Question

Do parts a and d of Exercise 10 for a quadratic form of n variables

Short answer

The system$\frac{d\overrightarrow{x}}{dt}=gradq$is linear by finding a matrix B in terms of the symmetric matrix A such that$gradq=B\overrightarrow{x}$is$2A=B$

The eigenvalues of A and B are negative definite.

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

Show that the systemdxdt=gradq  is linear by finding a matrix B in quadratic form

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

Assume that the system of differential equations as

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

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

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

Hence the solution

Step 3:Explanation for the relationship between the definiteness of the form q and the stability of the system dx→/dt= gradq

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,$d\overrightarrow{x}/dt=gradq$where A is gradq

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

Then the eigenvalues of A and B are all negative.

Thus the solution.