How do the solutions to the wave and heat equations converge in general?

Question

I would like to check my understanding with someone if possible.

When we cover the heat and wave equations, for instance, in "methods" courses at university, they normally restrict the initial functions to fairly nice functions, and the implication (though never quite stated) is that the solution to the PDE converges at least pointwise, if not uniformly.

Am I right in thinking that if the initial functions are merely arbitrary $L^2$ functions (and thus $L^2$-limits of partial sums of eigenfunctions of the equation with suitable coefficients) then the series solutions to these PDEs are themselves no more than $L^2$ limits?

I'm in need of a bit of reassurance - or correction - on this, since 'methods' courses don't discuss it and the abstract courses leave it, by my perception, slightly implicit! By the way, I'm not interested in stronger convergence, since I'm trying to understand the value of the full Hilbert space setting from the point of view of solving these and similar equations.

Many thanks if you can help.

Answer 1

Since you tagged it (spectral-theory), let me give a sort of a high-brow answer.

It the most general case, what you are looking for "can" be interpreted as a question generalizing the classical question about convergence of Fourier series. Given a compact domain $D$, you can (formally) decompose any $L^2$ function with vanishing boundary conditions in terms of the Dirichlet eigenfunctions of the Laplacian on the domain.

$$ f = \sum_{i = 0}^{\infty} f_i $$

with

$$ -\triangle f_i = \lambda_i f_i $$

Assuming the series converges absolutely in $L^2$, then you can write the $L^2$ solution to the heat equation as

$$ u(t,x) = \sum_k e^{-t\lambda_k}f_k $$

and to the wave equation as (roughly speaking; in reality for the wave equation you have to prescribe two initial values)

$$ u(t,x) = \sum_k e^{it\sqrt{\lambda_k}} f_k $$

Using that for $t > 0$, $e^{-t\lambda_k} < 1$ and $|e^{it\sqrt{\lambda_k}}| = 1$ you have that $u(t,x)$ will still be defined as an $L^2$ limit using the convergence at the initial time.

Here, however, you see a big difference. As long as the smallest eigenvalue of the Laplacian is non zero (which one can expect since harmonic functions on bounded domains with Dirichlet boundary must be the 0 function). We have that the Fourier multipliers in the heat equation case decreases exponentially in time. While the multipliers in the wave equation case is roughly constant.

So for the heat equation, you actually have the following nice fact:

Consider a solution $u(t,x)$. At any future time $t > 0$, the function $(\triangle)^M u(t,x)$ which we can formally write as $\sum (\lambda_k)^M e^{-t\lambda_k} f_k$ is absolutely convergent in $L^2$, using that exponential decay (in $\lambda$) faster than any polynomial growth. So using either elliptic regularity theory (that $\triangle v\in L^2 \implies v\in W^{2,2}$) or just plain Sobolev embedding, you actually get that the function $u$ is immediately smooth for any future time $t$.

Furthermore, you can transfer convergences. You have that the partial sums converges in all $L^2$ Sobolev spaces with the tail exponentially decaying. So the Sobolev embeddings also tell you that the partial sums converge in $C^k$ for any $k$.

This is an illustration of the infinite smoothing property of parabolic equations.

---

By contrast, the same is not true for the wave equation. In general you can not expect the solution to converge any better than the initial data, simply because the wave equation is $L^2$ conservative. (The solution operator is a unitary map on $L^2$.) (Notice that the heat equation is dissipative: the $L^2$ norm decays exponentially in time.)

The simplest illustration is the solution in the one-dimensional case. On a string, we can solve the wave equation using the method of characteristics and the principle of superposition to be just a left and a right traveling wave bouncing around in the box, with the same profile. So if you start with a square wave in a box, where you will pick up the Gibbs phenomenon on the edges, you will all have the same persist for all time.

Answer 2

In the case of the heat equation on a bounded domain, suppose the decomposition of the initial function is

$$u(x,0)=f(x)=\sum_n A_n e^{inx}$$

Then the solution at time $t>0$ is given by

$$u(x,t)=\sum_n A_n e^{inx-n^2 t}=\sum_n \frac{A_n e^{inx}}{(e^{-t})^{n^2}}$$

Note that, as for any $L^2$ function, we have $A_n\to 0$ as $\lvert n\rvert\to \infty$. Therefore the series converges uniformly, for all $t$ except $t=0$, to a continuous (in fact smooth) function, and this holds in all dimensions.

I'm not so sure about the wave equation, but I believe your initial conditions need to be continuous for uniform convergence to hold, or at least this is the case in dimensions 1, 2 and 3 (probably also in higher dimensions, but I make no guarantees, since the qualitative behaviour of the wave equation does change across dimensions). Otherwise the best you can hope for is $L^2$ convergence.