0
$\begingroup$

You have two solutions of an differential equation with the same eigenvalues. How can you see, that the phase space are lines:

$ y_1=c_1e^{\lambda t} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

$ y_2=c_2e^{\lambda t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

$\endgroup$
  • $\begingroup$ Exactly what solutions do you have? There are two rather different possibilities when you have a repeated eigenvalue. $\endgroup$ – amd Feb 16 at 23:17
  • $\begingroup$ The eigenvalues are greater than 0. $\endgroup$ – SvenMath Feb 16 at 23:21
  • $\begingroup$ That’s not the most interesting condition. A defective eigenvalue generates a very different phase portrait than if both its algebraic and geometric multiplicities are 2. $\endgroup$ – amd Feb 16 at 23:24
  • $\begingroup$ Sorry for that. The algebraic and geometric multiplicities are 2. $\endgroup$ – SvenMath Feb 16 at 23:33
  • $\begingroup$ $\dot y_1=\lambda_1 y_1$ and $\dot y_2=\lambda_2 y_2$ are lines, no? $\endgroup$ – Cosmas Zachos Aug 2 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.