# Classifaction of equilibrium points for a Hamiltonian System

## Homework Statement

For the system:
$$\frac{dx}{dt}=x\cos{xy} \: \: \frac{dy}{dt}=-y\cos{xy}$$
(a) is Hamiltonian with the function:
$$H(x,y)=\sin{xy}$$
(b) Sketch the level sets of H, and
(c) sketch the phase portrait of the system. Include a description of all equilibrium points and any saddle connections.

## The Attempt at a Solution

$$\frac{\partial H}{\partial y}=y\cos{xy}=-g \\ \frac{\partial H}{\partial x}=x\cos{xy}=f$$
So the function is Hamiltonian. I see that the equilibrium points are (0,0) and (±√π/2,±√π/2) by inspection. The problem I have is that the second set of equilibria have complex roots, but I don't see any of that behavior when I graph the phase portrait with pplane.

## Answers and Replies

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Mark44
Mentor

## Homework Statement

For the system:
$$\frac{dx}{dt}=x\cos{xy} \: \: \frac{dy}{dt}=-y\cos{xy}$$
(a) is Hamiltonian with the function:
$$H(x,y)=\sin{xy}$$
(b) Sketch the level sets of H, and
(c) sketch the phase portrait of the system. Include a description of all equilibrium points and any saddle connections.

## The Attempt at a Solution

$$\frac{\partial H}{\partial y}=y\cos{xy}=-g \\ \frac{\partial H}{\partial x}=x\cos{xy}=f$$
So the function is Hamiltonian. I see that the equilibrium points are (0,0) and (±√π/2,±√π/2) by inspection.
The equation $\cos(xy) = 0$ has many solutions of the form $xy = (2k + 1)\frac{\pi}2$, with k an integer. The point you show is just one equilibrium point.
rmiller70015 said:
The problem I have is that the second set of equilibria have complex roots
How are you getting this (the complex roots)?
rmiller70015 said:
, but I don't see any of that behavior when I graph the phase portrait with pplane.

I didn't approximate so I may have made a mistake typing it into my calculator. I'll approximate and see if I get something else. Maybe that's the problem.

Mark44
Mentor
I didn't approximate so I may have made a mistake typing it into my calculator.
I used plain old algebra, so I don't see how using approximations would be helpful.
rmiller70015 said:
I'll approximate and see if I get something else. Maybe that's the problem.

I get a Jacobian matrix of:
$$\begin{bmatrix} \cos{xy}-y\sin{xy} & -x^2\sin{xy} \\ xy\sin{xy} & xy\sin{xy}-\cos{xy} \end{bmatrix}$$
So then my matrix for that second point becomes:
$$\begin{bmatrix} -\sqrt{\frac{\pi}{2}} & -\frac{\pi}{2} \\ \frac{\pi}{2} & \frac{\pi}{2} \end{bmatrix}$$
Then using the trace and determinant I get:
$$\lambda^2-\Bigg(\frac{\pi}{2}-\sqrt{\frac{\pi}{2}}\Bigg)\lambda-\Bigg(\Big(\frac{\pi}{2}\Big)^\frac{3}{2}-\Big(\frac{\pi}{2}\Big)^2\Bigg)$$
So then the quadradic would be:
$$\frac{\Bigg(\frac{\pi}{2}-\sqrt{\frac{\pi}{2}}\Bigg)\pm\sqrt{\Bigg(\frac{\pi}{2}-\sqrt{\frac{\pi}{2}}\Bigg)^2+4\Bigg(\big(\frac{\pi}{2}\big)^\frac{3}{2}-\big(\frac{\pi}{2}\big)^2\Bigg)}}{2}$$

That is why I wanted to approximate and when I do, I still get a spiral, but this time a source. I used 1.51 as an approximation for π/2 and 1.25 for√π/2 then my approximation for b in the quadradic was 0.32 and c was 0.50. I hope I'm not completely missing something here.

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