product solutions for PDEs, physical motivation Given a boundary value problem with independent variables $x_1,x_2, \dots , x_n$ and a PDE say $U(x_i, y, \partial_j y,\partial_{ij} y, \dots )=0$ we typically begin constructing a general solution by making the ansatz $y = F_1(x_1)F_2(x_2) \cdots F_n(x_n)$ where each $F_i$ is a function of just one variable. Next, we plug this product solution into the given PDE and typically we obtain a family of ODEs for $F_1, F_2, \dots, F_n$ which are necessarily related by a characteristic constant. Often, boundary values are given which force a particular spectrum of possible constants. Each allowed value gives us a solution and the general solution is assembled by summing the possible BV solutions. (there is more for nonhomogenous problems etc... here I sketch the basic technique I learned in second semester DEqns as a physics undergraduate)
Examples, the heat equation, the wave equation, Laplace's equation for the electrostatic case, Laplace's equation as seen from fluids, Schrodinger's equation. With the exception of the last, these are not quantum mechanical. My question is simply this:

what is the physical motivation for proposing a product solution to the classical PDEs of mathematical physics?

I would ask this in the MSE, but my question here is truly physical. Surely the reason "because it works" is a reason, but, I also hope there is a better reason. At the moment, I only have some fuzzy quantum mechanical reason and I have to think there must be a classical physical motivation as well since these problems are not found in quantum mechanics. 
Thanks in advance for any insights ! 
 A: Whether or not a PDE allows separation of variables depends not only on the equation, but also the boundary conditions. The following conditions must be satisfied for the method of separation of variables to work:


*

*The differential operator should be separable. An example of the one which is not separable is
\begin{equation}
\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial x \partial y} + \frac{\partial^2u}{\partial y^2} 
\end{equation}

*Continuing examples in two dimensions, the initial and boundary conditions should be on lines $x = $ constant and $y=$ constant. An example of violation of this condition is if we are attempting to solve
\begin{equation}
\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0
\end{equation}
but the domain is a rectangle with sides oblique to the coordinate axes.

*If the linear operators defining the boundary conditions at $x=$ constant ($y=$ constant) must not involve partial derivatives of $u$ with respect to $y$ ($x$) and their coefficients should be independent of $y$($x$). For example, if we are solving
\begin{equation}
\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0
\end{equation}
on a square with sides parallel to the coordinate axis but the boundary condition at one of the vertices is
\begin{equation}
\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0
\end{equation}
These conditions suggest that the symmetry of the problem decides whether separation of variables will be useful.
