# Inverting an Equation Containing Elliptic Integrals

• I
• diegogarcia

#### diegogarcia

TL;DR Summary
Is it possible to analytically invert this elliptic integral equation?
Hello,

For my own amusement, I am deriving the eqations for various roulettes, i.e. a the trace of a curve rolling on another curve.

When considering rolling ellipses, I encounter equations containing elliptic integrals (of the second kind) that need to be inverted.

For example, here is one such equation:

t = a * elliptic_e(u, E)

where a, E are positive, real contants and t, u are the real variables of concern.

(The notation is from Maxima: https://maxima.sourceforge.io/docs/manual/maxima_91.html)

In other words, I need to express u as a function of t.

Can this equation be analytically inverted?

For specific values of t, I can easily find a value for u by using a numerical root finding method but an exact, analytical answer would be preferable.

Another such equation is:

elliptic_e(u, Er) = a/ar * elliptic_e(t, Ef)

Again, I need to invert this equation to find u as a function of t (a, ar, Er, Ef are all positive real constants).

I know that the inverse of an elliptic integral is an elliptic function but I don't know how to invert these equations.

The inverse of an incomplete elliptic integral of the first kind $F(\phi, \alpha)$ is an elliptic function. You seem to be dealing with the incomplete integral of the second kind $E(\phi, \alpha)$, which doesn't have that property.
Althougth $E(\phi,\alpha)$ is invertible with respect to $\phi$ for $\phi \in [0, \pi/2]$ at fixed $\alpha \in [0,1)$, this inverse is not, so far as I am aware, tabulated anywhere or implemented as a function in a scientific library, so in practise I suspect the answer is "no".
Rather than using a root-finding method for $t = aE(u,E)$ for multiple values of $t$, it is perhaps more efficient to solve the ODE $$\frac{du}{dt} = \frac1{a\sqrt{1 - E\sin^2u}},\qquad u(0)= 0$$ numerically. For $E(u,E_r) = (a_t/a_r)E(t,E_t)$ you instead need to solve $$\frac{du}{dt} = \frac{a_t}{a_r} \frac{\sqrt{1 - E_t\sin^2 t}}{\sqrt{1-E_r\sin^2 u}},\qquad u(0) = 0$$ numerically.