Question

Consider a quadratic form$\mathit{q}\left(\overrightarrow{x}\right)\mathbf{=}\overrightarrow{\mathbf{x}}\mathbf{·}\mathit{A}\overrightarrow{\mathbf{x}}$of two variables. Consider the system of differential equations$$\begin{gathered} \left|\frac{{\mathrm{dx}}_1}{\mathrm{dt}}=\frac{\partial q}{\partial x_1} \\ \frac{{\mathrm{dx}}_2}{\mathrm{dt}}=\frac{\partial q}{\partial x_2}\right| \end{gathered}$$

Or more sufficiently

$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{g}\mathit{r}\mathit{a}\mathit{d}\mathit{q}$

1. Show that the system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{g}\mathit{r}\mathit{a}\mathit{d}\mathit{q}$ is linear by finding a matrix B(in terms of the symmetric matrix A) such that$\mathit{g}\mathit{r}\mathit{a}\mathit{d}\mathit{q}\mathbf{=}\mathit{B}\overrightarrow{\mathbf{x}}$
2. When q is negative definite,draw a sketch showing possible level curves of q. On the same sketch draw a few trajectories of the system localid="1662089983392">dx→dt=gradq.What does your sketch suggest about the stability of the systemlocalid="1662089999708">dx→dt=gradq
3. Do the same as in part b for an indefinite quadratic form
4. Explain the relationship between the definiteness of the form q and the stability of the systemlocalid="1662090018926">dx→dt=gradq

Short answer

1. The system $\frac{d\overrightarrow{x}}{dt}=gradq$is linear by finding a matrixB such that $gradq=B\overrightarrow{x}$is
2. 
4. The zero state is a stable equilibrium of the systemlocalid="1659599650391" $\frac{d\overrightarrow{x}}{dt}=gradq$ if and only if q is negative definite.

Step-by-step solution

(a)Step 1: Show that the system  dx→dt=gradq is linear by finding a matrix B in quadratic form

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

Assume that the system of differential equations as

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

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

(b)Step 2:Explanation for the stability of the system  dx→dt=gradq using sketches in definite quadratic form

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

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

Then the eigenvalues of A and B are all negative.

Thus the diagram.

(c)Step 3: Explanation for the stability of the system  dx→dt=gradq using sketches in indefinite quadratic form

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

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

Then the eigenvalues of A and B are all negative.

Hence the diagram.

(d)Step 4: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$\frac{d\overrightarrow{x}}{dt}=gradq$,where A is gradq.

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

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