TITLE: An interesting definition of a structure-preserving map inspired by natural transformation
QUESTION [1 upvotes]: The book "Conceptual mathematics..." (Lawvere and Schanuel, 1997) introduces the category of "sets with endomaps" (that resemble a set theoretic representation of monoids). Then the book proceeds to give the definition of a "structure-preserving maps" between those structures that conceptually resemble functors. However, the definition of this map is actually quite different from the traditional functor definition and is actually closer to the definition of natural transformation (see the screenshot.)
Later the book also reviews the category of (set-theoretic representations of) graphs and the definition of maps between them is similar.
I was wondering what is the formal definition of such structure-preserving maps would be and if they are defined anywhere else.

REPLY [0 votes]: There are many versions in many abstraction levels of 'structure preserving maps'.
A 'set with endomap' in universal algebraic language is called a 'unary algebra': it is a set $X$ with a single unary operation $\alpha:X\to X$.
Further universal algebraic structures might have multiple finitary operations (an $n$-ary operation is simply a map $X^n\to X$).
Further, there are first order structures. Those themselves have multiple notions of 'structure preserving maps' because of the presence of relations.
A category theoretic approach is as follows:
Consider an endofunctor $T:\mathcal S\to\mathcal S$ on a category $\mathcal S$ (which is typically $\mathcal{Set}$).
Then a $T$-algebra is a pair $(x,\alpha)$ where $x\in Ob\,\mathcal S$ and $\alpha:T(x)\to x$.
A structure preserving map from $(x,\alpha)$ to $(y,\beta)$ is an arrow $f:x\to y$ in $\mathcal S$ which satisfies $\beta\circ T(f)=f\circ\alpha$.
Often, $T$ is assumed to be a monad and then the algebra structure also needs to be compatible with the monad structure (see section 'algebras over a monad').


TITLE: Electron phonon interaction : why $V(\overrightarrow{q}, \delta \epsilon)=0$ if $E-E_F>h \omega_D$ (superconductivity)
QUESTION [0 upvotes]: In my superconductivity course, we talked about the electron phonon interaction.
We said that this interaction is $0$ if $|E-E_F| > h \omega_D $
I would like to understand why ?
Why an electron at $E=E_F-h \omega_D$ doesn't feel the electron-phonon interaction for example ?
Same question for $E=E_F+h \omega_D$.

REPLY [0 votes]: You can justify this by using the Debye model of phonons which states that beyond a given frequency $\omega_D$, the density of states of phonons is zero.


text:
\begin{document}

\title{On the Extension of Onsager's Conjecture for General Conservation Laws}
\author{Claude Bardos\footnotemark[1] \quad Piotr Gwiazda\footnotemark[2] \quad Agnieszka \'Swierczewska-Gwiazda\footnotemark[3] \quad Edriss S.\, Titi\footnotemark[4] \, \and \,  Emil Wiedemann\footnotemark[5]}
\maketitle

\begin{abstract}

The aim of  this work   is to extend and prove the Onsager conjecture for a class of conservation laws that possess generalized entropy. One of the main findings of this work is the ``universality" of the Onsager exponent, $\alpha > 1/3$, concerning the regularity of the solutions, say in $C^{0,\alpha}$, that guarantees the conservation of the generalized entropy; regardless of the structure of the genuine nonlinearity in the underlying system.

\end{abstract}
{\bf Keywords:} {Onsager's conjecture, conservation laws, conservation of entropy.} \\
 {\bf MSC Subject Classifications:} {35Q31.}

\tableofcontents

\renewcommand{\thefootnote}{\fnsymbol{footnote}}

\footnotetext[1]{
Laboratoire J.-L. Lions, BP187, 75252 Paris Cedex 05, France. Email:
claude.bardos@gmail.com}

\footnotetext[2]{
Institute of Mathematics, Polish Academy of Sciences, \'Sniadeckich 8, 00-656 Warszawa, Poland.  Email: pgwiazda@mimuw.edu.pl}

\footnotetext[3]{
Agnieszka \'Swierczewska-Gwiazda:  Institute of Applied Mathematics and Mechanics, University of Warsaw, Banacha 2, 02-097 Warszawa, Poland.  Email: aswiercz@mimuw.edu.pl}

\footnotetext[4]{
Department of Mathematics,
                 Texas A\&M University, 3368 TAMU,
                 College Station, TX 77843-3368, USA. Also, Department of Computer Science and Applied Mathematics, The Weizmann Institute of Science, Rehovot 76100, Israel. Email: titi@math.tamu.edu \, and \, edriss.titi@weizmann.ac.il}

\footnotetext[5]{
 Institute of Applied Mathematics, Leibniz University Hannover, Welfengarten~1, 30167 Hannover, Germany. Email: wiedemann@ifam.uni-hannover.de}

\section{Introduction}

In this work  we aim at extending and proving  the Onsager conjecture for a class of conservation laws that admit a  generalized entropy. Roughly speaking, the Onsager conjecture \cite{ON} states that weak solutions of the three-dimensional Euler equations of inviscid incompressible flows conserve energy if the velocity field $u\in C^{0,\alpha}\,,$ for $\alpha > \frac13\,,$ and that the critical exponent $\alpha =\frac13$ is sharp. This conjecture has been the subject of intensive investigation for the last two decades. The sufficient condition direction was proved by Eyink \cite{GEY} for the case when  $\alpha > \frac12\,.$ Later, a complete proof was established by Constantin, E and Titi \cite{CET} (see also \cite{CCFS}) under slightly weaker regularity assumptions on the solution which involve a similar exponent $\alpha > \frac13\,.$  Duchon and Robert \cite{DuchonRobert} have shown, under similar sufficient conditions to those in \cite{CET}, a local version of the conservation of energy. It is worth mentioning that the above results are established in the absence of physical boundaries, i.e., periodic boundary conditions or the whole space. However, due to the well recognized  dominant role of the boundary in the generation of turbulence (cf.\, \cite{BT} and references therein) it seems very reasonable  to investigate the analogue of the Onsager conjecture in bounded domains. Indeed, for the three-dimensional Euler equations in a smooth bounded domain $\Omega$, subject to no-normal flow (slip) boundary conditions, it has been shown in \cite{BT2} that a weak solution conserves the energy provided the velocity field  $u\in C^{0,\alpha}(\overline{\Omega})\,,$ for $\alpha > \frac13\,,$    (see also \cite{RRS} for the case of the upper-half space under stronger conditions on the pressure term). A local version, analogue to that of \cite{DuchonRobert}, was established recently in \cite{BTW} under slightly weaker conditions to those in \cite{BT2}, but at the expense of additional sufficient conditions concerning the vanishing behavior of the energy flux near the boundary.

Showing the sharpness of the  exponent $\alpha = \frac13$ in Onsager's conjecture turns out to be much more subtle. This direction has been underlined  by a series of contributions
(cf.\, Isett \cite{Isett}, Buckmaster, De Lellis ,  Sz\'ekelyhidi and  Vicol  \cite{BDSV} and references therein) where  weak solutions,   $u\in C^{0,\alpha}\,,$ with $\alpha <\frac 13$, that  dissipate energy were constructed using the convex integration machinery. Notice, however,  that there exists a family of weak solutions to the three-dimensional Euler equations, that are not  more regular than $L^2$,  and which conserve the energy, cf. \cite{BT-Shear}.

It is most natural to ask whether the analogue of the Onsager conjecture is valid for other systems of conservation laws. Indeed, there has been some intensive  recent work  extending  the Onsager conjecture for other physical  systems, in the absence of physical boundaries, see, e.g.,  \cite{AW,DrEy,FGSW,GMS,Yu} and references therein. In this paper we consider systems of conservation laws with physical boundaries. We use the approach of \cite{DuchonRobert}, as it has been outlined and extended in \cite{BTW} in the presence of physical boundaries, to establish the local conservation of ``generalized entropies" (conserved quantities, which are not necessarily convex) for systems of conservation laws which possess such generalized entropies. This is accomplished provided the underlying weak solutions are locally in $C^{0,\alpha}\,,$ for $\alpha > \frac13\,.$ One of the  primary findings of this work  is the universality of the Onsager critical exponent $\alpha = \frac13\,,$ regardless of the structure of the genuine nonlinearity in the underlying system. Notably, in a forthcoming paper \cite{BGSTW} we will show the extension of these results, and provide additional explicit physical examples,  using more delicate harmonic analysis tools and function spaces. In particular, we will show similar results  under slightly weaker regularity assumptions of the solutions which are required to belong to some ``exotic" function spaces with exponent $\alpha = \frac13\,.$

\section{Local and global generalized entropies conservation}
In this section we state and prove our main result. In the first subsection we establish  the local entropy conservation for any weak solution that belongs to the  H\"older space $C^{0,\alpha}$ with $\alpha > 1/3\,.$ In the second subsection we state a fundamental Lemma concerning nonlinear commutator estimate of Friedrichs mollifier, and in the last subsection we state additional sufficient conditions for establishing the global entropy conservation.

\subsection{Local entropy conservation}
Let  $Q\subset\bR^{d+1}$ be an open set, and consider in $Q$ the following system of conservation laws:
\begin{equation}
\sum_{0\le i \le d} \del_{x_i} A_i(u)=0\,, \label{eq1}
\end{equation}

config: arxiv


text:
\begin{document}
\begin{frontmatter}

\title{A free boundary problem for a class of parabolic-elliptic type chemotaxis model
\tnoteref{sup}} \tnotetext[sup]{This work is supported by National Natural Science Foundation of China (Grant No. 11131005) and the Fundamental Research Funds for the Central Universities (Grant No. 2014201020202).}

\author[whu]{Hua Chen}
\ead{chenhua@whu.edu.cn}

\author[sxu]{Wenbin Lv}
\ead{lvwenbin@whu.edu.cn}

\author[whu]{Shaohua Wu}
\ead{wush8@sina.com}

\address[whu]{School of Mathematics and Statistics, Wuhan University,
              Wuhan 430072, China}

\address[sxu]{School of Mathematical Sciences, Shanxi University, Taiyuan 030006, China}

\begin{abstract}
In this paper, we study a free boundary problem for a class of
parabolic-elliptic type chemotaxis model in high dimensional symmetry domain $\Omega$. By
using the contraction mapping principle and operator semigroup approach, we
establish the existence of the solution for such kind of chemotaxis system in
the domain $\Omega$ with free boundary condition. Besides, we get the explicit formula for the free boundary and show the chemotactic collapse for the solution of the system.
\end{abstract}

\begin{keyword}{chemotaxis, free boundary problem, existence of solution.}

\MSC[2010] 35A01, 35K57, 35R35, 47D03.
\end{keyword}
\end{frontmatter}

\section{Introduction}

Understanding of the partially oriented movement of cells in response to chemical signals, chemotaxis, is of great significance in various contexts. This importance partly stems from the fact that when cells combined with the ability to produce the respective signal substance themselves, chemotaxis mechanisms are among the most primitive forms of intercellular communication. Typical examples include
aggregation processes such as slime mold formation in Dictyostelium Discoideum discovered by K.B. Raper \cite{K.B.R.}. A lot of mathematicians have made efforts to develop various models and investigated them from a viewpoint of mathematics. For a broad overview over various types of chemotaxis processes, we refer the reader to the surveys and papers \cite{C-W20071,C-W2008,C-W2011,C-Z2004,C-Z2005,C-Z2006,H-P,D.H.,K-S,W-Y,W-C-L2008,Y-C-L-S} and the references therein.

As we all know, in a standard setting for many partial differential equations, we usually assume that the process being described occurs in a fixed domain of the space. But in the real world, the following phenomenon may happen. At the initial state, a kind of amoeba occupy some areas. When foods become rare, they begin to secrete chemical substances on their own. Since the biological time scale is much slower than the chemical one, the chemical substances are full filled with whole domain and create a chemical gradient attracting the cells. In turn, the areas of amoeba may change according to the chemical gradient from time to time. In other words, a part of whose boundary is unknown in advance and that portion of the boundary is called a free boundary. In addition to the standard boundary conditions that are needed in order to solve the PDEs, an additional condition must be imposed at the free boundary. One then seeks to determine both the free boundary and the solution of the differential equations. The theory of free boundaries has seen great progress in the last thirty years; for the state of the field we refer to \cite{A.F.}.

In this paper, we consider the following high dimensional free boundary problem
of a chemotaxis model. Such kind of models can be found in \cite{C-L-W2015,C-L-W2016,C-W2007,C-W2012,W2010}, and we can
give more explanations in the appendix below.
\begin{equation}\label{19}
\begin{cases}
u_{t}=\nabla(\nabla u-u\nabla v),&\text{in}\quad\Omega_{t}\times(0,T),\\
u=0,&\text{in}\quad\Omega\times(0,T)\setminus\Omega_{t}\times(0,T),\\
\displaystyle -\nabla u\cdot\frac{\nabla\Phi}{|\nabla\Phi|}=ku,&\text{on}\quad\partial\Omega_{t}\times(0,T),\\
\displaystyle u\frac{\partial\Phi}{\partial t}=\nabla u\cdot\nabla\Phi-u\nabla v\cdot\nabla\Phi,&\text{on}\quad\partial\Omega_{t}\times(0,T),\\
u(x,0)=u_{0}(x),&\text{in}\quad\Omega_{0},\\
0=\Delta v+u-c,&\text{in}\quad\Omega\times(0,T),\\
\displaystyle \frac{\partial v}{\partial \mathbf{n}}=0,&\text{on}\quad\partial\Omega\times(0,T),
\end{cases}
\end{equation}
where
\begin{itemize}
  \item $\Omega\subset\mathbb{R}^n$ is a bounded open set with smooth boundary $\partial\Omega$ and $\mathbf{n}$ is unit outer normal vector of $\partial\Omega$. Besides, $\Omega$ is assumed as a symmetry domain, i.e. if $x\in\Omega$ then $-x\in\Omega$ as well;
  \item $c$ is a constant which will be determined in the following section;
  \item $k(x,t)=k(|x|,t)$ is radial symmetric and satisfying the Lipschitz condition on $|x|$, namely there exists a constant $L>0$, such that
\begin{equation}\label{103}
\big|k(|x_1|,t)-k(|x_2|,t)\big|\leqslant L\big||x_1|-|x_2|\big|,~x_1,x_2\in\Omega_t.
\end{equation}
Also, $k(x,t)$ is bounded on $t\in[0,+\infty)$. In other words, there exists a constant $C>0$ which depends on $x$, such that
\begin{equation}\label{105}
|k(x,\,t)|\leqslant C,~t\in[0,+\infty);
\end{equation}
  \item $u=u(x,\,t)$ is an unknown function of $(x,\,t)\in\Omega_t\times(0,\,T)$ and it stands for the density of cellular slime molds. In other words, the density $u(x,\,t)$ occupies the domain $\Omega_t$, an open subset of $\Omega$, in time $t$ and $u(x,t)=0$ in the outside of $\Omega_t$;
  \item $v=v(x,\,t)$ is an unknown function of $(x,\,t)\in\Omega\times(0,\,T)$ and it stands for the concentration of chemical substances secreted by the slime molds;
  \item $\Gamma_t:~\Phi(x,t)=0$ is an unknown free boundary.
\end{itemize}
For general smooth domain $\Omega$, the system (\ref{19}) is based on the well-known chemotaxis model with fixed boundary
\begin{equation}\label{20}
\begin{cases}
u_t=\nabla(\nabla u-u\nabla v),&\text{in}\quad\Omega\times(0,\,T),\\
0=\Delta v+u-c,&\text{in}\quad \Omega\times(0,\,T),\\
\displaystyle \frac{\partial u}{\partial \mathbf{n}}=\frac{\partial v}{\partial \mathbf{n}}=0,&\text{on}\quad\partial\Omega\times(0,T),\\
u(x,0)=u_{0}(x),&\text{in}\quad\Omega.
\end{cases}
\end{equation}
which was introduced by W. J\"{a}ger and S. Luckhaus~\cite{J-L} who first studied the conjectrues of Childress and Percus in 1992. Later, the problem \eqref{20} is intensively studied by many authors (see for instance
\cite{C-M,M.H.,H-M-V1997,H-M-V1998,J-L}). Concerning the problem, classical results state that
\begin{itemize}
  \item if n=2
   \begin{itemize}
     \item there exists a critical number $C(\Omega)$ such that $c\leqslant C(\Omega)$ implies that there exists a unique, smooth positive solution to \eqref{20} globally in time \cite{J-L};
     \item $\Omega$ is a disk, there exists a positive number $C$ such that $c\geqslant C$ implies radially symmetric positive initial values can be constructed such that explosion of the solution happens in the center of the disc in finite time \cite{J-L};
     \item $\Omega=\R^2$, then no radial, self-similar solutions of \eqref{20} exist such that $\int_{|x|\leqslant r}u(T,s)ds<\infty$ as $r\rightarrow0$ \cite{M.H.,H-M-V1997};
   \end{itemize}
  \item if $n\geqslant 3$
   \begin{itemize}
     \item $\Omega=\R^3$ for any $T>0$ and any constant $C>0$, the radial solution $(u,v)$ of \eqref{20} that is smooth for all times $0<t<T$, blows up at $r=0$ and $t=T$, and is such that $\int_{|x|\leqslant r}u(T,s)ds\rightarrow C$ \cite{M.H.,H-M-V1997,H-M-V1998};
     \item  $\Omega$ is a ball and some additional conditions, there exists a unique maximal classical solution with finite maximal existence time $T$ \cite{C-M}.
   \end{itemize}
\end{itemize}

In one dimensional case, if $k$ is a positive constant H. Chen and S.H. Wu \cite{C-W2007,C-W2012,W2010} studied the similar free boundary value problem (\ref{19}) and established the existence and uniqueness of the solution for the system (\ref{19}). However, to the best of our knowledge, high dimensional case for the free boundary value problem (\ref{19}) will be more important. In view of the biological relevance of the particular case $n=3$, we find it worthwhile to clarify these questions. In the present paper, we consider the system (\ref{19}) in a high dimensional symmetry domain $\Omega$. In addition, the condition that $k$ is a positive constant in \cite{C-W2007,C-W2012,W2010} seems too strict. So it is also worthwhile to consider the system with non-constant coefficient $k$.

This paper is arranged as follows. In section \ref{sec2}, we rewrite the model and present the main result of the paper. In section \ref{sec3}, we use the operator semigroup approach to establish some estimates which are essential in the proof of the main result. In section \ref{sec4}, we shall give the proof of the main result. In addition, we study the property of the free boundary in section \ref{sec5}.

\section{Main result}\label{sec2}

\subsection{Rewrite the model}

In this subsection, we rewrite the model with the form of radial symmetry. We assume that the environment and solution are radially symmetric. Without loss of generality, we assume that $\Omega=B_1(0)$ which represents a unite ball centered in origin and that $u$ and $v$ are radially symmetric with respect to $x=0$. The free boundary can be written as $r=|x|=h(t)$ and $h(0)=b$. Hence, the density $u(x,0)$ occupies the domain $\Omega_0=B_b(0)$ at the initial time $t=0$.

Let
$(\widetilde{u},\widetilde{v})$ denote the corresponding radial solution in $B_1(0)\times(0,T)$. In order to avoid confusion, we may write
\begin{equation*}
\widetilde{u}(r,t)=u(x,t),\quad \widetilde{v}(r,t)=v(x,t),
\end{equation*}
 with $r=|x|=(x_1^2+\cdots+x_n^2)^{\frac{1}{2}}\in (0,1)$ and
$$\frac{\partial r}{\partial x_i}=\frac{\partial}{\partial x_i}\left((x_1^2+\cdots+x_n^2)^{\frac{1}{2}}\right)=\frac{x_i}{r},\quad i=1,2,\cdots,n.$$
A simple calculation shows that
\begin{equation}\label{21}
u_{x_i}(x,t)=\widetilde{u}_r(r,t)\frac{x_i}{r},\quad v_{x_i}=\widetilde{v}_r(r,t)\frac{x_i}{r},
\end{equation}
\begin{equation}\label{24}
u_{x_ix_i}(x,t)=\widetilde{u}_{rr}(r,t)\frac{x_i^2}{r^2}+\widetilde{u}_r(r,t)\left(\frac{1}{r}-\frac{x_i^2}{r^3}\right),
\end{equation}
and
\begin{equation}\label{25}
v_{x_ix_i}(x,t)=\widetilde{v}_{rr}(r,t)\frac{x_i^2}{r^2}+\widetilde{v}_r(r,t)\left(\frac{1}{r}-\frac{x_i^2}{r^3}\right).
\end{equation}

\noindent $\bullet$ Reformulation of the boundary.

If $\Gamma_t: \widetilde{\Phi}(r,t)=0\Leftrightarrow r-h(t)=0$, then the condition of the free boundary converts into
\begin{equation*}
\widetilde{k}\big(h(t),t\big)\widetilde{u}\big(h(t),t\big)+\widetilde{u}_r\big(h(t),t\big)=0,
\end{equation*}
and
\begin{equation*}
-\widetilde{u}\big(h(t),t\big)h_t(t)=\widetilde{u}_r\big(h(t),t\big)-\widetilde{u}\big(h(t),t\big)\widetilde{v}_r\big(h(t),t\big).
\end{equation*}

Actually, substituting (\ref{21}) and
$$\Phi_{x_i}(x,t)=\frac{\partial \widetilde{\Phi}}{\partial r}\cdot\frac{\partial r}{\partial x_i}=\frac{x_i}{r},\quad i=1,2,\cdots,n,$$
into the third and forth equations of (\ref{19}), we can easily get
$$\widetilde{k}(r,t)\widetilde{u}(r,t)=-\nabla u\cdot\frac{\nabla\Phi}{|\nabla\Phi|}=-\sum_{i=1}^n\widetilde{u}_r(r,t)\frac{x_i^2}{r^2}=-\widetilde{u}_r(r,t),$$
and
\begin{equation}\label{26}
\begin{aligned}
-\widetilde{u}(r,t)h_t(t)=&u\frac{\partial\Phi}{\partial t}=\nabla u\cdot\nabla\Phi-u\nabla v\cdot\nabla\Phi=\widetilde{u}_r(r,t)-\widetilde{u}(r,t)\widetilde{v}_r(r,t),
\end{aligned}
\end{equation}
on $\Gamma_t$.

If $u>0$ on $\Gamma_t$, then (\ref{26}) is equivalent to
\begin{equation*}
h_t(t)=\widetilde{k}\big(h(t),t\big)+\widetilde{v}_r\big(h(t),t\big).
\end{equation*}

\noindent $\bullet$ Reformulation of the equation.

Substituting (\ref{21}), (\ref{24}) and (\ref{25}) into the first and sixth equations of (\ref{19}), we can easily obtain
\begin{align*}
\widetilde{u}_{t}(r,t)&=u_{t}(x,t)=\Delta u-\nabla u\nabla v-u\Delta v\\
&=\widetilde{u}_{rr}(r,t)+\frac{n-1}{r}\widetilde{u}_r(r,t)-\widetilde{u}_r(r,t)\widetilde{v}_r(r,t)\\
&\quad\quad-\widetilde{u}(r,t)\left(\widetilde{v}_{rr}(r,t)+\frac{n-1}{r}\widetilde{v}_r(r,t)\right)\\
&=r^{1-n}\left(r^{n-1}\widetilde{u}_r(r,t)\right)_r-r^{1-n}\left(r^{n-1}\widetilde{u}(r,t)\widetilde{v}_r(r,t)\right)_r,
\end{align*}
\begin{equation}\label{1-1}
\text{or}\quad r^{n-1}\widetilde{u}_{t}(r,t)=\big(r^{n-1}\widetilde{u}_r(r,t)-r^{n-1}\widetilde{u}(r,t)\widetilde{v}_r(r,t)\big)_r,
\end{equation}
and
\begin{align*}
0&=\Delta v+u-c\\
&=\widetilde{v}_{rr}(r,t)+\frac{n-1}{r}\widetilde{v}_r(r,t)+\widetilde{u}(r,t)-c\\
&=r^{1-n}\left(r^{n-1}\widetilde{v}_r(r,t)\right)_r+\widetilde{u}(r,t)-c,
\end{align*}
\begin{equation}\label{1-2}
\text{or}\quad 0=\big(r^{n-1}\widetilde{v}_r(r,t)\big)_r+r^{n-1}\widetilde{u}(r,t)-r^{n-1}c.
\end{equation}
Therefore, the model we are concerned here becomes
\begin{equation}\label{1}
\begin{cases}
\widetilde{u}_{t}(r,t)=\widetilde{u}_{rr}(r,t)+\dfrac{n-1}{r}\widetilde{u}_r(r,t)\\
\quad\quad\quad\quad-\widetilde{u}_r(r,t)\widetilde{v}_r(r,t)-\widetilde{u}(r,t)\left(\widetilde{v}_{rr}(r,t)\right.\\
\quad\quad\quad\quad\left.+\dfrac{n-1}{r}\widetilde{v}_r(r,t)\right),&0<r<h(t),\quad 0<t<T,\\
\widetilde{u}(r,t)=0,&h(t)<r<1,\quad 0<t<T,\\
\widetilde{u}_r(0,t)=0,&0<t<T,\\
\widetilde{u}_r(h(t),t)+\widetilde{k}(h(t),t)\widetilde{u}(h(t),t)=0,&0<t<T,\\
h_t(t)=\widetilde{k}(h(t),t)+\widetilde{v}_r(h(t),t),&0<t<T,\\
\widetilde{u}(r,0)=\widetilde{u}_{0}(r),&0<r<b,\\
0=\widetilde{v}_{rr}(r,t)+\dfrac{n-1}{r}\widetilde{v}_r(r,t)+\widetilde{u}(r,t)-c,&0<r<1,\quad 0<t<T,\\
\widetilde{v}_r(0,t)=0,&0<t<T,\\
\widetilde{v}_r(1,t)=0,&0<t<T,
\end{cases}
\end{equation}
which corresponds to the equation with normal coordinate
\begin{equation}\label{101}
\begin{cases}
u_{t}=\nabla(\nabla u-u\nabla v),&\text{in}\quad B_{h(t)}(0)\times(0,T),\\
u=0,&\text{in}\quad\big(B_1(0)\setminus \overline{B_{h(t)}(0)}\big)\times(0,T),\\
\displaystyle \frac{\p u}{\p \mathbf{n}}(x,t)+k(x,t)u(x,t)=0,&\text{on}\quad\partial B_{h(t)}(0)\times(0,T),\\
\displaystyle h_t(t)=k(x,t)+\frac{\p v}{\p \mathbf{n}}(x,t),&\text{on}\quad\partial B_{h(t)}(0)\times(0,T),\\
u(x,0)=u_{0}(x),&\text{in}\quad B_b(0),\\
0=\Delta v+u-c,&\text{in}\quad B_1(0)\times(0,T),\\
\displaystyle \frac{\partial v}{\partial \mathbf{n}}(x,t)=0,&\text{on}\quad\partial B_1(0)\times(0,T).
\end{cases}
\end{equation}

\subsection{Main result}

Now we introduce the following space notations, which will be used in the main result. For $T>0$, we define
\begin{gather*}
X_{u}^{1}(T)=L^\infty\left((0,T],H^{1,p}\big(B_{h(t)}(0)\big)\cap\left\{\frac{\p u}{\p \mathbf{n}}(x,t)+k(x,t)u(x,t)=0,x\in\p B_{h(t)}(0)\right\}\right),\\
X_{v}^{2}(T)=L^\infty\left((0,T],H^{2,p}\big(B_1(0)\big)\cap\left\{\frac{\p v}{\p \mathbf{n}}(x,t)=0,x\in\p B_1(0)\right\}\right),\\
Y_v(T)=\left\{v~\bigg|~\frac{1}{|B_1(0)|}\int_{B_1(0)} vdx=0\right\}.
\end{gather*}

Our main results are:
\begin{theorem}\label{th1}
Assume $k(x,t)$ is radial symmetric on $x$ and satisfies the conditions (\ref{103}) and (\ref{105}) and
$$c=nM,\quad M=\frac{\Gamma(\frac{n}{2})}{2\pi^{\frac{n}{2}}}\int_{B_b(0)}u_0(x)dx,$$
where $n$ is the dimension of the space. If
$$u_{0}(x)\in H^{1,p}\big(B_b(0)\big),~u_{0}>0,$$
are radial symmetric on $x$, where $0<b<1$ and $b$ is a constant. Then there exist $T>0$ small enough, a radial symmetric pair
$$\big(u(x,t),v(x,t)\big)\in (X_u^{1}(T))\times (X_{v}^{2}(T)\cap Y_v(T)),$$
and a curve $\Gamma_t:|x|=h(t)$, which are the solutions of \eqref{101} for $n<p$.
\end{theorem}

Concerning the free boundary, we have the following properties:

\begin{theorem}\label{th2}
Assume $1<n<p$, $k(x,t)=a|x|$ and $a>0$ is a constant. Then we have $$h(t)=\left[\left(b^n-\frac{M}{a+M}\right)e^{n(a+M)t}+\frac{M}{a+M}\right]^{\frac{1}{n}}.$$
Furthermore, if $M<\frac{ab^n}{1-b^n}$, then $h(t)$ is increasing and
\[h(t)\rightarrow 1~\text{when}~t\rightarrow \frac{1}{n(a+M)}\ln\frac{a}{(a+M)b^n-M}.\]
If $M>\frac{ab^n}{1-b^n}$, then $h(t)$ is decreasing and
\[h(t)\rightarrow0~\text{as}~t\rightarrow \frac{1}{n(a+M)}\ln\frac{M}{(1-b^n)M-ab^n}.\]
If $M=\frac{ab^n}{1-b^n}$, then $h(t)\equiv b$.
\end{theorem}

Concerning the global existence and blow-up in finite time for the solution, we have the following result:

\begin{theorem}\label{th3}
Assume $1<n<p$, $k(x,t)=a|x|$ and $a>0$ is a constant. If $M\leqslant\frac{ab^n}{1-b^n}$, then the solution $(u,v)$ is a global solution. If $M>\frac{ab^n}{1-b^n}$, then the system \eqref{101} collapses in a finite time $T^*=\frac{1}{n(a+M)}\ln\frac{M}{(1-b^n)M-ab^n}$. More precisely
$$u(x,t)\rightarrow \frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}M\delta(x)~\text{as}~t\rightarrow T^*.$$
\end{theorem}

\section{Some crucial estimates}\label{sec3}

In this section, we establish some crucial estimates, which will play the key roles in proving the local existence of radial symmetric solution of the system \eqref{101} in high dimensional case.

\subsection{Some basic properties of the solution}

\begin{lemma}
If $u_0(x)>0$ and $(u,v)$ is the radial symmetrical solution of the system \eqref{101}, then we have $u>0$ and
$$\int_{B_{h(t)}(0)}u(x,t)dx=\int_{B_b(0)}u_0(x)dx.$$
\end{lemma}

\begin{proof}
Since $\widetilde{u}_{0}(r)>0$, by standard maximal principle of the parabolic equation, it follows that $\widetilde{u}>0$. Integrating the equation (\ref{1-1}) over $(0,h(t))$, we have
\begin{equation*}
\begin{split}
\int_0^{h(t)} r^{n-1}\widetilde{u}_{t}(r,t)dr=&\int_0^{h(t)}\left(r^{n-1}\widetilde{u}_r(r,t)-r^{n-1}\widetilde{u}(r,t)\widetilde{v}_r(r,t)\right)_rdr\\
=&h^{n-1}(t)\left[\widetilde{u}_r\big(h(t),t\big)-\widetilde{u}\big(h(t),t\big)\widetilde{v}_r\big(h(t),t\big)\right]\\
=&-h^{n-1}(t)\widetilde{u}\big(h(t),t\big)h_t(t),
\end{split}
\end{equation*}
where the fifth equation of (\ref{1}) is used. Thus one has
$$\frac{d}{dt}\int_0^{h(t)}r^{n-1}\widetilde{u}(r,t)dr=\int_0^{h(t)} r^{n-1}\widetilde{u}_{t}(r,t)dr+h_t(t)h^{n-1}(t)\widetilde{u}\big(h(t),t\big)=0,$$
which implies that
\begin{equation*}
\int_0^{h(t)}r^{n-1}\widetilde{u}(r,t)dr=\int_0^{h(0)}r^{n-1}\widetilde{u}(r,0)dr.
\end{equation*}
By virtue of the fact
$$\int_{B_b(0)}u_0(x)dx=\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\int_0^br^{n-1}\widetilde{u}(r,0)dr,$$
we have
$$\int_{B_{h(t)}(0)}u(x,t)dx=\int_{B_b(0)}u_0(x)dx$$
as required. Hence, the proof of the lemma is completed.
\end{proof}

\begin{lemma}\label{lm4}
We have
$$\widetilde{v}_r(h(t),t)=Mh(t)-\frac{M}{h^{n-1}(t)}.$$
\end{lemma}

\begin{proof}
Integrating the equation (\ref{1-2}) over $(0,h(t))$, we have
$$h^{n-1}(t)\widetilde{v}_r(h(t),t)=nM\int_0^{h(t)}r^{n-1}dr-\int_0^{h(t)}r^{n-1}\widetilde{u}(r,t)dr.$$
Simple calculation yields
$$\widetilde{v}_r(h(t),t)=Mh(t)-\frac{M}{h^{n-1}(t)}.$$
This completes the proof of Lemma \ref{lm4}.
\end{proof}

\subsection{Some basic properties of the system}

Firstly, we define
$$B=\left\{h\in C(0,T]~\bigg|~h(0)=b,\left|\frac{h(t_1)-h(t_2)}{t_1-t_2}\right|\leqslant M_0,t_1,t_2\in(0,T),t_1\ne t_2\right\},$$
where $M_0<\frac{1}{2}\frac{\min\{b,1-b\}}{T}$ is a constant.

In this section, we shall establish some estimates which are important in the proof of the main result. For any fixed $h(t)\in B$, we consider the following problems
\begin{equation}\label{5}
\begin{cases}
u_{t}=\nabla(\nabla u-u\nabla v),&\text{in}\quad B_{h(t)}(0)\times(0,T),\\
u=0,&\text{in}\quad\big(B_1(0)\setminus \overline{B_{h(t)}(0)}\big)\times(0,T),\\
\displaystyle \frac{\p u}{\p \mathbf{n}}(x,t)+k(x,t)u(x,t)=0,&\text{on}\quad\partial B_{h(t)}(0)\times(0,T),\\
u(x,0)=u_{0}(x),&\text{in}\quad B_b(0),
\end{cases}
\end{equation}
and
\begin{equation}\label{6}
\begin{cases}
0=\Delta v+u-c,&\text{in}\quad B_1(0)\times(0,T),\\
\displaystyle \frac{\partial v}{\partial \mathbf{n}}(x,t)=0,&\text{on}\quad\partial B_1(0)\times(0,T).
\end{cases}
\end{equation}

Concerning the system \eqref{6}, we have the following result.

\begin{lemma}\label{lm2}
If $h(t)\in B$, then for $T>0$ small enough and $u\in X_u^1(T)$, the system \eqref{6} admits a unique solution $v\in X_v^2(T)\cap Y_v(T)$, and we have
\begin{equation}\label{9}
\sup_{0\leqslant t\leqslant T}\left\|v\right\|_{H^{2,p}(B_1(0))}\leqslant C\sup_{0\leqslant t\leqslant T}\|u\|_{L^p(B_{h(t)}(0))}+C.
\end{equation}
where $C$ is independent of $T$ and $h(t)\in B$.
\end{lemma}

\begin{proof}
It is obvious that the problem (\ref{6}) has a solution $v\in H^{1,p}(B_1(0))$ if and only if $\int_{B_1(0)}(u-c)dx=0$ i.e.
\begin{align*}
\int_{B_1(0)}(u-c)dx=&\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\int_0^{h(t)}r^{n-1}\widetilde{u}(r,t)dr-c|B_1(0)|\\
=&\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}M-c\frac{2\pi^{\frac{n}{2}}}{n\Gamma(\frac{n}{2})}\\
=&0.
\end{align*}
Moreover, we have
$$\sup_{0\leqslant t\leqslant T}\left\|v\right\|_{H^{2,p}(B_1(0))}\leqslant C\sup_{0\leqslant t\leqslant T}\|u\|_{L^p(B_{h(t)}(0))}+C.$$

From the above estimate, we can easily draw the conclusion.
\end{proof}

Now, we consider the system \eqref{5}. Let $y_i=\frac{x_i}{h(t)},~\mathfrak{u}(y,t)=u(x,t),~\mathfrak{v}(y,t)=v(x,t),~\mathfrak{k}(y,t)=k(x,t)$. Then the system \eqref{5} is equivalent to the following system
\begin{equation*}
\begin{cases}
\displaystyle\mathfrak{u}_t=\frac{1}{h^2(t)}\Delta\mathfrak{u}-\frac{1}{h^2(t)}\nabla\cdot(\mathfrak{u}\nabla\mathfrak{v})+\frac{ h'(t)}{h(t)}y\nabla\mathfrak{u},&\text{in}\quad y\in B_1(0),\quad0<t<T,\\
\displaystyle\frac{\partial\mathfrak{u}}{\partial \mathbf{n}}(y,t)+\mathfrak{k}\big(y,t\big)h(t)\mathfrak{u}(y,t)=0,&\text{on}\quad y\in\p B_1(0),\quad0<t<T,\\
\displaystyle\mathfrak{u}(y,0)=\mathfrak{u}_0(y),&\text{in}\quad y\in B_1(0).
\end{cases}
\end{equation*}
Hence, it can be reduced to the system of integral equation
\begin{equation*}
\mathfrak{u}(y,t)=e^{\alpha(t,0)\Delta}\mathfrak{u}(y,0)+\int_0^t e^{\alpha(t,s)\Delta}A_{\mathfrak{u}\mathfrak{v}}(s)ds,
\end{equation*}
where
$$\alpha(t,s)=\int_s^t \frac{1}{h^2(\tau)}d\tau,$$
\begin{equation*}
A_{\mathfrak{u}\mathfrak{v}}(s)=-\frac{1}{h^2(s)}\nabla\cdot(\mathfrak{u}\nabla\mathfrak{v})+\frac{ h'(s)}{h(s)}y\nabla\mathfrak{u}.
\end{equation*}
Let~$t\geqslant s>0$,~$e^{\alpha(t,s)\Delta}$~represents the operator semigroup on $L^{p}\big(B_1(0)\big)$ which is generated by
\begin{gather*}
L(\tau)=h^{-2}(\tau)\Delta,\\ D\big(L(\tau)\big)=H^{1,p}\big(B_1(0)\big)\cap\left\{\frac{\partial\mathfrak{u}}{\partial \mathbf{n}}(y,t)+\mathfrak{k}\big(y,t\big)h(t)\mathfrak{u}(y,t)=0,y\in\p B_1(0)\right\}.
\end{gather*}
and $e^{\alpha(t,s)\Delta}$ is an holomorphic semigroup on $L^{p}\big(B_1(0)\big)$. If~$h(t)\in B$, then
\begin{equation}\label{8}
0<b-M_0T\leqslant h(t)\leqslant b+M_0T<1,\quad 0\leqslant t\leqslant T.
\end{equation}
The operator-theoretic feature of $e^{\alpha(t,s)\Delta}$ necessary for the proof of this lemma is well known \cite{M.T.2013}. There exists a constant $C>0$ which depends on $M_0$ but independent of $t$ such that
\begin{equation}\label{12}
\|e^{\alpha(t,s)\Delta}f\|_{H^{\lambda+1,p}}\leqslant C(t-s)^{-\frac{1}{2}-\frac{n}{2}(\frac{1}{q}-\frac{1}{p})}\|f\|_{H^{\lambda,q}},\quad f\in H^{\lambda,q},
\end{equation}
where $1\leqslant q\leqslant p\leqslant+\infty$.

Concerning the system \eqref{5} and \eqref{6}, we have the following result.

\begin{lemma}\label{lm3}
If $h(t)\in B,~u_0\in H^{1,p}\big(B_b(0)\big)\cap\{\frac{\p u}{\p \bf{n}}(x,0)+k(x,0)u(x,0)=0,x\in \p B_b(0)\}$ and $v_0\in H^{2,p}\big(B_1(0)\big)\cap\{\frac{\p v}{\p \bf{n}}(x,0)=0,x\in\p B_1(0)\}$. Then for $T$ small enough, the system (\ref{5}) and (\ref{6}) admit a unique solution
$$u\in X_u^1(T),\quad v\in X_v^{2}(T)\cap Y_v(T).$$
Moreover, we have
$$\sup_{0\leqslant t\leqslant t_0}\|u\|_{H^{1,p}(B_{h(t)}(0))}\leqslant 2C\|u(\cdot,0)\|_{H^{1,p}(B_b(0))},$$
where $n<p$.
\end{lemma}

\begin{proof}
Firstly, we consider $w\in X_u^1(T)$ and $w(x,0)=u_0(x)$. Let $v=v(w)$ denote the corresponding solution of the equation
\begin{equation}\label{14}
\begin{cases}
0=\Delta v+w-c,&\text{in}\quad B_1(0)\times(0,T),\\
\displaystyle \frac{\partial v}{\partial \mathbf{n}}(x,t)=0,&\text{on}\quad\partial B_1(0)\times(0,T).
\end{cases}
\end{equation}
By Lemma \ref{lm2}, we have
$$\sup_{0\leqslant t\leqslant T}\left\|v\right\|_{H^{2,p}(B_1(0))}\leqslant C\sup_{0\leqslant t\leqslant T}\|w\|_{L^p(B_{h(t)}(0))}+C.$$

Secondly, for the solution $v$ of equation (\ref{14}), we define $u=u\big(v(w)\big)$ to be the corresponding solution of
\begin{equation*}
\begin{cases}
\displaystyle\mathfrak{u}_t=\frac{1}{h^2(t)}\Delta\mathfrak{u}-\frac{1}{h^2(t)}\nabla\cdot(\mathfrak{w}\nabla\mathfrak{v})+\frac{ h'(t)}{h(t)}y\nabla\mathfrak{w},&\text{in}\quad y\in B_1(0),\quad0<t<T,\\
\displaystyle\frac{\partial\mathfrak{u}}{\partial \mathbf{n}}(y,t)+\mathfrak{k}\big(y,t\big)h(t)\mathfrak{u}(y,t)=0,&\text{on}\quad y\in\p B_1(0),\quad0<t<T,\\
\displaystyle\mathfrak{u}(y,0)=\mathfrak{u}_0(y),&\text{in}\quad y\in B_1(0).
\end{cases}
\end{equation*}

Let
$$B(\mathcal{M},T)=\{w\in X_u^1\mid w(x,0)=u_0(x),~\sup_{0\leqslant t\leqslant T}\|\mathfrak{w}(\cdot,t)\|_{H^{1,p}(B_1(0))}\leqslant \mathcal{M}\},$$
and we define a mapping $Fw=u\big(v(w)\big)$ as follows
\begin{equation}\label{11}
\mathfrak{u}(y,t)=e^{\alpha(t,0)\Delta}\mathfrak{u}(y,0)+\int_0^t e^{\alpha(t,s)\Delta}A_{\mathfrak{w},\mathfrak{v}}(s)ds.
\end{equation}
Then local existence will be established via contraction mapping principle.

{\bf Step 1:} $F$ maps $B(\mathcal{M},T)$ into itself.

Noticing \eqref{11}, we have
\begin{equation}\label{13}
\begin{aligned}
\left\|\mathfrak{u}(\cdot,t)\right\|_{H^{1,p}}\leqslant&\left\|e^{\alpha(t,0)\Delta}\mathfrak{u}(\cdot,0)\right\|_{H^{1,p}}+\int_0^{t} \left\|e^{\alpha(t,s)\Delta}\frac{1}{h^2(s)}\nabla\cdot\big(\mathfrak{w}\nabla\mathfrak{v}\big)\right\|_{H^{1,p}} ds\\
&\quad\quad+\int_0^{t} \left\|e^{\alpha(t,s)\Delta}\frac{ h'(s)}{h(s)}y\nabla\mathfrak{w}\right\|_{H^{1,p}}ds\\
=&(S_1)+(S_2)+(S_3).
\end{aligned}
\end{equation}
Using the facts \eqref{8} and \eqref{12}, the terms of the right-hand side of \eqref{13} are estimated from above by
$$(S_1)\leqslant C\left\|\mathfrak{u}(\cdot,0)\right\|_{H^{1,p}}\leqslant C\left\|\mathfrak{u}_0\right\|_{H^{1,p}};$$
\begin{equation*}
\begin{aligned}
(S_2)&\leqslant C\int_0^{t}(t-s)^{-\frac{1}{2}}\left\|\mathfrak{w}\nabla\mathfrak{v}\right\|_{H^{1,p}}ds\\
&\leqslant Ct^{\frac{1}{2}}\sup_{0\leqslant s\leqslant T}\left\|\mathfrak{w}\right\|_{H^{1,p}}\cdot\sup_{0\leqslant s\leqslant T}\left\|\mathfrak{v}\right\|_{H^{2,p}};
\end{aligned}
\end{equation*}
and
$$(S_3)\leq C\int_0^{t}\left[1+(t-s)^{-\frac{1}{2}}\right]\left\|y\nabla\mathfrak{w}\right\|_{L^{p}}ds\leqslant C\left(t+t^{\frac{1}{2}}\right)\sup_{0\leqslant s\leqslant T}\left\|\mathfrak{w}\right\|_{H^{1,p}}.$$
For $w\in B(\mathcal{M},T)$ this implies
\begin{equation*}
\left\|\mathfrak{u}(\cdot,t)\right\|_{H^{1,p}}\leqslant C\left\|\mathfrak{u}_0\right\|_{H^{1,p}}+CT^{\frac{1}{2}}\mathcal{M}(\mathcal{M}+1)+C\left(T+T^{\frac{1}{2}}\right)\mathcal{M}.
\end{equation*}
If we take $\mathcal{M}>0$ as large as
$$C\left\|\mathfrak{u}_0\right\|_{H^{1,p}}\leqslant\frac{\mathcal{M}}{2},$$
and then take $T>0$ as small as
$$CT^{\frac{1}{2}}\mathcal{M}(\mathcal{M}+1)+C\left(T+T^{\frac{1}{2}}\right)\mathcal{M}\leqslant\frac{\mathcal{M}}{2},$$
it holds that
$$\sup_{0\leqslant s\leqslant T}\|\mathfrak{u}\|_{H^{1,p}}\leqslant \mathcal{M},$$
i.e. $F$ maps $B(\mathcal{M},T)$ into itself.

{\bf Step 2:} $F$ is a contract mapping.

For $w_1,w_2\in B(\mathcal{M},T)$, we can construct $v_1,v_2$ as above and it holds that
\begin{equation*}
\begin{cases}
0=\Delta (v_1-v_2)+(w_1-w_2),&\text{in}\quad B_1(0)\times(0,T),\\
\displaystyle \frac{\partial (v_1-v_2)}{\partial \mathbf{n}}(x,t)=0,&\text{on}\quad\partial B_1(0)\times(0,T).
\end{cases}
\end{equation*}
Similar to the proof of Lemma \ref{lm2}, we can obtain that for $T>0$ small enough
\begin{equation}\label{17}
\sup_{0\leqslant t\leqslant T}\left\|v_1-v_2\right\|_{H^{2,p}(B_1(0))}\leqslant C\sup_{0\leqslant t\leqslant T}\|w_1-w_2\|_{L^p(B_{h(t)}(0))}.
\end{equation}
Let $u_1$, $u_2$ denote the corresponding solution of \eqref{11} respectively. Then the difference $u_1-u_2$ satisfies
\begin{equation*}
\begin{cases}
\displaystyle(\mathfrak{u}_1-\mathfrak{u}_2)_t=\dfrac{1}{h^2(t)}\Delta(\mathfrak{u}_1-\mathfrak{u}_2)\\
\quad-\dfrac{1}{h^2(t)}\nabla\cdot[(\mathfrak{w}_1-\mathfrak{w}_2)\nabla\mathfrak{v}_1]\\
\displaystyle\quad-\frac{1}{h^2(t)}\nabla\cdot[\mathfrak{w}_2\nabla(\mathfrak{v}_1-\mathfrak{v}_2)]+\frac{ h'(t)}{h(t)}y\nabla(\mathfrak{w}_1-\mathfrak{w}_2),&\quad y\in B_1(0),\quad0<t<t_0,\\
\displaystyle\frac{\partial(\mathfrak{u}_1-\mathfrak{u}_2)}{\partial \mathbf{n}}(y,t)+\mathfrak{k}\big(y,t\big)h(t)(\mathfrak{u}_1-\mathfrak{u}_2)(y,t)=0,&\quad y\in\p B_1(0),\quad0<t<t_0,\\
\displaystyle(\mathfrak{u}_1-\mathfrak{u}_2)(y,0)=0,&\quad y\in B_1(0).
\end{cases}
\end{equation*}
We have
\begin{align*}
\|\mathfrak{u}_1-\mathfrak{u}_2\|_{H^{1,p}}\leqslant&\int_0^{t} \left\|e^{\alpha(t,s)\Delta}\frac{1}{h^2(s)}\nabla\cdot\left[(\mathfrak{w}_1-\mathfrak{w}_2)\nabla\mathfrak{v}_1\right]\right\|_{H^{1,p}} ds\\
&\quad+\int_0^{t} \left\|e^{\alpha(t,s)\Delta}\frac{1}{h^2(s)}\nabla\cdot\left[\mathfrak{w}_2\nabla(\mathfrak{v}_1-\mathfrak{v}_2)\right]\right\|_{H^{1,p}}\\
&\quad\quad+\int_0^{t} \left\|e^{\alpha(t,s)\Delta}\frac{ h'(s)}{h(s)}y\nabla(\mathfrak{w}_1-\mathfrak{w}_2)\right\|_{H^{1,p}}ds\\
\leqslant& C\int_0^{t}(t-s)^{-\frac{1}{2}}\left\|(\mathfrak{w}_1-\mathfrak{w}_2)\nabla\mathfrak{v}_1\right\|_{H^{1,p}}ds\\
&\quad+C\int_0^{t}(t-s)^{-\frac{1}{2}}\left\|\mathfrak{w}_2\nabla(\mathfrak{v}_1-\mathfrak{v}_2)\right\|_{H^{1,p}}ds\\
&\quad\quad+C\int_0^{t}\left[1+(t-s)^{-\frac{1}{2}}\right]\|y\nabla(\mathfrak{w}_1-\mathfrak{w}_2)\|_{L^{p}}ds.
\end{align*}
For $w_1,w_2\in B(\mathcal{M},T)$ this implies
\begin{align*}
\|\mathfrak{u}_1-\mathfrak{u}_2\|_{H^{1,p}}\leqslant& CT^{\frac{1}{2}}\sup_{0\leqslant s\leqslant T}\|\mathfrak{v}_1\|_{H^{2,p}}\sup_{0\leqslant s\leqslant T}\|\mathfrak{w}_1-\mathfrak{w}_2\|_{H^{1,p}}\\
&\quad+CT^{\frac{1}{2}}\sup_{0\leqslant s\leqslant T}\|\mathfrak{w}_2\|_{H^{1,p}}\sup_{0\leqslant s\leqslant T}\|\mathfrak{v}_1-\mathfrak{v}_2\|_{H^{2,p}}\\
&\quad\quad+C\left(T+T^{\frac{1}{2}}\right)\sup_{0\leqslant s\leqslant T}\|\mathfrak{w}_1-\mathfrak{w}_2\|_{H^{1,p}}\\
\leqslant& CT^{\frac{1}{2}}(\mathcal{M}+1)\sup_{0\leqslant s\leqslant T}\|\mathfrak{w}_1-\mathfrak{w}_2\|_{H^{1,p}}\\
&\quad+CT^{\frac{1}{2}}\mathcal{M}\sup_{0\leqslant s\leqslant T}\|\mathfrak{w}_1-\mathfrak{w}_2\|_{H^{1,p}}\\
&\quad\quad+C\left(T+T^{\frac{1}{2}}\right)\sup_{0\leqslant s\leqslant T}\|\mathfrak{w}_1-\mathfrak{w}_2\|_{H^{1,p}}.
\end{align*}
Taking $T>0$ as small as
$$CT^{\frac{1}{2}}(\mathcal{M}+1)+CT^{\frac{1}{2}}\mathcal{M}+C\left(T+T^{\frac{1}{2}}\right)\leqslant\frac{1}{2},$$
we obtain $F$ is a contract mapping. This completes the proof of the lemma.
\end{proof}

\section{The existence of the solution locally in time}\label{sec4}

Having at hand the preliminary material collected above, we are now prepared to prove our main
result on local in time existence of solutions. We prove Theorem \ref{th1} by using Lemma \ref{lm2} and Lemma \ref{lm3}.

For each $h(t)\in B$, by Lemma \ref{lm3} we know that there exists a pair
$$u\in X_u^1(T),\quad v\in X_v^{2}(T)\cap Y_v(T),$$
which is the solution of the system \eqref{5} and \eqref{6}.

The following, we will use the contraction mapping principle to show the existence of the free boundary $r=h(t)$.

Set
\begin{equation*}
g(t)=b+\int_0^tk(x,s)ds+\int_0^t\frac{\p v}{\p \mathbf{n}}(x,s)ds,\quad x\in \p B_{h(t)}(0),
\end{equation*}
then we have
\begin{equation}\label{18}
\begin{aligned}
\left|\frac{g(t_1)-g(t_2)}{t_1-t_2}\right|\leq&\left|\frac{1}{t_1-t_2}\right|\int_{t_1}^{t_2}\left|k\big(x,s\big)\right|ds+\left|\frac{1}{t_1-t_2}\right|\int_{t_1}^{t_2}\left|\frac{\p v}{\p \mathbf{n}}(x,s)\right|ds\\
\leqslant&\sup_{0\leqslant t\leqslant T}\|k(\cdot,t)\|_{L^\infty}+C\sup_{0\leqslant t\leqslant T}\|\nabla v(\cdot,t)\|_{L^\infty}\\
\leqslant&C+C\|u(\cdot,0)\|_{H^{2,p}(B_b(0))}.
\end{aligned}
\end{equation}
Let $M_1$ denote the constant at the right hand of (\ref{18}). If $T$ is small enough, then
$$\frac{1}{2}\frac{\min\{b,1-b\}}{T}>M_1.$$
Take $M_0=M_1$ in $B$, it is clear that $B\subset C[0,T]$.

Define $G:~h(t)\rightarrow g(t)$, therefore $G$ maps $B$ into itself. Next we will demonstrate
that $G$ is contractive. Then the fixed point theorem yields that there exist a pair $(u,v)$ and
a curve $\Gamma:~r=h(t)$ which are the solution of (\ref{1}).

For $h_1(t),~h_2(t)\in B$, let $(u_1,v_1),~(u_2,v_2)$ represent the corresponding solutions of \eqref{101} respectively, then
\begin{align*}
&G(h_1)-G(h_2)=g_1(t)-g_2(t)\\
=&\int_0^t\left(k(x,s)-k(y,s)\right)ds+\int_0^t\Big(\frac{\p v_1}{\p \mathbf{n}}(x,s)-\frac{\p v_2}{\p \mathbf{n}}(y,s)\Big)ds\\
=&(I_1)+(I_2),
\end{align*}
where $x\in\p B_{h_1(t)}(0),~y\in\p B_{h_2(t)}(0)$.

The first term $(I_1)$ on the right-hand side are estimated from above by
$$(I_1)\leqslant L\int_0^t\big|h_{1}(s)-h_{2}(s)\big|ds\leqslant Lt\sup_{0\leqslant s\leqslant T}|h_1-h_2|.$$
By Lemma \ref{lm4}, the second term $(I_2)$ on the right-hand side are estimated from above by
\begin{align*}
(I_2)=&\int_0^t\left|\frac{\p \widetilde{v_1}}{\p r}(h_1(s),s)-\frac{\p \widetilde{v_2}}{\p r}(h_2(s),s)\right|ds\\
\leqslant&\frac{c}{n}\int_0^t|h_1(s)-h_2(s)|ds+M\int_0^t\left|\frac{1}{h_1^{n-1}(s)}-\frac{1}{h_2^{n-1}(s)}\right|ds\\
\leqslant& Ct\sup_{0\leqslant s\leqslant T}|h_1-h_2|.
\end{align*}

Hence the fixed point theorem yields that there exist a pair $(u,v)$ and a curve $\Gamma:~r=h(t)\in B$ which are the solution of (\ref{1}).

\section{Proofs of Theorem \ref{th2} and Theorem \ref{th3}}\label{sec5}

In the section, we study the property of the free boundary and give the proofs for Theorem \ref{th2} and Theorem \ref{th3}.

\begin{proof}[The proof of Theorem \ref{th2}]
Let $(u,v)\in(X_u^{1}(T))\times (X_{v}^{2}(T)\cap Y_v(T))$ and a curve $\Gamma:~r=h(t)\in B$ be the solution of (\ref{1}). We have
\begin{equation*}
\begin{cases}
h_t(t)=\widetilde{k}(h(t),t)+\widetilde{v}_r(h(t),t),\\
h(0)=b.
\end{cases}
\end{equation*}
According to the assumption and Lemma \ref{lm4}, we get
\begin{equation*}
\begin{cases}
h_t(t)=(a+M)h(t)-\frac{M}{h^{n-1}(t)},\\
h(0)=b.
\end{cases}
\end{equation*}
Let $s(t)=h^{n}(t)$, then it holds that
\begin{equation*}
\begin{cases}
s_t(t)=n(a+M)s(t)-nM,\\
s(0)=b^n.
\end{cases}
\end{equation*}
A direct calculation gives the following unique solution:
$$s(t)=\left(b^n-\frac{M}{a+M}\right)e^{n(a+M)t}+\frac{M}{a+M}.$$
Hence, we obtain
$$h(t)=\left[\left(b^n-\frac{M}{a+M}\right)e^{n(a+M)t}+\frac{M}{a+M}\right]^{\frac{1}{n}}.$$
Thus, we can deduce all the results in Theorem \ref{th2} easily.
\end{proof}

\begin{proof}[The proof of Theorem \ref{th3}]
If $M=\frac{ab^n}{1-b^n}$, then $h(t)\equiv b$. Thus by the extensive method, we can get the global existence for the solution of the free boundary problem \eqref{101}.

If $M<\frac{ab^n}{1-b^n}$, then the result of Theorem \ref{th2} tells us that $h(t)$ is increasing and $h(t)\to 1$ when $t\to \frac{1}{n(a+M)}\ln\frac{a}{(a+M)b^n-M}$. Since we know that $0\leqslant h(t)\leqslant 1$, which means $h(t)\equiv 1$ when $t\geqslant\frac{1}{n(a+M)}\ln\frac{a}{(a+M)b^n-M}$. Thus, it is similar that
 we can get the global existence for the solution of the free boundary problem \eqref{101}.

Also from Theorem \ref{th2}, we know that if $M>\frac{ab^n}{1-b^n}$, then $0<h(t)<1$ for each $0<t<T^*=\frac{1}{n(a+M)}\ln\frac{M}{(1-b^n)M-ab^n}$. Then, applying the extensive method, we can assert that the problem \eqref{101} has a solution for each $t\in(0,T^*)$.

Notice that $h(t)\rightarrow 0$ as $t\rightarrow T^*$ and in the outside of $B_{h(t)}(0)$, the solution $u\equiv0$. Thus we have
$$u(x,t)\rightarrow \frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}M\delta(x)~\text{as}~t\rightarrow T^*,$$
which implies that the chemotaxis model \eqref{101} collapses in a finite time $t\to T^*=\frac{1}{n(a+M)}\ln\frac{M}{(1-b^n)M-ab^n}$. Theorem \ref{th3} is proved.
\end{proof}

\section{Appendix}

In this section, let us recall the construction of the system. All of the material here can be
found in \cite{C-W2012}.

Let $\Omega\subset \mathbb{R}^{n}$ be a bounded open domain and $\Omega_{0}\subset\Omega$ be an open sub-domain. Assume a population density $u(x,0)$ occupying the domain $\Omega_{0}$, and in the outside of $\Omega_{0}$ the population density $u(x,0)\equiv0$ and the external signal $v$ occupying $\Omega$. For $t>0$, $u(x,t)$ spreads to domain $\Omega_{t}\subset\Omega$. Let $\partial\Omega_{t}$ denote the boundary of $\Omega_{t}$ and $n_{t}$ denote the outer normal vector of $\partial\Omega_{t}$, then $\Gamma_{t}=\partial\Omega_{t}\times(0,T)$ is the free boundary.

The spatial diffusion of species is related to the free boundary of $\Omega_{t}$ at the time $t\geqslant 0$. Observe the flux is increasing with respect to the density of the species, so it would be reasonable to suppose that flux is proportional to the density. Thus we have following flux condition on $\partial\Omega_{t}$,
\begin{equation*}
\begin{split}
-\nabla u\cdot n_{t}=k(x,t)u&\qquad\text{on}\quad\partial\Omega_{t}
\end{split}
\end{equation*}
where $k(x,t)$ is positive function, and $\frac{1}{k(x,t)}>0$ is mass flow ratio.

On the other hand, noticing that the total flux on $\partial\Omega_{t}$ is
\begin{equation*}
\begin{split}
j=-\nabla u\cdot n_{t}+\chi u\nabla v\cdot n_{t},
\end{split}
\end{equation*}
By conservation of population, one has
\begin{equation}\label{1-6}
uv_{n_{t}}=-\nabla u\cdot n_{t}+\chi u\nabla v\cdot n_{t}\quad\text{on}\quad\partial\Omega_{t},
\end{equation}
where $v_{n_{t}}$ is the normal diffusion velocity of $\partial\Omega_{t}$. Assume $\Gamma_{t}:\Phi(x,t)=0$, then
\begin{equation}\label{1-3}
v_{n_{t}}=\left(\frac{dx_{1}}{dt},\frac{dx_{2}}{dt},\cdot\cdot\cdot,\frac{dx_{n}}{dt}\right)\cdot n_{t}=
\left(\frac{dx_{1}}{dt},\frac{dx_{2}}{dt},\cdot\cdot\cdot,\frac{dx_{n}}{dt}\right)\cdot\frac{\nabla\Phi}{|\nabla\Phi|},
\end{equation}
where $x=(x_{1},x_{2},\cdot\cdot\cdot,x_{n})$ and $\nabla=(\frac{\partial}{\partial x_{1}},\frac{\partial}{\partial x_{2}},\cdot\cdot\cdot,\frac{\partial}{\partial x_{n}})$.

Notice that
\begin{equation}\label{1-4}
\frac{\partial\Phi}{\partial t}+\frac{\partial\Phi}{\partial x_{1}}\cdot\frac{dx_{1}}{dt}+\frac{\partial\Phi}{\partial x_{2}}\cdot\frac{dx_{2}}{dt}+\cdot\cdot\cdot+\frac{\partial\Phi}{\partial x_{n}}\cdot\frac{dx_{n}}{dt}=0,
\end{equation}
thus \eqref{1-3} and (\ref{1-4}) give
\begin{equation}\label{1-5}
v_{n_{t}}=-\frac{1}{|\nabla\Phi|}\cdot\frac{\partial\Phi}{\partial t}.
\end{equation}
Combining (\ref{1-6}) and (\ref{1-5}),we obtain
\begin{equation*}
u\frac{\partial\Phi}{\partial t}=\nabla u\cdot\nabla\Phi-\chi u\nabla v\cdot\nabla\Phi\quad\text{on}\quad\partial\Omega_{t}.
\end{equation*}
At last we obtain the conditions of the free boundary $\Gamma_{t}$
\begin{equation*}
\begin{split}
-\nabla u\cdot\frac{\nabla\Phi}{|\nabla\Phi|}=ku&\quad\text{on}\quad\partial\Omega_{t},
\end{split}
\end{equation*}
and
\begin{equation*}
\begin{split}
u\frac{\partial\Phi}{\partial t}=\nabla u\cdot\nabla\Phi-\chi u\nabla v\cdot\nabla\Phi&\quad\text{on}\quad\partial\Omega_{t}.
\end{split}
\end{equation*}
Therefore the full free boundary problem reads
\begin{equation*}
\begin{cases}
u_{t}=\nabla(\nabla u-\chi u\nabla v)&\text{in}\quad\Omega_{t}\times(0,T),\\
u=0&\text{in}\quad\Omega\times(0,T)\setminus\overline{\Omega_{t}}\times(0,T),\\
-\nabla u\cdot\dfrac{\nabla\Phi}{|\nabla\Phi|}=ku&\text{on}\quad\partial\Omega_{t}\times(0,T),\\
u\dfrac{\partial\Phi}{\partial t}=\nabla u\cdot\nabla\Phi-\chi u\nabla v\cdot\nabla\Phi&\text{on}\quad\partial\Omega_{t}\times(0,T),\\
u(x,0)=u_{0}(x)&\text{in}\quad\Omega_{0},\\
0=\Delta v+u-c&\text{in}\quad\Omega\times(0,T),\\
\dfrac{\partial v}{\partial \mathbf{n}}=0&\text{on}\quad\partial\Omega\times(0,T),
\end{cases}
\end{equation*}
where $\Gamma_{t}:\Phi(x,t)=0$ is the free boundary.

\section*{Acknowledgments} The authors of this paper would like to thank the referee for the comments and helpful suggestions.


TITLE: Measuring sum of angles of triangles on different curvature
QUESTION [0 upvotes]: I was reading Richard Feynman’s Six Not so Easy pieces where I came across the concept of measuring geometric shapes drawn on different curved surfaces like the ones with negative and positive curvatures. So out of curiosity I drew a few triangles on a ball and using a protractor measured the sum to be approximately 210 degrees in all cases.
I wonder what would be the sum of angles of the triangle drawn on the interior of a hollow sphere. So, I took a bowl ( not an exact sphere) and drew a triangle on the inside but now I can’t measure it since the protractor won’t fit. Is there any other way to measure it? If not what would be the approximate summation? I also wonder how it would turn out on hyperboloids and more...

REPLY [2 votes]: Say you draw the triangle on the outside of a thin clear glass sphere. Now assume that the lines you drew were inside instead of outside. The angles would be the same, they just may be more easily measurable on the outside of the sphere.


TITLE: Prob. 6, Sec. 3.3, in Kreyszig's Functional Analysis Book: What is $Y^\perp$ if $Y = \ldots \subset \ell^2$?
QUESTION [8 upvotes]: Let $Y$ be the subset of $\ell^2$ given by
  $$ Y \colon= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N}} \in \ell^2 \ \colon \ \xi_{2n} = 0 \ \mbox{ for all } \ n \in \mathbb{N} \ \right\}. $$
  Then $Y$ is a closed subspace of the Hilbert space  $\ell^2$, and so $Y$ is also complete. What is $Y^\perp$?

Here is the relevant definition:

Let $X$ be an inner product space, and let $M$ be a non-empty subset of $X$. Then the orthogonal complement $M^\perp$ of $M$ is defined as follows:
  $$ M \colon= \left\{ \ x \in X \ \colon \  \langle x, y \rangle = 0 \ \mbox{ for all } \ y \in M \ \right\}. $$

My effort:

By definition,
  \begin{align}
& Y^\perp \\
&= \left\{ \ x \in \ell^2 \ \colon \ \langle x, y \rangle = 0 \ \forall y \in Y \ \right\} \\
&= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \left\langle \left( \xi_n \right)_{n \in \mathbb{N} }, \left( \eta_n \right)_{n \in \mathbb{N} } \right\rangle = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in Y \ \right\} \\
&= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty  \xi_n  \overline{  \eta_n} = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in Y \ \right\} \\
&= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty  \xi_n  \overline{  \eta_n} = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \mbox{ such that } \ \eta_{2n} = 0 \ \forall n \in \mathbb{N}  \ \right\} \\
&= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty  \xi_{2n-1}  \overline{  \eta_{2n-1} } = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \mbox{ such that } \ \eta_{2n} = 0 \ \forall n \in \mathbb{N}  \ \right\}.
\end{align}
  My intuition is that
  $$Y^\perp = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall n \in \mathbb{N}  \ \right\}.$$

Is it so? If yes, then how to prove this rigorously?
Another approach:

As $\ell^2$ is a Hilbert space and as $Y$ is a closed subspace of $\ell^2$, so $$ \ell^2 = Y \oplus Y^\perp, $$
  by Theorem 3.3-4 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig.
This means that every $x \in \ell^2$ has a unique representation
  $$ x = y+z, \ \mbox{ where } \ y \in Y \ \mbox{ and } \ z \in Y^\perp. $$
Now  if $x \colon= \left( \xi_n \right)_{n \in \mathbb{N}} \in \ell^2$, then, by definition, the series $\sum \left\lvert \xi_n \right\rvert^2$ of non-negative real numbers converges in $\mathbb{R}$.
Therefore, the sequence
  $$ \left( \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 \right)_{n \in \mathbb{N}} $$ of the partial sums of the series $\sum \left\lvert \xi_n \right\rvert^2$, which is also monotonically increasing, is convergent. Hence this sequence is bounded (from above). That is, for each  $n \in \mathbb{N}$, we have
  $$ \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 \leq \sum_{j=1}^{n+1} \left\lvert \xi_j \right\rvert^2, $$
  and
  $\lim_{n \to \infty} \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2$ exists in $\mathbb{R}$, and
$$ \lim_{n \to \infty} \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 = \sup \left\{ \ \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 \ \colon \ n \in \mathbb{N} \ \right\} < +\infty. $$
Now let $y \colon= \left( \eta_n \right)_{n \in \mathbb{N} }$ be ths sequence of (real or complex) numbers defined as follows: For each $n \in \mathbb{N}$, let
  $$ \eta_{2n-1} \colon= \xi_{2n-1} \ \mbox{ and } \eta_{2n} \colon= 0. $$
  Then the sequence
  $$ \left( \sum_{j=1}^n \left\lvert \eta_j \right\rvert^2 \right)_{n \in \mathbb{N} } $$
  of the partial sums of the series $\sum \left\lvert \eta_j \right\rvert^2$ is given by
  $$ \sum_{j=1}^{2n-1} \left\lvert \eta_j \right\rvert^2 = \sum_{j=1}^n \left\lvert \xi_{2j-1} \right\rvert^2, $$
  and
  $$ \sum_{j=1}^{2n} \left\lvert \eta_j \right\rvert^2 = \sum_{j=1}^n \left\lvert \xi_{2j-1} \right\rvert^2 $$
  for all $n \in \mathbb{N}$. And, this sequence, being a sequence of sums of non-negative real numbers, is also monotonically increasing and  bounded above by the sum of the convergent series $ \sum \left\lvert \xi_n \right\rvert^2$. So the sequence $$ \left( \sum_{j=1}^n \left\lvert \eta_j \right\rvert^2 \right)_{n \in \mathbb{N} } $$ is convergent, which means that the series
  $ \sum  \left\lvert \eta_j \right\rvert^2$ converges. Therefore,
   $y \colon= \left( \eta_n \right)_{n \in \mathbb{N} }$ is in $\ell^2$.
  So from the definition of $Y$ we can conclude that $y \in Y$.
Now let $z \colon= \left( \zeta_n \right)_{n \in \mathbb{N}}$ be the sequence of (real or complex) numbers defined as follows: For each $n \in \mathbb{N}$, let
  $$ \zeta_{2n-1} \colon= 0 \ \mbox{ and } \ \zeta_{2n} \colon= \xi_{2n}. $$
  Then $x = y+z$, and so $z = x-y \in \ell^2$, because $\ell^2$ is closed under addition and scalar multiplication.
Another way to show that $z \colon= \left( \zeta_n \right)_{n \in \mathbb{N}}$ is in $\ell^2$ is to show that the series $\sum \left\lvert \zeta_n \right\rvert^2$ of real numbers converges in $\mathbb{R}$.
Now the sequence
  $$ \left( \sum_{j=1}^n  \left\lvert \zeta_j \right\rvert^2 \right)_{n \in \mathbb{N} }$$
  of the partial sums of the series $\sum \left\lvert \zeta_n \right\rvert^2$ is  given by
  $$ \sum_{j=1}^{2n-1}  \left\lvert \zeta_j \right\rvert^2 = \sum_{j=1}^{2n-2}  \left\lvert \xi_{2j} \right\rvert^2,   $$
  and
  $$ \sum_{j=1}^{2n}  \left\lvert \zeta_j \right\rvert^2 = \sum_{j=1}^{2n}  \left\lvert \xi_{2j} \right\rvert^2   $$
  for all $n \in \mathbb{N}$. Thus this sequence, being a sequence of sums of non-negative real numbers, is monotonically increasing and bounded above by the sum of the series $\sum \left\lvert \xi_n \right\rvert^2$, and  so this sequence is convergent, which implies that $z \in \ell^2$.
Thus
  $$ y = \left( \eta_1, \eta_2, \eta_3, \ldots \right) = \left( \xi_1, 0, \xi_3, 0, \xi_5, 0, \ldots \right), $$
  and
  $$ z = \left( \zeta_1, \zeta_2, \zeta_3, \ldots \right) = \left( 0, \xi_2, 0, \xi_4, 0, \xi_6,  \ldots \right),  $$
  and so
  $$ \langle y, z \rangle = \sum_{n=1}^\infty \eta_n \overline{\zeta_n} = 0, $$
  and so $\langle y, z \rangle = 0$ also. In fact, we can also show that
  $\langle z, v \rangle = 0$ for all $v \in Y$. Thus $z \in Y^\perp$.
But so far I have only been able to show that the subspace
  $$ Z = \left\{  \ \left( \xi_n \right)_{n \in \mathbb{N} }  \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall \ n \in \mathbb{N} \ \right\}$$ is contained in $Y^\perp$.

Is what I have done so far correct, logically sound, and rigorous enough?
If so, then how to complete the above two arguments?
If not, then where do I lack?
P.S.:

In order to show that $Y$ is indeed closed in $\ell^2$, let us suppose that $x \colon= \left( \xi_n \right)_{n \in \mathbb{N} }$ is an element in the closure of $Y$ in $\ell^2$. Then, by Theorem 1.4-6 (a) in Kreyszig, there exists a sequence $\left(y_m \right)_{m \in \mathbb{N} }$, where $y_m \colon= \left( \eta_{mn} \right)_{n \in \mathbb{N} }$ for each $m \in \mathbb{N}$, converging to $x$ in $\ell^2$.
For each $m \in \mathbb{N}$, as $y_m = \left( \eta_{mn} \right)_{n \in \mathbb{N} } \in Y$, so we must have $\eta_{m, 2n} = 0$ for all $n \in \mathbb{N}$, that is, we must have $$ \eta_{m2} = \eta_{m4} = \eta_{m6} = \cdots = 0. $$
Now let us choose an arbitrary natural number $N$. We show that the sequence $\left( \eta_{mN} \right)_{m \in \mathbb{N} }$ converges (in $\mathbb{R}$ or $\mathbb{C}$ as the case may be) to the number $\xi_N$.
As the sequence $\left(y_m \right)_{m \in \mathbb{N} }$ converges in $\ell^2$ to the point $x$, so given any real number  $\varepsilon > 0$, there exists a natural number $M$ such that
  $$ d_{\ell^2} \left( y_m, x \right) < \varepsilon $$
  for all $m \in \mathbb{N}$ such that $m > M$.
  But we have $x = \left( \xi_n \right)_{n \in \mathbb{N} }$, and, for each $m \in \mathbb{N}$, we have $y_m = \left( \eta_{m n} \right)_{n \in \mathbb{N} }$. So we can conclude that
  $$ \sqrt{ \sum_{n = 1}^\infty \left\lvert \eta_{m n} - \xi_n \right\rvert^2 } < \varepsilon $$
  for all $m \in \mathbb{N}$ such that $m > M$.
And, so we obtain
  $$ \left\lvert \eta_{mN} - \xi_N \right\rvert = \sqrt{ \left\lvert \eta_{mN} - \xi_N \right\rvert  } \leq \sqrt{ \sum_{n = 1}^\infty \left\lvert \eta_{mn} - \xi_n \right\rvert^2 } < \varepsilon $$
  for all $m \in \mathbb{N}$ such that $m > M$. Thus it follows that the sequence $\left( \eta_{m N} \right)_{m \in \mathbb{N} }$ converges to $\xi_N$.
Now since for each $N \in \mathbb{N}$, we have
  $$ \xi_N = \lim_{m \to \infty} \eta_{m N}, $$
  and since for each $N \in \mathbb{N}$, we have $\eta_{m, 2N} = 0$,
  therefore we must have
  $$ \xi_{2N} = 0$$
  for all $N \in \mathbb{N}$, which shows that $x = \left( \xi_{n} \right)_{n \in \mathbb{N} }$ is in $Y$.
But this $x$ was an arbitrary element of the closure of $Y$ in $\ell^2$. Thus it follows that $\mbox{Cl}(Y) \subset Y \subset \mbox{Cl}(Y)$. Hence $Y$ is closed.

REPLY [2 votes]: You have the correct intuition.  We should indeed find that
$$
Y^\perp = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall n \in \mathbb{N}  \ \right\}.
$$
With the first approach, you've already showed that
$$
Y^\perp = \left\{ \ x \in \ell^2 \ \colon \ \langle x, y \rangle = 0 \ \forall y \in Y \ \right\}
\\= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty  \xi_{2n-1}  \overline{  \eta_{2n-1} } = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in \ell^2 \right\}
$$
(note that because we reindexed the sum, there is no need to say $\eta_{2n} = 0$ for all $n$).
Now, let $S = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall n \in \mathbb{N}  \ \right\}$.  It's easy to show that $S \subset Y^\perp$: if $\xi_{2n-1} = 0$ for all $n$, then $\sum_{n = 1}^\infty  \xi_{2n-1}  \overline{  \eta_{2n-1} } = 0$ for any $(\eta_n)$.
To show that $Y^\perp \subset S$, take any $(\xi_n) \in Y^\perp$.  define $\eta_n$
$$
\eta_n = \begin{cases}
\xi_n & n \text{ is even}\\
0 & n \text{ is odd}
\end{cases}
$$
We then have
$$
\langle (\xi_n),(\eta_n) \rangle = 0 \implies\\
\sum_{n=1}^\infty |\xi_n|^2 = 0 \implies (\xi_n) \in S
$$
as desired.

You might find that it's easier to do the problem, however, by first showing that the set
$$
\{e_{2n} : n \in \Bbb N\}
$$
(as in section 2.3) forms a Schauder-basis for the space $Y$.


TITLE: joint PDF of max and min of $n$ iid standard uniform random variables
QUESTION [3 upvotes]: Let $U_1, ... U_n$ be iid standard uniform variables.
Let $X = max(U_i)$ and $Y = min(U_i)$.
The goal is to compute the joint PDF of $X, Y$!
I have already computed the PDFs of $X$ and $Y$ separately.
But I am not sure if that is so useful because $X, Y$ are not independent (by definition $X \geq Y$) so it is not correct to just multiply the marginal PDFs of $X$ and $Y$.
I thought about starting from the CDF and taking advantage of the fact that all of the $U_i$'s are independent. So something like: $P(X \leq x, Y \leq y) = P(U_1, ...U_n \leq x, \text{at least one $U_i$ is less than }y)$ and you can already see the problem here -- I don't know how to express the relation for $Y$ in terms of all the $U_i$'s like that. when computing $Y$'s CDF, I did $1 - P(U_1 \geq y, ... U_n \geq y)$ and was able to take advantage of the independence of all the $U_i$'s there. But I'm not sure how to translate that to the joint PDF of $X, Y$.

REPLY [4 votes]: First, on $A=\{0\le y<x\le 1\}$,
\begin{align}
\mathsf{P}(X\le x,Y\le y)&=\mathsf{P}(X\le x)-\mathsf{P}(Y>y,X\le x) \\
&=\mathsf{P}\left(\bigcap_{1\le i\le n}(U_i\le x)\right)-\mathsf{P}\left(\bigcap_{1\le i\le n}(y<U_i\le x)\right) \\
&=x^n-(x-y)^n.
\end{align}
Therefore, on $A$,
$$
f_{X,Y}(x,y)=\frac{d^2}{dxdy}(x^n-(x-y)^n)=n(n-1)(x-y)^{n-2}.
$$


TITLE: Discuss the convergence or the divergence of the series
QUESTION [3 upvotes]: $x_n=($ln $n)^{-p}$ is the nth term of the series I am working on.
I have tried looking at the series for different ranges of $p$.
I also have noted that the ratio test is inconclusive.
My work:
If $p\leq0$, nth term does not go to 0 as n$\rightarrow\infty$. Thus the series diverges.
If $0<p<1$, the series diverges by the limit comparison test with $y_n=\sum\frac{1}{n(lnn)^p}$.
However, I don't know how to tackle the case for when $p\geq1$.
Could you give me an idea?

REPLY [2 votes]: Let $p\gt 1$. If we can show that for large enough $n$ we have $(\ln n)^p \lt n$, then Comparison with $\sum \frac{1}{n}$ will show that our series diverges.
Let $w=\ln n$. We want to show that for large $w$, we have $w^p \lt e^w$, or equivalently that $w\lt e^{w/p}$.  Note that in fact by L'Hospital's Rule we have
$$\lim_{w\to\infty} \frac{w}{e^{w/p}}=0.$$
It follows that if $w$ is large enough, then $w^p\lt e^w$.
If we don't want to use L'Hospital's Rule, we can use the series for $e^x$ to show that for all positive $w$ we have $e^{w/p}\gt 1+\frac{w}{p}+\frac{w^2}{2p^2}\gt \frac{w^2}{2p^2}$.  Thus if $w$ is positive then
$$\frac{w}{e^{w/p}} \lt \frac{2p^2}{w}.$$
This can be used to find an explicit $N$ such that if $n\ge N$ we have $(\ln n)^p \lt n$.


TITLE: Identifying free variables in row-reduction
QUESTION [0 upvotes]: For this matrix:
How do I know what the free variables are? Are they $x_4$ and $x_5$ ?

REPLY [3 votes]: We have, from the bottom up:

$x_5 = 1$
$x_3 = 2 + x_4$
$x_1 = 8 + 3 x_4 + 4 x_2$

So, our free variables are $x_2$ and $x_4$.
You are 'free' to choose any values you want for those, with the observation that two of the solution are dependent on those values.
Try some examples to see what is going on.


TITLE: dimension of subspace $V = \{A \in M_{2\times2}(\mathbb{C}) : \textrm{tr}(A) = 0, A^\textrm{T} = -\bar{A}\}$
QUESTION [3 upvotes]: Let V be the subspace defined by
$V = \{A\in M_{2\times2}(\mathbb{C}) : \textrm{tr}(A) = 0, A^\textrm{T} = -\bar{A}\}$ and note $V$ is a vector space over $\mathbb{R}$ with inner product $<A,B> = -\textrm{tr}(AB)$. Find an orthonormal basis for $V$ over $\mathbb{R}$ with respect to this inner product.
I was able to find one element of this basis
$$ \left[
  \begin{array}{ c c }
     0 & -\frac{1}{2} \\
     \frac{1}{2} & 0
  \end{array} \right]
$$
but I'm not sure what the other could be, seeing as we are finding a basis over $\mathbb{R}$. Further, I'm not sure what the dimension of the subspace ought to be, so I don't know how many elements to find. How would I figure this out?
Any help appreciated!

REPLY [4 votes]: We can write $V$ as
\begin{equation}
V = \{\begin{pmatrix}ia & -b+ic\\  b + ic & -ia\end{pmatrix}| a,b,c \in \mathbb{C}\}
\end{equation}
So we can see that the basis are
\begin{equation}
\{t_1, t_2, t_3\} = \{\frac{1}{{\sqrt{2}}}\begin{pmatrix}i & 0\\  0 & -i\end{pmatrix},\frac{1}{{\sqrt{2}}}\begin{pmatrix}0 & -1\\  1 & 0\end{pmatrix},\frac{1}{{\sqrt{2}}}\begin{pmatrix}0 & i\\  i & 0\end{pmatrix}\}
\end{equation}
It is then easy to see that the dimension is 3. Note that the vector space $V$ you described is also known as the lie algebra $\mathfrak{su}(2)$. The basis can easily be verified to be orthonormal $tr({t_i^\dagger t_j}) = \delta_{ij}$.


TITLE: Finding combination using generating function
QUESTION [4 upvotes]: Q. Find the no of ways of inserting r dolllars using 1 dollar, 2 dollar and 5 dollar tokens,when order doesn't matter and when order doesn't matter.
Ans.
When order doesn't matter..
$(1+x+x^2+x^3+..)(1+x^2+x^4+..)(1+x^5+x^{10}+..)$
Coefficient of $x^r$ in above generating function.
I clearly understand this. But I don't understand the process when order matters. Please somebody explain how (and why) to approach when order matters.

REPLY [6 votes]: Order Does Not Matter
If the order doesn't matter, then the coefficient of $x^n$ in
$$
\begin{align}
&\overbrace{\left(1+x+x^2+x^3+\dots\right)}^{\text{exponent = number of  $\$1$}}\overbrace{\left(1+x^2+x^4+x^6+\dots\right)}^{\text{exponent = 2 $\times$ number of $\$2$}}\overbrace{\left(1+x^5+x^{10}+x^{15}+\dots\right)}^{\text{exponent = 5 $\times$ number of $\$5$}}\\
&=\bbox[5px,border:2px solid #C0A000]{\frac1{1-x}\frac1{1-x^2}\frac1{1-x^5}}\\
&\small=1+x+2x^2+2x^3+3x^4+4x^5+5x^6+6x^7+7x^8+8x^9+10x^{10}+11x^{11}+13x^{12}\\
&\small\,+14x^{13}+16x^{14}+18x^{15}+20x^{16}+22x^{17}+24x^{18}+26x^{19}+29x^{20}+\dots
\end{align}
$$
is the number of ways to choose ones, twos, and fives that sum to $n$.
For example, suppose $n=12$. The case of $3$ $\$1$ tokens and $2$ $\$2$ tokens and $1$ $\$5$ token is counted by the $x^3$ term from the first sum times the $x^4$ term from the second sum times the $x^5$ term from the third sum.
Since the denominator of the generating function is $1-x-x^2+x^3-x^5+x^6+x^7-x^8$, the recursion for the coefficients is
$$
a_n=a_{n-1}+a_{n-2}-a_{n-3}+a_{n-5}-a_{n-6}-a_{n-7}+a_{n-8}
$$

Order Matters
If the order matters, break down the generating function into the sum of the generating functions for a given number of tokens:
$$
\begin{align}
&1+\overbrace{\left(x+x^2+x^5\right)\!\vphantom{\left(x^2\right)^2}}^{\text{one token}}+\overbrace{\left(x+x^2+x^5\right)^2}^{\text{two tokens}}+\overbrace{\left(x+x^2+x^5\right)^3}^{\text{three tokens}}+\dots\\
&=\bbox[5px,border:2px solid #C0A000]{\frac1{1-\left(x+x^2+x^5\right)}}\\
&=1+x+2x^2+3x^3+5x^4+9x^5+15x^6+26x^7+44x^8+75x^9\\
&\,+128x^{10}+218x^{11}+372x^{12}+634x^{13}+1081x^{14}+1843x^{15}+\dots
\end{align}
$$
Under the six tokens term, we would have a product like
$$
\small\left(x+x^2+x^5\right)\left(x+x^2+x^5\right)\left(x+x^2+x^5\right)\left(x+x^2+x^5\right)\left(x+x^2+x^5\right)\left(x+x^2+x^5\right)
$$
There is now a term to specify each possible ordering of the tokens. Using the example above, one such term for $3$ $\$1$ tokens and $2$ $\$2$ tokens and $1$ $\$5$ token would be
$$
\small\overbrace{\left(x\color{#C0C0C0}{+x^2+x^5}\right)}^{\$1}\overbrace{\left(\color{#C0C0C0}{x+}x^2\color{#C0C0C0}{+x^5}\right)}^{\$2}\overbrace{\left(\color{#C0C0C0}{x+x^2+}x^5\right)}^{\$5}\overbrace{\left(x\color{#C0C0C0}{+x^2+x^5}\right)}^{\$1}\overbrace{\left(x\color{#C0C0C0}{+x^2+x^5}\right)}^{\$1}\overbrace{\left(\color{#C0C0C0}{x+}x^2\color{#C0C0C0}{+x^5}\right)}^{\$2}
$$
Since the denominator of the generating function is $1-x-x^2-x^5$, the recursion for the coefficients is
$$
a_n=a_{n-1}+a_{n-2}+a_{n-5}
$$


TITLE: Determining the semigroup and relating it to the generator via the functional calculus
QUESTION [0 upvotes]: In the theory of Hille Yoshida, we have a (definitely real) banach space $L$ and on it a contractive, strongly continuous semigroup of operators. ($T_tT_s=T_{t+s}$ and $T_0=I$.)  We then define the generator $L$ as the strong derivative, defined where it exists, of the semigroup at $0$.  There is no possibility of a functional calculus relationship because we aren't even on a Hilbert space, but we still have that $e^{tL}=T_t$ in the sense that T_t is the strong limit of $(1-tL/n)^{-n}$ as $n\rightarrow \infty$.
Is there a relationship with how people use operator semigroups (such as $e^{it\Delta}$) in PDE?  It seems to me like Hille Yoshida can't apply in the complex world, and luckily nor would it need to because we often have Hilbert spaces in PDE situations, and thus could assign meaning via the functional calculus instead.  Whenever both assignments of meaning are possible, do they agree?
It seems reasonable to conjecture that $T_t$ is fully determined by $T_1$, at least in the Hille Yoshida setting. (I don't know much about how things are set up in PDE.)  Is this true?

REPLY [0 votes]: The semigroup solves Schrodinger or Heat types of equations
$$
           \frac{d}{dt} x(t) = Ax(t),\;\; x(0)=x_{0}
$$
There is a functional calculus that looks like the Laplace transform:
$$
         \mathscr{L}_{A}\{f\} = \int_{0}^{\infty}f(t)T(t)\,dt.
$$
So, for example,
$$
       \mathscr{L}_{A}\{f\star g\} = \mathscr{L}_{A}\{f\}\mathscr{L}_{A}\{g\},
$$
where $f\star g$ is the convolution of $f$ and $g$. This allows you to define powers of the generator under reasonable restrictions. And
$$
          \mathscr{L}_{A}\{e^{-st}\} = (sI-A)^{-1}.
$$
If $f \in L^{1}[0,\infty)$, then the ordinary Laplace transform $\mathscr{L}\{f\}$ is holomorphic in the right half-plane, and you can view $\mathscr{L}_{A}\{f\}$ as $(\mathscr{L}\{f\})(A)$, which is related to the holomorphic functional calculus. I'm not being specific about conditions. You just asked about general ideas. These techniques work nicely for studying operators in Banach spaces, especially general $L^{p}$.


TITLE: proof that $[0,1]$ is compact
QUESTION [0 upvotes]: I tried coming up with a proof of compactness of $[0,1]$ in $\mathbb{R}$ and thought of the following method. please let me know if it is correct or how it could be made more correct.
For any open cover of $[0,1]$ there exists an $\mathbb{\epsilon}$ such that  $[0,\mathbb{\epsilon})$ is contained in one open set of the cover. Using the Least upper bound property of reals, there exists $l_1$ such that $[0,l_1)$ is contained in one open set $U_1$ and for any $l_1<m, [0,m) $ is not contained in one open set.
Now there exists some $l_2$ such that $[l_1,l_2)$ is contained in one open set $U_2$ and for any $l_2<m, [l_2,m) $ is not contained in one open set.
this way one forms an increasing sequence of real numbers ${(0,l_1,l_2..)}$ this sequence must converge to some $l$.
if $l\ne1$, pick an $\mathbb{\epsilon}$ ball around $l,$ such that $  (l-\mathbb{\epsilon}, l+\mathbb{\epsilon})$ is covered in one open set.
 since $l$ was the limit point of sequence ${(0,l_1,l_2..)}$,  $  (l-\mathbb{\epsilon}, l+\mathbb{\epsilon})$ contains all but finitely many $l_n$'s.
let $l_m$ be the first entry in the sequence ${(0,l_1,l_2..)}$ which belongs to $  (l-\mathbb{\epsilon}, l+\mathbb{\epsilon})$,
form a new increasing sequence $(0,l_1,l_2....l_m,l..)$ similar to the previous method. This way one gets a increasing sequence not bounded by any $l_n<1$.
So there exists an increasing sequence of $(0,l_1,l_2...)$ getting arbitarily close to $1$ and their corresponding open set sequence,$(U_1,U_2,...)$  .
since there exists one open set containing $  (1-\mathbb{\epsilon}, 1]$. so all but finitely many $l_n$'s are contained in that open set. So finitely many open sets $(U_1,U_2...U_n)$ along with the open set covering $(1-\mathbb{\epsilon}, 1]$ cover $[0,1]$

REPLY [1 votes]: A proof along your lines is possible, but you have to be more greedy when choosing the $U_k$. Your algorithm could stop short long before the right end is reached, and you would have to restart with no guarantee of success.
We are given a family ${\cal U}$ of open sets $U\subset{\mathbb R}$ that together cover the interval $[0,1]$. Put $x_0:=0$ and choose recursively points $x_k\in\>]0,1]$ as follows:
Assume that $x_0$,$x_1$, $\ldots$, $x_m$ have been choosen.

If $x_m=1$, stop.
If $x_m<1$, put
$$x_{m+1}:=\sup\bigl\{x\leq 1\bigm| \exists\, U\in{\cal U}: \ [x_m,x]\subset U\bigr\}>x_m\ .$$ Since there is an open $U\in{\cal U}$ with $x_m\in U$ we can be sure that $x_{m+1}>x_m$.

I claim that this process will stop at a finite $m$. If not, consider the point $\xi:=\lim_{m\to\infty} x_m\leq1$. There is an open $U\in{\cal U}$ covering $\xi$
and therewith a point $x_m<\xi$. The inequalities $x_m<x_{m+1}<\xi$ then violate the choice of $x_{m+1}$.
We therefore may assume $x_m=1$ for some $m\geq1$. There is an $U_m\in{\cal U}$ covering $x_m=1$ and therewith an interval $J:=\>]1-\delta,1]$, $\>\delta>0$. By definition of $x_m$ we then can find an $U_{m-1}\in{\cal U}$ covering $[x_{m-1},x]$ for an $x\in J$. This $U_{m-1}$ will also cover an interval $J':=\>]x_{m-1}-\delta,x_{m-1}]$, $\>\delta>0$. By definition of $x_{m-1}$ we then can find an $U_{m-2}$, such that $\ldots$, etcetera. Proceeding in this way we obtain a finite sequence of open sets $U_k\in{\cal U}$ $\>(m\geq k\geq0)$ which together cover the interval $[0,1]$.


TITLE: Non-empty closed subset of the complex plane is the spectrum of a normal operator
QUESTION [1 upvotes]: This is an exercise in Chapter 13 of Rudin's Functional Analysis. The question is to show that every non-empty closed subset of $\mathbb{C}$ is the spectrum of some normal (not bounded) operator in a Hilbert space $H$ if dim $H=\infty$.
Does the question ask whether we find a normal operator for every infinite dimensional Hilbert space, or does it ask for the existence of some normal operator on some Hilbert space whose spectrum is equal to the given non-empty closed set?
I would be grateful for hints on this.

REPLY [3 votes]: All separable Hilbert spaces are isomorphic, so it doesn't really matter which Hilbert space you use.  That said, you might as well use one that is convenient: in this case, one that is connected in some way to the closed subset.  That's one hint.  Another hint: multiplication operators are normal.


TITLE: Condition for a submanifold to have a trivial normal bundle
QUESTION [2 upvotes]: Is the following assertion true?
Suppose $M$ is a smooth manifold of dimension $n$, and $S$ an embedded submanifold of dimension $k$. If there is an embedding $S\times \Bbb R^{n-k}\to M$ (with $S\times 0$ corresponding to $S$ in the obvious way), then the normal bundle of $S$ in $M$ is trivial.
I think this can be proved as follows, but I'm not sure about my argument: If $f$ is such an embedding, then the image of $f$ should be an open subset of $M$, so the normal bundle of $S$ in $M$ and the normal bundle of $S$ in $f(S\times \Bbb R^{n-k})$ are the same, but the latter is a trivial bundle.

REPLY [4 votes]: The sub-assertion

... the normal bundle of $S$ in $M$ and the normal bundle of $S$ in $f(S \times \mathbb R^{n-k})$ are the same...

is false as stated literally. They aren't the "same". But it contains a grain of truth which you can then use to turn into a proof.
The key thing to keep in mind is that whenever you are tempted to use the "s" word, you should instead ask yourself: What kind of isomorphism are we talking about here?
The true statement which you should have in place of the one above is

... the derivative $Df : T(S \times \mathbb R^{n-k}) \to TM$ restricts to an isomorphism between the normal bundle of $S \times 0$ in $S \times \mathbb R^{n-k}$ and the normal bundle of $f(S \times 0)$ in $M$...

And now you can go on to prove it.


TITLE: Help with developing equation with combinatorial numbers
QUESTION [0 upvotes]: How can I get from $$k^{4} = a*\binom{k}{1}+b*\binom{k}{2}+c*\binom{k}{3}+d*\binom{k}{4}$$ to
\begin{equation}
 k^{4} = \frac{(24a-12b+8c-6d)k+(12b-12c+11d)k^{2}+(4c-6d)k^{3}+dk^{4}}{4!}
\end{equation}
I know that for example, $$\binom{k}{1}== \frac{k!}{1!*(k-1)!}$$
but I can not figure what steps were exactly made. Thank you very much.

REPLY [1 votes]: Your definition of ${k\choose n}$ is correct, but we can write it in another way which is helpful for this question, namely: $${k \choose n} = \frac{k(k-1)\cdots(k-(n-1))}{n!}.$$
As such, we can then rewrite your equation as: $$k^4 = a\left(\frac{k}{1!}\right) + b\left(\frac{k(k-1)}{2!}\right) + c\left(\frac{k(k-1)(k-2)}{3!}\right) + d\left(\frac{k(k-1)(k-2)(k-3)}{4!}\right).$$
If we factorize the 4! out (to make life easier, and so we get it in the form you are after) that is the same as: $$k^4 = \left(\frac{1}{4!}\right)\left(24ak + 12bk(k-1) + 4ck(k-1)(k-2) + dk(k-1)(k-2)(k-3)\right).$$
Multiply out all the brackets: $$k^4 = \left(\frac{1}{4!}\right)\left(24ak + 12bk^2 - 12bk + 4ck^3 -12ck^2 + 8ck + dk^4 - 6dk^3 + 11dk^2 -6dk\right).$$
Then I'll just rearrange this line, so the coefficients of $k$ are together: $$k^4 = \left(\frac{1}{4!}\right)(24ak - 12bk + 8ck - 6dk + 12bk^2 - 12ck^2 + 11dk^2 + 4ck^3 - 6dk^3 + dk^4).$$
Then just factorize the $k$'s: $$k^4 = \left(\frac{1}{4!}\right)((24a - 12b + 8c -6d)k + (12b - 12c + 11d)k^2 + (4c - 6d)k^3 +dk^4.)$$
Which is what you were after: $$k^4 = \frac{(24a - 12b + 8c -6d)k + (12b - 12c + 11d)k^2 + (4c - 6d)k^3 +dk^4}{4!}.$$


TITLE: Dividing up a fortune
QUESTION [5 upvotes]: I am trying to solve this below problem:

At his death, a millionaire left his 10 children a million dollars in cash, all in $\$100$, $\$10$, $\$1$ bills,10-cent, and 1-cent coins.  Show that there is a way for them to split the fortune into ten stacks of equal value.  (Note that this would not be true if there were $\$3$ bills).

So the best I can do is set this up as solving an equation in five variables. I want to find $a,b,c,d,e$ so that
$$
10^6 = a \cdot 100 + b \cdot 10 + c \cdot 1 + d \cdot 0.1 + e \cdot .01.
$$
I can try to take congruences mod $10$. The LHS is clearly $0$, and any multiple of $10$ is certainly $0$, but $\frac{1}{10}$ is not even defined in $\mathbb{Z}/10\mathbb{Z}$, since that would amount to division by zero, so this strategy does not work.
Can anyone give me a hint on how to get started?

REPLY [2 votes]: First, prove that the number of 1-cent coins is divisible by 10. This follows from the fact that "cents place" of \$1 million is zero. Then, WLOG, exchange the 1-cent coins for an equivalent number of 10-cent coins.
Then, prove that the number of 10-cent coins is divisible by 10. This follows from the fact that there are no 1-cent coins, and the "10-cent place" of \$1 million is zero. Then, exchange the 10-cent coins for an equivalent number of \$1 coins.
...
Then, prove there must be a multiple of 10 \$100 bills. This is because all the currency now is in \$100 bills.