I am trying to understand the how Jacobian Elliptic functions are derived from the corresponding elliptic integrals.

From http://mathworld.wolfram.com/JacobiEllipticFunctions.html on Jacobi Elliptic Functions, an elliptic integral of the first kind is:

$$u = F(\phi,k) = \int_0^\phi \frac{d\theta} {\sqrt{(1-k^2\sin^2\theta)}}$$


$$ 0<k^2<1 $$ and $$ \phi = am(u,k) $$

k is the modulus and am is the amplitude.

I know that the Jacobian elliptic functions are found be inverting the elliptic integrals.

If $$ u=F(\phi,k) $$

then the inverse is

$$ F^{-1}(u,k) = F^{-1}[F(\phi,k)] = \phi $$


$$ F^{-1}(u,k) = \phi = am(u,k) $$

So $$ \sin(\phi) = \sin(am(u,k)) = sn(u,k) $$


$$ \cos(\phi) = \cos(am(u,k)) = cn(u,k) $$

and $$\sqrt{1-k^2\sin^2\phi} = \sqrt{1-k^2\sin^2(am(u,k))} = dn(u,k)$$

Am I correct in my reasoning for inverting the integral? It looks like the elliptic integrals are functions of two variables (k and $\phi$), so when you invert the integral, what happens to the k? If $$\phi=am(u,k)$$ why to you take the sine of $\phi$?

Much thanks for any insight!

  • $\begingroup$ May see another answer here. $\endgroup$ Apr 5, 2017 at 13:48

1 Answer 1


$k$ is normally regarded as a parameter in the integral, at least initially, with $\phi$ being the actual variable. The notation is odd for historic reasons, but one could equally write $\operatorname{am}_k(u)$ and so on. Think of what these functions are used for: solutions to differential equations like $$ u'' = -k^2\sin{u} $$ (although this is solved by am itself, rather than its sine or cosine), for example.

  • $\begingroup$ I am not all that familiar with how to solve nonlinear DEs. How does one solve $$u" = -k^2 sinu $$ as am(u)? Are there good references/books you recommend? $\endgroup$ Apr 5, 2017 at 14:48
  • $\begingroup$ Multiplying by $2u'$ and integrating gives $$ u'^2-u_0'^2 = k^2(\cos{u}-\cos{u_0}) $$ We can rearrange to the equation $$ dx = \frac{du}{\sqrt{u_0'^2-k^2\cos{u_0}+k^2\cos{u}}}, $$ and then you can use a double-angle formula to rewrite the second term so that it looks like a rescaling of the integrand of $F(,\kappa)$ for some $\kappa$ that is related to $k$, but not identical to it. Hence $ x = a F(bu,\kappa) $ for some $a,b,\kappa$ depending on $k$ and the initial conditions, and then you have to invert to get $u$ in terms of $x$, which is where am comes in. $\endgroup$
    – Chappers
    Apr 5, 2017 at 15:16
  • $\begingroup$ The notation for elliptic integrals and elliptic functions is indeed odd and confusing; that's exactly why the choice of delimiter is crucial. $\endgroup$ Mar 31, 2018 at 18:38

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