Hi everone,

Ok so to begin with I know that for Dirichlet, Neuman and Robyn boundary conditions you can use the method of fourier series to solve the heat and wave equations given Cauchy data on $[0,L]$. This essentially uses the fact that the bounary conditions are symmetric meaning that if I'm considering the operator $-d^2/dx^2$ with boundary conditions on $0$ and $L$ then I need $XX^{\prime}|_0^L = 0$ for the operator to be self adjoint so that I can use spectral theory on the solution operator (which is compact).

So when I'm given non-homogeneous boundary conditions like $u(0,t)=h(t)$ and $u(L,t)=g(t)$ then one method is to subtract off these boundary values by some function so that I can reduce it to solving $u_{tt} - u_{xx} = f$ with $u(0,t)=0=u(L,t)$. The same holds true for Neuman and Robyn boundary conditions. So my question becomes:

Question: Exactly what type of boundary conditions can I pose on the wave/heat equation on $[0,L]$ so that the method of Fourier series works to generate a solution? Is a necessary and sufficient condition that the boundary conditions be symmetric?

There are obvious compatibility conditions with the Cauchy data/boundary data but this is not my concern here. I'm simply aiming to understand when we can write out the solution was $u(t,x) = \sum_{k} c_k(t)e^{ikx}$. My feeling is that you need the boundary conditions to be symmetric for all times so that you may indeed write $c_k(t)$ in terms of the eigenvalues of $-d^2/dx^2$ for those given boundary conditions.

I hope this is clear. Best, Dorian


I don't have a complete answer, but just some preliminary thoughts: the idea behind using Fourier analysis to solve constant coefficient linear PDEs is to transform a partial differential equation into an ordinary differential equation. In symbols, suppose $\psi:I\times\mathbb{R}^n\to \mathbb{C}$ solves the PDE

$$ \sum_{0\leq i \leq N} P_i(\nabla) \partial_t^i \psi = 0 $$

where $P_i$ are constant coefficient polynomials, the Fourier transform "gives an equation"

$$ \sum_{0 \leq i \leq N} P_i(\xi) \partial_t^i \hat{\psi} = 0~. $$

The problem is: how do you interpret this equation? To treat it as an ODE, you need to treat $\hat{\psi}$ as a map $I \to X$ where $X$ is, say, the Hilbert space of $L^2$ functions over $\mathbb{R}^n$ or some such.

Now, in your case of prescribing boundary conditions for the interval $[0,1]$, if your boundary conditions were time independent, and if the boundary conditions plays well with the Fourier transform, then you can again recover the ODE formulation. (The solvability of the ODE, as you noted, depends on which Hilbert space you use and the properties of the polynomials $P_i$ on the Hilbert space.)

But if your boundary conditions are time dependent, then an immediate problem is that the Hilbert space in which $\hat{\psi}$ lives will be time-dependent. So the naive application of "Fourier" methods won't make sense. Geometrically the case where $X$ is a fixed Hilbert space is the analogue of solving an ODE on the trivial vector bundle $V$ over $I$ with trivial connection. The case where $X$ also varies with time can be thought of as having some sort of an attempt at writing down an ODE on an arbitrary vector bundle $V$ over $I$. Without specifying the connection, even the notion of an ODE is not well-defined.

To put it differently, since a connection over a curve is just an identification of the fibres (roughly speaking), what you need to use an analogue of the Fourier method is a collection of 1-parameter families of functions $\phi_i(t;x)$, such that

  • For each fixed $t$, the functions $\phi_i(t;x)$ forms an ON basis of some appropriate Hilbert space
  • Each $\phi_i(t;x)$ solves the PDE you are looking at

Just directly assuming the trace of $\phi_i$ on constant $t$ slices are the trigonometric functions is probably not the right way to go in general.

Not having thought about this problem in detail before, I don't have much more to say. But I suspect that the suitability of individual boundary conditions need to be examined on a case by case basis.


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