Finite propagation for the wave equation on a Riemannian manifold

Fix a nonempty closed $K\subseteq M$ and write $\rho (x)=d_g(x,K)$. By [F2], $\rho$ is 1-Lipschitz in the intrinsic metric. Local metric comparison [F3] bounds the length of sufficiently short coordinate segments above by a constant times their Euclidean length, so $\rho$ is locally Euclidean-Lipschitz in each chart.

For a smooth real- or complex-valued solution of $L_{q}u=0$, set $e=\frac{1}{2}(\mathrm{\mid }{\mathrm{\partial }}_{t}u{\mathrm{\mid }}^2+\mathrm{\mid }{\mathrm{\nabla }}_{g}u{\mathrm{\mid }}_g^2+\mathrm{\mid }u{\mathrm{\mid }}^2)$. Since $q$ is real, the product rule gives ${\mathrm{\partial }}_{t}e=\mathrm{Re}\text{\hspace{0.17em}⁣}({\mathrm{div}}_g(\overline{{\mathrm{\partial }}_{t}u}{\mathrm{\nabla }}_{g}u)+(1-q)\bar{u}{\mathrm{\partial }}_{t}u)\mathrm{.}$

For a fixed coordinate vector $a$, every sufficiently short segment satisfies $\mathrm{\mid }\rho (z+sa)-\rho (z)\mathrm{\mid }\le d_g(z+sa,z)\le {\int }_0^s\sqrt{g_{z+ra}(a,a)}dr\mathrm{.}$ Reversing the segment gives the same estimate for negative $s$, so wherever the line derivative exists its absolute value is at most $\sqrt{g_z(a,a)}$.

For each nonzero rational $a$, an invertible linear coordinate change with first axis parallel to $a$ and [F4]–[F5] identify the almost-everywhere line derivative from step 2.1 with $a\cdot D\rho$; the zero direction is immediate. Countably many rational vectors and a countable coordinate cover allow one common null exceptional set by Countable Choice [A1]. Outside it, the estimate holds for every rational vector, and continuity extends it to every real tangent vector. Thus $\mathrm{\mid }{\mathrm{\nabla }}_g\rho {\mathrm{\mid }}_g\le 1$ almost everywhere.

Fix $ϵ>0$. Define $b(r)=0$ for $r\le 0$ and $b(r)=e^{-1\mathrm{/}r}$ for $r>0$, and set ${\theta }_{ϵ}(s)=b(s)\mathrm{/}(b(s)+b(ϵ-s))$. Every derivative of $e^{-1\mathrm{/}r}$ tends to zero as $r\downarrow 0$, so $b$ is smooth; the denominator is positive everywhere. On $(0,ϵ)$ the ratio $b(s)\mathrm{/}b(ϵ-s)$ is increasing because its logarithmic derivative is $s^{-2}+(ϵ-s{)}^{-2}>0$, and outside this interval ${\theta }_{ϵ}$ is constant. Thus ${\theta }_{ϵ}$ is smooth nondecreasing, equals zero on $(-\mathrm{\infty },0]$ and one on $[ϵ,\mathrm{\infty })$. For $\eta (t,x)={\theta }_{ϵ}(\rho (x)-t)$, the one-dimensional chain rule on almost every coordinate line and [F4] give ${\mathrm{\partial }}_t\eta =-{\theta }_{ϵ}^{\mathrm{\prime }}(\rho -t),{\mathrm{\nabla }}_g\eta ={\theta }_{ϵ}^{\mathrm{\prime }}(\rho -t){\mathrm{\nabla }}_g\rho ,$ so ${\mathrm{\partial }}_t\eta +\mathrm{\mid }{\mathrm{\nabla }}_g\eta {\mathrm{\mid }}_g\le 0$ almost everywhere. With ${\rho }_{\mathrm{\partial }}=d_g(\cdot ,\mathrm{\partial }M)$, the cutoff ${\eta }_{\mathrm{\partial }}(t,x)={\theta }_{ϵ}({\rho }_{\mathrm{\partial }}(x)-(t-s_0))$ has the same inequality for $t\ge s_0$ and has zero boundary trace.

Multiply that identity by a nonnegative Lipschitz weight $\eta$ and integrate over $M$. In each chart the product of $\eta$ with a smooth coefficient is Lipschitz, so [F4] gives its weak derivative by difference quotients. Choose a boundary-compatible coordinate cover and subordinate partition $({\chi }_j)$ by [F7]. On a boundary half-chart, one-dimensional integration by parts in the normal half-line includes the endpoint value; [F6] identifies its coordinate density with $\eta \mathrm{Re}(\overline{{\mathrm{\partial }}_{t}u}{\mathrm{\partial }}_{{\nu }_{out}}u)d{S}_g$. Summing the partitioned fields ${\chi }_j\eta \overline{{\mathrm{\partial }}_{t}u}{\mathrm{\nabla }}_{g}u$ cancels the extra terms since ${\sum }_j{\mathrm{\nabla }}_g{\chi }_j=0$. For $E_{\eta }(t)={\int }_M\eta ed{V}_g$, this yields $E_{\eta }^{\mathrm{\prime }}(t)={\int }_M(({\mathrm{\partial }}_t\eta )e-\mathrm{Re}(\overline{{\mathrm{\partial }}_{t}u}g({\mathrm{\nabla }}_g\eta ,{\mathrm{\nabla }}_{g}u))+\eta \mathrm{Re}((1-q)\bar{u}{\mathrm{\partial }}_{t}u))d{V}_g+{\int }_{\mathrm{\partial }M}\eta \mathrm{Re}(\overline{{\mathrm{\partial }}_{t}u}{\mathrm{\partial }}_{{\nu }_{out}}u)d{S}_g\mathrm{.}$ The identity holds for each fixed $t$; the spatial weak-derivative calculation applies at every such time, while $\eta ,{\mathrm{\partial }}_t\eta ,e$ are bounded and continuous in $(t,x)$. Differentiation under the integral makes $E_{\eta }$ continuously differentiable.

The inequality $\mathrm{\mid }{\mathrm{\partial }}_{t}u\mathrm{\mid }\mathrm{\mid }{\mathrm{\nabla }}_{g}u{\mathrm{\mid }}_g\le e$ and step 4.1 give $({\mathrm{\partial }}_t\eta )e-\mathrm{Re}(\overline{{\mathrm{\partial }}_{t}u}g({\mathrm{\nabla }}_g\eta ,{\mathrm{\nabla }}_{g}u))\le 0.$ Since $q$ is smooth, the open sets $U_x=\{y:\mathrm{\mid }1-q(y)\mathrm{\mid }<\mathrm{\mid }1-q(x)\mathrm{\mid }+1\}$ cover compact $M$. A finite subcover gives $C<\mathrm{\infty }$ with $\mathrm{\mid }1-q\mathrm{\mid }\le C$. Thus $\mathrm{\mid }(1-q)\bar{u}{\mathrm{\partial }}_{t}u\mathrm{\mid }\le Ce$. The boundary term vanishes if $u=0$ on $\mathrm{\partial }M$ and also if $\eta =0$ there. In either case $E_{\eta }^{\mathrm{\prime }}(t)\le C{E}_{\eta }(t)$. If $E_{\eta }(s)=0$, the nonnegative differentiable function $H(t)=e^{-C(t-s)}{E}_{\eta }(t)$ satisfies $H(s)=0$ and $H^{\mathrm{\prime }}(t)\le 0$, so $H=E_{\eta }=0$ for all $t\ge s$.

For (i), take $\rho =d_g(\cdot ,K)$ and $\eta ={\theta }_{ϵ}(\rho -t)$. The homogeneous Dirichlet condition removes the boundary term. At $t=0$, $\eta$ vanishes on $K$; on $M\setminus K$, $u_0,u_1$ and the spatial gradient of $u_0$ vanish because their supports lie in $K$. Thus $E_{\eta }(0)=0$, and step 6.1 gives $E_{\eta }(t)=0$. Continuity and nonnegativity force $e(t,x)=0$ wherever $\eta (t,x)>0$, in particular where $d_g(x,K)>t+ϵ$. Given any $x$ with $d_g(x,K)>t$, choose $0<ϵ<d_g(x,K)-t$; then $u(t,x)=0$.

For (ii), first use $\eta \equiv 1$ on $[0,s_0]\times M$. The boundary flux vanishes because $h=0$ there, and the initial energy is zero; step 6.1 gives $u={\mathrm{\partial }}_{t}u={\mathrm{\nabla }}_{g}u=0$ up to $s_0$. With ${\rho }_{\mathrm{\partial }}=d_g(\cdot ,\mathrm{\partial }M)$, use ${\eta }_{\mathrm{\partial }}(t,x)={\theta }_{ϵ}({\rho }_{\mathrm{\partial }}(x)-(t-s_0))$. Its boundary trace is zero and its energy at $s_0$ is zero. Steps 5.1–6.1 yield $E_{{\eta }_{\mathrm{\partial }}}=0$; the continuity argument in step 7.1 then gives $u(t,x)=0$ whenever $d_g(x,\mathrm{\partial }M)>t-s_0$.

Reverse time by setting $\tilde{v}(s,x)=v(T-s,x)$. Since $q$ is time independent, $L_{q}v=0$ implies $L_q\tilde{v}=0$. Applying (ii) from the reversed initial time gives the terminal-data assertion.

Set $\delta =d_g(Q,\mathrm{\partial }M)>T\mathrm{/}2$. For every $t\in [0,T]$, either $t<\delta$ or $T-t<\delta$: if $t\le T\mathrm{/}2$ the first holds, and if $t\ge T\mathrm{/}2$ the second holds. In the first case (ii) with $s_0=0$ gives $u(t,\cdot )=0$ on $Q$; in the second case the terminal assertion gives $v(t,\cdot )=0$ on $Q$. Thus $(q_1-q_2)uv=0$ pointwise and its integral vanishes.