text:
\section{Introduction}

Let $G$ be a bounded domain in $\R^d$ ($d\geq 2$) with the smooth
boundary $\pt G$, $$L_k u(x)=\dfrac{1}{2}\sum\li_{i,j=1}^d
a^{(k)}_{ij}(x)\dfrac{\pt^2 u}{\pt x_i \pt x_j}+\sum\li_{i=1}^d
b_i^{(k)}(x)\dfrac{\pt u}{\pt x_i} \ , \ k=0,1 \ , \ x\in \R^d \ .$$

The coefficients are assumed to be smooth enough, say, in
$\contfunc^{(2)}(\R^d)$, i.e., having continuous second derivatives.

Boundary problems for the operator $L_\ve=L_0+\ve L_1$ in the domain
$G$ and corresponding initial-boundary problems for the equation
$\dfrac{\pt u^\ve(t,x)}{\pt t}=L_\ve u^\ve$, $t>0$, $x\in G$, are
considered. The operator $L_\ve$ is assumed to be elliptic for
$\ve>0$. One can study the limiting behavior of solutions of
stationary problems as $\ve\da 0$ and the limiting behavior of
solutions of initial-boundary problems as $\ve \da 0$ and $t\ra
\infty$.

If the operator $L_0$ is elliptic in $G\cup \pt G$, the problem is
simple: $u^\ve$ converges to the solution of corresponding problem
for the operator $L_0$. In the case of degenerate operator $L_0$,
situation is more complicated, and the question was considered in
numerous papers. First, the case of first order operator $L_0$ was
considered: $L_0=b^{(0)}(x)\cdot \grad$,
$b^{(0)}(x)=(b^{(0)}_1(x),...,b^{(0)}_d(x))$. N.Levinson
\cite{[Levinson]} showed in 1950-th that, if the characteristics of
$L_0$ (e.g., trajectories of the dynamical system
$\dot{X}_t=b^{(0)}(X_t)$ in $\R^d$) leave the domain $G$ in finite
time and cross the boundary in a regular way, then the solution of
the Dirichlet problem $L_\ve u^\ve=0$, $x\in G$, $u^\ve(x)|_{\pt
G}=\psi(x)$, converges as $\ve\da 0$ to the solution of degenerate
equation $L_0u^0(x)=0$, $x\in G$, with the boundary condition
$\psi(x)$ ($\psi(x)$ is assumed to be continuous) on the part of
$\pt G$ through which the characteristics leave the domain. Such a
solution $u^0(x)$ is unique.

Most of subsequent results concerning this problem were obtained by
probabilistic methods. With each operator $L_\ve$, $\ve\geq 0$, one
can (see \cite{[F Lemma]}, notice that the coefficients of
$a_{ij}^{(k)}(x)$ are in $\contfunc^{(2)}(\R^d)$) connect a
diffusion process $\widetilde{X}_t^\ve$ in $\R^d$ defined by the
stochastic differential equation

$$
\begin{array}{l}
\dot{\widetilde{X}}_t^\ve=b^{(0)}(\widetilde{X}_t^\ve)+\ve
b^{(1)}(\widetilde{X}_t^\ve)+
\sm^{(0)}(\widetilde{X}_t^\ve)\dot{W}_t^0+\sqrt{\ve}\sm^{(1)}(\widetilde{X}_t^\ve)\dot{W}_t^1
\ \ ,
\\
\widetilde{X}_0^\ve=x\in \R^d \ , \ t>0 \ , \
\sm^{(k)}(x)(\sm^{(k)}(x))^*=(a_{ij}^{(k)}(x))=a^{(k)}(x) \ , \ k=0
, 1 \ .
\end{array}$$

Here $W_t^0$ and $W_t^1$ are independent Wiener processes in $\R^d$.
Then the solution of the Dirichlet problem for the equation $L_\ve
u^\ve(x)=0$, $x\in G$, and of the initial boundary problem for
$\dfrac{\pt u^\ve(t,x)}{\pt t}=L_\ve u^\ve(t,x)$ can be represented
as expectations of corresponding functionals of
$\widetilde{X}_t^\ve$. The trajectories $\widetilde{X}_t^\ve$, in a
sense, play the same role as characteristics in the case of first
order operator $L_0$. Using these representations and studying
limiting behavior of process $\widetilde{X}_t^\ve$ one can describe
the limiting behavior of the boundary problems (see \cite{[F red
book]}, \cite{[FW book]}).

If problems with the Neumann boundary conditions are considered, one
can use the corresponding diffusion process with reflection on the
boundary (see, for instance, \cite[\S 2.5]{[F red book]}). Various
cases of first order operators $L_0$ not satisfying Levinson's
conditions were examined using the probabilistic approach (see
\cite{[F red book]}, \cite{[FW book]} and the references therein).

If the operator $L_0$ has terms with second derivatives, one can
introduce a generalized Levinson condition (\cite[\S 4.2]{[F red
book]}). Under this condition the equation $L_0 u^0(x)=0$, $x\in G$,
with appropriate Dirichlet type boundary conditions has a unique
solution, and the solution $u^\ve(x)$ of the Dirichlet problem for
equation $L_\ve u^\ve(x)=0$, $x\in G$ converges to $u^0(x)$ as $\ve
\da 0$. The difference with the classical Levinson case is just in
the rate of convergence: under mild additional assumptions
$|u^\ve(x)-u^0(x)|<\ve^\gm$ for some $\gm>0$ and $0<\ve<\!<1$, but
for any $\gm'>0$ one can find $L_0$ with infinitely differentiable
coefficients non-degenerating on $\pt G$ such that
$|u^\ve(x)-u^0(x)|$ is greater than $\ve^{\gm'}$ at a point $x\in G$
and $0<\ve<\!<1$.

A convenient way to specify the degeneration of $L_0$ is given by
the conservation laws. A function $H(x)$ is called a first integral
for the process $X_t^0$ corresponding to $L_0$ if $\Prob_x(X_t^0\in
S(H(x)))=1$ for all $t\geq 0$ and $x\in \R^d$, where $S(z)=\{y\in
\R^d: H(y)=z\}$; here and below the subscript $x\in \R^d$ in the
probability $\Prob_x$ or expected value $\E_x$ means that the
trajectory of the process starts at the point $x$.

We consider in this paper self-adjoint operators $L_0$ and $L_1$:
$$L_k u(x)=\dfrac{1}{2}\grad \cdot (a^{(k)}(x)\grad u(x)) \ .$$ Then
a smooth function $H(x)$ is a first integral for the process
$\widetilde{X}_t^0$ (for the corresponding operator $L_0$) if and
only if $\azero(x)\grad H(x) \equiv 0$. In general, the process
$\widetilde{X}_t^0$ can have several independent smooth first
integrals. To restrict ourselves to the case of one smooth first
integral we assume that $\e\cdot(\azero(x)\e)\geq
\underline{a}(x)|\e|_{\R^d}^2$ for each $\e\in \R^d$ such that
$\e\cdot \grad H(x)=0$: It is assumed that $\underline{a}(x)$ is
smooth and strictly positive if $x$ is not a critical point of
$H(x)$; if $x_0$ is a critical point, $\azero(x_0)=0$ and
$\underline{a}(x_0)=0$.

To be specific we consider the Neumann problem

$$\dfrac{1}{\ve}L_\ve u^\ve=\left(\dfrac{1}{\ve}L_0+L_1\right)u^\ve(x)=f(x) \ , \
\left.\dfrac{\pt u^\ve}{\pt \gm^\ve}(x)\right|_{\pt G}=0 \ ;
\eqno(1.1)$$ $\gm^\ve(x)$ here is the inward co-normal unit vector
to $\pt G$ corresponding to $L_\ve$. Let $X_t^\ve$ be the process in
$G\cup \pt G$ governed by the operator inside $G$ with reflection
along the co-normal to $\pt G$. Since $L_\ve$ is self-adjoint, the
Lebesgue measure is invariant for the process $X_t^\ve$, and the
problem (1.1) is solvable for continuous $f(x)$ such that
$\play{\int_{G}f(x)dx=0}$. Together with the last condition, we
assume that $L_1$ is not degenerate in $G \cup \pt G$, so that to
single out a unique solution of (1.1), we shall fix the value of
$u^\ve(x)$ at a point $x_{O}\in G\cup \pt G$ which is fixed the same
for all $\ve>0$. We let $u^\ve(x_O)=0$.

Then the solution of problem (1.1) can be written in the form (see,
for instance \cite{[F red book]})

$$u^\ve(x)=-\int_0^\infty \E_x f(X_t^\ve)dt+\int_0^\infty \E_{x_O} f(X_t^\ve)dt  \ . \eqno(1.2)$$

If the first integral $H(x)$ has in $G \cup \pt G$ no critical
points, one can describe the $\lim\li_{\ve\da 0}u^\ve(x)$ in the way
similar to \cite{[FW fish]}: One shall introduce a graph $\graph$
corresponding to the set of connected components of the
intersections of the level sets of $H(x)$ within $G$. A boundary
problem on $\graph$ with appropriate gluing conditions at the
vertices can be formulated, and the solution of this problem defines
$\lim\li_{\ve\da 0}u^\ve(x)$. If the function $H(x)$ has saddle
points inside $G$, additional branchings in the graph appear. The
gluing conditions at these new vertices can be calculated using the
results of \cite{[FWeb]}.

All mentioned above results concern the case when the rank of
$\azero(x)$ is constant and equal to $d-1$ for all $x\in G\cup \pt
G$ except the critical points of $H(x)$. In this paper, we consider
the case when $L_0$ is non-degenerate in a connected subdomain
$\cE\subset G$, and we let $H(x)$ be equal to a constant on $\cE$.
Outside $\cE$ the first integral $H(x)$ has a finite number of
critical points (see Fig.1). For convenience of presentation, we
shall then introduce several first integrals $H_k(x)$ ($k=1,...,r$)
for each of the connected components $U_1,...,U_r$ on which $L_0$ is
degenerate. We shall let $H(x)=H_k(x)$ for $x\in U_k$. A more
concrete setup of the problem is in Section 2. Existence of the
domain $\cE$ where the operator $L_0$ is not degenerate leads to
more general gluing conditions. The limiting process on the graph
spends a positive time at the vertex corresponding to $\cE$.

\begin{figure}
\renewcommand{\captionlabeldelim}{.}
\centering
\includegraphics[height=5cm, width=14cm , bb=11 82 399 217]{Figure_1_new.eps}
\caption{}
\end{figure}

Let $S(z)=\{x\in G\cup \pt G: H(x)=z\}$. The graph $\graph$ is the
result of identification of points of each connected component of
every level set $S(z)$. Let $\iiY: G\cup \pt G \ra \graph$ be the
identification mapping. We call $\iiY(x)$ the projection of $x$ onto
$\graph$. We consider the projection $Y_t^\ve=\iiY(X_t^\ve)$ of the
process $X_t^\ve$ on $\graph$ and prove that processes $Y_t^\ve$ on
$\graph$ converge weakly in the space of continuous functions $[0,T]
\ra \graph$ to a diffusion process $Y_t$ on $\graph$. The process
$Y_t$ is defined by a family of differential operators, one on each
edge, and by gluing conditions at the vertices. We calculate the
operators and the gluing condition. The function
$u^0(x)=\lim\li_{\ve\da 0}u^\ve(x)$ is constant on each connected
component of every level set of $H(x)$: $u^0(x)=v(\iiY(x))$. We
formulate a boundary problem for the function $v(y)$, $y\in \graph$,
which has a unique solution, and actually can be solved explicitly.

The organization of this paper is as follows: Section 2 sets up the
problem and gives the main results. Section 3 is devoted to the
proof of the main results in Section 2. Section 4 proves auxiliary
results needed in Section 3.

\section{Main results}

Let us first speak about our assumptions.

Suppose we have a bounded domain $G\subset \R^d$, with smooth
boundary $\pt G$. We assume that $d\geq 2$. Let $L_0$ be a
self-adjoint operator
$$L_0 u (x)=\dfrac{1}{2}\sum\li_{i,j=1}^d \dfrac{\pt}{\pt x_i}\left(\azero_{ij}(x) \dfrac{\pt u(x)}{\pt
x_j}\right)=\dfrac{1}{2}\grad\cdot (\azero(x)\grad u(x)) \ .$$

Let $U_1,...,U_r$ be several regions inside $G$. They are simply
connected open sets and their closure do not intersect each other.
Let us assume that the matrix
$\azero(x)\equiv(\azero_{ij}(x))_{1\leq i,j \leq d}$ is positive
definite on $\cE=[G]\backslash \left(\cup_{k=1}^r [U_k]\right)$
(here $[D]$ is the closure of a domain $D$). For $x\in [\cE]$, the
coefficients $\azero_{ij}(x), 1\leq i,j \leq d$ are assumed to be in
$\contfunc^{(3)}([\cE])$.

Let us discuss the case when $x\in \cup_{k=1}^r[U_k]$. For each
$k=1,...,r$ and $x\in [U_k]$, the coefficients $\azero_{ij}(x),
1\leq i,j \leq d$ are assumed to be in $\contfunc^{(3)}([U_k])$. We
assume that the matrix $(\azero_{ij}(x))_{1\leq i,j \leq d}$ is
degenerate on $\cup_{k=1}^r[U_k]$. To specify this degeneration, we
assume that within each $[U_k]$ there is a function $H_k$ which is a
first integral of the (degenerate) operator $L_0$, i.e.,
$\azero(x)\grad H_k(x)=0$ for $x\in [U_k]$. The function $H_k$ is
assumed to be in at least $\contfunc^{(4)}([U_k])$. Let $H_k$ have
only one minimum $m_k$ inside $U_k$. (We can always make this
assumption since if $m_k$ is a maximum we work with $-H_k$ instead
of $H_k$.) Let $x_k(m_k)$ be the point in $U_k$ corresponding to the
minimum $m_k$. This minimum is assumed to be non-degenerate, i.e.,
the matrix $\left(\dfrac{\pt^2 H_k}{\pt x_i\pt
x_j}(x_k(m_k))\right)_{1\leq i,j \leq d}$ is positive definite.
Since the choice of $H_k$ is up to a constant we can assume that
$H_k=0$ on $\pt U_k$. For $h\in (m_k,0]$ the level surfaces
$C_k(h)=\{x\in U_k: H_k(x)=h\}$ of the functions $H_k$ inside $U_k$
are closed surfaces of dimension $(d-1)$ and the operator $L_0$ is
non-degenerate on $C_k(h)$. Let $\gm_k=\pt U_k=C_k(0)$. A
non-degeneracy condition of $\azero(x)$ on $C_k(h)$ is assumed: for
any vector $\mathbf{e}\in \R^d$ such that $\mathbf{e}\cdot \grad
H_k=0$ we have $\mathbf{e}\cdot(\azero(x) \mathbf{e})\geq
\underline{a}(x)|\mathbf{e}|_{\R^d}^2$ for some $\underline{a}(x)>0$
and $x\neq x_k(m_k)$. We set $\azero(x_k(m_k))=0$ and
$C_k(m_k)=\{x_k(m_k)\}$. We assume that the level surfaces $C_k(h)$
for $h\in (m_k,0]$ divide $U_k\backslash \{x_k(m_k)\}$ into pieces
of $(d-1)$-dimensional surfaces.

For simplicity of presentation we will assume that $\grad H_k
(x)\neq 0$ for $x\in \gm_k$. One can introduce a global first
integral $H(x)$ on $[G]$ as in Section 1: $H(x)=H_k(x)$ for $x\in
U_k$ and $H(x)=0$ for $x\in [\cE]$. We notice that the function
$H(x)$ so defined as a global first integral is not smooth at
$\cup_{k=1}^r \gm_k$. However, this will not affect our analysis.

Let $\gm=\cup_{k=1}^r \gm_k$. We will define a unit vector field
$\e_d(x)$ in a small neighborhood of $\gm_k$ at the beginning of
Section 4. Roughly speaking, the vector field $\e_d(x)$ is the
direction in which the matrix $\azero(x)$ degenerates. We assume
that the order of degeneracy is given by the condition that for this
$\e_d(x)$ as $x$ belongs to the intersection of a small neighborhood
of $\cup_{k=1}^r \gm_k$ and $\cE$ we have
$$\text{const}_1\cdot\text{dist}^2(x,\gm)\leq
\e_d(x)\cdot(\azero(x)\e_d(x))\leq
\text{const}_2\cdot\text{dist}^2(x,\gm)$$ for some $\text{const}_1$,
$\text{const}_2>0$. The distance $\text{dist}(x,\gm)$ is the
Euclidean distance between $x$ and $\gm$. The vector field
$\e_d(x)=\dfrac{\grad H_k}{|\grad H_k|_{\R^d}}$ for $x\in \gm_k$.

In particular, our assumptions imply that the matrix $\azero(x)$ has
rank $d$ in $\cE$ and rank $(d-1)$ in $\cup_{k=1}^r ([U_k]\backslash
\{x_k(m_k)\})$. At the points $x_k(m_k)$ the matrix $\azero(x)$ is
just $0$. However, the coefficients $\azero_{ij}(x)$, $1\leq i,j\leq
d$ are only in $\contfunc^{(1)}$ for $x\in \gm$. We notice that in
this case results of \cite{[F Lemma]} do not apply. We shall then
assume that there is a decomposition
$\azero(x)=\sm^{(0)}(x)(\sm^{(0)}(x))^*$ for all $x\in [G]$. The
square matrix $\sm^{(0)}(x)$ has bounded Lipschitz continuous terms.

We shall assume that the operator $L_1$ governing the perturbation
is self-adjoint and strictly elliptic within $[G]$: $$L_1 u
(x)=\dfrac{1}{2}\sum\li_{i,j=1}^{d}\dfrac{\pt}{\pt
x_i}\left(\aone_{ij}(x)\dfrac{\pt u (x)}{\pt
x_j}\right)=\dfrac{1}{2}\grad\cdot (\aone(x)\grad u(x)) \ .$$

Again we denote the matrix $\aone(x)\equiv(\aone_{ij}(x))_{1\leq i,j
\leq d}$ and we assume that the terms $\aone_{ij}(x)$ are in class
$\contfunc^{(2)}(\R^d)$. In this case results of \cite{[F Lemma]}
apply and we have $\aone(x)=\sm^{(1)}(x)(\sm^{(1)}(x))^*$ for all
$x\in [G]$. The square matrix $\sm^{(1)}(x)$ have bounded Lipschitz
continuous terms.

Let us put a Neumann boundary condition with respect to co-normal
unit vector $\gm^\ve(x)$ pointing inward on $\pt G$. Let $X_t^\ve$
be the diffusion process in $[G]$, corresponding to the operator
$\dfrac{1}{\ve}L_0+L_1$ inside $G$ with co-normal reflection at $\pt
G$. We see that Lebesgue measure is invariant for the process
$X_t^\ve$.

\

Consider a simple example. We use polar coordinates for the plane
$\R^2$. Let $G=\{(r,\tht): 0\leq \tht\leq 2\pi, 0\leq r < 2\}$. Let
$U_1=\{(r,\tht): 0\leq \tht \leq 2\pi, 0\leq r <1\}$. Let
$H_1(r,\tht)=\dfrac{r^2-1}{2}$. Let $H(r,\tht)=H_1(r,\tht)$ for
$0\leq r<1$ and $H(r,\tht)=0$ for $1\leq r\leq 2$. Let
$\lb(r)=(r-1)^2$ for $1\leq r\leq 2$ and $\lb(r)=0$ for $0\leq r<
1$. Let
$$a^{(0)}(r,\tht)=\begin{pmatrix}\lb(r)\cos^2\tht+r\sin^2\tht & (\lb(r)-r)\sin\tht\cos\tht\\
(\lb(r)-r)\sin\tht\cos\tht &
\lb(r)\sin^2\tht+r\cos^2\tht\end{pmatrix} \ .$$ Let $\e_r=(\cos\tht,
\sin\tht)$ and $\e_\tht=(-\sin\tht, \cos\tht)$. Then it is easy to
check that $a^{(0)}(r,\tht)\e_r=\lb(r)\e_r$ and
$a^{(0)}(r,\tht)\e_\tht=r\e_\tht$. We have the decomposition
$a^{(0)}(x)=\sm^{(0)}(x)(\sm^{(0)}(x))^*$ where
$$\sm^{(0)}(r,\tht)=\begin{pmatrix}\sqrt{\lb(r)}\cos^2\tht+\sqrt{r}\sin^2\tht & (\sqrt{\lb(r)}-\sqrt{r})\sin\tht\cos\tht\\
(\sqrt{\lb(r)}-\sqrt{r})\sin\tht\cos\tht &
\sqrt{\lb(r)}\sin^2\tht+\sqrt{r}\cos^2\tht\end{pmatrix} \ .$$

\

Let us then speak about the results.

We construct a graph $\graph$ as follows. The graph $\graph$ has $r$
edges $I_1,...,I_r$ joined together at one vertex $O$. Let the other
endpoint of $I_k$ be $V_k$. Let us write $I_k=[m_k,0]$. The
coordinate $(k,H_k)$ is a global coordinate on $\graph$. The root
$O$ corresponds to all $(k,0)$ for $k=1,...,r$. Let us introduce an
identification map $\iY: [G] \ra \graph$: for $x\in [\cE]$ we have
$\iY(x)=O$ and for $x\in U_k$ we have $\iY(x)=(k,H_k(x))$. Let the
process $Y_t^\ve=\iY(X_t^\ve)$. We are going to prove, that as $\ve
\da 0$ the processes $Y_t^\ve$ converge weakly in the space
$\contfunc_{[0,T]}(\graph)$ to a Markov process $Y_t$ on $\graph$.

The process $Y_t$ is defined as follows. It is a diffusion process
on the graph $\graph$ with generator $A$ and the domain of
definition $D(A)$. Inside each $I_k$ it is governed by an operator
$\cL_k$ defined as $$\cL_k f(k,H_k)=
\dfrac{1}{2}M_k^{-1}(H_k)\dfrac{d}{dH_k}\left(M_k(H_k)\aonebar
(H_k)\dfrac{df}{dH_k}\right)\ .$$ Here
$$\aonebar(h)=M_k^{-1}(h)\int\li_{C_k(h)}\dfrac{(\aone(x)\grad H_k(x),\grad H_k(x))}
{|\grad H_k(x)|_{\R^d}}d\sm \ ,$$ and normalizing factor
$$M_k(h)=\int\li_{C_k(h)}\dfrac{d\sm}{|\grad H_k(x)|_{\R^d}} \
.$$ The notation $d\sm$ denotes the integral with respect to the
area element on $C_k(h)$.

We set $Af=\cL_k f$ for $(k,H_k)\in (I_k)$ ($(I_k)$ is the interior
of the interval $I_k$). Let the limit $\lim\li_{(k,H_k)\ra
O}Af(k,H_k)$ be finite and independent of $k$. This limit is set to
be $Af(O)$. The domain of definition $D(A)$ of the operator $A$
consists of those functions $f$ that are twice continuously
differentiable inside each $I_k$ having the limit $\lim\li_{H_k\ra
0}\dfrac{\pt f}{\pt H_k}(k,H_k)$. These functions satisfy the gluing
condition at the vertex $O$:
$$0=\Vol(\cE)\cdot Af(O)+\dfrac{1}{2}\sum\li_{k=1}^r p_k \cdot \lim\li_{H_k\ra
0}\dfrac{\pt f}{\pt H_k}(k,H_k) \ . \eqno(2.1)$$ Here $\Vol(\cE)$ is
the $d$-dimensional volume of the domain $\cE$ and
$$p_k=\int_{\gm_k}\dfrac{(a^{(1)}(x)\grad H_k(x), \grad H_k(x))}{|\grad H_k(x)|_{\R^d}}d\sm \
.$$

For the exterior vertices $V_1,...,V_r$ no additional assumptions
are to be imposed on the behavior of the function $f$ in the domain
$D(A)$.

It was proved in \cite{[FW Diffusion process on a graph]} the the
process $Y_t$ exists and is a strong Markov process on the graph
$\graph$.

We have

\

\textbf{Theorem 2.1.} \textit{As $\ve \da 0$ the processes $Y_t^\ve$
converge weakly to $Y_t$ in $\contfunc_{[0,T]}(\graph)$.}

\

Let $\mu_x^\ve$ be the distribution of the trajectory
$Y_t^\ve=\iiY(X_t^\ve)$ starting from a point $x\in [G]$ in the
space $\contfunc_{[0,T]}(\graph)$: for each Borel subset $B\subseteq
\contfunc_{[0,T]}(\graph)$ we set
$\mu_x^\ve(B)=\Prob_{X_0^\ve=x}(Y_{\bullet}^\ve\in B)$. Similarly,
for each $y\in \graph$ we let $\mu_y^0$ be the distribution of $Y_t$
in the space $\contfunc_{[0,T]}(\graph)$ with
$\mu_y^0(B)=\Prob_{Y_0=y}(Y_{\bullet}\in B)$. Theorem 2.1 can be
reformulated as

\

\textbf{Theorem 2.2.} \textit{For every $x\in [G]$ and every $T>0$
the distribution $\mu_x^\ve$ converges weakly to $\mu_{\iiY(x)}^0$
as $\ve \da 0$. For every bounded continuous functional $F$ on
$\contfunc_{[0,T]}(\graph)$ we have $$\E_{X_0^\ve=x}
F(Y_{\bullet}^\ve)\ra \E_{Y_0=\iiY(x)}F(Y_{\bullet})$$ as $\ve \da
0$.}

\

The process $Y_t^\ve$ can be viewed as the slow component of the
process $X_t^\ve$. The fast component $Z_t^\ve$ of $X_t^\ve$ is a
process governed by the operator $\dfrac{1}{\ve}L_0$. The process
$Z_t^\ve$ moves on $\iiY^{-1}(y)$ for each $y\in \graph$: it is
moving on $[\cE]$ when $y=O$ and it is moving on $C_k(H_k)$ when
$y=(k,H_k)$. Since Lebesgue measure is invariant for the process
$X_t^\ve$, the fast component $Z_t^\ve$, as $\ve>0$ is small, has,
approximately, a distribution with density
$\dfrac{1}{\text{Volume}(\cE)}$ on $[\cE]$ (with respect to Lebesgue
measure on $\R^d$) and $\dfrac{1}{M_k(H_k)}\dfrac{1}{|\grad
H_k|_{\R^d}}$ on $C_k(H_k)$ (with respect to the area element $d\sm$
on $C_k(H_k))$. Using this we can formulate the above two theorems
in terms of differential equations:

\

\textbf{Theorem 2.3.} \textit{Consider the Neumann problem}

$$\dfrac{1}{\ve}L_\ve u^\ve(x)=\left(\dfrac{1}{\ve}L_0+L_1\right)u^\ve(x)=f(x) \textit{\text{ for }} x\in G \ ,
\ \left.\dfrac{\pt u^\ve}{\pt \gm^\ve}(x)\right|_{x\in \pt G}=0$$
\textit{with a H\"{o}lder continuous function $f(x)$ satisfying
$\play{\int_G f(x)dx=0}$. Let $u^\ve(x_O)=0$ for some $x_O\in G$.
Then we have }$$\lim\li_{\ve\da 0}u^\ve(x)=v(\iiY(x))$$\textit{
where $v(y)$ is a continuous function on $\graph$ such that }$$\cL_k
v(y)=-\bar{f}(y) \textit{\text{ for }} y\in (I_k) \ , \ k=1,...,r \
.$$\textit{ Here}
$$\bar{f}(y)=\dfrac{1}{\text{Volume}(\cE)}\int_{\cE}f(x)dx$$\textit{ when $y=O$
and}
$$\bar{f}(y)=\dfrac{1}{M_k(H_k)}\int_{C_k(H_k)}f(x)\dfrac{d\sm}{|\grad
H_k(x)|_{\R^d}}$$\textit{ when $y=(k,H_k)$.} \textit{ The function
$v(y)$ satisfies the gluing condition (2.1) and $v(\iiY(x_O))=0$.
Such a function $v(y)$ is unique. }

\section{Proof of Theorem 2.1}

The \textbf{Proof} of Theorem 2.1 follows the arguments of \cite{[FW
book]}, \cite{[FW AMS]}, \cite{[DK1]} and \cite{[DK2]}.

Heuristically, the idea of \cite{[DK2]} can be explained as follows.
The process $X_t^\ve$ moves within $[G]$ and has Lebesgue measure as
its invariant measure. Since the process $X_t^\ve$ has a "fast"
component governed by the operator $\dfrac{1}{\ve}L_0$, it will
spend a positive amount of time proportional to $\Vol(\cE)$ within
$\cE$ as $\ve\da 0$. As we project $X_t^\ve$ onto the graph $\graph$
and the whole ergodic component $\cE$ corresponds to $O$, the
limiting process $Y_t$ has a boundary condition with a "delay" at
$O$. (We recommend a nice article \cite{[Ito-McKean 1963]} and a
brief summary \cite[\S 5.7]{[Ito-McKean book]} about this boundary
condition.) Our gluing condition (2.1) ensures that the process
$Y_t$ has an invariant measure on $\graph$ that agrees with the
Lebesgure measure on $[G]$. We also refer to \cite[Ch.8, pp.
347--350]{[FW book]} for an explanation of this.

Let us first introduce some notations. Below we will often suppress
the small parameter $\ve$ and it could be understood directly from
the context. Let $\gmin_k=C_k(-\ve^{1/2})$ and $\gmin=\cup_{k=1}^r
\gmin_k$. Let $\sm$ be the first time when the process $X_t^\ve$
hits $\gm$. Let $\tau$ be the first time when the process $X_t^\ve$
hits $\gmin$. Let $\sm_0=\sm$. Let $\tau_n$ be the first time
following $\sm_n$ when the process reaches $\gmin$. For $n\geq 1$
let $\sm_n$ be the first time after $\tau_{n-1}$ when the process
$X_t^\ve$ reaches $\gm$.

Let $\sm^*\in \{\sm_0,\sm_1,...\}$ and we denote by $m_{\sm^*}^x$
the measure on $\gm$ induced by $X_{\sm^*}^\ve$ starting at $x$.
That is,

$$m_{\sm^*}^x(A)=\Prob_x(X_{\sm^*}^\ve\in A) \ , \ A\in \cB(\gm) \ .$$

Let $\nu(\bullet)$ be the invariant measure of the induced chain
$X_{\sm_n}^\ve$ on $\gm$. The key lemma of \cite{[DK2]} is the
following

\

\textbf{Lemma 3.1.} \textit{Let $x\in [\cE]$. For each $\dt>0$ and
all sufficiently small $\ve$ there is a stopping time $\sm^*$ which
may depend on $\dt, \ve$ and $x$ and such that $$\E_x\sm^*\leq \dt \
, \eqno(3.1)$$
$$\sup\li_{x\in \gm}\text{Var} (m_{\sm^*}^x(dy)-\nu(dy))\leq \dt \ , \eqno(3.2)$$
where $\text{Var}$ is the total variation of the signed measure. }

\

Our proof of this lemma is similar while a bit simpler than that of
\cite{[DK2]}.

\

\textbf{Proof.} We will prove, in Lemma 4.11 that $X_{\sm_n}^\ve$
satisfies the Doeblin condition on $\gm$ uniformly in $\ve$. This
implies that one can choose an $N$ depending on $\dt$ but
independent of $\ve$ such that the distribution of $X_{\sm_N}^\ve$
is $\dt$-close to the invariant measure $\nu(\bullet)$ on $\gm$.
That is, as we set $\sm^*=\sm_N$ the condition (3.2) is satisfied.

We are going to prove in Lemmas 4.8, 4.9, 4.10, respectively, that
$$\lim\li_{\ve\da 0}\sup\li_{x\in [\cE]} \E_x\sm=0  \ , \eqno(3.3)$$
$$\lim\li_{\ve\da 0}\sup\li_{x\in \gm} \E_x\tau=0  \ , \eqno(3.4)$$
$$\lim\li_{\ve\da 0}\sup\li_{x\in\gmin} \E_x\sm=0  \ . \eqno(3.5)$$

We can write
$\sm_N=\sum\li_{i=1}^N[(\sm_i-\tau_{i-1})+(\tau_{i-1}-\sm_{i-1})]+\sm_0$.
For each $i=1,...,N$ the random variable $\sm_i-\tau_{i-1}$ has the
same distribution as $\sm$ for the process $X_t^\ve$ starting at
some point on $\gmin$; similarly, the random variable
$\tau_{i-1}-\sm_{i-1}$ has the same distribution as $\tau$ for the
process $X_t^\ve$ starting at some point on $\gm$. The results
(3.3), (3.4), (3.5) imply that as $\ve$ is small (notice that $N$ is
fixed at this stage) we can choose $\sm^*=\sm_N$ and the condition
(3.1) is also satisfied. $\square$

\

\textbf{Proof of Theorem 2.1.} The proof of Theorem 2.1 is the same
as the proof of Lemma 2.1 (including the proof of Lemma 3.4) stated
in \cite{[DK2]} using the above Lemma 3.1. For the sake of
completeness let us briefly repeat it here. Reasoning as in
\cite{[DK1]}, \cite{[DK2]}, \cite{[FW book]}, \cite{[FW AMS]}, it
suffices to prove that for a function $f\in D(A)$, for every $T>0$
and uniformly in $x\in [G]$ we have

$$\E_x\left[f(H(X_T^\ve))-f(H(X_0^\ve))-\int_0^T Af(H(X_s^\ve))ds\right]\ra 0$$
as $\ve \da 0$. Here $H(x)=H_k(x)$ if $x\in U_k$ and $H(x)=0$ if
$x\in [\cE]$. Let us replace the time interval $[0,T]$ by a larger
one $[0,\smtl]$, where $\smtl$ is the first of the stopping times
$\sm_n$ which is greater than or equal to $T$: $\smtl=\min\li_{n:
\sm_n>T}\sm_n$. Let $\smtl=\sm_{\widetilde{n}+1}$. We have

$$\begin{array}{l}
\play{\E_x\left[f(H(X_T^\ve))-f(H(X_0^\ve))-\int_0^T
Af(H(X_s^\ve))ds\right]}
\\
=\play{\E_x\left[f(H(X_\sm^\ve))-f(H(X_0^\ve))-\int_0^{\sm}
Af(H(X_s^\ve))ds\right]+\E_x\sum\li_{k=0}^{\ntl}\int_{\sm_k}^{\sm_{k+1}}Af(H(X_s^\ve))ds-}
\\
\play{ \ \ \ \ \ \
-\E_x\E_{X_T^\ve}\left[f(H(X_{\sm}^\ve))-f(H(X_0^\ve))-\int_0^{\sm}Af(H(X_s^\ve))ds\right]}
\\
=(I)+(II)-(III) \ .
\end{array}$$

If $x\in \cup_{k=1}^r U_k$ we have $|(I)|\ra 0$ uniformly in $x$ as
$\ve \da 0$ due to averaging principle. If $x\in [\cE]$ then
$|(I)|\ra 0$ uniformly in $x$ as $\ve \da 0$ due to Lemma 4.8. In a
similar way we see that $|(III)|\ra 0$ uniformly in $x\in [G]$ as
$\ve \da 0$.

Let $\play{\al_k=\int_{\sm_k}^{\sm_{k+1}}Af(H(X_s^\ve))ds}$ and let
$\play{\bt_k=\sum\li_{n=0}^\infty \E_x(\al_{k+n}|\cF_k)}$ ($\cF_k$
is the filtration generated by the process $X_t^\ve$ for $t\leq
\sm_k$). We have $\E_x(\al_k-\bt_k+\bt_{k+1}|\cF_k)=0$ and therefore
$\play{\left(\sum\li_{k=1}^n
(\al_k-\bt_k+\bt_{k+1}),\cF_{n+1}\right) \ , \ n\geq 0}$ is a
martingale. Using the optimal sampling theorem we have

$$\E_x\sum\li_{k=0}^{\ntl}\al_k=\E_x\sum\li_{k=0}^{\ntl}
(\al_k-\bt_k+\bt_{k+1})+\E_x(\bt_0-\bt_{\ntl+1})
=\E_x(\bt_0-\bt_{\ntl+1}) \ .$$

The above argument shows that for the proof of $|(II)|\ra 0$
uniformly in $x\in [G]$ as $\ve \da 0$ it suffices to prove
$\play{\sup\li_{x\in \gm}\left|\sum\li_{n=0}^\infty
\E_x\al_n\right|}\ra 0$ uniformly in $x\in \gm$ as $\ve \da 0$.

Let us first show that $\E_{\nu}\al_0=0$. By Lemma 4.11 the Markov
chain $X_{\sm_n}^\ve$, $n \geq 0$ on $\gm$ is ergodic and has
invariant measure $\nu$. Therefore we have $\play{\lim\li_{n\ra
\infty}\dfrac{\sm_n}{n}=\E_{\nu}\sm_1}$. By ergodicity of the
process $X_t^\ve$ and self-adjointness of $L_0$ and $L_1$ we also
have
$$\lim\li_{t\ra \infty}\E_{\nu}\dfrac{1}{t}\play{\int_0^t A
f(H(X_s^\ve))ds=\int_{[G]}A f(H(x))dx \ .}$$ These two equalities
imply that $\play{\E_{\nu}\al_0=\E_{\nu}\int_0^{\sm_1}A
f(H(X_s^\ve))ds=(\E_{\nu}\sm_1)\cdot \int_{[G]}A f(H(x))dx}$.

We have
$$\int_{[G]}Af(H(x))dx=\text{Volume}(\cE)\cdot Af(O)+\sum\li_{k=1}^r \int_{I_k}M_k(H_k)\cL_k f(H_k)dH_k \ .$$
(The notations agree with those in the definition of the process
$Y_t$.)

Since $$\int_{I_k}M_k(H_k)\cL_k
f(H_k)dH_k=\int_{I_k}\dfrac{d}{dH_k}\left(M_k(H_k)\overline{\aone}(H_k)\dfrac{df}{dH_k}\right)dH_k=
\dfrac{1}{2}p_k\cdot \lim\li_{H_k\ra 0}\dfrac{df}{dH_k}(k,H_k) \ ,$$
we can use our boundary condition (2.1) to have
$\play{\int_{[G]}Af(H(x))dx=0}$ and therefore $\E_{\nu}\al_0=0$.

From the fact that $\E_{\nu}\al_0=0$ one first derives that
$\sup\li_{x\in \gm}\E_x \al_n$ decays to $0$ exponentially fast and
therefore $\sup\li_{x\in \gm}\left|\sum\li_{n=0}^\infty \E_x
\al_n\right|<\infty$. It also gives, for $x\in \gm$, that, for
$\sm^*\in \{\sm_1,\sm_2,...\}$ we have

$$\left|\sum\li_{n=0}^\infty \E_x \al_n\right|\leq \|Af\|_{\infty}\cdot\E_x\sm^*+
\text{Var}(m_{\sm^*}^x-\nu)\cdot \sup\li_{x\in
\gm}\left|\sum\li_{n=0}^\infty \E_x \al_n\right| \ .$$

Using Lemma 3.1 we see that for any $\dt>0$ we can choose $\sm^*$
such that

$$\sup\li_{x\in \gm}\left|\sum\li_{n=0}^\infty \E_x \al_n\right|\leq \|Af\|_{\infty}\cdot\dt+
\dt\cdot \sup\li_{x\in \gm}\left|\sum\li_{n=0}^\infty \E_x
\al_n\right| \ , $$ which proves that $\play{\sup\li_{x\in
\gm}\left|\sum\li_{n=0}^\infty \E_x\al_n\right|}\ra 0$ uniformly in
$x\in \gm$ as $\ve \da 0$. This implies that $|(II)|\ra 0$ uniformly
in $x\in [G]$ as $\ve \da 0$ and Theorem 2.1 follows. $\square$

\

\section{Auxiliary results needed in the proof of Theorem 2.1}

We establish in this section all the auxiliary results needed in
Section 3 for the proof of Theorem 2.1.

Let us make some further geometric constructions. Since we assumed
that the closure of all these $U_k$'s do not intersect each other we
see that for sufficiently small neighborhoods of these $U_k$'s they
also do not intersect each other. Without loss of generality let us
speak about one of these $U_k$'s. We remind that the matrix
$\azero(x)=(\azero_{ij}(x))_{1\leq i,j\leq d}$ is non-negative
definite inside $[G]$ and has rank $d$ on $[G]\backslash
\cup_{k=1}^r[U_k]$ and rank $(d-1)$ on $\cup_{k=1}^r
([U_k]\backslash \{x_k(m_k)\})$. At the points $x_k(m_k)$ the matrix
$\azero(x)$ is just $0$. From our assumptions we see that
$\azero(x)\grad H_k=0$ on $C_k(h)$ and $\e\cdot (\azero(x)\e)\geq
\underline{a}(x)|\e|^2_{\R^d}$ for any unit vector $\e$ tangent to
$C_k(h)$. Here $\underline{a}(x)>0$ for $x\in C_k(h)$ and $h\in
(m_k,0]$. The eigenvalue $\lb(x)=0$ for $\azero(x)$, $x\in
C_k(0)=\gm_k$ is simple and is the smallest one in the spectrum of
$\azero(x)$. For $x\in \gm_k$ the family of eigen-polynomials
$p(\lb;x)=\det(\lb I-a^{(0)}(x))$ pass through the origin. They are
transversal (i.e. not tangent) to the axis $p=0$. The transversality
is preserved under a small perturbation. From here one can see that
the eigenvalue $\lb(x)$ will remain simple and is still the smallest
one in the spectrum of all the matrices $\azero(x)$ as $x$ belongs
to a small neighborhood of $\gm_k$. We then see from implicit
function theorem that this eigenvalue $\lb(x)$ is a
$\contfunc^{(3)}$ function in a small enough neighborhood of
$\gm_k$. As a consequence, the unit eigenvector $\e_d(x)$ (for
different $k$ it is different vector fields but for simplicity of
notation we ignore that $k$ in our notation) corresponding to this
smallest eigenvalue is a $\contfunc^{(3)}$ vector field in a
neighborhood of $\gm_k$, with $\e_d(x)=\dfrac{\grad H_k (x)}{|\grad
H_k (x)|_{\R^d}}$ for $x\in \gm_k$. Let $X^x(t)$ be the integral
curve of this vector field. We let $\dfrac{d
X^x(t)}{dt}=\e_d(X^x(t))$, $X^x(0)=x\in \gm_k$. As we are working
within a small neighborhood of $\gm_k$ and $\e_d(x)$ in this
neighborhood is a $\contfunc^{(3)}$ vector field, being transversal
to $\gm_k$ when $x\in \gm_k$, we see that for $t\in [0,
\overline{h}]$ with $\overline{h}$ sufficiently small the points
$X^x(t)$ for fixed $t$ and all $x\in \gm_k$ form a surface
$\contfunc^{(3)}$ diffeomorphic to $\gm_k$. In this way we obtain an
extension of $H_k$ to a neighborhood of $U_k$ by letting
$H_k(X^x(t))=t$ for $t\in [0, \overline{h}]$. The Euclidean distance
from a point $X^x(t)$ to $\gm_k$ is $\geq \underline{d} \cdot t$ for
some $\underline{d}>0$. Let us denote by $C_k(+t)$ the level surface
$\{H_k=+t\}$ for $t\in [0, \overline{h}]$. Let
$\gmout_k=C_k(+\ve^{1/4})$ and $\gmoout_k=C_k(+2\ve^{1/4})$. Let
$\gmout=\cup_{k=1}^r \gmout_k$ and $\gmoout=\cup_{k=1}^r \gmoout_k$.
We can take $\ve$ small such that all $\gmoout_k$'s do not intersect
each other and do not touch $\pt G$. We denote by $\cE(\ve^{1/4})$
those points of $x\in \cE$ which lie outside the union of the
neighborhoods of the $U_k$'s bounded by $\gmout_k$, and we denote
$\cE(2\ve^{1/4})$ in a similar way.

We shall denote, for $x\in \cE(\ve^{1/4})$, the stopping time
$\sm(\ve^{1/4})$ to be the time when $X_t^\ve$ first hits $\gmout$.
Notice that by our assumption, for a point $x\in \cE(\ve^{1/4})$ we
have $$\sum\li_{i,j=1}^d \azero_{ij}(x)\xi_i\xi_j\geq \text{const}
\cdot \ve^{1/2} \sum\li_{i=1}^d \xi_i^2$$ for all
$(\xi_1,...,\xi_d)\in \R^d$ and some $\text{const}>0$.

\

Within the rest of this section implied positive constants denoted
by $C_i$'s will not be explicitly pointed out unless necessary.
Also, sometimes we use the same symbol $C$ to denote different
implied positive constants which are not important.

\

\textbf{Lemma 4.1.} \textit{For any $0<\varkappa<1/2$, for any $\ve$
small enough we have
$$\sup\li_{x\in [\cE(\ve^{1/4})]}
\E_x\sm(\ve^{1/4})\leq C\ve^{1/2-\varkappa}$$ for some $C>0$.}

\

\textbf{Proof.} Let $u^\ve(x,t)=\Prob_x(\sm(\ve^{1/4})>t)$. Then
$u^\ve(x,t)$ solves the problem

$$\left\{\begin{array}{l}
\dfrac{\pt u^\ve}{\pt t}=\left(\dfrac{1}{\ve}L_0+L_1\right)u^\ve \ ,
\\
u^\ve(y,0)=1 \text{ for } y \in \cE(\ve^{1/4}) \ ,
\\
u^\ve(y,t)=0 \text{ for } y \in \gmout \text{ and } t>0 \ ,
\\
\dfrac{\pt u^\ve}{\pt \gm^\ve}(y,t)=0 \text{ for } y \in \pt G \ .
\end{array}\right.$$

Our argument is a slight modification of the standard estimates of
heat kernel temporal decay (see \cite{[Fabes-Strook 1986]}). We
first multiply the equation that $u^\ve$ satisfies and we integrate
with respect to $x$ in $\cE(\ve^{1/4})$:

$$\begin{array}{l}
\play{\dfrac{d}{dt}\int_{\cE(\ve^{1/4})}(u^\ve)^2dx}
\\
\play{=\int_{\cE(\ve^{1/4})}u^\ve \grad \cdot
\left[\left(\dfrac{1}{\ve}a^{(0)}(x)+a^{(1)}(x)\right)\grad
u^\ve\right]dx}
\\
\play{=\int_{\cE(\ve^{1/4})} \grad \cdot \left[ u^\ve
\left(\dfrac{1}{\ve}a^{(0)}(x)+a^{(1)}(x)\right)\grad
u^\ve\right]dx-\int_{\cE(\ve^{1/4})}\left<\left(\dfrac{a^{(0)}(x)}{\ve}+a^{(1)}(x)\right)\grad
u^\ve, \grad u^\ve\right>_{\R^d}dx}
\\
\play{=-\int_{\cE(\ve^{1/4})}\left<\left(\dfrac{a^{(0)}(x)}{\ve}+a^{(1)}(x)\right)\grad
u^\ve, \grad u^\ve\right>_{\R^d}dx} \ .
\end{array}$$

The last step in the above calculation makes use of our boundary
condition:

$$\begin{array}{l}
\play{\int_{\cE(\ve^{1/4})} \grad \cdot \left[ u^\ve
\left(\dfrac{1}{\ve}a^{(0)}(x)+a^{(1)}(x)\right)\grad
u^\ve\right]dx}
\\
\play{=\int_{\pt G}u^\ve
\left(\dfrac{1}{\ve}a^{(0)}(x)+a^{(1)}(x)\right)\grad u^\ve\cdot
\textbf{n} dS+\int_{\gmout}u^\ve
\left(\dfrac{1}{\ve}a^{(0)}(x)+a^{(1)}(x)\right)\grad u^\ve\cdot
\textbf{n} dS}
\\
=0 \ .
\end{array}$$

Therefore we see that we have, for some $C>0$, that

$$\dfrac{d}{dt}\int_{\cE(\ve^{1/4})}(u^\ve)^2dx\leq
-\dfrac{C}{\ve^{1/2}}\int_{\cE(\ve^{1/4})}|\grad u^\ve|^2dx\ .$$

We can extend the function $u^\ve$ to the whole domain of $G$ so
that $u^\ve=0$ on $\cE\backslash \cE(\ve^{1/4})$. We can then apply
a variant of Poincar\'{e} inequality (see \cite[Lemma 1]{[Lieb et
al]}) so that we have

$$\int_{\cE(\ve^{1/4})}|\grad
u^\ve|^2dx\geq C\left(\int_{\cE(\ve^{1/4})}(u^\ve)^qdx\right)^{1/q}
$$ for $1<q<\dfrac{2d}{d-2}$. The constant $C>0$ in the above inequality can be chosen
independent of $\ve$.

Now we use the H\"{o}lder inequality so that, for
$\dfrac{1}{\al}+\dfrac{1}{\bt}=1$, $\al>0$, $\bt>0$ we obtain

$$\int_{\cE(\ve^{1/4})}(u^\ve)^2dx\leq
\left(\int_{\cE(\ve^{1/4})}u^\ve dx\right)^{1/\al}
\left(\int_{\cE(\ve^{1/4})}(u^\ve)^{(2-1/\al)\bt}dx\right)^{1/\bt} \
.$$

Combining the above two inequalities we see that we have, for
$\al>\dfrac{d+2}{4}$, that

$$\int_{\cE(\ve^{1/4})}(u^\ve)^2dx\leq
C\left(\int_{\cE(\ve^{1/4})}u^\ve dx\right)^{1/\al}
\left(\int_{\cE(\ve^{1/4})}|\grad u^\ve|^2dx\right)^{1-1/2\al} \ .$$

Thus we have

$$\dfrac{d}{dt}\int_{\cE(\ve^{1/4})}(u^\ve)^2dx\leq
-\dfrac{C}{\ve^{1/2}}
\dfrac{\play{\left(\int_{\cE(\ve^{1/4})}(u^\ve)^2dx\right)^{\frac{2\al}{2\al-1}}}}
{\play{\left(\int_{\cE(\ve^{1/4})}u^\ve
dx\right)^{\frac{2}{2\al-1}}}} \ .$$

Integrating in $x$ the equation that $u^\ve$ satisfies and making
use of the parabolic maximum principle it is possible to see that we
have

$$\int_{\cE(\ve^{1/4})}u^\ve dx\leq C$$ for some $C>0$.

This shows that we have

$$\dfrac{d}{dt}\int_{\cE(\ve^{1/4})}(u^\ve)^2dx\leq
-\dfrac{C}{\ve^{1/2}}
\play{\left(\int_{\cE(\ve^{1/4})}(u^\ve)^2dx\right)^{\frac{2\al}{2\al-1}}}
\ .$$

We shall denote $S(t)$ the semigroup generated by the operator
$\dfrac{1}{\ve}L_0+L_1$ with the prescribed boundary conditions as
in the problem for $u^\ve$. From co-normal condition one can check
that $S(t)$ is self-adjoint. From the above inequality it is
standard to deduce
$$\|S(t)\|_{L^1(\cE(\ve^{1/4}))\ra L^2(\cE(\ve^{1/4}))}\leq \dfrac{C}{(\ve^{-1/2} t)^{q}}$$ for any
$q>d/4$. By self-adjointness of $S(t)$ we see that
$$\|S(t)\|_{L^2(\cE(\ve^{1/4}))\ra L^\infty(\cE(\ve^{1/4}))}\leq
\dfrac{C}{(\ve^{-1/2} t)^{q}} \ .$$ Since $S(t)=S(t/2)\circ S(t/2)$
we have $$\|S(t)\|_{L^1(\cE(\ve^{1/4}))\ra
L^\infty(\cE(\ve^{1/4}))}\leq \dfrac{C}{(\ve^{-1} t^2)^{q}}$$ for
any $q>d/4$. In particular, this means that we have
$$\sup\li_{x\in [\cE(\ve^{1/4})]}\Prob_x(\sm(\ve^{1/4})>\ve^{1/2-\kp})\leq C\ve^{2\kp q}$$
for $q>d/4$.

By strong Markov property of the process $X_t^\ve$ we see that
$$\begin{array}{l}
\E_x\sm(\ve^{1/4})
\\
=\play{\int_0^\infty \Prob_x(\sm(\ve^{1/4})>t)dt}
\\
\leq \ve^{1/2-\varkappa} \sum\li_{n=0}^\infty
\Prob_x(\sm(\ve^{1/4})>n\ve^{1/2-\varkappa})
\\
\leq \ve^{1/2-\varkappa} \sum\li_{n=0}^\infty (\sup\li_{x\in
[\cE(\ve^{1/4})]}\Prob_x(\sm(\ve^{1/4})>\ve^{1/2-\varkappa}))^n
\\
= \dfrac{\ve^{1/2-\varkappa}}{1-\sup\li_{x\in
[\cE(\ve^{1/4})]}\Prob_x(\sm(\ve^{1/4})>\ve^{1/2-\varkappa})} \leq
C\ve^{1/2-\varkappa}
\end{array}$$ for $\ve$ small enough. This implies the statement of the Lemma. $\square$

\

We shall denote by $\cS_k([0,\ve^{1/4}])$ the closed set bounded by
the surfaces $\gmout_k$ and $\gm_k$ and by
$\cS([0,\ve^{1/4}])=\cup_{k=1}^r\cS_k([0,\ve^{1/4}])$. We denote
$\cS_k([0,2\ve^{1/4}])$ and $\cS([0,2\ve^{1/4}])$ in a similar way
by replacing $\gmout_k$ by $\gmoout_k$.

Following the geometric construction stated before Lemma 4.1, for
$\ve>0$ small enough, and each $k=1,...,r$, at any point $x\in
\cS_k([0,2\ve^{1/4}])$ one can find $\e_d(x)=\dfrac{\grad
H_k(x)}{|\grad H_k(x)|_{\R^d}}$ and for any unite vector $\e$ such
that $\e\cdot \e_d(x)=0$ we have $\e\cdot (\azero(x)\e)\geq
\underline{a}|\e|_{\R^d}^2=\underline{a}$ for some
$\underline{a}>0$. Also $\azero(x)\e_d(x)=\lb(x)\e_d(x)$. The
eigenvalue $\lb(x)$ is in $\contfunc^{(3)}(\cS_k([0,2\ve^{1/4}]))$
with $\lb|_{\gm_k}=0$ and $\lb(x)>0$ for $x\in
\cS_k([0,2\ve^{1/4}])\backslash \gm_k$. Furthermore, for $\ve$ small
enough we have $C_1\cdot\text{dist}^2(x,\gm_k) \leq \lb(x)\leq
C_2\cdot\text{dist}^2(x,\gm_k)$ for some $C_1,C_2>0$ and $x\in
\cS([0,2\ve^{1/4}])$.

For each $x\in \cS_k([0, 2\ve^{1/4}])$, we can find a point
$\vphi^k(x)\in \gm_k$ such that $X^{\vphi^k(x)}(H_k(x))=x$ for the
flow $X^x(t)$ introduced in the geometric construction before Lemma
4.1. Let us introduce a local coordinate
$(\vphi^k_1,...,\vphi^k_{d-1}, H_k)$ in a small neighborhood of a
point $x\in \cS_k([0,2\ve^{1/4}])$. We take $H_k=H_k(x)$, which is
the extended first integral of $H_k$ to $\cS_k([0,2\ve^{1/4}])$; and
we take
$(\vphi^k_1,...,\vphi^k_{d-1})=(\vphi^k_1(x),...,\vphi^k_{d-1}(x))$
to be the local coordinate for the point $\vphi^k(x)$ on $\gm_k$. In
the more or less simpler case we can arrange the coordinate
$(\vphi^k_1(x),...,\vphi^k_{d-1}(x), H_k(x))$ in such a way that
this new coordinate system is an orthogonal curvilinear coordinate
system. (We will discuss the general case a bit later.) The metric
tensor of this new coordinate system is given by
$ds^2=E_1(x)(d\vphi^k_1(x))^2+...+E_{d-1}(x)(d\vphi^k_{d-1}(x))^2+E_d(x)(dH_k(x))^2$.
Here the functions $0<C_3<E_1(x),...,E_d(x)<C_4<\infty$ are in class
$\contfunc^{(3)}(\cS_k([0,2\ve^{1/4}]))$ with bounded derivatives.
We notice that by our geometric construction we have $C_5\cdot
H_k^2(x) \leq \lb(x)\leq C_6\cdot H_k^2(x)$. We shall let $\e_i(x)$
be the unit tangent vector on the axis curve corresponding to
$\vphi_i^k$ for $1\leq i\leq d-1$; $\e_d(x)$ be the unit tangent
vector on the axis curve corresponding to $H_k$. The vectors
$\e_1(x),...,\e_d(x)$ are basis vectors.

The theory of orthogonal curvilinear coordinate system (see, for
example, \cite[Ch.14]{[Zorich]}) tells us that for a differentiable
function $f$ on $\cS_k([0,2\ve^{1/4}])$ we have

$$\grad f (x)=\sum\li_{i=1}^{d-1}\dfrac{1}{\sqrt{E_i(x)}}\dfrac{\pt f}{\pt \vphi^k_i}(x)\e_i(x)
+\dfrac{1}{\sqrt{E_d(x)}}\dfrac{\pt f}{\pt H_k}(x) \e_d(x) \ ,$$ and
for a differentiable vector field $\mathbf{B}(x)=\sum\li_{i=1}^d
B^i(x)\e_i(x)$ on $\cS_k([0,2\ve^{1/4}])$ we have

$$\grad \cdot \mathbf{B}(x)= \dfrac{1}{\sqrt{\prod_{i=1}^d E_i(x)}}\left[\sum\li_{i=1}^{d-1}
\dfrac{\pt}{\pt \vphi_i^k}\left(\sqrt{\dfrac{\prod_{j=1}^d
E_j(x)}{E_i(x)}}B^i(x)\right)+\dfrac{\pt}{\pt
H_k}\left(\sqrt{\dfrac{\prod_{j=1}^d
E_j(x)}{E_d(x)}}B^d(x)\right)\right] \ .$$

Consider a function (so called "barrier function", see
\cite{[Hasminskii]} and \cite[Ch.3]{[F red book]}) $u_k(x)\in
\contfunc^{(2)}(\cS_k([0,2\ve^{1/4}]))$ which depends only on $H_k$
and is a constant on each level surface $\{H_k=\text{const}\}$. We
can write $u_k(x)=u_k(H_k)$ and we apply the above two formulas to
get

$$\begin{array}{l}
\left(\dfrac{1}{\ve}L_0+L_1\right)u_k(x)
\\
=\left(\dfrac{1}{2\ve}\grad \cdot (\azero(x)\grad
u_k(x))+\dfrac{1}{2}\grad \cdot (\aone(x)\grad u_k(x))\right)
\\
=\dfrac{1}{\sqrt{\prod_{i=1}^d
E_i(x)}}\left[\dfrac{1}{2}\dfrac{\pt}{\pt
H_k}\left(\sqrt{\dfrac{\prod_{i=1}^d
E_i(x)}{E_d^2(x)}}\left(\dfrac{\lb(x)}{\ve}+\mu_d(x)\right)\dfrac{du_k}{dH_k}(H_k)\right)\right.
\\
\left. \ \ \ \ \ +\sum\li_{i=1}^{d-1}\dfrac{1}{2}\dfrac{\pt}{\pt
\vphi_i^k}\left(\sqrt{\dfrac{\prod_{i=1}^d
E_i(x)}{E_d(x)E_i(x)}}\mu_i(x)\right)\cdot
\dfrac{du_k}{dH_k}(H_k)\right] \ .
\end{array}$$

Here the functions $\mu_1(x),...,\mu_d(x)$ are defined via the
relation $\aone(x) \e_d(x)=\mu_1(x)\e_1(x)+...+\mu_d(x)\e_d(x)$.
These functions are in $\contfunc^{(3)}(\cS([0,2\ve^{1/4}]))$ with
bounded derivatives. Notice that since $L_1$ is strictly elliptic,
the matrix $\aone(x)$ is positive definite, and therefore the
function $\mu_d(x)$ is uniformly bounded from below by a certain
positive constant.

For simplicity of notation let us define $A(x)=\sqrt{\prod_{i=1}^d
E_i(x)}$ and $A_i(x)=\dfrac{A(x)}{\sqrt{E_i(x)E_d(x)}}$ for
$i=1,...,d$. These functions are strictly positive (with uniform
lower bound) in $\contfunc^{(2)}(\cS([0,2\ve^{1/4}]))$ with bounded
derivatives. Under this notation we can write

$$\begin{array}{l}
\left(\dfrac{1}{\ve}L_0+L_1\right)u_k(x)
\\
=\dfrac{1}{A(x)}\left[\dfrac{1}{2}\dfrac{\pt}{\pt
H_k}\left(A_d(x)\left(\dfrac{\lb(x)}{\ve}+\mu_d(x)\right)\dfrac{du_k}{dH_k}(H_k)\right)
+\sum\li_{i=1}^{d-1}\dfrac{1}{2}\dfrac{\pt}{\pt
\vphi_i^k}\left(A_i(x)\mu_i(x)\right)\cdot
\dfrac{du_k}{dH_k}(H_k)\right] \ .
\end{array}$$

As a further simplification we shall define
$$\dfrac{1}{2}A_d(x)\lb(x)=K_1(x) \ ,$$
$$\dfrac{1}{2}A_d(x)\mu_d(x)=K_2(x) \ ,$$
$$\sum\li_{i=1}^{d-1}\dfrac{1}{2}\dfrac{\pt}{\pt
\vphi_i^k}(A_i(x)\mu_i(x))=K_3(x) \ .$$ We have
$$\play{\left(\dfrac{1}{\ve}L_0+L_1\right)u_k(x)=\dfrac{1}{A(x)}\left[\dfrac{\pt}{\pt
H_k}\left(\left(\dfrac{K_1(x)}{\ve}+K_2(x)\right)\dfrac{du_k}{dH_k}(H_k)\right)+K_3(x)\dfrac{du_k}{dH_k}(H_k)\right]}
\ . \eqno(4.1)$$

For a point $x\in \cS_k([0,2\ve^{1/4}])$ and $\ve$ small enough we
have $$C_7 H_k^2(x)\leq K_1(x)\leq C_8 H_k^2(x) \ ; \eqno(4.2)$$
$$C_9 H_k(x)\leq \dfrac{\pt}{\pt H_k}(K_1(x))\leq C_{10}H_k(x) \ ; \eqno(4.3)$$
$$0<C_{11}<K_2(x)<C_{12}<\infty \ ; \eqno(4.4)$$ $$\left|\dfrac{\pt}{\pt
H_k}(K_2(x))\right|\leq C_{13}<\infty  \ ; \eqno(4.5)$$
$$|K_3(x)|\leq C_{14}<\infty \ . \eqno(4.6)$$

We also notice that since we are working in $\cS_k([0,2\ve^{1/4}])$
and $\ve$ is small, the functions $A_d(x)=A_d(\vphi^k,H_k)$ and
$\lb(x)=\lb(\vphi^k,H_k)$ have Taylor expansions
$$A_d(\vphi^k,H_k)=A_d(\vphi^k,0)+O(H_k)\ ,$$
$$\lb(\vphi^k,H_k)=\dfrac{1}{2}\dfrac{\pt^2 \lb}{\pt H_k^2}(\vphi^k,0)H_k^2+O(H_k^3)\ .$$

Therefore we see that for $x\in \cS_k([0,2\ve^{1/4}])$ we have

$$K_1(x)=C_k(\vphi^k)H_k^2+O(H_k^3) \eqno(4.7)$$
with a certain positive function $C_k(\vphi^k)$.

\

In the general case the axis curve corresponding to $H_k$ will be
orthogonal to those corresponding to the $\vphi_i^k$'s, but the axis
curves corresponding to the $\vphi_i^k$'s are not necessarily
orthogonal. The calculation will be more bulky since the metric
tensor have cross terms with respect to the coordinate
$\vphi_i^k$'s, but the essence is the same as it is only important
to have the axis curves corresponding to $H_k$ being orthogonal to
those corresponding to the $\vphi_i^k$'s. To be more precise, let
$(g_{ij})_{1\leq i,j\leq d}$ be the metric tensor corresponding to
the (local) coordinate system $(\vphi_1^k,...,\vphi_{d-1}^k, H_k)$.
Let $\e_i(x)$ be the unit tangent vector on the axis curve
corresponding to $\vphi_i^k$ for $1\leq i\leq d-1$; $\e_d(x)$ be the
unit tangent vector on the axis curve corresponding to $H_k$. The
vectors $\e_1(x),...,\e_d(x)$ are basis vectors. We have
$g_{id}=g_{di}=0$ for $i=1,...,d-1$ and $g_{dd}>0$. Let
$(g^{ij})_{1\leq i,j \leq d}$ be the dual tensor, i.e.,
$(g^{ij})_{1\leq i,j\leq d}$ is the inverse matrix of
$(g_{ij})_{1\leq i,j\leq d}$. We have $g^{id}=g^{di}=0$ for $1\leq
i\leq d-1$ and $g^{dd}=\dfrac{1}{g_{dd}}$. For $u_k=u_k(H_k)$ we
have

$$\grad u_k(x)=\dfrac{1}{\sqrt{g_{dd}(x)}}\dfrac{du_k}{dH_k}\e_d(x) \ ,$$
and
$$\azero(x)\grad u_k(x)=\dfrac{\lb(x)}{\sqrt{g_{dd}(x)}}\dfrac{du_k}{dH_k}\e_d(x) \ ;$$
$$\aone(x)\grad u_k(x)=\dfrac{\mu_d(x)}{\sqrt{g_{dd}(x)}}\dfrac{du_k}{dH_k}\e_d(x)+
\dfrac{1}{\sqrt{g_{dd}(x)}}\dfrac{du_k}{dH_k}(\mu_1(x)\e_1(x)+...+\mu_{d-1}(x)\e_{d-1}(x))
\ .$$

Here, as before, we have
$\aone(x)\e_d(x)=\mu_1(x)\e_1(x)+...+\mu_d(x)\e_d(x)$. We shall then
apply a general formula that for a vector field
$\mathbf{B}(x)=\sum\li_{i=1}^dB^i(x)\e^i(x)$ we have $$\grad \cdot
\mathbf{B}(x)=\dfrac{1}{\sqrt{g(x)}}\sum\li_{i=1}^{d-1}\dfrac{\pt}{\pt
\vphi_i^k}(B^i(x)\sqrt{g^{ii}(x)}\sqrt{g(x)})+\dfrac{1}{\sqrt{g(x)}}\dfrac{\pt}{\pt
H_k}(B^d(x)\sqrt{g^{dd}(x)}\sqrt{g(x)}) \ .$$ Here
$g(x)=\det(g_{ij}(x))$. The basis $\e^1(x),...,\e^d(x)$ is the
reciprocal basis (normalized) dual to $\e_1(x),...,\e_d(x)$, i.e.,
$(\e_i,\e^j)_{(g_{ij})}=\dt_{ij}$ with respect to the inner product
$(\bullet,\bullet)_{(g_{ij})}$ defined by the metric tensor
$(g_{ij})$. By the fact that the metric tensor has no cross terms
between $H_k$ and $\vphi_i^k$'s, we actually have $\e^d(x)=\e_d(x)$
and
$\text{span}\{\e_1(x),...,\e_{d-1}(x)\}=\text{span}\{\e^1(x),...,\e^{d-1}(x)\}$.

We then see that the operator $\dfrac{1}{\ve}L_0+L_1$ applied to
$u_k(x)=u_k(H_k)$ will result in a formula which is the same as
(4.1). The functions $K_1(x)$, $K_2(x)$ and $K_3(x)$ will somehow be
different but they still satisfy the conditions (4.2)\,--\,(4.7).

\

Let $\zt([0,2\ve^{1/4}])$ be the first time when the process
$X_t^\ve$, starting from a point $x\in \cS([0,2\ve^{1/4}])$, hits
$\gm$ or $\gmoout$.

\

\textbf{Lemma 4.2.} \textit{We have }
$$\sup\li_{x\in
\cS([0,2\ve^{1/4}])} \E_x\zt([0,2\ve^{1/4}])\leq \ C\ve^{3/4}
$$ \textit{for some $C>0$.}

\

\textbf{Proof.} Let $$K_4(H_k)=\dfrac{H_k^2}{\ve}+1 \ $$ and
$$K_5(x)=\dfrac{1}{K_4(H_k(x))}\left(\dfrac{K_1(x)}{\ve}+K_2(x)\right) \ .$$

By (4.2) and (4.4) we can estimate
$$C_{15}\leq K_5(x)\leq
C_{16} \eqno(4.8)$$ for $C_{15}, C_{16}>0$, $x\in \cS_k([0,
2\ve^{1/4}])$ and $\ve$ small enough.

By (4.7) we see that

$$\begin{array}{l}
K_5(x)
\\
\play{=\dfrac{K_1(x)+\ve K_2(x)}{H_k^2+\ve}}
\\
\play{=\dfrac{C_k(\vphi^k(x))H_k^2+O(H_k^3)+\ve K_2(x)}{H_k^2+\ve}}
\\
\play{=C_k(\vphi^k(x))+H_k\cdot\dfrac{O(H_k^2)+\dfrac{\ve}{H_k}
(K_2-C_k(\vphi^k(x)))}{H_k^2+\ve}} \ .
\end{array}$$

Since $0\leq H_k\leq 2\ve^{1/4}$ we see that as $\ve$ is small we
have

$$\left|\dfrac{O(H_k^2)+\dfrac{\ve}{H_k}
(K_2-C_k(\vphi^k(x)))}{H_k^2+\ve}\right|\leq C_{17} \ .$$

Thus as $0\leq H_k\leq 2\ve^{1/4}$ we see that
$$\left|\dfrac{\pt}{\pt H_k}(K_5(x))\right|\leq C_{17} \ .$$

Let the barrier function $u_k^{(1)}(x)=u_k^{(1)}(H_k)$ be defined by

$$u_k^{(1)}(H_k)=\int_0^{H_k}\dfrac{K_6(\ve)-y}{K_4(y)}dy$$ with
$$K_6(\ve)=\left(\int_0^{2\ve^{1/4}}\dfrac{dy}{K_4(y)}\right)^{-1}
\left(\int_0^{2\ve^{1/4}}\dfrac{y dy}{K_4(y)}\right) \ .$$

It is easy to check that $$u_k^{(1)}(0)=u_k^{(1)}(2\ve^{1/4})=0 \
.$$

We can estimate $K_6(\ve)\leq 2\ve^{1/4}$ and we have

$$\play{\int_0^{H_k}\dfrac{dy}{K_4(y)}}=\int_{0}^{H_k}\dfrac{dy}{\dfrac{y^2}{\ve}+1}
=\ve^{1/2}\arctan(H_k\ve^{-1/2})\leq C_{18}\ve^{1/2} \ .
\eqno(4.9)$$

This gives the estimates $$0\leq u_k^{(1)}(H_k)\leq C_{19}\ve^{3/4}
\eqno(4.10)$$ and $$\left|\dfrac{du_k^{(1)}}{dH_k}(H_k)\right|\leq
C_{20}\ve^{1/4} \eqno(4.11)$$ for $0\leq H_k\leq 2\ve^{1/4}$. Apply
(4.1) to the function $u_k^{(1)}$ we can see, using (4.11), that,

$$\begin{array}{l}
\play{\left(\dfrac{1}{\ve}L_0+L_1\right)u_k^{(1)}(x)}
\\
\play{=\dfrac{1}{A(x)}\left[\dfrac{\pt}{\pt
H_k}\left(K_5(x)K_4(H_k(x))\dfrac{du_k^{(1)}}{dH_k}(H_k(x))\right)+K_3(x)\dfrac{du_k^{(1)}}{dH_k}(H_k(x))\right]}
\\
\play{\leq\dfrac{1}{A(x)}\left[\dfrac{\pt}{\pt
H_k}\left((K_6(\ve)-H_k(x))K_5(x)\right)+C_{21}\ve^{1/4}\right]}
\\
\play{=\dfrac{1}{A(x)}\left[-K_5(x)+\dfrac{\pt}{\pt H_k}(
K_5(x))(K_6(\ve)-H_k(x))+C_{21}\ve^{1/4}\right]\leq -C_{22}}
\end{array} \eqno(4.12)$$
for $x\in \cS([0,2\ve^{1/4}])$ and $\ve$ small enough.

We notice that this process $X_t^\ve$ before hitting $\gm$ or
$\gmoout$ is restricted to one of the $\cS_k([0,2\ve^{1/4}])$'s and
the bound (4.12) can be made uniform in $k$.

Now we apply It\^{o}'s formula to the function $u_k^{(1)}$
constructed above up to the stopping time $\zt([0,2\ve^{1/4}])$.
Taking expectation we get
$$u_k^{(1)}(x)=-\E_x\play{\int_0^{\zt([0,2\ve^{1/4}])}\left(\dfrac{1}{\ve}L_0+L_1\right)u_k^{(1)}
(X_s^\ve) ds}\geq C_{22}\E_x\zt([0,2\ve^{1/4}]) \ .\eqno(4.13) $$
From (4.10) and (4.13) we see that the statement of this Lemma
follows. $\square$

\

\textbf{Lemma 4.3.} \textit{For $x\in \gmout$ we have}
$$\Prob_x(X_{\zt([0,2\ve^{1/4}])}^\ve\in \gm)\geq C\ve^{1/4}$$ \textit{for some $C>0$.}

\

\textbf{Proof.} Let

$$K_7(H_k)=\max\li_{x\in \cS_k([0,2\ve^{1/4}]),
H_k(x)=H_k}\dfrac{\dfrac{\pt}{\pt
H_k}\left(\dfrac{K_1(x)}{\ve}+K_2(x)\right)+K_3(x)}{\dfrac{K_1(x)}{\ve}+K_2(x)}
\ .$$

Let, for a fixed $H_k\in[0,2\ve^{1/4}]$, the above maximum be
achieved at a point $(\vphi^k,H_k)$.

We will prove in Lemma 4.4 that we have the auxiliary estimate

$$\left|K_7(H_k)-\dfrac{2C_k(\vphi^k)H_k}{C_k(\vphi^k)H_k^2+\ve K_2(\vphi^k, H_k)}\right|\leq C_{23} \ . \eqno(4.14)$$

Let the barrier function $u_k^{(2)}(x)=u_k^{(2)}(H_k)$ be defined by

$$u_k^{(2)}(H_k)=1-\dfrac{\play{\int_0^{H_k}\exp\left(-\int_0^y
K_7(z)dz\right)dy}}{\play{\int_0^{2\ve^{1/4}}\exp\left(-\int_0^y
K_7(z)dz\right)dy}} \ .$$

It is easy to see that we have
$$u_k^{(2)}(0)=1 \ , \ u_k^{(2)}(2\ve^{1/4})=0 \ .$$

Apply formula (4.1) we can see that

$$\begin{array}{l}
\play{\left(\dfrac{1}{\ve}L_0+L_1\right)u_k^{(2)}(x)}
\\
\play{=\dfrac{1}{A(x)}\left[\left(\dfrac{K_1(x)}{\ve}+K_2(x)\right)\dfrac{d^2u_k^{(2)}}{dH_k^2}(H_k)
+\left(\dfrac{\pt}{\pt
H_k}\left(\dfrac{K_1(x)}{\ve}+K_2(x)\right)+K_3(x)\right)\dfrac{du_k^{(2)}}{dH_k}(H_k)\right]}
\\
\geq 0 \ .
\end{array} \eqno(4.15)$$

However, by (4.14), (4.4) and the property of $C_k(\vphi^k)$ in
(4.7) we can estimate

$$\int_{H_k}^{2\ve^{1/4}}\exp\left(-\int_0^y K_7(z)dz\right)dy\geq C_{24}
\int_{H_k}^{2\ve^{1/4}}\exp\left(-\int_0^y
\dfrac{2z}{z^2+C_{25}\ve}dz\right)dy \ , \eqno(4.16)$$

$$\int_0^{2\ve^{1/4}}\exp\left(-\int_0^y K_7(z)dz\right)dy\leq C_{26}
\int_0^{2\ve^{1/4}}\exp\left(-\int_0^y
\dfrac{2z}{z^2+C_{27}\ve}dz\right)dy \ . \eqno(4.17)$$

By (4.16) and (4.17), and taking into account the auxiliary
estimates made in Lemma 4.5 we see that
$$u_k^{(2)}(\ve^{1/4})\geq C_{28}\ve^{1/4} \ . \eqno(4.18)$$

This bound (4.18) can actually be made uniform in $k$. We can apply
It\^{o}'s formula to the function $u_k^{(2)}$ constructed above up
to the stopping time $\zt([0,2\ve^{1/4}])$. Taking expectation we
get
$$\Prob_x(X_{\zt([0,2\ve^{1/4}])}^\ve\in
\gm)-u_k^{(2)}(\ve^{1/4})=\E_x\play{\int_0^{\zt([0,2\ve^{1/4}])}\left(\dfrac{1}{\ve}L_0+L_1\right)u_k^{(2)}
(X_s^\ve) ds}\geq 0 \  \eqno(4.19)$$ for $x\in \gmout$. Now (4.18)
and (4.19) imply the statement of this Lemma. $\square$

\

\textbf{Lemma 4.4.} \textit{For a fixed $H_k\in [0,2\ve^{1/4}]$ and
the corresponding $\vphi^k$ defined as in the proof of Lemma 4.3, we
have}

$$\left|K_7(H_k)-\dfrac{2C_k(\vphi^k)H_k}{C_k(\vphi^k)H_k^2+\ve K_2(\vphi^k, H_k)}\right|\leq C$$
\textit{for some $C>0$.}

\

\textbf{Proof.} Using (4.7), we can write

$$K_7(H_k)=\dfrac{2C_k(\vphi^k)H_k+O(H_k^2)+
\ve K_{2,3}(\vphi^k,H_k)}{C_k(\vphi^k)H_k^2+O(H_k^3)+\ve
K_2(\vphi^k,H_k)} \ .$$

Here $K_{2,3}(\vphi^k,H_k)$ is a bounded function. We then have

$$\begin{array}{l}
\play{\left|K_7(H_k)-\dfrac{2C_k(\vphi^k)H_k}{C_k(\vphi^k)H_k^2+\ve
K_2(\vphi^k, H_k)}\right|}
\\
= \play{\left|\dfrac{2C_k(\vphi^k)H_k+O(H_k^2)+ \ve
K_{2,3}(\vphi^k,H_k)}{C_k(\vphi^k)H_k^2+O(H_k^3)+\ve
K_2(\vphi^k,H_k)}-\dfrac{2C_k(\vphi^k)H_k}{C_k(\vphi^k)H_k^2+\ve
K_2(\vphi^k, H_k)}\right|}
\\
\leq \play{\left|\dfrac{O(H_k^2)+\ve
K_{2,3}(\vphi^k,H_k)}{C_k(\vphi^k)H_k^2+O(H_k^3)+\ve
K_2(\vphi^k,H_k)}\right|+}
\\
\play{\ \ \ \ \ \ \
+\left|\dfrac{2C_k(\vphi^k)H_k}{C_k(\vphi^k)H_k^2+O(H_k^3)+\ve
K_2(\vphi^k,H_k)}-\dfrac{2C_k(\vphi^k)H_k}{C_k(\vphi^k)H_k^2+\ve
K_2(\vphi^k,H_k)}\right|}
\\
= \play{\left|\dfrac{O(H_k^2)+\ve
K_{2,3}(\vphi^k,H_k)}{C_k(\vphi^k)H_k^2+O(H_k^3)+\ve
K_2(\vphi^k,H_k)}\right|+}
\\
\play{\ \ \ \ \ \ \
+\left|\dfrac{2C_k(\vphi^k)H_k}{C_k(\vphi^k)H_k^2+\ve
K_2(\vphi^k,H_k)}\cdot
\dfrac{O(H_k^3)}{C_k(\vphi^k)H_k^2+O(H_k^3)+\ve
K_2(\vphi^k,H_k)}\right|\leq C \ .}
\end{array}$$ $\square$

\

\textbf{Lemma 4.5.} \textit{We have}

$$\int_{\ve^{1/4}}^{2\ve^{1/4}}\exp\left(-\int_0^y \dfrac{2z}{z^2+C\ve}dz\right)dy\geq C_{29}\ve^{3/4} \ ,$$

$$C_{31}\ve^{1/2}\geq \int_{0}^{C_{32}\ve^{1/4}}\exp\left(-\int_0^y \dfrac{2z}{z^2+C\ve}dz\right)dy\geq C_{30}\ve^{1/2} \ .$$

\

\textbf{Proof.} Evaluating the integrals, we have

$$\int_0^y \dfrac{2z}{z^2+C\ve}dz=\ln\left(\dfrac{y^2+C\ve}{C\ve}\right) \ ,$$

$$\int_a^b \exp\left(-\int_0^y\dfrac{2z}{z^2+C\ve}dz\right)dy=
\sqrt{C}\ve^{1/2}\left(\arctan(\dfrac{b}{\sqrt{C}\ve^{1/2}})-\arctan(\dfrac{a}{\sqrt{C}\ve^{1/2}})\right)
\ .$$

If $a=0$ and $b=C_{32}\ve^{1/4}$ we already get the second
inequality of this Lemma. Now suppose $a=\ve^{1/4}$ and
$b=2\ve^{1/4}$. We shall make use of an asymptotic expansion of
$\arctan(y)$ as $y\ra \infty$:

$$\arctan (y)=\dfrac{\pi}{2}-\dfrac{1}{y}+O(\dfrac{1}{y^2}) \ \text{ as } \ y \ra \infty \ .$$

This gives

$$\begin{array}{l}
\play{\int_{\ve^{1/4}}^{2\ve^{1/4}}\exp\left(-\int_0^y
\dfrac{2z}{z^2+C\ve}dz\right)dy}
\\
\geq
C_{33}\ve^{1/2}\left(-\dfrac{1}{2\ve^{-1/4}}+\dfrac{1}{\ve^{-1/4}}+O(\ve^{1/2})\right)\geq
C_{29}\ve^{3/4} \ .
\end{array}$$ $\square$

\

\textbf{Lemma 4.6.} \textit{We have}

$$\lim\li_{\ve\da 0}\sup\li_{x\in \gmout}\E_x\sm=0 \ .$$

\

\textbf{Proof.} Lemmas 4.1, 4.2 and 4.3 imply the statement of this
Lemma. For $x\in \gmout$ we have

$$\begin{array}{l}
\E_x\sm
\\
=\E_x\zt([0,2\ve^{1/4}])\1(X_{\zt([0,2\ve^{1/4}])}^\ve\in
\gm)+\E_x(\zt([0,2\ve^{1/4}])+\E_{X_{\zt([0,2\ve^{1/4}])}^\ve}\sm)\1(X_{\zt([0,2\ve^{1/4}])}^\ve\in
\gmoout)
\\
\leq \sup\li_{x\in
\cS([0,2\ve^{1/4}])}\E_x\zt([0,2\ve^{1/4}])+\E_x(\sup\li_{y\in
\gmoout}\E_y \sm(\ve^{1/4})+\sup\li_{x\in \gmout}\E_x
\sm)\1(X_{\zt([0,2\ve^{1/4}])}^\ve\in \gmoout)
\\
\leq \sup\li_{x\in
\cS([0,2\ve^{1/4}])}\E_x\zt([0,2\ve^{1/4}])+(\sup\li_{y\in
\gmoout}\E_y \sm(\ve^{1/4})+\sup\li_{x\in \gmout}\E_x
\sm)\Prob_x(X_{\zt([0,2\ve^{1/4}])}^\ve\in \gmoout) \ .
\end{array} \eqno(4.20) $$

Taking a sup over all $x\in \gmout$ we get

$$\sup\li_{x\in\gmout}\E_x\sm\leq \dfrac{\sup\li_{x\in
\cS([0,2\ve^{1/4}])}\E_x\zt([0,2\ve^{1/4}])+\sup\li_{y\in
\gmoout}\E_y \sm(\ve^{1/4})\cdot
\Prob_x(X_{\zt([0,2\ve^{1/4}])}^\ve\in
\gmoout)}{\Prob_x(X_{\zt([0,2\ve^{1/4}])}^\ve\in \gm)}\ . $$

Using Lemmas 4.1, 4.2 and 4.3 we see that the statement of this
Lemma follows. (We choose $\varkappa=1/8$ in Lemma 4.1.) $\square$

\

\textbf{Lemma 4.7.} \textit{We have}

$$\lim\li_{\ve\da 0}\sup\li_{x\in \cS([0,\ve^{1/4}])}\E_x\sm=0 \ .$$

\

\textbf{Proof.} This is a consequence of Lemmas 4.1, 4.2 and 4.6.
$\square$

\

\textbf{Lemma 4.8.} \textit{We have}

$$\lim\li_{\ve\da 0}\sup\li_{x\in [\cE]}\E_x\sm=0 \ .$$

\

\textbf{Proof.} This is a consequence of Lemmas 4.1 and 4.7.
$\square$

\

We shall denote by $\cS_k([-\ve^{1/2},\ve^{1/4}])$ the closed set
bounded by the surfaces $\gmin_k$ and $\gmout_k$ and by
$\cS([-\ve^{1/2},\ve^{1/4}])=\cup_{k=1}^r\cS_k([-\ve^{1/2},\ve^{1/4}])$.
We notice that by the same reason as before, a coordinate
$(\vphi_1^k,...,\vphi_{d-1}^k, H_k)$ exists in
$\cS_k([-\ve^{1/2},\ve^{1/4}])$. We denote $\cS_k([0,\ve^{1/4}])$,
$\cS([0,\ve^{1/4}])$ (replacing $\gmin_k$ by $\gm_k$) and
$\cS_k([-\ve^{1/2},0])$, $\cS([-\ve^{1/2},0])$ (replacing $\gmout_k$
by $\gm_k$) in a similar way.

\

\textbf{Lemma 4.9.} \textit{We have $$\lim\li_{\ve\da
0}\sup\li_{x\in \gm} \E_x\tau=0 \ .$$}

\

\textbf{Proof.} The proof of this lemma is very similar to and is a
bit simpler than that of Lemma 4.6. We shall construct two barrier
functions $u_k^{(3)}$ (for estimating the exit time from
$\cS_k([-\ve^{-1/2},\ve^{1/4}])$) and $u_k^{(4)}$ (for the
probability of hitting $\gmin$).

For the construction of $u_k^{(3)}$ all the arguments of Lemma 4.2
can be carried here with $\gm$ replaced by $\gmin$, $\gmout$
replaced by $\gm$ and $\gmoout$ replaced by $\gmout$. We are working
now with $\cS_k([-\ve^{1/2},\ve^{1/4}])$ and $H_k\in [-\ve^{1/2},
\ve^{1/4}]$. We apply formula (4.1) with the change of the estimates
(4.2)\,--\,(4.6) as follows: when $x\in \cS_k([0,\ve^{1/4}])$ there
is no change in the estimates; when $x\in \cS_k([-\ve^{1/2},0])$ we
replace (4.2) and (4.3) by $K_1(x)=\dfrac{\pt}{\pt H_k}(K_1(x))=0$
and (4.4)\,--\,(4.6) remain the same. The function $K_4(x)=1$ and
$K_5(x)$ is then defined in a same way as in Lemma 4.2 with an
estimate $0<C_{34}\leq K_5(x)\leq C_{35}<\infty$ for $x\in
\cS_k([-\ve^{1/2},0])$. Again we let
$$u_k^{(3)}(H_k)=\int_{-\ve^{1/2}}^{H_k}\dfrac{K_6(\ve)-y}{K_4(y)}dy$$
with
$$K_6(\ve)=\left(\int_{-\ve^{1/2}}^{\ve^{1/4}}\dfrac{dy}{K_4(y)}\right)^{-1}
\left(\int_{-\ve^{1/2}}^{\ve^{1/4}}\dfrac{ydy}{K_4(y)}\right) \ .$$

It is then checked that $u_k(-\ve^{1/2})=u_k(\ve^{1/4})=0$ and
$K_6(\ve)\leq 2\ve^{1/4}$. The estimate (4.9) is still working for
$H_k\in [-\ve^{1/2},\ve^{1/4}]$. The estimates (4.10), (4.11) and
(4.12) are still working. Let $\zt([-\ve^{1/2},\ve^{1/4}])$ be the
first time when the process $X_t^\ve$ starting from a point $x\in
\gm$, first hits $\gmout$ or $\gmin$. A similar statement of (4.13)
is then obtained. We have

$$\lim\li_{\ve \da 0}\sup\li_{x\in \cS([-\ve^{1/2},\ve^{1/4}])}\E_x
\zt([-\ve^{1/2},\ve^{1/4}])=0 \ . \eqno(4.21)$$

The estimate of the hitting probability is a bit simpler. We
construct a barrier function $u_k^{(4)}$ similarly as in Lemma 4.3.
The function $K_7(H_k)$ is defined as in Lemma 4.3. But now we have
the property that $|K_7(H_k)|\leq C$ for $H_k\in [-\ve^{1/2},0]$. We
let
$$u_k^{(4)}(H_k)=1-\dfrac{\play{\int_{-\ve^{1/2}}^{H_k}\exp\left(-\int_{-\ve^{1/2}}^y
K_7(z)dz\right)dy}}{\play{\int_{-\ve^{1/2}}^{\ve^{1/4}}\exp\left(-\int_{-\ve^{1/2}}^y
K_7(z)dz\right)dy}} \ .$$ We have $u_k^{(4)}(-\ve^{1/2})=1$,
$u_k^{(4)}(\ve^{1/4})=0$. We have

$$C_{36}\ve^{1/2}\leq
\int_{-\ve^{1/2}}^0\exp\left(-\int_{-\ve^{1/2}}^y
K_7(z)dz\right)dy\leq C_{37}\ve^{1/2} \ , $$ and by the second
inequality in Lemma 4.5 we see that

$$C_{38}\ve^{1/2}\leq
\int_{0}^{\ve^{1/4}}\exp\left(-\int_{-\ve^{1/2}}^y K_7(z)dz\right)dy
\leq C_{39}\ve^{1/2} \ .$$

These estimates ensure that an analogue of (4.19) works, but the
lower bound is a positive constant, and hence situation is a bit
simpler. We have

$$1\geq \Prob_x(X_{\zt([-\ve^{1/2},\ve^{1/4}])}^\ve\in \gmin)\geq u_k^{(4)}(0)\geq C_{40}>0 \ . \eqno(4.22)$$
uniformly in $x\in\gm$. The results (4.21), (4.22) and Lemma 4.8,
combined with a similar analysis of (4.20) in Lemma 4.6, give the
statement of this Lemma. $\square$

\

\textbf{Lemma 4.10.} \textit{We have $$\lim\li_{\ve\da
0}\sup\li_{x\in \gmin} \E_x\sm=0 \ .$$}

\

\textbf{Proof.} This is a result in the same essence of Lemma 3.2
(formula (10)) of \cite{[DK2]}. $\square$

\

\textbf{Lemma 4.11.} \textit{The process $X_{\sm_n}^\ve$ satisfies
the Doeblin condition on $\gm$.}

\

\textbf{Proof.} For each fixed $\ve>0$ we have the ergodicity of the
process $X_t^\ve$. The Doeblin condition is satisfied for the
process $X_t^\ve$ in $[\cE]$ and each of these $[U_k]$'s for
$k=1,...,r$. As we have Lemmas 4.8, 4.9 and 4.10, we see that the
statement of this Lemma follows. $\square$

\

\textbf{Acknowledgement.} We would first like to thank the anonymous
referee for the very careful reading of the manuscript and for
pointing out some errors that we did not notice, as well as giving
various important suggestions. We would also like to thank
Konstantinos Spiliopoulos for reading the first version of this
paper and to thank Rongrong Wang for a pleasant discussion on
problems about matrix perturbations.

\

redpajama_set_name: RedPajamaArXiv


Put our research in technology and innovative concepts to work for your specific needs and requirements. Whether you need an expert, an extra pair of capable hands or collaborator, we can help.
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Today, MainBoard, LLC operates as MainBoard Consulting.
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text:
News Outlets Call Trump's Attacks on Minority Congresswomen Anything But "Racist"
Fox News Was Duped by a Seth Rich Conspiracy Pushed by Russian Intelligence
Excited Local News Reporter Accidentally Loses His Microphone in a Tuba
This is one way to get the inside scoop.
A local news reporter in Phoenix named Cory McCloskey was doing a report from a pep rally and two things were very clear. First, Cory is a very excitable man. This is a guy who really gets into the pep spirit. And second, he specifically is excitable when it comes to marching bands. So excitable that he wanted to make sure the local viewers of Fox 10 not only got an overall sense of the band's sound, but also a sense of what each individual instrument sounded like. The only problem was, the tuba saw that mic near its mouth and decided it was hungry. The glorious moment of truth comes about 40 seconds into this video, but I'd watch the whole thing, because the energy of McCloskey's set-up really helps it out.
But as we established, Cory McCloskey is an excitable guy. The kind of guy who isn't going to let a little tuba/sousaphone action slow him down. Our man is going to finish his report by any means necessary. And he doesn't disappoint.
Cory McCloskey: Inspirational figure.
Related Stories for GQInternet
Jake Gyllenhaal Proves He Is Our Best Celebrity
The Knives Out Trailer Welcomes You to the Fucked-Up Murder Party
Aziz Ansari's New Netflix Special Right Now Wrestles with the Present
Max Cea

redpajama_set_name: RedPajamaCommonCrawl


A Gown can well deservedly boast of being one of the most popular attires among evening wear. A Gown is applied to any full-length woman's garment consisting of a fitted bodice and an attached skirt. It is exquisitely embroidered to give it a rich and elegant look. Associated with charm and luxury, designer evening gowns speak of indulgence in a subtle yet significant way.
Choose from a wide collection of formal evening gowns and wedding gowns made from luxurious fabric with alluring styles that exude an enticing appeal while defining your feminine elegance.


My blog beginnings: I first began blogging the Summer of 2011, thanks to a passionate & driven woman, Allie. Allie is the owner of Honey Bee Invites, a custom invitation company based out of Atlanta, GA. Allie gave me a shot in the wedding industry & mentored me while I learned the odds & ends of blogging, social media, & PR. I am forever grateful. Here is one of my posts on her blog, but feel free to stay up late and swoon over the rest of her blog and beautiful work!


text:
package org.zmlx.hg4idea.command;

import com.intellij.openapi.project.Project;
import org.jetbrains.annotations.Nullable;

import java.nio.charset.Charset;
import java.util.ArrayList;
import java.util.Collections;
import java.util.LinkedList;
import java.util.List;

public class HgCloneCommand {
  private final Project project;

  private String repositoryURL;
  private String directory;
  private final HgCommandAuthenticator authenticator = new HgCommandAuthenticator();

  public HgCloneCommand(Project project) {
    this.project = project;
  }

  public void setDirectory(String directory) {
    this.directory = directory;
  }

  public void setRepositoryURL(String repositoryURL) {
    this.repositoryURL = repositoryURL;
  }

  @Nullable
  public HgCommandResult execute() {
    final List<String> arguments = new ArrayList<String>(2);
    arguments.add(repositoryURL);
    arguments.add(directory);
    return authenticator.executeCommandAndAuthenticateIfNecessary(project, null, repositoryURL, "clone", arguments);
  }
}

redpajama_set_name: RedPajamaGithub


My muse seems to work best after midnight, following warm, cosy naps and bags of chocolate.
I think I am about halfway with the second draft now - that is 50% in sight, ladies and gents. Chapter 12 is finished, and 13 is breathing. This is good indeed, because some serious stuff is happening now. No, seriously.
This really is the best time of the day for me. Every one is asleep, the house is quiet apart from the scattered snores and I can really clear my head. The only problem is that I need to get up for work in just under 4 hours. gah!
This balancing work, life, family, being wife + mother and writing is hard sometimes. Invariably, one will suffer. I try and give each aspect a turn to suffer during the week - today, it is sleep's turn, obviously. Tomorrow, possibly my sanity.


text:
A Praia do Pinho é uma praia localizada na cidade de Balneário Camboriú no estado brasileiro de Santa Catarina. Ela é nacionalmente conhecida por ser destinada ao naturismo, sendo considerada uma das principais praias de nudismo do país  e a mais antiga para naturistas. Seu acesso se dá por conta da Rodovia Interpraias, através de um trevo da BR-101, sendo que compreende uma faixa de aproximadamente 500 m de extensão.

No início da década de 1980, a praia do Pinho começou a ser frequentada por praticantes do naturismo. A princípio, tiveram diversas brigas com outros frequentadores e habitantes do local, até que em 1984 a extinta Revista Manchete publicou um artigo sobre o Naturismo na Praia do Pinho, o qual serviu de base para as ideias naturistas após o fim do Regime militar no Brasil. As edições desta revistas se esgotaram rapidamente e ocorreu um sucesso avassalador da praia.

Nas areias da Praia do Pinho foi criada em 1988 a Federação Brasileira de Naturismo (FBrN), uma espécie de fusão da Associação Brasileira de Naturismo, feita por Celso Rossi. Em 2004, a praia passou a constar num guia das melhores praias naturistas do Mundo como a quinta praia mais bela do planeta neste quesito.

Em novembro de 2005 a praia do Pinho passa a ser afiliada à Federação Brasileira de Naturismo e no ano de 2007 foi a sede do 6º Encontro Brasileiro de Naturismo. Para se hospedar na praia é necessário possuir o Passaporte Naturista, também chamado de Cartão INF.

Em janeiro de 2013 a praia do Pinho sediou o primeiro campeonato de surf naturista do sul do Brasil.

Ligações externas

Pinho
Pinho

redpajama_set_name: RedPajamaWikipedia


NLC Recruitment 2018 Apprentice 460 Trainee Notification – Get Here Online NLC Important Dates, Eligibility Criteira, Pattern & Syllabus, Selection process & Vacancies details etc.
NLC Recruitment 2018: The Neyveli Lignite Corporation Limited has rolled out an official notification for the selection of 460 candidates at Technician Apprenticeship Training (TAT) & Graduate Apprenticeship Training (GAT) posts.
1. Candidates belonging to the States of Andhra Pradesh, Kerala, Karnataka, Tamil Nadu and the Union Territories of Puducherry and Laksha Deep only are eligible.
2. Candidates should have passed the qualifying exam on or after 01.04.2015.
3. Candidates should not have undergone or presently undergoing Apprenticeship Training in NLCI or elsewhere.


text:
namespace PS {

DEFINE_bool(enable_fault_tolerance, false, "enable fault tolerance feature");
DEFINE_int32(num_replicas, 0, "number of replica");
DEFINE_int32(num_servers, 1, "number of servers");
DEFINE_int32(num_workers, 1, "number of clients");
DEFINE_int32(num_unused, 0, "number of unused nodes");
DEFINE_int32(num_threads, 2, "number of computational threads");
DEFINE_string(app, "../config/rcv1_l1lr.config", "the configuration file of app");
DEFINE_string(node_file, "./nodes", "node information");

Postoffice::~Postoffice() {
  recving_->join();
  MessagePtr stop(new Message()); stop->terminate = true; queue(stop);
  sending_->join();
}

// TODO run a console if it is a Node::MANAGER
void Postoffice::run() {
  // omp_set_dynamic(0);
  // omp_set_num_threads(FLAGS_num_threads);
  yellow_pages_.init();
  recving_ = std::unique_ptr<std::thread>(new std::thread(&Postoffice::recv, this));
  sending_ = std::unique_ptr<std::thread>(new std::thread(&Postoffice::send, this));
  switch(myNode().role()) {
    case Node::SCHEDULER: {
      // get all node information
      yellow_pages_.add(myNode());
      nodes_are_ready_.get_future().wait();
      LI << "Scheduler connected " << FLAGS_num_servers << " servers and "
         << FLAGS_num_workers << " workers";

      // run the application
      AppConfig conf; readFileToProtoOrDie(FLAGS_app, &conf);
      AppPtr app = App::create(conf);
      yellow_pages_.add(app);
      app->run();
      app->stop();
      break;
    } default:
      // run as a daemon
      addMyNode("default", yellow_pages_.van().scheduler());
      while (!done_) usleep(300);
      break;
  }
}

void Postoffice::reply(
    const NodeID& recver, const Task& task, const string& reply_msg) {
  if (!task.request()) return;
  Task tk;
  tk.set_customer(task.customer());
  tk.set_request(false);
  tk.set_type(Task::REPLY);
  if (!reply_msg.empty()) tk.set_msg(reply_msg);
  tk.set_time(task.time());
  MessagePtr re(new Message(tk)); re->recver = recver; queue(re);
}

void Postoffice::queue(const MessageCPtr& msg) {
  if (msg->valid) {
    sending_queue_.push(msg);
  } else {
    // do not send, fake a reply mesage
    Task tk;
    tk.set_customer(msg->task.customer());
    tk.set_request(false);
    tk.set_type(Task::REPLY);
    tk.set_time(msg->task.time());
    MessagePtr reply(new Message(tk));
    reply->sender = msg->recver;
    reply->recver = msg->sender;
    yellow_pages_.customer(tk.customer())->exec().accept(reply);
  }
}

//  TODO fault tolerance, check if node info has been changed
void Postoffice::send() {
  MessageCPtr msg;
  while (true) {
    sending_queue_.wait_and_pop(msg);
    if (msg->terminate) break;
    Status stat = yellow_pages_.van().send(msg);
    if (!stat.ok()) {
      LL << "sending " << msg->debugString() << " failed. error: " << stat.ToString();
    }
  }
}

void Postoffice::recv() {
  while (true) {
    MessagePtr msg(new Message());
    auto stat = yellow_pages_.van().recv(msg);
    CHECK(stat.ok()) << stat.ToString();
    auto& tk = msg->task;
    if (tk.request() && tk.type() == Task::TERMINATE) {
      yellow_pages_.van().statistic();
      done_ = true;
      break;
    } else if (tk.request() && tk.type() == Task::MANAGE) {
      if (tk.has_mng_app()) manageApp(tk);
      if (tk.has_mng_node()) manageNode(tk);
    } else {
      if (tk.customer() == "default") continue;
      auto pt = yellow_pages_.customer(tk.customer());
      CHECK(pt) << "customer [" << tk.customer() << "] doesn't exist";
      pt->exec().accept(msg);
      continue;
    }
    auto ptr = yellow_pages_.customer(tk.customer());
    if (ptr != nullptr) ptr->exec().finish(msg);
    reply(msg->sender, msg->task);
  }
}

void Postoffice::manageApp(const Task& tk) {
  CHECK(tk.has_mng_app());
  auto& mng = tk.mng_app();
  if (mng.cmd() == ManageApp::ADD) {
    yellow_pages_.add(std::static_pointer_cast<Customer>(App::create(mng.app_config())));
  }
}

void Postoffice::manageNode(const Task& tk) {

  CHECK(tk.has_mng_node());
  auto& mng = tk.mng_node();
  std::vector<Node> nodes;
  for (int i = 0; i < mng.node_size(); ++i) {
    nodes.push_back(mng.node(i));
  }
  auto obj = yellow_pages_.customer(tk.customer());
  switch (mng.cmd()) {
    case ManageNode::ADD:
      for (auto n : nodes) yellow_pages_.add(n);
      if (yellow_pages_.num_workers() >= FLAGS_num_workers &&
          yellow_pages_.num_servers() >= FLAGS_num_servers) {
        nodes_are_ready_.set_value();
      }
      break;
    case ManageNode::INIT:
      for (auto n : nodes) yellow_pages_.add(n);
      if (obj != nullptr) {
        obj->exec().init(nodes);
        for (auto c : obj->children())
          yellow_pages_.customer(c)->exec().init(nodes);
      }
      break;
    case ManageNode::REPLACE:
      CHECK_EQ(nodes.size(), 2);
      obj->exec().replace(nodes[0], nodes[1]);
      for (auto c : obj->children())
        yellow_pages_.customer(c)->exec().replace(nodes[0], nodes[1]);
      break;

    default:
      CHECK(false) << " unknow command " << mng.cmd();
  }
}

void Postoffice::addMyNode(const string& name, const Node& recver) {
  Task task;
  task.set_request(true);
  task.set_customer(name);
  task.set_type(Task::MANAGE);
  auto mng_node = task.mutable_mng_node();
  mng_node->set_cmd(ManageNode::ADD);
  *(mng_node->add_node()) = myNode();

  MessagePtr msg(new Message(task));
  msg->recver = recver.id();
  queue(msg);
}

} // namespace PS


A picture is something that seizes an instant. A specific moment where you can cut off a feeling for the future. A picture is a place where you can completely hold an emotion. The Groovers are something like a picture. They're that place. Every song is the emption of a moment. An emotion that condenses within itself all the force it generates at its beginning. They're essential. Nothing is too much. Not a note. Not a word. Nothing is secondary. Everything is moulded on a feeling, on an emotion, on a moment opened up in time. It's the world of the Groovers. A world made of different musical traditions, but that somehow are similar. Traditions coming from the American county - from Springsteen to Morphine, passing through Replacements, Long Ryders, Mark Lanegan and Hoodoo Gurus - up to the provincial Italian reality. A reality made of workers clutched at faraway dreams, of hate and love for the little things of everyday. It's a sharp but indestructible power that holds up every song of the Groovers. It's not only words. It's not only music. But words and music together: that's their strength. The story of the Groovers is a long walk on a tortuous and complicated road. A road that took us to a place, walking with songs for the dreamers that live their own moments, their own emotions, their own feelings. That's the place where the Groovers took us after these ten years: the strength and the bravery of living a feeling.
" Thanks to my DJ friend in Italy, Massimo Ferro.........I recently discovered the entertaining & inventive music of The Groovers, also from Italy. Thru-out the tracks of the two cds I listenned to " September Rain " & " That's All Folks " .......I heard delightfull hot licks and heart heavy emotion. Combine this with a good time feel in the right spots......and the crafty songwriting of Michele Anelli. Call it roots rock, or Alt-Country......The Groovers hold their on.....& their future look bright in my opinion ".
The Groovers of Michele Anelli are back. It's one of the most important bands of the Italian roots-rock scene (if it can be defined). "That's all folks!!" is their fourth album. It is different than its predecessors. It is characterized by a clean prevail of acoustic and intimate atmospheres. Best worth of the formation is, to our notice, that of filtering influences typically American, like Bruce Springsteen (there are two covers, Factory and the ghost-track Stolen Car) or Steve Earle (but also the Velvet Underground!), with a sensibility more typically pop.
1991 7" Songs For The Dreamers Lost Ballads Soul Street September Rain That s All Folks!!


text:
Review - Battle Chess
Post by Trev » Sun Oct 05, 2008 10:49 pm
Most everyone knows what chess is, so what exactly puts the "battle" in this 3DO port of an old pc game? Well, the pieces in this game are represented by humans (with the exception of the rook who is some kind of strange rock monster ) And when a piece is captured from the board the game zooms in as the characters get into a short, but very amusing, fight for survival.
Of course the rules of chess still apply, so the captured piece will always lose. It is certainly worth seeing all the various combinations of fights though, as each piece reacts differently depending on the piece they confront. Again these are really quite funny, especially when the losing piece knows that they are done for (ex - check out the knight vs. knight for some classic Monty Python type humor ) After you have seen all the battle combos though, they do start to grow a bit stale. At that point you might think about switching to the 2d view which makes for a much quicker game. It also makes mistakes less likely thanks to the far less awkward viewing angle.
This game, though enhanced, is far from an example of 3DO power graphically, but the improved music and sound effects are noticeable. Fans of classical music will likely appreciate the pleasant background track (although the repetition has the potential to annoy during long games) Each piece has their own individual tune also which is a nice touch. The grunts of the combatants and the clashing of their weapons are quite clear too.
Along with a thorough instruction booklet, the game also includes a helpful tutorial which is great for newbies. A save option is available for lengthy game as well, and is a feature that will surely be made use of. All the usual chess options that have becomes standard over the years are present too, and I really could not think of anything in particular that was missing. One note ... even on the novice setting the cpu is quite challenging, so chess rookies may have a tough go at it.
The engine in Battle Chess is strong enough for the serious player, and the gimmicky but entertaining fights are enough to draw in the more casual ones. What more can I say … its chess. A pretty solid game for what it is. I suggest grabbing it in the Maniac Pak ... it's worth owning.
-Trev
zook1981
Post by zook1981 » Fri Oct 26, 2012 7:03 pm
Dude I love this game. Been playing it since the PC days back in the very early 90s!
Post by Trev » Fri Oct 26, 2012 7:31 pm
zook1981 wrote: Dude I love this game. Been playing it since the PC days back in the very early 90s!
Same here. In my computer class in middle school we had one day each week where we could play games. I would usually rush for Battle Chess.
Trev wrote:
Yea I can't stand Chess, and suck at it, but I can always play Battle Chess. To make it more enjoyable, I either play the computer or I play against another human but I control both player 1 and player 2 LOL. I just like the death moves.
JohnnyDude
Post by JohnnyDude » Mon Oct 29, 2012 3:58 am
zook1981 wrote:
Heck, I thought it was boring. I like regular chess "enough". Pawns die too frequently for this to hold interest! And the computer sometimes took forever to move!
Post by Trev » Mon Oct 29, 2012 4:03 am
JohnnyDude wrote: Heck, I thought it was boring. I like regular chess "enough". Pawns die too frequently for this to hold interest! And the computer sometimes took forever to move!
It's true that the cpu in Battle Chess isn't always the quickest. That said, having played several video game versions of chess, I didn't find it too bad. It allows you extra time to think ahead and plan your moves, and really chess is a slow paced game anyway.
Hmm. Well, come to think of it, I'm probably thinking of the Intellivision version I played. I remember that one even came with a printed warning about how the computer may take hours if set to the highest difficulty!
JohnnyDude wrote:
Ha ha, it's not often that I hear the 3DO being compared to the Intellivision.
But yeah, the game had so many ports who can keep track.
Martin III
Post by Martin III » Tue Oct 30, 2012 2:34 pm
Trev wrote: Ha ha, it's not often that I hear the 3DO being compared to the Intellivision.
I dunno, I find it easy to imagine Mr. Intellivision going on to become a spokesman for the 3DO. Can't you just see him comparing Total Eclipse to Star Fox and Cybermorph?
Post by Trev » Tue Oct 30, 2012 9:47 pm
Martin III wrote:
Post by goldenband » Sat Dec 15, 2012 7:32 am
I spent some time over the past week playing Battle Chess, though strictly with an eye towards its chessplaying abilities -- I used the 2D board, with animations off. I played three games against it, first with the White pieces on Level 1 (which I didn't record), then with White on Level 5, and finally with Black on Level 9. I won all three games, though the game on Level 9 took a long time -- over 20 hours! -- which made me grateful for the savegame option.
If anyone's interested, I've included the game scores below, as well as animated GIFs that replay the game. The Level 5 game was a tactical knockout, featuring a piece sacrifice that led to a mating attack and checkmate -- something that doesn't happen against computers very often -- while the Level 9 game was a slow positional crush, at the end of which the computer resigned because I would promote my pawn to a Queen and deliver checkmate within ~10 moves.
Overall I must say I'm not too impressed with the strength of the program. Compared to NES Chessmaster, it seems decidedly weaker to me, and it's not much better than Intellivision's USCF Chess program. I'd expect quite a bit more from the 3DO!
And other than the welcome ability to save one's game, it's also very short on options; for example, I wasn't able to spot a way to flip the board around. It doesn't offer any way to set up a timed game, or even have a clock for elapsed time, which the Intellivision game has. And its time usage is very simplistic -- it takes 20-25 minutes for every move on the highest difficulty, no matter how routine. That gets old fast, believe me.
I think all this is pretty typical of the Battle Chess series, though; it coasts by on gimmicks, but doesn't really have much under the hood. I'm a reasonably strong chessplayer, but I shouldn't be able to beat it this easily. Still, at least there's the option to allow the computer to resign, which was very much appreciated in the last game since it allowed me to finish the game tonight.
Anyway, here are the games! They're in PGN format, which is the standard format for computer-based chess. Sorry about the large diagrams, not sure how to shrink them on this board.
[White "goldenband"]
[Black "3DO Battle Chess, Level 5"]
[ECO "B30"]
1. e4 c5 2. Nf3 Nc6 3. b3 Nf6 4. e5 Ng4 5. Bb2 f6 6. Qe2 fxe5 7. Nxe5 Ngxe5 8. Bxe5 d6 9. Bb2 Nd4 10. Bxd4 cxd4 11. Qe4 e5 12. Bb5+ Kf7 13. Qd5+ Kg6 14. O-O Bf5 15. Qxb7 Bxc2 16. Na3 Qb8 17. Qf3 d5 18. Qxd5 Bxa3 19. Qe6+ Kg5 20. h4+ Kxh4 21. g3+ Kg5 22. f4+ exf4 23. gxf4+ Kh4 24. Kh2 Qc8 25. Bd7 Bf5 26. Qxf5 Qd8 27. Qh3# 1-0
[White "3DO Battle Chess, Level 9"]
[Black "goldenband"]
1. d4 Nf6 2. c4 c5 3. dxc5 Na6 4. Qd4 Qa5+ 5. Bd2 Qxc5 6. Nf3 Qxd4 7. Nxd4 e5 8. Nb5 d5 9. cxd5 Nxd5 10. e4 Ndb4 11. N1a3 Be6 12. Bc3 f6 13. b3 Bc5 14. Bxb4 Bxb4+ 15. Ke2 O-O 16. Nc2 Bc5 17. Rd1 Rfd8 18. f3 Rxd1 19. Kxd1 Rd8+ 20. Ke2 Rd7 21. Nc3 Nb8 22. Na4 Be7 23. Nc3 Nc6 24. Nb5 a6 25. Nc3 b5 26. Ne3 Nd4+ 27. Kf2 Bc5 28. Bd3 f5 29. exf5 Nxf5 30. Bxf5 Bxf5 31. Re1 Rd3 32. Ne4 Bb6 33. Ng5 h6 34. Ne4 Kf7 35. Re2 Ke6 36. Re1 Kd7 37. Re2 Bd4 38. Re1 Kc6 39. Re2 Kb6 40. Ng3 Bg6 41. Ne4 Bxe4 42. fxe4 Ka5 43. g4 Kb4 44. a4 Rxb3 45. axb5 axb5 46. h3 Rxe3 47. Rxe3 g5 48. Kf3 Bxe3 49. Kxe3 Kc3 50. Ke2 Kc2 51. h4 b4 0-1
Post by Trev » Sat Dec 15, 2012 7:21 pm
goldenband wrote: Overall I must say I'm not too impressed with the strength of the program. Compared to NES Chessmaster, it seems decidedly weaker to me, and it's not much better than Intellivision's USCF Chess program. I'd expect quite a bit more from the 3DO!
Keep in mind though that the 3DO version is essentially a port of a game from 1988. Outside of tweaks to improve the graphics & audio, it's still (for better or worse) the 88 engine .
Chessmaster, even at that time, had several games in the series already. Naturally they would keep tweaking it to make it better. By 1990 it was more advanced.
That said, I still think Battle Chess on 3DO compares favorably w/Chessmaster. I actually found the playstation version of chessmaster weaker challenge wise than Battle Chess on 3DO, especially on the lower levels.
Post by goldenband » Sun Dec 16, 2012 1:37 am
Trev wrote: Keep in mind though that the 3DO version is essentially a port of a game from 1988. Outside of tweaks to improve the graphics & audio, it's still (for better or worse) the 88 engine .
Fair point, though I still shake my finger at the idea that they'd re-release a 1988 engine untweaked!
Trev wrote: That said, I still think Battle Chess on 3DO compares favorably w/Chessmaster. I actually found the playstation version of chessmaster weaker challenge wise than Battle Chess on 3DO, especially on the lower levels.
Huh, that doesn't match my experience, at least with the NES version which I thought played a fairly good game. Maybe it's time to pit the two games against each other? People have done similar matches over at AtariAge, e.g. Atari 2600 chess vs. a 1980s electronic board.
Post by Trev » Sun Dec 16, 2012 5:27 am
I've done that also, playing chess games head to head. One of the games was CD-i's Sargon Chess, but I forgot the other.
I suppose it is possible that Battle Chess on 3DO is running a newer engine. I'm really just guessing it's not, based on the fact that (despite minor cosmetic changes) the game reaks of port.
As far as challenge goes, part of it may be that my memory is colored from playing Battle Chess when I was a kid. Obviously it seemed harder at that time as opposed to now when I am a solid player. I do think though, even at lower levels, it will challenge newbies.
Post by Martin III » Sun Dec 16, 2012 5:30 pm
Ha, I love it.
goldenband wrote: And other than the welcome ability to save one's game, it's also very short on options; for example, I wasn't able to spot a way to flip the board around. It doesn't offer any way to set up a timed game, or even have a clock for elapsed time, which the Intellivision game has. And its time usage is very simplistic -- it takes 20-25 minutes for every move on the highest difficulty, no matter how routine. That gets old fast, believe me.
25 minutes? Really? Heck, if a human player was taking 25 minutes on a move, I'd think something was wrong with them.
Post by goldenband » Sun Dec 16, 2012 7:31 pm
Martin III wrote: 25 minutes? Really? Heck, if a human player was taking 25 minutes on a move, I'd think something was wrong with them.
Well, I don't know if I've ever taken 25 minutes on a move, but I've definitely taken 15-20. But, that's always in the context of a timed game where I have between 1 hour and 3 hours (roughly speaking) to make all my moves. In other words, if I choose to spend 25 minutes on a single move, then it means I have that much less time for all my other moves -- so the opponent knows the torture will eventually end, one way or the other.
But when a game is untimed, then it's generally considered rude to spend more than a few minutes on a move. Introducing clocks was one of the big changes that led to the professionalization of chess, and timed games became the standard around the end of the 19th century (I might be a couple decades off, but thereabouts). Before that, there were players notorious for spending hours on a single move!
The longest time spent on a single move in a timed game is probably around 2 hours, since you almost never get more than 3 hours to make your first 40 moves (and after move 40, super-lengthy calculation is seldom required). Even today, you'll see grandmasters using up to an hour on a single move if they need to calculate really deeply, but in general that's fading away as time controls get faster. Most of the games I play these days are on the Internet, and are usually blitz (5 minutes total per player for the whole game), accelerated blitz (3 minutes), or even lightning (1 minute).
Anyway, the main thing is that the 3DO is taking the same rough amount of time on every move, and is doing so on things like routine recaptures where there's no subtlety to be discerned. That's a sign of poor programming, but it also reflects the fact that there's no time limit. I assume that on Level 9, it simply searches until it reaches a certain number of moves, or hits a RAM limitation.
mattyg
Post by mattyg » Mon Dec 17, 2012 5:53 am
Wasn't this port done by the Doom idiots? Or am I mistaken?
"purge with fire Bretheren"
3DO is dead , long live 3DO
http://www.youtube.com/watch?v=_K2kuP_anAc
"The greatest games in the world take seconds to learn but months to master. The worst take months to learn and minutes to finish."...MattyG


text:
Q: How to add Swagger related static files to Spring Boot + Jersey app? I am trying to add Swagger support to my REST API but I am confused how to add Swagger related static content (HTML, JS) files to my Spring Boot application.
I use the following dependencies:

*

*spring-boot-starter-parent:2.0.1.RELEASE

*spring-boot-starter-jersey:2.0.1.RELEASE

*swagger-jersey2-jaxrs:1.5.18
This is my swagger configuration:
@Configuration
public class SwaggerConfig {
    @Bean
    public BeanConfig swaggerConfiguration() {
        final BeanConfig beanConfig = new BeanConfig();
        beanConfig.setResourcePackage("a.b.c");
        beanConfig.setScan(true);
        beanConfig.setPrettyPrint(true);
        return beanConfig;
    }
}

And the jersey configuration:
@Component
public class JerseyConfig extends ResourceConfig {
    public JerseyConfig() {
        register(ImageResource.class);
        register(io.swagger.jaxrs.listing.ApiListingResource.class);
        register(io.swagger.jaxrs.listing.SwaggerSerializers.class);
    }
}

This part works like a charm, when I open http://localhost:8090/swagger.json then I can see the expected Swagger JSON content.
But I do not know, how to add the Swagger related static HTML content to my application. I can see that this content is in the springfox-swagger-ui.jar and I can add it to my project as a maven dependency, but how I can unpack the content from this jar?
And what is the proper way to overwrite the default swagger.json URL with my URL in the static Swagger file in order to Swagger show my REST API immediately when I open swagger-ui.html.

A: <dependency>
  <groupId>org.webjars</groupId>
  <artifactId>swagger-ui</artifactId>
  <version>${swagger-ui.version}</version>
</dependency>

Please, do not include springfox-swagger-ui.jar, it's meant to work with Spring's RestController.

A: You must have solved it now but it might help others so here's the complete procedure as I was also looking for a tutorial.
I am using Swagger V2 with Spring Boot 2 and it's straightforward 3 step process.
Step 1: Add required dependencies in pom.xml file. The second dependency is optional use it only if you need Swagger UI.
        <!-- https://mvnrepository.com/artifact/io.springfox/springfox-swagger2 -->
        <dependency>
            <groupId>io.springfox</groupId>
            <artifactId>springfox-swagger2</artifactId>
            <version>2.9.2</version>
        </dependency>
        <!-- https://mvnrepository.com/artifact/io.springfox/springfox-swagger-ui -->
        <dependency>
            <groupId>io.springfox</groupId>
            <artifactId>springfox-swagger-ui</artifactId>
            <version>2.9.2</version>
        </dependency>

Step 2: Add configuration class
@Configuration
@EnableSwagger2
public class SwaggerConfig {

     public static final Contact DEFAULT_CONTACT = new Contact("Usama Amjad", "https://stackoverflow.com/users/4704510/usamaamjad", "hello@email.com");
      public static final ApiInfo DEFAULT_API_INFO = new ApiInfo("Article API", "Article API documentation sample", "1.0", "urn:tos",
              DEFAULT_CONTACT, "Apache 2.0", "http://www.apache.org/licenses/LICENSE-2.0", new ArrayList<VendorExtension>());

    @Bean
    public Docket api() {
        Set<String> producesAndConsumes = new HashSet<>();
        producesAndConsumes.add("application/json");
        return new Docket(DocumentationType.SWAGGER_2)
                .apiInfo(DEFAULT_API_INFO)
                .produces(producesAndConsumes)
                .consumes(producesAndConsumes);

    }
}

Step 3: Setup complete and now you need to document APIs in controllers
    @ApiOperation(value = "Returns a list Articles for a given Author", response = Article.class, responseContainer = "List")
    @ApiResponses(value = { @ApiResponse(code = 200, message = "Success"),
            @ApiResponse(code = 404, message = "The resource you were trying to reach is not found") })
    @GetMapping(path = "/articles/users/{userId}")
    public List<Article> getArticlesByUser() {
       // Do your code
    }

Usage:
Swagger UI: You can access it via http://localhost:8080/swagger-ui.html

Postman: You can also access your Documentation JSON from http://localhost:8080/v2/api-docs and just copy paste it in Postman to use with it.

redpajama_set_name: RedPajamaStackExchange


Flooding can be a disaster for a business. As well as damage to premises or equipment you may also loose stock and supplies and will be unable to trade whilst the damage is being repaired. Your employees may be unable to get to work if flooding affects the transport links, and this may affect your business.
There are measures you can take to reduce the risk and impact of flooding, and make your business more resilient. It is a good idea to put these in place so that in the event of a flood or any other kind of emergency it enables you to offer business as usual in the shortest possible time. The plan does not have to be detailed and should not take up too much of your time.