I want to show that the null solution of the system $\frac{dx}{dt} = y$,$\frac{dy}{dt}= f(x)$, $((x(t), y(t)) ∈ \mathbb{R}^2)$ is stable, if the function f(x) satisfies the following
- f is continuously differentiable
- $f(x) = \left\{ \begin{array}{ll} > 0 \mbox{ if } x<0 \\ = 0 \mbox{ if } x = 0 \\ < 0 \mbox{ if } x >0 \end{array} \right. $
I already prove that the the stability of the null solution x(t) = ψ(t) of the system: $$\frac{dx_k}{dt} = \frac{\partial V}{\partial y_k} \mbox{ and } \frac{dy_k}{dt} = -\frac{\partial V}{\partial x_k}$$ For k = 1,...,n, $V=V(x,y) = V(x_1,...,x_n,y_1,...,y_n)$ which satisfies :
- V is a $C^2$-fonction
- V(0, 0) = 0.
- V(x, y) > 0 for (x,y) $\neq$ (0,0) ( or V(x,y) < 0 for (x,y) $\neq$ (0,0))
- ψ(t) = (x(t), y(t)) ≡ (0, 0) is a solution
For that, I proved that V(x,y) or -V(x,y) is a Lyapunov function.
So now I think that I have to use this to prove my question, does anybody can help me?