Stability Diagram Analysis

Consider the homogeneous two-dimensional linear system
\[ \mathbf{x}'=\mathbf{A}\mathbf{x}\]
where \(\mathbf{A}\) is a constant \(2\times 2\) real-valued matrix. Assume that \( 0\) is not an eigenvalue of \(\mathbf{A}\), i.e. \(\text{det}\mathbf{A}\neq 0\). Then \(\mathbf{0}=(0,0)\) is the only critical point. The characteristic equation of \(\mathbf{A}\) is

\[ \lambda^2-(\text{tr}\mathbf{A})\lambda+\text{det}\mathbf{A}=0.\]

Define the discriminant
\[\Delta=(\text{tr}\mathbf{A})^2-4\text{det}\mathbf{A}.\]
Let \(\lambda_1, \lambda_2\) be the eigenvalues of \(\mathbf{A}\). Then \(\lambda_1, \lambda_2\) are the roots of the characteristic equation of \(\mathbf{A}\), i.e.
\[(\lambda-\lambda_1)(\lambda-\lambda_2)=0.\]
Hence we get
\begin{aligned}
\text{tr}\mathbf{A} &= \lambda_1+\lambda_2 \\
\det\mathbf{A}&= \lambda_1\lambda_2.
\end{aligned}
Also,
\[\lambda_{1,2}=\frac{1}{2}(\text{tr}\mathbf{A}\pm\sqrt{\Delta}).\]
We determine the stability of the critical point by calculating the eigenvalues of \(\mathbf{A}\). From now on, we can directly determine it by observing \(\text{tr}\mathbf{A}\), \(\det\mathbf{A}\) and \(\Delta\).


Case 1. \(\text{det}\mathbf{A}<0\)
It must be real eigenvalues with opposite sign. In this case, \(\mathbf{0}\) is an unstable saddle point.

Case 2. \(\text{det}\mathbf{A}>0\)

• \(\text{tr}\mathbf{A}=0\)
  It must be pure imaginary eigenvalues. In this case, \(\mathbf{0}\) is a stable center.

• \(\text{tr}\mathbf{A}<0\)

It must be two negative real eigenvalues or complex conjugate eigenvalues with negative real part. In this case, \(\mathbf{0}\) is asymptotically stable.

• \(\text{tr}\mathbf{A}>0\) 
It must be two positive real eigenvalues or complex conjugate eigenvalues with positive real part. In this case, \(\mathbf{0}\) is unstable.

Case 3. \(\text{det}\mathbf{A}=0\)

One of eigenvalues must be zero.


Then we can portrait the following stability diagram.

You can check "Stability Diagram 學生作品" for the reference.