text:
\begin{document}
\title{
{\bf\Large  Positive solutions to some systems of coupled nonlinear Schr\"odinger equations}}
\author{{\bf\large Eduardo Colorado}\hspace{2mm}
{\bf\large}\vspace{1mm}\\
{\it\small Departamento de Matem\'aticas, Universidad Carlos III de Madrid}\\
{\it\small Avda. Universidad 30, 28911 Legan\'es (Madrid), Spain,}\\
{\it\small and}\\
{\it\small Instituto de Ciencias Matem\'aticas, ICMAT (CSIC-UAM-UC3M-UCM)}\\
{\it\small C/Nicol\'as Cabrera 15, 28049 Madrid, Spain.}\\
{\it\small e-mail: eduardo.colorado@uc3m.es, eduardo.colorado@icmat.es}}

\maketitle

\begin{center}
{\bf\small Abstract}

\vspace{3mm}
\hspace{.05in}\parbox{4.5in}
{{\small We study the existence of nontrivial bound state  solutions to the following system of coupled nonlinear time-independent Schr\"odinger equations
$$
- \D u_j+ \l_j u_j =\m_j u_j^3+ \sum_{k=1;k\neq j}^N\b_{jk}  u_ju_k^2,\quad u_j\in \Wan;\: j=1,\ldots,N
$$
where  $n=1,\,2,\, 3; \,\l_j,\, \m_j>0$ for $j=1,\ldots ,N$, the coupling parameters $\b_{jk}=\b_{kj}\in \mathbb{R}$ for $j,k=1,\ldots,N$, $j\neq k$.
Precisely, we  prove the existence of  nonnegative bound state solutions for suitable conditions on the coupling factors.
Additionally, with more restrictive conditions on the coupled parameters, we show that the bound states founded are positive.
}}
\end{center}

\vglue 0.5cm

\noindent{\it \footnotesize 2010 Mathematics Subject Classification}. {\scriptsize  35Q55, 35B38, 35J50, 47A55}.\\
{\it \footnotesize Key words}. {\scriptsize Nonlinear Schr\"odinger Equations,  Bound States, Critical Point Theory, Perturbation Theory}

\tableofcontents

\section{Introduction}

Problems on coupled Nonlinear Schr\"odinger Equations (NLSE for short) have been widely investigated in the last years. They arise
naturally in nonlinear Optics, and in the Hartree-Fock theory for Bose-Einstein condensates, among other physical problems.
For example, a planar light beam propagating in the $z$ direction in a non-linear medium can be described by a NLSE  of the form
$$
{\rm i}\, \bE_z+ \bE_{xx}+ \theta |\bE|^2 \bE=0,
$$
where ${\rm i}$, $\bE(x,z)$  denote the imaginary unit and  the complex envelope of an Electric field respectively, and $\theta>0$ is a normalization constant, corresponding to the fact that the medium is self-focusing.

Here we consider the following more general  system of $N$-coupled   NLSE
\be\label{eq:main-sys-time}
\left\{\begin{array}{ll}
&\dyle -{\rm i}\, \frac{\partial}{\partial t} E_j- \D E_j = \mu_j |E_j|^2E_j+\sum_{k\neq j;k=1}^N\b_{jk}  |E_k|^2E_j,\quad x\in\Rn,\quad t>0\\ & \\
& E_j=E_j(x,t)\in \mathbb{C},\quad E_j(x,t)\to 0  \mbox{ as }|x|\to \infty
\end{array}\right.
\ee
for $j=1,\ldots,N$, the coupled parameters $\b_{jk}=\b_{kj}\in \R$ for $j,k=1,\ldots,N$, $j\neq k$; $\l_j>0$ and  $\mu_j>0$ is for self-focusing in the $j$-th component of the beam. The solution $E_j$ represents the $j$-th component of the beam. The coupling constant $\b_{jk}$ means the interaction between the $j$-th and $k$-th component of the beam. If they are positive, the interaction is attractive while if they are negative we have a repulsive interaction. The mixed case involves different sign on the coupling parameters $\b_{jk}$. Here, we study  the attractive and mixed interactions.

When one looks for solitary wave solutions of the form $E_j(x,t)=e^{i\l_jt}u_j(x)$, with $\l_j>0$, $u_j$ are the real valued functions called standing wave solutions which solve the following system of $N$-coupled nonlinear time-independent Sch\-r\"o\-dinger equations
\be\label{eq:main-sys}
- \D u_j+ \l_j u_j = \mu_j u_j^3+\sum_{k\neq j;k=1}^N\b_{jk}  u_k^2u_j, \qquad u(x)\to 0 \mbox{ as }|x|\to \infty,
\ee
for $j=1,\ldots,N$. Solutions $u_j$  belong to the Sobolev space $\Wan$ ($n=1,\,2,\,3$) for all $j=1,\ldots,N$.

In the last years there has been a very active investigation on coupled systems of NLSE, from the physical point of view; see for instance \cite{hioe,me} or the book \cite{akanbook} and the references therein, as well as many works of mathematicians, from a more theoretical point of view,  dealing with the existence, multiplicity, uniqueness and qualitative properties of bound and ground state solutions; see the earlier works \cite{ac1,ac2,acr,bt,linwei,mmp,pomp,sirakov,ww}, and the more recent list (far from complete) \cite{chen-zou,dancer-wei,itanaka,liu-wang,soave,wei-yao}. This kind of systems have also been recently studied in the framework of the fractional and bi-harmonic Schr\"odinger operators in \cite{col} and \cite{acv} respectively.

Most of the existing works have been developed for systems with two equations, i.e., $N=2$. For a general $N$-system, in \cite{linwei} Lin-Wei  studied the attractive interaction case ($\b_{ij}>0$) in the dimensional case $n=1,\,2,\,3$ and small coupling factors, i.e., when the ground state could coincide with a semitrivial solution, which corresponds to the ground state of a single NLS equation in one component and zero the others. More recently, in \cite{liu-wang}  Liu-Wang gave a sufficient condition on large coupling coefficients for the existence of a nontrivial ground state solution in a system of $N$ ($N$-system for short) NLSE for the dimensional case $n=2,\, 3.$

The existence of positive bound states was firstly studied by Ambrosetti-C. in \cite{ac1,ac2}, where for a $2$-system in the dimensional case $n=2,3$  was proved, among other results, the existence of positive bound states for positive non-large coupling factors; see also \cite{col} for $N=2,\,n=1$. Even more, that existence result was also proved in \cite{ac2} for $N\ge 3$ in the attractive interaction case, provided the coupling factors are sufficiently small. Later in \cite{itanaka} with $N=2$, Ikoma-Tanaka shown that bound states obtained by Ambrosetti-C. in \cite{ac1,ac2} are indeed least energy (positive) solutions in some range of the small coupling factor. That result was recently improved by Chen-Zou in \cite{chen-zou} for a better suitable range of the parameters.

Here, we improve the only result (up to our knowledge) of the existence of bound state solutions for a system of coupled  NLSE such us \eqref{eq:main-sys} with $N\ge 3$. Precisely, we demonstrate the existence of new positive and non-trivial bound state solutions to \eqref{eq:main-sys} in dimensions $n=1,2,3$ different from all previously studied. We want to emphasize that the interaction  between the components of the beam  is analyzed here in the attractive case  and the mixed one.

The paper is organized as follows. In Section \ref{sec:not-prel} we introduce the functional setting, the notation and give  some definitions. Section \ref{sec:pr} contains some previous known results, while the last one, Section \ref{sec:bs} is devoted to prove the main results of the work.

\section{Functional setting and notation}\label{sec:not-prel}

In this section, as we just mentioned above, we will establish the corresponding functional framework, the notation and give some definitions. The functional setting comes from the variational structure of \eqref{eq:main-sys}, which can be derived from
$$
- \D u_j + \l_j u_j =   f_{u_j}(u_1,\ldots,u_n)
$$
where
$$
f(u_1,\ldots,u_n)=\dyle\sum_{j,k=1}^N\a_{jk}u_j^2u_k^2,
$$
with $\a_{jj}=\frac 14\mu_j$ for $j=1,\ldots,N$; the coupling parameters $\a_{jk}=\frac 14\b_{jk}$ are symmetric,  $\b_{jk}=\b_{kj}$, both for $k,j=1,\ldots,N$ with $k\neq j$.
We will work on the Sobolev  functional space $\Wan$, i.e., we will assume that $u_j\in \Wan$ for $j=1,\ldots,N$. We remember that the Sobolev space $E=\Wan$ can be defined as the completion of $\mathcal{C}_0^1(\Rn)$ with the norm given by
$$
\|u\|_{E}=\left(\intn ( |\n u|^2+ u^2)\, dx\right)^{\frac 12},
$$
which comes from the scalar product
$$
\langle u|v\rangle_E=\intn (\n u\cdot \n v+ uv)\, dx.
$$
We will denote the following equivalent norms in $E$,
\be\label{norma}
\|u\|=\|u\|_j=\left(\intn (|\n u|^2+\l_j u^2)\, dx\right)^{\frac 12},
\ee
with the corresponding scalar products
\be\label{p-escalar}
( u|v)=( u|v)_j=\intn (\n u\cdot \n v+ \l_j uv)\, dx; \quad j=1,\ldots,N,
\ee
dropping the sub-index $j$ in \eqref{norma}-\eqref{p-escalar} for short.
In this manner, solutions of \eqref{eq:main-sys} are the critical points $\bu=(u_1,\ldots,u_N)\in \E=E\times\stackrel{(N)}{\cdots}\times E$ of the  corresponding energy functional defined by
$$
\Phi (\bu)=\frac 12\|\bu\|^2- F(\bu),
$$
where
$$
F(\bu)= \intn f(\bu)\, dx\quad {and}\quad
\|\bu\|^2=\sum_{j=1}^N \|u_j\|^2.
$$
Also we define
$$
I_j(u)=\frac 12\intn (|\n u|^2+\l_j u^2)\, dx-\frac 14 \m_j\intn u_j^4\, dx\quad\mbox{for }j=1,\ldots,N
$$
With respect to the  coupling factors, we assume $m\in \mathbb{N}$, moreover $2\le 2m\le N$, and define $\b_k=\b_{2k-1,2k}$ and $\l_{2k-1}=\l_{2k}$ for $1\le k\le m$. If $2m<N$,  we define
 $\b_{m+\ell}^{(2k)}=\b_{2m+\ell,2k}$, $\b_{m+\ell}^{(2k-1)}=\b_{2m+\ell,2k-1}$  for any $1\le \ell \le N-2m$ and $k\neq 2m+\ell$. We also suppose that $\b_{m+\ell}^{(2k)}=\e\wt{\b}^{(2k)}_{m+\ell}$, $\b_{m+\ell}^{(2k-1)}=\e\wt{\b}^{(2k-1)}_{m+\ell}$.

 Let us denote $\Phi_\e=\Phi$ to emphasize its dependence on $\e$, then we can rewrite the functional $\Phi_\e$ as
$$
\Phi_{\e}(\bu)=\Phi_0-\e \wt{F} (\bu)
$$
where
$$
\Phi_0(\bu)=\sum_{j=1}^N I_j(u_j)-\frac 12\sum_{k=1}^m\b_{k}   \intn u_{2k-1}^2u_{2k}^2\,dx
$$
and
$$
\wt{F}(\bu)=\frac 12 \sum_{\ell=1}^{N-2m} \sum_{k=1}^m\intn \left(\wt{\b}_{k+\ell}^{(2k-1)}u_{2k-1}^2+\wt{\b}_{k+\ell}^{(2k)}u_{2k}^2\right) u_{2m+\ell}^2\,dx.
$$
For a function of two components we also set
\be\label{eq:Phi-k}
\Phi_k(u,v)=I_{2k-1}+I_{2k}-\frac 12 \b_k\intn u^2v^2\,dx\qquad 1\le k\le m.
\ee
We denote $\bo=(0,\ldots,0)$, then we say that a vector function (or a function to simplify) $\bu\in\E$ is positive (non-negative), namely $\bu>\bo$ ($\bu\ge \bo$) if every component of $\bu$ is positive (non-negative), i.e., $u_j>0$ ($u_j\ge 0$) for every $j=1,\ldots,N.$

\begin{Definition}
A bound state $\bu\in \E$ of (\ref{eq:main-sys}) is a critical point
of $\Phi$. A nontrivial bound state $\bu=(u_1,\ldots,u_N)$ is a bound state with all the components $u_j\not\equiv 0$, $j=1,\ldots,N$. Moreover, a positive bound state   $\bu\ge \bo$  such that its energy is minimal among {\it all the non-trivial bound states}, namely
\begin{equation}\label{eq:gr}
\Phi(\bu)=\min\{\Phi(\bv): \bv\in \E\setminus\{\bo\},\; \Phi'(\bv)=0\},
\end{equation}
is called a {\it ground state}  of \eqref{eq:main-sys}.
\end{Definition}

Note that a ground state can have some of the components equal zero, thus in some works the definition of a ground state corresponds to a positive ground state here.

We will focus on the existence of (non-trivial) non-negative and also on positive bound states.

Let us  denote $2^*=\frac{2n}{n-2}$ if $n\ge 3$ and $2^*=\infty$ if $n=1,\,2$, and remark that the functional make sense because $E\hookrightarrow L^4(\Rn)$ for $n=1,2,3$. Concerning the Palais-Smale $(PS)$ condition, one cannot expects a compact embedding neither in $E$ nor in the restriction on the even functions of $E$ into $L^q(\R)$ for any $1<q<\infty$.  If $n=2,\, 3$, the PS condition is easy to obtain because the space of radially symmetric functions of the Sobolev space  $E$, namely $H=E_{rad}$,  is compactly embedded into $L^{q}(\mathbb{R}^n)$ for $q<2^*$ (in particular for $q=4$), see for instance \cite{lions}.
Nevertheless, as we will see, the lack of compactness will not be a problem provided we will work on $H$  for all $n=1,2,3$. The reason is that we can obtain strong convergence in
$\mathbb{H}=H\times\stackrel{(N)}{\cdots}\times H$ thanks to the Local Inversion Theorem by a perturbation argument involving some non-degenerate critical points.

\section{Previous results}\label{sec:pr}

In order to prove the main result of the paper we introduce some known results we are going to use.
\begin{Remark}\label{rem:1}
Let $U$ be the unique positive solution in $H$ of the equation $- \D u+u=u^3$ (see the paper of Kwong \cite{kw}) then the scaled function
\be\label{eq:single}
U_j(x)=\sqrt{\frac{\l_j}{\mu_j}}\,U(\sqrt{\l_j}\,x),\quad j=1,\ldots,N
\ee
is the unique positive solution  in $H$ of  $- \D u+\l_ju=\m_j u^3$ for $j=1,\ldots,N$. We also recall that $U_j$ is the ground state, as well as a non-degenerate critical point of the corresponding functional $I_j$, $j=1,\ldots, N$.
\end{Remark}
Let us consider the following $2$-system of NLSE
\be\label{eq:2-sistema}
(S_2)\equiv\left\{\begin{array}{ll}
- \D u + \l u &= \mu_1 u^3+\b  v^2 u\\
- \D v + \l u &= \mu_2 v^3+\b u^2 v.
\end{array} \right.
\ee
Throughout this subsection we will maintain the notation of Section \ref{sec:not-prel} for a $2$-system  or a $3$-system of  NLSE with the natural meaning.

Let us  define
$$
 a_{k\b} = \sqrt{\frac{\mu_k(\mu_j-\b)}{\mu_k\mu_j
-\b^2}},\quad k\not= j,\;k=1,2.
$$
If $\l_1=\l_2=\l$,  we have the following explicit and positive solution of $(S_2)$, given by
\be\label{eq:2-ground}
(u^0,v^0)=(a_{1\b}U_1,a_{2\b}U_2)
\ee
with $U_1,\, U_2$ defined  in \eqref{eq:single}, see also \cite[Remarks 5.8-(b)]{ac2}.
\begin{Remark}\label{rem:non-degenerate}
  In  \cite[Lemma 2.2 and Theorem 3.1]{dancer-wei} Dancer-Wei proved that $(u^0,v^0)$ is a non-degenerate critical point to the corresponding energy functional of $(S_2)$.
\end{Remark}
Even more, Wei-Yao  proved in \cite{wei-yao} the following property of the critical point $(u^0,v^0)$.

\

\noindent
{\bf Theorem A.} {\it Assume that $\b>\max\{\m_1,\m_2\}$}, then $(u^0,v^0)$ is the unique positive solution of $(S_2).$
\begin{Remark}
According to \cite[Remark5.8-(b)]{ac2} and Theorem A one has that $(u^0,v^0)$ is a ground state of $(S_2)$, i.e.,
$$
\Phi(u^0,v^0)=\min\{\Phi(\bv): \bv\in \E\setminus\{\bo\},\; \Phi'(\bv)=0\},
$$
which also corresponds to
$$
\Phi(u^0,v^0)=\min_{\bu\in\cN}\Phi(\bu)
$$
where $\cN$ is the corresponding Nehari manifold, defined as
$$
\cN =\{ \bu\in \mathbb{E}\setminus\{{\bf 0}\}: (\Phi'(\bu)|\bu)=0\},
$$
that is  a natural constraint for $\Phi$; see \cite{ac2} for more details. Furthermore,
\be\label{eq:inf1}
\begin{array}{rcl}
 \|(u^0,v^0)\|& = &\dyle \inf_{\bu\in \mathbb{H}\setminus\{\bo\}} \frac{\|(u_1,u_2)\|^2}{\dyle\left(\m_1 \intn u_1^4\,dx+\m_2 \intn u_2^4\,dx+2\b\intn u_1^2u_2^2\,dx\right)^{\frac12}}\\ & &  \\
 & = & \dyle\inf_{\bu\in \mathbb{E}\setminus\{\bo\}} \frac{\|(u_1,u_2)\|^2}{\dyle\left(\m_1 \intn u_1^4\,dx+\m_2 \intn u_2^4\,dx+2\b\intn u_1^2u_2^2\,dx\right)^{\frac12}}.
\end{array}\ee
\end{Remark}

\section{Existence of bound states}\label{sec:bs}

In order to prove the existence  of a bound state solution to System \eqref{eq:main-sys}, first we demonstrate  a reduced result to both, clarify and also simplify the proof of the main result in Theorem \ref{th:ec2}. To do so we consider the following $3$-system of NLSE
$$
(S_3)\equiv \left \{
\begin{array}{ll}
- \D u_1+ \l_1 u_1 &= \mu_1 u_1^3+\b_{12}  u_1u_2^2+ \b_{13} u_1u_3^2\\
- \D u_2 + \l_2 u_2 &= \mu_2 u_2^3+\b_{21} u_2 u_1^2 + \b_{23}  u_2u_3^2\\
- \D u_3 + \l_3 u_3 &= \mu_3 u_3^3+\b_{31} u_3u_1^2 + \b_{32}u_3u_2^2
\end{array} \right.
$$
with $\l_j,\m_j>0$, $u_j\in H$, $j=1,\,2,\,3$ and $\b_{jk}=\b_{kj}$ for all $j,k=1,\,2,\, 3$; $j\neq k$.
\begin{Remark}\label{rem:sobolev-cte}
Let
  $$
\s_\l^2\equiv \inf_{\varphi\in E\setminus \{ 0\}}\frac{\|
\varphi\|_\l^2}{\| \varphi\|^2_{L^4}}=\inf_{\varphi\in E,\, \|
\varphi\|_{L^4}=1}\| \varphi\|_\l^2,
$$
denotes  the  best Sobolev constant in the embedding
of $(E,\|\cdot\|_{\l})$ into  $L^4(\Rn)$, where
$$
\| u\|^2_\l=\intn (|\n u|^2+\l u^2)\, dx.
$$
Easily one obtains that  $\s_\l$ is attained  at
$$
v_\l(x)=\s_\l^{-1}\sqrt{\l} U(\sqrt{\l}x),
$$
and one has
$$
\s_\l^4=\l^2\intn U^4(\sqrt{\l}x)dx=\l^{2-\frac n2}\intn U^4(x)dx.
$$
\end{Remark}
Let us set $\bu_0=(u^0,v^0,U_3)$, where $(u^0,v^0)$, $U_3$ are given by \eqref{eq:2-ground}, \eqref{eq:single} respectively. Then we have the following.
\begin{Theorem}\label{th:ec1}
If $\l_1=\l_2=\l$, $\b_{12}>\max\{ \m_1,\m_2\}$,
and $\b_{j3}=\e\wt{\b}_{j3}$, $j=1,2$, there exists $\e_0>0$ such that for $0<\e<\e_0$,  $(S_3)$ has a radial bound state $\bu_{\e}\ge \bo$ with  $\bu_{\e}\to \bu_0$ as $\e\to 0$. Furthermore, if both  $\b_{13}, \,\b_{23}\ge 0$ then $\bu_{\e}>\bo$.
\end{Theorem}
Notice that this result corresponds to $N=3$, $m=1$,  $\ell=1$.
\begin{Remark}\label{rem:2}
Note that hypothesis $\b_{12}>\max\{ \m_1,\m_2\}$ comes from Theorem A, on the contrary $(S_2)$ have no positive solutions. Multiplying the first (third) equation of $(S_3)$ by $u_3$ ($u_1$) and integrating on $\Rn$ we get
\be\label{eq:h1}
(\m_1-\b_{13})\intn \!\! u_1^3u_3\, dx+(\b_{12}-\b_{23})\intn\!\! u_1u_2^2u_3\, dx +(\b_{13}-\m_3)\intn\!\! u_1u_3^3\, dx=0.
\ee
Multiplying the second (third) equation of $(S_3)$ by $u_3$ ($u_2$) and integrating again on $\Rn$ we find
\be\label{eq:h2}
(\m_2-\b_{23})\intn\!\! u_2^3u_3\, dx+(\b_{12}-\b_{13})\intn\!\! u_1^2u_2u_3\, dx +(\b_{23}-\m_3)\intn\!\! u_2u_3^3\, dx=0.
\ee
As a consequence, if for instance $\mu_3<\b_{13},$ thanks to \eqref{eq:h1}, system $(S_3)$ has no positive solutions. Similarly, if $\m_3<\b_{23}$, due to   \eqref{eq:h1}, system  $(S_3)$ has no positive solutions. But $\m_3>\max\{\b_{13},\,\b_{23}\}$  provided $\e_0$ is sufficiently small.
\end{Remark}
\noindent {\it Proof of Theorem \ref{th:ec1}}.
Taking  $\e_0>0$ sufficiently small, we can suppose that $\m_3>\max\{\b_{13},\, \b_{23}\}$ and the compatibility conditions to have non-negative solutions given in \eqref{eq:h1}, \eqref{eq:h2} are satisfied. According to the notation of Section \ref{sec:not-prel} with $N=3$, $m=1$,  $\ell=1$, we  have
$$
\Phi_{\e}(\bu)=\Phi_0-\e \wt{F} (\bu)
$$
where
$$
\Phi_0(\bu)=\sum_{j=1}^3 I_j(u_j)-\frac 12\b \intn u_1^2u_2^2\,dx
$$
and
$$
\wt{F}(\bu)=\frac 12 \intn (\wt{\b}_{13}u_1^2u_3^2+\wt{\b}_{23}u_2^2u_3^2)\,dx.
$$
Let us consider the critical point $\bu_0$ of the unperturbed functional $\Phi_0$. We note that $U_3$ is a non-degenerate critical point of $I_3$ on $H$; see \cite{kw} and Remark \ref{rem:1}. Also, as we pointed out in Remark \ref{rem:non-degenerate}, $(u^0,v^0)$  is a non-degenerate critical point to the corresponding energy functional (of $(S_2)$ with $\l_1=\l_2=\l$ and $\b_{12}=\b$) acting on $\mathbb{H}$. As an immediate consequence, $\bu_0$ is a non-degenerate critical point of $\Phi_0$ on $\mathbb{H}$, then a straightforward application of the Local Inversion Theorem yields the existence of a critical point $\bu_\e$ of $\Phi_\e$ for any $0<\e<\e_0$ with $\e_0$ sufficiently small; see  \cite{a-m} for more details. Moreover, $\bu_\e\to \bu_0$ on $\mathbb{H}$ as $\e\to 0$. To complete the proof it remains to show that $\bu_\e\ge \bo$, and furthermore, in case $\b_{13},\,\b_{23}\ge 0$ then $\bu_\e>\bo$. We follow an argument of \cite{cing} with suitable modifications by separating the positive and negative parts and using energy type estimates.

Let us  denote the positive part
 $\bu_\e^+=(u_{1\e}^+,u_{2\e}^+,u_{3\e}^+)$ and  the negative part $\bu_\e^-=(u_{1\e}^-,u_{2\e}^-,u_{3\e}^-)$.
Notice that $(u^0,v^0)$ satisfies the identity \eqref{eq:inf1} and $U_3$ satisfies the following identity
\be\label{eq:inf2}
 \|U_3\|=\inf_{u\in H\setminus\{0\}} \frac{\|u\|^2}{\dyle\left(\mu_3 \intn u^4dx\right)^{\frac12}}.
\ee
As a consequence, from \eqref{eq:inf1}, resp. \eqref{eq:inf2}, it follows that
\be\label{eq:pm1}
\frac{ \|(u_{1\e}^\pm,u_{2\e}^\pm)\|^2}{ \dyle\left(\mu_1 \intn (u_{1\e}^\pm)^4\,dx+\mu_2 \intn (u_{2\e}^\pm)^4\,dx+2\b\intn (u_{1\e}^\pm)^2(u_{2\e}^\pm)^2\,dx\right)^{\frac12}} \ge  \|(u^0,v^0)\|,
\ee
resp.
\be\label{eq:pm2}
\frac{\|u_{3\e}^\pm\|^2}{\dyle\left(\mu_3 \intn (u_{3\e}^\pm)^4dx\right)^{\frac 12}} \geq  \|U_3\|.
\ee
Multiplying  the third equation of $(S_3)$ by $u_{3\e}^\pm$ and integrating on $\Rn$ one infers
\begin{eqnarray*}
\|u_{3\e}^\pm\|^2 &=& \mu_3 \intn (u_{3\e}^\pm)^4dx + \e \intn\left[ (u_{3\e}^\pm)^2(\wt{\b}_{13} u_{1\e}^2 +\wt{\b}_{23} u_{2\e}^2)   \right]dx\\
&\leq&\mu_3 \intn (u_{3\e}^\pm)^4dx \\ & & +\e \left( \intn (u_{3\e}^\pm)^4dx\right)^{1/2}
\left[ \wt{\b}_{13}\left( \intn u_{1\e}^4\right)^{1/2} +\wt{\b}_{23}\left(\intn u_{2\e}^4\right)^{1/2}   \right].
\end{eqnarray*}
This, jointly with \eqref{eq:pm2}, yields
$$
\|u_{3\e}^\pm\|^2 \le \frac{\|u_{3\e}^\pm\|^4}{ \|U_3\|^2} + \e\, \vartheta_\e\;\frac{\|u_{3\e}^\pm\|^2}{ \|U_3\|},
$$
where
$$
\vartheta_\e = \mu_3^{-1/2}\left[\wt{\b}_{13}\left( \intn u_{1\e}^4\right)^{1/2} +\wt{\b}_{23}\left(\intn u_{2\e}^4\right)^{1/2}   \right].
$$
Since $\bu_\e\to \bu_0$, clearly  $(u_{1\e},u_{2\e})\to (u^0,v^0)$, then one has $\e\vartheta_\e\to 0$ as $\e\to 0$.
Hence, if  $\|u_{3\e}^\pm\|>0$, one obtains
\begin{equation}\label{eq:pmm1}
\|u_{3\e}^\pm\|^2 \ge  \|U_3\|^2 +o(1),
\end{equation}
where $o(1)=o_\e (1)\to 0$ as $\e\to 0$.
Using again  $\bu_\e \to \bu_0$, then $u_{3\e}\to U_3>0 $ and as a consequence, for $\e$ small enough, $\|u_{3\e}^+\|>0$. Thus
\eqref{eq:pmm1} gives
\begin{equation}\label{eq:pm11}
\|u_{3\e}^+\|^2 \ge  \|U_3\|^2 +o(1).
\end{equation}
Multiplying now the first, resp. the second equation of
$(S_3)$ by $u_{1\e}^\pm$, resp.  $u_{2\e}^\pm$ and integrating on $\Rn$ one infers
\begin{eqnarray*}
\|(u_{1\e}^\pm,u_{2\e}^\pm)\|^2 &=& \mu_1 \intn (u_{1\e}^\pm)^4dx + \mu_2 \intn (u_{2\e}^\pm)^4dx\\
& & +\b  \intn \left[(u_{1\e}^\pm)^2u_{2\e}^2+ (u_{2\e}^\pm)^2u_{1\e}^2\right]dx \\
 &  &+ \e \intn u_{3\e}^2\left[ (\wt{\b}_{13} (u_{1\e}^\pm)^2 +\wt{\b}_{23} (u_{2\e}^\pm)^2)   \right]dx\\
&\le &\mu_1 \intn (u_{1\e}^\pm)^4dx + \mu_2 \intn (u_{2\e}^\pm)^4dx+ 2\b  \intn (u_{1\e}^\pm)^2(u_{2\e}^\pm)^2 dx \\
& & +\b  \intn \left[(u_{1\e}^\mp)^2(u_{2\e}^\pm)^2+ (u_{1\e}^\pm)^2(u_{2\e}^\mp)^2\right]dx\\
& & +\e \left( \intn u_{3\e}^4\right)^{\frac 12}\left[ \wt{\b}_{13} \left(\intn(u_{1\e}^\pm)^4\right)^{\frac 12} +\wt{\b}_{23}\left(\intn (u_{2\e}^\pm)^4\right)^{\frac 12}   \right]\! dx.
\end{eqnarray*}
This, jointly with \eqref{eq:pm1}, yields
\be
\begin{array}{rcl}\label{eq:masmenos1}
\|(u_{1\e}^\pm,u_{2\e}^\pm)\|^2 &  \leq  & \dyle\frac{\|(u_{1\e}^\pm,u_{2\e}^\pm)\|^4}{ \|(u^0,v^0)\|^2} + \b  \intn \left[(u_{1\e}^\mp)^2(u_{2\e}^\pm)^2+ (u_{1\e}^\pm)^2(u_{2\e}^\mp)^2\right]dx\\ & & \\
& &\dyle +\e\, \psi_\e\;\frac{\|(u_{1\e}^\pm,u_{2\e}^\pm)\|^2}{ \|(u^0,v^0)\|},
\end{array}
\ee
where
$$
\psi_\e =C\left( \intn u_{3\e}^4\right)^{1/2} \quad\mbox{for some constant}\quad C=C(\m_1,\m_2,\l,\b)>0.
$$
Using  that $u_{3\e}\to U_3$ we obtain
\be\label{eq:masmenos2}
\e\psi_\e\to 0\quad\mbox{as }\e\to 0,
\ee
and, from $(u_{1\e},u_{2\e})\to (u^0,v^0)>\bo$, we get $(u_{1\e}^-,u_{2\e}^-)\to \bo$ which implies
\be\label{eq:masmenos3}
\b  \intn \left[(u_{1\e}^\mp)^2(u_{2\e}^\pm)^2+ (u_{1\e}^\pm)^2(u_{2\e}^\mp)^2\right]dx\to 0\quad \mbox{as }\e\to 0.
\ee
Since $(u_{1\e}^+, u_{2\e}^+)\to (u^0,v^0)>\bo $, we have $\|(u_{1\e}^+, u_{2\e}^+)\|>0$ for $\e$ sufficiently small, thus
\eqref{eq:masmenos1}-\eqref{eq:masmenos3} gives
\be\label{eq:pm-fin11}
\|(u_{1\e}^+ ,u_{2\e}^+)\|^2 \geq  \|(u^0,v^0)\|^2 +o(1).
\ee
If  $\|(u_{1\e}^-, u_{2\e}^-)\|>0$, it is not so easy to obtain a similar estimate like \eqref{eq:masmenos3} with $\|(u_{1\e}^+ ,u_{2\e}^+)\|^2$ replaced by $\|(u_{1\e}^- ,u_{2\e}^-)\|^2$. To do so, we need to estimate more carefully the mixed term by \eqref{eq:masmenos3} since it has the same order of decay as  $\|(u_{1\e}^-, u_{2\e}^-)\|^2$.

By the Cauchy-Schwarz inequality, the convergence $\bu_\e\to \bu_0$ and Remark \ref{rem:sobolev-cte} we have
\be\begin{array}{rcl}\label{eq:new-estimate}
& &\dyle  \b \intn \left[(u_{1\e}^-)^2(u_{2\e}^+)^2+ (u_{1\e}^+)^2(u_{2\e}^-)^2\right]dx\\ & & \\
 & & \le \dyle
\b \left[\left( \intn (u_{1\e}^-)^4\right)^{1/2}\left( \intn (u_{2\e}^+)^4\right)^{1/2} + \left( \intn (u_{1\e}^+)^4\right)^{1/2}\left( \intn (u_{2\e}^-)^4\right)^{1/2}\right]\\ & & \\
& & \le  \dyle \b\frac{1}{\s_\l^2}\| u_{1\e}^{-}\|^2
\left(\l^{1-\frac{n}{4}}\left(\intn U^4\, dx\right)^\frac 12+o(1)\right)\frac{\b-\m_1}{\b^2-\m_1\m_2}
\\ & & \\
& &\quad \dyle + \b\frac{1}{\s_\l^2}\| u_{2\e}^{-}\|^2
\left(\l^{1-\frac{n}{4}}\left(\intn U^4\, dx\right)^\frac 12+o(1)\right)\frac{\b-\m_2}{\b^2-\m_1\m_2}\\
& & \\ & & \le  (\rho+o(1))\|(u_{1\e}^- ,u_{2\e}^-)\|^2,
\end{array}
\ee
where
\be\label{eq:rho}
\rho=\max\left\{ \frac{\b-\m_1}{\b^2-\m_1\m_2},\,\frac{\b-\m_2}{\b^2-\m_1\m_2}\right\}.
\ee
Even more, taking into account that $\b>\max\{\m_1,\m_2\}$ then $0<\rho<1$. As a consequence, if $\|(u_{1\e}^-, u_{2\e}^-)\|>0$, by  \eqref{eq:masmenos1}, \eqref{eq:masmenos2} and \eqref{eq:new-estimate}  we have the following estimate,
\be\label{eq:correjida}
\|(u_{1\e}^-, u_{2\e}^-)\|^2 \geq  (1-\rho)\|(u^0,v^0)\|^2 +o(1).
\ee
Now, suppose by contradiction that there exists $k\in \{1,2,3\}$ such that
$\|u_{k\e}^-\|>0$. Then we have two possibilities:
\begin{itemize}
\item
If $k=3$, as in \eqref{eq:pmm1} one obtains $\|u_{3\e}^-\|^2 \geq  \|U_3\|^2+o(1)$, hence
\begin{equation}\label{eq:pm12}
\|\bu_\e^-\|^2 =\|(u_{1\e}^- ,u_{2\e}^-)\|^2+\|u_{3\e}^-\|^2 \geq  \|U_3\|^2+o(1).
\end{equation}
Next, we evaluate the functional
\be\label{eq:eval-funct}
\Phi(\bu_\e)= \tfrac 14 \|\bu_\e\|^2 = \tfrac 14 \left[  \|\bu_\e^+\|^2 + \|\bu_\e^-\|^2\right].
\ee
On one hand, using \eqref{eq:pm11},  \eqref{eq:pm-fin11}  and \eqref{eq:pm12}, we infer
\begin{equation}\label{eq:pmm2}
\Phi(\bu_\e)\geq \tfrac 14 \|\bu_0\|^2 +\tfrac 14\|U_3\|^2+o(1).
\end{equation}
On the other hand, since $\bu_\e \to \bu_0$ we also find
\be\label{eq:final}
\Phi(\bu_\e)= \tfrac 14 \|\bu_\e\|^2\to \tfrac 14\|\bu_0\|^2,
\ee
which is in contradiction with \eqref{eq:pmm2}, proving that $u_{3\e}\geq 0$.
\item If $k\in \{1,2\}$, by \eqref{eq:correjida} we have
\begin{equation}\label{eq:pm122}
\|\bu_\e^-\|^2 =\|(u_{1\e}^- ,u_{2\e}^-)\|^2+\|u_{3\e}^-\|^2 \ge (1-\rho)\|(u^0,v^0)\|^2+o(1).
\end{equation}
Using \eqref{eq:pm-fin11}, \eqref{eq:correjida}, \eqref{eq:eval-funct} and \eqref{eq:pm122}, we get
$$
\Phi(\bu_\e)\geq \tfrac 14 \|\bu_0\|^2 +\tfrac 14(1-\rho)\|(u^0,v^0)\|^2+o(1).
$$
This is a contradiction with \eqref{eq:final}, proving that $(u_{1\e},u_{2\e})\ge \bo$.
\end{itemize}
In conclusion, we have proved that $\bu_\e\ge \bo$. Finally, if $\b_{12},\, \b_{23}\ge 0$, using once more that $\bu_\e\to \bu_0$ and applying the maximum principle it follows that  $\bu_\e>\bo$.
\rule{2mm}{2mm}

\

Let us set
\be\label{eq:limite}
\bz=((u_1^0,v_1^0),\ldots,(u_m^0,v_m^0),U_{2m+1},\ldots,U_{N}),
\ee
where $(u_k^0,v_k^0)$ is given by \eqref{eq:2-sistema} with the parameters $\l=\l_{2k}$, $\m_1=\m_{2k-1}$, $\m_2=\m_{2k}$, $\b=\b_{k}$, and $1\le k\le m$; see Theorem A.
And
$$
U_{2m+\ell}(x)=\sqrt{\dfrac{\l_{2m+\ell}}{\m_{2m+\ell}}}U(\sqrt{\l_{2m+\ell}}x)
$$
where $U$ is given by \eqref{eq:single}; see Remark \ref{rem:1}.

The main result of the paper is the following.
\begin{Theorem}\label{th:ec2}
If $\b_k>\max \{ \m_{2k-1},\m_{2k}\}$ for  $1\le k\le m$, there exists $\e_0>0$ such that for $0<\e<\e_0$, \eqref{eq:main-sys} has a radial bound state $\bu_{\e}\ge \bo$  with $\bu_{\e}\to \bz$ as $\e\to 0$. Furthermore, if  $\b_{m+l}\ge 0$ for all $1\le\ell\le N-2m$ then $\bu_{\e}>\bo$.
\end{Theorem}
\begin{Remark}
Notice that a condition on the coupling factors like in Remark \ref{rem:2} holds provided $\e_0$  is small enough.
\end{Remark}
\noindent {\it Proof of Theorem \ref{th:ec2}}.
We follow, with suitable modifications, the arguments in the proof of Theorem \ref{th:ec1}.

Let us consider the critical point $\bz$  (defined in \eqref{eq:limite}) of the unperturbed functional $\Phi_0$. We note that $U_{2m+\ell}$ is a non-degenerate critical point of $I_{2m+\ell}$ on $H$ for each $\ell$; see \cite{kw} and Remark \ref{rem:1}. Also, $(u_k^0,v_k^0)$ is a non-degenerate critical point of the corresponding functional $\Phi_{k}$ described in \eqref{eq:Phi-k}, for each $1\le k\le m$; see Remark \ref{rem:non-degenerate}.
As an immediate consequence, $\bu_0$ is a non-degenerate critical point of $\Phi_0$ on $\mathbb{H}$, then  the Local Inversion Theorem provides us with the existence of a critical point $\bu_\e$ of $\Phi_\e$ for any $\e<\e_0$ with $\e_0$ sufficiently small. Furthermore, $\bu_\e\to \bu_0$ on $\mathbb{H}$ as $\e\to 0$. To complete the proof it remains to show that $\bz\ge \bo$ and if  $\b_{m+l}\ge 0$ for all $1\le\ell\le N-2m$, then $\bu_{\e}>\bo$.

Following the notation of $\bz$ by \eqref{eq:limite}, let us denote
$$
\bu_\e=((u_1^\e,v_1^\e),\ldots,(u_m^\e,v_m^\e), u^\e_{2m+1},\ldots,u_N^\e).
$$
Using that $\bu_\e\to \bz>\bo$, as in \eqref{eq:pm11} and \eqref{eq:pm-fin11} it follows
\be\label{eq:pm-final1}
\|((u_{k}^\e)^+,(v_{k}^\e)^+)\|^2\ge \|(u_{k}^0,v_{k}^0)\|^2+o(1)\quad \mbox{for any }1\le k\le m,
\ee
 and
\be\label{eq:pm-final2}
 \|(u_{2m+\ell}^\e)^+\|^2\ge \|U_{2m+\ell}\|^2+o(1) \quad \mbox{for any }1\le\ell\le N-2m.
\ee
Suppose  by contradiction that there exists either $1\le k_0\le m$, or $1\le \ell_0\le N-2m$ such that either
\be\label{eq:either}
\|((u_{k_0}^\e)^-,(v_{k_0}^\e)^-)\|>0,
\ee
or
\be\label{eq:or}
\|(u_{2m+\ell_0}^\e)^-\|>0.
\ee
\begin{itemize}
\item If \eqref{eq:either} holds, by \eqref{eq:correjida}
\be\label{eq:pm-final1menos}
\|((u_{k_0}^\e)^- ,(v_{k_0}^\e)^-)\|^2 \ge  (1-\rho_{k_0})\|(u_{k_0}^0,v_{k_0}^0)\|^2 +o(1),
\ee
for some $0<\rho_{k_0}<1$ defined by \eqref{eq:rho} with with the parameters $\l=\l_{2{k_0}}$, $\m_1=\m_{2{k_0}-1}$, $\m_2=\m_{2k_0}$, $\b=\b_{k_0}$. Then we obtain
\be\label{eq:either2}
\|\bu_\e^-\|^2  \ge (1-\rho_{k_0})\|(u_{k_0}^0,v_{k_0}^0)\|^2 +o(1),
\ee
and by  \eqref{eq:pm-final1}, \eqref{eq:pm-final1menos} we have the following inequality
\be\label{eq:either22}
\Phi(\bu_\e)\ge \tfrac 14 \|\bz\|^2 +\tfrac 14 (1-\rho_{k_0}) \|(u_{k_0}^0,v_{k_0}^0)\|^2 +o(1).
\ee
\item If \eqref{eq:or} holds, by \eqref{eq:pm11}
\be\label{eq:pm-final2menos}
\|(u_{2m+\ell_0}^\e)^-\|^2 \geq  \|U_{2m+\ell_0}\|^2 +o(1),
\ee
which implies
\be\label{eq:or2}
\|\bu_\e^-\|^2  \ge \|U_{2m+\ell_0}\|^2 +o(1),
\ee
and by \eqref{eq:pm-final2} and \eqref{eq:pm-final2menos} the following inequality holds
\be\label{eq:or22}
\Phi(\bu_\e)\ge \tfrac 14 \|\bz\|^2 +\tfrac 14 \|U_{2m+\ell_0}\|^2 +o(1).
\ee
\end{itemize}
Since $\bu_\e\to \bz$ we have
$$
\Phi(\bu_\e)=\tfrac 14 \|\bu_\e\|^2\to \tfrac 14 \|\bz\|^2.
$$
This is a contradiction with \eqref{eq:either22} and also with \eqref{eq:or22}, proving that $\bu_\e\ge \bo$. To finish,  if  $\b_{m+l}\ge 0$ for all $1\le\ell\le N-2m$,  using once more that $\bu_\e\to \bz$ and applying the maximum principle it follows that  $\bu_\e>\bo$.
\rule{2mm}{2mm}

\

If $2m=N$, or equivalently $\ell=0$, we set $\bz_1=((u_1^0,v_1^0),\ldots,(u_m^0,v_m^0))$, and if $m=0$, or equivalently $\ell=N$, we set $\bz_2=(U_1,\ldots,U_N)$. Then as a consequence of Theorem \ref{th:ec2} we have the following.
\begin{corollary}\label{cor:ec1}
For $\e$  small enough,
\begin{itemize}
\item[(i)] if $2m=N\Leftrightarrow\ell=0$, then  \eqref{eq:main-sys} has a radial bound state $\bu_{\e}^{(1)}>\bo$  and  $\bu_{\e}^{(1)}\to \bz_1$ as $\e\to 0$.
\item[(ii)] if $m=0\Leftrightarrow \ell=N$,  then \eqref{eq:main-sys} has a radial bound state $\bu_{\e}^{(2)}\ge \bo$  and $\bu_{\e}^{(2)}\to \bz_2$ as $\e\to 0$. Furthermore, if  $\b_{\ell}\ge 0$ for all $1\le\ell\le N$ then $\bu_{\e}^{(2)}>\bo$.
\end{itemize}
\end{corollary}
We observe that Corollary \ref{cor:ec1}-(ii)  is nothing but \cite[Theorem 6.4]{ac2}.
\begin{Remark}
Note that in Theorem \ref{th:ec2}  some of the coupling factors $\b_k$, $1\le k\le m$ are big while the others $\b_{k+\ell}$, $1\le \ell \le N-2m$ are small, and  some of these can be negative giving rise to a non-negative new non-trivial bound states. Similar remark can be done for Corollary \ref{cor:ec1}-(ii).
\end{Remark}

\

\noindent {\bf Acknowledgements.}

The author thought about this work during a visit  at the Institut f\"ur Mathematik of the Goethe-Universit\"at Frankfurt in 2013, and he wishes to thank its hospitality. That visit was partially supported by Beca de Movilidad de Profesores de las Universidades P\'ublicas de Madrid 2012-2013 of Fundaci\'on Caja Madrid. The author also wants to thank Prof. Tobias Weth for useful discussions during that visit.

{\small

config: arxiv


TITLE: if $f: \mathbb{R}^n \rightarrow \mathbb {R}^m$ is differentiable at $a\in\mathbb{R}^n$ then it is continuous at $a$.
QUESTION [2 upvotes]: Prove that if $f: \mathbb{R}^n \rightarrow \mathbb {R}^m$ is differentiable at $a\in\mathbb{R}^n$ then it is continuous at $a$.

My attempt:
My intuition tells me that to show continuous, I need to show that $lim_{x \rightarrow a} f(x)= f(a)$.
By using the book "calculus on manifolds,"....
Let $f: \mathbb{R}^n \rightarrow \mathbb {R}^m$ be differentiable at $a \in \mathbb{R}^n$, then we can say that there exists a linear transformation $\lambda: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that $$ lim_{h \rightarrow0}= \frac {||f(a+h)-f(a)-\lambda(h)||}{||h||}=0$$
I am new to this and I am not sure where to go from here.

REPLY [0 votes]: Assume $f:\mathbb{R}^n\to \mathbb{R}^m$ is differentiable, then there exists a linear transformation $\lambda: \mathbb{R}^n\to \mathbb{R}^m $ such that $$\tag{1}\lim_{h\to 0}\frac{|f(a+h)-f(a) - \lambda (h)|}{|h|}=0.$$
Let $A=(a_{ij})$ be the Jacobian of $\lambda$, $\alpha = \max{|a_{ij}|}$, and $M=\sqrt{mn}\alpha$, then$^*$
$$\tag{2}|\lambda(h)|\leq M|h|$$ for all $h \in \mathbb{R}^n$.
We wish to show $\lim_{x \to a} f(x)=f(a)$. We can use $(2)$ to derive the inequality $$\frac{|f(x)-f(a) - \lambda(x-a)|}{|x-a|}\geq \frac{|f(x)-f(a)| - |\lambda(x-a)|}{|x-a|} \geq \frac{|f(x)-f(a)|}{|x-a|} - M.$$ Let $ \epsilon > 0 $ be given. Let $h=x-a$ in $(1)$, then for any $ \epsilon_1>0$ we can find $\delta >0$ such that $$\epsilon_1 > \frac{|f(x)-f(a) - \lambda(x-a)|}{|x-a|}$$ for all $x\in \mathbb{R}^n$ satisfying $|x-a|<\delta$. Since $\epsilon_1$ was arbitrary, we can choose the pair $(\epsilon_1, \delta)$ small enough that $$\epsilon > \delta( \epsilon_1 + M) > |x-a| (\epsilon_1 + M) > |f(x)-f(a)|.$$ $ \square$
$*$ Proof of (2):
By Cauchy-Schwarz, for any set of real numbers $x_1, ... , x_k$: $$\tag{3}(x_1+...+x_k)^2\leq k(x_1^2+...+x_k^2).$$
Consider
$$y^i= a_{i1} (h^1)+...+a_{in}(h^n),$$  then $$(y^i)^2= |(y^i)|^2\leq \left( |a_{i1} (h^1)|+...+|a_{in}(h^n)|\right)^2 \leq \alpha^2\left( \sum_{j=1}^{n} (h^j) \right)^2\leq n\alpha^2|h|^2, $$ where the last inequality follows from $(3)$. Thus
$$|\lambda(h)| ^2= \sum_{i=1}^{m} (y^i)^2 \leq mn\alpha^2 |h|^2= M^2 |h|^2.$$
$\square$


text:
(*
(C) Copyright Andreas Viktor Hess, DTU, 2020
(C) Copyright Sebastian A. Mödersheim, DTU, 2020
(C) Copyright Achim D. Brucker, University of Exeter, 2020
(C) Copyright Anders Schlichtkrull, DTU, 2020

All Rights Reserved.

Redistribution and use in source and binary forms, with or without
modification, are permitted provided that the following conditions are
met:

- Redistributions of source code must retain the above copyright
  notice, this list of conditions and the following disclaimer.

- Redistributions in binary form must reproduce the above copyright
  notice, this list of conditions and the following disclaimer in the
  documentation and/or other materials provided with the distribution.

- Neither the name of the copyright holder nor the names of its
  contributors may be used to endorse or promote products
  derived from this software without specific prior written
  permission.

THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
"AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR
A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT
OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT
LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
(INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
*)

(*  Title:      Keyserver.thy
    Author:     Andreas Viktor Hess, DTU
    Author:     Sebastian A. Mödersheim, DTU
    Author:     Achim D. Brucker, University of Exeter
    Author:     Anders Schlichtkrull, DTU
*)

section\<open>The Keyserver Protocol\<close>
theory Keyserver
  imports "../PSPSP"
begin

declare [[code_timing]]

trac\<open>
Protocol: keyserver

Types:
honest = {a,b,c}
server = {s}
agents = honest ++ server

Sets:
ring/1 valid/2 revoked/2

Functions:
Public sign/2 crypt/2 pair/2
Private inv/1

Analysis:
sign(X,Y) -> Y
crypt(X,Y) ? inv(X) -> Y
pair(X,Y) -> X,Y

Transactions:
# Out-of-band registration
outOfBand(A:honest,S:server)
  new NPK
  insert NPK ring(A)
  insert NPK valid(A,S)
  send NPK.

# User update key
keyUpdateUser(A:honest,PK:value)
  PK in ring(A)
  new NPK
  delete PK ring(A)
  insert NPK ring(A)
  send sign(inv(PK),pair(A,NPK)).

# Server update key
keyUpdateServer(A:honest,S:server,PK:value,NPK:value)
  receive sign(inv(PK),pair(A,NPK))
  PK in valid(A,S)
  NPK notin valid(_)
  NPK notin revoked(_)
  delete PK valid(A,S)
  insert PK revoked(A,S)
  insert NPK valid(A,S)
  send inv(PK).

# Attack definition
authAttack(A:honest,S:server,PK:value)
  receive inv(PK)
  PK in valid(A,S)
  attack.
\<close>\<open>
val(ring(A)) where A:honest
sign(inv(val(0)),pair(A,val(ring(A)))) where A:honest
inv(val(revoked(A,S))) where A:honest S:server
pair(A,val(ring(A))) where A:honest

occurs(val(ring(A))) where A:honest

timplies(val(ring(A)),val(ring(A),valid(A,S))) where A:honest S:server
timplies(val(ring(A)),val(0)) where A:honest
timplies(val(ring(A),valid(A,S)),val(valid(A,S))) where A:honest S:server
timplies(val(0),val(valid(A,S))) where A:honest S:server
timplies(val(valid(A,S)),val(revoked(A,S))) where A:honest S:server
\<close>

subsection \<open>Proof of security\<close>
protocol_model_setup spm: keyserver
compute_SMP [optimized] keyserver_protocol keyserver_SMP
manual_protocol_security_proof ssp: keyserver
  for keyserver_protocol keyserver_fixpoint keyserver_SMP
  apply check_protocol_intro
  subgoal by code_simp
  subgoal by code_simp
  subgoal by code_simp
  subgoal by code_simp
  subgoal by code_simp
  done

end

subset_name: curated


TITLE: uniqueness of solutions of $ax=b$ and $ya=b$ in a semigroup .
QUESTION [0 upvotes]: Suppose $G$ is a semigroup in which every equation of the form $ax=b$ or $ya=b$ has a solution. Does this solution have to be unique?

REPLY [0 votes]: If a semigroup has a left identity and any element has a left inverse, then it is a group (Hall M.,The theory of groups).
Your semigroup has these properties. Indeed, let $ea=a$ and $ax=b$. Then $eb=eax=b$, so $e$ is a common identity.


TITLE: Find $x(t)$ and $y(t)$ which satisfy the following differential equations
QUESTION [0 upvotes]: Find $x(t)$ and $y(t)$ which satisfy
$3\dot x + \dot y +5x-y=2e^{-t}+4e^{-3t}$,
$\dot x + 4\dot y -2x+7y=-3e^{-t}+5e^{-3t}$,
subject to $x=y=0$ at $t=0$.

This is how I tried it:
If we multiply the first equation by $4$ and then substract the second we get the nicer
$\dot x + 2x=e^{-3t}+e^{-t}+y$
Similarly, if we multiply the second equation by $3$ and substract the first, we get
$\dot y + 2y=e^{-3t}-e^{-t}+x$.
If now we try to solve the first one in the usual 'particular solution+general solution' way and substitude in for $\dot y$ from the second, we end up with a new equation for $x$ but it still involves $\dot x$. So it seems like we are just going in circles.

REPLY [1 votes]: derivative of
$$y=x'+2x-e^{-3t}-e^{-t}$$
is
$$y'=x''+2x'+3e^{-3t}+e^{-t}$$
substitute in the second equation to get
$$x''+2x'+3e^{-3t}+e^{-t}+2(x'+2x-e^{-3t}-e^{-t})=e^{-3t}-e^{-t}+x$$
$$x''+4x'+3x=0$$
$$m^2+4m+3=0$$
$$m_1=-1$$
$$m_2=-3$$
so
$$x=C_1e^{-t}+C_2e^{-3t}$$
$$x'=-C_1e^{-t}-3C_2e^{-3t}$$
so
$$y=-C_1e^{-t}-3C_2e^{-3t}+2(C_1e^{-t}+C_2e^{-3t})-e^{-3t}-e^{-t}$$
$$y=C_1e^{-t}-C_2e^{-3t}-e^{-3t}-e^{-t}$$


TITLE: Getting different answers for same problem on divergence and curl.
QUESTION [0 upvotes]: Given that $\vec{a}$ is a constant vector and $\vec{r}$ is a position vector.
We are asked to prove the following:
$$\nabla\times(\vec{a}\times\vec{r})=2\vec{a}$$
I tried two ways. Could prove it from one method but from another I am getting another result.
$$\rule{20cm}{0.4pt}$$
1st method(with correct result):
$$LHS=\nabla\times(\vec{a}\times\vec{r})$$
Let $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \,\,$ and $\,\,\vec{r}=x\hat{i}+y\hat{j}+z\hat{k},$ where $a_1,a_2,a_3$ are constants and $x,y,z$ is any arbitrary point.
$$Simplifying\,\,\,\, (\vec{a}\times\vec{r})$$
$$=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ x & y & z\end{vmatrix}$$
$$=\hat{i}(a_2z-a_3y)+\hat{j}(a_3x-a_1z)+\hat{k}(a_1y-a_2x)$$
Now performing the curl operation on the simplified result we obtained.
$$=\nabla\times(simplified\,result)$$
$$=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y}  & \frac{\partial}{\partial z} \\ a_2z-a_3y & a_3x-a_1z & a_1y-a_2x\end{vmatrix}$$
$$=2a_1x\hat{i}+2a_2y\hat{j}+2a_3z\hat{k}$$
$$=2\vec{a}$$
$$=RHS$$
$$\rule{20cm}{0.4pt}$$
2nd method with incorrect result.
We know that $$\nabla\times(\vec{a}\times\vec{r})=2[(\nabla.\vec{r})\vec{a}-(\nabla.\vec{a})\vec{r}]$$
Solving RHS,
$$Also,\nabla.\vec{r}=3\,and\, \nabla.\vec{a}=0$$
$$=2[3\vec{a}-0\vec{r}]$$
$$=6\vec{a}$$
which is incorrect. Where did I make mistake in this method?

REPLY [1 votes]: The "vector calculus identity" you use in your second calculation is just false. You actually have
\begin{align*}
\nabla \times (\vec{a} \times \vec{r}) &= \vec{a}(\nabla \cdot \vec{r}) - \vec{r}(\nabla \cdot \vec{a}) + (\vec{r} \cdot \nabla)\vec{a} - (\vec{a} \cdot \nabla) \vec{r} \\
&=
\vec{a}(\nabla \cdot \vec{r})  - (\vec{a} \cdot \nabla) \vec{r}.
\end{align*}
As you say, the first term on the RHS is then $3 \vec{a}$, but we need to calculate the other term. It is just the directional derivative of $\vec{r}$ in the direction of $\vec{a}$ multiplied by the magnitude of $\vec{a}$, so this is just $\vec{a}$. Thus
$$
\nabla \times (\vec{a} \times \vec{r}) = 3 \vec{a} - \vec{a} = 2 \vec{a},
$$
as desired.


TITLE: How to resolve a horizontal force acting on an object in an incline plane into a parallel force?
QUESTION [0 upvotes]: A particle of weight $W$ Newtons is supported on a smooth inclined plane by two forces each equal in
magnitude to half the weight, one acting horizontally and the other acting parallel to the plane.
We are asked to draw a diagram and here is the one provided by the answer:

First of all, why is the horizontal force ($\frac{w}{2}$) is acting to the right? Why can't it acts to the left instead?
Second, if the horizontal force is to the right, how can I resolve it into the parallel component, in order to get the form:

this is my diagram, why is it wrong?

REPLY [0 votes]: First of all, why is the horizontal force ($\frac{w}{2}$) is acting to the right? Why can't it acts to the left instead?

Let us suppose it to be acting towards the left. Now , what are the other forces or component of forces in that direction ? It is pretty clear that the component of gravity in that direction is zero . Now , the component of $N$ will be $N\sin \theta$ (you can get this using the alternate angle theorem for parallel lines). Also you will have a component of $\frac{w}{2}$ in the opposite direction and it will be $\frac{w}{2}\cos \theta$ (the reason for this is explained in your second question).
It is pretty obvious that in this situation the body can't be supported on the inclined (which is against the first line of your question) . So the force must be towards the right.

Second, if the horizontal force is to the right, how can I resolve it into the parallel component

From the picture, you can get that the angle between the force parallel to the Incline and the horizontal force is $\theta$.
Now
Here's the picture of resolution :

Vector resolution is a process in which we draw an imaginary right angled triangle with the vector as it's hypotenuse and then we find the value of that vector along the remaining two perpendicular sides of that triangle.
Since the angle between the actual force (green line) and the dotted blue line is $\theta$. So the component of that green line along the dotted line is $\frac{w}{2}\cos \theta$ using this
$$\cos \theta = \frac{base}{hypotenuse} =\frac{dotted \;blue\; line}{green \;line}$$
$dotted \; blue \; line = (green \; line)×(\cos \theta) =\frac{w}{2} \cos \theta$
and since you can move a vector on its line of support or along the line of support of a parallel vector. So , the bold blue line (along the incline) is also $\frac{w}{2}\cos \theta$.
Your last picture is useful only when you are resolving the force which is acting along the incline since the two axes along which it is being resolved are perpendicular. But if you are trying to resolve the horizontal force on those two lines then you can get its component on the hypotenuse only and not on the vertical line and this is against the definition of resolution. You need two mutually perpendicular axis for resolving a vector but your picture doesn't meet this criteria for the horizontal force and so it wrong to resolve that horizontal force along the two remaining sides.
Hope it helps .


TITLE: last non-zero digit sum
QUESTION [0 upvotes]: $
\sum_{n=p}^{n=q} F(n)
$
where
F(n) is last non-zero digit of n and $ 1 \leq p \leq q \leq 2^{31}$
For example,
F(1050) = 5
F(1008) = 8
F(2000) = 2
Can anyone help me how to get the result using number tricks?

REPLY [2 votes]: The last non-zero digit follow this pattern :
$\underbrace{123456789}_{u_1}$
$\underbrace{u_1 1 u_1 2 u_1 3 u_1 4 u_1 5 u_1 6 u_1 7 u_1 8 u_1 9 u_1}_{u_2}$
$\underbrace{u_2 1 u_2 2 u_2 3 u_2 4 u_2 5 u_2 6 u_2 7 u_2 8 u_2 9 u_1}_{u_3}$
So you just have to calculate the sum for $u_1$, $u_2$, $u_3$ , $\cdots$ and find where your sum begin and end. Then you can be smart, and calculate the number of each $u_n$ you'll need
For exemple, between 3456 and 11578, I have the sequence
$67896u_17u_18u_19u_1 5 u_2 6 u_2 7 u_2 8 u_2 9 u_2 4 u_3 5 u_3 6u_3 7u_3 8u_39u_31$ (we're at 10000)
$u_31 u_22u_23u_2 4u_2 5 u_1 1 u_1 2 u_13 u_1 4 u_1 5 u_1 6 u_1 7 12345678$
So, I need:

"6789" to get to the next 10,
"6789" + 4*$u_1$ to get to the next 100,
"56789" + 5*$u_2$ to get to the next 1000,
"45678911" + 8*$u_3$ to get to 11000
etc.

So by the same reasonning, if I want the sequence between 684551 and 767894, I'll will need :

"23456789" to get to 684560 (the numbers after the 1)
"6789" +4$u_1$ to get to 684600 (the numbers after the 5)
"6789" +4$u_2$ to get to 685000 (the numbers after the 5)
"56789" +5$u_3$ to get to 690000 (the numbers after the 4)
"9" +1$u_4$ to get to 700000 (the numbers after the 8)
"7" (we're at 700000)
"123456" + 6$u_4$ to get to 760000
"1234567" + 7$u_3$ to get to 767000
"12345678" + 8$u_2$ to get to 767800
"123456789" + 9 $u_1$ to get to 767890
"1234" to get to 767894


TITLE: Midpoint of two line segments in three dimensions
QUESTION [0 upvotes]: This might be an easy question, but since i'm new to solid shapes, i couldn't solve it. A= (7,1,3) B=(5,1,2) C=(4,-2,3) D=(6,m,n) I need to find m and n so that segments BD and AC have the smae midpoint. I wrote the midpoint formula for AC and found its midpoints and then tried to equate it with the midpoints of the segment DC, but i have 2 unknowns there and couldn't find m and n. Any help would be appreciated

REPLY [0 votes]: The midpoint of $AC$ has coordinates:
$$
M_{AC}=\left(\frac{x_A+x_C}{2},\frac{y_A+y_C}{2},\frac{z_A+z_C}{2}\right)=\left(\frac{11}{2},-\frac{1}{2},3 \right)
$$
In the same way,the midpoint of $BD$ has coordinates:
$$
M_{CD}=\left( \frac{11}{2},\frac{1+m}{2},\frac{2+n}{2}\right)
$$
so we want:
$$
\frac{1+m}{2}=-\frac{1}{2} \quad \land \quad \frac{2+n}{2}=3
$$
two equations for two unknowns.


TITLE: How to represent a network of chemical reactions?
QUESTION [1 upvotes]: I am trying to figure out a suitable representation for given set of chemical reactions, which happen over an exhaustive set of chemicals.
The chemicals are $A, B, C, D, E$ and the reactions are given below
$$A + A \rightarrow B$$
$$B + C \rightarrow D$$
$$B + D \rightarrow E$$
$$C + E \rightarrow D$$
$$E + E \rightarrow A$$
It's only obvious to form a matrix representaion of the reactions.
\begin{array}{X}
  & A & B & C & D & E \\
A & B & * & * & * & * \\
B & - & * & D & E & * \\
C & - & - & * & * & D \\
D & - & - & - & * & * \\
E & - & - & - & - & A \\
\end{array}
where $*$ means no product forms

I tried to label elements with unique integer IDs, so that the product (if any) can be treated as the weight of the edge, but I failed to incorporate this new property into a visualisation of the matrix.
Can any one tell me what form of graphs can be used to represent such data, where combinaiton of two nodes gives rise to another?

REPLY [1 votes]: For a matrix representation, you example seems to most naturally work as a 3D matrix, with the dimensions representing the first reactant, second reactant and product, and the value representing the yield, speed or some other property.
An alternative method might be to have a bipartite directed graph, with reactants/products in one part and the reactions themselves as nodes in the other part:

This might also work well as a 2D matrix, using a sentinel value of some sort to denote inputs and a yield or other value for the outputs.


TITLE: How to show that this function is integrable
QUESTION [0 upvotes]: Let $f\colon [0,1] \to \mathbb R  $ a function defined by
$$ f(x) =
\begin{cases}
1,  & \text{if $\exists p \in \mathbb N, x = \frac 1p$} \\
0, & \text{elsewhere}
\end{cases} $$
I have to show that $f$ is integrable in $[0,1]$.
I tried using Cauchy criterion but no luck so far. any hints are welcome.

REPLY [1 votes]: hint
Let $\epsilon>0$ be a small real.
$f $ is integrable at $[\frac {\epsilon}{2},1] $ as a stairs function, thus there exist a subdivision $s_1$ of $[\frac {\epsilon}{2},1] $ such that
$U (f,s1)-L (f,s1)<\frac {\epsilon}{2}$
put $s=s_1\cup \{0\} $
we have
$$U (f,s)-L (f,s)\leq $$
$$\frac {\epsilon}{2}+U (f,s_1)-L (f,s_1) <\epsilon$$


TITLE: Intuition behind gravitational potential
QUESTION [1 upvotes]: Gravitational potential at a point is equal to work done in bringing a unit mass from infinity to a particular point

That was the text book definition
$$V_{p} = -\frac{GM}{r}$$
If we calculate $V_p$ for the Earth surface with $R_e = 6.4 \times 10^6 \text{ m}$ and $M_e = 6 \times 10^{24} \text{ kg}$ we get
$$V_p = -6.35 \times 10^7 \text{ J/kg}$$
Q1) What does this number tell us? Does it mean I have to give a body of mass $1 \text{ kg}$ kinetic energy worth of $6.35 \times 10^7 \text{ J}$ in order for the body to leave this planet (escape Earth's gravitational influence) and after escaping Earth's influence its kinetic energy will be zero?
What does $-ve$ sign indicate?

Q2) Does higher magnitude of $V_{p}$ indicate higher potential energy of the body? If yes, we know that everything in this universe tries to reduce its potential so won't it make more sense if gravity doesn't attract but rather repel other mass as potential is zero at infinity, but it attracts? If higher magnitude of $V_{p}$ doesn't indicate higher potential energy of the body skip to next question.

Q3) We have two formulas to look at: $U = mgh$ and $V_p = -\frac{GM}{r}$. Let's deal with mass of $1 \text{ kg}$. Assuming that $-ve$ sign of gravitational potential has nothing to do with potential energy.
In the first formula we are multiplying by height, which means higher the body from the Earth surface higher is the potential which makes sense. Here $U$ and $h$ are directly proportional. The second formula doesn't make sense as we're dividing with height, $V_p = -\frac{G M_e}{R_e+h}$ here $h$ and potential are inversely related?
I'm confused regarding this, tell me where am I going wrong

REPLY [0 votes]: The equations in consideration are:
$$V = \frac{-GM}{r} \ = \frac{U_g}{m}$$
Here $M$ is the mass of the particle establishing the gravitational field. It follows that $U_g = -\frac{GMm}{r}$.

Q1) What does this number tell us?

The value you have obtained essentially means that $-6.35 \times 10^7 J$ of work needs to be done on an object of mass $1\text{kg}$ to bring it from $r = ∞$ to $r = 0$ at constant velocity.
The reason for it being negative is that the applied force would have to be directed opposite to the body's displacement. This is because the gravitational force is attractive and would otherwise cause the body to accelerate. Consequently, in order to maintain a constant velocity, a force must be applied in the opposite direction.

Q2) Does higher magnitude of Vp indicate higher potential energy of the body?

First, the gravitational potential energy is defined for the system of two bodies (or more), and not the body alone. The term "potential energy of the body" does not make sense.
As $$V = \frac{U_g}{m}$$ a higher value of $V$ would mean a higher value of $U_g$, the gravitational potential energy of the system.

so won't it make more sense if gravity doesn't attract but rather repel other mass as potential is zero at infinity, but it attracts?

Pay attention to the negative sign in the equation. As $r \rightarrow 0$ (as the force is attractive), $V \rightarrow -∞$, consistent with your claim of the potential decreasing.

...Here U and h are directly proportional. The second formula doesn't make sense as we're diving with height, $V_p=−\frac{GMe}{Re+h}$ here h and potential are inversely related?

Again, the negative sign plays a role here. $$V \propto -\frac{1}{r} \ \text{not} \ V \propto \frac{1}{r}$$
Hence, as $r$ increases, $V$ increases, consistent with the expression for the approximate formuala for the potential energy of a two mass system.
Hope this helps.


TITLE: Generating function for the number of choices $I,J\!\subseteq[n]$ such that $\max\,[n]\!\setminus\!(I\!\cup\!J) < \max I\!\cap\!J$
QUESTION [2 upvotes]: Suppose that each pair $I,J\!\subseteq[n]=\{1,\ldots,n\}$ for which
$$\max\,([n]\!\setminus\!(I\!\cup\!J)) < \max (I\!\cap\!J) \tag{1}$$ contributes $t^{|I|+|J|}$ to a generating function, and each pair $I,J\!\subseteq[n]$ for which
$$\max\,([n]\!\setminus\!(I\!\cup\!J)) > \max (I\!\cap\!J) \tag{2}$$ contributes $t^{|I|+|J|+1}$ to the same generating function. How can I show that this generating function is the one at the bottom of this calculation (taken from here):

When $I\cup J=[n]$, this is a special case (that you need not be concerned about) and only $I\!\cap\!J\!=\!\emptyset$ is allowed, so we are counting all choices of $I\!\subseteq\![n]$, of which there are $2^n$, hence we get $2^nt^{|I|+|J|}+2^nt^{|I|+|J|+1}= 2^nt^n+2^nt^{n+1} = (1+t)(2t)^n$, which is the first summand.
But when $I\cup J\subsetneq[n]$, is $L=I\cup J$ or $L=(I\!\cup\!J)\setminus(I\!\cap\!J)$? Where do $$t^3(2t)^{|L|}(1+t^2)^{n-|L|-1} \text{ and } (2t)^{|L|}(1+t^2)^{n-|L|-1} \text{ come from???}$$ I do not understand why $(1)$ should be equivalent to $(1')$ $\max([n]\!\setminus\!(I\!\cup\!J)) \notin K$. I suspect that $K\!=\!I\!\cap\!J$, and that in $(1)$ we have arbitrary $I,J$, but in $(1')$ we have $I=I\!\setminus\!I\!\cap\!J$ and $J=J\!\setminus\!I\!\cap\!J$.

REPLY [0 votes]: Instead of finding $a_k:=\big|\big\{I,\!J\!\subseteq\![n]; |I|\!+\!|J|\!=\!k, \max(I^C\!\cap\!J^C)\!>\!\max(I\!\cap\!J)\big\}$ $\cup$ $\big\{I,\!J\!\subseteq\![n]; |I|\!+\!|J|\!+\!1\!=\!k, \max(I^C\!\cap\!J^C)\!<\!\max(I\!\cap\!J)\big\}\big|$, we compute the generating function $\sum_{k=0}^{2n+1}\!a_kt^k$. A critical vertex $x_I\!\!\wedge\!y_J$ contributes $t^{|I|+|J|}\!=\!t^{|I\cup J|+|I\cap J|}$ and a critical vertex $x_I\!\!\wedge\!y_J\!\wedge\!z$ contributes $t^{|I|+|J|+1}\!=\!t^{|I\cup J|+|I\cap J|+1}$.
Let $u_k\!=\!x_k\!\wedge\!y_k$, so that we write any basis element as $x_{I'}\!\!\wedge\!y_{J'}\!\wedge\!u_{K}$ or $x_{I'}\!\!\wedge\!y_{J'}\!\wedge\!u_{K}\!\wedge\!z$, with $I'\!\cap\!J'\!=\! I'\!\cap\!K\!=\! J'\!\cap\!K\!=\!\emptyset$ and $L\!:=\!I'\!\cup\!J'$. If $L\!=\![n]$, then $K\!=\!\emptyset$, hence all $x_{I'}\!\!\wedge\!y_{J'}$ (cannot remove $u_i$) and $x_{I'}\!\!\wedge\!y_{J'}\!\wedge\!z$ (cannot add $u_i$) are critical, contributing $2^nt^n\!+\!2^nt^{n+1}$ (number of subsets $I'\!\subseteq\![n]$ is $2^n$). If $L\!\subsetneq\![n]$, then $x_{I'}\!\!\wedge\!y_{J'}\!\wedge\!u_{K}$ is critical iff $\max(L\!\cup\!K)^C\!>\!\max K$ iff $m\!:=\!\max L^C\!\notin\!K$, contributing
$$2^{|L|}\sum_{k=0}^{n-|L|-1}\!\!\binom{n\!-\!|L|\!-\!1}{k}t^{|L|+2k} \!=\! 2^{|L|}t^{|L|}\sum_{k=0}^{n-|L|-1}\!\!\binom{n\!-\!|L|\!-\!1}{k}(t^2)^k \!=\! (2t)^{|L|}(1\!+\!t^2)^{n-|L|-1}$$ (number of subsets $I'\!\subseteq\!L$ is $2^{|L|}$, number of subsets $K\!\subseteq\!(L\!\cup\!\{m\})^C$ with $k$ elements is $\binom{n-|L|-1}{k}$), and $x_{I'}\!\!\wedge\!y_{J'}\!\wedge\!u_{K}\!\wedge\!z$ is critical iff $\max(L\!\cup\!K)^C\!<\!\max K$ iff $\max L^C\!\in\!K$, contributing
$$2^{|L|}\sum_{k=0}^{n-|L|-1}\!\!\binom{n\!-\!|L|\!-\!1}{k}t^{|L|+2(k+1)+1} \!=\! 2^{|L|}t^{|L|+3}\sum_{k=0}^{n-|L|-1}\!\!\binom{n\!-\!|L|\!-\!1}{k}(t^2)^k \!=\! t^3(2t)^{|L|}(1\!+\!t^2)^{n-|L|-1}$$ (number of subsets $I'\!\subseteq\!L$ is $2^{|L|}$, number of subsets $K\!\setminus\!\{m\}\!\subseteq\!(L\!\cup\!\{m\})^C$ with $k$ elements is $\binom{n-|L|-1}{k}$).


TITLE: Proving that $-\frac{\cos x +\cos^2 x + \sin^2 x}{(1 + \cos x)^2} = -\frac{1}{1 + \cos x}$
QUESTION [0 upvotes]: Could anyone give me some details on how this happens?
$$-\frac{\cos x +\cos^2 x + \sin^2 x}{(1 + \cos x)^2} = -\frac{1}{1 + \cos x}$$

REPLY [6 votes]: $$-\frac{\cos x + \overbrace{\cos^2 x+ \sin^2 x}^{=1}}{(1+\cos x)^2} = -\frac{\cos x + 1}{(1+\cos x)^2} = -\frac{1}{1+\cos x} $$


TITLE: Extension of mapping of subset to homomorphism
QUESTION [0 upvotes]: Is the following proposition true?
Let $V$, $W\neq0$ be vector spaces over some field $F$ and let $S_v \subset V$.
Then if every mapping $f:S_v \to W$ can be uniquely extended to homomorphism $g:V\to W$, $S_v$ is a basis of $V$.
So far I tried to prove only that $S_v$ can contain only linear independent vectors.
I proceed as follows:
Let every mapping $f:S_v \to W$ be extendable to homomorphism $g:V\to W$. Let $B_v=(b_1,...,b_n)$ be a basis of V. Then it's true that $\forall v \in V$ with coordinates ($c_1,...,c_n$):
$g(v) = g(\sum_{ \substack{i\in \hat{n}}}c_ib_i)=\sum_{{i\in \hat{n}}}c_ig(b_i)$
Moreover let $S_v=(s_1,...,s_m)$ and as $g$ is extension of $f$ : $(\forall s \in S_v) (g(s) = f(s))$.
Then $S_v$ can't be linearly dependent because if it is then:
($\exists c_1,...c_m)((\exists i\in \hat{m})(c_i \neq 0 \wedge \sum_{ \substack{j\in \hat{m}}}c_js_j=0 ) \wedge g(\sum_{ \substack{i\in \hat{m}}}c_is_i)=\sum_{ \substack{i\in \hat{m}}}c_if(s_i) = 0)$
and that won't hold for all possible mappings $f$.
Can I somehow proceed further and prove that $S_v$ is basis?
I tried to define a homomorphism on $V$. Knowing that $S_v$ is linearly independent I can extend it to basis $B'_v = (s_1, ..., s_m, b'_{m+1}, ... , b'_n)$ and define $g$ also to map $(b'_{m+1}, ... , b'_n) \to W $.
At the end I will have:
$(\forall v \in V)(\exists c_1,...,c_n) (g(v) = g(\sum_{ {i\in \hat{m}}}c_is_i+\sum_{ {m<j\leq n}}c_jb'_i) = \sum_{ {i\in \hat{m}}}c_if(s_i)+\sum_{ {m<j\leq n}}c_jg(b'_i)))$.
But it won't be a unique extension because I can extend $S_v$ to basis in different ways. So the only option is to have $S_v$ as basis already.
Does this correctly prove that the proposition is true?
Corrections compared to the first version: forgot to mention uniqueness of extension, added restriction $W\neq0$.

REPLY [1 votes]: It works if you assume that $dim(W)\ge dim(V)$, as pointed out in the comments. I will assume that $W=V$ as it is the simplest case, and all other cases are a corollary of it.
Let $S_V=\{s_1, \dots \}$. Let $U=Span(S_V)$. Then we have that the map $f:S_V\to V$ given by $v\mapsto v$ can be extended to the identity map $V\to V$, and the map given by projection $V\to U$ both extend $f$, so they must coincide, and $U=Im(Pr_U)=Im(Id)=V$, so $S_V$ is a spanning set.
For independence, assume that $s_i=\sum_{i\neq j} a_js_j$. Then we have a map $f_i:S_V\to V$ given by $s_j\mapsto 0$ for $j\neq i$, and $s_i\mapsto s_i$. By assumption, this extends to $f:V\to V$. So that $s_i=f(s_i)=f(\sum_j a_js_j)=\sum_i a_jf(s_j)=0$. But if $s_i=0$, then the map $f:S_V\to V$ given by $s_i\mapsto 1$ has $1=f(s_i)=f(0)=f(2*0)=f(2s_i)=2$, a contradiction. Thus we have that the $s_j$ are independent, so that $S_V$ forms a basis as desired.