Let $ x = [x_1 \ x_2]^T$, $x\in\mathbb{R}^2$ be the state vector of the control system:
$$ \dot{x_1} = -x_1 + x_2\cdot x_1^2 $$ $$ \dot{x_2} = u $$
I want to check whether the function:
$$ V(x) = \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 $$
with derivative:
$$ \nabla{V(x)f(x)} + \nabla{V(x)g(x)}u = \dot{V}(x) = -x_1^2+x_1^3\cdot x_2+x_2\cdot u $$
is a Control Lyapunov Function (CLF).
Let $x \neq 0$, if $\nabla{V(x)g(x)} = 0 \Rightarrow x_2 = 0 $ and $\nabla{V(x)f(x)} = -x_1^2 <0 $ since $x\neq 0$ and $x_2 = 0$. So, $x_1$ can't be $0$.
The second property that needs to be satisfied states: $\forall\epsilon >0, \exists\delta>0$ such that for every $x\in \mathbb{R}^2$ with $0 < |x| < \delta$, $\exists u\in\mathbb{R}, \ |u|<\epsilon$ with $\nabla{V(x)f(x)} + \nabla{V(x)g(x)}u < 0$. This is what I came up with:
Define $\delta=\epsilon>0$ and let $0 < |x| < \delta$.
For $u = -x_1^3-x_2 \Rightarrow \dot{V}(x) = -x_1^2-x_2^2 < 0$.
Norm of $u$ is: $|u| = |x_1^3 + x_2| \leq |x_1|^3 + |x_2| \leq |x|^3 + |x| < |x|^3 + \delta$.
Since $|x|,\delta > 0 \rightarrow 0 < |x| \Rightarrow 0 < |x|^3 \ \text{and} \ |x| < \delta \Rightarrow |x|^3 < \delta^3$
Considering the above: $|u| < |x|^3 + \delta < \delta^3 + \delta$.
By defining $\delta^3 + \delta = \Delta > 0$ (obviously $\Delta = E = \epsilon^3 + \epsilon $), I concluded that:
$$ |u| < \Delta $$
and that $V(x)$ is a Control Lyapunov Function for this system. I am wondering if the procedure I implemented is correct or if I am missing something.