what to do with a system with a repeated zero eigenvalue, where you cannot impose a condition on one of the components of the generalised eigenvector?

Question

I have the non-linear system $$ \frac{dx}{dt}=ax-bxy \\ \frac{dy}{dt}=cx $$ This has infinitely many steady states of form $x^*=(0,y)$, but my problem is with steady state $x*=(0.\frac{a}{b})$ I linearised the system using the Jacobian to give $$ j= \begin{pmatrix} a-by & 0 \\ c & 0 \end{pmatrix} $$ which evaluated at the steady state of interest yields $$ j= \begin{pmatrix} 0 & 0 \\ c & 0 \end{pmatrix} $$ Evidently we have a repeated eigenvalue of 0, with an eigenvector of (0,1). My confusion comes when you search for the generalised eigenvector, which I find to be of form $$ \begin{pmatrix} \frac{1}{c} \\ x_2 \end{pmatrix} $$ where $x_2\in\mathbb{R}$. I don't really understand the consequences of this are, so any insight to what this particular steady state would be like, or what having infinitely many generalised eigenvectors with different slopes means would be greatly appreciated.

Answer 1

It turns out that it doesn’t matter what you pick for $x_2$. The general solution to the equation is $$C_1\pmatrix{0\\1} + C_2\left[t\pmatrix{0\\1}+\pmatrix{1/c\\x_2}\right] = \pmatrix{C_2/c\\C_2t+(C_1+C_2x_2)}.$$ Since $C_1$ and $C_2$ are arbitrary, we can adjust $C_1$ to produce any desired value for the constant term in the second component of the solution, so we may as well replace that entire expression by the arbitrary constant $C_1$. Assuming that $c\ne0$, we can also multiply through by $c$ (again, since the other two constants are arbitrary) and after relabeling the constants, we have the somewhat nicer-looking $$\pmatrix{\alpha\\\alpha ct+\beta} = \pmatrix{0\\\alpha ct}+\pmatrix{\alpha\\\beta}.$$ The trajectories are lines parallel to the $x_2$ axis, or, when $\alpha=0$, a single point on that axis. In general, when a $2\times2$ system has $0$ for a defective eigenvalue, the trajectories are lines parallel to the lone eigenvector.