What is the motivation behind a product solution?

Question

Let's consider the simple differential equation:

$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$

And let's assume we have some regular homogeneous boundary conditions like:

$$ u(a, y) = 0$$ $$ u(L, y) = 0$$ $$ u'(x, y) = 0$$

(I just made these up at the top of my head so feel free to improvise if they don't work with what you want to say.)

My question is: What is the intuition/motivation needed in taking a product solution of the form: $$\Psi = X(x)Y(y)$$ From what I understand, we also assume that the BCs apply to $\Psi$ too.

From what I've been told, there is no proof of this, it's just a reasonable trial that seems to work well in a lot of cases.

Answer 1

It is probably impossible to say for sure what brought Bernoulli and slightly later (thans @mkl314) Lagrange to consider a product of two functions as a solution of the wave equations. However, a good guess would be: the knowledge that the wave equation describes, well, waves. D'Alambert actually already written down his formula that provides a direct interpretation in terms of traveling waves, however, it was well known to all the great names I wrote above, that, given some conditions at the boundaries, what is observed in an experiment is the standing waves (Bernoulli in his paper from 1753 gives drawings of his father, who did a lot of experiments considering waves). And it also was well known that the simplest analytical representation of the standing waves is $$ A(t)\cos (\omega x+\phi), $$ hence a guess to look for a solution in the form $$ T(t)X(x). $$ And, of course, Fourier came later to claim that any function can be represented by its Fourier series to provide a (nonrigorous at the time) justification of the method for the linear equations for which the superposition principle holds.

Added: See here Bernoulli's paper with obvious "Fourier series" and especially with figures on standing waves.

Answer 2

This the PDE version of seperation of variables, it is a viable option, when the equation is linear, and the substitution produces seperable results, and the boundary conditions are homogeneous.

And in these nice cases, there is no reason that the solution cannot be of that form.

Answer 3

I don't know how you prove that something is reasonable. A simple answer might be: someone who came before us tried it, and it worked; thus it is reasonable.

Maybe you are looking for some motivation for trying a product solution rather than an argument for why it is reasonable? If so, I'd say because when $\Psi$ is put back into the PDE, the partial derivatives on $\Psi$ become ordinary derivatives on either $X$ or $Y$. And at this stage in a student's learning, solving two ODEs is simpler than tackling this new creature called a PDE.

Answer 4

First of all, what you mean by product of solutions is usually called method of separation of variables.

It is not always something reasonable to do as you could be missing out a lot of solutions.

However it happens that it could be useful especially if it yields a satisfactory solution and that you can prove that the solution is unique. In that case you've found it!

The idea behind it is that we don't know how to solve PDEs so by using a separation of variables, we transform the PDEs into ODEs, generally much simpler to deal with.

The wave and heat equations are good examples where you use them alongside Fourier series. I could give you some links if you haven't seen them yet.

Answer 5

With the proper selection of coordinates, any differential equation that is solvable in integrals (sequentially) separates. (This is the meat and potatoes of the Lie symmetry techniques in solving differential equations.) If your system has a lot of "obvious" symmetry, then the separation of variables corresponding to the obvious symmetry is a good thing to try first.

Solving in a box? Try $X(x,t)\cdot Y(y,t) \cdot Z(z,t)$.

Solving in a cylinder? Try $R(r,t) \cdot \Theta(\theta,t) \cdot Z(z,t)$.

Et c...