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:

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



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:


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

  • $\begingroup$ Maybe helpful: users.mai.liu.se/hanlu09/complex/elliptic (search for the word “pendulum”). $\endgroup$ Apr 18, 2018 at 6:46
  • $\begingroup$ 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. $\endgroup$ Apr 18, 2018 at 7:49

1 Answer 1


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






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, $c>h$.


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