Consider the system $$\left\{\begin{matrix} \frac{dx}{dt}=-2x\\ \frac{dy}{dt}=-2y-2z^2\sin y\cos y\\ \frac{dz}{dt}=-2z(\sin y)^2 \end{matrix}\right.$$ I want to show that the equilibrium point is stable.
To do that I want to construct the Lyapunov function.
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I have done the following:
We are looking for the function $v(x,y,z)$ such that $v(x,y,z)>0$ and $v(x,y,z)=0$ iff $x=y=z=0$ and $\frac{dv}{dt}\leq 0$.
$$\frac{dv}{dt}=v_x(-2x)+v_y(-2y-2z^2\sin y \cos y)-2z(\sin y)^2v_z$$
I thought to take $v_x=x, v_y=y, v_z=z$ then we would have $$\frac{dv}{dt}=-2x^2-2y^2-2yz^2\sin y \cos y-2z^2\sin^2 y$$
If we hadn't the term $2yz^2\sin \cos y$, we would have $\frac{dv}{dt}\leq 0$, right?
In this case we can't, can we?
There is no need for a Lyapunov function.
Notice that the origin is stable for the first and third equations (for the third note that $(\sin y)^2>0$ in a neighborhood of zero, outside zero).
So, it remains to consider the second equation: if $z$ is small (and you know that then it will be bounded), then the right-hand side is approximately $-2y+z^22y=(-2+2z^2)y$, since $\sin y\cos y/y\to1$ when $y\to0$.