The following is motivated by taking a product space $\Omega$ and splitting it into two parts via projections, whose subspaces, $T$ and $X$, are home to functions which satisfy a nonlinear PDE and a linear PDE respectively. I would like to understand if one can further take the Mellin followed by Fourier transforms of the nonlinear and linear PDE's respectively and arrive at the same resulting equation. I give some good evidence for this to be true. My question is at the end.

Begin by considering the nonlinear equation

$$c(t,x)\frac{\partial}{\partial t} \sqrt{g}=\Psi g \tag{1}$$

where $$\Psi g := \sum_{i=1}^n \sqrt{-\frac{\partial}{\partial x_i}g} \tag{2} $$

If we require that $g(t_1,x_i)$ satisfies $\forall i$

$$d(t_1,x_i)\frac{\partial^2}{\partial t_1^2}g(t_1,x_i)=- \frac{\partial}{\partial x_i} g(t_1,x_i) \tag{3} $$

and require that

$$ g:=g(t_1,x_1,x_2,\cdot\cdot\cdot,x_n) :=\prod_{i=1}^ng(t_1,x_i) \tag{4}$$

Then a particular solution can be found. For example let $c(t_1,x_1)=\frac{2\sqrt{t_1}}{\sqrt{x_1}}$ and let $d(t_1,x_i)={\frac{t_1}{x_i}}.$ We recover the function which leads to our solution $g$

$$ g(t_1,x_i)=e^{\frac{t_1}{\log x_i}} \tag{5} $$

for $x_i \ne 0,1$ and $t_1>0$.

We note that when $n=1,$ $g$ satisfies both the nonlinear $(1)$ and the linear $(3).$

We also note the following. If we suppose time is $n$-dimensional, then we have a generalization of $(3)$

$$ \bigg( \sum_{i=1}^n t_i \bigg) \Delta h=- n x_1 \frac{\partial}{\partial x_1}h \tag{6}$$

where we may write $$h:=h(x_1,t_1,t_2,\cdot\cdot\cdot,t_n)=\prod_{i=1}^n e^{\frac{t_i}{\log x_1}}$$

There is a projection at work here. We either collapse $n$-dimensional time down to one dimension and let space be $n$ dimensional, or vice versa. To see this we can let time and space have the same dimensions. Both will have multiple dimensions now. We form the $2n$-dimensional product space

$$ \Omega=\prod_{i=1}^n g(t_{i},x_i)=g(t_1,x_1)g(t_2,x_2)\cdot\cdot\cdot g(t_n,x_n) \tag{7} $$

Now $\forall i$ let $t_i=t_1.$ We recover $g$. Now $\forall i$ let $x_i=x_1$. We recover $h$. Note that $g$ satisfies $(1)$ and $h$ satisfies $(6)$.

In other words we have the time projection operator

$$ P_{T}: \Omega \to T $$

such that $$(t_1,t_2,\cdot\cdot\cdot,t_n,x_1,x_2,\cdot\cdot\cdot,x_n)\mapsto (t_1,x_1,x_2,\cdot\cdot\cdot,x_n) $$

and we have the space projection operator

$$ P_{X}: \Omega \to X $$

such that $$(t_1,t_2,\cdot\cdot\cdot,t_n,x_1,x_2,\cdot\cdot\cdot,x_n)\mapsto (t_1,t_2,\cdot\cdot\cdot,t_n,x_1) $$

I'm wondering why the time projection of the product space $\Omega$ yields a solution to a nonlinear PDE $(1)$ while the space projection of the product space $\Omega$ yields a solution to a linear PDE $(6).$

If $n=1$ then both $(1)$ and $(6)$ are simultaneously satisfied by $g$ and $h.$

One can transform $(4)$

$$ \Phi(t_1,z_1,z_2,\cdot\cdot\cdot, z_n)=\int_{(0,1)^n}x_1^{z_1}x_2^{z_2}\cdot\cdot\cdot x_n^{z_n}~ g~ \frac{dx_1dx_2\cdot\cdot\cdot dx_n}{x_1x_2 \cdot\cdot\cdot x_n} $$

and obtain

$$ \Phi(t_1,z_1,z_2,\cdot\cdot\cdot, z_n)= \prod_{i=1}^n \Phi(t_1,z_i)= \prod_{i=1}^n 2\sqrt{\frac{z_i}{t_1}}K_1\big(2\sqrt{z_i t_1}\big)$$

If $z_i=z_1$ $\forall i$ then we'll have a modified Bessel of the second kind to the $n$th power.

we can project down to

$$ \Phi(t_1,z_i)=2\sqrt{\frac{z_i}{t_1}}K_1\big(2\sqrt{z_i t_1}\big)$$

which individually satisfy the linear third order PDE

$$ z^2 \frac{\partial^3}{\partial z_i^3} \Phi(t_1,z_i)= t_1^2 \frac{\partial}{\partial t_1}\Phi(t_1,z_i) \tag{8}$$

We have a map given by the integral transform

$$ f: g \to \Phi $$

Here we know $f$ is linear, bijective, continuous and smooth.

**Question:**

Does $ \Phi(t_1,z_1,z_2,\cdot\cdot\cdot, z_n)$ satisfy some linear $n$-dimensional generalization of $(8)?$ (I would take a reference for this).

This will act like a bridge between $(1)$ and $(6).$

I formed the product space $$ \Xi = \prod_{i=1}^n \Phi(t_i,z_i) $$

and tried to investigate projections like I did with $\Omega.$ But I'm not aware of any $n$-dimensional linear PDE's that products of modified Bessel functions of the second kind satisfy, which are generalizations of $(8)$.

I think perhaps using a computer program could help here in terms of deducing the differential equation. But I'm also confident it can be done theoretically.