Understanding Fixed Points in Hamiltonian Systems

[Master1022]

Homework Statement

How can we use the potential function of a Hamiltonian system to determine the nature of the equilibrium

Relevant Equations

Hamiltonian system

Hi,

I was attempting a question about Hamiltonian systems from dynamic systems and wanted to ask a question that arose from it.

Homework Question: Given the system below:
\dot x_1 = x_2
\dot x_2 = x_1 - x_1 ^4
(a) Prove that the system is a Hamiltonian function and find the potential function
(b) Use the potential function to determine information about the fixed points of the system (along $x_2 = 0$)

My question: I don't understand how to do part (b). Specifically, I don't understand why (for potential function $V$):
"""
- If $V(\mathbf{x}_{fixed})$ is a minimum, the fixed point is a center
- If $V(\mathbf{x}_{fixed})$ is a maximum, the final point is a saddle point
"""

Attempt:
I know that a Hamiltonian system has the form:
\dot x_1 = \frac{\partial H}{\partial x_2} \dot x_2 = - \frac{\partial H}{\partial x_1}

and I can use these relations to calculate the Hamiltonian function:
H(x_1, x_2) = \frac{1}{2} x_2 ^2 + \frac{1}{5} x_1 ^5 - \frac{1}{2} x_1 ^2

By matching terms in $H(x_1, x_2) = KE + \text{Potential}$. So I can identify the potential function as: $V(x_1, x_2) = \frac{1}{5} x_1 ^5 - \frac{1}{2} x_1 ^2$

Now in terms of finding the equilibria:
- we can use $\frac{dV}{dx_1} = 0$ to find the equilibrium points which end up being: $(0, 0)$, $(1, 0)$

Then the solution says:
"""
- If $V(\mathbf{x}_{fixed})$ is a minimum, the fixed point is a center
- If $V(\mathbf{x}_{fixed})$ is a maximum, the fixed point is a saddle point
"""

Where do these come from?/Why is that the case?

Any help would be greatly appreciated.

 

[pasmith]

If H = \frac12x_2^2 + V(x_1) then the jacobian is <br /> J = \begin{pmatrix}<br /> \frac{\partial \dot x_1}{\partial x_1} &amp; \frac{\partial \dot x_1}{\partial x_2} \\<br /> \frac{\partial \dot x_2}{\partial x_1} &amp; \frac{\partial \dot x_2}{\partial x_2} \end{pmatrix} = <br /> \begin{pmatrix}<br /> 0 &amp; 1 \\<br /> - V&#039;&#039;(x_1) &amp; 0<br /> \end{pmatrix}.<br /> The eigenvalues \lambda of J are then given by <br /> \lambda^2 = -V&#039;&#039;(x_1).

 

[Master1022]

If H = \frac12x_2^2 + V(x_1) then the jacobian is <br /> J = \begin{pmatrix}<br /> \frac{\partial \dot x_1}{\partial x_1} &amp; \frac{\partial \dot x_1}{\partial x_2} \\<br /> \frac{\partial \dot x_2}{\partial x_1} &amp; \frac{\partial \dot x_2}{\partial x_2} \end{pmatrix} =<br /> \begin{pmatrix}<br /> 0 &amp; 1 \\<br /> - V&#039;&#039;(x_1) &amp; 0<br /> \end{pmatrix}.<br /> The eigenvalues \lambda of J are then given by <br /> \lambda^2 = -V&#039;&#039;(x_1).

Many thanks @pasmith ! Ah okay, this is very clear - now I can see the sign of the potential function evaluated at a point will determine whether the eigenvalues are have a non-zero imaginary component or not. Just to check, this rule wouldn't necessarily apply for other Hamiltonian systems because their potential functions may involve both variables and their KE may also have both variables?