Direction of the spiral trajectories for coupled differential equation system I am looking at the coupled system $x'=ax+by,$ $y'=cx+dy$ with non-real eigenvalues.
Do I know it correctly that the trajectories are clockwise if $b-c$ is positive and anticlockwise if $b-c$ is negative? I am quoting this claim from http://wwwf.imperial.ac.uk/metric/metric_public/differential_equations/second_order/qualitative_methods_1.html
Can you also please point me to an online resource where I can find the proof of the condition?
 A: Wew can diagonalize the matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \to \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix}$$
Then we work with $(X,Y)$ instead of $(x,y)$. Yhe eigenvalues are gicven by $$m_1,m_2=\frac{1}{2}[(a+d)\pm \sqrt{(a-d)^2+4bc}]$$ Suppose
Case (1): $m_1=1, m_2=2 \implies X=e^t, Y=e^{2t}\implies Y=X^2,$ we get parabolic trajectory.
Next, suppose
Case(2):$m=-1, m_2=-2 \implies X=e^{-t}, Y=e^{-2t}$, we again get get parabolic trajectory $Y=X^2$.
But the difference is that when time increases to $\infty$ in case (1) both $X,Y$ diverge away from origin (0,0). So the arrow goes out and the orbit is unstable. In case (2) the arrow will direct towards origin, orbit will be stable.
When eigenvalues are non-real (Complex) we get spiral trajectories if $\Re (m_1,m_2)>0$, $X,Y$ diverge, we get unstable spiral. Otherwise, we get stable spiral.
One may call unstable spiral as clockwise and stable one as anti-clockwise.
Any way the sign of real part of eigenvalues decides this.
When eigenvalues are real but of opposite sign the origin becomes a saddle point.
You find a good discussion on this in Problems in ODE by M.L. Krasnov et al (MIR Publication)
