 # Consider the system x &#x2032; </msup> = &#x2212;<!-- − --> x pipantasi4 2022-07-03 Answered
Consider the system
${x}^{\prime }=\frac{-x}{2};{y}^{\prime }=2y+{x}^{2}$
Show that this system is topologically conjugate to the linear system $D{F}_{\left(0,0\right)}$
a) Solve both the linear and nonlinear systems and express your answers as a flows ${\varphi }_{t}^{L}\left(x,y\right)$ and ${\varphi }_{t}^{N}\left(x,y\right)$ respectively.
b) As I found ${\varphi }_{t}^{L}\left({x}_{0},{y}_{0}\right)$ = $\left({x}_{0}{e}^{\frac{-1}{2}t},{y}_{0}{e}^{2t}\right)$ and ${\varphi }_{t}^{N}\left({x}_{0},{y}_{0}\right)$ = $\left({x}_{0}{e}^{\frac{-1}{2}t},\left({y}_{0}+\frac{1}{3}{x}_{0}^{2}\right){e}^{2t}-\frac{{x}_{0}^{2}}{3}{e}^{-t}\right)$ Do they look right to you?
c) Find the topological conjugacy that maps the flow of the nonlinear system to that of the linear system.
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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 $x↦x$. The second component is of the form $h\left(x,y\right)=\left(1+\frac{1}{3}\frac{{x}_{0}^{2}}{{y}_{0}}\right)y-\frac{1}{3}{x}^{2}$. You can easily check that ${h}^{\prime }=2h+{x}^{2}$ as it should be. Of course if you take $h\left({x}_{0}\mathrm{exp}\left(-\frac{1}{2}t\right),{y}_{0}\left(-2t\right)\right)$ you'll get the $y$ component of ${\varphi }_{t}^{N}$. 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 ${\varphi }_{t}^{L}$ to ${\varphi }_{t}^{N}$.
- You might also want to check the Hartman-Grobman theorem.