91Ó°ÊÓ

For a system,$\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$ 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

Consider a quadratic form $q\left(\stackrel{→}{x}\right)=\stackrel{→}{x}.A\stackrel{→}{x}$two variables$x_1   and  x_2$

Assume that the system of differential equations as

$$\begin{gathered} \left|\frac{d{x}_1}{dt}=\frac{∂q}{∂x_1} \\ \frac{d{x}_2}{dt}=\frac{∂q}{∂x_2} \\ \right| \end{gathered}$$

Also it can be termed as$\frac{d\stackrel{→}{x}}{dt}=gradq$ such that $gradq=B\stackrel{→}{x}$

$\begin{array}{rcl}\frac{d\stackrel{→}{x}}{dt}& =& B\stackrel{→}{x}\\ Assume  \frac{d{x}_1}{dt}& =& A\\ and      \frac{d{x}_2}{dt}& =& A\\ Since    \frac{d{x}_1}{dt}+ \frac{d{x}_2}{dt}& =& B\\ A+A& =& B\\ 2A& =& B\\ & & \end{array}$

Hence the solution

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

Similarly here,$d\stackrel{→}{x}/dt=gradq$where A is gradq

The zero state is a stable equilibrium of the system$d\stackrel{→}{x}/dt=gradq$if and only if q is negative definite.

Then the eigenvalues of A and B are all negative.

Thus the solution.