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 !



Does Huygens principle holds for heterogeneous media (variable coefficients)?
A pretty extensive reference on Huygens' principle is


[G] *Günther, Paul* MR 946226 [**Huygens’ principle and hyperbolic equations**](http://www.ams.org/mathscinet-getitem?mr=946226), *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.


