\section{Stability 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.