12

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.

turnip
  • 415

5 Answers5

6

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.

Artem
  • 14,847
  • Maybe Lagrange and even earlier Bernoulli? And how they both were doing without Euler coefficients, widely called now Fourier coefficients? And not both, but three! – mkl314 May 01 '14 at 03:15
  • @mkl314 Right, thanks! I remembered that Lagrange was reviewing a paper with separation of variables, and somehow thought that it was Bernoulli's. Actually Bernoulli did it first in 1753, and Lagrange slightly later, in his Analytical Mechanics. Bernoulli did not care about coefficients at all, and Lagrange considered a finite system of $n$ second order ODEs (points on springs) so he did not need Fourier's coefficients, because he already had a finite set of initial conditions. – Artem May 01 '14 at 03:32
5

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.

user88595
  • 4,579
  • I am familiar with the wave and heat equations (I am a physicist). I still don't really understand the intuition behind this however. For example, why isn't it a good idea to take $\Psi = XY - Y$ as my trial solution? – turnip Apr 30 '14 at 22:15
  • As they mention above, $\Psi = XY$ allows us to separate the PDE into two ODEs, one for each variable. The form $\Psi = XY - Y$ does not offer this advantage. – Alex G. May 01 '14 at 03:37
  • 1
    @PPG: What you are proposing is the same. $XY - Y = (X-1)Y = X_1Y_1$ with $Y_1 = Y$ and $X_1 = X-1$ both only being functions of $y$ and $x$ respectively. I think you meant just some other combination of $X$ and $Y$. I guess you could but once you replace the function, you then need to separate variables on each side of the equation and the more complicated the combination, the less likely you are to be able to separate them. That,s why we stay simple by setting $XY$. Of course you could recognise some kind of pattern and want to try a different combination of $X$ and $Y$ but not likely. – user88595 May 01 '14 at 14:59
  • @AlexGrounds: Actually it does ;-) Just shifts $X$ by one which makes the further algebra harder to solve. – user88595 May 01 '14 at 15:01
  • Interesting. I may have to experiment to see what sort of things come out at the end. – turnip May 01 '14 at 22:55
3

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...

Eric Towers
  • 70,953
2

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.

JohnD
  • 14,681
  • Yes, the motivation was exactly what I was wondering about. Imagine being the first person to encounter such a problem; what would make you think that a product solution is something worth trying? Maybe it all happened by chance... – turnip Apr 30 '14 at 22:18
  • I wouldn't say by chance, rather we only see the successful methodologies! Seeing that ${\partial^2\over \partial x^2}u(x,y)$ becomes $X''(x)Y(y)$ when $u(x,y)=X(x)Y(y)$ is both tractable and offers sufficient warrant or motivation to at least look for solutions of that form. – JohnD May 01 '14 at 02:58
0

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.

Ellya
  • 12,149
  • 2
    An equation's being linear seems absolutely unimportant. This substitution trick's staying a viable option is rooted in its being the cheapest tools to immediately produce explicit partial solutions to the nonlinear BVPs and IBVPs for PDEs with power-like nonlinearities. – mkl314 May 01 '14 at 04:44