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?
[Math] Does separation of variables in PDEs give a general solution
partial differential equations
Related Solutions
The basic equations for which separation of variables works are classical types of linear equations.
- Laplace equation $$ \nabla^{2}u = 0. $$
- Wave equation $$ \frac{\partial^{2}u}{\partial t^{2}} = c^{2}\nabla^{2}u + Vu $$
- Heat equation $$ \frac{\partial u}{\partial t} = k\nabla^{2}u + Vu $$
- Schrodinger Equation $$ i\hbar \frac{\partial u}{\partial t} = -\frac{\hbar^{2}}{2\mu}\nabla^{2}u + Vu $$
You have to be able to separate variables, which puts restrictions on $V$. You can separate variables in different coordinate systems, but normally only using orthogonal coordinate systems, of which there are a couple of dozen reasonable ones where the Laplacian separates. The domains can be infinite or semi-infinite in some cases, or bounded with faces that are part of a constant coordinate face. Boundary conditions must be consistent with separation of variables.
When all of these conditions are in place so that separation of variables can succeed, then basically you get what you want. The decompositions are complete in $L^{2}$ spaces. General solutions will either be built from discrete sums or integral sums of separated solutions with respect to separation parameters. (It's theoretically possible to end up with really pathological decompositions requiring general measures to build up full solutions, but not for Physically realistic functions $V$.) However, you may need both Fourier sums and integrals in the general case. $L^{2}$ theory comes from general Spectral Theory for unbounded selfadjoint linear operators on a Hilbert space, even though the heat equation is better suited for $L^{1}$. You'd be surprised how much it takes to justify the expansions for the Sturm-Liouville ODEs arising out of separation of variables problems.
Keep in mind that all bets are off if you don't choose the conditions correctly to end up with selfadjoint problems for the ODEs, but well-posed Physical problems can be expected to lead to well-posed Mathematical problems. Basically, if your conditions are correct, and the coordinate systems are separable, then everything works out, at least so far as $L^{2}$ theory is concerned. A thorough examination of pointwise results is probably not realistic, however. Pointwise convergence results for the ordinary trigonometric Fourier series would fill a small library.
For linear equations, the technique of separation of variables is used to find all separated solutions of the form $X_1(x_1)X_2(x_2)\cdots X_n(x_n)$. You find them all if your equation can be separated.
If you make a change of variables, then you will generally find a different set of separated solutions. For example, you might separate $X(x)Y(y)$, or instead $R(r)\Theta(\theta)$ where $r=\sqrt{x^2+y^2}$, etc. The separated solutions you end up with using a different set of coordinates will be different because the separated solutions $u(x,y)=X(x)Y(y)$ that vanish at some $x$ vanish for all $y$ at that $x$; so the zero sets are coordinate-aligned. The solutions $u(r,\theta)=R(r)\Theta(\theta)$ will vanish on circles and or rays starting at the origin. Note: just because you can separate in one set of coordinates does not mean you can separate in another.
The Laplacian separates in Cartesian coordinates, and one contributing factor is that there are no mixed derivatives. Coordinate changes that introduce mixed terms most often result in equations that are no longer separable. For example, try $u=X(x)Y(y)$ in the following $$ u_{xx}+ 2u_{xy} +u_{yy} = 0 \\ X''Y + 2 X'Y' + XY'' = 0 \\ \frac{X''}{X}+2\frac{X'}{X}\frac{Y'}{Y}+\frac{Y''}{Y}=0. $$ For the Laplacian, avoiding mixed terms essentially requires that you use orthogonal coordinate systems to separate variables, but that's not always enough to be able to separate. The spherical coordinate system and the cylindrical coordinate system are orthogonal because the coordinate surfaces $r=A$, $\theta=B$, $\phi=C$ are mutually orthogonal where they intersect. So, when you separate variables you don't end up with mixed derivative terms. As I recall, there are 24 orthogonal coordinate systems where the Laplacian separates. This limits you, because you can only impose conditions on surfaces where a coordinate is constant, but there are enough systems that significant problems could be solved exactly a century ago, enough to develop Quantum Mechanics, for example.
If you're going to use the technique on a system, then all of the equations need to separate, and maybe in the same coordinate system.
The basic reason you have enough separated solutions to build up a full solution for classical operators is that they are formally selfadjoint. Symmetry is a big part of what makes separation of variables work to give you a full range of solutions. For example, $$ \int_{\Omega} \nabla^2 f g dx = \int_{\Omega} f \nabla^2 g dx + \mbox{eval terms} $$ Fortunately, symmetry and Physics combine in a powerful way; and symmetry and math combine in a powerful way. It's difficult to do much in either field without symmetry or something close to it.
Best Answer
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:
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.