text:
\section{Introduction and results} \label{intro}

\noindent
 We consider the equation
\begin{equation} \label{eq}
-\Delta u + V(x)u = Q(x)u\log u^2 \quad\,\,\, \text{in $\mathbb{R}^N$},
\end{equation}
where the external potential $V$ and the term $Q$ are $1$-periodic functions of the variables
$x_1,\ldots,x_N$, $Q\in C^1(\mathbb{R}^N)$, $\min_{\mathbb{R}^N} Q >0$ and $\min_{\mathbb{R}^N}(V+Q)>0$.
The problem is formally associated with the energy functional $J:H^1(\mathbb{R}^N)\to\mathbb{R}\cup\{+\infty\}$ defined by
\begin{equation*}
J(u) = \frac12\int_{\rn}\left(|\nabla u|^2+(V(x)+Q(x))u^2\right)dx - \frac12\int_{\rn} Q(x)u^2\log u^2\, dx.
\end{equation*}
Problem~\eqref{eq} admits applications related to quantum mechanics,
quantum optics, nuclear physics, transport and diffusion
phenomena, open quantum systems,
effective quantum gravity, theory of
superfluidity and Bose-Einstein condensation
(see e.g.\ \cite{Zlosh} and the references therein).
We stress that, specifically, periodic potentials $V$ can play a significant r\^ ole in
crystals and in artificial crystals formed by light beams.
Although the logarithmic Schr\"odinger equation has been ruled out as a fundamental quantum wave equation by
very accurate experiments on neutron diffraction, it is currently under discussion if this equation can be adopted
as a simplified model for some physical phenomena.
We refer the reader to \cite{cazNA,caz1,cazhar}
for existence and uniqueness of solutions of the associated Cauchy problem
in a suitable functional framework and to a study of orbital stability, with respect to radial perturbations, of the ground state solution (see \cite{birula76,birula79,birula3}).
In light of a simple modification (see formula \eqref{logQ} in Section~\ref{sect2})
of the standard logarithmic Sobolev inequality \cite{LLoss}
\begin{equation}
\label{eq:logsob}
\int_{\rn} u^2 \log u^2 \leq \frac{a^2}{\pi}  \|\nabla u\|_2^2
+  (\log \| u\|_2^2 - N (1+\log a))\| u\|_2^2,
\quad
\hbox{for }u\in H^1(\mathbb{R}^N) \hbox{ and } a>0,
\end{equation}
it is easy to see that $J(u)>-\infty$ for all $u\in H^1(\mathbb{R}^N)$,
but there exist $u$ in $H^1(\mathbb{R}^N)$ with
$ \int_{\mathbb{R}^N} u^2 \log u^2 dx=-\infty $.
Thus, in general, $J$ fails to be finite and
$C^1$ smooth on $H^1(\mathbb{R}^N)$.

Due to this loss of smoothness, in order to study existence
of solutions, to the best of our knowledge, at least three approaches
were used so far in the literature.
On the one hand, in \cite{cazNA}, the idea is
to work in a suitable Banach space $W$ endowed with a Luxemburg type norm
in order to make the functional
$J:W\to\mathbb{R}$ well defined and $C^1$ smooth. On the other hand,
in \cite{guerrero} the authors
penalize the nonlinearity around the origin and try to obtain
a priori estimates to get a nontrivial solution
at the limit.
However, the drawback of these indirect approaches is that
the Palais-Smale condition cannot be obtained, due to a loss
of coercivity of the functional $J$,
and, in general, no multiplicity result can be obtained
by the Lusternik-Schnirelmann category theory.
Recently, in \cite{dams}, in the case of constant potentials $V$ and $Q$,
the existence of infinitely many weak solutions was obtained
by considering the functional  $J$ on $H^1_{\rm rad}(\mathbb{R}^N)$ as merely
lower semicontinuous and by applying the nonsmooth critical point theory of
\cite{DegZan2000}, originally formulated to tackle semilinear elliptic equations with one-sided
growth conditions, and based upon the general theory developed in \cite{CDM,CamDeg}. The restriction to the space of radially symmetric
functions in $H^1(\mathbb{R}^N)$ is related to having the Palais-Smale condition (in a suitable sense)
satisfied at an arbitrary energy level.

In this paper, we shall work in the unrestricted space $H^1(\mathbb{R}^N)$ and we exploit the
fact that the functional $J$, although being nonsmooth, can be decomposed into the
sum of a $C^1$ functional and a convex lower semicontinuous functional. If $u$ is a solution
to \eqref{eq}, so are the elements of its orbit under the action of $\mathbb{Z}^N,$
${\mathcal O}(u):=\{u(\cdot-k):k\in\mathbb{Z}^N\},$ and two solutions are said to be {\em geometrically distinct} if
${\mathcal O}(u_1)\cap {\mathcal O}(u_2)=\emptyset.$

\vskip2pt
By adapting some arguments in \cite{sw} and using tools from \cite{sz}, we prove the following result.

\begin{theorem} \label{thm}
Equation \eqref{eq} has infinitely many geometrically distinct solutions.
\end{theorem}

\vskip2pt
Furthermore, setting:
\begin{equation} \label{neh}
{\mathcal N}:=\Big\{u\in H^1(\mathbb{R}^N)\setminus\{0\}:
\int_{\rn}\left(|\nabla u|^2+V(x)u^2\right)dx=\int_{\rn} Q(x)u^2\log u^2\,dx\Big\},
\end{equation}
we have the following existence result for ground state solutions to \eqref{eq}.

\begin{theorem} \label{thm0}
The infimum $\inf_{{\mathcal N}} J>0$ is attained at a solution $u$ to equation \eqref{eq}. Moreover, $u(x)>0$ for all $x\in\mathbb{R}^N$ or $u(x)<0$ for all $x\in\mathbb{R}^N$.
\end{theorem}

\noindent
In the case of constant potentials, the ground state solution is known explicitly
(it is called the {\em Gausson} in the physical literature \cite{birula76,birula79,birula3}) and, as proved in \cite{dams}, it is nondegenerate, that is to say the dimension of the
nullspace of the linearized operator is $N$, i.e.\ smallest possible.

\vskip3pt

In what follows by a {\em solution} to \eqref{eq} we shall always mean a function $u\in H^1(\mathbb{R}^N)$ such that $u^2\log u^2\in L^1(\mathbb{R}^N)$ and
\[
\int_{\rn}(\nabla u\cdot \nabla v + V(x)uv)\,dx = \int_{\rn} Q(x)uv\log u^2\,dx, \quad \text{for all } v\in C_0^\infty(\mathbb{R}^N).
\]
Note that since $|u\log u^2|\le C(1+|u|^q)$, where $q>1$, we may use local estimates and standard bootstrap
to assert that $u$ is, in fact, a classical solution (cf.\ \cite[Appendix B]{struwe}).

\bigskip

\noindent \textbf{Notation.} $C, C_1, C_2$ etc. will denote positive constants whose exact values are inessential. $\langle .\, , . \rangle$
is the duality pairing between $E'$ and $E$, where $E$ is a Hilbert (more generally, Banach) space and $E'$ is its dual. $\|\cdot\|_p$ is
the norm of the space $L^p(\mathbb{R}^N)$.\  $2^*:=2N/(N-2)$ if $N\geq 3$ and $2^*:=\infty$ if $N=1$ or $2$.\
$B_R(x)$ denotes the open ball centered at $x$ and having radius $R$ and $S_R(x):=\partial B_R(x)$.
For a functional $J$, we set $J^b:=\{u\in E: J(u)\le b\}$, $J_a:=\{u\in E: J(u)\ge a\}$ as well as $J_a^b:=J_a\cap J^b$.

\bigskip
\noindent
{\bf Acknowledgements.} The first author would like to thank Prof.\ Iwo Bia\l ynicki-Birula
from the Center for Theoretical Physics PAN, Warsaw, for useful comments about
the relevance of equation \eqref{eq} from the physical point of view.
\bigskip

\section{Proof of Theorem~\ref{thm}}
\label{sect2}
\noindent
In order to prove Theorem~\ref{thm}, we first need to state some preliminary results.

\subsection{Preliminary results}
We shall denote by $E$ the Hilbert space of functions $u\in H^1(\mathbb{R}^N)$ and we endow it with the norm
$$
\|u\|:=\left(\int_{\rn}\left(|\nabla u|^2+(V(x)+Q(x))u^2\right)dx\right)^{1/2}\!\!\!\!\!\!\!.
$$
Furthermore, let us set:
\begin{align*}
&F_1(s) := \left\{
\begin{array}{ll}
-\frac12s^2\log s^2, & |s|<\delta, \\
\noalign{\vskip3pt}
-\frac12s^2(\log\delta^2+3)+2\delta|s|-\frac12\delta^2, & |s|>\delta,
\end{array}
\right.  \\
& F_2(s) := \left\{
\begin{array}{ll}
0, & |s|<\delta, \\
\noalign{\vskip3pt}
\frac12s^2\log(s^2/\delta^2)+2\delta|s|-\frac32s^2-\frac12\delta^2, & |s|>\delta.
\end{array}
\right.
\end{align*}
Then $F_2(s)-F_1(s)=\frac12s^2\log s^2$ and $F_1$ is convex, provided $\delta>0$ is sufficiently small
which we assume from now on. Moreover, $F_1,F_2\in C^1(\mathbb{R})$. Re-write the functional $J$ as
$J(u)=\Phi(u)+\Psi(u)$, $u\in E$, where we have set:
\begin{align*}
\Phi(u)&:=\frac12\|u\|^2 -\int_{\rn} Q(x)F_2(u)\,dx, \\
\Psi(u)&:=\int_{\rn} Q(x)F_1(u)\,dx.
\end{align*}
Choosing $p\in(2,2^*)$, we have $|F_2'(s)| \le C|s|^{p-1}$ for some $C>0$ and all $s\in\mathbb{R}$,
and hence it follows that  $\Phi\in C^1(E,\mathbb{R})$ \cite[Lemma 3.10]{wi}. Note that $\Psi$ is convex, $\Psi\ge 0$ and $\Psi(u) = +\infty$ for certain $u\in E$. Moreover, it is easy to see by Fatou's lemma that $\Psi$ (and therefore also $J$) is lower semicontinuous (cf.\ \cite[Proposition 2.9]{dams}).

\begin{remark}\rm
Theorems \ref{thm} and \ref{thm0} remain valid for any $F_1, F_2$ such that $F_1,F_2\in C^1(\mathbb{R}^N)$, $F_1$ is convex and nonnegative, $F_1(0)=0$, $|F_1'(s)|, |F_2'(s)| \le C(1+|s|^{p-1})$ for some $C > 0$ and  $p\in(2,2^*)$, and $F_2'(s)/s\to 0$ as $s\to 0$. Some additional conditions may also be needed in order to ensure that the corresponding functional $J$ satisfies the conclusion of Lemma \ref{bounded}. The proof is the same except that small modifications are necessary at some places because  $|F_2'(s)| \le C|s|^{p-1}$ may not hold. Instead, for each $\varepsilon>0$ there exists $C_\varepsilon>0$ such that $|F_2'(s)|\le \varepsilon|s|+C_\varepsilon|s|^{p-1}$.
\end{remark}

\begin{lemma} \label{domain}
If $\Omega\subset\mathbb{R}^N$ is a bounded domain, then $\Psi$ (and hence $J$) is of class $C^1$ in $H^1(\Omega)$.
\end{lemma}

\begin{proof}
Since $|F_1'(s)|\le C(1+|s|^{p-1})$, the conclusion follows from \cite[Lemma 2.16]{wi}. In \cite{wi} the result is stated in $H^1_0(\Omega)$ but the argument remains valid in $H^1(\Omega)$.
\end{proof}

We shall need the following definitions, essentially taken from \cite{sz}:

\begin{definitions} \rm
\label{def}
$(i)$ The set
\[
\partial J(u):=\{w\in E': \langle \Phi'(u),v-u\rangle+\Psi(v)-\Psi(u) \ge \langle w, v-u\rangle,\,\,\,\text{for all $v\in E$}
\]
is called the subdifferential of $J$ at $u$. \\
$(ii)$ $u\in E$ is a critical point of $J$ if $J(u)<+\infty$ and $0\in \partial J(u)$, i.e.\
\[
\langle \Phi'(u),v-u\rangle + \Psi(v)-\Psi(u) \ge 0, \quad\,\, \text{for all $v\in E$}.
\]
$(iii)$ $(u_n)$ is a  Palais-Smale sequence if $(J(u_n))$ is bounded and there exist $\varepsilon_n\to 0^+$ such that
\[
\langle \Phi'(u_n), v-u_n\rangle  + \Psi(v)-\Psi(u_n) \ge -\varepsilon_n\|v-u_n\|, \quad\,\, \text{for all $v\in E$}.
\]
$(iv)$ The set $D(J) := \{u\in E: J(u)<+\infty\}$ is called the effective domain of $J$.
\end{definitions}

\begin{lemma} \label{subdiff}
If $u\in D(J)$, then $\partial J(u)\ne\emptyset$, i.e.\ there exists $w\in E'$ such that
$$
\langle \Phi'(u),v-u\rangle +\Psi(v)-\Psi(u) \ge \langle w, v-u\rangle,\quad\,\,\text{for all $v\in E$}.
$$
Moreover, this $w$ is unique and satisfies
$$
\langle\Phi'(u),z\rangle+\int_{\rn} Q(x)F_1'(u)z\,dx = \langle w, z\rangle \quad \text{for all $z\in E$ such that $F_1'(u)z\in L^1(\mathbb{R}^N)$}.
$$
\end{lemma}

\begin{proof}
Assume by contradiction that for each $w\in E'$ there exists $v\in E$ such that
\[
\langle \Phi'(u), v-u\rangle+\Psi(v)-\Psi(u) - \langle w, v-u\rangle < 0.
\]
Let $u_n\in C_0^\infty(\mathbb{R}^N)$, $u_n\to u$ in $\|\cdot\|$ as $n\to\infty$. Then, by the lower semicontinuity of $\Psi$,
\begin{gather*}
\limsup_{n\to\infty}\left(\langle\Phi'(u_n),v-u_n\rangle+\Psi(v)-\Psi(u_n) - \langle w, v-u_n\rangle\right) \\
\le \langle\Phi'(u), v-u\rangle+\Psi(v)-\Psi(u) - \langle w, v-u\rangle < 0.
\end{gather*}
So $\partial J(u_n)=\emptyset$ for almost all $n\geq 1$ which is impossible because $u_n\in C_0^\infty(\mathbb{R}^N)$ and hence, using Lemma \ref{domain} and convexity of $\Psi$,
\[
\langle\Phi'(u_n),v-u_n\rangle  + \Psi(v)-\Psi(u_n) \ge \langle \Phi'(u_n)+\Psi'(u_n),v-u_n\rangle \quad \text{for all } v.
\]

Let now $v=u+tz$, where $z\in C_0^\infty(\mathbb{R}^N)$ and let $w\in\partial J(u)$. Then
\[
\langle\Phi'(u),z\rangle + \int_{\rn} Q(x)\,\frac{F_1(u+tz)-F_1(u)}t\,dx \ge \langle w,z\rangle.
\]
Since the integrand is 0 for $x\not\in\text{supp\,}z$, we can pass to the limit as $t\to 0$ (cf.\ Lemma \ref{domain}) and we obtain
\[
\langle\Phi'(u),z\rangle+\int_{\rn} Q(x)F_1'(u)z\,dx \ge \langle w,z\rangle.
\]
Since this also holds for $-z$,
\[
\langle\Phi'(u),z\rangle+\int_{\rn} Q(x)F_1'(u)z\,dx = \langle w,z\rangle, \quad \text{for all $z\in C_0^\infty(\mathbb{R}^N)$}.
\]
By density of $C_0^\infty(\mathbb{R}^N)$ in $E$, $w$ is unique and the equality above holds for all $z\in E$ such that $F_1'(u)z\in L^1(\mathbb{R}^N)$.
\end{proof}

\begin{definition} \rm
Let $u\in D(J)$. Then the unique element $w\in E'$ of $\partial J(u)$ introduced in Lemma~\ref{subdiff} will
be denoted by $J'(u)$.
\end{definition}

An immediate consequence of Lemma \ref{subdiff} is the following

\begin{corollary} \label{ps}
Suppose $(J(u_n))$ is bounded. Then  $(u_n)\subset E$ is a Palais-Smale sequence if and only if $J'(u_n)\to 0$ in $E'$
as $n\to\infty$, or equivalently,
\[
\lim_{n\to\infty} \sup\big\{\langle J'(u_n),v\rangle: v \in C_0^\infty(\mathbb{R}^N), \|v\| = 1\big\}= 0.
\]
\end{corollary}

It follows from Lemma \ref{subdiff} that if $u\in D(J)$, then
\[
\langle J'(u),v\rangle = \langle\Phi'(u),v\rangle+\int_{\rn} Q(x)F_1'(u)v\,dx \quad \text{for all $v\in E$ such that $F_1'(u)v\in L^1(\mathbb{R}^N)$}.
\]
Next we construct a vector field of pseudo-gradient type. Denote the set of critical points of the functional $J$ by $K$.

\begin{lemma} \label{pseudogradient}
There exist a locally finite  countable covering $(W_j)$ of $D(J)\setminus K$, a set of points $(u_j)\subset D(J)\setminus K$ and a locally Lipschitz continuous vector field $H: D(J)\setminus K\to E$ with the following properties: \\
(i) The diameter of $W_j$ and the distance from $u_j$ to $W_j$ tend to 0 as $j\to\infty$. \\
(ii) $\|H(u)\|\le 1$ and $\langle J'(u),H(u)\rangle > z(u)$, where $z(u):= \min\frac12\|J'(u_j)\|$ for all $j$ such that $u\in W_j$. \\
(iii) $H$ has locally compact support, i.e.\ for each $u_0\in D(J)\setminus K$ there
exists a neighbourhood $U_0$ of $u_0$ in $D(J)\setminus K$ and $R>0$ such that $\text{supp\,}H(u)\subset B_R(0)$ for all $u\in U_0$. \\
(iv) $J(u)>J(u_j)-\gamma_j$ for all $j$ such that $u\in W_j$, where $\gamma_j>0$ and  $\gamma_j\to 0$ as $j\to\infty$. \\
(v) $H$ is odd in $u$.
\end{lemma}

\begin{remark}\rm
The special properties of the covering $(W_j)$ and the field $H$ in Lemma~\ref{pseudogradient}
will be essential in the proofs of Lemmas \ref{limit} and \ref{critical}.
\end{remark}

\begin{proof}
Since $E$ is separable, there exists a countable dense set of points $(\widetilde u_k)\subset D(J)\setminus K$. For each $k$ we can choose $\widetilde v_k\in C_0^\infty(\mathbb{R}^N)$, $\|\widetilde v_k\|=1$, such that $\langle J'(\widetilde u_k),\widetilde v_k\rangle > \frac12\|J'(\widetilde u_k)\|$.
Since $\widetilde v_k$ has compact support, $u\mapsto \langle J'(u),\widetilde v_k\rangle$ is continuous according to Lemma \ref{domain} and hence
\begin{equation} \label{ineq}
\langle J'(u),\widetilde v_k\rangle > \frac12\|J'(\widetilde u_k)\|.
\end{equation}
 for all $u$ in some neighbourhood $W(\widetilde u_k)$ of $\widetilde u_k$ in $D(J)\setminus K$. Moreover, we may assume that the diameter of $W(\widetilde u_k)$ tends to 0 as $k\to\infty$ and  by the lower semicontinuity of $J$, $W(\widetilde u_k)$ may be chosen so that $J(u)>J(\widetilde u_k)-1/k$ in $W(\widetilde u_k)$. Clearly, $(W(\widetilde u_k))$ is a covering of $D(J)\setminus K$. Since $E$ is metric and hence paracompact, we can find a locally finite refinement $(W_j)$ of $(W(\widetilde u_k))$ and for each $W_j$ we choose $u_j:=\widetilde u_{k_j}$ for some $k_j$ such that $W_j\subset W(\widetilde u_{k_j})$ (note that $u_j$ may not be in $W_j$). So $(W_j)$ and $(u_j)$ satisfy (i) and by \eqref{ineq}, the inequality
\[
\langle J'(u),v_j\rangle > \frac12\|J'(u_j)\|
\]
holds for $u_j=\widetilde u_{k_j}$, $v_j=\widetilde v_{k_j}$ and all $u\in W_j\subset W(\widetilde u_{k_j})$.

Let $\widetilde \rho_j(u) := \text{dist}(u,E\setminus W_j)$,
\[
\rho_j(u) := \widetilde\rho_j(u) \Big/\sum_{j=1}^\infty \widetilde\rho_j(u) \quad \text{and} \quad H(u) := \sum_{j=1}^\infty\rho_j(u)v_j.
\]
It is easy to see that the properties (ii)-(iv) are satisfied  (in (iv) we take $\gamma_j=1/k_j$). Moreover, as $J$ is even, $H$ may be constructed so that $H(-u)=-H(u)$ (e.g.\ by taking $\pm W_j$, $ \pm u_j$ etc.). Hence also (v) holds.
\end{proof}

\begin{lemma} \label{bounded}
If $(u_n)$ is a sequence such that $(J(u_n))$ is bounded above and $J'(u_n)\to 0$, then $(u_n)$ is bounded. Moreover, since $\Psi\ge 0$, $(J(u_n))$ is also bounded below and hence it is a Palais-Smale sequence.
\end{lemma}

\begin{proof}
Let $(u_n)\subset E$ be a sequence with $(J(u_n))$ bounded above and $J'(u_n)\to 0$ as $n\to\infty$.
Then, choosing $v=u_n$ as test function, we end up with
\begin{equation}
\label{eq:boundL2}
C_1\| u_n\|_2^2 \le \int_{\rn} Q(x)u^2_n\,dx = 2 J(u_n) -  \langle J'(u_n),u_n\rangle
\leq C+o(1) \|u_n\|,\quad\text{as $n\to\infty$.}
\end{equation}
Replacing $u$ by $\sqrt{Q}u$ in \eqref{eq:logsob} yields
\[
\int_{\rn} Qu^2 \log(Qu^2)\,dx \leq \frac{2a^2}{\pi}  (\|\sqrt{Q}\nabla u\|_2^2 + \|u\nabla\sqrt{Q}\|_2^2)
+  (\log \| \sqrt{Q}u\|_2^2 - N (1+\log a))\| \sqrt{Q}u\|_2^2.
\]
This gives, taking $a>0$ small enough,
\begin{equation}
\label{logQ}
\int_{\rn} Q(x)u^2\log u^2\,dx \le \frac12\|\nabla u\|_2^2 + C_2(\log\|u\|_2^2+1)\|u\|_2^2.
\end{equation}
So using \eqref{eq:boundL2}, we obtain
\begin{align*}
C\ge 2J(u_n) & = \|u_n\|^2-\int_{\rn} Q(x)u_n^2\log u_n^2\,dx \ge \frac12\|u_n\|^2 - C_2(\log\|u_n\|_2^2+1)\|u_n\|_2^2 \\& \ge \frac12\|u_n\|^2-C_3(1+\|u_n\|^{1+\delta}),
\,\quad \text{as } n\to\infty,
\end{align*}
where we can take $\delta<1$.
Hence $(u_n)$ is bounded, proving the first assertion. Then the second
assertion immediately follows.
\end{proof}

Assume throughout the rest of this section that $J$ has only {\em finitely many} critical orbits and choose a finite set $\mathcal{F}\subset K$ such that $\mathcal{F} = -\mathcal{F}$ and each critical orbit has a unique representative in $K$.

\begin{lemma} \label{kappa}
$\kappa := \inf\{\|u-v\|: u,v\in K,\ u\ne v\} > 0$.
\end{lemma}
\begin{proof}
This follows by a straightforward modification of the proof of \cite[Lemma 2.13]{sw}.
\end{proof}

In the next three lemmas we adapt some arguments from \cite{sw}.

\begin{lemma}
\label{discreteness}
Let $(u_n^1), (u_n^2)\subset E$ be two Palais-Smale sequences,
Then either $\|u_n^1-u_n^2\|
\to 0$ as $n\to\infty$ or $\limsup_{n \to \infty}\|u_n^1-u_n^2\| \ge \kappa$.
\end{lemma}

\begin{proof}
By Lemma~\ref{bounded}, it follows that $(u_{n}^1)$ and $(u_{n}^2)$ are bounded in $E$. Choose $p\in(2,2^*)$.
Then $|F_2'(s)| \le C|s|^{p-1}$ for some $C>0$ and all $s\in\mathbb{R}$.

Suppose first $\|u_{n}^1-u_{n}^2\|_{p}\to 0$ as $n\to\infty$. Then, by Definition \ref{def}(iii), we obtain
\[
\langle\Phi'(u_n^1),u_n^2-u_n^1\rangle+\Psi(u_n^2)-\Psi(u_n^1)\ge -\varepsilon_n\|u_n^2-u_n^1\|,
\]
and a similar inequality holds with the roles of $u_n^1$ and $u_n^2$ interchanged. Hence
\begin{align*}
\|u_n^1-u_n^2\|^2 & =\langle\Phi'(u_n^1),u_n^1-u_n^2\rangle-\langle\Phi'(u_n^2),u_n^1-u_n^2\rangle \\
& \quad +
\int_{\rn} Q(x)(F_2'(u_n^1)-F_2'(u_n^2))(u_n^1-u_n^2)\,dx  \\
& \le 2\varepsilon_n \|u_n^1-u_n^2\| + C\int_{\rn} Q(x)(|u_n^1|^{p-1}+|u_n^2|^{p-1})|u_n^1-u_n^2|\,dx \\
& \le 2\varepsilon_n\|u_n^1-u_n^2\| + D\|u_n^1-u_n^2\|_p.
\end{align*}
So  $\|u_n^1-u_n^2\|\to 0$.

Suppose now $\|u_{n}^1-u_{n}^2\|_{p}\not\to 0$. By Lions' lemma \cite[Lemma I.1]{lions}, \cite[Lemma 1.21]{wi}, we can find  $\varepsilon>0$ and $(y_n)\subset\mathbb{R}^N$ with
$$
\int_{B_1(y_n)}(u_n^1-u_n^2)^2\,dx \ge \varepsilon,
$$
after passing to a subsequence. Since $J$ is invariant under
translations $u\mapsto u(\cdot-k)$, $k\in \mathbb{Z}^{N}$, we may assume the sequence $(y_n)$ is bounded. Hence, passing to a subsequence once more, $u_n^1\rightharpoonup u^1$, $u_n^2\rightharpoonup u^2$ and $u^1\ne u^2$.
The functional $\Psi$ is lower semicontinuous and hence weakly lower semicontinuous (by convexity). So $\Psi(u^1)<\infty$ and therefore $u^1\in D(J)$. Moreover, since $\langle J'(u_n^1),v\rangle\to 0$ for all $v\in C_0^\infty(\mathbb{R}^N)$, it is easy to see that $u^1\in K$. Similarly, $u^2\in K$. Hence
\[
\limsup_{n\to\infty}\|u_n^1-u_n^2\| \ge \|u^1-u^2\| \ge \kappa,
\]
concluding the proof.
\end{proof}

\begin{remark} \label{rem} \rm
For the purpose of the next section we note that if there are finitely many critical orbits below a certain level $d>0$, then the conclusions of Lemmas \ref{kappa}, \ref{discreteness} as well as of Lemmas \ref{limit} and \ref{critical} below remain valid on $J^d$. The proofs go through unchanged except that we need to show that $u^1, u^2\in J^d$ in the proof of Lemma \ref{discreteness}. Since $(u_n^1)$ is bounded and $J'(u^1)=0$,  we have
\begin{align*}
d & \ge J(u_n^1)  = J(u_n^1) - \frac12\langle J'(u_n^1),u_n^1\rangle + o(1) = \frac12\int_{\mathbb{R}^N}Q(x)(u_n^1)^2dx+o(1) \\
& \ge \frac12\int_{\mathbb{R}^N}Q(x)(u^1)^2dx+o(1) = J(u^1)-\frac12\langle J'(u^1),u^1\rangle +o(1) = J(u^1)+o(1).
\end{align*}
So $u^1\in J^d$ and similarly, $u^2\in J^d$.
\end{remark}

\noindent
Consider now the flow $\eta$ given by
\begin{equation}
\label{flow}
\left\{
\begin{array}{l}
\frac{d}{dt} \eta(t,u)=-H(\eta(t,u)), \\
\noalign{\vskip3pt}
\eta(0,u) = u, \ u\in D(J)\setminus K,
\end{array}
\right.
\end{equation}
and let $(T^-(u), T^+(u))$ be the maximal existence time for the trajectory $t\mapsto\eta(t, u)$.

\begin{lemma}
\label{limit}
Let $u \in D(J)\setminus K$. Then either $\lim_{t \to T^+(u)}\eta(t,u)$ exists and is a critical point of $J$ or $\lim_{t\to T^+(u)}J(\eta(t,u)) = -\infty$. In the latter case $T^+(u)=+\infty$.
\end{lemma}

\begin{proof}
Since $\eta(s,u) = u - \int_0^sH(\eta(\tau,u))\,d\tau$ and $H$ has locally compact support, $\tau\mapsto J(\eta(\tau,u))$ is continuously differentiable. To see this, consider
\[
\frac1h(\Psi(\eta(t+h,u))-\Psi(\eta(t,u)))  = \frac1h \int_{\rn} Q(x)(F_1(\eta(t+h,u)) - F_1(\eta(t,u)))\,dx.
\]
Since $\eta(s,u(x)) = u(x)$ for all $s\in[0,t+h]$ and $|x|$ large enough, we can pass to the limit as $h\to 0$ using Lemma \ref{domain} and we obtain, by Lemma \ref{pseudogradient},
\[
\frac d{dt}J(\eta(t,u)) = - \langle J'(\eta(t,u)), H(\eta(t,u))\rangle \le -z(\eta(t,u)) < 0.
\]
So $t\mapsto J(\eta(t,u))$ is decreasing.

Suppose $T^+(u)<\infty$ and let $0\le s<t<T^+(u)$. Then
\[
\|\eta(t,u)-\eta(s,u)\| \le \int_s^t \|H(\eta(\tau,u))\|\,d\tau \le t-s.
\]
Hence the limit exists and if it is not a critical point, then $\eta(\cdot,u)$ can be continued for $t>T^+(u)$.

Suppose $T^+(u)=+\infty$ and $J(\eta(t,u))$ is bounded below. It suffices to show that
for each $\varepsilon>0$ there exists $t_\varepsilon>0$ such that
$\|\eta(t_\varepsilon,u)-\eta(t,u)\|<\varepsilon$ whenever $t \ge t_\varepsilon$.
Assuming the contrary, we can find $\varepsilon\in(0,\kappa/2)$ and $(t_n)\subset\mathbb{R}^+$ with $t_n\to+\infty$ and $\|\eta(t_n,u)-\eta(t_{n+1},u)\|= \varepsilon$ for all $n\geq 1$. Choose the smallest $t_n^1\in(t_n,t_{n+1})$ such that $\|\eta(t_n,u)-\eta(t_n^1,u)\|= \varepsilon/3$ and let $\kappa_n := \min \{z(\eta(s,u)): s\in[t_n,t_n^1]\}$. Then $\kappa_n>0$ and
\begin{align*}
\frac\varepsilon 3 & = \|\eta(t_n^1,u)-\eta(t_n,u)\| \le \int_{t_{n}}^{t_n^1}  \|H(\eta(s,u))\| \,ds \le t_n^1-t_n \\
& \le \frac{1}{\kappa_n} \int_{t_n}^{t_n^1} \langle J'(\eta(s,u)),H(\eta(s,u))\rangle\,ds = \frac1{\kappa_n}\left(J(\eta(t_n,u))-J(\eta(t_n^1,u))\right).
\end{align*}
Since $J(\eta(t_n,u))-J(\eta(t_n^1,u))\to 0$, it follows that $\kappa_n\to 0$. Hence we can find $s_n^1\in [t_n,t_n^1]$ such that $z(\eta(s_n^1,u))\to 0$ as $n\to\infty$. So by Lemma \ref{pseudogradient} there exist $u_n^1$ (where $u_n^1=u_{j_n}$ for some $j_n$) such that $J'(u_n^1)\to 0$, $J(u_n^1)\le J(\eta(s_n^1,u))+\gamma_n^1$ and $\|u_n^1-\eta(s_n^1,u)\| \to 0$. Here it is important that the diameter of $W_{j_n}$ and the distance from $u_{j_n}$ to $W_{j_n}$ in Lemma \ref{pseudogradient} tend to 0 and that (iv) of this lemma gives a uniform bound from above for $J(u_n^1)$.  Similarly we can first find a largest $t_n^2\in[t_n^1,t_{n+1}]$ for which $\|\eta(t_{n+1},u)-\eta(t_n^2,u)\|= \varepsilon/3$ and then $s_n^2\in[t_n^1,t_{n+1}]$ and $u_n^2$ such that $J'(u_n^2)\to 0$, $J(u_n^2)\le J(\eta(s_n^2,u))+\gamma_n^2$ and $\|u_n^2-\eta(s_n^2,u)\| \to 0$. Since $(J(u_n^1))$, $(J(u_n^2))$ are bounded above, $(u_n^1)$ and $(u_n^2)$ are Palais-Smale sequences according to Lemma \ref{bounded}. Hence
\[
\frac\varepsilon 3 \le \limsup_{n\to\infty}\|u_n^1-u_n^2\|\le 2\varepsilon<\kappa,
\]
a contradiction to Lemma \ref{discreteness}. This completes the proof.
\end{proof}

\noindent
Let $d>0$ and choose $\varepsilon_0>0$ such that $J_{d-2\varepsilon_0}^{d+2\varepsilon_0}\cap K = K_d := \{u\in K: J(u)=d\}$. Denote
\[
U_\delta(K_d) := \{u\in E: \text{dist}(u,K_d)<\delta\}.
\]

\begin{lemma}
\label{critical}
For each $\delta>0$ there exists $\varepsilon\in(0,\varepsilon_0)$ such that
\[
\lim_{t\to T^+(u)}J(\eta(t,u))<d-\varepsilon, \quad\text{ whenever } u\in J^{d+\varepsilon}\setminus U_\delta(K_d).
\]
Moreover, $\eta(t,u)\notin U_{\delta/2}(K_d)$ for any $t\in[0,T^+(u))$.
\end{lemma}

\begin{proof}
Assume without loss of generality that $\delta<\kappa$. Let
\[
\tau := \inf\{z(u): u\in J^{d+2\varepsilon_0}_{d-2\varepsilon_0}\cap U_\delta(K_d)\setminus U_{\delta/2}(K_d)\}.
\]
We show that $\tau>0$. Arguing by contradiction, we find a sequence $w_n^1\in J^{d+2\varepsilon_0}_{d-2\varepsilon_0}\cap U_\delta(K_d)\setminus U_{\delta/2}(K_d)$ with $z(w_n^1)\to 0$ and then, using Lemma \ref{pseudogradient}, $u_n^1$ (where $u_n^1=u_{j_n}$ for some $j_n$) such that $J'(u_n^1)\to 0$, $\|u_n^1-w_n^1\|\to 0$ and $J(u_n^1)\le J(w_n^1)+\gamma_n^1$ ($\gamma_n^1\to 0$). Hence $J(u_n^1)$ is bounded above, so $(u_n^1)$ is  a Palais-Smale sequence by Lemma \ref{bounded}. Using Lemma \ref{kappa} and $\mathbb{Z}^N$-invariance of $J$ we may assume $w_n^1\in U_\delta(u_0)\setminus U_{\delta/2}(u_0)$ for some $u_0\in K_d$. Set $u_n^2:=u_0$ for all $n$. This is obviously a Palais-Smale sequence and we have
\[
\frac\delta 2 \le \limsup_{n\to\infty} \|u_n^1-u_n^2\|\le \delta < \kappa,
\]
a contradiction to Lemma \ref{discreteness}.  So $\tau>0$.
If the conclusion of the lemma is false, there exists $w\in K_d$ such that $\eta(t,u)$ will enter $U_{\delta/2}(w)$. Set
\begin{align*}
& t_1:= \sup \{t \in [0,T^+(u)): \eta(t,u) \not \in U_\delta(w)\}, \\
& t_2:= \inf \{t \in (t_1,T^+(u)): \eta(t,u) \in U_{\delta/2}(w)\}.
\end{align*}
Then
\[
\frac\delta 2 \le \|\eta(t_2,u)-\eta(t_1,u)\| \le \int_{t_1}^{t_2}\|H(\eta(s,u))\|\,ds \le t_2-t_1
\]
and therefore
\begin{equation*}
J(\eta(t_2,u)) - J(\eta(t_1,u))  = -\int_{t_1}^{t_2}\langle J'(\eta(s,u)),H(\eta(s,u))\rangle\,ds
 \le -\tau(t_2-t_1) \le -\frac{\tau\delta}2.
\end{equation*}
So $J(\eta(t_2,u))\le d+\varepsilon-\tau\delta/2 < d$ for $\varepsilon$ small. Hence $\eta(t,u)$ cannot enter $U_{\delta/2}(w)$.
\end{proof}

\begin{lemma} \label{mp}
There exist $\rho,b>0$ such that $J(u)\ge 0$ for all $u\in B_\rho(0)$ and $J(u)\ge b$ for all $u\in S_\rho(0)$.
\end{lemma}

\begin{proof}
Recalling that $\Psi\ge 0$ and $|F_2'(s)|\le C|s|^{p-1}$, we obtain $J(u) \ge \Phi(u) = \frac12\|u\|^2+o(\|u\|^2)$. Hence the conclusion.
\end{proof}

In the proof of Theorem \ref{thm} we shall need a variant of Benci's pseudoindex \cite{bbf, be} which we now introduce.  Let $\Sigma := \{A\subset D(J): A=-A \text{ and } A \text{ is compact}\}$ and
\begin{align*}
\mathcal{H} := \{h: D(J)\to E,\,\, \text{$h$ odd homeomorphism onto $h(D(J))$} \\ \text{and $J(h(u))\le J(u)$ for all $u\in D(J)$}\}.
\end{align*}
Denote Krasnoselskii's genus of $A\in\Sigma$ by $\gamma(A)$ \cite{struwe} and set
\[
i^*(A) := \min_{h\in\mathcal{H}} \gamma(h(A)\cap S_\rho(0)),
\]
where $\rho$ is as in Lemma \ref{mp}.

\begin{lemma} \label{index}
Let $A,B\in\Sigma$.\\
(i) If $A\subset B$, then $i^*(A)\le i^*(B)$. \\
(ii) $i^*(A\cup B)\le i^*(A)+\gamma(B)$. \\
(iii) If $g\in \mathcal{H}$, then $i^*(A)\le i^*(g(A))$. \\
(iv) Let $E_k$ be a $k$-dimensional subspace of $D(J)$. Then $i^*(E_k\cap\overline B_R(0))\ge k$ whenever $R$ is large enough.
\end{lemma}

\begin{proof}
(i) follows immediately from the properties of genus \cite{struwe}. \\
(ii) For each $h\in\mathcal{H}$,
\[
i^*(A\cup B) \le \gamma(h(A\cup B)\cap S_\rho(0)) = \gamma ((h(A)\cup h(B))\cap S_\rho(0)) \le \gamma(h(A)\cap S_\rho(0)) + \gamma(B).
\]
Taking the minimum over all $h\in\mathcal{H}$ on the right-hand side we obtain the conclusion. \\
(iii) Since $J(g(u))\le J(u)$ for all $u\in D(J)$, $h\circ g\in\mathcal{H}$ if $h\in\mathcal{H}$. Hence
\[
\min_{h\in\mathcal{H}}\gamma(h(A)\cap S_\rho(0)) \le \min_{h\in\mathcal{H}}\gamma((h\circ g)(A)\cap S_\rho(0)).
\]
(iv) It is easy to see that $J(u)\to-\infty$ uniformly for $u\in E_k$, $\|u\|\to\infty$. So $J(u)<0$ on $E_k\setminus B_R(0)$ if $R$ is large enough. Let $D := E_k\cap\overline B_R(0)$. Suppose $i^*(D)<k$, choose $h\in\mathcal{H}$ such that $\gamma(h(D)\cap S_\rho(0))<k$ and an odd mapping $f: h(D)\cap S_\rho(0) \to \mathbb{R}^{k-1}\setminus\{0\}$. Let $U:= h^{-1}(B_\rho(0))\cap E_k$. Since $J(h(u))\le J(u)< 0$ when $u\in E_k\setminus B_R(0)$, $U\subset D$ according to Lemma \ref{mp} and hence $U$ is an open and bounded neighbourhood of 0 in $E_k$. If $u\in \partial U$, then $h(u)\in S_\rho(0)$ and therefore $f\circ h: \partial U \to \mathbb{R}^{k-1}\setminus\{0\}$, contradicting the Borsuk-Ulam theorem \cite[Proposition II.5.2]{struwe}, \cite[Theorem D.17]{wi}. So $i^*(D)\ge k$.
\end{proof}

\subsection{Proof of Theorem \ref{thm} completed}
Let
\[
d_k := \inf_{i^*(A)\ge k} \sup_{u\in A}J(u).
\]
Since there exist sets of arbitrarily large pseudoindex $i^*$, $d_k$ is well defined for all $k\ge 1$ and it follows from  Lemma \ref{mp} that $d_k\ge b$.

 We shall show that $K_{d_k}\ne\emptyset$ and $d_k<d_{k+1}$ for all $k$, and this contradicts our assumption that there are only finitely many critical orbits.
Put $d\equiv d_k$. By Lemma \ref{kappa}, $\gamma(K_d)=0$ (if $K_d=\emptyset$) or 1. Let $U:=U_\delta(K_d)$ where $\delta$ is so small that $\gamma(\overline U)=\gamma(K_d)$ and choose $\varepsilon>0$ as in Lemma \ref{critical}. Choose $A\in\Sigma$ such that $i^*(A)\ge k$ and $\sup_{u\in A}J(u)\le d+\varepsilon$.
We need to modify the flow $\eta$. Let $\chi_1: E\to [0,1]$ be locally Lipschitz continuous and
such that $\chi_1=0$ on $J^{d-2\varepsilon_0}$, $\chi_1>0$ otherwise. Since $\{u\in E: J(u)>d+2\varepsilon_0\}$ is an open set, there exists a locally Lipschitz continuous function $\chi_2: E\to [0,1]$ such that $\chi_2=1$ on $J^{d+2\varepsilon_0}\setminus U_{\delta/2}(K_d)$ and $\chi_2 =0$ in a neighbourhood of $K\cap J_{d-2\varepsilon_0}$.  Put $\chi(u)=\chi_1(u)\chi_2(u)$. The flow $\widetilde\eta$ given by
\begin{equation*}
\left\{
\begin{array}{l}
\frac{d}{dt} \widetilde\eta(t,u)=-\chi(\widetilde\eta(t,u))H(\widetilde\eta(t,u)), \\
\widetilde\eta(0,u) = u, \ u\in D(J)
\end{array}
\right.
\end{equation*}
is defined for all $t>0$ an has the same flow lines on $J_{d-2\varepsilon_0}^{d+2\varepsilon_0}\setminus U_{\delta/2}(K_d)$ as $\eta$.
Choose $T$ so that $J(\widetilde\eta(T,u))<d-\varepsilon$ for all $u\in A\setminus U$. Such $T$ exists because $A$ is compact and $\lim_{t\to T^+(u)}J(\eta(t,u))\le d-2\varepsilon_0$. The properties of pseudoindex give
\[
k\le i^*(A) \le i^*(A\setminus U)+\gamma(\overline U) \le i^*(\widetilde\eta(T,A\setminus U))+\gamma(\overline U)\le k-1+\gamma(K_d).
\]
So $\gamma(K_d)\ne 0$ and hence $K_d\ne\emptyset$. If $d\equiv d_k=d_{k+1}$, then $i^*(A)\ge k+1$ and
therefore $\gamma(K_{d})\ge 2$ which is impossible. So $d_k<d_{k+1}$ for all $k$.  \qed
\smallskip

\section{Proof of Theorem \ref{thm0}}

\noindent
Let $u\in D(J)\setminus\{0\}$. Then the map
$s\mapsto J(su)$ admits a unique maximum point on $(0,\infty)$. In fact, if $\varphi:(0,\infty)\to\mathbb{R}$
is the map defined by
$$
\varphi(s):=J(su)=J(u) s^2-s^2\log s\int_{\mathbb{R}^N}Q(x)u^2dx,\quad\,\, s>0,
$$
we have $\varphi(s)>0$ for $s>0$ sufficiently small and
$\varphi(s)<0$ for all $s>0$ large enough. Moreover,
$\varphi'(s)=0$ with $s>0$ if and only if
$$
J(u)=\frac{2 \log s+1}{2} \int_{\mathbb{R}^N}Q(x)u^2dx,
$$
which proves the claim. Since $\varphi'(s)=\langle J'(su),u\rangle$,
the ray $\{su: s>0\}$ intersects the Nehari
manifold $\mathcal{N}$ (see definition \eqref{neh}) at exactly one point. Moreover, there exists $s_0>0$ such that
for all $u\in D(J)\cap S_1(0)$, $s\mapsto \Phi(su)$ is increasing when $0<s<s_0$. Since $s\mapsto \Psi(su)$ is increasing for all $s>0$ (by convexity),
$\mathcal{N}$ is bounded away from $0$.\
Alternatively one can observe that, if $u\in \mathcal{N}$, then inequality \eqref{logQ} yields
$$
\int_{\rn}\left(|\nabla u|^2+V(x)u^2\right)dx\le \frac12\|\nabla u\|_2^2 + C_2(\log\|u\|_2^2+1)\|u\|_2^2.
$$
Then, if $\|u\|$ is so small that $C_2(\log\|u\|_2^2+1)\leq \inf V$ one gets the contradiction $u=0$. Let
\[
\Gamma := \{\alpha\in C([0,1],E): \alpha(0)=0,\ J(\alpha(1))<0\}
\]
and
\[
c := \inf_{\alpha\in\Gamma} \sup_{s\in[0,1]} J(\alpha(s)).
\]
By Lemma \ref{mp}, $c\ge b>0$ (and $c$ is the Mountain Pass level). Clearly, $c \le c_{\mathcal{N}} := \inf_{\mathcal{N}} J$. Assume that, for some $\varepsilon_0>0$, there exists no nontrivial solution
 below the energy level $c+\varepsilon_0$. According to Remark \ref{rem}, we can use Lemma \ref{critical} with $U_\delta(K_c)=\emptyset$ and a
 sufficiently small $\varepsilon<\varepsilon_0$. Let $\chi: E\to [0,1]$ be a locally Lipschitz continuous function such that $\chi=0$ on $J^{c/2}$, $\chi>0$ otherwise, and consider the flow
 \begin{equation*}
\left\{
\begin{array}{l}
\frac{d}{dt} \widehat\eta(t,u)=-\chi(\widehat\eta(t,u))H(\widehat\eta(t,u)), \\
\widehat\eta(0,u) = u, \ u\in J^{c+\varepsilon}.
\end{array}
\right.
\end{equation*}
Choosing $\alpha\in \Gamma$ such that $\sup_{s\in[0,1]}J(\alpha(s))\le c+\varepsilon$ and setting $\beta(s) := \widehat\eta(T,\alpha(s))$, where $T$ is large enough,
we obtain $\sup_{s\in[0,1]}J(\beta(s)) < c$ which is a contradiction because $\beta\in\Gamma$. Hence there exists a sequence of nontrivial solutions $(u_n)$
with $\limsup_{n\to\infty}J(u_n)\le c$ (we do not exclude the possibility that $u_n=u$ for all $n$ and some $u$). Since $u_n\in\mathcal{N}$ and $c\le c_{\mathcal{N}}$,
it follows that $c=c_{\mathcal N}$ and thus $J(u_n)\to c$. Obviously, $(u_n)$ is a Palais-Smale sequence, hence it is bounded according to Lemma \ref{bounded} and we may assume that $u_n\rightharpoonup u$ in $E$ as $n\to\infty$.
As we have seen earlier, $u$ is a solution for \eqref{eq}. If $\|u_n\|_p\to 0$ for some $p\in(2,2^*)$, then
\begin{equation*}
 0=\langle J'(u_n),u_n\rangle \geq \|u_n\|^2-C\int_{\{|u_n|\geq 1\}}  |u_n|^{p}\,dx,
 \end{equation*}
 yielding $\|u_n\|\to 0$ as $n\to\infty,$ contrary to the fact that $(u_n)\subset {\mathcal N}$. Hence, by means of
 Lions' lemma \cite[Lemma I.1]{lions}, \cite[Lemma 1.21]{wi}, we have
$$
\int_{B_1(y_n)}u_n^2\,dx \ge \varepsilon,
$$
for some sequence $(y_n)\subset\mathbb{R}^N$ and some $\varepsilon>0$. As in the proof of Lemma \ref{discreteness} we may assume,
making translations if necessary, that $(y_n)$ is bounded. So for the (translated) sequence $(u_n)$ we have $u_n\rightharpoonup u\ne 0$ as $n\to\infty$.
Notice that $J(u)=\inf_{{\mathcal N}} J$. In fact, $J(u)\leq c$ by the same argument as in Remark \ref{rem} and obviously, $J(u)\ge c$.
Hence $u$ is a ground state solution.

Finally, the solution $u$ has constant sign. In fact, let us write $u = u^+ - u^-$. Then $J(u) = J(u^+)+J(u^-)$ and $u^+, u^-\in D(J)$.
Moreover, $0 = \langle J'(u),u^+\rangle = \langle J'(u^+),u^+\rangle $, so either $u^+\in {\mathcal N}$ or $u^+=0$.
A similar conclusion holds for $u^-$. Hence, either one of the functions $u^+$, $u^-$ is equal to $0$ or
$J(u) \ge 2c$, which yields a contradiction. Suppose $u=u^+$. Then, by a slight variant of the argument
in \cite[Section 3.1]{dams} it follows from the maximum principle (see \cite[Theorem 1]{va}) that $u(x)>0$, for a.e.\ $x\in\mathbb{R}^N$.   \qed

\bigskip

\section{A note on the $p$-Laplacian}

\noindent
Our arguments also allow to prove Theorems \ref{thm} and \ref{thm0} for the equation
\begin{equation} \label{plapl}
-\Delta_pu + V(x) |u|^{p-2}u = Q(x)|u|^{p-2}u\log |u|^p, \quad u\in W^{1,p}(\mathbb{R}^N),
\end{equation}
where $\Delta_pu:=\text{div}(|\nabla u|^{p-2}\nabla u)$ is the $p$-Laplacian ($1<p<N$) and $V,Q$ satisfy the conditions stated at the
beginning of Section \ref{intro}. Here one needs to make use of a $p$-logarithmic Sobolev inequality, see e.g.\ \cite{dD} and the references therein. The functional corresponding to \eqref{plapl} is
\[
J(u) = \frac1p\int_{\rn}(|\nabla u|^p+(V(x)+Q(x))|u|^p)\,dx - \frac1p\int_{\rn} Q(x)|u|^p\log|u|^p\,dx, \quad u\in W^{1,p}(\mathbb{R}^N).
\]
In order to show boundedness of the sequence $(u_n)$ with $J(u_n)$ bounded above and $J'(u_n)\to 0$, one needs to use
\cite[formula (3)]{dD} with $u$ replaced by $Q^{1/p}u$ and modify the proof of Lemma \ref{bounded} in a suitable way. We omit the easy but somewhat tedious details.

\bigskip
\bigskip

redpajama_set_name: RedPajamaArXiv


text:
Coronavirus worries prompt San Antonio employers to ban travel, work remotely and even close
H-E-B employees say new company policy makes them work sick
Business // Local Business
Madison Iszler, Staff writer
March 12, 2020 Updated: March 12, 2020 6:49 p.m.
USAA, which employs about 19,000 people in San Antonio, tried having 10 percent of its non-member contact staff work from home Monday.
San Antonio companies are sending employees home, conducting virtual meetings, restricting travel, deep-cleaning offices and, in some cases, temporarily closing their doors.
As coronavirus ricochets around the globe, employers are quickly embracing measures to keep the virus out of the workplace — and to reassure their workers and customers.
Coronavirus live updates: A timeline of COVID-19 in San Antonio
Coronavirus wreckage affects sports at every level
It's a "moment of inflection," said Kai Andrews, principal consultant at Point B, a Seattle management consulting and real estate development firm with an office in San Antonio.
Companies have talked for years about having employees work from home to cut costs and increase productivity. Many firms have resisted telecommuting, but the virus is forcing more of them to offer the option.
That means providing workers with the training and hardware to work remotely, Andrews said, and "getting people mentally prepared."
"Work will have to shift very quickly, but at the same time, (it has) to continue," Andrews said.
USAA, which employs about 19,000 people in San Antonio, directed some of its staff to work from home Monday — to gauge the effect telecommuting would have on its operations. Several days later, the company announced at least 25 percent of its employees, the ones who aren't in contact with customers, must work from home beginning next week.
"Employee health and safety, and our ability to continue serving members remain top priorities," said spokesman Matt Hartwig.
On ExpressNews.com: At some San Antonio companies, threat of coronavirus means no hugging or handshakes
USAA is taking additional steps, including banning international business travel, asking employees returning from international trips or cruises to work from home for 14 days and limiting participation in most outside conferences and training events.
Another local employer, technology company Rackspace, is also experimenting with work-from-home days, spokeswoman Michelle Doss said.
"This strategy will allow us to test and validate our tools to their full capacity," she said.
Small retailers generally don't have that luxury. They're also worried about customers staying away as the virus spreads. That's what's happening at La Villita Historic Arts Village downtown, said Deborah Sibley, who owns Capistrano Soap Co., which has a shop there.
Most of the village's visitors are in town for conventions and events, some of which have been canceled because of the coronavirus. More than 90 percent of customers at Sibley's shop are from another city, state or country.
"There's already an impact to traffic and revenue," she said.
Employees' exposure to the virus is another worry. Signs asking customers not to open bottles for sniffing or to use pumps and applicator sticks for sampling have been posted around the shop, and workers are wiping down surfaces.
"But that does not socially distance us," Sibley said. Starting Monday, she is temporarily closing the shop.
On ExpressNews.com: San Antonio loses conventions to coronavirus worries
H-E-B is suspending all in-store sampling and putting hand sanitizer at checkout stations for cashiers and customers, said spokeswoman Dya Campos.
Customers picking up orders or having items delivered won't be asked to sign a device to acknowledge the orders, unless they're buying alcohol, and can select a "leave at the door" option for delivery. About 204 of the company's pharmacies deliver, and the service is being offered for free under normal drop-off terms, Campos said.
As for non-store workers — such as headquarters employees — Campos said, "We are always prepared for remote working."
Employees at Valero Energy Corp. aren't being sent home, but the company has remote work systems in place that can be scaled up if necessary, said spokeswoman Lillian Riojas. Also, Valero's on-site health clinic is providing services for workers and their families.
"We are reviewing available information and facts with our medical director and evaluating precautionary actions, including potential work contingency plans similar to what we might implement during hurricanes," she said.
Limiting interactions by canceling events or putting more space between workers is another tack employers are taking.
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Independent Rough Terrain Center, a Cibolo business that produces industrial vehicles, hosted a barbecue lunch for employees Thursday. Managers provided hand sanitizer, and asked those waiting in line to keep their distance from each other and eat at their workplace instead of in a hall together.
That "disperses them across our campus," president and CEO Stephen Speakes said. IRTC is also banning travel and encouraging employees who feel they're at risk to work from home.
"Our first priority is to take care of our employees and their families," he said. "We want to ensure that we are mitigating risk wherever possible."
Having employees work remotely poses few challenges for Jungle Disk, a local cybersecurity business with a staff of about 25.
The company ran tests to make sure everyone can work from home, and its video conferencing and phone systems enable handling calls and meetings virtually, said CEO Bret Piatt. Some employees have elderly parents with health conditions, and Jungle Disk is giving them the option to work from home.
"This is one of the benefits we have in the tech industry," Piatt said. "As the situation continues to evolve, I suspect we'll see more than half of our employees work from home."
It's too early to tell what the long-term effects will be, but "it's inevitable that some companies are going to have to change business models," consultant Kai Andrews said.
madison.iszler@express-news.net
Lifestyle Brackenridge Park's low-water crossing now closed
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Lifestyle Austin Psych Fest 2023 announces lineup
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Home & Garden Texas turf: Lawn grasses compete with bluebonnets
Madison Iszler
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Madison Iszler covers real estate, retail, economic development, and other business topics for the San Antonio Express-News.
Reach Madison at 210-250-3242, madison.iszler@express-news.net and @madisoniszler.

redpajama_set_name: RedPajamaCommonCrawl


text:
Минский курьер — общественно-политическая, официальная минская газета (Белоруссия). Выходит три раза в неделю: вторник (16 стр.), среда (24 стр.), пятница (32 стр.).

Учредитель — Минский городской исполнительный комитет.

История
27 января 1908 года — первый номер (редактор — Е. А. Волина, издатель — В. И. Чаусов).

Первоначальная аудитория: местная интеллигенция. Держала читателей в курсе всех значительных событий в России и в мире, помещала остроумные отчеты о заседаниях Государственной думы, сообщения Петербургского телеграфного агентства (под рубрикой «Телеграммы»), знакомила с новинками русской и белорусской литературы, а также городскими новостями.

Формат: ежедневная, на четырёх полосах, русскоязычная.

Подписная цена на год: четыре с половиной рубля (совсем недешево, так, например, лошадь стоила около 45 рублей).

Проблем с распространением газеты не было, расходилась по подписке, продавалась разносчиками. В начале XX века всего вышло 148 номеров газеты (последний — 22 августа 1908 года).

Минск, 1919—1920 гг.
Время, которое образно можно охарактеризовать как demi-saison — «полу-время» или безвременье. В эти годы в Минске издавались газеты «Звон», «Беларусь», «Минскій курьер» (формат — ежедневная беспартийная газета, пыталась внешне продолжать традиции русскоязычной демократической печати, редактор — Александр Гзовский). По информационной насыщенности, репортерскому профессионализму «Минский курьер» — впереди всех в Минске (зеркало «демисезонной» эпохи).

2001 год
«Минский курьер» восстановлен в апреле 2001 года.

Подшивка старых номеров хранится в Национальном архиве Республики Беларусь.

2011 год
С 2011 года — в составе единого медиахолдинга, куда входят газеты «Вечерний Минск» и «Минский курьер», а также: информационный портал www.minsknews.by, УКВ-радиостанция «Радио-Минск» (92,4 — Минск, 98,1 — Могилёв, 100,4 — Брест, 105,6 — Гомель, 106,4 — Витебск, сайт www.radiominsk.by), ежемесячный журнал для детей и подростков «Качели». Все газеты медиахолдинга можно читать в электронном виде и оформить подписку.

Структура

Администрация
 Генеральный директор: Андрей Кривошеев
 Первый заместитель ген. директора: Андрей Басс
 Заместители: Дмитрий Партон, Наталья Вавилова, Татьяна Буйко, Кирилл Казаков
 Главный редактор: Кирилл Казаков

Отделы
 Отдел экономики
 Отдел городского хозяйства.
 Отдел общественных коммуникаций и торговли.
 Отдел социальной политики.
 Отдел медицины, физкультуры и туризма.
 Отдел образования и науки.
 Отдел культуры и истории.
 Отдел аналитической работы и обращений граждан.

Обратная связь с читателем
Краткие сообщения, свои пожелания и просто мысли вслух, поступающие на мобильный телефон редакции, дважды в неделю появляются в рубрике «Народные SMS-ки».

Имеется возможность задать вопрос на горячую линию.

Газета периодически проводит прямые линии, во время которых на вопросы читателей отвечают специалисты, руководители городских служб, представители районных администраций.

При этом разговор со специалистом попадает не только на страницы газеты, но и в эфир станции «Радио-Минск».

Примечания

См. также
 Официальный сайт газеты
 Средства массовой информации Белоруссии

Газеты Белоруссии
Первомайский район (Минск)

redpajama_set_name: RedPajamaWikipedia


Meet Your Neighbors with Front Yard Fridays!
Just how well do you know your neighbors? Is it an occasional hello when getting the mail or a cordial wave as you're coming and going? Or is it a "let's compare our yards conversation"? Perhaps you are close with your neighbors and have put a gate in-between your fences?
I can honestly say that I feel close to my immediate neighbors, I can even call them friends. Our kids are best friends with our next door neighbors' children and we often spend a Friday or Saturday night having laughs on the patio. We're also close to the neighbors on our other side, always willing to help each other or share a beer and talk about the neighborhood. We'll help them in a pinch and they'll help us. My kids love them and their dog, Duke — seriously, the best dog in the world.
But as far as everyone else goes, it's just a simple hi or wave, a friendly hello or check-in on the kids and their various activities if we happen to both be outside at the same time. No one is trying to be rude, it's just life. So what are we to do if we want to know our neighbors better?
There's a new trend spreading across the nation and now in The Meadows. Front Yard Fridays' goal is to promote close-knit communities starting with your neighbors. It's easy enough to participate and you will get to know your neighbors better, you may even make new friends.
Seriously, what happens on Friday at your house?
You're probably eating pizza or having a barbeque. At least that's every Friday at my house. We love our pizza nights. Well for Front Yard Fridays, all you have to do is take your pizza night or BBQ to the front yard and enjoy it with neighbors.
The Meadows is a great community, full of great neighbors from all over the country, even the world. We have plenty of events throughout the year. But why not take matters into your own hands?
Unwind and share your stories. Heck, even make it a weekly party! It'll be a great way to learn and meet new friends. I've recently learned that one of my neighbors is originally from Germany – pretty cool.
Does Your Block Know How to Party?
Why not kick off your Front Yard Friday with a bang? Turn your gathering into a block party. There's no better time or way to wrap up the summer. Share your summer stories, learn about a new camping spot or family vacation spot. You may find you have hobbies in common with your neighbors that you can try doing together next summer!
To help encourage your Front Yard Fridays, The Grange is even sweetening the pot for your first party. Simply submit your block party ideas/themes and if yours is the best, you'll win a $250 Safeway gift card to help buy food for your bash.
Who doesn't love a good party? Start thinking and email your ideas to Jen at The Grange, jreinhardt@meadowslink.com.


Lionel Messi opened the scoring with a brilliant free kick and laid on two goals as Argentina put their World Cup qualifying campaign back on track with a resounding 3-0 win over Colombia on Tuesday night.
The three points lifted Messi's team into fifth place in the South American group with 19 points, one behind Ecuador and Chile.
Brazil, away to Peru in Lima in a later match, top the table with 24 points while Uruguay are second with 23 after a 3-1 loss to Chile in Santiago. Colombia dropped to sixth on 18 points.
The top four go through to the 2018 finals in Russia with the fifth, at present Argentina, qualifying for an intercontinental playoff for another berth.
Argentina took a ninth-minute lead when Messi curled his free kick over the wall and into the top left corner of the leaping David Ospina's net off the underside of the bar.
Falcao came close for Colombia at the other end when he headed James Rodriguez's free kick just over Sergio Romero's bar after 18 minutes.
Five minutes later Messi's superb ball over the defence was met on the edge of the six-yard box with a firm header into the bottom far corner by Lucas Pratto.
Colombia pushed Argentina back in the second half though the home side threatened again when Messi fed Angel Di Maria but his shot from a tight angle hit the post.
Di Maria did score seven minutes from time after Messi cheekily stole the ball from a defender on the right wing, raced into the box and laid the ball square where the winger blasted it into the roof of the net.
Rodriguez hit the post with a free kick in stoppage time with Colombia's last chance of the match.
Meanwhile, Messi has led the Argentina squad in a media boycott following "grave" press reports about the team and their recent form. The 29-year-old announced the ban after scoring in the side's 3-0 victory over Colombia.
A radio report alleged that forward Ezequiel Lavezzi smoked marijuana after a training session, which he denies.
"We've received many accusations, a lot of lack of respect and we never said anything," Messi said.
Messi and the 25-man squad walked out of the news conference following the Colombia game.
It was also announced that Lavezzi is taking the radio reporter to court over the accusations.
"We know there are lots of you who are not in the game of not showing us respect, but getting into one's personal life is very grave," Messi added.
Argentina were criticised following their 3-0 defeat by Brazil on Friday but their win over Colombia moved them up to fifth place and back into qualifying contention.
Elsewhere in the group, Arsenal forward Alexis Sanchez scored twice as Chile beat Uruguay 3-1 to stay third.
Chile had been warned by Arsene Wenger that it would be "suicidal" to play Sanchez unless he was fully fit after he injured a muscle in training.


Viva la Foto is an open air photo booth company based out of Redding serving the greater North State of California. We custom tailor our services to your event. Transform your next event into a vibrant photo shoot with our photo booth rental!


Sharlet \sh(ar)-let\ as a girl's name is a variant of Charlotte (Old German), and the meaning of Sharlet is "free man".
The baby name Sharlet sounds like Sharlot, Sharlett and Charlet. Other similar baby names are Sharlee, Sharlene, Sharleen, Sharlena, Sharla, Charlee, Sharlie, Charley, Charlea, Sharlott, Sharlette, Sharel, Charlot, Charlett, Charle, Sharde, Shardei, Sharlyn, Sharly, Sharletta, Sharlan, Shirley, Shurlee, Shirlee, Shirlea and Sherlee.


text:
Q: JavaScript - why does this code log 'undefined' but show up in Chrome Console? (closures) I made this example:
(function(_stack_){
  var stack = {
    over: 'flow',
    run: function(){
      console.log(this);
    }
  };
  _stack_.app = stack;
  stack.run();
})(this);

app.foo = 'bar';

Output in Chrome console:

But how come the output of this:
(function(_stack_){
  var stack = {
    over: 'flow',
    run: function(){
      console.log(this.foo); // undefined
    }
  };
  _stack_.app = stack;
  stack.run();
})(this);

app.foo = 'bar';

Equals undefined, while it shows up in the Chrome console?
Check out the JS Bin
EDIT:
I understand that it becomes undefined because stack.run() is called before app.foo is assigned. The question is how come it shows up in the console?

A: this.foo isn't available at the time you run it
This is poor design, when using javascript like this you probably have more use of a custom constructor function and eventhandling. Your code and question suggests that the run method should be executed at the time you update the dynamically created variable app. That will never happened unless you implement some eventhandling for this.
Your code does however work if you call app.run(); after you set the app.foo variable.
// Antipattern ahead!
(function(_stack_){
  var stack = {
    over: 'flow',
    run: function(){
      // Prevent error messages
      if(typeof this.foo === 'undefined') {
          return;
      }
      console.log("log: " + this.foo); // undefined
    }
  };
  _stack_.app = stack;
})(this);

app.foo = 'bar';
app.run(); // log: bar

A: The reason is because stack.run() is executed before the creation of the foo property on the app object.
By the time you're looking in the debugger, the foo property has been created.

redpajama_set_name: RedPajamaStackExchange


Covenant's Connect activities are "inward focused", where we minister to individuals and families within Covenant's church body. Connect is geared to encourage each other through fellowship, connect teams, and membership.
Flocks. Covenant's body is distributed among groups called flocks, each ministered by a Connect Deacon.
Youth Group Rocks. Covenant's youth group is a dynamic group of young people that are excited about the church and meeting new people. At Covenant, the youth are encouraged to have fun, be themselves, and build relationships with other youth while growing closer to God. Fun events (Six Flags, Paintball, Bowling, Miniature Golf, etc.). Growing in Christ (30 Hour Famine, Fun In The Son, The Great Escape, Fall Retreat to Concord WV, Mission work, Leadership Group, etc.). High School Youth Group: Meets at Covenant on Wednesday nights from 8:00p.m. until 9:00p.m. Every week we relax, have fun, and chat about different topics. Middle School Youth Group: Meets at Covenant every week on Wednesdays from 6:30 p.m.to 7:30 p.m.
Other Connect Groups include Presbyterian Women, Book Club, Wednesday Walkers, and Knitting Club.


text:
The Allman Brothers Band is an American rock/blues band once based in Macon, Georgia. The band was formed in Jacksonville, Florida, in 1969 by brothersDuane Allman (slide guitar and lead guitar) and Gregg Allman (vocals, organ, songwriting), plus Dickey Betts (lead guitar, vocals, songwriting), Berry Oakley (bass guitar), Butch Trucks (drums), and Jai Johanny "Jaimoe" Johanson (drums).[2] While the band has been called the principal architects of Southern rock,[3] they also incorporate elements of blues, jazz, and country music, and their live shows have jam band-style improvisation and instrumentals.
The band achieved its artistic and commercial breakthrough in 1971 with the release of At Fillmore East, featuring extended renderings of their songs "In Memory of Elizabeth Reed" and "Whipping Post" and often considered one of the best live albums ever made.[4] George Kimball of Rolling Stone magazine hailed them as "the best damn rock and roll band this country has produced in the past five years."[5] A few months later, group leader Duane Allman was killed in a motorcycle accident. The group survived that and the death of bassist Oakley in another motorcycle accident a year later; with replacement members Chuck Leavell and Lamar Williams, the Allman Brothers Band achieved its peak commercial success in 1973 with the album Brothers and Sisters and the hit single "Ramblin' Man". Internal turmoil overtook the band soon after; the group dissolved in 1976, reformed briefly at the end of the decade with additional personnel changes, and dissolved again in 1982.
In 1989, the group reformed with some new members and has been recording and touring since. A series of personnel changes in the late 1990s was capped by the departure of Betts. The group found stability during the 2000s with bassist Oteil Burbridge and guitarists Warren Haynes and Derek Trucks (the nephew of their drummer), and became renowned for their month-long string of shows in New York City each spring. The band has been awarded eleven gold and five platinum albums between 1971 and 2005[6] and was inducted into the Rock and Roll Hall of Fame in 1995. Rolling Stone ranked them 52nd on their list of the 100 Greatest Artists of All Time in 2004.[7]
[hide] *1 Band history
1.2 Loss and triumph
1.3 Turmoil and dissolution
1.4 Revival
1.5 2000s onward
4.2 Live personnel
Band history[edit] Edit
Beginnings[edit] Edit
The group was formed in Jacksonville, FL, and consisted of Duane and Gregg Allman, Dickey Betts, Berry Oakley, Butch Trucks, and Jai Johanny Johanson.[2]Brothers Duane and Gregg Allman grew up in Daytona Beach, Florida, and had been playing music publicly since the early 1960s. They formed a garage bandcalled the Escorts in 1963, which evolved into the Allman Joys in 1965.[8] From there the brothers formed Hour Glass and moved to Los Angeles. The Hour Glass released two failed albums on Liberty Records in 1967 and 1968.[2][8] They were all released from the contract except Gregg, who Liberty thought might have some commercial potential.[8] Gregg and Duane had previously met Butch Trucks and his band The 31st of February while touring as the Allman Joys, and decided to record an album with them in September 1968, shortly after the breakup of Hour Glass. This album was eventually released as Duane & Greg Allman on the Bold Records label in 1972. Duane Allman played on Wilson Pickett's hit version of "Hey Jude" and became the primary session guitarist for FAME Studios in Muscle Shoals, Alabama, recording with Aretha Franklin, King Curtis, Percy Sledge, and others.[8] Allman started jamming with Dickey Betts, Butch Trucks and Berry Oakley in Jacksonville. Eddie Hinton, with whom Duane Allman had played in Muscle Shoals, was considered to play guitar, but Hinton refused in order to join the Muscle Shoals studio band.[citation needed] Duane brought in Jaimoe, a drummer he had played with in the past.
[1][2]Jaimoe, percussionist
Gregg was in Los Angeles, fulfilling the Hour Glass contract with Liberty Records. He was summoned back to Jacksonville.
[3][4]Duane Allman playing guitar at the Fillmore East, June 26, 1971 (late show)
The Allman Brothers Band played numerous shows in the South before releasing their debut album, The Allman Brothers Band in 1969 to critical acclaim. It featured future jam standards "Whipping Post" and a 12/8 time slide guitar tour de force "Dreams". A cult following began to build.
Idlewild South was released in 1970 to critical success and improved sales. Produced by Tom Dowd it featured the upbeat "Revival" and the moody-but-resolute "Midnight Rider". After completing the Idlewild South sessions Duane Allman joined Eric Clapton and his ad hoc Derek and the Dominos to record the classic Layla and Other Assorted Love Songs.
The Allman Brothers' Cincinnati, Ohio concerts on April 4 & 5, 1970 were recorded and later released (in 1991) as Live at Ludlow Garage. This album features a 44-minute version of "Mountain Jam," which sounds different compared to the version of this song recorded at the Fillmore East one year later.
1971 saw the release of a live album, At Fillmore East, recorded on Friday and Saturday March 12 and March 13 of that year at the legendary rock venue the Fillmore East. The album was another huge hit. Rolling Stone listed At Fillmore East as number 49 on of The 500 Greatest Albums of All Time.[9] It showcased the band's mix of jazz, classical music,hard rock, and blues, with arrangements propelled by Duane's and Betts' dual lead guitars, Oakley's long, melodic "third guitar" bass runs, the rhythm section's pervasively percussive yet dynamically flexible foundation, and Gregg Allman's gritty Ray Charles-like vocals and piano/organ play which all completed the band's wall of sound. The rendering of Blind Willie McTell's "Statesboro Blues" was a straight-ahead opener, the powerful "Whipping Post" (with its famous 11/4 bass opening) became the standard for an epic jam that never lost interest, while the ethereal-to-furious "In Memory of Elizabeth Reed" invited comparisons to John Coltrane and Miles Davis.[10][11]
The Allman Brothers were the last act to play the Fillmore East before it closed in June 1971. The final shows achieved legendary status, partly due to bands literally playing all night; in 2005 Gregg Allman would relate how the jamming musicians lost track of time, not realizing it was dawn until the side doors of the Fillmore were opened and the morning light poured in.
The band continued to tour; decades later, a special-order recording of one of their final concerts in this lineup, S.U.N.Y. at Stonybrook: Stonybrook, NY 9/19/71, would be released.[12] It reveals that Duane Allman's slide guitar playing on "Dreams" and other songs was touching the farthest reaches of both that instrument and his imagination.[12]
Loss and triumph[edit] Edit
[5][6]Dickey Betts on guitar
Duane Allman died not long after the Fillmore East album was certified gold, killed in a motorcycle accident on October 29, 1971 in Macon, Georgia, The popular version of the accident was that he collided with the rear of a flatbed truck that had turned in front of him. However, it has been revealed in recent years that the truck had turned before Allman was in view due to the road being on a slope. When Duane topped the crest of the hill it can be only assumed that he simply was going to swerve behind the truck. It is well known that Duane always broke speed limits on his motorcycle[citation needed] and on this day it would cost him his life. As he neared the truck and started the swerve the road dipped sharply and as he rode up out of the dip the motorcycle became airborne causing Duane to fly off the bike and land on the pavement. The motorcycle then landed on him with both sliding into a curb. Duane was taken to a hospital but never regained consciousness, dying from massive internal injuries. The group decided to carry on. The album continued to gain FM radio airplay, with stations even playing 13-minute and 23-minute selections.
Dickey Betts filled Duane's former role in completing the last album Duane participated in, Eat a Peach, released in February 1972. The album was often softer ("Blue Sky", "Little Martha") and wistful in tone ("Melissa", "Ain't Wastin' Time No More"), capped by the 34-minute "Mountain Jam" reverie taken from the Fillmore East concerts. Writer Greil Marcusdescribed parts of Eat a Peach as an "after-the-rain celebration... ageless, seamless... front-porch music stolen from the utopia of shared southern memory."
The group played some concerts as a five-man band, then decided to add Chuck Leavell, a pianist, to gain another lead instrument but without, however, directly replacing Duane. This new configuration debuted on November 2, 1972, on ABC's In Concert late-night television program.
Days later, on November 11, 1972, Berry Oakley died from head injuries he received in another motorcycle accident near Napier Avenue and Inverness Street, only three blocks from the site of Duane's accident the previous year. The common retelling that it was at exactly the same site as Duane's death is incorrect, as is the legend that the Eat a Peachalbum is named for what was being carried by the truck involved in Allman's accident.[13] Both Duane and Berry were 24 years old, yet another coincidence.
Oakley was replaced by Lamar Williams at the end of 1972, in time to finish the next album, Brothers and Sisters, released in August 1973.
Dickey Betts was becoming the group's unofficial leader.[8] Brothers and Sisters included the group's best known hits, "Ramblin' Man" and "Jessica", both written by Betts; the former reached No. 2 on the Billboard Hot 100 as a single, while the latter was a seven-minute instrumental hit.
The Allman Brothers Band had become one of the top concert draws in the country. Probably their most celebrated performance of the era took place on July 28, 1973 at the Summer Jam at Watkins Glen outsideWatkins Glen, New York, in a joint appearance with The Grateful Dead and The Band. An estimated 600,000 people made it to the racetrack where this massive outdoor festival took place.
In the wake of the Allman Brothers Band's success, many other Southern rock groups rose to prominence, including the Marshall Tucker Band (who played as the Allman Brothers Band's opening act for many shows on their 1973 tour) and Lynyrd Skynyrd.
Despite this success, Butch Trucks recalls of the time that, "celebrity is the worse thing that ever happened to the Allman Brothers." In an interview, Trucks recalled that after Allman's death in 1971, the band "got away from the music," producing "country-fried hit records," which created egos that "ripped [them] all apart." [14]
Another peak of the Allmans' success came on New Year's Eve, 1973, when promoter Bill Graham arranged for a nationwide radio broadcast of their concert from San Francisco's Cow Palace. New arrangements of familiar tunes such as "You Don't Love Me" went out over the airwaves, as the show stretched out over three sets, with Boz Scaggs sitting in, along with Grateful Dead members Jerry Garcia and Bill Kreutzmann(Allmans and Grateful Dead members guested at each other's shows multiple times in the early 1970s).
Turmoil and dissolution[edit] Edit
[7][8]Gregg Allman, playing organ and singing
Personality conflicts started to tear the band apart, however. Gregg Allman and Dickey Betts both began solo careers, while Allman married Cher, separated quickly, reconciled, and eventually separated again, all in a storm of publicity; drug abuse took its toll on the entire band. Musically, Betts and Leavell were pulling in opposite directions, with Allman trying to mediate.
The tension resulted in the uneven Win, Lose or Draw (1975), with some members not participating on all tracks or doing so only from afar. The few stand-out tracks included a stop-start take on Muddy Waters' "Can't Lose What You Never Had", Betts' instrumental "High Falls", and Allman's Jackson Browne-influenced title song.
The band managed to limp along until 1976, when Gregg Allman was arrested on federal drug charges and agreed to testify against a friend, tour manager, and bodyguard for the band, John "Scooter" Herring. Leavell, Johanson, and Williams formed Sea Level, while Betts worked on his solo career. All four swore that they would never work with Allman again.
Meanwhile, Capricorn Records released a compilation album, The Road Goes On Forever, and a poorly received live album, Wipe the Windows, Check the Oil, Dollar Gas. Neither sold very well.
The group reformed in 1978 and released the strong Enlightened Rogues (1979). It featured new members Dan Toler (guitar) and David Goldflies (bass), who replaced Leavell and Williams, both of whom concentrated on Sea Level instead. "Crazy Love" was a minor hit single, and the instrumental "Pegasus" got some airplay, but overall The Allman Brothers Band was no longer as popular as before, and financial woes plagued both the group and Capricorn Records, which collapsed in 1979. PolyGram took over the catalogue, and the Allman Brothers Band signed to Arista Records. The group released a pair of critically slammed albums, firing Jaimoe in the process, and then disbanded once again in early 1982.
Allman quickly formed the Gregg Allman Band with the Toler brothers Dan and David ("Frankie") (drums) in 1982 and began touring small venues and clubs. Betts, Leavell, Trucks and Goldflies formed the band Betts Hall Leavell Trucks (BHLT). Neither garnered attention from any record labels. BHLT would dissolve two years later.
The Allman Brothers Band reunited in 1986 for a pair of benefit concerts for promoter Bill Graham in New York and Macon. Allman, Betts, Trucks, Jaimoe, Leavell, and Dan Toler performed together but no subsequent reunion plans for the band were made. The following year, the Gregg Allman Band and the Dickey Betts Band co-headlined a theatre and club tour. After each band played a set of music, Betts, Allman and the Tolers performed a closing set of Allman Brothers music together.
In 1987, Epic Records signed both Allman and Betts to separate solo contracts. The Gregg Allman Band had a surprise FM hit single with the title track to the 1987 album I'm No Angel. Just Before the Bullets Flyquickly followed from Allman in 1988. The Dickey Betts Band, including Warren Haynes, was also formed during this time and released the album Pattern Disruptive in 1988. This series of collaboration among bandmembers and interest from a major label during the late 1980s laid the groundwork for the next era of Allman Brothers Band activity and success.
Revival[edit] Edit
[9][10]Warren Haynes, vocalist and guitarist
In 1989 The Allman Brothers reunited and returned to the popular consciousness of the American public, spurred by Gregg's recent FM radio success, the release of archival material by PolyGram, and the start of regular appearances on the American summer outdoor amphitheatre circuit. Warren Haynes (guitar, vocals), Johnny Neel (keyboards and harmonica), and Allen Woody (bass guitar) joined originals Allman, Betts, Jaimoe and Trucks. Leavell opted to go on tour again with the Rolling Stones, with whom he has been a touring member since 1982.
After the 20th Anniversary tour, the band signed to Epic Records and released Seven Turns (1990), which got excellent reviews. This was followed by Neel's departure and a series of moderately-selling, but critically well-received albums including Shades Of Two Worlds (1991) and Where It All Begins (1994, certified Gold by the RIAA 1998), both featuring new percussionist Marc Quiñones. Warren Haynes and Allen Woody formed their own side project Gov't Mule in 1994. In 1995, the band was inducted into the Rock and Roll Hall of Fame and in 1996 they won the Grammy Award for Best Rock Instrumental Performance for "Jessica". When Haynes and Woody decided to concentrate full-time on Gov't Mule in 1997, Haynes was replaced on guitar by Jack Pearson, while Woody was replaced on bass by Oteil Burbridge. Derek Trucks, nephew of original Brother Butch Trucks, replaced Pearson in 1999.
In 2000, the band forced Dickey Betts out for "personal and professional reasons." For this tour, he was replaced by Jimmy Herring. Betts then filed a lawsuit against the other three original members and the summer separation turned into a permanent divorce. Also in 2000, former bassist Allen Woody was found dead on August 26. The band did release the live CD Peakin' at the Beacon that year which chronicled the now-annual March tradition of a many-night stand at the Beacon Theater in New York City. The band has sold out the 2900-seat Upper West Side Manhattan theatre over 200 times since 1989. The tradition is known as the "Beacon Run" among fans, who travel from across the United States, Canada and Western Europe to see these annual March and April shows.
2000s onward[edit] Edit
[11][12]Derek Trucks playing guitar with The Allman Brothers Band, 2009
Warren Haynes began appearing with the Allmans again in 2000 and rejoined full-time in 2001, while also maintaining his active schedule with Gov't Mule. (Haynes also toured during this time and later in the decade with former members of the Grateful Dead). Haynes' return marked a new period of stability and productivity for the band after nearly four years of lineup shifts. The Michael Barbiero/Warren Haynes-produced Hittin' the Note was released in 2003 to popular and critical acclaim, as was the Live At the Beacon Theatre DVD film (2003, certified Platinum 2004). The live CD One Way Out (2004) also chronicled the Beacon concerts.
The Allman Brothers garnered back to back Grammy Award nominations in 2003 and 2004 in the category of Best Rock Instrumental for performances of "Instrumental Illness" from Hittin' The Note and One Way Out. In 2003, Rolling Stone magazine named Duane Allman, Warren Haynes, Dickey Betts, and Derek Trucks to their list of the 100 Greatest Guitarists of All Time,[15] with Allman coming in at No. 2 and Trucks being the youngest guitarist on their list.
The Warren Haynes and Derek Trucks lineup continued the band's connection with younger music fans via concert pairings with popular jam bands The String Cheese Incident,moe, and Dave Matthews Band among others. The Allman Brothers Band continue to be a major attraction at the Bonnaroo Music Festival since 2003. Since 2005, the Allmans have staged their own two day Wanee Music Festival at the Spirit of the Suwannee Music Park in Live Oak, Florida. The Allmans, Gov't Mule and The Derek Trucks Band perform on different stages along with younger roots artists including the North Mississippi Allstars, Robert Randolph and The Family Band, Medeski, Martin and Wood, Devon Allman's Honeytribe, Nickel Creek, Mofro and others.
Allman Brothers songs have been used in various advertising campaigns and television programs, with the most well-known use being that of "Jessica" used in both the 1977and 2002 formats of the BBC television series Top Gear. "Jessica" was also used in the film Field of Dreams. It was also the opening song of a famous radio show in the State of Bavaria, Germany during the 70s known as "Club 16" (http://peterhammer.de/sound4/br/club16/000000club16outro.html with an example while the show ends. )
[13][14]Oteil Burbridge, bass guitarist, 2005
The band cancelled their Beacon run for 2008 due to Gregg Allman recovering from hepatitis C treatments,[16] but they were back on the road that summer for the amphitheater circuit. On November 20, 2008, The Allman Brothers Band received the Legend Of Live Award at the 2008 Billboard Touring Awards ceremony in New York.[17] The award recognized "a touring professional who has had a significant and lasting impact on the concert industry."[18] At the ceremony, Gregg Allman talked about his brother Duane, saying: "It happens to be today would have been his 62nd birthday and I'm sure he's looking down on this and is really proud of us. We'll keep coming back until we can't come back no more."[17]
The Allman Brothers Band celebrated their 40th anniversary in 2009.[18] That year's Beacon run also marked the 20th anniversary of their appearances in that venue; and the band focused on paying tribute to Duane, inviting guest appearances from those who had played with him.[16] The 15-night stand was considered to be the best Beacon run of all,[19] highlighted by two nights of the first-ever stage appearance of Eric Clapton with the band and performances of numbers from Layla and Other Assorted Love Songs.[19] Guests on other nights included members of the Grateful Dead, Phish, Levon Helm and his band, and represented genres from bluesmanBuddy Guy to jazz great Lenny White to rap-rock-country eccentric Kid Rock.[19] On March 26, three days after the actual anniversary of the band's 1969 forming, what's been known as the "Legendary Jacksonville Jam", the band featured no guests but performed their first two albums in their entirety.[19]
In March 2010, The Allman Brothers Band's New York run changed venues from the Beacon Theater to the far uptown United Palace Theatre, but in March 2011, the Allman Brothers returned to the Beacon Theater, playing their 200th show there on March 26.
In 2012, the Allman Brothers Band was awarded the Grammy Lifetime Achievement Award.
In 2012, in cooperation with Live Nation Entertainment, the Allman Brothers Band started The Peach Music Festival. The festival is held annually at the Toyota Pavilion at Montage Mountain and Montage Mountain Ski Resort in Scranton, Pennsylvania, and takes place over the course of several days in mid-August. The Allman Brothers perform multiple times over the course of the festival, and it has featured other headliners such as Bob Weir & RatDog, Gov't Mule, The Black Crowes, and Zac Brown Band.
Skydog, a seven-CD box set tracing the virtuosity of Duane Allman on the slide guitar was released in 2013 with the help of his daughter, Galadrielle Allman. A March 16 interview with her on NPR's Weekend Edition Saturday by Scott Simon includes many details and is highlighted with clips of his playing.[20]
On January 8, 2014, guitarists Warren Haynes and Derek Trucks announced their intention to leave the band at the end of 2014, reportedly to work in their respective solo projects, Gov't Mule and Tedeschi Trucks Band[21][22] Gregg Allman revealed on January 28, 2014 that, following Haynes' and Trucks' announcement, The Allman Brothers Band are going to break up in 2014.[23][24]
Awards and recognition[edit] Edit
[15][16]Guest appearances during the "Beacon Run" are common. Here Eric Clapton joins the band in March 2009 to play songs from 1970's classic Layla and Other Assorted Love Songs, which had featured Duane Allman.*Grammy Award for Best Rock Instrumental Performance, 1996, "Jessica"
Grammy Lifetime Achievement Award, 2012
Inducted into the Rock and Roll Hall of Fame, 1995
Rolling Stone Magazine's "Greatest...of All Time" lists:
100 Greatest Artists of All Time (2004): No. 52[25]
500 Greatest Albums of All Time (2003): No. 49 for At Fillmore East[26]
100 Greatest Guitarists of All Time (2003):
No. 2 Duane Allman[27]
No. 23 Warren Haynes[28]
No. 58 Dickey Betts[29]
No. 81 Derek Trucks[30]
Main article: The Allman Brothers Band discography*The Allman Brothers Band (1969)
Idlewild South (1970)
At Fillmore East (1971)
Eat a Peach (1972)
Beginnings (1973)
Brothers and Sisters (1973)
Win, Lose or Draw (1975)
Wipe the Windows, Check the Oil, Dollar Gas (1976)
Enlightened Rogues (1979)
Reach for the Sky (1980)
Brothers of the Road (1981)
Seven Turns (1990)
Shades of Two Worlds (1991)
An Evening with the Allman Brothers Band: First Set (1992)
Where It All Begins (1994)
An Evening with the Allman Brothers Band: 2nd Set (1995)
Peakin' at the Beacon (2000)
Hittin' the Note (2003)
One Way Out (2004)[34]
Personnel[edit] Edit
Main article: List of The Allman Brothers Band members===Members[edit]===
Gregg Allman - organ, piano, guitar, vocals (1969-1976, 1978-1982, 1986, 1989-present)
Butch Trucks - drums, tympani (1969-1976, 1978-1982, 1986, 1989-present)
Jai Johanny "Jaimoe" Johanson - drums, percussion (1969-1976, 1978-1980, 1986, 1989-present)
Warren Haynes - guitar, slide guitar, vocals (1989-1997, 2000–present)
Marc Quiñones - drums, percussion, background vocals (1991–present)
Oteil Burbridge - bass, vocals (1997–present)
Derek Trucks - guitar, slide guitar (1999–present)
Duane Allman - guitar, slide guitar (1969-1971; died 1971)
Dickey Betts - guitar, slide guitar, vocals (1969-1976, 1978-1982, 1986, 1989-2000)
Berry Oakley - bass, vocals (1969-1972; died 1972)
Chuck Leavell - piano, synthesiser, background vocals (1972-1976, 1986)
Lamar Williams - bass, vocals (1972-1976; died 1983)
Dan Toler - guitar (1978-1982, 1986; died 2013)
David Goldflies - bass (1978-1982)
David "Frankie" Toler - drums (1980-1982; died 2011)
Mike Lawler - keyboards (1980-1982)
Allen Woody - bass, background vocals (1989-1997; died 2000)
Johnny Neel - keyboards, harmonica (1989-1990)
Jack Pearson - guitar, vocals (1997-1999)
Jimmy Herring - guitar (2000)
Live personnel[edit] Edit
Thom Doucette – harmonica, percussion (for the At Fillmore East concert)
Rudolph "Juici" Carter – saxophone (for the At Fillmore East concert)
Bobby Caldwell – percussion (for the At Fillmore East concert)
Elvin Bishop – vocals (for the At Fillmore East concert)
Steve Miller – piano (for the At Fillmore East concert)
Bruce Waibel – bass (1986 reunion concert; died 2003)
Retrieved from "https://men.fandom.com/wiki/The_Allman_Brothers_Band?oldid=3024"


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Read all instructions before use. The use of a mechanical pre-filter to remove particulate organic material from water prior to its interaction with this media will improve efficiency. Rinse Xport-BIO in purified water or low-TDS fresh water, then place media into a clean pail and fill with water from the aquarium into which the media is to be placed. Seed each 150g of Xport Bio Cubes used with 20-ml of MicroBacter7 (poured into the pail and dispersed by hand), then allow media to remain in this "bath" for 24-hours at room temperature. Remove Xport Bio Cubes from pail and pour water into the aquarium system, then place Xport Bio Cubes into canister filter, media reactor, or 800µm mesh filter bag; a flow-through rate of 1 - 3x the total aquarium volume per hour is recommended. Do not fluidize. Remove and rinse ~33% of the total Xport Bio Cubes employed in a pail of discarded aquarium water (when performing a water change) once or twice each month, and place media back into use. Apply 1-ml of MicroBacter7 per 20 US-gallons (~76 L) in system to aquarium in an area of high water flow to fortify media after rinsing.
Xport Bio Cubes is not a chemical filtration media; it must be seeded with appropriate bacteria to begin nitrification. Simply placing Xport Bio Cubes into an established aquarium will slowly seed the media, however seeding with MicroBacter7 as directed above will significantly decrease cycling time and therefore results in faster results. In all systems, begin with a ratio of 150 g Xport Bio Cubes per 300 US-gallons (1,136 L) in system. Nitrification will typically begin within one day; rate of nitrification increases with time (generally mature within 2 weeks) and increasing nitrogen content in system. If plants and/or zooxanthellate invertebrates (e.g. corals, clams, etc.) appear to react negatively to the use of Xport Bio Cubes, it may be an indication that nitrogen is being removed too aggressively; NeoNitro or FlorinGro may be employed to provide nitrogen as nitrate, and feeding with the appropriate foods is necessary for all zooxanthellate invertebrates. Note that the required amount of Xport Bio Cubes may change as the system matures, and as biomass and/or stocking density changes; high-nutrient systems may require a greater ratio of Xport Bio Cubes to water volume, evident as a persistent ammonia presence regardless of how much MicroBacter7 is employed.


text:
Q: Почему не работает код? Как исправить? код не может увидеть context и user, что делать?
main.dart
import 'dart:js';

import 'package:firebase_auth/firebase_auth.dart';
import 'package:flutter/material.dart';
import 'package:project_c/pages/Home.dart';

void main() {
  runApp(MyApp());
}

class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context){
    return new MaterialApp(
      debugShowCheckedModeBanner: false,
      home: new MyHomePage(),

    );
  }
}

class MyHomePage extends StatefulWidget {
  @override
  _MyHomePageState createState() => new _MyHomePageState();
}

class _MyHomePageState extends State<MyHomePage> {

  final GlobalKey<FormState> _formKey = GlobalKey<FormState>();
  String _email, _password;

  @override
  Widget build(BuildContext context) {
    return new Scaffold(
        resizeToAvoidBottomPadding: false,
        body: Column(
            children: <Widget>[
              Container(
                  child: Stack(
                      children: <Widget>[
                        Container(
                          padding: EdgeInsets.fromLTRB(15.0, 110.0, 0.0, 0.0),
                          child: Text(
                              'Hello',
                              style: TextStyle(
                                  fontSize: 80.0, fontWeight: FontWeight.bold)
                          ),
                        ),
                        Container(
                          padding: EdgeInsets.fromLTRB(15.0, 175.0, 0.0, 0.0),
                          child: Text(
                              'Gamer',
                              style: TextStyle(
                                  fontSize: 80.0, fontWeight: FontWeight.bold)
                          ),
                        ),
                        Container(
                          padding: EdgeInsets.fromLTRB(250.0, 175.0, 0.0, 0.0),
                          child: Text(
                              '.',
                              style: TextStyle(
                                  fontSize: 80.0,
                                  fontWeight: FontWeight.bold,
                                  color: Colors.green)
                          ),
                        ),
                        Container(
                            padding: EdgeInsets.only(
                                top: 330.0, left: 20.0, right: 20.0),

                            child: Form(
                              key: _formKey,
                              child: Column(
                                children: <Widget>[
                                  TextFormField(
                                    validator: (input) {
                                      if (input.isEmpty) {
                                        return 'Provide an email';
                                      }
                                    },
                                    decoration: InputDecoration(
                                        labelText: 'EMAIL',

                                        labelStyle: TextStyle(
                                            fontFamily: 'Montserrat',
                                            fontWeight: FontWeight.bold,
                                            color: Colors.grey
                                        ),
                                        focusedBorder: UnderlineInputBorder(
                                            borderSide: BorderSide(
                                                color: Colors.green)
                                        )

                                    ),
                                    onSaved: (input) => _email = input,

                                  ),
                                  SizedBox(height: 20.0),
                                  TextFormField(
                                    validator: (input) {
                                      if (input.length < 6) {
                                        return 'Longer password please';
                                      }
                                    },
                                    decoration: InputDecoration(
                                        labelText: 'PASSWORD',
                                        labelStyle: TextStyle(
                                            fontFamily: 'Montserrat',
                                            fontWeight: FontWeight.bold,
                                            color: Colors.grey
                                        ),
                                        focusedBorder: UnderlineInputBorder(
                                            borderSide: BorderSide(
                                                color: Colors.green)
                                        )

                                    ),
                                    onSaved: (input) => _password = input,
                                    obscureText: true,

                                  ),
                                  SizedBox(height: 5.0,),
                                  Container(
                                    alignment: Alignment(1.0, 0.0),
                                    padding: EdgeInsets.only(
                                        top: 15.0, left: 20.0),
                                    child: InkWell(
                                      child: Text('Forgot password',
                                          style: TextStyle(
                                              color: Colors.green,
                                              fontWeight: FontWeight.bold,
                                              fontFamily: 'Montserrat',
                                              decoration: TextDecoration
                                                  .underline)
                                      ),
                                    ),
                                  ),
                                  SizedBox(height: 40.0),
                                  Container(
                                    height: 40.0,
                                    child: Material(

                                      borderRadius: BorderRadius.circular(20.0),
                                      shadowColor: Colors.greenAccent,
                                      color: Colors.green,
                                      elevation: 7.0,
                                      child: RaisedButton(

                                        onPressed: signIn,
                                        child: Center(
                                          child: Text(

                                            'LOGIN',
                                            style: TextStyle(

                                                color: Colors.white,
                                                fontWeight: FontWeight.bold,
                                                fontFamily: 'Montserrat'

                                            ),

                                          ),
                                        ),

                                      ),

                                    ),
                                  ),

                                  SizedBox(height: 20.0),
                                  Container(
                                    height: 40.0,

                                    child: Material(

                                      borderRadius: BorderRadius.circular(20.0),
                                      shadowColor: Colors.greenAccent,
                                      color: Colors.white,
                                      elevation: 7.0,
                                      child: GestureDetector(

                                        onTap: () {},
                                        child: Center(
                                          child: Text(

                                            'REGISTER',
                                            style: TextStyle(

                                                color: Colors.black,
                                                fontWeight: FontWeight.bold,
                                                fontFamily: 'Montserrat'

                                            ),

                                          ),
                                        ),

                                      ),

                                    ),

                                  )

                                ],
                              ),
                            )
                        )

                      ]
                  )
              )

            ]

        )

    );

  }

  Future<void> signIn() async{
    final formState = _formKey.currentState;
    if(formState.validate()){
      formState.save();
      try{
        UserCredential user = await FirebaseAuth.instance.signInWithEmailAndPassword(email: _email, password: _password);
        Navigator.push(context, MaterialPageRoute(builder: (context) => Home(user: user)));
      }catch(e){
        print(e.message);
      }

    }

  }

  }

dart аналитика

A: Из-за того, что под капотом import 'dart:js' есть упоминания про свой context он перетягивает JsObject, а не  BuildContext, просто удали этот импорт(import 'dart:js';) или пометь его
import 'dart:js' as js;


Two spread opener for one of the most popular Polish magazines - Wysokie Obcasy Extra.
The text was about a couple who travelled around a world for more than three years. My map illustrates their itinerary and adventures.


Subject: Letter of Instruction -For the Turn in of Medical Materiel.
Class VIII Materiel for Redeploying Units, VII Corps.
MEDSOM by the individual units.
Document and have present Unit Destruction Officer and witnesses.