Consider the following system of partial differential equations:
$$ \begin{cases} 2\sqrt{s}\dfrac{\partial}{\partial s} \sqrt{\mp \Omega_s(x)}=\sqrt{x}\sqrt{\pm\dfrac{\partial}{\partial x}\Omega_s(x)} \\ s \dfrac{\partial^2}{\partial s^2}\Omega_s(x)=\pm x\dfrac{\partial}{\partial x}\Omega_s(x) \end{cases} $$
for Cauchy data given by $$\Omega_s(x)|_{s=0}$$
I discovered this system in the following way: I leveraged the ansatz that $\Omega_s(x)$ was a Cauchy foliation of a Lorentzian manifold $M,$ isometric to the classical Minkowski $(1+1)$ space in Dirac coordinates.
Working under these assumptions I derived that we should have $(M,g)$ with $g=\frac{dudv}{uv}.$
Next, I embedded $\Omega_s(x)$ in a Riemannian manifold, $\Bbb R^2$ with Cartesian coordinates and the usual metric. Then I showed that $\Omega_s(x)=e^{\frac{\mp s}{\log x}}$ for $x\ne 0,1$ satisfies the above system of equations.
There is a way to generalize and look at the $(n \times n)$ space and time lattice, which then the above system of equations corresponds to the lattice point $(1,1).$ In fact for any point on the lattice $(1,n)$ these lattice points correspond to solutions to an $n$-dim. version of the linear equation in the above system whilst any point on the lattice $(n,1)$ these lattice points correspond to solutions to an $n$-dim. version the the nonlinear equation in the above system.
And the point $(1,1)$ is special in that it corresponds to a solution to both the nonlinear and linear equation simultaneously.
The generalization of:
$$2\sqrt{s}\dfrac{\partial}{\partial s} \sqrt{\mp \Omega_s(x)}=\sqrt{x}\sqrt{\pm\dfrac{\partial}{\partial x}\Omega_s(x)}$$
is
$$\frac{2\sqrt{s}}{\sqrt{x_1}}\frac{\partial}{\partial s} \sqrt{\mp g}=\Psi g$$
for
$$\Psi g := \sum_{i=1}^n \sqrt{\pm \frac{\partial}{\partial x_i}g}$$
where
$$g:=\prod_{i=1}^n \Omega_s(x_i)=e^{\frac{\mp s}{\log x_i}}$$
Similarly,
$$\bigg( \sum_{i=1}^n s_i \bigg) \Delta h=\pm n x \frac{\partial}{\partial x}h$$
has solution given by a product
$$h:=\prod_{i=1}^n \Omega_{s_i}(x)$$
To conceptually think about this, we can think about $g$ and $h$ as specific projections of a more general product space, $\Phi$.
$$ \Phi:=\prod_{i=1}^n \Omega_{s_i}(x_i) $$
And realize that lattice points aside from $(1,n)$ and $(n,1)$ don't correspond to solutions to either of the PDE's.
Returning to the $(1+1)$ case and the initial system of equations at the very top of this post and taking the Mellin transform of $\Omega$ can I expect to recover a transformed version of the following nonlinear equation?
$$2\sqrt{s}\dfrac{\partial}{\partial s} \sqrt{\mp \Omega_s(x)}=\sqrt{x}\sqrt{\pm\dfrac{\partial}{\partial x}\Omega_s(x)}$$
Well the Mellin transform of $$s \dfrac{\partial^2}{\partial s^2}\Omega_s(x)=\pm x\dfrac{\partial}{\partial x}\Omega_s(x)$$
which is only valid for the $-$ sign,
returns
$$r^2 \frac{\partial^3}{\partial r^3} \chi_s(r)= s^2 \frac{\partial}{\partial s}\chi_s(r)$$
with particular solution $$ \chi_s(r)=2\sqrt{\frac{r}{s}}K_1(2\sqrt{rs}) $$
where $K_1$ is a modified Bessel function.
How do you take the Mellin transform of the nonlinear equation though?
I'm interested in this because then I could have $\chi_s(r)$ simultaneously satisfying a transformed version of this system:
$$ \begin{cases} 2\sqrt{s}\dfrac{\partial}{\partial s} \sqrt{\mp \Omega_s(x)}=\sqrt{x}\sqrt{\pm\dfrac{\partial}{\partial x}\Omega_s(x)} \\ s \dfrac{\partial^2}{\partial s^2}\Omega_s(x)=\pm x\dfrac{\partial}{\partial x}\Omega_s(x) \end{cases} $$
which would complete this system:
$$ \begin{cases} ???\\ r^2 \frac{\partial^3}{\partial r^3} \chi_s(r)= s^2 \frac{\partial}{\partial s}\chi_s(r) \end{cases} $$
for Cauchy data given by $$\chi_s(r)|_{r=0}$$