Trace-determinant plane (stability analysis)

Consider an ODE system described by the matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ . As we have seen, the eigenvalues of this matrix determine the solutions to the system.

\begin{align*} \det(A - I \lambda) &= 0 \\ \lambda^2 - (a+d)\lambda + ad-bc &= 0 \\ \lambda^2 - \mathrm{tr}(A)\lambda + \det(A) &= 0 \\ \frac{\mathrm{tr}(A)}{2} \pm \sqrt{\frac{\mathrm{tr}^2(A)}{4} - \det(A)} &= \lambda. \end{align*}

The discriminant of the quadratic equation formula will determine whether the eigenvalues are real or complex, and will determine the behavior of the system.

If $\mathrm{tr}(a)>0$ then the real part is positive, so the solution is unstable.
When $\det(A) > \mathrm{tr}^2(A)/4$ the eigenvalue is imaginary, so the solution is a spiral.
If $\det(A)<0$ then at least one eigenvalue is positive, so the solution is unstable.

Therefore stable solutions are only found in the top-left quadrant of the Poincaré diagram below.