Solving nonlinear system to change variable in a PDE

Question

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 $(r,s)$?

Answer 1

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!