Context:
In my recent answer I proved that a certain map $T$ is linear, and smooth. In that answer I work with a Cauchy foliation $\Omega_s(x,y,z)$ of a Lorentzian manifold. In this question I will take the same foliation of one lower dimension and embed it in Euclidean space with the usual metric, instead of Lorentzian space.
(2+1)-dim.
Let $x,y$ be space variables and let $s$ be a time variable i.e. $(2+1)$-dim. (later on I will reverse these dimensions).
With this embedding into Euclidean space it turns out that $\Omega$ now satisfies a nonlinear first order partial differential equation
$$2\sqrt{s}\frac{\partial}{\partial s}\sqrt{\Omega}=\sqrt{x}~\Psi\Omega$$
where $$\Psi \Omega := \sum_{i=1}^2 \sqrt{-\frac{\partial}{\partial x_i}\Omega}$$
let $x_1=x$ and $x_2=y$. Then equivalently we have
$$ 2\sqrt{s} \frac{\partial}{\partial s} \sqrt{\Omega_s(x,y)}=\sqrt{x}~\sqrt{-\frac{\partial}{\partial x}\Omega_s(x,y)}+ \sqrt{x}\sqrt{-\frac{\partial}{\partial y}\Omega_s(x,y)} \tag{1}$$
where we recover a particular solution using the ansatz:
$$\Omega_{s}(x,y)= \varphi_s(x)\varphi_s(y) \space\space\space\space\space{x,y\in(0,1)} \space\space\space{s >0}$$
where $\Omega_s(x,y)=e^{\frac{s}{\log x}+\frac{s}{\log y}}.$
This ansatz is leveraged from recognizing that $\varphi_s(x)$ is a solution to the linear parabolic diffusion equation:
$$ s\frac{\partial^2}{\partial s^2} \varphi_s(x)=-x\frac{\partial}{\partial x} \varphi_s(x) \tag{2} $$
and that $\varphi_s(y)$ is a solution to the linear parabolic diffusion equation:
$$s\frac{\partial^2}{\partial s^2} \varphi_s(y)=-y\frac{\partial}{\partial y} \varphi_s(y) \tag{3}$$
Probability theory:
If you're familiar with probability theory, if we think of $\varphi_s(x)$ and $\varphi_s(y)$ as unnormalized marginal distributions, then $\Omega_s(x,y)$ is just the joint distribution, for $x$ and $y$ independent random variables. This implies that the marginals satisfy linear PDE's but the joint satisfies a nonlinear PDE.
(1+2)-dim.
Now let $s_1$ and $s_2$ be time variables and let $x$ be the only space variable. We've reversed the above dimensions.
In this case, we have $$\psi_s(x)=e^{\frac{s_1}{\log x}}e^{\frac{s_2}{\log x}}=e^{\frac{s_1+s_2}{\log x}}$$
which solves the linear PDE
$$ \bigg(\sum_{i=1}^2 s_i\bigg)\Delta \psi=-2x\frac{\partial}{\partial x} \psi \tag{4} $$
Is there a theoretical reason why switching the dimensions of time and space here, results in a linear vs. nonlinear PDE?
Note that in one dimension both $\varphi$ and $\psi$ satisfy $(1)$ and $(4)$ making this nontrivial. In one dimension there is common ground between the linear and nonlinear PDE's but there is divergence when the dimensions increase.