# Why does an eigenvalue expansion 'work' for PDEs?

I understand the logic and rationale behind using a series of eigenfunctions to represent general solutions to simple partial differential equations with prescribed boundary values, such as the heat/diffusion equation, Laplace's equation, the wave equation, etc., and I can see why the series would converge to an acceptable solution of the problem through theory on Fourier series. For example, for the heat equation with Neumann boundary conditions in one dimension, we have:

$$u_t = ku_{xx}, 0<x<L$$ and ultimately a series solution:

$$\frac{A_0}{2} + \sum_{n=1}^\infty A_n e^{{(-n\pi/L)}^2kt}\mathrm{cos}(\frac{n\pi x}{L})$$

with the coefficients fitted using initial data.

However, what I'm having trouble understanding is why $\mathbf{all}$ solutions to the PDE and boundary value problem can be represented as a series of eigenfunctions - I can see that some solutions definitely can be, but don't see why it is guaranteed that $\mathbf{any}$ solution of the problem can be represented as a series of Eigenfunctions.

In the above problem it seems to me like the eigenfunctions are solutions to a special class of separable equations and although the sum of that series is a solution, I don't understand why all solutions can be written out like that.

Please forgive me if it is obvious or if I have made any conceptual errors - I feel like I'm missing something really big here.

• The more important thing at this level is that you've solved the problem for any initial condition in L2. Uniqueness is a separate issue, which is not proved by using this technique. – Ian Oct 6 '14 at 14:18
• Do you mean PDEs or just linear PDEs? – user_of_math Oct 6 '14 at 16:33

Broadly speaking, because of two things:

1. The eigenfunctions of $\frac{d^2}{dx^2}$ form a basis for a space of functions on $(0,L)$ which contains all functions we want as initial conditions.

2. A solution with given boundary and initial conditions is unique.

So, if we have some solution of PDE $u(x,t)$, then by expanding $u(x,0)$ into a series of eigenfunctions (using 1) and forming a series solution $v(x,t)=\sum\dots$ we get that $v$ satisfies the same boundary and initial conditions as $u$, and therefore $u=v$ (using 2).

Now, the above are not precisely worded because technicalities may obscure the picture. They include:

• What is the space of functions in 1? (Typical answer: those with square-integrable first derivatives, and with prescribed boundary conditions.)
• What kinds of boundary conditions we allow? What do we mean by a basis? This develops into the Sturm–Liouville theory.
• What do we mean by a solution, do functions actually have to have two continuous derivatives? And in what sense are boundary/initial conditions are satisfied, what kind of limit do we take as $t\to 0$, $x\to 0$ or $x\to L$?

So... I would not make a bold claim about all solutions until it is perfectly clear what a solution means. This clarity tends to arrive with a set of assumptions that replace the vagueness of any and all with the language of function spaces and operators on them. The possibility of writing solutions as eigenfunction series is essentially a property of the differential operators involved in the equation.