Given
$\dot x = x^2-y^4 $ and $ \dot y = y^2 -x^4 $
Find the index of the circle $x^2 + y^2 = a^2$ with $a < 1$
Attempt:
I employed linear analysis by finding all the fixed points. There are five total but the one at interest is the origin (0,0). Finding the trace and determinant of the Jacobian for this system classifies this fixed point as a non-isolated stable node which has an index of +1. Since a circle of a < 1 only contains this fixed point, the index of the circle is equal to +1.
What is another way of reaching the same conclusion? I was thinking to evaluate the index by evaluating the integral over the circle but it looked messy. Any thoughts?
Recall that the index remains unchanged under homotopies of the closed path that avoid the fixed points. In this case, we have $$ (x',y')=(x^2,y^2)-(y^4,x^4), $$ and so on the circle $x^2+y^2=a^2$ with $|a|$ very small the term $(x^2,y^2)$ determines the index, since $$ \lim_{(x,y)\to(0,0)}\frac{\|(y^4,x^4)\|}{\|(x^2,y^2)\|}=0. $$ In other words, for $|a|$ very small, the index is the same as that of $(x',y')=(x^2,y^2)$, namely $0$.