Question

Consider the system $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\left(\begin{array}{cc}0& 1\\ a& b\end{array}\right)\overrightarrow{\mathbf{x}}$where a and b are arbitrary constants for which values of a and b is the zero state a stable equilibrium solution?

Short answer

The value of arbitrary constants a and b in the zero state a stable equilibrium solution is$a=1$ and$b=-1$

Step-by-step solution

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.

Step 2:Solution for the values of k is the zero state a stable equilibrium solution

Consider the system $\frac{d\overrightarrow{x}}{dt}=\left(\begin{array}{cc}0& 1\\ a& b\end{array}\right)\overrightarrow{x}$where a and b are arbitrary constants

Here $A=\left(\begin{array}{cc}0& 1\\ a& b\end{array}\right)$which is the real $2\times 2$matrix.

$\begin{array}{rcl}A-\lambda I& =& 0\\ \left(\begin{array}{cc}0& 1\\ a& b\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)& =& 0\\ \left(\begin{array}{cc}0& 1\\ a& b\end{array}\right)-\left(\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right)& =& 0\\ \left(\begin{array}{cc}-\lambda & 1\\ a& b-\lambda \end{array}\right)& =& 0\\ \left(-\lambda \right)\left(b-\lambda \right)-a& =& 0\\ -\lambda b+{\lambda }^2-a& =& 0\\ {\lambda }^2-b\lambda -a& =& 0\\ & & \\ & & \end{array}$

Here$\lambda$is zero state according to the stability equilibrium condition because it is zero state in the asymptotically stable equilibrium.

So value for$b=-1$and$a=1$ and it is obtained from the zero state condition of asymptotically stable equilibrium

Thus, the value for the zero statesof a stable equilibrium solution is $a=1$and $b=-1$.