_+\text{ converges  a.e. in }\R^+\times\Omega;\label{convpr1_}
\end{align}
due to \cref{convv} and \cref{convpr1_},
\begin{align}
 {\text{for each }n,l\in\N\text{ a subsequence of }}\left(n-u_{ k}\right)_+\text{ converges a.e. in }\{v<l\};\label{connmu1}
\end{align}
due to \cref{connmu1}, $v\in L^1_{loc}(\R_0^+\times\overline{\Omega})$, and   a diagonal argument,
\begin{align}
 {\text{for each }n\in\N\text{ a subsequence of }}\left(n-u_{ k}\right)_+\text{ converges a.e. in }\R^+\times\Omega;\label{connmu2}
\end{align}
due to \cref{connmu2},
\begin{align}
 &{\text{for each }n\in\N\text{ a subsequence of }}n-\left(n-u_{ k}\right)_+\text{ converges to }{\eta}_n\in[0,n]\label{connmu3}\\
 &\text{a.e. in }\R^+\times\Omega\nonumber
\end{align}
for some measurable ${\eta}_n$;\\
due to \cref{connmu3},
\begin{align}
 {\text{for each }n\in\N\text{ a subsequence of }}u_{ k}\text{ { converges to} }{\eta}_n\qquad\text{a.e. in }\{{\eta}_n<n\};\label{connmu4}
\end{align}
due to \cref{connmu3} and $n\mapsto n-(n-u)_+$ increasing for all $u\in\R$,
\begin{align}
 {\eta}_n\underset{n\rightarrow\infty}{\nearrow}{ u}\in[0,\infty]\qquad\text{a.e. in }\R^+\times\Omega\label{etaincr}
\end{align}
for some measurable ${ u}$;\\
due to \cref{connmu4}-\cref{etaincr},
\begin{align}
 {\text{for each }n\in\N\text{ a subsequence of }}u_{ k}\text{ { converges to} }{\eta}_n{ =u}\ \ \text{a.e. in }{\{{ u}<n\}\subset\{{\eta}_n<n\}};\label{connmu5}
\end{align}
due to \cref{aproxiniu_}, \cref{massue}, \cref{etaincr}, $n-(n-u)_+\in[0,u]$, and the  monotone convergence theorem,  { $u\in L^{\infty}(\R^+;L^1(\Omega))$ and}
\begin{align}
 \|{ u}\|_{L^{\infty}(\R^+;L^1(\Omega))}\leq \|u_0\|_{L^1(\Omega)};\label{xiL1}
\end{align}
 due to $u\in L^{\infty}(\R^+;L^1(\Omega))$, \cref{connmu5}, and a diagonal argument,
\begin{align}
 &u_{k_m}
  \underset{m\rightarrow\infty}{\rightarrow}u\qquad\text{ a.e. in }\R^+\times\Omega.\label{convuae}
\end{align}

\subsection{Convergence in the variational reformulations}

In this sequel we prove that $u$ and $v$ satisfy variational reformulations \cref{superu} and \cref{prweakbc}-\cref{vweakbc}.
 From now on we fix arbitrary constants
\begin{align*}
 a,b>2
\end{align*}
and a function
\begin{align}
 &M:\N\times\N\rightarrow \R^+_0,\nonumber\\
 &{ M{ >} M_*{[c]}},\label{Ml}
\end{align}
with $M_*$ as defined in \cref{condM}. Thus, these parameters satisfy the requirements of {\cref{defsol}}. Further, throughout this section we assume $$n,l\in\N$$ to be arbitrary but fixed and make use of {\cref{Notation}} as well as {\cref{Notation2}\eqref{Notitem}}.

Consider an arbitrary sequence of functions
 \begin{align*}
&G_{{ m}}\in C_b(\R^+_0\times \R^+_0),
\end{align*}
such that for some $G\in C_b(\R^+_0\times \R^+_0)$
\begin{align*}
 G_{{ m}}\underset{m\rightarrow\infty}{\rightarrow}G\qquad\text{in } C_b(\R^+_0\times \R^+_0).
\end{align*}
Using the convergence results from \cref{SubSecv,SubSecue}, we deduce that:\\
due to \cref{convv} and \cref{convuae},
\begin{alignat}{3}
 &G_{{ m}}\left(u_{k_m},v_{k_m}\right)\underset{m\rightarrow\infty}{\rightarrow}G(u,v)&&\qquad\text{ a.e. in }\R^+\times\Omega;\label{convG}
\end{alignat}
due to \cref{convG} and the dominated convergence theorem,
\begin{align}
 &G_{{ m}}\left(u_{k_m},v_{k_m}\right)\underset{m\rightarrow\infty}{\rightarrow}G(u,v)\qquad\text{ in }L^p_{loc}(\R^+_0\times\Omega)\text{ for all }p>1;\label{convGLp}
\end{align}
due to \cref{convG},
\begin{align}
 G_{{ m}}(u_{k_m},v_{k_m})\underset{m\rightarrow\infty}{\rightarrow}{ G}(u,v)\qquad\text{ in }L^1_{loc}(\R^+_0\times\Omega);\label{convFL1}
\end{align}
due to \cref{convv}-\cref{convnv} and the chain rule,
\begin{align}
 \nabla \overline{v_{k_m}}^{\frac{b}{2}}\underset{m\rightarrow\infty}{\rightarrow}\nabla \overline{v}^{\frac{b}{2}}\qquad \text{in }L^s_{loc}(\R^+_0\times\Omega)\text{ and a.e. in }\R^+\times\Omega;\label{convnbv}
\end{align}
 due to \cref{est12new} and  \cref{convnbv},
\begin{align}
 \nabla \overline{v_{k_m}}^{\frac{b}{2}}\underset{m\rightarrow\infty}{\rightharpoonup}\nabla \overline{v}^{\frac{b}{2}}\qquad \text{in }L^2_{loc}(\R^+_0\times\Omega);\label{convweak}
\end{align}
due to \cref{convG} and \cref{convnbv},
\begin{align}
 G_{{ m}}\left(u_{k_m},v_{k_m}\right)\nabla \overline{v_{k_m}}^{\frac{b}{2}}\underset{m\rightarrow\infty}{\rightarrow}G\left(u,v\right)\nabla \overline{v}^{\frac{b}{2}}\qquad\text{ a.e. in }\R^+\times\Omega;\label{convGvae}
\end{align}
 due to \cref{est12new}, \cref{convGvae}, and the Lions lemma \cite[Lemma 1.3]{Lions}
\begin{align}
  G_{{ m}}\left(u_{k_m},v_{k_m}\right)\nabla \overline{v_{k_m}}^{\frac{b}{2}}\underset{m\rightarrow\infty}{\rightharpoonup}G\left(u,v\right)\nabla \overline{v}^{\frac{b}{2}}\qquad \text{in }L^2_{loc}(\R^+_0\times\Omega);\label{convGvweak}
\end{align}
 due to \cref{est11new}, \cref{convG}, and the Banach-Alaoglu theorem,
\begin{align}
 \nabla \left(\overline{u_{k_m}}^{\frac{a}{2}}\overline{v_{k_m}}^{\frac{b}{2}}\right)\underset{m\rightarrow\infty}{\rightharpoonup}\nabla \left(\overline{{ u}}^{\frac{a}{2}}\overline{{ v}}^{\frac{b}{2}}\right)\qquad \text{in }L^2_{loc}(\R^+_0\times\Omega).\label{convprw}
\end{align}
Let
\begin{align}
 0\leq\varphi\in C^1(\overline{\Omega}),\qquad 0\leq\psi\in C^1_0(\R_0^+)\nonumber
\end{align}
be arbitrary. Then:\\
 due to \eqref{Bconv} and \cref{convGvweak}-\cref{convprw},
\begin{align}
&\nabla \left(\overline{u_{k_m}}^{\frac{a}{2}}\overline{v_{k_m}}^{\frac{b}{2}}\right)+\Cr{B1}{[c_{k_m}]}(u_{k_m},v_{k_m})\nabla \overline{v_{k_m}}^{\frac{b}{2}}\underset{m\rightarrow\infty}{\rightharpoonup}\nabla \left(\overline{u}^{\frac{a}{2}}\overline{v}^{\frac{b}{2}}\right)+\Cr{B1}{[c]}(u,v)\nabla \overline{v}^{\frac{b}{2}},\label{convweak1}\\
 &\nabla\left(\overline{u_{k_m}}^{a}\overline{v_{k_m}}^{b} \right)+\Cr{B4}{[c_{k_m}]}(u_{k_m},v_{k_m})\nabla \overline{v_{k_m}}^{b}\underset{m\rightarrow\infty}{\rightharpoonup}\nabla\left(\overline{u}^{a}\overline{v}^{b} \right)+\Cr{B4}{[c]}(u,v)\nabla \overline{v}^{b}\label{convweak2}\\
 &\text{in }L^2_{loc}(\R^+_0\times\Omega);\nonumber
\end{align}
due to \cref{convweak1},
\begin{align}
 \underset{m\rightarrow\infty}{\lim\inf}\ &\int_0^{\infty}\psi\int_{\Omega}4\frac{a-1}{a}\left|\nabla \left(\overline{u_{k_m}}^{\frac{a}{2}}\overline{v_{k_m}}^{\frac{b}{2}}\right)+\Cr{B1}{[c_{k_m}]}(u_{k_m},v_{k_m})\nabla \overline{v_{k_m}}^{\frac{b}{2}}\right|^2\varphi\,dxds\nonumber\\
 \geq &\int_0^{\infty}\psi\int_{\Omega}4\frac{a-1}{a}\left|\nabla \left(\overline{u}^{\frac{a}{2}}\overline{v}^{\frac{b}{2}}\right)+\Cr{B1}{[c]}(u,v)\nabla \overline{v}^{\frac{b}{2}}\right|^2\varphi\,dxds;\label{conv9w}
\end{align}
  due to \cref{convM*}, \cref{Bpos}-\cref{Bconv},  \cref{Ml}, \cref{convGvae}, and Fatou's lemma,  taking $M_0:=M$
\begin{align}
 \underset{m\rightarrow\infty}{\lim\inf}\ &\int_0^{\infty}\psi\int_{\Omega}\left(4\frac{b-1}{b}(M-M_{*}{ [c_{k_m}]})+\Cr{B2}{[c_{k_m}]}(u_{k_m},v_{k_m})\right)\left|\nabla \overline{v_{k_m}}^{\frac{b}{2}}\right|^2\varphi\,dxds\nonumber\\
 \geq &\int_0^{\infty}\psi\int_{\Omega}\left(4\frac{b-1}{b}(M-M_{*}{ [c]})+\Cr{B2}{[c]}(u,v)\right)\left|\nabla \overline{v}^{\frac{b}{2}}\right|^2\varphi\,dxds\label{conv11w}
\end{align}
  due to \cref{Bbound2}-\cref{Bpos}, \cref{convv}, \cref{convuae}, \cref{convG}, and Fatou's lemma,
\begin{align}
 \underset{m\rightarrow\infty}{\lim\sup}\ &\int_0^{\infty}\psi\int_{\Omega}{\Cr{Bnew}}(u_{k_m},v_{k_m})\varphi-{\Cr{B5}}(u_{k_m},v_{k_m}){ u_{k_m}}\varphi\,dxds\nonumber\\
 \leq& \int_0^{\infty}\psi\int_{\Omega}{\Cr{Bnew}}(u,v)\varphi-{\Cr{B5}}(u,v){ u}\varphi\,dxds;\label{conv10w}
\end{align}
due to \cref{aproxiniu_}-\cref{aproxiniv_} and the dominated convergence theorem,
\begin{align}
 F\left(u_{k_m0},v_{k_m0}\right)\underset{m\rightarrow\infty}{\rightarrow}F(u_0,v_0)\qquad\text{ in }L^1_{loc}(\R^+_0\times\Omega).\label{convini}
\end{align}
Combining \cref{convFL1} and  \cref{conv9w}-\cref{convini}, we can pass to the limit superior on both sides of \cref{superue} and thus finally obtain that  $(u,v)$ satisfies \cref{superu}.

Let now
\begin{align}
 \varphi\in C^1(\overline{\Omega}),\qquad \psi\in C^1_0(\R_0^+),\nonumber
\end{align}
i.e., not necessarily nonnegative. Then:\\
due to {\eqref{Bconv},}  \cref{estintnabla_}, \cref{convweak2}, and the Banach-Alaoglu theorem,
\begin{align}
 &\nabla\cdot\int_0^{\infty}\psi\left(\nabla\left(\overline{u_{k_m}}^{a}\overline{v_{k_m}}^{b} \right)+\Cr{B4}{[c_{k_m}]}(u_{k_m},v_{k_m})\nabla \overline{v_{k_m}}^{b}\right)\,dt\nonumber\\
 \underset{m\rightarrow\infty}{\overset{*}{\rightharpoonup}}&\nabla\cdot\int_0^{\infty}\psi\left(\nabla\left(\overline{u}^{a}\overline{v}^{b} \right)+\Cr{B4}{[c]}(u,v)\nabla \overline{v}^{b}\right)\,dt\qquad\text{in }{\cal M}(\overline{\Omega}),\label{convbcpr}
\end{align}
and
\begin{align*}
\int_0^{\infty}\psi\left(\nabla\left(\overline{u}^{a}\overline{v}^{b} \right)+\Cr{B4}{[c]}(u,v)\nabla \overline{v}^{b}\right)\,dt\in {\cal DM}^2(\Omega);
\end{align*}
due to \cref{contitr}, \cref{uvebc}, \cref{convweak2}, and \cref{convbcpr}, $(u,v)$ satisfies  \cref{prweakbc};\\
due to \cref{estintDelta_}, \cref{convnv}, and the Banach-Alaoglu theorem,
\begin{align}
 \nabla\cdot\int_0^{\infty}\psi \nabla v_{k_m}\,dt\underset{m\rightarrow\infty}{\overset{*}{\rightharpoonup}}\nabla\cdot\int_0^{\infty}\psi\nabla v\,dt\qquad\text{in }{\cal M}(\overline{\Omega}),\label{convDelta}
\end{align}
and
\begin{align*}
\int_0^{\infty}\psi \nabla v\,dt\in {\cal DM}^{s}(\Omega);                                                              \end{align*}
due to \cref{contitr}, \cref{uvebc}, \cref{convnv}, and \cref{convDelta},  $v$ satisfies  \cref{vweakbc}.
\subsection{Convergence to \texorpdfstring{$\mu$}{}}
Finally, we establish \cref{superv}-\cref{massu0} for some element $\mu\in L^{\infty}_{w-*}(\R^+;{\cal M}_+(\overline{\Omega}))$. It holds that:\\
due to \cref{aproxiniu_} and  \cref{massue},
\begin{align}
 \int_{\Omega}u_{k_m}\,dx
 \phantom{\ }\equiv\phantom{\    }&\int_{\Omega}u_{k_m0}\,dx\nonumber\\
  \underset{m\rightarrow\infty}{\rightarrow}&\int_{\Omega}u_0\,dx.\label{conv9}
\end{align}
Now let
\begin{align*}
 { 0\leq}\varphi\in C(\overline{\Omega}).
\end{align*}
Then:\\
 due to \cref{aproxiniu_}, \cref{massue},
and the Banach-Alaoglu theorem (compare also {\cref{SecFSp}}), there exists some $\mu\in L^{\infty}_{w-*}(\R^+;{\cal M}(\overline{\Omega}))$, such that
\begin{align}
&\int_{\Omega} u_{k_m}\varphi\,dx
 \underset{m\rightarrow\infty}{\overset{*}{\rightharpoonup}}\int_{\Omega} u\varphi\,dx+\int_{\overline{\Omega}}\varphi\ (d\mu(\cdot))(x)\qquad\text{in }L^{\infty}(\R^+);\label{conv06}
\end{align}
due to \cref{convuae} and Fatou's lemma,
\begin{align}
 \underset{m\rightarrow\infty}{\lim}\int_{\Omega} u_{k_m}\varphi\,dx
 \geq\int_{\Omega}u\varphi\,dx\qquad\text{a.e. in }\R^+;\label{conv7}
\end{align}
due to \cref{conv06}  and  \cref{conv7},
\begin{align}
 \int_{\overline{\Omega}}\varphi\ (d\mu(t))(x)\geq0\qquad\text{for a.a. }t>0,\nonumber
\end{align}
so that $\mu(t)$ is for a.a. $t>0$ a non-negative measure;\\
due to \cref{conv9} and \cref{conv06} (set $\varphi\equiv 1$),  $u$ and $\mu$ satisfy  \cref{massu0}.

Finally, passing to the limit in \cref{superve} for $k=k_m$ as $m\rightarrow\infty$, and { using \cref{convv}-\cref{convnv}, we find} that $(u,v)$ satisfies \cref{superv}. {\cref{TheoEx}} is proved.

\section{Classical solutions to \texorpdfstring{\cref{KS}}{}: proof of {\texorpdfstring{\cref{classol}}{}}}\label{SecCl}
In this final section we prove  {\cref{classol}}.
Thus, we assume now that { the smoothness assumptions \eqref{chi1}-\eqref{uvsmooth} hold} and verify that in this case $(u,v)$ is, in fact, a classical solution to \cref{KS}.

Let $T_0\in(0,T)$ be arbitrary but fixed. Since both $c$ and $v$ are smooth, they are uniformly bounded in $[0,{ T_0}]\times\overline{\Omega}$. From now on we assume that
\begin{align*}
 n>\|u\|_{{ C}([0,{ T_0}]\times\overline{\Omega})}, \qquad l>\|v\|_{{ C}([0,{ T_0}]\times\overline{\Omega})}.
\end{align*}
Consequently,
\begin{align}
 \overline{u}, \overline{v}>0\qquad\text{ in }[0,{ T_0}]\times\overline{\Omega}.\label{posbar}
\end{align}

As $v$ is smooth, the weak boundary condition \cref{vweakbc} implies that $\partial_{\nu}v(t,\cdot)$ vanishes on the boundary of $\partial\Omega$ for all  $t\in(0,T_0]$. Further, since both $u$ and $v$ are smooth,  the weak boundary condition \cref{prweakbc} takes the form
\begin{align}
\left(\left(\nabla\left( \overline{u}^{a}\overline{v}^{b}\right)-\overline{u}^{a}\nabla \overline{v}^{b}\right)+\frac{1}{b}\left(b\left(M(n,l) +\overline{u}^{a}\right)-ac(u,v)\overline{u}^{a-1}\overline{v}\right)\nabla \overline{v}^{b}\right)\cdot\nu|_{\partial\Omega}=0.\nonumber
\end{align}
This means that
\begin{align}
-a\overline{u}^{a-1}\overline{v}^{b}\left(\partial_{\nu} u-c(u,v)\partial_{\nu} v\right)-b\left(M(n,l) +\overline{u}^{a}\right)\overline{v}^{b-1}\partial_{\nu}v=0\qquad\text{ { in} }(0,{ T_0]}\times \partial\Omega.\label{prclasbc}
\end{align}
Dividing  \cref{prclasbc} by $-a\overline{u}^{a-1}\overline{v}^{b}$ and plugging the boundary condition for $v$, we conclude that $\partial_{\nu}u(t,\cdot)$ vanishes on $\partial\Omega$ for all  $t\in(0,T_0]$ as well.

Next, exploiting the smoothness of $u$ and $v$, we integrate by parts in  \cref{superu} w.r.t. $t$ and $x$ and then apply the Du Bois-Reymond lemma. This results in the following differential inequality  in $(0,T_0]\times\Omega$:
\begin{align}
 &-a\overline{u}^{a-1}\overline{v}^{b}\left(\partial_tu -\nabla\cdot\left(\nabla u-c(u,v)\nabla v\right)\right)\nonumber\\
 &-b\left(M(n,l) +\overline{u}^{a}\right)\overline{v}^{b-1}\left(\partial_t v-(\Delta v-v+u)\right)\leq0.\label{Eq1ac}
\end{align}
Similarly, \cref{superv} implies that $\mu=\xi\,dx$ for some  density function $0\leq\xi\in  C((0, T_0]\times\overline{\Omega})$ and
\begin{align}
 \partial_t v=\Delta v-v+u+\xi\qquad\text{in }{(0,T_0]}\times\Omega.\label{vxi}
\end{align}
Dividing \cref{Eq1ac} by $-a\overline{u}^{a-1}\overline{v}^{b}$ and making use of \cref{vxi} we deduce that
\begin{align}
 &\partial_tu -\nabla\cdot\left(\nabla u-c(u,v)\nabla v\right)\geq-\frac{b}{a}\frac{\left(M(n,l) +\overline{u}^{a}\right)}{\overline{u}^{a-1}\overline{v}}\xi\qquad\text{in }{(0,T_0]}\times\Omega.\label{Eq1ineq}
 \end{align}
 Integrating \cref{Eq1ineq} by parts over $\Omega$  using the boundary conditions and, subsequently, integrating over $(0,t)$ for any  $t\in(0,T_0]$ we then have that
\begin{align}
 \|u(t,\cdot)\|_{L^1(\Omega)}\geq& \|u_0\|_{L^1(\Omega)}-\frac{b}{a}\int_0^t\int_{\Omega}\frac{\left(M(n,l) +\overline{u}^{a}\right)}{\overline{u}^{a-1}\overline{v}}\xi\,dxds.\label{mu1}
\end{align}
On the other hand,  \cref{massu0} implies that
\begin{align}
 \|u(t,\cdot)\|_{L^1(\Omega)}+\|\xi(t,\cdot)\|_{L^1(\Omega)}=\|u_0\|_{L^1(\Omega)}\qquad\text{ for all }{ t\in[0,T_0]}.\label{mu2}
\end{align}
Thus, combining \cref{posbar} and  \cref{mu1}-\cref{mu2}  we conclude that
\begin{align}
 \|\xi(t,\cdot)\|_{L^1(\Omega)}\leq&\frac{b}{a}\int_0^t\int_{\Omega}\frac{\left(M(n,l) +\overline{u}^{a}\right)}{\overline{u}^{a-1}\overline{v}}\xi\,dxds\nonumber\\
 \leq& \C({ T_0})\int_0^t\|\xi(s,\cdot)\|_{L^1(\Omega)}ds\qquad\text{ for all }{ t\in[0,T_0]}\nonumber
\end{align}
which yields that
\begin{align}
\xi\equiv0\label{xi0}
\end{align}
due to the Gronwall lemma. Plugging \cref{xi0} into \cref{vxi},  \cref{Eq1ineq}, and \cref{mu2} immediately yields that \cref{Eqv} is satisfied in the classical sense, and it holds that
\begin{align}
 &\partial_tu -\nabla\cdot\left(\nabla u-c(u,v)\nabla v\right)\geq0\qquad\text{in }{(0,T_0]}\times\Omega\label{Eq1ineq0}
 \end{align}
 and
 \begin{align}
  \|u(t,\cdot)\|_{L^1(\Omega)}=& \|u_0\|_{L^1(\Omega)}\qquad\text{for all }{ t\in[0,T_0]}.\label{mas}
 \end{align}
Since \cref{Eq1ineq0} is subject to the no-flux boundary conditions,   \cref{Eq1ineq0}  and \cref{mas} imply that equality holds in \cref{Eq1ineq0}. Thus, equation \cref{Equ} is satisfied in the classical sense, and {\cref{classol}} is proved.

\phantomsection
\printbibliography
\end{document}

config: arxiv


TITLE: Are there examples that suggest the Riemann Hypothesis might be false?
QUESTION [29 upvotes]: Are there examples that might suggest the Riemann hypothesis is false?
I mean, is there a zeta function $ \zeta (s,X) $ for some mathematical object $X$  with the properties

$ \zeta (1-s,X) $ and $ \zeta (s,X)$ are related by a functional equation.
$\zeta (s,X) $ can be expanded into an Euler product $ \zeta (s,X)= \prod _{i}(1-N(i)^{-s})^{-1}$.
the zeta function $\zeta (s,X) $ has zeroes of the form $ a+bi$ with $b\ne0$ for $a$ different from $\frac 12$.

That is, a zeta function with similar properties to the Riemann zeta but with zeroes off the critical line.

REPLY [11 votes]: There is the de-Bruijn Newman Constant $\Lambda $ . RH is equivalent that $\Lambda  \leq 0$. Up to now it is known  $-2.7 \cdot 10^{-9} < \Lambda  \leq 1/2$. See http://www.dtc.umn.edu/~odlyzko/doc/debruijn.newman.pdf


TITLE: When we use Lorenz gauge or Coulomb gauge, the result formula for electric $E$ and magnetic field $B$ is same or different?
QUESTION [1 upvotes]: Gauge condition can be chosen as you like or not?
is the Lorenz gauge is the only one correct? If  Coulomb gauge can obtained exactly same results as Lorenz gauge for the electromagnetic fields E and B, we can choose what we like, either Coulomb gauge or Lorenz gauge. If use different gauge we get different formula of electromagnetic fields $E$ and $B$, then we have to choose a corrected gauge to get the correct electromagnetic field formula. I am not clear that when we use Lorenz gauge or Coulomb gauge, the result formula for electric $E$ and magnetic field $B$  same or different?

REPLY [2 votes]: Gauge condition is any human imposed restriction on the functions $\varphi(\mathbf x,t), \mathbf A(\mathbf x,t)$ that does not change electric and magnetic field implied by those potential functions.
For any single physical situation, one can use either potential functions obeying the Coulomb gauge condition or those obeying the Lorenz gauge condition. Both choices give the same electric and magnetic field.


TITLE: order of integration cylindrical volume
QUESTION [0 upvotes]: I got the following integral $$\int_{h}^{R}\int_{0}^{2\pi}\int_{0}^{\sqrt{R^2-z^2}}A(2-\sin^2\theta)r dr d\theta dz$$
Can I solve it for: $$\int_{h}^{R}\int_{0}^{\sqrt{R^2-z^2}}r dr  dz\int_{0}^{2\pi}A(2-\sin^2\theta) d\theta $$?

REPLY [0 votes]: Yeah, you can solve it, except I would think the $$\int_0^{2\pi}A(2-sin^2\theta) d\theta$$ on the outside, before the other two integrals.


TITLE: Flat versus etale cohomology
QUESTION [22 upvotes]: Although the definition of etale ($\ell$-adic) cohomology is scary, I have at least some intuition for how it should behave: for instance, when it makes sense, I expect that it should be ``similar'' to the classical (singular) cohomology.
I'm in a situation in which I'm wondering if I can generalize a result in the realm of etale cohomology by using flat cohomology instead. (For instance, I want to be able to consider varieties over characteristic 2 and take cohomology with coefficients in the sheaf $\mu_2$, as an analogue of the case where one of the 2's is replaced by $p \neq 2$.)
Unfortunately, I have no intuition whatsoever for how flat cohomology should behave, so I am looking for mental models, slogans, etc. An answer of "don't use flat cohomology because it is impossible to work with" would also be helpful (although a little disappointing). For the sake of concreteness, here are some specific questions (but my interest is not limited to these)

Can I expect analogues of familiar nice things such as Poincare duality and cycle class maps?
Should this behave like (or even coincide with) etale cohomology in ``nice'' cases?
What are non-trivial examples where flat cohomology can be effectively computed? Especially examples where the computation is not by formally showing that it must be the same as etale cohomology, and then computing the latter.
Should I expect to have a good theory of characteristic classes?

REPLY [12 votes]: Here is a good example of some pathological behaviour that will shatter all your hopes and dreams.

Claim. Let $k$ be a perfect field, and let $X$ be a smooth, proper, integral $k$-scheme. Then the cohomology $H^1(X,\alpha_p)$ is a $k^p$-vector space. In particular, it is only finite if it is $0$ or $k$ is finite.

For a proof, see this MO post.
The reason this is disappointing is that $\alpha_p$ should be a constructible sheaf on the flat site, and from étale cohomology we expect the cohomology of a constructible sheaf to be finite. But it isn't!
On the other hand, for $\mu_p$-coefficients, we do have a fairly nice theory, thanks to Milne's arithmetic duality theorems. The first paper in this direction was his Duality in the flat cohomology of a surface, and this has been extended and rewritten many times. Some references were already given in the comments; I am not aware of any others (but they might well exist).
There are many other interesting sheaves to consider other than just sheaves represented by finite group schemes over the ground field. But I don't think there is a nice and unified theory in the same that there is for étale cohomology.
Remark. A nice conceptual remark is the fact that flat cohomology can be computed by the quasi-finite flat site. This is discussed in Milne's étale cohomology book, Example III.3.4. Thus, even though the flat site seems infinitely larger than the étale site (flat morphisms can have arbitrary relative dimension), it somehow isn't as big as you would think.


TITLE: How safe is to subtract or add equations in a system?
QUESTION [0 upvotes]: Let's take as example the following system:
$
\begin{cases}
2x + 1 = 0 \\
-x + 1 = 0
\end{cases}
$
I could take x from the first and check on the second which will give me no solution. Taking from one and replacing in other guarantees safety
But if I add them together I will get x + 2 = 0 => x = -2. Checking -2 will fail in both of them. More solutions than needed are shown, and for advanced problems is not easy if possible at all to check the solutions
a = 2, b = 3 => a + b = 5, but a + b = 5 does not mean a = 2, b = 3. Even if there are the same number of equations as the number of variables, since they are on the same domain, I guess some dirty tricks can be applied so the system does not result only true solutions
What rules shall be applied to the system in order to guarantee valid solutions?
Example (please mention the mistake):
We have 2 quadratric equations f(x) and g(x). By using f(x) = g(x) we get all x coordinates of the intersection points. Let's say we want to get all point with y = 2:
$
\begin{cases}
f(x) = 2 \\
g(x) = 2
\end{cases}
$
The system results f(x) = g(x) or f(x) - g(x) = 0, and we said earlier that this means we get all intersection points, contrary to our rule of y = 2. In a real example we may have $f(x) = x^2 + mx + 1$ (m parameter) and some conditions that our f shall accomplish (so we can determine m). Since the system will not result only true solutions, then the method is quite useless and other ways shall be taken

REPLY [1 votes]: When you have a system of equations, a linear combination of them produces another equation whose solutions include those of the original system. That’s an important nuance: nothing says that this single equation has exactly the same solution set as the original system, but the original system’s solutions are a subset of those of the new equation. That should make perfect sense if you think about it a bit. For example, if you have a pair of intersecting lines in the plane, every (nontrivial) linear combination of their equations describes some line that passes through the intersection point, but obviously that combined equation has other solutions as well—a whole line’s worth.
That’s what happened here. The solution set to the original system is a subset of the solutions to the combined equation $x+2=0$, just as it’s supposed to be: the empty set is a subset of any set, after all. The moral is that if you solve a derived equation in the process of solving a system of equations, you need to check those solutions for validity against the original system.


TITLE: Conditional probability density function for random variables $Y = X$
QUESTION [1 upvotes]: Lets say I have a uniform random variable $X$ in $[0, 1]$, and another random variable $Y = X$. Clearly, $Y$ is dependent on $X$.
The marginal pdf's for the random variables are
$$f_X(t) = f_Y(t) = \begin{cases}
1 && \text{if $t \in [0, 1]$} \\
0 && \text{otherwise}
\end{cases}$$
I am trying to figure out what the conditional pdf for $x$ conditioned on $y$ would be. The cdf seems pretty obvious.
$$F_{X|Y}(x|y) = P(X < x | Y = y) = P(y < x) = \begin{cases}
0 && \text{if $x \le y$} \\
1 && \text{if $x > y$}
\end{cases}$$
This is where I am confused. Because the pdf should be the derivative of this function, but we have a jump discontinuity. Thus we should have
$$f_{X|Y}(x|y) = 0 \quad\text{if } x \neq y$$
But this seems to break the property $$\int_{-\infty}^{\infty} f_{X|Y}(x|y) dx = 1$$
Can someone help clarify what I am misunderstanding about conditional pdf and cdfs?
Thanks.

REPLY [0 votes]: If you don't insist on being rigorous, you could say that the conditional pdf is the Dirac delta function,
$$ f_{X|Y}(x|y) = \delta( x - y).$$
Roughly speaking, this is an infinite "spike" at $x = y$: the function is zero when $x \neq y$, and takes the value infinity at $x = y$, and its integral over all $x$ is defined to be one.
This fits quite nicely with your discussion about differentiating the cumulative density function. The derivative of the cdf is zero when $x \neq y$, but it is infinite at $x = y$, and the integral of the derivative of the cdf should equal the jump in the cdf at $x = y$, which is indeed one. Again, this statement is not rigorous, because a function with a jump continuity is not really differentiable, and the delta function isn't really a function. (See here and here for more discussion.)


TITLE: Singularity structure of integrals of rational functions
QUESTION [8 upvotes]: Suppose I have a convergent integral of the form $\int_0^1dx_1\dots\int_0^1 dx_n \frac{P(x_i)}{Q(x_i)}$, where $P$ and $Q$ are polynomial functions of $n$ nonnegative real variables $x_i$.  Let the coefficients of the various monomials in $Q$ be $a_k$.  Depending on the structure of $Q$, the integral could develop singularities in the limit that some of the coefficients $a_k$ vanish.  For example, the integral $\int_0^1 dx\frac 1 {a+x}$ goes like $-\log a$ as $a\to 0$.  Is there a systematic way to extract the structure of these singularities for general multidimensional integrals?
The particular example I'm interested in is the following: I'd like to determine the singularities of
$\begin{eqnarray}
\int_0^1 d\lambda_1\int_0^1 d\lambda_2 \int_0^1d\alpha_1 \int_0^1d\alpha_2 \frac{\alpha_1\alpha_2}{(\lambda_2\alpha_1\alpha_2+\lambda_1\alpha_1\alpha_2+a\alpha_2\lambda_1^2+b\alpha_1\lambda_2^2+c\alpha_1\alpha_2)^2},
\end{eqnarray}$
as $a,b,$ and $c$ tend to zero.  If necessary, it's alright to assume they all tend to zero at the same rate.  I naively expect a leading singularity that looks like $\log a \log b \log c$, followed by sub-leading singularities that look like $\log a \log b$ times some convergent integral, $\log b \log c$ times some other convergent integral, etc..  For my purposes, these $\log$-squared terms suffice, but I'd be interested to know how to systematically go further.
Edit: The above integral is an example of a particular Feynman integral, written in terms of Feynman parameters.

REPLY [4 votes]: OK. It is going to be long and extremely boring, but, as promised, here goes. We'll get the precision $O(mL^2)$ where $m=\max(a,b,c), L=\max(\log\frac 1a,\log\frac 1b,\log\frac 1c)$, which should be enough for any reasonable interpretation of "$a,b,c$ going to $0$ at about the same rate".
We start with the approximation of an auxiliary integral:
$$
I(a,b)=\iint_{[a,+\infty)\times[b,+\infty)}\frac{1}{(1+x+y)^2}\frac{dx}{x}\frac{dy}{y}
$$
as $a,b\to 0+$.
First, we have
$$
\frac{1}{(1+x+y)^2}-\frac 1{(1+x)^2}\frac 1{(1+y)^2}=\frac{xy(2+2x+2y+xy)}{(1+x+y)^2(1+x)^2(1+y)^2}
\\
\le
\frac{2xy}{(1+x)^{2}(1+y)^{2}}\,,
$$
which means that $I(a,b)=J(a)J(b)+D_1+O(a+b)$ where
$$
J(a)=\int_{a}^\infty\frac{1}{(1+x)^2}\frac{dx}x\,,
$$
and
$$
D_1=\iint_{(0,\infty)^2}\frac{(2+2x+2y+xy)dx\,dy}{(1+x+y)^2(1+x)^2(1+y)^2}
$$
(below we will always denote by $D_k$ various numerical constants whose values are expressed by some convergent integrals. Some of them are easy to evaluate and some aren't, but that distinction won't bother us here).
Now
$$
J(a)=\int_a^\infty\left[\frac 1{(1+x)^2}-\chi_{[0,1]}(x)\right]\frac{dx}{x}+\log\frac 1a\,.
$$
Since for $x<1$, we have $\frac 1{1+x}-1=-\frac x{1+x}$, we conclude that
$$
J(a)=\log\frac 1a+D_2+O(a)
$$
with
$$
D_2=\int_0^\infty \left[\frac 1{(1+x)^2}-\chi_{[0,1]}(x)\right]\frac{dx}{x}
$$
Multiplying out, we get
$$
I(a,b)=\log\frac 1a\log\frac 1b+D_2(\log\frac 1a+\log\frac 1b)+D_3
\\
+O\left((a+b)(\log\frac 1b+\log\frac 1a)\right)\,.
$$
Now we can return to our quadruple integral. Let us rewrite the integrand in the form
$$
\frac{1}{(\lambda_1+\lambda_2+c)^2}\frac{1}{\left(1+\frac{a\lambda_1^2}{\lambda_1+\lambda_2+c}\frac 1{\alpha_1}+
\frac{b\lambda_2^2}{\lambda_1+\lambda_2+c}\frac 1{\alpha_2}\right)^2}
\,d\lambda_1\,d\lambda_2\,\frac{d\alpha_1}{\alpha_1}\,\frac{d\alpha_2}{\alpha_2}
$$
For fixed $\lambda_1,\lambda_2$, we can now happily integrate $\alpha_1,\alpha_2$ out using the computation above to reduce our integral to
$$
\iint_{[0,1]^2}
\frac 1{(\lambda_1+\lambda_2+c)^2}\left[\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}
\\+
D_2\left(\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}+\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}\right)+D_3\right]\,d\lambda_1\,d\lambda_2+E_1
$$
where the error term $E_1$ is controlled by
$$
\iint_{[0,1]^2}\frac{1}{(\lambda_1+\lambda_2+c)^2}\left[
\left(\frac{a\lambda_1^2}{\lambda_1+\lambda_2+c}+
\frac{b\lambda_2^2}{\lambda_1+\lambda_2+c}\right)
\\
\left(\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}+\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}\right)
\right]\,d\lambda_1\,d\lambda_2
$$
Now we do not need a special investigation of $E_1$ because the first factor in the brackets is  bounded by $a+b$ and if we look at the rest, it is just one of the terms in the main asymptotic integral. Next, once we open the parentheses and expand logarithms of products, we see that we just need to approximate three double integrals:
$$
I_2=\iint_{[0,1]^2}
\frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_1^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_2^2}\,,
$$
$$
I_1=\iint_{[0,1]^2}
\frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_1^2}\,,
$$
and
$$
I_0=\iint_{[0,1]^2}
\frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}
$$
IT is not hard to convince yourself that in all $3$ cases the integral over the domain $\lambda_1+\lambda_2>1$ tends to its value at $c=0$ and the deviation from that value is $O(c)$, so we can replace those parts by appropriate constants. In the other part it is convenient to make the change of variable $\lambda_1=v\lambda,\lambda_2=(1-v)\lambda$ ($0<v,\lambda<1$) and to integrate $v$ out (again splitting logarithms of products into sums). We see that we need to approximate
$$
I_2'=\int_0^1\frac{\lambda\,d\lambda}{(\lambda+c)^2}\log^2\frac{\lambda+c}{\lambda^2}\,,
$$
$$
I_1'=\int_0^1\frac{\lambda\,d\lambda}{(\lambda+c)^2}\log\frac{\lambda+c}{\lambda^2}
$$
$$
I_0'=\int_0^1\frac{\lambda\,d\lambda}{(\lambda+c)^2}
$$
To be continued...


TITLE: A white square has a red circle inscribed in it, which has a white square inscribed in it, and so on. Find the red area
QUESTION [2 upvotes]: A white  regular polygon with area $1$ has a red circle inscribed(the circle touches the edges of the square), with a white regular polygon with the same amount of sides as the first one(the polygon's corners touch the circle) and this pattern continues forever, creating a pattern like this:
(https://i.stack.imgur.com/IARUV.jpg)
1. What is the area that is coloured red if the polygon is a square?

If you knew the amount of sides, what would be a formula to find the area coloured red, in terms of $s$ (for sides)

REPLY [0 votes]: Consider the larger red area. It is the area of the circle having radius $r=1$ , $C=\pi$ less the square having area $A=\sqrt{2}^2=2$
So its area is $R_1=\pi-2$
Now consider the similitude ratio between the inscribed and the circumscribed squares. The inner square has side $l_1=\sqrt{2}$ and the outer is $L_1=2$
The ratio is $r_1=\frac{\sqrt 2}{2}$ then the ratio for areas is $r_1^2=\frac{1}{2}$
As the red part become smaller, the ratio between them is constant and is $\frac{1}{2}$
Therefore the sum of the red areas is
$$R=R_1+\frac{1}{2}R_1+\frac{1}{2}\frac{1}{2}R_1+\ldots=R_1\sum_{n=0}^{\infty}\frac{1}{2^n}=2R_1$$
so the red area is $R=2(\pi-2)$
Hope this helps


TITLE: Math learning after Differential Equations?
QUESTION [2 upvotes]: I made it up to DE(Pre-Calc, Calc I and Calc II with basic stats before that same school) in community college through a special program(I'm poor). I'll have to self-teach from now on but am still very in to math and engineering. What do people usually go to with math after DE?
I'm interested in topology but it looks like there is a lot of other things I need to do first like advanced algebra.

REPLY [2 votes]: If you haven't had the genuine treat of a basic course in complex variables, do that first. Topology is high in my list, but you really must get complex variables under your belt first. It will completely change your take on lots of stuff you have already done.


TITLE: Topologizing a free product G*H of discrete groups?
QUESTION [1 upvotes]: My -possibly flawed- mental picture of free products of groups certainly comes from the special case usually performed to illustrate the construction that proves the Banach-Tarski paradox. Thus I'm used to think of a free product as a certain kind of self-similar "fractal".
Although from the set theoretical or algebraic viewpoint it may be clear, one may ask if this "fractal" property is also expressible in metric/topological terms, and if it is possessed by free products of, say, discrete groups in general (not only $\mathbb{Z}*\mathbb{Z}$).

I would like to know if there is a more or less natural way to put a topology (or even a metric) on the free product $G*H$ of two discrete groups $G$ and $H$ so that the above "fractal" picture is retained.

I suppose the topology should satisfy:

compatibility with group structure: $G*H$ should be a topological group;
the subgroup of $G*H$ made of words with letters from $G$ (resp., from $H$), that we still call $G$ (resp., $H$), should inherit the discrete topology

In case we even look for a metric on $G*H$, I think it would be reasonable to require that

left (and right) translations should be homotheties: $d(gx,gy)=C\cdot d(x,y)$ for a constant $C=C(g)$ depending only on $g$,

and maybe:

$G*H$ should have fractionary Hausdorff dimension.

Perhaps a metric would be something like the metric one can put on the "space of words" used in the symbolic dynamics description (if I don't misremember - I'm no expert) of the horseshoe map...

Another random thought was originated by an answer to this MO question. It was noted that the modular group $PSL(2,\mathbb{Z})$ is isomorphic to the free product of cyclic groups $\mathbb{Z}/2*\mathbb{Z}/3$, and that it is not co-Hopf, i.e. it does have a proper subgroup isomorphic to itself (so, in a certain vein not dissimilar to the above one, it is "self similar").

May this fact have any interesting consequences about some properties of modular forms?

REPLY [3 votes]: I wouldn't agree with your interpretation of free groups, I think that your understanding comes from the free group on 2 letters with a particular action on a fractal.  That's not to say that there isn't much of interest to say about free products and actions on particular spaces.
But if you're just interested in free groups and how they are related to free products then I'd recommend reading up on Bass-Serre theory:
http://en.wikipedia.org/wiki/Bass%E2%80%93Serre_theory
This studies how groups act on simplicial trees.
Current research is focused on the automorphisms of free groups, if you want to learn about this you should search for articles on "Outer space", possibly by Karen Vogtmann.
[EDIT: As pointed out the last paragraph points to just one topic of active research, which by no means describes fully the current research environment. ]


TITLE: Using Abel's summation formula, prove f integrable if and only if the sum in n of the measures of the sets with $|f(x)| \geq n$ converges
QUESTION [1 upvotes]: The problem staetement is:
Let $(X, \mathcal{M}, \mu)$ be a finite measure space, and let $f:X \longrightarrow [-\infty,+\infty]$ be a measurable function. Prove that f is integrable if and only if $\sum_{n=1}^{\infty} \mu(\lbrace x \in X : |f(x)| \geq  n \rbrace) < \infty$.
Abel's formula: $a_1,...,a_n,b_1,...,b_{n+1} \in \mathbb{R}$, let's call $A_k=\sum_{j=1}^{k}a_j$ then $\sum_{k=1}^{n}a_kb_k=A_nb_{n+1}-\sum_{k=1}^{n}A_k(b_{k+1}-b_k)$.
I will call $f$ like its absolute value, since their integrability is equivalent, just to make the notation  a bit more readable. And also, now f is a nonnegative function, which is good.
I will define some sets at the beginning:
$A_k= \lbrace x \in X : f(x) \geq  n \rbrace$
$B_k= \lbrace x \in X :f(x) \in [k,k+1)\rbrace$. Remark: those are disjoint sets
It's obvious that $X=\cup_{k=0}^{\infty} B_k$, and $A_n=\cup_{k=n}^{\infty}$, so $\mu(X)= \sum_{k=1}^{\infty} \mu(B_k)$ and $\mu(A_n) = \sum_{k=n}^{\infty} \mu(B_k)$
Assume $\sum \mu(A_n) < \infty$, and let's see $f$ is integrable: (I think I have this part okay but please feel free to correct me)
Since $f$ is measurable and bounded in each $B_k$, is integrable in each $B_k$. So:
$\int_X f \,d\mu = \sum_{k=0}^{\infty} \int_{B_k}fd\mu \leq \lim_{n\to\infty} \sum_{k=0}^{n} (k+1)\mu(B_k)=\mu(X)+\mu(A_1)+\mu(A_2)+... < \infty$ beacuse $\mu$ is finite. This give us $f$ intergable, without using Abel's formula.
Now, my problem. How can I use Abel's formula to show that sum is convergent?, assuming $f$ integrable? I guess it's something like caling $a_k=1$ and $b_k=\mu(A_k)$ but I don't see how.
Thanks in advance.

REPLY [1 votes]: Ok, I see now a way to use Abel's summation formula. Set
$$A_n:=n,\quad  b_n:=-\mu(\{x: |f(x)|\geqslant n-1\})\\
\therefore\quad a_n=1,\quad b_{n+1}-b_n=\mu(\{x: n-1\leqslant |f(x)|< n\})$$
then from Abel's formula we have that
$$
\sum_{k=1}^n \mu(\{x: |f(x)|\geqslant k-1\})= \sum_{k=1}^n k\,\mu(\{x: k-1\leqslant |f(x)|<k\})+n\,\mu(\{x: |f(x)|\geqslant n\})
$$
Taking limits above we get
$$
\sum_{k\geqslant 1}k\,\mu(\{x:k-1\leqslant |f(x)|<k\})+\lim_{n\to\infty}n\,\mu(\{x: |f(x)|\geqslant n\})<\infty
$$
As both expressions on the LHS are non-negative we find that
$$
\sum_{k\geqslant 1}k\, \mu(\{x:k-1\leqslant |f(x)|<k\})<\infty
$$
Finally note that
$$
k\, \mu(\{x:k-1\leqslant |f(x)|<k\})\geqslant \int_{|f|^{-1}([k-1,k))}|f|\,d\mu\quad\text{ for each }k\in\{1,2,3,\ldots\}
\\
\therefore\quad \sum_{k\geqslant 1}k\,\mu(\{x:k-1\leqslant |f(x)|<k\})\geqslant \int_{X}|f|\mathop{}\!d \mu
$$
∎


TITLE: Prove $(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$
QUESTION [4 upvotes]: $x,y,z >0$, prove
$$(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$$
The term $\frac{4(x-y)^2}{xy+yz+zx}$ made this inequality tougher. It remains me of this inequality. I think one can prove it by expanding everything, but is there any elegant solution for this problem ?

REPLY [1 votes]: There is a very nice solution only using SOS, I was found it!
We need to prove:$$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) - 9 \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(\frac{x}{y}+\frac{y}{x}-2) + (\frac{y}{z}+\frac{z}{x}-2)+(\frac{z}{x}+\frac{x}{z}-2) \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$\frac{(x-y)^2}{xy}+\frac{(x-y)^2}{yz+zx}+(\frac{(y-z)^2}{yz}+\frac{(z-x)^2}{zx} -\frac{(x-y)^2}{yz+zx})\geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(x-y)^2(\frac{1}{xy}+\frac{1}{yz+zx})+\frac{(xz+yz-2xy)^2}{xyz(x+y)} \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(x-y)^2(\frac{1}{xy}+\frac{1}{yz+zx}-\frac{4}{xy+yz+zx})+\frac{(xz+yz-2xy)^2}{xyz(x+y)} \geq 0$$
or $$\frac{(x-y)^2(xz+yz-xy)^2}{xyz(x+y)(xy+yz+zx)}+\frac{(xz+yz-2xy)^2}{xyz(x+y)} \geq 0$$
Which is obvious!
Equality holds when $x=y=z$


TITLE: Importance of MHV amplitudes
QUESTION [8 upvotes]: Why are MHV amplitudes so important?  How/where are they used and why do people keep trying to rederive them in many different ways?

REPLY [4 votes]: MHV amplitudes are not really any more important than next-to maximal helicity violating amplitudes ($NMHV$) or $N^kMHV$ amplitudes. You need all of them to compute a general scattering amplitude.
Basically, scattering amplitudes for non-Abelian Yang Mills theories are very complicated to compute for more than 4 particles, so people work on formulating easier ways to do the computations. The first step is usually to strip off the color dependence of the amplitude, so that the object you really need to compute is the color ordered amplitude $\mathcal{A}(\pm,\pm,...\pm)$. These objects get multiplied by traces of color matrices at the end of the calculation.
At tree level, if all particles have the same helicity or only 1 helicity is different, the color ordered amplitude vanishes. So the MHV amplitude is the first non-vanishing color ordered amplitude. For $2 \to 2$ scattering you only need the MHV, since you can have at most 2 $+$ and 2 $-$ helicities. But if you were to compute say $3 \to 3$ scattering, you would also need something like $\mathcal{A}(+++---)$ which is not MHV.
MHV probably gets a lot of attention because the formulas are extremely simple (see: Parke-Taylor formula). Moreover, the discovery of the BCFW recursion relations allow you to generate higher order amplitudes from the lower order ones. In general, people are deriving new methods to calculate the amplitudes, and they can check their methods by matching onto the MHV amplitudes which have a simple form. But to compute a general amplitude you need more than just MHV.
Hope this helps.


TITLE: Why is Heisenberg's Uncertainty Principle not applicable to macroscopic objects?
QUESTION [0 upvotes]: Why is it that the uncertainty principle is not applicable to macroscopic objects? Specifically, at which scale does the uncertainty principle does not work? And also why? Why doesn't it work for macroscopic objects?

REPLY [0 votes]: First of all, it depends on what is meant by a macroscopic object. E.g., Bose-einstein condensates or a superconducting state are the examples of quantum phenomena occuring on macroscopic scale, and they do obey the uncertainty principle, such as, e.g., phase-number uncertainty relation for the supercondcuting state.
Secondly, we know of small but macroscopic objects, such as small proteins or fullerene molecules behaving as quantum object, tunneling through barriers, etc. The ucnertainty principle works here as well.
What the OP probably had in mind is the macroscopic obecets that we encounter every day, outside of a physics lab. Here the point is that they cannot be considered as a single object - due to the interactions the state of such objects is thermodynamic, i.e., the atoms/electrons/nuclei in such objects are typically localized on the length scales much smaller than the extent of the object.
The scales characterizing the lack of coherence in a macroscopic objects are thus coherence length of the particles in it (for mobile electrons it is usually known as a mean free path, which even in highest purity semiconductors is about hundreds of nanometers).


TITLE: labelled graph characteristic polynomial
QUESTION [4 upvotes]: Given the adjacency matrix $\mathbf{A}$ for a simple connected graph, the characteristic polynomial is defined as:
$$ p(\lambda) = \det(\lambda \mathbf{I} - \mathbf{A})$$
Now if an edge between vertex $i$ and $j$ is "labelled" with another variable $x$, then we could consider a bivariate polynomial:
$$ f(\lambda,x) = \det(\lambda \mathbf{I} + x (\mathbf{e}_{ij}+\mathbf{e}_{ji}) - \mathbf{A})$$
where $\mathbf{e}_{ij}$ is the matrix with all entries zero except row $i$, column $j$, which is 1.
Or similarly, if the vertex $i$ was labelled with the variable $x$, we could consider the polynomial:
$$ g(\lambda,x) = \det(\lambda \mathbf{I} + x \mathbf{e}_{ii} - \mathbf{A})$$
This of course could be extended to more labels, but I'm mostly interested in graphs with a single label at the moment.

Have these been studied before, and is there a name for these kinds of polynomials?

I'm particularly interested in learning about simple connected graphs which are non-isomorphic, and have the same "labelled characteristic polynomial", but differ only by a single labelled edge like
$$\det(\lambda \mathbf{I} + x(\mathbf{e}_{ij}+\mathbf{e}_{ji}) - A) = \det(\lambda \mathbf{I} + x(\mathbf{e}_{mn}+\mathbf{e}_{nm}) - A)$$
where $mn$ and $ij$ are distinct edges. I'm assuming that is possible given the difficulty of graph isomorphism, but I'm not sure how to go about constructing such graphs.

REPLY [1 votes]: OK, if you wish to obtain an answer here then I confirm my above comments that your should talk with a specialist. My google search extended my knowledge of an answer not far from zero.
I didn’t find exactly the same generalization of a characteristic polynomial of a graph as you proposed but I guess there should be some of them. See, for instance a paper “A Generalization of the Characteristic Polynomial of a Graph” by Richard J. Lipton and Nisheeth K. Vishnoi.
Personally I see two natural directions in which a further generalization may be useful. The first is a problem of an isomorphism of vertex or edge colored graphs. The second is a problem of a simultaneous isomorphism of two graphs with common vertex set (If I remember it right, a corresponding problem for the similarity of pairs of matrices is so called “wild problem” and when specialists in matrices encounter an equivalent problem they said with respect: “O-o, so it is the wild problem” and stop. Here is even written (in Russian) that it is not expected to obtain a reasonable answer to such a problem).

how to go about constructing non-isomorphic graphs with matching characteristic polynomial with a single "edge label".

I suggest such graphs exist, because the quest to find two non-isomorphic graphs with matching characteristic polynomials (so called cospectral graphs) and even more additional restrictions, looks elaborated. See, for instance, the following papers:

Anthony DiGenio “A Construction of Cospectral Graphs”.
Haruo Hosoya, Kyoko Ohta, Masaki Satomi “Topological twin graphs II. Isospectral polyhedral graphs with nine and ten vertices”.
M. Randić , W. R. Müller , J. V. Knop , and N. Trinajstić  “The Characteristic Polynomial as a Structure Discriminator”.


TITLE: example proof using predicate calculus with equality
QUESTION [1 upvotes]: I am trying to understand example 4 of section 21 of First Order Mathematical Logic by Angelo Margaris:

0 + 0 = 0 + 0 (=)
0 + y = y + 0
y' + 0 = y + 0'
y + 0' = (y + 0)'
y + 0' = (0 + y)'
0 + y' = (0 + y)'
0 + y' = y' + 0 (=, 6, 5, 3)

The only things introduced so far in this section have been the equality axioms and the following two theorems:
T1. $\forall x \forall y (x = y \to y = x)$
T2. $\forall x \forall y \forall z (x = y \land y = z \to x = z)$
It does not seem to me that steps 2-6 can be arrived at from these alone. Are they all assumptions? If they are all assumptions then why are steps 1, 2 and 4 stated?
What does the prime notation denote? Why is step 7 different from step 2?

REPLY [1 votes]: Margaris is a great book, but does require some patience getting to know his particular nomenclature. In addition, as Example 4 of Section 21 illustrates, he is sometimes not as clear as he could have been.
The purpose of this example is solely to show how three of the results introduced in the preceding two pages are used in an exemplary proof. These results are:

Equality Axiom (E1): $\forall x(x=x)$;
Theorem 1: $\vdash\forall x\forall y(x=y\rightarrow y=x)$;
Theorem 2: $\vdash\forall x\forall y\forall z(x=y\wedge y=z\rightarrow x=z)$

(E1) is used in step (1) of the proof.  Theorems 1 and 2 are used in step (7). These applications are properly explained in the paragraph following the example on page 106. Why no reasons are given for steps (2)-(6) is because the results required have yet to be introduced in the book. Indeed, Example 4 is exactly those first 8 (of 17) steps given as the proof for Theorem 4 of Section 24 (pages 129-130). Section 24 defines the first-order theory of natural numbers.

The set of proper symbols of $N$ is $\{\ =\ ,\ '\ ,\ +\ ,\ \cdot \,\ 0\ \}$, where $'$ is the first 1-place operation symbol, $+$ and $\cdot$ are the first two 2-place operation symbols, and $0$ is the first constant symbol. $r'$ is called the successor of $r$.

When you get to Section 24 and Theorem 4 therein, you will find that the justifications for steps (2)-(6) are based on the specific axioms of the natural numbers, an initial theorem of the number system, as well as some substitution theorems that actually immediately follow Example 4 in the book.
I wish this answer could be more interesting to those who do not have the Margaris text, but unfortunately the answer is a consequence of the book's pedantics and not really of a mathematical nature. But for partial completeness, see below for Theorem 4 and first 8 (of 17) steps of its proof covering the base induction step (from which the example above is taken).

Theorem 4 (Section 24/Margaris/First Order Number Theory): $\vdash
\forall x\forall y(x+y=y+x)$
Proof: Induction on $x$. $P(x)$ is $x+y=y+x$. First we prove $0+y=y+0$ by induction on $y$.

$0+0=0+0\qquad$ Equality Axiom $\forall x(x=x)$

$0+y=y+0\qquad\quad$ induction hypothesis
$y'+0=y+0'\qquad\ \ \ $ Number Theorem $\forall x\forall y(x'+y=x+y')$.
$y+0'=(y+0)'\qquad$ Number Axiom $\forall x\forall y(x+y'=(x+y)')$.
$y+0'=(0+y)'\qquad$ substitution of (2) and (4).
$0+y'=(0+y)'\qquad$ Number Axiom $\forall x\forall y(x+y'=(x+y)')$.
$0+y'=y'+0\qquad\ \ \ $ Theorems 1/2 from above.

$8$. $\forall y(0+y=y+0\rightarrow0+y'=y'+0)\qquad$ deduction from (2) to (7) and generalization


TITLE: Finding homomorphism between rings
QUESTION [3 upvotes]: I am asked to show that $\Bbb{R}[x]/((x+1)x)$ has zero divisors but no nipolent elements. So I decided to start with finding homomorphism $\phi\colon\Bbb{R}[x]\to\Bbb{R}$ such that $\ker(\phi)=((x+1)x)$ Normaly I would take evaluation homomorphism, but I cannot evaluate a polynomial at $x=0$ and $x=1$ at the same time. So what is the trick here?
Reamark:
((x+1)x) is an ideal generated by this polynomial

REPLY [6 votes]: We will denote by $\overline{f}$ the image of $f\in\mathbb{R}[x]$ under the canonical projection onto the quotient $A=\mathbb{R}[x]/\langle (x+1)x\rangle$.
Indeed, $A$ has nontrivial zero divisors e.g. $\overline{x+1}$ and $\overline{x}$.
Furthermore, $A$ has no nontrivial nilpotent elements. Indeed, let $\overline{f}$ be a nontrivial nilpotent element. Then $f^n$ is divisible by $(x+1)x$ for some positive integer $n$. Since $x+1$ and $x$ are both prime in $\mathbb{R}[x]$, and they have greatest common divisor $1$, $f$ must be divisible by $(x+1)x$.
Another way to see this is via the following isomorphisms.
$$A\cong \mathbb{R}[x]/\langle x+ 1\rangle\oplus\mathbb{R}[x]/\langle x\rangle\cong \mathbb{R}\oplus\mathbb{R}.$$
The first isomorphism is given by the Chinese remainder theorem, and the second isomorphism is induced by evaluation homomorphisms $\mathbb{R}[x]\to\mathbb{R}$ (at $-1$ for the first component and $0$ for the second).


TITLE: Is there a closed form expression for the following sum?
QUESTION [3 upvotes]: Is there a closed form expression for the following sum? $$\sum_{0\le i_1<i_2<\cdots<i_k\le n}r^{i_1+i_2+\cdots+i_k}$$
I can understand that there are $\binom{n}{k}$ such terms and the values that $i_1+\cdots+i_k$ can take vary from $\frac{k(k-1)}{2}$ to $k(n-k)+\frac{k(k+1)}{2}$. So it remains to find how often the term $\sum_{j=1}^k i_j=l$ is found as an exponent in the above sum. Can anyone give some idea? Thanks.

REPLY [1 votes]: In here we calculate the limit of $S_k(n)$ when $r$ goes to unity. It is easy to see that each of the terms in the sum over $j$ behaves like $O(\frac{1}{(1-r)^k})$ in that limit. Therefore in order to get the limit we need to extract the coefficient at $\frac{1}{(1-r)^k}$. In other words we need to compute:
\begin{eqnarray}
\left.\frac{(-1)^k}{k!} \frac{d^k}{d r^k} r^k \left( \left. S_k(n) \right|_{r \rightarrow 1-r} \right) \right|_{r\rightarrow 0} =
\frac{1}{k!} \sum\limits_{j=0}^k \sum\limits_{q_2=0}^k \binom{k}{q_2}
(-1)^{j+k+q_2} \left(\binom{k-j}{2} + j \cdot(n+1)\right)_{(k-q_2)} \cdot
{\mathcal A}^{(j,k-j)}_{q_2}
\end{eqnarray}
where
\begin{equation}
{\mathcal A}^{(j,k-j)}_{q_2} := \frac{d^{q_2}}{d r^{q_2}} \left.\left(\frac{r^k}{(1-r;1-r)_j(1-r;1-r)_{k-j}} \right)\right|_{r\rightarrow 0}
\end{equation}
Now the situation seems hopeless. On the face of it it will not be possible to find a closed form expression for the coefficients above and therefore it might seem we will never evaluate the sum. However, let us brace ourselves and make a step further. We use the Chu-Vandermonde identity to expand the right hand side. We have:
\begin{equation}
rhs = \frac{1}{k!}\sum\limits_{j=0}^k \sum\limits_{q_3=0}^k (j (n+1))_{(q_3)} \cdot \sum\limits_{q_2=0}^{k-q_3} (-1)^{k+q_2+j} \frac{k!}{q_2!q_3!(k-q_2-q_3)!} \cdot \binom{k-j}{2}_{(k-q_2-q_3)} \cdot {\mathcal A}^{(j,k-j)}_{q_2}
\end{equation}
Now a miracle occurs. We have checked using Mathematica that for fixed $q_3$ the sum of the right hand side is just equal to a power of $(1+n)$ multiplied by a Stirling number of the first kind. In other words we have:
\begin{equation}
\left(n+1\right)^{q_3} S^{(k)}_{q_3} = \binom{k}{q_3}
\frac{d^{q_3}}{d r_1^{q_3}} \frac{d^{k-q_3}}{d r^{k-q_3}}
\left.\left(
\sum\limits_{j=0}^k (-1)^j \frac{(1-r)^{\binom{k-j}{2}} (1-r_1)^{j (n+1)}}{(1-r;1-r)_j \cdot (1-r;1-r)_{k-j}} \cdot r^k
\right)
\right|_{r=0,r_1=0}
\end{equation}
for $k \in {\mathbb N}_+$ and $q_3=0,\cdots,k$.
Inserting the above to the right hand side of the equation in question gives:
\begin{equation}
rhs = \frac{1}{k!} \sum\limits_{q_3=0}^k (n+1)^{q_3} S^{(k)}_{q_3} = \frac{(n+1)_{(k)}}{k!} = \binom{n+1}{k}
\end{equation}
It would be interesting to prove the identity involving the Stirling numbers of the first kind.


TITLE: A question about Wall's construction for CW-complexes
QUESTION [4 upvotes]: For a given map $\phi :X\longrightarrow Y$, the mapping cylinder of $\phi$ is defined by $M_{\phi}:=Y\bigcup_{\phi} (X \times \{ 1\})$. Denote $\pi_n (M_{\phi},X \times \{ 1\} )$  by $\pi_n (\phi)$. The map $\phi$ is called $n$-connected if $X$ and $Y$ are connected and $\pi_i (\phi)=0$ for $1\leq i\leq n$.
Let $X$ be a CW-complex. Then conditions $\mathcal{F}_i$ and $\mathcal{D}_i$ on $X$ are defined inductively as follows:
$\mathcal{F}_1$: the group $\pi_1 (X)$ is finitely generated.
$\mathcal{F}_2$: the group $\pi_1 (X)$ is finitely presented, and for any  1-dimensional finite CW-complex $K$ and any map $\phi :K\longrightarrow X$ inducing an isomorphism of fundamental groups, $\pi_2 (\phi )$ is a finitely generated module over $\mathbb{Z}\pi_1 (X)$.
$\mathcal{F}_n$: the condition $\mathcal{F}_{n-1}$ holds, and for any $(n-1)$-dimensional finite CW-complex $K$ and any $(n-1)$-connected map $\phi :K\longrightarrow X$, $\pi_n (\phi )$ is a finitely generated $\mathbb{Z}\pi_1 (X)$-module.
$\mathcal{D}_n$: $H_i(\tilde{X})=0$ for $i>n$, and $H^{n+1}(X;\mathcal{B})=0$ for all coefficient bundles $\mathcal{B}$.
Proposition 3.3 in th paper "Finiteness conditions for CW-complexes" of C.T.C. Wall states that : If CW-complex $X$ satisfies the conditions $\mathcal{D}_2$ and $\mathcal{F}_2$, and $\pi_1 (X)$ is free, then $X$ has the homotopy of a finite bouquet of 1-spheres and 2-spheres.
Clearly, there exist a finite bouquet of circles $K$ and a map $\phi :K\longrightarrow X$ inducing an isomorphism of fundamental groups.
Wall proved that $\pi_2 (\phi)$ is a free $\mathbb{Z}\pi_1 (X)$-module. So we can attach a finite 2-cells to $K$, necessarily with trivial attaching map, to make a new complex $L$.  Since $\pi_2 (\phi)$ is free, we can extend the map $\phi$ to a $2$-connected map $\psi :L\longrightarrow X$.
My question is that:
Why   $\psi$ is a homotopy equivalence? Equivalently, why does $\psi$ is i-connected for all $i\geq 3$?

REPLY [2 votes]: The given conditions imply that the map $\psi:L\rightarrow X$ induces an isomorphism of fundamental groups, and so it lifts to a map $\tilde\psi: \widetilde L\rightarrow \widetilde X$.  The conditions also imply that $\tilde\psi$ is an isomorphism on homology.  Since these spaces are simply-connected, their homology groups being isomorphic implies that their homotopy groups are isomorphic.  But for $i>1$, $\pi_i(X)=\pi_i(\widetilde X)$ and similarly for $L$.  Hence $\psi$ is $i$-connected for all $i$.