Conversion to elliptic integrals

I have the following differential equation:

$$2\theta'' = h \sin\theta$$

After multiplying both sides by $\frac{\mathrm{d}\theta}{\mathrm{d}x}$, I obtain:

$$\frac{\mathrm{d}}{\mathrm{d}x} \left[\left(\frac{\mathrm{d}\theta}{\mathrm{d}x}\right)^2+h\cos\theta\right]=0$$

Therefore,

$\left(\frac{\mathrm{d}\theta}{\mathrm{d}x}\right)^2+h\cos\theta=c$

where $c$ is a constant that I need to determine via energy minimization.

I can take the square root of the expression to obtain:

$\mathrm{d}x=\frac{\mathrm{d}\theta}{\sqrt{c-h \cos\theta}}$.

Mathematica gives me a solution in terms of Jacobi Amplitudes. I am trying to express the right hand side as an elliptic integral of the first kind, but so far I have not succeeded. Can anyone give me some ideas on how to proceed?

Edit: I have already found a solution without the use of elliptic integrals in order to find the constant $c$. In that case, the constant $c$ is known and I have to solve the following differential equation:

$\left(\frac{\mathrm{d}\theta}{\mathrm{d}x}\right)^2=c-h\cos\theta$.

Could I find an expression for $\theta(x)$ without the use of elliptic integrals in that case?

• Maybe helpful: users.mai.liu.se/hanlu09/complex/elliptic (search for the word “pendulum”). – Hans Lundmark Apr 18 '18 at 6:46
• Thanks for the info Hans. The specific example you provided imposes a priori the initial conditions, whereas for my case, the constant c will be determined by energy minimization. I have already found a solution to my problem without the use of elliptic integrals, but it would be useful if I could also have it in that form. – Dimitris A. Apr 18 '18 at 7:49

I think an easy way to do it would be to write

$\cos\theta=1-2\sin\left(\theta/2\right)^2$

Then,

$c-h\cos\theta=c-h\left(1-2\sin\left(\theta/2\right)^2\right)=c-h+2h\sin\left(\theta/2\right)^2=c-h\left(1+\frac{2h}{c-h}\sin\left(\theta/2\right)^2\right)$

Consequently,

$\sqrt{c-h}\;\mathrm{d}x=\frac{\mathrm{d}\theta}{\sqrt{1+\frac{2h}{c-h}\sin\left(\theta/2\right)^2}}$

Then if we integrate, the right hand side is the elliptic integral of the first kind with modulus $k\equiv -2h/(c-h)$: $F\left(\theta/2,\frac{2h}{h-c}\right)$.

As a result, $\theta(x)=2\; \text{am}\left(\sqrt{c-h}\;x,\frac{2h}{h-c}\right)$

Edit: the $\sqrt{c-h}$ is legit since $c-h\cos\theta>0$ ,$\forall \theta$. Consequently, $c>h$.