text:
\begin{document}
\title  [Curvature estimates for the level sets]
{Curvature estimates for the level sets of spatial quasiconcave solutions to a class
of parabolic equations}
\author{Chuanqiang Chen}
\address{Department of Mathematics\\
         University of Science and Technology of China\\
         Hefei 230026, Anhui Province, CHINA.}
\email{cqchen@mail.ustc.edu.cn}
\author{Shujun Shi}
\address{Department of Mathematics\\
         University of Science and Technology of China\\
         Hefei 230026, Anhui Province, CHINA\\
         and
         School of Mathematical Sciences\\
         Harbin Normal University\\
         Harbin 150025, Heilongjiang Province, CHINA.}
\email{shjshi@mail.ustc.edu.cn}
\thanks{2000 Mathematics Subject Classification: 45B99, 35K10.}
\thanks{Keywords and phrases: curvature estimates, level sets, constant rank theorem, spatial quasiconcave solutions.}
\thanks{Research of the first author was supported by Grant 10871187 from the
National Natural Science Foundation of China. Research of the second
author was supported in part by the Science Research Program from
the Education Department of Heilongjiang Province (11551137).}

\begin{abstract} We prove a constant rank theorem for the second
fundamental form of the spatial convex level surfaces of solutions
to equations $u_t=F(\n^2u, \n u, u, t)$ under a structural
condition, and give a geometric lower bound of the principal
curvature of the spatial level surfaces.
\end{abstract}

\maketitle

\section{Introduction}

In this paper, we consider the convexity and principal curvature estimates of the spatial level surfaces of the spatial quasiconcave solutions to a
class of parabolic equations under some structural conditions. A
continuous function $u(x,t)$ on $\Omega \times [0,T]$ is called {\it
spatial quasiconcave} if its level sets $\{x \in \Omega| u(x,t)\ge
c\}$ are convex for each constant $c$ and any fixed $t \in [0,T]$.

The convexity of the level sets of the solutions to elliptic partial
differential equations has been studied extensively. For instance, Ahlfors \cite{AH} contains the well-known result that level curves of Green function on simply connected convex domain in the plane are the convex Jordan curves. In 1956, Shiffman \cite{Sh56} studied the minimal annulus in $\mathbb{R}^3$ whose
boundary consists of two closed convex curves in parallel planes $P_1, P_2$. He proved that the intersection of the surface with any parallel plane $P$, between $P_1$ and $P_2$, is a convex Jordan curve. In 1957, Gabriel \cite{Ga57} proved that the level sets of the Green function on a 3-dimensional bounded convex domain are strictly convex. In 1977, Lewis \cite{Le77} extended Gabriel's result to $p$-harmonic functions in higher dimensions.
Caffarelli-Spruck \cite{CS82} generalized the Lewis \cite{Le77} results to a class of semilinear elliptic partial differential equations. Motivated by the result of Caffarelli-Friedman \cite{CF85}, Korevaar \cite{Ko90} gave a new proof on the results of Gabriel and Lewis by applying the deformation process and the constant rank theorem of the second fundamental form of the convex level sets of $p$-harmonic function. A survey of this subject is given by Kawohl \cite{Ka85}. For more recent related extensions, please see the papers by Bianchini-Longinetti-Salani \cite{BLS}, Bian-Guan \cite{BG09}, Xu \cite{Xu08} and Bian-Guan-Ma-Xu \cite{BGMX}.

 There is also an extensive literature on the curvature estimates of the level sets of the solutions to elliptic partial differential equations. For 2-dimensional harmonic function and minimal surface with convex level curves, Ortel-Schneider \cite{OS}, Longinetti \cite{Lo83} and \cite{Lo87} proved that the curvature of the level curves attains its minimum on the boundary (see Talenti \cite{T83} for related
results). Longinetti also studied the precise relation between the curvature of the convex level curves and the height of 2-dimensional minimal surface in \cite{Lo87}.  Ma-Ou-Zhang \cite{MOZ09} got the Gaussian curvature estimates of the convex level sets on higher dimensional harmonic function, and Wang-Zhang \cite{WZ} got the similar curvature estimates of some quasi-linear elliptic equations under certain structure condition \cite{BLS}. Both of their test functions involved the Gaussian curvature of the boundary and the norm of the gradient on the boundary. Furthermore, for the $p$-harmonic function with strictly convex level sets, Ma-Zhang \cite{MZ} obtained that the curvature function introduced in it is concave with respect to the height of the $p$-harmornic function. For the principal curvature estimates in higher dimension, in terms of the principal curvature of the boundary and the norm of the gradient on the boundary, Chang-Ma-Yang \cite{CMY} obtained the lower bound estimates of principal curvature for the strictly convex level sets of higher dimensional harmonic functions and solutions to a class of semilinear elliptic equations under certain structure condition \cite{BLS}. Recently, in Guan-Xu \cite{GX}, they got a lower bound for the principal curvature of the level sets of solutions to a class of fully nonlinear elliptic equations in convex rings under the general structure condition \cite{BLS} via the approach of constant rank theorem.

Naturally, we hope to give a  characterization about the convexity and curvature of the level surfaces of the solutions to the corresponding parabolic equations. Borell \cite{Bo} showed the same property in \cite{Ga57} and \cite{Le77} for the solution of the corresponding heat conduction problem with zero initial data. In this paper, we will consider the following parabolic equations
\begin{equation}\label{1.1}
\frac{{\partial u}} {{\partial t}} = F(\n ^2u, \n u, u, t),
\quad~\text{in}~ \Omega  \times (0,T],
\end{equation}
where $\Omega$ is a domain in $\R^n$, and $\n ^2 u$, $\n u$ are the
spatial Hessian and spatial gradient of $u(x,t)$ respectively. Let
$\S^n$ denote the space of real symmetric $n \times n$ matrices,
$\Lambda\subset \S^n$ an open set,  $\mathbb S^{n-1}$ a unit sphere and $F=F(r,p,u,t)$  a
$C^{2,1}$ function in $\Lambda \times \R^n \times \R \times [0,T] $.
We will assume that $F$ satisfies the following conditions: there are
$\gamma_0>0$ and $c_0\in \mathbb R$,
\begin{equation}\label{1.2}
F^{\alpha\beta}:=\left(\frac{\partial F}{\partial
r_{\alpha\beta}}(r, p, u, t)\right) >0, \quad \forall \; (r, p, u,
t) \in \Lambda\times \mathbb R^{n}\times (-\gamma_0+c_0,
\gamma_0+c_0) \times [0,T],
\end{equation}
and for each $(\theta,u)\in \mathbb S^{n-1}\times \mathbb R$
fixed,
\begin{equation}\label{1.3}
F(s^2 A, s \theta, u, t) \text{ is locally concave in  } (A,s)
\text{ for each fixed }  t.
\end{equation}

Now we state our theorems.
\begin{theorem}\label{th1.1}
Suppose $u \in C^{3,1}(\Omega \times [0,T])$ is a spatial
quasiconcave solution to parabolic equation (\ref{1.1}) such that
$(\n^2u(x,t), \n u(x,t), u(x,t))\in \Lambda\times \mathbb R^n\times
(-\gamma_0+c_0, \gamma_0+c_0)$ for each $(x,t)\in \Omega \times [0,
T]$. Suppose that, $F$ satisfies conditions (\ref{1.2}) and
(\ref{1.3}),  $\n u \neq 0$ and the spatial level sets $\{x \in
\Omega |u(x,t) \ge c\}$ of $u$ are connected and locally convex for
all $c\in (-\gamma_0+c_0, \gamma_0+c_0)$ for some $\gamma_0>0$. Then
the second fundamental form of spatial level surfaces $\{x \in
\Omega |u(x,t) = c\}$ has the same constant rank for all $c\in
(-\gamma_0+c_0, \gamma_0+c_0)$. Moreover, let $l(t)$ be the minimal
rank of the second fundamental form in $\Omega$, then $l(s)
\leqslant l(t)$ for all $s \leqslant t \leqslant T$.
\end{theorem}

Inspired by \cite{GX}, we also consider to establish a geometric
lower bound for the principal curvature of the spatial level
surfaces of solutions to parabolic equation on the convex rings as
follows,
\begin{equation}\label{1.4}
\left\{\begin{array}{lcl} \frac{{\partial u}} {{\partial t}} =
F(\n ^2u, \n u, u, t)   &\text{in}& {\Omega  \times (0,T]}, \\
              u(x,0) = u_0 (x)  &\text{in}& \Omega,   \\
              u(x,t) =  0   &\text{on}&  \partial \Omega_0 \times (0,T],\\
              u(x,t) =  1   &\text{on}&  \partial \Omega_1 \times (0,T],
\end{array} \right.
\end{equation}
where $\Omega  = \Omega _0 \backslash \overline {\Omega _1 }$,
$\Omega_0$, $\Omega_1$ are two convex domains with $\overline
{\Omega _1 } \subset \Omega_0$, $F(\n ^2u_0, \n u_0, u_0, 0)>0$ and $u_0$ is quasiconcave and satisfies
\begin{equation}\label{1.5}
\left\{\begin{array}{lcl}
          u_0 =  0  & \text{on}&  \partial \Omega_0,\\
              u_0 =  1  & \text{on}&  \partial \Omega_1.
\end{array} \right.
\end{equation}
We denote $\kappa_s(x,t)$ the smallest principal curvature of
the spatial level set $\Sigma^{u(x_0,t)}=\{x \in \Omega |u(x,t) =
u(x_0,t)\}$ at $(x,t)$. For each $(x_0,t)$, set
\begin{equation}
\kappa^{u(x_0,t)}=\mathop {\inf }\limits_{x \in
\Sigma^{u(x_0,t)}}\kappa_s(x,t).
\end{equation}
We will assume that there exists $\lambda >0$, such that
\begin{equation}\label{1.7}
(F^{\alpha\beta}(\n ^2u, \n u, u, t)) \geq \lambda (\delta_{\a\b}),
\quad \forall (x,t) \in \overline{\Omega} \times [0,T].
\end{equation}

\begin{theorem}\label{th1.2}
Suppose $u \in C^{3,1}(\Omega \times [0,T])$ is a spatial
quasiconcave solution to parabolic equation (\ref{1.4}), and $F$
satisfies conditions (\ref{1.7}) and (\ref{1.3}),  $\n u \neq 0$,
then

\begin{equation}\label{1.8}
\kappa^{u(x,t)}\geq \min\{\kappa^0,\kappa^1e^{-A} \}e^{Au(x,t)}
\end{equation}
for some universal constant $A$ depending only on $ \left\| F
\right\|_{C^2 } $, $ n $, $ \lambda $, $ \mathop {\min
}\limits_{(x,t) \in \overline \Omega   \times [0,T]} \left| {\nabla
u} \right| $, $ \left\| u \right\|_{C^3 } $.
Moreover, if
$"="$ holds for some $u(x,t)\in (0, 1)$, then the $"="$ holds for all $u(x,t)\in [0, 1]$.

\end{theorem}

Theorem \ref{th1.1} and Theorem \ref{th1.2} may be looked as some parabolic versions for Theorem 1.1 in \cite{BGMX} and Theorem 1.5 in \cite{GX} respectively. The main idea to prove the main theorems in this paper can be found in the two literatures.

The rest of the paper is organized as follows. In section 2, we
prove Theorem \ref{th1.1}. In section 3, we prove Theorem
\ref{th1.2}.

\textbf{Acknowledgement} The authors would like to express sincere
gratitude to Prof. Xi-Nan Ma for his encouragement and many
suggestions in this subject.

\section{Proof of Theorem \ref{th1.1}}

Suppose $u(x,t) \in C^{3,1}(\Omega \times [0.T])$, and $u_n \ne 0$
for any fixed $(x,t)\in \Omega \times [0,T]$. It follows that the
upward inner normal direction of the spatial level sets $\{x \in
\Omega |u(x,t) = c\}$ is
\begin{eqnarray}\label{2.1}
\vec{n}= \frac{|u_n|}{|\n u|u_n}(u_1,u_2,...,u_{n-1},u_n),
\end{eqnarray}
where $\n u=(u_1,u_2,...,u_{n-1},u_n)$ is the spatial gradient of
$u$.

The second fundamental form $II$ of the spatial level surface of
function $u$ with respect to the upward normal direction (\ref{2.1})
is
\begin{equation}\label{2.2}
b_{ij} =  - \frac{|u_n|(u_n^2 u_{ij} + u_{nn}u_{i}u_j - u_nu_ju_{in}
- u_nu_iu_{jn})}{|\n u|u_n^3}.
\end{equation}
Set
\begin{equation}\label{2.3}
h_{ij} =u_n^2 u_{ij} + u_{nn}u_{i}u_j - u_nu_ju_{in} - u_nu_iu_{jn},
\end{equation}
we may write
\begin{equation}\label{2.4}
b_{ij} = - \frac{|u_n|h_{ij}}{|\n u|u_n^3}.
\end{equation}
Note that if $\Sigma ^{c,t} = \{x \in \Omega |u(x,t) = c\}$ is
locally convex, then the second fundamental form of $\Sigma^{c,t}$
is semipositive definite with respect to the upward normal direction
(\ref{2.1}). Let $a(x,t)=(a_{ij}(x,t))$ be the symmetric Weingarten
tensor of $\Sigma^{c,t}= \{x \in \Omega |u(x,t) = c \}$, then $a$ is
semipositive definite. As computed in \cite{BGMX}, if $u_n \ne 0$,
and the Weingarten tensor is
\begin{equation}\label{2.5}
a_{ij} =-\frac{|u_n|}{|\n u|{u_n}^3}\left\{h_{ij}
-\frac{u_iu_lh_{jl}}{W(1+W)u_n^2} -\frac{u_ju_lh_{il}}{W(1+W)u_n^2}
+ \frac{u_iu_ju_ku_l h_{kl}}{W^2(1+W)^2u_n^4}\right\}.
\end{equation}
With the above notations, at the point $(x,t)$ where $u_n(x,t)=|\nabla
u(x,t)|>0,\, u_i(x,t)=0$, $i=1, \cdots, n-1$, $a_{ij,k}$ is
commutative, that is, they satisfy the Codazzi property
$a_{ij,k}=a_{ik,j}, \; \forall i,j,k \le n-1$.

\subsection{Calculations on the test function}

Since Theorem \ref{th1.1} is of local feature, we may assume level
surface $\Sigma^{c,t} = \{x \in \Omega| u(x,t)=c\}$ is connected for
each $c\in (c_0-\gamma_0, c_0+\gamma_0)$. Suppose $a(x,t_0)$ attains
minimal rank $l=l(t_0)$ at some point $z_0 \in \Omega$. We may
assume $l\leqslant n-2$, otherwise there is nothing to prove.  And
we assume $u \in C^{3,1}(\Omega\times [0,T])$ and $u_n>0$ in the
rest of this paper. So there is a neighborhood $\mathcal {O}\times
(t_0-\delta, t_0+\delta]$ of $(z_0, t_0)$, such that there are $l$
"good" eigenvalues of $(a_{ij})$ which are bounded below by a
positive constant, and the other $n-1-l$ "bad" eigenvalues of
$(a_{ij})$ are very small. Denote $G$ be the index set of these
"good" eigenvalues and $B$ be the index set of "bad" eigenvalues.
And for any fixed point  $(x,t) \in \mathcal {O}\times (t_0-\delta,
t_0+\delta]$, we may express $(a_{ij})$ in a form of
(\ref{2.5}), by choosing $e_1,\cdots, e_{n-1},e_n$ such that

\begin{equation}\label{2.6}
 |\n u(x,t)|=u_n(x,t)>0\ \mbox{and}
(u_{ij}),i,j=1,..,n-1, \mbox{is diagonal at}\ (x,t).
\end{equation}

Without loss of generality we assume $ u_{11} \geq u_{22}\geq \cdots
\geq u_{n-1n-1} $. So, at $(x,t) \in \mathcal {O}\times (t_0-\delta,
t_0+\delta)$, from \eqref{2.5}, we have the matrix
$(a_{ij}),i,j=1,..,n-1,$ is also diagonal, and without loss of
generality we may assume $a_{11} \geq a_{22} \geq ... \geq a_{n-1,
n-1}$. There is a positive constant $C>0$ depending only on
$\|u\|_{C^{4}}$ and $\mathcal {O}\times (t_0-\delta, t_0+\delta]$,
such that $a_{11} \ge a_{22} \ge...\ge a_{ll} > C$ for all $(x,t)
\in \mathcal {O}\times (t_0-\delta, t_0+\delta)$. For convenience we
denote $ G = \{ 1, \cdots ,l \} $ and $ B = \{
l+1, \cdots, n-1 \} $ be the "good" and "bad" sets of indices
respectively. If there is no confusion, we also denote
\begin{eqnarray}\label{2.7}
\quad \quad \mbox{\it $G=\{a_{11},...,a_{ll}\}$ and
$B=\{a_{l+1,l+1},...,a_{n-1,n-1}\}$.}
\end{eqnarray}
Note that for any
$\delta>0$, we may choose $\mathcal {O}\times (t_0-\delta,
t_0+\delta]$ small enough such that $a_{jj} <\delta$ for all $j \in
B$ and $(x,t) \in \mathcal {O}\times (t_0-\delta, t_0+\delta]$.

For each $c$, let $a=(a_{ij})$ be the symmetric Weingarten tensor of
$\Sigma^{c,t}$. Set
\begin{eqnarray}\label{2.8}
 p(a)=\sigma_{l+1}(a_{ij}), \quad
q(a) &=& \left\{
 \begin{array}{llr}
\frac{\sigma_{l+2}(a_{ij})}{\sigma_{l+1}(a_{ij})},  &\mbox{if} \; \sigma_{l+1}(a_{ij})>0&\\
0,& \mbox{otherwise}. &
\end{array}
 \right.
\end{eqnarray}
Theorem \ref{th1.1} is equivalent to say $p(a)\equiv 0$ (defined in
(\ref{2.8}) ) in $\mathcal {O}\times (t_0-\delta, t_0]$. Since we
are dealing with general fully nonlinear equation (\ref{1.1}), as in
the case for the convexity of solutions in \cite{BG09}, there are
technical difficulties to deal with $p(a)$ alone. A key idea in
\cite{BG09} is the introduction of function $q$ as in (\ref{2.8})
and explore some crucial concavity properties of $q$. We consider
function
\begin{equation}
 \label{2.9}
 \phi(a)= p(a)+q(a),
\end{equation}
where $p$ and $q$ as in \eqref{2.8}. We will use notion $h=O(f)$ if
$|h(x,t)| \le Cf(x,t)$ for $(x,t) \in \O\times (t_0-\delta, t_0+\delta]$ with positive constant $C$ under
control.

To get around $p=0$, for $\varepsilon>0$ sufficiently small, we instead
consider
\begin{equation}\label{2.10}
\phi_\varepsilon (a)= \phi(a_\varepsilon),
\end{equation}
where $a_\varepsilon=a+\varepsilon I.$ We will also denote
$G_\varepsilon=\{a_{ii}+\varepsilon, i\in G\},$
$B_\varepsilon=\{a_{ii}+\varepsilon, i\in B\}.$

To simplify the notations, we will drop subindex $\varepsilon$
 with the understanding that all the estimates will be independent
 of $\varepsilon.$ In this setting, if we pick $\mathcal {O}\times (t_0-\delta,
t_0+\delta]$ small enough,
 there is $C>0$ independent of $\varepsilon$ such that
 \begin{equation}\label{2.11}
 \phi(a(x,t))\geq C\varepsilon, \quad \sigma_1(B)\geq C\varepsilon,
 ~\quad ~\mbox{ for all}\ (x,t) \in \mathcal {O}\times (t_0-\delta,
t_0+\delta].
 \end{equation}

In what follows, we will use $i, j, \cdots$ as indices run from $1$
to $n-1$ and use the Greek indices $\alpha, \beta, \cdots$ as
indices run from $1$ to $n$. Denote
\begin{align*}
& F^{\alpha \beta}  = \frac{{\partial F}} {{\partial u_{\alpha
\beta} }} , F^{p_\alpha } = \frac{{\partial F}} {{\partial u_\alpha
}} , F^u = \frac{{\partial F}}{{\partial u}} ,F^t  = \frac{{\partial F}}{{\partial t}} ,\\
&F^{\alpha \beta,\gamma \eta}  = \frac{{\partial ^2 F}} {{\partial
u_{\alpha \beta}\partial u_{\gamma \eta} }},  F^{\alpha
\beta,p_\gamma } = \frac{{\partial ^2 F}} {{\partial u_{\alpha
\beta}\partial u_\gamma }} , F^{\alpha \beta,u} = \frac{{\partial
^2 F}}{{\partial u_{\alpha \beta}\partial u}},\\
& F^{p_\alpha p_\beta }  = \frac{{\partial ^2 F}} {{\partial
u_\alpha\partial u_\beta}} ,  F^{p_\alpha, u}  = \frac{{\partial ^2
F}} {{\partial u_\alpha\partial u}} ,  F^{u,u}  = \frac{{\partial ^2
F}} {{\partial u}^2}.
\end{align*}
We also denote
\begin{equation}
\mathcal H_{\phi} = \sum_{i,j\in B}|\nabla a_{ij}|+\phi.
\end{equation}

\begin{lemma}\label{lem2.1}
For any fixed $(x,t) \in \mathcal {O}\times (t_0-\delta,
t_0+\delta]$, with the coordinate chosen as in (\ref{2.6}) and
(\ref{2.7}),
\begin{equation}\label{2.13}
\phi_t=-u_n^{-3}\sum_{j \in B} \left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}
{{\sigma}^2_1(B)}\right][u_n^2u_{jjt}-2u_nu_{jn}u_{jt} ]+O(\mathcal
H_{\phi})
\end{equation}
and
\begin{eqnarray}\label{3.9}
&&\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}\phi_{\alpha\beta} \notag
\\
&=&u_n^{-3}\sum_{j\in B}\left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)} {{\sigma}^2_1(B)}\right]
[-u_n^2\sum_{\alpha,\beta=1}^n F^{\alpha\beta}u_{\alpha\beta j
j}+2u_nu_{nj}\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta j}
\notag \\
&& \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad
+4u_nu_{nj}\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta
j}-6u_{nj}^2\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}u_{\alpha\beta
}]  \notag  \\
&&+2u_n^{-3}\sum_{j\in B,i\in G}\left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}
{{\sigma}^2_1(B)}\right]\sum_{\alpha,\beta=1}^{n}
F^{\alpha\beta}\frac{1}{u_{ii}}[u_nu_{ij\alpha}-2u_{i\a}u_{jn}][u_nu_{ij\b}-2u_{i\b}u_{jn}] \notag \\
&&-\frac{1}{{\sigma}^3_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\in
B}F^{\alpha\beta}[{\sigma}_1(B)a_{ii,\alpha}-a_{ii}\sum_{j\in
B}a_{jj,\alpha}][{\sigma}_1(B)a_{ii,\beta}-a_{ii}\sum_{j\in
B}a_{jj,\beta}]\nonumber\\
&& -\frac{1}{{\sigma}_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\neq j\in
B}F^{\alpha\beta}a_{ij,\alpha}a_{ij,\beta}+O(\mathcal H_{\phi}).
\notag
\end{eqnarray}
\end{lemma}

{\bf Proof:} For any fixed point $(x,t) \in \mathcal
{O}\times (t_0-\delta, t_0+\delta]$, choose a coordinate system as
in (\ref{2.6}) so that $|\n u|=  u_n >0$ and the matrix
$(a_{ij}(x,t))$ is diagonal for $1 \le i,j\le n-1$ and nonnegative.
From the definition of $\phi$,
\begin{eqnarray}\label{2.14}
a_{jj}=-\frac{h_{jj}}{u^3_n}=-\frac{u_{jj}}{u_n}=O(\mathcal
H_{\phi}) , \forall j\in B,
\end{eqnarray}
and
\begin{eqnarray}\label{2.15}
\phi_t &=& \sum_{j \in B} \left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)} {{\sigma}^2_1(B)}\right]
a_{jj,t}+O(\mathcal H_{\phi})  \notag  \\
&=& -u_n^{-3}\sum_{j \in B} \left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}
{{\sigma}^2_1(B)}\right][u_n^2u_{jj,t}-2u_nu_{jn}u_{jt} ]+O(\mathcal
H_{\phi})
\end{eqnarray}
Using relationship (\ref{2.14}), we have
\begin{eqnarray}\label{2.16}
\phi_{\alpha \beta}&=&\sum_{j\in B}\left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}
{{\sigma}^2_1(B)}\right]\Big[a_{jj,\alpha \beta}-2\sum_{i\in
G}\frac{a_{ij,\alpha}a_{ij,\beta}}{a_{ii}}\Big]
\nonumber\\
&&-\frac{1}{{\sigma}^3_1(B)}\sum_{i\in
B}\Big[{\sigma}_1(B)a_{ii,\alpha}-a_{ii}\sum_{j\in
B}a_{jj,\alpha}\Big]\Big[{\sigma}_1(B)a_{ii,\beta}-a_{ii}\sum_{j\in
B}a_{jj,\beta}\Big]\nonumber\\
&&-\frac{1}{{\sigma}_1(B)}\sum_{i\neq j\in
B}a_{ij,\alpha}a_{ij,\beta}+O(\mathcal H_{\phi}).
\end{eqnarray}

So far, we have followed standard calculations as in \cite{ GM03,
BGMX, BG09}. Since $u_k=0$ for $k=1,\cdots, n-1$, from
(\ref{2.5}),
\begin{eqnarray}\label{2.17}
u_n u_{ij\a}=-u_n^2 a_{ij,\a}+u_{nj}u_{i\a}
+u_{ni}u_{j\a}+u_{n\a}u_{ij}, \quad \forall\; i,j \le
n-1,\end{eqnarray}
and for each $j\in B$,
\begin{eqnarray}\label{2.18}
a_{jj,\alpha \beta}&=&-\frac{1}{u_n^3}h_{jj,\alpha \beta}+O(\mathcal
H_{\phi})  \notag  \\
&=&-\frac{1}{u_n^3}[u_n^2u_{jj\alpha\beta}+2u_{nn}u_{j\a}u_{j\b}+2u_{n\a}u_{nj}u_{j\b}+2u_{n\b}u_{nj}u_{j\a}
\notag   \\
&& \qquad \quad  -2u_nu_{nj}u_{\a\b
j}-2u_nu_{j\a}u_{nj\b}-2u_nu_{j\b}u_{nj\a}]+O(\mathcal H_{\phi}).
\end{eqnarray}
Hence, for $j\in B$,
\begin{eqnarray}\label{2.19}
\sum_{\alpha,\beta=1}^n F^{\alpha\beta}a_{jj,\alpha \beta} =&&
\sum_{\alpha,\beta=1}^n
\frac{F^{\alpha\beta}}{u_n^3}[-u_n^2u_{\alpha\beta
jj}-4u_{n\a}u_{nj}u_{j\b} +4u_nu_{j\a}u_{nj\b} \nonumber \\
&& \qquad \qquad \quad + 2u_nu_{nj}u_{\a\b j}-2u_{nn}u_{j\a}u_{j\b}]
+O(\mathcal H_{\phi}).
\end{eqnarray}
Using the fact that
$\sum_{\alpha=1}^n
F^{\alpha n}u_{n\a}=(\sum_{\alpha,\beta=1}^n
-\sum_{\b
=1}^{n-1}\sum_{\a=1}^n)F^{\alpha\beta}u_{\a\b}$, $\forall j\in B$,
\begin{eqnarray*}\label{3.24}
\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}u_{n\a}u_{j\b}=u_{nj}(\sum_{\alpha,\beta=1}^n
- \sum_{\b
=1}^{n-1}\sum_{\a=1}^n)F^{\alpha\beta}u_{\a\b} +O(\mathcal H_{\phi}),
\end{eqnarray*}
\begin{eqnarray*}\label{3.25}
\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}u_{j\a}u_{nj\b}
=u_{nj}(\sum_{\alpha,\beta=1}^n
-\sum_{\a=1}^{n-1}\sum_{\b=1}^n)F^{\alpha\beta}u_{\a\b
j}+O(\mathcal H_{\phi}),
\end{eqnarray*}
and
\begin{eqnarray*}\label{3.26}
&&-2u_{nn}\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}u_{j\a}u_{j\b}=-2u_{nn}F^{nn}u_{nj}^2+O(\mathcal H_{\phi})\\
&=&-2u_{nj}^2\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}u_{\a\b}+4u_{nj}^2\sum_{\a=1}^{n-1}F^{\a
n}u_{n\a}+2u_{nj}^2 \sum_{\alpha,\beta=1}^{n-1}
F^{\alpha\beta}u_{\a\b}+O(\mathcal H_{\phi})\nonumber.
\end{eqnarray*}
Put above to (\ref{2.19}),
\begin{eqnarray}\label{2.20}
&&\sum_{j\in B}\sum_{\alpha,\beta=1}^n F^{\alpha\beta}
u_n^3a_{jj,\alpha\beta}\nonumber\\
&=&-u_n^2\sum_{j\in B}\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}u_{\alpha\beta jj}+6u_n\sum_{j\in
B}u_{nj}\sum_{\alpha,\beta=1}^n F^{\alpha\beta}u_{\a\b j}\nonumber\\
&&-6\sum_{j\in B}u_{nj}^2\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}u_{\a\b}-4u_n\sum_{j\in
B}u_{nj}\sum_{\a=1}^{n-1}\sum_{\b=1}^nF^{\alpha\beta}u_{\a\b j}\nonumber\\
&&+8\sum_{j\in B}u_{nj}^2\sum_{\a=1}^{n-1}F^{\a
n}u_{n\a}+6\sum_{j\in B}u_{nj}^2 \sum_{\alpha,\beta=1}^{n-1}
F^{\alpha\beta}u_{\a\b}+O(\mathcal H_{\phi}).
\end{eqnarray}
By (\ref{2.17}), for $j\in B$,
\begin{eqnarray}\label{2.21}
&& u_n\sum_{\a=1}^{n-1}\sum_{\b=1}^nF^{\alpha\beta}u_{\a\b
j}=u_n\sum_{\a=1}^{n}\bigg(\sum_{i \in
B}F^{\alpha i}u_{ij\a}+\sum_{i \in G}F^{\alpha
i}u_{ij\a}\bigg)\nonumber\\
&=& \sum_{\a=1}^{n} \sum_{i \in G}F^{\alpha i}
(-u_n^2 a_{ij,\a}+u_{i\a}u_{jn} + u_{j\a}u_{in})\nonumber\\
&&+
\sum_{\a=1}^{n}\sum_{i \in B}F^{\alpha
i}(u_{i\a}u_{jn} + u_{j\a}u_{in})+O(\mathcal H_{\phi})\nonumber\\
&=&-u_n^2\sum_{\a=1}^{n} \sum_{i \in G}F^{\alpha i}
a_{ij,\a}+u_{nj}\sum_{i\in G}F^{ii}u_{ii}
+2u_{nj}(\sum_{i=1}^{n-1}F^{ni}u_{ni})+O(\mathcal H_{\phi}).
\end{eqnarray}
(\ref{2.20}) and (\ref{2.21}) yield
\begin{eqnarray}\label{3.29}
\sum_{\alpha,\beta=1}^n F^{\alpha\beta} u_n^3a_{jj,\alpha\beta}
&=&-u_n^2\sum_{\alpha,\beta=1}^n F^{\alpha\beta}u_{\alpha\beta
jj}+2u_n u_{nj}\sum_{\alpha,\beta=1}^n F^{\alpha\beta}u_{\a\b j} \notag  \\
&&+4u_n u_{nj}\sum_{\alpha,\beta=1}^n F^{\alpha\beta}u_{\a\b j}-6
u_{nj}^2\sum_{\alpha,\beta=1}^n F^{\alpha\beta}u_{\a\b} \notag \\
&&+4u_n^2 u_{nj}\sum_{\a=1}^{n} \sum_{i \in G}F^{\alpha i}
a_{ij,\a}+2 u_{nj}^2\sum_{i\in G}F^{ii}u_{ii} +O(\mathcal H_{\phi}).
\end{eqnarray}
So,
\begin{eqnarray}\label{3.9}
&&\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}\phi_{\alpha\beta} \notag
\\
&=&u_n^{-3}\sum_{j\in B}\left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)} {{\sigma}^2_1(B)}\right]
[-u_n^2\sum_{\alpha,\beta=1}^n F^{\alpha\beta}u_{\alpha\beta j
j}+2u_nu_{nj}\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta j}
\notag \\
&& \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad
+4u_nu_{nj}\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta
j}-6u_{nj}^2\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}u_{\alpha\beta
}]  \notag  \\
&&-2\sum_{j\in B,i\in G}\left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}
{{\sigma}^2_1(B)}\right]\left[\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}\frac{a_{ij,\alpha}a_{ij,\beta}}{a_{ii}}
-2\frac{u_{nj}}{u_n}\sum_{\a=1}^{n} F^{\alpha i}
a_{ij,\a}-\frac{u_{nj}^2}{u_n^3}F^{ii}u_{ii}\right]  \notag \\
&&-\frac{1}{{\sigma}^3_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\in
B}F^{\alpha\beta}[{\sigma}_1(B)a_{ii,\alpha}-a_{ii}\sum_{j\in
B}a_{jj,\alpha}][{\sigma}_1(B)a_{ii,\beta}-a_{ii}\sum_{j\in
B}a_{jj,\beta}]\nonumber\\
&& -\frac{1}{{\sigma}_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\neq j\in
B}F^{\alpha\beta}a_{ij,\alpha}a_{ij,\beta}+O(\mathcal H_{\phi}).
\notag
\end{eqnarray}
In fact, for any $i \in G, j \in B$,

\begin{eqnarray}\label{2.23}
&&\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}\frac{a_{ij,\alpha}a_{ij,\beta}}{a_{ii}}
-2\frac{u_{nj}}{u_n}\sum_{\a=1}^{n} F^{\alpha i}
a_{ij,\a}-\frac{u_{nj}^2}{u_n^3}F^{ii}u_{ii} \notag \\
&=&-\frac{1}{u_n^3}[\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}\frac{h_{ij,\alpha}h_{ij,\beta}}{h_{ii}}
-2\frac{u_{nj}}{u_n}\sum_{\a=1}^{n} F^{\alpha i}
h_{ij,\a}+u_{nj}^2F^{ii}u_{ii}]  \notag \\
&=&-\frac{1}{u_n^3}\left\{\sum_{\alpha,\beta=1}^{n-1}
F^{\alpha\beta}\frac{1}{u_n^2u_{ii}}[u_n^2u_{ij\alpha}-u_nu_{i\a}u_{jn}][u_n^2u_{ij\b}-u_nu_{i\b}u_{jn}]
\right. \notag  \\
&& \qquad \qquad  +2\sum_{\alpha=1}^{n-1}
F^{\alpha n}\frac{1}{u_n^2u_{ii}}[u_n^2u_{ij,\alpha}-u_nu_{i\a}u_{jn}][u_n^2u_{ijn}-2u_nu_{in}u_{jn}]  \notag  \\
&& \qquad \qquad + F^{nn}\frac{1}{u_n^2u_{ii}}[u_n^2u_{ijn}-2u_nu_{in}u_{jn}][u_n^2u_{ijn}-2u_nu_{in}u_{jn}]  \notag  \\
&& \qquad \qquad -2\sum_{\a=1}^{n-1} F^{\alpha
i}\frac{1}{u_n^2u_{ii}}[u_n^2u_{ij\a}-2u_nu_{i\a}u_{jn}][u_nu_{ii}u_{nj}] \notag \\
&& \qquad \qquad -2 F^{n i}\frac{1}{u_n^2u_{ii}} [u_n^2u_{ijn}-2u_nu_{in}u_{jn}][u_nu_{ii}u_{nj}]  \notag \\
&& \qquad \qquad
\left.+F^{ii}\frac{1}{u_n^2u_{ii}}(u_nu_{ii}u_{nj})^2\right\}
\notag \\
&=&-\frac{1}{u_n^3}\sum_{\alpha,\beta=1}^{n}
F^{\alpha\beta}\frac{1}{u_n^2u_{ii}}[u_n^2u_{ij\alpha}-2u_nu_{i\a}u_{jn}][u_n^2u_{ij\b}-2u_nu_{i\b}u_{jn}].
\end{eqnarray}
Obviously, we can get
\begin{eqnarray} \label{2.24}
\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}\frac{a_{ij,\alpha}a_{ij,\beta}}{a_{ii}}
-2\frac{u_{nj}}{u_n}\sum_{\a=1}^{n} F^{\alpha i}
a_{ij,\a}-\frac{u_{nj}^2}{u_n^3}F^{ii}u_{ii}\geq 0,
\end{eqnarray}
this is the Claim in \cite{BGMX}.

From the above formulas, Lemma \ref{2.1} holds. \qed

\subsection{Proof of Theorem \ref{th1.1}}

We start this section with a discussion on structure condition
\eqref{1.3}. For any function $F(r,p,u,t)$,  denote
$F^{\alpha\beta}=\frac{\partial F}{\partial r_{\alpha\beta}},
F^{p_l}=\frac{\partial F}{\partial u_l}, \cdots$ as partial
derivatives of $F$ with respect to corresponding arguments.
\begin{lemma}\label{lem2.2}
If $F$ satisfies condition \eqref{1.3}, then
\begin{eqnarray}\label{2.25}
Q(V,V)&=&F^{\alpha\beta,
\gamma \eta}X_{\alpha\beta}X_{\gamma \eta}+2F^{\alpha\beta,p_l}\theta_lX_{\alpha\beta}Y+F^{p_k, p_l}\theta_k\theta_l Y^2\nonumber\\
&&+4s ^{-1}F^{\alpha\beta}X_{\alpha\beta}Y-6F^{\alpha\beta}A_{\alpha\beta}Y^2\nonumber\\
&\leqslant& 0,
\end{eqnarray}
for every $(X_{\alpha\beta}, Y) = ((s^2\widetilde{X}_{\alpha\beta} +
2s A_{\alpha\beta}\widetilde{Y}),\widetilde{Y})$, with any
$\widetilde{V}=((\widetilde{X}_{\alpha\beta}),\widetilde{Y}) \in
\S^{n}\times \mathbb R$, where $F^{\alpha\beta,
rs},F^{\alpha\beta,u_l}, etc.$ are evaluated at $(s^{2}A, s\theta,
u, t)$, and the Einstein summation convention is used.
\end{lemma}
{\bf Proof:}
 Denoting $\widetilde{F}(A,s)=F(s^2 A, s\theta, u, t),$ condition
(\ref{1.3}) implies that $\widetilde{F}(A, s)$ is locally concave,
that is,
\begin{eqnarray}\label{2.26}
\widetilde{F}^{\alpha\beta, \gamma \eta} \widetilde{X}
_{\alpha\beta} \widetilde{X}_{\gamma \eta} + 2
\widetilde{F}^{\alpha\beta,s} \widetilde{X}_{\alpha\beta}
\widetilde{Y} + \widetilde{F}^{s,s} \widetilde{Y}^2 \leq 0,
\end{eqnarray}
for any $\widetilde{V}=((\widetilde{X}_{\alpha\beta}),\widetilde{Y})
\in \S^{n}\times \mathbb R$.

At $(A,s)$,
\begin{eqnarray*}
&& \widetilde{F}^{\alpha \beta ,\gamma \eta }  = F^{\alpha \beta
,\gamma \eta } s^2  \cdot s^2,  \\
&&\widetilde{F}^{\alpha \beta ,s}  = F^{\alpha \beta ,\gamma \eta }
s^2 \cdot 2sA_{\gamma \eta }  + F^{\alpha \beta ,p_l } s^2  \cdot
\theta _l  + F^{\alpha \beta } 2s,  \\
&&\widetilde{F}^{s,s}  = F^{\alpha \beta ,\gamma \eta } 2sA_{\alpha
\beta }  \cdot 2sA_{\gamma \eta }  + 2F^{\alpha \beta ,p_l }
2sA_{\alpha \beta }  \cdot \theta _l  + F^{p_k ,p_l } \theta _k
\cdot \theta _l  + F^{\alpha \beta } 2A_{\alpha \beta }.
\end{eqnarray*}
Set
\begin{eqnarray}
&& X_{\alpha \beta }  = s^2 \widetilde{X}_{\alpha \beta }  +
2sA_{\alpha \beta } \widetilde{Y},  \\
&&Y=\widetilde{Y},
\end{eqnarray}
so (\ref{2.26}) is equivalent to
\begin{eqnarray}\label{3.61}
&&F^{\alpha\beta,\gamma \eta}X_{\alpha\beta}X_{\gamma \eta}+2F^{\alpha\beta,p_l}\theta_lX_{\alpha\beta}Y+F^{u_k, p_l}\theta_k\theta_l Y^2\nonumber\\
&&+4s ^{-1}F^{\alpha\beta}s^2 \widetilde{X}_{\alpha\beta}\widetilde{Y}+2F^{\alpha\beta}A_{\alpha\beta}\widetilde{Y}^2\nonumber\\
&=&F^{\alpha\beta,\gamma \eta}X_{\alpha\beta}X_{\gamma \eta}+2F^{\alpha\beta,p_l}\theta_lX_{\alpha\beta}Y+F^{p_k, p_l}\theta_k\theta_l Y^2\nonumber\\
&&+4s ^{-1}F^{\alpha\beta}X_{\alpha\beta}Y-6F^{\alpha\beta}A_{\alpha\beta}Y^2\nonumber\\
&\leqslant& 0.\notag
\end{eqnarray}
Therefore, (\ref{2.25}) follows from above, and Lemma \ref{2.2} holds. \qed

Theorem~\ref{th1.1} is a direct consequence of the following
proposition and the strong maximum principle.

\begin{proposition}\label{thm2.3}
Suppose that the function $F, u$ satisfy assumptions in Theorem
\ref{th1.1}. If the second fundamental form $b_{ij}$ of
$\Sigma^{c,t_0}$ attains minimum rank $l=l(t_0)$ at certain point
$x_0\in \Omega$, then there exist a neighborhood $\mathcal {O}
\times(t_0-\delta_0,t_0+\delta_0]$ of $(x_0,t_0)$ and a positive
constant $C$ independent of $\phi$ (defined in (\ref{2.9})), such
that
\begin{equation}\label{2.29}
\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}\phi_{\alpha\beta}(x,t)-\phi_t\leq
C(\phi+|\nabla \phi|),~~\forall ~(x,t)\in \mathcal {O}
\times(t_0-\delta_0,t_0+\delta_0].
\end{equation}
\end{proposition}

{\bf Proof:}  Let $u\in C^{3,1}(\Omega
\times [0,T])$ be a spatial quasiconcave solution of equation
(\ref{1.1}) and $(u_{ij}) \in \S^n.$ Let $l=l(t_0)$ be the minimum
rank of the second fundamental forms $h_{ij}$ of $\Sigma^{c,t_0}$
($l \in \{0,1,...,n-1\}$) for every $c$ in $(-\gamma_0+c_0,
\gamma_0+c_0)$, suppose the minimum rank $l$ arrives at point $x_0
\in \Sigma^{c,t_0}$. We work on a small open neighborhood $\mathcal
{O} \times(t_0-\delta_0,t_0+\delta_0]$ of $(x_0,t_0)$. We may assume
$l \le n-2$. Lemma \ref{2.1} implies $\phi \in C^{1,1}(\mathcal {O}
\times(t_0-\delta_0,t_0+\delta_0]),$ $ \phi(x,t)\geq 0, ~\qquad~
\phi(x_0,t_0)=0 $. For $\epsilon>0$ sufficient small, let
$\phi_{\epsilon}$ defined as in (\ref{2.9}) and (\ref{2.10}), we
need to verify (\ref{2.29}) for each point $(x,t) \in \mathcal {O}
\times(t_0-\delta_0,t_0+\delta_0]$. For each fixed $(x,t)$, choose a
local coordinate $e_1,\cdots, e_{n-1}, e_n$ such that (\ref{2.6}) and
(\ref{2.7}) are satisfied. We want to establish differential
inequality (\ref{2.29}) for $\phi_\varepsilon$ defined in
(\ref{2.10}) with constant $C$ independent of $\varepsilon$. Note that we will omit the subindex $\varepsilon$ with the
understanding that all the estimates are independent of
$\varepsilon$.

By Lemma~\ref{lem2.1},
\begin{eqnarray}\label{2.30}
&&\sum_{\alpha,\beta=1}^nF^{\alpha\beta}\phi_{\alpha\beta}-\phi_t\nonumber\\
&\le& -u_n^{-3}\sum_{j\in
B}\left[\sigma_l(G)+\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}{{\sigma}^2_1(B)}\right]\Big[
u_n^2(\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{jj\alpha\beta}-u_{jjt})
\nonumber\\
&&-2u_nu_{jn}(\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}u_{j\alpha\beta}-u_{jt})-4u_nu_{jn}\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}u_{j\alpha\beta} +6u_{jn}^2\sum_{\alpha,\beta=1}^n
F^{\alpha\beta}u_{\alpha\beta}\Big]\nonumber\\
&&-\frac{1}{{\sigma}^3_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\in
B}F^{\alpha\beta}[{\sigma}_1(B)a_{ii,\alpha}-a_{ii}\sum_{j\in
B}a_{jj,\alpha}][{\sigma}_1(B)a_{ii,\beta}-a_{ii}\sum_{j\in
B}a_{jj,\beta}]\nonumber\\
&&-\frac{1}{{\sigma}_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\neq j,
i,j\in B}F^{\alpha\beta}a_{ij,\alpha}a_{ij,\beta}+O(\mathcal H_{\phi}).
\end{eqnarray}

For each $j\in B$, differentiating equation (\ref{1.1}) in $e_j$ direction at
$x$,
\begin{equation}\label{2.31}
u_{jt}=\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta
j}+F^{u_n}u_{jn}+O(\mathcal H_{\phi}),
\end{equation}
and
\begin{eqnarray}\label{2.32}
u_{jjt}&=&\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta
jj}+\sum_{\alpha,\beta,r,s=1}^nF^{\alpha\beta, rs}u_{\alpha\beta
j}u_{rsj} +2\sum_{\alpha,\beta,l=1}^nF^{\alpha\beta,
u_l}u_{\alpha\beta j}u_{lj}\nonumber\\
&&+2\sum_{\alpha,\beta=1}^nF^{\alpha\beta, u}u_{j\alpha\beta}u_j
+\sum_{l,s=1}^nF^{u_l, u_s}u_{lj}u_{sj}-2\sum_{l=1}^nF^{u_l,
u}u_{lj}u_j\nonumber\\
&&+F^{u,u}u_j^2+\sum_{l=1}^nF^{u_l}u_{ljj}+F^{u}u_{jj}.
\end{eqnarray}
It follows from (\ref{2.17}) that, at $(x,t)$
\begin{eqnarray}\label{2.33}
\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta
jj}-u_{jjt}&=&-\sum_{\alpha,\beta,r,s=1}^nF^{\alpha\beta,
rs}u_{\alpha\beta j}u_{rsj}
-2\sum_{\alpha,\beta=1}^nF^{\alpha\beta, u_n}u_{j\alpha\beta}u_{nj}\nonumber\\
&&-F^{u_n, u_n}u^2_{jn}-2\frac{F^{u_n}}{u_n}u^2_{jn}+O(\mathcal
H_{\phi}).
\end{eqnarray}
Since $u_{\alpha\beta jj}=u_{ jj\alpha\beta}$, (\ref{2.31}) and
(\ref{2.33}) yield
\begin{eqnarray}\label{2.34}
&&F^{\alpha\beta}\phi_{\alpha\beta}- \phi_t\nonumber\\
&\leq& \sum_{j\in
B}u_n^{-3}\left[\sigma_l(G)+\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}{{\sigma}^2_1(B)}\right]\left\{
\Big[\sum_{\alpha,\beta,r,s=1}^n
F^{\alpha\beta, rs}u_{\alpha\beta j}u_{rsj}\right.\nonumber\\
&&\qquad \qquad \qquad \qquad
+2\sum_{\alpha,\beta=1}^nF^{\alpha\beta,
u_n}u_{j\alpha\beta}u_{jn}+F^{u_n,u_n}u_{jn}^2\Big]u_n^2\nonumber\\
&&\qquad \qquad\qquad \qquad
+\left.4u_{jn}u_n\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{j\alpha\beta}
-6u_{jn}^2\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta}\right\}\nonumber\\
&&-\frac{1}{{\sigma}^3_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\in
B}F^{\alpha\beta}[{\sigma}_1(B)a_{ii,\alpha}-a_{ii}\sum_{j\in
B}a_{jj,\alpha}][{\sigma}_1(B)a_{ii,\beta}-a_{ii}\sum_{j\in
B}a_{jj,\beta}]\nonumber\\
&&-\frac{1}{{\sigma}_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\neq j,
i,j\in B}F^{\alpha\beta}a_{ij,\alpha}a_{ij,\beta}+O(\mathcal H_{\phi}).
\end{eqnarray}
For each $ j \in B$, set
\begin{eqnarray}\label{2.35}
S_j= && \Big[\sum_{\alpha,\beta,r,s=1}^n
F^{\alpha\beta, rs}u_{j\alpha\beta}u_{rsj}+2\sum_{\alpha,\beta=1}^nF^{\alpha\beta,
u_n}u_{j\alpha\beta}u_{jn}+F^{u_n,u_n}u_{jn}^2\Big]u_n^2\nonumber\\
&+&4\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{j\alpha\beta}u_{jn}u_n
-6\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta}u_{jn}^2
\end{eqnarray}
For each $j\in B$, set
\begin{eqnarray}\label{4.16}
&&X_{\alpha\beta}=u_{\alpha\beta j}u_n, \forall (\alpha, \beta), \\
&&Y=u_{jn}u_n.
\end{eqnarray}
In the coordinate system (\ref{2.6}),
\[(\n^2u(x), \n u(x), u(x), t)=(\n^2u,(0,...,0,|\n u|), u, t).\]
Equalize it to $(s^2A, s\theta, u, t)$, the components of
$\widetilde{V}$ defined in Lemma~\ref{lem2.2} are
\begin{eqnarray*}
&&\widetilde{X}_{\alpha\beta}=\frac{u_{\alpha\beta
j}}{u_n}-\frac{2u_{\alpha\beta}u_{jn}}{u^2_n}, \quad \forall
(\alpha,\beta),  \\
&&\widetilde{Y}=u_{jn}u_n.
\end{eqnarray*}
For $j \in B$, Lemma~\ref{lem2.2} implies
\begin{equation}\label{2.38}
S_j\le 0.
\end{equation}
Condition (\ref{1.2}) implies
\begin{equation}\label{2.39}
 (F^{\alpha\beta}) \ge \delta_0 I, \; \quad \mbox{for some $\delta_0>0$, and
$\forall x \in \O$.}
\end{equation}
Set
\[V_{i\alpha}={\sigma}_1(B)a_{ii,\alpha}-a_{ii}\sum_{j\in
B}a_{jj,\alpha}.\]
Combining (\ref{2.34}), (\ref{2.38}) and (\ref{2.39}),
\begin{eqnarray}\label{2.40}
F^{\alpha\beta}\phi_{\alpha\beta}\le  C(\phi+\sum_{i,j\in B}|\nabla
a_{ij}|)-\delta_0[\frac{\sum_{i\neq j\in
B, \alpha=1}^na^2_{ij\alpha}}{\sigma_1(B)}+\frac{\sum_{i\in
B,\alpha=1}^nV_{i\alpha}^2}{{\sigma}^3_1(B)}].
\end{eqnarray}
By Lemma~3.3 in \cite{BG09}, for each $M\ge 1$, for any $M\ge
|\gamma_i|\ge \frac{1}{M}$, there is a constant $C$ depending only
on $n$ and $M$ such that, $\forall \alpha$,
\begin{equation}\label{2.41}
\sum_{i,j\in B}|a_{ij\alpha}|\le C(1+\frac{1}{\delta_0^2})
(\sigma_1(B)+|\sum_{i\in B} \gamma_ia_{ii\alpha}|)
+\frac{\delta_0}{2}[\frac{\sum_{i\neq j \in B}|a_{ij\alpha}|^{2}}
{\sigma_{1}(B)}+
\frac{\sum_{i\in B}V_{i\alpha}^{2}}{\sigma_{1}^{3}(B)}].\end{equation}
Taking $\gamma_i=\sigma_l(G)+\frac{{\sigma}^2_1(B|i)-{\sigma}_2(B|i)}{{\sigma}^2_1(B)}$ for each $i\in B$,
the Newton-MacLaurine inequality implies
\[\sigma_l(G)+1 \ge
\sigma_l(G)+\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}{{\sigma}^2_1(B)}
\ge \sigma_l(G), \quad \forall j\in B.\]
Therefore we conclude from Lemma~\ref{lem2.1} and (\ref{2.41}) that
$\sum_{i,j \in B} |\nabla a_{ij}|$  can be controlled by the rest
terms on the right hand side in (\ref{2.40}) and $\phi + |\nabla
\phi|$. The proof is complete. \qed

\section{Proof of Theorem \ref{1.2}}

In this section, through modifying the proof of Theorem \ref{1.1},
we will give a proof of Theorem \ref{1.2}. Also it is a parabolic
equation case corresponding to \cite{GX}.

Suppose that $u(x,t)$ is a spatial quasiconcave solution of \eqref{1.4},
and assume that level surface $\Sigma^{u(x_0,t)} = \{x \in \Omega|
u(x,t)=u(x_0,t)\}$ is connected for each $(x_0,t) \in \mathcal
{O}\times [0, T]$.

Set
\begin{equation} \label{3.1}
\widetilde a = a - \eta _0 g I, \quad \eta _0 \geqslant 0, \quad
g(x,t) = e^{Au(x,t)},
\end{equation}
where $A > 0$ is a constant to be determined. We want to show $\widetilde{a}$ is
of constant rank. Theorem \ref{1.1} corresponds to the case
$\eta_0=0$. If $ \min \{ \kappa ^0 ,\kappa ^1 \}=0$, there is
nothing to prove instead of utilizing Theorem \ref{1.1}. We will
assume $ \min \{ \kappa ^0 ,\kappa ^1 \}>0$ in the rest of the
paper. Denote $\kappa_s(x,t)$ and $\widetilde \kappa_s(x,t)$ be the
minimum eigenvalue of matrix $a(x)$ and $\widetilde a(x)$
respectively. Since the spatial level sets are strictly convex, and
$\overline{\Omega}$ is compact, $a$ is strictly positive definite.
That is, $\kappa_s(x,t)$ has a positive lower bound.

For a positive constant $A$ to be determined, increasing $\eta_0$
from 0, such that $\widetilde a$ is degenerate at some points, i.e.
$\widetilde a$ is semi-positive with the rank is not full. \eqref{1.8}
follows easily if this happens only on the boundary. We want to show that, if the degeneracy happens
at an interior point of $\Omega$,  then $\widetilde a$ is degenerate through out $\Omega$
 with the same rank. This
implies that the "=" holds in \eqref{1.8} and Theorem \ref{th1.2} is proved.

Therefore, the main task is to prove constant rank theorem for
$\widetilde a$. Suppose $\widetilde a(x,t_0)$ attains minimal rank
$l=l(t_0)$ at some point $z_0 \in \Omega$. We may assume $l\leqslant
n-2$, otherwise there is nothing to prove. And we assume $u \in
C^{3,1}$ and $u_n>0$ in the rest of this paper. So there is
a neighborhood $\mathcal {O}\times (t_0-\delta, t_0+\delta]$ of
$(z_0, t_0)$, such that there are $l$ "good" eigenvalues of
$(\widetilde a_{ij})$ which are bounded below by a positive
constant, and the other $n-1-l$ "bad" eigenvalues of $(\widetilde
a_{ij})$ are very small. Denote $G$ be the index set of these "good"
eigenvalues and $B$ be the index set of "bad" eigenvalues. And for
any fixed point $(x,t) \in \mathcal {O}\times (t_0-\delta,
t_0+\delta]$, we may express $(\widetilde a_{ij})$ in a form
of \eqref{3.1} and (\ref{2.5}), by choosing $e_1,\cdots,
e_{n-1},e_n$ such that
\begin{equation}\label{3.2}
|\n u(x,t)|=u_n(x,t)>0\ \mbox{and}\
(u_{ij}),i,j=1,..,n-1,\ \mbox{is diagonal at}\ (x,t).
\end{equation}

Without loss of generality, we assume $ u_{11} \geq u_{22}\geq \cdots
\geq u_{n-1,n-1} $. So, at $(x,t) \in \mathcal {O}\times (t_0-\delta,
t_0+\delta)$, from \eqref{2.5}, we have the matrix
$(a_{ij}),i,j=1,..,n-1$, is also diagonal. And without loss of
generality we may assume $a_{11} \geq a_{22} \geq ... \geq a_{n-1,
n-1}$, then $\widetilde a_{11} \geq \widetilde a_{22} \geq ... \geq
\widetilde a_{n-1, n-1}$. There is a positive constant $C>0$
depending only on $\|u\|_{C^{4}}$ and $\mathcal {O}\times
(t_0-\delta, t_0+\delta]$, such that $\widetilde a_{11} \ge
\widetilde a_{22} \ge...\ge \widetilde a_{ll} > C$ for all $(x,t)
\in \mathcal {O}\times (t_0-\delta, t_0+\delta)$. For convenience we
denote  $ G = \{ 1, \cdots ,l \} $ and $ B = \{
l+1, \cdots, n-1 \} $ be the "good" and "bad" sets of indices
respectively. If there is no confusion, we also denote
\begin{eqnarray}\label{3.3}
\quad \quad \mbox{\it $G=\{\widetilde a_{11},...,\widetilde
a_{ll}\}$ and $B=\{\widetilde a_{l+1,l+1},...,\widetilde a_{n-1,
n-1}\}$.}
\end{eqnarray}
Note that for any $\delta>0$, we may choose $\mathcal {O}\times
(t_0-\delta, t_0+\delta]$ small enough such that $\widetilde a_{jj}
<\delta$ for all $j \in B$ and $(x,t) \in \mathcal {O}\times
(t_0-\delta, t_0+\delta]$.

For each $(x,t)$, let $a=(a_{ij})$ be the symmetric Weingarten
tensor of $\Sigma^{u(x,t)}$. Set
\begin{eqnarray}\label{3.4}
p(\widetilde a)=\sigma_{l+1}(\widetilde a_{ij}), \quad q(\widetilde
a) &=& \left\{
\begin{array}{llr}
\frac{\sigma_{l+2}(\widetilde a_{ij})}{\sigma_{l+1}(\widetilde a_{ij})},  &\mbox{if} \; \sigma_{l+1}(\widetilde a_{ij})>0,&\\
0,& \mbox{otherwise}. &
\end{array}
\right.
\end{eqnarray}
Theorem \ref{th1.2} is equivalent to say $p(\widetilde a)\equiv 0$
(defined in (\ref{3.4}) ) in $\mathcal {O}\times (t_0-\delta, t_0]$.
As in the description of the proof of Theorem \ref{th1.1}, we should consider the function
\begin{equation}\label{3.5}
\phi(\widetilde a)= p(\widetilde a)+q(\widetilde a),
\end{equation}
where $p$ and $q$ as in \eqref{3.4}. We will use notion $h=O(f)$ if
$|h(x,t)| \le Cf(x,t)$ for $(x,t) \in \O \times(t_0-\delta, t_0+\delta]$ with positive constant $C$ under
control.

To get around $p=0$, for $\varepsilon>0$ sufficiently small, we
instead consider
\begin{equation}\label{3.6}
\phi_\varepsilon (\widetilde a)= \phi(\widetilde a_\varepsilon),
\end{equation}
where $a_\varepsilon=\widetilde a+\varepsilon I.$ We will also
denote $G_\varepsilon=\{\widetilde a_{ii}+\varepsilon, i\in G\},$
$B_\varepsilon=\{\widetilde a_{ii}+\varepsilon, i\in B\}.$

To simplify the notations, we will drop subindex $\varepsilon$ with
the understanding that all the estimates will be independent of
$\varepsilon.$ In this setting, if we pick $\mathcal {O}\times
(t_0-\delta, t_0+\delta]$ small enough, there is $C>0$ independent
of $\varepsilon$ such that
\begin{equation}\label{2.11}
\phi(\widetilde a(x,t))\geq C\varepsilon, \quad \sigma_1(B)\geq
C\varepsilon, ~\quad ~\rm{ for ~all~ }(x,t) \in \mathcal {O}\times
(t_0-\delta, t_0+\delta].
 \end{equation}
We also denote
\begin{equation} \mathcal H_{\phi} = \sum_{i,j\in B}|\nabla \widetilde a_{ij}|+\phi.
\end{equation}

\begin{lemma}\label{lem3.3}
For any fixed $(x,t) \in \mathcal {O}\times (t_0-\delta,
t_0+\delta]$, with the coordinate chosen as in (\ref{3.2}) and
(\ref{3.3}),

\begin{eqnarray}\label{3.9}
&&\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}\phi_{\alpha\beta}-\phi_t
\notag  \\
&=&u_n^{-3}\sum_{j\in B}\left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)} {{\sigma}^2_1(B)}\right]
[-u_n^2(\sum_{\alpha,\beta=1}^n F^{\alpha\beta}u_{\alpha\beta j
j}-u_{jjt})+2u_nu_{nj}(\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta
j}-u_{jt})
\notag \\
&& \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad
+4u_nu_{nj}\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta
j}-6u_{nj}^2\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}u_{\alpha\beta
}]  \notag  \\
&&+2u_n^{-3}\sum_{j\in B,i\in G}\left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}
{{\sigma}^2_1(B)}\right]\sum_{\alpha,\beta=1}^{n}
F^{\alpha\beta}\frac{1}{u_{ii}}[u_nu_{ij\alpha}-2u_{i\a}u_{jn}][u_nu_{ij\b}-2u_{i\b}u_{jn}] \notag \\
&&+\eta _0 g \left[-A^2F^{nn}u_n^2+A O(1) + O (1)\right] \notag   \\
&&-\frac{1}{{\sigma}^3_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\in
B}F^{\alpha\beta}[{\sigma}_1(B)\widetilde a_{ii,\alpha}-\widetilde
a_{ii}\sum_{j\in B}\widetilde a_{jj,\alpha}][{\sigma}_1(B)\widetilde
a_{ii,\beta}-\widetilde a_{ii}\sum_{j\in B}\widetilde a_{jj,\beta}]\nonumber\\
&& -\frac{1}{{\sigma}_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\neq j\in
B}F^{\alpha\beta}\widetilde a_{ij,\alpha} \widetilde
a_{ij,\beta}+O(\mathcal H_{\phi}). \notag
\end{eqnarray}
\end{lemma}

{\bf Proof:} For any fixed $(x,t) \in \mathcal {O}\times (t_0-\delta,
t_0+\delta]$, we choose the coordinate  as in (\ref{3.2}) such that
$|\n u(x)|= u_n(x) >0$ and the matrix $(\widetilde a_{ij}(x))$ is
diagonal for $1 \le i,j\le n-1$ and nonnegative. From the definition
of $p$,
\begin{eqnarray}\label{3.9}
a_{jj}=-\frac{h_{jj}}{u^3_n}=-\frac{u_{jj}}{u_n}=O(\mathcal
H_{\phi}), \forall j\in B,
\end{eqnarray}
and
\begin{equation}\label{3.10}
\phi_t=\sum_{j \in B} \left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)} {{\sigma}^2_1(B)}\right]
\widetilde a_{jjt} +O(\mathcal H_{\phi}).
\end{equation}
Using relationship (\ref{3.9}), we have
\begin{eqnarray}\label{3.11}
\phi_{\alpha \beta}&=&\sum_{j\in B}\left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}
{{\sigma}^2_1(B)}\right]\Big[\widetilde a_{jj,\alpha
\beta}-2\sum_{i\in G}\frac{\widetilde a_{ij,\alpha}\widetilde
a_{ij,\beta}}{\widetilde a_{ii}}\Big] \nonumber\\
&&-\frac{1}{{\sigma}^3_1(B)}\sum_{i\in
B}\Big[{\sigma}_1(B)\widetilde a_{ii,\alpha}-\widetilde
a_{ii}\sum_{j\in B}\widetilde a_{jj,\alpha} \Big] \Big[{\sigma}_1(B)
\widetilde a_{ii,\beta} - \widetilde a_{ii}\sum_{j\in
B} \widetilde a_{jj,\beta} \Big]  \nonumber\\
&&-\frac{1}{{\sigma}_1(B)}\sum_{i\neq j\in B}\widetilde
a_{ij,\alpha}\widetilde a_{ij,\beta}+O(\mathcal H_{\phi}).
\end{eqnarray}

So,
\begin{eqnarray}\label{3.12}
&&\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}[\widetilde a_{jj,\alpha
\beta}-2\sum_{i\in G}\frac{\widetilde a_{ij,\alpha}\widetilde
a_{ij,\beta}}{\widetilde a_{ii}}] -\widetilde a_{jj,t} \notag  \\
&=&\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}a_{jj,\alpha
\beta}-a_{jj,t}+\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}(-\eta_0g_{\a\b})+\eta_0g_t
\notag \\
&&-2\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}\sum_{i\in
G}\frac{\widetilde a_{ij,\alpha}\widetilde a_{ij,\beta}}{\widetilde
a_{ii}}.
\end{eqnarray}
From the definition of $a_{ij}$, and $u_k = 0$ for $k = 1, \cdots
, n - 1$, we can get
\begin{eqnarray}\label{3.13}
u_n u_{ij\alpha }  =  - u_n ^2 a_{ij,\alpha }  + u_{nj} u_{i\alpha }
+ u_{ni} u_{j\alpha }  + u_{n\alpha } u_{ij}
\end{eqnarray}
and
\begin{eqnarray}\label{3.14}
u_n ^3 a_{jj,\alpha \beta } & = & - u_n ^2 u_{jj\alpha \beta }  +
2u_n u_{nj} u_{\alpha \beta j}  - 2u_n (u_{n\beta } u_{jj\alpha }  +
u_{n\alpha } u_{jj\beta } )   \notag \\
&& + 2u_n (u_{j\alpha } u_{nj\beta }  + u_{j\beta } u_{nj\alpha } )
+ 2u_{nj} (u_{n\alpha } u_{j\beta }  + u_{n\beta } u_{j\alpha } ) -
2u_{nn} u_{j\alpha } u_{j\beta }   \notag \\
&& - 2(u_{n\alpha } u_{n\beta }  + u_n u_{\alpha \beta n} )u_{jj}  -
2\eta _0 gu_{j\alpha } u_{j\beta } u_n  - 3\eta _0 u_n ^2
(u_{n\alpha } g_\beta   + u_{n\beta } g_\alpha  )  \notag \\
&& - \eta _0 g(3u_n ^2 u_{n\alpha \beta }  + 6u_{n\alpha } u_{n\beta
} u_n  + \sum\limits_{i = 1}^{n - 1} {u_{i\alpha } u_{i\beta } u_n }
) + O(\mathcal H_{\phi}).
\end{eqnarray}
Direct calculation and \eqref{3.13}, we can get
\begin{eqnarray}\label{3.15}
&&- a_{jj,t}  + \sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta }
( - \eta _0 g_{\alpha \beta } )}  + \eta _0 g_t  \notag  \\
& =& \frac{1} {{u_n ^3 }}[u_n ^2 u_{jjt}  - 2u_n u_{nj} u_{jt} ]
\notag  \\
&&+ \eta _0 g[ - A^2 F^{nn} u_n ^2  - A(\sum\limits_{\alpha \beta  =
1}^n {F^{\alpha \beta } u_{\alpha \beta } }  - u_t ) + \frac{{u_{nt}
}} {{u_n }}].
\end{eqnarray}
From \eqref{3.14},
\begin{eqnarray}
\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta } a_{jj,\alpha
\beta } }  &=& \sum\limits_{\alpha \beta  = 1}^n {\frac{{F^{\alpha
\beta } }} {{u_n ^3 }}[ - u_n ^2 u_{jj\alpha \beta }  + 2u_n
u_{nj}u_{\alpha \beta j} }   \notag  \\
&& - 4u_{nj} u_{n\alpha } u_{j\beta }  + 4u_{nj} u_{n\alpha }
u_{j\beta }  - 2u_{nn} u_{j\alpha } u_{j\beta }    \notag \\
&& - 2\eta _0 u_n ^2 u_{n\alpha } g_\beta   - \eta _0 g(u_n ^2
u_{n\alpha \beta }  + 2u_{j\alpha } u_{j\beta } u_n  +
\sum\limits_{i = 1}^{n - 1} {u_{i\alpha } u_{i\beta } u_n } )]
+O(\mathcal H_{\phi}),   \notag
\end{eqnarray}
so, as in \cite{GX}, we can get
\begin{eqnarray} \label{3.16}
&&u_n ^3 \sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta }
a_{jj,\alpha \beta } }   \notag  \\
& =&  - u_n ^2\sum\limits_{\alpha \beta = 1}^n {F^{\alpha \beta }
u_{jj\alpha \beta } }  + 2u_n u_{nj} \sum\limits_{\alpha \beta  =
1}^n {F^{\alpha \beta } u_{\alpha \beta
j} }   \notag   \\
&& + 4u_n u_{nj} \sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta
} u_{\alpha \beta j} }  - 6u_{nj} ^2 \sum\limits_{\alpha \beta  =
1}^n {F^{\alpha \beta } u_{\alpha \beta } }   \notag  \\
&& + 4u_n ^2 \sum\limits_{\alpha  = 1}^n {\sum\limits_{i \in G}
{F^{\alpha i} a_{ij,\alpha } } }  + 2u_{nj} ^2 \sum\limits_{i \in G}
{F^{ii} u_{ii} }   \notag  \\
&& + 2u_{nj} ^2 \sum\limits_{i \in B} {F^{ii} u_{ii} }  - 12u_{jn}
u_{jj} \sum\limits_{\alpha  = 1}^n {F^{j\alpha } u_{n\alpha } }  +
4u_n u_{jj} \sum\limits_{\alpha  = 1}^n {F^{j\alpha } u_{jn\alpha }
}  - 2u_{nn} F^{jj} u_{jj} ^2  \notag   \\
&& - \eta _0 g(u_n ^2 u_{n\alpha \beta }  + 2u_{j\alpha } u_{j\beta
} u_n  + \sum\limits_{i = 1}^{n - 1} {u_{i\alpha } u_{i\beta } u_n }
)   \notag \\
&& - 2\eta _0 \sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta }
u_{n\alpha } g_\beta  } u_n  + 4\eta _0 \sum\limits_{\alpha  = 1}^n
{F^{j\alpha } g_\alpha  u_{jn} u_n ^2 }  + O(\mathcal H_{\phi}),
\notag   \\
& =&  - u_n ^2\sum\limits_{\alpha \beta = 1}^n {F^{\alpha \beta }
u_{jj\alpha \beta } }  + 2u_n u_{nj} \sum\limits_{\alpha \beta  =
1}^n {F^{\alpha \beta } u_{\alpha \beta j} }   \notag  \\
&& + 4u_n u_{nj} \sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta
} u_{\alpha \beta j} }  - 6u_{nj} ^2 \sum\limits_{\alpha \beta  =
1}^n {F^{\alpha \beta } u_{\alpha \beta } }     \\
&& + 4u_n ^2 \sum\limits_{\alpha  = 1}^n {\sum\limits_{i \in G}
{F^{\alpha i} a_{ij,\alpha } } }  + 2u_{nj} ^2 \sum\limits_{i \in G}
{F^{ii} u_{ii} }   \notag  \\
&& +\eta _0 g \left[A O(1) + O (1)\right]  + O(\mathcal H_{\phi}).
\notag
\end{eqnarray}
Also, with the similar computations \eqref{2.23} in the Lemma \ref{lem2.1},
\begin{eqnarray} \label{3.17}
&&\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta }
\frac{{\widetilde a_{ij,\alpha } \widetilde a_{ij,\beta } }}
{{\widetilde a_{ii} }}}  - \frac{1} {{u_n ^3 }}[2u_n ^2 u_{nj}
\sum\limits_{\alpha  = 1}^n {F^{\alpha i} a_{ij,\alpha } }  + u_{nj}
^2 F^{ii} u_{ii} ]   \notag  \\
& =& \sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta }
\frac{{a_{ij,\alpha } a_{ij,\beta } }} {{a_{ii} }}}  + \eta _0
g\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta }
\frac{{a_{ij,\alpha } a_{ij,\beta } }} {{a_{ii} \widetilde a_{ii}
}}}   \notag  \\
&&- \frac{1} {{u_n ^3 }}[2u_n ^2 u_{nj} \sum\limits_{\alpha  = 1}^n
{F^{\alpha i} a_{ij,\alpha } }  + u_{nj} ^2 F^{ii} u_{ii} ]  \notag \\
& =& \eta _0 g\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta }
\frac{{a_{ij,\alpha } a_{ij,\beta } }} {{a_{ii} \widetilde a_{ii}
}}}  \notag  \\
&&- \frac{1} {{u_n ^3 }}\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha
\beta } \frac{1} {{u_{ii} }}} [ - u_n u_{ij\alpha }  + u_{nj}
u_{i\alpha } + u_{ni} u_{j\alpha } ][ - u_n u_{ij\beta }  + u_{nj}
u_{i\beta }  + u_{ni} u_{j\beta } ]   \notag  \\
&& - \frac{1} {{u_n ^3 }}[2u_{nj} \sum\limits_{\alpha  = 1}^n
{F^{\alpha i} ( - u_n u_{ij\alpha }  + u_{nj} u_{i\alpha }  + u_{ni}
u_{j\alpha } )} + u_{nj} ^2 F^{ii} u_{ii} ]   \notag  \\
& =& \eta _0 g\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta }
\frac{{a_{ij,\alpha } a_{ij,\beta } }} {{a_{ii} \widetilde a_{ii}
}}}  \notag  \\
&& - \frac{1} {{u_n ^3 }}\sum\limits_{\alpha \beta  = 1}^n
{F^{\alpha \beta } \frac{1} {{u_{ii} }}} [ - u_n u_{ij\alpha }  +
2u_{nj} u_{i\alpha } ][ - u_n u_{ij\beta }  + 2u_{nj} u_{i\beta } ]
\notag  \\
&& - \frac{1} {{u_n ^3 }}u_{jj} [\sum\limits_{\alpha  = 1}^{n - 1}
{F^{\alpha j} \frac{2} {{u_{ii} }}} u_{ni} ( - u_n u_{ij\alpha }  +
u_{nj} u_{i\alpha } ) + F^{ii} \frac{1} {{u_{ii} }}u_{jj} u_{ni} ^2
\notag  \\
&& \qquad \qquad + 2F^{jn} \frac{1} {{u_{ii} }}u_{ni} ( - u_n
u_{ijn}  + 2u_{nj} u_{in} ) + F^{ij} u_{ni} u_{nj} ]   \notag  \\
& =&  - \frac{1} {{u_n ^3 }}\sum\limits_{\alpha \beta  = 1}^n
{F^{\alpha \beta } \frac{1} {{u_{ii} }}} [ - u_n u_{ij\alpha }  +
2u_{nj} u_{i\alpha } ][ - u_n u_{ij\beta }  + 2u_{nj} u_{i\beta } ]
\notag   \\
&& + \eta _0 g\left[\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha
\beta } \frac{{a_{ij,\alpha } a_{ij,\beta } }} {{a_{ii} \widetilde
a_{ii}}}} +O(1). \right]
\end{eqnarray}

From the above calculations, the proof is complete.  \qed

Theorem~\ref{th1.2} is a direct consequence of the following
proposition and the strong maximum principle.

\begin{proposition}\label{thm4.1}
Suppose that the function $F, u$ satisfy assumptions in Theorem
\ref{th1.2}. If the second fundamental form $b_{ij}$ of
$\Sigma^{u(x,t_0)}$ attains minimum rank $l=l(t_0)$ at certain point
$x_0\in \Omega$, then there exist a neighborhood $\mathcal {O}
\times(t_0-\delta_0,t_0+\delta_0]$ of $(x_0,t_0)$ and a positive
constant $C$ independent of $\phi$ (defined in (\ref{3.5})), such
that
\begin{equation}\label{3.18}
\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}\phi_{\alpha\beta}(x,t)-\phi_t\leq
C(\phi+|\nabla \phi|)+\eta _0 g \left[-A^2F^{nn}u_n^2+A O(1) + O
(1)\right]
\end{equation}
holds for any $(x,t)\in \mathcal {O} \times (t_0-\delta_0,
t_0+\delta_0]$.
\end{proposition}

{\bf Proof:} Since
\begin{equation}\label{3.19}
u_{t}=F(\n ^2 u, \n u ,u, t),
\end{equation}
for each $j\in B$, differentiating the above equation in $e_j$
direction at $x$,
\begin{equation}\label{3.20}
u_{jt}=\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta
j}+F^{u_n}u_{jn}+O(\mathcal H_{\phi})
\end{equation}
and
\begin{eqnarray}\label{3.21}
u_{jjt}&=&\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta
jj}+\sum_{\alpha,\beta,r,s=1}^nF^{\alpha\beta, rs}u_{\alpha\beta
j}u_{rsj} +2\sum_{\alpha,\beta,l=1}^nF^{\alpha\beta,
u_l}u_{\alpha\beta j}u_{lj}\nonumber\\
&&+2\sum_{\alpha,\beta=1}^nF^{\alpha\beta, u}u_{j\alpha\beta}u_j
+\sum_{l,s=1}^nF^{u_l, u_s}u_{lj}u_{sj}-2\sum_{l=1}^nF^{u_l,
u}u_{lj}u_j\nonumber\\
&&+F^{u,u}u_j^2+\sum_{l=1}^nF^{u_l}u_{ljj}+F^{u}u_{jj}.
\end{eqnarray}
It follows from (\ref{3.9})and \eqref{3.13} that, at $(x,t)$
\begin{eqnarray}
 \sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta } u_{\alpha
\beta j} }  - u_{jt}  =  - F^{p_n } u_{nj}  + \eta _0 gF^{p_j } u_n
+ O(\mathcal H_{\phi})
\end{eqnarray}
and
\begin{eqnarray}
&&\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta } u_{\alpha
\beta jj} }  - u_{jjt}  \notag  \\
&=&  - \sum\limits_{\alpha \beta \gamma \eta = 1}^n {F^{\alpha \beta
,\gamma \eta } u_{\alpha \beta j} } u_{\gamma \eta j}  -
2\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta ,p_n } u_{\alpha
\beta j} u_{nj} } - F^{p_n ,p_n } u_{nj} u_{nj}  - 2\frac{{F^{p_n }
}} {{u_n }}u_{nj} ^2   \notag  \\
&& + \eta _0 g[ - AF^{p_n } u_n ^2 ]   \notag  \\
&& + \eta _0 g[2\sum\limits_{\alpha \beta  = 1}^n {F^{\alpha \beta
,p_j } u_{\alpha \beta j} u_n }  + F^{p_j ,p_j } u_{jj} u_n  +
2F^{p_n ,p_j } u_{nj} u_n  + F^{p_n } u_n  + 2F^{p_j } u_{jn}  +
F^{p_l } u_{nl} ]  \notag  \\
&&+ O(\mathcal H_{\phi}).   \notag
\end{eqnarray}
From lemma \ref{3.1},
\begin{eqnarray}\label{4.13}
&&F^{\alpha\beta}\phi_{\alpha\beta}- \phi_t\nonumber\\
&=& \sum_{j\in
B}u_n^{-3}\left[\sigma_l(G)+\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}{{\sigma}^2_1(B)}\right]\left\{
\Big[\sum_{\alpha,\beta,r,s=1}^n
F^{\alpha\beta, rs}u_{\alpha\beta j}u_{rsj}\right.\nonumber\\
&&\qquad \qquad \qquad \qquad
+2\sum_{\alpha,\beta=1}^nF^{\alpha\beta,
u_n}u_{j\alpha\beta}u_{jn}+F^{u_n,u_n}u_{jn}^2\Big]u_n^2\nonumber\\
&&\qquad \qquad\qquad \qquad
+\left.4u_{jn}u_n\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{j\alpha\beta}
-6u_{jn}^2\sum_{\alpha,\beta=1}^nF^{\alpha\beta}u_{\alpha\beta}\right\}\nonumber\\
&&+2u_n^{-3}\sum_{j\in B,i\in G}\left[ \sigma_l(G) +
\frac{{\sigma}^2_1(B|j)-{\sigma}_2(B|j)}
{{\sigma}^2_1(B)}\right]\sum_{\alpha,\beta=1}^{n}
F^{\alpha\beta}\frac{1}{u_{ii}}[u_nu_{ij\alpha}-2u_{i\a}u_{jn}][u_nu_{ij\b}-2u_{i\b}u_{jn}] \notag \\
&&+\eta _0 g \left[-A^2F^{nn}u_n^2+A O(1) + O (1)\right] \notag   \\
&&-\frac{1}{{\sigma}^3_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\in
B}F^{\alpha\beta}[{\sigma}_1(B)\widetilde a_{ii,\alpha}-\widetilde
a_{ii}\sum_{j\in B}\widetilde a_{jj,\alpha}][{\sigma}_1(B)\widetilde
a_{ii,\beta}-\widetilde a_{ii}\sum_{j\in B}\widetilde a_{jj,\beta}]\nonumber\\
&& -\frac{1}{{\sigma}_1(B)}\sum_{\alpha,\beta=1}^n\sum_{i\neq j\in
B}F^{\alpha\beta}\widetilde a_{ij,\alpha} \widetilde
a_{ij,\beta}+O(\mathcal H_{\phi}). \notag
\end{eqnarray}
So, following the argument in the proof of Proposition \ref{thm2.3},  we get,
\begin{equation}
\sum_{\alpha,\beta=1}^{n}F^{\alpha\beta}\phi_{\alpha\beta}(x,t)-\phi_t\leq
C(\phi+|\nabla \phi|)+\eta _0 g \left[-A^2F^{nn}u_n^2+A O(1) + O
(1)\right].
\end{equation}
The proof is completed. \qed

\bibliographystyle{amsplain}


TITLE: Finiteness and cardinality in abstract categories
QUESTION [5 upvotes]: My question is a very simple one.
What ways are there to generalize terms such as cardinality (or, more generally, the concept of finiteness) to abstract (and not concretizable) categories?
I have seen two ways to do so: One is to take a terminal object $\mathbf{T}$ and to count the numer of morphisms from $\mathbf{T}$ to an object $\mathbf{A}$ and define this number as the cardinality of $\mathbf{A}$. Clearly, this will coincide with the cardinality as defined in the category of sets as the number of morphisms from $\{1\}$ to another set is the cardinality of the letter. The downside of this approach is that it does not always work well: Even if a category has a terminal object, there is no garuantee that there are morphisms from this object to any other object and, thus, if we define cardinality in this way, it might give very misleading results: For instance, we can have a concrete category in which the number of morphisms from the terminal object does not reflect the finitness of an object that might be obtained by applying the forgetful functor to it.
Another way is to define an object as finite if it has a finite number of endomorphisms. Again, this coincides with the notional of finiteness in the category of sets. However, one can think of a category in which an object $\mathbf{A}$ only has the identity morphism as an endomorphism but an infinite number of morphisms exist between $\mathbf{A}$ and another object $\mathbf{B}$ (for which, again, only the identity morphism is an endomorphism). This creates a scenario in which it is hardly justified to think $\mathbf{A}$ and $\mathbf{B}$ as finite.
Are there other ways to define finitenes? In particular, is there a way to define finitness for a category such that, if this category is concretizable, it can be turned into a concrete category in which only the finite objects (abstractly defined) can have a finite image under the forgetful functor and this can be achieved for any abstractly defined finite object?

EDIT: In the comments, it was asked what categories I have in mind. Here is an answer to this:
The categories I have in mind are only required to have the following: There must exist an object $\mathbf{A}$ such that all finite, non-empty powers of $\mathbf{A}$ are also in the category (the product is of course defined in the usual category-theoretic way and the point of all this is to work with a generalized definition of clones in these categories). Apart from this condition, however, the category can have any possible structure. Although it would be nice to have a general definition of cardinality, all I am really interested in is to distinguish between finite and non-finite objects. The definition should satisfy the three conditions below.

Of course, it should coincide with the usual definition of finiteness if my category is Set.
If $\mathbf{A}$ and another object $\mathbf{B}$ is finite, then the number of morphisms from $\mathbf{B}$ to $\mathbf{A}$ should be finite (it would therefore be enough to have a notion of finiteness "with respect to $\mathbf{A}$", if such a thing makes sense to have).
The definition should give me as many finite objects as the "best" concretezation. By this, I mean the following: If my category is concretizable, then you should not be able to get more finite objects by defining finiteness over the image under the forgetful functor.

REPLY [4 votes]: In Pareigis' Categories and functors an object $x$ of an abelian category $A$ is called finite if $Hom(x,-) : A \to Ab$ commutes with coproducts. For example, in $R$-Mod, every finitely generated free module is finite. There it is also proved that every cocomplete abelian category, which has a finite projective generator, is equivalent to some $R$-Mod.
There are various notions of finitely generated objects. See for example also Pareigis' book, this question or nlab.


TITLE: Bijection between $ A ^ {B \cup C} $ and $ A ^ B \times A ^ C $, with $B\cap C=\varnothing$.
QUESTION [2 upvotes]: Let us consider $ f | _B $ the function restricted to $ B $ and analogously $ f | _C $. The formula for both is $ f | _B (b) = f (b) $ and $ f | _C (c) = f (c) $, for all $ b, c \ in B, C $ respectively.
Let's define $ F: A ^ {B \cup C} \to A ^ B \times A ^ C $, given by $$ F (f) = (f | _B, f | _C) $$.
To demonstrate injectivity, let $ f, f '\in A ^ {B \cup C} $, such that $ F (f) = F (f') $, that is $ (f | _B, f | _C) = (f '| _B, f' | _C) $. Hence, it follows that $ f (b) = f | _B = f '| _B = f' (b) $ and $ f (c) = f | _C = f '| _C = f' (c) $. This is that $ f = f '$. That way $ F $ is injective.
Some help demonstrating surjectivity.

REPLY [1 votes]: In general, given sets $B$ and $C$, we can form the disjoint union $B \sqcup C$ by choosing disjoint copies $B'$ of $B$ and $C'$ of $C$, and then $B \sqcup C = B' \cup C'$; for example $$B \sqcup C := (B \times \{0\}) \cup (C \times \{1\}).$$ Now, we claim that there is a bijection between $A^{B \sqcup C}$ and $A^B \times A^C$ (this is the reason of why the disjoint union $B \sqcup C$ is also denoted by $B + C$, or $B \oplus C$). Indeed, given functions $g: B \to A$ and $h : C \to A$, define $g \sqcup h : B \sqcup C \to A$ by $$(g \sqcup h) \circ j_B = g \textrm{ and } (g \sqcup h) \circ j_C = h$$ where $j_Z : Z \to B \sqcup C$ (for $Z \in \{B,C\}$) sends each element in $Z$ to its corresponding copy in $Z' \subseteq B \sqcup C$. Thus, the function $$A^B \times A^C \to A^{B \sqcup C}; \quad (g,h) \mapsto g \sqcup h$$ is a bijection, with inverse $$A^{B \sqcup C} \to A^B \times A^C; \quad f \mapsto (f \circ j_B,f \circ j_C).$$


TITLE: Is second derivative of a function related to curve smoothness?
QUESTION [1 upvotes]: If there exist a first derivative of a function at any point then the funtion is continuous at that point.
What if the second derivative of that function is also exist at that point ? Does this mean that function is smooth at that point ? Means instead of taking sharp bend the function is having curved shape ?

REPLY [1 votes]: It's easier to talk about discontinuity than continuity.
If a function has a discontinuity (i.e. is not continuous), then it typically means that there is "jump" in the function value, so its graph has a gap in it. This is not 100% true, though. There are weird functions that oscillate infinitely fast in a small region, and these are also not continuous. But let's ignore strange functions like that, for now.
If there is a discontinuity in the first derivative of a function, it means that its graph has a sharp corner -- a place where there is an abrupt change in direction.
If there is a discontinuity in second derivative, it means there is an abrupt change in curvature (or radius of curvature). Some people can see these discontinuities, and some people can't. Anyone with a background in graphics or design will certainly see them.
Things get a lot more interesting when you make surfaces from your curves, and you look at reflections in these surfaces, as you would when looking at a car, for example. Reflection "magnifies" discontinuities. A discontinuity in curvature will be clearly visible in a reflective surface, even to the untrained eye. Even a discontinuity in third derivative will be visible to a designer. That's why people who design cars don't use cubic splines (which you are studying, sounds like), because these have only continuity of first and second derivatives.


TITLE: Ring of Cauchy sequences.
QUESTION [2 upvotes]: Les $R$ be the ring of cauchy sequences in the field $\mathbb{R}$ of real numbers and $I=\{(x_n):(x_n) \rightarrow 0 \}$ be a subset of $R$. Then which of the following is correct:
$(1)$ For every real number $a$ their exists a $(x_n)$ in $R$ such that $x_n\rightarrow a$.
$(2)$ $I$ is prime ideal.
$(3)$ $I$ is maximal ideal.
$(4)$ $I$ is prime but not maximal.
$(5)$ $I$ is principal ideal.
$(6)$ Every real number has a unique representation in $R/I$.
I can see that for $a\in \mathbb{R}$ we have a sequence $(x_n)=(a,a,a,\ldots)$ such that $(x_n)\rightarrow a$. And if $x_n y_n\rightarrow 0$ then at least one of $x_n$ or $y_n$ must converse to $0$, so $I$ is prime ideal. But I am not able do anything for the rest options. Please help me

REPLY [2 votes]: Questions (1) - (4) and (6) are addressed in the comment.
For (5), we can show it is false by contradiction. Suppose that $t = (t_n)$ is a generator of this ideal. Note that no $t_n$ can be zero (otherwise no element with a non-zero entry in that spot could be a multiple of $t$).
Define $s = (t_n / n)$. Since $t$ is a generator, one has
$$
t = s\alpha
$$
for some $\alpha$. We compute $\alpha_n = n$ and therefore $\alpha$ is not a Cauchy sequence, a contradiction.
(We could probably word this in a less sloppy way to avoid proof by contradiction; the point is that nothing can be a generator because we can construct something such that the quotient isn't Cauchy.)


TITLE: Find the basis for kernel of a matrix transformation
QUESTION [2 upvotes]: $Let \ ψ\ :{Mat }_{ 2x2 }(ℝ)\ →\ { Mat }_{ 2x2 }(ℝ)\ be\ defined\ by$
$ ψ : \pmatrix{a&b\\c&d}→\pmatrix{a+b&a-c\\a+c&b-c}$ .
Find basis for ker ψ
I'm not sure how to do it for a matrix, I've set $\pmatrix{a+b&a-c\\a+c&b-c}$ to the zero matrix, and obtained 4 equations, but not sure how proceed from there.

REPLY [2 votes]: Easy Way
Assuming that $\text{Mat}_{2\times 2}(\mathbb{R})$ are being though of as additive groups, the kernel is made up of all matrices that are send to the zero matrix. So you need $a+b=0$, $a-c=0$, $a+c=0$ and $b-c=0$. Well, $a+b=0$ if and only if $a=-b$. Let's substitute that into the remaining three equations: $-b-c=0$, $-b+c=0$ and $b-c=0$. Well, $-b-c=0$ if and only if $b=-c$. Let's put that into the two remaining equations: $-(-c)+c=0$ and $-(-c)-c=0$. Well, $2c=0$ if and only if $c=0$. Hence $a=-b$, $b=-c$ and $c=0$ meaning that $a=b=c=0$.
You're free to choose $d$. So the matrix $\left[\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right]$ spans the kernel.

REPLY [1 votes]: HINT First identify $\text{Mat}_{2 \times 2}$ with $\mathbb{R}^4$, with basis matrices $$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, ...,$$ and write $\psi$ as a linear transformation $\mathbb{R}^4 \to \mathbb{R}^4$ with matrix $M$.
Then read the examples in this Wikipedia article, which give a method for finding a basis once you have a matrix.


TITLE: Definition of finite field with fixed characteristic
QUESTION [1 upvotes]: In this article Discrete logarithms in quasi-polynomial time in finite fields of fixed characteristic the term finite fields of fixed characteristic is not defined and I couldn't find it on the literature, too.

What is the definition of finite fields of fixed characteristic

REPLY [3 votes]: The term quasi-polynomial time means quasi-polynomial in...which parameters? What fixed characteristic says is that the problem is quasi-polynomial in the field size as the field size varies, as long as we keep the characteristic fixed.
So, for instance, if their algorithm takes $\le C2^pk^r$ steps over a field of characteristic $p$ and size $k$ for some constants $C,r$, then it is polynomial for fields of fixed characteristic, but exponential in the characteristic.


TITLE: Show $(m^2+n^2)(o^2+p^2)(r^2+s^2) \geq 8mnoprs$
QUESTION [2 upvotes]: Let $m,n,o,p,r,s$ be positive numbers. Show that:
$$(m^2+n^2)(o^2+p^2)(r^2+s^2) \geq 8mnoprs$$
I tested an example,
let $m,n,o,p,r,s=1,2,3,4,5,6$ respectively. Then:
$$(1+4)(9+16)(25+30)=(5)(25)(55) \geq 8123456$$
but I get $6875$ which is not greater than the right side.
So am i misunderstanding the problem? Any ideas?

REPLY [0 votes]: Yes, you misunderstood the problem, for $8mnoprs$ has tacit multiplication signs, it does not mean concatenation. Therefore, your test case is $$(1 + 4)(9 + 16)(25 + 30) = 6875 \geq 8 \times 1 \times 2 \times 3 \times 4 \times 5 \times 6.$$ But the first case I would test is the minimal case $m = n = o = p = r = s = 1$. Then $$2 \times 2 \times 2 \geq 8 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1.$$


TITLE: Why do pivot entries correspond to bound variables?
QUESTION [0 upvotes]: Why do the pivot entries of a matrix in row echelon form correspond to bound variables?
Take this matrix in row echelon form for example with variables $x_1, x_2, x_3, x_4$ (the last column is meant to be augmented).
$$A = \begin{bmatrix}
    1       & 2 & 0 &   0 &3\\
   0      & 0 & 1 &   0 &3\\
    0 & 0 & 0 &1&2
\end{bmatrix}$$

REPLY [0 votes]: The key idea here is that the rows of a matrix constitute a linear system of equations.
Since we have 3 equations and 5 unknowns (columns), our system has an infinitum of solutions, because we could choose two variables to be whatever we want.
It is somewhat an arbitrary rule that mathematicians always choose to say that the pivot entries are bound and the other ones ($x_2$ and $x_5$ in your example above) are "free".
I say that it is is arbitrary because if $x_1 + x_2 = 5$ then we could say $x_1 = 5 - x_2$ or equivalently $x_2 = 5 - x_1$. In any event, the number of pivots corresponds to the number of equations you actually have.
A good reference for learning about this is chapter 2 of Strang's Linear Algebra and It's Applications.


TITLE: $\int_{-\infty}^{\infty}e^{-x^n}dx$ converges $\iff n$ is even
QUESTION [3 upvotes]: I'd like to show that $$\int_{-\infty}^{\infty}e^{-x^n}dx \text{ converges} \iff n\text{ is even.}$$

My attempt:
Suppose $n$ is odd. Then $$\int_{-\infty}^{\infty}e^{-x^n}dx=\int_{0}^{\infty}e^{-x^n}dx+\underbrace{\int_{0}^{\infty}e^{x^n}dx}_{\text{diverges}}$$ The second integral diverges since $e^{x^n}\to\infty$ as $x\to\infty$.
If instead, $n$ is even, then $$\int_{-\infty}^{\infty}e^{-x^n}dx =2\int_{0}^{\infty}e^{-x^n}dx$$
I don't know how to show this is bounded.

I would like to know if there is any inequality I can use to bound this, and show that the integral is finite.

REPLY [0 votes]: Hint:
$$0\le\int_0^\infty e^{-x^n}\mathrm dx\le\sum_{k=0}^\infty e^{-k^n}$$
Applying the root test we have that
$$\limsup_{k\to\infty}\sqrt[k]{e^{-k^n}}=\limsup_{k\to\infty}e^{-k^{(n-1)}}=0,\quad n>1$$


TITLE: Ordering people with friendship constraints
QUESTION [19 upvotes]: $n$ people have to be ordered in a line, one after the other. We say that:

A person X sees a person Y, if X stands behind Y.
A person X hears a person Y, if X stands at most $k$ positions in front of Y.

Each person has two friends (friendship relation is not symmetric).
An order is good if each person either sees or hears both his/her friends

What is the largest value of $n$ (as a function of $k$) for which a good order always exists?

Here are some examples.
When $n\leq k+1$, a good order always exists, since each person can see or hear all other people.
When $k=1$ and $n\geq k+2$, a good order never exists, since the front person sees nobody and hears at most a single friend.
When $k=2$ and $n=4$. a good order always exists and can be found as follows. Put an arbitrary person (e.g. Alice) in the front. Put her two friends just behind her, so that she can hear both of them. Put the fourth person at the back.
When $k=2$ and $n=5$, things are getting interesting. Suppose we put Alice in the front, and her friends are Bob and Carl. So the order has to start as either Alice-Bob-Carl or Alice-Carl-Bob. But, if Bob's friends are David and Eva and Carl's friends are also David and Eva, then one of them (the one just behind Alice) will not hear both his friends.
We can try to put Bob at the front, then David and Eva. But, if both David's friends and Eva's friends are Alice and Carl, then one of them (the one just behind Bob) will not hear both his friends.
In that case, we can order the agents as: Carl-David-Eva-Alice-Bob; this order satisfies all the requirements.
Apparently, finding a good order becomes harder when $n$ is larger. Hence the question.
UPDATE 1: when $n\leq 2k$, a good order always exists. Put an arbitrary agent (say, Alice) in the front. Put one of her friends (say, Bob) just behind her. Put one of Bob's friends (say, Carl) just behind him. Continue like this until $k$ people are placed. Then, at position $k+1$, put Alice's second friend (if not already placed). At position $k+2$, put Bob's second friend (if not already placed), and so on. So the people at positions $1,\dots,k$ can hear all their friends. Since $n\leq 2k$, the people at positions $k+1,\dots,2k$ can hear everyone behind them. Hence the order is good.
This seems very not tight, since we just picked the front person arbitrarily. Surely we can do better by picking the front person more carefully, right?
UPDATE 2: Based on Kevin's probabilistic argument, I tried the following counting argument. The friendship relation can be roughly represented by a vector of size $2n$. For each 2 positions in the vector, there are $(n-1)(n-2)$ options, so the total number of vectors is $\approx n^{2n}$. On the other hand, the total number of orderings is $n!$. Therefore, if we could prove that each order is good for less than $n^{2n} / n!$ vectors, then this would imply that some vectors have no good orders.
Unfortunately, the latter claim is probably not true. Fix an order, and renumber the agents such that the order is $1,\dots,n$. To construct a vector for which this order is good, we have $k(k-1)$ options for the friends of person 1, $(k+1)k$ options for the friends of person 2, ..., $(n-2)(n-1)$ options for the friends of person $n-k$, and $(n-2)(n-1)$ options for the friends of the remaining people $n-k,\dots,n$. Multiplying all these numbers gives much more than $n^{2n} / n!$. So this counting argument does not work.
UPDATE 3: This question is related: How many nodes do you need? If our friendship graph if $k$-dense (as defined in that question), then a good order does not exist. Proof: regardless of who we put in the first $k$ locations, these $k$ people have together $k+1$ friends besides themselves. So, one of these people will not hear one of his friends. So a lower bound of $n$ in that question implies an upper bound of $n-1$ in the current question. In particular, for $k=2$, there is a 2-dense graph with 7 nodes; hence the largest number of people for which a good order exists is at most 6 (Peter showed a better upper bound: 5).

REPLY [1 votes]: Note: This is NOT a full answer; just two upper specific bounds.

Let $f(k)$ denote the greatest value of $n$ for which a good order is guaranteed to exist. Then:
We have $f(2)<6$, since the friendship graph consisting of three pairs of people: {0,1}, {2,3} and {4,5}, where each person from each pair is friends with each person from the next pair (with last pair wrapping back to the first one) admits no good order. More explicitly:
  0 => [2,3];  1 => [2,3];  2 => [4,5];  3 => [4,5];  4 => [0,1];  5 => [0,1]

This graph is nicely regular, which simplifies the analysis considerably. Without loss of generality, the first person in line is person $0$. Its two friends, $2$ and $3$ must be placed right behind (their relative order doesn't matter; it can be $2$ without loss of generality). However, this means at least one of their friends ($4$ and $5$) would be too far to be heard.

We also have $f(3)<9$, as demonstrated by the following friendship graph (with persons numbered $0$ to $8$) in which every person is friends with the next one (wrapping back from $8$ to $0$), plus the following friendship relations:
  0 => 5; 1 => 7; 2 => 0;   3 => 8; 4 => 1; 5 => 3;   6 => 2; 7 => 4; 8 => 6

Unfortunately, I have not managed to represent this graph in a very neat, regular structure like the previous one (although it can still look relatively nice if represented as a $3\times 3$ grid). As such, I have not managed to produce an easy proof of its desired no-good-order property; having only verified it via computer-based exhaustive search (which, of course, could have been implemented incorrectly; feel free to verify it independently).

A wild extrapolation might suggest $f(k)<3k$, although this is only a very wild guess.


TITLE: Financial Mathematics problem
QUESTION [2 upvotes]: $i^{(p)}$ is the nominal interest converted p-thly i.e the total interest per unit of time paid on a loan of amount 1 at time 0 where interest paid in p equal installments at the end of each p-th sub-interval.
I can't understand the statement given as follows :
Since $i^{(p)}$ is the total interest paid and each interest payment is of amount $\dfrac{i^{(p)}}{p}$ then the accumulated value at time 1 of the interest payments is :
$\dfrac{i^{(p)}}{p}(1+i)^{(p-1)/p} + \dfrac{i^{(p)}}{p}(1+i)^{(p-2)/p} +$.. ..... $+ \dfrac{i^{(p)}}{p} = i$
where $i$ is the effective rate of interest.
Can anyone explain the above statement and thus the equation ? Thanks in advance !

I tried and tried , finally I interpreted it in the following way , please correct me if wrong ,
We are adding the interest applicable per interval i.e in ( 1/p , 2/p , .... , 1) and equating it to the annual effective rate of interest $i$.
So the interest applicable per interval can be thought of as : $ \dfrac{i^{(p)}}{p} \times$ (Value accumulated just before that interval ) , so , interest applicable in the (1/p)th interval would be $\dfrac{i^{(p)}}{p} (1 + i) ^{(1 - \frac{1}{p})}$ , So ,
$\dfrac{i^{(p)}}{p}(1+i)^{(p-1)/p} + \dfrac{i^{(p)}}{p}(1+i)^{(p-2)/p} +$.. ..... $+ \dfrac{i^{(p)}}{p} = i$
Is this correct ??

REPLY [1 votes]: Let's look at this with an example.  Suppose $t = 1$ represents one year of time from $t = 0$.  Let $p = 12$, so that $i^{(12)}$ represents a nominal interest rate convertible monthly (12 times per year).  As this is a rate, it is also equivalent to an amount of interest paid per monetary unit.  The meaning of "convertible monthly" is that the interest accrues (i.e., is compounded) at the end of each month, but at an effective monthly rate of $i^{(12)}/12$.
We are interested in the equivalent effective annual rate $i$ corresponding to the nominal monthly rate $i^{(12)}$; that is to say, if interest were to be accrued at the end of one year, what would the required amount be in order to be equal to the total interest accrued if compounded monthly?
The idea, then, is to note that payments of $i^{(12)}/12$ are made at time points $t = 1/12, 2/12, 3/12, \ldots, 12/12$, and under the effective annual rate $i$, these payments have had time $1-t = 11/12, 10/12, \ldots, 0$ to accrue interest, respectively.  Therefore, the future value of these payments at time $t = 1$ are $$\frac{i^{(12)}}{12} (1 + i)^{11/12} + \frac{i^{(12)}}{12} (1 + i)^{10/12} + \cdots + \frac{i^{(12)}}{12} (1 + i)^0,$$ and this must equal $i$ itself, since the total interest accrued under both situations are equivalent.  Now, using the formula for a geometric series with common ratio $r = (1+i)^{1/12}$, we find that the LHS is $$\frac{(1+i)^{12/12} - 1}{(1+i)^{1/12} - 1} \frac{i^{(12)}}{12} = \frac{i^{(12)} i}{12((1+i)^{1/12} - 1)},$$ and equating this with $i$ and solving gives $$i = \left(1  + \frac{i^{(12)}}{12}\right)^{\!12} - 1,$$ and in general, it is easy to see that for a nominal rate of $i^{(p)}$ convertible $p$ times from $t = 0$ to $t = 1$, we have $$i = \left(1 + \frac{i^{(p)}}{p}\right)^{\!p} - 1.$$


TITLE: What is the difference between following approaches to line integrals?
QUESTION [0 upvotes]: What is the difference between following approaches concerning line integrals:

first approach is for complex function with parametrization $\gamma(t) = \cos t+i\sin t$ (Line_integral :: Example from Wikipedia).
Second approach is for $f(x,y)$ with parametrization $x=\cos t, y=\sin t$ (Line_integral :: Example from khanacademy.org).

Yes complex function $f(z)=1/z$ is $f: C \rightarrow C$ while the other is $f:R^2 \rightarrow R$, but what other differences there are or are they same thing said in different method?

REPLY [0 votes]: Let me add one more kind of line integration to make the difference clearer:

Line integral of a scalar function. This is your second approach. There is not much going on here. An intuitive description of the integral would be to sum the function values encountered along the curve.
Line integral of a vector function. Take a vector function with components $(f(x,y),g(x,y))$. Then the integral is defined by
$$
\int \left( f(x(t),y(t))x'(t)+g(x(t),y(t))y'(t) \right) {\mathrm{d}}t.
$$
A rough description would be to sum the tangential component of the vector field $(f,g)$ along the curve.
Line integral of a complex function. Take a complex function $f(x,y)+ig(x,y)$. Then the integral is
$$
\int \left( f(x(t),y(t))x'(t)-g(x(t),y(t))y'(t) \right) {\mathrm{d}}t + i\int \left( f(x(t),y(t))y'(t)+g(x(t),y(t))x'(t) \right) {\mathrm{d}}t.
$$
As you see, this is somewhat similar to the vector integral, in the sense that the integral depends on the interaction between the components of the function and the curve in a nontrivial way. However, this is "more scalar" than the vector integral because of the complex multiplication, in the same way complex numbers are "more scalar" than just two dimensional vectors.


TITLE: Need help in determining the volume of styrofoam used with dimensions $2.50ft + 1.50ft + 1.00ft$
QUESTION [0 upvotes]: So to elaborate on the title, the question is this:
The average density of Styrofoam is $1.00 \frac{kg}{m^3}$.  If a Styrofoam cooler is made with outside dimensions of $3.00ft$ $x$ $2.00ft$ $x$ $1.50ft$ and inside dimensions of $2.50ft$ $x$ $1.50ft$ $x$ $1.00ft$, determine the (a) volume of Styrofoam used in the cubic meters...
So the word that is throwing me off is used.  Am I to calculate the volume of the inside and outside and simply add them?
I was doing the following:
$$V=lwh$$
$$V=(3.00)(2.00)(1.50)$$
(Outer)$V=9ft^3$
$$V=(2.50)(1.50)(1.00)$$
(Inner)$V=3.75ft^3$
(Total)$V=12.75ft^3$
Am I on the right track at all?

REPLY [1 votes]: The outer volume gives you the amount of Styrofoam that it would take to make a cooler of the same size that was completely useless (because there was no space inside, as it was just a block of Styrofoam). The inner volume is the amount of Styrofoam that would have to be removed from such a block to make such a cooler. The amount of Styrofoam used, then, is the original/outer/useless volume minus the inner/removed volume. That is, the amount used is $5.25$ cubic feet.


TITLE: Hall's marriage Theorem and Tychonoff Theorem
QUESTION [4 upvotes]: I was reading this paper. In particular the second point. He proves the Hall's marriage Theorem for infinite family using the Tychonoff theorem on topological product of compact $T_2$ spaces and the Hall's marriage Theorem for finite family.
So assume that $Y$ is an infinite family of finite sets such that any $k$ sets in $Y$ have at least $k$ elements in their union ($k$ is a finite number). We want to prove that there is a $1-1$ function $f:Y\longrightarrow \underset{X\in Y}{\cup}X$.
Endow each $X\in Y$ with the discrete topology. Since $X$ is finite it is a compact $T_2$ space. The topological product $\underset{X\in Y}{\prod}X$ is a compact $T_2$ space.
We now take the $F$ to be the family of all sets $S_G$ of all $f\in \underset{X\in Y}{\prod}X$ such that their restriction to $G$ is injective; for all $G$ finite subsets of $Y$.
He claims that $S_G$ is closed and nonempty by the Hall's marriage theorem for finite family. I do not understand why. In particular I ask myself the following things:
As matter the nonemptiness I was thinking of finding an injective function using Hall's marriage theorem for the finite family $G$ and then to extend to all $Y$. But how? How can we say that  $\underset{X\in Y}{\prod}X$ is not empty without using the full strenght of the Axiom of Choice?
Where do we make (if we make) us of the fact that the product is a $T_2$ space?
The closeness of $S_G$ just came from the fact that the projection onto the factor is a continuus function and that the antiimages of closed set is a closed set?

REPLY [7 votes]: The paper shows that the infinite Hall theorem follows from the Tychonoff theorem for compact Hausdorff spaces (or even the Tychonoff theorem for finite discrete spaces). So we are allowed to use that fact.
Now the Tychonoff theorem for non-empty finite discrete spaces implies that the product of finite discrete spaces is non-empty (i.e. collections of finite sets have a choice function). This uses the same argument (essentially) as the fact that full Tychonoff implies full AC.
Proof: let $F_i, i \in I$ be a set of non-empty finite sets. Let $p$ be any point not in $\cup_i F_i$, and define $X_i = F_i \cup \{p\}$ for all $i$, and give this set the discrete topology. We know that $\prod_i X_i$ is compact by assumption. Define $F'_i = (\pi_i)^{-1}[F_i]$, where $\pi_i$ is the projection from $\prod_i X_i$ onto $X_i$. Then the $F'_i$ are closed in $\prod_i X_i$ and they have the finite intersection property: we can pick finitely many points in finitely many $F_j$ and $p$ for all other coordinates. We essentially added $p$ as the default, always present, choice point. The intersection of the $F'_i$ is thus non-empty by compactness, and this is what we needed.
The previous result let's us use that when the finite Hall theorem on a finite $G \subset Y$  gives us a choice function that is 1-1 there, and extend using choice; as said, not full choice, but choice for finite sets, which we are allowed to use, as we are working under the assumption of the Tychonoff theorem for finite discrete spaces. This covers the non-emptyness.
The closedness is just a product property: if $f \notin S_G$, it must be the case that there are $Y_1 \neq Y_2$ both in $G$, that have the same choice under $f$, i.e. $f(S_{Y_1}) = f(S_{Y_2})$. But then we can take the basic open product set $O$ that equals $\{f(S_{Y_1})\}$ at coordinate $Y_1$, and $\{f(S_{Y_2})\}$ at coordinate $Y_2$, and the full set elsewhere. All points in $O$ still have the property that injectiveness fail (at the same $Y_1$ and $Y_2$), so $O$ misses $S_G$. This shows that $S_G$ is closed. Note we use the discreteness here (to see the singletons are open).
And as the paper notes $S_{G_1 \cup G_2} \subset S_{G_1} \cap S_{G_2}$, so the set $S_G$, where $G$ ranges over the finite subsets of $Y$ has the finite intersection property, so has non-empty intersection (second application of the Tychonoff theorem for finite discrete spaces) and this is what we needed: a choice function for all of $Y$. It will clearly be injective, as failing to be injective already happens in a finite subset of $Y$.
So we do not need the full Tychonoff for compact Hausdorff spaces, but "only" the Tychonoff theorem for finite discrete spaces (which might or might not be weaker, I'd have to look it up). Edit: looked it up in "Consequences of the Axiom of Choice", and the Tychonoff theorem for compact Hausdorff, finite and even two point spaces are all equivalent (to the Boolean Prime Ideal theorem, among others). It's form 14 in the book, while Hall's theorem is form 107, so weaker.


TITLE: How to find eigenvalue of $\sigma$ and $\sigma^2$
QUESTION [0 upvotes]: How to find $\sigma^2$ for $\sigma^2= \sigma_x^2 + \sigma_y^2 + \sigma_z^2$ where $\sigma_x^2$, $\sigma_y^2$ and $\sigma_z^2$ are the identity matrix?
My attempt:
$\sigma^2 = 3 {\bf I}$ (identity matrix) so can I clarify that since only the number 3 lies diagonally in the matrix, its eigenvalue is 3?  Is it correct?
For another case where $\sigma = \sigma_x + \sigma_y + \sigma_z$ where
How to find the eigenvalue in this case?
My attempt:
I realized when I add them up together, it isn't a diagonal matrix so how to find its eigenvalue in this case?

REPLY [1 votes]: Use the definition of eigenvalue: the eigenvalues of a matrix $M$ are the solutions $\lambda$ of
$$|M-\lambda I|=0$$
This definition works for diagonal and non-diagonal matrices. In the specific case of diagonal matrices, the eigenvalues are indeed the diagonal elements, so the eigenvalues of $3I$ are indeed all $3$.


TITLE: The probability of getting at least one double six in a ten throws of a pair of fair dice.
QUESTION [1 upvotes]: I calculated the probability of obtaining a double six as 1/36 and not obtaining it as 35/36
But to get at least one double six, I multiplied the probability by power 10. Yet I can't seem to arrive at the answer. What am I doing wrong please

REPLY [1 votes]: Did you remember that you have to take the complement?  In other words, if $A$ is the event that there is at least one double-six observed in ten throws, and $B$ is the event that there are no double-sixes observed in ten throws, then $$\Pr[A] = 1 - \Pr[B].$$  Since $$\Pr[B] = \left(\frac{35}{36}\right)^{10},$$ the answer is not this number, but $1$ minus this number.


TITLE: Why is this ODE classified as non-autonomous?
QUESTION [2 upvotes]: In my textbook, the following
$$\frac{d^2{x}}{d{t^2}}-\alpha x\frac{d{x}}{d{t}}-x+x^3=\sin (\omega t) $$
is given as an example of a non-autonomous ODE, without explanation. From my understanding, an ODE is non-autonomous when the coefficient of the dependent variable and its derivatives depend on the independent variable. In this case does $1$ count as a derivative of the dependent variable and therefore it is being multiplied by $\sin (\omega t)$?

REPLY [3 votes]: A differential equation is called autonomous if it can be written as
$$
\frac{dx}{dt} = f(x).
$$
In other words, the RHS cannot depend on $t$ explicitly, but only through dependence on $x(t)$.
In your case, you have a $\sin(\omega t)$ term on the RHS, so having this form is not possible.
Here are examples of autonomous ODEs:
$$
\begin{split}
x'(t) &= \sin\left(x(t)^2\right) + e^{x(t)}\\
x''(t) - 3x'(t) &= \cos(3x(t) + 35)
\end{split}
$$


TITLE: Are there infinitely many primes of form $\underbrace{3\dots3}_n{}1$?
QUESTION [12 upvotes]: The following numbers are prime:
$ 31 $
 $ 331 $
 $ 3331 $
 $ 33331 $
 $ 333331 $
 $ 3333331 $
 $ 33333331 $
Which made me think, is there something we can use to prove/disprove the statement that there are infinitely many primes of this form?

More precisely, can we prove/disprove that there are infinitely many
  primes of form:
$$\frac{10^{n+1}-7}{3}$$

This is prime for $n=1,2,3,4,5,6,7,17,39,49,59\dots$ since I tested all $n\le60$

The only proofs for "infinitely many primes of form X" I know of are using the Dirichlet's theorem, but I don't see that it would be helpful in cases like this one.

REPLY [4 votes]: A proof that there are infinite many primes of this form seems to be out of reach, but since the sequence is not increasing very fast and the $gcd$ of the numbers seems to be $1$, there are probably infinite many. The first few $n$ giving proven primes , calculated with PARI/GP are :
? for(n=1,500,m=(10^(n+1)-7)/3;if(isprime(m,2)==1,print1(n," ")))
1 2 3 4 5 6 7 17 39 49 59 77 100 150 318 381
?

You can also look up the known primes in this superb factorization database :
http://factordb.com/index.php?query=%2810%5E%28n%2B1%29-7%29%2F3&use=n&n=1&sent=Show&VP=on&VC=on&EV=on&OD=on&PR=on&PRP=on&U=on&perpage=20&format=1


TITLE: Inequality-constrained quadratic program with non-negativity constraints
QUESTION [0 upvotes]: Given matrix $\mathbf{M} \in \mathbb{R}^{p \times p}$ and vector $\boldsymbol{\mu} \in \mathbb{R}^{p}$, I want to solve
$$\begin{array}{ll} \underset{\boldsymbol{\beta} \in \mathbb{R}^p}{\text{minimize}} & \boldsymbol{\beta}^T \mathbf{M} \, \boldsymbol{\beta}\\ \text{subject to} & \boldsymbol{\mu}^T \boldsymbol{\beta} \geq \sigma\\ & \boldsymbol{1}^T \boldsymbol{\beta} \leq 1\\ & \beta_i \geq 0, \quad \forall i \in \{ 1,\, 2,\, \dots, \, p\} \end{array}$$
where $\boldsymbol{1} \in\mathbb{R}^p$ is vector of ones.
If we only have the first and second inequality constraints, a closed form solution for $\boldsymbol{\beta}$ can be easily calculated with the Lagrangian method, e.g.,
$$L = \boldsymbol{\beta}^T \, \mathbf{M}\boldsymbol{\beta} + \lambda_1 (\boldsymbol{\mu}^T \boldsymbol{\beta} - \sigma) + \lambda_2 (1 - \boldsymbol{1}^T \boldsymbol{\beta})$$
and then we calculate $\frac{\partial L}{\partial \boldsymbol{\beta}} = 0$,  $\frac{\partial L}{\partial \lambda_1} = 0$, and $\frac{\partial L}{\partial \lambda_2} = 0$.
But now how can we deal with the non-negativity constraint and how can the Lagrangian be written? Do you think I have to solve the optimization problem with only the first two inequality constraints, and then after finding
$$\boldsymbol{\beta}^* = \begin{bmatrix} \beta_1^* & \beta_2^* & \cdots & \beta^*_p \end{bmatrix}^T$$
calculate $\max(\beta^*_i, \, 0)$ for $i \in [1,p]$? Any help will be very appreciated!

REPLY [1 votes]: The approach you suggested, using only (1) and (3) to find a candidate solution and then projecting onto nonnegative vectors will not generally work on its own. However, an iterative method using the same idea is called proximal gradient descent. Given a function $f(x)$ and the constraint that $x \in A$, the set of feasible solutions, as long as you can compute the projection operator $P_A$ you can use the projection of small updates to $x$ in the direction of the gradient as an iterative update.
$$ x \leftarrow P_A(x - \alpha\nabla f(x)) $$
Where $\alpha$ is a step size that can be chosen with a line search or heuristically. If too large, you can run the risk of moving too far away from the feasible set and getting slow convergence or divergence.
You have multiple, more complex constraints though, and it may be difficult to find a simple expression for a projection onto the intersection of the constraints. A potential solution is to use Dijkstra's algorithm for projections onto the intersection of sets. Naively cycling through the constraints and projecting onto them one at a time may not give the projection onto the intersection- it will be a point in the intersection, but it may not be the closest point to the input. Dijkstra's algorithm will with just a small amount of extra bookkeeping.
Another option is to simplify the problem- choose two elements of the current solution that violate their constraints, and solve for their optimal values holding all else constant. It will be a much simpler subproblem and can be very efficient for some problems. It will end up looking very much like the sequential minimal optimization method used to solve support vector machine learning- you may be able to adapt it nearly directly. This has the benefit of only needing to operate on the variables violating the constraints after each step, so if the solution path stays close to the interior of the constraints it will be close to unconstrained gradient descent in performance.
That said, if what you need is a solution and it's not important that you write it yourself, you will almost certainly have better performance with a general quadratic programming library. Setting up the problem in Python's cvxopt package takes less than 5 lines of code and an immense amount of work has been done on these kinds of solvers that you don't have to recreate.


TITLE: Using matrix of linear operator to solve a differential equation
QUESTION [0 upvotes]: Above is a matrix of a linear operator (on the vector space of polynomials of degree at most 3) relative to the basis,
{${1, x, x^2, x^3}$}
The operator is defined as I - L, where I is the identity matrix and L is the differentiation operator (d/dx).
How can I use this to solve the differential equation,
$y'-y=5x^2-3$
Any sort of help is appreciated.

REPLY [1 votes]: Write $y = y_0 + y_1 x + y_2 x^2 + y_3 x^3$, which becomes the vector $\vec{y} = [y_0, y_1, y_2, y_3]$. Then the LHS of your equation is $(L-I)\vec{y}$. The RHS of your equation is just $[-3, 0, 5, 0]$. So now you have a simple matrix equation to solve, which is equivalent to your differential equation.