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For a Hamiltonian system of odes on a plane, the eigenvalues of the linearisation matrix of fixed points are of the form $\pm \lambda$ for $\lambda \in \mathbb{R}$, (hyperbolic) or $\pm i \mu$ (elliptic) for $\mu \in \mathbb{R}$, or identically zero (parabolic stability).

Is it possible to have a $\mathbf{non}$-Hamiltonian system on a plane, such that the corresponding eigenvalues of all fixed points are of the above 3 types only? So essentially, as I understand, I am looking for a system of odes with forcing / dissipation (to obtain non-Hamiltonian character), but with only above type of behaviour, which would not exist?

Would appreciate an example of such a Hamiltonian (and more generally, how to construct it), or a prove of the assertion.

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Here's a counterexample for your conjecture: take \begin{align} \dot{x} &= x(1-x),\\ \dot{y} &= y\left(x-\frac{1}{2}\right). \end{align} Both equilibria of the system are saddles, but the system is not Hamiltonian, since $$ \frac{\partial}{\partial x}x(1-x) = 1-2x \neq \frac{\partial}{\partial y}y\left(\frac{1}{2}-x\right) = \frac{1}{2}-x, $$ so the system cannot be of the form \begin{align} \dot{x} &= \frac{\partial H}{\partial y},\\ \dot{y} &= -\frac{\partial H}{\partial x}. \end{align}

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