I have the following operator, defined for two twice-differentiable functions $f,g$:
$X(f,g):=\frac{(g')^3+fg'f''+g'(f')^2-ff'g''}{g'f''-f'g''}$
This operator has the following property: A curve defined by $x(t)=X(f(t),g(t)),y(t)=X(g(t),f(t))$ is the evolute of a curve $x(t)=f(t),y(t)=g(t)$. Now what I am trying to do is define an inverse operator such that
$X^{-1}(X(f,g),X(g,f))=f$
However, I have no idea how one would go about doing that. I tried assuming $X^{-1}$ is a ratio $\frac{A(f,g)}{B(f,g)}$ such that each "term" in $A,B$ satisfy the following:
Each "term" in $A,B$ is the product of three "components" picked from $[f,f',f'',g,g',g'' ]$
The sum of of the differential order of each term's "components" is $3$.
For example, the numerator of X consists of four "components" $[g',g',g'],[f,g',f''],[g',f',f'],[f,f',g'']$. I would let $A,B$ each consist of a linear combination of all possible "components", and attempt to solve $X^{-1}(X(f,g),X(g,f))=\frac{A(X(f,g),X(g,f))}{B(X(f,g),X(g,f))}=f$. I believe $X^{-1}$ has this form based on similar operators I have worked with and their inverses. However, this approach creates enormous equations to solve and is unmanageable. My question is how one would normally determine inverse operators of this form, and an approach that could work for this operator. Also, is there a way to define an involute of a curve as an envelope of lines, or limiting case of lines intersecting? This would solve the problem for me, as I have dealt a lot with envelopes of curves, and $x(t)=X^{-1}(f(t),g(t)),y(t)=X^{-1}(g(t),f(t))$ is the involute of $x(t)=f(t),y(t)=g(t)$. Any help is appreciated. Thank you!
I do not know how to solve the equations to find the following, but for any $t_0$, an involute of $X, Y$ can be calculated as $$ f(X,Y) = X - \frac{X'}{\def\abs#1{\left|#1\right|}\abs{(X,Y)'}}\cdot \int_{t_0}^t \abs{(X,Y)'}\, dt,\quad g(X,Y) = f(Y,X) $$ where $\abs\cdot$ denotes the standard absolute value on $\mathbb R^2$.