Question

Consider a system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$where A is a$\mathbf{2}\mathbf{\times }\mathbf{2}$matrix with$\mathbf{\text{tr}}\mathit{A}\mathbf{<}\mathbf{0}$. We are told that A has no real eigenvalue. What can you say about the stability of the system

Short answer

The given system $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$is stable.

Step-by-step solution

Step 1:Explanation for the continuous dynamical systems with eigen values p±iq

Consider the linear system$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$, where A is the real $2\times 2$matrix with complex eigenvalues$p\pm iq$and$q\ne 0$

Consider an eigenvector$\overrightarrow{v}+i\overrightarrow{w}$ with eigenvalue$p\pm iq$ . Then:

$\overrightarrow{x}\left(t\right)=e^{pt}S\left(\begin{array}{cc}\cos \left(qt\right)& -\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right)S^{-1}{\overrightarrow{x}}_0$, where$S=\left[\overrightarrow{w} \overrightarrow{v}\right]$

Recall that $S^{-1}{\overrightarrow{x}}_0$is the coordinate vector of $x_0$with respect to the basis $\overrightarrow{w},\overrightarrow{v}$.

Explanation of the stability of a continuous dynamical system

For a system,$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$ where A is the matrix form.

The zero state is an asymptotically stable equilibrium solution if and only if the real parts of all eigen values of A are negative.

Solution for the stability of the system

Consider A has no real eigen value, which means the real part of all the eigen values are negative.

Thus, the system $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$is stable