text:
\section{Introduction}
In recent years, the second order cone ${\cal L}^n_+:= \{(x_0,x^{n-1}) \in \mathbb R \times \mathbb R^{n-1}: x_0 \geq \|x^{n-1}\|\}$, also known as the Lorentz cone, attracted much attention of the researchers in optimization, particularly in conic optimization. Many optimization problems can be reformulated as the conic ones. There are computationally stable numerical algorithms for solving various such problems, including complementarity problems. The literature on the subject is vast and commonly accessible, see e.g., Alizadeh and Goldfarb \cite{alizadeh-goldfarb}, and a recent work \cite{ hao-wan-chi-jin}, by Hao et al., which is related to the Mixed Complementarity Problem over the second order cone.

The Lorentz cone has a particularly regular structure: it is a {\em self-dual} cone, whose base is isometric to the Euclidean unit ball in $\mathbb R^{n-1}$ and is {\em irreducible}. In the context of Euclidean Jordan algebras, ${\cal L}^n_+$ is a symmetric cone (of squares) in the spin algebra ${\cal L}^n$ it generates. We will not pursue this direction here.

There are several known important versions of the extended Lorentz cone, including {\em Bishop--Phelps cone} \cite{gowda-trott} and the Extended Second Order cone (ESOC), which was recently developed by S.Z. N\'emeth and his co-authors, see \cite{nemeth-zhang1,nemeth-zhang2,nemeth-xiao1,nemeth-xiao2,nemeth-xie-zhang}. The {\em Lyapunov rank} of a cone $K$, denoted by $\beta(K)$ (see its definition in the next section) is an invariant which shows that the Lorentz cone and ESOC are generally not linearly isomorphic. It was introduced and studied by F. Alizadeh et al. in \cite{alizadeh-etal} under the name of {\em bilinearity rank}. The Lyapunov rank of ${\cal L}^n_+$ was computed in \cite{alizadeh-etal} and \cite{gowda-tao}, and equals $\frac{n^2-n}{2}$. Orlitzky, in \cite{orlitzky}, showed that the latter quantity is the maximum value the Lyapunov rank can attain for a proper cone in $\mathbb R^n$. Sznajder, in \cite{sznajder}, showed that the ESOC is irreducible and computed its Lyapunov rank, which is generally lower than $\frac{n^2-n}{2}$.

In this article, we study another extension of the Lorentz cone, called the {\em Monotone Extended Second Order Cone} (MESOC)
\cite{ferreira-gao-nemeth}. There are three main results related to MESOC
here:

\vspace{-.3cm}
\begin{itemize}
  \item[$\circ$]\, computing its Lyapunov rank, which turns out, in general, is much lower than the minimal upper bound indicated in \cite{orlitzky},
  \item[\mbox{}]\, \vspace{-1.1cm}
  \item[$\circ$]\, proving that MESOC (in contrast to ESOC) is a {\em reducible} cone,
  \item[\mbox{}] \vspace{-1.1cm}
  \item[$\circ$]\, showing that a closed convex set is an {\em isotonic
	  projection set} with respect to MESOC if and only if it is a
	  {\em cylinder} (in an ambient space).
\end{itemize}

\vspace{-.3cm}
The paper is organized as follows: In Section 2, we collect the necessary definitions and provide examples of monotone cones. The main concept related to a cone $K$, on which the paper relies upon, is the {\em complementarity set of $K$}. In Section 3, for MESOC, we identify its dual space and investigate the structure of its complementarity set. We also formulate here and prove the results listed above. In Section 4, based on the work done in \cite{nemeth-zhang1} and \cite{nemeth-zhang2}, we study the properties of the Mixed Complementarity Problem (MiCP). By exploring the relationship of mixed complementarity problem and nonlinear complementarity problem derived in \cite{nemeth-zhang1}, and by using the isotonicity of MESOC obtained in Section 3, we generate a fixed point iteration sequence (called Picard iteration by some authors), which is convergent to a solution of the MiCP on a general closed and convex cone. The convergence of this iteration is order-based, rather than based on a usual contraction mapping principle, although the preprint \cite{banach-lorentz} and the example in the final section suggests that in certain situations it may be implicitly related to such a principle. This example is about a real MiCP example. We show the existence of a solution, in exact numbers, by using the above iteration.

\section{Preliminaries}

Denote the canonical unit vectors of $\mathbb R^n$ by $e^1,\ldots,e^n$ and let $e=e^1+\cdots +e^n$.
Any vector $z\in\mathbb R^n$ is considered to be a column vector and can be uniquely written as
$z=(z_1,\ldots, z_n)^\top:=z_1e^1+\cdots+z_ne^n$. In particular $e=(1,\ldots,1)^\top$.

The canonical {\em inner product} of any two vectors $x,y\in\mathbb R^n$ is defined as \[\langle
x,y\rangle:=x^\top y=x_1y_1+\cdots+x_ny_n.\] We identify $\mathbb R^p\times\mathbb R^q$ with $\mathbb R^{p+q}$ through $(x,y)=(x^\top,y^\top)^\top$.

We call the set
\begin{equation*
	\mathcal H(u,a):=\{x\in \mathbb R^n:\;\langle x-a,u\rangle=0\}
\end{equation*}
an \emph{affine hyperplane} through $a\in\mathbb R^m$ with the normal $u\in\mathbb R^n\setminus\{0\}$ and the corresponding sets
\[\mathcal H_-(u,a):=\{x\in \mathbb R^n:\langle x-a,u\rangle\le0\},\]
\[\mathcal H_+(u,a):=\{x\in \mathbb R^n:\langle x-a,u\rangle\ge0\},\]
\emph{closed half-spaces}.
An \emph{affine hyperplane} through the origin will be simply called \emph{hyperplane}.

A nonempty set $K\subseteq\mathbb R^n$ is a {\em cone} if for any $x\in K$ and $\forall \alpha > 0$, it holds $\alpha x \in K$. A set $K$ is a {\em convex cone} (i.e., cone $K$ is a convex set) if and only if for any $x,y\in K$ and $\forall \alpha,\beta > 0$, it holds $\alpha x + \beta y \in K.$

\vskip 0.1cm
\noindent A cone $K$ is called a {\em closed cone (pointed)} when it is a closed set ($K \cap -K =\{0\}$).

\vskip 0.1cm
\noindent The dual cone of a cone $K$ is given by
$$
K^* := \{y \in \mathbb R^n :\langle x, y\rangle\geq 0,  \forall x\in K\}.
$$

\noindent We define the following set, which is vital for our further considerations
$$
C(K):= \{(x, y): x\in K, y\in K^*,\, x \perp y\},
$$
called the {\em complementarity set of $K$}, where $x\perp y$ means $\langle x,y\rangle = 0$.

A cone $K\subseteq\mathbb R^n$ is called \emph{simplicial} if there is a basis $\{u^i:1\le i\le m\}$ of
$\mathbb R^n$ such that
\[
K=\left\{\alpha_1 u^1+\dots+\alpha_n u^n\,:\,\alpha_i\ge 0,\,1\le i\le n\right\}.
\]
The vectors $u^i$, $1\le i\le m$ are called the \emph{generators} of $K$. It is known that the
dual of a simplicial cone is also simplicial.

We present two examples of complementarity sets, the second will be used later.

\vskip 0.1cm
\noindent {\bf Example 1.
\, Define the {\em monotone cone} $\mathbb R_{\geq}^n$ as
$$
\mathbb R_{\geq}^n :=\{x\in\mathbb R^n:x_1\geq x_2\geq \cdots\geq x_n\}.
$$
It is easy to check that its dual cone $(\mathbb R_{\geq}^n)^*$ is given by
$$
(\mathbb R_{\geq}^n)^*=\left \{y\in\mathbb R^n: \sum_{i=1}^{j} y_j\geq 0,~ j=1,2,\ldots,n-1,~ \sum_{i=1}^{n}y_i=0 \right \}.
$$
It is an important object, also known as the {\em Schur cone} (see \cite{seeger-sossa}, Example 7.4), since it induces the so-called {\em Schur ordering}, which plays an important role in the theory of majorization, see \cite{marshall-olkin-arnold}.

\vskip 0.2cm
\noindent The complementarity set $C(\mathbb R_{\geq}^n)$ of the cone $\mathbb R_{\geq}^n$ is described as
$$
C(\mathbb R_{\geq}^n)=\left \{(x,y):x\in \mathbb R_{\geq},y\in (\mathbb R_{\geq}^n)^*,~ (x_i-x_{i+1})\sum_{j=1}^{i} y_j=0,~\forall i=1,2,\ldots,n-1\right \}.
$$

\noindent {\bf Example 2.} \label{example2}\,
We define the {\em monotone nonnegative cone} $\mathbb R_{\geq+}^n$ as:
$$\mathbb R_{\geq+}^n :=\{x\in\mathbb R^n: x_1\geq x_2 \geq\cdots\geq x_n\geq 0\}\,.$$
Its dual cone is given by:
$$(\mathbb R_{\geq+}^n)^*=\left \{y\in\mathbb R^n: \sum_{i=1}^{j}y_i\geq 0,\, j=1,2,\ldots,n\right \},$$
and the complementarity set of\, $\mathbb R_{\geq+}^n$ is equal to

\begin{eqnarray*}
C(\mathbb R_{\geq+}^n)=\left\{x\in\mathbb R_{\geq+}^n,y\in(\mathbb R_{\geq+}^n)^*: \left(x_j=x_{j+1} \textrm{ or } \sum_{i=1}^{j}y_j=0,\textrm{ }\forall j=1,2,\ldots,n-1\right)\, \right. \\ \left. \textrm{and}\left(x_n=0 \textrm{\, or } \sum_{i=1}^{n}y_i=0\right)\right\}.
\end{eqnarray*}
Both $\mathbb{R}^n_{\ge+}$ and $(\mathbb{R}^n_{\ge+})^*$ are simplicial cones.

\bigskip

Recall \cite{nemeth-zhang1} that the extended second order cone ({\rm ESOC}) is defined by
\[
\textrm{ESOC}=\{(x,u)\in\mathbb R^p\times\mathbb R^q:\,x\geq\|u\|e\}
\]
and its dual cone is given as
\[
\textrm{(ESOC)}^*=\{(x,u)\in\mathbb R^p\times\mathbb R^q:\,\langle x,e\rangle\geq\|u\|,\,x\geq 0\},
\]
where $p$ and $q$ are nonnegative integers.

\noindent A matrix $A\in\mathbb R^{n\times n}$ is called {\em Lyapunov-like} on $K$, if
\begin{equation} \label{lyap-like}
\langle Ax,y\rangle = 0, \text{ }\forall (x,y) \in C(K).
\end{equation}
Define a vector space $\LL(K)$ as the set of all Lyapunov-like matrices on $K$ and denote its dimension as $\beta(K)$, which we call the {\em Lyapunov rank} (or {\em bilinearity rank}) of $K$.

\vskip 0.1cm
\noindent For an arbitrary closed convex set $C \subseteq \mathbb{R}^m$, we define mapping $P_C$--{\em metric projection} onto $C$:
$$\mathbb{R}^m\ni x\mapsto P_Cx:=\arg\min\{\|y-x\| : y\in C\}.$$

\noindent Necessarily, $P_C$ is a point--to--point mapping, which is well defined from $\mathbb{R}^m$ onto $C$.
 We also indicate that the projection $P_C$ is {\em nonexpansive}, i.e., for any $x,y\in\mathbb{R}^m$,
\begin{equation}\label{nonexpansivity}
 	\|P_C(x)-P_{C}(y)\|\leq\|x-y\|.
\end{equation}

\noindent For any pointed closed convex cone $K\subset\mathbb{R}^m$, a mapping $F: \mathbb{R}^m \rightarrow \mathbb{R}^m$ is called
{\em $K$-isotone} if for any $x, y \in K$, $x\leq_{K} y$ implies $F(x)\leq_{K} F(y)$; here $x\leq_{K} y$ means $y-x \in K$.
If the projection $P_C$ is $K$-isotone, then the closed convex set $C \subseteq \mathbb{R}^m$ is called a {\em $K$-isotone projection set}.

Finally, for a proper closed convex cone $K \subset \mathbb{R}^m$ and a mapping $F: \mathbb{R}^m \rightarrow \mathbb{R}^m$ we define a {\em complementarity problem} CP$(K,F)$ as to find an $x^* \in K$ such that $F(x^*) \in K^*$ and $x^* \perp F(x^*)$. In other words, we seek an $x^*$ such that $(x^*,F(x^*)) \in C(K)$.

\section{Main Results}
The first topic we are interested in is the complementarity problem based on the monotone extended second order cone, which we introduce below.
Let $p$ and $q$ be two nonnegative integers.
{\em The monotone extended second order cone} (informally MESOC) is defined as follows:
 \begin{equation}
L:=\{(x,u)\in\mathbb{R}^p\times\mathbb{R}^q:x_1\geq x_2\geq\cdots\geq x_p\geq\|u\|\}\label{defmesoc}\,.
\end{equation}

\noindent In order to find solutions of a complementarity problem, first we need
to find the dual cone and the complementarity set of this cone. Although
a considerable part of the characterization has been already presented in
\cite{ferreira-gao-nemeth}, for the sake of completeness, we decided to
include it here.
\begin{proposition}\label{pd}
Let $p$ and $q$ be two nonnegative integers.  Then the dual cone of a monotone extended second order cone $L$ defined above is
\begin{equation} \label{dualmesoc}
	M:=\left \{(x,u)\in\mathbb{R}^p\times\mathbb{R}^q:\sum_{i=1}^{j}x_i\geq 0, \forall j\in\{1,\ldots,p-1\}, \sum_{i=1}^{p}x_i\geq\|u\| \right  \},
\end{equation}
that is, $M=L^*$.
\end{proposition}
\noindent {\bf Proof.}\, First, we show that $M\subseteq L^*$. Let $(x,u)\in L$ and $(y,v)\in M$. Using Abel's summation formula, we have
$$
\langle (x,u),(y,v)\rangle=
x^Ty+u^Tv= \sum_{i=1}^{p-1}(x_i-x_{i+1})\sum_{j=1}^{i}y_j+x_p\sum_{i=1}^{p}y_i+u^Tv\geq\|u\|\|v\|+u^Tv\geq 0.
$$
So, we have $M\subseteq L^*$. Now, we show that $L^*\subseteq M$. For, let $(y,v)\in L^*$ and $e=(1,1,\ldots,1) \in \mathbb R^p$. It is obvious that $(\|v\|e,-v) \in L$.
Suppose $v\neq 0$, then
$$ \langle (\|v\|e,-v), (y,v)\rangle \geq 0\Leftrightarrow \|v\|\sum_{i=1}^{p}y_i-\|v\|^2\geq 0.$$
Hence, $\displaystyle \sum_{i=1}^{p}y_i\geq \|v\|.$ When $v=0$, then\, $(e,0)\in L$ and $ (y,0)\in L^*$ imply that
$\displaystyle \sum_{i=1}^{p}y_i\geq 0=\|v\|.$

\vskip 0.1cm
\noindent We also have $\Big ((\underbrace{1,1,\ldots,1,0}_{i<p},\underbrace{0,\ldots,0}_{p-i}),(\underbrace{0,0,\ldots,0}_q)\Big)\in L$, and $(y,v)\in L^*$, which implies that
$$
\sum_{j=1}^{i}y_i\geq 0, \text{ }\forall i\in\{1,2,\ldots,p-1\}.
$$
Thus, $(y,v)\in M$, so $L^*\subseteq M$. Altogether, we have $L^* = M$. \hfill $\Box$

\vskip 0.1cm
\noindent After finding the dual of the monotone extended second order cone, we will describe the complementarity set of this cone. In order to do so, we need to use the inequality, introduced in Lemma \ref{p3} below.

\begin{lemma}\label{p3}
For every $(x,u) \in L$ and $(y,v) \in M$, we have
\begin{equation*}
	\langle x,y \rangle\geq\|u\|\sum_{i=1}^{p}y_i\geq\|u\|\|v\|\,.
\end{equation*}
\end{lemma}

\noindent {\bf Proof.}\, First, we prove that $\langle x,y\rangle\geq\|u\|\sum_{i=1}^{p}y_i$.
Since $(x,u)\in L, (y,v)\in M$, it follows that
$
x_1\geq x_2\geq\cdots\geq x_p\geq\|u\|,
$
$
\sum_{i=1}^{j}y_j\geq 0$ for all $j\in\{1,\ldots,p-1\}$ and  $\sum_{i=1}^{p}y_i\geq\|v\|\geq 0.
$
Thus, by using the backward induction,
\begin{equation*}
\begin{aligned}
\sum_{i=1}^{p}y_i &=y_1+y_2+\ldots+y_p \geq 0\\
&\implies (x_p-\|u\|)\sum_{i=1}^{p-1}y_i + (x_p-\|u\|)y_p\geq 0\\
&\implies (x_{p-1}-\|u\|)\sum_{i=1}^{p-2}y_i + (x_{p-1}-\|u\|)y_{p-1}+(x_p-\|u\|)y_p\geq 0\\
&\implies (x_{p-2}-\|u\|)\sum_{i=1}^{p-3}y_i +(x_{p-2}-\|u\|)y_{p-2}+ (x_{p-1}-\|u\|)y_{p-1}+(x_p-\|u\|)y_p\geq 0\\
&\cdots\\
&\implies (x_1-\|u\|)y_1+(x_2-\|u\|)y_2+\cdots+(x_p-\|u\|)y_p\geq 0\\
&\iff \langle x,y\rangle\geq\|u\|\sum_{i=1}^{p}y_i\,.
\end{aligned}
\end{equation*}
Finally, since $\langle x,y\rangle\geq\|u\|\sum_{i=1}^{p}y_i$ and $\sum_{i=1}^{p}y_i\geq\|v\|$, we have
$$
\langle x,y \rangle\geq\|u\|\sum_{i=1}^{p}y_i\geq\|u\|\|v\|\,.
$$

\vskip -1.1 cm \hfill $\Box$

\vskip 0.2cm
\vskip 0.2cm
\noindent By using Lemma \ref{p3}, we find the complementarity set of the monotone extended second order cone.

\begin{proposition}\label{cl}
Let $(x,y,u,v)\in C(L)$\footnote{By a slight abuse of the notation, we write $(x,u,y,v)$ instead of $((x,u),(y,v))$.}.
If $u\neq 0, v\neq 0$, then
\begin{eqnarray*}
C(L)=\left\{(x,u,y,v): (x,u)\in L,\textrm{ }(y,v)\in M,\textrm{ }\langle x,y\rangle=\|u\|\sum_{i=1}^{p}y_i,\textrm{ }~
\right.\\ \sum_{i=1}^{p}y_i=\|v\|,\textrm{\rm and }\exists\lambda> 0 ~\textrm{\rm such that } v=-\lambda u\Bigg \}\\=\left\{(x,u,y,v): (x,u)\in L, \textrm{ }(y,v)\in M,
\textrm{ }(x_i-x_{i+1})\sum_{j=1}^{i}y_j=0,\textrm{ }\forall i=1,\ldots,p-1, \textrm{ }\right.\\ x_p=\|u\|,\textrm{ }\sum_{i=1}^{p}y_i=\|v\|, \textrm{\rm and }\exists\lambda>0 ~\textrm{\rm such that } v=-\lambda u\Bigg\}\,.
\end{eqnarray*}
\end{proposition}

\noindent {\bf Proof.}\, Let
\begin{eqnarray*}
S:=\left\{(x,u,y,v): (x,u)\in L,\textrm{ }(y,v)\in M,\textrm{ }\langle x,y\rangle=\|u\|\sum_{i=1}^{p}y_i,\textrm{ }\sum_{i=1}^{p}y_i=\|v\|,\textrm { and}\right.\\\exists\lambda>0\textrm{ such that } v=-\lambda u\Bigg \}\,.
\end{eqnarray*}
Now, our task is to show that $C(L)=S$. First, we need to prove that $C(L)\subseteq S$.
For arbitrary $(x,u,y,v)\in C(L)$, by using Lemma \ref{p3}, we have
\begin{equation*}
\begin{aligned}
0 = \langle(x,u),(y,v)\rangle & = \langle x,y\rangle+\langle u,v\rangle\\
                              & \geq \|u\|\sum_{i=1}^{p}y_i + \langle u,v\rangle\\
                              & \geq \|u\|\|v\| + \langle u,v\rangle \geq 0\,.
\end{aligned}
\end{equation*}
Hence, all the inequalities above must be equalities, that is,
\begin{equation*}
\begin{aligned}
0=\langle x,y\rangle+\langle u,v\rangle & = \|u\|\sum_{i=1}^{p}y_i + \langle u,v\rangle \\
& = \|u\|\|v\| + \langle u,v\rangle = 0\,.
\end{aligned}
\end{equation*}
Thus,
\begin{equation}
\langle x,y\rangle=\|u\|\sum_{i=1}^{p}y_i=\|u\|\|v\|\label{e1}\,.
\end{equation}
Therefore,
\begin{equation*}
\|u\|\sum_{i=1}^{p}y_i=\|u\|\|v\
\end{equation*}
and
\begin{equation} \label{e3}
\|u\|\|v\|+\langle u,v\rangle=0\,.
\end{equation}

\noindent From \eqref{e1} we get $\langle x,y\rangle=\|u\|\sum_{i=1}^{p}y_i$ and, subsequently, $\sum_{i=1}^{p}y_i=\|v\|$. From the equality case in the Cauchy-Schwarz inequality, equation (\ref{e3}) implies that $\exists\lambda>0, v=-\lambda u$. Thus, $C(L)\subseteq S$.

\vskip 0.1cm
\noindent Now, for the converse inclusion $S\subseteq C(L)$. We have: $\forall (x,u,y,v)\in S$, $\exists\lambda>0$ such that $v = -\lambda u $, $(x,u)\in L, (y,v)\in M, x^Ty=\|u\|\sum_{i=1}^{p}y_i,\forall i=1,\ldots,p$, and $\sum_{i=1}^{p}y_i=\|v\|$. Thus
$$\langle(x,u),(y,v)\rangle = \langle x,y\rangle +\langle u,v\rangle = \|u\|\|v\| + \langle u,v\rangle = 0\,.$$
Therefore, $(x,u,y,v)\in C(L)$. Hence, $S\subseteq C(L)$.

\vskip 0.2cm
\noindent Finally, we have
\begin{equation}\label{cl1}
\begin{array}{r}
C(L)=S= \Bigg \{(x,u,y,v): (x,u)\in L,\textrm{ }(y,v)\in M,\textrm{ }\langle x,y\rangle=\|u\|\sum_{i=1}^{p}y_i,\textrm{  }\sum_{i=1}^{p}y_i=\|v\|,\\\\ \textrm{ and}~  \exists\lambda>0\textrm{ such that } v=-\lambda u\Bigg \}\,.
\end{array}
\end{equation}

\vskip -.3cm
\noindent Moreover,
$$
\begin{aligned}
\|u\|\sum_{i=1}^{p}y_i & = \langle x,y\rangle\\
& = y_1(x_1-x_2)+(y_1+y_2)(x_2-x_3)+\cdots+(y_1+y_2+\cdots+y_{p-1})(x_{p-1}-x_p)\\&+(y_1+y_2+\cdots+y_p)x_p \\
\end{aligned}
$$
if and only if

\begin{eqnarray*
(\|u\|-x_p)\sum_{i=1}^{p}y_i = y_1(x_1-x_2)+(y_1+y_2)(x_2-x_3)\\+\cdots+(y_1+y_2+\cdots+y_{p-1})(x_{p-1}-x_p)\,.
\end{eqnarray*}

In the equation above, it is obvious that the $\textrm{LHS} \leq 0$ and the $\mbox{RHS} \geq 0$, thus both the LHS and the RHS must be equal to 0. Since the components of the sum in the RHS are all nonnegative, each component must b      e equal to $0$. Hence, from equation \eqref{cl1} it follows that
\begin{eqnarray*}
C(L)=\Bigg\{(x,u,y,v): (x,u)\in L, \textrm{ }(y,v)\in M, \textrm{ }(x_i-x_{i+1})\sum_{j=1}^{i}y_j=0\, \textrm{ }\forall\, i=1,\ldots,p-1, \\x_p=\|u\|, \textrm{ }\sum_{i=1}^{p}y_i=\|v\|, \textrm{ and }\exists\lambda>0\, \textrm{ such that } v=-\lambda u\Bigg\}\,.
\end{eqnarray*}

\vskip -1.1cm \hfill $\Box$

\vskip 0.2cm
\vskip 0.2cm
\begin{lemma}\label{lyap_rank_lem}
	Let $A\in\mathbb R^{p\times p}$. Then, $A\in\LL(\mathbb R^p_{\ge+})$ if and only if it is of the form
	\begin{equation}\label{llmn}
		A=\left[
		\begin{array}{cccccc}
			a-\sum_{i=2}^pa_i & a_{2} & a_{3} & \cdots & \cdots & a_{p}\\
			& a-\sum_{i=3}^pa_i & a_{3} & \cdots & \cdots & a_{p}\\
			&  & a-\sum_{i=4}^pa_i & \cdots & \cdots & a_{p}\\
			&  &  &  \ddots& \vdots & \vdots\\
			& \emph{\textbf{0}} &  &  & a-a_p & a_{p}\\
			& & & & & a
		\end{array}
		\right],
	\end{equation}
	where $a,a_2,a_3\ldots,a_p\in\mathbb R$ are arbitrary.
\end{lemma}
\begin{proof}
	Let $e^i\in\mathbb R^p$, $1\le i\le p$ be the canonical unit vectors in $\mathbb R^p$ and
	$e^{p+1}$ be the zero vector in $\mathbb R^p$. Denote $u^i:=\sum_{k=1}^i e^k\in\mathbb{R}^n_{\ge+}$\, and\,
	$v^i:=e^i-e^{i+1}\in(\mathbb{R}^n_{\ge+})^*$, for\, $1\le i\le p$ (see Example \ref{example2}). Then,
	$\langle u^i,v^j\rangle=\delta_{ij}$, where
	$\delta_{ij}$ is the Kronecker symbol, that is, $\delta_{ii}=1$ and $\delta_{ij}=0$, for
	$i\ne j$. If follows that $(u^i,v^j)\in C(\mathbb{R}^n_{\ge+})$, whenever $i\ne j$ (as it can be seen
	from Example \ref{example2}, too). Hence, if
	$A\in\LL(\mathbb{R}^n_{\ge+})$ and $i\ne j$, then
	\begin{equation}\label{ij}
		\langle Au^i,v^j\rangle=\sum_{k=1}^i (a_{jk}-a_{j+1,k})=0,
	\end{equation}
	where we set $a_{p+1,k}:=0$. By using equation \eqref{ij}, we get
	\begin{equation}\label{ij2}
		\sum_{\ell=j}^p\langle Au^i,v^{\ell}\rangle=\sum_{k=1}^i a_{jk}=0,\quad \mbox{if}\quad
		j>i.
	\end{equation}
	By equation \eqref{ij2} we get
	\begin{equation}\label{ij3}
		a_{ji}=\sum_{k=1}^i a_{jk}-\sum_{k=1}^{i-1} a_{jk}=0, \quad \mbox{if}\quad
		j>i.
	\end{equation}
	By using again equation \eqref{ij}, we get
	\begin{equation}\label{ij4}
		a_{ji}-a_{j+1,i}=\sum_{k=1}^i (a_{jk}-a_{j+1,k})
		-\sum_{k=1}^{i-1} (a_{jk}-a_{j+1,k})=0,\quad \mbox{if}\quad j+1<i.
	\end{equation}
	Equations \eqref{ij}, \eqref{ij3} and \eqref{ij4} imply that $A$ is of the form \eqref{llmn}. Now,
	suppose that $A$ is of the form \eqref{llmn}. From Example \ref{example2}, any element
	$(x,y)\in C(\mathbb{R}^n_{\ge+})$ can be written in the form
	\begin{equation}\label{xy}
		(x,y)=\left(\sum_{i\in I}\alpha_i u^i,\sum_{j\in J}\beta_j v^j\right);\quad
		\alpha_i,\beta_j\ge 0,
	\end{equation}
	for some $I,J\subseteq\{1,2,\dots,n\}$ with $I\cup J=\{1,2,\dots,n\}$ and
	$I\cap J=\varnothing$, because $\{u^i:1\leq i\leq j\}\subseteq\mathbb{R}^n_{\ge+}$ and
	$\{v^i:1\leq i\leq j\}\subseteq(\mathbb{R}^n_{\ge+})^*$ are generators of the simplicial cones
	$\mathbb{R}^n_{\ge+}$ and $(\mathbb{R}^n_{\ge+})^*$, respectively, and $x\perp y$. As
	\(\langle Au^i,v^j\rangle=0,\) by considering the derivation of equations \eqref{ij},
	\eqref{ij2}, \eqref{ij3} and \eqref{ij4} above in the reverse order, equation \eqref{xy}
	implies that $\langle Ax,y\rangle=0$. Hence, $A\in\LL(\mathbb{R}^n_{\ge+})$.
\end{proof}
\begin{theorem
For the monotone extended second
order cone \eqref{defmesoc}, any Lyapunov like transformation $T$ is of the form
{\small
\begin{equation}\label{ee4}
T=\left[
\begin{array}{cccccc|ccc}
a-\sum_{j=2}^pa_{j} & a_{2} & a_{3} & \cdots & \cdots & a_{p} & c_1 & \cdots & c_q\\
 & a-\sum_{j=3}^pa_{j} & a_{3} & \cdots & \cdots & a_{p} & c_1 & \cdots & c_q \\
 &  & a-\sum_{j=4}^pa_{j} & \cdots & \cdots & a_{p} & c_1 & \cdots & c_q \\
 &  &  &  \ddots& \vdots & \vdots & \vdots & & \vdots\\
 & \emph{\textbf{0}} &  &  & a-a_{p} & a_{p} & c_1 & \cdots & c_q \\
 & & & & & a & c_1 & \cdots & c_q
\\ \hline
 &  &  & &  & c_1 & a & & *\\
 & \emph{\textbf{0}} &  & &  & \vdots & & \ddots & \\
 &  &  & &  & c_q & -* & & a\\
\end{array}
\right],
\end{equation}
}
where $a,a_2,a_3,\ldots,a_p,c_1,...,c_q\in\mathbb R$ are arbitrary.
Hence, its
Lyapunov rank is given by
\begin{equation*
\beta(L)=p+\frac{q(q+1)}2\,.
\end{equation*}
\end{theorem}

\noindent {\bf Proof.}\, Recall that the complementarity set for the monotone extended second order cone $L$ is
$$C(L)=\{((x,u),(y,v)) \in L\times M: (x,u)\perp (y,v)\}.$$
We partition the above set in the following way:
$$C(L):=C_1(L)\cup C_2(L)\cup C_3(L)\cup C_4(L),$$
where
\begin{equation*}
\begin{aligned}
	& C_1(L):=\{(x,0,y,0)\in C(L)\},\\
	& C_2(L):=\{(x,0,y,v)\in C(L):v\neq 0\},\\
	& C_3(L):=\{(x,u,y,v)\in C(L):u\neq 0\neq v\},\\
	& C_4(L):=\{(x,u,y,0)\in C(L):u\neq 0\}.
\end{aligned}
\end{equation*}

\noindent Since $x=0 \Rightarrow u=0$ and $y=0 \Rightarrow v=0$, for any Lyapunov-like transformation on
$L$ we only need to consider the case of\, $x\neq 0 \neq y$.  Let $T$ be any element of $\LL(L)$, so it has the following block form:

$$
\left[
\begin{array}{cc}
   A & B  \\
   C & D
  \end{array}
  \right]
  : \mathbb R^p\times\mathbb R^q\rightarrow\mathbb R^p\times\mathbb R^q,
$$
where $A\in\mathbb R^{p\times p},B\in\mathbb R^{p\times q}, C\in\mathbb R^{q\times p}$, and $D\in\mathbb R^{q\times q}$. Take any $(x,u,y,v) \in C(L)$. Then (\ref{lyap-like}) implies
$$
\langle Ax,y\rangle + \langle Bu,y\rangle + \langle Cx,v\rangle + \langle Du,v\rangle =0,
$$
$$
\langle Ax,y\rangle - \langle Bu,y\rangle - \langle Cx,v\rangle + \langle Du,v\rangle =0,
$$
where the latter equation comes from the former one by substituting $-u$ for $u$ and $-v$ for $v$. By adding and subtracting the above equations, we get
\begin{equation}
\begin{aligned}
\langle Ax,y\rangle + \langle Du,v\rangle=0,\\
\langle Bu,y\rangle + \langle Cx,v\rangle=0.\label{ee}
\end{aligned}
\end{equation}

\noindent By using an element $(x,0,y,0)\in L\times M$ in $C_1(L)$, with $x\in\mathbb R_{\geq+}^p$ and
$y\in(\mathbb R_{\geq+}^p)^*$, we get $\langle Ax,y\rangle = 0$, which implies that $A\in\LL(\mathbb R_{\geq+}^p)$.

Now, we will determine the structures of matrices $B$ and $C$. By using elements in $C_2(L)$, from the second equation in (\ref{ee}), we get
\begin{equation*
	\langle Cx,v\rangle = \langle B0,y\rangle + \langle Cx,v\rangle = 0.
\end{equation*}
Suppose that $Ca^i\neq 0$ for some $i<p$ and let $v:=\frac{Ca^i}{\|Ca^i\|}$, and $y:=e^j,(j>i)$, thus, $\langle y, e^j\rangle = 1 = \|v\|$. Hence, $(a^i,0,e^j,v)\in C_2(L)$. Then $0=\langle Cx,v\rangle=\langle Ca^i,v\rangle =\|Ca^i\|$, which leads to a contradiction. Hence, $Ca^i = 0$.
Then, for certain $c_1,\ldots,c_q\in\mathbb R$ we have
$$
C=\left[\begin{array}{c|c}\textbf{0}&\begin{array}{ccc}c_1\\ \vdots \\c_q\end{array}\end{array}\right]_{q\times p}.
$$
If $C=0$, the second equation in (\ref{ee}) demonstrates that $\langle Bu,y \rangle =0$ for all $(x,u,y,v)\in C_3(L)$. It is easy to verify that $(e,-v,e^i,v)\in C_3(L)$, where $v$ is an arbitrary unit vector in $\mathbb R^q$. Hence, $\langle B(-v),e^i\rangle = 0$, for all $1 \leq i \leq p$, thus $Bv=0$. In consequence, $B=0$.

If $C \neq 0$, first we need to find the structure of matrix $B$. We have $\langle Bu,y\rangle=0$ for any $(x,u,y,0)\in C_4(L)$.

Let $u^i$ denote the standard (canonical) unit vector in $\mathbb R^q$ and for any $n>m$, let\\
$y^{m,n}:=e^m-e^n \in\mathbb R^p$. Since $\left(e,u^i,y^{m,n},0\right)\in C_4(L)$,
$$
\langle Bu^i,y^{m,n}\rangle=0.
$$
Therefore,
$$
B=\begin{bmatrix}
b_1 & b_2 & \cdots &b_q \\
\vdots & \vdots &  & \vdots\\
 b_1 & b_2 & \cdots &b_q
\end{bmatrix}_{p\times q}\,.
$$
For $i=1,\ldots,q$ and $j=1,\ldots,p$, we have $(e,u^i,e^j,-u^i) \in C_3(L)$ and subsequently,
$$\langle Bu^i,e^j\rangle+\langle Ce,-u^i\rangle=0.$$
It readily implies $b_i=c_i$. Hence,
$$
B=\begin{bmatrix}
c_1 & c_2 & \cdots &c_q \\
\vdots & \vdots &  & \vdots\\
 c_1 & c_2 & \cdots &c_q
\end{bmatrix}_{p\times q}\,.
$$

\vskip 0.1cm
As $(e,u,\frac{1}{p}e, -u) \in C_3(L)$ for all $u$ with $\|u\|=1$, by using (\ref{ee}), we have
\begin{equation} \label{ee2}
\left\langle Ae,\frac{1}{p}e\right\rangle+\langle Du,-u\rangle =0.
\end{equation}

Let $\displaystyle a:=\frac{\langle Ae,e\rangle}{p}$. Then (\ref{ee2}) implies
$$\left\langle\left(\frac{D+D^T}{2}-aI\right)u,u\right\rangle=0,
$$
and hence
\begin{equation}
D+D^T=2aI.\label{ee3}
\end{equation}

\noindent Obviously, $(e,-u^1,e^1,u^1) \in C(L)$ and using the first equation in (\ref{ee}) gives
$$
\langle Ae,e^1\rangle -\langle Du^1,u^1\rangle=0,
$$
which implies that $d_{11}=\sum_{j}a_{1j}$. Thus, (\ref{ee3}) implies that $d_{11}=a$ and hence, $\sum_{j=1}^pa_{1j}=a$.

\noindent By changing $e^1$ to $e^2$ (yes, we can), we have $\sum_{j=2}^pa_{2j}=d_{22}=a$. By following this process, we obtain that $d_{ii}=\sum_{j=1}^pa_{ij}=a$, for all $1\leq i \leq p$.

Therefore, by equation (\ref{ee3}), $A\in\LL(\mathbb{R}^n_{\ge+})$ (shown above) and Lemma \ref{lyap_rank_lem},
any Lyapunov-like transformation on $L$ has the form \eqref{ee4}.

\vskip 0.1cm
Now, we want to show that any transformation $T$, which can be represented in the form (\ref{ee4}), is Lyapunov-like on $L$, so let $T$ be given as above. Then we have

\begin{equation}\label{ellt}
\langle T(x,u),(y,v)\rangle = \langle Ax,y\rangle+\langle Du,v\rangle+\langle Bu,y\rangle + \langle Cx,v\rangle.
\end{equation}

We wish to show that for any $(x,u,y,v) \in C(L)$, the RHS in the above equation is zero.
We will perform a case-by-case analysis.

\vskip 0.2cm
\noindent {\em Case 1.}\,
For any $(x,u,y,v):=(x,0,y,0)\in C_1(L)$, the RHS of \eqref{ellt} is equal to zero, as
$(x,y)\in\C\left(\mathbb R^n_{\ge+}\right)$ and we have already shown that $A\in\LL(\mathbb R^n_{\ge+})$, hence
it is enough to use again Lemma \ref{lyap_rank_lem}.

\vskip 0.2cm
\noindent {\em Case 2.}\, For any
$(x,u,y,v):=(x,0,y,v)\in C_2(L)$, the RHS of \eqref{ellt} is $(c_1v_1+...+c_qv_q)x_p$. Suppose
that $x_p\ne 0$. Then, since $(x,y)\in C\left(\mathbb R^n_{\ge+}\right)$, from Example \ref{example2} we
get
$y_1+\dots+y_p=0$. Hence, $(y,v)\in M$ and \eqref{dualmesoc} implies $v=0$, which contradicts
$(x,0,y,v)\in C_2(L)$. Thus, $x_p=0$ and therefore the RHS of \eqref{ellt} is zero.

\vskip 0.2cm
\noindent {\em Case 3.}\,
Take an arbitrary
$(x,u,y,v) \in C_3(L)$. Proposition \ref{cl} indicates that for some $\lambda >0$ one has $v=-\lambda u$, thus
\begin{equation}\label{ly}
\begin{aligned}
\langle Ax,y\rangle +\langle Du,v\rangle & = \langle Ax,y\rangle +\left\langle \frac{D+D^T}{2} u,v\right\rangle=\langle Ax,y\rangle + a\langle u,v\rangle\\
&=\langle z,y\rangle+a\langle u,v\rangle=\sum_{i=1}^{p-1}\left[(z_i-z_{i+1})\sum_{j=1}^iy^j\right]+z_p\sum_{i=1}^{p}y^i+a\langle u,v\rangle,
\end{aligned}
\end{equation}
where $z_i:=\sum_{j=1}^ia_jx_i+\sum_{k=i+1}^pa_kx_k$, for any $1\le i\le p-1$ and $z_p=\sum_{k=1}^pa_kx_p$. Then for any $1\le i\le p-1$, it is easy check that $z_i-z_{i+1}=\sum_{j=1}^ia_j(x_i-x_{i+1})$. By inserting these equalities and the formula for $z_p$ into equation \eqref{ly}, and by using Proposition \ref{cl}, we obtain $\langle Ax,y\rangle +\langle Du,v\rangle =0$.
We will show that
$$
\langle Bu,y\rangle + \langle Cx,v\rangle=0.
$$
By using the full power of Proposition \ref{cl}, including $v=-\lambda u$ for some $\lambda>0$, we have
$$
\langle Bu,y\rangle = \sum_{i=1}^q(c_iu_i)\cdot\sum_{i=1}^py_i=\|v\|\sum_{i=1}^q(c_iu_i)
$$
and
$$
\langle Cx,v\rangle = x_p\sum_{i=1}^q(c_iv_i)=\|u\|\sum_{i=1}^q(c_iv_i).
$$
Then
$$
\langle Bu,y\rangle + \langle Cx,v\rangle = \|v\|\sum_{i=1}^q(c_iu_i)+ \|u\|\sum_{i=1}^q(c_iv_i)
=\lambda\|u\|\sum_{i=1}^q(c_iu_i)-\lambda\|u\| \sum_{i=1}^q(c_iu_i)=0.
$$

\vskip 0.2cm
\noindent {\em Case 4.}\,
For any
$(x,y,u,v):=(x,u,0,v)\in C_4(L)$, the RHS of \eqref{ellt} is $(c_1u_1+...+c_qu_q)(y_1+...+y_q)$.
Suppose that $y_1+\dots+y_p\ne 0$. Then, since $(x,y)\in C(\mathbb R^n_{\ge+})$, from Example
\ref{example2} we get $x_p=0$.  Hence, $(x,u)\in M$ and \eqref{dualmesoc} implies $u=0$, which
contradicts $(x,u,y,0)\in C_4(L)$. Thus, $y_1+\dots+y_p=0$ and therefore the RHS of
\eqref{ellt} is zero.

In conclusion, the RHS of \eqref{ellt} is zero for any $(x,u,y,v)\in C_1(L)\cup C_2(L)\cup
C_3(L)\cup C_4(L)=C(L)$
Therefore, $T\in\LL(L)$. Following the definition of the Lyapunov rank, its value for the cone $L$ equals to the number of independent parameters in (\ref{ee4}), which is\, $p+\frac{q(q+1)}{2}$. \hfill $\Box$

\vskip 0.2cm
After calculating the Lyapunov rank of MESOC, we prove our second main result, namely that this cone is reducible. Recall that a cone $K$ in $\mathbb{R}^m$ is {\em reducible} if it can be expressed as a sum $K=K_1+K_2$, where $K_1, K_2 \neq \{0\}$ are cones with ${\rm span}(K_1) \cap {\rm span}(K_2)= \{0\}$. Otherwise, it is called {\em irreducible}.

\begin{theorem
For the monotone extended second order cone $L$ defined in (\ref{defmesoc}) one has $L=L_1+L_2$, where
$$L_1:={\rm cone}\,\Big\{ (\underbrace{1,\ldots,1}_p, m_1, \ldots, m_q): m_1^2+ \cdots+m_q^2 \leq 1 \Big\}~~{\rm and}$$
$$L_2:={\rm cone}\,\Big \{(\underbrace{1,0,\ldots,0}_p,\underbrace{0, \ldots,0}_q), (\underbrace{1,1,\ldots,0}_p,\underbrace{0, \ldots,0}_q), \ldots, (\underbrace{1,1,\ldots,1,0}_p,\underbrace{0, \ldots,0}_q) \Big\}.$$
As a by-product, we show that $L$ is a reducible cone.
\end{theorem}

\noindent {\bf Proof.}\, First, we show the inclusion $L\subseteq L_1+L_2$.

\noindent An arbitrary element $(x_1, \ldots, x_p, u_1, \ldots, u_q) \in L$, by the definition of $L$, can be represented as $(\sum_{i=1}^p a_i, \ldots, a_1+a_2, a_1, u_1, \ldots, u_q)$, where $a_i \geq 0$ for $i=2,\ldots,p$ and $a_1 \geq \|(u_1, \ldots,u_q)\|$. Hence,

\begin{align*}
 & (x_1, \ldots, x_p, u_1, \ldots, u_q)\\
& = \left(\sum_{i=1}^p a_i, \sum_{i=1}^{p-1} a_i, \ldots, a_1, u_1, \ldots, u_q\right) \\
& = (a_1, \ldots, a_1,u_1, \ldots, u_q) + (\underbrace{a_2, \ldots, a_2,0}_p,\underbrace{0, \ldots, 0}_q) + \cdots + (\underbrace{a_p,0, \ldots,0}_p, \underbrace{0, \ldots, 0}_q)\\
& = (a_1, \ldots, a_1,u_1, \ldots, u_q)+ a_2(\underbrace{1, \ldots, 1,0}_p,\underbrace{0, \ldots, 0}_q)+ \cdots + a_p(\underbrace{1,0, \ldots,0}_p, \underbrace{0, \ldots, 0}_q).
\end{align*}
Obviously, $a_2(\underbrace{1, \ldots, 1,0}_p,\underbrace{0, \ldots, 0}_q)+ \cdots + a_p(\underbrace{1,0, \ldots,0}_p, \underbrace{0, \ldots, 0}_q) \in L_2$.

\vspace{.2cm}
\noindent Now, we show that $(a_1, \ldots, a_1,u_1, \ldots, u_q) \in L_1$. It is trivial when $a_1=0$, so we assume that $a_1>0$. Thus, we have
$\displaystyle (a_1, \ldots, a_1,u_1, \ldots, u_q)=a_1\left (1, \ldots, 1,\frac{u_1}{a_1}, \ldots, \frac{u_q}{a_1}\right )$. As $a_1 \geq \|(u_1, \ldots,u_q)\|$, we get
$$a_1\geq \sqrt{u_1^2+ \cdots + u_p^2}\, \equiv\, 1 \geq \sqrt{\left ( \frac{u_1}{a_1}\right )^2 + \cdots + \left (\frac{u_q}{a_1}\right)^2  }, $$
which, by the definition of $L_1$, gives that $(a_1, \ldots, a_1,u_1, \ldots, u_q) \in L_1$.

\noindent Hence, we showed that an arbitrary element $(x_1, \ldots, x_p, u_1, \ldots, u_q) \in L$ can be represented as the sum of two elements\\
\centerline{$(a_1, \ldots, a_1,u_1, \ldots, u_q) \in L_1~ {\rm and}~ a_2(\underbrace{1, \ldots, 1,0}_p,\underbrace{0, \ldots, 0}_q)+ \cdots + a_p(\underbrace{1,0, \ldots,0}_p, \underbrace{0, \ldots, 0}_q) \in L_2\,.$}

\vskip 0.2cm
Now, for the inclusion $L_1+L_2 \subseteq L$. Observe that $L_1 \subseteq L$ and $L_2 \subseteq L$. From the convexity of the cone $L$, it follows that $L_1+l_2 \subseteq L+L=L$.

\vspace{.1cm}
\noindent

It concludes the proof of the equality $L=L_1+L_2$. Obviously, the cones $L_1, L_2 \neq \{0\}$ and ${\rm span}(L_1) \cap {\rm span}(L_2)= \{0\}$.     \hfill $\Box$

\vskip 0.2cm
The following three results will help us proving Theorem \ref{isotonicity-on-cylinder}, where a characterization of $K\subseteq\mathbb R^p\times\mathbb R^q$ as an L-isotone projection set is given.

\begin{theorem}[see \cite{nemeth-nemeth-2012}]\label{fooo}
	The closed convex set $C\subset\mathbb R^m$ with nonempty interior is a $K$-isotone projection set
	if and only if it is of the form
	\begin{equation*}
		C=\bigcap_{i\in \mathbb N} \mathcal H_-(u^i,a^i),
	\end{equation*}
	where each affine hyperplane $\mathcal H(u^i,a^i)$ is tangent to $C$ and it is a $K$-isotone projection set.
\end{theorem}

\noindent The following two lemmas are from \cite{nemeth-zhang1}.

\begin{lemma}\label{ti}
	Let $K\subset\mathbb R^m$ be a closed convex cone and $\mathcal H\subset\mathbb R^m$ be a hyperplane
	with a unit normal vector $a\in\mathbb R^m$. Then, $\mathcal H$ is a $K$-isotone projection set
	if and only if \[\langle x,y\rangle\ge\langle a,x\rangle\langle a,y\rangle,\] for any $x\in K$ and
	$y\in K^*$.
\end{lemma}

\begin{lemma}\label{leasy}
	Let $z\in\mathbb R^m$, $K\subset\mathbb R^m$ be a closed convex cone and $C\subset\mathbb R^m$ be a nonempty closed convex set. Then, $C$ is a $K$-isotone projection set if and only
	if $C+z$ is a $K$-isotone projection set.
\end{lemma}

\vskip 0.1cm
Finally, by using the above three results, we derive an isotonicity property of MESOC, which we will use to solve complementarity problems on the MESOC.
\begin{theorem}\label{isotonicity-on-cylinder}
Let $L$ be the MESOC corresponding to the dimensions $p$ and $q$, with  $q>1$. The closed convex set with nonempty interior $K\subseteq\mathbb R^p\times\mathbb R^q$ is an L-isotone projection set if and only if $K=\mathbb R^p\times C$, for some closed convex set with nonempty interior $C\subseteq\mathbb R^q$.
\end{theorem}

\noindent {\bf Proof.}\, First, suppose that $K=\mathbb R^p\times C$, where $C\subseteq\mathbb R^q$ is a nonempty closed convex set with nonempty interior. Let $(x,u),(y,v)\in\mathbb R^p\times\mathbb R^q$ be such that $(x,u)\leq_{L}(y,v)$, thus $(y-x,v-u)\in L$, i.e.,
\begin{equation}\label{defcon}
y_1-x_1\geq y_2-x_2\geq\cdots\geq y_p-x_p\geq\|v-u\|.
\end{equation}
Since $C$ is a closed and convex set in $\mathbb R^q$, by the nonexpansivity (\ref{nonexpansivity}) of
$P_C$, we have
\begin{equation*
\|v-u\|\geq\|P_Cv-P_Cu\|,
\end{equation*}
which together with (\ref{defcon}) yields
$$
y_1-x_1\geq y_2-x_2\geq\cdots\geq y_p-x_p\geq\|P_Cv-P_Cu\|.
$$
Thus,
$$
(y,P_Cv)-(x,P_Cu)\in L
$$
and therefore we have
$$
P_K(x,u)=(x,P_Cu)\leq_{L}(y,P_Cv)=P_K(y,v).
$$
In conclusion, $K$ is an $L$-isotone project set.

Conversely, suppose that the closed convex set $K\subseteq\mathbb R^p\times\mathbb R^q$ with nonempty interior  is an $L$-isotone project set. If $p=1$, then in \cite{nemeth-nemeth-2012} it has been proved that $K=\mathbb R^p\times C$, where $C$ is a nonempty, closed and convex subset with nonempty interior of $\mathbb R^q$. Therefore, assume that $p>1$. By Theorem \ref{fooo} and Lemma \ref{leasy}, we need to show that for any tangent hyperplane $\mathcal H$ of $K$ with unit normal $\gamma=(a,u)$, we have $a=0$. From Lemma \ref{ti}, we have
\begin{equation}\label{eau}
	\langle\zeta,\xi\rangle\ge\langle\gamma,\zeta\rangle\langle\gamma,\xi\rangle,
\end{equation}
for any $\zeta:=(x,v)\in L$ and $\xi:=(y,w)\in L^*$.
	By Lemma \ref{ti}, condition \eqref{eau} holds.
 	Let $x\in\mathbb R^p_+$ and $v\in\mathbb R^q$. Then, by equation \eqref{defmesoc}, and Proposition
	\ref{pd}, it is easy to check that $\zeta:=(\|v\|e,v)\in L$, $\xi:=(\|v\|x,-\langle e,x\rangle v)\in L^*$ and $\langle\zeta,\xi\rangle=0$. Hence,
	condition \eqref{eau} implies
	\begin{equation}\label{ea}
	    0\ge(\langle a,e\rangle\|v\|+\langle u,v\rangle)(\langle a,x\rangle\|v\|-\langle e,x\rangle\langle u,v\rangle).
	\end{equation}
	If in \eqref{ea} $x=e$ and we choose $v\neq 0$ such that $\langle u,v\rangle=0$, then $0\ge\langle a,e\rangle^2\|v\|^2$, and hence $\langle a,e\rangle=0$.
	Thus, \eqref{ea} becomes
	\begin{equation}\label{ea2}
	    0\ge\langle u,v\rangle(\langle a,x\rangle\|v\|-\langle e,x\rangle\langle u,v\rangle).
	\end{equation}
	First, suppose that $u\ne0$. Let $v^n\in\mathbb R^q$ be a sequence of points such that $\|v^n\|=1$, $\langle u,v^n\rangle>0$ and
	$\lim_{n\to+\infty}\langle u,v^n\rangle=0$. Let $n$ be an arbitrary positive integer.
	If in \eqref{ea2} we choose $\lambda>0$ sufficiently large such that $x:=a+\lambda e\ge0$ and $v=v^n$,
	we get \(0\ge\langle u,v^n\rangle(\|a\|^2-\lambda p\langle u,v^n\rangle),\) or equivalently $\|a\|^2\le\lambda p\langle u,v^n\rangle$. By letting
	$n\to+\infty$ in the last inequality, we obtain $\|a\|^2\le0$, or equivalently $a=0$.

    Next, suppose that $u=0$. Since $(a,u)$ is a unit vector, it follows that $a\ne 0$. Let $(x,y)\in C(\mathbb R^p_{\ge +})$ and $w\in\mathbb R^q$ such that
	$\langle y,e\rangle\ge\|w\|$. Then, by \eqref{defmesoc} and Proposition \ref{pd}, it is easy to check that
	$\zeta:=(x,0)\in L$, $\xi:=(y,w)\in L^*$ and $\langle\zeta,\xi\rangle=0$. Hence, inequality \eqref{eau} implies
	\begin{equation*
	    0\ge\langle a,x\rangle\langle a,y\rangle,
	\end{equation*}
	for any $(x,y)\in C(\mathbb R^p_{\ge +})$ with $\langle x,y\rangle=0$. From Example \ref{example2}, we can choose
	$x=e^1+\cdots+e^r$ and $y=e^s-e^{s+1}$, where $r, s\in\{1,\ldots,p\}$, and we set $e^{p+1}:=0$. Hence, $(a_1+\cdots +a_r)(a_s-a_{s+1})\le0$, where we set $a_{p+1}:=0$. Take now $r=1$ and for $s=1, \ldots,p$, add the inequalities $a_1(a_1-a_2)\leq 0, \ldots, a_1(a_p-a_{p+1})\leq 0$, to obtain (by the telescoping effect) $a_1\cdot a_1 \leq 0$, which gives $a_1=0$. Similarly, for $r=2$ and $s=2, \ldots,p$, add the inequalities $(0+a_2)(a_2-a_3) \leq 0, \ldots, a_2(a_p-a_{p+1}) \leq 0$, to get $a_2=0$. Acting similarly (with $r=3$, and so on), we get $a_3=0$, up to $a_p=0$. Thus, $a=0$. But this contradicts $a\ne 0$, so the case $u=0$ cannot hold. \hfill $\Box$

\vskip 0.1cm
It is a well-known that for the nonlinear complementarity problem NCP$(F,K)$, $x^*$ is its solution if and only if $x^*$ is a fixed point of the mapping $K\ni x\mapsto P_K(x-F(x))$. For an arbitrary sequence $\{x^n\}$ generated by the fixed point iteration process
\begin{equation}\label{picard iteration}
x^{n+1}=P_{K}(x^n-F(x^n)),
\end{equation}
if the mapping $F$ is continuous and the sequence $\{x^n\}$ is convergent to $x^*\in K$, then $x^*$ is a fixed point of the mapping $K\ni x\mapsto P_K(x-F(x))$, hence $x^*$ is a solution of the nonlinear complementarity problem NCP$(F,K)$.

\section{Mixed Complementarity Problem}
Facchinei and Pang defined the mixed complementarity problem (${\rm MiCP}$) on the nonnegative orthant (see Subsection 9.4.2 in \cite{facchinei-pang}). It is not only equivalent to a linearly constrained {\em variational inequality problem} (this relationship is also known as the {\em Karush-Kuhn-Tucker (KKT)} system of the variational inequality), but it can also be viewed as an {\rm NCP} for a particular non-pointed cone. N\'emeth and Zhang considered the MiCP defined on an arbitrary closed and convex cone. In Proposition \ref{isotonicity-on-cylinder}, we have already shown that the projection mapping onto a cylinder is an isotonic projection set with respect to MESOC. It is interesting to consider using isotonicity on MESOC as a tool to solve the MiCP.

\begin{lemma} (Lemma 4 in \cite{nemeth-zhang2}) \label{guohan1}
Let $K=\mathbb R^p\times C$, where $C$ is an arbitrary nonempty closed and convex cone in $\mathbb R^q$. Denote mapping $G: \mathbb R^p\times\mathbb R^q\mapsto\mathbb R^p$, mapping $H: \mathbb R^p\times\mathbb R^q\mapsto\mathbb R^q$ and mapping $F=(G;H): \mathbb R^p\times\mathbb R^q\mapsto\mathbb R^p\times\mathbb R^q$. Then the nonlinear complementarity problem {\rm NCP}$(F,K)$ is equivalent to the mixed complementarity problem {\rm MiCP}$(G,H,C,p,q)$ defined as
$$
G(x,u)=0,\,\,\,C\ni u\perp H(x,u)\in C^*.
$$
\end{lemma}

\noindent {\bf Proof.}\, It is standard and follows from the definition of the nonlinear complementarity problem NCP$(F,K)$, by noting that $K^* = \{0\}\times C^*$.  \hfill $\Box$

\vskip 0.2cm
By using the notations of Lemma \ref{guohan1}, the fixed point iteration (\ref{picard iteration}) can be rewritten as:
\begin{equation}\label{picard iteration 2}
\begin{cases}
x^{n+1}=x^n-G(x^n,u^n),  \\
u^{n+1}=P_C(u^n-H(x^n,u^n)).
\end{cases}
\end{equation}

\begin{proposition} (Proposition 2 in \cite{nemeth-zhang2}) \label{guohan2}
Let $L\subseteq \mathbb R^m$ be a pointed closed convex cone, $K \subseteq \mathbb R^m$ be a closed convex cone and $F: \mathbb R^m \rightarrow \mathbb R^m$ be a continuous mapping. Consider the sequence $\{x^n\}_{n\in\mathbb N}$ defined by (\ref{picard iteration}). Suppose that the mappings $P_K$ and $I - F$ are $L-isotone$ and $x^0 = 0 \leq_L x^1$. Let
$$
\Omega:=K\cap L\cap F^{-1}(L)=\{x\in K\cap L: F(x)\in L\}
$$
and
$$
\Gamma:=\{x\in K\cap L:P_K(x-F(x))\leq_L x\}.
$$
Then $\varnothing\ne\Omega\subset\Gamma$ and the sequence $\{x^n\}$ is convergent to $x^*$, which is a solution of {\rm NCP}$(F,K)$. Moreover, $x^*$ is a lower L-bound of $\Omega$ and the L-least element of $\Gamma$.
\end{proposition}

The following theorem provides sufficient conditions for the solvability of
the mixed complementarity problem \small MiCP$(G,H,C,p,q)$.
\begin{theorem}\label{MainT}
Let $L$ be the monotone extended second order cone corresponding to $p$ and $q$. For an arbitrary cone $K=\mathbb R^p\times C$, where $C$ be a closed convex cone, denote its dual cone by $K^*$. Let $F=(G;H): \mathbb R^p\times\mathbb R^q\mapsto\mathbb R^p\times\mathbb R^q$, such that $I-F$ is $L$-isotone, where I denotes the identical mapping, $G: \mathbb R^p\times\mathbb R^q\mapsto\mathbb R^p$ and $H: \mathbb R^p\times\mathbb R^q\mapsto\mathbb R^q$ are two continuous mappings. Consider a sequence $\{(x^n,u^n\}_{n\in\mathbb N}$ defined by $(\ref{picard iteration 2})$, where $x^0=0\in\mathbb R^p$ and $u^0=0\in\mathbb R^q$. Let $x,y\in\mathbb R^p$ and $u,v\in\mathbb R^q$. Suppose that the system of inequalities
\begin{equation}\label{si1}
y_i-x_i\geq y_{i+1}-x_{i+1}\geq\|v-u\|;\quad 1\le i\le p-1
\end{equation}
implies the system of inequalities
\begin{equation}\label{si2}
\begin{array}{rcl}
y_i-x_i-(G(y,v)_i-G(x,u)_i)&\geq& y_{i+1}-x_{i+1}-(G(y,v)_{i+1}-G(x,u)_{i+1})\\
&\geq&\|v-u-(H(y,v)-H(x,u))\|;
\end{array}
\end{equation}
$1\le i\le p-1$, and that $x^1_i\geq x^1_{i+1}\geq \|u^1\|$; $1\le i\le p-1$  (in particular, this holds when $-G(0,0)_i\geq-G(0,0)_{i+1}\geq\|H(0,0)\|$; $1\le i\le p-1$).
Let
$$
\Omega:=\{(x,u)\in\mathbb R^p\times C: x_1\geq\cdots\geq x_p\geq\|u\|,G(x,u)_1\geq \cdots \geq G(x,u)_p \geq\|H(x,u)\|\}
$$
and
\begin{eqnarray*}
\Gamma:=\{(x,u)\in\mathbb R^p\times C:x_1\geq\cdots\geq x_p\geq\|u\|,G(x,u)_1 \geq\cdots\geq G(x,u)_p\\\geq\|u-P_C(u-H(x,u))\|\}.
\end{eqnarray*}
Then $\varnothing\ne\Omega\subseteq\Gamma$, the sequence $\{(x^n,u^n)\}$ is convergent, and its limit $(x^*,u^*)$ is a solution of {\rm MiCP}$(G,H,C,p,q)$. Moreover, $(x^*,y^*)$ is a lower L-bound of $\Omega$ and the L-least element of $\Gamma$.
\end{theorem}

\noindent {\bf Proof.}\, Following the definition of the monotone extended second order cone, we have
$$
\Omega=K\cap L\cap F^{-1}(L)=\{z\in K\cap L: F(z)\in L\}
$$
and
$$
\Gamma=\{z\in K\cap L:P_K(z-F(z))\leq_Lz\},
$$
where \(z=(x,u)\).
Theorem \ref{isotonicity-on-cylinder} implies that $P_K$ is $L$-isotone. Since \eqref{si1}$\implies$\eqref{si2}, $I-F$ is $L$-isotone.
Meanwhile, $x_1^1\geq x_2^1\geq\cdots\geq x_p^1\geq\|u^1\|$ implies that $(x^0,y^0)=(0,0)\leq_L(x^1,y^1)$.
Then, by using Proposition \ref{guohan2}, we have that $\varnothing\ne\Omega\subset\Gamma$, the sequence $\{(x^n,u^n)\}$ is convergent, and its limit $(x^*,u^*)$ is a solution of MiCP$(G,H,C,p,q)$. Moreover, $(x^*,y^*)$ is a lower $L$-bound of $\Omega$ and the $L$-least element of $\Gamma$. \hfill $\Box$

\section{Numerical Example}
Let $L$ be the monotone extended second order cone, then suppose that $K=\mathbb R^2\times C$ where $C=\{(u_1,u_2)\in\mathbb R^2: u_1\geq u_2\geq 0\}$. Let $f_1(x,u)= \frac{1}{10}x_1-\frac{1}{20}x_2 +\frac{1}{20}\|u\|+1$ and $f_2(x,u)=\frac{1}{5}x_1-\frac{3}{20}x_2+\frac{1}{20}\|u\|-\frac{3}{5}$. Obviously, $f_1(x,u)$ and $f_2(x,u)$ are $L$-monotone. Define $\omega^1:=(2,1, \frac{1}{3},\frac{1}{6})$ and $\omega^2:=(2,1,\frac{1}{6},\frac{1}{3})$; it is easy to find out that\, $\omega^1, \omega^2 \in L$. Then, for two arbitrary vectors $(x,u), (y,v) \in \mathbb R^2\times\mathbb R^2$ such that $(x,u)\leq_L (y,v)$, by using the definition of the MESOC, we have that $y_1-x_1\geq y_2-x_2\geq\|v-u\|\geq\|u\|-\|v\|$. Hence,
$$
f_1(y,v)-f_1(x,u)=\frac{1}{10}(y_1-x_1)-\frac{1}{20}(y_2-x_2)-\frac{1}{20}(\|u\|-\|v\|)\geq 0,
$$
$$
f_2(y,v)-f_2(x,u)=\frac{1}{5}(y_1-x_1)-\frac{3}{20}(x_2-y_2)-\frac{1}{20}(\|u\|-\|v\|)\geq 0.
$$
Since $\omega^1, \omega^2, (y,v)-(x,u) \in L$, by using the convexity of $L$, if we have $(x,u)\leq_L(y,v)$, then
$$
(f_1(y,v)-f_1(x,u))\omega^1 + (f_2(y,v)-f_2(x,u))\omega^2\in L,
$$
which is equivalent to the following inequality:
$$
f_1(x,u)\omega^1+f_2(x,u)\omega^2\leq_L f_1(y,v)\omega^1+f_2(y,v)\omega^2.
$$
Thus, the mapping $f_1\omega^1+f_2\omega^2$ is $L$-isotone.
Now, we define functions $G$ and $H$ as follows:
$$
G(x,u):=\Bigg(\frac{2}{5}x_1+\frac{2}{5}x_2-\frac{1}{5}\|u\|-\frac{4}{5},-\frac{3}{10}x_1+\frac{6}{5}x_2-\frac{1}{10}\|u\|-\frac{2}{5}\Bigg),
$$
$$
H(x,u):=\Bigg(u_1-\frac{1}{15}x_1+\frac{1}{24}x_2-\frac{1}{40}\|u\|-\frac{7}{30},u_2-\frac{1}{12}x_1+\frac{7}{120}x_2-\frac{1}{40}\|u\|+\frac{1}{30}\Bigg).
$$
Hence, we get
$$
(x-G,u-H)=f_1\omega^1+f_2\omega^2=\Bigg(2f_1+2f_2,f_1+f_2,\frac{1}{3}f_1+\frac{1}{6}f_2,\frac{1}{6}f_1+\frac{1}{3}f_2\Bigg)
$$
is $L-$ isotone. Then we check that all the conditions in Theorem \ref{MainT} are satisfied.
Let us start at the initial condition. We have,
$$
-G(0,0,0,0)=\left(\frac{4}{5},\frac{2}{5}\right)\quad\mbox{\rm and}\quad
\|H(0,0,0,0)\|=\sqrt{\left(-\frac{7}{30}\right)^2+\left(\frac{1}{30}\right)^2}=\frac{\sqrt{2}}{6}.
$$
Evidently, $-G(0,0,0,0)_1\geq -G(0,0,0,0)_2\geq \|H(0,0,0,0)\|.$
Now, consider a vector $(\hat{x},\hat{u}):=(30,12,4,3)\in K$, which yields

$$
G(\hat{x},\hat{u})=\left(15,\frac{9}{2}\right)\quad\mbox{\rm and}\quad
H(\hat{x},\hat{u})=\left(\frac{257}{120},\frac{133}{120}\right).
$$
Moreover, we have that $G(\hat{x},\hat{u})_1\geq G(\hat{x},\hat{u})_2\geq\|H(\hat{x},\hat{u})\|$, which implies that $(\hat{x},\hat{u})\in\Omega$. Hence, $\Omega\neq\varnothing$.
Next, we solve the mixed complementarity problem MiCP$(G,H,C,p,q)$.
For an arbitrary element $(x,y)$, if it is a solution of MiCP$(G,H,C,p,q)$, then
\[
x-G(x,u) = (2f_1+2f_2,f_1+f_2)\text{ where } f_i=f_i(x,u),i=1,2,
\]
and $G(x,u)=0$. Thus, we have $x_1=2f_1+2f_2$ and $x_2=f_1+f_2$.
Moreover,
\begin{equation}\label{value-x1x2}
x_1=\frac{1}{3}\|u\|+\frac{4}{3}\text{ and }x_2=\frac{1}{6}\|u\|+\frac{2}{3}.
\end{equation}
Meanwhile, we have $u\perp H(x,u)$, which implies
\[
\langle u,H(x,u)\rangle=u_1\left(u_1-\frac{1}{3}f_1-\frac{1}{6}f_2\right)+u_2\left(u_2-\frac{1}{6}f_1-\frac{1}{3}f_2\right)=0.
\]
Then,
\begin{equation}\label{norm-u-u1u2}
\|u\|^2=u_1^2+u_2^2=\left(\frac{1}{3}u_1+\frac{1}{6}u_2\right)f_1+\left(\frac{1}{6}u_1+\frac{1}{3}u_2\right)f_2.
\end{equation}
We will figure out all the nonzero solutions on the boundary of $C$. For the first case, without loss of generality, suppose that $u_1=u_2 > 0$, so we have $\|u\|=\sqrt{2}u_1=\sqrt{2}u_2$ and, by using (\ref{norm-u-u1u2}),
\[
u_1=u_2=\frac{1}{4}\left(f_1+f_2\right).
\]
By using the definition of $f_1$ and $f_2$ as well as  (\ref{value-x1x2}), we get
\begin{equation*
u_1=u_2=\frac{48+2\sqrt{2}}{287}.
\end{equation*}
Thus, the solution of MiCP$(G,H,C,p,q)$ is
\[
(x,u)=\left(\frac{384+16\sqrt{2}}{287},\frac{192+8\sqrt{2}}{287},\frac{48+2\sqrt{2}}{287},\frac{48+2\sqrt{2}}{287}\right).
\]

For the second case, we consider $u_2=0$, which implies that $\|u\|=u_1$. Hence, the equation (\ref{norm-u-u1u2}) is equivalent to
\[
u_1^2-\left(\frac{1}{3}f_1+\frac{1}{6}f_2\right) u_1=0.
\]
Since $u_1\neq0$, we have
\[
u_1=\frac{1}{3}f_1+\frac{1}{6}f_2.
\]
By using the definition of $f_1$ and $f_2$, and (\ref{value-x1x2}) again, we have $u_1=\frac{212}{691}$, which implies that $u=\left(\frac{212}{691},0\right)$.
Thus,
\[
(x,u)=\left(\frac{992}{691},\frac{496}{691},\frac{212}{691},0\right).
\]
Consider $(0,0,0,0)$ as a starting point in the fixed point iteration process (\ref{picard iteration 2}). We have
\begin{equation}\label{Picard-applied}
\begin{cases}
x_{n+1}=x^n-G(x^n,u^n)=(2f_1(x^n,u^n)+2f_2(x^n,u^n),f_1(x^n,u^n)+f_2(x^n,u^n)),& \\
u_{n+1}=P_C(u^n-H(x^n,u^n))=P_C\left(\frac{1}{3}f_1(x^n,u^n)+\frac{1}{6}f_2(x^n,u^n),\frac{1}{6}f_1(x^n,u^n)+\frac{1}{3}f_2(x^n,u^n)\right).&
\end{cases}
\end{equation}
From the above equations we get $x^{n+1}_1\geq x^{n+1}_2$. Moreover, since as the starting point we set $(0,0,0,0)$, then for any arbitrary $i\in\mathbb N$, we have that $x_1^i\geq x_2^i\geq 0$.
Define the set $S$ as follows:
\[
S:=\left\{(x,u)\in\mathbb R^2\times\mathbb R^2:0\leq x_1<\frac{992}{691},0\leq x_2<\frac{496}{691},0\leq u_1<\frac{212}{691},u_2=0 \right\}.
\]
We want to show that for any $n\in\mathbb N$ we have $(x^n,u^n)\in S$. We will prove it by induction.
First, we have $(x^0,u^0)\in S$.
Suppose next\, $0\leq x_1^n < \frac{992}{691}$, $0\leq x_2^n < \frac{496}{691}$, $0\leq u_1^n < \frac{212}{691}$ and $u_2=0$, which is equivalent to $\|u^n\|=u_1^n$. Since $x_1^n\geq x_2^n$, we have
\[
\begin{aligned}
0<x_1^{n+1}=2f_1(x^n,u^n)+2f_2(x^n,u^n)
& = \frac{3}{5}x_1^n-\frac{2}{5}x_2^n+\frac{1}{5}u_1^n+\frac{4}{5}\\\\
& < \frac{3}{5}\cdot\frac{992}{691}-\frac{2}{5}\cdot\frac{496}{691}+\frac{1}{5}\cdot\frac{212}{691}+\frac{4}{5}= \frac{992}{691}.
\end{aligned}
\]
Similarly,
\[
\begin{aligned}
0<x_2^{n+1}=f_1(x^n,u^n)+f_2(x^n,u^n)
& = \frac{3}{10}x_1^n-\frac{1}{5}x_2^n+\frac{1}{10}u_1^n+\frac{2}{5}\\\\
& < \frac{3}{10}\cdot\frac{992}{691}-\frac{1}{5}\cdot\frac{496}{691}+\frac{1}{10}\cdot\frac{212}{691}+\frac{2}{5}=\frac{496}{691}.
\end{aligned}
\]
Meanwhile, we also have
\[
u^n-H(x^n,u^n)=\left(\frac{1}{3}f_1(x^n,u^n)+\frac{1}{6}f_2(x^n,u^n),\frac{1}{6}f_1(x^n,u^n)+\frac{1}{3}f_2(x^n,u^n)\right).
\]
Obviously, $(u^n-H(x^n,u^n))_1>0$, then
\[
\begin{array}{rcl}
(u^n-H(x^n,u^n))_1 & < & \displaystyle \frac{1}{3}\left(\frac{1}{10}\cdot\frac{992}{691}-\frac{1}{20}\cdot\frac{496}{691}+\frac{1}{20}\cdot\frac{212}{691}+1\right)\\\\
& & \displaystyle +\frac{1}{6}\left(\frac{1}{5}\cdot\frac{992}{691}-\frac{3}{20}\cdot\frac{496}{691}+\frac{1}{20}\cdot\frac{212}{691}-\frac{3}{5}\right)
=\frac{212}{691}
\end{array}
\]
and $0<(u^n-H(x^n,u^n))_2$.
It is easy to check that the projection of it onto $C$ such that $0\leq u^{n+1}_1 <\frac{691}{212}$ and $u_2^{n+1}=0$, must be given on the ray $\{(u_1,u_2):u_1\geq0,u_2=0\}$. It is equivalent to
\[
u^{n+1}=(u^{n+1}_1,u^{n+1}_2)=P_C\left(u^n-H(x^n,u^n)\right)= \left(\frac{1}{3}f_1(x^n,u^n)+\frac{1}{6}f_2(x^n,u^n),0\right).
\]

\noindent Thus, the system of equations (\ref{Picard-applied}) is equivalent to
\begin{equation}\label{iteration1}
\begin{cases}
x_1^{n+1}= \frac{3}{5}x_1^n-\frac{2}{5}x_2^n+\frac{1}{5}u_1^n+\frac{4}{5} , &   \\
x_2^{n+1}= \frac{3}{10}x_1^n-\frac{1}{5}x_2^n+\frac{1}{10}u_1^n+\frac{2}{5}, & \\
u_1^{n+1}= \frac{1}{15}x_1^n-\frac{1}{24}x_2^n+\frac{1}{40}u_1^n+\frac{7}{30}.
\end{cases}
\end{equation}
Moreover, we have $x_1^n=2x_2^n$, so (\ref{iteration1}) is equivalent to
\begin{equation}\label{iteration2}
\begin{cases}
x_1^{n+1}= 2x_2^{n+1}, &   \\
x_2^{n+1}= \frac{2}{5}x_2^n+\frac{1}{10}u_1^n+\frac{2}{5}, & \\
u_1^{n+1}= \frac{11}{120}x_2^n+\frac{1}{40}u_1^n+\frac{7}{30}.
\end{cases}
\end{equation}
The last two lines in (\ref{iteration2}) can be aggregated as follows
$$\left [\begin{array}{c} x_2^{n+1}\\ \mbox{} \vspace{-.3cm} \\ u_1^{n+1} \end{array} \right]=\left [ \begin{array}{rr} \frac{2}{5} & \frac{1}{10}\\ \mbox{} \vspace{-.3cm}\\
\frac{11}{120} & \frac{1}{40} \end{array} \right] \left [\begin{array}{c} x_2^n\\ \mbox{} \vspace{-.3cm}\\ u_1^n \end{array}\right]+
\left [\begin{array}{c} \frac{2}{5} \\ \mbox{} \vspace{-.3cm}\\ \frac{7}{30} \end{array}\right]. $$
One easily verifies that the above $2 \times 2$ matrix has both (real) eigenvalues whose absolute values are less than $1$, so it is a convergent matrix. Hence, the above process is convergent to the unique fixed point $\left[x_2^*~ u_1^*\right]'$ of the above equation, regardless of a starting point $\left[x_2^0~ u_1^0\right]' \in \mathbb R^2.$ Explicitly,
$$\left [\begin{array}{c} x_2^*\\ \mbox{} \vspace{-.3cm} \\ u_1^* \end{array} \right]=
\left [ \begin{array}{rr} \frac{3}{5} & -\frac{1}{10}\\ \mbox{} \vspace{-.3cm}\\
-\frac{11}{120} & \frac{39}{40} \end{array} \right]^{-1} \cdot \left [\begin{array}{c} \frac{2}{5} \\ \mbox{} \vspace{-.3cm}\\ \frac{7}{30} \end{array}\right]= \left [\begin{array}{c} \frac{496}{691} \\ \mbox{} \vspace{-.3cm}\\ \frac{212}{691} \end{array}\right]\,.$$
Bearing in mind that $x_1^{n+1}= 2x_2^{n+1}$ and $u_2^n=0$, we have the convergence:
$$(x^n,u^n)=(x_1^n,x_2^n,u_1^n, u_2^n) \rightarrow (x_1^*,x_2^*,u_1^*,0)= \left (\frac{992}{691}, \frac{496}{691}, \frac{212}{691}, 0 \right), $$
which is the same as one solution we have obtained on the boundary.
\begin{remark}
\emph{We remark that $f_1\omega^1+f_2\omega^2$ is not ESOC-isotone.
Indeed, if we assume that $f_1\omega^1+f_2\omega^2$ is $L$-isotone, then for any $(x,u)\leq_L(y,v)$ and $\omega^1,\,\omega^2\in L$, we have
\begin{equation} \label{f_1-f_2}
f_1(x,u)\omega^1+f_2(x,u)\omega^2\leq_L f_1(y,v)\omega^1+f_2(y,v)\omega^2
\end{equation}
and it is equivalent to
$$
(f_1(y,v)-f_1(x,u))\omega^1 + (f_2(y,v)-f_2(x,u))\omega^2\in L.
$$
For arbitrary $(x^*,u^*),(y^*,v^*)\in\mathbb R^p\times\mathbb R^q$, such that $y^*_1-x^*_1=\|u^*-v^*\|=\|u^*\|-\|v^*\|>0$, $y^*_2-x^*_2=2\|u^*-v^*\|=2(\|u^*\|-\|v^*\|)>0$, it is obvious that $(x^*,u^*)\leq_{\rm ESOC}(y^*,v^*)$. Since $f_1(x,u)= \frac{1}{10}x_1-\frac{1}{20}x_2 +\frac{1}{20}\|u\|+1$ and $f_2(x,u)=\frac{1}{5}x_1-\frac{3}{20}x_2+\frac{1}{20}\|u\|-\frac{3}{5}$,
\[
\begin{aligned}
f_1(y^*,v^*)-f_1(x^*,u^*)&=\frac{1}{10}(y^*_1-x^*_1)-\frac{1}{20}(y^*_2-x^*_2)-\frac{1}{20}(\|u^*\|-\|v^*\|)\\
&=\frac{1}{10}(\|u^*\|-\|v^*\|)-\frac{2}{20}(\|u^*\|-\|v^*\|)-\frac{1}{20}(\|u^*\|-\|v^*\|)\\
&=-\frac{1}{20}(\|u^*\|-\|v^*\|)< 0,
\end{aligned}
\]
\[
\begin{aligned}
f_2(y^*,v^*)-f_2(x^*,u^*)&=\frac{1}{5}(y^*_1-x^*_1)-\frac{3}{20}(x^*_2-y^*_2)-\frac{1}{20}(\|u^*\|-\|v^*\|)\geq 0\\
&=\frac{1}{5}(\|u^*\|-\|v^*\|)-\frac{6}{20}(\|u^*\|-\|v^*\|)-\frac{1}{20}(\|u^*\|-\|v^*\|)\\
&=-\frac{3}{20}(\|u^*\|-\|v^*\|)< 0,
\end{aligned}\textcolor{red}{,}
\]
contradicting (\ref{f_1-f_2}), so $f_1\omega^1+f_2\omega^2$ is not ESOC-isotone.
Let us recall that both $f_1$ and $f_2$ are MESOC-monotone (which has been proved in the numerical example) and not ESOC-monotone, which implies that $f_1(y,v)-f_1(x,u)$ and $f_2(y,v)-f_2(x,u)$ will not be nonnegative for all $(x,u)\leq_{\rm ESOC}(y,v)$. Since both $f_1(y^*,v^*)-f_1(x^*,u^*)$ and $f_2(y^*,v^*)-f_2(x^*,u^*)$ are negative, then by using convexity of ESOC, since $\omega^1,\,\omega^2\in \textrm{MESOC}\subseteq \textrm{ESOC}$, we have
$$
-(f_1(y^*,v^*)-f_1(x^*,u^*))\omega^1 - (f_2(y^*,v^*)-f_2(x^*,u^*))\omega^2\in \textrm{ESOC}.
$$
Meanwhile, if $f_1\omega_1+f_2\omega_2$ were ESOC-isotone, then
$$
(f_1(y^*,v^*)-f_1(x^*,u^*))w^1 + (f_2(y^*,v^*)-f_2(x^*,u^*))w^2\in \textrm{ESOC}.
$$
Since $\omega^1$, $\omega^2$ are linearly independent, it contradicts pointedness of ESOC.
Thus,
 $f_1\omega^1+f_2\omega^2$ is not ESOC-isotone.}
\end{remark}

\section*{Concluding remarks}

In this paper we study the Monotone Extended Second Order Cone (MESOC) as a new
generalization of the Lorentz cone $\mathcal{L}^n_+$. This cone is different both from $\mathcal{L}^n_+$ and the previously introduced Extended Second Order Cone (ESOC) \cite{nemeth-zhang1,nemeth-zhang2,nemeth-xiao1,nemeth-xiao2,nemeth-xie-zhang,sznajder} in many aspects, but bears similarities too. Both MESOC and ESOC are cones in $\mathbb R^p\times\mathbb R^q$. MESOC is sub-dual as is ESOC, but for $p>1$ it is not self-dual like $\mathcal{L}^n_+$. Both ESOC and MESOC become $\mathcal{L}^{q+1}_+$ when $p=1$.
Contrary to $\mathcal{L}^n_+$, for $q=1$ both ESOC and MESOC are polyhedral. MESOC and ESOC
are symmetric cones for $p=1$ only, that is, if and only if they are Lorentz cones. Contrary to
$\mathcal{L}^n_+$ and ESOC, MESOC is reducible. For both ESOC and MESOC the cylinders $\mathbb R^p\times C$, where $C$ is an arbitrary closed convex set with nonempty interior in $\mathbb R^q$, are isotone projection sets. In fact, these cylinders are isotone projection sets with respect to any intersection of ESOC with $U\times\mathbb R^q$, where $U$ is an arbitrary closed convex cone in $\mathbb R^p$ (the proof is similar to the first part of the proof of Theorem \ref{isotonicity-on-cylinder}). Contrary to ESOC, any isotone projection set with respect to MESOC is such a cylinder.

We determined the bilinearity rank of MESOC and used the MESOC-isotonicity of the projection onto the cylinder to solve general mixed complementarity problems. We illustrated the corresponding iterative method by using a numerical example with exact numbers. Although the iteration principle for the MESOC is similar to the corresponding one for ESOC,
we remark that there are mixed complementarity problems which can be solved iteratively by MESOC, but the same iterative scheme cannot be used via ESOC, because it does not satisfy the corresponding ESOC-isotonicity condition (merely the MESOC-isotonicity). This is due to the fact that although MESOC is a subset of ESOC, MESOC-isotonicity of mappings does not imply their ESOC-isotonicity. This idea is underlined in the preceding section.

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We posted about the astonishing Martini Mustang last year and gave you a run down of it's long specs list. Though what we neglected show you was this fantastic video from Petrolicious.
We may be late to the party here but regardless this video is too good not to document over here at 67mustangblog. If you've seen it you'll know what we're on about, if not, Enjoy!


I have mentioned before that I am an avid dry shampoo user – not only does it revive my hair when I'm feeling lazy but it gives me so MUCH VOLUME! What's not to love?! I love Batiste products, their dry shampoo works wonders.
Two weeks ago, while commuting home on the tube I spotted an advert for a new Batiste product that includes the usual dry shampoo component plus a conditioner! Hello, Batiste 2 in 1 Invisible Dry Shampoo & Conditioner! One of my fears was always that I was drying my hair and scalp out from using too much dry shampoo. So with added conditioner, surely I was helping it?!
But really, not hello, more GOODBYE. I've used it on three separate occasions now to test it out; the day after washing my hair, two days after washing my hair and STRAIGHT AFTER washing my hair. Every time my hair came out greasy and limp and worse than it was before. Am I doing it wrong? I've used it just like normal dry shampoo. Spraying near the roots and scrunching and brushing it through with my fingertips.
I feel like they have over-conditioned the stuff inside the can, and the ratio/consistency is not right. I'm on the verge of throwing the can out, because I can't use it.


BOSTON, Mass. /Massachusetts Newswire/ — A public memorial celebration will be held Saturday, October 10, at 2 p.m. ET. at the Museum of African American History, 46 Joy Street, Beacon Hill, Boston, MA to celebrate the life and legacy of Samuel Washington Allen – who also wrote under the pen name of Paul Vesey. The noted poet, professor, lawyer and translator passed away peacefully surrounded by family and friends at the age of 97 on June 27, 2015 in Norwood, Mass.
Samuel Allen's collections of poetry consist of the bilingual Elfenbeinzahne (Heidelberg, 1956); Ivory Tusks (New York, 1968); Paul Vesey's Ledger (London, 1975) and Every Round (Detroit, 1987). He translated Sartre's "Orphee noir" essay in 1951 while in Paris and was the editor and one of the translators of Poems from Africa (Crowell, 1973).
While majoring in sociology at Nashville's Fisk University, Allen studied writing with Harlem Renaissance poet, novelist and critic James Weldon Johnson. He graduated with high honors in 1938.
Allen then attended Harvard Law School where he earned his Doctor of Jurisprudence in 1941 – later becoming deputy assistant in the New York district attorney's office.
"While many African-American writers have sought to incorporate African elements into their works, few have realized that goal as elegantly and powerfully as Samuel W. Allen," stated James Manheim in Contemporary Black Biography, 2003.


Jenna Matthys was one of three winners for Word Sprout's Story Slam in on the theme "Thankful." For the first time, because of Word Sprout's new videorecording equipment and the talents of the awesome Allison Broeren, I can actually embed a YouTube version of the story Jenna told in this post. Due to spoilers in the interview, I'm going to suggest you listen to the story first.
My story was about being a nerdy kid, observing the secret life of parents, and growing up inside a conflicted marriage, thankful to have survived the occasional wars between them.
I had been thinking about the odd little blue house I grew up in on the East side of St. Paul, and about my father having secret toilet paper for a while, and every time I thought about it-it cracked me up. I started working on it a few months earlier, and dusted it off for the slam and finished it.
I start out with trying to think about the theme, looking the word up in the dictionary. Now I know what the word means without looking but sometimes there's a surprise. For instance at a recent storytelling event, the theme was Vice. Looking it up, I saw that the word "vice" comes from the Latin word vitae, meaning fault. For me, thinking about vices in terms of fault was a real jumping off point. Then I think about it, the theme, while brushing my teeth and taking a shower. all those little everyday activities. Then I start out writing and keep doing it until I have at least one page of…total crap. That first page is awful, almost always. So I rewrite it about 20 times. Vicious editing is a big part of my process. Then it's done, I'm happy with it-it's perfect, wouldn't change a thing, but it's too long. So then I really look at it, to figure what I can take out without losing the bones of the story.
The most difficult part is right before you go up there-cause the butterflies in my stomach are having an old school punk rock dance party and I am trying not to throw up. I get real nervous right before. Will I mess up? Will they get me? Will my Minnesota accent be so annoying people flee the room? That's the difficult part. The most rewarding part is when I can tell that the audience is really hearing me, relating to my story. I have also been blown away with how supportive the storytelling community has been to me; meeting so many other kind, supportive, artists has really been the best part.
Jenna Matthys is a poet and storyteller from Minneapolis, MN.
But I can tell you as well that she is not only a several-times winner of Story SlamMN! but she has won The Moth Story Slam at the Amsterdam Bar in St. Paul.
Because of that she will be participating in the first Moth Grand Slam at the Fitzgerald Theater on June 7th – along with myself, Annie Baxter, Ward Rubrecht, Christy Marie Kent, Taylor Arrell-Tower, Mike Fotis and a few other folks I do not know personally. You may know the Fitzgerald as the base from which Prairie Home Companion broadcasts. It seats over a thousand. Am I nervous? A little. Jenna? I doubt it.
I can't post the story I told at SlamMN this past Tuesday for "Provoked" – although it came in first. But it was not written down when I performed it. This seldom happens. In fact, where I generally fall flat is in getting off the page and onto the stage.
The story was a sensitive one for me – one that contains some difficult experiences that involve people I love. Despite numerous drafts, it was not coming together. I did not intend to perform at all, but turnout for the slam was light and there weren't enough names in the hat.
I put together a very rough structure on the back of an envelope.
Sometimes you discover enough in the process of doing that to make a story work in performance. Which apparently I did. But in this case what was ready for the stage is not yet ready for the page.
Next month are the semifinals for Story SlamMN! and the theme is "Passionate." There will be at least one brainstorming post for that. Meanwhile, if you'd like to read the story that qualified me for The Moth Grand Slam, it's right here. Also, in addition to the video for Jenna, I've now added the story Leif Wallin told for "Thankful" to his interview. Check it out!
I did laugh at Jenna's family arguments although I confess to 'missing' bits of what she says – I'm unfamiliar with her accent and it was a surprising 'throw' because she's clearly an articulate woman!
Meanwhile, perhaps a voice recorder for the next impromptu performance? A shame to not capture a gem..