text:
\begin{document}
\setlength{\abovedisplayshortskip}{2mm}
\setlength{\belowdisplayshortskip}{2mm}
\setlength{\abovedisplayskip}{2mm}
\setlength{\belowdisplayskip}{2mm}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\newtheorem{theoremalph}{Theorem}
 \renewcommand\thetheoremalph{\Alph{theoremalph}}

\newcommand{\bta}{\begin{theoremalph}}
\newcommand{\eeta}{\end{theoremalph}}
\newcommand{\bth}{\begin{theorem}}
\newcommand{\eeth}{\end{theorem}}
\newcommand{\ble}{\begin{lemma}}
\newcommand{\ele}{\end{lemma}}
\newcommand{\bco}{\begin{corollary}}
\newcommand{\eco}{\end{corollary}}
\newcommand{\bde}{\begin{definition}}
\newcommand{\ede}{\end{definition}}
\newcommand{\bpr}{\begin{proposition}}
\newcommand{\epr}{\end{proposition}}
\newcommand{\bre}{\begin{remark}}
\newcommand{\ere}{\end{remark}}
\newcommand{\beg}{\begin{example}}
\newcommand{\eeg}{\end{example}}

\newcommand{\beq}{\begin{equation}}
\newcommand{\eeq}{\end{equation}}
\newcommand{\ben}{\begin{equation*}}
\newcommand{\een}{\end{equation*}}
\newcommand{\beqn}{\begin{eqnarray}}
\newcommand{\eeqn}{\end{eqnarray}}
\newcommand{\be}{\begin{eqnarray*}}
\newcommand{\ee}{\end{eqnarray*}}
\newcommand{\ban}{\begin{align*}}
\newcommand{\ean}{\end{align*}}
\newcommand{\bal}{\begin{align}}
\newcommand{\eal}{\end{align}}
\newcommand{\bad}{\aligned}
\newcommand{\ead}{\endaligned}
\newcommand{\lan}{\langle}
\newcommand{\ran}{\rangle}

\newcommand{\na}{\nabla}
\newcommand{\vp}{\varphi}
\newcommand{\La}{\Lambda}
\newcommand{\la}{\lambda}
\newcommand{\Om}{\Omega}
\newcommand{\ta}{\theta}
\newcommand{\fr}{\frac}
\newcommand{\iy}{\infty}
\newcommand{\ve}{\varepsilon}
\newcommand{\pa}{\partial}
\newcommand{\al}{\alpha}
\newcommand{\mr}{\mathbb{R}^n}
\newcommand{\bu}{\bullet}
\newcommand{\si}{\sigma}
\newcommand{\Vol}{{\rm Vol}}
\newcommand{\Div}{{\rm div}}
\newenvironment{sequation}{\begin{equation}\small}{\end{equation}}
\newenvironment{tequation}{\begin{equation}\tiny}{\end{equation}}

\title
{$\mathcal{W}$-Entropy formulae  and  differential Harnack estimates for  porous medium equations on Riemannian manifolds}
\author{Yu-Zhao Wang}
\address{School of Mathematical Sciences, Shanxi University, Taiyuan, 030006, Shanxi, China}
\email{wangyuzhao@sxu.edu.cn}
\thanks{The author is supported by the National Science Foundation of China(NSFC, 11701347).}

\maketitle
\maketitle \numberwithin{equation}{section}
\maketitle \numberwithin{theorem}{section}
\setcounter{tocdepth}{2}
\setcounter{secnumdepth}{2}

\begin{abstract}
In this paper, we prove Perelman type $\mathcal{W}$-entropy formulae and global differential Harnack estimates  for positive solutions to porous medium equation on  closed Riemannian manifolds with Ricci curvature bounded below. As applications, we derive Harnack inequalities and Laplacian estimates.

\vspace{2mm}
\textbf{Mathematics Subject Classification (2010)}. Primary 58J35,	35K92; Secondary 35B40,35K55

\textbf{Keywords}. Porous medium equation, Perelman type entropy formula, differential Harnack  estimates, Bakry-\'Emery Ricci curvature.
\end{abstract}

\section{Introduction and main results}

Monotonicity formula and differential Harnack inequality are two  important tools in geometric analysis. The more spectacular one is the entropy monotonicity formula discovered by G.Perelman \cite{P} and related differential Harnack inequality for the conjugate heat equation under the Ricci flow. More precisely, let $(M,g(t))$ be the closed $n$-dimensional Riemannian manifolds along the Ricci flow and $(g(t), f(t), \tau(t))$ be a solution to conjugate heat equation coupled with Ricci flow
\begin{equation}\label{adjoint}
\partial_tg=-2{\rm Ric},\quad\partial_t f=-\Delta f+|\nabla f|^2+R+\frac n{2\tau},\quad\partial_t\tau=-1.
\end{equation}
Perelman \cite{P} introduced the following $\mathcal{W}$-entropy
$$
\mathcal{W}(g,f,\tau):=\int_M\Big(\tau(R+|\nabla f|^2)+f-n\Big)\frac{e^{-f}}{(4\pi\tau)^{\frac n2}}\,dV
$$
and proved its monotonicity
\begin{equation}\label{Pentropy}
\frac{d}{dt}\mathcal{W}(g,f,\tau)=2\tau\int_M\Big|R_{ij}+\nabla_i\nabla_jf-\frac1{2\tau}g_{ij}\Big|^2\frac{e^{-f}}{(4\pi\tau)^{\frac n2}}\,dV\ge0,
\end{equation}
where $\tau>0$ and $\int_M{(4\pi\tau)^{-\frac n2}}e^{-f}dV=1$. The static point of $\mathcal{W}$-entropy is called gradient shrinking Ricci soliton
\begin{equation}\label{soliton}
R_{ij}+\nabla_i\nabla_jf=\frac1{2\tau}g_{ij},
\end{equation}
which can be viewed as a model with singularity when studying the singularity formation of solutions of the Ricci flow.  Moreover, when $H(x,t)={(4\pi\tau)^{-\frac n2}}e^{-f}$ is a fundamental solution to the conjugate heat equation \eqref{adjoint}, Perelman's differential Harnack inequality holds \cite{P,NiLYH}
\begin{equation}\label{PLYH}
v_H:=\Big(\tau(R+2\Delta f-|\nabla f|^2)+f-n\Big)H\le0.
\end{equation}

A feature of Perelman's entropy monotonicity formula is that it holds at any dimension and without assumption of curvature condition. After Perelman's work, an interesting direction is that of finding entropy monotonicity formulae for other geometric evolution equations.  There were some development in this direction, L. Ni \cite{Nientropy}  derived the entropy monotonicity formula for the linear heat equation on Riemannian manifolds with nonnegative Ricci curvature,
\begin{equation}\label{Nientropy}
\frac{d}{dt}\mathcal{W}(f,\tau)=-2\tau\int_M\left(\Big|\nabla_i\nabla_jf-\frac1{2\tau}g_{ij}\Big|^2+R_{ij}f_if_j\right)u\,dV,
\end{equation}
where $u=(4\pi\tau)^{-\frac n2}e^{-f}$ is a positive solution to the heat equation $\partial_tu=\Delta u$ with $\int_Mu\,dV=1$, $\frac{d\tau}{dt}=1$ and $\mathcal{W}(f,\tau)$ is defined by
$$
\mathcal{W}(f,\tau):=\int_M\Big(\tau|\nabla f|^2+f-n\Big)u\,dV.
$$

When the Ricci curvature of $M$ is bounded below, Li-Xu \cite{LX} established some Perelman-Ni type entropy formulae for the linear heat equation and proved their monotonicity.

A natural question is how to establish the entropy formula for the nonlinear equation
on Riemannian manifolds.  Kotschwar-Ni \cite{KoNi} and Lu-Ni-Vazquez-Villani \cite{LNVV} obtained the entropy monotonicity formula for the $p$-Laplaican heat equation and the porous medium equation on compact Riemannian manifolds with nonnegative Ricci curvature respectively.
In \cite{WC2}, the author proved the entropy monotonicity formula for positive solution to the doubly nonlinear diffusion equation on the closed Riemannian manifolds with nonnegative Ricci curvature.

The first step in the study of the $\mathcal{W}$-entropy on Riemannian
manifolds with negative lower bound of Ricci curvature
is to find the suitable quantity to define the $\mathcal{W}$-entropy.
In \cite{LiLi2, LiLi3}, S. Li and X.-D. Li gave a new idea to introduce  the correct quantity for the definition of the $\mathcal{W}$-entropy for the heat equation of the Witten Laplacian on Riemannian manifolds with lower bound of the  infinity dimensional Bakry-Emery Ricci curvature. Motivated by their works, we can obtain the Perelman type $\mathcal{W}$-entropy monotonicity formula for the porous medium equation on closed Riemannian manifolds with Ricci curvature (or Bakry-Emery Ricci curvature) bounded below.

\begin{theorem}\label{KPMEentropy}
Let $(M,g)$ be a closed Riemannian manifold with  Ricci curvature bounded below by $-K(K\ge0)$. Suppose~$u$~be a smooth positive solution to the porous medium equation
\begin{equation}\label{PME}
\partial_tu=\Delta u^{\gamma}
\end{equation}
 and $v=\frac{\gamma}{\gamma-1}u^{\gamma-1}$ the pressure function. For any $\gamma>1$,
define the   Perelman-type $\mathcal{W}$-entropy
\begin{equation}\label{WKentropy}
\mathcal{W}_K(v,t):=\sigma_K\beta_K\int_M\left[\gamma\frac{|\nabla v|^2}{v}-\left(\frac{1}{\beta_K}+\frac{\dot{\sigma}_K}{\sigma_K}\right)\right]vu\,dV,
\end{equation}
Then we have
\begin{align}\label{WKDentropy}
\frac{d}{dt}\mathcal{W}_K(v,t)
\le&-2\sigma_K\beta_K\int_M (\gamma-1) \left|\nabla_i\nabla_jv+\frac{\eta_K}{n(\gamma-1)}g_{ij}\right|^2vu\,dV\\
&-2\sigma_K\beta_K\int_M\left[(\gamma-1)({\rm Ric}+Kg)(\nabla v,\nabla v )+((\gamma-1)\Delta v+\eta_K)^2\right]vu\,dV,\notag
\end{align}
where $a=\frac{n(\gamma-1)}{n(\gamma-1)+2}$, $\kappa=K\sup\limits_{M\times(0,T]}u^{\gamma-1}$, $\sigma_K=\left(\frac{e^{2{\kappa}t}-1}{2\kappa}\right)^{a}$, $\beta_K=\frac{\sinh(2\kappa t)}{2\kappa}$ and  $\eta_K=\frac{2a\kappa}{1-e^{-2\kappa t}}$.
Moreover, if ${\rm Ric}\ge-Kg$ for $K\ge0$, then  $\mathcal{W}_{K}(v,t)$ is monotone decreasing along the porous medium equation \eqref{PME}.
\end{theorem}
\begin{remark}
When $K=0$, $\sigma_0=t^{a}$, $\beta_0=t$ and $\eta_0=\frac{a}t$, the entropy formula \eqref{WKDentropy} in Theorem \ref{KPMEentropy} reduced the result of  Lu-Ni-Vazquez-Villani in \cite{LNVV},
\begin{align}\label{PMEentropy2}
\frac{d}{dt}\mathcal{W}_0(v,t)=&-2(\gamma-1)t^{a+1}\int_M\left[\Big|\nabla_i\nabla_jv
+\frac{a}{n(\gamma-1)t}g_{ij}\Big|^2+ {\rm Ric}(\nabla v,\nabla v)\right]vu\,dV\notag\\
&-2t^{a+1}\int_M\left[(\gamma-1)\Delta v+\frac {a}t\right]^2 vu\,dV,
\end{align}
where
\begin{equation}\label{PMEentropy1}
\mathcal{W}_0(v,t)= t^{a+1}\int_M\left(\gamma\frac{|\nabla v|^2}{v}-\frac{a+1}t\right)vu\,dV.
\end{equation}
\end{remark}

A weighted Riemannian manifold is a Riemannian manifold $(M,g)$ with a smooth measure $d\mu:= e^{-f}\,{dV}$, denoted by $(M,g, d\mu)$, where $f$ is a smooth function on $M$. The weighted Riemannian manifold carries a natural analog of the Ricci curvature, that is, the $m$-Bakry-\'Emery Ricci curvature, which is defined as
$$
{\rm Ric}_f^m:={\rm Ric}+\nabla\nabla f-\frac{\nabla f\otimes \nabla f}{m-n},\quad (n\leq m\leq \infty).
$$
In particular, when $m=\infty$, ${\rm Ric}^{\infty}_f={\rm Ric}_f:={\rm Ric}+\nabla\nabla f$ is the classical Bakry--\'Emery Ricci curvature, which was introduced in the study of diffusion processes and functional inequalities including the Poincar\'e and the logarithmic Sobolev inequalities(see \cite{BGL} for a comprehensive introduction),  then it is extensively investigated in the theory of the Ricci flow (for example, the gradient shrinking Ricci soliton equation \eqref{soliton} is precisely ${\rm Ric}_f =\frac{1}{2\tau} g$);  when $m=n$ if and only if $f$ is a constant function. There is also a natural analog of the Laplacian, namely, the so-called weighted Laplacian, denoted by $\Delta_f=\Delta-\nabla f\cdot\nabla$, which is a self-adjoint operator in $L^2(M, d\mu)$.

A nature question is which contents can be generalized to the weighted manifolds and what are the advantages and applications for the weighted case. There are a lot of progress in this direction, for instance, gradient estimates and Liouville theorems for symmetric diffusion operators $\Delta_f$ \cite{LiXD1}, some comparison geometry for the Bakry-Emery Ricci tensor \cite{WW} etc..  In view of the innovation and importance of above-mentioned Perelman $\mathcal{W}$-entropy formulae,  one tried to get the entropy formula for the weighted case, the first work is established by X.-D. Li for the weighted heat equation $\partial_t u=\Delta_fu$ in \cite{LiXD2,LiXD3},
  \begin{align}\label{Lientropy}
\frac{d}{dt}\mathcal{W}_f(v,\tau)=&-2t\int_M\left(\Big|\nabla_i\nabla_jv-\frac1{2t}g_{ij}\Big|^2+{\rm Ric}^m_f(\nabla v,\nabla v)\right)u\,d\mu\notag\\
&-\frac{2t}{m-n}\int_M\Big(\nabla f\cdot\nabla v+\frac{m-n}{2t}\Big)^2u\,d\mu,
\end{align}
where the $\mathcal{W}$-entorpy is defined by
$$
\mathcal{W}_f(v,t):=\int_M\Big(t|\nabla v|^2+v-m\Big)u\,d\mu,\quad u = \frac{e^{-v}}{(4\pi t)^{m/2}}.
$$
In particular, if the $m$-Bakry-\'Emery Ricci curvature is nonnegative, then $\mathcal{W}_f(v,\tau)$ is monotone decreasing along the weighed heat equation. When $m=n$, $f=const.$, \eqref{Lientropy} reduces to \eqref{Nientropy}.

In \cite{LiLi}, when $n\le m\in \mathbb{N}$,  S. Li and X.-D. Li gave a new proof of the $\mathcal{W}$-entropy formula \eqref{Lientropy} by using of the warped product approach and a natural geometric interpretation for the third term in \eqref{Lientropy}. Moveover, they extended the $\mathcal{W}$-entropy formula to the weighted heat equation on the weighted compact Riemannian manifolds with time dependent metrics and potentials under satisfying the curvature dimensional condition $CD(K,m)$ for some negative constant $K$. In \cite{LiLi2}, S. Li and X.-D. Li obtained the $\mathcal{W}$-entropy formula  on compact Riemannian manifolds with $(K,m)$-super Perelman Ricci flow, where $K\in \mathbb{R}$ and $m\in [n,\infty]$ are two constants. In their paper \cite{LiLi4}, S. Li and X.-D. Li pointed out that there is an essential and deep connection between the definition of the $W$-entropy and the Hamilton type Harnack inequality on complete Riemnnian manifolds with the $CD(K,m)$-condition. Recently, they \cite{LiLi3} introduced Perelman's $\mathcal{W}$-entropy  along geodesic flow on the Wasserstein space over Riemannian manifolds and proved a rigidity theorem. For further related study, see \cite{LiLi3, LiLi5,LiLi6}.

For the nonlinear case,  combining the analogous methods in \cite{KoNi}, \cite{LNVV} and \cite{LiXD2}, Wang-Yang-Chen \cite{WYC} and Huang-Li \cite{HL} obtained the entropy monotonicity formulae for the weighted $p$-Laplacian heat equation and the weighted porous medium equation with nonnegative $m$-Bakry-\'Emery Ricci curvature respectively. In \cite{WYZ}, the author expanded the entropy formulae of Kotschwar-Ni \cite{KoNi} and Wang-Yang-Chen \cite{WYC} to the case  where the $m$-dimensional
Bakry-\'Emery Ricci curvature is bounded from below.

 Inspired by above works, we can obtain entropy monotonicity formula for positive solution to the weighted porous medium equation with $m$-Bakry-\'Emery Ricci curvature bounded below,  which is a natural generalization of  Theorem \ref{KPMEentropy}.

\begin{theorem}\label{WKPMEentropy}
Let~$(M,g,d\mu)$ be a closed weighted Riemannian manifold with  $m$-Barky-Emery-Ricci curvature bounded below,  i.e. ${\rm Ric}_f^m\ge-Kg$, $K\ge0$. Suppose~$u$~be a smooth positive solution to equation \eqref{WPME}
\begin{equation}\label{WPME}
\partial_tu=\Delta_f u^{\gamma}
\end{equation}
 and $v=\frac{\gamma}{\gamma-1}u^{\gamma-1}$ the pressure function. For any $\gamma>1$,
the  Perelman-type $\mathcal{W}$-entropy is defined by
\begin{equation}\label{WKentropy}
\mathcal{W}_K(v,t):=\bar{\sigma}_K\bar{\beta}_K\int_M\left[\gamma\frac{|\nabla v|^2}{v}-\left(\frac{1}{\bar{\beta}_K}+\frac{\dot{\bar{\sigma}}_K}{\bar{\sigma}_K}\right)\right]vu\,d\mu,
\end{equation}
Then we have
\begin{small}
\begin{align}\label{WKPMEent}
&\frac{d}{dt}\mathcal{W}_K(v,t)
\le-2\bar{\sigma}_K\bar{\beta}_K\int_M (\gamma-1) \left[\left|\nabla_i\nabla_jv+\frac{\bar{\eta}_K}{n(\gamma-1)}g_{ij}\right|^2+({\rm Ric}^m_f+Kg)(\nabla v,\nabla v )\right]vu\,d\mu\notag\\
&-2\bar{\sigma}_K\bar{\beta}_K\int_M\left[\Big((\gamma-1)\Delta_f v+\bar{\eta}_K\Big)^2+\frac{\gamma-1}{m-n}\left(\langle\nabla v,\nabla f\rangle-(m-n)\frac{\bar{\eta}_K}{m(\gamma-1)}\right)^2\right]vu\,d\mu.
\end{align}
\end{small}
where $\bar{a}=\frac{m(\gamma-1)}{m(\gamma-1)+2}$, ${\kappa}=K\sup\limits_{M\times(0,T]}u^{\gamma-1}$, $\bar{\sigma}_K=\left(\frac{e^{2{\kappa}t}-1}{2\kappa}\right)^{\bar{a}}$, $\bar{\beta}_K=\frac{\sinh(2{\kappa}t)}{2{\kappa}}$, $\bar{\eta}_K=\frac{2\bar{a}{\kappa}}{1-e^{-2{\kappa}t}}$.
Moreover, if ${\rm Ric}_f^m\ge-Kg$ for $K\ge0$, then  entropy $\mathcal{W}_{K}(v,t)$ is monotone decreasing along the weighted porous medium equation \eqref{WPME}.
\end{theorem}

In the second part of this paper, we study the differential Harnack inequality for the porous medium equation. Such inequality was first proved by  Li and Yau \cite{LY} for solutions to the heat equation on Riemannian manifolds, i.e. if $u$ is a positive solution to $\partial_tu=\Delta u$ on $M$ with nonnegative Ricci curvature, then
\begin{equation}\label{LY}
\frac{|\nabla u|^2}{u^2}-\frac{u_t}{u}\le\frac{n}{2t}.
\end{equation}
Later on, it has been extensively investigated in other geometric evolution equations, such as Hamilton's estimates for the Ricci flow and the mean curvature flow, the corresponding results for the K\"ahler Ricci flow and the Gauss curvature flow were proved by H. Cao and B. Chow, Perelman's differential Harnack inequality \eqref{PLYH} for the conjugate heat equaiton under Ricci flow, etc.. For more progress in this direction, see the survey \cite{Ni} and references therein.

It is a long time question of proving the sharp Li-Yau Harnack inequality for positive solution of the heat equation on Riemannian manifolds with negative Ricci curvature bound (see P.393 in \cite{CLN}). There are many works in this direction. Let ${\rm Ric}\ge-K$ for some $K>0$, the original result of Li-Yau \cite{LY} is,
$$
\frac{|\nabla u|^2}{u^2}-\alpha\frac{u_t}{u}\le\frac{\alpha^2}{2(\alpha-1)}nK+\alpha^2\frac{n}{2t},
$$
where $\alpha$ is a constant and  $\alpha>1$.
In \cite{Ham93}, Hamilton  proved
$$
\frac{|\nabla u|^2}{u^2}-e^{2Kt}\frac{u_t}{u}\le e^{4Kt}\frac{n}{2t}.
$$
Recently, Li-Xu \cite{LX} obtained some new Li-Yau type estimates,
\begin{align}\label{LiXu}
\frac{|\nabla u|^2}{u^2}-\left(1+\frac{2}3Kt\right)\frac{u_t}{u}\le&\frac{n}{2t}+\frac{nK}{2}\Big(1+\frac13Kt\Big),\\
\frac{|\nabla u|^2}{u^2}-\left(1+\frac{\sinh (Kt)\cosh (Kt)-Kt}{\sinh^2( Kt)}\right)\frac{u_t}{u}\le&\frac{nK}{2}\Big(1+\coth(Kt)\Big).\notag
\end{align}
In \cite{LiLi,LiLi3}, S. Li and X.-D. Li proved an analog of the Harnack inequality \eqref{LiXu} for the weighted heat equation on weighted complete Riemannian manifolds with  ${\rm Ric}^m_f\ge-K$ . B.Qian \cite{QianB} gave a further generalization under the  proper assumptions of $\alpha(t)$ and $\varphi(t)$,
\begin{equation}\label{QLY}
\frac{|\nabla u|^2}{u^2}-\alpha(t)\frac{u_t}{u}\le\varphi(t).
\end{equation}
There are some results on Li-Yau, Hamilton and Li-Xu type differential Harnack estimates for the porous medium equation \cite{LNVV, HHL, WC1}, the $p$-heat equation \cite{KoNi,WYC,WYZ} and the doubly nonlinear diffusion equation \cite{WC2} on Riemannian manifolds with Ricci curvature bounded below.

In the second part of this paper, we obtain  Qian type differential Harnack estimate for the porous medium  equation on the closed Riemannian manifold with Ricci curvature bounded below, which can also be generalized to the case of the weighted Riemanian manifolds with the $m$-Bakry-\,Emery Ricci curvature bound below.

Now we first give two assumptions, let $\sigma(t)\in C^1(M)$ and satisfy (See \cite{QianB}).
\begin{description}
  \item[(A1)] For a $t>0$, $\sigma(t)>0$, $\sigma'(t)>0$, $\lim\limits_{t\to0}\sigma(t)=0$ and $\lim\limits_{t\to0}\frac{\sigma(t)}{\sigma'(t)}=0$;
  \item[(A2)] For any $T>0$, $\frac{(\sigma')^2}{\sigma}$ is continuous and integrable on the interval $[0,T)$.
\end{description}

\begin{theorem}\label{pmeGK}
Let $(M^n,g)$ be a closed Riemannian manifold with ${\rm Ric}\ge-Kg$ for $K\ge0$. Suppose  $u(x,t)$ be a positive solution to equation \eqref{PME} and $v$ the  pressure function. For any $\gamma>1$, we have
\begin{equation}\label{PMELYH1}
\frac{|\nabla v|^2}{v}-\alpha(t)\frac{v_t}{v}\le\varphi(t).
\end{equation}
Here
\begin{align}\label{PMEalphavarphi}
\alpha(t)=1+\frac{2{\kappa}}{\sigma}\int^t_0\sigma(s)ds,\quad
\varphi(t)={\kappa}a
+\frac{{\kappa}^2a}{\sigma}\int_0^t\sigma(s)ds+\frac{a}{4\sigma}\int^t_0\frac{(\sigma'(s))^2}{\sigma(s)}ds,
\end{align}
and
$\sigma(t)$ satisfies the assumptions $\mathbf{(A1)}$ and $\mathbf{(A2)}$,  $a=\frac{n(\gamma-1)}{n(\gamma-1)+2}$ and  ${\kappa}=K\sup\limits_{M\times(0,T]}u^{\gamma-1}$.
\end{theorem}

\begin{remark}
\begin{enumerate}
      \item When $K=0$, $\alpha(t)=1$, $\sigma(t)=t^2$ and $\varphi(t)=\frac at$, the estimate \eqref{PMELYH1} in Theorem \ref{pmeGK} reduces to the Aronson-Benilan's estimate in \cite{LNVV}, i.e.
          $$
          \frac{|\nabla v|^2}{v}-\frac{v_t}{v}\le\frac at.
          $$
      \item When $K>0$, $\sigma(t)=t^2$,  $\alpha(t)=1+\frac{2\kappa t}3$ and $\varphi(t)=\frac at+a\kappa\left(1+\frac{\kappa t}3\right)$, the estimate \eqref{PMELYH1} in Theorem \ref{pmeGK} reduces to the Li-Xu type estimate in \cite{HHL}
          $$
          \frac{|\nabla v|^2}{v}-\left(1+\frac{2\kappa t}3\right)\frac{v_t}{v}\le\frac at+a\kappa\left(1+\frac{\kappa t}3\right).
          $$
      \item When $K>0$, $\sigma(t)=\sinh^2(\kappa t)$,  $\alpha(t)=1+\frac{\sinh(\kappa t)\cosh(\kappa t)-\kappa t}{\sinh^2(\kappa t)}$ and $\varphi(t)=a\kappa(1+\coth(\kappa t))$, the estimate \eqref{PMELYH1} in Theorem \ref{pmeGK} reduces to another Li-Xu type estimate in \cite{HHL}
          $$
          \frac{|\nabla v|^2}{v}-\left(1+\frac{\sinh(\kappa t)\cosh(\kappa t)-\kappa t}{\sinh^2(\kappa t)}\right)\frac{v_t}{v}\le a\kappa(1+\coth(\kappa t)).
          $$
         \item All of above gradient estimates are valid for the weighted case by similar method.
    \end{enumerate}
\end{remark}

There are some applications for differential Harnack estimates, including Harnack inequalities and Laplacian estimates.
Integrating the estimate \eqref{PMELYH1} in Theorem \ref{pmeGK} along a
minimizing path between two points, we can get the Harnack inequalities for positive solutions to the porous medium equation \eqref{PME}.

\begin{corollary}\label{Harnack}
For any $(x_1,t_1)$ and $(x_2,t_2)$ with $0<t_1\le t_2<T$, we have
$$
v(x_1,t_1)-v(x_2,t_2)\le v_{max}\int^{t_2}_{t_1}\frac{\varphi(t)}{\alpha(t)}dt
+\frac{1}{4}\frac{d(x_2,x_1)^{2}}{(t_2-t_1)^{2}}\int^{t_2}_{t_1}\alpha(t)dt
$$
and
$$
\frac{v(x_1,t_1)}{v(x_2,t_2)}\le\exp\left(\int^{t_2}_{t_1}\frac{\varphi(t)}{\alpha(t)}dt
+\frac{1}{4}\frac1{v_{max}}\frac{d(x_2,x_1)^{2}}{(t_2-t_1)^{2}}\int^{t_2}_{t_1}\alpha^{\frac1{p-1}}(t)dt\right),
$$
where $v_{max}=\sup_{M\times[0,T)}v$.
\end{corollary}

\begin{corollary}\label{pfLaEst}
Under the same assumptions as in Theorem \ref{pmeGK}, define $\beta(t)$ by $\frac{\alpha(t)-1}{\alpha(t)}=(\gamma-1)(\beta(t)-1)$ for $\alpha(t)>1$ and such that $1<\beta(t)<\frac{\gamma}{\gamma-1}$. Assume that ${\rm Ric}\ge-K$ for some $K\ge0$, then
\begin{equation}\label{pfLaE}
\Delta(v^{\beta})\ge-\frac{\beta}{\alpha(\gamma-1) } v_{max}^{\beta-1}\varphi(t),
\end{equation}
where $v_{max}:=\sup_{M\times[0,T]}v$.
\end{corollary}

This paper is organized as follows. In Section 2, we establish some evolution equations and then prove the entropy monotonicity formulae,  i.e. Theorem \ref{KPMEentropy} and Theorem \ref{WKPMEentropy}. In section 3, we obtain the Qian type differential Harnack estimate. Finally, Harnack inequalities and Laplacian estimates are derived as applications.

\section{Entropy monotonicity formulae}

Let~$(M,g)$ be a closed Riemannian manifold. Suppose~$u$~be a smooth solution to \eqref{PME} and $v=\frac{\gamma}{\gamma-1}u^{\gamma-1}$ be the pressure function, define the parabolic operator
$$
\square:=\partial_t-(\gamma-1)v\Delta.
$$
Thus $v$ satisfies the equation
$$
\label{pressure}\square v =|\nabla v|^2.
$$

\begin{lemma}\label{pmeBochner} Let  $\alpha,\beta$ be two constants and $w=|\nabla v|^2$, $F_{\alpha}=\alpha\frac{v_t}{v}-\frac{|\nabla v|^2}{v}$,
then we have the following evolution equations(See \cite{LNVV}),
\begin{align}
\label{pmeBochner0}\square v_t=&(\gamma-1) v_t\Delta v+2\langle\nabla v,\nabla v_t\rangle,\\
\label{pmeBochner2}\square v^{\beta}=&\beta\big(\beta+\gamma-\beta\gamma\big)v^{\beta-1}w,\\
\label{pmeBochner3}\square w=&2\langle\nabla v,\nabla w\rangle
+2(\gamma-1)w\Delta v
-2(\gamma-1)v\Big(|\nabla\nabla v|^2+{\rm Ric}(\nabla v,\nabla v)\Big)\\
\label{pmeBochner4}\square F_{\alpha}=&2\gamma \left\langle \nabla v ,\nabla F_{\alpha}\right\rangle+2(\gamma-1)\Big(|\nabla\nabla v|^2+{\rm Ric}(\nabla v,\nabla v)\Big)
+\Big(F_1^2+(\alpha-1)\Big(\frac{v_t}v\Big)^2\Big).
\end{align}
\end{lemma}

By means of the  Bochner-type formulae in Lemma \ref{pmeBochner}, we have the following integral formulae, which  are useful for the proof of the $W$-entropy
formulae (See\cite{LNVV}).
\begin{lemma}\label{pmeint}
Let $u$ be a positive solution to \eqref{PME}, we have
\begin{align}\label{pmeint1}
\frac{d}{dt}\int_Mvu\,dV=&(\gamma-1)\int_M(\Delta v)vu\,d\mu=-\gamma\int_M|\nabla v|^2u\,dV.\\
\label{pmeint2}
\frac{d^2}{dt^2}\int_Mvu\,dV=&2(\gamma-1)\int_M\left(|\nabla\nabla v|^2+{\rm Ric}(\nabla v,\nabla v)+(\gamma-1)(\Delta v)^2\right)vu\,dV.
\end{align}
\end{lemma}

Applying integral formulae in Lemma \ref{pmeint}, we can deduce the $\mathcal{W}$-entropy formula for the porous medium equation on the closed Riemannian manifolds with  Ricci curvature lower bound.

\begin{proof}[\bf{Proof of Theorem \ref{KPMEentropy}}] Firstly, Boltzmann-Nash type entropy is defined by
$$
\mathcal{N}_K(t):=-\sigma_K(t)\int_Mvu\, dV,
$$
where $\sigma_K(t)$ is a function of $t$, then \eqref{pmeint1}  implies that
\begin{align}\label{Kpment1}
\frac{d}{dt}\mathcal{N}_K(t)=&-\dot{\sigma}_K\int_Mv u\,d\mu-\sigma_K\frac{d}{dt}\int_Mvu\,dV\notag\\
=&-\sigma_K\int_M\left((\gamma-1)\Delta v+(\log\sigma_K)'\right)vu\,dV
\end{align}

By \eqref{pmeint2}, \eqref{Kpment1} and the assumption ${\kappa}=K\sup_{M\times[0,T)}u^{\gamma-1}=\frac{\gamma-1}{\gamma}K\sup_{M\times[0,T)}v$, one has
\begin{small}
\begin{align}\label{Kpment3}
\frac{d^2}{dt^2}\mathcal{N}_K(t)
=&-\sigma_K\frac{d^2}{dt^2}\int_Mvu\,dV
-2\dot{\sigma}_K\frac{d}{dt}\int_Mvu\,dV-\ddot{\sigma}_K\int_Mv u\,dV\notag\\
=&-2(\gamma-1)\sigma_K\int_M\left[|\nabla\nabla v|^2+{\rm Ric}(\nabla v,\nabla v)+(\gamma-1)(\Delta v)^2\right]vu\,dV\notag\\
&+2\frac{\dot{\sigma}_K}{\sigma_K}\frac{d}{dt}\mathcal{N}_K
+\left(\frac{\ddot{\sigma}_K}{\sigma_K}
-\frac{2\dot{\sigma}_K^2}{\sigma_K^2}\right)\mathcal{N}_K\notag\\
=&-2(\gamma-1)\sigma_K\int_M\left[|\nabla\nabla v|^2+({\rm Ric}+Kg)(\nabla v,\nabla v)+(\gamma-1)(\Delta v)^2\right]vu\,dV\notag\\
&+2(\gamma-1)\sigma_K\int_MK|\nabla v|^2vu\,dV+2\frac{\dot{\sigma}_K}{\sigma_K}\frac{d}{dt}\mathcal{N}_K
+\left(\frac{\ddot{\sigma}_K}{\sigma_K}
-\frac{2\dot{\sigma}_K^2}{\sigma_K^2}\right)\mathcal{N}_K\notag\\
\le&-2(\gamma-1)\sigma_K\int_M \left[|\nabla\nabla v|^2+({\rm Ric}+Kg)(\nabla v,\nabla v )+(\gamma-1)(\Delta v)^2\right]vu\,dV\notag\\
&+2\left(\frac{\dot{\sigma}_K}{\sigma_K}+{\kappa}\right)\frac{d}{dt}\mathcal{N}_K
+\left(\frac{\ddot{\sigma}_K}{\sigma_K}
-\frac{2\dot{\sigma}_K^2}{\sigma_K^2}-2{\kappa}\frac{\dot{\sigma}_K}{\sigma_K}\right)\mathcal{N}_K.
\end{align}
\end{small}
Inspired by S. Li and X.-D. Li \cite{LiLi2}(see also \cite{LiLi3} for a survey),
we define the Perelman type $\mathcal{W}$-entropy by
\begin{small}
\begin{align}\label{KWentropy}
\mathcal{W}_K(t):=&\frac1{\dot{\alpha}_K(t)}\frac d{dt}(\alpha_K(t)\mathcal{N}_K(t))
=\mathcal{N}_K+\beta_K(t)\frac{d}{dt}\mathcal{N}_K\notag\\
=&-\sigma_K\int_M\Big[\beta_K(\gamma-1)\Delta v+\big(1+(\log\sigma_K)'\beta_K\big)\Big]vu\,dV,\notag\\
=&\sigma_K\beta_K\int_M\left[\gamma\frac{|\nabla v|^2}{v}-\left(\frac{1}{\beta_K}+\frac{\dot{\sigma}_K}{\sigma_K}\right)\right]vu\,dV,
\end{align}
\end{small}
where $\beta_K(t)=\frac{\alpha_K}{\dot{\alpha}_K}$, then
$$
\frac{d}{dt}\mathcal{W}_K(t)=\beta_K\left(\frac{d^2}{dt^2}\mathcal{N}_K
+\frac{1+\dot{\beta}_K}{\beta_K}\frac{d}{dt}\mathcal{N}_K\right).
$$
Combining \eqref{Kpment1} and \eqref{Kpment3},
we have
\begin{small}
\begin{align}\label{wkpment1}
\frac{d}{dt}\mathcal{W}_K(t)\le&
-2(\gamma-1)\sigma_K\beta_K\int_M \left[|\nabla\nabla v|^2+({\rm Ric}+Kg)(\nabla v,\nabla v )+(\gamma-1)(\Delta v)^2\right]vu\,dV\notag\\
&+2\beta_K\left(\frac{\dot{\sigma}_K}{\sigma_K}+\frac{1+\dot{\beta}_K}{2\beta_K}
+{\kappa}\right)\frac{d}{dt}\mathcal{N}_K
+\beta_K\left(\frac{\ddot{\sigma}_K}{\sigma_K}
-\frac{2\dot{\sigma}_K^2}{\sigma_K^2}
-2{\kappa}\frac{\dot{\sigma}_K}{\sigma_K}\right)\mathcal{N}_K.
\end{align}
\end{small}
On the other hand,
\begin{align}\label{wkpment2}
(\gamma-1)\left|\nabla_i\nabla_jv+\frac{\eta_K(t)}{n(\gamma-1)}g_{ij}\right|^2
=&(\gamma-1)|\nabla\nabla v|^2+\frac{2\eta_K}{n}\Delta v+\frac{\eta_K^2}{n(\gamma-1)}.
\end{align}
Putting \eqref{wkpment2} into \eqref{wkpment1}, we get
\begin{small}
\begin{align}\label{wkpment3}
&\frac{d}{dt}\mathcal{W}_K(t)\notag\\
\le&
-\sigma_K\beta_K\int_M 2 (\gamma-1)\left[\left|\nabla_i\nabla_jv+\frac{\eta_K}{n(\gamma-1)}g_{ij}\right|^2+({\rm Ric}+Kg)(\nabla v,\nabla v )\right]vu\,dV\notag\\
&-\sigma_K\beta_K\int_M\left[2(\gamma-1)^2(\Delta v)^2+2(\gamma-1)\left((\log\sigma_K)'+\frac{1+\dot{\beta}_K}{2\beta_K}+{\kappa}
-\frac{2\eta_K}{n(\gamma-1)}\right)\Delta v\right]vu\,dV\notag\\
&-\sigma_K\beta_K\int_M\left[(\log\sigma_K)''+ ((\log\sigma_K)')^2+\frac{1+\dot{\beta}_K}{\beta_K}(\log\sigma_K)'-\frac{2\eta_K^2}{n(\gamma-1)}\right]vu\,dV.
\end{align}
\end{small}
Set $\lambda=(\log\sigma_K)'$ and choose a proper function $\eta_{K}(t)$ such that
\begin{equation}\label{etabetak}
\left\{
  \begin{array}{l}
    2\eta_K=\lambda+\frac{1+\dot{\beta}_K}{2\beta_K}+{\kappa}
-\frac{2}{n(\gamma-1)}\eta_K \\
   2\eta_K^2=\lambda'+ \lambda^2+\frac{1+\dot{\beta}_K}{\beta_K}\lambda-\frac{2}{n(\gamma-1)}\eta_K^2,
  \end{array}
\right.
\end{equation}
which is equivalent to
\begin{align}\label{etak}
0=&\eta^2_K-2\lambda\eta_K+\frac{a}{a+1}
\left(\lambda^2-\lambda'+2{\kappa}\lambda\right)\notag\\
=&(\eta_K-\lambda)^2-\frac{1}{a+1}\left(\lambda^2
+a\left(\lambda'-2{\kappa}\lambda\right)\right),
\end{align}
where $a=\frac{n(\gamma-1)}{n(\gamma-1)+2}$. Solving the equation \eqref{etak}, we get a special solution
\begin{equation}\label{lambdaetak}
\lambda=\eta_K=\frac{2a{\kappa}}{1-e^{-2{\kappa}t}}.
\end{equation}
Putting \eqref{lambdaetak} back to system \eqref{etabetak}, we have
$$
\frac{1+\dot{\beta}_K}{\beta_K}=2\kappa\coth (kt),\quad \beta_K=\frac{\sinh(2{\kappa}t)}{2{\kappa}}
$$
and
$$
\alpha_K={\kappa}\tanh({\kappa}t),\quad
\sigma_K=\left(e^{{\kappa}t}\frac{\sinh({\kappa}t)}{\kappa}\right)^{a}=\left(\frac{e^{2{\kappa}t}-1}{2\kappa}\right)^{a}.
$$

Therefore, from \eqref{wkpment3}, we obtain the entropy monotonicity formula,
\begin{small}\begin{align}\label{wkpment4}
\frac{d}{dt}\mathcal{W}_K(t)\le&
-\sigma_K\beta_K\int_M 2 (\gamma-1)\left[\left| \nabla_i\nabla_jv+\frac{\eta_K}{n(\gamma-1)}g_{ij}\right|^2+({\rm Ric}+Kg)(\nabla v,\nabla v )\right]vu\,dV\notag\\
&-\sigma_K\beta_K\int_M2\left[(\gamma-1)\Delta v+\eta_K\right]^2vu\,dV.
\end{align}
\end{small}
Thus, when ${\rm Ric}\ge-Kg$, $K\ge0$, $\sigma_K>0,\eta_K>0$, Perelman type entropy \eqref{KWentropy} is monotone decreasing  along the porous medium equation \eqref{PME}.
\end{proof}

In \cite{LiLi,LiLi2}, when $n\le m\in \mathbb{N}$, Li-Li~gave another proof for entropy formula of Witten Laplacian and explained its geometric meaning by warped product, so we can also obtain the entropy formula  \eqref{WKPMEent} for the weighted porous medium equation by the analogous method. In fact, once one can obtain the entropy formula on Riemannian manifold, the weighted version is a direct result by this warped product approach when $m>n$ is a positive integer.

\begin{proof}[\bf{Proof of Theorem \ref{WKPMEentropy}}]
Let $\overline{M}=M\times N$ be a warped product manifold equipped the metric
$$
g_{\overline{M}}=g_M+e^{-\frac{2f}q}g_N,\quad q={m-n},
$$
where $m,n,q$ are the dimensions of $\overline{M}, M, N$ respectively,
then the volume measure satisfies
\begin{equation}\label{volmeasure}
\,\,dV_{\overline{M}}=e^{-f}\,\,dV_M\otimes \,\,dV_N=\,d\mu\otimes \,\,dV_N.
\end{equation}
where $(N^q,g_N)$ is a compact Riemannian manifold. Let $\pi:M\times N\to M$ be a nature projection map, $\overline{X}$ and $X$ are vector fields on $\overline{M}$ and $M$ respectively, then \cite{Besse}
$$
{\rm Ric}_{\overline{M}}(\overline{X},\overline{X})=\pi^*\left({\rm Ric}_{M}(X,X)-qe^{\frac fq}\nabla\nabla e^{-\frac fq}(X,X)\right),
$$
that is
\begin{equation}\label{mBER}
{\rm Ric}_f^m(X,X)=\pi_*\left({\rm Ric}_{\overline{M}}(\overline{X},\overline{X})\right).
\end{equation}
Assume~$V_N(N)=1$, $\overline{\nabla}$~denotes Levi-Civita connection on $(\overline{M},g_{\overline{M}})$, for any ~$v\in C^2(M)$, by S.Li and X.-D.Li \cite{LiLi}, we know
\begin{equation}\label{warped}
\overline{\nabla}_i\overline{\nabla}_jv=\nabla_i\nabla_jv,\quad
\overline{\nabla}_{\alpha}\overline{\nabla}_{\beta}v=-\frac 1q g_{\alpha\beta}g^{kl}\partial_kv\partial_lf,\quad
\overline{\nabla}_i\overline{\nabla}_{\alpha}v=0.
\end{equation}
where $i,j=1,2,\cdots,n$, and $\alpha,\beta=n+1,n+2,\cdots,m$, moreover,
\begin{equation}\label{warpedLaplacian}
\Delta_{\overline{M}}=\Delta_f+e^{-\frac{2f}{q}}\Delta_N.
\end{equation}
Thus, if~$u:M\to[0,\infty)$~ is a positive solution to ~\eqref{PME},
then~$u$~ satisfies the following equation
$$
\frac{\partial u}{\partial t}=\bar{\Delta}(u^{\gamma}).
$$
Moreover, in term of \eqref{volmeasure}, the weighted entropy functional~\eqref{WKentropy}~is indeed the functional on ~$(\overline{M},g_{\overline{M}})$
$$
\overline{\mathcal{W}}_K(v,t)=\sigma_K\beta_K\int_{\overline{M}}\left[\gamma\frac{|\nabla v|^2}{v}-\left(\frac{1}{\beta_K}+\frac{\dot{\sigma}_K}{\sigma_K}\right)\right]vu\,dV_{\overline{M}},
$$
Applying the entropy formula on $(\overline{M},g_{\overline{M}})$ in Theorem \ref{KPMEentropy}, we have
\begin{small}
\begin{align}\label{KPMEnt}
&\frac{d}{dt}\overline{\mathcal{W}}_K(t)\notag\\
\le&
-2(\gamma-1)\bar{\sigma}_K\bar{\beta}_K\int_{\overline{M}}\left(\left| \bar{\nabla}_i\bar{\nabla}_jv+\frac{\bar{\eta}_K}{m(\gamma-1)}\bar{g}_{ij}\right|_{\bar{g}}^2
+(\overline{{\rm Ric}}+K\bar{g})(\bar{\nabla} v,\bar{\nabla} v )\right)vu\,dV_M dV_N\notag\\
&-2\bar{\sigma}_K\bar{\beta}_K\int_{\overline{M}}\left((\gamma-1)\bar{\Delta} v+\eta_K\right)^2vu\,dV_M dV_N.
\end{align}
\end{small}
where
$$
\bar{\eta}_K=\frac{2\bar{a}{\kappa}}{1-e^{-2{\kappa}t}},\quad\bar{\beta}_K=\frac{\sinh(2{\kappa}t)}{2{\kappa}},\quad
\bar{\sigma}_K=\left(\frac{e^{2{\kappa}t}-1}{2\kappa}\right)^{{\bar{a}}},\quad \bar{a}=\frac{m(\gamma-1)}{m(\gamma-1)+2}.
$$
An analogous calculation in \cite{LiLi} gives,
\begin{small}
\begin{align}\label{WHessian}
\Big|\overline{\nabla}_i\overline{\nabla}_jv+\frac{\bar{\eta}_K}{m(\gamma-1)}\bar{g}_{ij}\Big|^2
=&\Big|\nabla_i\nabla_jv+\frac{\bar{\eta}_K}{m(\gamma-1)}g_{ij}\Big|^2+\Big|\overline{\nabla}_{\alpha}\overline{\nabla}_{\beta}v+\frac{\bar{\eta}_K}{m(\gamma-1)}g_{\alpha\beta}\Big|^2\\
=&\Big|\nabla_i\nabla_jv+\frac{\bar{\eta}_K}{m(\gamma-1)}g_{ij}\Big|^2+\frac{1}{m-n}\Big|\nabla v\cdot\nabla f-\frac{\bar{\eta}_K(m-n)}{m(\gamma-1)}\Big|^2
\end{align}
\end{small}
Combining ~\eqref{mBER}~with~\eqref{warpedLaplacian}~, one has
\begin{equation}\label{BERic}
{\rm \overline{Ric}}(\overline{\nabla} v,\overline{\nabla} v)={\rm Ric}^m_f(\nabla v,\nabla v),
\end{equation}
and
\begin{equation}\label{WLaplacian}
\bar{\Delta}v=\Delta_{f}v.
\end{equation}
Put \eqref{WHessian}, \eqref{BERic} and \eqref{WLaplacian} into \eqref{KPMEnt}, we can get the  desired result~\eqref{WKPMEent}.
\end{proof}

\section{Differential Harnack estimates and applications}
In this section, we prove a Qian type differential Harnack estimates \cite{QianB} for the porous medium equation on the compact Riemannian manifolds with lower bound of Ricci curvature, which generalize the Lu-Ni-Vazquez-Villani's estimate \cite{LNVV} and Huang-Huang-Li's estimate \cite{HHL}, Harnack inequalities and Laplacian estimate are derived as applications.
\begin{proposition} Let  $\sigma(t)$ be a functions of $t$ and satisfy the assumption (A1)(A2) and $\alpha(t), \varphi(t)$ are defined in \eqref{PMEalphavarphi}.  Suppose~$u$~be a smooth solution to \eqref{PME} and $v=\frac{\gamma }{\gamma-1}u^{\gamma-1}$ the pressure funciton. Define
$$
F_{\alpha}:=\alpha(t)\frac{v_t}{v}-\frac{|\nabla v|^2}{v}+\varphi(t).
$$
Set ${\kappa}=K\sup\limits_{M\times(0,T]}(u^{\gamma-1})$, for any $\gamma>1$, we have
\begin{small}
\begin{align}\label{GWDBochner1}
\square F_{\alpha}\ge&2\gamma \left\langle \nabla v ,\nabla F_{\alpha}\right\rangle+\frac1a\left((\gamma-1)\Delta v+\frac{a\sigma'}{2\sigma}+{a\gamma\kappa}\right)^2-\frac{\sigma'}{\sigma}F_{\alpha}
+(\alpha-1)\Big(\frac{v_t}v\Big)^2,
\end{align}
\begin{align}\label{GWDBochner2}
\square(\sigma F_{\alpha})\ge&2\gamma\sigma \left\langle \nabla v ,\nabla F_{\alpha}\right\rangle+\frac{\sigma}a\left((\gamma-1)\Delta v+\frac{a\sigma'}{2\sigma}+{a\gamma\kappa}\right)^2
+(\alpha-1)\sigma\Big(\frac{v_t}v\Big)^2.
\end{align}
\end{small}
\end{proposition}

\begin{proof} By using of \eqref{pmeBochner4}, we have
 \begin{align}\label{WDBochner4}
\square F_{\alpha}=&2\gamma\left\langle \nabla v ,\nabla F_{\alpha}\right\rangle+2(\gamma-1)\Big(|\nabla\nabla v|^2+{\rm Ric}(\nabla v,\nabla v)\Big)\notag\\
&+\left(((\gamma-1)\Delta v)^2+(\alpha-1)\Big(\frac{v_t}v\Big)^2\right)
+\alpha'\Big(\frac{v_t}{v}\Big)+\varphi'.
\end{align}
 Applying Cauchy-Schwartz inequality and \eqref{WDBochner4}, when ${\rm Ric}\ge-Kg$, we get
\begin{small}
\begin{align}\label{pmegradientest}
&\square F_{\alpha}-2\gamma \left\langle \nabla v ,\nabla F_{\alpha}\right\rangle\notag\\
\ge&\frac{1}{a}((\gamma-1)\Delta v)^2-2(\gamma-1)K|\nabla v|^2+(\alpha-1)\Big(\frac{v_t}v\Big)^2
+\alpha'\Big(\frac{v_t}{v}\Big)+\varphi'\notag\\
=&\frac{(\gamma-1)^2}{a}(\Delta v+\eta)^2-\frac{2 (\gamma-1)^2}{a}\eta\Delta v-\frac{(\gamma-1)^2}{a}\eta^2-2(\gamma-1)K|\nabla v|^2\notag\\
&+(\alpha-1)\Big(\frac{v_t}v\Big)^2
+\alpha'\Big(\frac{v_t}{v}\Big)+\varphi'\notag\\
\ge&\frac{(\gamma-1)^2}{a}(\Delta v+\eta)^2+\Big(\alpha'-\frac{2 (\gamma-1)}{a}\eta\Big)\frac{v_t}{v}-\Big(2\gamma{\kappa}
-\frac{2 (\gamma-1)}{a}\eta\Big)\frac{|\nabla v|^2}{v}\notag\\
&-\frac{(\gamma-1)^2}{a}\eta^2+(\alpha-1)\Big(\frac{v_t}v\Big)^2
+\varphi'\notag\\
=&\frac{(\gamma-1)^2}{a}(\Delta v+\eta)^2+\Big(2\gamma{\kappa}-\frac{2 (\gamma-1)}{a}\eta\Big)\left(\frac{\alpha'-\frac{2(\gamma-1)}{a}\eta}{2\gamma{\kappa}-\frac{2(\gamma-1)}{a}\eta}\frac{v_t}v- \frac{|\nabla v|^2}{v}+\varphi\right)\notag\\
&+(\alpha-1)\Big(\frac{v_t}v\Big)^2+\varphi'-\Big(2\gamma{\kappa}-\frac{2 (\gamma-1)}{a}\eta\Big)\varphi-\frac{(\gamma-1)^2}{a}\eta^2.
\end{align}
\end{small}

Choosing the proper functions $\sigma(t)$ and $\eta(t)$ such that $\alpha(t)$ and $\varphi(t)$ satisfy the following system
\begin{equation}\label{sigeta}
\left\{
  \begin{array}{rl}
\frac{\sigma'}{\sigma}&=\frac{2(\gamma-1)}{a}\eta-2\gamma{\kappa} \\
\alpha&=\frac{\alpha'-\frac{2(\gamma-1)}{a}\eta}{2\gamma{\kappa} -\frac{2 (\gamma-1)}{a}\eta} \\
\eta^2&=\frac{a}{(\gamma-1)^2}\left(\varphi'-\Big(2\gamma{\kappa}-\frac{2 (\gamma-1)}{a}\eta\Big)\varphi\right).
  \end{array}
\right.
\end{equation}
Plugging \eqref{sigeta} into \eqref{pmegradientest}, we get
$$
\square F_{\alpha}\ge2\gamma\left\langle \nabla v ,\nabla F_{\alpha}\right\rangle
+\frac{(\gamma-1)^2}{a}(\Delta v+\eta)^2-\frac{\sigma'}{\sigma}F_{\alpha}
+(\alpha-1)\Big(\frac{v_t}v\Big)^2,
$$
that is \eqref{GWDBochner1}.
Inequality \eqref{GWDBochner2} is a direct result of \eqref{GWDBochner1} and
$$
\square G=\square (\sigma F_{\alpha})=\sigma\square F_{\alpha}+\sigma' F_{\alpha}.
$$
In fact, the first equation in \eqref{sigeta} is equivalent to
$
\frac{2(\gamma-1)}{a}\eta(t)=\frac{\sigma'}{\sigma}+2\gamma{\kappa}.
$
Put this into  the last two equations in \eqref{sigeta}, we have
$$
\left\{
  \begin{array}{rl}
(\sigma\alpha)'=&\sigma'+2\gamma{\kappa}\sigma\\
(\sigma\varphi)'=&\frac{a\sigma}{4}\Big(\frac{\sigma'}{\sigma}+2\gamma{\kappa}\Big)^2,
  \end{array}
\right.
$$
Integral above identities on $[0,t]$, we can obtain the explicit expressions of $\alpha(t)$ and $\varphi(t)$ in \eqref{PMEalphavarphi}.
\end{proof}

\begin{proof}[\bf{Proof of Theorem \ref{pmeGK}}]
Since $\gamma>1$, $a>0$ and $M$ is closed,  the standard parabolic maximum principle in \eqref{GWDBochner2} implies $F_{\alpha}\ge0$, that is \eqref{PMELYH1} in Theorem \ref{pmeGK}.
\end{proof}

\begin{proof}[\bf{Proof of Corollary \ref{Harnack}}]
Let $\varsigma(t)$ be a constant speed geodesic with $\varsigma(t_1)=x_1$ and $\varsigma(t_2)=x_2$ such that $|\dot{\varsigma}(t)|=\frac{d(x_2,x_1)}{t_2-t_1}$. Using differential Harnack estimate \eqref{PMELYH1} and Young inequality, we have
\begin{small}
\begin{align*}
v(x_2,t_2)-v(x_1,t_1)=&\int^{t_2}_{t_1}v_t+\langle\nabla v,\dot{\varsigma}(t)\rangle dt\\
\ge&\int^{t_2}_{t_1}\left(\frac1{\alpha(t)}|\nabla v|^2-\frac{\varphi(t)}{\alpha(t)}v-\frac1{\alpha(t)}|\nabla v|^2-\frac{1}{4}\alpha(t)|\dot{\varsigma}(t)|^{2}\right)dt\\
\ge&-v_{max}\int^{t_2}_{t_1}\frac{\varphi(t)}{\alpha(t)}dt
-\frac{1}{4}\frac{d(x_2,x_1)^{2}}{(t_2-t_1)^{2}}\int^{t_2}_{t_1}\alpha(t)dt
\end{align*}
\end{small}
and
\begin{small}
\begin{align*}
\log\frac{v(x_2,t_2)}{v(x_1,t_1)}
=&\int^{t_2}_{t_1}\left(\frac{d}{dt}\log v(x,t)+\nabla\log v\cdot\dot{\varsigma}(t)\right)dt\\
\ge&\int^{t_2}_{t_1}\left(\frac{1}{\alpha(t)}\Big(|\nabla v|^2-\varphi(t)\Big)
-\frac1{\alpha(t)}|\nabla v|^2-\frac{1}{4}\frac{|\dot{\varsigma}(t)|^{2}}{v_{max}}\alpha(t)\right)dt\\
\ge&-\int^{t_2}_{t_1}\frac{\varphi(t)}{\alpha(t)}dt
-\frac{1}{4}\frac1{v_{max}}\frac{d(x_2,x_1)^{2}}{(t_2-t_1)^{2}}\int^{t_2}_{t_1}\alpha(t)dt.
\end{align*}
\end{small}
This finishes  the proof of Corollary \ref{Harnack}.
\end{proof}

\begin{proof}[\bf{Proof of Corollary \ref{pfLaEst}}]
Since $\alpha(t)>1$, direct calculation implies that
\begin{small}
\begin{align*}
\Delta(v^{\beta})=&\beta v^{\beta-1}\left(\Delta v+(\beta-1)\frac{|\nabla v|^2}{v}\right)\\
=&\frac{1}{\alpha (\gamma-1) }\beta v^{\beta-1}\left(\alpha (\gamma-1)\Delta v+\alpha(\gamma-1)(\beta-1)\frac{|\nabla v|^2}{v}\right)\\
=&\frac{1}{\alpha (\gamma-1) }\beta v^{\beta-1}\left(\alpha \frac{v_t}{v}-\frac{|\nabla v|^2}{v}\right).
\end{align*}
\end{small}
The estimate \eqref{pfLaE} follows from \eqref{PMELYH1}.
\end{proof}

\section*{Acknowledgments} The author would like to thank Professor Xiang-Dong Li and Dr. Songzi Li for their interest and illuminating discussions. The author is also thankful to the anonymous reviewers for their constructive comments and suggestions on the earlier version for this paper.

config: arxiv


text:
\section{Two Thirds as Pandigital Fraction}
Tags: Pandigital Fractions

\begin{theorem}
$\dfrac 2 3$ cannot be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]].
\end{theorem}

\begin{proof}
Can be verified by brute force.
[[Category:Pandigital Fractions]]
5itdiplla5qizaah8gv9ev3z2sozpk9
\end{proof}

config: wiki


TITLE: Simulating the Electric Field inside a Hollow Sphere of Uniform Charge
QUESTION [1 upvotes]: I've recently been studying Gauss's Law and have come across some results that I want to verify. For one, I am trying to verify that the electric field inside a conductor (in this case it could be hollow or solid as the charges will be uniformly distributed along the surface no matter what) is indeed 0. I found that this must be true in order for the charges to remain stationary and due to the symmetry created with Gauss's Law, especially with those problems that involve finding the electric field in a cavity with a point charge in it. Because the charge in the cavity is dependent on only the point charge (according to Gauss's Law), this must imply that all the other charges along the inside wall of the cavity and the outside wall of the sphere must create a net electric field of 0. I tried to simulate this to verify it.
So I used the algorithm describe here in order to generate a set of uniformly ditributed points on a sphere. Here is a visual plot of the resulting points (the red dot is the test point):

Then, I made the following method to calculate the electric field from each of these points at some test point:
import math
def calc_efield(point, testpoint):
    k = 1
    diffx = testpoint[0] - point[0]
    diffy = testpoint[1] - point[1]
    diffz = testpoint[2] - point[2]
    dist = math.sqrt((diffx ** 2) + (diffy ** 2) + (diffz ** 2))
    alpha = math.acos(abs(diffx/dist))
    beta = math.acos(abs(diffy/dist))
    gamma = math.acos(abs(diffz/dist))

    eField = k/(dist**2)
    eFieldx = math.cos(alpha) * eField
    eFieldy = math.cos(beta) * eField
    eFieldz = math.cos(gamma) * eField

    if abs(diffx) < .001:
        eFieldx = 0
    elif diffx < 0:
        eFieldx = eFieldx * -1
    if abs(diffy) < .001:
        eFieldy = 0
    elif diffy < 0:
        eFieldy = eFieldy * -1
    if abs(diffz) < .001:
        eFieldz = 0
    elif diffz < 0:
        eFieldz = eFieldz * -1
    return eFieldx, eFieldy, eFieldz

However, after iterating through each of these points and summing the electric field in all directions, I only arrive at a relatively net zero result when the test point is at the origin (0,0,0) (and even then it is not exactly zero).
Is there something I'm missing either in physical predictions made in the first paragraph or in my code? I made N=2e3 -- is that sufficiently large?

REPLY [2 votes]: You will never get exactly zero for two reasons.

You are making a discrete approximation to a continuous distribution,
and
computers have limited precision. They get the absolutely
correct answer to floating point operations only very rarely (such as the multiplication of two integers.)

You need to compare the value that you calculated to something else in order to determine if the answer is small.  I'd suggest comparing to the the magnitude of the electric field at the origin due to one of your charges.


TITLE: Question on doing the integral for Fermi golden rule
QUESTION [6 upvotes]: Today in the lecture, my professor did something which confused me

As an example, we consider the photoelectric effect, in which an
  electron bound in a Coulomb potential is ionized after interacting
  with an external electromagnetic field. [...] The absorption rate is given
  by
  $$ \Gamma_{i\rightarrow
f}=\frac{4a_{0}^{3}e^{2}}{m^{2}\pi\hbar^{4}c^{2}}\frac{(\mathbf{A}_{0}\cdot\mathbf{p}_{f})^{2}}{(1+(p_{f}a_{0}/\hbar^{2})^{4}}\delta(p_{f}^{2}/2m-E_{i}-\hbar\omega)
$$
This gives us the rate for a precise final momentum $p_{f}$ .
  Typically, what we want to know is the rate of electrons detected in a
  solid angle $\mathrm{d}\Omega$
  $$
\frac{\mathrm{d}\Gamma}{\mathrm{d}\Omega}=\int_{0}^{\infty}p_{f}^{2}\,\mathrm{d}p_{f}\,\Gamma_{i\rightarrow
p_{f}} $$
Doing the $p_{f}$ integral, we have $$
\frac{\mathrm{d}\Gamma}{\mathrm{d}\Omega}=\frac{4a_{0}^{3}e^{2}p_{f}}{m\pi\hbar^{4}c^{2}}\frac{(\mathbf{A}_{0}\cdot\mathbf{p}_{f})^{2}}{(1+(p_{f}a_{0}/\hbar^{2})^{4}}
$$
where $p_{f}=\sqrt{2m(E_{i}+\hbar\omega)}$.

but I don't get how the integral just multiplies a term $m \, p_f$ to the first expression. I thought that the $\delta$ function replace $p_f$ with $\sqrt{2m(E_{i}+\hbar\omega)}$.

REPLY [8 votes]: The delta function $\delta(x)$ has unit area, but the function $\delta(2x)$ is "half as wide" and thus has half as much area; thus you can pick up extra factors from 'how fast' you cross the peak of the delta function. The general identity is
$$\delta(f(x)) = \sum \frac{\delta(x-x_i)}{\big| df/dx|_{x=x_i} \big|}$$
where the $x_i$ are the roots of $f$. In this case the factor you pick up is
$$p_f^2 \left(\frac{d}{dp_f}(p_f^2/2m - E_i - \hbar \omega) \right)^{-1} = p_f^2 (m/p_f) = m p_f.$$


TITLE: Are the measurement outcomes of an observable gaussian distributed?
QUESTION [1 upvotes]: Suppose in an experiment we perform $n$ independent measurements to find the true value of an observable $X$. Let the outcomes of $n$ measurement are denoted by $x_1,x_2,...x_n$. If $n$ is sufficiently large, will these measured outcomes $\{x_i\}$ be Gaussian distributed?
Please note that I am not asking whether the means of the measurements are Gaussian distributed. I know they are.

REPLY [1 votes]: It's certainly possible to identify an observable with a non-Gaussian distribution.
An example from my professional life was a detector which collected Cherenkov photons from relativistic electrons. We needed the same detector to be sensitive to a macroscopic electron current (nanoamps) but also, with a gain change, to be able to trigger on single electrons. In the counting mode we determined that each electron sent about ten photons to the photomultiplier --- a number small enough that we had to account for the asymmetry in the Poisson distribution ($10\pm\sqrt{10}$) rather than using the Gaussian approximation.


TITLE: Congruence modulo a power of two
QUESTION [1 upvotes]: let $a \geq 3$. Let $2^a \mid c^2-1$.
By simple divisibility rules I get
$$c \equiv 1 \mod 2^a \quad \text{or} \quad c \equiv -1 \mod 2^a  $$
or
$$c \equiv 1 \mod 2^{a-1},\quad  c \equiv  -1 \mod 2 $$
or
$$c \equiv -1 \mod 2^{a-1},\quad  c \equiv  1 \mod 2 $$
Can you explain me how to obtain the following congruence:
$$
c \; \equiv \; \pm 1 + 2^{a-1} \mod 2^a
$$
??

REPLY [2 votes]: We want to solve the congruence $(c-1)(c+1)\equiv 0\pmod{2^a}$, where $a\ge 3$. Note that $c$ must be odd, and that $c-1$ and $c+1$ are two consecutive even numbers.
Since $c-1$ and $c+1$ are consecutive even numbers, one of them is congruent to $2$ modulo $4$, and therefore only has one $2$ to contribute to the product.
Case 1: [$c+1\equiv 2\pmod{4}$]  Since $2^a$ divides $(c-1)(c+1)$, $2^{a-1}$ must divide $c-1$. So $c\equiv 1\pmod{2^{a-1}}$, and therefore $c=1+k2^{a-1}$ for some integer $k$. If $k$ is even, then $s\equiv 1\pmod{2^a}$. If $k$ is odd then $c\equiv 1+2^{a-1}\pmod{2^{a}}$.
It is clear that $c\equiv 1\pmod{2^a}$ and $c\equiv 1+2^{a-1}\pmod{2^a}$ are solutions.
Case 2: [$c-1\equiv 2\pmod{4}$] Then $2^{a-1}$ must divide $c+1$, and therefore $c=-1+k2^{a-1}$ for some integer $k$. If $k$ is even then $c\equiv -1+2^{a-1}\pmod{2^{a}}$, and if $k$ is odd then $c\equiv -1+2^{a-1}\pmod{2^{a}}$.
If $a\ge 3$, then the four solutions are distinct modulo $2^a$.


TITLE: Solving equations involving permutations and combinations
QUESTION [0 upvotes]: I need to solve for $n$ if $$P^n_4=60\times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$

REPLY [1 votes]: By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$.  Positive solution is $n=8$.

REPLY [0 votes]: You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)\Big[(n-2)(n-3)-30\Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\\\implies n^2-5n-24=(n-8)(n+3)=0\\\therefore n=0,1,-3,8$$If you define $P^n_4=\displaystyle\frac{n!}{(n-4)!}$, you need $n\ge4$. This leaves you with the answer $n=8$.


TITLE: Proof of relationship $S^2−S(a+b+c+d+e)+ab+bc+cd+de+ea=0$ between areas connected to a pentagon
QUESTION [8 upvotes]: So recently I've been looking around at some other problems to see if they could help me solve an ongoing problem, and I found a theorem that was mentioned that I feel that might be useful to my attempt at solving the ongoing problem. The problem goes a little something like this:

The area of a convex pentagon $ABCDE$ is denoted by $S$. Let $a$, $b$, $c$, $d$, $e$ be the areas of $ΔABC$,$ΔBCD$,$ΔCDE$,$ΔDEA$,$ΔEAB$; then prove that $S^2−S(a+b+c+d+e)+ab+bc+cd+de+ea=0$.

Now there wasn't a proof mentioned in that post for this theorem mentioned, and I looked around the web a bit for some context on this theorem, but I couldn't find who made or how to prove it, even after a few hours of searching. I figured I could ask for a bit of help here on solving this, as I haven't been able to make much progress on this problem yet. I initially was trying to find the area of each triangle in terms of two segments and then tried relating that back to the pentagon's area, but I just couldn't make much progress going that way. I would really appreciate it if anyone could give some ideas on proving this problem.

REPLY [7 votes]: Here is a proof using barycentric coordinates.
Let us use compact notation $[MNP]$ for the oriented area of triangle $MNP$ (positive is $M,N,P$ in this order has the direct orientation, negative otherwise).

Let us consider $ABC$ as the reference triangle.
The barycentric coordinates (b.c. in short) of points $D$ and $E$ are
$$b.c.(D)=(\underbrace{\dfrac{[\color{red}{D}BC]}{[ABC]}}_{(+)},\underbrace{\dfrac{[A\color{red}{D}C]}{[ABC]}}_{(-)},\underbrace{\dfrac{[AB\color{red}{D}]}{[ABC]}}_{(+)})=(\dfrac{b}{a},\dfrac{a+d-S}{a},\dfrac{S-b-d}{a})$$
(notations (+), resp. (-), for positive, resp. negative orientation) and:
$$b.c.(E)=(\dfrac{[\color{red}{E}BC]}{[ABC]},\dfrac{[A\color{red}{E}C]}{[ABC]},\dfrac{[AB\color{red}{E}]}{[ABC]})=(\dfrac{S-c-e}{a},\dfrac{a+c-S}{a},\dfrac{e}{a}).$$
Let us consider as well $A$ with $b.c.(A)=(1,0,0)$ evidently.
Let us now express a classical property: the determinant of the barycentrical coordinates of 3 points $M,N,P$ is equal to the ratio $[MNP]/[ABC]$ of the area of the triangle to the area of the reference triangle giving here for triangle $MNP \equiv ADE$:
$$\begin{vmatrix}1&b/a&(S-c-e)/a\\0&(a+d-S)/a&(a+c-S)/a\\0&(S-b-d)/a&e/a\end{vmatrix}=\dfrac{d}{a}$$
factorizing $1/a$ in the second and third columns, we obtain the equivalent form:
$$\begin{vmatrix}(a+d-S)&(a+c-S)\\(S-b-d)&e\end{vmatrix}=ad \tag{1}$$
Expanding (1) gives the looked for relationship:
$$S^2−S(a+b+c+d+e)+(ab+bc+cd+de+ea)=0\tag{2}$$
Important edit: I have finally found a very interesting reference to formula (2) in [this answer] (https://mathoverflow.net/q/151316) [and also used here which is another answer to the same question]. More than that, the author of the answer establishes a "mirror formula", surprisingly, identical to (2):
$$S^2−S(a'+b'+c'+d'+e')+(a'b'+b'c'+c'd'+d'e'+e'a')=0\tag{3}$$
with the complementary areas:
$$a':=S-a, \ b':=S-b, \, c':=S-c, \ d':=S-d, \ e':=S-e$$
i.e. areas of quadrilaterals $CDEA, DEAB,....$ !
Remark 1: It is explained in the given reference that by solving (2) as a quadratic one obtains a formula for $S$ as a function of $a,b,c,d,e$ under the condition to take the largest root, whereas, doing the same in (3), one has to take the smallest root.
Remark 2: There is a connection of (2) with the shoelace formula.


TITLE: Proof dot product identity
QUESTION [3 upvotes]: From the dot product properties we can write:
$$
\langle v, u \rangle = v^{T} u = \sum_{j}^{} u_j v_j
$$
I want to prove that:
$$
\langle v, u \rangle u = (u u^{T}) v
$$
I tried rewriting the first expression as
$$
\langle v, u \rangle u  = (v^{T} u)u = \cdots =  (u u^{T}) v
$$
But I don't know how to continue to reach the right hand side. Is this approach the right way?

REPLY [1 votes]: We have$\, \langle v, u \rangle u = u \langle u, v \rangle =
u (u^{T} v) = (u u^{T})v \,$ where the first equality is commutativity
of multiplication of a scalar times a vector. The second equality by
definition of dot (or inner) product. The third equality by associativity of
multiplicaton of matrices and vectors.


TITLE: How can I compute $\mathbb P\{T_n<T_0\}$?
QUESTION [4 upvotes]: Let $(X_n)_{n}$ a random walk over $\mathbb Z$ starting at $0$, i.e. $\mathbb P\{X_0=0\}=1$. I denote $T_k=\inf\{n\geq 1\mid X_n=k\}$. I suppose that $$\mathbb P\{X_{n+1}=X_n+1\mid X_n,...,X_0\}=p\quad \text{and}\quad \mathbb P\{X_{n+1}=X_n-1\mid X_n,...,X_0\}=q.$$
Remark that $q=1-p$. How can I compute $\mathbb P\{T_n<T_0\}$, i.e. the probability to touch $n$ before touching $0$ ? In fact I have problem to interpret $\{T_n<T_0\}$ using $(X_n)_n$.

I tired as follow :

$\mathbb P\{T_1<T_0\}=\mathbb P\{X_1=1\mid X_0=0\}=p$.
$\mathbb P\{T_2<T_0\}=\mathbb P\{X_1=1\mid X_0=0\}+\mathbb P\{X_2=2\mid X_1=1\}=2p$
$\mathbb P\{T_3<T_0\}=\mathbb P\{X_1=1\mid X_0=0\}+\mathbb P\{X_2=2\mid X_1=1\}+\mathbb P\{X_3=3\mid X_2=2\}+\mathbb P\{X_3=2\mid X_2=2\}(\mathbb P\{X_4=3\mid X_3=2\}+\mathbb P\{X_4=1\mid X_3=2\})$

I know that the last one is not clear at all, but I don't know how to interpret that fact that the walker can past many time between 1 and 2 many times before arriving at $3$.
Any explanation would be appreciated.

REPLY [3 votes]: Good question! For your three examples, I think you shouldn't condition on $X_0 =0$ because it does not change anything ($P(X_0=0)=1$). Instead we should condition on the value of $X_1$ (for $n\geq 2$) and use Bayesian formula. $P(T_1<T_0) = P(X_1 = 1) = p$ is obvious. For $T_2$, we have
$$
P(T_2<T_0) = P(T_2<T_0|X_1=1) \cdot P(X_1=1)+P(T_2<T_0|X_1=-1) \cdot P(X_1=-1).
$$
Since $P(T_2<T_0|X_1=1) = P(X_2=2|X_1=1) = p$, and $P(T_2<T_0|X_1=-1)=0$ (starting at $-1$, you must pass $0$ before arrive at $2$), we have $P(T_2<T_0) = p^2$.
For general $n>2$, we can still do
$$
P(T_n<T_0) = P(T_n<T_0|X_1=1) \cdot P(X_1=1)+P(T_n<T_0|X_1=-1) \cdot P(X_1=-1)
$$
and get $P(T_n<T_0) = P(T_n<T_0|X_1=1) \cdot P(X_1=1)$ because $P(T_n<T_0|X_1=-1) = 0$. So it remains to get $P(T_n<T_0|X_1=1)$.
Now it turns to a special case of the Gambler's ruin problem: we start at $1$, and we want to find the probability that we hit $n$ before hitting $0$. Doing a shift, it's equivalent to the probability that we start at $0$ and hit $n-1$ before hitting $-1$.
Thus by the well-known result (see at the bottom of page 3) for Gambler's ruin problem, where $a = n-1$ and $b=1$ for our problem, we get
$$
P(T_n<T_0|X_1=1) = \begin{cases}
\frac{1-(q/p)^{n-1}}{1-(q/p)^{n}},\ p\neq 0.5\\
\frac{1}{n},\ p=0.5.
\end{cases}
$$
In conclusion, we have
$$
P(T_n<T_0) = P(T_n<T_0|X_1=1) \cdot P(X_1=1) = \begin{cases}
p\frac{1-(q/p)^{n-1}}{1-(q/p)^{n}},\ p\neq 0.5\\
\frac{1}{2n},\ p=0.5.
\end{cases}
$$


TITLE: Which is more accurate, a quantum gyroscope or ring laser gyroscope?
QUESTION [1 upvotes]: I can't find any source which says whether quantum or laser gyroscopes are more accurate. I'm also wondering what the advantages and disadvantages are of each technology.

REPLY [1 votes]: Many use the terms accuracy, precision, and sensitivity interchangeably when they are different but most often taken to mean sensitivity. I will assume by accurate you mean sensitivity. I will also assume you are referring to atom gyroscopes as quantum as light-based gyroscopes can, in principle also be quantum.
This is not a straightforward question to answer, simply because the devices are so different. For example, we shall look at an all things equal comparison of light and atom-based gyros. The operational principle behind these gyros is the Sagnac effect, for light and atoms propagating around a single loop the phase accumulated due to a rotation is
$$\Phi_L=\frac{8\pi\Omega A}{\lambda c}, \qquad\text{and}\qquad\Phi_A=\frac{4m\Omega A}{\hbar},$$
where $\Omega$ is the rotation being measured $A$ the area enclosed $\lambda$ the wavelength of light and $m$ the mass of the atom. For optical gyroscopes operating with visible light and an atom gyro using $^{87}$Rb, the ratio of the two accumulated phases is $\Phi_A/\Phi_L\approx10^{10}$. This is a huge increase in sensitivity, and this direct comparison is the current drive behind the interest of making atom-based interferometers in general. However, some things have been assumed in this comparison that aren't appropriate for actual applications. The area enclosed was equated, for optical gyroscopes the area enclosed can order of magnitude larger than atom-based gyros. The reason for this is maintaining coherence and achieving the propagation. Another is particle flux, light-based gyros will have a far higher flux of photons than atom-based gyros.
All that said, there is a drive to get atom-based gyros up and running. This is because, in the long run with advancing atom trapping and coherence control techniques, atom-gyros are set to outperform light-based gyros. Although in practice it might be beneficial to have a hybrid gyroscope as atom-gyros have a dead-time (atoms need to be trapped and cooled first before the rotation measurement), the superior short-term sensitivity of an atom-based gyro can be used together with the continuous-mode operation of light-based gyro.


text:
\begin{document}

\onehalfspacing

\maketitle

\begin{abstract}
We investigate the distribution of the function $\omega(n)$, the number of distinct prime divisors of $n$, in residue classes modulo $q$ for natural numbers $q$ greater than 2. In particular we ask `prime number races' style questions, as suggested by Coons and Dahmen in their paper `On the residue class distribution of the number of prime divisors of an integer'.
\end{abstract}

\section{Introduction}

Let $q \geq 2$ be an integer, $a\in\{0,1,\ldots, q-1\}$ represent some residue class modulo $q$ and $\omega(n)$ denote the number of distinct prime divisors of $n$. Define
$$N_{a,q}(x):=\#\{n \leq x : \omega(n)\equiv a \mod q\}.$$
Seeing no reason why $\omega(n)$ should favour any particular residue class, we expect that for all $a$,
\begin{equation}
N_{a,q}(x)\sim \frac{x}{q} \:\:\:\: \text{ as } \:\:\: x \rightarrow \infty.
\end{equation}

In fact, it was proved in~\cite{Add} that
$$N_{a,q}(x) - \frac{x}{q} = O\left(\frac{x}{(\log x)^{c(q)}}\right)$$
with $c(q) = 1-\cos(\frac{2\pi}{q})$. It was also proved that for $q>2$ the error term here is best possible, since it was also determined that for $q>2$
$$N_{a,q}(x) - \frac{x}{q} = \Omega_{\pm}\left(\frac{x}{(\log x)^{c(q)}}\right).$$ This is in stark contrast to the case $q=2$ for which we expect ``square-root cancellation". Indeed,
$$
N_{a,2}(x) - \frac{x}{2} =O(x^{1/2+o(1)}) \:\: \text{  for  }  a=0 \text{  and  } 1
$$
is equivalent to the Riemann Hypothesis. For $q=2$, it is well known that (1) is equivalent to the prime number theorem.

In~\cite{Coons-Dah}, the authors suggest that, in the spirit of prime number races, it would be interesting to investigate the sign changes of $N_{a,q}(x)-N_{b,q}(x).$ The traditional prime number races concern the popularity of residue classes for prime numbers rather than for the values of $\omega$, that is, sign changes of $\pi(x;a,q)-\pi(x;b,q)$ where $\pi(x;a,q)$ is the number of primes less than or equal to $x$ which are congruent to $a$ modulo $q$. Rubinstein and Sarnak~\cite{Rub-Sar} proved under certain reasonable assumptions that the set
$$\{x \in \mathbb{N} \: : \:\pi(x;3,4)<\pi(x;1,4)\},$$
for example, does not have a natural density in the integers but does have a \textit{logarithmic density}, defined for a subset $E\subset \mathbb{N}$, if the limit exists, to be $\lim_{X\rightarrow \infty}\frac{1}{\log X}\sum_{\substack{ x\leq X \\ x \in E }}\frac{1}{x}$. Loosely speaking, the reason for this is that the difference $\pi(x;3,4) - \pi(x;1,4)$ can be written  as a sum of terms of the form $\sin(\gamma \log x)/\gamma$, where $\gamma$ ranges over the imaginary parts of the zeros of certain Dirichlet $L$-functions. For an introduction to this topic see~\cite{Gran-Martin}.

For our investigations it is natural to consider mean values of the multiplicative functions $n\mapsto z^{\omega(n)}$ where $z$ is taken to be a complex $q$-th root of unity. By applying a classical result first due to Selberg concerning such mean values we will establish an asymptotic formula for $N_{a,q}(x)$ with main term and next highest order term in the case $q>2$.  This will be used to prove our main theorem. The formula will contain an expression of the form $\cos(A \log \log x + B)$ and so in our case we have neither natural nor logarithmic density, but instead need to go further and define the notion of \emph{loglog density}. We say a subset $E \subset \mathbb{N}$ has \emph{loglog density} $\delta$ if $$\frac{1}{\log \log X}\sum_{\substack{ x\leq X \\ x \in E }}\frac{1}{x\log x} \rightarrow \delta \text{  as  } X \rightarrow \infty.$$ With this we can now state our main theorems.

\begin{theorem}\label{thm1}
Let $q>2$ be an integer and $a,b\in \{0,1,\ldots, q-1\}$ with $a\neq b$. The set $$E_{a,b}:=\{x \in \mathbb{N} \: :\: N_{a,q}(x) < N_{b,q}(x) \}$$ has no natural density, in fact $$\liminf_{X\rightarrow \infty}\frac{1}{X}[1,X]\cap E_{a,b} = 0 \:\:\text{  and  }\:\:  \limsup_{X\rightarrow \infty}\frac{1}{X}[1,X]\cap E_{a,b} = 1.$$
\end{theorem}

\begin{theorem}\label{thm2}
The set $E_{a,b}$, defined in Theorem~\ref{thm1}, has no natural or logarithmic density, but has \emph{loglog density} equal to $1/2$.
\end{theorem}

Given a complete ordering on the residue classes, we can also ask how often the different `competitors' in our race are in that order.

\begin{theorem}\label{thm3}
Let $q>2$ be an integer and $a \in \{0, 1, \ldots, q-1\}$. Each of the following sets has \emph{loglog density} $\frac{1}{2q}$
$$U_{a,q}:=\{x \in \mathbb{N} \: :\: N_{a, q}(x)>N_{a-1, q}(x)>N_{a+1, q}(x)>\cdots>N_{a-i,q}(x)>N_{a+i,q}(x)>\cdots \}$$
$$V_{a, q}:=\{x \in \mathbb{N} \: :\: N_{a, q}(x)>N_{a+1, q}(x)>N_{a-1, q}(x)>\cdots>N_{a+i,q}(x)>N_{a-i,q}(x)>\cdots \}.$$ Therefore, since there are $2q$ of them, these are the only permutations which appear with non-zero \emph{loglog densities.}
\end{theorem}

\begin{exmp}
When $q=6$ and $a=4$ we get
$$\lim_{X\rightarrow \infty}\frac{1}{\log \log X}\sum_{\substack{x \leq X \\ N_{4, 6}(x)>N_{3, 6}(x)>N_{5, 6}(x)>N_{2, 6}(x)>N_{0, 6}(x)>N_{1, 6}(x)}}\frac{1}{x\log x}=\frac{1}{12}$$
$$\lim_{X\rightarrow \infty}\frac{1}{\log \log X}\sum_{\substack{x \leq X \\ N_{4, 6}(x)>N_{5, 6}(x)>N_{3, 6}(x)>N_{0, 6}(x)>N_{2, 6}(x)>N_{1, 6}(x)}}\frac{1}{x\log x}=\frac{1}{12}.$$
\end{exmp}

Theorem~\ref{thm1} follows easily from the proofs of Proposition~\ref{prop7} and Theorem~\ref{thm3}. Theorem~\ref{thm2} follows from Theorem~\ref{thm3} because, of the $2q$ permutations with non-zero loglog density, there are $1/q$ in which $N_{a,q}(x)<N_{b,q}(x)$.

We also prove that certain orderings can occur only a finite number of times.

\begin{theorem}\label{thm4}
Any permutation $\sigma$ of $\{0,1,\ldots, q-1\}$ for which the set $$\{x\in \mathbb{N}\: :\: N_{\sigma(0),q}(x)\geq N_{\sigma(1), q}(x) \geq \ldots \geq N_{\sigma (q-1),q}(x) \}$$ is infinite is such that $\sigma(0) = \sigma (1) \pm 1 \mod q$.
\end{theorem}

Notice that there are $2q!/(q-1)$ such permutations so if $q$ is large, a vanishingly small proportion of the possible permutations occur infinitely often.

\begin{exmp}
If $q=4$ then only 16 out of the 24 orderings can occur for arbitrarily large $x$. There is a point after which, if ``0 is in the lead", then 2 cannot be second and vice versa. Similarly, they cannot simultaneously hold positions 3rd and 4th, and the same goes for the pair of residue classes 1 and 3.
\end{exmp}

Let us look at the start of the mod 4 race before moving on to the proofs. For a better view, the mean has been subtracted and the points are plotted on a loglog scale.

\begin{figure}[H]
{
\centering
\includegraphics[scale=0.45]{omgracesgraph.pdf}
\caption{$\omega$ mod 4 race}
}
\end{figure}

The plotted points along with their colours are as follows:
\begin{align*}
&\text{ Red: }\{\;\big(\log\log x, (\frac{1}{x}N_{0,4}(x)-\frac{1}{4})\log x\big) \::\:10\leq x \leq 10^8\},\\
&\text{ Blue: }\{\;\big(\log\log x, (\frac{1}{x}N_{1,4}(x)-\frac{1}{4})\log x\big) \::\:10\leq x \leq 10^8\},\\
&\text{ Orange: }\{\;\big(\log\log x, (\frac{1}{x}N_{2,4}(x)-\frac{1}{4})\log x\big) \::\:10\leq x \leq 10^8\},\\
&\text{ Green: }\{\;\big(\log\log x, (\frac{1}{x}N_{3,4}(x)-\frac{1}{4})\log x\big) \::\:10\leq x \leq 10^8\}.
\end{align*}

This data strongly suggests, at least for the $q=4$ race, that any ordering not of the form stated in Theorem~\ref{thm4} can $\emph{never}$ occur. It may not be unreasonable to conjecture that this is the case for all $q$.

We remark that similar results can be proved for the sets $$\{n\leq x \: : \: \Omega(n) \equiv a \mod q\} \text{   and   } \{n, \text{ square-free }\leq x \: : \: \omega(n) \equiv a \mod q\}$$ where $\Omega(n) = \sum_{p^k|n}1$ counts prime divisors with multiplicities.

\section{Preliminaries}

To save space, we will use $\log_2 x$ and $e_2(x)$ to denote $\log\log x$ and $\exp(\exp(x))$ respectively. We start by proving an asymptotic formula for $N_{a,q}(x)$.

\begin{prop}\label{prop5}
For $x\geq 3$ we have $$N_{a,q}(x) = \frac{x}{q}\left(1+2|g(\phi)|\cos\Big(\sin\Big(\frac{2\pi}{q}\Big)\log_2 x + \theta -\frac{2\pi a}{q}\Big)(\log x)^{\cos(\frac{2\pi}{q})-1} + O((\log x)^{\cos(\frac{4\pi}{q})-1})\right)$$

where $g(z):=\frac{1}{\Gamma(z)}\prod_p \Big(1+\frac{z}{p-1}\Big)(1-p^{-1})^z$ and $\phi = \phi(q) = e^{2\pi i /q}$ and $\theta = \arg g(\phi)$.
\end{prop}

\begin{proof}

The main result we will make use of is [1, Theorem 2] from which it follows that for $z$ a complex variable bounded in absolute value by 1, and $g(z)$ defined as in the proposition we have
\begin{equation}A_z(x):=\sum_{n\leq x}z^{\omega(n)}=g(z)x(\log x)^{z-1} + O\big(x(\log x)^{\Re z -2})\big).\end{equation}
These mean values satisfy
$$A_{\phi^j}(x)=\sum_{n\leq x}\phi^{j\omega(n)}=\sum_{k=0}^{q-1}\phi^{jk}N_{k,q}(x)\:\:\text{  for all } j = 0, \ldots, q-1.$$
We can isolate the $N_{a,q}(x)$ as follows
$$\frac{1}{q} \sum_{j = 0}^{q-1}\overline{\phi}^{aj} A_{\phi^j} = \frac{1}{q} \sum_{j=0}^{q-1} \overline{\phi}^{aj} \sum_{k=0}^{q-1} \phi^{jk} N_{k,q}(x)
= \sum_{k=0}^{q-1}  N_{k,q}(x)  \frac{1}{q} \sum_{j=0}^{q-1}  \phi^{j(k-a)} = N_{a,q}(x).$$
Substituting (2) into the line above gives
$$N_{a,q}(x)=\frac{x}{q}\left(1+\sum_{j=1}^{q-1}\Big(\overline{\phi}^{aj}g(\phi^j)e^{i\sin(\frac{2\pi j}{q})\log_2 x}(\log x)^{\cos (\frac{2\pi j}{q})-1} + O((\log x)^{\cos (\frac{2\pi j}{q})-2})\Big) \right).$$
For sufficiently large $x$, the terms in this sum with largest absolute value are those with $j=1$ and $j=q-1$. Each of the others is $\ll (\log x)^{\cos(\frac{4\pi}{q})-1}.$ Combining this observation with the fact that $g(\overline{z})=\overline{g(z)}$ we get

\begin{align*}
N_{a,q}(x)&=\frac{x}{q}\left(1+2\Re( \overline{\phi}^a g(\phi)e^{i \sin(\frac{2\pi}{q})\log_2 x})(\log x)^{\cos(\frac{2\pi}{q})-1} + O((\log x)^{\cos(\frac{4\pi}{q})-1})\right) \\
&=\frac{x}{q}\left(1+2|g(\phi)|\cos\Big(\sin\Big(\frac{2\pi}{q}\Big)\log_2 x + \theta -\frac{2\pi a}{q}\Big)(\log x)^{\cos(\frac{2\pi}{q})-1} + O((\log x)^{\cos(\frac{4\pi}{q})-1})\right)
\end{align*}
where $\theta = \arg g(\phi).$

\end{proof}

Notice that if $q=2$ then $\phi=-1$ and $g(-1)=0$. It is for this reason we cannot say any more in this most interesting case. Indeed we actually suspect a much stronger error term of $O(x^{1/2+o(1)})$ in this case. For $q>2$ though, $g(\phi)\neq 0$, as $\Gamma$ has no pole there and the product has only non-zero terms.

From this, we see immediately that
$$N_{a,q}(x) - \frac{x}{q} = O\left(\frac{x}{(\log x)^{1-\cos(\frac{2\pi}{q})}}\right)$$
and also
$$N_{a,q}(x) - \frac{x}{q} = \Omega_{\pm}\left(\frac{x}{(\log x)^{1-\cos(\frac{2\pi}{q})}}\right).$$

Before trying to understand when $N_{a,q}(x)$ is less than this average value of $x/q$, we start with the simpler, but related, question of when the secondary term is negative. That is, when $$\cos\Big(\sin\Big(\frac{2\pi}{q}\Big)\log_2 x + \theta -\frac{2\pi a}{q}\Big)<0.$$ Now cosine is negative ``about half the time" which might suggest that $N_{a,q}(x)$ is less than its average value ``about half the time" too. To be precise, the set $\{n\in \mathbb{N}:\cos n <0\}$ has \textit{natural density} 1/2 in $\mathbb{N}$. That is, $\lim_{x\rightarrow \infty}\frac{1}{x}\sum_{\substack{ n\leq x \\ \cos n<0}}1 =\frac{1}{2}.$ It is not true though that $\{n\in \mathbb{N} : \cos (\log_2 n) < 0\}$ has natural density 1/2. In fact this set has no natural density, as we shall see below. The presence of the $\cos(\log_2 x)$ type term is why we ought to be looking at the $\log\log$ density.

The property of possessing a natural density is stronger than that of possessing a logarithmic density, which is stronger still than having a loglog density. In fact, a straightforward application of partial summation proves that if a set $E\subset \mathbb{N}$ has a natural density then it also has a logarithmic density and the two are equal, and if $E$ has a logarithmic density then it has a loglog density and the two are equal. The following lemma, which we do not prove here, reassures us that our notions of logarithmic and loglog density are sound.

\begin{lemma}\label{lem} There exist constants $\gamma$, $\mu$ such that
\begin{align*}
\sum_{1\leq n \leq x}\frac{1}{n} &= \log x + \gamma + O\Big(\frac{1}{x}\Big).\\
\sum_{2\leq n \leq x}\frac{1}{n\log n} &= \log_2 x + \mu + O\Big(\frac{1}{x \log x}\Big).
\end{align*}
\end{lemma}

\begin{prop}\label{prop7}
Let $A, B \in \mathbb{R}$ with $A > 0$. The set $$E:=\{n\in \mathbb{N} : \cos(A\log_2(n) + B)<0\}$$ has \emph{loglog density} 1/2 but no natural or logarithmic densities.
\end{prop}

\begin{proof}
For $N\in \mathbb{N}$, let $x_N:= e_2((2N\pi-\pi/2-B)/A) $ and $y_N:=e_2((2N\pi+\pi/2-B)/A)$ so that
\begin{align*}
\cos(A\log_2 x+B) > 0 &\Leftrightarrow x\in (x_N,y_N) \text{ for some } N \text{ and }\\
\cos(A\log_2 x+B) < 0 &\Leftrightarrow x\in (y_{N-1},x_{N}) \text{ for some }N.
\end{align*}
Writing $[\alpha]$ for the largest integer at most $\alpha$, we therefore have
$$\frac{1}{x_N}\sum_{\substack{n\leq x_N \\ n\in E}}1 \geq \frac{[x_N]-[y_{N-1}]}{x_N} \rightarrow 1 \text{ as } N\rightarrow \infty$$
and
$$\frac{1}{y_N}\sum_{\substack{n\leq y_N \\ n\in E}}1 \leq 1 -  \frac{[y_N]-[x_N]}{y_N} \rightarrow 0 \text{ as } N\rightarrow \infty,$$
so $E$ certainly doesn't have a natural density. When we look at the logarithmic density, we get, since $\log y_N = e^{\pi/A}\log x_N,$
$$\frac{1}{\log x_N}\sum_{\substack{n\leq x_N \\ n\in E}}\frac{1}{n} = \frac{e^{\pi/A}}{\log y_N}\sum_{\substack{n\leq y_N \\ n\in E}}\frac{1}{n},$$
so for the limit to exist as $N\rightarrow \infty$ we would need $e^{\pi/A} = 1$ which is impossible.

When we look at the loglog density however, we get, using Lemma~\ref{lem}
\begin{align*}
\limsup_{x\rightarrow \infty}\frac{1}{\log_2 x}\sum_{\substack{n\leq x \\ n\in E}}\frac{1}{n\log n} &= \lim_{N\rightarrow \infty}\frac{1}{\log_2 x_N}\sum_{m=2}^{N}\sum_{n\in(y_{m-1}, x_m)}\frac{1}{n\log n} \\
&= \lim_{N\rightarrow \infty}\frac{1}{\log_2 x_N}\sum_{m=2}^{N}\left(\log_2(x_{m})-\log_2(y_{m-1})+\mu - \mu +O\Big(\frac{1}{m\log m}\Big) \right)\\
&= \lim_{N\rightarrow \infty}\frac{1}{(2N\pi-\pi/2-B)/A}\sum_{m=2}^{N}\left(\pi/A + O\Big(\frac{1}{m\log m}\Big) \right)\\
&= \frac{1}{2}.
\end{align*}
A similar calculation shows that $\liminf_{x\rightarrow \infty}\frac{1}{\log_2 x}\sum_{\substack{n\leq x \\ n\log n\in E}}\frac{1}{n\log n} =\frac{1}{2}$ and the result follows.
\end{proof}

It is tempting to conclude that the set $$\{x \in \mathbb{N} \::\: N_{a, q}(x)<\frac{x}{q}\}$$ has no natural or logarithmic densities but has loglog density 1/2. Unfortunately, to prove this rigorously we will need to account for the error introduced by the terms we have left out. We will do this shortly. If we forget about error terms for the moment though (which we can only really do when $\sin(2\pi/q)\log_2 x +\theta-\frac{2\pi a}{q}$ is not too close to a zero of $\cos$), then asking ``for which value of $a$ is $N_{a,q}(x)$ largest" is tantamount to asking ``for which value of $a$ is $\frac{2\pi}{q}\Big(\frac{q}{2\pi}(\sin(2\pi/q)\log_2 x +\theta)-a\Big)$ closest to some $2n\pi \in 2\pi\mathbb{Z}$". The answer is the closest integer to $\frac{q}{2\pi}(\sin(2\pi/q)\log_2 x +\theta)$ modulo $q$ which clearly depends on $x$. Any given $a$ will therefore produce the most values of $n\leq x$ such that $\omega(n)\equiv a \mod q$ when there exists some $m\in \mathbb{Z}$ such that, $$a-\frac{1}{2} < \frac{q}{2\pi}(\sin(2\pi/q)\log_2 x +\theta)+mq < a +\frac{1}{2}.$$ A similar calculation to that in the proof of Proposition~\ref{prop7} shows that for each $a$, the set of such $x$ values has loglog density $1/q$.

We end this section with a picture of the curves
$$\cos\Big(\sin\Big(\frac{2\pi}{6}\Big)\log_2 x + \theta -\frac{2\pi a}{6}\Big)$$
for $a = 0 , 1, \ldots, 5$ plotted on a scale which makes the oscillations visible.

\begin{figure}[H]
{
\centering
\includegraphics[scale=0.37]{sinsketch.pdf}
\caption{Shifted sinusoidal curves}
}
\end{figure}

Although this picture isn't to be taken too seriously it serves as a useful illustration to have in mind for comparing the secondary terms.

\section{Proof of Theorem 3}

\begin{proof}
Let $q\geq 3$ be some fixed integer, $a \in \{0,1,\ldots, q-1\}$ and $\theta$ be defined as in Proposition~\ref{prop5}. Let $\epsilon>0$ be small and define $$U_{a,q}^{\epsilon-}:=\{x\in\mathbb{N}\::\: -\frac{\pi}{q}+\sqrt{\epsilon}<\sin(\frac{2\pi}{q})\log_2 x + \theta - \frac{2\pi a}{q} + 2n\pi < -\sqrt{\epsilon} \text{ for some }n\in \mathbb{Z}\},$$
$$ U_{a,q}^{\epsilon+}:=\{x\in\mathbb{N}\::\: -\frac{\pi}{q}-\sqrt{\epsilon}<\sin(\frac{2\pi}{q})\log_2 x + \theta - \frac{2\pi a}{q} + 2n\pi < \sqrt{\epsilon} \text{ for some }n\in \mathbb{Z}\}.$$
First let us see how, for small $\epsilon$, these sets approximate $U_{a,q}$. Our formula for $N_{a,q}(x)$ gives
\begin{multline*}
\frac{N_{a,q}(x)-N_{b,q}(x)}{(\log x)^{\cos(\frac{2\pi}{q})-1}}=\\2|g(\phi)|\Big(\cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi a}{q}\Big)- \cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi b}{q}\Big)\Big) +o(1)
\end{multline*}
as $x \rightarrow \infty$ and for $q\geq3$ we have $g(\phi)\neq0$. Therefore, for all $\epsilon>0$ there exists some $X_0(\epsilon)$ such that for $x\geq X_0$ and for each $a, b \in \{0, \ldots, q-1\}$, $$\cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi a}{q}\Big)- \cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi b}{q}\Big)>\epsilon$$
which implies that $N_{a,q}(x)>N_{b,q}(x).$ We will use this fact to prove that for $x\geq X_0$ and $\epsilon$ sufficiently small we have
\begin{equation}
x\in U^{\epsilon-}_{a,q} \Rightarrow x \in U_{a,q}
\end{equation}
and
\begin{equation}
x \in U_{a,q} \Rightarrow x \in U^{\epsilon+}_{a,q}.
\end{equation}
It follows that $$ \sum_{\substack{ x\leq X \\ x\in U_{a, q}^{\epsilon-}}}\frac{1}{x \log x} + O_\epsilon(1) \leq \sum_{\substack{ x\leq X \\ n\in U_{a, q}}}\frac{1}{x \log x} \leq \sum_{\substack{ x\leq X \\ x\in U_{a, q}^{\epsilon+}}}\frac{1}{x \log x}+O_\epsilon(1).$$
After showing that each of $U_{a,q}^{\epsilon\pm}$ has loglog density $\frac{1}{2q}+o_\epsilon(1)$ for arbitrarily small $\epsilon$, where $o_\epsilon(1)$ is a quantity that tends to 0 as $\epsilon \rightarrow 0$, we will have shown that $U_{a,q}$ has loglog density $\frac{1}{2q}$. The result for $V_{a,q}$ is proved in much the same way.\\

\par

\textit{Proof of (3) and (4)}.

Suppose $x\geq X_0$ and $x\in U^{\epsilon-}_{a,q}$ and $\epsilon$ is small enough so that $\sin(\frac{\pi}{q})\sin(\sqrt{\epsilon})>\epsilon.$ For example, $\epsilon < 1/q^2$ will do. In order to show that $x \in U_{a,q}(x)$ we need to show
\begin{enumerate}[label=(\alph*)]
\item        $N_{a-i,q}(x)>N_{a+i,q}(x)$ for all $i \in \{ 1, 2, \ldots, [\frac{q-1}{2}] \}$
\item        $N_{a+i,q}(x)>N_{a-i-1,q}(x)$ for all $ i \in \{ 0, 1, \ldots, [\frac{q-2}{2}] \}$.
\end{enumerate}
To do so we will use the identity
\begin{equation}\cos(\xi + A) - \cos(\xi + B) = -2\sin\left(\frac{A-B}{2}\right)\sin\left(\xi + \frac{A+B}{2}\right).\end{equation}
For (a), let $i \in \{ 1, 2, \ldots, \text{ or } [\frac{q-1}{2}] \}$, then
\begin{multline*}
\cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi (a-i)}{q}\Big)- \cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi (a+i)}{q}\Big)\\=-2\sin(\frac{2\pi i}{q})\sin(\xi)
\end{multline*}
where $\xi = \sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi a}{q} \in \left(-\frac{\pi}{q}+ \sqrt{\epsilon}, -\sqrt{\epsilon} \right)$ which is $>\epsilon.$ This proves (a). For (b), let  $ i \in \{ 0, 1, \ldots, \text{ or } [\frac{q-2}{2}] \}$, then
\begin{multline*}\cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi (a+i)}{q}\Big)- \cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi (a-i-1)}{q}\Big)\\
=2\sin(\frac{\pi(2i+1)}{q})\sin(\xi+\frac{\pi}{q})
\end{multline*}
where again $\xi \in \left(-\frac{\pi}{q}+ \sqrt{\epsilon}, -\sqrt{\epsilon} \right)$ which is again $>\epsilon.$ This proves (b) and that $x \in U_{a,q}(x).$

Now suppose $x\geq X_0$ and $x \in U_{a,q}$. To show that $x\in U_{a,q}^{\epsilon+}$ we need to find some $n\in \mathbb{Z}$ such that $$-\frac{\pi}{q}-\sqrt{\epsilon}<\sin(\frac{2\pi}{q})\log_2 x + \theta - \frac{2\pi a}{q} + 2n\pi < \sqrt{\epsilon}.$$ Suppose this is not the case. Then either we can find some $n$ such that $$-\pi-\sqrt{\epsilon} \leq \sin(\frac{2\pi}{q})\log_2 x + \theta - \frac{2\pi a}{q} + 2n\pi \leq  -\frac{\pi}{q}-\sqrt{\epsilon}.$$ In which case
\begin{multline*}
\cos\Big(\sin(\frac{2\pi}{q})\log_2 x+\theta -\frac{2\pi (a-1)}{q}\Big)-\cos\Big(\sin(\frac{2\pi}{q})\log_2 x+\theta -\frac{2\pi a}{q}\Big) \\
=-2 \sin(\frac{\pi}{q})\sin(\xi + \frac{\pi }{q})
\end{multline*} for some $ \xi\in\Big[-\pi-\sqrt{\epsilon}, -\frac{\pi}{q}-\sqrt{\epsilon}\Big]$. This is then $>\epsilon$ and so $N_{a-1, q}(x)>N_{a, q}(x)$, contrary to our assumption on $x$.
Or else we can find some $n$ such that $$\sqrt{\epsilon} \leq \sin(\frac{2\pi}{q})\log_2 x + \theta - \frac{2\pi a}{q} + 2n\pi  \leq \pi-\sqrt{\epsilon}.$$ In which case
\begin{multline*}
\cos\Big(\sin(\frac{2\pi}{q})\log_2 x+\theta -\frac{2\pi (a+1)}{q}\Big)-\cos\Big(\sin(\frac{2\pi}{q})\log_2 x+\theta -\frac{2\pi (a-1)}{q}\Big) \\
=-2\sin( -\frac{2\pi}{q} ) \sin(\xi   )
\end{multline*} for some $ \xi\in (\sqrt{\epsilon}, \pi-\sqrt{\epsilon})$. Again making the difference $>\epsilon$ and so $N_{a+1, q}(x)>N_{a-1,q}(x)$, again contrary to our assumption on $x$. We must therefore have $x \in U_{a,q}^{\epsilon+}$.\\

It remains to calculate the loglog densities of $U^{\epsilon-}_{a,q}$ and $U^{\epsilon+}_{a,q}$. This is similar to the proof of Proposition~\ref{prop7}. For $N\in\mathbb{N}$ define $x_N:=e_2((2N\pi+2a\pi/q-\theta-\pi/q+\sqrt{\epsilon})/\sin(2\pi/q))$ and $y_N:=e_2((2N\pi+2a\pi/q-\theta-\sqrt{\epsilon})/\sin(2\pi/q))$, then for $x\geq X_0$ we have $x \in U^{\epsilon-}_{a, q}$ if and only if $x\in (x_N, y_N)$ for some $N$ and we therefore have
\begin{align*}
\liminf_{X\rightarrow \infty}\frac{1}{\log_2 X} \sum_{ \substack{n\leq x \\ n\in U^{\epsilon-}_{a, q} }}\frac{1}{n \log n} &= \lim_{N \rightarrow \infty}\frac{1}{\log_2 x_N}\sum_{m=1}^{N-1}\sum_{n\in (x_m, y_m)}\frac{1}{n\log n} \\
&=  \lim_{N \rightarrow \infty}\frac{1}{\log_2 x_N}\sum_{m=1}^{N-1}\left(\frac{(\pi/q)-2\sqrt{\epsilon}}{\sin(2\pi/q)}+O\Big(\frac{1}{m \log m}\Big)\right) \\
&= \lim_{N \rightarrow \infty}\frac{\sin(2\pi/q)}{2N\pi+2a\pi/q-\theta-\pi/q+\sqrt{\epsilon}}\Big((N-1)\frac{(\pi/q)-2\sqrt{\epsilon}}{\sin(2\pi/q)}+O(\log N) \Big)\\
&=\frac{1}{2q} - \frac{\sqrt{\epsilon}}{\pi}.
\end{align*}
Also,
\begin{align*}
\limsup_{X\rightarrow \infty}\frac{1}{\log_2 X} \sum_{ \substack{n\leq x \\ n\in U^{\epsilon-}_{a, q} }}\frac{1}{n\log n} &= \lim_{N \rightarrow \infty}\frac{1}{\log_2(y_N))}\sum_{m=1}^{N}\sum_{n\in (x_m, y_m)}\frac{1}{n\log n}=\frac{1}{2q} - \frac{\sqrt{\epsilon}}{\pi}
\end{align*}

Hence the loglog density of $U^{\epsilon-}_{a,q}$ exists and is equal to $\frac{1}{2q}-\frac{\sqrt{\epsilon}}{\pi}$. A very similar calculation shows that the loglog density of $U^{\epsilon+}_{a,q}$ is $\frac{1}{2q}+\frac{\sqrt{\epsilon}}{\pi}$ and so by (4) and the fact that $\epsilon$ can be taken arbitrarily small we can conclude that $U_{a,q}$ has loglog density $\frac{1}{2q}$.

\end{proof}

\section{Proof of Theorem 4}

As in the previous proof, let $\epsilon$ be small enough and $X_0$ large enough so that $\sin(\frac{\pi}{q})\sin(\sqrt{\epsilon})>\epsilon$ and that for $x\geq X_0$ we have $$\cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi a}{q}\Big)- \cos\Big(\sin(\frac{2\pi}{q})\log_2 x + \theta -\frac{2\pi b}{q}\Big)>\epsilon \Rightarrow N_{a,q}(x)>N_{b,q}(x)$$
and
$$x\in U^{\epsilon-}_{a,q} \Rightarrow x \in U_{a,q}$$
and
$$x \in U_{a,q} \Rightarrow  x \in U^{\epsilon+}_{a,q}.$$

We will show that only the permutations stated in Theorem~\ref{thm4} can occur for $x\geq X_0$. Suppose that we have some $x\geq X_0$ for which $N_{a,q}(x)$ is leading, that is, $\max_{c \:mod\: q}N_{c,q}(x) = N_{a,q}(q)$. It follows that $x\in U_{a,q}^{\epsilon+} \cup V_{a,q}^{\epsilon+}$ since otherwise there would be some integer $n$ such that
$$\frac{\pi}{q}+\sqrt{\epsilon} \leq \sin(\frac{2\pi}{q})\log_2 x + \theta - \frac{2\pi a}{q} + 2n\pi  \leq 2\pi-\frac{\pi}{q}-\sqrt{\epsilon}$$
and hence
\begin{multline*}
\cos\Big(\sin(\frac{2\pi}{q})\log_2 x+\theta -\frac{2\pi b}{q}\Big)-\cos\Big(\sin(\frac{2\pi}{q})\log_2 x+\theta -\frac{2\pi a}{q}\Big) \\
=-2\sin( \frac{\pi(a-b)}{q} ) \sin(\xi +\frac{\pi(a-b)}{q}  )
\end{multline*}
for some $\xi \in (\frac{\pi}{q}+\sqrt{\epsilon}\:,\:\: 2\pi -\frac{\pi}{q} - \sqrt{\epsilon}).$ But then this is $>\epsilon$ for either $b=a+1 \mod q$ or $b=a-1 \mod q$ contradicting the assumption that $N_{a,q}(x)$ was leading.

To prove Theorem~\ref{thm4} it suffices to prove that $\max _{\pm 1}N_{a \pm 1, q}(x) > N_{b,q}(x)$ for all $b \neq a, a\pm1 \mod q$. This follows, in a by now familiar fashion, from the fact that there exists an integer $n$ such that
$$-\frac{\pi}{q}-\sqrt{\epsilon} < \sin(\frac{2\pi}{q})\log_2 x + \theta - \frac{2\pi a}{q} + 2n\pi  < \frac{\pi}{q}+\sqrt{\epsilon}$$
since then
\begin{align*}
&\max _{\pm} \:\:\:\cos\Big(\sin(\frac{2\pi}{q})\log_2 x+\theta - \frac{2\pi(a \pm 1)}{q}\Big)-\cos\Big(\sin(\frac{2\pi}{q})\log_2 x+\theta -\frac{2\pi b}{q}\Big) \\
&=\max_{\pm} \:\:\:-2\sin( \pi\frac{b-(a \pm 1)}{q}) \sin(\xi -\frac{\pi(b-a\pm 1)}{q}  )\\
&=\max_{\pm} \:\:\: 2\sin( \pi\frac{b-(a \pm 1)}{q}) \sin(\pi\frac{(b-a\pm 1)}{q} -\xi )
\end{align*}
where $\xi \in (-\frac{\pi}{q}-\sqrt{\epsilon}\:,\:\: \frac{\pi}{q} + \sqrt{\epsilon})$, so this is $>\epsilon$ for $b \neq a, a\pm 1 \mod q$ which proves the claim.

\section*{Acknowledgements}

The author would like to thank his supervisor, Andrew Granville, for suggesting this question. This work was supported by the Engineering and Physical Sciences Research Council EP/L015234/1 via the EPSRC Centre for Doctoral Training in Geometry and Number Theory (The London School of Geometry and Number Theory), University College London.

\nocite{Mont-Vau}
\nocite{Sel-Sat}


TITLE: prove or disprove: $\lim_{x\to \infty} \frac{f(x)}{g(x)}=\lim \frac{f'(x)}{g'(x)}=0 \implies \lim \frac{\frac{f(x)}{g(x)}}{\frac{f'(x)}{g'(x)}}\ne 0$
QUESTION [4 upvotes]: my attempt $$\lim_{x\to \infty}\frac{\frac{f(x)}{g(x)}}{\frac{f'(x)}{g'(x)}}\text{ yields }\frac{0}{0}$$ then use l'hopital on this
$$\lim_{x\to \infty}\frac{\frac{f(x)}{g(x)}}{\frac{f'(x)}{g'(x)}} =\lim_{x\to \infty} \frac{\frac{f'}{g}-\frac{fg'}{g^2}}{\frac{f''}{g'}-\frac{f'g''}{g'^2}} $$
$$\lim_{x\to \infty}\frac{\frac{f(x)}{g(x)}}{\frac{f'(x)}{g'(x)}} =\lim_{x\to \infty} \frac{\frac{f'}{f}\frac{f}{g}-\frac{f}{g}\frac{g'}{g}}{\frac{f''}{f'}\frac{f'}{g'}-\frac{f'}{g'}\frac{g''}{g'}} $$
$$\lim_{x\to \infty}\frac{\frac{f}{g}}{\frac{f'}{g'}} =\lim_{x\to \infty} \frac{\frac{f}{g} \left( \frac{f'}{g'}-\frac{f}{g}\right)}{\frac{f'}{g'} \left( \frac{f''}{f'}-\frac{g''}{g'}\right)} $$ assuming such limits exists, and equal to $L$,  leads to
$$\lim_{x\to \infty}\frac{\frac{f}{g}}{\frac{f'}{g'}} =\lim_{x\to \infty} \frac{\frac{f}{g}  }{\frac{f'}{g'}  }\cdot \lim \frac{  \left( \frac{f'}{g'}-\frac{f}{g}\right)}{ \left( \frac{f''}{f'}-\frac{g''}{g'}\right)} $$
$$L =L\cdot \lim_{x\to \infty} \frac{  \left( \frac{f'}{g'}-\frac{f}{g}\right)}{ \left( \frac{f''}{f'}-\frac{g''}{g'}\right)}  $$
I am stuck here..
I would really like to prove that $L$ is not zero and not $\infty$ the proof is  obvious for polynomials functions f, and g, and I can not find any counter examples.. any help would be much appreciated..
we can assume the initial condition comes from appropriately using l'hopital

REPLY [1 votes]: Just saw Barry Cipra had given the same answer...not sure how I could have missed that.
If $f=e^{-x^2}$ and $g=e^{-x}$, you get
$$
\lim_{x\rightarrow\infty}\frac{f}{g}
=\lim_{x\rightarrow\infty}\frac{f'}{g'}
=\lim_{x\rightarrow\infty}\frac{f/f'}{g/g'}
=0
$$
which is a counter-example to your hypothesis.
For the alternative hypothesis ($L=\infty$), you can set $f=e^x$, $g=e^{x^2}$ to get
$$
\lim_{x\rightarrow\infty}\frac{f}{g}
=\lim_{x\rightarrow\infty}\frac{f'}{g'}
=0,
\quad
\lim_{x\rightarrow\infty}\frac{f/f'}{g/g'}
=\infty.
$$