TITLE: When does Separation of Variables yield basis of solution set?
QUESTION [14 upvotes]: In mathematical physics, one often employs the technique 'Separation of Variables' to find the full solution set to some linear partial differential equation.
For instance, consider the differential equation (1D heat equation):
$$\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0$$
The driving premise behind 'separation of variables' is that solutions of the following form make up a basis (are total in the sense that infinite, converging sums are allowed) for the entire solution space:
$$u(x,t) = X(x)T(t)$$
Let this set of solutions be denoted $U$. Let the full set of solutions be denoted $S$.
Is there a proof that $\overline{\mathrm{lin}}(U) = S$?
What is the complete classification of PDEs for which this is true?

I have another way of wording this question which may give some insight.
Let $X$ be the vector space of $L^2$ functions of $x$, and $T$ the vector space of $L^2$ functions of $t$. I think that $X \otimes T$ is the $L^2$ space of functions of $(x,t)$.
The Separation of Variables technique assumes that the subspace of $X \otimes T$ corresponding to the solution set is given by some $X_1 \otimes T_1$ where $X_1$ and $T_1$ are subspaces of $X$ and $T$ respectively, as these spaces admit bases of form $\{f_1 \otimes g_1, f_2 \otimes g_2 , \dots\}$.
However, not every subspace of $X \otimes T$ is of this form, as I found in this question. Given a PDE, how do we know when the solution set will be of the special form $X_1 \otimes T_1$?

REPLY [4 votes]: Here is a large class of counterexamples. Let $F(x,t)$ be a smooth nonvanishing function $\mathbb{R}^2 \to \mathbb{R}$ which $F$ does not factor as $g(x) h(t)$. As an example, I'll take $\exp(-x^2-xt-t^2)$. Consider the differential equation
$$\frac{\partial u}{\partial t} = \frac{\partial F/\partial t}{F} u.$$
So, in my example, $\tfrac{\partial u}{\partial t} = -(x+2t) u$. 
The solutions to this differential equation are of the form $F(x,t) v(x)$. So, if $F(x,t)$ does not factor, then no solution of this differential equation factors.

The second part of this answer is going to be very imprecise because I am terrible at functional analysis, but I am hoping someone else will write up an answer that fills in the gaps. Separation of variables should work when the system has time symmetry. This includes every example that DisintegratingByParts gives.
Let $X$ be a manifold and let's say we have a differential operator $D$ on $C^{\infty}(X \times \mathbb{R})$ which commutes with $\partial/(\partial t)$. 
Then the kernel of $D$ (space of solutions to the equation) is closed under $\partial/\partial t$. Therefore, this follows from proving some statement along the lines of "a nice subspace $V$ of $C^{\infty}(X \times \mathbb{R})$ which is closed under $\partial/\partial t$ obeys $V=\bigoplus_{\omega \in \mathbb{R}} V \cap C^{\infty}(X)  e^{i \omega t} $." This direct sum should presumably be in the sense of integrating against some sort of measure on $\mathbb{R}$. 
The finite dimensional analogue (which is true) is that, if $H$ is a finite dimensional vector space,  $A: H \to H$ is a diagonalizable operator such that $H = \bigoplus H_{\lambda}$ with $A$ acting by $\lambda$ on $H_{\lambda}$ and $V \subset H$ is an $A$-invariant subspace, then $V = \bigoplus_{\lambda} (V \cap H_{\lambda})$. The relevance is that the $i \omega$ eigenspace of $\partial/\partial t$ on $C^{\infty}(X \times \mathbb{R})$ is $C^{\infty}(X) e^{i \omega t}$.
As I said, I am terrible at functional analysis, so I don't know what the careful formulation of this statement would be. But I hope someone will come along to record it and give a reference for it.