\chapter{Hopfield Network Stability Analysis} 
\section{Introduction}

In this chapter, we apply the second method of Lyapunov to analyze stability in the sense of Lyapunov of the Hopfield neural network. The neural network will be viewed as a large interconnected dynamic system, that is as an interconnection of many neurons. We first analyze the stability of a single neuron, and then proceed with the stability analysis of the whole network in terms of the properties of the single neurons. The Hopfield type neural network can be viewed as an associative memory. This is because it has stable states that attract the neighboring states to develop in time. This behavior can be interpreted as an evolution of an imperfect, corrupted pattern towards the correct, stored pattern---hence the term associative memory.

We first review some, not particularly well known, results from matrix theory that we use in our presentation. In particular, we review selected properties of M-matrices that are useful in the analysis of large scale interconnected systems. Next, we present the Comparison Theorem which we use when estimating trajectory bounds of neural networks. Having reviewed the relevant background material, we start with the analysis of a single neuron. After that, we discuss the stability and instability of the whole network in terms of the properties of the individual neurons. We conclude this chapter with remarks in the closing section.

\section{Properties of M-Matrices}

We start with the definition of the M-matrix.
\begin{defn}
A real $n \times n$ matrix $\mb{\bm$A$}$ is an M-matrix if all its off-diagonal elements are non-positive, that is, $a_{ij} \le 0$, $i \ne j$, and all its principal minors are positive.
\end{defn}
Note that if $\mb{\bm$A$}$ is an M-matrix, then its diagonal elements are positive since its principal minors are positive.\\
In establishing the needed properties of the M-matrices, we use the following technical lemma due to Fiedler and Pt\'{a}k~\cite{fiedler62}.
\begin{lemma}
\label{lemma:lu}
The following properties af an arbitrary matrix $\mb{\bm$A$} \in \mb{\Bb R}^{n \times n}$ are equivalent:
\begin{itemize}
\item[(i)] The leading principal minors of $\mb{\bm$A$}$ are positive,
\item[(ii)] There exists a lower triangular matrix $\mb{\bm$L$} \in \mb{\Bb R}^{n \times n}$ and an upper triangular matrix $\mb{\bm$U$} \in \mb{\Bb R}^{n \times n}$ both with positive diagonal elements such that $\mb{\bm$A$}=\mb{\bm$L$}\mb{\bm$U$}$
\end{itemize}
\end{lemma}
{\bf Proof}\quad We first prove the implication (i)$ \Rightarrow $(ii). The proof is by induction. Obviously the implication holds for $n=1$. Suppose the statement holds for all real matrices of order $n-1$. Let us represent $\mb{\bm$A$} \in \mb{\Bb R}^{n \times n}$ as follows:
\[
\mb{\bm$A$} = \left[\begin{array}{cc}
\mb{\bm$A$}_{n-1} & \mb{\bm$a$}\\
\mb{\bm$b$} & a_{nn}
\end{array} \right],
\]
where $\mb{\bm$A$}_{n-1} \in \mb{\Bb R}^{(n-1) \times (n-1)}$, $\mb{\bm$a$} \in \mb{\Bb R}^{(n-1) \times 1}$, $\mb{\bm$b$} \in \mb{\Bb R}^{1 \times (n-1)}$, and $a_{nn} \in \mb{\Bb R}$. From the inductive hypothesis, the matrix $\mb{\bm$A$}_{n-1}$ has positive leading principal minors, $\mb{\bm$A$}_{n-1}=\mb{\bm$L$}_{n-1}\mb{\bm$U$}_{n-1}$, where $\mb{\bm$L$}_{n-1} \in \mb{\Bb R}^{(n-1) \times (n-1)}$ is a lower triangular matrix with positive diagonal elements, and $\mb{\bm$U$}_{n-1} \in \mb{\Bb R}^{(n-1) \times (n-1)}$ is an upper triangular matrix with positive diagonal elements. Observe that
\[
\left[\begin{array}{cc}
\mb{\bm$I$}_{n-1} & \mb{\bm$0$}\\
-\mb{\bm$b$}\mb{\bm$A$}_{n-1}^{-1} & 1
\end{array} \right] \mb{\bm$A$} = \left[\begin{array}{cc}
\mb{\bm$A$}_{n-1} & \mb{\bm$a$}\\
\mb{\bm$0$} & a_{nn} - \mb{\bm$b$}\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}
\end{array}\right],
\]
where from the assumptions $\det \mb{\bm$A$}_{n-1}>0$, and $\det \mb{\bm$A$}>0$. Hence,
\[
0< \det \mb{\bm$A$}=\left( a_{nn} - \mb{\bm$b$}\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$} \right) \det \mb{\bm$A$}_{n-1}.
\]
Thus we have
\[
a_{nn} - \mb{\bm$b$}\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}>0.
\]
Let
\[
\mb{\bm$L$}=\left[\begin{array}{cc}
\mb{\bm$L$}_{n-1} & \mb{\bm$0$}\\
\mb{\bm$b$}\mb{\bm$U$}_{n-1}^{-1} & 1
\end{array} \right],
\]
and
\[
\mb{\bm$U$}=\left[\begin{array}{cc}
\mb{\bm$U$}_{n-1} & \mb{\bm$L$}_{n-1}^{-1}\mb{\bm$a$}\\
\mb{\bm$0$} & a_{nn} - \mb{\bm$b$}\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}
\end{array}\right].
\]
One can check that
\[
\mb{\bm$A$}=\mb{\bm$L$}\mb{\bm$U$},
\]
and that the matrices $\mb{\bm$L$}$ and $\mb{\bm$U$}$ satisfy (ii).

We now prove the implication (i)$ \Leftarrow $(ii). From the Generalized Cauchy's Theorem it follows that the leading principal minors of $\mb{\bm$A$}=\mb{\bm$L$}\mb{\bm$U$}$ are just the products of the corresponding leading principal minors of $\mb{\bm$L$}$ and $\mb{\bm$U$}$. This is because the matrix $\mb{\bm$L$}$ is lower triangular with positive diagonal elements and the matrix $\mb{\bm$U$}$ is upper triangular with positive diagonal elements. This completes our proof of the lemma.\\
\hspace*{5.5in} $\Box$

The following theorem, also due to Fiedler and Pt\'{a}k~\cite{fiedler62}, establishes some properties of the M-matrices in terms of their LU-decomposition.
\begin{thm}
\label{thm:m_matlu}
Let $\mb{\bm$A$} \in \mb{\Bb R}^{n \times n}$ be an M-matrix. Then there is a lower triangular matrix $\mb{\bm$L$} \in \mb{\Bb R}^{n \times n}$ and an upper triangular matrix $\mb{\bm$U$} \in \mb{\Bb R}^{n \times n}$ both with positive diagonal elements. Their off-diagonal elements are non-positive. Furthermore, both $\mb{\bm$L$}^{-1}$ and $\mb{\bm$U$}^{-1}$ are non-negative which implies that $\mb{\bm$A$}^{-1}=\mb{\bm$U$}^{-1}\mb{\bm$L$}^{-1}$ is non-negative.
\end{thm}
{\bf Proof}\quad We prove this theorem by induction. The theorem holds for $n=1$. Suppose the theorem holds for all matrices of order $n-1$. Consider now the factorization from the proof of Lemma~\ref{lemma:lu}. We have
\[
\mb{\bm$A$}=\mb{\bm$L$}\mb{\bm$U$} = \left[\begin{array}{cc}
\mb{\bm$L$}_{n-1} & \mb{\bm$0$}\\
\mb{\bm$b$}\mb{\bm$U$}_{n-1}^{-1} & 1
\end{array} \right]\left[\begin{array}{cc}
\mb{\bm$U$}_{n-1} & \mb{\bm$L$}_{n-1}^{-1}\mb{\bm$a$}\\
\mb{\bm$0$} & a_{nn} - \mb{\bm$b$}\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}
\end{array}\right],
\]
where $\mb{\bm$L$}_{n-1}$, $\mb{\bm$U$}_{n-1}$, $\mb{\bm$L$}_{n-1}^{-1}$, $\mb{\bm$U$}_{n-1}^{-1}$ and $\mb{\bm$A$}_{n-1}^{-1}=\mb{\bm$U$}_{n-1}^{-1}\mb{\bm$L$}_{n-1}^{-1}$ satisfy the inductive hypotheses. We see that $l_{ij} \le 0$ for $i>j$ and $u_{kl} \le 0$ for $k<l$ which means that the off-diagonal elements of $\mb{\bm$L$}$ and $\mb{\bm$U$}$ are non-positive. Their diagonal elements are positive. The inverse of $\mb{\bm$L$}$ is
\[
\mb{\bm$L$}^{-1}=\left[\begin{array}{cc}
\mb{\bm$L$}_{n-1}^{-1} & \mb{\bm$0$}\\
-\mb{\bm$b$}\mb{\bm$L$}_{n-1}^{-1} & 1
\end{array}\right].
\]
From the inductive assumption $\mb{\bm$L$}_{n-1}^{-1}$ and $\mb{\bm$A$}_{n-1}^{-1}$ are non-negative and hence $\mb{\bm$L$}_{n-1}^{-1}$ is non-negative.\\
The inverse of $\mb{\bm$U$}$ is
\[
\mb{\bm$U$}^{-1} = \left[\begin{array}{cc}
\mb{\bm$U$}_{n-1}^{-1} & -\mb{\bm$A$}_{n-1}^{-1} \frac{\mb{\bm$a$}}{ a_{nn} - \mb{\bm$b$} \mb{\bm$A$}_{n-1}^{-1} \mb{\bm$a$}}\\
\mb{\bm$0$} & \left( a_{nn}-\mb{\bm$b$}\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$} \right) ^{-1}
\end{array} \right].
\]
By the inductive assumptions and the fact that $a_{nn}-\mb{\bm$b$}\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}>0$, the matrix $\mb{\bm$U$}^{-1}$ is non-negative, and the proof is complete.\\
\hspace*{5.5in} $\Box$

We now consider a more general class of matrices called Z-matrices.
\begin{defn}
A real $n \times n$ matrix $\mb{\bm$A$}$ whose off-diagonal elements are non-positive is called a Z-matrix.
\end{defn}
Note that M-matrices are Z-matrices.\\
To simplify our presentation, we use the notation $\mb{\bm$B$}>0$, to mean that all the elements of $\mb{\bm$B$}$ are positive, where $\mb{\bm$B$}$ can be a matrix or a vector. Respectively, $\mb{\bm$B$} \ge 0$ means that all the elements of $\mb{\bm$B$}$ are non-negative. We are now ready to state and prove the following theorem concerning properties of Z-matrices. The theorem can be found in Fiedler and Pt\'{a}k~\cite{fiedler62}.
\begin{thm}
\label{thm:z_mat}
Let $\mb{\bm$A$} \in \mb{\Bb R}^{n \times n}$ be a Z-matrix. Then, the following conditions are equivalent:
\begin{itemize}
\item[(i)] There exists a vector $\mb{\bm$x$} \ge 0$ such that $\mb{\bm$Ax$}>0$;
\item[(ii)] There exists a vector $\mb{\bm$y$}>0$ such that $\mb{\bm$Ay$}>0$;
\item[(iii)] There exists a diagonal matrix $\mb{\bm$D$}$ with positive diagonal elements such that $\mb{\bm$ADe$}>0$, where $\mb{\bm$e$}=\left[\begin{array}{ccc}
1, & \ldots, & 1
\end{array}\right]^T \in \mb{\Bb R}^n$;
\item[(iv)] There exists a diagonal matrix $\mb{\bm$D$}$ with positive diagonal elements such that the matrix $\mb{\bm$W=AD$}$ has a dominant positive main diagonal, that is for $i=1,2,\ldots,n$, we have $w_{ii}>\sum_{\stackrel{j=1}{j \ne i}}^n w_{ij}$.
\end{itemize}
\end{thm}
{\bf Proof}\quad We prove this theorem using the following scheme:
\[
\mb{(i)} \Rightarrow \mb{(ii)} \Rightarrow \mb{(iii)} \Rightarrow \mb{(iv)} \Rightarrow \mb{(i)}.
\]
We begin with the implication (i)$ \Rightarrow $(ii). Thus suppose that there exists a vector $\mb{\bm$x$} \ge 0$ such that $\mb{\bm$Ax$} > 0$. Let $\mb{\bm$e$}=\left[\begin{array}{ccc}
1, & \ldots, & 1
\end{array}\right]^T \in \mb{\Bb R}^n$, $\epsilon > 0$, and $\mb{\bm$y$} = \mb{\bm$x$} + \epsilon \mb{\bm$e$} > 0$. Then
\[
\mb{\bm$Ay$} = \mb{\bm$A(x$} + \epsilon \mb{\bm$e$}) = \mb{\bm$Ax$} + \epsilon \mb{\bm$Ae$} > 0
\]
for a sufficiently small $\epsilon > 0$.

We next prove that (ii)$ \Rightarrow $(iii). For this let $\mb{\bm$y$}=\left[\begin{array}{ccc}
y_1, & \ldots, & y_n
\end{array}\right]^T > 0$ be such that $\mb{\bm$Ay$} > 0$. Let
\[
\mb{\bm$D$} = \left[\begin{array}{ccccc}
y_1 & 0 & 0 & \ldots & 0\\
0 & y_2 & 0 & \ldots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \ldots & y_n
\end{array} \right].
\]
It then follows that $\mb{\bm$ADe$} = \mb{\bm$Ay$} > 0$.

To prove (iii)$ \Rightarrow $(iv), let $\mb{\bm$D$} > 0$ be a diagonal matrix with positive diagonal elements such that $\mb{\bm$ADe$} > 0$. We denote
\[
\mb{\bm$AD$}=\mb{\bm$W$} = \left[\begin{array}{cccc}
w_{11} & -w_{12} & \ldots & -w_{1n}\\
-w_{21} & w_{22} & \ldots & -w_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-w_{n1} & -w_{n2} & \ldots & w_{nn}
\end{array} \right],
\]
where $w_{ij} \ge 0$ for $i \ne j$. Note that $\mb{\bm$ADe$}=\mb{\bm$We$}>0$ which is equivalent to
\[
w_{ii} > \sum_{\stackrel{j=1}{j \ne i}}^{n} w_{ij}, \qquad i=1,2,\ldots ,n.
\]
Since $w_{ij} \ge 0$, we have
\[
w_{ii} > \sum_{\stackrel{j=1}{j \ne i}}^{n} |w_{ij}|.
\]
The above means that $\mb{\bm$W$}$ is a row diagonal dominant matrix with positive main diagonal. Note $a_{ii} > 0$ since $0 < w_{ii} = a_{ii} d_i$ and $d_i > 0$.

To complete the proof of the theorem we now show that (iv)$ \Rightarrow $(i). For this let $\mb{\bm$D$}$ be a diagonal matrix with positive diagonal elements such that $\mb{\bm$W$} = \mb{\bm$AD$}$ has a positive dominant main diagonal. Then
\[
0 < \mb{\bm$We$} = \mb{\bm$ADe$} = \mb{\bm$Ax$},
\]
where
\[
\mb{\bm$x$}=\left[\begin{array}{ccc}
d_1 & \ldots & 0\\
\vdots & \ddots & \vdots\\
0 & \ldots & d_n
\end{array} \right] \left[\begin{array}{c}
1\\
\vdots\\
1
\end{array} \right]=\left[\begin{array}{c}
d_1\\
\vdots\\
d_n
\end{array}\right] > 0.
\]
Hence $(i)$ is satisfied.\\
\hspace*{5.5in} $\Box$

To proceed further, we need the following two technical results.
\begin{lemma}
\label{lemma:hadamard}
An $n \times n$ matrix $\mb{\bm$A$}$ with complex coefficients such that
\[
|a_{ii}| > \sum_{\stackrel{j=1}{j \ne i}}^{n} |a_{ij}|\quad\mb{for}\quad i=1,2,\ldots,n,
\]
is nonsingular, that is, $\det \mb{\bm$A$} \ne 0$.
\end{lemma}
{\bf Proof}\quad We prove the above lemma by contraposition. Suppose that $\det \mb{\bm$A$} = 0$. Then, there exists a nontrivial solution $\mb{\bm$x$} \ne \mb{\bm$0$}$ to $\mb{\bm$Ax=0$}$, that is for $k=1,2,\ldots,n$,
\[
\sum_{j=1}^{n} a_{kj} x_j = 0.
\]
Let
\[
x_i = \max_{1 \le k \le n} |x_k|.
\]
Rearranging $\sum_{j=1}^{n} a_{ij} x_j = 0$ yields
\[
-a_{ii}x_i=a_{i1}x_1 + \cdots + a_{i,i-1}x_{i-1} + a_{i,i+1}x_{i+1} + \cdots + a_{in}x_n.
\]
Taking the absolute values of the both sides of the above equation gives
\[
|a_{ii}||x_i|=|a_{i1}x_1 + \cdots + a_{i,i-1}x_{i-1} + a_{i,i+1}x_{i+1} + \cdots + a_{in}x_n| \le \sum_{\stackrel{j=1}{j \ne i}}^{n} |a_{ij}||x_j|.
\]
Thus
\[
|a_{ii}| \le \sum_{\stackrel{j=1}{j \ne i}}^{n} |a_{ij}| \frac{|x_j|}{|x_i|} \le  \sum_{\stackrel{j=1}{j \ne i}}^{n} |a_{ij}|
\]
which contradicts the assumption that $|a_{ii}| > \sum_{\stackrel{j=1}{j \ne i}}^{n} |a_{ij}|$.\\
\hspace*{5.5in} $\Box$

The above lemma says that if a matrix $\mb{\bm$A$}$ is row diagonal dominant then it is nonsingular. Following the arguments of the proof of the lemma, we can show that if a square, real or complex, matrix is column diagonal dominant then it is also nonsingular. We state this result in the following corollary.
\begin{corollary}
A real or complex $n \times n$ matrix $\mb{\bm$A$}$ such that 
for $j=1,2,\ldots,n$,
\[
|a_{jj}| > \sum_{\stackrel{k=1}{k \ne j}}^{n} |a_{kj}|
\]
is nonsingular. 
\end{corollary}

We now use Lemma~\ref{lemma:hadamard} to prove the following theorem due to Gershgorin.
\begin{thm}
\label{thm:gershgorin}
Given a real or complex $n \times n$ matrix $\mb{\bm$A$}$. Let $D_i$, $i=1,2,\ldots,n$, denotes the $i$-th closed disk in the complex plane centered at $a_{ii}$ whose radius is $r_i=\sum_{\stackrel{j=1}{j \ne i}}^{n} |a_{ij}|$. Then the eigenvalues of $\mb{\bm$A$}$ are located in the union $\bigcup_{i=1}^{n}D_i$. 
\end{thm}
{\bf Proof}\quad The eigenvalues of $\mb{\bm$A$}$ are the roots of the characteristic equation
\[
\det [\lambda \mb{\bm$I$} - \mb{\bm$A$}] = \det \left[\begin{array}{cccc}
\lambda - a_{11} & -a_{12} & \ldots & -a_{1n}\\
-a_{21} & \lambda - a_{22} & \ldots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots \\
-a_{n1} & -a_{n2} & \ldots & \lambda -a_{nn}
\end{array} \right]=0.
\]
By Lemma~\ref{lemma:hadamard}, the characteristic polynomial is nonzero if for all $i=1,2,\ldots,n$,
\[
|\lambda -a_{ii}| > \sum_{\stackrel{j=1}{j \ne i}}^{n} |a_{ij}|,
\]
that is, for all $i=1,2,\ldots,n$
\[
|\lambda -a_{ii}| > r_i.
\]
Thus the eigenvalues of $\mb{\bm$A$}$ must be in the set $\bigcup_{i=1}^{n} D_i$.\\
\hspace*{5.5in} $\Box$

In the following theorem, due to Fiedler and Pt\'{a}k~\cite{fiedler62}, we collect a number of equivalent conditions characterizing the M-matrices.
\begin{thm}
\label{thm:m_matsum}
Let $\mb{\bm$A$}$ be a Z-matrix. Then the following conditions are equivalent
\begin{itemize}
\item[(i)] There exists a vector $\mb{\bm$x$} \ge 0$ such that $\mb{\bm$Ax$} > 0$;
\item[(ii)] There exists a diagonal matrix $\mb{\bm$D$}$ with positive diagonal such that the matrix $\mb{\bm$W=AD$}$ has a positive main diagonal and is row diagonal dominant;
\item[(iii)] Eigenvalues of $\mb{\bm$W$}$ have positive real parts and principal minors of $\mb{\bm$W$}$ are positive;
\item[(iv)] All principal minors of $\mb{\bm$A$}$ are positive, that is, $\mb{\bm$A$}$ is an M-matrix;
\item[(v)] All leading principal minors of $\mb{\bm$A$}$ are positive;
\item[(vi)] There exists a lower triangular matrix $\mb{\bm$L$}$ and an upper triangular matrix $\mb{\bm$U$}$ both with positive diagonals such that $\mb{\bm$A=LU$}$. Furthermore, $l_{ij} \le 0$ for $i>j$, and $u_{kl} \le 0$ for $k<l$;
\item[(vii)] $\mb{\bm$A$}^{-1}$ exists and $\mb{\bm$A$}^{-1} \ge 0$.
\end{itemize}
\end{thm}
{\bf Proof}\quad The implication (i)$ \Rightarrow $(ii) follows from Theorem~\ref{thm:z_mat}. To prove (ii)$ \Rightarrow $(iii), we represent the matrix $\mb{\bm$W=AD$}$ as follows
\[
\mb{\bm$W=AD$}=\left[\begin{array}{cccc}
w_{11} & -w_{12} & \ldots & -w_{1n}\\
-w_{21} & w_{22} & \ldots & -w_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-w_{n1} & -w_{n2} & \ldots & w_{nn}
\end{array} \right],
\]
where $w_{ij} \ge 0$ for $i \ne j$, and $w_{ii} > 0$, as follows from Theorem~\ref{thm:z_mat}. Furthermore
\[
w_{ii} > \sum_{\stackrel{j=1}{j \ne i}}^{n}w_{ij} \quad \mbox{for} \quad i=1,2,\ldots,n.
\]
The above implies that each principal minor matrix of $\mb{\bm$W$}$ is also row diagonal dominant. By the Theorem of Gershgorin, the eigenvalues of $\mb{\bm$W$}$ as well as its principal minor matrices are located on the disks with centers at $w_{ii}$ and radii $r_i < w_{ii}$. Hence they must have real positive parts.

We shall now show that (iii)$ \Rightarrow $(iv) holds. First note that since the determinant of a square matrix is the product of its eigenvalues and the product of complex eigenvalues of a real matrix is positive, it follows that all principal minors of the matrix $\mb{\bm$W$}$ are positive. Next, the matrix $\mb{\bm$W$}$ has the form
\[
\mb{\bm$W=AD$}=\left[\begin{array}{cccc}
a_{11} & -a_{12} & \ldots & -a_{1n}\\
-a_{21} & a_{22} & \ldots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-a_{n1} & -a_{n2} & \ldots & a_{nn}
\end{array} \right]
\left[\begin{array}{cccc}
d_{11} & \ldots & {} & 0\\
{} & d_{22} & {} & {}\\
\vdots & {} & \ddots & \vdots\\
0 & \ldots & {} & d_{nn}
\end{array} \right],
\]
where $a_{ij} \ge 0$ for $i \ne j$, $a_{ii} > 0$, and $d_{ii} > 0$. Thus if the principal minors of $\mb{\bm$W$}$ are positive, the principal minors of $\mb{\bm$A$}$ must be positive, and vice versa. This is because
\[
\mb{\bm$W$}\left[\begin{array}{cccc}
i_1 & i_2 & \ldots & i_k\\
i_1 & i_2 & \ldots & i_k
\end{array} \right]=
\mb{\bm$A$}\left[\begin{array}{cccc}
i_1 & i_2 & \ldots & i_k\\
i_1 & i_2 & \ldots & i_k
\end{array} \right]
\left[\begin{array}{cccc}
d_{i_1 i_1} & \ldots & {} & 0\\
{} & d_{i_2 i_2} & {}& {}\\
{} & {} & \ddots & {}\\
0 & \ldots & {} & d_{i_k i_k}
\end{array}\right],
\]
where
\[
\mb{\bm$W$}\left[\begin{array}{cccc}
i_1 & i_2 & \ldots & i_k\\
i_1 & i_2 & \ldots & i_k
\end{array} \right]
\]
is a matrix formed from $i_1,i_2,\ldots,i_k$ columns and $i_1,i_2,\ldots,i_k$ rows of the matrix $\mb{\bm$W$}$.

The implication (iv)$ \Rightarrow $(v) follows from the observation that the leading principal minors are also principal minors.

The implications (v)$ \Rightarrow \mb{(vi)} \Rightarrow $(vii) follow from Theorem~\ref{thm:m_matlu}.

Finally, to complete the proof we show that (vii)$ \Rightarrow $(i). For this note that if $\mb{\bm$A$}^{-1}$ exists and $\mb{\bm$A$}^{-1} \ge \mb{\bm$0$}$, then
\[
\mb{\bm$x=A$}^{-1}\mb{\bm$e$} \ge \mb{\bm$0$},
\]
where $\mb{\bm$e$}=\left[1,1,\ldots,1\right]^T \in \mb{\Bb R}^n$. Hence
\[
\mb{\bm$Ax=e$} > \mb{\bm$0$}
\]
which implies that (i) is satisfied.\\
\hspace*{5.5in} $\Box$


\section{A Comparison Theorem}
We now state and prove a theorem that allows one to reduce the study of some problems involving differential equations to the study of corresponding problems using differential inequalities instead.

We consider the following differential inequality
\[
\dot{x}(t) \le \omega(t,x(t)),\qquad x(t_0)=x_0,
\]
where $\omega : \mb{\Bb R}^2 \to \mb{\Bb R}$ is a continuous function in a domain $\Delta \subset \mb{\Bb R}^2$. We assume that $\omega$ satisfies the following Lipschitz condition in $\Delta$:
\[
|\omega (t,x) - \omega (t,y)| \le L|x-y|,
\]
where $L\ge 0$ is a Lipschitz constant.\\
Consider now an associated differential equation
\[
\dot{\tilde{x}}(t) = \omega(t,\tilde{x}(t)),\qquad \tilde{x}(t_0)=\tilde{x}_0.
\]
We refer to the above equation as the {\em equation of comparison}.\\
Denote the solution to the equation of comparison by
\[
\tilde{x}(t)=\tilde{x}(t;t_0,\tilde{x}_0).
\]
The following theorem, that we found, for example, in Corduneanu~\cite[page 49]{corduneanu71}, relates the solutions of the differential inequality and the associated differential equation.
\begin{thm}
\label{thm:comparison}
If $x(t)$ is a differentiable solution to the differential inequality and $\tilde{x}(t)$ is the solution to the associated differential equation with the corresponding initial conditions such that
\[
x_0 \le \tilde{x}_0,
\]
then
\[
x(t) \le \tilde{x}(t)
\]
for all $t \ge t_0$.
\end{thm}
{\bf Proof}\\
We prove the theorem by contradiction. We suppose that $x_0 \le \tilde{x}_0$ but $x(t) \le \tilde{x}(t)$ does not hold. Thus there exists $t_1 > t_0$ such that
\[
x(t_1) > \tilde{x}(t_1).
\]
Let
\[
y(t)=x(t)-\tilde{x}(t).
\]
Note that $y(t_0)=x(t_0)-\tilde{x}(t_0) \le 0$, and $y(t_1)=x(t_1)-\tilde{x}(t_1) > 0$. Denote by $\tau$ the largest $t \in \left[ t_0, \, t_1 \right]$ such that $y(\tau) = 0$. Such a value of $t$ exists by virtue of the continuity of $y$. Thus
\[
y(t) > 0 \quad \mbox{for all} \quad t \in \left( \tau, t_1 \right].
\]
On the other hand, for $t \in \left[ \tau, t_1 \right]$, we have
\begin{eqnarray*}
y(t) &=& x(t)-\dot{\tilde{x}}(t)\\
&\le& \omega (t,x(t)) - \omega (t, \tilde{x}(t))\\
&\le& |\omega (t,x(t)) - \omega (t, \tilde{x}(t))|\\
&\le& L|x(t)-\tilde{x}(t)|.
\end{eqnarray*}
In the interval $\left[ \tau, t_1 \right]$ we have
\[
y(t)=x(t)-\tilde{x}(t) \ge 0.
\]
Hence
\begin{eqnarray*}
\dot{y}(t) &\le& L|x(t)-\tilde{x}(t)|\\
&=& L(x(t)-\tilde{x}(t))\\
&=& Ly(t).
\end{eqnarray*}
Therefore, in the interval $\left[ \tau, t_1 \right]$, the function $y$ satisfies
\[
y(t) \le y( \tau )e^{L(t - \tau )} = 0
\]
since $y(\tau) = 0$. But this is a contradiction to the previous conclusion that $y(t) > 0$ for all $t \in \left( \tau, t_1 \right]$. Hence the proof is complete.\\
\hspace*{5.5in} $\Box$

\section{Lyapunov Analysis of the Hopfield Network}
Recall the equations describing the Hopfield neural net. They are
\[
C_i\frac{du_i}{dt}+\frac{u_i}{r_i}=\sum_{j=1}^{N}T_{ij}x_j + I_i,\quad i=1,2,\ldots,N,
\]
where $x_i=g_i(u_i)$. We assume that the functions $g_i:\mb{\Bb R} \to (-1,1)$, $i=1,2,\ldots,N$, are continuously differentiable and strictly monotonically increasing. We also assume that
\[
u_ig_i(u_i)>0\quad \mbox{for} \quad u_i \ne 0,
\]
and 
\[
g_i(0)=0.
\]
To proceed further we divide both sides of the equations describing the Hopfield network by $C_i$ to get
\[
\dot{u}_i=-b_iu_i+\sum_{j=1}^{N}A_{ij}g_j(u_j) + U_i,\quad i=1,2,\ldots,N,
\]
where $b_i=\frac{1}{r_iC_i}$, $A_{ij}=\frac{T_{ij}}{C_i}$, and $U_i=\frac{I_i}{C_i}$.\\
We are interested in the behavior of solutions of the above model near its equilibrium points. If $U_i=0$, the equilibrium points are obtained by solving the set of algebraic equations
\[
-b_iu_i+\sum_{j=1}^{N}A_{ij}g_j(u_j)=0,\quad i=1,2,\dots,N,
\]
for $\mb{\bm$u$}=\left[u_1,u_2,\ldots,u_N \right]^T \in \mb{\Bb R}^N$.\\
In some applications the external inputs $U_i$ are held constant, that is, $U_i=c_i$. In this case the equilibrium points are obtained by solving for $u_i$ the set of the following algebraic equations
\[
-b_iu_i+\sum_{j=1}^{N}A_{ij}g_j(u_j) + c_i=0,\quad i=1,2,\dots,N.
\]
However, we know that we can always transform the algebraic equations with non-zero constant external inputs into the algebraic equations where the external inputs are set to zero. Indeed, let
\[
u_i=y_i+\frac{c_i}{b_i},
\]
and
\[
G_i(y_i)=g_i\left(y_i+\frac{c_i}{b_i}\right).
\]
Substituting the above into the algebraic equations in which the $c_i$ terms are present yields
\[
\dot{y}_i=-b_iy_i+\sum_{j=1}^{N}A_{ij}G_j(y_j),\quad i=1,2,\ldots,N.
\]
Observe that the right side of the above differential equations vanishes at the points $\mb{\bm$y$}$ where
\begin{eqnarray*}
&{}&-b_iy_i+\sum_{j=1}^{N}A_{ij}G_j(y_j)\\
&=&-b_iy_i+\sum_{j=1}^{N}A_{ij}g_j(y_j + \frac{c_j}{b_j})=0,\quad i=1,2,\ldots,N.
\end{eqnarray*}
Thus, after the transformation, the algebraic equations with external constant inputs used for computing the equilibrium points assume the same form as the algebraic equations corresponding to zero external inputs. Hence, without loss of generality we can assume that the external inputs are zero.

Throughout this section, we assume that an isolated equilibrium point of interest is located at the origin of the state-space $\mb{\Bb R}^N$. If this is not the case then by an appropriate change of variables we can arrange for the equilibrium point of interest to be translated to the origin of $\mb{\Bb R}^N$. Indeed, suppose $\mb{\bm$u$}^{*} \ne \mb{\bm$0$}$ is an isolated equilibrium point of interest. Let for $i=1,2,\ldots,N$
\begin{eqnarray*}
x_i &=& u_i-u^{*}_i\\
h_i(u_i) &=& h_i(x_i + u^{*}_i)=g_i(u_i)-g_i(u^{*}_i).
\end{eqnarray*}
Substituting the above into $\dot{u}_i=-b_iu_i+\sum_{j=1}^{N}A_{ij}g_j(u_j) + U_i$ yields
\begin{eqnarray*}
\dot{x}_i &=&-b_ix_i - b_iu_i^{*} + \sum_{j=1}^{N}A_{ij}\left(h_j(x_j+u^{*}_j)+g_j(u^{*}_j) \right)+U_i\\
&=&-b_ix_i+\sum_{j=1}^{N}A_{ij}h_j(x_j+u^{*}_j)+\left\{-b_iu^{*}_i+\sum_{j=1}^{N}A_{ij}g_j(u^{*}_j)\right\} + U_i.
\end{eqnarray*}
Note that, by the definition of the equilibrium point, the term in the curly bracket vanishes. Thus we have
\[
\dot{x}_i =-b_ix_i+ \sum_{j=1}^{N}A_{ij}h_j(x_j+u^{*}_j)+U_i.
\]
Let
\[
G_j(x_j)=h_j(x_j+u^{*}_j)=g_j(x_j+u^{*}_j)-g_j(u^{*}_j).
\]
Then the above differential equations can be written as
\[
\dot{x}_i =-b_ix_i+ \sum_{j=1}^{N}A_{ij}G_j(x_j)+U_i, \quad i=1,2,\ldots,N,
\]
where $G_j:\mb{\Bb R} \to (c_1,c_2) \subseteq (-2,2)$, $c_1 < 0 < c_2$, $G_j(0)=0$, $G_j$ is continuously differentiable and strictly monotonically increasing in $x_j$. Furthermore $x_jG_j(x_j) >0$ for all $x_j \ne 0$. Observe that the above system of differential equations has an equilibrium point at the origin of $\mb{\bm$x$}^{*}=\mb{\bm$0$}$, $\mb{\bm$U$}^{*}=\mb{\bm$0$}$. We assume that there is no other equilibrium point in a neighborhood of the origin, that is, $\mb{\bm$x^{*}=0$}$ is an isolated equilibrium state of the above system corresponding to $\mb{\bm$U$}=\mb{\bm$0$}$.\\
We now represent the above model of the Hopfield neural network in the form
\[
\dot{x}_i =-b_ix_i+A_{ii}G_i(x_i)+ \sum_{\stackrel{j=1}{j \ne i}}^{N}A_{ij}G_j(x_j)+U_i, \quad i=1,2,\ldots,N,
\]
We shall view the terms
\[
\sum_{\stackrel{j=1}{j \ne i}}^{N}A_{ij}G_j(x_j)
\]
as the interconnections between the $i$-th neuron and the remaining neurons in the network. We shall refer to the equations
\[
\dot{x}_i =-b_ix_i+A_{ii}G_i(x_i)+U_i,\quad i=1,2,\ldots,N
\]
as models of free isolated neurons. In what follows we analyze the stability of an isolated, single, neuron.

\subsection{Single Neuron Stability Analysis}
A single neuron in the Hopfield network, when the external input is set to zero, is modeled by a first order differential equation
\[
\dot{x}_i =-b_ix_i+A_{ii}G_i(x_i).
\]
Assume that for some $r_i>0$ there exist constants $\sigma _{i1} > 0$ and $\sigma _{i2} > 0$ such that for all $x_i \in (-r_i,r_i)$, $x_i \ne 0$, we have
\[
\sigma_{i1}x_i^2 < x_i G_i (x_i) < \sigma_{i2} x_{i2}^2,
\]
or equivalently
\[
\sigma_{i1} < \frac{G_i (x_i)}{x_i} < \sigma_{i2}.
\]
Consider now, as in Michel et al.\ ~\cite{michel89}, the following positive definite Lyapunov function candidate
\[
V_i(x_i)=\frac{1}{2} x_{i}^2.
\]
Its time derivative evaluated along the trajectories of the single neuron model is
\begin{eqnarray*}
\dot{V}_i (x_i(t)) &=& x_i \dot{x}_i\\
&=& x_i \left( -b_ix_i+A_{ii}G_i(x_i) \right).
\end{eqnarray*}
If there exists an $r_i > 0$ such that for all $x_i \in (-r_i,r_i)$
\[
\left\{ \begin{array}{ccc}
-b_ix_i+A_{ii}G_i(x_i)>0 & \mbox{if} & x_i < 0\\
-b_ix_i+A_{ii}G_i(x_i)=0 & \mbox{if} & x_i=0\\
-b_ix_i+A_{ii}G_i(x_i)<0 & \mbox{if} & x_i > 0,
\end{array} \right.
\]
then for $x_i \ne 0$
\[
\dot{V}_i(x_i(t)) < 0
\]
and $\dot{V}_i(0)=0$. Hence the null solution $x_i=0$ is uniformly asymptotically stable. Figure~\ref{fig:st_nen} illustrates this case.
\begin{figure}
\vspace{3in}
\hskip1in\special{bmp:c:/mybook/st_nen.bmp x=4in y=3in}
\caption{Characteristics of a single neuron model whose null solution is uniformly asymptotically stable.}
\label{fig:st_nen}
\end{figure}
Now, if there exists an $r_i > 0$ such that for $x_i \in (-r_i,r_i)$ we have
\[
\left\{ \begin{array}{ccc}
-b_ix_i+A_{ii}G_i(x_i)<0 & \mbox{if} & x_i < 0\\
-b_ix_i+A_{ii}G_i(x_i)=0 & \mbox{if} & x_i=0\\
-b_ix_i+A_{ii}G_i(x_i)>0 & \mbox{if} & x_i > 0,
\end{array} \right.
\]
then in addition to $V_i$ being positive definite,
\[
\dot{V}_i(x_i(t)) < 0
\]
for $x_i \ne 0$, and $\dot{V}_i(0)=0$. Thus the null solution $x_i=0$ is unstable. Figure~\ref{fig:ust_nen} illustrates the case when the null solution of a single neuron is unstable.
\begin{figure}
\vspace{3in}
\hskip1in\special{bmp:c:/mybook/ust_nen.bmp x=4in y=3in}
\caption{Characteristics of single neuron model whose null solution is unstable.}
\label{fig:ust_nen}
\end{figure}
If in addition to the sufficient conditions for asymptotic stability, shown above, we also have
\[
-b_i+A_{ii} \delta_i < 0,
\]
where
\[
\delta_i = \left\{ \begin{array}{ccc}
\sigma_{i1} & \mbox{if} & A_{ii} < 0\\
\sigma_{i2} & \mbox{if} & A_{ii} > 0,
\end{array} \right.
\]
then the null solution of the single neuron model is exponentially stable. Indeed, for $x_i \ne 0$ we have
\begin{eqnarray*}
\dot{V}_i(x_i(t)) &=& x_i\dot{x}_i\\
&=&  x_i \left( -b_ix_i+A_{ii}G_i(x_i) \right)\\
&=& \left( -b_i+A_{ii} \frac{G_i(x_i)}{x_i} \right)x_i^2\\
&\le& \left(-b_i + A_{ii} \delta_i \right) x_i^2\\
&=& \left( \frac{-b_i + A_{ii} \delta_i}{2} \right)V_i(x_i(t)).
\end{eqnarray*}
Applying Theorem~\ref{thm:comparison} yields
\begin{eqnarray*}
V_i(x_i(t)) &=& \frac{x_i^2(t)}{2}\\
&\le& \exp \left\{\frac{-b_i + A_{ii} \delta_i}{2}(t-t_0) \right\} V_i(x_i(t_0))\\
&=& \exp \left\{\frac{-b_i + A_{ii} \delta_i}{2}(t-t_0) \right\}\frac{x_i(t_0)}{2}.
\end{eqnarray*}
Hence
\[
|x_i(t)| \le \exp \left\{\frac{-b_i + A_{ii} \delta_i}{4}(t-t_0) \right\}|x_i(t_0)|.
\]
From the above we conclude that the null solution is not only asymptotically stable but even more, it is exponentially stable.

\subsection{Stability Analysis of the Network}
We now investigate the stability of the null solution of the whole neural network. As we mentioned before, without loss of generality, we can assume that the external inputs $U_i$ are set to zero. Thus the neural network model we analyze in this section is
\[
\dot{x}_i=-b_ix_i+\sum_{j=1}^{N}A_{ij}G_j(x_j), \quad i=1,2,\ldots,N.
\]
We impose the following assumptions on the interconnections:\\
{\bf A1}: There exist positive constants $r_i$, $r_j$, and $a_{ij}$ such that for $|x_i| < r_i$ and $|x_j| < r_j$, $i,j=1,2,\ldots,N$, we have
\[
x_iA_{ij}G_j(x_j) \le |x_i|a_{ij}|x_j|.
\]
{\bf A2}: There exists a vector $\mb{\bm$\alpha$}=\left[\alpha_1, \alpha_2,\ldots,\alpha_N \right]^T$, where $\alpha_i >0$ for $i=1,2,\ldots,N$ such that the matrix $\mb{\bm$S$}\in \mb{\Bb R}^{N \times N}$ is negative definite, where the elements of the matrix $\mb{\bm$S$}$ are defined as
\[
s_{ij}=\left\{\begin{array}{ccc}
\alpha_i \left(-b_i + a_{ii} \right) & \mbox{if} & i=j\\
\frac{1}{2}\left(\alpha_i a_{ij} + \alpha_j a_{ji}\right) & \mb{if}& i \ne j.
\end{array}\right.
\]
We now state and prove a theorem concerned with the exponential stability of the neural network model. This Theorem can be found in Michel et al.\ ~\cite{michel89}.
\begin{thm}
The null solution of the neural network model
\[
\dot{x}_i=-b_ix_i+\sum_{j=1}^{N}A_{ij}G_j(x_j),\quad i=1,2,\ldots,N,
\]
is exponentially stable if the assumptions {\bf A1} and {\bf A2} are satisfied.
\end{thm}
{\bf Proof}\quad Consider the following Lyapunov function candidate
\begin{eqnarray*}
V(\mb{\bm$x$}) &=& \frac{1}{2}\sum_{i=1}^{N}\alpha_i x_i^2\\
&=&\frac{1}{2}\mb{\bm$x$}^T \left[\begin{array}{cccc}
\alpha_1 & {} & \ldots& 0\\
{} & \alpha_2 & {} & {}\\
\vdots & {} & \ddots & \vdots\\
0 & {} & \ldots & \alpha_N
\end{array}\right] \mb{\bm$x$},
\end{eqnarray*}
where $\alpha_i > 0$, $i=1,2,\ldots,N$ are defined in assumption {\bf A2}. Note that $V$ is real symmetric positive definite. The time derivative of $V$ evaluated along the trajectories of the neural model is
\begin{eqnarray*}
\dot{V}(\mb{\bm$x$}(t)) &=& \sum_{i=1}^{N} \alpha_i x_i \dot{x}_i\\
&=& \sum_{i=1}^{N} \alpha_i x_i \left( -b_ix_i + \sum_{j=1}^{N} A_{ij} G_j(x_j)\right)\\
&\le& \sum_{i=1}^{N} \alpha_i \left( -b_i x_i^2 + |x_i|\sum_{j=1}^{N}a_{ij}|x_j| \right)\\
&=& |\mb{\bm$x$}|^T \mb{\bm$R$} |\mb{\bm$x$}|,
\end{eqnarray*}
where $|\mb{\bm$x$}|=\left[|x_1|,|x_2|,\ldots,|x_N|\right]^T$, and the matrix $\mb{\bm$R$} \in \mb{\Bb R}^{N \times N}$ has its elements defined as
\[
r_{ij}= \left\{\begin{array}{ccc}
\alpha_i \left( -b_i + a_{ii} \right) & \mb{if} & i=j\\
\alpha_i a_{ij} & \mb{if} & i \ne j.
\end{array}\right.
\]
Note that
\begin{eqnarray*}
|\mb{\bm$x$}|^T \mb{\bm$R$} |\mb{\bm$x$}| &=& |\mb{\bm$x$}|^T \left( \frac{\mb{\bm$R$}+\mb{\bm$R$}^T}{2} \right)|\mb{\bm$x$}|\\
&=& |\mb{\bm$x$}|^T \mb{\bm$S$}|\mb{\bm$x$}|\\
&\le& \lambda_{max}(\mb{\bm$S$})\|\mb{\bm$x$}\|^2,
\end{eqnarray*}
where the matrix $\mb{\bm$S$}$ is defined in {\bf A2}. The last inequality follows from Rayleigh's Principle. By assumption {\bf A2} the matrix $\mb{\bm$S$}$ is real symmetric negative definite. Hence $\lambda_{max}(\mb{\bm$S$})$ is real and negative. Thus
\[
\dot{V}(\mb{\bm$x$}(t)) \le  \lambda_{max}(\mb{\bm$S$})\|\mb{\bm$x$}(t)\|^2 < 0
\]
in a neighborhood of $\mb{\bm$x$}=\mb{\bm$0$}$ of radius
\[
r=\min_{1 \le i \le N}r_i.
\]
Since $V$ is a quadratic Lyapunov function, the null solution is exponentially stable.\\
\hspace*{5.5in} $\Box$

Note that in our stability analysis of the Hopfield neural network we do not require that the parameters $A_{ij}$, and hence the interconnections $T_{ij}$, be symmetric. The symmetry requirement on the interconnections was usually imposed in the early stability analyses of this network.

Note also that a necessary condition for the matrix $\mb{\bm$S$}$ to be negative definite is that its diagonal elements be negative, that is,
\[
-b_i+a_{ii} < 0, \quad i=1,2,\ldots,N.
\]
However this condition is not sufficient. The above necessary condition, though, can shed some light on the influence of the stability properties of the individual neurons on the stability of the whole network. To show how the stability of the individual neurons may affect the stability of the whole network, we set $i=j$ in {\bf A1} to obtain
\[
x_iA_{ii}G_i(x_i) \le a_{ii} |x_i|^2.
\]
Applying the sector condition $\sigma_{i1}x_i^2 < x_i G_i (x_i) < \sigma_{i2} x_{i2}^2$ to the above inequalities yields
\[
a_{ii}=A_{ii} \delta_i,
\]
where
\[
\delta_i=\left\{\begin{array}{ccc}
\sigma_{i2} & \mb{if} & A_{ii} > 0\\
\sigma_{i1} & \mb{if} & A_{ii} < 0
\end{array} \right.
\]
Therefore for $i=1,2,\ldots,N$, we have
\[
-b_i + A_{ii} \delta_i < 0
\]
since $-b_i+a_{ii} < 0$ and $a_{ii}=A_{ii} \delta_i$. Thus {\bf A2} can only be satisfied if the null solution of each individual neuron is exponentially stable.\\
We now present another sufficiency condition for the exponential stability of the Hopfield neural network model. Before we state the condition we need to perform some algebraic manipulations. We first combine the assumption {\bf A2} with the sector condition $\sigma_{i1}x_i^2 < x_i G_i (x_i) < \sigma_{i2} x_{i2}^2$. It follows from the sector condition that for $x_j \ne 0$
\[
\frac{G_j(x_j)}{x_j} < \sigma_{j2}.
\]
Applying the above when $i \ne j$ yields
\begin{eqnarray*}
|x_i A_{ij} G_j(x_j)| &\le& |A_{ij}||x_iG_j(x_j)|\\
&=&|A_{ij}||x_i \frac{G_j(x_j)}{x_j} x_j|\\
&\le& |A_{ij}||x_i||x_j|\, \sigma_{j2}.
\end{eqnarray*}
When $i=j$ we obtain
\[
x_iA_{ii}G_i(x_i) \le \delta_i A_{ii} x_i^2,
\]
where $\delta_i$ is defined as above. We use the results of the above manipulations in the following assumption:\\
{\bf A3:} There exist positive constants $r_i$ and $r_j$ such that for $|x_i| < r_i$ and $|x_j| < r_j$ the interconnections of the neural network model $\dot{x}_i=-b_ix_i+\sum_{j=1}^{N}A_{ij}G_j(x_j)$, $i=1,2,\ldots,N$ satisfy the conditions
\begin{eqnarray*}
x_i A_{ii} G_i(x_i) &\le& \delta_i A_{ii} x_i^2 \quad \mb{for} \quad i=j\\
|x_i A_{ij} G_j (x_j)| &\le& |x_i||A_{ij}||x_j| \sigma_{j2} \quad \mb{for} \quad i \ne j.
\end{eqnarray*}
We shall use another assumption.\\
{\bf A4:} There exists a vector $\mb{\bm$\alpha$}=\left[\alpha_1,\alpha_2,\ldots,\alpha_N \right]^T \in \mb{\Bb R}^N$, $\alpha_i > 0$, such that the matrix $\mb{\bm$S$} \in \mb{\Bb R}^{N \times N}$ is negative definite, where the elements of $\mb{\bm$S$}$ are defined as
\[
s_{ij} = \left\{ \begin{array}{ccc}
\alpha _i (-b_i + \delta _i A_{ii} )& \mb{if} & i=j\\
\frac{1}{2} \left(\alpha _i |A_{ij}| \sigma_{j2} + \alpha _j |A_{ji}| \sigma_{j2} \right) & \mb{if} & i \ne j.
\end{array} \right.
\]
We are now ready to state and prove an alternative sufficiency condition for the Hopfield neural network to be exponentially stable. This theorem can be found in Michel et al.\ ~\cite{michel89}.
\begin{thm}
The null solution of the neural network model
\[ 
\dot{x}_i=-b_ix_i+\sum_{j=1}^{N}A_{ij}G_j(x_j),\quad i=1,2,\ldots,N,
\]
is exponentially stable if the assumptions {\bf A3} and {\bf A4} are satisfied.
\end{thm}
{\bf Proof}\quad Consider the following quadratic positive definite Lyapunov function candidate
\[
V(\mb{\bm$x$})=\frac{1}{2} \sum_{i=1}^{N} \alpha_i x_i^2,
\]
where $\alpha_i > 0$, $i=1,2,\ldots,N$, are defined in {\bf A4}. Taking into account the assumptions {\bf A1} and {\bf A2} we evaluate the time derivative of $V$ along the trajectories of the neural network to get
\begin{eqnarray*}
\dot{V}(\mb{\bm$x$}(t)) &=& \sum_{i=1}^{N} \alpha_i x_i \left(-b_ix_i+\sum_{j=1}^{N}A_{ij}G_j(x_j) \right)\\
&=& \sum_{i=1}^{N} \alpha_i x_i \left(-b_i x_i+ A_{ii} G_i(x_i) + \sum_{\stackrel{j=1}{j \ne i}}^{N}A_{ij} G_j(x_j) \right)\\
&\le& \sum_{i=1}^{N} \alpha_i \left( \left(-b_i + \delta_i A_{ii} \right) x_i^2 +\sum_{\stackrel{j=1}{j \ne i}}^{N} |A_{ij}||x_i||x_j| \, \sigma_{j2} \right)\\
&=& \frac{1}{2}|\mb{\bm$x$}|^T\mb{\bm$S$}|\mb{\bm$x$}|\\
&<& 0 \quad \mb{for} \quad \mb{\bm$x$} \ne \mb{\bm$0$}.
\end{eqnarray*}
Thus the null solution of the neural network model is exponentially stable since $V$ is a quadratic Lyapunov function.\\
\hspace*{5.5in} $\Box$

\section{M-Matrices in Stability Analysis}
An alternative approach to the stability analysis of the Hopfield neural network is by using properties of M-matrices given in Theorem~\ref{thm:m_matsum}. This approach was proposed by Michel at al.\ ~\cite{michel89}. To proceed we need the following assumption:\\
{\bf A5:} The matrix $\mb{\bm$D$} \in \mb{\Bb R}^{N \times N}$ with its elements defined as
\[
d_{ij}=\left\{\begin{array}{ccc}
b_i - \delta_i A_{ii} & \mb{if} & i=j\\
-|A_{ij}| \sigma _{j2} & \mb{if} & i \ne j
\end{array} \right.
\]
is an M-matrix.

We now prove a sufficiency condition, using certain properties of M-matrices, for an equilibrium state of the Hopfield neural network model to be asymptotically stable. Without loss of generality we state and prove the condition for $\b x=\b 0$.
\begin{thm}
\label{thm:stab_m_mat}
The null solution of the neural network model 
\[
\dot{x}_i=-b_ix_i+\sum_{j=1}^{N}A_{ij}G_j(x_j),\quad i=1,2,\ldots,N,
\]
is asymptotically stable if the assumptions {\bf A3} and {\bf A5} are satisfied.
\end{thm}
{\bf Proof}\quad Consider the following positive definite Lyapunov function candidate
\[
V(\mb{\bm$x$}) = \sum_{i=1}^{N} \alpha_i |x_i|,
\]
where $\alpha_i > 0$, for $i=1,2,\ldots,N$, are weighting factors to be specified later in the proof. The time derivative of $V(\mb{\bm$x$}(t))$ along the solutions of the neural network when $x_i \ne 0$ evaluates to
\[
\dot{V}(\mb{\bm$x$}(t))=\sum_{i=1}^{N} \alpha_i (\mb{sign}\,x_i) \dot{x}_i,
\]
where
\[
\mb{sign}\,x_i = \frac{|x_i|}{x_i} = \frac{x_i}{|x_i|} = \left\{ \begin{array}{ccc}
1 & \mb{if} & x_i > 0\\
-1 & \mb{if} & x_i < 0.
\end{array} \right.
\]
Hence
\begin{eqnarray*}
\dot{V} &=& \sum_{i=1}^{N} \alpha_i (\mb{sign}\, x_i) \left\{-b_ix_i+A_{ii}G_i(x_i) + \sum_{\stackrel{j=1}{j \ne i}}^{N}A_{ij}G_j(x_j) \right\}\\
&=& \sum_{i=1}^{N} \alpha_i \left\{-b_i|x_i|+\frac{x_iA_{ii}G_i(x_i)}{|x_i|} + \sum_{\stackrel{j=1}{j \ne i}}^{N}\frac{x_iA_{ij}G_j(x_j)}{|x_i|} \right\}\\
&\le& \sum_{i=1}^{N} \alpha_i \left\{(-b_i +\delta_i A_{ii})|x_i| + \sum_{\stackrel{j=1}{j \ne i}}^{N}|A_{ij}| \sigma_{j2}|x_j|  \right\}\\
&=& -\mb{\bm$\alpha $}^T \mb{\bm$D$}|\mb{\bm$x$}|,
\end{eqnarray*}
where $|\mb{\bm$x$}|=\left[|x_1|,|x_2|,\ldots,|x_N| \right]^T$, and the matrix $\mb{\bm$D$}$ is defined in {\bf A5}. Since $\mb{\bm$D$}$ is an M-matrix, by Theorem~\ref{thm:m_matsum}, $\mb{\bm$D$}^{-1}$ exists and $\mb{\bm$D$}^{-1} \ge \mb{\bm$0$}$. Let
\[
\mb{\bm$y$}^T=\mb{\bm$\alpha$}^T \mb{\bm$D$}.
\]
Then
\[
\mb{\bm$\alpha$}=\left(\mb{\bm$D$}^{-1}\right)^T\mb{\bm$y$}.
\]
If we choose $\mb{\bm$y$}>\mb{\bm$0$}$ then $\mb{\bm$\alpha$}>\mb{\bm$0$}$ and
\[
\dot{V}(\mb{\bm$x$}(t)) \le -\mb{\bm$y$}^T |\mb{\bm$x$}|
\]
is negative definite in some neighborhood of $\mb{\bm$x$}=\mb{\bm$0$}$.\\
The case when one or more $x_i=0$ on an open time interval can be dealt with in a similar fashion as the previous case when all $x_i \ne 0$. In particular, if $x_k=0$ then we consider the following Lyapunov function candidate
\[
V_{(k)}(\mb{\bm$x$})=\sum_{\stackrel{i=1}{i \ne k}}^{N} \alpha_i |x_i|.
\]
The time derivative of $V_{(k)}(\mb{\bm$x$}(t))$ along the solutions of the neural network evaluates to
\[
\dot{V}(\mb{\bm$x$}(t)) \le -\mb{\bm$\alpha$}_{(k)}^T \mb{\bm$D$}_{(k)}| \mb{\bm$x$}_{(k)}|,
\]
where 
\begin{eqnarray*}
\mb{\bm$\alpha$}_{(k)} &=& \left[\alpha_1,\ldots,\alpha_{k-1},\alpha_{k+1},\ldots,\alpha_N \right]^T \in \mb{\Bb R}^{N-1},\\
|\mb{\bm$x$}_{(k)}| &=& \left[ |x_1|,\ldots,|x_{k-1}|,|x_{k+1}|,\ldots,|x_N| \right]^T,
\end{eqnarray*}
and $\mb{\bm$D$}_{(k)} \in \mb{\Bb R}^{(N-1) \times (N-1)}$ is obtained by deleting the $k$th row and the $k$th column from $\mb{\bm$D$}$.\\
Notice that $\mb{\bm$D$}_{(k)}$ is an M-matrix. Hence we can find a vector $\mb{\bm$y$}_{(k)} \in \mb{\Bb R}^{N-1}$, $\mb{\bm$y$}_{(k)} > \mb{\bm$0$}$ that results in
\[
\dot{V}_{(k)}(\mb{\bm$x$}(t)) \le -\mb{\bm$y$}_{(k)}^T |\mb{\bm$x$}_{(k)}| <0
\]
in some neighborhood of the origin of $\mb{\Bb R}^N$.\\
The case when some $x_i=0$ for time instances of measure zero does not pose any problems.\\
In summary, the null solution of the neural network is asymptotically stable.\\
\hspace*{5.5in} $\Box$

We give yet another sufficiency condition, in terms of M-matrices, for asymptotic stability of an equilibrium state of the neural network. This theorem is also due to Michel et al.\ ~\cite{michel89}. This time instead of using the assumption {\bf A5}, we use the following assumption:\\
{\bf A6:} The matrix $\mb{\bm$D$} \in \mb{\Bb R}^{N \times N}$ with its elements defined as
\[
d_{ij}=\left\{\begin{array}{ccc}
\frac{b_i}{\sigma_{i2}} - A_{ii} & \mb{if} & i=j\\
-|A_{ij}| & \mb{if} & i \ne j
\end{array}\right.
\]
ia an M-matrix.
\begin{thm}
\label{thm:a3_a6}
The null solution of the neural network model 
\[
\dot{x}_i=-b_ix_i+\sum_{j=1}^{N}A_{ij}G_j(x_j),\quad i=1,2,\ldots,N,
\]
is asymptotically stable if the assumptions {\bf A3} and {\bf A6} are satisfied.
\end{thm}
{\bf Proof}\quad We consider a similar Lyapunov function candidate as in the previous Theorem, that is,
\[
V(\mb{\bm$x$}) = \sum_{i=1}^{N} \alpha_i |x_i|,
\]
where $\alpha_i > 0$, $i=1,2,\ldots,N$, are to be determined. The time derivative of $V$ along the solutions of the neural network, when $x_i \ne 0$, $i=1,2,\ldots,N$, evaluates to
\begin{eqnarray*}
\dot{V}(\mb{\bm$x$}(t)) &=& \sum_{i=1}^{N} \alpha_i (\mb{sign}\, x_i) \left\{-b_ix_i+A_{ii}G_i(x_i) + \sum_{\stackrel{j=1}{j \ne i}}^{N}A_{ij}G_j(x_j) \right\}\\
&=& \sum_{i=1}^{N} \alpha_i \left\{-b_i|x_i|+\frac{A_{ii}G_i(x_i)|x_i|}{x_i} + \sum_{\stackrel{j=1}{j \ne i}}^{N}\frac{A_{ij}G_j(x_j)x_i}{|x_i|}\right\}\\
&\le& \sum_{i=1}^{N} \alpha_i \left\{\left(-b_i \frac{x_i}{G_i(x_i)}+A_{ii} \right) \frac{G_i(x_i)}{x_i}|x_i|+ \sum_{\stackrel{j=1}{j \ne i}}^{N}|A_{ij}||G_j(x_j)|  \right\}.
\end{eqnarray*}
From the sector condition $\sigma_{i1} < \frac{G_i (x_i)}{x_i} < \sigma_{i2}$, we obtain
\[
\frac{1}{\sigma_{i1}} > \frac{x_i}{G_i(x_i)} > \frac{1}{\sigma_{i2}}.
\]
Furthermore, since $\sigma_{i1} > 0$  in the sector condition, and we assumed $x_i \ne 0$, for $i=1,2,\ldots,N$, we can write
\[
|G_j(x_j)|=\frac{G_j(x_j)}{x_j}|x_j|.
\]
Taking the above into account yields
\begin{eqnarray*}
\dot{V}(\mb{\bm$x$}(t)) &\le& \sum_{i=1}^{N} \alpha_i \left\{ \left( -\frac{b_i}{\sigma_{i2}} + A_{ii} \right) \frac{G_i(x_i)}{x_i} |x_i| + \sum_{\stackrel{j=1}{j \ne i}}^{N}|A_{ij}|\frac{G_j(x_j)}{x_j}|x_j|\right\}\\
&=& -\left[\alpha_1,\alpha_2,\ldots,\alpha_N \right]^T \b D \left[\begin{array}{c}
\frac{G_1(x_1)}{x_1}|x_1|\\
\frac{G_2(x_2)}{x_2}|x_2|\\
\vdots\\
\frac{G_N(x_N)}{x_N}|x_N|
\end{array}\right]\\
&=& -\mb{\bm$\alpha$}^T \mb{\bm$D$} \mb{\bm$w$},
\end{eqnarray*}
where
\[
\b D=\left[\begin{array}{cccc}
\frac{b_1}{\sigma_{12}}-A_{11} & -|A_{12}| & \ldots & -|A_{1N}|\\
-|A_{21}| & \frac{b_2}{\sigma_{22}}-A_{22} & \ldots & -|A_{2N}|\\
\vdots & \vdots & \ddots & \vdots \\
-|A_{N1}| & -|A_{N2}| & \ldots & \frac{b_N}{\sigma_{N2}}-A_{NN}\\
\end{array}\right].
\]
Note that
\[
\mb{\bm$w$}=\left[\frac{G_1(x_1)}{x_1}|x_1|,\frac{G_2(x_2)}{x_2}|x_2|,\ldots,\frac{G_N(x_N)}{x_N}|x_N| \right]^T > \mb{\bm$0$},
\]
and the matrix $\mb{\bm$D$}$ is an M-matrix. Hence there exists an $\mb{\bm$\alpha$} > \mb{\bm$0$}$ such that $\mb{\bm$y$} = \mb{\bm$\alpha$}^T \mb{\bm$D$}> \mb{\bm$0$}$ and thus
\[
\dot{V}(\mb{\bm$x$}(t)) \le \mb{\bm$y$}^T\mb{\bm$w$}<0.
\]
If some $x_i=0$ then we proceed similarly as in the second part of the proof of the previous Theorem.

In summary, the null solution of the neural network is asymptotically stable.\\
\hspace*{5.5in} $\Box$

\section{Estimating Trajectory Bounds}
Our next task is to estimate trajectories's bounds near an equilibrium point. The bounds obtained can serve as parameters characterizing qualitative behavior of the network. In the subsequent development we do not require the external inputs $U_i$, $i=1,2,\ldots,N$ to be set to zero. Thus the network we analyze is described by
\[
\dot{x}_i =-b_i x_i+ \sum_{j=1}^{N}A_{ij}G_j(x_j)+U_i, \quad i=1,2,\ldots,N.
\]
We use the assumption {\bf A6} that states that the matrix $\mb{\bm$D$}$ is an M-matrix. Since $\mb{\bm$D$}$ is an M-matrix there exist positive constants $\lambda_i$, $i=1,2,\ldots, N$, and $\varepsilon > 0$ such that
\[
\left( \frac{b_i}{\sigma_{i2}} - A_{ii} \right) - \sum_{\stackrel{j=1}{j \ne i}}^{N} \left( \frac{\lambda_j}{\lambda_i} \right) |A_{ij}| \ge \varepsilon .
\]
We also use another assumption:\\
{\bf A7:} For the neural network model $\dot{x}_i =-b_ix_i+ \sum_{j=1}^{N}A_{ij}G_j(x_j)+U_i$, $i=1,2,\ldots,N$, there exists a positive constant $k$ such that for $t \ge 0$
\[
\sum_{i=1}^{N} \lambda_i |U_i(t)| \le k.
\]
We use the following notation. Let
\[
c = \varepsilon \min_{1 \le i \le N} \lambda_i \sigma_{i1},
\]
and
\[
\mb{\bm$\Phi$} (t;t_0,\mb{\bm$x$}_0)=\left[ \Phi_1 (t;t_0,\mb{\bm$x$}_0),\Phi_2 (t;t_0,\mb{\bm$x$}_0),\ldots,\Phi_N (t;t_0,\mb{\bm$x$}_0) \right]^T
\]
denotes the solution to the neural network model subject to the initial condition $\mb{\bm$x$}_0$ at $t_0 \ge 0$.
\begin{thm}
If for $i=1,2,\ldots,N$ we have
\[
\left( \frac{b_i}{\sigma_{i2}} - A_{ii} \right) - \sum_{\stackrel{j=1}{j \ne i}}^{N} \left( \frac{\lambda_j}{\lambda_i}\right) |A_{ij}| \ge \varepsilon
\]
and {\bf A7} are satisfied then
\begin{eqnarray*}
\|\mb{\bm$\Phi$} (t;t_0,\mb{\bm$x$}_0) \| &\stackrel{\rm def}{=}& \sum_{i=1}^{N} \lambda_i |\Phi_i (t;t_0,\mb{\bm$x$}_0)|\\
&\le& \left( \gamma -\frac{k}{c} \right) \exp (-c(t-t_0)) + \frac{k}{c},\quad t \ge t_0 \ge 0,
\end{eqnarray*}
provided
\[
\gamma > \frac{k}{c},
\]
and
\begin{eqnarray*}
\| \mb{\bm$x$}_0 \| &\stackrel{\rm def}{=}& \sum_{i=1}^{N} \lambda_i |x_{i0}|\\
&\le& \gamma.
\end{eqnarray*}
\end{thm}
{\bf Proof}\quad Consider the following positive definite Lyapunov function candidate
\[
V(\mb{\bm$x$})=\sum_{i=1}^{N} \lambda_i |x_i|,
\]
where $x_i \ne 0$, $i=1,2,\ldots, N$. We evaluate the time derivative of $V(\mb{\bm$x$}(t))$ along the solutions of the neural network model in a similar fashion as in the proof of the previous theorem to get
\[
\dot{V}(\mb{\bm$x$}(t)) \le -\mb{\bm$\lambda$}^T \mb{\bm$D$}\mb{\bm$w$} + \sum_{i=1}^{N}\lambda_i |U_i(t)|.
\]
Note that $\mb{\bm$w$} > \mb{\bm$0$}$. It follows from the assumptions that for $i=1,2,\ldots,N$,
\[
\mb{\bm$\lambda$}^T \left[\begin{array}{c}
-|A_{1i}|\\
\vdots\\
\frac{b_i}{\sigma_{i2}} - A_{ii}\\
\vdots\\
-|A_{Ni}|
\end{array}\right] \frac{G_i(x_i)}{x_i}|x_i| \ge \lambda_i \, \varepsilon \,\sigma_{i1} |x_i|.
\]
Hence, using the notation introduced before we stated the theorem, we obtain
\[
\dot{V} \le -cV + u,
\]
where $u=\sum_{i=1}^{N}\lambda_i |U_i(t)|$. Applying now Theorem~\ref{thm:comparison} yields
\begin{eqnarray*}
\| \mb{\bm$\Phi$}(t;t_0,\mb{\bm$x$}_0) \| &\stackrel{\rm def}{=}& \sum_{i=1}^{N} \lambda_i |x_i|\\
&=& \sum_{i=1}^{N} \lambda_i |\Phi_i (t;t_0,\mb{\bm$x$}_0)|\\
&\le& \left(\gamma - \frac{k}{c}\right) \exp(-c(t-t_0) +\frac{k}{c}
\end{eqnarray*}
provided
\[
\| \mb{\bm$\Phi$}(t_0;t_0,\mb{\bm$x$}_0) \| = \sum_{i=1}^{N} \lambda_i |x_{i0}| \le \gamma,
\]
and
\[
\gamma > \frac{k}{c}.
\]
The case when some $x_i=0$ can be dealt with in similar fashion as in the proof of Theorem~\ref{thm:stab_m_mat}.\\
\hspace*{5.5in} $\Box$

\section{Instability of the Hopfield Network}
Of course not all of the equilibrium states of the Hopfield neural network model have to be stable. We mow present a sufficiency condition for an equilibrium state to be unstable. In the forthcoming discussion we use the following assumption:\\
{\bf A8:} The matrix $\mb{\bm$D$} \in \mb{\Bb R}^{N \times N}$ with its elements defined as
\[
d_{ij}=\left\{\begin{array}{ccc}
\sigma_i & \mb{if} & i=j\\
-|A_{ij}| & \mb{if} & i \ne j
\end{array}\right.
\]
is an M-matrix, where
\begin{eqnarray*}
\sigma_i &=& \frac{b_i}{\sigma_{i2}} - A_{ii} \quad \mb{if} \quad i \in F_s,\\
\sigma_i &=& \frac{b_i}{\sigma_{i1}} - A_{ii} \quad \mb{if} \quad i \in F_u,\\
F &=& F_s \bigcup F_u,\\
F &=& \{1,2,\ldots,N\}.
\end{eqnarray*}
By Theorem~\ref{thm:m_matsum}, the matrix $\mb{\bm$D$}$ is an M-matrix if all its principal minors are positive. Thus for {\bf A8} to hold we have to have
\[
\begin{array}{ccc}
\frac{b_i}{\sigma_{i2}} - A_{ii} > 0 & \mb{if} & i \in F_s,\\
-\frac{b_i}{\sigma_{i1}} + A_{ii} > 0 & \mb{if} & i \in F_u,
\end{array}
\]
that is, all its first order principal minors must be positive. Observe that if $i \in F_u$ then $b_i/ \sigma_{i1} < A_{ii}$. Since $b_i/ \sigma_{i1} > 0$, hence $A_{ii} > 0$ for $i \in F_u$. One can verify that an individual neuron for which $A_{ii} > 0$ and $b_i/ \sigma_{i1} < A_{ii}$ is unstable. Indeed, let us consider the following Lyapunov function candidate
\[
v_i(x_i)=|x_i|
\]
for the single neuron model $\dot{x}_i =-b_ix_i+A_{ii}G_i(x_i)$. The time derivative of $v_i$ evaluated along the trajectories of the single neuron model, when $x_i \ne 0$, is
\begin{eqnarray*}
\dot{v}_i &=& \frac{x_i}{|x_i|}\dot{x}_i\\
&=& \frac{x_i}{|x_i|}\left( -b_i x_i+A_{ii}G_i(x_i) \right)\\
&\ge& -b_i|x_i| + \frac{1}{|x_i|}\sigma_{i1} A_{ii} x_i^2\\
&=& \sigma_{i1} \left( -\frac{b_i}{\sigma_{i1}} + A_{ii} \right)|x_i|\\
&>& 0.
\end{eqnarray*}
Therefore using the Test for Instability we conclude that the single neuron with the index $i \in F_u$ is unstable. On the other hand, using the same Lyapunov function candidate as above, we can show that the single neurons with the indices $i \in F_s$ are asymptotically stable. We now generalize the above results to the whole network. The following theorem is due to Michel et al.\ ~\cite{michel89}.
\begin{thm}
The null solution of the neural network model 
\[
\dot{x}_i=-b_ix_i+\sum_{j=1}^{N}A_{ij}G_j(x_j),i=1,2,\ldots,N,
\]
is unstable if {\bf A3} and {\bf A8} are satisfied, where $F_u \ne \emptyset$, ($\emptyset$ denotes an empty set)
\end{thm}
{\bf Proof}\quad We use the following Lyapunov function candidate
\[
V(\mb{\bm$x$})=\sum_{i \in F_u} \alpha_i \left( -|x_i| \right)+\sum_{i \in F_s} \alpha_i |x_i|,
\]
where $\alpha_i > 0$, $i=1,2,\ldots,N$, are to be determined. Note that in every neighborhood of $\mb{\bm$x$}=\mb{\bm$0$}$ there is at least one $\mb{\bm$x$}$ such that $V(\mb{\bm$x$}) < 0$. We now evaluate the time derivative of $V$ for $x_i \ne 0$, $i=1,2,\ldots,N$, along the solutions of the neural network model to get
\begin{eqnarray*}
\dot{V}(\mb{\bm$x$}(t)) &=& -\sum_{i \in F_u} \alpha_i \mb{sign}(x_i)\dot{x}_i +\sum_{i \in F_s} \alpha_i \mb{sign}(x_i)\dot{x}_i\\
&=& \sum_{i \in F_u} \alpha_i \left( b_i|x_i| - \frac{A_{ii} G_i(x_i)|x_i|}{x_i} - \sum_{\stackrel{j=1}{j \ne i}}^{N} \frac{A_{ij}G_j(x_j)|x_i|}{x_i} \right)\\
&+& \sum_{i \in F_s} \alpha_i \left(- b_i|x_i| + \frac{A_{ii} G_i(x_i)|x_i|}{x_i} + \sum_{\stackrel{j=1}{j \ne i}}^{N} \frac{A_{ij}G_j(x_j)|x_i|}{x_i} \right)\\
&\le& \sum_{i \in F_u} \alpha_i \left( \left(\frac{b_i}{\sigma_{i1}}-A_{ii} \right) \frac{G_i(x_i)}{x_i}|x_i| + \sum_{\stackrel{j=1}{j \ne i}}^{N}|A_{ij}|\frac{G_j(x_j)}{x_i}|x_i| \right)\\
&-& \sum_{i \in F_s} \alpha_i \left( \left(\frac{b_i}{\sigma_{i2}}-A_{ii} \right) \frac{G_i(x_i)}{x_i}|x_i| + \sum_{\stackrel{j=1}{j \ne i}}^{N}|A_{ij}|\frac{G_j(x_j)}{x_i}|x_i| \right)\\
&=& -\mb{\bm$\alpha$}^T \mb{\bm$D$}\mb{\bm$w$}\\
&<&0
\end{eqnarray*}
for some $\mb{\bm$\alpha$}>\mb{\bm$0$}$. We can find an appropriate $\mb{\bm$\alpha$}$ using the method of the proof of Theorem~\ref{thm:a3_a6}. Using the Test for Instability we conclude that the null solution of the neural network is unstable.\\
\hspace*{5.5in} $\Box$

\section{Summary}
In this chapter we applied the qualitative theory of large scale interconnected dynamic systems to the analysis of the Hopfield neural network model. We viewed the Hopfield neural network as an interconnection of many single neurons. We then expressed the properties of the network in terms of the properties of the individual neurons and interconnections between them. This approach was originally proposed by Michel, Farrell, and Porod~\cite{michel89}. The above presented approach to the analysis of the Hopfield neural network constitutes a useful tool in a the associative memory design toolbox.

\section{Exercises}
\begin{exercise}\rm
Show
\end{exercise}