Does Huygens principle holds for heterogeneous media (variable coefficients)?

Question

I'm having trouble to find references on that. Consider for instance a very simple model of a wave equation with variable coefficients: $$\partial_{tt}^2 u(x,t) - \nabla \cdot( a(x) \nabla u(x,t)) = f(x,t), \quad x \in \mathbb R^3,\ t>0,$$ with some initial conditions $u(\cdot,0) = u_0$ and $\partial_t u(\cdot,0) = u_1$, where $a$ is a non-constant function with the usual elliptic property.

My question is: does the Huygens principle hold for this equation ? This is well-known that it is true when $a$ is constant. I can't really find any good references, either these are some physics paper without any rigour, or these are very abstract papers on general hyperbolic models on manifolds and I can't really tell if it's applying to my problem or not because I'm having trouble to understand them.

Any help would be great !

Answer 1

A pretty extensive reference on Huygens' principle is

[G] Günther, Paul MR 946226 Huygens’ principle and hyperbolic equations, Perspectives in Mathematics ISBN: 0-12-307330-8.

First, some definitions. Denote by $A\cdot B = \delta^{ij} A_i B_j$ the usual Euclidean inner product and by $\nabla_i = \frac{\partial}{\partial x^i}$ the usual Euclidean gradient. Let $(\Sigma,h) = (\mathbb{R}^3,a^{-1}\delta_{ij} dx^i dx^j)$ be the spatial metric, where $a=a(x)$. The corresponding Laplace operator is $\Delta_h = \frac{1}{\det h} \nabla \cdot \sqrt{\det h} h^{-1}\cdot \nabla = \nabla \cdot a(x) \nabla - \frac{3}{2} (\nabla a) \cdot \nabla$. Note that the spatial scalar curvature is $$R_h = -a [-2\nabla\cdot\nabla \log a + \frac{1}{2} (\nabla\log a)\cdot(\nabla\log a)].$$ Let also the static spacetime $(M,g) = (\mathbb{R}\times \Sigma, -dt^2+h)$. The operator $P = \partial_{tt} - \Delta_h - 2 h^{ij} B_i \nabla_j - C$, with $B_i = \frac{3}{4} \nabla_i \log a$ and $C = 0$, is your wave operator.

Result: Thm.VII-3.1 of [G] implies that for a static spacetime $(M,g)$ such that $\Delta_h R_h$ is $\mu_h$-integrable and $$\tag{$*$} 0 \le \int_{\mathbb{R}^3} (\Delta_h R_h) \mu_h < \infty,$$ with $\mu_h = a^{-3/2} d^3x$ the volume element of $h$, Huygens' principle holds for $P$ if and only if it is a "trivial operator." The hardest part is chasing down the definition of "trivial operator" in the book and various corresponding properties. Prop.VI-3.6(b) implies that the necessary and sufficient condition for $P$ to be "trivial" is that the Weyl tensor $W_g = 0$, or equivalently $(M,g)$ is conformally flat, and some other conditions on the coefficients on $B_i$ and $C$. These conditions are examined in the discussion following the statement of Thm.VII-3.1 and in this case, I believe, come down to the requirement that $C = -\frac{1}{6} R_h$ and the strongest constraints of all, that $(\Sigma,h)$ is a space of constant curvature. A 3-space of constant curvature is locally isometric to either the Euclidean plane (zero curvature), the 3-sphere (positive curvature) or the 3-hyperbolic space (negative curvature). The sign of the curvature is determined by the sign of $R_h$. Thus, since your $C=0$, the only way that the Huygens' principle could be satisfied is if $(\Sigma,h)$ were a subset of flat Euclidean space, meaning that $a(x)$ is a constant.

Note that the integrand in $(*)$ is an exact form, $(\Delta_h R_h) \mu_h = d({*_h}d R_h)$. Hence, the result is automatically zero as long as Stokes' theorem could be applied, which happens precisely when the gradient $\nabla_h R_h$ goes to zero sufficiently quickly at infinity on $\mathbb{R}^3$. This of course depends on the asymptotics of $a(x)$ at infinity.