We are searching for an analytic solution to the given equation for $f_text{n}(u)$, for $u in (0, d/2)$ (this problem is a snippet from this paper here)

$$-partial^2_{u} f_text{n} + leftlbrack 1 – f_text{n} ^{2} +frac{j_{u}^2}{f_text{n} ^4}rightrbrack

f_text{n} = 0$$

With a little manipulation (integrate from 0 to $x$, use that $f’$ is minimal at $x=0$…):

$$left(frac{partial f_n^2}{partial u}right)^2 = 2 (f_n^2 – f^2_0) left( f_n^2 (2 – f_n^2 – f^2_0) + frac{2j_u^2}{f^2_0}right)$$

or more usefully:

$$ 2left(frac{partialPsi}{partial u}right)^2 = left( (Psi^2 + f_0^2) (2 – Psi^2 – 2f^2_0) + frac{2j_u^2}{f^2_0}right)$$.

I know from extensive hand-calculation that this has a Jacobi elliptic function solution.

$$Psi_n^2 = f_n^2 – f_0^2 = frac{b_n^2 a_n^2}{a_n^2 + b_n^2} operatorname{sd}^2left(u sqrt{frac{a_n^2 + b_n^2}{2}}, sqrt{frac{b_n^2}{a_n^2 + b_n^2}}right)$$

where

$$

a_n^2 = – left(1-frac{3}{2}f_0^2right) + sqrt{2left(1-frac{1}{2}f_0^2right)^2 + frac{2j_u^2}{f_0^2}}

\

b_n^2 = + left(1-frac{3}{2}f_0^2right) + sqrt{2left(1-frac{1}{2}f_0^2right)^2 + frac{2j_u^2}{f_0^2}}

$$

When I try and replicate this in Mathematica, I obtain a different solution:

```
an^2 == - ( 1 - (3/2) f0^2) + Sqrt(2 (1 - f0^2/2)^2 + 2 ju^2/f0^2)
bn^2 == + ( 1 - (3/2) f0^2) + Sqrt(2 (1 - f0^2/2)^2 + 2 ju^2/f0^2)
eqn = {Sqrt(2)*Psi'(x) == Sqrt((an^2 + Psi(x)^2)*(bn^2 - Psi(x)^2))}
sol = DSolve(eqn, Psi(x), x)
```

Output:

{{Psi(x) -> bn*JacobiSN((Sqrt(2) anx + 2*an*C1)/2, -(bn^2/an^2))}}

It is not clear to me that these solutions are equivalent, and moreover, the form looks completely different, let alone being a completely different JEF solution. I have only looked into `DSolve`

in the past, but I am not too sure how to use `Reduce`

to find a better solution.

Any advice would be gratefully accepted. My end result is that I want to find the analytic solution, ideally able to find the results for $a_n$ and $b_n$ as well, but I appreciate mathematical manipulation is somewhat necessary!