Inverting an Equation Containing Elliptic Integrals

[diegogarcia]

TL;DR

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.

 

[pasmith]

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 <br /> \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 <br /> \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.