<p>The fact that it's a Lyapunov function means that
<span class="math-container">$$
\frac{dL}{dt} = \frac{\partial L}{\partial x} \cdot \frac{dx}{dt} \leq 0
$$</span>
If <span class="math-container">$I-W$</span> is positive semidefinite, <span class="math-container">$L$</span> is bounded from below by <span class="math-container">$-\frac{1}{2} b^T (I-W)^{-1} b $</span>.
This means that the network will end up in a minimum of <span class="math-container">$L$</span> and cannot keep moving for ever.</p>

CC BY-SA 4.0
2019-04-11T08:25:05.437Z
2019-04-11T08:25:05.437Z
Subhaneil Lahiri
<p>For a given Hopfeild network : 
$\tau\frac{dx}{dt} = -x + f(Wx + b) $</p>

<p>a general Lyapunov stability function for Hopfeild networks is :</p>

<p>$L(x) = -\frac{1}{2}x^TWx + b^Tx + F(x)$  </p>

<p>where $F(x) = \int^{x} f^{-1}(y)dy$</p>

<p>In the case of linear networks, $f(x) = x$, the Lyapunov function becomes: </p>

<p>$L(x) = \frac{1}{2}x^T(I - W)x + b^Tx$ </p>

<p>It is asserted that if $(I-W)$ is positive semi definite, then the $L(x)$ is the Lyapunov function and the system is stable. 
Why is that a sufficient condition?</p>

CC BY-SA 3.0
2017-11-05T06:03:45.587Z
can you link the study you are reading from, preferably with a direct quote?
honi
<theoretical-neuroscience>
Why is positive semi definiteness a sufficient condition for Lyapunov stability of the Hopfield networks?
Anonymous
# Why is positive semi definiteness a sufficient condition for Lyapunov stability of the Hopfield networks?

(Asked by: username_0 on 2017-11-05)

For a given Hopfeild network :
$\tau\frac{dx}{dt} = -x + f(Wx + b) $

a general Lyapunov stability function for Hopfeild networks is :

$L(x) = -\frac{1}{2}x^TWx + b^Tx + F(x)$

where $F(x) = \int^{x} f^{-1}(y)dy$

In the case of linear networks, $f(x) = x$, the Lyapunov function becomes:

$L(x) = \frac{1}{2}x^T(I - W)x + b^Tx$

It is asserted that if $(I-W)$ is positive semi definite, then the $L(x)$ is the Lyapunov function and the system is stable.
Why is that a sufficient condition?

**Tags:** theoretical-neuroscience

> **Comments:**
> **username_1**: can you link the study you are reading from, preferably with a direct quote?  

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## 1 answers

### Answer by username_2 on 2019-04-11. Score: 1

The fact that it's a Lyapunov function means that
$$
\frac{dL}{dt} = \frac{\partial L}{\partial x} \cdot \frac{dx}{dt} \leq 0
$$
If $I-W$ is positive semidefinite, $L$ is bounded from below by $-\frac{1}{2} b^T (I-W)^{-1} b $.
This means that the network will end up in a minimum of $L$ and cannot keep moving for ever.


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