text:
\begin{document}

\title{\textbf{Existence of multiple solutions to a class of  nonlinear Schr\"{o}dinger system with external sources terms}
 }
\author{Zexin Qi$^a$,~~ Zhitao Zhang$^{b,}$\thanks{Corresponding author, supported by NSFC 11325107, 11271353, 11331010.} \\
{\small $^a$ College of Mathematics and Information Science,}\\
{\small Henan Normal University, Xinxiang 453007, China}\\
 {\small E-mail: qizedong@126.com}\\
{\small $^b$ Academy of Mathematics and Systems Science,}\\
{\small the Chinese Academy of Sciences, Beijing 100190, China}\\
{\small E-mail: zzt@math.ac.cn} \ }
\date{}
\maketitle
\begin{abstract}
We  study a class of nonlinear schr\"{o}dinger system with  external sources terms  as perturbations in order to obtain existence of multiple solutions,
 this system arises from Bose-Einstein condensates etc..
 As these  external sources   terms are positive functions and small in some sense, we use Nehari manifold to get the existence of a positive ground state solution and a positive  bound state solution.
\end{abstract}

\noindent {\sl Keywords:} nonlinear schr\"{o}dinger system;
multiple solutions; ground state; Nehari manifold \

\vskip 0.2cm

\noindent {\sl AMS Subject Classification (2010):} 35B20, 35J47,
35J50.\

\renewcommand{\theequation}{\thesection.\arabic{equation}}
\setcounter{equation}{0}

\section{Introduction}
We are concerned with the following   nonlinear schr\"{o}dinger system with external sources terms
\begin{equation}\label{QZ1.1}
     \left\{
   \begin{array}{ll}
      -\Delta u+\lambda_{1}u=\mu_{1}u^{3}+\beta uv^{2}+f(x), & x\in  \Omega, \\
      -\Delta v+\lambda_{2}v=\mu_{2}v^{3}+\beta u^{2}v+g(x), & x\in \Omega, \\
     u=v=0, & x\in\partial \Omega,
   \end{array}
 \right.
\end{equation}
where $\Omega$ is a bounded domain in~$\mathbb{R}^{N}, N\leq3$,
$\lambda_{1},\lambda_{2},\mu_{1},\mu_{2}$ and $\beta$ are positive
constants, $f(x),g(x)$ are external sources terms. \par (1.1) is a
perturbed version of the following system
\begin{equation}\label{1.2}
     \left\{
   \begin{array}{ll}
      -\Delta u+\lambda_{1}u=\mu_{1}u^{3}+\beta uv^{2}, &  x\in  \Omega, \\
      -\Delta v+\lambda_{2}v=\mu_{2}v^{3}+\beta u^{2}v, &  x\in \Omega, \\
     u=v=0, &  x\in\partial \Omega.
   \end{array}
 \right.
\end{equation}
 \par
 System (1.2) arises from many physical problems, especially in describing some phenomenon in nonlinear optics (\cite{2},\cite{9}). It also models the Hartee-Fock theory for a double
  condensate, i.e., a binary mixture of  Bose-Einstein condensates in two different hyperfine states $|1\rangle$ and $|2\rangle$ (\cite{15}). We refer the reader to
   \cite{3,1,10,ChenZ,CTV2,CTV3,41,42,43,LW,18,NTTV1,NTTV2,34}, and the references therein for  interesting  existence
    of solutions or properties of solutions. The parameters $\mu_i$ and $\beta$ are the intraspecies and interspecies scattering lengths respectively. The sign of the scattering length $\beta$ determines whether the interaction of states $|1\rangle$ and $|2\rangle$ are repulsive or attractive. When $\beta<0$, the interactions of states $|1\rangle$ and $|2\rangle$ are repulsive. In contrast, when $\beta>0$ the interactions of states $|1\rangle$ and $|2\rangle$ are attractive. For atoms of the single state $|j\rangle$, when $\mu_j>0$, the interactions of the single state $|j\rangle$ are attractive.

 \par
  Naturally, people concern nontrivial solutions(solutions with both components nonzero) of the system. In recent years, many interesting works have been devoted to searching ground states and bound states
  for this system, see \cite{3,1,10,ChenZ,LW,18,34} etc. and the references therein.\par
     \textit{\bf A positive ground state solution} we mean a solution of a schr\"{o}dinger system which has the least
      energy among all nonzero solutions, and both of its components are
positive. Note here we call a function positive if it is
  nonnegative and nonzero. A \textit{\bf bound state solution} refers to limited-energy solution.
  As for system (1.2), well-known results indicate the existence of positive ground state is closely related to
   the parameters, see \cite{1},\cite{34} etc. and a remark for the bounded-domain case in \cite{17}.

 In this paper, multiplicity result is established when the perturbations are sufficiently small. If $f(x)$ and $g(x)$ are both positive,
  we  can  find a positive ground state.
  \par
   To be precise, let $S_{4}$ be the best
  Sobolev constant of the embedding: $ H_{0}^{1}(\Omega)\hookrightarrow L^{4}(\Omega)$, then we have\\
 \\
 \textbf{Theorem~1.1}~~\textit{Assume that}~ $f(x),g(x)\in L^{\frac{4}{3}}(\Omega)$,~\textit{both nonzero. Then there exists a positive constant}~$\Lambda =\Lambda(\lambda_{1},\lambda_{2},\mu_{1},\mu_{2}, \beta,S_{4})$, \textit{such that whenever
$\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda $, system}~(1.1)~\textit{has two nontrivial solutions.~Furthermore,~if~$f$ and $g$~are both positive, system}~(1.1)~
\textit{has one positive ground state solution and one positive bound state solution}.\\

Let $H:=H_{0}^{1}(\Omega)\times H_{0}^{1}(\Omega)$ with  the norm

\begin{center}
$\|(u,v)\|:=\left(\displaystyle\int_{\Omega}(|\nabla u|^{2}+\lambda_{1}u^{2})+\displaystyle\int_{\Omega}(|\nabla v|^{2}+\lambda_{2}v^{2})\right)^{\frac{1}{2}}.$
\end{center}
\par An element ${(}u,v{)}\in H$ is called a weak solution of~(1.1),~if the equality

  $$\displaystyle\int_{\Omega}(
\nabla u \nabla \varphi+\nabla v\nabla \psi-\lambda_{1}u\varphi-\lambda_{2}v\psi-\mu_{1}u^{3}\varphi-\mu_{2}v^{3}\psi$$
$$-\beta uv^{2}\varphi-\beta u^{2}v\psi-f\varphi-g\psi)dx=0~~~~~~~~~~~~$$

holds for all $(\varphi,\psi)\in H$. A weak solution of (1.1) corresponds to a critical point of the following $C^{1}$-functional
\begin{equation}
J(u,v)=\frac{1}{2}\|(u,v)\|^{2}-\frac{1}{4}(\mu_{1}\|u\|_{4}^{4}+\mu_{2}\|v\|_{4}^{4}+2\beta\int_{\Omega}u^{2}v^{2})-\int_\Omega(fu+gv).
\end{equation}
\par Denote the Nehari manifold associated with the functional by
 \begin{center}
 $\mathcal{N}:=\{(u,v)\in H : \langle J^{'}(u,v),(u,v)\rangle=0\}.$
\end{center}
\par It is well-known that all critical points lie in
the~Nehari~manifold, and it is usually  effective to consider
the existence of critical points in this smaller subset of the Sobolev space. For fixed
~$(u,v)\in H \setminus\{(0,0)\}$, denote
\begin{equation}\label{}
\phi(t)=\phi_{(u,v)}(t):=J(tu,tv),~t>0
\end{equation}
the so called fibering map in the direction $(u,v)$.~Such maps are often used to investigate Nehari manifolds for
 various semi-linear problems(\cite{27,Poh,32,40}). Our method is, roughly speaking, to figure out two non-degenerate
  parts of the Nehari manifold, and then consider minimization problems in the two parts
  respectively to obtain two nontrivial solutions, especially to obtain two positive solutions.  Under some other assumptions, there may exist more solutions,
   after we  submitted this manuscript, we thank the referee to tell us
   that existence of infinitely many nontrivial
  solutions of systems related to \eqref{QZ1.1}  are recently obtained  in \cite{YueZ} by different methods under
  different conditions, they   assume  the coefficient matrix is either positive
definite when $N=1,2$ or anti-symmetric when $N=3$. But here for
\eqref{QZ1.1}, the coefficient matrix
  $\left(\begin{array}{lll}
  \mu_1~ \beta\\
  \beta~ \mu_2\end{array}
  \right)$
 usually doesn't satisfy those assumptions, especially the anti-symmetric assumption implies that $\beta=0$ when
 $N=3$, then \eqref{QZ1.1} becomes two independent differential equations without any
 couplings. Our assumption here on $\beta$  can be $\beta>-\sqrt{\mu_1\mu_2}$(notice the remark at the end
 of the paper).
  \par
Our paper is organized as follows. In section 2, we  use fibering
maps to divide the Nehari manifold into three parts. A basic work
related can be found in \cite{Poh},\cite{32}.
 The number $\Lambda$ in Theorem 1.1 is determined to ensure a satisfactory partition. In Section 3 we set up a critical
  point lemma   to consider two minimization problems.
  In the last section, we give the proof of Theorem 1.1.
\section{Partition of the Nehari manifold}
\numberwithin{equation}{section}
 \setcounter{equation}{0}
\par
Denote~$\Phi(u,v):= \langle J^{'}(u,v),(u,v)\rangle$. By (1.3) one has
\begin{center}
$\langle J^{'}(u,v),(u,v)\rangle=\|(u,v)\|^{2}-(\mu_{1}\|u\|_{4}^{4}+\mu_{2}\|v\|_{4}^{4}+2\beta\int_{\Omega}u^{2}v^{2})-\int_\Omega(fu+gv),$
\end{center}
\begin{center}
$\langle \Phi^{'}(u,v),(u,v)\rangle=2\|(u,v)\|^{2}-4(\mu_{1}\|u\|_{4}^{4}+\mu_{2}\|v\|_{4}^{4}+2\beta\int_{\Omega}u^{2}v^{2})-\int_\Omega(fu+gv).$
\end{center}
\par Now we divide the Nehari manifold into three parts:
\begin{center}
   $ \mathcal{N}^{+}:=\{(u,v)\in \mathcal{N} : \langle \Phi^{'}(u,v),(u,v)\rangle>0\};$
\end{center}
\begin{center}
   $ \mathcal{N}^{0}:=\{(u,v)\in \mathcal{N} : \langle \Phi^{'}(u,v),(u,v)\rangle=0\};$
\end{center}
\begin{center}
   $ \mathcal{N}^{-}:=\{(u,v)\in \mathcal{N} : \langle \Phi^{'}(u,v),(u,v)\rangle<0\}.$
\end{center}
\par Obviously, only $\mathcal{N}^{0}$ contains the element $(0,0)$, and it is easy to see $\mathcal{N}^{+}\cup \mathcal{N}^{0}$ and $\mathcal{N}^{-}\cup \mathcal{N}^{0}$ are both closed subsets of $H$.

In order to make an explanation of such partition, for $(u,v)\in H\setminus\{(0,0)\}$,  let us consider the fibering map defined in (1.4).
Since
\begin{center}
 $\phi'(t)=\dfrac{1}{t}\langle J'(tu,tv),(tu,tv)\rangle,$
\end{center}
we know $(u,v)\in \mathcal{N}$ if and only if $\phi'(1)=0$. Moreover, $(tu,tv)\in \mathcal{N}$ with $t>0$ if and only if $\phi'(t)=0$. \par
Thus for a fixed direction $(u,v)\in H \setminus\{(0,0)\}$, we can obtain all elements on the Nehari manifold which lie in this direction if we can find all stationary points of the fibering map. As a result, one would obtain all nonzero elements of the Nehari manifold if one could find stationary points of fibering maps in all directions. We remark that the number of roots of the equation $\phi'(t)=0$ does not depend on the norm of $(u,v)$, once this direction is fixed. Indeed, for $\delta>0$ we have
\begin{center}
$\phi_{(\delta u,\delta v)}^{'}\left(\dfrac{t}{\delta}\right)=\delta\phi_{(u,v)}^{'}(t)$.
\end{center}
 \par Thus
\begin{center}
 $\phi_{(u,v)}^{'}(t)=0,~t>0\Longleftrightarrow\phi_{(\delta u,\delta v)}^{'}\left(\dfrac{t}{\delta}\right)=0,~t>0.$
\end{center}
\par Furthermore,
\begin{center}
$\phi_{(\delta u,\delta v)}^{''}\left(\dfrac{t}{\delta}\right)=\delta^{2}\phi_{(u,v)}^{''}(t)$,
\end{center}
which means it is also irrelevant with the norm of $(u,v)$ if we consider the sign of second derivative of the fibering map at its stationary points.
Moreover, stationary points can be classified into three types, namely local minimum, local maximum and turning point, according to the sign of second derivative of the fibering map at these points.\par
 We now disclose the relationship between such classification and the former partition of the Nehari manifold. By direct calculation, we get
\begin{center}
 $\phi^{'}(t)=\dfrac{1}{t} \langle J^{'}(tu,tv),(tu,tv)\rangle=\dfrac{1}{t}\Phi(tu,tv),$
\end{center}
\begin{center}
 $\phi^{''}(t)=\dfrac{1}{t^{2}}[\langle\Phi^{'}(tu,tv),(tu,tv)\rangle-\Phi(tu,tv)].$
\end{center}
\par Thus if~$\phi^{'}(t)=0$,~then $(tu,tv)\in \mathcal{N}$, and $\Phi(tu,tv)=0$, which yields
\begin{center}
 $\phi^{''}(t)=\dfrac{1}{t^{2}} \langle\Phi^{'}(tu,tv),(tu,tv)\rangle.$
\end{center}
\par Now it is easy to check:
\begin{center}
$t(u,v)\in \mathcal{N}^{+},~t>0 \Longleftrightarrow \phi^{'}(t)=0, \phi^{''}(t)>0;$
\end{center}
\begin{center}
$t(u,v)\in \mathcal{N}^{0},~t>0 \Longleftrightarrow \phi^{'}(t)=0, \phi^{''}(t)=0;$
\end{center}
\begin{center}
$t(u,v)\in \mathcal{N}^{-},~t>0 \Longleftrightarrow \phi^{'}(t)=0, \phi^{''}(t)<0.$
\end{center}
\par We can explore the Nehari manifold through fibering maps. In
fact,  we give the following important lemma to show
when  the degenerate part of the Nehari manifold is clear and
simple. \par
To simplify the calculation, let us introduce some notations
that will be used repeatedly in the rest. For $(u,v)\in H$, define
\begin{equation}\label{}
A=A(u,v):=\mu_{1}\|u\|_{4}^{4}+\mu_{2}\|v\|_{4}^{4}+2\beta\int_{\Omega}u^{2}v^{2},
\end{equation}
\begin{equation}\label{}
B=B(u,v):=\int_\Omega(fu+gv).
\end{equation}
\textbf{Lemma~2.1}~~\textit{Suppose that $f(x),g(x)\in L^{\frac{4}{3}}(\Omega)$,~both nonzero,~then there exists a positive constant~$\Lambda =\Lambda(\lambda_{1},\lambda_{2},\mu_{1},\mu_{2}, \beta,S_{4})$
 such that  $\mathcal{N}^{0}=\{(0,0)\}$ when
$\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda$.\\}
\\
\textbf{Proof.}~~By the analysis above, we only need to prove for~each $(u,v)\in H $ with $\|(u,v)\|=1$,~the fibering map~$\phi(t)=\phi_{(u,v)}(t)$~has no stationary point that is a turning point. By the notations  (2.1) and (2.2), we can write
\begin{align}\label{}
    \phi(t)&=J(tu,tv)\notag\\
           &=\dfrac{t^{2}}{2}\|(u,v)\|^{2}-\dfrac{t^{4}}{4}(\mu_{1}\|u\|_{4}^{4}+\mu_{2}\|v\|_{4}^{4}+2\beta\displaystyle\int_{\Omega}u^{2}v^{2})-t\int_\Omega(fu+gv)\notag\\
           &=\dfrac{t^{2}}{2}-\dfrac{At^{4}}{4}-Bt.\notag
\end{align}
\par Thus
\begin{center}
   $\phi'(t)=t-At^{3}-B$,\;\;\;\;\;\;\;\;$\phi''(t)=1-3At^{2}.$
\end{center}
\par Notice~$A=A(u,v)=\mu_{1}\|u\|_{4}^{4}+\mu_{2}\|v\|_{4}^{4}+2\beta\int_{\Omega}u^{2}v^{2}>0$.~Define $\psi(t)=t-At^{3},~t>0$, then $\phi'(t)=0\Longleftrightarrow \psi(t)=B$.
\par
Let us consider the graph of $\psi(t)$: $\psi''(t)=-6At<0$,~so~$\psi(t)$~is strictly concave;~$\psi'(t)=0\Longleftrightarrow t=\sqrt{\frac{1}{3A}}$,~thus~$\psi(t)$~takes its maximum at~$t=\sqrt{\frac{1}{3A}}$, the value of which is~$\frac{2}{3}\sqrt{\frac{1}{3A}}$. Also, $\lim_{t\rightarrow0^{+}} \psi(t)=0$,~$ \psi(\infty)=-\infty$. All the above implies if~$0<B<\frac{2}{3}\sqrt{\frac{1}{3A}}$, the equation $\phi'(t)=0$~has exactly two roots $t_{1}$,~$t_{2}$ satisfying $0<t_{1}<\sqrt{\frac{1}{3A}}<t_{2}$; if~$B\leq0$ the equation $\phi'(t)=0$ has exactly one root denoted $t_{2}^{'}$, which satisfies $\sqrt{\frac{1}{3A}}<t_{2}^{'}$. Since~$\phi''(t)=1-3At^{2}$,~considering the above two cases we have $\phi''(t_{1})>0$,~$\phi''(t_{2})<0$, and $\phi''(t_{2}^{'})<0$. So when $0<B<\frac{2}{3}\sqrt{\frac{1}{3A}}$, one has $t_{1}(u,v)\in \mathcal{N}^{+}$,~$t_{2}(u,v)\in \mathcal{N}^{-}$;~
when $B\leq0$ one has $t_{2}^{'}(u,v)\in \mathcal{N}^{-}$. Since $f(x),g(x)$ are both nonzero, it is easy to check that the sets $\{(u,v)\in H :\|(u,v)\|=1,0<B<\frac{2}{3}\sqrt{\frac{1}{3A}}\}\not=\emptyset$ and $\{(u,v)\in H :\|(u,v)\|=1,B\leq0 \}\not=\emptyset$, so $\mathcal{N}^{+}\not=\emptyset$ and $\mathcal{N}^{-}\not=\emptyset$. To finish the proof, we are in a position to determine a number $\Lambda$ such that whenever
$\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda$ we have
\begin{equation}\label{}
     B<\frac{2}{3}\sqrt{\frac{1}{3A}}.
\end{equation}
\par In fact, since $(u,v)$ lies on the unit sphere of $H$, by Sobolev inequality and~H\"{o}lder~inequality one obtains an upper bound for
~$A$, which yields the existence of a positive constant $\alpha=\alpha(\lambda_{1},\lambda_{2},\mu_{1},\mu_{2},\beta,S_{4})$ such that
\begin{equation}\label{}
        0<\alpha\leq\frac{2}{3}\sqrt{\frac{1}{3\sup\limits_{\|(u,v)\|=1}A(u,v)}}.
\end{equation}
\par On the other hand, by Sobolev inequality and~H\"{o}lder inequality, we obtain that
 \begin{center}
  $B=\displaystyle\int_\Omega(fu+gv)
   \leq \|f\|_{\frac{4}{3}}\|u\|_{4}+\|g\|_{\frac{4}{3}}\|v\|_{4}
    \leq \sqrt{2}S_{4}\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}\|(u,v)\|.$
\end{center}
\par That is,
\begin{equation}\label{}
B\leq \sqrt{2}S_{4}\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}.
\end{equation}
\par Now take $\Lambda:=\Lambda(\lambda_{1},\lambda_{2},\mu_{1},\mu_{2},\beta,S_{4}) =\frac{\alpha}{\sqrt{2}S_{4}}$,

by (2.4),(2.5) we know when
$\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda$,~(2.3) holds.\hfill{$\square$}

\section{Reduction to minimization problems}
\numberwithin{equation}{section}
 \setcounter{equation}{0}
Here we reduce our discussion into two minimization problems through
a critical point lemma. As to constraint minimization problems, we
refer the reader to \cite{8}. Let $X$,$Y$ be real Banach spaces,
$U\subset X$ be an open set. Suppose that $f:U\rightarrow
\mathbb{R}^{1}$, $g:U\rightarrow Y$ are $C^{1}$ mappings in this
section. Let
\begin{center}
   $ M=\{x\in U:g(x)=\theta\}.$
\end{center}
\par For the following minimization problem
\begin{equation}\label{}
\min_{x\in M}f(x),
  \end{equation}
we have\\
\\
\textbf{Lemma~3.1}~(\cite{8},~Theorem 4.1.1)~~\textit{Suppose that~$x_{0}\in M$~solves}~(3.1),~\textit{and that~$Im~g'(x_{0})$~is closed.~Then there exists~$(\lambda,y^{\ast})\in \mathbb{R}^{1}\times Y^{\ast}$~such that~$(\lambda,y^{\ast})\neq (0,\theta)$,~and
\begin{center}
$\lambda f'(x_{0})+ g'(x_{0})^{\ast}y^{\ast}=0.$
\end{center}
Furthermore,~if~$Im~g'(x_{0})=Y$,~then~$\lambda\neq 0$.}\\

Let us introduce two minimization problems
\begin{equation}\label{}
 \theta^{+}= \inf_{(u,v)\in \mathcal{N}^{+}}J(u,v);
  \end{equation}
\begin{equation}\label{}
 \theta^{-}= \inf_{(u,v)\in \mathcal{N}^{-}}J(u,v).
  \end{equation}
\textbf{Lemma ~3.2}~~\textit{If~$(u_{1},v_{1})$~solves}~(3.2),~\textit{then~$(u_{1},v_{1})$~is a nontrivial weak solution of}~(1.1);~
\textit{if~$(u_{2},v_{2})$~solves}~(3.3),~\textit{then~$(u_{2},v_{2})$~is a nontrivial weak solution of}~(1.1).\\
\\
\textbf{Proof.}~~We prove the first assertion, since the second is similar. \par
Recall~$\Phi(u,v)= \langle J^{'}(u,v),(u,v)\rangle$~and
\begin{center}
   $ \mathcal{N}^{+}=\{(u,v)\in \mathcal{N} : \langle \Phi^{'}(u,v),(u,v)\rangle>0\}.$
\end{center}
\par Let~$U=\{(u,v)\in H :\langle \Phi^{'}(u,v),(u,v)\rangle>0\}$,~and rewrite
\begin{center}
   $ \mathcal{N}^{+}=\{(u,v)\in U:\Phi(u,v)=0 \}.$
\end{center}
\par Now we use Lemma 3.1 to consider problem (3.2) by taking
$X=H,~Y=\mathbb{R}^{1},~U=\{(u,v)\in H :\langle
\Phi^{'}(u,v),(u,v)\rangle>0\},~f=J,~g=\Phi,~M=\mathcal{N}^{+}$.
Then there exists a real pair~$(\lambda_{1},\lambda_{2})\neq
(0,0)$,~satisfying
\begin{center}
    $\lambda_{1}J'(u_{1},v_{1})=\lambda_{2}\Phi'(u_{1},v_{1}).$
\end{center}
\par Since $\langle \Phi^{'}(u_{1},v_{1}),(u_{1},v_{1})\rangle>0$, we
have~$\Phi^{'}(u_{1},v_{1})\neq 0$,~again by Lemma 3.1,~one may
assume~$\lambda_{1}\neq0$.~In other words, there exists a real
number~$\mu$ such that
\begin{equation}\label{}
    J'(u_{1},v_{1})=\mu\Phi'(u_{1},v_{1}).
\end{equation}
\par Since
\begin{center}
   $\mu \langle \Phi^{'}(u_{1},v_{1}),(u_{1},v_{1})\rangle=\langle J^{'}(u_{1},v_{1}),(u_{1},v_{1})\rangle=0$
\end{center}
and~$\langle \Phi^{'}(u_{1},v_{1}),(u_{1},v_{1})\rangle>0$,~we
obtain~$\mu =0$. By (3.4)~we get~$J'(u_{1},v_{1})=0$.~
Obviously~$(u_{1},v_{1})\neq (0,0)$, noticing that (1.1)~doesn't
admit semi-trivial (nonzero but one component being trivial)
solutions, we know that ~$(u_{1},v_{1})$~is a nontrivial weak
solution of (1.1).\hfill{$\square$}
\section{Proof of the main result}
\numberwithin{equation}{section}
 \setcounter{equation}{0}
When $\mathcal{N}^{0}=\{(0,0)\}$, we show the solvability of the two
problems (3.2) and (3.3), and give the proof of Theorem 1.1.

The following lemma implies that~$\theta^{+}$~and~$\theta^{-}$~are both finite,~and any minimizing sequence for (3.2) or (3.3) are bounded.\\
\\
\textbf{Lemma~4.1}~~\textit{Assume that $f(x),g(x)\in L^{\frac{4}{3}}(\Omega)$,~then~ $J$~ is coercive and bounded from below on~$\mathcal{N}$~}(\textit{thus on~$\mathcal{N}^{+}$~and~$\mathcal{N}^{-}$}).\\
\\
\textbf{Proof.}~~Let~$(u,v)\in \mathcal{N}$, from the definition of the~Nehari~manifold we have
\begin{center}
   $
   \|(u,v)\|^{2}-\displaystyle\int_\Omega(fu+gv)=(\mu_{1}\|u\|_{4}^{4}+\mu_{2}\|v\|_{4}^{4}+2\beta\displaystyle\int_{\Omega}u^{2}v^{2})$,
\end{center}
this equality together with~(1.3)~yield
\begin{equation}
J(u,v)=\frac{1}{4}\|(u,v)\|^{2}-\frac{3}{4}\int_\Omega(fu+gv).
\end{equation}
\par Note that
\begin{equation}\label{}
   \displaystyle\int_\Omega(fu+gv)
   \leq \|f\|_{\frac{4}{3}}\|u\|_{4}+\|g\|_{\frac{4}{3}}\|v\|_{4}
    \leq \sqrt{2}S_{4}\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}\|(u,v)\|,
  \end{equation}
combining (4.1) and (4.2) we obtain
\begin{equation}
J(u,v)\geq \frac{1}{4}\|(u,v)\|^{2}-\frac{3\sqrt{2}}{4}S_{4}\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}\|(u,v)\|.
\end{equation}
\par Since the right hand side of the inequality (4.3) is a quadratic function of $\|(u,v)\|$, it is easy to know that $J$ is coercive and bounded from below on~$\mathcal{N}$.\hfill{$\square$}\\

From now on, $\Lambda$ refers to the number in Lemma 2.1. Although
the lemma below is valid under weaker conditions, we would assume
$f(x),g(x)\in L^{\frac{4}{3}}(\Omega)$, both nonzero, and
$\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda$ in order
to ensure $\mathcal{N}^{0}=\{(0,0)\}$ by Lemma 2.1, because our discussion bases on a good partition of the
Nehari manifold. We  will often use the proof
of Lemma 2.1 to analyze some properties of the manifold as well as
associated fibering maps.

 For the first minimization problem, we establish:\\
\\
\textbf{Lemma~4.2}~~\textit{Assume that $f(x),g(x)\in L^{\frac{4}{3}}(\Omega)$, both nonzero, and that $\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda$. Then $\theta^{+} < 0$}.\\
\\
\textbf{Proof.}~~By the proof of Lemma 2.1, we may take one $(u,v)\in H $,~$\|(u,v)\|=1$ such that $0<B<\frac{2}{3}\sqrt{\frac{1}{3A}}$. Now the derivative  $\phi'(t)$ of the
 fibering map in this direction has exactly two positive zeros: $t_{1}$,~$t_{2}$, satisfying~$0<t_{1}<\sqrt{\frac{1}{3A}}<t_{2}$, and $t_{1}\cdot(u,v)\in \mathcal{N}^{+}$.

By $\phi'(t)=t-At^{3}-B$ we have $\lim_{t\rightarrow0^{+}}
\phi'(t)=-B<0$, and $\phi''(t)>0$, $\forall
t\in(0,\sqrt{\frac{1}{3A}})$. Since $\phi'(t_{1})=0$, we know that
~$\phi(t_{1})<\lim_{t\rightarrow0^{+}} \phi(t)=0$. Since
~$\phi(t_{1})=J(t_{1}u,t_{1}v)\geq\theta^{+}$, we obtain that $\theta^{+} < 0$.\hfill{$\square$}\\

For the second minimization problem we need the following lemma, from which we know $\mathcal{N}^{-}$~stays away from the origin.\\
\\
\textbf{Lemma~4.3}~~\textit{Assume that $f(x),g(x)\in L^{\frac{4}{3}}(\Omega)$, both nonzero, and that $\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda$. Then $\mathcal{N}^{-}$ is closed}.\\
\\
\textbf{Proof.} Under the  assumptions one has
\begin{center}
    $cl~({{\mathcal{N}^{-}}})\subset\mathcal{N}^{-}\cup\mathcal{N}^{0}=\mathcal{N}^{-}\cup \{(0,0)\},$
\end{center}
where we denote $cl~({{\mathcal{N}^{-}}})$ the closure of
$\mathcal{N}^{-}$. Thus we only need to prove $(0,0) \not\in
cl~({{\mathcal{N}^{-}}})$, which is equivalent to prove that
\begin{center}
    ~$dist((0,0),\mathcal{N}^{-})>0.$
\end{center}
\par Take~$(u,v)\in\mathcal{N}^{-}$,~and denote
\begin{center}
$(\widetilde{u},\widetilde{v})=\dfrac{(u,v)}{\|(u,v)\|},$
\end{center}
then~$\|(\widetilde{u},\widetilde{v})\|=1$. Under the assumptions,
by the proof of Lemma 2.1, we obtain that
$A:=A(\widetilde{u},\widetilde{v}),~B:=B(\widetilde{u},\widetilde{v})$
satisfy $B<\frac{2}{3}\sqrt{\frac{1}{3A}}$.~Furthermore,
if~$0<B<\frac{2}{3}\sqrt{\frac{1}{3A}}$,~the
equation~$\phi'_{(\widetilde{u},\widetilde{v})}(t)=0$~has exactly
two roots also denoted by~$t_{1}$, $t_{2}$,~which
satisfy~$t_{2}\cdot(\widetilde{u},\widetilde{v})\in
\mathcal{N}^{-}$,~$t_{1}\cdot(\widetilde{u},\widetilde{v})\in
\mathcal{N}^{+}$, we have $t_{2}\cdot(\widetilde{u},\widetilde{v})=
(u,v)$,~so~$t_{2}= \|(u,v)\|$.~If~$B\leq 0$,~then the equation
~$\phi'_{(\widetilde{u},\widetilde{v})}(t)=0$~has exactly one
root~still denoted by $\widetilde{t}_{2}$,~thus we get
~$\widetilde{t}_{2}\cdot(\widetilde{u},\widetilde{v})= (u,v)$, then
$\widetilde{t}_{2}= \|(u,v)\|$. Since~${t_{2}}>\sqrt{\frac{1}{3A}}$,
$\widetilde{t}_{2}>\sqrt{\frac{1}{3A}}$ in the proof of Lemma
2.1,~so no matter which of the above two cases happens,~ we always
obtain~$\|(u,v)\|>\sqrt{\frac{1}{3A}}$. Noticing that $A$ is bounded
from above,~ we know that there exists~$\tau>0$~such that
\begin{center}
    $\|(u,v)\|>\tau.$
\end{center}
\par We obtain
\begin{center}
    $dist((0,0),\mathcal{N}^{-})=\inf\limits_{(u,v)\in\mathcal{N}^{-}}\|(u,v)\|\geq\tau>0,$
\end{center}
which completes the proof.\hfill{$\square$}\\

In order to abstract a $(PS)_{\theta^{+}}$~sequence from the minimizing sequence for problem (3.2), we use the idea of \cite{32} to obtain the following lemma. \\
\\
\textbf{Lemma~4.4}~~\textit{Assume that $f(x),g(x)\in
L^{\frac{4}{3}}(\Omega)$, both nonzero. Then for $(u,v)\in
\mathcal{N}^{+}$, there exists~$\epsilon=\epsilon(u,v)>0$~and a
differentiable
function~$\xi^{+}:B_{\epsilon}(0,0)\rightarrow\mathbb{R}_{+}:=(0,+\infty)$
such that}
\begin{itemize}
  \item $\xi^{+}(0,0)=1;$
  \item $\xi^{+}(w,z)(u-w,v-z)\in \mathcal{N}^{+},\forall(w,z)\in B_{\epsilon}(0,0);$
  \item $ \langle (\xi^{+})'(0,0),(w,z)\rangle=[\|(u,v)\|^{2}-3A(u,v)]^{-1}[2\displaystyle\int_{\Omega}(\nabla u\nabla w+\lambda_{1}uw+\nabla v\nabla z+\lambda_{2}vz)-4\displaystyle\int_{\Omega}(\mu_{1}u^{3}w+\mu_{2}v^{3}z+\beta uv^{2}w+\beta u^{2}vz)-\displaystyle\int_{\Omega}(fw+gz)].$
\end{itemize}
\textbf{Proof.}~~Define a $C^{1}$-mapping~$F:\mathbb{R}_{+}\times H \rightarrow \mathbb{R}$ as follows:
\begin{center}
    $F(t,(w,z))=t\|(u-w,v-z)\|^{2}-t^{3}A(u-w,v-z)-\displaystyle\int_\Omega[f\cdot(u-w)+g\cdot(v-z)].$
\end{center}
\par
 By (2.1), we know that
$A(u-w,v-z)=\mu_{1}\|u-w\|_{4}^{4}+\mu_{2}\|v-z\|_{4}^{4}+2\beta\displaystyle\int_{\Omega}(u-w)^{2}(v-z)^{2}$.
Since $(u,v)\in \mathcal{N}^{+}$~we have
\begin{center}
    $F(1,(0,0))=0.$
\end{center}
\par Consider the fibering map~$\phi(t)=\phi_{(u,v)}(t):=J(tu,tv)$,~since
\begin{center}
   $F(t,(0,0))=t\|(u,v)\|^{2}-t^{3}(\mu_{1}\|u\|_{4}^{4}+\mu_{2}\|v\|_{4}^{4}+2\beta\displaystyle\int_{\Omega}u^{2}v^{2})-\displaystyle\int_\Omega(fu+gv),$
\end{center}
we have
\begin{center}
   $F(t,(0,0))=\phi'(t).$
\end{center}
\par Since $(u,v)\in \mathcal{N}^{+}$, we get $\phi''(1)>0$,~thus
\begin{center}
   $\dfrac{\partial F}{\partial t}(1,(0,0))=\phi''(1)>0.$
\end{center}
\par We apply the implicit function theorem at point (1,(0,0)) to obtain the existence of~$\epsilon=\epsilon(u,v)>0$~and differentiable
function~$\xi^{+}(u,v)(i.e.,~t(u,v)):B_{\epsilon}(0,0)\rightarrow\mathbb{R}_{+}$
such that
\begin{itemize}
  \item $\xi^{+}(0,0)=1;$
  \item $\xi^{+}(w,z)\cdot(u-w,v-z)\in \mathcal{N},~\forall(w,z)\in B_{\epsilon}(0,0).$
\end{itemize}
\par Besides, we obtain the third conclusion of this lemma by calculation.~To finish the second one, we only need to choose $\epsilon=\epsilon(u,v)>0$ small,~such that~$\xi^{+}(w,z)(u-w,v-z)\in \mathcal{N^{+}},~\forall(w,z)\in B_{\epsilon}(0,0)$.~Indeed,~since
$\mathcal{N}^{-}\cup \mathcal{N}^{0}$~is
closed,~$dist((u,v),\mathcal{N}^{-}\cup \mathcal{N}^{0})>0$. Since
~$\xi^{+}(w,z)\cdot(u-w,v-z)$~is continuous with~respect to
$(w,z)$,~when~$\epsilon=\epsilon(u,v)>0$~is small enough, we know

\begin{center}
   $ \|\xi^{+}(w,z)\cdot(u-w,v-z)-(u,v)\| < \dfrac{1}{2}dist((u,v),\mathcal{N}^{-}\cup \mathcal{N}^{0}),\forall(w,z)\in B_{\epsilon}(0,0).$
\end{center}
That is,~$\xi^{+}(w,z)\cdot(u-w,v-z)$~does not belong to~$
\mathcal{N}^{-}\cup \mathcal{N}^{0}$.~Thus
~$\xi^{+}(w,z)\cdot(u-w,v-z)\in \mathcal{N^{+}}$ and our proof is completed.\hfill{$\square$}\\

Similarly, we can establish the following lemma, which will be used to abstract a $(PS)_{\theta^{-}}$~sequence from the minimizing sequence for problem (3.3).\\
\\
\textbf{Lemma~4.5}~~\textit{Assume that $f(x),g(x)\in
L^{\frac{4}{3}}(\Omega)$, both nonzero. Then for $(u,v)\in
\mathcal{N}^{-}$, there exists~$\epsilon=\epsilon(u,v)>0$~and a
differentiable
function~$\xi^{-}:B_{\epsilon}(0,0)\rightarrow\mathbb{R}_{+}$ such
that}
\begin{itemize}
  \item $\xi^{-}(0,0)=1;$
  \item $\xi^{-}(w,z)(u-w,v-z)\in \mathcal{N}^{-},\forall(w,z)\in B_{\epsilon}(0,0);$
  \item $ \langle (\xi^{-})'(0,0),(w,z)\rangle=[\|(u,v)\|^{2}-3A(u,v)]^{-1}[2\displaystyle\int_{\Omega}(\nabla u\nabla w+\lambda_{1}uw+\nabla v\nabla z+\lambda_{2}vz)-4\displaystyle\int_{\Omega}(\mu_{1}u^{3}w+\mu_{2}v^{3}z+\beta uv^{2}w+\beta u^{2}vz)-\displaystyle\int_{\Omega}(fw+gz)].$
\end{itemize}

We are in a position to give:\\
\\
\textbf{Lemma~4.6}~~\textit{Assume that $f(x),g(x)\in
L^{\frac{4}{3}}(\Omega)$, both nonzero, and that
$\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda$. Then
there exists a sequence~$\{(u_{n},v_{n})\}\subset
\mathcal{N}^{+}$~such that ($n \rightarrow \infty$)}:
\begin{enumerate}
  \item $J(u_{n},v_{n})\rightarrow \theta^{+};$
  \item $J'(u_{n},v_{n})\rightarrow 0.$
\end{enumerate}
\textbf{Proof.}~~Notice~$\mathcal{N}^{+}\cup \{(0,0)\}$~is closed
in~$H$,~we use the Ekeland's variational principle(\cite{14}) on
$\mathcal{N}^{+}\cup \{(0,0)\}$ to obtain a minimizing sequence
$\{(u_{n},v_{n})\}\subset \mathcal{N}^{+}\cup \{(0,0)\}$ such that
\begin{description}
  \item[(a)] $J(u_{n},v_{n})< \inf\limits_{(u,v)\in \mathcal{N}^{+}\bigcup \{(0,0)\}}J(u,v)+\dfrac{1}{n}$;
  \item[(b)] $J(w,z)\geq J(u_{n},v_{n}) -\dfrac{1}{n}\|(w-u_{n},z-v_{n})\|,\;\;\forall (w,z)\in \mathcal{N}^{+}\cup \{(0,0)\}$.
\end{description}
\par
By Lemma~4.2, we know $\theta^{+} < 0$; since $ J(0,0)=0$, we get
that $$ \inf\limits_{(u,v)\in \mathcal{N}^{+}\bigcup
\{(0,0)\}}J(u,v)=\theta^{+}.$$ \par
Thus $J(u_{n},v_{n})\rightarrow
\theta^{+}$,   we may assume $\{(u_{n},v_{n})\}\subset
\mathcal{N}^{+}$, then the first assertion holds.

For the second assertion, Firstly we have   that
$\inf\limits_n\|(u_{n},v_{n})\|\geq m>0$, where $m$ is a constant. Indeed, if
not, then $J(u_{n},v_{n})$~would converge to~zero. Moreover, by
Lemma~4.1 we know that $J$~is coercive~on $\mathcal{N}^{+}$,~ then
$\{\|(u_{n},v_{n})\|\}$ is bounded.  i.e., $\exists M>0$ such that
\begin{equation}\label{}
    0<m\leq \|(u_{n},v_{n})\| \leq M.
\end{equation}
\par Now by contradiction we assume $\|J'(u_{n},v_{n})\|\geq C>0$ as $n$
is large, otherwise we may extract a subsequence to get the
conclusion.

Now let us take~$(u,v)=(u_{n},v_{n})$ in Lemma 4.4,~and define a differentiable function~$\xi_{n}^{+}:(-\epsilon,\epsilon)\rightarrow \mathbb{R}_{+}$~as:
\begin{center}
   $\xi_{n}^{+}(\delta ):= \xi^{+}\left(\dfrac{\delta J'(u_{n},v_{n})}{\|J'(u_{n},v_{n})\|}\right),$
\end{center}
\par then by  Lemma 4.4, we know that $\xi_{n}^{+}(0)=
\xi^{+}(0,0)=1$,~and for~$ \delta \in (-\epsilon,\epsilon)$~we have
\begin{center}
   $(w,z)_{\delta}:=\xi_{n}^{+}(\delta )\cdot[(u_{n},v_{n})-\dfrac{\delta J'(u_{n},v_{n})}{\|J'(u_{n},v_{n})\|}]\in\mathcal{N}^{+}.$
\end{center}
\par
Since~$(u_{n},v_{n})$~satisfies (b),~one has
\begin{equation}\label{}
J(u_{n},v_{n})-J((w,z)_{\delta})\leq \dfrac{1}{n}\|(w,z)_{\delta}-(u_{n},v_{n})\|.
\end{equation}
\par Expanding the left hand side of (4.5) we get
\begin{align}\label{}
    J(u_{n},v_{n})-J((w,z)_{\delta}) =& (1-\xi_{n}^{+}(\delta ))\langle J'((w,z)_{\delta}),(u_{n},v_{n})\rangle\notag\\
                                     &+\delta\xi_{n}^{+}(\delta )\langle J'((w,z)_{\delta}),
     \frac {J'(u_{n},v_{n})}{\|J'(u_{n},v_{n})\|}\rangle\notag\\
                                     &+o(\|(w,z)_{\delta}-(u_{n},v_{n})\|).
\end{align}
\par Combining~(4.6)~with (4.5) we obtain

     $$~~~~(1-\xi_{n}^{+}(\delta ))\langle J'((w,z)_{\delta}),(u_{n},v_{n})\rangle+\delta\xi_{n}^{+}(\delta )\langle J'((w,z)_{\delta}),
     \dfrac {J'(u_{n},v_{n})}{\|J'(u_{n},v_{n})\|}\rangle$$
     $$\leq o(\|(w,z)_{\delta}-(u_{n},v_{n})\|)+\dfrac{1}{n}\|(w,z)_{\delta}-(u_{n},v_{n})\|.~~~~~~~~~~~~~~~~~~~~~~~~$$

\par
Divide~the above inequality by $\delta$ for~$\delta\neq0$~and
let~$\delta\rightarrow 0$, then we get

    $$ -(\xi_{n}^{+})'(0)\langle J'(u_{n},v_{n}),(u_{n},v_{n})\rangle+
     \|J'(u_{n},v_{n})\| $$
    $$ \leq (o(1)+\dfrac{1}{n})(1+|(\xi_{n}^{+})'(0)|\cdot\|(u_{n},v_{n})\|).~~~~~~~~~~~$$

\par That is,
\begin{center}
     $\|J'(u_{n},v_{n})\| \leq (o(1)+\dfrac{1}{n})\cdot(1+|(\xi_{n}^{+})'(0)|\cdot\|(u_{n},v_{n})\|).$
\end{center}
\par
By (4.4), we only need to show $|(\xi_{n}^{+})'(0)|$~is
uniformly bounded with respect to $n$. Noticing
that
\begin{center}
    $(\xi_{n}^{+})'(0)=\langle(\xi^{+})'(0,0),\dfrac {J'(u_{n},v_{n})}{\|J'(u_{n},v_{n})\|}\rangle,$
\end{center}
by~(4.4) and the third assertion of Lemma 4.4,  we can get that there exists $C>0$ such that
\begin{center}
    $|(\xi_{n}^{+})'(0)|\leq\dfrac{C}{|\|(u_{n},v_{n})\|^{2}-3A(u_{n},v_{n})|}.$
\end{center}
\par Thus we only need to prove that $|\|(u_{n},v_{n})\|^{2}-3A(u_{n},v_{n})|$~has a positive lower bound.~Assume the contrary, then up to a subsequence,
\begin{equation}\label{}
    \|(u_{n},v_{n})\|^{2}-3A(u_{n},v_{n})=o(1).
\end{equation}
\par Since~$\{(u_{n},v_{n})\}\subset \mathcal{N}^{+}$,
\begin{equation}\label{}
    \|(u_{n},v_{n})\|^{2}-A(u_{n},v_{n})=B(u_{n},v_{n}).
\end{equation}
\par From (4.7) and (4.8) we have,
\begin{equation}\label{}
   B(u_{n},v_{n})=\frac{2}{3} \|(u_{n},v_{n})\|^{2}+o(1).
\end{equation}
\par For fixed~$f(x),g(x)$, since
$\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda$,~there must exist a small positive~$\tau$~such that
\begin{equation}\label{}
    \max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<(1-\tau)\Lambda.
\end{equation}
\par By the proof of Lemma~2.1, one knows from the derivation of (2.3) that more accurate inequality will occur once (4.10) holds. That is, one has for~$\|(u,v)\|=1$,
\begin{center}
     $B(u,v)<\dfrac{2}{3}(1-\tau)\sqrt{\dfrac{1}{3A(u,v)}}.$
\end{center}
\par Thus by homogeneity,
\begin{equation}\label{}
    B(u_{n},v_{n})<\frac{2}{3}(1-\tau)\sqrt{\frac{\|(u_{n},v_{n})\|^{2}}{3A(u_{n},v_{n})}}\|(u_{n},v_{n})\|^{2}.
\end{equation}
\par Dividing~(4.11)~by~$\|(u_{n},v_{n})\|^{2}$ and letting $n \rightarrow \infty$, we reach a contradiction from (4.4),~(4.7)~and~(4.9), that
\begin{center}
    ${\dfrac{2}{3}}\leq\dfrac{2}{3}(1-\tau).$
\end{center}
This completes the proof.\hfill{$\square$}\\

On the other hand, we have\\
\\
\textbf{Lemma~4.7}~~\textit{Assume~$f(x),g(x)\in L^{\frac{4}{3}}(\Omega)$, both nonzero, and $\max\{\|f\|_{\frac{4}{3}},\|g\|_{\frac{4}{3}}\}<\Lambda$. Then there exists a sequence~$\{(u_{n},v_{n})\}\subset \mathcal{N}^{-}$,~such that when~$n \rightarrow \infty$, it holds}:
\begin{enumerate}
  \item $J(u_{n},v_{n})\rightarrow \theta^{-};$
  \item $J'(u_{n},v_{n})\rightarrow 0.$
\end{enumerate}
\textbf{Proof.}~~By Lemma~4.3, $\mathcal{N}^{-}$~is closed in~$H$.~We use Ekeland's variational principle on~$\mathcal{N}^{-}$~to obtain a minimizing sequence~$\{(u_{n},v_{n})\}\subset \mathcal{N}^{-}$, such that
\begin{enumerate}
  \item $J(u_{n},v_{n})<\inf\limits_{(u,v)\in \mathcal{N}^{-}}J(u,v)+\dfrac{1}{n};$
  \item $J(w,z)\geq J(u_{n},v_{n}) -\dfrac{1}{n}\|(w-u_{n},z-v_{n})\|$~holds for all~$(w,z)\in \mathcal{N}^{-}$.
\end{enumerate}
\par From Lemma 4.1 and the remark before Lemma 4.3 we obtain estimates similar to (4.4):
\begin{center}\label{}
    $0<\tilde{m}\leq \|(u_{n},v_{n})\| \leq \tilde{M}$,
\end{center}
where~$\tilde{m}$~and~$\tilde{M}$~are positive constants. By Lemma 4.5, the rest of the proof is similar to that of Lemma 4.6,  we omit it.\hfill{$\square$}\\

Since system (1.1) is  subcritical, it is not difficult to obtain compactness conditions for the functional $J$. That is,\\
\\
\textbf{Lemma~4.8}~~\textit{Assume~$f(x),g(x)\in L^{\frac{4}{3}}(\Omega)$, then $J$ satisfies $(PS)$ condition, i.e., for any $c\in \mathbb{R}$,
any sequence $\{u_{n}\}\subset H$ for which $J(u_{n})\rightarrow c,J'(u_{n})\rightarrow 0$ as $n\rightarrow+\infty$ possesses a convergent subsequence}.\\

 We are now able to give\\

  \par
  \textbf{Proof of Theorem 1.1.}~ Firstly let us consider the minimization problem~(3.2).~By Lemma 4.6,~there exists~$\{(u_{n},v_{n})\}\subset \mathcal{N}^{+}$~such that as~$n\rightarrow\infty$
\begin{center}
  $J(u_{n},v_{n})\rightarrow \theta^{+}$,~~~~
   $J'(u_{n},v_{n})\rightarrow 0$.
\end{center}
\par Since $J$ satisfies $(PS)$condition by Lemma 4.8, we find a $(w_{1},z_{1})\in cl~(\mathcal{N}^{+})\subset\mathcal{N}^{+}\cup \{(0,0)\}$ such that $J(w_{1},z_{1})= \theta^{+}$,~$J'(w_{1},z_{1})=0 $. By Lemma 4.2, $J(w_{1},z_{1})<0$. Thus
$(w_{1},z_{1})\neq(0,0)$,~which implies~$(w_{1},z_{1})\in \mathcal{N}^{+}$.~We see~$(w_{1},z_{1})$~is a nontrivial weak solution of~(1.1) by Lemma 3.2.

Furthermore,~if~$f$ and $g$~are both positive,~we show the minimizer can be chosen to be a multiple of $(|w_{1}|,| z_{1}|)$. Indeed,~$\|(w_{1},z_{1})\|=\|(|w_{1}|,| z_{1}|)\|$.~Let
  $$(|w_{0}|, |z_{0}|):=\dfrac{(|w_{1}|,| z_{1}|)}{\|(|w_{1}|,| z_{1}|)\|},\;\;\;(w_{0}, z_{0}):=\dfrac{(w_{1}, z_{1})}{\|(w_{1}, z_{1})\|}.$$
 \par Since $(w_{1},z_{1})\in \mathcal{N}^{+}$,~from the proof of Lemma 2.1 we know
$$B(\dfrac{w_{1}}{\|(w_{1},z_{1})\|},\dfrac{z_{1}}{\|(w_{1},z_{1})\|})>0,$$
 thus $B(|w_{1}|,| z_{1}|)\geq B(w_{1},z_{1})>0$, which yields $B(|w_{0}|, |z_{0}|)>0$. By the proof of Lemma 2.1, there exists $t_{1}>0$ such that
$t_{1}\cdot(|w_{1}|,| z_{1}|)\in \mathcal{N}^{+}$.~Since
 $t_{1}\|(w_{1},z_{1})\|\cdot(|w_{0}|, |z_{0}|)=t_{1}\|(|w_{1}|,|z_{1}|)\|\cdot(|w_{0}|, |z_{0}|)=t_{1}\cdot(|w_{1}|,| z_{1}|)\in \mathcal{N}^{+}$,
 we know~$t_{1}\|(w_{1},z_{1})\|$~is the first stationary point of the fibering map in the direction $(|w_{0}|, |z_{0}|)$.
  \par
  Moreover, $(w_{1},z_{1})\in \mathcal{N}^{+}$ is equivalent to $\|(w_{1},z_{1})\|(w_{0}, z_{0})\in \mathcal{N}^{+}$, so $\|(w_{1},z_{1})\|$ is the first stationary point of the fibering map in the direction $(w_{0}, z_{0})$. Since $B(|w_{0}|,| z_{0}|)\geq B(w_{0},z_{0})>0$ and $A(|w_{0}|,| z_{0}|)=A(w_{0},z_{0})$, we can compare the above two roots of the associated fibering map to infer $t_{1}\|(w_{1},z_{1})\|\geq\|(w_{1},z_{1})\|$. That is
\begin{equation}\label{}
    t_{1}\geq 1.
\end{equation}
Taking account of the graph of the fibering map in direction~$(|w_{1}|,| z_{1}|)$, one has from (4.12) and~the fact $t_{1}\cdot(|w_{1}|,| z_{1}|)\in \mathcal{N}^{+}$ that,
\begin{center}
    $J(t_{1}|w_{1}|,t_{1}| z_{1}|)\leq J(|w_{1}|,| z_{1}|)$.
\end{center}
Thus
\begin{center}
    $\theta ^{+}\leq J(t_{1}|w_{1}|,t_{1}| z_{1}|)\leq J(|w_{1}|,| z_{1}|)\leq J(w_{1}, z_{1})=\theta ^{+}$,
\end{center}
from which we know~$(t_{1}|w_{1}|,t_{1}| z_{1}|)$~solves problem~(3.2).~By Lemma 3.2 we see~$(t_{1}|w_{1}|,t_{1}| z_{1}|)$~is a weak solution of system (1.1).

Now we consider the minimization problem~(3.3). By Lemma 4.7, there exists~$\{(u_{n},v_{n})\}\subset \mathcal{N}^{-}$, such that  as~$n\rightarrow\infty$
\begin{center}
  $J(u_{n},v_{n})\rightarrow \theta^{-}$,~~~~
   $J'(u_{n},v_{n})\rightarrow 0$.
\end{center}
Since~$J$~satisfies~$(PS)$~condition by Lemma 4.8, we find a~$\{(w_{2},z_{2})\}\in cl~(\mathcal{N}^{-})=\mathcal{N}^{-}$
such that $J(w_{2},z_{2})= \theta^{-}$,~$J'(w_{2},z_{2})=0 $, and $(w_{2},z_{2})$ is a nontrivial weak solution of (1.1) by Lemma 3.2.

Furthermore,~if~$f$ and $g$~are both positive,~we show the minimizer for (3.3) can be chosen to be a multiple of $(|w_{2}|,| z_{2}|)$. By the proof of Lemma 2.1, there exists~$t_{2}>0$~such that
$t_{2}\cdot(|w_{2}|,| z_{2}|)\in \mathcal{N}^{-}$. Moreover, one calculates that the two fibering maps in direction $(w_{2}, z_{2})$ and $(|w_{2}|,| z_{2}|)$ has the same turning point denoted $t_{0}=\sqrt{\frac{1}{3A(w_2,z_2)}}=\sqrt{\frac{1}{3A(|w_2|,|z_2|)}}$. Thus~$t_{2}>t_{0}$, and by investigating the graph of the fibering map in direction $(w_{2},z_{2})$~one gets
\begin{center}
    $J(t_{2}w_{2},t_{2} z_{2})\leq J(w_{2}, z_{2})$.
\end{center}
\par Now we have
\begin{center}
    $\theta ^{-}\leq J(t_{2}|w_{2}|,t_{2}| z_{2}|)\leq J(t_{2}w_{2},t_{2} z_{2})\leq J(w_{2}, z_{2})=\theta ^{-}$,
\end{center}
so~$(t_{2}|w_{2}|,t_{2}| z_{2}|)$~solves~(3.3).~By Lemma 3.2 we see~$(t_{2}|w_{2}|,t_{2}| z_{2}|)$~is a weak solution of system (1.1).

We finish the proof by showing $\theta_{+}<\theta_{-}$. In fact, if $(w_{2},z_{2})$   is the minimizer of (3.3) satisfying $B(w_{2},z_{2})\leq0$, then
the associated fibering map has only one stationary point, which implies $\theta_{-}>0$ by the proof of Lemma 2.1.
 So $\theta_{+}<\theta_{-}$ by Lemma 4.2. On the other hand, if $(w_{2},z_{2})$ satisfies $B(w_{2},z_{2})>0$, then
 the associated fibering map $\phi(t)$ has two stationary points: $t_{1},t_{2}(=1)$. Thus from the graph of this
  fibering map, we get immediately that $\theta_{+}\leq \phi(t_{1}) <\phi(t_{2})=\theta_{-}$. \hfill{$\square$}
 \vskip 0.4cm
 {\bf Remark} ~~~ For the definition $A(u,v)$ (2.1), we have by H\"{o}lder inequality $$\int_{\Omega}u^2v^2\leq(\int_{\Omega}u^4)^{\frac{1}{2}}(\int_{\Omega}v^4)^{\frac{1}{2}}=\|u\|_4^2\cdot\|v\|_4^2,$$
 then
 $$A(u,v)\geq\mu_1\|u\|_4^4+\mu_2\|v\|_4^4-2|\beta|\|u\|_4^2\|v\|_4^2\geq 2(\sqrt{\mu_1\mu_2}-|\beta|)\|u\|_4^2\cdot\|v\|_4^2,$$
 and the equality  is satisfied for the second inequality if and only if $\mu_1\|u\|_4^4=\mu_2\|v\|_4^2$.
 Thus as $\|(u,v)\|=1, \beta>-\sqrt{\mu_1\mu_2}$, we have $$A(u,v)>0.$$
 Therefore, as $\beta>-\sqrt{\mu_1\mu_2}$, all the proofs are valid, Theorem 1.1  above is still true.\hfill{$\square$}
\par
{\bf Acknowledgements:} The authors thank the referees for their
careful reading and helpful suggestion.


TITLE: What's the meaning of $\mathbb{N}_0,\wedge$
QUESTION [2 upvotes]: What's the meaning of the $\wedge$ in $\mathbb{N}_0,\wedge$?
The context is an exercise about the properties of mathematical operations.
I'm aware of $\wedge$ as a logic operator, but I'm quite sure that's not meant here.
I've found lots and lots of pages about a $\wedge$- or external product on vectorspaces, but I don't think I should interpret $\mathbb{N}_0$ as a vectorspace?
I've also found $a\wedge b = \min(a,b)$ (in this question).  That would make sense to me.  Is there a logical explanation why this should the correct interpretation?

REPLY [1 votes]: In some programming languages, $\wedge$ means the bitwise xor function.  That is, $a\wedge b$ means computing the sum of $a$ and $b$ in base two without carry: so $2^k \wedge 2^k = 0$, $2^k \wedge 2^l = 2^k + 2^l$ if $k \ne l$, and $\wedge$ is commutative and associative.


TITLE: Can this be converted into a polynomial equation for $x$?
QUESTION [2 upvotes]: I came upon this monster equation while fiddling with the area of pentagons:
$$\sqrt{(a+b+c-x)(a+b-c+x)(a+c-b+x)(b+c-a+x)}+\sqrt{(c+d+x)(c+d-x)(c-d+x)(d-c+x)}=4T$$
Where $a,b,c,d,e,T$ are known quantities. So my question is:

Can this be converted into a polynomial equation for $x$?
If so how can we do it, and what would be its degree?

Any help is appreciated :)

REPLY [2 votes]: If you have an equation of the form $\sqrt a + \sqrt b = c$ then by taking squares we get $2\sqrt {ab} = c^2-a-b$, and then $4ab = (c^2-a-b)^2$.
You can also develop the product in the equation $(\sqrt a + \sqrt b - c)(\sqrt a - \sqrt b - c)(- \sqrt a + \sqrt b - c)(- \sqrt a - \sqrt b - c) = 0$, which gives you (again) $4ab = (c^2-a-b)^2$.


TITLE: $2^W = (2a_1+1)^2 \cdot 2^{M_1}-a_1$ - is there a nontrivial diophantine solution in $M_1,a_1$ given possible?
QUESTION [3 upvotes]: Update/error: I had a small error in the $(2a+1)$-term in the formula which should have been $(a+1)$ . Because I added now my derivations in long and broad at the end of the question I also rewrite a bit the unknowns $a$ by a shift of $a+1$ to $a$ now reading as $a_1$ and $M_1=M-1$ instead of $M$
I'm looking at whether the rhs in the following equation
$$2^W = (2a_1+1)^2 \cdot 2^{M_1}-(a_1)$$
can ever be a perfect power of $2$ for integer $W>0$, $M_1 \ge 0$ and integer (positive or negative) $a_1$.
I know that for $a_1=1$ and $M_1=0$ we have one solution:
$$2^W = (3)^2 \cdot 2^0-(+1) = 9 - 1 = 2^3 $$

Q : Is there another or no other solution?
  Q : how to do a proof?

I fiddled with this today and didn't get it. Maybe this is easy and I'm only dense at the moment...

Motivation: I'm considering the possibility of 2-step-cycles in the generalized Collatz-problem $ b={m \cdot a +1 \over 2^A} \to a={m \cdot b +1 \over 2^B}$ for odd $m$ and $a,b$ (note: both $a,m$ in the positive and negative).
Letting $a,b$ both be unknown I first checked numerically up to very large positive $m$ and verified the sparseness of known cycles up to some $m \approx 10^{1400}$. But a proof of the nonexistence of further 2-step-cycles seemed out of reach for me.
So I tried with a simplified version, where $b=1$ and only $a$ and $m$ is unknown.
This is what I found so far:
A useful ansatz for the analysis of cycles is to write the equality of products
$$ b \cdot a ={m \cdot a +1 \over 2^A} \cdot {m \cdot b +1 \over 2^B}$$ and reorder to arrive at
$$ 2^{A+B} = (m+{1 \over a}) \cdot (m+{1 \over b})$$
For convenience I always write $S$ for the sum $A+B$ and to simplify the problem we set here $b=1$ and get the

Main formula:
$$ 2^S = (m+1)\cdot(m+1/a) \tag 1$$
Because the second parenthese has a denominator $a$ it follows by the first parenthese, that $m$ must have the structure with some integer $m_1$
$$m=m_1 \cdot a - 1 \tag {1.1}$$
The equation changes to
$$ 2^S = (a\cdot m_1)\cdot(a\cdot m_1-1+1/a) \\
= m_1\cdot(a^2\cdot m_1-a+1) \tag 2$$
Because both factors must be perfect powers of $2$ it follows $m_1$ must be a perfect power of $2$ (possibly $2^0=1$), say $m_1=2^M$ so $ m=2^M \cdot a-1$ and then
$$ 2^S = 2^M\cdot(a^2\cdot 2^M-a+1) \tag {2.1}$$
Because by definition all $a,b$ in the generalized Collatz-problem with $b={m \cdot a +1\over 2^A} $ are odd we can write the structure of $a$ by some integer $a_1$ as $a=2a_1+1$.
Beforehand, we exclude $a=1$ by the problem definition (we want discuss the existence of a two-step-cycle $1 \to a \to 1$) so a solution $a_1=0$ is not wanted - it would be the trivial one-step-cycle which is existent for many $m$ repeated two times).
What we have now is
$$ 2^S = 2^M\cdot((2a_1+1)^2\cdot 2^M-2a_1-1+1)\\
   2^{S-M-1} = (2a_1+1)^2\cdot 2^{M-1}-a_1 \tag 3$$
or simplified for the actual original question
$$   2^W = (2a_1+1)^2\cdot 2^{M_1}-a_1 \tag {3.1 for OP} $$
Small solutions are
 a_1   a   (2a_1+1)^2    rhs                M       m     S     comment
------------------------------------------------------------------------
  1    3        9       9* 2^{M-1}-1        1      5      5    (well known) solution 1
  0    1        1       2^{M-1}            any    -1      2M   trivial, not wanted
 -1   -1        1       2^{M-1}+1           1     -3      3     new (?)
-23  -45     2025  2025*2^{M-1}+23          1    -91     13     new (?)
 ??? ???      ???                                       .

The likely full set of 2-step-cycles with $b$ varying over the odd integers and $a \ne b$ seems to be
   S    m:   a    b   a ...
   ------:-----------------
   3,   3:  -7,  -5,  -7
   3,  -3:  -1,   1,  -1
   5,   5:   1,   3,   1
   7, -11:  -1,   3,  -1
  13, -91: -45,   1, -45
  15, 181:  27, 611,  27
  15, 181:  35,  99,  35

But this was taken using brute force up to $m \approx 10^{300}$, no analytical evaluation about the possibility of 2-step-cycles with $m$ of larger absolute values has been done.

REPLY [1 votes]: We want to show that $$(2a+1)^2\cdot 2^M-a=2^W$$ with integers $W,a,M$, $W>0$ , $a\ne 0$ , $M\ge 0$ has the only solutions $M=0$ and $a=-23,-1,1$
In the case $M=0$, we have $$\frac{(8a+3)^2+7}{16}=2^W$$ so we can use Nagell's equation to show that $-23,-1,1$ are the only solutions.
For $M>0$ We can rewrite the left side as $$\frac{[(2a+1)\cdot 2^{M+2}-1]^2+2^{M+3}-1}{2^{M+4}}$$
The given link about Nagell's equation mentions that $u^2+2^m-1=2^n$ has at most $2$ positive solutions for $m \ge 4$. It is easy to see that $u=1$ and $u=2^{m-1}-1$ are solutions, and there are no more.
The expression $(2a+1)\cdot 2^{M+2}-1$ cannot be $\pm1$.
Assume the expression is $\pm(2^{M+2}-1)$. In the case of the plus-sign we get $a=0$. The minus-sign also leads easily to a contradiction. Therefore, there are no solutions. This completes the proof.


TITLE: Does it make any sense to prove $0.999\ldots=1$?
QUESTION [0 upvotes]: I have read this post which contains many proofs of $0.999\ldots=1$.
Background
The main motivation of the question was philosophical and not mathematical. If you read the next section of the post then you will see that I have asked for a "meaning" of the symbol $0.999\ldots$ other than defining it to be $1$. Now here is a epistemological problem and this is mainly the problem from which the question arose. Suppose you know that $1$ is a real number. Now I give you a symbol, say $0.999\ldots$ which from now on I will denote as $x$. Now I ask you whether $x$ is a real number. To answer this, if you define $x=1$ then you are already attributing the properties of $1$ to $x$ among which one is it being a real number without proving whether to $x$ we can indeed attribute the properties of $1$.
A common response to this question has been to define the symbol $x$ as the limit of the sequence $\left(\displaystyle\sum_{i=1}^n \dfrac{9}{10^i}\right)_{n\in\mathbb{N}}$ and then prove that the limit of this sum is indeed $1$. But again the problem is that you are defining the symbol $x$ to be a real number and hence are assuming a priori that the symbol $x$ denotes a real number.
As per the discussion that has been conducted with Simply Beautiful Art let me state my position again in brief,

Also let me say that I do not disallow $0.999…$ to be a real number. My impression that if you assume $0.999…$ to be a real number then there is no sense in proving that $0.999…$ is indeed equal to $1$ because either you define it to be $1$ or you prove the equality as a theorem. But if you are going to use the limit definition of $0.999…$ then what you are a priori assuming it to be a real number which is an assumption that I don't allow. What can be allowed is that $0.999…$ is a number (but not necessarily a real number).

Question
My question is,

Does it make any sense to prove this equality?

Can one give any "meaning" of the symbol $0.999\ldots$ other than defining it to be $1$?

REPLY [0 votes]: From epistemological point of view, there is no difference in that question and the question if some two objects are both elephants. If they share the same set of properties that defines an object being an elephant, they are. One elephant might have a leg missing or very short trunk, or broken tusk, and still be an elephant.
The same for $0.9999\cdots$ and $1$.
When you talk about $1$ you usually consider a unique element, known as identity element, that is required to exist within multiplication of real numbers, we will mark it as $e$. It satisfies $a \cdot e=a, a \neq 0$. That is what we are looking for. $e$ not $1$ or $0.999\cdots$. Why $e$? Because it gives as the set of properties to check if $1$ and $0.999\cdots$ are both elephant.
From that point of view,  $1$ or $0.999\cdots$ are two equivalent representations of $e$.
So there is nothing to prove. All proofs are actually quite artificial and actually make it more complicated, because any proof that $0.999\cdots=1$ already contains the axiom that they are equal. It must contain it. So it does not say anything new or moves us into a direction of some sort of clarification.
Sure, if you understand $0.999\cdots$ as a series, then you need to prove that its limit is $1$, but that is way advanced and no different than singing an English lullaby for no reason suddenly in Chinese.
To "prove" that $0.999\cdots$ and $1$ belong to the same class of equivalence, all you need is the defined set of properties of real numbers. This is always implied, since if we are not looking into reals, they do not have to be the same whatsoever. For example, lexicographically, obviously, they are not the same words and so on. It is the concept of real numbers that is making them equivalent words, words with the same meaning: potayto, potahto. It is an edible plant. And that is what I am concerned of, since I am hungry. The same, I need something that will represent an element that does nothing during the multiplication. And for that I can use either.


TITLE: What kind of equations are differentiable?
QUESTION [1 upvotes]: F(n) = F(n-1) + F(n-2)

Is this differentiable? How or why not?
I understand how to differentiate equations like x**3 + 2x**2 + 3x + 5. But this is a recurrence relation and I don't know how it is differentiated.
I'm trying to understand the rate of growth of this equation. Can you tell me what would be the rate of growth at n = 987654?

REPLY [3 votes]: Equations can never be differentiable, but functions may or may not be.
Without any context, it looks like your equation is actually a recurrence equation, and $n$ is assumed to be an integer. Functions that are only defined for integer values are never differentiable.
On the other hand, your equation defines the so called Fibonacci numbers (or something similar, depending on the initial values), and a resonable interpretation of "rate of growth" would be $F(n+1)-F(n)$, which can be computed. If $F(0) = F(1) = 1$, then
$$F(n) = \frac{1}{\sqrt 5}\left( \frac{1+\sqrt5}{2}\right)^n - \frac{1}{\sqrt 5}\left( \frac{1-\sqrt5}{2}\right)^n.$$
I'll leave it to you to compute $F(987655)-F(987654)$.


TITLE: What is the a matrix decomposition to swap out an eigenvalue?
QUESTION [0 upvotes]: I have the following problem: let $A$ be an $m \times m$ matrix that is not symmetric and real. Then, I am interested in its eigenspaces that correspond to the largest eigenvalue, $\lambda_{\text{max}}$. Without affecting the column span of the eigenspace associated with the largest root, say $\text{sp}\,r_{\text{max}}$, what can I do to $A$ to effect the following change: 1. remove $\lambda_{\text{max}}$, 2. insert $\mu$ in its stead, 3. obtain the relation $A\, r_{\text{max}} = \mu r_{\text{max}}$ so that the eigenvalue associated with $\text{sp}\,r_{\text{max}}$ is now $\mu$. I was thinking of possibly using the real Schur decomposition, thanks to its numerical stability. Any help is much appreciated!
Follow-up: suppose that the eigenvalue is potentially complex but the matrix is real. Then, as suggested, calculating $A + I \left(\mu -\lambda_{\text{max}} \right)$ would cause a problem because $\lambda_{\text{max}}$ comes in a conjugate pair. We (re-)define the notion of 'max' to be maximal by modulus, so $\lambda_{\text{max}}$ is such that $||\lambda_{\text{max}}||$ is maximal. (This arises from the fact that I am estimating $A$ using least squares. A few references, avenues I have considered is applying the method of Gershgorin discs or the Bauer-Fike theorem just to bound $||\lambda_{\text{max}}||$ using the (real) entries of $A$ but have not made much progress.
Moreover, it would be interesting to know if one could use the derivative of $A$ with respect to its eigenvalues to 'slide it' towards a matrix that would have the desired root, very much one uses the first derivative of a function to attain a linear approximation to.

REPLY [3 votes]: The "obvious" thing to do is to change each $\lambda_{max}$ on the diagonal of the Schur decomposition to $\mu$, but that may change the eigenspace.  For example, if
$$A = \pmatrix{1 & 1\cr 0 & 2\cr}$$ (already upper triangular) the eigenspace for $2$ is
spanned by $\pmatrix{1\cr 1\cr}$, but if you change the $2$ to $\mu$ the eigenspace for $\mu$ becomes
$\pmatrix{1/(\mu-1)\cr 1\cr}$.
Instead, why not just add $(\mu - \lambda_{max}) I$ to the matrix?  The eigenspaces stay the same, just that all eigenvalues are shifted by $\mu - \lambda_{max}$.
EDIT: Another possibility is to use the functional calculus:
for any polynomial $p(X)$, $p(A)$ is a matrix whose eigenvalues are $p(\lambda)$ for eigenvalues $\lambda$ of $A$. If $v$ is an eigenvector of $A$ for $\lambda$, then $p(A) v = p(\lambda) v$. Moreover, if $p$ has real coefficients, $p(A)$ will still be real.  Choose a polynomial such that
$p(\lambda_{max}) = \mu$.


TITLE: How can I parametize the intersection of sphere and plane?
QUESTION [1 upvotes]: I am doing Stokes' theorem and have to parametize the the curve of intersection of the sphere $x^2+y^2+z^2=a^2$ and plane $x+y+z=0$
I can't think of any parametization..
I let $z=-y-x$ and plug in sphere equation and got nothing

REPLY [1 votes]: This curve will be a circle centered at the origin (why?).   Any circle in $\mathbb{R}^3$ centered at the origin admits a parametrization of the form $\alpha(t) = \mathbf{v_1}\cos(t) + \mathbf{v_2}\sin(t)$, where $\mathbf{v_1}$ and $\mathbf{v_2}$ are orthogonal vectors of equal magnitude.
One way of going about finding appropriate vectors $\mathbf{v_1}$ and $\mathbf{v}_2$ would be to choose an arbitrary $\mathbf{v_1}$ on the boundary of the circle: say $x = 0, y=-a/\sqrt{2}$ and $z = a/\sqrt{2}$.  Then find the unit normal vector $N$ to the plane and take $\mathbf{v}_2 = N \times \mathbf{v}_1$.


TITLE: Why does it seem to decrease probability to roll more times?
QUESTION [2 upvotes]: I am looking how to find the probability of rolling 10 sixes in 60 rolls. After reading Calculate probabilities. 40 sixes out of a 100 fair die rolls & 20 sixes out of a 100 fair die roll, it seems that I can use the following formula: $60 \choose 10$$(\frac{1}{6})^{10}$$(1-\frac{1}{6})^{60-10}$
That comes to about 13.7%. So far, so good.
But then I decided to test it. What about getting 10 sixes in 120 rolls?
$120 \choose 10$$(\frac{1}{6})^{10}$$(\frac{5}{6})^{120-10}$
That's only 0.37%! I'm getting something wrong, but what?

REPLY [4 votes]: The formula you quote is for getting exactly (not at least) $10\ 6$'s in $60$ rolls.  The factor $(1-\frac 16)^{60-10}$ ensures that the other dice roll something other than $6$.  Your extension is correct for getting exactly (not at least) $10\ 6$'s in $120$ rolls.  The expected number is $20$, so it is not surprising that the probability has decreased-you are farther from the expected number.

REPLY [1 votes]: What you've calculated (correctly) are the probabilities that you get exactly $10$ sixes in $60$ rolls and $120$ rolls, respectively. If you think about this, it makes sense that the probability of the latter should be lower. Intuitively, there are more possibilities for how many sixes show up in the second case.


text:
\subsection{Review on the bi-algebraic system}\label{SubsectionReviewOnBiAlgSystem}
Let $M = \Gamma \backslash \cX^+$ be a connected mixed Shimura variety of Kuga type associated with $(P,\cX^+)$. Let $\bu \colon \cX^+ \rightarrow M$ be the uniformization.

Recall the realization of $\cX^+$ in $\mathsection$\ref{SubsectionRealizationOfX}. By Proposition~\ref{PropositionRealizationOfX} $\cX^+$ can be realized as an open (in the usual topology) semi-algebraic subset of a complex algebraic variety $\cX^\vee$.

\begin{defn}
\begin{enumerate}
\item[(i)] A subset $\tilde{Y}$ of $\cX^+$ is said to be \textbf{irreducible algebraic} if it is a complex analytic irreducible component of $\cX^+ \cap W$, where $W$ is an algebraic subvariety of $\cX^\vee$.
\item[(ii)] A subset $\tilde{Y}$ of $\cX^+$ is said to be \textbf{irreducible bi-algebraic} if it is an irreducible algebraic subset of $\cX^+$ and $\bu(\tilde{Y})$ is an algebraic subvariety of $M$.
\item[(iii)] A closed irreducible subvariety $Y$ of $M$ is said to be \textbf{bi-algebraic} if one (and hence any) complex analytic irreducible subvariety of $\bu^{-1}(Y)$ is irreducible algebraic in $\cX^+$. A closed subvariety $Y$ of $M$ is said to be bi-algebraic if all of its irreducible components are bi-algebraic.
\end{enumerate}
\end{defn}
The following result is not hard to prove.
\begin{lemma}
If $F_1$ and $F_2$ are bi-algebraic subvarieties of $M$, then every irreducible component of $F_1 \cap F_2$ is also bi-algebraic.
\end{lemma}
In view of this, we can introduce the following notation.
\begin{notation}\label{NotationBiAlg}
\begin{enumerate}
\item[(i)] Let $\tilde{Z}$ be any complex analytic irreducible subset of $\cX^+$. Use $\tilde{Z}^{\Zar}$ to denote the smallest irreducible algebraic subset of $\cX^+$ which contains $\tilde{Z}$, and use $\tilde{Z}^{\biZar}$ to denote the smallest irreducible bi-algebraic subset of $\cX^+$ which contains $\tilde{Z}$.
\item[(ii)] Let $Z$ be any subset of $M$ (not necessarily a subvariety). Use $Z^{\biZar}$ to denote the smallest bi-algebraic subvariety of $M$ which contains $Z$ (hence $Z^{\biZar}$ contains $Z^{\Zar}$).
\end{enumerate}
\end{notation}

\subsection{Statement of the Ax-Schanuel theorem}
Let $M = \Gamma \backslash \cX^+$ be a connected mixed Shimura variety of Kuga type associated with $(P,\cX^+)$. Let $\bu \colon \cX^+ \rightarrow M$ be the uniformization. The typical case is when $M= \mathfrak{A}_{g,D}$ is the universal abelian variety.

\begin{thm}[Ax-Schanuel for mixed Shimura varieties of Kuga type]\label{ThmASUnivAbVar}
Let $\mathscr{Z}$ be a complex analytic irreducible subvariety of $\mathrm{graph}(\bu) \subseteq \cX^+ \times M$, and denote by $\tilde{Z}$ the image of $\mathscr{Z}$ under the natural projection $\cX^+ \times M \rightarrow \cX^+$. Then
\[
\dim \mathscr{Z}^{\Zar} - \dim \tilde{Z} \ge \dim \tilde{Z}^{\biZar},
\]
where $\mathscr{Z}^{\Zar}$ means the Zariski closure of $\mathscr{Z}$ in $M \times \cX^+$. Moreover the equality holds if and only if $\tilde{Z}$ is a complex analytic irreducible component of $\tilde{Z}^{\Zar} \cap \bu^{-1}(\bu(\tilde{Z})^{\Zar})$.
\end{thm}

We also state the weak Ax-Schanuel theorem for mixed Shimura varieties of Kuga type to make the statement more clear. It follows directly from Theorem~\ref{ThmASUnivAbVar}.
\begin{thm}[weak Ax-Schanuel for mixed Shimura varieties of Kuga type]\label{ThmWASUnivAbVar}
Let $\tilde{Z}$ be a complex analytic irreducible subset of $\cX^+$. Then
\[
\dim (\bu(\tilde{Z}))^{\Zar} + \dim \tilde{Z}^{\Zar} \ge \dim \tilde{Z} + \dim \tilde{Z}^{\biZar}.
\]
Moreover the equality holds if and only if $\tilde{Z}$ is a complex analytic irreducible component of $\tilde{Z}^{\Zar} \cap \bu^{-1}(\bu(\tilde{Z})^{\Zar})$.
\end{thm}

Before moving on, we point out that the ``Moreover'' part of Theorem~\ref{ThmWASUnivAbVar} immediately follows from the main part and the Intersection Dimension Inequality.

\begin{rmk}
Our proof of Theorem~\ref{ThmASUnivAbVar} uese the work of Mok-Pila-Tsimerman \cite{MokAx-Schanuel-for} on the Ax-Schanuel theorem for pure Shimura varieties and extends their proof. As a statement itself, Theorem~\ref{ThmASUnivAbVar} implies the pure Ax-Schanuel theorem.
\end{rmk}

\subsection{Geometric description of bi-algebraic subvarieties}\label{SubsectionGeomDescriptionBiAlg}
By \cite[Theorem~8.1]{GaoTowards-the-And}, bi-algebraic subvarieties of $\mathfrak{A}_{g,D}$ are precisely the weakly special subvarieties defined by Pink \cite[Definition~4.1.(b)]{PinkA-Combination-o}. The definition of weakly special subvarieties will be given in $\mathsection$\ref{SubsectionWeaklySpecial}. For the moment we present the geometric description of these weakly special subvarieties. Recall the projection $\pi \colon \mathfrak{A}_{g,D} \rightarrow \A_{g,D}$, which is proper. We have the following result.
\begin{prop}[{\cite[Proposition~1.1]{GaoA-special-point}}]\label{PropGeomDescriptionOfBiAlgebraic}
A closed irreducible subvariety $Y$ of $\mathfrak{A}_{g,D}$ is bi-algebraic (or equivalently weakly special) if and only if the following conditions hold:
\begin{enumerate}
\item[(i)] Its projection $\pi(Y)$ is a weakly special subvariety of $\A_{g,D}$;
\item[(ii)] Up to taking a finite covering of $\pi(Y)$, we have that $Y$ is the translate of an abelian subscheme of $\pi^{-1}(\pi(Y)) \rightarrow \pi(Y)$ by a torsion section and then by a constant section.
\end{enumerate}
\end{prop}
Let us explain condition~(ii) in more details. We have that $\pi(Y)$ is a closed irreducible subvariety of $\A_{g,D}$, hence $\pi^{-1}(\pi(Y)) = \mathfrak{A}_{g,D}|_{\pi(Y)}$ is an abelian scheme over $\pi(Y)$. Condition~(ii) means: there exists a finite covering $B' \rightarrow \pi(Y)$ such that under the base change $\mathfrak{A}' := B' \times_{\pi(Y)} \pi^{-1}(\pi(Y))$ and the natural projection $p' \colon \mathfrak{A}' \rightarrow \pi^{-1}(\pi(Y))$, we have $Y = p'(\mathfrak{B}+\sigma + \sigma_0)$, where $\mathfrak{B}$ is an abelian subscheme of $\mathfrak{A}' \rightarrow B'$, $\sigma$ is a torsion section of $\mathfrak{A}'\rightarrow B'$, and $\sigma_0$ is a constant section of (the largest constant abelian subscheme of) $\mathfrak{A}'\rightarrow B'$.

Proposition~\ref{PropGeomDescriptionOfBiAlgebraic} is proven as a consequence of the following result.
\begin{prop}[{\cite[Proposition~3.3]{GaoA-special-point}}]\label{PropGeomOfsgInUnivAbVar}
Let $B$ be an irreducible subvariety of $\A_{g,D}$. Then
\[
\begin{array}{c}
\{\text{Up to taking a finite covering of }B,\text{ the translates of an abelian subscheme} \\
\text{of }\pi^{-1}(B) \rightarrow B\text{ by a torsion section and then by a constant section}\} \\
= \{\text{irreducible components of }\pi^{-1}(B) \cap F : F\text{ weakly special in }  \mathfrak{A}_{g,D} \text{ with } B \subseteq \pi(F)\}.
\end{array}
\]
\end{prop}

\subsection{Review on weakly special subvarieties}\label{SubsectionWeaklySpecial}
Now let us give the definition of weakly special subvarieties following Pink. Let $(P,\cX^+)$ and $M$ be as above Theorem~\ref{ThmASUnivAbVar}.
\begin{defn}
\begin{enumerate}
\item[(i)] A subset $\tilde{Y}$ of $\cX^+$ is said to be \textbf{weakly special} if there exist a connected mixed Shimura subdatum of Kuga type $(Q,\cY^+)$ of $(P,\cX^+)$, a normal subgroup $N$ of $Q^{\der}$, and a point $\tilde{y} \in \cY^+$ such that $\tilde{Y} = N(\R)^+\tilde{y}$.
\item[(ii)] A subvariety $Y$ of $M$ is said to be \textbf{weakly special} if $Y = \bu(\tilde{Y})$ for some weakly special subset $\tilde{Y}$ of $\cX^+$.
\end{enumerate}
\end{defn}
Our formulation is slightly different from \cite[Definition~4.1.(b)]{PinkA-Combination-o}, but it is not hard to show that they are equivalent; see \cite[$\mathsection$5.1]{GaoTowards-the-And}.

config: arxiv


TITLE: Congruences and Legendre
QUESTION [0 upvotes]: I am trying to solve a Legendre symbol problem and have got it down to the following:

When $p \equiv 1\mod4$ and a prime such that $p \neq 2,7$, $\left(\frac{7}{p}\right) = \left(\frac{p}{7}\right) \iff p \equiv 1,2,4\mod7 \iff  p \equiv 1,9,25\mod28$.

Why does this last 'if and only if' hold?

REPLY [1 votes]: It is not helpfully phrased. If $p=1\bmod4$, then $p=1\bmod7$ iff $p=1\bmod28$. Similarly, if $p=1\bmod4$, then $p=2\bmod7$ iff $p=9\bmod28$. Finally, if $p=1\bmod4$, then $p=4\bmod7$ iff $p=25\bmod28$.
You can easily check these. $p=1\bmod4$ implies $p=1,5,9,13,17,21$ or 25 mod 28. But the only one of these which is 1 mod 7 is 1. Similarly for the other cases.


TITLE: Extension of map $f$
QUESTION [2 upvotes]: Let $X$ be a compact Hausdorff space,$C \subseteq X$ closed and $f:C \rightarrow S^n$ a continuous map.Is it correct to say that there exists an open neighbourhood $U \supseteq C$ in $X$ and a continuous map $F:U \rightarrow S^n$ such that $ F|_C=f$.
I'm wondering if I can use Tietze extension theorem to show it's correct.If so,what would be formal proof then?

REPLY [2 votes]: Yes. Let $f_i : C \to \mathbb R$ be the coordinate functions of $f : C \to S^n  \hookrightarrow \mathbb R^{n+1}$. By the Tietze extension theorem each $f_i$ has an extension $F_i : X \to \mathbb R$ which gives an extension $F : X \to R^{n+1}$ of $f$. Now let $V = R^{n+1} \setminus \{ 0\}$. There exists a retraction $r : V \to S^n, r(x) = x/\lVert x\rVert$. Then $U = F^{-1}(V)$ is an open neigbourhood of $C$ in $X$ and
$$f ' : U \stackrel{F}{\rightarrow} V \stackrel{r}{\rightarrow} S^n$$
is an extension of $f$.
Note that this works for any normal $X$.


TITLE: How does one find how many nilpotent groups are of any particular order?
QUESTION [1 upvotes]: For example, a past exam question asks to find all the nilpotent groups of order 70. How many are there and is there a general method to finding the answer?

REPLY [1 votes]: The paper 'Nilpotent Numbers' of Pakianathan and Shankar (http://www2.math.ou.edu/~shankar/papers/nil2.pdf), shows that every group of order $n$ is nilpotent if and only if $p^k\not\equiv 1$ mod $q$, whenever $p,q$ are distinct primes with $p^k$ and $q$ dividing $n$.


TITLE: Simplifying trig expression cos and sin.
QUESTION [0 upvotes]: I'm reading a solution right now, and I am stuck.
$$\frac{\cos\left(\frac{\theta}{2}\right) - \cos\left(n + \frac{1}{2}\right)\theta}{2} \sin\left(\frac{\theta}{2}\right) =\frac{\sin(n\frac{\theta}{2}))\sin\left(\left[n+1\right]\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}$$
How could these two equations be equal.

REPLY [0 votes]: Not sure about the accuracy/format of the problem as
using $\displaystyle \cos C-\cos D=2\sin\frac{C+D}2\sin\frac{D-C}2$
$$\cos\frac\theta2-\cos\left(n+\frac12\right)\theta=2\sin\frac{n\theta}2\sin\frac{(n+1)\theta}2 $$


TITLE: Applying Fourier series for $|\sin x|$
QUESTION [1 upvotes]: Why when we apply Fourier series for $|\sin x|$ from $0 < x < \pi$ , we set $2L = 2\pi$?
Shouldn't it be $2L = \pi$?
In Schaum's Outline of Advanced Calculus book, there's a question that says:

"Expand $f(x) = \sin x, 0 < x < \pi$, in a Fourier cosine series.
A Fourier series consisting of cosine terms alone is obtained only for an even function. Hence, we extend the definition of $f(x)$ so that it becomes even. With this extension, $f(x)$ is then defined in an interval of length $2\pi$. Taking the period as $2 \pi$, we have $2L = 2\pi$ so that $L = \pi$."

REPLY [1 votes]: Given a function $f\!: x\mapsto f(x)$ on some interval $I$ of length $L>0$ one obtains its Fourier expansion by extending $f$ to all of ${\mathbb R}$ periodically using the given $L$ as period length. The coefficients of the Fourier series should then be computed using the standard formulae for period length $L$ and integrating over the interval $I$ where the function $f$ was given in the first place.
In the case at hand we  have $I=[0,\pi]$ and $L=\pi$, and the standard formulae give
$$a_k={2\over\pi}\int_0^\pi \sin x \cos(2 k x)\ dx,\qquad
b_k={2\over\pi}\int_0^\pi \sin x \sin(2 k x)\ dx.$$
Now there is the extra condition that we want only $\cos$-terms. In order to enforce this we must make sure that the periodically extended function is even. But here we are lucky: As $\sin(\pi -x)\equiv \sin x$ the extended function is even automatically. This means that in the above formulae the $b_k$ are all zero, and there remains nothing to be done apart from calculating the integrals for $a_k$.
It would be another matter if the given function would not have this symmetry with respect to $x\mapsto \pi -x$, as in the case of $g(x):=\sin{x\over 2}$ $(0 < x <\pi)$. In this case it would be necessary to extend $g$ first  to an even function on the interval $[-\pi,\pi]$, whereupon the standard formulae for period length $2\pi$ can be applied to compute the $a_k$.


TITLE: Intervals that are free of primes
QUESTION [21 upvotes]: How can I prove that exists intervals as large as I want that are free of primes?

I mean, $\forall \ k \in \mathbb{N}, \exists \ k$ consecutive positive integers none of which is a prime.

REPLY [3 votes]: I would have formulated it another way. Using the contraposit of what we want to prove, there is a k for wich there is no sequence of k consecutive non-prime number. This lead to number of prime at least $\frac{n}{k}$ ie, $O(n)$. This is in contradiction with the prime number theorem suggested by Harold.


TITLE: How can $1/V$ be equal to $0$ in Boyle's Law?
QUESTION [1 upvotes]: In relation to ideal gases, Boyle's Law states that pressure is inversely proportional to volume under constant temperature. In other words,
$$P \propto 1/V$$
Below is a graph that plots pressure, $P$, against inverse volume, $1/V$.

How can $1/V$ ever equal zero? How is this possible?

REPLY [2 votes]: You've got to start with the ideal gas law
$$pV=mRT$$
Where, in this version, $m$ is the mass of the gas, $R$ is the specific gas constant and $T$ is the temperature. If $m$ and $T$ are constant, then $mRT$ = constant = $C$.
$$p=\frac{C}{V}$$
It is important to note that Boyle's law only applies to a closed system, i.e., a system where $m$ is constant. So even if the volume increases there are always the same number of gas molecules within that volume. Although the density of the gas (molecules per unit volume) keeps decreasing with increasing volume, it never becomes zero. And as long as there are gas molecules, there will be collisions between the molecules and any surfaces within or bounding that volume resulting in pressure proportional to the collision rate.
So if the volume were infinite, it would simply mean that an infinite amount of time would be required for a collision to occur. The rate of collisions approaches zero, but can never actually become zero as long as there are gas molecules in the volume.
Hope this helps.


TITLE: What is $1 + 999999...$ (an infinite string of $9$s)?
QUESTION [7 upvotes]: I'm doing a programming assignment in Haskell, and it involves adding "infinite" lists. At the bottom of the assignment, our lecturer has written ... "in some sense an infinite string of $9$s is the additive inverse of the number $1$, so we might say that it is equal to $-1$". I find this to be very confusing, is there actually any sense in this statement?

REPLY [1 votes]: Indeed! $\ldots 999999=-1$! I will try to elucidate it.
$$\cdots + 1000x + 100x + 10x + 1x + \frac{x}{10} + \frac{x}{100} + \frac{x}{1000} + \cdots = 0$$
Why is the equation above true? Try multiplying the left-hand side by $10$. As you see, multiplying it by $10$ doesn't change it, therefore it can only be $0$.
Let's plug $x = 9$ into the equation:
$$\cdots + 9000 + 900 + 90 + 9 + \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots = 0$$
or
$$\dots 999999,999999 \ldots = 0$$
or
$$\ldots999999 = -0,999999\ldots$$
We know the famous equation:
$$0,999999\ldots = 1$$
Therefore
$$\ldots 999999 = -1$$


TITLE: Using Separation of Variables to solve $\vartriangle u = 0, \; u(1,\theta) = g(\theta)$
QUESTION [0 upvotes]: Let $\Omega = \{(x,y) \in \mathbb{R}^2 \mid x^2 + y^2< 1\} = \{(r,\theta) \mid 0 \leq r < 1, \; 0 \leq \theta < 2\pi\}$. Use seperation of variables $(r,\theta)$ to solve the Dirichlet problem $$\begin{cases}
      \vartriangle u  = 0 \text{ in }\Omega \\
      u(1,\theta) = g(\theta)
   \end{cases}$$
Assuming $u(r,\theta) = X(r)Y(\theta)$ then we get $X''(r)Y(\theta) = -X(r)Y''(\theta)$. Dividing both sides of the equation by $X(r)Y(\theta)$, we obtain $$\frac{X''(r)}{X(r)} = \frac{Y''(\theta)}{Y(\theta)} = - \lambda$$.
Since $u(1,\theta) = g(\theta)$, then $X(1)Y(\theta) = g(\theta)$.
But, I do not know what to do from here. Can anyone provide some help?

REPLY [1 votes]: The Laplacian in polar coordinates is
$$
        \Delta f(r,\theta)= \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}f\right)+\frac{1}{r^2}\frac{\partial^2 f}{\partial\theta^2}.
$$
Let $f(r,\theta)=R(r)\Theta(\theta)$, divide $\Delta f=0$ by $f$ to obtain the separated equations:
$$
    \frac{r}{R}\frac{d}{dr}r\frac{dR}{dr}=\lambda=-\frac{\Theta''}{\Theta} \\
    r(rR')'-\lambda R=0,\;\;\Theta''+\lambda\Theta =0.
$$
Periodicity in $\theta$ requires $\lambda=n^2,\;\;n=0,1,2,3,\cdots$, and $\Theta_n(\theta)=A_n\cos n\theta + B_n\sin n\theta$, which reduces to $A_n$ for $n=0$. The corresponding solutions in $r$ are $C_n r^n+Dr^{-n}$ for $n\ne 0$ and $C_0 + D_0 \ln(r)$ for $n=0$. The solution that is non-singular at $r=0$ has the series form
$$
      f(r,\theta) = E_0+\sum_{n=1}^{\infty}E_nr^n\cos n\theta+F_nr^{n}\sin n\theta.
$$
The condition $f(1,\theta)=g(\theta)$ determines the coefficients $E_n,F_n$ through the Fourier series
$$
           g(\theta)=E_0+\sum_{n=1}^{\infty}E_n\cos n\theta+F_n\sin n\theta.
$$


TITLE: Contradiction in Abbott's Understanding Analysis or my confusion?
QUESTION [1 upvotes]: Here are two statements that the textbook considers:
(1) For any two real numbers $a, b$
$$a \leq b \quad iff \quad a < b + \epsilon, \quad \forall \epsilon > 0$$
(2) Given an upper bound $s$ of a subset $A \subseteq \mathbb{R},$
$$s = \operatorname{sup}A \quad iff \quad \exists a \in A \quad s.t. \quad s - \epsilon < a, \quad \forall \epsilon > 0.$$
However, is it not the case that, by the first statement, if
$$s - \epsilon < a, \quad \forall \epsilon > 0,$$
then
$$s \leq a.$$
However, as $s$ is given to be an upper bound of the set $A,$ it is also the case that
$$s \geq a.$$
Therefore, it is necessarily the case that
$$s = a.$$
However, it is obviously not "necessarily" the case that the supremum of a set $A$ must be included in that set $A$ (that is, that the supremum of a set $A$ must be equal to some number in that set $A$).
Where am I confused? Or, where is Abbott confused?

REPLY [4 votes]: Your attempt to rephrase Abbott resulted in an ambiguous statement:
$$s = \sup A \quad \mathrm{iff} \quad \exists a \in A \quad \mathrm{s.t.} \quad s - \epsilon < a, \quad \forall \epsilon > 0.$$
Does this mean
$$s = \sup A \quad \mathrm{iff} \quad ((\exists a \in A \quad \mathrm{s.t.} \quad s - \epsilon < a), \quad \forall \epsilon > 0), \tag1$$
or does it mean
$$s = \sup A \quad \mathrm{iff} \quad (\exists a \in A \quad \mathrm{s.t.} \quad (s - \epsilon < a, \quad \forall \epsilon > 0))? \tag2$$
Indeed $(2)$ would be incorrect in this context, because it is indeed equivalent to
$s = \sup A \iff (\exists a \in A:s \leq a).$
Note that it is true that $\sup A \leq \max A$ when $\max A$ exists (that is, when $A$ has a greatest element), but the idea of the supremum is that it is defined even when $A$ does not have a greatest element.
But $(1)$ is OK (apart from the fact that is an abominable mangling of the notation of predicate logic).
Note that if we write
$$a \leq b \iff (\forall \epsilon > 0 : a < b + \epsilon),$$
we fix the numbers $a$ and $b$ first, and then we say that a certain inequality involving those two already-chosen numbers must be true for
every value of $\epsilon$ that we might put in that inequality.
But in the right-hand side of $(1),$ which (more conventionally written) is
$$\forall \epsilon > 0 : \exists a \in A : s - \epsilon < a,$$
the $\forall$ symbol means the rest of the formula has to be evaluated for every possible positive value of $\epsilon,$
but each time we choose $\epsilon$ first and then say something about $a$ and $s.$
The way this works when there is no greatest element in $A$ is that we are allowed to choose different values of $a$ for different values of $\epsilon,$
and indeed we will have to do so.


text:
\section{Statistical phase transition}\label{sec:matrixWinfotheorresults}

The phase transition manifests itself as a singularity (more precisely a discontinuous first order derivative) in the mutual information
$I(\bX\otimes \bX;\bW)=H(\bW)-H(\bW|\bX\otimes \bX)$. Note that because the data $\bW$ depends on $\bX$ only through $\bX\otimes \bX$ we have $H(\bW|\bX\otimes \bX)=H(\bW|\bX)$ and therefore $I(\bX\otimes \bX; \bW)=I(\bX; \bW)$. From now on we use the form $I(\bX; \bW)$.

To state the result, we define the {\it potential function}:

\begin{align}\label{26}
i_n^{\rm pot}(q, \lambda,\rho) \equiv \frac{\lambda}{4} (q-\rho)^2+I_n(X;\sqrt{\lambda q}X+Z)\, ,
\end{align}

where $I_n(X;\sqrt{\lambda q}X+Z)$ is the mutual information for a scalar gaussian channel, with $X\sim P_{X,n}$ and $Z\sim{\cal N}(0,1)$. The mutual information $I_n$ is indexed by $n$ because of its dependence on $P_{X,n}$.

\begin{thm}[Mutual information for the sparse spiked Wigner model]\label{thm:ws}
Let the sequences $\lambda_n$ and $\rho_n$ verify \eqref{mainregime} with $\beta\in [0, 1/6)$ and $\gamma >0$. There exists $C>0$ independent of $n$ such that
\begin{align}\label{mainbound}
\frac{1}{\rho_n|\ln\rho_n|}\Big|\frac{1}{n}I(\bX;\bW) - \inf_{q\in [0,\rho_n]} i^{\rm pot}_n(q,\lambda_n,\rho_n)\Big|
\le C\frac{(\ln n)^{1/3}}{n^{(1-6\beta)/7}}\,.
\end{align}
\end{thm}

\begin{figure}[t!]
\centering
\includegraphics[trim={0 0 0 1.2cm},clip,width=0.45\linewidth]{./figures/allornothing_MI}
\includegraphics[trim={0 0 0 1.2cm},clip,width=0.45\linewidth]{./figures/allornothing_matrixMMSE.pdf}

 \caption{\footnotesize A sequence of suitably normalized asymptotic mutual information $(\rho |\ln \rho|)^{-1}\inf_{q\in [0,\rho]} i^{\rm pot}_n(q,\lambda,\rho)$ (left) and associated minimum mean-square error (MMSE) $\rho^{-2} \frac{d}{d\lambda}\inf_{q\in [0,\rho]} i^{\rm pot}_n(q,\lambda,\rho)$ (right) curves as a function of $\lambda/\lambda_c(\rho)$ with $\lambda_c(\rho) = 4\vert \ln \rho\vert/\rho$ for the model $X_i \sim {\rm Ber}(\rho)$ and various $\rho=\rho_n$ values (that can be converted to signal sizes through $\rho_n=\Omega(n^{-\beta})$ given a sparsity scaling $\beta$) using the potential function defined in \eqref{26}. These are the curves towards which, respectively, the finite size mutual information $(n\rho|\ln\rho|)^{-1}I(\bX;\bW)$ and minimum mean-square error $(n\rho)^{-2}{\rm MMSE}((X_iX_j)_{i<j}|\bW)$, converge: see theorem~\ref{thm:ws} and corollary~\ref{Cor1:MMSEwigner}. In the sparse limit $\rho\to 0$, the MMSE curves approach a $0$--$1$ phase transition with the discontinuity at $\lambda = \lambda_c(\rho)$. This corresponds to an angular point for the mutual information (by the I-MMSE relation).}
\label{fig:MMSEandMI}
\end{figure}

The mutual information is thus given, to leading order, by a one-dimensional variational problem
$$
I(\bX;\bW) = n\rho_n \vert \ln \rho_n \vert \inf_{q\in [0,\rho_n]} i^{\rm pot}_n(q,\lambda_n,\rho_n) + {\rm correction\,\,terms}\,.
$$
The factor $\rho_n|\ln \rho_n|$ is related to the entropy (in nats) of the support of the signal given by
$- n (\rho_n\ln\rho_n + (1-\rho_n)\ln(1-\rho_n)),$ which behaves like $n \rho_n\vert \ln\rho_n\vert$ for $\rho_n\to 0_+$. In particular, for both the {\it Bernoulli} and {\it Bernoulli-Rademacher} distributions an analytical solution of the variational problem, given in appendix F, shows that $(\rho_n |\ln \rho_n|)^{-1} \inf_{q\in [0,\rho_n]} i^{\rm pot}_n(q,\lambda_n,\rho_n)$ tends to the singular function $\gamma \mathbb{I}(\gamma \le 1)+\mathbb{I}(\gamma \geq 1)$
as $n\to +\infty$ and $\rho_n\to 0$, where we recall that $\lambda_n = 4\gamma \vert \ln \rho_n\vert\rho_n^{-1}$. See figure~\ref{fig:MMSEandMI}. {Let us mention that $\beta<1/6$ is probably not a fundamental limit to the validity of the result but rather is an artefact of the sub-optimality of our proof technique.}

We now turn to the consequences for the MMSE. It is convenient to work with the ``matrix'' MMSE defined as
${\rm MMSE}((X_iX_j)_{i<j}|\bW) \equiv \mathbb{E} \Vert (X_iX_j)_{i<j} - \mathbb{E}[(X_iX_j)_{i<j} \vert \bW]\Vert_{\rm F}^2$.
This quantity satisfies the I-MMSE relation \cite{GuoShamaiVerdu_IMMSE, guo2011estimation} (see also appendix I for a self-contained derivation),

$$
\frac{d}{d\lambda_n} \frac{1}{n} I(\bX; \bW) = \frac{1}{2n^2}{\rm MMSE}((X_iX_j)_{i<j}|\bW)\,.
$$

In appendix J we prove:

\begin{corollary}[Minimum mean-square error for the sparse spiked Wigner model]\label{Cor1:MMSEwigner}
Let $\frac{1}{2}{m}_n(\lambda,\rho_n)\equiv \rho_n^{-2} \frac{d}{d\lambda}\inf_{q\in [0,\rho_n]} i^{\rm pot}_n(q,\lambda,\rho_n)$. Let $\epsilon > 0$ and sequences $\lambda_n$ and $\rho_n$ verifying \eqref{mainregime} with $\beta\in [0, 1/13)$. There exists $C'>0$ independent of $n$ such that
\begin{align*}
 {m}_n(\lambda_n+\epsilon,\rho_n) \! -\ \! \frac{C^\prime}{\epsilon} \frac{(\ln n)^{4/3}}{n^{(1-13\beta)/7}}
 \le
 \frac{{\rm MMSE}((X_iX_j)_{i<j}\vert \bW)}{(n\rho_n)^2}
 \le{m}_n(\lambda_n-\epsilon,\rho_n) \! + \! \frac{C^\prime}{\epsilon} \frac{(\ln n)^{4/3}}{n^{(1-13\beta)/7}}.
\end{align*}
\end{corollary}

Concretely the derivative $(d/d{\lambda_n})\inf_{q\in [0,\rho_n]} i^{\rm pot}_n(q,\lambda_n,\rho_n)$ is computed, using the envelope theorem \cite{milgrom2002envelope}, as $(\partial/\partial{\lambda_n}) i^{\rm pot}_n(q_n^*,\lambda_n,\rho)$ where $q_n^*=q_n^*(\lambda_n,\rho_n)$ is the solution of the variational problem, which is unique almost everywhere (except at the phase transition point, see e.g.\ \cite{barbier2017phase} for such proofs). For Bernoulli and Bernoulli-Rademacher distributions, we easily compute the limiting behavior $m_n(\lambda_n,\rho_n)$ from the solution of the variational problem stated above, and find that $(n\rho_n)^{-2}{\rm MMSE}((X_iX_j)_{i<j}\vert \bW)$ tends to $\mathbb{I}(\gamma \le 1)$ as $n\to +\infty$.

Figure \ref{fig:MMSEandMI} shows the mutual information and MMSE computed from the numerical solution of the variational problem for a sequence of ${\rm Ber}(\rho_n)$ distributions. We check that the limiting curves are indeed approached as $\rho_n \to 0$ and, in particular, the suitably rescaled MMSE displays the all-or-nothing transition at $\lambda / \lambda_c(\rho_n) =1$ as $n\to +\infty$ with $\lambda_c(\rho_n) = 4\vert\ln\rho_n\vert/ \rho_n$. For the Bernoulli-Rademacher distribution the transition location is the same, suggesting that the hardness of the inference is only related, for discrete priors, to the recovery of the support. For more generic distributions than these two cases the situation is richer.
Although one generically observes phase transitions {\it in the same scaling regime}, the limiting curves appear to be more complicated than the simple staircase shape and the jumps are not necessarily located at $\gamma=1$. A classification of these transitions is an interesting problem that is out of the scope of this paper.


TITLE: Find the function $f$ such that $g_{x}(z)=f(x,z)$, $g_{y}(z) =f(z,y)$
QUESTION [0 upvotes]: Find the function $f\colon\mathbb{R}^{2}\rightarrow \mathbb{R}$ such that $g_{x}(z)=f(x,z)$, $g_{y}(z)=f(z,y)$ are continuous, but $f$ is not continuous($\forall x,y\in \mathbb{R}$).

REPLY [1 votes]: Hint: Consider the map, $f:\Bbb R^2 \to \Bbb R$ defined by,
$$f(x,y) =
\begin{cases}
\frac{xy}{x^2+y^2},  & \text{if}\;(x,y)\not=(0,0) \\[2ex]
0, & \text{if } (x,y)=(0,0)
\end{cases}
$$
You can verify that $f$ is not continuous and that it is continuous in each variable seperately.


TITLE: Which spheres are fiber bundles?
QUESTION [11 upvotes]: The Hopf fibration is a fiber bundle with total space $S^3$, and there are similar constructions for $S^7$ and $S^{15}$.  Are there any other ways to regard a sphere as a nontrivial fiber bundle?
My guess is no.  Perhaps it would help to look at the long exact sequence of homotopy groups.
I'm not requiring that the base and fiber be manifolds, smooth or otherwise.  As a side note, I'm also curious to know whether the base and fiber are generally manifolds if the total space is a manifold -- I'm guessing yes.

REPLY [0 votes]: I've been confusing myself into thinging the below constructions would answer the question.
I'll leave them in case they might interest anyone and give ideas.

For $p,q>1$, describe points $X\in S^{pq-1}$ as $p\times q$ matrices with norm 1: i.e. $X=(x_{ij})$ where $\sum_{i,j}x_{ij}^2=1$. Let $U\in SO(p)$ act on $S^{pq-1}$ by $X\mapsto UX$. This gives us an action of $SO(p)$ on $S^{pq-1}$, and by letting $B=S^{pq-1}/SO(p)$, we get a fibration
$$SO(p)\rightarrow S^{pq-1}\rightarrow S^{pq-1}/SO(p).$$
Problem: The action is not free!

If I'm not overly confusing myself here, there's always
$$F\rightarrow S^{pq-1}\rightarrow S^{q-1}$$
where you describe points $X\in S^{pq-1}$ as $p\times q$ matrices with norm 1, i.e. $\sum_{i,j}x_{ij}^2=1$, which map to $Y\in S^{q-1}$ by $y_j=\sum_i x_{ij}^2$ and on which matrices in $U\in SO(p)$ act on the fibers by $X\mapsto UX$.
Problem: Fibres depend on the number of coordinates in $S^{q-1}$ are zero. The action is not free (see above) and irrelevant for the argument.


TITLE: Problem on improper integrals
QUESTION [1 upvotes]: I'm having some trouble evaluating the following integrals:
$$
I=\int_{-\infty}^0\frac{dx}{\sqrt{1-x^3}},\quad
J=\int_0^1\frac{dx}{\sqrt{1-x^3}}.
$$
Any help will be appreciated.

REPLY [4 votes]: The $J$ integral is reduced to the Euler integral of the first kind using $u$-substitution $u=x^3$:
$$
  J = \frac{1}{3} \int_0^1 \left(1-u\right)^{1/2-1} u^{1/3-1} \mathrm{d}u = \frac{1}{3} \operatorname{Beta}\left(\frac{1}{3}, \frac{1}{2} \right)
$$
The $I$ integral can be reduced to the beta integral using $x^3 = \frac{u}{1-u}$, i.e. $u=\frac{x^3}{1+x^3}$:
$$
  I = \int_0^\infty \frac{\mathrm{d}x}{\sqrt{1+x^3}} =\frac{1}{3} \int_0^1 u^{1/3-1} \left(1-u\right)^{1/6-1} \mathrm{d}u = \frac{1}{3} \operatorname{Beta}\left(\frac{1}{3}, \frac{1}{6} \right)
$$