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 $$\dot{V}(x) = 2x$$ ?

• 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. – Kwin van der Veen Aug 29 at 16:13

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 $$t \mapsto \left(\frac{d}{dx} V \right)(x(t)) \frac{d}{dt} x(t) = \left(\frac{d}{dx} V \right)(x(t))\dot x(t)$$.