Question

Determine the stability of the system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ -1& -1& -2\end{array}\right)\overrightarrow{\mathbf{x}}$

Short answer

The stability of the system $\frac{d\overrightarrow{x}}{dt}=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ -1& -1& -2\end{array}\right)\overrightarrow{x}$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, $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$here 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 dx→dt=(000001-1-1-2)x→

Consider the stability of the system is$\frac{d\overrightarrow{x}}{dt}=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ -1& -1& -2\end{array}\right)\overrightarrow{x}$

Here A is the real matrix:

$A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& -1& -2\end{array}\right)$

$$\begin{gathered} \frac{d\overrightarrow{x}}{dt}=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& -1& -2\end{array}\right)\overrightarrow{x} \\ A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& -1& -2\end{array}\right) \end{gathered}$$

Find $A-\lambda I$as:

$\begin{array}{rcl}A-\lambda I& =& 0\\ \left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& -1& -1\end{array}\right)-\lambda \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)& =& 0\\ \left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& -1& -1\end{array}\right)-\left(\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right)& =& 0\\ \left(\begin{array}{ccc}-\lambda & 1& 0\\ 0& -\lambda & 1\\ -1& -1& -2-\lambda \end{array}\right)& =& 0\end{array}$

Find characteristics polynomial as:

$\begin{array}{rcl}-\lambda \left[\left(-\lambda \right)\left(-2-\lambda \right)-\left(-1\right)\right]-1\left(0-\left(-1\right)\right)& =& 0\\ -\lambda (2\lambda +{\lambda }^2+1)-\left(1\right)& =& 0\\ -2{\lambda }^2-{\lambda }^3-\lambda -1& =& 0\\ {\lambda }^3+2{\lambda }^2+\lambda +1& =& 0\\ & & \lambda is negative\\ & & \\ & & \end{array}$

The stability of the system is negative for all eigen values.

Hence the system is asymptotically stable.