Consider the two systems of ODEs $$ \begin{cases}\dot{x_1}=-\lambda_1x_1\\\dot{x_2}=-\lambda_2x_2\end{cases}\text{ and }\begin{cases}\dot{y_1}=ay_1+by_2\\\dot{y_2}=-by_1+ay_2\end{cases}. $$ The question is to determine whether these systems are topologically conjugate or not.

I was given the following definition of topologically conjugate dynamical systems:

Two systems $\dot{x}= f(x_1,x_2)$, $\dot{y}= f(y_1,y_2)$ are topologically conjugate if there exists a continuous, invertible function (or "map") $F :(x_1,x_2) -> (y_1,y_2)$ which takes values in $(x_1,x_2)$ and maps them over to $(y_1,y_2)$ coordinates. So in other words, we have a continuously invertible $F$ such that $g\circ F = F\circ f$.

This is my attempt:

Notice that the solution to the ODEs on the left is

The system on the left has solution:

\begin{equation*} {\mathbf x}(t) = c_1 e^{-\lambda_1 t} {\mathbf v}_1 + c_2 e^{-\lambda_2 t} {\mathbf v}_2. \end{equation*}

Here , ${\mathbf v}_1$, ${\mathbf v}_2$ are both eigenvectors of the system on the left.

The one on the right has solution:

\begin{equation*} {\mathbf y}(t) = c_1 e^{\alpha t} \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} + c_2 e^{\alpha t} \begin{pmatrix} \sin \beta t \\ \cos \beta t \end{pmatrix}. \end{equation*}

The question is if we can now explicitly find a map $F$ in $(x_1,x_2)$ such that it changes to $(y_1,y_2)$ coordinates.

So, using the solution the first system, we may write:

\begin{equation*} {\mathbf y}(t) = \begin{pmatrix} \cos \beta t\ & \sin \beta t \\ -\sin \beta t & \cos \beta t \end{pmatrix} \begin{pmatrix} c_1x_1(-\frac{at}{\lambda_1}) \\ c_2x_2(-\frac{at}{\lambda_2}) \end{pmatrix} \end{equation*}

We easily notice that the matrix of transformation is invertible, as determinant is equal to 1. So, using this logic, we can say that there exists an invertible, continuous map such that the two dynamical systems are topologically conjugate.

Is this correct? I feel I am missing something crucial.



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