1

I need to plot the phase portrait of the following gradient system, $x'=-\nabla V$.

$$V=x^2+y^2$$

Here are my workings so far,

$$x'=-\nabla V = \begin{bmatrix}-2x \\ -2y \end{bmatrix} = \begin{bmatrix}-2 & 0 \\ 0 & -2\end{bmatrix} \begin{bmatrix}v_1 \\ -v_2 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$$

$\lambda ^2+4\lambda+4=0$
$\lambda=-2,-2$

But how do I find the fixed points and draw the phase portrait?

fr14
  • 717
  • 3
  • 10
  • 23
  • A fixed point is when $\mathbf{x}'=(-2x,-2y)=(0,0)$... – Ian Mar 08 '17 at 20:55
  • One way to draw the phase portraits is to draw the equipotential surfaces and then their gradient lines. In your example, the equipotential surfaces are all circles around the origin. So the phase portrait is simply the straight lines passing through the origin. – polfosol Mar 12 '17 at 13:38

0 Answers0