Let $a>0 $. Show that $V(x, y)=x^{2}+2 y^{2}$ is a strict Lyapunov function for the system $$ x^{\prime}(t)=a y^{2}-x, \quad y^{\prime}(t)=-y-a x^{2} $$
The function is clearly continuous, positive everywhere except at the origin. We need to show that $V'(x, y) \lt 0$.
We have: $$ \begin{aligned} \nabla V &=\langle 2 x, 4 y\rangle \\ \bf{f'} &=\left\langle a y^{2}-x,-y-a x^{2}\right\rangle \end{aligned} $$
Taking the dot product of these two vectors gives $-2x^2-4y^2-4yax^2+2axy^2$. The notes indicate to look at the region ($\frac{-1}{a}$,$\frac{1}{a}$) $\times$ ($\frac{-1}{4a} $,$\frac{1}{4a}$), but how does one know where the function attains a maximum in this region? Do I just find all critical points and then check boundaries?
Here $V(x,y) = x^2+2y^2$ should be used locally to study the near-origin stability.
We have
$$ \dot V = -(2x^2+4y^2+4ayx^2-2axy^2) $$
$\dot V$ has a stationary point at $(0,0)$ because it is a solution for
$$ \nabla\dot V = \cases{-2 x - 4 a x y + a y^2=0\\ -2 a x^2 - 4 y + 2 a x y =0 } $$
and also is a maximum because the Hessian at $(0,0)$ is
$$ H = \left(\begin{array}{cc} -2 & 0\\ 0 &-4 \end{array}\right) $$
having negative eigenvalues.
Concluding, there is a open set $U(x,y)$ containing the origin such that for $(x,y)\in U(x,y)\Rightarrow\cases{V(x,y) \ge 0\\ \dot V(x,y) \le 0}$ with the equality at $(0,0)$
Attached for $a=1$ the level curves for $V$ first and for $\dot V$ last.