text:
\section{Introduction}\label{intro}

In this paper we consider existence and regularity for Poiseuille flow of nematic liquid crystals. Liquid crystals have many forms and a particular form is the nematic whose molecules can be viewed as
rod-like/thread-like.  Macroscopically,  the state of a nematic liquid crystal is  characterized by its velocity field ${\bf u}$ for the flow and its director field  ${\bf n}\in \mathbb S^2$ for the alignment of the rod-like feature.  These two characteristics interact with each other so that any distortion of the director ${\bf n}$ causes
a motion ${\bf u}$  and, likewise,  any flow ${\bf u}$ affects the alignment ${\bf n}$. Using the convention to denote $\dot{f}=f_t+{\bf u}\cdot \nabla f$ the material derivative, the full Ericksen-Leslie model for nematics  is given as follows
\begin{equation}\label{wlce}
\begin{cases}
\rho\dot {\bf u}+\nabla P=\nabla\cdot\sigma-\nabla\cdot\left(\frac{\partial W}{\partial\nabla {\bf n}}\otimes\nabla {\bf n}\right),
\\
\nabla\cdot {\bf u}=0,\\
\nu{\ddot {\bf n}}=\lambda{\bf n}-\frac{\partial W}{\partial  {\bf n}}-{\bf g}+\nabla\cdot\left(\frac{\partial W}{\partial\nabla {\bf n}}\right), \\
|{\bf n}|=1.\\
\end{cases}
\end{equation}
In  (\ref{wlce}),  $P$ is the pressure, $\lambda$ is the Lagrangian multiplier of the constraint $|{\bf n}|=1$, $\rho$ is the density, $\nu$ is the inertial coefficient of the director ${\bf n}$, and $W$,
${\bf g}$  and $\sigma$ are the Oseen-Frank energy,  the kinematic transport and the viscous stress tensor, respectively (see, e.g., \cites{DeGP,ericksen62, frank58, liucalderer00,leslie68,Les, lin89} for details).

\subsection{Poiseulle flow of nematics} For Poiseulle flows of  nematics with a choice of coordinates system, ${\bf u}$ and ${\bf n}$ take the form
\[{\bf u}(x,t)= (0,0, u(x,t))^T \;\mbox{ and }\;{\bf n}(x,t)=(\sin\theta(x,t),0,\cos\theta(x,t))^T,\]
where the motion ${\bf u}$ is along the $z$-axis and   the director ${\bf n}$ lies in the $xz$-plane with angle $\theta$ made from the $z$-axis.   For this case of Poiseulle flows, taking $\rho=\nu=1$ for simplicity, the Ericksen-Leslie  model is reduced to system (\ref{sysf}) and (\ref{sysw}) below, whose  derivation is available in the literature (see, e.g., \cite{CHL20}),
\begin{align}\label{sysf}
    \displaystyle u_t&=\left(g(\theta)u_x+h(\theta)\theta_t\right)_x,\\\label{sysw}
\theta_{tt}+\gamma_1\theta_t&=c(\theta)\big(c(\theta)\theta_x\big)_x-h(\theta)u_x.
\end{align}
We will be interested in Cauchy problem of the Poiseulle flow
 with initial data
\beq\label{initial}
u(x,0)=u_0(x)\in H^1(\mathbb {R}),\; \theta(x,0)=\theta_0(x)\in H^1,
\; \theta_t(x,0)=\theta_1(x)\in L^2(\mathbb {R}).
\eeq
We further impose an assumption
\beq\label{tecinitial}\begin{split}
u'_0(x)+\frac{h(\theta_0(x))}{g(\theta_0(x))}&\theta_1(x)\in H^1(\mathbb R)\cap L^1(\mathbb {R}).
\end{split}
\eeq

 The functions $c(\theta)$, $g(\theta)$ and $h(\theta)$ in the model  are given by
\begin{align}\label{fgh}\begin{split}
 g(\theta):=&\alpha_1\sin^2\theta\cos^2\theta+\frac{\alpha_5-\alpha_2}{2}\sin^2\theta+\frac{\alpha_3+\alpha_6}{2}\cos^2\theta+\frac{\alpha_4}{2},\\
  h(\theta):=&\alpha_3\cos^2\theta-\alpha_2\sin^2\theta=\frac{\gamma_1+\gamma_2\cos(2\theta) }{2},\\
  c^2(\theta):=&K_1\cos^2\theta+K_3\sin^2\theta,
 \end{split}
 \end{align}
 where, depending on the nematic material and temperature, $K_1>0$ and $K_3>0$ are Frank's coefficients for splay and bending energies,  and $\alpha_j$'s $(1 \leq j \leq 6)$ are the Leslie dynamic coefficients. The following relations are assumed in the literature.
\begin{align}\label{a2g}
\gamma_1 =\alpha_3-\alpha_2,\quad \gamma_2 =\alpha_6 -\alpha_5,\quad \alpha_2+ \alpha_3 =\alpha_6-\alpha_5.
\end{align}
The first two relations are compatibility conditions, while the third relation is called a Parodi's relation, derived from Onsager reciprocal relations expressing the equality of certain relations between flows and forces in thermodynamic systems out of equilibrium (cf. \cite{Parodi70}).
They also satisfy the following empirical relations (p.13, \cite{Les})
\begin{align}\label{alphas}
&\alpha_4>0,\quad 2\alpha_1+3\alpha_4+2\alpha_5+2\alpha_6>0,\quad \gamma_1=\alpha_3-\alpha_2>0,\\
&  2\alpha_4+\alpha_5+\alpha_6>0,\quad 4\gamma_1(2\alpha_4+\alpha_5+\alpha_6)>(\alpha_2+\alpha_3+\gamma_2)^2\notag.
\end{align}
Note that the fourth relation  is implied by the third together with the last relation and the last can be rewritten as
$\gamma_1(2\alpha_4+\alpha_5+\alpha_6)> \gamma_2^2$.
 {Very importantly, relations (\ref{a2g}) and (\ref{alphas}) imply that (see formula (2.4) in \cite{CHL20}),  for some constant $ \overline C>0$,
\begin{align}\label{positiveDamping}
g(\theta)\ge g(\theta)-\frac{h^2(\theta)}{\gamma_1}>\overline C.
\end{align}
}

It is known that the solution of system \eqref{sysf} and \eqref{sysw} generically form finite time cusp singularity \cites{CS22,CHL20}.{ The goal of this paper is to establish a global existence of H\"older continuous solutions for the Cauchy problem  \eqref{sysf}-\eqref{tecinitial} beyond singularity formation as stated in Theorem \ref{main} below.}

\subsection{Directly relevant results and our results.}

 In \cite{CHL20}, a special case of Poiseuille flow was treated.  More precisely, the authors chose the parameters as
\[\rho=\nu=1,\; \alpha_1=\alpha_5=\alpha_6=0,\;\alpha_2=-1,\;\alpha_3=\alpha_4=1,\]
which result in $\gamma_1=2$, $\gamma_2=0$, and $g=h=1$. With this special choice of parameters, system (\ref{sysf}) and (\ref{sysw}) becomes
\begin{align}\label{simeqn0}
\begin{split}
 u_t=&(u_x+\theta_t)_x,\\
\theta_{tt}+2\theta_t=&c(\theta)(c(\theta)\theta_{x})_x
 -u_x.
\end{split}
\end{align}
 In \cite{CHL20},  on one hand, the authors constructed solutions for \eqref{simeqn0} with smooth initial data that produce, in finite time, cusp singularities—blowups of $u_x$ and $\theta_t$. The method directly extends  that of \cites{CZ12, GHZ} for  variational wave equations.   On the other hand, the global existence of weak solutions, which are H\"older continuous, of system \eqref{simeqn0} were established for general initial data similar to (\ref{initial}).
 The latter resolved satisfactorily the physical concerns from application point of view about what happens after the finite time singularity formation.
  A crucial ingredient for existence beyond singularity formation is the  identification of the quantity
\[J(x,t)=u_x(x,t)+\theta_t(x,t)\]
 and the reveal of a singularity cancellation—the quantity $J$ remains bounded and H\"older continuous while  its components $u_x$ and $\theta_t$ approach infinity at formations of cusp singularities.

The change of coordinates framework in \cite{BZ}  for the variational wave equations was used to cope with the wave part $\eqref{simeqn0}_2$, and will be used in this paper too for \eqref{sysw}.
The detailed idea will be given in Section \ref{idea}. See other works on the global well-posedness of H\"older continuous solutions for variational wave equations \cites{BC,BC2015,BH,BHY,BCZ, CCD, ZZ03,ZZ10,ZZ11,GHZ}.

 In a recent paper \cite{CS22}, the singularity formation for the general system \eqref{sysf} and \eqref{sysw} was established.
 As mentioned above, we are concerning with the global existence of the Cauchy problem for the general system \eqref{sysf} and \eqref{sysw}. {It should be pointed out that the generalization is far beyond straightforward.
One apparent trouble  is that the diffusion coefficient $g(\theta(x,t))$ in the parabolic equation \eqref{sysf} is only H\"older continuous, which creates difficulties in handling the quantity $J$ for the singularity (see system (\ref{J_eq}) and the discussion followed for details). This leads the authors to introduce and work with the potential $A$ of $J$ in (\ref{VarA}).  Another difficulty is caused by rough (worse than H\"older continuous) non-homogeneous terms in the parabolic equation for $A$, in addition to the diffusion coefficient $g(\theta(x,t))$ being only H\"older continuous. This difficulty is overcome with a careful analysis in \cite{MS} that relies on but goes beyond treatments in \cite{Fri}. The work in \cite{MS} has a much more broad interest besides a direct application to the present work.}

For the statement of our result, we need the following definition of weak solutions.
\begin{definition}\label{def1}
For any fixed time $T\in (0,\infty)$, we say that $(u(x,t),\theta(x,t))$ is a weak solution of \eqref{sysf}, \eqref{sysw}, and \eqref{initial} over $\mathbb {R}\times [0,T]$ if
\begin{itemize}

 \item
 For any test function $\phi\in H^1_0(\mathbb {R}\times (0,T)),$
\begin{align}\label{thetaweak1}
    \int_0^T\int_\mathbb {R}\theta_t\phi_t-\gamma_1\theta_t\phi\,dx\,dt=\int_0^T\int_\mathbb {R} (c(\theta)\theta_x)(c(\theta)\phi)_x+hu_x\phi\,dx\,dt,
\end{align}
\begin{align}\label{uweak}
    \int_0^T\int_R u\phi_t-(gu_x+h\theta_t)\phi_x\,dx\,dt=0.
\end{align}

\item The initial data
\begin{align}\label{data}
    u(x,0)=u_0(x),\; \theta(x,0)=\theta_0(x),\; \text{and}\; \theta_t(x,0)=\theta_1(x)
\end{align}
hold point-wise for $u(x,0)$ and $\theta(x,0),$ and in $L^p$ sense for $\theta_t(x,0)$ for any $p\in[1,2).$
\end{itemize}
\end{definition}

 Throughout the paper, we will always assume relations \eqref{a2g} and \eqref{alphas} for the Leslie coefficients $\alpha_j$'s and refer to $g$, $h$ and $c$ as the functions given in \eqref{fgh}.
\begin{thm}[Global Existence]\label{main}  For any fixed time $T\in (0,\infty)$, the Cauchy problem of system \eqref{sysf} and \eqref{sysw} with  the initial conditions $u_0(x)$, $\theta_0(x)$ and $\theta_1(x)$ given in \eqref{initial} and \eqref{tecinitial} has a weak solution $(u(x,t),\theta(x,t))$ defined on $\mathbb {R}\times [0,T]$ in the sense of Definition \ref{def1}.
    Moreover,
    \[u(x,t)\in L^2([0,T],H^1(\mathbb {R}))\cap L^\infty(\mathbb {R}\times [0,T])\] and
    \[\theta(x,t)\in C^{1/2}(\mathbb {R}\times [0,T])\cap L^2([0.T],H^1(\mathbb {R})),\]
and,     for any $t\in[0,T]$, the associated energy
\begin{align}\label{calEdef}
\mathcal{E}(t):=\int_\mathbb {R}\theta_t^2+c^2(\theta)\theta_x^2+u^2\,dx
\end{align}
    satisfies
    \begin{align}
        \mathcal{E}(t)\leq \mathcal{E}(0)-\int_0^t\int_\mathbb {R} (u_x+\frac{h}{g}\theta_t)^2+\theta_t^2\,dx\,dt.
    \end{align}
\end{thm}

\medskip

The rest of this paper is organized as follows. In Section \ref{idea}, we introduce the main idea of this work. In Section \ref{J2theta}, as the first step in carrying out the main idea, we analyze the wave equation for $\theta$ with a prescribed forcing term.
 In Section \ref{HK}, we recall some basic properties from \cite{Fri} on parabolic differential operators with only H\"older continuous diffusion coefficients. In Sections \ref{J2v}, we apply the formulation in Section \ref{HK} and results in \cite{MS} to analyze $u$-component.  In Sections \ref{FixedPoint}-\ref{bdE} we prove the  existence of weak solution for the Cauchy problem \eqref{sysf} and \eqref{sysw}.

\section{Main idea of this work}\label{idea}
 The approach developed in \cite{CHL20} for the special case with $g=h=1$ provides a framework for the general system \eqref{sysf} and \eqref{sysw}. There are, however, a number of crucial issues in this generalization.

Note that singularity formation is unavoidable in general \cite{CS22}. Similar to \cite{CHL20}, we introduce a new variable
\[v=\int_{-\infty}^x u\,dx\]
 and obtain, from \eqref{sysf},
\beq\label{veq}
v_t=g(\theta)v_{xx}+h(\theta) \theta_t=g(\theta) u_x+h(\theta) \theta_t.
\eeq
Motivated by singularity cancellation revealed in \cite{CHL20},  we also introduce
\begin{align}\label{VarJ}
J=\frac{v_t}{g(\theta)}=u_x+\frac{h(\theta)}{g(\theta)}\theta_t,
\end{align}
which agrees with the function $J$  in \cite{CHL20} for the special case. It follows from \eqref{veq} and \eqref{VarJ} that $J$ satisfies
\beq\label{J_eq}
(g(\theta)J)_t=g(\theta) (g(\theta)J)_{xx}+h(\theta) \theta_{tt}+h'(\theta) \theta_t^2+g'(\theta)\theta_t
J-g'(\theta)\frac{h(\theta)}{g(\theta)}\theta_t^2.\eeq
However, it turns out the coefficient $g(\theta)$ in $(g(\theta) J)_{xx}$ is  only H\"older continuous in general.
And for the general case, we do not have an explicit formula using the heat kernel as in the special case.
The work in \cite{Fri} provides an implicit  expression for the kernel which can be used to treat the nonhomogeneous terms in (\ref{J_eq}) in an indirect way (see Section \ref{HK} for more details). In order to follow the framework in \cite{Fri} for the nonhomogeneous parabolic equation \eqref{J_eq},
we introduce a new auxiliary function $A,$

\begin{align}\label{VarA}
A(x,t)=\hat A(x,t)-A_0(x)=\int_{-\infty}^xJ(z,t)\,dz-\int_{-\infty}^xJ(z,0)\,dz,
\end{align}
that is $A_x=J-J_0,$ with $J_0=J(x,0)$ determined by \eqref{tecinitial}.

 By comparing equation (\ref{J_eq}) with equation (\ref{AAeq}) for $A$ given below, we avoid some complication in handling the rough term $\theta_{xx}$.

Instead of working directly on system \eqref{sysf} and \eqref{sysw}, we will treat the equivalent system in terms of the quantities $(v, \theta,A)$ as
\begin{align}
  v_t=g(\theta)v_{xx}&+h(\theta) \theta_t,\label{ueq}\\
   \theta_{tt}+\big(\gamma_1-\frac{h^2(\theta)}{g(\theta)}\big)\theta_t&=c(\theta)(c(\theta)\theta_x)_x-h(\theta)\hat A_x,\label{thetaeq}\\
    A_t=g(\theta)A_{xx}-&\gamma_1A+g'(\theta)\theta_xA_x+g'(\theta)\theta_xJ_0+F(\theta, v), \label{AAeq}
   \end{align}
  where
\begin{align}\label{Fun}\begin{split}
F&=G+f,\\
f&=[\gamma_1-\frac{h^2}{g}]v_x+\frac{h(\theta)c^2(\theta)}{g}\theta_x+g(\theta)J'_0,\\
G&=\int_{-\infty}^x[\frac{h'}{g}-\frac{g'h}{g^2}]\theta_t^2-[\gamma_1-\frac{h^2}{g}]'\theta_zv_z-(\frac{h(\theta)c(\theta)}{g})'c(\theta)\theta_z^2\,dz-\gamma_1 A_0,
    \end{split}
\end{align}
and
\beq\label{a0a0x}
A_0(x)=\int_{-\infty}^xJ_0(z)\,dz=\int_{-\infty}^xJ(z,0)\,dz,
\eeq
Note $A(x,0)=0.$

  The splitting of $F=f+G$ is based on different regularities of each term, as we will see later.
  A derivation for (\ref{AAeq}) from  \eqref{J_eq} is provided in Appendix \ref{App-A}.\\

Roughly, Theorem \ref{main} will be proved in the following steps.
 \bigskip

\paragraph{\bf Step 1:} For any given $J(x,t)\in C^\alpha\cap L^2\cap L^\infty$ for some $\alpha>0$, we consider the wave equation (\ref{thetaeq}) with $A_x$ being replaced by $J$
\begin{align}\label{waveJ}
    \theta_{tt}+\big(\gamma_1-\frac{h^2(\theta)}{g(\theta)}\big)\theta_t&=c(\theta)(c(\theta)\theta_x)_x-h(\theta)J.
    \end{align}\\
    Using very similar method as in \cite{CHL20},  the existence of a $C^{1/2}$ solution $\theta^J$ of (\ref{waveJ}) will be shown in Section \ref{J2theta}.
     \bigskip

\paragraph{\bf Step 2:}   With $\theta^J$ obtained from Step 1,  we then solve $v^J$  from the equation
    \begin{align}\label{vJ}
    v_t=g(\theta^J)v_{xx}+h(\theta^J)\theta_t^J,
    \end{align}
 and show that both $v^J$ and $u^J=v_x^J$ are in $ C^\alpha\cap L^2\cap L^\infty$ in Section \ref{J2v}.
    \bigskip

\paragraph{\bf Step 3:} With $(v^J,\theta^J)$ from the above steps, we will solve for $A^J$ from
\[A_t=g(\theta^J)A_{xx}-\gamma_1A+g'\theta_x^JJ+F(\theta^J,v^J).\]
An expression of $A^J=\mathcal{N}(J)$ by the so called parametrix method in \cite{Fri} is very helpful. Recall that $A_x+J_0=J$. After setting $\mathcal{M}(J)=\left(\mathcal{N}(J)\right)_x+J_0$, we then have  a fixed point problem for the map $J\mapsto \mathcal{M}(J)$ that will be analyzed by the Schauder fixed point theory in Section \ref{FixedPoint}.

\begin{center}
\begin{figure}[h]
\begin{tikzcd}
J  \arrow[dr,leftrightarrow] \arrow[r, mapsto,"A=\mathcal{N}(J)"] & A \arrow[d, "\mathcal{M}(J)=A_x+J_0"] \\
& \mathcal{M}(J)
\end{tikzcd}
\caption{A diagram explaining the relation between the quantities $A,J$ and the map $\mathcal{M}.$}
\label{diagram}
\end{figure}
\end{center}

\section{The solution $\theta$ of (\ref{waveJ}) with a fixed  $J$}\label{J2theta}

Given any function $J(x,t)\in L^2\cap L^\infty\cap C^\alpha$ for some $\alpha>0,$ we consider (\ref{waveJ}) recast below
\begin{align}\label{waveAx}
    \theta_{tt}+\big[\gamma_1-\frac{h^2(\theta)}{g(\theta)}\big]\theta_t&=c(\theta)(c(\theta)\theta_x)_x-h(\theta)J.
\end{align}
It is crucial that   $\gamma_1-\frac{h^2(\theta)}{g(\theta)}>C_*$  for some constant $C_*>0$ (see (\ref{positiveDamping})).

Using the change of coordinates method in \cites{BZ,CHL20}, we can prove the global existence of weak solution for \eqref{waveAx} in a very similar way as for the simplified system. To make this paper self-contained, we include the proof.\\

Given a point $(x_0,t_0)$, we define the characteristic curves
    \[x^\pm(s)\equiv x^\pm(s;\, x_0,\, t_0)\]
    by the solutions $x^\pm(s)$ of
    \[\frac{d x^\pm(s)}{d s}=\pm c\bigl(\theta(x^\pm(s),s)\bigr)\;\mbox{ with }\; x^\pm(t_0)=x_0.\]
 Note that $x^\pm(0)$ are the intersections of the characteristic curves $x^\pm(s)$ with the $x$-axis.
      With the help of the following gradient variables
      \begin{align}\label{RS}\begin{split}
          S(x,t)= &\theta_t(x,t) - c(\theta(x,t)) \theta_x(x,t),\\
 R(x,t)=& \theta_t(x,t)+c(\theta(x,t)) \theta_x(x,t),
\end{split}
      \end{align}
we make the change of coordinates $(x,t)\mapsto (X,Y)$:
\begin{align}\label{charaXY}\begin{split}
X\equiv X(x,t):=&	\int_1^{x^-(0;\, x,\, t)} (1+ R^2(x',0))\,d x',\\
 Y\equiv Y(x,t):=&	\int_{x^+(0;\, x,\, t)}^1 (1+ S^2(x',0))\,d x'.
 \end{split}
\end{align}
 Note that, for any differentiable function $f$, one has
\begin{align}\label{forward}\begin{split}
f_t(X,Y) +c(\theta) f_x(X,Y)&=2c X_x f_X,\\
f_t(X,Y)-c(\theta) f_x(X,Y)&=-2c Y_x f_Y.
\end{split}
\end{align}

In order to complete the system, we introduce several variables:
\beq\label{defwz}
w:=2\arctan R,\quad z:=2\arctan S,
\eeq
and
\beq\label{defpq}
p:=\frac{1+R^2}{X_x},\quad q:=\frac{1+S^2}{-Y_x}.
\eeq

We write the system with respect to the new coordinates to obtain
\begin{align}\label{semisys11}
&\theta_X=\frac{\sin w}{4c(\theta)}p,\quad \theta_Y=\frac{\sin z}{4c(\theta)}q
\end{align}
and to close the system we derive the equations for $z,\omega,p$ and $q$
\begin{align}
&\textstyle z_X=p
\Big\{\frac{c'}{4c^2}(\cos^2\frac{w}{2}-\cos^2\frac{z}{2})+\frac{b(\theta)}{4c}(\sin w\cos^2\frac{z}{2}+\sin z\cos^2\frac{w}{2})-\frac{h(\theta)}{c}J\cos^2\frac{z}{2}\cos^2\frac{w}{2} \Big\},\\
&\textstyle w_Y=q
\Big\{\frac{c'}{4c^2}(\cos^2\frac{z}{2}-\cos^2\frac{w}{2})+\frac{b(\theta)}{4c}(\sin w\cos^2\frac{z}{2}+\sin z\cos^2\frac{w}{2})-\frac{h(\theta)}{c}J\cos^2\frac{z}{2}\cos^2\frac{w}{2}\Big\}, \\
&\textstyle p_Y=pq\Big\{\frac{c'}{8c^2}(\sin z-\sin w)
+\frac{b(\theta)}{2c}(\frac{1}{4}\sin w\sin z+\sin^2 \frac{w}{2}\cos^2\frac{z}{2})
-\frac{h(\theta)}{2c}J\sin w\cos^2\frac{z}{2}\Big\},\\\label{semisys55}
&\textstyle q_X=pq\Big\{\frac{c'}{8c^2}(\sin w-\sin z)
+\frac{b(\theta)}{2c}(\frac{1}{4}\sin w\sin z+\sin^2 \frac{z}{2}\cos^2\frac{w}{2})
-\frac{h(\theta)}{2c}J\sin z\cos^2\frac{w}{2}\Big\},
\end{align}

and

\beq
\left\{
\begin{array}{ll}\label{eqnxt}
\displaystyle x_X=\frac{1}{2X_x}=\frac{1+\cos w}{4}p,\\
\\
\displaystyle x_Y=\frac{1}{2Y_x}=-\frac{1+\cos z}{4} q,
\end{array}
\right.
\qquad
\left\{
\begin{array}{ll}
\displaystyle t_X=\frac{1}{2cX_t}=\frac{1+\cos w}{4c}p,\\
\\
\displaystyle t_Y=-\frac{1}{2cY_t}=\frac{1+\cos z}{4c} q,
\end{array}
\right.
\eeq
where $b(\theta)=\frac{h^2(\theta)}{g(\theta)}-\gamma_1.$ See \cite{CHL20} for the derivation of \eqref{semisys11}-\eqref{eqnxt}.\\

Comparing this system with that in \cite{CHL20} for the special case $g=h=1$ and $\gamma_1=2$, one can see that the appearance of the bounded and smooth functions $g$ and $h$ does not create any difficulties applying the process in \cite{CHL20}.

\begin{prop}[Local Existence]\label{localwave}
There exists $ T>0$ sufficiently small such that system \eqref{semisys11}-\eqref{semisys55} has a solution $(\theta,z,\omega, p,q)(X,Y)$ defined on
\[\Omega_ T:=\{(X,Y)\in\mathbb {R}^2: d\big((X,Y),\gamma\big)\leq T\},\] where $\gamma$ is the curve in the $(X,Y)$ plane corresponding to the line $t=0$ on the $(x,t)$ plane and $d(\cdot,\gamma)$ is the distance between the curve and a point.
\end{prop}
\begin{proof}
The proof is similar to that in \cite{CHL20} and is outlined in Appendix \ref{app-semilinear}.
\end{proof}

Now, to extend the solution globally, meaning to any arbitrary time $0<T<\infty,$ we need some a prior uniform bound on $p$ and $q$.

\begin{lemma}\label{lemma3.3}Consider any solution of \eqref{semisys11}-\eqref{semisys55} constructed in the local existence result with $t\in [0, T]$. Then, we have
\beq\label{bdspq}
0<A_1\leq \max_{(X,Y)\in \Omega_ T }\left\{p(X,Y),\, q(X,Y)\right\}\leq A_2,
\eeq
for some constants $A_1$ and $A_2$ independent of $ T.$
\end{lemma}
\begin{proof}
We skip the proof since it is entirely similar to that for Lemma 6.2 in \cite{CHL20}.
\end{proof}
Now we need to transfer the solution back to the original coordinate system $(x,t)$ using \eqref{eqnxt}. We note that in general, due to the lack of enough regularity of the solution $\theta,$ we might well lose the  uniqueness of the characteristic curves. So $x(X,Y),t(X,Y)$ might not be well defined. Instead, we will show that $\theta(x,t)$ is always well defined. In fact, a possible scenario in this case is that we might have the following
    \[X_1:=X(x_1,t_1)=X(x_2,t_2)=:X_2\] and \[Y_1:=Y(x_1,t_1)=Y(x_2,t_2)=:Y_2.\] But, one can show that $\theta(X_1,Y_1)=\theta(X_2,Y_2)$ and hence $\theta$ remains well defined (\cite{BZ}).\\

Finally, we can prove the existence of weak solution in $(x,t)$ coordinates.

\begin{prop}\label{propth} Given a function  $J(x,t)\in (L^2\cap L^\infty\cap C^\alpha)([0,T],\mathbb {R})$ and initial data $\theta_0(x)$ and $\theta_1(x)$ as in \eqref{data}. Then
\eqref{waveAx} has a weak solution $\theta(x,t)=\theta^J(x,t)$ in the following sense
\begin{align}
    \int_0^T\int_\mathbb {R}\theta_t\phi_t-\big[\gamma_1-\frac{h^2(\theta)}{g(\theta)}\big]\theta_t\phi\,dx\,dt=\int_0^T\int_\mathbb {R} (c(\theta)\theta_x)(c(\theta)\phi)_x+h(\theta)J\phi\,dx\,dt,
\end{align}
for any function $\phi\in H^1_0(\mathbb {R}\times (0,T))$. Moreover, we have
\[\theta(x,t)\in C^{1/2}([0,T],\mathbb {R})\cap L^\infty([0,T],H^1(\mathbb {R})).\]
The initial data is satisfied in the sense that $\displaystyle\theta(x,0)=\theta_0(x)$ point-wise and $\displaystyle \theta_t(x,0)=\theta_1(x)$ in $L^p_{loc}$ for $p=[1,2).$
\end{prop}

\begin{proof}

We show that the solution constructed as a fixed point of the map in Appendix \ref{app-semilinear} satisfies the weak formulation,
\begin{align*}
    \int_0^T\int_\mathbb {R}\theta_t\phi_t+b(\theta)\theta_t\phi\,dx\,dt=\int_0^T\int_\mathbb {R} (c(\theta)\theta_x)(c(\theta)\phi)_x+hJ\phi\,dx\,dt,
\end{align*}
where $b(\theta)=\frac{h^2(\theta)}{g(\theta)}-\gamma_1.$
The calculations are similar to the one in \cite{BZ}. Rewrite the above equation in terms of the variables $R$ and $S$ defined in \eqref{RS} to get
\begin{align*}
  \int_0^T\int_\mathbb {R}\big[\phi_t-c\phi_x\big]R+\big[\phi_t+c\phi_x\big]S+c'\theta_x(S-R)\phi+b(\theta)(R+S)\phi-2hJ\phi\,dx\,dt=0.
\end{align*}
Using \eqref{forward} and $\displaystyle dxdt=\frac{pq}{2c(1+R^2)(1+S^2)}dXdY,$
we obtain
\begin{align*}
 &   \int_0^T\int_\mathbb {R}\bigg\{\big(-2cY_x\phi_Y\big)R+\big(2cX_x\phi_X\big)S+c'\big[\theta_XX_x+\theta_YY_x\big](S-R)\phi\\
    &\qquad -\big[\gamma_1-\frac{h^2(\theta)}{g(\theta)}\big](R+S)\phi-2hJ\phi\bigg\}\frac{pq}{2c(1+R^2)(1+S^2)}dXdY=0.
\end{align*}
Apply (\ref{defpq}) and (\ref{semisys11}) to get
\begin{align*}
  &  \int_0^T\int_\mathbb {R}\frac{R}{1+R^2}p\phi_Y+\frac{S}{1+S^2}q\phi_X+\frac{c'pq}{8c^2}\bigg(\frac{\sin\omega}{1+S^2}-\frac{\sin z}{1+R^2}\bigg)(S-R)\phi\\
    &\qquad -\frac{pq}{2c}\big[\gamma_1-\frac{h^2(\theta)}{g(\theta)}\big]\frac{R+S}{(1+R^2)(1+S^2)}\phi-\frac{pqhJ}{c(1+R^2)(1+S^2)}\phi\,dX\,dY=0.
\end{align*}
Noticing $\displaystyle \frac{R}{1+R^2}=\frac{\sin \omega}{2}\;\mbox{and}\; \frac{S}{1+S^2}=\frac{\sin z}{2}$ from (\ref{defwz}), we get
\begin{align*
  &  \int_0^T\int_\mathbb {R}\frac{\sin\omega}{2}p\phi_Y+\frac{\sin z}{2}q\phi_X\\
 & \qquad  +\frac{c'pq}{8c^2}\big(\sin\omega\sin z-\sin\omega \cos^2\frac{z}{2}\tan \frac{\omega}{2}-\sin z \cos^2\frac{\omega}{2}\tan \frac{z}{2}\big)\phi\\
    &\qquad -\frac{pq}{2c}\big[\gamma_1-\frac{h^2(\theta)}{g(\theta)}\big]\big(\cos\frac{\omega}{2}\cos^2\frac{z}{2}\sin \frac{\omega}{2}+\cos^2\frac{\omega}{2}\cos\frac{z}{2}\sin \frac{z}{2}\big)\phi\\
    &\qquad -\frac{pq}{c}h\cos^2\frac{\omega}{2}\cos^2\frac{z}{2}J\phi\,dX\,dY=0.

\end{align*}

Integrating the first two terms by parts and using the equations for $p,q,\omega$ and $z$, we prove that the weak formulation \eqref{thetaweak1} is satisfied.\\

It remains to show the H\"older continuity of $\theta(x,t)$ with exponent $1/2.$ This follows from the fact that
\begin{align*}
    \int_0^t[\theta_t\pm c(\theta)\theta_x]^2\,dt\leq C
\end{align*}
for any $t\in [0,T],$ where the constant $C$ depends only on $t.$ Using the change of coordinates \eqref{forward} (See Appendix \ref{app-semilinear}), we obtain
\begin{align*}
     \int_0^t[\theta_t+c(\theta)\theta_x]^2\,dt&=
    \int_{X_0}^{X_t}(2cX_x\theta_X)^2\frac{1}{2X_t}\,dX\\
    &=\int_{X_0}^{X_t}\big(\frac{4c}{p(1+\cos \omega}\frac{\sin \omega}{4c}p\big)^2\frac{1+\cos \omega}{4c}p\,dX\\
    &=\int_{X_0}^{X_t}\frac{\sin^2\omega}{1+\cos \omega}p\,dX\leq C.
\end{align*}
Similar calculations for $\theta_t-c(\theta)\theta_x$ gives a similar bound. The two bounds together imply the square integrability of $\theta_t$ and $\theta_x$ hence Sobolev embedding implies the H\"older continuity of $\theta(x,t)$ with exponent $1/2.$\\

Finally we show a bound for the energy $E(t)$ defined as
\begin{align}
  E(t):=\frac{1}{2}\int_\mathbb {R}\theta_t^2+c(\theta)^2\theta_x^2\,dx.
\end{align}
For any fixed $0<T<\infty,$ let $\Omega_T:=\mathbb {R}\times [0,T].$ For any function $J(x,t)\in L^\infty\cap L^2\cap C^\alpha(\Omega_T),$ the energy of a weak solution $\theta$ of \eqref{waveAx} satisfies a prior bound. More precisely,
the energy satisfies the following bound
\begin{align}
    E(t)\leq C_E,
\end{align}
for some $C_E$ depending on $E(0)$ and $J.$ The proof is similar to the one in \cite{CHL20} and is provided in Appendix \ref{Appenergy}.

The estimate obtained,
\beq\label{ET}
\frac{1}{2}\max_{0\leq t
\leq T}E(t)\leq E(0)+C\int_0^T\int_{-\infty}^\infty|J|^2\,dxdt
\eeq
for some constant $C$. This implies that $\theta_t(\cdot, t)$ and $\theta_x(\cdot, t)$ are both square integrable functions in $x$, so are $R$ and $S$.

The proof that the solution satisfies the initial condition follows by the same argument in \cite[Theoerem 1]{BZ}. We omit the proof here.
\end{proof}

\section{A brief review of parabolic differential operators with non-constant H\"older coefficients.}\label{HK}
In this section we summarise relevant results from first chapter in \cite{Fri} in terms of the specific form of equations appeared in this paper for direct future usage.

 Let the differential operator $\mathcal{L}$ be defined as
\[\mathcal{L}:=\partial_t-g(\theta)\partial_{xx}+\gamma_1,\]
where $g(\theta)$ is a strictly positive smooth function and $\theta$ is H\"older continuous with exponent $1/2$ with respect to $x$ and $t,$ and $\gamma_1$ is a positive constant. Consider the  differential equation
\begin{align}\label{diff}
    \mathcal{L}\,\omega&=0.
\end{align}

\par

Note that, for any fixed $(\xi,\tau)$,  the heat kernel of the operator
 \[\mathcal{L}_0^{\xi,\tau}:=\partial_t-g(\theta(\xi,\tau))\partial_{xx}\]
 is
\begin{align}\label{heat}
    H^{\xi,\tau}(x-\xi,t-\tau)=\frac{1}{2\sqrt{g(\theta(\xi,\tau))}\sqrt{t-\tau}}e^{-\frac{(x-\xi)^2}{4g(\theta(\xi,\tau))(t-\tau)}}.
\end{align}

\begin{remark}
    The superscripts in $\mathcal{L}_0^{\xi,\tau}$ and $H^{\xi,\tau}(x-\xi,t-\tau)$ indicate the dependence  on $(\xi,\tau)$ via $g(\theta(\xi,\tau)).$
\end{remark}

 Several results established in Chapter 1 of  \cite{Fri} will be recalled and used. These include Theorems 8-11, displays (4.15), (6.12) and (6.13) in \cite{Fri}.
\begin{prop}
There exists a function $\Phi$ such that $\Gamma$ given by
\begin{align}\label{GGGamma}
\Gamma(x,t,\xi,\tau)&=H^{\xi,\tau}(x-\xi,t-\tau)
+\int_\tau^t\int_\mathbb{R}H^{y,s}(x-y,t-s)\Phi(y,s;\xi,\tau)\,dy\,ds
\end{align}
 satisfies \eqref{diff}. Moreover, one has
    \begin{align}\label{phiest}
|\Phi(y,s;\xi,\tau)|\leq \frac{const}{(s-\tau)^{5/4}}e^{\frac{-d(y-\xi)^2}{4(s-\tau)}},
\end{align}
where $d$ is a constant depending on $\|g\|_{L^\infty(\mathbb {R})}$ and $\gamma_1.$
\end{prop}

Set $\displaystyle \Omega_T:=\mathbb {R}\times (0,T]$ for some $T>0$ and consider the Cauchy problem
\begin{align}\label{chy}
    &\mathcal{L}\,\omega(x,t)=f(x,t),\quad \text{on}\quad\Omega_T\\\label{chyinit}
    &\omega(x,0)=\phi(x),\quad \text{at} \quad t=0
\end{align}
where $f$ is H\"older continuous on $ \overline\Omega_T$ and $\phi$ is continuous on $\mathbb {R}$. It is shown in \cite[Theorem 12]{Fri} that the function
 \begin{align}
     \omega(x,t)=\int_R\Gamma(x,t,\xi,0)\phi(\xi)\,d\xi+\int_0^t\int_R\Gamma(x,t;\xi.\tau)f(\xi,\tau)\,d\xi\,d\tau
 \end{align}
is a {\em classical} solution of the Cauchy problem \eqref{chy} and \eqref{chyinit}. Moreover,
\begin{align}\label{Gammaest}
    |\Gamma(x,t;\xi,\tau)|\lesssim \frac{1}{\sqrt{t-\tau}}e^{-\frac{d(x-\xi)^2}{4(t-\tau)}}\approx H(x-\xi,t-\tau),
\end{align}
\begin{align}\label{Gamma_xest}
    |\Gamma_x(x,t;\xi,\tau)|\lesssim\frac{1}{{t-\tau}}e^{-\frac{d(x-\xi)^2}{4(t-\tau)}}\approx\frac{1}{\sqrt{t-\tau}}H(x-\xi,t-\tau),
\end{align}
where $d$ is a constant depending on $g$ and $\lesssim$ and $\approx$ mean $\leq$ up to a constant and $=$ up to a constant, respectively. In both cases the constant is uniform in $(x,t,\xi,\tau).$  \\

We now apply the above results from \cite{Fri} to get a preliminary result for later usage.
Set
\begin{align*}
M_f(x,t)&:=\int_0^t\int_\mathbb {R}\Gamma(x,t;\xi,\tau)f(\xi,\tau)\,d\xi\,d\tau,\\
    M_{f,x}(x,t)&:=\int_0^t\int_\mathbb {R}\Gamma_x(x,t;\xi,\tau)f(\xi,\tau)\,d\xi\,d\tau.
\end{align*}
\begin{prop}\label{GammaL2}
If $f(x,t)\in L^2(\Omega_T),$
then
\begin{align}\label{LinfMf}
\|M_f\|_{L^\infty(\Omega_T)}\lesssim T^{1/4}\|f\|_{L^2(\Omega_T)},\;&\; \|M_{f,x}\|_{L^\infty(\Omega_T)}\lesssim  T^{1/4}\|f\|_{L^\infty((0,T),L^2(\mathbb {R}))},\\\label{L2Mf}
\|M_f\|_{L^2(\Omega_T)}\lesssim T\|f\|_{L^2(\Omega_T)},\;&\mbox{ and }\; \|M_{f,x}\|_{L^2(\Omega_T)}\lesssim \sqrt{T}\|f\|_{L^2(\Omega_T)}.
\end{align}
\end{prop}
\begin{proof}
To estimate the $L^\infty$ norm we use \eqref{Gammaest}, \eqref{Gamma_xest} and Cauchy–Schwarz inequality,
\begin{align*}
    |M_f|&\lesssim \bigg(\int_0^t\int_\mathbb {R} \frac{1}{{t-\tau}}e^{-\frac{d(x-\xi)^2}{2(t-\tau)}}\,dx\,d\tau\bigg)^{1/2}\|f\|_{L^2(\Omega_T)}\lesssim T^{1/4}\, \|f\|_{L^2(\Omega_T)}.
\end{align*}
Taking the sup over $\Omega_T,$ we get the first estimate in \eqref{LinfMf}. Similarly,
\begin{align*}
    |M_{f,x}|&\lesssim \bigg(\int_0^t\int_\mathbb {R} \frac{1}{{|t-\tau|}^{2-2r}}e^{-\frac{d(x-\xi)^2}{2(t-\tau)}}\,dx\,d\tau\bigg)^{1/2}\bigg(\int_0^t\int_\mathbb {R} \frac{1}{{|t-\tau|}^{2r}}f^2\,dx\,d\tau\bigg)^{1/2}\\
    &\lesssim \bigg(\int_0^t \frac{1}{{t-\tau}^{\frac{3}{2}-2r}}\,d\tau\bigg)^{1/2} \bigg(\int_0^t \frac{1}{{|t-\tau|}^{2r}}\,dx\,d\tau\bigg)^{1/2}\|f\|_{L^\infty((0,T),L^2(\mathbb {R}))}.
\end{align*}
For $r=\frac{3}{8},$ we get the second estimate in \eqref{LinfMf}.\\

To estimate the $L^2$ norm we use \eqref{Gammaest}, \eqref{Gamma_xest} and the Young's convolution inequality with $r=2$, $p=1$, and $q=2$. On $\Omega_T,$
\begin{align*}
\|M\|_{L^2}\lesssim & \bigg\|\int_0^T\int_R\frac{1}{\sqrt{t-\tau}}e^{-\frac{d(x-\xi)^2}{4(t-\tau)}}|f(\xi,\tau)|\,d\xi\,d\tau\bigg\|_{L^2}\\
&\lesssim\|H\ast f\|_{L^2}\leq \|H\|_{L^{1}}\|f\|_{L^2}=C\, T\,\|f\|_{L^2},\\
\|M_x\|_{L^2}\lesssim & \bigg\|\int_0^T\int_R\frac{1}{{t-\tau}}e^{-\frac{d(x-\xi)^2}{4(t-\tau)}}|f(\xi,\tau)|\,d\xi\,d\tau\bigg\|_{L^2}\\
&\lesssim\|\frac{1}{\sqrt{t}}H\ast f\|_{L^2}\leq \|\frac{1}{\sqrt{t-\tau}} H\|_{L^{1}}\|f\|_{L^2}=C\, \sqrt{T}\,\|f\|_{L^2}.
\end{align*}
This completes the proof.
\end{proof}

\section{Existence of a solution $v^J$ for $\eqref{veq}$.}\label{J2v}
Recall that $\theta=\theta^J$ is the solution of wave equation (\ref{waveJ}) depending on $J$.
We now consider Cauchy problem of   \eqref{vJ}
\begin{align}\label{vveq}
& v_t=g(\theta^J)v_{xx}+h(\theta^J)\theta^J_t,\\
 \label{stvinit}
  &  v(x,0)=v_0(x)
\end{align}
and denote the solution by $v^J$.

\begin{prop}\label{propv}
Let  $v_0(x)$ be defined as $v'_0(x)=u_0(x).$ For any $T\in (0,\infty),$ there exists a function $$v^J(x,t)\in L^2((0,T),H^1(\mathbb {R}))$$ that satisfies \eqref{vveq} and \eqref{stvinit} in the sense that
\begin{align}\label{vweak}
     \int_0^T\int_R v^J\phi_t-v^J_x(g(\theta^J)\phi)_x+h(\theta^J)\theta^J_t\phi\,dx\,dt=0
\end{align}
for any $\phi\in H^1_0((0,T],\mathbb {R})$ and, as $t\to 0^+$,
\begin{align}\label{limv}
v^J(x,t)&\to v_0(x) \mbox{ point-wise, }  ,\\\label{limv_x}
v_x^J(x,t)&\to v'_{0}(x)=u_0(x) \mbox{ almost everywhere. }
\end{align}
 Moreover,
\[v^J, v^J_x\in C^\alpha((0,T],\mathbb {R})\cap L^\infty((0,T],\mathbb {R}) \mbox{ for any } 0<\alpha<1/4.\]
\end{prop}

\begin{proof} { For simplicity, we will drop the subscript $J$ in the proof.\\

Since $\theta_t$ is generally not H\"older continuous, to apply the results in Section \ref{HK}, we let $\theta_t^\epsilon$ be the mollification of $\theta_t$ for $\epsilon>0$ small.} It is  known that $\theta_t^\epsilon\in C^\infty_c(\Omega_T)$ and, as $\epsilon\to 0$, $\theta_t^\epsilon\to \theta_t$  in $L^2(\Omega_T).$ Denote the solution of
\begin{align}\label{stv}
&v_t=g(\theta)v_{xx}+h(\theta)\theta_t^\epsilon
\end{align}
with the same initial condition (\ref{stvinit})
by $v^\epsilon(x,t).$ As discussed in Section \ref{HK}, $v^\epsilon(x,t)$ is
a classical solution and can be written explicitly as
\begin{align}\label{vepsilon}
    v^\epsilon(x,t)=\int_\mathbb {R} \Gamma^0(x,t,\xi,0)v_0(\xi)\,d\xi+\int_0^t\int_\mathbb {R}\Gamma^0(x,t;\xi,\tau)h(\theta)\theta_\tau^\epsilon(\xi,\tau)\,d\xi\,d\tau,
\end{align}
{where $\Gamma^0$ is the  kernel of the operator ${\mathcal L}_0:=\partial_t-g(\theta(x,t))\partial_{xx}$.
Note that the operator ${\mathcal L}_0$ is the same as  ${\mathcal L}$ in (\ref{diff}) with $\gamma_1=0$. We comment that all estimates in Section \ref{HK} for $\Gamma$ still hold true for $\Gamma^0$ with possibly different constants.}\\

Clearly, $v^\epsilon$ in \eqref{vepsilon} satisfies the weak formulation \eqref{vweak} since it is a classical solution to \eqref{stv} and \eqref{stvinit}. We have
\begin{align}\label{epsilonweak}
   \int_0^T\int_R v^\epsilon\phi_t-v^\epsilon_x(g\phi)_x+h\theta_t^\epsilon\phi\,dx\,dt=0
\end{align}
for all $\phi\in H^1_0(\mathbb {R}\times (0,T)).$
At this point, we claim that the expressions
\begin{align}\label{compv}
    v(x,t)=\int_\mathbb {R} \Gamma^0(x,t,\xi,0)v_0(\xi)\,d\xi+\int_0^t\int_\mathbb {R}\Gamma^0(x,t;\xi,\tau)h(\theta)\theta_\tau(\xi,\tau)\,d\xi\,d\tau
\end{align}
and
\begin{align}\label{compv_x}
    v_x(x,t)=\int_\mathbb {R} \Gamma^0_x(x,t,\xi,0)v_0(\xi)\,d\xi+\int_0^t\int_\mathbb {R}\Gamma^0_x(x,t;\xi,\tau)h(\theta)\theta_\tau(\xi,\tau)\,d\xi\,d\tau
\end{align}
are the limits of $v^\epsilon(x,t)$ and $v_x^\epsilon(x,t),$ respectively, in $L^2(\Omega_T)$ sense, and hence, $v(x,t)$
 is a weak solution of \eqref{vveq}.\\

 Subtract (\ref{compv}) from (\ref{vepsilon}) and apply Proposition \ref{GammaL2} and estimate \eqref{Gammaest} to get
 \begin{align*}
    |v^\epsilon-v|\leq \int_0^T\int_\mathbb {R}|\Gamma^0||h||\theta^\epsilon_\tau-\theta_\tau|\,d\xi\,d\tau
\end{align*}
and
\begin{align*}
    \|v^\epsilon-v\|_{L^2(\Omega_T)}&\lesssim \|H\ast (|h||\theta^\epsilon_\tau-\theta_\tau|)\|_{L^2(\Omega_T)}\\
    &\lesssim \|H\|_{L^1(\Omega_T)}\|\theta^\epsilon_\tau-\theta_\tau\|_{L^2(\Omega_T)}\to 0\;\mbox{ as }\; \epsilon\to 0.
\end{align*}
 Similarly, by Proposition \ref{GammaL2} and estimate \eqref{Gamma_xest} we have
 \begin{align*}\begin{split}
    \|v_x^\epsilon-v_x\|_{L^2(\Omega_T)}&\lesssim \|\frac{1}{\sqrt{t}}H\ast (|h||\theta^\epsilon_\tau-\theta_\tau|)\|_{L^2(\Omega_T)}\\
    &\lesssim
    \|\frac{1}{\sqrt{t-\tau}}H\|_{L^1(\Omega_T)}\|\theta^\epsilon_\tau-\theta_\tau\|_{L^2(\Omega_T)}\to 0\;\mbox{ as }\; \epsilon\to 0.
    \end{split}
\end{align*}
Taking $\epsilon\to 0$ in \eqref{epsilonweak} we obtain \eqref{vweak}. Hence, the weak formulation \eqref{vweak} is satisfied as a limit of the classical solution $v^\ve$ to the initial value problem \eqref{stv},\eqref{stvinit}.\\

For the initial data, the first limit \eqref{limv} follows from \cite{Fri}.
The second limit \eqref{limv_x} can be shown by considering the equation satisfied by the first term of \eqref{compv_x} that is by letting $u^0(x,t)=v^0_x(x,t):=\int_\mathbb {R} \Gamma^0_x(x,t,\xi,0)v_0(\xi)\,d\xi.$ Then
\[u^0_t-(g(\theta)u^0_x)_x=0\]
\[u^0(x,0)=u_0(x)\in H^1(\mathbb {R}).\]

It is known that the solution $u^0\in C([0,T],L^2(\mathbb {R}))$ and $u^0(x,0)=u_0(x)$ almost everywhere \cite{Evans}. Hence, we obtain
\[v_x(x,0)=v'_0(x), \mbox{ almost everywhere }\]

Finally, to show $v, v_x\in L^\infty\cap C^\alpha(\mathbb {R}\times (0,T]),$ we use Propositions \ref{GammaL2} and \ref{mainlemma}, which are established in \cite{MS} for more general equation was studied and H\"older estimates were established.
\end{proof}
\section{Existence of a solution $J.$}\label{FixedPoint}
{ Recall in Sections \ref{J2theta} and \ref{J2v}, for any $J(x,t)\in C^\alpha\cap L^2\cap L^\infty$ for some $\alpha>0$, we solve $\theta^J$ from system (\ref{thetaeq}) with $A_x$ replaced by $J$,  then solve $v^J$ (and hence $u^J$) from system (\ref{ueq}) with $\theta$ replaced by $\theta^J$, and we show that
\[
v^J, u^J \in L^2\cap L^\infty\cap C^\alpha(\mathbb {R}\times (0,T))\;\mbox{ and }\;
\theta^J_t, \theta^J_x\in L^\infty((0,T),L^2(\mathbb {R})).\]

We now solve $A^J$ from system (\ref{AAeq}) with $(\theta,v,A_x+A_{0,x})$ replaced by $(\theta^J,v^J,J)$. With all these preparations, we will then define a mapping $\mathcal M(J)$ so that its fixed point gives rise to a solution of our original Cauchy problem.}

In view of system (\ref{AAeq}), define the operator
\[\mathcal{L}^J:=\partial_t-g(\theta^J)\partial_{xx}+\gamma_1.\]
The function $A$, introduced in (\ref{VarA}), satisfies \eqref{AAeq} recast below
\begin{align}\label{Aeq}
    \mathcal{L}^J A=&F(\theta^J,\theta^J_t,\theta^J_x, v^J)+g'(\theta^J)\theta^J_x J,
\end{align}
where $F(\theta,\theta_t,\theta_x, v)=f(\theta,\theta_x, v_x)+G(\theta,\theta_t,\theta_x, v_x)$ with
$f$ and $G$ given in from \eqref{Fun}, along with the initial data\begin{align}\label{A_0}
    A(x,0)=0.
\end{align}

Formally, following the discussion in Section \ref{HK}, $A$ can be expressed as
\begin{align}\label{Aexpression}
    A(x,t)=\int_0^t\int_\mathbb {R}\Gamma(x,t;\xi,\tau)\big[F(\theta^J,v^J)+g'(\theta^J)\theta^J_{\xi}J\big](\xi,\tau)\,d\xi\,d\tau.
\end{align}
We now define a mapping $\mathcal{M}$ by
\begin{align}\begin{split}\label{5.6}
    &\mathcal{M}( J)(x,t):=A_x(x,t)+J_{0}\\
&=
\int_0^t\int_\mathbb {R}\Gamma_x(x,t;\xi,\tau)\big[F(\theta^J,v^J)+g'(\theta^J)\theta^J_\xi J\big]\,d\xi\,d\tau+J_0.
\end{split}
\end{align}

The goal is to show the existence of a fixed point $J^*=\mathcal{M}(J^*)$ in a suitable space. This will lead to a weak solution $(\theta^{J^*},u^{J^*})$ for \eqref{sysf} and \eqref{sysw}.

We first give a uniform a priori energy estimate for $\mathcal{E}(t)$ defoined in \eqref{calEdef}.
\begin{thm}\label{lemma3.2} For any fixed $T>0$ and for any weak solution $(u(x,t),\theta(x,t))$ of system \eqref{sysf} and \eqref{sysw}, we have, for $t\in[0,T]$,
\beq\label{engineq2}
\mathcal{E}(t)\leq \mathcal{E}(0) -\iint_{\mathbb {R}\times [0,t]}(\frac{v_t^2}{g^2}+\theta^2_t)\,dxdt.
\eeq
\end{thm}
The proof is the same as the one in \cite{CHL20}. One can find it in the appendix \eqref{appE}.
\begin{cor}
For any weak solution of system \eqref{sysf} and \eqref{sysw}, there exists a constant $C_0$ depending on $\mathcal{E}(0),\|J'_0\|_{L^2(\mathbb {R})},$ and $\|J_0\|_{L^1(\mathbb {R})}$, such that
\beq\label{Gfbound}
\|G\|_{L^\infty(\Omega_T)}\leq C_0,\qquad \|f\|_{L^\infty([0,T],L^2(R))}\leq C_0,\eeq
\end{cor}
The proof is straightforward using \eqref{calEdef}, \eqref{tecinitial}, \eqref{positiveDamping},
and definition of $G$ and $f$ in \eqref{Fun}.

Now we fix an arbitrary $T>1$ once and for all and
consider the following spaces over $\overline{\Omega}_T:=\mathbb {R}\times [0,T].$ Denote
\begin{align*}
    L_*:=\cap_{p\in[a+2,\infty)}L_*^p,
\end{align*}
for fixed $a>0$, and
\begin{align*}
    N_T:=C_*^\alpha\cap L_*^2\cap L_*^\infty(\overline{\Omega}_T),
\end{align*}
with
\begin{align*}
 \|S\|_{N_T(\overline\Omega_T)}=\max\left\{\|S\|_{L_*^\infty(\overline\Omega_T)}, \|S\|_{L_*^2(\overline\Omega_T)}, \|S\|_{C_*^\alpha(\overline\Omega_T)}\right\},
\end{align*}
where $\alpha\in (0,1/4)$ and
\[\|S\|_{L_*^\infty(\overline\Omega_T)}=\|e^{-\lambda t}S(x,t)\|_{L^\infty(\overline\Omega_T)},\]

\[\|S\|_{L_*^p(\overline\Omega_T)}=\|e^{-\lambda t}S(x,t)\|_{L^p(\overline\Omega_T)},\]
and

\[\|S\|_{C_*^\alpha(\overline\Omega_T)}=\sup_{\|(h_1,h_2)\|>0}e^{-\lambda t}\frac{|S(x+h_1,t+h_2)-S(x,t)|}{\|(h_1,h_2)\|^\alpha},\]
for some $\lambda=\lambda(\mathcal{E}(0),T)>0$ sufficiently large that will be determined later. Let
\beq\label{kT}
k_T=2(\|J_0\|_{C^\alpha\cap L^2\cap L^\infty}+\max\{C_0,C_0^2\}T^2).
\eeq
We define
\begin{align*}
    K_T&:=\big\{\mathcal{S}(x,t)\in N_T:\|\mathcal{S}\|_{N_T}\leq k_T,\, \mathcal{S}(x,0)=J_0(x)\big\}.

\end{align*}
By \eqref{tecinitial} and the Sobolev embedding, it is clear that $\|J_0(x)\|_{N_T(\mathbb {R})}<\|J_0(x)\|_{C^\alpha\cap L^2\cap L^\infty}.$

Furthermore, for any fixed $T$, it is easy to show that the $N_T$ norm and the $C^\alpha\cap L^2\cap L^\infty$ norm are equivalent. In \cite{BZ}, a similar norm was used to prove the existence.
\begin{cor}
    $K_T$ is compact in $L_*$ on any $\overline\Omega_T.$
\end{cor}
\begin{proof} This can be proved using  a very similar method as in Section 6.3 of \cite{CHL20}.
\end{proof}

We now recall the Schauder Fixed Point Theorem that will be applied to complete our analysis.

\begin{thm}[Schauder Fixed Point Theorem]\label{sch}
Let $E$ be a Banach space, and let $K$ be a convex set in $E$. Let  $\mathcal T: K\to K$ be a continuous map such that $\mathcal T(K) \subset K$, where
$K$ is a compact subset of $E$. Then $\mathcal T$ has a fixed point in $K$.
\end{thm}

The main step  is to verify the two assumptions of Theorem \ref{sch}, that is
\bigskip
\begin{enumerate}[i.]
    \item \emph{The continuity of the map $\mathcal{M}:K_T\to K_T.$}\smallskip\\
    This can be verified using the same argument as in \cite{CHL20}. The idea is to use the change of coordinates and the semi-linear system introduced previously along with the regularity of the transformation that preserves the continuity of the map. We refer the reader to \cite{CHL20}.
    \bigskip
    \item \emph{The inclusion $\mathcal{M}(K_T)\subset K_T.$
    }
 \end{enumerate}
 Now we prove (ii).\\

 The following proposition, with the help of \cite[Propositions 1.1 and 1.2]{MS}, is the key estimate to show that for any $T>0$ we have the inclusion \[\mathcal{M}(J):K_T\to K_T.\]

To state the proposition, let $G\in L^\infty(\Omega_T)$ and $f\in$ $L^\infty((0,T),L^2(\mathbb {R}))$ and define
\begin{align*}
    M_{G}(x,t):=\int_0^t\int_{\mathbb {R}}\Gamma(x,t;\xi,\tau)G(\xi,\tau)\,d\xi\,d\tau,
\end{align*}
\begin{align*}
M_{f}(x,t):=\int_0^t\int_{\mathbb {R}}\Gamma(x,t;\xi,\tau)f(\xi,\tau)\,d\xi\,d\tau.
\end{align*}
and
\begin{align*}
    M_{G,x}(x,t):=\int_0^t\int_{\mathbb {R}}\Gamma_x(x,t;\xi,\tau)G(\xi,\tau)\,d\xi\,d\tau,
\end{align*}
\begin{align*}
M_{f,x}(x,t):=\int_0^t\int_{\mathbb {R}}\Gamma_x(x,t;\xi,\tau)f(\xi,\tau)\,d\xi\,d\tau.
\end{align*}
By the upper bounds established in \cite[Sections 2 and 4]{MS} and then by \eqref{Gfbound},
\begin{align}\label{infbound}
\begin{split}
\|M_{G,x}\|_{L^\infty(\Omega_ T)}&\lesssim  T^{1/2}\|G\|_{L^\infty(\Omega_T)}\leq C_0T^{1/2},\\
\|M_{f,x}\|_{L^\infty(\Omega_ T)}&\lesssim  T^{1/4}\|f\|_{L^\infty((0,T),L^2(R))}\leq C_0T^{1/4},
\end{split}
\end{align}
\begin{align}\label{alphbound}
\begin{split}
\|M_{G,x}\|_{C^\alpha(\Omega_ T)}&\lesssim  T^{r}\|G\|_{L^\infty(\Omega_T)}\leq C_0 T^r,\\
\|M_{f,x}\|_{L^\alpha(\Omega_ T)}&\lesssim  T^{s}\|f\|_{L^\infty((0,T),L^2(R))}\leq C_0 T^s,
\end{split}
\end{align}
for some fixed $0<r,s<1.$ And
\begin{align}\label{L2bound}
\begin{split}
 \|M_{G,x}\|_{L^2(\Omega_ T)}&\leq \frac{1}{2} T^{3/2} \|G\|_{L^\infty(\Omega_T)}\|G_x\|_{L^\infty((0,T),L^1(\mathbb {R}))}\leq C^2_0 T^{3/2},\\
    \|M_{f,x}\|_{L^2(\Omega_ T)}&\leq  T^{1/2}\|f\|_{L^\infty((0,T),L^2(R))}\leq C_0T^{1/2}.
\end{split}
\end{align}

Using these estimates, \eqref{Gfbound} and the equivalence of $N_T$ norm and $C^\alpha\cap L^2\cap L^\infty$ norm when $T$ is given, it is easy to prove the following proposition.
\begin{prop}\label{mainlemma} Assume $f$ and $G$ satisfy the bounds in \eqref{Gfbound}. For any given $T>1$, we have
$$\displaystyle \max\{\|M_{G,x}(x,t)\|_{N_ T},\|M_{f,x}(x,t)\|_{N_ T}\}\leq \max\{C_0,C_0^2\}T^2.$$

\end{prop}

The map $\mathcal{M}$ in \eqref{5.6} contains two terms in the integration. The first term is $F^J:=G+f$ and the second term is $\displaystyle g'(\theta^J)\theta^J_\xi J\in L^\infty((0,T),L^2(\mathbb {R})).$ Proposition \eqref{mainlemma} gives a uniform bound on the first term, where the bound is less than $k_T,$ chosen in \eqref{kT}.

To control the second term, a special treatment needed due to the extra explicit dependence on $J.$
Denote
\[Q(x,t):=\int_0^t\int_\mathbb {R} \Gamma_x(x,t,\xi,\tau)g'\theta_\xi(\xi,\tau)J(\xi,\tau)\,d\xi d\tau.\]
We write
\begin{align*}
    \big|\int_0^t\int_\mathbb {R}& e^{\lambda\tau} \Gamma_x(x,t,\xi,\tau)g'\theta_\xi(\xi,\tau)e^{-\lambda\tau}J(\xi,\tau)\,d\xi d\tau\big|\\
    &\lesssim \|e^{-\lambda\tau} J\|_{L^\infty(\Omega_T)}\,\mathcal{E}(0) \bigg[\int_0^t\int_\mathbb {R}(t-\tau)^{3/4} \Gamma_x^2(x,t,\xi,\tau)\,d\xi\,d\tau\bigg]^{1/2}\bigg[\int_0^t\frac{e^{2\lambda\tau}}{(t-\tau)^{3/4}}\,d\tau\bigg]^{1/2}\\
    &\leq \|e^{-\lambda\tau} J\|_{L^\infty(\Omega_T)}\,\mathcal{E}(0)\,t^{1/8}\bigg[\int_0^t\frac{e^{2\lambda\tau}}{(t-\tau)^{3/4}}\,d\tau\bigg]^{1/2}\\
    &=\|e^{-\lambda\tau} J\|_{L^\infty(\Omega_T)}\,\mathcal{E}(0)\,t^{1/8}\bigg[\int_0^t\frac{e^{2\lambda(t-\tau)}}{\tau^{3/4}}\,d\tau\bigg]^{1/2},
\end{align*}
where we used \eqref{Gamma_xest}.
Multiplying by $e^{-\lambda t},$ we get
\begin{align*}
    \big|e^{-\lambda t}\int_0^t\int_\mathbb {R}& e^{\lambda\tau} \Gamma_x(x,t,\xi,\tau)g'\theta_\xi(\xi,\tau)e^{-\lambda\tau}J(\xi,\tau)\,d\xi d\tau\big|\\
    &\leq \|e^{-\lambda\tau} J\|_{L^\infty(\Omega_T)}\mathcal{E}(0)\,t^{1/8}\bigg[\int_0^t\frac{e^{-2\lambda\tau}}{\tau^{3/4}}\,d\tau\bigg]^{1/2}\\
    &=\|e^{-\lambda\tau} J\|_{L^\infty(\Omega_T)}\mathcal{E}(0)\,t^{1/8}\bigg[\int_0^{1/\lambda}\frac{e^{-2\lambda\tau}}{\tau^{3/4}}\,d\tau+\int_{1/\lambda}^{t}\frac{e^{-2\lambda\tau}}{\tau^{3/4}}\,d\tau\bigg]^{1/2}\\
    &\leq \|e^{-\lambda\tau} J\|_{L^\infty(\Omega_T)}\mathcal{E}(0)\,t^{1/8}\bigg[\frac{1}{\lambda^{1/4}}+\lambda^{3/4}\frac{1}{\lambda}(e^{-1}-e^{-\lambda T})\bigg]^{1/2}\\
    &\leq \sqrt{2} \|e^{-\lambda\tau} J\|_{L^\infty(\Omega_T)}\mathcal{E}(0)\,T^{1/8}\frac{1}{\lambda^{1/8}},
\end{align*}
which yields, for $J\in K_T,$
\begin{align}
    \|Q\|_{L_*^\infty(\Omega_T)}\leq C_T \frac{1}{\lambda^{1/8}} \|J\|_{L_*^\infty(\Omega_T)}\mathcal{E}(0)\leq \sqrt{2}\,T^{1/8} \frac{1}{\lambda^{1/8}} k_T\mathcal{E}(0).
\end{align}
Hence, choosing $\lambda>(2\sqrt{2})^8T\mathcal{E}^8(0),$ we obtain the bound
\begin{align}\label{Linf*}
  \|Q\|_{L_*^\infty(\Omega_T)}\leq k_T/2.
\end{align}
For the $L^2_*$ estimate, using \eqref{Gamma_xest} and \eqref{L2bound}

\begin{align*}
  e^{-\lambda t}|Q(x,t)|&\leq e^{-\lambda t}\int_0^t\int_\mathbb {R} \frac{1}{{t-\tau}}e^{-\frac{d(x-\xi)^2}{4(t-\tau)}}|g'||\theta_\xi(\xi,\tau)||J(\xi,\tau)|\,d\xi d\tau\\
  &\leq \|g'\|_{L^\infty} \|J\|_{L_*^\infty}\int_0^t\int_\mathbb {R} \frac{1}{{t-\tau}}e^{-\frac{d(x-\xi)^2}{4(t-\tau)}}|\theta_\xi(\xi,\tau)|e^{\lambda(\tau-t)}\,d\xi d\tau\\
&=\|g'\|_{L^\infty}\|J\|_{L_*^\infty}\int_0^t\int_\mathbb {R} \frac{1}{{\tau}}e^{-\frac{d\,\xi^2}{4\tau}}|\theta_\xi(x-\xi,t-\tau)|e^{-\lambda\tau}\,d\xi d\tau.
\end{align*}
Taking the $L^2$ in $x,$
\begin{align*}
    \|e^{-\lambda t}Q(x,t)\|_{L^2(\mathbb {R})}&\leq \|g'\|_{L^\infty} \|J\|_{L_*^\infty(\Omega_T)} \|\theta_\xi\|_{L^\infty((0,T),L^2(\mathbb {R}))}\int_0^t\frac{1}{\sqrt{\tau}}e^{-\lambda \tau}\,d\tau\\
    &\leq \|g'\|_{L^\infty}\|J\|_{L_*^\infty} \mathcal{E}(0)\frac{3}{\sqrt{\lambda}}.
\end{align*}
This gives
\begin{align}
    \|Q\|_{L^2_*(\Omega_T)}\leq \|g'\|_{L^\infty}k_T \mathcal{E}(0)\frac{3}{\sqrt{\lambda}}\,\sqrt{T}.
\end{align}
Hence, for $\lambda>36\,\mathcal{E}(0)^2\|g'\|_{L^\infty}^2T$ we have
\begin{align}\label{L2*}
    \|Q\|_{L^2_*(\Omega_T)}\leq k_T/2.
\end{align}
By a similar argument one can show the existence of $\lambda>0$ depending only on $\mathcal{E}(0),\,T$ and bound of $g$ and $g'$ such that
\begin{align}\label{alpha*}
    \|Q\|_{C^\alpha_*(\Omega_T)}\leq k_T/2.
\end{align}
\begin{remark}
    The weight $e^{-\lambda t}$ introduced in the norm $N_T$ helps to get the inclusion of the map $\mathcal{M},$ particularly, for the term $Q$ treated above. One can see, for example, the $L^\infty_*$ estimate; we got \[\|Q\|_{L^\infty_*}\leq C_T\frac{1}{\lambda^{1/8}} \|J\|_{L^\infty_*}\mathcal{E}(0).\]
    This allows us to choose $\lambda=\lambda(T,\mathcal{E}(0))$ such that $\|Q\|_{L^\infty_*}<\frac{1}{2}  \|J\|_{L^\infty_*}.$ In other words, the size of $\|Q\|_{N_T}$ shrinks faster than the size of $\|J\|_{N_T}$ as $\lambda$ gets large.
\end{remark}

Now, Proposition \ref{mainlemma} with \eqref{Linf*},\eqref{L2*} and \eqref{alpha*} show that for any fixed $T>1,$
there exists $\lambda>0$ large enough (depending only on $T, \mathcal{E}(0), \|g\|_{L^\infty}$ and $\|g'\|_{L^\infty}$)
such that $$\mathcal{M}(K_T)\subset K_T.$$ Hence, by the Schauder fixed point theorem, we have a fixed point $J^*=\mathcal{M}(J^*)\in K_T.$ Since for a fixed finite time $T$ the $N_T$ norm and the $L^2\cap L^\infty\cap C^\alpha(\Omega_T)$ norm are equivalent,  $J^*\in L^2\cap L^\infty\cap C^\alpha.$ Therefore, we have the existence of $A(x,t)$ by \eqref{Aexpression}.

The next proposition shows that the weak formulation is well-defined. Before that we will introduce $A^\ve(x,t)$ to be the solution to
\begin{align}\label{Aepsilon}
    A^\ve_t-g(\theta)A^\ve_{xx}+\gamma_1A^\ve=F^{\epsilon}+g'\theta_x^\epsilon J.
\end{align}
Here $F^\epsilon,$ defined in \eqref{Fun}, and $\theta_x^\epsilon$ are smooth mollification where the mollification of $F$ is acting only on $\theta_x$ and $\theta_t$. Precisely, $F^\epsilon=F(\theta,\theta_t^\epsilon,\theta_x^\epsilon,v_x)$. The mollification is not needed for $v_x$ since we have shown $\theta\in C^{1/2}(\Omega_T)$ and $v_x\in C^\alpha(\Omega_T)$. Hence, the classical theory discussed in Section \ref{HK} applies.

\begin{prop}\label{propA}
For any $T\in (0,\infty),$ there exists a function $A(x,t)$ such that $$A_x\in L^2\cap L^\infty\cap C^\alpha([0,T],\mathbb {R})$$
and
\begin{align}\label{Aweak}
    \int_0^T\int_\mathbb {R} A\phi_t-g(\theta) A_x\phi_x-\gamma_1 A\phi\,dx\,dt=-\int_0^T\int_\mathbb {R} (F+g'(\theta)\theta_x J_0)\phi\,dx\,dt,
\end{align}
for any $\phi\in C^\infty_c(\mathbb {R}\times (0,T)).$ More precisely, the weak formulation \eqref{Aweak} is satisfied as a limit of $A^\ve.$
\end{prop}
\begin{proof}
First, for short, we denote $F^{J^*}$ by $F$ and $\theta^{J^*}$ by $\theta.$
The fixed point $J^*$ gives
\begin{align}\label{Express4A}
A(x,t)=\int_0^t\int_\mathbb {R}\Gamma(x,t;\xi,\tau)\big[F+g'(\theta)\theta_\xi J^*\big]\,d\xi\,d\tau.
\end{align}
Now we show that $A$ satisfies \eqref{Aweak}.
We can write the classical solution $A^\epsilon$ of \eqref{Aepsilon} explicitly as
\begin{align}
    A^\epsilon(x,t)=\int_0^t\int_\mathbb {R}\Gamma(x,t;\xi,\tau)\big[F^{\epsilon}+g'\theta_\xi^{\epsilon}J^*\big]\,d\xi\,d\tau.
\end{align}
The weak formulation of the solution to \eqref{Aepsilon} is defined as the following
\begin{align}\label{eweak}
    \int_0^T\int_\mathbb {R} A^\epsilon \phi_t- A^\epsilon_x(g\phi)_x-\gamma_1 A^\epsilon\phi\,dx\,dt=-\int_0^T\int_\mathbb {R} \big[F^{\epsilon}+g'\theta_x^{\epsilon}J^*\big]\phi\,dx\,dt
\end{align}
since $A^\epsilon$ is a classical solution.
The same argument as in Section \ref{J2v} can be applied here to show that $A^\ve, A_x^\ve\to A,A_x$ in the $L^2_{loc}$ sense. This means
taking $\epsilon\to 0$ in \eqref{eweak} and using $J^*-J_0=A_x$ we obtain \eqref{Aweak}.

An application of Proposition \ref{mainlemma} shows

$$\displaystyle A_x\in L^\infty\cap L^2\cap C^\alpha([0,T],\mathbb {R}).$$
\end{proof}
\begin{cor}\label{hatweak}
The function $\hat A(x,t):=A(x,t)+A_0(x)$ satisfies the following identity
  \begin{align}
    \int_0^T\int_\mathbb {R} \hat A \phi_t-g(\theta)\hat A_x\phi_x-\gamma_1 \hat A\phi\,dx\,dt=&-\int_0^T\int_\mathbb {R} \hat F\phi\,dx\,dt
\end{align}
where
\begin{align}\begin{split}
\hat F&=\hat f+\hat G,\\
\hat f&=[\gamma_1-\frac{h^2}{g}]v_x+\frac{h(\theta)c^2(\theta)}{g}\theta_x,\\
\hat G&=\int_{-\infty}^x[\frac{h'}{g}-\frac{g'h}{g^2}]\theta_t^2-[\gamma_1-\frac{h^2}{g}]'\theta_zv_z-(\frac{h(\theta)c(\theta)}{g})'c(\theta)\theta_z^2\,dz.
    \end{split}
\end{align}
\end{cor}
\begin{proof}
Using $A=\hat A(x,t)-A_0$ and \eqref{Aweak}, we have
\begin{align}
    \int_0^T\int_\mathbb {R} \hat A \phi_t-g(\theta)\hat A_x\phi_x-\gamma_1 \hat A\phi\,dx\,dt=&\int_0^T\int_\mathbb {R} g(\theta) J'_0\phi-\gamma_1 A_0\phi-F\phi\,dx\,dt\nonumber\\
    =&-\int_0^T\int_\mathbb {R} \hat F\phi\,dx\,dt\nonumber
\end{align}
\end{proof}

\section{Weak formulations}\label{bdE}
To summarize, from Propositions \ref{propth}, \ref{propv}, \ref{propA} and Corollary \ref{hatweak} we have proved that $(\theta,v,A)$ defines a weak solution in the sense that, for any test function $\phi\in H_0^1((0,T)\times\mathbb {R})$,
     \begin{align}\label{thetaweakk}
    \int_0^T\int_\mathbb {R}\theta_t\phi_t-\big(\gamma_1-\frac{h^2}{g}\big)\theta_t\phi\,dx\,dt=\int_0^T\int_\mathbb {R} (c(\theta)\theta_x)(c(\theta)\phi)_x+h\hat A_x\phi\,dx\,dt,
\end{align}
\begin{align}\label{vvweak}
    \int_0^T\int_\mathbb {R} v\phi_t-v_x(g\phi)_x+h\theta_t\phi\,dx\,dt=0,
\end{align}
\begin{align}\label{AAweak}
    \int_0^T\int_\mathbb {R} \hat A \phi_t-g(\theta)\hat A_x\phi_x-\gamma_1 \hat A\phi\,dx\,dt=-\int_0^T\int_\mathbb {R} \hat F\phi\,dx\,dt.
\end{align}

Recall,
\begin{align}\begin{split}
\hat F&=\hat f+\hat G,\\
\hat f&=[\gamma_1-\frac{h^2}{g}]v_x+\frac{h(\theta)c^2(\theta)}{g}\theta_x,\\
\hat G&=\int_{-\infty}^x[\frac{h'}{g}-\frac{g'h}{g^2}]\theta_t^2-[\gamma_1-\frac{h^2}{g}]'\theta_zv_z-(\frac{h(\theta)c(\theta)}{g})'c(\theta)\theta_z^2\,dz.
    \end{split}
\end{align}

Now we show that $(u,\theta)$ with $u=v_x$ satisfies the requirement in definition \eqref{def1} for a weak solution.
By \eqref{vvweak}, for any test function $\eta\in C_c^\infty(\Omega_T)$, choose $\phi=\eta_x$,  we have
\begin{align}
    \int_0^T\int_\mathbb {R} u\eta_t-(gu_x+h\theta_t)\eta_x\,dx\,dt=0.
\end{align}

Next we establish the relation between $v_t$ and $J$. Precisely, we show that
\begin{align}
    J-\frac{v_t}{g}=0\;\mbox{ almost everywhere.}
\end{align}
Taking $\displaystyle\phi=\frac{\psi_t}{g}$ in \eqref{vweak}, we have
\begin{align}\label{0}
    \int_0^T\int_\mathbb {R} -v_t\frac{\psi_t}{g}-v_t \psi_{xx}+h\theta_t \frac{\psi_t}{g}\,dx\,dt=0,
\end{align}
which is
\begin{align}\label{1}
    \int_0^T\int_\mathbb {R} \frac{v_t}{g}(\psi_t+g\psi_{xx})-(\frac{h\theta_t }{g}) \psi_t\,dx\,dt=0.
\end{align}
By \eqref{thetaweakk}, if we choose $\displaystyle \phi=\frac{h}{g}\psi,$ then
 \begin{align}\label{2}\begin{split}
  &  \int_0^T\int_\mathbb {R}\theta_t\frac{h}{g}\psi_t+(\frac{h}{g})'\theta^2_t\psi-\big(\gamma_1-\frac{h^2}{g}\big)\theta_t\frac{h}{g}\psi\,dx\,dt\\
  &\qquad  =\int_0^T\int_\mathbb {R} (c\theta_x)(c\frac{h}{g}\psi)_x+hJ\frac{h}{g}\psi\,dx\,dt.
  \end{split}
\end{align}
Finally, for \eqref{AAweak}, we choose $\phi=\psi_x$ to get
\begin{align}\label{3}
    &\iint -J\psi_t-gJ\psi_{xx}+\gamma_1J\psi\,dx\,dt\\
    &\qquad =\iint (\frac{h}{g})'\theta_t^2\psi-[\gamma_1-\frac{h^2}{g}] v_{xx}\psi
    -(\frac{hc}{g})'c\theta_x^2\psi-\frac{hc^2}{g}\theta_x\,dx\,dt.\nonumber
\end{align}
Adding \eqref{1}-\eqref{3}, and using that
\[
\iint (\frac{h^2}{g}-\gamma_1)\frac{1}{g}\psi(v_t-gv_{xx}-\theta_t h)\,dx\,dt=0,
\]
we obtain
\begin{align}\label{identity}
 \iint (J-\frac{v_t}{g})(\psi_t+g\psi_{xx}+(\frac{h^2}{g}-\gamma_1)\psi)\,dx\,dt=0.
\end{align}
Denote $\displaystyle a:=J-\frac{v_t}{g}.$ Now we show that the weak solution of
\[
a_t-(ag)_{xx}-\beta a=0\; \mbox{ where }\; \beta=\frac{h^2}{g}-\gamma_1
\]
with $a(x,0)=0$, has only zero solution almost everywhere. To prove that, let $\displaystyle m=ga$, so the problem becomes
\[
m_t-g m_{xx}-(\beta+\frac{g_t}{g})m=0
\]
with $m(x,0)=0$. We need to prove that $m=0$ almost everywhere is the only solution.
    Applying same arguments from Section \ref{HK}, we have an integral formula of the solution and we write the map of solutions as
\[\hat m(x,t)=\int_0^t\int_\mathbb {R}\Gamma(x,t;\xi,\tau)\big(\big[\beta+\frac{g_t}{g}\big]m\big)(\xi,\tau)\,d\xi\,d\tau.\]
Following the same fixed point arguments in Section \ref{FixedPoint}, we have the following estimate for some short time $t$ and some constant $r>0$
\[\|m\|_{L^\infty}<t^r \|m\|_{L^\infty}\]
since $m(x,0)=0$, so $m=0$, a.e. when $t$ is small enough, then for any $t$. Hence,
\[
J=\frac{v_t}{g} \quad \hbox{almost everywhere.}
\]
Furthermore, by \eqref{vvweak} and choosing $\phi=\frac{\varphi}{g}$, we have
\begin{align}\label{4}
    \int_0^T\int_\mathbb {R} \frac{v_t}{g}\varphi+u_x\varphi+\frac{h\theta_t }{g}\varphi\,dx\,dt=0.
\end{align}
So
\[
J=\frac{v_t}{g}=u_x+\frac{h\theta_t}{g} \quad \hbox{almost everywhere.}
\]
This also shows that $u_x\in L^\infty([0,T],L^2(\mathbb {R}))$.

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Q: IdentityServer4 storing Client Secrets I'm currently trying to build an single sign-on Server for a couple of clients to use. Because I don't exactly know, how many clients that will be, I planned to make it so I can add clients at runtime using the EntityFramework Configuration Store.
My problem is now how to set the client secrets. I tried generating a new GUID and using that as a secret. The problem now is, that the Configuration Store just wants to save the hashed version of the secret and I would need to access the plain secret to add it to the actual client application.
I assume that this is on purpose and that it is discouraged to save the plain version of the secret? What would be the go-to solution for saving secrets?

A: Use following algorithm to generate sha256 hash. This is the same algorithm used in IdentityServer4.Models.HashExtensions class.
using System.Security.Cryptography;

static class Extentions
{

    public static string Sha256(this string input)
    {

        using (SHA256 shA256 = SHA256.Create())
        {
            byte[] bytes = Encoding.UTF8.GetBytes(input);
            return Convert.ToBase64String(((HashAlgorithm)shA256).ComputeHash(bytes));
        }
    }
}

void Main()
{
    Console.WriteLine( "secret-as-guid".Sha256());
}

A: You should not store the client secret in plain text.
Always assume that your configuration database gets compromised - and then those secrets can be used to impersonate your clients.
This might be slightly inconvenient for you - but it is a best practice (and also in-line with how other token services deal with that).
If you have other means of protecting the secret at rest - you can add the the plain text based secret validator to DI
https://github.com/IdentityServer/IdentityServer4/blob/dev/src/IdentityServer4/Validation/PlainTextSharedSecretValidator.cs

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Properties and Applications: blue purple, red light purple.
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X-Factor (2nd series) #2
Jeff Jensen (writer), Arthur Ranson (artist), Paul Mounts (colors), Paul Tutrone (letters), Franco and Yoshii (assistant editors), Powers and Lis (editors), Joe Quesada (editor in chief), Bill Jemas (president)
Eighteen months after having his right arm desiccated by a mutant arms dealer, Agent Kearse finally has an operation that replaces his useless limb with a fully functional artificial one. After a period of recovery, he and his partner, Catherine Gray, are called into their new assignment; investigating death threats being directed at baseball homerun star Tony Robb. While secretly a mutant, Robb has been receiving threats from someone identifying him as such, and instructing him not to take the homerun record. With heightened security, Robb continues his playing, intending to reveal his mutant nature, and retire, after the game where he breaks the record. As he nears his goal of a 73rd homerun, Robb panics and instinctively uses his mutant power to alter his apparent location. This moment actually saves his life, as the would-be assassin misses Robb's vital organs. Agents Gray and Kearse corner the shooter, but find that he has already been subdued by the X-Man, Nightcrawler, who disappears in a puff of brimstone smelling smoke. Robb quickly learns that he is paralyzed from the waist down, and calls off revealing his secret. The secret still comes out, ruining his intended status as a "goodwill ambassador to mutantkind". As Catherine Gray deals with her mother, a rabid mutant-rights advocate with conspiracy theories, and her husband's questions about their status as possible future parents, Kearse deals with the fallout of his son, an avid Robb fan, learning Robb's mutant nature. As he throws out his son's Robb memorabilia collection, Kearse thinks of his own father teaching his to play baseball, and how after 18 months, it will be good to feel strong again.
As he stands in the batter's box, awaiting his pitch, Tony Robb wonders how they will come for him. A rifle short, high and hard like a fastball at his head? Or a snaking curve, sneaking out of the corner of his eye? Or will it be something impossible to anticipate, something sinking and wicked, split-fingered and dirty? With a fluid motion, the pitcher releases his pitch, sending the ball toward the strike zone. With skill honed by countless practicing, Robb swings his bat, hitting the ball squarely. As the ball soars into the night sky, he thinks that however they come at him, he knows this for certain: he will keep swinging. They greatest home run hitter of his generation will not be deterred from making history. Then, he will be through with the damn game.
As Robb's 32nd home run of the year arcs into the third deck of Baltimore's Levinson Field, a horde of fans tries to catch the ball. Out of the legion of arms reaching skyward, it is the arm of Aaron Kearse, his only good arm no less, that reaches high enough to make the catch. With pride and great pleasure, Kearse sits back down and hands his prized catch to his son, Matthew. Matthew continues holding fast to the ball, even as his father carries him back to their car. Rambling on, Matthew asks his father if he knew that Tony Robb reached 500 home runs faster than anyone else in history? Or that he's the only player to be A.L. MVP five years in a row?
Kearse smiles as he basks in his son's excitement of baseball; a passion Kearse himself inherited from his own father. For one blissful moment, he thinks to himself, there is no such thing as mutants. After a quick drive home, Kearse picks up his sleeping son, and takes him upstairs to his room. After being tucked in by him, Matthew asks his father if he is scared of the operation. Kearse replies that he is a little. Matthew promptly offers to pray, if his father wants, to which Kearse replies that he would like that. After thanking his son for the kind thought, Kearse tells him good night, and turn off the light, bringing darkness to the boy's bedroom. It is in the semi-darkness of an operation room sometime later that the surgeon asks Aaron Kearse to begin to count backwards from ten. Kearse begins the countdown, but before he can complete the number seven, the anesthetic takes effect, bringing darkness to his world as well.
(18 months ago)
In Las Vegas, parked along a darkened street, Agents Kearse and Green stand outside their car. Seeing an approaching van, Kearse tells Green that it is he. Leaning against their parked car, Green asks Duncan if they are rolling tape. The electronically transmitted voice of Agent Duncan replies 10-4, and recommends that they be careful. After parking a close distance away, the van's driver gets out and lights a cigarette. He greets the two undercover agents as gentlemen and tells them he hopes that he wasn't too difficult to find. Approaching the newcomer, Kearse replies that his directions were crystal clear. The van driver laughs and says he wishes that all his customers were as capable. Some of them couldn't find the Statue of Liberty even if they flew into it. Not getting a reaction from the two men, the newcomer tells them it was a joke and swiftly offers a cigarette. Green replies no thanks and asks if he is alone. The smoking man asks Green if he is trying to make him nervous. Addressing the smoking man as Mr. Wither, Green replies that he's just making sure that there'll be no surprises. Wither replies that he's sure and, opening the van, asks if they would like to inspect the merchandise.
The double doors of the van's back opens up. Inside, the agents see a pile of prosthetic arms. Dryly, Green tells Kearse that this is funny. He's an arms dealer. Get it? An arms dealer. Responding as equally dry, Kearse states that the problem is that they're there for rocket launchers, not prosthetic limbs. Standing behind the two agents, Wither drops his cigarette and then looks at the two agents with glowing eyes. "Perhaps, not yet," he says, responding to Kearse's statement.
The right arms of both Agents Kearse and Green explode in pain. Wither tells them that the pain they are now experiencing is the muscle tissue of their right arms shriveling and dying on the bone. Coldly, he says to them that they didn't know he could do that, did they? As the two men
collapse to their knees and then roll onto the ground, Wither calmly walks to his van and picks up two prosthetic arms and drops them on the ground. In calm contrast to the agents' agony, Wither tells them that intelligence is as invaluable to his line of work as it is to theirs. Remarking that they are with the Federal Bureau of Investigation, he tells them that he did his homework. As he closes the van doors, Wither points out that if they had done their homework too, they might have learned that he possessed some genetic advantages most human do not. After getting in the van's driver seat and closing the door, Wither tells Green that this is why he can work alone and why the norms will never catch him. Finally, he bids the gentlemen a good evening and
tells them to enjoy the new arms. With this he drives away, leaving the two men still in agony. Through electronic transmissions, Kearse hears Agent Duncan ask what is happening? What is happening?
(present)
Sitting on his bed, Agent Kearse rubs lotion on his right arm. No longer the shriveled husk that it was 18 months ago, the arm is still not the healthy match of Kearse's left limb. From behind, Matthew asks his father if he can touch it. When his father responds with a course no, Matthew asks please, but is rebuked again. After a second please, Kearse tells his son he can't until it heals and begins to bandage the arm. A voice at the door tells the two that it will be a few days more. Kearse looks and, to his surprise, sees his old partner Kendall Green. Amy Kearse, Matthew's mother, who is also standing in the doorway, calls for Matthew to come with her; they have to get him to baseball practice. After reaching his mother, Matthew, decked out in his Eagles little league uniform, turns to his dad and suggests that he and Mr. Green can go to the batting cages next week. They can set their arms at super strength ... Matthew is interrupted by his mother, who says that it is enough and they should go.
Putting on his shirt, Kearse turns to Green and asks if super strength is even possible. Green responds that he wishes, although he can enhance his strength if knows how to "jimmy" the synthetics arm. However, Green continues, being a by-the-book Christian that he is, he will probably want to do it the old-fashioned way. Presenting his old partner a box, Green tells Kearse that he will need this. Opening the box, Kearse finds a "speed bag". As Kearse examines the gift, Green tells him that it is something that helped him get through his physical therapy. Recently, he
adds, he got one of the body-sized ones. For him, it's all about hand-eye at this point.
Placing the bag back inside the box, Kearse tells Green that he would like it if they could do it together. He notes that they haven't really hung out together much since ... Hearing Kearse trailing off, Green continues. He tells Kearse that he's surprised that he waited so long to have the procedure done. For him, it was a no-brainer. Feeling his recently acquired limb, Kearse replies that there was something about it that felt... unnatural. He just wanted to make sure it was the right thing to do, with his faith, with Amy, with Matthew. Green replies that he wouldn't worry about Matthew; it looks like his father just bumped Tony Robb off as his hero; not that he wasn't before, of course. After a slight chuckle from Kearse, Green says speaking of Robb, strangely
enough, he wants Kearse and Catherine to take a look at something when have a moment. As the two shake with their artificial hands, Green tells Kearse that he wants them on it as soon as they're ready. He then adds that he thinks they will find it most interesting.
That night, Kearse and his wife, Amy, sit in bed together reading. Amy reads The Divine Conspiracy, while Kearse leafs through the documents that Green gave earlier. One of the papers is a statement from Louis P. Handlesman, Manager/Lawyer for Tony Robb. It is regarding
Robb's mutant background and death threats. As he reads on, Kearse learns that Robb discovered he was a mutant at age five, when he awoke one night to find his skin shimmering with gray oil. Apparently inherited from his father, Robb found that he would alter how light
reflected off of his skin, allowing him to literally control how other people perceived him. A man of principle, Robb's father, Wesley, had never used his power other than to hide the oily secretion that made his ability possible. As this control required incessant concentration, Wesley taught his son mental toughness through athletics, beginning with baseball. Tony quickly took to the sport and became very good at it, turning it into a professional career.
Standing behind the batter's box on the baseball field, Louis Handlesman, Tony's agent, speaks with Agents Gray and Kearse. She explains to them that Tony's father kept Tony very sheltered and managed nearly every aspect of Tony's career and life in order to protect him from the mutant hating world. Driving home the point, Handlesman points out that he himself didn't know, and he's been the Robb family lawyer for 30 years. He, however, was told last summer by Tony after the death of his father. Probing the very reason why they are there, Agent Gray states that it is apparently not a secret anymore. Holding out a piece of paper, Handlesman tells Gray that the first one came in two months ago. He states that it could be some crank who's closer to the truth than he knows, but given the new strategy, they thought that they should get the FBI involved. Taking the paper, Gray unfolds it and reads the contents. In large letters cut from various magazines and newspapers, the message reads "break 70 and you die mutie scum". Seeing this, Kearse brings asks what Handlesman meant by new strategy. Handlesman tells the agents that Tony has grown tired of living in the closet and has decided to go public. Further, he has decided that it is his responsibility to live openly and champion mutant rights. Putting the plan into perspective, Handlesman states that Robb could be to mutants what Muhammed Ali was to blacks.
Continuing, Handlesman says that they've got a marketing plan as well as endorsements from the X-Men and the Minority Action Network. When Agent Gray recognizes the network, referring to them as TMAN, Handlesman acknowledges that they are a bunch of kooky 60's radicals, but they carry weight with the baby boomers and make them remember when they actually cared about civil rights. After Kearse asks when they announce, Handlesman replies after Robb breaks the record. Afterward, Robb will announce his immediate retirement from baseball so he can pursue his new calling as "the goodwill ambassador for mutantkind". Promptly, Gray tells the agent that they're going to need the names of everyone in the loop. Obviously, she says, the investigation begins there. After Handlesman replies done, he asks if there is anything else. Gray
replies, not at the moment. Looking over at Kearse who is looking out into the field, Gray asks if he has any questions. Not turning around, he tells Gray that he has nothing.
As they walk to their cars, Agent Gray asks why she gets the feeling the only cause getting championed is the growth of Robb's wallet. After Kearse replies with an absentminded, right, Gray asks what is bugging him. Kearse replies that he's just trying to figure out how he's going
to tell Matthew. He practically worships Robb. Now that he's a mutant ... As Kearse trails off, Gray asks why would that crush his kid? Is this, like a bad thing? Kearse responds that it is not; it's just ... surprising. With this he tells her that he will see her tomorrow and gets into his car. Unconvinced, Gray mutters sure, whatever, and gets into her own car.
As she sits behind the wheel, Gray picks up the Planned Parenthood brochure in the passenger's seat and begins to read it. Almost inaudibly, a voice from the back seat asks if he is gone. Is it safe? As an older black woman sits upright and into view of the rearview mirror, Agent Gray tells her that she was wondering if she would show up. Following this, Gray asks her mother what the hell she wants? Innocently, Gray's mother asks if a mother has to have a reason to see her daughter. In turn, Gray coldly repeats her question. Turning her demeanor to the serious, Gray's mother asks her daughter if Handlesman told her about their involvement? About how crucial this could be for the advancement of tolerance? Visibly uninterested, Agent Gray mutters, yeah, yeah, Muhammed Ali, blah, blah, blah.
Undeterred by her daughter's statement, Gray's mother asks if it wouldn't be a real tragedy if instead if becoming the next Muhammed Ali, he became the next Dr. King... if she knows what she means. Gray tells her mother that it is her job to make sure he doesn't. Incredulously, Gray's mother responds that her sources say that the FBI mutant task force is a sham, a front for something insidious. Angrily, Gray asks her mother when she will let this conspiracy crap go? Gray's mother retorts that it isn't crap; it's true. She tried to warn her in Los Angeles through her new associates. Gray responds that she figured it was from her, so she promptly ignored it... because it was from her. Looking out her side window, Gray points out to her mother that she didn't speak to her for five years and finally says hello through a vigilante telepath. Gray's mother tries to explain her actions; her work, it's a crucial time... Tearing up, Gray points out that yes, work is important... but so is she, so was dad. Wiping her tears away, Gray tells her mother to get out. Seconds later, she is gone.
In Matthew's room, filled with memorabilia of Tony Robb, Amy Kearse asks her husband when he is going to tell their son. After he flippantly replies, how about never, Amy points to the room's décor and says that they need to tell him. But how, Kearse asks? While Matthew knows there are mutants, he doesn't know the how and why of it. Speaking almost from his own thoughts, Kearse points out that the Bible doesn't really cover this; it goes against everything they've been taught. Looking to the ground, Kearse tells his wife that the world is changing so fast, faster than he can keep up with. Taking the basket of laundry she has just filled, Amy tells Kearse that she thinks that it's about time she showed her if it's true what they say about men with bionic arms. Walking away, she tells him she will be in the bedroom. With a little better demeanor, Kearse rises and asks rhetorically how he could resist an offer like that?
As time passes, Robb becomes more of the celebrity. After more home runs are added to his personal record, Sports Illustrated gives him a nickname that sticks: Tony Terrific. Meanwhile, Handlesman secretly negotiates long-term contracts with Nike, Coke and McDonald's, all eager to use him to target the burgeoning mutant market. Robb himself spends his free time in "media manners" classes who teach him to use more emotion, be less detatched, more outrageous and outspoken, and not introverted like Pete Samprasy. Through it all, though, Robb thinks if only they knew how little he really cared. All he wants is out. On a balmy summer night, after his 72nd home run, Robb returns home to find a final death threat is there to greet him. In the previous cut-and-pasted letters, the message reads, "it will happen when you round third". As he crumbles the paper in his hand, Robb thinks that he would like to see them try.
In the locker room before the game, Kearse finds Robb sitting alone and somber. Breaking the ice with a joke, Kearse asks Robb if he can trouble him for an autograph. Seeing that the newcomer is Agent Kearse, Robb asks how the Tony Robb deathwatch is going. To this, Kearse replies simply that nobody's dying tonight. Sitting down next to Robb, Kearse tells Robb that he has a question that he could help wondering... has he ever used his mutant power to cheat? Examining a baseball in his hand, Robb replies only once, when he was a kid. It was during a Little League championship, it was the bottom of the ninth, with two outs. The team was down 1 to nothing, and it was his turn at bat. The other team's pitcher, Killer Cody, was pitching the game of his life, and Robb says he knew that he wasn't going to hit off of him. So instead of being the "goat", Robb used his power to make Cody misjudge his location and hit him. His father, Robb says, sat there giving him a look that he will never forget. They lost the game, and he never cheated again. After pausing a moment, Kearse tells Robb that he doesn't have to come out. Kearse begins to snap, but decides against it. A moment later, Kearse is gone.
Later, in front of the 47,312 spectators waiting to see history, Tony Robb steps up to bat. The pitcher sends in a fastball, low and inside ... but not low enough. As he has done 72 times before, Robb sends the ball into left field. Though everyone else believes that Robb has hit #73, Robb knows better and begins to sprint. True to his instinct, the ball has ricocheted off the wall and as Robb rounds third, the left fielder, Tim Hernandez, picks up the ball. Suddenly, the death threat flashes through Robb's mind, and on instinct, Robb's power activates, changing his "strike zone". The first shot, aimed at his chest, passes through his right thigh. The second shot, aimed at his head, smashes into his lower back. It will be years before Tony Robb will walk again.
In the security control room, Agents Gray and Kearse move into action. As they move into the upper deck for the shooter, pointed out by a boy witness, the agents prepare themselves for anything. Closing into the area they believe the shooter to be, Agent Gray detects the smell of
sulfur, which is followed by the sound of a multiple bamfs, and the cries and pleas of someone in pain. Gray and Kearse turn the corner, bearing their guns, and, identifying themselves as FBI, tell whom they have surprised to freeze. Standing over the apparent gunman, is a man... of sorts. Clad in a dark leather uniform, is a creature with blue furry skin, yellow glowing eyes, and a prehensile tail that ends in a triangular point. Before the two agents can react to this apparent mutant's presence, he disappears in puff of smoke and an audible bamf.
The shooter turns out to be Earl Whitacker, a hardcore San Francisco Giants fan who also turns out to be a paranoid schizophrenic. Already believing there was a conspiracy to rob Barry Bonds of his homerun record, Whitacker was told that Robb was a mutant, and he believed it. While all of this is learned in the subsequent FBI investigation, it is never learned how Whitacker managed to sneak a high-powered rifle into Levinson Field. Whitacker himself would only smile and joke: "I am just a patsy. I am just a patsy".
After spending a week in the hospital, trying in vain to feel his legs, Robb realizes that being a "goodwill ambassador to mutantkind" is not what he wants, a normal life is. Greeted by the fans and reporters upon his release from the hospital, Robb makes a short statement to thank his fans. His statement, however, is interrupted by multiple questions from reporters asking him to comment on rumors, ranging from high levels of mutagenic steroids reportedly in his bloodstream to reports of "third species" literature being found in his apartment. As Robb's agent, Handlesman, tries to end the press conference, Agent Gray looks at Kearse and jokingly says, "Say it ain't so Joe. Say it ain't so."
Not immediately going home, Gray drives for hours, consumed by memories of her beautiful baby girl, engulfed in flames of power manifesting too early. Next she thinks of a young boy, crucified on the Hollywood sign and a man too scared to live openly as a mutant. No, she decides, it
would be cruel to bring another life into this world. Entering her apartment, Gray finds her mother sitting on the front porch, in the dark. She tells her daughter that they have learned since her day. The killing fails, they assassinate the character instead. As her mother continues to try to convince her that they will stop at nothing, Gray silently opens the door, enters, and closes the door behind her.
Now leaning against the closed door, Gray takes a moment to herself. Her personal moment ends when a voice tells her that her mother has been out there for hours. Ramon, her husband, stands up from his chair and walks toward her. He tells Catherine that he didn't let her mother in,
assuming that she wouldn't want him to, but did offer a blanket and a cookie. He says that he thought that they would at least be hospitable in their bitterness. As Catherine walks up the steps, Ramon tells her that they need to talk about the baby, and what they are going to do. Still walking, Catherine tells him that she knows they do, but not tonight.
Kearse and his wife never tell Matthew about Robb's secret, but Matthew still learns it in time from television. Not taking it well, he throws out all of his Robb memorabilia, but not before writing "cheater" over the posters. As he throws out his son's formerly prized collection, Kearse thinks about the first time he hit a baseball, when he was five. His father had tried so many times to teach him how to hit, but Kearse swung and missed every time. Although the Louisville Slugger was a bit cumbersome for his size, Kearse finally got he hang of the bat. After many times of closing his eyes, young Kearse had kept them open and watched the ball the whole way.
As the modern Kearse recreated his memory with his son's bat, he recalls the charge of power that traveled down through the wood into his slight frame, and how his father laughed in delight and cheered him. As the ball he hits today soars into the sky, it ruptures and unravels, trailing twine. Basking in the moment, and the memories of moments passed, Kearse thinks of how it will be good to feel strong again.
Agent Catherine Gray, Agent Aaron Kearse (both FBI's Mutant Civil Rights Task Force)
Ramon Gray, Catherine's husband
Catherine Gray's mother
Amy Kearse, Aaron's wife
Matthew Kearse, Aaron's son
Tony Robb, baseball pro
Louis P. Handlesman, Robb's manager/lawyer
Nightcrawler (X-Men)
FBI MCRTF Field Director Kendall Green
Earl Whitacker
Levinson Park players and attendees
Hospital surgeons
In flashbacks:
Wesley Robb, Tony Robb's father
Wither, mutant arms dealer
The title of the story, the Natural, undoubtedly refers to the Robert Redford film where he played a baseball player with exceptional skill.
As of this issue's publication only 17 major leaguers have hit 500 home runs. The record for fastest player to reach this mark is currently Mark McGuire, who reached it in 1999 after 5,487 at-bats.
No major leaguer has ever been named MVP five years in a row, in either the American League or the National League. The closest player ever was Barry Bonds, who was named MVP in the National League in 1990, 1992, 1993 & 2001.
The arms dealer Wither's name is most likely an alias. Given that he is extremely careful in his dealings, it is unlikely he would use his real name. Also, the word wither is an apt description of Wither's mutant power.
Pete Sampras is a tennis player who has repeatedly won championships at Wimbledon and the US Open. He is known for his straightforward nature and his easy-going personality.
When Robb is trying to make his 73rd home run, he is trying to beak the record currently held by Barry Bonds, who hit 73 home runs in one season in 2001.
Catherine's mocking statement of "Say it ain't so, Joe" is a reference to baseball legend Shoeless Joe Jackson. Jackson was an out-fielder in Chicago's team, the White Sox, who made it to the World Series in 1919. This particular World Series was mired in scandal, however, when it was
learned that gamblers bribed eight of the team's players with deliberately losing the series. While Jackson did accept the money, it was never conclusively proven that he did anything to sabotage the team's chances of winning. Nevertheless, he was banned from professional baseball forever. According to legend, after the scandal broke, a disheartened boy approached Jackson upon hearing the news and asked him to "Say it ain't so, Joe."
The reference to 3rd species literature refers to John Sublime's attempt at transplanting organs from mutants into humans and creating, what he refered to, as U-Men. (New X-Men #118 – 120)
Louisville Slugger is a company that has made baseball bats since 1884.
Douglas Mangum.
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Infinity War #2
Wolverine is responsible for killing all of his 6 known children.


text:
Placide Massey, né le  à Tarbes et mort dans la même ville le , était un botaniste et jardinier français du .

Biographie

Enfance et formation
Placide Massey, le grand bienfaiteur de la ville de Tarbes, y naquit le  de Anne Marmouget et de Jean Massey, un maître cordonnier.

Dès 1778, il demeure au numéro 9 de la rue des Grands-Fossés. Il fait ses études à l'École centrale de Tarbes, l'actuel lycée lycée Théophile-Gautier. Il est élève et aide-pharmacien chez Lécussan, au numéro 11 de la place Marcadieu.

À l'âge de 18 ans, il abandonne cet emploi et s'engage dans l'armée. Quelques années après, il quitte l'armée et devient l'adjoint de Louis Ramond (1755-1827).

Botaniste et jardinier reconnu
En 1808, il est nommé intendant des jardins de la reine Hortense (en France et au Royaume de Hollande). Il exécute, sur l'ordre du roi Louis-Philippe, de très importantes plantations d'arbres, dans la plaine de Trianon, de Chèvreloup et au parc de Saint-Cloud.

Il assure aussi, dès 1819, la direction du potager du roi de Versailles jusqu'en 1848.

Installation à Tarbes
En 1829 et 1852, il achète, à Tarbes, des terrains pour créer un grand jardin et y plante des arbres.

En 1850, il prend sa retraite et fait bâtir sa maison et continue la plantation d'arbres aux essences rares. Il veut également doter sa ville d'un muséum d'histoire naturelle, et fait construire à cette fin un bâtiment de style oriental, dominé par une tour d'observation sur les Pyrénées, œuvre de l'architecte Jean-Jacques Latour. C'est dans ce bâtiment, inachevé à sa mort, que fut créé le musée Massey.

Mort
Placide Massey meurt à Tarbes le , à l'âge de 76 ans, et est inhumé au cimetière Saint-Jean. Il a légué son jardin, le Jardin Massey, et presque tous ses biens à la ville de Tarbes.

Son testament
« Je donne à la commune de Tarbes, ma ville natale, tous les immeubles que je possède sur son territoire et consistant en jardin d'ornement, pépinière, maisons, prairies, le jardin d'ornement pour servir de promenade ; la pépinière pour continuer la culture d'arbres fruitiers, les maisons et prairies pour employer leurs produits à l'entretien du jardin d'ornement et de la pépinière.
Comme les produits ne peuvent pas être suffisants pour l'entretien du jardin d'ornement à cause du jardin d'hiver que j'y fais construire, je donne en outre, à la commune de Tarbes, soixante actions du Chemin de fer du nord qui produiront environ deux mille francs par an. »

Notes et références

Voir aussi

Bibliographie
 .

Articles connexes
 Jardin Massey
 Musée Massey
 Rue Massey

Liens externes

Naissance à Tarbes
Personnalité française du XIXe siècle
Botaniste français
Personnalité liée aux Hautes-Pyrénées
Naissance en octobre 1777
Décès en novembre 1853
Décès à Tarbes
Décès à 76 ans

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