Lemma: Lemma B: witness forcing mismatch

On the same slab $M_T$, with the transport (eq:htransport) and equal ordinary zero Cauchy data, VI's first coordinate equation becomes

The prepared candidate instead satisfies (eq:candidate-PDE). Their normalized forcing difference is

For every $\lambda >0$, its time component is nonzero as a smooth function. It is supported in the temporal preparation layer $[0,\epsilon ]$, but not in a compact spatial set.

There are a phase $\zeta \in \mathbb{C}$, $\left|\zeta \right|=1$, and a real nonnegative $f\in {C_c}^{\infty }({M_T}^{\circ })$ such that

Thus a real smooth witness detects the forcing mismatch.

Proof. Equation (eq:VI-first) and Lemma lem:transport give

Multiplying by $-1$ gives (eq:VI-normalized). Subtracting its forcing from the candidate forcing gives

which proves (eq:forcing-residual). The sign normalization is essential: the native candidate has ${\square }_{-}{X}_{\mathrm{cand}}=+F_{\mathrm{co}}$, not $-F_{\mathrm{co}}$.

Suppose ${{\mathfrak{r}}_{\lambda }}^0$ were identically zero. Then $(\sigma -1){V_{\lambda }}^{00}$ would be constant. Its value at $0$ is zero, so

Choose a nonempty open interval $J\subset (0,\epsilon )$ with $0<\sigma <1$. On $J$, the last equality implies ${V_{\lambda }}^{00}=0$, hence ${S_{\lambda }}^{00}=({V_{\lambda }}^{00}{)}^{\prime \prime }=0$. But the coefficient

in (eq:S00) has strictly nonzero real part on $J$. The exponential never vanishes. This is a contradiction.

The residual vanishes for $t\le 0$, because $V_{\lambda }=0$, and for $t\ge \epsilon$, because $\sigma =1$. At a time where it is nonzero it is nonzero at every spatial point; its support is therefore not spatially compact.

Choose $t_c\in (0,\epsilon )$ with ${{\mathfrak{r}}_{\lambda }}^0(t_c)\ne 0$, and choose

By continuity, $\mathrm{Re}(\zeta {{\mathfrak{r}}_{\lambda }}^0)>0$ on a sufficiently small interval about $t_c$. Take a nonnegative nonzero time bump $b$ supported there and a nonnegative nonzero spatial bump $c$. Then $f(t,\mathbf{x})=b(t)c(\mathbf{x})$ proves (eq:forcing-test). All operations are real linear after taking $\mathrm{Re}(\zeta \cdot )$.