Outgoing exterior Dirichlet uniqueness for the smooth combined-field geometry

Set $w(x,t)=e^{kt}u(x)$ on the connected open set ${\mathrm{\Omega }}^{+}\times \mathbb{R}\subset {\mathbb{R}}^4$. Distributional differentiation gives ${\mathrm{\Delta }}_{x,t}w=e^{kt}({\mathrm{\Delta }}_{x}u+k^{2}u)=0$. By [F4] it has a smooth harmonic representative. This implies $u$ has a smooth Helmholtz representative: the distributional identity ${\mathrm{\partial }}_{t}w=kw$ is also a classical identity for this representative, so $w(x,t)=e^{kt}w(x,0)$, and comparison as distributions gives $u(x)=w(x,0)$.

Choose $a$ so large that $\overline{{\mathrm{\Omega }}^{-}}\subset B_a$, and fix $r>a$. The annular domain $B_r\setminus \overline{{\mathrm{\Omega }}^{-}}$ has inner boundary $\mathrm{\Gamma }$ and outer sphere. We derive the Green formula for this arbitrary weak solution from the trace/lift part of [F1]. A zero-trace $H^1$ function supported in a boundary chart is approximable by smooth functions supported on its own side: flatten the chart and use $v(z,t)={\int }_0^t{\mathrm{\partial }}_{s}v(z,s)ds$, which gives ${ϵ}^{-2}{\int }_{0<t<2ϵ}\mathrm{\mid }v{\mathrm{\mid }}^2\le C{\int }_{0<t<2ϵ}\mathrm{\mid }{\mathrm{\partial }}_{t}v{\mathrm{\mid }}^2\to 0$. A cutoff removing $t<ϵ$ therefore converges in $H^1$, and subsequent interior mollification, followed by the finite chart partition, proves the claim. Hence the weak PDE applies to every $H^1$ test with zero trace on both boundaries. For a compact inner trace lift $Lg$, define $⟨q^{+},g⟩=-\int (\mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }Lg}-k^{2}u\overline{Lg})$. The preceding approximation makes this independent of the lift, and [F1] bounds it in $H^{-1\mathrm{/}2}$. For any test subtract its inner trace lift and an outer-sphere lift supported away from $\mathrm{\Gamma }$; the remaining test has zero trace on both boundaries. The weak PDE kills that remainder, and classical integration by parts near the outer sphere is valid by step 1.1. Testing the resulting identity with $u$ gives ${\int }_{B_r\setminus \overline{{\mathrm{\Omega }}^{-}}}(\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2-k^2\mathrm{\mid }u{\mathrm{\mid }}^2)dx={\int }_{\mathrm{\mid }x\mathrm{\mid }=r}{\mathrm{\partial }}_{r}u\bar{u}dS-⟨q^{+},{\gamma }^{+}u⟩\mathrm{.}$ The last term is zero by the zero trace. The left side is real, hence $J(r)=r^2\mathrm{Im}{\int }_{{\mathbb{S}}^2}{\mathrm{\partial }}_{r}u(r\omega )\overline{u(r\omega )}d\omega =0$. Only the original obstacle boundary has zero trace; no zero trace is assumed on the enclosing sphere.

At this fixed $r$, both $u(r,\cdot )$ and ${\mathrm{\partial }}_{r}u(r,\cdot )$ are smooth, so their ordered harmonic projections converge in $L^2$ by [F2]. Cauchy–Schwarz passes the inner product to the limit of finite orthogonal sums. By [F2]–[F3], this gives $J(r)={\lim }_{N\to \mathrm{\infty }}{\sum }_{\mathrm{\ell }=0}^N\mathrm{\parallel }{c}_{\mathrm{\ell }}{\mathrm{\parallel }}_2^2\mathrm{/}k\mathrm{.}$ Every summand is nonnegative and step 2.1 gives zero, so all $c_{\mathrm{\ell }}=0$. Completeness in [F2] now makes $u(r,\cdot )=0$ in $L^2$, hence pointwise by smoothness. This holds for every $r>a$, so $u=0$ on $\mathrm{\mid }x\mathrm{\mid }>a$. No degree sum was interchanged with the radiation limit at infinity.

The harmonic lift $w$ of step 1.1 vanishes on $({\mathbb{R}}^3\setminus \overline{B_a})\times \mathbb{R}$, a nonempty open subset of ${\mathrm{\Omega }}^{+}\times \mathbb{R}$. The latter is connected because ${\mathrm{\Omega }}^{+}$ is connected by hypothesis. [F4] gives $w=0$ everywhere there, hence $u=w(\cdot ,0)=0$ on ${\mathrm{\Omega }}^{+}$.