I would like to find the inverse of function of 2 variables in then analytic form. Unfortunately, the functions $f,g$ have a "complex" form. Suppose that
\begin{align*} x=F(u,v) & ,\quad y=G(u,v)\\ u=f(x,y), & \quad v=g(u,v), \end{align*}
where $f\equiv F^{-1}$ and $g\equiv G^{-1}$. Differentiating the equations leads to
\begin{align*} \left[\begin{array}{c} dx\\ dy \end{array}\right] & =\left[\begin{array}{cc} x_{u} & x_{v}\\ y_{u} & y_{v} \end{array}\right]\left[\begin{array}{c} du\\ dv \end{array}\right],\\ \left[\begin{array}{c} du\\ dv \end{array}\right]=\left[\begin{array}{cc} u_{x} & u_{y}\\ v_{x} & v_{y} \end{array}\right]\left[\begin{array}{c} dx\\ dy \end{array}\right]\Rightarrow & \left[\begin{array}{c} dx\\ dy \end{array}\right]=\frac{1}{\delta^{\prime}}\left[\begin{array}{cc} v_{y} & -u_{y}\\ -v_{x} & u_{x} \end{array}\right]\left[\begin{array}{c} du\\ dv \end{array}\right], \end{align*}
where $\delta^{\prime}=u_{x}v_{y}-v_{x}u_{y}$, and $\delta=x_{u}y_{v}-x_{v}y_{u}=1/\delta^{\prime}$. Then, differential equations for the inverse are
\begin{align*} \left[\begin{array}{cc} x_{u} & x_{v}\\ y_{u} & y_{v} \end{array}\right] & =\delta\left[\begin{array}{cc} v_{y} & -u_{y}\\ -v_{x} & u_{x} \end{array}\right]. \end{align*}
Let us start with the relatively simple function,
\begin{align*} x=\cos v\tan u, & y=\frac{\sin v}{\cos u},\\ \end{align*} then, $h=(\cos^{2}v+\sin^{2}v\sin^{2}u)/\cos^{3}u$, and \begin{align*} \frac{\cos v}{\cos^{2}u}=h\frac{dv}{dy},\quad & \sin v\tan u=h\frac{du}{dy},\\ -\frac{\sin u\sin v}{\cos^{2}u}=h\frac{dv}{dx},\quad & \frac{\cos v}{\cos u}=h\frac{du}{dx}. \end{align*}
For example, the last equation leads to
\begin{align*} dx & =\frac{\cos^{2}v+\sin^{2}v\sin^{2}u}{\cos^{2}u\cos v}du\Rightarrow x=\int\frac{\cos^{2}v+\sin^{2}v\sin^{2}u}{\cos^{2}u\cos v}du. \end{align*}
Do you find these equations helpful for practical finding the inverse or not? In my opinion, they are not. The "classic" approach leads to the simple substitution
\begin{align*} x^{2} & =(1-\sin^{2}v)\tan^{2}u\Rightarrow x^{2}=(\tan^{2}u-y^{2}\sin^{2}u). \end{align*}
Let us move to the more complex function representing the "engineering product" that I am not able to solve using any substitution. Since its Jacobian matrix at $u=0, v=0$ is not singular, its inversion at this points exists. However, how to find it...
\begin{align*} b & =\frac{\pi}{2v}-\frac{2v}{\pi},\quad c=\frac{2}{\pi}u,\quad d=\frac{1-c^{2}}{\sin u-c},\\ x= & \dfrac{\frac{b}{d}\sin u-0.5b}{1+\frac{b^{2}}{d^{2}}}v,\quad y=\dfrac{\frac{d^{2}}{b^{2}}\sin u+0.5d}{1+\frac{d^{2}}{b^{2}}}\cos v. \end{align*}
Questions:
- Is there any more advanced method for finding the inverse in the analytic form or not, or, only numerical methods like Gauss-Newton are available?
- Is it useful to find some polynomial approximation of the function and find the inverse from the polynomial approximation?
Thank you very much for your answers.