Toronto Math Forum
MAT2442018F => MAT244Lectures & Home Assignments => Topic started by: Victor Ivrii on November 20, 2018, 07:37:04 AM

When considering system
\begin{equation*}
\mathbf{x}'=A\mathbf{x} \qquad \text{with } \
A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}
\end{equation*}
and discovering that it has a repeated root $\mu\ne 0$, but only one eigenvector, we know that $\mu>0$ corresponds to an unstable node, and $\mu<0$ t corresponds to a stable node. Furthermore, on the pictures below a straight line is directed along this single eigenvector. However, how to distinguish between two upper pictures and two lower pictures?
Observe that, $\mu$ is a double root of the equation $\lambda ^2 (a+d)\lambda + ad bc=0$ with the discriminant $D:=(a+d)^24(ad bc))=(ad)^2+4bc$, and we consider the case $D=0\implies bc <0$ (except $a=d$). Thus $b$ and $c$ are not $0$ and have opposite signs.
Then, if $b<0$ (and $c>0$) "rotation" is counterclockwise, and if $b>0$ (and $c<0$) "rotation" is clockwise. I say "rotation" because it is not a real rotation as in the case of complexconjugate roots (http://forum.math.toronto.edu/index.php?topic=1525.0), only a halfturn rotation by $\pm \pi$, but it works! Indeed, if we slightly increase $b$ or $c$, we get $D<0$ and we will have a focal point and the picture needs to be consistent.
If $a=d=\mu$ then either $b=0$ or $c=0$ but not both (otherwise there would be 2 linearly independent eigenvectors) and we use the same sign criteria, looking at $b$ or $c$ which is not $0$.