# sn(u), Jacobi elliptic function, for simple pendulum of any amplitude

#### mishima

Problem Statement
Boas Chapter 11, Section 12, Problem 22. Show that the exact solution when $\alpha$ is not small is
$$sin \frac \theta 2=sin\frac \alpha 2 sn \sqrt \frac g l t$$
where k = $sin\frac \alpha 2$ is the modulus of the elliptic function.
Relevant Equations
$$sn(u) = \int_0^\phi \frac{d\theta}{\sqrt{1-k^{2}sin^{2}\theta}}=sin\phi$$
I understand how to reach

$$\int_0^\phi \frac{d\theta}{\sqrt{1-k^{2}sin^{2}\theta}}=\sqrt \frac g l t$$

from physics but from there I don't get how to turn that into this new (for me) sn(u) form.

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#### mishima

Anything I can do to make this more answerable?

I've read the sn, cn, and dn functions were widely used in the past but not so much these days...hard to find some worked examples.

#### George Jones

Staff Emeritus
Gold Member
Problem Statement: Boas Chapter 11, Section 12, Problem 22.
Have you done problem 17 from the same section. If not, then first do problem 17.

• mishima and dextercioby

#### mishima

Thanks, I did that one earlier, and could recognize the same $\frac {sin\frac \theta 2} { sin \frac \alpha 2}$ substitution in the exact solution, but I don't understand how this makes the sn(u) function.

#### George Jones

Staff Emeritus
Gold Member
Thanks, I did that one earlier, and could recognize the same $\frac {sin\frac \theta 2} { sin \frac \alpha 2}$ substitution in the exact solution, but I don't understand how this makes the sn(u) function.
Can you show your solution to question 17?

• mishima

#### mishima

Alright.

The section narrative provided the starting point of
$$\int_0^\alpha \frac {d\theta} {\sqrt {cos\theta-cos\alpha}}=\sqrt {\frac {2g} l} \frac {T_\alpha} 4$$
where $\alpha$ is the amplitude of the pendulum's swing, and $T_\alpha$ is the period for a swing from $\alpha$ to $-\alpha$. $~T_\alpha/4$ then represents a swing through $\pi/2$ and is the t in problem 22.

I can change the cosines into sines using the double angle trig identity:

$$\int_0^\alpha \frac {d\theta} {\sqrt {2sin^2\frac {\alpha} 2-2sin^2\frac \theta 2}}=\sqrt {\frac {2g} l} \frac {T_\alpha} 4$$

Now I can make the denominator more like an elliptic integral by pulling out the following factor:

$$\int_0^\alpha \frac {d\theta} {\sqrt 2 sin\frac {\alpha} 2\sqrt {1-\frac {sin^2\frac {\theta} 2 }{sin^2\frac {\alpha} 2}}}=\sqrt {\frac {2g} l} \frac {T_\alpha} 4$$

Now the substitution $x=\frac {sin^2\frac {\theta} 2 }{sin^2\frac {\alpha} 2}$, its differential, and corresponding limits of integration changes it from Legendre to Jacobi form. This step is of course making my spider senses tingle with problem 22, but again I don't quite understand the sn(u) notation:

$$\sqrt 2 \int_0^1 \frac {dx} {cos\frac \theta 2 \sqrt {1-x^2}}$$

Then its just a matter of using the pythagorean formula for sines and cosines to turn the cosine in the denominator into the Jacobi form, then one last substitution to get the k in the right place:

$$\sqrt 2 \int_0^1 \frac {dx} {\sqrt {1-x^2sin^2\frac \alpha 2} \sqrt {1-x^2}}$$

But thats just

$$\sqrt 2 K(sin\frac \alpha 2)$$

#### George Jones

Staff Emeritus
Gold Member
Good.

This step is of course making my spider senses tingle with problem 22, but again I don't quite understand the sn(u) notation
To link up with sn, let' s redo problem 17, but with different limits of integration. Suppose that $\theta=0$ at $t=0$, and that the angular position at general $t$ is $\theta$. Then, equation (12.6) of the 3rd edition gives

$$\sqrt{\frac{2g}{l}} \int_0^t dt' = \int_0^\theta \frac{d \theta'}{\sqrt{\cos \theta' - \cos \alpha}} .$$

Now, use the same substitutions that are used in problem 17. What happens?

"sn(u), Jacobi elliptic function, for simple pendulum of any amplitude"

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