• Support PF! Buy your school textbooks, materials and every day products Here!

Classifaction of equilibrium points for a Hamiltonian System

  • #1

Homework Statement


For the system:
[tex]
\frac{dx}{dt}=x\cos{xy}
\: \:
\frac{dy}{dt}=-y\cos{xy}[/tex]
(a) is Hamiltonian with the function:
[tex]
H(x,y)=\sin{xy}[/tex]
(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.

Homework Equations




The Attempt at a Solution


[tex]\frac{\partial H}{\partial y}=y\cos{xy}=-g \\
\frac{\partial H}{\partial x}=x\cos{xy}=f[/tex]
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

  • #2
33,267
4,966

Homework Statement


For the system:
[tex]
\frac{dx}{dt}=x\cos{xy}
\: \:
\frac{dy}{dt}=-y\cos{xy}[/tex]
(a) is Hamiltonian with the function:
[tex]
H(x,y)=\sin{xy}[/tex]
(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.

Homework Equations




The Attempt at a Solution


[tex]\frac{\partial H}{\partial y}=y\cos{xy}=-g \\
\frac{\partial H}{\partial x}=x\cos{xy}=f[/tex]
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.
 
  • #3
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.
 
  • #4
33,267
4,966
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.
 
  • #5
I get a Jacobian matrix of:
[tex]
\begin{bmatrix}
\cos{xy}-y\sin{xy} & -x^2\sin{xy} \\
xy\sin{xy} & xy\sin{xy}-\cos{xy}
\end{bmatrix}
[/tex]
So then my matrix for that second point becomes:
[tex]
\begin{bmatrix}
-\sqrt{\frac{\pi}{2}} & -\frac{\pi}{2} \\
\frac{\pi}{2} & \frac{\pi}{2}
\end{bmatrix}
[/tex]
Then using the trace and determinant I get:
[tex]
\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)[/tex]
So then the quadradic would be:
[tex]
\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}[/tex]

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.
 
Last edited:

Related Threads on Classifaction of equilibrium points for a Hamiltonian System

Replies
2
Views
1K
Replies
1
Views
3K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
935
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
983
  • Last Post
Replies
0
Views
993
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
9
Views
2K
Top