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…
[Math] Finite speed of propagation of wave equation
ap.analysis-of-pdes
Related Solutions
You can approach this dually using that for classical solutions the finite propagation speed holds. This argument is similar in spirit to this answer of mine for low regularity uniqueness for the linear wave equation.
1
First write $w = u-v$. Then you see that $w$ solves a linear wave equation of the form
$$ \Box w = G(u,v) w $$
where the potential $G(u,v) \lesssim |u|^{p-1} + |v|^{p-1}$. Pretend now that $G$ is a smooth function of $(t,x)$.
2
Suppose for convenience that $x_0 = 0$. You want to prove that $w(t,\cdot) |_{B(0,t_0 - t)} = 0$ for $t\in (0,t_0)$. So it is enough to show that $\int w(t,x) f(x) ~dx = 0$ for every $f\in C^\infty_c(B(0,t_0-t))$. We will try to do so using a duality argument.
Fix $T\in (0,t_0)$, and take $f\in C^\infty_c(B(0,t_0-T))$. Solve the wave equation $$ \Box \varpi = G \varpi $$ with initial data $\varpi(0,T) = 0$ and $\partial_t\varpi(0,T) = f$. Under the assumption that $G$ is smooth, we have that $\varpi$ is a smooth function with compact support for all time. In fact, for $t\in (0,T)$ we have that $\mathrm{supp}~ \varpi \subset B(0,t_0 - t)$ using the finite speed of propagation for classical solutions.
3
That $w$ is an energy solution implies that the following identity holds for any test function $\varpi$:
$$ \int_0^T \int_{\mathbb{R}^d} \Box w \varpi - w \Box \varpi = \int_{\mathbb{R}^d} w \partial_t\varpi - \partial_t w \varpi \Big|_{t = 0}^T $$
Using the support property for $\varpi$ as derived above, you have that the right hand integral vanishes at $t = 0$ since $w,\partial_t w$ vanishes on $B(0,t_0)$. The integral at $t = T$ has only one term, and that is $\int w f$.
The left hand side however vanishes: since the two functions solve the same (linear) equation you have $LHS = \int Gw~v - w ~Gv = 0$.
4
The above works assuming that $G$ is smooth. In general, $G$ is not. Replace $G$ by $G_\epsilon$ through mollification. And replace $\varpi$ correspondingly by $\varpi_\epsilon$. Then it suffices to show that
$$ \int_0^T \int_{\mathbb{R}^d} (G - G_\epsilon) w \varpi_\epsilon \to 0 $$
Let me do the critical case for convenience; the argument should be similar for the subcritical cases with some adjustment of the exponents.
Noting that $p-1 = \frac{4}{d-2}$, by Strichartz inequality, $G$ belongs to space-time norm $L^1([0,T]; L^q_x)$ for any $q\in [\frac{d}{2},d]$. And the mollification will converge in that norm. Since $w$ is energy class you have that it is uniformly bounded in $L^\infty_t L^{2d/(d-2)}_x$.
The Strichartz estimates together with Gronwell's inequality can be used to show that if $G_\epsilon \in L^1_t L^d_x$, then
$$ \|\varpi_\epsilon(t)\|_{\dot{H}^1} \lesssim e^{C(T-t) \|G_\epsilon\|_{L^1_t L^d_x}} \|f\|_{L^2} $$
This shows that $\|\varpi_\epsilon(t)\|_{L^\infty_t L^{2d/(d-2)}}$ is uniformly bounded on the time interval $[0,T]$. And hence the fact that $G-G_\epsilon$ converges to zero in $L^1_t L^{d/2}$ gives that $|\int w(T,x) f(x) ~dx | = 0$ after taking the limit.
Strichartz estimates on domains is a difficult problem!
First: on bounded domains you cannot have any global in time Strichartz estimates. This is because of the presence of standing waves. (Set initial data to be an eigenfunction of the Laplacian.)
On the other hand, there is still the possibility of local in time Strichartz. But recall that Strichartz estimates capture dispersive phenomenon, where two wave packets starting out at the same location with different velocities will separate spatially. When you work on a domain, the wave packet may now hit the boundary and reflect back. So you will expect some degree of losses due to such "singularities".
There is a lot of research on how to understand this. I would suggest starting by looking up papers by Oana Ivanovici and tracing through the literature.
(The problem on "exterior" domains are somewhat easier, especially when the domains have nice boundaries such that each wave packet can only hit it at most once and reflect. [For example, when the domains are convex.] In those cases Strichartz estimates have been proven.)
Finally: Strichartz estimates for variable coefficient wave equations (on the whole space) have also been previously studied. Look up papers by Hart Smith and Daniel Tataru (together and separately).
Best Answer
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 $E\equiv 0$.