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.