My book has the given example:
The quadratic system \begin{align*}\ \dot{x}&=\mu x-y+x^2\\ \dot{y}&=x+\mu y+x^2 \end{align*} has a weak focus of multiplicity at the origin for $\mu=0$ since the Liapunov number $\sigma=-3\pi\neq 0$\
I was wondering if someone could help explain to me how they arrive at the $\sigma=-3\pi$?
My book has a formula for the liapunov number, which I am assuming they used for the calculation. It is as follows: $$\sigma=\frac{3\pi}{2b}\{[3(a_{30+b_{03}})+(a_{12}+b_{21})]-\frac{1}{b}[2(a_{20}b_{20}-a_{02}b_{02})-a_{11}(a_{02}+a_{20})+b_{11}(b_{02}+b_{20})]\}$$ Where $$p(x,y)=\sum_{i+j\geq 2}a_{ij}x^iy^j \quad \text{and} \quad q(x,y)=\sum_{i+j\geq 2}b_{ij}x^iy^j$$ But I am having diffuctly understanding how to use this formula?
All of the examples I have seen are similar to this, where they only give the result and not the calculation. Can someone help me understand, how exactly I am supposed to use the formula for the calculation?
I am also interested in the difference of calculating the liapunov number in 2d and 3d?
Any help is appreciated! :)
You have $p(x,y)=x^2$, so $a_{2,0}=1$, all other $a_{i,j}=0$. Likewise $q(x,y)=x^2$ giving $b_{2,0}=0$, the rest zero.
With $b=1$ the formula evaluates directly to $\sigma=-3\pi$, which presumably means that the solutions spiral inwards, as that would be the natural calibration of $σ$.
Empirically $r(t)^{-2}-r_0^{-2}\approx 0.5(t-t_0)$, so the motion is indeed a spiral inwards. I can't imagine what the connection to $\sigma$ would be.