text:
\begin{document}

\newcommand{\e}{\epsilon}
\renewcommand{\a}{\alpha}
\renewcommand{\b}{\beta}
\newcommand{\om}{\omega}
\newcommand{\La}{\Lambda}
\newcommand{\la}{\lambda}
\newcommand{\p}{\partial}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\id}{\int_D}
\newcommand{\na}{\nabla}

\newtheorem{corollary}{Corollary}
\newtheorem{proposition}{Proposition}
\newtheorem{lemma}{Lemma}
\newtheorem{theorem}{Theorem}

\def\sqg{G(\sqrt{(x-\xi)^2+(y-\eta)^2})}
\def\kernel{G(\sqrt{x^2+y^2})}
\def\be{\begin{eqnarray}}
\def\ee{\end{eqnarray}}
\def\qed{\hbox{\vrule width 6pt height 6pt depth 0pt}}

\title{Asymptotic Dynamical Difference between the Nonlocal and Local
	Swift-Hohenberg Models
\footnote{Author for correspondence: Professor Jinqiao Duan,
Department of Mathematical Sciences,
Clemson University, Clemson, South Carolina 29634, USA.
$\;$ E-mail:  duan@math.clemson.edu; $\;$ Fax: (864)656-5230.} }

\author{Guoguang Lin$^1$, Hongjun Gao$^2$,
Jinqiao Duan$^3$ and Vincent J. Ervin$^3$
   \\
   \\
\\1. Graduate School, Chinese Academy of Engineering Physics \\
P. O. BOX 2101, Beijing  100088, and
Department of Mathematics \\
Yunnan University, Kunming  650091,  China.
\\
\\2. Laboratory of Computational Physics\\
Institute of Applied Physics and Computational Mathematics\\
Beijing, 100088, China.
\\
\\3. Department of Mathematical Sciences\\
Clemson University, Clemson, South Carolina 29634, USA.  }

\date{June 11, 1998 }

\maketitle

\begin{abstract}

In this paper the difference in the asymptotic dynamics between
the nonlocal and local two-dimensional Swift-Hohenberg models
is investigated. It is shown that the bounds  for the dimensions
of the global attractors for the nonlocal
and local Swift-Hohenberg models differ by an absolute constant,
which depends only on the Rayleigh number,
and upper and lower bounds of the kernel of the nonlocal
nonlinearity.  Even when this kernel of the nonlocal operator
is a constant function, the dimension bounds  of the global attractors
still differ by an absolute constant depending on the Rayleigh number.

\bigskip
\bigskip
{\em Running Title:} Nonlocal Swift-Hohenberg Model

\bigskip
{\em Key Words:}  asymptotic behavior, nonlocal nonlinearity, global attractor,
		dimension estimates

\bigskip
{\em PACS Numbers:}   02.30, 03.40, 47.20

\end{abstract}

\newpage

   \section{Introduction}

Fluid convection due to density gradients arises in
geophysical fluid flows in the atmosphere, oceans
and the earth's mantle.
The Rayleigh-Benard convection is a prototypical model
for fluid convection, aiming at predicting spatio-temporal
convection patterns.
The mathematical model for the
Rayleigh-Benard convection involves nonlinear Navier-Stokes
partial differential equations coupled with the temperature
equation. When the Rayleigh number is near the onset of the
convection, the Rayleigh-Benard convection model may be approximately
reduced to an amplitude or order parameter equation, as derived
by Swift and Hohenberg (\cite{Swift}).

In the current literature, most work on the
Swift-Hohenberg model deals with the following one-dimensional equation
for $w(x,t)$, which is a
localized, one-dimensionalized version of the model
originally derived by Swift and Hohenberg (\cite{Swift}),
\begin{eqnarray}
	w_t & = & \mu w -(1+\p_{xx})^2 w - w^3.
\label{sh2}
\end{eqnarray}
The cubic term $w^3$ is used as an approximation of
a nonlocal integral term.
For the (local) one-dimensional Swift-Hohenberg
equation (\ref{sh2}), there has been some recent research on
propagating or steady patterns (e.g., \cite{Eckmann_Wayne}, \cite{Hilali},
\cite{Lerman}).
Mielke and Schneider(\cite{Mielke})  proved the existence of
the global attractor
in a weighted Sobolev space on the whole real line.
Hsieh et al. (\cite{Hsieh}, \cite{Hsieh2})
remarked that the elemental instability mechanism is
the negative diffusion term $-w_{xx}$.

Roberts (\cite{Roberts}, \cite{Roberts2})
recently re-examined  the rationale for using the
Swift-Hohenberg model as a reliable model of the spatial
pattern evolution in specific physical systems.
He   argued  that,
although the localization
approximation used in (\ref{sh2})
makes some sense in the one-dimensional case,
this approximation is deficient
in the two-dimensional convection problem and
one should use the nonlocal Swift-Hohenberg model
(\cite{Swift}, \cite{Roberts}, \cite{Roberts2}):
\begin{eqnarray}
 u_t  = \mu u -(1+\De)^2 u
	- u \int_D G(\sqrt{(x-\xi)^2+(y-\eta)^2}) u^2(\xi, \eta, t)d\xi d\eta ,
\label{sh}
\end{eqnarray}
where $u=u(x,y,t)$ is the unknown amplitude function,
$\mu$ measures the difference of the Rayleigh number
from its critical onset value,
$\De = \p_{xx} + \p_{yy}$ is the Laplace operator, and
$G(r)$ is a given radially symmetric function ($r=\sqrt{x^2+y^2}$).
The equation is defined for
$t>0$ and $(x,y) \in D$, where $D$ is a bounded planar domain with
smooth boundary $\p D$.

The two-dimensional version of the local  Swift-Hohenberg equation
for $u(x, y, t)$ is
\begin{eqnarray}
	u_t & = & \mu u -(1+\De)^2 u - u^3.
\label{sh3}
\end{eqnarray}
Here $u^3$ is used to approximate the nonlocal term in (\ref{sh}).

Roberts (\cite{Roberts}, \cite{Roberts2}) noted that the range of Fourier
harmonics generated by the nonlinearities is fundamentally
different in two-dimensions than in one-dimension. This difference
requires a more sophisticated treatment of two-dimensional
convection problem, which leads to nonlocal nonlinearity
in the Swift-Hohenberg model. He also argued
that nonlocal operators naturally appear in systematic derivation of
simplified models for pattern evolution, and nonlocal operators
also permit symmetries which are consisitent with
physical considerations.

In this paper, we discuss the difference between
nonlocal and local two-dimensional
Swift-Hohenberg models (\ref{sh}), (\ref{sh3}),
from a viewpoint of asymptotic dynamics.
We show that the bounds  for the dimensions
of the global  attractors for the nonlocal
and  local Swift-Hohenberg models differ by an absolute
constant, which depends only on
the the Rayleigh number, and upper and
lower bounds of the kernel of the nonlocal
nonlinearity.  Even when this kernel is a constant function,
the dimension bounds  of the global attractors
still differ by a constant depending on the Rayleigh number.
In \S 2 and \S 3, we will consider the
nonlocal and local Swift-Hohenberg models, respectively.
Finally in \S 4, we summarize the results.

\section{Nonlocal Swift-Hohenberg Model}

In this section, we discuss the global attractor
and its dimension estimate for the nonlocal
Swift-Hohenberg model (\ref{sh}).
In the following we use the abbreviations $L^2=L^2(D)$,
$L^{\infty}=L^{\infty}(D)$,  $H^k = H^k(D)$ and $H^k_0 =H^k_0(D)$  ($k$ is
a non-negative integer)
for the standard Sobolev spaces.
Let $(\cdot, \cdot)$,
$\| \cdot \| \equiv \| \cdot \|_2$ denote the standard
inner product and norm in $L^2$,
respectively. The norm for $H^k_0$  is
$\| \cdot \|_{H^k_0}$. Due to the Poincar\'e inequality,
$\|D^k u\|$ is an equivalent norm in $ H^k_0$.

We rewrite the two-dimensional nonlocal Swift-Hohenberg equation (\ref{sh})
as
\begin{eqnarray}
u_t + \a u +2\De  u + \De^2 u
  + u \int_D G(\sqrt{(x-\xi)^2+(y-\eta)^2}) u^2(\xi, \eta, t)d\xi d\eta=0,
\label{sh4}
\end{eqnarray}
where  $\a = 1 - \mu$.
This equation is supplemented with the initial condition
 \be
 u(x,y,0)=u_0(x,y),
	\label{sh5}
 \ee
and the  boundary conditions
 \be
 u|_{\p D}   =    0, \;\;  \frac{\p u}{\p n} |_{\p D} = 0,
	\label{sh6}
 \ee
where $n$ denotes the unit outward normal vector of the boundary $\p D$.

In this paper, we assume the following conditions
for every $t \geq 0 $ and $(x,y) \in D$,
 \be
  0 < b  \leq  \kernel  \leq  a,  \; \mbox{and}  \;
 G, \; \na G, \; \De G \in L^{\infty}(D),
 \label{sh7}
 \ee
where $a, b > 0$ are some positive constants
and $\na = (\p_x, \p_y)$ is the gradient operator.
Denote $K_1= \|\na G\|_{\infty}$ and $K_2=\|\De G\|_{\infty}$.

To study the global attractor, we need to derive
some a priori estimates about solutions.

\begin{lemma}
 Suppose $u$ is a solution of (\ref{sh4})-(\ref{sh6}).
 Then  $u$ is uniformly (in time)
 bounded,  and  the following estimates hold for $t>0$

 \be
 \|u(x,y,t)\|^2 \leq \|u_0(x,y)\|^2 exp(-2\mu t)+ \frac{\mu}{b}, \;
  \label{sh8}
 \ee
 and thus
\be
 {\lim \sup}_{t\rightarrow +\infty} \|u(x,y,t)\|    \leq  \sqrt{\frac{\mu}{b}} \equiv R ,
\label{sh9}
\ee
where $R = \sqrt{\frac{\mu}{b}}$.
\label{bound}

\end{lemma}

{\bf Proof.}  Taking the inner product of (\ref{sh4}) with $u$, we have

\be
\frac{1}{2}\frac{d}{dt}\|u\|^2 + \|\De u\|^2 + 2(\De u,u) + \a \|u\|^2
				\nonumber  \\
  + (u^2,\id \sqg u^2 (\xi,\eta)d\xi d\eta)   = 0.
\label{sh10}
\ee
Note that
$$  2|(\De u,u)|\leq 2\|\De u\|\|u\|  \leq  \|\De u\|^2 + \|u\|^2,$$

$$  (u^2,\id \sqg u^2(\xi,\eta)d\xi d\eta) $$
$$ = \id u^2(\id \sqg u^2(\xi,\eta)d\xi d\eta)dxdy$$

$$ \geq b\id u^2(x,y)dxdy \id u^2(\xi,\eta)d\xi d\eta = b\|u\|^4.
$$
Then from (\ref{sh10}) we get
\be
 \frac{d}{dt}\|u\|^2 + 2(\a-1)\|u\|^2 + 2b\|u\|^4 \leq 0.
 \label{L2norm}
 \ee
It is easy to see that if $\a \geq 1$, i.e., $\mu \leq 0$,
then  all solutions
approach zero in $L^2$. We will not consider this simple dynamical
case. In the rest of this paper we assume that $\mu > 0$, i.e.,
$\a < 1$.

Thus we have, for any constant $\e >0$,
\be
\frac{d}{dt}\|u\|^2 +  2\e \|u\|^2 +2(\a-1-\e)\|u\|^2+ 2b\|u\|^4 \leq  0,
\ee
or
\be
\frac{d}{dt}\|u\|^2 +  2\e \|u\|^2
 + [ \frac{(\a-1-\e)}{\sqrt{2b}} + \sqrt{2b}\|u\|^2 ]^2
 	\leq \frac{(\a-1-\e)^2}{2b}.
\ee
 So
\be
\frac{d}{dt}\|u\|^2 +  2\e \|u\|^2  \leq  \frac{(\a-1-\e)^2}{2b}.
\ee
By the usual Gronwall inequality (\cite{Temam}) we obtain
\be
 \|u\|^2 \leq \|u_0\|^2 exp(-2\e t)+ \frac{(\a-1-\e)^2}{4b\e}.
\ee
When $\e = 1-\a = \mu$, we get the optimal or tight estimate
\be
 \|u\|^2 \leq \|u_0\|^2 exp(-2\mu t) + \frac{\mu}{b}.
\ee
This completes the proof of
Lemma \ref{bound}.   $\hfill \qed$

Moreover, higher order derivatives of $u$ are also
uniformly bounded.

\begin{lemma}
 Suppose $u$ is a solution of (\ref{sh4})-(\ref{sh6}). Then $\|\na u\|$
 and $\|\De u\|$ are uniformly (in time)  bounded.
\label{highbound}
\end{lemma}

In order to prove this lemma, we recall a few useful inequalities.

{\em Uniform Gronwall inequality} (\cite{Temam}).
Let $g,h,y$ be three positive locally
integrable functions on $[t_0,+\infty)$ satisfying the inequalities
$$
\frac{dy}{dt}\leq gy + h ,
$$
with  $\int_t^{t+1}gds \leq a_1,$  $\int_t^{t+1}hds\leq a_2$
and $\int_t^{t+1}yds \leq a_3$ for $t\geq t_0,$
where the $a_i$(i=1,2,3) are positive constants. Then
$$
y(t+1) \leq (a_2+a_3)exp(a_1), \mbox{for} \; t \geq t_0 .
$$

{\em Gagliardo-Nirenberg inequality} (\cite{Pazy}).
Let $w \in L^q\cap W^{m,r}(D)$, where $1 \leq q,r \leq \infty$.
For any integer $j$, $0\leq j \leq m$, $\frac{j}{m}\leq \la \leq 1.$
$$
\|D^j w\|_p \leq C_0 \|w\|_q^{1-\la}\|D^m w\|_r^{\la}
\label{GN}
$$
provided
$$\frac{1}{p} = \frac{j}{n} + \la (\frac{1}{r} - \frac{m}{n})
+ \frac{1-\la}{q},$$
and $m-j-\frac{n}{r}$ is not a nonnegative integer
If $m-j-\frac{n}{r}$ is a nonnegative integer, then the inequality
(\ref{GN}) holds for
$\la = \frac{j}{m}$.

{\em Poincar\'e inequality} (\cite{Friedman}).
For $w \in H_0^1(D)$,
$$ \la_1 \| w \|^2 \leq \|\na w \|^2 , $$
where $\la_1$ is the first eigenvalue of $-\De$ on the domain $D$, with zero
Dirichlet boundary condition
on $\p D$.

{\bf Proof of Lemma \ref{highbound}}.
Due to the boundary condition (\ref{sh6}) on $ \na u$
and the Poincar\'e  inequality, we get
$ \| \na u \|^2 \leq \la_1^{-1} \|\De u \|^2$. Hence it is
sufficient to prove that $ \|\De u\|$ is
bounded.   We first show that $\int_t^{t+1}\|\De u\|^2ds$
is bounded. In fact, using
 $$  2|(\De u,u)| \leq 2\|\De u\|\|u\|  \leq \frac12 \|\De u\|^2 + 2\|u\|^2,$$
in (\ref{sh10}), we get
\be
 \frac{d}{dt} \|u\|^2 + \|\De u\|^2 + 2(\a-2)\|u\|^2 + 2b\|u\|^4 \leq 0.
\ee
Since
$$2b\|u\|^4 + 2(\a-2)\|u\|^2 = b\|u\|^2 +
2\b(\|u\|^4 + \frac{2\a-4-\b}{2\b}\|u\|^2)$$
$$=b\|u\|^2 + 2\b(
\|u\|^2 + \frac{2\a-4-\b}{4\b})^2 -
\frac{(2\a -4 -\b)^2}{8\b}$$
$$\geq b\|u\|^2 - \frac{(2\a -4 -\b)^2}{8\b},$$
we conclude
\be
\frac{d}{dt} \|u\|^2 + \|\De u\|^2 + b\|u\|^2 \leq
\frac{(2\a -4 -\b)^2}{8\b} =\frac{(2 + 2\mu + \b)^2}{8\b}.
\label{sh11.5}
\ee
Integrating (\ref{sh11.5}) with respect to $t$ from $t$ to $t+1$ and noting
Lemma \ref{bound}, we see that $\int_t^{t+1}\|\De u\|^2ds$ is bounded.

Now, multiplying (\ref{sh4}) by $\De^2 u$ and integrating over $D$,
it follows that

$$
\frac{1}{2}\frac{d}{dt}\|\De u\|^2 + \|\De^2 u\|^2 +
2\id \De u\De^2udxdy + \a \|\De u\|^2$$
\be
+\id (u \id \sqg u^2(\xi,\eta)d\xi d\eta)\De^2 udxdy = 0.
\label{sh12}
\ee
Note that
\be
2|\id\De u\De^2 udxdy |  \leq \frac{1}{2}\|\De^2u\|^2 + 2\|\De u\|^2,
\ee
and
$$
 | \id ( u \id \sqg u^2(\xi,\eta)d\xi d\eta)  \De^2  udxdy |
$$
$$
  =   | \id (\De u)^2(\id \sqg  u^2(\xi,\eta)d\xi d\eta)dxdy
$$
$$
       +   \id u\De u (\id(\De \sqg))u^2(\xi,\eta)d\xi d\eta)dxdy
$$
$$
   +   2\id \na u\De u(\id(\na \sqg ))u^2(\xi,\eta)d\xi d\eta)dxdy|
$$
$$
   \leq    (a\|u\|^2+2\la_1^{-\frac12} \|\na G \|_{\infty}\|u\|^2
   + \frac12 \|\De G \|_{\infty}\|u\|^2 ) \|\De u\|^2
	       + \frac12 \|\De G \|_{\infty} \|u\|^4
$$
$$
   \leq    (a +2\la_1^{-\frac12} K_1  + \frac12 K_2 )\|u\|^2 \| \De u \|^2
   			+ \frac12 K_2  \|u\|^4,
$$
where $a, K_1, K_2$ are various upper bounds of $G$ defined
in (\ref{sh7}), and $R$ is the $L^2$ bound of the
solution $u$ as in Lemma 1.
Hence by (\ref{sh12}) we get
\be
\frac{d}{dt}\|\De u\|^2 \leq 2 [(a +2\la_1^{-\frac12} K_1 + \frac12 K_2)\|u\|^2
	 -\a +2]  \|\De u\|^2   +  K_2  \|u\|^4.
\label{sh13}
\ee
Finally, applying the uniform Gronwall inequality  (\ref{sh13})
and noting Lemma \ref{bound}, we conclude
 that $\| \De u\|^2$ is uniformly
bounded for all $t\geq 0.$
This proves Lemma \ref{highbound}.  $\hfill \qed$

We now have the following global existence and uniqueness
result.

\begin{theorem}
Let $u_0(x,y) \in L^2(D)$ and $G$ satisfies (\ref{sh7}),
then the initial-boundary value problem
$(\ref{sh}), (\ref{sh5}), (\ref{sh6})$ has a unique global
solution $u\in L^{\infty}(0,\infty; H^2_0(D))$.  Moreover,
the corresponding solution semigroup $S(t)$, defined by
$$
u = S(t) u_0,
$$
has a bounded absorbing set
$$
B_0 = \{ u  \in H^2_0(D):  (\|u\|^2+\|\na u\|^2
   +\|\De u\|^2)^{\frac{1}{2}}  \leq  \tilde{R}  \},
$$
where $\tilde{R} $ is a postive constant which depending on the
uniform  bound of $\|u\|, \|\na u\|, \|\De u\|$.
Finally, the solution semigroup $S(t)$, when restricted on
$H^2_0(D)$, is continuous from
$H^2_0(D)$ into $H^2_0(D)$ for $t > 0$.
\label{global}
\end{theorem}

{\bf Proof.}
The global existence, uniqueness and absorbing property
follow from standard arguments (e.g., \cite{Temam})
together with Lemmas \ref{bound}, \ref{highbound} above.
The absorbing property also follows from these two lemmas.

We now prove that $S(t)$ is continuous in $H^2(D)\cap H_0^1(D)$.
Suppose that $u_0,v_0\in H^2_0(D)$ with $\|\De u_0\|,
\|\De v_0\|\leq 2R_1,$
we denote by $u(t), v(t)$ the corresponding solutions, i.e.,
$u(t) = S(t)u_0, v(t) = S(t)v_0$.  Let
$w(t)=u(t)-v(t).$ Then $w(t)$ satisfies
$$
w_t + \De^2 w + 2 \De w + \a w + w\id \sqg u^2(\xi,\eta)
d\xi d\eta+$$
\be
v\id \sqg (u(\xi,\eta)+v(\xi,\eta))w(\xi,\eta)d\xi d\eta = 0.
\label{sh14}
\ee
Applying the Gagliardo-Nirenberg inequality

 $$ \|u\|_{\infty} \leq C_0\|\De u\|,$$
and the Poincar\'e inequality

$$ \|w\| \leq \frac{1}{\la_1}\|\De w\|,$$
we obtain (similar to the proof of Lemma \ref{highbound}),

$$
\frac{d}{dt}\|\De w\|^2 \leq C_1 \|\De w\|^2,
$$
which implies that $\|\De w(t)\|^2 \leq \|\De w_0\|^2 exp(C_1 t)$
for some positive constant $C_1$. This shows that $S(t)$ is continuous.
						$\hfill \qed$

This theorem implies that  (\ref{sh4})-(\ref{sh6})
defines an infinite dimensional nonlocal dynamical system.

In the rest of this section, we consider the global attractor
for the nonlocal dynamical system (\ref{sh4})-(\ref{sh6}).
We will establish the following result about the global attractor.

\begin{theorem}
There exists a global attractor ${\cal A}$
for the nonlocal dynamical system
$(\ref{sh}), (\ref{sh5}), (\ref{sh6})$.
The global attractor is the $\omega-$limit set
of the absorbing set $B_0$ (as in Theorem \ref{global}), and it
has the following properties:

(i) $A$ is compact and $S(t){\cal A} = {\cal A},$ for $t > 0$;

(ii) for every bounded set $B\subset H^2_0(D)$,
$\lim\limits_{t\to \infty} d(S(t)B,{\cal A})=0;$

(iii)${\cal A}$ is connected in $H^2_0(D),$
where
$d(X,Y)=\sup\limits_{x\in X}\inf\limits_{y\in  Y}\|x-y\|_{H^2_0(D)}$
is the Hausdorff distance.

Moreover, the global attractor ${\cal A}$ has finite Hausdorff dimension
$d_H({\cal A}) \leq m$,
where
$$
m \sim  C (1+\sqrt{ \mu + (2a-b)\frac{\mu}{b} }),
\label{attractor}
$$
where $ C >0$ is a constant depending only on the domain $D$,
and $a>0, b>0$ are the upper, lower bounds of the kernel $G$,
respectively.
\end{theorem}

{\bf Proof.}
The existence and properties of ${\cal A}$ are
quite standard now (see \cite{Temam} and references therein).
We omit this part, and only estimate the dimensions below.

As in  \cite{Temam}, we may use
the so-called Constantin-Foias-Temam trace formula (which works for
the semiflow $S(t)$ here) to estimate
the sum of the global Lyapunov exponents of
${\cal A}$. The sum of these Lyapunov exponents can then
 be used to estimate the upper bounds of ${\cal A}$'s
Hausdorff dimension, $d_H({\cal A})$.
To this end,
we linearize equation (\ref{sh4}) about a solution $u(t)$ in the
global attractor to obtain an equation for $v(t)$ and then
use the trace formula to estimate the sum of
the global Lyapunov exponents. Doing so, we obtain
\be
v_t + L(u(t))  v = 0,
\label{sh15}
\ee
where
$$
L(u(t))v = \De^2 v + 2\De v + \a v +
v\id \sqg u^2(\xi,\eta)d\xi d\eta
$$
$$+2u\id \sqg u(\xi,\eta) v(\xi,\eta)d\xi d\eta.
$$
This equation is supplemented with
$v(x,y , 0) = \xi(x,y) \in  H^2_0(D)$.
Denote by $\xi_1(x,y), \ldots , \xi_m(x,y)$,
$m$ linearly independent functions
in $H^2_0(D)$, and $v_{i}(x,y , t)$ the solution of (\ref{sh15})
satisfying $v_{i}(x,y , 0) = \xi_{i}(x,y)$, $i = 1, \ldots, m$.
Let $Q_m (t)$ represent the orthogonal projection of $H^2_0(D) $
onto the subspace spanned by $\{v_1(x,y,t), \ldots , v_m(x,y, t) \}$.

We need to estimate the lower bound of
$Tr(L(u(t)Q_m(t)))$, which gives bounds on the sum of
global Lyapunov exponents.
Note that in \cite{Temam},
the linearized equation like (\ref{sh15}) is written
as $v_t = L(u(t))v$ and in that case one needs to
estimate the upper bound of $Tr(L(u(t)Q_m(t)))$.
Suppose that $\phi_1(t),...,\phi_m(t)$ is
an orthonormal basis  ($\|\phi_j\| =1$)
of the subspace $Q_m(t) H_0^2(D)$
for any $t>0$.

Now we estimate  the lower bound of
$Tr(L(u(t)Q_m(t)))$. It is easy to see that

$$Tr(L(u(t)Q_m))$$

$$
=\sum\limits^m_{j=1}(\De^2\phi_j+2\De\phi_j+\a \phi_j+\phi_j\id
	\sqg u^2(\xi,\eta)d\xi d\eta,\phi_j)
$$
$$
+ \sum\limits^m_{j=1}(2u\id \sqg u(\xi,\eta) \phi_j d\xi d\eta,\phi_j).
$$
Since
$(2\De\phi_j, \phi_j) \geq -(\frac1{\e}\|\De \phi_j\|^2 +\e \|\phi_j\|^2)$
for any constant $\e > 1$, we get
$$
Tr(L(u(t)Q_m)) \geq \sum\limits^m_{j=1} [(1-\frac1{\e})\|\De \phi_j\|^2+
b\|\phi_j\|^2\|u\|^2 + (\a - \e) \|\phi_j\|^2 ]
$$
$$
+ \sum\limits^m_{j=1} 2\id u\phi_jdxdy (\id \sqg u(\xi,\eta) \phi_j d\xi d\eta)
$$
$$
\geq \sum\limits^m_{j=1} (1-\frac1{\e}) \|\De \phi_j\|^2
   + \sum\limits^m_{j=1} (b\|u\|^2+ \a-\e -2 a\|u\|^2)
$$
\be
 = \sum\limits^m_{j=1} (1-\frac1{\e}) \|\De \phi_j\|^2
  + [1-\mu-\e + (b-2 a) \|u\|^2 ] m.
\label{sh18}
\ee

We introduce notation $f(x,y) = \sum\limits_{j=1}^m | \phi_j|^2$.
Note that $m = \id f(x,y)dxdy$.
By the generalized Sobolev-Lieb-Thirring
inequality (\cite{Temam}, page 462),

$$\id f^3(x,y)dxdy \leq K_0 \sum\limits^m_{j=1} \|\De \phi_j\|^2,$$
where $K_0>0$ depending only on the domain $D$.
Moreover, due to the fact that $L^3 (D) \hookrightarrow L^1(D)$,
$$m^3 = (\id f(x,y)dxdy)^3 \leq C_2 \id f^3(x,y)dxdy $$

$$\leq K_0 C_2 \sum\limits^m_{j=1} \|\De \phi_j\|^2 $$

$$ = C\sum\limits^m_{j=1} \|\De \phi_j\|^2 $$
for some constants $C_2>0, C>0$ depending only on the domain $D$.

Thus
\be
(1-\frac1{\e}) \sum\limits^m_{j=1} \|\De \phi_j\|^2
  \geq    (1-\frac1{\e}) \frac1{C} m^3.
     \label{sh19}
\ee

Therefore, by (\ref{sh18})-(\ref{sh19}) we have
\be
Tr (L(u(t)Q_m))
& \geq & \frac{1-\frac1{\e}}{C} m^3 - (\mu-1 +\e + (2a-b) \|u\|^2) m
		 		\nonumber   \\
& \geq & \frac{1-\frac1{\e}}{C} m^3 - (\mu-1 +\e + (2a-b)\frac{\mu}{b}) m
				\nonumber   \\
& > & 0
\ee
whenever
\be
m >\sqrt{[ \mu-1 +\e + (2a-b) \frac{\mu}{b} ]
	\frac{C}{1-\frac1{\e}}}.
\label{mm}
\ee
The right hand side of (\ref{mm}) has the minimal value of
\be
   m \sim    C (1+\sqrt{\mu + (2a-b)\frac{\mu}{b}})
   \label{dimension}
\ee
when
$ \e = 1+\sqrt{ \mu + (2a-b)\frac{\mu}{b} }$.

As in \cite{Temam},  we conclude that the Hausdorff dimension of
${\cal A}$ is  estimated as in  (\ref{dimension}).
This proves Theorem \ref{attractor}.       $\hfill \qed$

 \section{Local Swift-Hohenberg Model}

Similarly, for the two-dimensional local Swift-Hohenberg
equation (\ref{sh3}), we can obtain the existence of
the global attractor $\tilde {\cal A}$. We omit this part
and will only estimate the dimension
of $\tilde {\cal A}$.

\begin{theorem}
There exists the  global attractor $\tilde {\cal A}$  for the
local dynamical system (\ref{sh3}), (\ref{sh5}), (\ref{sh6}).
The Hausdorff dimension  of  $\tilde {\cal A}$ is finite, and
$d_H(\tilde {\cal A} )  \leq  m_1 \sim  C (1+\sqrt{ \mu }) $,
where $ C $ is a constant depending only on the domain $D$.
 \label{attractor2}
\end{theorem}

{\bf Proof.} As in the proof of Ttheorem \ref{attractor},
 we consider the linearized equation   of (\ref{sh3}), defined by
$$ v_t + L_1(u(t))v = 0,$$
where
$$L_1(u(t))v = \De^2v + 2\De v + \a v + 3u^2v.  $$

Then we estimate
$$ Tr(L_1(u(t)Q_m)) $$

$$
=\sum\limits^m_{j=1}(\De^2\phi_j+2\De\phi_j+\a \phi_j+3u^2\phi_j ,\phi_j)
$$
$$
= \sum\limits^m_{j=1}[\|\De \phi_j\|^2 + 2(\De \phi_j, \phi_j)
	+\a \|\phi_j\|^2 + 3(u^2\phi_j, \phi_j)]
$$
$$
\geq \sum\limits^m_{j=1} (1-\frac1{\e})\|\De \phi_j\|^2
  + \sum\limits^m_{j=1}(\a -\e),
$$
where we have used the fact that  $3(u^2\phi_j, \phi_j) \geq 0$.
Noting   again that
$m^3 \leq  C\sum\limits^m_{j=1} \|\De \phi_j\|^2$ and $\a =1-\mu$,
we have
\be
Tr (L_1(u(t)Q_m))
\geq \frac{1-\frac1{\e}}{C} m^3 -(\mu -1+\e) m >0
\ee
whenever
\be
m >\sqrt{( \mu-1 +\e ) \frac{C}{1-\frac1{\e}} }.
\label{mmm}
\ee
The right hand side of (\ref{mmm}) has the minimal value of
\be
   m \sim  C (1+\sqrt{ \mu })
\ee
when
$ \e = 1+\sqrt{ \mu }$.
This completes the proof.      		$\hfill \qed$

\section{Discussions}

In this paper, we have discussed the Hausdorff dimension estimates
for the global attractors of the two-dimensional nonlocal and local
Swift-Hohenberg model for Rayleigh-Benard convection.

The Hausdorff dimension for the global attractor of the
nonlocal model is estimated as
$$
   m \sim    C (1+\sqrt{\mu + (2a-b)\frac{\mu}{b}}),
$$
while for the local model this estimate is
$$
   m \sim  C (1+\sqrt{ \mu }),
$$
where $C>0$ is an absolute constant depending only on
the fluid convection domain, and $\mu>0$ measures the
difference of the Rayleigh number from its critical
convection onset value.
Note that $a, b > 0$ are the upper and lower
bounds, respectively, of the kernel $G$ of the nonlocal nonlinearity
in (\ref{sh}).

The two dimension estimates above
differ by an absolute constant $(2a-b)\frac{\mu}{b}$,
which depends only on
the the Rayleigh number through $\mu$, and upper and
lower bounds of the kernel $G$ of the nonlocal
nonlinearity.  Moreover, if the kernel $G$ is a constant function
(thus, $a=b=G$), then the dimension estimate for the
nonlocal model becomes
$$
   m \sim    C (1+\sqrt{ 2 \mu}),
$$
which still differs from the dimension
estimate for the local model by a constant depending on the Rayleigh number
through $\mu$.

\bigskip

{\bf Acknowledgement.}
Part of this work was done while Jinqiao Duan was visiting the
Institute for Mathematics and its Applications (IMA), Minnesota,
and the Center for Nonlinear Studies, Los Alamos National
Laboratory.
    This work was supported by the Nonlinear Science Program of China,
the National Natural Science Foundation of China Grant 19701023,
the Science Foundation of Chinese
Academy of Engineering Physics  Grant 970682, and the USA National Science
Foundation Grant DMS-9704345.


TITLE: How to construct an $AF$-algebra $N$ with $K_0(N)=\mathbb Q$?
QUESTION [1 upvotes]: On the paper i'm reading, it says letting $N=\lim M_2\otimes ...\otimes M_{n!}$ will do.
I wonder why it doesn't simply let $N=\lim M_2\otimes ...\otimes M_n$. Take arbitary $p\in N\otimes M_n$, there is $p'\in M_{m!}\otimes M_n$ such that $|p'-p|<1/2$ so that $[p']=[p]$ in $K_0(N)$. Therefore, for any $r\in\mathbb N$, letting $rs>m$ and mapping $p'$ to $p'\otimes 1_{(rs)!/m!}\in M_{(rs)!}\otimes M_n$, we see $q=p\otimes 1_{(rs)!/m!/r}$ satisfies that $r[q]=[p]$. This shows $K_0(N)=\mathbb Q$.
Am I right?

REPLY [1 votes]: You're right, if you take $A_n = \otimes_1^n M_i \simeq M_{n!}$ with connecting maps $A_n \ni a \mapsto a \otimes 1_{n+1} \in A_{n+1}$, then
$$A = \underset{\to}{\lim} A_n $$
has $K_0(A) \simeq \mathbb{Q}$. Indeed, just use continuity of $K_0$. Applying $K_0$ to the sequence we get
$$ \mathbb{Z} \to^2 \mathbb{Z} \to^3 \mathbb{Z} \to \cdots, $$
and its not too hard to convince yourself that the resulting limit is $\mathbb{Q}$. Note that the algebra $A$ is actually the universal UHF algebra in this case.
The same argument applies to more general inductive limits which give UHF algebras (see Direct limit of $\mathbb{Z}$-homomorphisms). As long as in the induced sequence on $K$-theory
$$ \mathbb{Z} \to^{a_1} \mathbb{Z} \to^{a_2} \mathbb{Z} \to \cdots, $$
each prime divides an $a_i$ for some $i$, you should get all of $\mathbb{Q}$ as your limit. So the limit you described works as well (although I'm sure of why one might prefer one limit to another in this specific case). For example, you could take $\underset{\to}{\lim} M_2 \otimes M_3 \otimes \cdots \otimes M_{n!} \simeq M_{(n!)!}$, which intuitively is a subsequence of the one you wrote down. (I think I wrote a wrong limit before, edited for corectness)
There is also a classification of UHF algebras via their corresponding supernatural numbers, one can show that the supernatural numbers of the two sequences end up being the same.


TITLE: $\Bbb{R}_\ell$ is Lindelöf...Why is this Map Well-defined?
QUESTION [0 upvotes]: Here is Munkres' proof that $\Bbb{R}_\ell$, the Sorgenfrey line ($\Bbb{R}$ with the lower limit topology), is Lindelöf (example 3 in section 30):

Let $C = \bigcup_{i \in I}(a_i,b_i)$. . .Let $x$ be a point of $\Bbb{R}-C$. We know that $x$ belongs to no open interval $(a_i,b_i)$; therefore $x = a_i$ for some index $i$. Choose such a $i$ and then choose $q_x$ to be a rational number belonging to the interval $(a_i,b_i)$. It follows that if $x$ and $y$ are two points of $\Bbb{R}-C$ with $x < y$, then $q_x < q_y$. . .Therefore the map $x \mapsto q_x$ of $\Bbb{R}-C$ into $\Bbb{Q}$ is injective, so that $\Bbb{R}-C$ is countable.

My question is pretty simple: why is this map--well--a map? It doesn't appear to be well-defined, since there are may rationals in $(a_i,b_i)$.

REPLY [0 votes]: Being very explicit:
Enumerate the rationals as $\mathbb{Q} = \{q_1, q_2 ,q_3, \ldots\}$.
Now, if $X$ is is in $\mathbb{R} - C$, we know $I_x = \{i \in I: x_i \in [a_i, b_i)\}$ is a non-emty set, because the $[a_i, b_i), i \in I$ form a cover of $\mathbb{R}$. So we pick one such $i(x) \in I_x$ for each $x$, by the Axiom of Choice. Indeed this means that $x= a_{i(x)}$, because it must be in $[a_{i(x)}, b_{i(x)})$ but cannot be in $(a_{i(x)}, b_{i(x)}) \subseteq C$, as $x \notin C$.
The open interval $(a_{i(x)}, b_{i(x)})$ must contain some rational number so let $n(x)$ be the smallest $n$ such that $q_{n(x)} \in (a_{i(x)}, b_{i(x)})$. Note that by the enumeration of the rationals we can avoid choice here. This $q_{n(x)}$ is then denoted by $q_x$. It's clear such a $q_x$ exists (if we assume choice for the picking of the $i(x)$). It's not a "map" as some formula, but a "choice function" for some simultaneous choices; this happens a lot in topology.
Now, if $x \neq y$ are both in $\mathbb{R} - C$ we must have $x < y$ or $y < x$, so assume WLOG that $x < y$. We know $x =a_{i(x)}$ and again $y \notin (a_{i(x)}, b_{i(x)}) \subseteq C$, so that in fact $y \ge b_{i(x)}$.
Hence we have the situation $$x = a_{i(x)} < q_x = q_{n(x)} < b_{i(x)} \le y = a_{i(y)} < q_{n(y)} = q_y$$ so indeed $x < y$ implies $q_x < q_y$, as claimed.


TITLE: Showing $\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n b_i \right)^\frac1{n}\le\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n a_i \right)^\frac1{n}$?
QUESTION [12 upvotes]: If $a_1\ge a_2 \ge a_3 \ldots $ and if $b_1,b_2,b_3\ldots$ is any rearrangement of the sequence $a_1,a_2,a_3\ldots$ then for each $N=1,2,3\ldots$ one has
$$\sum^N_{n=1}\left(\prod_{i=1}^n b_i \right)^{\frac{1}{n}}\le \sum^N_{n=1}\left(\prod_{i=1}^n a_i \right)^{\frac{1}{n}}$$
This comes from page 177 of "The Cauchy-Schwarz Master Class".
The solution in the back argues that, by hypothesis, $b_1\le a_1,b_2\le a_2,b_3\le a_3\dots$  Therefore, it follows that $(b_1b_2\cdots b_n)^{1/n}\le (a_1a_2\cdots a_n)^{1/n}$.
It seems to me that for $N=3$, with a sequence $a_1=3$,$a_2=2$ and $a_3=1$, and it's rearrangement $b_1=1$,$b_2=2$ and $b_3=3$, this is not the case.
Am I missing something obvious?

In order to provide the context, here is the relevant portion from the book (Steele J.M. The Cauchy-Schwarz master class, CUP 2004, p.273):
Solution for Exercise 11.7. This observation is painfully obvious,
but it seems necessary for completeness. The hypothesis gives us the
bounds $b_1 \le a_1, b_2 \le a_2, \dots , b_N \le a_N$; thus, for all $1 \le n \le N$ we have
$(b_1b_2\dots b_n)^{1/n} \le (a_1a_2\dots a_n)^{1/n}$, which is more than we need. There
are questions on infinite rearrangements which are subtle, but this is not
one of them.

REPLY [7 votes]: I think you are correct with your observation.
Maybe the author wanted to say that for each $n$ such that $1\le n\le N$ we have
$$a_1a_2\dots a_n \ge b_1b_2 \dots b_n,$$
which follows from the fact that if we reorder $b_1,b_2,\ldots,b_n$ from the largest element $c_1\ge c_2\ge\ldots\ge c_n$, then $c_1\le a_1,c_2\le a_2,\ldots,c_n\le a_n$ and $b_1b_2\dots b_n = c_1c_2\dots c_n$.
Although this observation seems to be easy, I have trouble writing a simple and clear proof of it :-(


TITLE: Are partial derivatives of parametric surfaces always orthogonal?
QUESTION [0 upvotes]: For a surface $\mathbf{r} = \mathbf{r}(s, t)$ are the partial derivatives $\mathbf{r}_s$ and $\mathbf{r}_t$ in general orthogonal?
I was thinking the surface of a sphere and in that case indeed these vectors are orthogonal at every point. We can think the partial derivatives as the tangent vectors of the curves $\mathbf{r} (s, t_0)$ and $\mathbf{r} (s_0, t)$ respectively. Can the above fact for the surface of a sphere be generalized?

REPLY [0 votes]: Your example is apparently $$\mathbf r(\theta, \phi) = (\cos \theta\sin\phi, \sin\theta\sin\phi,\cos\phi)$$
where it is indeed true that $\mathbf r_\theta \cdot \mathbf r_\phi = 0$. But notice that this a standard coordinate parametrization of the sphere. This coordinate system was chosen exactly for the property that the $\theta$ and $\phi$ coordinate directions are perpendicular to each other.
In general, this is not true. Another familiar way of parametrizing half the sphere is $$\mathbf r(x, y) = (x, y,\sqrt{1 - x^2 - y^2})$$
for which we find
$$\mathbf r_x = (1, 0, \frac {-x}{\sqrt{1 - x^2 - y^2}})\\
\mathbf r_y = (0, 1, \frac {-y}{\sqrt{1 - x^2 - y^2}})$$
and thus
$$\mathbf r_x \cdot \mathbf r_y = \frac {xy}{1 - x^2 - y^2}$$
which is only $0$ when either $x$ or $y$ is $0$.


TITLE: Finding Number of cases
QUESTION [1 upvotes]: Suppose we have 20 cards numbered 1 through 20 and we want to pick 5 of them at the same time
a.How large is our total sample space ?
b.What is the probability of numbers 1 and 2 are not in our sample?
c.What is the probability of at least one card number is 3 , 4 ?
for a I think C(20,5)
I have no idea for b and c

REPLY [2 votes]: A natural sample space is the collection of all "hands" of $5$ numbers. There are $\dbinom{20}{5}$ such hands, and they are all equally likely.
For b), there are $\dbinom{18}{5}$ ways to choose $5$ cards, none of which is $1$ or $2$. So our probability is
$$\frac{\binom{18}{5}}{\binom{20}{5}}.$$
If desired, considerable simplification is possible.
For c), the simplest way is to subtract the answer to b) from $1$. This is because the event "at least one from $1$ or $2$" happens precisely if the event described in b) doesn't happen.
We can also solve the problem in a way close to the idea you mentioned in a comment. If we want a $1$ or a $2$ or both, there are three ways this can happen: (i) $1$ but not $2$; (ii) $2$ but not $1$; both $1$ and $2$.
The number of choices for (i) is $\dbinom{18}{4}$. The number of choices for (ii) is also $\dbinom{18}{2}$. And the number of choices for (iii) is $\dbinom{18}{3}$. Add up, and for the probability divide by $\dbinom{20}{5}$.

REPLY [1 votes]: Your answer of $\binom{20}5$ for (a) is correct.
(b) To get the probability that $1$ and $2$ are not in the sample, you must count the $5$-card sets that don’t contain either $1$ or $2$. These are the $5$-card sets chosen from $\{3,4,5,\dots,20\}$; how many of them are there? This is really a problem just like (a). Once you have that, what fraction must you form to get the desired probability?
(c) Now observe that the events in (b) and (c) are complementary, so their probabilities add up to a known amount; what is that amount? Use it and your answer to (b) to get the answer to (c).


TITLE: Three inequalities with a common constraint
QUESTION [0 upvotes]: A question struck me today and to be able to answer it I wrote it down as a mathematical expression and have been able to simplify the question to the following:
Given $$x > 1,y > 1,z > 1$$ $$x+y+z > S$$ what is the smallest $S$ that guarantees
$$Ax > A+B+C$$
$$By > A+B+C$$
$$Cz > A+B+C$$
if $A,B,C$ are chosen correctly after $x,y,z$ become known, where $A,B,C$ are positive numbers?
I do not even know if there exists a value of $S$ that satisfies this. Furthermore, after I find out the value of $S$, how would I go about choosing the numbers $A, B$, and $C$? I would guess that by finding $S$, you would also find out how to choose the numbers, but perhaps that is not the case.
Do not refrain from answering even if you do not have a complete answer. Partial answers are certainly not unwanted.

REPLY [1 votes]: No value of $S$ guarantees that this is possible.
Note that choosing positive $A,B,C$ is equivalent to choosing positive $\lambda,\mu,\nu$ such that $\lambda+\mu+\nu=1$ and $\lambda x>1,\mu y>1,\nu z>1$. [Just divide the inequalities by $A+B+C$ to see this.]
Now suppose that $x>S,y=z=1.5$. Then we certainly have $x+y+z>S$. But to ensure that $\mu y>1$ and $\nu z>1$, we need $$\mu>\frac{2}{3},\nu>\frac{2}{3},\mu+\nu<1$$ which is impossible.


TITLE: Multivariable critical points
QUESTION [1 upvotes]: I have the following function:
$\ f(x,y) = (x^3-y^2)(x-1) $
Expanding the brackets gives:
$\ f(x,y) = x^4 - x^3 - xy^2 + y^2 $
I understand to find the first derivative of both variables and set to zero:
$\ f_x ==> 4x^3 - 3x^2 - y^2 = 0.....[1] $
$\ f_y ==> y - 2xy = 0.....[2] $
$\ y = 2xy.....[3] $
Substituting [3] into [1]:
$\ 4x^3 - 3x^2 - 4x^2y^2 = 0 $
$\ x^2(4x - 3 - y^2) = 0$
I am stuck at this point and can not factorise further to find the critical points. Any hints or clues would be helpful. Thank you.

REPLY [1 votes]: The second partial derivative is $-2xy+2y$ which gives $-xy+y=y(-x+1)=0$, $y=0$ or $x=1$. By replacing $y=0$ you have  $4x^3-3x^2=0$  or $x=1$,  $4-3-y^2=0$.


TITLE: How to calculate the focal spot size of concentrated sunlight
QUESTION [0 upvotes]: I know that concentrated photovoltaic systems use Fresnel Lenses to concentrate incident sunlight to a point. I also know that this point isn’t infinitely small and has an actual size since the lenses don’t concentrate to infinity.
My question is how do I calculate the true spot focus size of the concentrated sunlight?
Is there an equation that relates distance, focal length and other properties perhaps?

REPLY [1 votes]: The sun is not a point source, so you can't get a point image. You get an image of the Sun. You can use basic lens maker's formulas with the angular size of the Sun as seen from the Earth (same as the angular size of the Moon) and the focal length of your lens.


TITLE: Branch cut for $\sqrt{1-z^2}$ - Can I use the branch cut of $\sqrt{z}$?
QUESTION [12 upvotes]: I was trying to clarify some questions I had about elliptic integrals using
http://websites.math.leidenuniv.nl/algebra/ellcurves.pdf
There they define the map
$$\phi\colon w\mapsto \int_0^w\frac{\mathrm{d}z}{\sqrt{1-z^2}}$$
on $\mathbb{C}\setminus[-1,1]$ to get $\phi$ well-defined up to periods of the integral. The choice of the interval $[-1,1]$ is made so that $\sqrt{1-z^2}$ admits a single-valued branch.
Now, I know that the principal branch of the square root $\sqrt{z}$ is discontinuous on the half-line $(-\infty,0)$, so to get a holomorphic map we restrict to $\mathbb{C}\setminus (-\infty,0]$. Substituting $1-z^2$ for $z$ we get that the appropriate branch cuts for the above mapping $\sqrt{1-z^2}$ would be $(-\infty,-1]$ and $[1,\infty)$, which is somewhat the opposite of the suggested interval $[-1,1]$.
From that I conclude that they didn't choose the principal branch, otherwise for e.g. $z=2$ the map would be discontinuous.
My question is: Are both choices possible? Then there must be some way to choose another branch of $\sqrt{1-z^2}$. Is there a good way to see how to choose "elegant" branch cuts and the corresponding holomorphic branches?
A thought of my own: It should be possible to instead integrate on the Riemann sphere, using $\infty$ and not $0$ as a starting point. Then the two intervals would "swap roles". But I don't see how to formalize this.

REPLY [5 votes]: If we choose a branch of $\sqrt{z}$ with branch cut along the positive axis $[0,\infty)$, then we won't need to integrate on the Riemann sphere, since $1-z^2\ge 0$ if and only if $1\ge z^2$, which happens if and only if $z\in [-1,1]$.
This is equivalent to Potato's answer, which is a nice geometrical interpretation of the branch cut.


TITLE: What is the significance of the branch cut in renormalization group logarithms?
QUESTION [9 upvotes]: What is the physical significance of the branch cut in renormalization group logarithms?
(Is this just an avatar of the optical theorem, or is there something to be understood about these logarithms from the point of view of taking the scale as a complex parameter? Usually non-analyticity is a signal of something of physical importance, right?)

REPLY [13 votes]: Let me take a stab at answering this (somewhat vague) question. You said you are interested in the analytic structure of QFT. But you also mentioned the RG, which is somewhat different. I will  try to address the analytic structure of QFT  and then emphasize that the renormalization group can be thought of as merely a trick to improve perturbation theory.
So if one promotes all momenta in a scattering process to complex variables, then the complex function one obtains is not single valued, in other words, the amplitude lives in a multi-cover of some $C^n$. All the branch cuts must live on the real axis.
The significance of the coefficients of the brach cuts is that some "virtual particles can become on shell," or more abstractly, that some subset of the external legs can create an on-shell state with the prescribed kinematics. (These are the famous unitarity cuts.)
There could be poles on the real axis too, but these should be thought of as a special case of a  branch cut, one where the discontinuity is a delta function. This means that the state in the Hilbert space is “isolated” and it is thus interpreted as a bound state.
RG is quite a different set of ideas altogether.  Perhaps it is best to illustrate why we need the equations of RG by an example. Consider some two point function of currents in a QCD-like theory. One removes the divergences with dimensional regularization and so the answer should be $\sim g^2$, where $g$ is the gauge coupling. However, since there are virtual particles in the loop this cannot be the right answer because there must be a branch cut. And indeed, one finds $g^2\log(p^2/\mu^2)$ where $\mu$ is an arbitrary fixed scale and $p$ is the momentum.
The branch cut is independent of $\mu$ because the imaginary part must be finite, scheme independent.
A result like $g^2\log(p^2/\mu^2)$ is highly unsatisfying practically because the expansion in $g$ breaks down at large enough momentum. This is where the RG equations come in.  Basically to understand the large $p^2$ regime  (which is ought to be weakly coupled in AF theories!!) one needs to re-sum  perturbation theory, but one needs to worry only about the leading large logarithms which endanger the $g$ expansion. One finds that the answer can be written as $g^2(p^2)$ and $g^2(p^2)$ satisfies the RG equation, which, in QCD, guarantees that, at large $p$, $g^2(p^2)$ is small. So perturbation theory is saved. Another way to think about RG is that there is a family of possible regularizations of the theory, parameterized by some scale $\mu$, and we can compute any process we want with any choice of $\mu$ , these are all equivalent, but to change $\mu$ to $\mu'$ one needs to redefine the coupling constants. And indeed, to avoid large logarithms it is best to choose $\mu$ around the momentum scale. This is why $g^2(p^2)$ effectively becomes $g^2(\mu)$.
So to summarize, the RG's claim to fame is that it re-sums the dominant terms in perturbation theory to all orders in theories where ordinary perturbation theory is jeopardized by the presence of parametrically large pre-factors in front of the coupling constants. The presence of these large pre-factors can be understood from the presence of branch cuts in momentum space.


TITLE: Central Difference Method
QUESTION [0 upvotes]: Solve the following using the central difference method:
$y(x)= y'+ y + 2x$  where $0 < x < 4$  with $n=4$ subintervals (thus $h=1$).
Given that $y(0)=0$ and $y(3)=1$, find $y(1)$.
Really stuck on this question. Feel like there is an error in the question, shouldn't $y(0)$ and $y(2)$ be given to find $y(1)$ using the central difference method?

REPLY [0 votes]: I really think that your equation is
$$
y''=y'+y+2x.
$$
The central difference quotients for stepsize $h$ are
$$
2h·y'(x)\approx y(x+h)-y(x-h)\\
h^2·y''(h)\approx y(x+h)-2y(x)+y(x-h)
$$
With $h=1$ you get
$$
2y(x+1)-4y(x)+2y(x-1)=y(x+1)-y(x-1)+2y(x)+4x\\
\iff y(x+1)-6y(x)+3y(x-1)=4x
$$
Evaluating the boundary conditions, the resulting linear system
\begin{aligned}
y(2)-6y(1)+3·0&=4\\
1-6y(2)+3y(1)&=8
\end{aligned}
can easily be solved for the values of $y(1)$ and $y(2)$.


TITLE: How to prove that these partial binomial sums are zero?
QUESTION [4 upvotes]: I am trying to prove that the following equation is equal to zero.
$$
0=
 \sum_{j=J+1}^N \Big(j (1-q)+ (j-J) (q N-j) \Big) \cdot q^{j} (1-q)^{N -j}   \binom{N}{j}   \label{zero1}$$
Where
$J,N \in \mathbb{Z}^+$ and $J<N$ and
$0<q<1$ is a probability.
Numerical simulations (see link) suggest that this is true.
Tips and hints (or solutions) are very welcome!

REPLY [7 votes]: Write $p=1-q$, $qN-j=(N-j)q-jp$, the $j$-th summand is $$(j(1-j+J)p+(j-J)(N-j)q)q^jp^{N-j} {N\choose j}.$$
Expand the brackets and consider it as a homogeneous polynomial in $p$ and $q$ of degree $N+1$. The coefficient of $q^{j+1}p^{N-j}$ equals
$$
(j-J)(N-j){N\choose j}+(j+1)(J-j){N\choose j+1}=0.
$$


TITLE: Average acceleration and centripetal acceleration
QUESTION [0 upvotes]: What is the difference between centripetal acceleration and average acceleration  in this worked example?
If $\dfrac{\Delta v}{\Delta t}=\dfrac{v^2}{r}$ and $v=3$ m/s change in time is $1$ second  and $r$ is $1$ m why change in tangential velocity is not $9$ m/s$^2$ but $6$ m/s?

REPLY [0 votes]: Here another way to solve this problem.
As explained in some of the previous answers, the centripetal acceleration is a vector pointing to the center of the circle or normal to the velocity vector. As the toy car is moving down the circle, counterclockwise, from top position (π/2) to the bottom position (-π/2), the acceleration vector is moving counterclockwise as well, from -π/2 to π/2, as shown below:

The average acceleration would be a vector some of all these vectors divided by the total angle π/2 –(-π/2)= π.
As can be seen from the picture, the vertical components of these vectors (blue and red) will cancel each other, while horizontal components (green, pointing to the right) will add up. Therefore, the sum of all vectors will be pointing to the right.
The horizontal component of a vector with an absolute value 9 could be calculated as 9*cos(x), where the x is angle, therefore the average acceleration magnitude could be calculated as follows:

This is not exactly 6, but close:)


TITLE: Realation between direct sum and Cartesian product
QUESTION [0 upvotes]: Let $X_1, X_2$ be such that $X_1 = A_1\oplus B_1$ and $X_2 = A_2\oplus B_2$.
What happens if I consider the Cartesian product $X_1\times X_2$?
Could I write
$$X_1\times X_2 = A\oplus B?$$
There exist some relations between $A, A_1, A_2$ and $B, B_1, B_2$?
Could anyone please help or give some references?
Thank you in advance!

REPLY [0 votes]: We have $X_1 \times X_2 \cong (A_1 \times A_2) \oplus (B_1 \times B_2)$. To prove this, it suffices to show that each element $x \in X_1 \times X_2$ can be written in the form $v_1 + v_2$ with $v_1 \in A_1 \times A_2$ and $v_2 \in B_1 \times B_2$, and that $(A_1 \times A_2 )\cap(B_1 \times B_2) = \{0\}$.
Indeed, by the definition of the Cartesian product we have $x = (x_1,x_2)$ for $x_1 \in X_1,x_2 \in X_2$. It follows that $x_1 = a_1 + b_1$ and $x_2 = a_2 + b_2$ with $a_i \in A_i$ and $b_i \in B_i$, so that
$$
x = (x_1,x_2) = (a_1,b_1) + (a_2,b_2).
$$
For the second part, we note that
$$
(A_1 \times A_2 )\cap(B_1 \times B_2) = (A_1 \cap B_1) \times (A_2 \cap B_2) = \{0\}\times\{0\} = \{0\}.
$$


TITLE: Can I prove that $A\Rightarrow (B\wedge C)$ by proving that $\neg B\Rightarrow\neg A$?
QUESTION [1 upvotes]: Let $A$, $B$ and $C$ be propositional variables. I need to prove a statement of the form $A\Rightarrow (B\wedge C)$ and I’m thinking of proving it by showing the equivalent contrapositive statement
$$\neg(B\wedge C)\Rightarrow \neg A$$
Specifically, I’m thinking of simply showing that
$$\neg B\Rightarrow \neg A$$
However, I’m not 100% sure my approach is correct. My question is: can I prove that $A\Rightarrow(B\wedge C)$ by proving $\neg B\Rightarrow\neg A$? My attempt at answering my own question is below.
Statement: $(\neg B\Rightarrow\neg A)\Rightarrow(A\Rightarrow(B\wedge C))$
Proof:
Suppose that
$$\neg (B\wedge C)\Rightarrow \neg A$$
Because
$$\neg B\Rightarrow\neg(B\wedge C)$$
it follows that
$$\neg B\Rightarrow \neg A$$
EDIT: I understand from the answers that my reasoning is false. Suppose nonetheless that I still want I to show
$$\neg(B\wedge C)\Rightarrow \neg A$$
How do I do it?

REPLY [1 votes]: If $A$ and $B$ are true while $C$ is false, then $\lnot B\to\lnot A$ is true and $\lnot (B\land C)$ is true, meaning $\lnot (B\land C)\to \lnot A$ is false; a true statement cannot imply a false statement.


TITLE: If neutrinos were their own antiparticles, would be observe a photon background from $\nu$-$\nu$ annihilations?
QUESTION [0 upvotes]: If neutrinos were their own antiparticles, would we observe a photon background from neutrino-neutrino annihilations?
As neutrinos are so abundant, why don't we observe such a background? Is the cross-section just too small? Or is some conservation law preventing such annihilations?

REPLY [2 votes]: SM model neutrinos cannot annihilate into the photons, because neutrinos have non-zero lepton number. To remedy this, you need to introduce additional lepton-number breaking interaction to the SM. A way to do it is to introduce a mass term for neutrinos. Introducing it directly poses a theoretical problem (to not break the SU(2) electroweak symmetry it would require a non-renormalizable interaction with the Higgs field), so it's usually done indirectly, via neutrino mixing with heavy right-handed neutrinos. Then the mass term for SM neutrinos appears in an effective theory.
In such a model the annihilation is possible, but the cross section would still be extremely low. Neutrinos do not directly interact with photons, which means that to create a photon pair they could only do it via intermediate virtual particles. For example, the process could be represented by the following Feynman diagram (among others, but they are all similar):

The amplitude for this process would be very low, mostly because of the neutrino mass term, but also because of the fact that it requires a virtual W boson. As a consequence, the process would be undetectable with our current capabilities.


TITLE: Distribution of Brownian Motion squared
QUESTION [0 upvotes]: Could anyone please help me resolve the following paradox?
(i) Brownian Motion squared (i.e. $(B(t)^2$) can be written as:
$$2*[Ito Integral(B(t))] - t$$
(simple, using Ito's lemma). The distribution of Ito integral of a process is Normal, where the mean is zero and the variance can be computed using Ito Isometry. Therefore, using this argumentation, the distribution of Brownian motion squared should be normal with mean $-t$ and variance equal to $2t^2$.
(ii) Standard Brownian motion $B(t)$ at a point in time is, by definition, normally distributed with mean zero and variance $t$. Therefore, it would seem logical that Brownian Motion squared has a (non-centered) Chi-squared distribution with one degree of freedom.
Which one is correct?

REPLY [2 votes]: Your second reasoning is correct, and the first one is wrong. The stochastic integral
$$\int_0^t f(s) \, dB_s$$
is Gaussian if $f$ is a deterministic function. You are interested in the stochastic integral
$$\int_0^t B_s \, dB_s = \frac{B_t^2}{2}-\frac{t}{2}$$
i.e. $f(s) = B_s$ which is clearly not deterministic.
Remark: Note that $B_t^2$ is a non-negative random variable - this means, in particular, that it cannot be Gaussian!


TITLE: Universal Chevalley group associated to $D_l$
QUESTION [1 upvotes]: Consider the simple Lie algebra $D_l$.  Consider the universal Chevalley group $G$ over a field $K$ associated to it. Then $G$ is a subgroup of the orthogonal group $O_{2l}(K, f)$ where $f$ is the quadratic form $x_1x_{-1}+\cdots +x_lx_{-l}$. Then it is shown in the book 'Simple groups of Lie type' by Carter p. 185 that $G$ is isomorphic to the commutator subgroup of $O_{2l}(K, f)$.
My question is :
What is the maximal torus of $G$? Can we write it explicitly? What is its dimension?. I think that its true that $G$ is contained in $SO_{2l}(K, f)$, which is the special orthogonal group. I know that the maximal torus of $SO_{2l}(K, f)$ is $Diag(t_1,t_2,\cdots , t_l, t_1^{-1},\cdots , t_l^{-1})$. So the maximal torus of $G$ is a subgroup of the above torus of $SO_{2l}(K, f)$. I want to write the maximal torus of $G$ explicitly like above. Thanks for help and reference. It will be also nice if someone also describes $G$  more clearly to me.

REPLY [3 votes]: The question shows some confusion but also illustrates the difficulty of working with the Spin groups (the universal groups of types $B_\ell$ and $D_\ell$).  It's easier to work with the single Lie algebra of type $D_\ell$, of course, but the associated groups are essential for many purposes including applications to particle physics.    Chevalley (followed by Steinberg, Borel-Tits, and others) appreciated the possibility of working over arbitrary fields in the spirit of classical Lie group theory.  But the literature and notation can easily cause some confusion.
First of all, type $D_\ell$ (of rank $\ell \geq 4$) leads to an algebraic or topological fundamental group of order 4, which is cyclic for $\ell$ odd (as in type $A_3$) but a Klein 4-group for $\ell$ even.    So several simple algebraic groups of type $D_\ell$ tend to arise, or their analogues  over smaller fields including finite fields.   Each of the algebraic groups is equal to its derived group, and is generated by root groups corresponding to simple roots and their negatives; this much is true for any connected semisimple algebraic group.    But over a field which is not algebraically closed, the abstract group structure may get more complicated: this was studied in detail by Dieudonne and others.
The most familiar algebraic group of type $D_\ell$ is $SO(2\ell)$, which in turn leads to the adjoint group $PSO(2 \ell)$ by factoring out a center of order 2.   Here you use the first fundamental weight attached to the Lie algebra, to get a faithful irreducible representation in dimension $2 \ell$.  However, your assertion (in the third sentence) that the "universal" group lives in the same dimension is false.   So this confusion propagates to other fields as well.   Instead, the universal covering group (simply connected algebraic group) of type $D_\ell$ is more elusive.   To get a faithful representation of it you typically need dimension $2^\ell$, which involves study of an associated Clifford algebra.   In turn, there is a "half-spin" group, arising from either of the fundamental weights attached to the two branched endpoints of the Dynkin diagram.   Like the special orthogonal group this group has the Spin group as a two-fold universal covering group and projects onto the adjoint group.  So even over an algebraically closed field the group theory gets complicated; it's more so over other fields.
Having said all of this, I think your main question has an easy answer: the diagonal subgroup of the universal (or other) group of type $D_\ell$ over an arbitrary field has a "maximal torus" which is a direct product of $\ell$ copies of the multiplicative group of the field.  Of course, it doesn't have a "dimension" unless the field is algebraically closed (then the dimension is $\ell$), but if you start with the standard $2^\ell$-dimensional realization of the Spin group, such a torus (up to conjugacy) can be realized in the group of diagonal matrices.   It's just not a convenient group to work with in matrix form.  Working "downstairs" in $SO(2\ell)$ is usually more practical.


TITLE: The evaluation map of Brauer-Manin obstruction is zero?
QUESTION [3 upvotes]: Let $X$ be a proper $\mathbb{Z}_p$ scheme. The point $x: \mathbb{Z}_p \to X$ induces the map $(A \mapsto A(x))$, $\operatorname{Br}(X) \to \operatorname{Br}(\mathbb{Z}_p)$. Since $\operatorname{Br}(\mathbb{Z}_p) = 0$ we have $A(x)=0$. Now the valuative criterion of properness implies that the natural map $X(\mathbb{Z}_p) \to X(\mathbb{Q}_p)$ is bijective. This would make the Brauer-Manin obstruction vacuous, which isn't true. What went wrong?

REPLY [2 votes]: There is no contradiction, which you can see by spelling things out a little bit more. If $A \in \operatorname{Br}(X)$, then indeed the evaluation map $\operatorname{ev}_A:X(\mathbb{Q}_p)\to \operatorname{Br}(\mathbb{Q}_p)$ vanishes for the reason you give. However, the Brauer-Manin obstruction is mostly used for Brauer classes $A' \in \operatorname{Br}(X_{\mathbb{Q}_p})$ which come from the generic fiber, so do not necessarily extend to the whole of $X$. The evaluation map for $A'$ no longer factors via $\operatorname{Br}(\mathbb{Z}_p)$, so the contradiction vanishes.


TITLE: Estimating the rank of a large sparse matrix
QUESTION [8 upvotes]: Consider a large sparse n by n matrix.  Are there any methods to estimate its rank in time roughly proportional the number of elements in the matrix?

REPLY [5 votes]: This has also been studied from the property testing / query complexity point of view. For example:

Li, Wang, & Woodruff. Improved Testing of Low Rank Matrices, KDD '14. (Freely available version from author here.)

One thing they show is that one can test (with high probability) if an $n \times n$ matrix has rank $\leq r$ or requires an $\varepsilon$ fraction of its entries to be changed in order to get rank $\leq r$ by querying only $O(r^2 / \varepsilon)$ entries of the matrix.