# Understanding Fixed Points in Hamiltonian Systems

• Master1022
In summary, the conversation was about a question involving Hamiltonian systems and finding information about fixed points using the potential function. The Hamiltonian function and potential function were derived, and the process for finding equilibria was explained. The conversation also discussed the relationship between the potential function and the eigenvalues of the Jacobian matrix. Finally, it was clarified that the rules for determining the type of fixed point based on the potential function may not apply to other Hamiltonian systems due to differences in their potential and kinetic energy functions.
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.

Last edited:
If $H = \frac12x_2^2 + V(x_1)$ then the jacobian is $$J = \begin{pmatrix} \frac{\partial \dot x_1}{\partial x_1} & \frac{\partial \dot x_1}{\partial x_2} \\ \frac{\partial \dot x_2}{\partial x_1} & \frac{\partial \dot x_2}{\partial x_2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ - V''(x_1) & 0 \end{pmatrix}.$$ The eigenvalues $\lambda$ of $J$ are then given by $$\lambda^2 = -V''(x_1).$$

Master1022
pasmith said:
If $H = \frac12x_2^2 + V(x_1)$ then the jacobian is $$J = \begin{pmatrix} \frac{\partial \dot x_1}{\partial x_1} & \frac{\partial \dot x_1}{\partial x_2} \\ \frac{\partial \dot x_2}{\partial x_1} & \frac{\partial \dot x_2}{\partial x_2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ - V''(x_1) & 0 \end{pmatrix}.$$ The eigenvalues $\lambda$ of $J$ are then given by $$\lambda^2 = -V''(x_1).$$

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?

## 1. What are fixed points in Hamiltonian systems?

Fixed points, also known as equilibrium points or stationary points, are points in a Hamiltonian system where the system remains at rest or in a steady state. In other words, the derivative of the system at these points is zero, and the system does not change over time.

## 2. How do fixed points affect the behavior of a Hamiltonian system?

Fixed points play a critical role in determining the stability and dynamics of a Hamiltonian system. The behavior of the system near a fixed point is determined by the eigenvalues of the Jacobian matrix at that point. If all eigenvalues are negative, the fixed point is stable, and the system will tend towards it. If at least one eigenvalue is positive, the fixed point is unstable, and the system will move away from it.

## 3. Can a Hamiltonian system have more than one fixed point?

Yes, a Hamiltonian system can have multiple fixed points. The number and location of fixed points depend on the specific system and its parameters. Some systems may have only one fixed point, while others may have an infinite number of fixed points.

## 4. How can one determine the stability of a fixed point in a Hamiltonian system?

The stability of a fixed point can be determined by examining the eigenvalues of the Jacobian matrix at that point. If all eigenvalues are negative, the fixed point is stable. If at least one eigenvalue is positive, the fixed point is unstable. In some cases, the stability of a fixed point may also depend on the system's parameters.

## 5. Are fixed points always present in Hamiltonian systems?

No, not all Hamiltonian systems have fixed points. Some systems may have a constant energy surface, which means there are no equilibrium points. In such cases, the system may exhibit other types of behavior, such as periodic or chaotic motion.

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