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Consider the system $x'= \frac{-x}{2}; y' = 2y + x^2 $

Show that this system is topologically conjugate to the linear system $\overrightarrow {y'}$ = $DF_{(0,0)}$ $\overrightarrow {y}$

a) Solve both the linear and nonlinear systems and express your answers as a flows $\phi_t^L (x,y)$ and $\phi_t^N (x,y)$ respectively.

b) As I found $\phi_t^L (x_0,y_0)$ = $(x_0 e^{\frac{-1}{2}t}, y_0 e^{2t})$ and $\phi_t^N (x_0,y_0)$ = $(x_0 e^{\frac{-1}{2}t}, (y_0 + \frac{1}{3}x_0^2)e^{2t} - \frac{x_0^2}{3}e^{-t})$ Do they look right to you? Also, I am wondering do I need them to show part a? If so, how? If not, what is the right approach?

c) Find the topological conjugacy that maps the flow of the nonlinear system to that of the linear system.

How should I approach this?

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You are almost done, once you solve both systems (which in principle is not always so easy or even possible) you have all you need to find the conjugacy.

  • I think your b) is the answer of a).

  • The conjugacy is just a map obtained from the flows. The first component is the identity as $x\mapsto x$. The second component is of the form $h(x,y)=\left(1+\frac{1}{3}\frac{x_0^2}{y_0}\right)y-\frac{1}{3}x^2$. You can easily check that $h'=2h+x^2$ as it should be. Of course if you take $h\left(x_0\exp\left( -\frac{1}{2}t\right),y_0\left( -2t\right) \right)$ you'll get the $y$ component of $\phi_t^N$. This conjugacy $h$ maps trajectories from the linear system to the nonlinear systems, however it is smooth and therefore $h^{-1}$ gives the answer to your question c), you just need to solve for $y$ and check that indeed it is the required conjugacy from $\phi_t^L$ to $\phi_t^N$.

-You might also want to check the Hartman-Grobman theorem.

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