text:
\section{Introduction}

Let $U=U(t)$ denote the \emph{primary} velocity of the fluid, parallel to the walls of the channel, and let  $v=v(t,x)$ denote the \emph{secondary} velocity of the turbulent motion. According to \cite{Burgers}, they satisfy the following system of equations
\begin{align}
\frac{{\rm d} U }{{\rm d} t} ( t)&=P-\nu U( t) -\int_0^1v^2( t,y) {\rm d} y,   \label{1} \\
\frac{\partial v }{\partial t}( t,x) &=\nu \frac{\partial ^2v }{\partial x^2}( t,x)+U( t) v( t,x) -\frac \partial {\partial x}( v^2( t,x))
\label{2}
\end{align}
for $t>0$ and $x\in (0,1)$. The system is considered with the initial and boundary conditions
\begin{equation} \label{3}
\begin{aligned}
U( 0) &=U_{0},\ v( 0,x) =v_0( x) ,\quad  x\in (0,1)\\
v( t,0) &=v( t,1) =0,\quad  t>0.
\end{aligned}
\end{equation}

In \eqref{1} and \eqref{2}, $P$ is  a constant representing an exterior force, analogous to the mean pressure gradient in the hydrodynamic case, and $\nu =\frac{\mu }{\rho }>0$, where $\rho$ is the density and  $\mu$ is the viscosity of the fluid. We assume that $\rho$ and $\mu$ are constant.   The system is derived from the theory of turbulent fluid motion and has similar properties to  the Navier--Stokes equation, but is much simpler to study.

The existence and uniqueness of the global solution  was established  by D\l otko  \cite{Dlotko}, using the Galerkin method. The necessary and sufficient conditions on $P$ and $\nu $ such that the solutions to \eqref{1}--\eqref{3} satisfy $U(t)\rightarrow 0$ and $v(t)\rightarrow 0$ as $t\rightarrow +\infty $ are given in \cite{Cholewa-Dlotko} and \cite{Dlotko}.

In order to obtain nontrivial limiting behaviour  of solutions as $t\rightarrow +\infty$, the   following stochastic perturbation was  proposed
\begin{equation}\label{4}
{\rm d} U(t) =\left(P-\nu U( t) -\int_0^1  v^2(t,y){\rm d} y \right){\rm d} t+g_0\left(U(t),v(t)\right){\rm d} W_0(t),
\end{equation}
\begin{equation}\label{5}
\begin{aligned}
{\rm d} v(t,x) &=\left(\nu \frac{\partial ^2v }{\partial x^2}( t,x)+U( t) v(t,x)-\frac \partial {\partial x}( v^2( t,x) ) \right){\rm d} t    \\
&\quad+g_1\left(U(t),v( t,\cdot ) \right)(x){\rm d} W_1(t,x)
\end{aligned}
\end{equation}
with the initial and boundary conditions \eqref{3}. Above $W_0$ is a real-valued Wiener process,  $W_1$ is an independent of $W_0$  cylindrical Wiener process in $L^2:= L^2(0,1)$, see the next section,  $g_0\colon \mathbb{R}\times L^2\mapsto \mathbb{R}$  and $g_1\colon \mathbb{R}\times L^2\mapsto L^2 $ are Lipschitz continuous functions.

The existence and uniqueness of the global solution  to \eqref{3}--\eqref{5}  in the space $H:= \mathbb{R}\times L^2$ was established  by Twardowska and Zabczyk  \cite{Twardowska-Zabczyk1, Twardowska-Zabczyk2,Twardowska-Zabczyk3}. The existence and uniqueness of the solution for the classical one dimensional stochastic Burgers equation driven by cylindrical Wiener process, as well as the existence of an invariant measure was established by Da Prato, Debussche and Temam \cite{DaPrato-Debussche-Temam}.

As the main result, we prove the existence of an invariant measure to system \eqref{3}--\eqref{5}. In the proof we use  the Krylov--Bogolyubov theorem adapting   the method od Da Prato and Gatarek  \cite{DaPrato-Gatarek}, see also \cite{DaPrato-Zabczyk}. The main difficulty is  caused  by  the fact that since the noise is not an $\mathbb{R}\times L^2$-valued process we cannot use the It\^o formula.  Due to existence of  nonlinear diffusion terms and different structure of our system we cannot apply the Hopf--Cole transformation as for example in  \cite{E-Khanin-Mazel-Sinai}.

A similar problem of an hydrodynamic  equation coupled with heat equation was studied by Ferrario  \cite{Ferrario}. In the considered there B\'ennard  problem:  a  viscous  fluid,  in  a  rectangular container, is heated from below and the top surface is taken at constant temperature. Heating the fluid, its  density changes and the gradient of density gives rise  to a motion of particles from the bottom to the top of the container.

\section{Preliminaries and formulation of the main result}

Let $L^2:=L^2(0,1)$ be the Hilbert space equipped with the  scalar product  $\langle \varphi ,\psi \rangle =\int_0^1\varphi(x)\psi (x){\rm d} x$ and the corresponding norm $\| \cdot \|$.

Let $(\Omega ,\mathcal{F},(\mathcal{F}_t)_{t\ge 0 },\mathbb{P})$ be a filtered probability space. We assume that $(\Omega, \mathcal{F},\mathbb{P})$ is complete, the filtration is right-continuous and each $\mathcal{F}_t$ contains  all $\mathbb{P}$-null sets in $\mathcal{F}$. We consider the one-dimensional Wiener process $W_0(t)$,  $t\ge 0$, and a  cylindrical Wiener process $W_1(t,\cdot )$, $t\ge 0$,  in $L^2$. It can be defined as the sum
$$
W_1(t,x)=\sum_{k=1}^\infty W^k(t)e_k(x),
$$
where $(e_k)$ is any orthonormal basis of $L^2$ and $(W^k)$ are independent real-valued Wiener processes. The sum converges however not in $L^2$ but in any Hilbert space $V$ such that the imbedding  $L^2\hookrightarrow V$ is  Hilbert--Schmidt.  We assume that for all $t>s>0$, $k\in \mathbb{N}$, the increments  $W_0(t)-W_0(s)$ , $W^k(t)-W^k(s)$ are independent of the $\sigma$-field $\mathcal{F}_s$.

From now on,  we assume that  $(e_k)$ is the  orthonormal basis of the eigenvectors of the Laplace operator on $L^2$ with Dirichlet boundary conditions, that is
\begin{equation}\label{6}
e_k(x)=\sqrt{\frac 2\pi }\sin k\pi x,\qquad x\in (0,1),\  k=1,2,\dots .
\end{equation}

\begin{definition}
A pair of processes $(U,v)$ is \emph{a weak solution} to problem \eqref{3}--\eqref{5} if and only if they are adapted  with  continuous trajectories  in $\mathbb{R}$ and $L^2$, respectively, and  such that  for arbitrary $t\geq 0$,
\begin{align*}
U(t)&=U_0+tP-\nu \int_0^tU(s){\rm d} s-\int_0^t\| v(s)\|^2{\rm d} s  \\
&+\int_0^tg_0\left(U(s),v(s)\right) {\rm d} W_0(s),\quad \mathbb{P}\text{-a.s.,}
\end{align*}
and for arbitrary $\varphi \in C_0^\infty (0,1)$,
\begin{align*}
\langle v(t),\varphi \rangle  &= \langle v_0,\varphi \rangle +\int_0^tU( s) \langle v(s),\varphi \rangle {\rm d} s\\
&\quad +\int_0^t\left\langle v(s),\nu \frac{{\rm d}  ^2\varphi }{{\rm d}  x^2} \right\rangle {\rm d} s+\int_0^t\left\langle v^2(s),\frac{{\rm d}  \varphi }{{\rm d}  x}\right \rangle {\rm d} s
\\
&\quad +\int_0^t\left\langle \varphi ,g_1\left(U(s),v(s, \cdot)\right) {\rm d} W_1(s)\right\rangle ,\qquad \mathbb{P}\text{-a.s.}
\end{align*}

\end{definition}

We introduce now an  equivalent concept of the \emph{integral or mild solution}, see e.g.  \cite{DaPrato-Zabczyk1, DaPrato-Zabczyk}. Namely, let $(e_k)$ be the orthonormal basis of $L^2$ given by \eqref{6}, and let $S(t)$, $t\geq 0$, be the classical heat semigroup on $L^2$. Then, for any $\psi\in L^2$,
$$
S(t)\psi=\sum_{k=1}^\infty {\rm e}^{-\frac{\pi ^2}\nu k^2t}\langle \psi ,e_k\rangle e_k
$$
with the convergence of the series in $L^2$. It is well known that the generator $A$ of the semigroup $S(t)$, $t\geq 0$, is identical with the second derivative operator $\frac{{\rm d}  ^2}{{\rm d}  x^2}$ on the domain $D(A)$ consisting of functions $\psi $ such that $\psi $ and $\frac{{\rm d} \psi }{{\rm d}  x}$ are absolutely continuous, $\psi,\ \frac{{\rm d} \psi }{{\rm d}  x},  \  \frac{{\rm d}  ^2\psi }{{\rm d}  x^2}\in L^2$,  and  $\psi(0)=\psi(1)=0$.

A proof of the following lemma can be find e.g.  in \cite{DaPrato-Debussche-Temam, Twardowska-Zabczyk1, Twardowska-Zabczyk2, Twardowska-Zabczyk3}.
\begin{lemma}\label{Lemma1}
The operators $S(t)$, $t> 0$, can be extended linearly to the space of all distributions of the form $\frac {{\rm d} \psi}  {{\rm d}  x}$, $\psi\in L^1(0,1)$, in such a way that they  take values in $L^2$ and for all $\psi\in L^2$ and $t\ge 0$,
$$
\left\| S(t)\frac {{\rm d}  \psi }{{\rm d}  x}\right\| \leq \| \psi\| _{L^1(0,1)}\left(\sum_{k=1}^\infty \frac{2\pi }{\sqrt{\nu }}k^2{\rm e}^{- \frac{2\pi ^2}\nu k^2t}\right)^{1/2}
\le \frac{C}{\sqrt{t}}\| \psi\| _{L^1(0,1)}.
$$
\end{lemma}

\begin{definition}
A pair of continuous adapted processes $(U,v)$ with values in $\mathbb{R}$ and $L^2$, respectively, is said to be \emph{an integral or mild solution to problem} \eqref{3}--\eqref{5} if
$$
\begin{aligned}
U(t)&={\rm e}^{-\nu t}U_0+\int_0^t{\rm e}^{-\nu (t-s)}\left(P-\| v(s)\| ^2\right){\rm d} s
 \\
&\quad +\int_0^t{\rm e}^{-\nu (t-s)}g_0\left(U(s),v(s)\right){\rm d} W_0(s),
\end{aligned}
$$
and
$$
\begin{aligned}
v(t)&=S(t)v_0+\int_0^tS(t-s)\left( U( s) v(s) - \frac {{\rm d} v^2}{{\rm d}  x}(s)\right){\rm d} s  \\
&\quad +\int_0^tS(t-s)g_1\left(U(s),v(s)\right){\rm d} W_1(s).
\end{aligned}
$$
\end{definition}

In the integral
$$
\int_0^tS(t-s)\frac {{\rm d} v^2} {{\rm d} x}(s){\rm d} s
$$
we use the extension of the operator $S(t-s)$ described in Lemma \ref{Lemma1}.

\bigskip
Let $Z_T^p$, $p>1$, denote the space of all continuous adapted processes $X = (U,v)$  on $[0,T]$ with values on $\mathbb{R}\times L^2$ such that
$$
\left\| (U, v)\right\| _{p,T}=\| U\| _{1,p,T}+\| v\| _{2,p,T}<+\infty,
$$
where
$$
\begin{aligned}
\| U\| _{1,p,T}&=\left(\mathbb{E}\, \sup_{t\in [0,T]}\vert U(t)\vert  ^p\right)^{1/p},\\
\|v\|_{2,p,T}&= \left(\mathbb{E}\, \sup_{t\in [0,T]}\| v(t)\| ^p\right)^{1/p}.
\end{aligned}
$$

The first part of the theorem below concerning the existence of the solution was proven  in \cite{Twardowska-Zabczyk1,Twardowska-Zabczyk2,Twardowska-Zabczyk3} and it is stated here only for the sake of completeness.  The second and the third parts concerning existence and uniqueness of the invariant measure are the main result of the present paper. Their proofs are presented in Sections \ref{SInvariant} and \ref{SSFeller-Ireducibility}, respectively.
\begin{theorem}
Assume that the mappings  $g_0\colon \mathbb{R}\times L^2\mapsto \mathbb{R}$ and $g_1\colon \mathbb{R}\times L^2\mapsto L^2$ are bounded and Lipschitz  continuous. Then the following assertions are true:\\
$(i)$  For any initial values $U_0\in \mathbb{R}$ and $v_0\in L^2$,  system \eqref{3}--\eqref{5} has a unique weak solution in any $Z_T^p$-space, $T>0$, $p\ge 1$.  Moreover, \eqref{3}--\eqref{5} defines Feller Markov process on the Hilbert space $H:= \mathbb{R}\times L^2$. \\
$(ii)$ There is an invariant measure $\mu$  for the Markov family defined by system \eqref{3}--\eqref{5}. \\
$(iii)$ If $g_0$ and $g_1$ are separated from zero; i.e. there is a constant $C>0$ such that
$$
\left\vert \frac{1}{g_0(r,\psi)}\right\vert  \le C, \quad \left\| \frac{1}{g_1(r,\psi)}\right\|\le  C \qquad \text{for all $r\in \mathbb{R}$ and $\psi \in L^2$},
$$
then the corresponding transition semigroup $(P_t)$ is strong Feller and irreducible, and the invariant measure $\mu$  is unique. Finally, for any initial value $(U_0,v_0)$ the law $\mathcal{L}(U(t),v(t))$ of the solution at time $t$ converges in the total variation norm to $\mu$ as $t\to +\infty$.
\end{theorem}
\section{Auxiliary results}

To prove the existence of an invariant measure we shall apply the classical Krylov--Bogo\-lyubov criterion  in the form of the following proposition, see \cite{DaPrato-Zabczyk}.

\begin{proposition}\label{P1} \textbf{(Krylov--Bogo\-lyubov)}  Assume that  $X$ is a Feller Markov process on a Hilbert space $H$ and $H_0\subset H$ is a Hilbert space contained in $H$ with compact embedding. If for some X(0) and for  each $\varepsilon >0$ there exists a constant $M$ such that for all $t>0$,
$$
\frac 1t\int_0^tP\left(\| X(s)\| _{H_0}\geq M\right){\rm d} s\leq \varepsilon ,
$$
then there exists an invariant measure for $X$ in $H$.
\end{proposition}

Let us recall that  $\nu$ is the viscosity parameter. The eigenvalues of $A$ are $-\lambda_m = {\frac{\pi^2 m^2}{\nu}}$,  $m=1,\ldots$.   The operator $-A$ is self-adjoint and positive definite and for arbitrary $\alpha >0$ the fractional power $(-A)^{\alpha}$ is given by the formulae,
$$
(-A)^{\alpha}\psi = \sum_{m=1}^{\infty} \lambda_m^{\alpha}\psi_me_m\quad \text{for}\quad  \psi=\sum_{m=1}^{\infty} \psi_m e_m \in D(-A)^{\alpha},
$$
where $\psi \in D(-A)^{\alpha}$ if and only if
$$
\|(-A)^{\alpha}\psi \| =\left(\sum_{m=1}^{\infty} \lambda_m^{ 2\alpha} \psi _m^2 \right) ^{1/2} < +\infty.
$$
We set $H^{2\alpha} = D(-A)^{\alpha}$.

In our  proof of the existence of an invariant measure to \eqref{3}--\eqref{5} we will show that the  assumptions of the proposition above are satisfied  for    $H=\mathbb{R}\times L^2$ and  $H_0=\mathbb{R}\times H^{\frac 14}$.

Let $X(t)=\left (U(t), v(t)\right)$ be the solution to problem \eqref{3}--\eqref{5}.   We assume that $U(0)=0$ and $v_0=0$. Given  $L>0$ set
\begin{equation} \label{7}
\begin{aligned}
Z_L (t)&:=\int_0^t{\rm e}^{-L(t-s)} S(t-s)g_1\left(U(s),v(s)\right){\rm d} W_1(s),\qquad t>0, \\
Y_L(t)&:= \int_0^t{\rm e}^{-(L+\nu) (t-s)}g_0\left(U(s),v(s)\right){\rm d} W_0(s),\qquad t>0.
\end{aligned}
\end{equation}
\begin{lemma}\label{L2}
Let $p\ge 2$ and $\varepsilon >0$. There exists $L_0\ge 0$ such that for $L\ge L_0$ we have
$$
\sup_{0\leq t<+\infty }\mathbb{E}\left( \left\| Z_L (t)\right\|_{H^\frac 14} ^p+ \left\vert Y_L(t)\right\vert ^p\right)
\le \varepsilon .
$$
\end{lemma}
\begin{proof} Let $\| B \|_{(HS)}$ denote the Hilbert--Schmidt norm of a linear operator $B\colon L^2\mapsto L^2$. In \eqref{7} we identify an element   $g_1\left(U(s),v(s)\right)$ of $L^2$ with the  multiplication operator
$$
g_1\left(U(s),v(s)\right)\colon L^2\ni \psi \mapsto g_1\left(U(s),v(s)\right)\psi \in L^2.
$$
Clearly, its operator norm
$$
\left\| g_1\left(U(s),v(s)\right)\right\|_{L(L^2,L^2)} \le \sup_{r\in \mathbb{R}, \psi \in L^2} \|g_1(r,\psi)\|:= K.
$$
Hence using the Burkholder  inequality  and denoting by $S_L(t) ={\rm e}^{-Lt}S(t)$,  we obtain
\begin{align*}
\mathbb{E}\left\| Z_L (t)\right\|_{H^\frac 14} ^p &\le c_p \mathbb{E} \left( \int_0^t\left\|(-A)^{\frac 18}S_L(t-s) g_1\left(U(s),v(s)\right)\right \|_{(HS)}^2 {\rm d} s \right)^{\frac{p}{2}}\\
&\le c_pK^p \left( \int_0^{+\infty} \left\|(-A)^{\frac 18}S_L(s)\right \|_{(HS)}^2 {\rm d} s \right)^{\frac{p}{2}}\\
&\le c_pK^p \left( \int_0^{+\infty} {\rm e}^{-2Ls} \left\|(-A)^{\frac 18}S (s /2)S(s/2) \right \|_{(HS)}^2 {\rm d} s \right)^{\frac{p}{2}}.
\end{align*}
Since
\begin{align*}
\left\|(-A)^{\frac 18}S( s/ 2)S( s/2) \right \|_{(HS)}&\le \left\|(-A)^{\frac 18}S( s/ 2) \right \|_{L(L^2,L^2)}\left\|S( s/2) \right \|_{(HS)}\\
&\le C s^{-\frac 1 8} \left\|S( s/2) \right \|_{(HS)},
\end{align*}
we have
\begin{align*}
\mathbb{E}\left\| Z_L (t)\right\|_{H^\frac 14} ^p &\le C' \left( \int_0^{+\infty} {\rm e}^{-2Ls } s^{-\frac 14} \sum_{k=1}^\infty {\rm e}^{-\frac{\pi ^2}\nu k^2s}
 {\rm d} s \right)^{\frac{p}{2}}.
\end{align*}
Therefore the desired conclusion follows from the Lebesgue dominated convergence theorem and fact that
$$
\int_0^{+\infty}  s^{-\frac 14} \sum_{k=1}^\infty {\rm e}^{-\frac{\pi ^2}\nu k^2s}{\rm d} s =
 \int_0^{+\infty} s^{-\frac 14} {\rm e}^{-s}{\rm d} s \sum_{k=1}^\infty \left(\frac{\pi ^2}\nu k^2\right)^{-1 + \frac 14} <+\infty.
$$
Clearly, in the same way one can obtain the desired estimate for $Y_L$.
\end{proof}
\begin{corollary}\label{C1}
Taking into account the  Sobolev embedding $H^{\frac14}\hookrightarrow L^4$,  see Theorem 3, p. 8 in \cite{Runst-Sickel}, we obtain
$$
\lim_{L\to +\infty} \sup_{t\ge 0} \mathbb{E}\, \|Z_L(t)\|_{L^4}^p=0.
$$
\end{corollary}

Let
$$
\begin{aligned}
V_L(t)&:= v(t)-Z_L(t),\\
U_L(t)&:= U(t)-Y_L(t).
\end{aligned}
$$
Note that $V_L$ and $U_L$ are solutions to the following system of equations with random coefficients:
$$
\begin{aligned}
\frac{{\rm d} U_L}{{\rm d} t}(t)&= P-\nu U_L(t)- \|V_L(t)+Z_L(t)\|^2 + L Y_L(t),\\
U_L(0)&=U_0
\end{aligned}
$$
and
$$
\begin{aligned}
\frac{\partial  V_L}{\partial t}(t,x)&= \nu \frac{\partial ^2V_L}{\partial x^2} (t,x) + \left(U_L(t)+Y_L(t)\right) \left( V_L(t,x)+ Z_L(t,x)\right) \\
&\quad - \frac{\partial }{\partial x} \left( V_L(t,x)+ Z_L(t,x)\right)^2 +LZ_L(t,x),\\
V_L(0,x)&=v_0(x),\\
V_L(t,0)&=0=V_L(t,1),\qquad t>0.
\end{aligned}
$$

Recall, see e.g. \cite{Runst-Sickel},  that $H^1_0$ is the Sobolev space  consisting of absolutely continuous functions $\psi $ such that $ \frac{{\rm d} \psi }{{\rm d}  x}\in L^2$,  and  $\psi(0)=\psi(1)=0$.  We can now state  the main result of this section.
\begin{lemma} \label{L3}
Given $L$ write
$$
y_L(t):= \|V_L(t)\|^2 + U_L^2(t), \qquad t\ge 0.
$$
Then  there are independent of $L$  constants $\beta >0$ and  $ C>0$ such that
$$
\frac{{\rm d} }{{\rm d} t} y_L(t)+ \beta \left[ y_L(t)+ \|V_L(t)\|_{H^1_0}\right] \le C y_L(t) \xi_L(t) + C\eta_L(t),
$$
where
$$
\begin{aligned}
\xi_L(t)&:=  \vert  Y_L(t)\vert  +  \|Z_L(t)\|^2   +\|Z_L(t)\| + \| Z_L(t) \| _{L^4}^{\frac 8 4},\\
\eta_L(t)&:= (1+L^2) \left( Y_L ^4(t)+   \|Z_L(t)\|_{L^4}^4\right)  +1.
\end{aligned}
$$
\end{lemma}
\begin{proof} We have
\begin{align*}
\left\langle \frac{\partial V_L }{\partial t},V_L \right\rangle  &=\nu \left \langle  \frac{\partial ^2V_L }{\partial x^2},V_L \right\rangle  +U\left\langle V_L
+Z_L ,V_L \right\rangle \\
&-\left\langle \frac \partial {\partial x}(V_L +Z_L )^2,V_L\right\rangle
+L \left\langle Z_L ,V_L \right\rangle.
\end{align*}
Therefore
\begin{align*}
\frac 12\frac {{\rm d}}{{\rm d} t}\| V_L\|^2 &= -\nu \left\langle \frac{\partial V_L }{\partial x},\frac{\partial V_L }{\partial x}\right\rangle  +U\langle V_L ,V_L\rangle +U\langle Z_L ,V_L\rangle \\
&+\left\langle V_L ^2,\frac{\partial V_L }{\partial x}\right\rangle +2\left\langle V_L Z_L ,\frac{\partial V_L }{\partial x}\right\rangle  +\left\langle Z_L ^2,\frac{\partial V_L }{\partial x}\right\rangle  +L \langle  Z_L ,V_L \rangle\\
&= -\nu \left\langle \frac{\partial V_L }{\partial x},\frac{\partial V_L }{\partial x}\right\rangle  +U\langle V_L ,V_L\rangle +U\langle Z_L ,V_L\rangle \\
&+2\left\langle V_L Z_L ,\frac{\partial V_L }{\partial x}\right\rangle  +\left\langle Z_L ^2,\frac{\partial V_L }{\partial x}\right\rangle  +L \langle Z_L ,V_L \rangle,
\end{align*}
since $\left \langle V_L ^2,\frac{\partial V_L }{\partial x}\right\rangle = 0$.  In other words we have
\begin{align*}
& \frac 12\frac {{\rm d}}{{\rm d} t} \| V_L \| ^2+\nu \| V_L \| _{H_0^1}^2 \\
&=U\| V_L \| ^2+U\langle V_L ,Z_L\rangle   +2\int_0^1  V_LZ_L \frac{\partial V_L }{\partial x}{\rm d} x+\int_0^1  Z_L ^2\frac{\partial V_L }{\partial x}{\rm d}  x+L \langle  Z_L ,V_L \rangle.
\end{align*}
Further we have
\begin{align*}
\left\vert \int_0^1Z_L ^2\frac{\partial V_L }{\partial x}{\rm d} x\right\vert  &\le \| Z_L ^2\|^{\frac 12}\left\| \frac{\partial V_L }{\partial x}\right\| =\| Z_L \| _{L^4}^2\| V_L \| _{H_0^1}  \\
&\le \frac \nu 4 \| V_L \| _{H_0^1}^2+\frac 1{\nu }\| Z_L \| _{L^4}^4.
\end{align*}
Using an   interpolation estimate, see e.g.  \cite{Lions-Magenes},  we obtain
\begin{align*}
2\left\vert  \int_0^1  V_L Z_L\frac{\partial V_L }{\partial x}{\rm d} x\right\vert
&\le 2C_1 \| V_L \| ^{\frac 3 4} \| V_L \| _{H_0^1}^{\frac 54} \|Z_L\|_{L^4} \le \frac \nu 4 \| V_L \| _{H_0^1}^2 + C_2\| V_L \| ^2  \|Z_L\|_{L^4}^{\frac 83}
\end{align*}
Hence
\begin{align*}
\frac 12\frac {{\rm d}}{{\rm d} t} \| V_L \| ^2+\frac {\nu}{2}  \| V_L \| _{H_0^1}^2&\le  U\| V_L \| ^2+U\langle V_L ,Z_L \rangle +  L \|Z_L\| \|V_L\|   \\
&+C_2  \|V_L \| ^2\| Z_L \| _{L^4}^{\frac 8 4}  +\frac 1 \nu \| Z_L \|_{L^4}^4.
\end{align*}
Recall that $U= U_L+Y_L$ and
\begin{align*}
\frac 12\frac{{\rm d} U_L^2}{{\rm d} t}+\nu U_L^2&=U_L\left(P-\| V_L +Z_L \| ^2 + LY_L\right)\\
&= U_L\left( P-\| V_L \| ^2-2\langle V_L ,Z_L \rangle -\| Z_L \| ^2\right) +LU_LY_L
\\
&\le \frac{\nu}{2} U_L^ 2 + \frac{1}{2\nu} P^2 -U_L\| V_L \| ^2-2U_L\langle V_L ,Z_L\rangle -U_L\| Z_L \| ^2 \\
&\qquad +LU_LY_L.
\end{align*}
Therefore adding the above inequality we obtain
\begin{align*}
&\frac 12\frac {{\rm d}}{{\rm d} t}\left[  \| V_L \| ^2+U_L^2\right ]+\frac{\nu}{2} \left[ \| V_L \|
_{H_0^1}^2+ U_L^2\right]  \\
&\le Y_L \| V_L\| ^2+(Y_L-U_L) \langle V_L ,Z_L \rangle +L \|Z_L\| \|V_L\|  \\
&+C_2  \|V_L \| ^2\| Z_L \| _{L^4}^{\frac 8 4}  +\frac 1 \nu \| Z_L \|_{L^4}^4 + \frac{1}{2\nu} P^2 -U_L\| Z_L \| ^2 +LU_LY_L.
 \end{align*}
Thus, using  the Young inequality $|ab|\le a^p/p+ b^q/q$, we obtain
\begin{align*}
&\frac {{\rm d}}{{\rm d} t}\left [ \| V_L\| ^2+U_L^2\right]+\frac{\nu}{2}  \left[  \| V_L \|_{H_0^1}^2+U_L^2\right ]  \\
&\le  2 \vert Y_L\vert  \| V_L\| ^2+2 (Y_L-U_L) \langle V_L ,Z_L \rangle +C_3 L^2  \|Z_L\|^2   \\
&\quad +2C_2  \|V_L \| ^2\| Z_L \| _{L^4}^{\frac 8 4}  +\frac 2 \nu \| Z_L \|_{L^4}^4 + \frac 1\nu P +C_3\| Z_L \| ^4 +C_3L^2  Y_L^2\\
&\le 2 \vert Y_L \vert \| V_L\| ^2+2 \vert Y_L\vert \|Z_L\|\|V_L\| +   2 \vert U_L\vert \|V_L\| \|Z_L \|   +C_3 L^2  \|Z_L\|^2   \\
&\quad +2C_2  \|V_L \| ^2\| Z_L \| _{L^4}^{\frac 8 4}  +\frac 2 \nu \| Z_L \|_{L^4}^4 + \frac{1}{\nu}P +C_3 \| Z_L \| ^4 +C_3L^2  Y_L^2\\
&\le 2 \vert Y_L \vert \| V_L\| ^2+ \vert Y_L\vert ^2 +  \|Z_L\|^2 \|V_L\|^2  +   U_L^2   \|Z_L \| + \|V_L\|^2 \|Z_L\| \\
&\quad  +C_3 L^2  \|Z_L\|^2  +2C_2  \|V_L \| ^2\| Z_L \| _{L^4}^{\frac 8 4}  +\frac 2 \nu \| Z_L \|_{L^4}^4 \\
&\le + \frac{1}{\nu} P +C_3 \| Z_L \| ^4 +C_3L^2  Y_L^2\\
&\le C_4 \left[ \| V_L\| ^2+U_L^2\right] \left(Y_L(t) +  \|Z_L\|^2   +\|Z_L\| + \| Z_L \| _{L^4}^{\frac 8 4} \right) \\
&\quad +C_4\left( (1+L^2) \vert Y_L\vert ^2 + (1+L^2)  \|Z_L\|^2 +\| Z_L \|_{L^4}^4 + P\right)
\end{align*}
and the desired estimate follows from the Poincar\'e inequality
$$
\pi ^2\| V_L \| ^2\le \| V_L \| _{H_0^1}^2
$$
and from the fact that $\|Z_L\|+\|Z_L\|^2 \le 2 \|Z_L\|^4_{L^4} +2$.
 \end{proof}
 \begin{corollary}\label{C2}
 Let $X=(U,v)$ be the solution to problem \eqref{3}--\eqref{5} with initial conditions $U(0)=U_0\in \mathbb{R}$ and $v(0)=v_0\in L^2$. Then
 $$
 \lim_{M\to +\infty} \limsup_{T\to +\infty} \frac 1T \int_0^T \mathbb{P}\left( \vert U(t)\vert + \|v(t)\|\ge M\right){\rm d} t=0.
 $$
 \end{corollary}
 \begin{proof} We use the idea of Da Prato and Gatarek  \cite{DaPrato-Gatarek}, see also Da Prato and Zabczyk \cite{DaPrato-Zabczyk}. Namely, let us fix $M\ge 2$.  Recall, see Lemma \ref{L3},  that $ y_L(t):= \|V_L(t)\|^2 + U_L^2(t)$. Given $L>0$ set
 $$
 \zeta_L(t):= \log \left( y_L(t) \vee \frac{M}{2}\right).
 $$
 Then
 $$
 \frac{{\rm d}}{{\rm d} t}\zeta_L(t)= \chi_{\{y_L(t)\ge M/2\}}\frac {1}{y_L(t)}\frac{{\rm d}}{{\rm d} t} y_L(t).
 $$
Recall, see Lemma \ref{L3} that
 $$
 \frac{{\rm d} }{{\rm d} t} y_L(t)+ \beta y_L(t)\le C y_L(t) \xi_L(t) + C\eta_L(t).
$$
Multiplying both sides of the inequality above, by
$$
\chi_{\{y_L(t)\ge M/2\}}\frac {1}{y_L(t)}\le \frac 2M,
$$
we obtain
$$
\frac{{\rm d} }{{\rm d} t} \zeta_L(t)+ \beta \chi_{\{y_L(t)\ge M/2\}}\le C\xi_L(t)+ \frac{2C}{M}\eta_L(t).
$$
Hence, after integrating over $t$ and taking expectation, we get
\begin{align*}
&\mathbb{E}\left(\zeta_L(T)-\zeta_L(0)\right) + \beta \int_0^T \mathbb{P}\left( y_L(t)\ge \frac{M}{2}\right){\rm d} t\\
&\quad \le C\int_0^T \left( \mathbb{E}\,\xi_L(t)+ \frac{2 \, \mathbb{E}\, \eta_{L}}{M}\right){\rm d} t.
\end{align*}
Since $\zeta_L(T)-\zeta_L(0)\ge 0$ because $M\ge 2$, we have
$$
\frac{1}{T} \int_0^T \mathbb{P}\left( y_L(t)\ge \frac{M}{2}\right){\rm d} t\le \frac{C}{\beta T}\int_0^T \left( \mathbb{E}\,\xi_L(t)+ \frac{2\, \mathbb{E}\, \eta_{L}}{M}\right){\rm d} t.
$$
Let us fix an $\varepsilon >0$. Then, by Lemma \ref{L2} there is a  constant $L_0$  such that for any $L\ge L_0$,
$$
\sup_{t\ge 0}  \mathbb{E}\,\xi_L(t)\le \varepsilon \qquad \text{and}\qquad \sup_{t\ge 0}  \mathbb{E}\,\eta_L(t)\le (1+L^2)\varepsilon +1.
$$
Therefore, for $L\ge L_0$,
$$
\frac{1}{T} \int_0^T \mathbb{P}\left( y_L(t)\ge \frac{M}{2}\right){\rm d} t\le \frac{C }{\beta }\left[ \varepsilon + \frac{(1+L^2)\varepsilon +1}{M}\right].
$$
Since
$$
U^2(t) + \|v(t)\|^2 = y_L(t)+ Y^2_L(t) + \|Z_L(t)\|^2,
$$
we have by the Chebyshev inequality
\begin{align*}
&\frac{1}{T} \int_0^T \mathbb{P}\left( U^2(t) + \|v(t)\|^2 \ge M \right){\rm d} t\\
&\quad \le \frac{1}{T} \int_0^T\left[  \mathbb{P}\left( y_L(t) \ge \frac{M}{2} \right)+\mathbb{P}\left(  Y^2_L(t) + \|Z_L(t)\|^2
\ge \frac{M}{2} \right)\right] {\rm d} t\\
&\quad \le \frac{C }{\beta }\left[ \varepsilon + \frac{(1+L^2)\varepsilon +1}{M}\right] + \frac{2}{M} \sup_{t\ge 0} \mathbb{E}\left[ Y^2_L(t) + \|Z_L(t)\|^2\right],
\end{align*}
and the desired conclusion follows since by Lemma \ref{L2},
$$
\lim_{L\to +\infty} \sup_{t\ge 0} \mathbb{E}\left[ Y^2_L(t) + \|Z_L(t)\|^2\right] =0.
$$
To be more precise, given $\varepsilon >0$ we first find $L$ such that
$$
\sup_{t\ge 0} \mathbb{E}\left[ Y^2_L(t) + \|Z_L(t)\|^2\right]  \le \frac{\varepsilon}{2},
$$
and than $M>2$ such that
$$
\frac{C }{\beta }\left[ \varepsilon + \frac{(1+L^2)\varepsilon +1}{M}\right] \le \frac{\varepsilon }{2}.
$$
\end{proof}
\section{Proof of the existence of an  invariant measure}\label{SInvariant}
Taking into account the Krylov-Bogolyubow criterion (see Proposition \ref{P1}) and Corollary \ref{C2} it is enough to show that the family of  laws $(\mathcal{L}(v(t))$, $t>0$ is tight in the space $H^{\frac 14}$.  Recall that  $v(t)=V_L(t) +Z_L(t)$. On the other hand by Lemma \ref{L3} we have  the inequality
\begin{align*}
\beta \|V_L(t)\|^2_{H^1_0}\le C y_L(t) \xi_L(t) + C\eta_L(t)\le C\left[ y^2_L(t) + \xi^2_L(t) +\eta _L(t)\right].
\end{align*}
Since
$$
y_L(t) = \|V_L(t)\|^2 + U_L^2(t)\le 2\left(  \|v(t)\|^2 + U^2(t)\right) + 2\left( \|Z_L(t)\|^2 + Y^2_L(t)\right),
$$
and $\|\cdot \|_{H^{\frac 1 4}}\le C_ 1 \|\cdot \|_{H^1_0}$ we have
\begin{align*}
\|v(t)\|_{H^{\frac 14}}&\le  \|Z_L(t)\|_{H^{\frac 14}} + C_2 \left(  \|v(t)\|^2 + U^2(t)\right)  \\
&\quad+ C_2 \left( \|Z_L(t)\|^2 + Y^2_L(t)\right)+ C_2 \left[ \xi^2_L(t) +\eta _L(t)\right]^{\frac 12}\\
&\le C_2 \left(  \|v(t)\|^2 + U^2(t)\right)  + R_L(t),
\end{align*}
where, by Lemma \ref{L2},
$$
R_L(t):=  \|Z_L(t)\|_{H^{\frac 14}}+ C_2 \left( \|Z_L(t)\|^2 + Y^2_L(t)\right)+ C_2 \left[ \xi^2_L(t) +\eta _L(t)\right]^{\frac 12}
$$
satisfies
$$
\sup_{t>0}  \mathbb{E}\, R_L(t)<+\infty.
$$
Hence
\begin{align*}
\mathbb{P}\left( \|v(t)\|_{H^{\frac 14}}\ge M\right) &\le \mathbb{P}\left(\|v(t)\|^2 + U^2(t)  \ge \frac{M}{2C_2}\right)+ \mathbb{P}\left( R_L(t)\ge\frac{M}{2} \right).
\end{align*}
By Corollary \ref{C2},
$$
\lim_{M\to +\infty} \limsup_{T\to +\infty} \frac 1 T \int_0^T  \mathbb{P}\left(\|v(t)\|^2 + U^2(t)  \ge \frac{M}{2C_2}\right){\rm d} t=0.
$$
By Chebyshev's inequality
\begin{align*}
&\lim_{M\to +\infty} \limsup_{T\to +\infty} \frac 1 T \int_0^T  \mathbb{P}\left(R_L(t) \ge \frac{M}{2}\right){\rm d} t\\
&\quad \le
\lim_{M\to +\infty}\frac{2\sup_{t>0}  \mathbb{E}\, R_L(t)}{M}=0. \qquad \square
\end{align*}
\section{Strong Feller and Irreducibility}\label{SSFeller-Ireducibility}

In this section we will show that if $g_0$ and $g_1$ are separated from $0$, then the semigroup defined by problem \eqref{3}--\eqref{5} is strong Feller and irreducible.

Let us recall first some basic definitions and theorems. Assume that $P_t(z,\Gamma)$, $t\ge 0$, $z\in E$, $\Gamma \in \mathcal{B}(E)$ is a transition probability   on a Polish space $E$. Let $(P_t)$ be the corresponding transition semigroup.
\begin{definition}
We call a transition probability  $P$ \emph{regular}  if the measures $P_t(u,\cdot)$ and $P_t(z,\cdot)$,  $u,z\in E$ and $t,s>0$  are mutually absolutely continuous. \\
We call $P$ \emph{strong Feller} if the semigroup $P_t\colon B_b(E)\mapsto C_b(E)$ for $t> 0$. We call it \emph{irreducible} if $P(t,z,\mathcal{O})>0$ for all $t>0$, $z\in E$ and non-empty open set $\mathcal{O}\subset E$.
\end{definition}
The first part of the theorem below is due to Khas'minskii \cite{Khas'minskii}. The second part is due to Doob \cite{Doob}. For the proofs of thees two parts we refer the reader also to \cite{DaPrato-Zabczyk}. The third part contains results of Seidler \cite{Seidler} and Stettner \cite{Stettner}.
\begin{theorem}
(i) If a transition probability  is strong Feller and irreducible then it is regular. (ii) If $P$ is stochastically continuous and regular transition probability and $\mu$ is an invariant measure with respect to $P$, than $\mu$ is unique, equivalent to all measures $P_t(z,\cdot)$ and for all $z\in E$ and $\Gamma \in \mathcal{B}(E)$,
$$
\lim_{t\to +\infty} P_t(z,\Gamma) = \mu(\Gamma),
$$
(iii) Assume that the Markov process $X$ with a transition function $P$ is right-continuous, strong Markov and  regular.  Assume that $\mu$ is an invariant-measure for $X$. Then  for any probability measure $\tilde \mu$ on $E$ we gave
$$
\lim_{t\to +\infty} \|P_t^*\mu-\mu\|_{\mathrm  TV}=0.
$$
\end{theorem}

Recall that the \emph{total variation norm} of  two probability measures $\mu$ and $\tilde \mu$ on $E$ is given as follows
$$
\|\mu-\tilde \mu\|_{\mathrm  TV} = \sup_{\Gamma\in \mathcal{B}(E)} \left\vert \mu(\Gamma)-\tilde \mu(\Gamma)\right\vert.
$$
\subsection{Proof of strong Feller property} \label{SSFeller} In this part we are going to apply some standard techniques. Therefore we present here only a sketch of the proof. First we need to approximate $g_0$ and $g_1$ by sequence of $C^\infty$ mappings  $g_0^n$ and $g_1^n$ such that the corresponding Lipschitz constants converge; i.e.
$$
\|g_0^n\|_{\mathrm {Lip}}\to \|g_0\|_{\mathrm {Lip}}\quad \text{and} \quad   \|g_1^n\|_{\mathrm {Lip}}\to \|g_1\|_{\mathrm {Lip}}.
$$
We require also point convergence and the upper and lower bonds:
$$
\vert g_0^n(r,\psi)- g_0(r,\psi)\vert \to 0\  \text{and} \  \| g_1^n(r,\psi)- g_1(r,\psi)\| \to 0,\  \forall\, r\in \mathbb{R}, \psi\in L^2.
$$
and there is a constant $C\in (0,+\infty)$ such that for all $r\in \mathbb{R}$ and $\psi\in L^2$ all the following numbers
$$
  \vert g_0^n(r,\psi)\vert , \quad  \frac{1}{\vert g_0^n(r,\psi)\vert } ,\quad \left \|g_1^n(r,\psi)\right \|,\quad  \left \|\frac{1}{g_1(r,\psi)}\right \|
 $$
 are less or equal to $C$. In \cite{Peszat-Zabczyk} one can find a construction of such sequences.

We need also to regularize (in fact localize) the mappings $\mathcal{N}(\psi)=\|\psi \|^2$ and $B(\psi)= -\frac{\partial }{\partial x}\psi^2$. We set
\begin{equation}\label{N}
\mathcal{N}_n(\psi) = \mathcal{N}(\psi) \phi_n(\|\psi\|^2),
\end{equation}
where $\phi_n\in C^\infty(\mathbb{R})$ is such that $0\le \phi_n(r)\le 1$, $|\phi'(r)|\le 2$ for $r\in \mathbb{R}$, $\phi_{n}(r)=1$ for $r\le n+1$ and $\phi_n(r)= r $ for $r\le n$.  In the same fashion we localize $B$,
\begin{equation}\label{B}
B_n(\psi)= B\left(\psi \phi_n(\|\psi\|^2)\right).
\end{equation}
Let $X_{n,m}^z(t)=(U^z_{n,m}(t), v^z_{n,m}(t))$ be the value at time $t$ of the solution to the problem obtained from \eqref{3}--\eqref{5} by replacing $g_i$ by $g_i^m$,  and $\mathcal{N}$ and $B$ by $\mathcal{N}_n$ and $B_n$. We denote   by $z=(U(0), v(0)$ the initial data. We take  for granted the existence and uniqueness of the solution to the modify problem on the state space $H=\mathbb{R}\times L^2$.

Let
$$
P^{n,m}_tf(z)= \mathbb{E}\, f(X^z_{n,m}(t)), \qquad z=(r,\psi)\in \mathbb{R}\times L^2,
$$
be the transition semigroup to the modify problem. As in \cite{Peszat-Zabczyk} one can derive  the Bismut--Elworthy--Li formula
\begin{align*}
&D_r   P^{n,m}_tf((r,\psi))a  + D_\psi P^{n,m}_tf\left((r,\psi)\right)[\varphi] \\
&\quad = \mathbb{E} \, f(X^z_{n,m}(t)) Y_{n,m}(r,\psi,t, a, \varphi),
\end{align*}
where: $a\in \mathbb{R}$ and $\varphi\in L^2$ are directions,  $D_r$ and  $D_\psi $ denote the derivatives in $r$ and $\psi$,
\begin{align*}
Y_{n,m}(r,\psi,t, a, \varphi)&= Y^1_{n,m}(r,\psi,t)a +Y^2_{n,m}(r,\psi,t, \varphi),\\
Y^1_{n,m}(r,\psi,t)&:= \frac {1}{t} \int  _0^t \frac{1}{g_0(X^{(r,\psi)}_{n,m}(s))} \frac{\partial }{\partial r} U^{(r,\psi)}_{n,m}(s){\rm d} W_0(s),\\
Y^2_{n,m}(r,\psi,t,\varphi)&:= \frac {1}{t} \int  _0^t \left \langle \frac{1}{g_1(X^{(r,\psi)}_{n,m}(s))} D_\psi U^{(r,\psi)}_{n,m}(s)[\varphi] , {\rm d} W_0(s)\right\rangle.
\end{align*}
We left to the reader the verification of the fact  that there is an independent of $m$, $r$,  $\psi$, $a$,   and $\varphi$ constant $C(n, t)$ such that
$$
\mathbb{E}\left \vert Y_{n,m}(r,\psi,t,a,\varphi)\right\vert \le C(n,t)\left[ \vert a\vert ^2 + \|\varphi\|^2\right]^{\frac{1}{2}} .
$$
This ensures not  only the strong Feller property of $(P^{n,m}_t)$ but also gives the estimate
$$
\|DP^{n,m}_tf(z)\|_{L(H,H)}\le C(n,t)\sup_{z\in H}\vert f(z)\vert.
$$
Therefore $P_tf$ is Lipschitz continuous function on $H$, and
$$
\|P^{n,m}_tf\|_{\textrm{Lip}}\le C(n,t).
$$
Since $C(n,t)$ does not depend on $m$ it enables us to pass with $n\to \infty$ and prove  the strong Feller property of the transition semigroup $(P^n_t)$ for the problem, say problem $\Pi_n$,  obtained from \eqref{3}--\eqref{5} by replacing only $\mathcal{N}$ and $B$ by $\mathcal{N}_n$ and $B_n$. In fact, $P^nf$ is Lipschitz continuous and for any bounded measurable function $f$ we have
$$
\|P^{n}_tf\|_{\textrm{Lip}}\le C(n,t).
$$

We can now conclude by using a standard localization procedure. Namely let $X^z_n$ be the solution to problem $\Pi_n$ and let $X^z$ be the solution to problem \eqref{3}--\eqref{5}. Then, from the uniqueness of the solutions and since for $\psi\colon \|\psi\|\le n$ we have $\mathcal{N}_n(\psi)=\mathcal{N}(\psi)$ and $B_n(\psi)=B(\psi)$, we infer that
\begin{equation}\label{Lok}
X^z_n(s)= X^z(s) \qquad \text{for $s<\tau_{z}^n$},
\end{equation}
where
$$
\tau_z^n:= \inf\{ s\colon \|X^z(s)\|\ge n\}.
$$
We can now repeate arguments from \cite{DaPrato-Gatarek, DaPrato-Zabczyk}. Namely, let $f$ be a bounded measurable function on $H=\mathbb{R}\times L^2$. We have
\begin{align*}
\left\vert P_t^n f(z)-P_tf(z)\right\vert &= \left\vert \mathbb{E}\left[ f(X^z_n(t))-f(X^z(t))\right]\chi_{\{t\ge \tau_z^n\}}\right\vert \\
&\le 2 \sup_{z\in H}\vert f(z)\vert \, \mathbb{P}\left( t\ge \tau_z^n\right).
\end{align*}
Since functions $P^n_tf$ are   continuous, it is enough to show that for  all $M>0$ and $t>0$ we have
$$
\lim_{n\to +\infty} \sup_{\|z\|_H\le M} \mathbb{P} \left( t\ge \tau_z^n\right)=0.
$$
This can be written equivalently as
$$
\lim_{n\to +\infty} \sup_{\|z\|_H\le M} \mathbb{P} \left( \sup_{s\le t} \|X^z(s)\|_H \ge n\right)=0,
$$
which can be easily deuced from Lemma \ref{L3},  see \cite{DaPrato-Gatarek} for more details, or from the existence of solution in the spaces  $Z_T^p$, $p>1$, established in \cite{Twardowska-Zabczyk1, Twardowska-Zabczyk2, Twardowska-Zabczyk3}.

\subsection{Proof of irreducibility} We will follow \cite{DaPrato-Gatarek, Peszat-Zabczyk}. Namely, from \eqref{Lok} we infer that it is enough to prove the irreducibility for the problem $\Pi_n$  obtained from the original problem \eqref{3}--\eqref{5} by replacing $\mathcal{N}$ and $B$ by $\mathcal{N}_n$ and $B_n$, see formulae \eqref{N} and \eqref{B} from Section \ref{SSFeller}. The idea is to use the Girsanov theorem. Let $\mathcal{O}\not =\emptyset$ be an open set in $H$ and let $T>0$. Let $z= (U_0, v_0)$ be a fixed initial datum.  We are going to show that $P^n_T(z, \mathcal{O})>0$.  Let $\overline z=(\overline U, \overline v)\in \mathcal{O}$. Since the domain $D(A)$ is dense in $L^2$ we can assume that $\overline v\in D(A)$. We will show that for any $\varepsilon>0$, $P^n_T(z, B(\overline z,\varepsilon))>0$, where  $B(\overline z,\varepsilon)$ is the ball in $H$ of radius $\varepsilon $ and center at $\overline z$.  To explain the idea it will be convenient for us to  write our problem in the following form
$$
{\rm d} X_n =\left[ \mathcal{A}X_n + F_n(X_n)\right]{\rm d} t + \sigma(X_n){\rm d} W,
$$
with with properly chosen  $\mathcal A$, $F_n$ and $\sigma$. In particular  $\mathcal{A}= (A,-\nu )$. Given  $0<h<T$ define
$$
f^{h}(s)= \begin{cases} 0&\text{for $s\le T-h$}, \\
h^{-1}{\rm e}^{\mathcal{A}(s-(T-h)}\left( \overline z- X_n(T-h)\right) - \mathcal{A}z &\text{for  $s> T-h$}.
\end{cases}
$$
Note that the mapping $f^{h}$ is such that
\begin{equation}\label{Poy}
X_n(T-h) + \int_{T-h}^T {\rm e} ^{\mathcal{A}(T-s)}f^{h,m}(s){\rm d} s = \overline z.
\end{equation}
Let
$$
\zeta_{h,m}(s):= \sigma(X_n(s))^{-1} \left[ f^{h}(s)\right] \chi_{\{\|X_n(T-h)\|_H\le m\}}.
$$
Then under the probability measure
$$
{\rm d} \mathbb{P}^* = \exp\left\{-\int_0^T  \zeta_{h,m}(s){\rm d} W(s)-\frac 12 \int_0^T \|\zeta_{h,m}(s)\|^2 {\rm d} s\right\} {\rm d} \mathbb{P},
$$
$$
W^*(t)= W(t)+ \int_0^t \zeta_{h,m}(s){\rm d} s, \qquad t\le T,
$$
is a cylindrical Wiener process on $H$.  Therefore the law of $X_n$ is equivalent to the law of the solution $\tilde X_n$ to the following equation
$$
{\rm d} \tilde X_n=\left[  \mathcal{A}\tilde X_n + F_n(\tilde X_n)+ f^h\chi_{\{\|\tilde X_n\|_H\le m\}}\right] {\rm d} t +  \sigma(X_n){\rm d} W.
$$
Taking into account, \eqref{Poy}, we have
$$
\tilde X_n(T)= \overline z + R_h(T)\quad \text{if $\|X_n(T-h)\|_H\le m$,}
$$
where
$$
R_h(T):= \int_{T-h}^T {\rm e}^{\mathcal{A}(T-s)}F_n(\tilde X_n(s)){\rm d} s+ \int_{T-h}^T {\rm e}^{\mathcal{A}(T-s)}\sigma(\tilde X_n(s)){\rm d} W(s).
$$
Obviously we have
$$
\lim_{m\to +\infty}\mathbb{P}\left( \|X_n(T-h)\|_H\le m\right)=1.
$$
Thus the proof will be completed as soon as we can show that for any $\varepsilon >0$ we have
$$
\lim_{h\downarrow 0} \mathbb{P}\left( \|R_h(T)\|_{H}<\varepsilon\right)>0.
$$
Since  it can be done in the same way as in Da Prato and Gatarek \cite{DaPrato-Gatarek} and we leave it to the reader.

redpajama_set_name: RedPajamaArXiv


Myriad Paths – Box. Wrestle. Fence.
Handwork has certain character to it. In Valkyrie, our handwork is mostly modern boxing, with legwork thrown in from Savate and Shotokan. It's a good blend, workable and enjoyable. Very practical. But it's not a complete game.
Modern boxing is a distillation of things that work, but much as we see in modern fencing, it's a distillation that has thrown out workable things for the sake of consistency, whether for rules or safety. In fencing, the fight between the French and Italian schools, and the fights within the Italians schools, left us with a winner and many losers. It's worth remembering that the losers once stood shoulder to shoulder with the winner, and earned equal respect. That shouldn't be forgotten. Best is often only a marginal distinction.
Boxing is an open art. The feet and hips swing, drive the shoulders, and the hands arc to the target. The arc might be wide or short, but the arc is the birth of all boxing techniques. If it were a Chinese art, I would call it soft, or internal. It looks hard and external, but the reality is quite different (like many arts). A good boxer floats, and moves like wind or water around the opponent.
Harder forms of boxing, Wing Chun for example, lock in the hips and shoulders square to the opponent. Power is generated through tight motions. Internal tensions allow for rapid rotations on small scale, and good bracing to ground and gravity for extra drive. Punches can drive deep and hard, like a sledgehammer. Hard control of lines allow for angulation work to break down the opponents structure and overpower their defense with superior lines. In our method of fencing, we call this the transition from the open to the wide engagement.
For all of it's apparent muscularity and hardness, the third form of boxing is the precision style I learned in Shotokan. It's a combination of the open and the wide. Like modern boxing, you move and flow to find the opening to land your shot, and like Wing Chun, you use tight and hard mechanics to transition arcs into linear motion. The difference is that the motion carries forward and deep. The body extends and penetrates deep into the opponents guard, landing a precision blow that rapidly retreats. The hips and shoulders are square, but the limbs extend far more. In our fencing, this can be seen as part of the narrow engagement.
One isn't better than the other, if the training is equal. But it is true that for the average martial artist, a lifetime is only enough time to master one aspect.
Can you expand upon the idea of going from open to wide engagement? My guess is that the concept has to do with more than just measure, but the fencing vocabulary that I currently have at my disposal is having a hard time placing it into an understandable category.
We use an aggressively modern shorthand for discussing and teaching swordplay. Engagements incorporate measure, line and tempo for us, and describe common encounters as opposed to principles. A narrow engagement is blades close and inline, wide is an engagement that involves angulation, and open is an engagement with invitations.
Going from open to wide would be the action of closing out the invitation without decreasing measure. For example, I arrive at measure in Marozzo's Coda Lunga e larga against an opponent in Capo Ferro's Terza, but neither I nor more my opponent act for whatever reason. Seeing his inaction, and still being confident in my ability to react to his intentions, I move my blade towards parity with his guard, but continue passing my tip over his blade so that I have greater control with the strong. That would complete the change of engagement, and I should follow up with some technique that I had planned to trigger on attaining the engagement.


text:
You Ain't Never Been No Little Girl, Taylor Townsend
Danielle Parhizkaran/USA TODAY Sports
Taylor Townsend
Women's Tennis Association
It's the summer of 2012, and here's me: I'm a 16-year-old girl from the South Side of Chicago. I'm a Wiz Khalifa stan, my favorite athlete is Martina Navratilova, and I like to dance. I'm Taylor….. but my friends call me Tay. Oh, yeah — and I'm the No. 1 ranked junior tennis player in the world.
Here's what I'm not:
I'm not from some rich family…... I'm not white…….
I'm not your typical tennis kid.
And one more thing: I'm not thin.
That last part, it's important to me. It's important to this story. My body — it's just a part of who I am. And if that makes you uncomfortable, then I don't know what to say. This might not be the article for you.
I'm not thin, and I've never been thin — that's just the truth about it, straight up. Like, forreal, I used to be out there on the court with my lil rolls hanging out in my tight tank tops. Why?? Because why not?? I liked the way I looked. I liked the way I felt, and I wasn't ashamed.
Anyway, now you can picture me at 16.
That's where this story starts. I'm 16, I'm coming off some great results, and I'm only a few weeks from the U.S. Open.
And then…….. I get this phone call.
It was an official from the USTA. (That's like being called to the principal's office.) They said, "Taylor, you need to come to Florida — now. We're putting you on an eight-week block of fitness training."
I didn't even have to do the math. I heard "eight weeks" and my stomach just dropped.
Eight weeks meant missing the U.S. Open.
Now, this wasn't the first time my fitness had come up. It was something I'd been working on with my coaches….. and I'll be the first to admit that conditioning wasn't an area of strength for me that summer. But I was still getting results where it mattered most, you know what I mean?? 2012 Australian Open juniors in singles and doubles. 2012 Wimbledon juniors in doubles. I was still winning grand slams.
It was frustrating!! Like, here I was, flying back to Florida to start my fitness "hiatus" — while all the other juniors I knew (girls I was ranked higher than) were on their way to New York to start getting ready for the Open.
And it only got worse from there. Pretty much right when I got back, I was in a fitness session — and all of a sudden the campus physio runs up to me. He's like, "Taylor, you need to get over here. Stop everything."
I'm thinking, like, Dang, I just can't catch a break. What's wrong now??
And he says, "We just got your blood work back."
I had gotten some routine physicals and blood work done as part of my fitness training, just basic stuff, you know, and ... plot twist!!
My blood work came back showing that I was anemic.
As in — I'd been under serious cardiovascular stress for who knows how long. I'd been playing sick, man. That's what had been going on with me.
Clive Brunskill/Getty Images
Figuring that out was scary….. but in a way, it was also a relief at the same time??
Because it was kind of just like — okay, now we know. It's not a fitness thing. It's a health thing. And now we're aware of it, and we're about to take care of it, and….. yeah, I'll say it: Now I can go to the Open.
They send me to a hematologist, and I ask him straight up like, "What's gonna make me feel good enough to play in 17 days?"
He then explains this process of liquid iron injections.
He said, "You're going to feel like superwoman."
So I called up the folks at USTA and let them know what the hematologist said. Told them I was good to go for New York. Put me in the juniors draw, put me in the main draw, put me in doubles, singles, sign me up, the whole deal.
What you think happened next?
Yeah — they still said no.
They couldn't stop me from playing juniors (I'd qualified automatically with my ranking), but they were like, "Yeah….. no main draw. We're not giving you a wild card. It's like we told you: You're simply not fit to play."
Okay, now might be a good time to stop and ask yourself something. Just take a second, and think about all this, and ask yourself: What do you think "fit to play" really means?
Like — what are we actually talking about here, right??
I was the No. 1 ranked junior in the world — and that was with my anemia, y'all. Now the doctor is telling me that if I get my iron levels up, I'm going to feel 100%. And yet….. they're still trying to keep me out.
Seriously, what's the problem??
What's sad is, looking back, I really trusted those people. Like, at first — I really believed their "fit to play" stuff was about looking out for me. As frustrated as I was, I'd cooperated with the USTA's "eight weeks off" decision no problem, no questions asked. But that was before I got my anemia diagnosis.
Once I got the diagnosis, I figured the original conversation about my fitness was over….. or at least clearly different now. Right? I mean, that diagnosis probably saved my life, let alone my tennis career. I'm so grateful for it. But like I said: It also proved this was a health problem. And to me, that meant it was no longer something the USTA and a bunch of fitness trainers should be speaking on. Now it was something that should be left up to doctors.
And my doctor said I was good to go!!
So why is the USTA basically saying, Nah. We're sticking with our original decision??? Why is my ranking saying I'm the best in the world….. and my doctor is saying I'm okay to play….. and meanwhile the USTA is saying I need to see a nutritionist….. and lose some weight???
It made no sense. It was confusing as hell. And it hurt — it hurt really bad.
Sixteen years old, and getting to No. 1 in juniors as a Black girl from the South Side? I was so proud of that. I was so proud of who I was, and what I'd achieved. And I think I had it in my head, like, alright — I know I might be an outsider in this sport. I know I might not be like all these other tennis kids. But once I got to No. 1?? Once I climbed that mountain?? Now they'll be proud to have me. Now I'll be treated like a part of American tennis. Now I'll be one of them.
Doesn't exactly work that way, though, does it.
As a matter of fact….. it worked pretty much exactly the opposite.
It worked the way things usually work in a country that hates fat Black women.
I don't think that's a controversial opinion, by the way. To me, America hating fat Black women — it's just part of life. It's in the culture. It's in the health-care system. You see it in Hollywood, you see it in sports. You don't have to look around very hard. It's everywhere.
And it's especially everywhere in the world of tennis. I mean, think about it: They didn't just alienate me for not fitting the "mold" of what a tennis player should look like — they punished me. They took away something I'd earned.
I was fat, and I was Black, so they took away my dream.
Or at least they tried.
America hating fat Black women — it's just part of life. It's in the culture. It's in the health-care system. You see it in Hollywood, you see it in sports. You don't have to look around very hard.
- Taylor Townsend
I knew the USTA could stop me from getting funding….. and they could stop me from getting a wild card into the main draw….. but like I said: They couldn't actually stop me. My spots in the juniors draws, those were automatically mine. So I got some money together and paid my own way to New York.
It was worth every penny :)
Even with all that chaos leading up to the tournament, I still made the quarters in singles — and won in doubles. Yeah, that's right. Me and Gabby.
We WON.
I mean….. y'all. Can you even imagine how good that felt??? To lift a freaking U.S. Open trophy on Arthur Ashe, with a mouth full of braces, after they tried to play me like they did??
Chris Trotman/Getty Images for USTA
In the post-match presser, all of these media people were asking, you know — "Why aren't you playing with the pros? Why didn't you get a wild card?" And I just remember taking this pause in my head, and thinking about it. Thinking about saying something easy, and diplomatic….. something that wouldn't ruffle any feathers. But then I thought, You know what? F*** it.
I'm gonna be real.
I decided I wasn't gonna let myself be embarrassed anymore — I wasn't gonna let myself be humiliated by this rich, white tennis world that I had spent my entire childhood scraping and crawling and bending over backwards to fit into.
I took a deep breath….. and aired everything out to the press. I gave them the real. Told them what actually happened.
I spoke my truth.
…..and lived happily ever after, right?? That's the end of the article, right???
Nah, come on.
Mr. and Mrs. Young, they were really my start in tennis.
They were good friends of my parents, and had their own academy in Chicago. My mom and Mr. Young had known each other since they were teenagers — that's pretty much how the Black tennis world operated back then. It was small, like a family. Everybody just knew each other.
After Mr. Young got married, my mom started playing doubles with his wife, Illona. She played until she was eight months (!!) pregnant with me. I was probably about four when they handed me my first racket.
Mr. Young's academy was over by the Museum of Science and Industry at Washington Park — even just writing that down brings back memories. I remember running around, mimicking what all the older kids were doing, and how I'd just be, you know…… out there, carefree. Having fun and being four. I didn't know anything yet about the game or its rules.
And when I say "rules"..... I don't just mean understanding how points work and all that. I'm also talking about the rules of the game that we might not say out loud. The rules of "tennis etiquette," and respectability politics, and all the other things that go along with it. When you're just starting out, you don't think about any of that.
But you learn to — quick.
I was barely old enough to know how to play when I first started hearing about how I looked.
I can still remember the first time I was playing and heard one of the moms say, "Look at her. She's too….. big." I was probably four or five when that happened.
The thing is — you don't even really understand what people are saying around you at that age. I mean, you gotta put yourself in the mind of a child. When you hear something like that and you're that young, it's really just like this split second of a moment in your head, where you're feeling any type of way about it. Where maybe you're kind of scared or self-conscious. And you don't even really know why, if that makes sense.
Like, you know that there's something "different" about you….. and that it's not good. You know that the grown-ups can sense it….. and they don't like it. As a child, you have this amazing ability to shrug stuff like that off — but also to carry it forever.
And yeah, this was tennis, this was a very white world. But if I'm being honest, I wasn't only hearing things like that from white people. I heard about my weight in my own community — even in my own house. My mom was still a tennis mom, you know?? And sometimes that just came before everything else.
Those little moments, they get under your skin and they stick with you.
You grow up, you move on — but you never really forget.
Courtesy of Taylor Townsend
When the Youngs moved their academy to Atlanta, my family moved too. And now being in the South….. I definitely started to notice this vibe.
Just this weird kind of vibe.
It's like that very specific feeling you get when you walk into a room, and you can tell you're not really supposed to be there. Or that people don't really want you there. There's just this tension.
It's heavy.
I mean, when me and my big sister, Symone, started having success….. how do you think those parents reacted to these two little Black girls taking over tennis in Georgia, of all places?
Lemme tell you: People had a lot of issues with that.
Oh, they made that known.
It's not like anyone was throwing things at us on the court or anything. No one was outwardly saying anything like racist racist. But even still, man. When you walk into a room, or when you're in a big crowd….. you know. You feel it.
And it just felt like there was a lot of resentment.
It felt like I was getting criticism that other players weren't getting. Parents started complaining before every match, coming out with all these excuses about why I shouldn't be playing against their kids and whatnot. Sometimes they'd go to the tournament director about me, asking for my birth certificate.
No — for real!! They'd ask for my birth certificate.
Parents would be out there, like, "Oh, no, she can't be 11. She's too big. She's too developed."
My mom was the one who shielded me from all that. She was always like, "You're a child — you should stay in a child's place. Let me deal with the grown-up stuff." So for the most part, me and Symone, we just went out and played. But we still knew what was going on. We'd overhear them talking about it, see the looks on their faces, stuff like that. And it was tough. Seeing how you're making things hard on your family? That's tough at any age.
And the thing about those parents who would say sh*t about me — it was like they felt their kids were the ones getting robbed of an opportunity. Like they weren't getting the "country club tennis experience" they had signed up for. Tennis was supposed to be "their" sport. And yet here was their kid, getting beat by a big Black girl.
I'd always try to downplay it like I thought it was funny — like, Haha, all I can do is laugh at how mad these other parents are.
But there wasn't really anything funny about it.
It was actually pretty f*cked up.
David Gray/AFP via Getty Images
I cried sometimes late at night.
I was starting to understand some of what was going on, and it hurt like hell. But starting to is only starting to — and there was still so much going on that I couldn't wrap my head around. So much that was just leaving me feeling bad and confused.
I think that's the one major thing that's difficult to explain to people who've never experienced racism themselves. How it's not the stuff you understand that hurts the most.
It's the stuff you don't understand.
When you're young and you're Black, and you're in a situation where, for whatever reason, you're getting singled out in a negative way — whether it's by another parent, or a teacher, or a store manager, or a cop, or whoever?? You really don't get it.
When all this racist sh*t is going down — 9 times out of 10?? You think it's just you.
And like I said: That's heavy.
You walk around carrying this invisible weight — this pressure on your whole spirit — because the worlds you're trying to fit into are rejecting you in so many ways. And you think it's your fault. Then you add sports into the mix, competition and all of that, winning and losing….. and it just becomes that much more complicated.
And that much more ugly.
I mean, of course I wondered where I fit in. Of course I cared what people thought, even if I said I didn't.
I'd go through different phases over the years. For a while, I even tried to make myself smaller — I tried to act more in line with how I thought people wanted me to. I minimized myself and kind of dimmed myself, dimmed my light and my shine, for other people's comfort. But it just felt like no matter what I did, it was never enough.
And then eventually you stop caring.
That's the truth of it, if I'm being real — that's what comes next. Not some amazing epiphany, or some big moment of enlightenment, or empowerment.
What comes next, is just……. what comes next.
You accept that you can't hop out of your own skin. You accept that — whether you want to or not, and whether it's hurtful or not, and whether you understand why or not — you can't change, not in that way.
This is how it is.
This is how it's gonna be.
In the beginning, there was no Taylor Townsend.
There was Taylor and Symone.
Seriously, in those early days, you didn't say Taylor without mentioning Symone — we were a package deal.
And you already know they made the "Venus and Serena" comparisons….. which was a big deal, believe me. Venus and Serena are household names anywhere. But where me and Symone were from?? They were larger than life. They were almost like an idea. And I've definitely looked up to them — and what their success represented — ever since I started playing.
But at the same time, I also remember just……. wanting to be my own person. You know what I mean?? I was like, I'm not the "next" ANYBODY!! I'm Taylor Townsend!!! From the South Side of Chicago!!!
And I think Symone, in her own way, felt the same.
Actually — let me tell you something about sisters, man.
It's so funny. Like, as much as we love each other to death?? We could not stand playing with each other. I'm not even kidding. There was this one time, we're playing doubles together. And I remember we're playing and playing, it's deep into the match….. and I kinda start to notice something: Symone keeps missing the same shot.
God bless Symone: She is the sweetest person you will ever meet in your life. I'm the fiery one. (Fiery, that's probably code for some other words, but we'll leave it there.) So anyway, Symone keeps missing this same…. damn…. shot. And internally I'm trying to calm myself down, like, Taylor. This is okay. This is fine. But she just keeps making that same mistake. And now in my head I'm starting to trip a lil bit, like, Uhhhhh. Hmm. I need to see some URGENCY, sis!!!
But she just keeps missing lol. And eventually I can't even help myself anymore. Now I'm straight-up cussing her. I'm like, "What the f*ck are you doing, girl???" I mean, she's my older sister, so you have to keep that in mind. No matter what, it's still love. But I'm just like — y'all. I'm going in on her, in the middle of the match, like it might as well have been the two of us at home.
Next point I served, I made sure I hit Symone square in the back with it.
And of course the next point she served, she made sure she hit me just the same.
Now we're literally going back and forth with it like that, like — forget about the actual MATCH right now. We're at war. We're in our own little "game within the game"…... of using our first serves to try to tag each other. Mom and Dad literally had their heads in their hands, they were so embarrassed. We lost (of course we lost), and they were rushing us out the stadium all angry, whispering like — "This is over." HA.
They were right about that. It was over. Symone and I weren't allowed to play doubles together ever again. But it was hilarious, though. Like….. we just were Those Sisters. Each wanting to be our own person, so damn bad. And at the same time?? Loving each other — needing each other — pretty much to death.
What me and Symone have, I wouldn't trade it for anything.
You know, sometimes I think to myself, like — What if Symone had been there with me, all the way back then, when everything was going down with the USTA?? Would it maybe have gone any differently? Sh*t….. if Symone had been there, would EVERYTHING have been different??
It's crazy how life works out, though. No lie: Symone was way better than me. She used to beat me all the time, for years. She would just Big Sister the hell out of me on the court. And there's no question in my mind — a lot of the player I turned into, it came from chasing Symone for all those years. Came from chasing the level that she was at.
But then one day something horrible happened.
Symone was about 15, and I was about 13 — and we were coaches at this tennis camp for the summer.
Symone had the four-year-old kids, and like….. if you know how it is with four-year-olds and tennis, there's only about so much you can do with them. Their little attention spans get tired quick. So it's the end of practice, and Symone has them playing Duck Duck Goose.
I remember this clear as day. She had them all in a circle, you know, criss cross applesauce, and she was going around, like: Duck…... duck…... duuuck….… GOOSE!!!! And now she's running the circle, with some little kid coming after her.
And then all of a sudden, outta nowhere this one little boy jerks his body back out the circle.
Like — he stuck his whole body out.
Symone, just on instinct (and being a great athlete), she jumped over him, so that she didn't hit him. And when she jumped, it was like……. man. It was like, her foot just got caught in the net and twisted, somehow. She tore all the ligaments in her knee. And when I say all I mean all of those damn things: ACL, MCL, PCL. She messed up her ankles bad. She ruptured a nerve in her foot. You hear about "freak accidents" — that's what this was. It was just this freak, shocking thing.
Symone being Symone, she still went on to play at a high level and get a tennis scholarship to college….. but nothing was really the same after that injury. She pursued other things, and grew in a different direction. And I'm so unbelievably proud of her for it. (We don't have to be on a tennis court for that girl to be Big Sistering the hell out of me.) But at the same time, I won't lie: While I would never in a million years compare what Symone went through to what I went through….. after she got injured, nothing was really the same for me either.
You have to understand that tennis….. it wasn't just my thing growing up. It was our thing.
Like I said — it wasn't Taylor Townsend. It was Taylor and Symone.
I didn't care that she was better than me….. I didn't care that she was thinner than me….. I didn't care that she was the sweet one and I was the (lol) not as sweet one. That's just my BIG SISTER, man. That's my ace. My everything. And tennis was our escape from the world, right?? But we were each other's escape from tennis.
And looking back, I mean….. that's part of why I wanted to tell y'all that story about us tagging each other with our serves playing doubles. Because it's almost like, when you're talking about that relationship you have with a sister or brother — it's not all about these movie-scene heart-to-hearts necessarily. It's not only about those Hallmark moments. So much of it is also about the opposite of those moments. So much of it is about how, with Symone, I didn't have to worry about any of these "outside" forces — no tennis parents or country club managers or USTA whatevers. I could just be myself.
I could be myself no matter what.
Whether it meant I was being that jealous little sister, walking around with my tiny fist all clenched up, big mad that I was too young to get lessons, picking up Symone's loose balls and throwing them back on the court, just so I could be a part of it….. or whether it meant I was acting a fool as her doubles partner and cussing her out for no good reason and reaching new levels of being a brat, with those first serves hitting her SPLAT in the ass (sorry Symone)….. or whether it meant….. man, I mean, yeah: whether it meant I was feeling like sh*t from all the antiblackness and fatphobia that I was internalizing on a daily basis, and just wanting a steady shoulder to cry on, just needing someone to let it all out with, who not only loved me — but who had experienced those same traumas.
Whether it was good or bad or embarrassing or depressing or anywhere in between.
Whether it was silly-silly or realer than real.
It was Taylor and Symone.
Venus and Serena…... nah, lol. Not quite. But we were two Black sisters trying to make it in this tennis thing — and we were in it together.
Then sh*t happened.
Life happened.
And it was like….. trying to make it suddenly got a whole lot harder.
Suddenly I wasn't just struggling with being a Black girl in the tennis world.
Now I was doing it alone.
Coming off that U.S. Open win, all I wanted was to move forward. I wanted to have everything with the USTA in my rearview mirror, so I could just focus on my training. I was planning to go pro the next year, so that was the only thing that was important to me. No more soap opera — just grind.
But it turns out that 16-year-old Black girls can't take public shots at the biggest organization in American tennis and then simply go back to their business.
Reporters started coming around for interviews. USTA officials came out to "clear things up." (Translation: They basically denied all of it.) Somehow they got my mom involved. It's like stuff just kept spiraling out of control. Like it went from not a thing, to barely a thing, to a THING, to a situation, to almost this like National Incident, so damn fast.
And it sucks, man, because I wasn't asking for a fight. Like, my God — I wasn't looking for beef with the freaking USTA. People understand that, right?? I wasn't trying to "grab the spotlight," or "put the whole system on trial," or represent this or that.
I was just trying to speak my truth.
All I wanted — pretty much all I ever wanted, my whole life — was to play tennis.
But I had clearly put some powerful people in a hot pile of fiery sh*t, and brought them scrutiny they didn't want. And that decision came with painful consequences. I mean, it's not like anyone could stop me from playing….. but they could slowly squeeze out every drop of what I loved about playing.
I left the USTA in 2013, then turned professional in 2014 — which should have been one of the happiest moments of my life. Not a lot of people ever get to go pro at tennis….. let alone people from where I'm from. But all I can really remember about that time now is sadness. All I can really remember is this feeling that tennis wasn't tennis anymore. It was like it had turned into this Other Bullsh*t, that just kept on collecting more and more bullsh*t.
It was like — now, for me, tennis was the feeling of getting betrayed by a bunch of people who I thought were on my team. Tennis was the feeling of being stuck with a life where answering questions about my weight in public had literally become part of my job. Tennis was the feeling of having this permanent cloud hanging over my career….. before my career had even gotten started.
Michael Dodge/Getty Images
I had a rough first couple of years as a pro.
My ranking climbed steadily for a while, but in 2015, it slipped from 94 in February to like….. outside of the top 300 by November. In less than three years, I went from being the No. 1 junior in the world to, "Man, what happened to Taylor Townsend?"
Off the court, those years were just as bad. My parents split up. My relationship with my mom got harder, as she made decisions about my career that I didn't agree with. I actually quit for a minute. I really missed Symone. I think I had developed this defense mechanism where I was just, like, shutting people further and further out — to the point where I probably even ended up shutting out myself.
And maybe the worst part of all of it? Is that for the first time in my life, I started to worry about the way I looked.
That confident and carefree girl who let her lil rolls hang out of her tank tops….. she was gone. I started to feel down about my size and my appearance. I started to cover up more. I didn't want people to see me at all. I lost sight of who I was, if that makes sense — I lost sight of the version of myself that made me happy.
I lost sight of that girl from the South Side who just wanted to PLAY.
And to be real with you? I was messed up.
I don't want to beat around the bush about that. I don't care who you are — you go through something like that, and you're going to be messed up from it. I honestly still can't really talk about the depression aspect of it all, because it's just hard to articulate. It's hard to describe something that's taking place inside your head, you know?? But that's how it was for me mostly. I was still "me," on the surface. I wasn't missing practices. I wasn't missing matches. I was acting pretty much the same.
But I wasn't the same.
I was different after. Of course I was.
I'll be honest: Deep in my heart, as crazy as it might sound to people, I think there's a universe out there where tennis fans are saying Serena, they're saying Venus — and they're saying Taylor in that same breath. They're saying Taylor Townsend, right up there with all the other greats of the game.
If you think I'm tripping, that's fine.
But if you were there — I mean really, really there — then you know.
You know what it was.
Those "what if"s….. they're hard, man. They're just f*cking hard. Because you never wanna make excuses. You never wanna blame anyone for what's happened to you. For the places that you didn't go. Or the heights that you didn't reach.
But at the same time….. how do you stop blaming yourself?
And I guess that's what I'm still trying to figure out, in some ways. I know that tennis wasn't taken away from me. But I also know that it was. I know that the things I've had to carry — they're more than just a tennis career. And the way that I was treated, it's less than I deserved.
But it's what I have to live with, and I'm going to live with it.
Back when I was first starting to take tennis seriously, around like seven years old, the most prestigious thing they had for my age group was this tournament called Little Mo. To qualify, you had to play in your local sectional, then a regional, and then, if you placed in your region, you got to go to the national tournament in Austin.
Oh yeah and one more thing about Little Mo: They gave out the biggest trophies.
So when the time came, and I actually made it to nationals? You couldn't tell me nothin'. We got to the tournament, they had the trophies all lined up for display, and it was just like….. whew. I was a kid in a candy store, you know what I mean?? Once I saw the size of that second-place trophy, it was a wrap, man. In my head, I had already made it. I was just like, Oh my gosh, I get a big-ass trophy — that's it. I'm living the dream!! And I kind of went out there and just didn't really care what happened.
I played terrible, and got second place.
But then I came off the court….. and there was no high fives or nothing. Mr. Young wasn't even cross, he was just matter-of-fact. He looked at me and he was like, "You just wasted your time. You're so happy to get that second-place trophy, you don't even realize you should've won first place."
Then he asks me to give him the trophy — and he hands it to Donald Jr., his son who's with us.
And he tells Donald Jr. to go out to the parking lot and smash it. (My mom was there, but she knew what the deal was. She gave Donald Jr. the nod, like, "Go ahead.") So Donald Jr. took my trophy out to the lot, and — with like all the strength in his little body — he threw it on the concrete. It shattered into I don't even know how many pieces. I remember just standing there, like….. mouth open. Mostly mad about losing my big-ass trophy, and about how Donald Jr. did me like that, more than learning any lesson really.
But it was a moment that's really stuck with me. It's something I'll never forget.
And looking back on it now….. I think I know why. I think that for as much as Mr. Young having my trophy smashed doesn't seem like it has anything to do with what I've been talking about here, with my body and my Blackness and my struggling as a pro and all of that — it really has everything to do with it. Because you can say what you want to about Mr. Young's methods, or whatever else. But I'll tell you one thing I know for sure, as far as that moment goes: It was about tennis.
It was about tennis, man.
It wasn't about how I looked — it was about how I played. It was about someone believing in me, and investing in me, and pushing me to be excellent. Not whiter, or sweeter, or quieter, or thinner. Excellent.
Thick and Black and proud and excellent.
It was about someone wanting me to be me.
Tim Clayton
In 2015, I did something so simple that I keep asking myself why I didn't do it sooner: I started working with Mr. Young again.
I'll never forget our first practice back together.
At this point, I was still kind of focused on my body — still kind of focused on losing weight as a part of my training. But Mr. Young….. he just wasn't having it. He was like, "Look, I know where you came from. And you ain't never been no little girl, Taylor Townsend. You ain't never been little. And that's okay!! We both know what you're capable of. So why start trying to look some other way now??"
Damn, good point.
We worked together for the next five years — and won my first-ever WTA title together. I also went on a crazy singles run to the Round of 16 at the 2019 U.S. Open, and then made it to the semifinals of women's doubles there in 2020.
When I went on pregnancy hiatus at the end of last year, I felt better about my game than I had in years. I also felt more at peace with my place in the world than I had in….. ever. I'm in a place now where I can finally pull back a little and start to see the big picture. Where I can see that everything I've been through — it didn't just give me hurt.
It gave me strength.
And now I don't only have my strength — I have my reason.
Aydn Aubrey Johnson.
Born March 14, 2021. Six pounds and 14 ounces.
My baby boy.
Honestly, I'm still in shock. I mean, my body — my body!! — created a real human being. It just freaks me out sometimes. To think about what Taylor Townsend from the South Side has done in such a short time on this earth is pretty crazy.
And I've obviously changed a lot physically from all that crazy. Going through pregnancy and childbirth….. y'all, I've never been this big or this heavy. I have stretch marks in places I never had 'em. Certain stuff hanging lower than it used to hang lol. The body is an adventure, man. It's doing what it's doing.
But I love being a mom. I'm pouring everything into Adyn right now — I'm giving that boy all I've got, forreal.
Being a mom….. it also really makes you think. It's like it has this magical way of forcing you to look outside yourself. And one of the ways I've caught myself doing that a lot lately, has been in figuring out how I want to perceive my own career. Like, I've been thinking a lot about how I want to define my successes and my failures. I've been thinking about what is the right way to look forward, and the right way to look back.
And I'll tell you what I've realized.
I've realized that for as much as Venus and Serena have been these amazing legends for me to look up to, and these standard-bearers for me to measure myself against, and this essential part of my journey as a Black woman trying to make it in the tennis world….. they're not actually my story. They're an inspiration, they're the pinnacle, they're two absolute icons, on and off the court. And you can't tell the story of Black women in tennis without them.
But the full story?? The full story of Black women in tennis is a whole mess of stories.
There's the Williams sisters, of course, overcoming everything they overcame to become a pair of damn GOATs.
But on the other side of the coin, there's also hundreds (thousands?!) of stories you'll probably never even hear about, of Black girls who just didn't get a shot. Whether that's because of money….. or racism….. or lack of support….. or gatekeeping bullsh*t….. or because the system just kind of failed them, the way it fails so many Black women, all the f*cking time….. and whether their dream ended in the pros, or in college, or in high school or grade school or — real talk — maybe they never got to pick up a racket at all. There's just got to be so many stories out there of Black women that never got their chance.
And then somewhere in there, somewhere else within that mess of stories, you've got mine.
You've got Taylor Townsend, now 25 years OLD, from the South Side of Chicago.
You may have read a thing or two about her.
She made some noise a while back in juniors. Won a few big tournaments. Got a dope ranking. Then there was an issue with the USTA?? Or some sh*t. And I think she struggled turning pro. Found her game, though, I heard. Been hitting the hell out of the ball. Winning again. Has a handsome-ass kid, too.
Honestly — it seems like she's doing alright.
Wasn't easy, no doubt….. but she's living, she's breathing.
She's building a future, you know what I mean??
I love stories like that.
Good for her.
Not Here to Dance
Ada Hegerberg
I Just Wanna Say My Sisters' Names
Reggie Bullock is using National Gun Violence Awareness Day to remember his sisters, Mia and Keiosha.
Reggie Bullock
"You Know What I Heard About Kordell Stewart???"
Back then, I'm at the top of the mountain. I'm 26 years old. I'm the quarterback of the greatest franchise in the NFL.
Kordell Stewart
Dear Black Women
As Black women, there's so much pressure on us to be everything to everyone.
A'ja Wilson

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: jquery animate, can't get it to work I'm trying to animate a height property, when hovering, but I can't get it to work.
I am getting into the hover function when hovering, but it just doesn't animate the height.
if (!Modernizr.csstransitions) { // Test if CSS transitions are supported
if ($('#belt figure').length) {
    $('#belt figure figcaption').hover(function () {
        $(this).animate({ height: '7em;' }, { queue: false, duration: 500 });
    }, function () {
        $(this).animate({ height: '3.8em;' }, { queue: false, duration: 500 });
    });
}

}

A: this is html code for my example

Hello!

and this jquery for animate div height:
$("#block").hover(function(){
  $(this).animate({
    height: "100px"

  }, 500);

},function(){
    $(this).animate({height: "50px"
  }, 500);
});

You can also see example this following linkhttp://jsfiddle.net/jzRPa/29/

A: Have you tried
$('#belt figure figcaption').hover(function () {
    $(this).animate({ height: '7em' }, { queue: false, duration: 500 });
}, function () {
    $(this).animate({ height: '3.8em' }, { queue: false, duration: 500 });
});

withouth the ;

redpajama_set_name: RedPajamaStackExchange


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Contact our scrap metal merchants today, in Lincoln, Lincolnshire, to discover more about our reliable waste management services.


text:
Q: Window does not open new window or tab on external url link click I need to open an URL using pyQt5. The page has several links that open a new window. pyQt5 opens a windows for the URL but does not do anything after clicking on a link that should open a new window.
P.S I'm using pyQt5.6
I have tried it on Linux centOs but nothing works.
from PyQt5.QtCore import QUrl
from PyQt5.QtGui import QDesktopServices
from PyQt5.QtWidgets import QApplication
from PyQt5.QtWebEngineWidgets import QWebEngineView, QWebEnginePage

class WebEnginePage(QWebEnginePage):
    def acceptNavigationRequest(self, url,  _type, isMainFrame):
        if _type == QWebEnginePage.NavigationTypeLinkClicked:
            return True
        return QWebEnginePage.acceptNavigationRequest(self, url,  _type,      isMainFrame)

class HtmlView(QWebEngineView):
    def __init__(self, *args, **kwargs):
        QWebEngineView.__init__(self, *args, **kwargs)
        self.setPage(WebEnginePage(self))

if __name__ == '__main__':
    import sys

    app = QApplication(sys.argv)
    w = HtmlView()
    w.load(QUrl("https://gmail.com"));
    w.show()
    sys.exit(app.exec_())

I expect it to open a new window on click of target='_blank' on any webpage.

A: You have to override the createWindow method and return a QWebEngineView, but for the object not to be distruded it must be the child of another window or be part of a container that has a longer life cycle.
from PyQt5 import QtCore, QtGui, QtWidgets, QtWebEngineWidgets

class WebEnginePage(QtWebEngineWidgets.QWebEnginePage):
    def acceptNavigationRequest(self, url,  _type, isMainFrame):
        if _type == QtWebEngineWidgets.QWebEnginePage.NavigationTypeLinkClicked:
            return True
        return super(WebEnginePage, self).acceptNavigationRequest(url, _type, isMainFrame)

class HtmlView(QtWebEngineWidgets.QWebEngineView):
    def __init__(self, windows, *args, **kwargs):
        super(HtmlView, self).__init__(*args, **kwargs)
        self.setPage(WebEnginePage(self))
        self._windows = windows
        self._windows.append(self)

    def createWindow(self, _type):
        if QtWebEngineWidgets.QWebEnginePage.WebBrowserTab:
            v = HtmlView(self._windows)
            v.resize(640, 480)
            v.show()
            return v

if __name__ == '__main__':
    import sys

    app = QtWidgets.QApplication(sys.argv)
    windows = []
    w = HtmlView(windows)
    w.load(QtCore.QUrl("https://gmail.com"));
    w.show()
    sys.exit(app.exec_())


text:
Q: Location-based app to match users - do I need precise geolocation? Imagine an app like Tinder but not for dating. It connects people in the same city/town/country, not based on exact location. So just one broad location check e.g. on app start-up is enough.
If I only need these 3 levels of location (city/town/country), do I still need to go with usual APIs like Geolocation API or Core Location? Or is there a way to simplify it, since no such granularity is needed.
Or would trying to simplify it only lead to more custom work?


text:
Composers » David Dubery
David Dubery
30 performances.
An August Midnight
Pitville Pump Room, Cheltenham
Swans in Flight
Huntingdon Hall
Home is so sad
West Seattle Library
Three Songs to Poems...
One Night in Decembe...
Threesome for 2 play...
Drama Studio, Middlewich High School
Middleton Hall
Chapter House, Gloucester Cathedral, Gloucester
Music Center of the Northwest, Seattle WA 98113
Cuarteto Ib�rico: (L...
Westbourne Community Hall, West Kirby
Cuarteto Iberico: (L...
St Peter's, Oxted
St Oswald's Church, Durham
Ushaw College, Durham
Bridgewater Hall, Manchester
St Edmund's Church, Southwold
Sonata for Cello and...
St Mary's Church, Aylesbury
Mrs. Harris in Paris...
Doncaster Museum Art Gallery
Wagtails on the Dyfi
Y Tabernacl, Machynlleth
Tanto gentile e tant...
Saint Luke Church, Chicago
Wesley's Chapel and Leysian Mission, London
St Pancras Parish Church
St Olave Church
Sarabande & Gavotte
Temple Methodist Church
Pinchbelly Park
St James', Gatley
Cuarteto Iberico
Cuarteto Iberico: (Los fantasmas de los tiempos pasados)
Cuarteto Ib�rico: (Los fantasmas de los tiempos pasados)
Mrs. Harris in Paris (Valse Temptation)
One Night in December (Away in a Manger)
Sonata for Cello and Piano
Tanto gentile e tanto onesta (Sonetti d'amore)
Three Songs to Poems by Robert Graves
Threesome for 2 players (Sonatina for oboe and piano)
David Dubery Biography
David Dubery was born in Durban, South Africa, in 1948, and studied piano from the age of five. The rugged sea and landscapes there instilled in him a love of nature. To further his musical and artistic skills the family emigrated to his mother's home town of Manchester, England in 1961, before South Africa became a Republic. He started composing in 1962, continued with piano lessons developing his skills in sight reading and improvisation and had many opportunities during his teens to perform his own piano compositions at school speech days and concerts, and as a junior student at the Northern School of Music (NSM) [1964-67]. After winning the 1965 Composition Prize he was awarded a course of composition lessons with Dorothy Pilling, best known for Harmonization of Melodies at the Keyboard, as well as delightful Vignettes for piano solo and a number of attractive songs. His insatiable thirst to discover music was so strong he accepted the offer of a three year degree course at the NSM. His piano tutors were Eileen Chadwick and the distinguished concert pianist Kendall Taylor CBE. The principal, Ida Carroll kept him busy with many recitals in College, Music Clubs and Societies. Studying with Albert Haskayne he developed a useable baritone voice and sang in Choral Works and Operas with the Hall� and NSM Orchestras under the direction of Sir Charles Groves, James Robertson, and Maurice Handford. Most memorable were Sir John Barbirolli's glorious farewell performances of Verdi's Otello.
In 1969 he was awarded a Hesse Scholarship to Aldeburgh where he came under the influences of Benjamin Britten, Peter Pears and Imogen Holst and where he formed friendships with Peter Holman and David Matthews. Time spent with them resulted in a collaborative private recording for Imogen of her father's opera Sita [1970].
His first BBC broadcast was Castles in the Air in 1971 following a successful performance at NSM sung by Diane Marshall. Other broadcasts followed; a recital with violinist Philip Sutton [BBC Northern S.O.], Rhymes for Children [1973] and Two Canticles for voice & piano I am my own and not my own: Christ on the Cross [1980].
Since 1972 he has enjoyed a multi-faceted music career as solo pianist, accompanist, composer, vocal coach, musical director and teacher of piano and voice. During the 1970's he wrote several musicals that received a number of highly praised staged performances and during these years he partnered in concert some outstanding singers at the start of their professional careers including Jane Eaglen, Barry Banks, Joan Rogers, and Claire Moore and Jeffrey Lawton. Throughout the 1980's he toured as part of the Verlaine Duo with oboist Jonathan Tobutt which gave him the chance to write a Sonata Since dawn is breaking [1981], Sonatina [1986] and solo Elegy for E.S. [1983]. The duo made its BBC broadcast debut for the Manchester Midday Concerts Society in 1982 and London debut at St Martin in the Fields in 1986.
He coached actors from musical theatre, TV, stage and film, taught cast members from over 25 West-End and national-tour productions, and worked in the departments of Dance, Drama and Music in Colleges of Further Education, The Manchester School of Music and The Actors Centre. From 1985-2003 at the Northern Ballet School he was staff pianist, vocal tutor, and Musical Director for their Showcase Productions at the Dance House Theatre.
Being largely self-taught he works in a traditional language preferring miniature to medium sized projects. His music is tonal, rooted in past traditions, and working with dancers and actors has ensured rhythm and movement play an integral part in its composition. European travel, visiting Italian cities, coasts and islands, and in particular Lake Como, has allowed him to indulge his passion for photography, capturing images he often uses as a resource and inspiration for his music creation. He enjoys writing for specific performers including Craig Ogden; guitar, James Bowman: counter-tenor, John Turner; recorders, Bernadette Greevy, Adrienne Murray: mezzo-sopranos, Graham Salvage:bassoon, Richard Simpson::Oboe.
His music is published by Roberton/Goodmusic, Forsyth, Peacock Press, Sunshine Music/Spartan Press and Emerson. Recordings include --
British Recorder Concertos Dutton CDLX 7154
http://www.duttonvocalion.co.uk/proddetail.asp?prod=CDLX7154
Songs & Chamber Music Metier MSV 28523
http://www.divineartrecords.com/CD/28523info.htm
Observations. Seventeen Songs and a String Quartet. Metier msv 28548
Antony Hopkins A Portrait Divine art dda21217
http://www.divine-art.co.uk/CD/21217info.htm
Anthony Burgess - The man and his music. Metier msv 77202
Sheet music by David Dubery


I started painting ski trail maps in 1988 and since then over 200 resorts use or have used my illustrations for their base art. Clients are from Australia, Canada, Chile, New Zealand, Fiji, Japan, Scotland, the Caribbean, China, and Serbia as well as the United States.
Like James Niehues' art? Become a fan!


Circa Early 1900's. This beautiful ring has black enamel along the outside of the design. Inside the enamel are 4 seed pearls securely set in 6 prong heads. In the very center is an old European cut diamond which helps to date this ring between 1900 – 1930. The shank of the ring loops into the sides of the top with a sweet curve east and west of the top. The top measures 7/8″ knuckle to knuckle. Pictured as received from a local estate. Very good estate condition with minimal wear indicative of age. Cleaning and/or polishing by request at no additional cost.
SIZE: Fits like an 8 1/4 but cannot be measured on a mandrel because of the flat top. Ring can be sized for an additional fee. Please inquire if needed.
COMMENTS: Circa Early 1900's. This beautiful ring has black enamel along the outside of the design. Inside the enamel are 4 seed pearls securely set in 6 prong heads. In the very center is an old European cut diamond which helps to date this ring between 1900 – 1930. The shank of the ring loops into the sides of the top with a sweet curve east and west of the top. The top measures 7/8″ knuckle to knuckle. Pictured as received from a local estate. Very good estate condition with minimal wear indicative of age. Cleaning and/or polishing by request at no additional cost.


text:
El saladar de Lo Poyo es un espacio protegido situado en la ribera del Mar Menor, entre Los Urrutias y Los Nietos, en el municipio de Cartagena en la comunidad autónoma de la Región de Murcia en España.Como lugar protegido se encuentra incluidos dentro de los denominados Espacios abiertos e islas del Mar Menor con la categoría de Parque natural, LIC y ZEPA.

El espacio protegido de Lo Poyo se conforma como un criptohumedal ubicado al sur del litoral del Mar Menor. Está constituido por unas antiguas salinas, en la actualidad abandonadas, que se ejecutaron sobre una primitiva laguna interior.

Vegetación

La vegetación se distribuye en tres zonas bien diferenciadas. Por la singularidad y diversidad de las especies presentes en este espacio, las zonas de arenales y saladares han sido propuestas como microrreserva botánica.

Carrizales
Formado mayoritariamente por carrizos (Phragmites australis), los cuales ocupan las zonas desecadas de las antiguas salinas.

Arenales
Situados sobre en la banda de arena que separa las antiguas salinas del mar. En estos arenales, pueden encontrarse especies propias de la vegetación dunar como la zanahoria marítima (Echinophora spinosa), la siempreviva amarilla (Helichrysum stoechas), la hierba pincel (Coris monspeliensis), la azucena de mar (Pancratium maritimum) o el cambrón (Lycium intricatum). Destaca especialmente la presencia de algunos ejemplares de la esparraguera del Mar Menor (Asparagus macrorrhizus), una especie endémica del entorno de la laguna y en peligro crítico de extinción.

Saladares

Un saladar se caracteriza por poseer un suelo cargado de sales (principalmente cloruros) y que, además, suele poseer un nivel freático próximo a la superficie. La vegetación que crece sobre estos suelos, capaz de tolerar la sal, se denomina halófita.

En el caso del saladar de Lo Poyo, esta vegetación está compuesta mayoritariamente por sosa (Suaeda vera) sosa alacranera (Sarcocornia fruticosa), siempreviva morada (Limonium caesium), lechuga de mar (Limonium cossonianum) y taray (Tamarix boveana).

Problemas medioambientales

Algunos lugares de este espacio protegido se encuentran especialmente alterados por la presencia de residuos con altos contenidos en metales pesados - plomo y zinc principalmente- así como arsénico, producto del lavado del mineral procedente de la cercanas minas de La Unión, que han llegado a este espacio a través de las ramblas que desembocan en el Mar Menor.

Recientes estudios realizados por la Universidad Politécnica de Cartagena proponen, como medio para limpiar estos espacios contaminados, la inundación de los antiguos humedales y la reforestación con diversas especies vegetales

Por otro lado, los arenales se encontraban profundamente degradados debido al paso incontrolado de vehículos y la presencia de especies alóctonas. Recientemente, en 2014, los caminos que cruzaban los arenales han sido cortados, la vegetación alóctona eliminada, y se ha realizado una campaña de reforestación con especies autóctonas, entre las que destaca la sabina de las dunas (Juniperus turbinata), especie en peligro de extinción en la Región de Murcia.

Galería fotográfica

Referencias

 Murcia, en clave ambiental. El Mar Menor y su entorno: un verdadero tesoro botánico.
 Catálogo de Recursos Naturales del Mar Menor.

Zonas de especial protección para aves de la Región de Murcia
Espacios naturales de Cartagena

redpajama_set_name: RedPajamaWikipedia


This engagement session was an amazing opportunity for us. We photographed Adam and Sarah on a private 650 acre piece of land in the breathtaking Santa Ynez Valley. The land is only accessible by 4x4 vehicles and is beautiful and wild. We had a great time getting to know Adam and Sarah and scoping out the ceremony location for their sunrise wedding ceremony later this Summer. They are some of the coolest people we've ever met. Adam is the first legit cowboy we've ever really hungout with - he shoes horses for a living and is a really nice guy. In all, it was a great time meeting some new friends and getting some seriously killer pics.


text:
Q: DFS with stack to find length of longest path from a certain node in a directed, acyclic, unweighted graph I am trying to figure out a way to implement a DFS search to find the length of the longest path from a given node on a directed graph represented as an adjacency list by using a stack, and not using recursion.  Specifically, I want to implement the DFS search so that as it runs, the stack gets populated as shown in the picture below..
If that isn't clear, this video is sort of how I want the stack to be built up as my program runs (DFS starts at around 12:45 : https://www.youtube.com/watch?v=pcKY4hjDrxk
But im struggling to find a way to achieve this, as I am still pretty new to programming in general. My current code represents the graph as an unordered map, with each entry containing a vector of all the nodes that point to it. i.e:
std::unordered_map<long, std::vector<long>> nodes;

And basically, I want to implement a DFS search from all nodes with a key value of -1 in that unordered_map as shown in the picture and in the video- with the stack getting allocated as shown. I was thinking, that way I can just record when the stack reaches its maximum size, and that would be the longest path.
One thing to note is the graphs in this specific problem is that each edge will only have one outgoing degree, as shown in the picture.. Is this possible, or will I have to use some sort of recursion to do what I want? Thanks in advance for any help.

A: You can probably use a task list instead of recursion.  If you use the task list in FIFO order like a queue, you get a breadth-first search; if you use it LIFO like a stack, you get depth-first behavior.
However, note that it is possible for a DAG with N nodes to have O(2^(N/2)) possible paths!  You should not need to evaluate all possible paths to solve your problem, though, so be careful not to write an algorithm that can take exponential time.
In order to do that, you will need to mark which nodes you have processed.  Also, since you are looking for the longest path, you'll need to track per-node information about the longest path found so far.

A: Although it's possible to achieve this without recursion, as an alternative, I would suggest you to design a function this way, which requires you to write less code and which will provide the nice intuition to understand this algorithm for you're a beginner. And you'll not need to think about creating stack by your own
const int n = 100;
vector< int > graph[n];
int ans = 0, level = 0;
int vis[n];
void dfs(int src) {
  level++;
  ans = max(level, ans);
  for (int x: graph[src]) {
    if(vis[x]) continue;
    vis[x] = 1;
    dfs(x);
    level--;
  }
}

I hard coded the value of n and graph you can change it as you need as required structure.
Where we take the advantages of the stack created for the recursion tree by the program.
This function will work in O(V+E) for a given graph of V nodes and E edges.
Note that, if your Graph is so large that the stack created default by the program can't handle, then you still have to write your own stack to handle the recursion.