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1. The problem statement, all variables and given/known data

[tex]-t[/tex] + [tex]\frac{P(\alpha)}{4}[/tex] = [tex]\sqrt{\frac{l}{g}}[/tex][tex]\int[/tex][tex]\frac{ds}{\sqrt{1-k^{2}sin^{2}s}}[/tex] = [tex]\sqrt{\frac{l}{g}}[/tex]F(k, [tex]\phi[/tex]).

It is integrated from 0 to [tex]\phi[/tex]

For fixed k, F(k, [tex]\phi[/tex]) has an "inverse," denoted by sn(k, u), that satisfies u = F(k, [tex]\phi[/tex]) if and only if sn(k,u) = sin [tex]\phi[/tex]. The function sin [tex]\phi[/tex] is called a Jacobi elliptic function and has many properties that resemble those of the sine function. Using the Jacobi elliptic function sin [tex]\phi[/tex], express the equation of motion for the pendulum in the form

[tex]\theta[/tex] = 2 arcsin [tex]\left \{[/tex]k sn[tex]\left[[/tex]k, [tex]\sqrt{\frac{g}{l}}[/tex][tex]\left([/tex][tex]-t[/tex] + [tex]\frac{P(\alpha)}{4}[/tex])]}

0 [tex]\leq[/tex] t [tex]\leq[/tex] [tex]\frac{P(\alpha)}{4}[/tex]

where k is defined as sin [tex]\alpha[/tex]/2)

3. The attempt at a solution

I looked at the answer and tried to backtrack it. I see that [tex]\sqrt{\frac{l}{g}}[/tex] is next to [tex]-t[/tex] + [tex]\frac{P(\alpha)}{4}[/tex], so this means that [tex]\sqrt{\frac{l}{g}}[/tex] was divided into both sides.

Looking through my integral table, I found that

[tex]\int\frac{du}{\sqrt{a^{2} - u^{2}}}[/tex] = arcsin [tex]\frac{u}{a}[/tex]

I set a = 1 and u = k^{2}sin^{2}s. With that I got arcsin [tex]\frac{ksins}{1}[/tex]. So where did the 2 come from? Was I wrong to set k part of 2? And where did the [tex]\theta[/tex] comes from?

Nevermind, I think I got it now. The 2 and [tex]\theta[/tex] came from some earlier part of the problem which is not posted here. Thank you, whoever stops by :D

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# Homework Help: Jacobi Elliptic Equations

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