\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{graphicx,epic}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 243, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/243\hfil Semi-hyperbolic patches for UTSD equations]
{Properties of solutions in semi-hyperbolic patches for
unsteady transonic small disturbance equations}

\author[I. Jegdi\'c, K. Jegdi\'c \hfil EJDE-2015/243\hfilneg]
{Ilija Jegdi\'c, Katarina Jegdi\'c}

\address{Ilija Jegdi\'c \newline
Department of Mathematics and Physics,
Houston Baptist University,
7502 Fondren Rd, Houston, TX 77074, USA}
\email{i\_jegdic@yahoo.com}

\address{Katarina Jegdi\'c \newline
Department of Mathematics and Statistics,
University of Houston - Downtown,
1 Main St, Houston, TX 77002, USA}
\email{jegdick@uhd.edu}

\thanks{Submitted April 27, 2015. Published September 22, 2015.}
\subjclass[2010]{35L65}
\keywords{Unsteady transonic small disturbance equation;
\hfill\break\indent mixed type system; semi-hyperbolic patch; Goursat-type problem}

\begin{abstract}
 We consider a two-dimensional Riemann problem for the unsteady transonic small
 disturbance equation resulting in diverging rarefaction waves.
 We write the problem in self-similar coordinates and we obtain a mixed
 type (hyperbolic-elliptic) system.
 Resolving the one-dimensional discontinuities in the far field, where
 the system is hyperbolic, and using characteristics, we formulate the
 problem in a semi-hyperbolic patch that is between the hyperbolic and
 the elliptic regions. A semi-hyperbolic patch is known as a region where
 one family out of two nonlinear families of characteristics starts on a
 sonic curve and ends on a transonic shock. We obtain existence of a smooth
 local solution in this semi-hyperbolic patch and we prove various properties
 of global smooth solutions based on a characteristic decomposition using
 directional derivatives.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

\subsection*{Motivation}
Two-dimensional Riemann problems for systems of
hyperbolic conservation laws have been studied extensively during
the last several decades. Various systems have been considered:
steady/unsteady transonic small disturbance equation, nonlinear
wave system, pressure gradient system, potential flow equations and Euler
gas dynamics equations (see \cite{CKK0} by \v{C}ani\'c, Keyfitz and Kim,
and \cite{K} by Keyfitz).

Riemann problems for the unsteady transonic small disturbance (UTSD) equation
resulting in shock reflection were studied by Brio and Hunter in \cite{BH}
(Mach reflection), by \v{C}ani\'c, Keyfitz and Kim in \cite{CKK1}, and
by Jegdi\'c, Keyfitz and \v{C}ani\'c in \cite{J2} (transonic or strong
regular reflection) and by \v{C}ani\'c, Keyfitz and Kim in \cite{CKK2}
(supersonic or weak regular reflection). Solution to a Riemann problem
for the steady case with shocks was obtained by \v{C}ani\'c, Keyfitz and
 Lieberman in \cite{CKL}.

In this article we revisit the Riemann problem for the UTSD equation
studied by Chen, Christoforou and Jegdi\'c in \cite{CCJ}.
The initial data is chosen to be constant in three sectors, symmetric
about $x$-axis, resulting in diverging rarefaction waves. As in the study
of general two-dimensional Riemann problems for systems of hyperbolic
conservation laws, we rewrite the problem in self-similar coordinates
and we obtain a mixed type system.
It is well-known that when this system is linearized about a constant state,
it changes type across a sonic parabola and it is hyperbolic
(elliptic) outside (inside).
Since the problem is symmetric with respect to the horizontal axis,
we focus our attention on the upper half plane.
The one dimensional discontinuity in the hyperbolic region becomes a
rarefaction wave that stretches down towards the origin. It interacts
with the sonic parabola for one of the constant
states and gives rise to a very complicated free boundary problem.
Our preliminary local analysis in \cite{CCJ} suggested that there are a
degenerate hyperbolic region and a degenerate elliptic region coupled
via a sonic curve near the point of intersection of the
left-border of the rarefaction wave with the above mentioned sonic
parabola. The same Riemann problem for the UTSD equation was studied by
Tesdall and Keyfitz \cite{TK}.
They linearized the problem and solved it exactly, and they presented numerical
results for the nonlinear problem indicating that the rarefaction wave reflects
off a sonic line and forms a transonic shock. Such
regions where one family out of two nonlinear families of characteristics
starts at on a sonic curve and ends on a transonic shock are called
semi-hyperbolic patches. Existence of solutions
in semi-hyperbolic patches was obtained for the pressure gradient
system by Song and Zheng in \cite{SZ}, for the Euler equations
by Li and Zheng in \cite{LZ} and by Zheng in \cite{Z1},
and for the nonlinear wave system (NLWS)
by Hu and Wang in \cite{HW}. They derive characteristics decomposition
for the problem and use it to
\begin{enumerate}
\item prove that a hyperbolic state adjacent to a constant state is a simple wave,
\item formulate a Goursat-type boundary value problem,
\item prove existence of a local solution to the Goursat problem,
\item derive various properties of global smooth solutions, and
\item use those properties and
results by Dai and Zhang in \cite{Dai} for gas dynamics equations,
to obtain a global solution to the degenerate Goursat problem.
\end{enumerate}
We follow ideas in \cite{SZ, HW}
to derive characteristic decomposition of the
problem using directional derivatives along characteristics.
\v{C}ani\'c and Keyfitz proved in \cite{CK} that for $2 \times 2$
quasi-one-dimensional systems, a
hyperbolic state adjacent to a hyperbolic constant state is a simple wave,
completing the above mentioned step (1).
In this article, we complete steps (2)-(4) and we believe
that techniques from \cite{Dai}
could be extended to the UTSD equation to prove existence of a global
solution to the Goursat problem. We leave that study for the future work.


Numerical studies of the full gas dynamics equations involving interaction
of rarefaction waves also indicate existence of semi-hyperbolic regions
(see \cite{Glimm} by Glimm et al. and
\cite{Sheng} by Sheng, Wang and Zhang).


\subsection*{Summary of results}
This article is organized as follows.
In \S 2 we formulate the Riemann problem of interest
for the UTSD equation and we rewrite it using self-similar coordinates.
 We recall the structure of the solution in the hyperbolic region
(see \cite{CCJ, TK}) and, using analysis via characteristics,
we state the problem in the semi-hyperbolic patch.
To analyze the problem, we define directional derivatives along characteristics.

In \S 3 we rewrite the problem, again, using parabolic coordinates and we
formulate the Goursat-type boundary value problem
with two boundary curves given by two characteristics.

Section \S 4 deals with existence
of a local solution to this Goursat problem (Theorem \ref{thm4.1})
and various properties of smooth solutions in the entire
semi-hyperbolic patch.
We establish the signs of directional derivatives in the domain and
their monotonicity along the characteristics curves.
We prove that positive/negative characteristics are convex/concave.
We also establish that the global minimum of the directional derivatives
is achieved along the boundary which consists of the characteristic curves
and we obtain existence of the sonic boundary along which the two directional
derivatives coincide. Finally, we derive an estimate of the $C^1$
norm of the solution away from the sonic curve.

In \S 5 we remark that the solution in the semi-hyperbolic patch,
in the region outside of the Goursat problem, is a simple wave using the
result by \v{C}ani\'c and Keyfitz in \cite{CK}.
We conclude by showing that one family of characteristics forms an envelope
before they become sonic, indicating existence of a transonic shock.


\section{Formulation of the problem}

In this section, we pose our Riemann problem and rewrite it using self-similar
coordinates. We formulate the problem in the semi-hyperbolic patch.


\subsection*{Formulation of the initial value problem}

We consider the unsteady transonic small disturbance equation
\begin{equation} \label{utsd}
\begin{gathered}
u_t + uu_x + v_y = 0, \\
-v_x+u_y = 0,
\end{gathered}
\end{equation}
where $(x,y) \in \mathbb{R}^2$, $t \in [0,\infty)$, and
 $u,v: [0, \infty) \times \mathbb{R}^2 	o \mathbb{R}$ denote
velocities in $x$- and $y$-directions, respectively. Given parameter $a>0$, we
consider Riemann initial data posed in three
sectors (see Figure \ref{fig1})
\begin{equation}\label{initial}
(u,v)(0,x,y)=\begin{cases}
(-1,a), & x/a<y<0, \\
(-1,-a), & 0<y<-x/a, \\
(0,0), & \text{otherwise}.
\end{cases}
\end{equation}

\begin{figure}[htb]
\begin{center} % Riemann.eps
\setlength{\unitlength}{1mm}
\begin{picture}(60,40)(0,0)
\put(0,20){\vector(1,0){60}}
\put(30,0){\vector(0,1){40}}
\put(10,0){\line(1,1){20}}
\put(10,40){\line(1,-1){20}}
\put(26,38){$y$}
\put(56,16){$x$}
\put(0,34){$x=-ay$}
\put(0,24){$(-1,-a)$}
\put(0,14){$(-1,-a)$}
\put(0,4){$x=ay$}
\end{picture}
\end{center}
\caption{Riemann initial data} \label{fig1}
\end{figure}

\subsection*{Formulation of the problem in self-similar coordinates}

We formulate the problem in self-similar coordinates $\xi=x/t$ and
$\eta=y/t$ and we obtain the system
\begin{equation} \label{mixed}
\begin{gathered}
(u-\xi) u_{\xi} - \eta u_{\eta} + v_{\eta} = 0, \\
-v_{\xi} + u_{\eta} = 0,
\end{gathered}
\end{equation}
whose eigenvalues are
\begin{equation} \label{ev_xieta}
\Lambda_{\pm} = \frac{\eta \pm \sqrt{\eta^2 + 4(\xi - u)}}{2(\xi-u)}.
\end{equation}
We also recall that by eliminating $v$ in \eqref{mixed}, we obtain
the following second order equation for $u$
\begin{equation}\label{2nd}
(u-\xi)u_{\xi\xi} - \eta u_{\xi\eta} + u_{\eta \eta} + (u_{\xi}-1)u_{\xi}=0.
\end{equation}
Therefore, when the above system is linearized about a constant state,
it changes type across the sonic parabola
\begin{equation} \label{Pu}
P_u: \xi + \frac{\eta^2}{4}=u,
\end{equation}
and the system is hyperbolic if $\xi + \eta^2/4 >u$ and it is elliptic
if $\xi + \eta^2/4 <u$.
Since the problem is symmetric with respect to $\xi$-axis, we consider
the problem in the upper-half plane. The initial condition \eqref{initial}
becomes a boundary condition in the far field and results in a rarefaction
wave in the upper-half plane given by
\begin{equation} \label{R}
u = \xi + a \eta - a^2, \quad v = au, \quad a^2-1\leq \xi+a \eta \leq a^2.
\end{equation}
Hence, in a time-like direction, the left border of the rarefaction wave
is given by the line $\xi+a \eta =a^2$ along which $(u,v)=(0,0)$, and the
right border of the rarefaction wave is given by $\xi+a\eta=a^2-1$ along
which $(u,v)=(-1,-a)$. The rarefaction wave stretches down until it
becomes sonic. We compute the sonic line to be $\eta=2a$ using \eqref{Pu}
and \eqref{R}. However, it turns that the problem is much more complicated
due to the interaction of the rarefaction wave with the sonic parabola
\[
P_0 : \xi + \frac{\eta^2}{4}=0,
\]
corresponding to the state $(0,0)$.
More precisely, the left border of
the rarefaction wave intersects parabola $P_0$ at the point $A(-a^2, 2a)$.
We find the two characteristics at the point $A$, by solving the equations
\[
\frac{d \eta}{d \xi} = \Lambda_{\pm},
\]
and by substituting $\xi-u=a^2-a\eta$ in \eqref{ev_xieta} and noting that
$\eta \geq 2a$. We obtain the line
\[
\xi+a\eta=a^2 \quad \text{(positive characteristics)}
\]
and the parabola
\[
\xi = -\frac{(\eta -a)^2}{2}-\frac{a^2}{2} \quad \text{(negative characteristics)} .
\]

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig2} \\ % xieta.eps
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)(0,0)
\put(40,4){$\xi$}
\put(20,60){$\eta$}
\put(18,17){$P_0$}
\put(-7,17){$P_{-1}$}
\put(18,26){$A$}
\put(-15,48){$B$}
\put(-1,30){$C$}
\put(-38,45){$D$}
\put(15,53){$(0,0)$}
\put(-30,53){$(-1,-a)$}
\end{picture}
\end{center}
\caption{Semi-hyperbolic patch in $(\xi,\eta)$-plane}
\label{fig2}
\end{figure}

An easy calculation shows that the negative characteristics through $A$
intersects the right border of the rarefaction
wave at the point $B(-a^2-1-a\sqrt{2}, 2a+\sqrt{2})$ which is below the
projected parabola $P_0$ (see Figure \ref{fig2}). Therefore,
the information from the point $A$ travels back into the hyperbolic region and,
as a consequence the rarefaction wave is curved below the curve $AB$
(which will from now on denote the negative characteristics at the point $A$).
 We observe that this negative characteristics
continues as a straight line and intersects the sonic parabola
$P_{-1}: \xi+\eta^2/4=-1$, corresponding to the state $(-1,-a)$, tangentially
at the point $D$. The problem we are interested in is
formulated as follows:
\begin{quote}
 Given a positive characteristics $BC$, where $C$ is a sonic point, find the
solution in the domain
$ABDCA$, where the curve $ABD$ is a negative
characteristics, $AC$ is a sonic curve and $CD$ is the envelope of
negative characteristics.
\end{quote}


\subsection*{Directional derivatives along $\Lambda_{\pm}$ characteristics}

We denote the directional derivatives along $\Lambda_{\pm}$ characteristics by
\begin{equation} \label{xiDpm}
\partial^{\pm} = \partial_{\xi} + \Lambda_{\pm} \partial_{\eta}.
\end{equation}
Using \eqref{ev_xieta} we find
\[
u = \xi - \frac{\eta \Lambda_{\pm}+1}{\Lambda_{\pm}^2}.
\]
Differentiating this expression, and using $\partial^{\pm}\xi=1$ and
$\partial^{\pm}\eta=\Lambda_{\pm}$, we obtain
\begin{equation}
\begin{aligned}
\partial^{\pm} u
& = 1 - \frac{(\Lambda_{\pm}^2 + \eta \partial^{\pm} \Lambda_{\pm}) \Lambda_{\pm}
 -(\eta \Lambda_{\pm} +1) 2 \Lambda_{\pm} \partial^{\pm} \Lambda_{\pm}}
 {\Lambda_{\pm}^4}  \\
 & = \frac{\eta \Lambda_{\pm}+2}{\Lambda_{\pm}^3} \partial^{\pm} \Lambda_{\pm},
\end{aligned}\label{L}
\end{equation}
or, equivalently,
\begin{equation}\label{pmL}
\partial^{\pm} \Lambda_{\pm}
= \frac{\Lambda_{\pm}^3}{\eta \Lambda_{\pm}+2} \partial^{\pm} u.
\end{equation}

\section{Formulation of the Goursat problem in parabolic coordinates}

It is convenient to formulate the problem in parabolic coordinates $(\rho, \eta)$, 
where $\rho=\xi+\eta^2/4$. The system \eqref{mixed} becomes
\begin{gather*}
(u-\rho) u_{\rho}-\frac{\eta}{2}u_{\eta} + v_{\eta}  =  0, \\
\frac{\eta}{2} u_{\rho} - v_{\rho} + u_{\eta}  =  0,
\end{gather*}
with eigenvalues
\begin{equation} \label{ev_rhoeta}
\lambda_{\pm} = \pm \frac{1}{\sqrt{\rho-u}}.
\end{equation}
We eliminate variable $v$ in the above system and we obtain the second order 
equation for $u$,
\begin{equation}
\label{rho2nd}
(u-\rho) u_{\rho\rho} + u_{\eta \eta} - \frac{u_{\rho}}{2}+u_{\rho}^2=0.
\end{equation}
We find the left and the right borders of the rarefaction wave to be the parabolas
\[
\rho=\left( \frac{\eta}{2}-a\right)^2 \quad \text{and} \quad 
\rho=\left( \frac{\eta}{2}-a\right)^2-1,
\]
respectively (see Figure \ref{fig3}).
Furthermore, we find points $A(0,2a)$, $B(-1/2, 2a+\sqrt{2})$ and the negative
characteristics at $A$ to be the parabola
\begin{equation}\label{l-p}
\rho=-\left( \frac{\eta}{2}-a\right)^2.
\end{equation}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig3} \\ % rhoeta
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)(0,0)
\put(40,8){$\rho$}
\put(-4,72){$\eta$}
\put(2,28){$A$}
\put(-20,56){$B$}
\put(-25,36){$C$}
\end{picture}
\end{center} 
\caption{Region $ABC$ in the $(\rho,\eta)$-plane}
\label{fig3}
\end{figure}

\subsection*{Directional derivatives along $\lambda_{\pm}$ characteristics}

We use the notation
\begin{equation}\label{l-1}
\lambda^{-1} = \sqrt{\rho-u},
\end{equation}
and we denote the directional derivatives along
$\lambda_{\pm}$ characteristics by
\begin{equation}
\label{rhoDpm}
\partial_{\pm} = \partial_{\eta} \pm \lambda^{-1} \partial_{\rho}
 = \partial_{\eta} \pm \sqrt{\rho-u} \, \partial_{\rho}.
\end{equation}
Hence,
\begin{align*}
\partial_{\pm} u 
& = \partial_{\eta}u(\xi, \eta) \pm \lambda^{-1} \partial_{\rho} u (\xi, \eta) \\
 & = u_{\xi}\xi_{\eta} + u_{\eta} \pm \lambda^{-1} u_{\xi} \xi_{\rho} \\
 & = u_{\xi} \Big(-\frac{\eta}{2} \pm \lambda^{-1} \Big)+u_{\eta}\\
 & = \Big( -\frac{\eta}{2} \pm \lambda^{-1}\Big) \Big\{u_{\xi} +
 \Big( -\frac{\eta}{2} \pm \lambda^{-1}\Big)^{-1}u_{\eta} \Big\}.
\end{align*}
An easy calculations shows
\[
\left(-\frac{\eta}{2} \pm \lambda^{-1}\right)^{-1} = \left(-\frac{\eta}{2} \pm \sqrt{\rho-u}\right)^{-1} =
\Lambda_{\pm},
\]
implying
\[
\partial_{\pm} u = \Lambda_{\pm}^{-1} \partial^{\pm} u.
\]
Recalling \eqref{L}, we obtain
\begin{equation}\label{important}
\partial_{\pm} u = \frac{\eta \Lambda_{\pm}+2}{\Lambda_{\pm}^4} 
\partial^{\pm} \Lambda_{\pm}.
\end{equation}
We use this relationship between $\partial_{\pm} u$ and 
$\partial^{\pm} \Lambda_{\pm}$ in
\S 4 to investigate the convexity/concavity of $\Lambda_{\pm}$ characteristics 
in the region $ABC$.

\subsection*{Characteristics decomposition}

Using the second order equation \eqref{rho2nd}, we compute
\begin{align*}
\partial_+ \partial_- u 
& = \partial_+ \left( u_{\eta} - \sqrt{\rho-u} \, u_{\rho}\right) \\
 & = \frac{u_{\rho}}{2 \sqrt{\rho-u}} (u_{\eta} - \sqrt{\rho-u} \, u_{\eta})\\
 & = \frac{u_{\rho}}{2 \sqrt{\rho-u}} \partial_-u.
\end{align*}
We obtain, from \eqref{rhoDpm}, that
\begin{equation}\label{urho}
u_{\rho} = \frac{\partial_+u - \partial_-u}{2 \sqrt{\rho-u}}.
\end{equation}
Therefore,
\[
\partial_+\partial_- u
 = \frac{1}{4(\rho-u)}\left( \partial_+ u - \partial_- u\right)\partial_-u.
\]
We define
\begin{equation}\label{Q}
Q:= \frac{1}{4(\rho-u)},
\end{equation}
and note that $Q>0$ in the hyperbolic region.
Hence, we obtain
\begin{equation}\label{plusminus}
\partial_+\partial_-u = Q(\partial_+u - \partial_-u)\partial_-u.
\end{equation}
Similarly,
\begin{equation}
\label{minusplus}
\partial_-\partial_+u = Q(\partial_-u - \partial_+u)\partial_+u.
\end{equation}
Equations \eqref{plusminus}-\eqref{minusplus} are analogous to the 
characteristic decomposition equations obtained for the pressure-gradient 
system in \cite{SZ} and the nonlinear wave system
in \cite{HW}. We note that
\begin{equation} \label{ueta}
\frac{1}{2}(\partial_+u + \partial_-u)=u_{\eta}.
\end{equation}

\subsection*{Formulation of the Goursat problem in the domain $ABC$}

Using \eqref{plusminus}, \eqref{minusplus} and \eqref{ueta},
the second order equation \eqref{rho2nd} for $u$ can be written as the system
\begin{equation}\label{system}
\begin{bmatrix}
\partial_+ u \\
\partial_- u \\
u
\end{bmatrix}_{\eta}
+\begin{bmatrix}
-\lambda^{-1} & 0 & 0 \\
\quad 0 & \lambda^{-1} & 0 \\
\quad 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
\partial_+ u \\
\partial_- u \\
u
\end{bmatrix}_{\rho}
=\begin{bmatrix}
Q(\partial_-u - \partial_+ u)\partial_+ u \\
Q(\partial_+u - \partial_- u)\partial_- u \\
\frac{1}{2}(\partial_+ u + \partial_- u)
\end{bmatrix}.
\end{equation}
We define
\[
W=(\partial_+ u, \partial_- u, u)^T,
\]
and we specify $W$ along characteristics $AB$ and $CB$. We denote
\begin{equation}\label{condition}
W_-=W|_{AB} \quad \text{and} \quad W_+=W|_{CB}.
\end{equation}

First, we recall from \eqref{l-p} that
\begin{equation}
\label{AB}
AB = \{ (\rho, \eta) : \rho=-\left( \frac{\eta}{2}-a\right)^2, \;
2a \leq \eta \leq 2a+\sqrt{2} \}.
\end{equation}
Further, along $AB$ we have
\begin{equation}
\label{uAB}
u = \xi+a\eta-a^2
 = \rho-\frac{\eta^2}{4} + a \eta - a^2
 = \rho -\left( \frac{\eta}{2}-a\right)^2
 = 2 \rho.
\end{equation}
Hence, along $AB$ we have
\begin{equation}\label{u-AB}
\partial_-u=\partial_{\eta}u - \sqrt{\rho-u}\, \partial_{\rho} u =-2\sqrt{-\rho},
\end{equation}
implying $\partial_-u<0$ on $AB \setminus \{ A\}$ and $\partial_-u(A)=0$.

Secondly, let the positive characteristics $CB$ be given by the curve 
$\rho=f(\eta)$. Note that this curve is increasing, implying $f'>0$. 
Recall that the characteristics $AB$ is given by the parabola \eqref{l-p}, 
which has vertex at $(-1,2a)$ and the second derivative $1/2$. 
Since the point $C$ is below this parabola, but above the line $\eta=2a$, it is
reasonable to expect that $f''>1/2$. For the rest of the paper we assume 
that $f''>1/2$. Since the positive characteristics $CB$ satisfies the equation
\[
f'=\frac{d \rho}{d\eta}=\sqrt{\rho-u} = \sqrt{f-u},
\]
we obtain
\begin{equation}\label{uCB}
u = f - (f')^2.
\end{equation}
Therefore, along $CB$ we have
\begin{equation}\label{u+CB}
\partial_+ u = \partial_{\eta}u + \sqrt{\rho-u}
 \partial_{\rho} u = f'-2f'f''=f'(1-2f'')<0.
\end{equation}

Next, we specify directional derivatives $\partial_+u$ along $AB$ 
and $\partial_-u$ along $CB$.
To obtain $\partial_+ u$ along $AB$, we rewrite \eqref{minusplus} as
\[
-\partial_-\partial_+u+Q \partial_-u \partial_+u = Q(\partial_+u)^2,
\]
and, after diving by $(\partial_+u)^2$, we obtain
\begin{equation}
\label{first}
\partial_- \Big(\frac{1}{\partial_+u}\Big)+\frac{Q \partial_-u}{\partial_+u}=Q.
\end{equation}
We recall the expression for $\rho$ along the characteristics $AB$ in \eqref{AB} 
and expressions for $Q$, $u$ and $\partial_-u$ from \eqref{Q}, \eqref{uAB} 
and \eqref{u-AB}, respectively. Hence,
along $AB$, \eqref{first} becomes
\begin{equation}\label{ODE1}
\frac{d}{d\eta} \frac{1}{\partial_+u} - \frac{1}{\eta-2a} \frac{1}{\partial_+u} 
= \frac{1}{(\eta-2a)^2},
\end{equation}
which is a linear ODE for $1/\partial_+u$. Using \eqref{u+CB}, we impose the 
condition at the point $B$ to be
\begin{equation}
\label{C1}
\frac{1}{\partial_+u}\left(2a+\sqrt{2}\right)
=\frac{1}{f'(1-2f'')}\left(2a+\sqrt{2} \right).
\end{equation}
The solution of \eqref{ODE1} is given by
\[
\frac{1}{\partial_+u}=-\frac{1}{2(\eta-2a)} + C(\eta - 2a),
\]
where constant $C$ is determined from the condition \eqref{C1}.

To obtain $\partial_-u$ along $CB$, similarly as above, we 
rewrite \eqref{plusminus} as
\[
\partial_+\Big( \frac{1}{\partial_-u}\Big) + \frac{Q \partial_+u}{\partial_-u}=Q.
\]
We recall that $\rho=f(\eta)$ along $CB$, and using expressions for
$u$, $\partial_+u$ and $Q$ from \eqref{uCB}, \eqref{u+CB} and
\eqref{Q}, respectively, the above equation along $CB$ becomes
\[
\frac{d}{d \eta} \Big(\frac{1}{\partial_-u}\Big)
 +\frac{1-2f''}{4f'}\frac{1}{\partial_-u}=
\frac{1}{4(f')^2}.
\]
Using \eqref{u-AB}, we impose a condition for $1/\partial_-u$ at the point $B$ to be
\[
\frac{1}{\partial_-u}\big(2a+\sqrt{2} \big)=-2\sqrt{-f( 2a+\sqrt{2})}.
\]
Solution of this linear ODE problem gives $\partial_-u$ along $CB$.

\section{Existence of a local smooth solution to the Goursat problem 
and properties of smooth solutions}

In this section we prove existence of a smooth local solution st the point $B$
to the Goursat problem \eqref{system}
with conditions \eqref{condition} along characteristics $AB$ and $CB$.
In the series of Lemmas that follow, we establish
the signs of the directional derivatives $\partial_{\pm} u$ in the domain $ABC$, 
their monotonicity along the characteristic curves,
global minimum of $\partial_{\pm}u$, convexity/concavity of positive/negative 
characteristics, existence of a sonic curve $AC$ and an estimate of the $C^1$ norm of
$W$ away from the sonic curve.

\begin{theorem} \label{thm4.1}
The Goursat problem \eqref{system}-\eqref{condition} has a smooth local 
solution at the point $B$.
\end{theorem}

\begin{proof}
The eigenvalues of \eqref{system} are
$\lambda_1 = -\lambda^{-1}$, $\lambda_2 = \lambda^{-1}$  and $\lambda_3 =0$,
and the corresponding eigenvectors are
\[
\vec{l}_1 = (1,0,0), \quad \vec{l}_2=(0,1,0) \quad \text{and} \quad \vec{l}_3=(0,0,1).
\]
To prove the theorem, as in \cite{HW, SZ}, it suffices to check compatibility 
conditions at the point $B$ (see \cite[\S 4.3]{LectureZ})
\[
W_+(B)=W_-(B)
\]
and
\[
\frac{\vec{l}_3 (B) W_-'(B) - \frac{\partial_+u(B) 
+ \partial_-u(B)}{2}}{\lambda_1(B)-\lambda_3(B)}
= \frac{\vec{l}_3 (B) W_+'(B) - \frac{\partial_+u(B) 
 + \partial_-u(B)}{2}}{\lambda_2(B)-\lambda_3(B)}.
\]
The first condition is clearly satisfied. Noting that 
$\vec{l}_3(B)W_-'(B) = \partial_-u(B)$ and
$\vec{l}_3(B)W_+'(B) = \partial_+u(B)$, and recalling the definition 
of $\lambda^{-1}$ in \eqref{l-1},
implies that the second condition is also satisfied.
\end{proof}

\begin{lemma} \label{lem4.2}
Let $u$ be a smooth solution of the Goursat problem for 
\eqref{rho2nd} in the domain $ABC$
with conditions \eqref{uAB} and
\eqref{uCB} along characteristics $AB$ and $CB$, respectively. Then
\begin{itemize}
\item[(a)] $\partial_-u <0$ and $\partial_+u < 0$ in the interior of $ABC$,

\item[(b)] $\partial_-u$ is increasing (decreasing) function along $\lambda_+$ characteristics if
\[
\partial_+u \geq \partial_-u \quad (\partial_+u \leq \partial_-u),
\]

\item[(c)] $\partial_+u$ is decreasing (increasing) function
along $\lambda_-$ characteristics if
\[
\partial_+u \geq \partial_-u \quad (\partial_+u \leq \partial_-u).
\]
\end{itemize}
\end{lemma}

\begin{proof}
Proof of (a) was partially given in \cite{CCJ} where we analyzed the 
problem only in a neighborhood of the point $A$,
and for completeness we recall it here. Let $P$ be a point in the
domain $ABC$ and let $P_1$ and $P_2$ be intersections of positive characteristics 
through $P$ with $AB$ and negative characteristics through $P$ with $BC$, 
respectively (Figure \ref{fig4}).

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig4} \\ % rhoeta
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)(0,0)
\put(38,6){$A$}
\put(-10,67){$B$}
\put(-41,14){$C$}
\put(-7,40){$P$}
\put(13,55){$P_1$}
\put(-32,52){$P_2$}
\put(32,35){$\lambda_-$}
\put(-43,36){$\lambda_+$}
\end{picture}
\end{center} 
\caption{Interior point $P$ with its $\lambda_{\pm}$ characteristics}
\label{fig4}
\end{figure}

We recall the characteristic decomposition \eqref{plusminus}-\eqref{minusplus} 
and that $\partial_-u<0$ and
$\partial_+u<0$ hold on $AB \setminus \{A \}$ and $CB$, respectively, 
from \eqref{u-AB} and \eqref{u+CB}.
Integrating equation \eqref{plusminus} along positive characteristics $PP_1$, 
we obtain
\begin{equation} \label{PP1}
\partial_-u(P)=\partial_-u(P_1) e^{{\int_P^{P_1}} Q (\partial_+u - \partial_-u) ds}.
\end{equation}
Since $\partial_-u(P_1)<0$, we obtain $\partial_-u(P)<0$. 
Similarly, integrating \eqref{minusplus}
along negative characteristics $PP_2$, we obtain
\begin{equation} \label{PP2}
\partial_+u(P)=\partial_+u(P_2) e^{{\int_P^{P_2}} Q (\partial_-u - \partial_+u) ds}.
\end{equation}
Using that $\partial_+u(P_2)<0$, we obtain $\partial_+u(P)<0$.

Parts (b) and (c) follow immediately from \eqref{PP1} and \eqref{PP2}, 
noting that the sign of the expressions in the
exponent depends on the sign of $\partial_+u-\partial_-u$, since $Q>0$.
\end{proof}

\begin{lemma} \label{lem4.3}
Let $u$ be a smooth solution of the Goursat problem for the equation 
\eqref{rho2nd} in the domain $ABC$ with conditions \eqref{uAB} and
\eqref{uCB} along characteristics $AB$ and $CB$, respectively.
If the parameter $a$ in \eqref{initial} is chosen so that $a>1$, then
$\Lambda_-$ characteristics are concave and $\Lambda_+$ characteristics 
are convex in the region $ABC$.
\end{lemma}

\begin{proof}
We remark that the maximum of $\xi$ is at the point $A$ and, that (according to the
part (a) of the previous Lemma, the minimum of $u$ is at the point $B$. Therefore
\[
\xi-u \leq -a^2 + 1 <0,
\]
if $a>1$.

To complete the proof, we recall the relationship \eqref{important} 
between $\partial_{\pm} u$ and
$\partial^{\pm} \Lambda_{\pm}$.
First, we claim that
\begin{equation}\label{claim1}
\Lambda_->-\frac{2}{\eta}.
\end{equation}
Using the expression for $\Lambda_-$ from \eqref{ev_xieta} and rearranging 
the terms in \eqref{claim1},
since $\xi-u<0$, we obtain that proving \eqref{claim1} is equivalent to proving
\begin{equation}\label{step1}
-\eta \sqrt{\eta^2 + 4(\xi-u)} < -\eta^2 - 4(\xi-u).
\end{equation}
Note that the right-hand side in the above inequality is $-(\eta^2 + 4 (\xi-u))$, 
which is negative by hyperbolicity.
Hence, to show \eqref{step1} we need to show
\[
\left(\eta \sqrt{\eta^2 + 4(\xi-u)}\right)^2 > (\eta^2 + 4(\xi-u))^2.
\]
After simple calculations, the above inequality is equivalent to
\[
\eta^2 + 4(\xi-u)>0,
\]
which is, again, satisfied by hyperbolicity.
Since $\partial_-u<0$ in $ABC$, using claim the \eqref{claim1}, 
from \eqref{important}, we obtain that
\[
\partial^-\Lambda_-<0,
\]
implying that $\Lambda_-$ characteristics are concave in $ABC$.

Similar calculations show that
\[
\Lambda_+ < -\frac{2}{\eta},
\]
implying, from \eqref{important} and the fact that $\partial_+u<0$ in $ABC$, that
\[
\partial_+ \Lambda^+ >0,
\]
i.e., $\Lambda^+$ characteristics are convex in $ABC$.
\end{proof}


\begin{lemma} \label{lem4.4}
If $u$ is a smooth solution of the Goursat problem for  \eqref{rho2nd} in the 
domain $ABC$ with conditions \eqref{uAB} and
\eqref{uCB} along characteristics $AB$ and $CB$, respectively,
then $\lambda_+$ characteristics are convex and $\lambda_-$ characteristics 
are concave in the domain $ABC$.
\end{lemma}

\begin{proof}
Consider a positive characteristics
\[
\frac{d \rho}{d \eta} = \sqrt{\rho -u}.
\]
Then
\[
\frac{d^2 \rho}{d \eta^2} 
= \frac{\sqrt{\rho-u}-\partial_+u}{2 \sqrt{\rho-u}}>\frac{1}{2}>0,
\]
by Lemma \ref{lem4.2}, implying that positive characteristics are convex.

Next, let a negative characteristics be given by $\rho=g(\eta)$. Then
\[
g'=\frac{d \rho}{d \eta} = -\sqrt{\rho -u}=-\sqrt{g-u},
\]
implying, after solving for $u$ and differentiating along this characteristics, 
that
\begin{equation}\label{f''}
\frac{\partial_-u}{g'}=1-2g''.
\end{equation}
We claim that
\begin{equation}\label{claim2}
\frac{\partial_-u}{g'}> 2.
\end{equation}
Using \eqref{plusminus} and expression for $Q$ in \eqref{Q}, we obtain
\begin{equation}
\begin{aligned}
\partial_+\Big(\frac{\partial_-u}{g'} \Big)
&= - \partial_+ \Big(\frac{\partial_-u}{\sqrt{\rho-u}} \Big) \\
& = - \frac{Q (\partial_+u - \partial_-u)\partial_-u \, \sqrt{\rho-u}
- \frac{\partial_-u}{2 \sqrt{\rho-u}}(\sqrt{\rho-u}-\partial_+u)}{\rho-u}  \\
& = -\frac{1}{4} \frac{\partial_-u}{\sqrt{\rho-u}}\Big(\frac{\partial_+u}{\rho-u}
-\frac{\partial_-u}{\rho-u} \Big)+\frac{\partial_-u}{2(\rho-u)}-\frac{\partial_-u \, \partial_+u}
{2 \sqrt{\rho-u}(\rho-u)}  \\
& = \frac{\partial_-u}{g'} \Big\{\frac{1}{4} \Big(\frac{\partial_+u}{\rho-u}
-\frac{\partial_-u}{\rho-u} \Big)-\frac{1}{2\sqrt{\rho-u}}
+ \frac{\partial_+u}{2(\rho-u)}
\Big\}.
\end{aligned} \label{expression}
\end{equation}
We consider two cases.

$\bullet$ If
\[
\frac{1}{4} \Big(\frac{\partial_+u}{\rho-u}
-\frac{\partial_-u}{\rho-u} \Big)-\frac{1}{2\sqrt{\rho-u}}
 + \frac{\partial_+u}{2(\rho-u)} < 0,
\]
then \eqref{expression} implies
\[
\partial_+\Big(\frac{\partial_-u}{g'} \Big)<0,
\]
since $\partial_-u <0$ and $g'<0$. Hence,
$\frac{\partial_-u}{g'}$ decreases along positive characteristics.
For an interior point $P$, let $S$ be the intersection of positive characteristics through $P$ and
the curve $AB$. Then
\[
\frac{\partial_-u}{g'}(P) 
 > \frac{\partial_-u}{-\sqrt{\rho-u}}(S) 
  = \frac{2 \sqrt{-\rho}}{\sqrt{\rho-u}}(S) 
  = 2,
\]
using that along $AB$ we have $u=2\rho$ and $\partial_-u=-2\sqrt{-\rho}$ from \eqref{uAB} and \eqref{u-AB},
respectively.
Hence, the claim \eqref{claim2} is proved.

$\bullet$ If
\[
\frac{1}{4} \Big(\frac{\partial_+u}{\rho-u}
-\frac{\partial_-u}{\rho-u} \Big)-\frac{1}{2\sqrt{\rho-u}}
 + \frac{\partial_+u}{2(\rho-u)} \geq 0,
\]
then
\[
\frac{1}{4}\frac{\partial_-u}{g'} 
\geq \Big( \frac{1}{2\sqrt{\rho-u}} - \frac{\partial_+u}{2(\rho-u)}
-\frac{1}{4} \frac{\partial_+u}{\rho-u}\Big) \sqrt{\rho-u}> \frac{1}{2},
\]
since $\partial_+u<0$ inside $ABC$. Hence,
\[
\frac{\partial_-u}{g'} >2,
\]
and the claim \eqref{claim2} is proved.

Therefore, \eqref{f''} implies
\[
g''<-\frac{1}{2}<0,
\]
and, as a consequence, $\lambda_-$ characteristics are concave in the region $ABC$.
\end{proof}


\begin{lemma} \label{lem4.5}
Let $u$ be a smooth solution of the Goursat problem \eqref{rho2nd} 
in the domain $ABC$ with conditions \eqref{uAB} and
\eqref{uCB} along characteristics $AB$ and $CB$. Then
\begin{itemize}
\item[(a)] $\rho-u$ is increasing along positive characteristics,

\item[(b)] $\rho-u$ is increasing along negative characteristics,

\item[(c)] level curves of $\rho-u$ are non-characteristic, and

\item[(d)] $AC$ is a sonic curve and $\partial_+u=\partial_-u$ on $AC$.
\end{itemize}
\end{lemma}


\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig5} \\ % PQRS.eps
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)(0,0)
\put(38,6){$A$}
\put(-10,67){$B$}
\put(-42,14){$C$}
\put(-17,10){$P$}
\put(1,8){$R$}
\put(14,55){$Q$}
\put(-32,51){$S$}
\put(32,35){$\lambda_-$}
\put(-43,36){$\lambda_+$}
\end{picture}
\end{center} 
\caption{$\lambda_{\pm}$ characteristics inside the region $ABC$}
\label{fig5}
\end{figure}

\begin{proof}
To prove part (a), let $PQ$ be a positive characteristics (Figure \ref{fig5}).
Then, along $PQ$, we have
\[
\partial_+(\rho-u) = \partial_+\rho - \partial_+u >0,
\]
since $\partial_+\rho=\sqrt{\rho-u}>0$ and $\partial_+u<0$, by the previous Lemma.

To prove part (b), let $RS$ be a negative characteristics (Figure \ref{fig5}), 
given by $\rho=g(\eta)$.
Since $g$ is decreasing and concave, we have $g'<0$ and $g''<0$. Then along $RS$ we have
\[
g'=\frac{d\rho}{d\eta}=-\sqrt{\rho-u}=-\sqrt{g-u},
\]
and, therefore,
$u= g-(g')^2$,
yielding
$\partial_-u = g' - 2 g'g''$.
Hence
\[
\partial_-(\rho-u)=2g'g''>0,
\]
implying $\rho-u$ is increasing along $RS$.

For part (c), clearly, the level curve of the function
$\rho-u$, by (a) and (b), must be noncharacteristic.

Finally, to prove part (d),
we remark that since $\rho-u$ is increasing along both characteristic families,
the level curve $AC$ ($\rho-u=0$) consists of sonic points only.
Further, using the definition of derivatives $\partial_{\pm}$ 
from \eqref{rhoDpm}, since
along the sonic curve $AC$ we have $\rho=u$, we obtain
$\partial_+ u = \partial_-u$.
\end{proof}

\begin{lemma} \label{lem4.6}
For a smooth solution $u$ of the Goursat problem \eqref{rho2nd} in the
 domain $\Omega=ABC$, with conditions \eqref{uAB} and
\eqref{uCB} along characteristics $AB$ and $CB$, we have
\begin{equation} \label{m}
m:=\min_{AB \cup CB} \partial_{\pm} u \leq \min_{\Omega}\partial_{\pm}u \leq 0.
\end{equation}
\end{lemma}

\begin{proof}
As in \cite{HW, SZ}, we prove \eqref{m} in two cases.

$\bullet$  First, assume $\partial_+ u \geq \partial_-u$ in the entire domain 
$\Omega$. Then, by (b) and (c) of Lemma \ref{lem4.2}, we have that $\partial_-u$ increases 
along positive characteristics and $\partial_+u$ decreases along negative 
characteristics. Let $T \in \Omega$ be an arbitrary point and let $S$ be the
intersection of negative characteristics through $T$ with the curve $CB$ 
(Figure \ref{fig6}). We have
\[
\partial_+ u (T) \geq \partial_+u(S) \geq m.
\]
Also, let $P$ be the intersection of positive characteristics through $T$ 
with the sonic curve $AC$.
Then
\[
\partial_-u(T) \geq \partial_-u(P)=\partial_+u (P),
\]
by Lemma \ref{lem4.5}, part (d). Now consider the negative characteristics at $P$ and 
assume it intersects $CB$ at the point $S_1$. Since $\partial_+u$ decreases 
along negative characteristics we have
\[
\partial_+u(P) \geq \partial_+u(S_1) \geq m.
\]
Hence $\partial_{\pm} u (T) \geq m$.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig6} \\ % m1.eps
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)(0,0)
\put(38,6){$A$}
\put(-10,67){$B$}
\put(-42,14){$C$}
\put(-14,10){$P$}
\put(-14,10){$P$}
\put(-2,34){$T$}
\put(-38,42){$S_1$}
\put(32,35){$\lambda_-$}
\put(-42,30){$\lambda_+$}
\end{picture}
\end{center} 
\caption{Case 1: $\partial_+ u \geq \partial_-u$ in $\Omega$}
\label{fig6}
\end{figure}


$\bullet$ Secondly, we assume that there is a point in $\Omega$ such that 
$\partial_+u < \partial_-u$ at that point.
By the continuity, let $\mathcal N$ be the largest neighborhood of this point, 
inside $\Omega$, such that $\partial_+u < \partial_-u$ in $\mathcal N$. 
Let $Z \in \overline{\mathcal N}$.
By Lemma \ref{lem4.2}, $\partial_-u$ decreases along positive characteristics in $\mathcal N$. 
Let $Z_1 \in \partial N$ be the  point on positive
characteristics through $Z$, as in Figure \ref{fig7}. Then
\[
\partial_-u(Z) \geq \partial_-u(Z_1)=\partial_+u(Z_1).
\]

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig7} \\ % m2.eps
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)(0,0)
\put(38,6){$A$}
\put(-10,67){$B$}
\put(-42,14){$C$}
\put(1,51){$Z_3$}
\put(-19,43){$Z_2$}
\put(1,36){$Z_1$}
\put(-6,26){$Z$}
\put(24,44){$\lambda_-$}
\put(-35,48){$\lambda_+$}
\end{picture}
\end{center} 
\caption{Case 2: $\partial_+ u < \partial_-u$ in $\mathcal N$}
\label{fig7}
\end{figure}

In the neighborhood of $Z_1$ outside of $\mathcal N$ we have 
$\partial_+u \geq \partial_-u$ and we draw
negative characteristics at $Z_1$ in the direction of increasing $\eta$.
For any point $Z_2$ on this characteristics, we have
\[
\partial_+u(Z_1) \geq \partial_+u(Z_2),
\]
by Lemma \ref{lem4.2}. If this negative characteristics intersects the curve $CB$, 
\eqref{m} is proved for $\partial_-u(Z)$. Otherwise, we draw this characteristics 
as long as $\partial_+u \geq \partial_-u$ and assume that $Z_2$ is the ending point.
 Next, we draw positive characteristics starting at the point
$Z_2$, in the direction of increasing $\eta$, along which 
$\partial_+u < \partial_-u$. For any point $Z_3$ along this curve, we have
\[
\partial_+u(Z_2) = \partial_-u(Z_3) \geq \partial_-u(Z_3).
\]
If this positive characteristics intersects the curve $AB$, \eqref{m} is proved for
$\partial_-u(Z)$. Otherwise, we draw this positive characteristics as long as
$\partial_+u < \partial_-u$ ending at the point $Z_3$. From the point $Z_3$, we
continue by following negative characteristics. We repeat the procedure, 
by following positive or negative characteristics, depending on the sign of 
$\partial_+u-\partial_-u$, until we reach either $AB$ or $CB$. 
Hence, $\partial_-u(Z) \geq m$.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig8} \\ % m3.eps
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)(0,0)
\put(38,6){$A$}
\put(-10,67){$B$}
\put(-42,14){$C$}
\put(-14,45){$Z$}
\put(-1,34){$Z_1$}
\put(24,44){$\lambda_-$}
\put(-35,48){$\lambda_+$}
\end{picture}
\end{center} 
\caption{Case 2: $\partial_+ u < \partial_-u$ in $\mathcal N$}
\label{fig8}
\end{figure}

To prove $\partial_+u(Z) \geq m$, we recall that $\partial_+u$ increases 
along negative characteristics that are in $\mathcal N$.
 Let $Z_1 \in \partial \mathcal N$ be the point on the negative
characteristics through $Z$ as in Figure \ref{fig8}. Then
\[
\partial_+u(Z) \geq \partial_+u(Z_1) = \partial_-u(Z_1).
\]
Since $Z_1 \in {\overline{\mathcal N}}$, by the previous part of the proof, we have
\[
\partial_-u(Z_1) \geq m.
\]
Therefore, $\partial_{\pm}u(Z) \geq m$.
\end{proof}


\begin{lemma} \label{lem4.7}
Let $\Omega_{\epsilon}=\{(\rho, \eta) \in ABC : \rho-u > \epsilon\}$ and let
$W \in C^1(\Omega_{\epsilon})$ be a solution of the Goursat problem \eqref{system}
 with conditions \eqref{condition} along characteristics $AB$ and $CB$. Then
\[
\|W\|_{C^1(\Omega_{\epsilon})} \leq \frac{C}{\epsilon^{5/2}},
\]
for a constant $C$ independent of $\epsilon$.
\end{lemma}

\begin{proof}
From equations \eqref{urho} and \eqref{ueta} and Lemma \ref{lem4.5}, we have
\[
|u_{\rho}| = \Big|\frac{\partial_+u - \partial_-u}{2 \sqrt{\rho -u}}\Big|
\leq \frac{|m|}{\sqrt{\epsilon}}
\]
and
\[
|u_{\eta}|= \Big|\frac{\partial_+u + \partial_- u}{2}\Big| \leq |m|,
\]
where $m$ is as in \eqref{m}. Also, \eqref{plusminus}-\eqref{minusplus} imply
\begin{equation}\label{pmmp}
|\partial_{\pm}\partial_{\mp}u|=|Q(\partial_{\pm} u- \partial_{\mp}u)\partial_{\mp}u|
=\Big| \frac{1}{4(\rho-u)} (\partial_{\pm} u- \partial_{\mp}u)\partial_{\mp}u \Big|
\leq \frac{|m|^2}{2 \epsilon}.
\end{equation}
Next, we obtain estimates on $\partial_{\pm}\partial_{\pm}u$. As in \cite{HW, SZ},
we use the commutator relation from
\cite[Proposition 2.1]{LectureZ}
\[
\partial_-\partial_+I - \partial_+\partial_-I 
= \frac{\partial_-\lambda^{-1} - \partial_+(-\lambda)^{-1}}
{(-\lambda)^{-1}-\lambda^{-1}} (\partial_-I - \partial_+I),
\]
where $I$ is any quantity, and $\partial_{\pm}$ are directional derivatives 
along characteristics $d\rho/d\eta=(\pm \lambda)^{-1}$.
We use this relation for $\partial_-u$ and obtain
\begin{equation} \label{com}
\partial_-\partial_+\partial_- u - \partial_+\partial_-\partial_-u 
= \frac{\partial_-\lambda^{-1} 
 - \partial_+(-\lambda)^{-1}}{(-\lambda)^{-1}-\lambda^{-1}}
(\partial_- \partial_- u - \partial_+\partial_-u).
\end{equation}
On the other hand, by differentiating expression \eqref{plusminus} along 
negative characteristics,
we obtain
\[
\partial_-\partial_+\partial_-u = \partial_-Q (\partial_+u - \partial_-u) \partial_-u
+ Q (\partial_-\partial_+u - \partial_-\partial_-u) \partial_-u
+Q (\partial_+u - \partial_-u) \partial_-\partial_-u.
\]
By substituting this in \eqref{com}, we obtain
\begin{equation} \label{+--}
\begin{aligned}
\partial_+\partial_-\partial_-u
& = \big\{-\frac{\partial_-\lambda^{-1} - \partial_+(-\lambda)^{-1}}
 {(-\lambda)^{-1}-\lambda^{-1}}
+ Q (\partial_+u - 2 \partial_-u) \big\}
\partial_-\partial_-u \\
&\quad +
\frac{\partial_-\lambda^{-1} - \partial_+(-\lambda)^{-1}}
{(-\lambda)^{-1}-\lambda^{-1}}
Q (\partial_+u - \partial_-u)\partial_-u \\
&\quad +
\partial_-Q (\partial_+u - \partial_-u)\partial_-u \\
&\quad  +Q^2 (\partial_-u - \partial_+u)
\partial_+u \, \partial_-u.
\end{aligned}
\end{equation}
We recall that
\[
\lambda^{-1}=\sqrt{\rho-u}, \quad (-\lambda)^{-1}=-\sqrt{\rho-u}, \quad
(-\lambda)^{-1}-\lambda^{-1}=-2\sqrt{\rho-u},
\]
and we compute
$\partial_-\rho=-\sqrt{\rho-u}$  and $\partial_+\rho = \sqrt{\rho-u}$.
Therefore,
\[
\partial_-\lambda^{-1}=\frac{1}{2\sqrt{\rho-u}}(\partial_-\rho-\partial_-u) =
- \frac{\sqrt{\rho-u}+\partial_-u}{2 \sqrt{\rho-u}},
\]
and, similarly,
\[
\partial_+(-\lambda)^{-1} = -\frac{\sqrt{\rho-u}-\partial_+u}{2 \sqrt{\rho-u}}.
\]
Hence,
\[
\frac{\partial_-\lambda^{-1}
- \partial_+(-\lambda)^{-1}}{(-\lambda)^{-1}-\lambda^{-1}} =
\frac{\partial_-u + \partial_+ u }{4(\rho-u)}.
\]
Also, we recall the expression for $Q$ in \eqref{Q}, and find
\[
\partial_-Q = -\frac{1}{4(\rho-u)^2} (\partial_-\rho - \partial_-u) =
\frac{\sqrt{\rho-u}+ \partial_-u}{4(\rho-u)^2}.
\]
By substituting these expressions in \eqref{+--}, we obtain
\[
\partial_+\partial_-\partial_- u =
-\frac{3}{4(\rho-u)^2}\partial_-u\, \partial_-\partial_-u + \frac{\partial_-u
(\partial_+u - \partial_-u)}{4(\rho-u)^2}
\Big\{ \frac{5 \partial_-u}{4}+\sqrt{\rho-u}\Big\},
\]
which is a differential equation for $\partial_-\partial_- u$ along
positive characteristics.
By integrating this equation along positive characteristics, we obtain
\[
|\partial_-\partial_-u| \leq \frac{C}{\epsilon^2}\quad \text{on } \Omega_{\epsilon}.
\]
Similarly,
\[
|\partial_+\partial_+u| \leq \frac{C}{\epsilon^2} \quad\text{on } \Omega_{\epsilon}.
\]
The last two estimates together with \eqref{pmmp}, \eqref{ueta} and \eqref{urho},
give
\begin{gather*}
|(\partial_{\pm} u)_{\eta}| = |\partial_{\pm}(u_{\eta})|
= \Big|\partial_{\pm}\Big( \frac{\partial_+u - \partial_-u}{2}\Big)\Big|
\leq \frac{C}{\epsilon^2},
\\
|(\partial_{\pm} u)_{\rho}|=|\partial_{\pm}(u_{\rho})|=
\Big| \partial_{\pm} \Big(\frac{\partial_+u-\partial_-u}{2 \sqrt{\rho-u}}
\Big)\Big|
\leq \frac{C}{\epsilon^{5/2}}.
\end{gather*}
\end{proof}

Analogous properties for solutions to the pressure-gradient system and
the nonlinear wave system to those obtained in Lemmas \ref{lem4.2}--\ref{lem4.7}, 
are used to extend the local smooth
solution at the point $B$ to the whole domain $ABC$ using techniques
from \cite{Dai} by Dai and Zhang for gas dynamics equations.
We believe that these techniques could be extended for the UTSD 
equation as well and we leave that study
for the future work.

\section{The envelope of $\lambda_-$ characteristics and existence of a shock}

As in \cite{HW, SZ}, we conclude by showing that one family of characteristics 
(in this case, negative characteristics) forms an envelope before they become sonic. 
We use \cite[Theorem 3.2]{CK}, which states that for $2 \times 2$ 
quasi-one-dimensional systems (such as \eqref{mixed}), a hyperbolic state 
adjacent to a hyperbolic constant state must be a simple wave
(also, see the argument using Riemann invariants in \S 2 in \cite{LZZ}).

Hence, the solution in the region below the curve $BD$, to the left of $CB$ 
(see Figure \ref{fig2}), is a simple wave. In our case,
the family of negative characteristics are straight lines along which $u$ is
constant.
Therefore, in this region $\partial_-u=0$ and
\begin{equation}\label{shock}
\partial_-\Big( \frac{1}{\partial_+u}\Big)
=-\frac{\partial_-\partial_+u}{(\partial_+u)^2}
=-\frac{Q(\partial_-u - \partial_+u)\partial_+u}{(\partial_+u)^2}
=Q = \frac{1}{4(\rho-u)}>0,
\end{equation}
by \eqref{minusplus}. Hence, $\frac{1}{\partial_+u}$ is increasing along 
negative characteristics in the direction from $B$ to $D$. 
We recall from \eqref{u+CB} that $\frac{1}{\partial_+u}<0$ along the
curve $CB$. Also note from \eqref{shock} that $\frac{1}{\partial_+u}$ 
becomes infinite at the sonic points, since $\rho=u$ there. 
Hence, $\frac{1}{\partial_+u}$ must become zero before
the negative characteristics reach their sonic points. Consequently, 
$\partial_+u$ becomes infinite before negative characteristics become sonic 
indicating that negative characteristics form an envelope before their sonic points. 
As in \cite{SZ} and in \S 9 in \cite{Z1},
this indicates existence of a shock.


\subsection*{Conclusion}

In this article we consider a Riemann problem for the unsteady transonic 
small disturbance equation resulting in diverging rarefaction waves. 
We write the problem in self-similar and parabolic coordinates,
and obtain a system that changes type from hyperbolic to elliptic. 
We find the solution in the hyperbolic region and we formulate a problem 
in a semi-hyperbolic patch where one family out of
two families of characteristic curves starts at a sonic curve and ends on
 a transonic shock. We prove existence of a smooth local solution
and we derive various properties of the global solution in the semi-hyperbolic 
patch adopting the ideas of characteristic decomposition for the nonlinear 
wave system and the pressure gradient system in \cite{HW, SZ}.

Our future work consists of three main steps: (1) computing the
numerical approximations of the global solution to the problem using the 
numerical method developed in \cite{I}, (2) using the approach in \cite{Dai} 
to prove existence of the solution in the entire semi-hyperbolic patch, 
and (3) establishing the iteration scheme to prove existence of a global solution, 
as outlined for the nonlinear wave system in \cite{IK}.

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\end{document}