Phase Plane

Consider a first order LCCODE system containing $n=2$ variables:

$$\dot{\bf x}=\left[\begin{array}{c}\dot{x}_1 \dot{x}_2\end{... ...gin{array}{c}\dot{x}_1 \dot{x}_2\end{array}\right] ={\bf Ax}$$

The general solution is

$${\bf x}=c_1e^{\lambda_1t}{\bf v}_1+c_2e^{\lambda_2t}{\bf v}_2$$

where $\lambda_i$ and ${\bf v}_i$ are the eigenvalue and corresponding eigenvectors of ${\bf A}$ satisfying ${\bf Av}_i=\lambda_i{\bf v}_i$. Specifically, these eigenvalues can be found by solving the following characteristic polynomial:

$$\det(\lambda{\bf I}-{\bf A})=(\lambda-a_{11})(\lambda-a_{22}... ... =\lambda^2-(a_{11}+a_{22})\lambda+a_{11}a_{22}-a_{12}a_{21}=0$$

where $a_{11}+a_{22}=tr({\bf A})$ and $a_{11}a_{22}-a_{12}a_{21}=det({\bf A})$ are the trace and determinant of ${\bf A}$ respectively. Solving the equadratic equation, we get the two roots:

$$\lambda_{1,2}=\frac{1}{2}(a_{11}+a_{22}\pm \sqrt{\Delta}) =\frac{1}{2}(a_{11}+a_{22} \pm j \sqrt{-\Delta})$$

where $\Delta=(a_{11}+a_{22})^2-4(a_{11}a_{22}-a_{12}a_{21})$.

$$\left[\begin{array}{rr}6&2\\ 3&7\end{array}\right],\;\;\; \l... ...],\;\;\; \left[\begin{array}{rr}-5&-2\\ 4&-3\end{array}\right]$$

$$\begin{array}{c\vert\vert c\vert c\vert c\vert c\vert c\vert ... ...3.5\pm j\,4.0 & 0\pm j\,4.0 & -4\pm j\,2.7\\ \hline \end{array}$$

1. $\lambda_1>0,\;\lambda_2 > 0$ unstable
2. $\lambda_1>0,\;\lambda_2 < 0$ saddle point
3. $\lambda_1<0,\;\lambda_2 < 0$ stable
4. $\lambda=a\pm j\,b,\\ ;\;a>0$ unstable, spiral out
5. $\lambda=a\pm j\,b,\\ ;\;a=0$ marginal stable
6. $\lambda=a\pm j\,b,\\ ;\;a<0$ stable, spiral in