Proposition: Explicit edge-zone defect

Let $\alpha \in {C_c}^{\infty }((0,\epsilon ))$ be nonnegative and not identically zero, let $\beta \in {C_c}^{\infty }({\mathbb{R}}^3)$ with $\int \beta {\mathrm{d}}^{3}x=1$, and let $\xi =\alpha (t)\beta (\mathbf{x}){\partial }_1$, that is ${\xi }^1=\alpha \beta$ and ${\xi }^0={\xi }^2={\xi }^3=0$. Then for every $t\in [\epsilon ,T]$,

Consequently ${\mathsf{X}}_1[K\xi ]\ne \xi$. Moreover, with ${{\Delta }^{(1)}}_{\epsilon ,\xi }:=(\xi -{\mathsf{X}}_1[K\xi ]{)}^{\mu }{\partial }_{\mu }{\varphi }_0$, there are a smooth ${\varphi }_0$ and an $f\in {C_c}^{\infty }({M_T}^{\circ })$ with $\int f{{\Delta }^{(1)}}_{\epsilon ,\xi }{\mathrm{d}}^{4}x\ne 0$. This $\xi$ is not in ${\mathfrak{g}}_U$.

Proof. The source $-{\partial }_{\alpha }({\sigma }_{\epsilon }{a_1}^{\alpha \mu }[K\xi ])$ is smooth and compactly supported, so $u^{\mu }:={{\mathsf{X}}_1}^{\mu }[K\xi ]$ has compact spatial support on each slice and the spatial integrals ${\mathcal{X}}^{\mu }(t)=\int u^{\mu }(t,\mathbf{x}){\mathrm{d}}^{3}x$, ${\Xi }^{\mu }(t)=\int {\xi }^{\mu }(t,\mathbf{x}){\mathrm{d}}^{3}x$ are smooth in $t$. Integrating the coefficient equation over space, the Laplacian term and spatial divergences drop out:

From ${a_1}^{0\mu }[K\xi ]=({\partial }_{\rho }{\xi }^{\rho }){\eta }^{0\mu }-{\partial }^0{\xi }^{\mu }-{\partial }^{\mu }{\xi }^0$, using ${\eta }^{0\mu }={{\delta }^{\mu }}_0$, ${\partial }^0={\partial }_0$, and the vanishing of integrated spatial derivatives,

Hence ${\ddot{\mathcal{X}}}^{\mu }=({\sigma }_{\epsilon }{\dot{\Xi }}^{\mu }{)}^{\cdot }$. Both ${\dot{\mathcal{X}}}^{\mu }$ and ${\sigma }_{\epsilon }{\dot{\Xi }}^{\mu }$ vanish at $t=0$, so ${\dot{\mathcal{X}}}^{\mu }={\sigma }_{\epsilon }{\dot{\Xi }}^{\mu }$ and ${\mathcal{X}}^{\mu }(t)={{\int }_0}^t{\sigma }_{\epsilon }{\dot{\Xi }}^{\mu }\mathrm{d}s$. For $\mu =1$, ${\Xi }^1=\alpha$, and for $t\ge \epsilon$,

while ${\xi }^1(t,\cdot )=0$ for $t\ge \epsilon$. The integral is strictly positive because ${\sigma }_{\epsilon }^{\prime }>0$ on $(0,\epsilon )$ and $\alpha \ge 0$ is not identically zero. This proves (eq:edgegap).

On $\{t\ge \epsilon \}$ the function $u^1$ solves the homogeneous wave equation and has nonzero spatial integral, so $u^1(t_1,{\mathbf{x}}_1)\ne 0$ at some $(t_1,{\mathbf{x}}_1)\in {M_T}^{\circ }$. Take ${\varphi }_0=x^1$; then ${{\Delta }^{(1)}}_{\epsilon ,\xi }=(\xi -u{)}^1=-u^1$ near that point, and a nonnegative bump $f$ supported where $u^1$ has fixed sign gives a nonzero smeared value. A free Klein–Gordon solution with ${\partial }_1{\varphi }_0\ne 0$ at a point where $u^1\ne 0$ works as well. Finally $\mathrm{supp}\xi \subset \{0<t<\epsilon \}$ is disjoint from $U$.