# Fundamental Solutions for the Klein-Gordon Equation
in de Sitter Spacetime

Karen Yagdjian, Anahit Galstian

###### Abstract

In this article we construct the fundamental solutions for the Klein-Gordon
equation in de Sitter spacetime. We use these fundamental solutions to
represent solutions of the Cauchy problem and to prove $L^{p}-L^{q}$ estimates
for the solutions of the equation with and without a source term.

Department of Mathematics, University of Texas-Pan American, 1201 W.
University Drive,
Edinburg, TX 78541-2999, USA

##  0 Introduction and Statement of Results

In this paper we construct the fundamental solutions for the Klein-Gordon
equation in the de Sitter spacetime and use these fundamental solutions to
find representations of the solutions to the Cauchy problem as well as
$L^{p}-L^{q}$ estimates for them.

After averaging on a suitable scale, our universe is homogeneous and
isotropic; therefore, the properties of the universe can be properly described
by treating the matter as a perfect homogeneous fluid. In the models of the
universe proposed by Einstein [12] and de Sitter [11] the line element is
connected with the proper mass density and the proper pressure in the universe
by the field equations for a perfect fluid. There are two alternatives which
lead to the solutions of Einstein and de Sitter, respectively [22, Sec.132].

In the models proposed by Einstein [12] and de Sitter [11] the universe is
assumed to be a static system, which means that we can introduce a system of
coordinates $x^{i}=(r,\theta,\phi,ct)$ in which the line element has the
static and spherically symmetric form

$ds^{2}=-b(r)c^{2}dt^{2}+a(r)dr^{2}+r^{2}(d\theta^{2}+\sin^{2}d\phi^{2}),$

where $a$ and $b$ are functions of $r$ only. On account of the assumed
homogeneity of the universe any point in the space may be taken as the origin
$r=0$ of the spatial system of coordinates. The functions $a(r)$ and $b(r)$
are connected with the proper mass density $\mu$ and the proper pressure $p$
in the universe by the field equations for a perfect fluid

$\displaystyle(\mu c^{2}+p)b^{\prime}=0,$ (0.1)
$\displaystyle\frac{b^{\prime}}{abr}-\frac{1}{r^{2}}\left(1-\frac{1}{a}\right)+\Lambda=\kappa
p,$ (0.2)
$\displaystyle\frac{a^{\prime}}{a^{2}r}+\frac{1}{r^{2}}\left(1-\frac{1}{a}\right)-\Lambda=\kappa\mu
c^{2},$ (0.3)

where $\Lambda$ is cosmological constant, while $p$ and $\mu$ are constants.
The general solution of the equation (0.3) is

$a(r)=\left(1-\frac{2M_{bh}}{r}-\frac{\Lambda r^{2}}{3}\right)^{-1}.$ (0.4)

The constant of integration $M_{bh}$ may have a meaning of the “mass of the
black holes”. There are two alternatives $b^{\prime}=0$ or $\mu c^{2}+p=0$,
which lead to the solutions of Einstein and de Sitter, respectively. In the
case of de Sitter universe $ab=constant$. By a trivial change of scale of the
time variable, this constant can, of course, always be made equal to 1, which
means

$b(r)=1-\frac{2M_{bh}}{r}-\frac{\Lambda r^{2}}{3}\,.$

The corresponding metric with the line element

$ds^{2}=-\left(1-\frac{2M_{bh}}{r}-\frac{\Lambda
r^{2}}{3}\right)c^{2}\,dt^{2}+\left(1-\frac{2M_{bh}}{r}-\frac{\Lambda
r^{2}}{3}\right)^{-1}dr^{2}+r^{2}(d\theta^{2}+\sin^{2}\theta\,d\phi^{2})$

is called the Schwarzschild - de Sitter metric. The Cauchy problem for the
linear wave equation without source term on the maximally extended
Schwarzschild - de Sitter spacetime in the case of non-extremal black-hole
corresponding to parameter values

$0<M_{bh}<\frac{1}{3\sqrt{\Lambda}},$

is considered by Dafermos and Rodnianski [8]. They proved that in the region
bounded by a set of black/white hole horizons and cosmological horizons,
solutions converge pointwise to a constant faster than any given polynomial
rate, where the decay is measured with respect to natural future-directed
advanced and retarded time coordinates.

There is an important question of local energy decay for the solution of the
wave equation and Klein-Gordon equation in black hole spacetime. Results on
the decay of local energy can provide a proof of the global nonlinear
stability of the spacetime. The global nonlinear stability we believe is only
known for Minkowski spacetime. Bony and Hafner [4] describe an expansion of
the solution of the wave equation in the Schwarzschild - de Sitter metric in
terms of resonances. The resonances correspond to the frequencies and rates of
dumping of signals emitted by the black hole in the presence of perturbations
(see [9, Chapter 4.35]). The main term in the expansion obtained in [4] is due
to a zero resonance. The error term decays polynomially if one permits a
logarithmic derivative loss in the angular directions and exponentially if one
permits a small derivative loss in the angular directions.

In the present paper we set $M_{bh}=0$ to exclude black holes. The case of the
Klein-Gordon equation in the presence of the black holes will be discussed in
the forthcoming paper. Thus, the de Sitter line element has the form

$ds^{2}=-\left(1-\frac{r^{2}}{R^{2}}\right)c^{2}\,dt^{2}+\left(1-\frac{r^{2}}{R^{2}}\right)^{-1}dr^{2}+r^{2}(d\theta^{2}+\sin^{2}\theta\,d\phi^{2})\,.$

The transformation

$\displaystyle r^{\prime}=\frac{r}{\sqrt{1-r^{2}/R^{2}}}e^{-ct/R}\,,\quad
t^{\prime}=t+\frac{R}{2c}\ln\left(1-\frac{r^{2}}{R^{2}}\right)\,,\quad\theta^{\prime}=\theta\,,\quad\phi^{\prime}=\phi$

leads to the following form for the line element:

$ds^{2}=-c^{2}\,d{t^{\prime}}^{2}+e^{2ct^{\prime}/R}(d{r^{\prime}}^{2}+r^{\prime
2}\,d{\theta^{\prime}}^{2}+r^{\prime
2}\sin^{2}\theta^{\prime}\,d{\phi^{\prime}}^{2})\,.$ (0.5)

Finally, defining new space coordinates $x^{\prime}$, $y^{\prime}$,
$z^{\prime}$ connected with $r^{\prime}$, $\theta^{\prime}$, $\phi^{\prime}$
by the usual equations connecting Cartesian coordinates and polar coordinates
in a Euclidean space, (0.5) may be written [22, Sec.134]

$ds^{2}=-c^{2}\,d{t^{\prime}}^{2}+e^{2ct^{\prime}/R}(d{x^{\prime}}^{2}+d{y^{\prime}}^{2}+d{z^{\prime}}^{2})\,.$
(0.6)

The new coordinates $x^{\prime}$, $y^{\prime}$, $z^{\prime}$, $t^{\prime}$ can
take all values from $-\infty$ to $\infty$. Here $R$ is the “radius” of the
universe. The de Sitter model allows us to get an explanation of the actual
red shift of spectral lines observed by Hubble and Humanson [22]. The de
Sitter model also enjoys the advantage of being the only time-dependent
cosmological model for which both particle creation and vacuum stress have
been explicitly evaluated by all known techniques [3]. In a certain sense all
solutions look like the de Sitter solution at late times [16].

The homogeneous and isotropic cosmological models possess highest symmetry
that makes them more amenable to rigorous study. Among them we mention FLRW
(Friedmann-Lematre-Robertson-Walker) models. The simplest class of
cosmological models can be obtained if we assume additionally that the metric
of the slices of constant time is flat and that the spacetime metric can be
written in the form

$ds^{2}=-dt^{2}+a^{2}(t)(d{x}^{2}+d{y}^{2}+d{z}^{2})$

with an appropriate scale factor $a(t)$. Although on the made assumptions, the
spatially flat FLRW models appear to give a good explanation of our universe.
The assumption that the universe is expanding leads to the positivity of the
time derivative $\frac{d}{dt}a(t)$. A further assumption that the universe
obeys the accelerated expansion suggests that the second derivative
$\frac{d^{2}}{dt^{2}}a(t)$ is positive. A substantial amount of the
observational material can be satisfactorily interpreted in terms of the
models which take into account existing acceleration of the recession of
distant galaxies.

The time dependence of the function $a(t)$ is determined by the Einstein field
equations for gravity. The Einstein equations with the cosmological constant
$\Lambda$ have form

$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=-8\pi GT_{\mu\nu}-\Lambda g_{\mu\nu},$

where term $\Lambda g_{\mu\nu}$ can be interpreted as an energy-momentum of
the vacuum. Even a small value of $\Lambda$ could have drastic effects on the
evolution of the universe. Under the assumption of FLRW symmetry the equation
of motion in the case of positive cosmological constant $\Lambda$ leads to the
solution

$a(t)=a(0)e^{\sqrt{\frac{\Lambda}{3}}t},$

which produces models with exponentially accelerated expansion. The model
described by the last equation is usually called the de Sitter model.

The unknown of principal importance in the Einstein equations is a metric $g$.
It comprises the basic geometrical feature of the gravitational field, and
consequently explains the phenomenon of the mutual gravitational attraction of
substance. In the presence of matter these equations contain a non-vanishing
right hand side $-8\pi GT_{\mu\nu}$. In general the matter fields described by
the function $\phi$ must satisfy equations of motion and in the case of the
massive scalar field the equation of motion is that $\phi$ should satisfy the
Klein-Gordon equation generated by the metric $g$. In the de Sitter universe
the equation for the scalar field with mass $m$ and potential function $V$
written out explicitly in coordinates is (See, e.g. [14, 26].)

$\phi_{tt}+nH\phi_{t}-e^{-2Ht}\bigtriangleup\phi+m^{2}\phi=-V^{\prime}(\phi)\,.$
(0.7)

Here $x\in{\mathbb{R}}^{n}$, $t\in{\mathbb{R}}$, and $\bigtriangleup$ is the
Laplace operator on the flat metric,
$\bigtriangleup:=\sum_{j=1}^{n}\frac{\partial^{2}}{\partial x_{j}^{2}}$, while
$H=\sqrt{\Lambda/3}$ is Hubble constant. If we introduce new unknown function
$u=e^{-\frac{n}{2}Ht}\phi$, then the semilinear Klein-Gordon equation for $u$
on de Sitter spacetime takes the form

$u_{tt}-e^{-2Ht}\bigtriangleup
u+M^{2}u=-e^{\frac{n}{2}t}V^{\prime}(e^{-\frac{n}{2}t}u),$ (0.8)

where $M^{2}:=m^{2}-\frac{n^{2}}{4}H^{2}$. Henceforth the quantity $M$, with
nonnegative real part $\Re M\geq 0$, defined by

$M^{2}:=m^{2}-\frac{n^{2}}{4}H^{2}\,,$

will be called the “curved mass” of particle. We extract a linear part of the
equation (0.8) as an initial model that must be treated first:

$u_{tt}-e^{-2Ht}\bigtriangleup u+M^{2}u=0\,.$ (0.9)

The fundamental solutions and the Cauchy problem for the equation with $M=0$
in the backward direction of time are considered in [33].

The de Sitter line element in the higher dimensional analogue of de Sitter
space is

$ds^{2}=-dt^{2}+e^{2Ht}\big{(}(dx^{1})^{2}+\ldots+(dx^{n})^{2}\big{)}\,.$

It is a simplified version of the multidimensional cosmological models with
the metric tensor given by

$g=-e^{2\gamma(t)}dt^{2}+e^{2\phi^{1}(t)}g_{1}+\ldots+e^{2\phi^{n}(t)}g_{n}\,,$

and can be chosen as a starting point for the study. The multidimensional
cosmological models have attracted a lot of attention during recent years in
constructing mathematical models of an anisotropic universe (see, e.g. [7, 16]
and references therein).

The equation (0.9) is strictly hyperbolic. That implies the well-posedness of
the Cauchy problem for (0.9) in the different functional spaces. The
coefficient of the equation is an analytic function and Holmgren’s theorem
implies a local uniqueness in the space of distributions. Moreover, the speed
of propagation is finite, namely, it is equal to $e^{-Ht}$ for every
$t\in{\mathbb{R}}$. The second-order strictly hyperbolic equation (0.9)
possesses two fundamental solutions resolving the Cauchy problem. They can be
written microlocally in terms of the Fourier integral operators [17], which
give a complete description of the wave front sets of the solutions. The
distance between two characteristic roots $\lambda_{1}(t,\xi)$ and
$\lambda_{2}(t,\xi)$ of the equation (0.9) is
$|\lambda_{1}(t,\xi)-\lambda_{2}(t,\xi)|=e^{-Ht}|\xi|$, $t\in{\mathbb{R}}$,
$\xi\in{\mathbb{R}}^{n}$. It tends to zero as $t$ approaches $\infty$. Thus,
the operator is not uniformly (that is for all $t\in{\mathbb{R}}$) strictly
hyperbolic. Moreover, the finite integrability of the characteristic roots,
$\int_{0}^{\infty}|\lambda_{i}(t,\xi)|dt<\infty$, leads to the existence of
so-called “horizon” for that equation. More precisely, any signal emitted from
the spatial point $x_{0}\in{\mathbb{R}}^{n}$ at time $t_{0}\in{\mathbb{R}}$
remains inside the ball $|x-x_{0}|<\frac{1}{H}e^{-Ht_{0}}$ for all time
$t\in(t_{0},\infty)$. The equation (0.9) is neither Lorentz invariant nor
invariant with respect to usual scaling and that brings additional
difficulties. In particular, it can cause a nonexistence of the $L^{p}-L^{q}$
decay for the solutions in the direction of time. In [30] it is mentioned the
model equation with permanently bounded domain of influence, power decay of
characteristic roots, and without $L^{p}-L^{q}$ decay for the solutions that
illustrates that phenomenon. The above mentioned $L^{p}-L^{q}$ decay estimates
are one of the important tools for studying nonlinear equations (see, e.g.
[25, 27]).

The time inversion transformation $t\to-t$ reduces the equation (0.9) to the
mathematically equivalent equation

$u_{tt}-e^{2Ht}\bigtriangleup u+M^{2}u=0\,.$ (0.10)

The wave equation, that is equation (0.10) with $M=0$, was investigated in
[15] by the second author. More precisely, in [15] the resolving operator for
the Cauchy problem

$u_{tt}-e^{2t}\bigtriangleup u=0,\qquad u(x,0)=\varphi_{0}(x),\quad
u_{t}(x,0)=\varphi_{1}(x)\,,$ (0.11)

is written as a sum of the Fourier integral operators with the amplitudes
given in terms of the Bessel functions and in terms of confluent
hypergeometric functions. In particular, it is proved in [15] that for $t>0$
the solution of the Cauchy problem (0.11) is given by

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle-i\frac{2}{(2\pi)^{n}}\int_{{\mathbb{R}}^{n}}\Big{\\{}e^{i[x\cdot\xi+(e^{t}-1)|\xi|]}H_{+}\big{(}\frac{1}{2};1;2ie^{t}|\xi|\big{)}H_{-}\big{(}\frac{3}{2};3;2i|\xi|\big{)}$
$\displaystyle\hskip
51.21504pt-\,e^{i[x\cdot\xi-(e^{t}-1)|\xi|]}H_{-}\big{(}\frac{1}{2};1;2ie^{t}|\xi|\big{)}H_{+}\big{(}\frac{3}{2};3;2i|\xi|\big{)}\Big{\\}}|\xi|^{2}\hat{\varphi}_{0}(\xi)d\xi$
$\displaystyle-i\frac{1}{(2\pi)^{n}}\int_{{\mathbb{R}}^{n}}\Big{\\{}e^{i[x\cdot\xi+(e^{t}-1)|\xi|]}H_{+}\big{(}\frac{1}{2};1;2ie^{t}|\xi|\big{)}H_{-}\big{(}\frac{1}{2};1;2i|\xi|\big{)}$
$\displaystyle\hskip
51.21504pt-\,e^{i[x\cdot\xi-(e^{t}-1)|\xi|]}H_{-}(\frac{1}{2};1;2ie^{t}|\xi|)H_{+}\big{(}\frac{1}{2};1;2i|\xi|\big{)}\Big{\\}}\hat{\varphi}_{1}(\xi)d\xi\,.$

In the notations of [2] the last functions are
$H_{-}(\alpha;\gamma;z)=e^{i\alpha\pi}\Psi(\alpha;\gamma;z)$ and
$H_{+}(\alpha;\gamma;z)=e^{i\alpha\pi}\Psi(\gamma-\alpha;\gamma;-z)$, where
function $\Psi(a;c;z)$ is defined in [2, Sec.6.5]. Here $\hat{\varphi}(\xi)$
is a Fourier transform of $\varphi(x)$.

The typical $L^{p}-L^{q}$ decay estimates obtained in [15] by dyadic
decomposition of the phase space contain some derivative loss. More precisely,
it is proved that for the solution $u=u(x,t)$ to the Cauchy problem (0.11)
with $n\geq 2$, $\varphi_{0}(x)\in C_{0}^{\infty}({\mathbb{R}}^{n})$ and
$\varphi_{1}(x)=0$ for all large $t\geq T>0$, the following estimate is
satisfied

$\|u(x,t)\|_{L^{q}({\mathbb{R}}^{n})}\leq
C(1+e^{t})^{-\frac{1}{2}(n-1)(\frac{1}{p}-\frac{1}{q})}\|\varphi_{0}\|_{W^{N}_{p}({\mathbb{R}}^{n})},$
(0.12)

where $1<p\leq 2$, $\frac{1}{p}+\frac{1}{q}=1$, and
$\frac{1}{2}(n+1)(\frac{1}{p}-\frac{1}{q})\leq
N<\frac{1}{2}(n+1)(\frac{1}{p}-\frac{1}{q})+1$ and
$W^{N}_{p}({\mathbb{R}}^{n})$ is the Sobolev space. In particular, the
derivative loss, $N$, is positive, unless $p=q=2$. This derivative loss
phenomenon exists for the classical wave equation as well. Indeed, it is well-
known (see, e.g., [19, 20, 24]) that for the Cauchy problem
$u_{tt}-\bigtriangleup u=0$, $u(x,0)=\varphi(x)$, $u_{t}(x,0)=0$, the estimate
$\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x}^{n})}\leq
C\|\varphi(x)\|_{L^{q}({\mathbb{R}}_{x}^{n})}$ fails to fulfill even for small
positive $t$ unless $q=2$. The obstacle is created by the distinguishing
feature of the (different from translation) Fourier integral operators of
order zero, which compose a resolving operator.

According to Theorem 1 [15], for the solution $u=u(x,t)$ to the Cauchy problem
(0.11) with $n\geq 2$, $\varphi_{0}(x)=0$ and $\varphi_{1}(x)\in
C_{0}^{\infty}({\mathbb{R}}^{n})$ for all large $t\geq T>0$ and for any small
$\varepsilon>0$, the following estimate is satisfied

$\|u(x,t)\|_{L^{q}({\mathbb{R}}^{n})}\leq
C_{\varepsilon}(1+t)(1+e^{t})^{r_{0}-n(\frac{1}{p}-\frac{1}{q})}\|\varphi_{1}\|_{W^{N}_{p}({\mathbb{R}}^{n})},\,$

where $1<p\leq 2$, $\frac{1}{p}+\frac{1}{q}=1$,
$r_{0}=\max\\{\varepsilon;\frac{(n+1)}{2}(\frac{1}{p}-\frac{1}{q})-\frac{1}{q}\\}$,
$\frac{n+1}{2}(\frac{1}{p}-\frac{1}{q})-\frac{1}{q}\leq
N<\frac{n+1}{2}(\frac{1}{p}-\frac{1}{q})+\frac{1}{p}$.

The nonlinear equations (0.7) and (0.8) are those we would like to solve, but
the linear problem is a natural first step. Exceptionally efficient tool for
the studying nonlinear equations is a fundamental solution of the associate
linear operator.

The fundamental solutions for the operator of the equation (0.11) are
constructed in [33] and the representations of the solutions of the Cauchy
problem

$u_{tt}-e^{2t}\bigtriangleup u=f(x,t),\qquad u(x,0)=\varphi_{0}(x),\quad
u_{t}(x,0)=\varphi_{1}(x)\,,$ (0.13)

are given in the terms of the solutions of wave equation in Minkowski
spacetime. Then in [33] for $n\geq 2$ the following decay estimate

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}\\!\\!$
$\displaystyle\\!\\!\leq$
$\displaystyle\\!\\!\\!\\!Ce^{t(2s-n(\frac{1}{p}-\frac{1}{q}))}\int_{0}^{t}(1+t-b)\|f(x,b)\|_{{L}^{p}({\mathbb{R}}^{n})}\,db$
$\displaystyle+C(e^{t}-1)^{2s-n(\frac{1}{p}-\frac{1}{q})}\left\\{\|\varphi_{0}(x)\|_{{L}^{p}({\mathbb{R}}^{n})}+\|\varphi_{1}(x)\|_{{L}^{p}({\mathbb{R}}^{n})}(1+t)(1-e^{-t})\right\\}$

is proven, provided that $s\geq 0$, $1<p\leq 2$, $\frac{1}{p}+\frac{1}{q}=1$,
$\frac{1}{2}(n+1)\left(\frac{1}{p}-\frac{1}{q}\right)\leq 2s\leq
n\left(\frac{1}{p}-\frac{1}{q}\right)<2s+1$. Moreover, this estimate is
fulfilled for $n=1$ and $s=0$ as well as if $\varphi_{0}(x)=0$ and
$\varphi_{1}(x)=0$. Case of $n=1$, $f(x,t)=0$, and non-vanishing
$\varphi_{1}(x)$ and $\varphi_{1}(x)$ also is discussed in Section 8 [33].

In the construction of the fundamental solutions for the operator (0.9) we
follow the approach proposed in [29] that allows us to represent the
fundamental solutions as some integral of the family of the fundamental
solutions of the Cauchy problem for the wave equation without source term. The
kernel of that integral contains the Gauss’s hypergeometric function. In that
way, many properties of the wave equation can be extended to the hyperbolic
equations with the time dependent speed of propagation. That approach was
successfully applied in [31, 32] by the first author to investigate the
semilinear Tricomi-type equations.

Thus, in the present paper we consider Klein-Gordon operator in de Sitter
model of the universe, that is

${\mathcal{S}}:=\partial_{t}^{2}-e^{-2t}\bigtriangleup+M^{2}\,,$

where $M\geq 0$ is the reduced mass, $x\in{\mathbb{R}}^{n}$,
$t\in{\mathbb{R}}$. We look for the fundamental solution
$E=E(x,t;x_{0},t_{0})$,

$E_{tt}-e^{-2t}\Delta E-M^{2}E=\delta(x-x_{0},t-t_{0}),$

with a support in the “forward light cone” $D_{+}(x_{0},t_{0})$,
$x_{0}\in{\mathbb{R}}^{n}$, $t_{0}\in{\mathbb{R}}$, and for the fundamental
solution with a support in the “backward light cone” $D_{-}(x_{0},t_{0})$,
$x_{0}\in{\mathbb{R}}^{n}$, $t_{0}\in{\mathbb{R}}$, defined as follows

$\displaystyle D_{\pm}(x_{0},t_{0})$ $\displaystyle:=$
$\displaystyle\Big{\\{}(x,t)\in{\mathbb{R}}^{n+1}\,;\,|x-x_{0}|\leq\pm(e^{-t_{0}}-e^{-t})\,\Big{\\}}\,.$
(0.14)

In fact, any intersection of $D_{-}(x_{0},t_{0})$ with the hyperplane
$t=const<t_{0}$ determines the so-called dependence domain for the point
$(x_{0},t_{0})$, while the intersection of $D_{+}(x_{0},t_{0})$ with the
hyperplane $t=const>t_{0}$ is the so-called domain of influence of the point
$(x_{0},t_{0})$. The equation (0.9) is non-invariant with respect to time
inversion. Moreover, the dependence domain is wider than any given ball if
time $const>t_{0}$ is sufficiently large, while the domain of influence is
permanently, for all time $const<t_{0}$, in the ball of the radius
$e^{t_{0}}$.

Define for $t_{0}\in{\mathbb{R}}$ in the domain $D_{+}(x_{0},t_{0})\cup
D_{-}(x_{0},t_{0})$ the function

$\displaystyle E(x,t;x_{0},t_{0})$ $\displaystyle=$
$\displaystyle(4e^{-t_{0}-t})^{iM}\Big{(}(e^{-t}+e^{-t_{0}})^{2}-(x-x_{0})^{2}\Big{)}^{-\frac{1}{2}-iM}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-t_{0}}-e^{-t})^{2}-(x-x_{0})^{2}}{(e^{-t_{0}}+e^{-t})^{2}-(x-x_{0})^{2}}\Big{)},$

where $F\big{(}a,b;c;\zeta\big{)}$ is the hypergeometric function (See, e.g.
[2].). Let $E(x,t;0,b)$ be function ( ‣ 0 Introduction and Statement of
Results), and set

${\mathcal{E}}_{+}(x,t;0,t_{0}):=\cases{E(x,t;0,t_{0})\quad\mbox{\rm
in}\,\,D_{+}(0,t_{0}),\cr 0\hskip 56.9055pt\mbox{\rm
elsewhere}}\,,\qquad{\mathcal{E}}_{-}(x,t;0,t_{0}):=\cases{E(x,t;0,t_{0})\quad\mbox{\rm
in}\,\,D_{-}(0,t_{0}),\cr 0\hskip 56.9055pt\mbox{\rm elsewhere}}\,.$

Since function $E=E(x,t;0,t_{0})$ is smooth in $D_{\pm}(0,t_{0})$ and is
locally integrable, it follows that ${\mathcal{E}}_{+}(x,t;0,$ $t_{0})$ and
${\mathcal{E}}_{-}(x,t;0,$ $t_{0})$ are distributions whose supports are in
$D_{+}(0,t_{0})$ and $D_{-}(0,t_{0})$, respectively. The next theorem gives
our first result.

###### Theorem 0.1

Suppose that $n=1$. The distributions ${\mathcal{E}}_{+}(x,t;0,t_{0})$ and
${\mathcal{E}}_{-}(x,t;0,t_{0})$ are the fundamental solutions for the
operator ${\mathcal{S}}=\partial_{t}^{2}-e^{-2t}\bigtriangleup+M^{2}$ relative
to the point $(0,t_{0})$, that is

${\mathcal{S}}{\mathcal{E}}_{\pm}(x,t;0,t_{0})=\delta(x,t-t_{0}),$

or

$\frac{\partial^{2}}{\partial
t^{2}}{\mathcal{E}}_{\pm}(x,t;0,t_{0})-e^{-2t}\frac{\partial^{2}}{\partial
x^{2}}{\mathcal{E}}_{\pm}(x,t;0,t_{0})+M^{2}{\mathcal{E}}_{\pm}(x,t;0,t_{0})=\delta(x,t-t_{0}).$

To motivate one construction for the higher dimensional case $n\geq 2$ we
follow the approach suggested in [29] and represent fundamental solution
${\mathcal{E}}_{+}(x,t;0,t_{0})$ as follows

$\displaystyle{\mathcal{E}}_{+}(x,t;0,t_{0})$ $\displaystyle=$
$\displaystyle\int_{e^{-t}-e^{-t_{0}}}^{e^{-t_{0}}-e^{-t}}{\mathcal{E}}^{string}(x,r)(4e^{-t_{0}-t})^{iM}\Big{(}(e^{-t_{0}}+e^{-t})^{2}-r^{2}\Big{)}^{-\frac{1}{2}-iM}$
$\displaystyle\hskip 71.13188pt\times
F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-t_{0}}-e^{-t})^{2}-r^{2}}{(e^{-t_{0}}+e^{-t})^{2}-r^{2}}\Big{)}\,dr,\quad
t>t_{0},$

where the distribution ${\mathcal{E}}^{string}(x,t)$ is the fundamental
solution of the Cauchy problem for the string equation:

$\frac{\partial^{2}}{\partial
t^{2}}{\mathcal{E}}^{string}-\frac{\partial^{2}}{\partial
x^{2}}{\mathcal{E}}^{string}=0,\qquad{\mathcal{E}}^{string}(x,0)=\delta(x),\,\,\,{\mathcal{E}}^{string}_{t}(x,0)=0\,.$

Hence, ${\mathcal{E}}^{string}(x,t)=\frac{1}{2}\\{\delta(x+t)+\delta(x-t)\\}$.
The integral makes sense in the topology of the space of distributions. The
fundamental solution ${\mathcal{E}}_{-}(x,t;0,t_{0})$ for $t<t_{0}$ admits a
similar representation.

We appeal to the wave equation in Minkowski spacetime to obtain in the next
theorem very similar representations of the fundamental solutions of the
higher dimensional equation in de Sitter spacetime with $n\geq 2$.

###### Theorem 0.2

If $x\in{\mathbb{R}}^{n}$, $n\geq 2$, then for the operator
${\mathcal{S}}=\partial_{t}^{2}-e^{-2t}\bigtriangleup+M^{2}$ the fundamental
solution ${\mathcal{E}}_{+,n}(x,t;x_{0},t_{0})$
$(={\mathcal{E}}_{+,n}(x-x_{0},t;0,t_{0}))$ with a support in the forward cone
$D_{+}(x_{0},t_{0})$, $x_{0}\in{\mathbb{R}}^{n}$, $t_{0}\in{\mathbb{R}}$,
supp$\,{\mathcal{E}}_{+,n}\subseteq D_{+}(x_{0},t_{0})$, is given by the
following integral $(t>t_{0})$

$\displaystyle{\mathcal{E}}_{+,n}(x-x_{0},t;0,t_{0})$ $\displaystyle=$
$\displaystyle
2\int_{0}^{e^{-t_{0}}-e^{-t}}{\mathcal{E}}^{w}(x-x_{0},r)(4e^{-t_{0}-t})^{iM}\Big{(}(e^{-t_{0}}+e^{-t})^{2}-r^{2}\Big{)}^{-\frac{1}{2}-iM}$
(0.16) $\displaystyle\hskip 71.13188pt\times
F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-t_{0}}-e^{-t})^{2}-r^{2}}{(e^{-t_{0}}+e^{-t})^{2}-r^{2}}\Big{)}\,dr.$

Here the function ${\mathcal{E}}^{w}(x,t;b)$ is a fundamental solution to the
Cauchy problem for the wave equation

${\mathcal{E}}^{w}_{tt}-\bigtriangleup{\mathcal{E}}^{w}=0\,,\quad{\mathcal{E}}^{w}(x,0)=\delta(x)\,,\quad{\mathcal{E}}^{w}_{t}(x,0)=0\,.$

The fundamental solution ${\mathcal{E}}_{-,n}(x,t;x_{0},t_{0})$
$(={\mathcal{E}}_{-,n}(x-x_{0},t;0,t_{0}))$ with a support in the backward
cone $D_{-}(x_{0},t_{0})$, $x_{0}\in{\mathbb{R}}^{n}$, $t_{0}\in{\mathbb{R}}$,
supp$\,{\mathcal{E}}_{-,n}\subseteq D_{-}(x_{0},t_{0})$, is given by the
following integral $(t<t_{0})$

$\displaystyle{\mathcal{E}}_{+,n}(x-x_{0},t;0,t_{0})$ $\displaystyle=$
$\displaystyle-2\int_{e^{-t_{0}}-e^{-t}}^{0}{\mathcal{E}}^{w}(x-x_{0},r)(4e^{-t_{0}-t})^{iM}\Big{(}(e^{-t_{0}}+e^{-t})^{2}-r^{2}\Big{)}^{-\frac{1}{2}-iM}$
(0.17) $\displaystyle\hskip 71.13188pt\times
F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-t_{0}}-e^{-t})^{2}-r^{2}}{(e^{-t_{0}}+e^{-t})^{2}-r^{2}}\Big{)}\,dr.$

In particular, the formula (0.16) shows that Huygens’s Principle is not valid
for the waves propagating in de Sitter model of the universe. Fields
satisfying a wave equation in de Sitter model of the universe can be
accompanied by tails propagating inside the light cone. This phenomenon will
be discussed in the spirit of [28] in the forthcoming paper.

Next we use Theorem 0.1 to solve the Cauchy problem for the one-dimensional
equation

$u_{tt}-e^{-2t}u_{xx}+M^{2}u=f(x,t)\,,\qquad t>0\,,\quad x\in{\mathbb{R}}\,,$
(0.18)

with vanishing initial data,

$u(x,0)=u_{t}(x,0)=0\,.$ (0.19)

###### Theorem 0.3

Assume that the function $f$ is continuous along with its all second order
derivatives, and that for every fixed $t$ it has a compact support,
supp$f(\cdot,t)\subset{\mathbb{R}}$. Then the function $u=u(x,t)$ defined by

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,f(y,b)(4e^{-b-t})^{iM}\Big{(}(e^{-t}+e^{-b})^{2}-(x-y)^{2}\Big{)}^{-\frac{1}{2}-iM}$
$\displaystyle\hskip 71.13188pt\times
F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-(x-y)^{2}}{(e^{-b}+e^{-t})^{2}-(x-y)^{2}}\Big{)}$

is a $C^{2}$-solution to the Cauchy problem for the equation (0.18) with
vanishing initial data, (0.19).

The representation of the solution of the Cauchy problem for the one-
dimensional case ($n=1$) of the equation (0.9) without source term is given by
the next theorem.

###### Theorem 0.4

The solution $u=u(x,t)$ of the Cauchy problem

$u_{tt}-e^{-2t}u_{xx}+M^{2}u=0\,,\qquad u(x,0)=\varphi_{0}(x)\,,\qquad
u_{t}(x,0)=\varphi_{1}(x)\,,$ (0.20)

with $\varphi_{0},\varphi_{1}\in C_{0}^{\infty}({\mathbb{R}})$ can be
represented as follows

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\frac{1}{2}e^{\frac{t}{2}}\Big{[}\varphi_{0}(x+1-e^{-t})+\varphi_{0}(x-1+e^{-t})\Big{]}+\int_{0}^{1-e^{-t}}\big{[}\varphi_{0}(x-z)+\varphi_{0}(x+z)\big{]}K_{0}(z,t)\,dz$
$\displaystyle+\,\,\int_{0}^{1-e^{-t}}\,\Big{[}\varphi_{1}(x-z)+\varphi_{1}(x+z)\Big{]}K_{1}(z,t)dz\,,$

where the kernels $K_{0}(z,t)$ and $K_{1}(z,t)$ are defined by

$\displaystyle K_{0}(z,t)$ $\displaystyle:=$
$\displaystyle-\left[\frac{\partial}{\partial b}E(z,t;0,b)\right]_{b=0}$
$\displaystyle=$
$\displaystyle(4e^{-t})^{iM}\big{(}(e^{-t}+1)^{2}-z^{2}\big{)}^{-iM}\frac{1}{[(1-e^{-t})^{2}-z^{2}]\sqrt{(e^{-t}+1)^{2}-z^{2}}}$
$\displaystyle\times\Bigg{[}\big{(}e^{-t}-1-iM(e^{-2t}-1-z^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(1-e^{-t})^{2}-z^{2}}{(1+e^{-t})^{2}-z^{2}}\Big{)}$
$\displaystyle\hskip
28.45274pt+\big{(}1-e^{-2t}+z^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(1-e^{-t})^{2}-z^{2}}{(1+e^{-t})^{2}-z^{2}}\Big{)}\Bigg{]}$

and

$\displaystyle K_{1}(z,t)$ $\displaystyle:=$ $\displaystyle E(z,t;0,0)$
$\displaystyle=$
$\displaystyle(4e^{-t})^{iM}\big{(}(1+e^{-t})^{2}-z^{2}\big{)}^{-\frac{1}{2}-iM}F\left(\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(1-e^{-t})^{2}-z^{2}}{(1+e^{-t})^{2}-z^{2}}\right),\,0\leq
z\leq 1-e^{-t},$

respectively.

The kernels $K_{0}(z,t)$ and $K_{1}(z,t)$ play leading roles in the derivation
of $L^{p}-L^{q}$ estimates. Their main properties are listed and proved in
Section 8.

Next we turn to the higher-dimensional equation with $n\geq 2$.

###### Theorem 0.5

If $n$ is odd, $n=2m+1$, $m\in{\mathbb{N}}$, then the solution $u=u(x,t)$ to
the Cauchy problem

$u_{tt}-e^{-2t}\Delta u+M^{2}u=f,\quad u(x,0)=0,\quad u_{t}(x,0)=0,$ (0.21)

with $f\in C^{\infty}({\mathbb{R}}^{n+1})$ and with the vanishing initial data
is given by the next expression

$\displaystyle u(x,t)$ $\displaystyle=$ $\displaystyle
2\int_{0}^{t}db\int_{0}^{e^{-b}-e^{-t}}dr_{1}\,\left(\frac{\partial}{\partial
r}\Big{(}\frac{1}{r}\frac{\partial}{\partial
r}\Big{)}^{\frac{n-3}{2}}\frac{r^{n-2}}{\omega_{n-1}c_{0}^{(n)}}\\!\\!\int_{S^{n-1}}f(x+ry,b)\,dS_{y}\right)_{r=r_{1}}$
(0.22)
$\displaystyle\times(4e^{-b-t})^{iM}\left((e^{-t}+e^{-b})^{2}-r_{1}^{2}\right)^{-\frac{1}{2}-iM}F\left(\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-r_{1}^{2}}{(e^{-b}+e^{-t})^{2}-r_{1}^{2}}\right)\\!\\!,$

where $c_{0}^{(n)}=1\cdot 3\cdot\ldots\cdot(n-2)$. Constant $\omega_{n-1}$ is
the area of the unit sphere $S^{n-1}\subset{\mathbb{R}}^{n}$.

If $n$ is even, $n=2m$, $m\in{\mathbb{N}}$, then the solution $u=u(x,t)$ is
given by the next expression

$\displaystyle u(x,t)$ $\displaystyle=$ $\displaystyle
2\int_{0}^{t}db\int_{0}^{e^{-b}-e^{-t}}dr_{1}\,\left(\frac{\partial}{\partial
r}\Big{(}\frac{1}{r}\frac{\partial}{\partial
r}\Big{)}^{\frac{n-2}{2}}\frac{2r^{n-1}}{\omega_{n-1}c_{0}^{(n)}}\\!\\!\int_{B_{1}^{n}(0)}\frac{f(x+ry,b)}{\sqrt{1-|y|^{2}}}\,dV_{y}\right)_{r=r_{1}}$
(0.23)
$\displaystyle\times(4e^{-b-t})^{iM}\left((e^{-t}+e^{-b})^{2}-r_{1}^{2}\right)^{-\frac{1}{2}-iM}F\left(\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-r_{1}^{2}}{(e^{-b}+e^{-t})^{2}-r_{1}^{2}}\right)\\!\\!.$

Here $B_{1}^{n}(0):=\\{|y|\leq 1\\}$ is the unit ball in ${\mathbb{R}}^{n}$,
while $c_{0}^{(n)}=1\cdot 3\cdot\ldots\cdot(n-1)$.

Thus, in both cases, of even and odd $n$, one can write

$\displaystyle u(x,t)$ $\displaystyle=$ $\displaystyle
2\int_{0}^{t}db\int_{0}^{e^{-b}-e^{-t}}dr\,v(x,r;b)(4e^{-b-t})^{iM}\left((e^{-t}+e^{-b})^{2}-r^{2}\right)^{-\frac{1}{2}-iM}$
(0.24) $\displaystyle\hskip 71.13188pt\times
F\left(\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-r^{2}}{(e^{-b}+e^{-t})^{2}-r^{2}}\right),$

where the function $v(x,t;b)$ is a solution to the Cauchy problem for the wave
equation

$v_{tt}-\bigtriangleup v=0\,,\quad v(x,0;b)=f(x,b)\,,\quad v_{t}(x,0;b)=0\,.$

The next theorem gives representation of the solutions of equation (0.9) with
the initial data prescribed at $t=0$.

###### Theorem 0.6

The solution $u=u(x,t)$ to the Cauchy problem

$u_{tt}-e^{-2t}\bigtriangleup u+M^{2}u=0\,,\quad u(x,0)=\varphi_{0}(x)\,,\quad
u_{t}(x,0)=\varphi_{1}(x)\,,$ (0.25)

with $\varphi_{0}$, $\varphi_{1}\in C_{0}^{\infty}({\mathbb{R}}^{n})$, $n\geq
2$, can be represented as follows:

$\displaystyle u(x,t)$ $\displaystyle=$ $\displaystyle
e^{\frac{t}{2}}v_{\varphi_{0}}(x,\phi(t))+\,2\int_{0}^{1}v_{\varphi_{0}}(x,\phi(t)s)K_{0}(\phi(t)s,t)\phi(t)\,ds$
(0.26)
$\displaystyle+\,2\int_{0}^{1}v_{\varphi_{1}}(x,\phi(t)s)K_{1}(\phi(t)s,t)\phi(t)\,ds,\quad
x\in{\mathbb{R}}^{n},\,\,t>0\,,$

$\phi(t):=1-e^{-t}$, by means of the kernels $K_{0}$ and $K_{1}$ have been
defined in Theorem 0.4. Here for $\varphi\in C_{0}^{\infty}({\mathbb{R}}^{n})$
and for $x\in{\mathbb{R}}^{n}$, $n=2m+1$, $m\in{\mathbb{N}}$,

$\displaystyle v_{\varphi}(x,\phi(t)s):=\Bigg{(}\frac{\partial}{\partial
r}\Big{(}\frac{1}{r}\frac{\partial}{\partial
r}\Big{)}^{\frac{n-3}{2}}\frac{r^{n-2}}{\omega_{n-1}c_{0}^{(n)}}\int_{S^{n-1}}\varphi(x+ry)\,dS_{y}\Bigg{)}_{r=\phi(t)s}$

while for $x\in{\mathbb{R}}^{n}$, $n=2m$, $m\in{\mathbb{N}}$,

$v_{\varphi}(x,\phi(t)s):=\Bigg{(}\frac{\partial}{\partial
r}\Big{(}\frac{1}{r}\frac{\partial}{\partial
r}\Big{)}^{\frac{n-2}{2}}\frac{2r^{n-1}}{\omega_{n-1}c_{0}^{(n)}}\int_{B_{1}^{n}(0)}\frac{1}{\sqrt{1-|y|^{2}}}\varphi(x+ry)\,dV_{y}\Bigg{)}_{r=s\phi(t)}\,.$

The function $v_{\varphi}(x,\phi(t)s)$ coincides with the value
$v(x,\phi(t)s)$ of the solution $v(x,t)$ of the Cauchy problem

$v_{tt}-\bigtriangleup v=0,\quad v(x,0)=\varphi(x),\quad v_{t}(x,0)=0\,.$

As a consequence of the above theorems we obtain in Sections 9-10 for $n\geq
2$ and for the particles with “large” mass $m$, $m\geq n/2$, that is, with
nonnegative curved mass $M\geq 0$, the following $L^{p}-L^{q}$ estimate

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}\\!\\!$
$\displaystyle\\!\\!\leq\\!\\!$
$\displaystyle\\!\\!C\int_{0}^{t}\|f(x,b)\|_{{L}^{p}({\mathbb{R}}^{n})}e^{-b}\left(e^{-b}-e^{-t}\right)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}\left(1+t-b\right)db$
(0.27)
$\displaystyle+C(1+t)(1-e^{-t})^{2s-n(\frac{1}{p}-\frac{1}{q})}\Big{\\{}e^{\frac{t}{2}}\|\varphi_{0}(x)\|_{{L}^{p}({\mathbb{R}}^{n})}+(1-e^{-t})\|\varphi_{1}\|_{{L}^{p}({\mathbb{R}}^{n})}\Big{\\}}$

provided that $s\geq 0$, $1<p\leq 2$, $\frac{1}{p}+\frac{1}{q}=1$,
$\frac{1}{2}(n+1)\left(\frac{1}{p}-\frac{1}{q}\right)\leq 2s\leq
n\left(\frac{1}{p}-\frac{1}{q}\right)<2s+1$. Moreover, according to Theorem
7.1 the estimate (0.27) is valid for $n=1$ and $s=0$ as well as if
$\varphi_{0}(x)=0$ and $\varphi_{1}(x)=0$. Case of $n=1$, $f(x,t)=0$, and non-
vanishing $\varphi_{1}(x)$ and $\varphi_{1}(x)$ is discussed in Section 8. The
case of particles with small mass $m<n/2$ will be discussed in the forthcoming
paper.

The paper is organized as follows. In Section 1 we construct the Riemann
function of the operator of (0.9) in the characteristic coordinates for the
case of $n=1$. That Riemann function used in Section 2 to prove Theorem 0.1.
Then in Section 3 we apply the fundamental solutions to solve the Cauchy
problem with the source term and with the vanishing initial data given at
$t=0$. More precisely, we give a representation formula for the solutions. In
that section we also prove several basic properties of the function
$E(x,t;y,b)$. In Sections 4-5 we use formulas of Section 3 to derive and to
complete the list of representation formulas for the solutions of the Cauchy
problem for the case of one-dimensional spatial variable. The higher-
dimensional equation with the source term is considered in Section 6, where we
derive a representation formula for the solutions of the Cauchy problem with
the source term and with the vanishing initial data given at $t=0$. In the
same section this formula is used to derive the fundamental solutions of the
operator and to complete the proof of Theorem 0.6. Then in Sections 7-10 we
establish the $L^{p}-L^{q}$ decay estimates. Applications of all these results
to the nonlinear equations will be done in the forthcoming paper.

##  1 The Riemann Function

In the characteristic coordinates $l$ and $m$,

$l=x+e^{-t},\qquad m=x-e^{-t},$ (1.1)

one has

$\displaystyle\frac{\partial^{2}}{\partial t^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(l-m)\left(\frac{\partial}{\partial
l}-\frac{\partial}{\partial
m}\right)+\frac{1}{4}(l-m)^{2}\left(\frac{\partial^{2}}{\partial
l^{2}}-2\frac{\partial^{2}}{\partial l\,\partial
m}+\frac{\partial^{2}}{\partial m^{2}}\right)$

and

$\displaystyle e^{-2t}\frac{\partial^{2}}{\partial
x^{2}}=\frac{1}{4}(l-m)^{2}\left(\frac{\partial^{2}}{\partial
l^{2}}+2\frac{\partial^{2}}{\partial l\,\partial
m}+\frac{\partial^{2}}{\partial m^{2}}\right).$

Then the operator ${\mathcal{S}}$ of the equation (0.18) reads

${\mathcal{S}}:=\frac{\partial^{2}}{\partial
t^{2}}-e^{-2t}\frac{\partial^{2}}{\partial
x^{2}}+M^{2}=-(l-m)^{2}\Bigg{\\{}\frac{\partial^{2}}{\partial l\,\partial
m}-\frac{1}{2(l-m)}\Big{(}\frac{\partial}{\partial l}-\frac{\partial}{\partial
m}\Big{)}-\frac{1}{(l-m)^{2}}M^{2}\Bigg{\\}}\,.$ (1.2)

In particular, in the new variables the equation

$\left(\frac{\partial^{2}}{\partial t^{2}}-e^{-2t}\frac{\partial^{2}}{\partial
x^{2}}+M^{2}\right)u=0\quad\mbox{\rm
implies}\quad\Bigg{\\{}\frac{\partial^{2}}{\partial l\,\partial
m}-\frac{1}{2(l-m)}\Big{(}\frac{\partial}{\partial l}-\frac{\partial}{\partial
m}\Big{)}\Bigg{\\}}u-\frac{1}{(l-m)^{2}}M^{2}u=0\,.$

We need the following lemma with $\displaystyle\gamma=\frac{1}{2}+iM$.

###### Lemma 1.1

The function

$V(l,m;a,b)=(l-b)^{-\gamma}(a-m)^{-\gamma}F\Big{(}\gamma,\gamma;1;\frac{(l-a)(m-b)}{(l-b)(m-a)}\Big{)}$

solves the equation

$\Bigg{\\{}\frac{\partial^{2}}{\partial l\,\partial
m}-\frac{1}{(l-m)}\gamma\Big{(}\frac{\partial}{\partial
l}-\frac{\partial}{\partial m}\Big{)}\Bigg{\\}}V(l,m;a,b)=0\,.$ (1.3)

Proof. We denote the argument of the hypergeometric function by $z$, and
evaluate its derivatives,

$z:=\frac{(l-a)(m-b)}{(l-b)(m-a)}\,,\qquad\frac{\partial}{\partial
l}z=\frac{(a-b)(b-m)}{(l-b)^{2}(a-m)}\,,\quad\frac{\partial}{\partial
m}z=-\frac{(a-b)(a-l)}{(b-l)(a-m)^{2}}\,.$

Further, we obtain

$\frac{\partial}{\partial
l}V(l,m;a,b)=(a-m)^{-\gamma}(l-b)^{-\gamma-1}\Bigg{\\{}-\gamma
F\Big{(}\gamma,\gamma;1;z\Big{)}+\frac{(a-b)(b-m)}{(l-b)(a-m)}F^{\prime}_{z}\Big{(}\gamma,\gamma;1;z\Big{)}\Bigg{\\}}\,.$

Next

$\frac{\partial}{\partial
m}V(l,m;a,b)=(a-m)^{-\gamma}(l-b)^{-\gamma}\Bigg{\\{}\gamma(a-m)^{-1}F\Big{(}\gamma,\gamma;1;z\Big{)}-\frac{(a-b)(a-l)}{(b-l)(a-m)^{2}}F_{z}^{\prime}\Big{(}\gamma,\gamma;1;z\Big{)}\Bigg{\\}}\,.$

Then

$\displaystyle\left(\frac{\partial}{\partial l}-\frac{\partial}{\partial
m}\right)V(l,m;a,b)$ $\displaystyle=$
$\displaystyle(a-m)^{-\gamma}(l-b)^{-\gamma}\left(-\gamma\right)F\big{(}\gamma,\gamma;1;z\big{)}\Big{\\{}\frac{1}{(l-b)}+\frac{1}{(a-m)}\Big{\\}}$
$\displaystyle+(a-m)^{-\gamma-1}(l-b)^{-\gamma-1}F_{z}^{\prime}\big{(}\gamma,\gamma;1;z\big{)}(a-b)\left\\{\frac{(b-m)}{(l-b)}-\frac{(a-l)}{(a-m)}\right\\}\,.$

Furthermore,

$\displaystyle\frac{\partial^{2}}{\partial l\,\partial m}V(l,m;a,b)$
$\displaystyle=$
$\displaystyle(a-m)^{-\gamma-1}\Bigg{[}-\gamma(l-b)^{-\gamma-1}\Bigg{\\{}\gamma
F\Big{(}\gamma,\gamma;1;z\Big{)}-\frac{(a-b)(a-l)}{(b-l)(a-m)}F_{z}^{\prime}\Big{(}\gamma,\gamma;1;z\Big{)}\Bigg{\\}}\Bigg{]}$
$\displaystyle+\,(a-m)^{-\gamma-1}\Bigg{[}(l-b)^{-\gamma}\frac{\partial}{\partial
l}\Bigg{\\{}\gamma
F\Big{(}\gamma,\gamma;1;z\Big{)}-\frac{(a-b)(a-l)}{(b-l)(a-m)}F_{z}^{\prime}\Big{(}\gamma,\gamma;1;z\Big{)}\Bigg{\\}}\Bigg{]}.$

We calculate

$\displaystyle\frac{\partial}{\partial l}\Bigg{\\{}\gamma
F\Big{(}\gamma,\gamma;1;z\Big{)}-\frac{(a-b)(a-l)}{(b-l)(a-m)}F_{z}^{\prime}\Big{(}\gamma,\gamma;1;z\Big{)}\Bigg{\\}}$
$\displaystyle=$ $\displaystyle\gamma
F^{\prime}_{z}\Big{(}\gamma,\gamma;1;z\Big{)}\frac{(a-b)(b-m)}{(l-b)^{2}(a-m)}-\frac{(a-b)^{2}}{(b-l)^{2}(a-m)}F_{z}^{\prime}\Big{(}\gamma,\gamma;1;z\Big{)}$
$\displaystyle-\frac{(a-b)(a-l)}{(b-l)(a-m)}F_{zz}{}^{\prime}{}^{\prime}\Big{(}\gamma,\gamma;1;z\Big{)}\frac{(a-b)(b-m)}{(l-b)^{2}(a-m)}\,.$

Here $\frac{\partial}{\partial
l}\frac{(a-b)(a-l)}{(b-l)(a-m)}=\frac{(a-b)^{2}}{(b-l)^{2}(a-m)}$. Finally,

$\displaystyle\frac{\partial^{2}}{\partial l\,\partial m}V(l,m;a,b)$
$\displaystyle=$
$\displaystyle(a-m)^{-\gamma-1}\Bigg{[}-\gamma(l-b)^{-\gamma-1}\Bigg{\\{}\gamma
F\Big{(}\gamma,\gamma;1;z\Big{)}-\frac{(a-b)(a-l)}{(b-l)(a-m)}F_{z}^{\prime}\Big{(}\gamma,\gamma;1;z\Big{)}\Bigg{\\}}\Bigg{]}$
$\displaystyle+\,(a-m)^{-\gamma-1}(l-b)^{-\gamma}\Bigg{\\{}\gamma
F^{\prime}_{z}\Big{(}\gamma,\gamma;1;z\Big{)}\frac{(a-b)(b-m)}{(l-b)^{2}(a-m)}$
$\displaystyle-\frac{(a-b)^{2}}{(b-l)^{2}(a-m)}F_{z}^{\prime}\Big{(}\gamma,\gamma;1;z\Big{)}-\frac{(a-b)(a-l)}{(b-l)(a-m)}F_{zz}{}^{\prime}{}^{\prime}\Big{(}\gamma,\gamma;1;z\Big{)}\frac{(a-b)(b-m)}{(l-b)^{2}(a-m)}\Bigg{\\}}.$

The coefficients of the derivatives of the hypergeometric function,
$F_{zz}{}^{\prime}{}^{\prime}$, $F_{z}{}^{\prime}$, and of $F$ in the
expression for $\displaystyle\frac{\partial^{2}}{\partial l\,\partial
m}V(l,m;a,b)$ are

$\displaystyle(a-m)^{-\gamma-3}(l-b)^{-\gamma-3}(a-b)^{2}(a-l)(b-m)\,,$
$\displaystyle(a-m)^{-\gamma-2}(l-b)^{-\gamma-2}(a-b)\Big{\\{}\gamma(b-m-a+l)-(a-b)\Big{\\}}\,,$
$\displaystyle(a-m)^{-\gamma-1}\left(-\gamma\right)(l-b)^{-\gamma-1}\\{-\left(-\gamma\right)\\}\,=\,-(a-m)^{-\gamma-1}(l-b)^{-\gamma-1}\gamma^{2}\,,$

respectively. The coefficients of $F_{z}{}^{\prime}$ and $F$ in the expression
for $\displaystyle\frac{1}{(l-m)}\gamma\left(\frac{\partial}{\partial
l}-\frac{\partial}{\partial m}\right)V(l,m;a,b)$ are

$\frac{1}{(l-m)}\gamma(a-m)^{-\gamma-1}(l-b)^{-\gamma-1}(a-b)\left\\{\frac{(b-m)}{(l-b)}-\frac{(a-l)}{(a-m)}\right\\}$

and

$-\,\frac{1}{(l-m)}\gamma(a-m)^{-\gamma}(l-b)^{-\gamma}\gamma\Big{\\{}\frac{1}{(l-b)}+\frac{1}{(a-m)}\Big{\\}}\,.$

Now we turn to the equation (1.3). The coefficients of $F$ and
$F_{z}{}^{\prime}$ in that equation are

$\frac{1}{(m-l)}(a-b)(a-m)^{-\gamma-1}(l-b)^{-\gamma-1}(-\gamma^{2}),$

and

$\displaystyle\frac{1}{(m-l)}(a-m)^{-\gamma-1}(l-b)^{-\gamma-1}(a-b)$
$\displaystyle\times\Bigg{[}\gamma\frac{(m-l)}{(a-m)(l-b)}(b-m-a+l)+\gamma\left\\{\frac{(b-m)}{(l-b)}-\frac{(a-l)}{(a-m)}\right\\}-\frac{(m-l)}{(a-m)(l-b)}(a-b)\Bigg{]}.$

The first two terms in the brackets can be written as follows

$\displaystyle\gamma\frac{(m-l)}{(a-m)(l-b)}(b-m-a+l)+\gamma\left\\{\frac{(b-m)}{(l-b)}-\frac{(a-l)}{(a-m)}\right\\}=-2\gamma
z\,,$

while the last term can be transformed to

$\displaystyle-\frac{(m-l)}{(a-m)(l-b)}(a-b)=1-z\,.$

Thus, the coefficient of $F_{z}{}^{\prime}$ in the equation (1.3) is

$\displaystyle\frac{1}{(m-l)}(a-m)^{-\gamma-1}(l-b)^{-\gamma-1}(a-b)\left[1-(1+2\gamma)z\right]\,.$

Finally, the coefficient of $F_{zz}{}^{\prime}{}^{\prime}$ in the equation
(1.3) is

$\displaystyle\frac{1}{(m-l)}(a-m)^{-\gamma-1}(l-b)^{-\gamma-1}(a-b)\Big{[}\frac{(a-b)(a-l)(b-m)(m-l)}{(a-m)^{2}(l-b)^{2}}\Big{]}\,,$

where

$\displaystyle\frac{(a-b)(a-l)(b-m)(m-l)}{(a-m)^{2}(l-b)^{2}}=z(1-z)\,.$

Hence, the left-hand side of (1.3) reads

$\displaystyle\Bigg{\\{}\frac{\partial^{2}}{\partial l\,\partial
m}-\frac{1}{(l-m)}\gamma\Big{(}\frac{\partial}{\partial
l}-\frac{\partial}{\partial m}\Big{)}\Bigg{\\}}V(l,m;a,b)$ $\displaystyle=$
$\displaystyle\frac{1}{(m-l)}(a-m)^{-\gamma-1}(l-b)^{-\gamma-1}(a-b)\Big{[}z(1-z)F_{zz}+\big{(}1-(1+2\gamma)z\big{)}F_{z}-\gamma^{2}F\Big{]}=0\,,$

and vanishes, since $F$ solves the Gauss hypergeometic equation with $c=1$,
$a=\gamma$, and $b=\gamma$. Lemma is proven. $\square$

###### Lemma 1.2

For $\gamma\in{\mathbb{C}}$ such that $F\big{(}\gamma,\gamma;1;z\big{)}$ is
well defined, the function

$\displaystyle E(l,m;a,b)$ $\displaystyle:=$
$\displaystyle(a-b)^{\gamma-\frac{1}{2}}(l-m)^{\gamma-\frac{1}{2}}V(l,m;a,b)$
$\displaystyle=$
$\displaystyle(a-b)^{\gamma-\frac{1}{2}}(l-m)^{\gamma-\frac{1}{2}}(l-b)^{-\gamma}(a-m)^{-\gamma}F\Big{(}\gamma,\gamma;1;\frac{(l-a)(m-b)}{(l-b)(m-a)}\Big{)}$

solves the equation

$\Bigg{\\{}\frac{\partial^{2}}{\partial l\,\partial
m}-\frac{1}{2(l-m)}\Big{(}\frac{\partial}{\partial l}-\frac{\partial}{\partial
m}\Big{)}\Bigg{\\}}E(l,m;a,b)+\frac{1}{(l-m)^{2}}\left(\frac{1}{2}-\gamma\right)^{2}E(l,m;a,b)=0\,.$
(1.4)

Proof. Indeed, straightforward calculations show

$\displaystyle(a-b)^{-\gamma+\frac{1}{2}}\Bigg{\\{}\frac{\partial^{2}}{\partial
l\,\partial m}-\frac{1}{2(l-m)}\Big{(}\frac{\partial}{\partial
l}-\frac{\partial}{\partial
m}\Big{)}+\frac{1}{(l-m)^{2}}\left(\frac{1}{2}-\gamma\right)^{2}\Bigg{\\}}E$
$\displaystyle=$
$\displaystyle(l-m)^{\gamma-\frac{1}{2}}\left[V_{lm}-\frac{1}{(l-m)}\gamma\left(V_{l}-V_{m}\right)\right]=0\,.$

Lemma is proven. $\square$

Consider now the operator

${\mathcal{S}}_{ch}^{*}:=\frac{\partial^{2}\,}{\partial l\,\partial
m}+\frac{1}{2(l-m)}\Big{(}\frac{\partial\,}{\partial
l}-\frac{\partial\,}{\partial m}\Big{)}-\frac{1}{(l-m)^{2}}(M^{2}+1)\,,$

which is a formally adjoint to the operator

${\mathcal{S}}_{ch}:=\frac{\partial^{2}\,}{\partial l\,\partial
m}-\frac{1}{2(l-m)}\Big{(}\frac{\partial\,}{\partial
l}-\frac{\partial\,}{\partial m}\Big{)}-\frac{1}{(l-m)^{2}}M^{2}\,.$

In the next lemma the Riemann function is presented.

###### Proposition 1.3

The function

$\displaystyle R(l,m;a,b)$ $\displaystyle=$ $\displaystyle(l-m)E(l,m;a,b)$
$\displaystyle=$
$\displaystyle(a-b)^{iM}(l-m)^{1+iM}(l-b)^{-\frac{1}{2}-iM}(a-m)^{-\frac{1}{2}-iM}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(l-a)(m-b)}{(l-b)(m-a)}\Big{)}$

defined for all $l$, $m$, $a$, $b\in{\mathbb{R}}$, such that $l>m$, is a
unique solution of the equation ${\mathcal{S}}_{ch}^{*}R=0$, which satisfies
the following conditions:
$(\mbox{\rm i})$ $\displaystyle{R_{l}=\frac{1}{2(l-m)}R\quad}$ along the line
$m=b$;
$(\mbox{\rm ii})$ $\displaystyle{R_{m}=-\frac{1}{2(l-m)}R\quad}$ along the
line $l=a$;
$(\mbox{\rm iii})$ $R(a,b;a,b)=1$.

Proof. Indeed, if we denote $\gamma=\frac{1}{2}+iM$, then for the Riemann
function we have

$R(l,m;a,b)=(a-b)^{\gamma-\frac{1}{2}}(l-m)^{\gamma+\frac{1}{2}}V(l,m;a,b)=(l-m)E(l,m;a,b)\,.$

The operators ${\mathcal{S}}_{ch}$ and ${\mathcal{S}}_{ch}^{*}$ can be written
as follows:

$\displaystyle{\mathcal{S}}_{ch}$ $\displaystyle=$
$\displaystyle\frac{\partial^{2}\,}{\partial l\,\partial
m}-\frac{1}{2(l-m)}\Big{(}\frac{\partial\,}{\partial
l}-\frac{\partial\,}{\partial
m}\Big{)}+\frac{1}{(l-m)^{2}}\left(\gamma-\frac{1}{2}\right)^{2}\,,$
$\displaystyle{\mathcal{S}}_{ch}^{*}$ $\displaystyle=$
$\displaystyle\frac{\partial^{2}\,}{\partial l\,\partial
m}+\frac{1}{2(l-m)}\Big{(}\frac{\partial\,}{\partial
l}-\frac{\partial\,}{\partial
m}\Big{)}-\frac{1}{(l-m)^{2}}\left(1-\left(\gamma-\frac{1}{2}\right)^{2}\right)\,.$

The direct calculations show that, if function $u$ solves the equation
${\mathcal{S}}_{ch}u=0$, then the function $v=(l-m)u$ solves the equation
${\mathcal{S}}_{ch}^{*}v=0$, and vice versa. Then Lemma 1.2 completes the
proof. Lemma is proven. $\square$

##  2 Proof of Theorem 0.1

Next we use Riemann function $R(l,m;a,b)$ and function $E(x,t;x_{0},t_{0})$
defined by ( ‣ 0 Introduction and Statement of Results) to complete the proof
of Theorem 0.1, which gives the fundamental solution with a support in the
forward cone $D_{+}(x_{0},t_{0})$, $x_{0}\in{\mathbb{R}}^{n}$,
$t_{0}\in{\mathbb{R}}$, and the fundamental solution with a support in the
backward cone $D_{-}(x_{0},t_{0})$, $x_{0}\in{\mathbb{R}}^{n}$,
$t_{0}\in{\mathbb{R}}$, defined by (0.14) with plus and minus, respectively.

We present a proof for ${\mathcal{E}}_{+}(x,t;0,b)$ since for
${\mathcal{E}}_{-}(x,t;0,b)$ it is similar. First, we note that the operator
${\mathcal{S}}$ is formally self-adjoint, ${\mathcal{S}}={\mathcal{S}}^{*}$.
We must show that

$<{\mathcal{E}}_{+},{\mathcal{S}}\varphi>=\varphi(0,b)\,,\qquad\mbox{\rm for
every}\quad\varphi\in C_{0}^{\infty}({\mathbb{R}}^{2})\,.$

Since $E(x,t;0,b)$ is locally integrable in ${\mathbb{R}}^{2}$, this is
equivalent to showing that

$\int\\!\\!\int_{{\mathbb{R}}^{2}}{\mathcal{E}}_{+}(x,t;0,b){\mathcal{S}}\varphi(x,t)\,dx\,dt=\varphi(0,b),\quad\mbox{\rm
for every}\quad\varphi\in C_{0}^{\infty}({\mathbb{R}}^{2}).$ (2.1)

In the mean time ${D(x,t)}/{D(l,m)}=(l-m)^{-1}$ is the Jacobian of the
transformation (1.1). Hence the integral in the left-hand side of (2.1) is
equal to

$\displaystyle\int\\!\\!\int_{{\mathbb{R}}^{2}}{\mathcal{E}}_{+}(x,t;0,b){\mathcal{S}}\varphi(x,t)\,dx\,dt=\int_{b}^{\infty}dt\int_{e^{-t}-e^{-b}}^{e^{-b}-e^{-t}}E(x,t;0,b){\mathcal{S}}\varphi(x,t)\,dx$
$\displaystyle=$
$\displaystyle-\int_{-e^{-b}}^{\infty}\int_{-\infty}^{e^{-b}}R(l,m;e^{-b},-e^{-b})\,dl\,dm\Bigg{\\{}\frac{\partial^{2}}{\partial
l\,\partial m}-\frac{1}{2(l-m)}\Big{(}\frac{\partial}{\partial
l}-\frac{\partial}{\partial
m}\Big{)}-\frac{1}{(l-m)^{2}}M^{2}\Bigg{\\}}\varphi.$

We consider the first term of the right hand side, and integrate it by parts

$\displaystyle\int_{-e^{-b}}^{\infty}dm\int_{-\infty}^{e^{-b}}dl\,\,R(l,m;e^{-b},-e^{-b})\frac{\partial^{2}}{\partial
l\,\partial m}\varphi$ $\displaystyle=$
$\displaystyle\int_{-e^{-b}}^{\infty}R(e^{-b},m;e^{-b},-e^{-b})\frac{\partial\varphi}{\partial
m}\Bigg{|}_{l=e^{-b}}\,dm-\Bigg{[}-\int_{-\infty}^{e^{-b}}\,dl\left(\frac{\partial}{\partial
l}R(l,-e^{-b};e^{-b},-e^{-b})\right)\varphi\Bigg{|}_{m=-e^{-b}}$
$\displaystyle\hskip
56.9055pt-\int_{-e^{-b}}^{\infty}dm\int_{-\infty}^{e^{-b}}\,dl\left(\frac{\partial^{2}}{\partial
l\,\partial m}R(l,m;e^{-b},-e^{-b})\right)\varphi\Bigg{]}$ $\displaystyle=$
$\displaystyle-\varphi(e^{-b},-e^{-b})-\int_{-e^{-b}}^{\infty}dm\,\left(\frac{\partial}{\partial
m}R(e^{-b},m;e^{-b},-e^{-b})\right)\varphi(e^{-b},m)$
$\displaystyle+\int_{-\infty}^{e^{-b}}\,dl\left(\frac{\partial}{\partial
l}R(l,-e^{-b};e^{-b},-e^{-b})\right)\,\varphi(l,-e^{-b})+\int_{-e^{-b}}^{\infty}dm\int_{-\infty}^{e^{-b}}\,dl\left(\frac{\partial^{2}}{\partial
l\,\partial m}R(l,m;e^{-b},-e^{-b})\right)\varphi.$

Then, for the second term we obtain

$\displaystyle-\int_{-e^{-b}}^{\infty}dm\int_{-\infty}^{e^{-b}}\,dl\,R(l,m;e^{-b},-e^{-b})\frac{1}{2(l-m)}\Big{(}\frac{\partial}{\partial
l}-\frac{\partial}{\partial m}\Big{)}\varphi$ $\displaystyle=$
$\displaystyle-\int_{-\infty}^{e^{-b}}R(l,-e^{-b};e^{-b},-e^{-b})\frac{1}{2(l+e^{-b})}\varphi(l,-e^{-b})\,dl$
$\displaystyle-\int_{-e^{-b}}^{\infty}R(e^{-b},m;e^{-b},-e^{-b})\frac{1}{2(e^{-b}-m)}\varphi(e^{-b},m)\,dm$
$\displaystyle-\int_{-e^{-b}}^{\infty}dm\int_{-\infty}^{e^{-b}}dl\,\frac{1}{(l-m)^{2}}R(l,m;e^{-b},-e^{-b})\varphi(l,m)\,$
$\displaystyle+\int_{-e^{-b}}^{\infty}dm\int_{-\infty}^{e^{-b}}\,dl\,\Big{[}\frac{1}{2(l-m)}\Big{(}\frac{\partial}{\partial
l}-\frac{\partial}{\partial
m}\Big{)}R(l,m;e^{-b},-e^{-b})\Big{]}\varphi(l,m)\,.$

Using properties of the Riemann function we derive

$\displaystyle\int_{-e^{-b}}^{\infty}dm\int_{-\infty}^{e^{-b}}R(l,m;e^{-b},-e^{-b})\,dl\,\Bigg{\\{}\frac{\partial^{2}}{\partial
l\,\partial m}-\frac{1}{2(l-m)}\Big{(}\frac{\partial}{\partial
l}-\frac{\partial}{\partial
m}\Big{)}-\frac{1}{(l-m)^{2}}M^{2}\Bigg{\\}}\varphi$ $\displaystyle=$
$\displaystyle-\varphi(e^{-b},-e^{-b})-\int_{-e^{-b}}^{\infty}\left(\frac{\partial}{\partial
m}R(e^{-b},m;e^{-b},-e^{-b})\right)\varphi(e^{-b},m)\,dm$
$\displaystyle+\int_{-\infty}^{e^{-b}}\left(\frac{\partial}{\partial
l}R(l,-e^{-b};e^{-b},-e^{-b})\right)\,\varphi(l,-e^{-b})\,dl$
$\displaystyle-\int_{-\infty}^{e^{-b}}R(l,-e^{-b};e^{-b},-e^{-b})\frac{1}{2(l+e^{-b})}\varphi(l,-e^{-b})\,dl$
$\displaystyle-\int_{-e^{-b}}^{\infty}R(e^{-b},m;e^{-b},-e^{-b})\frac{1}{2(e^{-b}-m)}\varphi(e^{-b},m)\,dm$
$\displaystyle=$ $\displaystyle-\varphi(e^{-b},-e^{-b})\,.$

Theorem is proven. $\square$

##  3 Application to the Cauchy Problem: Source Term and $n=1$

Consider now the Cauchy problem for the equation (0.18) with vanishing initial
data (0.19). For every $(x,t)\in D_{+}(0,b)$ one has $e^{-t}-e^{-b}\leq x\leq
e^{-b}-e^{-t}$, so that

$\displaystyle E(x,t;0,b)$ $\displaystyle=$
$\displaystyle(4e^{-b-t})^{iM}\Big{(}(e^{-t}+e^{-b})^{2}-x^{2}\Big{)}^{-\frac{1}{2}-iM}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-x^{2}}{(e^{-b}+e^{-t})^{2}-x^{2}}\Big{)}.$

The coefficient of the equation (0.9) is independent of $x$, therefore
${\mathcal{E}}_{+}(x,t;$ $y,b)$ $={\mathcal{E}}_{+}(x-y,t;0,b)$. Using the
fundamental solution from Theorem 0.1 one can write the convolution

$u(x,t)=\int_{-\infty}^{\infty}\\!\int_{-\infty}^{\infty}{\mathcal{E}}_{+}(x,t;y,b)f(y,b)\,db\,dy=\int_{0}^{t}\\!db\\!\int_{-\infty}^{\infty}{\mathcal{E}}_{+}(x-y,t;0,b)f(y,b)\,dy,$
(3.1)

which is well-defined since supp$f\subset\\{t\geq 0\\}$. Then according to the
definition of the distribution ${\mathcal{E}}_{+}$ we obtain the statement of
the Theorem 0.3. Thus, Theorem 0.3 is proven.

###### Remark 3.1

The argument of the hypergeometric function is nonnegative and bounded,

$0\leq\frac{(e^{-b}-e^{-t})^{2}-z^{2}}{(e^{-b}+e^{-t})^{2}-z^{2}}<1\quad\mbox{
for all}\quad b\in(0,t),\,\,z\in(e^{-t}-e^{-b},e^{-b}-e^{-t})\,.$

The following corollary is a manifestation of the time-speed transformation
principle introduced in [29]. It implies the existence of an operator
transforming the solutions of the Cauchy problem for the string equation to
the solutions of the Cauchy problem for the inhomogeneous equation with time-
dependent speed of propagation. One may think of this transformation as a
“two-stage” Duhamel’s principal, but unlike the last one, it reduces the
equation with the time-dependent speed of propagation to the one with the
speed of propagation independent of time.

###### Corollary 3.2

The solution $u=u(x,t)$ of the Cauchy problem (0.18)-(0.19) can be represented
by (0.24), where the functions
$v(x,t;\tau):=\frac{1}{2}(f(x+t,\tau)+f(x-t,\tau))$, $\tau\in[0,\infty)$, form
a one-parameter family of solutions to the Cauchy problem for the string
equation, that is, $v_{tt}-v_{xx}=0$, $v(x,0;\tau)=f(x,\tau)$,
$v_{t}(x,0;\tau)=0$.

Proof. From the convolution (3.1) we derive

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}db\int_{-(e^{-b}-e^{-t})}^{e^{-b}-e^{-t}}dz\,f(z+x,b)(4e^{-b-t})^{iM}\Big{(}(e^{-t}+e^{-b})^{2}-z^{2}\Big{)}^{-\frac{1}{2}-iM}$
$\displaystyle\times
F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-z^{2}}{(e^{-b}+e^{-t})^{2}-z^{2}}\Big{)}$
$\displaystyle=$ $\displaystyle
2\int_{0}^{t}db\int_{0}^{e^{-b}-e^{-t}}dz\,\frac{1}{2}\\{f(x+z,b)+f(x-z,b)\\}(4e^{-b-t})^{iM}\Big{(}(e^{-t}+e^{-b})^{2}-z^{2}\Big{)}^{-\frac{1}{2}-iM}$
$\displaystyle\times
F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-z^{2}}{(e^{-b}+e^{-t})^{2}-z^{2}}\Big{)}\,.$

Corollary is proven. $\square$

Some Properties of the Function ${\mathbf{E}(x,t;y,b)}$.

In this section we collect some elementary auxiliary formulas in order to make
the proofs of main theorems more transparent.

###### Proposition 3.3

Let $E(x,t;x_{0},t_{0})$ be function defined by ( ‣ 0 Introduction and
Statement of Results). One has

$\displaystyle E(x,t;y,b)$ $\displaystyle=$ $\displaystyle E(y,b;x,t),\,\,$
(3.2) $\displaystyle E(x,t;y,b)=E(x-y,t;0,b)$ , $\displaystyle
E(x,t;0,b)=E(-x,t;0,b),$ (3.3) $\displaystyle E(x,t;0,-\ln(x+e^{-t}))$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{1}{\sqrt{e^{-t}}\sqrt{x+e^{-t}}},$ (3.4)
$\displaystyle\frac{\partial}{\partial
b}\Big{(}e^{-b}E(e^{-b}-e^{-t},t;0,b)\Big{)}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}e^{t/2}e^{-b/2},$ (3.5)
$\displaystyle\frac{\partial}{\partial
b}\Big{(}be^{-b}E(-e^{-b}+e^{-t},t;0,b)\Big{)}$ $\displaystyle=$
$\displaystyle\frac{\partial}{\partial
b}\Big{(}be^{-b}E(e^{-b}-e^{-t},t;0,b)\Big{)}=\frac{1}{4}e^{t/2}e^{-b/2}(2-b),$
(3.6) $\displaystyle\lim_{y\rightarrow
x+e^{-b}-e^{-t}}\frac{\partial}{\partial x}E(x-y,t;0,b)$ $\displaystyle=$
$\displaystyle-\frac{1}{16}(1+4M^{2})e^{t/2}e^{b/2}(e^{t}-e^{b}),$ (3.7)
$\displaystyle\lim_{y\rightarrow x-e^{-b}+e^{-t}}\frac{\partial}{\partial
x}E(x-y,t;0,b)$ $\displaystyle=$
$\displaystyle\frac{1}{16}(1+4M^{2})e^{t/2}e^{b/2}(e^{t}-e^{b}),$ (3.8)
$\displaystyle\left[\frac{\partial}{\partial
b}E(x,t;0,b)\right]_{b=-\ln(x+e^{-t})}$ $\displaystyle=$
$\displaystyle\frac{1}{16}e^{t}\frac{4+e^{t}x(1+4M^{2})}{\sqrt{1+e^{t}x}},$
(3.9) $\displaystyle\Bigg{[}\frac{\partial}{\partial
b}E(x,t;0,b)\Bigg{]}_{b=0}\\!\\!$ $\displaystyle\\!\\!=\\!\\!$
$\displaystyle-(4e^{-t})^{iM}\big{(}(e^{-t}+1)^{2}-x^{2}\big{)}^{-iM}\frac{1}{[(1-e^{-t})^{2}-x^{2}]\sqrt{(e^{-t}+1)^{2}-x^{2}}}$
(3.10)
$\displaystyle\times\Bigg{[}\big{(}e^{-t}-1-iM(e^{-2t}-1-x^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(1-e^{-t})^{2}-x^{2}}{(1+e^{-t})^{2}-x^{2}}\Big{)}$
$\displaystyle\hskip
14.22636pt+\big{(}1-e^{-2t}+x^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(1-e^{-t})^{2}-x^{2}}{(1+e^{-t})^{2}-x^{2}}\Big{)}\Bigg{]}.$

Proof. The properties (3.2),(3.3), and (3.4) are evident. To prove (3.5) and
(3.6) we write

$E(e^{-b}-e^{-t},t;0,b)=(4e^{-b-t})^{-\frac{1}{2}},$

that implies both (3.5) and (3.6). To prove (3.7) we denote

$\displaystyle
z:=\frac{(e^{-b}-e^{-t})^{2}-(x-y)^{2}}{(e^{-b}+e^{-t})^{2}-(x-y)^{2}}\,,$

so that,

$\displaystyle\frac{\partial z}{\partial x}$ $\displaystyle=$
$\displaystyle-2(x-y)\frac{4e^{-b}e^{-t}}{[(e^{-b}+e^{-t})^{2}-(x-y)^{2}]^{2}}\,,\qquad\frac{\partial
z}{\partial
b}=-\frac{4e^{-b}e^{-t}\left(-e^{-2t}+e^{-2b}+(x-y)^{2}\right)}{\left[(e^{-b}+e^{-t})^{2}-(x-y)^{2}\right]^{2}}\,.$

Then we obtain

$\displaystyle\frac{\partial}{\partial x}E(x-y,t;0,b)$ $\displaystyle=$
$\displaystyle(4e^{-b-t})^{iM}\Bigg{[}(-2(x-y))(-\frac{1}{2}-iM)\Big{(}(e^{-t}+e^{-b})^{2}-(x-y)^{2}\Big{)}^{-\frac{3}{2}-iM}$
(3.11) $\displaystyle\times F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;z\Big{)}$
$\displaystyle+\Big{(}(e^{-t}+e^{-b})^{2}-(x-y)^{2}\Big{)}^{-\frac{1}{2}-iM}\Big{(}\frac{\partial
z}{\partial
x}\Big{)}F_{z}\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;z\Big{)}\Bigg{]}\,.$

It is easily seen that

$\lim_{y\to x+e^{-b}-e^{-t}}z=0,\qquad\lim_{y\to
x+e^{-b}-e^{-t}}\frac{\partial z}{\partial x}=\frac{1}{2}(e^{t}-e^{b}),$

while according to (20) [2, Sec.2.8 v.1], we have

$\displaystyle\lim_{y\to
x+e^{-b}-e^{-t}}\partial_{z}F\left(\frac{1}{2}+iM,\frac{1}{2}+iM;1;z\right)$
$\displaystyle=$
$\displaystyle\frac{\Gamma^{2}(\frac{3}{2}+iM)}{\Gamma^{2}(\frac{1}{2}+iM)}=\left(\frac{1}{2}+iM\right)^{2}\,.$
(3.12)

Consequently, from (3.11) we obtain

$\displaystyle\lim_{y\to x+e^{-b}-e^{-t}}\frac{\partial}{\partial
x}E(x-y,t;0,b)$ $\displaystyle=$
$\displaystyle(4e^{-b-t})^{iM}\Bigg{[}2(e^{-b}-e^{-t})\left(-\frac{1}{2}-iM\right)\Big{(}4e^{-t}e^{-b}\Big{)}^{-\frac{3}{2}-iM}+\Big{(}4e^{-t}e^{-b}\Big{)}^{-\frac{1}{2}-iM}\frac{1}{2}(e^{t}-e^{b})\left(\frac{1}{2}+iM\right)^{2}\Bigg{]}$
$\displaystyle=$
$\displaystyle-e^{\frac{t}{2}}e^{\frac{b}{2}}\frac{1}{16}(e^{t}-e^{b})\Big{[}1+4M^{2}\Big{]}\,.$

The proof of (3.8) is similar. To prove (3.9) we write

$\displaystyle\frac{\partial}{\partial b}E(x,t;0,b)$ (3.13) $\displaystyle=$
$\displaystyle(-iM)(4e^{-b-t})^{iM}\Big{(}(e^{-t}+e^{-b})^{2}-x^{2}\Big{)}^{-\frac{1}{2}-iM}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-x^{2}}{(e^{-b}+e^{-t})^{2}-x^{2}}\Big{)}$
$\displaystyle-\Big{(}\frac{1}{2}+iM\Big{)}(4e^{-b-t})^{iM}(-e^{-b})2(e^{-t}+e^{-b})\Big{(}(e^{-t}+e^{-b})^{2}-x^{2}\Big{)}^{-\frac{3}{2}-iM}$
$\displaystyle\times
F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-x^{2}}{(e^{-b}+e^{-t})^{2}-x^{2}}\Big{)}$
$\displaystyle+(4e^{-b-t})^{iM}\Big{(}(e^{-t}+e^{-b})^{2}-x^{2}\Big{)}^{-\frac{1}{2}-iM}\frac{\partial}{\partial
b}\Bigg{(}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-x^{2}}{(e^{-b}+e^{-t})^{2}-x^{2}}\Big{)}\Bigg{)}.$

On the other hand (3.12) and (3.13) imply

$\displaystyle\left[\frac{\partial}{\partial
b}E(x,t;0,b)\right]_{b=-\ln(x+e^{-t})}$ $\displaystyle=$
$\displaystyle(4e^{-b-t})^{iM}\Big{(}4e^{-b-t}\Big{)}^{-\frac{1}{2}-iM}\Bigg{[}(-iM)+2\Big{(}\frac{1}{2}+iM\Big{)}e^{-b}(e^{-t}+e^{-b})\Big{(}4e^{-b-t}\Big{)}^{-1}$
$\displaystyle\hskip 142.26378pt+\frac{\partial}{\partial
b}\Bigg{(}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-x^{2}}{(e^{-b}+e^{-t})^{2}-x^{2}}\Big{)}\Bigg{)}\Bigg{]}_{b=-\ln(x+e^{-t})}$
$\displaystyle=$
$\displaystyle\frac{1}{2}e^{\frac{1}{2}b+\frac{1}{2}t}\Bigg{[}(-iM)+\frac{1}{2}\Big{(}\frac{1}{2}+iM\Big{)}(e^{-t}+e^{-b})e^{t}-\frac{4e^{-b}e^{-t}\left(e^{-2b}-e^{-2t}+x^{2}\right)}{\left[(e^{-b}+e^{-t})^{2}-x^{2}\right]^{2}}\Big{(}\frac{1}{2}+iM\Big{)}^{2}\Bigg{]}_{b=-\ln(x+e^{-t})}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{1}{\sqrt{1+xe^{t}}}e^{t}\Bigg{[}-iM+\frac{1}{2}\Big{(}\frac{1}{2}+iM\Big{)}(2+xe^{t})-\frac{xe^{t}}{2}\Big{(}\frac{1}{2}+iM\Big{)}^{2}\Bigg{]}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{1}{\sqrt{1+xe^{t}}}e^{t}\frac{4+xe^{t}(1+4M^{2})}{8}\,.$

Thus, (3.9) is proven. To prove (3.10) we appeal to (23)[2, v.1, Sec.2.8],
that reads with $a=b=\frac{1}{2}+iM$, $c=1$,

$\displaystyle F_{z}\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;z\Big{)}$
$\displaystyle=$
$\displaystyle\frac{1}{z(1-z)}\Bigg{\\{}\Big{[}-\Big{(}\frac{1}{2}-iM\Big{)}+\Big{(}\frac{1}{2}+iM\Big{)}z\Big{]}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;z\Big{)}$
$\displaystyle\hskip
56.9055pt+\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;z\Big{)}\Bigg{\\}}\,.$

Then we plug the last relation in (3.13) and obtain

$\displaystyle\Bigg{[}\frac{\partial}{\partial b}E(x,t;0,b)\Bigg{]}_{b=0}$
$\displaystyle=$ $\displaystyle-
iM(4e^{-t})^{iM}\Big{(}(e^{-t}+1)^{2}-x^{2}\Big{)}^{-\frac{1}{2}-iM}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}$
$\displaystyle-\Big{(}\frac{1}{2}+iM\Big{)}(4e^{-t})^{iM}(-1)2(e^{-t}+1)\Big{(}(e^{-t}+1)^{2}-x^{2}\Big{)}^{-\frac{3}{2}-iM}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}$
$\displaystyle+(4e^{-t})^{iM}\Big{(}(e^{-t}+1)^{2}-x^{2}\Big{)}^{-\frac{1}{2}-iM}\left[\frac{\partial}{\partial
b}\left(F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta\Big{)}\right)\right]_{b=0}\,,$

where we have denoted

$\zeta:=\frac{(e^{-b}-e^{-t})^{2}-x^{2}}{(e^{-b}+e^{-t})^{2}-x^{2}}\,,\qquad\zeta_{0}:=\frac{(1-e^{-t})^{2}-x^{2}}{(1+e^{-t})^{2}-x^{2}}\,,\qquad\zeta_{0}(1-\zeta_{0})=\frac{4e^{-t}[(1-e^{-t})^{2}-x^{2}]}{[(1+e^{-t})^{2}-x^{2}]^{2}}\,,$

with

$\frac{\partial}{\partial
b}\zeta=-\frac{4e^{-b}e^{-t}\left(e^{-2b}-e^{-2t}+x^{2}\right)}{\left[(e^{-b}+e^{-t})^{2}-x^{2}\right]^{2}}\,,\qquad\frac{\partial\zeta}{\partial
b}\Big{|}_{b=0}=-\frac{4e^{-t}\left(1-e^{-2t}+x^{2}\right)}{\left[(1+e^{-t})^{2}-x^{2}\right]^{2}}\,.$

Hence, due to (20) [2, v.1, Sec.2.8], we obtain

$\displaystyle\Bigg{[}\frac{\partial}{\partial
b}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-x^{2}}{(e^{-b}+e^{-t})^{2}-x^{2}}\Big{)}\Bigg{]}_{b=0}$
$\displaystyle=$
$\displaystyle-\frac{1-e^{-2t}+x^{2}}{(1-e^{-t})^{2}-x^{2}}\Bigg{\\{}\Big{[}-\Big{(}\frac{1}{2}-iM\Big{)}+\Big{(}\frac{1}{2}+iM\Big{)}\zeta_{0}\Big{]}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}$
$\displaystyle\hskip
170.71652pt+\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}\Bigg{\\}}\,.$

Hence,

$\displaystyle\Bigg{[}\frac{\partial}{\partial b}E(x,t;0,b)\Bigg{]}_{b=0}$
$\displaystyle=$ $\displaystyle-
iM(4e^{-t})^{iM}\Big{(}(e^{-t}+1)^{2}-x^{2}\Big{)}^{-\frac{1}{2}-iM}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}$
$\displaystyle-\Big{(}\frac{1}{2}+iM\Big{)}(4e^{-t})^{iM}(-1)2(e^{-t}+1)\Big{(}(e^{-t}+1)^{2}-x^{2}\Big{)}^{-\frac{3}{2}-iM}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}$
$\displaystyle+(4e^{-t})^{iM}\Big{(}(e^{-t}+1)^{2}-x^{2}\Big{)}^{-\frac{1}{2}-iM}$
$\displaystyle\times\Bigg{[}-\frac{1-e^{-2t}+x^{2}}{(1-e^{-t})^{2}-x^{2}}\Bigg{\\{}\Big{[}-\Big{(}\frac{1}{2}-iM\Big{)}+\Big{(}\frac{1}{2}+iM\Big{)}\zeta_{0}\Big{]}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}$
$\displaystyle+\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}\Bigg{\\}}\Bigg{]}$
$\displaystyle=$
$\displaystyle(4e^{-t})^{iM}\Big{(}(e^{-t}+1)^{2}-x^{2}\Big{)}^{-\frac{1}{2}-iM}\Bigg{[}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}\Big{\\{}-iM+(1+2iM)\frac{e^{-t}+1}{(e^{-t}+1)^{2}-x^{2}}$
$\displaystyle-\frac{1-e^{-2t}+x^{2}}{(1-e^{-t})^{2}-x^{2}}\Big{[}-\Big{(}\frac{1}{2}-iM\Big{)}+\Big{(}\frac{1}{2}+iM\Big{)}\frac{(1-e^{-t})^{2}-x^{2}}{(1+e^{-t})^{2}-x^{2}}\Big{]}\Big{\\}}$
$\displaystyle-\frac{1-e^{-2t}+x^{2}}{(1-e^{-t})^{2}-x^{2}}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}\Bigg{]}.$

Finally,

$\displaystyle\Bigg{[}\frac{\partial}{\partial b}E(x,t;0,b)\Bigg{]}_{b=0}$
$\displaystyle=$
$\displaystyle(4e^{-t})^{iM}\big{(}(e^{-t}+1)^{2}-x^{2}\big{)}^{-\frac{1}{2}-iM}$
$\displaystyle\times\Bigg{[}\frac{e^{t}-iM+e^{2t}\big{(}iM(1+x^{2})-1\big{)}}{e^{t}\big{(}2+e^{t}(x^{2}-1)\big{)}-1}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}$
$\displaystyle\hskip
28.45274pt-\frac{1-e^{-2t}+x^{2}}{(1-e^{-t})^{2}-x^{2}}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta_{0}\Big{)}\Bigg{]}\,.$

The formula (3.10) and, consequently, the proposition are proven. $\square$

##  4 The Cauchy Problem: Second Datum and $n=1$

In this section we prove Theorem 0.4 in the case of $\varphi_{0}(x)=0$. More
precisely, we have to prove that the solution $u(x,t)$ of the Cauchy problem
(0.20) with $\varphi_{0}(x)=0$ and $\varphi_{1}(x)=\varphi(x)$ can be
represented as follows

$u(x,t)=\int_{0}^{1-e^{-t}}\,\Big{[}\varphi(x+z)+\varphi(x-z)\Big{]}K_{1}(z,t)dz=\int_{0}^{1}\,\Big{[}\varphi(x+\phi(t)s)+\varphi(x-\phi(t)s)\Big{]}K_{1}(\phi(t)s,t)\phi(t)ds,$
(4.1)

where $\phi(t)=1-e^{-t}$. The proof of the theorem is splitted into several
steps.

###### Proposition 4.1

The solution $u=u(x,t)$ of the Cauchy problem (0.20) with $\varphi_{0}(x)=0$
and $\varphi_{1}(x)=\varphi(x)$ can be represented as follows

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}db\,\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}(2-b)+\frac{1}{16}be^{-3b/2}e^{t/2}(e^{b}-e^{t})(1+4M^{2})\Big{]}$
(4.2) $\displaystyle\hskip
28.45274pt\times\Big{[}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{]}$
$\displaystyle+\int_{0}^{t}db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi(y)b\Big{[}e^{-2b}\,\left(\frac{\partial}{\partial
y}\right)^{2}E(x-y,t;0,b)-M^{2}E(x-y,t;0,b)\Big{]}\,\,.$

Proof. We look for the solution $u=u(x,t)$ of the form
$u(x,t)=w(x,t)+t\varphi(x)$. Then (0.20) implies

$\displaystyle
w_{tt}-e^{-2t}w_{xx}+M^{2}w=te^{-2t}\varphi^{(2)}(x)-M^{2}t\varphi(x),\qquad
w(x,0)=0,\quad w_{t}(x,0)=0\,.$

We set $f(x,t)=te^{-2t}\varphi^{(2)}(x)-M^{2}t\varphi(x)$ and due to Theorem
0.3 obtain

$w(x,t)=\widetilde{w(x,t)}-M^{2}\int_{0}^{t}b\,db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi(y)E(x-y,t;0,b),$

where

$\displaystyle\widetilde{w(x,t)}$ $\displaystyle:=$
$\displaystyle\int_{0}^{t}be^{-2b}\,db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi^{(2)}(y)E(x-y,t;0,b)\,.$

Then we integrate by parts:

$\displaystyle\widetilde{w(x,t)}\\!\\!$ $\displaystyle\\!\\!=\\!\\!$
$\displaystyle\\!\\!\int_{0}^{t}be^{-2b}\,db\Bigg{[}\varphi^{(1)}(x+e^{-b}-e^{-t})E(-e^{-b}+e^{-t},t;0,b)-\varphi^{(1)}(x-e^{-b}+e^{-t})E(e^{-b}-e^{-t},t;0,b)\Bigg{]}$
$\displaystyle-\int_{0}^{t}be^{-2b}\,db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi^{(1)}(y)\frac{\partial}{\partial
y}E(x-y,t;0,b)\,.$

But

$\varphi^{(1)}(x+e^{-b}-e^{-t})=-e^{b}\frac{\partial}{\partial
b}\varphi(x+e^{-b}-e^{-t})\quad\mbox{\rm
and}\quad\varphi^{(1)}(x-e^{-b}+e^{-t})=e^{b}\frac{\partial}{\partial
b}\varphi(x-e^{-b}+e^{-t})\,.$

Then, $E(e^{-b}-e^{-t},t;0,b)=E(-e^{-b}+e^{-t},t;0,b)$ due to (3.3), and we
obtain

$\displaystyle\widetilde{w(x,t)}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}be^{-2b}\,db\Bigg{[}-e^{b}\frac{\partial}{\partial
b}\varphi(x+e^{-b}-e^{-t})-e^{b}\frac{\partial}{\partial
b}\varphi(x-e^{-b}+e^{-t})\Bigg{]}E(e^{-b}-e^{-t},t;0,b)$
$\displaystyle-\int_{0}^{t}be^{-2b}\,db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi^{(1)}(y)\frac{\partial}{\partial
y}E(x-y,t;0,b)\,.$

One more integration by parts leads to

$\displaystyle\widetilde{w(x,t)}$ $\displaystyle=$
$\displaystyle-2te^{-t}\varphi(x)E(0,t;0,t)$
$\displaystyle+\int_{0}^{t}\,db\Big{(}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{)}\frac{\partial}{\partial
b}\Big{(}be^{-b}E(e^{-b}-e^{-t},t;0,b)\Big{)}$
$\displaystyle-\int_{0}^{t}be^{-2b}\,db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi^{(1)}(y)\frac{\partial}{\partial
y}E(x-y,t;0,b)\,.$

Since $E(0,t;0,t)=e^{t}/2$ we use (3.6) of Proposition 3.3 to derive the next
representation

$\displaystyle\widetilde{w(x,t)}+t\varphi(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\frac{1}{4}e^{t/2}e^{-b/2}(2-b)\Big{(}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{)}\,db$
$\displaystyle-\int_{0}^{t}be^{-2b}\,db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi^{(1)}(y)\frac{\partial}{\partial
y}E(x-y,t;0,b)\,.$

The integration by parts in the second term leads to

$\displaystyle\widetilde{w(x,t)}+t\varphi(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\frac{1}{4}e^{t/2}e^{-b/2}(2-b)\Big{[}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{]}\,db$
$\displaystyle-\int_{0}^{t}be^{-2b}\,db\,\varphi(x+e^{-b}-e^{-t})\left[\frac{\partial}{\partial
y}E(x-y,t;0,b)\right]_{y=x+e^{-b}-e^{-t}}$
$\displaystyle+\int_{0}^{t}be^{-2b}\,db\,\varphi(x-e^{-b}+e^{-t})\left[\frac{\partial}{\partial
y}E(x-y,t;0,b)\right]_{y=x-e^{-b}+e^{-t}}$
$\displaystyle+\int_{0}^{t}be^{-2b}\,db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi(y)\left(\frac{\partial}{\partial
y}\right)^{2}E(x-y,t;0,b)\,.$

The application of (3.7) and (3.8) of Proposition 3.3 and
$\frac{\partial}{\partial y}E(x-y,t;0,b)=-\frac{\partial}{\partial
x}E(x-y,t;0,b)$ imply

$\displaystyle\widetilde{w(x,t)}+t\varphi(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}(2-b)+\frac{1}{16}be^{-3b/2}e^{t/2}(e^{b}-e^{t})(1+4M^{2})\Big{]}\Big{[}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{]}\,db$
$\displaystyle+\int_{0}^{t}be^{-2b}\,db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,\varphi(y)\left(\frac{\partial}{\partial
y}\right)^{2}E(x-y,t;0,b)\,.$

Thus, for the function $u(x,t)=w(x,t)+t\varphi(x)$ we have obtained

$\displaystyle u(x,t)$ (4.3) $\displaystyle=$
$\displaystyle\int_{0}^{t}\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}(2-b)+\frac{1}{16}be^{-3b/2}e^{t/2}(e^{b}-e^{t})(1+4M^{2})\Big{]}\Big{[}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{]}\,db$
$\displaystyle+\int_{0}^{t}be^{-2b}\,db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,\varphi(y)\left(\frac{\partial}{\partial
y}\right)^{2}E(x-y,t;0,b)$
$\displaystyle-M^{2}\int_{0}^{t}b\,db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,\varphi(y)E(x-y,t;0,b)\,\,.$

The proposition is proven. $\square$

###### Corollary 4.2

The solution $u=u(x,t)$ of the Cauchy problem (0.20) with $\varphi_{0}(x)=0$
and $\varphi_{1}(x)=\varphi(x)$ can be represented as follows

$\displaystyle u(x,t)$ (4.4) $\displaystyle=$
$\displaystyle\int_{0}^{t}\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}(2-b)+\frac{1}{16}be^{-3b/2}e^{t/2}(e^{b}-e^{t})(1+4M^{2})\Big{]}\Big{[}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{]}\,db$
$\displaystyle+\int_{0}^{t}db\int_{0}^{e^{-b}-e^{-t}}dz\,\Big{[}\varphi(x-z)+\varphi(x+z)\Big{]}b\left[e^{-2b}\,\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}E(z,t;0,b)\right]\,,$

as well as by (4.1), where

$\displaystyle K_{1}(z,t)$ $\displaystyle=$
$\displaystyle\Big{[}\frac{1}{4}e^{t/2}\big{(}2+\ln(z+e^{-t})\big{)}+\frac{1}{16}(1+4M^{2})e^{3t/2}z\ln(z+e^{-t})\Big{]}\frac{1}{\sqrt{z+e^{-t}}}$
(4.5)
$\displaystyle+\int_{0}^{-\ln(z+e^{-t})}b\left[e^{-2b}\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}E(z,t;0,b)\right]db\,.$

Proof of Corollary. By means of the statement (4.2) of Proposition 4.1 and
(3.3) we obtain

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}db\,\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}(2-b)+\frac{1}{16}be^{-3b/2}e^{t/2}(e^{b}-e^{t})(1+4M^{2})\Big{]}$
$\displaystyle\hskip
28.45274pt\times\Big{[}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{]}$
$\displaystyle+\int_{0}^{t}db\int_{0}^{-(e^{-b}-e^{-t})}(-1)dz\,\varphi(x-z)\left[be^{-2b}\,\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}bE(z,t;0,b)\right]$
$\displaystyle+\int_{0}^{t}db\int_{-(e^{-b}-e^{-t})}^{0}dz\,\varphi(x+z)\left[be^{-2b}\,\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}bE(z,t;0,b)\right]\,.$

To prove (4.1) with $K_{1}(z,t)$ defined by (4.5) we apply (3.3) and write

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}db\,\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}(2-b)+\frac{1}{16}be^{-3b/2}e^{t/2}(e^{b}-e^{t})(1+4M^{2})\Big{]}$
$\displaystyle\hskip
28.45274pt\times\Big{[}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{]}$
$\displaystyle+\int_{0}^{t}\,db\int_{0}^{e^{-b}-e^{-t}}dz\,\Big{[}\varphi(x+z)+\varphi(x-z)\Big{]}\left[be^{-2b}\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}bE(z,t;0,b)\right].$

Next we make change $z=e^{-b}-e^{-t}$, $dz=-e^{-b}db$,
$db=-(z+e^{-t})^{-1}dz$, and $b=-\ln(z+e^{-t})$ in the first integral:

$\displaystyle\int_{0}^{t}\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}(2-b)+\frac{1}{16}be^{-3b/2}e^{t/2}(e^{b}-e^{t})(1+4M^{2})\Big{]}\Big{[}\varphi(x+e^{-b}-e^{-t})+\varphi(x-e^{-b}+e^{-t})\Big{]}\,db$
$\displaystyle=$
$\displaystyle\int_{0}^{1-e^{-t}}\Big{[}\frac{1}{4}e^{t/2}\big{(}2+\ln(z+e^{-t})\big{)}+\frac{1}{16}(1+4M^{2})e^{3t/2}z\ln(z+e^{-t})\Big{]}\frac{1}{\sqrt{z+e^{-t}}}$
$\displaystyle\hskip
56.9055pt\times\Big{[}\varphi(x+z)+\varphi(x-z)\Big{]}\,dz\,.$

Then

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int_{0}^{1-e^{-t}}\Big{[}\frac{1}{4}e^{t/2}\big{(}2+\ln(z+e^{-t})\big{)}+\frac{1}{16}(1+4M^{2})e^{3t/2}z\ln(z+e^{-t})\Big{]}\frac{1}{\sqrt{z+e^{-t}}}$
$\displaystyle\hskip
56.9055pt\times\Big{[}\varphi(x+z)+\varphi(x-z)\Big{]}\,dz$
$\displaystyle+\int_{0}^{t}\,db\int_{0}^{e^{-b}-e^{-t}}dz\,\Big{[}\varphi(x+z)+\varphi(x-z)\Big{]}b\left[e^{-2b}\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}E(z,t;0,b)\right].$

In the last integral we change the order of integration,

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int_{0}^{1-e^{-t}}\Big{[}\frac{1}{4}e^{t/2}\big{(}2+\ln(z+e^{-t})\big{)}+\frac{1}{16}(1+4M^{2})e^{3t/2}z\ln(z+e^{-t})\Big{]}\frac{1}{\sqrt{z+e^{-t}}}\Big{[}\varphi(x+z)+\varphi(x-z)\Big{]}\,dz$
$\displaystyle+\int_{0}^{1-e^{-t}}\,dz\Big{[}\varphi(x+z)+\varphi(x-z)\Big{]}\int_{0}^{-\ln(z+e^{-t})}db\,b\left[e^{-2b}\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}E(z,t;0,b)\right],$

and obtain (4.1), where $K_{1}(z,t)$ is defined by (4.5). Corollary is proven.
$\square$

Proof of Theorem 0.4 with $\varphi_{0}=0$. The next lemma completes the proof
of Theorem 0.4.

###### Lemma 4.3

The kernel $K_{1}(z,t)$ defined by (4.5) coincides with one given in Theorem
0.4.

Proof. We have due to Lemma 1.2, (3.3), and by integration by parts

$\displaystyle\int_{0}^{-\ln(z+e^{-t})}b\left[e^{-2b}\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}E(z,t;0,b)\right]\,db=\int_{0}^{-\ln(z+e^{-t})}b\Big{(}\frac{\partial}{\partial
b}\Big{)}^{2}E(z,t;0,b)db$ $\displaystyle=$
$\displaystyle(-\ln(z+e^{-t}))\Big{[}\frac{\partial}{\partial
b}E(z,t;0,b)\Big{]}_{b=-\ln(z+e^{-t})}-E(z,t;0,-\ln(z+e^{-t}))+E(z,t;0,0)\,.$

On the other hand, (3.4) and (3.9) of Proposition 3.3 imply

$\displaystyle\int_{0}^{-\ln(z+e^{-t})}db\,b\left[e^{-2b}\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}E(z,t;0,b)\right]$ $\displaystyle=$
$\displaystyle-\ln(z+e^{-t})\frac{1}{16}\frac{4+e^{t}z(1+4M^{2})}{\sqrt{e^{-t}}\sqrt{z+e^{-t}}}-\frac{1}{2}\frac{1}{\sqrt{e^{-t}}\sqrt{z+e^{-t}}}+E(z,t;0,0).$

Thus, for the kernel $K_{1}(z,t)$ defined by (4.5) we have

$\displaystyle K_{1}(z,t)$ $\displaystyle=$
$\displaystyle\Big{[}\frac{1}{4}e^{t/2}\big{(}2+\ln(z+e^{-t})\big{)}+\frac{1}{16}(1+4M^{2})e^{3t/2}z\ln(z+e^{-t})\Big{]}\frac{1}{\sqrt{z+e^{-t}}}$
$\displaystyle-\ln(z+e^{-t})\frac{1}{16}\frac{4+e^{t}z(1+4M^{2})}{\sqrt{e^{-t}}\sqrt{z+e^{-t}}}-\frac{1}{2}\frac{1}{\sqrt{e^{-t}}\sqrt{z+e^{-t}}}+E(z,t;0,0)$
$\displaystyle=$ $\displaystyle
e^{t/2}\Big{[}\frac{1}{4}\big{(}2+\ln(z+e^{-t})\big{)}+\frac{1}{16}(1+4M^{2})e^{t}z\ln(z+e^{-t})$
$\displaystyle-\ln(z+e^{-t})\frac{1}{16}(4+e^{t}z(1+4M^{2}))-\frac{1}{2}\Big{]}\frac{1}{\sqrt{z+e^{-t}}}+E(z,t;0,0)$
$\displaystyle=$ $\displaystyle E(z,t;0,0)\,.$

The last line coincides with $K_{1}(z,t)$ of Theorem 0.4. Lemma is proven.
$\square$

##  5 The Cauchy Problem: First Datum and $n=1$

In this section we prove Theorem 0.4 in the case of $\varphi_{1}(x)=0$. Thus,
we have to prove the representation given by Theorem 0.4 for the solution
$u=u(x,t)$ of the Cauchy problem (0.20) with $\varphi_{1}(x)=0$, that is
equivalent to

$u(x,t)=\frac{1}{2}e^{\frac{t}{2}}\Big{[}\varphi_{0}(x+1-e^{-t})+\varphi_{0}(x-1+e^{-t})\Big{]}+\,\int_{0}^{1}\big{[}\varphi_{0}(x-\phi(t)s)+\varphi_{0}(x+\phi(t)s)\big{]}K_{0}(\phi(t)s,t)\phi(t)\,ds,$

where $\phi(t)=1-e^{-t}$. The proof of this case consists of the several
steps.

###### Proposition 5.1

The solution $u=u(x,t)$ of the Cauchy problem (0.20) can be represented as
follows

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\frac{1}{2}e^{\frac{t}{2}}\Big{[}\varphi_{0}(x+1-e^{-t})+\varphi_{0}(x-1+e^{-t})\Big{]}$
$\displaystyle-\int_{0}^{t}\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}+\frac{1}{16}(1+4M^{2})e^{t/2}e^{-3b/2}(e^{t}-e^{b})\Big{]}\Big{[}\varphi_{0}(x+e^{-b}-e^{-t})+\varphi_{0}(x-e^{-b}+e^{-t}))\Big{]}\,db$
$\displaystyle+\int_{0}^{t}db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi_{0}(y)\left[e^{-2b}\left(\frac{\partial}{\partial
y}\right)^{2}E(x-y,t;0,b)-M^{2}E(x-y,t;0,b)\right]\,.$

Proof. We set $u(x,t)=w(x,t)+\varphi_{0}(x)$, then

$w_{tt}-e^{-2t}w_{xx}+M^{2}w=e^{-2t}\varphi_{0,xx}-M^{2}\varphi_{0}(x)\,,\qquad
w(x,0)=0\,,\qquad w_{t}(x,0)=0\,.$

Next we plug $f(x,t)=e^{-2t}\varphi_{0,xx}(x)-M^{2}\varphi_{0}(x)$ in the
formula given by Theorem 0.3 and obtain

$\displaystyle w(x,t)$ $\displaystyle=$
$\displaystyle\widetilde{w(x,t)}-\int_{0}^{t}db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,M^{2}\varphi_{0}(y)E(x-y,t;0,b),$
(5.2)

where we have denoted

$\displaystyle\widetilde{w(x,t)}$ $\displaystyle:=$
$\displaystyle\int_{0}^{t}db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,e^{-2b}\varphi_{0}^{(2)}(y)E(x-y,t;0,b)\,.$

Next we integrate by parts and apply (3.3):

$\displaystyle\widetilde{w(x,t)}$ $\displaystyle=$
$\displaystyle\int_{0}^{t}e^{-2b}\Big{[}\varphi_{0}^{(1)}(x+e^{-b}-e^{-t})-\varphi_{0}^{(1)}(x-e^{-b}+e^{-t})\Big{]}E(e^{-b}-e^{-t},t;0,b)\,db$
$\displaystyle-\int_{0}^{t}db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,e^{-2b}\varphi_{0}^{(1)}(y)\frac{\partial}{\partial
y}E(x-y,t;0,b)\,.$

On the other hand,

$\varphi_{0}^{(1)}(x+e^{-b}-e^{-t})=-e^{b}\frac{\partial}{\partial
b}\varphi_{0}(x+e^{-b}-e^{-t})\quad\mbox{\rm
and}\quad\varphi_{0}^{(1)}(x-e^{-b}+e^{-t})=e^{b}\frac{\partial}{\partial
b}\varphi_{0}(x-e^{-b}+e^{-t})$

imply

$\displaystyle\widetilde{w(x,t)}$ $\displaystyle=$
$\displaystyle-\int_{0}^{t}e^{-b}\Big{[}\frac{\partial}{\partial
b}\varphi_{0}(x+e^{-b}-e^{-t})+\frac{\partial}{\partial
b}\varphi_{0}(x-e^{-b}+e^{-t})\Big{]}E(e^{-b}-e^{-t},t;0,b)\,db$
$\displaystyle-\int_{0}^{t}db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,e^{-2b}\varphi_{0}^{(1)}(y)\frac{\partial}{\partial
y}E(x-y,t;0,b)\,.$

One more integration by parts leads to

$\displaystyle\widetilde{w(x,t)}+\varphi_{0}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{2}e^{\frac{t}{2}}\Big{[}\varphi_{0}(x+1-e^{-t})+\varphi_{0}(x-1+e^{-t})\Big{]}$
$\displaystyle+\int_{0}^{t}\Big{[}\varphi_{0}(x+e^{-b}-e^{-t})+\varphi_{0}(x-e^{-b}+e^{-t})\Big{]}\frac{\partial}{\partial
b}\Big{(}e^{-b}E(e^{-b}-e^{-t},t;0,b)\Big{)}\,db$
$\displaystyle-\int_{0}^{t}db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,e^{-2b}\varphi_{0}^{(1)}(y)\frac{\partial}{\partial
y}E(x-y,t;0,b)\,,$

where $E(0,t;0,t)=\frac{1}{2}e^{t}$ and
$E(1-e^{-t},t;0,0)=\frac{1}{2}e^{\frac{t}{2}}$ have been used. Next we apply
(3.5) of Proposition 3.3 and an integration by parts to obtain

$\displaystyle\widetilde{w(x,t)}+\varphi_{0}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{2}e^{\frac{t}{2}}\Big{[}\varphi_{0}(x+1-e^{-t})+\varphi_{0}(x-1+e^{-t})\Big{]}$
$\displaystyle-\int_{0}^{t}\frac{1}{4}e^{t/2}e^{-b/2}\Big{[}\varphi_{0}(x+e^{-b}-e^{-t})+\varphi_{0}(x-e^{-b}+e^{-t})\Big{]}\,db$
$\displaystyle+\int_{0}^{t}db\,e^{-2b}\Big{[}\varphi_{0}(y)\frac{\partial}{\partial
x}E(x-y,t;0,b)\Big{]}_{y=x-e^{-b}+e^{-t}}^{y=x+e^{-b}-e^{-t}}$
$\displaystyle+\int_{0}^{t}db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,\varphi_{0}(y)e^{-2b}\left(\frac{\partial}{\partial
y}\right)^{2}E(x-y,t;0,b)\,.$

We have due to (3.7) and (3.8) of Proposition 3.3

$\displaystyle\widetilde{w(x,t)}+\varphi_{0}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{2}e^{\frac{t}{2}}\Big{[}\varphi_{0}(x+1-e^{-t})+\varphi_{0}(x-1+e^{-t})\Big{]}$
$\displaystyle-\int_{0}^{t}\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}+\frac{1}{16}(1+4M^{2})e^{t/2}e^{-3b/2}(e^{t}-e^{b})\Big{]}\Big{[}\varphi_{0}(x+e^{-b}-e^{-t})+\varphi_{0}(x-e^{-b}+e^{-t}))\Big{]}\,db$
$\displaystyle+\int_{0}^{t}db\int_{x-e^{-b}+e^{-t}}^{x+e^{-b}-e^{-t}}dy\,\varphi_{0}(y)e^{-2b}\left(\frac{\partial}{\partial
y}\right)^{2}E(x-y,t;0,b)\,.$

Then the last equation together with (5.2) proves the desired representation.
Proposition is proven. $\square$

Completion of the proof of Theorem 0.4. We make change $z=e^{-b}-e^{-t}$,
$dz=-e^{-b}db$, and $b=-\ln(z+e^{t})$ in the second term of the representation
given by the previous proposition:

$\displaystyle\int_{0}^{t}\Big{[}\frac{1}{4}e^{t/2}e^{-b/2}+\frac{1}{16}(1+4M^{2})e^{t/2}e^{-3b/2}(e^{t}-e^{b})\Big{]}\Big{[}\varphi_{0}(x+e^{-b}-e^{-t})+\varphi_{0}(x-e^{-b}+e^{-t}))\Big{]}\,db$
$\displaystyle=$
$\displaystyle\int^{1-e^{-t}}_{0}\Big{[}\frac{1}{4}e^{t/2}\sqrt{z+e^{-t}}+\frac{1}{16}(1+4M^{2})e^{t/2}(\sqrt{z+e^{-t}})^{3}\frac{ze^{t}}{z+e^{-t}}\Big{]}\Big{[}\varphi_{0}(x+z)+\varphi_{0}(x-z)\Big{]}\frac{1}{z+e^{-t}}\,dz$
$\displaystyle=$
$\displaystyle\int^{1-e^{-t}}_{0}e^{t}\Big{[}\frac{1}{4}+\frac{1}{16}(1+4M^{2})ze^{t}\Big{]}\frac{1}{\sqrt{e^{t}z+1}}\Big{[}\varphi_{0}(x+z)+\varphi_{0}(x-z)\Big{]}\,dz\,.$

Next we apply (3.3) to the last term of that representation, and then we
change the order of integration:

$\displaystyle\int_{0}^{t}db\int_{x-(e^{-b}-e^{-t})}^{x+e^{-b}-e^{-t}}dy\,\varphi_{0}(y)\left[e^{-2b}\left(\frac{\partial}{\partial
y}\right)^{2}E(x-y,t;0,b)-M^{2}E(x-y,t;0,b)\right]$ $\displaystyle=$
$\displaystyle\int_{0}^{t}db\int_{0}^{e^{-b}-e^{-t}}dz\,\Big{[}\varphi_{0}(x+z)+\varphi_{0}(x-z)\Big{]}\left[e^{-2b}\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}E(z,t;0,b)\right]$ $\displaystyle=$
$\displaystyle\int_{0}^{1-e^{-t}}dz\Big{[}\varphi_{0}(x+z)+\varphi_{0}(x-z)\Big{]}\int_{0}^{-\ln(z+e^{-t})}db\,\left[e^{-2b}\left(\frac{\partial}{\partial
z}\right)^{2}E(z,t;0,b)-M^{2}E(z,t;0,b)\right]\,.$

On the other hand, since the function $E(z,t;0,b)$ solves Klein-Gordon
equation, the last integral is equal to

$\displaystyle\int_{0}^{1-e^{-t}}dz\Big{[}\varphi_{0}(x+z)+\varphi_{0}(x-z)\Big{]}\int_{0}^{-\ln(z+e^{-t})}db\,\left[\left(\frac{\partial}{\partial
t}\right)^{2}E(z,t;0,b)\right]$ $\displaystyle=$
$\displaystyle\int_{0}^{1-e^{-t}}dz\Big{[}\varphi_{0}(x+z)+\varphi_{0}(x-z)\Big{]}\left[\frac{\partial}{\partial
b}E(z,t;0,b)\right]_{b=0}^{b=-\ln(z+e^{-t})}\,.$

Application of (3.9) and (3.10) completes the proof. Theorem 0.4 is proven.
$\square$

##  6 n-Dimensional Case, $n\geq 2$

Proof of Theorem 0.5. Let us consider the case of $x\in{\mathbb{R}}^{n}$,
where $n=2m+1$, $m\in{\mathbb{N}}$. First for the given function $u=u(x,t)$ we
define the spherical means of $u$ about point $x$:

$\displaystyle I_{u}(x,r,t)$ $\displaystyle=$
$\displaystyle\frac{1}{\omega_{n-1}}\int_{S^{n-1}}u(x+ry,t)\,dS_{y}\,,$

where $\omega_{n-1}$ denotes the area of the unit sphere
$S^{n-1}\subset{\mathbb{R}}^{n}$. Then we define an operator $\Omega_{r}$ by

$\Omega_{r}(u)(x,t):=\Big{(}\frac{1}{r}\frac{\partial}{\partial
r}\Big{)}^{m-1}r^{2m-1}I_{u}(x,r,t)\,.$

One can show that there are constants $c_{j}^{(n)}$, $j=0,\ldots,m-1$, where
$n=2m+1$, with $c_{0}^{(n)}=1\cdot 3\cdot 5\cdots(n-2)$, such that

$\Big{(}\frac{1}{r}\frac{\partial}{\partial
r}\Big{)}^{m-1}r^{2m-1}\varphi(r)=r\sum_{j=0}^{m-1}c_{j}^{(n)}r^{j}\frac{\partial^{j}}{\partial
r^{j}}\varphi(r)\,.$

One can recover the functions according to

$\displaystyle u(x,t)$ $\displaystyle=$ $\displaystyle\lim_{r\to
0}I_{u}(x,r,t)=\lim_{r\to 0}\frac{1}{c_{0}^{(n)}r}\Omega_{r}(u)(x,t)\,,$ (6.1)
$\displaystyle u(x,0)$ $\displaystyle=$ $\displaystyle\lim_{r\to
0}\frac{1}{c_{0}^{(n)}r}\Omega_{r}(u)(x,0)\,,\quad u_{t}(x,0)=\lim_{r\to
0}\frac{1}{c_{0}^{(n)}r}\Omega_{r}(\partial_{t}u)(x,0)\,.$ (6.2)

It is well known that
$\Delta_{x}\Omega_{r}h=\frac{\partial^{2}}{\partial\,r^{2}}\Omega_{r}h$ for
every function $h\in C^{2}({\mathbb{R}}^{n})$. Therefore we arrive at the
following mixed problem for the function $v(x,r,t):=\Omega_{r}(u)(x,r,t)$:

$\displaystyle\cases{v_{tt}(x,r,t)-e^{-2t}v_{rr}(x,r,t)+M^{2}v(x,r,t)=F(x,r,t)\quad\mbox{\rm
for all}\quad t\geq 0\,,\,\,r\geq 0\,,\,\,x\in{\mathbb{R}}^{n}\,,\cr
v(x,0,t)=0\quad\mbox{\rm for all}\quad t\geq 0\,,\quad
x\in{\mathbb{R}}^{n},\cr v(x,r,0)=0\,,\quad v_{t}(x,r,0)=0\quad\mbox{\rm for
all}\quad r\geq 0\,,\quad x\in{\mathbb{R}}^{n}\,,\cr
F(x,r,t):=\Omega_{r}(f)(x,t)\,,\quad F(x,0,t)=0\,,\quad\mbox{\rm for all}\quad
x\in{\mathbb{R}}^{n}\,.}$

It must be noted here that the spherical mean $I_{u}$ defined for $r>0$ has an
extension as even function for $r<0$ and hence $\Omega_{r}(u)$ has a natural
extension as an odd function. That allows replacing the mixed problem with the
Cauchy problem. Namely, let functions $\widetilde{v}$ and $\widetilde{F}$ be
the continuations of the functions $v$ and $F$, respectively, by

$\widetilde{v}(x,r,t)=\cases{\,v(x,r,t),\,\,if\,\,r\geq
0\cr-v(x,-r,t),\,\,if\,\,r\leq
0}\,,\quad\widetilde{F}(x,r,t)=\cases{\,F(x,r,t),\,\,if\,\,r\geq
0\cr-F(x,-r,t),\,\,if\,\,r\leq 0}\,.$

Then $\widetilde{v}$ solves the Cauchy problem

$\displaystyle\cases{\widetilde{v}_{tt}(x,r,t)-e^{-2t}\widetilde{v}_{rr}(x,r,t)+M^{2}\widetilde{v}(x,r,t)=\widetilde{F}(x,r,t)\quad\mbox{\rm
for all}\quad t\geq 0\,,\quad r\in{\mathbb{R}}\,,\quad
x\in{\mathbb{R}}^{n}\,,\cr\widetilde{v}(x,r,0)=0\,,\quad\widetilde{v}_{t}(x,r,0)=0\quad\mbox{\rm
for all}\quad r\in{\mathbb{R}}\,,\quad x\in{\mathbb{R}}^{n}.}$

Hence, according to Theorem 0.3, one has the representation

$\displaystyle\widetilde{v}(x,r,t)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}db\int_{r-(e^{-b}-e^{-t})}^{r+e^{-b}-e^{-t}}dr_{1}\,\widetilde{F}(x,r_{1},b)(4e^{-b-t})^{iM}\left((e^{-t}+e^{-b})^{2}-(r-r_{1})^{2}\right)^{-\frac{1}{2}-iM}$
$\displaystyle\hskip 85.35826pt\times
F\left(\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-(r-r_{1})^{2}}{(e^{-b}+e^{-t})^{2}-(r-r_{1})^{2}}\right).$

Since $u(x,t)=\lim_{r\to 0}\big{(}\widetilde{v}(x,r,t)/(c_{0}^{(n)}r)\big{)}$,
we consider the case of $r<t$ in the above representation to obtain:

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\frac{1}{c_{0}^{(n)}}\int_{0}^{t}db\int_{0}^{e^{-b}-e^{-t}}dr_{1}\,\lim_{r\to
0}\frac{1}{r}\left\\{\widetilde{F}(x,r+r_{1},b)+\widetilde{F}(x,r-r_{1},b)\right\\}$
$\displaystyle\times(4e^{-b-t})^{iM}\left((e^{-t}+e^{-b})^{2}-r_{1}^{2}\right)^{-\frac{1}{2}-iM}F\left(\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-r_{1}^{2}}{(e^{-b}+e^{-t})^{2}-r_{1}^{2}}\right)\,.$

Then by definition of the function $\widetilde{F}$, we replace $\lim_{r\to
0}\frac{1}{r}\Big{\\{}\widetilde{F}(x,r-r_{1},b)+\widetilde{F}(x,r+r_{1},b)\Big{\\}}$
with
$2\Big{(}\frac{\partial}{\partial r}F(x,r,b)\Big{)}_{r=r_{1}}$ in the last
formula. The definitions of $F(x,r,t)$ and of the operator $\Omega_{r}$ yield:

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\frac{2}{c_{0}^{(n)}}\int_{0}^{t}db\int_{0}^{e^{-b}-e^{-t}}dr_{1}\,\left(\frac{\partial}{\partial
r}\Big{(}\frac{1}{r}\frac{\partial}{\partial
r}\Big{)}^{m-1}r^{2m-1}I_{f}(x,r,t)\right)_{r=r_{1}}$
$\displaystyle\times(4e^{-b-t})^{iM}\left((e^{-t}+e^{-b})^{2}-r_{1}^{2}\right)^{-\frac{1}{2}-iM}F\left(\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{-b}-e^{-t})^{2}-r_{1}^{2}}{(e^{-b}+e^{-t})^{2}-r_{1}^{2}}\right)\,,$

where $x\in{\mathbb{R}}^{n}$, $n=2m+1$, $m\in{\mathbb{N}}$. Thus, the solution
to the Cauchy problem is given by (0.22). We employ the method of descent to
complete the proof for the case of even $n$, $n=2m$, $m\in{\mathbb{N}}$.
Theorem 0.5 is proven. $\square$

Proof of (0.16) and (0.17). We set $f(x,b)=\delta(x)\delta(t-t_{0})$ in (0.22)
and (0.23), and we obtain (0.16) and (0.17), where if $n$ is odd, then

$E^{w}(x,t):=\frac{1}{\omega_{n-1}1\cdot 3\cdot
5\ldots\cdot(n-2)}\frac{\partial}{\partial
t}\Big{(}\frac{1}{t}\frac{\partial}{\partial
t}\Big{)}^{\frac{n-3}{2}}\frac{1}{t}\delta(|x|-t)\,,$

while for $n$ even we have

$E^{w}(x,t):=\frac{2}{\omega_{n-1}1\cdot 3\cdot
5\ldots\cdot(n-1)}\frac{\partial}{\partial
t}\Big{(}\frac{1}{t}\frac{\partial}{\partial
t}\Big{)}^{\frac{n-2}{2}}\frac{1}{\sqrt{t^{2}-|x|^{2}}}\chi_{B_{t}(x)}\,.$

Here $\chi_{B_{t}(x)}$ denotes the characteristic function of the ball
$B_{t}(x):=\\{x\in{\mathbb{R}}^{n};\,|x|$ $\leq t\\}$. Constant $\omega_{n-1}$
is the area of the unit sphere $S^{n-1}\subset{\mathbb{R}}^{n}$. The
distribution $\delta(|x|-t)$ is defined by
$<\delta(|\cdot|-t),f(\cdot)>=\int_{|x|=t}f(x)\,dx$ for $f\in
C^{\infty}_{0}({\mathbb{R}}^{n})$.

Proof of Theorem 0.6. First we consider case of $\varphi_{0}(x)=0$. More
precisely, we have to prove that the solution $u(x,t)$ of the Cauchy problem
(0.25) with $\varphi_{0}(x)=0$ can be represented by (0.26) with
$\varphi_{0}(x)=0$. The next lemma will be used in both cases.

###### Lemma 6.1

Consider the mixed problem

$\displaystyle\cases{v_{tt}-e^{-2t}v_{rr}+M^{2}v=0\quad\mbox{\rm for all}\quad
t\geq 0\,,\quad r\geq 0\,,\quad\cr v(r,0)=\tau_{0}(r)\,,\quad
v_{t}(r,0)=\tau_{1}(r)\quad\mbox{\rm for all}\quad r\geq 0\,,\cr
v(0,t)=0\quad\mbox{\rm for all}\quad t\geq 0\,,}$

and denote by $\widetilde{\tau}_{0}(r)$ and $\widetilde{\tau}_{1}(r)$ the
continuations of the functions $\tau_{0}(r)$ and $\tau_{1}(r)$ for negative
$r$ as odd functions: $\widetilde{\tau}_{0}(-r)=-\tau_{0}(r)$ and
$\widetilde{\tau}_{1}(-r)=-\tau_{1}(r)$ for all $r\geq 0$, respectively. Then
solution $v(r,t)$ to the mixed problem is given by the restriction of (4.1) to
$r\geq 0$:

$\displaystyle v(r,t)$ $\displaystyle=$
$\displaystyle\frac{1}{2}e^{\frac{t}{2}}\Big{[}\widetilde{\tau}_{0}(r+1-e^{-t})+\widetilde{\tau}_{0}(r-1+e^{-t})\Big{]}+\,\int_{0}^{1}\big{[}\widetilde{\tau}_{0}(r-\phi(t)s)+\widetilde{\tau}_{0}(r+\phi(t)s)\big{]}K_{0}(\phi(t)s,t)\phi(t)\,ds$
$\displaystyle+\int_{0}^{1}\Big{[}\widetilde{\tau}_{1}\Big{(}r+\phi(t)s\Big{)}+\widetilde{\tau}_{1}\Big{(}r-\phi(t)s\Big{)}\Big{]}K_{1}(\phi(t)s,t)\phi(t)\,ds\,,$

where $K_{0}(z,t)$ and $K_{1}(z,t)$ are defined in Theorem 0.4 and
$\phi(t)=1-e^{-t}$.

Proof. This lemma is a direct consequence of Theorem 0.4. $\square$

Now let us consider the case of $x\in{\mathbb{R}}^{n}$, where $n=2m+1$. First
for the given function $u=u(x,t)$ we define the spherical means of $u$ about
point $x$. One can recover the functions by means of (6.1), (6.2), and

$\displaystyle\varphi_{i}(x)$ $\displaystyle=$ $\displaystyle\lim_{r\to
0}I_{\varphi_{i}}(x,r)=\lim_{r\to
0}\frac{1}{c_{0}^{(n)}r}\Omega_{r}(\varphi_{i})(x)\,,\quad i=0,1\,.$

Then we arrive at the following mixed problem

$\displaystyle\cases{v_{tt}(x,r,t)-e^{-2t}v_{rr}(x,r,t)+M^{2}v(x,r,t)=0\quad\mbox{\rm
for all}\quad t\geq 0\,,\,\,r\geq 0\,,\,\,x\in{\mathbb{R}}^{n}\,,\cr
v(x,0,t)=0\quad\mbox{\rm for all}\quad t\geq 0\,,\quad
x\in{\mathbb{R}}^{n}\,,\cr v(x,r,0)=0\,,\quad
v_{t}(x,r,0)=\Phi_{1}(x,r)\quad\mbox{\rm for all}\quad r\geq 0\,,\quad
x\in{\mathbb{R}}^{n}\,,}$

with the unknown function $v(x,r,t):=\Omega_{r}(u)(x,r,t)$, where

$\displaystyle\Phi_{i}(x,r):=\Omega_{r}(\varphi_{i})(x)=\Big{(}\frac{1}{r}\frac{\partial}{\partial
r}\Big{)}^{m-1}r^{2m-1}\frac{1}{\omega_{n-1}}\int_{S^{n-1}}\varphi_{i}(x+ry)\,dS_{y}\,,$
(6.3) $\displaystyle\Phi_{i}(x,0)=0\,,\quad i=0,1,\quad\quad\mbox{\rm for
all}\quad x\in{\mathbb{R}}^{n}\,.$ (6.4)

Then, according to Lemma 6.1 and to $u(x,t)=\lim_{r\to
0}\big{(}v(x,r,t)/(c_{0}^{(n)}r)\big{)}$, we obtain:

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\frac{1}{c_{0}^{(n)}}\lim_{r\to
0}\frac{1}{r}\int_{0}^{1}\Big{[}\widetilde{\Phi}_{1}\big{(}x,r+\phi(t)s\big{)}+\widetilde{\Phi}_{1}\big{(}x,r-\phi(t)s\big{)}\Big{]}K_{1}(\phi(t)s,t)\phi(t)\,ds\,.$

The last limit is equal to

$\displaystyle 2\int_{0}^{1}\left(\frac{\partial}{\partial
r}\Phi_{1}(x,r)\right)_{r=\phi(t)s}K_{1}(\phi(t)s,t)\phi(t)\,ds$
$\displaystyle=$ $\displaystyle 2\int_{0}^{1}\left(\frac{\partial}{\partial
r}\Big{(}\frac{1}{r}\frac{\partial}{\partial
r}\Big{)}^{\frac{n-3}{2}}\frac{r^{n-2}}{\omega_{n-1}}\int_{S^{n-1}}\varphi_{1}(x+ry)\,dS_{y}\right)_{r=\phi(t)s}K_{1}(\phi(t)s,t)\phi(t)\,ds\,.$

Thus, Theorem 0.6 in the case of $\varphi_{0}(x)=0$ is proven.

Now we turn to the case of $\varphi_{1}(x)=0$. Thus, we arrive at the
following mixed problem

$\displaystyle\cases{v_{tt}(x,r,t)-e^{-2t}v_{rr}(x,r,t)+M^{2}v(x,r,t)=0\quad\mbox{\rm
for all}\quad t\geq 0\,,\,\,r\geq 0\,,\,\,x\in{\mathbb{R}}^{n}\,,\cr
v(x,r,0)=\Phi_{0}(x,r)\,,\quad v_{t}(x,r,0)=0\quad\mbox{\rm for all}\quad
r\geq 0\,,\quad x\in{\mathbb{R}}^{n}\,,\cr v(x,0,t)=0\quad\mbox{\rm for
all}\quad t\geq 0\,,\quad x\in{\mathbb{R}}^{n}\,,}$

with the unknown function $v(x,r,t):=\Omega_{r}(u)(x,r,t)$ defined by (6.3),
(6.4). Then, according to Lemma 6.1 and to $u(x,t)=\lim_{r\to
0}\big{(}v(x,r,t)/(c_{0}^{(n)}r)\big{)}$, we obtain:

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\frac{1}{c_{0}^{(n)}}e^{\frac{t}{2}}\lim_{r\to
0}\frac{1}{2r}\Big{[}\widetilde{\Phi}_{0}(x,r+e^{t}-1)+\widetilde{\Phi}_{0}(x,r-e^{t}+1)\Big{]}$
$\displaystyle+\,\frac{2}{c_{0}^{(n)}}\int_{0}^{1}\lim_{r\to
0}\frac{1}{2r}\big{[}\widetilde{\Phi}_{0}(x,r-\phi(t)s)+\widetilde{\Phi}_{0}(x,r+\phi(t)s)\big{]}K_{0}(\phi(t)s,t)\phi(t)\,ds\,,$
$\displaystyle=$
$\displaystyle\frac{1}{c_{0}^{(n)}}e^{\frac{t}{2}}\left(\frac{\partial}{\partial
r}\Phi_{0}(x,r)\right)_{r=\phi(t)}+\,\frac{2}{c_{0}^{(n)}}\int_{0}^{1}\left(\frac{\partial}{\partial
r}\Phi_{0}(x,r)\right)_{r=\phi(t)s}K_{0}(\phi(t)s,t)\phi(t)\,ds$
$\displaystyle=$ $\displaystyle
e^{\frac{t}{2}}v_{\varphi_{0}}(x,\phi(t))+\,2\int_{0}^{1}v_{\varphi_{0}}(x,\phi(t)s)K_{0}(\phi(t)s,t)\phi(t)\,ds\,.$

Theorem 0.6 is proven. $\square$

##  7 $L^{p}-L^{q}$ and $L^{q}-L^{q}$ Estimates for the Solutions of One-
dimensional Equation, $n=1$

Consider now the Cauchy problem for the equation (0.18) with the source term
and with vanishing initial data (0.19). In this and next sections we restrict
ourselves to the particles with “large” mass $m\geq n/2$, that is, with
nonnegative curved mass $M\geq 0$, to make presentation more transparent. The
case of $m<n/2$ will be discussed in the forthcoming paper.

###### Theorem 7.1

For every function $f\in C^{2}({\mathbb{R}}\times[0,\infty))$ such that
$f(\cdot,t)\in C_{0}^{\infty}({\mathbb{R}}_{x})$ the solution $u=u(x,t)$ of
the Cauchy problem (0.18), (0.19) satisfies inequality

$\displaystyle\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x})}$ $\displaystyle\leq$
$\displaystyle
C_{M}e^{t(1-1/\rho)}\int_{0}^{t}(1+t-b)(e^{t-b}-1)^{1/\rho}(e^{t-b}+1)^{-1}\|f(x,b)\|_{L^{p}({\mathbb{R}}_{x})}\,db$

for all $t>0$, where $1<p<\rho^{\prime}$,
$\frac{1}{q}=\frac{1}{p}-\frac{1}{\rho^{\prime}}$, $\rho<2$,
$\frac{1}{\rho}+\frac{1}{\rho^{\prime}}=1$.

Proof. Using the fundamental solution from Theorem 0.1 one can write the
convolution

$u(x,t)=\int_{-\infty}^{\infty}\\!\int_{-\infty}^{\infty}{\mathcal{E}}_{+}(x,t;y,b)f(y,b)\,db\,dy=\int_{0}^{t}\\!db\\!\int_{-\infty}^{\infty}{\mathcal{E}}_{+}(x-y,t;0,b)f(y,b)\,dy\,.$

Due to Young’s inequality we have

$\displaystyle\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x})}$ $\displaystyle\leq$
$\displaystyle
c_{k}\int_{0}^{t}db\Bigg{(}\int_{-(\phi(t)-\phi(b))}^{\phi(t)-\phi(b)}|E(x,t;0,b)|^{\rho}dx\Bigg{)}^{1/\rho}\|f(x,b)\|_{L^{p}({\mathbb{R}}_{x})},$

where $1<p<\rho^{\prime}$, $\frac{1}{q}=\frac{1}{p}-\frac{1}{\rho^{\prime}}$,
$\frac{1}{\rho}+\frac{1}{\rho^{\prime}}=1$, $\phi(t)=1-e^{-t}$. The integral
in parentheses can be transformed as follows

$\displaystyle\int_{-(\phi(t)-\phi(b))}^{\phi(t)-\phi(b)}|E(x,t;0,b)|^{\rho}dx$
$\displaystyle=$ $\displaystyle
2e^{-t+t\rho}\int_{0}^{e^{t-b}-1}\Big{(}(e^{t-b}+1)^{2}-y^{2}\Big{)}^{-\frac{\rho}{2}}\left|F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{t-b}-1)^{2}-y^{2}}{(e^{t-b}+1)^{2}-y^{2}}\Big{)}\right|^{\rho}dy.$

On the other hand, due to integral representation for the hypergeometric
function (1)[2, v.1, Sec.2.12] for $\zeta\in[0,1)$ one has

$\displaystyle\left|F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\zeta\Big{)}\right|$
$\displaystyle\leq$
$\displaystyle\frac{1}{|\Gamma\left(\frac{1}{2}+iM\right)|^{2}}\pi
F\Big{(}\frac{1}{2},\frac{1}{2};1;\zeta\Big{)}\,.$ (7.1)

Thus,

$\int_{-(\phi(t)-\phi(b))}^{\phi(t)-\phi(b)}|E(x,t;0,b)|^{\rho}dx\leq
C_{M}e^{-t+t\rho}\int_{0}^{e^{t-b}-1}\\!\left((e^{t-b}+1)^{2}-y^{2}\right)^{-\frac{\rho}{2}}\left|F\Big{(}\frac{1}{2},\frac{1}{2};1;\frac{(e^{t-b}-1)^{2}-y^{2}}{(e^{t-b}+1)^{2}-y^{2}}\Big{)}\right|^{\rho}\\!dy.$

###### Lemma 7.2

[33] For all $z>1$ the following estimate

$\int_{0}^{z-1}((z+1)^{2}-r^{2})^{-\frac{\rho}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\right)^{\rho}dr\leq
C(1+\ln
z)^{\rho}(z-1)(z+1)^{-\rho}F\Big{(}\frac{1}{2},\frac{\rho}{2};\frac{3}{2};\frac{(z-1)^{2}}{(z+1)^{2}}\Big{)}$

is fulfilled, provided that $1<p<\rho^{\prime}$,
$\frac{1}{q}=\frac{1}{p}-\frac{1}{\rho^{\prime}}$,
$\frac{1}{\rho}+\frac{1}{\rho^{\prime}}=1$. In particular, if $\rho<2$, then

$\displaystyle\int_{0}^{z-1}((z+1)^{2}-r^{2})^{-\frac{\rho}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\right)^{\rho}dr$
$\displaystyle\leq$ $\displaystyle C(1+\ln z)^{\rho}(z-1)(z+1)^{-\rho}\,.$

Completion of the proof of Theorem 7.1. Thus for $\rho<2$ and $z=e^{t-b}$ we
have

$\displaystyle\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x})}$ $\displaystyle\leq$
$\displaystyle
C_{M}e^{t(1-1/\rho)}\int_{0}^{t}(1+t-b)(e^{t-b}-1)^{1/\rho}(e^{t-b}+1)^{-1}\|f(x,b)\|_{L^{p}({\mathbb{R}}_{x})}\,db\,.$

The last inequality implies the estimate of the statement of theorem. Theorem
7.1 is proven. $\square$

###### Proposition 7.3

The solution $u=u(x,t)$ of the Cauchy problem

$u_{tt}-e^{-2t}u_{xx}+M^{2}u=0\,,\qquad u(x,0)=\varphi_{0}(x)\,,\qquad
u_{t}(x,0)=\varphi_{1}(x)\,,$

with $\varphi_{0}$, $\varphi_{1}\in C_{0}^{\infty}({\mathbb{R}})$ satisfies
the following estimate

$\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x})}\leq
C(1+t)\Big{(}e^{\frac{t}{2}}\|\varphi_{0}(x)\|_{L^{q}({\mathbb{R}}_{x})}+(e^{t}-1)e^{-t}\|\varphi_{1}(x)\|_{L^{q}({\mathbb{R}}_{x})}\Big{)}\,\quad\mbox{for
all}\,\quad t\in(0,\infty).$ (7.2)

Proof. First we consider the equation without source term but with the second
datum that is the case of $\varphi_{0}=0$. For the convenience we drop
subindex of $\varphi_{1}$. Then we apply the representation given by Theorem
0.4 for the solution $u=u(x,t)$ of the Cauchy problem with $\varphi_{0}=0$,
and obtain

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\int_{0}^{1-e^{-t}}\,\Big{[}\varphi_{1}(x-z)+\varphi_{1}(x+z)\Big{]}K_{1}(z,t)dz\,,$

where the kernel $K_{1}(z,t)$ is defined in Theorem 0.4. Hence, we arrive at
inequality

$\|u(x,t)\|_{{L}^{q}({\mathbb{R}})}\leq
C_{M}\|\varphi(x)\|_{{L}^{q}({\mathbb{R}})}\int_{0}^{1-e^{-t}}\,|K_{1}(r,t)|dr\,.$

To estimate the last integral we denote it by $I_{1}$,

$\displaystyle I_{1}(z)$ $\displaystyle:=$
$\displaystyle\int_{0}^{1-e^{-t}}\,|K_{1}(r,t)|dr\,,$

and with $z=e^{t}>1$ due to (7.1) we write

$\displaystyle I_{1}(z)$ $\displaystyle\leq$ $\displaystyle
C\int_{0}^{z-1}\,\frac{1}{\sqrt{(1+z)^{2}-y^{2}}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{y^{2}-(1-z)^{2}}{y^{2}-(1+z)^{2}}\right)dy\,.$
(7.3)

Then, according to Lemma 7.2 (the case of $\rho=1$) we have for that integral
the following estimate

$I_{1}(e^{t})\leq C(1+t)(e^{t}-1)(e^{t}+1)^{-1}\,.$ (7.4)

Finally, (7.3) and (7.4) imply the $L^{q}-L^{q}$ estimate (7.2) for the case
of $\varphi_{0}=0$.

Next we consider the equation without source but with the first datum, that
is, the case of $\varphi_{1}=0$. We apply the representation given by Theorem
0.4 for the solution $u=u(x,t)$ of the Cauchy problem with $\varphi_{1}=0$,
and obtain

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle\frac{1}{2}e^{\frac{t}{2}}\Big{[}\varphi_{0}(x+1-e^{-t})+\varphi_{0}(x-1+e^{-t})\Big{]}+\int_{0}^{1-e^{-t}}\big{[}\varphi_{0}(x-r)+\varphi_{0}(x+r)\big{]}K_{0}(r,t)\,dr\,,$

where the kernel $K_{0}(r,t)$ is defined in Theorem 0.4. Then we easily obtain
the following two estimates:

$\displaystyle\|u(x,t)-\int_{0}^{1-e^{-t}}\big{[}\varphi_{0}(x-r)+\varphi_{0}(x+r)\big{]}K_{0}(r,t)\,dr\|_{{L}^{q}({\mathbb{R}})}$
$\displaystyle\leq$ $\displaystyle
e^{\frac{t}{2}}\|\varphi_{0}(x)\|_{{L}^{q}({\mathbb{R}})}$

and

$\displaystyle\|u(x,t)\|_{{L}^{q}({\mathbb{R}})}$ $\displaystyle\leq$
$\displaystyle
e^{\frac{t}{2}}\|\varphi_{0}(x)\|_{{L}^{q}({\mathbb{R}})}+2\|\varphi_{0}(x)\|_{{L}^{q}({\mathbb{R}})}\int_{0}^{1-e^{-t}}\left|K_{0}(r,t)\right|dr\,.$

Finally, the following lemma completes the proof of proposition.

###### Lemma 7.4

The kernel $K_{0}(r,t)$ has an integrable singularity at $r=e^{t}-1$, more
precisely, one has

$\displaystyle\int_{0}^{1-e^{-t}}\left|K_{0}(r,t)\right|dr\leq
C(e^{t}-1)e^{-\frac{1}{2}t}(1+t)\quad\mbox{for all}\quad t\in[0,\infty)\,.$

Proof. For the integral we obtain

$\displaystyle\int_{0}^{1-e^{-t}}\left|K_{0}(r,t)\right|dr$
$\displaystyle\leq$
$\displaystyle\int_{0}^{z-1}\frac{1}{[(z-1)^{2}-y^{2}]\sqrt{[(z+1)^{2}-y^{2}]}}$
$\displaystyle\times\Bigg{|}\big{(}z-z^{2}-iM(1-z^{2}-y^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}$
$\displaystyle\hskip
19.91684pt+\big{(}z^{2}-1+y^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}\Bigg{|}dy$

for all $z:=e^{t}>1$. We divide the domain of integration into two zones,

$\displaystyle Z_{1}(\varepsilon,z)$ $\displaystyle:=$
$\displaystyle\left\\{(z,r)\,\Big{|}\,\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\leq\varepsilon,\,\,0\leq
r\leq z-1\right\\},$ (7.5) $\displaystyle Z_{2}(\varepsilon,z)$
$\displaystyle:=$
$\displaystyle\left\\{(z,r)\,\Big{|}\,\varepsilon\leq\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}},\,\,0\leq
r\leq z-1\right\\},$ (7.6)

and split the integral into conformable two parts,

$\displaystyle\int_{0}^{e^{t}-1}\left|K_{0}(r,t)\right|dr$ $\displaystyle=$
$\displaystyle\int_{(z,r)\in
Z_{1}(\varepsilon,z)}\left|K_{0}(r,t)\right|dr+\int_{(z,r)\in
Z_{2}(\varepsilon,z)}\left|K_{0}(r,t)\right|dr\,.$

In the first zone we have

$\displaystyle
F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}\\!\\!$
$\displaystyle\\!\\!=\\!\\!$
$\displaystyle\\!\\!1+\left(\frac{1}{2}+iM\right)^{2}\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}+O\left(\left(\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\right)^{2}\right),$
(7.7) $\displaystyle
F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}\\!\\!$
$\displaystyle\\!\\!=\\!\\!$
$\displaystyle\\!\\!1-\left(\frac{1}{4}+M^{2}\right)\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}+O\left(\left(\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\right)^{2}\right).$
(7.8)

We use the last formulas to estimate the term containing the hypergeometric
functions:

$\displaystyle\Bigg{|}\big{(}z-z^{2}-iM(1-z^{2}-r^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\Big{)}$
(7.9) $\displaystyle\hskip
28.45274pt+\big{(}z^{2}-1+r^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\Big{)}\Bigg{|}$
$\displaystyle\leq$ $\displaystyle\frac{1}{2}\big{[}(z-1)^{2}-r^{2}\big{]}$
$\displaystyle+\left|\big{(}z-z^{2}-iM(1-z^{2}-r^{2})\big{)}\left(\frac{1}{2}+iM\right)^{2}-\big{(}z^{2}-1+r^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}\left(\frac{1}{4}+M^{2}\right)\right|\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}$
$\displaystyle+\Big{(}\left|z-z^{2}-iM(1-z^{2}-r^{2})\right|+\left|z^{2}-1+y^{2}\right|\Big{)}O\left(\left(\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\right)^{2}\right)$
$\displaystyle=$ $\displaystyle\frac{1}{2}\big{[}(z-1)^{2}-r^{2}\big{]}$
$\displaystyle+\frac{1}{8}\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\left|(1-2iM)(-1+4M^{2})(y^{2}+z^{2}-1)+2(1+2iM)^{2}(-z^{2}+z+iM(y^{2}+z^{2}-1))\right|$
$\displaystyle+\Big{(}\left|z-z^{2}-iM(1-z^{2}-r^{2})\right|+\left|z^{2}-1+y^{2}\right|\Big{)}O\left(\left(\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\right)^{2}\right).$

Hence, we have to consider the following three integrals, which can be easily
evaluated and estimated,

$\displaystyle A_{1}$ $\displaystyle:=$ $\displaystyle\int_{(z,r)\in
Z_{1}(\varepsilon,z)}\frac{1}{\sqrt{(z+1)^{2}-r^{2}}}dr\leq\mbox{\rm
Arctan}\left(\frac{z-1}{2\sqrt{z}}\right)\leq\frac{\pi}{2}\,,$ $\displaystyle
A_{2}$ $\displaystyle:=$ $\displaystyle\int_{(z,r)\in
Z_{1}(\varepsilon,z)}\frac{z^{2}}{((z+1)^{2}-r^{2})\sqrt{(z+1)^{2}-r^{2}}}dr\leq(z+1)^{-1/2}(z-1),$

and

$A_{3}:=\int_{(z,r)\in
Z_{1}(\varepsilon,z)}\frac{\left|z-z^{2}-iM(1-z^{2}-r^{2})\right|+\left|z^{2}-1+r^{2}\right|}{\sqrt{(z+1)^{2}-r^{2}}}\frac{(z-1)^{2}-r^{2}}{((z+1)^{2}-r^{2})^{2}}dr\leq
C_{M}(z+1)^{-1/2}(z-1)$

for all $z\in[1,\infty)$. Finally, for the integral over the first zone we
have obtained

$\displaystyle\int_{(z,r)\in
Z_{1}(\varepsilon,z)}dr\frac{1}{[(z-1)^{2}-r^{2}]\sqrt{[(z+1)^{2}-r^{2}]}}$
$\displaystyle\times\Bigg{|}\big{(}z-z^{2}-iM(1-z^{2}-r^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\Big{)}$
$\displaystyle\hskip
28.45274pt+\big{(}z^{2}-1+r^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\Big{)}\Bigg{|}\,\leq\,C_{M}(z+1)^{-1/2}(z-1)$

for all $z\in[1,\infty)$. In the second zone we have

$\varepsilon\leq\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\leq 1\quad\mbox{\rm
and}\quad\frac{1}{(z-1)^{2}-r^{2}}\leq\frac{1}{\varepsilon[(z+1)^{2}-r^{2}]}\,.$
(7.10)

According to the formula 15.3.10 of [2, Ch.15] the hypergeometric functions
obey the estimate

$\left|F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;x\Big{)}\right|\leq
C\,\,\mbox{\rm
and}\,\,\left|F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;x\Big{)}\right|\leq
C\big{(}1-\ln(1-x))\,\,\,\forall x\in[\varepsilon,1).$ (7.11)

This allows to estimate the integral over the second zone:

$\displaystyle\int_{(z,r)\in
Z_{2}(\varepsilon,z)}dr\frac{1}{[(z-1)^{2}-r^{2}]\sqrt{(z+1)^{2}-r^{2}}}$
(7.12)
$\displaystyle\times\Bigg{|}\big{(}z-z^{2}-iM(1-z^{2}-r^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\Big{)}$
$\displaystyle\hskip
14.22636pt+\big{(}z^{2}-1+r^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\Big{)}\Bigg{|}\leq
C_{M}(z+1)^{-1/2}(z-1)$

for all $z\in[1,\infty)$. Indeed, for the argument of the hypergeometric
functions we have

$\varepsilon\leq\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}=1-\frac{4z}{(z+1)^{2}-r^{2}}<1,\quad\frac{4z}{(z+1)^{2}-r^{2}}<1-\varepsilon\quad\mbox{\rm
for all}\quad(z,r)\in Z_{2}(\varepsilon,z)\,.$ (7.13)

Hence,

$\left|F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\Big{)}\right|\leq
C\left(1+\ln z\right),\,(z,r)\in Z_{2}(\varepsilon,z).$ (7.14)

To prove (7.12) we estimate the following integral

$\displaystyle\int_{(z,r)\in
Z_{2}(\varepsilon,z)}\frac{z^{2}}{((z-1)^{2}-r^{2})\sqrt{(z+1)^{2}-r^{2}}}dr$
$\displaystyle\leq$ $\displaystyle
C_{\varepsilon}z^{2}\int_{0}^{z-1}\frac{1}{((z+1)^{2}-r^{2})^{3/2}}dr\,\leq\,C_{\varepsilon}\frac{(z-1)}{\sqrt{z}}\,.$

Thus, the lemma is proven. $\square$

##  8 $L^{p}-L^{q}$ Estimates for Equation with $n=1$ and without Source Term.
Some Estimates of Kernels $K_{0}$ and $K_{1}$

###### Theorem 8.1

Let $u=u(x,t)$ be a solution of the Cauchy problem

$u_{tt}-e^{-2t}u_{xx}+M^{2}u=0\,,\qquad u(x,0)=\varphi_{0}(x)\,,\qquad
u_{t}(x,0)=\varphi_{1}(x)\,,$

with $\varphi_{0}$, $\varphi_{1}\in C_{0}^{\infty}({\mathbb{R}})$. If
$\rho\in(1,2)$, then

$\displaystyle\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x})}$ $\displaystyle\leq$
$\displaystyle
e^{\frac{t}{2}}\|\varphi_{0}(x)\|_{L^{q}({\mathbb{R}}_{x})}+C_{\rho}(1+t)(e^{t}-1)^{\frac{1}{\rho}}e^{t[\frac{1}{2}-\frac{1}{\rho}]}\|\varphi_{0}(x)\|_{L^{p}({\mathbb{R}}_{x})}$
$\displaystyle+\,C_{\rho}(1+t)(e^{t}-1)^{\frac{1}{\rho}}e^{-\frac{t}{\rho}}\|\varphi_{1}(x)\|_{L^{p}({\mathbb{R}}_{x})}\,,$

for all $t\in(0,\infty)$. Here $1<p<\rho^{\prime}$,
$\frac{1}{q}=\frac{1}{p}-\frac{1}{\rho^{\prime}}$,
$\frac{1}{\rho}+\frac{1}{\rho^{\prime}}=1$. If $\rho=1$, then

$\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x})}\leq
C(1+t)\Big{(}e^{\frac{t}{2}}\|\varphi_{0}(x)\|_{L^{q}({\mathbb{R}}_{x})}+(e^{t}-1)e^{-t}\|\varphi_{1}(x)\|_{L^{q}({\mathbb{R}}_{x})}\Big{)}\,,$
(8.1)

for all $t\in(0,\infty)$.

Proof. For $\rho=1$ we just apply Proposition 7.3. To prove this theorem for
$\rho>1$ we need some auxiliary estimates for the kernels $K_{0}$ and $K_{1}$.
We start with the case of $\varphi_{0}=0$, where the kernel $K_{1}$ appears.
The application of Theorem 0.4 and Young’s inequality lead to

$\displaystyle\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x})}$ $\displaystyle\leq$
$\displaystyle
2\Bigg{(}\int_{0}^{1-e^{-t}}|K_{1}(x,t)|^{\rho}dx\Bigg{)}^{1/\rho}\|\varphi_{1}(x)\|_{L^{p}({\mathbb{R}}_{x})},$

where $1<p<\rho^{\prime}$, $\frac{1}{q}=\frac{1}{p}-\frac{1}{\rho^{\prime}}$,
$\frac{1}{\rho}+\frac{1}{\rho^{\prime}}=1$. Now we have to estimate the last
integral.

###### Proposition 8.2

We have

$\displaystyle\left(\int_{0}^{1-e^{-t}}|K_{1}(x,t)|^{\rho}dx\right)^{1/\rho}$
$\displaystyle\leq$ $\displaystyle C(1+t)(1-e^{-t})^{1/\rho}\quad\mbox{for
all}\,\quad t\in(0,\infty)\,.$

Proof. One can write

$\left(\int_{0}^{1-e^{-t}}|K_{1}(x,t)|^{\rho}dx\right)^{1/\rho}\leq
C_{M}e^{t(1-1/\rho)}\Bigg{(}\int_{0}^{e^{t}-1}\big{(}(e^{t}+1)^{2}-y^{2}\big{)}^{-\frac{\rho}{2}}\Big{|}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(e^{t}-1)^{2}-y^{2}}{(e^{t}+1)^{2}-y^{2}}\right)\Big{|}^{\rho}dy\Bigg{)}^{1/\rho}.$

Denote $z:=e^{t}>1$ and consider the integral
$\displaystyle\int_{0}^{z-1}\left|\frac{1}{\sqrt{(1+z)^{2}-x^{2}}}F\Big{(}\frac{1}{2},\frac{1}{2};1;\frac{(z-1)^{2}-x^{2}}{(z+1)^{2}-x^{2}}\Big{)}\right|^{\rho}dx$
of the right-hand side. Then we apply Lemma 7.2 and obtain

$\left(\int_{0}^{1-e^{-t}}|K_{1}(x,t)|^{\rho}dx\right)^{1/\rho}\leq
Ce^{t(1-1/\rho)}(1+\ln e^{t})(e^{t}-1)^{1/\rho}(e^{t}+1)^{-1}\leq
C(1+t)(1-e^{-t})^{1/\rho}\,.$

Proposition is proven. $\square$

Thus, the theorem in the case of $\varphi_{0}=0$ is proven.

Now we turn to the case of $\varphi_{1}=0$, where the kernel $K_{0}$ appears.
The application of Theorem 0.4 leads to

$\displaystyle\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x})}$ $\displaystyle\leq$
$\displaystyle
e^{\frac{t}{2}}\|\varphi_{0}(x)\|_{L^{q}({\mathbb{R}}_{x})}+\,\left\|\int_{0}^{1-e^{-t}}\left[\varphi_{0}(x-z)+\varphi_{0}(x+z)\right]K_{0}(z,t)\,dz\right\|_{L^{q}({\mathbb{R}}_{x})}\,.$

Similarly to the case of the second datum we arrive at

$\displaystyle\|u(x,t)\|_{L^{q}({\mathbb{R}}_{x})}$ $\displaystyle\leq$
$\displaystyle
e^{\frac{t}{2}}\|\varphi_{0}(x)\|_{L^{q}({\mathbb{R}}_{x})}+\,\|\varphi_{0}(x)\|_{L^{p}({\mathbb{R}}_{x})}\left(\int_{0}^{1-e^{-t}}|K_{0}(r,t)|^{\rho}dr\right)^{1/\rho}\,.$

The next proposition gives an estimate for the integral of the last
inequality.

###### Proposition 8.3

Let $1<p<\rho^{\prime}$, $\frac{1}{q}=\frac{1}{p}-\frac{1}{\rho^{\prime}}$,
$\frac{1}{\rho}+\frac{1}{\rho^{\prime}}=1$, and $\rho\in[1,2)$. We have

$\displaystyle\left(\int_{0}^{1-e^{-t}}|K_{0}(r,t)|^{\rho}dr\right)^{1/\rho}$
$\displaystyle\leq$ $\displaystyle
C_{\rho}(1+t)(e^{t}-1)^{\frac{1}{\rho}}e^{t(\frac{1}{2}-\frac{1}{\rho})}\quad\mbox{
for all}\quad t\in(0,\infty)\,.$

Proof. We turn to the integral ($z:=e^{t}>1$)

$\displaystyle\left(\int_{0}^{1-e^{-t}}|K_{0}(r,t)|^{\rho}dr\right)^{1/\rho}\\!\\!$
$\displaystyle\\!\\!=\\!\\!$
$\displaystyle\\!\\!\Bigg{(}\int_{0}^{z-1}dy\left(\frac{1}{[(z-1)^{2}-y^{2}]\sqrt{(z+1)^{2}-y^{2}}}\right)^{\rho}$
$\displaystyle\times\Bigg{|}\big{(}z-z^{2}-iM(1-z^{2}-y^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}$
$\displaystyle+\big{(}z^{2}-1+y^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}\Bigg{|}^{\rho}\Bigg{)}^{1/\rho}.$

The formulas (7.7) and (7.8) describe the behavior of the hypergeometric
functions in the neighbourhood of zero. Consider therefore two zones,
$Z_{1}(\varepsilon,z)$ and $Z_{2}(\varepsilon,z)$, defined in (7.5) and (7.6),
respectively. We split integral into two parts:

$\displaystyle\int_{0}^{1-e^{-t}}\left|K_{0}(r,t)\right|^{\rho}dr$
$\displaystyle=$ $\displaystyle\int_{(z,r)\in
Z_{1}(\varepsilon,z)}\left|K_{0}(r,t)\right|^{\rho}dr+\int_{(z,r)\in
Z_{2}(\varepsilon,z)}\left|K_{0}(r,t)\right|^{\rho}dr\,.$

In the proof of Lemma 7.4 the relation (7.9) was checked in the first zone. If
$1\leq z\leq N$ with some constant $N$, then the argument of the
hypergeometric functions is bounded,

$\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\leq\frac{(z-1)^{2}}{(z+1)^{2}}\leq\frac{(N-1)^{2}}{(N+1)^{2}}<1\quad\mbox{for
all}\quad r\in(0,z-1),$ (8.2)

and we obtain with $z=e^{t}$,

$\displaystyle\Bigg{(}\int_{0}^{1-e^{-t}}\left|K_{0}(r,t)\right|^{\rho}dr\Bigg{)}^{1/\rho}$
$\displaystyle\leq$ $\displaystyle
C_{M,N}\Bigg{(}\int_{0}^{z-1}\Bigg{[}\frac{1}{[(z-1)^{2}-y^{2}]\sqrt{(z+1)^{2}-y^{2}}}$
$\displaystyle\hskip
2.84544pt\times\Bigg{\\{}\frac{1}{2}[(z-1)^{2}-y^{2}]+z^{2}\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}+z^{2}\left(\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\right)^{2}\Bigg{\\}}\Bigg{]}^{\rho}dy\Bigg{)}^{1/\rho}$
$\displaystyle\leq$ $\displaystyle
C_{M,N}\Bigg{(}\int_{0}^{z-1}\Bigg{[}\frac{1}{\sqrt{(z+1)^{2}-y^{2}}}\Bigg{\\{}1+z^{2}\frac{1}{(z+1)^{2}-y^{2}}\Bigg{\\}}\Bigg{]}^{\rho}dy\Bigg{)}^{1/\rho}$
$\displaystyle\leq$ $\displaystyle C_{M,N}(z-1)^{1/\rho}(z+1)^{-1}\,.$

Thus, we can restrict ourselves to the case of large $z\geq N$ in both zones.
Consider therefore for $\rho\in(1,2)$ the following integrals over the first
zone

$\displaystyle A_{4}$ $\displaystyle:=$ $\displaystyle\int_{(z,r)\in
Z_{1}(\varepsilon)}\left(\frac{1}{\sqrt{(z+1)^{2}-r^{2}}}\right)^{\rho}dr\,\,\leq\,\,\int_{0}^{z-1}\left(\frac{1}{\sqrt{(z+1)^{2}-r^{2}}}\right)^{\rho}dr$
$\displaystyle\leq$ $\displaystyle
C(z-1)(z+1)^{-\rho}F\Big{(}\frac{1}{2},\frac{\rho}{2};\frac{3}{2};\frac{(z-1)^{2}}{(z+1)^{2}}\Big{)}$
$\displaystyle\leq$ $\displaystyle C(z-1)(z+1)^{-\rho}\,,$ $\displaystyle
A_{5}$ $\displaystyle:=$ $\displaystyle\int_{(z,r)\in
Z_{1}(\varepsilon)}\left(\frac{z^{2}}{[(z+1)^{2}-r^{2}]\sqrt{(z+1)^{2}-r^{2}}}\right)^{\rho}dr\leq
Cz^{2\rho}(z-1)(z+1)^{-3\rho}F\Big{(}\frac{1}{2},\frac{3\rho}{2};\frac{3}{2};\frac{(z-1)^{2}}{(z+1)^{2}}\Big{)}.$

Then, according to (15.3.6) of Ch.15[1] and [2],

$\displaystyle F\left(a,b;c;\zeta\right)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}F\left(a,b;a+b-c+1;1-\zeta\right)$
$\displaystyle+(1-z)^{c-a-b}\frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}F\left(c-a,c-b;c-a-b+1;1-\zeta\right)$

for all $\zeta\in{\mathbb{C}}$, $|\arg(1-\zeta)|<\pi$. We use (8) with

$\zeta=\frac{(z-1)^{2}}{(z+1)^{2}},\qquad 1-\zeta=\frac{4z}{(z+1)^{2}}$

to obtain for $\rho<2$ and large $z\geq N$ the following estimate for the
hypergeometric function,

$F\Big{(}\frac{1}{2},\frac{3\rho}{2};\frac{3}{2};\frac{(z-1)^{2}}{(z+1)^{2}}\Big{)}\leq
C(z+1)^{-1+\frac{3\rho}{2}}\,.$ (8.4)

Thus,

$\displaystyle A_{5}$ $\displaystyle\leq$ $\displaystyle
C(z-1)(z+1)^{-1+\frac{\rho}{2}}\,.$

For the next term we obtain a similar estimate,

$\displaystyle A_{6}:=\int_{(z,r)\in
Z_{1}(\varepsilon)}\left|\frac{z^{2}}{((z-1)^{2}-r^{2})\sqrt{(z+1)^{2}-r^{2}}}\left(\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\right)^{2}\right|^{\rho}dr$
$\displaystyle\leq$ $\displaystyle C(z-1)(z+1)^{-1+\frac{\rho}{2}}\,.$

Hence,

$\displaystyle\int_{(z,r)\in
Z_{1}(\varepsilon,z)}\left|K_{0}(r,t)\right|^{\rho}dr\leq
C(z-1)(z+1)^{-1+\frac{\rho}{2}}\,.$

In the second zone $Z_{2}(\varepsilon,z)$ for the argument of the
hypergeometric functions we have (7.10), (7.13), and (7.14). We have to
estimate the following integral

$\displaystyle A_{7}$ $\displaystyle:=$ $\displaystyle\int_{(z,r)\in
Z_{2}(\varepsilon,z)}\left|\frac{z^{2}\left(1+\ln
z\right)}{((z-1)^{2}-r^{2})\sqrt{(z+1)^{2}-r^{2}}}\right|^{\rho}dr\,.$

We apply (7.10) and (8.4) to obtain

$\displaystyle A_{7}$ $\displaystyle\leq$ $\displaystyle C\left(1+\ln
z\right)^{\rho}(z-1)(z+1)^{-1+\frac{\rho}{2}}\,.$

Hence,

$\displaystyle\int_{(z,r)\in
Z_{2}(\varepsilon,z)}\left|K_{0}(r,t)\right|^{\rho}dr\leq C\left(1+\ln
z\right)^{\rho}(z-1)(z+1)^{-1+\frac{\rho}{2}}\qquad\mbox{\rm for all}\quad
z\geq N\,.$

Proposition is proven. $\square$

##  9 $L^{p}-L^{q}$ Estimates for the Equation with Source, $n\geq 2$

For the wave equation the Duhamel’s principle allows to reduce the case of
source term to the case of the Cauchy problem without source term and
consequently to derive the $L^{p}-L^{q}$-decay estimates for the equation. For
(0.9) the Duhamel’s principle is not applicable straightforward and we have to
appeal to the representation formula of Theorem 0.5. In fact, one can regard
that formula as an expansion of the two-stage Duhamel’s principle. In this
section we consider the Cauchy problem (0.21) for the equation with the source
term with zero initial data.

###### Theorem 9.1

Let $u=u(x,t)$ be solution of the Cauchy problem (0.21). Then for $n>1$ one
has the following decay estimate

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}$
$\displaystyle\leq$ $\displaystyle
C\int_{0}^{t}db\,\|f(x,b)\|_{{L}^{p}({\mathbb{R}}^{n})}\int_{0}^{e^{-b}-e^{-t}}dr\,r^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{\sqrt{(e^{-t}+e^{-b})^{2}-r^{2}}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(e^{-b}-e^{-t})^{2}-r^{2}}{(e^{-b}+e^{-t})^{2}-r^{2}}\right)$

provided that $s\geq 0$, $1<p\leq 2$, $\frac{1}{p}+\frac{1}{q}=1$,
$\frac{1}{2}(n+1)\left(\frac{1}{p}-\frac{1}{q}\right)\leq 2s\leq
n\left(\frac{1}{p}-\frac{1}{q}\right)<2s+1$.

Proof. In both cases, of even and odd $n$, one can write the representation
(0.24). Due to the results of [5, 23] for the wave equation, we have

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}$
$\displaystyle\leq$ $\displaystyle
C\int_{0}^{t}db\int_{0}^{e^{-b}-e^{-t}}\|(-\bigtriangleup)^{-s}v(x,r;b)\|_{{L}^{q}({\mathbb{R}}^{n})}\frac{1}{\sqrt{(e^{-t}+e^{-b})^{2}-r^{2}}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(e^{-b}-e^{-t})^{2}-r^{2}}{(e^{-b}+e^{-t})^{2}-r^{2}}\right)dr$
$\displaystyle\leq$ $\displaystyle
C\int_{0}^{t}db\,\|f(x,b)\|_{{L}^{p}({\mathbb{R}}^{n})}\int_{0}^{e^{-b}-e^{-t}}r^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{\sqrt{(e^{-t}+e^{-b})^{2}-r^{2}}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(e^{-b}-e^{-t})^{2}-r^{2}}{(e^{-b}+e^{-t})^{2}-r^{2}}\right)dr\,.$

The theorem is proven. $\square$

We are going to transform the estimate of the last theorem to more cosy form.
To this aim we estimate for $n(\frac{1}{p}-\frac{1}{q})<2s+1$ the last
integral of the right hand side. If we replace $e^{-b}/e^{-t}>1$ with
$z:=e^{-b}/e^{-t}>1$, then the integral will be simplified.

$\displaystyle\int_{0}^{e^{-b}-e^{-t}}r^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{\sqrt{(e^{-t}+e^{-b})^{2}-r^{2}}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(e^{-b}-e^{-t})^{2}-r^{2}}{(e^{-b}+e^{-t})^{2}-r^{2}}\right)dr$
$\displaystyle=$ $\displaystyle
e^{-t[2s-n(\frac{1}{p}-\frac{1}{q})]}\int_{0}^{z-1}y^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{\sqrt{(z+1)^{2}-y^{2}}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\right)dy$

###### Lemma 9.2

[33, Lemma 9.2] Assume that $0\geq 2s-n(\frac{1}{p}-\frac{1}{q})>-1$. Then

$\int_{0}^{z-1}r^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{\sqrt{(z+1)^{2}-r^{2}}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\right)\,dr\leq
Cz^{-1}(z-1)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}(1+\ln z),$

for all $z>1$.

###### Corollary 9.3

Let $u=u(x,t)$ be solution of the Cauchy problem (0.21). Then for $n\geq 2$
one has the following decay estimate

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}$
$\displaystyle\leq$ $\displaystyle
C\int_{0}^{t}\|f(x,b)\|_{{L}^{p}({\mathbb{R}}^{n})}e^{-b}\left(e^{-b}-e^{-t}\right)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}\left(1+t-b\right)\,db$
(9.1)

provided that $s\geq 0$, $1<p\leq 2$, $\frac{1}{p}+\frac{1}{q}=1$,
$\frac{1}{2}(n+1)\left(\frac{1}{p}-\frac{1}{q}\right)\leq 2s\leq
n\left(\frac{1}{p}-\frac{1}{q}\right)<2s+1$.

Proof. Indeed, we apply Lemma 9.2 with $z=e^{t-b}$ to the right-hand side of
the estimate given by Theorem 9.1 :

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}$
$\displaystyle\leq$ $\displaystyle
C\int_{0}^{t}db\,\|f(x,b)\|_{{L}^{p}({\mathbb{R}}^{n})}e^{-t[2s-n(\frac{1}{p}-\frac{1}{q})]}z^{-1}(z-1)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}(1+\ln
z)$ $\displaystyle\leq$ $\displaystyle
C\int_{0}^{t}\|f(x,b)\|_{{L}^{p}({\mathbb{R}}^{n})}e^{-b}\left(e^{-b}-e^{-t}\right)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}\left(1+t-b\right)\,db\,.$

Corollary is proven. $\square$

##  10 $L^{p}-L^{q}$ Estimates for Equation without Source, $n\geq 2$

The $L^{p}-L^{q}$-decay estimates for the energy of the solution of the Cauchy
problem for the wave equation without source can be proved by the
representation formula, $L_{1}-L_{\infty}$ and $L_{2}-L_{2}$ estimates, and
interpolation argument. (See, e.g., [25, Theorem 2.1].) There is also a proof
of the $L^{p}-L^{q}$-decay estimates that is based on the microlocal
consideration and dyadic decomposition of the phase space. (See, e.g., [5,
23].) To avoid the derivative loss and obtain more sharp estimates we appeal
to the representation formula provided by Theorem 0.6.

###### Theorem 10.1

The solution $u=u(x,t)$ of the Cauchy problem (0.25) satisfies the following
$L^{p}-L^{q}$ estimate

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}$
$\displaystyle\leq$ $\displaystyle
C(1+t)(1-e^{-t})^{2s-n(\frac{1}{p}-\frac{1}{q})}\Big{\\{}e^{\frac{t}{2}}\|\varphi_{0}(x)\|_{{L}^{p}({\mathbb{R}}^{n})}+(1-e^{-t})\|\varphi_{1}\|_{{L}^{p}({\mathbb{R}}^{n})}\Big{\\}}$

for all $t\in(0,\infty)$, provided that $s\geq 0$, $1<p\leq 2$,
$\frac{1}{p}+\frac{1}{q}=1$,
$\frac{1}{2}(n+1)\left(\frac{1}{p}-\frac{1}{q}\right)\leq 2s\leq
n\left(\frac{1}{p}-\frac{1}{q}\right)<2s+1$.

Proof. We start with the case of $\varphi_{0}=0$. Due to Theorem 0.6 for the
solution $u=u(x,t)$ of the Cauchy problem (0.25) with $\varphi_{0}=0$ and to
the results of [5, 23] we have:

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}$
$\displaystyle\leq$ $\displaystyle
C\|\varphi_{1}\|_{{L}^{p}({\mathbb{R}}^{n})}\int_{0}^{1-e^{-t}}r^{2s-n(\frac{1}{p}-\frac{1}{q})}\left|K_{1}(r,t)\right|\,dr$
$\displaystyle\leq$ $\displaystyle
C\|\varphi_{1}\|_{{L}^{p}({\mathbb{R}}^{n})}e^{-t[2s-n(\frac{1}{p}-\frac{1}{q})]}\int_{0}^{e^{t}-1}y^{2s-n(\frac{1}{p}-\frac{1}{q})}\big{(}(e^{t}+1)^{2}-y^{2}\big{)}^{-\frac{1}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(e^{t}-1)^{2}-y^{2}}{(e^{t}+1)^{2}-y^{2}}\right)\,dy\,.$

To continue we apply Lemma 9.2 and obtain

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}$
$\displaystyle\leq$ $\displaystyle
C\|\varphi_{1}\|_{{L}^{p}({\mathbb{R}}^{n})}(1+t)(1-e^{-t})^{1+2s-n(\frac{1}{p}-\frac{1}{q})}\,.$

Thus, in the case of $\varphi_{0}=0$ the theorem is proven.

Next we turn to the case of $\varphi_{1}=0$. Due to Theorem 0.6 for the
solution $u=u(x,t)$ of the Cauchy problem (0.25) with $\varphi_{1}=0$ and to
the results of [5, 23] we have:

$\displaystyle\|(-\bigtriangleup)^{-s}u(x,t)\|_{{L}^{q}({\mathbb{R}}^{n})}$
$\displaystyle\leq$ $\displaystyle
C\Bigg{(}e^{\frac{t}{2}}(1-e^{-t})^{2s-n(\frac{1}{p}-\frac{1}{q})}+\int_{0}^{1-e^{-t}}r^{2s-n(\frac{1}{p}-\frac{1}{q})}|K_{0}(r,t)|\,dr\Bigg{)}\|\varphi_{0}(x)\|_{{L}^{p}({\mathbb{R}}^{n})}.$

One can estimate the last integral

$\displaystyle\int_{0}^{1-e^{-t}}r^{2s-n(\frac{1}{p}-\frac{1}{q})}|K_{0}(r,t)|\,dr$
$\displaystyle\leq$ $\displaystyle
e^{-t[2s-n(\frac{1}{p}-\frac{1}{q})]}\int_{0}^{e^{t}-1}y^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{[(e^{t}-1)^{2}-y^{2}]\sqrt{(e^{t}+1)^{2}-y^{2}}}$
$\displaystyle\times\Bigg{|}\big{(}e^{t}-e^{2t}-iM(1-e^{2t}-y^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{t}-1)^{2}-y^{2}}{(e^{t}+1)^{2}-y^{2}}\Big{)}$
$\displaystyle\hskip
28.45274pt+\big{(}e^{2t}-1+y^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(e^{t}-1)^{2}-y^{2}}{(e^{t}+1)^{2}-y^{2}}\Big{)}\Bigg{|}\,dy\,.$

The following proposition gives the remaining estimate for that integral and
completes the proof of the theorem.

###### Proposition 10.2

If $2s-n(\frac{1}{p}-\frac{1}{q})>-1$, then

$\displaystyle\int_{0}^{z-1}y^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{[(z-1)^{2}-y^{2}]\sqrt{(z+1)^{2}-y^{2}}}$
$\displaystyle\times\Bigg{|}\big{(}z-z^{2}-iM(1-z^{2}-y^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}$
$\displaystyle\hskip
28.45274pt+\big{(}z^{2}-1+y^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}\Bigg{|}\,dy$
$\displaystyle\leq$ $\displaystyle
Cz^{-\frac{1}{2}}(z-1)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}\left(1+\ln
z\right)\quad\mbox{ for all}\quad z>1.$

Proof. We follow the arguments have been used in the proof of Proposition 8.3.
If $1\leq z\leq N$ with some constant $N$, then the argument of the
hypergeometric functions is bounded (8.2), and the integral can be estimated
by:

$\displaystyle\int_{0}^{z-1}y^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{[(z-1)^{2}-y^{2}]\sqrt{(z+1)^{2}-y^{2}}}$
$\displaystyle\times\Bigg{|}\big{(}z-z^{2}-iM(1-z^{2}-y^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}$
$\displaystyle\hskip
28.45274pt+\big{(}z^{2}-1+y^{2}\big{)}\Big{(}\frac{1}{2}-iM\Big{)}F\Big{(}-\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}\Bigg{|}\,dy$
$\displaystyle\leq$ $\displaystyle
C_{M}\int_{0}^{z-1}y^{2s-n(\frac{1}{p}-\frac{1}{q})}\Bigg{[}\frac{1}{\sqrt{(z+1)^{2}-y^{2}}}\Bigg{\\{}1+z^{2}\frac{1}{(z+1)^{2}-y^{2}}\Bigg{\\}}\Bigg{]}\,dy$
$\displaystyle\leq$ $\displaystyle
C_{M}z^{-1}(z-1)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}\quad\mbox{\rm for all}\quad
z\in[1,N]\,.$

Thus, we can restrict ourselves to the case of large $z\geq M$ in both zones
$Z_{1}(\varepsilon,z)$ and $Z_{2}(\varepsilon,z)$, defined in (7.5) and (7.6),
respectively. In the first zone we have (7.9). Consider therefore the
following inequalities,

$\displaystyle A_{8}$ $\displaystyle:=$ $\displaystyle\int_{(z,r)\in
Z_{1}(\varepsilon,z)}r^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{\sqrt{(z+1)^{2}-r^{2}}}\,dr$
$\displaystyle\leq$ $\displaystyle
Cz^{-1}(z-1)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}\quad\mbox{\rm for all}\quad
z\in[N,\infty)\,.$

For $0\geq a>-1$ and $z\geq N$ the following integral can be easily estimated:

$\displaystyle\int_{0}^{z-1}r^{a}\frac{1}{((z+1)^{2}-r^{2})^{3/2}}dr$
$\displaystyle=$
$\displaystyle\int_{0}^{z/2}r^{a}\frac{1}{((z+1)^{2}-r^{2})^{3/2}}dr+\int_{z/2}^{z-1}r^{a}\frac{1}{((z+1)^{2}-r^{2})^{3/2}}dr$
$\displaystyle\leq$
$\displaystyle\frac{16}{9}z^{-3}\int_{0}^{z/2}r^{a}dr+\frac{z^{a}}{4^{a}}\int_{z/2}^{z-1}\frac{1}{((z+1)^{2}-r^{2})^{3/2}}dr$
$\displaystyle\leq$ $\displaystyle Cz^{a-3/2}\quad\mbox{\rm for all}\quad
z\in[N,\infty)\,.$

Hence,

$\displaystyle A_{9}$ $\displaystyle:=$ $\displaystyle z^{2}\int_{(z,r)\in
Z_{1}(\varepsilon,z)}r^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{\sqrt{(z+1)^{2}-r^{2}}}\frac{1}{(z+1)^{2}-r^{2}}dr$
$\displaystyle\leq$ $\displaystyle
z^{2}\int_{0}^{z-1}r^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{\sqrt{(z+1)^{2}-r^{2}}}\frac{1}{(z+1)^{2}-r^{2}}dr$
$\displaystyle\leq$ $\displaystyle
Cz^{-\frac{1}{2}}(z-1)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}\quad\mbox{\rm for
all}\quad z\in[N,\infty)\,,$

and

$\displaystyle A_{10}$ $\displaystyle:=$ $\displaystyle z^{2}\int_{(z,r)\in
Z_{1}(\varepsilon,z)}r^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{((z-1)^{2}-r^{2})\sqrt{(z+1)^{2}-r^{2}}}\left(\frac{(z-1)^{2}-r^{2}}{(z+1)^{2}-r^{2}}\right)^{2}dr$
$\displaystyle\leq$ $\displaystyle z^{2}\int_{(z,r)\in
Z_{1}(\varepsilon,z)}r^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{\sqrt{(z+1)^{2}-r^{2}}}\frac{1}{(z+1)^{2}-r^{2}}dr$
$\displaystyle\leq$ $\displaystyle
Cz^{-\frac{1}{2}}(z-1)^{1+2s-n(\frac{1}{p}-\frac{1}{q})}\quad\mbox{\rm for
all}\quad z\in[N,\infty)\,.$

Finally,

$\displaystyle\int_{(z,y)\in
Z_{1}(\varepsilon,z)}\,y^{2s-n(\frac{1}{p}-\frac{1}{q})}\frac{1}{[(z-1)^{2}-y^{2}]\sqrt{(z+1)^{2}-y^{2}}}$
$\displaystyle\times\Bigg{|}\big{(}z-z^{2}-iM(1-z^{2}-y^{2})\big{)}F\Big{(}\frac{1}{2}+iM,\frac{1}{2}+iM;1;\frac{(z-1)^{2}-y^{2}}{(z+1)^{2}-y^{2}}\Big{)}$