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 !

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/90779/2451 $\endgroup$
    – Qmechanic
    Jun 4, 2014 at 4:33
  • $\begingroup$ While I have no evidence to back me up, I'd like to think that it's due to orthogonality of space & time that we can assume $U(x,y,t)=X(x)Y(y)T(t)$. Note also that separation of variables only works for linear homogeneous PDEs with linear homogeneous boundary conditions. Without that, you have to resort to other techniques (e.g., Green's function) $\endgroup$
    – Kyle Kanos
    Jun 4, 2014 at 14:27
  • $\begingroup$ I had come across a reference which "proved" separation of variable tricks using Lie algebras, it was one of the things that motivated the discovery of Lie algebras by Lie. This is detailed in the book: ima.umn.edu/~miller/separationofvariables.html $\endgroup$
    – user7757
    Jun 19, 2014 at 5:32
  • $\begingroup$ @ramanujan_dirac that text is fascinating. However, so far as I see, math. I hold out hope for a more physical answer. Perhaps it's hidden somewhere in ima.umn.edu/~miller/sepofvariablestalk.pdf The problem of really understanding separation of variables in general is much harder than I first suspected. $\endgroup$ Jun 20, 2014 at 15:08

1 Answer 1


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:

  1. 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}

  2. 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.

  3. 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.

  • $\begingroup$ But, is this physical motivation? Moreover, are you certain there is not some coordinate system in which some of the problems you list here do not become separable? $\endgroup$ Oct 12, 2014 at 16:05
  • $\begingroup$ @James, you are right. You always choose a coordinate system that suits the symmetry of your problem. For example, if your domain is circular, you would write the Laplacian in polar coordinates, in which it could be separable. $\endgroup$
    – Amey Joshi
    Oct 12, 2014 at 16:09
  • $\begingroup$ I don't think we'll be able to carefully describe the extent to which separation of variables applies mathematically, see the linked work in my earlier comments. It's a deep question. That said, I am still hopeful there is some physical motivation... $\endgroup$ Oct 12, 2014 at 16:13

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