10
$\begingroup$

When a partial differential equation is solved using the separation of variables method, is the produced solution the most general one that satisfies the equation or have we lost some forms of the solution because of the assumption that it is in the form of separated variables?

$\endgroup$
3
$\begingroup$

Here go some thoughts:

  • Separation of variables is a powerful technique which may be particularly useful for boundary value problems and, generally speaking, when the equation is linear. Generally speaking again, the solution is fully recovered in terms of eigenfunctions (read about Sturm-Liouville theory for further understanding).

  • If no boundary conditions are specified, but a condition on the solution is required (for example passing through a curve) the method of characteristics if often used to write 2nd order PDEs into canonical form or turn 1st order PDEs into a set of ODEs defined over certain curves called characteristics.

An example, related to my question, is the following:

Consider the PDE:

$$z_{xx} - z_{yy} = 0, \quad (x,y) \in \mathbb{R}^2, \quad z = z(x,y). $$

Then, by separation of variables, i.e., assuming solutions $z = P(x)Q(y)$ we obtain:

$$P'' - \lambda P = 0, \quad Q''-\lambda Q = 0, \quad \lambda \in \mathbb{R}.$$

Then, for $\lambda < 0$ (note that we have no restrictions on $\lambda$), the solution is given by:

$$P(x) = A_1 \cos \mu x + A_2 \sin \mu x, \quad Q(y) = B_1 \cos \mu y + B_2 \sin \mu y, \quad \mu = \sqrt{|\lambda|}.$$ Then, the solution can be integrated over all possible values of $\lambda \in (-\infty,0)$ to make the solution independent from $\lambda$*.

Cheers!


$*$ This step is beyond my knowledge about PDEs and still remains unknown for me.

$\endgroup$
  • $\begingroup$ I think I understand now. The integration you mention in the last step can tranform the function from the form X(x)*T(t) to a more general form so we haven't lost possible solutions by assuming that the solution was seperable. $\endgroup$ – patatahooligan Sep 9 '14 at 17:08
  • $\begingroup$ That was precisely what I was wondering when I (tried) answered your question. $\endgroup$ – Dmoreno Sep 9 '14 at 17:10
  • $\begingroup$ In the case of sine/cosine function I think the integral (or series if there are boundary conditions restricting the values of μ) can indeed yield any function form. What I'm wondering is whether this would remain true for other forms of eigenfunctions (stemming from other forms of PDEs). I'll look it up and consider asking it as a seperate question because it's an issue of its own. $\endgroup$ – patatahooligan Sep 9 '14 at 17:16
  • $\begingroup$ I'm in a hurry now, but IMO this is a question which deserves a deeper discussion, so I will be looking after for any more elaborated answer or thoughts of yours. Cheers! $\endgroup$ – Dmoreno Sep 9 '14 at 17:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.