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The method of separation of variables is introduced in every textbook on mathematical physics.

A basic question is rarely addressed: does this method exhaust all the solutions?

Is there any fundamental theorem available for answering this question? At least for some kinds of equation?

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Commented Feb 25, 2015 at 22:29
  • $\begingroup$ Posted there too. $\endgroup$
    – Beamer
    Commented Feb 25, 2015 at 22:35
  • $\begingroup$ I believe if the individual functions you use in separation techniques form a complete set of functions then separation of variables will exhaust all solutions. We skimmed over completeness in our book as well on the basis that it's difficult to prove, although we did get a reference to outside texts. $\endgroup$
    – MonkeysUncle
    Commented Feb 25, 2015 at 22:41
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    $\begingroup$ Dear Jiang-min Zhang. In general, it is frown upon to cross-post simultaneously, because it may waste potential answerer's time. As a minimum you should mention the cross-post (on both sites!). The preferred procedure is to not cross-post, and if the post hasn't received an acceptable answer after, say, a couple of days, then you could flag for migration. $\endgroup$
    – Qmechanic
    Commented Feb 25, 2015 at 22:50
  • $\begingroup$ This is called Sturm-Liouville theory and this is a math question. $\endgroup$
    – DanielSank
    Commented Feb 26, 2015 at 7:39

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What do you mean exactly by "exhaust all the solutions"? If you mean: "Are there solutions to the equation that cannot be separated?" then the answer is yes. It's a tool for solving problems by constraining the form of possible solutions. By adding this constraint you're helping yourself solve the problem in some cases by ruling out other solutions, but that doesn't make those solutions impossible.

If instead you mean: "Can you construct all solutions by combining the solutions found from separation of variables in various ways?" then you're asking a more difficult question. In general the answer is no. For instance, in the 1-D case of Schrodinger's equation, you can use separation of variables to get a set of solutions and then use linear combinations of those solutions to construct any solution if you force the potential to not depend on time. So all of those solutions to Schrodinger's equation are separable. But there are, of course, cases where the potential does change with time and you will not in general be able to use your previous solutions to construct a solution that satisfies the equation.

Put another way, it's like the adage: all dimes are coins, but not all coins are dimes. All separable solutions are solutions, but not all solutions are separable. There are surely cases where separable solutions can be shown to span all of the possible solutions, but that is not always the case.

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I'll give you an example that shows that that one non-separable solution of a PDE, could have been obtained a sum of separable solutions.

$\beta x \frac{\partial \psi}{\partial x} - y \frac{\partial \psi}{\partial y} = 0$

Using separation of variables you find a set of solutions parametrized by $\alpha$

$X_\alpha(x)Y_\alpha(y) = (xy^\beta)^{\alpha/2}$

However there exists an inseparable set of solutions parametrized by $\gamma$

$\text{exp}(\gamma xy^\beta)$

However in this particular example the inseparable solution can be cast as an infinite sum of separable ones (owing to the linearity of the PDE) and the specific form of the answers

$\text{exp}(\gamma xy^\beta) = \sum_{\alpha\in2\mathbb{N}}C_\alpha X_\alpha(x)Y_\alpha(y) =\sum_{\alpha\in2\mathbb{N}} \frac{\gamma^{\alpha/2}}{(\alpha/2)!}(xy^\beta)^{\alpha/2}$

so that $C_\alpha \equiv \frac{\gamma^{\alpha/2}}{(\alpha/2)!}$

This does not say in generality whether we can do the same thing in all situations. But as has been mentioned earlier, it does if the basis of solutions is a complete basis

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