TITLE: Mittag-Leffler function recurrence relation
QUESTION [2 upvotes]: The general Mittag-Leffler function $$E_{a,b}(z)=\sum_{h=0}^{\infty}\frac{z^h}{\Gamma(ha+b)}$$ satifies the recurrence $$E_{a,b}(z)=zE_{a,b+a}(z)+\frac1{\Gamma(b)}.$$
I am having a hard time in proving this recurrence. Does it follow immediately from definition. I am not getting it.I think the property of Gamma function is used here. Any easy proofs? Thanks beforehand.

REPLY [3 votes]: Well $\;\displaystyle zE_{a,b+a}(z)=\sum_{h=0}^{\infty}\frac{z^{h+1}}{\Gamma((h+1)a+b)}$ so yes it is immediate.
For other interesting properties of the Mittag-Leffler function see this thread.


TITLE: Commutative Diagram and Essential Extension
QUESTION [0 upvotes]: I have a question about commutative diagram.

is a commutative diagram of right R-modules and right R-module homomorphisms. Suppose that f and g
are isomorphisms, and also rows are exact. I want to show  if $\phi'(M) < M_2$ is an essential extension, then $h$ must be injective.

REPLY [1 votes]: Hint:
Since $M_2$ is an essential extension of $\phi'(M)$, to show $\ker h=\{0\}$, you only have to show that $\phi'(M)\cap \ker h=\{0\}$, i.e. that for all $m\in M$,
$$h\bigl(\phi'(m)\bigr)=0\implies \phi'(m)=0.$$
Set $n=g(m)$ and do some diagram hunting.
Some more details. Here's how you can start a-hunting:
As the right square is commutative, you can check that $\psi'(n)=0$, hence, as the bottom row is exact, there exists $n_1\in N_1$ such that $n=\psi(n_1)$. Now $f$ is an isomorphism, so there's a unique $m_1\in M_1$ such that $f(m_1)=n_1$. What can you say about $\phi(m_1)$, given that $g$ is an isomorphism?


TITLE: Experimental results of Higgs Boson
QUESTION [2 upvotes]: In 2012, ATLAS and CMS posted results of the mass of the Higgs boson, $126.0 \pm 0.4$ (stat) $\pm 0.4$ (sys) $\ \mathrm{GeV}/c^2$ and $125.3 \pm 0.4$ (stat) $\pm 0.5$ (sys) $\mathrm{GeV}/c^2$.
What do these kind of uncertainty mean? Why do they use two types of uncertainty?

REPLY [0 votes]: The 'stat' part is statistical error, so you have some number of events that form a peak, sitting on some number of events that form a smooth background. Estimating the centroid of that peak involves a statistical error.
You can get idea for how that works by randomly generating a Gaussian on top of some uniform distribution: how well do you estimate the known centroid? (It's possible the Higgs peak would be Lorentzian (Cauchy) on a smooth monotonic background, but the idea is the same).
The systematic error probably has 2 main sources:
(1) How well does the detector measure the mass (which may involve 2 main sources: measuring energy (via calorimeters) and measuring momentum (via bending in a magnetic field). Calorimeters capture electromagnetic showers and count photons. Photon may be absorbed before detection; energy may escape the calorimeter. Calibrating these effects have systematic biases.
Momentum measurement involves particle ID and tracking. Tracking involves systematic errors from wire/strip position inside the track-detector, and survey errors--as one must align the sub-detector element within the whole detector apparatus, all with respect to the beam axis. There is also magnetic field calibration and uniformity concerns.
Particle ID (e.g. Cherenkov and Transition Radiation detection) can also introduce systematic errors.
(2) Modeling the background: The Higgs mass peak sits on some smooth background that must me modeled by putting the Standard Model into a massive Monte Carlo an computing the events with no Higgs Boson.
Of course, the Standard Model deals with quarks, and the experimenter has protons (in the initial state), and pions/kaons/and so on, in the final state. It is a huge modeling effort to understand the quantitative connection between the 2--one needs nucleon structure functions/generalized Parton distributions) to understand the initial state, and hardronization models to understand the final state.
When speaking with laymen, one thing they find disappointing is that there is no single event that we can say: that is a Higgs Boson decaying. Rather we say: that is what a Higgs Boson decay looks like; now this event may be a background event that doesn't involve the Higgs at all. In fact, looking at some preliminary data:

you can say, "This event is most likely is not a Higgs Boson, but it could be. Maybe... But the Higgs is there, really."
As to why this is done--the statistical error is obviously random. The systematic error may have a random component, and it may be an overall bias for all measurements. When comparing the Higgs mass to other measured masses, that can be important information.
Also: when you are systemic error dominated, more beam-time will not improve the measurement, only a better apparatus (or Standard Model Monte Carlo) will.


TITLE: Generating function for product of binomial coefficients
QUESTION [5 upvotes]: In general, if $a(n)$ is an integer sequence with generating function $A(t)$ and $b(n)$ is an integer sequence with generating function $B(t)$, it is not easy to find the generating function $C(t)$ for $c(n)=a(n)b(n)$ in terms of $A$ and $B$, i.e., to find the Hadamard product of $A$ and $B$. However, it is not impossible to do so in some special cases.
I am interested in the case where $a(n) = \binom{N}{n}$ and $b(n) = \binom{M}{n}$, i.e., where the sequences are binomial coefficients. In this case $A(t) = (1+t)^N$ and $B(t) = (1+t)^M$. But what is $C$?
Thanks in advance for any advice or references!

REPLY [0 votes]: Here is a "solution" that might be useful; but it is not wholly satisfactory.
(as usual read carefully before using (sigh!))
We use the “Snake Oil Method” from Wilf's book “generatingfunctionology” (second edition) Amazon
or free: Free version
A good book in either form. Chapter 4, page 125 and following the example on 126.  We get:
$\left(\begin{array}{c}
N\\
n
\end{array}\right)\left(\begin{array}{c}
M\\
n
\end{array}\right)= \left[t^{n}\right]\left[s^{N}\right]\frac{1}{1-s}\cdot\left(1+\frac{\left(t\cdot s\right)}{\left(1-s\right)}\right)^{M}$
Let
$F(N,M,n;t)=\displaystyle \sum_{n=0}\left(\begin{array}{c}
N\\
n
\end{array}\right)\left(\begin{array}{c}
M\\
n
\end{array}\right)t^{n}$
We “dodge the bullet” and sum on an independent parameter N and variable s. In this case it's easy to undo afterwards.
$f(M,N;s,t)={\displaystyle \sum_{n=0}\left(\begin{array}{c}
i\\
n
\end{array}\right)\left(\begin{array}{c}
M\\
n
\end{array}\right)t^{n}s^{i}={\displaystyle {\displaystyle \sum_{n=0}}}}\left(\begin{array}{c}
M\\
n
\end{array}\right)t^{n}\cdot{\displaystyle \sum_{i=0}}\left(\begin{array}{c}
i\\
n
\end{array}\right)s^{i}$
$={\displaystyle {\displaystyle \sum_{n=0}}}\left(\begin{array}{c}
M\\
n
\end{array}\right)\cdot t^{n}\cdot\frac{s^{n}}{\left(1-s\right)^{n+1}}=\frac{1}{1-s}\cdot{\displaystyle {\displaystyle \sum_{n=0}}}\left(\begin{array}{c}
M\\
n
\end{array}\right)\cdot\frac{\left(t\cdot s\right)^{n}}{\left(1-s\right)^{n}}=\frac{1}{1-s}\cdot\left(1+\frac{\left(t\cdot s\right)}{\left(1-s\right)}\right)^{M}$
Now it's obvious how to pick out $t^{n}$
$\left[t^{n}\right]\frac{1}{1-s}\cdot\left(1+\frac{\left(t\cdot s\right)}{\left(1-s\right)}\right)^{M}=\frac{1}{1-s}\cdot\left(\begin{array}{c}
M\\
n
\end{array}\right)\cdot\left(\frac{\left(s\right)}{\left(1-s\right)}\right)^{n}$
And  $s^{N}$
$\left[s^{N}\right]\frac{1}{1-s}\left(\frac{\left(s\right)}{\left(1-s\right)}\right)^{n}=\left[s^{N-n}\right]\frac{1}{\left(1-s\right)^{n+1}}=\left(\begin{array}{c}
n+\left(N-n\right)\\
N-n
\end{array}\right)=\left(\begin{array}{c}
N\\
N-n
\end{array}\right)=\left(\begin{array}{c}
N\\
n
\end{array}\right)$
But not what we would prefer.
$\left[s^{N}\right]\frac{1}{1-s}\cdot\left(1+\frac{\left(t\cdot s\right)}{\left(1-s\right)}\right)^{M}$ to leave $\left[t^{n}\right]$
Which I can't see how to do.


TITLE: What does the following equation say about the conditional distribution $P(ab|xy)$?
QUESTION [1 upvotes]: What does the equation below say about the conditional distribution $P(ab|xy)$?
$$ P(ab|xy) = \int{ d\lambda \ \rho(\lambda)\ P(a|x\lambda)\ P(b|y\lambda) }$$
What does $\lambda$ denote here? Could you point me to some resources where I could learn about the semantics of the expression on the right hand side?
I can provide more context if needed.

REPLY [1 votes]: It sounds like you have random variables $A$, $B$, $X$, $Y$, $\Lambda$ for which
$$\Pr(A=a, B=b\mid X=x, Y=y) = \mathbb E_\lambda\left[\Pr(A=a\mid X=x, \Lambda=\lambda)\cdot\Pr(B=b\mid Y=y, \Lambda=\lambda) \right]$$
Here $\mathbb E$ is expectation and $\Pr$ is probability (I'm assuming discrete random variables $A$, $B$, $X$, $Y$ here but it would be similar for continuous ones.)
In words, it is not quite true that

$A$ and $B$ are independent given $X$ and $Y$,
$A$ only depends on $X$, and
$B$ only depends on $Y$.

But it is true when given $\Lambda$.
For instance let $A$=daughter's hairstyle, $B$=son's hairstyle, $X$=mom's hairstyle, $Y$=dad's hairstyle, and $\Lambda$=the age difference between son and daughter. Something like that.


TITLE: Any idea on how WolframAlpha did this sum?
QUESTION [0 upvotes]: So I've been trying to work out this sum for quite a while:

WolframAlpha unfortunately won't supply step by step proofs for this for some reason...
As for how i tried to prove this:
I looked at sin((2n-1)*a) as the imaginary part of cis((2n-1)a), which is a geometric series (a_n = a_1q^(n-1), where a_1 = cis(a) and q=cis(2a)), therefore its sum from n=1 to n=k should be:

Meaning the sum is:
(Note: I used De Moivre's Theorem to put the n power inside the cis)
But no matter how i tried continuing from here i always arrived at a huge trigonometric expression which i did not see how it (the imaginary part) could become what WolframAlpha is saying it is...
Any directions/solutions would be appreciated.
Thanks

REPLY [0 votes]: (i) $$\sin a+ \sin 3a+...+\sin ((2k-1)a)$$ $$=\sin a+\sin 2a+...+\sin 2ka-(\sin  2a+\sin 4a+...+\sin 2ka)$$ (ii) Use forward differences $\Delta f(x)=f(x+1)-f(x)$ and $\sum_{k=1}^b \Delta f(k)=f(b+1)-f(1)$ $$\text{(iii)   }\sin k \theta=(\Delta (\cos ((k-1/2)\theta)) )/(-2 \cos (\theta/2))$$ ()iv  Put in (i) $\theta=a$ and $\theta =2a$ and subtract.


text:
\begin{document}

\title{Factorizations of Matrices Over Projective-free Rings}

\author{H. Chen \thanks{Department of Mathematics, Hangzhou Normal University, Hangzhou,
310036, People's Republic of China, e-mail: huanyinchen@yahoo.cn},
H. Kose\thanks{Department of Mathematics, Ahi Evran University,
Kirsehir, Turkey, handankose@gmail.com}, Y. Kurtulmaz
\thanks{Department of Mathematics, Bilkent University, Ankara, Turkey, yosum@fen.bilkent.edu.tr}}

\maketitle

\begin{abstract} An element of a ring $R$ is called strongly $J^{\#}$-clean
provided that it can be written as the sum of an idempotent and an
element in $J^{\#}(R)$ that commute. We characterize, in this
article, the strongly $J^{\#}$-cleanness of matrices over
projective-free rings. These extend many known results on strongly
clean matrices over commutative local rings.

\vspace{2mm} \noindent {\bf 2010 Mathematics Subject Classification :} 15A13, 15B99, 16L99.\\
\noindent {\bf Key words}: strongly $J^{\#}$-matrix,
characteristic polynomial, projective-free ring.
\end{abstract}

\section{Introduction}

Let $R$ be a ring with an identity. We say that $x\in R$ is
strongly clean provided that there exists an idempotent $e\in R$
such that $x-e\in U(R)$ and $ex=xe$. A ring $R$ is strongly clean
in case every element in $R$ is strongly clean (cf. [9-10]). In
[2, Theorem 12], Borooah, Diesl, and Dorsey provide the following
characterization: Given a commutative local ring $R$ and a monic
polynomial $h\in R[t]$ of degree $n$, the following are
equivalent: $(1)$ $h$ has an $SRC$ factorization in $R[t]$; $(2)$
every $\varphi\in M_n(R)$ which satisfies $h$ is strongly clean.
It is demonstrated in [6, Example 3.1.7] that statement $(1)$ of
the above can not weakened from $SRC$ factorization to $SR$
factorization. The purpose of this paper is to investigate a
subclass of strongly clean rings which behave like such ones but
can be characterized by a kind of $SR$ factorizations, and so get
more explicit factorizations for many class of matrices over
projective-free rings.

Let $J(R)$ be the Jacobson radical of $R$. Set
$$J^{\#}(R)=\{ x\in R~\mid~\exists ~n\in {\Bbb N}~\mbox{such
that}~x^n\in J(R)\}.$$ For instance, let $R=M_2({\Bbb Z}_2)$. Then
$$J^{\#}(R)=\{ \left(
\begin{array}{cc}
0&0\\
0&0
\end{array}
\right), \left(
\begin{array}{cc}
0&1\\
0&0
\end{array}
\right), \left(
\begin{array}{cc}
0&0\\
1&0
\end{array}
\right) \},$$ while $J(R)=0$. Thus, $J^{\#}(R)$ and $J(R)$ are
distinct in general. We say that an element $a\in R$ is strongly
$J^{\#}$-clean provided that there exists an idempotent $e\in R$
such that $a-e\in J^{\#}(R)$ and $ea=ae$. If $R$ is a commutative
ring, then $a\in R$ is strongly $J^{\#}$-clean if and only if
$a\in R$ is strongly $J$-clean (cf. [3]). But they behave
different for matrices over commutative rings. A Jordan-Chevalley
decomposition of $n\times n$ matrix $A$ over an algebraically
closed field (e.g., the field of complex numbers), then $A$ is an
expression of it as a sum: $A=E+W$, where $E$ is semisimple, $W$
is nilpotent, and $E$ and $W$ commute. The Jordan-Chevalley
decomposition is extensively studied in Lie theory and operator
algebra. As a corollary, we will completely determine when an
$n\times n$ matrix over a filed is the sum of an idempotent matrix
and a nilpotent matrix that commute. Thus, the strongly
$J^{\#}$-clean factorizations of matrices over rings is also an
analog of that of Jordan-Chevalley decompositions for matrices
over fields.

We characterize, in this article, the strongly $J^{\#}$-cleanness
of matrices over projective-free rings. Here, a commutative ring
$R$ is projective-free provided that every finitely generated
projective $R$-module is free. For instances, every commutative
local ring, every commutative semi-local ring, every principal
ideal domain, every B$\acute{e}$zout domain (e.g., the ring of all
algebraic integers) and the ring $R[x]$ of all polynomials over a
principal domain $R$ are all projective-free. We will show that
strongly $J^{\#}$-clean matrices over projective-free rings are
completely determined by a kind of $``SC"$-factorizations of the
characteristic polynomials. These extend many known results on
strongly clean matrices to such new factorizations of matrices
over projective-free rings (cf. [1-2] and [5]).

Throughout, all rings with an identity and all modules are unitary
modules. Let $f(t)\in R[t]$. We say that $f(t)$ is a monic
polynomial of degree $n$ if $f(t)=t^n+a_{n-1}t^{n-1}+\cdots
+a_1t+a_0$ where $a_{n-1},\cdots ,a_1,a_0\in R$. We always use
$U(R)$ to denote the set of all units in a ring $R$. If
$\varphi\in M_n(R)$, we use $\chi (\varphi)$ to stand for the
characteristic polynomial $det( tI_n-\varphi)$.

\section{Full Matrices Over Projective-free Rings}

Let $A=\left(
\begin{array}{cc}
1&1\\
1&0
\end{array}
\right)\in M_2({\Bbb Z}_2)$. It is directly verified that $A\in
M_2({\Bbb Z}_2)$ is not strongly $J^{\#}$-clean, though $A$ is
strongly clean. It is hard to determine strongly cleanness even
for matrices over the integers, but completely different situation
is in the strongly $J^{\#}$-clean case. The aim of this section is
to characterize a single strongly $J^{\#}$-clean $n\times n$
matrix over projective-free rings. Let $M$ be a left $R$-module.
We denote the endomorphism ring of $M$ by $end(M)$.

\begin{lem} Let $M$ be a left $R$-module, and let $E=end(M)$, and let $\alpha\in E$. Then the following are equivalent:
\begin{enumerate}
\item [(1)]{\it $\alpha\in E$ is strongly $J^{\#}$-clean.}
\vspace{-.5mm}
\item [(2)]{\it There exists a left $R$-module decomposition $M=P\oplus Q$ where $P$ and $Q$ are $\alpha$-invariant, and
$\alpha|_P\in J^{\#}\big(end(P)\big)$ and $(1_M-\alpha)|_Q\in
J^{\#}\big(end(Q)\big)$.}
\end{enumerate}
\end{lem}
\begin{proof} $(1)\Rightarrow (2)$ Since $\alpha$ is strongly $J^{\#}$-clean in $E$, there exists an idempotent $\pi\in E$ and a $u\in J^{\#}(E)$
such that $\alpha=(1-\pi )+u$ and $\pi u=u\pi$. Thus,
$\pi\alpha=\pi u\in J^{\#}\big(\pi E\pi\big)$. Further, $1-\alpha
=\pi +(-u)$, and so $(1-\pi)(1-\alpha)=(1-\pi )(-u)\in J^{\#}\big(
(1-\pi)E(1-\pi)\big)$. Set $P=M\pi $ and $Q=M(1-\pi )$. Then
$M=P\oplus Q$. As $\alpha\pi=\pi\alpha$, we see that $P$ and $Q$
are $\alpha$-invariant. As $\alpha\pi\in J^{\#}\big(\pi
E\pi\big)$, we can find some $t\in {\Bbb N}$ such that
$(\alpha\pi)^t\in J\big(\pi E\pi\big)$. Let $\gamma\in end(P)$.
For any $x\in M$, it is easy to see that $(x)\pi \big(1_P-\gamma
\big(\alpha|_P\big)^t\big)=(x)\pi
\big(\pi-(\pi\overline{\gamma}\pi) (\pi\alpha\pi)^t\big)$ where
$\overline{\gamma}: M\to M$ given by
$(m)\overline{\gamma}=(m)\pi\gamma$ for any $m\in M$. Hence,
$1_P-\gamma \big(\alpha|_P\big)^t\in aut(P)$. Hence
$\big(\alpha|_P\big)^t\in J\big(end(P)\big)$. This implies that
$\alpha|_P\in J^{\#}\big(end(P)\big)$. Likewise, we verify that
$(1-\alpha)|_Q\in J^{\#}\big(end(Q)\big)$.

$(2)\Rightarrow (1)$ For any $\lambda\in end(Q)$, we construct an
$R$-homomorphism $\overline{\lambda}\in end(M)$ given by
$\big(p+q\big)\overline{\lambda}=(q)\lambda$. By hypothesis,
$\alpha|_P\in J^{\#}\big(end(P)\big)$ and $(1_M-\alpha)|_Q\in
J^{\#}\big(end(Q)\big)$. Thus, $\alpha
=\overline{1_{Q}}+\overline{\alpha
|_{P}}-\overline{(1_M-\alpha)|_{Q}}$. As $P$ and $Q$ are
$\alpha$-invariant, we see that
$\alpha\overline{1_{Q}}=\overline{1_{Q}}\alpha$. In addition,
$\overline{1_{Q}}\in end(M)$ is an idempotent. As
$\big(\overline{\alpha
|_{P}}\big)\big(\overline{(1_M-\alpha)|_{Q}}\big)=0=\big(\overline{(1_M-\alpha)|_{Q}}\big)\big(\overline{\alpha
|_{P}}\big)$, we show that $\overline{\alpha
|_{P}}-\overline{(1_M-\alpha)|_{Q}}\in J^{\#}\big(end(M)\big)$, as
required.
\end{proof}

\begin{lem} Let $R$ be a ring, and let $M$ be a left $R$-module. Suppose that $x,y,a,b\in end(M)$ such that $xa+yb=1_M, xy=yx=0, ay=ya$ and $xb=bx$.
Then $M=ker(x)\oplus ker(y)$ as left $R$-modules.
\end{lem}
\begin{proof} Straightforward. (cf. [6, Lemma 3.2.6]).
\end{proof}
\begin{lem} Let $R$ be a commutative ring, and let $\varphi\in M_n(R)$. Then the following are equivalent:
\begin{enumerate}
\item [(1)]{\it $\varphi\in J^{\#}\big(M_n(R)\big)$.}
\item [(2)]{\it $\chi (\varphi)\equiv t^n \big(mod~ J(R)\big)$, i.e., $\chi (\varphi)-t^n\in J(R)[t]$.}
\item [(3)]{\it There exists a monic polynomial $h\in R[t]$ such that $h\equiv t^{degh} \big(mod~J(R)\big)$ for which $h(\varphi)=0$.}
\end{enumerate}
\end{lem}
\begin{proof} $(1)\Rightarrow (2)$ Since $\varphi\in J^{\#}\big(M_n(R)\big)$, there exists some $m\in {\Bbb N}$ such that $\varphi^m\in
J\big(M_n(R)\big)$. As $J\big(M_n(R)\big)=M_n\big(J(R)\big)$, we
get $\overline{\varphi}\in N\big(M_n(R/J(R))\big)$. In view of [6,
Proposition 3.5.4], $\chi\big(\overline{\varphi}\big)\equiv
t^n\big(mod~N(R/J(R))\big)$. Write
$\chi(\varphi)=t^n+a_1t^{n-1}+\cdots +a_n$. Then
$\chi\big(\overline{\varphi}\big)=t^n+\overline{a_1}t^{n-1}+\cdots
+\overline{a_n}$. We infer that each $a_i^{m_i}+J(R)=0+J(R)$ where
$m_i\in {\Bbb N}$. This implies that $a_i\in J^{\#}(R)$. That is,
$\chi(\varphi)\equiv t^n \big(mod~ J^{\#}(R)\big)$. Obviously,
$J(R)\subseteq J^{\#}(R)$. For any $x\in J^{\#}(R)$, then there
exists some $m\in {\Bbb N}$ such that $x^n\in J(R)$. For any
maximal ideal $M$ of $R$, $M$ is prime, and so $x\in M$. This
implies that $x\in J(R)$; hence, $J^{\#}(R)\subseteq J(R)$.
Therefore $J^{\#}(R)=J(R)$, as required.

$(2)\Rightarrow (3)$ Choose $h=\chi (\varphi)$. Then $h\equiv
t^{degh} \big(mod~J(R)\big)$. In light of the Cayley-Hamilton
Theorem, $h(\varphi)=0$, as required.

$(3)\Rightarrow (1)$ By hypothesis, there exists a monic
polynomial $h\in R[t]$ such that $h\equiv t^{degh}
\big(mod~J(R)\big)$ for which $h(\varphi)=0$. Write
$h=t^n+a_1t^{n-1}+\cdots +a_n$. Choose
$\overline{h}=t^n+\overline{a_1}t^{n-1}+\cdots +\overline{a_n}\in
\big(R/J(R)\big)[t]$. Then $\overline{h}\equiv t^n \big( mod~
N(R/J(R))\big)$ for which
$\overline{h}\big(\overline{\varphi}\big) = 0$. According to [6,
Proposition 3.5.4], there exists some $m\in {\Bbb N}$ such that
$\big(\overline{\varphi}\big)^m=\overline{0}$ over $R/J(R)$.
Therefore $\varphi^m\in M_n\big(J(R)\big)$, and so $\varphi\in
J^{\#}\big(M_n(R)\big)$.
\end{proof}

\begin{df} For $r\in R$, define
$${\Bbb J}_r=\{ f\in R[t]~|~f~\mbox{monic, and}~f\equiv (t-r)^{^{degf}} \big(mod~ J^{{\#}}(R)\big)\}.$$
\end{df}
\begin{lem} Let $R$ be a projective-free ring, let $\varphi\in M_n(R)$, and let $h\in R[t]$ be a monic polynomial of degree $n$. If $h(\varphi)=0$ and
there exists a factorization $h=h_0h_1$ such that $h_0\in {\Bbb
J}_0$ and $h_1\in {\Bbb J}_1$, then $\varphi$ is strongly
$J^{\#}$-clean.
\end{lem}
\begin{proof} Suppose that $h=h_0h_1$ where $h_0\in {\Bbb J}_0$ and $h_1\in {\Bbb J}_1$. Write $h_0=t^p+a_1t^{p-1}+\cdots +a_p$
and $h_1=(t-1)^q+b_1t^{q-1}+\cdots +b_q$. Then each $a_i,b_j\in
J^{\#}(R)$. Since $R$ is commutative, we get each $a_i,b_j\in
J(R)$. Thus, $\overline{h_0}=t^p$ and
$\overline{h_1}=(t-\overline{1})^q$ in $\big(R/J(R)\big)[t]$.
Hence, $\big(\overline{h_0},\overline{h_1}\big)=\overline{1}$, In
virtue of [6, Lemma 3.5.10], we have some $u_0, u_1\in R[t]$ such
that $u_0h_0+u_1h_1=1$. Then
$u_0(\varphi)h_0(\varphi)+u_1(\varphi)h_1(\varphi)=1_{nR}$. By
hypothesis,
$h(\varphi)=h_0(\varphi)h_1(\varphi)=h_1(\varphi)h_0(\varphi)=0$.
Clearly, $u_0(\varphi)h_1(\varphi)=h_1(\varphi)u_0(\varphi)$ and
$h_0(\varphi)u_1(\varphi)=u_1(\varphi)h_0(\varphi)$. In light of
Lemma 2.2, $nR=ker\big(h_0(\varphi)\big)\oplus
ker\big(h_1(\varphi)\big)$. As $h_0t=th_0$ and $h_1t=th_1$, we see
that $h_0(\varphi)\varphi=\varphi h_0(\varphi)$ and
$h_1(\varphi)\varphi=\varphi h_1(\varphi)$, and so
$ker\big(h_0(\varphi)\big)$ and $ker\big(h_1(\varphi)\big)$ are
both $\varphi$-invariant. It is easy to verify that
$h_0\big(\varphi~|_{ker(h_0(\varphi))}\big)=0$. Since $h_0\in
{\Bbb J}_0$, we see that $h_0\equiv t^{degh_0}
\big(mod~J^{\#}(R)\big)$; hence, $\varphi~|_{ker(h_0(\varphi))}\in
J^{\#}\big(end(kerh_0(\varphi))\big)$.

It is easy to verify that
$h_1\big(\varphi~|_{ker(h_1(\varphi))}\big)=0$. Set
$g(u)=(-1)^{degh_1}h_1(1-u)$. Then
$g\big((~1-\varphi~)|_{ker(h_1(\varphi))}\big)=0$. Since $h_1\in
{\Bbb J}_1$, we see that $h_1\equiv
(t-1)^{degh_1}\big(mod~J^{\#}(R)\big)$. Hence, $g(u)\equiv
(-1)^{degh_1}(-u)^{degg}\big(mod~J(R)\big)$. This implies that
$g\in {\Bbb J}_0$. By virtue of Lemma 2.3,
$(1-\varphi)~|_{ker(h_1(\varphi)})\in
J^{\#}\big(end(ker(h_1(\varphi))\big))$. According to Lemma 2.1,
$\varphi\in M_n(R)$ is strongly $J^{\#}$-clean.
\end{proof}

The matrix
$$\varphi =\left(
\begin{array}{ccccc}
0&0&\cdots &0&-a_{0}\\
1&0&\cdots &0&-a_{1}\\
\vdots&\vdots&\ddots &\vdots&\vdots\\
0&0&\cdots &1&-a_{n-1}
\end{array}
\right)\in M_n(R)$$ is called the companion matrix $C_h$ of $h$,
where $h=t^n+a_{n-1}t^{n-1}+\cdots +a_1t+a_0\in R[t]$.

\begin{thm} Let $R$ be a projective-free ring and let $h\in R[t]$ be a monic polynomial of degree $n$. Then the following are equivalent:
\begin{enumerate}
\item [(1)]{\it Every $\varphi\in M_n(R)$ with $\chi (\varphi)=h$ is strongly $J^{\#}$-clean.}
\item [(2)]{\it The companion matrix $C_h$ of $h$ is strongly $J^{\#}$-clean.}
\item [(3)]{\it There exists a factorization $h=h_0h_1$ such that
$h_0\in {\Bbb J}_0$ and $h_1\in {\Bbb J}_1$.}
\end{enumerate}
\end{thm}
\begin{proof} $(1)\Rightarrow (2)$ Write $h=t^n+a_{n-1}t^{n-1}+\cdots +a_1t+a_0\in R[t]$. Choose $$C_h=\left(
\begin{array}{ccccc}
0&0&\cdots &0&-a_{0}\\
1&0&\cdots &0&-a_{1}\\
\vdots&\vdots&\ddots &\vdots&\vdots\\
0&0&\cdots &1&-a_{n-1}
\end{array}
\right)\in M_n(R).$$ Then $\chi (C_h)=h$. By hypothesis, $C_h\in
M_n(R)$ is strongly $J^{\#}$-clean.

$(2)\Rightarrow (3)$ In view of Lemma 2.1, there exists a
decomposition $nR=A\oplus B$ such that $A$ and $B$ are
$\varphi$-invariant, $\varphi~|_{A}\in J^{\#}\big(end_R(A)\big)$
and $(1-\varphi)~|_{B}\in J^{\#}\big(end_R(B)\big)$. Since $R$ is
a projective-free ring, there exist $p,q\in {\Bbb N}$ such that
$A\cong pR$ and $B\cong qR$. Regarding $end_R(A)$ as $M_p(R)$, we
see that $\varphi~|_{A}\in J^{\#}\big(M_p(R)\big)$. By virtue of
Lemma 2.3, $\chi (\varphi~|_{A})\equiv t^p \big(mod~
J^{\#}(R)\big)$. Thus $\chi (\varphi~|_{A})\in {\Bbb J}_0$.
Analogously, $(1-\varphi)~|_{B}\in J^{\#}\big(M_q(R)\big)$. It
follows from Lemma 2.3 that $\chi
\big((1-\varphi)~|_{B}\big)\equiv
t^{q}~\big(~\mbox{mod}~J^{\#}(R)\big)$. This implies that
$det\big(\lambda I_q- (1-\varphi)~|_{B}\big)\equiv
\lambda^{q}~\big(~\mbox{mod}~J^{\#}(R)\big)$. Hence,
 $det\big((1-\lambda)I_q- \varphi~|_{B}\big)\equiv (-\lambda)^{q}~\big(~\mbox{mod}~J^{\#}(R)\big)$. Set $t=1-\lambda$. Then
 $det\big(tI_q- \varphi~|_{B}\big)\equiv (t-1)^{q}~\big(~\mbox{mod}~J^{\#}(R)\big)$. Therefore we get $\chi (\varphi~|_{B})
 \equiv (t-1)^{q}~\big(~\mbox{mod}~J^{\#}(R)\big)$. We infer that $\chi (\varphi~|_{B})\in {\Bbb J}_1$.
 Clearly, $\chi (\varphi)=\chi (\varphi~|_{A})\chi (\varphi~|_{B})$. Choose $h_0=\chi (\varphi~|_{A})$ and $h_1=\chi (\varphi~|_{B})$.
 Then there exists a factorization $h=h_0h_1$ such that $h_0\in {\Bbb J}_0$ and $h_1\in {\Bbb J}_1$, as desired.

$(3)\Rightarrow (1)$ For every $\varphi\in M_n(R)$ with $\chi
(\varphi)=h$, it follows by the Cayley-Hamilton Theorem that
$h(\varphi)=0$. Therefore $\varphi$ is strongly $J^{\#}$-clean by
Lemma 2.5.
\end{proof}

\begin{cor} Let $F$ be a field, and let $A\in M_n(F)$. Then the following are equivalent:
\begin{enumerate}
\item [(1)]{\it $A$ is the sum of an idempotent matrix and a nilpotent matrix that
commute.}
\item [(2)]{\it $\chi(A)=t^s(t-1)^t$ for some $s,t\geq 0$.}
\end{enumerate}
\end{cor}
\begin{proof} As
$J\big(M_n(F)\big)=0$, we see that a $n\times n$ matrix contains
in $J^{\#}\big(M_n(F)\big)$ if and only if $A$ is a nilpotent
matrix. So $A\in M_n(F)$ is strongly $J^{\#}$-clean if and only if
$A$ is the sum of an idempotent matrix and a nilpotent matrix that
commute. By virtue of Theorem 2.6, we see that $A\in M_n(F)$ is
the sum of an idempotent matrix and a nilpotent matrix that
commute if and only if $\chi(A)=h_0h_1$, where $h_0\in {\Bbb J}_0$
and $h_1\in {\Bbb J}_1$. Clearly, $h_0\in {\Bbb J}_0$ if and only
if $h_0\equiv t^{degh_0}(mod~J^{\#}(F))$. But $J^{\#}(F)=0$, and
so $h_0=t^s$, where $s=degh_0$. Likewise, $h_1=(t-1)^t$, where
$t=degh_1$. Therefore we complete the proof.\end{proof}

For matrices over integers ,we have a similar situation. As
$J\big(M_n({\Bbb Z})\big)=0$, we see that an $n\times n$ matrix
contains in $J^{\#}\big(M_n({\Bbb Z})\big)$ if and only if it is a
nilpotent matrix. Likewise, we show that $A\in M_n({\Bbb Z})$ is
the sum of an idempotent matrix and a nilpotent matrix that
commute if and only if $\chi(A)=t^s(t-1)^t$ for some $s,t\geq 0$.
For instance, choose $A=\left(
\begin{array}{ccc}
-2&2&-1\\
-4&4&-2\\
-1&1&0
\end{array}
\right)\in M_3({\Bbb Z})$. Then $\chi(A)=t(t-1)^2$. Thus, $A$ is
the sum of an idempotent matrix and an nilpotent matrix that
commute. In fact, we have a corresponding factorization $A=\left(
\begin{array}{ccc}
-1&1&0\\
-2&2&0\\
0&0&1
\end{array}
\right)+\left(
\begin{array}{ccc}
-1&1&-1\\
-2&2&-2\\
-1&1&-1
\end{array}
\right)$.

\begin{cor} Let $R$ be a projective-free ring, and let $\varphi\in M_2(R)$. Then $\varphi$ is strongly $J^{\#}$-clean if and only if
\begin{enumerate}
\item [(1)]{\it $\chi (\varphi )\equiv t^2 \big(mod~ J(R)\big)$; or}
\vspace{-.5mm}
\item [(2)]{\it $\chi (\varphi )\equiv (t-1)^2 \big(mod~ J(R)\big)$; or}\vspace{-.5mm}
\item [(3)]{\it $\chi (\varphi )$ has a root in $J(R)$ and a root in $1+J(R)$.}\vspace{-.5mm}
\end{enumerate}
\end{cor}
\begin{proof} Suppose that $\varphi$ is strongly $J^{\#}$-clean. By virtue of Theorem 2.6, there exists a factorization $\chi(\varphi )=h_0h_1$ such
that $h_0\in {\Bbb J}_0$ and $h_1\in {\Bbb J}_1$.

Case I. $deg(h_0)=2$ and $deg(h_1)=0$. Then $h_0=\chi(\varphi )=
t^2-tr(\varphi)t+det(\varphi)$ and $h_1=1$. As $h_0\in {\Bbb
J}_0$, it follows from Lemma 2.3 that $\varphi\in
J^{\#}\big(M_2(R)\big)$ or $\chi (\varphi )\equiv t^2 \big(mod~
J(R)\big)$.

Case II. $deg(h_0)=1$ and $deg(h_1)=1$. Then $h_0=t-\alpha$ and
$h_1=t-\beta$. Since $R$ is commutative, $J^{\#}(R)=J(R)$. As
$h_0\in {\Bbb J}_0$, we see that $h_0\equiv t (mod~ J(R))$, and
then $\alpha\in J(R)$. As $h_1\in {\Bbb J}_1$, we see that
$h_1\equiv t-1 (mod~ J(R))$, and then $\beta\in 1+J(R)$. Therefore
$\chi (\varphi )$ has a root in $J(R)$ and a root in $1+J(R)$.

Case III. $deg(h_0)=0$ and $deg(h_1)=2$. Then
$h_1(t)=det\big(tI_2-\varphi\big)\equiv (t-1)^2 (mod~ J(R))$. Set
$u=1-t$. Then $det\big(uI_2-(I_2-\varphi)\big)\equiv u^2 \big(mod~
J(R)\big)$. According to Lemma 2.3, $I_2-\varphi\in
J^{\#}\big(M_2(R)\big)$ or $\chi (\varphi )\equiv (t-1)^2
\big(mod~ J(R)\big)$.

We will suffice to show the converse. If $\chi (\varphi )\equiv
t^2 \big(mod~ J(R)\big)$ or $\chi (\varphi )\equiv (t-1)^2
\big(mod~ J(R)\big)$, then $\varphi\in J^{\#}\big(M_2(R)\big)$ or
$I_2-\varphi \in J^{\#}\big(M_2(R)\big)$. This implies that
$\varphi$ is strongly $J^{\#}$-clean. Otherwise, $\varphi,
I_2-\varphi\not\in J\big(M_2(R)\big)$. In addition, $\chi (\varphi
)$ has a root in $J(R)$ and a root in $1+J(R)$. According to [4,
Theorem 16.4.31], $\varphi$ is strongly $J$-clean, and therefore
it is strongly $J^{\#}$-clean.
\end{proof}

Choose $A=\left(
\begin{array}{cc}
\overline{0}&\overline{2}\\
\overline{1}&\overline{3}
\end{array}
\right)\in M_2\big({\Bbb Z}_4\big)$. It is easy to check that $A,
I_2-A\in M_2\big({\Bbb Z}_4\big)$ are not nilpotent. But
$\chi(A)=t^2+t+2$ has a root $\overline{2}\in J({\Bbb Z}_4)$ and a
root $\overline{1}\in 1+J({\Bbb Z}_4)$. As $J({\Bbb Z}_4)=\{
\overline{0},\overline{2}\}$ is nil, we know that every matrix in
$J^{\#}\big(M_2({\Bbb Z}_4)\big)$ is nilpotent. It follows from
Corollary 2.8 that $A$ is the sum of an idempotent matrix and a
nilpotent matrix that commute. Let ${\Bbb Z}_{(2)}=\{
\frac{m}{n}~|~m,n\in {\Bbb Z}, 2~\nmid~n\}$, and let $A=\left(
\begin{array}{cc}
1&1\\
\frac{2}{9}&0
\end{array}
\right)\in M_2({\Bbb Z}_{(2)})$. Then $J({\Bbb Z}_{(2)})=\{
\frac{2m}{n}~|~m,n\in {\Bbb Z}, 2~\nmid~n\}$. As
$\chi(A)=t^2-t+\frac{2}{9}$ has a root $\frac{1}{3}\in 1+J({\Bbb
Z}_{(2)})$ and a root $\frac{2}{3}\in J({\Bbb Z}_{(2)})$. In light
of Corollary 2.8, $A$ is strongly $J$-clean.

\begin{cor} Let $R$ be a projective-free ring, and let $f(t)=t^2+at+b\in R[t]$ be degree $2$ polynomial with $1+a\in J(R),
b\not\in J(R)$. Then the following are equivalent:
\begin{enumerate}
\item [(1)]{\it Every $\varphi\in M_2(R)$ with $\chi (\varphi )=f(t)$ is strongly $J^{\#}$-clean.}
\vspace{-.5mm}
\item [(2)]{\it There exist $r_1\in J(R)$ and $r_2\in 1+J(R)$ such that $f(r_i)=0$.}
\item [(3)]{\it There exists $r\in J(R)$ such that $f(r)=0$.}
\end{enumerate}
\end{cor}
\begin{proof} $(1)\Rightarrow (2)$ Since every $\varphi\in M_2(R)$ with $\chi (\varphi )=f(t)$ is strongly $J^{\#}$-clean, it follows by Corollary
2.8 that $f(t)=(t-r_1)(t-r_2)$ with $r_1\in J(R), r_2\in 1+J(R)$.

$(2)\Rightarrow (3)$ is trivial.

$(3)\Rightarrow (1)$ As $r^2+ar+b=0$, we see that
$f(t)=(t-r)(t+a+r)$. Clearly, $t-r\in {\Bbb J}_0$. As $1+a+r\in
J(R)$, we see that $t+a+r\in {\Bbb J}_1$. According to Theorem
2.6, we complete the proof.
\end{proof}

Let $\varphi$ be a $3\times 3$ matrix over a commutative ring $R$.
Set $mid(\varphi)=det(I_3-\varphi)-1+tr(\varphi)+det(\varphi).$

\begin{cor} Let $R$ be a projective-free ring, and let $\varphi\in M_3(R)$. Then $\varphi$ is strongly $J^{\#}$-clean if and
only if

{\rm (1)} $\chi (\varphi )\equiv t^3 \big(mod~ J(R)\big)$; or

{\rm (2)} $\chi (\varphi )\equiv (t-1)^3 \big(mod~ J(R)\big)$; or

{\rm (3)} $\chi(\varphi )$ has a root in $1+J(R)$,$tr(\varphi)\in
1+J(R)$,mid$(\varphi)\in J(R), det(\varphi)\in J(R)$;or

{\rm (4)} $\chi (\varphi )$ has a root in $J(R)$, $tr(\varphi)\in
2+J(R)$, mid$(\varphi)\in 1+J(R), det(\varphi)\in J(R)$.

\end{cor}

\begin{proof} Suppose that $\varphi$ is strongly $J^{\#}$-clean. By virtue of Theorem 2.6, there exists a factorization $\chi(\varphi )=h_0h_1$ such that $h_0\in {\Bbb J}_0$ and $h_1\in {\Bbb J}_1$.

Case I. $deg(h_0)=3$ and $deg(h_1)=0$. Then $h_0=\chi(\varphi )$
and $h_1=1$. As $h_0\in {\Bbb J}_0$, it follows from Lemma 2.3
that $\varphi\in J^{\#}\big(M_3(R)\big)$.

Case II. $deg(h_0)=0$ and $deg(h_1)=3$. Then
$h_1(t)=det\big(tI_3-\varphi\big)\equiv (t-1)^3 (mod~ J(R))$. Set
$u=1-t$. Then $det\big(uI_3-(I_3-\varphi)\big)\equiv u^3(mod~
J(R))$. According to Lemma 2.3, $I_3-\varphi\in
J^{\#}\big(M_3(R)\big)$.

Case III. $deg(h_0)=2$ and $deg(h_1)=1$. Then $h_0=t^2+at+b$ and
$h_1=t-\alpha$. As $h_0\in {\Bbb J}_0$, we see that $h_0\equiv t^2
(mod~ J(R))$; hence, $a,b\in J(R)$. As $h_1\in {\Bbb J}_1$, we see
that $h_1\equiv t-1 (mod~ J(R))$; hence, $\alpha\in 1+J(R)$. We
see that $a-\alpha=-tr(\varphi),b-a\alpha=mid(\varphi)$ and
$-b\alpha=-det(\varphi)$. Therefore $tr(\varphi)\in 1+J(R),
mid(\varphi)\in J(R)$ and $det(\varphi)\in J(R)$.

Case IV. $deg(h_0)=1$ and $deg(h_1)=2$. Then $h_0=t-\alpha$ and
$h_1=t^2+at+b$. As $h_0\in {\Bbb J}_0$, we see that $h_0\equiv t
(mod~ J(R))$; hence, $\alpha\in J(R)$. As $h_1\in {\Bbb J}_1$, we
see that $h_1\equiv (t-1)^2 (mod~ J(R))$, and then $a\in -2+J(R)$
and $b\in 1+J(R)$. Obviously,
$\chi(\varphi)=t^3-tr(\varphi)t^2+mid(\varphi)t-det(\varphi)$, and
so $a-\alpha=-tr(\varphi),b-a\alpha=mid(\varphi)$ and
$-b\alpha=-det(\varphi)$. Therefore $tr(\varphi)\in 2+J(R),
mid(\varphi)\in 1+J(R)$ and $det(\varphi)\in J(R)$.

Conversely, if $\chi (\varphi )\equiv t^3 \big(mod~ J(R)\big)$ or
$\chi (\varphi )\equiv (t-1)^3 \big(mod~ J(R)\big)$, then
$\varphi\in J^{\#}\big(M_3(R)\big)$ or $I_3-\varphi \in
J^{\#}\big(M_3(R)\big)$. Hence, $\varphi$ is strongly
$J^{\#}$-clean. Suppose $\chi (\varphi )$ has a root $\alpha\in
1+J(R)$ and $tr(\varphi)\in 1+J(R), det(\varphi)\in J(R)$. Then
$\chi (\varphi)=(t^2+at+b)(t-\alpha)$ for some $a,b\in R$. This
implies that $a-\alpha=-tr(\varphi), -b\alpha=-det(\varphi)$.
Hence, $a,b\in J(R)$. Let $h_0=t^2+at+b$ and $h_1=t-\alpha$. Then
$\chi(\varphi)=h_0h_1$ where $h_0\in {\Bbb J}_0$ and $h_1\in {\Bbb
J}_1$. According to Theorem 2.6, $\varphi$ is strongly
$J^{\#}$-clean.

Suppose $\chi (\varphi )$ has a root $\alpha\in J(R)$ and
$tr(\varphi)\in 2+J(R), mid(\varphi)\in 1+J(R)$ and
$det(\varphi)\in J(R)$. Then $\chi (\varphi)=(t-\alpha)(t^2+at+b)$
for some $a,b\in R$. This implies that $a-\alpha=-tr(\varphi),
b-a\alpha=mid(\varphi)$. Hence, $a\in -2+J(R)$,$b\in 1+J(R)$. Let
$h_0=t-\alpha$ and $h_1=t^2+at+b$. Then $\chi(\varphi)=h_0h_1$
where $h_0\in {\Bbb J}_0$ and $h_1\in {\Bbb J}_1$. According to
Theorem 2.6, $\varphi$ is strongly $J^{\#}$-clean, and we are
done.
\end{proof}

\section{Matrices Over Power Series Rings}

The purpose of this section is to extend the preceding discussion
to matrices over power series rings. We use $R[[x]]$ to stand for
the ring of all power series over $R$. Let
$A(x)=\big(a_{ij}(x)\big)\in M_n\big(R[[x]]\big)$. We use $A(0)$
to stand for $\big(a_{ij}(0)\big)\in M_n(R)$.

\begin{thm} Let $R$ be a projective-free ring, and let $A(x)\in M_2\big(R[[x]])$. Then the following are equivalent:
\begin{enumerate}
\item [(1)]{\it $A(x)\in M_2\big(R[[x]])$ is strongly $J^{\#}$-clean.}
\vspace{-.5mm}
\item [(2)]{\it $A(0)\in M_2(R)$ is strongly $J^{\#}$-clean.}\vspace{-.5mm}
\end{enumerate}
\end{thm}
\begin{proof} $(1)\Rightarrow (2)$ Since $A(x)$ is strongly $J^{\#}$-clean in $M_2\big(R[[x]]\big)$, there exists an $E(x)=E^2(x)\in
M_2\big(R[[x]]\big)$ and a $U(x)\in J^{\#}\big(M_2(R[[x]])\big)$
such that $A(x)=E(x)+U(x)$ and $E(x)U(x)=U(x)E(x)$. This implies
that $A(0)=E(0)+U(0)$ and $E(0)U(0)=U(0)E(0)$ where
$E(0)=E^2(0)\in M_2(R)$ and $U(0)\in J^{\#}\big(M_2(R)\big)$. As a
result, $A(0)$ is strongly $J^{\#}$-clean in $M_2(R)$.

$(2)\Rightarrow (1)$ Construct a ring morphism $\varphi: R[[x]]\to
R, f(x)\mapsto f(0)$. Then $R\cong R[[x]]/kerf$, where $kerf=\{
f(x)~|~f(0)=0\}\subseteq J\big(R[[x]]\big)$. For any finitely
generated projective $R[[x]]$-module $P$,
$P\bigotimes\limits_{R}\big(R[[x]]/kerf\big)$ is a finitely
generated projective $R[[x]]/kerf$-module; hence it is free. Write
$P\bigotimes\limits_{R}\big(R[[x]]/kerf\big)\cong
\big(R[[x]]/kerf\big)^m$ for some $m{\Bbb N}$. Then
$P\bigotimes\limits_{R}\big(R[[x]]/kerf\big)\cong
\big(R[[x]]\big)^m\bigotimes\limits_{R}\big(R[[x]]/kerf\big)$.
That is, $P/P\big(kerf\big)$ $\cong
\big(R[[x]]\big)^m/\big(R[[x]]\big)^m\big(kerf\big)$
wit$kerf\subseteq J\big(R[[x]]\big)$. By Nakayama Theorem, $P\cong
\big(R[[x]]\big)^m$ is free. Thus, $R[[x]]$ is projective-free.
Since $A(0)$ is strongly $J^{\#}$-clean in $M_2(R)$, it follows
from Corollary 2.8 that $A(0)\in J^{\#}\big(M_2(R)\big)$, or
$I_2-A(0)\in J^{\#}\big(M_2(R)\big)$, or the characteristic
polynomial $\chi\big(A(0)\big)=y^2+\mu y+\lambda$ has a root
$\alpha\in 1+J(R)$ and a root $\beta\in J(R)$. If $A(0)\in
J^{\#}\big(M_2(R)\big)$, then $A(x)\in
J^{\#}\big(M_2(R[[x]])\big)$. If $I_2-A(0)\in
J^{\#}\big(M_2(R)\big)$, then $I_2-A(x)\in
J^{\#}\big(M_2(R[[x]])\big)$. Otherwise, we write
$y=\sum\limits_{i=0}^{\infty}b_ix^i$  and $\chi(A(x))=y^2-\mu
(x)y-\lambda (x)$. Then $y^2=\sum\limits_{i=0}^{\infty}c_ix^i$
where $c_i=\sum\limits_{k=0}^{i}b_kb_{i-k}$. Let
$\mu(x)=\sum\limits_{i=0}^{\infty}\mu_ix^i,
\lambda(x)=\sum\limits_{i=0}^{\infty}\lambda_ix^i\in R[[x]]$ where
$\mu_0=\mu$ and $\lambda_0=\lambda$. Then, $y^2-\mu (x)y-\lambda
(x)=0$ holds in $R[[x]]$ if the following equations are satisfied:
$$\begin{array}{c}
b_0^2-b_0\mu_0-\lambda_0=0;\\
(b_0b_1+b_1b_0)-(b_0\mu_1+b_1\mu_0)-\lambda_1=0;\\
(b_0b_2+b_1^2+b_2b_0)-(b_0\mu_2+b_1\mu_1+b_2\mu_0)-\lambda_2=0;\\
\vdots
\end{array}$$ Obviously, $\mu_0=\alpha + \beta\in U(R)$ and $\alpha - \beta\in U(R)$ . Let $b_0=\alpha$. Since
$R$ is commutative, there exists some $b_1\in R$ such that
$$b_0b_1+b_1(b_0-\mu_0)=\lambda_1+b_0\mu_1.$$ Further, there exists some $b_2\in R$ such that
$$b_0b_2+b_2(b_0-\mu_0)=\lambda_2-b_1^2+b_0\mu_2+b_1\mu_1.$$ By
iteration of this process, we get $b_3,b_4,\cdots $. Then $y^2-\mu
(x)y -\lambda (x)=0$ has a root $y_0(x)\in 1+J\big(R[[x]]\big)$.
If $b_0=\beta\in J(R)$, analogously, we show that $y^2-\mu (x)y
-\lambda (x)=0$ has a root $y_1(x)\in J\big(R[[x]]\big)$. In light
of Corollary 2.8, the result follows.
\end{proof}

\begin{cor} Let $R$ be a projective-free ring, and let $A(x)\in M_2\big(R[[x]]/(x^{m})\big)$ $(m\geq 1)$. Then the following
are equivalent:
\begin{enumerate}
\item [(1)]{\it $A(x)\in M_2\big(R[[x]]/(x^m)\big)$ is strongly $J^{\#}$-clean.}
\vspace{-.5mm}
\item [(2)]{\it $A(0)\in M_2(R)$ is strongly $J^{\#}$-clean.}\vspace{-.5mm}
\end{enumerate}
\end{cor}
\begin{proof} $(1)\Rightarrow (2)$ is obvious.

$(2)\Rightarrow (1)$ Let $\psi: R[[x]]\to R[[x]]/(x^m),
\psi(f)=\overline{f}$. Then it reduces a surjective ring
homomorphism $\psi^*: M_2\big(R[[x]]\big)\to
M_2\big(R[[x]]/(x^m)\big)$. Hence, we have a $B\in
M_2\big(R[[x]]\big)$ such that $\psi^*\big(B(x)\big)=A(x)$.
According to Theorem 3.1, we complete the proof.
\end{proof}

\begin{ex} {\rm Let $R={\Bbb Z}_4[x]/(x^2)$, and let $A(x)=\left( \begin{array}{cc}
\overline{2}&\overline{2}+\overline{2}x\\
\overline{2}+x&\overline{3}+\overline{3}x \end{array} \right)$
$\in M_2(R)$. Obviously, ${\Bbb Z}_4$ is a projective-free ring,
and that $R={\Bbb Z}_4[[x]]/(x^2)$. Since we have the strongly
$J^{\#}$-clean decomposition $A(0)=\left(
\begin{array}{cc}
\overline{0}&\overline{2}\\
\overline{2}&\overline{1} \end{array} \right)+\left(
\begin{array}{cc}
\overline{2}&\overline{0}\\
\overline{0}&\overline{2} \end{array} \right)$ in $M_2({\Bbb
Z}_4)$, it follows by Corollary 3.2 that $A(x)\in M_2(R)$ is
strongly $J^{\#}$-clean.}
\end{ex}

\begin{thm} Let $R$ be a projective-free ring, and let $A(x)\in M_3\big(R[[x]])$. Then the following are equivalent:
\begin{enumerate}
\item [(1)]{\it $A(x)\in M_3\big(R[[x]])$ is strongly $J^{\#}$-clean.}
\vspace{-.5mm}
\item [(2)]{\it $A(x)\in M_3\big(R[[x]]/(x^m)\big)(m\geq 1)$ is strongly $J^{\#}$-clean.}
\item [(3)]{\it $A(0)\in M_3(R)$ is strongly $J^{\#}$-clean.}
\vspace{-.5mm}
\end{enumerate}
\end{thm}
\begin{proof} $(1)\Rightarrow (2)$ and $(2)\Rightarrow (3)$ are clear.

$(3)\Rightarrow (1)$ As $A(0)$ is strongly $J^{\#}$-clean in
$M_3(R)$, it follows from Corollary 2.10 that $A(0)\in
J^{\#}\big(M_3(R)\big)$, or $I_3-A(0)\in J^{\#}\big(M_3(R)\big)$,
or $\chi\big(A(0)\big)$ has a root in $J(R)$ and
$tr\big(A(0)\big)\in 2+J(R), mid\big(A(0)\big)\in 1+J(R),
det\big(A(0)\big)\in J(R)$, or $\chi\big(A(0)\big)$ has a root in
$1+J(R)$ and $tr\big(A(0)\big)\in 1+J(R), mid\big(A(0)\big)\in
J(R), det\big(A(0)\big)\in J(R)$. If $A(0)\in
J^{\#}\big(M_3(R)\big)$ or $I_3-A(0)\in J^{\#}\big(M_3(R)\big)$,
then $A(x)\in J^{\#}\big(M_3(R[[x]])\big)$ or $I_3-A(x)\in
J^{\#}\big(M_3(R[[x]])\big)$. Hence, $A(x)\in M_3\big(R[[x]]\big)$
is strongly $J^{\#}$-clean. Assume that
$\chi\big(A(0)\big)=t^3-\mu t^2-\lambda t-\gamma $ has a root
$\alpha\in J(R)$ and $tr\big(A(0)\big)\in
2+J(R),mid\big(A(0)\big)\in 1+J(R),det\big(A(0)\big)\in J(R)$.
Write $y=\sum\limits_{i=0}^{\infty}b_ix^i$. Then
$y^2=\sum\limits_{i=0}^{\infty}c_ix^i$ where
$c_i=\sum\limits_{k=0}^{i}b_kb_{i-k}$. Further,
$y^3=\sum\limits_{i=0}^{\infty}d_ix^i$ where
$d_i=\sum\limits_{k=0}^{i}b_kc_{i-k}$. Let
$\mu(x)=\sum\limits_{i=0}^{\infty}\mu_ix^i,
\lambda(x)=\sum\limits_{i=0}^{\infty}\lambda_ix^i,
\gamma(x)=\sum\limits_{i=0}^{\infty}\gamma_ix^i\in R[[x]]$ where
$\mu_0=\mu,\lambda_0=\lambda$ and $\gamma_0=\gamma$. Then,
$y^3-\mu (x)y^2-\lambda (x)y-\gamma(x)=0$ holds in $R[[x]]$ if the
following equations are satisfied:
$$\begin{array}{c}
b_0^3-b_0^2\mu_0-b_0\lambda_0-\gamma_0=0;\\
(3b_0^2-2b_0\mu_0-\lambda_0)b_1=\gamma_1+b_0^2\mu_1+b_0\lambda_1;\\
(3b_0^2-2b_0\mu_0-\lambda_0)b_2=\gamma_2+b_0^2\mu_2+b_1^2\mu_0+2b_0b_1\mu_1+b_0\lambda_2+b_1\lambda_0-3b_0b_1^2;\\
\vdots
\end{array}$$ Let $b_0=\alpha\in J(R)$. Obviously, $\mu_0=trA(0)\in 2+J(R)$ and $\lambda_0=-midA(0)\in U(R)$.
Hence, $3b_0^2-2b_0\mu_0-\lambda_0\in U(R)$. Thus, we see that
$b_1=(3b_0^2-2b_0\mu_0-\lambda_0)^{-1}(\gamma_1+b_0^2\mu_1+b_0\lambda_1)$
and
$b_2=(3b_0^2-2b_0\mu_0-\lambda_0)^{-1}(\gamma_2+b_0^2\mu_2+b_1^2\mu_0+2b_0b_1\mu_1+b_0\lambda_2+b_1\lambda_0-3b_0b_1^2)$.
By iteration of this process, we get $b_3,b_4,\cdots $. Then
$y^3-\mu (x)y^2 -\lambda (x)y-\gamma(x)=0$ has a root $y_0(x)\in
J\big(R[[x]]\big)$. It follows from $trA(0)\in 2+J(R)$ that
$trA(x)\in 2+J\big(R[[x]]\big)$. Likewise, $midA(x)\in
1+J\big(R[[x]]\big)$. According to Corollary 2.10, $A(x)\in
M_3\big(R[[x]])$ is strongly $J^{\#}$-clean.

Assume that $\chi\big(A(0)\big)$ has a root $1+\alpha\in J(R)$ and
$tr\big(A(0)\big)\in 1+J(R), mid\big(A(0)\big)\in J(R), detA(0)\in
J(R)$. Then $det\big(I_3-A(0)\big)=1-trA(0)+midA(0)-detA(0)\in
J(R)$. Set $B(x)=I_3-A(x)$. Then $\chi\big(B(0)\big)$ has a root
$\alpha\in J(R)$ and $tr\big(B(0)\big)\in 2+J(R), detB(0)\in
J(R)$. This implies that $midB(0)=detA(0)-1+trB(0)+detB(0)\in
1+J(R)$. By the preceding discussion, we see that $B(x)\in
M_3\big(R[[x]])$ is strongly $J^{\#}$-clean, and then we are done.
\end{proof}

From this evidence above, we end this paper by asking the
following question: Let $R$ be a projective-free ring, and let
$A(x)\in M_n\big(R[[x]]) (n\geq 4) $. Do the strongly
$J^{\#}$-cleanness of $A(x)\in M_3\big(R[[x]])$ and $A(0)\in
M_3(R)$ coincide with each other?\\

{\bf\large Acknowledgements}\ \ This research was supported by the
Scientific and Technological Research Council of Turkey (2221
Visiting Scientists Fellowship Programme) and the Natural Science
Foundation of Zhejiang Province (Y6090404).

config: arxiv


TITLE: Jordan-Brouwer theorem for closed surfaces
QUESTION [1 upvotes]: The following is Exercise (6) of Chapter 4 of Curves and Surfaces, 2nd edition, by Montiel and Ros, in which the reader is asked to prove the Jordan-Brouwer Theorem for closed surfaces

Theorem(Jordan-Brouwer): If $S$ is a closed connected surface, then $\Bbb{R}^3 \setminus S$ has exactly two connected components whose common boundary is $S$.

My question is in itens b and c below. How to prove them?
Thanks in advance and kind regards.

EDIT Why isn't the following a counterexample to b?

REPLY [3 votes]: First, an important observation: Montiel and Ros seem to diverge from the more commonly accepted terminology of "closed manifold" to mean "compact and without boundary". By closed they really seem to mean closed in the topological sense, and manifold/surface is already assumed to be boundaryless.
They prove the Jordan-Brouwer separation theorem (for closed surfaces in $\mathbb{R}^3$) assuming compactness, and leave the closed case as an exercise. This is the exercise in the question.

With that said, for $(b)$, you can perturb the triangle generated by $p_1,p_2,p_3$ a little in order for it to be transverse to the surface and not change the number of intersections of the lines and the surface. Now, the intersection of such a triangle with the surface is a compact $1$-manifold with boundary and its boundary is precisely the intersections of the lines $[p_1,p_2], [p_2,p_3]$ and $[p_1,p_3]$ with the surface. Of course, $[p_2,p_3]$ does not intersect the surface, so it counts only the intersections of the lines $[p_1,p_2]$ and $[p_1,p_3]$. Since a compact $1$-manifold with boundary is a union of circles and closed intervals, the boundary has an even number of points and the result follows.
For $(c)$, you can pretty much follow the argument he gives for the compact case and define the two sets $A_{even}$ (resp. $A_{odd}$) as the set of points $x$ of $\mathbb{R}^3-S$ such that there exists a segment starting at $x$, ending in $B$ and intersecting $S$ an even (resp. odd) amount of times. Both sets are open by basic facts of transversality, and both sets are disjoint by $(b)$. $A_{even}$ is non-empty since you can pick a point in the ball and have empty intersection. That $A_{odd}$ is non-empty follows from the fact that if you consider a minimum of the function $x \mapsto \lVert x-p_0\rVert$ on the surface  then you just need to extend the segment connecting $p_0$ to the minimum a bit in order to get an element of $A_{odd}$. By the way, such a minimum exists because the minimum of the function restricted to the intersection of the surface with an appropriately large enough closed ball centered on $p_0$, thus a compact set where the existence of the point achieving a minimum is guaranteed, is the same as the minimum of the function on the entire surface.
Now you know that $\mathbb{R}^3-S$ is not connected. Then Proposition $4.4$ in the book concludes what is requested at the end. For reference, the proposition states:

Let $S$ be a closed connected surface. Then $\mathbb{R}^3-S$ has at most two connected components. Moreover, each connected component of $\mathbb{R}^3-S$ has $S$ as its boundary.

Since the arguments therein are not relevant to the exercise and take up a page, I don't see it being much useful to transfer them here. But this ends the exercise.


text:
\begin{document}
\thispagestyle{empty}
\title{\bf Boundary feedback stabilization of a chain of serially connected strings}
\author{
Ka\"{\i}s Ammari
\thanks{UR Analysis and Control of Pde, UR 13ES64, Department of Mathematics, Faculty of Sciences of Monastir, University of Monastir, Tunisia, e-mail: kais.ammari@fsm.rnu.tn}
 \, and \, Denis Mercier \thanks{UVHC, LAMAV, FR CNRS 2956, F-59313 Valenciennes, France, email: denis.mercier@univ-valenciennes.fr}}
\date{}
\maketitle

\begin{quotation}
{\bf Abstract.} {\small We consider $N$ strings connected to one another and forming a particular network which is a chain of strings. We study a stabilization problem and precisley we prove that the energy of the solutions of the dissipative system decay exponentially to zero when the time tends to infinity, independently of the densities of the strings. Our technique is based on a
frequency domain method and a special analysis for the resolvent. Moreover,
by same appraoch, we study the transfert function associated to the chain of strings and the stability of the Schr\"odinger system.}
\end{quotation}
2010 Mathematics Subject Classification. 35L05, 35M10, 35R02, 47A10, 93D15, 93D20.\\
Key words and phrases. Network, wave equation, resolvent method, transfert function, boundary feedback stabilization.

\section{Introduction} \label{secintro}

\setcounter{equation}{0}
We consider the evolution problem $(P)$  described by the following system of $N$ equations:
\begin{equation*}
\leqno(P)
\left \{
\begin{array}{l}
(\partial_t^2 u_{j}- \rho_j \partial_x^2u_{j})(t,x)=0,\, x\in(j,j+1),\, t\in(0,\infty),\, j = 0,...,N-1, \\
\rho_0 \, \partial_x u_0(t,0) = { \bf \partial_t u_0(t,0)},\ u_{N-1}(t,N)=0,\, t\in(0,\infty),\\
u_{j-1}(t,j)=u_{j}(t,j),\, t\in(0,\infty),\, j = 1,...,N-1,\\
- \rho_{j-1} \partial_x u_{j-1}(t,j)+ \rho_j \partial_x u_{j}(t,j)= 0,\, t\in(0,\infty),\, j = 1,...,N-1, \\
u_j(0,x)=u_j^0(x),\ \partial_t u_j(0,x)=u_j^1(x), \,x \in (j,j+1),\,   j=0,...,N-1,
\end{array}
\right.\\
\end{equation*}

where $\rho_j > 0, \, \forall \, j=0,...,N-1$.

We can rewrite the system (P) as a first order hyperbolic system, by putting
$$
V_j = \left(
\begin{array}{ll}
\partial_t u_j \\ \rho_j \, \partial_x u_j
\end{array}
\right), \; \hbox{and} \; V_j^0 = \left(
\begin{array}{c}
u^1_j \\ \rho_j \, \partial_x u^0_j
\end{array}
\right), \, 0 \leq j \leq N-1,
$$

\begin{equation*}
\leqno(P^\prime)
\left \{
\begin{array}{l}
(\partial_t V_{j}- B_j  \, \partial_x V_{j})(t,x)=0,\, x\in(j,j+1),\, t\in(0,\infty),\, j = 0,...,N-1, \\
C_0 V_0(t,0) = 0, \, C_{N-1} V_{N-1} (t,N) = 0,\, t\in(0,\infty),\\
V_{j-1}(t,j)=V_{j}(t,j),\, t\in(0,\infty),\, j = 1,...,N-1,\\
V_j(0,x)= V_j^0 (x), \,x\in(j,j+1),\,   j=0,...,N-1,
\end{array}
\right.\\
\end{equation*}
where
\be
\label{C0CN}
B_j = \left(\begin{array}{ll} 0 & 1 \\ \rho_j & 0\end{array} \right), C_0 = \left(\begin{array}{lclc} 1 & - 1 \\ 0 & 0 \end{array} \right), \, C_{N-1} = \left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array} \right), \, 0 \leq j \leq N -1.
\ee

Models of the transient behavior of some or all of the state variables describing the motion of flexible structures have been of great interest
in recent years, for details about physical motivation for the models, see \cite{dagerzuazua}, \cite{lagnese} and the references therein. Mathematical
analysis of transmission partial differential equations is detailed in \cite{lagnese}.

Let us first introduce some notation and definitions which will be used throughout the rest of the paper, in particular some which are linked to the notion of $C^{\nu }$- networks,
$\nu \in \nline$ (as introduced in \cite{jvb}). \\
Let $\Gamma$ be a connected topological graph embedded in $\rline$, with $N$ edges ($N \in \nline^{*}$).
Let $K=\{k_{j}\, :\, 0\leq j\leq N-1\}$ be the set of the edges of $\Gamma$. Each edge $k_{j}$ is a Jordan curve in $\rline$ and is assumed to be parametrized by its arc length $x_{j}$ such that
the parametrization $\pi _{j}\, :\, [j,j+1]\rightarrow k_{j}\, :\, x_{j}\mapsto \pi _{j}(x_{j})$ is $\nu$-times differentiable, i.e. $\pi _{j}\in C^{\nu }([j,j+1],\rline)$ for all $0\leq j\leq N-1$. The density of the edge $k_j$ is $\rho_j>0$.
The $C^{\nu}$- network $R$  associated with $\Gamma$ is then defined as the union $$R=\bigcup _{j=0}^{N-1}k_{j}.$$

We study a feedback stabilization problem for a wave and a Schr\"dinger equations in networks, see \cite{ammari1}-\cite{amjellk}, \cite{lagnese} and Figure \ref{fig}.
\begin{figure}[ht]
\begin{center}
\input{maillagecordes}
\caption{Serially connected strings}
\end{center}
\label{fig}
\end{figure}

More precisely, we study a linear system modelling the vibrations of a chain of strings. For each edge $k_{j}$, the scalar function $u_j(t,x)$ for $x \in R$ and $t > 0$
contains the information on the vertical displacement of the string, $0 \leq j \leq N-1$.

Our aim is to study the behaviour of the resolvent of the spatial operator
which is defined in Section \ref{resolvent} and to obtain stability result for $(P)$.

We define the natural energy $E(t)$ of a solution $\underline{u} = (u_0,...,u_{N-1})$ of $(P)$ and the natural energy of a solution $V$ of $(P^\prime)$, respectively, by
\be
\label{energy1}
E(t)=\frac{1}{2} \ds \sum_{j=0}^{N-1} \left( \int_{j}^{j+1} \left(|\partial_t u_{j}(t,x)|^2+ \rho_j \, |\partial_x u_{j}(t,x)|^2\right){\rm d}x \right),
\ee
\be
\label{energy1b}
e(t)=\frac{1}{2} \, \ds \sum_{j=0}^{N-1} \left\|V_{j}\right\|^2_{L^2_{\rho_j}(j,+,j+1) \times L^2(j,j+1)}, \, 0 \leq j \leq N-1,
\ee
where $L^2_{\rho_j} (j,j+1) = L^2_{\rho_j} ((j,j+1),dx) = L^2((j,+j+1),\rho_j \, dx)$.

We note that $E(t)  \asymp e(t), \, \forall \, t \geq 0$.

We can easily check that every sufficiently smooth solution of $(P)$ satisfies the following dissipation law
\begin{equation}\label{dissipae1}
E^\prime(t) = - \ds \bigl|\partial_t u_{0}(t,0)\bigr|^2\leq 0, \,
e^\prime(t) = - \ds \bigl|C_0 V_{0}(t,0)\bigr|^2\leq 0,
\end{equation}
and therefore, the energy is a nonincreasing function of the time variable $t$.

The result concerns the well-posedness of the solutions of $(P)$ and the exponential decay of the energy $E(t)$ of the solutions of $(P)$.

\medskip

The main result of this paper then concerns the precise asymptotic behaviour of the solutions of $(P)$.
Our technique is based on a frequency domain method and a special analysis for the resolvent.

\medskip

This paper is organized as follows:
In Section \ref{wellposed}, we give the proper functional setting for system $(P)$ and prove that the system is well-posed.
In Section \ref{resolvent}, we then show that the energie of system $(P)$ tends to zero.
We study, in Section \ref{resolvent}, the stabilization result for $(P)$ by the frequency domain technique
and give the explicit decay rate of the energy of the solutions of $(P)$. Finally, in the last sections,
we study the transfert function associated to a string network and the exponential stability of the Schr\"odinger system.

\section{\label{wellposed}Well-posedness of the system}

In order to study system $(P)$ we need a proper functional setting.
We define the following spaces
$$
H = \ds \prod_{j=0}^{N-1} (L^2_{\rho_j}(j,j+1) \times L^2(j,j+1))
$$
and
$$
V= \bigg \{\underline{u}=(u_0,...,u_{N-1}) \in \ds \prod_{j=0}^{N-1} H^1(j,j+1),\\
u_{N-1}(N)=0, \, u_{j-1}(j) = u_{j} (j), \, j=1, \ldots ,N-1 \bigg \},
$$
equipped with the inner products
\begin{equation}\label{ipVb}
<V,\tilde{V}>_{H}=\sum_{j=0}^{N-1}\left(\int_j^{j+1} \rho_j \,  u_{j}(x) \,  \overline{\tilde{u}_{j}(x)} + v_{j}(x) \,  \overline{\tilde{v}_{j}(x)}
dx\right), \, V = \left(\begin{array}{c} \underline{u} \\ \underline{v} \end{array} \right),
\tilde{V} = \left(\begin{array}{c} \underline{\tilde{u}} \\ \underline{\tilde{v}} \end{array} \right),
\end{equation}
\begin{equation}\label{ipV}
<\underline{u},\,\underline{\tilde{u}}>_{V}=\sum_{j=0}^{N-1}\left(\int_j^{j+1} \rho_j \partial_x u_{j}(x)\partial_x  \overline{\tilde{u}_{j}(x)}
dx\right).
\end{equation}

It is well-known that system $(P)$ may be rewritten as the first order evolution equation
\begin{equation} \left\{
\begin{array}{l}
U^\prime =\mathcal{A} U,\\
U(0)=(\underline{u}^{0},\,\underline{u}^{1})=U_0,\end{array}\right.\label{pbfirstorder}\end{equation}
where $U$ is the vector $U=(\underbar{u},\,\partial_t \underbar{u})^t$ and the operator $\mathcal{A} : {\cal D}({\cal A}) \rightarrow {\cal H} = V\times \ds \prod_{j=0}^{N-1} L^2(j,j+1)$ is defined by
\[\mathcal{A} (\underline{u},\underline{v})^t:=(\underline{v}, (\rho_j \partial_x^2u_{j})_{0 \leq j\leq N-1})^t,\]
with
\begin{multline*}
{\cal D}({\cal A}):=\left\{(\underline{u},\,\underline{v})\in \prod_{j=0}^{N-1} H^2(j,j+1) \times V \,:
\mbox {\textrm{satisfies }} \,(\ref {e2}) \; \mbox{\textrm{to}} \; (\ref {e5}) \; \mbox{\textrm{hereafter}}
\right\},\end{multline*}
\begin{equation}\label{e2}
\rho_0 \, \partial_x u_{0}(0) =  v_{0}(0)
\end{equation}
\begin{equation}\label{e5}
- \rho_j \partial_x u_{j}(j) + \rho_{j-1} \partial_x u_{j-1}(j)= 0, \quad j = 1,...,N-1.
\end{equation}

It is clear that $\mathcal{H}$ is a Hilbert space,
equipped with the usual inner product
\begin{multline*}
\left\langle\left(\begin{array}{c}\underline{u}\\\underline{v}\end{array}\right),
\left(\begin{array}{c}\underline{\tilde{u}}\\ \underline{\tilde{v}}\end{array}\right)\right\rangle_{{\cal H}} =
\sum_{j=0}^{N-1}\left(\int_{j}^{j+1}\left(v_{j}(x)\overline{\tilde{v}_{j}(x)}
+ \rho_j \partial_x u_{j}(x)\partial_x\overline{\tilde{u}_{j}(x)}\right){\rm d}x. \right.
\end{multline*}

By the same way we define the operator $A$ as following:
$$
A : {\cal D}(A) \subset H \rightarrow H, A V = B \, \partial_x \, V, \, \forall \,  V \in {\cal D}(A),
$$
where $${\cal D}(A) = \left\{V = (V_j)_{0 \leq j \leq N-1} \in H, \, V_j \in (H^1(j,j+1))^2, \,  V_{j-1} (j) = V_j (j), 1 \leq j \leq N-1, \right.
$$
$$
C_0 V_0 (0 ) = 0, \, C_{N-1} V_{N-1} (N) = 0  $$ and
$B = (B_j)_{0 \leq j \leq N-1}.$

Now we can prove the well-posedness of system $(P)$ and that the solution of $(P)$ satisfies the dissipation law (\ref{dissipae1}).

\begin{proposition}\label{3exist1}
(i) For an initial datum $U_{0}\in \mathcal{H}$, there exists a unique solution $U\in C([0,\,+\infty),\, \mathcal{H})$
to  problem (\ref{pbfirstorder}). Moreover, if $U_{0}\in \mathcal{D}(\mathcal{A})$, then
$$U\in C([0,\,+\infty),\, \mathcal{D}(\mathcal{A}))\cap C^{1}([0,\,+\infty),\, \mathcal{H}).$$

(ii) The solution $\underline{u}$ of $(P)$ with initial datum in $\mathcal{D}(\mathcal{A})$ satisfies \rfb{dissipae1}.
Therefore the energy is decreasing.
\end{proposition}

\begin{proof}
(i) By Lumer-Phillips' theorem (see \cite{Pazy, tucsnakbook}), it suffices to show that
$\mathcal{A}$ is dissipative and  maximal.

We first prove that $\mathcal{A}$ is dissipative. Take $U=(\underline{u},\underline{v})^{t}\in \mathcal{D}(\mathcal{A})$.
Then
\begin{multline*}
\left\langle\mathcal{A}U,\, U \right\rangle_{\mathcal{H}}=\sum_{j=0}^{N-1}\left(\int_{j}^{j+1}\left(\rho_j \partial_x^2u_{j}(x)\overline{v_{j}(x)}
+ \rho_j \partial_x v_{j}(x)\partial_x \overline{u_{j}(x)}\right){\rm d}x\right ).
\end{multline*}
By integration by partsand by using the transmission and boundary conditions, we have

\begin{equation}\label{dissipativeness}
\Re\left(\left\langle\mathcal{A}U,\, U \right\rangle_{\mathcal{H}}\right)=- \left|v_{0}(0)\right|^2\leq 0.
\end{equation}
This shows the dissipativeness of $\mathcal{A}$.

\medskip

Let us now prove that $\mathcal{A}$ is maximal, i.e. that
$\lambda I-\mathcal{A}$ is surjective for some $\lambda>0$.

Let $(\underline{f}, \underline{g})^{t}\in \mathcal{H}$. We look for $U=(\underline{u}, \underline{v})^{t}\in \mathcal{D}(\mathcal{A})$ solution of
\begin{equation}\label{eqmaxmon}
(\lambda I-\mathcal{A})\left(\begin{array}{c}
\underline{u}\\\underline{v}\end{array}\right)=\left(\begin{array}{c}
\underline{f}\\\underline{g}\end{array}\right),\end{equation}
or equivalently
\begin{equation} \left\{
\begin{array}{ll}
\lambda u_{j}-v_{j}=f_{j} & \forall j\in\{0,...,N-1\},\\
\lambda v_{j}- \rho_j \partial^{2}_xu_{j}=g_{j} & \forall j\in\{0,...,N-1\}.\end{array}\right.\label{eqmaxmon2}\end{equation}

Suppose that we have found $\underline{u}$ with the appropriate regularity.
Then for all  $j\in\{0,...,N-1\},$ we have
\begin{equation} v_{j}:=\lambda u_{j}-f_{j}\in V.\label{maxmonv}\end{equation}
It remains to find $\underline{u}$. By (\ref{eqmaxmon2}) and (\ref{maxmonv}), $u_{j}$ must
satisfy, for all $j=0,...,N-1$,
$$
\lambda^{2}u_{j}- \rho_j \partial^{2}_xu_{j}=g_{j}+\lambda f_{j}.
$$
Multiplying these identities by a test function $\underline{\phi}$, integrating in space and using integration by
parts, we obtain
\begin{multline*}
\sum_{j=0}^{N-1}\int_j^{j+1}
\left(\lambda^2u_{j}\overline{\phi_{j}}+ \rho_j \partial_x u_{j}\partial_x \overline{\phi_{j}}\right)dx
-\sum_{j=0}^{N-1}\left[\rho_j \partial_xu_{j}\overline{\phi_{j}}\right]_j^{j+1}  \\
=\sum_{j=0}^{N-1}\int_j^{j+1}\left( g_j+\lambda f_j \right)\overline{\phi_j} \, dx.
\end{multline*}
Since $(\underline{u},\underline{v})\in \mathcal{D}(\mathcal{A})$ and $(\underline{u},\underline{v})$ satisfies (\ref{maxmonv}), we then have
\begin{multline}\label{maxmoneq1}
\sum_{j=0}^{N-1}\int_j^{j+1}\left(\lambda^2u_{j}\overline{\phi_{j}}+\rho_j \partial_xu_{j}\partial_x\overline{\phi_{j}}\right)dx +
\\
\left(\lambda u_0(0) - f_0(0) \right) \overline{\phi_0}(0) =\sum_{j=0}^{N-1}\int_j^{j+1}\left(g_j+\lambda f_j\right)\overline{\phi_j}dx.
\end{multline}
This problem has a unique solution $\underline{u}\in V$ by Lax-Milgram's lemma, because the left-hand side  of (\ref{maxmoneq1}) is coercive on $V$.
If we consider $\underline{\phi}\in \ds \prod_{j=0}^{N-1}\mathcal{D}(j,j+1)\subset V$, then $\underline{u}$
satisfies
$$\begin{array}{c}
\displaystyle{\lambda^{2}u_{j}-\rho_j \partial_x^{2}u_{j}=g_{j}+\lambda f_{j} \quad\hbox{ in } \mathcal{D}^\prime (j,j+1),\quad  j=0,\cdots,N-1.}
\end{array}$$
This directly implies that  $\underline{u}\in \ds \prod_{j=0}^{N-1}H^{2}(j,j+1)$ and then
$\underline{u}\in V\cap \ds \prod_{j=0}^{N-1}H^{2}(j,j+1)$.
Coming back to (\ref{maxmoneq1}) and by integrating by parts, we find
$$\begin{array}{ll}
- \ds \sum_{j=0}^{N-1}\left(\rho_j \partial_x u_{j}(j) \overline{\phi_{j}}(j) - \rho_{j} \partial_x u_{j}(j+1)\overline{\phi_{j}}(j+1) \right)\\
+ \left(\lambda u_0(0) - f_0(0) \right)\overline{\phi_0}(0)
= 0.
\end{array}
$$
Consequently, by taking particular test functions  $\underline{\phi}$, we obtain
$$\begin{array}{c}
\rho_0 \partial_x u_{0}(0)= v_0(0) \quad \hbox{and}\quad \rho_j \partial_x u_{j}(j) - \rho_{j-1} \partial_x u_{j-1}(j) =0, \, j=1,\cdots,N-1.
\end{array}$$
In summary we have found $(\underline{u},\underline{v})^{t}\in \mathcal{D}(\mathcal{A})$ satisfying (\ref{eqmaxmon}), which finishes the proof of (i).

\medskip

(ii) To prove (ii), it suffices to derivate the energy (\ref{energy1}) for regular solutions and to use system $(P)$.
The calculations are analogous to those of the proof of the dissipativeness of $\mathcal{A}$ in (i), and then, are left to the reader.
\end{proof}

\begin{remark} \label{opha}
By the same we can prove that the operator $A$ is a m-dissipatif operator of $H$ and generates a $C_0-$ semigroup of contractions of $H$.
\end{remark}

\bigskip

\section{Exponential stability} \label{resolvent}

We prove a decay result of the energy of system $(P)$, independently of $N$ and of the densities, for all initial data in the energy space. Our technique is based on a frequency domain method and a special analysis for the resolvent.
\begin{theorem} \label{lr}
There exists a constant $C, \omega >0$ such that, for all $(\underline{u}^0,\underline{u}^1)\in {\cal H}$, the solution of system $(P)$ satisfies the following estimate
\BEQ{EXPDECEXP3nb}
E(t) \le C \, e^{- \omega \,t} \, \left\Vert (\underline{u}^0,\underline{u}^1) \right\Vert_{{\cal H}}^2,
\FORALL t > 0.
\EEQ
\end{theorem}

{\it Proof.}
By classical result (see Huang \cite{huang} and Pr\"{u}ss \cite{pruss}) it suffices to show that ${\cal A}$ satisfies the
following two conditions:) of a $C_0$ semigroup
of contractions on a Hilbert space:

\be
\rho ({\cal A})\supset \bigr\{i \beta \bigm|\beta \in \rline \bigr\} \equiv i \rline, \label{1.8w} \ee
and \be \limsup_{|\beta |\to \infty }  \|(i\beta -{\cal A})^{-1}\|_{{\cal L}(\caH)} <\infty,
\label{1.9}
\ee
where $\rho({\cal A})$ denotes the resolvent set of the operator ${\cal A}$.

Then the proof of Theorem \ref{lr} is based on the following two lemmas.

\begin{lemma} \label{condsp}
The spectrum of ${\cal A}$ contains no point on the imaginary axis.
\end{lemma}

\begin{proof}
Since ${\cal A}$ has compact resolvent,
its spectrum $\s({\cal A})$ only consists of eigenvalues of ${\cal A}$. We will
show that the equation
\be
{\cal A} Z = i \beta \, Z
\label{1.10}
\ee
with $Z= \left(\begin{array}{l} \underline{y} \cr \underline{v} \end{array} \right) \in {\cal D}({\cal A})$ and $\beta \neq 0$ has only the trivial solution.

By taking the
inner product of (\ref{1.10}) with $Z \in {\cal H}$ and using
\be
\label{1.7}
\Re <{\cal A}Z,Z>_{{\cal H}} = - \, \left| v_0 (0)\right|^2,
\ee
we obtain
that $v_0(0)=0$. Next, we eliminate $\underline{v}$ in (\ref{1.10})
to get
a second order ordinary differential system:
\be
\left\{ \begin{array}{l}
\rho_j \, \frac{d^2 y_j}{dx^2} + \beta^2 \, y_j = 0, \, (j,j+1), \, j = 0,...,N-1,\\
y_0(0) = \frac{d y_0}{dx}(0)=0, \, y_{N-1} (N) = 0, \\
y_{j-1}(j) = y_{j}(j),\, j = 1,...,N-1.
\end{array}
\right.
\label{1.11}
\ee
The above system has only trivial solution.

\end{proof}

\begin{lemma}\label{lemresolvent}
The resolvent operator of $\mathcal{A}$ satisfies condition \rfb{1.9}.
\end{lemma}

\begin{proof}
In order to prove \rfb{1.9} or by equivalence the following
\be
\limsup_{|\beta |\to \infty }  \|(i\beta -A)^{-1}\|_{{\cal L}(H)} <\infty,
\label{1.9n}
\ee
we  will compute and estimate the resolvent of the operator $A$ associated to the problem $(P^\prime)$. \\
More precisely, let $\lambda=i\lb,\lb \in \R,$  $G=(G_0,...,G_{N-1}) \in H,$ we look for $W=(W_0,...,W_{N-1})\in \caD(A)$  solution of
\be \label{respb} (i \lb - B \partial_x ) W =G, \ee
where $B=(B_0,...,B_{N-1})$ and $$B_j=\left(\begin{array}{ll}
0 & 1  \\
\rho_j & 0
\end{array} \right), \;j=0,...N-1.
$$
We want to prove that there exists a constant $C$ independent of $\lb$ such that
\be \label{estresolv} \norm{W}{H} \leq C \norm{G}{H}\ee

{\bf First step :} Computation of the resolvent\\
From (\ref{respb}) we have
$$\partial_x W_j = i\lb B_j^{-1} W_i - B_j^{-1}G_j,j=0,...,N-1$$
therefore

\be\label{W0} W_0(x)=e^{i \lb (x-1) B_0^{-1} }F_0-\int_1^x e^{i \lb (x-s) B_0^{-1} } B_0^{-1} G_0(s)\; ds, \forall x\in [0,1],\ee
and
\be \label{Wj} W_j(x)=e^{i \lb (x-j) B_j^{-1} }F_j-\int_j^x e^{i \lb (x-s) B_j^{-1} } B_j^{-1} G_j(s) \; ds, \forall x\in [j,j+1],j=1,2,...,N-1,\ee
where $F_0=W_0(1)$ and $F_j=W_j(j), j=0,...,N-1.$
For simplification we set
\be\label{Gjt} \tilde{G_j} (x)=\int_j^x e^{i \lb(x-s)  B_j^{-1} } B_j^{-1} G_j(s) \; ds ,j=0,...,N-1.\ee
Using the transmission conditions at nodes $j=1,...,N-1$ we have
\be\label{F1} F_1=F_0, \mbox{ and } F_j=W_{j-1}(j), j=2,...,N-1,\ee
which implies that
\be\label{Fj0} F_j=\left( \prod_{k=j-1}^{k=1} e^{i \lb B_k^{-1}}\right)F_0- \left(\sum_{p=2}^{j-1}\left( \prod_{k=j-1}^{k=p} e^{i \lb B_k^{-1}}\right)
\tilde{G}_{p-1}(p)
+ \tilde{G}_{j-1}(j)\right) ,j=2,...,N-1.\ee
For all $j=1,...,N-1$ we set $$M_j(\lb)=\left( \prod_{k=j-1}^{k=1} e^{i \lb B_k^{-1}}\right)$$ and
\be \label{gammaj}\Gamma_j(\lb)=\sum_{p=2}^{j-1}\left( \prod_{k=j-1}^{k=p} e^{i \lb B_k^{-1}}\right) \tilde{G}_{p-1}(p)
 +\tilde{G}_{j-1}(j),\ee
hence  \be \label{Fj} F_j=M_j(\lb) F_0-\Gamma_j(\lb).\ee
Note that the solution $W$ is completely determined if $F_0$ is known. Indeed, it suffices to insert the identity (\ref{Fj}) in  (\ref{Wj}).

Thus, we give the equation satisfied by  $F_0.$ The boundary conditions at nodes $x=0$ and $x=N$ are respectively $C_0 W_0(0)=\left(\begin{array}{l}
0  \\
0
\end{array} \right),$ and
$C_{N-1} W_{N-1}(N)=\left(\begin{array}{l}
0  \\
0
\end{array} \right),$
 where $C_0, C_{N-1}$ are  the matrices  given in (\ref{C0CN}).
Since the second lines of $C_0$ and $C_{N-1}$ vanish the previous equations may be written as
$$(1,-1). W_0(0)=0,\;  \mbox{\rm and }  (1,0). W_{N-1}(N)=0,$$
where "$.$" represents the matrix-product of a vector line by a vector column.
These equations are  equivalent to
\be\label{eq1} (1,-1).e^{-i\lb B_0^{-1} }F_0 = (1,-1),\tilde{G_0}(0) \ee
and
$$(1,0).e^{i\lb B_{N-1}^{-1} }F_{N-1} = (1,0).\tilde{G_{N-1}}(N). $$
Inserting (\ref{Fj}) in the previous equation we get
\be\label{eq2}(1,0).e^{i\lb B_{N-1}^{-1} }M_{N-1}(\lb)F_0 = (1,0).\left( \tilde{G}_{N-1}(N)+e^{i\lb B_{N-1}^{-1} } \Gamma_{N-1}(\lb)\right). \ee
If we denote by $H_{N-1}$ the $2\times 2$ matrix whose the first line is the vector line is $(1,-1).e^{-i\lb B_0^{-1} }$
and the second line is $(1,0).e^{i\lb B_{N-1}^{-1} }M_{N-1}(\lb)$ i.e

\be\label{HN} H_{N-1}=\left(\begin{array}{ll}
(1,-1).e^{-i\lb B_0^{-1} }\\
(1,\;\;0).e^{i\lb B_{N-1}^{-1}}
\end{array} \right)\ee
and $Y_{N-1}$ the $2\times 1$ vectors columns by

\be \label{YN} Y_{N-1}=\left(\begin{array}{c}
(1,-1).\tilde{G_0}(0)\\
(1,0).\left( \tilde{G_{N-1}}(N)+e^{i\lb B_{N-1}^{-1} } \Gamma_{N-1}(\lb)\right)
\end{array} \right)\ee
then equation (\ref{eq1}) and (\ref{eq2}) are equivalent to the following system:

\be \label{eq3}  H_{N-1} F_0= Y_{N-1}. \ee

{\bf Second  step: estimate of $F_0$} \\

We first start by given an estimation of $\tilde{G}=(\tilde{G_0},...,\tilde{G}_{N-1})$ where $\tilde{G_j}$ are defined in (\ref{Gjt}).

For all $j=0,...,N-1,$ the matrix is $B_j$ is invertible:
$$B_j^{-1}=\left(
\begin{array}{cc}
 0 & \frac{1}{\rho_j} \\
 1 & 0
\end{array}
\right)$$

\ and we  easily find after some computation that
\be\label{expbj} e^{i\lb x  B_j^{-1}}=\left(
\begin{array}{cc}
 \cos(\dfrac{\lb x}{\sqrt{\rho_j}}) & \dfrac{i \sin(\dfrac{\lb x}{\sqrt{\rho_j}})}{\sqrt{\rho_j}} \\
 i \sqrt{\rho_j}\sin(\dfrac{\lb x}{\sqrt{\rho_j}}) & \cos(\dfrac{\lb x}{\sqrt{\rho_j}})
\end{array}
\right).
\ee
Since $\lb \in \R,$ from the previous identity, we directly get the following estimates

\be \label{est1}
|\tilde{G_j}(j)|
\lesssim \norm{G}{H},\;|\tilde{G_j}(j+1)|
\lesssim \norm{G}{H}, j=0,...,N-1 \ \ee

\be \label{est2}
\norm{\tilde{G}}{H}
\lesssim \norm{G}{H}. \ee

From  the definition of $\Gamma_j$ in (\ref{gammaj}) we also get
\be \label{est3} |\Gamma_j(\lb) | \lesssim  \norm{G}{H}, j=1,...,N-1. \ee

It follows that
\be \label{est4} \|Y_{N-1}\| \lesssim \norm{G}{H}. \ee

Note that from (\ref{expbj}) the entries of $H_{N-1}$ are bounded and so it is for the entries of the matrix $( \mbox{\rm com} H_{N-1})^T.$  Assume for the moment that
there exists a constant $\gamma_{N-1}>0$ such that
\be \label{det} \forall \lb \in\R, |\det(H_{N-1})|\geq \gamma_{N-1},\ee
then it follows with (\ref{est4}) that
\be\label{estF0} \|F_0\|=\dfrac{1} {\det H_{N-1}} ( \mbox{\rm com} H_{N-1})^T Y_{N-1} \lesssim  \norm{G}{\caH}.\ee
It remains to prove (\ref{det}).

The idea of the proof is that (\ref{det}) is well known for $N=1$ and that this property spreads by iteration.

First, similarly to (\ref{HN}) we define for all $N\in \N^*$ the matrix
\be\label{HNt} \tilde{H}_{N-1}=\left(\begin{array}{ll}
(1,-1).e^{-i\lb B_0^{-1} }\\
(0,\;\;1).e^{i\lb B_{N-1}^{-1}}
\end{array} \right)\ee
and we set
$$D_{N-1}=\det(H_{N-1}),\;\; \tilde{D}_{N-1}=\det(\tilde{H}_{N-1}), \forall N\in \N^*.$$
Particularly, for $N=1$ we have
$$H_0=\left(
\begin{array}{cc}
 \cos(\dfrac{\lb}{\sqrt{\rho_0}})+i \sqrt{\rho_0} \sin(\dfrac{\lb}{\sqrt{\rho_0}}) & -\cos(\dfrac{\lb}{\sqrt{\rho_0}})-i\dfrac{ \sin(\dfrac{\lb}{\sqrt{\rho_0}})}{\sqrt{\rho_0}} \\
 \cos(\dfrac{\lb}{\sqrt{\rho_0}}) & -i\dfrac{ \sin(\dfrac{\lb}{\sqrt{\rho_0}})}{\sqrt{\rho_0}}
\end{array}
\right),$$
and
$$\tilde{H}_0=\left(
\begin{array}{cc}
 \cos(\dfrac{\lb}{\sqrt{\rho_0}})+i \sqrt{\rho_0} \sin(\dfrac{\lb}{\sqrt{\rho_0}}) & -\cos(\dfrac{\lb}{\sqrt{\rho_0}})-i\dfrac{ \sin(\dfrac{\lb}{\sqrt{\rho_0}})}{\sqrt{\rho_0}} \\
i \sqrt{\rho_0} \sin(\dfrac{\lb}{\sqrt{\rho_0}}) &  \cos(\dfrac{\lb}{\sqrt{\rho_0}})
\end{array}
\right).$$
Thus $$D_0= \cos(\dfrac{2\lb}{\sqrt{\rho_0}}) +  i \dfrac{\sin(\dfrac{2\lb}{\sqrt{\rho_0}})}{\sqrt{\rho_0}},\; \tilde{D}_0=
\cos(\dfrac{2\lb}{\sqrt{\rho_0}}) +  i \sqrt{\rho_0} \sin(\dfrac{2\lb}{\sqrt{\rho_0}}),$$
and  $$|D_0|^2=\cos^2(\dfrac{2\lb}{\sqrt{\rho_0}})+\dfrac{\sin^2(\dfrac{2\lb}{\sqrt{\rho_0}})}{\rho_0}\geq \min(1,\dfrac{1}{\rho_0})>0,$$
 $$|\tilde{D}_0|^2=\cos^2(\dfrac{2\lb}{\sqrt{\rho_0}})+\rho_0\sin^2(\dfrac{2\lb}{\sqrt{\rho_0}})\geq \min(1,\rho_0)>0.$$
It is useful for the sequel to remark that
$$\Re(D_0 \overline{\tilde{D}_0)}=1.$$
Using (\ref{expbj}) we have the following identity
$$\left\{\begin{array}{lll}
D_{N-1} =  \cos(\dfrac{\lb}{\sqrt{\rho_{N-1}}})D_{N-2}+\dfrac{i}{\sqrt{\rho_{N-1}}} \sin(\dfrac{\lb }{\sqrt{\rho_{N-1}}})\tilde{D}_{N-2}, \\
\tilde{D}_{N-1} = i  \sqrt{\rho_{N-1}} \sin(\dfrac{\lb }{\sqrt{\rho_{N-1}}})D_{N-2}+\cos(\dfrac{\lb}{\sqrt{\rho_{N-1}}})\tilde{D}_{N-2}.
\end{array} \right.
$$
A simple computation shows that
$$\Re(D_{N-1}\overline{\tilde{D_{N-1}}})=\Re(D_{N-2}\overline{\tilde{D}_{N-2}})$$ consequently,
$$\forall N \in \N^*, \;\Re(D_{N-1}\overline{\tilde{D}_{N-1}})=1.$$

Now, since
$$|D_{N-1}|^2=
$$
$$
(\cos(\dfrac{\lb}{\sqrt{\rho_{N-1}}}),\sin(\dfrac{\lb}{\sqrt{\rho_{N-1}}}))
\left(\begin{array}{ll}
|D_{N-2}|^2 &\dfrac{1}{\sqrt{\rho_{N-1}}} \Im(D_{N-2}\overline{\tilde{D}_{N-2}}) \\
\dfrac{1}{\sqrt{\rho_{N-1}}}\Im(D_{N-2}\overline{\tilde{D}_{N-2}}) & |\dfrac{1}{\sqrt{\rho_{N-1}}}\tilde{D}_{N-2}|^2
\end{array} \right)
\left(\begin{array}{l}
\cos(\dfrac{\lb}{\sqrt{\rho_{N-1}}}) \\
\sin(\dfrac{\lb}{\sqrt{\rho_{N-1}}})
\end{array} \right),
$$
it follows that
\be \label{mu}|D_{N-1}|^2 \geq \mu_{min,N-2},\ee
 where $\mu_{min,N-2}$ is the smallest eigenvalue of the matrix in the previous identity.
The determinant of this matrix is:
$$\dfrac{1}{\rho_{N-1}}\Re(D_{N-2}\overline{\tilde{D}_{N-2}})^2=\dfrac{1}{\rho_{N-1}}.$$
Since $D_{N-2}$ and $\tilde{D}_{N-2}$ are clearly bounded the trace of this matrix is bounded, i.e
$$\exists C'_{N-1}>0, |D_{N-2}|^2+ |\dfrac{1}{\sqrt{\rho_{N-1}}}\tilde{D}_{N-2}|^2 \leq C'_{N-1}.$$
It follows that $$\mu_{min,N-2}\geq \dfrac{1}{\rho_{N-1} C'_{N-1}}.$$
Setting $C_{N-1}=\sqrt{\dfrac{1}{\rho_{N-1} C'_{N-1}}},$ then (\ref{mu}) implies (\ref{det}). Consequently we have prove the estimate (\ref{estF0}) for $F_0.$

Finally, using estimates (\ref{estF0}), (\ref{est1}), (\ref{est2}) in (\ref{F1}) and (\ref{Fj0}) and the fact that the matrices involved in (\ref{Fj0}) are uniformly bounded we get
$$\|F_j\| \lesssim  \norm{G}{\caH}, \forall j=1,...N-1.$$
Using (\ref{estF0}) and the previous estimates in (\ref{W0}) and (\ref{Wj}), we get (\ref{estresolv}).

Which implies \rfb{1.9n} and thereafter \rfb{1.9}, and end the proof of Theorem \ref{lr}.

\end{proof}

\section{Comments and related questions}
The same strategy can be applied to stabilize the following models and to verify and compute the transfer function.

\subsection{Transfer function}
We can use the same strategy to verify that the operator $H(\lambda) = \lambda \, C^*(\lambda^2 I +\underline{A}  )^{-1} \, C \in {\cal L}(U), \, \lambda \in \cline_+,$ satisfies the property (\ref{Hest}) of the following lemma,
where here $\underline{A}$ is the self-adjoint operator corresponding to the conservative problem associated  to problem (P), namely we replace in (P) the boundary feedback condition by
\begin{equation}\label{newe2}
\rho_0 \, \partial_x u_{0}(t,0) =0,
\end{equation}
i.e.,

$\underline{A} : {\cal D}(\underline{A}) \subset \underline{H} = \ds \prod_{j=0}^{N-1} L^2(j,j+1) \rightarrow \underline{H}$ is defined by
\[\underline{A} (\underline{u}):=(- \, \rho_j \partial_x^2u_{j})_{0 \leq j\leq N-1},
\]
with
\begin{multline*}
{\cal D}(\underline{A}):=\left\{\underline{u} \in \prod_{j=0}^{N-1} H^2(j,j+1)\,:
\mbox {\textrm{satisfies }} \,(\ref {e2bn}) \; \mbox{\textrm{to}} \; (\ref {e5bn}) \; \mbox{\textrm{hereafter}}
\right\},\end{multline*}
\begin{equation}\label{e2bn}
\rho_0 \, \partial_x u_{0}(0) =  0
\end{equation}
\begin{equation}\label{e5bn}
- \rho_j \partial_x u_{j}(j) + \rho_{j-1} \partial_x u_{j-1}(j)= 0, \quad j = 1,...,N-1.
\end{equation}
$$
C \in {\cal L}(\mathbb{C}, V^\prime = {\cal D}(\underline{A}^\half)^\prime), \, Ck = \sqrt{\rho_0} \, \underline{A}_{-1} \underline{N} k = k \, \left( \begin{array}{c} \frac{1}{\sqrt{\rho_0}} \,  \delta_0 \\ . \\ . \\. \\ 0 \end{array} \right), \, \forall \, k \in \mathbb{C},
$$
$$
C^* \underline{u} = \left( \frac{1}{\sqrt{\rho_0}} \, u_0(0) \ 0 \; ... \; 0 \right), \, \forall \, \underline{u} \in V,
$$
where $\underline{A}_{-1}$ is the extension of $\underline{A}$ to
$({\cal D}(\underline{A}))^\prime$ (the duality is in the sense of $\underline{H}$) and $\underline{N}$ is the Neumann map,
$$
\left\{
\begin{array}{lll}
\rho_j \, \partial_{x}^2 (\underline{N}k)_j = 0, \, (j,j+1), \, 0 \leq j \leq N -1, \\ \partial_{x} (\underline{N} k)_0 (0) = k, \, (\underline{N}k)_{N -1} (N) = 0,\\
\rho_{j-1} \, \partial_{x} (\underline{N}k)_{j-1} (j) = \rho_{j} \, \partial_{x} (\underline{N}k)_{j} (j), \, 1 \leq j \leq N-1.
\end{array}
\right.
$$

\begin{lemma} The transfer function $H$ satisfies the following estimate:
\be\label{Hest}
\sup_{\Re \lambda = \gamma} \left\| \lambda \, C^*(\lambda^2 I
+\underline{A})^{-1} \, C \right\|_{{\cal L}(U)} < \infty,
\ee
for $\gamma > 0.$
\end{lemma}
\begin{proof}

In the same way as the proof of Lemma \ref{lemresolvent}, we give an equivalent formulation of the function $H.$ For that purpose, we consider  $W=(W_j)_{0 \leq j \leq N-1} \in H,W_j \in (H^1(j,j+1))^2$ solution of
\be \label{respbb}
\begin{array}{l}
(\lambda - B \partial_x ) W =0, \\
W_{j-1} (j) = W_j (j), 1 \leq j \leq N-1, \\
\tilde{C}_0 W_0 (0 ) =  \left(\begin{array}{l}
z \\
0
\end{array} \right)
,\, C_{N-1} W_{N-1} (N) = 0
\end{array}
\ee where $z \in \cline,$ $B$ is defined as in the proof of Lemma \ref{lemresolvent}, $C_{N-1}$ is the matrix given in (\ref{C0CN}) and
$\tilde{C}_0  =\left(\begin{array}{ll}
0 &1  \\
0&0
\end{array} \right).$
Therefore, for $\lambda \in \cline, \Re(\lambda)=\gamma>0,$  the transfer function is
$$ H(\lambda) : z\in \cline \mapsto (1,0). W_0(0)\in \cline.$$
Consequently to prove (\ref{Hest}) it suffices to check that for  a fixed  $\gamma>0,$ there exists a constant $c_\gamma>0$ such that
\be \label{448} \forall \lambda \in \cline, \Re(\lambda)= \gamma,\;\; |(1,0). W_0(0)| \leq c_\gamma |z| .
\ee

Using (\ref{W0}), (\ref{Wj}), (\ref{Fj0}), we
have
$$W_0(x)=e^{\lambda (x-1) B_0^{-1}} F_0, W_j(x)=e^{\lambda (x-j)B_j^{-1}} F_j,j =1,...,N-1,$$
where
$$ F_0=W_0(1),\; F_1=F_0,\; \mbox{\rm and } F_j=\left( \prod_{k=j-1}^{k=1} e^{\lambda B_k^{-1}}\right)F_0, \; j=2,...,N-1.
$$
Therefore, from the boundary conditions at $x=0$ and $x=N,$ we find that $F_0$ is the solution of
$$ H_{N-1}F_0= \left(\begin{array}{l}
z \\
0
\end{array} \right),$$
where
$H_{N-1}$ is the $2\times2 $ matrix
$$\left(\begin{array}{ll}
(0,1).e^{-\lambda B_0^{-1}} \\
(1,0 ). \left(\displaystyle \prod_{k=N-1}^{k=1} e^{\lambda B_k^{-1}}\right)
\end{array} \right),
$$
with the convention that $\left(\displaystyle \prod_{k=N-1}^{k=1} e^{\lambda B_k^{-1}}\right)
$ is the identity matrix if $N=1.$

{\bf Estimate of $F_0$}\\
Since for all $j$ $$e^{\lambda  x B_j^{-1} }=\left(
\begin{array}{cc}
 \cosh(\frac{\lambda x}{c_j}) & \frac{\sinh(\frac{ \lambda x}{c_j})}{c_j} \\
 c_j \sinh(\frac{\lambda x }{c_j}) & \cosh(\frac{ \lambda x}{c_j})
\end{array}
\right),
$$
it is clear that there exists a constant $c'_\gamma>0$ sucht that
\be \label{up} \forall \lambda \in \cline, \Re(\lambda)=\gamma,\; \|H_{N-1}\|\leq c'_\gamma.\ee
We need a similar estimate for $H_{N-1}^{-1};$ this will be done by giving a lower uniform bound of $|D_{N-1}|$ on the line $\Re(\lambda)=\gamma,$ where
we have set $D_{N-1}=\det(H_{N-1}).$

Thus we introduce the matrix
$$\tilde{H}_{N-1}= \left(\begin{array}{ll}
(0,1).e^{-\lambda B_0^{-1}} \\
(0,1). \left( \prod_{k=N-1}^{k=1} e^{\lambda B_k^{-1}}\right)
\end{array} \right), \; N\geq 1,
$$
and set $\tilde{D}_{N-1}=\det(\tilde{H}_{N-1}).$
Now, we prove by iteration that
$$ \Re (D_{N-1} \overline{\tilde{D}_{N-1}})\geq k_{N-1} >0.$$
For $N=1$ we have
$$H_0=\left(
\begin{array}{cc}
-\sinh(\frac{\lambda }{c_0}) &\cosh(\frac{\lambda }{c_0}) \\
 1 & 0
\end{array}
\right),\;
\tilde{H}_0=\left(
\begin{array}{cc}
-\sinh(\frac{\lambda }{c_0}) &\cosh(\frac{\lambda }{c_0}) \\
 0 & 1
\end{array}
\right),
$$
thus $$\Re (D_0 \overline{\tilde{D}_0})=\dfrac{1}{2} \sinh(\frac{2 \gamma}{c_0} )>0.$$
Assume that there exists a constant $k_{N-2}>0$ such that
\be \label{rere} \Re (D_{N-2} \overline{\tilde{D}_{N-2}})\geq k_{N-2} >0.\ee

We have the following easily checked identities
$$D_{N-1}= \cosh(\frac{\lambda}{c_{N-1}}) D_{N-2}+\dfrac{1}{c_{N-1}}\sinh(\frac{\lambda}{c_{N-1}})\tilde{D}_{N-2} $$
$$\tilde{D}_{N-1}=c_{N-1} \sinh(\frac{\lambda}{c_{N-1}}) D_{N-2}+\cosh(\frac{\lambda}{c_{N-1}})\tilde{D}_{N-2}. $$
A computation leads to
$$\begin{array}{lll}
\Re (D_{N-1} \overline{\tilde{D}_{N-1}})&=&\Re(\cosh(\frac{\lambda}{c_{N-1}})\overline{\sinh(\frac{\lambda}{c_{N-1}})})
(c_{N-1}|D_{N-2}|^2+\dfrac{1}{c_{N-1}}|\overline{\tilde{D}_{N-2}}|^2)\\
&+&(|\cosh(\frac{\lambda}{c_{N-1}})|^2+|\sinh(\frac{\lambda}{c_{N-1}})|^2)\Re (D_{N-2} \overline{\tilde{D}_{N-2}})\\
&=& \dfrac{1}{2} \sinh(\frac{2 \gamma}{c_{N-1}} )(c_{N-1}|D_{N-2}|^2+\dfrac{1}{c_{N-1}}|\overline{\tilde{D}_{N-2}}|^2)\\
&+&\cosh(\frac{2 \gamma}{c_{N-1}} )\Re (D_{N-2} \overline{\tilde{D}_{N-2}})\\
&\geq& \cosh(\frac{2 \gamma}{c_{N-1}} ) k_{N-2}\\
&=& k_{N-1}>0.
\end{array}
$$
We have proved (\ref{rere}). It follows
$$|D_{N-1} \tilde{D}_{N-1}|\geq k_{N-1}>0.$$ But $|\tilde{D}_{N-1}|$  is obviously upper bounded on the line $\Re(\lambda)=\gamma,$ consequently there exists a constant $k'_{N-1}>0$ such that
$$\forall \lambda, \Re(\lambda)=\gamma,\; |D_{N-1}|\geq k'_{N-1}>0.$$
Finally, with (\ref{up}) we deduce that $H_{N-1}^{-1}$ is bounded on the line $\Re(\lambda)=\gamma$ and it follows that there exits $c_\gamma >0$ such that
$$\forall z \in \cline, \forall \lambda :  \Re(\lambda)=\gamma,\; |F_0|\leq c_\gamma z.$$

Conclusion: (\ref{448}) is a direct consequence of the previous estimate. The proof is complete

\end{proof}
As application is that the open loop system associated to $(P)$ is satisfies a regularity property.
\begin{corollary}
Let $T > 0$. Then, for all $v \in L^2(0,T)$ the following problem
\begin{equation*}
\left \{
\begin{array}{l}
(\partial_t^2 \psi_{j}- \rho_j \partial_x^2 \psi_{j})(t,x)=0,\, x\in(j,j+1),\, t\in(0,\infty),\, j = 0,...,N-1, \\
\rho_0 \, \partial_x \psi_0(t,0) = v(t),\ \psi_{N-1}(t,N)=0,\, t\in(0,\infty),\\
\psi_{j-1}(t,j)=\psi_{j}(t,j),\, t\in(0,\infty),\, j = 1,...,N-1,\\
- \rho_{j-1} \partial_x \psi_{j-1}(t,j)+ \rho_j \partial_x \psi_{j}(t,j)= 0,\, t\in(0,\infty),\, j = 1,...,N-1, \\
\psi_j(0,x)=0,\ \partial_t \psi_j(0,x)=0, \,x \in (j,j+1),\,   j=0,...,N-1.
\end{array}
\right.
\end{equation*}
admits a unique solution $(\psi,\partial_t \psi) \in C(0,T; {\cal H})$ which satisfies the following regularity property (says also open loop admissibility): there exists a constant $C > 0$ such that
$$
\int_0^T \left| \partial_t \psi_0 (t,0)\right|^2 \, dt \leq C \, \left\|v\right\|^2_{L^2(0,T)}, \, \forall \, v \in L^2(0,T).
$$
\end{corollary}

Moreover, according to \cite[Theorem 2.2]{ammari}, we have that:

\begin{corollary}
The system $(P)$ is exponentially stable in the energy space if and only if there exists $T, C > $ such that
$$
\int_0^T \left| \partial_t \varphi_0 (t,0)\right|^2 \, dt \geq C \, \left\|(\varphi^0,\varphi^1)\right\|^2_{{\cal H}}, \, \forall \, (\varphi^0,\varphi^1) \in {\cal D}({\cal A}),
$$
where
\begin{multline*}
{\cal D}({\cal A}):=\left\{(\underline{u},\,\underline{v})\in \prod_{j=0}^{N-1} H^2(j,j+1) \times V \,:
\mbox {\textrm{satisfies }} \,(\ref {e2sd}) \; \mbox{\textrm{to}} \; (\ref {e5sd}) \; \mbox{\textrm{hereafter}}
\right\},\end{multline*}
\begin{equation}\label{e2sd}
\rho_0 \, \partial_x u_{0}(0) =  0
\end{equation}
\begin{equation}\label{e5sd}
- \rho_j \partial_x u_{j}(j) + \rho_{j-1} \partial_x u_{j-1}(j)= 0, \quad j = 1,...,N-1.
\end{equation}
and
$\varphi = (\varphi_0,...,\varphi_{N-1})$ satisfies the following problem
\begin{equation*}
\left \{
\begin{array}{l}
(\partial_t^2 \varphi_{j}- \rho_j \partial_x^2 \varphi_{j})(t,x)=0,\, x\in(j,j+1),\, t\in(0,\infty),\, j = 0,...,N-1, \\
\rho_0 \, \partial_x \varphi_0(t,0) = 0,\ \varphi_{N-1}(t,N)=0,\, t\in(0,\infty),\\
\varphi_{j-1}(t,j)=\varphi_{j}(t,j),\, t\in(0,\infty),\, j = 1,...,N-1,\\
- \rho_{j-1} \partial_x \varphi_{j-1}(t,j)+ \rho_j \partial_x \varphi_{j}(t,j)= 0,\, t\in(0,\infty),\, j = 1,...,N-1, \\
\varphi_j(0,x)=\varphi_j^0(x),\ \partial_t \varphi_j(0,x)=\varphi_j^1(x), \,x \in (j,j+1),\,   j=0,...,N-1.
\end{array}
\right.
\end{equation*}
\end{corollary}

\section{Schr\"{o}dinger system} \label{Schr}

\setcounter{equation}{0}
We consider the evolution problem $(S)$  described by the following system of $N$ equations:
\begin{equation*}
\leqno(S)
\left \{
\begin{array}{l}
(\partial_t u_{j}+i \rho_j \partial_x^2u_{j})(t,x)=0,\, x\in(j,j+1),\, t\in(0,\infty),\, j = 0,...,N-1, \\
\rho_0 \, \partial_x u_0(t,0) = { \bf i u_0(t,0)},\ u_{N-1}(t,N)=0,\, t\in(0,\infty),\\
u_{j-1}(t,j)=u_{j}(t,j),\, t\in(0,\infty),\, j = 1,...,N-1,\\
- \rho_{j-1} \partial_x u_{j-1}(t,j)+ \rho_j \partial_x u_{j}(t,j)= 0,\, t\in(0,\infty),\, j = 1,...,N-1, \\
u_j(0,x)=u_j^0(x),\,   j=0,...,N-1,
\end{array}
\right.\\
\end{equation*}

where $\rho_j > 0, \, \forall \, j=0,...,N-1$.

We define the natural energy $E(t)$ of a solution $u = (u_0,...,u_{N-1})$ of $(S)$ by
\be
\label{schrenergy1}
E(t)=\frac{1}{2} \ds \sum_{j=0}^{N-1}  \int_{j}^{j+1} | u_{j}(t,x)|^2 {\rm d}x,
\ee
We can easily check that every sufficiently smooth solution of $(S)$ satisfies the following dissipation law
\begin{equation}\label{schrdissipae1}
E^\prime(t) = - \ds \bigl| u_{0}(t,0)\bigr|^2\leq 0, \,
\end{equation}
and therefore, the energy is a nonincreasing function of the time variable $t$.

In order to study system $(S)$ we introduce the following Hilbert space

$$
\mathcal{H}= \bigg \{u=(u_0,...,u_{N-1}) \in  \ds \prod_{j=0}^{N-1} L^2(j,j+1)) \bigg \},
$$
equipped with the inner product
$$
<u,\tilde{u}>_{\mathcal{H}}=\sum_{j=0}^{N-1} \int_j^{j+1}  u_{j}(x) \, \overline{\tilde{u}_{j}(x)} dx.
$$

The system $(S)$ is a first order evolution equation which as the form
\begin{equation} \left\{
\begin{array}{l}
u' =\mathcal{A} u,\\
u(0)=u^{0},\,\end{array}\right.\label{schrfirstorder}\end{equation}

where $u^{0}=(u_0^0,u_1^0,...,u_{N-1}^0) \in \cH $  and the operator
$\mathcal{A} : {\cal D}({\cal A}) \rightarrow \mathcal{H}$ is defined by
\[\mathcal{A} u:=(-i \rho_j \partial_x^2u_{j})_{0 \leq j\leq N-1},\]
with
\begin{multline*}
{\cal D}({\cal A}):=\left\{u\in \prod_{j=0}^{N-1} H^2(j,j+1) \,:
\mbox {\textrm{satisfies }} \,(\ref {schre2}) \; \mbox{\textrm{to}} \; (\ref {schre4}) \; \mbox{\textrm{hereafter}}
\right\},\end{multline*}
\begin{equation}\label{schre2}
\rho_0 \, \partial_x u_0(0) = { \bf i u_0(0)}, u_{N-1}(N)=0,
\end{equation}
\begin{equation}\label{schre3}
u_{j-1}(j)=u_{j}(j),\, j = 1,...,N-1,\\
\end{equation}
\begin{equation}\label{schre4}
-\rho_{j-1} \partial_x u_{j-1}(j)+ \rho_j \partial_x u_{j}(j)= 0,\, j = 1,...,N-1. \\
\end{equation}

Now we can prove the well-posedness of system $(S)$ and that the solution of $(S)$ satisfies the dissipation law (\ref{schrdissipae1}).

\begin{proposition}\label{schr3exist1}
(i) For an initial datum $u^{0}\in \mathcal{H}$, there exists a unique solution $u\in C([0,\,+\infty),\, \mathcal{H})$
to  problem (\ref{schrfirstorder}). Moreover, if $u^{0}\in \mathcal{D}(\mathcal{A})$, then
$$u\in C([0,\,+\infty),\, \mathcal{D}(\mathcal{A}))\cap C^{1}([0,\,+\infty),\, \mathcal{H}).$$

(ii) The solution $u$ of $(S)$ with initial datum in $\mathcal{D}(\mathcal{A})$ satisfies \rfb{schrdissipae1}.
Therefore the energy is decreasing.
\end{proposition}

\begin{proof}
(i) By Lumer-Phillips' theorem, it suffices to show that
$\mathcal{A}$ is dissipative and  maximal.

$\mathcal{A}$ is clearly dissipative. Indeed, by integration by parts and by using the transmission and boundary conditions, we have

\begin{equation}\label{schrdissipativeness}
\forall  u \in \mathcal{D}(\mathcal{A}),\; \Re\left(\left\langle\mathcal{A}u,\, u \right\rangle_{\mathcal{H}}\right)=- \left|u_{0}(0)\right|^2\leq 0.
\end{equation}

\medskip

Now, let $f\in \mathcal{H}$. We look for $u\in \mathcal{D}(\mathcal{A})$ solution of
\begin{equation}\label{schrbijection1}
-\mathcal{A} u = f.
\end{equation}
or equivalently
\begin{equation}
i \rho_j \partial_x^2u_{j}=f_{j} ,\;  \forall j\in\{0,...,N-1\},
\label{schrbijection2}\end{equation}
and $u$ satisfies the boundary and transmission conditions (\ref{schre2})-(\ref{schre4}).

The general solution of (\ref{schrbijection2}) is
$$u_j(x)=\dfrac{1}{i\rho_j} \int_{j}^{x}\left( \int_{j}^{u} f_j(s) ds \right ) du  + p_i(x),\; j=0,...,N-1,\; x\in[j,j+1],$$
where each  $p_j$ is a polynomial of degree 1. It remains to find $p_j,\;j=0,...,N-1$ such that the equations  (\ref{schre2})-(\ref{schre4}) are satisfied. It is equivalent to solve a linear system with $2N$ equations and $2N$ unknowns.  This system admits an unique solution if and only if the corresponding homogeneous system admits only the trivial solution.

So we suppose that $p_j, j=0,...,N-1$ are polynomials of degree 1 and satisfy (\ref{schre2})-(\ref{schre4}). Then by integrations by parts and using (\ref{schre2})-(\ref{schre4}) we get
$$
\ds \sum_{j=0}^{N-1} \rho_j \int_j^{j+1} |p_j'(x)|^2 dx=  -\sum_{j=0}^{N-1} \rho_j \int_j^{j+1} p_j''(x) \overline{p_j(x) }dx=0.
$$
Consequently, the polynomials $p_j$ are constant and finally vanish from the continuity conditions and the
right Dirichlet condition.
Hence we have proved that (\ref{schrbijection1}) admits an unique solution. Consequently $\mathcal{A}$ is maximal.

\medskip

(ii) To prove (ii), we use the same argument as in the proof of Theorem \ref{3exist1}.
\end{proof}

\bigskip

\subsection{Exponential stability of the Schr\"{o}dinger system} \label{scrresolvent}

The stability result of system $(S)$ is given by
\begin{theorem} \label{schlr}
There exist constants  $C>0$ and $\omega >0$ such that, for all $u^0\in {\cal H}$, the solution of system $(S)$ satisfies the following estimate
\BEQ{schrEXPDECEXP3nb}
E(t) \le C \, e^{- \omega \,t} \, \left\Vert u^0 \right\Vert_{{\cal H}}^2,
\FORALL t > 0.
\EEQ
\end{theorem}

{\it Proof.} As in the proof of Theorem (\ref{lr}) the result is based on the following two lemmas.

\begin{lemma} \label{schrcondsp}
The spectrum of ${\cal A}$ contains no point on the imaginary axis.
\end{lemma}

\begin{proof}
Since ${\cal A}$ has compact resolvent,
its spectrum $\s({\cal A})$ only consists of eigenvalues of ${\cal A}$. We will
show that the equation
\be
{\cal A} u  = i \beta \,u
\label{schr1.9nn}
\ee
with $u = (u_0,...,u_{N-1}) \in {\cal D}({\cal A})$ and $\beta \neq 0, \beta \in \R$ has only the trivial solution.

By taking the
inner product of (\ref{schr1.9nn}) with $u  \in {\cal H}$ and using (\ref{schrdissipativeness}) we get that $u_0(0)=0$. From the left boundary condition we deduce also that $\partial_x u_0(0)=0.$ Therefore we get that $u_0=0$ since $\rho_0 \partial^2_x u_0=\beta u_0.$ Therefore by iteration we easily find $u_j=0, j=1,...,N-1.$

The system (\ref{schr1.9nn})  has only trivial solution.

\end{proof}

\begin{lemma}\label{schrlemresolvent}
The resolvent operator of $\mathcal{A}$ satisfies
\be
\limsup_{|\beta |\to \infty }  \|(i\beta -\mathcal{A})^{-1}\|_{{\cal L}(H)} <\infty,
\label{schr1.9n}
\ee
\end{lemma}

\begin{proof}
In order to prove \rfb{schr1.9n}  we look for $u=(u_0,...,u_{N-1})\in \caD(\cA)$  solution of
\be\label{schrL2} (i \beta -  \cA ) u =g,\ee  where $\beta \in \R$ and $g=(g_0,...,g_{N-1}) \in \mathcal{H}.$

We will consider two cases since for each case the method is different.

{\bf First case : $\beta>0$. }

{\bf First step :} Computation of the resolvent

The solution of( \ref{schrL2}) satisfies
\be \label{schrequation} i \beta u_j +i \rho_i \partial_x^2  u_j =g_j,\; i=0,...,N-1.\ee An easy calculation shows that
\be\label{defuj} u_j(x)=G_i(x), + c_j^1 \cos (\dfrac{\sqrt{\beta }x}{\sqrt{\rho_j}}) + c_j^2\dfrac{1}{\beta} \sin (\dfrac{\sqrt{\beta }x}{\sqrt{\rho_j}}), x\in[j,j+1],\; j=0,...,N-1,\ee
where
\be \label{defGJ}G_j(x)=\int_j^{x} \dfrac{\sin (\dfrac{\sqrt{\beta }(x-s)}{\sqrt{\rho_j}})}{ i\sqrt{\beta} \sqrt{\rho_j}}
 g_i(s) ds,\ee
and $c_j^1,c_j^2, \in \C.$ Note that $c_j^1=u_j(j)$ and $\rho_j (\partial_x u_j)(j)=c_j^2.$

Now, let $F_j=\left(
\begin{array}{l}
c_j^1\\
c_j^2\\
\end{array} \right),\;
j=0,...,N-1,$
then using the transmission conditions (\ref{schre3})-(\ref{schre4}) we have
$$\begin{array}{lll}
F_{j+1}&=&\left(
\begin{array}{c}
u_j(j+1)\\
\rho_j(\partial_x u_j)(j+1)\\
\end{array} \right)\\
&=&
\left(\begin{array}{cc}
\cos(\dfrac{\sqrt{\beta}}{\sqrt{\rho_j}}) &\dfrac{\sin(\dfrac{\sqrt{\beta}}{\sqrt{\rho_j}})}{\sqrt{\beta}\sqrt{\rho_j}}  \\
-\sqrt{\beta}\sqrt{\rho_j}\sin(\dfrac{\sqrt{\beta}}{\sqrt{\rho_j}}) & \cos(\dfrac{\sqrt{\beta}}{\sqrt{\rho_j}}) \\
\end{array} \right)F_j+
\left(
\begin{array}{l}
G_j(j+1)\\
(\partial_x G_j)(j+1)\\
\end{array} \right).
\end{array}
$$

For simplification we introduce the matrix $M_j$ and the vector $W_j$ as
\be\label{defMJWJ} M_j=\left(\begin{array}{cc}
\cos(\dfrac{\sqrt{\beta}}{\sqrt{\rho_j}}) &\dfrac{\sin(\dfrac{\sqrt{\beta}}{\sqrt{\rho_j}})}{\sqrt{\beta}\sqrt{\rho_j}}  \\
-\sqrt{\beta}\sqrt{\rho_j}\sin(\dfrac{\sqrt{\beta}}{\sqrt{\rho_j}}) & \cos(\dfrac{\sqrt{\beta}}{\sqrt{\rho_j}}) \\
\end{array} \right),\; W_j=\left(
\begin{array}{l}
G_j(j+1)\\
(\partial_x G_j)(j+1)\\
\end{array} \right),\ee
hence the transmission conditions are
\be \label{defFj} F_{j+1}=M_j F_j +W_j, j=0,...,N-1.\ee

It follows that
\be\label{schrfn} F_N=(\prod_{j=N-1}^0 M_j) F_0 +\sum_{k=0}^{N-2}(\prod_{j=N-2}^{k+1} M_j)W_k+W_{N-1}.\ee

From the first boundary condition  (\ref{schre2}) we have $F_0=c_0^1\left(\begin{array}{c}
1 \\
i
\end{array} \right),
$ (i.e, $c_0^2=i c_0^1)$) therefore we now compute $c_0^1$ by using the second boundary condition  (\ref{schre2}). For that we set

\be\label{schalphagamma} \prod_{j=N-1}^0 M_j=
\left(\begin{array}{cc}
\alpha_{N,1} & \gamma_{N,1} \\
\alpha_{N,2} & \gamma_{N,2}
\end{array} \right),\ee
and

\be\label{schrsecondmembre} -\sum_{k=0}^{N-2}(\prod_{j=N-2}^{k+1} M_j)W_k+W_{N-1}=
\left(\begin{array}{c}
\omega_{N,1} \\
\omega_{N,2}
\end{array} \right).\ee
Thus we find that
$$c_0^1=\dfrac{\omega_{N,1}}{\alpha_{N,1}+i \gamma_{N,1}}.$$

This last identity completely determine the solution $u$ of (\ref{schrL2}).

{\bf Second  step :} Estimate of $F_0$ for $\beta$ large.

From one hand, since the order of each matrix $M_j$ is
\be\label{schorderM}\left(\begin{array}{cc}
O(1) & O(\dfrac{1}{\sqrt{\beta}} )\\
O(\sqrt{\beta}) & O(1)
\end{array} \right)
\ee then it is easy to see that all the matrices involved in (\ref{schrsecondmembre}) have the same order.

On the other hand, from (\ref{defGJ})  we have clearly
\be\label {schest1}  |G_j(j+1)| \lesssim \dfrac{1}{\sqrt{\beta}}\, \| g_j\|\lesssim \dfrac{1}{\sqrt{\beta}}\, \| g\|,\; j=0,....,N-1\ee
and
\be\label {schest2}  |(\partial_x G_j)(j+1)| \lesssim  \| g_j\|\lesssim \| g\|,\; j=0,....,N-1.\ee

Therefore using the order (\ref{schorderM}) and estimate (\ref{schest1})-(\ref{schest2}) for $W_k$
in (\ref{schrsecondmembre}) we get
\be\label{schestomega1}
\omega_{N,1} \lesssim \dfrac{1}{\sqrt{\beta}}\, \| g\|.
\ee

Now, we remark that for all $j=0,...,N-1,\; \det M_j=1,$ which implies from (\ref{schalphagamma}) that
$$ \alpha_{N,1} \gamma_{N,2}-\alpha_{N,2} \gamma_{N,1}=1.$$
Thus
$$|\alpha_{N,1}+i \gamma_{N,1}||\alpha_{N,2}+i \gamma_{N,2}|\geq Re[(\alpha_{N,1}+i \gamma_{N,1})(\alpha_{N,2}+i \gamma_{N,2})]=1,$$

implies  with (\ref{schorderM}) and (\ref{schalphagamma}) that
$$\dfrac{1}{ |\alpha_{N,1}+i \gamma_{N,1}|}\leq |\alpha_{N,2}+i \gamma_{N,2}|\leq O(\sqrt{\beta}).$$
The previous estimate and (\ref{schestomega1}) lead to
\be\label{schfinal1} |c_0^1| \lesssim \|g\| \mbox{ and } |c_0^2| \lesssim \|g\| . \ee

 {\bf Last step :} Estimate of $u.$

 First, from (\ref{defGJ}) we have
$$\|G_j\|\lesssim \|g_j\| \lesssim \|g\|, \; j=0,...,N-1.$$

Then, using (\ref{defFj}), (\ref{schorderM}) and (\ref{schfinal1}) we get by iteration the  components of $F_j,j=0,...,N-1$ satisfy :
$$c_j^1 \lesssim \|g\| \mbox { and } c_j^1 \lesssim \sqrt{\beta} \|g\|.$$

Consequently, using the two previous estimates in (\ref{defuj}) we directly obtain that the solution of (\ref{schrL2}) satisfies

\be \label{schrfinal2} \|u\| \lesssim \|g\|  \;\; ( \beta \rightarrow +\infty  ).\ee

{\bf Second case : $\beta<0$. }

If $\beta<0$ the previous procedure doesn't work but fortunately, in this case,
we can get the estimate (\ref{schrfinal2}) directly. Indeed, multiplying (\ref{schrequation}) by $\dfrac{\overline{u_j}}{i},$ integrating by parts, summing from $j=0$ to $N-1$ and using the boundary-transmision conditions we have
$$\begin{array}{lll}
 -\beta \ds \sum_{j=0}^{N-1} \int_j^{j+1}|u_j(x)|^2dx  &+&\ds\sum_{j=0}^{N-1} \rho_j \int_j^{j+1}|\partial_x u_j(x)|^2dx \\
&+ & i|u_0(0)|^2\\
&=&\dfrac{1}{i} \ds \sum_{j=0}^{N-1} \int_j^{j+1} g_j(x) \overline{u_j(x)} dx.
\end{array}
$$

Therefore,
$$ -\beta \sum_{j=0}^{N-1} \int_j^{j+1}|u_j(x)|^2dx \leq \Re[\dfrac{1}{i} \sum_{j=0}^{N-1} \int_j^{j+1} g_j(x) \overline{u_j(x)} dx ] \leq \|u\| \|g\|,
$$
and we find
\be \label{schrfinal3} \|u\| \lesssim \|g\|  \;\; ( \beta \rightarrow -\infty  ).\ee
Finally the result follows from (\ref{schrfinal2})-(\ref{schrfinal3}).
\end{proof}


TITLE: For $ x^3+y^2= 2x^2y+2 $ Find $\frac{dy}{dx}$ and the slope of the tangent to the curve at $(-1,-1)$
QUESTION [0 upvotes]: For $ x^3+y^2= 2x^2y+2 $ Find $\frac{dy}{dx}$ and the slope of the tangent to the curve at $(-1,-1)$

Having a little problem finding the derivative in respect to $\frac{dy}{dx}$ for $2x^2y+2.$
To my understandings I need to use implicit differentiation; thus, wherever there is a $y$, after I take the derivative of $y$, I need to attach $\frac{dy}{dx}$. However, how does one do that for $2x^2y+2$?

REPLY [0 votes]: let $$f(x,y)=x^3+y^2-2x^2y-2$$
$$\frac{df}{dx}=3x^2-4xy$$
$$\frac{df}{dy}=2y-2x^2$$
$$\frac{dy}{dx}=\frac{\frac{df}{dx}}{\frac{df}{dy}}=\frac{3x^2-4xy}{2y-2x^2}$$


TITLE: Find all the infinite cardinals $a$ such that $a^4 + 16^a = 16^a$
QUESTION [0 upvotes]: Find all the infinite cardinals $a$ such that  $a^4 + 16^a = 16^a$.
I think this is equivalent to $a^4 + (2^a)^4 = (2^a)^4$, but I don't know how to prove the latter.
Any help would be appreciated.

REPLY [1 votes]: For all infinite cardinals under AC, $a^4=a, 16^a=2^a \gt a, a+b=\max(a,b)$