text:
\section{Introduction}
Free boundary problems for incompressible, inviscid flows and for active scalars are mathematically challenging and physically interesting. Moreover, their applications are really spread, from geothermal reservoirs (see \cite{CF}) to tumor growth (see \cite{F}), passing through weather forecasting (see \cite{majda1996two, constantin1994singular}).

In particular, the evolution of a fluid in a porous medium is important in the Applied Sciences and Engineering (see \cite{bear}) but also in Mathematics (see, for instance, \cite{c-c-g10}). The effect of the medium has important consequences and the usual equations for the conservation of momentum, \emph{i.e.} the Euler or Navier-Stokes equations, must be replaced with an empirical law: Darcy's Law
\begin{equation}
\frac{\mu}{\kappa}v=-\nabla p -g(0,\rho),
\label{eq1}
\end{equation}
where $\mu$ is the dynamic viscosity of the fluid, $\kappa$ is the permeability of the porous medium, $g$ is the acceleration due to gravity, $v$ is the velocity of the fluid, $\rho$ is the density and $p$ is the pressure (see \cite{bear}). In our favourite units, we can assume $g=\mu=1.$

A very important part of the theory of flow in porous media studies the coexistence of two immiscible fluids with different qualities in the same volume. The case of two immiscible and incompressible fluids is known as the Muskat o Muskat-Leverett problem (see \cite{Muskat} and also \cite{SCH}). In this case the density is given by
\begin{equation}\label{density}
\rho=\rho^2\textbf{1}_{\{y<f(x,t)\}}+\rho^1\textbf{1}_{\{y>f(x,t)\}},
\end{equation}
where
\begin{equation}\label{gamma}
\Gamma(t)=\{(x,f(x,t)): \: x\in\mathbb R\},
\end{equation}
is the interface between both phases. This interface is an unknown in the evolution. If $\rho^2>\rho^1$, the system is in the so-called \emph{stable} (or Rayleigh-Taylor stable) regime.

Given the depth of the porous medium, $l>0$, we define the following dimensionless parameter (see \cite{bona2008asymptotic} and references therein)
\begin{equation}\label{dimensionless}
\mathcal{A}=\frac{\|f_0\|_{L^\infty}}{l}.
\end{equation}

If the porous medium has infinite depth ($\mathcal{A}=0$), the equation for the interface is
\begin{equation}\label{full}
\partial_t f=\frac{\rho^2-\rho^1}{2\pi}\text{P.V.}\int_\mathbb R \frac{(\partial_x f(x)-\partial_x f(x-\eta))\eta}{\eta^2+(f(x)-f(x-\eta))^2}d\eta,
\end{equation}
where $\text{P.V.}$ denotes principal value. This situation is known as the \emph{deep water regime}. This case has been extensively studied (see \cite{ambrose2004well, castro2012breakdown, ccfgl, ccgs-10, ccgs-13, c-c-g10, c-g07, c-g09, escher2011generalized, e-m10, KK} and references therein).

If the initial data verifies $|f_0|<l$, in the presence of impervious boundaries (see Figure \ref{bandafig}), the system is in the regime $0<\mathcal{A}<1$. This is known as the \emph{confined Muskat problem}. The equation for the interface corresponding to this situation when $l=\pi/2$ is
\begin{multline}
\partial_t f(x,t) = \frac{\rho^2-\rho^1}{4\pi}\text{P.V.}\int_\mathbb R\bigg{[}\frac{ (\partial_x f (x)-\partial_x f (x-\eta ))\sinh\left(\eta\right)}{\cosh \left(\eta\right)-\cos\left(f(x)-f(x-\eta)\right)}\\
+\frac{(\partial_x f(x)+\partial_x f(x-\eta))\sinh\left(\eta\right)}{\cosh \left(\eta\right)+\cos\left(f(x)+f(x-\eta)\right)}\bigg{]}d\eta.
\label{eq0.1}
\end{multline}
This case has been studied in \cite{BCG, CGO, GG, G}. Notice that the second kernel becomes singular when $f$ reaches the boundaries.

\begin{figure}[h!]
		\begin{center}
		\includegraphics[scale=0.3]{FH.eps}
		\end{center}
		\caption{Physical situation for an interface $\Gamma$ in the strip $\mathbb R\times (-l,l)$.}
\label{bandafig}
\end{figure}

It has been proved that $\mathcal{A}$ plays an important role on the evolution of $\|f(t)\|_{C^1}$. For instance, if $0<\mathcal{A}<1$, $\|f(t)\|_{L^\infty}$ decays slower than in the deep water regime. Moreover, to ensure that $\|\partial_x f(t)\|_{L^\infty}<1$ for every time, one needs to impose conditions on the amplitude and the slope of the initial data related to the depth $l$ (see \cite{CGO}). Notice that in the deep water case, the condition is only on $\|\partial_x f_0\|_{L^\infty}$ (see \cite{c-g09}). Finally, we also mention that, using a computer assisted proof, the authors in \cite{GG} proved that there exists curves, $z=(z_1(\alpha),z_2(\alpha))$, solutions of of the confined Muskat problem ($0<\mathcal{A}<1$) corresponding to the initial data $z_0$, such that $\partial_x z_1(0,\delta)<0$ for a sufficiently small time $\delta>0$ (\emph{i.e.} the wave \emph{breaks}). The same initial curve $z_0$ when plugged into the infinitely deep Muskat problem ($\mathcal{A}=0$) verifies $\partial_x z_1(0,\delta)>0$ (\emph{i.e.} the wave becomes a smooth graph).

The last case corresponds to $\mathcal{A}=1$. This case is known as the \emph{large amplitude regime}. In this situation, the initial interface reach the impervious walls at least in one point.

Let's write $\Lambda u=\sqrt{-\partial_x^2}u$ for the square root of the laplacian. In \cite{c-g-o08}, the authors proposed the problem
\begin{equation}\label{modelodiego}
\partial_t f(x)=-\frac{1}{1+(\partial_x f(x))^2}\Lambda f(x),
\end{equation}
as a model of the dynamics of an interface in the two phase, deep water Muskat problem \eqref{full}. If we define $g = \partial_x f$ and we take the derivative $\partial_x$ to equation \eqref{modelodiego}, we get
\begin{equation}\label{modelodiegoderiv}
g_t = -\Lambda g + \partial_x\left(\frac{g^2Hg}{1+g^2}\right).
\end{equation}
This latter equation will be helpful because it has divergence form.

In this paper, to study the effect of the boundaries in the large amplitude regime, we propose and study a model of \eqref{eq0.1}. Assume $l=\pi/2$ and $f(\tilde{x})=\pm l$ for some $\tilde{x}$. Then, the second term in \eqref{eq0.1} reduces to
$$
\frac{(\partial_x f(\tilde{x})+\partial_x f(\tilde{x}-\eta))\sinh\left(\eta\right)}{\cosh \left(\eta\right)+\cos\left(f(\tilde{x})+f(\tilde{x}-\eta)\right)}=\frac{\partial_x f(\tilde{x}-\eta)\sinh\left(\eta\right)}{\cosh \left(\eta\right)-\cos\left(f(\tilde{x})-f(\tilde{x}-\eta)\right)}\approx H\partial_x f(\tilde{x}),
$$
where $H$ denotes the Hilbert transform. Therefore, using \eqref{eq0.1}, $\partial_t f(\tilde{x})=0$ and the diffusion degenerates.

To capture this crucial fact when $f(\tilde{x})=\pm l$, we introduce the equation
\begin{equation}\label{model1}
\partial_t f(x)=\left(-\frac{1}{1+(\partial_x f(x))^2}+\frac{1}{1+l^2-\left(f(x)\right)^2}\right)\Lambda f(x),
\end{equation}
as a model of \eqref{eq0.1} in the large amplitude regime. Let's point out that, if $l=\infty$, formally, we recover \eqref{modelodiego}.

Notice that
\begin{equation}\label{eqRayleigh}
\frac{1}{1+(\partial_x f(x))^2}>\frac{1}{1+l^2-\left(f(x)\right)^2},
\end{equation}
guarantees that the model \eqref{model1} is in the so-called stable regime. In other words, if \eqref{eqRayleigh} holds, the model has a nonlocal, non-degenerate diffusion. Notice that the set of functions verifying \eqref{eqRayleigh} is not empty.

To bound the stability condition, we define
\begin{equation}\label{sigma}
\sigma(t)=\max_x\left\{\frac{1}{1+l^2-\left(f(x)\right)^2}-\frac{1}{1+\left(\partial_x f(x)\right)^2}\right\}.
\end{equation}
Using this function, the stability condition is equivalent to $\sigma(t)<0$.

\begin{remark}
Let us mention that
$$
f(x)=\pm l\sin(x),\pm l\cos(x),
$$
are steady solutions in the unstable case.
\end{remark}

\subsection{Notation and functional framework}
We write $Hu$ for the Hilbert transform of the function $u$ and $\Lambda u=\sqrt{-\partial_x^2}u=\partial_x Hu$ for the Zygmund operator, \emph{i.e.}
$$
\widehat{Hu}(\xi)=-i\frac{\xi}{|\xi|}\hat{u}(\xi)\text{ and }\widehat{\Lambda u}(\xi)=|\xi|\hat{u}(\xi),
$$
where $\hat{\cdot}$ denotes the usual Fourier transform.

We write $\Omega$ for our spatial domain. From this point onwards, we consider either $\Omega=\mathbb R$ or $\Omega=\mathbb T$, in particular our domain is always onedimensional. We write $H^s(\Omega)$ for the usual $L^2$-based Sobolev spaces with norm
$$
\|f\|_{H^s}^2=\|f\|_{L^2}^2+\|f\|_{\dot{H}^s}^2, \|f\|_{\dot{H}^s}=\|\Lambda^s f\|_{L^2},
$$
We denote $l>0$ the depth and
$$
H^s_l=\left\{f\in H^s\text{ s.t. }\|f\|_{L^\infty}< l\right\}.
$$
The fractional $L^p$-based Sobolev spaces, $W^{s,p}(\Omega)$, are
$$
W^{s,p}=\left\{f\in L^p(\Omega), \partial_x^{\lfloor s\rfloor} f\in L^p, \frac{|\partial_x^{\lfloor s\rfloor}f(x)-\partial_x^{\lfloor s\rfloor}f(y)|}{|x-y|^{\frac{1}{p}+(s-\lfloor s\rfloor)}}\in L^p(\Omega\times\Omega)\right\},
$$
with norm
$$
\|f\|_{W^{s,p}}^p=\|f\|_{L^p}^p+\|f\|_{\dot{W}^{s,p}}^p,
$$
$$
\|f\|_{\dot{W}^{s,p}}^p=\|\partial_x^{\lfloor s\rfloor} f\|^p_{L^p}+\int_\Omega\int_\Omega\frac{|\partial_x^{\lfloor s\rfloor}f(x)-\partial_x^{\lfloor s\rfloor}f(y)|^p}{|x-y|^{1+(s-\lfloor s \rfloor)p}}.
$$
Notice that $W^{k+s,\infty}$, $k\in\mathbb N$, $1>s\geq0$, reduces to the usual H\"older continuous space $C^{k+s}$. We define
$$
\mathcal{C}_l=\frac{l^2}{1+l^2}.
$$
This is the constant appearing in the linear problem.

\subsection{Statement of the results for \eqref{model1}}
In this section we collect the statement of the results concerning the equation \eqref{model1}. We start with local well-posedness of classical solutions and a continuation criterion when the initial data is in the stable regime:
\begin{teo}\label{teo1}
Let $f_0\in H^s_l(\Omega)$, $s\geq3$, $l>0$, $\epsilon>0$ and $\Omega=\mathbb T,\mathbb R$, be an initial data satisfying the stability condition \eqref{eqRayleigh}. Then there exists a unique solution $f(x,t)$ to \eqref{model1} such that
$$
f(x,t)\in C([0,T(f_0)],H^3)\cap L^2([0,T(f_0)],H^{3.5}).
$$
Moreover, if $T^*$ is the maximum lifespan of the solutions, then, $T^*=\infty$ or
\begin{equation}\label{continuationcriteria}
\sup_{0\leq t\leq {T^*}}\|f(t)\|_{W^{2+\epsilon,\infty}}=\infty.
\end{equation}
\end{teo}

Notice that the continuation criteria \eqref{continuationcriteria}, as is written in the previous result, doesn't deal with the possibility of reaching the unstable regime. However, in Proposition \ref{prop2} below we address this question.

We also prove that there is a unique local smooth solution even when the Rayleigh-Taylor condition \eqref{eqRayleigh} is not satisfied but our initial data is analytic. We prove this result complexifying the equation and using a Cauchy-Kowalevski Theorem (see \cite{nirenberg1972abstract} and \cite{nishida1977note}).

We define the complex strip $\mathcal S_r=\{x+i\xi,|\xi|<r\},$ and $\gamma=x\pm ir$, $\gamma'=x\pm ir'$ for $x\in\Omega$. We consider the Hardy-Sobolev spaces (see \cite{bakan2007hardy})
\begin{equation}\label{Xdef}
H^3(\mathcal S_r)=\{f(x+i\xi) \text{ analytic on } \mathcal S_r \text{ s.t. }f(x\pm ir)\in H^3(\Omega) \text{ and } f(x)\in\mathbb R\},
\end{equation}
with norm
$$
\|f(\gamma)\|^2_{H^3(\mathcal S_r)}=\sum_{\xi=\pm r}\|f(\gamma)\|^2_{H^3(\Omega)}.
$$
These spaces form a Banach scale with respect to the parameter $r$. In the same way we define $\|f\|^2_{L^2(\mathcal S_r)}=\sum_{\xi=\pm r}\|f(\gamma)\|^2_{L^2(\Omega)}.$ We also have, for $0<r'<r$,
\begin{equation}\label{cauchy2}
\|\partial_x \cdot\|_{L^2(\mathcal S_{r'})}\leq\frac{C}{r-r'}\|\cdot\|_{L^2(\mathcal S_r)}.
\end{equation}
The complex extension of the equation can be written as
\begin{equation}\label{eqmodelIA}
\partial_t f(\gamma)=\left(-\frac{1}{1+(\partial_x f(\gamma))^2}+\frac{1}{1+l^2-\left(f(\gamma)\right)^2}\right)\Lambda f(\gamma),
\end{equation}
where
$$
\Lambda u(\gamma)=\frac{1}{\pi}\text{P.V.}\int_\mathbb R\frac{u(\gamma)-u(\gamma-\eta)}{\eta^2}d\eta,\text{ }
\frac{1}{2\pi}\text{P.V.}\int_\mathbb T\frac{u(\gamma)-u(\gamma-\eta)}{\sin^2\left(\frac{\eta}{2}\right)}d\eta.
$$
Notice that the variable $\eta$ is a real number: $\eta\in\Omega$. Given a positive $\tau<1$, we define
\begin{equation*}
d_1[f](\gamma)=\frac{1}{\tau+l^2-\left(\text{Re} f(\gamma)\right)^2}, d_2[f](\gamma)=\frac{1}{\tau-\left(\text{Im} \partial_x f(\gamma)\right)^2},
\end{equation*}
Given $R>0$, we define the open set
\begin{equation*}
\mathcal{O}^\tau_R=\{f\in H^3(\mathcal S_r), \text{ s.t. }0<d_1[f]<R,0<d_2[f]<R, \|f\|_{H^3(\mathcal S_r)}<R\}.
\end{equation*}
We remark that in this set we have
$$
(\text{Re} f(\gamma))^2<l^2+\tau, (\text{Im} \partial_x f(\gamma))^2<\tau.
$$

\begin{teo}\label{teoCK}Let $f_0\in \mathcal{O}^{\tau_0}_{R_0}$ for some $0<\tau_0<1$, $R_0>0$ be the initial data for \eqref{model1}. Then, there exists $T(f_0)$ and a unique solution $f(x,t)\in C([-T,T],H^3(\mathbb R))$.
\end{teo}

This result is interesting because there exist functions such that $f(\tilde{x})=l$ in this set $O^\tau_R$. For instance one can consider $f^a_0(x)=a\cos(x)+l-a$ for a small enough $a$. In particular, this case is analogous to the case where the initial data reaches the boundary, \emph{i.e.} the large amplitude regime.

We study the decay of some lower order norms and other qualitative properties:
\begin{prop}\label{prop1}
Given $f_0\in H^3_l(\Omega),$ $l>0$, $\Omega=\mathbb R,\mathbb T$, in the stable regime, then the solution of \eqref{model1} verifies:
\begin{itemize}
\item
The even/odd symmetry of the initial data propagates.
\item
$$
\|f(t)\|_{L^\infty(\Omega)}\leq\|f_0\|_{L^\infty(\Omega)},
$$
\item Assume that $f_0\in H^3_l(\mathbb T)$ is odd, then
\begin{equation}\label{decay}
\|f(t)\|_{L^\infty(\mathbb T)}\leq \frac{e^{-Ct}\|f_0\|_{L^\infty(\mathbb T)}}{1+\mathcal{A}\left(e^{-Ct}-1\right)},
\end{equation}
where $\mathcal{A}$ is defined in \eqref{dimensionless}.
\item As long as the solution remains in the stable case, the solution is in $L^\infty_t\dot{H}^{0.5}_x\cap L^2_t\dot{H}^{1}_x$ and we have the following energy balance
$$
\|f(t)\|^2_{\dot{H}^{0.5}(\Omega)}-2\int_0^t \sigma(s)\|f(s)\|^2_{\dot{H}^{1}(\Omega)}ds\leq \|f_0\|^2_{\dot{H}^{0.5}(\Omega)},
$$
where $\sigma(t)<0$ is defined in \eqref{sigma}. Moreover, if the solution is in the stable regime up to time $T$, then
$$
\sup_{0\leq t\leq T}\|f(t)\|_{L^2}\leq c(\|f_0\|_{H^{0.5}},T).
$$

\item Assume that $f_0\in H^3_l(\mathbb T)$, then
\begin{equation}\label{decayh0.5}
\|f(t)\|_{\dot{H}^{0.5}(\mathbb T)}\leq e^{2\int_0^t\sigma(s)ds}\|f_0\|_{\dot{H}^{0.5}(\mathbb T)}.
\end{equation}
\item Given $f_0\in H^3_l(\mathbb R)$ and assuming that the solution is in the stable regime in the time interval $[0,T]$, we obtain
$$
\|f(t)\|_{L^\infty(\mathbb R)}\leq \left(\left(\frac{\frac{1-\mathcal{A}}{\mathcal{A}}}{1+l^2}\right)\frac{3t}{\mathfrak{C}(f_0,T)}+\frac{1}{\|f_0\|^3_{L^\infty(\mathbb R)}}\right)^{-1/3},
$$
where
$$
\mathfrak{C}(f_0,T)=\|f_0\|_{L^2}+ \frac{2\sqrt{T}\|f_0\|_{\dot{H}^{0.5}}}{\min_{0\leq s\leq T}\sqrt{|\sigma(s)|}},
$$
is a bound for $\|f(t)\|_{L^2(\mathbb R)}$.

\end{itemize}
\end{prop}

Recall that \eqref{decay} in Proposition \ref{prop1} gives us that, in the case
$$
\mathcal{A}=\frac{2}{\pi}\|f_0\|_{L^\infty(\mathbb T)}\approx 1,
$$
(the interface is close to the boundary)
$$
\|f(t)\|_{L^\infty(\mathbb T)}\leq \frac{e^{-\frac{\mathcal{C}_l}{l}t}\|f_0\|_{L^\infty(\mathbb T)}}{1+\mathcal{A}\left(e^{-\frac{\mathcal{C}_l}{l}t}-1\right)}\approx \|f_0\|_{L^\infty(\mathbb T)},
$$
and our decay estimate degenerates. This fact has been observed for equation \eqref{eq0.1} in \cite{CGO}. Moreover, it has also been observed in the numerical simulations in Section \ref{secdec} (see Figure \ref{deccase1}).

Notice that there is not a $L^2$ maximum principle, but we can use backward bootstrapping to bound the $L^2$ norm once that we now a bound for $\dot{H}^{0.5}$.

We prove that if the initial data is small, then there exists a global-in-time solution. Furthermore, we obtain some decay estimates in a lower norm. Thus, these results complement the decay rates proved in Proposition \ref{prop1}. We will use the approach in \cite{cheng2012global, shatahnormalforms}. Notice that, given $f_0\in H^3$, there exists a time of existence $T_0=T(f_0)$ and the solution is on the stable regime. For any $T<T_0$, we define the total norm
\begin{equation}\label{totalnorm}
\vertiii{f}_T=\sup_{0<t<T}\{\|f(t)\|_{X}+\mathcal{D}(t)\|f(t)\|_{Y}\},
\end{equation}
where $X\subset Y$ are Banach spaces. The function $\mathcal{D}(t)\rightarrow\infty$ as $t\rightarrow\infty$ and gives us the decay in the lower order norm. Using Duhamel's principle we write the expression for the mild solution
\begin{equation}\label{Duhamel}
f(x,t)=e^{-t\mathcal{C}_l\Lambda}f_0+\int_0^te^{-(t-s)\mathcal{C}_l\Lambda}NL(s)ds,
\end{equation}
where
\begin{equation}\label{nonlinear}
NL=\Lambda f\left[\frac{(\partial_x f)^2\left(1+2l^2+l^4-l^2f^2\right)+f^2}{(1+(\partial_x f)^2)(1+l^2-f^2)(1+l^2)}\right].
\end{equation}
\begin{teo}\label{teo3}
Let $f_0\in H^3(\mathbb T)$ be an odd initial data for equation \eqref{model1} in the stable regime.  Then, there exist $\delta>0$, such that if $\|f_0\|_{H^3(\mathbb T)}\leq \delta$ the corresponding solution is global in time and the solution verifies
$$
\|f(t)\|_{C^1(\mathbb T)}\leq \frac{l}{4} e^{-\mathcal{C}_lt}.
$$
\end{teo}

\begin{remark}
The oddness assumption is related to the decay estimate. We know that the odd character of the initial data propagates, so the solution will have zero mean and then the equilibrium solution is $f_\infty\equiv0$. However, as the mean is not preserved, it is not clear, and in general it is not true, that the mean will propagate for general initial data with zero mean.
\end{remark}

There are several results with limited regularity for \eqref{full} (see \cite{ccgs-13}). In particular, the authors in this paper proved the global existence of smooth solution corresponding to initial data with small derivative in the Wiener algebra. We prove that \eqref{model1} also captures these features. In particular, we study the equation \eqref{model1} when the initial data is only $H^2$ and we prove local existence for small initial data in both spatial domains, the real line and the torus.
\begin{teo}\label{teo4}
Let $f_0\in H^2(\Omega)$, $\Omega=\mathbb T,\mathbb R$, be the initial data for equation \eqref{model1} in the stable regime. We assume that $\|f_0\|_{H^{2}(\Omega)}\leq \delta$ for a small enough $\delta>0$. Then, there exists at least one local solution
$$
f\in C([0,T(f_0)],H^2(\Omega))\cap L^2([0,T(f_0)],H^{2.5}(\Omega)).
$$
\end{teo}
Notice that the solution is classical, but if the initial data is only $H^2$ the well-posedness for arbitrary data can not be achieved by standard energy methods. In the case where the initial data is odd and periodic, we can improve the previous local-in-time result:

\begin{teo}\label{teo5}
Let $f_0\in H^2(\mathbb T)$ be an odd initial data for equation \eqref{model1} in the stable regime.  Then, there exist $\delta>0$, such that if $\|f_0\|_{H^{2}(\mathbb T)}\leq \delta$ there exists at least one global in time solution. This solution verifies
$$
\|f(t)\|_{H^1(\mathbb T)}\leq \frac{l}{4} e^{-\mathcal{C}_lt},\text{ and }\|f(t)\|_{C^1(\mathbb T)}\leq\|f_0\|_{C^1(\mathbb T)}.
$$
\end{teo}

The two main possibilities for finite time blow up seem to be
\begin{enumerate}
\item To reach the unstable regime,
\item a blow up of the curvature for the case $f(\tilde{x})=l$.
\end{enumerate}

To reach the unstable regime is similar to the \emph{turning} singularities presents for \eqref{full} and \eqref{eq0.1} in \cite{ccfgl} and \cite{BCG, CGO, GG}. We discard this situation for \eqref{model1}. In particular we prove that, if the solution reaches the unstable case, the $W^{2+\epsilon,\infty}$ blows up first. The second source of singularity, a blow up of the curvature when the initial data reaches the boundaries may take two different forms: a corner-type singularity (blow up of the second derivative while the first derivative is bounded) and a cusp-type singularity (blow up of the first and second derivatives). We prove that, if the second derivative blows up, then the norm $W^{1+\epsilon,\infty}$ blows up first. Notice that, as a consequence of our proof, we get that if the initial data reaches the boundary, then the solution corresponding to this initial data reaches the boundary as long as it remains smooth.

We collect these two results in the next proposition:

\begin{prop}\label{prop2}
Let $f_0\in H^3(\Omega)$ be the initial data for equation \eqref{model1} and $0<T<\infty$ be an arbitrary parameter. We assume that the corresponding solution is $f(x,t)\in C([0,T),H^3(\Omega))$. Then,
\begin{itemize}
\item If $f_0$ is in the stable regime and $T$ is the first time where the solution leaves the stable regime, then
$$
\limsup_{t\rightarrow T}\int_0^t\|f(s)\|^2_{\dot{W}^{2+\epsilon,\infty}}ds=\infty.
$$
\item If $f_0$ is analytic and there exists $\tilde{x}$ such that $f_0(\tilde{x})=l$, then
$$
f(\tilde{x},t)=l,\,\forall 0\leq t\leq T
$$
and
$$
\max_{0\leq t\leq T}|\partial_x^2 f(\tilde{x},t)|\leq c(l,f_0)e^{l\int_0^T\Lambda f(\tilde{x},s)ds}.
$$
Consequently, the curvature can not blow up for a $W^{1+\epsilon,\infty}$ solution.
\end{itemize}
\end{prop}

There are three main questions that remain open for this model: an existence theory for initial data in $H^{1.5}\cap W^{1,\infty}$, a proof of finite time singularities where the curvature blows up and the existence of a geometry (instead of a flat strip) that enhances the similarities between the Muskat problem and the model introduced in this paper.

\subsection{Statement of the results for \eqref{modelodiego}}
We obtain a new energy balance for \eqref{modelodiego}. To do that, we consider the evolution of the entropy
$$
\int_\Omega \partial_x f(t)\log(1+(\partial_x f(t))^2)dx.
$$
\begin{prop}\label{prop3}
Given $f_0\in W^{1,3}(\Omega)\cap W^{1,1}(\Omega),$ $\Omega=\mathbb R,\mathbb T$, then the solution of \eqref{modelodiego} verifies the following energy balance
\begin{multline}\label{energybalance}
\int_\Omega \partial_x f(t)\log(1+(\partial_x f(t))^2)dx\\
+2\int_\Omega \arctan(\partial_x f(t))-\int_0^t\int_\Omega   \frac{\Lambda \partial_x f}{1+(\partial_x f)^2}dxds\\
=\int_\Omega \partial_x f_0\log(1+(\partial_x f_0)^2)dx+2\int_\Omega \arctan(\partial_x f_0).
\end{multline}
\end{prop}

Furthermore, under a positiveness hypothesis for $\partial_x f_0$, we can use this energy balance to obtain global existence of weak solutions with rough initial data. This energy balance fully exploits the diffusive character of the equation \eqref{modelodiego}. Notice also that, due to the positiveness of $g$, we can not recover a smooth, periodic $f$ from this $g$. Now we define our notion of weak solution:

\begin{defi} \label{definition 1}
$g(x,t)$ is a global weak solution of \eqref{modelodiegoderiv} if
$$
g(x,t)\in L^\infty([0,T],L^\infty(\mathbb T))\cap L^2([0,T],\dot{H}^{0.5}(\mathbb T))
$$
and \eqref{modelodiegoderiv} holds in the sense of distributions: for any $\psi\in C^{\infty}\left([0,T)\times\mathbb{T}\right)$, periodic in space and with compact support in time,
\[
\int^{T}_{0}\int_{\mathbb{T}} -\Lambda\psi g+\partial_t\psi g - \partial_x\psi\left(\frac{g^2Hg}{1+g^2}\right)dxdt-\int_\mathbb T \psi(x,0)g_0(x)dx=0,
\]
for every $T<\infty$.
\end{defi}

We state now our result:
\begin{teo}\label{teoweak}
Let $g_0\in L^\infty(\mathbb T)$ be a positive initial data for equation \eqref{modelodiegoderiv}. Then, there exist at least one global weak solution
$$
g(x,t)\in L^\infty([0,T],L^\infty(\mathbb T))\cap L^2([0,T],\dot{H}^{0.5}(\mathbb T)),
$$
satisfying the bounds
$$
\|g(t)\|_{L^\infty(\mathbb T)}\leq\|g_0\|_{L^\infty(\mathbb T)},\;\min_x g(x,t)>\min_x g_0(x)\;\forall\,t>0,
$$
and
$$
\int_0^t\|g(s)\|_{\dot{H}^{0.5}}^2\leq\frac{(1+\|g_0\|_{L^\infty})^2}{\min_x g_0}\left(\int_\mathbb T g_0\log(1+g_0^2)dx+2\int_\mathbb T \arctan(g_0)\right).
$$
\end{teo}

\begin{remark}
Notice that if we study the evolution of the energy
$$
\int (1 + \frac{1}{6}(\partial_x f (x, t))^2 )(\partial_x f (x, t))^2 dx,
$$
under \eqref{modelodiego}, we find
$$
\|\partial_x f\|_{L^2}^2+\frac{1}{6}\|\partial_x f\|_{L^4}^4+\int_0^t\|\partial_x f(s)\|^2_{\dot{H}^{0.5}}ds=\|\partial_x f_0\|_{L^2}^2+\frac{1}{6}\|\partial_x f_0\|_{L^4}^4.
$$
We thank the anonymous referee for pointing out this energy. This energy balance can be used to extend Theorem \ref{teoweak} to arbitrary (non necessarily positive) $g\in L^2\cap L^4$.
\end{remark}

\subsection{Plan of the paper}

The structure of the paper is as follows: In section \ref{secmodelodiego}, we prove the energy balance \eqref{energybalance} for the solutions of equation \eqref{modelodiego} and we use it to prove global existence of weak solutions of \eqref{modelodiegoderiv}. The results concerning \eqref{model1} are contained from Section \ref{secmodel} to Section \ref{seclarge}. In Section \ref{secmodel} we obtain well-posedness in Sobolev spaces and in an analytical framework for equation \eqref{model1}. In Section \ref{secqual} we study the qualitative properties of the solutions and we get some maximum principles for different lower order norms. In this Section, using the same scheme as in \cite{cheng2012global}, we also prove a global existence and decay in $C^1$ for the mild solution corresponding to small initial data in $H^3$. In Section \ref{seclim} we obtain existence and decay in $H^1$ for the mild solution corresponding to initial data small in $H^2$. In Section \ref{secdec} we present some numerics comparing the solutions to equations \eqref{full} and \eqref{eq0.1}. We present these simulations for the sake of completeness and to bring into comparison with the simulations corresponding to equation \eqref{model1}. In Section \ref{secnum} we present some numerics for equation \eqref{model1}. In particular we compute the evolution of a family of initial data reaching the boundary. In the last Section we study analytically some properties of the solutions when the initial data reaches the boundary. Notice that these solutions exist due to the well-posedness result in the analytical framework.

\section{A new energy balance and global weak solutions for \eqref{modelodiego}}\label{secmodelodiego}
Now we show a new energy balance for the derivative of \eqref{modelodiego}.
\begin{proof}[Proof of Proposition \ref{prop3}]
We consider the equation for the derivative of the interface evolving in the infinite depth regime \eqref{modelodiegoderiv}.

Now consider the evolution of the following quantity,
\begin{align*}
\frac{d}{dt}\int_\Omega g\log(1+g^2) &= \int g_t\log(1+g^2) + \int_\Omega\frac{2g^2g_t}{1+g^2}\\
&= \int_\Omega g_t\log(1+g^2) - 2\int_\Omega\frac{g_t}{1+g^2}\\
&= \int_{\Omega} g_t\log(1+g^2) - 2\int_\Omega \partial_t(\arctan(g))
\end{align*}
since $\int_\Omega g_t = 0$.

Let's look at the first term in the right hand side,
\begin{align}
&\int_\Omega g_t\log(1+g^2) = \int_\Omega -\Lambda g\log(1+g^2) + \partial_x \left(\frac{g^2Hg}{1+g^2}\right)\log(1+g^2)\label{E:aux2}
\end{align}
The second term,
\begin{align*}
&\int_\Omega\partial_x \left(\frac{g^2Hg}{1+g^2}\right)\log(1+g^2) = \int_\Omega \partial_x(Hg - \frac{Hg}{1+g^2})\log(1+g^2)\\
& = \int_\Omega \Lambda g\log(1+g^2) - \int_\Omega \partial_x\left(\frac{Hg}{1+g^2}\right)\log(1+g^2)\\
&= \int_\Omega \Lambda g\log(1+g^2) + \int_\Omega  Hg \frac{2gg_x}{(1+g^2)^2}\\
&=\int_\Omega \Lambda g\log(1+g^2) + \int_\Omega   \frac{\Lambda g}{1+g^2}
\end{align*}
Putting this back together into \eqref{E:aux2},
\begin{equation*}
\frac{d}{dt}\int_\Omega g\log(1+g^2)+2\frac{d}{dt}\int_\Omega \arctan(g) = \int_\Omega   \frac{\Lambda g}{1+g^2}.
\end{equation*}
\end{proof}

\begin{remark}
Now we can symmetrize the \emph{extra} term in \eqref{energybalance},
\begin{equation*}
\int_\mathbb T\frac{\Lambda g}{1+g^2} = -\frac{1}{4\pi}\iint_{\mathbb T\times\mathbb T} \frac{(g(x)-g(y))^2(g(x)+g(y))}{\sin^2(\frac{x-y}{2})(1+(g(x))^2)(1+(g(y))^2)}dxdy,
\end{equation*}
in the periodic case and
\begin{equation*}
\int_\mathbb R\frac{\Lambda g}{1+g^2} = -\frac{1}{2\pi}\iint_{\mathbb R\times\mathbb R} \frac{(g(x)-g(y))^2(g(x)+g(y))}{(x-y)^2(1+(g(x))^2)(1+(g(y))^2)}dxdy,
\end{equation*}

which is negative if we assume that $g\geq 0$. This observation will allow us to gain half a derivative from this energy identity.
\end{remark}

We fix $\Omega=\mathbb T$ to simplify and we consider $f_0\in W^{1,\infty}$. Consequently $g_0\in L^\infty$. We also assume $g_0>0$. In particular this implies that
$$
\|g(t)\|_{L^\infty(\mathbb T)}\leq\|g_0\|_{L^\infty(\mathbb T)},\;\min_x g(x,t)>\min_x g_0(x)\;\forall\,t>0.
$$

We use the previous energy identity to get compactness and to construct weak solutions:

\begin{proof}[Proof of Theorem \ref{teoweak}]
We define the approximate problems
\begin{equation}\label{E:smoothg}
g^\epsilon_t = -\Lambda g^\epsilon + \partial_x\left(\frac{(g^\epsilon)^2Hg^\epsilon}{1+(g^\epsilon)^2}\right) + \epsilon \partial_x^2g^\epsilon,\;g^\epsilon(x,0)=\rho_\epsilon*g_0,
\end{equation}
where $\rho_\epsilon$ is a standard mollifier.

Multiplying equation \eqref{E:smoothg} by $-\partial_x^2g^\epsilon$, and integrating over the torus, we obtain,

\begin{align*}
-\int_\mathbb T g^\epsilon_t\partial_x^2g^\epsilon &= \int_\mathbb T\Lambda g^\epsilon g^\epsilon_{xx} - \int_\mathbb T \left(\frac{(g^\epsilon)^2Hg^\epsilon}{1+(g^\epsilon)^2}\right)_x g^\epsilon_{xx} - \epsilon\|g^\epsilon\|^2_{\dot{H}^2}\\
\frac{1}{2}\frac{d}{dt}\|g^\epsilon\|^2_{\dot{H}^1} &=\int_\mathbb T\frac{\Lambda g^\epsilon}{1+(g^\epsilon)^2}g^\epsilon_{xx} - \int_\mathbb T \frac{Hg^\epsilon 2g^\epsilon g^\epsilon_x}{(1+(g^\epsilon)^2)^2}g_{xx} -\epsilon\|g_{xx}\|^2_{\dot{H}^2}.
\end{align*}
To estimates the remaining terms, we will use the following inequalities which are a direct consequence of Gagliardo-Nirenberg,
\begin{equation}\label{GN2}
\|u\|_{\dot{W}^{1,4}}^2\leq C\|u\|_{\dot{H}^{2}}\|u\|_{L^\infty}\text{ and }\|u\|_{\dot{W}^{1,3}}^3\leq C\|u\|^2_{\dot{H}^{1.5}}\|u\|_{L^\infty}.
\end{equation}
These inequalities are valid for zero mean, periodic functions, but as the $L^1$ norm of our solution propagates with the evolution, we can adapt the argument straightforwardly. Using this into our estimate,
\begin{align*}
\frac{1}{2}\frac{d}{dt}\|g_x\|^2_0 + \epsilon\|g_{xx}\|^2_0&\leq \|g_{xx}\|_0\|\Lambda g\|_0 + 2\|g_{xx}\|_0\|g_x\|_{L^4}\|Hg\|_{L^4}\|g\|_{L^\infty}\\
&\leq C( \lambda \|g_{xx}\|^2_0 + \frac{1}{\lambda}\|g\|^{10}_{L^\infty}).
\end{align*}
Choosing $\lambda = \epsilon/2C$, we absorb the second derivative into the left side, and integrating in time we obtain,
\begin{equation}
\|g^\epsilon_x(t)\|^2_0 + \epsilon\|g^\epsilon_{xx}\|^2_{L^2_t L^2_x} \leq \|g^\epsilon_x(0)\|^2_0 + \frac{C}{\epsilon}t\|g^\epsilon\|^{10}_{L^\infty_t L^\infty_x}.
\end{equation}
Since the $L^\infty$-norm of $g^\epsilon$ is uniformly bounded, we have a global estimate for the $H^1$ norm of $g^\epsilon$ for every $\epsilon>0$.

We study the evolution of $\int_\mathbb T g^\epsilon\log(1+(g^\epsilon)^2)$. We find
\begin{multline*}
\int_\mathbb T g^\epsilon(t)\log(1+(g^\epsilon(t))^2)dx+2\int_\mathbb T \arctan(g^\epsilon)\\
-\int_0^t\int_\mathbb T   \frac{\Lambda g^\epsilon}{1+(g^\epsilon)^2}dxds+2\epsilon\int_0^t\int_\mathbb T\frac{g^\epsilon(\partial_x g^\epsilon)^2}{1+(g^\epsilon)^2}dxdt\\
=\int_\mathbb T g^\epsilon_0\log(1+(g^\epsilon_0)^2)dx+2\int_\mathbb T \arctan(g^\epsilon_0).
\end{multline*}
This implies the uniform-in-$\epsilon$ bound
$$
\int_0^t\|g^\epsilon(s)\|_{\dot{H}^{0.5}}^2\leq\frac{(1+\|g_0\|_{L^\infty})^2}{\min_x g_0}\left(\int_\mathbb T g_0\log(1+g_0^2)dx+2\int_\mathbb T \arctan(g_0)\right).
$$
Banach-Alaoglu Theorem implies $g\in L^\infty([0,T],L^\infty)\cap L^2([0,T],H^{0.5})$. Using \eqref{modelodiegoderiv} we get a uniform bound for $\partial_t g^\epsilon$ in $L^2_tH^{-1}_x$. Thus, we can apply Lemma \ref{lemma:2.1} with
$$
X_0=H^{0.5},\;X=L^2,\;Y=H^{-1},
$$
to get strong convergence $g^\epsilon\rightarrow g$ in $L^2([0,T]\times\mathbb T)$. This compactness implies the convergence of the weak formulations.
\end{proof}

\section{Well-posedness}\label{secmodel}
\subsection{Well-posedness in Sobolev spaces}
First, we prove local well-posedness in the stable regime:
\begin{proof}[Proof of Theorem \ref{teo1}]
We proof the case $s=3, l=\pi/2$ being the other cases analogous. We define the energy
\begin{equation}\label{eqenergy}
E(t)=\|f(t)\|_{H^3}+\|d[f]\|_{L^\infty}+\|D[f]\|_{L^\infty},
\end{equation}
where
\begin{equation}\label{eqdistance}
d[f]=\frac{1}{\left(\frac{\pi}{2}\right)^2-\left(f(x)\right)^2-(\partial_x f(x))^2},
\end{equation}
and
\begin{equation}\label{eqDistance}
D[f]=\frac{1}{\left(\frac{\pi}{2}\right)^2-\left(f(x)\right)^2}.
\end{equation}
The quantity $d[f]$ controls the stability condition \eqref{eqRayleigh} for our model. Indeed, if initially $d[f_0]>0$, then, as long as the energy remains bounded, $d[f]>0$. This implies that the dynamics is in the stable regime \eqref{eqRayleigh}. The quantity $D[f]$ ensures that we don't leave the set $H^s_l$.

\textbf{Estimates for $\partial_t f,\partial_t\partial_x f$:} By the basic properties of the Hilbert transform and the Sobolev embedding, we get
\begin{equation}\label{patf}
\|\partial_t f\|_{L^\infty}\leq 2\|H\partial_x f\|_{L^\infty}\leq C\|f\|_{H^{3}},
\end{equation}
and
\begin{multline}\label{patpaxf}
\|\partial_t \partial_x f\|_{L^\infty}\leq 2\|H\partial_x^2 f\|_{L^\infty}\\
+C\left(\|\partial_x f\|_{L^\infty}\|\partial_x^2 f\|_{L^\infty}+\|f\|_{L^\infty}\|\partial_x f\|_{L^\infty}\right)\|\Lambda f\|_{L^\infty}\\
\leq C\|f\|_{H^3}(\|f\|_{H^3}^2+1).
\end{multline}

\textbf{Estimates for $d[f]$:} We compute
$$
\frac{d}{dt}d[f]=(d[f])^2\left(2f(x)\partial_t f(x)+2\partial_x f(x)\partial_t\partial_x f(x)\right).
$$
Using the definition of the energy \eqref{eqenergy}, we get
$$
\frac{d}{dt}d[f]\leq Cd[f]E\left(E+1\right)^4,
$$
thus, integrating in $(t,t+h)$
$$
d[f](t+h)\leq d[f](t)e^{C\int_t^{t+h} E\left(E+1\right)^4}ds.
$$
We have
\begin{multline}\label{patdf}
\frac{d}{dt}\|d[f]\|_{L^\infty}=\lim_{h\rightarrow0}\frac{\|d[f](t+h)\|_{L^\infty}-\|d[f](t)\|_{L^\infty}}{h}\\
\leq \|d[f](t)\|_{L^\infty}\lim_{h\rightarrow0}\frac{e^{C\int_t^{t+h} E\left(E+1\right)^4}ds-1}{h}\leq CE^2\left(E+1\right)^4.
\end{multline}

\textbf{Estimates for $D[f]$:}
In the same way,
\begin{equation}\label{patDf}
\frac{d}{dt}\|D[f]\|_{L^\infty}\leq CE^2\left(E+1\right)^4.
\end{equation}

\textbf{Estimates for the higher order terms:}
The higher order terms are
$$
I_1=\int_\Omega \left(\frac{1}{1+\left(\frac{\pi}{2}\right)^2-\left(f(x)\right)^2}-\frac{1}{1+\left(\partial_x f(x)\right)^2}\right)\Lambda \partial_x^3 f(x)\partial_x^3f(x)dx,
$$
$$
I_2=\int_\Omega \frac{2\partial_x f(x)\partial_x^4 f(x)}{\left(1+\left(\partial_x f(x)\right)^2\right)^2}\Lambda f(x)\partial_x^3f(x)dx.
$$

Notice that, due to \eqref{patdf}, $\sigma(t)<0$ for sufficiently small time. To estimate $I_1$ we use the pointwise inequality \cite{cordoba2003pointwise, cor2}
$$
2\theta\Lambda\theta\geq \Lambda\theta^2.
$$
This inequality and the self-adjointness of the operator allow us \emph{to integrate by parts} in the stable regime (which is guaranteed for a short time by \eqref{patdf}). We get
$$
I_1=J_1+J_2,
$$
with
\begin{eqnarray}\label{eqhigher}
J_1&\leq& \frac{1}{2}\int_\Omega \Lambda\left(\frac{1}{1+\left(\frac{\pi}{2}\right)^2-\left(f(x)\right)^2}-\frac{1}{1+\left(\partial_x f(x)\right)^2}-\sigma(t)\right)\nonumber\\
&&\times (\partial_x^3f(x))^2dx\nonumber\\
&\leq& C\|f\|^2_{H^3}\left\|\Lambda \left[\frac{1}{1+\left(\frac{\pi}{2}\right)^2-\left(f(x)\right)^2}-\frac{1}{1+\left(\partial_x f(x)\right)^2}\right]\right\|_{H^{0.6}}\nonumber\\
&\leq& \left(\left\|\frac{2f\partial_x f}{(1+\left(\frac{\pi}{2}\right)^2-\left(f(x)\right)^2)}\right\|_{H^{1}}+\left\|\frac{2\partial_x f \partial_x^2f}{(1+\left(\partial_x f(x)\right)^2)}\right\|_{H^{1}}\right)\nonumber\\
&&\times C\|f\|^2_{H^3},
\end{eqnarray}
and
\begin{equation}\label{eqhigher2}
J_2= \sigma(t)\int_\Omega (\Lambda^{0.5}\partial_x^3f(x))^2dx<0.
\end{equation}
The term $I_2$ can be bounded as in \cite{c-g-o08}. With the same ideas, we can handle the lower order terms. We conclude
$$
\frac{d}{dt}\|f(t)\|_{H^3}\leq C(E+1)^7.
$$

\textbf{Obtaining uniform estimates:} Collecting the estimates (see \eqref{patdf}, \eqref{patDf}, \eqref{eqhigher}, \eqref{eqhigher2}), we get
$$
\frac{d}{dt}E\leq C(E+1)^7.
$$
Using Gronwall's inequality, we obtain
$$
E(t)\leq C(f_0),\text{ if }0\leq t \leq T(f_0).
$$
With this a priori estimate we can obtain the local existence of smooth solutions using the standard arguments (see \cite{bertozzi-Majda}). Moreover, \eqref{eqhigher} and \eqref{eqhigher2} give us
$$
\frac{d}{dt}\|f(t)\|^2_{H^3}-\sigma(t)\|f(t)\|^2_{\dot{H}^{3.5}}\leq C(E+1)^k.
$$
Integrating in time, we conclude $f\in L^2_tH^{3.5}_x$.

\textbf{Uniqueness:} Let's assume that there exists $f_1,$ $f_2$, two different solutions corresponding to the same initial data and denote $f=f_1-f_2$. Then, with the same ideas, we get
$$
\frac{d}{dt}\|f\|_{L^2}^2\leq C(\|f_1\|_{H^3},\|f_2\|_{H^3})\|f\|_{L^2}^2,
$$
and using Gronwall inequality, we conclude the uniqueness.

\textbf{Continuation criterion:} We use Lemma \ref{lemaaux} ($\alpha=1,d=1,\gamma=0.6$) in \eqref{eqhigher} to get
\begin{equation}\label{eqbound}
\Lambda\left[\frac{1}{1+l^2-(f(x))^2}\right]\leq C(l)\|f\|_{W^{0.6,\infty}}^2+\frac{2f(x)}{(1+l^2-(f(x))^2)^2}\Lambda f(x),
\end{equation}
\begin{equation}\label{eqbound2}
\Lambda\left[\frac{-1}{1+(\partial_x f(x))^2}\right]\leq C\|\partial_x f\|_{W^{0.6,\infty}}^2+\frac{2\partial_x f(x)}{(1+(\partial_x f(x))^2)^2}\Lambda \partial_x f(x).
\end{equation}
From here we conclude the result.

\end{proof}

\subsection{Well-posedness in the analytical framework}
We start with a useful Lemma:
\begin{lem}\label{CKprop}
Consider $0\leq r'<r$ and the set $\mathcal{O}^\tau_R$. Then, for $f,g\in \mathcal{O}^\tau_R$, the spatial operator in \eqref{eqmodelIA}, $F:\mathcal{O}^\tau_R\rightarrow H^3(\mathcal S_{r'})$ is continuous. Moreover, the following inequalities holds:
\begin{enumerate}
 \item $\|F[f]\|_{H^3(\mathcal S_{r'})}\leq\frac{C^\tau_R}{r-r'}\|f\|_{H^3(\mathcal S_r)},$
 \item $\|F[f]-F[g]\|_{H^3(\mathcal S_{r'})}\leq\frac{C^\tau_R}{r-r'}\|f-g\|_{H^3(\mathcal S_r)}.$
\end{enumerate}
\end{lem}
\begin{proof}
For the sake of brevity, we only prove the first part. The second one is analogous. Notice that
$$\|\Lambda f\|_{L^2(\mathcal S_r')}\leq\frac{C}{r-r'}\|Hf\|_{L^2(\mathcal S_r)}\leq\frac{C}{r-r'}\|f\|_{L^2(\mathcal S_r)}$$
By definition:
$$
\|F[f]\|^2_{H^3(\mathcal S_r')}=I+II
$$
where
$$
I=\left\|\left(\frac{-1}{1+(\partial_x f)^2}+\frac{1}{1+l^2-f^2} \right)\Lambda f\right\|_{L^2(\mathcal S_r')}^2
$$
$$
II=\left\|\partial_x^3\left(\left(\frac{-1}{1+(\partial_x f)^2}+\frac{1}{1+l^2-f^2} \right)\Lambda f\right)\right\|_{L^2(\mathcal S_r')}^2
$$
To estimate $I$, we use H\"older and the fact that we're working in the open set $\mathcal{O}^\tau_R$ to get
$$
1-\tau\leq|1+(\partial_x f)^2|^2,
$$
and, as a consequence,
$$
\left|\frac{1}{1+(\partial_x f)^2}\right|
\leq \frac{1}{\sqrt{1-\tau}}.
$$
A similar bound holds for the term $\frac{1}{1+l^2-f^2}$. Hence, we have:
$$
\left\|\left(\frac{-1}{1+(\partial_x f)^2}+\frac{1}{1+l^2-f^2} \right)\Lambda f\right\|_{L^2(\mathcal S_r')}\leq\frac{C^\tau_R}{r-r'}\|f\|_{L^2(\mathcal S_r)}
$$
In $II$ we compute the third derivative. The terms involving $1$,$2$ and $3$ derivatives can be bounded using the previous ideas, the open set definition and the Banach scale property. For the terms involving $4$ derivatives we use \eqref{cauchy2}. In particular
\begin{multline*}
\|(1+(\partial_x f)^2)^{-2}\partial_x f\partial_x^4 f\|_{L^2(\mathcal S_r')}\\
\leq\|(1+(\partial_x f)^2)^{-2}\partial_x f\|_{L^\infty(\mathcal S_r')}\|\partial_x(\partial_x^3 f)\|_{L^2(\mathcal S_r')}\leq \frac{C^\tau_R}{r-r'}\|f\|_{H^3(\mathcal S_r)},
\end{multline*}
\begin{multline*}
\left\|\left(\frac{-1}{1+(\partial_x f)^2}+\frac{1}{1+l^2-f^2} \right)\partial_x^3\Lambda f\right\|_{L^2(\mathcal S_r')}\\
\leq\left\|\left(\frac{-1}{1+(\partial_x f)^2}+\frac{1}{1+l^2-f^2} \right)\right\|_{L^\infty(\mathcal S_r')}\|\partial_x^3(\Lambda f)\|_{L^2(\mathcal S_r')}^2\\
\leq\frac{C^\tau_R}{r-r'}\|f\|_{H^3(\mathcal S_r)}.
\end{multline*}
This concludes the proof.
\end{proof}

The former Lemma is used in the proof of Theorem \ref{teoCK}

\begin{proof}[Proof of Theorem \ref{teoCK}]
The proof follows the ideas in \cite{nirenberg1972abstract, nishida1977note} (see also \cite{ccfgl, CGO, marchioro1994mathematical}. We fix $R>R_0>0$ and $1>\tau>\tau_0$ and we consider the following Picard's iteration scheme
$$
f^n(t)=f_0+\int_0^tF[f^{n-1}]ds.
$$
By induction hypothesis we have $f^{j}\in \mathcal{O}^\tau_R$ for $1<j<n$. Using Lemma \ref{CKprop} and the ideas in \cite{nirenberg1972abstract, nishida1977note} we can find $T_0>0$ such that $\|f^n\|_{H^3(\mathcal S_r)}<R$, consequently, we need to find $T_1,T_2$ such that
$$
0<d_1[f^n]<R,\;0<d_2[f^n]<R.
$$
We obtain $T_2$ for $d_2$, being $T_1$ similar. We have
$$
\tau-\text{Im}\partial_x f^n\geq\tau-\text{Im}\partial_x f^0-tC^\tau_R=\tau-\tau_0+\tau_0-\text{Im}\partial_x f^0-tC^\tau_R\geq \frac{1}{R_0}-tC^\tau_R>\frac{1}{R},
$$
by taking $0\leq t\leq T_2=T_2(R,\tau)$ small enough. We define $T=\min\{T_0,T_1,T_2\}>0$ and we conclude.
\end{proof}

\section{Qualitative theory}\label{secqual}

\subsection{Decay estimates for the lower norms}

\begin{proof}[Proof of Proposition \ref{prop1}] We assume $l=\pi/2$ without losing generality.

\textbf{Step 1:} The proof of this part is straightforward.

\textbf{Step 2:} We denote $M(t)=\max_x f(x,t)=f(X_t,t)$ and $m(t)=\min_x f(x,t)=f(x_t,t)$. Using Rademacher Theorem as in \cite{cor2} and \cite{c-g09}, we obtain
$$
\frac{d}{dt}f(X_t)=-\frac{\left(\frac{\pi}{2}\right)^2-\left(f(X_t)\right)^2}{1+\left(\frac{\pi}{2}\right)^2-\left(f(X_t)\right)^2}\Lambda f(X_t).
$$
Using the kernel representations
$$
\Lambda u(x)=\frac{1}{\pi}\text{P.V.}\int_\mathbb R\frac{u(x)-u(x-\eta)}{\eta^2}d\eta,\text{ }
\frac{1}{2\pi}\text{P.V.}\int_\mathbb T\frac{u(x)-u(x-\eta)}{\sin^2\left(\frac{\eta}{2}\right)}d\eta,
$$
for the flat at infinity case and for the periodic case, respectively. We conclude $M'(t)<0$. With the same approach we get $m'(t)>0$.

\textbf{Step 3:} The solution remains odd, so, as in \cite{AGM}, we have
\begin{multline*}
\Lambda f(X_t)=\frac{1}{2\pi}\text{P.V.}\int_\mathbb T\frac{f(X_t)-f(X_t-\eta)}{\sin^2(\eta/2)}d\eta\\
\geq\frac{1}{2\pi}\text{P.V.}\int_\mathbb T f(X_t)-f(X_t-\eta)d\eta=f(X_t).
\end{multline*}
Thus, we have
$$
\frac{\left(\frac{\pi}{2}\right)^2-\left(f(X_t)\right)^2}{1+\left(\frac{\pi}{2}\right)^2-\left(f(X_t)\right)^2}\Lambda f(X_t)\geq \frac{\left(\frac{\pi}{2}\right)^2-\left(f(X_t)\right)^2}{1+\left(\frac{\pi}{2}\right)^2}f(X_t)\geq \frac{\left(\frac{\pi}{2}-f(X_t)\right)}{1+\left(\frac{\pi}{2}\right)^2}f(X_t).
$$
We conclude
$$
\frac{d}{dt}f(X_t)\leq -\frac{\left(\frac{\pi}{2}-f(X_t)\right)}{1+\left(\frac{\pi}{2}\right)^2}f(X_t),
$$
therefore
$$
\|f(t)\|_{L^\infty(\mathbb T)}\leq \frac{\pi \|f_0\|_{L^\infty(\mathbb T)}}{\pi e^{\frac{\pi}{2}\frac{t}{1+\left(\frac{\pi}{2}\right)^2}}+2\|f_0\|_{L^\infty(\mathbb T)}\left(1-e^{\frac{\pi}{2}\frac{t}{1+\left(\frac{\pi}{2}\right)^2}}\right)}.
$$

\textbf{Step 4:} We test the equation \eqref{model1} against $\Lambda f$, integrate in space and use the self-adjointness. Recalling the definition of $\sigma(t)$ \eqref{sigma}, we have
$$
\|f(t)\|^2_{\dot{H}^{0.5}}-2\int_0^t \sigma(s)\|f(s)\|^2_{\dot{H}^{1}}\leq \|f_0\|^2_{\dot{H}^{0.5}}.
$$
Assume again that the solution doesn't leave the stable regime up to time $T$, then testing equation \eqref{model1} against $f$, we get
$$
\frac{d}{dt}\|f\|_{L^2}^2\leq 4\|\Lambda f\|_{L^2}\|f\|_{L^2},
$$
and integrating in time,
\begin{multline*}
\|f(t)\|_{L^2}\leq \|f_0\|_{L^2}+ \frac{4}{\min_{0\leq s\leq T}\sqrt{|\sigma(s)|}}\int_0^t\sqrt{-\sigma(s)}\|f(s)\|_{\dot{H}^1}ds\\
\leq \|f_0\|_{L^2}+ \frac{2\sqrt{T}\|f_0\|_{\dot{H}^{0.5}}}{\min_{0\leq s\leq T}\sqrt{|\sigma(s)|}}.
\end{multline*}

\textbf{Step 5:} Using Poincar\'e inequality and recalling that fractional derivatives have zero mean, we get
$$
\frac{1}{2}\frac{d}{dt}\|f(t)\|^2_{\dot{H}^{0.5}(\mathbb T)}\leq \sigma(t)\|f(t)\|^2_{\dot{H}^{0.5}(\mathbb T)}.
$$
Using Gronwall inequality we conclude the result.

\textbf{Step 6:} Taking $T$ such that the solution is in the stable regime and using a previous step, we have
$$
\|f(t)\|_{L^2}\leq \mathfrak{C}(f_0,T).
$$
Using Rademacher Theorem and the decay of the amplitude, we obtain
\begin{eqnarray*}
\frac{d}{dt}\|f(t)\|_{L^\infty(\mathbb R)}&=&-\left(\frac{\left(\frac{\pi}{2}\right)^2-\|f(t)\|_{L^\infty(\mathbb R)}^2}{1+\left(\frac{\pi}{2}\right)^2-\|f(t)\|_{L^\infty(\mathbb R)}^2}\right)\Lambda f(x_t)\\
&\leq& -\left(\frac{\frac{\pi}{2}-\|f(t)\|_{L^\infty(\mathbb R)}}{1+\left(\frac{\pi}{2}\right)^2}\right)\frac{\|f(t)\|_{L^\infty(\mathbb R)}^{3}}{\mathfrak{C}(f_0,T)}\\
&\leq& -\left(\frac{\frac{\frac{\pi}{2}}{\|f(t)\|_{L^\infty(\mathbb R)}}-1}{1+\left(\frac{\pi}{2}\right)^2}\right)\frac{\|f(t)\|_{L^\infty(\mathbb R)}^{4}}{\mathfrak{C}(f_0,T)}\\
&\leq& -\left(\frac{\frac{1-\mathcal{A}}{\mathcal{A}}}{1+\left(\frac{\pi}{2}\right)^2}\right)\frac{\|f(t)\|_{L^\infty(\mathbb R)}^{4}}{\mathfrak{C}(f_0,T)},
\end{eqnarray*}
thus, for $0\leq t\leq T$ we conclude
$$
\|f(t)\|_{L^\infty(\mathbb R)}\leq \frac{1}{\sqrt[3]{\left(\frac{\frac{1-\mathcal{A}}{\mathcal{A}}}{1+\left(\frac{\pi}{2}\right)^2}\right)\frac{3t}{\mathfrak{C}(f_0,T)}+\frac{1}{\|f_0\|^3_{L^\infty(\mathbb R)}}}}
$$
\end{proof}

\subsection{Global existence and decay estimates in $C^1$}
\begin{proof}[Proof of Theorem \ref{teo3}]
In this case we have $X=H^3$, $Y=C^1$ and $\mathcal{D}(t)=e^{\mathcal{C}_lt}$. We use the estimate
\begin{equation}\label{normlinearoperator}
\|e^{-t\mathcal{C}_l\Lambda}\|_{L^\infty\rightarrow L^\infty}\leq e^{-\mathcal{C}_lt},
\end{equation}
and get
\begin{equation}\label{boundduhamel}
\|f(t)\|_{Y}\leq\|f_0\|_{Y}e^{-\mathcal{C}_lt}+\int_0^te^{-\mathcal{C}_l(t-s)}\|NL(s)\|_{Y}ds.
\end{equation}
Using Sobolev embedding, we have
\begin{equation*
\|NL(s)\|_{L^\infty}\leq C(l)\|f\|_{H^{1.6}}\left(\|\partial_x f\|^2_{L^\infty}+\|f(t)\|_{L^\infty}^2\right)\leq
C(l)\frac{\vertiii{f}_T^3}{(\mathcal{D}(t))^{2}}.
\end{equation*}
and
\begin{multline*}
\|\partial_x NL(s)\|_{L^\infty}\leq C(l)\left[\|f\|_{H^{2.6}}\left(\|\partial_x f\|^2_{L^\infty}+\|f(t)\|_{L^\infty}^2\right)\right.\\
+\|f\|_{H^{1.6}}\left(\|\partial_x f\|_{L^\infty}\|\partial_x^2 f\|_{L^\infty}\left(\|\partial_x f\|^2_{L^\infty}+\|f(t)\|_{L^\infty}^2\right)\right.\\
+\|\partial_x f\|_{L^\infty}\|f\|_{L^\infty}\left(\|\partial_x f\|^2_{L^\infty}+\|f(t)\|_{L^\infty}^2\right)\\
\left.\left.+\|\partial_x f\|_{L^\infty}\|\partial_x^2 f\|_{L^\infty}+\|\partial_x f\|_{L^\infty}\|f\|_{L^\infty}\right)\right].
\end{multline*}
Recalling the following inequalities
\begin{equation}\label{GN}
\|u\|_{H^{s}}\leq C\|u\|^{1-s/r}_{L^2}\|u\|^{s/r}_{H^r}\text{ and }\|u\|_{L^\infty}\leq C\|\partial_x u\|^{0.5}_{L^2}\|u\|^{0.5}_{L^2},
\end{equation}
and using $\|\partial_x f\|_{L^2}\leq C\|\partial_x f\|_{L^\infty}$, we get
$$
\|f\|_{H^{1.6}}\leq C\left(\|f\|_{L^\infty}+\|\partial_x f\|^{0.7}_{L^2}\|\partial_x f\|^{0.3}_{H^2}\right)\leq C\vertiii{f}_Te^{-\mathcal{C}_lt0.7},
$$
$$
\|\partial_x^2 f\|_{L^\infty}\leq C\|\partial_x f\|^{0.5}_{L^2}\|\partial_x^3 f\|^{0.5}_{L^2}\leq C\vertiii{f}_Te^{-\mathcal{C}_lt0.5}.
$$
Putting all the estimates together, we conclude the following estimate
\begin{equation*
\|NL(s)\|_{C^1}\leq C(l)\left(\vertiii{f}_T^3+\vertiii{f}_T^5\right)e^{-2\mathcal{C}_ls}.
\end{equation*}
Inserting this estimate in \eqref{boundduhamel}, we obtain
$$
e^{\mathcal{C}_lt}\|f(t)\|_{C^1}\leq\|f_0\|_{C^1}+C(l)\left(\vertiii{f}_T^3+\vertiii{f}_T^5\right)\int_0^\infty e^{-\mathcal{C}_ls}ds.
$$
With the energy estimates in Theorem \ref{teo1} and the definition of \eqref{totalnorm}, we get
$$
\frac{d}{dt}\|f(t)\|_{H^3}^2\leq C\|f(t)\|_{H^3}\mathcal{P}(\vertiii{f}_T)e^{-\mathcal{C}_lt},
$$
for a polynomial $\mathcal{P}$ with powers bigger than one. Integrating in time and collecting all the estimates together, we conclude
$$
\vertiii{f}_T\leq C\|f_0\|_{H^3}+\|f_0\|_{C^1}+\mathcal{Q}(\vertiii{f}_T)\leq C\delta+\mathcal{Q}(\vertiii{f}_T),
$$
where $\mathcal{Q}$ is a polynomial with high powers. From this latter inequality,  by a standard continuation argument, we obtain the global existence if $\delta$ is small enough. Moreover, if we take $\delta$ small enough, we can ensure that
$$
e^{\mathcal{C}_lt}\|f(t)\|_{C^1}\leq \frac{l}{4}.
$$
\end{proof}

\section{Limited regularity}\label{seclim}
\begin{proof}[Proof of Theorem \ref{teo4}]
We explain how to obtain the good bounds, then, using mollifiers for the initial data the result follows.

\textbf{Step 1: Case $\Omega=\mathbb T$} Given $\delta>0$, we define the energy
$$
E(t)=\|f(t)\|_{H^2}+\|f(t)\|_{\dot{W}^{1,\infty}}+\frac{1}{2\delta-\|f(t)\|_{H^2}}.
$$
We define $M(t)=\max_x \partial_x f(x,t)=\partial_x f(x_t)$ as in Proposition \ref{prop1}. Then, using Rademacher's Theorem, we have
$$
M'=\left(\frac{1}{1+l^2-(f(x_t))^2}-\frac{1}{1+(\partial_x f(x_t))^2}\right)\Lambda\partial_x f(x_t)+\Lambda f(x_t)\frac{2f(x_t)\partial_x f(x_t)}{(1+l^2-(f(x_t))^2)^2},
$$
thus, using Proposition \ref{prop1},
$$
M'\leq \left(\frac{1+l^2-\|f_0\|_{L^\infty}+2\|f(t)\|_{H^2}\|f_0\|_{L^\infty}}{(1+l^2-\|f_0\|_{L^\infty}^2)^2}-\frac{1}{1+(\partial_x f(x_t))^2}\right)M(t).
$$
If $\|f_0\|_{H^2}\leq \delta,$ due to the form of the energy, there exist a time $T^*$ such that $\max_{0\leq t\leq T^*}\|f(t)\|_{H^2}\leq 2\delta$. At this step in the proof, this time may depend on the regularization parameter, but we are going to bound it uniformly. Consequently,
$$
M'\leq \left(\frac{1}{1+l^2-C\delta^2}+\frac{C\delta^2}{(1+l^2-\|f_0\|_{L^\infty})^2}-\frac{1}{1+C\delta^2}\right)M(t),
$$
and, if $\delta<<1$ we obtain $M(t)\leq M(0),\text{ for }0\leq t\leq T^*.$ In the same way we obtain reverse inequality for $m(t)=\min_x \partial_x f(x,t)=\partial_x f(x_t)$. So,
$$
\|\partial_x f(t)\|_{L^\infty}\leq\|\partial_x f_0\|_{L^\infty}\leq C\delta\text{ for }0\leq t\leq T^*.
$$
From this decay we obtain that the solution relies in the stable regime. Recalling the definition \eqref{sigma}, we compute
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}\|f\|^2_{\dot{H}^2}&\leq& \left((4\|f_0\|_{L^\infty}+8)\|f(t)\|_{\dot{W}^{1,\infty}}+\|f_0\|_{L^\infty}\right)\|f\|^2_{\dot{H}^2}\\
&&+(3+\|f(t)\|_{H^2}2(1+2\|f(t)\|_{\dot{W}^{1,\infty}}))\|f\|^3_{\dot{W}^{2,3}}\\
&&+\left(8(\|f_0\|^2_{L^\infty}+\|f\|_{\dot{W}^{1,\infty}})\|f(t)\|_{\dot{W}^{1,4}}^2\right)\|f\|_{\dot{H}^2}\\
&&+\frac{1}{2}\int_\Omega\Lambda\left[\frac{1}{1+l^2-(f(x))^2}-\frac{1}{1+(\partial_x f(x))^2}\right](\partial_x^2f(x))^2dx\\
&&+\sigma(t)\|f(t)\|^2_{\dot{H}^{2.5}}.
\end{eqnarray*}
We use interpolation \eqref{GN2} to obtain
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}\|f\|^2_{\dot{H}^2}&\leq& C\delta\|f\|^2_{\dot{H}^2}+\sigma(t)\|f(t)\|^2_{\dot{H}^{2.5}}+(1+\delta)C\|f\|^2_{\dot{H}^{2.5}}\|f\|_{\dot{W}^{1,\infty}}\\
&&+\frac{1}{2}\int_\Omega\Lambda\left[\frac{1}{1+l^2-(f(x))^2}-\frac{1}{1+(\partial_x f(x))^2}\right](\partial_x^2f(x))^2dx.
\end{eqnarray*}
We use \eqref{eqbound} and \eqref{eqbound2}. Thus, using the $L^p$-boundedness of the singular integral operators,
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}\|f\|^2_{\dot{H}^2}&\leq& C\delta\|f\|^2_{\dot{H}^2}+\sigma(t)\|f(t)\|^2_{\dot{H}^{2.5}}+(1+\delta)C\|f\|^2_{\dot{H}^{2.5}}\|f\|_{\dot{W}^{1,\infty}}\\
&& +C(l)\|f\|_{W^{0.6,\infty}}^2\|f\|_{\dot{H}^2}^2+\|f_0\|_{L^\infty}\|\Lambda f\|_{L^\infty}\|f\|_{\dot{H}^2}^2\\
&&+C\|\partial_x f\|_{W^{0.6,\infty}}^2\|f\|_{\dot{H}^2}^2+\|f\|_{\dot{W}^{1,\infty}}\|f\|_{\dot{W}^{2,3}}^3\\
&\leq& C\delta\|f\|^2_{\dot{H}^2}+\sigma(t)\|f(t)\|^2_{\dot{H}^{2.5}}+C\|f\|^2_{\dot{H}^{2.5}}\|f\|_{\dot{W}^{1,\infty}}\\
&&+C\|f\|_{H^{2.1}}^2\|f\|_{\dot{H}^2}^2\\
&\leq& C\delta\|f\|^2_{\dot{H}^2}+\sigma(t)\|f(t)\|^2_{\dot{H}^{2.5}}+C\|f\|^2_{\dot{H}^{2.5}}\|f\|_{\dot{W}^{1,\infty}}\\
&&+C\|f\|^{0.32}_{L^2}\|f\|_{\dot{H}^2}^2\|f\|^{1.68}_{H^{2.5}}\\
&\leq& C\delta\|f\|^2_{\dot{H}^2}+\sigma(t)\|f(t)\|^2_{\dot{H}^{2.5}}+C\|f\|^2_{\dot{H}^{2.5}}\|f\|_{\dot{W}^{1,\infty}}\\
&&+C(\epsilon)\delta^2\|f\|_{\dot{H}^2}^{12.5}+C\epsilon\|f\|^{2}_{H^{2.5}}
\end{eqnarray*}
where we have used the continuous embedding $H^{2.1}\subset W^{1.6,\infty}$, \eqref{GN} and Young's inequality with $p=2/1.68$ and $q=6.25$. Notice that, if $\delta$ is small enough,
$$
\sigma(t)\leq \frac{1}{1+l^2-C\delta^2}-\frac{1}{1+C\delta^2}<0.
$$
Let $K\in\mathbb N$ be a fixed number. Inserting the latter bound we get
\begin{multline*}
\frac{d}{dt}\|f\|^2_{\dot{H}^2}+\frac{1}{K(1+C\delta^2)}\|f\|^2_{\dot{H}^{2.5}}\leq C(l,\epsilon)\delta\|f\|^2_{\dot{H}^2}\\
+\|f\|^2_{\dot{H}^{2.5}}\left(\frac{1}{1+l^2-C\delta^2}-\frac{K-1}{K(1+C\delta^2)}+C\epsilon+C\delta\right).
\end{multline*}
Thus, taking $0<\delta,\epsilon<<1$ small enough and $K=K(l)$ large enough, we obtain
$$
\|f\|_{\dot{H}^2}\leq\|f_0\|_{\dot{H}^2}e^{C(\epsilon,l)\delta t}.
$$
Putting all together, we obtain
$$
\frac{d}{dt}E\leq C(l)E(1+E^2)\leq C(l,\epsilon)\delta (1+E)^3.
$$
This bound doesn't depends on the regularization parameter, so using Gronwall inequality, we  obtain a time $T^*=T^*(l,\epsilon,\delta)$ where the solution remain in a ball with radius $2\delta$ in $H^2$. Moreover, due to the evolution of $M(t),m(t)$ and the Proposition \ref{prop1}, the solution doesn't leave the stable regime. This concludes the result in the periodic case.

\textbf{Step 2: Case $\Omega=\mathbb R$} Given $u(x)$, we define $X$ such that $u(X)=\max_x u(x)$. Then
\begin{multline*}
\Lambda u(X)=\frac{1}{\pi}\text{P.V.}\int_\mathbb R\frac{u(X)-u(X-y)}{y^2}dy\geq \frac{2u(X)}{\pi}\int_1^\infty y^{-2}dy\\
-\frac{1}{\pi}\int_{B^c(0,1)}\frac{u(X-y)}{y^2}dy\geq \frac{2}{\pi}u(X)-\frac{2}{3\pi}\|u\|_{L^2}
\end{multline*}
We define $M(t)$ as before. Then, using Rademacher's Theorem, we have
\begin{eqnarray*}
M'&\leq& \left(\frac{2\|f(t)\|_{H^2}\|f_0\|_{L^\infty}}{(1+l^2-\|f_0\|_{L^\infty}^2)^2}\right.\\
&&\left.+\frac{2}{\pi(1+l^2-\|f_0\|^2_{L^\infty})}-\frac{2}{\pi(1+(\partial_x f(x_t))^2)}\right)M(t)\\
&&-\left(\frac{1}{\pi(1+l^2-\|f_0\|^2_{L^\infty})}-\frac{1}{\pi(1+(\partial_x f(x_t))^2)}\right)\frac{2}{3}\|f\|_{\dot{H}^1}\\
&\leq& \left(\frac{C\delta^2}{(1+l^2-C\delta^2)^2}+\frac{2}{\pi(1+l^2-C\delta^2)}-\frac{2}{\pi(1+C\delta^2)}\right)M(t)\\
&&\left(\frac{-1}{\pi(1+l^2-C\delta^2)}+\frac{1}{\pi(1+C\delta^2)}\right)\frac{2}{3}\|f\|_{\dot{H}^1}
\end{eqnarray*}
and, if $\delta$ is taken small enough, integrating in time and using Proposition \ref{prop1}, we have
$$
\left(\frac{-1}{1+l^2-C\delta^2}+\frac{1}{1+C\delta^2}\right)\|f\|_{L^2_t\dot{H}^1}^2\leq-\int_0^t\sigma(s)\|f(s)\|_{\dot{H}^1}^2ds\leq \|f_0\|_{\dot{H}^{0.5}}^2,
$$
thus,
$$
\int_0^t\|f(s)\|_{\dot{H}^1}ds\leq \sqrt{t}\|f\|_{L^2_t\dot{H}^1}\leq \sqrt{t}C(l,\delta)\|f_0\|_{\dot{H}^{0.5}},
$$
and
$$
M(t)\leq M(0)+\sqrt{t}C(l,\delta)\|f_0\|_{\dot{H}^{0.5}}.
$$
With the same ideas we obtain the appropriate bound for $m(t)$ and we get
$$
\|f(t)\|_{\dot{W}^{1,\infty}}\leq \|f_0\|_{\dot{W}^{1,\infty}}+\sqrt{t}C(l,\delta)\|f_0\|_{\dot{H}^{0.5}}.
$$
With this bound and Proposition \ref{prop1} we conclude the existence of a time $T_1=T_1(\delta,l)$ such that the solution doesn't leave the stable regime. We define the energy
$$
E(t)=\|f(t)\|_{H^2}+\frac{1}{2\delta-\|f(t)\|_{H^2}}.
$$
With the same ideas as in the periodic case we get a second time $T_2=T_2(\delta,l,\epsilon)$ such that the solution remains in the ball with center the origin and radius $2\delta$ in $H^2$. We take $T^*=\min(T_1,T_2)$ the time of existence and we conclude the result.
\end{proof}
If we add a symmetry hypothesis for the initial data we can improve Theorem \ref{teo4}:
\begin{proof}[Proof of Theorem \ref{teo5}]
We have $X=H^{2.5}$ with norm $\|f(t)\|_X=\|f(t)\|_{H^2}+\int_0^t\|f(s)\|^2_{H^{2.5}}ds,$
$Y=H^1$ and $\mathcal{D}(t)=e^{\mathcal{C}_lt}$. Since Theorem \ref{teo4}, there exist a local solution on the interval $[0,T]$. We define the total norm $\vertiii{f}_T$ as in \eqref{totalnorm}. We use
$$
\|e^{-\mathcal{C}_lt\Lambda}\|_{L^2\rightarrow L^2}\leq e^{-\mathcal{C}_lt},
$$
Due to the interpolation inequality \eqref{GN} and using the expression \eqref{nonlinear}, we get
$$
\|NL(s)\|_{L^2}\leq C(l)\|\partial_x f\|_{L^2}\left(\|\partial_x f\|_{L^\infty}^2+\|f\|_{L^\infty}^2\right)\leq C(l)\frac{\vertiii{f}_T^3}{(\mathcal{D}(s))^{2}}.
$$
and, using Bochner Theorem,
$$
e^{\mathcal{C}_lt}\|f(t)\|_{L^2}\leq\|f_0\|_{L^2}+C(l)\vertiii{f}_T^3\int_0^t\frac{1}{e^{\mathcal{C}_ls}}ds\leq \|f_0\|_{L^2}+C(l)\vertiii{f}_T^3.
$$
With the same ideas, we get
$$
\|\partial_x NL(s)\|_{L^2}\leq C(l)\left(\frac{\vertiii{f}_T^{7/3}\|f(t)\|^{2/3}_{\dot{H}^{2.5}}}{(\mathcal{D}(t))^{4/3}}+\frac{\vertiii{f}_T^{3}+\vertiii{f}_T^{5}}{(\mathcal{D}(t))^{2}}\right).
$$
Consequently,
\begin{eqnarray*}
e^{\mathcal{C}_lt}\|f(t)\|_{\dot{H}^1}&\leq&\|f_0\|_{\dot{H}^1}+C(l)\left(\vertiii{f}_T^{3}+\vertiii{f}_T^{5}\right)\int_0^t\frac{1}{e^{\mathcal{C}_ls}}ds\\
&&+C(l)\vertiii{f}_T^{7/3}\int_0^t\frac{\|f(s)\|^{2/3}_{\dot{H}^{2.5}}ds}{e^{\mathcal{C}_ls(1/3)}}\\
&\leq& \|f_0\|_{\dot{H}^1}+C(l)\left(\vertiii{f}_T^{3}+\vertiii{f}_T^{5}\right)\\
&&+C(l)\vertiii{f}_T^{7/3}\left(\int_0^t\frac{1}{e^{\mathcal{C}_ls/2}}ds\right)^{2/3}\left(\int_0^t\|f(s)\|^{2}_{\dot{H}^{2.5}}ds\right)^{1/3}\\
&\leq&\|f_0\|_{\dot{H}^1}+C(l)\left(\vertiii{f}_T^{3}+\vertiii{f}_T^{5}+\vertiii{f}_T^{8/3}\right)
\end{eqnarray*}

We need to obtain \emph{bona fide} a priori estimates on the $H^2$ seminorm for small initial data. With the same estimates as in Theorem \ref{teo4} we get
$$
\frac{d}{dt}\|f\|^2_{\dot{H}^2}+\|f\|^2_{\dot{H}^{2.5}}\leq C(l,\|f_0\|_{H^2})\mathcal{P}(\vertiii{f}_T)e^{-\mathcal{C}_ls/4},
$$
where $C(l,\|f_0\|_{H^2})\rightarrow 1$ as $\|f_0\|_{H^2}\rightarrow0$ and $\mathcal{P}$ is a polynomial with high powers. Thus, adding both estimates,
$$
\vertiii{f}_T\leq \|f_0\|_{H^2}+ C(l,\|f_0\|_{H^2})\mathcal{Q}(\vertiii{f}_T).
$$
If $\|f_0\|_{H^2}<<1$ is choosen small enough, this nonlinear Gronwall-type inequality and the fact $\|f(t)\|_{W^{1,\infty}}\leq \|f_0\|_{W^{1,\infty}}$ (again, for a small enough $H^2$ initial data) give us the global existence by means of a classical continuation argument.
\end{proof}

\section{Numerical simulations for the confined Muskat problem}\label{secdec}
In this section we perform numerical simulations for equations \eqref{full} and \eqref{eq0.1} to study the decay of $\|f(t)\|_{L^ \infty}$. The main purpose of these simulations is to compare the behaviour when the depth is finite (equation \eqref{eq0.1}) with the case where the depth is infinite (equation \eqref{full}). We consider equations \eqref{eq0.1} and \eqref{full} where $\rho^2-\rho^1=4\pi$. For each initial datum we approximate the solutions of \eqref{eq0.1} and \eqref{full} with the same numerical and physical parameters.

To perform the simulations we follow the ideas in \cite{c-g-o08}. The interface is approximated using cubic splines with $N$ spatial nodes. The spatial operator is approximated with Lobatto quadrature (using the function \emph{quadl} in Matlab). Then, three different integrals appear for a fixed node $x_i$: the integral between $x_{i-1}$ and $x_i$, the integral between $x_i$ and $x_{i+1}$ and the nonsingular ones. In the two first integrals we use Taylor series to remove the singularity. In the nonsingular integrals the integrand is made explicit using the splines. We use a classical explicit Runge-Kutta method of order 4 to integrate in time. In the simulations we take $N=300$ and $dt=10^{-3}$. In what follows we change slightly the notation and write $f^{\pi/2}(x,t)$ for the solution of \eqref{eq0.1} and $f^{\infty}(x,t)$ for the solution of \eqref{full}. Notice that the superscript denotes the depth in each situation. Then, given an initial datum $f(x,0)=f_0(x)$, which is the same for both evolution problems, we are computing a numerical approximation for $f^{\pi/2}(x,t)$ and $f^ \infty(x,t)$. The initial datum considered is
\begin{equation}\label{caso1confmus}
f_0(x)=\left(\frac{\pi}{2}-0.0001\right)e^{-x^6}.
\end{equation}
We obtain Figures \ref{deccase1}. We can see that the decay is slower in the finite depth case and the existence of a \emph{big} time interval with a very small decay.

\begin{figure}[h!]
		\begin{center}
		\includegraphics[scale=0.35]{RealMuskat.eps}
		\end{center}
		\caption[Numerics for the confined Muskat problem]{a) Dynamics for $f^{\pi/2}(x,t)$, b) dynamics for $f^\infty(x,t)$, c) $f^{\pi/2}(x,t)$ (blue) and $f^ \infty(x,t)$ (red) for the same times $t_i$ and initial datum given by \eqref{caso1confmus}.}
\label{deccase1}
\end{figure}

\section{Numerical simulations for \eqref{model1}}\label{secnum}
In this section we take $l=\pi/2$. To approximate the solutions to \eqref{model1} we use a Fourier collocation method with a explicit Runge-Kutta scheme for the time integration. We write $N$ for the number of spatial nodes. Then the operator $\Lambda$ can be easily discretized using the Fast Fourier Transform routine. To perform the multiplications we jump to the physical space. To advance in time we use a Runge-Kutta (4,5) scheme.

\subsection{Decay of $\|f(t)\|_{L^\infty}$}
We consider $l=\pi/2$ and $f_0=(l-0.001)\cos(x)$ and we study $\|f(t)\|_{L^\infty}$. We show the results on Figure \ref{decaymodelfig}. We see that the evolution is qualitatively similar to the dynamics of the same quantity for equation \eqref{eq0.1} (see Figure \ref{deccase1}).
\begin{figure}[h!]
		\begin{center}
		\includegraphics[scale=0.2]{Decaymodel.eps}
		\end{center}
		\caption{Evolution of $\|f(t)\|_{L^\infty}$}
\label{decaymodelfig}
\end{figure}

\subsection{Reaching the boundary} We consider $a>0$ and define the family of initial data
$$
f^a_0(x)=\cos(x)*a+\pi/2-a.
$$
Notice that $f^a_0(0)=\pi/2=l$, thus, the equation is in the unstable regime.

\begin{figure}[h!]
		\begin{center}
		\includegraphics[scale=0.25]{onedimevolution.eps}
		\end{center}
		\caption{Evolution of $f(x,t)$ for the different cases $a=0.1,0.2,0.3,0.4$ with $N=2^{16}$. The wide line corresponds to the initial data.}
\label{evofig}
\end{figure}

\begin{figure}[h!]
		\begin{center}
		\includegraphics[scale=0.25]{onedimevolutiondx.eps}
		\end{center}
		\caption{Evolution of $\partial_x f(x,t)$ for the different cases $a=0.1,0.2,0.3,0.4$ with $N=2^{17}$. The wide line corresponds to the initial data.}
\label{evodxfig}
\end{figure}

\begin{figure}[h!]
		\begin{center}
		\includegraphics[scale=0.25]{onedimevolutionLambda.eps}
		\end{center}
		\caption{Evolution of $\Lambda f(x,t)$ for the different cases $a=0.1,0.2,0.3,0.4$ with $N=2^{17}$. The wide line corresponds to the initial data.}
\label{evolambdafig}
\end{figure}

We see in Figure \ref{evofig} that the second derivative at $x=0$ grows. In Figures \ref{evodxfig} and \ref{evolambdafig}, we observe that there exists two points where $\partial_x^2 f$ and $\partial_x\Lambda f$ are large.

\section{Large time dynamics}\label{seclarge}
In this section we show that the solution never leaves the stable regime and that, for H\"older solutions, the curvature is bounded at the point where the initial data reach the boundary. These statement excludes the two main candidates for finite time singularities.

\begin{proof}[Proof of Proposition \ref{prop2}]
\textbf{Step 1:} We define
$$
\Sigma(x,t)=\frac{1}{1+l^2-(f(x,t))^2}-\frac{1}{1+(\partial_x f(x,t))^2}.
$$
We compute
\begin{equation}\label{eqsigma}
\partial_t \Sigma=\frac{-2f\Sigma\Lambda f}{(1+l^2-f^2)^2}+\frac{-2\partial_x f\left(\Sigma\Lambda\partial_x f+\partial_x \Sigma\Lambda f\right)}{(1+\partial_x f^2)^2}.
\end{equation}
Using \eqref{sigma} and Rademacher Theorem, we get
$$
\frac{d}{dt}\sigma(t)=-2\sigma(t)\left(\frac{f(x_t,t)\Lambda f(x_t,t)}{(1+l^2-(f(x_t,t))^2)^2}+\frac{\partial_x f(x_t,t)\Lambda\partial_x f (x_t,t)}{(1+(\partial_x f(x_t,t))^2)^2}\right),
$$
thus,
$$
\sigma(t)=\sigma(0)\exp\left(-2\int_0^t \left(\frac{f(x_s,s)\Lambda f(x_s,s)}{(1+l^2-(f(x_s,s))^2)^2}+\frac{\partial_x f(x_s,s)\Lambda\partial_x f (x_s,s)}{(1+(\partial_x f(x_s,s))^2)^2}\right)ds\right).
$$
From this equation we conclude the first statement.

\textbf{Step 2:} We consider an initial data such that $f(\tilde{x})=l$ for some $\tilde{x}$. Evaluating \eqref{eqsigma} at $\tilde{x}$ we get
$$
\Sigma(\tilde{x},t)=\Sigma(\tilde{x},0)\exp\left(-2\int_0^t\frac{f(\tilde{x},s)\Lambda f(\tilde{x},s)}{(1+l^2-(f(x_s,s))^2)^2}ds\right).
$$
Assuming $f(x,t)\in C([0,T],H^3(\Omega))$ we obtain $\Sigma(\tilde{x},t)=0$ for all $0\leq t\leq T$. We compute
\begin{eqnarray*}
\partial_t\partial_x^2f&=&\Lambda\partial_x^2 f\left(\frac{1}{1+l^2-f^2}-\frac{1}{1+\partial_x f^2}\right)\\
&&+2\Lambda\partial_x f\left(\frac{2f\partial_x f}{(1+l^2-f^2)^2}+\frac{2\partial_x f\partial_x^2 f}{(1+\partial_x f^2)^2}\right)\\
&&+\Lambda f\left(\frac{2\partial_x f^2+2f\partial_x^2f}{(1+l^2-f^2)^2}+\frac{2\partial_x f\partial_x^3 f+2\partial_x^2f^2}{(1+\partial_x f^2)^2}\right)\\
&&+2\Lambda f\left(\frac{(2f\partial_x f)^2}{(1+l^2-f^2)^3}-\frac{(2\partial_x f\partial_x^2 f)^2}{(1+\partial_x f^2)^3}\right).
\end{eqnarray*}
If we evaluate at $x=\tilde{x}$ and we use the fact that $f(\tilde{x})=l$ is the maximum, we obtain
\begin{eqnarray*}
\partial_t\partial_x^2f(\tilde{x})&=&2\partial_x^2f(\tilde{x})\Lambda f(\tilde{x})\left(l+\partial_x^2f(\tilde{x})\right).
\end{eqnarray*}
From this ODE we conclude the result.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Q: Press left button, press right button, press left and right buttons simultaneously using mousedown handler I realise minesweeper game using JavaScript. I need to defer 3 events (press left button, press right button, press left and right simultaneously) using mousedown handler.
let isLeftMouseDown = false;
let isRightMouseDown = false;
let timer,
timeout = 100;
function clickCell(event){
    let $target = event.target;
    let clickCell = getCellById($target.id);
    let isRightMB;
    let isLeftMB;
    event = event || window.event;
    if ("which" in event){
        isRightMB = event.which == 3;
        isLeftMB = event.which == 1;
    }
    else if ("button" in event){
        isRightMB = e.button == 2;
        isLeftMB = e.button == 0;
    }

    if(isLeftMB){
        isLeftMouseDown = true;
    }
    else if(isRightMB){
        isRightMouseDown = true;
    }
    if(!timer){
        timer=setTimeout(function(){
            if(isLeftMouseDown){
                if(!clickCell.flag){
                    $($target).removeClass("close").addClass("open0");
                }
            }
            else if(isRightMouseDown){
                toggleFlag(clickCell)
            }
        }, timeout)
    }
    else{
        clearTimeout(timer);
        timer = null;
        if(isLeftMouseDown && isRightMouseDown){
            openCellEnviroment($target);
        }
    }
}
$field.on('mousedown', clickCell);

I used setTimeOut but it works not perfect. I want to know anything else way.

A: Look for these
element.addEventListener('click', (e) => {//do something on click on element})

redpajama_set_name: RedPajamaStackExchange


Kingston – St. Lawrence College will have newly renovated libraries but no college librarian to support students, staff, and faculty, says the Ontario Public Service Employees Union (OPSEU).
OPSEU faculty members from all 24 public community colleges elected a new Divisional Executive on the weekend. In voting June 5, delegates chose JP Hornick (Local 556, George Brown), Kevin MacKay (L. 240, Mohawk), RM Kennedy (L. 558, Centennial), Martin Devitt (L.


The school is very pleased to be awarded the judgement of GOOD in all four areas by a team of three inspectors in September 2013.
The school was further inspected in February 2018 and this reported that the school continues to be Good.
This inspection is undertaken under the remit from the National Society for church schools.


Kat Von D Beauty has a range of foundation called 'Lock It Foundation' and today, I am going to be talking about it. As always with my foundation and concealer posts, I am going to go over main headings including available shades, consistency/coverage/best applied with, shelf life/amount, lasting power and wear test, chemical (formula) analysis, marketing accuracy and an overall review and summary.
The Kat Von D Lock It Foundation comes in 33 different shades. These shades range from lights to mediums to tan shades to dark shades. I think the Kat Von D Lock It Foundation shade range is pretty versatile and I'm pretty sure there will be a shade for everyone. One thing about a large shade range is that it can make it more confusing to pick what shade you are, but then again a large shade range is a really good thing.
The Consistency of the foundation is creamy and very thick (as you can see in the swatch above). The Coverage of the foundation is full and you don't need much at all to build up to high coverage (more on this in the wear test below). The foundation is best applied with a Flat Kabuki Foundation Brush and can be applied with a Beauty Blender also, but if your using a Beauty Blender beware of how much product you apply because that could turn into a mess very easily.
The Shelf Life of the Kat Von D Lock It Foundation is 12 months (1 year) and each bottle comes with 30 ml (1.0 fluid ounces) of product. This means you have 12 months to use 30 ml of product, which is easy to do because foundation is a product used everyday for most.
Second Check In- 4.00 PM– four hours later– The foundation is still fully on and no coverage has been lost. There is still creasing under the eyes but no more than before. The finish and overall look is still really nice despite the tiny bit of creasing.
Third Check In- 6.00 PM– six hours later– The coverage is still full and none has come off after 6 hours. There is a bit more settling around the mouth and eye area but nothing too crazy. The settling is because of the thick consistency and it does feel a bit cakey, which doesn't worry me but will others. The finish is still flawless and I still love the finish.
Fourth Check In- 8.00 PM– eight hours later– The foundation hasn't worn off and it's still fully on. The finish is still beautiful and is providing full coverage still. Again, there is creasing but nothing too crazy. The foundation has stayed on all day with good coverage and I like the foundation a lot.
The Kat Von D Lock It Foundation has long-lasting power and passed my 8-hour lasting power test. It does crease a little bit, which is due to the thickness of the foundation, but the creasing is nothing too crazy. I didn't put concealer over the foundation so I could test it alone, but if you put concealer under the eyes it might not crease as much there, but at the end of the day (after 8 hours) there is slight settling. Again, this is nothing crazy and I really love the coverage and finish of the foundation. The wear of the foundation did pass my 8-hour wear test, even though it does crease and settle it's nothing crazy.
The Kat Von D Lock It Foundation has 28 ingredients in the formula and is ranked by the Cosmetic Analysis Database as an 'Average' formula. The Cosmetic Analysis Database ranks each individual chemical into 5 categories based on safety and functioning- highly recommended, recommended, average, not recommended and to be avoided- and then returns a formula reading based on this.
This 'Average' formula ranking is a result of 15 recommended chemicals, 9 average chemicals, 4 not recommended chemicals and NO to be avoided chemicals. The fact that there are 15 recommended chemicals in the formula means that the bulk of the formula contains good, recommended, high quality ingredients.
I am going to go over the 4 not recommended chemicals first and then I will go over the recommended chemicals in the formula. The 4 not recommended chemicals are Aluminium Starch Octenylsuccinate, Isododecane, Disodium EDTA and PEG/PPG-18/18 Dimethicone. Aluminium Starch Octenylsuccinate is an absorbent, anticaking and viscosity controlling agent that is ranked as 'not recommended' because it has skin irritating characteristics. Isododecane is a solvent and emollient agent in the formula and is ranked as 'not recommended' because it has negative characteristics such as building a waterproof barrier, promoting comedone construction (clogged pore construction), drying the skin out, leading to heat build-up and preventing the skin from breathing, among more. The fact that the foundation contains Isododecane means it is a matte foundation.
Disodium EDTA is a chelating and viscosity controlling agent in the formula and is ranked as 'not recommended' because it is not environmentally friendly. PEG/PPG-18/18 Dimethicone acts as an emulsifying agent and is ranked as 'not recommended' because it has characteristics such as irritating the skin, being irrelevant to the environment and making the skin porous thereby allowing the absorption of hazardous materials.
It's important to note there are NO to be avoided chemicals, which are the chemicals you want to avoid in formulas, so that is really good. Also note that all products no matter what have not recommended products, I have never came across a product that contains only recommended chemicals.
Now, let's get onto the good chemicals in the formula, which outweigh the bad. The formula contains 7 moisturizing chemicals including Trimethylsiloxysilicate, Trihydroxystearin and Isononyl Isononanoate which soften and smooth the skin and maintain it in good condition, Butylene Glycol which holds and retains moisture and Glyceryl Caprylate, Dimer Dilinoleyl Dimer Diilnoleate and Glycerin which also soften and smooths the skin.
Other than that, the preservative chemical used to extend shelf life is Potassium Sorbate, meaning the formula is paraben free. All in all, the Kat Von D Lock It Foundation has a good formula that contains mostly good, high quality ingredients.
Airless chamber prevents contamination, Pump system for precise dose measurements, Paraben-free, sulfate-free and phthalate-free, Synthetic fragrances.
What emollients (moisturizers) are used in the foundation?? Do they provide a velvety texture for perfectly smooth, hydrated skin?
There are a variety of emollient chemicals in the formula. Emollient chemicals have a moisturizing effect by preventing water loss from the skin, which gives the skin moisture (water is what moisturizes the skin, so preventing the loss of water from the skin moisturizes the skin). Emollient chemicals in the formula include Trimethylsiloxysilicate, Isononyl Isononanoate, Glyceryl Caprylate and Trihydroxystearin, along with a few more chemicals that also have other purposes.
These emollient chemicals all smooth and soften the skin and contribute to skin moisturizing. Do these emollient chemicals provide a velvety texture for perfectly, hydrated skin like Kat Von D Beauty claims ?? Yes, that is one way- a fancy way- to put it. These chemicals have moisturizing effects so do have some benefits in regards to providing moisture to ensure that the product glides on smoothly. This claim is correct but as I said, is explained in a fancy type of way, which is expected for product marketing. Overall, Kat Von D has accurately marketed this product in regards to their emollient and moisturizing benefit claims.
What siliconic elastomers and siliconic esters are in the foundation and what are their proper chemical name? Do these chemicals provide 24-hour transfer resistance? Aka, do they keep the foundation on and stop it from transferring?
Silicone chemicals in the formula include Trimethylsiloxysilicate, Polysilicone-11, Phenyl Trimethicone, Dimethicone Crosspolymer, PEG/PPG-18/18 Dimethicone and Silica Dimethyl Silicate.
Silicone chemicals have a variety of functions, including giving products transfer proof properties and helping products to stay on the skin for long periods of time as well as preventing any loss of colour. This means that Kat Von D has accurately marketed the use and transfer proof abilities of silicone chemicals in the formula. However, while Silicone chemicals do have transfer proof characteristics, this can also be classed as a negative characteristic because you have to make sure you remove silicone based foundations properly because they really do make the product stick to the skin, which can be bad if not removed and can cause breakouts, etc. So, while Kat Von D has marketed the benefits of silicone chemicals accurately, she also hasn't mentioned the downside to Silicone chemicals, which is expected with marketing and I don't have a problem with it.
Are there any parabens, sulfates or phlataletes in the formula?
There are no parabens in the formula and Potassium Sorbate is used instead, so Kat Von D has marketed this correctly. There are also no sulfates or phlataletes so this is marketed properly too.
The foundation has a very thick consistency and provides full coverage. Kat Von D hasn't lied when she says this foundation is full coverage, it sure is!
All in all, the marketing of the Kat Von D Lock It Foundation is accurate and all claims can be backed up. While there are a few claims that are over exaggerated, there are no straight up lies and over exaggeration is typical with brand marketing.
The Kat Von D Lock It Foundation comes in 33 different shades. It has a thick, creamy consistency and is full coverage. It is best applied with a Beauty Blender or a Flat Kabuki Foundation brush, which packs it nicely into the skin. The foundation has a shelf life of 12 months and contains 30 ml (1.0 fluid ounces) of product.
The foundation has a long lasting power and passed my 8-hour lasting power test because it stayed put all day. It also passed my 8-hour wear test, although there was a little bit of creasing and settling towards the end because of how thick the foundation is. The marketing strategies and claims are accurate and there are no straight out lies or claims that don't back up, only over exaggeration which is typical with any brand.
All in all, I do really like the Kat Von D Lock It Foundation but as you will see below in the ranking list, it isn't my go to or holy grail foundation. It is good though and I really like it's coverage.
• Shelf Life of foundation is 12 months and each bottle comes with 30 ml (1.0 fluid ounces) of product.
• The Wear of the foundation is good even though there is a bit of creasing and settling towards the end of the day. It still passed my 8-hour wear test because the creasing wasn't anything too bad.
• Formula contains 15 recommended chemicals, 9 average chemicals, 4 not recommended chemicals and no to be avoided chemicals.
I like the Kat Von D Lock It Foundation better than the Milani 2-in-1 Foundation and Too Faced Born This Way Foundation, but not better than the Tarte Amazonian Clay Foundation, Urban Decay All Nighter Foundation or Ella Bache Sunguard Foundation. This is just because the Kat Von D Foundation wears worse and creases/settles more than the other top foundations.
Where To Buy Kat Von D Foundation???


Thank you for this amazing website, it helped me a lot with my last stay in Belgrade, I discovered so many spots thanks to you.
I am Ilef from Tunisia, and I am in love with Serbia, I have been twice there. Once in summer 2016 and the other in summer 2017.
It would be nice from you to introduce us to winter life in Belgrade; the best spots to go to, the main activities, etc… as I am going back there in December and that most of the blogs about Belgrade on Internet are about the summer life there.
I will write about some winter acitvities soon!


Armed with the artifact, Manuela returned to the goddess. Her trident ended the attacks on her people and answered the calls for Tsovinar's Wrath. Later, in the betrayal of Tsovinar and the overthrow of her priests the trident was lost to the seas.
Base Damage: 2d6+3 (see properties below base damage does not include abilities), with an additional 2d8 when thrown against enemies of Tsovinar.
Dimensions: 12 inch blade, 72 inches long total, 4lbs.
Requires Attunement?: Yes, 24 hour dedication on the water in a boat.
The Trident of the Storm (Tsovinar's Blessing) is a magic weapon that starts with a 2d6 damage that grants a +3 bonus to attack and damage rolls made with it. An additional 2d8 happens against enemies of tsovinar. The trident also has other functions that resemble belt of dwarvenkind in granting a constitution bonus only, a dwarven thrower granting the return of the weapon, and a sword of sharpness. The difference is a limb is skewered and removed on a roll of 15-20, not just a 20 when a critical hit is taken.
It is unknown of more abilities come to bear if or when Tsovinar is freed.
Waterborne. This item floats on water and other liquids. Its bearer has advantage on Strength (Athletics) checks to swim.
Metamorphic. The item periodically and randomly alters its appearance in slight ways. The bearer has no control over these minor alterations, which have no effect on the item's use.
Muttering. The item grumbles and mutters. A creature who listens carefully to the item might learn something useful.
You have immunity to poison damage.
The range of your darkvision increases by 60 feet (or you gain dark vision 60′ if you didn't have it).
You gain proficiency with sailor's tools related to navigation, trident and trade (haggling).
If you are holding the trident, you can use your action to cast the conjure elemental spell from it, summoning a water elemental. You can't use this property again until the next dawn.
You can use an action to touch the trident to a fixed pool of water and cast the teleport spell from the trident. If your intended destination is in water, there is no chance of a mishap or arriving somewhere unexpected. You can't use this property again until 3 days have passed.
The only way to destroy the tridentis to melt it down on Anbar The Black Forge, where it was created.
Sentience and Personality. It is not known if the trident is sentient, no records remain of this at this point.
This entry was posted in Item of Wonder and tagged artifact, asim, weapon. Bookmark the permalink.


text:
The Great Seal Bug (also known as the Thing) was a Soviet mission in which a listening device was implanted within a copy of the Great Seal of the United States which was presented as a plaque to the the United States Ambassador to the Soviet Union W. Averell Harriman.
Henry Cabot Lodge displays the bug in front of the United Nations.
The plaque was presented On August 4, 1945 by a delegation of the Young Pioneer Organization of the Soviet Union, a youth organization for Soviet youth aged 10 to 15. It was presented in friendship as a peaceful gesture towards the Soviet's World War II ally. Unbeknownst to the ambassador, the plaque contained a listening device that was hidden in holes drilled under the beak of the eagle in the plaque. The device was extremely thin and powered remotely by an electromagnet energy source making it very difficult to detect. It was designed by Leon Theremin, a famed Russian inventor. The microphone hidden inside was passive and was only activated when the Soviets decided to turn it on. They usually did so from a van parked outside of the ambassador's house in which they ultra-high frequency beams at the house. Activated, the microphone would transmit any conversations coming from within range of the bug. When they turned the beams off, the bug was virtually undetectable.
It was hung in the Ambassadors' residential study (in Spasso House) and remained there until 1952. At that point, George Kennan was the ambassador. The bug was detected during a technical surveillance countermeasures inspection when a technician(Joseph Jacob Bezjian) with an untuned video receiver discovered it. A The existence of the bug was accidentally discovered by a British radio operator had overheard American conversations on an open radio channel and the Americans were alerted.
Ambassador Kennan later revealed in his memoirs that the Spasso House had been redecorated under Soviet supervision without and American presence. Surveillance sweeps were done routinely but revealed nothing. He also said that technicians conducting the surveillance asked him to pretend to be working out of the house one evening instead of his office. He therefore summoned his secretary to the Spasso House and began dictating a fictitious letter. At one point, one of the technicians encouraged him to continue with the dictation and began sweeping the room quietly. Eventually his sweep led to the Great Seal plaque which e took apart and found the bug.
The Soviets were undeterred and continued planting bugs throughout the rest of the Cold War.
Agnes Smedley 5 views
Operation Gold Whittaker Chambers

redpajama_set_name: RedPajamaCommonCrawl


There are a lot of masks, avatars, talking robots and even disembodied speaking computers in geek multimedia, and behind every one of those iconic characters there is a voice – sometimes one that becomes as much a part of pop culture as Luke Skywalker or the Iron Throne. That means that casting a voice actor can be as important as casting … well, any other actor. Take, for example, Cobra Commander, infamous for his shrill, grating tones in the '80s G.I. Joe cartoon. But when it came time to cast the new G.I. Joe: Retaliation, director Jon M. Chu decided to go in a different direction. "I mean, it [was] high-pitched screaming," Chu said. "We couldn't go there, it was just too annoying." But even after he tapped Robert Baker – an actor friend from his college days – for the man behind the chrome, it still took a while to get exactly right. "Ultimately what we found was that his breathing was something very important," Chu said. "Every time he breathes in, you hear this hiss. Not a hiss – it's a rattle. You could get personality out of that rattle." And with voicing a villain, particularly one who had a previous cartoon incarnation, knowing his personality is half the battle. Whether it's Johnny Five fretting about "No disassemble!" or the passive-aggression of rogue AI GLaDOS in Portal – not to mention the overthrow-plotting sass machine that is Siri – the voices behind some of the most interesting characters in media have become as recognizable to us as their faces (or lack thereof). To honor their righteousness, we here at Wired compiled some of our favorite voices from movies, TV, videogames and music.
The Cobra Commander featured in the G.I. Joe: A Real American Hero cartoon of the early and mid-1980s was – despite being the leader of a dangerous terrorist cell – not really all that threatening. Sure he had the über-militant royal blue uniform and faceless, steely mask, but also the ear-piercing wail of a male version of an evil witch going through puberty. Voiced by Chris Latta – who was the voice of the equally shrill Starscream on the Transformers cartoon of that time – Cobra Commander back then was all hiss and no bite.


text:
Six States to Go Green with $145M from DOE
July 9, 2009 • Staff • Bookmark +
WASHINGTON - More than $145 million in Recovery Act funding has been awarded to six states to support their green efforts, according to the U.S. Department of Energy (DOE).
Under the DOE's State Energy Program (SEP), the following states have proposed statewide plans that prioritize energy savings, create or retain jobs, increase the use of renewable energy, and reduce greenhouse gas emissions: Arkansas, Georgia, Kentucky, Mississippi, Montana, and New York.
With the initial 10 percent of total funding previously available to states to support planning activities, the July 6 awards bring the states' current funding to 50 percent. The remaining 50 percent of funds will be released once states meet reporting, oversight, and accountability milestones required by the Recovery Act.
The latest funds were distributed July 6:
Arkansas - $15.7 million.
Georgia - $32.9 million.
Kentucky - $21 million.
Mississippi - $16.1 million.
Montana - $10.3 million.
New York - $49.2 million.
For detailed information on these statewide plans, view the full release from the DOE.
In addition to these states, the DOE also awarded $8.2 million to the U.S. Virgin Islands to advance energy efficiency initiatives and renewable energy projects on the islands.
Read more about New York Mississippi Montana Recovery Act Georgia U.S. Department of Energy


text:
Asmussen Enters Five in Competitive Remington Springboard Mile
December 13, 2018 December 13, 2018 rowan wardLeave a comment
by Melissa Bauer-Herzog
Remington Park, home of the Springboard Mile. (Photo: Melissa Bauer-Herzog)
A four-time winner of the $400,000 Springboard Mile, Steve Asmussen hasn't seen the winner's circle for the race since 2014 when Bayerd beat Shotgun Kowboy by a head, but Remington's leading trainer looks to break that streak on Sunday.
The trainer has five horses entered in the one-mile 2-year-old feature from those who have spent their whole career at the track to those shipping in for the first time.
The leader of the group according to the morning line odds is Bankit, a Central Banker colt who is coming into the race off a victory in the Sleepy Hollow Stakes at Belmont Park and broke his maiden at Saratoga over the summer.
Originally pointed to the Remsen, co-owner Winchell Thoroughbreds' manager David Fiske said for a variety of reasons they changed course to this race.
"We had some other horses for the Kentucky Jockey Club, we seem to have a lot of stakes winning- and stakes placed-two year old colts this year," Fiske said. "Not that I'm complaining and we're very fortunate to be in that position but trying to find spots to run them and not run them over the top of one another and that's just ours. Steve's got a bunch of other nice 2-year-old colts as well. The $400,000 [purse] is attractive, Steve's got a division at Remington, Darren [Fleming's] there. All that works out, it's not really out of the way for us."
While Asmussen's string is formidable, a field of 12 horses searching for Kentucky Derby points makes this anything but a slam dunk for the trainer on paper.
Kelly Breen ships in Epic Dreamer, who has been named the morning line favorite at odds of 8/5. Coming into this race off a 2 ½ length maiden victory at Belmont Park on October 26, Epic Dreamer was making his second start. The month previously at the same track, the Orb colt had finished second by 1 ¾ lengths behind next-start Grade 3 Nashua Stakes winner Vekoma.
Another shipper coming in for the race is Brad Cox's Marquee Prince, who has already tested stakes company when finishing fifth in the Street Sense Stakes at Churchill Downs behind future Grade 1 winner Improbable.
Brad Cox trainee Marquee Prince, before the 2018 Street Sense Stakes. (Photo: Melissa Bauer-Herzog)
"He's doing really well," Cox said. "After the Street Sense he had a few really nice works at Churchill and we shipped him over to Oaklawn from Churchill, which is obviously a little closer to Remington. He had a good breeze here on Monday. He's a 2-year-old who has won two races and you don't have many options with those type of horses this time of year. We wanted to get him back around two turns. The Street Sense was just a race where we were obviously at Churchill and he was doing well. I took a shot there, he ran respectable and we're excited about getting him back around two turns."
Of the local challengers, Danny Pish's D Toz and Asmussen's Long Range Toddy seem to have the two best opportunities for a fully Southwest region-raced horse to get back to the winner's circle for the first time since Louies Flower in 2013.
Second in a maiden race at Remington on September 30 behind Cowboy Mischief, D Toz broke his maiden next out by 1 ¾ lengths. The gelding's most recent start came in the Don C. McNeill Stakes on November 24, where he beat the field by two lengths – including Cowboy Mischief. With a field that looks full of speed, D Toz won't get the lead easily, which is a downside but the gelding has proven to like this track and has regular rider Lane Luzzi aboard for this contest.
One of two horses owned by Willis Horton Racing along with Bankit, Kentucky-bred Long Range Toddy was also bred by Horton, who is still looking for his first victory in this race after Will Take Charge finished second to 128-to-1 longshot Texas Bling in 2012.
Fourth first out in an August 31 maiden going 5 ½ furlongs here, Long Range Toddy has put that beat behind him in a big way. Next out, Long Range Toddy proved too good for the field when romping by 5 ¾ lengths at the one mile distance of the Springboard Mile. In the local prep for this stakes race, he wasn't as dominant but proved to be too good for the field when pressing the pace and winning the Listed Clever Trevor Stakes by a length over seven furlongs. The Take Charge Indy colt seems to enjoy more distance – something that could be expected from his breeding – so the return to a mile should suit him well.
Picks and Ponderings: Fort Lauderdale Stakes Preview and Saturday Gulfstream Stakes Selections
2018 Remington Springboard Mile Picks and Analysis


Travelling to Buenos Aires? Here is a list of must-see Buenos Aires museums.
There are plenty of museums in Buenos Aires. The official tourism agency lists more than 160! So that you won't have to go over the whole list, let us point out for you the most interesting ones.


Gross has a message for Janet Yellen: Take a lesson from monopoly .. Gross likens the economy to the board game where players collect $200 every time they pass the "Go" square. That payment -- the game's version of credit creation -- ensures expansion at the start, but it also dooms players because the amount is fixed & doesn't grow, eventually leading to stagnation & bankruptcies ..


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Home / News / Auto Shows
2011 Jaguar C-X16 Concept: 2011 Frankfurt Auto Show
Martin Padgett September 13, 2011 Comment Now!
We've gone from teasers to previews to mega-galleries, and we're still not tired of showing you the stunning Jaguar C-X16 concept, on the floor this week at the 2011 Frankfurt Auto Show and hopefully, on a winding, kinky piece of tarmac within the next 18 months.
The C-X16 makes chief designer Ian Callum smile widely, and it's easy to see why: the two-seat derivative of the next-generation Jag XK platform is stubby and swift-looking, more dynamic from most angles than the larger, less lithe XK. The side view reads especially well in the C-X16's silver show paint, but after seeing it person, we're not sure the taillamps are quite up to the rest of the car.
There's no lack of excitement in the "production concept" specs. A supercharged, 3.0-liter V-6 breathes out 375 horsepower and 332 pound-feet of torque, which could push this sportscar to 0-60 times in the 4.4-second range, and up to a top speed of 186 mph. A "push to pass" button controls something dubbed a "hybrid boost function," which can send more power to the rear wheels when it's wanted or needed. It's all very Formula 1 meets the Jetsons.
More conventional-sounding are a ZF-bred eight-speed automatic transmission with stop-start technology wedded to it, seats formed from composite materials, and sheets upon sheets of red leather upholstery, which the C-X16 concept wears without a shred of shame.
A production version could bow before the end of next year, and it may as well wear XE badging--since, as Jaguar promises, this is the real successor to the E-Type celebrating its 50th anniversary this year.
If you're still unsatiated, check out our mega-gallery of the C-X16, and an accompanying video teaser.
High-Res Gallery: 2011 Jaguar C-X16 Concept
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