i have attempted this question and done as much as i possibly could, any help regarding this question would be very helpful and appreciated.
a) show that the second-order differential equation for the van der pol oscillator, that is
$$\frac{d^2x}{dt^2}+\beta (x^2-1) \frac{dx}{dt}+x=0 $$
where $\beta$ is a constant parameter, is equivalent to the following first order system in the plane
$$\frac{dx}{dt}=y, \qquad \frac{dy}{dt}=-x+\beta (1-x^2)y. $$
b) find the unique steady state for the system, and perform linearization around this steady state in order to determine its stability.
c) show in the case $\beta <0$ that $V(x,y)=x^2+y^2$ is a weak liapunov function for the system, in the sense that $\frac{dV}{dt}$ is negative definate. demonstrate also that in the case,
$$W(x,y)=(2+\beta^2)x^2-2\beta xy+2y^2, $$
is a strong liapunov function by showing there exists $\epsilon>0$ such that in the region when $|x|,|y|<\epsilon$, $\frac{dW}{dt}$ is negative definate. what about the case $\beta>0$? can you apply the poincare-bendixson theorem in the latter case?
This is what i have tried to attempt so far:
a) do i have to substitue $\frac{dx}{dt}=y$ here? i have tried this but dont really know how it gets to the equations above..,
b)steady state = $(0,0)$ with associated linear system:
$$\frac{dx}{dt}=y, \qquad \frac{dy}{dt} = -x + \beta y $$
and so
$$A = \begin{pmatrix} 0 & 1 \\ -1 & \beta \end{pmatrix} $$
which gives:
$$\lambda= \frac{\beta \pm \sqrt{(\beta^2-4)}}{2} $$
thus the origin is an unstable spiral if $0<\beta<2$ and an unstable node if $\beta>2$
c) $$\frac{dV}{dt} = xy+y(-x+\beta (1-x^2)y) $$
$$ = 1-x^2$$
hence $\frac{dV}{dt} \leq 0$ if $x^2<1$ and $\beta <0$ and is stable
and $\frac{dV}{dt} \geq 0$ if $x^2<1$ and $\beta >0$ and is unstable
taking $V(x,y)=x^2+y^2$ and the domain $x^2+y^2<1$ then the origin is stable if $\beta <0$ and unstable if $\beta >0$
is this right so far? not sure on how to do the $W(x,y)$ part.
many thanks in advance for any help.