TITLE: Prove that $A^TD-C^TB=I$
QUESTION [3 upvotes]: Let A,B,C,D be complex matrices $n \times n$ such that $AB^T,CD^T$ are symmetric and $AD^T-BC^T=I$. Prove that $A^TD-C^TB=I$. Can anyone give me any idea? Thank you.

REPLY [9 votes]: Hint: The given condition says that
$$
\pmatrix{A&-B\\ -C&D}\pmatrix{D^T&B^T\\ C^T&A^T} = \pmatrix{I&0\\ 0&I}.
$$
Now, note that $XY=I$ implies that $YX=I$ and in turn $X^TY^T=I$.


TITLE: A question about the Yoneda embedding
QUESTION [0 upvotes]: Write $y:\mathcal{C}\to[\mathcal{C}^{\text{op}},\mathbf{Set}]$ for the Yoneda embedding on a small category $\mathcal{C}$.
If there is a category $\mathcal{D}$ and functors $G:\mathcal{C}\to\mathcal{D}$ and $F:\mathcal{D}\to[\mathcal{C}^{\text{op}},\mathbf{Set}]$ such that $y=F\circ G$, is it necessary that $\mathcal{D}$ becomes a subcategory of $[\mathcal{C}^{\text{op}},\mathbf{Set}]$?

REPLY [0 votes]: No, just adjoin a disjoint object to $C$ with however many endomorphisms, and factor Yoneda through the resulting $D$ by sending the new object, say, to the terminal preheaf.
Or better yet, let $D$ be the disjoint union of two copies of the presheaf category.


TITLE: Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$
QUESTION [1 upvotes]: Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$

I have tried to express $x$ in terms of $a$ and $b$, namely: $x = a+bi, |x|= \sqrt{a^2+b^2}$ and to evaluate this but this simply led me nowhere. I guess there must be a clever way to solve that I don't see. I'd be thankful if you could give me some hints (not the actual solution) on how to proceed with this problem.

REPLY [1 votes]: A purely geometric proof, based on the inscribed angle theorem:


TITLE: How do you find the mass involved in fusion reactions in the center of a star?
QUESTION [1 upvotes]: How have scientists found the amount of a stars mass that is in the central region where nuclear fusion occurs? For example, our Sun has about 10% of its mass in the center, how have scientists found that?

REPLY [3 votes]: The equations Countto10 is alluding to are the equations of stellar structure, which describe how the pressure ($P$), luminosity ($L$), temperature ($T$), and enclosed mass ($M$) change at a radius $r$ in the star1:
$$\frac{\mathrm{d}M_r}{\mathrm{d}r}=4\pi r^2\rho(r)\tag{Mass}$$
$$\frac{\mathrm{d}P}{\mathrm{d}r}=-\frac{GM_r\rho(r)}{r^2}\tag{Pressure}$$
$$\frac{\mathrm{d}L}{\mathrm{d}r}=4\pi r^2\epsilon(r)\tag{Luminosity}$$
$$\frac{\mathrm{d}T}{\mathrm{d}r}=\frac{3\rho(r)\kappa(r)}{64\pi r^2\sigma[T(r)]^3}L(r)\tag{Temperature}$$
$\epsilon$, $\rho$ and $\kappa$ are the rate of energy generation, density and opacity, respectively. All other symbols are constants. They're coupled differential equations, and so they must be solved by numerical methods, not analytical ones like we might use for "normal" differential equations. One of the simplest common methods is the fitting method. We know that $M(r=0)=0$ and $L(r=0)=0$, but we don't know $P(r=0)$ and $T(r=0)$. However, we can figure out the values of $P$ and $T$ at the surface.
We can guess what $L$ and $M$ will be at the surface and what $P$ and $T$ will be at the center of the star. We then "integrate" both inwards and outwards until we reach a "fitting point", and check to see whether the equations match up. If they don't, we can still make some adjustments, and redo the process. There are other tricks that make the process more interesting and certainly quicker. At the end of it all, though, you have a description of how all these variables change with radius inside the star. You can then figure out how much mass, for instance, is contained within a certain region, and if we can figure out the outer radius of the region where the primary nuclear fusion takes place, we can find $M$ at that radius. That gives you the result you're looking for.
This is all very complicated, and I probably didn't make things seem any simpler. That's okay; the process really isn't simple. Something more hands-on might convince you of the power of some of these assumptions. We can approximate the star by assuming that it is a polytrope, that is, that its pressure and density are related by a certain simple formula. After doing some mathematical playing around, we eventually arrive at a simple differential equation, called the Lane-Emden equation:
$$\frac{1}{\xi^2}\frac{\mathrm{d}}{\mathrm{d}\xi}\left(\xi^2\frac{\mathrm{d}\theta}{\mathrm{d}\xi}\right)=-\theta^n$$
Ack! More variables! $\xi$ is known as the dimensionless radius, because it scales linearly with $r$. $\theta$ is a function of $\xi$, and $P$, $\rho$ and $T$ are all functions of $\theta$. So if we solve this differential equation, we can come up with some really interesting estimates for the structure of the star!2
I had some old Python code for a fourth-order Runge-Kutta method, and I used it to numerically integrate the Lane-Emden equation (although I did not use a large step size). I then plotted the temperature, density, and pressure of the star divided by their central values:

Look at how quickly they drop off! $\theta=0$ at just over $\xi=6$, so about halfway through the star, conditions are much different than at the center. This seems to confirm the assertion that the core contains about 10% of the Sun's mass.

1 The equation for temperature is really only valid under certain conditions.

2 The assumption fails near the surface, but that's okay. We only really care about the core here.


TITLE: Calculating Laurent series for log function
QUESTION [2 upvotes]: Suppose I have the function:
$ f(z) = \frac{log(z+1)}{z^3} $
I need to find the Laurent series of this function to determine its order.
In this case we take $log(z)$ to be $ln|z|+i\arg (z)$
So we want to find the power series for $\frac{ln|z+1|}{z^3} + \frac{i \arg (z+1)}{z^3}$
$ = \sum_{n=1}^{\infty}[\frac{(-1)^{n+1}z^{n-3}}{n} ]+ \frac{i \arg(z+1)}{z^3}$
I'm not sure how to incorporate the arg into the series though, any ideas?
EDIT: Log defined in my notes

REPLY [1 votes]: One may recall that, as $z \to 0$, by using the Taylor series expansion,
$$
\log(1+z)=z-\frac{z}2+\frac{z^3}3+O(z^4)
$$ giving the Laurent series expansion
$$
\frac{\log(1+z)}{z^3}=\frac1{z^2}-\frac1{2z}+\frac13+O(z)
$$as $z \to 0$.


TITLE: How can one find the x value that gives the largest possible value to the equation (100floor(10000/(floor(100000/(2x))+1)))/x?
QUESTION [0 upvotes]: How can one find the x value that gives the largest possible value to the equation
$$\frac{100\left\lfloor\frac{10000}{\left\lfloor \frac{100000}{2x}\right\rfloor+1}\right\rfloor}{x}$$
where $x$ is a whole number, and 100000 is divisible by $x$.

REPLY [0 votes]: $\frac{\mathrm{10}^{\mathrm{5}} }{{x}}\in{Z}^{+} ,\frac{\mathrm{10}^{\mathrm{2}} \left[\frac{\mathrm{10}^{\mathrm{4}} }{\left[\frac{\mathrm{10}^{\mathrm{5}} }{\mathrm{2}{x}}\right]+\mathrm{1}}\right]}{{x}}\left({MAX}\right) \\ $
$=\frac{\mathrm{10}^{\mathrm{2}} \left(\frac{\mathrm{10}^{\mathrm{4}} }{\frac{\mathrm{10}^{\mathrm{5}} }{\mathrm{2}{x}}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}}\right)}{{x}}=\frac{\mathrm{2}.\mathrm{10}^{\mathrm{6}} }{\mathrm{10}^{\mathrm{5}} +{x}} \\ $
$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\neq\mathrm{0}\Rightarrow{x}=\mathrm{1} \\ $


text:
\begin{document}

\large

\title[On a boundary value problem]
{On a boundary value problem in subsonic aeroelasticity and
the cofinite Hilbert transform.}
\author[Peter L. Polyakov]{Peter L. Polyakov}
\address{Department of Mathematics, University of Wyoming, Laramie, WY 82071, USA}
\email{ polyakov@@uwyo.edu}
\subjclass{45B05,45E05}
\date{\today}
\keywords{Hilbert transform, Fredholm determinant, Resolvent}

\begin{abstract}
We study a boundary value problem in subsonic aeroelasticity and introduce the
{\it cofinite Hilbert transform} as a tool in solving an auxiliary linear integral
equation on the complement of a finite interval on the real line $\R$.
\end{abstract}
\maketitle

\centerline{\DATE}

\bigskip

\section{Introduction.}\label{Introduction}
\indent
We consider the linearized subsonic inviscid compressible flow equation in 2D
(\cite{BAH}, \cite{Ba2})
\begin{equation}\label{LinearEquation}
a^2_{\infty}\left(1-M^2\right)\frac{\partial^2\phi}{\partial x^2}
+a^2_{\infty}\frac{\partial^2\phi}{\partial z^2}
=\frac{\partial^2\phi}{\partial t^2}
+2Ma_{\infty}\frac{\partial^2\phi}{\partial t\partial x},
\end{equation}
where $a_{\infty}$ is the speed of sound,
${\dis M=\frac{U}{a_{\infty}}<1}$ - the Mach number, $U$ - free stream velocity,
$\phi(x,z,t)$ - small disturbance velocity potential, considered on
$$\R^2_{+}\times\overline{\R_{+}}=\left\{(x,z,t):-\infty<x<\infty,\ 0<z<\infty,\
0\leq t<\infty\right\},$$
with boundary conditions:
\begin{itemize}
\item
flow tangency condition
\begin{equation}\label{Flow_Tangency}
\frac{\partial\phi}{\partial z}(x,0,t)=w_a(x,t),\ |x|<b,
\end{equation}
where $b$ is the "half-chord", and $w_a$ is the given normal velocity of the wing,
without loss of generality we will assume in what follows that $b=1$,
\item
{\it "strong Kutta-Joukowski condition"} for the acceleration potential
$$\psi(x,z,t):=\frac{\partial\phi}{\partial t}+U\frac{\partial\phi}{\partial x},$$
\begin{equation}\label{Kutta}
\psi(x,0,t)=0\ \mbox{for}\ 1<|x|<A\ \mbox{for some}\ A>1,
\end{equation}
\item
far field condition
$$\phi(x,z,t)\to 0,\ \mbox{as}\ |x|\to\infty,\ \mbox{or}\ z\to\infty.$$
\end{itemize}
\indent
Boundary condition (\ref{Kutta}), though being motivated by one of the "auxiliary boundary
conditions" from (\cite{BAH}, p. 319), is weaker, because it requires
that $\psi(x,0,t)=0$ not on the whole $\R\setminus[-1,1]$, but only on finite intervals
adjacent to the interval $[-1,1]$. On the other hand this change in the boundary condition
allows application of some new mathematical tools different from tools in \cite{BAH} and
\cite{Ba2}.\\

\indent
In order to formulate our main result we introduce the following notations. We
denote by $\widehat{w_a}$ the Laplace transform of the function $w_a$ with respect
to time variable
$$\widehat{w_a}(x,z,\lambda)=\int_0^{\infty}e^{-\lambda t}w_a(x,z,t)dt$$
for $\mbox{Re}\lambda>\sigma_a>0.$ We also denote
$$\begin{array}{ll}
{\dis r(\lambda)=\frac{\lambda M}{U\sqrt{1-M^2}},}
\vspace{0.1in}\\
{\dis d(\lambda)=\frac{\lambda M^2}{U(1-M^2)}.}
\end{array}$$
\indent
In sections \ref{Solvability} and \ref{GLambdaResolvent} we construct a function
${\cal D}_{N}(\lambda)$ (equation (\ref{D_N})), analytic in the half-plane
$\mbox{Re}\lambda>\sigma_a>0$, and depending only on the function
$K_0$ - the modified Bessel function of the third kind.

\indent
The following theorem represents the main result of the paper.

\begin{theorem}\label{Main}\ Let function ${\cal D}_{N}(\lambda)$ from equation (\ref{D_N}),
mentioned above, have no zeros in the strip $\left\{\mbox{Re}\lambda\in [\sigma_1,\sigma_2]\right\}$,
where $\sigma_1>\sigma_a$. Let $I(1)=[-1,1]$, and let
$w_a(\cdot,t) \in L^2\left(I(1)\right)$ be such that for some $\epsilon>0$
\begin{equation}\label{wCondition}
{\dis \left\|\widehat{w}_a(\cdot,\sigma+i\eta)\right\|_{L^2\left(I(1)\right)}
<\exp\left\{-e^{|\eta|}\cdot(1+|\eta|)^{2+\epsilon}\right\}\ \mbox{for}\
\sigma\in[\sigma_1,\sigma_2] }
\end{equation}
\indent
Then equation (\ref{LinearEquation}) has a solution of the form
\begin{equation}\label{Solution}
\begin{array}{ll}
{\dis \phi(x,z,t)=-\frac{1}{2\pi\sqrt{1-M^2}}
\int_{\sigma^{\prime}-i\infty}^{\sigma^{\prime}+i\infty}e^{d(\sigma^{\prime}+i\eta)x} }
\vspace{0.1in}\\
{\dis \times\left[\int_{-\infty}^\infty K_0\left(r(\sigma^{\prime}+i\eta)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)
h_a(y,\sigma^{\prime}+i\eta)dy\right]e^{(\sigma^{\prime}+i\eta)t}d\eta.}
\end{array}
\end{equation}
This solution is independent of $\sigma^{\prime}\in [\sigma_1,\sigma_2]$,
satisfies boundary conditions above, and function $h_a$ satisfies the estimate
$$\int_{-\infty}^\infty(1+|x|)^{p-2}\left|h_a(x,\sigma^{\prime}+i\eta)\right|^pdx
<\frac{C}{\left(1+|\eta|\right)^m}$$
for arbitrary $m>0$, ${\dis p<\frac{4}{3} }$, and $C>0$ independent of $\lambda$.
\end{theorem}

\indent
The author thanks A.V. Balakrishnan for suggesting and explaining the problem considered
here, and for his hospitality during author's visits to UCLA.

\section{"General" solution.}
\label{General}

\indent
We are seeking a solution of the equation (\ref{LinearEquation}) of the form
\begin{equation}\label{SolutionForm}
\phi(x,z,t)=\int_{\sigma-i\infty}^{\sigma+i\infty}
\xi(x,z,\lambda)e^{(\sigma+i\eta)t}d\eta,
\end{equation}
where $\lambda=\sigma+i\eta$, $\sigma>\sigma_a$ and $\xi(x,z,\lambda)\in L^1\eta(\R)$.
Then, substituting the expression above into equation (\ref{LinearEquation}),
we obtain the following auxiliary equation for $\xi$
\begin{equation}\label{LaplaceEquation}
a^2_{\infty}\left(1-M^2\right)\frac{\partial^2 \xi}{\partial x^2}
+a^2_{\infty}\frac{\partial^2 \xi}{\partial z^2}
-\lambda^2 \xi-2M\lambda a_{\infty}\frac{\partial \xi}{\partial x}=0.
\end{equation}
\indent
To describe the general solution of equation (\ref{LaplaceEquation}) satisfying
the far field condition we consider, following \cite{Ba2}
$$D(\omega,\lambda)=M^2\left(\frac{\lambda}{U}\right)^2+2i\frac{\lambda}{U}M^2\omega
+\left(1-M^2\right)\omega^2,$$
and prove two lemmas below.\\

\begin{lemma}\label{RootD}\ There exists a function $\sqrt{D(\omega,\lambda)}$, analytic
with respect to complex variable $\frac{\lambda}{U}+i\omega\ (\frac{\lambda}{U}
\in\C,\ \omega\in \R)$ in the half-plane ${\dis \mbox{Re}\lambda>\sigma_a}$,
and such that $\mbox{Re}\sqrt{D(\omega,\lambda)}>0$.
\end{lemma}

\indent
{\bf Proof.}\ Representing $D(\omega,\lambda)$ as
$$D(\omega,\lambda)=M^2\left(\frac{\lambda}{U}\right)^2+2i\frac{\lambda}{U}M^2\omega
+\left(1-M^2\right)\omega^2=M^2\left(\frac{\lambda}{U}+i\omega\right)^2+\omega^2,$$
we obtain that the image of the half-plane
${\dis \mbox{Re}\lambda>\sigma_a}$
under the map $D(\omega,\lambda)$ is contained in the domain
$\C\setminus{\R^{-}}$. Then the branch of the function $\sqrt{}$ considered
on the complex plane with the cut along the negative part of the real axis is well
defined and analytic on the image of $D$, and its real part satisfies condition
of the Lemma. Therefore, the composition $\sqrt{D}$ is also analytic and
satisfies the same condition.
\qed

\begin{lemma}\label{Fourier}\ The following equality holds
$$\frac{e^{d(\lambda)x}}{\sqrt{1-M^2}}
K_0\left(r(\lambda)\left(\frac{x^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)
={\cal F}\left[\frac{e^{\dis -z\sqrt{D(\omega,\lambda)} }}
{2\sqrt{D(\omega,\lambda)}}\right],$$
where ${\cal F}$ denotes the Fourier transform, or
$$\frac{e^{d(\lambda)x}}{\sqrt{1-M^2}}
K_0\left(r(\lambda)\left(\frac{x^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)$$
$$=\int_{-\infty}^\infty e^{ix\omega}
\frac{e^{\dis -z\left((1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)}}d\omega.$$
\end{lemma}

\indent
{\bf Proof.}\ First, we represent $D\left(\omega,\lambda\right)$ as
$$D\left(\omega,\lambda\right)=\left(1-M^2\right)\omega^2+2i\frac{\lambda}{U}M^2\omega
+M^2\left(\frac{\lambda}{U}\right)^2$$
$$=\left(\omega\sqrt{1-M^2}+i\frac{\lambda M^2}{U\sqrt{1-M^2}}
\right)^2+\left(\frac{\lambda M^2}{U\sqrt{1-M^2}}\right)^2
+M^2\left(\frac{\lambda}{U}\right)^2$$
$$=(1-M^2)(\omega+id(\lambda))^2+r^2(\lambda).$$
\indent
Changing variables in equality (\cite{EMOT})
$$K_0\left(r(x^2+z^2)^{\frac{1}{2}}\right)
=\int_{-\infty}^\infty e^{ixu}
\frac{e^{\dis -z\left(u^2+r^2\right)^{\frac{1}{2}}} }{2\sqrt{u^2+r^2}}du,$$
we obtain
$$K_0\left(r(x^2+z^2)^{\frac{1}{2}}\right)
=\int_{-\infty}^\infty e^{ix\sqrt{1-M^2}\omega}
\frac{e^{\dis -z\left((1-M^2)\omega^2+r^2\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)\omega^2+r^2}}d\left(\sqrt{1-M^2}\omega\right),$$
and then
$$\frac{1}{\sqrt{1-M^2}}K_0\left(r\left(\frac{x^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)
=\int_{-\infty}^\infty e^{ix\omega}
\frac{e^{\dis -z\left((1-M^2)\omega^2+r^2\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)\omega^2+r^2}}d\omega.$$
\indent
We transform the equality above by integrating function
$$g(x,w)=e^{ixw}\frac{e^{\dis -z\left((1-M^2)w^2+r^2\right)^{\frac{1}{2}}} }
{\sqrt{(1-M^2)w^2+r^2}},\hspace{0.1in}w\in\C,$$
analytic with respect to $w$, over the piecewise linear contour
$$\left[-C, C, C+id, -C+id\right]\in \C,\hspace{0.1in}\mbox{with}\hspace{0.05in}
C\in\R, C>0,\ d\in\C,\ \mbox{Re}d>0.$$
Then we obtain
\begin{equation}\label{ContourIntegral}
\int_{-C}^Cg(x,w)dw+\int_{C}^{C+id}g(x,w)dw+\int_{C+id}^{-C+id}g(x,w)dw
+\int_{-C+id}^{-C}g(x,w)dw=0.
\end{equation}
\indent
For $C$ large enough we have the following estimates for $w=u+iv \in \left[-C, -C+id\right]$,
and $w \in \left[C, C+id\right]$
$$\left|e^{ix(u+iv)}\right|<e^{|x|\cdot\mbox{\scriptsize Re}d},
\hspace{0.1in}\left|\sqrt{(1-M^2)w^2+r^2}\right|>\sqrt{1-M^2}\frac{C}{2},$$
$$\left|e^{\dis -z\left((1-M^2)w^2+r^2\right)^{\frac{1}{2}}}\right|
<e^{-z\sqrt{1-M^2}\frac{C}{2}},$$
and therefore for $z>0$
$$\left|\int_{C}^{C+id}g(x,w)dw\right|,\left|\int_{-C}^{-C+id}g(x,w)dw\right|\to 0
\hspace{0.1in}\mbox{as}\hspace{0.05in}C\to\infty.$$
\indent
Using the last estimate in (\ref{ContourIntegral}) we obtain equality
$$\int_{-\infty}^\infty e^{ix\omega}
\frac{e^{\dis -z\left((1-M^2)\omega^2+r^2(\lambda)\right)^{\frac{1}{2}}} }
{\sqrt{(1-M^2)\omega^2+r^2(\lambda)}}d\omega$$
$$=\int_{-\infty}^\infty e^{ix(\omega+id(\lambda))}
\frac{e^{\dis -z\left((1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)\right)^{\frac{1}{2}}} }
{\sqrt{(1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)}}d(\omega+id(\lambda)),$$
and finally
\begin{equation}\label{SFormula}
\frac{e^{d(\lambda)x}}{\sqrt{1-M^2}}
K_0\left(r(\lambda)\left(\frac{x^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)
\end{equation}
$$=\int_{-\infty}^\infty e^{ix\omega}
\frac{e^{\dis -z\left((1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)}}d\omega.$$
\qed

\indent
Using now Lemmas ~\ref{RootD} and ~\ref{Fourier} we consider a special representation
of the general solution of (\ref{LaplaceEquation}). Namely, using notations of
Lemma~\ref{Fourier}, and denoting
$$S(x,z,\lambda)=-\frac{e^{d(\lambda)x}}{\sqrt{1-M^2}}
K_0\left(r(\lambda)\left(\frac{x^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right),$$
we consider
\begin{equation}\label{LapSolution}
\begin{array}{ll}
{\dis \xi(x,z,\lambda)=\int_{-\infty}^\infty S(x-y,z,\lambda)v_a(y,\lambda)dy }
\vspace{0.1in}\\
{\dis =-\frac{e^{d(\lambda)x}}{\sqrt{1-M^2}}\int_{-\infty}^\infty e^{-d(\lambda)y}
K_0\left(r(\lambda)\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)v_a(y,\lambda)dy. }
\end{array}
\end{equation}

\begin{proposition}\label{LaplaceSolution}\ Function $\xi$ defined by formula
(\ref{LapSolution}) satisfies equation (\ref{LaplaceEquation}).
If
$$\xi(x,z,\lambda),\ \frac{\partial^2 \xi}{\partial x^2}(x,z,\lambda),\
\frac{\partial^2 \xi}{\partial z^2}(x,z,\lambda),\ |\eta|^2\xi(x,z,\lambda),\
|\eta|\frac{\partial\xi(x,z,\lambda)}{\partial x}\in L^1(\R_{\eta}),$$
where $\lambda=\sigma+i\eta$,
then the inverse Laplace transform of $\xi$, defined by the formula (\cite{Boc})
\begin{equation}\label{InverseLaplace}
\phi(x,z,t)=\frac{1}{2\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}e^{(\sigma+i\eta)t}
\xi(x,z,\sigma+i\eta)d\eta
\end{equation}
satisfies equation (\ref{LinearEquation}).
\end{proposition}

\indent
{\bf Proof.}\  To prove that $\xi$ defined above satisfies
equation (\ref{LaplaceEquation})
it suffices to prove that function $S$ satisfies the same equation. For $S$ we have
$$a^2_{\infty}\left(1-M^2\right)\frac{\partial^2 S}{\partial x^2}
+a^2_{\infty}\frac{\partial^2 S}{\partial z^2}
-\lambda^2 S-2M\lambda a_{\infty}\frac{\partial S}{\partial x}$$
$$=a^2_{\infty}\left[\frac{\partial^2 S}{\partial z^2}
+\left(1-M^2\right)\frac{\partial^2 S}{\partial x^2}
-M^2\left(\frac{\lambda}{U}\right)^2 S-2M^2\frac{\lambda}{U}\frac{\partial S}{\partial x}\right].$$
\indent
Using then formula (\ref{SFormula}), we obtain
$$\frac{\partial^2 S}{\partial z^2}
+\left(1-M^2\right)\frac{\partial^2 S}{\partial x^2}
-M^2\left(\frac{\lambda}{U}\right)^2 S-2M^2\frac{\lambda}{U}\frac{\partial S}{\partial x}$$
$$=-\int_{-\infty}^\infty e^{ix\omega}\left((1-M^2)(\omega+id)^2+r^2\right)
\frac{e^{\dis -z\left((1-M^2)(\omega+id)^2+r^2\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)(\omega+id)^2+r^2}}d\omega$$
$$+\int_{-\infty}^\infty e^{ix\omega}\left(1-M^2\right)\omega^2
\frac{e^{\dis -z\left((1-M^2)(\omega+id)^2+r^2\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)(\omega+id)^2+r^2}}d\omega$$
$$+\int_{-\infty}^\infty e^{ix\omega}M^2\left(\frac{\lambda}{U}\right)^2
\frac{e^{\dis -z\left((1-M^2)(\omega+id)^2+r^2\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)(\omega+id)^2+r^2}}d\omega$$
$$+\int_{-\infty}^\infty e^{ix\omega}2M^2\frac{\lambda}{U}i\omega
\frac{e^{\dis -z\left((1-M^2)(\omega+id)^2+r^2\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)(\omega+id)^2+r^2}}d\omega=0.$$
\indent
To prove that function $\phi$ defined by formula (\ref{InverseLaplace}) satisfies
equation (\ref{LinearEquation}) we apply the inverse Laplace transform to equality
$$a^2_{\infty}\left(1-M^2\right)\frac{\partial^2 \xi}{\partial x^2}
+a^2_{\infty}\frac{\partial^2 \xi}{\partial z^2}
-\lambda^2 \xi-2M\lambda a_{\infty}\frac{\partial
\xi}{\partial x}=0$$
and obtain equation (\ref{LinearEquation}) for $\phi$.
\qed

\section{Boundary Conditions.}
\label{Boundary}

\indent
In this section we reformulate the boundary conditions of section ~\ref{Introduction}
in terms of function $v_a(y,\lambda)$ from formula (\ref{LapSolution}).\\
\indent
To check the flow tangency condition (\ref{Flow_Tangency}) we use formulas (\ref{LapSolution})
and (\ref{SFormula}), and obtain
\begin{equation}\label{inversion}
\frac{\partial}{\partial z}\xi(x,z,\lambda)\Big|_{z=0}
=\frac{\partial}{\partial z}\int_{-\infty}^\infty
S(x-y,z,\lambda)v_a(y,\lambda)dy\Big|_{z=0}
\end{equation}
$$=-\frac{\partial}{\partial z}\int_{-\infty}^\infty v_a(y,\lambda)dy
\int_{-\infty}^\infty e^{i(x-y)\omega}
\frac{e^{\dis -z\left((1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)}}d\omega\Big|_{z=0}$$
$$=\frac{1}{2}\int_{-\infty}^\infty e^{ix\omega}d\omega\int_{-\infty}^\infty
e^{-iy\omega}v_a(y,\lambda)dy=\pi\cdot v_a(x,\lambda),$$
which, after comparison with equality (\ref{Flow_Tangency}) leads to a unique choice
\begin{equation}\label{v=w}
v_a(x,\lambda)=\frac{1}{\pi}\widehat{w_a}(x,\lambda)\ \mbox{for}\ |x|<1.
\end{equation}
\indent
To satisfy the Kutta-Joukowski boundary condition (\ref{Kutta}) we should have
$${\dis \left(\frac{\partial\phi}{\partial t}
+U\frac{\partial\phi}{\partial x}\right)\Big|_{z=0}
=0\hspace{0.1in}\mbox{for}\ 1<|x|<A,}$$
or equality
$$\lambda\xi(x,0,\lambda)
+U\frac{\partial\xi}{\partial x}(x,0,\lambda)
=0\hspace{0.1in}\mbox{for}\ 1<|x|<A$$
for function $\xi$.\\
\indent
Substituting $\xi$ from formula (\ref{LapSolution}) into equality above
we obtain the following condition for $1<|x|<A$:
\begin{equation}\label{ZeroCondition}
\begin{array}{ll}
{\dis 0=\left(\lambda+U\frac{\partial}{\partial x}\right)\xi(x,0,\lambda) }
\vspace{0.1in}\\
{\dis =-\left(\lambda+U\frac{\partial}{\partial x}\right)
\frac{e^{d(\lambda)x}}{\sqrt{1-M^2}}\int_{-\infty}^\infty e^{-d(\lambda)y}
K_0\left(r(\lambda)\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)
v_a(y,\lambda)dy\Big|_{z=0}. }
\end{array}
\end{equation}
\indent
To reformulate the last condition as an integral equation we
use condition (\ref{v=w}), and define
$$g_a(x,\lambda)=\frac{e^{d(\lambda)x}}{\pi}\int_{-1}^1 e^{-d(\lambda)y}
R(x-y,\lambda)\widehat{w_a}(y,\lambda)dy\ \mbox{for}\ 1<|x|<A,$$
with kernel $R(x,\lambda)$ defined by the formula
\begin{equation}\label{RKernel}
R(x,\lambda)=\left[\left(\lambda+Ud(\lambda)\right)K_0
\left(\frac{r(\lambda)|x|}{\sqrt{1-M^2}}\right)
+U\frac{\partial}{\partial x}K_0\left(\frac{r(\lambda)|x|}{\sqrt{1-M^2}}\right)\right].
\end{equation}
Then condition (\ref{ZeroCondition}) will be satisfied if $v_a$ will
satisfy the following integral equation
$$e^{d(\lambda)x}\int_{|y|>1} e^{-d(\lambda)y}
R(x-y,\lambda)v_a(y,\lambda)dy=-g_a(x,\lambda)\ \mbox{for}\ 1<|x|<A.$$
\indent
Further simplifying the equation above we consider
$h_a(y,\lambda):=e^{-d(\lambda)y}\cdot v_a(y,\lambda)$ as an
unknown function, and rewrite it as
\begin{equation}\label{MainEquation}
\int_{|y|>1}R(x-y,\lambda)h_a(y,\lambda)dy
=f_a(x,\lambda)\hspace{0.1in}\mbox{for}\ 1<|x|<A,
\end{equation}
where $f_a(x,\lambda)=-e^{-d(\lambda)x}\cdot\chi_A(x) g_a(x,\lambda)$
is defined for
$$\left\{(x,\lambda)\in \R\times\C:
|x|>1, \mbox{Re}\lambda\in[\sigma_1,\sigma_2]\right\}$$
by the formula
\begin{equation}\label{fDefinition}
f_a(x,\lambda)=-\frac{\chi_A(x)}{\pi}\int_{-1}^1 e^{-d(\lambda)y}
R(x-y,\lambda)\widehat{w_a}(y,\lambda)dy
\end{equation}
with
$$\chi_A(x)=\left\{\begin{array}{ll}
1\hspace{0.1in}\mbox{if}\ x\in [-A,A]\setminus[-1,1],\vspace{0.1in}\\
0\hspace{0.1in}\mbox{otherwise}.
\end{array}\right.$$

\section{Cofinite Hilbert transform.}
\label{CofiniteTransform}

\indent
As a first step in the analysis of solvability of (\ref{MainEquation}) we prove
solvability for the operator, closely related to operator
${\cal R}_{\lambda}$ from (\ref{MainEquation}), and which in analogy with the Tricomi's
definition of the finite Hilbert transform \cite{Tr} we call the
{\it cofinite Hilbert transform}.\\
\indent
We define the cofinite Hilbert transform on the set of functions on
$$I^c(1)=\R\setminus[-1,1]$$
by the formula
\begin{equation}\label{POperator}
{\cal P}[h](x)=\frac{1}{\pi}\int_{|y|>1}\frac{h(y)}{y-x}dy\
\mbox{for}\ |x|>1,
\end{equation}
where the integral
$$\int_{|y|>1}=\int_{-\infty}^{-1}+\int_1^{\infty}$$
is understood in the sense of Cauchy's principal value.\\
\indent
In the proposition below we prove solvability for the nonhomogeneous integral
equation with operator ${\cal P}$ in weighted spaces
$${\cal L}^p\left(I^c(1)\right)=\left\{f:\int_{|x|>1}|x|^{p-2}\left|f(x)\right|^pdx
<\infty\right\}$$
with
$$\|f\|_{{\cal L}^p\left(I^c(1)\right)}
=\left(\int_{|x|>1}|x|^{p-2}\left|f(x)\right|^pdx\right)^{1/p}.$$

\begin{proposition}\label{Cofinite}\ For any function $f \in {\cal L}^q\left(I^c(1)\right)$
with ${\dis q>\frac{4}{3} }$ there exists a solution $h$ of equation
\begin{equation}\label{Pequation}
{\cal P}[h]=f,
\end{equation}
such that $h \in {\cal L}^p\left(I^c(1)\right)$ for any ${\dis p<\frac{4}{3} }$.
\end{proposition}

\indent
{\bf Proof.}\ We consider the following diagram of transformations
\begin{equation}\label{Diagram}
\begin{array}{ccc}
L^p\left(I(1)\right)&\mapright{-\cal T}&L^p\left(I(1)\right)
\vspace{0.1in}\\
\mapdown{\Theta}&&\mapdown{\Theta}\vspace{0.1in}\\
{\cal L}^p\left(I^c(1)\right)&\mapright{\cal P}&{\cal L}^p\left(I^c(1)\right),
\end{array}
\end{equation}
where ${\cal T}$ is the finite Hilbert transform, ${\cal P}$ is the
cofinite Hilbert transform, and
$$\Theta:L^p\left(I(1)\right)\to {\cal L}^p\left(I^c(1)\right)$$
is defined by the formula
\begin{equation}\label{ThetaFormula}
\Theta[f](x)=\frac{1}{x}f\left(\frac{1}{x}\right).
\end{equation}
\indent
To prove that the maps in diagram (\ref{Diagram}) are well defined we use
equality
$$\left\|\Theta[f]\right\|_{{\cal L}^p}^p
=\int_{|x|>1}|x|^{p-2}\left|\Theta[f](x)\right|^pdx
=\int_{|x|>1}|x|^{p-2}\frac{\left|f\left(\frac{1}{x}\right)\right|^p}{|x|^p}dx$$
$$=-\int_1^{-1}\left|f\left(t\right)\right|^pdt
=\|f\|_p^p,$$
and notice that for
$$\Theta^*:{\cal L}^p\left(I^c(1)\right)\to L^p\left(I(1)\right)$$
defined by the same formula
$$\Theta^*[f](x)=\frac{1}{x}f\left(\frac{1}{x}\right),$$
we have
\begin{equation}\label{Thetaand*}
\Theta\circ\Theta^*\left[f\right](x)=\Theta
\left[\frac{1}{y}f\left(\frac{1}{y}\right)\right](x)
=\frac{1}{x}\cdot xf(x)=f(x).
\end{equation}
\indent
Diagram (\ref{Diagram}) is commutative, as can be seen from equality
$${\cal P}\left[\Theta[f]\right](x)=\frac{1}{\pi}\int_{|y|>1}
\frac{f\left(\frac{1}{y}\right)}{y(y-x)}dy
=\frac{1}{\pi}\int_{-1}^1t\frac{f\left(t\right)}{(\frac{1}{t}-x)t^2}dt$$
$$=\frac{1}{\pi x}\int_{-1}^1\frac{f\left(t\right)}{\frac{1}{x}-t}dt
=\Theta\left[-{\cal T}[f]\right].$$
\indent
To "invert" operator ${\cal P}$ we use commutativity of diagram (\ref{Diagram}),
relation (\ref{Thetaand*}), and operator (\cite{So},\cite{Tr})
$${\cal T}^{-1}:L^{\frac{4}{3}+}\left(I(1)\right)\to L^{\frac{4}{3}-}\left(I(1)\right),$$
defined by the formula
$${\cal T}^{-1}[g](x)=-\frac{1}{\pi}\int_{-1}^{1}\sqrt{\frac{1-y^2}{1-x^2}}
\frac{g(y)}{y-x}dy,$$
and satisfying
$${\cal T}\circ{\cal T}^{-1}[f]=f.$$
\indent
Namely, we define operator
$${\cal P}^{-1}:{\cal L}^{\frac{4}{3}+}\left(I^c(1)\right)\to
{\cal L}^{\frac{4}{3}-}\left(I^c(1)\right)$$
by the formula
$${\cal P}^{-1}[f]=-\Theta\circ{\cal T}^{-1}\circ\Theta^*[f].$$
\indent
Then
$${\cal P}\circ{\cal P}^{-1}[f]=-{\cal P}\circ\Theta\circ{\cal T}^{-1}\circ\Theta^*[f]
=\Theta\circ{\cal T}\circ{\cal T}^{-1}\circ\Theta^*[f]
=\Theta\circ\Theta^*[f]=f,$$
and we obtain the statement of the proposition for
$$h={\cal P}^{-1}[f].$$
\indent
To find an explicit formula for ${\cal P}^{-1}$ we use explicit formulas for $\Theta$
and ${\cal T}^{-1}$, and obtain
\begin{equation}\label{P-1Forfula}
\begin{array}{ll}
{\dis {\cal P}^{-1}[f](x)=\frac{1}{\pi x}\int_{-1}^{1}\sqrt{\frac{1-y^2}{1-1/x^2}}
\cdot\frac{f(1/y)}{y(y-1/x)}dy }\vspace{0.1in}\\
{\dis =\frac{|x|}{\pi}\int_{-1}^{1}\sqrt{\frac{1-y^2}{x^2-1}}
\left[\frac{1}{y}f\left(\frac{1}{y}\right)\right]\frac{dy}{xy-1}. }
\end{array}
\end{equation}
\qed

\indent
{\bf Remark.}\hspace{0.1in}Following \cite{Tr} we notice that solution of equation
(\ref{Pequation}) is unique in ${\cal L}^2\left(I^c(1)\right)$, but is not unique in larger
spaces. Namely, function
$$h(x)=\frac{1}{\sqrt{x^2-1}}$$
is a solution, and the only one in ${\cal L}^{2-}\left(I^c(1)\right)$ up to the
linear dependence, of the homogeneous equation
$${\cal P}[h]=0.$$
\qed

\section{Solvability of equation (\ref{MainEquation}).}
\label{Solvability}

\indent
From the asymptotic expansions of $K_0(\zeta)$
(see \cite{EMOT}, \cite{GR}) we obtain the following representations of the function
$R(x,\lambda)$ for $\lambda$ such that $\mbox{Re}\lambda\in[\sigma_1,\sigma_2]$
with $\sigma_1>\sigma_a$:
\begin{equation}\label{RRepresentation}
\begin{array}{ll}
{\dis R(x,\lambda)=-\frac{U}{x}+\lambda\log\left(\lambda|x|\right)\alpha(\lambda|x|)
+\lambda\beta\left(\lambda|x|\right)+\gamma\left(\lambda|x|\right)
\hspace{0.1in}\mbox{for}\ |\lambda x|\leq B, }\vspace{0.1in}\\
{\dis R(x,\lambda)=\lambda\delta\left(\lambda|x|\right)\frac{e^{-(\sigma+i\eta)|x|}}
{\sqrt{|\lambda|\cdot|x|}}\hspace{0.1in}\mbox{for}\ |\lambda x|> B, }
\end{array}
\end{equation}
where $\alpha(\zeta)$, $\beta(\zeta)$, $\gamma(\zeta)$, and $\delta(\zeta)$ are bounded
analytic functions on $\mbox{Re}\zeta>\epsilon>0$ and $B>0$ is some constant.\\
\indent
Using representations (\ref{RRepresentation}) we introduce function $M(x,\lambda)$,
analytic with respect to $\lambda\in \left\{\mbox{Re}\lambda>\sigma_a\right\}$,
uniquely defined by (\ref{RRepresentation}), and such that
$${\dis R(x,\lambda)=-\frac{U}{x}+M(x,\lambda). }$$
We consider then operators
$${\cal M}_{\lambda}[f](x)=\int_{|y|>1}\chi_A(x)M(x-y,\lambda)f(y)dy,$$
and
$${\cal R}_{\lambda}=\pi U\cdot{\cal P}+{\cal M}_{\lambda}.$$
\indent
In the next proposition we prove compactness of the operator
${\dis \frac{1}{\pi U}{\cal M}_{\lambda}\circ{\cal P}^{-1} }$
on ${\cal L}^{2}\left(I^c(1)\right)$.\\

\begin{proposition}\label{RFredholm}\ For any fixed $\lambda\in\C$ operator
${\dis {\cal N}_{\lambda}=\frac{1}{\pi U}{\cal M}_{\lambda}\circ{\cal P}^{-1} }$
is compact on ${\cal L}^{2}\left(I^c(1)\right)$, and therefore operator
\begin{equation}\label{GOperator}
{\cal G}_{\lambda}={\cal R}_{\lambda}\circ{\cal P}^{-1}
=\left(\pi U\cdot{\cal P}+{\cal M}_{\lambda}\right)\circ{\cal P}^{-1}
=\pi U\left({\cal I}+{\cal N}_{\lambda}\right),
\end{equation}
where ${\cal I}$ is the identity operator, is a Fredholm operator on
${\cal L}^{2}\left(I^c(1)\right)=L^{2}\left(I^c(1)\right)$. In addition, kernel
$N(x,y,\lambda)$ of the operator ${\cal N}_{\lambda}$ admits estimate
\begin{equation}\label{NEstimate}
\int_{\R^2}|N(x,y,\lambda)|^2dxdy<C|\lambda\log{\lambda}|^2
\end{equation}
with constant $C$ independent of $\lambda$.
\end{proposition}

\indent
{\bf Proof.}\ Using formula (\ref{P-1Forfula}) for ${\cal P}^{-1}$, we obtain
$${\cal N}_{\lambda}[g](x)={\cal M}_{\lambda}
\left[\frac{|x|}{\pi^2 U}\int_{-1}^{1}\sqrt{\frac{1-u^2}{x^2-1}}
\left[\frac{1}{u}g\left(\frac{1}{u}\right)\right]\frac{du}{xu-1}\right]$$
$$={\cal M}_{\lambda}\left[\frac{|x|}{\pi^2 U}\int_{|y|>1}\sqrt{\frac{1-\frac{1}{y^2}}{x^2-1}}
y^2g(y)\frac{dy}{y^2(x-y)}\right]$$
$$=\frac{\chi_A(x)}{\pi^2 U}\int_{|u|>1}M(x-u,\lambda)du
\int_{|y|>1}\frac{|u|\sqrt{y^2-1}}{|y|\sqrt{u^2-1}}g(y)\frac{dy}{(u-y)}$$
$$=\frac{\chi_A(x)}{\pi^2 U}\int_{|y|>1}g(y)dy
\int_{|u|>1}M(x-u,\lambda)\frac{|u|\sqrt{y^2-1}}{|y|\sqrt{u^2-1}}
\frac{du}{(u-y)}=\int_{|y|>1}N(x,y,\lambda)g(y)dy$$
with kernel
$$N(x,y,\lambda)=\frac{\chi_A(x)}{\pi^2 U}
\int_{|u|>1}M(x-u,\lambda)\frac{|u|\sqrt{y^2-1}}{|y|\sqrt{u^2-1}}
\frac{du}{(u-y)}.$$
\indent
To prove compactness of operator ${\cal N}_{\lambda}$ we use representation
$$N(x,y,\lambda)=\frac{1}{\pi^2 U}\left[N_1(x,y,\lambda)+N_2(x,y,\lambda)\right],$$
with
$$N_1(x,y,\lambda)=\frac{\chi_A(x)}{|y|}\int_{|u|>1}M(x-u,\lambda)|u|
\frac{du}{(u-y)},$$
and
$$N_2(x,y,\lambda)=\frac{\chi_A(x)}{|y|}\int_{|u|>1}M(x-u,\lambda)
\frac{|u|\left(\sqrt{y^2-1}-\sqrt{u^2-1}\right)}{\sqrt{u^2-1}}
\frac{du}{(u-y)}$$
$$=-\frac{\chi_A(x)}{|y|}\int_{|u|>1}M(x-u,\lambda)
\frac{|u|\left(y+u\right)du}{\left(\sqrt{y^2-1}+\sqrt{u^2-1}\right)\sqrt{u^2-1}},$$
and prove Hilbert-Schmidt property (cf.\cite{L})
of kernels $N_1(x,y,\lambda)$ and $N_2(x,y,\lambda)$.\\
\indent
For $N_1(x,y,\lambda)$ we notice that for fixed $x$ satisfying $1<|x|<A$
$$\int_{|u|>1}M(x-u,\lambda)|u|\frac{du}{(u-y)}$$
is a multiple of the Hilbert transform of an $L^{2}\left(I^c(1)\right)$ - function
$M(x-u,\lambda)|u|$ with
$$\|M(x-u,\lambda)|u|\|_{L^{2}_u}<\infty.$$
Therefore we have
\begin{equation}\label{N1Estimate}
\begin{array}{lll}
{\dis \int_{1<|x|<A}dx\int_{|y|>1}dy\left|N_1(x,y,\lambda)\right|^2 }\vspace{0.1in}\\
{\dis =\int_{1<|x|<A}dx\int_{|y|>1}dy\left|\frac{1}{|y|}\int_{|u|>1}M(x-u,\lambda)|u|
\frac{du}{(u-y)}\right|^2 }\vspace{0.1in}\\
{\dis < C\int_{1<|x|<A}dx\left\|M(x-u,\lambda)|u|\right\|_{L^{2}_u}^2<\infty. }
\end{array}
\end{equation}
\indent
For $N_2(x,y,\lambda)$ we have
\begin{equation}\label{N2Integral}
\begin{array}{llll}
{\dis \int_{1<|x|<A}dx\int_{|y|>1}dy\left|N_2(x,y,\lambda)\right|^2 }\vspace{0.1in}\\
{\dis =\int_{1<|x|<A}dx\int_{|y|>1}\frac{dy}{|y|^2}\left|\int_{|u|>1}M(x-u,\lambda)
\frac{|u|\left(y+u\right)du}{\left(\sqrt{y^2-1}+\sqrt{u^2-1}\right)\sqrt{u^2-1}}\right|^2 }
\vspace{0.1in}\\
{\dis \leq 2\int_{1<|x|<A}dx\int_{|y|>1}\frac{dy}{|y|^2}
\left|\int_0^{\infty}M(x-\sqrt{t^2+1},\lambda)
\frac{\left(y+\sqrt{t^2+1}\right)dt}{\left(\sqrt{y^2-1}+t\right)}\right|^2 }
\vspace{0.1in}\\
{\dis +2\int_{1<|x|<A}dx\int_{|y|>1}\frac{dy}{|y|^2}
\left|\int_0^{\infty}M(x+\sqrt{t^2+1},\lambda)
\frac{\left(y-\sqrt{t^2+1}\right)dt}{\left(\sqrt{y^2-1}+t\right)}\right|^2, }
\end{array}
\end{equation}
where we changed variable to $t=\sqrt{u^2-1}$.\\
\indent
Both integrals of the right hand side of (\ref{N2Integral}) are estimated analogously,
therefore we will present an estimate of the first of them only.\\
\indent
For $1<|x|<A$ and $|y|>2$ we have inequality
\begin{equation}\label{y-tEstimate}
\left|\frac{y+\sqrt{t^2+1}}{\sqrt{y^2-1}+t}\right|<C
\end{equation}
for some $C$ independent of $y$, and therefore, using representations (\ref{RRepresentation}),
we obtain
\begin{equation}\label{N2Estimate1}
\begin{array}{ll}
{\dis \int_{1<|x|<A}dx\int_{|y|>2}\frac{dy}{|y|^2}
\left|\int_0^{\infty}M(x-\sqrt{t^2+1},\lambda)
\frac{\left(y+\sqrt{t^2+1}\right)dt}{\left(\sqrt{y^2-1}+t\right)}\right|^2 }\vspace{0.1in}\\
{\dis \leq C^2\int_{1<|x|<A}dx\int_{|y|>2}\frac{dy}{|y|^2}
\left|\int_0^{\infty}M(x-\sqrt{t^2+1},\lambda)dt\right|^2<\infty. }
\end{array}
\end{equation}
\indent
For $1<|x|<A$, $1<|y|<2$, and $t>A+B$ we again use inequality (\ref{y-tEstimate}) and obtain
\begin{equation}\label{N2Estimate2}
\begin{array}{ll}
{\dis \int_{1<|x|<A}dx\int_{1<|y|<2}\frac{dy}{|y|^2}
\left|\int_{A+B}^{\infty}M(x-\sqrt{t^2+1},\lambda)
\frac{\left(y+\sqrt{t^2+1}\right)dt}{\left(\sqrt{y^2-1}+t\right)}\right|^2 }\vspace{0.1in}\\
{\dis \leq C^2\int_{1<|x|<A}dx\int_{|y|>2}\frac{dy}{|y|^2}
\left|\int_0^{\infty}M(x-\sqrt{t^2+1},\lambda)dt\right|^2<\infty. }
\end{array}
\end{equation}
\indent
For $1<|x|<A$, $1<|y|<2$, and $t<A+B$ we have
$$\int_{1<|x|<A}dx\int_{1<|y|<2}\frac{dy}{|y|^2}\left|\int_0^{A+B}M(x-\sqrt{t^2+1},\lambda)
\frac{\left(y+\sqrt{t^2+1}\right)dt}{\left(\sqrt{y^2-1}+t\right)}\right|^2$$
$$< C\int_{1<|x|<A}dx\int_{1<|y|<2}dy\left|\int_0^{A+B}
\frac{M(x-\sqrt{t^2+1},\lambda)dt}{\left(\sqrt{y^2-1}+t\right)}\right|^2$$
$$< C|\lambda\log{\lambda}|^2\int_{1<|x|<A}dx\int_{1<|y|<2}dy\left|\int_0^{A+B}
\frac{dt}{\left(\sqrt{y^2-1}+t\right)}\right|^2$$
$$+ C|\lambda|^2\int_{1<|x|<A}dx\int_{1<|y|<2}dy\left|\int_0^{A+B}
\frac{\log{|x-\sqrt{t^2+1}|}}{\left(\sqrt{y^2-1}+t\right)}dt\right|^2$$
$$<C|\lambda|^2\left(|\log{\lambda}|^2+\int_{1<|x|<A}dx\int_{1<|y|<2}dy\left|\int_0^{A+B}
\frac{\log{|x-\sqrt{t^2+1}|}}{\left(\sqrt{y^2-1}+t\right)}dt\right|^2\right),$$
where we used representation
$$M(x-\sqrt{t^2+1},\lambda)=\lambda\left(\log{\lambda}+\log{|x-\sqrt{t^2+1}|}\right)
\alpha(\lambda|x-\sqrt{t^2+1}|)$$
$$+\lambda \beta(\lambda|x-\sqrt{t^2+1}|)+\gamma(\lambda|x-\sqrt{t^2+1}|)$$
for $1<|x|<A$ and $0<t<A+B$, which is a corollary of (\ref{RRepresentation}).\\
\indent
To estimate the last integral we represent it as
$$\int_{1<|x|<A}dx\int_{1<|y|<2}dy\left|\int_0^{A+B}
\frac{\log{|x-\sqrt{t^2+1}|}}{\left(\sqrt{y^2-1}+t\right)}dt\right|^2$$
$$=\int_{1<|x|<A}dx\int_{1<|y|<2}dy\left|\int_{S(x,y)}
\frac{\log{|x-\sqrt{t^2+1}|}}{\left(\sqrt{y^2-1}+t\right)}dt\right|^2$$
$$+\int_{1<|x|<A}dx\int_{1<|y|<2}dy
\left|\int_{\left\{\left[0,A+B\right]\setminus S(x,y)\right\}}
\frac{\log{|x-\sqrt{t^2+1}|}}{\left(\sqrt{y^2-1}+t\right)}dt\right|^2$$
where $S(x,y)=\left\{t:\left|x-\sqrt{t^2+1}\right|
\geq\frac{1}{2}\left(x-1\right)\sqrt{y^2-1}\right\}$.\\
\indent
Then for $S(x,y)$ we have
$$\int_{1<|x|<A}dx\int_{1<|y|<2}dy\left|\int_{S(x,y)}
\frac{\log{|x-\sqrt{t^2+1}|}}{\left(\sqrt{y^2-1}+t\right)}dt\right|^2$$
$$\leq C\int_{1<|x|<A}dx\int_{1<|y|<2}dy\left|\int_0^{A+B}
\frac{\log{|x-1|}+\log{(\sqrt{y^2-1})}}{\left(\sqrt{y^2-1}+t\right)}dt\right|^2$$
$$\leq C\int_{1<|x|<A}dx\int_{1<|y|<2}dy\log^2{\sqrt{y^2-1}}\cdot\left(\log{|x-1|}
+\log{\sqrt{y^2-1}}\right)^2<C<\infty.$$
\indent
For $t\in \left[0,A+B\right]\setminus S(x,y)$ we have
$$1+\frac{t^2}{2}\geq \sqrt{t^2+1}> x-\frac{1}{2}\left(x-1\right)\sqrt{y^2-1},$$
and therefore
$$\frac{t^2}{2}\geq \left(x-1\right)\left[1-\frac{1}{2}\sqrt{y^2-1}\right],$$
or
$$t\geq C\sqrt{x-1}.$$
\indent
Using the last inequality we obtain
$$\left|dt\right|\leq\left|\frac{\sqrt{t^2+1}}{t}du\right|
\leq C\left|\frac{1}{\sqrt{x-1}}du\right|,$$
and switching to variable $u=\sqrt{t^2+1}$ for $\left[0,A+B\right]\setminus S(x,y)$,
we obtain
$$\int_{1<|x|<A}dx\int_{1<|y|<2}dy
\left|\int_{\left\{\left[0,A+B\right]\setminus S(x,y)\right\}}
\frac{\log{|x-\sqrt{t^2+1}|}}{\left(\sqrt{y^2-1}+t\right)}dt\right|^2$$
$$\leq C\int_{1<|x|<A}dx\int_{1<|y|<2}dy
\left|\int_{x-\frac{1}{2}(x-1)\sqrt{y^2-1}}^{x+\frac{1}{2}(x-1)\sqrt{y^2-1}}\
\frac{\log{|u-x|}}{\sqrt{y^2-1}\cdot\sqrt{x-1}}du\right|^2$$
$$\leq C\int_{1<|x|<A}dx\int_{1<|y|<2}dy
\left|\frac{(x-1)\sqrt{y^2-1}\cdot\left(\log{(x-1)}+\log{\sqrt{y^2-1}}\right)}
{\sqrt{y^2-1}\cdot\sqrt{x-1}}\right|^2<C<\infty.$$
\indent
Combining the last two estimates above we obtain
$$\int_{1<|x|<A}dx\int_{1<|y|<2}dy\left|\int_0^{A+B}
\frac{\log{|x-\sqrt{t^2+1}|}}{\left(\sqrt{y^2-1}+t\right)}dt\right|^2<C<\infty,$$
and therefore
\begin{equation}\label{N2Estimate}
\int_{1<|x|<A}dx\int_{1<|y|<2}\frac{dy}{|y|^2}\left|\int_0^{A+B}M(x-\sqrt{t^2+1},\lambda)
\frac{\left(y+\sqrt{t^2+1}\right)dt}{\left(\sqrt{y^2-1}+t\right)}\right|^2
<C|\lambda\log{\lambda}|^2.
\end{equation}
\indent
To prove estimate (\ref{NEstimate}) we use the following lemma.\\

\begin{lemma}\label{MEstimates}\ The following estimates hold for $1<|x|<A$
and $\mbox{Re}\lambda\in [\sigma_1,\sigma_2]$
\begin{equation}\label{MNorm}
\begin{array}{ll}
{\dis \int_{\R}|M(x-u,\lambda)|^2u^2du<C|\lambda|^{1+\epsilon}\ \mbox{for arbitrary}\
\epsilon>0, }\vspace{0.1in}\\
{\dis \left|\int_0^{\infty}M(x-\sqrt{t^2+1},\lambda)dt\right|<C \sqrt{|\lambda|}. }
\end{array}
\end{equation}
\end{lemma}

\indent
{\bf Proof.}\ Using representation (\ref{RRepresentation})
for $1<|x|<A$ and $|\lambda(x-u)|\leq B$ we obtain
$$M(x-u,\lambda)\cdot u=\left[\lambda\log\left(\lambda|x-u|\right) \alpha(\lambda|x-u|)
+\lambda \beta\left(\lambda|x-u|\right)+\gamma\left(\lambda|x-u|\right)\right]u,$$
and therefore
$$\int_{|x-u|\leq B/|\lambda|}|M(x-u,\lambda)|^2u^2du$$
$$<C\int_{|x-u|\leq B/|\lambda|}\left[|\lambda|^2\left(|\log{\lambda}|^2+\log^2|x-u|\right)
|\alpha(\lambda|x-u|)|^2\right.$$
$$\left.+|\lambda|^2 |\beta\left(\lambda|x-u|\right)|^2
+|\gamma\left(\lambda|x-u|\right)|^2\right]u^2du
<C|\lambda||\log{\lambda}|^2.$$
\indent
For $1<|x|<A$ and $|\lambda(x-u)|\geq B$ from (\ref{RRepresentation}) we have
$$M(x-u,\lambda)=\lambda \delta\left(\lambda|x-u|\right)\frac{e^{-(\sigma+i\eta)|x-u|}}
{\sqrt{|\lambda|\cdot|x-u|}},$$
and therefore
$$\int_{|x-u|\geq B/|\lambda|}|M(x-u,\lambda)|^2u^2dx$$
$$<C\int_{\R}|\lambda|^{1+\epsilon}|\delta\left(\lambda|x-u|\right)|^2
\frac{e^{-2\sigma|x-u|}u^2du}{\left(|\lambda(x-u)|\right)^{\epsilon}|x-u|^{1-\epsilon}}
<C|\lambda|^{1+\epsilon}$$
for any $\epsilon>0$.\\
\indent
Combining the estimates above we obtain the first estimate from (\ref{MNorm}).\\
\indent
For the second integral in (\ref{MNorm}) we use representation (\ref{RRepresentation}),
and obtain for $1<|x|<A$ and $|\lambda(x-\sqrt{t^2+1})|\leq B$
$$\left|\int_{|x-\sqrt{t^2+1}|\leq B/|\lambda|}M(x-\sqrt{t^2+1},\lambda)dt\right|$$
$$\leq \int_{|x-\sqrt{t^2+1}|\leq B/|\lambda|}
\Big|\lambda\log\left(\lambda|x-\sqrt{t^2+1}|\right)\alpha(\lambda|x-\sqrt{t^2+1}|)$$
$$+\lambda \beta\left(\lambda|x-\sqrt{t^2+1}|\right)
+\gamma\left(\lambda|x-\sqrt{t^2+1}|\right)\Big|dt\leq C|\log{\lambda}|,$$
where we used the fact that the length of the interval of integration
is bounded by ${\dis \frac{C}{|\lambda|}}$ for some $C>0$.\\
\indent
For $1<|x|<A$ and $|\lambda(x-\sqrt{t^2+1})|> B$ using representation (\ref{RRepresentation})
we obtain
$$\left|\int_{|x-\sqrt{t^2+1}|> B/|\lambda|}M(x-\sqrt{t^2+1},\lambda)dt\right|$$
$$< C\int_{\R}\lambda \delta\left(\lambda|x-\sqrt{t^2+1}|\right)
\frac{e^{-(\sigma+i\eta)|x-\sqrt{t^2+1}|}dt}{\sqrt{|\lambda|\cdot|x-\sqrt{t^2+1}|}}
<C \sqrt{|\lambda|}.$$
\indent
Combining the two estimates above we obtain the second estimate of (\ref{MNorm}).
\qed

\indent
Using now estimates (\ref{MNorm}) from the lemma above in estimates (\ref{N1Estimate}),
(\ref{N2Estimate1}), and (\ref{N2Estimate2}) and combining them with estimate (\ref{N2Estimate})
we obtain estimate (\ref{NEstimate}) of Proposition~\ref{RFredholm}.
\qed

\indent
Proposition~\ref{RFredholm} allows us to reduce the question of solvability of
(\ref{MainEquation}) to the solvability of corresponding equation for ${\cal G}_{\lambda}$.
Namely, calling those $\lambda$ for which operator ${\cal G}_{\lambda}$ is not invertible
by {\it characteristic values of} ${\cal G}_{\lambda}$, we have

\begin{proposition}\label{RSolvability}\ If $\lambda_0$ is not a characteristic value
of ${\cal G}_{\lambda}$, then for arbitrary function $f \in {\cal L}^2\left(I^c(1)\right)$
and $\lambda=\lambda_0$ there exists a solution $h$ of equation (\ref{MainEquation}) such
that $h \in {\cal L}^p\left(I^c(1)\right)$ for any ${\dis p<\frac{4}{3} }$.
\end{proposition}

\indent
{\bf Proof.}\ Considering  a solution of
$${\cal G}_{\lambda}[g]={\cal R}_{\lambda}\circ{\cal P}^{-1}[g]=f$$
we define $h={\cal P}^{-1}[g]$, which satisfies equation (\ref{MainEquation})
and belongs to ${\cal L}^p\left(I^c(1)\right)$ for any
${\dis p<\frac{4}{3} }$ according to Proposition~\ref{Cofinite}.

\qed

\section{ The resolvent of operator ${\cal G}_{\lambda}$.}
\label{GLambdaResolvent}

\indent
In this section we construct the resolvent of the operator ${\cal G}_{\lambda}$
and show that it is a Fredholm operator also analytically depending on
$\lambda \in \left\{\mbox{Re}\lambda>\sigma_1\right\}$.\\
\indent
Let ${\cal T}:L^2(\R)\to L^2(\R)$ be an integral operator with kernel $T(x,y)$ satisfying
Hilbert-Schmidt condition. Following \cite{C}, we consider for operator ${\cal T}$
Hilbert's modification of the original Fredholm's determinants:
$${\cal D}_{T,m}\left(t_1,\dots,t_m\right)
=\left|
\begin{array}{cccc}
0&T(t_1,t_2)&\cdots&T(t_1,t_m)\\
T(t_2,t_1)&0&\cdots&T(t_2,t_m)\\
\vdots&&&\vdots\\
T(t_m,t_1)&\cdots&T(t_m,t_{m-1})&0
\end{array}
\right|,$$
\begin{equation}\label{D_T}
{\cal D}_{T}=1+\sum_{m=1}^{\infty}\delta_m
=1+\sum_{m=1}^{\infty}\frac{1}{m!}\int_{\R}\cdots\int_{\R}
{\cal D}_{T,m}\left(t_1,\dots,t_m\right)dt_1\cdots dt_m,
\end{equation}
$${\cal D}_{T,m}\left(\begin{array}{c}
x\\
y
\end{array}t_1,\dots,t_m\right)
=\left|
\begin{array}{cccc}
T(x,y)&T(x,t_1)&\cdots&T(x,t_m)\\
T(t_1,y)&0&\cdots&T(t_1,t_m)\\
\vdots&&&\vdots\\
T(t_m,y)&\cdots&T(t_m,t_{m-1})&0
\end{array}
\right|,$$
and
\begin{equation}\label{D_Tfunction}
\begin{array}{ll}
{\dis {\cal D}_{T}\left(\begin{array}{c}
x\\
y
\end{array}\right)
=T(x,y)+\sum_{m=1}^{\infty}\delta_m\left(\begin{array}{c}
x\\
y
\end{array}\right) }\vspace{0.1in}\\
{\dis =T(x,y)+\sum_{m=1}^{\infty}\frac{1}{m!}\int_{\R}\cdots\int_{\R}
{\cal D}_{T,m}\left(
\begin{array}{c}
x\\
y
\end{array}t_1,\dots,t_m\right)dt_1\cdots dt_m. }
\end{array}
\end{equation}
\indent
We start with the following proposition, which summarizes the results from \cite{C}
(cf. also \cite{M}), that will be used in the construction of the resolvent of
${\cal G}_{\lambda}$.

\begin{proposition}\label{Resolvent} (\cite{C})\ Let function
$T(x,y):\R^2\to \C$ satisfy Hilbert-Schmidt condition
$$\|T\|^2=\int_{\R^2}\left|T(x,y)\right|^2dxdy<\infty.$$
\indent
Then function
${\cal D}_{T}\left(\begin{array}{c}
x\\
y
\end{array}\right)\in L^2(\R^2)$ is well defined, and the following estimates hold:
\begin{equation}\label{DEstimate1}
\left|\delta_m\right|\leq\left(\frac{e}{m}\right)^{m/2}\|T\|^m,\
\left|{\cal D}_{T}\right|\leq e^{\frac{\|T\|^2}{2}},
\end{equation}
\begin{equation}\label{DEstimate2}
\left|{\cal D}_{T}\left(\begin{array}{c}
x\\
y
\end{array}\right)\right|
\leq e^{\frac{\|T\|^2}{2}}
\left(|T(x,y)|+\sqrt{e}\alpha(x)\beta(y)\right),
\end{equation}
where
$$\alpha^2(x)=\int_{\R}\left|T(x,t)\right|^2dt,\hspace{0.1in}
\beta^2(y)=\int_{\R}\left|T(t,y)\right|^2dt.$$
\indent
If ${\cal D}_{T}\neq 0$
then kernel
\begin{equation}\label{HResolvent}
H(x,y)=\left[{\cal D}_{T}\right]^{-1}
\cdot{\cal D}_{T}\left(\begin{array}{c}
x\\
y
\end{array}\right)
\end{equation}
defines the resolvent of operator ${\cal I}-{\cal T}$, i.e. it satisfies the
following equations
\begin{equation}\label{ResolventEquations}
\begin{array}{ll}
{\dis H(x,y)+\int_{\R}T(x,t)\cdot H(t,y)dt=T(x,y), }
\vspace{0.1in}\\
{\dis H(x,y)+\int_{\R}T(t,y)\cdot H(x,t)dt=T(x,y), }
\end{array}
\end{equation}
and therefore operator ${\cal I}-{\cal H}$ is the inverse of
operator ${\cal I}+{\cal T}$.
\end{proposition}

\qed

\indent
Using Proposition~\ref{Resolvent}, we construct the resolvent of
operator ${\cal G}_{\lambda}=\pi U\left({\cal I}+{\cal N}_{\lambda}\right)$,
defined in (\ref{GOperator}), and prove the estimate that will be necessary in the proof
of Theorem~\ref{Main}.\\

\begin{proposition}\label{GResolvent}\ The set of characteristic values of operator
${\cal G}_{\lambda}$ coincides with the set
$$E({\cal G})=\left\{\lambda\in \C:\ \mbox{Re}\lambda>\sigma_1,\ {\cal D}_{N_{\lambda}}=0\right\}$$
and consists of at most countably many isolated points.\\
\indent
For $\lambda \notin E({\cal G})$ there exists an operator ${\cal H}_{\lambda}$ with kernel
$H(x,y,\lambda)$ satisfying the Hilbert-Schmidt condition and such that
operator ${\cal I}-{\cal H}_{\lambda}$ is the inverse of operator ${\cal I}+{\cal N}_{\lambda}$,
and therefore operator ${\dis \frac{1}{\pi U}\left({\cal I}-{\cal H}_{\lambda}\right) }$
is the inverse of operator ${\cal G}_{\lambda}$.\\
\indent
If function ${\cal D}_{N}(\lambda)={\cal D}_{N_{\lambda}}$ has no zeros in a strip
$\left\{\lambda:\ \sigma_1<\mbox{Re}\lambda<\sigma_2\right\}$, then operator ${\cal H}_{\lambda}$
admits estimate
\begin{equation}\label{HNormEstimate}
\left||{\cal H}_{\lambda}\right\|
<\exp\left\{e^{|\eta|}\cdot(1+|\eta|)^{2+\epsilon}\right\}
\end{equation}
for $\lambda\in \left\{\sigma_1+\gamma <\mbox{Re}\lambda <\sigma_2-\gamma\right\}$
and arbitrary $\epsilon>0$.
\end{proposition}

\indent
{\bf Proof.}\ Applying Proposition~\ref{Resolvent} to operator ${\cal N}_{\lambda}$
we obtain the existence of functions
\begin{equation}\label{D_N}
{\cal D}_{N}(\lambda)={\cal D}_{N_{\lambda}}
\end{equation}
and
$${\cal D}_{N}\left(\left.\begin{array}{c}
x\\
y
\end{array}\right|\lambda\right)
={\cal D}_{N_{\lambda}}\left(\begin{array}{c}
x\\
y
\end{array}\right)$$
such that for any fixed $\lambda$, satisfying ${\cal D}_{N}(\lambda)\neq 0$, kernel
$$H(x,y,\lambda)=\left[{\cal D}_{N}(\lambda)\right]^{-1}
\cdot{\cal D}_{N}\left(\left.\begin{array}{c}
x\\
y
\end{array}\right|\lambda\right)\in L^2(\R^2),$$
and operator ${\cal I}-{\cal H}_{\lambda}$ is the inverse of
operator ${\cal I}+{\cal N}_{\lambda}$.\\
\indent
Terms of the series (\ref{D_T}) for ${\cal N}_{\lambda}$ analytically depend on $\lambda$,
and according to estimates (\ref{DEstimate1}) this series converges uniformly with respect
to $\lambda$ on compact subsets of $\left\{\lambda\in\C:\ \mbox{Re}\lambda>\sigma_1\right\}$.
Therefore, ${\cal D}_{N}(\lambda)$ is an analytic function on
$\left\{\lambda\in\C:\ \mbox{Re}\lambda>\sigma_1\right\}$, and the set $E({\cal G})$
consists of at most countably many isolated points.\\
\indent
Analyticity of ${\cal I}-{\cal H}_{\lambda}$ with respect to $\lambda$ on
$$\left\{\lambda\in\C:\ \mbox{Re}\lambda>\sigma_1\right\}\setminus E({\cal G})$$
follows from the Theorem VI.14 in \cite{RS}. It is proved by approximation of the kernel by
degenerate kernels and by the argument that can be traced back to at least \cite{M}.\\
\indent
To prove estimate (\ref{HNormEstimate}) we use the well known estimate (\cite{L})
$$\left\|T\right\|^2\leq \int_{\R}|T(x,y)|^2dxdy$$
for integral operators. Using this estimate, estimates (\ref{DEstimate2}) and
(\ref{NEstimate}) we obtain
$$\left\|{\cal D}_{N}\left(\left.\begin{array}{c}
x\\
y
\end{array}\right|\lambda\right)\right\|
< \exp{\left\{C(1+|\eta|)^2\cdot\log^2{|\eta|}\right\}}
(1+|\eta|)^4\cdot\log^4{|\eta|}.$$
\indent
To estimate function $\left[{\cal D}_{N}(\lambda)\right]^{-1}$ for
$\lambda\in \left\{\sigma_1+\gamma <\mbox{Re}\lambda <\sigma_2-\gamma\right\}$
we use the following lemma.

\begin{lemma}\label{Estimate-ofD-1}\ If function ${\cal D}_{N}(\lambda)={\cal D}_{N_{\lambda}}$
has no zeros in the strip
$\left\{\lambda:\ \sigma_1<\mbox{Re}\lambda<\sigma_2\right\}$, then estimate
\begin{equation}\label{D-1Estimate}
\left|1/{\cal D}_{N}\left(\lambda\right)\right|
<\exp\left\{e^{|\eta|}\cdot(1+|\eta|)^{2+\epsilon}\right\}
\end{equation}
holds for $\lambda\in \left\{\sigma_1+\gamma <\mbox{Re}\lambda <\sigma_2-\gamma\right\}$
with fixed $\gamma>0$ and arbitrary $\epsilon>0$.
\end{lemma}

\indent
{\bf Proof.}\ We consider a biholomorphic map
$$\Psi:\ \left\{\lambda:\ \sigma_1<\mbox{Re}\lambda<\sigma_2\right\}\to
\D(1)=\left\{z\in\C: |z|<1\right\},$$
defined by the formula
$$\Psi(\lambda)=\frac{e^{i(\lambda-\sigma_1)\frac{\pi}{\sigma_2-\sigma_1}}-i}
{e^{i(\lambda-\sigma_1)\frac{\pi}{\sigma_2-\sigma_1}}+i}.$$
\indent
Denoting
$$w=u+iv=e^{i(\lambda-\sigma_1)\frac{\pi}{\sigma_2-\sigma_1}},$$
we obtain for the circle $C(r)=\left\{z:\ |z|=r\right\}$
$$\Psi^{-1}\left(C(r)\right)=\left\{\sigma+i\eta:\
\left|e^{i(\lambda-\sigma_1)\frac{\pi}{\sigma_2-\sigma_1}}-i\right|
=r\left|e^{i(\lambda-\sigma_1)\frac{\pi}{\sigma_2-\sigma_1}}+i\right|\right\}$$
$$=\left\{u+iv:\ \left(u^2+v^2-2v+1\right)
=r^2\left(u^2+v^2+2v+1\right)\right\}$$
$$=\left\{u+iv:\ u^2+\left(v-\frac{1+r^2}{1-r^2}\right)^2
=\frac{4r^2}{(1-r^2)^2}\right\}.$$
\indent
Introducing coordinates
$$t=\mbox{Re}\frac{\pi(\lambda-\sigma_1)}{\sigma_2-\sigma_1},\
s=\mbox{Im}\frac{\pi(\lambda-\sigma_1)}{\sigma_2-\sigma_1},$$
such that
$$w=u+iv=e^{i(\lambda-\sigma_1)\frac{\pi}{\sigma_2-\sigma_1}}
=e^{it-s}=e^{-s}\left(\cos{t}+\sin{t}\right),$$
we can rewrite the last condition as a quadratic equation with respect
to $e^{-s}$ for fixed $t$
$$\left(e^{-s}-\sin{t}\frac{1+r^2}{1-r^2}\right)^2
+\cos^2{t}\left(\frac{1+r^2}{1-r^2}\right)^2-\frac{4r^2}{(1-r^2)^2}=0.$$
\indent
Solving equation above we obtain
$$e^{-s}=\sin{t}\frac{1+r^2}{1-r^2}
\pm\sqrt{\frac{4r^2}{(1-r^2)^2}-\cos^2{t}\left(\frac{1+r^2}{1-r^2}\right)^2}$$
with solutions existing for $t$ such that
$$|\cos{t}|\leq \frac{2r}{1-r^2}\frac{1-r^2}{1+r^2}=\frac{2r}{1+r^2}.$$
\indent
The maximal value for $e^{-s}$ is achieved at $t=\frac{\pi}{2}$ and it is
$$e^{-s}=\frac{1+r^2}{1-r^2}+\frac{2r}{1-r^2}
=\frac{1+r^2+2r}{1-r^2}=\frac{(1+r)^2}{1-r^2}=\frac{1+r}{1-r}.$$
Therefore the maximal value for $|s|$ is achieved at $t=\frac{\pi}{2}$, is
equal to $|s|=\log\left(\frac{1+r}{1-r}\right)$, and for $r=1-\delta$
we have the maximal value
\begin{equation}\label{s_max}
\max|s|=\log\left(\frac{1+r}{1-r}\right)=-\log{\delta}+\log{(2-\delta)}.
\end{equation}
\indent
Since function ${\cal D}_{N}\left(\lambda\right)$ has no zeros
in $\left\{\lambda:\ \sigma_1<\mbox{Re}\lambda<\sigma_2\right\}$ we can consider
analytic function $\log\left({\cal D}_{N}\left(\lambda\right)\right)$
in this strip, and using estimates (\ref{DEstimate1}) and (\ref{NEstimate}),
and equality (\ref{s_max}), we obtain the following estimate for
$z=(1-\delta)e^{i\theta}$
$$\log\left|{\cal D}_{N}\left(\Psi^{-1}(z)\right)\right|
\leq \frac{\left\|N_{\Psi^{-1}(z)}\right\|^2}{2}
\leq C\left|\Psi^{-1}(z)\cdot\log\left(\Psi^{-1}(z)\right)\right|^2$$
$$\leq C|\log{\delta}\cdot\log\left(\log{\delta}\right)|^2.$$
\indent
Using then the Borel-Caratheodory inequality (\cite{Ti1}, \cite{Boa}) on disks with radii
$$1-2\delta=r<R=1-\delta,$$
we obtain
$$\left|\log\left({\cal D}_{N}\left(\Psi^{-1}(z)\right)\right)
\right|_{\{|z|=1-2\delta\}}$$
$$\leq\frac{2-4\delta}{\delta}\max_{|z|=R}
{\mbox{Re}\left\{\log\left({\cal D}_{N}\left(\Psi^{-1}(z)\right)\right)\right\}}
+\frac{1-\delta+1-2\delta}{\delta}|\log\left({\cal D}_{N}\left(\Psi^{-1}(0)\right)\right)|$$
$$< \frac{C}{\delta}\log^2{\delta}\cdot\log^2\left(\log{\delta}\right),$$
or
$$-\frac{C}{\delta}\log^2{\delta}\cdot\log^2\left(\log{\delta}\right)
<\mbox{Re}\left\{\log\left({\cal D}_{N}\left(\Psi^{-1}(z)\right)\right)
\right\}\Big|_{\{|z|=1-2\delta\}}
<\frac{C}{\delta}\log^2{\delta}\cdot\log^2\left(\log{\delta}\right).$$
\indent
From the last estimate we obtain an estimate for
the function $\left|1/{\cal D}_{N}\left(\Psi^{-1}(z)\right)\right|$ in the disk
$\D(1-2\delta)$:
\begin{equation}\label{D-lowestimate}
\left|1/{\cal D}_{N}\left(\Psi^{-1}(z)\right)\right|\Bigg|_{\{|z|\leq 1-2\delta\}}
< \exp\left\{\frac{|\log{\delta}|^{2+\epsilon}}{\delta}\right\}
\end{equation}
for arbitrary $\epsilon>0$.\\
\indent
For a fixed $t\in\left(0,\pi\right)$ and arbitrary $s$ we have that
$t+is\in\Psi^{-1}\left(\D(r)\right)$ with $r=1-2\delta$ if
$$e^{|s|}\leq \sin{t}\cdot\frac{1+r^2}{1-r^2}
+\sqrt{\frac{4r^2}{(1-r^2)^2}-\cos^2{t}\cdot\left(\frac{1+r^2}{1-r^2}\right)^2}$$
$$=\sin{t}\cdot\frac{2-4\delta+4\delta^2}{2\delta(2-2\delta)}
+\frac{\sqrt{4(1-2\delta)^2-\cos^2{t}\cdot(2-4\delta+4\delta^2)^2}}
{2\delta(2-2\delta)},$$
and therefore for any interval $\left[\gamma^{\prime},\pi-\gamma^{\prime}\right]$
there exist constants $C_1,\ C_2$ such that conditions
$$t\in \left[\gamma^{\prime},\pi-\gamma^{\prime}\right],\hspace{0.1in}
\frac{C_1}{\delta}<e^{|s|}<\frac{C_2}{\delta}$$
imply that $t+is \in \Psi^{-1}\left(\D(1-2\delta)\right)$.\\
\indent
Using then estimate (\ref{D-lowestimate}) we obtain for $\lambda$ with
$\mbox{Re}\lambda \in \left[\sigma_1+\frac{\gamma^{\prime}(\sigma_2-\sigma_1)}{\pi},
\sigma_2-\frac{(\pi-\gamma^{\prime})(\sigma_2-\sigma_1)}{\pi}\right]$ the estimate
$$\left|1/{\cal D}_{N}\left(\lambda\right)\right|
<\exp\left\{e^{|s|}\cdot(1+|s|)^{2+\epsilon}\right\}$$
for arbitrary $\epsilon>0$, which leads to estimate (\ref{D-1Estimate}).
\qed

\indent
Combining now estimate for ${\dis \left\|{\cal D}_{N}\left(\left.\begin{array}{c}
x\\
y
\end{array}\right|\lambda\right)\right\| }$ with (\ref{D-1Estimate}) we obtain
estimate (\ref{HNormEstimate}).
\qed

\section{Proof of Theorem~\ref{Main}.}
\label{Proof}

\indent
Before proving Theorem~\ref{Main} we will prove two lemmas, that will be used in
the proof of this theorem.\\
\indent
In order to assure applicability of Proposition~\ref{RSolvability} to
$f_a$, defined in (\ref{fDefinition}), we have to prove that
$$f_a \in {\cal L}^2\left(I^c(1)\right)$$
for $\widehat{w_a}$ satisfying (\ref{wCondition}). In the lemma below we
prove the necessary property of $f_a$.

\begin{lemma}\label{fCondition}\ If $\widehat{w_a}$ satisfies condition
(\ref{wCondition}) then $f_a(x,\lambda)$ defined by the formula (\ref{fDefinition})
is a function in ${\cal L}^2\left(I^c(1)\right)$ for any fixed $\lambda$,
which satisfies the estimate
\begin{equation}\label{f_aEstimate}
\left\|f_a(\cdot,\sigma+i\eta)\right\|_{{\cal L}^2\left(I^c(1)\right)}
<C\exp{\left\{-e^{|\eta|}\cdot(1+|\eta|)^{2+\epsilon}\right\}}
\end{equation}
with some $\epsilon>0$ for $\sigma\in[\sigma_1,\sigma_2]$.
\end{lemma}

\indent
{\bf Proof.}\ For a fixed $\lambda=\sigma+i\eta$ with $\sigma\in[\sigma_1,\sigma_2]$ we choose
$B>1$, and using second representation from (\ref{RRepresentation}) of $R(x,\lambda)$
for $|\lambda x|>B$, obtain an estimate
$$\left|R(x-y,\lambda)\right|<C\frac{|\lambda|^{1/2}e^{-\lambda|x-y|}}{\sqrt{|x-y|}}.$$
Using then condition (\ref{wCondition}), we have
\begin{equation}\label{BigxEstimate}
\left(\int_{|x|>B/|\lambda|}\left|f_a(x,\lambda)\right|^2dx\right)^{1/2}
\end{equation}
$$=\frac{1}{\pi^2}\left|\int_{|x|>B/|\lambda|}\left|\int_{-1}^1 e^{-d(\lambda)y}
R(x-y,\lambda)\widehat{w_a}(y,\lambda)dy\right|^2dx\right|^{1/2}$$
$$< C|\lambda|^{1/2}\left(\int_{|x|>B/|\lambda|}\left(\int_{-1}^1
e^{-\sigma|x|}\left|\widehat{w_a}(y,\lambda)\right|dy\right)^2dx\right)^{1/2}$$
$$< C|\lambda|^{1/2}\int_{-1}^1\left|\widehat{w_a}(y,\lambda)\right|dy
<C|\lambda|^{1/2}\int_{-1}^1\left|\widehat{w_a}(y,\lambda)\right|^2dy$$
$$<C\exp{\left\{-e^{|\eta|}\cdot(1+|\eta|)^{2+\epsilon}\right\}}.$$
\indent
For $|\lambda x|<B$ we use the first representation from (\ref{RRepresentation})
for $R(x-y,\lambda)$. Since the Hilbert transform is a bounded linear operator from
$L^q$ into $L^q$ (see \cite{Ti2}, \cite{Tr}), and kernels $\alpha\left(\lambda(x-y)\right)$,
$\beta\left(\lambda(x-y)\right)$, and  $\gamma\left(\lambda(x-y)\right)$ from
(\ref{RRepresentation}) are bounded, we obtain
\begin{equation}\label{SmallxEstimate}
\left(\int_{|x|<|B/\lambda|}\left|f_a(x,\lambda)\right|^2dx\right)^{1/2}
<C\left|\lambda\log{\lambda}\right|
\cdot\left\|\widehat{w_a}(y,\lambda)\right\|_{L^2\left(I(1)\right)}
\end{equation}
$$<C\exp{\left\{-e^{|\eta|}\cdot(1+|\eta|)^{2+\epsilon}\right\}},$$
where in the last inequality we used condition (\ref{wCondition}).\\
\indent
Combining estimates (\ref{BigxEstimate}) and (\ref{SmallxEstimate}) we obtain
(\ref{f_aEstimate}).
\qed

\begin{lemma}\label{LaplaceInverse}\ If a function $h(y,\lambda)$ satisfies
estimate
\begin{equation}\label{hCondition}
\int_{-\infty}^{\infty}e^{-\sigma_1\cdot|y|}|h(y,\sigma+i\eta)|dy
<\frac{C}{(1+|\eta|)^{\frac{5}{2}+\epsilon}}
\end{equation}
for some $\epsilon>0$ and $\sigma_1<\mbox{Re}\lambda<\sigma_2$, then function
\begin{equation}\label{xi_inL}
\xi(x,z,\lambda)=e^{d(\sigma+i\eta)x}\left[\int_{-\infty}^\infty
K_0\left(r(\sigma+i\eta)\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)
h(y,\sigma+i\eta)dy\right]\in{L^1_{\eta}(\R)}
\end{equation}
for $\sigma\in [\sigma_1,\sigma_2]$, and satisfies conditions
\begin{equation}\label{xiDerivatives}
\frac{\partial^2\xi(x,z,\sigma+i\eta)}{\partial x^2},\
\frac{\partial^2\xi(x,z,\sigma+i\eta)}{\partial z^2},\
|\eta|^2\xi(x,z,\sigma+i\eta),\
|\eta|\frac{\partial\xi(x,z,\sigma+i\eta)}{\partial x}\in L^1(\R_{\eta}).
\end{equation}
\indent
Function
$$\phi(x,z,t)=-\frac{1}{2\pi\sqrt{1-M^2}}
\int_{\sigma-i\infty}^{\sigma+i\infty}e^{d(\sigma+i\eta)x}$$
$$\times\left[\int_{-\infty}^\infty K_0\left(r(\sigma+i\eta)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)
h(y,\sigma+i\eta)dy\right]e^{(\sigma+i\eta)t}d\eta$$
is then well defined for $z>0$, and doesn't depend on $\sigma\in [\sigma_1,\sigma_2]$.
\end{lemma}

\indent
{\bf Proof.}\ To prove inclusion (\ref{xi_inL}) of the Lemma it suffices to prove that
under conditions above estimate
\begin{equation}\label{LaplaceEstimate1}
\left\|e^{d(\sigma+i\eta)x}\left[\int_{-\infty}^\infty K_0\left(r(\sigma+i\eta)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)
h(y,\sigma+i\eta)dy\right]\right\|_{L^1_{\eta}(\R)}<C(M,z)
\end{equation}
holds uniformly with respect to $\sigma \in \left[\sigma_1,\sigma_2\right]$
for fixed $x$, fixed $z>0$, and for some $\sigma_1>\sigma_a$. Applying then Theorem 47 from
\cite{Boc} we will obtain the second part of the Lemma.\\
\indent
Using asymptotics of $K_0(\zeta)$ for large and for small $|\zeta|$ (\cite{EMOT}) we obtain
the existence for fixed $z>0$ of a constant $A(z)>0$, large enough, such that estimates
\begin{equation}\label{LargeAEstimate}
\left|K_0\left(r(\sigma+i\eta)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)\right|
<\frac{C(M,z)e^{-\sigma|x-y|}}{\sqrt{|\sigma+i\eta|\cdot|x-y|}}\ \mbox{for}\ |x-y|>A(z),
\end{equation}
and
\begin{equation}\label{SmallAEstimate}
\left|K_0\left(r(\sigma+i\eta)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)\right|
<\frac{C(M,z)}{\sqrt{|\sigma+i\eta|}}\ \mbox{for}\ |x-y|<A(z),
\end{equation}
hold uniformly for $\sigma \in \left[\sigma_1,\sigma_2\right]$,
with a constant $C$ depending on $M$ and $z$.\\
\indent
Combining estimates (\ref{LargeAEstimate}) and (\ref{SmallAEstimate}) with the
estimate for $h(y,\lambda)$ we obtain
$$\left|e^{d(\sigma+i\eta)x}\left[\int_{-\infty}^\infty K_0\left(r(\sigma+i\eta)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)
h(y,\sigma+i\eta)dy\right]\right|$$
$$<\frac{C(M,z)}{\sqrt{|\sigma+i\eta|}}
\int_{-\infty}^\infty e^{-\sigma_1\cdot|y|}|h(y,\sigma+i\eta)|dy
<\frac{C(M,z)}{\left(1+|\eta|\right)^{3+\epsilon}},$$
for $z>0$, which leads to estimate (\ref{LaplaceEstimate1}).\\
\indent
Again using estimates (\ref{LargeAEstimate}) and (\ref{SmallAEstimate}) and analogous
estimates for
$$\frac{\partial}{\partial x}K_0\left(r(\sigma+i\eta)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right),\
\frac{\partial^2}{\partial x^2}K_0\left(r(\sigma+i\eta)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right),$$
and
$$\frac{\partial^2}{\partial z^2}K_0\left(r(\sigma+i\eta)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)$$
we obtain inclusions (\ref{xiDerivatives}).
\qed\\

\indent
To prove Theorem~\ref{Main} we consider $w_a$ satisfying condition (\ref{wCondition}),
and define $f_a$ by the formula (\ref{fDefinition}). Using Lemma~\ref{fCondition} we obtain
that $f_a$ satisfies estimate (\ref{f_aEstimate}). Applying Proposition~\ref{RSolvability}
to $f_a$ and using estimate (\ref{HNormEstimate}) from Proposition~\ref{GResolvent}
we obtain the existence of $h_a$ satisfying equation (\ref{MainEquation}) and such that
$$\left\|h_a(\cdot,\sigma+i\eta)\right\|_{{\cal L}^p\left(I^c(1)\right)}
<\exp\left\{e^{|\eta|}\cdot(1+|\eta|)^{2+\epsilon}\right\}
\cdot\left\|f_a(\cdot,\sigma+i\eta)\right\|_{{\cal L}^2\left(I^c(1)\right)}
<\frac{C}{\left(1+|\eta|\right)^m}$$
for arbitrary $m$, arbitrary $p<\frac{4}{3}$, and $\sigma\in[\sigma_1,\sigma_2]$,
with $\sigma_a<\sigma_1$.\\
\indent
Using the estimate above for $p=1$, we obtain
\begin{equation}\label{h_aFarEstimate}
\int_{|x|>1}\left|h_a(x,\sigma+i\eta)\right|\cdot|x|^{-1}dx
<\frac{C}{\left(1+|\eta|\right)^m}.
\end{equation}
\indent
From the definition of $h_a$ on $[-1,1]$ as
$$h_a(x,\lambda)=\frac{e^{-d(\lambda)x}\cdot \widehat{w_a}(x,\lambda)}{\pi}$$
and from condition (\ref{wCondition}) we obtain
$$\left\|h_a(x,\sigma+i\eta)\right\|_{L^p\left(I(1)\right)}
=\left\|e^{-d(\lambda)x}\cdot\widehat{w_a}(x,\sigma+i\eta)\right\|_{L^p\left(I(1)\right)}$$
$$<C\left\|\widehat{w_a}(\cdot,\sigma+i\eta)\right\|_{L^2\left(I(1)\right)}
<\frac{C}{\left(1+|\eta|\right)^m}\
\mbox{for}\ p<\frac{4}{3},\ \sigma\in[\sigma_1,\sigma_2]\ \mbox{with}\ \sigma_a<\sigma_1,$$
and therefore
\begin{equation}\label{h_aCloseEstimate}
\left\|h_a(\cdot,\sigma+i\eta)\right\|_{L^1\left(I(1)\right)}
<\frac{C}{\left(1+|\eta|\right)^m}
\end{equation}
for arbitrary $m>0$.\\
\indent
From the estimates (\ref{h_aFarEstimate}) and (\ref{h_aCloseEstimate}) we conclude
that function $h_a$ satisfies estimate (\ref{hCondition}), and therefore, applying
Lemma~\ref{LaplaceInverse} and Proposition~\ref{LaplaceSolution}, we obtain that function
$\phi(x,z,t)$ in formula (\ref{Solution})
is well defined and satisfies equation (\ref{LinearEquation}).\\
\indent
To prove that $\phi(x,z,t)$ satisfies boundary condition (\ref{Flow_Tangency}) we
fix $x\in[-1,1]$ and denote $\delta=\min\left\{x+1,1-x\right\}$. Then we have
\begin{equation}\label{Tangency1}
\lim_{z\to 0}\frac{\partial}{\partial z}\xi(x,z,\lambda)
=\lim_{z\to 0}\frac{\partial}{\partial z}\int_{-\infty}^\infty
S(x-y,z,\lambda)e^{d(\lambda)y}h_a(y,\lambda)dy
\end{equation}
$$=\lim_{z\to 0}\frac{\partial}{\partial z}\int_{x-\frac{\delta}{2}}^{x+\frac{\delta}{2}}
S(x-y,z,\lambda)e^{d(\lambda)y}h_a(y,\lambda)dy$$
$$+\lim_{z\to 0}\frac{\partial}{\partial z}
\int_{\R\setminus[x-\frac{\delta}{2},x+\frac{\delta}{2}]}
S(x-y,z,\lambda)e^{d(\lambda)y}h_a(y,\lambda)dy.$$
For the first integral in the right hand side of (\ref{Tangency1}) we obtain
using Lemma~\ref{Fourier}
$$\lim_{z\to 0}\frac{\partial}{\partial z}\int_{x-\frac{\delta}{2}}^{x+\frac{\delta}{2}}
S(x-y,z,\lambda)e^{d(\lambda)y}h_a(y,\lambda)dy$$
$$=-\lim_{z\to 0}\frac{\partial}{\partial z}\int_{x-\frac{\delta}{2}}^{x+\frac{\delta}{2}}
e^{d(\lambda)y}h_a(y,\lambda)dy\int_{-\infty}^\infty e^{i(x-y)\omega}
\frac{e^{\dis -z\left((1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)\right)^{\frac{1}{2}}} }
{2\sqrt{(1-M^2)(\omega+id(\lambda))^2+r^2(\lambda)}}d\omega$$
$$=\widehat{w_a}(x,\lambda).$$
\indent
For the second integral in the right hand side of (\ref{Tangency1}) we have
$$\lim_{z\to 0}\frac{\partial}{\partial z}
\int_{\R\setminus[x-\frac{\delta}{2},x+\frac{\delta}{2}]}
S(x-y,z,\lambda)e^{d(\lambda)y}h_a(y,\lambda)dy$$
$$=-\frac{e^{d(\lambda)x}}{\sqrt{1-M^2}}\lim_{z\to 0}
\int_{\R\setminus[x-\frac{\delta}{2},x+\frac{\delta}{2}]}
\left[\frac{\partial}{\partial z}K_0\left(r(\lambda)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)\right]h_a(y,\lambda)dy.$$
\indent
Using then estimate (\ref{h_aFarEstimate}) and equality
$$\lim_{z\to 0}\left[\frac{\partial}{\partial z}K_0\left(r(\lambda)
\left(\frac{(x-y)^2}{1-M^2}+z^2\right)^{\frac{1}{2}}\right)\right]=0$$
for $y\in \R\setminus[x-\frac{\delta}{2},x+\frac{\delta}{2}]$ we obtain that
the second integral in the right hand side of (\ref{Tangency1}) is equal to zero.\\
\indent
From equalities above we conclude that
$$\lim_{z\to 0}\frac{\partial}{\partial z}\xi(x,z,\lambda)=\widehat{w_a}(x,\lambda)$$
for $x\in [-1,1]$ and $\mbox{Re}\lambda\in [\sigma_1,\sigma_2]$, and therefore
$$\lim_{z\to 0}\frac{\partial}{\partial z}\phi(x,z,t)=w_a(x,t).$$
\indent
Straightforward substitution of $v_a(x,\lambda)=e^{d(\lambda)x}h_a(x,\lambda)$
into the formula (\ref{LapSolution}), with $h_a(x,\lambda)$ defined as
$$h_a(x,\lambda)=\left\{
\begin{array}{ll}
{\dis \frac{1}{\pi}e^{-d(\lambda)x}\cdot \widehat{w_a}(x,\lambda)\
\mbox{for}\ x\in[-1,1], }\vspace{0.1in}\\
\mbox{solution of equation}\ (\ref{MainEquation})\ \mbox{for}\ x\in\R\setminus[-1,1],
\end{array}\right.$$
shows that $\xi(x,z,\lambda)$ defined by this formula satisfies equation (\ref{ZeroCondition})
for $1<|x|<A$. Then for $\phi(x,z,t)$ defined by formula (\ref{InverseLaplace}) we will have
$$\frac{\partial\phi(x,0,t)}{\partial t}+U\frac{\partial\phi(x,0,t)}{\partial x}$$
$$=\frac{1}{2\pi}\left(\frac{\partial}{\partial t}+U\frac{\partial}{\partial x}\right)
\int_{\sigma-i\infty}^{\sigma+i\infty}e^{\lambda t}\xi(x,0,\lambda)d\eta$$
$$=\frac{1}{2\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}e^{\lambda t}
\left(\lambda+U\frac{\partial}{\partial x}\right)\xi(x,0,\lambda)d\eta=0$$
for $1<|x|<A$.

\qed

config: arxiv


TITLE: A doubt about Theorem 22 in textbook Algebra by Saunders MacLane and Garrett Birkhoff
QUESTION [3 upvotes]: I'm reading Theorem 22 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.

It follows that $\phi_*S = \phi[S] := \{\phi(x) \mid x \in S\}$ and $\phi^*T = \phi^{-1}[T] := \{x \in G \mid \phi(x) \in T\}$.

and its proof

Here is Proposition 10:

Due to the properties of the set-valued functions $\phi[\cdot], \phi^{-1}[\cdot]$ induced from $\phi (\cdot)$, we always have $\phi_{*}\left(S_{1} \cap S_{2}\right) \subseteq \phi_{*} S_{1} \cap \phi_{*} S_{2}$. One sufficient condition for the equality to hold is that $\phi$ is injective.
Could you please elaborate on how Proposition 10 lead to the equality?

REPLY [2 votes]: Clearly, $\phi [S_{1} \cap S_{2}] \subseteq \phi [S_{1}] \cap \phi [S_{2}]$. Below is my use of kernel to obtain $\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi [S_{1} \cap S_{2}]$:

For $y \in \phi[S_1] \cap \phi[S_2]$, $y=\phi(x_1)=\phi(x_2)$ for some $x_1 \in S_1, x_2 \in S_2$. Then $\phi(x_1 x_2^{-1}) = \phi(x_1) \phi(x_2)^{-1} =1$. Hence $x_1 x_2^{-1} \in \operatorname{ker} \phi \subseteq S_1 \cap S_2$ and thus $x_1 x_2^{-1} \in S_1$. Because $S_1$ is a subgroup, $x_2 \in S_1$. Hence $x_2 \in S_1 \cap S_2$. The result then follows.

Here is a lemma suggested by @user1729.

Lemma: If $\phi:G \to H$ is morphism of groups and $S$ is a subgroup of $G$, then $\phi^{-1}[\phi[S]] = S \operatorname{ker} \phi = S$.

Proof: Notice that $\phi^{-1}[\phi[S]] = \{x \in G \mid \exists y\in S: \phi(x) = \phi (y)\} \overset{(\star)}{=} \{x \in G \mid \exists y\in S: xy^{-1} \in \operatorname{ker} \phi\}$. It follows that $\phi^{-1}[\phi[S]] = S \operatorname{ker} \phi$. Notice that $S \subseteq\phi^{-1}[\phi[S]]$. With similar reasoning in my above approach, we get $(x,y) \in S \times G$ and $\phi(x) = \phi(y)$ implies $y \in S$. Hence $\phi^{-1}[\phi[S]] = S$.
$(\star)$: This is because $\phi$ is a morphism of groups.
Then we use this lemma to obtain the latter inclusion as follows:
We have $\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi [S_{1}]$ and thus $\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq \phi^{-1}[\phi [S_{1}]] \color{red}{=} S_1$. Similarly, $\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq S_2$. Hence $\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq S_1 \cap S_2$ and thus $\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi[S_1 \cap S_2]$.


TITLE: Why is the force of gravity not equal to the normal force on an inclined plane?
QUESTION [1 upvotes]: When an object is at rest the force of gravity and normal force cancel out. However, when place on an inclined plane the downward force of gravity is not equal to the normal force that is perpendicular to the ramp. Why is that?
My second question is why is normal force perpendicular to the ramp?

REPLY [0 votes]: One more thing to note, is that normal force is force of constraint.
We know, that the object cannot move under the surface and there must be some force that prevents this from happening. Such force must be perpendicular to the surface, because it only prohibits things to go under it and does not influence anything else. This is how normal force is defined.
But of course, the true contact force from the ramp might be more complicated than that. We call this additional complication friction.


TITLE: Proof of Quotient-Remainder Theorem by induction
QUESTION [0 upvotes]: I am trying to prove the quotient-remainder theorem by mathematical induction:

Quotient-Remainder Theorem:
$\forall a,b \in \mathbb{Z}, b> 0: \ \exists! q, \ r \in \mathbb{Z}, \
a = bq \ + \  r, \ r \in [0,b) $

Proof that $ a \in \mathbb{Z} $ exists:
           Divide into 2 cases: $ a \ge 0 $ or $ a < 0 $

  Case 1: $ a \ge 0 $

             Base Case: Let $ a = 0 $.

                        It is
  clear that if we let $ q = r = 0 $, then $ a = b(0) + (0) = 0 $.

                        Hence,
  base case is true.

  Inductive Step: Assume $ a = k \in \mathbb{Z} $ exists

  Hence, $ k = bq + r $

  Proof: Let $ a = k + 1\in \mathbb{Z} $

  By inductive hypothesis, $ k = bq + r $

  Hence, $ k + 1 = bq + r + 1 $ and $ k + 1 = bq + (r + 1) $ where
  $ r + 1 \in \mathbb{Z} $

                        by
  closure property of integers

  Thus, $ \forall a \in \mathbb{Z}, a \ge 0 \ $, $ a = bq + r $

  Case 2: $ a < 0 $

             Base Case: Let $ a = -1 $.

  It is clear that if we let $ q = 0 $ and $ r = -1 $, then $ a = b(0) + (-1) = -1 $.

  Hence, base case is true.

  Inductive Step: Assume $ a = k \in \mathbb{Z} $ exists

  Hence, $ k = bq + r $

  Proof: Assume $ a = -(k + 1) \in \mathbb{Z} $

                        By
  inductive hypothesis, $ k = bq + r $

  Hence, $ k + 1 = bq + r + 1 $ and $ -(k + 1) = b(-q) + (-r - 1) $ where

  $ -r - 1 \in \mathbb{Z} $ by closure property of integers

  Thus, $ \forall a \in \mathbb{Z}, a < 0 \ $, $ a = bq + r $

I am not sure if my proof is correct. Also, I do not know how to prove $ r \in [0,b) $ and the uniqueness of $ q $ and $ r $. Could someone please advise me?

REPLY [1 votes]: Your induction case didn't take $r+1=b $ into account and in the end all you proved was $a=0b+a $.
Do this: assume $a=qb+r;0\le r <b $.  If $r <b-1$ then $a+1 = qb + (r+1) $.  And if $r=b-1$ then $a+1 =qb +r+1 =qb+b= (q+1)b+0$.
So if $a =qb +r;r\in [0,b) $ has solution the $a+1=pb+s;s\in [0,b) $ has solution.
One can do the same for $a-1$.  If $a=qb+r$ and $r > 0$ then $a-1= qb +(r-1) $ is a solution.  If $r=0$ then $a-1= (q-1)b +(b-1) $ is a solution.
But I haven't shown they are unique.  You can show uniqueness via induction and I spent a long time in my rough draft showing just that.  But in the end it was unnecessarily complicated and confusing and, frankly bizarre.  (I'll leave it in as a post script.)  It's much better to show uniqueness directly.
Suppose $a=qb+r=pq+s $ then $(r-s)=(p-q)b $.  If $r,s \in [0,b) $ then $-b < r-s =(p-q)b <b $.  So $-1 < p-q < 1$ so $p-q=0;p=q $ and $r=s $.
So solutions are unique.
=== postscript: bizarre rough draft==
Base step.  Need to show $0=0b + 0$ is unique.
Proof: (assume all variables in this post are integers.)
If $0=qb +r$ then $r = -qb $.  If $q \le -1$ the $r \ge b $ which isn't acceptable.  If $q >0$ then $r < 0$ which isn't acceptable.  So $0,0$ is unique solution.
Induction case.
$a=qb+r $ is unique solution.
Let $a+1 = pb +s;0\le s <b $ be an acceptable solution (if there is one, which we need to be aware there might not be).
Then $a = pb + s-1;-1\le s-1 < b-1 $.  If $s -1 \ge 0$ then $p=q $ and $s=r+1$ (and $r < b-1$) as $p,r $ was a unique solution.  So $p=q;s=r+1$ would be a solution and it would be unique.
But it is a solution.  So it is unique
Otherwise if our hypothetical solution had $s=0$ then it would follow $r= b-1$. We would have $a+1 = pq +0 = qb +r+1= qb+b =(q+1)b $ and so $p=q+1$ and $s=0$.  That would have to be the solution and it would be unique.
But it is a solution.  So it is unique.


TITLE: How to find a rectangle which is formed from the lines?
QUESTION [2 upvotes]: I have Cartesian coordinates  $A$ and $B$.
Line $AB$ is the axis (center) of the rectangle.
And I have $H$ (height).
I need Cartesian  coordinates blue rectangle.

REPLY [2 votes]: To get the coordinates of the corner points, we need to displace $A$ and $B$ by $\pm h$ along the normal of the line connecting them (i.e. along a vector pointing 90° away from the line).
Let $A = (A_x, A_y)$ and $B = (B_x, B_y)$.  Then the vector from $A$ to $B$ is
$$v = B-A = (v_x, v_y),$$
where
$$v_x = B_x-A_x \text{ and } v_y = B_y-A_y.$$
Rotating this vector by 90° is equivalent to swapping its coordinates and negating one of them, giving the normal vector
$$n = (v_y, -v_x) = (B_y-A_y,\ A_x-B_x).$$
However, the length of $n$ is the same as the length of $v$; we need to scale it down to $1$ before multiplying it with $\pm h$.  To do this, we can divide $n$ by its length $|n| = |v| = \sqrt{v_x^2+v_y^2}$ to get the unit normal vector
$$\hat n = \frac{n}{|n|} = \left({\frac{v_y}{\scriptstyle \sqrt{v_x^2+v_y^2}},\ \frac{-v_x}{\scriptstyle \sqrt{v_x^2+v_y^2}}}\right).$$
We can then calculate the coordinates of the four corners as $A + h \hat n$, $A - h \hat n$, $B + h \hat n$ and $B - h \hat n$.


TITLE: least squares regression in 3space
QUESTION [2 upvotes]: robjohn is giving me a hand with this, but in case anybody else knows...
I need to do a least-squares regression for linearity on a set of coordinates in 3space. If the dataset is linear, I need to see if it is close to vertical or horizontal. How could I do this?
Many thanks in advance
Joe Stavitsky

REPLY [1 votes]: I posted this this morning in chat to Joe, but since the question is here, too, I will post it here.
Uniquely, up to the sign of $u$, specify a line as $p+ut$ where $|u|=1$, $p\cdot u=0$, and $t\in\mathbb{R}$. The square of the distance of $x$ from this line is
$$
|x-p|^2-((x-p)\cdot u)^2 = |x-p|^2-(x\cdot u)^2\tag{1}
$$
Thus, for a given $\{x_k\}$, we wish to minimize
$$
\Delta=\frac1n\sum_{k=1}^n\left[|x_k-p|^2-(x_k\cdot u)^2\right]\tag{2}
$$
over all $p$ and $u$ so that
$$
|u|=1\text{ and }p\cdot u=0\tag{3}
$$
Thus, we need that for any variation of $u$ and $p$ that maintain $(3)$, the variation of $(2)$ must be $0$.  That is, whenever
$$
u\cdot\delta u=0\quad\text{and}\quad p\cdot\delta u+u\cdot\delta p=0\tag{4}
$$
we want the variation of $\Delta$ to vanish
$$
0=\sum_{k=1}^n\left[(x_k-p)\cdot\delta p+u\cdot x_kx_k\cdot\delta u\right]\tag{5}
$$
Define
$$
\bar{x}=\frac1n\sum_{k=1}^nx_k\quad\text{and}\quad X=\frac1n\sum_{k=1}^nx_kx_k^T\tag{6}
$$
and $(5)$ becomes
$$
0=(\bar{x}-p)\cdot\delta p+(Xu)\cdot\delta u\tag{7}
$$
Considering $(4)$, if we hold $u$ constant and vary $p$, we get that for any $\delta p$ so that $u\cdot\delta p=0$
$$
0=(\bar{x}-p)\cdot\delta p\tag{8}
$$
Because $(8)$ holds for any $\delta p$ which is perpendicular to $u$, we get that $\bar{x}-p\,||\,u$, which leads to
$$
\bar{x}-p=(\bar{x}-p)\cdot uu=\bar{x}\cdot uu\tag{9}
$$
Combining $(4)$ and $(9)$ yields
$$
\bar{x}\cdot\delta u+u\cdot\delta p=0\tag{10}
$$
Plugging $(9)$ into $(7)$ and applying $(10)$ yields
$$
\begin{align}
0
&=\bar{x}\cdot uu\cdot\delta p+(Xu)\cdot\delta u\\
&=-\bar{x}\cdot u\bar{x}\cdot \delta u+(Xu)\cdot\delta u\\
&=\left(\left(X-\bar{x}\bar{x}^T\right)u\right)\cdot\delta u\tag{11}
\end{align}
$$
Since $(11)$ holds for any $\delta u$ perpendicular to $u$, we must have that
$$
\left(X-\bar{x}\bar{x}^T\right)u\,||\,u\tag{12}
$$
That is, $u$ is an eigenvector of $X-\bar{x}\bar{x}^T$.
Plugging $(9)$ into $(2)$ yields $\Delta$ as a function of $u$,
$$
\begin{align}
\Delta
&=\frac1n\sum_{k=1}^n\left[|x_k-p|^2-(x_k\cdot u)^2\right]\\
&=\frac1n\sum_{k=1}^n|(x_k-\bar{x})+\bar{x}\cdot uu|^2-u^TXu\\
&=\frac1n\sum_{k=1}^n\left(|x_k-\bar{x}|^2+2(x_k-\bar{x})\cdot u\bar{x}\cdot u+(\bar{x}\cdot u)^2\right)-u^TXu\\
&=\operatorname{Var}[\{x_k\}]-u^T\left(X-\bar{x}\bar{x}^T\right)u\tag{13}
\end{align}
$$
and since
$$
X-\bar{x}\bar{x}^T=\frac1n\sum_{k=1}^n(x_k-\bar{x})(x_k-\bar{x})^T\tag{14}
$$
we see that $X-\bar{x}\bar{x}^T$ is positive semi-definite. Therefore, the maximum eigenvalue corresponds to the minimum of $\Delta$.
Once we have found $u$, a unit eigenvector of $X-\bar{x}\bar{x}^T$ with the greatest eigenvalue, we can use $(9)$ to get $p=\bar{x}-\bar{x}\cdot uu$.


TITLE: Upward continued fractions
QUESTION [5 upvotes]: Has anybody seen "upward continued fractions", such as
$$
\frac{1+\large{\frac{1+\large{\frac{1+...}{a_2}}}{a_1}}}{a_0} \quad?
$$
These can be formed, for any real number $x$ with $0<x\le 1$, by defining $x_0:=x$ and inductively
$$
a_n:=\lfloor{x_n}\rfloor+1\qquad\text{and}\qquad x_{n+1}:=a_nx_n-1.
$$
It is easy to check that $1\le a_1\le a_2\le a_3\le\dots$, and that the sequence of $a_i$'s is eventually constant if and only if $x$ is rational.  This procedure yields expressions like the one displayed above, since
$$
x = \frac{1+x_1}{a_0}=\frac{1+\large{\frac{1+x_2}{a_1}}}{a_0}
=\frac{1+\large{\frac{1+\Large{\frac{1+x_3}{a_2}}}{a_1}}}{a_0} = \cdots
$$
I looked for these in google, wikipedia, and standard texts like Perron's, but couldn't find them.  Have they been studied?  I ask because a high school student invented these before my eyes today, and I'd like to tell him what he's rediscovered (assuming that in fact these have been studied before).

REPLY [0 votes]: You can find plenty on information on those special continued fractions at
the following link (American Mathematical Monthly):
https://www.facebook.com/AmerMathMonthly/photos/a.250425975006394.53155.241224542593204/1055084257873891/?type=3&theater
Also at:
https://youtu.be/mjSHenvXsEs


TITLE: Identity proofs
QUESTION [2 upvotes]: I'm strugling with these two combinatorial identities:
$$  \binom{n+2}{3}=\sum_{i=1}^{n} i(n+1-i) $$
and
$$\binom{n+1}{2}^2=\sum_{i=1}^{n} i^3$$
Please give me some footholds and hints

REPLY [0 votes]: For the first problem split the RHS:
$$\sum_{i=1}^n i(n+1-i) = \sum_{i=1}^n in +i - i^2 = n\sum_{i=1}^n i + \sum_{i=1}^n i - \sum_{i=1}^n i^2$$
Now we know that the some for consecutive integer from $1$ to $n$ is $\frac{n(n+1)}{2}$ and for the squares of consecutive integers from $1$ to $n$ is $\frac{n(n+1)(2n+1)}{6}$. So we substitute and we have:
$$n\sum_{i=1}^n i + \sum_{i=1}^n i - \sum_{i=1}^n i^2 = n \times \frac{n(n+1)}{2} + \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$$
$$n \times \frac{n(n+1)}{2} + \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{6} \times (3n + 3 - 2n - 1) = \frac{n(n+1)(n+2)}{3!} = \frac{(n+2)!}{3!(n-1)!} = \binom{n+2}{3}$$
And for the second problem it's a direct consequence from:
$$\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 = \binom{n+1}{2}^2$$
Note that this substitutions for the summations are well known, but they are proven using induction.


TITLE: Combination with repetition with limit
QUESTION [0 upvotes]: I have a following problem:
Having n boxes, and k balls where each box can contain maximum of m balls, we randomly insert the balls to the boxes. What is the expected number of non-empty boxes after the operation? If anyone would point me to sources when that problem was studied or solved I would be grateful, below I present what I managed  to get myself.
The formula I got is:
$\frac{\sum\limits_{i=\lceil \frac{k}{m}\rceil}^{k} i{n \choose i} \sum\limits_{h=0}^{\lfloor \frac{k-i}{m}\rfloor} (-1)^{h}{i \choose h} {k-1-hm \choose k-i-hm}}{\sum\limits_{i=0}^{\lfloor \frac{k}{m+1} \rfloor} (-1)^{i}{n \choose i}{n+k-1-i(l+1) \choose k-i(m+1)}} $
(By a computer program I wrote I already checked it on many input values, and it always worked. Explanation of the formula (the sum limits may be largely ignored, they just say outside where addends would be 0 anyway): In denominator I count all combinations that keep the max limit given. First component (for i = 0) is just a regular combination with repetition, then I subtract all that have at least 1 box overfilled, then I need to add those that have at least 2 box overfilled (since earlier I counted
them too many times) and so on according to inclusion-exclusion principle. In numerator situation is similar, yet every time I choose just i boxes that can be used (i.e. only them can be non-empty)).
This formula can be written in a little nicer form as follows:
$\frac{\sum\limits_{i=\lceil \frac{k}{m}\rceil}^{k} i{n \choose i} \sum\limits_{h=0}^{\lfloor \frac{k-i}{m}\rfloor} (-1)^{h}{i \choose h} {k-1-hm \choose i-1}}{\sum\limits_{i=0}^{\lfloor \frac{k}{m+1} \rfloor} (-1)^{i}{n \choose i}{n+k-1-i(m+1) \choose n-1}}$
But apart from that I don't know how to convert it to a digestible (i.e. reasonably efficiently computable) one. At this point I am not considering approximating, yet if someone had a good advice there, I would be grateful too (approximating every binomial separately doesn't seem to be a good idea).
Additionally those equations may help:
$\sum\limits_{i} (-1)^{i}{n \choose i}{n+k-1-i \choose n-1} = 0$
,
$\sum\limits_{i} (-1)^{i}{n \choose i}{n+k-1-2i \choose n-1} = {n \choose k}$
,
$\sum\limits_{i} (-1)^{i}{n \choose i}{n+k-1-(k+1)i \choose n-1} = {n + k -1 \choose k}$
(they come from taking the denominator, and setting max limit to respectively 0, 1, k, which respectively means no such combinations, combinations without repetition, combinations with repetition. Same can be done with nominator. Those equations may be actually new, since I didn't see them at http://en.wikipedia.org/wiki/Binomial_coefficient , Knuth's "The Art of Computer Programming" or anywhere else that I looked up to now).
The most problematic part is for sure the "hm" in the sum.
Any help or advice will be greatly appreciated.

REPLY [2 votes]: I didn't get the chance to work out the precise formula, but maybe this will help. It will give you a shorter answer anyway. In a nutshell: linearity of expectation is your friend.
Let $X$ be the number of empty boxes, and let $X_i$ be the random variable which is $0$ if box #$i$ is not empty and $1$ otherwise, so that
$ X=X_1 + \cdots +X_n.$
Then,
$$\mathbb{E}(X)=\mathbb{E}(X_1)+\cdots+\mathbb{E}(X_n)=n\mathbb{E}(X_1).$$
Now, $\mathbb{E}(X_1)=\mathbb{P}(X_1=1)$ which is easy to compute since it's just the probability of distributing the balls so that box #1 is empty, which is a classical application of inclusion-exclusion and the combination with repetition formula.
This way, you will get something much simpler. There is still an alternating sum, and there is no way around that, but there only one level of summation rather than the two nested levels in your formulas.
I hope this helps!