# Finite propagation speed of second order operators

I was reading about the finite propagation speed of the wave equation on a Riemannian manifold. I was wondering if instead of the Laplace-Beltrami operator $\Delta$ we consider the equation $$\partial_t^2 u = Lu$$ where $L$ is a negative-definite self-adjoint elliptic operator of order 2, would we still have finite speed of propagation property? Thanks a lot for your help.

• Searching for "finite speed" and "elliptic" on MathSciNet turns up ams.org/mathscinet-getitem?mr=1205177 Section 1.3 of which is "1.3. Finite speed propagation and essential selfadjointness". – Willie Wong Feb 3 '14 at 9:22
• I have the impression that any textbook on linear hyperbolic PDE's discusses this. – Deane Yang Feb 3 '14 at 20:57

Your example is a special case of an operator with a wave-like principal symbol on a Lorentzian manifold. Namely, if the coefficients of the second order derivatives in $L$ are $L = h^{ij} \partial_i \partial_j + \cdots$, then $h^{ij}$ is an inverse Riemannian metric and it is always possible to write $L = \Delta_h + B$, with $\Delta_h$ the $h$-Laplacian and $B$ a first order differential operator. Your equation then becomes $\Delta_g \psi = B\psi$, where $\Delta_g = \partial_t^2 - \Delta_h$ is the $g$-Laplacian and $g=\mathrm{diag}(1,-h)$ is a Lorentzian metric. In other words, $\Delta_g$ is a wave operator. Thus, you get the usual finite speed of propagation results for your equation, because the lower order operator $B$ does not affect them.