Does separation of variables in PDEs give a general solution?

Question

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?

Answer 1

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!

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