# Finite speed of propagation of wave equation

Let us consider the wave equation $$(\partial_t^2 - \Delta)u = 0$$ on a domain $U$ with coercive homogeneous boundary conditions $$Bu|_{\partial U} = 0$$ that make $-\Delta$ self-adjoint. My question is, how can we construct such boundary conditions that the wave equation does not have finite speed of propagation any more? Can it be done at all? Thanks in advance...

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I don't see how a boundary condition would help you. You can show finite propagation speed locally, with an energy estimate. – Christian Remling Jun 18 '14 at 23:17
@ChristianRemling I am not sure that the local derivation works always. For example, consider the boundary condition $Bu|_{\partial U} = g$, $g$ non-zero. Here the solution has to have infinite speed of propagation necessarily. I feel that the usual proof works fine up to the Dirichlet or Neumann condition, but not sure about other boundary conditions. – student Jun 19 '14 at 0:55
@ChristianRemling If the boundary condition is $Bu|_{\partial U} = g, g$ non-zero, then $u(t, x) = 0$ will not even solve the wave equation. As far as I can see, the boundary condition matters. – student Jun 19 '14 at 1:38

The propagation speed is still finite because the following standard argument works independently of what happens at the boundary:

Assume that $u\in C^2$ solves the wave equation and $u(t=0,x)=0$ on $B_r(0)$ for some $r>0$, such that $\overline{B}_r\subset U$. Then $u(t,x)=0$ on $B_{r-|t|}$. If we apply this to the difference $u=u_2-u_1$ of two solutions that agree on some open set, we obtain finite propagation speed.

To prove this claim (for $t\ge 0$), consider $$E(t) = \int_{B_{r-t}} (u_t^2 + |\nabla u |^2)\, dx .$$ Then $$E'(t) = -\int_{S_{r-t}} (u_t^2 + |\nabla u |^2)\,d\sigma + 2\int_{B_{r-t}} (u_t u_{tt} + \nabla u_t \cdot \nabla u )\, dx .$$ Integration by parts lets me rewrite $$\int_{B_{r-t}}\nabla u_t \cdot \nabla u\, dx = -\int_{B_{r-t}} u_t\Delta u\, dx + \int_{S_{r-t}} u_t\, n\cdot \nabla u\, d\sigma ,$$ where $n$ denotes the outer normal unit vector on the sphere. Thus, using the equation, $$E'(t) = \int_{S_{r-t}}(-u_t^2-|\nabla u|^2+ 2 u_t\, n\cdot \nabla u )\, d\sigma \le 0 ,$$ so if $E(0)=0$, then we must have that $E\equiv 0$.

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Sorry for the late reply. This seems convincing enough. I have two questions: 1. Can you give a reference to the differentiation under the integration formula where the boundary is dependent on the variable? I know the formula you state is right, I am just looking for a reference. 2. I think the calculation you give should work even when the center of the ball $B_{r−t}$ is on the boundary. Is that correct? – student Jun 26 '14 at 20:31
(1) I think an easy argument is to write the integral in spherical coordinates, then the radius of $B$ only affects the integration in the radial variable and the claim becomes the fundamental theorem. (2) I don't think this can work, except perhaps in special situations (such as boundary is a half plane locally). – Christian Remling Jun 26 '14 at 21:28
But can you claim finite propagation speed if your argument does not work on balls centered at the boundary? – student Jun 26 '14 at 21:48
I am more interested in the physical significance of the term FPS. In other words, I want to make sure that if the support of the initial condition is away from the boundary, for some small time $t$ the solution will still be zero on the boundary. Can you please shed some light on this? This was the basic reason behind my question.... – student Jun 27 '14 at 1:17
Oh, so it seems I somewhat misunderstood the question (sorry). But can't we just argue as follows: Suppose that $u(t=0)$, $u_t(t=0)$ have compact support inside $\Omega$. Solve the wave equation in $\mathbb R^n$ with these initial conditions (and extended to the whole space in the obvious way, by setting them equal to zero outside $\Omega$). Then $u=0$ on $\partial\Omega$ for a while by FPS on the whole space. But this function, restricted to $\Omega$, solves your problem as well (as long as $u=0$ on $\partial\Omega$). – Christian Remling Jun 27 '14 at 1:57

The wave propagation speed can be infinite if your boundaries have a fractal shape, because of the different scaling of space and time. This idea is due to Strichartz, see Laplacians on fractals (2005).

For application to a Sierpinski gasket, see Infinite propagation speed for wave solutions on some post-critically-finite fractals (2011).

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from best I can tell the links you gave says infinite propagation speed is possible for wave equations on fractals, not wave equations on sets whose boundaries are fractals. – Willie Wong Jun 19 '14 at 14:32
hmm, I didn't intend to make a subtle distinction here, but if a fractal, say the Sierpinski gasket, is embedded in three-dimensional space, doesn't its boundary have a fractal dimension? – Carlo Beenakker Jun 19 '14 at 14:46
I am more wondering about the reverse: how bad must the interior be if the boundary is a fractal. And to your reply, no, not quite: for the Sierpinski gasket, for the purpose of its Laplacian, the boundary is only the three vertices of the big triangle. So the boundary is not a fractal. (Note that this boundary is not a boundary in the topological sense, but is the sense for which the boundary value problem makes sense; this has to do with how the fractal Laplacian is defined.) – Willie Wong Jun 19 '14 at 15:50