91Ó°ÊÓ

Consider a quadratic form $q\left(\stackrel{→}{x}\right)=\stackrel{→}{x}·A\stackrel{→}{x}$of 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$\frac{d\stackrel{→}{x}}{dt}=gradq$,where A is gradq

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

Then the eigenvalues of A and B are all negative.

Thus the diagram.

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,$\frac{d\stackrel{→}{x}}{dt}=gradq$where A is gradq

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

Then the eigenvalues of A and B are all negative.

Hence the diagram.

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$\frac{d\stackrel{→}{x}}{dt}=gradq$,where A is gradq.

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

Then the eigenvalues of real parts ofAand B are all negative.