TITLE: Complex Inner Product of Vectors
QUESTION [2 upvotes]: In the case of vectors $a,b \in \mathbb{R}^n$ with inner product $\langle a,b\rangle=a^Tb$ we do not need to worry about whether we mean to project $a$ onto $b$ or $b$ onto $a$ because $\langle a,b\rangle=\langle b,a\rangle$.
However, for $a,b \in \mathbb{C}^n$ we now need to be careful about what we mean since $\langle a, b\rangle=a^Hb \neq \langle b, a\rangle$. If I want to project $a$ onto $b$, does that correspond to $\langle a,b\rangle$ or $\langle b,a\rangle$? How would I show this to myself?

REPLY [3 votes]: (Your convention with the conjugate on the first vector is far from universal, by the way.) :)
Well, you want to write $a = cb + v$, where $c$ is a scalar and $v$ is orthogonal to $b$. So, taking the hermitian inner product with $b$, we have
$$\langle b,a\rangle = \langle b,cb+v\rangle = c\|b\|^2 + \langle b,v\rangle,$$
so, not surprisingly,
$$c = \frac{\langle b,a\rangle}{\|b\|^2}.$$
Note that if you did the inner product "on the other side," here's what you'll get:
$$\langle a,b\rangle = \langle cb,b\rangle + \langle v,b\rangle.$$
The last term still vanishes, of course, but what about the first term on the right? It will become $\bar c\|b\|^2$, so now you have
$$\bar c = \frac{\langle a,b\rangle}{\|b\|^2}.$$
The answers are in agreement.


TITLE: minimum number of neighbors based minimum degree of vertices
QUESTION [0 upvotes]: Let G be a graph of $n \geq 4 $
Where $\delta (G) \geq \frac {2n+1}{3} $
Let $(u,v)$ be an edge, and let $N(u)$ and $N(v)$ be the neighbours of $u$ and $v$ amongst $V(G)\setminus \{u,v\}.$ Show that $$\left|N(u)\cap N(v)\right|\geq \frac{n+2}{3}\geq 2.$$
Any idea how they are coming up with the $\frac{n+2}{3} $ lower bound from $\delta (G) \geq \frac {2n+1}{3} $ ?

REPLY [1 votes]: This is just PIE. We know that $|N(A)\cup N(B)| \leq n$. So we have $$n\geq |N(A)\cup N(B)|= |N(A)|+| N(B)| -|N(A)\cap N(B)| \geq 2\cdot {2n+1\over 3}-|N(A)\cap N(B)| $$
and thus $|N(A)\cap N(B)|\geq  {n+2\over 3}$


text:
\begin{document}\thispagestyle{empty}
\title{\bf Pointwise controllability as limit of internal controllability for the beam equation}\maketitle

\begin{tabular}{c}
\sc Mama Abdelli $^{1}$, Akram Ben Aissa $^{2}$  \ \\
\small $^{1}$ Universit\'e Djillali Liab\'es,
\small   Laboratoire de Math\'ematiques,\qquad \\
\small B. P. 89, Sidi Bel Abb\'es 22000, ALGERIA.  \\e-mail:
abdelli.mama@gmail.com\\ \small $^{2}$ D\'epartement de
Math\'ematiques, Facult\'e des Sciences de
Monastir,\qquad  \\
\small 5019 Monastir, TUNISIA.\\e-mail: akram.benaissa@fsm.rnu.tn \\

\end{tabular}

\vskip20pt

 \renewcommand{\abstractname} {\bf Abstract}
 \begin{abstract}
This work is devoted to prove the pointwise controllability of the
  Bernoulli-Euler beam equation. It is obtained as a limit of internal controllability of  the same type of equation.
Our approach is based on the techniques used in {\bf\cite{FP1}}.
\end{abstract}
\bigskip\noindent
\\

{\small \noindent {\bf Keywords}: Beam equation, Internal control,
Exact controllability, Pointwise control.\\
{\small \noindent {\bf AMS Subject Classifications}: 35E15, 93D05, 93B07, 93D20.
\section {Introduction}
In this paper, we are interested in  the passage from internal  exact controllability of beam equation to pointwise exact controllability.
We consider the following initial
and boundary value problem
\begin{equation}\label{p1}
\Frac{\partial ^2 u}{\partial
t^2}(x,t)+\Frac{\partial^{4}u}{\partial x^4}(x,t)= g_n(x,t),\,\,\,
0<x<1,\,\,\,\, t>0,
\end{equation}
\begin{equation}\label{p2}
u(0,t)= \Frac{\partial u}{\partial x}(1,t)= \Frac{\partial^2
u}{\partial x^2}(0,t)= \Frac{\partial^3 u}{\partial
x^3}(1,t)=0,\quad t>0,
\end{equation}
\begin{equation}\label{p3}
u(x,0)=u^{0}(x),\,\,\,\,\ \Frac{\partial u}{\partial
t}(x,0)=u^{1}(x),\,\,\ 0<x<1,\end{equation} where $g_n,\,\,u^0,\,\,
u^1$ are in suitable spaces with
$\mathrm{supp}\,(g_n)=[\xi,\xi+\frac{1}{n}]$, $n\in\mathbb{N}^*$ and
$\xi \in (0,1)$.\\
 Here $u$ denotes the transverse displacement of
the beam, we suppose that the length of the beam is equal to $1$ and
the control depends on a parameter $n\in\N^*$. Recall that this
model describes the transversal vibrations of the Bernoulli-Euler
beam.

The problem of internal exact  controllability  was studied by Haraux {\bf\cite{HA1}}, Jaffard {\bf\cite{ja}} and Lions {\bf\cite{lio}}. The pointwise exact controllability for a strategic point was studied by Haraux and Jaffard   {\bf\cite{HA2}} and Lions {\bf\cite{lio}}. However, the convergence of the internal exact controllability of equation (\ref{p1})-(\ref{p3}) to the pointwise exact controllability has apparently not yet been studied.

The aim of this paper is to describe what happens when $n$ tends to
infinity, we can't hope to get a pointwise control for the limit
problem for any strategic  point in $(0,1)$,
 we use the same  techniques introduced in
{\bf\cite{FP1}}.

Our purpose in this paper is to  prove  the pointwise controllability of the
  Bernoulli-Euler beam equation. It is obtained as a limit of internal controllability of  the same type of equation.
Our approach based on the techniques used in {\bf\cite{FP1}}. This result can be proved by the standard HUM method
(Hilbert uniqueness method) by J.L. Lions {\bf\cite{lio}}. As n tends to infinity, we obtain the
solution of an exact internal controllability problem which converges towards to the
solution of an exact pointwise controllability  problem.

The plan of the paper is as follows. In section $2$ we show the regularity of weak solutions
of problem (\ref{p1})-(\ref{p3}) for a strategic point in $(0,1)$
and we  study the behavior of these solutions in an interval of length  $\frac{1}{n}$.
The  exact controllability results are given in section $3$ . In section $4$ we prove an
inverse inequality which will give us the estimates on the internal
controls in the case of  a strategic point. Finally, in section
$5$ as  n tends to infinity we prove that the pointwise
exact controllability problem is obtained as limit of exact internal controllability
problem of the beam equation.
 \section {Estimation and regularity results near a point}
Now introducing the Hilbert spaces
$$
V=\{u\in H^2(0,1),\,\,\, u(0)=0,\,\ \frac{d u}{d
x}(1)= 0\}.
$$
$V'$ is the dual space of $V$ with respect to the pivot space
$L^2(0,1)$, where the duality is in the sense of $L^2(0,1)$.\\
and
$${\mathcal D}(\partial_x^4) = \left\{u\in H^4(0,1),\,\,\, u(0)=\frac{d u}{d x}(1)=0,\,\,\,
 \frac{d^2 u}{d x^2}(0)=\frac{d^3 u}{d x^3}(1)=0\right\}.$$
 Consider two given functions $(u^0,u^1)$ in $
L^2(0,1)\times V'$,\,\ $g_n\in L^2(0,T,L^2(0,1))$ and
$\mathrm{supp}\,(g_n)=[\xi,\xi+\frac{1}{n}]$,\,\ $\xi \in (0,1)$ and
 we will take $\frac{1}{n} < 1-\xi$. Let
$u$ be the solution of (\ref{p1})-(\ref{p3}).
\begin{proposition}\label{pro}
Assume $g_n \in L^2(0,T,L^2(0,1))$ and $(u^0,u^1)$ in  $
L^2(0,1)\times V'$. Then for any $T> 0$, problem
(\ref{p1})-(\ref{p3}) admits a unique solution
$$
u  \in{\mathcal C}\left(0,T,L^2(0,1)\right) \cap {\mathcal
C}^1\left(0,T,V'\right).
$$
Moreover,
\begin{equation}\label{mi5}
u(\xi,t)\in L^2(0,T),
\end{equation}
and there exists a constant $C> 0$ (independent on $n$ and T), such that
\begin{equation}\label{mama}
n\Int_{\xi}^{\xi+\frac{1}{n}}\Int_0^T
\Big|u(x,t)\Big|^2\,dx\,dt
 \leq
C\Big(\|g_n\|^2_{L^{2}(0,T,L^2(0,1))} + \|u^0\|^2_{L^2(0,1)} +
\|u^1\|^2_{V'}\Big),
\end{equation}
for all $n>0$.
\end{proposition}
\begin{proof}
In order to prove (\ref{mama}) we put
$$
u^0(x)=\sum_{m=0}^\infty a_m\sin\Big(\frac{2m+1}{2}\pi x\Big),\,\,\,\,
u^1(x)=\sum_{m=0}^\infty \frac{b_m}{(\frac{2m+1}{2}\pi)^2}\sin\Big(\frac{2m+1}{2}\pi x\Big),
$$
and
$$
g_n(x,t)=\sum_{m=0}^\infty g_m(t)\sin\Big(\frac{2m+1}{2}\pi x\Big),
$$
with $(a_m), \Big(\frac{b_m}{(\frac{2m+1}{2}\pi)^2}\Big)\in l^2(\R)$ and for $t$ fixed $(g_m(t))\in l^2(\R)$.\\
\\
\\
The solution of (\ref{p1})-(\ref{p3}) is given by
{\small\begin{equation}\label{CV}
\begin{split}
u(x,t)&=\sum_{m=0}^\infty\left\{ a_m\cos\Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big] +
\frac{b_m}{(\frac{2m+1}{2}\pi)^4}\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]\right.\\&+\left.
\frac{1}{(\frac{2m+1}{2}\pi)^4}\Int_0^t
\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2(t-s)\Big]g_m(s)\,ds\right\}\sin\Big(\frac{2m+1}{2}\pi x\Big).
\end{split}
\end{equation}}
Which implies that
{\small\begin{equation*}
\begin{split}
u(\xi,t)&=\sum_{m=0}^\infty\left\{ a_m\cos\Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big] +
\frac{b_m}{(\frac{2m+1}{2}\pi)^4}\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]\right.\\&+\left.
\frac{1}{(\frac{2m+1}{2}\pi)^4}\Int_0^t
\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2(t-s)\Big]g_m(s)\,ds\right\}\sin\Big(\frac{2m+1}{2}\pi \xi\Big).
\end{split}
\end{equation*}}
We see that
$$
\|u^0\|^2_{L^2(0,1)}+ \|u^1\|^2_{V'}= \frac{1}{2}\sum_{m=0}^\infty\Big[a_m^2+
\frac{b_m^2}{(\frac{2m+1}{2}\pi)^8}\Big],
$$
and
$$
\|g_n\|^2_{L^2(0,1)}=\frac{1}{2}\sum_{m=0}^\infty g^2_m(t).
$$
Integrating (\ref{CV}) over $(0,1)$, we get
{\small\begin{equation}\label{BBN}
\begin{split}
\Int_0^1u^2(x,t)\,dx&=\Frac{1}{2}\sum_{m=0}^\infty\left\{a_m^2\cos^2\Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]+
\frac{b^2_m}{(\frac{2m+1}{2}\pi)^8}\sin^2\Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]\right.\\&+\left.
\frac{1}{(\frac{2m+1}{2}\pi)^8}\Big(\Int_0^t
\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2(t-s)\Big]g_m(s)\,ds\Big)^2\right.\\&+\left.2a_m \frac{b_m}{(\frac{2m+1}{2}\pi)^4}\cos\Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]\sin \Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]
\right.\\&+\left.2\frac{b_m}{(\frac{2m+1}{2}\pi)^8}\sin \Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]\Int_0^t
\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2(t-s)\Big]g_m(s)\,ds
\right.\\&+\left.\frac{2a_m}{(\frac{2m+1}{2}\pi)^4} \cos \Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]\Int_0^t
\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2(t-s)\Big]g_m(s)\,ds
\right\}
\end{split}
\end{equation}}
we shall estimate the third term of right hand side of (\ref{BBN}), we used H\"{o}lder inequality, we get
{\small\begin{equation}\label{BBN1}
\begin{split}
\sum_{m=0}^\infty \Frac{1}{(\frac{2m+1}{2}\pi)^8}\Big(\Int_0^t
\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2(t-s)\Big]g_m(s)\,ds\Big)^2
&\leq \sum_{m=0}^\infty \Int_0^t \sin^2\Big[\Big(\frac{2m+1}{2}\pi\Big)^2(t-s)\Big]\,ds \Int_0^t g^2_m(s)\,ds
\\&\leq C(T)\sum_{m=0}^\infty \Int_0^Tg^2_m(t)\,dt.
\end{split}
\end{equation}}
By Young's inequality, we get
{\small\begin{equation}\label{BBN2}
\begin{split}
\sum_{m=0}^\infty
2a_m \frac{b_m}{(\frac{2m+1}{2}\pi)^4}\cos\Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]\sin \Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]
\leq c_1\sum_{m=0}^\infty a_m^2 +  c'_1\sum_{m=0}^\infty \frac{b^2_m}{(\frac{2m+1}{2}\pi)^8},
\end{split}
\end{equation}}
and  using Young's and H\"{o}lder inequalities, we have
{\small\begin{equation}\label{BBN3}
\begin{split}
\sum_{m=0}^\infty 2\frac{b_m}{(\frac{2m+1}{2}\pi)^8}\sin \Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]&
\Int_0^t\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2(t-s)\Big]g_m(s)\,ds
\\&\leq C(T)\sum_{m=0}^\infty \frac{b_m}{(\frac{2m+1}{2}\pi)^8}\Big(\Int_0^T g^2_m(t)\,dt\Big)^{\frac{1}{2}}
\\&\leq
 c_2 \sum_{m=0}^\infty\frac{b^2_m}{(\frac{2m+1}{2}\pi)^{16}} +
  c'_2 \sum_{m=0}^\infty \Int_0^T g^2_m(t)\,dt
  \\& \leq
 c_2 \sum_{m=0}^\infty\frac{b^2_m}{(\frac{2m+1}{2}\pi)^8} +
  C(T)\sum_{m=0}^\infty \Int_0^T g^2_m(t)\,dt,
\end{split}
\end{equation}}
and
{\small\begin{equation}\label{BBN4}
\begin{split}
\sum_{m=0}^\infty \frac{2a_m}{(\frac{2m+1}{2}\pi)^4}\cos \Big[\Big(\frac{2m+1}{2}\pi\Big)^2t\Big]&\Int_0^t
\sin\Big[\Big(\frac{2m+1}{2}\pi\Big)^2(t-s)\Big]g_m(s)\,ds\\&\leq
 c_3 \sum_{m=0}^\infty a_m^2 +
  C(T)\sum_{m=0}^\infty \Int_0^T g^2_m(t)\,dt.
\end{split}
\end{equation}}
Integrating (\ref{BBN}) in $(0,T)$ and using (\ref{BBN1})-(\ref{BBN4}), we obtain from (\ref{BBN}) that
{\small\begin{equation*}
\begin{split}
\Int_0^1 \Int_0^Tu^2(x,t)\,dt\,dx&\leq \frac{C(T)}{2}\sum_{m=0}^\infty\left\{a_m^2+\frac{b_m^2}{(\frac{2m+1}{2}\pi)^8}
+\Int_0^Tg^2_m(t)\,dt\right\}.
\end{split}
\end{equation*}}
Then
{\small\begin{equation*}
\begin{split}
\Int_\xi^{\xi+\frac{1}{n}}\Int_0^Tu^2(x,t)\,dt\,dx \leq \frac{C(T)}{n}\Big(\|u^0\|^2_{L^2(0,1)} + \|u^1\|^2_{V'}
+\|g_n\|^2_{L^2(0,T,L^2(0,1))}\Big).
\end{split}
\end{equation*}}
This completes the proof of proposition \ref{pro}.\end{proof}
\section{Internal exact controllability of the beams
equation} We consider now the following homogenous problem
\begin{equation}\label{p4o}
\Frac{\partial ^2 \phi}{\partial
t^2}(x,t)+\Frac{\partial^{4}\phi}{\partial x^4}(x,t)= 0,\,\,\,
0<x<1,\,\,\,\ t>0,
\end{equation}
\begin{equation}\label{p5o}
\phi(0,t)= \Frac{\partial \phi}{\partial x}(1,t)= \Frac{\partial^2
\phi}{\partial x^2}(0,t)= \Frac{\partial^3 \phi}{\partial
x^3}(1,t)=0,
\end{equation}
\begin{equation}\label{p6o}
\phi(x,0)=\phi^{0}(x),\,\,\,\,\ \Frac{\partial \phi}{\partial
t}(x,0)=\phi^{1}(x),\,\,\ 0<x<1.\end{equation} where $(\phi^0,
\phi^1) \in L^2(0,1) \times V'$.
\begin{lemma}\label{lem2}
Let $\xi \in (0,1)$, then for any natural integer $m$ we have
{\small\begin{equation}\label{tu111}
\begin{split} \Inf_{m\geq 0} \Int_{\xi}^{\xi
+\frac{1}{n}}\sin^2\left(\frac{2m+1}{2}\pi x\right)\,dx &\geq
\frac{1}{2n} - \Frac{1}{\pi}\sin\Big( \frac{\pi}{2n}\Big)
\\& \geq c_{\pi}o\Big(\frac{1}{n^3}\Big).
\end{split}
\end{equation}}
\end{lemma}
\begin{proof} For $m \geq 0$, it is sufficient to note that for
any $x\in(0,1)$, we have {\small\begin{equation*}
\begin{split} \Int_{\xi}^{\xi
+\frac{1}{n}}\sin^2\left(\frac{2m+1}{2}\pi x\right)\,dx &=
\frac{1}{2m+1}\Sum_{k=0}^{2m}\Int_{(2m+1)\xi+\frac{k}{n}}^{(2m+1)\xi+\frac{k+1}{n}}\sin^2
\Big(\frac{\pi}{2}y\Big)\,dy \\& = \frac{1}{2m+1}\Int_{(2m+1)\xi}
^{(2m+1)(\xi+\frac{1}{n})}\sin^2\Big(\frac{\pi}{2}y\Big)\,dy\\& = \frac{1}{2m+1}\Int_{(2m+1)\xi}
^{(2m+1)(\xi+\frac{1}{n})}\Big(\Frac{1}{2}-\Frac{1}{2}\cos(\pi y)\Big)\,dy
\\& = \frac{1}{2n}-
\frac{1}{\pi(2m+1)}\sin\Big[\frac{\pi}{2n}(2m+1)\Big]\cos\Big[\Big(\xi
+\frac{1}{2n}\Big) \Big(2m+1\Big)\Big],
\end{split}
\end{equation*}}
it is clear that
$$
\sin\Big( \frac{\pi}{2n}(2m+1)\Big)= \frac{\pi}{2n}(2m+1) +c_{\pi,m}
o\Big(\frac{1}{n^3}\Big).
$$
Therefore, we get {\small\begin{equation*}
\begin{split}\Int_{\xi}^{\xi
+\frac{1}{n}}\sin^2\left(\frac{2m+1}{2}\pi x\right)\,dx & \geq
\frac{1}{2n} - \Frac{1}{\pi}\Big[\frac{\pi}{2n}
+o\Big(\frac{1}{n^3}\Big)\Big] \\& \geq
c_{\pi}o\Big(\frac{1}{n^3}\Big)
 .\end{split}
\end{equation*}}
The proof of lemma \ref{lem2} is now completed.
\end{proof}
The previous lemma is an essential tool  to show the following
Proposition.
\begin{proposition}\label{pro2}
Let $T\geq 2$, then we have the following. For almost all $\xi \in
(0,1)$ the solution $\phi$ of (\ref{p4o})-(\ref{p6o}) satisfies
\begin{equation}\label{obser1}
\Int_0^T\Int_\xi ^{\xi +\frac{1}{n}}\Big|\phi(x,t)
\Big|^2\,dx\,dt\geq
 c_{\pi,n}(\|u^0\|^2_{L^2(0,1)} + \|u^1\|^2_{V'}),
\end{equation}
$$
\forall (u^0,u^1) \in  L^2(0,1)\times V',
$$
where $c_{\pi,n} = c_{\pi}o(\frac{1}{n^3})$.\\
\end{proposition}
\begin{proof}
 The solution of (\ref{p4o})-(\ref{p6o})  is given by
$$
\phi(x,t)=
\sum_{m=0}^{\infty}\left\{a_m\cos\Big[\left(\frac{2m+1}{2}\pi\right)^2
t\Big]+
\frac{b_m}{\left(\frac{2m+1}{2}\pi\right)^4}\sin\Big[\left(\frac{2m+1}{2}\pi\right)^2
t\Big]\right\}\sin\left(\frac{2m+1}{2}\pi x\right).
$$
A simple calculation shows that
 {\small\begin{equation}\label{amma}
 \begin{split} \Int_0^2\Big|\phi(x,t)\Big|^2\,dt
 =\Sum_{m=0}^{\infty}\left\{a_m^2+\frac{b_m^2}{\left(\frac{2m+1}{2}\pi\right)^8}\right\}
 \sin^2\left(\frac{2m+1}{2}\pi
x\right),
\end{split}
\end{equation}}
we have {\small\begin{equation*}
\begin{split}\Int_{\xi}^{\xi +\frac{1}{n}} \Int_0^2\Big|\phi(x,t)\Big|^2\,dx
\,dt \geq
\Sum_{m=0}^{\infty}\left\{a_m^2+\frac{b_m^2}{\left(\frac{2m+1}{2}\pi\right)^8}\right\}\Inf_{m\geq 0} \Int_{\xi}^{\xi
+\frac{1}{n}}\sin^2\left(\frac{2m+1}{2}\pi x\right)\,dx,
\end{split}
\end{equation*}}
for every $T \geq 2$, we get
\begin{equation}\label{tu11}
\Int_{\xi}^{\xi +\frac{1}{n}} \Int_0^T\Big|\phi(x,t)\Big|^2\,dx \,dt
\geq
\Sum_{m=0}^{\infty}\left\{a_m^2+\left(\frac{b_m}{\left(\frac{2m+1}{2}\pi\right)^4}\right)^2\right\}\Inf_{m\geq
0} \Int_{\xi}^{\xi +\frac{1}{n}}\sin^2\left(\frac{2m+1}{2}\pi
x\right)\,dx.
\end{equation}
 Consequently, by  lemma \ref{lem2} and  using (\ref{tu111}) and
(\ref{tu11}), we obtain (\ref{obser1}).\\
 This achieve the proof of proposition \ref{pro2}.
\end{proof}
Let $(y^0,y^1)\in
 V \times L^2(0,1)$ and  $\psi_n$ be the solution of
\begin{equation}\label{23}
\Frac{\partial ^2 \psi_n}{\partial
t^2}(x,t)+\Frac{\partial^{4}\psi_n}{\partial x^4}(x,t)=
\chi_n(x)\widetilde{\phi}_n(x,t) ,\,\,\,\ 0<x<1,\,\,\,\,\ t>0,
\end{equation}
\begin{equation}\label{p5oo}
\psi_n(0,t)= \Frac{\partial \psi_n}{\partial x}(1,t)= \Frac{\partial^2
\psi_n}{\partial x^2}(0,t)= \Frac{\partial^3 \psi_n}{\partial x^3}(1,t)=0,
\end{equation}
\begin{equation}\label{p6oo}
\psi_n(x,0)=y^{0}(x),\,\,\,\,\ \Frac{\partial \psi_n}{\partial
t}(x,0)=y^{1}(x),\,\,\ 0<x<1,
\end{equation}
\begin{equation}\label{p7oo}
\psi_n(x,T)=\frac{\partial\psi_n}{\partial t}(x,T)=0,
\end{equation}
where $\chi_n$ is the characteristic function of
$(\xi,\xi+\frac{1}{n})$ and  $\widetilde{\phi}_n(x,t)= n\phi(x,t)$,
$\phi$ is the solution of (\ref{p4o})-(\ref{p6o}).
\begin{lemma}{\bf\cite{HA1}}\label{mo}
We suppose that $(y^0,y^1) \in V \times L^2(0,1)$, then
$$
\Big(y^0,
\widetilde{\phi}_n^1\Big)-\Big(y^1,\widetilde{\phi}_n^0\Big)=\Int_\xi^{\xi+\frac{1}{n}}\Int_0^T
\Big|\widetilde{\phi}_n(x,t)\Big|^2\,dx\,dt.
$$
\end{lemma}
\section{An inverse inequality}
 In this section we suppose  that the point  $\xi$ is strategic, that's
\begin{equation}\label{H}
  \sin\Big(\frac{2m+1}{2}\pi \xi\Big)\neq 0,\quad\forall\,m\in\mathbb{N}.
  \end{equation}
   The quantity
$$
\Big(\Int_0^T \phi^2(\xi,t)\,dt\Big)^{(1/2)}
$$
where $\phi$ is a solution of (\ref{p4o})-(\ref{p6o}) defines a norm
on the space ${\mathcal D}(0,1)\times {\mathcal D}(0,1)$ and the
initial data $\phi^0$ and $\phi^1$  are given by
$$
\phi^0= \Sum_{m=0}^{\infty} a_m \sin\Big(\frac{2m+1}{2}\pi
x\Big),\,\,\,\,\, \phi^1= \Sum_{m=0}^{\infty}
\frac{b_m}{(\frac{2m+1}{2}\pi)^2}\sin\Big(\frac{2m+1}{2}\pi x\Big).
$$
Let $F$ be a real Hilbert space
$$
(\phi^0,\phi^1)\in F\Leftrightarrow
\Sum_{m=0}^{\infty}\Big\{a_m^2+\frac{b_m^2}{(\frac{2m+1}{2}\pi)^8}\Big\}\sin^2\Big(\frac{2m+1}{2}\pi
\xi\Big) <\infty.
$$
We denote by $F$ the completion of ${\mathcal D}(0,1)\times {\mathcal D}(0,1)$ for this norm and we denote
 by $\|.\|_F$ the following quantity:
$$
\|\phi\|_{F}=\Big(\Int_0^T \phi^2(\xi,t)\,dt\Big)^{1/2}.
$$
Therefore
$$
L^2(0,1)\times V'\subset F.
$$
If
$$
y^0(x)=\Sum_{m=0}^{\infty}a_m\sin\Big(\frac{2m+1}{2}\pi x\Big),\,\,\,\,\,\,\,\,\
y^1(x)=\Sum_{m=0}^{\infty}\frac{b_m}{(\frac{2m+1}{2}\pi)^2}\sin\Big(\frac{2m+1}{2}\pi x\Big),
$$
 and therefore, its dual
$$
(y^0,y^1)\in F'\Leftrightarrow \Sum_{m=0}^{\infty}\frac{a_m^2+\frac{b_m^2}{(\frac{2m+1}{2}\pi)^4}}{\sin^2\Big(\frac{2m+1}{2}\pi \xi\Big)}
<\infty.
$$
\begin{remark}
If $\xi\in (0,1)$ satisfying (\ref{H}),  then there exists a
constant $C> 0$ such that $|\sin\Big(\frac{2m+1}{2}\pi\xi\Big)|\geq
C,\,\,\, \forall m\in \N$
 and therefore  $L^2(0,1)\times V'= F$.\\
  For the proof, see {\bf\cite{AmTu2}},{\bf\cite{AmTu1}}.
\end{remark}
 The main result of this section is the following:
\begin{theorem}
For $T\geq 2$, there exists $c> 0$, such that for $(\phi^0,\phi^1)
\in F$, the solution $\phi$ of (\ref{p4o})-(\ref{p6o}) satisfies
\begin{equation}\label{ZZ}
\|(\phi^0,\phi^1)\|_F^2 \leq c \Big(n
\Int_0^T\Int_\xi^{\xi+\frac{1}{n}} \phi^2(x,t)\,dx\,dt\Big).
\end{equation}
\end{theorem}
\begin{proof}
For $T\geq 2$. Using (\ref{amma}), we have
{\small\begin{equation*}
\begin{split}
\Int_0^T\Int_\xi^{\xi+\frac{1}{n}}\phi^2(x,t)\,dx\,dt&\geq
\Int_0^2\Int_\xi^{\xi+\frac{1}{n}}\phi^2(x,t)\,dx\,dt\\&=
\sum_{m=0}^{\infty}\left\{a_m^2+\left(\frac{b_m}{\left(\frac{2m+1}{2}\pi\right)^4}\right)^2\right\}
\Int_\xi^{\xi+\frac{1}{n}}\sin^2\left(\frac{2m+1}{2}\pi
x\right)\,dx.
\end{split}
\end{equation*}}
Now, we have to prove that there exists $c> 0$ independent on $n$
such that for every integer $m\in \N$, we have
$$
n\Int_\xi^{\xi+\frac{1}{n}}\sin^2\left(\frac{2m+1}{2}\pi x\right)\,dx\geq
 c \sin^2\left(\frac{2m+1}{2}\pi \xi\right).
$$
For $b\geq 0,\,\, t\geq 0$, we set
$$
I(b,t)= \Int_0^1\sin^2(\pi(b+tz))\,dz.
$$
As
$$
n\Int_\xi^{\xi+\frac{1}{n}}\sin^2\left(\frac{2m+1}{2}\pi x\right)\,dx =
 I\Big(\frac{2m+1}{2}\xi,\frac{2m+1}{2n}\Big),
$$
it is sufficient to prove that there exists $c> 0$ such that
\begin{equation}\label{tt}
\forall t\geq 0,\,\, I(b,t) \geq c\sin^2(\pi b).
\end{equation}
We have the formula
{\small\begin{equation*}
\begin{split}
\forall t\geq 0,\,\, I(b,t)&= \Frac{1}{2}\Big(1-\Frac{\sin (2\pi (b+t))-\sin (2\pi b)}{2\pi t}\Big)\\&=
\Frac{1}{2}\Big(1-\Frac{\sin (2\pi b)[\cos (2\pi t)-1] +\sin (2\pi t)\cos (2\pi b)}{2\pi t}\Big)
\\&=
\Frac{1}{2}\Big(1-\Frac{-2\sin (2\pi b)\sin^2(\pi t) +2\cos (\pi t)\sin (\pi t)\cos (2\pi b)}{2\pi t}\Big)\\&=
\frac{1}{2}\left(1-\frac{\sin (\pi t)}{\pi t}\cos(\pi(2b+t))\right)
\end{split}
\end{equation*}}
If $t \geq \frac{1}{2}$, then $I(b,t)\geq \frac{1}{2}(1-\frac{2}{\pi})$\\
If $t< \frac{1}{2}$, we distinguish two cases:\\
{\bf{Case 1.}} $]b,b+t[\subset ]p,(p+1)[$. It is then enough to
consider the case $p=0$ and as $\sin(\pi .)$ is concave on $[0,1]$, we
obtain
{\small\begin{equation*}
\begin{split}
\forall z \in (0,1),\,\,\, \sin ((1-z)b\pi + (b+t)z\pi)&=\sin ((1-z)\pi b+ z\pi(t+b))\\&\geq (1-z)\sin(\pi b)+z\sin (\pi(t+b))\\&\geq
(1-z)\sin(\pi b).
\end{split}
\end{equation*}}
Then
$$
\forall z \in (0,1),\,\,\, |\sin(\pi( b+tz))|\geq (1-z)|\sin (\pi b)|.
$$
Hence
\begin{equation*}
 \begin{split}
I(b,t)&\geq \sin^2 (\pi b)\Int_0^1(1-z)^2\,dz\\&
\geq \frac{1}{3}\sin^2 (\pi b).
\end{split}
\end{equation*}
{\bf{Case 2.}} $p-1\leq b\leq p\leq b+t \leq p+1$. It is enough
here to consider the case $p=1$ and we write $1= b+z_0t$ writh
$z_0 \in (0,1)$. We have
$$
|\sin (\pi (b+tz))|\geq (1-z)|\sin (\pi b)|\,\,\,\,\mbox{for}\,\,\,\, z\leq z_0,
$$
and
$$
|\sin (\pi (b+tz))|\geq z|\sin (\pi( b+t))|\,\,\,\,\mbox{for}\,\,\,\, z>z_0.
$$
Now, if $z_0 \geq \frac{1}{2}$, we find
\begin{equation*}
 \begin{split}
I(b,t)&\geq \sin^2 (\pi b) \Int_0^{z_0} (1-z)^2\,dz+\sin^2 (\pi b) \Int_{z_0}^1 (1-z)^2\,dz\\& \geq \sin^2 (\pi b) \Int_0^{z_0} (1-z)^2\,dz \\& \geq \frac{7}{24}\sin^2(\pi b).
\end{split}
\end{equation*}
If $z_0< \frac{1}{2}$, then
{\small\begin{equation*}
\begin{split}b+t-1= b+t-(b+z_0t)& = 1-2z_0t +t -b\\&= 1-b +t(1-2z_0)\\&> 1-b
\end{split}
\end{equation*}}
and
$$
\sin(\pi(b+t-1))= -\sin (\pi (b+t))
$$
 and we have
\begin{equation*}
 \begin{split}
I(b,t)&\geq\sin^2(\pi(b+t-1))\Int_0^{z_0} z^2\,dz
+\sin^2(\pi(b+t-1))\Int_{z_0}^1 z^2\,dz
\\&\geq \sin^2(\pi(b+t-1))\Int_{z_0}^1 z^2\,dz
 \\&\geq  \frac{1}{3}(1-z_0^3)\sin^2(\pi(1-b))\\&\geq \frac{7}{24}\sin^2(\pi b).
\end{split}
\end{equation*}
The proof of the theorem \ref{pro2} is complete.
\end{proof}
\section{Estimates on the controls}
For $T\geq 2$ and $\frac{1}{n}\widetilde{\phi}_n(x,t) =\phi(x,t)$
where $\phi$ is the solution of (\ref{p4o})-(\ref{p6o}), we have
\begin{theorem}\label{RR}
1. If $(y^0,y^1)\in V\times L^2(0,1)$, we have
\begin{equation}\label{a}
\|\widetilde{\phi}_n^0\|_{L^2(0,1)}+\|\widetilde{\phi}_n^1\|_{V'}=o(n^3),
\end{equation}
and
\begin{equation}\label{aa}
\Int_\xi^{\xi+\frac{1}{n}}\Int_0^T\Big
|\widetilde{\phi}_n(x,t)\Big|^2\,dx\,dt=o(n^3).
\end{equation}
2. If $\xi$ is strategic and $(y^0,y^1)\in F'$, we have
\begin{equation}\label{aaa}
\|(\widetilde{\phi}_n^0,\widetilde{\phi}_n^1)\|_{F}=o(n),
\end{equation}
and
\begin{equation}\label{aaaa}
\Int_\xi^{\xi+\frac{1}{n}}\Int_0^T\Big
|\widetilde{\phi}_n(x,t)\Big|^2\,dx\,dt=o(n).
\end{equation}
\end{theorem}
\begin{proof} 1. Applying H\"{o}lder and Young's inequalities. Hence, we see from (\ref{obser1}) and using lemma \ref{mo}, we have
 {\small\begin{equation}\label{oo}
 \begin{split}
\Big(\|\widetilde{\phi}_n^0\|_{L^2(0,1)}+\|\widetilde{\phi}_n^1\|_{V'}\Big)^2&\leq
c\Big(\|\widetilde{\phi}_n^0\|^2_{L^2(0,1)}+\|\widetilde{\phi}_n^1\|^2_{V'}\Big)\\&\leq
c n^2\Big(\|\phi^0\|^2_{L^2(0,1)}+\|\phi^1\|^2_{V'}\Big)
\\&\leq
c n^3 \Int_\xi^{\xi+\frac{1}{n}}\Int_0^T\Big
|\widetilde{\phi}_n(x,t)\Big|^2\,dx\,dt
\\& \leq c n^3
(\|\widetilde{\phi}_n^0\|_{L^2(0,1)}\|y^1\|_{L^2(0,1)}+\|\widetilde{\phi}_n^1\|_{V'}\|y^0\|_{V})
\\& \leq c n^3
(\|\widetilde{\phi}_n^0\|_{L^2(0,1)}+\|\widetilde{\phi}_n^1\|_{V'})
\end{split}
\end{equation}}
2. When the point $\xi$ is strategic and the initial data $(y^0,y^1)\in F'$.
Hence, we see from (\ref{ZZ}), that
 {\small\begin{equation*}
 \begin{split}
\|(\widetilde{\phi}_n^0,\widetilde{\phi}_n^1)\|^2_{F}&\leq c n
\Int_\xi^{\xi+\frac{1}{n}}\Int_0^T\Big
|\widetilde{\phi}_n(x,t)\Big|^2\,dx\,dt\\& \leq c n
\|(\widetilde{\phi}_n^0,
\widetilde{\phi}_n^1)\|_F\|(y^0,y^1)\|_{F'}\\& \leq c n
\|(\widetilde{\phi}_n^0, \widetilde{\phi}_n^1)\|_F.
\end{split}
\end{equation*}}
The proof of Theorem \ref{RR} is now complete.
\end{proof}
\section{Controllability limit as $n\rightarrow \infty$}
We study here the possibility of convergence of the solution of the
controllability  problems defined by (\ref{p5oo})-(\ref{p7oo}).
This convergence depends on the nature of the point $\xi$ and on
the space of the initial data $y^0$ and $y^1$.\\
If the point (\ref{H}) holds, we have the following theorem.
\begin{theorem}
Suppose that $T\geq 2$, if $\xi$ checks (\ref{H}), $y^0$ and $y^1$
belong to $F'$.\\
 Then, the solution of (\ref{p5oo})-(\ref{p7oo})
converges for the weak* topology of $L^{\infty}(0,T,V)$ to the
solution of the following pointwise system
\begin{equation}\label{p44oo}
\Frac{\partial ^2 \psi}{\partial
t^2}(x,t)+\Frac{\partial^{4}\psi}{\partial x^4}(x,t)=
v(t)\delta_\xi,\,\,\, 0<x<1,\,\,\,\ t>0,
\end{equation}
\begin{equation}\label{p55oo}
\psi(0,t)= \Frac{\partial \psi}{\partial x}(1,t)= \Frac{\partial^2
\psi}{\partial x^2}(0,t)= \Frac{\partial^3 \psi}{\partial x^3}(1,t)=0,
\end{equation}
\begin{equation}\label{p66oo}
\psi(x,0)=y^{0}(x),\,\,\,\,\ \Frac{\partial \psi}{\partial
t}(x,0)=y^{1}(x),\,\,\ 0<x<1,\end{equation}
\begin{equation}\label{p77oo}
\psi(x,T)=\frac{\partial \psi}{\partial t}(x,T)=0,\,\,\ 0<x<1,
\end{equation}
where $v\in L^2(0,T)$ and $ \phi(\xi,t)+\frac{1}{2n} \frac{\partial
\phi}{\partial x}(\xi,t)$ converges for the weak* topology of to
$v(t)$ in $H^{-1}(0,T)$.
\end{theorem}
\begin{proof} Multiplying (\ref{p1}) by $\psi_n(x,t)$ and
integrating by parts on  $(0,T)\times(0,1)$, we have\\

$
\forall (u^0,u^1,g_n) \in L^2(0,1)\times V'\times L^2(0,T,L^2(0,1)),
$
{\small\begin{equation}\label{AQZ}
 \begin{split}
&\Int_0^T\Int_0^{1}g_n(x,t)\psi_n(x,t)\,dx\,dt\\&=
\Int_0^T\Int_\xi^{\xi+\frac{1}{n}}\widetilde{\phi}_n(x,t)u(x,t)\,dx\,dt-
\Int_0^1y^0(x)u^1(x)\,dx +\Int_0^1 y^1(x)u^0(x)\,dx
\\&=n
\Int_0^T\Int_\xi^{\xi+\frac{1}{n}}\phi(x,t)u(x,t)\,dx\,dt-
\Int_0^1y^0(x)u^1(x)\,dx +\Int_0^1 y^1(x)u^0(x)\,dx.
\end{split}
\end{equation}}
Now, we prove that $(\psi_n)$  and $(g_n)$ are  bounded in
$L^{\infty}(0,T,L^2(0,1))$.\\Define
$$
\begin{array}{ll}
K_n: L^2(0,1)\times V'\times L^2(0,T,L^2(0,1))\rightarrow \R\\
\hskip4.1cm (u^0,u^1,g_n)\mapsto n
\Int_0^T\Int_\xi^{\xi+\frac{1}{n}}\phi(x,t)u(x,t)\,dx\,dt.
\end{array}
$$
Using H\"{o}lder inequality, we have
\begin{equation}\label{A}
|K_n(u^0,u^1,g_n)|^2 \leq \Big(n
\Int_0^T\Int_\xi^{\xi+\frac{1}{n}}\Big|\phi(x,t)\Big|^2\,dx\,dt\Big)\Big(
n\Int_0^T\Int_\xi^{\xi+\frac{1}{n}}|u(x,t)|^2\,dx\,dt\Big).
\end{equation}
Replacing (\ref{mama}) in (\ref{A}) and from (\ref{aaaa}), we have
$$
|K_n(u^0,u^1,g_n)|^2 \leq c\Big(\|u^0\|^2_{L^2(0,1)}+\|u^1\|^2_{V'} + \|g_n\|^2_{L^{2}(0,T,L^2(0,1))}
 \Big),
$$
which proves that the linear forms $K_n$ are bounded in $L^2(0,1) \times V' \times L^\infty(0,T,L^2(0,1))$.\\
Therefore, $(\psi_n)_n$ and $(g_n)_n$ are bounded in $L^\infty(0,T,L^2(0,1))$
after extraction of a subsequence of $(\psi_n)_n$ and $(g_n)_n$ still denoted by
$(\psi_n)_n$ and $(g_n)_n$, such that
$$
\psi_n \rightharpoonup \psi\,\,\,\mbox{ weakly* in }\,\,\,L^\infty(0,T,L^2(0,1)),
$$
and
$$
g_n \rightharpoonup g\,\,\,\mbox{ weakly* in }\,\,\,L^\infty(0,T,L^2(0,1)).
$$
The limit of $K_n$ is given  in the following lemma.
\begin{lemma}\label{50}
 The linear forms $K_n$ converge in $ L^2(0,1)\times V \times
L^\infty(0,T,L^2(0,1))$ weakly* to the $K$ defined by
\begin{equation}\label{sv}
K(u^0,u^1,g)=\Int_0^T v(t)u(\xi,t)\,dt,
\end{equation}
where $v\in L^2(0,T)$ and
$$
 \phi(\xi,t)+\frac{1}{2n}
\Frac{\partial \phi}{\partial x}(\xi,t)\rightharpoonup
 v(t)\,\,\,\mbox{weakly* in}\,\,\,\
 H^{-1}(0,T).
$$
\end{lemma}
In order to prove the previous lemma, we  need  the following
result.
\begin{lemma}\label{mm1}
Let $(\phi^0,\phi^1)\in L^2(0,1)\times V'$ and the solution
$\phi(x,t)$ of the problem (\ref{p4o})-(\ref{p6o}) satisfies
\begin{equation}\label{R}
\Int_0^T\Int_\xi^{\xi+\frac{1}{n}} |\phi(x,t)|^2\,dx\,dt =
o\Big(\frac{1}{n}\Big).
\end{equation}
Then, after extraction of a subsequence
$$
\phi(\xi,t)+\frac{1}{2n} \frac{\partial \phi}{\partial x}(\xi,t)
\rightharpoonup  v(t)\,\,\,\mbox{weakly* in}\,\,\,\
  H^{-1}(0,T),
$$ where  $v\in L^{2}(0,T)$.
\end{lemma}
{\bf{Proof of lemma \ref{mm1}.}} In order to prove  lemma \ref{mm1}
we suppose that $w = \Big(\frac{\partial^4}{\partial
x^4}\Big)^{-1}u$ \\
such that $w$ is the solution of
\begin{equation}\label{pp1}
\Frac{\partial ^2 w}{\partial t^2}(x,t)+\Frac{\partial^{4}w}{\partial x^4}(x,t)=
f_n(x,t),\,\,\, 0<x<1,\,\,\,\ t>0,
\end{equation}
\begin{equation}\label{pp2}
w(0,t)= \Frac{\partial w}{\partial x}(1,t)= \Frac{\partial^2
w}{\partial x^2}(0,t)= \Frac{\partial^3 w}{\partial x^3}(1,t)=0,
\end{equation}
\begin{equation}\label{pp3}
w(x,0)=w^{0}(x),\,\,\,\,\ \Frac{\partial w}{\partial
t}(x,0)=w^{1}(x),\,\,\ 0<x<1.
\end{equation}
with initial data
\begin{equation}\label{ggg}
 \left\{
\begin{array}{ll}
w^0=\Big(\frac{\partial^4}{\partial x^4}\Big)^{-1}u^0
\in {\mathcal D}(\partial_x^4)\\
w^1=\Big(\frac{\partial^4}{\partial x^4}\Big)^{-1}u^1\in V\\
f_n=\Big(\frac{\partial^4}{\partial x^4}\Big)^{-1}g_n \in L^2(0,T;{\mathcal D}(\partial_x^4)).
\end{array} \right.
\end{equation}
The trace regularity for (\ref{pp1})-(\ref{pp3}) is given in the
theorem below.
\begin{theorem}\label{h}
 Suppose that $f_n\in L^2(0,T;{\mathcal D}(\partial_x^4))$ and
$(w^0,w^1)\in {\mathcal D}(\partial_x^4)\times V$ the solution $w$
of (\ref{pp1})-(\ref{pp3})
 verifies
\begin{equation}\label{gg}
\Frac{\partial^4 w}{\partial x^4}(\xi,t)\in L^2(0,T),
\end{equation}
and the mapping
\begin{equation}\label{ggg}
\begin{array}{ll}
L^2(0,T;{\mathcal D}(\partial_x^4))\times {\mathcal
D}(\partial_x^4)\times V\rightarrow L^2(0,T)\\ \hskip2.75cm
(f_n,w^0,w^1)\mapsto \Frac{\partial^4 w}{\partial x^4}(\xi,t),
\end{array}
\end{equation}
is linear and continuous.\\
 Furthermore, we have
\begin{equation}\label{mamo}
n\Int_{\xi}^{\xi+\frac{1}{n}}\Int_0^T
\Big|\Frac{\partial^4 w}{\partial x^4}(x,t)\Big|^2\,dx\,dt
 \leq
C\Big(\|f_n\|^2_{L^2(0,T;{\mathcal D}(\partial_x^4))} +
\|w^0\|^2_{{\mathcal D}(\partial_x^4)} + \|w^1\|^2_{V}\Big).
\end{equation}
\end{theorem}
{\bf{Proof of Theorem \ref{h}.}} The proof of (\ref{gg}) and  (\ref{mamo}) can be done by using obvious adaptations of the
proof of (\ref{mi5}) and (\ref{mama}), so it is omitted.\df
 From (\ref{ZZ}) and (\ref{R})
it follows that $\phi(\xi,t)$ is bounded in $F$, after extraction of
a subsequence, $\phi(\xi,t)$ converges in
$L^2(0,T)$ weakly.\\
On the other hand, from (\ref{obser1}), (\ref{aaa}) and
(\ref{aaaa}) we have
 $$\|\phi^0\|_{L^2(0,1)} +
\|\phi^1\|_{V'}=o(n).
$$
Using (\ref{gg}) and (\ref{ggg}) we can easily prove that the mapping
$$
(\phi^0,\phi^1)\in  L^2(0,1)\times V'\rightarrow \Frac{\partial
\phi}{\partial x}(\xi,t) \in H^{-1}(0,T),
$$
is linear and  continuous.\\
Furthermore, we have
$$
\Big\|\Frac{\partial \phi}{\partial x}(\xi,t)\Big\|_{H^{-1}(0,T)} =
o(n).
$$
Now, we prove that $v\in L^{2}(0,T)$ that is
$$
\forall u \in {\mathcal D}(0,T),\,\,\,\,\, |(v,u)|_{{\mathcal
D'}\times {\mathcal D}}\leq c\|u\|_{L^2(0,T)}.
$$
We define the following functions
$$
\Phi(x,t)=\Int_0^t\phi(x,\tau)\,d\tau-\Big(\frac{\partial^4}{\partial
x^4}\Big)^{-1}\phi^1(x),
$$
and
$$
S_n(x,t)= \Int_0^t
\Phi(x,\tau)\,d\tau-\Big(\frac{\partial^4}{\partial
x^4}\Big)^{-1}\phi^0(x).
$$
The functions $\Phi$ and  $S_n$ are solutions of
(\ref{p4o})-(\ref{p6o}) with initial data in $V\times L^2(0,1)$ and
${{\mathcal D}(\partial_x^4)} \times V$
$$
\|\Phi^0\|_{V}+\|\Phi^1\|_{L^2(0,1)}  = o(n),
$$
and
$$
\|S_n^0\|_{{{\mathcal D}(\partial_x^4)} }+\|S_n^1\|_{V} = o(n).
$$
For $u\in {\mathcal D}(0,T)$, we have
{\small\begin{equation*}
 \begin{split}
n\Int_0^T\Int_\xi^{\xi + \frac{1}{n}}\phi(x,t)u(t)\,dx\,dt&=
n\Int_0^T\Int_\xi^{\xi + \frac{1}{n}}S_n(x,t)\frac{\partial^2 u}{
\partial t^2}(t)\,dx\\&= \Int_0^T\Big(S_n(\xi,t)+\frac{1}{2n}
\frac{\partial S_n}{\partial
x}(\xi,t)\Big)\frac{\partial^2u}{\partial t^2}(t)\,dt\\&
+n\Int_0^T\Int_\xi^{\xi+\frac{1}{n}}\frac{\partial^2 u} {\partial
t^2}(t)\Int_\xi^x\Int_\xi^y \frac{\partial^2 S_n}{\partial
z^2}(z,t)\,dz\,dy\,dx\,dt.
\end{split}
\end{equation*}}
Then
\begin{equation}\label{bb}
\Big(S_n(\xi,t)+\frac{1}{2n} \frac{\partial S_n}{\partial
x}(\xi,t),\frac{\partial^2 u}{\partial t^2}(t)\Big)_{{\mathcal
D}',{\mathcal D}}=n\Int_0^T\Int_\xi^{\xi+\frac{1}{n}}
\phi(x,t)u(t)\,dx\,dt-R_n,
\end{equation}
where
$$
R_n=n\Int_0^T\Int_\xi^{\xi+\frac{1}{n}}\frac{\partial^2
u}{\partial t^2}(t)\Int_\xi^x\Int_\xi^y \frac{\partial^2
S_n}{\partial z^2}(z,t)\,dz\,dy\,dx\,dt,
$$
then, we prove that
$$
\Lim_{n\rightarrow \infty}R_n =0.
$$
Using H\"{o}lder's inequality, we have
{\small\begin{equation*}
 \begin{split}
 |R_n|&\leq n\Big\|\frac{\partial^2 u}{\partial t^2}\Big\|_{L^2(0,T)}
 \Big[\Int_0^T\frac{1}{n}\Int_\xi^{\xi+\frac{1}{n}}(x-\xi)
\Int_\xi^x(y-\xi)\Int_\xi^y\Big|\frac{\partial^2 S_n}{\partial
z^2}(z,t)\Big|^2\,dz\,dy\,dx\,dt\Big]^{1/2}\\&
\leq\Frac{1}{\sqrt{8}\sqrt{n}}\Big\|\frac{\partial^2 u}{\partial
t^2}\Big\|_{L^2(0,T)}\|(S_n^0, S_n^1)\|_{{\mathcal
D}(\partial_x^4)\times V}.
\end{split}
\end{equation*}}
Thus
$$
\Lim_{n\rightarrow \infty}R_n =0.
$$
Integrating by part, we get
$$
\Big(S_n(\xi,t)+\frac{1}{2n} \frac{\partial S_n}{\partial
x}(\xi,t),\frac{\partial^2 u}{\partial t^2}(t)\Big)_{{\mathcal
D}',{\mathcal D}} = \Big(\phi(\xi,t)+\frac{1}{2n}\frac{\partial
\phi}{\partial x}(\xi,t),u(t)\Big)_{{\mathcal D}',{\mathcal D}}.
$$
Then
$$
\Big|\Big(\phi(\xi,t)+\frac{1}{2n}\frac{\partial \phi}{\partial
x}(\xi,t),u(t)\Big)_{{\mathcal D}',{\mathcal D}}\Big|\leq
c\|u\|_{L^2(0,T)}+|R_n|.
$$
Passing to the limit as $n$ tends to infinity, we obtain
$$
\Big|(v,u)\Big|\leq c\|u\|_{L^2(0,T)},
$$
which proves that $v$ belongs to $L^2(0,T)$. The proof of lemma
\ref{mm1} is now complete.
\end{proof}
{\bf{Proof of lemma \ref{50}}.}
Passing to the limit in (\ref{AQZ}), we have\\

$
\forall (u^0,u^1,g)\in L^2(0,1)\times V'\times L^2(0,T,L^2(0,1)),
$
\begin{equation*}
\Int_0^T\Int_0^1 g(x,t)\psi(x,t)\,dx\,dt=\Int_0^T v(t)u(\xi,t)\,dt
-\Int_0^1y^0(x)u^1(x)\,dx+\Int_0^1 y^1(x)u^0(x)\,dx,
\end{equation*}
where $u$ is the solution of (\ref{p1})-(\ref{p3}) and $\psi$ is the solution of (\ref{p44oo})-(\ref{p77oo}).\\
 Since the linear form $K$ defined in (\ref{sv}) is meaningful  on $L^2(0,1)\times V'\times L^2(0,T,L^2(0,1))$, it is sufficient to prove that $(K_n)_n$  converges to $K$ on a dense subspace of $L^2(0,1)\times V'\times L^2(0,T,L^2(0,1))$ and, for example,   we consider
 $(u^0,u^1,g_n)\in   L^2(0,1)\times V'\times L^2(0,T,L^2(0,1))$
 $$
 \begin{array}{ll}
 L_n: L^2(0,T, {\mathcal D}(\partial_x^4))\rightarrow \R\\
 \hskip3.50cm u \mapsto n\Int_0^T \Int_\xi^{\xi+
 \frac{1}{n}} \phi(x,t)u(x,t)\,dx\,dt,
 \end{array}
 $$
 are defined and bounded on $L^2(0,T,{\mathcal D}(\partial_x^4))$.\\
 They converge for the weak topology $L^2(0,T,({\mathcal D}(\partial_x^4))')$ to an element $L$ of
 $L^2(0,T,({\mathcal D}(\partial_x^4))')$. In order to determine $L$, we write for $u\in {\mathcal D}(0,T;C^\infty(0,1))$:
 {\small\begin{equation*}
\begin{split}
L_n(u)&= n\Int_0^T\Int_\xi^{\xi+
 \frac{1}{n}}\phi(x,t)u(\xi,t)\,dx\,dt
 \\&+n\Int_0^T \Int_\xi^{\xi+
 \frac{1}{n}}\phi(x,t)\Big(\Int_\xi^x
 \frac{\partial u}{\partial y}(y,t)\,dy\Big)\,dx\,dt.
\end{split}
\end{equation*}
We have already seen that
 $$
 \lim_{n\rightarrow \infty}n\Int_{\xi}^{\xi +
 \frac{1}{n}}\Int_0^T \phi(x,t)u(\xi,t)\,dx\,dt = \Int_0^T  v(t)u(\xi,t)\,dt.
 $$
On the other hand, it is easy to prove, using H\"{o}lder inequality, that for every $u_n\in {\mathcal D}(0,T;C^\infty(0,1))$, we have
$$
 \lim_{n\rightarrow \infty}n\Int_\xi^{\xi+
 \frac{1}{n}}\Int_0^T \phi(x,t)\Big(\Int_\xi^x
 \frac{\partial u}{\partial y}(y,t)\,dy\Big)\,dx\,dt = 0.
 $$
This completes the proof.\df
\begin{remark}

By the same method  we can obtain the pointwise controllability
of the  Kirchhoff beam equation
\begin{equation*}
\Frac{\partial ^2 u}{\partial
t^2}(x,t)-\Frac{\partial^{4}u}{\partial t^2\partial x^2}(x,t)+
\Frac{\partial^{4}u}{\partial x^4}(x,t)= v(t)\delta_{\xi},\,\,\,
0<x<1,\,\,\,\ t>0,
\end{equation*}
\begin{equation*}
u(0,t)= \Frac{\partial u}{\partial x}(1,t)= \Frac{\partial^2
u}{\partial x^2}(0,t)= \Frac{\partial^3 u}{\partial x^3}(1,t)=0,
\end{equation*}
\begin{equation*}
u(x,0)=u^{0}(x),\,\,\,\,\ \Frac{\partial u}{\partial
t}(x,0)=u^{1}(x),\,\,\ 0<x<1,\end{equation*}
as a limit of internal exact controllability of
\begin{equation*}
\Frac{\partial ^2 u}{\partial t^2}(x,t)-\Frac{\partial^{4}u}{\partial t^2\partial x^2}(x,t)+
\Frac{\partial^{4}u}{\partial x^4}(x,t)=
g_n(x,t),\,\,\, 0<x<1,\,\,\,\ t>0,
\end{equation*}
\begin{equation*}
u(0,t)= \Frac{\partial u}{\partial x}(1,t)= \Frac{\partial^2
u}{\partial x^2}(0,t)= \Frac{\partial^3 u}{\partial x^3}(1,t)=0,
\end{equation*}
\begin{equation*}
u(x,0)=u^{0}(x),\,\,\,\,\ \Frac{\partial u}{\partial
t}(x,0)=u^{1}(x),\,\,\ 0<x<1,
\end{equation*}

\end{remark}

config: arxiv


TITLE: Factor complex equation
QUESTION [0 upvotes]: Im having some difficulty in factoring the following complex equation. The image bellow is taken from WolframAlpha, can anyone explain how I can factor this equation.
In the task I am told one solution has a real part of 1.

REPLY [0 votes]: $z^4-2z^3+12z^2-14z+35=0$
Rewrite the term $12z^2$ as $5z^2+7z^2$:
$z^4-2z^3+5z^2+7z^2-14z+35=0$
Take out the common factors:
$z^2(z^2-2z+5)+7(z^2-2z+5)=0$
Factored form:
$(z^2+7)(z^2-2z+5)=0$
The solutions to this equation are:
$x=\pm\sqrt{7}i$ and $x=1\pm2i$.


TITLE: Amicable chains of numbers
QUESTION [4 upvotes]: Let amicable chain of $n$ numbers be (a$_{1}$, a$_{2}$, a$_{3},\dotsc$, a$_{n}$) such that  for $i < n$ sum of proper divisors of a$_{i}$  is a$_{i+1}$. For sum of proper divisors of a$_{n}\rightarrow$ a$_{1}$. All numbers in the chain are different.
Case $n = 1$  has small solutions ($\{6\}, \{28\},\dotsc$).
Case $n=2$ has ($\{220, 284\},\{1184,1210\},\dotsc$) examples below $1000$.
What about $n=3$? It doesn't seem to have any examples below $364$ million. Are examples known?
Note that this is different from definition given in  "Amicable Number Triples" L. E. Dickson.

Update:
I checked up to Amicable$_{5}$ (up to $50$M) and up to Amicable$_{32}$ (up to $10$M). Here are some Amicable$_{n}$ with the smallest number in a cycle  shown:
Amicable$_{12}$: $4256$
Amicable$_{5}$: $12496$
Amicable$_{28}$: $14316$
Amicable$_{4}$ are common:
$1264460$; $2115324$; $2784580$; $4938136$; $7169104$; $18048976$; $18656380$; $28158165$; $46722700$
Amicable$_{3}$ - no hits.

REPLY [0 votes]: You can read this link: https://www.jstor.org/stable/2973750. It is very informative on the generalisation of these numbers.


TITLE: How to express $\cos\left(\frac{3\pi}{2} + x\right)$ in terms of $\sin x$ and $\cos x$
QUESTION [1 upvotes]: Express $$\cos \Bigg(\frac{3\pi}{2} + x\Bigg)$$ in terms of $\sin x$ and $\cos x$

My try:
$$\cos \Big(\frac{3\pi}{2} + x\Big)=\cos \frac{3\pi}{2} \cos x-\sin \frac{3\pi}{2} \sin x=0-(-1)\sin x=\sin x$$
Is this the correct way to do it?

REPLY [1 votes]: Recall the identities for sine and cosine in the four quadrants. One of the QIV identities is the following.
$$\cos\big(\frac{3\pi}{2}+x\big) = \sin x$$


TITLE: Vector spaces and infinite cyclic linear transformations II
QUESTION [0 upvotes]: Let $V$ be the vector space of infinite dimension on the field $\mathbb{Z}_2$. Let's say $$ V=\langle v_1\rangle+\langle v_2\rangle+\dots+ \langle v_n\rangle+\dots$$ where each $v_i$ has order $2$ and form a basis of the vector space V.
Let $\alpha$ be the linear transformation of $V$ such that $$v_i\mapsto v_1+v_2+\dots+v_i, \text{ for each } i\geq 1$$
Is there a proper $\alpha$-invariant infinite subspace of $V$?

REPLY [1 votes]: No there is no proper $\alpha$-invariant infinite dimensional subspace.
For $k \ge 0$, let $V_k$ be the subspace of $V$ spanned by $v_1,\ldots,v_k$. Then $V_k$ is $\alpha$-invariant, and we claim that these are the only proper $\alpha$-invariant subspaces.
Let $W$ be an $\alpha$-invariant nonzero subspace of $V$, let $w \in W$,and let $k$ be the highest $v_k$ occurring in the sum of the basis elements for $w$. We claim that $V_k \le W$. Since $w + \alpha(w) \in W$, and has highest term $v_{k-1}$, this is a straightforward induction of $k$.
So either there is a maximal $v_k$ occurring as highest term in the elements of $W$, in which case $W=V_k$, or there is no such highest term, in which case $W$ contains $V_k$ for all $k$, so $W=V$.


text:
\begin{document}

\title{The Secrecy Rate Region of the Broadcast Channel}

\author{Ghadamali Bagherikaram, Abolfazl S. Motahari, Amir K. Khandani\\
Coding and Signal Transmission Laboratory,\\
 Department of
Electrical
and Computer Engineering,\\
 University of Waterloo, Waterloo, Ontario,
 N2L 3G1\\
 Emails: \{gbagheri,abolfazl,khandani\}@cst.uwaterloo.ca}

\maketitle
\begin{abstract}
In this paper, we consider a scenario where a source node wishes to
broadcast two confidential messages for two respective receivers,
while a wire-tapper also receives the transmitted signal. This model
is motivated by wireless communications, where individual secure
messages are broadcast over open media and can be received by any
illegitimate receiver. The secrecy level is measured by equivocation
rate at the eavesdropper. We first study the general (non-degraded)
broadcast channel with confidential messages. We present an inner
bound on the secrecy capacity region for this model. The inner bound
coding scheme is based on a combination of random binning and the
Gelfand-Pinsker bining. This scheme matches the Marton's inner bound
on the broadcast channel without confidentiality constraint. We
further study the situation where the channels are degraded. For the
degraded broadcast channel with confidential messages, we present
the secrecy capacity region. Our achievable coding scheme is based
on Cover's superposition scheme and random binning. We refer to this
scheme as Secret Superposition Scheme. In this scheme, we show that
randomization in the first layer increases the secrecy rate of the
second layer. This capacity region matches the capacity region of
the degraded broadcast channel without security constraint. It also
matches the secrecy capacity for the conventional wire-tap channel.
Our converse proof is based on a combination of the converse proof
of the conventional degraded broadcast channel and Csiszar lemma.
Finally, we assume that the channels are Additive White Gaussian
Noise (AWGN) and show that secret superposition scheme with Gaussian
codebook is optimal. The converse proof is based on the generalized
entropy power inequality.
\end{abstract}
\section{Introduction}
The notion of information theoretic secrecy in communication systems
was first introduced by Shannon in \cite{1}. The information
theoretic secrecy requires that the received signal of the
eavesdropper does not provide even a single bit information about
the transmitted messages. Shannon considered a pessimistic situation
where both the intended receiver and the eavesdropper have direct
access to the transmitted signal (which is called ciphertext). Under
these circumstances, he proved a negative result showing that
perfect secrecy can be achieved only when the entropy of the secret
key is greater than or equal to the entropy of the message. In
modern cryptography, all practical cryptosystems are based on
Shannnon's pessimistic assumption. Due to practical constraints,
secret keys are much shorter than messages. Therefore, these
practical cryptosystems are theoretically susceptible of breaking by
attackers. However, the goal of designing such practical ciphers is
to guarantee that there exists no efficient algorithm for breaking
them.

Wyner in \cite{2} showed that the above negative result is a
consequence of Shannon's restrictive assumption that the adversary
has access to precisely the same information as the legitimate
receiver. Wyner considered a scenario in which a wire-tapper
receives the transmitted signal over a degraded channel with respect
to the legitimate receiver's channel. He further assumed that the
wire-tapper has no computational limitations and knows the codebook
used by the transmitter. He measured the level of ignorance at the
eavesdropper by its equivocation and characterized the
capacity-equivocation region. Interestingly, a non-negative perfect
secrecy capacity is always achievable for this scenario.

The secrecy capacity for the Gaussian wire-tap channel is
characterized by Leung-Yan-Cheong in \cite{3}. Wyner's work then is
extended to the general (non-degraded) broadcast channel with
confidential messages (BCC) by Csiszar and Korner \cite{4}. They
considered transmitting confidential information to the legitimate
receiver while transmitting common information to both the
legitimate receiver and the wire-tapper. They established a
capacity-equivocation region of this channel.

The BCC is further studied recently in \cite{5,6,7}, where the
source node transmits a common message for both receivers, along
with two additional confidential messages for two respective
receivers. The fading BCC is investigated in \cite{8,9} where the
broadcast channels from the source node to the legitimate receiver
and the eavesdropper is corrupted by multiplicative fading gain
coefficients, in addition to additive white Gaussian noise terms.
The Channel State Information (CSI) is assumed to be known at the
transmitter. In \cite{10}, the perfect secrecy capacity is derived
where the channels are slow fading. Moreover, the optimal power
control policy is obtained for different scenarios regarding
availability of CSI. In \cite{11}, the wire-tap channel is extended
to the parallel broadcast channels and the fading channels with
multiple receivers. Here, the secrecy constraint is a perfect
equivocation for each messages, even if all the other messages are
revealed to the eavesdropper. The secrecy sum capacity for a reverse
broadcast channel is derived for this restrictive assumption. The
notion of the wire-tap channel is also extended to multiple access
channels \cite{12,13,14,15}, relay channels \cite{16,17,18,19},
parallel channels \cite{20} and MIMO channels
\cite{21,22,23,24,25,26}. Some other related works on communication
of confidential messages can be found in \cite{27,28,29,30,31}.

In this paper, we consider a scenario where a source node wishes to
broadcast two confidential messages for two respective receivers,
while a wire-tapper also receives the transmitted signal. This model
is motivated by wireless communications, where individual secure
messages are broadcast over shared media and can be received by any
illegitimate receiver. In fact, we simplify the restrictive
constraint imposed in \cite{11} and assume that the eavesdropper
does not have access to the other messages. We first study the
general broadcast channel with confidential messages. We present an
achievable rate region for this channel. Our achievable coding
scheme is based on the combination of the random binning and the
Gelfand-Pinsker bining \cite{32}. This scheme matches the Marton's
inner bound \cite{33} on the broadcast channel without
confidentiality constraint. We further study the situation where the
channels are physically degraded and characterize the secrecy
capacity region. Our achievable coding scheme is based on Cover's
superposition coding \cite{34} and the random binning. We refer to
this scheme as Secret Superposition Coding. This capacity region
matches the capacity region of the degraded broadcast channel
without security constraint. It also matches the secrecy capacity of
the wire-tap channel.

The rest of the paper is organized as follows. In section II we
introduce the system model. In Section III, we provide an inner
bound on the secrecy capacity region when the channels are not
degraded. In section IV, we specialize our channel to the physically
degraded and establish the secrecy capacity region. In Section V, we
conclude the paper.

\section{Preliminaries}

In this paper, a random variable is denoted by a capital letter
(e.g. X) and its realization is denoted by a corresponding lower
case letter (e.g. x). The finite alphabet of a random variable is
denoted by a script letter (e.g. $\mathcal{X}$) and its probability
distribution is denoted by $P(x)$. Let  $\mathcal{X}$ be a finite
alphabet set and denote its cardinality by $|\mathcal{X}|$. The
members of  $\mathcal{X}^{n}$ will be written as
$x^{n}=(x_{1},x_{2},...,x_{n})$, where subscripted letters denote
the components and superscripted letters denote the vector. The
notation $x^{i-1}$ denotes the vector $(x_{1},x_{2},...,x_{i-1})$. A
similar notation will be used for random variables and random
vectors.

Consider a Broadcast Channel with Confidential messages as depicted
in fig.\ref{f1}.
\begin{figure}
\centerline{\includegraphics[scale=.6]{alf1.eps}} \caption{Broadcast
Channel with Confidential Messages} \label{f1}
\end{figure}
 In this confidential setting, the transmitter ($X$) wants to broadcast some secret messages to the
legitimated receivers ($Y_{1}$,$Y_{2}$), and prevent the
eavesdropper ($Z$) from having any information about the messages. A
discrete memoryless broadcast channel with confidential messages is
described by finite sets $\mathcal{X}$,
$\mathcal{Y}_{1}$,$\mathcal{Y}_{2}$,$\mathcal{Z}$, and a conditional
distribution $P(y_{1},y_{2},z|x)$. The input of the channel is
$x\in\mathcal{X}$ and the outputs are
$(y_{1},y_{2},z)\in(\mathcal{Y}_{1}\times\mathcal{Y}_{2}\times\mathcal{Z})$
for receiver $1$, receiver $2$, and the eavesdropper, respectively.
The transmitter wishes to send independent messages $(W_{1},W_{2})$
to the respective receivers in $n$ uses of the channel while
insuring perfect secrecy. The channel is discrete memoryless in the
sense that
\begin{equation}
P(y_{1}^{n},y_{2}^{n},z^{n}|x^{n})=\prod_{i=1}^{n}P(y_{1,i},y_{2,i},z_{i}|x_{i}).
\end{equation}
A $((2^{nR_{1}},2^{nR_{2}}),n)$ code for a broadcast channel with
confidential messages consists of a stochastic encoder
\begin{equation}
f:(\{1,2,...,2^{nR_{1}}\}\times\{1,2,...,2^{nR_{2}}\})\rightarrow
\mathcal{X}^{n},
\end{equation}
and two decoders,
\begin{equation}
g_{1}:\mathcal{Y}_{1}^{n}\rightarrow \{1,2,...,2^{nR_{1}}\}
\end{equation}
and
\begin{equation}
g_{2}:\mathcal{Y}_{2}^{n}\rightarrow \{1,2,...,2^{nR_{2}}\}.
\end{equation}
The average probability of error is defined as the probability that
the decoded messages are not equal to the transmitted messages; that
is,
\begin{equation}
P_{e}^{(n)}=P(g_{1}(Y_{1}^{n})\neq W_{1}\cup g_{2}(Y_{2}^{n})\neq
W_{2}).
\end{equation}

The knowledge that the eavesdropper gets about $W_{1}$ and $W_{2}$
from its received signal $Z^{n}$ is modeled as
\begin{IEEEeqnarray}{lr}
I(Z^{n},W_{1})=H(W_{1})-H(W_{1}|Z^{n}),\\
I(Z^{n},W_{2})=H(W_{2})-H(W_{2}|Z^{n}),
\end{IEEEeqnarray}
and
\begin{equation}
I(Z^{n},(W_{1},W_{2}))=H(W_{1},W_{2})-H(W_{1},W_{2}|Z^{n}).
\end{equation}
Perfect secrecy revolves around the idea that the eavesdropper
cannot get even a single bit information about the transmitted
messages. Perfect secrecy thus requires that
\begin{IEEEeqnarray}{lr}
I(Z^{n},W_{1})=0\Leftrightarrow H(W_{1})=H(W_{1}|Z^{n}),\\
\nonumber I(Z^{n},W_{2})=0\Leftrightarrow H(W_{2})=H(W_{2}|Z^{n}),
\end{IEEEeqnarray}
and
\begin{equation}
I(Z^{n},(W_{1},W_{2}))=0\Leftrightarrow
H(W_{1},W_{2})=H(W_{1},W_{2}|Z^{n}).
\end{equation}
 The secrecy levels of confidential messages $W_{1}$ and
$W_{2}$ are measured at the eavesdropper in terms of equivocation
rates which are defined as follows.
\begin{defi}
The equivocation rates $R_{e1}$, $R_{e2}$ and $R_{e12}$
 for the
Broadcast channel with confidential messages are:
\begin{IEEEeqnarray}{lr}
R_{e1}=\frac{1}{n}H(W_{1}|Z^{n}),\\
\nonumber R_{e2}=\frac{1}{n}H(W_{2}|Z^{n}), \\ \nonumber
R_{e12}=\frac{1}{n}H(W_{1},W_{2}|Z^{n}).
\end{IEEEeqnarray}
\end{defi}
The perfect secrecy rates $R_{1}$ and $R_{2}$ are the amount of
information that can be sent to the legitimate receivers not only
reliably but also confidentially.
\begin{defi}
A secrecy rate pair $(R_{1},R_{2})$ is said to be achievable if for
any $\epsilon>0$, there exists  a sequence of
$((2^{nR_{1}},2^{nR_{2}}),n)$ codes, such that for sufficiently
large $n$, we have:
\begin{IEEEeqnarray}{rl}
\label{l0}P_{e}^{(n)}&\leq \epsilon,\\
\label{l1}
 R_{e1}&\geq R_{1}-\epsilon_{1},\\
\label{l2} R_{e2}&\geq R_{2}-\epsilon_{2},\\
\label{l3}
 R_{e12}&\geq R_{1}+R_{2}-\epsilon_{3}.

\end{IEEEeqnarray}
\end{defi}
In the above definition, the first condition concerns the
reliability, while the other conditions guarantee perfect secrecy for each individual message and both messages as well.

 The capacity region is defined as follows.
\begin{defi}
The capacity region of the broadcast channel with confidential
messages is the closure of the set of all achievable rate pairs
$(R_{1},R_{2})$.
\end{defi}

\section{General BCCs}
In this section, we consider the general broadcast channel with
confidential messages and present an achievable rate region. Our
achievable coding scheme is based on a combination of the random
binning and the Gelfand-Pinsker bining schemes \cite{32}. The
following theorem illustrates the achievable rate region for this
channel.

\begin{thm}
Let $\mathbb{R}_{I}$ denote the union of all non-negative rate pairs
$(R_{1},R_{2})$ satisfying
 \begin{IEEEeqnarray}{rl}
    R_{1}&\leq I(V_{1};Y_{1})-I(V_{1};Z), \\
    R_{2}&\leq I(V_{2};Y_{2})-I(V_{2};Z), \\
    R_{1}+R_{2}&\leq
    I(V_{1};Y_{1})+I(V_{2};Y_{2})-I(V_{1},V_{2};Z)-I(V_{1};V_{2}).
  \end{IEEEeqnarray}
over all joint distributions
$P(v_{1},v_{2})P(x|v_{1},v_{2})P(y_{1},y_{2},z|x)$. Then any rate
pair $(R_{1},R_{2})\in \mathbb{R}_{I}$ is achievable for the
broadcast channel with confidential messages.
\end{thm}

\begin{rem}
If we remove the secrecy constraints by setting
$\mathcal{Z}=\emptyset$, then the above rate region reduces to
Marton's achievable region for the general broadcast channel.
\end{rem}
\begin{rem}
If we remove one of the users by setting e.g.,
$\mathcal{Y}_{2}=\emptyset$, then we get the Csiszar and Korner's
secrecy capacity for the other user.
\end{rem}

\begin{proof}

1) \textit{Codebook Generation}:
 The structure of the encoder is
depicted in Fig.\ref{f2}.
\begin{figure}
\centerline{\includegraphics[scale=.6]{alf2.eps}} \caption{The
Stochastic Encoder} \label{f2}
\end{figure}
Fix $P(v_{1})$, $P(v_{2})$ and  $P(x|v_{1},v_{2})$. The stochastic
encoder generates $2^{n(I(V_{1};Y_{1})-\epsilon)}$ independent and
identically distributed sequences $v_{1}^{n}$ according to the
distribution $P(v_{1}^{n})=\prod_{i=1}^{n}P(v_{1,i})$. Next,
randomly distribute these sequences into $2^{nR_{1}}$ bins such that
each bin contains $2^{n(I(V_{1};Z)-\epsilon)}$ codewords. Similarly,
it generates $2^{n(I(V_{2};Y_{2})-\epsilon)}$ independent and
identically distributed sequences $v_{2}^{n}$ according to the
distribution $P(v_{2}^{n})=\prod_{i=1}^{n}P(v_{2,i})$. Next,
randomly distribute these sequences into $2^{nR_{2}}$ bins such that
each bin contains $2^{n(I(V_{2};Z)-\epsilon)}$ codewords. Index each
of the above bins by $w_{1}\in\{1,2,...,2^{nR_{1}}\}$ and
$w_{2}\in\{1,2,...,2^{nR_{2}}\}$ respectively.

2) \textit{Encoding}: To send messages $w_{1}$ and $w_{2}$, the
transmitter looks for $v_{1}^{n}$ in bin $w_{1}$ of the first bin
set and looks for $v_{2}^{n}$ in bin $w_{2}$ of the second bin set,
such that $(v_{1}^{n},v_{2}^{n})\in
A_{\epsilon}^{(n)}(P_{V_{1},V_{2}})$ where
$A_{\epsilon}^{(n)}(P_{V_{1},V_{2}})$ denotes the set of jointly
typical sequences $v_{1}^{n}$ and $v_{2}^{n}$ with respect to
$P(v_{1},v_{2})$. The rates are such that there exist more than one
joint typical pair, the transmitter randomly chooses one of them and
then generates $x^{n}$ according to
$P(x^{n}|v_{1}^{n},v_{2}^{n})=\prod_{i=1}^{n}P(x_{i}|v_{1,i},v_{2,i})$.
This scheme is equivalent to the scenario in which each bin is
divided into subbins and the transmitter randomly chooses one of the
subbins of bin $w_{1}$ and one of the subbins of bin $w_{2}$. It
then looks for a joint typical sequence $(v_{1}^{n},v_{2}^{n})$ in
the corresponding subbins and generates $x^{n}$.

3) \textit{Decoding}: The received signals at the legitimate
receivers, $y_{1}^{n}$ and $y_{2}^{n}$, are the outputs of the
channels $P(y_{1}^{n}|x^{n})=\prod_{i=1}^{n}P(y_{1,i}|x_{i})$ and
$P(y_{2}^{n}|x^{n})=\prod_{i=1}^{n}P(y_{2,i}|x_{i})$, respectively.
The first receiver looks for the unique sequence $v_{1}^{n}$ such
that $(v_{1}^{n},y_{1}^{n})$ is jointly typical and declares the
index of the bin containing $v_{1}^{n}$ as the message received. The
second receiver uses the same method to extract the message $w_{2}$.

4) \textit{Error Probability Analysis}: Since the region of
(\ref{l0}) is a subset of Marton's region then, error probability
analysis is the same as \cite{33}.

5) \textit{Equivocation Calculation}: The proof of secrecy
requirement for each individual message (\ref{l1}) and (\ref{l2}) is straightforward and may therefore be omitted.

To prove the requirement of (\ref{l3}) consider $H(W_{1},W_{2}|Z^{n})$, we have
\begin{eqnarray}\nonumber \label{l4}
nR_{e12}&=&H(W_{1},W_{2}|Z^{n})\\ \nonumber  &\geq& H(W_{1},W_{2},Z^{n})-H(Z^{n})\\
\nonumber &=&
H(W_{1},W_{2},V_{1^{n}},V_{2}^{n},Z^{n})-H(V_{1}^{n},V_{2}^{n}|W_{1},W_{2},Z^{n})-H(Z^{n})\\
\nonumber &=&H(W_{1},W_{2},V_{1}^{n},V_{2}^{n})+ H(Z^{n}|W_{1},W_{2},V_{1}^{n},V_{2}^{n})-H(V_{1}^{n},V_{2}^{n}|W_{1},W_{2},Z^{n})-H(Z^{n})\\
\nonumber &\stackrel{(a)}{\geq}&
H(W_{1},W_{2},V_{1}^{n},V_{2}^{n})+H(Z^{n}|W_{1},W_{2},V_{1}^{n},V_{2}^{n})-n\epsilon_{n}-H(Z^{n})\\
\nonumber &\stackrel{(b)}{=}&H(W_{1},W_{2},V_{1}^{n},V_{2}^{n})+H(Z|V_{1}^{n},V_{2}^{n})-n\epsilon_{n}-H(Z^{n}) \\
\nonumber&\stackrel{(c)}{\geq}&
H(V_{1}^{n},V_{2}^{n})+H(Z^{n}|V_{1}^{n},V_{2}^{n}) -n\epsilon_{n}-H(Z^{n}) \\
\nonumber &\stackrel{(d)}{=}&
H(V_{1}^{n})+H(V_{2}^{n})-I(V_{1}^{n};V_{2}^{n})-I(V_{1}^{n},V_{2}^{n};Z^{n})-
n\epsilon_{n}\\
\nonumber &\stackrel{(e)}{=}&
I(V_{1}^{n};Y_{1}^{n})+I(V_{2}^{n};Y_{2}^{n})-I(V_{1}^{n};V_{2}^{n})-I(V_{1}^{n},V_{2}^{n};Z^{n})- n\epsilon_{n}\\
\nonumber &\geq& nR_{1}+nR_{2}-n\epsilon_{n},
\end{eqnarray}

where $(a)$ follows from Fano's inequality which states that for
sufficiently large $n$ we have
$H(V_{1}^{n},V_{2}^{n}|W_{1},W_{2},Z^{n})$ $\leq h(P_{we}^{(n)})$
$+nP_{we}^{n}I(V_{1},V_{2};Z)\leq n\epsilon_{n}$. Here $P_{we}^{n}$
denotes the wiretapper's error probability of decoding
$(v_{1}^{n},v_{2}^{n})$ in the case that the bin numbers $w_{1}$ and
$w_{2}$ are known to the eavesdropper. Since the sum rate is less
than $I(V_{1},V_{2};Z)$, then $P_{we}^{n}\rightarrow 0$ for
sufficiently large $n$. $(b)$ follows from the following Markov
chain: $(W_{1},W_{2})\rightarrow (V_{1},V_{2})\rightarrow$ $ Z$.
Hence, we have
$H(Z^{n}|W_{1},W_{2},V_{1}^{n},V_{2}^{n})=H(Z^{n}|V_{1}^{n},V_{2}^{n})$.
$(c)$ follows from the fact that
$H(W_{1},W_{2},V_{1}^{n},V_{2}^{n})\geq H(V_{1}^{n},V_{2}^{n})$.
$(d)$ follows from that fact that
$H(V_{1}^{n})=I(V_{1}^{n};Y_{1}^{n})$ and
$H(V_{2}^{n})=I(V_{2}^{n};Y_{2}^{n})$.

\end{proof}

\section{The secrecy capacity Region of the Degraded BCCs}
In this section, we consider the degraded broadcast channel with
confidential messages and establish its secrecy capacity region.
\begin{defi}
A broadcast channel with confidential messages is said to be
physically degraded, if $X\rightarrow Y_{1} \rightarrow
Y_{2}\rightarrow Z$ forms a Markov chain. In the other words, we
have
\begin{equation}
P(y_{1},y_{2},z|x)=P(y_{1}|x)P(y_{2}|y_{1})P(z|y_{2}).
\end{equation}
\end{defi}
\begin{defi}
A broadcast channel with confidential messages is said to be
stochastically degraded if its conditional marginal distributions
are the same as that of a physically degraded broadcast channel,
i.e., if there exist two distributions $P^{'}(y_{2}|y_{1})$ and
$P^{'}(z|y_{2})$ such that
\begin{eqnarray}
P(y_{2}|x)=\sum_{y_{1}}P(y_{1}|x)P^{'}(y_{2}|y_{1})\\ \nonumber
P(z|x)=\sum_{y_{2}}P(y_{2}|x)P^{'}(z|y_{2})
\end{eqnarray}
\end{defi}
\begin{lem}
The secrecy capacity region of a broadcast channel with confidential
messages depends only on the conditional marginal distributions
$P(y_{1}|x)$, $P(y_{2}|x)$ and $P(z|x)$.
\end{lem}
\begin{proof}
The proof is very similar to \cite{34} and may therefore be omitted
here.
\end{proof}
In the following theorem, we fully characterize the capacity region
of the physically degraded broadcast channel with confidential
messages.
\begin{thm}
The capacity region for transmitting independent secret information
over the degraded broadcast channel is the convex hull of the
closure of all $(R_{1},R_{2})$ satisfying
\begin{IEEEeqnarray}{rl}\label{l7}
    R_{1}&\leq I(X;Y_{1}|U)+I(U;Z)-I(X;Z), \\
    R_{2}&\leq I(U;Y_{2})-I(U;Z).
\end{IEEEeqnarray}
for some joint distribution $P(u)P(x|u)P(y_{1}, y_{2},z|x)$.
\end{thm}
\begin{rem}
If we remove the secrecy constraints by setting
$\mathcal{Z}=\emptyset$, then the above theorem reduces to the
capacity region of the degraded broadcast channel.
\end{rem}
\begin{proof}

\textit{Achievability}: The coding scheme is based on Cover's
superposition coding and the random bining. We refer to this scheme
as Secure Superposition Coding scheme. The available resources at
the encoder are used for two purposes: to confuse the eavesdropper
so that perfect secrecy can be achieved for both layers, and to
transmit the messages in the main channels. To satisfy
confidentiality, the randomization used in the first layer is again
used in the second layer. This makes a shift of $I(U;Z)$ in the
bound of $R_{1}$. The formal proof of the achievability is as
follows:

1) \textit{Codebook Generation}: The structure of the encoder is
depicted in Fig.\ref{f3}.
\begin{figure}
\centerline{\includegraphics[scale=.6]{alf3.eps}} \caption{Secret
Superposition structure}  \label{f3}
\end{figure}
Let us fix $P(u)$ and $P(x|u)$. The stochastic encoder generates
$2^{n(I(U;Y_{2})-\epsilon)}$ independent and identically distributed
sequences $u^{n}$ according to the distribution
$P(u^{n})=\prod_{i=1}^{n}P(u_{i})$. Next, we randomly distribute
these sequences into $2^{nR_{2}}$ bins such that each bin contains
$2^{n(I(U;Z)-\epsilon)}$ codewords. We index each of the above bins
by $w_{2}\in\{1,2,...,2^{nR_{2}}\}$. For each codeword of $u^{n}$,
it also generates $2^{n(I(X;Y_{1}|U)-\epsilon)}$ independent and
identically distributed sequences $x^{n}$ according to the
distribution $P(x^{n}|u^{n})=\prod_{i=1}^{n}P(x_{i}|u_{i})$. We
randomly distribute these sequences into $2^{nR_{1}}$ bins such that
each bin contains $2^{n(I(X;Z)-I(U;Z)-\epsilon)}$ codewords. We
index each of the above bins by $w_{1}\in\{1,2,...,2^{nR_{1}}\}$.

2) \textit{Encoding}: To send messages $w_{1}$ and $w_{2}$, the
transmitter randomly chooses one of the codewords in bin $w_{2}$,
say $u^{n}$. Then given $u^{n}$, the transmitter randomly chooses
one of $x^{n}$ in bin $w_{1}$ of the second layer and sends it.

3) \textit{Decoding}: The received signal at the legitimate
receivers, $y_{1}^{n}$ and $y_{2}^{n}$, are the outputs of the
channels $P(y_{1}^{n}|x^{n})=\prod_{i=1}^{n}P(y_{1,i}|x_{i})$ and
$P(y_{2}^{n}|x^{n})=\prod_{i=1}^{n}P(y_{2,i}|x_{i})$, respectively.
Receiver $2$ determines the unique $u^{n}$ such that
$(u^{n},y_{2}^{n})$ are jointly typical and declares the index of
the bin containing $u^{n}$ as the message received. If there is none
of such or more than of one such, an error is declared. Receiver $1$
looks for the unique $(u^{n},x^{n})$ such that
$(u^{n},x^{n},y_{1}^{n})$ are jointly typical and declares the
indexes of the bins containing $u^{n}$ and $x^{n}$ as the messages
received. If there is none of such or more than of  one such, an
error is declared.

4) \textit{Error Probability Analysis}: Since each rate pair of
(\ref{l7}) is in the capacity region of the degraded broadcast
channel without confidentiality constraint, then it can be readily
shown that the error probability is arbitrarily small, c.f.
\cite{34}.

5) \textit{Equivocation Calculation}: To prove the secrecy
requirement of (\ref{l1}), we have
\begin{eqnarray}\nonumber
nR_{e1}&=&H(W_{1}|Z^{n})\\ \nonumber
&\geq& H(W_{1}|Z^{n},U^{n})\\ \nonumber &=&
H(W_{1},Z^{n}|U^{n})-H(Z^{n}|U^{n}) \\ \nonumber &=&
H(W_{1},X^{n},Z^{n}|U^{n})-H(Z^{n}|U^{n})-H(X^{n}|W_{1},Z^{n},U^{n})\\
\nonumber
&\stackrel{(a)}{=}&H(W_{1},X^{n}|U^{n})+H(Z^{n}|W_{1},U^{n},X^{n})-H(Z^{n}|U^{n})-n\epsilon_{n}\\
\nonumber &\stackrel{(b)}{\geq}&
H(X^{n}|U^{n})+H(Z^{n}|X^{n})-H(Z^{n}|U^{n})-n\epsilon_{n}\\
\nonumber &\stackrel{(c)}{=}&
H(X^{n};Y_{1}^{n}|U^{n})+I(U^{n};Z^{n})-I(X^{n};Z^{n})-n\epsilon_{n}\\
\nonumber &\geq & nR_{1}-n\epsilon_{n},
\end{eqnarray}
where $(a)$ follows from Fano's inequality which states that
$H(X^{n}|W_{1},Z^{n},U^{n})\leq
h(P_{we}^{(n)})+nP_{we}^{n}I(X;Z)\leq n\epsilon_{n}$ for
sufficiently large $n$. Here $P_{we}^{n}$ denotes the wiretapper's
error probability of decoding $x^{n}$ given the bin number and the
codeword $u^{n}$ are known to the eavesdropper. Since the rate is
less than $I(X;Z)$, then $P_{we}^{n}\rightarrow 0$ for sufficiently
large $n$. $(b)$ follows from the  fact that $(W_{1},U)\rightarrow X
\rightarrow Z$ forms a Markov chain. Thus we have
$I(W_{1},U^{n};Z^{n}|X^{n})=0$, where it is implied that
$H(Z^{n}|W_{1},U^{n},X^{n})=H(Z^{n}|X^{n})$. $(c)$ follows from two
identities: $H(X^{n}|U^{n})=I(X^{n};Y_{1}^{n}|U^{n})$ and
$H(Z^{n}|X^{n})-H(Z^{n}|U^{n})=I(U^{n};Z^{n})-I(X^{n};Z^{n})$. Since the proof of the requirement
(\ref{l2}) is straightforward, we need to prove the requirement of (\ref{l3}).
\begin{eqnarray}\nonumber\label{l6}
nR_{e12}&=&  H(W_{1},W_{2}|Z^{n})\\ \nonumber &\geq& H(W_{1},W_{2},Z^{n})-H(Z^{n})\\
\nonumber &=&
H(W_{1},W_{2},U^{n},X^{n},Z^{n})-H(U^{n},X^{n}|W_{1},W_{2},Z^{n})-H(Z^{n})\\
\nonumber &=&
H(W_{1},W_{2},U^{n},X^{n})\\ \nonumber &+& H(Z^{n}|W_{1},W_{2},U^{n},X^{n})-H(U^{n},X^{n}|W_{1},W_{2},Z^{n})-H(Z^{n})\\
\nonumber &\stackrel{(a)}{\geq}& H(W_{1},W_{2},U^{n},X^{n})+H(Z^{n}|W_{1},W_{2},U^{n},X^{n})-n\epsilon_{n}-H(Z^{n})\\
\nonumber &\stackrel{(b)}{=}&H(W_{1},W_{2},U^{n},X^{n})+H(Z|U^{n},X^{n})-n\epsilon_{n}-H(Z^{n})\\
\nonumber&\stackrel{(c)}{\geq}&
H(U^{n},X^{n})+H(Z^{n}|U^{n},X^{n})-n\epsilon_{n}-H(Z^{n})\\
\nonumber &=&
H(U^{n})+H(X^{n}|U^{n})-I(U^{n},X^{n};Z^{n})- n\epsilon_{n}\\
\nonumber &\stackrel{(d)}{=}&
I(U^{n};Y_{2}^{n})+I(X^{n};Y_{1}^{n}|U^{n})-I(X^{n};Z^{n})-I(U^{n};Z^{n}|X^{n})- n\epsilon_{n}\\
\nonumber &\geq& nR_{1}+nR_{2}-n\epsilon_{n},
\end{eqnarray}
where $(a)$ follows from Fano's inequality that
$H(U^{n},X^{n}|W_{1},W_{2},Z^{n})\leq
h(P_{we}^{(n)})+nP_{we}^{n}I(U,X;Z)\leq n\epsilon_{n}$ for
sufficiently large $n$. Here $P_{we}^{n}$ denotes the wiretapper's
error probability of decoding $(u^{n},x^{n})$ in the case that the
bin numbers $w_{1}$ and $w_{2}$ are known to the eavesdropper. The
eavesdropper first looks for the unique $u^{n}$ in bin $w_{2}$ of
the first layer, such that it is jointly typical with $z^{n}$. Since
the number of candidate codewords is less than $I(U;Z)$, then the
probability of error is arbitrarily small for a sufficiently large
$n$. Next, given $u^{n}$, the eavesdropper looks for the unique
$x^{n}$ in bin $w_{1}$ which is jointly typical with $z^{n}$.
Similarly, since the number of available candidates is less than
$I(X;Z)$, then the probability of error decoding is arbitrarily
small.  $(b)$ follows from the  fact that $(W_{1},W_{2})\rightarrow
U\rightarrow X\rightarrow Z$ forms a Markov chain. Therefore, we
have $I(W_{1},W_{2};Z^{n}|U^{n},X^{n})=0$, where it is implied that
$H(Z^{n}|W_{1},W_{2},U^{n},X^{n})=H(Z^{n}|U^{n},X^{n})$. $(c)$
follows from the fact that $H(W_{1},W_{2},U^{n},X^{n})\geq
H(U^{n},X^{n})$. $(d)$ follows from that fact that
$H(U^{n})=I(U^{n};Y_{2}^{n})$ and
$H(X^{n}|U^{n})=I(X^{n};Y_{1}^{n}|U^{n})$.

 \textit{Converse}: The
transmitter sends two independent secret messages $W_{1}$ and
$W_{2}$ to receiver $1$ and receiver $2$ respectively. Let us define
$U_{i}=(W_{2},Y_{1}^{i-1})$. The following chain of inequality
clarifies the proof:
\begin{eqnarray}\nonumber
nR_{1} &\stackrel{(a)}{\leq}&
\sum_{i=1}^{n}I(W_{1};Y_{1,i}|W_{2},Z_{i},Y_{1}^{i-1},\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{3}\\
\nonumber &=&
 \sum_{i=1}^{n}I(W_{1};Y_{1,i}|U_{i},Z_{i},\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{3}\\
\nonumber &\stackrel{(b)}{\leq}& \sum_{i=1}^{n} I(X_{i};Y_{1,i}|U_{i},Z_{i},\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{3}\\
\nonumber &\stackrel{(c)}{=}&
\sum_{i=1}^{n}I(X_{i};Y_{1,i},U_{i},Z_{i}|\widetilde{Z}^{i+1})-I(X_{i};Z_{i}|\widetilde{Z}^{i+1})-I(X_{i};U_{i}|Z_{i},\widetilde{Z}^{i+1}) +n\delta_{1}+n\epsilon_{3}\\
\nonumber&\stackrel{(d)}{=}& \sum_{i=1}^{n} I(X_{i};Y_{1,i}|U_{i},\widetilde{Z}^{i+1})+I(X_{i};U_{i}|\widetilde{Z}^{i+1})-I(X_{i};Z_{i}|\widetilde{Z}^{i+1})-I(X_{i};U_{i}|Z_{i},\widetilde{Z}^{i+1}) +n\delta_{1}+n\epsilon_{3}\\
\nonumber &\stackrel{(e)}{=}& \sum_{i=1}^{n} I(X_{i};Y_{1,i}|U_{i},\widetilde{Z}^{i+1})-I(X_{i};Z_{i}|\widetilde{Z}^{i+1})+I(Z_{i};U_{i}|\widetilde{Z}^{i+1})-I(Z_{i};U_{i}|X_{i},\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{3}\\
\nonumber &\stackrel{(f)}{=}& \sum_{i=1}^{n}I(X_{i};Y_{1,i}|U_{i},\widetilde{Z}^{i+1})-I(X_{i};Z_{i}|\widetilde{Z}^{i+1})+I(Z_{i};U_{i}|\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{3}\\
\nonumber,
\end{eqnarray}
 $(a)$ follows from the
following lemma (\ref{ll1}). $(b)$ follows
from the data processing theorem. $(c)$ follows from the chain rule.
$(d)$ follows from the fact that
$I(X_{i};Y_{1,i},U_{i},Z_{i}|\widetilde{Z}^{i+1})=I(X_{i};U_{i}|\widetilde{Z}^{i+1})+I(X_{i};Y_{1,i}|U_{i},\widetilde{Z}^{i+1})+I(X_{i};Z_{i}|Y_{1,i},U_{i},\widetilde{Z}^{i+1})$
 and from the fact that $\widetilde{Z}^{i+1}U_{i}\rightarrow X_{i} \rightarrow Y_{1,i} \rightarrow Y_{2,i} \rightarrow
Z_{i}$ forms a Markov chain, which means that
$I(X_{i};Z_{i}|Y_{1,i},U_{i},\widetilde{Z}^{i+1})=0$. $(e)$ follows from the fact that
$I(X_{i};U_{i}|\widetilde{Z}^{i+1})-I(X_{i};U_{i}|Z_{i},\widetilde{Z}^{i+1})=I(Z_{i};U_{i}|\widetilde{Z}^{i+1})-I(Z_{i};U_{i}|X_{i},\widetilde{Z}^{i+1})$.
$(f)$ follows from the fact that $\widetilde{Z}^{i+1} U_{i}\rightarrow
X_{i} \rightarrow Z_{i}$ forms a Markov chain. Thus
$I(Z_{i};U_{i}\widetilde{Z}^{i+1}|X_{i})=0$ which implies that $I(Z_{i};U_{i}|X_{i},\widetilde{Z}^{i+1})=0$.
\begin{lem}: \label{ll1}
For the broadcast channel with confidential messages of
$(W_{1},W_{2})\rightarrow X^{n} \rightarrow
Y_{1}^{n}Y_{2}^{n}Z^{n}$, the perfect secrecy rates are bounded as
follows,
\begin{IEEEeqnarray}{rl}
    nR_{1}&\leq \sum_{i=1}^{n}I(W_{1};Y_{1i}|W_{2},Z_{i},Y_{1}^{i-1},\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{3}, \\
    nR_{2}&\leq \sum_{i=1}^{n}I(W_{2};Y_{2i}|Z_{i},Y_{2}^{i-1},\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{2}.
\end{IEEEeqnarray}
\end{lem}
\begin{proof}
We need to prove the first bound. The second bound can similarly be
proven. According to the above discussion $nR_{1}$ is bounded as
follows:
\begin{eqnarray}\nonumber
nR_{1} &\stackrel{(a)}{\leq}& H(W_{1}|W_{2},Z^{n})+n\epsilon_{3} \\
\nonumber &\stackrel{(b)}{\leq}&
H(W_{1}|W_{2},Z^{n})-H(W_{1}|Y_{1}^{n},W_{2})+n\delta_{1}+n\epsilon_{3}\\
\nonumber &=& I(W_{1};Y_{1}^{n}|W_{2})-I(W_{1};Z^{n}|W_{2})+n\delta_{1}+n\epsilon_{3}

\end{eqnarray}
where $(a)$ follows from the secrecy constraint that
$H(W_{1},W_{2}|Z^{n})\geq H(W_{1},W_{2})-n\epsilon_{3}$, the fact
that $H(W_{2}|Z^{n})\leq H(W_{2})$ and the fact that two messages
are independent. $(b)$ follows from Fano's inequality that
$H(W_{1}|Y_{1}^{n},W_{2}) \leq n\delta_{1}$.

Next, we expand $I(W_{1};Y_{1}^{n}|W_{2})$ and $I(W_{1};Z^{n}|W_{2})$
starting with $I(W_{1};Y_{1}|W_{2})$ and $I(W_{1};\widetilde{Z}^{n}|W_{2})$,
respectively.
\begin{eqnarray}\nonumber
I(W_{1};Y_{1}^{n}|W_{2}) &=&\sum_{i=1}^{n}I(W_{1};Y_{1i}|W_{2},Y_{1}^{i-1}) \\
\nonumber
&=&\sum_{i=1}^{n}I(W_{1},\widetilde{Z}^{i+1};Y_{1i}|W_{2},Y_{1}^{i-1})-I(\widetilde{Z}^{i+1};Y_{1i}|W_{1},W_{2},Y_{1}^{i-1})\\
\nonumber &=&\sum_{i=1}^{n}
I(W_{1};Y_{1i}|W_{2},Y_{1}^{i-1},\widetilde{Z}^{i+1})+I(\widetilde{Z}^{i+1};Y_{1i}|W_{2},Y_{1}^{i-1})-I(\widetilde{Z}^{i+1};Y_{1i}|W_{1},W_{2},Y_{1}^{i-1})\\
\nonumber
&=&\sum_{i=1}^{n}I(W_{1};Y_{1i}|W_{2},Y_{1}^{i-1},\widetilde{Z}^{i+1})+\Delta_{1}-\Delta_{2},
\end{eqnarray}
where,
$\Delta_{1}=\sum_{i=1}^{n}I(\widetilde{Z}^{i+1};Y_{1i}|W_{2},Y_{1}^{i-1})$ and
$\Delta_{2}=\sum_{i=1}^{n}I(\widetilde{Z}^{i+1};Y_{1i}|W_{1},W_{2},Y_{1}^{i-1})$.
Similarly, we have,
\begin{eqnarray}\nonumber
I(W_{1};Z^{n}|W_{2}) &=&\sum_{i=1}^{n}I(W_{1};Z_{i}|W_{2},\widetilde{Z}^{i+1}) \\
\nonumber
&=&\sum_{i=1}^{n}I(W_{1},Y_{1}^{i-1};Z_{i}|W_{2},\widetilde{Z}^{i+1})-I(Y_{1}^{i-1};Z_{i}|W_{1},W_{2},\widetilde{Z}^{i+1})\\
\nonumber &=&\sum_{i=1}^{n}
I(W_{1};Z_{i}|W_{2},Y_{1}^{i-1},\widetilde{Z}^{i+1})+I(Y_{1}^{i-1};Z_{i}|W_{2},\widetilde{Z}^{i+1})-I(Y_{1}^{i-1};Z_{i}|W_{1},W_{2},\widetilde{Z}^{i+1})\\
\nonumber
&=&\sum_{i=1}^{n}I(W_{1};Z_{i}|W_{2},Y_{1}^{i-1},\widetilde{Z}^{i+1})+\Delta_{1}^{*}-\Delta_{2}^{*},
\end{eqnarray}
where,
$\Delta_{1}^{*}=\sum_{i=1}^{n}I(Y_{1}^{i-1};Z_{i}|W_{2},\widetilde{Z}^{i+1})$
and
$\Delta_{2}^{*}=\sum_{i=1}^{n}I(Y_{1}^{i-1};Z_{i}|W_{1},W_{2},\widetilde{Z}^{i+1})$.
According to lemma $7$ of \cite{4}, $\Delta_{1}=\Delta_{1}^{*}$ and
$\Delta_{2}=\Delta_{2}^{*}$. Thus, we have,
\begin{eqnarray}\nonumber
nR_{1}&\leq& \sum_{i=1}^{n}I(W_{1};Y_{1i}|W_{2},Y_{1}^{i-1},\widetilde{Z}^{i+1})-I(W_{1};Z_{i}|W_{2},Y_{1}^{i-1},\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{3}
\\ \nonumber &=&\sum_{i=1}^{n}H(W_{1}|W_{2},Z_{i},Y_{1}^{i-1},\widetilde{Z}^{i+1})-H(W_{1}|W_{2},Y_{1i},Y_{1}^{i-1},\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{3}\\ \nonumber
&\stackrel{(a)}{\leq}&\sum_{i=1}^{n}H(W_{1}|W_{2},Z_{i},Y_{1}^{i-1},\widetilde{Z}^{i+1})-H(W_{1}|W_{2},Y_{1i},Z_{i},Y_{1}^{i-1},\widetilde{Z}^{i+1})+n\delta_{1}+n\epsilon_{3}\\
\nonumber
&=&\sum_{i=1}^{n}I(W_{1};Y_{1i}|W_{2},Z_{i},Y_{1}^{i-1},\widetilde{Z}^{i+1})+n\delta_{1}\nonumber+n\epsilon_{3},
\end{eqnarray}
where $(a)$ follows from the fact that conditioning always decreases
the entropy.
\end{proof}
For the second receiver, we have
\begin{eqnarray}\nonumber
nR_{2} &\stackrel{(a)}{\leq}&\sum_{i=1}^{n}I(W_{2};Y_{2,i}|Y_{2}^{i-1},Z_{i},\widetilde{Z}^{i+1})+n\delta_{2}+n\epsilon_{1}\\
\nonumber
&=&\sum_{i=1}^{n}H(Y_{2,i}|Y_{2}^{i-1},Z_{i},\widetilde{Z}^{i+1})-H(Y_{2,i}|W_{2},Y_{2}^{i-1},Z_{i},\widetilde{Z}^{i+1})+n\delta_{2}+n\epsilon_{1}\\
\nonumber &\stackrel{(b)}{\leq}&\sum_{i=1}^{n}H(Y_{2,i}|Z_{i},\widetilde{Z}^{i+1})-H(Y_{2,i}|W_{2},Y_{1}^{i-1},Y_{2}^{i-1},Z_{i},\widetilde{Z}^{i+1})+n\delta_{2}+n\epsilon_{1}\\
\nonumber &\stackrel{(c)}{=}&\sum_{i=1}^{n}H(Y_{2,i}|Z_{i},\widetilde{Z}^{i+1})-H(Y_{2,i}|U_{i},Z_{i},\widetilde{Z}^{i+1})+n\delta_{2}+n\epsilon_{1}\\
\nonumber &=&\sum_{i=1}^{n}I(Y_{2,i};U_{i}|Z_{i},\widetilde{Z}^{i+1})+n\delta_{2}+n\epsilon_{1}\\
\nonumber &=&\sum_{i=1}^{n}I(Y_{2,i};U_{i}|\widetilde{Z}^{i+1})+I(Y_{2,i};Z_{i}|U_{i},\widetilde{Z}^{i+1})-I(Y_{2,i};Z_{i}|\widetilde{Z}^{i+1})+n\delta_{2}+n\epsilon_{1}\\
\nonumber &=&\sum_{i=1}^{n}I(Y_{2,i};U_{i}|\widetilde{Z}^{i+1})-I(Z_{i};U_{i}|\widetilde{Z}^{i+1})+I(Z_{i};U_{i}|Y_{2,i},\widetilde{Z}^{i+1})+n\delta_{2}+n\epsilon_{1}\\
\nonumber
&\stackrel{(d)}{=}&\sum_{i=1}^{n}I(Y_{2,i};U_{i}|\widetilde{Z}^{i+1})-I(Z_{i};U_{i}|\widetilde{Z}^{i+1})+n\delta_{2}+n\epsilon_{1}\nonumber,
\end{eqnarray}
where $(a)$ follows from the lemma (\ref{ll1}). $(b)$
follows from the fact that conditioning always decreases the entropy.
$(c)$ follows from the fact that $Y_{2}^{i-1}\rightarrow W_{2}\widetilde{Z}^{i+1}Y_{1}^{i-1}\rightarrow Y_{2i}\rightarrow Z_{i}$ forms a Markov chain. $(d)$ follows from the fact that
$\widetilde{Z}^{i+1}U_{i}\rightarrow Y_{2,i} \rightarrow Z_{i}$ forms a
Markov chain. Thus $I(Z_{i};U_{i}\widetilde{Z}^{i+1}|Y_{2i})=0$ which implies that $I(Z_{i};U_{i}|Y_{2i},\widetilde{Z}^{i+1})=0$. Now, following \cite{34}, let us define the time
sharing random variable $Q$ which is uniformly distributed over
$\{1,2,...,n\}$ and independent of
$(W_{1},W_{2},X^{n},Y_{1}^{n},Y_{2}^{n})$. Let us define
$U=U_{Q},~ V=(\widetilde{Z}^{Q+1},Q),~ X=X_{Q},~ Y_{1}=Y_{1,Q},~ Y_{2}=Y_{2,Q},~ Z=Z_{Q}$,
then we can bound $R_{1}$ and $R_{2}$ as follows
\begin{IEEEeqnarray}{rl}
    R_{1}&\leq I(X;Y_{1}|U,V)+I(U;Z|V)-I(X;Z|V), \\
    R_{2}&\leq I(U;Y_{2}|V)-I(U;Z|V).
\end{IEEEeqnarray}
Since Conditional mutual informations are average of unconditional
ones, the maximum region is achieved when $V$ is a constant. This
proves the converse part.
\end{proof}
\section{Gaussian BCCs}
In this section we consider the physically degraded AWGN broadcast channel with confidential messages. We show that secret superposition coding with
 Gaussian codebook is optimal. At time $i$ the received signals are $Y_{1i}=X_{i}+n_{1i}$,  $Y_{2i}=X_{i}+n_{2i}$ and  $Z_{i}=X_{i}+n_{3i}$, where $n_{1i}$'s,
$n_{2i}$'s and $n_{3i}$'s are each independent identically distributed Gaussian random variables with zero means and $Var(n_{ji})=N_{j}$, j=1,2,3. All noises are independent
of $X_{i}$ and $N_{1}\leq N_{2}\leq N_{3}$. Assume that transmitted power is limited to $E[X^{2}]\leq P$. Since the channels are degraded, at time $i$,
$Y_{1i}=X_{i}+n_{1i}$,  $Y_{2i}=Y_{1i}+n_{2i}^{'}$ and  $Z_{i}=Y_{2i}+n_{3i}^{'}$, where $n_{1i}$'s are i.i.d $\mathcal{N}(0,N_{1})$,
$n_{2i}^{'}$'s are i.i.d $\mathcal{N}(0,N_{1}-N_{2})$, and $n_{3i}^{'}$'s are i.i.d $\mathcal{N}(0,N_{2}-N_{3})$. Fig.\ref{f4} shows the equivalent channels for the physically
degraded AWGN-BCCs. The following theorem illustrates the secrecy capacity region of AWGN-BCCs.
\begin{thm}
The secrecy capacity region of the AWGN broadcast channel with confidential messages is given by the set of rates pairs $(R_{1},R_{2})$
such that
\begin{IEEEeqnarray}{rl}
    R_{1}&\leq C\left(\frac{\alpha P}{N_{1}}\right)+C\left(\frac{(1-\alpha) P}{\alpha P +N_{3}}\right)-C\left(\frac{P}{N_{3}}\right), \\
    R_{2}&\leq C\left(\frac{(1-\alpha) P}{\alpha P +N_{2}}\right)-C\left(\frac{(1-\alpha) P}{\alpha P +N_{3}}\right).
\end{IEEEeqnarray}
for some $\alpha \in [0,1]$.
\end{thm}
\begin{figure}
\centerline{\includegraphics[scale=.7]{alf4.eps}} \caption{equivalent channels for the AWGN-BCCs}  \label{f4}
\end{figure}
\begin{proof}

\textit{Achievability}: Let $U\sim \mathcal{N}(0,(1-\alpha)P)$ and $X^{'}\sim \mathcal{N}(0,\alpha P)$ be independent and $X=U+X^{'} \sim \mathcal{N}(0,P)$.
Therefore, the amount of $I(X;Y_{1}|U)$, $I(U;Z)$, $I(X;Z)$, and $I(U;Y_{2})$ can be easily evaluated.
Now consider the following secure superposition coding scheme:

1) \textit{Codebook Generation}: Generate $2^{nI(U;Y_{2})}$ i.i.d Gaussian codewords $u^{n}$ with average power $(1-\alpha)P$
and randomly distribute these codewords into $2^{nR_{2}}$ bins. Then index each bin by $w_{2}\in\{1,2,...,2^{nR_{2}}\}$.
Generate an independent set of $2^{nI(X^{'};Y_{1}})$ i.i.d Gaussian codewords $x^{'n}$ with average power $\alpha P$. Then, Randomly distribute
them into $2^{nR_{1}}$ bins. Index each bin by $w_{1}\in\{1,2,...,2^{nR_{1}}\}$.

2) \textit{Encoding}: To send messages $w_{1}$ and $w_{2}$, the
transmitter randomly chooses one of the codewords in bin $w_{2}$,
(say $u^{n}$) and one of the codewords in bin $w_{1}$ (say $x^{'n}$ ). Then, simply transmits
$x^{n} =u^{n}+ x^{'n}$.

3) \textit{Decoding}: The received signal at the legitimate
receivers are $y_{1}^{n}$ and $y_{2}^{n}$ respectively.
Receiver $2$ determines the unique $u^{n}$ such that
$(u^{n},y_{2}^{n})$ are jointly typical and declares the index of
the bin containing $u^{n}$ as the message received. If there is none
of such or more than of one such, an error is declared. Receiver $1$
uses successive cancelation method; first decodes $u^{n}$ and subtracts off $y_{1}^{n}$ and then
looks for the unique $x^{'n}$ such that
$(x^{'n},y_{1}^{n})$ are jointly typical and declares the
indexes of the bin containing $x^{'n}$ as the message
received.

The error probability analysis and equivocation calculation is straightforward and may therefor be omitted.

\textit{Converse}: According to the previous section, $R_{2}$ is bounded as follows:

\begin{eqnarray}\label{l10}
R_{2} \leq I(Y_{2};U|Z) = H(Y_{2}|Z)-H(Y_{2}|U,Z)
\end{eqnarray}
The classical entropy power inequality states that:
\begin{eqnarray}\nonumber
2^{\frac{2}{n}H(Y_{2}+n_{3}^{'})}\geq 2^{\frac{2}{n}H(Y_{2})}+2^{\frac{2}{n}H(n_{3}^{'})}
\end{eqnarray}
Therefore, $H(Y_{2}|Z)$ can be written as follows:
\begin{eqnarray}\nonumber
H(Y_{2}|Z)&=&H(Z|Y_{2})+H(Y_{2})-H(Z)\\ \nonumber
&=& \frac{n}{2}\log(N_{3}-N_{2})+H(Y_{2})-H(Y_{2}+n_{3}^{'})\\ \nonumber
&\leq&\frac{n}{2}\log(N_{3}-N_{2})+H(Y_{2})-\frac{n}{2}\log(2^{\frac{2}{n}H(Y_{2})}+N_{3}-N_{2})
\end{eqnarray}
On the other hand, for any fixed $a\in \mathcal{R}$, the function
\begin{eqnarray}\nonumber
f(t,a)=t-\frac{n}{2}\log(2^{\frac{2}{n}t}+a)
\end{eqnarray}
is concave in $t$ and has a global maximum at $t=t_{max}$. Thus, $H(Y_{2}|Z)$ is maximized when $Y_{2}$ (or equivalently $X$) has Gaussian distribution. Hence,
\begin{eqnarray} \label{l8}
H(Y_{2}|Z)&\leq& \frac{n}{2}\log(N_{3}-N_{2})+\frac{n}{2}\log(P+N_{2})-\frac{n}{2}\log(P+N_{3})\\ \nonumber
&=&\frac{n}{2}\log\left(\frac{(N_{3}-N_{2})(P+N_{2})}{P+N_{3}}\right)
\end{eqnarray}
Now consider the term $H(Y_{2}|U,Z)$. This term is lower bounded with $H(Y_{2}|U,X,Z)=\frac{n}{2}\log(N_{2})$ which is greater
 than $\frac{n}{2}\log(\frac{N_{2}(N_{3}-N_{2})}{N_{3}})$.
Hence,
\begin{eqnarray}\label{l9}
\frac{n}{2}\log(\frac{N_{2}(N_{3}-N_{2})}{N_{3}})\leq H(Y_{2}|U,Z)\leq H(Y_{2}|Z)
\end{eqnarray}
Inequalities (\ref{l8}) and (\ref{l9}) imply that there exists a $\alpha \in [0,1]$ such that
\begin{eqnarray}\label{l11}
H(Y_{2}|U,Z)=\frac{n}{2}\log\left(\frac{(N_{3}-N_{2})(\alpha P+N_{2})}{\alpha P+N_{3}}\right)
\end{eqnarray}
Substituting (\ref{l11}) and (\ref{l8}) into (\ref{l10}) yields the desired bound
\begin{eqnarray}
R_{2} &\leq&  H(Y_{2}|Z)-H(Y_{2}|U,Z) \\ \nonumber
&\leq&\frac{n}{2}\log\left(\frac{(P+N_{2})(\alpha P + N_{3})}{(P+N_{3})(\alpha P + N_{2})}\right)\\ \nonumber
&=& C\left(\frac{(1-\alpha) P}{\alpha P +N_{2}}\right)-C\left(\frac{(1-\alpha) P}{\alpha P +N_{3}}\right)
\end{eqnarray}
To bound the rate $R_{1}$, we need the following generalized entropy power inequality which is proven in \cite{35}.
\begin{lem} \cite{35}:
Let $n_{1}$, $n_{2}$ be two gaussian random variables. Let $U$ be a random variable independent of $n_{1}$ and $n_{2}$. Consider
The optimization problem
\begin{eqnarray}
\max_{P(X|U)}&& H(X+n_{1}|U)-H(X+n_{2}|U)\\
\text{subject to}&&  Var(X|U)\leq s
\end{eqnarray}
where the maximization is over all distribution of $X$ given $U$ independent of $n_{1}$ and $n_{2}$. A Gaussian $P(x|u)$ with same variance for each $u$ is an optimal solution
for this optimization problem.
\end{lem}
The rate $R_{1}$ is bounded as follows
\begin{eqnarray}
R_{1}&\leq& I(X;Y_{1}|U)-I(X;Z)+I(U;Z)\\ \nonumber
&=&H(Y_{1}|U)-H(Y_{1}|X,U)+H(Z|X)-H(Z|U) \\ \nonumber
&=&H(Y_{1}|U)-H(Z|U)+\frac{n}{2}\log(\frac{N_{3}}{N_{1}})\\ \nonumber
&=&H(X+n_{1}|U)-H(X+n_{3}|U)+\frac{n}{2}\log(\frac{N_{3}}{N_{1}})
\end{eqnarray}
On the other hand using (\ref{l11}), when $Z=0$ and $n_{2}=0$ then $Var(X|U)=\alpha P$. Therefore, According to the above lemma the Gaussian distribution is
optimum and $R_{1}$ is bounded as
\begin{eqnarray}
R_{1}&\leq& \frac{n}{2}\log\left( \frac{\alpha P + N_{1}}{\alpha P + N_{3}}\frac{N_{3}}{N_{1}} \right)\\ \nonumber
&=& C\left(\frac{\alpha P}{N_{1}}\right)+C\left(\frac{(1-\alpha) P}{\alpha P +N_{3}}\right)-C\left(\frac{P}{N_{3}}\right)
\end{eqnarray}
\end{proof}
\section{Conclusion}
A generalization of the wire-tap channel to the case of two
receivers and one eavesdropper is considered. We established an
inner bound for the general (non-degraded) case. This bound matches
Marton's bound on broadcast channels without security constraint.
Furthermore, we considered the scenario in which the channels are
degraded. We established the perfect secrecy capacity region for
this case. The achievability coding scheme is a secret superposition
scheme where randomization in the first layer helps the secrecy of
the second layer. The converse proof combines the converse proof for
the degraded broadcast channel without security constraint and the
perfect secrecy constraint. We proved that the secret superposition
scheme with Gaussian codebook is optimal in AWGN-BCCs. The converse
proof is based on the generalized entropy power inequality and
Csiszar lemma.


TITLE: Hyper-Virial Theorem for Central potentials
QUESTION [1 upvotes]: I saw an exercise on the hyper virial theorem for central potential and, as I was looking for some clues, I found it hasn't been solved here, so after finding the solution I thought I may include it as well in case anyone finds it useful (for some reason, I only encountered it until graduate school).
Consider a central potential of the form $V(r)=kr^{-s}$. Show that the hypervirial theorem satisfies:
$$\langle p_r r^s\rangle=-\frac{i\hbar s}{2}\langle r^{s-1}\rangle$$
Where $p_r=-i\hbar\partial/\partial_r$ and $s\in Z$

REPLY [1 votes]: The simplest way to prove it is by considering the commutator:
$[p_r,r^s]=r[p_r,r^{s-1}]+[p_r,r^{s-1}]r=r^2[p_r,r^{s-2}]+2r[p_r,r^{s-2}]+[p_r,r^{s-2}]r^2+...+=r^{s-1}[p_r,r]+...+[p_r,r]r^{s-1}=r^{s-1}(-i\hbar)+.. (\textrm{other} \ (s-2) \ \textrm{terms with} \ r^{s-1})..+(-i\hbar)r^{s-1}=-i\hbar sr^{s-1}$
Taking the expectation value:
$\langle[p_r,r^s]\rangle=-i\hbar\langle r^{s-1}\rangle$
On the other hand:
$\langle[p_r,r^s]\rangle=\langle p_r r^s-r^s p_r \rangle=\langle p_r r^s\rangle-\langle r^s p_r \rangle=\langle p_r r^s\rangle+\langle p_r r^s\rangle$
For stationary states. Therefore:
$\langle[p_r,r^s]\rangle=2\langle p_r r^s\rangle$
$$\langle p_r r^s\rangle=-\frac{i\hbar s}{2}\langle r^{s-1}\rangle$$
Which completes the proof. Feel free to add any other method you may consider useful.


TITLE: Calculating coefficients in a differential equation
QUESTION [0 upvotes]: It's a pretty open-minded exercise I found online. It says, you're advising a social network company and they're trying to model an equation for $u(t)$, being this the amount of active users in the network.
They know their equation will have this following "shape".
$$\frac{\partial^2 u}{\partial t^2} + c_{1} \frac{\partial u}{\partial t}  + c_{2} (1 + t)u = K$$
It says, how would you gather the data to make research easier (seems you have that social network, or a similar one you can sample data from). What mathematical tools would you use to estimate the coefficients $c1$, $c2$, and $K$. Finally, how would you verify and validate the final model you'd obtain?

REPLY [0 votes]: If the experimental data $(u_1,t_1)$ , $(u_2,t_2,)$ , ... , $(u_k,t_k)$, ... , $(u_n,t_n)$ are sufficiently numerous and well distributed, it is possible at each $t_k$ to compute approximates of $\frac{du}{dt}$ and $\frac{d^2u}{dt^2}$ thanks to numerical methods. Then a simple linear regression will lead to approximates of $c_1$ , $c_2$ and $K$.
But it is not the more accurate method, especially in case of not regulary distributed $t_k$. Better use numerical integration and the equation :
$$u +c_1\int udt +c_2\int\left(\int (1+t)udt \right)dt=K\left(\frac{t^2}{2}+at+b\right)$$
A linear regression leads to $c_1$, $c_2$, $K$, $a$, $b$.
The additional constants $a$ and $b$ depend on the lower bound of the integral in the numerical process. They are not used latter.
The general principle of this method is explained in the paper : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales
Of course, this would require a few preliminary studies and tests in order to develop a reliable algorithm. If you need more specific information and help about this method, you are welcome to contact me directly.


TITLE: $\frac{1}{n}\int\limits_0^1 f(x) dx=\int\limits_0^{\theta(n)} f(x) dx+\int\limits_{1-\theta(n)}^1 f(x) dx $
QUESTION [1 upvotes]: Let f be a positive and continuous function on the interval $[0, 1]$. It can be shown
that there exists, for every natural number $n$, a real number $\theta(n) \in [0, 1]$ such that
$$\frac{1}{n}\int\limits_0^1 f(x) dx=\int\limits_0^{\theta(n)} f(x) dx+\int\limits_{1-\theta(n)}^1 f(x) dx $$
What is
$\lim\limits_{n\rightarrow \infty} n\theta(n)$?
Please give some hint?

REPLY [2 votes]: Use the mean value theorem for integrals, there exists $x_n$ and $y_n$ such that:
$$\int_0^{\theta(n)}f(x)dx=f(x_n)\theta(n)\ \ \ \ \ \ \  \int_{1-\theta(n)}^1=f(y_n)\theta(n)
\tag 1$$
now prove the following:

$x_n\to 0$ and $y_n\to 1$ how : LHS tends to zero,hence RHS tends to zero and is a sum of positive terms because $f$ positive so every term tends to zero hence $\theta(n)\to 0$ and use $0\leq x_n\leq \theta(n)$ the same method for $y_n$.
$f$ is continue replace the formulas $(1)$ in the original equality and prove that:
$$n\theta(n)\to\frac{1}{f(1)+f(0)}\int\limits_0^1 f(x) dx $$

the limit is $+\infty$ if $f(0)=f(1)=0$


TITLE: Evaluate the integral $\int \frac{1}{x^3\sqrt{x^2+x-1}}\mathrm dx$
QUESTION [2 upvotes]: What substitution to use in this integral? I tried to factorize $x^2+x-1$ and use a substitution $u=\sqrt{\frac{a(x-\alpha)}{x-\beta}}$.

REPLY [2 votes]: $$I=\int \frac{1}{x^3\sqrt{x^2+x-1}}dx$$
Euler Substitution:
Let $x=\frac{u^2+1}{2u+1}$, $$I=2\int \frac{(2u+1)^2}{(u^2+1)^3}du$$
Partial fraction get $I=2\int \frac{4u-3}{(u^2+1)^3}du+8\int\frac{1}{(u^2+1)^2}du$


TITLE: Proof of w obeys Chi-square distribution with r degrees of freedom
QUESTION [0 upvotes]: In my econometrics book, it has a footnote saying:
Let $\mathbf{x}$ be an $m$ dimensional random vector, if $\mathbf{x} \sim N(\mu, \Sigma)$ where $\Sigma$ is nonsingular, then $ (\mathbf{x} - \mu)'\Sigma^{-1} (\mathbf{x} - \mu) \sim \chi^2(m)$ with $m$ degrees of freedom.
How to prove this?
Thanks a lot.

REPLY [2 votes]: It is not correct and should be: $$(\mathbf x-\mu)^T\Sigma^{-1}(\mathbf x-\mu)\sim\chi^2(m)$$
You can write $\mathbf x-\mu=A\mathbf u$ where $\mathbf u$ has standard normal distribution and $\Sigma=AA^T$.
Then:  $$(\mathbf x-\mu)^T\Sigma^{-1}(\mathbf x-\mu)=\mathbf u^T\mathbf u$$
Here $\mathbf u^T\mathbf u$ has chi-square distribution.


TITLE: Finding the percentage of water by volume of a solution in containers.
QUESTION [1 upvotes]: Two containers A and B contain solutions made up of only alcohol and water. The solution in container A has 30 percent alcohol by volume. They are both poured into an empty container which resulted in a solution having 25 percent water by volume. assume. Assume that the volume of the solution in container A is 3 times as large as that of container B. What is the percentage of water by volume of the solution in container B?
I have given this a try, but it doesn't seem to work.
$ a_C = \text{amount of alcohol in container} \, C$
$w_c = \text{amount of water in container} \, C$
$V_c = \text{volume of solution in container} \, C$
$a_A = 0.3 V_A$
$w_a = 0.7 V_A$ (shouldn't it be?)
$V_A = 3 V_B$
Then I tried this, but I am getting a negative value which doesn't seem to make sense unless you know what it means.
$w_A + w_b = 0.25 (V_a + V_b)$
$0.7 V_a + w_b = 0.25 ( 3 V_b + V_b ) $
$2.1 V_b + w_b = 0.25 (4 V_b) $
$2.1 V_b + w_b = V_b$
$w_b = -1.1 V_b$
I also tried assuming that the amount of water inside the new container is not $w_A + w_B$, say $W_{AB}$, but I don't think it is solvable then.
$W_{AB} = 0.25 (V_A + V_B) $
$W_{AB} = 0.25 (3 V_B + V_B) $
$W_{AB} = 0.25 (4 V_B)$
$W_{AB} = V_B$
This makes it seem that the amount of water inside the new container is just the amount of water inside container B. Thus, the percentage of water by volume of the solution in container B is 100%?
It is a multiple-choice question too.
a) 8
b) 9
c) 10
d) 12
e) none of these
Please help. Thank you.

REPLY [0 votes]: They mean to say the final mix is $25\%$ alcohol by volume, not water.  You should then get $10\%$ for an answer.  The final mix is $4$ times the volume of $B$, so it needs the volume of $B$ of alcohol.  $A$ supplies $3 \cdot 30\%$ of the volume of $B$ of alcohol, so you need $10\%$ of the volume of $B$ of alcohol to come from $B$.


TITLE: Number of real roots to $4^{\sin^2{x}}-2^{\cos{2x}}+1=0.$
QUESTION [2 upvotes]: Problem: Find the number of real roots to the equation $$4^{\sin^2{x}}-2^{\cos{2x}}+1=0 \ , \ \ \ \ \ \ \ \ \ 0<x\leq\pi.$$
Attempt: Simplifying i get $$2^{2\sin^2{x}}-2^{\cos{2x}}=2^{2(1-\cos{2x})}-2^{\cos{2x}}=\frac{2^2}{\left(2^{\cos{2x}}\right)^2}-2^{\cos{2x}}=-1.$$
Setting $t=2^{\cos{2x}}$ gives the third degree equation $$\frac{4}{t^2}-t+1=0 \Longleftrightarrow t^3-t^2-4=(t-2)(t^2+t+2)=0 \Longleftrightarrow t_{\mathbb{R}}=2.$$
So
$$2^{\cos{2x}}=t=2=2^1 \Longleftrightarrow \cos{2x}=1 \Longleftrightarrow x=\pi k \ , \ k\in \mathbb{Z}. $$
So, $k=0\Rightarrow x=0$ and $k=1\Rightarrow x=\pi.$ The only solution of $x$ that lies in the interval $(0,\pi]$ is given when $k=1$, thus the equation only has one solution in the given interval.
Questions:
1) I got the correct answer, but is my reasoning correct? Just wan't to make sure that the correct answer isn't just an accident/coincidence.
2) During the simplification step, I used that $\cos{2\theta} = 1-2\sin^2{\theta} \Leftrightarrow 2\sin^2{\theta}=1-\cos{2\theta}.$ Is there a more efficient way of using any other identity?
3) I factored the equation $t^3-t^2-4 = 0$ using a computer, but how can I solve it without knowing that one root is 2, in order to factor?

REPLY [3 votes]: Though you have made a mistake in your first line, but seeing the triviality of the mistake and the fact that you are aware of it now, let us answer your questions one by one.

The reasoning is correct and the answer is certainly not a fluke (for a moment let us shun out your mistake). I am talking generally about the procedural method of finding the answer.
Here is one more efficient way (as far as I am concerned)
$$4^{\sin^2{x}}-2^{\cos{2x}}+1=0 \ , \ \ \ \ \ \ \ \ \ 0<x\leq\pi.$$
$$(2^{\sin^2{x}})^2-2^{1-2\sin^{2}{x}}+1=0$$
$$(2^{\sin^2{x}})^2- \frac{2}{(2^{\sin^{2}{x}})^2}+1=0$$
Let us take $(2^{\sin^2{x}})^2=t$
$$\Rightarrow t^2-2+t=0 $$
This can be further solved easily and afterwards, you can put the value of $t$ in the equation and get the answer.
Though this is not much simpler, but still I thought of giving this way. Your own method is completely sufficient and right.

3.As far as cubic equations are concerned, there is no single formula(upto high school and undergraduate level) and hence most of the questions that you are going to encounter would require you to check some integral values and will give the right result also. But if you are not much satisfied with my explanation, you can check this website: https://math.vanderbilt.edu/schectex/courses/cubic/
Hope I have covered all the points and there is no doubt left.