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 $X_1 \otimes T_1$?

  • $\begingroup$ Note, your heat equation has a wrong sign. $\endgroup$ Feb 24, 2015 at 22:27
  • $\begingroup$ Right you are, thank you. $\endgroup$
    – Myridium
    Feb 24, 2015 at 22:39

2 Answers 2


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.

  • $\begingroup$ This is beyond me at the moment, I'm afraid. When you speak of separating variables, are you referring to after you've assumed a solution of the form $X(x)T(t)$? I would like to know how we know in the first place that we can make a basis from these solutions. $\endgroup$
    – Myridium
    Feb 24, 2015 at 22:55
  • $\begingroup$ @Myridium : Separating the variables means you start by assuming separated solutions $T(t)X(x)Y(y)\cdots$, plug them in, and you're able to separate out the variables one at a time, until you find end up with separate ordinary differential equations for the individual factors. For example, you get $X''+\lambda X=0$ and $T'+\lambda T=0$, etc. $\endgroup$ Feb 24, 2015 at 23:15
  • $\begingroup$ Even if all works out well and we can obtain separated solutions to the equations, how can we be sure that these form a basis? $\endgroup$
    – Myridium
    Feb 24, 2015 at 23:21
  • 2
    $\begingroup$ @Myridium : Perhaps I should be a little more clear. If you can separate variables, and if you pose the endpoint conditions correctly, then you end up with enough solutions to build up everything. As I mentioned, there is a tremendous body of Math devoted to justifying everything that Fourier originally did, but it does work out. You get a complete set (total set) of solutions to such problems. I noticed someone -1'd this answer, but it is a correct answer. $\endgroup$ Feb 25, 2015 at 23:14
  • 1
    $\begingroup$ I don't doubt that we end up with enough solutions to build up everything as you say. It's just that I'm a conscientious chap and I feel a need to prove something myself before I feel justified in using it. You mention that Fourier proved this originally? $\endgroup$
    – Myridium
    Feb 25, 2015 at 23:18

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .