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The task is to find and classify the fixed (equilibrium) points of the system :

$$ \begin{cases} \dot x = (2x - y) (x - 2) \\ \dot y = xy - 2\end{cases}$$

Solving

$$ \begin{cases} \dot x = (2x - y) (x - 2) = 0 \\ \dot y = xy - 2 = 0\end{cases}$$

gives three points $(1,2), (2,1)$ and $(-1,-2)$. I have already classified the first two so no problems with those. However, for $(-1, -2)$, if we shift the origin by changing the variables

$$ \begin{cases} x = p - 1 \\ y = q - 2\end{cases}$$

we have $$ \begin{cases} \dot p = (2x - y) (x - 2) = (2(p - 1) - q + 2 )(p - 3) = -6p + 3q -pq + 2p^2\\ \dot q = xy - 2 =(p-1)(p-2) + 2 = -2p - q + pq\end{cases}$$

Now, to find the linear approximation around the origin, I know that the higher-order terms can be ignored. What about the $pq$ terms? If I ignore them too, then I'll have

$$\begin{cases} \dot p \approx -6p + 3q\\ \dot q \approx -2p - q \end{cases} \implies \lambda_{1,2} = \{-3, -4\}$$

Which means the equilibrium point is a sink or stable non-degenerate node. Is that right?

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    $\begingroup$ Yes that's right $pq$ is also a higher order term. $\endgroup$ May 25, 2020 at 21:56

1 Answer 1

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Indeed, the $pq$ terms can be treated as higher-order terms and the equilibrium point turns out to be a nodal sink.

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