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\begin{document}

\title[Regularity and stability for solutions]{Regularity and stability for solutions to elliptic equations and systems arising from high-contrast composites}

\author[Z. Zhao]{Zhiwen Zhao}

\address[Z. Zhao]{Beijing Computational Science Research Center, Beijing 100193, China.}
\email{zwzhao365@163.com}

\date{\today}

\maketitle

\begin{abstract}
The main objective of this paper is to establish the regularity and stability for solutions to the conductivity problems with degenerate coefficients in the presence of two rigid conductors, as one conductor keeps motionless and another conductor moves in some direction by
a sufficiently small translational distance. Our results contain the following three cases: two perfect conductors, two insulators, a perfect conductor and an insulator. Further, we extend the results to the elasticity problem.

\end{abstract}

\maketitle

\section{Introduction and main results}

This investigation is stimulated by the problem of damage and fracture in high-contrast fiber-reinforced composites. The material fracture occurs in the thin gaps between fibers, since there always appear high concentrated fields including extreme electric and stress fields in these zones. These physical fields can be described by the gradients of solutions to elliptic equations and systems with discontinuous coefficients. Much effort has been devoted to quantitative analysis for the behavior of the gradients in the narrow regions between rigid inclusions since Babu\u{s}ka et al's famous numerical work \cite{BASL1999}. In the case of nondegenerate finite coefficients, it has been proved in \cite{DL2019,LV2000,LN2003,BV2000} that the gradients of solutions remain bounded independent of the distance $\varepsilon$ between inclusions. Especially in \cite{LN2003}, Li and Nirenberg established stronger $C^{1,\alpha}$ estimates for general second-order elliptic systems with piecewise H\"{o}lder continuous coefficients, which completely demonstrates the numerical observation in \cite{BASL1999}.

However, the situation becomes significantly different when the coefficients degenerate to be zero or infinity. Let $D\subseteq\mathbb{R}^{n}\,(n\geq3)$ be a bounded domain with $C^{2,\alpha}$ boundary, whose interior contains two $C^{2,\alpha}\,(0<\alpha<1)$ inclusions $D_{1}$ and $D_{2}$ with $\varepsilon$ apart, where $\varepsilon$ is a sufficiently small positive constant. Suppose further that $D_{i}$, $i=1,2$ are far away from the external boundary $\partial D$. When the conductivity $k$ degenerates to be $0$ or $\infty$, we obtain the following three types of boundary value problems (see \cite{BLY2009,BLY2010}):
\begin{align}\label{con006}
\begin{cases}
\Delta u=0,&\mathrm{in}\;D\setminus\overline{D_{1}\cup D_{2}},\\
u|_{+}=u|_{-},&\mathrm{on}\;\partial{D}_{i},\,i=1,2,\\
u=C_{i}^{0},&\mathrm{on}\;\overline{D}_{i},\,i=1,2,\\
\int_{\partial D_{i}}\frac{\partial u}{\partial\nu}\big|_{+}=0,&i=1,2,\\
u=\varphi,&\mathrm{on}\;\partial D,
\end{cases}
\end{align}
and
\begin{align}\label{PRO002}
\begin{cases}
\Delta u=0,&\mathrm{in}\;D\setminus(\partial D_{1}\cup\partial D_{2}),\\
u|_{+}=u|_{-},&\mathrm{on}\;\partial{D}_{i},\,i=1,2,\\
\frac{\partial u}{\partial\nu}\big|_{+}=0,&\mathrm{on}\;\partial{D}_{i},\,i=1,2,\\
u=\varphi,&\mathrm{on}\;\partial D,
\end{cases}
\end{align}
and
\begin{align}\label{PRO09}
\begin{cases}
\Delta u=0,&\mathrm{in}\;D\setminus(\partial D_{1}\cup\overline{D}_{2}),\\
u|_{+}=u|_{-},&\mathrm{on}\;\partial{D}_{i},\,i=1,2,\\
\frac{\partial u}{\partial\nu}\big|_{+}=0,&\mathrm{on}\;\partial{D}_{1},\\
u=C^{0},&\mathrm{on}\;\overline{D}_{2},\\
\int_{\partial D_{2}}\frac{\partial u}{\partial\nu}\big|_{+}=0,\\
u=\varphi,&\mathrm{on}\;\partial D,
\end{cases}
\end{align}
where $u\in H^{1}(D)$, $\varphi\in C^{2,\alpha}(\overline{D})$, $\varphi\not\equiv0$ on $\partial D$, $C_{i}^{0}$, $i=1,2,$ are the free constants determined by the fourth line of \eqref{con006}, the free constant $C^{0}$ is determined by the fifth line of \eqref{PRO09}, $\nu$ denotes the unit outer normal to the domain. Here and throughout this paper the subscript $\pm$ shows the limit from outside and inside the domain, respectively. It is worth pointing out that problems \eqref{con006} and \eqref{PRO002} are, respectively, called the perfect and insulated conductivity problems and their gradients of solutions always blow up with respect to the distance $\varepsilon$ between inclusions, as the distance $\varepsilon$ tends to zero. By the elliptic regularity theory, we know that $u\in C^{2,\alpha}(\overline{D\setminus D_{1}\cup D_{2}})$ for the perfect conductivity problem \eqref{con006}, $u\in C^{2,\alpha}(\overline{D\setminus D_{1}\cup D_{2}})\cap C^{2,\alpha}(\overline{D_{1}\cup D_{2}})$ for the insulated conductivity problem \eqref{PRO002}, and $u\in C^{2,\alpha}(\overline{D\setminus D_{1}\cup D_{2}})\cap C^{2,\alpha}(\overline{D}_{1})$ for the conductivity problem \eqref{PRO09} with a perfect conductor and an insulator.

With regard to the perfect conductivity case, there is a long list of literature involving different techniques and cases based on shapes of inclusions, dimensions, and applied boundary conditions. In the presence of two strictly convex inclusions, the gradient blow-up rates for the perfect conductivity problem have been proved to be $\varepsilon^{-1/2}$ \cite{AKLLL2007,BC1984,BLY2009,AKL2005,Y2007,Y2009,K1993} for $n=2$, $|\varepsilon\ln\varepsilon|^{-1}$ \cite{BLY2009,LY2009,BLY2010,L2012} for $n=3$ and $\varepsilon^{-1}$ \cite{BLY2009} for $n\geq4$, respectively. For more precise characterizations on the singular behavior of the electric field concentration, see \cite{ACKLY2013,LLY2019,BT2013,KLY2013,KLY2014} and the references therein. In the case when the current-electric field relation is nonlinear, we refer to \cite{G2012,CS2019,CS201902}. In addition, similar results have also been extended to the Lam\'{e} system with partially infinite coefficients, see \cite{BLL2015,BLL2017,KY2019} and the references therein. Different from the perfect conductivity case, it has been recently shown in \cite{DLY2022} that the gradient blow-up rate for the insulated conductivity problem also depends on the principal curvature of the surfaces of inclusions besides the dimension. For more related studies on the insulated conductivity problem, see \cite{AKL2005,AKLLL2007,BLY2010,Y2016,W2021,LY2021,DLY2021,LY202102}. As for problem \eqref{PRO09}, Dong and Yang \cite{DY2022} recently proved that there appears no blow-up for gradient of the solution, which partially answers a question raised by Kang \cite{K2022}.

The above-mentioned work mainly focus on the establishment of gradient estimates and asymptotics in the thin gaps between inclusions and aim to reveal the dependence on the distance between inclusions. By contrast, the distance between two inclusions considered in this paper is of constant order but not a infinitely small quantity. In this case, we dedicate to studying the regularity and stability of solutions, as one inclusion moves in some direction for a sufficiently small distance and another inclusion keeps motionless. We will prove that the solutions are smooth and stable with respect to the small translational distance.

\subsection{Main results}
Although the results of this paper can be extended to any finite number of inclusions of general convex shapes, we restrict to two regular curvilinear squares and cubes for the convenience of presentation. For $x^{0}\in\mathbb{R}^{n}$, $r>0$ and $m,n\geq2$, let $\mathcal{B}_{r}^{(m)}(x^{0})$ represent the curvilinear squares and cubes centered at $x^{0}$ with the radius $r$ as follows:
\begin{align*}
\sum^{n}_{i=1}|x_{i}-x_{i}^{0}|^{m}=r^{m}.
\end{align*}
In particular, when $m=2$, it is a circle or ball, that is, $\mathcal{B}_{r}^{(2)}(x^{0})=B_{r}(x^{0})$. In the following, let
\begin{align}\label{DOM001}
D=\mathcal{B}_{10}^{(m)}(0),\;\, D_{1}^{h}=\mathcal{B}^{(m)}_{1}((3+h)e_{n}),\;\, D_{2}=\mathcal{B}^{(m)}_{1}(0),\;\,\Omega_{h}=D\setminus\overline{D_{1}^{h}\cup D_{2}},
\end{align}
where $0<|h|\ll1/2$ and $e_{n}=(0',1)$. Here and throughout this paper, we use superscript prime to denote $(n-1)$-dimensional variables. For $m\geq2$, define
\begin{align}\label{alpha}
\gamma=
\begin{cases}
\min\{\alpha,m-2\},&2<m<3,\\
\alpha,&m=2\;\mathrm{or}\;m\geq3.
\end{cases}
\end{align}

The regularity and stability results for the perfect conductivity problem are stated as follows.
\begin{theorem}\label{thm001}
For $n\geq2$ and $0<|h|\ll1/2$, let $u_{h}\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{h})$ be the solution of perfect conductivity problem \eqref{con006} with $D,D_{1}^{h},D_{2},\Omega_{h}$ defined by \eqref{DOM001} and $\gamma$ given by \eqref{alpha}. Then there exist a small constant $h_{0}>0$ and a unique function $v(\cdot,x)\in C^{\infty}((-h_{0},h_{0}))$ such that $v(h,\cdot)\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{h})$ solves equation \eqref{con006} and $v(0,x)=u_{0}(x)$. Moreover, $\|v(h,\cdot)-v(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(|D_{1}^{h}\Delta D_{1}^{0}|)$, where $D_{1}^{h}\Delta D_{1}^{0}=(D_{1}^{h}\setminus D_{1}^{0})\cup(D_{1}^{0}\setminus D_{1}^{h})$.
\end{theorem}
\begin{remark}
Observe that $\partial D$ and $\partial D_{i}$, $i=1,2$ are $C^{\infty}$ if $m\geq2$ is an even number. Then in the case when $\varphi\in C^{\infty}(\overline{D})$ and $m\geq2$ is an even number, by applying the proofs of Theorems \ref{thm001}--\ref{thm005} with a slight modification, we can obtain that for any $k\geq0$, $\|v(h,\cdot)-v(0,\cdot)\|_{C^{k}(\overline{D\setminus D_{1}^{0}\cup D_{1}^{h}\cup D_{2}})}=O(h)\rightarrow0$, as $h\rightarrow0$. In addition, when these two inclusions move in some directions simultaneously by a small distance, the similar results can be also obtained by slightly modifying the proofs. For example, if $D_{2}$ is replaced by $D_{2}^{h}:=\mathcal{B}^{(m)}_{1}(\pm h e_{n})$, we can obtain that $\|v(h,\cdot)-v(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(|(D_{1}^{h}\Delta D_{1}^{0})\cup(D_{2}^{h}\Delta D_{2}^{0})|)$.

\end{remark}

Now we list the corresponding results for the insulated conductivity problem as follows.
\begin{theorem}\label{thm002}
For $n\geq2$ and $0<|h|\ll1/2$, let $u_{h}\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{h})\cap C^{2,\gamma}(\overline{D_{1}^{h}\cup D_{2}})$ be the solution of insulated conductivity problem \eqref{PRO002} with $D,D_{1}^{h},D_{2}$, $\Omega_{h}$ defined by \eqref{DOM001} and $\gamma$ given by \eqref{alpha}. Then there exist a small constant $h_{0}>0$ and a unique function $v(\cdot,x)\in C^{\infty}((-h_{0},h_{0}))$ such that $v(h,\cdot)\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{h})\cap C^{2,\gamma}(\overline{D_{1}^{h}\cup D_{2}})$ verifies equation \eqref{PRO002} and $v(0,x)=u_{0}(x)$. Furthermore, $\|v(h,\cdot)-v(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(|D_{1}^{h}\Delta D_{1}^{0}|)$, where $D_{1}^{h}\Delta D_{1}^{0}=(D_{1}^{h}\setminus D_{1}^{0})\cup(D_{1}^{0}\setminus D_{1}^{h}).$
\end{theorem}

With regard to the conductivity problem with an insulator and a perfect conductor, we have
\begin{theorem}\label{thm003}
For $n\geq2$ and $0<|h|\ll1/2$, let $u_{h}\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{h})\cap C^{2,\gamma}(\overline{D_{1}^{h}})$ be the solution of problem \eqref{PRO09} with $D,D_{1}^{h},D_{2},\Omega_{h}$ defined by \eqref{DOM001} and $\gamma$ given by \eqref{alpha}. Then there exist a small constant $h_{0}>0$ and a unique function $v(\cdot,x)\in C^{\infty}((-h_{0},h_{0}))$ such that $v(h,\cdot)\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{h})\cap C^{2,\gamma}(\overline{D_{1}^{h}})$ solves equation \eqref{PRO09} and $v(0,x)=u_{0}(x)$. Moreover, $\|v(h,\cdot)-v(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(|D_{1}^{h}\Delta D_{1}^{0}|)$, where $D_{1}^{h}\Delta D_{1}^{0}=(D_{1}^{h}\setminus D_{1}^{0})\cup(D_{1}^{0}\setminus D_{1}^{h}).$

\end{theorem}

The aforementioned results can be also extended to the elasticity problem modeled by the Lam\'{e} system with partially infinite coefficients. Specifically, consider the following boundary value problem:
\begin{align}\label{La.002}
\begin{cases}
\partial_\alpha(A_{ij}^{\alpha\beta}\partial_\beta u_{h}^j)=0,\quad&\mathrm{in}\;D\setminus\overline{D_{1}^{h}\cup D_{2}},\\
u_{h}|_{+}=u_{h}|_{-},&\mathrm{on}\ \partial{D}_{1}^{h}\cup\partial D_{2},\\
u_{h}=\sum^{\frac{n(n+1)}{2}}_{\alpha=1}C_{1\alpha}^{h}\phi_{\alpha},&\mathrm{on}~\overline{D_{1}^{h}},\\
u_{h}=\sum^{\frac{n(n+1)}{2}}_{\alpha=1}C_{2\alpha}^{h}\phi_{\alpha},&\mathrm{on}~\overline{D}_{2},\\
\int_{\partial{D}_{1}^{h}}A_{ij}^{\alpha\beta}\partial_\beta u_{h}^j\nu_{\alpha}\phi_{k}^{i}=0,&k=1,2,...,\frac{n(n+1)}{2},\\
\int_{\partial{D}_{2}}A_{ij}^{\alpha\beta}\partial_\beta u_{h}^j\nu_{\alpha}\phi_{k}^{i}=0,&k=1,2,...,\frac{n(n+1)}{2},\\
u_{h}=\varphi,&\mathrm{on}\ \partial{D},
\end{cases}
\end{align}
where $u_{h}=(u_{h}^{1},...,u_{h}^{n})\in H^{1}(D;\mathbb{R}^{n})$, $\varphi\in C^{2,\alpha}(\overline{D};\mathbb{R}^{n})$, the free constants $C_{i\alpha}^{h}$, $i=1,2,\,\alpha=1,2,...,\frac{n(n+1)}{2}$ are determined by the fifth and sixth lines of \eqref{La.002}, $\nu=(\nu_{1},...,\nu_{n})$ represents the unit outer normal vector of $\partial D_{1}^{h}$ and $\partial D_{2}$, $\{\phi_{\alpha}\}_{\alpha=1}^{\frac{n(n+1)}{2}}$ is a basis of the linear space of rigid displacement $\Phi:=\{\phi\in C^1(\mathbb{R}^{n}; \mathbb{R}^{n})\,|\,\nabla\phi+(\nabla\phi)^T=0\}$, whose elements are given by $\{e_{i},\,x_{k}e_{j}-x_{j}e_{k}\,|\,1\leq i\leq n,\,1\leq j<k\leq n\}$, and
\begin{align*}
A_{ij}^{\alpha\beta}=\lambda\delta_{i\alpha}\delta_{j\beta}+\mu(\delta_{i\beta}\delta_{\alpha j}+\delta_{ij}\delta_{\alpha\beta}),\quad i,j,k,l=1,2,...,n.
\end{align*}
Here $\mu,n\lambda+2\mu\in(0,\infty)$, $\delta_{ij}$ is the kronecker symbol, that is, $\delta_{ij}=0$ if $i\neq j$, $\delta_{ij}=1$ if $i=j$. Similarly, we have

\begin{theorem}\label{thm005}
For $n\geq2$ and $0<|h|\ll1/2$, let $u_{h}\in H^{1}(D;\mathbb{R}^{n})\cap C^{2,\gamma}(\overline{\Omega}_{h};\mathbb{R}^{n})$ be the solution of the elasticity problem \eqref{La.002} with $D,D_{1}^{h},D_{2},\Omega_{h}$ defined by \eqref{DOM001} and $\gamma$ given by \eqref{alpha}. Then there exist a small constant $h_{0}>0$ and a unique function $v(\cdot,x)\in C^{\infty}((-h_{0},h_{0});\mathbb{R}^{n})$ such that $v(h,\cdot)\in H^{1}(D;\mathbb{R}^{n})\cap C^{2,\gamma}(\overline{\Omega}_{h};\mathbb{R}^{n})$ satisfies problem \eqref{La.002} and $v(0,x)=u_{0}(x)$. Furthermore, $\|v(h,\cdot)-v(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(|D_{1}^{h}\Delta D_{1}^{0}|)$, where $D_{1}^{h}\Delta D_{1}^{0}=(D_{1}^{h}\setminus D_{1}^{0})\cup(D_{1}^{0}\setminus D_{1}^{h}).$
\end{theorem}

In the following, we will establish the regularity and stability of solutions to the conductivity problems with degenerate coefficients and the elasticity problem with respect to small translational distance in Sections \ref{SEC02} and \ref{SEC03}, respectively.

\section{Regularity and stability for the conductivity problem with degenerate coefficients}\label{SEC02}

Choose a cutoff function $\eta\in C^{\infty}(\mathbb{R}^{n})$ satisfying that $\eta=1$ in $\mathcal{B}_{1+|h|}^{(m)}(3e_{n})$, $\eta=0$ in $\mathbb{R}^{n}\setminus\mathcal{B}_{3/2}^{(m)}(3e_{n})$ and $|\nabla\eta|\leq C(m,n).$ Let $y=\psi(x)=(x',x_{n}-h\eta(x))$. Then $\psi$ is a diffeomorphism from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$, as $|h|$ is sufficiently small. In fact, since $\det(\partial_{x}y)=1-h\partial_{x_{n}}\eta$, it suffices to pick $\delta_{0}:=\delta_{0}(m,n)=2^{-1}(\sup\limits_{x\in\mathbb{R}^{n}}|\partial_{x_{n}}\eta|)^{-1}$ and let $h\in(-\delta_{0},\delta_{0})$. Remark that this diffeomorphism carry out translation and split joint in the interior of $D$.

Rewrite the original problems \eqref{con006}, \eqref{PRO002} and \eqref{PRO09} as follows:
\begin{align}\label{con002}
\begin{cases}
\mathrm{div}(a_{h}^{k}(x)\nabla u_{h}^{k})=0,&\mathrm{in}\; D,\\
u_{h}^{k}=\varphi, &\mathrm{on}\;\partial D,
\end{cases}
\end{align}
where the coefficients $a_{k}(x)$, $k=1,2,3$ are, respectively, given by
\begin{align}\label{QKW001}
a_{h}^{1}(x)=&
\begin{cases}
\infty,&x\in D_{1}^{h}\cup D_{2},\\
1,&x\in \Omega_{h},
\end{cases}\quad a_{h}^{2}(x)=
\begin{cases}
0,&x\in D_{1}^{h}\cup D_{2},\\
1,&x\in \Omega_{h},
\end{cases}
\end{align}
and
\begin{align}\label{QKW00100}
a_{h}^{3}(x)=&
\begin{cases}
0,&x\in D_{1}^{h},\\
\infty,&x\in D_{2},\\
1,&x\in\Omega_{h}.
\end{cases}
\end{align}
Denote $\tilde{u}_{h}^{k}(y)=u_{h}^{k}(x)$, $k=1,2,3$. By the above change of variables, problem \eqref{con002} turns into
\begin{align}\label{ZWQ001Z66}
\begin{cases}
\mathrm{div}(\tilde{A}_{h}^{k}(y)\nabla \tilde{u}_{h}^{k})=0,&\mathrm{in}\; D,\\
\tilde{u}_{h}^{k}=\varphi, &\mathrm{on}\;\partial D,
\end{cases}
\end{align}
where $\tilde{A}_{h}^{1}(y)=\infty I$ in $\overline{D_{1}^{0}\cup D_{2}}$, $\tilde{A}_{h}^{2}(y)=0I$ in $\overline{D_{1}^{0}\cup D_{2}}$, $\tilde{A}_{h}^{3}(y)=0I$ in $\overline{D_{1}^{0}}$, $\tilde{A}_{h}^{3}=\infty I$ in $\overline{D}_{2}$, $I$ is the identity matrix, and for $y\in \Omega_{0}$, $k=1,2,3,$
\begin{gather}
\begin{align*}
\tilde{A}_{h}^{k}(y)=&(\tilde{a}_{ij})=\frac{(\partial_{x}y)(\partial_{x}y)^{t}}{\det(\partial_{x}y)}
=\frac{1}{b_{n}}
\begin{pmatrix}1&0&\cdots&0&b_{1} \\ 0&1&\cdots&0&b_{2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\cdots&1&b_{n-1}\\ b_{1}&b_{2}&\cdots&b_{n-1}&\sum^{n-1}_{i=1}b_{i}^{2}+b_{n}^{2}
\end{pmatrix},
\end{align*}
\end{gather}
and
\begin{align*}
b_{i}=-h\partial_{x_{i}}\eta(\psi^{-1}(y)),\;i=1,...,n-1,\quad b_{n}=1-h\partial_{x_{n}}\eta(\psi^{-1}(y)).
\end{align*}

Introduce the following function spaces:
\begin{align*}
\begin{cases}
X^{1}=\{\tilde{v}\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{0})\,|\,\text{$\nabla \tilde{v}=0$ in $\overline{D_{1}^{0}\cup D_{2}}$, $\tilde{v}=\varphi$ on $\partial D$}\},\\
Y^{1}=\{f\in H^{-1}(D)\cap C^{0,\gamma}(\overline{\Omega}_{0})\,|\,\text{$f=0$ in $\overline{D_{1}^{0}\cup D_{2}}$}\},
\end{cases}
\end{align*}
and
\begin{align*}
\begin{cases}
X^{2}=\{\tilde{v}\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{0})\cap C^{2,\gamma}(\overline{D_{1}^{0}\cup D_{2}})\,|\,\text{$\tilde{v}=\varphi$ on $\partial D$}\},\\
Y^{2}=\{f\in H^{-1}(D)\cap C^{0,\gamma}(\overline{\Omega}_{0})\cap C^{0,\gamma}(\overline{D_{1}^{0}\cup D_{2}})\},
\end{cases}
\end{align*}
and
\begin{align*}
\begin{cases}
X^{3}=\{\tilde{v}\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{0})\cap C^{2,\gamma}(\overline{D_{1}^{0}})\,|\,\text{$\nabla \tilde{v}=0$ in $\overline{D}_{2}$, $\tilde{v}=\varphi$ on $\partial D$}\},\\
Y^{3}=\{f\in H^{-1}(D)\cap C^{0,\gamma}(\overline{\Omega}_{0})\cap C^{0,\gamma}(\overline{D_{1}^{0}})\,|\,\text{$f=0$ in $\overline{D}_{2}$}\},
\end{cases}
\end{align*}
with the corresponding norms as $\|\tilde{v}\|_{X^{k}}:=\|\tilde{v}\|_{H^{1}(D)}+\|\nabla\tilde{v}\|_{L^{\infty}(\frac{1}{2}D)}$ and $\|f\|_{Y^{k}}:=\|f\|_{H^{-1}(D)}=\sup\{\langle f,w\rangle\,|\,w\in H^{1}_{0}(D),\,\|w\|_{H^{1}_{0}(D)}\leq1\}$, $k=1,2,3,$ where $\gamma$ is defined by \eqref{alpha}, $\langle\, ,\rangle$ denotes the pairing between $H^{-1}(D)$ and $H_{0}^{1}(D)$. It is easy to verify that $X^{k}$ and $Y^{k}$ are Banach spaces. Denote $K:=(-\delta_{0},\delta_{0})$ and $F^{k}(h,\tilde{v}):=\mathrm{div}(\tilde{A}_{h}^{k}(y)\nabla \tilde{v})$, $k=1,2,3$.

\begin{prop}\label{QAT001}
For each $k=1,2,3,$ the map $F^{k}$ is in $C^{\infty}(K\times X^{k},Y^{k})$ in the sense that $F^{k}$ possesses continuous Fr\'{e}chet derivatives of any order.

\end{prop}

\begin{proof}
Observe first that for any $s\geq0$,
\begin{align}\label{MZ001}
\partial_{h}^{s}F^{1}(h,\tilde{v})=
\begin{cases}
\partial_{i}(\partial_{h}^{s}\tilde{a}_{ij}\partial_{j}\tilde{v}),&\mathrm{in}\;\Omega_{0},\\
0,&\mathrm{in}\;D\setminus\Omega_{0},
\end{cases}
\end{align}
and
\begin{align}\label{MZ006}
F^{2}(h,\tilde{v})=
\begin{cases}
\partial_{i}(\tilde{a}_{ij}\partial_{j}\tilde{v}),&\mathrm{in}\;\Omega_{0},\\
\Delta \tilde{v},&\mathrm{in}\;D\setminus\Omega_{0},
\end{cases}\;F^{3}(h,\tilde{v})=
\begin{cases}
\partial_{i}(\tilde{a}_{ij}\partial_{j}\tilde{v}),&\mathrm{in}\;\Omega_{0},\\
\Delta \tilde{v},&\mathrm{in}\;\overline{D_{1}^{0}},\\
0,&\mathrm{in}\;\overline{D}_{2},
\end{cases}
\end{align}
and, for $s\geq1$, $k=2,3,$
\begin{align}\label{MZ009}
\partial_{h}^{s}F^{k}(h,\tilde{v})=
\begin{cases}
\partial_{i}(\partial_{h}^{s}\tilde{a}_{ij}\partial_{j}\tilde{v}),&\mathrm{in}\;\Omega_{0},\\
0,&\mathrm{in}\;D\setminus\Omega_{0}.
\end{cases}
\end{align}
Note that for each $h\in K$, $\partial_{h}^{s}F^{k}(h,\tilde{v})$ is linear with respect to $\tilde{v}$, where $s\geq0,\,k=1,2,3$. According to the definition of $\tilde{a}_{ij}$, it follows from a direct computation that $|\partial_{h}^{s}\tilde{a}_{ij}|\leq C(s,m,n)$ in $\overline{K\times\Omega_{0}}$. This, together with the fact that $\tilde{v}\in X^{k}$, shows that $\partial_{h}^{s}F^{k}(h,\tilde{v})\in L^{2}(D)$. Hence, for any $w\in H_{0}^{1}(D)$, $\|w\|_{H^{1}_{0}(D)}\leq1$,

$(i)$ if $k=1$ and $s\geq0$, we have from \eqref{MZ001}, integration by parts, H\"{o}lder's inequality and Trace Theorem that
\begin{align*}
|\langle \partial_{h}^{s}F^{1},w\rangle|=&\left|\int_{D}\partial_{h}^{s}F^{1}(h,\tilde{v})w\,dy\right|=\left|\int_{\Omega_{0}}\partial_{h}^{s}F^{1}(h,\tilde{v})w\,dy\right|\notag\\
=&\left|\int_{\partial D_{1}^{0}\cup\partial D_{2}}\partial_{h}^{s}\tilde{a}_{ij}\partial_{j}\tilde{v}\nu_{i}w-\int_{\Omega_{0}}\partial_{h}^{s}\tilde{a}_{ij}\partial_{j}\tilde{v}\partial_{i}wdy\right|\notag\\
\leq&C(\|\nabla\tilde{v}\|_{L^{\infty}(\frac{1}{2}D)}\|w\|_{L^{2}(\partial D_{1}^{0}\cup\partial D_{2})}+\|\nabla \tilde{v}\|_{L^{2}(\Omega_{0})}\|\nabla w\|_{L^{2}(D)})\notag\\
\leq&C(\|\nabla\tilde{v}\|_{L^{\infty}(\frac{1}{2}D)}+\|\nabla\tilde{v}\|_{L^{2}(\Omega_{0})})\|w\|_{H^{1}_{0}(D)}\leq C\|\tilde{v}\|_{X^{1}};
\end{align*}

$(ii)$ for $k=2,3$, if $s=0$, it follows from \eqref{MZ006}--\eqref{MZ009}, integration by parts, Trace Theorem and H\"{o}lder's inequality again that
\begin{align*}
|\langle F^{2},w\rangle|=&\left|\int_{D}F^{2}(h,\tilde{v})w\,dy\right|=\left|\int_{\Omega_{0}}\partial_{i}(\tilde{a}_{ij}\partial_{j}\tilde{v})w\,dy+\int_{D_{1}^{0}\cup D_{2}}\Delta\tilde{v}w\right|\notag\\
=&\left|\int_{\partial D_{1}^{0}\cup\partial D_{2}}\left(\frac{\partial\tilde{v}}{\partial\nu}\Big|_{+}+\frac{\partial\tilde{v}}{\partial\nu}\Big|_{-}\right)w-\int_{\Omega_{0}}\tilde{a}_{ij}\partial_{j}\tilde{v}\partial_{i}w-\int_{D_{1}^{0}\cup D_{2}}\nabla\tilde{v}\nabla w\right|\notag\\
\leq& C(\|\nabla\tilde{v}\|_{L^{\infty}(\frac{1}{2}D)}+\|\nabla\tilde{v}\|_{L^{2}(D)})\|w\|_{H^{1}_{0}(D)}\leq C\|\tilde{v}\|_{X^{2}},
\end{align*}
and
\begin{align*}
|\langle F^{3},w\rangle|=&\left|\int_{D}F^{3}(h,\tilde{v})w\,dy\right|=\left|\int_{\Omega_{0}}\partial_{i}(\tilde{a}_{ij}\partial_{j}\tilde{v})w\,dy+\int_{D_{1}^{0}}\Delta\tilde{v}w\right|\notag\\
=&\left|\int_{\partial D_{1}^{0}\cup\partial D_{2}}\frac{\partial\tilde{v}}{\partial\nu}\Big|_{+}\cdot w+\int_{\partial D_{1}^{0}}\frac{\partial\tilde{v}}{\partial\nu}\Big|_{-}\cdot w-\int_{\Omega_{0}}\tilde{a}_{ij}\partial_{j}\tilde{v}\partial_{i}w-\int_{D_{1}^{0}}\nabla\tilde{v}\nabla w\right|\notag\\
\leq& C(\|\nabla\tilde{v}\|_{L^{\infty}(\frac{1}{2}D)}+\|\nabla\tilde{v}\|_{L^{2}(\Omega_{0}\cup D_{1}^{0})})\|w\|_{H^{1}_{0}(D)}\leq C\|\tilde{v}\|_{X^{3}},
\end{align*}
and, if $s\geq1$, using the fact that $\partial_{h}^{s}\tilde{a}_{ij}=0$ on $\partial D_{1}^{0}\cup\partial D_{2}$,
\begin{align*}
|\langle\partial_{h}^{s}F^{k},w\rangle|=&\left|\int_{D}\partial_{h}^{s}F^{k}(h,\tilde{v})w\,dy\right|=\left|\int_{\Omega_{0}}\partial_{i}(\partial_{h}^{s}\tilde{a}_{ij}\partial_{j}\tilde{v})w\,dy\right|=\left|\int_{\Omega_{0}}\partial_{h}^{s}\tilde{a}_{ij}\partial_{j}\tilde{v}\partial_{i}w\right|\notag\\
\leq& C\|\nabla\tilde{v}\|_{L^{2}(\Omega_{0})}\|\nabla w\|_{L^{2}(\Omega_{0})}\leq C\|\tilde{v}\|_{X^{k}},
\end{align*}
Therefore, we deduce that for $k=1,2,3,$ $\|\partial_{h}^{s}F^{k}(h,\tilde{v})\|_{Y^{k}}\leq C\|\tilde{v}\|_{X^{k}}$ for all $\tilde{v}\in X^{k}$. Then for any $h\in K$, $\partial_{h}^{s}F^{k}(h,\cdot):X^{k}\rightarrow Y^{k}$ is a bounded linear operator with uniformly bounded norm on $K$. Hence we have from standard theories in functional analysis that for $k=1,2,3,$ $F^{k}$ is a $C^{\infty}$ map from $K\times X^{k}$ to $Y^{k}$.

\end{proof}

Note that for $k=1,2,3,$ $F^{k}(h,\tilde{u}+\tilde{v})-F^{k}(h,\tilde{u})=F^{k}(h,\tilde{v}):=\mathcal{L}^{h,k}_{\tilde{u}}\tilde{v}$, where $h\in K$ and $\tilde{u},\tilde{v}\in X^{k}$. Then we obtain that the linear bounded operator $\mathcal{L}^{h,k}_{\tilde{u}}:X^{k}\rightarrow Y^{k}$ is the Fr\'{e}chet derivative of $F^{k}$ with respect to $\tilde{u}$ at $(h,\tilde{u}).$ When $h=0$, $y=\psi(x)=x$. Then we have $\mathcal{L}^{0,k}_{u_{0}^{k}}\tilde{v}=\mathrm{div}(A_{0}^{k}\nabla\tilde{v})$, where $A_{0}^{k}=a_{0}^{k}(x)I$ with $a_{0}^{k}(x)$ defined by \eqref{QKW001}--\eqref{QKW00100} with $h=0$. In particular, $\mathcal{L}^{0,k}_{u_{0}^{k}}u_{0}^{k}=\mathrm{div}(A_{0}^{k}\nabla u_{0}^{k})=0.$

\begin{prop}\label{QAT002}
For every $k=1,2,3,$ the operator $\mathcal{L}^{0,k}_{u_{0}^{k}}:X^{k}\rightarrow Y^{k}$ is an isomorphism.
\end{prop}

\begin{proof}
For $k=1,2,3$, if $\tilde{v}_{i}\in X^{k}$, $i=1,2$ satisfy $\mathcal{L}^{0,k}_{u_{0}^{k}}\tilde{v}_{1}=\mathcal{L}^{0,k}_{u_{0}^{k}}\tilde{v}_{2}$, then $\mathrm{div}(A_{0}^{k}\nabla(\tilde{v}_{1}-\tilde{v}_{2}))=0$. Since $\tilde{v}_{1}-\tilde{v}_{2}=0$ on $\partial D$, it then follows from the uniqueness of weak solution that $\tilde{v}_{1}=\tilde{v}_{2}$ in $D$. Then for $k=1,2,3,$ the map $\mathcal{L}^{0,k}_{u_{0}^{k}}:X^{k}\rightarrow Y^{k}$ is injective. In the following we divide into three subcases to verify that for $k=1,2,3,$ $\mathcal{L}^{0,k}_{u_{0}^{k}}:X^{k}\rightarrow Y^{k}$ is surjective.

{\bf Case 1.} Consider the case when $k=1.$ Choose $g\in C^{2}(\overline{D})$ such that $g=C_{1}^{0}$ in $D_{1}^{0}$, $g=C_{2}^{0}$ in $D_{2}$, and $g=\varphi$ on $\partial D$, where the constants $C_{i}^{0}$, $i=1,2$ are defined by the fourth line of problem \eqref{con006}. Then $\mathrm{div}(A_{0}^{1}\nabla g)\in H^{-1}(D)$. For any $f\in Y^{1}$, we have $f-\mathrm{div}(A_{0}^{1}\nabla g)\in H^{-1}(D)\cap C^{0,\gamma}(\overline{\Omega}_{0})$ and thus $f-\mathrm{div}(A_{0}^{1}\nabla g)\in H^{-1}(\Omega_{0})\cap C^{0,\gamma}(\overline{\Omega}_{0})$. Since $\Delta:H^{1}_{0}(\Omega_{0})\rightarrow H^{-1}(\Omega_{0})$ is an isomorphism, there exists a unique weak solution $w\in H^{1}_{0}(\Omega_{0})$ such that $\mathrm{div}(A_{0}^{1}\nabla w)=f-\mathrm{div}(A_{0}^{1}\nabla g)$ in $\Omega_{0}$. In view of $f-\mathrm{div}(A_{0}^{1}\nabla g)\in C^{0,\gamma}(\overline{\Omega}_{0}),$ we deduce from the elliptic regularity theories that $w\in C^{2,\gamma}(\overline{\Omega}_{0}).$ Set
\begin{align*}
\tilde{w}=
\begin{cases}
w,&\mathrm{in}\;\Omega_{0},\\
0,&\mathrm{in}\;D\setminus\Omega_{0},
\end{cases}\quad \tilde{u}=\tilde{w}+g,\,\;\mathrm{in}\;D.
\end{align*}
Therefore, $\tilde{u}\in X^{1}$ satisfies $\mathcal{L}^{0,1}_{u_{0}^{1}}\tilde{u}=f$.

{\bf Case 2.} Consider the case when $k=2.$ For any $f\in Y^{2}$, we have $f\in C^{0,\gamma}(\overline{\Omega}_{0})$. Then by the classical elliptic theories, we obtain that there exists a unique solution $\tilde{u}_{1}\in C^{2,\gamma}(\overline{\Omega}_{0})$ for equation $\mathrm{div}(A_{0}^{2}\nabla \tilde{u}_{1})=f$ satisfying that $\tilde{u}_{1}=\varphi$ on $\partial D$ and $\frac{\partial\tilde{u}_{1}}{\partial\nu}|_{+}=0$ on $\partial D_{1}^{0}\cup D_{2}$. Since $\tilde{u}_{1}\in C^{2,\gamma}(\partial D_{1}^{0}\cup \partial D_{2})$ and $f\in C^{0,\gamma}(\overline{D_{1}^{0}\cup D_{2}})$, it then follows from the elliptic theories again that there exists unique solutions $\tilde{u}_{2}\in C^{2,\gamma}(\overline{D_{1}^{0}})$ and $\tilde{u}_{3}\in C^{2,\gamma}(\overline{D}_{2})$ such that
\begin{align*}
\begin{cases}
\Delta\tilde{u}_{2}=f,&\mathrm{in}\; D_{1}^{0},\\
\tilde{u}_{2}=\tilde{u}_{1},&\mathrm{on}\;\partial D_{1}^{0},
\end{cases}\quad
\begin{cases}
\Delta\tilde{u}_{3}=f,&\mathrm{in}\; D_{2},\\
\tilde{u}_{3}=\tilde{u}_{1},&\mathrm{on}\;\partial D_{2}.
\end{cases}
\end{align*}
Take
\begin{align*}
\tilde{u}=
\begin{cases}
\tilde{u}_{1},&\mathrm{in}\;\overline{\Omega}_{0},\\
\tilde{u}_{2},&\mathrm{in}\;D_{1}^{0},\\
\tilde{u}_{3},&\mathrm{in}\; D_{2}.
\end{cases}
\end{align*}
Then $\tilde{u}\in X^{2}$ solves $\mathcal{L}^{0,2}_{u_{0}^{2}}\tilde{u}=f$.

{\bf Case 3.} Consider the case of $k=3$. Pick $g\in C^{2}(\overline{D})$ satisfying that $g=C^{0}$ in $D_{2}$, and $g=\varphi$ on $\partial D$, where $C^{0}$ is determined by the fifth line of problem \eqref{PRO09}. For any $f\in Y^{3}$, we have $f-\mathrm{div}(A_{0}^{3}\nabla g)\in C^{0,\gamma}(\overline{\Omega}_{0})$. It then follows from the classical elliptic theories that there exists a unique solution $\tilde{u}_{1}\in C^{2,\gamma}(\overline{\Omega}_{0})$ for equation $\mathrm{div}(A_{0}^{3}\nabla \tilde{u}_{1})=f-\mathrm{div}(A_{0}^{3}\nabla g)$ such that $\tilde{u}_{1}=0$ on $\partial D\cup\partial D_{2}$ and $\frac{\partial\tilde{u}_{1}}{\partial\nu}|_{+}=0$ on $\partial D_{1}^{0}$. Since $\tilde{u}_{1}\in C^{2,\gamma}(\partial D_{1}^{0})$, then we have from the classical elliptic theories again that there exists a unique solution $\tilde{u}_{2}\in C^{2,\gamma}(\overline{D_{1}^{0}})$ satisfying that
\begin{align*}
\begin{cases}
\Delta\tilde{u}_{2}=f,&\mathrm{in}\; D_{1}^{0},\\
\tilde{u}_{2}=\tilde{u}_{1},&\mathrm{on}\;\partial D_{1}^{0}.
\end{cases}
\end{align*}
Let
\begin{align*}
\tilde{u}=
\begin{cases}
\tilde{u}_{1}+g,&\mathrm{in}\;\overline{\Omega}_{0},\\
\tilde{u}_{2},&\mathrm{in}\;D_{1}^{0},\\
g,&\mathrm{in}\; D_{2}.
\end{cases}
\end{align*}
Hence $\tilde{u}\in X^{3}$ verifies $\mathcal{L}^{0,3}_{u_{0}^{3}}\tilde{u}=f$. The proof is complete.

\end{proof}

We are now ready to give the proofs of Theorems \ref{thm001}, \ref{thm002} and \ref{thm003}.

\begin{proof}[Proofs of Theorems \ref{thm001}, \ref{thm002} and \ref{thm003}]
Combining Propositions \ref{QAT001} and \ref{QAT002}, we deduce from the implicit function theorem (see \cite{N2001}) that for $k=1,2,3,$ there exist a small positive constant $h_{k}$ and a unique function $\tilde{v}^{k}(\cdot,y)\in C^{\infty}((-h_{k},h_{k}))$ such that $\tilde{v}^{k}(h,\cdot)\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{0})$ satisfies equation \eqref{ZWQ001Z66} and $\tilde{v}^{k}(0,y)=u_{0}^{k}(y)$. By taking $v^{k}(h,x)=\tilde{v}^{k}(h,y)$ with $y=\psi(x)$, we obtain that $v^{k}(\cdot,x)\in C^{\infty}((-h_{k},h_{k}))$, $v^{k}(0,x)=u_{0}^{k}(x)$, and $v^{k}(h,\cdot)\in H^{1}(D)\cap C^{2,\gamma}(\overline{\Omega}_{h})$ satisfies equation \eqref{con002}. It remains to establish the stability of $v^{k}(h,x)$ with respect to $h$.

{\bf Case 1.} Consider the case of $k=1$. For any fixed $h\in(-h_{1},h_{1})$, the solution $v^{1}(h,x)$ of problem \eqref{con002} can be split as follows:
\begin{align}\label{DEC001}
v^{1}(h,x)=C_{1}^{h}v_{1}(h,x)+C_{2}^{h}v_{2}(h,x)+v_{0}(h,x),\quad\mathrm{in}\;\Omega_{h},
\end{align}
where $v_{i}$, $i=0,1,2$, respectively, solve
\begin{align*}
\begin{cases}
\Delta_{x} v_{0}=0,&\mathrm{in}\;\Omega_{h},\\
v_{0}=0,&\mathrm{on}\;\overline{D_{1}^{h}\cup D_{2}},\\
v_{0}=\varphi,&\mathrm{on}\;\partial D,
\end{cases}
\end{align*}
and
\begin{align*}
\begin{cases}
\Delta_{x} v_{1}=0,&\mathrm{in}\;\Omega_{h},\\
v_{1}=1,&\mathrm{on}\;\overline{D_{1}^{h}},\\
v_{1}=0,&\mathrm{on}\;\overline{D}_{2}\cup\partial D,
\end{cases}\quad\begin{cases}
\Delta_{x} v_{2}=0,&\mathrm{in}\;\Omega_{h},\\
v_{2}=1,&\mathrm{on}\;\overline{D}_{2},\\
v_{2}=0,&\mathrm{on}\;\overline{D_{1}^{h}}\cup\partial D.
\end{cases}
\end{align*}
Substituting \eqref{DEC001} into the fourth line of \eqref{con006}, we have
\begin{align*}
\begin{cases}
a_{11}^{h}C_{1}^{h}+a_{12}^{h}C_{2}^{h}+b_{1}^{h}=0,\\
a_{21}^{h}C_{1}^{h}+a_{22}^{h}C_{2}^{h}+b_{2}^{h}=0,
\end{cases}
\end{align*}
where $a_{ij}^{h}$ and $b_{i}^{h}$, $i,j=1,2,$ are defined by
\begin{align*}
a_{1j}^{h}=\int_{\partial D_{1}^{h}}\frac{\partial v_{j}}{\partial\nu},\quad a_{2j}^{h}=\int_{\partial D_{2}}\frac{\partial v_{j}}{\partial\nu},\quad b_{1}^{h}=\int_{\partial D_{1}^{h}}\frac{\partial v_{0}}{\partial\nu},\quad b_{2}^{h}=\int_{\partial D_{2}}\frac{\partial v_{0}}{\partial\nu}.
\end{align*}
By Cramer's rule, we derive
\begin{align*}
C_{1}^{h}=\frac{a_{12}^{h}b_{2}^{h}-b_{1}^{h}a_{22}^{h}}{a_{11}^{h}a_{22}^{h}-a_{12}^{h}a_{21}^{h}},\quad C_{2}^{h}=\frac{a_{21}^{h}b_{1}^{h}-b_{2}^{h}a_{11}^{h}}{a_{11}^{h}a_{22}^{h}-a_{12}^{h}a_{21}^{h}}.
\end{align*}
Integrating by parts, we obtain
\begin{align*}
a_{ij}^{h}=\int_{\Omega_{h}}\nabla v_{i}\nabla v_{j},\quad b_{i}^{h}=\int_{\Omega_{h}}\nabla v_{0}\nabla v_{i},\quad i,j=1,2.
\end{align*}

For $x\in\partial D_{1}^{h}\setminus D_{1}^{0}$, we have from the mean value theorem and the boundary estimates for elliptic equations that for some $\theta_{i}\in(0,1)$, $i=0,1,2,$
\begin{align*}
|v_{i}(h,x)-v_{i}(0,x)|=&|v_{i}(0,x',x_{n}-h)-v_{i}(0,x',x_{n})|\notag\\
=&|\partial_{x_{n}} v_{i}(0,x',x_{n}-\theta_{i}h)|h\leq Ch.
\end{align*}
In exactly the same way, we deduce that for $x\in\partial D_{1}^{0}\setminus D_{1}^{h}$,
\begin{align*}
|v_{i}(h,x)-v_{i}(0,x)|=|v_{i}(h,x',x_{n})-v_{i}(h,x',x_{n}+h)|\leq Ch.
\end{align*}
These two relations, in combination with the fact that $v_{i}(h,x)-v_{i}(0,x)=0$ on $\partial D_{2}\cup\partial D$, gives that
\begin{align*}
|v_{i}(h,x)-v_{i}(0,x)|\leq Ch,\quad\mathrm{on}\;\partial(D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}).
\end{align*}
By the maximum principle, we further have
\begin{align*}
|v_{i}(h,x)-v_{i}(0,x)|\leq Ch,\quad\mathrm{in}\;D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}.
\end{align*}
For $i=0,1,2,$ denote
\begin{align*}
V_{i}(h,x)=\frac{v_{i}(h,x)}{h},\quad h\in(-h_{1},h_{1}).
\end{align*}
Then we have $|V_{i}(h,x)-V_{i}(0,x)|\leq C\,\;\mathrm{in}\;D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}.$ This, together with the standard elliptic estimates, shows that
\begin{align*}
\|V_{i}(h,\cdot)-V_{i}(0,\cdot)\|_{C^{2}(\overline{D\setminus (D_{1}^{0}\cup D_{1}^{h}\cup D_{2})})}\leq C,
\end{align*}
and thus
\begin{align*}
\|v_{i}(h,\cdot)-v_{i}(0,\cdot)\|_{C^{2}(\overline{D\setminus (D_{1}^{0}\cup D_{1}^{h}\cup D_{2})})}\leq Ch,
\end{align*}
which leads to that for $i,j=1,2,$
\begin{align*}
a_{ij}^{h}=&\int_{\Omega_{h}}\nabla v_{i}(h,x)\nabla v_{j}(h,x)=\int_{D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}}\nabla v_{i}(h,x)\nabla v_{j}(h,x)+O(h)\notag\\
=&\int_{D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}}[\nabla v_{i}(0,x)\nabla v_{j}(0,x)+(\nabla v_{i}(h,x)-\nabla v_{i}(0,x))\nabla v_{j}(h,x)]\notag\\
&+\int_{D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}}\nabla v_{i}(0,x)(\nabla v_{j}(h,x)-\nabla v_{j}(0,x))+O(h)\notag\\
=&a_{ij}^{0}+O(h).
\end{align*}
By the same argument, we also have $b_{i}^{h}=b_{i}^{0}+O(h)$, $i=1,2.$ Hence we obtain
\begin{align}\label{DZQAM001}
C_{i}^{h}=C_{i}^{0}+O(h),\quad i=1,2.
\end{align}
Observe that for $i=0,1,2,$ $v_{i}(h,x)=v_{i}(0,x)$ on $\overline{D_{1}^{0}\cap D_{1}^{h}}$. Then for $x\in D_{1}^{h}\setminus\overline{D_{1}^{0}}$, we deduce from the standard elliptic estimates that
\begin{align*}
&|v_{i}(h,x',x_{n})-v_{i}(0,x',x_{n})|\notag\\
&=\bigg|v_{i}(0,x',x_{n})-v_{i}\Big(0,x',3+\mathrm{sgn}(h)\Big(1-\sum^{n-1}_{i=1}|x_{i}|^{m}\Big)^{1/m}\Big)\bigg|\leq Ch,
\end{align*}
where $\mathrm{sgn}$ is the sign function satisfying that $\mathrm{sgn}(h)=1$ if $h>0$, and $\mathrm{sgn}(h)=-1$ if $h<0$. In exactly the same way, we obtain that for $x\in D_{1}^{0}\setminus\overline{D_{1}^{h}}$,
\begin{align*}
&|v_{i}(h,x',x_{n})-v_{i}(0,x',x_{n})|\notag\\
&=\bigg|v_{i}(h,x',x_{n})-v_{i}\Big(h,x',3+h-\mathrm{sgn}(h)\Big(1-\sum^{n-1}_{i=1}|x_{i}|^{m}\Big)^{1/m}\Big)\bigg|\leq Ch.
\end{align*}
Combining these above facts, we obtain that for $h\in(-h_{1},h_{1})$,
\begin{align}\label{ZMQ006}
\|v_{i}(h,\cdot)-v_{i}(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(h),\quad i=0,1,2.
\end{align}
Observe that
\begin{align}\label{ZMQ009}
v^{1}(h,x)-v^{1}(0,x)=&\sum^{2}_{i=1}\left(C_{i}^{h}(v_{i}(h,x)-v_{i}(0,x))+(C_{i}^{h}-C_{i}^{0})v_{i}(0,x)\right)\notag\\
&+v_{0}(h,x)-v_{0}(0,x).
\end{align}
Then inserting \eqref{DZQAM001}--\eqref{ZMQ006} into \eqref{ZMQ009}, we deduce
\begin{align*}
\|v^{1}(h,\cdot)-v^{1}(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(h),\quad h\in(-h_{1},h_{1}).
\end{align*}

{\bf Case 2.} Consider the case of $k=2$. To begin with, for $h\in(-h_{2},h_{2})$, in view of $\frac{\partial v^{2}(0,x)}{\partial\nu}|_{+}=0$ on $\partial D_{1}^{0}\cup \partial D_{2}$ and $\frac{\partial v^{2}(h,x)}{\partial\nu}|_{+}=0$ on $\partial D_{1}^{h}\cup\partial D_{2}$, it follows from the standard elliptic estimates that $x\in \partial D_{1}^{h}\setminus D_{1}^{0}$,
\begin{align*}
\left|\frac{\partial (v^{2}(h,x)-v^{2}(0,x))}{\partial\nu}\Big|_{+}\right|=&\left|\frac{\partial v^{2}(0,x',x_{n}-h)}{\partial\nu}\Big|_{+}-\frac{\partial v^{2}(0,x',x_{n})}{\partial\nu}\Big|_{+}\right|\notag\\
\leq&|\nabla(v^{2}(0,x',x_{n}-h)-v^{2}(0,x',x_{n}))|\leq Ch,
\end{align*}
and, for $x\in\partial D_{1}^{0}\setminus D_{1}^{h}$,
\begin{align*}
\left|\frac{\partial (v^{2}(h,x)-v^{2}(0,x))}{\partial\nu}\Big|_{+}\right|=&\left|\frac{\partial v^{2}(h,x',x_{n})}{\partial\nu}\Big|_{+}-\frac{\partial v^{2}(h,x',x_{n}+h)}{\partial\nu}\Big|_{+}\right|\notag\\
\leq&|\nabla(v^{2}(h,x',x_{n})-v^{2}(h,x',x_{n}+h))|\leq Ch.
\end{align*}
Then we have
\begin{align}\label{NED001}
\frac{\partial (v^{2}(h,x)-v^{2}(0,x))}{\partial\nu}\Big|_{+}=
\begin{cases}
O(h),& \mathrm{on}\;(\partial D_{1}^{h}\setminus D_{1}^{0})\cup(\partial D_{1}^{0}\setminus D_{1}^{h}),\\
0,&\mathrm{on}\;\partial D_{2}.
\end{cases}
\end{align}
For simplicity, denote $w(h,x):=v^{2}(h,x)-v^{2}(0,x)$ and $\widetilde{\Omega}_{h}:=D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}$. Then $w(h,x)$ solves
\begin{align*}
\begin{cases}
\Delta_{x} w(h,x)=0,&\mathrm{in}\;\widetilde{\Omega}_{h},\\
w(h,x)=0,&\mathrm{on}\;\partial D.
\end{cases}
\end{align*}
Multiplying this equation by $|w(h,x)|^{p-2}w(h,x)$ with $p\geq2$, it then follows from \eqref{NED001}, integration by parts, H\"{o}lder's inequality, Trace Theorem and Poincar\'{e}'s inequality that for $h\in(-h_{2},h_{2}),$
\begin{align*}
&(p-1)\int_{\widetilde{\Omega}_{h}}|\nabla w(h,x)|^{2}|w(h,x)|^{p-2}=\int_{\partial\widetilde{\Omega}_{h}}\frac{\partial w(h,x)}{\partial\nu}\Big|_{+}\cdot w(h,x)|w(h,x)|^{p-2}\notag\\
&\leq Ch\int_{(\partial D_{1}^{h}\setminus D_{1}^{0})\cup(\partial D_{1}^{0}\setminus D_{1}^{h})}|w(h,x)|^{p-1}\leq Ch\||w(h,\cdot)|^{p-1}\|_{W^{1,1}(\widetilde{\Omega}_{h})}\notag\\
&\leq Ch\|\nabla |w(h,\cdot)|^{p-1}\|_{L^{1}(\widetilde{\Omega}_{h})}=Ch(p-1)\int_{\widetilde{\Omega}_{h}}|\nabla w(h,x)||w(h,x)|^{p-2}\notag\\
&\leq Ch(p-1)\left(\int_{\widetilde{\Omega}_{h}}|\nabla w(h,x)|^{2}|w(h,x)|^{p-2}\right)^{\frac{1}{2}}\left(\int_{\widetilde{\Omega}_{h}}|w(h,x)|^{p-2}\right)^{\frac{1}{2}}\notag\\
&\leq Ch(p-1)\left(\int_{\widetilde{\Omega}_{h}}|\nabla w(h,x)|^{2}|w(h,x)|^{p-2}\right)^{\frac{1}{2}}\|w(h,\cdot)\|_{L^{p}(\widetilde{\Omega}_{h})}^{\frac{p-2}{2}},
\end{align*}
which implies that
\begin{align}\label{ZWQA05}
\left(\int_{\widetilde{\Omega}_{h}}\big|\nabla|w(h,x)|^{\frac{p}{2}}\big|^{2}\right)^{\frac{1}{2}}\leq Cph\|w(h,\cdot)\|_{L^{p}(\widetilde{\Omega}_{h})}^{\frac{p-2}{2}}.
\end{align}
Applying Sobolev-Poincar\'{e} inequalities to \eqref{ZWQA05}, we obtain
\begin{align}\label{ZWQA08}
\|w(h,\cdot)\|_{L^{tp}(\widetilde{\Omega}_{h})}\leq& (Cp^{2})^{1/p}h^{2/p}\|w(h,\cdot)\|_{L^{p}(\widetilde{\Omega}_{h})}^{\frac{p-2}{p}},
\end{align}
where $t:=t(n)$ is given by
\begin{align*}
\begin{cases}
t>1,&n=2,\\
t=\frac{n}{n-2},&n>2.
\end{cases}
\end{align*}
Let
\begin{align*}
p_{k}=2t^{k},\quad k\geq0,\,n\geq2.
\end{align*}
By iteration with \eqref{ZWQA08}, we deduce that for $k>2$,
\begin{align*}
\|w(h,\cdot)\|_{L^{p_{k}}(\widetilde{\Omega}_{h})}\leq&\prod^{k-1}_{i=0}(Cp_{i}^{2})^{1/p_{i}}h^{\frac{2}{p_{k-1}}+\sum\limits^{k-2}_{j=0}\frac{2}{p_{j}}\prod\limits^{k-1}_{l=j+1}(1-\frac{2}{p_{l}})}\notag\\
\leq& Ch t^{\sum\limits^{k-1}_{i=0}\frac{i}{t^{i}}}\leq Ch,
\end{align*}
where $C$ is independent of $k$. By sending $k\rightarrow\infty$, we obtain
\begin{align}\label{AMYW001}
\|v^{2}(h,\cdot)-v^{2}(0,\cdot)\|_{L^{\infty}(\overline{D\setminus(D_{1}^{0}\cup D_{1}^{h}\cup D_{2})})}\leq Ch.
\end{align}
It is worthwhile to emphasize that \eqref{AMYW001} is proved by using Moser's iteration argument. For $x=(x',x_{n})\in D_{1}^{h}\setminus D_{1}^{0}$, denote
\begin{align*}
I_{1}:=&v^{2}\Big(h,x',3+h+\mathrm{sgn}(h)\Big(1-\sum^{n-1}_{i=1}|x_{i}|^{m}\Big)^{1/m}\Big)\notag\\
&-v^{2}\Big(0,x',3+h+\mathrm{sgn}(h)\Big(1-\sum^{n-1}_{i=1}|x_{i}|^{m}\Big)^{1/m}\Big),\\
I_{2}:=&v^{2}(h,x',x_{n})-v^{2}\Big(h,x',3+h+\mathrm{sgn}(h)\Big(1-\sum^{n-1}_{i=1}|x_{i}|^{m}\Big)^{1/m}\Big),\\
I_{3}:=&v^{2}\Big(0,x',3+h+\mathrm{sgn}(h)\Big(1-\sum^{n-1}_{i=1}|x_{i}|^{m}\Big)^{1/m}\Big)-v^{2}(0,x',x_{n}).
\end{align*}
From \eqref{AMYW001}, we have $|I_{1}|\leq Ch$. Applying the standard elliptic estimates for $v^{2}(h,x)$ in $\overline{D_{1}^{h}}$ and $v^{2}(0,x)$ in $\overline{D\setminus(D_{1}^{0}\cup D_{2})}$, we obtain that $|I_{2}|\leq Ch$ and $I_{3}\leq Ch$. Then we obtain
\begin{align*}
\|v^{2}(h,\cdot)-v^{2}(0,\cdot)\|_{L^{\infty}(\overline{D_{1}^{h}\setminus D_{1}^{0}})}\leq Ch.
\end{align*}
In exactly the same way, we also have $\|v^{2}(h,x)-v^{2}(0,h)\|_{L^{\infty}(\overline{D_{1}^{0}\setminus D_{1}^{h}})}\leq Ch.$ Hence we have
\begin{align}\label{ZZW006}
\|v^{2}(h,\cdot)-v^{2}(0,\cdot)\|_{L^{\infty}(\overline{(D_{1}^{h}\setminus D_{1}^{0})\cup(D_{1}^{0}\setminus D_{1}^{h})})}\leq Ch.
\end{align}
Then applying the maximum principle for $v^{2}(h,x)-v^{2}(0,x)$ in $\overline{D_{1}^{0}\cap D_{1}^{h}}$, we derive
\begin{align*}
\|v^{2}(h,\cdot)-v^{2}(0,\cdot)\|_{L^{\infty}(\overline{D_{1}^{0}\cap D_{1}^{h}})}\leq Ch,
\end{align*}
which, together with \eqref{AMYW001}--\eqref{ZZW006}, reads that
\begin{align}\label{QLAZM006}
\|v^{2}(h,\cdot)-v^{2}(0,\cdot)\|_{L^{\infty}(\overline{D})}\leq Ch,\quad h\in(-h_{2},h_{2}).
\end{align}

{\bf Case 3.} Consider the case when $k=3$. Similarly as before, for any given $h\in(-h_{3},h_{3})$, we decompose the solution $v^{3}(h,x)$ of problem \eqref{PRO09} as follows:
\begin{align}\label{DEC006}
v^{3}(h,x)=C^{h}v_{1}(h,x)+v_{2}(h,x),\quad\mathrm{in}\;\Omega_{h},
\end{align}
where $v_{i}$, $i=1,2$, respectively, satisfy
\begin{align*}
\begin{cases}
\Delta_{x} v_{1}(h,x)=0,&\mathrm{in}\;D\setminus(\partial D_{1}^{h}\cup\overline{D}_{2}),\\
\frac{\partial v_{1}(h,x)}{\partial\nu}\big|_{+}=0,&\mathrm{on}\;\partial D_{1}^{h},\\
v_{1}(h,x)=1,&\mathrm{on}\;\overline{D}_{2},\\
v_{1}(h,x)=0,&\mathrm{on}\;\partial D,
\end{cases}\quad\begin{cases}
\Delta_{x} v_{2}(h,x)=0,&\mathrm{in}\;D\setminus(\partial D_{1}^{h}\cup\overline{D}_{2}),\\
\frac{\partial v_{2}(h,x)}{\partial\nu}\big|_{+}=0,&\mathrm{on}\;\partial D_{1}^{h},\\
v_{2}(h,x)=0,&\mathrm{on}\;\overline{D}_{2},\\
v_{2}(h,x)=\varphi,&\mathrm{on}\;\partial D.
\end{cases}
\end{align*}
Inserting \eqref{DEC006} into the fifth line of \eqref{PRO09}, we obtain
\begin{align*}
a_{21}^{h}C^{h}=Q^{h}[\varphi],\quad\text{and thus } C^{h}=\frac{Q^{h}[\varphi]}{a_{21}^{h}},
\end{align*}
where $a_{21}^{h}$ and $Q^{h}[\varphi]$ are given by
\begin{align*}
a_{21}^{h}=\int_{\partial D_{2}}\frac{\partial v_{1}(h,x)}{\partial\nu},\quad Q^{h}[\varphi]=-\int_{\partial D_{2}}\frac{\partial v_{2}(h,x)}{\partial\nu}.
\end{align*}
From integration by parts, we see
\begin{align*}
a_{21}^{h}=\int_{\Omega_{h}}|\nabla v_{1}(h,x)|^{2},\quad Q^{h}[\varphi]=-\int_{\Omega_{h}}\nabla v_{1}(h,x)\nabla v_{2}(h,x).
\end{align*}
By the same arguments as in \eqref{QLAZM006}, it follows from Moser's iteration argument, the standard elliptic estimates and the maximum principle that for $i=1,2,$
\begin{align*}
\|v_{i}(h,\cdot)-v_{i}(0,\cdot)\|_{L^{\infty}(\overline{D})}\leq Ch,
\end{align*}
which, in combination with the rescale argument and the standard elliptic estimates, reads that
\begin{align*}
\|v_{i}(h,\cdot)-v_{i}(0,\cdot)\|_{C^{2}(\overline{D\setminus (D_{1}^{0}\cup D_{1}^{h}\cup D_{2})})}\leq Ch.
\end{align*}
This leads to that
\begin{align*}
a_{21}^{h}=a_{21}^{0}+O(h),\quad Q^{h}[\varphi]=Q^{0}[\varphi]+O(h),
\end{align*}
and we thus have $C^{h}=C^{0}+O(h)$. Note that
\begin{align*}
v^{3}(h,x)-v^{3}(0,x)=&C^{h}(v_{1}(h,x)-v_{1}(0,x))+(C^{h}-C^{0})v_{1}(0,x)\notag\\
&+v_{2}(h,x)-v_{2}(0,x).
\end{align*}
Therefore, combining these above facts, we deduce
\begin{align*}
\|v^{3}(h,\cdot)-v^{3}(0,\cdot)\|_{L^{\infty}(\overline{D})}\leq Ch.
\end{align*}
The proof is complete.

\end{proof}

\section{Regularity and stability for the elasticity problem}\label{SEC03}

As shown in the appendix of \cite{BLL2015}, problem \eqref{La.002} is the limiting case of the following problem
\begin{align}\label{ZWE006}
\begin{cases}
\partial_\alpha(A_{ij}^{\alpha\beta}\partial_\beta u_{h}^j)=0,&\mathrm{in}\; D,\\
u_{h}=\varphi, &\mathrm{on}\;\partial D,
\end{cases}
\end{align}
with
\begin{align*}
A_{ij}^{\alpha\beta}(x)=
\begin{cases}
\lambda_{1}\delta_{i\alpha}\delta_{j\beta}+\mu_{1}(\delta_{i\beta}\delta_{\alpha j}+\delta_{ij}\delta_{\alpha\beta}),&\mathrm{in}\;D_{1}^{h}\cup D_{2},\\
\lambda\delta_{i\alpha}\delta_{j\beta}+\mu(\delta_{i\beta}\delta_{\alpha j}+\delta_{ij}\delta_{\alpha\beta}),&\mathrm{in}\;\Omega_{h},
\end{cases}
\end{align*}
as $\mu_{1},n\lambda_{1}+2\mu_{1}\rightarrow\infty$. So problem \eqref{La.002} can be rewritten as \eqref{ZWE006} with $(\lambda_{1},\mu_{1})$ and $(\lambda,\mu)$ satisfying that $\mu_{1}=n\lambda_{1}+2\mu_{1}=\infty$, $\mu,n\lambda+2\mu\in(0,\infty)$.

Denote $\tilde{u}_{h}(y)=u_{h}(x)$, where $y=\psi(x)$. Then problem \eqref{ZWE006} becomes
\begin{align}\label{ZWQ001}
\begin{cases}
\partial_\alpha(\tilde{A}_{ij}^{\alpha\beta}\partial_\beta \tilde{u}_{h}^j)=0,&\mathrm{in}\; D,\\
\tilde{u}_{h}=\varphi, &\mathrm{on}\;\partial D,
\end{cases}
\end{align}
where, for $i,j,\alpha,\beta=1,2,...,n,$ $\tilde{A}_{ij}^{\alpha\beta}|_{D_{1}^{0}\cup D_{2}}=A_{ij}^{\alpha\beta}|_{D_{1}^{h}\cup D_{2}}$, and for $i,j=1,2,...,n,$
\begin{align*}
(\tilde{A}_{ij}^{\alpha\beta}(y))=\frac{(\partial_{x}y)(A_{ij}^{\alpha\beta})(\partial_{x}y)^{t}}{\det(\partial_{x}y)}=(\lambda\delta_{i\alpha}\delta_{j\beta}+\mu(\delta_{i\beta}\delta_{\alpha j}+\delta_{ij}\delta_{\alpha\beta}))+O(h),\quad\mathrm{in}\;\Omega_{0}.
\end{align*}
Similarly as above, define
\begin{align*}
X=&\{\tilde{v}\in H^{1}(D;\mathbb{R}^{n})\cap C^{2,\gamma}(\overline{\Omega}_{0};\mathbb{R}^{n})\,|\,\text{$\nabla\tilde{v}+(\nabla\tilde{v})^{T}=0$ in $\overline{D_{1}^{0}\cup D_{2}}$, $\tilde{v}=\varphi$ on $\partial D$}\},\notag\\
Y=&\{f\in H^{-1}(D;\mathbb{R}^{n})\cap C^{0,\gamma}(\overline{\Omega}_{0};\mathbb{R}^{n})\,|\,\text{$f=0$ in $\overline{D_{1}^{0}\cup D_{2}}$}\},
\end{align*}
with their norms as $\|\tilde{v}\|_{X}:=\|\tilde{v}\|_{H^{1}(D;\mathbb{R}^{n})}+\|\nabla\tilde{v}\|_{L^{\infty}(\frac{1}{2}D;\mathbb{R}^{n})}$ and
$$\|f\|_{Y}:=\|f\|_{H^{-1}(D;\mathbb{R}^{n})}=\sup\{\langle f,w\rangle\,|\,w\in H^{1}_{0}(D;\mathbb{R}^{n}),\,\|w\|_{H^{1}_{0}(D;\mathbb{R}^{n})}\leq1\},$$
where $\gamma$ is given by \eqref{alpha}, $\langle\, ,\rangle$ denotes the pairing between $H^{-1}(D;\mathbb{R}^{n})$ and $H_{0}^{1}(D;\mathbb{R}^{n})$. It is not difficult to demonstrate that $X$ and $Y$ are Banach spaces. Define $K:=(-\delta_{0},\delta_{0})$ and $F(h,\tilde{v}):=(F_{1}(h,\tilde{v}),...,F_{n}(h,\tilde{v}))=(\partial_\alpha(\tilde{A}_{1j}^{\alpha\beta}\partial_\beta \tilde{v}^j),...,\partial_\alpha(\tilde{A}_{nj}^{\alpha\beta}\partial_\beta \tilde{v}^j)).$

\begin{lemma}\label{QAT001V2}
The map $F$ is in $C^{\infty}(K\times X,Y)$ in the sense that $F$ possesses continuous Fr\'{e}chet derivatives of any order.

\end{lemma}

\begin{proof}
To begin with, for any $k\geq0$, $h\in K$ and $\tilde{v}\in X$, we have
\begin{align*}
\partial_{h}^{k}F(h,\tilde{v})=
\begin{cases}
(\partial_\alpha(\partial_{h}^{k}\tilde{A}_{1j}^{\alpha\beta}\partial_\beta \tilde{v}^j),...,\partial_\alpha(\partial_{h}^{k}\tilde{A}_{nj}^{\alpha\beta}\partial_\beta \tilde{v}^j)),& \mathrm{in}\; \Omega_{0},\\
(0,...,0),&\mathrm{in}\;D\setminus\Omega_{0}.
\end{cases}
\end{align*}
Observe that for each $h\in K$, $\partial_{h}^{k}F$ is linear in $\tilde{v}$. By a direct calculation, we obtain that for $i,j,\alpha,\beta=1,2,...,n,$ $|\partial_{h}^{k}\tilde{A}_{ij}^{\alpha\beta}|\leq C(k,m,n)$ in $\overline{K\times\Omega_{0}}$. This, in combination with the fact that $\tilde{v}\in X$, implies that $\partial_{h}^{k}F(h,\tilde{v})\in L^{2}(D;\mathbb{R}^{n})$. Then we deduce from integration by parts, H\"{o}lder's inequality and Trace Theorem that for all $w\in H_{0}^{1}(D;\mathbb{R}^{n})$, $\|w\|_{H^{1}_{0}(D;\mathbb{R}^{n})}\leq1$,
\begin{align*}
|\langle \partial_{h}^{k}F,w\rangle|=&\left|\int_{D}\partial_{h}^{k}F(h,\tilde{v})w\,dy\right|=\left|\int_{\Omega_{0}}\partial_{h}^{k}F(h,\tilde{v})w\,dy\right|\notag\\
=&\left|\int_{\partial D_{1}^{0}\cup\partial D_{2}}\partial_{h}^{k}\tilde{A}_{ij}^{\alpha\beta}\partial_\beta \tilde{v}^j\nu_{\alpha}w^{i}-\int_{\Omega_{0}}\partial_{h}^{k}\tilde{A}_{ij}^{\alpha\beta}\partial_\beta \tilde{v}^j\partial_{\alpha}w^{i}dy\right|\notag\\
\leq&C(\|\nabla\tilde{v}\|_{L^{\infty}(\frac{1}{2}D;\mathbb{R}^{n})}\|w\|_{L^{2}(\partial D_{1}^{0}\cup\partial D_{2})}+\|\nabla \tilde{v}\|_{L^{2}(\Omega_{0};\mathbb{R}^{n})}\|\nabla w\|_{L^{2}(D;\mathbb{R}^{n})})\notag\\
\leq& C(\|\nabla\tilde{v}\|_{L^{\infty}(\frac{1}{2}D;\mathbb{R}^{n})}+\|\nabla\tilde{v}\|_{L^{2}(\Omega_{0};\mathbb{R}^{n})})\|w\|_{H^{1}_{0}(D;\mathbb{R}^{n})}\leq C\|\tilde{v}\|_{X}.
\end{align*}
This yields that $\|\partial_{h}^{k}F(h,\tilde{v})\|_{Y}\leq C(k,m,n)\|\tilde{v}\|_{X}$ for all $\tilde{v}\in X$. Hence, for any $h\in K$, $\partial_{h}^{k}F(h,\cdot):X\rightarrow Y$ is a bounded linear operator with uniformly bounded norm on $K$. Then it follows from standard theories in functional analysis that $F$ is a $C^{\infty}$ map from $K\times X$ to $Y$.

\end{proof}

Observe that $F(h,\tilde{u}+\tilde{v})-F(h,\tilde{u})=F(h,\tilde{v}):=\mathcal{L}^{h}_{\tilde{u}}\tilde{v}$, where $h\in K$ and $\tilde{u},\tilde{v}\in X$. Therefore, we know that the linear bounded operator $\mathcal{L}^{h}_{\tilde{u}}:X\rightarrow Y$ is the Fr\'{e}chet derivative of $F$ with respect to $\tilde{u}$ at $(h,\tilde{u}).$ Moreover, we have $\mathcal{L}^{0}_{u_{0}}\tilde{v}=(\partial_\alpha(A_{1j}^{\alpha\beta}\partial_\beta \tilde{v}^j),...,\partial_\alpha(A_{nj}^{\alpha\beta}\partial_\beta \tilde{v}^j))$ and $\mathcal{L}^{0}_{u_{0}}u_{0}=0.$

\begin{lemma}\label{QAT002V2}
The operator $\mathcal{L}^{0}_{u_{0}}:X\rightarrow Y$ is an isomorphism.
\end{lemma}

\begin{proof}
Similar to Proposition \ref{QAT002}, we deduce from the uniqueness of weak solution that $\mathcal{L}^{0}_{u_{0}}:X\rightarrow Y$ is injective. It remains to show that $\mathcal{L}^{0}_{u_{0}}:X\rightarrow Y$ is also surjective. Take $g\in C^{2}(\overline{D};\mathbb{R}^{n})$ such that $g=\sum^{\frac{n(n+1)}{2}}_{\alpha=1}C_{1\alpha}^{0}\phi_{\alpha}$ in $D_{1}^{0}$, $g=\sum^{\frac{n(n+1)}{2}}_{\alpha=1}C_{2\alpha}^{0}\phi_{\alpha}$ in $D_{2}$, and $g=\varphi$ on $\partial D$, where the constants $C_{i\alpha}^{0}$, $i=1,2,\,\alpha=1,2,...,\frac{n(n+1)}{2}$ are given by \eqref{La.002}. Then $\partial_\alpha(A_{ij}^{\alpha\beta}\partial_\beta g^j)=0$ in $D_{1}^{0}\cup D_{2}$, $i=1,2,...,\frac{n(n+1)}{2}$, and $(\partial_\alpha(A_{1j}^{\alpha\beta}\partial_\beta g^j),...,\partial_\alpha(A_{nj}^{\alpha\beta}\partial_\beta g^j))\in H^{-1}(D;\mathbb{R}^{n})$. For any $f\in Y$, we have $f-(\partial_\alpha(A_{1j}^{\alpha\beta}\partial_\beta g^j),...,\partial_\alpha(A_{nj}^{\alpha\beta}\partial_\beta g^j))\in H^{-1}(D;\mathbb{R}^{n})\cap C^{0,\gamma}(\overline{\Omega}_{0};\mathbb{R}^{n})$ and then it also belongs to $H^{-1}(\Omega_{0})\cap C^{0,\gamma}(\overline{\Omega}_{0})$. Hence by using the regularity theories for elliptic systems, we obtain that there exists a unique solution $w\in H^{1}_{0}(\Omega_{0};\mathbb{R}^{n})\cap C^{2,\gamma}(\overline{\Omega}_{0})$ such that $\partial_\alpha(A_{ij}^{\alpha\beta}\partial_\beta w^j)=f^{i}-\partial_\alpha(A_{ij}^{\alpha\beta}\partial_\beta g^j)$ in $\Omega_{0}$. Let
\begin{align*}
\tilde{w}=
\begin{cases}
w,&\mathrm{in}\;\Omega_{0},\\
0,&\mathrm{in}\;\overline{D_{1}^{0}\cup D_{2}},
\end{cases}\quad \tilde{u}=\tilde{w}+g,\,\;\mathrm{in}\;D.
\end{align*}
Then $\tilde{u}\in X$ solves $\mathcal{L}^{0}_{u_{0}}\tilde{u}=f$. The proof is finished.

\end{proof}

Based on the above facts, we now present the proof of Theorem \ref{thm005}.
\begin{proof}[Proof of Theorem \ref{thm005}]
A combination of the implicit function theorem and Lemmas \ref{QAT001V2} and \ref{QAT002V2} shows that there exist a small constant $h_{0}>0$ and a unique function $\tilde{v}(\cdot,y)\in C^{\infty}((-h_{0},h_{0});\mathbb{R}^{n})$ such that $\tilde{v}(h,\cdot)\in H^{1}(D;\mathbb{R}^{n})\cap C^{2,\gamma}(\overline{\Omega}_{0};\mathbb{R}^{n})$ is the solution for problem \eqref{ZWQ001} and $\tilde{v}(0,y)=u_{0}(y)$. By letting $v(h,x)=\tilde{v}(h,y)$ with $y=\psi(x)$, we derive that $v(\cdot,x)\in C^{\infty}((-h_{0},h_{0});\mathbb{R}^{n})$, $v(0,x)=u_{0}(x)$, and $v(h,\cdot)\in H^{1}(D;\mathbb{R}^{n})\cap C^{2,\gamma}(\overline{\Omega}_{h};\mathbb{R}^{n})$ solves equation \eqref{La.002}. We now proceed to study the stability of $v(h,x)$ with respect to $h$.

For later convenience, we first rewrite the original problem \eqref{La.002} as follows:
\begin{align}\label{AAE001}
\begin{cases}
\nabla\cdot(\mathbb{C}^0e(v))=0,&\mathrm{in}\;\Omega_{h},\\
v|_{+}=v|_{-},&\mathrm{on}\ \partial D_{1}^{h}\cup\partial D_{2},\\
v=\sum^{\frac{n(n+1)}{2}}_{\alpha=1}C_{1\alpha}^{h}\phi_{\alpha},&\mathrm{on}\;\overline{D_{1}^{h}},\\
v=\sum^{\frac{n(n+1)}{2}}_{\alpha=1}C_{2\alpha}^{h}\phi_{\alpha},&\mathrm{on}\;\overline{D}_{2},\\
\int_{\partial{D}_{1}^{h}}\frac{\partial v}{\partial \nu_0}\big|_{+}\cdot\phi_{\alpha}=0,&\alpha=1,2,...,\frac{n(n+1)}{2},\\
\int_{\partial{D}_{2}}\frac{\partial v}{\partial \nu_0}\big|_{+}\cdot\phi_{\alpha}=0,&\alpha=1,2,...,\frac{n(n+1)}{2},\\
v=\varphi,&\hbox{on}\ \partial{D},
\end{cases}
\end{align}
where $e(v)=\frac{1}{2}(\nabla v+(\nabla v)^{T})$, $\mathbb{C}^{0}=(C^{0}_{ijkl})$, $C_{ijkl}^{0}=\lambda\delta_{ij}\delta_{kl} +\mu(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})$, $(\mathbb{C}^{0}e(v))_{ij}=\sum_{k,l=1}^{n}C_{ijkl}^{0}(e(v))_{kl}$, and
\begin{align*}
\frac{\partial v}{\partial \nu_0}\Big|_{+}&:=(\mathbb{C}^0e(v))\nu=\lambda(\nabla\cdot v)\nu+\mu(\nabla v+(\nabla v)^T)\nu.
\end{align*}
For $h\in(-h_{0},h_{0})$, we split the solution $v(h,x)$ of problem \eqref{AAE001} as follows:
\begin{align}\label{DEC002}
v(h,x)=\sum^{\frac{n(n+1)}{2}}_{\alpha=1}\sum^{2}_{i=1}C_{i\alpha}^{h}v_{i\alpha}(h,x)+v_{0}(h,x),\quad\mathrm{in}\;\Omega_{h},
\end{align}
where $v_{0}$ and $v_{i\alpha}$, $i=1,2,\,\alpha=1,2,...,\frac{n(n+1)}{2}$, respectively, satisfy
\begin{align*}
\begin{cases}
\nabla\cdot(\mathbb{C}^0e(v_{0}))=0,&\mathrm{in}\;\Omega_{h},\\
v_{0}=0,&\mathrm{on}\;\overline{D_{1}^{h}\cup D_{2}},\\
v_{0}=\varphi,&\mathrm{on}\;\partial D,
\end{cases}
\end{align*}
and
\begin{align*}
\begin{cases}
\nabla\cdot(\mathbb{C}^0e(v_{1\alpha}))=0,&\mathrm{in}\;\Omega_{h},\\
v_{1\alpha}=\phi_{\alpha},&\mathrm{on}\;\overline{D_{1}^{h}},\\
v_{1\alpha}=0,&\mathrm{on}\;\overline{D}_{2}\cup\partial D,
\end{cases}\;\,\begin{cases}
\nabla\cdot(\mathbb{C}^0e(v_{2\alpha}))=0,&\mathrm{in}\;\Omega_{h},\\
v_{2\alpha}=\phi_{\alpha},&\mathrm{on}\;\overline{D}_{2},\\
v_{2\alpha}=0,&\mathrm{on}\;\overline{D_{1}^{h}}\cup\partial D.
\end{cases}
\end{align*}
For $i=1,2$ and $\alpha,\beta=1,2,...,\frac{n(n+1)}{2}$, denote
\begin{align*}
a_{i1\alpha\beta}^{h}:=&\int_{\partial{D}_{1}^{h}}\frac{\partial v_{i\alpha}}{\partial \nu_0}\Big|_{+}\cdot\phi_{\beta},\quad a_{i2\alpha\beta}^{h}:=\int_{\partial{D}_{2}}\frac{\partial v_{i\alpha}}{\partial \nu_0}\Big|_{+}\cdot\phi_{\beta} ,\notag\\ b_{1\beta}^{h}:=&-\int_{\partial D_{1}^{h}}\frac{\partial v_{0}}{\partial \nu_0}\Big|_{+}\cdot\phi_{\beta},\quad b_{2\beta}^{h}:=-\int_{\partial D_{2}}\frac{\partial v_{0}}{\partial \nu_0}\Big|_{+}\cdot\phi_{\beta}.
\end{align*}
Then substituting \eqref{DEC002} into the fifth and sixth lines of \eqref{AAE001}, we obtain that for $\beta=1,2,...,\frac{n(n+1)}{2},$
\begin{align}\label{AHNTW009}
\begin{cases}
\sum\limits_{\alpha=1}^{\frac{n(n+1)}{2}}\sum\limits^{2}_{i=1}C_{i\alpha}^{h} a_{i1\alpha\beta}^{h}=b_{1\beta}^{h},\\
\sum\limits_{\alpha=1}^{\frac{n(n+1)}{2}}\sum\limits^{2}_{i=1}C_{i\alpha}^{h} a_{i2\alpha\beta}^{h}=b_{2\beta}^{h}.
\end{cases}
\end{align}
For brevity, denote $\mathbb{A}^{h}_{ij}=(a_{ij\alpha\beta}^{h})_{\frac{n(n+1)}{2}\times\frac{n(n+1)}{2}}$, $i,j=1,2,$ and
\begin{align*}
X_{i}=\big(C_{i1}^{h},C_{i2}^{h},...,C_{i\frac{n(n+1)}{2}}^{h}\big)^{T},\;\,Y_{i}=\big(b_{i1}^{h},b_{i2}^{h},...,b_{i\frac{n(n+1)}{2}}^{h}\big)^{T},\quad i=1,2.
\end{align*}
Therefore, \eqref{AHNTW009} becomes
\begin{gather*}
\begin{pmatrix} \mathbb{A}_{11}^{h}&\mathbb{A}_{21}^{h} \\  \mathbb{A}_{12}^{h}&\mathbb{A}_{22}^{h}
\end{pmatrix}
\begin{pmatrix}
X_{1}\\
X_{2}
\end{pmatrix}=
\begin{pmatrix}
Y_{1}\\
Y_{2}
\end{pmatrix}.
\end{gather*}
For $i,j=1,2,\,\alpha=1,2,...,\frac{n(n+1)}{2}$, by replacing the elements of $\alpha$-th column in the matrix $\mathbb{A}_{ij}^{h}$ by column vector $(Y^{1},Y^{2})^{T}$, we obtain new matrix $\mathbb{A}_{ij\alpha}^{h}$ as follows:
\begin{gather*}
\mathbb{A}_{ij\alpha}^{h}=
\begin{pmatrix}
a_{ij11}^{h}&\cdots&b_{j1}^{h}&\cdots&a_{ij1\frac{n(n+1)}{2}}^{h} \\\\ \vdots&\ddots&\vdots&\ddots&\vdots\\\\a_{ij\frac{n(n+1)}{2}1}^{h}&\cdots&b_{j\frac{n(n+1)}{2}}^{h}&\cdots&a_{ij\frac{n(n+1)}{2}\frac{n(n+1)}{2}}^{h}
\end{pmatrix}.
\end{gather*}
For $\alpha=1,2,...,\frac{n(n+1)}{2}$, write
\begin{align*}
\mathbb{F}^{h}_{1\alpha}=\begin{pmatrix} \mathbb{A}^{h}_{11\alpha}&\mathbb{A}^{h}_{21} \\  \mathbb{A}^{h}_{12\alpha}&\mathbb{A}^{h}_{22}
\end{pmatrix},\;\,\mathbb{F}^{h}_{2\alpha}=\begin{pmatrix} \mathbb{A}^{h}_{11}&\mathbb{A}^{h}_{21\alpha} \\  \mathbb{A}^{h}_{12}&\mathbb{A}^{h}_{22\alpha}
\end{pmatrix},\;\, \mathbb{F}^{h}=\begin{pmatrix} \mathbb{A}^{h}_{11}&\mathbb{A}^{h}_{21} \\  \mathbb{A}^{h}_{12}&\mathbb{A}^{h}_{22}
\end{pmatrix}.
\end{align*}
It then follows from Cramer's rule that
\begin{align*}
C_{i\alpha}^{h}=\frac{\det\mathbb{F}^{h}_{i\alpha}}{\det\mathbb{F}^{h}},\quad i=1,2,\,\alpha=1,2,...,\frac{n(n+1)}{2}.
\end{align*}

The following proof is similar to that in the scalar case above. First, it follows from the boundary estimates for elliptic systems that for $x\in\partial D_{1}^{h}\setminus D_{1}^{0}$,
\begin{align*}
|v_{i\alpha}(h,x)-v_{i\alpha}(0,x)|\leq|v_{i\alpha}(0,x',x_{n}-h)-v_{i\alpha}(0,x',x_{n})|+h\leq Ch,
\end{align*}
and for $x\in\partial D_{1}^{0}\setminus D_{1}^{h}$,
\begin{align*}
|v_{i\alpha}(h,x)-v_{i\alpha}(0,x)|\leq|v_{i\alpha}(h,x',x_{n})-v_{i\alpha}(h,x',x_{n}+h)|+h\leq Ch.
\end{align*}
In view of the fact that $v_{i\alpha}(h,x)-v_{i\alpha}(0,x)=0$ on $\partial D_{2}\cup\partial D$, we then have
\begin{align*}
|v_{i\alpha}(h,x)-v_{i\alpha}(0,x)|\leq Ch,\quad\mathrm{on}\;\partial(D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}).
\end{align*}
From the maximum modulus principle, we derive
\begin{align*}
|v_{i\alpha}(h,x)-v_{i\alpha}(0,x)|\leq Ch,\quad\mathrm{in}\;D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}.
\end{align*}
Then we deduce from the rescale argument and the standard interior and boundary estimates for elliptic systems that
\begin{align}\label{ZKA001}
\|v_{i\alpha}(h,\cdot)-v_{i\alpha}(0,\cdot)\|_{C^{2}(\overline{D\setminus (D_{1}^{0}\cup D_{1}^{h}\cup D_{2})})}\leq Ch.
\end{align}
Since $v_{i\alpha}(h,x)=v_{i\alpha}(0,x)$ in $\overline{D_{1}^{0}\cap D_{1}^{h}}$, it then follows from the standard elliptic estimates that for $x\in D_{1}^{h}\setminus\overline{D_{1}^{0}}$,
\begin{align*}
&|v_{i\alpha}(h,x',x_{n})-v_{i\alpha}(0,x',x_{n})|\notag\\
&\leq\bigg|v_{i\alpha}(0,x',x_{n})-v_{i\alpha}\Big(0,x',3+\mathrm{sgn}(h)\Big(1-\sum^{n-1}_{i=1}|x_{i}|^{m}\Big)^{1/m}\Big)\bigg|+h\leq Ch,
\end{align*}
and for $x\in D_{1}^{0}\setminus\overline{D_{1}^{h}}$,
\begin{align*}
&|v_{i\alpha}(h,x',x_{n})-v_{i\alpha}(0,x',x_{n})|\notag\\
&\leq\bigg|v_{i\alpha}(h,x',x_{n})-v_{i\alpha}\Big(h,x',3+h-\mathrm{sgn}(h)\Big(1-\sum^{n-1}_{i=1}|x_{i}|^{m}\Big)^{1/m}\Big)\bigg|+h\leq Ch,
\end{align*}
where $\mathrm{sgn}(h)$ is the sign function in $h$. A consequence of these above facts shows that for $h\in(-h_{0},h_{0})$, $i=1,2,\,\alpha=1,2,...,\frac{n(n+1)}{2}$, $\|v_{i\alpha}(h,\cdot)-v_{i\alpha}(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(h)$. By the same argument, we also have $\|v_{0}(h,\cdot)-v_{0}(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(h)$.

Observe from integration by parts that for $i,j=1,2$ and $\alpha,\beta=1,2,...,\frac{n(n+1)}{2}$,
\begin{align*}
a_{ij\alpha\beta}^{h}=\int_{\Omega_{h}}(\mathbb{C}^{0}e(v_{i\alpha}),e(v_{j\beta})),\quad b_{i\alpha}^{h}=-\int_{\Omega_{h}}(\mathbb{C}^{0}e(v_{0}),e(v_{i\alpha})).
\end{align*}
From \eqref{ZKA001}, we obtain that for $i,j=1,2,\,\alpha,\beta=1,2,...,\frac{n(n+1)}{2}$,
\begin{align*}
a_{ij\alpha\beta}^{h}=&\int_{D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}}(\mathbb{C}^{0}e(v_{i\alpha}(h,x)),e(v_{j\beta}(h,x)))+O(h)\notag\\
=&\int_{D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}}(\mathbb{C}^{0}e(v_{i\alpha}(h,x)-v_{i\alpha}(0,x)),e(v_{j\beta}(h,x)))\notag\\
&+\int_{D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}}(\mathbb{C}^{0}e(v_{i\alpha}(0,x)),e(v_{j\beta}(h,x)-v_{j\beta}(0,x)))\notag\\
&+\int_{D\setminus\overline{D_{1}^{0}\cup D_{1}^{h}\cup D_{2}}}(\mathbb{C}^{0}e(v_{i\alpha}(0,x)),e(v_{j\beta}(0,x)))+O(h)\notag\\
=&a_{ij\alpha\beta}^{0}+O(h).
\end{align*}
In exactly the same way, we obtain $b_{i\alpha}^{h}=b_{i\alpha}^{0}+O(h)$, $i=1,2,\,\alpha=1,2,...,\frac{n(n+1)}{2}$. Combining these above facts, we deduce that $C_{i\alpha}^{h}=C_{i\alpha}^{0}+O(h)$, $i=1,2,\,\alpha=1,2,...,\frac{n(n+1)}{2}$. Since
\begin{align*}
v(h,x)-v(0,x)=&\sum^{2}_{i=1}\sum^{\frac{n(n+1)}{2}}_{\alpha=1}\left(C_{i\alpha}^{h}(v_{i\alpha}(h,x)-v_{i\alpha}(0,x))+(C_{i\alpha}^{h}-C_{i\alpha}^{0})v_{i\alpha}(0,x)\right)\notag\\
&+v_{0}(h,x)-v_{0}(0,x),
\end{align*}
it then follows from the above facts that
\begin{align*}
\|v(h,\cdot)-v(0,\cdot)\|_{L^{\infty}(\overline{D})}=O(h),\quad h\in(-h_{0},h_{0}).
\end{align*}

\end{proof}

\noindent{\bf{\large Acknowledgements.}}
The author is deeply grateful to Professor JinGang Xiong for asking this question and giving thorough guidance and numerous supports. The author was partially supported by CPSF (2021M700358).

\bibliographystyle{plain}

\def$'${$'$}

redpajama_set_name: RedPajamaArXiv


WAKEFIELD — Chief Michael Sullivan reports that the Wakefield Fire Department is responding to a low-tier hazardous materials incident in town.
A state hazmat team has been dispatched to 19 Fairmount Ave. to respond to Tier 1 activation, following a report by a resident that he dropped a mercury thermometer in his home.
Upon assembly of the team, two of the five team members will don hazmat suits and enter the home, where they will conduct air sampling tests and test the contents of the bag where the thermometer was disposed. The team will then further isolate the substance and prepare it for proper disposal.
Following an initial investigation, Wakefield Fire determined that the resident double-bagged the damaged thermometer and has since obtained medical attention for an unrelated matter. The homeowner reported to firefighters that he did not suffer mercury contamination as a result of the spill, but may obtain further tests as a precaution.


text:
Contaminated retention pond ordered closed at Clean Harbors Colfax
Richard Sharkey
rsharkey@thetowntalk.com, (318) 487-6490
The retention pond at Clean Harbors Colfax in Grant Parish will be closed following tests that showed it had levels of toxins that exceeded standards.
Water from the contaminated pond is released into tributaries that ultimately carry it to the Red River.
State Department of Environmental Quality Secretary Chuck Carr Brown said Clean Harbors Colfax has been directed to get rid of the pond and replace it with a system "that will provide greater protections to human health and the environment."
The pond, which holds rainwater with residue from chemicals on the facility's burn pads, will be replaced by a storage tank/treatment system.
"We're talking about a maximum of six to nine months" to replace the pond, Phillip Retallick told The Town Talk on Thursday. He is a senior vice president of Clean Harbors, the parent company of the Colfax-area facility.
"We're still designing the replacement system. It is going to have a containment system. It's just deciding what type of containment system to use," he said
Brenda Vallee, one of the leaders of the Central Louisiana Coalition for A Safe and Healthy Environment, said the contaminated pond should be shut down quickly.
"We want it, like, yesterday," Vallee said of closing the pond. "We don't want to drag it out. We want it closed now."
Vallee lives about 2 miles from Clean Harbors Colfax.
The ordered closure of the pond was noted in two letters from Louisiana Department of Environmental Quality obtained by The Town Talk, including one from DEQ Secretary Brown.
"… the LDEQ directed Clean Harbors to: close the Pond (as soon as practicable) in accordance with applicable regulatory standards and to replace the Pond with alternate wastewater treatment system that will provide greater protections to human health and the environment …," Secretary Brown wrote in a Sept. 16 letter to state Rep. Terry Brown.
Terry Brown, I-Colfax, had expressed concerns about the pond and the burn pad in a letter to DEQ earlier in September.
That came after tests conducted in June and reported to DEQ in August found higher-than-allowable amounts of arsenic, lead, perchlorate and explosive compound RDX in the pond, documents show.
Clean Harbors Colfax conducts open burning of more than 500,000 pounds of munitions, explosives and other waste annually at its facility located about 5 miles from the town of Colfax.
Central Louisiana Coalition for A Safe and Healthy Environment, a citizens group, was formed to oppose open burning at Clean Harbors Colfax.
Vallee and other CLHCC members have said they are not looking to close the facility down, but they would like to see a closed-container system used instead of open burning.
The company says the open burning is not a threat to public health, but it is researching alternative methods of disposal.
Wilma Subra, a New Iberia chemist who provides technical assistance to Louisiana Environmental Action Network, said closed-container burning could be used for most of the waste stream at Clean Harbors Colfax.
But some of the materials there should be disposed of through methods other than open burning or closed- container burning, she said.
"Not all of the material that they are able to accept would be appropriate for the contained-burn category …," Subra told The Town Talk. "But there are other methods of dealing with the other components that you could not do in some type of contained burn."
Subra was at Vallee's house on Wednesday and took photos of a large, black smoke plume coming from the Clean Harbors Colfax site.
DEQ Secretary Brown, in his letter, said measures are underway to mitigate the pond contamination.
"… As directed by DEQ, Clean Harbors has begun implementing and is continuing to evaluate interim measures that will reduce the volume and/or concentration of contaminated storm water entering the Pond prior to the Pond's closure," Brown said in the letter.
He said measures being implemented or under consideration "include, but are not limited to, use of explosion-proof vacuum(s) to collect ash residues from the burn pans and the Pad, the immediate removal of all sludges/sediments currently located in the Pond, daily inspection of storm water filters, and frequent power washing of the Pad (with muriatic acid)."
The pad is where open burning takes place on burn pans. Among the materials openly burned at Clean Harbors Colfax are high explosives, warheads, fireworks, rocket motors, munitions, propellants, shaped charges, detonating cord, nitro-related compounds and undeployed airbags.
In addition to Secretary Brown's letter to Rep. Brown, Vallee received a letter dated Sept. 23 from DEQ's Elliott Vega, assistant secretary, Office of Environmental Services.
The letter assured Vallee that DEQ is "working to address the issues associated with exceedances of Risk Evaluation/Corrective Action Program (RECAP) screening standards in the recently sampled groundwater wells at the facility. In addition to directing Clean Harbors to close the Retention Pond as soon as is practicable, the Department is weighing its options to address the collections of groundwater samples surrounding the Retention Pond."
Open-burn bill withdrawn; compromise resolution offered
Subra said samples were taken from the pond and the sediment in the pond, which is "primarily the debris" from the burn pad and burn pans.
Although the pond showed higher-than-allowed levels of certain toxins, the water from the pond can be discharged, she said.
"There were allowed, when the pond got to a certain level, to discharge it into the surface streams. The discharge only required them to monitor for oil and grease, and chemical oxygen demand and pH. …
"So even if they had those things in it, they could discharge because there wasn't a limit that it had to be below," Subra said.
The pond is a "source of surface contamination and a source of groundwater contamination," which she believes led to DEQ's directive to close the pond.
When the facility replaces the pond with a storage tank, Subra hopes that water, even if treated, won't be released into streams. She would like to see Clean Harbors "haul it off to some other location."
However, Retallick said the replacement system is expected to be a "combination of the containment system and a treatment system, which would have a discharge."
The company will meet all criteria set by DEQ, Retallick said, adding, "We're committed to making the change."
Vallee said burning at Clean Harbors Colfax had stopped for a while a couple of weeks ago, and now it appears the burning is somewhat reduced from the norm.
Retallick said the company halted burning at times during the past 30 days while it performed "some maintenance and upgrades to our burn pad," including pouring new cement slabs.
Regular burning operations at the site have resumed.

redpajama_set_name: RedPajamaCommonCrawl


Caring for our children is a high priority at Forest Baptist Church. Our First Steps Preschool leadership team works extra hard so that members and visitors alike feel at home leaving their children in our care. The clean, modern, preschool facilities are designed with children's needs in mind and provide spacious room and adequate resources for those from birth to five years of age.


text:
Q: Import maps + Stimulus: error in JS "Unknown Content-Type text/html" I'm having a JS issue with my first Rails app, which I suspect is related to my using import maps instead of Webpack. I've searched and searched but haven't found any discussions of this.
It's a Rails 6 app with some JS via Stimulus, which I installed by adding importmap-rails and then stimulus-rails. It works fine locally, but in production on Heroku the JS doesn't work and I see errors like this in the browser console: Uncaught (in promise) Error: Unknown Content-Type "text/html; charset=utf-8" doFetch https://plainreading.herokuapp.com/assets/es-module-shims-424762490b07cc7f9a37745e8b53356bca11997a8e313f47d780e29009df7e8e.js:580
I'm wondering if it's because I removed Webpack from my app, using How to completely remove webpack and all its dependencies from Rails App. I removed it because I was getting Webpack-related build errors in Heroku, and it's my understanding that I don't need Webpack if I'm using import maps.
A while ago I fixed a similar issue in a static site on Netlify by including this in its netlify.toml config:
[[headers]]
  for = "/*.js"
  [headers.values]
    Content-Type = "text/javascript; charset=utf-8"

I couldn't find any similar config for Heroku. So then I tried customizing the response headers in the app, but I couldn't find a way to do that for my JS files, only for the main HTML response and for public assets.
Here's the repo in case it helps: https://github.com/fpsvogel/plainreading

A: In the end I solved the issue by doing the following:

*

*Install Rails 7 alpha.

*Create a test project with import maps: rails _7.0.0.alpha2_ new importmap-test. (You can use -j to set a different JS bundling option, such as -j esbuild, but the default is import maps.)

*Upgrade my main project to Rails 7 alpha.

*In my main project, make all JS-related code identical to the corresponding code in the test project. For me this meant overwriting app/javascript/controllers/index.js.

Then my JS worked without errors.

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text:
Telangana EAMCET Details 2015
After the Intermediate examinations both the Telugu States are heading for another showdown on the conduct of the Engineering Agriculture and Medical Common Entrance Test (Eamcet) with the Telangana government planning to conduct its own test and the A.P. government likely to challenge it.
Notification Issued on : Feb 25th, 2015
Online Application Start from : Feb 28th, 2015
Online Application Last Date : April 9th, 2015
With Rs.500/- Fine Last Date : April 15th, 2015
With Rs.1000/- Fine Last Date : April 22nd, 2015
With Rs.5000/- Fine Last Date : May 5th, 2015
With Rs.10,000 /- Fine Last Date : May 12th, 2015
Data Correction from : April 15 to 20, 2015
Hall Tickets Download From : May 8th to 12th, 2015
EAMCET EXAM : May 14th, 2015
Engineering : 10am to 1 pm
Medical and Agr : 2.30pm to 5.30pm
Preliminary Key : May 16th, 2015
Objections Receive Last Date : May 23rd, 2015
Declaration of Ranks : May 28th, 2015
Telangana officials have indicated the seriousness of the government to go ahead with its own test for entry into engineering and medical courses. "A meeting would be convened next week and a final decision would be taken," said T. Papi Reddy, Chairman of the Telangana State Council of Higher Education (TSCHE). But the decision may face some legal hurdles if the AP government opposes it, particularly with regard to the 15 per cent non-local (open merit) quota that has to continue for 10 years as per the A.P. Reorganisation Act.
The 15 per cent open merit quota is open to non-local candidates where they compete with the local meritorious students. The better of them will be allotted the seat irrespective of the region in both States. A separate Eamcet for both States means separate merit lists. Can candidates appearing in two separate tests compete for the same reserved seats to be filled by common merit is the question. Officials say students might challenge such admissions, and more so in medical and dental seats, where competition is high and every mark counts. The same case applies to top engineering colleges also.


RT: RT is our abbreviation for Real time. Data synchronization jobs that take place when an action occurs. Click here for more information about Real-Time.
What are webhooks and crons? What is the difference and which systems are which?
Can I set a real time (RT) job to run at a certain time?


The Biennial Project presents an open call to artists worldwide for The Biennial Project Biennial 2019 – an online juried biennial, with accepted work to be displayed on our website. In addition, all accepted work will be shown via a large screen digital presentation at a champagne reception to be held at the CENTRO CULTURALE DON ORIONE ARTIGIANELLI (donorione-venezia.it) in Venice during the Opening Press Week of The Venice Biennale 58.
– Accepted work will be displayed permanently on the very busy Biennial Project Website.
– Accepted work will also be shown on a large screen digital presentation in Venice during the Opening Week of The Venice Biennale 58.
– Online galleries with accepted work will be promoted on The Biennial Project Blog posting reaching over 40,000 art lovers.
– Work will be juried by The world renowned Biennial Project Artists, with Special Prize Winners chosen by by jurors to be announced.
– Hundreds of entries are highlighted on our well-trafficked social media sites during the months leading up to the entry deadline.
Now we know that artists love to wait until the last possible moment to respond to calls for entry.
But there are great reasons to break from this time-honored tradition and enter the The Biennial Project Biennial 2019 early.
You will have a much better chance of having your work promoted as part of the publicity for The Biennial Project Biennial 2019.
We feature an entry of the day every day during the entire entry period – sending a chosen image out on Facebook to our following of over 5,000 people – and linking to the entrant's website or facebook page.
Additionally, we send out Entry Sampler Email Blasts several times during the lead up to the contest, highlighting some of the most interesting work that we have received.
Near the end of the entry period we are swamped with the number of replies, whereas at the beginning the pace of entries is less frenzied – greatly increasing your chances of being included in one of these publicity vehicles.
Not to mention that your image could be selected to appear on a postcard or a poster for The Biennial Project Biennial 2019.


text:
Global Force Wrestling Lucha Underground Pro Wrestling Top Stories
Impact Wrestling & AAA Announce Cross-Promotion, Co-Branded Events & Distribution
Posted By: William B. West 0 Comment AAA, GFW
IMPACT Wrestling, a subsidiary of Anthem Sports & Entertainment Corp., today unveiled details about the landmark alliance announced last month with Promociones Antonio Peña S.A. de C.V., the owner of Lucha Libre AAA, Mexico's preeminent professional wrestling organization founded in 1992.
The alliance coincides with a mutual milestone, as this year marks the 15th anniversary of IMPACT Wrestling and 25th anniversary of Lucha Libre AAA.
"This is an exciting time of celebration for both IMPACT Wrestling and AAA Lucha Libre," said Ed Nordholm, president of IMPACT Wrestling. "By combining our efforts on talent, touring and distribution, we will increase the global footprint of our respective companies and create the best product for our passionate fan bases."
The previously announced talent exchange between the organizations will kick into full gear on July 2 when IMPACT Wrestling presents Slammiversary 2017 from The IMPACT Zone in Orlando, Florida. The blockbuster pay-per-view event will showcase top talent from the high-flying AAA roster.
Meanwhile, AAA will present its 25th Anniversary Triplemania XXV on August 26, 2017 in Mexico City. The annual extravaganza will feature top IMPACT Wrestling stars Bobby Lashley, Eddie Edwards and Davey Richards, and an IMPACT championship belt will be defended on Mexican soil.
The two organizations also intend to launch co-branded live tours in the fall.
In addition to these cross-promotional initiatives, Anthem Sports & Entertainment Corp. will work with Lucha Libre AAA to expand the distribution of Fight Network, the world's premier combat sports channel and home of IMPACT Wrestling in Canada. AAA will also facilitate distribution of IMPACT Wrestling in new territories throughout Latin America and South America, while Anthem will facilitate the expansion of the Lucha Libre AAA footprint through Fight Network's diverse distribution channels.
"The best way to celebrate these 25 years of uninterrupted work is sharing with all of you the news that we presented today – our alliance with IMPACT Wrestling and the announcement of our international tour that includes not only the North American stops that we will complete with IMPACT Wrestling but also countries like Colombia and Japan, in addition to other activities that will help us to celebrate this important anniversary," said Dorian Roldan, Executive VP of Enterprise Development for AAA.
Jeff Jarrett, Chief Creative Officer of IMPACT Wrestling, added: "I am excited to see the stars of IMPACT Wrestling and AAA on both anniversary shows and eager to start the co-branded live events as we continue to build promotional bridges on a global basis."
For more information, please visit www.impactwrestling.com.
Rey Mysterio At MCW In Joppa, MD This Saturday
Longtime WWE Network Subscribers Could Win WrestleMania 34 Tickets
Gail Kim Training For Comeback
May 20, 2017 William B. West 0
TNA Wrestling
This Week's TNA Impact Free Online
Paige Offers More Details On Albert El Patron Incident
July 12, 2017 William B. West 0


Station Hollywood: Happy birthday to Shilpa Shetty.
Happy birthday to Shilpa Shetty. The viewers of bollywood films wish happy birthday to Shilpa Shetty on 8th June, 2018. She had maintained her romantic image and brilliant figure for over 15 years. She had remained indispensable for the last 15 years. Starting her journey with Shahrukh Khan in the film Baazigaar, in the early 90s, she had acted in different types of films. She acted in romantic films like Dhadkan opposite Akshay Kumar and Sunil Shetty. The film had successful songs like "tum dil ki dhadkan me". She also acted in action thrillers like Prithvi, opposite Sunil Shetty. She was also sublime in off-beat films like Tarqueib. Nana Patekar played the role of the crime-investigator in the film. She was the friend of Tabu, the girl killed in the film. She also had speaking terms with Ashutosh Rana who was initially suspected to have murdered Tabu. Nana Patekar at the end of the film found out the murderer. But Shilpa Shetty showed her acting skills while relying to interrogation of Nana Patekar in the film. Her performance in the film Auzaar opposite Salman Khan and Sanjay Kapoor, also require special mention. Salman Khan was a CBI investigator in the film and was given the task of investigation the unlawful business affairs of Sanjay Kapoor. In fact the underworld involvement of Paresh Rawal, the father of Sanjay Kapoor was at issue. Shilpa Shetty bridged the gap between Salman Khan and Sanjay Kapoor, after Salman convinced her that he concealed his identity for the benefit of Sanjay Kapoor, his friend. Shilpa gave brilliant lips and danced well in the songs "oye oye oye oyeoye", and other songs composed by Anu Malik. Shilpa Shetty had shown her comedy instincts in a number of films like Chor Machaye Shor. She played the role of the daughter of Paresh Rawal, and got attached with Bobby Deol, who was a subordinate of Paresh Rawal. He pretended to be twin brother, one having a moustache, the other without a moustache. In that comedy drama, Shilpa Shetty should her innocence and comedy instinct along with Bipasha Basu, who also played a significant role in the film. Even at her middle age she had shown dance numbers which could have made young actresses envious about her. For example, in Dostana, released in 2008, she performed in a dance number opposite John Abraham, which amazed the viewers. After marrying Raj Kundra she had reduced signing in as many number of films as she used to do earlier. She also had bought major share of Rajasthan Royals team which played IPL cricket in 2010. In 2011 her team Rajasthan Royals did not fare well in IPL, so she should concentrate in films. In 2013, she had been accused of being part of match-fixing scandal. Both her husband Mr. Kundra and herself faced charges of match fixing. She should act in more films. She should be followed by the new generation of actresses who want to stay fit and contribute in bollywood films for more than 20 years.


HEARTLAND HERO: Willi Huber, left, and his wife, Edna.
Canterbury mountain man was a decorated German war hero. MIKE CREAN reports.
The Iron Cross saved Willi Huber's life. The Austrian mountain boy who fought in the German Army in World War II was awarded the highest military honour of Iron Cross First Class for his actions in the great tank battle at Kursk.
As the war neared its close Huber was seriously wounded and taken to a military hospital. Though not fully recovered, he discharged himself and headed for home in Austria.
Chaos ruled on the roads, railways and canals. He risked arrest and instant execution as a deserter. He says the people's respect for the medal he wore assured him rides and he got almost home, to Schladming in the Steria district.
Huber has lived in New Zealand since 1953, mostly in Christchurch. He became prominent in skiing, especially in establishing the Mt Hutt ski area, and in mountain climbing and guiding. He has been honoured at Mt Hutt for his contribution and by the Austrian Government for forging good relations between New Zealand and Austria.
Huber and his wife, Edna, retired to Geraldine after their Cashmere house burned down after the 2011 earthquake.
"I love New Zealand. I never would live anywhere else," he says.
Born in 1923, Huber enjoyed farm life in the Alps. He loved taking the cows and sheep into the high pastures in summer. He would stay in the mountains, climbing for weeks on end. His love of mountains moved him to seek a career in mountain guiding, against his father's wishes that he stay on the farm.
When war came, he realised he would be conscripted so he volunteered in the hope he would have some say in where he served. He was assigned to a motorcycle reconnaissance unit, as machine gunner, and saw action in the invasion of Russia. He got close enough to Moscow to see its city spires and domes. Then the Germans were turned back. He speaks of being stuck in snow and unbelievable cold, with no petrol and meagre supplies, for two months. As part of the general retreat, he was brought back to Germany for re-training in tanks. He then served as gunner in a Panzer tank on the Russian front.
The Iron Crosses (he was awarded a Second Class cross as well) recognised the hits he scored against Russian tanks at Kursk. Asked how many tanks he hit, he replies: "I can't remember and it doesn't matter".
His division was recalled to help shore up German defences in France against Allied invasion. Again, petrol supplies ran short, making battle hopeless. He recalls hearing Allied planes overhead every night, dropping supplies to French Resistance fighters, while the Germans had nothing. He knew the war was lost but fought on. The troops had been trained to believe they could still win and that Hitler's secret weapons would turn things around, he says.
Forced back into Germany, Huber was attacked by a Russian officer who stabbed him in the chest and leg before he could draw his pistol and shoot him. His knife wounds were life-threatening and he needed hospital care.
From hospital he could see boats going up the Danube River. He walked out and began a series of boat rides.
About 90km short of Schladming he had to go into hospital again. While he was there, the war ended. He completed his journey home and recuperated for three months.
Having volunteered for the German Army put him off-side with many Austrians, he says. One betrayed him to the American occupation force. He was arrested and held as a "political prisoner" for 16 months.
Huber believes his background as a country farm boy helped him survive the war.
"I had a logic of survival. I knew what to do in any circumstances - find food etc. I could deal with the country and with people."
Huber next trained and became qualified in mountain guiding and ski instructing. He made many climbs in the alps, including the first ascent of Mt Steinweg. His guiding of an English doctor who had studied in Otago led to an interest in New Zealand. The doctor told him he would love New Zealand and should climb Mt Cook. He urged Huber to apply for a scheme to go to New Zealand with a team of carpenters to assemble Austrian-built kitset houses.
Huber knew nothing of carpentry but applied anyway. He was called to an interview and told the panel he was not a carpenter. They said they knew that, but an English doctor had recommended him as someone who would do well in New Zealand. He was accepted.
He worked on State houses at Porirua for 18 months before being released from his contract. Then he bought a Volkswagen car and, with two mates, headed for Mt Cook. His plan, he says, was to climb the mountain, then return to Austria.
Bad weather made climbing impossible. While waiting for cloud to clear, he met Edna. The former Londoner had emigrated to Australia in 1948 and was at Mt Cook on a tour of New Zealand. They married in 1956 and had four children.
Huber settled his family on Christchurch's Cashmere Hills. He took many jobs to provide a good education for the children. While working at Tisdalls sports and hunting shop on Cashel St he set up the store's skiing department. He was also a commercial traveller in outdoor sports clothes and equipment.
Being in Christchurch allowed Huber to climb and ski in the Southern Alps. He lost count of his ascents of Mt Cook, many of them while guiding foreign climbers.
He was an instigator of Mt Hutt ("Just say I was a pioneer of Mt Hutt," he says modestly.) He did investigations of sites for ski runs and found the route for the access road. He negotiated with European firms for ski lift systems and equipment. He attracted Austrian Olympic ski teams for training camps.
Still there was time for Himalayan climbing expeditions, for teaching skiing and for frequent trips home.
At 90 years old, Huber still has niggles from his war wounds, but remains as lively as anyone 20 years younger.


You are here: Home / Dinner Recipes / Quick & Easy Vegan Tacos!
Quick & Easy Vegan Tacos!
Lentils, walnuts, chipotles OH MY! Yes, this "meat" is the perfect filling for vegan tacos! Add all the toppings and turn dinner into a fun plant-based taco night!
What makes these vegan tacos so great?
All wrapped in a romaine lettuce cup! Sounds delicious right?
I don't need to tell you this but tacos are one of the easiest, most fun, kid-approved meals out there. This vegan taco recipe fits the bill too. The vegan taco "meat" is beyond filling and is super healthy.
That just makes more room for extra chips, salsa, and guacamole!
Try a fun salsa as a topping like this jicama mango salsa!
If you tried these vegan tacos, let me know in the comments below and share a photo using #mindfulavocado. I'd love to hear from you! Want to see more recipes? Let's get social! CONNECT WITH ME on Facebook, Twitter, Instagram, or Pinterest to see what I'm currently cooking up!
This plant-based and gluten-free taco recipe is perfect for weeknight meals. Vegan taco filling nestled in a lettuce cup and topped with tons of veggies! This healthy meal is a MUST TRY!
Rinse uncooked lentils and add to water. Cover and reduce heat to low-medium. Let cook for 15 minutes.
Remove lentils from the burner and set aside for 3-5 minutes covered. Remove cover and fluff with a fork.
Add cooked lentils, diced tomatoes, walnuts, garlic, and seasonings to the food processor. Pulse until desired consistency is reached (about 10-12 times).
Place a medium skillet over medium heat and add 1 Tablespoon of olive oil. Add taco meat and let cook 5-8 minutes until tomatoes are cooked.
I love your clean layout. Your photos are beautiful too… and I am SO all over this dish!! Kusod to you Amanda!
That sounds really tasty, I make a delicious lentil 'bolognese' for pasta and love the texture.
These tacos scream out loud: Tuck in!! These are so much my kind of quick food, so nice and easy to make especially on summer days when cooking can be a bit of a drag. Gorgeous !
Thanks for sharing such a great recipe. My vegan cuisine will come and visit soon and will try your recipe.
Lentil and walnut?! Wow, what an interesting combo! But I LOVE both separately so how could this be bad!? I love meat but I am always looking to take a break for it every now and then and this would be so perfect!


The growth of packaged cider in recent years has been impressive to say the least. Cider has grown from having little presence in the Australian market just five years ago, to a popular trend that is suited to almost any drinking occasion today. Against a backdrop of an overall sluggish packaged liquor market, cider continues to buck the trend – posting strong sales growth and market share gains.
While the top five cider brands have not changed position, innovation has been instrumental to driving the category's stellar performance. A consistent flow of newcomers and further explosion of flavours including mango/raspberry and strawberry/kiwi – are just some of the top new product developments generating incremental dollars.
While Australians' love affair with cider continues, manufacturers and brands will be competing for their share of cider's success. To do this, marketers need to know who their consumers are in order to attract cider faithfuls and newcomers to their brand i.e. who is the cider drinker, what is the best way to reach them, when and where they are drinking, what alternate segments are in their repertoire, and what other liquor categories hold the greatest threat and opportunity for cider?
It will be those brands that have a true understanding of their target consumer that will cut through the clutter and provide a clear growth strategy to succeed in his increasingly crowded arena.
You can learn about the secrets behind cider's success in Nielsen's 2013 Cider Report.
Click here for purchase and download information.


text:
Каньйон Диявола () — природна визначна пам'ятка, яка знаходиться в штаті Аризона, США. Це глибокий каньйон, який розташований в Парку Диявола, на північ від міста Флагстафф.

Каньйон Диявола сформувався більше ніж 60 мільйонів років тому в результаті діяльності річки Диявола. Довжина каньйону становить близько 24 км, а глибина — до 300 метрів. У каньйоні можна побачити вражаючі кольорові формації скель, печери, водоспади та інші природні чудеса.

Каньйон Диявола є популярним місцем для туристів, які люблять природу та аутдор-активності. Тут можна займатися піших прогулянках, скелелазінні, рафтингу та іншими видами спорту

Примітки

Каньйони США

redpajama_set_name: RedPajamaWikipedia


In the Artists' Footsteps In the Artists' Footsteps Dumfries and Galloway can boast a rich artistic history. Landscape and light have combined to lure painters to live and work in the area for at least 200 years. The Artists' Footsteps project documents the landscape paintings, their artists and the places that inspired their work. For inclusion the painting must be of a Dumfries and Galloway subject and by a deceased artist.
Collection Collection The Kirkcudbright Artists' Collection comprises over 600 works of fine art, illustrations, books, decorative art, craftwork, silverware and furniture. In addition, it includes an extensive collection of archives and photographs.
Café Café Kirkcudbright Galleries Café offers a light, varied menu using fresh local produce, situated on the first floor of the gallery, overlooking St Cuthbert's church and flooded in natural light to create a most relaxing, modern cafe environment. Serving the leading Italian coffee "Segafredo" alongside Edinburgh's famous tea merchant Brodies. The visit to the cafe is an essential point of your tour of Kirkcudbright Galleries.


Citrouilles [H1055] - $10.00 : Yarn Tree, Your wholesale source for cross stitch supplies.
Cross stitch pattern by Camille Colje-Camps. A tower of teacups, pumpkins, black cats and mice. Hearts and spiders accent. Stitch count 81w x 163h. Stitched on #3029 Aida 20ct sand.