\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{amssymb}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 165, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/165\hfil Singularities of solutions]
{Singularities of solutions to partial differential
equations in a complex domain}

\author[N. Bentiba \hfil EJDE-2016/165\hfilneg]
{Naouel Bentiba}

\address{Naouel Bentiba \newline
Department of Mathematics,
Faculty of Sciences,
University Badji Mokhtar Annaba,
 23000 Annaba, Algeria}
\email{nabn\_math@hotmail.com}

\thanks{Submitted May 4, 2016. Published June 28, 2016.}
\subjclass[2010]{35A20, 35C10, 33C05, 33B15}
\keywords{Cauchy problem; singular solutions; hypergeometric equation}

\begin{abstract}
 We give an explicit representation of the solution of the following singular
 Cauchy problem with analytic data,
 \[
 u_{tt}-xu_{xx}+Cu_{x}-B(t^2-4x)^{-1}u=0.
 \]
 We also study the singularities of this solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks



\section{Introduction}

In this article, we study the singularities of the solution to the
following Cauchy problem with analytic data in the complex domain
\begin{equation} \label{e1.1}
\begin{gathered}
LU=\partial_t^2U-x\partial_{x}^2U+C\partial_{x}U-B(t^2
-4x)^{-1}U=0, \\
U(0,x)=U_0 x)  ,\\
 U_t(0,x) =U_1(x)  ,
\end{gathered}
\end{equation}
where $C$ and $B$ are real numbers. The variables $t,x$ and the unknown
$U$ are complex numbers.
Our aim is to give an explicit representation of the solution in terms of
Gauss hypergeometric functions and study its singularities.

Equation \eqref{e1.1} arises naturally by linearizing the
nonlinear partial differential equation with double characteristics in
involution
\[
\partial_t^2U-x\partial_{x}^2U+C\partial_{x}U=aU^{p},
\]
around its self-similar solution $U=(t^2-4x)^{\frac{-1}{p-1}}$, where $a$
is real number, $p>1$ and $B=ap$. 

In the complex domain, the study of the singularities of the solutions of the
nonlinear Cauchy problem is very complicated. Especially if the roots of the
characteristic polynomial are double and not holomorphic at the origin (as is
in our case). Indeed the technical difficulties are such that even in the
linear case there are no methods for obtaining general theorems.

However in the case of second order equations, many authors showed that the
solutions of certain evolution or degenerate linear equations can be expressed
in terms of hypergeometric functions. Since these hypergeometric functions
have intrinsic singularities, they permit the analysis of the structure of the
solutions and therefore to describe their singularities; see 
\cite{a2,b1,b2,h1,r1}.

We will show that the solution, depending on various parameters, might have
singularities on the characteristic surfaces:
\begin{equation} \label{e1.2}
K_1:x=0,\quad K_2:4x-t^2=0.
\end{equation}
Before developing the theory and to get some insight to the problem,
we present four concrete examples.

\begin{example} \label{examp1} \rm
In $\mathbb{C}^2$ consider the Cauchy problem
\begin{gather*}
\partial_t^2U-x\partial_{x}^2U+2\partial_{x}U-28(t^2-4x) ^{-1}U=0,\\
U(0,x)  =x^{4},\\
U_t(0,x)  =0.
\end{gather*}
The solution is
\[
U(t,x)=\frac{1}{4^2}(4x-t^2)^2(x^2-t^2
x+\frac{t^{4}}{6})  .
\]
We observe that $U$ is polynomial.
\end{example}

\begin{example} \label{examp2} \rm
The Cauchy problem
\begin{gather*}
\partial_t^2U-x\partial_{x}^2U-2\partial_{x}U-12(t^2-4x)^{-1}U=0,\\
U(0,x)  =x,\\
U_t(0,x)  =0,
\end{gather*}
has solution
\[
U(t,x)=\frac{1}{16}x^{-1}(4x-t^2)^2.
\]
Thus it is singular only on $K_1:x=0$.
\end{example}

\begin{example} \label{examp3} \rm
In $\mathbb{C}^2$ consider the Cauchy problem
\begin{gather*}
\partial_t^2U-x\partial_{x}^2U+\partial_{x}U+6(t^2-4x)^{-1}U=0,\\
U(0,x)  =x,\\
U_t(0,x)  =0.
\end{gather*}
Its solution is
\[
U(t,x)=4x^2(4x-t^2)^{-1}.
\]
Thus, $U(t,x)  $ is singular only on $K_2:4x-t^2=0$.
\end{example}

\begin{example} \label{examp4} \rm
Consider the  Cauchy problem
\begin{gather*}
\partial_t^2U-x\partial_{x}^2U-\partial_{x}U-4(t^2-4x)^{-1}U=0,\\
 U(0,x)  =x,\\
 U_t(0,x)  =0.
\end{gather*}
Its solution is 
\[
U(t,x)=\frac{1}{4}\big[  t\sqrt{(4x-t^2)  }\arcsin(
\frac{t}{2\sqrt{x}})  +(4x-t^2)  \big]  .
\]
Thus, $U(t,x)  $ is singular both on $K_1:x=0$ and on
$K_2:4x-t^2=0$.
\end{example}

\section{Statement of the problem}

In a neighborhood $\Omega$ of the origin of $\mathbb{C}^2$,
we consider the Cauchy problem with analytic data:
\begin{equation} \label{e2.1}
\begin{gathered}
LU=\partial_t^2U-x\partial_{x}^2U+C\partial_{x}U-B(t^2 -4x)^{-1}U=0,\\
U(0,x)  =u_0(x)  ,\\
U_t(0,x)  =u_1(x)  ,
\end{gathered}
\end{equation}
where 
\[
u_0(x) =\sum_{l=0}^\infty a_{l}x^{l}, \quad
u_1(x) =\sum_{l=0}^\infty b_{l}x^{l}
\]
 are analytic and $C$, $B$ are real numbers.

Our purpose is to construct an exact solution in terms of hypergeometric
functions and to show that the solution might have singularities on the
characteristic surfaces \eqref{e1.2}. We denote 
\begin{gather*}
\Omega_{R} =\{  (t,x)  \in\mathbb{C}^2:| t| <R,\; | x| <R\}  ,\\
K_1  = \{  (t,x)  \in\Omega_{R}: x=0\}  ,\\
K_2  = \{  (t,x)  \in\Omega_{R}:4x-t^2=0\}  .
\end{gather*}
Using these notation, we have the following theorems.

\begin{theorem} \label{thm1}
The Cauchy problem \eqref{e2.1} has a unique solution of the form :
\[
U(t,x)  =\sum_{l\geq0}  (a_{l}V_{l,\beta}+b_{l}W_{l,\beta'})  ,
\]
with
\begin{equation} \label{e2.2}
\begin{aligned}
V(t,x)
& =V_{l,\beta}(t,x) \\
& =(4)^{-\beta}(x)^{l-\beta}(4x-t^2)^{\beta}
F\big(\beta-l,C+\beta-l+1,\frac{1}{2};\frac
{t^2}{4x}\big)  ,
\end{aligned}
\end{equation}
where $\beta$ satisfies
\[
\beta(2l-\beta-C-\frac{1}{2})  =\frac{1}{4}B,
\]
\begin{equation} \label{e2.3}
\begin{aligned}
W(t,x)  & =W_{l,\beta'}(t,x)=\\
  & =(4)^{-\beta'}(x)^{l-\beta'}(4x-t^2)^{\beta'}t
  F\big(\beta'-l,C+\beta'-l+1,\frac{3}{2};\frac{t^2}{4x}\big)  ,
\end{aligned}
\end{equation}
where
\[
\beta'(2l-\beta'-C+\frac{1}{2})  =\frac{1}{4}B,
\]
and $F$ denotes the Gauss hypergeometric function.
 This solution might have singularities on $K=$ $K_1\cup K_2$.
\end{theorem}

\subsection{Construction of the solutions}

According to the principle of superposition, it is sufficient to study the
following two Cauchy problems:
\begin{equation}  \label{e2.4}
\begin{gathered}
LV=0,\\
V(0,x)  =x^{l},\\
V_t(0,x)  =0,
\end{gathered}
\end{equation}
and
\begin{equation} \label{e2.5}
\begin{gathered}
LW=0,\\
W(0,x)  =0,\\
W_t(0,x)  =x^{l}.
\end{gathered}
\end{equation}

First we solve \eqref{e2.4}.
Setting the Characteristics equations $x=\xi_1$ and $4x-t^2=\xi_2$.
Let $V(t,x)  =V_{l}(t,x)  =\xi_1^{l}v(z)  $ with 
$z=1-\frac{\xi_2}{4\xi_1}$. Substituting 
$\xi_1^{l}v(z)  $ for $V$ , $LV=0$  becomes
\begin{align*}
LV&=\frac{1}{2}\xi_1^{l-1}V'+z\xi_1^{l-1}V''-z^2
\xi_1^{l-1}V''-2z\xi_1^{l-1}V'+lz\xi_1^{l-1}V'\\
&\quad +zl\xi_1^{l-1}V'-l(l-1)  \xi_1^{l-1}V
+C(l\xi_1^{l-1}V-z\xi_1^{l-1}V')  +\frac{B}{\xi_2
}\xi_1^{l}V=0.
\end{align*}
Simplifying this equation and replacing  
$\frac{\xi_1}{\xi_2}$ by
$\frac{1}{4(1-z) }$, we have
\begin{equation} \label{e2.6}
\begin{aligned}
&z(1-z)  v''+\big(\frac{1}{2}-(C+2(1-l)  )  z\big)  v'\\
&+(l(C-l+1)  -\frac{1}{4}B(\frac{1}{z-1}))  v=0.
\end{aligned}
\end{equation}
The substitution $v=(1-z)^{\beta}y$ leads to
\begin{equation} \label{e2.7}
\begin{aligned}
&z(1-z)  y''+\big(\frac{1}{2}-(C+2(
1-l+\beta)  )  z\big)  y' \\
&+\big(l(C-l+1)  +\beta(\beta+C+1-2l)  \big)y=0.
\end{aligned}
\end{equation}
Therefore \eqref{e2.7} 
 is equivalent to a Gauss differential equation with parameters
\[
\big(\beta-l,C+\beta-l+1,\frac{1}{2}\big)  ,
\]
if and only if
\begin{equation} \label{e2.8}
\beta\big(2l-\beta-C-\frac{1}{2}\big)  =\frac{1}{4}B.
\end{equation}

According to hypergeometric equation theory, we have:
A first solution of Gauss equation for $| z| <1$ is 
\[
y_1(z)  =F(a,b;c;z)  
=F\big(\beta
-l,C+\beta-l+1;\frac{1}{2};z\big)  .
\]
A second solution is
\begin{align*}
y_2(z)  
&=z^{1/2}F\big(a-c+1,b-c+1;2-c;z\big) \\
&=z^{1/2}F\big( \beta-l+\frac{1}{2},C+\beta-l+\frac{3}{2},\frac{3}{2};z\big)  .
\end{align*}
A complete solution of the Gauss equation is
\begin{equation} \label{e2.9}
\begin{aligned}
y  & =DF\big( \beta-l,C+\beta-l+1,\frac{1}{2};z\big)  \\
&\quad+ Ez^{1/2}F\big(\beta-l+\frac{1}{2},C+\beta-l+\frac{3}{2},\frac
{3}{2};z\big)  ,
\end{aligned}
\end{equation}
with $z=\frac{t^2}{4x}$, for $| z| <1$, where $D$ and
$E$ are constants.

It follows that $V=x^{l}(1-z)^{\beta}y$ is a solution of
$LV=0$. Taking into account the Cauchy data
\[
V(0,x)  =x^{l},\quad V_t(0,x)  =0.
\]
we have to choose $D=1$ and $E=0$. Hence $V(t,x)$ reduces to
\begin{equation} \label{e2.10}
V(t,x)  =V_{l\beta}(t,x)  
=x^{l}\big(1-\frac{t^2}{4x}\big)^{\beta}
F\big(\beta-l,C+\beta-l+1,\frac{1}{2};\frac{t^2}{4x}\big)  .
\end{equation}
In a similar way, we solve the second Cauchy problem \eqref{e2.4}
 by setting
\[
W(t,x)  =tx^{l}\big( 1-\frac{t^2}{4x}\big)^{\beta'}y;
\]
we obtain
\begin{equation} \label{e2.11}
\begin{aligned}
W(t,x)
& =W_{l,\beta'}(t,x)  \\
& =t(4)^{-\beta'}x^{l-\beta'}(
4x-t^2)^{\beta'}F\big(\beta'-l,C+\beta^{\prime
}-l+1,\frac{3}{2};\frac{t^2}{4x}\big)  ,
\end{aligned}
\end{equation}
where $\beta'$ and $l$ satisfy
\[
\beta'(\frac{1}{2}+2l-C-\beta')  =\frac{B}{4}.
\]

\begin{remark} \label{rmk1} \rm
When $\beta-l=-n$ or $ C+\beta-l+1=-n,\ n\in \mathbb{N}$,
the solution $V$ of \eqref{e2.4} is 
\begin{align*}
V(t,x)  
& =(4)^{l-n}(x)^{n}(4x-t^2)^{l-n}
\Big(\sum_{i=0}^n \alpha _{i}(\frac{t^2}{4x})^{i}\Big)  ,\\
& =( 4)^{l-n}(4x-t^2)^{l-n}
\Big( \sum_{i=0}^n \alpha_{i}x^{n-i}(\frac{t^2}{4})^{i}\Big);
\end{align*}
its last term is
\[
c_{n}t^{2n}(4x-t^2)^{l-n},
\]
where $c_{n}$ depends on parameters $C,l$ and $\beta$.
\end{remark}

 Therefore, we have some results for $V$.
\begin{itemize}
\item[(1)] When $l-n\geq0$, the solution $V(t,x) $ is a polynomial.

\item[(2)] When $l-n<0$, the solution is singular on the surface $K_2:4x-t^2=0$.

\item[(3)] When $C+\beta-l+1=-n$, we have the following results:
 \begin{itemize}
\item[(i)] For $C<-n-1$, the solution is singular on the surface $K_1$.

\item[(ii)] For $C>-n-1$ \ and $l-(C+n+1)  <0$, the solution
is singular on the surface $K_2$.

\item[(iii)] For $C>-n-1$ and $l-(C+n+1)  >0$, the solution
is polynomial.

\end{itemize}
\end{itemize}

\subsection{Singularities}
In this section, we study the singularities of the solution. The mapping
\[
z=\frac{t^2}{4x},
\]
transforms
\begin{gather*}
\text{ }t   =0,\text{ into }z=0,\\
K_2 :t^2-4x=0,\text{ into }z=1,\\
K_1  :x=0,\text{ into }z=\infty.
\end{gather*}


By construction, the solution $U$ is composed of a hypergeometric function
which is holomorphic on $D-(0,1,\infty)  $, where $D$ is the
Riemann sphere. So, the study of the singularities of the solution is reduced
to those corresponding well known properties of Gauss functions. It follows
that $U$ is ramified around $K_1\cup K_2$.

\subsection{Convergence of the solution}
The study of the convergence of this solution is reduced to estimate 
the Gauss functions.
For this, we apply the following result.

\begin{lemma} \label{lem1}
If $a\geq b>c>0$ and $d=a+b-c$, then for $| z| <1$,
\begin{equation} \label{e2.12}
F(a,b;c;z)  \ll(1-z)^{-d}\frac{\Gamma(
c)  \Gamma(d)  }{\Gamma(a)  \Gamma(b)  }.
\end{equation}
\end{lemma}

For the proof of this lemma, see \cite{p1}.

\begin{theorem} \label{thm2}
The series $\sum_{l\geq0} a_{l}V_{l}$ and 
$\sum_{l\geq0} \sum b_{l}W_{l}$ converge for $| x| <\frac{R}{4}$,
where $R$ is the radius of convergence of $u_0$ and $u_1$.
\end{theorem}

\begin{proof}
We have
\[
V(t,x)=V_{l}(t,x)=(x)^{l}\big( 1-\frac{t^2}{4x}\big)
^{\beta}F\big( \beta-l,C+\beta-l+1;\frac{1}{2};\frac{t^2}{4x}\big)  .
\]
Let 
\[
\beta=\beta_1=\frac{2l-C-\frac{1}{2}-\sqrt{\Delta}}{2}
\]
be one of the roots of the equation \eqref{e2.8}, where
\begin{equation} \label{e2.13}
\Delta=\big( 2l-C-\frac{1}{2}\big)^2-B
\end{equation}
Putting : $a=\beta-l$, $b=C+\beta-l+1$, and $c=\frac{1}{2}$, we have
\[
d=a+b-c=2\beta-2l+C+\frac{1}{2}=-\sqrt{\Delta}<0.
\]
In this case, we recall the Euler transformation
\[
F(\alpha,\delta;\gamma;z)  
=(1-z)^{\gamma -\alpha-\delta}F(\gamma-\alpha,\gamma-\delta;\gamma;z)  .
\]
Applying this transformation to $F(\beta-l,C+\beta-l+1;\frac{1}
{2};\frac{t^2}{4x})$, we obtain
\begin{align*}
&F\big(\beta-l,C+\beta-l+1;\frac{1}{2};\frac{t^2}{4x}\big)  \\
&=\big(
1-\frac{t^2}{4x}\big)^{\sqrt{\Delta}}
F\big(l-\beta+\frac{1}
{2},l-C-\beta-\frac{1}{2};\frac{1}{2};\frac{t^2}{4x}\big)  .
\end{align*}
For $l$ large, $l-\beta=l+o(1)$, and so by applying Lemma \ref{lem1}
to $F\big( l-\beta+\frac{1}{2},l-C-\beta-\frac{1}{2};\frac{1}
{2};\frac{t^2}{4x}\big)  $ we have 
\[
F\big( l-\beta+\frac{1}{2},l-C-\beta-\frac{1}{2};\frac{1}{2};\frac{t^2
}{4x}\big)  \ll\big( 1-\frac{t^2}{4x}\big)^{-\sqrt{\Delta}}M,
\]
where $y=\frac{\sqrt{\Delta}}{2}$ and 
\[
M=\frac{\Gamma(\frac{1}{2})  \Gamma(2y)  }{\Gamma(y-\frac
{2C+1}{4})  \Gamma(y+\frac{2C+3}{4})  }.
\]
Stirling's formula gives
\[
M\sim 2^{2y-1}\frac{\sqrt{2\pi}(y)^{y-\frac{1}{2}}e^{-y}
\sqrt{2\pi}(y)^{y}e^{-y}}
{\Big(\sqrt{2\pi}(y)
^{y-(\frac{2C+3}{4})  }e^{-y}\Big)  \Big(\sqrt{2\pi}(
y)^{y+(\frac{2C+1}{4})  }e^{-y}\Big)  };
\]
then
$M\sim 2^{2y-1}$.
Therefore,
\begin{align*}
| V(t,x)  |  
& \leqslant 2^{2y-1}| x| ^{l}| 1-\frac{t^2}{x}|
^{\beta_1}| 1-\frac{t^2}{x}| ^{\sqrt{\Delta}}|
1-\frac{t^2}{x}| ^{-\sqrt{\Delta}}\\
& \leqslant2^{2y-1}| x| ^{l}| 1-\frac{t^2}
{x}| ^{\beta_1}.\text{ }
\end{align*}
As $\beta_1\to 0$ for $l$ large, we deduce that
\[
\underset{l\to \infty}{\lim\sup}| V_{l}| ^{1/l}\leq4| x| .
\]
It follows that $\sum_{l\geq0} a_{l}V_{l}$ converges for
\[
| x| <\frac{R}{4}.
\]
 In the similar way, we show that $\sum b_{l}W_{l}$  converges for
\[
| x| <\frac{1}{4}R.
\]
This completes the proof.
\end{proof}

\begin{thebibliography}{00}

\bibitem{a1} M. Abramowitz, I. A. Stegun, (Eds.);
 \emph{Handbook of Mathematical
Function}, National Bureau of Standards, Washington, Tenth printing,
1972.

\bibitem{a2} Akhmadjon K. Urinov, Akhrorjon I. Ismoilov, Azizbek O. Mamanazarov;
\emph{A Cauchy-Goursat problem for the generalized Euler-Poisson-Darboux
equation}; Contemp. Anal. Appl. Math, Vol. 4, No.1, 1-22, 2016.

\bibitem{b1} A. Bentrad; 
\emph{Exact solutions for a different version of the
nonhomogeneous E-P-D equation.} Complex Var. Elliptic Equ., vol. 51. No.3
March 2006, pp. 243-253.

\bibitem{b2} D. W. Bresters; 
\emph{On the equation of Euler-Poisson-Darboux, }Siam
J.Math.Anal.1973 no.1 pp. 31-41.

\bibitem{h1} A. Hasanov;
 \emph{The solution of the Cauchy problem for generalized
Euler-Poisson-Darboux equation.}
Int. J. Appl. Math. Stat., vol. 8, M07, 2007, pp. 30-44.

\bibitem{p1} S. Ponnusamy, M.\ Vuorinen; 
\emph{Asymptotic expansions and inequalities for hypergeometric functions}; 
Mathematika, 44 (1997), pp. 278-301.

\bibitem{r1} Rakhila B. Seilkhanova, Anvar H. Hasanov;
\emph{Particular Solutions of Generalized Euler-Poisson-Darboux equation}; 
Electron. J. Differential Equations, Vol. 2015 (2015), No. 09, pp. 1-10.

\end{thebibliography}

\end{document}