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From here, there are (at least) two different wave equations with variable wave speed.

Either $c^2(x)$ is outside the Laplacian: $$ \begin{cases}u_{tt} - c^2(x) \Delta u = 0 \quad \textrm{ in } \mathbb R \times \mathbb R^n \\ u(0,x) = f(x); \quad u_t(0,x)= g(x). \end{cases} $$ or $c^2(x)$ is between the two nablas: $$ \begin{cases}u_{tt} - \nabla\cdot(c^2(x)\nabla u) = 0 \quad \textrm{ in } \mathbb R \times \mathbb R^n \\ u(0,x) = f(x); \quad u_t(0,x)= g(x). \end{cases} $$ My question is, what do the differences in the location of $c^2(x)$ actually represent? What is the physical meaning distinguishing these two equations?

Wapiti
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    Consider the second equation (called "divergence form"). If for each $t$, $x\mapsto u(t,x)\in C^\infty_c(\mathbb R^n)$, then $\frac{d^2}{dt^2}\int_{\mathbb R^n}u(t,x)dx=0$, so that $\int_{\mathbb R^n}udx=\int_{\mathbb R^n} f(x)+t\int_{\mathbb R^n} g(x)$. So conservation of the center-of-mass holds in the divergence case. Note that energy conservation holds in both cases, but the definition of energy is $\int u_t^2+c^2(x)|\nabla u|^2dx$ in the second case. – user254433 May 09 '18 at 01:12
  • I don't understand how you deduce conservation of center-of-mass. Also, the definition of energy only has one time derivative in the divergence form equation? – Wapiti May 09 '18 at 01:45
  • If you integrate $u_{tt}-\nabla\cdot(c^2\nabla u)$ over each $B_k(0)$, apply the divergence theorem, and let $k\to\infty$, then $\int_{\mathbb R^n} u_{tt}=\lim\int_{\partial B_k}c^2(x)\frac{\partial u}{\partial n}d\sigma(x)=0$ since $x\mapsto u(t,x)\in C^\infty_c$.

    In divergence form, the energy density is $u_t^2+c^2|\nabla u|^2$, but it's $c^{-2}u_t^2+|\nabla u|^2$ in the nondivergence case (your first equation).

     – user254433 May 12 '18 at 05:00

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