= Solution
Put $s=\sqrt b$. The two <fixed points> on the invariant line $x=0$ are $P_+=(0,s)$ and $P_-=(0,-s)$. There are also $Q_\pm=(\pm\sqrt{b-a^2/4},-a/2)$ when $a^2<4b$. At equality they coalesce with $P_-$.
The <Jacobian matrix> is
$$
J(x,y)=\begin{pmatrix}-a-2y&-2x\\2x&2y\end{pmatrix}.
$$
At $P_+$ its <eigenvalues> are $-a-2s$ and $2s$, so it is always a <saddle equilibrium>. At $P_-$ they are $2s-a$ and $-2s$, so it is a <saddle equilibrium> for $a<2s$ and a <stable node> for $a>2s$, with a zero <eigenvalue> at $a=2s$.
At $Q_\pm$ the characteristic equation is $\lambda^2+a\lambda+4b-a^2=0$, giving
$$
\lambda=\frac{-a\pm\sqrt{5a^2-16b}}2.
$$
For $0<a^2<16b/5$ these points are <stable spirals>; for $16b/5<a^2<4b$ they are <stable nodes>. At $a^2=16b/5$ the double negative <eigenvalue> has only one independent eigenvector, so these are improper <stable nodes>. When $a=0$ they are <center equilibria>, as established in part (iii). Thus the collision at $\boxed{a^2=4b>0}$ changes both the number and stability of <fixed points>; it is the <bifurcation> classified below.
Back to article page
