Is this an elliptic integral or not?

Question

I am working with the following differential equation: $$4\left(\frac{dz}{dx}\right)^2+z^4=4$$ On rearrangement, this yields $$\frac{dz}{dx}=\frac{\sqrt{4-z^4}}{{2}}$$ Using $z=\sqrt{2}\tan \theta$, we further get $$\frac{d\theta}{\sqrt{1-2\sin^2\theta}}=\frac{dx}{\sqrt{2}}$$

Now, as per this, this, this and other links, the term on the left is an elliptic integral. But Mathworld says that, for an elliptic integral of the form $\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}$, the bound on $k$ is given by $$0<k^2<1$$

But for my integral, $k^2=2>1$, which is also causing me difficulty in numerically trying to integrate the problem.

Can somebody tell me what is correct and what not and how I should proceed to integrate the LHS of the equation, numerically (or if possible analytically)?

Answer 1

$$\frac{dz}{dx}=\pm\frac{\sqrt{4-z^4} }{{2}}$$ $$dx=\pm 2\frac{dz}{\sqrt{4-z^4}}$$ $$x(z)=\pm 2\int\frac{dz}{\sqrt{4-z^4}}+c$$ This is an elliptic integral of first kind.

The inverse function $z(x)$ is a Jacobi elliptic function : $$z(x)=\pm \sqrt{2}\text{ sn}\left(\frac{x-c}{\sqrt{2}} \: \Bigg| \: -1 \right)$$ sn$(u|m)$ is the Jacobi elliptic sine function, with $u=\frac{x-c}{\sqrt{2}}$ and $m=-1$. See Wikipedia.

Answer 2

This equation is interesting in that the points $(z, z') = (\sqrt{2}, 0)$ and $(-\sqrt{2},0)$ are equilibria. So whenever $z$ hits $\pm\sqrt{2}$ the solution does not extend in a natural way. I guess that any physically meaningful solution will immediately escape this equilibria. One such solution is

$$ z(x) = \sqrt{2} \operatorname{dn} \left( \frac{x}{\sqrt{2}} \, \middle| \, 2\right), $$

which satisfies $z(0) = \sqrt{2}$ and $z'(0) = 0$. Here, $\operatorname{dn}(\cdot \mid m)$ is the Jacobi delta amplitude and we are adopting the convention that $m = k^2$ is the parameter and $k$ is the elliptic modulus. This at least provides a way of computing $z(x)$, but I am not sure if $\operatorname{dn}$ can be easily computed in a numerical way.

Finally, here are some computations using Mathematica.

Answer 3

The solutions of $$a\left(\frac{dz}{dx}\right)^2+z^4=b$$ are given in terms of the Jacobi elliptic function $$z=\pm\sqrt[4]{b}\, \text{sn}\left(\left.\frac{\sqrt[4]{b} }{\sqrt{a}}x+c\right|-1\right)$$

Answer 4

As a supplement to the fine answers already provided by Sangchul and M. Jacquelin:

The elliptic functions (not elliptic integrals!) that solve the nonlinear ODE presented in the OP satisfy what are called the reciprocal modulus and imaginary modulus identities, which are useful in the respective cases of $k>1$ and purely imaginary $k$.

(At this juncture, I would like to remind everybody to mind their argument conventions; the state of affairs is already confusing as it is.)

For example, with M'sieur Jacquelin's solution, applying this imaginary modulus identity yields

$$\sqrt{2}\,\operatorname{sn}\left(\frac{x-c}{\sqrt{2}}\middle|-1\right)=\operatorname{sd}\left(x-c\middle|\frac12\right)$$

(or $\operatorname{sd}\left(x-c,\frac1{\sqrt 2}\right)$ if you prefer using moduli instead of parameters).

(A similar operation can be done for the answer of M. Leibovici.)

Sangchul's solution, OTOH, admits a reciprocal modulus transformation:

$$\sqrt{2} \operatorname{dn} \left( \frac{x}{\sqrt{2}} \, \middle| \, 2\right)=\sqrt{2}\operatorname{cn}\left(x\middle| \frac12\right)$$

Of interest is that all the transformed answers now have $m=\frac12$, which is now more amenable to AGM-type computations unlike the forms with parameters outside $[0,1)$.