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

• Math100
Math100
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?

The fixed points are at $(\dot x, \ddot x) = (0,0)$. You have correctly found these to be at $(x, \dot x) = ((n + \frac12)\pi,0), n \in \mathbb{Z}$. 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 $C$ at a saddle point, which will give you the equation of the phase path(s) connecting adjacent saddles.

pasmith said:
The fixed points are at $(\dot x, \ddot x) = (0,0)$. You have correctly found these to be at $(x, \dot x) = ((n + \frac12)\pi,0), n \in \mathbb{Z}$. 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 $C$ 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?

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.

Math100

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