Solving nonlinear system to change variable in a PDE I want to change variables of the PDE
\begin{align*}
\left(-\partial_x^2 - \frac{1}{x}\partial_x + \frac{t^2}{x^2}\partial_t^2\right)\psi = \lambda \psi
\end{align*}
to variables
\begin{align*}
r &:= \sqrt{(x+2bt)(x-2c/t)} \\
s &:= te^{\tau}\sqrt{\frac{1-\frac{2c}{xt}}{1+\frac{2bt}{x}}}
\end{align*}
$\tau, b,c$ are complex constants, which can be used to give some freedom in choosing the variable transformation.
Using the chain rule, I found that I needed to solve for $x,t$ in terms of $r$ and $s$ in order to define the operator in $(r,s)$ space. I plugged this into Mathematica, and got an indecipherable stream of characters. How can I solve for $(x,t)$ in terms of $(r,s)$?
 A: Denote
\begin{align}
A&=\sqrt{x-\frac{2c}{t}},\\
B&=\sqrt{x+2bt},
\end{align}
and your change-of-variable reads
\begin{align}
r&=BA,\\
s&=te^{\tau}\frac{A}{B}.
\end{align}
These two relations yield
\begin{align}
rs&=te^{\tau}A^2=te^{\tau}\left(x-\frac{2c}{t}\right)=e^{\tau}\left(xt-2c\right),\\
\frac{r}{s}&=\frac{1}{te^{\tau}}B^2=\frac{1}{te^{\tau}}\left(x+2bt\right)=e^{-\tau}\left(\frac{x}{t}+2b\right),
\end{align}
or equivalently,
\begin{align}
xt&=e^{-\tau}rs+2c,\\
\frac{x}{t}&=e^{\tau}\frac{r}{s}-2b.
\end{align}
These last relations would lead to
\begin{align}
x^2&=xt\cdot\frac{x}{t}=\left(e^{-\tau}rs+2c\right)\left(e^{\tau}\frac{r}{s}-2b\right),\\
t^2&=xt\cdot\frac{t}{x}=\frac{e^{-\tau}rs+2c}{e^{\tau}r/s-2b}.
\end{align}
Therefore,
\begin{align}
x&=\sqrt{\left(e^{-\tau}rs+2c\right)\left(e^{\tau}\frac{r}{s}-2b\right)},\\
t&=\sqrt{\frac{e^{-\tau}rs+2c}{e^{\tau}r/s-2b}}.
\end{align}
By the way, I am interested in how you come up with this change-of-variable? This is quite technical to me!
