product solutions for PDEs, physical motivation

Question

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 !

This post imported from StackExchange Physics at 2014-06-06 20:04 (UCT), posted by SE-user James S. Cook

Comments

Related: physics.stackexchange.com/q/90779/2451

This post imported from StackExchange Physics at 2014-06-06 20:04 (UCT), posted by SE-user Qmechanic

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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)

This post imported from StackExchange Physics at 2014-06-06 20:04 (UCT), posted by SE-user Kyle Kanos

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Does BV stand for Boundary Value? Related post http://physics.stackexchange.com/questions/109750/when-can-we-assume-that-the-wavefunction-is-separable/109787#109787

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Hi @user26143 nice to see you here :-). I there are on topic SE posts (your own or the ones of others) you would like to see on PhysicsOverflow, you can request them to be imported here.

Answer 1

The product ansatz just assumes that each variable can be varied independently, with the effect on the solution being independent relative perturbations. Such an ansatz is useful for linear PDEs whenever (and only when) the differential operator in the problem is a sum of differential operators depending one one variable only. 

Comments

thanks for this answer, I would have upvoted it long ago, but I've been a little slow to gain rep around here.

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Hi @JamesSCook if you have SE posts you would like to have on PhysicsOverflow too, you could suggest them to be imported and then you would have more (well deserved!) rep too ... ;-)