text:
Endothelial damage due to ischemia and reperfusion is prevented with SIN-1
D. Johnson, J. A. Freischlag, R. Lesniak, H. Kelly, J. H. Mudaliar, R. A. Cambria, M. D G R Seabrook, J. B. Towne
Acute ischemia followed by reperfusion results in direct endothelial damage characterized by cell swelling, increased permeability and loss of acetylcholine-mediated vasorelaxation. Ischemia followed by reperfusion in a New Zealand white rabbit hindlimb has been shown to result in loss of acetylcholine-induced relaxation of superficial femoral arteries. This loss of relaxation in response to acetylcholine is a reflection of the decreased nitric oxide availability that occurs with reperfusion injury. The purpose of this investigation was to evaluate the effect of SIN-1, a direct nitric oxide donor, on this endothelial injury. New Zealand white rabbits underwent complete ischemia of the right hindlimb for 3 h followed by 2 h of reperfusion. Aliquots of 20 ml of either 0.88-mM SIN-1 or normal saline was infused via a lateral branch of the right common iliac artery during the first 20 min of reperfusion. Sham vessels were subjected to the S-h operative intervention to control for anesthetic effect. Control vessels were harvested from rabbits not exposed to ischemia or reperfusion. Superficial femoral artery rings were evaluated in vitro for endothelial cell-mediated relaxation. Rings were contracted with potassium chloride and norepinephrine and then exposed to standardized incremental doses of acetylcholine to measure percent relaxation. Artery sections were sent for hematoxylin and eosin staining. No significant differences were seen in contraction caused by either potassium chloride or norepinephrine in all four experimental groups. Saline infused vessel rings relaxed a mean of 8.42 ± 2.39% and 49.57 ± 8.65% in response to acetylcholine doses of 3 × 10−8 M and 1 × 10−7M, respectively. In contrast, SIN-1 infused vessels relaxed a mean of 57.82 ± 2.65% and 100.23 ± 1.53% to the same doses of acetylcholine. Control and sham arteries showed a similar relaxation response as compared with SIN-1 infused vessels. Differences in relaxation when comparing saline infused vessels with SIN-1 infused, sham and control arteries, were significantly different at each dose of acetylcholine from 3 × 10−8M to 1 × 10−7M (P < 0.05, ANOVA). Histologic examination of the vessels revealed no morphologic differences among the experimental groups. All vessels werestructurally normal with an intact endothelium. In this model of rabbit hindlimb ischemia, preservation of endothelial cell-mediated vasorelaxation occurs with administration of intra-arterial SIN-1 during reperfusion. This preservation of endothelial function cannot be explained by histologic changes in the arterial wall or attributed to altered arterial contractility in response to potassium chloride or norepinephrine.
endothelium. SIN-1
Dive into the research topics of 'Endothelial damage due to ischemia and reperfusion is prevented with SIN-1'. Together they form a unique fingerprint.
Acetylcholine Medicine & Life Sciences 100%
Reperfusion Medicine & Life Sciences 94%
Ischemia Medicine & Life Sciences 81%
Potassium Chloride Medicine & Life Sciences 57%
Hindlimb Medicine & Life Sciences 52%
Rabbits Medicine & Life Sciences 43%
Norepinephrine Medicine & Life Sciences 39%
Arteries Medicine & Life Sciences 34%
Johnson, D., Freischlag, J. A., Lesniak, R., Kelly, H., Mudaliar, J. H., Cambria, R. A., Seabrook, M. D. G. R., & Towne, J. B. (1998). Endothelial damage due to ischemia and reperfusion is prevented with SIN-1. Vascular, 6(5), 527-532. https://doi.org/10.1177/096721099800600518
Endothelial damage due to ischemia and reperfusion is prevented with SIN-1. / Johnson, D.; Freischlag, J. A.; Lesniak, R.; Kelly, H.; Mudaliar, J. H.; Cambria, R. A.; Seabrook, M. D G R; Towne, J. B.
In: Vascular, Vol. 6, No. 5, 1998, p. 527-532.
Johnson, D, Freischlag, JA, Lesniak, R, Kelly, H, Mudaliar, JH, Cambria, RA, Seabrook, MDGR & Towne, JB 1998, 'Endothelial damage due to ischemia and reperfusion is prevented with SIN-1', Vascular, vol. 6, no. 5, pp. 527-532. https://doi.org/10.1177/096721099800600518
Johnson D, Freischlag JA, Lesniak R, Kelly H, Mudaliar JH, Cambria RA et al. Endothelial damage due to ischemia and reperfusion is prevented with SIN-1. Vascular. 1998;6(5):527-532. https://doi.org/10.1177/096721099800600518
Johnson, D. ; Freischlag, J. A. ; Lesniak, R. ; Kelly, H. ; Mudaliar, J. H. ; Cambria, R. A. ; Seabrook, M. D G R ; Towne, J. B. / Endothelial damage due to ischemia and reperfusion is prevented with SIN-1. In: Vascular. 1998 ; Vol. 6, No. 5. pp. 527-532.
@article{6df4b3aaecc84b718e12d457cd358ad2,
title = "Endothelial damage due to ischemia and reperfusion is prevented with SIN-1",
abstract = "Acute ischemia followed by reperfusion results in direct endothelial damage characterized by cell swelling, increased permeability and loss of acetylcholine-mediated vasorelaxation. Ischemia followed by reperfusion in a New Zealand white rabbit hindlimb has been shown to result in loss of acetylcholine-induced relaxation of superficial femoral arteries. This loss of relaxation in response to acetylcholine is a reflection of the decreased nitric oxide availability that occurs with reperfusion injury. The purpose of this investigation was to evaluate the effect of SIN-1, a direct nitric oxide donor, on this endothelial injury. New Zealand white rabbits underwent complete ischemia of the right hindlimb for 3 h followed by 2 h of reperfusion. Aliquots of 20 ml of either 0.88-mM SIN-1 or normal saline was infused via a lateral branch of the right common iliac artery during the first 20 min of reperfusion. Sham vessels were subjected to the S-h operative intervention to control for anesthetic effect. Control vessels were harvested from rabbits not exposed to ischemia or reperfusion. Superficial femoral artery rings were evaluated in vitro for endothelial cell-mediated relaxation. Rings were contracted with potassium chloride and norepinephrine and then exposed to standardized incremental doses of acetylcholine to measure percent relaxation. Artery sections were sent for hematoxylin and eosin staining. No significant differences were seen in contraction caused by either potassium chloride or norepinephrine in all four experimental groups. Saline infused vessel rings relaxed a mean of 8.42 ± 2.39% and 49.57 ± 8.65% in response to acetylcholine doses of 3 × 10−8 M and 1 × 10−7M, respectively. In contrast, SIN-1 infused vessels relaxed a mean of 57.82 ± 2.65% and 100.23 ± 1.53% to the same doses of acetylcholine. Control and sham arteries showed a similar relaxation response as compared with SIN-1 infused vessels. Differences in relaxation when comparing saline infused vessels with SIN-1 infused, sham and control arteries, were significantly different at each dose of acetylcholine from 3 × 10−8M to 1 × 10−7M (P < 0.05, ANOVA). Histologic examination of the vessels revealed no morphologic differences among the experimental groups. All vessels werestructurally normal with an intact endothelium. In this model of rabbit hindlimb ischemia, preservation of endothelial cell-mediated vasorelaxation occurs with administration of intra-arterial SIN-1 during reperfusion. This preservation of endothelial function cannot be explained by histologic changes in the arterial wall or attributed to altered arterial contractility in response to potassium chloride or norepinephrine.",
keywords = "endothelium. SIN-1, ischemia, nitric oxide, reperfusion",
author = "D. Johnson and Freischlag, {J. A.} and R. Lesniak and H. Kelly and Mudaliar, {J. H.} and Cambria, {R. A.} and Seabrook, {M. D G R} and Towne, {J. B.}",
journal = "Vascular",
publisher = "SAGE Publications Ltd",
T1 - Endothelial damage due to ischemia and reperfusion is prevented with SIN-1
AU - Johnson, D.
AU - Freischlag, J. A.
AU - Lesniak, R.
AU - Kelly, H.
AU - Mudaliar, J. H.
AU - Cambria, R. A.
AU - Seabrook, M. D G R
AU - Towne, J. B.
N2 - Acute ischemia followed by reperfusion results in direct endothelial damage characterized by cell swelling, increased permeability and loss of acetylcholine-mediated vasorelaxation. Ischemia followed by reperfusion in a New Zealand white rabbit hindlimb has been shown to result in loss of acetylcholine-induced relaxation of superficial femoral arteries. This loss of relaxation in response to acetylcholine is a reflection of the decreased nitric oxide availability that occurs with reperfusion injury. The purpose of this investigation was to evaluate the effect of SIN-1, a direct nitric oxide donor, on this endothelial injury. New Zealand white rabbits underwent complete ischemia of the right hindlimb for 3 h followed by 2 h of reperfusion. Aliquots of 20 ml of either 0.88-mM SIN-1 or normal saline was infused via a lateral branch of the right common iliac artery during the first 20 min of reperfusion. Sham vessels were subjected to the S-h operative intervention to control for anesthetic effect. Control vessels were harvested from rabbits not exposed to ischemia or reperfusion. Superficial femoral artery rings were evaluated in vitro for endothelial cell-mediated relaxation. Rings were contracted with potassium chloride and norepinephrine and then exposed to standardized incremental doses of acetylcholine to measure percent relaxation. Artery sections were sent for hematoxylin and eosin staining. No significant differences were seen in contraction caused by either potassium chloride or norepinephrine in all four experimental groups. Saline infused vessel rings relaxed a mean of 8.42 ± 2.39% and 49.57 ± 8.65% in response to acetylcholine doses of 3 × 10−8 M and 1 × 10−7M, respectively. In contrast, SIN-1 infused vessels relaxed a mean of 57.82 ± 2.65% and 100.23 ± 1.53% to the same doses of acetylcholine. Control and sham arteries showed a similar relaxation response as compared with SIN-1 infused vessels. Differences in relaxation when comparing saline infused vessels with SIN-1 infused, sham and control arteries, were significantly different at each dose of acetylcholine from 3 × 10−8M to 1 × 10−7M (P < 0.05, ANOVA). Histologic examination of the vessels revealed no morphologic differences among the experimental groups. All vessels werestructurally normal with an intact endothelium. In this model of rabbit hindlimb ischemia, preservation of endothelial cell-mediated vasorelaxation occurs with administration of intra-arterial SIN-1 during reperfusion. This preservation of endothelial function cannot be explained by histologic changes in the arterial wall or attributed to altered arterial contractility in response to potassium chloride or norepinephrine.
AB - Acute ischemia followed by reperfusion results in direct endothelial damage characterized by cell swelling, increased permeability and loss of acetylcholine-mediated vasorelaxation. Ischemia followed by reperfusion in a New Zealand white rabbit hindlimb has been shown to result in loss of acetylcholine-induced relaxation of superficial femoral arteries. This loss of relaxation in response to acetylcholine is a reflection of the decreased nitric oxide availability that occurs with reperfusion injury. The purpose of this investigation was to evaluate the effect of SIN-1, a direct nitric oxide donor, on this endothelial injury. New Zealand white rabbits underwent complete ischemia of the right hindlimb for 3 h followed by 2 h of reperfusion. Aliquots of 20 ml of either 0.88-mM SIN-1 or normal saline was infused via a lateral branch of the right common iliac artery during the first 20 min of reperfusion. Sham vessels were subjected to the S-h operative intervention to control for anesthetic effect. Control vessels were harvested from rabbits not exposed to ischemia or reperfusion. Superficial femoral artery rings were evaluated in vitro for endothelial cell-mediated relaxation. Rings were contracted with potassium chloride and norepinephrine and then exposed to standardized incremental doses of acetylcholine to measure percent relaxation. Artery sections were sent for hematoxylin and eosin staining. No significant differences were seen in contraction caused by either potassium chloride or norepinephrine in all four experimental groups. Saline infused vessel rings relaxed a mean of 8.42 ± 2.39% and 49.57 ± 8.65% in response to acetylcholine doses of 3 × 10−8 M and 1 × 10−7M, respectively. In contrast, SIN-1 infused vessels relaxed a mean of 57.82 ± 2.65% and 100.23 ± 1.53% to the same doses of acetylcholine. Control and sham arteries showed a similar relaxation response as compared with SIN-1 infused vessels. Differences in relaxation when comparing saline infused vessels with SIN-1 infused, sham and control arteries, were significantly different at each dose of acetylcholine from 3 × 10−8M to 1 × 10−7M (P < 0.05, ANOVA). Histologic examination of the vessels revealed no morphologic differences among the experimental groups. All vessels werestructurally normal with an intact endothelium. In this model of rabbit hindlimb ischemia, preservation of endothelial cell-mediated vasorelaxation occurs with administration of intra-arterial SIN-1 during reperfusion. This preservation of endothelial function cannot be explained by histologic changes in the arterial wall or attributed to altered arterial contractility in response to potassium chloride or norepinephrine.
KW - endothelium. SIN-1
KW - ischemia
KW - nitric oxide
KW - reperfusion
JO - Vascular
JF - Vascular


text:
Trevor Lawrence Junior (ur. 17 lutego 1974), znany również jako TrevBeats, to amerykański perkusista rhythmandbluesowy. Lawrence współpracował z takimi wykonawcami jak Snoop Dogg, Alicia Keys, Mariah Carey, Lionel Richie, Stevie Wonder, Dizzy Gillespie, Ashanti, Jennifer Lopez, Taj Mahal, Dr. Dre, Macy Gray czy The Temptations. Jest współautorem piosenki I'm So Excited.

Syn saksofonisty jazzowego Trevora Lawrencea Seniora i wokalistki Lyndy Laurence.

Przypisy

Amerykańscy perkusiści jazzowi
Amerykańscy muzycy bluesowi
Urodzeni w 1974

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}

	\hskip 0.5cm Following the seminal work of J.~Kigami~\cite{Kigami1989},~\cite{Kigami1993},
	~\cite{Kigami2001},~\cite{Kigami2003} in the field of analysis on fractals, the natural step was to explore the numerical related areas. \\

	It has been initiated, in the case of he Sierpi\'{n}ski gasket, by~K.~Dalrymple, R.~S.~Strichartz, and J.~Vinson~\cite{DalrympleFDM}, who gave an equivalent method for the finite difference approximation. More precisely, the authors use the spectral shape of the solution (heat kernel), which involves eigenvalues and eigenvectors, an therefore calls for an approximation of the eigenvalues. This work has been followed by the one of M.~Gibbons, A.~Raj and R.~S.~Strichartz~\cite{GibbonsFEM}, where they describe how one can build approximate solutions, by means of piecewise harmonic, or biharmonic, splines, again in the case of~$\mathcal SG$.  They go so far as giving theoretical error estimates, through a comparison with experimental numerical data. \\

	After our work~\cite{RianeDavidM}, where we built a Laplacian on the Minkowski curve, we went so far as implementing the resulting finite difference scheme, for which one cannot find any equivalent in the existing literature.  \\

	The novelty of our contribution layed in defining the discretization of the considered PDE's (heat and wave equation), by taking into account the recursive construction of the matrix related to the sequence of graph Laplacians. Contrary to the aforementioned work, we thus did not call for approximations of the eigenvalues. This enabled us not only to compute the consistency error, but, alos, to set stability conditions of Courant-Friedrichs-Lewy type, and, then, to prove the  convergence of the scheme.\\

	In the sequel, we extend this method to Sierpi\'{n}ski simplices (Gasket and Tetrahedron), in the case of the heat equation. First, we build the finite difference scheme. Then, we give a theoretical study of the error, compute the scheme error, give stability conditions, and prove the convergence of the scheme.

	\vskip 1cm

	\section{The Sierpi\'{n}ski simplices}

	In the sequel, we place ourselves in the Euclidean space of dimension~$d-1$ for a strictly positive integer~$d$, referred to a direct orthonormal frame. The usual Cartesian coordinates will be denoted by~$(x_1,x_2,...,x_{d-1})$.\\

	\noindent Let us introduce the family of contractions$f_i$,~\mbox{$1\leq i \leq d$}, of fixed point~$P_{i-1}$ such that, for any~$X\,\in\,\R^{d-1}$, and any integer~$i$ belonging to~\mbox{$\left \lbrace 1, \hdots,d \right \rbrace $}:

	$$f_i(X)=\frac{1}{2}(X+P_{i-1})$$

	\vskip 1cm

	\noindent According to~\cite{Hutchinson1981}, there exists a unique subset $\mathfrak{SS} \subset \R^{d-1}$ such that:

	\[\mathfrak{SS} = \underset{  i=1}{\overset{d}{\bigcup}}\, f_i(\mathfrak{SS})\]
	\noindent which will be called the Sierpi\'{n}ski simplex.\\

	\vskip 1cm

	\noindent We will denote by~$V_0$ the ordered set, of the points:

	$$\left \lbrace P_{0},\hdots,P_{d-1}\right \rbrace$$

	\noindent The set of points~$V_0$, where, for any~$i$ of~\mbox{$\left \lbrace  0,...,d-1  \right \rbrace$}, every point~$P_i$ is linked to the others, constitutes an complete oriented graph, that we will denote by~$ {\mathfrak {SS}}_0$.~$V_0$ is called the set of vertices of the graph~$ {\mathfrak {SS}}_0$.\\

	\noindent For any strictly positive integer~$m$, we set:
	$$V_m =F \left (V_{m-1}\right )$$

	\noindent The set of points~$V_m$, where the points of an~$m^{th}$-order cell are linked in the same way as ${\mathfrak {SS}}_0$, is an oriented graph, which we will denote by~$ {\mathfrak {SS}}_m$.~$V_m$ is called the set of vertices of the graph~$ {\mathfrak {SS}}_m$. We will denote, in the following, by~${\mathcal N}_m$ the number of vertices of the graph~$ {\mathfrak {SS}}_m$.

	\vskip 1cm

	\begin{proposition}
		Given a natural integer~$m$, we will denote by~$\mathcal{N}_m$ the number of vertices of the graph ${\mathfrak {SS}}_m$. One has:

		$$\mathcal{N}_0  =d$$

		\noindent and, for any strictly positive integer~$m$:

		$$\mathcal{N}_m   =d\,\mathcal{N}_{m-1}- \frac{d\,(d-1)}{2}$$
	\end{proposition}

	\vskip 1cm

	\begin{proof} 	The graph $\mathfrak{SS}_m$ is the union of~$d$ copies of the graph $\mathfrak{SS}_{m-1}$. Each copy sharesa vertex with the other ones. So, one may consider the copies as the vertices of a complete graph~$K_d$, the number of edges is equal to~$\displaystyle\frac{d\,(d-1)}{2}$, which leads to~$\displaystyle\frac{d\,(d-1)}{2}$ vertices to take into account.

	\end{proof}

	\vskip 1cm

	\begin{remark}
		\noindent One may check that~$\mathcal{N}_m=\displaystyle\frac{d^{m+1}+d}{2}$.
	\end{remark}

	\vskip 1cm

	\section{The finite difference method on the Minkowski curve}

	In the sequel, we will denote by~$T$ a strictly positive real number, by~$\mathcal{N}_0$ the cardinal of $V_0$, and by~$\mathcal{N}_m$ the cardinal of $V_m$.\\

	\subsection{The heat equation}

	\subsubsection{Formulation of the problem}

	\noindent We may now consider a solution~$u$ of the problem:

	$$ \left \lbrace  \begin{array}{ccccc}
	\displaystyle \frac{\partial u}{\partial t}(t,x)-\Delta u(t,x)&=&0 & \forall \, (t,x)\,\in \,\left]0,T\right[  \times {\mathfrak{SS}}\\
	u(t,x)&=&0 & \forall \, ( x,t) \,\in \,\partial {\mathfrak{SS} } \times \left[0,T\right[\\
	u(0,x)&=&g(x)  & \forall \,x\,\in \,{\mathfrak{SS}}
	\end{array}\right.$$

	\noindent In order to define a numerical scheme, one may use a first order forward difference scheme to approximate the time derivative~$\displaystyle \frac{\partial u}{\partial t}$. The Laplacian is approximated by means of the graph Laplacians~$\Delta_m \,u$, defined on the sequence of graphs~$\left ( {\mathfrak{SS}}_m\right)_{m\in\N^{\star}}$. \\

	\noindent To this purpose, we fix a strictly positive integer~$N $, and set:

	$$ \displaystyle{h =\frac{T}{N}}$$

	\noindent One has, for any integer~$k$ belonging to~$\left \lbrace 0, \hdots, N-1 \right \rbrace$:

	$$   \forall \,X\,\in \, {\mathfrak{SS}} \,:   \quad
	\displaystyle\frac{\partial u}{\partial t}(kh,x)=\displaystyle\frac{1}{h}\,\left( u((k+1)\, h,X)-u(kh,X)\right)+{\mathcal O}(h)
	$$

	\noindent According to~\cite{StrichartzLivre2006}, the Laplacian on Sierpi\'{n}ski simplices~$\mathfrak{SS}$ is given by:

	$$ \forall \, X\,\in \,{\mathfrak{SS}} \,:   \quad
	\Delta u(t,X)= 	\displaystyle\lim_{m\rightarrow +\infty} r^{-m}\left(\int_{\mathfrak{GSG}} \psi^{(m)}_{X_m} \, d\mu\right)^{-1} \, \left( \sum_{X_m \underset{m}\sim Y}  u(t,Y)- u(t,X_m)\right)
	$$

	\noindent where, for any naturel integer~$m$,~\mbox{$X_m\, \in\, V_m\setminus V_0$}, $\psi^{(m)}_{X_m}$ a piecewise harmonic function, and:

	$$	\displaystyle\lim_{m\rightarrow \infty}X_m=X$$

	\noindent This enables one to approximate the Laplacian, at a~$m^{th}$ order,~$m \,\in\,\N^\star$, using the graph normalized Laplacian as follows:

	$$ \forall \, k\, \left \lbrace 0, \hdots, N-1 \right \rbrace,\,\forall \,X\,\in \,\mathfrak{SS}  \,: \quad
	\Delta u(t,X)\approx r^{-m}\left(\int_{\mathfrak{SS}} \psi^{(m)}_{X_m} d\mu\right)^{-1} \,\left( 	\displaystyle\sum_{X_m \underset{m}\sim Y}  u(kh,Y)- u(k\, h,X_m)\right)
	$$

	\noindent By combining those two relations, one gets the following scheme, for any integer~$k$ belonging to~$\left \lbrace 0, \hdots, N-1 \right \rbrace$, any point~$P_j$ of~$V_0$,~$0\leq j \leq N_0-1$, and any~$X$ in the set~\mbox{$V_m\setminus V_0$} :

	$$\left ({\mathcal S}_{\mathcal H}\right) \quad \left \lbrace \begin{array}{cccc}
	\displaystyle \frac{u^m_h((k+1)\,h,X)-u^m_h(k\,h,X)}{h}&=& r^{-m}\left(\int_{\mathfrak{GSG}} \psi^{(m)}_{X_m} d\mu\right)^{-1} \,\left(\displaystyle \sum_{X \underset{m}\sim Y}  u^m_h(k\,h,Y)- u^m_h(kh,X)\right) &   \\
	u^m_h(k\,h,P_j)&=&0 &     \\
	u^m_h(0,X)&=&g(X) &
	\end{array} \right.$$

	\noindent Let us define the approximate equation as:

	$$
	u^m_h((k+1)\,h,X)=u^m_h(k\, h,X) + h\, r^{-m}\left(\int_{\mathfrak{GSG}} \psi^{(m)}_{X_m} d\mu\right)^{-1} \, \left( \sum_{X \underset{m}\sim Y}  u^m_h(k\, h,Y)- u^m_h(k\,h,X)\right) \quad
	\forall \, k\, \in\,\left \lbrace 0, \hdots, N-1 \right \rbrace,  \, \forall \,X\,\in \,V_m\setminus V_0
	$$

	\noindent We now fix~$m\,\in \,\N$, and denote any~$X\,\in \,V_m\setminus V_0$ as~$X_{w,P_i}$, where~\mbox{$w\,\in\,\{1,\dots,d\}^m$} is a word of length $m$, and where $P_i$,~$0 \leq i \leq d-1$ belongs to~$V_0$. We also set:

	$$n=\#\left \lbrace  w \, \in\,\{1,\dots,N\}^m \right \rbrace $$

	\noindent This enables one to introduce, for any integer~$k$ belonging to~$ \left \lbrace 0, \hdots, N-1 \right \rbrace$, the solution vector $U(k)$ as:

	\begin{align*}
	U^m_h(k) &=\left(
	\begin{matrix}
	u^m_h(k\,h,X_1)\\
	\vdots\\
	u^m_h(k\,h,X_{\mathcal{N}_m-d})\\
	\end{matrix}
	\right)\\
	\end{align*}

	\noindent which satisfies the recurrence relation:

	$$U^m_h(k+1) = A \, U^m_h(k)$$

	\noindent where:

	$$
	A=I_{\mathcal{N}_m -d}-h\,\tilde{\Delta}_{m}
	$$

	\noindent  and where~$I_{\mathcal{N}_m -d}$ denotes the~$ {(\mathcal{N}_m -d)}\times  {(\mathcal{N}_m -d)}$ identity matrix, and $\tilde{\Delta}_{m}$ the~$ {(\mathcal{N}_m -d)}\times  {(\mathcal{N}_m -d)}$ normalized Laplacian matrix.

	\vskip 1cm

	\subsubsection{Theoretical study of the error, for H\"{o}lder continuous functions}

	\noindent \emph{i}. \underline{General case}\\

	In the spirit of the work of~R.~S.~Strichartz~\cite{Strichartz1999},~\cite{Strichartz2000}, it is interesting to consider the case of H\"{o}lder continuous functions. Why ? First, H\"{o}lder continuity implies continuity, which is a required condition for functions in the domain of the Laplacian (we refer to our work~\cite{RianeDavidM} for further details).\\
	\noindent Second, a H\"{o}lder condition for such a function will result in fruitful estimates for its Laplacian, which is a limit of difference quotients.

	\vskip 1cm

	\noindent Let us thus consider a function~$u$ in the domain of the Laplacian, and a nonnegative real constant~$\alpha$ such that:

	$$\forall\, (X,Y)\,\in \mathfrak{SS}^2,\forall\,t>0\, : \quad |u(t,X)-u(t,Y)| \leq C(t) \, |X-Y|^\alpha$$

	\noindent where~$C$ denotes a positive function of the time variable~$t$.\\

	\noindent Given a strictly positive integer~$m$, due to:

	$$\Delta_m u(t,X) = \displaystyle\sum_{Y \in V_m,\,Y\underset{m}{\sim} X} \left (u(t,Y)-u(t,X)\right)  \quad \forall\,t>0,\, \forall\, X\,\in\, V_m\setminus V_0 $$

	\noindent Given a strictly positive integer~$m$, its Laplacian is defined as the limit :

	$$\Delta_{\mu} u(t,X)=\displaystyle\lim_{m\rightarrow+\infty} r^{-m} \left( \int_{\mathfrak{SS}} \psi_{X_m}^{(m)} d\mu\right)^{-1}\, \Delta_m u(t, X_m)  \quad \forall\,t>0,\, \forall\, X\,\in\, \mathfrak{SS}$$

	\noindent where $\left (X_m\,\in\,V_m\setminus V_0 \right)_{m\in\N}$ is a sequence a points such that:
	$$\displaystyle\lim_{m\rightarrow+\infty}X_m = X$$

	\noindent and where~$r$ denotes the normalization ratio,~$\psi_{X_m}^{(m)}$ a harmonic spline function, and where:

	$$\Delta_m u(t,X) = \displaystyle\sum_{Y \in V_m,\,Y\underset{m}{\sim} X} \left (u(t,Y)-u(t,X)\right)  \quad \forall\,t>0,\, \forall\, X\,\in\, V_m\setminus V_0 $$

	\noindent Let us now introduce a strictly positive number~$\delta_{ij}:=|P_i-P_j|$, for any~$P_i$ belonging to the set~\mbox{$V_0$}, and any~$P_j$ such that $P_j{\sim} P_i$. We set: $\delta_i=\max_{j}\delta_{ij}$.\\

	\noindent In the other hand, define $R$ to be the contraction ratio of the similarity $f_i$, and $R=\frac{1}{2}$.\\

	\noindent One has then, for any~$X$ belonging to the set~\mbox{$ V_m\setminus V_0$}, any integer~$k$ belonging to~$\left \lbrace 0, \hdots, N-1 \right \rbrace$, and any strictly positive number~$h$:

	$$\begin{array}{ccc} \left |r^{-m}\, \left( \displaystyle\int_{\mathfrak{SS}} \psi_{X_m}^{(m)} d\mu\right)^{-1} \right | \, h \, \left |\Delta_m u(k\,h,X)\right | &\leq & \left | r^{-m} \, \left( \displaystyle\int_{\mathfrak{SS}} \psi_{x_m}^{(m)} d\mu\right)^{-1} \right |\, h \, \displaystyle\sum_{Y \in V_m,\,Y\underset{m}{\sim} X} \left |u(k\,h,Y)-u(k\,h,X)\right| \\
	&\leq & \left | r^{-m} \,\left( \displaystyle\int_{\mathfrak{SS}} \psi_{X_m}^{(m)} d\mu\right)^{-1} \right | \,h\,C(k\,h) \,\displaystyle\sum_{Y \in V_m,\,Y\underset{m}{\sim} X}   |X-Y|^\alpha  \\
	&\leq & \left | r^{-m}\,  \left( \displaystyle\int_{\mathfrak{SS}} \psi_{X_m}^{(m)} d\mu\right)^{-1} \right | \,h\,C(k\,h) \,\displaystyle\sum_{m \, \mid \, Y \in V_m,\,Y\underset{m}{\sim} X}   \displaystyle \delta^{\alpha} \,R^{m\,\alpha}  \\
	&\leq  & \left | r^{-m} \left( \int_{\mathfrak{SS}} \psi_{X_m}^{(m)}\, d\mu\right)^{-1} \right | \,h\,C(k\,h) \,\displaystyle\sum_{p=0}^{+ \infty}   \displaystyle \delta^{\alpha}\, R^{p\, \alpha }  \\
	&=  & \delta^{\alpha} \,  \displaystyle  \frac{ \left |r^{-m}\, \left( \displaystyle\int_{\mathfrak{SS}} \psi_{X_m}^{(m)} d\mu\right)^{-1} \right | \,h\,C(k\,h)}{(1-R^{ \alpha })}  \\
	\end{array}$$

	\noindent We used the fact that, for $X\underset{m}{\sim} Y$, $X$ and $Y$ have addresses such that:

	$$X=f_w(P_i) \quad , \quad Y=f_w(P_j)$$

	\noindent for some $P_i$ and $P_j$ in $V_0$ and $w\in \{1,\hdots ,d\}^m$. May one set:

	$$R(w) =R_{w_1}R_{w_2} \hdots R_{w_m}$$

	\noindent one gets:

	$$\left |X-Y \right | =\left |fw(P_i)-f_w(P_j) \right |= R(w) \left |P_i-P_j \right |  \leq R^m \,\delta $$

	\noindent The scheme~\mbox{$\left ({\mathcal S}_{\mathcal H}\right) $} allow us to write:

	$$\begin{array}{ccccc}  \left |u((k+1)\,h,X)-u(k\,h,X) \right | &\leq &
	\delta^{\alpha} \,  \displaystyle  \frac{ \left |r^{-m}\, \left( \displaystyle\int_{\mathfrak{SS}} \psi_{X_m}^{(m)} d\mu\right)^{-1} \right | \,h\,C(k\,h)}{(1-{\frac{1}{2}}^{ \alpha })}
	\end{array}$$

	\noindent One may note that a required condition for the convergence of the scheme is:
	$$ \displaystyle \lim_{m\to + \infty,\, h \to 0^+} \left | r^{-m} \,\left( \displaystyle\int_{\mathfrak{SS}} \psi_{X_m}^{(m)}\, d\mu\right)^{-1} \right | \,h\,C(k\,h)=0$$

	\noindent In the case where~$C$ is a constant function, it reduces to:
	$$ \displaystyle \lim_{m\to + \infty,\, h \to 0^+} \left | r^{-m} \,\left( \displaystyle\int_{\mathfrak{SS}} \psi_{X_m}^{(m)}\, d\mu\right)^{-1} \right | \,h =0$$

	\vskip 1cm

	\subsubsection{Consistency, stability and convergence}

	\paragraph{The scheme error}$\,$\\

	\noindent Let us consider a continuous function~$u$ defined on $\mathfrak{SS}$. For all~$k$ in~$\left \lbrace 0, \hdots, N-1 \right \rbrace$ :

	$$   \forall \,X\,\in \, {\mathfrak{SS}} \,:   \quad
	\displaystyle\frac{\partial u}{\partial t}(k\,h,X)=\displaystyle\frac{1}{h}\,\left( u((k+1)\, h,X)-u(k\, h,X)\right)+{\mathcal O}(h)
	$$

	\noindent In the other hand, given a strictly positive integer~$m$,~$X\in V_m\setminus V_0$, and a harmonic function~$\psi_X^{(m)}$ on the~$m^{th}$-order cell, taking the value $1$ on $X$ and $0$ on the others vertices (see \cite{Strichartz1999}):

	\begin{align*}
	\displaystyle\int_{\mathfrak{SS}} \psi_X^{(m)} (y) \, ( \Delta u(X) - \Delta u(Y))\,  d\mu(Y) &= \displaystyle\frac{2}{d} \, d^{-m}\Delta u(X) - \left(\frac{d+2}{d}\right)^{m}\,  \Delta_m u(X)
	\end{align*}

	\noindent  Then:

	\begin{align*}
	\Delta u(X) -\displaystyle  \frac{d}{2}\, (d+2)^{m} \Delta_m u(X) &= \frac{d}{2} \, d^{m} \int_{\mathfrak{SS}} \psi_X^{(m)} (Y) ( \Delta u(X) - \Delta u(Y)) \, d\mu(Y)
	\end{align*}

	\noindent Let us now consider the case of H\"older continuous functions, as in the above:

	$$\forall\, (X,Y)\,\in \mathfrak{SS}^2 : \quad |u( X)-u( Y)| \leq C  \, |X-Y|^\alpha$$
	\noindent where~$C$ and~$\alpha$ arenonnegative real constants.\\

	\noindent Given a strictly positive integer~$m$, due to:

	$$\Delta_m u( X) = \displaystyle\sum_{Y \in V_m,\,Y\underset{m}{\sim} X} \left (u( Y)-u( X)\right)  \quad   \forall\, X\,\in\, V_m\setminus V_0 $$

	\noindent this yields:

		$$\left \vert \Delta_m u( X) \right \vert \lesssim  \left \vert Y- X\right\vert ^\alpha \quad  \forall\, X\,\in\, V_m\setminus V_0 $$

		\noindent and thus:

		$$\left \vert \Delta u( X) \right \vert \lesssim  \left \vert  Y - X\right\vert ^\alpha \quad  \forall\, X\,\in\, {\mathfrak SS} \setminus V_0 $$

	\noindent One may note that:
	$$\displaystyle\frac{d}{2} d^{m} \int_{\mathfrak{SS}} \psi_X^{(m)} (Y) \, ( \Delta u(X) - \Delta u(Y))\, d\mu(Y)$$
	\noindent is the mean value of~$\Delta u(X) - \Delta u(Y)$ over the~$m^{th}$-order cell containing $X$, and $\Delta u$ is a continuous function, so we can apply the mean value formula for integrals; there exists $c_m$ in the $m^{th}$-order cell containing $X$ such that :

		\begin{align*}
		\left |\Delta u(x) - \displaystyle\frac{d}{2}(d+2)^{m}\,  \Delta_m u(X)\right | &= \Delta u(X) - \Delta u(c_m)\\
		&\lesssim  |X-c_m|^{\alpha}\\
		&\lesssim   \left(\displaystyle\frac{1}{2}\right)^{m\,\alpha}
		\end{align*}

	\noindent  Finally:

	\begin{align*}
	\Delta u(x) &= \displaystyle \frac{d}{2}\, (d+2)^{m} \, \Delta_m u(x) + \mathcal{O}(2^{-m\,\alpha})
	\end{align*}

	\vskip 1cm
	\paragraph{Consistency}

	\begin{definition}
		The scheme is said to be \textbf{consistent} if the consistency error go to zero when $h\rightarrow 0$ and $m \rightarrow+ \infty$, for some norm.
	\end{definition}

	\noindent The consistency error of our scheme is given by :

	\[
	\varepsilon^m_{k,i} = \mathcal{O}(h) + \mathcal{O}(2^{-m \alpha}) \quad 0\leq k \leq N-1,\, 1\leq i \leq \mathcal{N}_m-d.
	\]

	\noindent One may check that

	$$ \displaystyle \lim_{h\rightarrow 0, \, m\rightarrow+\infty} \varepsilon^m_{k,i}=0$$

	\noindent The scheme is then consistent.

	\paragraph{Stability}

	\begin{definition}
		\noindent Let us recall that the \textbf{spectral norm}~$\rho$ is defined as the induced norm of the norm $\parallel \cdot \parallel_2$. It is given, for a square matrix~$A$, by:

		\[ \rho(A) =\sqrt{\lambda_{\max} \, \left( A^T \, A\right)}\]

		\noindent where $\lambda_{\max}$ stands for the spectral radius.
	\end{definition}

	\vskip 1cm

	\begin{proposition}

		\noindent Let us denote by~$\Phi$ the function such that:

		$$ \forall\, x \neq 0 :\quad  \Phi(x)=x\, (d+2-x).$$
		\noindent According to~\cite{Fukushima1992}, the eigenvalues~$\lambda_{m}$,~$m\,\in\,\N$, of the Laplacian are related recursively:

		$$ \forall\, m\geq 1\, :\quad \lambda_{m-1}=\Phi(\lambda_{m}).$$

	\end{proposition}

	\vskip 1cm

	\noindent We deduce that, for any strictly positive integer~$m$:

	\[\lambda^{\pm}_{m}=\displaystyle\frac{(d+2)\pm \sqrt{(d+2)^2-4\, \lambda_{m-1}}}{2}\]

	\noindent Let us introduce the functions~$\phi^-$ and~$\phi^+$ such that, for any~$x$ in~\mbox{$ \left]-\infty,\displaystyle\frac{(d+2)^2}{4}\right]$} :

	$$
	\phi^-(x) =\displaystyle\frac{(d+2)- \sqrt{(d+2)^2-4\, x}}{2}\quad , \quad
	\phi^+(x)=\displaystyle\frac{(d+2) + \sqrt{(d+2)^2-4\, x}}{2}
	$$

	\noindent $\phi^+(0)=d+2$, $\phi^-\left(\frac{(d+2)^2}{4}\right)=\displaystyle\frac{d+2}{2}$, $\phi^-(0)=0$, and $\phi^+\left(\frac{(d+2)^2}{4}\right)=\displaystyle\frac{d+2}{2}$.\\

	\noindent The function~$\phi^-$ is increasing. Its fixed point is~$x^{-,\star}=0$.\\

	\noindent The function $\phi^+$ is non increasing. Its fixed point is~$x^{+,\star}=(d+2)-1$.\\
	\noindent One may also check that the following two maps are contractions, since:

	$$
	\left |\displaystyle \frac{d}{dx}\phi^-(0)\right | =\frac{1}{\sqrt{(d+2)^2}}
	=\frac{1}{d+2}<1$$

	\noindent and:

	$$
	\left |\displaystyle \frac{d}{d\, x}\phi^+\left ((d+2)-1\right )\right |=\frac{1}{\sqrt{(d+2)^2-4\, (d+2)+4}}
	=\displaystyle\frac{1}{d}<1.$$

	\noindent In~\cite{Shima},~T.~Shima shows that the Laplacien on $V_1$ has Dirichlet eigenvalues $d+2$ with multiplicity $d-1$, and $2$ with multiplicity $1$, and gives the complete spectrum for $m\geq 1$.\\

	\noindent The complete Dirichlet spectrum, for $m\geq 2$, is generated by the recurrent stable maps (convergent towards the fixed points) $\phi^+$ and $\phi^-$ with initial values $2$, $d+2$ and $2\, d$.\\

	\noindent One may finally conclude that, for any naural integer~$m$:

	\begin{align*}
	0 \leq \lambda_m \leq 2\, d\\
	\end{align*}

	\vskip 1cm

	\begin{definition}
		The scheme is said to be:
		\begin{itemize}
			\item unconditionally stable if there exist a constant~$C < 1$ independent of $h$ and $m$ such that:

			$$\rho(A^{k})\leq C \quad \forall \, k\, \in \, \{1,\hdots,N\}$$

			\item conditionally stable if there exist three constants $\alpha> 0$, $C_1 > 0$ and $C_2 < 1$ such that:

			$$ h \leq C_1 \, ((d+2)^{-m} )^{\alpha} \Longrightarrow \rho(A^{k})\leq C_2  \quad \forall \, k\, \in \, \{1,\hdots,N\}$$

		\end{itemize}
	\end{definition}

	\vskip 1cm

	\begin{proposition}

		\noindent Let us denote by~$\gamma_i$,~\mbox{$i=1,\hdots,\mathcal{N}_m -d$}, the eigenvalues of the matrix~$A$. Then:

		$$\forall\, i=1,\hdots,\mathcal{N}_m -d\,: \quad h\, (d+2)^{m}\leq \displaystyle\frac{2}{d^2} \Longrightarrow |\gamma_i|\leq 1.$$
	\end{proposition}

	\vskip 1cm

	\begin{proof}

		\noindent Let us recall our scheme writes, for any integer~$k$ belonging to~\mbox{$\left \lbrace 1, \hdots, N\right \rbrace $}:
		{
			$$\left ({\mathcal S}_{\mathcal H}\right) \quad \left \lbrace \begin{array}{cccc}
			\displaystyle \frac{u^m_h((k+1)\,h,X_i)-u^m_h(k\,h,X_i)}{h}&=&	\displaystyle \frac{d}{2}(d+2)^{m}  \,\displaystyle \sum_{X_i \underset{m}\sim Y}  \left ( u^m_h(k\,h,Y)- u^m_h(kh,X_i)\right) &  \quad \forall \,  1\leq i \leq \mathcal{N}_m-d \\
			u^m_h(k\,h,P_j)&=&0 &     \\
			u^m_h(0,X_i)&=&g(X_i) & \quad 1\leq i \leq \mathcal{N}_m-d \\
			\end{array} \right.$$

			\noindent i.e., under matrix form:

			$$
			U^m_h(k) =
			\left(
			\begin{matrix}
			u^m_h(k\,h,X_{1})\\
			\vdots\\
			u^m_h(k\,h,X_{\mathcal{N}_m-d})\\
			\end{matrix}
			\right)
			\qquad \forall\,  k \, \in \left \lbrace 1, \hdots, N\right \rbrace.$$
		}
		\noindent It satisfies the recurrence relation:

		$$U^m_h(k+1) = A \, U^m_h(k) \qquad \forall\,  k \, \in \left \lbrace 1, \hdots, N\right \rbrace$$

		\noindent where:

		$$
		A=I_{\mathcal{N}_m -d}-h\,\tilde{\Delta}_{m}.
		$$

		\noindent One may use the recurrence to find:

		$$U^m_h(k)=A^{k}\, U^m_h(0) \qquad \forall\,  k \, \in \, \left \lbrace 1, \hdots, N\right \rbrace.$$

		\noindent The eigenvalues~$\gamma_i$,~\mbox{$i=1,\hdots,\mathcal{N}_m -d$}, of $A$ are such that:

		$$\gamma_i = 1 - h\, ( \frac{d}{2}(d+2)^{m} )\lambda_i$$

		\noindent One has, for any integer~$i$ belonging to~\mbox{$\left \lbrace1,\hdots,\mathcal{N}_m -d\right \rbrace$} :

		$$ 1 - h\, \displaystyle \frac{d}{2}(d+2)^{m}\, (2\, d) \leq\gamma_i\leq 1$$

		\noindent which leads to:

		$$h\, (d+2)^{m}\leq \displaystyle\frac{2}{d^2} \Longrightarrow |\gamma_i|\leq 1.$$

	\end{proof}
	\vskip 1cm

	\paragraph{Convergence}

	\begin{definition}
		\noindent
		\begin{itemize}
			\item The scheme is said to be convergent for the matrix norm $\| \cdot  \|$ if :
			{
				$$\displaystyle\lim_{h\rightarrow 0, \, m\rightarrow +\infty} \left  \| \left( u(kh,X_i)- u^m_h(k\,h,X_i) \right)_{0\leq k\leq N,\, 1\leq i \leq \mathcal{N}_m} \right  \|=0$$
			}
			\item The scheme is said to be conditionally convergent for the matrix norm $ \| \cdot  \|$ if there exist two real constants $\alpha$ and $C$ such that :

			$$\displaystyle \lim_{h\leq C\left((d+2)^{-m}\right)^{\alpha}, \, m\rightarrow +\infty} \left  \| \left( u(k\,h,X_i)- u^m_h(k\,h,X_i) \right)_{0\leq k\leq N,\, 1\leq i \leq \mathcal{N}_m}\right  \|=0$$

		\end{itemize}
	\end{definition}

	\vskip 1cm

	\begin{theorem}
		{	\noindent If the scheme is stable and consistent, then it is also convergent for the norm~$\| \cdot  \|_{2,\infty}$, such that:

			$$	\left \| \left( u^m_h(k\, h,X_i) \right)_{0\leq k\leq N, x\in V_m\setminus V_0} \right  \|_{2,\infty}=\displaystyle \max_{0\leq k\leq N} \left( d^{-m}\sum_{1\leq i\leq \mathcal{N}_m}  \left  |u^m_h(k\,h,X_i) \right |^2)\right)^{\frac{1}{2}}$$
		}
	\end{theorem}

	\vskip 1cm

	\begin{proof}

		\noindent Let us set:

		$$ w^k_i  = u(k\,h,X_i)-u^h_m(k\,h,X_i), \quad 0\leq k \leq N,\, 1\leq i\leq \mathcal{N}_m.$$

		\noindent One may check that:

		\begin{align*}
		w^k_{\mathcal{N}_m-d+1}=\dots=w^k_{\mathcal{N}_m}&=0 \qquad 0\leq k \leq N\\
		w^0_i&=0 \qquad 1\leq i \leq \mathcal{N}_m -d
		\end{align*}

		\noindent Let us now introduce, for any integer~$k$ belonging to~\mbox{$\left \lbrace 0, \hdots,   N \right \rbrace$}:

		$$W^k=
		\left(
		\begin{matrix}
		w^k_1\\
		\vdots\\
		w^k_{\mathcal{N}_m-d}
		\end{matrix}
		\right)
		\quad , \quad  E^k=
		\left(
		\begin{matrix}
		\varepsilon^m_{k,1}\\
		\vdots\\
		\varepsilon^m_{k,\mathcal{N}_m-d}
		\end{matrix}
		\right)
		$$

		\noindent One has then~$W^0=0$, and, for any integer~$k$ belonging to~\mbox{$\left \lbrace 1, \hdots,   N-1\right \rbrace$}:

		$$
		W^{k+1} =A\, W^k+h\, E^k  $$

		\noindent One finds recursively, for any integer~$k$ belonging to~$\left \lbrace 0, \hdots,   N-1\right \rbrace$:

		$$
		W^{k+1}=A^k W^0+h\, \displaystyle\sum_{j=0}^{k-1} A^j\,  E^{k-j-1}=h\, \displaystyle\sum_{j=0}^{k-1} A^j \, E^{k-j-1}
		$$

		\noindent Since the matrix~$A$ is a symmetric one, the~\textbf{CFL} stability condition $h\,(d+2)^{m}\leq \displaystyle\frac{2}{d^2}$ yields, for any integer~$k$ belonging to~\mbox{$\left \lbrace  0,\hdots,N\right \rbrace $}:

		\begin{align*}
		|W^k|&\leq h\, \left(\sum_{j=0}^{k-1}\parallel A \parallel^j \right)\, \left(\displaystyle\max_{0\leq k \leq j-1} | E^{k} | \right)\\
		&\leq h\, k\, \left(\max_{0\leq k \leq j-1} | E^{k} | \right)\\
		&\leq h\, N\, \left(\max_{0\leq k \leq j-1} | E^{k} | \right)\\
		&\leq T\, \left(\max_{0\leq k \leq j-1} \left(\sum_{i=1}^{\mathcal{N}_m-d} \, |\varepsilon^m_{k,i}|^2 \right)^{\frac{1}{2}} \right)\\
		\end{align*}

		\noindent One deduces then:

		\begin{align*}
		\max_{0\leq k\leq N} \left( d^{-m}\sum_{i=1}^{\mathcal{N}_m-d}   |w^k_i|^2)\right)^{\frac{1}{2}}&=d^{-\frac{m}{2}}\max_{1\leq k\leq N}|\, W^k|\\
		&\leq \left(d^{-\frac{m}{2}}\right)\,  T\left(\max_{0\leq k \leq N-1} \left(\displaystyle\sum_{i=1}^{\mathcal{N}_m-d} |\varepsilon^m_{k,i}|^2 \right)^{\frac{1}{2}} \right)\\
		&\leq \left(d^{-\frac{m}{2}}\right) \, T\left((\mathcal{N}_m-d)^{\frac{1}{2}}\max_{0\leq k \leq N-1,\, 1\leq i \leq \mathcal{N}_m-d} |\varepsilon^m_{k,i}| \right)\\
		&=\sqrt{\left(d^{-m}\, \displaystyle\frac{d^{m+1}-d}{2}\right)} \, T\left(\displaystyle\max_{0\leq k \leq N-1,\, 1\leq i \leq \mathcal{N}_m-d} |\varepsilon^m_{k,i}| \right)\\
		&=   \mathcal{O}(h) + \mathcal{O}(2^{-m \alpha})  \\
		&=   \mathcal{O}((d+2)^{-m}) + \mathcal{O}(2^{-m\alpha})  \\
		&= \mathcal{O}(2^{-m\alpha}) .\\
		\end{align*}

		\noindent The scheme is thus convergent.
	\end{proof}

	\vskip 1cm

	\begin{remark}
		\noindent One has to bear in mind that, for piecewise constant functions~$u$ on the~$m^{th}$-order cells:
		{
			$$\left \|  \left( u^m_h(k\, h,X_i) \right) \right \| _{2}= \left( d^{-m}\, \displaystyle \sum_{1\leq i\leq \mathcal{N}_m}   |u^m_h(kh,X_i)|^2)\right) ^{\frac{1}{2}}
			=\left \|  \left( u^m_h(k\, h,X_i) \right) \right \| _{L^2(\mathfrak{SS})}.$$

		}
	\end{remark}

	\vskip 1cm

	{
		\subsubsection{The specific case of the implicit Euler Method}

		\noindent Let us consider the implicit Euler scheme, for any integer~$k$ belonging to~$\left \lbrace 0, \hdots, N-1 \right \rbrace$, any point~$P_j$ of~$V_0$,~$0\leq j \leq N_0-1$, and any~$X$ in the set~\mbox{$V_m\setminus V_0$} :

		$$\left ({\mathcal S}_{\mathcal H}\right) \quad \left \lbrace \begin{array}{cccc}
		\displaystyle \frac{u^m_h(k\,h,X)-u^m_h((k-1)\,h,X)}{h}&=& r^{-m}\left(\int_{\mathfrak{GSG}} \psi^{(m)}_{X_m} d\mu\right)^{-1} \,\left(\displaystyle \sum_{X \underset{m}\sim Y}  u^m_h(k\,h,Y)- u^m_h(kh,X)\right) &   \\
		u^m_h(k\,h,P_j)&=&0 &     \\
		u^m_h(0,X)&=&g(X) &
		\end{array} \right.$$

		\noindent Let us define the approximate equation as:

		$$
		u^m_h(k\,h,X) - h\times r^{-m}\left(\int_{\mathfrak{GSG}} \psi^{(m)}_{X_m} d\mu\right)^{-1} \, \left( \sum_{X \underset{m}\sim Y}  u^m_h(k\, h,Y)- u^m_h(k\,h,X)\right)  = u^m_h((k-1)h,X) \quad
		\forall \, k\, \in\,\left \lbrace 0, \hdots, N-1 \right \rbrace,  \, \forall \,X\,\in \,V_m\setminus V_0
		$$

		\noindent As before, we fix~$m\,\in \,\N$, and denote any~$X\,\in \,V_m\setminus V_0$ as~$X_{w,P_i}$, where~\mbox{$w\,\in\,\{1,\dots,d\}^m$} denotes a word of length $m$, and where $P_i$,~$0 \leq i \leq d-1$ belongs to~$V_0$. Let us also set:

		$$n=\#\left \lbrace  w \, \in\,\{1,\dots,N\}^m \right \rbrace $$

		\noindent We get, for any integer~$k$ belonging to~$ \left \lbrace 0, \hdots, N-1 \right \rbrace$, the solution vector $U(k)$ as before:

		\begin{align*}
		U^m_h(k) &=\left(
		\begin{matrix}
		u^m_h(k\,h,X_1)\\
		\vdots\\
		u^m_h(k\,h,X_{\mathcal{N}_m-d})\\
		\end{matrix}
		\right)\\
		\end{align*}

		\noindent It satisfies the recurrence relation:

		$$\tilde{A} \, U^m_h(k) = U^m_h(k-1)$$

		\noindent where:

		$$
		\tilde{A}=I_{\mathcal{N}_m -d} + h\times\tilde{\Delta}_{m}
		$$

		\noindent  and where~$I_{\mathcal{N}_m -d}$ denotes the~$ {(\mathcal{N}_m -d)}\times  {(\mathcal{N}_m -d)}$ identity matrix, and $\tilde{\Delta}_{m}$ the~$ {(\mathcal{N}_m -d)}\times  {(\mathcal{N}_m -d)}$ normalized Laplacian matrix.

		\vskip 1cm

		\paragraph{Consistency, stability and convergence}

		\subparagraph{\emph{i}. The scheme error}

		\noindent Let $u$ a function defined on $\mathfrak{SS}$. For all~$k$ in~$\left \lbrace 0, \hdots, N-1 \right \rbrace$ :

		$$   \forall \,X\,\in \, {\mathfrak{SS}} \,:   \quad
		\displaystyle\frac{\partial u}{\partial t}(kh,X)=\displaystyle\frac{1}{h}\,\left( u(kh,X)-u((k-1)h,X)\right)+{\mathcal O}(h)
		$$

		\noindent In the other hand, for $X\in V_m\setminus V_0$:

		\begin{align*}
		\Delta u(x) &= \frac{d}{2}(d+2)^{m} \Delta_m u(x) + \mathcal{O}(2^{-m\alpha})
		\end{align*}

		\subparagraph{\emph{ii}. Consistency}

		\noindent The consistency error of the implicit Euler scheme is given by :

		\[
		\varepsilon^m_{k,i} = \mathcal{O}(h) + \mathcal{O}(2^{-m\alpha}) \quad 0\leq k \leq N-1,\, 1\leq i \leq \mathcal{N}_m-d
		\]

		\noindent We can check that

		\[ \lim_{h\rightarrow 0, m\rightarrow\infty} \varepsilon^m_{k,i}=0\]

		\noindent The scheme is then consistent.

		\paragraph{Stability}

		\begin{definition}
			The scheme is said to be :
			\begin{itemize}
				\item unconditionally stable for the norm $\parallel .\parallel_{\infty}$ if there exist a constant $C >0$ independent of $h$ and $m$ such that :

				$$ \parallel U^m_h(k)\parallel_{\infty} \leq C \parallel U^m_h(0) \parallel_{\infty} \quad \forall k\in \{1,...,N\}$$

				\item conditionally stable if there exist three constants $\alpha> 0$, $C_1 > 0$ and $C_2 < 1$ such that :

				$$ h \leq C_1 ((d+2)^{-m} )^{\alpha} \Longrightarrow \parallel U^m_h(k)\parallel_{\infty} \leq C_2 \parallel U^m_h(0) \parallel_{\infty} \quad \forall k\in \{1,...,N\}$$

			\end{itemize}
		\end{definition}

		\noindent Let us recall that our scheme writes:
		\footnotesize

		$$\left ({\mathcal S}_{\mathcal H}\right) \quad \left \lbrace \begin{array}{cccc}
		\displaystyle \frac{u^m_h(k\,h,X_i)-u^m_h((k-1)\,h,X_i)}{h}&=& \displaystyle \frac{d}{2}(d+2)^{m}  \,\left(\displaystyle \sum_{X_i \underset{m}\sim Y}  u^m_h(k\,h,Y)- u^m_h(kh,X_i)\right) &  \quad \forall\,  1\leq k\leq N,\, 1\leq i \leq \mathcal{N}_m-d \\
		u^m_h(k\,h,P_j)&=&0 &   \quad \forall \, 1\leq k\leq N  \\
		u^m_h(0,X_i)&=&g(X_i) & \quad 1\leq i \leq \mathcal{N}_m-d \\
		\end{array} \right.$$
		\normalsize

		\noindent i.e., under matrix form :

		\[
		U^m_h(k) =
		\left(
		\begin{matrix}
		u^m_h(k\,h,X_{1})\\
		\vdots\\
		u^m_h(k\,h,X_{\mathcal{N}_m-d})\\
		\end{matrix}
		\right)
		\]

		\noindent which satisfies the recurrence relation:

		$$\tilde{A} \, U^m_h(k) = U^m_h(k-1)$$

		\noindent where :

		$$
		\tilde{A}=I_{\mathcal{N}_m -d} + h\times\tilde{\Delta}_{m}
		$$

		\noindent One has:
		$$ \parallel \tilde{A}^{-1} \parallel_{\infty} \leq 1\quad \text{and thus} \quad
		\parallel \tilde{A}^{-n} \parallel_{\infty} \leq 1$$

		\noindent This enables us to conclude that the scheme is unconditionally stable :

		$$ U^m_h(k) \leq  U^m_h(0)$$

		\vskip 1cm

		\subparagraph{\emph{iii}. Convergence}

		\begin{theorem}
			\noindent The implicit Euler scheme is convergent for the norm $\parallel . \parallel_{2,\infty}$.

		\end{theorem}

		\vskip 0.5cm

		\begin{proof}

			\noindent Let:

			$$ w^k_i = u(kh,X_i)-u^h_m(kh,X_i), \quad 0\leq k \leq N,\, 1\leq i\leq \mathcal{N}_m$$

			\noindent One may check that:

			\begin{align*}
			w^k_{\mathcal{N}_m-d+1}=\dots=w^k_{\mathcal{N}_m}&=0 \qquad 0\leq k \leq N\\
			w^0_i&=0 \qquad 1\leq i \leq \mathcal{N}_m -d
			\end{align*}

			\noindent We set:

			$$ W^k=
			\left(
			\begin{matrix}
			w^k_1\\
			\vdots\\
			w^k_{\mathcal{N}_m-d}
			\end{matrix}
			\right)
			\quad , \quad
			E^k=
			\left(
			\begin{matrix}
			\varepsilon^m_{k,1}\\
			\vdots\\
			\varepsilon^m_{k,\mathcal{N}_m-d}
			\end{matrix}
			\right)
			$$

			\noindent Thus,~$W^0=0$, and,  for~$0\leq k \leq N-1 $:

			\begin{align*}
			W^{k+1}&=\tilde{A}^{-1} W^k+h\, E^k \quad 0\leq k \leq N-1\\
			\end{align*}

			\noindent We find, by induction, for~$0\leq k \leq N-1 $:

			\begin{align*}
			W^{k+1}&=\tilde{A}^{-k} W^0+h\, \displaystyle \sum_{j=0}^{k-1} \tilde{A}^{-j} E^{k-j-1}  \\
			&=h\, \displaystyle \sum_{j=0}^{k-1} \tilde{A}^{-j} E^{k-j-1}
			\end{align*}

			\noindent Due to the stability of the scheme, we have, for~$k=0,\hdots,N$:

			\begin{align*}
			|W^k|&\leq h\,\left(\displaystyle \sum_{j=0}^{k-1}\parallel \tilde{A}^{-1} \parallel^j \right)\left(\max_{0\leq k \leq j-1} | E^{k} | \right)\\
			&\leq h\, k\,\left(\max_{0\leq k \leq j-1} | E^{k} | \right)\\
			&\leq h\,N\,\left(\max_{0\leq k \leq j-1} | E^{k} | \right)\\
			&\leq T\,\left(\max_{0\leq k \leq j-1} \left(\sum_{i=1}^{\mathcal{N}_m-d} |\varepsilon^m_{k,i}|^2 \right)^{1/2} \right)\\
			\end{align*}

			\newpage
			\noindent One deduces then:

			\begin{align*}
			\max_{0\leq k\leq N} \left( d^{-m}\sum_{i=1}^{\mathcal{N}_m-d}   |w^k_i|^2)\right)^{\frac{1}{2}}&=(d)^{-\frac{m}{2}}\max_{1\leq k\leq N}|W^k|\\
			&\leq \left(d^{-\frac{m}{2}}\right) T\left(\max_{0\leq k \leq N-1} \left(\sum_{i=1}^{\mathcal{N}_m-d} |\varepsilon^m_{k,i}|^2 \right)^{1/2} \right)\\
			&\leq \left(d^{-\frac{m}{2}}\right) T\left((\mathcal{N}_m-d)^{\frac{1}{2}}\max_{0\leq k \leq N-1,\, 1\leq i \leq \mathcal{N}_m-d} |\varepsilon^m_{k,i}| \right)\\
			&=\sqrt{\left(d^{-m}\frac{d^{m+1}-d}{2}\right)} T\left(\max_{0\leq k \leq N-1,\, 1\leq i \leq \mathcal{N}_m-d} |\varepsilon^m_{k,i}| \right)\\
			&= \left( \mathcal{O}(h) + \mathcal{O}(2^{-m\alpha}) \right)\\
			\end{align*}

			\noindent The scheme is thus convergent.
		\end{proof}
	}

	\vskip 1cm

	\subsubsection{Numerical results - Gasket and Tetrahedron}

	\paragraph{Recursive construction of the matrix related to the sequence of graph Laplacians}$\,$\\

	In the sequel, we describe our recursive algorithm used to construct matrix related to the sequence of graph Laplacians, in the case of Sierpi\'{n}ski Gasket and Tetrahedron.\\

	\subparagraph{\emph{i}. The Sierpi\'{n}ski Gasket.}

	\begin{center}
		\includegraphics[scale=0.5]{SG.eps}
		\captionof{figure}{$m^{th}$-order cell of the Sierpi\'{n}ski Gasket.}
		\label{fig1}
	\end{center}

	\noindent One may note, first, that, given a strictly positive integer~$m$,  a~$m^{th}$-order triangle has three corners, that we will denote by~$C1$,~$C2$ and~$C3$ ; the $(m+1)^{th}$-order triangle is then constructed by connecting three $m$ copies $T(n)$ with $n=1,\,2,\, 3$.\\

	\noindent The initial triangle is labeled such that $C1 \sim 1$, $C2 \sim 2$ and $C3 \sim 3$~(see figure~1).

	\begin{figure}[!htb]

		\minipage{0.32\textwidth}
		\includegraphics[width=\linewidth]{SG1.eps}
		\caption{The first copy $T(1)$}
		\endminipage\hfill
		\minipage{0.32\textwidth}
		\includegraphics[width=\linewidth]{SG2.eps}
		\caption{The second copy $T(2)$}
		\endminipage\hfill
		\begin{center}
			\minipage{0.32\textwidth
			\includegraphics[width=\linewidth]{SG3.eps}
			\caption{The third copy $T(3)$}
			\endminipage
		\end{center}
	\end{figure}

	\noindent The fusion is done by connecting $C2(1,m) \sim C1(2,m)$, $C3(1,m) \sim C1(3,m)$, and $C3(2,m) \sim C2(3,m)$~(see figures~2,~3,~4).\\

	\noindent The label of the corner vertex can be obtained by means of the following recursive sequence, for any strictly positive integer~$m$:

	\begin{align*}
	C1(n,m)&=1+(n-1)\,\mathcal{N}_{m-1}\\
	C2(n,m)&=I2(m)+(n-1)\,\mathcal{N}_{m-1}\\
	C3(n,m)&=n \,\mathcal{N}_{m-1}\\
	\end{align*}

	\noindent where:

	\begin{align*}
	\mathcal{N}_{-1}&=3 \quad , \quad
	I2(0)=0\\
	I2(m)&=I2(m-1)+\mathcal{N}_{m-2}-1.\\
	\end{align*}
	\newpage
	\begin{enumerate}

		\item One may start with the initial triangle with the set of vertices~$V_0$. The corresponding matrix is given by:

		$$A_0 = \left(
		\begin{matrix}
		2 & -1 & -1 \\
		-1 & 2 & -1 \\
		-1 & -1 & 2 \\
		\end{matrix}
		\right)$$

		\item If $m=0$, the Laplacian matrix is $A_0$, else, $A_m$ is constructed recursively from three copies of the Laplacian matrices $A_{m-1}$ of the graph $V_{m-1}$. First, we build, for any strictly positive integer~$m$, the block diagonal matrix:

		$$B_m = \left(
		\begin{matrix}
		A_{m-1} & 0 & 0 \\
		0 & A_{m-1} & 0 \\
		0 & 0 & A_{m-1} \\
		\end{matrix}
		\right)$$

		\item One may then introduce, for any strictly positive integer~$m$, the connection matrix as in~\cite{UtaFreiberg2004}:

		$$C_m = \left(
		\begin{matrix}
		C2(1,m) & C3(1,m) & C3(2,m) \\
		C1(2,m) & C1(3,m) & C2(3,m) \\
		\end{matrix}
		\right)$$

		{	\item One has then to sum the rows (resp. the columns) $C_m(2,j)$ and $C_m(1,j)$, and delete the row and the column $C_m(2,j)$.}

	\end{enumerate}

	\vskip 1cm

	\subparagraph{\emph{ii}. The Sierpi\'{n}ski Tetrahedron.}

	\vskip 1cm

	\begin{center}
		\includegraphics[scale=0.7]{ST.eps}
		\captionof{figure}{$m^{th}$-order cell of the Sierpi\'{n}ski Tetrahedron.}
		\label{fig1}
	\end{center}

	One may note, first, that, given a strictly positive integer~$m$,  a~$m^{th}$-order tetrahedron has four corners~$C1$,~$C2$,~$C3$ and~$C4$~(see figure 5), and that the $(m+1)^{th}$-order triangle is constructed by connecting four~$m$ copies $T(n)$, with $n=1,2,3,4$~(see figure 6, 7, 8, 9).\\

	As in the case of the triangle, the initial tetrahedron is labeled such that $C1 \sim 1$, $C2 \sim 2$, $C3 \sim 3$ and $C4 \sim 4$.

	\begin{figure}[!htb]

		\minipage{0.32\textwidth}
		\includegraphics[width=\linewidth]{ST1.eps}
		\caption{The first copy $T(1)$.}
		\endminipage\hfill
		\minipage{0.32\textwidth}
		\includegraphics[width=\linewidth]{ST2.eps}
		\caption{The second copy $T(2)$.}
		\endminipage\hfill
		\minipage{0.32\textwidth}
		\includegraphics[width=\linewidth]{ST3.eps}
		\caption{The third copy $T(3)$.}
		\endminipage\hfill

		\begin{center}
			\minipage{0.32\textwidth
			\includegraphics[width=\linewidth]{ST4.eps}
			\caption{The fourth copy $T(4)$.}
			\endminipage
		\end{center}
	\end{figure}

	\noindent The fusion is done by connecting $C2(1,m) \sim C1(2,m)$, $C3(1,m) \sim C1(3,m)$, $C4(1,m) \sim C1(4,m)$, $C3(2,m) \sim C2(3,m)$, $C4(2,m) \sim C2(4,m)$, $C4(3,m) \sim C3(4,m)$.\\

	\newpage
	\noindent The number of corners can be obtained by means of the following recursive sequence, for any strictly positive integer~$m$:

	\begin{align*}
	C1(n,m)&=1+(n-1)\,\mathcal{N}_{m-1}\\
	C2(n,m)&=I2(m)+(n-1)\,\mathcal{N}_{m-1}\\
	C3(n,m)&=I3(m)+(n-1) \,\mathcal{N}_{m-1}\\
	C4(n,m)&=n \,\mathcal{N}_{m-1}\\
	\end{align*}

	\noindent where:
	{
		\begin{align*}
		\mathcal{N}_{-1}&=3\\
		I2(0)&=0\\
		I2(m)&=I2(m-1)+\mathcal{N}_{m-2}-1\\
		I3(1)&=3\\
		I3(m)&=I3(m-1)+2\times\mathcal{N}_{m-2}-3\\
		\end{align*}
	}
	\newpage
	\begin{enumerate}

		\item One starts with initial tetrahedron with the set of vertices $V_0$. The corresponding matrix is given by:

		$$A_0 = \left(
		\begin{matrix}
		3 & -1 & -1 & -1\\
		-1 & 3 & -1 & -1\\
		-1 & -1 & 3 & -1 \\
		-1 & -1 & -1 & 3 \\
		\end{matrix}
		\right)$$
		\vskip 0.5cm

		\item If $m=0$ the Laplacian matrix is $A_0$, else, for any strictly positive integer~$m$, $A_m$ is constructed recursively from three copies of the Laplacian matrices $A_{m-1}$ of the graph $V_{m-1}$. Thus, we build the block diagonal matrix:

		$$B_m = \left(
		\begin{matrix}
		A_{m-1} & 0 & 0 & 0 \\
		0 & A_{m-1} & 0 & 0 \\
		0 & 0 & A_{m-1} & 0 \\
		0 & 0 & 0 & A_{m-1} \\
		\end{matrix}
		\right)$$

		\vskip 0.5cm
		\item We then write the connection matrix:
		{
			$$C_m = \left(
			\begin{matrix}
			C2(1,m) & C3(1,m) & C3(2,m) & C4(1,m) & C4(2,m) & C4(3,m)\\
			C1(2,m) & C1(3,m) & C2(3,m) & C1(4,m) & C2(4,m) & C3(4,m)\\
			\end{matrix}
			\right)$$

			\vskip 0.5cm

			\item One then has to sum the rows (resp. the columns) $C_m(2,j)$ to $C_m(1,j)$, and delete the row and the column $C_m(2,j)$.}

	\end{enumerate}

	\vskip 1cm

	\newpage
	\paragraph{Numerical results}

	\subparagraph{\emph{i}. The Sierpi\'{n}ski Gasket}$\,$\\

	\noindent In the sequel (see figures 10 to 14), we present the numerical results for~$m=6$, $T=1$ and $N=2\times 10^5$.

	\begin{center}
		\includegraphics[scale=1.5]{SGk0.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=0$.}
		\label{fig1}
	\end{center}

	\begin{center}
		\includegraphics[scale=1.5]{SGk10.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=10$.}
		\label{fig1}
	\end{center}

	\begin{center}
		\includegraphics[scale=1.5]{SGk100.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=100$.}
		\label{fig1}
	\end{center}

	\begin{center}
		\includegraphics[scale=1.5]{SGk500.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=500$.}
		\label{fig1}
	\end{center}

	\begin{center}
		\includegraphics[scale=1.5]{SGk1000.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=1000$.}
		\label{fig1}
	\end{center}

	\newpage
	\subparagraph{\emph{ii}. The Sierpi\'{n}ski Tetrahedron}$\,$\\

	In the sequel (see figures~15 to~19), we present the numerical results for~$m=5$, $T=1$ and $N=10^5$.  \\

	Our heat transfer simulation consists in a propagation scenario, where the initial condition is a harmonic spline~$g$, the support of which being a~$m$-cell, such that it takes the value~$1$ on a vertex $x$, and~$0$ otherwise.\\

	The color function is related to the gradient of temperature, high values ranging from red to blue.  \\

	\begin{center}
		\includegraphics[scale=1]{STk0.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=0$.}
		\label{fig1}
	\end{center}

	\begin{center}
		\includegraphics[scale=1]{STk10.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=10$.}
		\label{fig1}
	\end{center}

	\begin{center}
		\includegraphics[scale=1]{STk50.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=50$.}
		\label{fig1}
	\end{center}

	\begin{center}
		\includegraphics[scale=1]{STk100.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=100$.}
		\label{fig1}
	\end{center}

	\begin{center}
		\includegraphics[scale=1]{STk500.eps}
		\captionof{figure}{The graph of the approached solution of the heat equation for $k=500$.}
		\label{fig1}
	\end{center}

	\vskip 1cm

	An interesting feature in our work is that, contrary to existing ones, we do not rely on heat kernel estimates. Using a direct method has thus enabled us to discuss the choices of parameters as the integer~$m$, the step~$h$, and the convergence.\\

	As expected, the numerical scheme is unstable and diverges until one respects the stability condition between~$h$ and~$m$. Also, the propagation process evolves with time, directed from hot regions, towards cold ones.\\

	In order to go further, we have also studied the evolution in time of the temperature of a point $x_0$. Figures~20 and~21 respectively display the graph and log-log graph of the temperature as a function of time.\\

	\begin{center}
		\includegraphics[scale=1]{Temperature.eps}
		\captionof{figure}{The graph of the temperature as a function of time.}
		\label{fig1}
	\end{center}

	\begin{center}
		\includegraphics[scale=1]{LoglogTemperature.eps}
		\captionof{figure}{The log-log graph of the temperature as a function of time.}
		\label{fig1}
	\end{center}

	We find, numerically that the temperature follows the law :

	$$\ln \left  (u(t,x_0)\right ) \approx -1.01275 - 1.44972 \, \ln t$$

	It is interesting to note that the  slope is close of the spectral dimension~$d_S=\displaystyle \frac{\ln 3}{\ln 2}$, which yields a power law of the form:

	$$u(t,x_0)\approx C\, t^{d_S}$$

	\noindent where~$C$ is a strictly positive real constant. This results holds for different values of~$m$.\\

	\noindent Such a result suggests that the spectral dimension belongs to the spectrum of the Laplacian, which is in accordance with theoretical results (see section~3, and~\cite{Fukushima1992}).

	\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


text:
Spiritual Awakening Process
5th chakra
Cultivating a spiritual awakening
spiritual attention
the art of listening
voice chakra
Part and parcel to finding your voice is practicing the art of listening. Lots of people hear, but they don't listen. There's a Chinese proverb that says, "To listen well, is as powerful a means of influence as to talk well, and is as essential to all true conversation."
It sounds simple enough, but how many of us really do this? How many times are you simply waiting for the person to stop talking so you can say what you want to say? How many times are you being silent so as to not make waves or upset someone? And in that silence is a discord in you that is resisting whatever the person is saying, so you aren't really listening at all.
Deaf to the World
Our egos have managed to turn themselves into megaphones. Blasting out opinions on the 24hour news cycle, in boardrooms and business meetings, and in relationships, our ego voices have been filling ears with constant nonsense. So what has happened? We've tuned out. We've shut down. We've decided that it's all crap and that we shouldn't pay attention. This shows up in so many ways. In politics, it means people celebrated a 60% plus voter turn out for the 2008 Presidential election. That means four out of ten REGISTERED voters didn't vote (and probably didn't pay attention to the election). But since 61.7% was a high water mark not seen since 1968, everyone was super excited.
Political apathy is one element of tuning out, and it's a symptom of tuning out in the rest of our lives. If you think that you don't have any power and you tune out, you've created a self-fulfilling prophecy. When you aren't listening and paying attention, you don't know what's going on. You make yourself that much more powerless. The same principle applies to raising a family. If your children are upset about something and you don't listen, then how can you help them and resolve the problem? If your husband has had a hard day, tuning out because he's just "complaining again" doesn't help him to feel heard and validated (and he's probably felt unheard and unvalidated for most of the day, which is why you're getting to hear it now). You can see the cycle can you not? You don't have to solve other people's problems for the most part. Generally speaking, you just have to listen.
Listening with Intention
It's a strange thing, but because so many people have never been fully listened to, the experience can be very intense at first. When I'm talking with a close friend and we're chest deep in a conversation, we're super focused on each other. There are a variety of levels of listening that happen between us as well as other people. A lot of it has to do with body orientation, so let's start there:
Eye contact. This is a no brainer, right? Strong, steady eye contact confirms that someone is with you. You can look away and blink, though. This isn't a staring contest. You don't need to bore a hole through someone's skull. Soft, but steady gaze is the way to go.
Chest direction. You can definitely feel the difference if someone is listening to you with their head turned toward you and the rest of their body pointed else where. The body does tell you where most of someone's attention is.
Hip/Lower torso direction. When someone's whole body is pointed at you, that's a whole other level of connection. Some of the deepest conversations for me end up with my friend and me having our whole bodies directly opposite to each other. That's whole body listening.
Making Space in the Mind
Now that you've got your body pointed in the right direction, let's talk about clearing some of the debris between the ears. Don't worry. You don't have to get it all at once, and you will be spending a lot of time clearing out this space. In many respects the spiritual path is like a big yard sale--you start getting rid of what you don't want or need (a spiritual awakening is like a store clearance/going out of business sale, btw, but I'm getting ahead of myself). To start, simply notice the noise in your mind. Here are some questions:
What stories are running through your head when you talk to this person?
Why are you distracted?
What are your expectations about this relationship and conversation?
These are just a few with which to begin. Ask these questions of yourself. Notice the noise. Now, start to put it aside as you're listening to someone. Really focus on their words and on what they're saying. You don't even necessarily have to understand the conversation to be present to it. Reciprocating and discussion are important to active listening, but for this blog post, I'm just talking about listening and making space to hear the person and where they're coming from. I'm also talking about noticing the filters and projections you put on someone that don't allow you to hear them for who they are. There are so many things in the way to listening, which should be such a simple thing.
Find a Listening Partner
I've been around meditation groups and other gatherings where there's an activity in which one person talks and the other just listens. They don't comment on what the other person has said. They simply listen. This, of course, is the start of developing a space in yourself as a witness. The witness (which I'll also talk about in later blogs) is a powerful place. This is where you can step back to see things as they really are and to hold space for whatever is occurring without ego judgment. For now, it's enough to start to practicing listening.
I encourage you to find a friend with whom you can listen to with active eye contact and your bodies directed at each other. Spend ten minutes doing this, and then you can talk about what you noticed inside yourselves. At this point, it's important not to be critiquing the other person. There's enough critique of the "other" in our society. You're at the beginning of the process, and you need to start to turn your attention inwards to you (and even if you're not at the start of a spiritual path, this is a good reminder). As you do this, you're going to notice some changes and become aware of how a lot of relationships do and don't work for you. You're going to start to see a lot more clearly, and in turn, you'll going to start to be able to voice your truth more easily because you started to really listen.
Next blog: Using Your Intuition
This picture is gift from Becky Stiller. You can see more of her photography on this Flickr link.
Jim Tolles November 04, 2010 Email Post
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text:
Strange Case of Dr Jekyll and Mr Hyde
Dr. Jekyll and Mr Hyde transforming potion | Strange Case of Dr Jekyll and Mr Hyde
Halloween· Mixers· Robert Louis Stevenson· Strange Case of Dr Jekyll and Mr Hyde
Wonder what makes this Dr Jekyll and Mr Hyde transforming potion special? There's two reasons. While butterfly pea flower extract…


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← There was a clear call today for a NI Referendum
Proportional Representation is needed for the UK →
A very disturbing attack onto the National Trust
Posted on October 13, 2021 by ianchisnall
In the past I have written ac couple of times about the challenges being presented to the National Trust by parts of the Conservative Party. They seem unwilling to accept that some of the premises which the National Trust now owns on behalf of our Nation should be disclosed with their genuine history. Almost a year ago there was this piece and it was a reflection of the statement by John Hayes who is a very extreme MP on a number of themes. Back on the 12th November 2020 he stated
Britain's heritage is under attack, ironically from those missioned to be the guardians of it. The National Trust, while losing money and sacking staff, has commissioned an expensive review of its properties' links with colonialism, including Churchill's Chartwell; unheroic characters at the National Maritime Museum are re-evaluating Nelson's heroic status; and the custodians of the Churchill War Rooms are claiming that they need to look again at Churchill's legacy. Can we have a debate on how these charitable organisations' purpose is being perverted by political posturing, as they all seem to be in the thrall of the militant Black Lives Matter movement? Mr Deputy Speaker, defending our history and heritage is our era's battle of Britain.
There now appears to be another challenge being set out to attack the National Trust by a group of people seeking to become a new group of Trustees. The most disturbing concept if accurate in this piece in the Guardian is that this group of people who are working together and have got funding to campaign for some of their group to succeed. As the article states
a range of "extreme" positions taken by individuals involved in a group called Restore Trust, which is backing a slate of candidates in elections for the NT's governing council.
The group, which has been heavily promoted in the pages of the rightwing press after its foundation and has spoken of building up a "fighting fund" of tens of thousands of pounds, is also now using paid-for social media adverts to try to influence the election.
This is deeply concerning on several fronts but what is equally disturbing is that one of the people who has been identified within the group is a person who I have spoken to on one occasion and who tragically has a very strong view about themes that he wishes to change in our nation and beyond based on the article. He he has set himself up as though he was able to speak out on behalf of Christianity under his charitable agency called Christian Voice which is not promoting views that all Christians agree with. It raises many challenges, and when on the one occasion we spoke about his public criticism of someone else who had been a board member of the BBC he took my view as to determine that I am not a Christian. The irony was that the person he was criticising was someone who I had met and who was a Christian and his anger was the way in which the BBC was behaving. The fact is that board members don't control every detail operated by their agency but perhaps he will have a different expectation if he is appointed as a Trustee. He has also visited Brighton on several occasions to campaign against the Pride events which creates a whole range of challenges for our area. Inevitably some other people agree with his views and may will support him in his new approach. However we need to enable the approach taken in the past by the National Trust to help explain what some of their premises have been used for in the past. Given that Slavery is still being identified today as a challenge from a range of churches, it seems acceptable that our history is also being displayed by groups like the National Trust. Equally if people in the past were LGBT, it would be very strange for the National Trust to be unable to disclose this in their settings.
About ianchisnall
I am passionate about the need for public policies to be made accessible to everyone, especially those who want to improve the wellbeing of their communities. I am particularly interested in issues related to crime and policing as well as health services and strategic planning.
View all posts by ianchisnall →
This entry was posted in Charities, Church Teaching, Parliament and Democracy and tagged John Hayes, National Trust. Bookmark the permalink.


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Home » News » UK GRANTS K310 MILLION FOR WILDLIFE CRIME INVESTIGATION
UK GRANTS K310 MILLION FOR WILDLIFE CRIME INVESTIGATION
March 15, 2016 2:20 pm Comments Off on UK GRANTS K310 MILLION FOR WILDLIFE CRIME INVESTIGATION Lilongwe Wildlife Trust
A specialized Wildlife Crime Investigations Unit is being introduced in Malawi, thanks to funding from the British Government.
Read the Daily Times coverage below, the British High Commission press release here and read the LWT/IFAW joint press release here. Photo from left to right: H.E. Michael Nevin, British High Commissioner, Chiza Manda, Deputy Director, DNPW and Jonny Vaughan, Executive Director, Lilongwe Wildlife Trust.
By Gladys Kamakanda
Lilongwe — British Government has announced grant amounting to £297, 000 (about K310 million) to New Malawi Wildlife Crime Intelligence Unit (WCIU) project to tackle the illegal wildlife trade.
British High Commissioner Michael Nevin made the announcement Tuesday during a press briefing at Lilongwe Wildlife Centre.
He said that this will build on previous British government funding to support law enforcement work through Royal Society for the Prevention of Cruelty to Animals (RSPCA) with the Malawi authorities and the support his government is providing locally through the British High Commission.
"Malawi is marking a positive name for itself with a progressive attitude towards tackling wildlife crime. It is important that we work collectively and regionally to prevent Malawi and its neighbours from being used as source and transit points for this damaging criminal activity," said Nevin.
He said in tandem with efforts to address issues such as corruption and professionalism, updating of legal framework and work with communities, Malawi can become a model for other countries to emulate.
Nevin said the project will help to develop sustainable livelihoods for communities affected by illegal wildlife trade through the introduction of Community Enforcement Networks whilst at the same time strengthening law enforcement in the country.
He said project partners Malawi's Department of National Parks and Wildlife (DNPW), International Fund for Animal Welfare (IFAW) and the Lilongwe Wildlife Trust will focus on law enforcement to provide a swift and effective response to serious wildlife crimes, particularly those involving elephants and illegal ivory trade using a multi-agency approach.
Lilongwe Wildlife Trust Executive Director, Jonathan Vaughan, said the extent and nature of the illegal wildlife trade globally has escalated significantly in recent years to involve organized criminal syndicates and Malawi has fallen victim as a trafficking hotspot and a distribution and transit hub for illicit ivory.
"A special WCIU will help to strengthen law enforcement, both in terms of the role of the criminal justice system and that of communities whilst also facilitating cooperation with our regional partners.
"In conjunction with other government-led initiatives, we hope to see improved interception rates which should dater would-be wildlife criminals who currently see Malawi as a soft target, particularly when it comes to the illicit ivory trade," said Vaughan.
He explained that Malawi has become a hotspot targeted by criminal groups for ivory trafficking, due to most its location between some of the country's worst hit by poaching – Tanzania, Mozambique and Zambia.
Deputy Director for DNPW, Chiza Manda, said this will go a long way in the fight against illegal wildlife trade and poaching that the country is currently going through.
"On behalf of the Malawi government I would like to extend my gratitude to the Challenge Fund for awarding the grant and the British High Commission for their ongoing support. We look forward to working with our NGO partners, IFAW and Lilongwe Wildlife Trust, on its delivery," said Manda.


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Q: Regression and contrast codings with multiple categorical variables In regression with multiple explanatory categorical variables, how should I model the problem to compare the effects of the categorical variables with each other?
Most contrast coding schemes (e.g. also on Wikipedia) seem to be designed (or at least, that's how they are described) to compare the effects of different levels within a given categorical variable (e.g. German vs British nationality). But what if I want to compare contributions between categorical variables?
For example, say that we want to estimate the contribution of two explanatory categorical variables (dress size and color) to a single dependent variable (dress price). How can I design my coding to measure interpretable contributions of size and color to price?

A: Reference cell coding:
There are P categorical variables with N levels each. This will generate P*(N-1) dummies where some combination of levels of categories is selected as baseline and others are estimated in relation to it. Baseline could be S and Green for example. Coding is done by having 0/1 indicator for N-1 levels each.
Deviation from the mean coding is such that sum of codes is zero for N levels.

redpajama_set_name: RedPajamaStackExchange


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Q: Laravel: disallow direct access to Post method by users I have a website which is developed in Laravel.
Problem:
I have a route method (POST)
Route::post('/profile/edit/save', 'ProfileController@save');

if I enter this url "mywebsite.com/profile/edit/save" I get an error

    Symfony \ Component \ HttpKernel \ Exception \ MethodNotAllowedHttpException
my .env file:
 APP_NAME=Laravel
 APP_ENV=production
 APP_DEBUG=false
 APP_LOG_LEVEL=debug
 APP_URL=http://localhost

I get this error while running this code on my server,in Localhost I don't get any errors.
I have cleared the cache but it's not working
How can i solve this Problem?

A: I have solved it by adding this in app\Exceptions\handler.php
    public function render($request, Exception $exception)
        {

             if($exception instanceof \Symfony\Component\HttpKernel\Exception\MethodNotAllowedHttpException){
                 return redirect('/');
             }

        return parent::render($request, $exception);
    }

A: You cannot access to post url like you tried above
You can only access by submitting form like this...
<form action="route('post_insert')" method="post">
  {{csrf_field()}}
   <input type="text" />
<input type="submit"/>
</form>

your route must include this name...
Route::post('/profile/save', 'ProfileController@save')->name('post_insert');


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Q: Pass value of selected country (dropdown) in session and then reload current page (Rails 4) In my Rails app, I'm trying to ask the user in the footer which country he'd like to see (France/United States/Germany...) and then store this in session and reload the page he's currently on.
Note that: I just want the current page to reload but I don't want to add a params in the URLs of my app(ie I don't want example.com/browser_country=France). I want them to stay as they are (example.com).
How can I do that?
This is my code below but it's not working (i'm a Rails newbie):
<footer class="footer">

    <small>
        MySuper Website <br/>
        <select name= "browser_country", :onchange => "location.href = '#{current_path}'"  >
           <option value="">Select Country</option>
              <option value="1">France</option>
              <option value="2">United States</option>
        </select>

    </small>

Here if the controller that loads the page (homepage)on  /app/controllers/static_pages_controller.rb
class StaticPagesController < ApplicationController

  def home
  end

  def set_country_into_session
    session[:country] = :browser_country # i'm putting in session the country selected in the dropdown
  end

end

A: You will need to add an ajax call to this process.
# your js file
$(function() {
    $('select[name="browser_country"]').change(function() {
        $.ajax({
            url: 'static_pages/set_country_into_session',
            data: 'browser_country=' + this.value,
            success: function() {
                location.reload();
            }
        });
    });
});

# static_pages_controller.rb
def set_country_into_session
  session[:country] = params[:browser_country]

  render text => :ok
end


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M666: How long has On A Pale Horse been together?
Jerry: About 2 years as a whole.
M666: What is the meaning of On A Pale Horse and how did you get it?
Jerry: Have you ever read the book? Read it!
M666: Who founded On A Pale Horse originally?
Jerry: Nobody founded On A Pale Horse. The band was a collective idea of
what we really believed in. The music was a style of music that I really wanted
to play back in the Painface days, but didn't come out right. We decided as
a band to go on a quest to find our own original style and let everyone in the
band have free roam of their musical talents.
M666: What is your favorite song to play live?
Jerry: Every single song. Each song has it's own personality and things
that I enjoy playing. I look forward to playing every song during the set.
M666: What were some of your influences growing up?
Jerry: I grew up listening to Jimi Hendrix, Old AC/DC, Led Zepplin, Mountain
and any other classic rock. I love the true meaning of being a musician.
M666: What is your favorite band in the world?
Jerry: Any band that has Scott "Wino" Weinrich in. That dude is the true
guitarist. Every band that he has ever been in was genius. He is my absolute
idol.
M666: What other bands have you been in?
Jerry: I was in Painface (Jas), and several other stupid teenage bands back
in the day.
M666: What was your first job?
Jerry: My first job was a cook for a local restaurant in my hometown. I was
thirteen.
M666: What is your opinion on Slipknot?
Jerry: I think every member of that band is truely gifted. They are all
geniuses in their own way. I consider all of them my brothers and friends.
M666: What is your favorite movie?
Jerry: My favorite movie of all time would have to be "Slingblade". When I
first sat down to watch it I didn't think it would end up the way it did.
Great movie in my terms. My second favorite would have to be "Gummo".
M666: How many albums do you think On A Pale Horse will make?
Jerry: Do you know what the winning Lotto Numbers are? Who knows? Everyone
hopes they will be together forever. Let's just jam as long as we can and
see what happens.
M666: How would you describe Aaron?
Jerry: Aaron is one of my closest friends. He has always stuck by my band
and I have always stuck by his. No matter what we did in music, we always
talked about being in a band together. When the position became available.
approached me with the idea of him on vocals. I couldn't say no. Aaron is a
great guy.
M666: What music genre does On A Pale Horse belong to?
Jerry: I hope we don't fit a genre…..I know we will start one!
M666: Last question….what do you think of my website, "Metabolic666"?
Jerry: Great informational website, but there needs to be more Pale Horse on
here! Ha Ha Ha!!!
M666: Thank you Jerry
Jerry: No problem
-Interview by Finnur/Master of Crowz of MFKR1.com & MFKRboard.com


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- National Wind Watch: Wind Energy News - https://www.wind-watch.org/news -
Study says wind turbines affect naval aviation negatively
Posted By Wind Watch Highlight On December 7, 2016 @ 7:35 am In Texas | Comments Disabled
A new study on the effects of those giant wind turbines on naval aviation in South Texas has been completed. Local officials were briefed on the results Tuesday afternoon.
According to those at the meeting, the U.S. Navy has concluded that wind turbines do have a negative impact on Doppler radar and the safety of pilots at the local training facility.
Congressman Blake Farenthold's office said the study showed that the wind turbines from the proposed Chapman Ranch wind farm can reduce detection of planes near wind turbines, cause false echoes of planes and false echoes in areas adjacent to wind farms. They added that some mitigation efforts were discussed including a second radar to improve plane detection and legislation that would create a 30-mile wind turbine buffer zone around air installations.
Other proposals include implementation of a special air traffic rule in South Texas operating areas and upgrading exisiting radar facilities.
The study said the U.S. Navy will work with wind developers to minimize impacts through the best location for turbines and agreements to curtail operations in certain situations. No other details on the impact on operations at the Naval Air Station in Corpus Christi were immediately available.
3News reached out to Enbridge, the new owners of the Chapman Ranch wind farm project, for comment but did not hear from them by news time.
Article printed from National Wind Watch: Wind Energy News: https://www.wind-watch.org/news
URL to article: https://www.wind-watch.org/news/2016/12/07/study-says-wind-turbines-affect-naval-aviation-negatively/


On Friday, February 8th, 2019 cryptocurrency litecoin (LTC) succeeded to surge 30.91% (seven-month high) just within the one-day timeframe. With such a sharp increase, litecoin has become the 4th largest crypto by market cap, according to data from CoinMarketCap.
Following litecoin, almost all of the first 100 cryptocurrencies by market cap see a powerful rise in price.
At press time, litecoin is trading at $43.46 (+30.91%).
Earlier, litecoin reached a key milestone — 60 million LTC in circulation.


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Tuition Waivers and Scholarships
Tuition Waivers and Scholarships Bobby G. Gaines High School Equivalency Waiver Program
Bobby G. Gaines High School Equivalency Waiver Program
Did you recently receive your State of Oklahoma Recognized High School Equivalency Diploma (GED, HiSET, or TASC)? Did you complete OCCC's High School Equivalency (HSE) classes? If so, you may be eligible for this wonderful tuition waiver opportunity!
The Bobby G. Gaines High School Equivalency Waiver Program is available to students who have earned their State of Oklahoma recognized High School equivalency Diploma (GED, HiSet, or TASC) and who completed Oklahoma City Community College's Preparation Program. Students must be enrolled in 12 hours a semester to receive a $900.00 tuition waiver per year. Students are responsible for fees and books.
Students must submit the application by February 1, 2022. Students must enter OCCC the fall semester immediately after receiving the GED, HiSet or TASC. Only U.S. citizens and permanent residents are eligible to apply for this scholarship.
Students must have earned their GED, HiSet, or TASC
Students must be enrolled in 12 hours a semester to get up to $900.00 ($450 fall & spring) tuition waiver
Must complete the FAFSA
The application deadline for this scholarship is February 1, 2022


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Q: Using AXIOS to POST to external server API to retrieve SSOID In short: Here is some Python code posting to Betfair API. I would like to use Axios to do the same thing.
resp = requests.post('https://identitysso-cert.betfair.com/api/certlogin',
                     data=payload, cert=('TestApp.crt', 'client-2048.key'), headers=headers)

I'm reading through AXIOS docs, and am curious how to apply the cert=('TestApp.crt', 'cient-2048.key') field.
In detail: Currently, I have this:
axios({
  method: "POST",
  headers: headers,
  url: "https://identitysso-cert.betfair.com/api/certlogin",
  data: payload,
});

Would I use the form-data library replacing cert=('TestApp.crt', 'cient-2048.key') with form<FormData>
const FormData = require("form-data");

const form = new FormData();
form.append("my_field", "my value");
form.append("my_buffer", new Buffer(10));
form.append("my_file", fs.createReadStream("/foo/bar.jpg"));

axios.post("https://example.com", form, { headers: form.getHeaders() });

EDIT:
Scrapped the FormData route, and am using HTTPS for node js.
I add this along with the options I provide to Axios.
const httpsAgent = new https.Agent({
  cert: fs.readFileSync("certificat.crt"),
  ca: fs.readFileSync("key.pem"),
});

I in turn get this error:

Error: "Error: SSL Error: SELF_SIGNED_CERT_IN_CHAIN"

A: Digging a little deeper,
The error SELF_SIGNED_CERT_IN_CHAIN means that you cannot use self-signed certificates.
I ended up using Python to accomplish what I needed.


17 Jan 2019 How Much Do Crowns Cost?
Most of us, at some point in our lives, are going to need a crown. Crowns are tooth shaped caps that are put on teeth sometimes after a root canal, after several cavities on the same too, or after infection or injury. Crowns can help preserve a tooth that is weakened or compromised. A crown is a much better option than having a tooth pulled and then having to pay for an implant or deal with a bridge. Crowns can look very similar to real teeth and can function as real teeth, too.
The question that all of us want answered before we go to the dentist is how much a crown is going to cost. Unfortunately, it's difficult to estimate the cost of crowns because they are a custom product. Each one is custom-made to fit perfectly in a person's mouth. Several factors can influence the cost of a dental crown: the material the crown is made of, how big the crown is, the location of the tooth being crowned, the experience of the dentist, and the "going" rates of dentistry in your area.
Other factors that can affect the cost of a crown include whether or not you have insurance that will pay part of the bill and the structure of the tooth itself. Do you need a root canal before a crown is placed? Or does your tooth need to be built up in order to be fit with a crown? Additional procedures will increase the cost of getting a crown. Meeting with your dentist and having him or her look at your tooth is the only way to figure out what you really need. If you meet with one dentist and worry that the price is too high, know that you can get a second opinion.
All Ceramic or Porcelain Crowns – These crowns look the most like real teeth. Many people prefer them although they can break under force.
Zirconia or Milled Crown – Crown made from zirconia are ideal for those who have broken their natural teeth or porcelain crowns through bruxing or grinding.
Porcelain-fused-to-metal crowns are stronger than ceramic crowns. The crown has a metal structure that is covered with porcelain.
All Resin – All resin crowns are less expensive than other crown types but are far weaker. Typically all-resin crowns are only used as a temporary solution while a more permanent option such as one listed above is being made.
Gold alloys are strong and resistant to chipping. They will be gold in appearance.
Base metal alloys are also strong. They will be silver in appearance. These are typically only used for small children when a crown is only needed on a baby tooth for a year or two.
If your dentist decides you need a crown, he or she will discuss with you the best type of crown for your particular problem. If the tooth being repaired is in the front of your mouth, most likely your dentist will recommend a ceramic crown because the look the best. If you are repairing a molar, a different type of crown like a porcelain-fused-to-metal may be recommended because it is stronger and could withstand pressure from grinding better. Costs for all of these types of crowns can vary depending on many different factors.
Prices or cost for a crown varies depending on where you live, the type of materials used, and even the experience level of the lab technician who makes the crown and the training and experience level of your dentist. Some dental offices have in-house machines that make your crown while you wait. These can add cost to your crown as well. There are many factors to consider when deciding what type of crown is the best for you. Read more about the different types of dental crowns in Salt Lake City and how they're made as well as what to expect during your crown appointment.
Whatever crown you choose, your mouth will thank you for years to come! At Holladay Family Dental we take the time to educate our patients on the many different options and prices of crowns available to them. Dr. Greaves is one the best dentists in Salt Lake City and he would love to help you decide on the best type of crown for you that will also fit your budget.


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Q: object mapper conversion issue when string contains'\/' patterns I have following type of token that as to be send as it is in request.
//token contains following value
token = "ABCD/saljljlkljljljl";
class Token{
        public String token;
    public String getToken() {
        return token;
    }

    public void setToken(String token) {
        this.token = token;
    }

}

when I try to convert it to string using object mapper class, it converts it as follows.
ObjectMapper mapper = new ObjectMapper();
mapper.writeValueAsString(token)
{"token":"ABCD\/saljljlkljljljl"}
Note one more '\' added in front of '/' due to which my calls are failing.Don't no why its not reflecting even in stackoverflow reader pane also.
Please help out on this.


This is a moment when Cody the Pug was content to just watch the hens, and Bo the cat was content to lay in the hay watching Cody bond with the hens.
The latest additions began as three ducks hanging out together. There was an "incident" at the waterfront over the weekend involving Bo the cat, and the Indian Runner has been missing, without a trace, since.
This is the beautiful Indian Runner, last seen taunting Bo the cat at the waterfront several days ago, then escaping, we thought, beneath the dock. If you see him, tell him to contact his buddies, they miss him.
Bo looking innocent, days before the "episode".