TITLE: Finding $\text{Res}(f,0)$ where $f(z)=\frac{1}{z^2\sin(z)}$
QUESTION [2 upvotes]: I am trying to determine the residue of $z=0$ where $f(z)=\frac{1}{z^2\sin(z)}$.
I have determined that $z=0$ is a pole of order $3$. Hence to compute the residue, I use $$\text{Res}(f,0)=\frac{1}{2}\lim_{z\to 0}\frac{\partial^2}{\partial z^2}\left(\frac{z}{\sin(z)}\right).$$
I am convinced I am doing something incorrect, as the calculations become very convoluted. I have computed that $$\frac{\partial^2}{\partial z^2}=\frac{z\sin^3(z)-2\sin^2(z)\cos(z)+2z\sin(z)\cos^2(z)}{\sin^4(z)},$$ but am unsure of how to further compute the residue, as I'm unable to solve this limit.
edit
I have solved the problem via the repeated use of L'Hopitals rule, but this took the work of computing some ugly limits. Is there another way?

REPLY [0 votes]: $\frac{\partial^2}{\partial z^2}(\frac z{\sin z})=\frac{\partial}{\partial z}\frac{\sin z-z\cos z}{\sin^2 z}=\frac{(\sin z-z\cos z)2\sin z\cos z+z\sin^3 z}{\sin^4 z}=\frac{2\cos z(\sin z-z\cos z)}{\sin^3 z}+\frac z{\sin z}=\frac{2\cos z}{\sin^2 z}-\frac{2z\cos^2 z}{\sin^3 z}+\frac z{\sin z}$.
Now for the middle term (as the other two are easier) , we use L'Hôpital's:  $\frac{2\cos^2 z-4 z\cos z\sin z}{3\sin^2 z\cos z}=\frac{2\cos z}{3\sin^2 z}-\frac{4z}{3\sin z}$
Now for the first term (as the other is easy), L'Hôpital's again:  $-\frac1{3\cos z}\to-\frac13$.
The limit of each term is now easy:
$-1-(-\frac13-\frac43)+1=\frac13$.
I get a limit of $\frac13$.


text:
\begin{document}
\baselineskip=18pt

\title[]{On Caccioppoli's inequalities of Stokes equations and Navier-Stokes equations near boundary}
\author{Tongkeun Chang and Kyungkeun Kang}

\thanks{}

\begin{abstract}

We study Caccioppoli's inequalities of the non-stationary Stokes
equations and Navier-Stokes equations. Our analysis is local near
boundary and we prove that, in contrast to the interior case, the
Caccioppoli's  inequalities of the Stokes equations and the
Navier-Stokes equations, in general, fail near boundary.

\end{abstract}

\maketitle

\section{Introduction}
\setcounter{equation}{0}

We consider first non-stationary Stokes equations near flat boundary
\begin{equation}\label{Stokes-10}
w_t - \De w + \na \pi =0\qquad \mbox{div } w =0 \quad \mbox{ in }
\,Q^+_{(0,1),1}:=B^+_{0,1}\times (0, 1),
\end{equation}
where $B^+_{0,r}:=\{ x=(x', x_n )\in \R: |x|<r, x_n >0\}$ for $r>0$. Here
no-slip boundary condition is given on the flat boundary, i.e.
\begin{equation}\label{Stokes-20}
w=0 \quad \mbox{ on } \,\Sigma:=(B_{0,1}\cap\{x_n=0\})\times (0,1),
\end{equation}
where $B_{0,r}:=\{ x\in \R: |x|<r \}$ for $r>0$. We emphasize that boundary
conditions are prescribed only on flat boundary of $Q^+_{(0,1),1}$, not on the
rounded boundary, $\{ x\in \R: |x|=1, x_n> 0 \}\times (0,1)$. From
now on, we denote $Q^+_{(0,1),1}$ and $B^+_{0,1}$ by $Q^+_1$ and
$B^+_1$, respectively, unless any confusion is to be expected.

We can compare similar situation to the heat equation, i.e.
\begin{align*}
v_t - \De v =0\quad \mbox{ in } \,Q^+_1
\end{align*}
with homogeneous boundary condition
\begin{align*}
v=0 \quad \mbox{ on } \,\Sigma.
\end{align*}
It is then well-known that the following  priori estimate, so called
Caccioppoli's inequality, is available:
\begin{equation}\label{Heat-50}
\norm{\nabla v}_{L^2(Q^+_{\frac{1}{2}})}\le c\norm{v}_{L^2(Q^+_1)},
\end{equation}
where $Q^+_{\frac{1}{2}}:=B^+_{\frac{1}{2}}\times (\frac{3}{4}, 1)$
and $c$ is independent of $v$. For the Stokes equations
\eqref{Stokes-10}-\eqref{Stokes-20}, an energy estimate shows
\begin{equation}\label{Stokes-55}
\norm{\nabla w}_{L^2(Q^+_{\frac{1}{2}})}\le
c\bke{\norm{w}_{L^2(Q^+_1)}+\norm{\pi}_{L^2(Q^+_1)}},
\end{equation}
and, to the authors' knowledge, the Caccioppoli's inequality as in
\eqref{Heat-50} is not known for the Stokes equations
\eqref{Stokes-10}-\eqref{Stokes-20}, namely it is unknown whether or
not the following inequality is available;
\begin{equation}\label{Stokes-57}
\norm{\nabla w}_{L^2(Q^+_{\frac{1}{2}})}\le c\norm{w}_{L^2(Q^+_1)},
\end{equation}
where $c$ is independent of $w$.

We remark that it was shown in \cite{Kang05} that the maximum of
normal derivatives of tangential components for Stokes equations are
not controlled by the righthand side of \eqref{Stokes-57}. More
precisely, an example of the Stokes equations
\eqref{Stokes-10}-\eqref{Stokes-20} was constructed such that
$\displaystyle\sup_{Q^+_{1/2}}|D_{x_3} w|$ is not bounded, although
$\int_{Q^+_1}|w|^2$ is finite. Furthermore,  solutions of the Stokes
equations \eqref{Stokes-10}-\eqref{Stokes-20} may not be even H\"older
continuous, unless corresponding pressures are  integrable (see \cite[Remark
6]{Kang05} and compare to \cite{Seregin00}).

We can also consider the Navier-Stokes equations near flat boundary,
i.e.
\begin{align*}
u_t - \De u + \na p =-\mbox{div}(u\otimes u), \qquad \mbox{div } u
=0 \quad \mbox{ in } \,Q^+_1
\end{align*}
with no-slip boundary condition
\begin{align*}
u=0 \quad \mbox{ on } \,\Sigma.
\end{align*}
Again, we emphasize that homogeneous boundary conditions are
assigned only on flat boundary of $Q_1^+$. We can ask whether or not the
following Caccioppoli type's inequality of the Navier-Stokes
equations is satisfied:
\begin{equation}\label{NSE-80}
\norm{\nabla u}^2_{L^2(Q^+_{\frac{1}{2}})}\le
c\bke{\norm{u}^2_{L^2(Q^+_1)}+\norm{u}^3_{L^3(Q^+_1)}},
\end{equation}
where $c$ is independent of $u$. The main point of the above
inequality is that pressure does not appear in the righthand side.

Our main goal is to show that it is not, in general, true to obtain
near boundary the Caccioppoli's inequality \eqref{Stokes-57} of the
Stokes equations and Caccioppoli type's inequality \eqref{NSE-80} of
the Navier-Stokes equations. More precisely, we construct  sequences
of smooth solutions (for example $W^{2,1}_2 (Q^+_1)$-solutions) of the Stokes equations and the Navier-Stokes equations
such that righthand-sides of \eqref{Stokes-57} and \eqref{NSE-80}
are uniformly bounded but left-hand sides are not uniformly bounded, i.e.
$L^2-$norms of gradient of velocities, say $\norm{\nabla u_n}^2_{L^2(Q^+_{\frac{1}{2}})}$, tend to infinity
by passing them to the limit (see Theorem
\ref{maintheorem}).

For the interior case, it was independently shown in  \cite{CSYT08},
\cite{J} and \cite{Wolf15} by different method of proofs that the
Caccioppoli's inequality \eqref{Stokes-57} of the Stokes equations is
valid in the interior.

The Caccioppoli type's inequality \eqref{NSE-80} was also extended in the interior
to {\it suitable} weak solutions of some nonlinear fluid equations,
e.g. the Navier-Stokes equations \cite{Wolf15}, Magnetohydrodynamics
equations \cite{Choe-Yang15} and non-Newtonian fluid flow
\cite{Jin-Kang17}.

We remark that the Caccioppoli type's inequality of the stationary
case was obtained in the interior or near boundary (refer to e.g.
\cite{Giaquinta-Modica}, \cite{Kang04} and \cite{Sverak-Tsai98}).

Our main result reads as follows:
\begin{theo}\label{maintheorem}
The Caccioppoli's inequality \eqref{Stokes-57} of the Stokes equations
and Caccioppoli type's inequality \eqref{NSE-80} of the
Navier-Stokes equations, in general, fail  near boundary.
\end{theo}

We remark that our analysis is only local near boundary. Our
construction of sequences of solutions failing \eqref{Stokes-57} or
\eqref{NSE-80} are caused by non-local behavior of solutions and
singular boundary conditions away from flat boundary. Therefore,
such construction would not be applicable to the Stokes and the
Navier-Stokes equations in domains with no-slip conditions
on boundaries everywhere.

We briefly give the mainstream of
how we show Theorem \ref{maintheorem}.

Firstly, we construct, in a half-space, a very weak solution of the Stokes equations
whose normal derivatives of tangential components is not locally square integrable near boundary.
More precisely, we use the explicit formula \eqref{rep-bvp-stokes-w} of the Stokes equations with a
boundary condition given in \eqref{boundarydata} and \eqref{0502-6}.
In particular, among all split terms of the normal derivative of tangential component, it turns out that the following integral (see \eqref{0411-v}) is not square integrable in a local neighborhood near boundary (see Proposition \ref{lemma0406}).
\[
D_{x_n} \int_0^t \int_{\Rn} B_{in}(x'- y', x_n, t -s) g_n(y', s) dy'ds,\qquad i=1,2, \cdots n-1,
\]
where $B_{in}$ is given in \eqref{B-tensor}.
Indeed, one crucial estimate
is
\[
\int_{\Rn} e^{-\frac{|X'-z'|^2}{t-s}} \frac{z_1}{|z'|^n} dz' \gtrsim (t-s)^{\frac{n-1}2}\bke{1-e^{-\frac{c}{t-s}}}, \quad |X'| \geq \frac12,
\]
which causes the unbounded $L^2$-norm of normal derivative in $B_r(0)\times (t_0, t_0+{r^2})$
(see \eqref{est-J1J2J3}, \eqref{grad-w1} and \eqref{L2-blowup}). On the other hand, we can see that
the solution is integrable in $L^4 (0, \infty; L^p(\R_+))$ for $n/(n-1)<p<\infty$ (see Proposition \ref{lemma0406}).

Next step is to regularize the boundary data that is originally singular so that corresponding solutions of the Stokes equations
are regular. If the Caccioppoli's inequality near boundary is valid, then the regular solutions
should satisfy the inequality. If this is the case, by passing to the limit, the gradient of the limit solution must
be square integrable near boundary, which leads to a contradiction (see for details Subsection 4.1 in Section 4).

For the Navier-Stokes equations, we consider the similar situation as in the Stokes equations.
If we denote by $w$ the solution of the Stokes equations mentioned above, we look for a solution $u$
of the form $u=w+v$ such that $v$ solving the following perturbed Navier-Stokes equations:
\[
v_t-\Delta v+\nabla\pi=-{\rm div}\,( (v+w)\otimes (v+w)), \qquad {\rm div}\, v=0
\]
with zero boundary and zero initial data.
In fact, we construct a weak solution $v$ whose gradient is square integrable, in case that the size of
$w$ is assumed to be a sufficiently small.
Therefore, similar argument as in Stokes equations yields that the Caccioppoli's inequality near boundary
fails in general for the Navier-Stokes equations as well (see for details Subsection 4.2 in Section 4).

This paper is organized as follows. In Section 2, we recall some
known results and introduce new estimates that are useful for our purpose. Section 3 is devoted
to constructing a very weak solution in a half space such that
$L^2-$norm of its gradient is not bounded. In Section 4, we present
the proof of Theorem \ref{maintheorem}.

\section{Preliminaries}
\setcounter{equation}{0}

For notations,  we denote $x=(x^{\prime},x_n)$, where the symbol
${\prime}$ means the coordinate up to $n-1$, that is,
$x^{\prime}=(x_1,x_2,\cdot\cdot\cdot,x_{n-1})$. We write $D_{x_i} u$
  as the partial derivative of $u$ with respect to $x_i, \,\, 1 \leq i \leq n$, i.e., $D_{x_i} u(x) =
\frac{\pa }{\pa x_i}u(x)$. Throughout this paper we denote by $c$ various generic positive constant and by $c(*,\cdots,*)$ depending on the quantities appearing in the
parenthesis.

Next, we introduce notions of {\it very weak solutions} for the
Stokes equations and the Navier-Stokes equations with non-zero boundary
values in a half-space $\hR$. To be more precise, we consider first
the following Stokes equations in $\hR$ with non-zero boundary values:
\begin{equation}\label{Stokes-bvp-200}
w_t - \De w + \na \pi = {\rm div} \,F, \qquad \mbox{div } w =0,\quad
\mbox{ in }
 \hR \times (0,\infty)
\end{equation}
with
\begin{equation}\label{Stokes-bvp-210}
\rabs{w}_{t=0}= 0, \qquad  \rabs{w}_{x_n =0} = g.
\end{equation}
We now define very weak solutions of
\eqref{Stokes-bvp-200}-\eqref{Stokes-bvp-210}.
\begin{defin}\label{stokesdefinition}
Let $g \in L^1_{\rm{loc}} (\pa \hR\times (0, T))$ and $ F \in
L^1_{\rm{loc}}(\hR\times (0, T))$. A vector field $u\in
L^1_{\rm{loc}}( \hR \times (0, T))$ is called a very weak solution
of the Stokes equations \eqref{Stokes-bvp-200}-\eqref{Stokes-bvp-210},
if the following equality is  satisfied:
\begin{align*}
-\int^T_0\int_{\hR }u\cdot \Delta \Phi
dxdt=\int^T_0\int_{\hR}\bke{u\cdot \Phi_t-F:\nabla \Phi} dxdt
-\int^T_0 \int_{\pa \hR} g \cdot D_{x_n}\Phi  dx' dt
\end{align*}
for each $\Phi\in C^2_c(\overline{\hR}\times [0,T])$ with
\begin{align}\label{0528-1}
\mbox{\rm div } \Phi=0,\quad \Phi\big|_{\pa \hR \times (0, T)}=0, \quad
\Phi(\cdot,T) =0.
\end{align}
In addition, for each $\Psi\in C^1_c(\overline{\hR})$ with
 $\Psi \big|_{\pa \hR }=0$
\begin{equation}\label{Stokes-bvp-2200}
\int_{\hR} u(x,t) \cdot \na \Psi(x) dx =0  \quad  \mbox{ for all}
\quad 0 < t< T.
\end{equation}
\end{defin}

We recall  existence results of the Stokes equations
\eqref{Stokes-bvp-200}-\eqref{Stokes-bvp-210}.
\begin{prop}\label{thm-stokes}(\cite[Theorem 1.2]{CJ})\,\,
Let  $1<p, q<\infty$. Assume that $g\in
L^q(0,\infty;\dot{B}^{-\frac{1}{p}}_{pp}(\Rn))$ and $F\in
L^{q_1}(0,\infty; L^{p_1}(\hR)) $, where $p_1, q_1$ satisfy
$n/p_1+2/q_1=n/p+2/q+1$ with $q_1\leq q$ and $p_1 \leq p$. Then, there is the unique
very weak solution $w\in L^q(0,\infty;L^p(\R_+))$ of the Stokes equations
\eqref{Stokes-bvp-200}-\eqref{Stokes-bvp-210} such that $w$
satisfies
\[
\| w\|_{L^q(0,\infty;L^p(\R_+))}\leq c \bke{\|F\|_{L^{q_1}(0,\infty; L^{p_1}(\R_+))} +\|
g\|_{L^q(0,\infty;\dot{B}^{-\frac{1}{p}}_{pp}(\Rn))}}.
\]
\end{prop}
Here, $\dot{B}^{\al}_{pp}(\Rn)$ is homogeneous Besov space in $\Rn$ (see \cite{BL} for definition and properties of homogeneous Besov spaces).

The estimate of the following proposition may be known to the experts, but we couldn't find it in the literature.
Since it will be used to prove Theorem \ref{maintheorem}, we give its details in Appendix \ref{appendixa}.
\begin{prop}\label{theo0503}
Let $1< p, q< \infty$, $F \in L^q (0,\infty; L^p (\R_+))$, $\rabs{F}_{x_n =0} \in L^q (0, \infty; \dot B^{-\frac1p}_{pp} (\Rn))$ and  $g =0$. Then, there is the unique
very weak solution $w $ of the Stokes equations
\eqref{Stokes-bvp-200}-\eqref{Stokes-bvp-210} such that $w$
satisfies
\begin{align}\label{0504-1}
\|  \na w\|_{L^q (0, \infty;   L^p (\R_+)} \leq c \bke{\| F\|_{L^{q} (0,\infty;  L^p (\R_+))} +  \| \rabs{F}_{x_n =0}\|_{L^q (0, \infty; \dot B^{-\frac1p}_{pp}(\Rn))}}.
\end{align}
\end{prop}
In \cite{GGS}, the authors showed \eqref{0504-1} with additional condition  ${\rm div}\, F =0$ and $F_{in}|_{x_n =0}=0, \, i =1, \cdots , n$. In \cite{KS}, the authors    showed \eqref{0504-1} with condition $p =q$, then the second term of the right-hand side is dropped.

We also consider the boundary value problem of the
Navier-Stokes equations in a half-space, namely
\begin{equation}\label{NSE-bvp-230}
u_t - \De u + \na p =-\mbox{div}(u\otimes u), \qquad \mbox{div } u
=0 \qquad\mbox{ in } \hR \times (0,\infty)
\end{equation}
with
\begin{equation}\label{NSE-bvp-240}
\rabs{u}_{t=0}= 0, \qquad  \rabs{u}_{x_n =0} = g.
\end{equation}
We mean by very weak solutions of the Navier-Stokes equations
\eqref{NSE-bvp-230}-\eqref{NSE-bvp-240} as follows:

\begin{defin}
Let $g \in L^1_{\rm{loc}}(\pa \hR\times (0, T))$. A vector field
$u\in L^2_{loc}( \hR \times (0, T))$ is called a very weak solution
of the non-stationary Navier-Stokes equations
\eqref{NSE-bvp-230}-\eqref{NSE-bvp-240}, if the following equality
is  satisfied:
\begin{align*}
-\int^T_0\int_{\hR}u\cdot \Delta \Phi dxdt=\int^T_0\int_{\hR}(u\cdot
\Phi_t+(u\otimes u):\nabla \Phi) dxdt -\int^T_0 \int_{\pa \hR} g
\cdot D_{x_n} \Phi dx' dt
\end{align*}
for each $\Phi\in C^2_c(\overline{\hR}\times [0,T])$ satisfying \eqref{0528-1}.
In addition, for each $\Psi\in C^1_c(\overline{\hR})$ with
 $\Psi \big|_{\pa \hR }=0$,  $u$ satisfies \eqref{Stokes-bvp-2200}.
\end{defin}

\section{Stokes equations with boundary data in a half-space}
\setcounter{equation}{0}

 We let $N$ and $\Gamma$  be the fundamental solutions to the Laplace equation
 and the heat equation, respectively, i.e.
\begin{align*}
N(x) = \left\{\begin{array}{ll}\vspace{2mm} -\frac{1}{(n-2)\om_n} \frac{1}{|x|^{n-2}},& n \geq 3\\
\frac1{2\pi} \ln |x|, & n =2
\end{array}
\right. \qquad
\Gamma(x,t)  = \left\{\begin{array}{ll} \vspace{2mm}
\frac{1}{\sqrt{4\pi t}^n} e^{ -\frac{|x|^2}{4t} }, &   t > 0,\\
0, & t < 0,
\end{array}
\right.
\end{align*}
where $\om_n$ is the measure of the unit sphere in $\R$. For
convenience, we introduce a tensor $L_{ij}$ and a scalar function
$A$ defined by
\begin{align}\label{L-tensor}
{L}_{ij} (x,t) & =  D_{x_j}\int_0^{x_n} \int_{\Rn}   D_{z_n}
\Ga(z,t) D_{x_i}   N(x-z)  dz,\quad
i,j=1,2,\cdots,n,\\
\notag A(x,t) & =\int_{\Rn}\Ga(z^{\prime},0,t)N(x^{\prime}-z^{\prime},x_n)dz^{\prime}.
\end{align}

The Poisson kernel $(K,\pi ) $ of the Stokes equations is  given as
\[
K_{ij}(x'-y',x_n,t) =  -2 \delta_{ij} D_{x_n}\Ga(x'-y', x_n,t)
-L_{ij} (x'-y',x_n,t)
\]
\begin{equation}\label{Poisson-tensor-K}
+  \de_{jn} \de(t)  D_{x_i} N(x'-y',x_n),
\end{equation}
\[
\pi_j (x'-y',x_n,t )=-2\delta(t) D_{x_j}D_{x_n}N(x'-y',x_n)+4
D_{x_n}D_{x_n}A(x'-y',x_n,t)
\]
\begin{equation}\label{Poisson-tensor-P}
+4D_{t} D_{x_j}A(x'-y',x_n,t),
\end{equation}
where $\delta(t)$ is the Dirac delta function and $\delta_{ij}$ is
the Kronecker delta function.

We recall the following relations on $L$  (see \cite{So}):
\begin{align} \label{1006-3}
\sum_{i=1}^{n} L_{ii} = -2D_{x_n} \Ga, \quad\qquad  L_{in} =
L_{ni}  + B_{in} \,\, \mbox{ if }\, i \neq n,
\end{align}
where
\begin{equation}\label{B-tensor}
B_{in}(x,t) = -\int_{\Rn}D_{x_n} \Ga(x^\prime -z^\prime ,
x_n, t) D_{z_i} N( z^{\prime},0) dz^\prime.
\end{equation}
Furthermore, we remind estimates of $L_{ij}$ defined in
\eqref{L-tensor} (see \cite{So}).
\begin{equation}\label{est-L-tensor}
|D^{l_0}_{x_n} D^{k_0}_{x'} D_{t}^{m_0} L_{ij}(x,t)| \leq
\frac{c}{t^{m_0 + \frac12} (|x|^2 +t )^{\frac12 n + \frac12 k_0}
(x_n^2 +t)^{\frac12 l_0}},
\end{equation}
where $ 1 \leq  i \leq n$ and $1 \leq j \leq n-1$.

It is shown in \cite{So} that the solution $(w,\pi)$ of the Stokes
equations \eqref{Stokes-bvp-200}-\eqref{Stokes-bvp-210} with $ F =0$ is expressed by
\begin{align}\label{rep-bvp-stokes-w}
w_i(x,t) = \sum_{j=1}^{n}\int_0^t \int_{\Rn} K_{ij}(
x^{\prime}-y^{\prime},x_n,t-s)g_j(y^{\prime},s) dy^{\prime}ds,\\
\label{rep-bvp-stokes-p}
\pi(x,t) = \sum_{j=1}^{n}\int_0^t \int_{\Rn}
\pi_j(x^{\prime}-y^{\prime},x_n,t-s) g_j(y^{\prime},s)
dy^{\prime}ds,
\end{align}
where $(K_{ij}, \pi_j)$ is Poisson kernel of the Stokes equations given
in \eqref{Poisson-tensor-K} and \eqref{Poisson-tensor-P}.

Next, we will construct a solution $w$ of Stokes equations via
\eqref{rep-bvp-stokes-w} and \eqref{rep-bvp-stokes-p} for a certain $g$ such that
$L^2$-norm of $\nabla w$ is not bounded. For convenience, we denote
\[
A = \{ y'=(y_1, y_2, \cdots, y_{n-1})\in \Rn \, | \, 1 < |y'| < 2, \,  -2 < y_i
< -1, \, \, 1 \leq i \leq n-1 \}.
\]
We note that if $y' \in A$, then $|y'| \leq  2< -2y_i$ for all $1\leq i\leq n-1$. Let
\begin{align}\label{boundarydata}
g^1_n(y') = \chi_{A}(y'),\qquad  g^2_n(s) =
\bke{s-\frac1{16}}^{-\frac14} \abs{\ln \bke{s-\frac1{16}}}^{-\frac14
-\ep}\chi_{\frac1{16} < s< \frac14}(s),
\end{align}
where $\ep$ is a fixed constant with $0 <\ep \leq \frac14$ and $\chi$ is characteristic function. We introduce
a non-zero boundary data $g:\mathbb R^{n-1}\times \mathbb
R_+\rightarrow  \R$ with only $n-$th component defined
by
\begin{align}\label{0502-6}
g(y', s) = (0, \cdots, 0,  g_n (y',s))= (0, \cdots, 0,\al g^1_n (y')g^2_n(s)),
\end{align}
where $\al> 0$ is determined late on.

The following is two dimensional cartoon for $A$ that is the support of $g_n^1$.

\begin{tikzpicture}[scale=3]

\draw (0,0) ellipse (8.5pt and 8.5pt);

\draw (0,0) ellipse (17pt and 17pt);

\draw (0,0) ellipse (34pt and 34pt);

\draw (-0.8, -0.75) node {$A$};

\draw(0.13,1.5) node {$y_2$};

\draw (0,1.5) --(-0.05, 1.45);
\draw (0,1.5) --(0.05, 1.45);

\draw (-2,0) -- (2,0);

\draw (-2,-1.2) -- (2,-1.2);

\draw (-2,-0.6) -- (2,-0.6);

\draw (0,-1.5) -- (0, 1.5);

 \draw (-0.6, 1.5) -- (-0.6,-1.5);

\draw (-1.2, 1.5) -- (-1.2,-1.5);

\draw (-0.65, -0.7) node {$ \bullet$};

\draw (-0.55, -0.75) node {$  y'$};

\draw (-0.32, -0.1) node {$-\frac12$};

\draw (-0.62, -0.1) node {$-1$};

\draw (-1.22, -0.1) node {$-2$};

\draw (2.05,-0.099) node {$y_1$};

\draw (2,0) --(1.95, 0.05);
\draw (2,0) --(1.95, -0.05);

\draw (-0.2, -0.15) node {$ \bullet$};
\draw (0.0, 0.0) node {$ \bullet$};
\draw (0.05, -0.05) node {$ 0$};

\draw (-0.1,-0.1 ) node {$x'$};

\draw (-0.68,-0.98) -- (-0.6,-0.9);

\draw (-0.743,-0.945) -- (-0.6,-0.8);

\draw (-0.799,-0.899) -- (-0.6,-0.7);

\draw (-0.85,-0.85) -- (-0.6,-0.6);

\draw (-0.9,-0.8) -- (-0.7,-0.6);

\draw (-0.95,-0.74) -- (-0.8,-0.6);

\draw (-0.98,-0.68) -- (-0.9,-0.6);

\draw (0.0, -1.7) node {Figure; A is the support of $g_n^1$ in $\mathbb R^2$};

\end{tikzpicture}

\begin{rem}\label{rem0101-1}
We note that
$g \in L^4(0,T;\dot{B}^{-\frac{1}{p}}_{pp}(\Rn))$ for $\frac{n}{n-1}  <p <
\infty$. In fact, for $ 1 < r <\infty$ satisfying $ -\frac{n-1}r = -\frac1p -\frac{n-1}p$, we can see that (refer to e.g. \cite[Theorem 6.5.1]{BL})
\begin{align*}
\| g(t)\|_{ \dot{B}^{-\frac{1}{p}}_{pp}(\Rn) } \leq c \| g(t)\|_{L^r(\Rn)} < \infty.
\end{align*}

\end{rem}

\begin{prop}\label{lemma0406}
Let $w$ be a solution of the Stokes equations
\eqref{Stokes-bvp-200}-\eqref{Stokes-bvp-210} with $F=0$ defined by
\eqref{rep-bvp-stokes-w}, where the boundary data $g$ is given in
\eqref{boundarydata} and \eqref{0502-6}. Let $ \frac{n}{n-1} < p < \infty$, $t_0 = \frac1{16}$ and
$r \leq \frac12$. Then, $w$ satisfies
\begin{equation}\label{L4Lp-w}
\| w\|_{L^4(0,\infty;L^p(\R_+))}<\infty,
\end{equation}
\begin{equation}\label{L2-grad-w}
\int_{t_0}^{ t_0 + r^2} \int_{B(0, r)} |D_{x_n} w_i(x,t) |^2 dxdt =
\infty \qquad  i =1,\cdots, n-1,
\end{equation}
\begin{equation}\label{L2-grad-w-n}
\int_{t_0}^{ t_0 + r^2} \int_{B(0, r)} |D_{x} w_n(x,t) |^2 dxdt <\infty.
\end{equation}
\end{prop}

\begin{proof}
We recall  Proposition \ref{thm-stokes} and Remark \ref{rem0101-1}, which implies that for $\frac{n}{n-1} <
p < \infty$
\begin{align*}
\| w\|_{L^4(0,\infty;L^p(\R_+))}\leq c_p \| g\|_{L^4(0,\infty;\dot{B}^{-\frac{1}{p}}_{pp}(\Rn))}  \leq c \| g\|_{L^4(0, \infty;L^r(\Rn))} < \infty.
\end{align*}
Hence, the first inequality \eqref{L4Lp-w} is immediate.

Next, we show the estimate \eqref{L2-grad-w}.  Since the arguments are similar,
we only prove the case of $i =1$. It follows from \eqref{Poisson-tensor-K} and
\eqref{rep-bvp-stokes-w} that
\begin{align*}
w_1(x,t) & =    \int_0^t   \int_{A} L_{1 n}(x'- y', x_n, t
-s) g_n(y', s) dy' ds -   \int_{A} D_{x_1} N(x' -y', x_n)g_n(y',t) dy'\\
&:=  w^1_{1}(x,t) + w^2_{1}(x,t).
\end{align*}
We note that $|x' -y'| \geq \frac12$ for $|x'| \leq \frac12  $ and $
y' \in A$, and so, for $(x, t) \in B(0, \frac12) \times (t_0,
\frac14) $ we have
\begin{align}\label{0411-w}
|\na w_1^2(x,t)| \leq c (t-t_0)^{-\frac14} |\ln ( t -t_0)|^{-\frac14 -\ep}.
\end{align}
Since $L_{1n} = L_{n1} + B_{1n}$ by the second equality of \eqref{1006-3}, we divide $w_1^1= w_1^{11} + w_1^{12}$ by
\begin{align}\label{0411-v}
\notag w^1_1(x,t) &=  \int_0^t \int_{\Rn} L_{n1}(x'- y', x_n, t -s) g_n(y',
s)dy'ds\\
\notag & \qquad  +\int_0^t \int_{\Rn} B_{1n}(x'- y', x_n, t -s) g_n(y', s)dy'ds\\
&: =
w^{11}_1 (x,t) + w^{12}_1 (x,t).
\end{align}

Now, we estimate of  $w^{11}_1$.  We assume that $ t -t_0 < x_n^2$. Note that $|x' -y'| \geq \frac12$ for $x \in B(0, \frac12) $ and $y' \in A$. Due to \eqref{est-L-tensor},
for $(x,t) \in B(0, \frac12) \times (t_0, \frac14)$, we have
\begin{align}

\notag |D_x w^{11}_1 (x,t)|

 & \leq   c\int_{t_0}^t \int_{1 \leq |y'| \leq 2 } \frac1{ (t
-s)^\frac12 ( x_n^2 + t -s)^\frac12  ( |x' -y'|^2 + x_n^2 + t
-s)^\frac{n}2 }\\
\notag &\qquad \qquad \qquad \qquad \times  (s -t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep} dy'
ds\\
\notag  &\leq   c
  x_n^{-1} \int_{t_0}^{t}  \frac1{ (t
-s)^\frac12  }
 (s-  t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep} ds\\
\notag & = c x_n^{-1} \int_{t_0}^{\frac{t_0 +t}2}   \frac1{ (t
-s)^\frac12  }
 (s-  t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep} ds\\
\notag&  \qquad  +  cx_n^{-1} \int^{t}_{\frac{t_0 +t}2}   \frac1{ (t
-s)^\frac12  }
 (s-  t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep} ds\\
\label{w11-0}
 & = I_1 + I_2.
\end{align}
Since $t - t_0 < x_n^2$, we have
\begin{align} \label{0409I2}
\notag I_2 & \leq c (t-  t_0)^{-\frac34} |\ln (t- t_0)|^{-\frac14 -\ep} \int^{t}_{\frac{t_0 +t}2}   \frac1{ (t
-s)^\frac12  }   ds\\
 &  =
c   (t-  t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}.
\end{align}
On the other hand, noting that for $0 < a < \frac12$, we observe
\begin{align}\label{0409-1}
 c_1  a^{\frac34} |\ln a|^{-\frac14 -\ep} \leq   \int_0^a s^{-\frac14 } |\ln s|^{-\frac14 -\ep} ds \leq c_2  a^{\frac34} |\ln a|^{-\frac14 -\ep}.
\end{align}
Indeed, via Hospital's Theorem, we have
\[
\lim_{a \ri 0} \frac{ \int_0^a s^{-\frac14 } |\ln s|^{-\frac14 -\ep} ds }{ a^{\frac34} |\ln a|^{-\frac14 -\ep}} = \lim_{a \ri 0} \frac{ a^{-\frac14 } |\ln a|^{-\frac14 -\ep}  }{ \frac34 a^{-\frac14} |\ln a|^{-\frac14 -\ep} -(\frac14 +\ep)  a^{-\frac14} |\ln a|^{-\frac54 -\ep}} = \frac43.
\]
Since $a^{\frac34} |\ln a|^{-\frac14 -\ep}, \,    \int_0^a s^{-\frac14 } |\ln s|^{-\frac14 -\ep} ds > 0$ for $0 < a < \frac12$, which implies \eqref{0409-1}.

Due to \eqref{0409-1}, since $t -t_0 < x_n^2$, we obtain
\begin{align}
\notag I_1 & \leq  c (t -t_0)^{-1}  \int_{t_0}^{\frac{t_0 +t}2}
 (s-  t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep} ds\\
\label{0409-2}
& \leq   c    (t-t_0 )^{-1}  \int_0^{\frac{t -t_0 }2}
 s^{-\frac14} |\ln s|^{-\frac14 -\ep} ds
 \leq    c   (t-  t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}.
\end{align}
Summing up \eqref{w11-0}, \eqref{0409I2} and \eqref{0409-2},
for $(x,t) \in B(0, \frac12) \times (t_0, \frac14)$ and $ t -t_0 < x_n^2$,  we have
\begin{align}\label{w11-1}
|D_x w^{11}_1 (x,t)| \leq  c   (t-  t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}.
\end{align}

Next, we estimate $w_1^{12}$. Reminding \eqref{B-tensor}, we note that
\begin{align}\label{w12}
\notag D_{x_n} w^{12}_1(x,t) &= c_n \int_{t_0}^t \int_{\Rn} g_n(y', s)  (t-s)^{-\frac{n+2}2}    \bke{-2 + \frac{4 x_n^2}{t-s}}\\
 & \qquad \qquad \times  e^{-\frac{x_n^2}{t-s}}
\int_{\Rn}  e^{-\frac{|x' -y'-z'|^2}{t-s}}   \frac{z_1}{|z'|^n}   dz' dy' ds.
\end{align}
For fixed $X' = x' -y'$, we divide $\Rn$ by three disjoint sets
$D_1, D_2$ and $D_3$ defined by
\[
D_1=\bket{z'\in\Rn: |X'-z'| \leq \frac1{10} |X'|},
\]
\[
D_2=\bket{z'\in\Rn: |z'| \leq \frac1{10} |X'|}, \qquad D_3=\Rn\setminus
(D_1\cup D_2).
\]
We then split the following integral into three terms as follows:
\begin{align*}
\int_{\Rn} e^{-\frac{|X'-z'|^2}{t-s}} \frac{z_1}{|z'|^n} dz' =
\int_{D_1}\cdots + \int_{D_2} \cdots+ \int_{D_3}\cdots := J_1 + J_2
+J_3.
\end{align*}
Since $\int_{D_2}\frac{z_1}{|z'|^n} dz' =0$,  we have $\int_{D_2} \frac{z_1}{|z'|^n} e^{-\frac{|X'-z'|^2}{t-s}}   dz' = \int_{D_2} \frac{z_1}{|z'|^n} \big(
e^{-\frac{|X'-z'|^2}{t-s}} -  e^{-\frac{|X'|^2}{t-s}} \big) dz'$.  Using the Mean-value Theorem, we  have
\begin{align}\label{est-J2}
\notag |J_2| & =\abs{\int_{D_2}   \frac{z_1}{|z'|^n} \big(
e^{-\frac{|X'-z'|^2}{t-s}} -  e^{-\frac{|X'|^2}{t-s}} \big) dz'
}\leq   c(t-s)^{-1 }|X'| e^{-c\frac{|X'|^2}{t-s}}  \int_{D_2}
\frac{1}{|z'|^{n-2}}   dz'\\
&\leq  c (t-s)^{-1 }|X'|^2 e^{-c\frac{|X'|^2}{t-s}} \leq  c
e^{-c\frac{|X'|^2}{t-s}}.
\end{align}
Since $\int_{|z'| > a} e^{-|z'|^2} dz' \leq c_1 e^{-c_2 a^2}, \, a > 0$, we have
\begin{align}
\notag|J_3| &\leq    \frac{c}{|X'|^{n-1}}\int_{D_3} e^{-\frac{|z' -X'|^2}{t-s}}
dz' \leq   \frac{c}{|X'|^{n-1}} \int_{\{|z'-X'| \geq \frac1{10} |X'|\}} e^{-\frac{|z'-X'|^2}{t-s}} dz'\\
\notag & =   \frac{c(t-s)^{\frac{n-1}2}}{|X'|^{n-1}} \int_{\{|z'| \geq \frac1{10}\frac{ |X'|}{\sqrt{t-s}}\}} e^{- |z'|^2} dz'\\
\label{est-J3}
&\leq   \frac{c(t-s)^{\frac{n-1}{2}}}{|X'|^{n-1}}  e^{-c\frac{ |X'|^2}{t-s}}\leq   c
e^{-c \frac{|X'|^2}{t-s}}.
\end{align}
Since    $ \frac12 \leq |X'| \leq 2$, from \eqref{est-J2} and \eqref{est-J3}, we have
\begin{align}\label{0604}
|J_2|, \, |J_3| \leq   c e^{-c\frac1{ ( t-s)}}
\end{align}

Now, we estimate $J_1$. We note that for $|x'| < \frac12$ and $y'
\in A$ we see that
$X_1 = x_1 - y_1 \geq -\frac12 + 1 =\frac12$ and
$\frac15 |X'| \leq \frac15 (|x'| + | y'|) \leq \frac15 \cdot \frac52
\leq X_1$. Then, for $|X'-z'| \leq \frac1{10} |X'| $,  we have
\[
z_1 = z_1 -X_1 + X_1 \geq X_1 - |z_1 - X_1| \geq X_1 - |X' -z'|
\]
\[
\geq X_1 - \frac1{10} |X'| \geq \frac15|X'| - \frac1{10} |X'| = \frac1{10} |X'|.
\]
Therefore,  we obtain
\begin{align}\label{est-J1}
\notag J_1 & = \int_{D_1} e^{-\frac{|X'-z'|^2}{t-s}} \frac{z_1}{|z'|^n} dz'
\geq \frac{c}{|X'|^{n-1}} \int_{D_1} e^{-\frac{|X'-z'|^2}{t-s}}
dz'\\
\notag & = \frac{c(t-s)^{\frac{n-1}2}}{|X'|^{n-1}}
\int_{\{|z'| \leq \frac1{10}\frac{ |X'|}{\sqrt{t-s}}\}}  e^{-|z'|^2}
dz'\\
& \geq \frac{c(t-s)^{\frac{n-1}2}}{|X'|^{n-1}} \geq c(t-s)^{\frac{n-1}2} .
\end{align}
Combining \eqref{0604} and \eqref{est-J1}, we
obtain
\begin{equation}\label{est-J1J2J3}
\int_{\Rn} e^{-\frac{|x'-y'-z'|^2}{(t-s)}} \frac{z_1}{|z'|^n} dz'
\geq   c  \Big(   (t -s)^{\frac{n-1}2}  -
e^{-c\frac{1}{(t-s)}} \Big) \geq   c (t -s)^{\frac{n-1}2} \Big(1     -
e^{-c\frac{1}{(t-s)}} \Big).
\end{equation}
Noting that $ -2 +  \frac{4 x_n^2}{t-s} > 2$
for $t-t_0  < x_n^2$ and   $t_0 < s< t$,  it follows from \eqref{w12} and \eqref{est-J1J2J3} that
\begin{align*}
|D_{x_n} w^{12}_1(x,t)| &\geq c \int_{t_0}^t \int_{A } (s-t_0)^{-\frac14}|\ln ( s -t_0)|^{-\frac14 -\ep}  (t-s)^{-\frac32}
  e^{-\frac{x_n^2}{t-s}}
\big( 1 - e^{-c\frac1{t-s}}  \big) dy' ds\\
&\geq c \int_{t_0}^t   (s-t_0)^{-\frac14}  |\ln (s-t_0)|^{-\frac14 -\ep} (t-s)^{-\frac32}     e^{-\frac{x_n^2}{t-s}} ds -c.
\end{align*}
Splitting the above integral over intervals $(t_0, \frac12 (t +
t_0))$ and $(\frac12 (t + t_0), t)$, we first compute
\begin{align*}
&\int_{t_0}^{\frac12 (t + t_0) }  (s-t_0)^{-\frac14} |\ln (
s-t_0)|^{-\frac14 -\ep} (t -s)^{-\frac32} e^{-\frac{x_n^2}{t -s}}    ds\\
&
\geq  c (t-t_0)^{-\frac32} e^{-\frac{x_n^2}{t -t_0}}
\int_{t_0}^{\frac12 (t + t_0)}       (s-t_0)^{-\frac14} |\ln
(s-t_0)|^{-\frac14 -\ep}  ds\\
&
\geq  c (t-t_0)^{-\frac32} e^{-\frac{x_n^2}{t -t_0}}
\int_0^{\frac12 (t - t_0)}       s^{-\frac14} |\ln
s|^{-\frac14 -\ep}  ds\\
&
\geq  c(t-t_0)^{-\frac34}|\ln (t-t_0)|^{-\frac14 -\ep}
e^{-\frac{x_n^2}{t-t_0}}.
\end{align*}
For the third inequality, we use $\int_0^a   s^{-\frac14} |\ln
s|^{-\frac14 -\ep}  ds \geq c   a^{\frac34} |\ln
a|^{-\frac14 -\ep}  $ for $ a \leq \frac12$ (see \eqref{0409-1}).

On the other hand, since $\int_a^\infty s^\frac12 e^{-s} ds \geq c
a^\frac12 e^{- a}$ for $ a > 1$, we have
\begin{align*}
&\int_{ \frac12 ( t + t_0)}^t (s-t_0)^{-\frac14} |\ln
(s-t_0)|^{-\frac14 -\ep} (t -s)^{-\frac32} e^{-\frac{x_n^2}{t -s}}
   ds\\
&\geq  c(t-t_0)^{-\frac14}|\ln (t-t_0)|^{-\frac14 -\ep} \int_{\frac12
(t+t_0)}^t (t -s)^{-\frac32} e^{-\frac{x_n^2}{t -s}}       ds\\
&=   c(t-t_0)^{-\frac14}|\ln (t-t_0)|^{-\frac14 -\ep} \int_0^{\frac12
( t-t_0)} s^{-\frac32} e^{-\frac{x_n^2}s}        ds\\
&\geq  c(t-t_0)^{-\frac14}|\ln (t-t_0)|^{-\frac14 -\ep} x_n^{-1}
\int^\infty_{\frac{2x_n^2}{t-t_0}} s^{\frac12} e^{-s}         ds\\
&\geq c(t-t_0)^{-\frac34} |\ln (t-t_0)|^{-\frac14 -\ep}
e^{-c\frac{x_n^2}{t-t_0}}.
\end{align*}
Adding the estimates above,  for $(x,t) \in B (0,\frac12) \times (t_0, \frac12)$ and  $t -t_0 \leq x_n^2$,  we have
\begin{align}\label{0411-v-2}
|D_{x_n} w^{12}_1(x,t)| &\geq c (t-t_0)^{-\frac34} |\ln (t-t_0)|^{-\frac14 -\ep} e^{-\frac{x_n^2}{t-t_0}} -c.
\end{align}
From \eqref{0411-w}, \eqref{w11-1} and \eqref{0411-v-2},
for $t -t_0 \leq x_n^2$ and $(x,t) \in B(0,\frac12) \times (t_0, \frac12)$, we get
\begin{equation}\label{grad-w1}
|D_{x_n} w_1(x,t)|
  \geq  c(t-t_0)^{-\frac34}|\ln (t-t_0)|^{-\frac14 -\ep}  e^{-\frac{x_n^2}{t-t_0}}
  -c (t-t_0)^{-\frac14} |\ln ( t -t_0)|^{-\frac14 -\ep}.
\end{equation}
Since
\begin{align*}
\notag & \int_{t_0}^\frac14 \int_{   \{(t-t_0)^\frac12 \leq x_n \leq
1 \}}  (t-t_0)^{-\frac12 } |\ln (t-t_0)|^{-\frac12 -2\ep}  dx_n dt <
\infty,
\end{align*}
from  \eqref{grad-w1}, for any $\ep$ with $0<\ep \leq  \frac14$,  we obtain
\begin{align}
\notag & \int_{t_0}^{t_0 + r^2} \int_{B_r(0)} |D_{x_n} w_1(x,t)|^2 dxdt \\
\notag
&\geq c\int_{t_0}^{t_0 + r^2}  \int_{\sqrt{t-t_0}}^r
\int_{|x'| < r}  (t-t_0)^{-\frac32} |\ln (t-t_0)|^{-\frac12 -2\ep}
e^{-\frac{x_n^2}{t-t_0}}  dx'dx_n dt-c\\
\notag
&\geq cr^{n-1} \int_0^{r^2}  \int_1^{\frac{r}{\sqrt{t}}}
t^{-1 } |\ln t|^{-\frac12 -2\ep}  e^{-x_n^2} dx_n dt-c\\
\label{L2-blowup}
&\geq cr^{n-1} \int_0^{\frac14 r^2}      t^{-1 } |\ln t|^{-\frac12
-2\ep} dt -c= \infty.
\end{align}
Therefore, we complete the proof of \eqref{L2-grad-w}.

It remains to prove \eqref{L2-grad-w-n}. It follows from \eqref{Poisson-tensor-K} and
\eqref{rep-bvp-stokes-w} that
\begin{align}\label{0212-1}
\notag w_n(x,t) & =   \int_0^t \int_{A} D_{x_n} \Ga(x'- y', x_n, t
-s)  g_n( y', s)  dy' ds\\
\notag  & \qquad    +    \int_0^t   \int_{A} L_{n n}(x'- y', x_n, t
-s) g_n( y', s) dy' ds
  -    \int_{A} D_{x_n} N(x' -y', x_n) g_n( y', t) dy'\\
& :=  w^1_n(x,t) + w^2_n(x,t ) + w^3_n (x,t).
\end{align}
As the same reason with \eqref{0411-w},  we obtain
\begin{align}\label{0502-5}
\int_{t_0}^{t_0 + r^2} |D_x w^3_n (x,t)|^2 dxdt < \infty.
\end{align}
We note that  by the first equality  \eqref{1006-3}, the kernel of $w^2_n$ also satisfies  \eqref{est-L-tensor}. With the same estimate \eqref{w11-1}, for $(x,t) \in B(0, \frac12) \times (t_0, \frac14)$ and $ t -t_0 < x_n^2$, we have
\begin{align}\label{wn1-1}
 |D_x w^2_n (x,t)|\leq   c   (t-  t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}.
\end{align}

Now, we assume that $ x_n^2 < t -t_0  $. From \eqref{est-L-tensor},
for $(x,t) \in B(0, \frac12) \times (t_0, \frac14)$, we have
\begin{align}
\notag |D_x w_n^2 (x,t)|
&\leq   c\int_{t_0}^t  \frac1{ (t
-s)^\frac12 ( x_n^2 + t -s)^\frac12  }  (s -t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep}ds\\
\notag&\leq   c\int_{t_0}^{t-x_n^2}  \frac1{ (t
-s)^\frac12 ( x_n^2 + t -s)^\frac12  }  (s -t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep}ds\\
\notag&\qquad +  c \int_{t -x_n^2}^t \frac1{ (t
-s)^\frac12 ( x_n^2 + t -s)^\frac12  }  (s -t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep} ds\\
\label{w11-2}
& = II_1 + II_2.
\end{align}
Let  $2 x_n^2 < t-t_0$ ($\frac{t+t_0}2 < t - x_n^2$). Note that for $t -x_n^2 < s<t$, we have  $(s -t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep} \leq c (t -t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}$. Hence, we have
\begin{align}\label{0501-2}
\notag II_2 &\leq  cx_n^{-1}  (t -t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}\int_{t -x_n^2}^t  \frac1{ (t
-s)^\frac12   }  ds\\
& = c   (t -t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}.
\end{align}
Since $\frac{t+t_0}2 < t - x_n^2$, from \eqref{0409-1}, we have
\begin{align} \label{0409I2-2}
\notag II_1 & \leq c (t -t_0)^{-1} \int_{t_0}^{\frac{t +t_0}2}      (s -t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep}ds\\
 \notag & \qquad  + c (t -t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}\int_{\frac{t+t_0}2}^{t-x_n^2}   (t-s)^{-1}  ds\\
\notag   &  = c (t -t_0)^{-1} \int_{0}^{\frac{t -t_0}2}      s^{-\frac14} |\ln s|^{-\frac14 -\ep}ds
   + c (t -t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}\ln(\frac{2x_n^2}{t-t_0})\\
  &  \leq c (t -t_0)^{-\frac14  } |\ln (t-t_0)|^{-\frac14 -\ep}
   + c (t -t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}\ln(\frac{2x_n^2}{t-t_0}).
\end{align}
Hence, summing up \eqref{w11-2}, \eqref{0501-2} and \eqref{0409I2-2},
for $(x,t) \in B(0, \frac12) \times (t_0, \frac14)$ and $ 2x_n^2 <  t -t_0 $, we have
\begin{align}\label{0502w11-2}
|D_x  w_n^2 (x,t)| \leq  c   (t-  t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep} \big( 1  + \ln(\frac{2x_n^2}{t-t_0}) \big).
\end{align}

Next, we assume that $ x_n^2 < t -t_0<2x_n^2  $ ( $\frac{t+t_0}2 > t - x_n^2> t_0$). From \eqref{0409-1}, we have
\begin{align}
\notag II_1
\notag &\leq   c (t -t_0)^{-1} \int_{t_0}^{t-x_n^2}   (s -t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep}ds\\
\notag & \leq   c (t -t_0)^{-1} \int_0^{t-t_0}   s^{-\frac14} |\ln s|^{-\frac14 -\ep}ds\\
& \leq  c   (t-  t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep},
\label{w11-3}
\end{align}
\begin{align} \label{0501-3}
\notag II_2&\leq c  (t -t_0)^{-1}  \int_{t -x_n^2}^{\frac{t +t_0}2}      (s -t_0)^{-\frac14} |\ln (s- t_0)|^{-\frac14 -\ep}ds\\
 &\qquad   + c  (t -t_0)^{-\frac34} |\ln (t- t_0)|^{-\frac14 -\ep} \int_{\frac{t+t_0}2}^{t} \frac1{ (t
-s)^\frac12  }   ds\\
\notag    &\leq  c    (t-  t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}.
\end{align}

From the  estimates \eqref{w11-2},  \eqref{w11-3} and  \eqref{0501-3},  for $(x,t) \in B_{\frac12} \times (t_0, \frac14)$ and $ x_n^2 < t -t_0<2x_n^2  $
\begin{align}\label{0502-2}
|D_x  w_n^2 (x,t)| \leq  c   (t-  t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}.
\end{align}

From \eqref{wn1-1}, \eqref{0502w11-2} and \eqref{0502-2}, we have
\begin{align}\label{0502-3}
\notag \int_{t_0}^{t_0 + r^2} \int_{B_r} | \na w^2_n (x,t)|^2 dxdt
 &  \leq c\int_{t_0}^{t_0 + r^2} \int_0^{   \sqrt{\frac{t -t_0}2}}    (t -t_0)^{-\frac12} |\ln (t- t_0)|^{-\frac12 -2\ep}\ln^2(\frac{2x_n^2}{t-t_0}) dx_ndt\\
&\qquad  + \int_{t_0}^{t_0 + r^2} \int_0^r (t-  t_0)^{-\frac12} |\ln (t- t_0)|^{-\frac12 -2\ep} dx_ndt
 < \infty.
\end{align}
Similarly , we have
\begin{align}\label{0502-4}
 \int_{t_0}^{t_0 + r^2} \int_{B_r} | \na w^1_n (x,t)|^2 dxdt
 & < \infty.
\end{align}

Hence,  from \eqref{0212-1}, \eqref{0502-5}, \eqref{0502-3} and \eqref{0502-4}, we complete the proof of \eqref{L2-grad-w-n}, and thus we deduce the proposition.
\end{proof}

\begin{rem}
From the estimates of the proof of Proposition \ref{lemma0406} (in particular $w^{12}_1$), for $(x,t) \in B_{\frac12} \times (t_0, \frac14)$, we can  obtain
\begin{align*}
| D_{x_n} w_i(x,t)| & \leq c (t-t_0)^{-\frac34} |\ln (t-t_0)|^{-\frac14 -\ep} e^{-\frac{x_n^2}{t-t_0}}\\
 &\qquad      + c(t -t_0)^{-\frac14} |\ln (t- t_0)|^{-\frac14 -\ep}\ln(\frac{2x_n^2}{t-t_0}) \chi_{x_n^2\leq t -t_0} \quad i = 1, \cdots, n-1
\end{align*}
and which implies  $D_{x_n} w_i\in L^4(t_0, \frac14; L^p( B_\frac12))$ for any $1\le p<2$.
\end{rem}

\section{Proof of Theorem \ref{maintheorem} }
\label{proofStokes}
\setcounter{equation}{0}

\subsection{Stokes equations}
We take a sequence  $
\{ g_{n,k}| k=1,2,\cdots\}$ with $ g_{n,k}\in C^\infty_c (A \times
(t_0, \frac14)) $ such that $g_{n,k}$ goes to $ g_n $ in $L^4(0,
\infty; L^p(\Rn) )$ as $k \ri \infty$, where $g_n$ is defined in \eqref{0502-6}. Let $\tilde{g}_k =(0,
g_{nk})$ and $w_k$ be solution of the Stokes equations
\eqref{Stokes-bvp-200}-\eqref{Stokes-bvp-210} defined by
\eqref{rep-bvp-stokes-w}. Since the boundary data $\tilde g_{k}$ are functions in $C^\infty_c(\Rn \times (0, \infty))$, we can obtain that $w_k $ is a classical
function, e.g.  $w_k \in \dot C^{l, \frac{l }2}(\overline \R_+
\times [0, \infty))\cap \dot W_p^{l, \frac{l}{2}} (\R_+\times [0, \infty))$
for any $l \in {\mathbb N}$. Note that $w_k =0$ on $B_\frac12 \times (t_0,\frac14) $.

Suppose that the Caccioppoli's inequality holds for smooth solutions
of the Stokes equations, that is,  $w_k$ satisfies the following
inequality for $k \in {\mathbb N}$;
\begin{equation}\label{h-cacciopoli}
\int_{t_0}^{t_0 + \frac14 r^2} \int_{B^+_{\frac12 r}(0)}  | \na  w_k(x,t)|^2 dxdt
\leq c r^{-2} \int_{t_0}^{t_0 + r^2} \int_{B^+_r(0)} | w_k(x,t)|^2 dxdt,
\end{equation}
where $c$ is independent of solutions.
Due to  Proposition \ref{thm-stokes}, we note that $w_k$ converges
to $u$ in $L^4(0, \infty; L^2 (\R_+ ))$ and on the other hand, by
the above inequality \eqref{h-cacciopoli}, $\na w_k$ converges to
$\na w$  in $L^2 (B^+_{\frac12 r}(0)  \times (t_0, t_0+ r^2))$. Hence, we
also get
\begin{align*}
\int_{t_0}^{t_0 + \frac14 r^2} \int_{B^+_{\frac12 r}(0)}  | \na  w(x,t)|^2 dxdt
\leq c r^{-2} \int_{t_0}^{t_0 + r^2} \int_{B^+_r(0)} | w(x,t)|^2 dxdt \leq C.
\end{align*}
This is, however, contrary to Proposition \ref{lemma0406}. Therefore, the
Caccioppoli's inequality is not true for the Stokes equations near
boundary. We complete the proof of the Theorem \ref{maintheorem} for
the case of Stokes equations.

\subsection{Navier-Stokes equations}
Let $g$ be a boundary data
defined in \eqref{0502-6} and $w$ be a solution of the Stokes
equations \eqref{Stokes-bvp-200}-\eqref{Stokes-bvp-210} defined by
\eqref{rep-bvp-stokes-w}. By the result of  Proposition \ref{lemma0406}, for $\frac{n}{n-1} <p < \infty$, we have
\begin{align}
\label{uc1-1}  \| w\|_{L^4(0,\infty;L^p (\R_+))} \leq c\| g\|_{L^4(0,\infty;\dot{B}^{-\frac{1}{p}}_{pp}(\Rn))} \leq c\al, \quad
 \| \na w\|_{L^2(B_r \times (t_0, t_0 + r^2))} = \infty
\end{align}
for all $ r>0$, where $\al >0$ is defined in \eqref{0502-6}.

Next, we consider the following perturbed Navier-Stokes equations in
$\R_+\times (0,\infty)$:
\begin{equation}\label{CCK-Feb7-10}
v_t-\Delta v+\nabla q+{\rm div}\,\left(v\otimes v+v\otimes
w+w\otimes v\right)=-{\rm div}\,(w\otimes w), \quad {\rm
div} \, v =0
\end{equation}
with homogeneous initial and boundary data, i.e.
\begin{equation}\label{pnse-bdata-20}
v(x,0)=0,\qquad v(x,t)=0 \,\,\mbox{on} \,\,\{x_n=0\}.
\end{equation}
Our aim is to establish the existence of solution  $v$ for
\eqref{CCK-Feb7-10} satisfying $v \in L^4(0, \infty; L^{2n}(\R_+))$
and $\na v \in L^2(0, \infty; L^{n}(\R_+))$. In order to do that, we
consider the iterative scheme for \eqref{CCK-Feb7-10}, which is
given as follows: For a positive integer $m\ge 1$
\begin{align*}
&v^{m+1}_t-\Delta v^{m+1}+\nabla q^{m+1}=-{\rm
div}\,\left(v^{m}\otimes v^{m}+v^{m}\otimes w+w\otimes
v^{m}+w\otimes w \right),\\
& \qquad \qquad \qquad \qquad \qquad {\rm div} \, v^{m+1} =0
\end{align*}
with homogeneous initial and boundary data, i.e. $v^{m+1}(x,0)=0$ and
$v^{m+1}(x,t)=0$ on $\{x_n=0\}$.
We set $v^1=0$. We then have, due to Proposition \ref{theo0503}, we have
\begin{align}\label{est-v2-10}
\notag \|\na  v^2\|_{L^2(0, \infty; L^{n}(\R_+))}  &\leq c \big(   \||w|^2\|_{L^2(0, \infty; L^{n}(\R_+)) } + \|w\otimes w|_{x_n =0}\|_{L^2(0, \infty; \dot B^{-\frac1n}_{nn} (\Rn)) } \big)\\
\notag & \leq c \big(  \|w\|^2_{L^4(0, \infty; L^{2n}(\R_+)) } + \|w\otimes w|_{x_n =0}\|_{L^2(0, \infty; L^{n-1} (\Rn)) }    \big)\\
& \leq c \big(  \|w\|^2_{L^4(0, \infty; L^{2n}(\R_+)) } + \|g\|^2_{L^4(0, \infty; L^{2(n-1)} (\Rn)) }    \big).
\end{align}
On the other hand, from Proposition \ref{thm-stokes},   we have
\begin{equation}\label{est-v2-20}
\| v^2\|_{L^4(0, \infty; L^{2n}(\R_+))}\leq c \|w \otimes
w\|_{L^2(0, \infty; L^{n}(\R_+))} \leq c   \|w\|^2_{L^4(0, \infty;
L^{2n}(\R_+)) }.
\end{equation}
By \eqref{uc1-1}, we have  $ A :=    \|w\|_{L^4(0, \infty; L^{2n}(\R_+)) } +  \|g\|_{L^4(0, \infty; L^{2(n-1)} (\Rn)) } \leq c\al$, where $\al>0$ is defined in \eqref{0502-6}. Taking $\al >0$ small such that
$A<\frac{1}{4c}$, where $c$ is the constant in
\eqref{est-v2-10}-\eqref{est-v2-20} such that
\begin{align*}
\|\na  v^2\|_{L^2(0, \infty; L^{n}(\R_+))}  +   \| v^2\|_{L^4(0, \infty; L^{2n}(\R_+))} < A.
\end{align*}
Then, iterative arguments show that
\begin{align}\label{0531-1}
\notag&\| \na v^{m+1}\|_{L^2(0, \infty; L^{n}(\R_+)) }\\
 \notag & \quad \leq c \big(
\||v^m|^2+|v^mw|+|w|^2\|_{L^2(0, \infty; L^{n}(\R_+))} + \|w\otimes w|_{x_n =0}\|_{L^2(0, \infty; \dot B^{-\frac1n}_{nn} (\Rn)) } \big) \\
\notag &\quad \leq  2c\left(\|v^m\|^2_{L^4(0, \infty; L^{2n}(\R_+))  }+\|w\|^2_{L^4(0, \infty; L^{2n}(\R_+))  } +  \|g\|^2_{L^4(0, \infty; L^{2(n-1)} (\Rn)) }\right)\\
 & \quad   \leq 4c A^2 <A.
\end{align}
Similarly, we note that
\begin{align}\label{0531}
\notag\| v^{m+1}\|_{L^4(0, \infty; L^{2n}(\R_+))} \leq c
\||v^m|^2+|v^mw|+|w|^2\|_{L^2(0, \infty; L^{n}(\R_+))}\\
\leq  2c\left(\|v^m\|^2_{L^4(0, \infty; L^{2n}(\R_+))  }+\|w\|^2_{L^4(0, \infty; L^{2n}(\R_+))  }\right)   \leq 4c A^2 <A.
\end{align}

We denote $V^{m+1}:=v^{m+1}-v^{m}$ and $Q^{m+1}:=q^{m+1}-q^{m}$ for $m\ge
1$. We then see that $(V^{m+1}, Q^{m+1})$ solves
\[
V^{m+1}_t-\Delta V^{m+1}+\nabla Q^{m+1}=-{\rm
div}\,\left(V^{m}\otimes v^{m}+v^{m-1}\otimes V^{m}+V^{m}\otimes
w+w\otimes V^{m}\right),
\]
\[
{\rm div} \, V^{m+1} =0,
\]
with homogeneous initial and boundary data, i.e. $V^{m+1}(x,0)=0$
and $V^{m+1}(x,t)=0$ on $\{x_n=0\}$. Taking sufficiently small
$\al>0$ such that $A < \frac1{6c}$, from \eqref{0531-1} and  \eqref{0531}, we obtain
\begin{align*}
\| \na V^{m+1}\|_{L^{2}(0, \infty; L^{n}(\R_+))}+\| V^{m+1}\|_{L^4 (0, \infty; L^{2n}(\R_+))}\\
\leq c \norm{|V^mv^m|+|V^mv^{m-1}|+|V^mw|}_{L^2(0, \infty; L^{n}(\R_+)) }\\
\leq 3c A \norm{V^m}_{L^4(0, \infty; L^{2n}(\R_+))}
<\frac12 \norm{V^m}_{L^4(0, \infty; L^{2n}(\R_+))}.
\end{align*}

Therefore, $(v^m, \na v^m)$ converges to $(v, \na v)$ in $L^4(0, \infty;
L^{2n}(\R_+))\times L^2(0, \infty; L^{n}(\R_+))$ such that $v$
solves in the sense of distributions
\[
v_t-\Delta v+\nabla \Pi=-{\rm div}\,\left(v\otimes v+v\otimes
w+v\otimes w+w\otimes w\right),
\]
\[
{\rm div} \, v =0,
\]
with homogeneous initial and boundary data, i.e. $v(x,0)=0$ and
$v(x,t)=0$ on $\{x_n=0\}$.

We then set $u:=v+w$ and $p =\pi + q$, which becomes a very weak solution of the
Navier-Stokes equations in $\R_+ \times (0, \infty)$, namely
\[
u_t-\Delta u+\nabla p=-{\rm div}\,\left(u\otimes u\right),\qquad
{\rm div} \, u=0,
\]
with boundary data $u(x,t)=g(x)$ on $\{x_n=0\}$ and homogeneous
initial data $u(x,0)=0$ such that
\begin{align}\label{0502-7}
 \| u \|_{L^4(0, \infty;  L^{2n}(\R_+))} \leq c,\qquad \| \na u\|_{L^2(B_r \times (t_0, t_0 + r^2))} = \infty \qquad \forall r >0.
\end{align}

Similarly, as in case of Stokes equations, we take  $  g_{n,k} \in
C^\infty_c (A \times (t_0, \frac14)) $  such that $g_{n,k}$ goes to
$g_n$ in $L^4(0,\infty; L^{2n}(\Rn ))$ as $k \ri \infty$. We denote
$\tilde{g}_k =(0,  g_{nk})$. Let $w_k$ be a solution of Stokes equations
with boundary data $\tilde{g}_k$ and $v_k$ be a solution of
\eqref{CCK-Feb7-10}-\eqref{pnse-bdata-20} with replacement of $w$ by
$w_k$. We recall that $w_k$ is smooth vector field such that $w_k $
converges to $w$ in $L^4 (0,\infty ; L^{2n}(\R_+ ))$.

We also observe that $v_k$ is smooth vector field such that
\[
\| \na v_k\|_{L^2(0, \infty;L^{n}(\R_+))}+\| v_k \|_{L^4(0,\infty;
L^{2n}(\R_+))} \leq c.
\]
By weak compactness, there is a subsequence of $\{ v_k\}$, redefined
as $v_k$, and $v\in L^4(0, \infty;L^{2n}(\R_+))$ such that $v_k$
weakly converges in $L^4(0, \infty;L^{2n}(\R_+))$ to $v$.   This
implies that $u_k = w_k + v_k$ weakly converges to $u = w + v$ in
$L^4(0,\infty; L^{2n}(\R_+))$.

Suppose that the Caccioppoli's inequality holds for smooth solutions
for the Navier-Stokes equations, that is, for $k \in {\mathbb N}$,
$u_k$ is assumed to satisfy the following inequality;
\begin{align*}
\int_{t_0}^{t_0 + \frac14 r^2} \int_{B^+_{\frac12 r}(0)}  | \na  u_k(x,t)|^2 dxdt
&\leq c r^{-2} \int_{t_0}^{t_0 + r^2} \int_{B^+_r(0)} | u_k(x,t)|^2 dxdt\\
&\qquad  + c r^{-1} \int_{t_0}^{t_0 + r^2} \int_{B^+_r(0)} | u_k(x,t)|^3 dxdt,
\end{align*}
where $c>0$ is independent of $k$. Since $\| u_k \|_{L^4(0,
\infty;L^{2n}(\R_+))} \leq c$ for all $k$, this leads to
\begin{align*}
\int_{t_0}^{t_0 + \frac14 r^2} \int_{B^+_{\frac12 r}(0)}  | \na  u(x,t)|^2 dxdt
\leq c <\infty.
\end{align*}
This is, however, contrary to \eqref{0502-7}, and therefore,
the Caccioppoli's inequality is not true for the Navier-Stokes
equations near boundary. This completes the proof of Theorem \ref{maintheorem}. \qed

\appendix
\setcounter{equation}{0}

\section{Proof of Proposition \ref{theo0503}}
\label{appendixa}

\subsection{Helmholtz projection  in half space} \label{projection}
\setcounter{equation}{0}

It is well known that the Helmholtz projection   ${\mathbb P}$ in half space $\R_+$ is given  by
\begin{align}\label{Hprojection}
{\mathbb P} f = f - \na {\mathbb Q}_1f - \na {\mathbb Q}_2 f = f -\na {\mathbb Q}f,
\end{align}
where ${\mathbb Q}_1 f$ and ${\mathbb Q}_2f $ satisfy the following equations;
\begin{align*}
\De {\mathbb Q}_1 f ={\rm div}\, f,\qquad
{\mathbb Q}_1 f|_{x_n =0} =0
\end{align*}
and
\begin{align*}
\De{\mathbb Q}_2f =0,\qquad
D_{x_n} {\mathbb Q}_2f|_{x_n =0} = \big(f_n -D_{x_n}{\mathbb Q}_1f\big)|_{x_n =0}.
\end{align*}
Note that ${\mathbb Q}_1f$ and ${\mathbb Q}_2f$ are represented by
\begin{align}\label{0427-1}
{\mathbb Q}_1f(x)& =- \int_{\R_+}  \na_y \big(N(x- y) - N (x - y^* ) \big) \cdot  f(y) dy,\\
\label{0427-2} {\mathbb Q}_2f (x) & = \int_{\Rn}   N(x'-y',x_n)  \big(f_n(y', 0)  - D_{y_n} {\mathbb Q}_1f (y', 0) \big) dy',
\end{align}
where $y^*= (y', -y_n)$.
Note that $({\mathbb P}f)_n|_{x_n =0} =0$.

\begin{lemm}\label{lemma0523-3}

Let  $f_i = {\rm div} \, F_i $ and   $f = ( f_1, \cdots, f_n)$.   Then,
\begin{align*}
{\mathbb Q}_1  f(x) &=\sum_{ k \neq n} D_{x_k} {\mathbb Q}_1 F_k(x) +  D_{x_n} A(x),\\
{\mathbb Q}_2 f(x) &= \sum_{  k \neq  n} D_{x_k}  {\mathbb Q}_2 F_k(x) - \sum_{  k \neq  n} D_{x_k} B^1_k(x)  - \sum_{ k \neq n} D^2_{x_k} B^2(x),\\
&\qquad ,
\end{align*}
where
\begin{align*}
A(x) & = -\int_{\R_+}\na_y \big( N(x-y) + N(x-y^*) \big)\cdot F_n(y) dy +2 B^1_n(x),\\
B^1_k(x) & =  -  \int_{\Rn} N(x' -y', x_n) F_{kn} (y', 0) dy', \qquad 1 \leq k \leq n,\\
B^2(x) & =  \int_{\Rn} N(x'-y',x_n) A(y',0) dy'.
\end{align*}

\end{lemm}

\begin{proof}

From \eqref{0427-1}, we have
\begin{align}\label{0424-1}
\notag {\mathbb Q}_1  f(x)
\notag &= - \sum_{ k  \neq n} D_{x_k}\int_{\R_+} \na_y \big( N(x-y) - N(x-y^*) \big) \cdot F_k(y) dy\\
\notag & \quad  -  D_{x_n} \int_{\R_+}\na_y \big( N(x-y) + N(x-y^*) \big)\cdot F_n(y) dy\\
\notag &\quad +2  D_{x_n} \int_{\Rn} N(x'-y', x_n) F_{nn} (y', 0) dy'\\
&:=\sum_{ k \neq n} D_{x_k} {\mathbb Q}_1 F_k(x) +  D_{x_n} A(x).
\end{align}

Since $\De A = {\rm div} \, F_n = f_n$, from \eqref{0424-1}, we have  $D_{y_n}{\mathbb Q}_1f (y) = D_{y_n} \sum_{k \neq n} D_{y_k} {\mathbb Q}_1 F_k(y)  +  f_n (y) - \De' A(y) $. Hence, we have
\begin{align}\label{0501-1}
\notag \int_{\Rn} N(x' -y', x_n) D_{y_n} {\mathbb Q}_1f (y',0) dy'
\notag & = \sum_{  k \neq  n}  D_{x_k} \int_{\Rn} N(x'-y',x_n)   D_{y_n} {\mathbb Q}_1 F_k(y',0)  dy'\\
\notag &\quad  + \int_{\Rn} N(x'-y',x_n)f_n(y',0) dy'\\
&\quad  + \sum_{ k \neq n} D_{x_k} \int_{\Rn} N(x'-y',x_n) D_{y_k}A(y',0) dy'.
\end{align}

Hence, from \eqref{0427-2} and  \eqref{0501-1}, we have
\begin{align*}
{\mathbb Q}_2 f(x) &=
-\sum_{  k \neq  n}  D_{x_k} \int_{\Rn} N(x'-y',x_n)  D_{y_n} {\mathbb Q}_1 F_k(y',0)  dy'\\
&\quad  - \sum_{ k \neq n} D_{x_k} \int_{\Rn} N(x'-y',x_n) D_{y_k}A(y',0) dy'.
\end{align*}
We complete the proof.
\end{proof}

\subsection{Proof of Proposition \ref{theo0503}}
To prove Proposition \ref{theo0503}, we use the following Proposition.
\begin{prop}(\cite[Proposition 3.2]{CJ})
\label{0503prop2}
 Let $1<p,q<\infty$. Let $\Ga *' g(x,t) =  \int_{-\infty}^t \int_{\Rn} \Ga(x' -y', x_n, t-\tau)g(y',\tau) dy' d\tau $.
Then
\begin{align*}
\| D_x \Ga *' g\|_{L^q({\mathbb R};L^p(\R_+))}  \leq c \|  g\|_{L^q({\mathbb R};\dot{B}^{-\frac{1}{p}}_{pp}(\Rn))}.
\end{align*}
\end{prop}

We consider the Stokes equations
\begin{align}\label{maineq-stokesh=0}
\begin{array}{l}\vspace{2mm}
v_t - \De v + \na \Pi =f, \qquad \mbox{div } v =0, \mbox{ in }
 \R_+\times (0,\infty),\\
\hspace{30mm}v|_{t=0}= 0, \qquad  v|_{x_n =0} = 0,
\end{array}
\end{align}
where $f = \mbox{div}\, F$.

Let  $f = {\mathbb P}f + \na {\mathbb Q} f$ be  a  decomposition of $f$ defined \eqref{Hprojection}. Note that $({\mathbb P} f)_n|_{x_n =0} =0$.  We define  $(v, \Pi_0)$  by
\begin{equation}\label{expression-v}
v_i (x,t) =\int_0^t \int_{{\mathbb R}^n_+} G_{ij}(x,y, t-\tau)
({\mathbb P} f)_j(y,\tau) dyd\tau,
\end{equation}
\begin{equation*}
\Pi_0(x,t) =\int_0^t \int_{{\mathbb R}^n_+} P(x,y, t-\tau)
\cdot  ({\mathbb P} f) (y,\tau)dyd\tau,
\end{equation*}
where $G$ and $P$ are defined by
\begin{align}\label{formulas-v}
G_{ij} &= \de_{ij} (\Ga(x-y, t) - \Ga(x-y^*,t )) + 4(1 -\de_{jn})
\frac{\pa}{\pa x_j} \int_0^{x_n} \int_{{\mathbb R}^{n-1}}
            \frac{\pa N(x-z)}{\pa x_i} \Ga(z -y^* , t) dz,
\end{align}
\begin{align*}
\notag P_j(x,y,t) & =4 (1 - \de_{jn}) \frac{\pa }{\pa x_j}\Big[ \int_{{\mathbb
R}^{n-1}} \frac{\pa N(x' - z', x_n)}{\pa x_n} \Ga(z' -y', y_n,t) dz'\\
& \quad +\int_{{\mathbb R}^{n-1}} N(x' -z',x_n) \frac{\pa \Ga(z'-y', y_n,
t)}{\pa y_n} dz'\Big].
\end{align*}
From \cite{So}, $(v, \Pi_0)$ satisfies
\begin{align*}
\begin{array}{l}\vspace{2mm}
v_t - \De v + \na \Pi_0 = {\mathbb P}\, f, \qquad \mbox{div } v =0, \mbox{ in }
 \R_+\times (0,\infty),\\
\hspace{30mm}v|_{t=0}= 0, \qquad  v|_{x_n =0} = 0.
\end{array}
\end{align*}
Let $\Pi = \Pi_0 + {\mathbb Q}f$. Then, $(v, \Pi)$ is solution of \eqref{maineq-stokesh=0}.

Let $1<p, \, q <\infty$.  In Section 3 in  \cite{CK}, the authors showed that
 $v $ defined by
\eqref{expression-v}   has the following estimate;
\begin{align}\label{0417-2}
\|  \na v \|_{L^q(0, \infty;  L^{p}(\R_+)) } &
   \leq c \big( \| \na   \Gamma* {\mathbb P} f\|_{ L^q(0, \infty;  L^{p}(\R_+)) } +
  \|  \na \Gamma^* * {\mathbb P} f \|_{ L^q(0, \infty; L^{p}(\R_+)} \big),
\end{align}
where $\Gamma^* *  f (x,t) = \int_0^t \int_{\R_+} \Ga(x-y^*, t-\tau) f(y, \tau) dyd\tau$.

\begin{lemm}\label{0929-1}
Let $1 <  p,\, q < \infty$.
Let  $F \in L^{q} (0, \infty, L^p(\R_+))$. Then,
\begin{align*}
\notag \| \na  \Ga* {\mathbb P}({\rm div} \, F)\|_{L^q (0, \infty; L^{p}(\R_+)} + \| \na \Ga^* * {\mathbb P}({\rm div} \, F)\|_{L^q(0, \infty;  L^{p}(\R_+))}\\
 \leq c \big(\| F\|_{L^{q} (0,\infty;  L^p (\R_+))} +  \| F|_{x_n =0}\|_{L^q (0, \infty; \dot B^{-\frac1p}_{pp}(\Rn))}\big).
\end{align*}

\end{lemm}
\begin{proof}
Since the proofs are exactly same, we only prove the case of $ \Ga^* * {\mathbb P}({\rm div} \, F)$.

Since $({\mathbb P}\, {\rm div}\, F)_j (t) = {\rm div}\, F_j(t) - D_{x_j} {\mathbb Q} \, {\rm div}\, F(t)$,   for $1 \leq j \leq n-1$, we have
\begin{align*}
\notag (\Ga^* * ({\mathbb P}\, {\rm div}\, F) )_j (x,t) &= \int_0^t \int_{{\mathbb R}^n_+}
 \na_y \Ga(x-y^*, t-\tau) \cdot F_{j}(y,\tau) dyd\tau \\
\notag &\quad  + \int_0^t \int_{\Rn}
   \Ga(x'-y', x_n, t-\tau) F_{jn}(y',0,\tau) dy' d\tau \\
&\quad +  \int_0^t \int_{{\mathbb R}^n_+}
D_{y_j} \Ga(x-y^*, t-\tau) {\mathbb Q}({\rm div} \, F )(y,\tau) dyd\tau.
\end{align*}
Note that $D_{x_n}^2 A(t) = -\De' A (t)+ {\rm div }\, F_n(t)$. From Lemma \ref{lemma0523-3}, we have
\begin{align*}
\notag \Ga^* * ({\mathbb P}\, {\rm div}\, F)_n (x,t)
&= \int_0^t \int_{{\mathbb R}^n_+}  \na_y \Ga(x-y^*, t-\tau) \cdot F_n(y,\tau) dyd\tau dz'\\
\notag &\quad  + \int_0^t \int_{\Rn}
   \Ga(x'-y', x_n, t-\tau)  F_{nn}(y',0,\tau) dy' d\tau \\
\notag &\quad + \sum_{1 \leq k \leq n-1} D_{x_k}  \int_0^t \int_{{\mathbb R}^n_+}  \Ga(x-y^*, t-\tau)D_{y_n} {\mathbb Q}  F_k(y,\tau) dyd\tau\\

\notag &\quad +  \int_0^t \int_{{\mathbb R}^n_+} \na' \Ga(x-y^*, t-\tau)\cdot \na'A(y,\tau) dyd\tau\\

\notag &\quad +  \int_0^t \int_{{\mathbb R}^n_+} \na' \Ga(x-y^*, t-\tau)\cdot \na' D_{y_n}B^1_k(y,\tau) dyd\tau\\
&\quad +  \int_0^t \int_{{\mathbb R}^n_+} \na' \Ga(x-y^*, t-\tau)\cdot \na' D_{y_n}B^2(y,\tau) dyd\tau.
\end{align*}

Due to parabolic type's Calderon-Zygmun Theorem and Proposition \ref{0503prop2},  for  $ 1 < p, q< \infty$, we have
\begin{align*}
\|\na \Ga^* * ({\mathbb P}\, {\rm div}\, F)\|_{L^q(0, \infty; L^p(\R_+)) }
& \leq   \big(
    \| D_x {\mathbb Q} F  \|_{ L^q(0,   \infty; L^p(\R_+))} + \|D_x A\|_{ L^q(0,   \infty; L^p(\R_+))}\\
     & \quad  +  \|D_x B^1_k F_k \|_{ L^q(0,   \infty; L^p(\R_+))} +   \|D^2_x B^2 \|_{ L^q(0,   \infty; L^p(\R_+))}\\
     & \quad + \| F\|_{L^q(0,   \infty; L^p(\R_+)) }    + \| F|_{x_n =0}\|_{L^q (0, \infty; \dot B^{-\frac1p}_{pp}(\Rn))}   \big).
\end{align*}
It is well known (see e.g. \cite{St})  that  $D B^1_k(t)$ are bounded from $\dot{B}^{l-\frac{1}{p}}_{pp}(\Rn)$ to $\dot W^l_p(\R_+)$, $l \in {\mathbb N} \cup \{ 0\}$ so that
\begin{align}\label{Poisson}
  \| D_x B^1_k(t)\|_{\dot W^l_p(\R_+)}\leq c\|F(t)|_{x_n =0}\|_{\dot B^{l-\frac{1}{p}}_{pp}(\Rn)}.
 \end{align}

By Calderon-Gygmund Theorem and \eqref{Poisson}, we have
\begin{align*}
\| D_x {\mathbb Q}_1 F(t)\|_{L^p (\R_+)} & \leq c \|  F(t)\|_{ L^p (\R_+)},\\
 \| D_x A(t) \|_{ L^p(\R_+)}  & \leq c \|  F(t)\|_{ L^p(\R_+)} + c\|F(t)|_{x_n =0}\|_{\dot B^{k-\frac{1}{p}}_{pp}(\Rn)},\\
\|D^2_x B^2(t) \|_{L^p (\R_+)} & \leq  c \| A(t)|_{x_n =0} \|_{\dot B^{1  -\frac1p}_{pp} (\Rn)} \leq  c \| D_xA(t) \|_{L^p (\R_+)}\\
 &\leq c \| F(t)\|_{L^p (\R_+)} + c\|F(t)|_{x_n =0}\|_{\dot B^{-\frac{1}{p}}_{pp}(\Rn)}.
\end{align*}
Finally, from \eqref{Poisson}, we have
  \begin{align*}
 \| D_x  {\mathbb Q}_2 F_k (t)\|_{L^p (\R_+)} & \leq c  \|\big( F_k (t)- \na {\mathbb Q}_1F_k(t) \big)_n|_{x_n =0} \|_{\dot B^{-\frac1p}_{pp} (\Rn)}.
  \end{align*}
Since $F_k(t) - \na {\mathbb Q}_1F_k(t)$ is in $ \in L^p (\R_+)$ and  divergence free in $\R_+$, its normal component  has trace  $ \big( F_k(t) - \na {\mathbb Q}_1F_k (t) \big)_n|_{x_n =0} \in \dot B^{ -\frac1p}_{pp} (\Rn)$  (see \cite{galdi}). Hence,  we have
 \begin{align*}
\|\big( F_k(t) - \na {\mathbb Q}_1F_k (t) \big)_n|_{x_n =0} \|_{\dot B^{ -\frac1p}_{pp} (\Rn)} \leq  \|F_k(t) - \na {\mathbb Q}_1F_k (t) \|_{L^p (\R_+)}\leq c \|F_k(t)   \|_{ L^p (\R_+)}.
 \end{align*}
Therefore, we complete the proof of Lemma \ref{0929-1}.
\end{proof}
With the aid of the estimate \eqref{0417-2} and Lemma \ref{0929-1}, Proposition  \ref{theo0503} is immediate.

\section*{Acknowledgements}
T. Chang is partially supported by NRF-2017R1D1A1B03033427 and K. Kang is partially
supported by NRF-2017R1A2B4006484 and NRF-2015R1A5A1009350.

config: arxiv


TITLE: existence of Lebesgue integral
QUESTION [3 upvotes]: Hi, I have a question on my homework.
For each positive integer $n$, let $f_n:\mathbb{R}\to\mathbb{R}$ be integrable, $ ~f_n\geq 0$ and $f_n(x)\to f$ pointwise. I need to show that if $\int f_n$ converges to some finite $c\geq 0$, then $\int f$ exists and $0\leq\int f\leq c$.
I am thinking that for an arbitrary function $f$ , the Lebesgue integral exists iff $f$ is Lebesgue integrable or $\int f$ is infinite (is this correct?). However, for a nonnegative function $f$, the Lebesgue integral always exists and $\int f = \sup\{\int g：0\leq g\leq f, ~~g$ bounded and supported on a set of finite measure$\}$.
If what I am thinking is correct, then the question seems quite straigtforward. We have $f_n(x)$ converges to $f(x)$ for every $x$ and $f_n\geq 0$ for every $n$. So $f$ is nonnegative everywhere and $\int f$ exists since the Lebesgue integral exists for all nonnegative functions. That $0\leq\int f\leq c$ just follows from the fact that $f \leq 0$ and Fatou's Lemma.
Can someone tell me under what condition the Lebesgue integral exists? Is my attempt correct?

REPLY [1 votes]: Depending on your course, showing that "the Lebesgue integral exists" could include showing that the integral (which can be defined for any non-negative measurable function) is finite. But it doesn't matter here: for any non-negative measurable (you didn't mention this; see also Juan's comment) function $g$, $\int g$ can be defined, and Fatou's lemma holds in that context, so that finiteness of the integral is established by afterwards by $$\int f\leq c.$$
In short: Your approach is correct.


TITLE: With respect to inertial observer standing at the starting point of why is the velocity of ejected mass of rocket $v + v_0$
QUESTION [0 upvotes]: We derived the equation of motion of a rocket this way: all the velocities are taken with respect to inertial observer standing where the rocket starts. As we take upwards direction to be positive.  The initial momentum of rocket is mv. That's fine. Now at a certain time $\delta t$ later the momentum becomes $(m - \delta m) (v + \delta v) - \delta m (v + \delta v_0) $ {where $\delta m$ = amount of decrease of mass of rocket or amount of gas emitted. $\delta v$ = increase of velocity of rocket due to mass loss and conservation of momentum.  Now my question is why we take the velocity of the gas to be $v + v_0$ . As we measuring with respect to inertial observer standing at the starting point of rocket. And if we are measuring with respect to the rocket then the initial momentum should be zero as well as the velocity of the rocket relative to itself is zero. So in my opinion the relative velocity of the gas with respect to the inertial observer should be $v_0 -v$ ;$v_0$ = velocity of the gas.

REPLY [1 votes]: The proper derivation says that we are in an inertial reference frame, and at some time $t$, we see the rocket traveling at speed $v(t)$ with total mass $m(t)$ ejecting mass backwards at a constant exhaust speed $s$ relative to the rocket. A momentum balance therefore says,
$$
m(t) v(t) = m(t+dt) v(t+dt) + [m(t)-m(t + dt)](v(t) - s).
$$
The term on the left is the total momentum of the rocket at time $t$, the first term on the right is the new momentum of the rocket at time $t+dt$, and the second term on the right is a chunk of fuel which has been ejected at speed $s$ backwards relative to the rocket, which is traveling forward at speed $v$.
Expanding we have$f(t+dt)=f(t) + f'(t) dt + \dots$. Thus, this becomes$$
m~ v = (m + m'~ dt)(v+v'~dt) - m'~ dt(v - s).
$$
Discarding the $dt^2$ term on the right yields
$$
0= m'~v+ v'~m - m'~v +m'~s.
$$
And so finally,
$$
v'~m = -m'~s.
$$This integrates directly to $$
v(t) - v(0) = - s\ln\big(m(t)/m(0)\big).
$$
You can get there somewhat more easily by just using a reference frame that happens to be moving forward at speed $v$ but that always seemed like cheating for me: it is important to understand that the two $m'~v $ terms cancel each other out.


TITLE: Prove that a function is holomorphic
QUESTION [1 upvotes]: I want to prove that for $f:\mathbb{R}\to\mathbb{R}$ a continuous function such that $\int_{\mathbb{R}}\frac{|f(t)|}{1+t^{2}}dt<\infty$ ,for $z=x+iy$,
$$u(z)=\frac{y}{\pi}\int_{\mathbb{R}}\frac{f(t)}{(x-t)^{2}+y^{2}}dt$$
is harmonic in the upper half plane $U=\{z\in\mathbb{C}:Im(z)>0\}$. For this I want to see that $u(z)$ is the real part of a holomorphic function, I obtain that
$$u(z)= Re\left(\frac{1}{\pi}\int_{\mathbb{R}}\frac{i}{z-t}f(t)dt\right).$$
How can I prove that $\int_{\mathbb{R}}\frac{i}{z-t}f(t)dt$ is holomorphic?

REPLY [0 votes]: Let $g(z)=\int_\mathbb R \frac 1 {z-t} f(t) dt$. The integral exists for $Im(z) >0$. [ Use the hypothesis and the fact that $\frac {1+t^{2}} {(x-t)^{2}+y^{2}}$ is a bounded function on $\mathbb R$. (Here x and y are the real and imaginary parts of z). Now consider $g(z+h)-g(z)= \int_\mathbb R (\frac 1 {z+h-t}-\frac 1 {z-t}) f(t) dt$ $= \int_\mathbb R (\frac {-h} {(z+h-t)(z+t)}f(t) dt$. Divide by $h$ and take the limit. Once again use the fact that $\frac {1+t^{2}} {(z+h-t)(z-t)}$ is uniformly bounded in $t$ for $|h| <y/2$ so Dominated Convergence Theorem can be applied to show that $g$ is differentaiable at each point of the upper half plane.


TITLE: Limit of the Schröder numbers ratio
QUESTION [3 upvotes]: I have been playing around with interesting integer sequences and came across Schröder number which defines the number of lattice paths of n x n grid.
The recurrence formula to calculate these numbers is as follows:
$$
S_n = 3S_{n-1} + \sum_{k=1}^{n-2}S_kS_{n-k-1} \text{  for  }n\ge2 \text{ with } S_0 =1, S_2=2
$$
The first few numbers in the sequence are:
$$
1, 2, 6, 22, 90, 394, 1806, 8558
$$
I was curious about the following ratio and its limit:
$$
\lim_{n\rightarrow\infty}\frac{S_{n-1}}{S_{n}} = ?
$$
So I wrote a small C++ program with mixed-precision to handle the large numbers and surprisingly there appears to be an asymptote that is not zero. So far my little program has reached n = 20,000, but it's slowing down significantly as the numbers are getting truly large but it has given a preliminary result of
$$
\lim_{n\rightarrow\infty}\frac{S_{n-1}}{S_{n}} = 0.17158...
$$
My question now is, is this simply a ghost limit for not having gone far enough, or does a limit exist that is not zero? My math expertise is lacking somewhat to attempt to find that limit analytically (can it even be done?).
If a limit exists would it imply that, for large n, the number of additional lattice paths of a n x n grid only increases by about 17% by expanding the grid by 1?
Code for anyone who wants to try it for themselves:
#include <iostream>
#include <gmp.h>
#include <gmpxx.h>
#include <vector>

void schroder(std::vector<mpz_class>& n_vec);

int main(int argc, char **argv)
{
    int n=2;
    int n_max;
    std::cout << "Enter n_max: ";
    std::cin >> n_max;
    std::vector<mpz_class> s = {mpz_class(1), mpz_class(2)};
    while (n<n_max){
        mpq_class div (s[s.size()-2],s.back());
        gmp_printf ("%d\t %.*Ff\n", n, 100, mpf_class(div,500));
        schroder(s);
        n++;
    }
    return 0;
}

void schroder(std::vector<mpz_class>& n_vec){
    int n = n_vec.size();
    mpz_class sum = mpz_class(0);
    for(int k = 1; k<=(n-2); k++){
        sum += n_vec[k]*n_vec[n-k-1];
    }
    sum += 3*n_vec[n-1];
    n_vec.push_back(sum);
}

REPLY [2 votes]: Pietro has given the answer, and the easiest way to derive it. I'd like to add that if you type the decimal approximation you have into Wolfram Alpha, it lists the correct answer, $3-\sqrt{8}=3-2\sqrt{2}$, as a possible closed-form.
With a few more correct digits, the OEIS will also suggest the closed-form, although you have to be a bit careful because the first digit is different.


TITLE: $y \leq x \leq y \implies x = y$
QUESTION [3 upvotes]: For a proof, I need to use the fact that if an integer, say $x$ is bounded on both sides by another integer $y$, then $x = y$.
Is this something that needs to be proven, or is this part of the definition of equality of two integers? My gut instinct says it is the latter, but I would like a source if possible as I cannot find one.

REPLY [4 votes]: A possible (semi-formal) proof.
We have that : $y \le x \le y$ is ; $(y \le x) \text { and } (x \le y)$.
But $(x \le y)$ means : $(x < y) \text { or } (x=y)$.
And $(y \le x)$ means : $(y < x) \text { or } (y=x)$.
Four cases : (1) $(x < y)$ and $(y < x)$ : contradiction. The same for (2) $(x < y)$ and $(y = x)$ and (3) $(y < x)$ and $(x=y)$.
Why contradiction ? Because $x < y$ is defined as : $y=k+x$ for some $k \ne 0$.
Thus, we are left with :

$(x=y)$.

Alternatively, we may start defining directly $x \le y$ as : $y=k+x$, where in this case $k$ may be $0$.
Thus, from $(x \le y)$ and $(y \le x)$ we have : $y=k+x$ and $x=l+y$.
Now, using property of $=$, we have :

$y=k+x=k+(l+y)=(k+l)+y$

But this holds only if $k=l=0$.
Thus, we have $y=k+x=0+x=x$.


TITLE: Monotonic Function and the opposite monotonicity
QUESTION [2 upvotes]: Does there exists $f: \mathbb{R} \to \mathbb{R},g: \mathbb{R} \to \mathbb{R}$, such that $f,g$ are onto function and satisfies:

$f(g(x))$ strictly monotonically increasing and $g(f(x))$ strictly monotonically decreasing.

This question occur when I realize that  if $g,f$ are monotonic functions with same monotonicity, $g(f(x))$ will increase, if they are  monotonic functions with opposite monotonicity, $g(f(x))$ will decrease. But what about $g,f$ are not monotonic?
I haven't got any idea about it, thanks alot for your help.

REPLY [2 votes]: This is not an answer, but can lead to one. Here are a few remarks on your problem:

since $f \circ g$ and $g \circ f$ are both injective it follows that $f,g$ are injective functions, which coupled with the fact that $f,g$ are onto prove that $f,g$ are bijections of $\Bbb{R}$ in $\Bbb{R}$.
note that $f,g$ cannot be monotonic on the same interval, because then $f\circ g$ and $g \circ f$ would have the same monotonicity on that interval.
an injective, continuous function is monotonic. Therefore $f,g$ cannot be both continuous on an interval $I$.

These being said, if there are some examples of functions like in your problem, then they will look very nasty. If one is continuous on an interval $I$ then the other one will not be continuous on any subinterval of $I$.


TITLE: counter example for $\mathbb{E}[\log(x)] <\mathbb{E}[\log(y)]$ when $\mathbb{E}[x]< \mathbb{E}[y]$ and vice versa
QUESTION [1 upvotes]: I am looking for an example that $\mathbb{E}[x]< \mathbb{E}[y]$ but not $\mathbb{E}[\log(x)] <\mathbb{E}[\log(y)]$ and an example that $\mathbb{E}[\log(x)] <\mathbb{E}[\log(y)]$ but not $\mathbb{E}[x]< \mathbb{E}[y]$.
Is there any such example? Or are they equivalent conditions?
Please note we can assume both $x,y\geq 0$.

REPLY [4 votes]: Suppose $x=1$ with certainty and $y$ is equal to $0.4$ or $2$ with equal probability. Then
$$ \mathbb E[x] = 1 \lt 1.2 = \mathbb E[y],$$
and
$$ \mathbb E[\log(x)] = 0 \gt -0.11157 \approx \mathbb E[\log(y)].$$
Flip $x$ and $y$ and you get your other example.


TITLE: Please find the quadratic form solution
QUESTION [0 upvotes]: Let
$q(x) =x_1^2+x_2^2+3x_3^2-2x_1x_3+2x_2x_3+x_1+x_3+5 =x^TQx+b^Tx+5$
$ \begin{matrix}
  x=(    x_1 & x_2 & x_3 )^T\\
  \end{matrix}
$
the first question was find a symmetric matrix Q and a vector b. so I found that matrix by using way to solve quadratic form
$$
  Q=     \begin{matrix}
        1 & 0 & -1 \\
        0 & 1 & 1 \\
       -1 & 1 & 3 \\
        \end{matrix}
$$
and $$ b =\begin{matrix}
        (1 & 0 & 1)^T\\
       \end{matrix}$$
The second question was 'find, in terms of Q and b and find the minimum of q(x)'
If $ \partial q(x)/∂x=
\begin{matrix}
        (∂q/∂x_1& ∂q/∂x_2&∂q/∂x_3)^T\\
       \end{matrix}$
how can I solve second question?
how can I approach in good way?

REPLY [1 votes]: I just wanna share my try,
I guess we can find the minimum of this function by differentiation. Let's find our critical points, that is
$$ \exists x_{0} \in R^3, / \bigtriangledown  q(x_{0})=0?$$
note that $q(x): R^3 \rightarrow R$
$$\bigtriangledown  q(x)=(2x-2z+1,2y+2z,6z-2x+2y+1)$$
Now we should solve the following system
$2x-2z+1=0$
$2y+2z=0$
$6z-2x+2y+1=0$
it's east to find $x_{o}=(-3/2 , 1,-1)$ as its solution. Thus we just need to check if the Hessian is a positive matrix (cause it implies this point is a minimum). If you derivative again, you'll find:
$ Hq(x_{o})=
\begin{matrix}
        2 & 0 & -2 \\
        0 & 2 & 2 \\
       -2 & 2 & 6 \\
        \end{matrix}
$
As far as I remember there is a theorem which relate Hessian with the quadratic form, check your notes cause it will help. However, in our case $Hq(x_{0})=2Q$, then we just need to check if Q is a positive matrix.
$A_{1} >0, A_{2}>0, A_{3}>0$ so Q is a positive matrix thus $x_{0}$ is the minimum of q(x) $\forall b$


TITLE: Probability for Communication Networks
QUESTION [0 upvotes]: A computer communication channel transmits words of $n$ bits using an error-correcting code which is
capable of correcting errors in up to $k$ bits. Here each bit is either a $0$ or a $1$. Assume each bit is
transmitted correctly with probability $p$ and incorrectly with probability $q$ independently of all other
bits.

(a) Find a formula for the probability that a word is correctly transmitted.
(b) Calculate the probability of correct transmission for $n = 8$, $k = 2$, and $p = 0:01$.

REPLY [1 votes]: This is a typical Binomial Distribution calculation. The probability that $k$ or fewer bits are transmitted incorrectly is
$$\sum_{i=0}^k \binom{n}{i}q^ip^{n-i}.$$
Remark: In the formula, I have used your assertion that the probability that a bit is transmitted incorrectly is called $q$. There may be a typo in the question, since in the concrete example you use $p=0.01$, meaning that the probability $q=1-p$ that a bit is transmitted incorrectly is $0.99$. If there is such a high probability of incorrect transmission of a bit, we would be much better off reversing the bit received!
If there really is a typo, and the probability of incorrect transmission is $p$, then in the formula above, just interchange $q$ and $p$.


TITLE: Compute Elementary set
QUESTION [0 upvotes]: Define f : N×N → N (Zero counted as natural number) by letting f(x,y) = 7x+3y.
Compute f^-1({16}) and f^-1 ({11})
I understand if f(x)= x^2 for some R you can just go
f^-1({16}) = {x|f(x) E {16} }
           = {x|x^2 = 16} = {-4,4}
But what confuses me about my original question is when it has "3y" I don't understand how to move forward with that.

REPLY [1 votes]: $f^{-1}(16)$ by definition is $\{(x,y)\in \mathbb N \times \mathbb N| 7x + 3y = 16\}$
If $7x + 3y = 16$
$6x + 3y + x = 15 + 1$
$2x + y + \frac x3 = 5 + \frac 13$
So $x$ must have a remainder $1$ when divided by $3$.  In other words $x = 3k + 1; k \in \mathbb N$
$2x + y + \frac x3 = 6k + 2 + y + k + \frac 13 = 5 + \frac 13$
$7k + y = 3$
$6k + k + y = 3$
$2k + \frac {k + y}3 = 1$;  So $3$ divides $k+ y$..i.e. $k + y = 3j$ for some natural $j$
So $2k + j = 1$.  There is only one natural solution for that.
$k = 0; j = 1\implies y=3; x = 1$.
So $f^{-1}(16)  = (1,3)$
Notice if we weren't restricted to natural numbers but to any integers then would be infinite solutions:
$2k = 1-j$ implies $j$ is odd.  So $j= 2m+1$ and $k = -m$ and so
$x = 1 + 3k$ for any $k$ and $k+ y = 3(-2k + 1); y = 3-7k $ for the same $k$.


TITLE: How to solve this kind of questions translation of $3D$
QUESTION [1 upvotes]: Translate $\Delta PQR$ with vertices $P(2, 4, -7), Q(3, 7, 18), R(-5, 12, 8)$  by  $(-3, 2, 5)$.
This is my question , it came in our test and I am afraid to come in the final exam so kindly, I want to know how to solve such questions, our lecturer tough us only about $2D$ and he did not teach us about $3D$ so I understand $2D$ but I could not understand $3D$ .
Please, can you explain the solution step by step because I want to know how you solve it?
Thank in advance!

REPLY [0 votes]: Denote by $\tau(PQR)=\tau P \tau Q \tau R $ translated triangle from conditions we have that
$$\tau P=\tau(2,4,-7)=(2,4,-7)+(-3,2,5)=(2-3,4+2,-7+5)=(-1,6,-2)$$
$$\tau P=\tau(3,7,18)=(3,7,18)+(-3,2,5)=(3-3,7+2,18+5)=(0,9,33)$$
$$\tau P=\tau(-5,12,8)=(-5,12,8)+(-3,2,5)=(-5-3,12+2,8+5)=(-8,14,13)$$
$$\tau(PQR=(-1,6,-2),(0,9,33),(-8,14,13)$$


text:
\begin{document}

\begin{abstract}
We prove that the $C^1$ interior of the set of all topologically stable $C^1$ incompressible flows is contained in the set of Anosov incompressible flows. Moreover, we obtain an analogous result for the discrete-time case.
\end{abstract}

\maketitle

\noindent\emph{MSC 2000:} primary 37C10;37C15; secondary 37C05, 37C27.\\
\emph{keywords:} Volume-preserving flows and diffeomorphisms, topological stability, Anosov systems.\\

\begin{section}{Introduction: basic definitions and statement of the results}

We consider an $n$-di\-men\-sio\-nal ($n\geq 3$) closed and connected $C^\infty$ Riemaniann manifold $M$ endowed with a volume-form $\omega$. Let $\mu$ denote the measure associated to $\omega$, that we call Lebesgue measure, and let $d(\cdot,\cdot)$ denote the metric induced by the Riemannian structure. We say that a vector field $X\colon M\rightarrow TM$ is \emph{divergence-free} if $\nabla\cdot X=0$ or equivalently if the measure $\mu$ is invariant for the associated flow, $X^t\colon M\rightarrow M$, $t \in \mathbb{R}$. In this case we say that the flow is \emph{incompressible} or \emph{volume-preserving}. We denote by $\mathfrak{X}_\mu^r(M)$ ($r\geq 1$) the space of $C^r$ divergence-free vector fields on $M$ topologized with the usual $C^r$ Whitney topology.

Given $X\in\mathfrak{X}_\mu^{1}(M)$ let $Sing(X)$ denote the set of \emph{singularities} of $X$ and $\mathpzc{R}=\mathpzc{R}(M):=M\setminus Sing(X)$ the set of \emph{regular} points. We say that $\sigma\in Sing(X)$ is a \emph{hyperbolic singularity} if $DX_\sigma$ has no eigenvalue with null real part.

Let $X,Y\in\mathfrak{X}_\mu^1(M)$; $Y$ is \emph{semiconjugated} to $X$ if the flow associated to $Y$ is semiconjugated to the one of $X$, i.e., there exists a continuous and onto map $h\colon M\rightarrow M$ and a continuous real map $\tau \colon M\times\mathbb{R}\rightarrow\mathbb{R}$ such that
\begin{enumerate}
 \item [a)] for any $x\in M$, $\tau_x \colon \mathbb{R}\rightarrow\mathbb{R}$ is an orientation preserving homeomorphism where $\tau(x,0)=0$ and
 \item [b)] for all $x\in M$ and $t\in\mathbb{R}$ we have $h(Y^t(x))=X^{\tau(x,t)}(h(x))$.
\end{enumerate}

We say that $X\in\mathfrak{X}_\mu^{1}(M)$ is \emph{topologically stable} in $\mathfrak{X}_\mu^{1}(M)$ if for any $\epsilon>0$, there exists $\delta>0$ such that for any $Y\in \mathfrak{X}_\mu^{1}(M)$ $\delta$-$C^0$-close to $X$, there exists a semiconjugacy from $Y$ to $X$, i.e., there exists $h\colon M\rightarrow M$ and $\tau \colon M\times\mathbb{R}\rightarrow\mathbb{R}$ satisfying a) and b) above, and $d(h(x),x)<\epsilon$ for all $x\in M$. It is worth to emphasize that our definition of topological stability is restricted to the conservative setting and not to the  broader space of dissipative flows. We denote the set of topological stable incompressible flows by $\mathpzc{TS}_\mu(M)$.

A vector field is said to be \emph{Anosov} if the tangent bundle $TM$
splits into three continuous $DX^t$-invariant nontrivial subbundles
$E^0\oplus E^1\oplus E^2$ where $E^0$ is the flow direction, the
sub-bundle $E^2$ is uniformly contracted and the
sub-bundle $E^1$ is uniformly expanded by $DX^t$
for $t>0$. Of course that, for an Anosov flow, we have $Sing(X)=\emptyset$ which follows from the fact that the dimensions of the subbundles are constant on the entire manifold.

For analogous definitions in the volume-preserving diffeomorphisms context see \S\ref{diff}.

The concept of topological stability was first introduced by Walters. In (\cite{W}) he proved that Anosov diffeomorphisms are topologically stable. In (\cite{Ni}) Nitecki proved that topological stability was a necessary condition to get Axiom A plus strong transversality. Later, in (\cite{Ro2}), Robinson proved that Morse-Smale flows are topologically stable. In the mid 1980's, Hurley obtained necessary conditions for topological stability (see \cite{H,H2,H3}). About ten years ago it was proved by Moriyasu, Sakai and Sumi  (see~\cite{MSS}) that, if $X$ is a vector field in the $C^1$ interior of the set of topologically stable vector fields (in the broader space of dissipative flows) then $X$ satisfies  the Axiom A and the strong transversality properties. Our main result (Theorem~\ref{teo1}) is a generalization of the main theorem in (\cite{MSS}) for divergence-free vector fields. Although this result is expectable, its proof uses perturbations techniques that only recently become available.

Given a set $A\subset \mathfrak{X}^1_\mu(M)$ let $int_{C^1}(A)$ denote the interior of $A$ in $\mathfrak{X}^1_\mu(M)$ with respect to the $C^1$-topology.

\begin{maintheorem}\label{teo1}
If $X\in int_{C^1}(\mathpzc{TS}_\mu(M))$ then $X$ is Anosov.
\end{maintheorem}

Due to well-know results (see~\cite{PT}) about the restrictions of the existence of manifolds supporting Anosov flows, we obtain that, for general manifolds, the $C^1$-interior of topological stable incompressible flows must be empty. It is also interesting to note that, in the geodesic flow context, Anosov systems are not necessarily topological stable (see~\cite{R}).

\bigskip

Given $x\in \mathpzc{R}(X)$ we consider its normal bundle $N_{x}=X(x)^{\perp}\subset T_{x}M$ and define the \emph{linear Poincar\'{e} flow} by $P_{X}^{t}(x):=\Pi_{X^{t}(x)}\circ DX^{t}_{x}$ where $\Pi_{X^{t}(x)}:T_{X^{t}(x)}M\rightarrow N_{X^{t}(x)}$ is the projection along the direction of $X(X^{t}(x))$. Let $\Lambda \subset R$ be an $X^t$-invariant and compact set. We say that $\Lambda$ is a \emph{hyperbolic} set for the linear Poincar\'{e} flow if there exist constants $\lambda\in (0,1)$, $C>0$ and a splitting $N_x=N^u_x\oplus N^s_x$ such that for all $x\in \Lambda$ we have:
$$\|(P_{X}^{t}(x))^{-1}|_{N^{u}_{x}}\|<C\lambda^{t}\text{ and }\|P_{X}^{t}(x)|_{N^{s}_{x}}\|<C\lambda^{t}.$$

We say that $X \in \mathcal{G}_{\mu}^1(M)$ if there exists a neighborhood $\mathcal{V}$ of $X$ in $\mathfrak{X}_\mu^{1}(M)$ such that any $Y\in\mathcal{V}$, has  all the closed orbits and all the singularities of hyperbolic-type.

The following result, which is important \emph{per se}, will be crucial to obtain Theorem~\ref{teo1}.

\begin{maintheorem}\label{teo2}
 If $X\in int_{C^1}(\mathpzc{TS}_\mu(M))$ then $X\in \mathcal{G}^1_{\mu}(M)$.
\end{maintheorem}

The next result was proved recently by Ferreira (\cite{F}) and is a generalization of a  three-dimensional theorem by the authors (\cite{BeRo3}).
\begin{maintheorem}(Ferreira~\cite{F})\label{Ferreira}
If $X\in \mathcal{G}^1_{\mu}(M)$ then $X$ is Anosov.
\end{maintheorem}

Theorem~\ref{teo1} is a direct consequence of Theorem ~\ref{teo2} and Theorem~\ref{Ferreira} and for this reason we just have to concentrate on the proof of Theorem~\ref{teo2}.

\end{section}

\begin{section}{Perturbation lemmata}
\begin{subsection}{Perturbations near singularities}

Some key results to perform perturbations in the conservative setting are available (see~\cite{AM}). Nevertheless, neither \cite[Theorem 3.1]{AM} nor \cite[Theorem 3.2]{AM} are adequate to go on with the proof of our result. Therefore, we need to obtain an upgrade of these pasting lemmas and this is the content of the Lemma~\ref{am}. We would like to thank Carlos  Matheus for a valuable suggestion for the proof of this lemma.

\begin{lemma}\label{am} Let $M$ be a compact and boundaryless Riemannian manifold of dimension $\geq 2$. Given $\epsilon>0$, $X\in\mathfrak{X}^1_{\mu}(M)$, a compact $\mathcal{K}\subset M$ and an open neighborhood $\mathcal{U}$ of $\mathcal{K}$,  there are $\delta>0$  and an open set $\mathcal{K} \subset \mathcal{V} \subset \mathcal{U}$ such that, if  $Y\in\mathfrak{X}^2_{\mu}(M)$ is $\delta$-$C^1$-close to $X$ in $\mathcal{U}$, then there exists $Z\in \mathfrak{X}^1_{\mu}(M)$  satisfying
\begin{enumerate}
 \item [a)] $Z=Y$ in $\mathcal{V}$,
 \item [b)] $Z$ is $\epsilon$-$C^1$-close to $X$ and
 \item [c)] $Z=X$ outside $\mathcal{U}$.
\end{enumerate}
\end{lemma}

\begin{proof}
We consider $\mathcal{V}\supset \mathcal{K}$ such that $\partial \mathcal{V}$ is $C^\infty$, $\overline{\mathcal{V}}\subset \mathcal{U}$ and $\{\mathcal{U},\text{int}(M\setminus \overline{\mathcal{V}})\}$ is an open covering of $M$. Let $\alpha\colon M\rightarrow[0,1]$ be a $C^\infty$ function such that $\alpha=1$ in $\mathcal{V}$, $\alpha=0$ outside $\mathcal{U}$ and $|\nabla\alpha|\leq K$, where $K$ is a positive constant depending only on $\mathcal{U}$ and $\mathcal{V}$. Define
\begin{equation}
 Z_0(x) := \alpha(x) Y(x) + (1-\alpha(x)) X(x),
\end{equation}
where $Y\in\mathfrak{X}^2_{\mu}(M)$ is $\delta$-$C^1$-close to $X$ on a small open neighborhood $\mathcal{U}$ of $\mathcal{K}$, where $\delta>0$ will be fixed in (\ref{delta}). We observe that $Z_0=Y$ inside $\mathcal{K}$ and $Z_0=X$ outside $\mathcal{U}$, and thus $\nabla\cdot Z_0=0$ in the closed set $\overline{\mathcal{V}}\cup (M\setminus\mathcal{U})$. However, although $\nabla\cdot Z_0$ is close to zero in $\mathcal{U}\setminus\mathcal{V}$, in general $\nabla\cdot Z_0\not=0$. Actually,

\begin{eqnarray*}
\nabla\cdot Z_0&=& (\nabla\alpha)\cdot Y+\alpha(\nabla\cdot Y)  -(\nabla\alpha)\cdot  X+(1-\alpha)\nabla\cdot Y\\
&=& (\nabla\alpha)\cdot Y -(\nabla\alpha)\cdot X= (\nabla\alpha)\cdot (Y- X),
\end{eqnarray*}
and we have $|\nabla\cdot Z_0|<K\delta$.

Now we will make use of \cite[Theorem 2]{DM} in order to obtain $Z_1\in\mathfrak{X}^2_{\mu}(M)$ (supported in $\mathcal{U}\setminus\mathcal{V}$) such that $\nabla\cdot Z_1=-\nabla\cdot Z_0$ and $Z_1=0$ in $\partial(\mathcal{U}\setminus\mathcal{V})$.

Finally, we define
\begin{equation}
Z:=Z_0+Z_1.
\end{equation}
Of course that, by construction, $\nabla\cdot Z=0$ and c) holds.

In $\mathcal{U}\setminus\mathcal{V}$ we have
\begin{eqnarray*}
 \|Z-Y\|_{C^1}&=&\|Z_0+Z_1-Y\|_{C^1}\leq \|Z_1\|_{C^1}+\|Z_0-Y\|_{C^1}\\
&\leq& C\|\nabla\cdot Z_1\|_{C^0}+\|Z_0-Y\|_{C^1},
\end{eqnarray*}
where $C>0$ is a constant given in Dacorogna-Moser theorem (see~\cite[Theorem 2.3]{AM}) that depends only on $\mathcal{U}$.

Going back to the beginning of the proof, we take
\begin{equation}\label{delta}
\delta<\min\left\{\epsilon,\frac{\epsilon}{2CK}\right\}.
\end{equation}

Now, using $|\nabla\cdot Z_0|<K\delta$ we get,
\begin{eqnarray*}
\|Z-Y\|_{C^1}&\leq& C\|\nabla\cdot Z_1\|_{C^0}+\|Z_0-Y\|_{C^1} \leq CK\delta+\|\alpha Y + (1-\alpha) X-Y\|_{C^1}\\
&\leq& \frac{\epsilon}{2}+\| (1-\alpha) X-(1-\alpha) Y\|_{C^1}\leq \frac{\epsilon}{2}+|1-\alpha| \|X- Y\|_{C^1}\\
&\leq& \frac{\epsilon}{2}+\|X- Y\|_{C^1}\leq \frac{\epsilon}{2}+\delta<\epsilon.
\end{eqnarray*}

\end{proof}

The next lemma is the divergence-free vector fields version of \cite[Lemma 1.1]{MSS}.

\begin{lemma}\label{linear}
Let $\sigma$ be a singularity of $X \in \mathfrak{X}_\mu^1(M)$. For any $\epsilon >0$ there exist $\delta_0>0$ and $\epsilon_0>0$ such that if $Y_\delta \colon T_\sigma M \rightarrow T_\sigma M$ is a traceless linear map $\frac{\delta}{2}$-$C^0$-close to $DX_\sigma$ (with $\delta<\delta_0$) then there exists $Z_\delta \in  \mathfrak{X}_\mu^{1}(M)$, such that $Z_\delta=Y_\delta$ in $B_{\epsilon_0/4}(\sigma)$, $Z_\delta$ is $\epsilon$-$C^1$-close to $X$ and $Z_\delta=X$ outside the set $B_{\epsilon_0}(\sigma)$.
\end{lemma}
\begin{proof}

Let $(U,\phi)$ be a conservative chart given by Moser's Theorem (\cite{Mo}) such that $\sigma \in U$ and $\phi(\sigma)=\textbf{0}$. Now we will work on the euclidean space $\mathbb{R}^n$.

We fix $\delta\in(0,\delta_0)$ where $\delta_0>0$ will be defined in the sequel. For simplicity we assume that $Y_\delta \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ is written in the canonical coordinates. Now, we consider the divergence-free linear vector field in $\mathbb{R}^n$ defined (in the canonical coordinates) by;
\begin{equation}\label{Y}
 (\dot{x}_1,...,\dot{x}_n)=Y_\delta (x_1,...,x_n).
\end{equation}

Now, let $\mathcal{K}=\overline{B_{\epsilon_0/4}(\textbf{0})}$ and $\mathcal{U}=B_{\epsilon_0}(\textbf{0})$. Since, by hypothesis, the map $Y_\delta$ is $\frac{\delta}{2}$-$C^0$-close to $DX_\textbf{0}$, if we choose $\epsilon_0$ very small, then $X$ is $\frac{\delta_0}{2}$-$C^1$-close to $DX_{\textbf{0}}$ when restricted to $\mathcal{U}$.  Therefore $Y_\delta$ (defined in (\ref{Y})) and $X$ are $\delta_0$-$C^1$-close in $\mathcal{U}$. This conditions gives $\epsilon_0$ depending on $\delta_0$. So, by Lemma~\ref{am}, fixed $\epsilon$, there exist $\delta_0$ and there are an open set $\mathcal{V}$; $\mathcal{K} \subset \mathcal{V} \subset \mathcal{U}$ and $Z_\delta\in \mathfrak{X}^1_{\mu}(M)$  such that $Z_\delta=Y_\delta$ in $\mathcal{V}$, $Z_\delta$ is $\epsilon$-$C^1$-close to $X$ and $Z_\delta=X$ outside $\mathcal{U}$ for $\delta<\delta_0$. The lemma is proved.
\end{proof}

\end{subsection}

\begin{subsection}{Perturbations near closed orbits}

Lemma~\ref{mpl} below is a \emph{Franks' lemma} for incompressible flows.

Define $\Gamma(p, \tau)=\{X^t(p);\,\, t \in [0, \tau]\}$. Let $V, \tilde{V} \subset N_p$, $dim(V)=j$, $2 \leq j \leq n-1$, and  $N_p=V \oplus \tilde{V}$. A \emph{one-parameter linear family} $\{A_t\}_{t\in \mathbb{R}}$ associated to $\Gamma(p, \tau)$ and $V$  is defined as follows:
\begin{itemize}
\item $A_t\colon N_p \rightarrow N_p$ is a linear map, for all $t\in \mathbb{R}$,
\item $A_t=id$, for all $t\leq 0$, and $A_t=A_{\tau}$, for all $t\geq \tau$,
\item $A_t|_V \in sl(j, \mathbb{R})$, and $A_t|_{\tilde{V}}\equiv id$, $\forall t \in [0, \tau]$, in particular we have  $\det(A_t)=1$, for all $t\in \mathbb{R}$, and
\item the family $A_t$ is $C^\infty$ on the parameter $t$.
\end{itemize}

\begin{lemma}\label{mpl}(\cite[Lemma 3.2]{BeRo2}) Given $\epsilon>0$ and a vector field $X \in \mathfrak{X}_\mu^4(M)$ there exists $\theta_0=\theta_0(\epsilon,X)$ such that $\forall \tau \in [1,2]$, for any periodic point $p$ of period greater than $2$, for any sufficient small flowbox  $\mathcal{T}$ of $\Gamma(p, \tau)$ and for any one-parameter linear family $\{A_t\}_{t \in [0, \tau]}$ such that $\|  \dot{A}_t A_t^{-1}\|<\theta_0$, $\forall t \in [0, \tau]$, there exists $Y \in \mathfrak{X}_\mu^1(M)$ satisfying the following properties
\begin{enumerate}
\item [(A)] $Y$ is $\epsilon$-$C^1$-close to $X$;
\item [(B)] $Y^t(p)=X^t(p)$, for all $t \in \mathbb{R}$;
\item [(C)] $P_Y^\tau(p)=P_X^\tau(p) \circ A_{\tau}$, and
\item [(D)] $Y|_{\mathcal{T}^c}\equiv X|_{\mathcal{T}^c}$.
\end{enumerate}
\end{lemma}
\end{subsection}
\end{section}

\begin{section}{Proof of Theorem~\ref{teo2}}

In order to go on with our proof we observe that the main steps are based on the arguments in \cite[\S 2]{MSS}. From \S\ref{sing} and \S\ref{closed} below it follows that any $X\in int_{C^1}(\mathpzc{TS}_\mu(M))$ has all its singularities and closed orbits of hyperbolic-type. Therefore, $int_{C^1}(\mathpzc{TS}_\mu(M))\subset \mathcal{G}^1_\mu(M)$.

\begin{subsection}{Singularities}\label{sing}
We are going to prove that, if $X\in int_{C^1}(\mathpzc{TS}_\mu(M))$, then any singularity of $X$ is hyperbolic. By contradiction let us assume that there exists a non-hyperbolic singularity $\sigma \in Sing(X)$. According to Lemma~\ref{linear} we consider a family of divergence-free vector fields $\{Z_\delta\}_{\delta\geq 0}$, where $Z_{\delta}$ is linear and $Z_0=DX_{\sigma}$ in $B_{\frac{\epsilon_0}{4}}(\sigma)$, and $\delta$ is sufficiently small to assure that this family is contained in $int_{C^1}(\mathpzc{TS}_\mu(M))$. We observe that
we can chose $\delta^\prime$ arbitrarily small such that $\sigma$ is a hyperbolic singularity for $Z_{\delta^\prime}$ and such that $Z_{\delta^{\prime}}$ is semiconjugated to $Z_0$, that is, there exists a continuous and onto map $h\colon M\rightarrow M$ (arbitrarily close to the identity depending on $\delta^{\prime}$) and a continuous real map $\tau \colon M\times\mathbb{R}\rightarrow\mathbb{R}$ such that for any $x\in M$, $\tau_x \colon \mathbb{R}\rightarrow\mathbb{R}$ is an orientation preserving homeomorphism where $\tau(x,0)=0$ and for all $x\in M$ and $t\in\mathbb{R}$ we have
\begin{equation}\label{h}
 h(Z_{\delta^\prime}^t(x))=Z_{0}^{\tau(x,t)}(h(x)).
\end{equation}

As $\sigma$ is non-hyperbolic for $Z_0$, there exists $z\in M$ such that $$\sigma\notin \{B_{\epsilon}(Z_{0}^{t}(z))\colon t\in\mathbb{R}\} \text{ and } \{B_{\epsilon}(Z_{0}^{t}(z))\colon t\in\mathbb{R}\}\subset B_{\frac{\epsilon_0}{8}}(\sigma)$$ and $\epsilon<\frac{\epsilon_0}{16}$. Let $w\in h^{-1}(z)$. From (\ref{h}) we get that $h(Z_{\delta^\prime}^t(w))=Z_0^{\tau(w,t)}(z)$ and, since $h$ is arbitrarily close to the identity, we obtain that $$\{Z_{\delta^\prime}^{t}(w)\colon t\in\mathbb{R}\}\subset \{B_{\epsilon}(Z_{0}^{t}(z))\colon t\in\mathbb{R}\}\subset B_{\frac{\epsilon_0}{8}}(\sigma),$$
which is a contradiction because $\sigma$ is a hyperbolic singularity of $Z_{\delta^\prime}$ and, when restricted to $B_{\frac{\epsilon_0}{8}}(\sigma)$, the vector field $Z_{0}$ is linear.
\end{subsection}

\begin{subsection}{Closed orbits}\label{closed}
Fix $X\in int_{C^1}(\mathpzc{TS}_\mu(M))$. Now we are going to prove that all the closed orbits of $X$ are hyperbolic.

Assume that $X$ has a non-hyperbolic closed orbit $p$ of period $\pi(p)$. In order to proceed with the arguments in \cite[\S 2]{MSS} we need to $C^1$-approximate the vector field $X$ by a new one which is \emph{linear} in a neighborhood of the closed orbit $p$. To perform this task, in the conservative setting, it is required more differentiability of the vector field (cf. Lemma~\ref{mpl}).

If $X$ is of class $C^\infty$, take $Z=X$, otherwise we use \cite{Z} in order to obtain a $C^\infty$ incompressible vector field $Y\in int_{C^1}(\mathpzc{TS}_\mu(M))$, arbitrarily $C^1$-close to $X$, and such that $Y$ has a closed orbit\footnote{Notice that $q$ may not be the analytic continuation of $p$. This is precisely the case when $1$ is an eigenvalue of $P_X^{\pi(p)}(p)$.} $q$, close to $p$, and with period $\pi(q)$ close to $\pi(p)$. If $q$ is not hyperbolic take $Z=Y$. If $q$ is hyperbolic for $P_Y^{\pi(q)}(q)$, then, since $Y$ is $C^1$-arbitrarily close to $X$, the distance between the spectrum of $P_Y^{\pi(q)}(q)$ and $\mathbb{S}^1$ can be taken arbitrarily close to zero (weak hyperbolicity). Now, we are in position to apply Lemma~\ref{mpl} to obtain a new vector field $Z\in\mathfrak{X}^\infty_\mu(M)\cap int_{C^1}(\mathpzc{TS}_\mu(M))$, $C^1$-close to $Y$ and having a non-hyperbolic closed orbit.

Now, we argue as in \cite[\S 3]{BeRo1} in order to obtain $L\in int_{C^1}(\mathpzc{TS}_\mu(M))$ such that $L$ is linear (equal to $P_Z^t$) in a neighborhood of the closed non-hyperbolic orbit, $\Gamma$.

Finally, we $C^1$-approximate $L$ by $W\in int_{C^1}(\mathpzc{TS}_\mu(M))$ such that $\Gamma$ is hyperbolic (for $W$). This is a contradiction because $W$ is semiconjugated to $L$, although there is an $L^t$-orbit (different from $\Gamma$) contained in a small neighborhood of $\Gamma$ and the same cannot occur for $W^t$ because $\Gamma$ is a hyperbolic closed orbit for $W^t$.

\end{subsection}

\end{section}

\begin{section}{The volume-preserving diffeomorphisms case}\label{diff}
Let $\text{Diff}_\mu^{\,\,1}(M)$ denote the set of volume-preserving (or conservative) diffeomorphisms defined on $M$, and consider this space endowed with the $C^1$ Whitney topology. In this section we assume that $\dim(M)\geq 2$. We say that a diffeomorphism $f$ is \emph{Anosov} if, there exist $\lambda\in(0,1)$ and $C>0$ such that the tangent vector bundle over $M$ splits into two $Df$-invariant subbundles $TM=E^u\oplus E^s$, with $\|Df^n|_{E^s}\|\leq C\lambda^n$ and $\|Df^{-n}|_{E^u}\|\leq C\lambda^n$.

We say that $f \in \mathcal{F}_{\mu}^1(M)$ if there exists a neighborhood $\mathcal{V}$ of $f$ in $\text{Diff}_\mu^{\,\,1}(M)$ such that any $g\in\mathcal{V}$ has all the periodic orbits  hyperbolic. In~\cite{AC} Arbieto and Catalan proved the following result.

\begin{maintheorem}\label{AC}
If $f\in \mathcal{F}^1_{\mu}(M)$ then $f$ is Anosov.
\end{maintheorem}

Given $f,g\in\text{Diff}_\mu^{\,\,1}(M)$ we say that $g$ is \emph{semiconjugated} to $f$ if there exists a continuous and onto map $h\colon M\rightarrow M$ such that for all $x\in M$ one has $h(g(x))=f(h(x))$.

We say that $f$ is \emph{topologically stable} in $\text{Diff}_\mu^{\,\,1}(M)$ if, for any $\epsilon>0$, there exists $\delta>0$ such that for any $g\in \text{Diff}_\mu^{\,\,1}(M)$ $\delta$-$C^0$-close to $f$, there exists a semiconjugacy from $g$ to $f$, i.e., there exists $h\colon M\rightarrow M$ satisfying $h(g(x))=f(h(x))$ and $d(h(x),x)<\epsilon$, for all $x\in M$. Once again we emphasize that our definition of topological stability is restricted to the conservative setting and not to the  broader space of dissipative diffeomorphisms. We denote the set of topological stable incompressible flows by $\emph{\textbf{TS}}_\mu(M)$.

In this section we obtain the discrete-time version of Theorem~\ref{teo1}.

\begin{maintheorem}
If $f\in int_{C^1}(\textbf{TS}_\mu(M))$ then $f$ is Anosov.
\end{maintheorem}

The proof is similar to the one done in \S\ref{closed} and, for this reason, we present a brief highlight of it. As before, given $f\in int_{C^1}(\textbf{TS}_\mu(M))$, we prove that all its  periodic orbits are hyperbolic; from this it follows that $int_{C^1}(\textbf{TS}_\mu(M))\subset \mathcal{F}^1_{\mu}(M)$. Then, using Theorem~\ref{AC}, we obtain that any $f\in int_{C^1}(\textbf{TS}_\mu(M))$ is Anosov.

Once again we assume, by contradiction, that there is some non-hyperbolic orbit. Now, to argue as in the flow case, we make use of the following two ingredients:
\begin{enumerate}
 \item a way to linearize the diffeomorphism in a neighborhood of a periodic point and
 \item a ``perturbation of the derivative'' result in the vein of Lemma~\ref{mpl}.
\end{enumerate}

The item (2) is available in the literature (see~\cite[Proposition 7.4]{BDP}).

For (1) we just have to approximate $f\in int_{C^1}(\emph{\textbf{TS}}_\mu(M))$ by a diffeomorphism $g\in int_{C^1}(\emph{\textbf{TS}}_\mu(M))\cap \text{Diff}_\mu^{\,\,\infty}(M)$ using a recent result by Avila~(\cite{A}), and then we use the Pasting lemma~\cite[Theorem 3.6]{AM} to obtain $h\in int_{C^1}(\emph{\textbf{TS}}_\mu(M))\cap \text{Diff}_\mu^{\,\,\infty}(M)$ such that $h=Dg_p$ in a neighborhood of the periodic orbit $p$. This is precisely what we need to obtain a contradiction as we did in \S\ref{closed}.

\end{section}

\section*{Acknowledgements}

The authors were partially supported by the FCT-Funda\c{c}\~ao para a Ci\^encia e a Tecnologia, project PTDC/MAT/099493/2008.
MB was partially supported by FCT (SFRH/BPD/20890/2004).


TITLE: Equation of an $(n-2)-$sphere in $\mathbb{R}^n$.
QUESTION [0 upvotes]: I am looking for the equation of an $(n-2)$-sphere in $\mathbb{R}^{n}$ generated from the intersection of the $(n-1)$- sphere $x_1^2 + x_2^2 + \cdots + x_n^2 = r^2$, and the hyperplane perpendicular to a given vector $ \mathbf{v} \in \mathbb{R}^{n}$ and containing the point $(0,0,\ldots,0)$.
I just need some hint to solve this problem. Thanks for any help!

REPLY [0 votes]: The additional condition is $\mathbf{v}\cdot (x_1,\dots,x_n) = 0$.


TITLE: Proposition Question
QUESTION [2 upvotes]: I am trying to translate this into propositional symbols but (for me) it's so complicated. Can someone help me figure this out.
"If it rains then I will carry a sharp object and I will start laughing while everyone catches their breath; otherwise I will not play outside and pretend to be invisible."
Thanks in advance.
Sorry for the newbie question.

REPLY [1 votes]: It is all done with sentences and connectives: A sentence is usually  denoted by Capital letters like $A,B$, etc., the connectives are  or:=$ \lor$; and:=$\land$, and if-then:= $ \rightarrow$, and negation connective is $\lnot$ so that, e.g., If I go, then I will see a movie is transcribed as : say $A$ is " If I go" , and $B$ is , "I will see a movie" is represented as $ A\rightarrow B$. Then you can use additional connectives to make longer sentences. Does that help?


TITLE: Can any number of squares sum to a square?
QUESTION [46 upvotes]: Suppose
$$a^2 = \sum_{i=1}^k b_i^2$$
where $a, b_i \in \mathbb{Z}$, $a>0, b_i > 0$ (and $b_i$ are not necessarily distinct).
Can any positive integer be the value of $k$?

The reason I am interested in this: in a irreptile tiling where the smallest piece has area $A$, we have $a^2A = \sum_{i=1}^k b_i^2A$, where we have $k$ pieces scaled by $b_i$ to tile the big figure, which is scaled by $a$. I am wondering what constraints there are on the number of pieces.
Here is an example tiling that realizes $4^2 = 3^2 + 7 \cdot 1^2$, so $k = 8$.

REPLY [1 votes]: Many of the other answers (based on extending smaller values of $k$) yield a fairly large value for the LHS.  It’s worth pointing out that we actually know a great deal about the set of integers representable as the sum of $k$ nonzero squares.
See, for instance, this other answer, where it is shown that every integer $\ge 34$ is the sum of five positive squares.  Immediately, by padding with $1$s, we can extend this to

For $k\ge 5$, every integer $> k+28$ can be written as the sum of $k$ nonnegative squares.

Therefore, we can just choose $a^2$ to be the first square greater than $k+28$, which makes $a^2 \le k + 2\sqrt k + O(1)$, which makes $$a \le \sqrt k + 1 + O(k^{-1/2}).$$ This basically optimal as clearly $a \ge \lceil \sqrt k\rceil^2$.
We might be able to shave down  the constant $28$ somewhat by increasing the lower bound on $k$, but it can’t be reduced all the way to $0$ because of the gap between the first two nonzero squares $1$ and $4$.  There must be some constant $C>0$ that is the optimal replacement for $28$ for all $k\ge k_0$ (I wouldn’t be surprised if it is already optimal).