How should I find the equilibrium points and the general equation?

  • Thread starter Math100
  • Start date
  • Tags
    Equilibrium
In summary, to find equilibrium points in a system, first set the derivatives of the system’s equations equal to zero to identify where the system is at rest. Then, solve the resulting equations simultaneously to determine the specific values of the variables at equilibrium. The general equation can be derived from the original system by substituting these equilibrium values back into the equations, allowing for a comprehensive understanding of the system's behavior at those points.
  • #1
Math100
802
222
Homework Statement
Find the equilibrium points and the general equation for the phase paths of ## \ddot{x}+cos(x)=0 ##. Obtain the equation of the phase path joining two adjacent saddles.
Relevant Equations
For the general equation ## \ddot{x}=f(x, \dot{x}) ##, equilibrium points lie on the ## x ## axis, and are given by all solutions of ## f(x, 0)=0 ##, and the phase paths in the plane ## (x, y) (y=\dot{x}) ## are given by all solutions of the first-order equation ## \frac{dy}{dx}=\frac{f(x, y)}{y} ##.
Consider the differential equation ## \ddot{x}+cos(x)=0 ##.
Note that ## \ddot{x}=f(x, \dot{x}) ##, so we have ## f(x, y)=-cos(x) ##.
Then ## f(x, 0)=-cos(x)=0 ##.
This gives ## x=n\pi-\frac{\pi}{2} ## for some ## n\in\mathbb{Z} ##.
Since the differential equation for the phase paths is given by ## \frac{dy}{dx}=-\frac{cos(x)}{y} ##,
it follows that ## y dy=-cos(x)dx\implies \int y dy=-\int cos(x)dx\implies \frac{y^2}{2}=-sin(x)+C ##.
Thus, ## y^2=-2sin(x)+C\implies y=\pm\sqrt{C-2sin(x)} ## where ## C ## is an arbitrary constant.

Above is my work for this problem. However, I've only found the general equation/solution for the phase paths of ## \ddot{x}+cos(x)=0 ## but how should I find the equilibrium points based on the equation I've found above in my work ## x=n\pi-\frac{\pi}{2} ## for some ## n\in\mathbb{Z} ##? Also, how should I obtain the equation of the phase path joining two adjacent saddles from here?
 
Physics news on Phys.org
  • #2
The fixed points are at [itex](\dot x, \ddot x) = (0,0)[/itex]. You have correctly found these to be at [itex](x, \dot x) = ((n + \frac12)\pi,0), n \in \mathbb{Z}[/itex]. But which are saddles? You need to find the eigenvalues of the Jacobian at these fixed points to determine that.

Then you can find the value of [itex]C[/itex] at a saddle point, which will give you the equation of the phase path(s) connecting adjacent saddles.
 
  • #3
pasmith said:
The fixed points are at [itex](\dot x, \ddot x) = (0,0)[/itex]. You have correctly found these to be at [itex](x, \dot x) = ((n + \frac12)\pi,0), n \in \mathbb{Z}[/itex]. But which are saddles? You need to find the eigenvalues of the Jacobian at these fixed points to determine that.

Then you can find the value of [itex]C[/itex] at a saddle point, which will give you the equation of the phase path(s) connecting adjacent saddles.
Why is it ## (x, \dot{x})=((n+\frac{1}{2})\pi, 0) ## for some ## n\in\mathbb{Z} ## instead of ## (x, \dot{x})=((n-\frac{1}{2})\pi, 0) ## for some ## n\in\mathbb{Z} ##? Also, what's the Jacobian in this problem and how to find the eigenvalues of this Jacobian in order to find those saddles?
 
  • #4
Math100 said:
Why is it ## (x, \dot{x})=((n+\frac{1}{2})\pi, 0) ## for some ## n\in\mathbb{Z} ## instead of ## (x, \dot{x})=((n-\frac{1}{2})\pi, 0) ## for some ## n\in\mathbb{Z} ##?

Do you think these are different?

Also, what's the Jacobian in this problem and how to find the eigenvalues of this Jacobian in order to find those saddles?

If you don't recall what the jacobian of a system of ODEs is or how to find it, you need to reread your notes.
 
  • Informative
Likes Math100

FAQ: How should I find the equilibrium points and the general equation?

What are equilibrium points in a system of equations?

Equilibrium points are the values of the variables in a system of equations where the system is in a state of balance, meaning that the rates of change of the variables are zero. In a graphical context, these points correspond to the intersections of the curves or lines representing the equations.

How do I find equilibrium points for a system of differential equations?

To find equilibrium points for a system of differential equations, you set the derivatives equal to zero and solve the resulting algebraic equations. This involves finding the values of the variables that satisfy the system when the rates of change are zero.

What is the general equation for a system of equations?

The general equation for a system of equations can vary depending on the context, but it typically refers to a set of equations that describe the relationships between multiple variables. For linear systems, it can be represented in matrix form as Ax = b, where A is a matrix of coefficients, x is a vector of variables, and b is a vector of constants.

Can equilibrium points be stable or unstable?

Yes, equilibrium points can be classified as stable or unstable. A stable equilibrium point is one where small perturbations will decay over time, returning the system to equilibrium. An unstable equilibrium point, on the other hand, will lead to the system moving away from equilibrium when perturbed.

What methods can I use to visualize equilibrium points?

To visualize equilibrium points, you can use phase portraits, which plot the trajectories of the system in a state space, or you can graph the equations themselves to see where they intersect. Software tools like MATLAB or Python libraries such as Matplotlib can help create these visualizations effectively.

Back
Top