I have the system $$\dot{x}=y+2xy\\\dot{y}=-x+x^2-y^2$$ Which is Hamiltonian with $$H(x,y)=\frac{1}{2}x^2-\frac{1}{3}x^3+xy^2+\frac{1}{2}y^2$$
Now I want to plot the phase portrait for the system so I seek to classify the equilibrium points.
We have the following equilibrium points:
$(0,0),(1,0),(-1/2,\sqrt3/2),(-1/2,-\sqrt3/2)$
To classify these we can linearise the system and calculate the eigenvalues of the resulting Jacobian matrix which only depends on the sign of $H_{xy}^2-H_{xx}H_{yy}$ at the equilibrium point for Hamiltonian systems.
Here we get the point $(0,0)$ is a centre of the system, and the rest are saddle points I want to figure out the straight line paths for the saddle points for the $3$ equilibrium points but I'm not sure how to do this, I know we can search for eigenvalues but this seems quite long and maybe not what is expected perhaps there is a more efficient methodology?
I guess this mainly boils down to two questions, is what I have done so far correct? And how do I proceed efficiently to find the straight line paths.
One last point, the example asks me to determine which region of the phase plane are the phase paths periodic which I have absolutely no idea how to proceed with any hints for that part would be great also.
Many thanks.
Here's the phase plane sketched using PPlane:
The paths connecting the three saddle points certainly seem to be straight lines (note that this is in general not the case!). I would say that the best way to check this, is to find a formula for each of these lines, for example of the form $y = a x + b$, and substitute it in the Hamiltonian system to see if it's indeed a solution.
On your second point, you have to think about how orbits associated to periodic solutions would look like in the phase plane. For example, take your favourite periodic solution, like $x(t) = \sin t$, and plot its orbit in the phase plane. What do you notice?