Maybe this is due to the fact that I never had a dedicated course about PDE but here is the situation. $$ \frac{\partial^2 u}{\partial \xi \partial \tau} = -e^u $$
And a suggested substitution $y = e^u$, $x = k \xi + \Omega \tau + \varphi$. However I am unsure how to actually apply it. Also does this mean that some information about the main equation is lost because we have only $x$ instead of $\xi$ and $\tau$?
The supposed result is then $$ k \Omega y y'' - k \Omega (y')^2+y^3=0 $$
If $x=\xi+\tau$ and $t=\xi-\tau$, then $\xi=\frac{1}{2}(x+t)$, $\tau=\frac{1}{2}(x-t)$, and $$ \frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+ \frac{\partial u}{\partial \tau}\frac{\partial \tau}{\partial x}=\frac{1}{2}\left(\frac{\partial u}{\partial \xi}+\frac{\partial u}{\partial \tau}\right), $$ and similarly $$ \frac{\partial u}{\partial t}=\frac{1}{2}\left(\frac{\partial u}{\partial \xi}-\frac{\partial u}{\partial \tau}\right). $$ Hence $$ \frac{\partial^2 u}{\partial^2t}-\frac{\partial^2 u}{\partial^2x}=\frac{1}{4}\left(\frac{\partial}{\partial \xi}-\frac{\partial}{\partial \tau}\right)^2u-\frac{1}{4}\left(\frac{\partial}{\partial \xi}+\frac{\partial}{\partial \tau}\right)^2u=-\frac{\partial^2u}{\partial\xi\partial \tau} $$
Let $\begin{cases}x=\dfrac{\xi+\tau}{2}\\t=\dfrac{\xi-\tau}{2}\end{cases}$ ,
Then $\dfrac{\partial u}{\partial\tau}=\dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial\tau}+\dfrac{\partial u}{\partial t}\dfrac{\partial t}{\partial\tau}=\dfrac{1}{2}\dfrac{\partial u}{\partial x}-\dfrac{1}{2}\dfrac{\partial u}{\partial t}$
$\dfrac{\partial^2u}{\partial\xi\partial\tau}=\dfrac{\partial}{\partial\xi}\left(\dfrac{1}{2}\dfrac{\partial u}{\partial x}-\dfrac{1}{2}\dfrac{\partial u}{\partial t}\right)=\dfrac{\partial}{\partial x}\left(\dfrac{1}{2}\dfrac{\partial u}{\partial x}-\dfrac{1}{2}\dfrac{\partial u}{\partial t}\right)\dfrac{\partial x}{\partial\xi}+\dfrac{\partial}{\partial t}\left(\dfrac{1}{2}\dfrac{\partial u}{\partial x}-\dfrac{1}{2}\dfrac{\partial u}{\partial t}\right)\dfrac{\partial t}{\partial\xi}=\dfrac{1}{2}\biggl(\dfrac{1}{2}\dfrac{\partial^2u}{\partial x^2}-\dfrac{1}{2}\dfrac{\partial^2u}{\partial x\partial t}\biggr)+\dfrac{1}{2}\biggl(\dfrac{1}{2}\dfrac{\partial^2u}{\partial x\partial t}-\dfrac{1}{2}\dfrac{\partial^2u}{\partial t^2}\biggr)=\dfrac{1}{4}\dfrac{\partial^2u}{\partial x^2}-\dfrac{1}{4}\dfrac{\partial^2u}{\partial t^2}$
$\therefore\dfrac{1}{4}\dfrac{\partial^2u}{\partial x^2}-\dfrac{1}{4}\dfrac{\partial^2u}{\partial t^2}=-e^u$
$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial^2u}{\partial x^2}+4e^u$
This is a PDE of the form http://eqworld.ipmnet.ru/en/solutions/npde/npde2103.pdf.
It has the general solution $u(x,t)=f(x-t)+g(x+t)-2\ln\left(k\int^{x-t}e^{f(r)}~dr-\dfrac{1}{2k}\int^{x+t}e^{g(s)}~ds\right)$
$\therefore u(\xi,\tau)=f(\tau)+g(\xi)-2\ln\left(k\int^\tau e^{f(r)}~dr-\dfrac{1}{2k}\int^\xi e^{g(s)}~ds\right)$
