# Jacobi Elliptic Equations

1. Nov 8, 2007

### sourlemon

[Solved] Jacobi Elliptic Equations

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

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

It is integrated from 0 to $$\phi$$

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

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

0 $$\leq$$ t $$\leq$$ $$\frac{P(\alpha)}{4}$$

where k is defined as sin $$\alpha$$/2)

3. The attempt at a solution
I looked at the answer and tried to backtrack it. I see that $$\sqrt{\frac{l}{g}}$$ is next to $$-t$$ + $$\frac{P(\alpha)}{4}$$, so this means that $$\sqrt{\frac{l}{g}}$$ was divided into both sides.

Looking through my integral table, I found that

$$\int\frac{du}{\sqrt{a^{2} - u^{2}}}$$ = arcsin $$\frac{u}{a}$$

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

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

Last edited: Nov 8, 2007