text:
\section{Introduction}

Let $\mathbb{G} = (\mathbb{R}^n, \star)$ be a stratified Lie group and $\{X_1, \dots, X_m\}$ be a system of generators for $\mathbb{G}$, i.e. a basis for the first strata of $\mathbb{G}$.
For the self-adjoint subelliptic operator $$\Delta_\mathbb{G} = - \sum_{i=1}^m X_i^2,$$
and $p > 1$, consider the semilinear subelliptic heat equation
\begin{equation}\label{MainProblem}
\begin{cases}
u_t(t,x) - \Delta_\mathbb{G} u(t,x) = |u(t,x)|^p + f(x), & (t,x) \in (0,T) \times \mathbb{G}, \\
u(0, x) = u_0(x), & x \in \mathbb{G},
\end{cases}
\end{equation}
with a non-zero forcing term $f$ depending only upon the space variable $x \in \mathbb{G}$ where the given functions $f, u_0 \in L^1_{loc}(\mathbb{G})$ are non-negative.
A critical exponent is the largest real number $p_c$ such that (\ref{MainProblem}) has no global solution whenever $p < p_c$.
In the supercritical case of $p > p_c$, there always exists some $f$ and $u_0$ for which (\ref{MainProblem}) has a global solution.

The importance of studying PDEs on stratified Lie groups is apparent from the lifting theorem given by Rothschild and Stein in their celebrated paper \cite{RothschildStein}.
Even though a generic H\"ormander operator which is sum of squares of vector fields does not become a sub-Laplacian, it can be, roughly speaking, approximated by a sub-Laplacian on some stratified Lie group.
This makes it crucial to study the second order partial differential equations on stratified Lie groups.

A vast literature is available on problems similar to (\ref{MainProblem}) and this pursuit started from the seminal paper \cite{Fujita} of Fujita.
For $\mathbb{G} = (\mathbb{R}^n, +)$ and $f \equiv 0$, it was proved that $p_c = \frac{n+2}{n}$.
The critical case $p= p_c$ was then dealt with in \cite{Hayakawa} where it was proved that a global solution does not exist.
In \cite{Bandle} the problem (\ref{MainProblem}) was studied for $\mathbb{G} = (\mathbb{R}^n, +)$ and the authors proved that  $p_c = \frac{n}{n-2}$ when $\int_{\mathbb{R}^n} f(x) dx > 0$.
Then $p_c = \frac{Q+2}{Q}$ was documented by Pascucci in \cite{Pascucci} which provided a natural generalization of Fujita's work on stratified Lie groups $\mathbb{G} = (\mathbb{R}^n, \star)$.
Thus, an essential question was to develop Pascucci's result for the inhomogeneous case.

The main goal of the present paper is to seek an answer to this question.
In this paper, thanks to the available Lie group dilation structure, we extend some results proved in \cite{Bandle} from the abelian case $\mathbb{G} = (\mathbb{R}^n, +)$ to an arbitrary stratified Lie group $\mathbb{G} = (\mathbb{R}^n, \star)$.
While attending to the supercritical case we could not use the classical technique of finding a solution of the form $\epsilon (1 + d(x))^{k}$ ($d$ being a homogeneous norm) since $|\nabla_\mathbb{G} d|$ may not be identically one on $\mathbb{G} \setminus \{0\}$.
This is where the techniques from \cite{Zhang} turned out to be useful.
Employment of the test function method is inspired from \cite{Torebek, JleliKawakamiSamet, GeorgievPalmieri, MitidieriPokhozhaev} and we refer the reader to \cite{RuzhanskyYessirkegenov, Pohozaev, Pascucci, Levine} for further discussions on similar problems.

This paper is organized as follows.
In \Cref{Preliminary} we collect some definitions and results to provide the necessary background and aid smooth reading of the paper.
The main results are proved in \Cref{Results}.
We show in \Cref{subcritical} and \Cref{Critical} that the nonexistence results derived for the Heisenberg group $\mathbb{H}^n$ in \cite[Section 4]{Torebek} can be generalized to $\mathbb{G}$.
In Section 2 of our recent article \cite{SuraganTalwar}, we have already proved that all the results presented in \cite[Section 3]{Torebek} are valid in a more generalized setting, which includes the case of stratified Lie groups.
An existence result is proved in \Cref{Supercritical} which helps us conclude that for (\ref{MainProblem})  $p_c = \frac{Q}{Q-2}$ when $Q > 2$ and $p_c = \infty$ when $Q \in \{ 1,2 \}$.

\section{Preliminaries}\label{Preliminary}
We refer to \cite[Chapter 1]{Bonfiglioli} for the fundamentals of the theory of stratified Lie group $\mathbb{G} = (\mathbb{R}^n, \star)$.
There are $n$ positive integers $1 \leq \sigma_1 \leq  \dots \leq \sigma_n$ associated to $\mathbb{G}$ such that for every $\lambda > 0$ the maps $$\delta_\lambda(x_1, \dots x_n) = (\lambda^{\sigma_1} x_1, \dots \lambda^{\sigma_n} x_n)$$ are group automorphisms known as the dilations on $\mathbb{G}$.
There exists $m \leq n$ linearly independent vector fields $X_1, \dots, X_m$ such that $\{ X_k : 1 \leq k \leq m  \}$ along with its $r \in \mathbb{N}$ commutators generate the Lie algebra of $\mathbb{G}$.
In this case $\sigma_1 =\sigma_2 = \dots \sigma_m = 1$ and we say that $\mathbb{G}$ has step $r$.
The integer $Q = \sum_{i=1}^n  \sigma_i$ is known as the homogeneous dimension of $\mathbb{G}$ and $Q = n$ if and only if  $\mathbb{G}$ is the same as  $(\mathbb{R}^n, +)$.

For $l \in \mathbb{R}$, a vector field $X$ on $\mathbb{G}$ is said to be $\delta_\lambda$-homogeneous of degree $l$ if for every smooth function $f$ on $\mathbb{G}$ the equation $$X(f(\delta_\lambda(x)))= \lambda^l (X f) (\delta_\lambda(x))$$ is satisfied for every $x \in \mathbb{G}$.
Likewise, a real valued function $\phi$ on $\mathbb{G}$ is said to be $\delta_\lambda$-homogeneous of degree $l \in \mathbb{R}$ if $$\phi(\delta_\lambda(x)) = \lambda^l \phi(x)$$ for every $x \in \mathbb{G}$.
As one would expect, if $X$ is $\delta_\lambda$-homogeneous of degree $l_1$ and $\phi$ is $\delta_\lambda$-homogeneous of degree $l_2$, then $X \phi$ is $\delta_\lambda$-homogeneous of degree $l_2 - l_1$ and $X^k$ is $\delta_\lambda$-homogeneous of degree $kl_1$ for every $k \in \mathbb{N}$.

The Carnot-Carath\'eodory (or control) distance as well as the $\Delta_\mathbb{G}$-gauge prove to be of great use in our computations.
Recall that all the homogeneous norms on $\mathbb{G}$ are equivalent.
Till \Cref{Critical} is proved, we use $d$ to denote the Carnot-Carath\'eodory distance.
It gives rise to a homogeneous norm $d(x) := d(x,0)$ on $\mathbb{G}$ satisfying the triangle inequality \cite[p. 4]{Pascucci}.
Then $d$ (and every other homogeneous norm on $\mathbb{G}$) and the smooth vector fields  $X_k$ are $\delta_\lambda$-homogeneous of degree 1 for $1 \leq k \leq m$.
It is known (see e.g. \cite{VSC}) that for the heat kernel $h_t(x,y)$, i.e. the fundamental solution of $\frac{\partial}{\partial t} - \Delta_\mathbb{G}$, the global estimate  \begin{equation}\label{BoundsForHeatKernel}
\frac{1}{C t^{\frac{Q}{2}}} \exp^{\frac{-C d(x,y)^2}{t}} \leq h_t(x,y) \leq  \frac{C}{t^{\frac{Q}{2}}} \exp^{\frac{-d(x,y)^2}{C t}}\end{equation} holds for every $t \in (0, \infty)$.

After proving \Cref{Critical} we use $d$ for $\Delta_\mathbb{G}$-gauge which is by definition a smooth out of origin symmetric  homogeneous  norm satisfying the equation $$\Delta_\mathbb{G}(d^{2-Q}) = 0$$ in $\mathbb{G} \setminus \{0\}$.
Then there exists $c > 0$ such that  $d$ satisfies the pseudo-triangle inequality $$d(x,y) \leq c(d(x) + d(y))$$ for every $x,y \in \mathbb{G}$.
From \cite{Folland} it is  known that for the $\Delta_\mathbb{G}$-gauge  $d$ there exists $\beta_d >0$ such that $$ H(x,y):= \frac{\beta_d}{d(x,y)^{Q-2}}$$ is the fundamental solution for the sub-Laplacian $\Delta_\mathbb{G}$.
One should not confuse $\delta_\lambda$ with $\delta$ which is used as a positive real number at times.
Recall from \cite{Bonfiglioli} that  if $B(x,t)$ denotes the ball in $\mathbb{G}$ of radius $a$ and center at $x$ with respect to $d$, then $|B(x,a)| = a^Q |B(x,1)|$ ($|E|$ being the Haar measure of the subset $E \subseteq \mathbb{G}$) and for every measurable function $f$ we have (see \cite[Proposition 5.4.4]{Bonfiglioli}) \begin{equation}\label{IntegrationOfRadialFunctions}
\int_{B(0,a)} f(d(x)) dx = Q |B(0,1)| \int_0^a f(\delta) \delta^{Q-1} d\delta.
\end{equation}

Let us fix some notations.
We denote the space of all continuous functions on $\mathbb{G}$ which vanish at infinity by $C_0(\mathbb{G})$.
For a Banach space $(X, \norm{\cdot})$, the space of all $X$-valued continuous and vanishing at infinity functions on a locally compact Hausdorff topological space $S$ is denoted by $C_0(S,X)$.
This space is a Banach space when the norm of $w \in  C_0(S,X)$ is given by $\norm{w}_{\sup} = \sup \{ \norm{w(s)} : s \in S \}$.
We set $x = (x_1, x_2, \dots, x_n) \in \mathbb{G}$ and similarly $x' = (x_1', x_2', \dots, x_n')$.
Although we use $C$ to denote a constant whose value may keep on changing from line to line $2C$ is also used at some places to make the inequalities easier to understand.

\section{Fujita exponent is $\frac{Q}{Q-2}$}\label{Results}

Let us fix $\mathbb{G}^{\ast} = \mathbb{G} \setminus \{0\}$ and $\frac{1}{p} + \frac{1}{p'} = 1$.

\begin{defin}\label{DefinitionLocalWeakSolution}
	Let $u_0 \in L^1_{loc}(\mathbb{G})$.
	Then $u \in L^p_{loc}((0,T), L^p_{loc}(\mathbb{G}))$ is called a local in time weak solution of (\ref{MainProblem}) if for every non-negative test function $\psi \in C^1_0( (0,T); C_0^2(\mathbb{G}^{\ast}) )$ satisfying $\psi(T, \cdot) = 0$ we have
\begin{eqnarray}
	\int_0^T \int_{\mathbb{G}^{\ast}} |u|^p \psi + \int_{\mathbb{G}^{\ast}} u_0 \psi(0, x) + \int_0^T \int_{\mathbb{G}^{\ast}} f \psi + \int_0^T \int_{\mathbb{G}^{\ast}} u \psi_t + \int_0^T \int_{\mathbb{G}^{\ast}} u \Delta_\mathbb{G} \psi = 0.
\end{eqnarray}
When $T = \infty$, such $u$ is called a global in time weak solution.
\end{defin}

Let us start with providing a result  of independent interest in relation to which we refer the reader to \cite[Theorem 2.5]{MontiSerra} where a similar result is proved in a different setting.
It is the idea of the proof of \Cref{NabladIsBounded} which helps us to obtain the non-existence results later in the paper.
\begin{prop}\label{NabladIsBounded}
For the control distance $d$ on $\mathbb{G}$, the function $|\nabla_\mathbb{G} d|$ is bounded on $\mathbb{G}^{\ast}$.
\end{prop}
\begin{proof}
Let us put to use the fact that the vector fields $\{ X_k: 1 \leq k \leq m \}$ and $d$ are  $\delta_\lambda$-homogeneous of degree $1$.
This tells us that $X_k d$ is $\delta_\lambda$-homogeneous of degree $0$ which means that for every $x \in \mathbb{G}$ and $\lambda > 0$ $$(X_kd) (\delta_{\lambda}(x)) = (X_kd)(x).$$
Thus, $$\sup_{x \in \mathbb{G}^{\ast}} \{ (X_kd)(x) \} = \sup_{x \in \mathbb{G} : d(x) = 1} \{ (X_kd)(x) \}.$$
As $X_kd$ is continuous on the compact sphere $\{x \in \mathbb{G} : d(x) = 1\}$, we obtain that  this supremum is finite for every $1 \leq k \leq m$.
Hence, $$|\nabla_\mathbb{G} d|^2 = \sum_{k=1}^m (X_k d)^2$$ is bounded on $\mathbb{G}^{\ast}$.
\end{proof}

Let us now discuss the subcritical case.
\begin{theorem}\label{subcritical}
Let $Q \geq 3$, $p < \frac{Q}{Q-2}$, $u_0 \geq 0$ and $\int_{\mathbb{G}^\ast} f(x) dx > 0$.
Then (\ref{MainProblem}) does not admit a global in time weak solution.
\end{theorem}
\begin{proof}
We adapt the proof of \cite[Theorem 3.2]{SuraganTalwar} and use some delicate reasoning.
Let us suppose on the contrary that $u$ is a global in time weak solution.
Then for every test function $\psi$, $$\int_0^T \int_{\mathbb{G}^{\ast}} |u|^p \psi + \int_{\mathbb{G}^{\ast}} u_0 \psi(0, x) + \int_0^T \int_{\mathbb{G}^{\ast}} f \psi + \int_0^T \int_{\mathbb{G}^{\ast}} u \psi_t + \int_0^T \int_{\mathbb{G}^{\ast}} u \Delta_\mathbb{G} \psi = 0.$$
As both $\psi$ and $u_0$ are non-negative, we obtain
\begin{eqnarray}\label{Inequality}
\int_{0}^T \int_{\mathbb{G}^{\ast}} f \psi + \int_{0}^T \int_{\mathbb{G}^{\ast}} |u|^p \psi &\leq& - \int_{0}^T \int_{\mathbb{G}^{\ast}} u \psi_t   - \int_{0}^T \int_{\mathbb{G}^{\ast}} u \Delta_\mathbb{G} \psi \nonumber  \\
& \leq &  \int_{0}^T \int_{\mathbb{R}^n} | u || \psi_t| + |u|| \Delta_\mathbb{G} \psi|.
\end{eqnarray}
Set $P(x) = \frac{d(x)}{\sqrt{T}} $.
Then $P(x) = (p \circ d)(x)$ where  $p: \mathbb{R} \to \mathbb{R}$ is given by $$p(w)= \frac{w}{\sqrt{T}}.$$
Define a test function (see \Cref{DefinitionLocalWeakSolution}) with separated variables as $$\psi(t,x) = \Phi(P(x)) \Phi(2t/T),$$ where \begin{equation*}
\Phi(z) = \begin{cases}
1, & z \in [0,1],\\
\searrow, & z \in [1, 2],\\
0, & z \in (2, \infty),
\end{cases}
\end{equation*}
is a function in $C^2_0(\mathbb{R}^+)$ satisfying, for $$(\mathbb{G}^{\ast}) := \delta_{\frac{1}{\sqrt{T}}} \mathbb{G}^{\ast} := \left\{\delta_{\frac{1}{\sqrt{T}}}(x) : x \in \mathbb{G}^{\ast} \right\},$$ the estimates $$\int_{(\mathbb{G}^{\ast})} \Phi(P(x'))^{\frac{-1}{p-1}} dx' < \infty$$ and $$\int_{0}^2    |\Phi(t')|^{\frac{-1}{p-1}}|  \Phi'(t')|^{\frac{p}{p-1}} dt' < \infty$$  where $t' = \frac{2t}{T}$ and $x' = \delta_{\frac{1}{\sqrt{T}}} x$.

Note that $$\text{supp}(\psi) \subseteq \left\{ (t,x) \in \mathbb{R}^+ \times \mathbb{G}^{\ast} : 0 \leq 2t/T \leq 2, 0 \leq P(x) \leq 2 \right\}. \smallskip$$
Now $\psi_t(x,t) = \frac{\partial}{\partial t}(\Phi(P(x)) \Phi(2t/T)) = \frac{2}{T}  \Phi(P(x)) \Phi'(2t/T)$ implies that $$\text{supp}(\psi_t) \subseteq \left\{ (t,x) \in \mathbb{R}^+ \times \mathbb{G}^{\ast} : 1 \leq 2t/T \leq 2, 0 \leq P(x) \leq 2 \right\}.$$
To keep the computations for $\Delta_\mathbb{G} \psi(t,x) = \Phi(2t/T) \Delta_\mathbb{G} \Phi(P(x))$  clear, we define $\Psi = \Phi \circ p$.
Then $\Psi$ is a smooth function from $\mathbb{R}$ to $\mathbb{R}$.
Note that
$$\Delta_\mathbb{G} \psi(t,x) = \Phi(2t/T) \Delta_\mathbb{G} \Phi(P(x)) = \Phi(2t/T) \left(\Delta_\mathbb{G}  (\Psi \circ d)\right)(x),$$
$$\Psi' = (\Phi \circ p)' =  \Phi'(p) p'$$ and $$\Psi'' = (\Phi \circ p)'' = (\Phi'(p) p')' = (\Phi'(p))' p' =  \Phi''(p) (p')^2.$$
Then
\begin{eqnarray*}
	\left(\Delta_\mathbb{G}  (\Psi \circ d)\right)(x) &=&   \sum_{k=1}^m  \left(X_k^2 (\Psi \circ d)\right) (x)\\
	&=&  \sum_{k=1}^m \left( \Psi^{''}(d(x)) (X_k(d(x)))^2 + \Psi'(d(x)) X_k^2(d(x)) \right)\\
	&=&  \sum_{k=1}^m \left( \Phi^{''}(P(x)) \frac{1}{T} \left( X_k(d(x)) \right)^2
	+ \frac{\Phi^{'}(P(x))}{\sqrt{T}} \left( X_k^2(d(x))  \right) \right).
\end{eqnarray*}
implies that \[\text{supp}(\Delta_\mathbb{G} \psi) \subseteq \left\{ (t,x) \in \mathbb{R}^+ \times \mathbb{G}^{\ast} : 0 \leq 2t/T \leq 2, 1 \leq P(x) \leq 2 \right\}.\]

For every $1 \leq k \leq m$, as $d$ and $X_k$ are $\delta_\lambda$-homogeneous of degree $1$ and $X_k^2$ is $\delta_\lambda$-homogeneous of degree $2$, we obtain \begin{eqnarray*}
\left(\Delta_\mathbb{G}  (\Psi \circ d)\right)(\delta_{\sqrt{T}} (x))
&=& \sum_{k=1}^m \left( \Phi^{''}(P(\delta_{\sqrt{T}} (x))) \frac{1}{T} \left( X_k(d(\delta_{\sqrt{T}} (x))) \right)^2 \right.  \\
&+& \left. \frac{\Phi^{'}(P(\delta_{\sqrt{T}} (x)))}{\sqrt{T}} \left( X_k^2(d(\delta_{\sqrt{T}} (x)))  \right) \right)\\
	&=& \sum_{k=1}^m \left( \Phi^{''}(P(\delta_{\sqrt{T}} (x))) \frac{1}{T} \left( X_k(d(x)) \right)^2
	+ \frac{\Phi^{'}(P(\delta_{\sqrt{T}} (x)))}{T} \left( X_k^2(d(x)) \right) \right).
\end{eqnarray*}
Consider the compact sphere  $$S = \left\{ x \in \mathbb{G} : d(x) = 1 \right\}.$$
Every element of $\mathbb{G}$ can be written as $\delta_{\frac{1}{\sqrt{T}}} (x)$ for some $x \in S$ and $T > 0$.
Also, since the vector fields under discussion are smooth $X_k^2(d(x))$ and $X_k(d(x))$ must be bounded for $x \in S$.
Finally, as $\Phi$ is smooth and compactly supported, there exists $C > 0$ such that $$ |\Delta_\mathbb{G}  (\Psi \circ d)| \leq \frac{C}{T}.$$

Using inequality 3.2, $\frac{p}{2}$-Young inequality and the similar arguments as in \cite[Theorem 3.2]{SuraganTalwar} one obtains
\[\int_{0}^T \int_{\mathbb{G}^{\ast}} f \psi \leq \frac{1}{p' (\frac{p}{2})^{p'-1}} \Big( \int_{0}^T \int_{\mathbb{G}^{\ast}} |\psi|^{\frac{-1}{p-1}}|\Delta_\mathbb{G} \psi|^{\frac{p}{p-1}} + |\psi|^{\frac{-1}{p-1}}|\psi_t|^{\frac{p}{p-1}} \Big).\]

Now, \begin{eqnarray*}
\int_{0}^T \int_{\mathbb{G}^{\ast}} |\psi|^{ \frac{-1}{p-1}}|\Delta_\mathbb{G} \psi|^{\frac{p}{p-1}} & \leq  & \int_{0}^T \int_{\mathbb{G}^{\ast}} |\Phi(P(x)) \Phi(2t/T)|^{\frac{-1}{p-1}} \left(\frac{C}{T}\right)^{\frac{p}{p-1}} \\
& =& \left(\frac{C}{T}\right)^{\frac{p}{p-1}} \int_{0}^T \int_{\mathbb{G}^{\ast}} |\Phi(P(x)) \Phi(2t/T)|^{\frac{-1}{p-1}} \\
&= & \left(\frac{C}{T}\right)^{\frac{p}{p-1}} T^{\frac{Q}{2}} \frac{T}{2} \int_{0}^2 \int_{(\mathbb{G}^{\ast})} |\Phi(P(x')) \Phi(t')|^{\frac{-1}{p-1}} \\
&= & \left(\frac{C}{T}\right)^{\frac{p}{p-1}} T^{\frac{Q}{2}} T \int_{(\mathbb{G}^{\ast})} |\Phi(P(x'))|^{-1/(p-1)} \leq C_1 T^{\frac{Q}{2}+1 - \frac{p}{p-1}}
\end{eqnarray*}
and similarly,
\begin{eqnarray*}
\int_{0}^T \int_{\mathbb{G}^{\ast}}  |\psi|^{\frac{-1}{p-1}}|\psi_t|^{\frac{p}{p-1}} &=&  \int_{0}^T \int_{\mathbb{G}^{\ast}}  |\Phi(P(x)) \Phi(2t/T)|^{\frac{-1}{p-1}} \left|\frac{2}{T}  \Phi(P(x)) \Phi'(2t/T)\right|^{\frac{p}{p-1}} \\
& = &  \left(\frac{2}{T}\right)^{{\frac{p}{p-1}}} \int_{0}^T \int_{\mathbb{G}^{\ast}}  |\Phi(P(x))|  |\Phi(2t/T)|^{\frac{-1}{p-1}}|  \Phi'(2t/T)|^{\frac{p}{p-1}}\\
& = & \left(\frac{2}{T}\right)^{{\frac{p}{p-1}}} \frac{T}{2} T^{\frac{Q}{2}} \int_{0}^2 \int_{(\mathbb{G}^{\ast})}  |\Phi(P(x'))|  |\Phi(t')|^{\frac{-1}{p-1}}|  \Phi'(t')|^{\frac{p}{p-1}}.
\end{eqnarray*}

As, \begin{equation*}
\int_{0}^T \int_{\mathbb{G}^{\ast}} f \psi = \int_{0}^T \Phi(2t/T) \int_{\mathbb{G}^{\ast}} f \Phi(P(x))  =  \frac{T}{2} \int_{0}^2 \Phi(t') \int_{\mathbb{G}^{\ast}} f \Phi(P(x)) \geq T \int_{\mathbb{G}^{\ast}} f \Phi(P(x))
\end{equation*}
we obtain $$T \int_{\mathbb{G}^{\ast}} f \Phi(P(x)) \leq \int_{0}^T \int_{\mathbb{G}^{\ast}} f \psi \leq C_1 T^{\frac{Q}{2}+1 - \frac{p}{p-1}}.$$
By dominated convergence theorem, $ \int_{\mathbb{G}^{\ast}} f \Phi \to \int_{\mathbb{G}^{\ast}} f $ as $T \to \infty$.
Using the fact that $p < \frac{Q}{Q-2}$ implies $\frac{Q}{2} < \frac{p}{p-1}$ we obtain $\int_{\mathbb{G}^{\ast}} f \leq 0$ by taking $T \to \infty$.
This contradicts the hypothesis that $\int_{\mathbb{G}^{\ast}} f > 0$.
So, no such $u$ exists and the statement of the theorem is proved.
\end{proof}

\begin{remark}
Note from the proof of \Cref{subcritical} that the assumption $p < \frac{Q}{Q-2}$ was used near the end.
Thus, we have the inequality, $$T \int_{\mathbb{G}^{\ast}} f \Phi(P(x)) \leq \int_{0}^T \int_{\mathbb{G}^{\ast}} f \psi \leq C_1 T^{\frac{Q}{2}+1 - \frac{p}{p-1}}$$ for every $p > 1$ and $Q \in \mathbb{N}$.
Since for $Q \in  \{ 1,2\}$ and $p > 1$ we have $\frac{Q}{2} - \frac{p}{p-1} < 0$, one can take $T \to \infty$ and conclude that $\int_{\mathbb{G}^{\ast}} f \leq 0$.
Hence, $p_c = \infty$ when $Q \in  \{ 1,2\}$, $u_0 \geq 0$ and $\int_{\mathbb{G}^\ast} f(x) dx > 0$.
\end{remark}

We now discuss the critical case.
\begin{theorem}\label{Critical}
Let $p = \frac{Q}{Q-2}$, $u_0 \geq 0$ and $\int_{\mathbb{G}^\ast} f(x) dx > 0$.
Then (\ref{MainProblem}) does not admit a global in time weak solution.
\end{theorem}
\begin{proof}
Let us suppose on the contrary that $u$ is a global in time weak solution for (\ref{MainProblem}).
For any $0 < R < \infty$, set $P(x) := \frac{ \ln\left(\frac{d(x)}{\sqrt{R}}\right) }{\ln(\sqrt{R})}$ and define a test function (see \Cref{DefinitionLocalWeakSolution}) with separated variables as $\psi(t,x) = \Phi(P(x)) \Phi(t/T)$ where \begin{equation*}
\Phi(z) = \begin{cases}
1, & z \in (- \infty ,0],\\
	\searrow, & z \in [0, 1],\\
	0, & z \in (1, \infty),
	\end{cases}
\end{equation*}
is a function in $C^\infty(\mathbb{R})$ satisfying, for $$(\mathbb{G}^{\ast}) := \delta_{\frac{1}{\sqrt{R}}} \mathbb{G}^{\ast} := \left\{\delta_{\frac{1}{\sqrt{R}}}(x) : x \in \mathbb{G}^{\ast} \right\},$$ the estimates $$\int_{(\mathbb{G}^{\ast})} \Phi(P(x'))^{\frac{-1}{p-1}} dx' < \infty$$ and $$\int_{0}^2    |\Phi(t')|^{\frac{-1}{p-1}}|  \Phi'(t')|^{\frac{p}{p-1}} dt' < \infty$$  where $t' = \frac{t}{T}$ and $x' = \delta_{\frac{1}{\sqrt{R}}} (x)$.
Note that $$\text{supp}(\psi) \subseteq \left\{ (t,x) \in \mathbb{R}^+ \times \mathbb{G}^{\ast} : 0 \leq t \leq T, - \infty < P(x) \leq 1 \right\}.$$
As $- \infty \leq P(x) \leq 1$ if and only if $d(x) \leq R$, the set $\text{supp}(\psi)$ is compact.
The fact that $$\psi_t(x,t) = \frac{\partial}{\partial t}(\Phi(P) \Phi(t/T)) = \frac{1}{T}  \Phi(P) \Phi'(t/T)$$ implies the embedding   $$\text{supp}(\psi_t) \subseteq \left\{ (t,x) \in \mathbb{R}^+ \times \mathbb{G}^{\ast} : 0 \leq t \leq T, - \infty < P(x) \leq 1 \right\}.$$

To keep the computations for $\Delta_\mathbb{G} \psi(t,x) = \Phi(2t/T) \Delta_\mathbb{G} \Phi(P(x))$ clear, let $$p(w) = \frac{\ln(w)}{\ln({\sqrt{R}})}.$$
Then $p'(w) = \frac{1}{w \ln({\sqrt{R}})}$ and $p''(w) = -\frac{1}{w^2 \ln({\sqrt{R}})}$ for every $w \in \mathbb{R}^+$.
For $\Psi = \Phi \circ p$, which is a smooth function from $\mathbb{R}^{+}$ to $\mathbb{R}$, we have
$$\Psi' = (\Phi \circ p)' =  \Phi'(p) p'$$ and $$\Psi'' = (\Phi \circ p)'' = (\Phi'(p) p')' = (\Phi'(p))' p' + \Phi'(p) p''  =  \Phi''(p) (p')^2 + \Phi'(p) p''$$ which implies that
\begin{eqnarray*}
\left(\Delta_\mathbb{G}  (\Psi \circ d)\right)(x) &=&   \sum_{k=1}^m  \left(X_k^2 (\Psi \circ d)\right) (x)\\
&=&  \sum_{k=1}^m \left( \Psi^{''}(d(x)) (X_k(d(x)))^2 + \Psi'(d(x)) X_k^2(d(x)) \right)\\
&=&  \sum_{k=1}^m \left( \left( \Phi^{''}(P(x)) \frac{1}{d(x)^2 \ln(\sqrt{R})^2} - \frac{\Phi^{'}(P(x))}{d(x)^2 \ln(\sqrt{R})} \right) \left( X_k(d(x)) \right)^2  \right. \\
& + & \left.  \frac{\Phi^{'}(P(x))}{d(x) \ln(\sqrt{R})}   \left( X_k^2(d(x))  \right)  \right).
\end{eqnarray*}

Hence $$\text{supp}(\Delta_\mathbb{G}(\psi)) \subseteq \left\{ (t,x) : 0 \leq P \leq 1, 0 \leq t \leq T \right\}.\smallskip$$
For large $R$  we have $\ln(\sqrt{R}) \geq 0$.
Thus,
\begin{eqnarray*}
0 \leq P(x) \leq 1 &\iff&   0 \leq  \ln(\frac{d(x)}{\sqrt{R}}) \leq \ln(\sqrt{R})\\
& \iff&  \exp{(0)} \leq  \exp{\left(\ln\left(\frac{d(x)}{\sqrt{R}}\right)\right)} \leq \exp{(\ln(\sqrt{R}))}     \\
&   \iff & 1 \leq \frac{d(x)}{\sqrt{R}} \leq \sqrt{R} \iff \sqrt{R} \leq d(x) \leq R,
\end{eqnarray*}
implying that $$\text{supp}(\Delta_\mathbb{G}(\psi)) \subseteq \left\{ ( t,x) : \sqrt{R} \leq d(x) \leq R, 0 \leq t \leq T \right\}.$$
For $(t,x) \in \mathbb{R}^+ \times \mathbb{G}^{\ast}$, just as in the subcritical case, we have
 \begin{eqnarray*}
\left(\Delta_\mathbb{G}  (\Psi \circ d)\right)(\delta_{\sqrt{R}} (x))
&=&  \sum_{k=1}^m \left( \frac{\Phi^{'}(P(\delta_{\sqrt{R}} (x)))}{d(\delta_{\sqrt{R}} (x)) \ln(\sqrt{R})}   \left( X_k^2(d(\delta_{\sqrt{R}} (x)))  \right) \right. \\
& -& \left.   \frac{\Phi^{'}(P(\delta_{\sqrt{R}} (x)))}{d(\delta_{\sqrt{R}} (x))^2 \ln(\sqrt{R})} \left( X_k(d(\delta_{\sqrt{R}} (x))) \right)^2    \right. \\
& + & \left.  \frac{ \Phi^{''}(P(\delta_{\sqrt{R}} (x)))}{d(\delta_{\sqrt{R}} (x))^2 \ln(\sqrt{R})^2} \left( X_k(d(\delta_{\sqrt{R}} (x))) \right)^2    \right)\\
&=&  \sum_{k=1}^m \left( \frac{\Phi^{'}(P(\delta_{\sqrt{R}} (x)))}{\sqrt{R} d(x) \ln(\sqrt{R})}   \left( \frac{1}{\sqrt{R}} X_k^2(d(x))  \right) \right. \\
& -& \left.  \frac{\Phi^{'}(P(\delta_{\sqrt{R}} (x)))}{R d(x)^2 \ln(\sqrt{R})} \left( X_k(d(x)) \right)^2    \right. \\
& + & \left.  \frac{ \Phi^{''}(P(\delta_{\sqrt{R}} (x)))}{R d(x)^2 \ln{(\sqrt{R})}^2} \left( X_k(d(x)) \right)^2    \right).
\end{eqnarray*}
Considering the sphere $S$ as in \Cref{subcritical}, we obtain that $$|\Delta_\mathbb{G}  (\Psi \circ d)| \leq  \frac{C}{R \ln{(\sqrt{R})}}$$ and $$\int_{0}^T \int_{\mathbb{G}^{\ast}} f \psi \leq \frac{1}{p' (\frac{p}{2})^{p'-1}} \Big( \int_{0}^T \int_{\mathbb{G}^{\ast}} |\psi|^{\frac{-1}{p-1}}|\Delta_\mathbb{G} \psi|^{\frac{p}{p-1}} + |\psi|^{\frac{-1}{p-1}}|\psi_t|^{\frac{p}{p-1}} \Big).$$

With change of the variables $x \to x'$ and $t \to t'$ we obtain  \begin{eqnarray*}
\int_{0}^T \int_{\mathbb{G}^{\ast}} |\psi|^{\frac{-1}{p-1}}|\Delta_\mathbb{G} \psi|^{\frac{p}{p-1}} & \leq  & \int_{0}^T \int_{\mathbb{G}^{\ast}} |\Phi(P(x)) \Phi(t/T)|^{\frac{-1}{p-1}} \left(\frac{C}{\ln{(\sqrt{R})}  R }\right)^{\frac{p}{p-1}} \\
& \leq & \left(\frac{C}{\ln{(\sqrt{R})}  R }\right)^{\frac{p}{p-1}} \int_{0}^T \int_{\mathbb{G}^{\ast}} |\Phi(P(x)) \Phi(t/T)|^{\frac{-1}{p-1}} \\
&\leq & \left(\frac{C}{\ln{(\sqrt{R})}  R }\right)^{\frac{p}{p-1}} R^{\frac{Q}{2}} T \int_{0}^2 \int_{(\mathbb{G}^{\ast})} |\Phi(P(x')) \Phi(t')|^{\frac{-1}{p-1}} \\
&\leq & \left(\frac{C}{\ln{(\sqrt{R})}  R } \right)^{\frac{Q}{2}} R^{\frac{Q}{2}} 2 T \int_{(\mathbb{G}^{\ast})} |\Phi(P(x'))|^{\frac{-1}{p-1}} \\
&\leq & C_1 T  \ln{(\sqrt{R})^{- \frac{Q}{2}}}
\end{eqnarray*}
and similarly,
\begin{eqnarray*}
\int_{0}^T \int_{\mathbb{G}^{\ast}}  |\psi|^{\frac{-1}{p-1}}|\psi_t|^{\frac{p}{p-1}} &\leq&  \int_{0}^T \int_{\mathbb{G}^{\ast}}  |\Phi(P(x)) \Phi(t/T)|^{\frac{-1}{p-1}} \left|\frac{1}{T}  \Phi(P(x)) \Phi'(t/T) \right|^{\frac{p}{p-1}} \\
& \leq &  \left(\frac{1}{T}\right)^{\frac{p}{p-1}} \int_{0}^T \int_{\mathbb{G}^{\ast}}  |\Phi(P(x))|  |\Phi(t/T)|^{\frac{-1}{p-1}}|  \Phi'(t/T)|^{\frac{p}{p-1}}\\
& \leq & \left(\frac{1}{T}\right)^{\frac{Q}{2}} T R^{\frac{Q}{2}} \int_{0}^2 \int_{(\mathbb{G}^{\ast})}  |\Phi(P'(x))|  |\Phi(t')|^{\frac{-1}{p-1}}|  \Phi'(t')|^{\frac{p}{p-1}}\\
& \leq & C_1 T^{\frac{-Q}{2}} T R^{\frac{Q}{2}}.
\end{eqnarray*}

	Moreover, \begin{eqnarray*}
		\int_{0}^T \int_{\mathbb{G}^{\ast}} f \psi &=& \int_{0}^T \Phi(t/T) \int_{\mathbb{G}^{\ast}} f \Phi(P(x))  =  T \int_{0}^1 \Phi(t') \int_{\mathbb{G}^{\ast}} f \Phi(P(x)) \geq  C_2 T \int_{\mathbb{G}^{\ast}} f \Phi(P(x))
	\end{eqnarray*}
	implies $$\int_{\mathbb{G}^{\ast}} f \Phi(P(x)) \leq C' (  \ln{(\sqrt{R})^{- \frac{Q}{2}}} +  T^{\frac{-Q}{2}} R^{\frac{Q}{2}} ).$$
Putting $T^2 = R$ and then taking $R \to \infty$ we obtain $\int_{\mathbb{G}^{\ast}} f \leq 0$ which  contradicts the hypothesis that $\int_{\mathbb{G}^{\ast}} f > 0$.
So, no such $u$ exists and the statement is proved.
\end{proof}

One major difficulty that we encounter is that for a $\Delta_\mathbb{G}$-gauge $d$, the function $\nabla_\mathbb{G} d$ is identically one on $\mathbb{G}^{\ast}$ if and only if $\mathbb{G}$ is $(\mathbb{R}^n, +)$ \cite[p. 466]{Bonfiglioli}.
This is what makes it difficult to use the classical  technique of finding a function of the form $u(x) = \epsilon (1 + d(x))^{k}$ for suitable $k$ and $\epsilon$ such that $-\Delta u - u^p \geq 0$.
However, we now show that the ideas of \cite{Zhang} work on $\mathbb{G}$ when we take $d$ to be a $\Delta_\mathbb{G}$-gauge satisfying $d(x,y) \leq c (d(x) + d(y))$ for every $x,y \in \mathbb{G}$.
For this we study (\ref{MainProblem}) using an integral equation.
\begin{defin}(Mild solution)
A local in time mild solution of (\ref{MainProblem}) is a function $u \in C([0,T], C_0(\mathbb{G}))$ such that $$u(t) = e^{-t \Delta_\mathbb{G}} u(0) + \int_{0}^t e^{-(t-s) \Delta_\mathbb{G}} (|u(s)|^p + f) ds,$$ for any $ t \in [0,T)$.
The function $u$ is called a global mild solution of (\ref{MainProblem}) if $T = + \infty$.
\end{defin}

It is not difficult to see that the proof of \cite[Lemma 3.4]{Torebek} works on $\mathbb{G}$ and hence every mild solution is a weak solution as well.
To be able to do some analysis about existence of global positive mild solutions, we need a couple of indispensable lemmas.

\begin{lemma}\label{Proposition1OfZhang}
	For $\delta > 0$, there exists a constant $C$ depending only upon $\delta, Q$ and $w \geq 2$ such that
	$$\sup_{x \in \mathbb{G}} \int_\mathbb{G} \frac{1}{d(x,y)^{Q-2}} \frac{1}{1 + d(y)^{w+ \delta}} dy < C.$$
\end{lemma}
\begin{proof}
Since the Haar measure on $\mathbb{G}$ is invariant under translation, the proof is same as that of \cite[Proposition 2.1]{Zhang} except that we need to use \Cref{IntegrationOfRadialFunctions} to obtain
\begin{eqnarray*}
		\int_\mathbb{G} \frac{1}{d(y)^{Q-2}} \frac{1}{1 + d(y)^{w + \delta}} dy &=&  \lim_{s \to \infty} \int_{\{ y \in \mathbb{G}: d(y) \leq s \}} \frac{1}{d(y)^{Q-2}} \frac{1}{1 + d(y)^{w + \delta}} dy \\
		&  = &  C \lim_{s \to \infty} \int_0^s \frac{a^{Q-1}}{(1 + a^{w + \delta})a^{Q-2}} da.
	\end{eqnarray*}
	Since $w \geq 2$ and $\delta> 0$ this integral is finite, giving the result.
\end{proof}

\begin{lemma}\label{Proposition2OfZhang}
For $\delta > 0$ and every $x,y \in \mathbb{G}$ we have
	$$\int_\mathbb{G} \frac{1}{d(x,y)^{Q-2}} \frac{1}{1 + d(y)^{Q+ \delta}} dy \leq \frac{C}{1+ d(x)^{Q-2}}$$ for some constant $C$ which is independent of $x$.
\end{lemma}
\begin{proof}
The proof is refinement of the proof of \cite[Proposition 2.2]{Zhang} and we provide an elaborate reasoning for the sake of completeness.
When $d(x) \leq 1$, with $w = Q$ in \Cref{Proposition1OfZhang} we obtain $$ \int_\mathbb{G} \frac{1}{d(x,y)^{Q-2}} \frac{1}{1 + d(y)^{Q+ \delta}} dy \leq C = \frac{2C}{2} \leq \frac{2 C}{1+ d(x)^{Q-2}}.$$

Let us assume that $d(x) > 1$.
If $d(y) \leq \frac{d(x)}{2c}$ then the pseudo-triangle inequality  $d(x) \leq c(d(x,y) + d(y))$ tells us that  $\frac{d(x)}{2c} \leq d(x,y)$.
Moreover, if $d(y) \geq \frac{d(x)}{2c}$, then using the assumption that  $d(x) \geq 1$ we obtain
\begin{eqnarray*}
\frac{1}{1 + d(y)^{Q+ \delta}} &=& \frac{1}{1 + d(y)^{Q - 2} d(y)^{\delta + 2}} \leq  \frac{C}{1 + d(x)^{Q - 2} d(y)^{\delta + 2}} \\
&\leq&  \frac{2 C}{(1 + d(x)^{Q - 2}) (1 + d(y)^{2 + \delta}) }.
\end{eqnarray*}
With $w = 2$ in \Cref{Proposition1OfZhang}, we have
\begin{eqnarray*}
\int_\mathbb{G} \frac{1}{d(x,y)^{Q-2}} \frac{1}{1 + d(y)^{Q+ \delta}} dy&=& \int_{d(y) \geq \frac{d(x)}{2c}} \frac{1}{d(x,y)^{Q-2}} \frac{1}{1 + d(y)^{Q+ \delta}} dy \\
&+& \int_{d(y) \leq \frac{d(x)}{2c}} \frac{1}{d(x,y)^{Q-2}} \frac{1}{1 + d(y)^{Q+ \delta}} dy \\
&\leq& \int_{d(y) \geq \frac{d(x)}{2c}} \frac{C}{(1 + d(x)^{Q - 2}) (1 + d(y)^{2 + \delta})} \frac{1}{d(x,y)^{Q-2}} dy \\
&+& \frac{C}{d(x)^{Q-2}} \quad \text{(From \Cref{IntegrationOfRadialFunctions})} \\
&\leq& \frac{C}{1 + d(x)^{Q-2}}  \int_{d(y) \geq \frac{d(x)}{2c}} \frac{1}{d(x,y)^{Q-2}} \frac{1}{1 + d(y)^{2+ \delta}} dy \\
&+& \frac{C}{d(x)^{Q-2}} \\
		&\leq& \frac{C}{1 + d(x)^{Q-2}}  + \frac{C}{d(x)^{Q-2}} \leq \frac{C}{d(x)^{Q-2}} \\
		& \leq &  \frac{2 C}{1 + d(x)^{Q-2}} \ \ (\text{as } d(x) > 1).
	\end{eqnarray*}
	This completes the proof.
\end{proof}

\begin{theorem}\label{Supercritical}
	For $p > \frac{Q}{Q-2}$, there exists $f$ and $u_0$ such that a positive global mild solution of (\ref{MainProblem}) exists.
\end{theorem}
\begin{proof}
We intend to use the Banach fixed point theorem.
	For $M >0$, let
	$$ S_M = \left\{ u \in C([0, \infty), C_0(\mathbb{G}) ) : 0 \leq u(t)(x) \leq \frac{M}{1 + d(x)^{Q-2}} \text{ for every } t \in [0, \infty) \right\}.$$
Then $S_M$ is a complete metric space and we will be able to apply the Banach fixed point theorem to any contraction from $S_M$ to itself.
Define $$T_M : S_M \to S_M$$ by $$T_M(u)(t) = e^{-t \Delta_\mathbb{G}} u(0) + \int_{0}^t e^{-(t-s) \Delta_\mathbb{G}} (u(s)^p + f) ds.$$
The result will  follow if we can prove that $T_M$ is a well-defined contraction for some $M$.

As $p > \frac{Q}{Q-2}$, there exists $\delta > 0$ such that $\frac{1}{(1 + d(x)^{Q-2})^p} \leq \frac{C}{(1 + d(x)^{Q+ \delta})}$ for every $x \in \mathbb{G}$.
For $\epsilon > 0$,  fix $$u(0)(x) = f(x) = \frac{\epsilon}{1+ d(x)^{Q + \delta}}.$$
To prove that $T_M$ is well-defined, we show that there exists $C > 0$ for which $$e^{-t \Delta_\mathbb{G}} u(0)(x) \leq \frac{C}{1 + d(x)^{Q-2}}$$ and $$\left (\int_{0}^t e^{-(t-s) \Delta_\mathbb{G}} (u(s)^p + f) ds \right )(x) \leq \frac{C}{1 + d(x)^{Q-2}}.$$
Using the pseudo-triangle inequality $d(x) \leq c(d(y) + d(x,y))$ for every $x,y \in \mathbb{G}$, we obtain
\begin{eqnarray*}
e^{-t \Delta_\mathbb{G}} u(0)(x) &=& \int_\mathbb{G} h_t(x,y) u(0)(y) dy = \epsilon \int_\mathbb{G}  \frac{h_t(x,y)}{1+ d(y)^{Q + \delta}} dy \\
&\leq &  \frac{\epsilon C}{t^{\frac{Q}{2}}}  \int_\mathbb{G}  \frac{1}{1+ d(y)^{Q + \delta}} \exp^{\frac{-d(x,y)^2}{C t}} dy\\
&=& \frac{\epsilon C}{t^{\frac{Q}{2}}}  \left[  \int_{d(y) > \frac{d(x)}{2c}}  \frac{1}{1+ d(y)^{Q + \delta}} \exp^{\frac{-d(x,y)^2}{C t}} dy \right. \\
&+& \left. \int_{d(y) \leq \frac{d(x)}{2c}}  \frac{1}{1+ d(y)^{Q + \delta}} \exp^{\frac{-d(x,y)^2}{C t}} dy \right] \\
&\leq& \frac{\epsilon C}{t^{\frac{Q}{2}}}  \left[  \frac{1}{1+ (\frac{d(x)}{2c})^{Q + \delta}}  \int_{d(y) > \frac{d(x)}{2c}}  \exp^{\frac{-d(x,y)^2}{C t}} dy \right. \\
&+& \left. \int_{d(y) \leq \frac{d(x)}{2c}}  \frac{1}{1+ d(y)^{Q + \delta}} \exp^{\frac{-d(x,y)^2}{C t}} dy \right].
\end{eqnarray*}
Since $\frac{d(x,y)^2}{C t} = d\left(\delta_{\frac{1}{\sqrt{Ct}}}x,\delta_{\frac{1}{\sqrt{Ct}}}y\right)^2$, with change of variable $x' \leftrightarrow \delta_{\frac{1}{\sqrt{Ct}}}x$ we obtain $$\int_\mathbb{G} \exp^{\frac{-d(x,y)^2}{C t}} dy =  \int_\mathbb{G} \sqrt{Ct}^{Q}\exp^{-d(x',y')^2} dy' = C t^{\frac{Q}{2}} \int_\mathbb{G} \exp^{-d(x',y')^2} dy' \leq C t^{\frac{Q}{2}}$$ where $C$ is independent of $x$ and the integral $\int_\mathbb{G} \exp^{-d(x',y')^2} dy'$ is finite because (\ref{IntegrationOfRadialFunctions}) applies.

We separately deal with the two possible cases.
Let us first assume that $d(x) > 1$.
Using the fact that $d(x) \leq c( d(y) + d(x,y))$ one obtains
	\begin{eqnarray*}
e^{-t \Delta_\mathbb{G}} u(0)(x) &\leq& \frac{\epsilon C}{t^{\frac{Q}{2}}}  \left[  \frac{1}{1+ (\frac{d(x)}{2c})^{Q + \delta}}  \int_{d(y) > \frac{d(x)}{2c}}  \exp^{\frac{-d(x,y)^2}{C t}} dy \right. \\
&+& \left. \int_{d(y) \leq \frac{d(x)}{2c}}  \frac{1}{1+ d(y)^{Q + \delta}} \exp^{\frac{-d(x,y)^2}{C t}} dy \right] \\
& \leq & \frac{\epsilon C}{t^{\frac{Q}{2}}}  \left[  \frac{t^{\frac{Q}{2}}}{1+ (\frac{d(x)}{2c})^{Q + \delta}} + \int_{d(y) \leq \frac{d(x)}{2c}}  \frac{1}{1+ d(y)^{Q + \delta}} \exp^{\frac{-d(x,y)^2}{C t}} dy \right] \\
& = & \frac{\epsilon C}{t^{\frac{Q}{2}}}  \left[  \frac{t^{\frac{Q}{2}}}{1+ (\frac{d(x)}{2})^{Q + \delta}} \right. \\
&+& \left. \int_{d(y) \leq \frac{d(x)}{2c}}  \frac{d(x,y)^{Q}}{(1+ d(y)^{Q + \delta}) d(x,y)^{Q}} \exp^{\frac{-d(x,y)^2}{C t}} dy \right] \\
& \leq & \frac{\epsilon C}{t^{\frac{Q}{2}}}  \left[  \frac{t^{\frac{Q}{2}}}{1+ (\frac{d(x)}{2})^Q} \right. \\
&+& \left. \frac{(2c)^Q}{d(x)^Q} \int_{d(y) \leq \frac{d(x)}{2}}  \frac{d(x,y)^{Q}}{(1+ d(y)^{Q + \delta})} \exp^{\frac{-d(x,y)^2}{2 C t}} \exp^{\frac{-d(x,y)^2}{2 C t}} dy \right] \\
& \leq & \frac{\epsilon C}{1+ d(x)^Q} + \frac{\epsilon C}{ d(x)^Q} \leq \frac{\epsilon  2 C}{1+ d(x)^{Q-2}} + \frac{\epsilon C}{ d(x)^Q}\\
&\leq&   \frac{\epsilon  2 C}{1+ d(x)^{Q-2}},
\end{eqnarray*}
	where the third last inequality comes from H\"older's inequality since $\int_\mathbb{G} \exp^{\frac{-d(x,y)^2}{2 C t}} dy \leq C t^{\frac{Q}{2}}$ and $\frac{d(x,y)^{Q}}{(1+ d(y)^{Q + \delta})} \exp^{\frac{-d(x,y)^2}{2 C t}} \leq C$, and the last inequality is true whenever $d(x) > 1$.

To attend to the other case, let $$S = \left\{ x \in \mathbb{G}: d(x) \leq 1 \right\}.$$
Then $S$ is a compact and hence the continuous function $x  \to \int_{d(y) \leq \frac{d(x)}{2}}  \frac{1}{1+ d(y)^{Q + \delta}} \exp^{\frac{-d(x,y)^2}{C t}} dy$ is bounded by $t^{\frac{Q}{2}} C$ on $S$.
In this case,
\begin{eqnarray*}
e^{-t \Delta_\mathbb{G}} u(0)(x) &\leq& \frac{\epsilon C}{t^{\frac{Q}{2}}}  \left[  \frac{1}{1+ (\frac{d(x)}{2c})^{Q + \delta}}  \int_{d(y) > \frac{d(x)}{2c}}  \exp^{\frac{-d(x,y)^2}{C t}} dy \right. \\
&+& \left. \int_{d(y) \leq \frac{d(x)}{2c}}  \frac{1}{1+ d(y)^{Q + \delta}} \exp^{\frac{-d(x,y)^2}{C t}} dy \right] \\
& \leq & \frac{\epsilon C}{t^{\frac{Q}{2}}}  \left[  \frac{t^{\frac{Q}{2}}}{1+ (\frac{d(x)}{2c})^{Q + \delta}} + t^{\frac{Q}{2}} C \right] \\
& \leq & \frac{\epsilon C}{t^{\frac{Q}{2}}}  \left[  \frac{t^{\frac{Q}{2}}}{1+ (\frac{d(x)}{2c})^{Q}} + \frac{2 t^{\frac{Q}{2}} C}{1+ d(x)^Q} \right] (\text{as } d(x) \leq 1) \\
&\leq& \frac{\epsilon C}{1+ d(x)^Q} \leq   \frac{\epsilon  2 C}{1+ d(x)^{Q-2}}.
\end{eqnarray*}
Hence, $e^{-t \Delta_\mathbb{G}} u(0)(x) \leq \frac{C}{1 + d(x)^{Q-2}}$ for every $x \in \mathbb{G}$ and $t \in [0, \infty)$.

Now that we have the sought after bound for $e^{-t \Delta_\mathbb{G}} u(0)$, let us work with $\left (\int_{0}^t e^{-(t-s) \Delta_\mathbb{G}} (u(s)^p + f) ds \right )(x)$.
From the choice of $\delta$, we obtain
\begin{eqnarray*}
 \left (\int_{0}^t e^{-(t-s) \Delta_\mathbb{G}} (u(s)^p + f) ds \right )(x)  &=& \int_0^t e^{-(t-s) \Delta_\mathbb{G}} (u(s)^p(x) + f(x)) ds \\
&=& \int_0^t \int_\mathbb{G} h_{(t-s)}(x,y) (u(s)^p(y) + f(y)) dy ds \\
&\leq&  \int_{0}^t \int_\mathbb{G} h_{(t-s)}(x,y) \left(\frac{M^p}{(1 + d(y)^{Q-2})^p} + \frac{\epsilon}{1+ d(y)^{Q + \delta}}\right) dy ds\\
&\leq & \int_{0}^t \int_\mathbb{G} h_{(t-s)}(x,y) \left(\frac{C M^p}{1 + d(y)^{Q+ \delta}} + \frac{\epsilon}{1+ d(y)^{Q + \delta}}\right) dy ds\\
&=& \int_{0}^t \int_\mathbb{G}  h_{(t-s)}(x,y)\frac{C M^p + \epsilon}{1 + d(y)^{Q+ \delta}} dy ds.
\end{eqnarray*}

If $H$ denotes the fundamental solution for $\Delta_\mathbb{G}$, then $\Delta_\mathbb{G} (H) = 0$ in $\mathbb{G}^{\ast}$.
It follows from the definition of the heat kernel $h_t(x,y)$ that for every fixed $y \in \mathbb{G}$, $\frac{\partial h_t(x,y)}{\partial t} - \Delta_\mathbb{G} h_t(x,y) = 0$.
Integrating this equation over time and using (\ref{BoundsForHeatKernel})  we obtain $$0=\lim_{s \to \infty} \int_0^s \left(\frac{\partial h_t(x,y)}{\partial t} - \Delta_\mathbb{G} h_t(x,y) \right) dt =  - \lim_{s \to \infty}  \Delta_\mathbb{G} \int_0^s  h_t(x,y) dt.$$
Non-negativity of the heat kernel now gives  $$\int_0^s h_{s-t}(x,y) dt \leq \lim_{s \to \infty} \int_0^s h_{s-t}(x,y) dt = H(x,y).$$
Since $H(x,y) = \frac{\beta_d}{d(x,y)^{Q-2}}$ for some $\beta_d > 0$, using \Cref{Proposition2OfZhang} we obtain
\begin{eqnarray*}
\left (\int_{0}^t e^{-(t-s) \Delta_\mathbb{G}} (u(s)^p + f) ds \right )(x)  &\leq& \int_{0}^t \int_\mathbb{G}  h_{(t-s)}(x,y)\frac{C M^p + \epsilon}{1 + d(y)^{Q+ \delta}} dy ds \\
& \leq & \int_\mathbb{G}  H(x,y)\frac{C M^p + \epsilon}{1 + d(y)^{Q+ \delta}} dy\\
& \leq & \int_\mathbb{G} \frac{\beta_d}{d(x,y)^{Q-2}} \frac{C M^p + \epsilon}{1 + d(y)^{Q+ \delta}} dy\\
&\leq&  \frac{C (C M^p + \epsilon)(\beta_d)}{1+ d(x)^{Q-2}}.
\end{eqnarray*}
Thus, there exists small $M$ and $\epsilon$ such that  $T_M$ is well-defined on $S_M$.

To prove that $T_M$ is a contraction, note that $(Q-2)(p-1) - 2 = \delta > 0$  and hence \begin{eqnarray*}
\norm{ T_M(u_1) - T_M(u_2) }_{\sup} &=& \sup_{t \in [0, \infty)} \left\| \int_{0}^t \exp^{-(t-s)\Delta_\mathbb{G}}( u_1(s)^p - u_2(s)^p ) \right\|\\
& \leq & p \norm{u_1 - u_2}_{\sup} \sup_{t \in [0, \infty)}  \int_\mathbb{G} \int_{0}^t h_{(t-s)}(x,y) \left(\frac{M}{1 + d(y)^{Q-2}}\right)^{p-1} dy \\
& \leq & p M^{p-1} \norm{u_1 - u_2}_{\sup}  \sup_{x \in \mathbb{G}}  \int_\mathbb{G} H(x,y) \frac{1}{1 + d(y)^{2 + \delta}} dy \\
&\leq & C p M^{p-1} \norm{u_1 - u_2}_{\sup}
\end{eqnarray*}
where $\int_\mathbb{G} H(x,y) \frac{1}{1 + d(x,0)^{2 + \delta}} dy < \infty$ can be seen by using $w = 2$ in \Cref{Proposition1OfZhang}.
This completes the proof.
\end{proof}

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ES Chronicles Home > Jenny Nakai
Jenny Nakai, Geophysicist
Rio Grande Rift Project - 5:25
Jenny describes her research, which involves using data from the EarthScope Transportable Array to study seismicity in the Rio Grande Rift region.
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The Rio Grande Rift appears to slowly be spreading apart, so Jenny is looking at whether there are earthquakes associated with this extension and whether these earthquakes are broadly distributed from east to west or are concentrated along normal fault zones. Her work includes poring through extensive seismic data to find events that might be actual earthquakes and to distinguish them from false detections caused by external events such as mine blasts. The goal is to build an earthquake catalog that identifies when, where, and at what depth earthquakes happened in this region.
Jenny's work has been supported by the National Science Foundation's Graduate Research Fellowship Program and by two EarthScope projects: Crustal Deformation Measurements and a Multidisciplinary Geophysical Investigation of the Rio Grande Rift (EAR-0454541) and Rio Grande Rift II: Kinematics and Dynamics of Continental Deformation in Low Strain-Rate Environments (EAR-1053596). Principal investigators for the two EarthScope projects are Anne Sheehan, Robert Nerem, and Anthony Lowry.
Meet Jenny Nakai - 4:41
Jenny talks about her interest in engineering and science as a means to be useful and solve problems and the importance of education in the Navajo culture in general. She explains how resource extraction such as coal mining has had a negative environmental impact on the Navajo land and how, in the past, few tribal members have had the knowledge and training to deal with these problems or to educate the public about ways to solve them.
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Vicky the hygienist is awesome!
Always great service and they listen to my concerns and perform only what I need not to just make sell service.
A good experience. Staff knowledgeable. Friendly. Clean office.
Polite, professional on time appointments.


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A look at the guest list from Dwyane Wade and Gabrielle Union's wedding
Inquiring Heat Nation minds want to know...who made it to the Summer of Love celebration.
By Eileen Navarrete Sep 1, 2014, 4:08pm EDT
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Dwyane Wade Instagram
By now, more details have emerged from the Dwyane Wade and Gabrielle Union 1930's themed wedding.
Fun facts that we have learned include the hiring of hip-hop violinist Lee England Jr. to provide music, Gabrielle Union's BFF Essence Atkins officiating the ceremony, and guests being bused in from Fontainebleau Miami Beach to the private Chateau Artisan located in rural Redlands near Homestead (a nearly 45-minute drive from South Beach). Also, John Legend provided live music for the newlyweds as well as the guests.
A fun detail the couple added to their party was the use of themed bars that included the Battier Bar (Shane Battier) with a moonshine bar, Bar Melo (Carmelo Anthony), Bosh's Brewery (Chris Bosh) and Ray's Retirement Home (Ray Allen). There was also a "Cleveland, this Way!" bar, but unfortunately no details were shared on what this bar included.
There was no shortage of food with plenty of late night snacks like waffles, cupcakes and cookies. The Wade Union is now official and whether we receive more details or not is not important. The love between Wade and Union is reason enough to celebrate.
Take a look at the guests who attended:
Headed to #thewadeunion pic.twitter.com/qDeiNZMtwV
— Dorell Wright (@DWRIGHTWAY1) August 30, 2014


"Abolish Profit" is an actual platform point being proposed by a major socialist organization…is it a good idea? Does it make sense?
Paul Mladjenovic, the author of books such as Stock Investing For Dummies and the instructor of courses such as Home Business Goldmine will analyze this point in detail.
Fantastic presentation that even a liberal dem could understand.


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Roy Hodgson
Roy Hodgson does not want another Fulham last game moment with Crystal Palace
And he's not predicting how many points will be required to survive
Lee Wilmot
Roy Hodgson was in no doubt when asked if he would take being in with a shot of survival on the last day of the season - no.
Crystal Palace are currently in the bottom three, but things are so tight at the bottom that during Monday night's defeat to Manchester United, when they were 2-0 up, the Eagles were in 13th place.
And Hodgson does not want survival to go down to the final day, he is hoping Palace are out of trouble before then.
"I don't like to think about the last game against West Brom," he said. "I've had one of those games in my career, only one and I don't really want another one, I'd be happier if we could extricate ourselves from the bottom three before that."
Roy Hodgson's press conference recap: Crystal Palace boss discusses injuries, Chelsea and the relegation battle
And Hodgson is not predicting how many points it will take for a team to get safe either.
He added: "I don't know. I don't guess, that's the point I've been trying to make for a long time, I don't second guess, I don't read into things, I don't try and work out what [Antonio] Conte or Chelsea or Man Utd are thinking, I just get on with the day-to-day work, it must sound pretty boring but that's what being a football coach is.
A good time to play Chelsea? Crystal Palace manager denies suggestions, saying it could be the worst time to face the champions
"There will be plenty of people out there who would have decided the number of points necessary, I'm just waiting to see how many that will be.
"When I was at Fulham we survived on 36, my friend Trevor Brooking went down on 42 with West Ham, so I would say somewhere between 36 and 42 would be a decent answer."


In 2017-18 Emmanuel Denguessi will enter his third season with the Capilano Blues with a promising young roster. Denguessi previously played for the Blues in 2007-08, before moving on to play for Columbia Bible College. With CBC he helped lead the Bearcats to a CCAA National Championship and PACWEST silver medal 2010-11. During his PACWEST career he was awarded CCAA and PACWEST First and Second Team All-Star honours, voted BCCAA Rookie of the Year while with Capilano in 2007-08, and named Team MVP at both CBC and Capilano. Denguessi was also named a CCAA Academic All-Canadian and CCAA National Scholar.
Previously, Denguessi has travelled, competed and coached all over the world for over 14 years as part of the Cameroonian national team. He received a Bachelor in Counselling and Diploma in Theological Studies from Columbia Bible College, and he has a Certificate in Life Coaching.


I'm originally from Minnesota. In the first verse, he talks about the snow and a famous club called First Ave., where I've been and can picture so clearly in my mind. So that makes me feel extremely nostalgic right off the bat.
And then there's the fact that I know I'm currently living the 'good old days' right now.
I just married my best friend. I get to ride a horse that I have built a great partnership with. I've reached a point in my life where I'm pretty comfortable with who I am and what I'm doing. I feel as though I've truly found a riding 'home' – in the barn I'm at, the friends I've made, and the sport I'm doing.
I feel like this is a golden time in my life that I will always remember. But thinking that way makes me sad, because it reminds me that everything has an end. That's not a bad thing – the good things come to an end just as the bad things do.
I said all of this to Trainer B as we drove down the highway to the show. Her response was so lovely. She said, in her experience, that every phase of life becomes the good old days – even the hard ones. There's beauty and joy and precious moments in every time of life that we will look back on fondly and wish to return to.
I don't want to miss the magic of these days just because I'm already sad knowing that they will eventually come to an end. I want to feel gratitude for these magical days.
I feel grateful for the fact that Duke now nickers to me every time I approach his stall.
I feel grateful for every cross country school.
I feel grateful for the moments when we achieve the feeling of suppleness in Dressage.
I feel grateful for the days spent building jumps with friends.
I feel grateful for every braid I put in Duke's mane.
I feel grateful for the smell of fresh grass and the sound of a horse grazing.
I feel grateful for the special conversations that only seem to happen when hauling horses down the road.
I feel grateful for the knowledge I am gaining about this sport and about horsemanship in general.
I feel grateful for hay breath and a soft nose.
I think the magic actually comes from the fact that I'm feeling gratitude in combination with the knowledge that nothing stays the same forever. I like to call it pre-nostalgia. Or maybe it's nostalgia-adjacent. Whatever it is, I'm just grateful to be living it.
Love. What you said, and what B said.
You are right nothing can stay the same but it is so good to be grateful and enjoy every minute of the present. There may be change ahead but it will not necessarily be a change for the worst. Even if it is there will be sunshine again after a tough patch and you will be in another time of loving your life and all the wonder in it. Thanks for this great post.


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UCF to Face USF with the AAC Title at Stake
ORLANDO, Fla. (UCFKnights.com) – The No. 28 UCF men's tennis team will face the No. 46 USF Bulls in the American Athletic conference championship tomorrow at 1 p.m. This will be the second time the two teams have met in the last eight days.
The Knights will be seeking their first conference championship in the American Athletic Conference in it's sixth year of existence. The last time the Knights won a conference championship was in 2005 when they were a part of the A-Sun. UCF has three conference championships to their name, all three coming in a row from 2003-2005.
On Sunday, there are a few story lines to follow. One of them being UCF's top doubles tandem consisting of Bogdan Pavel and Trey Hilderbrand, who are ranked No. 16 in the nation. They have won eight straight matches, nine of their last ten and have defeated seven out of eight ranked opponents they've faced. Additionally, Mikhail Sokolovskiy has been on a tear of his own, winning eight out of his last ten matches and coming out victorious in his last five. He is 28-9 on the 2018-2019 season, good enough for a tie of third place on the single season wins list for UCF. He has picked up two singles victories in this tournament and looks to add one more in the championship match.
As a whole, the Knights will be looking to win their ninth straight doubles point and earn their sixth consecutive shutout. UCF has won 23 consecutive points, which we're pretty sure is a record. The Knights swept five straight opponents earlier in the season, all of them by a score of 4-0. Of their fifteen wins, the Knights have earned sweeps in 11 of them.
USF has won four championships in the American, winning it the first four seasons of the conference's existence. Tulane won it last year to snap USF's streak. The Knights will look to become the third team to win the American Athletic Conference Championship.


For anyone starting on the rewarding journey into computer networking, Packet Tracer is an indispensable learning tool.
If you're planning to take the Cisco Certified Network Associate (CCNA) ) exam, you must practice on the Packet Tracer networking simulation software from Cisco.
On Cisco's Networking Academy web site, the latest 64-bit version of Packet Tracer is available for both Linux and Windows for free.
So I headed to Cisco's Packet Tracer Download web site and downloaded the Packet Tracer 7.2.1 tarball for Linux.
Before you can download Packet Tracer, you must register with the site.
For the purpose of this post, I downloaded Packet Tracer 7.2 to a 10-year-old Dell 780 desktop PC running Linux Mint 18.3.
I extracted Packet Tracer 7.2.1 into the same folder in which I had downloaded the file.
Now that you have downloaded the Packet Tracer software, fire up your terminal and extract the software (see command below).
Now that we completed the extraction, let's take a quick look at the folder.
So let's move to the next step: Running the install script.
I did not encounter any major issues in installing Packet Tracer except for a minor bump when the process stopped momentarily and asked for my sudo password to copy the files into opt/pt.
ALL OR USE THE SOFTWARE.
I.	DISTRIBUTION RIGHTS.  The terms in this Section I (Distribution Rights) of the Agreement apply solely to any Cisco Networking Academy instructor, administrator or other person or entity approved in writing by Cisco Systems, Inc. (each, an "Instructor") to distribute and use the Cisco Packet Tracer software and related documentation (collectively, the "Software") in accordance with the terms and conditions of this Agreement.
Cisco, Cisco Systems, and the Cisco Systems logo are registered trademarks of Cisco Systems, Inc. in the U.S. and certain other countries.  Any other trademarks mentioned in this document are the property of their respective owners.
Voila, Packet Tracer 7.2 is now installed on my Linux Mint PC.
After rebooting the system, I was able to launch Packet Tracer and get it to work without any issues.
But if you're like me and want to do more with less, you can set up an alias by opening the .bashrc file with nano or vim and inserting alias pt="packettracer" at the end of the file, save and close it. Don't forget to run source ~/.bashrc so that you don't have to close the terminal for the new alias to work. The next time you want to launch Packet Tracer, just type pt instead of the longer packettracer on the terminal.
Overall, I did not encounter any showstoppers in the process of downloading, installing or using Packet Tracer 7.2.1 on my Linux PC.
As you can see from the Packet Tracer image at the top of this post, I even moved a PC and switch into the logical work area and hooked them up with a console/rollover cable.
Getting Packet Tracer 7.2 on my Linux desktop PC took about five-minutes.
Since Linux Mint is based on Ubuntu, Packet Tracer should work on Ubuntu too.
One of these days, I'll test Packet Tracer on my CentOS 7 desktop and see how the process goes.


The internet is buzzing about "Running Man" star Lee Kwang Soo who has recently confirmed he's been dating actress Lee Sun Bin for the past five months.
An exclusive report by TV Report, a Korean news outlet, reported that the couple has been dating for five months. Shortly after their report was published, both actors' agencies confirmed the news.
According to TV Report and Lee Kwang Soo's own agency, the actor wasn't shy in expressing his feelings for his girlfriend. He usually presents her as his lover in gatherings with friends and family.
Lee Kwang Soo is a huge star in Asia; he's known for his awkward comedy in "Running Man" and has made a name for himself as an actor appearing in various productions in through the years.
But who is Lee Sun Bin? and why does she seems so familiar?
Today, I'd like to introduce you to the rising star Lee Sun Bin!
Lee Sun Bin is a 24-year-old rising actress but she actually got her start as an idol, she had debuted back in 2011 as part of a Kpop group called JQT, she joined them in 2011 after a member departed the group. She was a vocalist in that group.
However, shortly after and in 2012, the group's contract with their label at the time expired and the girls didn't renew, thus they disbanded.
She left the agency and joined another one; she joined as an acting trainee and began promoting again in 2014 snagging CF roles.
In 2016, her agency revealed to news outlets that she will be debuting in a new girl group in 2017, but those plans fell through.
Although she started out as an idol, many people don't know she was actually a part of a Kpop group which is because of her rising popularity as an actress.
She snagged her first role in the 2014 Chinese drama "Saint Wang Xizhi" and got her breakthrough role in OCN's hit drama "38 Squad" alongside Seo In Guk and Ma Dong Suk.
In 2018, Lee Sun Bin hit the jackpot, she was finally able to get her first EVER leading role in JTBC's "Sketch" opposite Singer-actor Rain and Lee Dong Gun.
She was also cast to play the leading role in the upcoming short drama "Has Already Begun" which will air on January 17, 2019.
Who is lucky enough to go on TV and say a celebrity is their ideal style only to end up dating him in real life?
Lee Sun Bin is an example of an extremely lucky person!
Lee Sun Bin had previously stated that Lee Kwang Soo was her ideal type, not once, but twice.
Also, on an episode of "Running Man" she appeared as a guest and even took a liar detector test to prove she was interested in Lee Kwang Soo and that he was for real her ideal type.
She was asked the question by Yoo Jae Suk and put her hand on the infamous liar detector test, the test results were, you guessed it, positive!
Lee Sun Bin has appeared in many music videos across the years, most notably in YG Entertainment group iKON. She appeared in the music video of their track "Apology," where she played one of the members' girlfriend, she can be seen crying and shouting in the MV. It was a song about a breakup.


Varsity wins AVCTL-Division 2 – undefeated league champions at 6-0-0. Finishes with best record in school history at 16 wins, 2 losses, 0 ties.
Kwame Parry: 1st Team All AVCTL-Division 2 Forward, AVCTL-Division 2 League MVP (unanimous choice), All-Metro Area Team.
Tyson Rey: 1st Team All AVCTL-Division 2 Defender.
Ian Young: 1st Team All AVCTL-Division 2 Midfielder.
Ethan Doud: 2nd Team All AVCTL-Division 2 Midfielder.
Christian Raffinengo: 2nd Team All AVCTL-Division 2 Forward.
Fernando Vidrio: 2nd Team All AVCTL-Division 2 Forward.
Cory Gleason: Honorable Mention All AVCTL-Division 2 Midfielder.
Trey Greening: Honorable Mention All AVCTL-Division 2 Goalkeeper.


Whether or not it's raining today where you are, I hope you're finding lots of bright, happy little things to make your world more colorful!
Unfortunately, a few days before that, some jerk dumped a dog in the parking lot and zoomed off in a pickup truck (with kids in there screaming and crying, ugh) before anyone could get the license plate number. Abandoned pets often don't survive for long because of not knowing how to find food. We called Animal Control, who came pretty quickly, but the dog already had run off and they couldn't locate the poor beast.
Most people would never be so cruel to either their dog or their children, of course; but I believe that we all could be kinder. Although animals and young children often don't understand the details of what's going on around them, usually they can tell whether or not they are being treated kindly; and those small acts of kindness can work much greater changes than we know. Just as a few more warm days made a big difference in how many young herons lived through the winter, a few kind acts can much improve the life of a pet or a child.
I'm posting this Nurturing Thursday entry on Friday because I felt somewhat distracted this week and didn't get around to writing a post earlier, but it's all okay. The work presentation that I mentioned in last week's post went very smoothly; there was no need to worry.
Often that's the way it goes—when we're doing things that we haven't tried before and developing our skills, it seems scary at the time. Then we find that life has settled back down into a pattern that comfortably includes something more than we used to know. That happened all the time back when we were kids, and it can be just as true when we're adults exploring new challenges!
Next week I'm scheduled to present a continuous improvement project, which I mentioned in a post in April, to my boss's boss. It is just a small beginner-level project, there is a standard template for presenting the results, and it should only take about a half-hour. I thought it was helpful to learn the basic concepts of continuous improvement and to gain more experience with programs such as Excel and PowerPoint, which I don't regularly use in my work. Still, there's always a bit of stress involved when doing anything new.
I did a practice presentation today with some of my coworkers, and they were positive and encouraging. When I asked whether I sounded nervous, they said no, I was fine. It's always good to have a bit of gentle encouragement to expand those comfort zones!
Along with my husband and son, I went to the rowing club to watch the city's Independence Day fireworks show. The boathouse is just up the river and has a great view. Each year the club has a picnic before we set up folding chairs on the dock to watch the fireworks. Everyone takes turns cooking hamburgers and other food on the grill, and we all bring snacks and sweet treats for our friends to share. There are old-fashioned games such as water balloon toss and three-legged race. For prizes, the winners get chocolate bars.
Although we didn't have any rowing races or other events on the water this year because of nearby bridge construction, everyone had a good time anyway. I'm wishing my readers plenty of fun with friends too!


If Terrorism has no Religion what is the logic behind ramzan ceasefire ?
Radical elements among Rohingya Muslims are a serious threat to national security says central govt in Apex Court.
Is India ready to handle a Nepal-like situation?


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Q: How can I get GADM level 0 (countries) shapefiles? When downloading GADM data from their website (https://gadm.org), It is highly suggested to download it country by country. I would need the whole world at once with only level 0 geometries.
The global dataset provided (https://gadm.org/data.html) does not include all the level 0 geometries but only the smallest subsets of each country.
Is there a simple way to get the level 0 data for all countries?

redpajama_set_name: RedPajamaStackExchange


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excerpt from article originally posted in : FEROCIOUS URBANITIES
By Foteini Vergidou.
What people often find most fascinating in generative art is not the replacement of physical tools, like brushes, paint and clay, by computer applications, but the fact that the computer works in such way that it seems to directly create art on its own.
The artist behind the work is the architect who has conceived, designed and created the simulation that allows the computer to act this way.
Still, the main difference from all other artistic tools throughout art history is the significant independence of the machine in the art-making process.
In his latest series of generative images and audio-reactive visuals, Constantine Nisidis works with minimal and abstract forms and structures, while showcasing a great deal of code mastery. The Greek artist is well known in the vvvv community for his live coding performances and interactive works that destroy and reconstruct themselves randomly. Both sound and image interact in real time and space, continuously alternating roles between static and motion, towards an unknown direction.
Nisidis' focal point of practice is on creative coding, preliminary design, interactive media and projection mapping, while his study in augmented reality (AR), virtual reality (VR) and mixed reality (MR) have influenced his body of work heavily.
His solo show in Paris in 2010 WWW+DOWNLOADJESUSNOW+COM, showcased the possibilities of a computer to randomly produce a great variety of visuals through structures and tools independently, while many shows and collaborations have followed in the recent years. In 2015, his collaboration with international artist Hiwa K. titled "Blind as the Mother Tongue" was awarded the Cinema award in Vienna Contemporary.
Back to generative art, the nature of this movement defines our approach to understand it. The viewer has to analyse the work into its two dimensions. The first level of the work should be studied for its development, how the artist is matching the algorithmic and the unpredictable elements into one coherent generative system. While this demonstrates technique and delicacy, the other level of the work lies on its conceptual performance and its way to allow the viewer to engage with its procedural aesthetics and its symbolic representation of the world.
Generative art uses code to explore the cult of contemporary neo beauty by creating avatars, acclaimed worlds or just abstract forms, in the same way, the Symbolist movement of the 19th century used metaphorical forms and indirect methods to represent reality and talk about the Ideal against the norm.
Stéfane Mallarmé's manifesto to "depict not the thing, but the effect it produces" applies in the practice of generative art, where artists wants us not to focus on the code, which is the actual work, but on the result it produces at an output, i.e. a computer screen, a projected surface, a sound speaker or a 3D printed object.
Our aesthetics change because we change. And defining the New Aesthetic is necessarily problematic because it is still elusive.
It's a vibe, a sensibility associated with the future.
[..] continue reading here
Categories: article, concept, creative coding, procedural animation
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Home » Effective Communications and Deliberate Indifference Revisited
Effective Communications and Deliberate Indifference Revisited
By William Goren on December 22, 2020
Posted in ACAA, ADA, Federal Cases, Rehabilitation Act, Title II, Title III
Today's blog entry comes out of the Northern District of Georgia, and it involves the question of what is deliberate indifference in effective communication cases. We have talked about effective communication and deliberate indifference numerous times before in the blog. The case of the day is Nix v. Advanced Urology Institute of Georgia198199. By way of full disclosure, I have previously on another matter consulted with local counsel on this case previously but have never consulted with the lead counsel. As usual, blog entry is divided into categories and they are: facts; court's reasoning introductory matters; court's reasoning effective communication; court's reasoning deliberate indifference;; the case is being appealed and should be; and thoughts/takeaways. Of course, the reader is free to concentrate on any or all of the categories.
Facts Lifted Directly from the Opinion (citations are internal to the opinion)
Nix is deaf and primarily communicates using American Sign Language (ASL).2 Nix also reads and writes in English.3 Advanced Urology is a medical 1 The following facts are undisputed by the parties or otherwise supported by undisputed evidence in the record. 2 ECF 88, ¶ 1. 3 Id. ¶ 2. practice specializing in urological health.4 Prior to its involvement with Nix, Advanced Urology had treated deaf patients, but did not have an ongoing agreement with an interpreting agency.5 On February 5, 2018, Nix contacted Advanced Urology through a video relay service to make an appointment at its facility in Snellville, Georgia.6 Due to the emergency nature of Nix's symptoms, the parties scheduled the appointment for February 7.7 Nix did not request an in-person interpreter during this initial call.8 The next morning, Nix called back and requested an interpreter.9 That request was forwarded to Missy Sherling—Advanced Urology's VicePresident of Clinical Strategy—who, in turn, initiated a call with Steve Karasick and Kelly Brauer, Advanced Urology's then-Chief Executive Officer and Surgery Center Director, respectively.10 Brauer, Karasick, and Sherling discussed Nix's request, concluded they did not have sufficient time to procure an interpreter 4 Id. ¶ 3; ECF 89-1, ¶ 16. 5 ECF 89-1, ¶ 18. 6 Id. ¶ 18. 7 Id. ¶ 24; ECF 72 (Nix Dep. Tr. 78:11–21). 8 ECF 72 (Nix Dep. Tr. 81:10–20). 9 ECF 89-1, ¶ 25. 10 Id. ¶ 27. through their usual interpreting agency, and decided to look for other ways to fulfill Nix's request.11 While attempting to locate an interpreter, Sherling learned that an Advanced Urology employee, Samantha Fazzolare, had a friend, Dalton Belew, who "could do basic signing."12 Fazzolare did not inform Sherling of Belew's profession or represent Belew as a trained or professional interpreter.13 Fazzolare provided Sherling with Belew's contact information.14 Sherling did not, however, conduct any investigation into Belew's background or qualifications as an interpreter.15 Based on Sherling's subjective belief that Belew previously interpreted for another medical practice, Sherling and Karasick made the decision to ask Belew to interpret Nix's appointment.16 Belew agreed.17 Contrary to Sherling's beliefs, Belew was not certified in ASL,18 had never interpreted in a 11 ECF 66 (Sherling 30(b)(6) Dep. Tr. 15:2–23). 12 Id. at 17:18–18:3. 13 ECF 76 (Fazzolare Dep. Tr. 18:6–19:3). 14 ECF 89-1, ¶ 33. 15 ECF 66 (Sherling 30(b)(6) Dep. Tr. 27:3–21). 16 Id. at 74:18–25. 17 ECF 89-1, ¶ 38. 18 ECF 68 (Belew Dep. Tr. 12:14–16). medical setting,19 and characterized his own skills as "intermediate."20 Belew had instead been previously employed as a video editor, floor manager at a news company, and at a dental office in an administrative position and as a sterilizer.21 After securing Belew's agreement, Sherling called Nix and informed her an interpreter had been secured for her appointment.22 On February 7, Nix and Belew arrived at Advanced Urology for Nix's appointment. The parties dispute many of the specific details of the appointment. However, the evidence is undisputed that Nix and Belew experienced significant difficulties communicating with each other through ASL. Due to Belew's struggles, and his wearing of scrubs bearing Advanced Urology's insignia, Nix became convinced Belew worked as a nurse, not an actual interpreter.23 At some point, Nix abandoned communication with Belew entirely and began writing notes and using gestures to communicate directly with the medical staff.24 Belew nonetheless remained in the room while Nix partially undressed and 19 Id. at 13:2–5. 20 Id. at 14:11. 21 Id. at 10:2–12:13. 22 ECF 89-1, ¶ 41. 23 E.g., ECF 72 (Nix Dep. Tr. 118:1–119:3). 24 ECF 89-1, ¶¶ 67–68. underwent two non-invasive ultrasounds.25 At the conclusion of the appointment, Nix and Belew filled out various forms affirming that Belew acted as an interpreter during the entire appointment.26 Advanced Urology ultimately issued Belew a check for $100 for his services.27 After her appointment, Nix and her husband complained to Advanced Urology regarding its decision to use Belew as an interpreter; Advanced Urology apologized and refunded their $40 co-pay for the appointment.2
plaintiff filed suit alleging violations of title III of the ADA, §504 to Rehabilitation Act, §1557 of the patient protection and Affordable Care Act, fraud, negligence, and intentional infliction of emotional distress. Of course, summary judgment motions were filed. She sought both injunctive relief and damages.
Court's Reasoning Introductory Matters
The Affordable Care Act, the ADA, and the Rehabilitation Act all use the same substantive standards.
Plaintiff withdrew her claims for injunctive relief in her response to defendant's motion for summary judgment.
Plaintiff is a qualified person with a disability.
Plaintiff utilizes ASL at her primary means of communication.
Court's Reasoning Effective Communication
With respect to whether communication is effective, the question is whether plaintiff experienced an impairment in her ability to communicate medically relevant information with hospital staff with a focus on the effectiveness of the communication and not on the medical success of the outcome.
While perfect communication is not required, communication must still be effective.
Ample evidence could lead a reasonable juror to find the communication between the plaintiff and the interpreter the practice hired to be ineffective. Examples include: the interpreter struggled to communicate even the most basic information, including his own name, during her appointment; the interpreter used incorrect signs for certain words, and the interpreter did not know what to sign or could not understand what plaintiff was attempting to communicate to him.
Frustrated with the interpreter's performance, plaintiff ultimately abandoned communication through ASL and resorted to self-help by exchanging handwritten notes in English directly with the medical staff.
The 11th Circuit has held that the use of written notes may not be appropriate in the medical setting because such matters involve more complexity, such as communication about medical history or diagnoses, conversations about medical procedures and treatment decision, or communication of instruction for care at home or elsewhere. Written notes are better suited for more basic everyday interactions, such as the purchasing of an item in the hospital gift shop.
The interpreter's strained efforts in ASL interpretation combined with a mere four pages of handwritten notes created during the course of a three hour appointment, which included two noninvasive ultrasounds, does not constitute effective communications as a matter of law. So, genuine issues of material fact exists as to whether the limited auxiliary aids employed by the defendant impaired plaintiff's ability to effectively communicate medically relevant information with that staff.
Court's Reasoning Deliberate Indifference
Deliberate indifference is an exacting standard requiring showing more than gross negligence.
No requirement exists in federal or Georgia law requiring a medical facility to provide an ASL certified interpreter for deaf patients.
The mere failure to provide a patient with an interpreter is not enough to support a finding of deliberate indifference.
Defendant did attempt to accommodate plaintiff's last-minute request for an in person interpreter and procured one for her appointment. It is true that the interpreter constituted a poor choice and perhaps the defendant should have more rigorously vetted his qualifications as an interpreter before hiring them.
No evidence exists that the defendant knew that it would be substantially likely that the interpreter could not effectively communicate with the plaintiff and yet did nothing about it.
There is no dispute that the plaintiff ever specifically informed the defendant's medical staff that she needed or wanted a different interpreter.
Plaintiff did not raise any grievance concerning the interpreter's performance until after the appointment.
The interpreter believed that he and the plaintiff adequately communicated during the appointment, with the exception of certain medical terminology.
Plaintiff's medical providers willingly engaged in the exchange of written notes to facilitate communication.
At best, the evidence demonstrates that the defendant should have done more to ascertain the interpreter's competency in ASL before hiring him, but that is not enough to demonstrate deliberate indifference.
No evidence exists creating a triable issue of fact that the defendant has knowledge that plaintiff's rights would substantially likely to be violated and yet failed to act.
This Case Is Being Appealed and It Should
While the court found that effective communication would not present here, it didn't mention the standard in Silva I, which is: "To be ineffective communication, it is sufficient if the patient experiences a real hindrance, because of her disability, which affects her ability to exchange material medical information with her health care providers. This standard is consistent with the requirement that hospitals afford a level of communicationto a deaf patient about medically relevant information that is substantially equal to that afforded to non-disabled patients."
To show deliberate indifference in the 11th Circuit, Liese v. Indian River Community Hospital District180185200, which we discussed here181186201, a plaintiff has to show that the defendant: (1) knew they had failed to provide plaintiff with appropriate auxiliary aids necessary to ensure effective communication; (2) had the authority to order that aid be provided; and (3) was deliberately indifferent as to defendant's failure to provide aid. A deliberate refusal to provide the auxiliary aid and service does constitute deliberate indifference under Silva I, discussed here182187202.
The court frequently cite to Silva II, here183188203, which vacated the summary judgment on deliberate indifference but threw out the injunctive relief claim due to policy changes of the Defendant (the policy changes required: the provision of live in person interpreters upon request by a patient or guest; provided for VRI while waiting for a live interpreter; and provided for the scheduling of live interpreters for regularly scheduled appointments), as a basis for their decision. However, the impression created is not right. In Silva II, here184189204, the 11th Circuit said that deliberate indifference by itself is not the exacting standard, per Nix's understanding for example, rather deliberate indifference requires that the indifference be a deliberate choice, which is the exacting standard.
Silva II said that deliberate indifference occurs when the defendant knew that the harm to a federally protected right would substantially likely and failed to act on that likelihood. In other words, plaintiff must show ineffective communication done with knowledge that it would substantially likely to occur.
Thoughts/Takeaways
I, with co-counsel, have previously represented culturally deaf individuals, Deaf. From my experience, the culturally deaf can be a little too trusting of the hearing community at times. If you are representing a culturally deaf individual seeking medical care, you absolutely have to insist that he or she not go through with the appointment the minute they realize that a qualified interpreter is not there. They just have to walk away. You have to tell the culturally deaf individual to not even engage in the exchange of written notes except for the purposes of saying that he or she need a qualified interpreter.
The court creates a higher standard for deliberate indifference than what was created in Silva I and Silva II. It resembles more the deliberate indifference standard adopted by the Seventh Circuit in a case we discussed here185190205.
While the court says effective communication did not occur, they didn't bring up the Silva I standard, which is whether communication was hindered.
If you are representing a culturally deaf individual in a case like this, it becomes really important to have a table interpreter, an interpreter that acts very much as part of the legal team, who really understands how interpreting can go awry when a qualified interpreter is not involved. Also, be prepared to have an expert talk about the English ability levels of the plaintiff and the importance of ASL given that English ability (with respect to a culturally deaf individual, it is not unusual for such individuals to have English reading skills not higher than fourth grade if that).
Prior to a culturally deaf individual walking away from the appointment, it is really helpful if the culturally deaf individual makes clear that the interpreter is not qualified and that they will not take any medical care offered until a qualified interpreter is present.
Not sure why the plaintiff withdrew injunctive relief claims. I can say that one reason might be that the plaintiff had no intention of ever returning to that practice for reasons that are perfectly understandable.
A culturally deaf individual need to immediately stop the appointment the minute he or she realizes the interpreter is not qualified. He or she should not wait till the end of the appointment to express those concerns.
Depending upon your state, there may be informed consent concerns as well when a qualified interpreter is not provided for a culturally deaf individual. In the case that I served as co-counsel on, we had both informed consent claims as well as ADA claims in our complaint.
Since the interpreter was not a qualified interpreter and bound by an interpreter code of ethics, there may be a breach of confidentiality claims against the physician as well in this kind of scenario.
The 11th Circuit has been extremely progressive with respect to the rights of persons with disabilities. In light of Silva I, II and Liese, I like the chances of the plaintiff on appeal, though one never knows.
Tags: deliberate indifference, effective communications, Liese v. Indian River County Hospital District, Nix v. advanced urology Institute of Georgia, silva v. Baptist health South Florida I, Silva v. Baptist health South Florida II


text:
[Opinion] Each athlete should have the option not to stand during the national anthem
Eli Harold (58), Colin Kaepernick (7) and Eric Reid (35) of the San Franciso 49ers kneel during the national anthem before a game against the Dallas Cowbowy on October 2, 2016, at Levi's Stadium, in Santa Clara, Calif. (Nhat V. Meyer/Bay Area News Group/TNS)
Benjamin Tichauer, Writer
Our country's national anthem has been at the forefront of controversy over the last several years. There are groups of people who feel that the Star-Spangled Banner does not fairly represent their culture and therefore have chosen not to stand while it's being played.
In 2016, NFL quarterback Colin Kaepernick of the San Francisco 49ers was the first professional athlete to kneel during the pre-game national anthem in an effort to call attention to social injustice. This led many people to question the song's meaning and caused a rift between players and their sports leagues across the United States. The national anthem should not be required before every professional sporting event since our country has changed in many ways and people want their voices to be heard more than ever.
Francis Scott Key wrote the Star-Spangled Banner in 1814. At first, theStar-Spangled Banner trailed in popularity behind other 19th century patriotic tunes such as Yankee Doodle and Hail Columbia. But during and after the Civil War, the Star-Spangled Banner was used as a song to boost morale.
In the late 19th century, live military bands occasionally played the Star-Spangled Banner at baseball games, but the song wasn't yet a tradition at sporting events. That changed on Sep. 5, 1918, during Game 1 of the World Series between the Boston Red Sox and the Chicago Cubs during World War I. The song became even more popular during the second world war as patriotism continued to surge.
In 2016, when Kaepernick decided not to stand during the anthem, he didn't do so out of disrespect for the military, rather choosing to take a knee to protest police brutality. This sparked many activists and civil rights groups to agree with the athlete's decision.
During a very turbulent 2020, there were many incidents that racially divided our country. This brought the subject of social injustice and standing during the anthem at sporting events to the forefront of controversy again.
But Kaepernick wasn't the only athlete to protest. In August of 2020, players from the National Basketball Association, Women's National Basketball Association, Major League Baseball and Major League Soccer decided to go on strike, walking off courts and playing fields in protest of systemic racism and police brutality. These strikes, which occurred in the wake of protests after the deaths of unarmed black people at the hands of law enforcement, allowed black athletes to use their platform to protest racial injustice in our country.
The NBA's Dallas Mavericks did not play the national anthem in their first 13 preseason and regular season home games to start the 2021 season, breaking with tradition in professional sports. This, once again, sparked many people to come out and question if it should really be a requirement for all the teams to play the anthem before each game. The NBA then mandated that the Mavericks start playing the anthem again as it was a longstanding league requirement.
Some may say that sports and politics shouldn't be mixed, however, sports, politics and racial injustice are intertwined in today's day and age. Overall, the anthem should not be required to be played at every game, as some traditions are meant to be changed. Athletes have the right to protest or not stand while the Star-Spangled Banner is being played.
Ben Tichauer
Marjory Stoneman Douglas High
Benjamin Tichauer is a sophomore at Marjory Stoneman Douglas High School. He enjoys playing tennis and plays for the school team. He likes coaching kids...
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Aaron	• Apr 12, 2021 at 6:29 pm
You should always stand for National Anthem that
represents freedom is disrespect the National Anthem or the flag then you disrespect the country and the men and women you gave there live so we could be free and this whole Black Lives Matter crap is dumb cause all lives matter not just black lives
https://eagleeye.news/19565/opinion/opinion-each-athlete-should-have-the-option-not-to-stand-during-the-national-anthem/#comment-379
adviser	• May 7, 2021 at 9:52 am
While we do appreciate you engaging with our content Aaron, we would like to note that your comment appears to lack the full understanding of the Black Lives Matter movement. Of course all lives matter however, due to systemic oppression and the racist history of our country, it has been made abundantly clear that marginalized communities do not matter, especially the Black community. We can provide you with ample resources that highlight the historical struggle the Black community has faced and hopefully that will broaden your mindset. In regards to the BLM movement, the whole purpose of it is to promote accountability for our law enforcement. Black people are killed at a disproportionately higher rate by police officers than other race/ethnic group, and despite there being statistics and real life examples to prove this, there has been little to no change to eliminate this problem. To make our response short, think of it as "Black Lives Matter Too" – The Eagle Eye Editorial Board


text:
Q: Firebase cloud functions throws timeout exception but standalone works fine I am trying to call a third party API using my Firebase cloud functions. I have billing enabled and all my other function are working fine.
However, I have one method that throws Timeout exception when it tries to call third API. The interesting thing is, when I run the same method from a standalone nodeJS file, it works fine. But when I deploy it on Firebase cloud or start the function locally, it shows timeout error.
Following is my function:
exports.fetchDemo = functions.https.onRequest(async (req, response) =>
{
    var res = {};
    res.started = true;
    await myMethod();
    res.ended = true;
    response.status(200).json({ data: res });
});

async function myMethod() {
     var url = 'my third party URL';
     console.log('Line 1');
     const res = await fetch(url);
     console.log('Line 2'); // never prints when run with cloud functions
     var data = await res.text();
     console.log(`Line 3: ${data}`);
}

Just now I also noticed, when I hit the same URL in the browser it gives the following exception. It means, it works only with standalone node.
<errorDTO>
  INTERNAL_SERVER_ERROR
 <uid>c0bb83ab-233c-4fe4-9a9e-3f10063e129d</uid>
</errorDTO>

Any help will be appreciated...

A: It turned out that one of my colleague wrote a new method with the name fetch. I was not aware about it. So when my method was calling to the fetch method, it was actually calling his method he wrote down the file. I just took git update and did not notice he wrote this method.


Lots of people have encountered the different DirectX errors while playing the Battlefield 4 or Battlefield 1. DirectX Function errors can appear both at BF start, and in its process. In this guide, we'll take a look at typical DirectX errors in different versions of Battlefield and possible solutions.
This error can happen any time, from a few seconds in, to minutes after opening the game, regardless of settings, and only seems to crash Battlefield 4. Fortunately there are a few ways to get around this, which we will describe below.
Hold down [Windows] + [R] to open a prompt window.
Type msconfig in the prompt window, then click OK. You should see the System Configuration dialog box.
Click the Selective startup option.
Uncheck the Load startup items box.
Check the Hide all Microsoft services option you see in the bottom-left part of the tab. This will allow you to keep all the essential programs Windows needs to function.
Make sure Auto Login is Off.
Go to System preferences in the Apple menu.
Open the Users and groups panel.
Click the lock at the bottom corner and, if necessary, enter your password.
Click on Login Options, find Automatic Login, and select Off (this can be turned back on after the clean boot process is complete).
After logging in, go to the Apple menu and restart.
Once you've done that, type your username and password, without immediately hitting Enter.
Hold down the Shift key as you click the login button.
You user account will now load without additional start-up items or applications. Release Shift after your icons and desktop disappear.
If you can now play the game without the crash occurring, more likely than not one of the programs you've disabled was responsible for the issue. Try to find out which one (through trial and error, maybe beginning with some of the more recent programs you've installed just before the issue began) and then reinstall it or remove it altogether.
Overclocking is a way of making your microprocessor run faster than the speed it was optimally designed for. Sometimes the graphics card is overclocked by default, in which case it might be causing problems without you even knowing it. There are a number of ways to downclock your processor, but one of the easiest is by using the (free) EVGA Precision software.
The white arrow shows the clock speed of your graphics card. You should find the standard clock speed of your card (look for that information on the manufacturer's website) and if that figure is lower than what the software is telling you the speed currently is, use the GPU Clock Offset slider to bring that figure down.
Errors of DirectX Function in Battlefield 1 can be related to different factors. Let's analyze the most important of them.
Perhaps your GPU does not support the required version of DirectX. The successful installation of DirectX does not mean that a new version of the DX libraries will be used (the video chip simply cannot use them and when the game is trying to call a new DX function, an error occurs).
In the event that you own the old video card (Nvidia 600 or 700 series etc.), make sure that your graphics card supports DirectX 11 (or above).
Disabling DX12 in the game menu can also fix the problem. Press Esc, then go to the Options -> Video -> Advanced. Switch the option DX12 Enabled to Off.
Try to reduce the GPU settings in the Nvidia Control Panel.
Maximum pre-rendered frames — 3.
Try to uninstall the current version of DirectX and install the older version. Also you can try to update your video driver to the last version.
Also try to delete the entire contents of the folder C:\Users\%username%\Documents\Battlefield 1\settings.
Please note that without an overlay, you will not be able to view the personal profiles of other players. All other options (invitation of friends to the squad and others) will be available without restrictions.
Sometimes repairing of the Microsoft Visual C++ libraries might help. Go to Control panel -> Programs -> Programs and Features. In the list of installed aps find Microsoft Visual C++ 2012, 2013, 2015 Redistributable, right click on it and select Repair option.
Also you can use to repair original Microsoft Visual C++ Redistributable packages from the directory C:\Program Files (x86)\Origin Games\Battlefield 1\__Installer\vc.
There is an opinion that the overheating of one of the video card's (VRM) components is caused by DirectX function errors, so a slight decrease in the frequency of the graphics chip and its voltage helps.
If you are facing with DirectX errors after installing Windows updates (upgrades), you can try to uninstall them, reboot your computer and check the Battlefield operation.
We hope the fixes outlined in this post have been helpful and have allowed you to enjoy Battlefield 4 without the game crashing on you. If you are aware of other solution to the problem, please let us know in the comments.


From the moment its first 385-case lot was labeled, The Prisoner has been an outlier in the California wine industry.
It was an oxygen epiphany. Could oxygen be the key to making wines that shimmer with life? I, for one, have now added it to the long list of factors that may determine whether a wine seems dead or alive.


Basic broadband is a Local Loop Unbundled Service* and is available at most exchanges. As a result, we are able to offer ADSL to our clients at significantly reduced prices with potentially faster speeds.
This product is an inclusive Broadband, Line and Call product.
* Local Loop Unbundling (LLU) is the process where BT makes its local network (the copper cables that run from customers premises to the telephone exchange) available to other companies at the local exchange.


text:
Mink found to have coronavirus on two Dutch farms - ministry
Sunday, 26 Apr 2020
AMSTERDAM (Reuters) - Two mink farms in the Netherlands have been put into quarantine after animals were found to be infected with the new coronavirus, the agriculture ministry said on Sunday, urging people to report any other likely cases in the animals.
The mink, which were tested after showing signs of having trouble breathing, were believed to have been infected by employees who had the virus, the ministry said in a statement.
The possibility that they could further spread the virus to humans or other animals on the farms was "minimal", the ministry said, citing advice from national health authorities.
However movement of the ferret-like mammals and their manure was banned and the ministry said it was studying the outbreak carefully, including testing the air and soil. People were advised not to travel within 400 meters of the farms.
They were the first reported cases in animals in the Netherlands of the disease, which has been found in some pets and zoo animals around the world after spreading among people.
The towns where the farms are located, Germert-Bakel and Laarbeek, are both in the southern Noord Brabant province of the Netherlands which has seen the country's worst coronavirus outbreak.
The mink are bred for their fur, which is sold in China, Korea, Greece and Turkey. After pressure from animal rights activists, the Dutch government banned new mink farms in 2013 and said existing ones would have to close by 2024.
The World Health Organization has said bats in China, where the new coronavirus emerged last year, were a likely reservoir of COVID-19 and that an intermediate animal host that is yet to be identified had then infected humans.
(Reporting by Toby Sterling; editing by Philippa Fletcher)


text:
Western Montana sees multiple crashes due to winter weather
On behalf of Terrazas Henkel, P.C. posted in car accidents on Thursday, February 21, 2019.
Although people in Montana are familiar with snow, ice and other winter weather conditions, not everyone in the state takes the roadway hazards these conditions can create as seriously as they should, and accidents can result. Due to storms that passed through the area approximately a week and a half ago, emergency crews in western Montana have had their hands full with a number of crashes that have caused either partial or complete highway closings on a temporary basis.
Several of these crashes occurred along Interstate 90. In the Bonner area, near mile marker 110, there was an accident with injuries late last Wednesday morning that blocked one of the eastbound lanes, prompting the issuance of a MEANS alert throughout Missoula County. At about the same time, though further west near mile marker 90, a one-vehicle crash also necessitated the closure of one lane of the interstate. A jackknifed semi truck blocked the left lane 3 1/2 miles west of St. Regis, and there were also crashes reported at mile markers 27, 37, 50, 99 and 105.
However, the accidents were not confined to the interstate. Though details are not available, it is known that there was a crash in the Ronan area on Round Butte Road near mile marker 3. On U.S. Highway 93, five miles north of Dayton in Lake County, a semi-truck slid off the road. A head-on collision approximately 20 miles west of Kalispell blocked U.S. Highway 2, according to state officials, and another wreck on U.S. Highway 93, this time between Missoula and Lolo, blocked the northbound lanes, bringing traffic to a standstill.
Details enumerating how many people these crashes involved are not available. Unfortunately, the ones who fail to pay attention to winter road conditions, or to take them seriously, are not necessarily the ones who pay the price. For this reason, those injured in winter car accidents may find it helpful to consult an attorney.
Related Posts: What kind of burns can you suffer in a car crash?, Tips on driving safely in the winter, Snow removal, plows and traffic accidents, What should you know to stay safe on an icy road?


text:
Q: "require": cannot load such file-- open-url(LoadError) Here's my .rb code. Kindly help please, thank you so much!
require 'open-uri'
require 'nokogiri'

url = "http://en.wikipedia.org/wiki/List_of_current_NBA_team_rosters"
page = Nokogiri::HTML(open(uri))

puts page

This is the error appear in my command prompt:

C:\Sites>ruby wikipedia_scraper.rb
  C:/RailsInstaller/Ruby2.3.0/lib/ruby/2.3.0/rubygems/core_ext/kernel_require.rb:55:in require': cannot load such file -- open-url (LoadError from C:/RailsInstaller/Ruby2.3.0/lib/ruby/2.3.0/rubygems/core_ext/kernel_require.rb:55:inrequire'
          from wikipedia_scraper.rb:1:in `'


First copy!! I always do a "launch" of every novel and novella of mine. A ritual of buying one print copy off Amazon. Well, another launch. USPS just delivered the Amazon copy so it's "official" We have a launch. A very bright spot in otherwise crappy day.
So now the print version is shipping on Amazon. Yeah!!
From Prairie Rose Publications -- Cheryl Moss Pierson and Livia Washburn Reasoner publishers, and Liva did the gorgeous cover.