0
$\begingroup$

I am figuring out (trying) to find an explicit topological conjugation between two ODEs.

$$\frac{dx}{dt} = -x \quad (\text{flow: } e^{-t})$$

and

$$\frac{dx}{dt} = Ax, \qquad A = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}$$ (a $2 \times 2$ matrix with hyperbolic eigenvalues)

My question at hand is it is easy to find a homeomorphism between two matrices, but a matrix and a non-matrix has me a bit confused at the moment. I got stuck after finding the matrix exponential of the $2 \times 2$ matrix because I do not know what to do afterwards...

$\endgroup$
  • $\begingroup$ I know very little about this subject, but I am assuming $x$ is a vector of length 2 here? In which case, $-x = -Ix$ where I is the identity matrix, so you really do have two matrices. $\endgroup$ – Steven Gubkin Feb 5 '14 at 20:44
  • $\begingroup$ Got something from the answer below? $\endgroup$ – Did Mar 17 '14 at 8:17
3
$\begingroup$

Assuming that you agree with the comment of @Steven Gubkin, which has to be ... OK, and maybe this is known to the OP ... but here it goes.

First note that by a similarity matrix you can transform

$$ \dot y=\begin{bmatrix} -2 & 1\\ 1 & -2 \end{bmatrix}y $$ to $$ \dot z=Bz=\begin{bmatrix} -3 & 0\\ 0 & -1 \end{bmatrix}z. $$

The previous transformation is done by defining $z=Cy$ with $C=\begin{bmatrix} -1 & 1\\ 1 & 1\end{bmatrix}$, and noting that $CAC^{-1}=B$.

Next you already have the second equation as you want it, so $x_2=z_2$. For the first coordinate just define $x_1=z_1^{1/3}$ and perform the required computations.

Side note: It is important to note that you cannot require a smooth conjugacy, as in that case you'd need that the spectrum of both matrices are the same.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.