Consider the system following system of differential equations \begin{equation} \dot{x} = y - x^{3} \\ \dot{y} = -2x -y^{3}(1-y^{2}) \end{equation} Construct a (strict) Lyapunov function of the form $\alpha x^{2} + \beta y^{2}$, where $\alpha$ and $\beta$ are real valued constants, in order to show that the origin is asymptotically stable.
I've made the following attempt at this problem.
Let $L(x, y) = \alpha x^{2} + \beta y^{2}$. Then clearly $L(x, y) > L(0, 0) = 0$, regardless of the values for $\alpha$ and $\beta$ (of course assuming they are non-zero). We continue by computing the derivative $\dot{L}$, \begin{align*} & \dot{L} = \frac{\partial L}{\partial x}\cdot \dot{x} + \frac{\partial L}{\partial y} \cdot \dot{y} \\ & = 2\alpha x(y-x^{3}) + 2\beta y(-2x-y^{3}(1-y^{2})) \\ & = 2\alpha xy - 2\alpha x^{4} - 4\beta xy - 2\beta y^{4} + 2\beta y^{6} \\ & = (2\alpha - 4\beta)xy + 2\beta(-y^{4} + y^{6}) - 2\alpha x^{4} \end{align*}
We want this derivative to be strictly negative, as such I either wanted to let $2\alpha - 4\beta = 0$, or perhaps choose a region around the origin for which the derivative is strictly negative and proceed to find possible values for $\alpha$ and $\beta$.
I've made an attempt at the former, $2\alpha - 4\beta = 0$ implies $2\alpha = 4\beta$. Inserting this into the derivative gives us the following: $\dot{L} = 2\beta(-y^{4} + y^{6}) - 8\alpha x^{4}$.
I'm unsure of the continuation of my attempt from this point, clearly $y^{6} > y^{4}$ except when $y \in [0, 1]$. As such I considered $y$ outside this region, which would mean we'd want a negative value for $\beta$. But this would make $\alpha$ positive and we wouldn't have certainty over the sign of the derivative.
Would anyone be able to help me with this problem, or have any tips on how to continue?
The question is only asking about showing asymptotic stability of the origin and not global asymptotic stability. And your analysis for $\alpha=2\beta>0$ is sufficient for proving this, i.e. $L(x,y)>0\ \forall\ x,y\neq0$ and $\dot{L}(x,y)<0\ \forall\ x,y\neq0$ and $-1<y<1$.
This can also be verified by looking at the stream plot:
It can be noted that the basin of attraction of the origin is not equal to the values of $(x,y)$ for which $\dot{L}(x,y)\leq0$. Instead, a lower bound of the border of the basin of attraction is given by the level set of the Lyapunov function, i.e. $L(x,y)=c$, such that for all points in that set satisfy $\dot{L}(x,y)\leq0$.