TITLE: exponential inequality via calculus
QUESTION [1 upvotes]: I am wondering if an inequality such as
$$
\exp(x+y)\leq \exp(x) + \exp(y)
$$
for certain values of $x$ and $y$. According to wolfram alpha, it should hold for the region $\mathcal{R}=\{(x,y)\in\mathbb{R}^2\colon x\leq0,y\leq 0 \}$ which for me works. My calculus is a bit rusty, but what I tried is defining the function $f(x,y)=\exp(x+y)- \exp(x) - \exp(y)$ and try to minimise/maximise it. The gradient $\nabla f(x,y)$ vanishes just at $(0,0)$, and the determinant of the Hessian matrix is negative at this point, so this is a saddle point, which makes sense since according to WA the reverse inequality holds in the complement of the region $\mathcal{R}$. How can I search for the maximum of $f$ in the desired region?
Thanks in advance.
PS: I have the feeling that I should be able to say something even stronger, but with this it would suffice.

REPLY [1 votes]: For any $y\in\mathbb{R}$, and $x\leq 0$ then $0<\exp(y)$ and $0<\exp(x)\leq 1$ which imply that
$$\exp(x+y)=\exp(x)\cdot \exp(y)\leq \exp(y) \leq \exp(x)+\exp(y).$$
By symmetry the same holds when  $y\leq 0$.
P.S. If $x>0$ then it is easy to verify that $\exp(x+y)\leq \exp(x)+\exp(y)$ holds if and only if
$$y\leq x-\ln(\exp(x)-1).$$
Take a look here.


TITLE: Is there a continuous surjection from $\mathbb R$ to $C_p([0,1])$?
QUESTION [7 upvotes]: Let $C_p([0,1])$ denote the set of continuous functions $[0,1]\to\mathbb R$ with the subspace topology coming from the product $\mathbb R^{[0,1]}.$ The "p" stands for pointwise. Is this space a continuous image of $\mathbb R$?

Some observations:

$C_p([0,1])$ is not locally compact
To be an image of $\mathbb R$ a space must be $\sigma$-compact
$C_p([0,1])$ is not Baire
It would be equivalent to consider maps $[0,1]\to [0,1].$ (In one direction we can "clip" a function to $[0,1],$ and in the other direction every real-valued continuous function is $N$ times a $[0,1]$ valued function, and we can use interleaving to encode all functions hitting each multiple of $N$ into a single surjection from $\mathbb R.$)
There is a nice diagonalization argument showing that $C_p(\mathbb R)$ is not a continuous image of $\mathbb R.$ See this answer and use the obvious continuous $C_p(\mathbb R)\to C_p(\mathbb N)$ coming from restriction to $\mathbb N$.

REPLY [4 votes]: Henno Brandsma's answer give a reference for a more general question. Here I will just record a mostly self-contained proof for this specific case.
Given a function $\phi:\mathbb R\to C_p([0,1]),$ let $K_m$ be the
image of $[-m,m]$ for each integer $m\geq 1.$ Inductively define a sequence of closed sets by $X_0=C_p([0,1])$ and for each $n\geq 1:$

$m_n=\min\{m\geq 1\mid K_m\cap X_{n-1}\neq \emptyset\},$ or $m_n=\infty$ if there is no such $m$
$X_n=\{f\in X_{n-1}\mid f(1/n)=(-1)^n/m_n\},$ where $1/\infty=0.$

So $m_n$ is a weakly increasing sequence, and $X_n$ is a weakly decreasing sequence of sets.
If $\max_n(m_n)=m<\infty,$ then $K_m\cap X_{n-1}\neq\emptyset$ never becomes empty for any $n.$ If $K_m$ were compact, then $(K_m\cap X_{n-1})_{n\geq 1}$ would be a non-increasing sequence of non-empty closed sets, so there would be some $f\in\bigcap_{n\geq 1}(K_m\cap X_{n-1}).$ But that gives $\liminf_{n\to \infty} f(1/n)\leq -1/m$ and $\limsup_{n\to\infty} f(1/n)\geq 1/m,$ which is not possible because $f$ must be continuous at $0.$ So in this case $K_m$ is not compact.
If $m_n\to\infty$ then $\bigcap_{n\geq 1} X_n$ contains the function $f:[0,1]\to\mathbb R$ that linearly interpolates between the points $(1/n,(-1)^n/m_n),$ setting $f(0)=0$ (note $1/m_n\to 0$). But for each $m$ we have $m_n>m$ for some $n,$ which gives $K_m\cap X_{n-1}=\emptyset,$ which implies $f\not\in K_m.$ So in this case $f\not\in\bigcup_{m\geq 1} K_m.$
In either case we have shown that $\{K_m\}$ is not a cover of $C_p([0,1])$ by compact sets. This means the original function $\phi$ is either not continuous or not surjective.


TITLE: Show that; $\displaystyle\int_0^\infty\frac{f(bx)-f(ax)}{x}dx=(l-f(0))\ln\frac{b}{a}$
QUESTION [5 upvotes]: Assume that $f\in C^1([0,\infty))$, $f'$ is monotonic on $[0,\infty)$ and $\lim\limits_{x\to\infty} f(x)=l$ is finite. If $a,b>0$, prove that; $$\displaystyle\int_0^\infty\frac{f(bx)-f(ax)}{x}dx=(l-f(0))\ln\frac{b}{a}$$

The trick is clear for me, I have to write is as double integral, i.e. the function $\frac{f(bx)-f(ax)}{x}$ as an integral, obviously with limits $a$ and $b$. For example
$\displaystyle\int_a^bf'(xy)dy$$\quad$(Here is $f$ differentiated with respect to $y$)
works.
Now if I assume that, $\displaystyle\int_0^\infty\Big(\displaystyle\int_a^bf'(xy)dy\Big)dx=\displaystyle\int_a^b\Big(\displaystyle\int_0^\infty f'(xy)dx\Big)dy$
and $\displaystyle\int_0^\infty f'(xy)dx=\frac{l-f(0)}{y}$ $(\bigstar)$
then $\displaystyle\int_a^b\frac{l-f(0)}{y}=(l-f(0))\ln\frac{b}{a}$ gives the correct result.
MY QUESTIONS
1)Why is the line with $(\bigstar)$ correct ? Because $f'(xy)$ is differentiated w.r. to $y$ but if we integrate w.r.to $x$ why do we get $f$ again ?
2)Why can we change the order of integration, We have to make use of the monotonicity of $f'$, I guess $\lim\limits_{x\to\infty}f'(x)=0$, but this is not sufficient.

REPLY [4 votes]: Ad 1): $f'(xy)$ is the derivative of $f$, evaluated at the point $xy$. Since $f$ is a function of a single variable, "differentiated w.r.t" some variable is not a distinction to be made, you can differentiate $f$ only with respect to its single real argument. You can differentiate the composition $F(x,y) = f(xy)$ with respect to $x$ or $y$, and the results are $\frac{\partial F}{\partial x}(x,y) = f'(xy)\cdot y,\; \frac{\partial F}{\partial y}(x,y) =f'(xy)\cdot x$. From which you can see that
$$f'(xy)= \frac{1}{y}\frac{\partial F}{\partial x}(x,y)= \frac{1}{x}\frac{\partial F}{\partial y}(x,y),$$
which gives you the respective results when integrating with respect to $x$ or $y$.
Ad 2): The monotonicity of $f'$ together with $\lim\limits_{x\to\infty} f'(x) = 0$ (which follows from $\lim\limits_{x\to\infty} f(x) = l$) imply that $f'$ has constant sign, either it is always non-negative, or always non-positive. Tonelli's theorem then assures that the interchange of order of integration is allowed.


TITLE: Algebras with Calabi-Yau derived categories
QUESTION [3 upvotes]: Let $A$ be a finite dimensional algebra over a field $k$ and assume that global dimension of $A$ is finite. I want to describe such algebras $A$ with Calabi-Yau derived category.
A triangulated category with a Serre functor $S$ is called Calabi-Yau if $S(-) \cong (-)[n]$, where $n$ is an integer.
For the triangulated category $D^b(A)$ Serre's functor is Nakayama's functor
$$
S(M) = D(A) \otimes^{\mathbb{L}} M,
$$
where $M \in D^b(A)$ and $D(-)=\operatorname{Hom}_k(-,k)$
If we assume that $S(M)=M[n]$ and take $M=A$ we get $D(A) \cong A[n]$, so Calabi-Yau dimension $n=0$ and $D(A) \cong A$ as left modules. So, $A$ is a Frobenius algebra.
The conclusion is: algebra $A$ is Frobenius if and only if $D^b(A)$ is Calabi-Yau. In this case Calabi-Yau dimension is $0$.
Is all the above correct?

REPLY [3 votes]: Almost correct, except that "Frobenius" is not sufficient. In order that $D(A)\otimes -$ is isomorphic to the identity functor (as functors from left $A$-modules to left $A$-modules) you need not only that $D(A)\cong A$ as left $A$-modules, but that they are isomorphic as $A$-bimodules. Algebras with this stronger property are called "symmetric".
Edit: Actually, I answered in a hurry and misread exactly what you asked, so let me clarify.
Unless it is semisimple, a symmetric algebra has infinite global dimension, and the derived category $D^b(A)$ is not Calabi-Yau. So the only examples are the semisimple symmetric $k$-algebras, which are precisely the separable $k$-algebras.
The sense in which more general symmetric algebras are $0$-Calabi-Yau is that the category $K^b(P_A)$ of perfect complexes (i.e., those isomorphic to bounded complexes of finitely generated projective modules) is $0$-Calabi-Yau. More generally, there is a natural isomorphism $\text{Hom}(P,X)\cong D\text{Hom}(X,P)$ whenever $P$ and $X$ are objects of $D^b(A)$ and $P$ is perfect.


TITLE: Differentiability of $f$ if $f^{-1}$ is differentiable
QUESTION [0 upvotes]: Suppose that $f$ is a one-to-one function and that $f^{−1}$ has a derivative which is nowhere $0$. Prove that $f$ is differentiable.

REPLY [2 votes]: Since $f$ is one-to-one, $f^{−1} $ is a function whose inverse is $f$.
Since $f^{−1} $ is differentiable, $f^{−1} $ is continuous.
Let $y = f(x)$ , so $f^{−1} (y) = x$.
$$\lim_{h\to0}\frac{f(x+h)−f(x)}{h}=\lim_{h\to0}\frac{f(x+h)−y}{h}.$$
Now since $f^{−1}$ is continuous, we can write $x + h = f^{−1} (y + k)$ where $k \to 0 $ as $  h \to 0.$
Now
$$\lim_ {h\to0} \frac {f(x+h)−y}{h} = \lim_{h\to0}  \frac {f(f^{−1} (y+k))−y}{(x+h)−x} = \lim_{h\to0}
\frac {(y+k)−y}{f^{−1} (y+k)−f^{−1} (y)} = \lim _{k\to0} \frac {k}{f^{−1} (y+k)−f^{−1} (y)} $$
as "$h \to 0 \implies k \to 0$".
Hence
$$ \lim_{h\to0} \frac {f(x+h)−y}{h} = \lim _ {k\to0}\frac {k} { f^{−1}(y+k) − f^{−1}(y) }  =  \frac  {1} {(f^{−1})′(y) } = \frac  {1} {(f^{−1})′(f(x)) }.$$
So f is differentiable.


TITLE: Where did I go wrong with the Gram-Schmidt orthogonalisation process?
QUESTION [1 upvotes]: Problem: Let $\alpha = \left\{(1,2,0), (1,0,1), (2,3,1)\right\}$ be a basis vor $\mathbb{R}^3$. Apply the Gram-Schmidt orthogonalisation process to turn $\alpha$ into an orthonormal basis for $\mathbb{R}^3$ with respect to the standard innerproduct.
Attempt at solution: I change this set to an orthogonal set first, and normalize afterwards.
Define: \begin{align*} v_1 &= (1,2,0) \\ v_2 &= (1,0,1) - \frac{ \langle (1,0,1), (1,2,0) \rangle}{\langle (1,2,0), (1,2,0) \rangle} (1,2,0) \\ v_3 &= (2,3,1) - \frac{ \langle (2,3,1), (1,0,1) \rangle}{ \langle (1,0,1), (1,0,1) \rangle} (1,0,1) - \frac{ \langle (2,3,1), (1,2,0) \rangle}{\langle (1,2,0), (1,2,0) \rangle} (1,2,0) \end{align*}
Then this gives \begin{align*} v_2 = (1,0,1) - \frac{1}{1+4} (1,2,0) = (1,0,1) - (\frac{1}{5}, \frac{2}{5}, 0) = (\frac{4}{5}, - \frac{2}{5}, 1)
\end{align*} and \begin{align*} v_3 &= (2,3,1) - \frac{(2+1)}{(1+1)} (1,0,1) - \frac{(2+6)}{(1+4)}(1,2,0) \\ &= (2,3,1) - (\frac{3}{2}, 0, \frac{3}{2}) - (\frac{8}{5}, \frac{16}{5}, 0) \\ &= (- \frac{11}{10}, - \frac{1}{5}, - \frac{1}{2}). \end{align*}
Now I wanted to check my work. So I compute the innerproduct of $v_2$ with $v_3$, i.e. \begin{align*} \langle (\frac{4}{5}, - \frac{2}{5}, 1), (- \frac{11}{10}, - \frac{1}{5}, - \frac{1}{2}) \rangle \end{align*} but I don't get zero? When I compute $\langle v_1, v_2 \rangle$, however, I do get zero. So something must be wrong with my $v_3$. I triple checked my work and I can't find the mistake. Please help!

REPLY [1 votes]: The problem is that you projected two non-orthogonal vectors, $v_1$ and $v_2$, out of $v_3$. Each of those orthogonalizations kills the other. In orthogonalizing $v_3$, you need to use the result of orthogonalizing $v_2$ against $v_1$, not $v_2$ itself.


TITLE: Find simple path with the largest sum of weights
QUESTION [1 upvotes]: In the longest path problem, given a weighted graph G and a starting node v0, find the simple path starting with v0 with the largest sum of weights.
Show that the longest path problem is NP-hard in general. Assume an efficient algorithm to solve the longest path and show how an efficient algorithm which decides the existence of a Hamiltonian path can be derived from that.
Please help :( Thank you!

REPLY [2 votes]: The main method to show a problem is NP-hard is via the process of reduction. Reduction means taking some unknown problem $p$, and showing that if you had a way to solve $p$ in polynomial time, then you can solve a known NP-hard problem $q$ in polynomial time.
You are given the fact that the problem Hamiltonian Path is NP-hard already (so it's our $q$ in this case). The goal is to show that the longest path problem (which is our $p$) is harder than the Hamiltonian Path problem. You can do this by finding an instance of the longest path problem, which when solved, also solves the Hamiltonian Path problem.
Here's a framework for the proof of this problem. Your job is to figure out what $G'$, the weights of each edge, and $W$ should be.
Let $G = (V,E)$ be a graph, and $v_{0} \in V$. We want to know if $G$ has a Hamiltonian path starting from $v_{0}$. Create a graph $G' = (V', E')$ with weights (INSERT WEIGHTS HERE). Now suppose $G'$ has a longest path of total weight $W$. Then G (DOES / DOES NOT) have a Hamiltonian path, because (REASON). This shows that Longest Path reduces to Hamiltonian Path, and hence the former is NP-hard.
Note that there are two cases you have to cover for: $G$ does or does not have a Hamiltonian path, and in the positive case, you do have to explain how to "convert" the longest path in $G'$ to the Hamiltonian path in $G$.
Hint: $G'$ is not too different from $G$. Good luck!


TITLE: Find center of mass and moment applied on beam structure
QUESTION [0 upvotes]: I have a simple physics problem to solve but it is giving me a slightly difficult time to figure out. The problem:

I have a beam structure with same cross section. It consists of three beam. First in vertical with length 1meter. Second beam joined on top with a angle of 20degrees to the right with lenth 1.1meter and on top of the second beam, a third beam parallel to the ground with 1meter length.
I need to find the center of gravity and how much moment I have on the base of the structure so I would add a mass that would be sufficient to avoid the structure to fall over.

any ideas how?

REPLY [1 votes]: I need to find the center of gravity

Center of mass (CoM) is found by adding up all particles $i$:
$$\vec r_{cm}=\frac{\sum m_i \vec r_i}{\sum m_i}$$
$\vec r_{cm}$ is the coordinate vector of the CoM.
Since you have straight beams (assumingly cylinder shaped), start out with symmetry considerations for each of them. The CoM for each beam must be at the exact center (assuming constant density throughout).  Put these three CoMs into the formula above to find the combined CoM of the whole structure.
A side note: If the gravitational acceleration $\vec g$ is constant throughout, then the CoM is equal to the center of gravity.

and how much moment i have on the base of the structure so I would add a mass that would be sufficient to avoid the structure to fall over.

Formula for torque about an axis (the tip of the base point):
$$\vec \tau=\vec r_{cm} \times \vec F$$
In your 2D-case simplified to:
$$\tau=r_{cm,\perp} F=r_{cm} F \sin \theta$$
where $r_{cm} \sin \theta$ must be $r_{cm}$'s horizontal component (since gravity $\vec F$ is vertical).
Find this torque of gravity on the structure, and then place an object on the base in whatever manner the problem allows so it gives the opposite torque equal in magnetude.


TITLE: A question about the proof of Schwarz inequality
QUESTION [5 upvotes]: There are many proofs of the Cauchy-Schwarz inequality, here's one of them: Consider the following quadratic polynomial: $$f(x)=\left(\sum_{i=1}^{n} a_i^2  \right)x^2-2\left(\sum_{i=1}^{n} a_ib_i  \right)x+\sum_{i=1}^{n} b_i^2=\sum_{i=1}^{n} (a_ix-b_i)^2.  $$ Since $f(x)\geqslant0$ for all $x\in\Bbb R$, it follows that the discriminant of $f(x)$ is negative, i.e., $$
\left(\sum_{i=1}^{n}a_ib_i\right)^2-\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{i=1}^{n}b_i^2\right)\leqslant0
$$
Therefore: $$\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{i=1}^{n}b_i^2\right)\geqslant\left(\sum_{i=1}^{n}a_ib_i\right)^2.$$
the question is:
why are we able to say :"Since $f(x)\geqslant0$ for all $x\in\Bbb R$, it follows that the discriminant of $f(x)$ is negative" ??

REPLY [1 votes]: We are not able to say that the discriminant of $f(x)$ is negative.  Look at exactly what you wrote:"Since $f(x) \ge 0$ for all x∈R, it follows that the discriminant of f(x) is negative, i.e.,  $$\left(\sum_{i=1}^{n}a_ib_i\right)^2-\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{i=1}^{n}b_i^2\right)\leqslant0$$.  This inequality does not follow from what you said (i.e. it is not a strict inequality, which your statement implies).  We don't know the discriminant is negative, it could be $0$.  We know it can't be positive, for then the quadratic would have two real roots, but since $f(x) \ge$  $0$ $\forall$ $x$ $ \in \mathbb{R}$, it can have at most 1 real root.


TITLE: "Tricks" for solving the determinant of a matrix
QUESTION [15 upvotes]: I've got an exam coming up, and I'm worried that (as I usually tend to do) I will mess up with some minor algebraic calculation in a large problem like finding the determinant of a 4x4 or larger matrix. I'm likely to be asked to use Laplace Expansion, and I'd like to know if there are any other quick and dirty ways I can try to make sure my answer was correct. Considering I'm not alowed to use a calculator, is there some other non-standard method I can use to try to verify a correct solution?
I know how to find the determinant no problem, but I'm very prone to minor mistakes and I don't want to have to spend too much time reading over every problem in fine detail to make sure I didn't omit a negative sign somewhere.

REPLY [2 votes]: You might consider Pivotal Condensation.
PC reduces an $n\times n$ determinant to an $(n-1)\times(n-1)$ determinant whose entries happen to be $2\times 2$ determinants. Simply iterate until your determinant gets to reasonable size. (You can/should stop at $3\times 3$, at which point it's easy enough to compute the final result manually.)
Web-searching "pivotal condensation" will give you lots of explanations and examples. This document even includes a simple proof. This lengthy YouTube video purports to talk through the process, but I didn't view the whole thing.
Pivotal condensation can be extremely tedious; it may-or-may-not be time-effective in an exam situation. However, at every stage of the process, you compute mere $2\times 2$ determinants, which are super-easy and super-quick. So there's a trade-off. (Of course, there's another trade-off: with so many to compute, the danger of stupid arithmetic mistakes accumulates.)
I've actually found pivotal condensation helpful with certain symbolic matrices (in particular, ones with trigonometric terms), since it gives me a systematic way to (attempt to) combine elements. In that regard, it's one of my favorite "tricks".


text:
\begin{document}
\maketitle

\thanks{Fakult\"at f\"ur Mathematik, Universit\"at Wien,
\\
Nordbergstrasse 15, A-1090 Vienna,  Austria. \\
al$_-$sakhnov@yahoo.com}

\begin{abstract}
A factorization formula for wave functions, which is basic in the inverse spectral transform approach
to initial-boundary value problems, is proved in greater generality than before.
Applications follow. Related compatibility  questions for the GBDT version of   B\"acklund-Darboux
transformation are treated too.
\end{abstract}

{MSC(2010):}  37K15; 46N20; 47A68; 35C08; 46E40.

{\it Keywords: } Wave function, fundamental solution, factorization,
compatibility, inverse
spectral transform,  integrable equation, initial-boundary value problem,
B\"acklund-Darboux transformation.

\section{Introduction} \label{intro}
\setcounter{equation}{0}
Zero curvature representation of the integrable nonlinear equations
is a well known approach (see \cite{AKNS, TF, Nov, ZM} and references in \cite{TF}),
which was developed
soon after the seminal Lax pairs appeared in \cite{Lax}. Namely, many
integrable nonlinear equations admit representation
(zero curvature representation)
\begin{align}&      \label{0.1}
G_t(x,t,z)-F_x(x,t,z)+[G(x,t,z),F(x,t,z)]=0, \\
& \nonumber
 G_t:=\frac{\p}{\p t}G, \quad [G,F]:=GF-FG,
\end{align}
which is the compatibility condition of the auxiliary linear systems
\begin{equation}      \label{0.2}
\frac{\p}{\p x}w(x,t,z)=G(x,t,z)w(x,t,z), \quad \frac{\p}{\p t}w(x,t,z)=F(x,t,z)w(x,t,z).
\end{equation}
Here $G$ and $F$ are $m \times m$ matrix functions, and
$z$ is the spectral parameter, which will be omitted sometimes in our notations.

Solution of integrable nonlinear equations is closely related to Lax pairs
and  zero curvature representations, which have been mentioned above, and has been
a great breakthrough in the second half of the 20th century. An active study
of the cases, which are close to integrable in a certain sense,  followed
(see, for instance, some references in \cite{Bo, Persp, Ka}).
Initial-boundary value problems  for  integrable nonlinear equations
can be considered as an important example, where integrability is
"spoiled" by the boundary conditions. These problems are
of great current interest, and inverse spectral transform (ISpT) method
\cite{Ber, BerG, KvM, SaA1, SaA2, SaA4, SaA5, SaL1, SaL2, SaL3, SaL4}
is one of the fruitful approaches in this domain. Further we assume that
$x, \, t$ belong to a semi-strip
\begin{equation}      \label{0.3}
{\mathcal D}=\{(x,\, t):\,0 \leq x <\infty, \,\,0\leq t<a\}.
\end{equation}
Normalize fundamental solutions of the auxiliary systems by the initial conditions
\begin{align} \label{0.4}
&\frac{d}{d x}W(x,t,z)=G(x,t,z)W(x,t,z), \quad W(0,t,z)=I_m;
\\ \label{0.5} &
\frac{d}{d t}R(x,t,z)=F(x,t,z)R(x,t,z), \quad R(x,0,z)=I_m,
\end{align}
where $I_m$ is the identity matrix of order $m$.
If condition \eqref{0.1} holds, the fundamental solution  of \eqref{0.4}
 admits factorization
\begin{equation} \label{0.6}
W(x,t,z)=R(x,t,z)W(x,0,z)R(t,z)^{-1}, \quad R(t,z):=R(0,t,z).
\end{equation}
Formula \eqref{0.6} is one of the basic and actively used formulas in the
 inverse spectral transform method (see \cite{SaA1, SaA2, SaA4, SaA5, SaL1, SaL2, SaL3, SaL4}
and references therein). It was derived in \cite{SaL1, SaL2} under some smoothness conditions
(conitinuous differentiability of $G$ and $F$, in particular): see formulas (1.6) in \cite{SaL1}, p.22
and in \cite{SaL2}, p. 39.

Here we prove \eqref{0.6}  under weaker conditions and in much greater detail, which is important
for applications.  Namely, we prove the following theorem.
\begin{Tm} \label{TmM} Let $m \times m$ matrix functions $G$ and $F$ and their derivatives $G_t$ and
$F_x$  exist on the semi-strip $\mathcal{D}$,
let $G$, $G_t$, and $F$ be continuous with respect to $x$ and $t$ on $\mathcal{D}$,
 and let \eqref{0.1}
hold. Then the equality
\begin{equation} \label{0.6'}
W(x,t,z)R(t,z)=R(x,t,z)W(x,0,z), \quad R(t,z):=R(0,t,z),
\end{equation}
is true.
 \end{Tm}
 Note that constructions similar to \eqref{0.6} appear also in the theory of
 Knizhnik-Zamolodchikov equation (see Theorem 3.1 in \cite{SaLKZ2} and see also \cite{SaLKZ}).

Theorem \ref{TmM} is proved in Section \ref{Pr}.
Section \ref{some} is dedicated to  applications to initial-boundary value
problems, and Theorem \ref{evol} on the evolution of the Weyl function
for the "focusing"  modified Korteweg-de Vries (mKdV) equation is proved
there as an example.

Related questions of the equality of mixed derivatives and application
of this equality  to the GBDT version (see \cite{GKS2, GKS6, KaSa, MST, SaA2, SaA2', SaAJFA, SaA3, SaA6}
and references therein)
of the B\"acklund-Darboux transformation are
treated in Section \ref{Darb}.

As usual, by $\BN$ we denote the set of positive integers, by $\BC$ we denote the complex plane,
and by $\BC^m$ is denoted the $m$-dimensional coordinate space
over $\BC$.  By $\Im z$ is denoted  the imaginary part of $z \in \BC$,
and $\arg z$ is the argument of $z$.
By $C^k({\mathcal D})$ we denote functions and matrix functions, which are $k$ times continuously differentiable on ${\cal D}$.

\section{Proof of Theorem \ref{TmM}} \label{Pr}
\setcounter{equation}{0}
The spectral parameter $z$ is non-essential for the formulation of
Theorem \ref{TmM} and for its proof and we shall omit it in this section.
We shall need the proposition below.
\begin{Pn} \label{PM} Let the $m \times m$ matrix function $W$ be given on the
 semi-strip $\mathcal{D}$ by  equation \eqref{0.4}, where $G(x,t)$
and $G_t(x,t)$ are continuous matrix functions in $x$ and $t$.

(i)Then the derivative $W_t$ exists and matrix functions $W$ and $W_t$
are continuous with respect to $x$ and $t$
on the semi-strip $\mathcal{D}$.

(ii)Moreover, the mixed derivative $W_{tx}$ exists and the equality
$W_{tx}=W_{xt}$ holds on $\mathcal{D}$.
 \end{Pn}
\begin{proof}.
Consider system
 \begin{align}      \label{1.2}
&\frac{d}{d x}y=\wh G(x,y)y, \quad
 \wh G(x,y)=\wh G(x,y_{m+1}):=\left[
\begin{array}{cc}
G(x,y_{m+1}) &0\\ 0&0
\end{array}
\right],
\end{align}
where $\wh G$ is an $m+1 \times m+1$ matrix function and $y_{m+1}$ is the last entry of the column vector $y\in \BC^{m+1}$.
Denote by $W_j$ and $e_j$ the $j$-th columns of $W$ and $I_m$,
respectively ($1 \leq j \leq m$).
 It easily follows from \eqref{0.4} that the solution of \eqref{1.2} with the initial condition
 \begin{align}      \label{1.3}
& y(0)=g=\left[
\begin{array}{c}e_j \\ t
\end{array}
\right]
\end{align}
has the form
\begin{equation}      \label{1.4}
y(x,g)=\left[
\begin{array}{c}W_j(x,t) \\ t
\end{array}
\right].
\end{equation}
Putting $G(x,t)=G(0,t)$ for $-\ve \leq x \leq 0$ whereas $t \geq 0$, and putting $G(x,t)=G(x,0)+tG_t(x,0)$
for $-\ve \leq t \leq 0$ ($\ve>0$) we extend $G$ so that $G$ and $G_t$ remain continuous on the
rectangles
\begin{equation}      \label{1.5}
\mathcal{D}(a_1,\, a_2)=\{(x,\, t): \,  -\ve\leq x \leq a_1,  \, \, -\ve \leq t \leq a_2<a\},
\quad a_1, a_2 \, \in \BR_+.
\end{equation}
Hence, it follows from the definition
of $\wh G$ in \eqref{1.2} that $\wh G(x,y)$ and, as a consequence, the vector function $\wh G(x,y)y$ are continuous  on $\mathcal{D}(a_1,\, a_2)$ together
with their derivatives  with respect to the entries of $y$. Thus, according  to the classical theory
of ordinary differential equations (see, for instance, theorem on pp. 305-306 in \cite{Sm})
the
partial first derivatives of $y(x,g)$ with respect to the entries of $g$
exist in the interior
$\mathcal{D}_i(a_1,\, a_2)$ of  $\mathcal{D}(a_1,\, a_2)$.
Moreover,
$y$ and its partial  derivatives  with respect to the entries of $g$ are continuous.  In particular,
since by
 \eqref{1.3} we have $g_{m+1}=t$,
the functions $y$ and $y_t$ are continuous in all rectangles $\mathcal{D}_i(a_1,\, a_2)$.
Taking into account \eqref{1.4}, we see that $W$ and $W_t$
are continuous  in the rectangles $\mathcal{D}_i(a_1,\, a_2)$, and the statement (i) is true.

In view of \eqref{0.4} and considerations above the derivatives $W_x$, $W_{xt}$, and $W_t$ exist and are
continuous in the rectangles $\mathcal{D}_i(a_1,\, a_2)$. Hence, by a  stronger formulation (see, for instance,
\cite{See, Aks} or p. 201 in \cite{Mar}) of the well-known theorem on mixed derivatives, $W_{tx}$ exists in
$\mathcal{D}_i(a_1,\, a_2)$  and $W_{tx}=W_{xt}$. Thus,  the statement (ii) follows.
\end{proof}
Now, we can follow the scheme from Chapter 3 in \cite{SaL2}  (see also Chapter  12 in \cite{SaL4}).
\begin{proof} of  Theorem \ref{TmM}.
According to statement (i) in Proposition  \ref{PM} the matrix function $W_t$ exists
and is continuous. Introduce $U(x,t)$ by the equality
\begin{equation}      \label{1.6}
U:=W_t-FW.
\end{equation}
By \eqref{0.4},  \eqref{1.6}, and  statement (ii) in Proposition  \ref{PM}  we have
\begin{equation}      \label{1.7}
U_x=W_{tx}-F_xW-FW_x=W_{xt}-F_xW-FGW.
\end{equation}
It  is immediate also from \eqref{0.4}  that
\begin{equation}      \label{1.8}
 W_{xt}=\big(GW\big)_t=G_tW+GW_t.
\end{equation}
Formulas  \eqref{1.7} and   \eqref{1.8} imply
\begin{equation}      \label{1.9}
 U_{x}=G_tW+GW_t-F_xW-FGW=(G_t-F_x+GF-FG)W+GW_t-GFW.
\end{equation}
It follows from \eqref{0.1}, \eqref{1.9}, and definition  \eqref{1.6}  that $U_{x}=GU$,
that is, $U$ and $W$ satisfy the same equation. Taking into account $W(0,t)=I_2$,
we derive $W_t(0,t)=0$, and so by \eqref{1.6} we have $U(0,t)=-F(0,t)$.  Finally, as
\[
U_{x}=GU, \quad W_x=GW, \quad U(0,t)=-F(0,t), \quad W(0,t)=I_2,
\]
we have $U(x,t)=-W(x,t)F(0,t)$ or, equivalently,
\begin{equation}      \label{1.10}
W_t(x,t)-F(x,t)W(x,t)= -W(x,t)F(0,t).
 \end{equation}
 Put
 \begin{equation}      \label{1.11}
Y(x,t)=W(x,t)R(t), \quad Z(x,t)=R(x,t)W(x,0).
 \end{equation}
 Recall that $R(t)=R(0,t)$. Therefore
 \eqref{0.5}, \eqref{1.10}, and \eqref{1.11}  imply that
\begin{align}     \nonumber
Y_t(x,t)&=\big(F(x,t)W(x,t)-W(x,t)F(0,t)\big)R(t)+W(x,t)F(0,t)\big)R(t)
\\ &  \label{1.12}
=F(x,t)Y(x,t), \quad Y(x,0)=W(x,0).
 \end{align}
 Formulas \eqref{0.5} and \eqref{1.11}  imply that
 \begin{equation}      \label{1.13}
Z_t(x,t)=F(x,t)Z(x,t), \quad Z(x,0)=W(x,0).
 \end{equation}
 By \eqref{1.12} and \eqref{1.13} $Y=Z$, that is, \eqref{0.6'} holds.
\end{proof}
\begin{Rk} \label{Dom0}
Though the case of continuous $F$ is more convenient for applications,
it is immediate from the proof that the statement of Theorem  \ref{TmM}
is true, when $F$ is differentiable with respect to $x$, and measurable
and summable  with respect to $t$ on all  finite intervals from $\BR_+$ .
\end{Rk}
According to the proof of
Theorem \ref{TmM} the following remark is also true.
\begin{Rk} \label{Dom}
Theorem \ref{TmM} holds on the domains more general than $\cal D$.
In particular, it holds
if we consider  $(x,\, t)\in {\cal I}_1 \times {\cal I}_2$,
where ${\cal I}_k$
 $(k=1,2)$  is the interval $[0, \,b_k)$  $\,(0<b_k\leq \infty)$.
\end{Rk}

Another interesting case of matrix factorizations related to boundary value
problems is treated in \cite{BHLW, KL}.

\section{Some applications} \label{some}
\setcounter{equation}{0}
The {\it matrix "focusing" mKdV equation} has the form
\begin{align}  \label{2.1} &
4v_t=v_{xxx}+3\big(v_xv^*v+vv^*v_x\big),
\end{align}
where $v(x,t)$ is a $p \times p$ matrix function.
 Equation \eqref{2.1} is equivalent
(see \cite{CD, TF, Wad} and references therein)
to zero curvature
equation \eqref{0.1}, where the $m \times m$ ($m=2p$)
matrix functions $G(x,t,z)$ and $F(x,t,z)$ are given by the formulas
\begin{align}  \label{2.2} &
G=izj+V, \quad
j=\left[
\begin{array}{lr}
I_p& 0 \\ 0 & -I_p
\end{array}
\right],
\quad
 V=\left[
\begin{array}{lr}
0& v \\ - v^* & 0
\end{array}
\right],
\\ \label{2.3} &
F=-iz^3j-z^2V-\frac{iz}{2}\big(V^2+V_xj\big)+\frac{1}{4}
\big(V_{xx}-2V^3-V_xV+VV_x\big).
\end{align}
At first we omit the variable $t$ in $V$ and $v$.
The Weyl theory of  the skew-self-adjoint Dirac system (also called
Zakharov-Shabat or AKNS system)
 \begin{equation}      \label{2.4}
\frac{d}{dx}w(x,z)=\big(izj+V(x)\big)w(x,z), \quad x \geq 0
 \end{equation}
 was treated in \cite{CG, FKS, SaA1, SaA2} (see also preliminaries
 in \cite{SaA7}).

 For the case of measurable matrix function $v$ such that
 \begin{equation} \label{2.5}
\sup_{0<x<\infty}\|v(x)\| \leq M,
\end{equation}
the Weyl matrix function $\vp$ of system \eqref{2.4} is  uniquely  defined in the semi-plane
$\Im z<-M$ by the inequality
\begin{equation} \label{2.6}
\int_0^\infty \left[ \begin{array}{lr}   \varphi (z)^* &
I_p
\end{array} \right]
  W(x, z)^*
W(x, z)
\left[ \begin{array}{c}
  \varphi (z) \\ I_p \end{array} \right] dx < \infty, \quad  \Im z<-M<0,
\end{equation}
where $W$ is the normalized by $W(0,z)=I_m$  fundamental solution of \eqref{2.4}.
Weyl functions are constructed using pairs of
meromorphic $p \times p$  matrix functions $P_1(z), \, P_2(z)$,  which are nonsingular
and have property-$j$, that is,
 \begin{equation} \label{2.6'}
\clp(z)^*\clp(z)>0, \quad \clp(z)^*j\clp(z) \leq 0, \quad \clp: \left[
\begin{array}{c}
P_1 \\ P_2
\end{array}
\right].
\end{equation}
 \begin{Tm}\label{DP} \cite{SaA1} There is a unique Weyl function of such a system
 \eqref{2.4}  that \eqref{2.5} holds. This Weyl function is holomorphic
 in the semi-plane $\Im z<-M$.
 It is
 given by the equality
 \begin{align} \label{2.7}
\vp(z)=& \lim_{r \to \infty}\big({\cal A}_{11}(r,z)P_1(r,z)+{\cal A}_{12}(r,z)P_2(r,z)\big)
\\ \nonumber & \times
\big({\cal A}_{21}(r,z)P_1(r,z)+{\cal A}_{22}(r,z)P_2(r,z)\big)^{-1} \quad (\Im z<-M),
\\ \label{2.8} &
{\cal A}(r,z)=\big\{{\cal A}_{kp}(r,z)  \big\}_{k,p=1}^2:=W(r,\ov z)^*,
\end{align}
 where the pairs $\{P_1, \, P_2\}$ are arbitrary  pairs satisfying \eqref{2.6'}.
 \end{Tm}
 Our next theorem on the evolution of the Weyl function
 in the case of the focusing mKdV follows from Theorems \ref{TmM} and \ref{DP}.
 The case of the defocusing mKdV was earlier treated in \cite{SaL1, SaL2, SaL4}.
  \begin{Tm}\label{evol} Let a $p \times p$ matrix function $v \in C^1({\mathcal D})$
have a continuous partial second derivative  $v_{xx}$, and let $v_{xxx}$  exist.
 Assume that $v$  satisfies mKdV \eqref{2.1}  and that the inequalities
  \begin{equation} \label{2.8'}
  \sup_{(x,t)\in {\mathcal D}}\|v(x,t)\| \leq M, \quad
\sup_{(x,t)\in {\mathcal D}}\big( \|v_x(x,t)\|+ \|v_{xx}(x,t)\| \big) <\infty
\end{equation}
hold.

 Then the evolution $\vp(t,z)$ of the Weyl function  of the skew-self-adjoint Dirac system
 \eqref{0.4}, where $G$ has the form \eqref{2.2}, is given by the equality
 \begin{equation} \label{2.9}
\vp(t,z)=\big(R_{11}(t,z)\vp(0,z)+R_{12}(t,z)\big)
\big(R_{21}(t,z)\vp(0,z)+R_{22}(t,z)\big)^{-1}
\end{equation}
in the semi-plane $\Im z <-M<0$.
Here the block matrix function
 \begin{equation} \label{2.9'}
R(t,z)=\{R_{kn}(t,z)\}_{k,n=1}^2=R(0,t,z)
\end{equation}
is defined
by the boundary values $v(0,t)$, $v_x(0,t)$, and $v_{xx}(0,t)$ via formulas
\eqref{0.5} and \eqref{2.3}.
  \end{Tm}
  \begin{proof}. As $V^*=-V$ and $(V_x j)^*=V_xj$, it is immediate from \eqref{2.3}
  that $F(x,t, \ov z)^*+F(x,t,z)=0$. Hence, it follows from
  \eqref{0.5}  that
\[
  \frac{\p}{\p t}\big(R(x,t, \ov z)^*R(x,t,z)\big)=0.
\]
  Therefore, using equalities  $R(x,0,z)=I_m$ and \eqref{2.9'}, we get
 \[
 R(x,t, \ov z)^*R(x,t,z)=I_m, \qquad R(t, \ov z)^*R(t,z)=I_m,
 \]
 or, equivalently,
\begin{equation} \label{2.10}
R(x,t, \ov z)^*=R(x,t,z)^{-1}, \qquad R(t, \ov z)^*=R(t,z)^{-1}.
\end{equation}
In view of  \eqref{2.10} rewrite \eqref{0.6'} in the form
\begin{equation} \label{2.11}
{\cal A}(x,t, z)R(x,t,z)=R(t,z){\cal A}(x,0, z),
\end{equation}
where  ${\cal A}(x,t, z):=W(x,t, \ov z)^*$ (compare with \eqref{2.8}).
Let ${\cal P}(x,z)$ satisfy \eqref{2.6'} and put
\begin{equation} \label{2.12}
\wt\clp(x,t,z)=\left[
\begin{array}{c}
\wt P_1(x,t,z) \\ \wt P_2(x,t,z)
\end{array}
\right]=R(x,t,z)\clp(x,z).
\end{equation}
By \eqref{2.11} and \eqref{2.12} we have
\begin{equation} \label{2.12'}
{\cal A}(x,t, z)\wt\clp(x,t,z)=R(t,z){\cal A}(x,0, z)\clp(x,z).
\end{equation}

Now, taking into account that $\clp(x,z)$ is a nonsingular pair with property-$j$,
we show that $\wt\clp(x,t,z)$ is a nonsingular pair with property-$j$ too.
According to  \eqref{0.5}, \eqref{2.3}, and \eqref{2.8'} we get
\begin{equation} \label{2.13}
 \frac{\p}{\p t}\big(R(x,t, z)^*jR(x,t,z)\big)=R(x,t, z)^*\big(i(\ov z^3-z^3)I_m+O(z^2)\big)R(x,t,z)
\end{equation}
for $z \to \infty$. Formula \eqref{2.13} implies that for some
\begin{equation} \label{2.14}
M_1>M>0 \quad (M \geq \sup_{(x,t)\in {\mathcal D}}\|v(x,t)\|),
\end{equation}
and for all $z$ from the domain
\begin{equation} \label{2.15}
{\mathcal D}_1=\{z: \,z\in \BC, \quad \Im z <-M_1,  \quad 0>\arg z > - \pi/4\}
\end{equation}
 we have
 \[
  \frac{\p}{\p t}\big(R(x,t, z)^*jR(x,t,z)\big) \leq 0,
 \]
 and so
 \begin{equation} \label{2.16}
 R(x,t, z)^*jR(x,t,z)\leq j.
 \end{equation}
 Relations \eqref{2.6'}, \eqref{2.12}, and \eqref{2.16} imply that
  \begin{equation} \label{2.17}
\wt \clp(x,t,z)^* \wt \clp(x,t,z)>0, \quad \wt \clp(x,t,z)^*j \wt \clp(x,t,z) \leq 0 \quad (z\in {\cal D}_1).
\end{equation}
Clearly, it suffices to prove \eqref{2.9} for values of $z$ from ${\cal D}_1$.
(According to \eqref{2.14} and \eqref{2.15} the domain ${\cal D}_1$ belongs to the semi-plane
$\Im z<-M$.)

In a way similar to the proofs of \eqref{2.10} and \eqref{2.16} we derive
 \begin{equation} \label{2.17'}
 \cla(x,t, z)=W(x,t,z)^{-1}, \qquad    W(x,t, z)^*jW(x,t,z)\geq j \quad (\Im z<-M).
 \end{equation}
It is immediate from \eqref{2.17'} that
\begin{align} \label{o0}&
\cla(x,t, z)^*j\cla(x,t,z)\leq j \qquad (\Im z<-M).
 \end{align}
Hence, inequalities
\eqref{2.6'} and \eqref{2.17} imply
\begin{align} \label{o1}&
\det\big({\cal A}_{21}(x,0,z) P_1(x,z)+{\cal A}_{22}(x,0,z)P_2(x,z)\big)\not=0 \quad (\Im z<-M), \\
\label{o2}&
\det\big({\cal A}_{21}(x,t,z)\wt P_1(x,t,z)+{\cal A}_{22}(x,t,z)\wt P_2(x,t,z)\big)\not=0
\quad (z\in {\cal D}_1).
 \end{align}
 In view of \eqref{o1} rewrite \eqref{2.12'} as
 \begin{align}\nonumber
{\cal A}(x,t, z)\wt\clp(x,t,z)=&R(t,z)
\left[
\begin{array}{c}
\phi(x,0,z) \\ I_p
\end{array}
\right]
\\ \label{o3} &
\big({\cal A}_{21}(x,0,z) P_1(x,z)+{\cal A}_{22}(x,0,z)P_2(x,z)\big),
\end{align}
\begin{align}
\nonumber
\phi(x,0,z):=& \big({\cal A}_{11}(x,0,z) P_1(x,z)+{\cal A}_{12}(x,0,z)P_2(x,z)\big)
\\ & \label{o4} \times
\big({\cal A}_{21}(x,0,z) P_1(x,z)+{\cal A}_{22}(x,0,z)P_2(x,z)\big)^{-1}.
\end{align}
According to \eqref{o1}-\eqref{o3} we get
\begin{align} \nonumber&
\big({\cal A}_{11}(x,t,z) \wt P_1(x,t,z)+{\cal A}_{12}(x,t,z)\wt P_2(x,t,z)\big)
\\ & \label{o5}  \times
\big({\cal A}_{21}(x,t,z) \wt P_1(x,t, z)
+{\cal A}_{22}(x,t,z)\wt P_2(x,t,z)\big)^{-1}
\\ & \nonumber
=
\big(R_{11}(t,z) \phi(x,0,z)+R_{12}(t,z)\big)
\big(R_{21}(t,z) \phi(x,0,z)+R_{22}(t,z)\big)^{-1}.
\end{align}
As $\wt \clp(x,t,z)$ satisfies \eqref{2.17} for $z \in {\cal D}_1$, using \eqref{2.7} we derive
\begin{align} \nonumber
\vp(t,z)=& \lim_{x \to \infty} \big({\cal A}_{11}(x,t,z) \wt P_1(x,t,z)+{\cal A}_{12}(x,t,z)\wt P_2(x,t,z)\big)
\\ & \label{o6}  \times
\big({\cal A}_{21}(x,t,z) \wt P_1(x,t, z)
+{\cal A}_{22}(x,t,z)\wt P_2(x,t,z)\big)^{-1} \quad (z \in {\cal D}_1).
\end{align}
In a similar way we derive from \eqref{2.7} and  \eqref{o4} that
\begin{align}  & \label{o7}
\vp(0,z)=\lim_{x \to \infty}\phi(x,0,z) \quad (\Im z <-M).
\end{align}
Let us show that
 \begin{equation} \label{o8}
\det
\big(R_{21}(t,z)\vp(0,z)+R_{22}(t,z)\big)\not=0  \quad (z \in {\cal D}_1).
\end{equation}
Indeed, it follows from \eqref{2.6'}, \eqref{o0}, and  \eqref{o4} that
 \begin{align}\label{o9}
&
[\phi(x,0,z)^* \quad I_p] j
\left[
\begin{array}{c}
\phi(x,0,z) \\ I_p
\end{array}
\right]\leq 0.
\end{align}
By \eqref{o7} and \eqref{o9} the inequality
\begin{align}\label{o10}
&
[\vp(0,z)^* \quad I_p] j
\left[
\begin{array}{c}
\vp(0,z) \\ I_p
\end{array}
\right]\leq 0.
\end{align}
is true. Finally, inequalities \eqref{2.16} and \eqref{o10} imply
\begin{align}\label{o11}
&
[\vp(0,z)^* \quad I_p] R(t,z)^*jR(t,z)
\left[
\begin{array}{c}
\vp(0,z) \\ I_p
\end{array}
\right]\leq 0 \quad (z\in {\cal D}_1).
\end{align}
It is immediate from \eqref{o11} that  \eqref{o8} holds.
Relations \eqref{o5}-\eqref{o8} imply  \eqref{2.9} in the domain ${\cal D}_1$.
Hence, by analyticity  equality \eqref{2.9} holds  in the semi-plane $\Im z <-M$.
\end{proof}
In a way similar to \cite{Kr} and to more general constructions for self-adjoint systems in \cite{SaL2,
SaL4} (see also some references therein), one can use structured operators to solve inverse
problem for system \eqref{2.4} too. Namely, to recover $v$,  which satisfies condition \eqref{2.5},
from the Weyl function $\vp$ we use operators $S_l$ (acting in $L_p^2(0,l)$, $0<l<\infty$) of the form
\begin{equation}\label{2.18}
S_lf=f(x)+ \frac{1}{2}\int_0^l
\int^{x+r}_{|x-r|} s^{\prime}\Big(
\frac{\la+x-r}{2} \Big) s^{\prime} \Big( \frac{\la+r-x}{2} \Big)^*d zf(r) dr.
\end{equation}
Here $s^{\prime}(x):
=\frac{d}{dx}s(x)$.  Below we give the procedure from \cite{SaA1} modified in accordance with \cite{FKS, SaA2}.

First, we recover
a $p \times p$ matrix function $s(x)$  with
the entries from $L^2(0,l)$ (i.e., $s(x) \in L^2_{p \times p}(0,l)$) via the
Fourier transform. That is, we put
\begin{equation}\label{2.19}
\displaystyle{s(x)= \frac{i}{2 \pi}e^{- \eta x}{\mathrm{
l.i.m.}}_{a \to \infty} \int_{- a}^{a}e^{i \xi
x} z^{-1} \vp(z /2) d \xi \quad (z= \xi +i \eta , \quad \eta
<-2M),}
\end{equation}
the limit l.i.m. being the limit in $L^2_{p \times p}(0,l)$.  Formula (\ref{2.19}) has sense for any $l<\infty$,
and so
the matrix function $s(x)$ is defined on the non-negative real  semi-axis $x\geq 0$.  Moreover,
 $s$ is absolutely continuous, it does not depend on the choice of $\eta <-
2M$, $s^{\prime}$ is bounded on any finite interval, and $s(0)=0$.
To define the operator $S_l$ we substitute $s^{\prime}(x)$
into  (\ref{2.18}).

Next, denote the $p\times 2p$ block rows of $W$ by $\om_1$ and $\om_2$:
\begin{equation}\label{2.20}
\om_1(x)=[ I_p \quad 0]W(x, 0), \quad  \om_2(x)=[0 \quad I_p]W(x, 0).
\end{equation}
It follows from (\ref{2.2}) and (\ref{2.4}) that $W(x,0)^*W(x,0)=I_{m}$. Hence, by
(\ref{2.2}) , (\ref{2.4}),  and  (\ref{2.20}) we have
\begin{equation}\label{2.21}
v(x)=\om_1^{\prime}(x) \om_2(x)^*,
\end{equation}
and $\om_1$, $\om_2$ satisfy the equalities
\begin{equation} \label{2.22}
\om_1(0)=[I_p \quad 0], \quad \om_1 \om_1^* \equiv I_p, \quad
\om_1^{\prime}
\om_1^* \equiv 0, \quad
\om_1 \om_2^* \equiv 0.
\end{equation}
It is immediate that $\om_1$ is uniquely recovered from $\om_2$ using  (\ref{2.22}).

Finally, we obtain $\om_2$ via the formula
\begin{equation}\label{2.23}
\om_2(l)=[0 \quad I_p]- \int_0^l \Big(S_l^{-1} s'(x)\Big)^*[I_p
\quad s(x)]dx \quad (0<l< \infty),
\end{equation}
where $S_l^{-1}$ is applied to $ s^{\prime}$ columnwise.
\begin{Tm}\label{IP}
Assume that $\vp$ is the Weyl function of system \eqref{2.4}, where $j$ and $V$ have the form
\eqref{2.2} and $v$ satisfies \eqref{2.5}. Then $v$ is recovered from $\vp$ via formulas
\eqref{2.21}-\eqref{2.23}, where $s$ and $S_l$ are given by equalities \eqref{2.18} and \eqref{2.19}.
All the mentioned above relations are well-defined and the inequalities $S_l \geq I$ hold.
\end{Tm}
Another inverse problem, where condition \eqref{2.5} on $v$ is substituted by a condition
on $\vp$, is also solved in \cite{FKS, SaA2, SaA7} using the same procedure.
\begin{Rk} \label{step} One can apply Theorems \ref{evol} and \ref{IP}
to recover solutions of mKdV. Theorems on the evolution of the Weyl functions
constitute also the first step in proofs of uniqueness and existence
of the solutions of nonlinear equations via  ISpT method (see, for instance, \cite{SaA7}).
\end{Rk}

\section{Factorization of the fundamental solution via Darboux matrix} \label{Darb}
\setcounter{equation}{0}
Various versions of B\"acklund-Darboux transformation and commutation \\
methods are widely used in spectral theory, differential equations
and nonlinear integrable equations (see, for instance,
\cite{CC, Ci, Cr, D, Gu, GeT, Kr2, March, MS, ZM2}
and numerous references therein). In this section we consider a so called GBDT version
of the B\"acklund-Darboux transformation (see references in Introduction
and some basic notations and results in Appendix).
The statement of Theorem \ref{WF} is formulated and proved here in greater generality than before.

One can apply Theorem \ref{GBDTrd} on GBDT to construct solutions  and
  wave functions of nonlinear integrable equations.
For this purpose we use auxiliary linear systems for integrable nonlinear
equation, namely, linear systems :
\begin{eqnarray}       \label{g15}
&&w_x=Gw, \quad  w_t=Fw;
\\ && \label{g16}
G(x,t,z)=-\sum_{k=0}^r  z^k q_k(x,t)-\sum_{s=1}^l \sum_{k=1}^{r_s}(z-c_s)^{-k}
q_{sk}(x,t),
\\ && \label{g17}
F(x,t,z)=-\sum_{k=0}^R  z^k Q_k(x,t)-\sum_{s=1}^L \sum_{k=1}^{R_s}(z-C_s)^{-k}
Q_{sk}(x,t),
\end{eqnarray}
and zero curvature (compatibility condition) representation (\ref{0.1}) of the
integrable nonlinear equation itself. We consider nonlinear equations on the domain
$(x,t) \in  {\cal I}_1 \times {\cal I}_2$, where ${\cal I}_k$
 ($k=1,2$)   is the interval $[0, \,b_k)$,  $\,(0<b_k\leq \infty)$.
By  Theorem \ref{TmM} and Remark \ref{Dom} the following corollary is true.
\begin{Cy}\label{Rk3.3}
Let coefficients $\{q_k(x,t)\}$ and $\{q_{sk}(x,t)\}$ be
differentiable with respect to  $t$ and let coefficients $\{Q_k(x,t)$ and $\{Q_{sk}(x,t)\}$ be
differentiable with respect to  $x$ on the domain $ {\cal I}_1 \times {\cal I}_2$.
Assume also that matrix functions $\{q_k(x,t)$, $\{q_{sk}(x,t)\}$,
$\{Q_k(x,t)$, $\{Q_{sk}(x,t)\}$, $\big\{\frac{\p}{\p t}q_k(x,t)\big\}$, and
$\big\{\frac{\p}{\p t}q_{sk}(x,t)\big\}$
are continuous with respect to $x$ and $t$, and that zero curvature
equation
\eqref{0.1}, where $G$ and $F$
are given by \eqref{g16} and \eqref{g17},
holds. Then there is the fundamental solution of \eqref{g15}  normalized
by the condition
\begin{align} \label{ap1}
w(0,0,z)=I_{m}.
\end{align}
\end{Cy}
\begin{proof}. Put
\begin{align} \label{ap2}
w(x,t,z)=W(x,t,z)R(t,z).
\end{align}
By \eqref{0.4}, \eqref{0.5}, and \eqref{0.6'} we see that  $w$ given by \eqref{ap2}
satisfies \eqref{g15}. According to the second relations in \eqref{0.4} and \eqref{0.5}
equality \eqref{ap1} holds too.
\end{proof}

Further asume that $G$, $F$, and coefficients in \eqref{g16} and \eqref{g17}
satisfy conditions of Corollary \ref{Rk3.3}, and that $w$ is given by \eqref{ap2}.

When we deal with two auxiliary linear systems,
we fix $n \in \BN$, three $n \times n$ parameter matrices,
namely $A_1$, $A_2$, and $S(0,0)$, and two $n\times m$ parameter matrices,
namely, $\Pi_1(0,0)$ and $\Pi_2(0,0)$. These matrices are chosen so
that they satisfy the matrix identity
\begin{equation}       \label{g18}
A_1S(0,0)-S(0,0)A_2=\Pi_1(0,0)\Pi_2(0,0)^*.
\end{equation}
Compare \eqref{g18} with a similar matrix identity   (\ref{g2}) for parameter matrices
$A_k$, $\Pi_k(0)$, and $S(0)$  in Appendix. Matrix functions $\Pi_1(x,t)$, $\Pi_2(x,t)$, and $S(x,t)$
are determined by the initial values $\Pi_1(0,0)$, $\Pi_2(0,0)$, and $S(0,0)$,
respectively, differential equations (\ref{g3})--(\ref{g5})  with respect to derivatives in  $x$
and similar equations with respect to derivatives in $t$. That is,
$\Pi_1$, $\Pi_2$, and $S$ satisfy  equations:
\begin{align}       \label{ng3}
&\big(\Pi_1 \big)_x=\sum_{k=0}^r A_1^k \Pi_1 q_k+\sum_{s=1}^l \sum_{k=1}^{r_s}(A_1 -c_s I_n)^{-k}
\Pi_1 q_{sk},
\end{align}
\begin{align}
      \label{ng4}
&\big(\Pi_2^* \big)_x=-\Big(\sum_{k=0}^r  q_k\Pi_2^*A_2^k+\sum_{s=1}^l \sum_{k=1}^{r_s}q_{sk}\Pi_2^*(A_2 -c_s I_n)^{-k}
\Big),
\\   \label{ng5}
S_x=&\sum_{k=1}^r\sum_{j=1}^k A_1^{k-j} \Pi_1 q_k\Pi_2^*A_2^{j-1}-\sum_{s=1}^l \sum_{k=1}^{r_s}
\sum_{j=1}^k
(A_1 -c_s I_n)^{j-k-1} \\ \nonumber & \times
\Pi_1 q_{sk}\Pi_2^*(A_2 -c_s I_n)^{-j},
\end{align}
which coincide with (\ref{g3})--(\ref{g5}), and additional equations with respect to $t$:
\begin{align}       \label{g3'}
&\big(\Pi_1 \big)_t=\sum_{k=0}^R A_1^k \Pi_1 Q_k+\sum_{s=1}^L \sum_{k=1}^{R_s}(A_1 -C_s I_n)^{-k}
\Pi_1 Q_{sk},\\
      \label{g4'}
&\big(\Pi_2^* \big)_t=-\Big(\sum_{k=0}^R  Q_k\Pi_2^*A_2^k+\sum_{s=1}^L \sum_{k=1}^{R_s}Q_{sk}\Pi_2^*(A_2 -C_s I_n)^{-k}
\Big),
\\     \label{g5'}
S_t=&\sum_{k=1}^R \sum_{j=1}^k A_1^{k-j} \Pi_1 Q_k\Pi_2^*A_2^{j-1}-\sum_{s=1}^L \sum_{k=1}^{R_s}
\sum_{j=1}^k
(A_1 -C_s I_n)^{j-k-1} \\ \nonumber & \times
\Pi_1 Q_{sk}\Pi_2^*(A_2 -C_s I_n)^{-j}.
\end{align}
We require
\begin{equation}       \label{r30}
\{c_s\}\cap \s(A_k)=\emptyset, \quad \{C_s\}\cap \s(A_k)=\emptyset \quad (k=1,2),
\end{equation}
where $\s(A)$ is the spectrum of $A$.
Then, Theorem \ref{GBDTrd}  provides expessions for derivatives
$\big(w_A(x,t,z)\big)_x$ and $\big(w_A(x,t,z)\big)_t$:
\begin{align}& \label{r30'}
\big(w_A\big)_x=\wt G w_A-w_AG, \quad \big(w_A\big)_t=\wt F w_A-w_A F,
\end{align}
where $\wt G$ has the same structure as $G$ and is given by formulas
\eqref{g9}-\eqref{g13}. Similarly $\wt F$ has the same structure as $F$, namely,
\begin{align}&
 \label{g43}
\wt F(x,t,z)=-\sum_{k=0}^R  z^k \wt Q_k(x,t)-\sum_{s=1}^L \sum_{k=1}^{R_s}(z-C_s)^{-k}
\wt Q_{sk}(x,t),
\end{align}
where coefficients are given by  \eqref{g10}-\eqref{g13} after substitution $Q_k$, $Q_{sk}$,
$\wt Q_k$, $\wt Q_{sk}$, $R$, $R_s$, $L$, and $L_s$ instead of
$q_k$, $q_{sk}$,
$\wt q_k$, $\wt q_{sk}$, $r$, $r_s$, $l$, and $l_s$, respectively, in those formulas.
The matrix function $w_A$ in \eqref{r30'}  has the  form
\begin{equation}       \label{g27}
w_A(x, t, z)=  I_m-\Pi_2(x,t)^*S(x,t)^{-1}(A_1-z I_n)^{-1}\Pi_1(x,t)
\end{equation}
(compare with   \eqref{g7}).
By  \eqref{g15} and \eqref{r30'} we have
\begin{equation}       \label{g35}
\wt w _x=\wt G \wt w, \quad  \wt w_t=\wt F\wt w, \quad \wt w(x,t,z):=w_A(x,t,z)w(x,t,z) .
\end{equation}

The following theorem shows that $\wt G$ and $\wt F$ satisfy zero curvature
equation
\begin{align}& \label{g42}
\wt G_t- \wt F_x+[\wt G, \wt F]= 0
\end{align}
on  the domain $\cld_S$ of the points of invertibility
of $S$:
\begin{align}& \label{g40}
\cld_S={\cal I}_1 \times {\cal I}_2 \setminus \mclo_S, \quad \mclo_S=
\{(x,t)\, :   \det S(x,t)=0\}.
\end{align}
\begin{Tm} \label{WF}
Let $G$, $F$, and coefficients in \eqref{g16} and \eqref{g17}
satisfy conditions of Corollary \ref{Rk3.3}.
Assume that $\Pi_1$, $\Pi_2$, and $S$ satisfy \eqref{g18}-\eqref{g5'}.
Then,  the coefficients
$\{\wt q_k\}$, $\{\wt q_{sk}\}$ and $\{\wt Q_k\}$, $\{\wt Q_{sk}\}$
of  $\wt G$ and $\wt F$, respectively,  are  continuous on $\cld_S$
together with derivatives $\{ \frac{\p}{\p t}\wt q_k\}$ and $\{\frac{\p}{\p t} \wt q_{sk}\}$.
Moreover, zero curvature
equation
 \eqref{g42} holds on $\cld_S$.
\end{Tm}
\begin{proof}. By  \eqref{g10}-\eqref{g13}
and \eqref{ng3}-\eqref{g5'} the differentiability and continuity  statements
of our theorem for the coefficients of $\wt G$ and $\wt F$ are true.
Moreover, it follows from \eqref{0.1} that coefficients $\{ \frac{\p}{\p x} Q_k\}$
and $\{\frac{\p}{\p x}  Q_{sk}\}$, and hence also coefficients
$\{ \frac{\p}{\p x}\wt Q_k\}$ and $\{\frac{\p}{\p x} \wt Q_{sk}\}$, are continuous
 on $\cld_S$.
Therefore, the matrix functions $\wt G$, $\wt G_t$,   $\wt F$, and $\wt F_x$
are continuous on the domain $\cld_S$.

Thus, taking into account \eqref{g35} we see that
$\wt w$, $\wt w_x$, $\wt w_t$, and $\wt w_{xt}$ exist and are continuous
on $\cld_S$. Hence, the conditions of the stronger formulation
of the theorem on mixed derivative, which is already used in the proof of Theorem
\ref{TmM}, are fulfilled and $\wt w_{xt}=\wt w_{tx}$ in the interior $(\cld_S)_i$ of
$\cld_S$. Using \eqref{g35} rewrite the equality $\wt w_{xt}=\wt w_{tx}$ in the form
\begin{align}& \label{g41}
(\wt G_t+\wt G \wt F)\wt w= (\wt F_x+\wt F \wt G)\wt w.
\end{align}
It follows from \eqref{g15} and \eqref{ap1} that $w$ is invertible on ${\cal I}_1 \times {\cal I}_2 $,
and it follows from \eqref{ap3} that $w_A$ is invertible on $\cld_S$. Thus
$\wt w =w_A w$ is invertible on $\cld_S$. Now,  it follows from \eqref{g41} that
\eqref{g42} holds  in $(\cld_S)_i$. By continuity, \eqref{g42} holds  on $\cld_S$.
\end{proof}
\begin{Rk}\label{F2}
Theorem \ref{WF} is basic to construct solutions and wave functions
of  integrable equations via GBDT. The corresponding normalized wave functions $\breve w$
(see  \cite{GKS2, GKS6, SaA2, SaA2', SaA3, SaA4, SaA6}) have the form :
\begin{align}& \label{g44}
\breve w(x,t,z)=\wt w(x,t,z)w_A(0,0,z)^{-1}=w_A(x,t,z)w(x,t,z)w_A(0,0,z)^{-1}.
\end{align}
\end{Rk}
 Compatibility of the equations \eqref{ng3}-\eqref{g5'} is a separate question.
In full generality it will be addressed elsewhere, and here we consider an important
and characteristic example of the compatibility of equations \eqref{ng3} and \eqref{g3'},
when $G$ and $F$   are polynomials.
\begin{Pn} Let $G$ and $F$ be polynomials
\begin{align} \label{g45}&
G(x,t,z)=-\sum_{k=0}^r  z^k q_k(x,t), \quad F(x,t,z)=-\sum_{s=0}^R  z^s Q_s(x,t),
 \end{align}
such that the conditions of  Corollary \ref{Rk3.3} are satisfied.
 Then, the corresponding equations \eqref{ng3} and \eqref{g3'}, which determine  $\Pi_1$, are compatible.
\end{Pn}
\begin{proof}. When $G$ and $F$ are polynomials, equations
  \eqref{ng3} and \eqref{g3'} take the form
\begin{align} \label{g46}&
\big(\Pi_1\big)_x=\sum_{k=0}^r  A_1^k\Pi_1 q_k, \quad \big(\Pi_1\big)_t=\sum_{s=0}^R  A_1^s \Pi_1Q_s.
 \end{align}
Denote the $p$-th column of $\Pi_1$ by $\big(\Pi_1\big)_p$ and introduce a block vector
with $\big(\Pi_1\big)_p$ as its blocks:
\begin{align} \label{g47}&
\ora{\Pi}_1=\left[
\begin{array}{c}
\big(\Pi_1\big)_1\\ \cdots \\ \big(\Pi_1\big)_m
\end{array}
\right] \in \BC^{mn}.
 \end{align}
Equations \eqref{g46} can be rewritten in an equivalent form in terms of  $\ora{\Pi}_1$:
\begin{align} \label{g48}&
\big(\ora{\Pi}_1\big)_x=\g\ora{\Pi}_1 , \quad
\big(\ora{\Pi}_1\big)_t=\G \ora{\Pi}_1, \\ \label{g49}&
\g(x.t):=\sum_{k=0}^r q_k(x,t)^T \otimes A_1^k, \quad \G(x,t):=\sum_{s=0}^R Q_s(x,t)^T \otimes A_1^s,
 \end{align}
where $q^T$ denotes the transpose of a matrix $q$, and $q\otimes A$ is the Kronecker product
of matrices $q$ and $A$. As $\{q_k\}$ and $\{Q_k\}$ satisfy conditions of  Corollary \ref{Rk3.3},
so $\g$ and $\G$ satisfy the differentiability and continuity conditions of Theorem \ref{TmM},
and to prove the compatibility it remains only to show that zero curvature equation
\begin{align} \label{g50}&
\g_t-\G_x+[G,f]=0
 \end{align}
holds. For that purpose consider  the block in the $i$-th block row and in the $p$-th block column
in \eqref{g50}.  We get an equality
\begin{align} \nn&
\sum_{k=0}^r \Big(\frac{\p}{\p t}q_k\Big)_{pi}A_1^k-\sum_{s=0}^R \Big(\frac{\p}{\p x}Q_s\Big)_{pi}A_1^s
+\Big(\sum_{k=0}^r  (q_k)_i^T\otimes A_1^k \Big)\Big(\sum_{s=0}^R(Q_s^T)_p\otimes A_1^s  \Big)
\\ \label{g51} &
-
\Big(\sum_{s=0}^R (Q_s)_i^T\otimes A_1^s \Big)\Big(\sum_{k=0}^r(q_k^T)_p\otimes A_1^k  \Big)=0,
 \end{align}
 where $ (q_k)_i^T$ is the transpose of the $i$-th column of $q_k$.  Rewrite \eqref{g51} as
 \begin{equation} \nn
\sum_{k=0}^r \Big(\frac{\p}{\p t}q_k\Big)_{pi}A_1^k-\sum_{s=0}^R \Big(\frac{\p}{\p x}Q_s\Big)_{pi}A_1^s
+\sum_{k=0}^r\sum_{s=0}^R\Big((Q_s q_k)_{pi}-(q_kQ_s)_{pi}\Big)A_1^{k+s}=0.
\end{equation}
It is immediate that independently from the choice of $A_1$ the equality above
follows from the equality
\begin{align} \label{g52}&
\sum_{k=0}^r \Big(\frac{\p}{\p t}q_k\Big)_{pi}z^k-\sum_{s=0}^R \Big(\frac{\p}{\p x}Q_s\Big)_{pi}z^s
+\sum_{k=0}^r\sum_{s=0}^R\Big((Q_s q_k)_{pi}-(q_kQ_s)_{pi}\Big)z^{k+s}=0.
\end{align}
In other words equation \eqref{g50} follows from
\begin{align} \label{g53}&
\sum_{k=0}^r z^k\frac{\p}{\p t}q_k-\sum_{s=0}^R z^s\frac{\p}{\p x}Q_s
+\sum_{k=0}^r\sum_{s=0}^Rz^{k+s}\Big(Q_s q_k-q_kQ_s\Big)=0.
\end{align}
Notice that in view of  \eqref{g45}  formula \eqref{g53} is equivalent to
\eqref{0.1}.
Thus, \eqref{g53} holds, and so \eqref{g50} holds too.
\end{proof}

{\bf Acknowledgement.}
The work of A.L. Sakhnovich was supported by the Austrian Science Fund (FWF) under
Grant  no. Y330.

\appendix

\section{Appendix. GBDT for system  depending rationally on
spectral  parameter}\label{RD}
\setcounter{equation}{0}

In this appendix we consider the  GBDT version of the
B\"acklund-Darboux transformation (BDT) for a general case of first order
system  depending rationally on the spectral parameter $z$:
\begin{equation}       \label{g1}
w_x=Gw, \quad G(x,z)=-\Big(\sum_{k=0}^r  z^k q_k(x)+\sum_{s=1}^l \sum_{k=1}^{r_s}(z-c_s)^{-k}
q_{sk}(x)\Big),
\end{equation}
where $x \in {\cal I}$,   and the coefficients
$q_k(x)$ and $q_{sk}(x)$ are $m \times m$ locally integrable matrix functions. To simplify notations we assume that
${\cal I}$ is either interval $[0, \,b]$ $(0<b<\infty)$ or interval $[0, \,b)$ $(0<b\leq \infty)$.
In our presentation of GBDT  we follow Section 3 of the review \cite{SaA6}. Further references one can find
in Introduction and  \cite{SaA6}.

As GBDT is a so called iterated BDT we fix an integer $n>0$. Next, we fix five  matrices,
namely, $n \times n$ matrices $A_k$ ($k=1,2$) and $S(0)$, and $n \times m$ matrices
$\Pi_k(0)$ ($k=1,2$). It is required that these matrices form an $S$-node, that is,
the identity
\begin{equation}       \label{g2}
A_1S(0)-S(0)A_2=\Pi_1(0)\Pi_2(0)^*
\end{equation}
holds. Matrix functions $\Pi_k(x)$ are introduced via initial values $\Pi_k(0)$ and linear differential
equations:
\begin{eqnarray}       \label{g3}
&&\big(\Pi_1 \big)_x=\sum_{k=0}^r A_1^k \Pi_1 q_k+\sum_{s=1}^l \sum_{k=1}^{r_s}(A_1 -c_s I_n)^{-k}
\Pi_1 q_{sk},\\
      \label{g4}
&&\big(\Pi_2^* \big)_x=-\Big(\sum_{k=0}^r  q_k\Pi_2^*A_2^k+\sum_{s=1}^l \sum_{k=1}^{r_s}q_{sk}\Pi_2^*(A_2 -c_s I_n)^{-k}
\Big),
\end{eqnarray}
where $\{q_k\}$ and $\{q_{sk}\}$ are  coefficients from $G$.
Compare  (\ref{g1}) with (\ref{g4}) to see that $\Pi_2^*$ can be viewed
as a generalized eigenfunction of the system $u_x=Gu$.

Matrix function $S(x)$ is introduced via $\frac{d}{dx}S$ by the equality
\begin{eqnarray}       \label{g5}
S_x&=&\sum_{k=1}^r\sum_{j=1}^k A_1^{k-j} \Pi_1 q_k\Pi_2^*A_2^{j-1}-\sum_{s=1}^l \sum_{k=1}^{r_s}
\sum_{j=1}^k
(A_1 -c_s I_n)^{j-k-1} \\ \nonumber && \times
\Pi_1 q_{sk}\Pi_2^*(A_2 -c_s I_n)^{-j}.
\end{eqnarray}
Equality   (\ref{g5}) is chosen so that the identity
$\Big(A_1S-SA_2\Big)_x=\Big(\Pi_1\Pi_2^*\Big)_x$ holds.
Hence, taking into account (\ref{g2}) we have
\begin{equation}       \label{g6}
A_1S(x)-S(x)A_2=\Pi_1(x)\Pi_2(x)^*, \quad x\in {\cal I}.
\end{equation}
By Theorem \ref{GBDTrd} below, the Darboux matrix for system (\ref{g1}) has the form (\ref{0.2}) :
\begin{equation}       \label{g7}
w_A(x, z)=  I_m-\Pi_2(x)^*S(x)^{-1}(A_1-zI_n)^{-1}\Pi_1(x).
\end{equation}
In other words, $w_A$ satisfies the equation
\begin{equation}       \label{g8}
\frac{d}{d x}w_A(x, z)=\wt G(x,z)w_A(x, z)-w_A(x, z)G(x,z),
\end{equation}
where $\wt G$ has the same structure as $G$:
\begin{equation}       \label{g9}
\wt G(x,z)=-\Big(\sum_{k=0}^r  z^k \wt q_k(x)+\sum_{s=1}^l \sum_{k=1}^{r_s}(z-c_s)^{-k}
\wt q_{sk}(x)\Big).
\end{equation}
The transformed coefficients $\wt q_k$ and $\wt q_{sk}$ are given by the formulas
\begin{equation}       \label{g10}
\wt q_k=q_k-\sum_{j=k+1}^r\Big(q_jY_{j-k-1}-X_{j-k-1}q_j+\sum_{i=k+2}^jX_{j-i}q_j
Y_{i-k-2}\Big),
\end{equation}
\begin{equation}
  \label{g11}
\wt q_{sk}=q_{sk}
+\sum_{j=k}^{r_s}\Big(q_{sj}Y_{s,k-j-1}-X_{s,k-j-1}q_{sj}-\sum_{i=k}^jX_{s,i-j-1}q_{sj}
Y_{s,k-i-1}\Big),
\end{equation}
where $X_k(x)$, $Y_k(x)$, $X_{sk}(x)$, and $Y_{sk}(x)$ are expressed in terms of the
matrices $A_k$ and matrix functions $S(x)$ and $\Pi_k(x)$:
\begin{eqnarray}       \label{g12}
&&X_k=\Pi_2^*S^{-1}A_1^k\Pi_1, \quad X_{sk}=\Pi_2^*S^{-1}(A_1-c_s I_n)^k\Pi_1, \\
&&  \label{g13}
Y_k=\Pi_2^*A_2^kS^{-1}\Pi_1,  \quad Y_{sk}=\Pi_2^*(A_2-c_s I_n)^kS^{-1}\Pi_1.
\end{eqnarray}
Denote the spectrum of matrix $A$ by $\s(A)$.
\begin{Tm} \label{GBDTrd} \cite{SaA3} Let first order system   (\ref{g1}) and five
matrices $S(0)$, $A_k$, and $\Pi_k$  $(k=1,2)$ be given. Assume that
the identity  (\ref{g2}) holds and that $\{c_s\}\cap \s(A_k)=\emptyset$ $(k=1,2)$. Then, in the
points of invertibility of $S$, the transfer matrix function $w_A$ given by
(\ref{g7}), where $S$ and $\Pi_k$ are determined by (\ref{g3})--(\ref{g5}),
 satisfies equation (\ref{g8}), where $\wt G$ is determined
 by the formulas (\ref{g9})--(\ref{g13}).
 \end{Tm}
\begin{Rk} \label{det}
The matrix function $w_A$ is invertible, since it can be derived from
\eqref{g6} that
\begin{equation}       \label{ap3}
w_A(x, z)^{-1}=  I_m+\Pi_2(x)^*(A_2-zI_n)^{-1}S(x)^{-1}\Pi_1(x).
\end{equation}
\end{Rk}

config: arxiv


TITLE: A water drop in a falling lift
QUESTION [1 upvotes]: Consider a lift, which is at rest in an homogeneous gravity field. There is a thin layer(with thickness $h$) of water on the floor of the lift. At some moment a single cable, supporting the lift, breaks and the lift begins free falling(forever). It is easy to describe qualitatively what happens with water, i think: the formation of a drop begins. During this process the drop jumps up from the floor and once the total kinetic energy of the drop is dissipated into the heat, the center of the drop stands at some height $H$ from the floor. (All the kinetic energy of the drop is coming from the difference in surface tension energy of water between initial and end moments.)
Question: How to determine(approximately) $H$ ?
For simplicity let's assume that the cross-section of the lift is a disc-shaped with radius $R$.  $(h<<R)$
Remark: A key point is to determine how the total kinetic energy distributes between internal kinetic energy and translational kinetic energy of the drop, i think. With this the problem will be solved.
So it is convenient at the early stage of formation of the drop ignore viscosity effects of water and air resistance.
Edit:
I would like to clarify that by internal kinetic energy i mean the kinetic energy of macroscopic motion of water inside the drop. At early stage there is no energy loss assumed, for simplicity.

REPLY [2 votes]: The only reason the water would form into a drop is the surface tension. You can work out the reduction in the energy stored in the surface tension in being a drop rather than a flat surface using a simple calc of the surface area. There's no particular reason I can see that the drop would float off the floor at all.

REPLY [0 votes]: Unless there is some inital temperature difference, its total kinetic energy is just its translational kinetic energy, which is the energy gained from its change in shape due to surface tension, from which you can get its temperature and also its velocity $v_0$.
You have the surface area, temperatures, conductivities, so you can get the heat transfer per time from fourier's law.
From which you can get the energy E(t), which gives you v(t), a solution to v(t)=0, if it exists, is T, then integrate v(t) from 0 to T to get H.
Most likely it wont come to a full stop, it could be in thermal equilibrium and still have translation kinetic energy, and even more likely it approaches velocity 0 asymptotically.


TITLE: Prove parallel line is tangent to second circle
QUESTION [0 upvotes]: Two circles $\Gamma_1,\Gamma_2$ have centers $O_1,O_2$. Let $\Gamma_1\cap\Gamma_2=A,B$, with $A\neq B$. An arbitrary line through $B$ intersects $\Gamma_1$ at $C$ and $\Gamma_2$ at $D$. The tangents to $\Gamma_1$ at $C$ and to $\Gamma_2$ at $D$ intersect at $M$. Let $N=AM\cap CD$. Let $l$ be a line through $N$ parallel to $CM$, and let $l\cap AC=K$. Prove that $BK$ is tangent to $\Gamma_2$.

$\qquad\quad$

Here is some progress I have:
We are looking to prove $\angle O_2BP=90^{\circ}$, and since $\angle O_2DP=90^{\circ}$, if we could prove $BP=PD$, we would be done by congruent triangles. So we are looking to prove $\dfrac{\sin \angle BDP}{\sin \angle DBP}=1$. Let $AM\cap BK=l$. We have $\angle BDP=\angle QBN$. By the law of sines on  $\triangle DNM,\triangle BNQ$, we have $\sin \angle BDP=\sin \angle NDM=\dfrac{NM}{DM}\sin \angle DNM$ and $\sin\angle QBN=\dfrac {NQ}{BQ}\sin\angle BNQ$. Dividing, the sines cancel (since they are supplementary), and we are left with $\dfrac{NM\times BQ}{DM\times NQ}$, so it remains to prove $\dfrac{NM}{DM}=\dfrac{NQ}{BQ}$.
I'm not sure what to do from here. We would be done if we could prove $\triangle NBQ\sim\triangle NDM$, but this would imply $\angle QNB=\angle MND=90^{\circ}$, but from drawing multiple diagrams, it looks like this is not always the case.
As always, any ideas are appreciated!

REPLY [1 votes]: Let us temporarily remove some of the distracting lines first so that we can use the following diagram to show ADMC is cyclic.

That claim is true because $x + (y_1) + [z_1] = x + (y_2) + [z_2] = 180^0$.
Now we add back the lines KB, ANM, KN.

From the fact that “$z_3 = z_1 = z_ 2$”, we can say $KN // CM$.
From the fact that "$\angle 1 = \angle 2 = \angle 3 = \angle 4$”, we can conclude that KB is tangent to the circle ADB.


TITLE: Sequence of Measurable Functions (Unsigned and Complex-Valued)
QUESTION [2 upvotes]: I am having several difficulties in solving the following problem about measurable functions:
Let $(X,\mathcal{B})$ be a measurable space.
If $f_n : X \to [0,+\infty]$ are a sequence of measurable functions that converge pointwise to a limit $f:X \to [0,+\infty]$, then show that $f$ is also measurable. Obtain the same claim if $[0,+\infty]$ is replaced by $\mathbb{C}.$
The given definition of a measurable function is analogous to that of a continuous function in Topology, according to Terrence Tao which is the author of the textbook I am using:
Let $(X,\mathcal{B})$ be a measurable space, and let $f: X \to [0,+\infty]$ or $f:X \to \mathbb{C}$ be an unsigned or complex-valued function. We say $f$ is measurable if $f^{-1}(U)$ is $\mathcal{B}-$measurable for every open subset $U$ of $[0,+\infty]$ or $\mathbb{C}$.
This is my failed attempt to a solution since I think that I am just writing definitions and not actually proving anything.
$(X,\mathcal{B})$ is a measurable space if $\mathcal{B}$ is a $\sigma-$algebra of subsets of $X$. We wish to show that $f:X\to [0,+\infty]$ is measurable if $f_n : X \to [0,\infty]$ are a sequence of measurable functions converging pointwise to $f$. Since we are in a $\sigma-$algebra, then we get $\forall E_1,E_2 \cdots \in \mathcal{B}, \quad \cup_{n=1}^{\infty}E_n$, call this countable union $U$.
Because the $f_n$ sequence are measurable functions, then $f_n^{-1}(U)$ is $\mathcal{B}-$measurable for every open subset $U$ of $[0,+\infty]$. Our goal is to use this to show that $f$ is measurable.
Here is where I am stuck. How could I go to the next step from here? Additionally, how would this solution extend to $\mathbb{C}$?
Update: I have been reviewing some lecture notes and the textbook and noticed that some of the proofs make a jump from using the quantifiers $\exists$ and $\forall$ to switch with $\cup$ and $\cap$ respectively.
Just in case anyone is interested in reading up this problem, it is coming from Terrence Tao's Introduction to Measure Theory. This is Exercise 1.4.29 (vi).
Here is the link to free pdf version:
http://terrytao.files.wordpress.com/2011/01/measure-book1.pdf

REPLY [0 votes]: Since $f_n$ converges pointwise to $f$, we have $f(x)=\lim_{n \to \infty}f_n(x) = \lim \sup_{n\to \infty}f_n(x) = \inf_{N>0}\sup_{n\geq N}f_n(x)$.
This implies that for some $\lambda$, the set $\{x \in [0,+\infty]:f(x)> \lambda\}$ is equal to $\cup_{M>0}\cap_{N>0}\{x\in [0,+\infty]: \sup_{n \geq N} f_n(x) > \lambda + \frac{1}{M}\}$ outside of a set with measure zero. Actually, this set is
$\cup_{M>0}\cap_{N>0}\cup_{n \geq N}\{x \in [0,+\infty]: f_n(x)> \lambda + \frac{1}{M}\}$ outside of a set with measure zero. Because each $f_n$ is measurable, the set $\{x\in [0,+\infty]: f_n(x)> \lambda + \frac{1}{M}\}$ is measurable. Since countable unions and intersections are measurable in $\sigma-$algebras, which is what is given since $(X,\mathcal{B})$ is a measurable space, then $f$ is measurable.
This is what I have so far, although I am unsure if I can simply change the domain from $[0,+\infty]$ to $\mathbb{C}$ and then use the same argument. Any suggestions would be greatly appreciated since I have spent a lot of time on this problem already.


TITLE: Forms on vector bundles: vertically compactly supported
QUESTION [0 upvotes]: Definition: Let $\pi:V \rightarrow M$ be a vector bundle.
 $\Omega^p_{cv}(V)$ is the sections of $p$ form on $V$, such that $\pi^{-1}(K) \cap supp \, (w) $ for all $K \subseteq M$ compact.
Definition: $\Omega^p_c(V)$ is the set of sections $p$ forms on $V$ such that its support is compact.

It is claimed that

$\Omega^p_c(V) = \Omega^p_{cv}(V)$  when $M$ is compact.

How is this true?  $\subseteq $ is clear.

Source, Pg 84, Proof of Theorem 6.2.10, line +3

REPLY [2 votes]: Let $\omega$ be a form on $V$ which is compactly supported on each fiber $V_x$. Consider $\text{supp}(\omega) \subset V$; this is a closed subset of $V$ whose intersection with each fiber is compact. The goal is to show all such sets are compact themselves.
Put a Riemannian metric over V (just so that I may measure "size of vectors"). Then if $X \subset V$ is closed, with the property that each $X \cap V_x$ is compact, then we may define a function $\sigma: M \to \Bbb R_{\geq 0}$ given by sending $\sigma(x)$ to the largest $\|v\|$ of any $v \in X \cap V_x$. This is well-defined because $X \cap V_x$ is compact. The map $\sigma$ may be seen to be continuous by brute-force; I do not want to do it here. (Intuitively, the nearby sets $X \cap V_x$ are 'close', and hence should have 'nearby' size bounds.)
Thus $\sigma$ is a continuous map from a compact space to $\Bbb R$, and hence has an absolute maximum, say $L$. Thus the set $X$ is contained in the closed $L$-disc bundle of $V$, which is a compact space (given any sequence $(m_n, v_n)$ of norm bounded by $L$, choose a subsequence so that $m_n$ is convergent to $m$; working on a chart to identify all nearby fibers with $V_x$, use compactness of the $L$-disc ball to see that a further subsequence may be chosen so that $v_n$ converges). $X$, being a closed subspace of a compact Hausdorff space, is compact.
So any form with compact vertical support on a vector bundle $V$ over compact base $M$ has compact support, as desired.


TITLE: Find the expected value of $F(X+a)$ where $X\sim\mathcal{N}(0,1)$ with CDF $F$
QUESTION [5 upvotes]: Let $X\sim\mathcal{N}(0,1)$ and denote the standard normal CDF by $F$. It is easy to see that $F(X)\sim\mathcal{U}(0,1)$. Namely, for $u\in(0,1)$ we have
\begin{align*}
P(F(X)\leq u)=P(X\leq F^{-1}(u)) = F(F^{-1}(u))= u,
\end{align*}
so the expected value is
\begin{align*}
E(F(X))=0.5.
\end{align*}
Now consider the random variable $F(X+a)$ where $a>0$ is a constant. How can we find the expected value
\begin{align*}
E(F(X+a))=\;?
\end{align*}
I suspect the answer to be $F(a)$. Additionally, can we find a simple form for the density or distribution function of the random variable $F(X+a)$?

REPLY [3 votes]: Your conjectured value for $E[F(X+a)]$ is close but not quite correct.
$$
E[F(X+a)] = \int_{-\infty}^\infty F(x+a)f(x)\, \mathrm dx
= \int_{-\infty}^\infty \left[\int_{-\infty}^{x+a}f(y)\, \mathrm dy\right]f(x)\, \mathrm dx
$$
Now suppose that $Y\sim N(0,1)$ is independent of $X$. Then, the integral on the right can be interpreted as calculating
$P(Y \leq X+a) = P(Y-X \leq a)$. But, $Y-X \sim N(0,2)$ and so $$E[F(X+a)] = F(a/\sqrt 2)$$
and not $F(a)$ as you have conjectured.  See also this answer of mine on stats.SE for a more
general version of the argument used here.


TITLE: How do you find the equation of a plane which is parallel to another plane which is at a unit distance of from some point
QUESTION [0 upvotes]: Ok I know how to solve the problem if you just assume the plane's equation is $x-2y+2z+k = 0$ ($k$ is some constant) and then use the distance formula to find the value of k and then finally obtaining the required equation ...
But is there some other way of doing this which involves something like: $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c1}{c2}=$constant (parallelism condition)?
Thank you!

REPLY [0 votes]: Let's say you have a plane
$$a x + b y + c z = d \tag{1}\label{NA1}$$
The normal of the plane is
$$\vec{n} = ( a , b , c )$$
The signed distance from point $(x, y, z)$ to the plane is
$$s(x, y, z) = \frac{ a x + b y + c z - d}{\sqrt{ a^2 + b^2 + c^2 }} \tag{2}\label{NA2}$$
Signed distance is positive in the same direction as the normal, negative in the opposite direction.
OP wants to find a plane with normal $\vec{n} = (a, b, c)$, that is at signed distance $\rho$ from point $(\chi, \gamma, \zeta)$, i.e. $s(\chi, \gamma, \zeta) = \rho$. (Planes are parallel if their normals are parallel, i.e. $\vec{n}_1 = k \, \vec{n}_2$, $k \in \mathbb{R}$, $k \ne 0$. OP can simply choose $k = 1$, and reuse the normal vector from the original plane.)
Substituting into $\eqref{NA2}$ yields
$$\rho = \frac{a \chi + b \gamma + c \zeta - d}{\sqrt{ a^2 + b^2 + c^2 }}$$
which we can solve for d, getting
$$\bbox[1em, border:3px double lightgray]{d = a \chi + b \gamma + c \zeta - \rho \sqrt{ a^2 + b^2 + c^2 } } \tag{3}\label{NA3}$$
Note that $\eqref{NA3}$ works for all planes having normal $(a, b, c)$, any points $(\chi, \gamma, \zeta)$, and all signed distances $\rho$.

REPLY [0 votes]: There's a solution which involves the vector equation of the planes: if the given plane $H_0$ has equation $ax+by+cz=k$, let $\vec n=(a,b,c)$ be the normal vector to $H_0$, $M_0(_0,y_0,z_0)$ a point in $H_0$. Denote $H$ a plane parallel to $H0$ and at distance $1$.
This means the point $N_0=M_0\pm\dfrac{\vec n}{\bigl|\mkern2mu\vec n\mkern2mu\bigr|}$ must be in $H$ so that its vector equation is
$$\vec n\cdot M=\vec n\cdot N_0=\vec n\cdot M_0 \pm\frac{\vec n\cdot\vec n}{\bigl|\mkern2mu\vec n\mkern2mu\bigr|}=\vec n\cdot M_0\pm\bigl|\mkern2mu\vec n\mkern2mu\bigr|$$
But as $M_0$ is in $H_0$, $\;\vec n\cdot M_0=ax_0+by_0+cz_0=k $, so an equation of $H$ is
$$ax+by+cz=k\pm\bigl|\mkern2mu\vec n\mkern2mu\bigr|=k\pm\sqrt{a^2+b^2+c^2}.$$


text:
\subsection{Dynamics, Evolution of Info Measures, and Simulation}

The material here is taken mainly from the books by Reif, Kittel, and
F.~P.~Kelly, {\it Reversibility and Stochastic Networks},
(Chaps 1--3), J.~Wiley \& Sons, 1979.

\subsubsection{Markovian Dynamics, Global Balance and Detailed Balance}

So far we discussed only physical systems in equilibrium. For these systems,
the Boltzmann--Gibbs distribution is nothing but the stationary distribution of
the microstate $x$ at every given time instant $t$. However, this is merely
one part of the picture. What is missing is the temporal probabilistic behavior, or
in other words, the laws that underly the
evolution of the microstate with time. These are dictated by
dynamical properties of the system, which constitute
the underlying physical laws in the microscopic level. It is customary then to model
the microstate at time $t$ as a random process
$\{X_t\}$, where $t$ may denote either discrete time or continuous time,
and among the various models, one of the most common ones is the Markov
model. In this section, we discuss a few of the properties of these processes
as well as the evolution of information measures, like entropy, divergence
(and more) associated with them.

We begin with an isolated system in continuous time, which
is not necessarily assumed to have reached (yet) equilibrium. Let us suppose that
$X_t$, the
microstate at time $t$, can take on values in a discrete set $\calX$. For
$r,s\in\calX$, let
\begin{equation}
W_{rs}=\lim_{\delta\to
0}\frac{\mbox{Pr}\{X_{t+\delta}=s|X_t=r\}}{\delta}~~~~r\ne s
\end{equation}
in other words, $\mbox{Pr}\{X_{t+\delta}=s|X_t=r\}=W_{rs}\cdot \delta
+o(\delta)$. Letting $P_r(t)=\mbox{Pr}\{X_t=r\}$, it is easy to see that
\begin{equation}
P_r(t+\mbox{d}t)=\sum_{s\ne r}P_s(t)W_{sr}\mbox{d}t+P_r(t)\left(1-\sum_{s\ne
r}W_{rs}\mbox{d}t\right),
\end{equation}
where the first sum describes the probabilities of all possibile transitions from other states to
state $r$ and the second term describes the probability of not leaving state
$r$. Subtracting $P_r(t)$ from both sides and dividing by $\mbox{d}t$, we
immediately obtain the following set of differential equations:
\begin{equation}
\frac{\mbox{d}P_r(t)}{\mbox{d}t}=\sum_{s}[P_s(t)W_{sr}-P_r(t)W_{rs}],~~~r\in\calX,
\end{equation}
where $W_{rr}$ is defined in an arbitrary manner, e.g., $W_{rr}=0$ for all
$r$.
These equations are called the {\it master equations}.\footnote{Note that the
master equations apply in discrete time too, provided that the derivative at
the l.h.s.\ is replaced by a simple difference, $P_r(t+1)-P_r(t)$, and
$\{W_{rs}\}$ designate one--step state transition probabilities.}
When the process
reaches stationarity, i.e., for all $r\in\calX$, $P_r(t)$ converge to some $P_r$ that is
time--invariant, then
\begin{equation}
\sum_s[P_sW_{sr}-P_rW_{rs}]=0,~~~\forall~r\in\calX .
\end{equation}
This is called {\it global balance} or {\it steady state}. When the system is
isolated (microcanonical ensemble), the steady--state distribution must be
uniform, i.e., $P_r=1/|\calX|$ for all $r\in\calX$.
From quantum mechanical considerations, as well as considerations
pertaining to time reversibility in the microscopic level,\footnote{Think,
for example, of an isolated system
of moving particles, obeying the differential
equations $m\mbox{d}^2\br_i(t)/\mbox{d}t^2=\sum_{j\ne i}F(\br_j(t)-\br_i(t))$, $i=1,2,\ldots,n$, which
remain valid if the time variable $t$ is replaced by $-t$ since
$\mbox{d}^2\br_i(t)/\mbox{d}t^2=\mbox{d}^2\br_i(-t)/\mbox{d}(-t)^2$.}
it is customary to
assume $W_{rs}=W_{sr}$ for all pairs $\{r,s\}$. We then observe that, not
only, $\sum_s[P_sW_{sr}-P_rW_{rs}]=0$, but moreover, each individual term in
the sum vanishes, as
\begin{equation}
P_sW_{sr}-P_rW_{rs}=\frac{1}{|\calX|}(W_{sr}-W_{rs})=0.
\end{equation}
This property is called {\it detailed balance}, which is stronger than global
balance, and it means equilibrium, which is stronger than steady state. While
both steady--state and equilibrium refer to a situation of time--invariant
state probabilities $\{P_r\}$, a steady--state still allows cyclic flows of
probability. For example, a Markov process with cyclic deterministic
transitions $1\to 2\to 3\to 1\to 2\to 3\to \cdot\cdot\cdot$ is in steady state
provided that the probability distribution of the initial state is uniform
$(1/3,1/3,1/3)$, however, the cyclic flow among the states is in one
direction. On the other hand, in detailed balance ($W_{rs}=W_{sr}$ for an
isolated system),
which is equilibrium, there
is no net flow in any cycle of states. All the net cyclic
probability fluxes vanish, and
therefore, time reversal would not change the probability law, that is,
$\{X_{-t}\}$ has the same probability law as $\{X_t\}$. For example,
if $\{Y_t\}$ is a Bernoulli process, taking values equiprobably in
$\{-1,+1\}$, then $X_t$ defined
recursively by
\begin{equation}
X_{t+1}=(X_t+Y_t)\mbox{mod} K,
\end{equation}
has a symmetric state--transition probability matrix $W$, a uniform stationary
state distribtuion, and it satisfies detailed balance.

\subsubsection{Evolution of Information Measures}

Returning to the case where the process
$\{X_t\}$ pertaining to our isolated system has not necessarily reached
equilibrium,
let us take a look at the entropy of the state
\begin{equation}
H(X_t)=-\sum_rP_r(t)\log
P_r(t).
\end{equation}
We argue that $H(X_t)$ is monotonically non--decreasing, which is in
agreement with the second law (a.k.a.\ the H--Theorem). To this end,
we next show that
\begin{equation}
\frac{\mbox{d}H(X_t)}{\mbox{d}t}\ge 0,
\end{equation}
where for convenience, we denote
$\mbox{d}P_r(t)/\mbox{d}t$ by $\dot{P}_r(t)$.
\begin{eqnarray}
\frac{\mbox{d}H(X_t)}{\mbox{d}t}&=&-\sum_r[\dot{P}_r(t)\log
P_r(t)+\dot{P}_r(t)]\nonumber\\
&=&-\sum_r\dot{P}_r(t)\log
P_r(t)~~~~~~~~~~~~~~\sum_r\dot{P}_r(t)=0\nonumber\\
&=&-\sum_r\sum_sW_{sr}[P_s(t)-P_r(t)]\log
P_r(t))~~~~~~~~~~~W_{sr}=W_{rs}\nonumber\\
&=&-\frac{1}{2}\sum_{r,s}W_{sr}[P_s(t)-P_r(t)]\log P_r(t)-\nonumber\\
& &\frac{1}{2}\sum_{s,r}W_{sr}[P_r(t)-P_s(t)]\log
P_s(t)\nonumber\\
&=&\frac{1}{2}\sum_{r,s}W_{sr}[P_s(t)-P_r(t)]\cdot[\log P_s(t)-\log
P_r(t)]\nonumber\\
&\ge& 0.
\end{eqnarray}
where the last inequality is due to the increasing monotonicity of the logarithmic
function: the product $[P_s(t)-P_r(t)]\cdot[\log P_s(t)-\log P_r(t)]$ cannot be
negative for any pair $(r,s)$, as the two factors of this product are either both negative,
both zero, or
both positive. Thus, $H(X_t)$ cannot decrease with time.

This result has a discrete--time analogue: If a finite--state Markov process
has a symmetric transition probability matrix, and so, the stationary
state distribution is uniform, then $H(X_t)$ is a monotonically
non--decreasing sequence.

A considerably more general result is the following: If $\{X_t\}$
is a Markov process with a given state transition probability matrix
$W=\{W_{rs}\}$ (not necessarily symmetric) and $\{P_r\}$ is a stationary state
distribution, then the function
\begin{equation}
U(t)=\sum_rP_r\cdot V\left(\frac{P_r(t)}{P_r}\right)
\end{equation}
is monotonically strictly increasing provided that
$V(\cdot)$ is strictly concave.
To see why this is true, we use the fact that $P_s=\sum_rP_rW_{rs}$ and define
$\tilde{W}_{sr}=P_rW_{rs}/P_s$. Obviously, $\sum_r\tilde{W}_{sr}=1$ for all
$s$, and so,
\begin{equation}
\frac{P_r(t+1)}{P_r}=\sum_s\frac{P_s(t)W_{sr}}{P_r}=\sum_s\frac{\tilde{W}_{rs}P_s(t)}{P_s}
\end{equation}
and so, by the concavity of $V(\cdot)$:
\begin{eqnarray}
U(t+1)&=&\sum_rP_r\cdot V\left(\frac{P_r(t+1)}{P_r}\right)\nonumber\\
&=&\sum_rP_r\cdot V\left(\sum_s\tilde{W}_{rs}\frac{P_s(t)}{P_s}\right)\nonumber\\
&>&\sum_r\sum_sP_r\tilde{W}_{rs}\cdot V\left(\frac{P_s(t)}{P_s}\right)\nonumber\\
&=&\sum_r\sum_sP_sW_{sr}\cdot V\left(\frac{P_s(t)}{P_s}\right)\nonumber\\
&=&\sum_sP_s\cdot V\left(\frac{P_s(t)}{P_s}\right)=U(t).
\end{eqnarray}
Here we required nothing except the existence of a stationary distribution.
Of course in the above derivation $t+1$ can be
replaced by $t+\tau$ for any positive real $\tau$ with the appropriate
transition probabilities, so the monotonicity of $U(t)$ applies to
continuous--time Markov processes as well.

Now, a few interesting choices of the function $V$ may be considered:
\begin{itemize}
\item
For $V(x)=-x\ln x$, we have $U(t)=-D(P(t)\|P)$. This means that
the divergence between $\{P_r(t)\}$ and the steady state distribution $\{P_r\}$
is monotonically strictly decreasing, whose physical interpretation could be
the decrease of the free energy, since we have already seen that the free
energy is the physical counterpart of the divergence.
This is a more general rule, that
governs not only isolated systems, but any Markov process with a stationary
limiting distribution (e.g., any Markov process whose distibution converges
to that of the Boltzmann--Gibbs distribution).
Having said that, if we now particularize this result to the case where
$\{P_r\}$ is the uniform distribution (as in an isolated system), then
\begin{equation}
D(P(t)\|P)=\log|\calX|-H(X_t),
\end{equation}
which means that the decrease of divergence is equivalent to the increase in
entropy, as before. The difference, however, is that
here it is more general as
we only required a uniform steady--state distribution, not
necessarily detailed balance.\footnote{For the uniform distribution to be a
stationary distribution, it is sufficient (and necessary) that $W$ would be a
doubly stochastic matrix, namely, $\sum_rW_{rs}=\sum_rW_{sr}=1$. This
condition is, of course, weaker than detailed balance, which means that $W$ is
moreover symmetric.}
\item
Another interesting choice of $V$ is $V(x)=\ln
x$, which gives $U(t)=-D(P\|P(t))$. Thus, $D(P\|P(t))$ is also monotonically
decreasing. In fact, both this and the monotonicity result of the previous item, are
in turn, special cases of a more general result concerning the divergence (see
also the book by Cover and Thomas, Section 4.4). Let
$\{P_r(t)\}$ and $\{P_r'(t)\}$ be two time--varying state--distributions
pertaining to the same Markov chain, but induced by two different initial state
distributions, $\{P_r(0)\}$ and $\{P_r'(0)\}$. Then $D(P(t)\|P'(t))$ is
monotonically non--increasing. This happens because
\begin{eqnarray}
D(P(t)\|P'(t))&=&\sum_rP_r(t)\log\frac{P_r(t)}{P_r'(t)}\nonumber\\
&=&\sum_{r,s}P_r(t)P(X_{t+\tau}=s|X_t=r)
\log\frac{P_r(t)P(X_{t+\tau}=s|X_t=r)}{P_r'(t)P(X_{t+\tau}=s|X_t=r)}\nonumber\\
&=&\sum_{r,s}P(X_t=r,~X_{t+\tau}=s)
\log\frac{P(X_t=r,~X_{t+\tau}=s)}{P'(X_t=r,~X_{t+\tau}=s)}\nonumber\\
&\ge&D(P(t+\tau)\|P'(t+\tau))
\end{eqnarray}
where the last inequality follows from the data processing theorem of the
divergence: the divergence between two joint distributions of $(X_t,X_{t+\tau})$
is never smaller than the
divergence between corresponding marginal distributions of $X_{t+\tau}$.
\item
Yet another choice is $V(x)=x^s$, where $s\in[0,1]$ is a
parameter. This would yield the increasing monotonicity of
$\sum_r P_r^{1-s}P_r^{s}(t)$, a metric that plays a role in
the theory of asymptotic
exponents of error probabilities pertaining to the optimum
likelihood ratio test between two probability distributions. In particular,
the choice $s=1/2$ yields balance between the two kinds of error and it is
intimately related to the Bhattacharyya distance. Thus, we obtained some sorts
of generalizations of the second law to information measures other than
entropy.
\end{itemize}

For a general Markov process, whose steady state--distribution is not
necessarily uniform, the condition of detailed balance, which means
time--reversibility, reads
\begin{equation}
P_sW_{sr}=P_rW_{rs},
\end{equation}
both in discrete time and continuous time (with the corresponding meaning of
$\{W_{rs}\}$). The physical interpretation is that now our system is (a small)
part of a large isolated system, which obeys detailed balance w.r.t.\ the
uniform equilibrium distribution, as before.
A well known example of a process that obeys detailed balance in its more
general form
is an M/M/1 queue with an arrival rate $\lambda$ and service rate $\mu$
($\lambda < \mu$). Here, since all states are arranged along a line, with
bidirectional transitions between neighboring states only (see Fig.\
\ref{mm1}), there cannot be
any cyclic probability flux. The steady--state distibution is well--known
to be geometric
\begin{equation}
P_r=\left(1-\frac{\lambda}{\mu}\right)
\cdot\left(\frac{\lambda}{\mu}\right)^r, ~~~~~~r=0,1,2,\ldots,
\end{equation}
which indeed satisfies the detailed
balance $P_r\lambda=P_{r+1}\mu$ for all $r$.
Thus, the Markov process $\{X_t\}$,
designating the number of customers in the queue at time $t$,
is time--reversible.

It is interesting to point out that in order to check for the detailed balance property, one does
not necessarily have to know the equilibrium distribution $\{P_r\}$ as above.
Applying detailed balance to any $k$ pairs of states in a cycle, $(s_1,s_2)$,
$(s_2,s_3),\ldots, (s_k,s_1)$,
and multiplying the respective detailed balance equations, the steady state
probabilities cancel out and one easily obtains
\begin{equation}
W_{s_1s_2}W_{s_2s_3}\cdot\cdot\cdot W_{s_{k-1}s_k}W_{s_ks_1}=
W_{s_ks_{k-1}}W_{s_{k-1}s_{k-2}}\cdot\cdot\cdot W_{s_2s_1}W_{s_1s_k},
\end{equation}
so this is clearly a necessary condition for detailed balance. One can show
conversely, that if this equation applies to any finite cycle of states, then
the chain satisfies detailed balance, and so this is also a sufficient condition.
This is true both in discrete time and continuous time, with the corresponding
meanings of $\{W_{rs}\}$
(see Kelly's book, pp.\ 22--23).

\begin{figure}[ht]
\hspace*{2cm}\input{mm1.pstex_t}
\caption{\small State transition diagram of an M/M/1 queue.}
\label{mm1}
\end{figure}

In the case of detailed balance, there is another
interpretation of the approach to equilibrium and the growth of $U(t)$.
We can write the master equations as follows:
\begin{equation}
\frac{\mbox{d}P_r(t)}{\mbox{d}t}=\sum_s\frac{1}{R_{sr}}\left(\frac{P_s(t)}{P_s}-\frac{P_r(t)}{P_r}\right)
\end{equation}
where $R_{sr}=(P_sW_{sr})^{-1}=(P_rW_{rs})^{-1}$. Imagine now an electrical
circuit where the indices $\{r\}$ designate the nodes. Nodes $r$ and $s$ are connected by
a wire with resistance $R_{sr}$ and every node $r$ is grounded via a capacitor
with capacitance $P_r$ (see Fig.\ \ref{circuit}).
If $P_r(t)$ is the charge at node $r$ at time $t$,
then the master equations are the Kirchoff equations of the currents at each
node in the
circuit. Thus, the way in which probability spreads across the circuit is
analogous to the way charge spreads across the circuit and probability fluxes
are now analogous to electrical currents.
If we now choose
$V(x)=-\frac{1}{2}x^2$, then $-U(t)=\frac{1}{2}\sum_r\frac{P_r^2(t)}{P_r}$,
which means that the energy stored in the capacitors dissipates as heat in the
wires until the system reaches equilibrium, where all nodes have the same
potential, $P_r(t)/P_r=1$, and hence detailed balance corresponds to
the situation where all individual currents vanish (not only their algebraic
sum).

\begin{figure}[ht]
\hspace*{2cm}\input{circuit.pstex_t}
\caption{\small State transition diagram of a Markov chain
(left part) and the electric circuit that emulates the
dynamics of $\{P_r(t)
\}$ (right part).}
\label{circuit}
\end{figure}

We have seen, in the above examples, that various choices of the function $V$
yield various `metrics' between $\{P_r(t)\}$ and $\{P_r\}$, which are both
marginal distributions of a single symbol. What about joint distributions of
two or more symbols? Consider, for example, the function
\begin{equation}
J(t)=\sum_{r,s}P(X_0=r,~X_t=s)\cdot V\left(\frac{P(X_0=r)P(X_t=s)}{P(X_0=r,~X_t=s)}\right),
\end{equation}
where $V$ is concave as before. Here, by the same token, $J(t)$ is a `metric'
between the joint probability distribution $\{P(X_0=r,~X_t=s)\}$ and the
product of marginals $\{P(X_0=r)P(X_t=s)\}$, namely, it a measure of the
amount of statistical dependence between  $X_0$ and $X_t$. For $V(x)=\ln x$,
we have, of course, $J(t)=-I(X_0;X_t)$. Now, using
a similar chain of inequalities as before, we get the non--decreasing
monotonicity of $J(t)$ as follows:
\begin{eqnarray}
J(t)&=&\sum_{r,s,u}P(X_0=r,~X_t=s,~X_{t+\tau}=u)\cdot V\left(\frac{P(X_0=r)P(X_t=s)}{P(X_0=r,~X_t=s)}
\cdot\frac{P(X_{t+\tau}=u|X_t=s)}{P(X_{t+\tau}=u|X_t=s)}\right)\nonumber\\
&=&\sum_{r,u}P(X_0=r,~X_{t+\tau}=u)\sum_sP(X_t=s|X_0=r,~X_{t+\tau}=u)\times\nonumber\\
& &V\left(\frac{P(X_0=r)P(X_t=s,~X_{t+\tau}=u)}
{P(X_0=r,~X_t=s,~X_{t+\tau}=u)}\right)\nonumber\\
&\le&\sum_{r,u}P(X_0=r,~X_{t+\tau}=u)\times\nonumber\\
& &V\left(\sum_s P(X_t=s|X_0=r,~X_{t+\tau}=u)\cdot
\frac{P(X_0=r)P(X_t=s,~X_{t+\tau}=u)}
{P(X_0=r,~X_t=s,~X_{t+\tau}=u)}\right)\nonumber\\
&=&\sum_{r,u}P(X_0=r,~X_{t+\tau}=u)\cdot V\left(\sum_s
\frac{P(X_0=r)P(X_t=s,~X_{t+\tau}=u)}
{P(X_0=r,~X_{t+\tau}=u)}\right)\nonumber\\
&=&\sum_{r,u}P(X_0=r,~X_{t+\tau}=u)\cdot V\left(
\frac{P(X_0=r)P(X_{t+\tau}=u)}
{P(X_0=r,~X_{t+\tau}=u)}\right)=J(t+\tau).
\end{eqnarray}
This time, we assumed nothing beyond Markovity (not even homogeneity). This is
exactly the generalized data processing theorem of Ziv and Zakai (J.~Ziv and
M.~Zakai, ``On functionals satisfying a data-processing
theorem,'' {\em IEEE Trans.~Inform.~Theory\/},
vol.~IT--19, no.~3, pp.~275--283, May 1973), which yields the ordinary data
processing theorem (of the mutual information) as a special case.
Thus, we see
that the second law of thermodynamics is (at least indirectly) related to the data processing
theorem via the fact that they both stem from some more general principle
concerning monotonic evolution of `metrics' between probability distributions
defined using convex functions.
In a very
similar manner, one can easily show that the generalized conditional entropy
\begin{equation}
\sum_{r,s}P(X_0=r,~X_t=s)\cdot V\left(\frac{1}{P(X_0=r|X_t=s)}\right)
\end{equation}
is monotonically non--decreasing with $t$ for any concave $V$.

\subsubsection{Monte Carlo Simulation}

Returning to the realm of Markov processes
with the detailed balance property,
suppose we want to simulate a
physical system, namely, to sample from the Boltzmann--Gibbs
distribution
\begin{equation}
P_r=\frac{e^{-\beta E_r}}{Z(\beta)}.
\end{equation}
In other words, we wish to
generate a discrete--time Markov process
$\{X_t\}$, possessing the detailed balance property,
whose marginal converges to the Boltzmann--Gibbs distribution.
This approach is called {\it dynamic Monte Carlo} or {\it Markov chain Monte Carlo}
(MCMC).
How should we select
the state transition probability matrix $W$ to this end?
Substituting $P_r=e^{-\beta E_r}/Z(\beta)$ into the detailed balance
equation, we readily see that a necessary condition is
\begin{equation}
\frac{W_{rs}}{W_{sr}}=e^{-\beta(E_s-E_r)}.
\end{equation}
The {\it Metropolis algorithm} is one popular way to implement such a Markov
process in a rather efficient manner. It is based on the concept of factoring $W_{rs}$ as a
product $W_{rs}=C_{rs}A_{rs}$, where $C_{rs}$ is the conditional probability of selecting
$X_{t+1}=s$ as a {\it candidate} for the next state, and $A_{rs}$
designates the
probability of {\it acceptance}. In other words, we first choose a candidate
according to $C$, and then make a final decision whether we accept this
candidate or stay in state $r$. The Metropolis algorithm pics $C$ to
implement a uniform distribution among $n$ states `close' to $r$ (e.g.,
flipping one spin of a $n$--spin configuration). Thus,
$W_{rs}/W_{sr}=A_{rs}/A_{sr}$, and so, it remains to choose $A$ such that
\begin{equation}
\frac{A_{rs}}{A_{sr}}=e^{-\beta(E_s-E_r)}.
\end{equation}
The Metropolis algorithm defines
\begin{equation}
A_{rs}=\left\{\begin{array}{ll}
e^{-\beta(E_s-E_r)} & E_s > E_r\\
1 & \mbox{otherwise}\end{array}\right.
\end{equation}
In simple words, the algorithm works as follows:
Given that $X_t=r$,
first randomly select one candidate $s$ for $X_{t+1}$ among $n$ possible (neighboring)
states. If $E_s < E_r$ always accept $X_{t+1}=s$ as the next state. If $E_s\ge
E_r$, then
randomly draw a RV $Y\in\mbox{Unif}[0,1]$. If $Y <
e^{-\beta(E_s-E_r)}$, then again, accept $X_{t+1}=s$ as the next state.
Otherwise, stay in state $r$, i.e., $X_{t+1}=r$.
To see why this choice of $A$ works, observe that
\begin{equation}
\frac{A_{rs}}{A_{sr}}=\left\{\begin{array}{ll}
e^{-\beta(E_s-E_r)} & E_s > E_r\\
\frac{1}{e^{-\beta(E_r-E_s)}} & E_s \le E_r\end{array}\right.
= e^{-\beta(E_s-E_r)}.
\end{equation}
There are a few nice things about this algorithm:
\begin{itemize}
\item Energy differences between neighboring states, $E_s-E_r$, are normally
easy to calculate. If $r$ and $s$ differ by a single component of the
microstate $\bx$, and the if the Hamiltonian structure consists of
short--range interactions only,
then most terms of the Hamiltonian are the same for $r$ and $s$, and only a
local calculation is required for evaluating the energy difference.
\item Calculation of $Z(\beta)$ is not required, and
\item Chances are that you don't get stuck in the same state for too long.
\end{itemize}
The drawback, however, is that aperiodicity is not guaranteed. This depends
on the Hamiltonian.

The {\it heat bath} algorithm (a.k.a.\ {\it Glauber
dynamics}) alleviates this shortcoming
and although somewhat slower than Metropolis to equilibrate, it
guarantees all the good properties of a Markov chain: irreducibility,
aperiodicity, and convergence to stationarity. The only difference is that
instead of the above choice of $A_{rs}$, it is redefined as
\begin{eqnarray}
A_{rs}&=&\frac{1}{2}\left[1-\tanh\left(\frac{\beta(E_s-E_r)}{2}\right)\right]
\nonumber\\
&=&\frac{e^{-\beta(E_s-E_r)}}{1+e^{-\beta(E_s-E_r)}}\nonumber\\
&=&\frac{P_s}{P_s+P_r},
\end{eqnarray}
which is also easily shown to satisfy the detailed balance condition.
The heat bath algorithm generalizes
easily to sample from any distribution $P(\bx)$ whose configuration space is
of the form $\calX^n$. The algorithm can be described by the following
pseudocode:
\begin{enumerate}
\item Select $\bX_0$ uniformly at random across $\calX^n$.
\item {\bf For} $t=1$ {\bf to} $t=T$:
\item Draw an integer $i$ at random with uniform distribution across
$\{1,2,\ldots,n\}$.
\item For each $x\in\calX$, calculate
\begin{equation}
P(X^i=x|\bX^{\sim i}=\bx_t^{\sim i})=\frac{P(X^i=x,\bX^{\sim i}=\bx_t^{\sim
i})}{\sum_{x'\in\calX}P(X^i=x',\bX^{\sim i}=\bx_t^{\sim i})}.
\end{equation}
\item Set $x_{t+1}^j=x_t^j$ for all $j\ne i$ and $x_t^i=X^i$,
where $X^i$ is drawn
according to
$$P(X^i=x|\bX^{\sim i}=\bx_t^{\sim i}).$$
\item {\bf end}
\item {\bf Return} the sequence $\bX_t$, $t=1,2,\ldots,T$.
\end{enumerate}
It can be easily seen that the resulting Markov chain satisfies detailed
balance and that in the case of binary alphabet (spin array) it implements
the above expression of $A_{rs}$. One can also easily generalize the
Metropolis algorithm, in the same spirit, as $e^{-\beta(E_s-E_r)}$ is nothing
but the ratio $P_s/P_r$.


TITLE: Characteristic polynomial of projection
QUESTION [1 upvotes]: Let $V$ be a $n$ dimensional vector space over a field $\mathbb{k}$, and let $P:V \rightarrow V$ be a linear map such that $P^2=P$. i.e.,  A linear map is a projection.
I want to find all possible characteristic polynomials and their corresponding minimal polynomial.

I know the form of characteristic polynomial $\phi_P(t) = \operatorname{det}(P-tI)$, and minimal polynomial divides characteristic polynomial.
From the assumption $P^2-P=0$ so the minimal polynomial $m_P(x)$ is of the following form $x, (x-1), x(x-1)$. i.e., eigenvalues of $P$ are $0$ or $1$.
But I have a problem with analyzing characteristic polynomial. Here is my trial.
Since $\phi_P(t) = \operatorname{det}(P-tI) = (-1)^n t^n + \cdots$, my guess of the form of characteristic polynomial is $n$ degree polynomial. Since minimal polynomial divides characteristic polynomial $t(t-1)$ should be a factor of $\phi_P(t)$... but what about other $n-2$ degrees of polynomials? The only eigenvalues of projection matrix are $0$ and $1$ so my naive guess is $\phi_P(t) = (-1)^n t^{m} (t-1)^{n-m}$ with $ 1< m <n$. Is this correct? If so how one can prove this?

REPLY [1 votes]: You have a small error: you mention the minimal polynomial could be $x$ or $x-1$, but only consider the case $x(x-1)$ in the last paragraph.
After correcting this mistake, you will have shown that the characteristic polynomial must be of the form $(-1)^n t^m (t-1)^{n-m}$ for $0 \le m \le n$.
It remains to check that all of these polynomials are the characteristic polynomial of some projection; to show this it suffices to consider diagonal matrices with $0$s and $1$s on the diagonal.