text:
\section{Introduction}
\subsection{General Overview}
We will be concerned with the one-parameter family of partial differential equations
\begin{equation}\label{GPJa}
\begin{cases}
&u_{txx}+uu_{xxx}-au_xu_{xx}=0,\\
&\left.u\right|_{t=0}=u^0,
\end{cases}
\end{equation}
the \textit{generalized Proudman--Johnson equation} (henceforth abbreviated as GPJ equation), for an unknown, scalar function $u$ with initial condition $u^0$ and a parameter $a\in\mathbb{R}$.
We will study equation \eqref{GPJa} together with either \textit{Dirichlet boundary conditions}, i.e., the function $u:[0,1]\times [0,t^*)\to\mathbb{R}$ satisfies
\begin{equation}\label{DirichletBC}
u(0,t)=u(1,t)=0,\qquad t\in [0,t^*),
\end{equation}
or together with \textit{periodic boundary conditions} (normalizing $u$ to have zero mean), i.e., the function $u:\mathbb{R}\times[0,t^*)\to\mathbb{R}$ satisfies
\begin{equation}\label{periodicBC}
\begin{split}
&u(x+1,t)=u(x,t),\qquad (x,t)\in\mathbb{R}\times [0,t^*),\\
&\int_0^1u(x,t)\, dx=0,\qquad t\in[0,t^*).
\end{split}
\end{equation}
Throughout, we denote the maximal existence time of a solution to equation \eqref{GPJa} as $t^*$.\\
Equation \eqref{GPJa} was introduced in \cite{okamoto2000} for specific parameter values of $a$ as a model derived from high-dimensional Navier--Stokes equations with certain symmetries. Mainly, however, equation \eqref{GPJa} has been introduced as a mathematical extension to the inviscid Proudman--Johnson equation ($a=1$), cf. \cite{childress_ierley_spiegel_young_1989}, which models inviscid, incompressible fluids close to a wall.\\
Indeed, defining the mock vorticity as $\omega:=u_{xx}$, equation \eqref{GPJa} becomes
\begin{equation}
\omega_t+u\omega_x=a\omega u_x,
\end{equation}
which can be interpreted as an $a$-weighted scalar toy-model for the three-dimensional, inviscid, incompressible vorticity equation, cf. \cite{chorin2012mathematical}. For different values of the parameter $a$, either the transport term $u\omega_x$ or the stretching term $\omega u_x$ becomes dominant. Therefore, understanding global existence and blow-up scenarios for equation \eqref{GPJa} is expected to shed light on the role and interplay of nonlinear transport and nonlinear stretching terms in global well-posedness of solutions.\\
Local well-posedness of equation \eqref{GPJa} in the periodic regime was established in \cite{okamoto2000} (cf. also \cite{Okamoto2009}):
\begin{theorem}[Theorem 2.1 in \cite{Okamoto2009}]
For any $u_{xx}^0\in L^2(0,1)/\mathbb{R}$ there exists $T>0$  and a solution of \eqref{GPJa}, satisfying periodic boundary conditions, unique in the class
\begin{equation}
u_{xx}\in C\Big([0,T];L^2(0,1)/\mathbb{R}\Big)\cap C^1_w\Big([0,T];H^{-1}(0,1)/\mathbb{R}\Big),
\end{equation}
where the subscript $w$ implies weak topology. If, in addition, $u_{xx}^0\in H^m(0,1)/\mathbb{R}$ with $m=1,2,...,$ then $u_{xx}\in C^0\Big([0,T];H^m(0,1)/\mathbb{R}\Big)$.
\end{theorem}
Different global-existence and blow-up scenarios for equation \eqref{GPJa} have been analyzed. The following theorem gives an overview of some criteria for periodic solutions.

\begin{theorem}[Theorem 3.3 in \cite{Okamoto2009}]\label{Okamototheorem}
	The following statements hold true:
\begin{enumerate}
	\item Suppose that $a<-2$ and that $\int_0^1\Big(u_x^0(s)\Big)^3\,ds<0$. Then $\|u_x(t)\|_{L^2(0,1)}$ blows up in finite time.
	\item Suppose that $-2\leq a<-1$. Then $u_x$ remains bounded in $L^2$-norm but blows up in finite time in $L^\infty$ norm unless $u\equiv 0$.
	\item Suppose that $-1\leq a<0$ and that $u_{xx}^0\in L^{-\frac{1}{a}}(0,1)/\mathbb{R}$. Then the solution exists globally in time.
	\item Suppose that $0\leq a<1$ and that $u_{xxx}^0\in L^{\frac{1}{1-a}}(0,1)/\mathbb{R}$. Then the solution exists globally in time.
\end{enumerate}
\end{theorem}

In \cite{CHO2010392}, existence of global weak solutions to equation \eqref{GPJa} for the parameter range $a=-\frac{n+3}{n+1},\quad n\in\mathbb{N}$, and in \cite{cho2012global} for a general $a\in [-2,-1)$ has been proved by the method of characteristics, devised in \cite{bressan2005global} in the context of the Hunter--Saxton equation.\\
In \cite{Sarria2013,SarriaSaxton2013} different global existence and blow-up scenarios based on a representation formula for the derivative of the velocity field along trajectories where proved for $u_x$. In \cite{Sarria2013,SarriaSaxton2013} it is first noted that $u_x$ along trajectories satisfies a Riccati differential equation, which is equivalent to a second-order ODE for which then, in turn, knowledge of a special solution implies a representation formula for the general solution.\\
In \cite{ConstantinWunsch2009,Wunsch2011} blow-up conditions for $a=1$ and general $a$ were given by a method based on the time evolution of suprema and infima. For the special case $a=1$ (Proudman--Johnson equation), different, more specific blow-up criteria can be obtained, cf. \cite{childress_ierley_spiegel_young_1989} for various methods related to trajectories.\\
For parameters $a>1$, comparably little is known about the formation of a singularity. In \cite{Okamoto2009} special, non-smooth self-similar solutions of the form $u(x,t)=\frac{F(x)}{T-t}$ with blow-up time $T$ where constructed by solving the ODE
\begin{equation}\label{ODeselfsimilar}
F''+FF'''-aF'F''=0,
\end{equation}
for $a>1$.  For more special blow-up solutions, we also refer to \cite{nagayama2002blow}. Equation \eqref{ODeselfsimilar}, however, only admits non-smooth solution for $a>1$, while numerical computations suggest that also smooth initial conditions become singular in finite time for this parameter range, cf. \cite{Okamotopresentation}.

\subsection{Results of the paper}
In this paper, we prove new criteria on global existence and formation of a finite-time singularity of equation \eqref{GPJa} in the class of $x$-Lipschitz continuous velocity fields. We remark that the presented methods also apply to lower regularity solutions.\\
First, the GPJ equation is reformulated in Lagrangian variables (Section \ref{Reform}), where it takes the form of an inhomogeneous Liouville equation. Subsequently, along the same lines as in Liouville's original paper \cite{liouville1853equation}, we solve the equation for the flow map and present an explicit solution formula - explicit up to the solution of an ordinary differential equation (Section \ref{SolutionFormula}). This allows us to infer several global existence and singularity criteria (Section \ref{Criteria}). Specifically, we can report the following improvements and novelties compared to the results in \cite{okamoto2000}, i.e., the assumptions of Theorem \ref{mainthm} compare to the assumptions of our main theorem, Theorem \ref{Okamototheorem}, as follows.
\begin{enumerate}
		\item  The assumptions in Theorem \ref{mainthm} for the parameter regime $a<-3$ only involve integrability constraints of the second derivative of $u^0$ or a comparison of the minimum and the $L^2$-norm of $u_x^0$, which are weaker as compared to the sign assumption on $\int_0^1\Big(u_x^0(s)\Big)^3\, ds$ in Theorem \ref{Okamototheorem}.\\

		\item The regime where a singularity forms for any sufficiently smooth initial condition is extended from $[-2,-1)$ in Theorem \ref{Okamototheorem} to $[-3,-1)$ in Theorem \ref{mainthm}.\\

		\item Theorem \ref{mainthm} provides assumptions that guarantee global existence for $a>-1$ and assumptions that guarantee the formation of a finite-time singularity for $a\geq 1$, even for smooth initial conditions, for which, to the knowledge of the author, no results have been established so far.\\
\end{enumerate}
 The main theorems are illustrated on several examples. A special emphasis is put on solutions with a parabolic initial velocity field, for which the derivative of the flow map satisfies the constant curvature Liouville equation (Section \ref{constantcurvaturechapter}).\\
 Finally, in Section \ref{Appendix}, we derive the GPJ equation for general parameter values $a\in\mathbb{R}$ from the inviscid, compressible Euler equations.

\subsection{Reformulation in Lagrangian Variables}\label{Reform}
In this section, we will reformulate equation \eqref{GPJa} in Lagrangian variables and, subsequently, solve it explicitly in terms of particle trajectories. \\
First, we define the flow associated to the one-dimensional, time dependent vector field $u$, solving equation \eqref{GPJa}, as
\begin{equation}\label{flowu}
\begin{cases}
&F_t(\xi,t)=u(F(\xi,t).t)=:\hat{u}(\xi,t),\quad (\xi,t)\in[0,1]\times [0,t^*),\\
&F(\xi,0)=\xi, \quad \xi\in [0,1].
\end{cases}
\end{equation}
Here, $t^*$ denotes again the maximal existence time for the flow map. We will assume that the initial condition $u^0$ is sufficiently smooth to guarantee existence and uniqueness of solutions to the ODE \eqref{flowu}. In particular, we will assume local well-posedness in the space of $C^1$-functions with Lipschitz continuous derivative, i.e.,
\begin{equation}
u\in C\Big(0,t^*; C^{1,Lip}(0,1)\Big).
\end{equation}
This allows us to infer that the flow map will have the same regularity as the velocity field in $x$ and, in particular, we can take second derivatives with respect to $x$ almost everywhere. Clearly, also less restrictive assumptions on the velocity field $u$ guarantee the subsequent formulas to hold true.\\
We denote the first and second $x$-derivatives of $u$ along trajectories as
\begin{equation}
\hat{u}_x(\xi,t):=u_x(F(\xi,t),t),\qquad \hat{u}_{xx}(\xi,t):=u_{xx}(F(\xi,t),t),
\end{equation}
where $F$ is a solution to \eqref{flowu}.
Given that the vector field $u$ is $C^1$ with Lipschitz continuous derivative  in $x$ and continuous in $t$, it follows from standard existence and uniqueness theory of ordinary differential equations that the map $F$ defines a local $C^1$-diffeomorphism at any time $t\in [0,\infty)$, which implies that
\begin{equation}\label{posFxi}
F_\xi(\xi,t)> 0, \quad (\xi,t)\in[0,1]\times [0,t^*).
\end{equation}
Indeed, a sign change of $F_\xi$ at some time $t^*$ indicates a break-down of a solution to equation \eqref{flowu}.\\
Assuming Dirichlet boundary conditions \eqref{DirichletBC} the flow map $F$ associated to $u$ has to fixed-points at $\xi=0$ and $\xi=1$, i.e.,
\begin{equation}\label{DirichletF}
F(0,t)=0,\qquad F(1,t)=1,\qquad t\in[0,t^*).
\end{equation}
On the other hand, assuming mean-free periodic boundary conditions \eqref{periodicBC}, the flow map $F(.,t):\mathbb{R}\to\mathbb{R}$ defines a diffeomorphism for every $t\in[0,t^*)$ and we obtain that
\begin{equation}
F(\xi,t)=F(\xi+1,t)-1,\qquad (\xi,t)\in \mathbb{R}\times[0,\infty),
\end{equation}
thanks to the periodicity of $u$ and the uniqueness of the flow map, or, to put it differently, the map $G(\xi,t):=F(\xi,t)-\xi$ is periodic for $(\xi,t)\in\mathbb{R}\times [0,t^*)$. The mean-free condition on $u$ translates to the following constraint for the flow map $F$:
\begin{equation}\label{flowperiodic}
\begin{split}
0&=\int_0^1u(x,t)\, dx=\int_{F^{-1}(0,t)}^{F^{-1}(0,t)+1}u(F(\xi,t),t)F_\xi(\xi,t)\,d\xi\\
&=\int_{F^{-1}(0,t)}^{F^{-1}(0,t)+1}F_t(\xi,t)F_\xi(\xi,t)\, d\xi,\qquad t\in[0,t^*),
\end{split}
\end{equation}
where we have used the definition of $F$ in \eqref{flowu},  the non-degeneracy condition \eqref{posFxi} and the flow property of $F$.\\

By taking a $\xi$-derivatives in \eqref{flowu}, we obtain expressions for the $x$-derivatives of $u$ along trajectories as
\begin{equation}\label{hatux}
\hat{u}_x(\xi,t)=\frac{F_{t\xi}(\xi,t)}{F_{\xi}(\xi,t)}=\frac{\partial}{\partial t}\log F_\xi(\xi,t),
\end{equation}
where we have used \eqref{posFxi} in the definition of the logarithmic derivative. Taking another $\xi$-derivative in \eqref{hatux}, we obtain
\begin{equation}\label{hatuxx}
\hat{u}_{xx}(\xi,t)=\frac{1}{F_{\xi}(\xi,t)}\left(\frac{F_{t\xi}(\xi,t)}{F_{\xi}(\xi,t)}\right)_{\xi}=\frac{1}{F_{\xi}(\xi,t)}\frac{\partial^2}{\partial t\partial \xi}\log F_\xi(\xi,t).
\end{equation}
The time derivative of $\hat{u}_{xx}$ along trajectories is given by
\begin{equation}\label{hatuxxt}
\begin{split}
(\hat{u}_{xx})_t(\xi,t)&=u_{xxt}(F(\xi,t),t)+\hat{u}_{xxx}(\xi,t)F_t(\xi,t)\\
&=u_{xxt}(F(\xi,t),t)+\hat{u}_{xxx}(\xi,t)\hat{u}(\xi,t).
\end{split}
\end{equation}
Evaluating \eqref{GPJa} along $x=F(\xi,t)$ and using \eqref{hatuxxt}, we find that
\begin{equation}
\frac{d}{dt}\hat{u}_{xx}=a\hat{u}_x\hat{u}_{xx},
\end{equation}
which can be solved explicitly to
\begin{equation}\label{solvehatuxx}
\hat{u}_{xx}(\xi,t)=u_{xx}^0(\xi)\exp\left\{a\int_0^t\hat{u}_x(\xi,s)\, ds\right\},
\end{equation}
where we have used that $\hat{u}_{xx}(\xi,0)=u_{xx}(F(\xi,0),0)=u_{xx}(\xi,0)=u_{xx}^0(\xi)$. At this point, we note that formula \eqref{solvehatuxx} for $a=1$ appears in \cite{ConstantinWunsch2009} with details of how it was derived.\\
Inserting the expressions for $\hat{u}_{x}$ and $\hat{u}_{xx}$ in terms of the flow map $F$ and its derivatives as derived in \eqref{hatux} and \eqref{hatuxx}, equation \eqref{solvehatuxx} becomes
\begin{equation}\label{reformF}
\begin{split}
\frac{1}{F_{\xi}(\xi,t)}\frac{\partial^2}{\partial t\partial \xi}\log F_\xi(\xi,t)&=u_{xx}^0(\xi)\exp\left\{a\int_0^t\frac{d}{ds}\log F_\xi(\xi,s)\, ds\right\}\\
&=u_{xx}^0(\xi)F_\xi(\xi,t)^a,
\end{split}
\end{equation}
for $(\xi,t)\in [0,1]\times [0,t^*)$, where we have used that $\log F_{\xi}(\xi,0)=\log(1)=0$, by the definition of $F$ in \eqref{flowu}. Multiplying equation \eqref{reformF} by $F_\xi$ and introducing
\begin{equation}
f_a(\xi,t)=F_\xi(\xi,t)^{a+1},
\end{equation}
for $a\in\mathbb{R}\setminus\{-1\}$, we arrive at the equation
\begin{equation}\label{Liouvilleinhom}
\frac{\partial^2}{\partial t\partial \xi}\log f_a(\xi,t)=(a+1)u_{xx}^0(\xi)f_a(\xi,t),\qquad (a\neq-1)
\end{equation}
for $(\xi,t)\in [0,1]\times [0,t^*)$ and $a\neq -1$, while for $a=-1$, we simply obtain
\begin{equation}\label{a=-1}
\frac{\partial^2}{\partial t\partial \xi}\log F_\xi(\xi,t)=u_{xx}^0(\xi),\qquad (a=-1),
\end{equation}
for $(\xi,t)\in [0,1]\times [0,t^*)$.\\
\begin{remark}
Equation \eqref{Liouvilleinhom} is an inhomogeneous Liouville equation with  $(a+1)u^0_{xx}(\xi)$ as a curvature term. For $(a+1)u^0_{xx}(\xi)=const.$,  the solution $f_a$ describes the conformal factor of a metric on a surface with constant curvature, cf. \cite{liouville1853equation}. In Section \ref{constantcurvaturechapter}, we solve the constant curvature equation for two values of $a$ explicitly, exemplifying this remarkable property of the GPJ equation.
\end{remark}

\subsection{Solution formula for the flow map $F$}\label{SolutionFormula}
First, we will start with the explicit solution to equation \eqref{a=-1}. Integrating with respect to $x$ and $t$ in \eqref{a=-1}, we find that
\begin{equation}
\log F_\xi(\xi,t)=\Big[u_x^0(\xi)-u_x^0(0)\Big]t+\theta(t)+\rho(\xi),
\end{equation}
for two functions $\rho:[0,1]\to\mathbb{R}$ and $\theta:[0,t^*)\to\mathbb{R}$. Denoting $\theta_0:=\theta(0)$, it follows from the definition of $F$ in \eqref{flowu} that $0=\log (1)=\log F_\xi(\xi,0)=\theta_0+\rho(\xi)$ and hence, after taking an exponential,
\begin{equation}
F_\xi(\xi,t)=\exp\left\{\theta(t)-\theta_0+\Big[u_x^0(\xi)-u_x^0(0)\Big]t\right\}.
\end{equation}
Integrating once more with respect to $\xi$ we arrive at the expression
\begin{equation}
F(\xi,t)=\int_0^\xi \exp\left\{\theta(t)-\theta_0+\Big[u_x^0(l)-u_x^0(0)\Big]t\right\}\, dl+\mu(t),
\end{equation}
for some function $\mu:[0,t^*)\to\mathbb{R}$.\\
Assuming Dirichlet boundary conditions, it follows that $\mu(t)\equiv0$, since $F(0,t)=0$ for all $t\in[0,t^*)$, and also that

\begin{equation}
e^{\theta(t)-\theta_0}=\left[\int_0^1 \exp\left\{\Big[u_x^0(l)-u_x^0(0)\Big]t\right\}\, dl\right]^{-1},
\end{equation}

since $F(1,t)=1$ for all $t\in[0,t^*)$. The flow map then takes the from
\begin{equation}
F(\xi,t)=\frac{\int_0^\xi \exp\left\{\Big[u_x^0(l)-u_x^0(0)\Big]t\right\}\, dl}{\int_0^1 \exp\left\{\Big[u_x^0(l)-u_x^0(0)\Big]t\right\}\, dl},
\end{equation}
or, after cancellation of the term $e^{-u^0_x(0)t}$,
\begin{equation}\label{flowa=0}
F(\xi,t)=\frac{\int_0^\xi \exp\left\{u_x^0(l)t\right\}\, dl}{\int_0^1 \exp\left\{u_x^0(l)t\right\}\, dl}.
\end{equation}
\medskip

We now turn to the case $a\neq -1$. To facilitate notation and to emphasize the connection to the geometric nature of equation \eqref{Liouvilleinhom}, we set
\begin{equation}
K_a(\xi)=(a+1)u_{xx}^0(\xi),\qquad  \xi\in[0,1].
\end{equation}
In our analysis, we follow the original approach of Liouville as outlined in \cite{liouville1853equation} (in the cited paper, however, only the constant curvature case $K_a=const.$ is considered).\\
Let $f_a=\frac{\partial g_a}{\partial t}$, for some function $g_a:[0,1]\times [0,t^*)\to\mathbb{R}$. As $K_a$ is $t$-independent, equation \eqref{Liouvilleinhom} can be integrated with respect to $t$:
\begin{equation}\label{solveLiouville1}
\frac{\partial }{\partial\xi}\log\frac{\partial g_a}{\partial t}=\left(\frac{\partial g_a}{\partial t}\right)^{-1}\frac{\partial^2 g_a}{\partial t\partial \xi}=K_ag_a+\rho_1,
\end{equation}
for some function $\xi\mapsto\rho_1(\xi), \rho_1:[0,1]\to\mathbb{R}$. Multiplying equation \eqref{solveLiouville1} with $\frac{\partial g_a}{\partial t}$ and integrating once more with respect to $t$, we have that
\begin{equation}\label{solveLiouville2}
\frac{\partial g_a}{\partial \xi}=\frac{K_a}{2}g_a^2+\rho_1g_a+\rho_2,
\end{equation}
for some function $\xi\mapsto\rho_2(\xi), \rho_2:[0,1]\to\mathbb{R}$.\\
 Equation \eqref{solveLiouville2} is a Riccati differential equation in the $\xi$-variable and can be solved by quadrature, once a particular solution is known. So, let $\xi\mapsto h_a(\xi), h:[0,1]\to\mathbb{R}$ be a particular (t-independent) solution to equation \eqref{solveLiouville2} and define
 \begin{equation}\label{solveLiouville4}
 g_a(\xi,t)=h_a(\xi)+\frac{1}{L_a(\xi,t)},\qquad (\xi,t)\in [0,1]\times [0,t^*).
 \end{equation}
 Then $L_a$ satisfies the linear ordinary differential equation
 \begin{equation}\label{solveLiouville3}
 \begin{split}
 \frac{\partial L_a}{\partial\xi}&=-(K_ah_a+\rho_1)L_a-\frac{K_a}{2}\\
 &=:-\rho_3L_a-\frac{K_a}{2}.
 \end{split}
 \end{equation}
 Equation \eqref{solveLiouville3} can be integrated explicitly to
 \begin{equation}
 \begin{split}
 L_a(\xi,t)&=\exp\left\{-\int_0^\xi\rho_3(s)\,ds\right\}\left(\phi(t)-\int_0^\xi\frac{K_a(s)}{2}\exp\left\{\int_0^s\rho_3(l)\,dl\right\}\, ds\right)\\
 &=:\frac{1}{R(\xi)}\left(\phi(t)-\int_0^\xi\frac{K_a(s)}{2}R(s)\, ds\right),
 \end{split}
 \end{equation}
for some function $\phi:[0,t^*)\mapsto \mathbb{R}$ and some function $R:[0,1]\to(0,t^*)$. Setting $\psi(\xi):=-\frac{1}{2}\int_0^\xi K_a(s)R(s)\, ds$, we can express \eqref{solveLiouville4} as
\begin{equation}
g_a(\xi,t)=h_a(\xi)-\frac{2\psi'(\xi)}{K_a(\xi)[\phi(t)+\psi(\xi)]},\qquad (\xi,t)\in[0.1]\times[0,t^*),
\end{equation}
and hence, $f_a$ becomes
\begin{equation}
   f_a(\xi,t)=\frac{2\psi'(\xi)\phi'(t)}{K_a(\xi)[\phi(t)+\psi(\xi)]^2},\qquad (\xi,t)\in[0.1]\times[0,t^*),
\end{equation}
for two arbitrary functions $\phi:[0,t^*)\to\mathbb{R}$ and $\psi:[0,1]\to\mathbb{R}$. Since $f_a> 0$, we have to choose the functions $\phi$ and $\psi$ such that
\begin{equation}\label{condposphipsi}
\frac{\psi'(\xi)\phi'(t)}{K_a(\xi)}> 0,\qquad (\xi,t)\in [0,1]\times [0,t^*).
\end{equation}
In particular, $\phi$ is either increasing or decreasing. Finally, since $f_a>0$, we arrive at
\begin{equation}\label{solveLiouville5}
F_\xi(\xi,t)=\left(\frac{2\psi'(\xi)\phi'(t)}{K_a(\xi)[\phi(t)+\psi(\xi)]^2}\right)^\frac{1}{a+1},\qquad (\xi,t)\in[0.1]\times[0,t^*).
\end{equation}
We will determine the functions $\phi$ and $\psi$ in \eqref{solveLiouville5} to match the boundary conditions of the flow map $F$. First, evaluating \eqref{solveLiouville5} at $t=0$ and using the definition of $F$ in \eqref{flowu}, it follows that
\begin{equation}\label{solvepsi}
\frac{2\psi'(\xi)\phi'_0}{K_a(\xi)[\phi_0+\psi(\xi)]^2}=1,
\end{equation}
where we have denoted $\phi(0)=\phi_0$ and $\phi'(0)=\phi'_0$. Equation \eqref{solvepsi} can be integrated explicitly to
\begin{equation}\label{solvepsi2}
\psi(\xi)=-\frac{1+\phi_0\left(\overline{\kappa}+\frac{1}{2\phi'_0}\int_0^\xi K_a(s)\,ds\right)}{\overline{\kappa}+\frac{1}{2\phi'_0}\int_0^\xi K_a(s)\,ds},\qquad \xi\in[0,1],
\end{equation}
for some constant $\overline{\kappa}\in\mathbb{R}$. Since $K_a(\xi)=(a+1)u_{xx}^0(\xi)$, equation \eqref{solvepsi2} reduces to
\begin{equation}
\psi(\xi)=-\frac{1+\phi_0\left(\kappa+\frac{a+1}{2\phi'_0}u^0_x(\xi)\right)}{\kappa+\frac{a+1}{2\phi'_0}u^0_x(\xi)},\qquad \xi\in[0,1],
\end{equation}
where we have denoted $\kappa=\overline{\kappa}-\frac{a+1}{2\phi_0'}u_x^0(0)$, and $F_\xi$ becomes
\begin{equation}\label{solveLiouville6}
F_\xi(\xi,t)=\left(\frac{\phi'(t)}{\phi_0'}\right)^\frac{1}{a+1}\left[1-(\phi(t)-\phi_0)\left(\kappa+\frac{a+1}{2\phi'_0}u^0_x(\xi)\right)\right]^{-\frac{2}{a+1}},
\end{equation}
for $(\xi,t)\in [0,1]\times [0,t^*).$
Rescaling
\begin{equation}
\phi(t)\mapsto\frac{1}{\phi_0'}(\phi(t)-\phi_0),
\end{equation}
we can assume that $\phi_0=0$ as well as $\phi_0'=1$ and hence, equation \eqref{solveLiouville6} simplifies to
\begin{equation}\label{Fxiphi}
F_\xi(\xi,t)=\left(\phi'(t)\right)^\frac{1}{a+1}\left[1-\phi(t)\left(\kappa+\frac{a+1}{2}u^0_x(\xi)\right)\right]^{-\frac{2}{a+1}},\qquad  (\xi,t)\in [0,1]\times [0,t^*),
\end{equation}
for some $\kappa\in\mathbb{R}$. We can give an expression of the constant $\kappa$ in terms of the function $\phi$ as follows.  Taking a $t$-derivative in equation \eqref{Fxiphi}, it follows from the initial condition $F_{t\xi}(\xi,0)=u_{x}^0(\xi)$, $\xi\in[0,1]$, that
\begin{equation}
\frac{1}{a+1}\phi_0''+\frac{2}{a+1}\left(\kappa+\frac{a+1}{2}u_x^0(\xi)\right)=u_x^0(\xi),\quad \xi\in[0,1],
\end{equation}
where we have used that $\phi(0)=0$, as well as $\phi'(0)=1$ and where we have abbreviated $\phi''_0:=\phi''(0)$. Hence, $\kappa$ is given as
\begin{equation}\label{kappaphipp}
\kappa=-\frac{\phi_0''}{2}.
\end{equation}
Sine $\kappa$ as given by equation \eqref{kappaphipp} is a second-derivative property of the function $\phi$, the flow map $F$ should not depend upon $\kappa$. Indeed, we can define another function $\eta$ through the M\"obius transform
\begin{equation}
\eta(t)=\frac{\phi(t)}{1-\kappa\phi(t)},\qquad \phi(t)=\frac{\eta(t)}{1+\kappa\eta(t)},\qquad t\in (0,t^*),
\end{equation}
which also satisfies $\eta(0)=0$ and $\eta'(0)=1$ such that the derivative of the flow map becomes
\begin{equation}\label{Fxi}
F_\xi(\xi,t)=\left(\eta'(t)\right)^\frac{1}{a+1}\left[1-\eta(t)\frac{a+1}{2}u^0_x(\xi)\right]^{-\frac{2}{a+1}},\qquad  (\xi,t)\in [0,1]\times [0,t^*),
\end{equation}
i.e., we have eliminated the dependence upon $\kappa$ while keeping the first-derivative properties of the unknown function $\eta$.\\
Integrating equation \eqref{Fxi} with respect to $\xi$ gives
\begin{equation}\label{flowmapDirichlet}
F(\xi,t)=\mu(t)+\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^\xi\left[1-\eta(t)\frac{a+1}{2}u^0_x(s)\right]^{-\frac{2}{a+1}}\, ds,\qquad  (\xi,t)\in [0,1]\times [0,t^*),
\end{equation}
for some function $\mu:[0,t^*)\to\mathbb{R}$.\\
Assuming Dirichlet boundary conditions \eqref{DirichletF}, it immediately follows that $\mu\equiv 0$ and hence
\begin{equation}\label{FDirichlet}
F(\xi,t)=\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^\xi\left[1-\eta(t)\frac{a+1}{2}u^0_x(s)\right]^{-\frac{2}{a+1}}\,ds,\qquad  (\xi,t)\in [0,1]\times [0,t^*).
\end{equation}
To match the boundary condition $F(1,t)=1,\quad t\in[0,t^*)$, the function $\eta$ has to satisfy the implicit ordinary differential equation
\begin{equation}\label{eqeta}
\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^1\left[1-\eta(t)\frac{a+1}{2}u^0_x(s)\right]^{-\frac{2}{a+1}}\,ds=1,\qquad  t\in [0,t^*),
\end{equation}
which, in general, cannot be solved explicitly.\\
On the other hand, assuming periodic mean-free boundary conditions \eqref{periodicBC}, which imply that $F(\xi,t)=F(\xi+1,t)-1,\quad t\in[0,t^*)$, we obtain that $\eta$ has to satisfy equation \eqref{eqeta} as well. By the mean-free constraint \eqref{flowperiodic}, the function $\mu$ is then given as the solution to the differential equation
\begin{footnotesize}
\begin{equation}
\begin{split}
\mu'(t)=-\int_{F^{-1}(0,t)}^{F^{-1}(0,t)+1}\left(\eta'(t)\right)^\frac{1}{a+1}&\left[1-\eta(t)\frac{a+1}{2}u^0_x(\xi)\right]^{-\frac{2}{a+1}}\times\\
&\times\frac{d}{dt}\left(\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^\xi\left[1-\eta(t)\frac{a+1}{2}u^0_x(s)\right]^{-\frac{2}{a+1}}\, ds\right)\, d\xi,
\end{split}
\end{equation}
\end{footnotesize}

\noindent
with initial condition $\mu(0)=0$, since $F(0,\xi)=\xi$.\\
Note that formula  \eqref{flowmapDirichlet} only holds as long as $\eta'(t)\geq 0$ and $1-\eta(t)\frac{a+1}{2}u^0_x(\xi)>0$ for all $\xi\in(0,1)$.

\begin{example}[$a=-3$, Burger's equation]
Assume Dirichlet boundary conditions and let $a=-3$. For this parameter value of $a$, the GPJ equation reduces to Burger's equation, for which the flow map can be computed easily as
\begin{equation}
F_{Burgers}(\xi,t)=u^0(\xi)t+\xi,\qquad (\xi,t)\in [0,1]\times [0,t^*).
\end{equation}
Now, consider the flow map \eqref{flowmapDirichlet} with $a=-3$:
\begin{equation}\label{FBurgers}
\begin{split}
F(\xi,t)&=\sqrt{\frac{1}{\eta'(t)}}\int_0^\xi 1+\eta(t)u^0_x(s)\, ds\\
&=\sqrt{\frac{1}{\eta'(t)}}\Big(\xi +\eta(t)u^0(\xi) \Big),
\end{split}
\end{equation}
where we have used that $u^0(0)=0$ in the second step. Note that formula \eqref{FBurgers} holds only for $t\in[0,t^*)$, where $1+\eta(t)u^0_x(s)> 0$. From the second boundary condition, $F(1,t)=1, t\in[0,t^*)$ and $u^0(1)=0$, it follows that $\eta$ satisfies the ordinary differential equation
\begin{equation}\label{eqphiBurgers}
\sqrt{\frac{1}{\eta'(t)}}=1,\qquad t\in[0,t^*],
\end{equation}
which, remembering that $\eta(0)=0$ as well as $\eta'(0)=1$, implies that
\begin{equation}
\eta(t)=t.
\end{equation}
The flow map then takes the form
\begin{equation}
F(\xi,t)=u^0(\xi)t+\xi,\quad (\xi,t)\in [0,1]\times [0,t^*),
\end{equation}
which is exactly the flow map of Burger's equation. In particular, we recover the explicit formula for the blow-up time,
\begin{equation}
t^*=-\frac{1}{\min_{\xi\in[0,1]}u_x^0(\xi)}.
\end{equation}
\end{example}

\begin{example}[$a=-2$, Hunter--Saxton equation]
Assume Dirichlet boundary conditions and let $a=-2$. For this parameter value of $a$, the GPJ equation reduces to the Hunter--Saxton equation.
Consider now the flow map \eqref{flowmapDirichlet} with $a=-2$:
\begin{equation}\label{flowHS}
\begin{split}
F(\xi,t)&=\frac{1}{\eta'(t)}\int_0^\xi\left[1+\eta(t)\frac{1}{2}u^0_x(\xi)\right]^2\, ds\\
&=\frac{1}{\eta'(t)}\left[\xi+ u^0(\xi)\eta(t)+\frac{\eta^2(t)}{4}\int_0^\xi(u_x^0(s))^2\,ds\right],
\end{split}
\end{equation}
for $(\xi,t)\in [0,1]\times [0,t^*)$. We remark again that the representation formula \eqref{flowHS} only holds as long as
\begin{equation}\label{condHS}
1+\eta(t)\frac{1}{2}u^0_x(\xi)> 0.
\end{equation}
Evaluating equation \eqref{flowHS} at $\xi=1$ and using the boundary condition $F(1,t)=1, t\in[0,t^*)$, we obtain the following ordinary differential equation for $\eta$:
\begin{equation}\label{etaeqHS}
\eta'(t)=1+\frac{1}{4}\|u_x^0\|_{L^2(0,1)}^2\eta^2(t),\qquad t\in[0,t^*].
\end{equation}
Equation \eqref{etaeqHS} can be integrated to
\begin{equation}
\eta(t)=\frac{2}{\|u_x^0\|_{L^2(0,1)}}\tan\left(\frac{t}{2}\|u_x^0\|_{L^2(0,1)}\right),\qquad  t\in[0,t^*],
\end{equation}
where we have used that $\eta(0)=0$ and $\eta'(0)=1$.
Finally, the flow map takes the form
\begin{equation}
\begin{split}
F(\xi,t)&=\cos^{2}\left(\frac{t}{2}\|u_x^0\|_{L^2(0,1)}\right)\left[\xi+\frac{2}{\|u_x^0\|_{L^2(0,1)}}\tan\left(\frac{t}{2}\|u_x^0\|_{L^2(0,1)}\right)u^0(\xi)\right.\\
&\qquad\left.+\frac{1}{\|u_x^0\|_{L^2(0,1)}^2}\tan^2\left(\frac{t}{2}\|u_x^0\|_{L^2(0,1)}\right)\int_0^\xi(u_x^0(s))^2\,ds\right],\quad (\xi,t)\in[0,1]\times [0,t^*).
\end{split}
\end{equation}
Since $\eta$ blows up in finite time, we find that condition \eqref{condHS} will be violated at the blow-up time
\begin{equation}
t^*=-\frac{2}{\|u_x^0\|_{L^2(0,1)}}\arctan\left(\frac{\|u_x^0\|_{L^2(0,1)}}{\min_{\xi\in[0,1]}u^0_x(\xi)}\right).
\end{equation}
\end{example}

\subsection{Constant Curvature Solutions}\label{constantcurvaturechapter}

Consider the GPJ equation with Dirichlet boundary conditions and initial condition $u^0(x)=\gamma x(x-1)$, cf. Figure \ref{Parabola}.

\begin{figure}
	\includegraphics[scale=0.8]{Parabola}
	\caption{The initial condition $u^0(x)=-x(x-1)$, satisfying Dirichlet boundary conditions.}
	\label{Parabola}
\end{figure}

 Since $u_x^0(x)=\gamma (2x-1)$, the representation formula \eqref{flowmapDirichlet} gives
\begin{equation}\label{flowconstant}
\begin{split}
F(\xi,t)&=(\eta'(t))^\frac{1}{a+1}\int_0^\xi\Big[1-\gamma\frac{a+1}{2}(2s-1)\eta(t)\Big]^{-\frac{2}{a+1}}\, ds\\
&=(\eta'(t))^\frac{1}{a+1}\frac{a+1}{-2A(t)(a-1)}\left([1+(1-2\xi)A(t)]^{\frac{a-1}{a+1}}-[1+A(t)]^\frac{a-1}{a+1}\right),
\end{split}
\end{equation}
for $(\xi,t)\in[0,1]\times [0,t^*)$ and $a\neq 1$. (The case $a=1$ will be treated below.)
 where we have abbreviated
\begin{equation}
A(t)=\gamma\frac{a+1}{2}\eta(t).
\end{equation}
Using the boundary condition $F(1,t)=1,\quad t\in[0,t^*)$, the formula \eqref{flowconstant} leads to the following ordinary differential equation for the function $\eta$:
\begin{equation}\label{ODEconstantcurvature}
1=(\eta'(t))^\frac{1}{a+1}\frac{1}{\gamma(1-a)\eta(t)}\left([1-\gamma\frac{a+1}{2}\eta(t)]^{\frac{a-1}{a+1}}-[1+\gamma\frac{a+1}{2}\eta(t)]^\frac{a-1}{a+1}\right),
\end{equation}
for $t\in (0,t^*)$, which, for general $a$, cannot be solved in terms of elementary functions.\\

\begin{example}[Constant Curvature for $a=0$]
Consider the constant curvature flow \eqref{flowconstant} for $a=0$. Equation \eqref{ODEconstantcurvature} becomes
\begin{equation}
1=\frac{\eta'(t)}{1-\frac{\gamma^2}{4}\eta^2(t)},\quad t\in(0,t^*),
\end{equation}
which, using $\eta_0=0$ and $\eta'_0=1$, can be solved to $\eta(t)=\frac{2}{\gamma}\tanh\left(\frac{\gamma}{2} t\right),\quad t\in(0,t^*)$. Hence, the flow map takes the form
\begin{equation}\label{Flowa1}
F(\xi,t)=\frac{\xi\left(\coth\left(\frac{\gamma t}{2}\right)-1\right)}{1-2\xi+\coth\left(\frac{\gamma t}{2}\right)},\quad (\xi,t)\in [0,1]\times (0,\infty).
\end{equation}
In particular, the flow map exists for all times. The time evolution of the $\xi$ of \eqref{Flowa1} now describes a surface of constant curvature, cf. Figure \ref{3Dconstantcurvature}.
\begin{figure}
	\includegraphics[scale=0.5]{3Dconstantcurvature}
	\caption{The $\xi$-derivative of \eqref{Flowa1} for the parameter value $\gamma=2$ and for $(\xi,t)\in[0,1]\times [0,1]$.}
	\label{3Dconstantcurvature}
\end{figure}

\end{example}

\begin{example}[Constant  Curvature for $a=1$]\label{examplea=1}
	For $a=1$, equation \eqref{GPJa} becomes the Proudman--Johnson equation. We revisit the example $u^0(x)=\gamma x(x-1)$, which was already treated in \cite{childress_ierley_spiegel_young_1989}, from a flow map point of view. The ordinary differential equation \eqref{eqeta} takes the form
	\begin{equation}\label{eqlog}
	\begin{split}
	1&=\sqrt{\eta'(t)}\int_0^1\frac{1}{1-\eta(t)\gamma(2s-1)}\, ds\\
	&=-\sqrt{\eta'(t)}\frac{1}{2\gamma\eta(t)}\log\left|\frac{1-\gamma\eta(t)}{1+\gamma\eta(t)}\right|\, ds,
	\end{split}
	\end{equation}
for $t\in[0,t^*)$. By comparing the signs of $\gamma$ and by noting that $\eta(t)$ will be positive for, at least, small $t$, equation \eqref{eqlog} can be integrated implicitly to
\begin{equation}\label{defPsi}
\begin{split}
t&=\frac{1}{4\gamma^2}\int_0^\eta\frac{1}{y^2}\log^2\left|\frac{1-\gamma y}{1+\gamma y}\right|\, dy\\
&=:\Psi(\eta).
\end{split}
\end{equation}
Since the function $y\mapsto \frac{1}{y}\log\left|\frac{1-\gamma y}{1+\gamma y}\right|$ is square integrable over $\mathbb{R}$, as can be seen by integration by parts, the function $\eta\mapsto \Psi(\eta)$ does not blow up and hence, the map $t\mapsto\eta(t)$ is unbounded, in fact, even blows up in finite time cf. Figure \ref{FigPsieta}. This implies, that the condition $1+\eta(t)\frac{1}{2}u^0_x(\xi)> 0$ will be violated at a finite time $t^*<\infty$ and hence, no global solution exists.
\begin{figure}
	\includegraphics[scale=0.8]{Psiofeta}
	\caption{The function $\eta\mapsto\Psi(\eta)$ as defined in \eqref{defPsi} for the parameter value $\gamma=2$.}
	\label{FigPsieta}
\end{figure}
\end{example}

\section{Criteria for formation of a singularity and global existence}\label{Criteria}

In this section, we present some singularity criteria and global well-posedness criteria for different values of $a$.\\
We note that the representation formula \eqref{flowmapDirichlet} only holds true as long as $1-\eta(t)\frac{a+1}{2}u_x^0(\xi)>0$. If, for some $(\xi^*,t^*)\in[0,1]\times(0,\infty)$, we have that $0=1-\eta(t^*)\frac{a+1}{2}u_x^0(\xi^*)=\Big(\eta'(t^*)\Big)^{-\frac{1}{a+1}}F_\xi(\xi^*,t^*)$, then the flow map $F$ ceases to be a diffeomorphism of the interval $[0,1]$ and hence, the velocity field $u$ can no longer belong to the class $u\in C\Big(0,t^*; C^{1,Lip}(0,1)\Big)$. Therefore, global existence of solutions $u\in C\Big(0,t^*; C^{1,Lip}(0,1)\Big)$ is equivalent to preservation of the inequality $1-\eta(t)\frac{a+1}{2}u_x^0(\xi)>0$ for all $(\xi,t)\in[0,1]\times [0,\infty)$.\\
Throughout, we write
\begin{equation}
-u_{min}\leq u_x^0(\xi)\leq u_{max},\quad \xi\in[0,1],
\end{equation}
for the negative minimal value and the positive maximal value of $u_x^0$. In the proof of our main theorem, we will need the following inequality.

\begin{lemma}[Reverse quadratic Bernoulli inequality]
	Let $x>-1$ and let $0<\alpha<1$. Then
	 \begin{equation}\label{reverseBernoulli}
	 (1+x)^\alpha\geq \frac{\alpha(\alpha-1)}{2}x^2+\alpha x+\frac{\alpha(3-\alpha)}{2}.
	 \end{equation}
\end{lemma}
\begin{proof}
Since $\alpha-1<0$, we can apply Bernoulli's inequality to obtain
\begin{equation}\label{Bernoulli1}
(1+x)^{\alpha-1}\geq 1+(\alpha-1)x,\quad x>-1.
\end{equation}
Integrating the inequality \eqref{Bernoulli1} from $-1$ to $x$ gives
\begin{equation}
\frac{1}{\alpha}(1+x)^\alpha\geq x+1+\frac{\alpha-1}{2}x^2-\frac{\alpha-1}{2},\quad x>-1,
\end{equation}
which, since $\alpha>0$, gives \eqref{reverseBernoulli}.
\end{proof}

\begin{remark}
	Note that it is only possible to obtain a quadratic inequality of the form \eqref{reverseBernoulli} for $0<\alpha<1$, as for this parameter regime, $\alpha-1<0$ holds, allowing to apply the usual Bernoulli inequality to $(1+x)^{\alpha-1}$, while at the same time $\lim_{x\to-1}(1+x)^\alpha=0$.
\end{remark}

The following theorem gives conditions under which the solution exists for all times or becomes singular in finite time, depending on the parameter $a$.

\begin{theorem}\label{mainthm}
Let $u\in C\Big(0,t^*; C^{1,Lip}(0,1)\Big)$ be a solution to equation \eqref{GPJa}, either with Dirichlet boundary conditions \eqref{DirichletBC} or with periodic boundary conditions and zero mean \eqref{periodicBC}, and let  $t^*\in (0,\infty]$ be its maximal existence time.
\begin{itemize}

	\item For $a<-3$, the solution exists only locally in time, i.e., $t^*<\infty$, provided that
	\begin{equation}\label{Assa-3}
	\|u_x^0\|_{L^2(0,1)}<\sqrt{\frac{3a+5}{a+3}}u_{min},
	\end{equation}
	or provided that
	\begin{equation}\label{Ass-3alternative}
	u_{xx}^0\in L^{\frac{2}{2+(a+1)q}}(0,1),
	\end{equation}
	for some $0<q<-\frac{2}{a+1}$.\\

	\item For $-3\leq a<-1$, the solution exists only locally in time, i.e., $t^*<\infty$ and the maximal existence time can be estimated as
	\begin{equation}
	0<t^*<-\frac{2}{(a+1)u_{min}}.
	\end{equation}

	\item For $a=-1$, the solution exists globally in time, i.e., $t^*=\infty$.\\

	\item For $-1<a<0$, the solution exists globally in time, i.e., $t^*=\infty$, provided that

	\begin{equation}\label{Assuxx}
	u_{xx}^0\in L^{-\frac{1}{a}}(0,1).
	\end{equation}\\

	\item For $-1<a$, the solution exists globally in time, i.e., $t^*=\infty$, provided that
	\begin{equation}\label{Assa-10}
	u_{max}< \frac{1}{\sqrt{1+a}}\|u_x^0\|_{L^2(0,1)},
	\end{equation}

	\item For $1\leq a$, the solution exists only locally in time, i.e., $t^*<\infty$, provided that the function
	\begin{equation}\label{defPsiThm}
	\psi(\eta):=\int_0^1\frac{ds}{1-\frac{a+1}{2}\eta u_x^0(s)},
	\end{equation}
	is square-integrable over the interval $[0,\frac{2}{(a+1)u_{max}}]$.
\end{itemize}
\end{theorem}

\begin{remark}
		It is not immediately clear that assumptions \eqref{Assa-10} and the square-integrability of expression \eqref{defPsiThm} are mutually exclusive. We refer to Example \ref{example} for the special case of a piece-wise linear initial condition. For initial conditions which oscillate equally around zero, i.e., initial conditions with $u_{max}=u_{min}$, there is a upper bound on $a$ such that \eqref{Assa-10} can be satisfied.
		Indeed, by Jensen's inequality
		\begin{equation}
		u_{max}<\frac{1}{\sqrt{1+a}}\|u_x^0\|_{L^2(0,1)}<\frac{1}{\sqrt{1+a}}u_{max},
		\end{equation}
		implying that
		\begin{equation}
		-1<a<0.
		\end{equation}
\end{remark}

\begin{proof}
We divide the proof of Theorem \ref{mainthm} in different section, each one employing different techniques to deduce the claims. To facilitate notation, we introduce the parameter
\begin{equation}
\alpha=-\frac{2}{a+1},
\end{equation}
whose values in dependence upon $a$ are shown in Table \ref{table}.\\
\begin{center}
	\begin{table}[h!]
\begin{tabular}{ |l|l| }
	\hline
	\multicolumn{2}{|c|}{Parameter values $a$ and $\alpha$} \\
	\hline
	$ -\infty<a<-3$ & $\,\,\,\,\,0<\alpha<1$ \\
	$-3\leq a<-1$ & $\,\,\,\,\,1\leq\alpha<\infty$ \\
	$-1<a<0$ & $-\infty<\alpha<-2$ \\
	$\,\,\,\,\,0<a<1$ & $-2<\alpha<-1$\\
	$\quad1<a,\infty$ & $-1<\alpha<0$\\
	\hline
\end{tabular}
\caption{}
\label{table}
\end{table}
\end{center}

\textbf{The case $-\infty<a<-3$ or $0<\alpha<1$}\\
Assume, to the contrary, that $1+\eta (t)\frac{u^0_x(\xi)}{\alpha}>0$, while $\eta'(t)>0$, for all $(\xi,t)\in[0,1]\times[0,\infty)$. First, let us assume \eqref{Assa-3}. This assumption allows us to apply the reversed quadratic Bernoulli inequality \eqref{reverseBernoulli} to equation \eqref{eqeta}, which in turn implies that
\begin{equation}\label{inequa-3}
\begin{split}
1&=\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^1\left[1+\eta(t)\frac{u^0_x(s)}{\alpha}\right]^\alpha\,ds\\
&\geq\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^1\frac{(\alpha-1)}{2\alpha}\Big(u^0_x(s)\eta(t)\Big)^2+u^0_x(s)\eta(t)+\frac{\alpha(3-\alpha)}{2}\, ds\\
&=\left(\eta'(t)\right)^\frac{1}{a+1}\left(\frac{\alpha(3-\alpha)}{2}+\frac{(\alpha-1)}{2\alpha}\|u_x^0\|^2_{L^2(0,1)}\eta^2(t)\right),
\end{split}
\end{equation}
for all $t\in(0,\infty)$, where in the last equality, we have used either Dirichlet or periodic boundary conditions. Since $a+1<0$, inequality \eqref{inequa-3} is equivalent to
\begin{equation}
1\leq \eta'(t)\left(\frac{\alpha(3-\alpha)}{2}+\frac{(\alpha-1)}{2\alpha}\|u_x^0\|^2_{L^2(0,1)}\eta^2(t)\right)^{a+1},
\end{equation}
for $t\in (0,\infty)$, as long as the expression in the brackets is non-negative. Therefore, $\eta$ can be bounded from below by a monotonically increasing function which goes to
\begin{equation}
\eta^*=\alpha\sqrt{\frac{3-\alpha}{1-\alpha}}\frac{1}{\|u_x^0\|_{L^2(0,1)}},
\end{equation}
as $t\to\infty$. Therefore, for any $\varepsilon>0$, there exists a time $t^*(\varepsilon)$, such that, at the minimum of the expression $1+\eta (t)\frac{u^0_x(\xi)}{\alpha}$ in $\xi$, we find that
\begin{equation}
\begin{split}
1-\eta(t^*)\frac{u_{min}}{\alpha}&\leq 1-\frac{u_{min}}{\alpha}\left(\alpha\sqrt{\frac{3-\alpha}{1-\alpha}}\frac{1}{\|u_x^0\|_{L^2(0,1)}}-\varepsilon\right)\\
&< 1-\sqrt{\frac{3-\alpha}{1-\alpha}}\frac{u_{min}}{\|u_x^0\|_{L^2(0,1)}}+\frac{u_{min}}{\alpha}\varepsilon.
\end{split}
\end{equation}
By assumption \eqref{Assa-3} and since $\varepsilon$ was arbitrary, we obtain a contradiction.\\

Now, let us assume \eqref{Ass-3alternative}. For any $0<q<\alpha$ and any two numbers $\xi_1,\xi_2\in[0,1]$, we can estimate
\begin{equation}\label{inequa-3alternative}
\begin{split}
\frac{\alpha}{\eta(q+1)}\left[\left(1+\eta\frac{u_x^0(\xi)}{\alpha}\right)^{q+1}\right]_{\xi_1}^{\xi_2}&=\frac{\alpha}{\eta(q+1)}\int_{\xi_1}^{\xi_2}\frac{d}{ds}\left(1+\eta\frac{u_x^0(s)}{\alpha}\right)^{q+1}\, ds\\
&=\int_{\xi_1}^{\xi_2}u_{xx}^0(s)\left(1+\eta\frac{u_x^0(s)}{\alpha}\right)^{q}\, ds\\
&\leq \|u_{xx}^0\|_{L^\frac{\alpha}{\alpha-q}}\left\|\left(1+\eta\frac{u^0_{x}}{\alpha}\right)^q\right\|_{L^{\frac{\alpha}{q}}(0,1)}\\
&=\|u_{xx}^0\|_{L^{\frac{\alpha}{\alpha-q}}(0,1)}\left(\int_0^1\left[1+\eta\frac{u^0_{x}(s)}{\alpha}\right]^\alpha\, ds\right)^{\frac{q}{\alpha}},
\end{split}
\end{equation}
for $0<\eta<\frac{|\alpha|}{u_{min}}$, where we have applied H\"older's inequality with the pair of conjugated exponents  $\left(\frac{\alpha}{q},\frac{\alpha}{\alpha-q}\right)$, using assumption \eqref{Ass-3alternative}. Since $u_x^0$ is mean free for both Dirichlet and periodic boundary conditions, we can find $\xi_1<\xi_2\in [0,1]$ such that $u_x^0(\xi_1)<0$, while $u_x^0(\xi_2)>0$ (possibly after multiplying the first equality in \eqref{inequa-3alternative} by $-1$). This implies that we can bound
\begin{equation}\label{inequxi1xi2}
\begin{split}
\frac{\alpha}{\eta(q+1)}\left[\left(1+\eta\frac{u_x^0(\xi)}{\alpha}\right)^{q+1}\right]_{\xi_1}^{\xi_2}&=\frac{\alpha}{\eta(q+1)}\left[\left(1+\eta\frac{u_x^0(\xi_2)}{\alpha}\right)^{q+1}-\left(1+\eta\frac{u_x^0(\xi_1)}{\alpha}\right)^{q+1}\right]\\
&\geq \frac{\alpha}{\eta(q+1)}\left[\left(1+\eta\frac{u_x^0(\xi_2)}{\alpha}\right)^{q+1}-1\right],
\end{split}
\end{equation}
since, for this choice of $\xi_1$, we have that $0<\left(1+\eta\frac{u_x^0(\xi_1)}{\alpha}\right)^{q+1}<1$.
 Since also $\frac{q}{\alpha}>0$, we can therefore estimate
 \begin{equation}
 \begin{split}
1&=(\eta'(t))^{\frac{1}{a+1}}\left(\int_0^1\left[1+\eta(t)\frac{u^0_{x}(s)}{\alpha}\right]^\alpha\, ds\right)\\
&\geq(\eta'(t))^{\frac{1}{a+1}}\left( \frac{\alpha}{\|u_{xx}^0\|_{L^{\frac{\alpha}{\alpha-q}}(0,1)}}\frac{1}{\eta(t)(q+1)}\left[\left(1+\eta(t)\frac{u_x^0(\xi)}{\alpha}\right)^{q+1}\right]_{\xi_1}^{\xi_2}\right)^{\frac{\alpha}{q}}\\
&\geq (\eta'(t))^{\frac{1}{a+1}} \left( \frac{\alpha}{\|u_{xx}^0\|_{L^{\frac{\alpha}{\alpha-q}}(0,1)}}\frac{1}{\eta(t)(q+1)}\left[\left(1+\eta(t)\frac{u_x^0(\xi_2)}{\alpha}\right)^{q+1}-1\right]\right)^{\frac{\alpha}{q}},
\end{split}
 \end{equation}
for $t\in(0,t^*)$, where we have used inequality \eqref{inequa-3alternative} in the first step and inequality \eqref{inequxi1xi2} in the last step. Since now $a+1<0$, it follows that $\eta'$ can be bounded from below as

 \begin{equation}\label{help}
 \begin{split}
 1&\leq \left( \frac{\alpha}{\|u_{xx}^0\|_{L^{\frac{\alpha}{\alpha-q}}(0,1)}}\frac{1}{(q+1)}\right)^{\frac{\alpha}{q}(a+1)}\eta'(t)[\eta(t)]^{-(a+1)\frac{\alpha}{q}}\left[\left(1+\eta(t)\frac{u_x^0(\xi_2)}{\alpha}\right)^{q+1}-1\right]^{(a+1)\frac{\alpha}{q}}\\
 &= \left( \frac{\alpha}{\|u_{xx}^0\|_{L^{\frac{\alpha}{\alpha-q}}(0,1)}}\frac{1}{(q+1)}\right)^{-\frac{2}{q}}\eta'(t)[\eta(t)]^{\frac{2}{q}}\left[\left(1+\eta(t)\frac{u_x^0(\xi_2)}{\alpha}\right)^{q+1}-1\right]^{-\frac{2}{q}},
 \end{split}
 \end{equation}

by the definition of $\alpha$. Integrating both sides of the inequality \eqref{help} with respect to $t$, we have that
 \begin{equation}
t\leq C\int_0^\eta y^{\frac{2}{q}}\left[\left(1+y\frac{u_x^0(\xi_2)}{\alpha}\right)^{q+1}-1\right]^{-\frac{2}{q}}\, dy,
 \end{equation}
for some constant $C>0$.
We will now show that
 \begin{equation}\label{intcond}
\int_0^{\frac{\alpha}{u_{min}}} y^{\frac{2}{q}}\left[\left(1+y\frac{u_x^0(\xi_2)}{\alpha}\right)^{q+1}-1\right]^{-\frac{2}{q}}\, dy<\infty,
\end{equation}
to obtain a contradiction.
First, we note that the integrand in \eqref{intcond} is bounded  for $\eta\in(0,\infty)$ (since $0<q\leq\alpha <1$):
 \begin{equation}
 \begin{split}
 \lim_{y\to 0}y^{\frac{2}{q}}\left[\left(1+y\frac{u_x^0(\xi_2)}{\alpha}\right)^{q+1}-1\right]^{-\frac{2}{q}}&=\lim_{y\to 0}y^{\frac{2}{q}}\left[\sum_{n=1}^\infty{q+1 \choose n}\left(y\frac{u_x^0(\xi_2)}{\alpha}\right)^n\right]^{-\frac{2}{q}}\\
 &=\lim_{y\to 0}\left[\sum_{n=1}^\infty{q+1 \choose n}\left(\frac{u_x^0(\xi_2)}{\alpha}\right)^ny^{n-1}\right]^{-\frac{2}{q}}\\
 &=\lim_{y\to 0} \left(\frac{u_x^0(\xi_2)}{\alpha}\right)^{-\frac{2}{q}}\left[\sum_{n=0}^\infty{q+1 \choose n+1}\left(y\frac{u_x^0(\xi_2)}{\alpha}\right)^n\right]^{-\frac{2}{q}}\\
 &=\left(\frac{u_x^0(\xi_2)}{\alpha}\right)^{-\frac{2}{q}}(q+1)^{-\frac{2}{q}}<\infty,
 \end{split}
 \end{equation}
 where we have expanded the expression in a binomial series, using that $y$ is small enough. Since $u_x^0(\xi_2)>0$, the integrand in \eqref{intcond}  does not have any pole for $y\geq 0$. Since
\begin{equation}
y^{\frac{2}{q}}\left[\left(1+y\frac{u_x^0(\xi_2)}{\alpha}\right)^{q+1}-1\right]^{-\frac{2}{q}}=\mathcal{O}(y^{-2}), \quad \text{ for } y\to\infty,
\end{equation}
the inequality \eqref{intcond} is indeed satisfied - a contradiction.

 This proves the claim.\\

\textbf{The case $-3\leq a<-1$ or $1\leq\alpha<\infty$}\\
Assume again, to the contrary, that $1+\eta (t)\frac{u^0_x(\xi)}{\alpha}>0$, while $\eta'(t)>0$, for all $(\xi,t)\in[0,1]\times[0,\infty)$. For this range or parameter values, the function $x\mapsto x^\alpha$ is convex, while the function $x\mapsto x^\frac{1}{a+1}$ is monotonically decreasing. \footnote{Note that this is the maximal parameter range with these properties.} We can therefore apply Jensen's inequality to \eqref{eqeta} to obtain
\begin{equation}\label{inequa-3-1}
\begin{split}
1&=\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^1\left[1+\eta(t)\frac{u^0_x(s)}{\alpha}\right]^\alpha\, ds\\
&\geq \left(\eta'(t)\right)^\frac{1}{a+1}\left(\int_0^11+\eta(t)\frac{u^0_x(s)}{\alpha}\, ds\right)^\alpha\\
&= \left(\eta'(t)\right)^\frac{1}{a+1},
\end{split}
\end{equation}
for $t\in[0,\infty)$, where in the last step, we have again used either Dirichlet or periodic boundary conditions. Inverting and integrating inequality \eqref{inequa-3-1} gives
\begin{equation}
t\leq \eta(t),\quad t\in[0,\infty),
 \end{equation}
where we have used the initial data for $\eta$ - again a contraction to $1+\eta (t)\frac{u_{min}}{\alpha}>0$ at $t^*=\frac{\alpha}{u_{min}}$.\\

\textbf{The case $a=-1$ or $\alpha=-\infty$}\\
In this case, we have derived the explicit formula \eqref{flowa=0} flow map, which shows that the solution exists for all times.\\

\textbf{The case $-1<a<0$ or $-\infty<\alpha<-2$}\\
For any two numbers $\xi_1,\xi_2\in [0,1]$, $\xi_1<\xi_2$, we can calculate
\begin{equation}\label{inequ-1a1}
\begin{split}
\frac{-\alpha}{\eta}\left[\left(1+\eta\frac{u_x^0(\xi)}{\alpha}\right)^{-1}\right]^{\xi_2}_{\xi_1}&=\frac{-\alpha}{\eta}\int_{\xi_1}^{\xi_2}\frac{d}{ds}\left(1+\eta\frac{u_x^0(s)}{\alpha}\right)^{-1}\, ds\\
&=\int_{\xi_1}^{\xi_2}u_{xx}^0(s)\left(1+\eta\frac{u_x^0(s)}{\alpha}\right)^{-2}\, ds\\
&\leq \|u_{xx}^0\|_{L^{\frac{\alpha}{\alpha+2}}(0,1)}\left\|\left(1+\eta\frac{u^0_{x}}{\alpha}\right)^{-2}\right\|_{L^{-\frac{\alpha}{2}}(0,1)}\\
&=\|u_{xx}^0\|_{L^{\frac{\alpha}{\alpha+2}}(0,1)}\left(\int_0^1\left[1+\eta\frac{u^0_{x}(s)}{\alpha}\right]^\alpha\, ds\right)^{-\frac{2}{\alpha}},
\end{split}
\end{equation}
for $0<\eta<\frac{|\alpha|}{u_{max}}$, where we have applied H\"older's inequality with the pair of conjugated exponents  $\left(-\frac{\alpha}{2},\frac{\alpha}{\alpha+2}\right)$, using assumption \eqref{Assuxx}.\\
Since $a+1>0$, it follows from \eqref{eqeta} that we can bound
\begin{equation}
\begin{split}
1&\geq \eta'\left(\frac{|\alpha|}{\eta\|u_{xx}^0\|_{L^{\frac{\alpha}{\alpha+2}}(0,1)}}\right)^{-\frac{\alpha}{2}(a+1)}\left(\left[\left(1+\eta\frac{u_x^0(\xi)}{\alpha}\right)^{-1}\right]^{\xi_2}_{\xi_1}\right)^{-\frac{\alpha}{2}(a+1)}\\
&\geq \eta'\left(\frac{|\alpha|}{\|u_{xx}^0\|_{L^{\frac{\alpha}{\alpha+2}}(0,1)}}\right)\left(\left[\left(1+\eta\frac{u_x^0(\xi)}{\alpha}\right)^{-1}\right]^{\xi_2}_{\xi_1}\right)\eta^{-1},
\end{split}
\end{equation}
for any choice of $\xi_1,\xi_2\in [0,1]$, by the definition of $\alpha$. Choosing $\xi_1,\xi_2$ such that
\begin{equation}\label{xi1xi2}
0<-\frac{\alpha}{u_x^0(\xi_2)}<-\frac{\alpha}{u_x^0(\xi_1)},
\end{equation}
or, equivalently, since $\alpha<0$,
\begin{equation}\label{xi1xi2explicit}
0<u_x^0(\xi_1)<u_x^0(\xi_2),
\end{equation}
we can bound $t\mapsto \eta(t)$ from above by a positive multiple of the inverse of the monotonically increasing function
\begin{equation}
\Psi(\eta):=\int_0^\eta y^{-1}\left(\left[\left(1+y\frac{u_x^0(\xi)}{\alpha}\right)^{-1}\right]^{\xi_2}_{\xi_1}\right)\, dy.
\end{equation}
First, we note that $\Psi$ is well-defined around $\eta=0$ since
\begin{equation}
\begin{split}
 \lim_{y\to 0}y^{-1}\left(\left[\left(1+y\frac{u_x^0(\xi)}{\alpha}\right)^{-1}\right]^{\xi_2}_{\xi_1}\right)

&=\left(\frac{u_x^0(\xi_2)-u_x^0(\xi_1)}{|\alpha|}\right)<\infty.
\end{split}
\end{equation}
In fact, by partial fraction decomposition, we can write $\Psi$ explicitly as
\begin{equation}
\begin{split}
\Psi(\eta)&=\int_0^\eta y^{-1}\left(\left[\left(1+y\frac{u_x^0(\xi)}{\alpha}\right)^{-1}\right]^{\xi_2}_{\xi_1}\right)\, dy\\
&=\int_0^\eta\frac{1}{y}\frac{1+y\frac{u_x^0(\xi_1)}{\alpha}-\left(1+y\frac{u_x^0(\xi_2)}{\alpha}\right)}{\left(1+y\frac{u_x^0(\xi_1)}{\alpha}\right)\left(1+y\frac{u_x^0(\xi_2)}{\alpha}\right)} \,dy\\
&=\int_0^\eta \frac{u_x^0(\xi_2)-u_x^0(\xi_1)}{|\alpha|\left(1+y\frac{u_x^0(\xi_1)}{\alpha}\right)\left(1+y\frac{u_x^0(\xi_2)}{\alpha}\right)}\, dy\\
&=\int_0^\eta -\frac{u_x^0(\xi_2)}{\alpha\left(1+y\frac{u_x^0(\xi_2)}{\alpha}\right)}+\frac{u_x^0(\xi_1)}{\alpha\left(1+y\frac{u_x^0(\xi_1)}{\alpha}\right)}\, dy\\
&=\left[-\log\left(1+y\frac{u_x^0(\xi_2)}{\alpha}\right)+\log\left(1+y\frac{u_x^0(\xi_1)}{\alpha}\right)\right]_0^\eta\\
&=\log\left(\frac{1+\eta\frac{u_x^0(\xi_1)}{\alpha}}{1+\eta\frac{u_x^0(\xi_2)}{\alpha}}\right),
\end{split}
\end{equation}
which shows that $\Psi\to\infty$ as $\eta\to\eta*:=\frac{|\alpha|}{u_x^0(\xi_2)}$.\\
As $t\mapsto\eta(t)$ is the inverse of the monotonically increasing function $\Psi$, the minimal instance of blow-up for $\Psi$ is the smallest upper bound on $\eta$.
The instance of blow up for $\Psi$ is minimal for any choice of $\xi_2$ such that $u_x^0(\xi_2)=u_{max}$. Indeed, we have that
\begin{equation}
\min_{\xi_2\in[0,1]}\eta^*=\min_{\xi_2\in[0,1]}-\frac{\alpha}{u_x^0(\xi_2)}=\frac{|\alpha|}{u_{max}}.
\end{equation}
Therefore, the function $\eta$ is monotonically increasing and bounded from above by $\frac{|\alpha|}{u_{max}}$ and hence, the solution exists for all times. This proves the claim.\\

\textbf{The case $-1<a$ or $-\infty<\alpha<0$}\\
Applying Bernoulli's inequality to equation \eqref{eqeta} after squaring gives
\begin{equation}\label{inequa-10}
\begin{split}
1&=\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^1\left[1+\eta(t)\frac{u^0_x(s)}{\alpha}\right]^{2\frac{\alpha}{2}}\,ds\\
&=\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^1\left[1+2\eta(t)\frac{u^0_x(s)}{\alpha}+\frac{1}{\alpha^2}\eta^2(t)\Big(u_x^0(s)\Big)^2\right]^{\frac{\alpha}{2}}\,ds\\
&\geq\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^1 1+\eta(t)u^0_x(s)+\frac{1}{2\alpha}\eta^2(t)\Big(u_x^0(s)\Big)^2\,ds\\
&=\left(\eta'(t)\right)^\frac{1}{a+1}\left(1+\frac{1}{2\alpha}\|u_x^0\|_{L^2(0,1)}^2\eta^2(t)\right),
\end{split}
\end{equation}
for all $t\in(0,t^*)$, where in the last step, we have also used Dirichlet or periodic boundary conditions. Since $a+1>0$, the function $x\mapsto x^{a+1}$ is monotonically increasing and it follows from \eqref{inequa-10} that $\eta$ can be bounded from above by a monotonically increasing function which goes to
\begin{equation}
\eta^*= \frac{\sqrt{2|\alpha|}}{\|u_x^0\|_{L^2(0,1)}},
\end{equation}
as $t\to\infty$. Now, we can estimate
\begin{equation}
\begin{split}
1+\eta(t)\frac{u^0_x(s)}{\alpha}&\geq 1-\eta(t)\frac{u_{max}}{|\alpha|}\\
&\geq 1-\frac{\sqrt{2|\alpha|}}{\|u_x^0\|_{L^2(0,1)}}\frac{u_{max}}{|\alpha|}\\
&>0,
\end{split}
\end{equation}
by assumption \eqref{Assa-10}. This proves the claim.\\

\textbf{The case $1\leq a$ or $-1\leq\alpha<0$}\\
Assume, to the contrary, that $1+\eta(t)\frac{u^0_x(\xi)}{\alpha}>0$, while $\eta'(t)>0$, for all $(\xi,t)\in[0,1]\times[0,\infty)$. Since $-1<\alpha<0$ for this parameter range, we can apply Jensen's inequality with the convex function $y\mapsto -y^{-\alpha}$ to equation \eqref{eqeta} to obtain
\begin{equation}
\begin{split}
1&=\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^1\left[1+\eta(t)\frac{u^0_x(s)}{\alpha}\right]^{\alpha}\,ds\\
&\leq \left(\eta'(t)\right)^\frac{1}{a+1}\left(\int_0^1\left[1+\eta(t)\frac{u^0_x(s)}{\alpha}\right]^{-1}\,ds\right)^{-\alpha},\end{split}
\end{equation}
or, equivalently, after taking an $(a+1)$-power,
\begin{equation}\label{inequa1}
1\leq \eta'(t)\left(\int_0^1\left[1+\eta(t)\frac{u^0_x(s)}{\alpha}\right]^{-1}\,ds\right)^2.
\end{equation}

Integrating both sides of the inequality \eqref{inequa1} with respect to $t$ gives
\begin{equation}
t\leq \int_0^{\eta(t)}\left(\int_0^1\left[1+y\frac{u_x^0(s)}{\alpha}\right]^{-1}\right)^2\, dy\leq \int_0^{\frac{|\alpha|}{u_{max}}}\left(\int_0^1\left[1+y\frac{u_x^0(s)}{\alpha}\right]^{-1}\right)^2\, dy<\infty,
\end{equation}
which is a contraction. Here, we have used that $1+\eta(t)\frac{u_x^0(\xi)}{\alpha}>0$ for all $(\xi,t)\in [0,1]\times [0,\infty)$ for the second inequality. This proves the claim.

\end{proof}

\begin{remark}
Generally, the integral expression on the left-hand side of equation \eqref{eqeta} can be expanded in a Binomial series around $\eta=0$ as
\begin{equation}
\int_0^1\left[1+\eta\frac{u^0_x(s)}{\alpha}\right]^{\alpha}\,ds=\sum_{n=0}^\infty {\alpha\choose n} \alpha^{-n}\left[\int_0^1\Big(u_x^0(s)\Big)^n\, ds\right]\eta^n,
\end{equation}
which, since
\begin{equation}
\lim_{n\to\infty}\left|{\alpha \choose n}\right|^{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{\Gamma(-\alpha)^{\frac{1}{n}}n^{\frac{\alpha+1}{n}}}=1,
\end{equation}
for $\alpha\notin \mathbb{N}$ (where the Gamma function has poles), defines an analytic function with radius of convergence
\begin{equation}\label{radiusofconvergence}
\begin{split}
R&=|\alpha|\left\{\limsup_{n\to\infty}\left|\int_0^1\Big(u_x^0(s)\Big)^n\, ds\right|^{\frac{1}{n}}\right\}^{-1}\\
&\geq|\alpha|\|u_x^0\|_{L^\infty(0,1)},
\end{split}
\end{equation}
where the inequality in \eqref{radiusofconvergence} can be deduced by passing to the subsequence of even $n's$.
Since $u_{max}$ and $u_{min}$ can be very different, we see that there is no hope on a general global existence or singularity criterion solely relying upon the failure of analyticity, i.e., of $\eta$ approaching $\|u_x^0\|_{L^\infty(0,1)}$.
\end{remark}

The following theorem presents more refined conditions which guarantee global existence in the parameter range $a>-1$.

\begin{theorem}[Improved Global Existence for $a>-1$]\label{improvedthm}
Let $u\in C\Big(0,t^*; C^{1,Lip}(0,1)\Big)$ be a solution to equation \eqref{GPJa} with $-1<a$, either with Dirichlet boundary conditions \eqref{DirichletBC} or with periodic boundary conditions and zero mean \eqref{periodicBC}, and let  $t^*\in (0,\infty]$ be its maximal existence time.
\begin{itemize}
	\item For any $n\geq2$, let $\eta^*$ be the minimal positive root (if such a root exists) of the polynomial
	\begin{equation}
	p_n(\eta)=1+\frac{1}{n}\sum_{k=2}^n{n \choose k}\left(\frac{-2}{a+1}\right)^{1-k}\left[\int_0^1\Big(u_x^0(s)\Big)^k\, ds\right]\eta^k,
	\end{equation}
	and assume that
	\begin{equation}
	\eta^*u_{max}<\frac{2}{a+1}.
	\end{equation}
	Then, the solution exists globally in time, i.e., $t^*=\infty$.
\end{itemize}
\end{theorem}
\begin{proof}
The proof is completely analogous to the proof of the third case in Theorem \ref{mainthm}. Indeed, we can apply Bernoulli's inequality after taking an $n^{th}$ power in the integral of equation \eqref{eqeta} to obtain
\begin{equation}
\begin{split}
1&=\left(\eta'(t)\right)^\frac{1}{a+1}\int_0^1\left[1+\eta(t)\frac{u^0_x(s)}{\alpha}\right]^{n\frac{\alpha}{n}}\,ds\\
&\geq \left(\eta'(t)\right)^\frac{1}{a+1}\left(1+\frac{1}{n}\sum_{k=2}^n{n \choose k}\alpha^{1-k}\left[\int_0^1\Big(u_x^0(s)\Big)^k\, ds\right]\eta^k\right),
\end{split}
\end{equation}
implying, since $x\mapsto x^{a+1}$ is monotonically increasing, that $\eta$ can be bounded from above by a monotonically increasing function which can be bounded from above by $\eta^*$. Since, by assumption $\eta^*u_{max}<-\alpha$, the claim readily follows.
\end{proof}

\begin{remark}
If, in addition to the assumptions of Theorem \ref{improvedthm}, the initial velocity field $u^0$ is even, i.e.,
\begin{equation}
u^0\left(\frac{1}{2}-x\right)=u^0\left(\frac{1}{2}+x\right),\quad x\in\left[0,\frac{1}{2}\right],
\end{equation}
or odd, i.e.,
\begin{equation}
u^0\left(\frac{1}{2}-x\right)=-u^0\left(\frac{1}{2}+x\right),\quad x\in\left[0,\frac{1}{2}\right],
\end{equation}
then less restrictive assumptions can be made to guarantee global existence.
\end{remark}

\begin{example}\label{example}
Let us consider the initial condition $u_0(x)=\gamma x(x-1)$, for which finite-time blow up was proved in \cite{childress_ierley_spiegel_young_1989}, once again and contrast it to assumption \eqref{Assa-10}. We calculate
\begin{equation}
\begin{split}
\frac{1}{\sqrt{a+1}}\|u_x^0\|_{L^2(0,1)}&=\sqrt{\frac{1}{2}\int_0^1\gamma^2(2x-1)^2\,dx}\\
&=\frac{\gamma}{\sqrt{3}},
\end{split}
\end{equation}
while $u_{max}=\gamma$, which shows that the condition \eqref{Assa-10} is not satisfied for this initial condition.
\end{example}
\begin{example}[$a\geq1$, Proudman--Johnson-type singularity]
Consider the GPJ equation with $a\geq1$. In their paper \cite{childress_ierley_spiegel_young_1989}, the authors analyzed different blow-up scenarios for different initial conditions, cf. also Example \ref{examplea=1}. Here, we consider the initial condition
\begin{equation}\label{ua=1}
u_x^0(x)=
\begin{cases}\gamma-4\gamma x\qquad&\text{ for } 0<x<\frac{1}{2}\\
4\gamma x-3\gamma\qquad&\text{ for } \frac{1}{2}<x<1,
\end{cases}
\end{equation}
for $\gamma\in\mathbb{R}$, such that $u^0$ satisfies either Dirichlet boundary conditions or is mean-free. The Fourier series of $u_x^0$ is given by
\begin{equation}\label{Fourierexample}
u_x^0(x)=\frac{\gamma}{2\pi^2}\sum_{n\in\mathbb{Z}^*}\frac{(e^{\pi\ri n}-1)[2(e^{\pi\ri n}-1)-\pi\ri n(e^{\pi\ri n}+1)]}{n^2}e^{2\pi\ri nx},
\end{equation}
and since $u_x^0$ is Lipschitz continuous, we find that the Fourier series \eqref{Fourierexample} converges uniformly, cf. Figure \ref{FigAbsandmode3}. Hence, we can choose a sufficiently high truncation of \eqref{Fourierexample} to obtain a smooth (even analytic) initial condition that satisfies the singularity criterion in Theorem \ref{mainthm}.\\
\begin{figure}
	\includegraphics[scale=0.6]{Absandmode3}
	\caption{Initial condition \eqref{ua=1} for $\gamma=0.5$ together with its three-mode truncation.}
	\label{FigAbsandmode3}
\end{figure}
Indeed, equation \eqref{eqeta} can be integrated explicitly to
\begin{equation}\label{intexample}
\begin{split}1&=\sqrt{\eta'(t)}\int_0^1\frac{1}{1-\frac{a+1}{2}\eta(t)u^0_x(s)}\,ds\\
&=\sqrt{\eta'(t)}\left(\int_0^{\frac{1}{2}}\frac{ds}{1-\frac{a+1}{2}\gamma\eta(t)+2(a+1)\gamma\eta(t)s}+\int_{\frac{1}{2}}^1\frac{ds}{1+3\frac{a+1}{2}\gamma\eta(t)-2(a+1)\gamma\eta(t)s}\right)\\
&=\sqrt{\eta'(t)}\left(\frac{1}{2(a+1)\gamma \eta(t)}\left[\log\left(1-\frac{a+1}{2}\gamma\eta(t)+2(a+1)\gamma\eta(t)s\right)\right]_{0}^{\frac{1}{2}}\right.\\
&\qquad\qquad\qquad\qquad\left.-\frac{1}{2(a+1)\gamma \eta(t)}\left[\log\left(1+3\frac{a+1}{2}\gamma\eta(t)-2(a+1)\gamma\eta(t)s\right)\right]_{\frac{1}{2}}^1\right)\\
&=\frac{\sqrt{\eta'(t)}}{2(a+1)\gamma \eta(t)}\left[\log\left(1+\frac{a+1}{2}\gamma \eta(t)\right)-\log\left(1-\frac{a+1}{2}\gamma \eta(t)\right)\right.\\
&\qquad\qquad\qquad\qquad\left.-\log\left(1-\frac{a+1}{2}\gamma \eta(t)\right)+\log\left(1+\frac{a+1}{2}\gamma \eta(t)\right)\right]\\
&=\frac{\sqrt{\eta'(t)}}{2(a+1)\gamma\eta(t)}\log\left|\frac{1+\frac{a+1}{2}\gamma\eta(t)}{1-\frac{a+1}{2}\gamma\eta(t)}\right|,
\end{split}
\end{equation}
and, since the logarithmic expression in \eqref{intexample} is square-integrable, it follows that $\eta$ becomes unbounded and hence, the solution becomes singular in finite time. The blow-up time depends on $a$ and $\gamma$.\\
For a sufficiently high modular truncation, we have also found a smooth (in fact real analytic) initial condition which leads to formation of a singularity for the parameter value $a>1$. This answers a question by Okamoto \cite{Okamotopresentation}.\\
Note, however, that for the one-mode truncation
\begin{equation}\label{initialcos}
u_x^0=\frac{8\gamma}{\pi^2}\cos(2\pi x),
\end{equation}
equation \eqref{eqeta} takes the form
\begin{equation}\label{intexample2}
\begin{split}1&=\sqrt{\eta'(t)}\int_0^1\frac{1}{1-\eta(t)u^0_x(s)}\,ds\\
&=\sqrt{\eta'(t)}\int_0^1\frac{ds}{1-\eta(t)\frac{8\gamma}{\pi^2}\cos(2\pi s)}\\
&=\sqrt{\eta'(t)}\frac{1}{\pi}\int_0^\infty\frac{ds}{1-\eta(t)\frac{8\gamma}{\pi^2}+\left(1+\eta(t)\frac{8\gamma}{\pi^2}\right)s^2}\\
&=\sqrt{\eta'(t)}\frac{1}{1-\eta(t)\frac{8\gamma}{\pi^2}},
\end{split}
\end{equation}
and hence, $\eta(t)< \frac{8\gamma}{\pi^2}$ for all $t\in(0,\infty)$. In particular, the solution with initial condition \eqref{initialcos} exists for all times.\\
Also, we can check condition \eqref{Assa-10} for the initial condition \eqref{initialcos}. Indeed, we find that
\begin{equation}
u_{max}=\frac{8\gamma}{\pi^2}>\frac{8\gamma}{\pi^2}\frac{1}{\sqrt{2}\sqrt{a+1}}=\frac{1}{\sqrt{a+1}}\|u_x^0\|_{L^2(0,1)},
\end{equation}
which shows that condition \eqref{Assa-10} is violated for $a$ sufficiently large.
\end{example}

\section{Summary and Further Perspectives}
We proved finite-time existence criteria as well as global existence criteria for the GPJ equation in dependence of the parameter $a$. We strengthened existing conditions on the formation of a singularity and provided new conditions depending on properties of the initial velocity profile. In particular, we showed that for $a>1$  solutions with sufficiently well behave initial conditions will exist for all times, while sufficiently singular - but still smooth - initial conditions will lead to the formation of a finite-time singularity. This behavior is exemplified for a truncation of a zigzag function. In the appendix, we also gave a physical derivation of the GPJ equation from the compressible, two-dimensional Euler equations.\\
While the finite-time existence results in, e.g., \cite{Sarria2013, SarriaSaxton2013} also specified the nature of the singularity, such as blow-up in the $L^p$-norm of $u_x$, Theorem \ref{mainthm} only guarantees that the solution cannot be extended, but does not give further information on the nature of the singularity. Does the solution blow up or does the solution lose its (Lipschitz) regularity, i.e., is there blow-up in its derivatives? In \cite{Sarria2013}, the authors showed, for $a=1$, that there exist smooth initial data such that the solutions exist globally. Since their results were obtained with quite different techniques, it would be interesting to compare the singularity and global existences results in Theorem \ref{mainthm} for $a\geq 1$ and $a>-1$ respectively to the case of $a=1$ in \cite{Sarria2013}\\
Also, it would be interesting to extend the flow map approach laid out in the present paper to also give information on how a solution becomes singular. This will, however, also require information on the second derivative of $t\mapsto\eta(t)$, which has to be derived from new estimates.\\

\textbf{Acknowledgments}. \\
	The author would like to thank the anonymous reviewer for several useful comments and suggestions.

\section{Appendix: Physical Derivation of the GPJ Equation}\label{Appendix}
In this section, we provide a physical derivation of the GPJ equation for a general value of $a$. For special values of $a$, the GPJ equation is already known to have physical applications, cf. \cite{okamoto2000}. We will find that the GPJ equations model the dynamics of an inviscid, compressible fluid close to a wall. This is an immediate generalization of the initial motivation of the Proudman--Johnson equation, modeling inviscid, incompressible fluid motion close a wall, cf. \cite{childress_ierley_spiegel_young_1989}.\\
Consider the three-dimensional Euler equations for an isentropic, compressible fluid in vorticity form, \cite[p.24]{chorin2012mathematical},
\begin{equation}\label{Euler}
\frac{\partial}{\partial t}\left(\frac{\boldsymbol{\omega}}{\rho}\right)+(\mathbf{u}\cdot\nabla)\left(\frac{\boldsymbol{\omega}}{\rho}\right)=\left(\frac{\boldsymbol{\omega}}{\rho}\cdot\nabla\right)\mathbf{u},
\end{equation}
together with the equation of mass conservation
\begin{equation}
\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{u})=0,
\end{equation}
for a three-dimensional velocity field $\mathbf{u}:\mathbb{R}^3\times (0,t^*)\to\mathbb{R}^3,\quad (x,y,z:t)\mapsto\mathbf{u}(x,y,z;t)$, the three-dimensional vorticity $\boldsymbol{\omega}=\nabla\times\mathbf{u}$ and the non-negative mass density $\rho:\mathbb{R}^3\times(0,t^*)\to\mathbb{R}$. Here, we denoted the minimum of the maximal existence times of $\mathbf{u}$ and $\rho$ as $t^*$.\\
After differentiating the product expressions, equation \eqref{Euler} can be written equivalently in terms of the material derivative $\frac{D}{Dt}:=\frac{\partial}{\partial t}+\mathbf{u}\cdot\nabla$
as
\begin{equation}\label{Eulermaterial}
\frac{1}{\rho}\frac{D\boldsymbol{\omega}}{Dt}-\frac{1}{\rho^2}\frac{D\rho}{Dt}\boldsymbol{\omega}=\left(\frac{\boldsymbol{\omega}}{\rho}\cdot\nabla\right)\mathbf{u}.
\end{equation}
Since the mass density $\rho$ is non-negative, there exists a scalar function $R:\mathbb{R}^3\times(0,t^*)\to\mathbb{R}$ such that
\begin{equation}\label{exprho}
\rho(x,y,z;t)=e^{R(x,y,z;t)}.
\end{equation}
Assuming a velocity field of the form
\begin{equation}\label{ansatzu}
\mathbf{u}(x,y,z;t)=\Big(u(x,t),\beta u_x(x,t)y,0\Big),
\end{equation}
for $\beta\in\mathbb{R}$ a parameter, an unknown scalar function $u:\mathbb{R}\times(0,t^*)\to\mathbb{R}$ and $y\geq 0$, the three-dimensional vorticity becomes
\begin{equation}
\boldsymbol{\omega}(x,y,z;t)=\Big(0,0,-\beta u_{xx}(x,t)y\Big),
\end{equation}
while the divergence of $\mathbf{u}$ reads
\begin{equation}\label{divu}
\nabla\cdot\mathbf{u}=(1+\beta)u_x.
\end{equation}
Introducing the scalar vorticity
\begin{equation}
\omega(x,y;t)=-\beta u_{xx}(x,t)y,
\end{equation}
equation \eqref{Eulermaterial} simplifies to the scalar equation
\begin{equation}
\frac{1}{\rho}\frac{D\omega}{Dt}-\frac{1}{\rho^2}\frac{D\rho}{Dt}\omega=0,
\end{equation}
or equivalently, in terms of the exponential density \eqref{exprho}, after multiplication by $e^R$,
\begin{equation}\label{Eulerred}
\frac{D\omega}{Dt}=\frac{DR}{Dt}\omega.
\end{equation}
Since, by \eqref{exprho} and by \eqref{divu}, the equation of mass conservation becomes
\begin{equation}
\frac{DR}{Dt}=-\nabla\cdot\mathbf{u}=-(1+\beta)u_x,
\end{equation}
the Euler equations \eqref{Eulerred} take the form
\begin{equation}
\omega_t+u\omega_x+\beta u_x\omega_y=-(1+\beta)u_x\omega,
\end{equation}
or, in terms of the scalar velocity $u$, after division by $-\beta y$,
\begin{equation}\label{GPJxx}
u_{xxt}+uu_{xxx}+(1+2\beta)u_xu_{xx}=0.
\end{equation}
Setting
\begin{equation}
a:=-(1+2\beta),
\end{equation}
we obtain the GPJ equation
\begin{equation}\label{GPJapendix}
u_{xxt}+uu_{xxx}-au_xu_{xx}=0,
\end{equation}
as defined in \eqref{GPJa}.\\
If we now assume periodic boundary conditions in the ansatz \eqref{ansatzu}, or if we define the scalar velocity $u$ on the interval $[0,1]$ together with Dirichlet boundary conditions, we obtain the according boundary conditions for the GPJ equation.\\
As the velocity field \eqref{ansatzu} becomes unbounded for $y\to\infty$, the GPJ equations give a model for, e.g., near-wall dynamics for compressible, inviscid fluids only for small enough $y$. \\

\bibliographystyle{abbrv}

redpajama_set_name: RedPajamaArXiv


text:
Q: Firebase promises-based Auth flow always redirects I'd like to have the copy-paste "Firebase Authorize using Google" flow for my app -- and I do, with one problem: it always redirects to Google, even when it is already authorized and bounces right back.  Is there any way to have it not bounce through Google when the user has recently authenticated and is a returning user?
Promise.resolve().then(function(){
  return new Promise(function(resolve) {
    console.group('init');

    firebase.initializeApp({
      apiKey: "MY KEY",
      authDomain: "MINE.firebaseapp.com",
      databaseURL: "https://MINE.firebaseio.com",
      storageBucket: "MINE.appspot.com",
      messagingSenderId: "ALSOMINE"
    });

    // Using this to make sure not in intermediate state
    firebase.auth().onAuthStateChanged(function(user) {
      console.log('User state stablilized.');
      resolve();
    });
  });
}).then(function() {
  // All the various login states
  return new Promise(function(resolve, reject) {
    if (firebase.auth().currentUser) {
      console.log('You were already logged in - great!')
      resolve(firebase.auth().currentUser);
    }

    firebase.auth().getRedirectResult().then(function(result) {
      if (result.user) {
        console.log('Welcome back and congratulations on your successful login!', result.user);
        resolve(result.user);
      } else {
        console.log('Not logged in, not returning - time to send you on your way.');
        var provider = new firebase.auth.GoogleAuthProvider();
        provider.addScope('https://www.googleapis.com/auth/plus.login');
        firebase.auth().signInWithRedirect(provider);
        console.log('You should never see this line because it redirected!');
      }
    }).catch(function(error) {
      console.error('Returning from redirect login error:', error);
      reject(error);
    });
  });

}).then(function(user) {
  console.log('Signed in user', firebase.auth().currentUser.email);
  // LETS ROCK

A: After you call:
resolve(firebase.auth().currentUser);
Call:
return;

redpajama_set_name: RedPajamaStackExchange


text:
Q: Edit function but still call original function I am trying to edit every function under the UI "class" to return their first argument but still run their native code.
Here is what I have tried:
for(var i in UI) {
    var cur = UI[i];
    UI[i] = function() {
        cur.call(this, arguments);
        return arguments[0];
    }
}

This returns my arguments but no longer runs the original function correctly as the UI element is not called.
Thanks in advance.

A: Typical JavaScript closure inside loops – simple practical example problem. There is only a single cur variable in your scope and it will have the value assigned to it in the last iteration of the loop.
Use let and pass the arguments along properly (via rest args and apply):
for(let i in UI) {
    let cur = UI[i];
    UI[i] = function(...args) {
        cur.apply(this, args);
        return args[0];
    }
}

If you are stuck with ES5:
for(var i in UI) {
  (function(cur) {
    UI[i] = function() {
        cur.apply(this, Array.prototype.slice.call(arguments));
        return arguments[0];
    }
  }(UI[i]));
}


A glowing blue gateway swirls inside an open doorway. Portal steps in front of the entrance to another dimension and drapes his trench coat over himself, slipping an arm into each sleeve. He's valiantly preparing to depart on another adventure in the multiverse.
Angel of Justice stretches her wings beside him in their shared cubicle, ready to see him off. She observes the special key dangling from a string around Portal's neck. "I've never noticed how nice your necklace looks," she tells him.
Portal looks up at Angel and their eyes meet. He is completely taken by her angelic aurora and forgets to reply. Angel smiles and laughs. "Portal…?" He snaps out of her trance and blushes embarrassed.
Angel confidently leans in close to Portal and takes the key in her hand. Portal gulps, unsure of how to deal with her close proximity. Angel examines the key while Portal studies her face, lost in her splendor.
Portal places one foot into the blue whirling doorway. "Well! It's time for me to go on another adventure! Wish me luck!" His trench coat blows in the wind created by the portal as he enters the blue light. I must look so awesome right now, he thinks to himself, before realizing that he'll need the key to get back!
"Oh no! I can't be separated from my key or…!" Portal exclaims as he loses his balance and falls through the gateway. The door swings closed behind him and collapses with a pop!
Check back here next Tuesday for Chapter 5!


Round up your favorite board games, grab a pizza, and have a game night! Take it a step further by creating a custom Monopoly-style game filled with places that are meaningful to you as a couple (with your wedding venue as Boardwalk, naturally).
Love cooking? Painting? Dancing? Sign up for a class together to expand your horizons. Bonus: You'll have built-in date nights for as long as the class runs.
Explore your area like you did when you were kids—on bikes. If you're feeling adventurous, rent a tandem bike for two. Have a picnic in the park while you're out and about.
A full vacation may be (way) out of the budget right now, but an overnight trip to a bed & breakfast can give you the R&R you need without breaking the bank.
Baseball fan? Find out where your favorite team does their spring training. You can score great seats for a fraction of the regular-season tickets. Plus, you can get much closer to your favorite players than you can at a regular-season game, and you might even snag an autograph.


It isn't easy photographing dogs but over our 15 years of studio experience, we have developed a method that guarantees stunning results every time!
Our team are extensively trained to work with dogs in a studio environment and understand how different each and every dog is. That is why we aim to capture your dog's unique personality.
Cheeky? Adorable? Lounger? Or energetic? Tell us a little about your pooch and we will decide on a theme for the photoshoot. Our Pawtrait shoots are all about having fun and capturing that unique character in your dog, so why not bring along some of their favourite toys or encourage them to show off their special tricks!
We will take a wide range of photos during the session, and in a couple of days we will ask you along to view the best images in your own slideshow.
Once you have selected your favourite shots your photographer will take you through all of the wonderful and amazing options of how to best display them in your home. Whether you want clean and modern frames or a large canvas – whatever your tastes, we will help you get that WOW factor!
You can build your own print package with your Pawtrait Shots starting at just £34.95 for 30 minutes. Add as many or as few prints, frames, canvases and optional extras thereafter as you like. You can even take all your photos away with you on a CD for just £50.
It's Time to Fluff your Tails and Comb your Coats!
Of course you want your pooch to look their best for their Photoshoot, so why not bring them along for a visit to our Grooming Spa before? If you do, we'll give you an extra £10 OFF our Spritz & Go service.
Simply click on the coupon below and present the PDF (printed or digitally) to a member of staff when you book your studio shoot.


On the basis of this icchA -- the first wish, as it may be called, and the Upanishad also says: 'akAmayata' - the kriyA-shakti (the power of Action) begins the pancha-kRtya-leelA. Thus, what was the paraBrahman by itself in itself, willed to 'become'. It is for this divine will that the Upanishad uses the word 'kAma', meaning desire.
This 'desire' is not to be taken in any derogatory sense. It is pure Divine Will from Being to Becoming. Thus the first evolute from Brahman is this divine kAma. So the Shakti that is the origin of this is called KAmeshvari and the ShivaM in which this kAma sprouted is therefore called KAmeshwara.


Raquel André is a collector of rare things.
In Lisbon, Ponta Delgada, Rio de Janeiro, Loulé, Minde, Paredes de Coura, Sever do Vouga, Ovar, Manaus and Barreiro, she has collected over 137 meetings and counting (up til November 2017). People of all nationalities, genders and ages have accepted to meet her at someone's flat, and for an hour, to construct a fictional intimacy to be captured in memory and photographs. In each city that the performance travels, it collects more lovers and accumulates new meetings. The photographs and details of these meetings are the contents of a performance that tells a story about what this collection of relationships may mean. Just what are we looking for when we meet someone? In the age of e-mail, facebook, instagram, tinder, and grinder, we have all become experts at faking intimacy.
We post what we eat, who we kiss, where we go, what we're thinking and reading, what we like and dislike – all translated into views, likes, and comments.
Raquel's collection is the results of an obsessive fascination with the terabytes of information that exist in each minuscule movement of another person. It is a reflection on intimacy that is explored one-to-one and amplified for the stage, all real and all fake. Each time the door opens for a new lover, Raquel André dives into an abyss that is the other, and reality and fiction merge. Each encounter is real. The flirtation is real. The intimacy may feel more real than fiction. And Raquel, the obsessive collector, holds on to the moments of each meeting, the rare objects of her peculiar collection, ephemeral and infinite.
This work is part of your project Collection of People: Collection of Lovers, Collection of Collectors, Collection of Artists and Collection of Spectators.


TY Overnight trip to North Coast & Belfast.
Tuesday 22nd May-Wednesday 23rd May.
Cost (travel & accommodation) covered. Dinner on Tuesday and Breakfast on Wednesday also included.
Lunch money is needed for Tuesday & Wednesday - spending money optional.
Permission slip due Tuesday morning latest.
This surprise trip to the North is a truly cross curricular extravaganza: culture, geography, history, religion, politics and more. It is a wonderful opportunity to see the beautiful North Coast and fascinating Belfast pre-Brexit! The week of the trip is Ecumenical Bible Week in Dublin - how fitting to have the opportunity to stay near Ballymena in a Prebyterian Church hall as well as visiting some key historical and religious sites of significance, Catholic & Protestant.
North Coast: Giant's Causeway, Carrick-a-rede Bridge etc.
Late afternoon drive to accommodation in Ballymena via Dark Hedges (as seen in Game of Thrones).
Accommodation is Second Broughshane Presbyterian Church, 5 Raceview Road, Broughshane, Ballymena.
Sleeping bag, sleeping mat & pillow are required. There are shower and toilet facilities in the hall.
Arrival back at school approx 5pm.


text:
One-Year Anniversary of New Yorker-Inspired Portraits of Women Surgeons
UCSF Department of Surgery - by Richard Barg - May 26, 2017
Surgeons, from left, Lucy Kornblith, Rita Mukhtar and Carter Lebares pose for a photograph in the lobby of the UCSF Clinical Science Building in San Francisco, Calif., on Wednesday, April 19, 2017. (Courtesy Anda Chu/Bay Area News Group)
The April 3, 2017 cover of the New Yorker depicting four women in blue doctors' scrubs over an operating table has gone viral. Reflecting the power of that image, female surgeons throughout the world have been posting group selfie photos of themselves recreating the pose. Malika Favre, who has designed a number of covers for the magazine, told CNN that "none has ever taken off like this one."
The Mercury News reports on this phenomenon in the Bay Area, speaking to women surgeons at all three major academic medical centers in the region. That group included UCSF surgeons Lucy Kornblith, MD, Rita Mukhtar, M.D. and Carter Lebares, M.D., all of whom offered their thoughts in a video accompanying the story. (Note: Video appears below author line and date on Mercury News site does not display as a video in all browsers).
What does a female surgeon look like?
Obscured by her cap, mask and gown, it might be hard to tell for patients waking up in the operating room to the face of progress.
But a captivating New Yorker magazine cover this month of four women surgeons — dressed head to toe in surgical attire, only brows and eyes visible, gazing down on their patient — has sparked a surgeon selfie movement across social media from the Bay Area to Saudi Arabia.
Sister surgeons at Stanford University, UC San Francisco, UC Davis and many more are following the lead of a University of Wisconsin endocrine surgeon who re-created the image with three colleagues and posted the selfie on Twitter.
Dr. Lucy Kornblith thinks women surgeons in the Bay Area are already seeing the results. At UC San Francisco, after all, a number of leading female surgeon mentors, including Dr. Nancy Ascher, have been leading the way for years.
So when the New Yorker arrived in Kornblith's mailbox with the four female surgeons staring her in the face, "I didn't think anything of it, to be perfectly honest,'' Kornblith said. "It didn't strike me as being abnormal.''
Read full story in The Mercury News
Tags: Women in Surgery
People: Lucy Kornblith, M.D. Carter Lebares, M.D. Rita Mukhtar, M.D.
The Muriel Steele Society Invites You to Celebrate International Women's Day at UCSF
Lucy Kornblith, M.D. Recipient of 5-Year $1M NIH K23 Award for Study of Post-Injury Platelet Biology
Rachael Callcut, MD, MSPH Named Chair of Research and Scholarship Committee for EAST
Rebecca Plevin, MD Named Director of San Francisco Wraparound Project at ZSFG
Trauma Surgeon Peggy Knudson, MD Leads First Large-scale Investigation of Pulmonary Clots after Injury
CIRM Awards Tippi MacKenzie, MD $12.1M to Treat Fetuses in the Womb Alpha Thalassemia Major
Translating Regenerative Surgery on the Liver From Concept to Therapeutic Reality
Lauren Eyler and Catherine Juillard Develop Algorithm to Facilitate Health Disparities Research
Amanda Sammann Launches The Better Lab to Drive Innovation in Healthcare Delivery
Trauma Surgeon Rachael Callcut Participates in Fleet Week Disaster Training Aboard USS San Diego
Laura Esserman Named to TIME 100 List of Most Influential People in the World
Nancy Ascher Receives Chancellor Diversity Award for the Advancement of Women
Madhulika Varma Awarded Maurice Galante, MD Distinguished Professorship in Surgical Oncology
Dr. M. Margaret Knudson Named Medical Director of ACS-MHS Partnership
Avoiding Surgery in the Elderly
Postpartum Moms Turning to Plastic Surgery
Being Small Has Its Advantages, Especially With Minimally Invasive Lung Surgery

redpajama_set_name: RedPajamaCommonCrawl


text:
White House begins talks with lawmakers on COVID-19 relief
(AP) — Top aides to President Joe Biden on Sunday began talks with a group of moderate Senate Republicans and Democrats on a $1.9 trillion coronavirus relief package as Biden faces increasing headwinds in his effort to win bipartisan backing for the initial legislative effort of his presidency.
Lawmakers on the right question the wisdom of racking up bigger deficits while those on the left are urging Biden not to spend too much time on bipartisanship when the pandemic is killing thousands of Americans each day and costing more jobs amid tightening restrictions in many communities.
At least a dozen senators met for an hour and 15 minutes in a virtual call with White House National Economic Council director Brian Deese and other senior White House officials. Many hope to approve a relief package before former President Donald Trump's trial, which is set to begin in two weeks, overtakes Washington's attention.
Sen. Angus King, an independent from Maine, called the opening talks a "serious effort."
"There was not a hint of cynicism or lack of commitment to at least trying to work something out," King said. "If they were just trying to jam this through, I don't think it would have interrupted the Packers game."
King told reporters that there was "absolute consensus" among the group that the No. 1 priority was to speed up the distribution of vaccinations and expanding COVID-19 testing and tracing.
The White House did not seem to budge on breaking up the package or reducing the overall price tag, even as it pushes for bipartisan support. There was also no discussion of breaking the package apart or pushing it through on a procedural move that could be done without Republicans, King said.
Senators from both parties raised questions about the economic aid provisions, particularly making direct $1,400 payments to Americans more tailored to recipients based on need.
Senators also wanted more data on how the White House reached the $1.9 trillion figure.
Sen. Jeanne Shaheen, D-N.H., told The Associated Press that no red lines were drawn. But she added there was consensus among the call's participants "that the more targeted the aid is the more effective it can be."
Overall, "it was a conversation and it was not about drawing lines in the sand," Shaheen said. "It was about how can we work together to help the people of this country."
Many of the senators are from a bipartisan group that struck the contours of the last COVID-19 deal.
The two leaders of the House's Problem Solvers Caucus, Reps. Josh Gottheimer, D-N.J., and Tom Reed, D-N.Y., as well as White House coronavirus response coordinator Jeff Zients and White House legislative affairs director Louisa Terrell joined the call.
Out of the gate, Biden has made clear that quickly passing another round of coronavirus relief is a top priority as he seeks to get the surging pandemic and the related economic crisis under control, while demonstrating he can break the gridlock that has ailed Congress for much of the last two presidencies.
Biden and his aides in their public comments have stressed that his plan is a starting point and that finding common ground on relief should be attainable considering the devastating impact the pandemic is exacting on Democratic and Republican states alike. With more than 412,000 dead and the economy again losing jobs, Biden has argued there is no time to lose.
"We're going to continue to push because we can't wait," said White House principal deputy press secretary Karine Jean-Pierre. "Just because Washington has been gridlocked before doesn't mean it needs to continue to be gridlocked
Central to Biden's campaign pitch, beyond healing the wounds created by Trump's presidency, was that he was a proven bipartisan dealmaker, one who would draw upon his decades in the Senate and deep relationships with Republicans to bridge partisan divides.
India expects more clashes with Chinese troops in Himalayas
(Reuters) – A security assessment by Indian police in the Himalayan region of Ladakh says
(Reuters) – More than 160 people have died from the cold in Afghanistan this month
Seven shot dead in another mass shooting in California
(Reuters) – Seven people were killed in a mass shooting at two locations in the


Twelve stories of wayward travelers. Twelve stamps on the passport of the soul.
Twelve stories of wayward travelers. Twelves stamps on the passport of the soul. Walter struggles to see past his disfigurement, but a trip to Thailand and a mysterious woman may challenge his quest f... or perfection. Married to a workaholic, Leslie is accustomed to solitude until her overbearing Indian mother-in-law turns her kitchen into a verbal battlefield. A woman looks for the perfect souvenir in the Turkish Bazaar and instead finds a passionate love affair. From ice storms in Maine to swimming pools in Vera Cruz, errant souls must grapple with past baggage to forge profound connections and a new sense of direction. Bonds of Love & Blood is a poignant collection of prize-winning stories featuring life's unexpected arrivals and departures. If you like intimate accounts of lives at the crossroads, vibrant portrayals of global locales, and elegantly-crafted prose, then you'll love Marylee MacDonald's compelling tales of travel. Buy Bonds of Love & Blood to embark on a heart-wrenching journey of love, loss, and redemption today!


text:
Home › News › Keeping records, crunching numbers – for your health
Keeping records, crunching numbers – for your health
The Report: April / May 2005 vol.26 num.2
by DAN KEETON
hen you look around the health care field, youd be hard pressed to find a more specialized combination of skills than that held by Gerald Yu, a Health Records Administrator who oversees the collection and management of clinical data for the Fraser Health Authority.
Gerald Yu
Health Records Administrator
Steward, Eagle Ridge Hospital
Hospital Records Administrators must possess advanced computer skills and be proficient in statistics, classification systems, and pathophysiology. They must know how to interpret health records, and be able to pick out whats important ... because its based on the reading and interpretation of clinical data which ultimately become statistics that decisions are made about the management of health care.
Yus skills are particularly in demand now that health records are being transferred to electronic health records (EHR) and that data elements are being standardized across the health regions in Canada. The task is to use the same standards and be able to compare data across the region, the province, and the rest of Canada. Its an important job and one that could have potential for high stress. Yu, based at Eagle Ridge Hospital in Port Moody, enjoys the challenge.
The advantages of modernizing health information management are numerous, he says. It helps target the right kind of information to the right user, enhance cost savings, and make unified information available for governments, health administrators, doctors, and other stakeholders in the health care system.
For the past year and a half he has been the Coordinator of Data Quality for the region, working to streamline how facilities keep track of their health information. -Before this, every single hospital would do something different in collecting health information. Were looking at a system where we all collect the same set of data elements for a patient with 'appendicitis so that for example, an 'appendicitis profile can be created for research, program management, CQI (continuous quality improvement), etc."
Quality of data is measured by what Yu calls six -dimensions." -One is relevancy: is the data valuable to the systems stakeholders? Another is accuracy: does the data reflect the reality of the subject?"
Then theres consistency, which asks, -Is the data free from variation and contradiction? We collect two types of data, one for inpatient and the other for what is called surgical daycare," Yu explains.
-We report it to the ministries of health, federal and provincial, and the Canadian Institute of Health Information, which keeps a kind of national health score card. Internally, we use the data to manage various programs. For example, well compare lengths of stay at different hospitals and perhaps be able to realize some cost savings through benchmarking and CQI," Yu says.
Theres also the comparability dimension. Is the data comparable to data across the region, British Columbia, and the rest of Canada? And theres timeliness: can such information be made available promptly?
And finally, theres the question of accessibility. -We ask ourselves, 'Is it user friendly? What we want is to have a lot of the data available on the Internet, Intranet, and Internal Network Drive, readily accessible to the various users."
And in this age of protection of private information, privacy is closely guarded by Health Records Administrators. -We provide general statistics. No patients personal information is attached to those numbers. Mostly people just want to know, for example, how many pneumonia cases we see, and how long they stay in the hospital.
-Its such powerful data because its clinically based. It helps us to better manage [treatment] programs. If we see a spike in, say, respiratory cases, we see a trend and can gear towards doing some proactive things in the future. Maybe we can learn from past practices and address the issue at an ambulatory setting instead of having all the people come into the acute facility."
Yus job has evolved over time. HRAs were once known as health records librarians because the job involved keeping paper records. -We still have paper records but a lot of the information we have is now on computer. Its more than just dealing with health records ... its the whole gamut of information.
-We still have charts," Yu says. -Some people still prefer them, and there arent always enough computers. But eventually well all be using electronic health records and information."
Why did Yu opt for what he says is considered more a -geek" than a -glamour" profession? -I always wanted to get into health care, but I didnt want to be something like a nurse. I like statistics, research, and data reporting."
In BC, there are two routes to health information management (also known as health records administration). A two-year diploma program is offered at Douglas College in Greater Vancouver or online through Canadian Healthcare Association. And theres a four-year bachelor program at institutes such as the University of Victoria, University of Western Ontario in London, and Ryerson University in Toronto ... this route encompasses more in depth knowledge of management, information system, and research . Yu originally obtained a diploma and went on to complete a Masters of Health Administration at the University of BC.
There are few men in the profession, says Yu. -They are maybe only about three out of every 100 HRAs. Im hoping more men will get into it." He thinks his professional organizations, the Canadian Health Information Management Association and Health Record Association of BC, should more actively promote the profession.
An HSA steward for the past five years, Yus commitment to health care and his co-workers is clear. -We see health care as our number-one priority. Wed rather work with management, but we wont roll over. Well stand up for what we believe in. Its a two-way street."
RT @FHcareer: "My mom's a #healthcare worker and I look up to her. When I used to visit her at work I would think that it would b… https://t.co/duMWZFZP5s — 1 week 1 hour ago


I think that's when I began to have a bit of a crush on him, it's probably what got that watchband broken! Everyday at lunch his girl friend gave him such a hard time. He'd come to class, and we'd talk for a few minutes before it began. He canada goose outlet black friday was always so nice, I couldn't understand why she felt the need to treat him so badly..
canada goose clearance Do I in Evolution like I believe in the Resurrection of Jesus Christ? Absolutely not! I accept the Theory of Evolution like I use the Quadratic Formula; they are both useful for a certain class of problems that I sometimes have to solve. I certainly do not place my eternal life and soul in the care of a scientific theory or a mathematical formula. I am no atheist canada goose clearance.


Kinetic and adaptive assemblies can play an important role in future resilient and sustainable architecture. In the projects below, students investigate strategies for transformative designs.
Variations on Theo Kansen's walking mechanism.
Developing and testing variations of Theo Jansen's walking mechanism for new types of movements.
Using the kit-of-parts to generate variation and test for their feasibility.
Parametric studies of apertures using Grasshopper plugin for Rhino. This allowed for better understanding of constraints and possible solutions/arrangements.
Extending 2-dimensional geometry into 3-dimensional kinetic form.