I'm working with a non-linear second order ODE which has an analytical solution in terms of the Jacobi elliptical function $sn(u|k^2)$. The equation is $y''=y(\gamma - \frac{y^2}{2})$ where $\gamma$ is a constant and should be less than 0 for the specific problem I am dealing with.

I want to solve the equation by hand so I am using a first integral to make it a first order non linear ODE because it does not depend on $y'$. Doing that I get:

$$\frac{dy}{dx}= \pm \sqrt{y^2 \left( y-\frac{y^2}{4} \right) + C_1}$$

Where $C_1$ is an integration constant. I can separate both sides of this equation and I get the next integral:

$$\int_0^y \frac{dr}{\sqrt{r^2(\gamma-\frac{r^2}{4}) + C_1}}$$

Which I know that can be transformed into an elliptic integral because that is the result I get if I introduce it into Mathematica (the elliptic integral of the first kind with $\phi=i arcsinh(\alpha y)$) where $i$ is the imaginary unit and $\alpha$ is some constant which depends upon $\gamma$ and $C_1$.

What I want to know is which changes of variables should I perform to get the corresponding elliptic integral:

$$\int_0^{iarcsinh(\alpha y)} \frac{dt}{\sqrt{1 - k^2 sin^2(t)}}$$

I saw this thread: Evaluating the elliptic integral $\int_{-\pi}^\pi\frac{dx}{\sqrt{(t-2\cos x)^2-4}}$ and I tried to perform some change similar to the ones that are proposed there but with no result.


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    $\begingroup$ Which system are you trying to handle? If it's a known one (e.g. a simple pendulum) then there are good refs available. $\endgroup$ – Semiclassical Jul 30 '14 at 14:39
  • $\begingroup$ Thanks! I think I can get it from here using some change like $t=i \cdot arcsinh(r)$ and taking into account that $sinh(ir)=i \cdot sin(r)$. Anyway, I'll try it tomorrow and I'll let you know if I get stucked. Thanks again! $\endgroup$ – A. A. Jul 30 '14 at 17:32
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    $\begingroup$ Good good. When you complete it, post the answer/accept to close the question off :)! $\endgroup$ – Chinny84 Jul 30 '14 at 18:53

Rewrite the expression under the square root as $$r^2\left(\gamma-\frac{r^2}{4}\right)+C_1=\frac{1}{4}\left(a^2-r^2\right)\left(b^2+r^2\right),$$ with some $a,b$, then set $r=a\sin t$.

  • $\begingroup$ I removed my comment in light of this nice answer and a +1 :). $\endgroup$ – Chinny84 Jul 30 '14 at 20:11
  • $\begingroup$ Thanks! I got the result I was looking for :) $\endgroup$ – A. A. Jul 31 '14 at 9:19

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