0
$\begingroup$

I'm having trouble to undertsand a basic example of the use of lyapunov stability. I'm trying to determine if the equilibrium point of a differential equation is stable.

Let a dynamic system defined by a first order differential equation : $\dot{x} = -x - x^3$. We have $x = 0$ for only equilibrium point.

Let a function $V(x) = x^2$ ; strictly positive except on the equilibrium point $x = 0$. The derivative of this function is $\dot{V}(x) = \dfrac{\partial V}{\partial x} \dot{x}$ with $\dot{x} = -x - x^3$.

$\dot{V}(x) = 2x (-x - x^3) = -2x^2 - 2x^4$. Thus, $\dot{V}(x) < 0$ for $x \neq 0$

From the Lyapunov theorem, the state of the system shoud converge to 0 from any initial state.


I don't understand where $\dot{x}$ come from in $\dot{V}(x) = \dfrac{\partial V}{\partial x} \dot{x}$. Why are we doing a partial derivative in this example and why $\dot{x}$ is here ?

Can we say that the derivative of $V(x)$ is $\dfrac{\partial V}{\partial x} \dot{x}$ ? Shouldn't it be only $\dot{V}(x) = 2x$ ?

$\endgroup$
  • $\begingroup$ Technically $x=\pm i$ would also be also be equilibria of the system. However if $x(t_0)$ is real then indeed $x=0$ would be the only equilibrium the system could converge to. $\endgroup$ – Kwin van der Veen Aug 29 at 16:13
2
$\begingroup$

The notation $\dot x$ stands for the function given by $t \mapsto \frac{d}{dt}x(t)$, and this rule extends to functions depending on $x$ and $t$ similarly: $\dot V$ stands for the function given by $t \mapsto \frac{d}{dt} \left( V(x(t)) \right)$, which, by the chain rule, expands to $t \mapsto \left(\frac{d}{dx} V \right)(x(t)) \frac{d}{dt} x(t) = \left(\frac{d}{dx} V \right)(x(t))\dot x(t)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.