Existence and uniqueness for the exterior vector Helmholtz problem with prescribed tangential trace and outgoing radiation

The test space and the variational form. Define $X:=\{v\in H^1(A;{\mathbb{C}}^3):v_{\text{tangential}}=0\text{ on }\mathrm{\partial }O\}$, the tangential part being taken with respect to the trace of $v$ on the smooth boundary component $\mathrm{\partial }O$; by the boundedness of the trace operator [L10] applied componentwise, $X$ is a closed subspace of the complex Hilbert space $H^1(A;{\mathbb{C}}^3)$ of [L21], hence complete. Define the sesquilinear form $a(E,v):={\int }_A(\mathrm{\nabla }E:\overline{\mathrm{\nabla }v}-k^{2}E\cdot \bar{v})dx+{\int }_{\mathrm{\partial }O}\kappa E_n\overline{v_n}dS-⟨T_{R}E,v{⟩}_{S_R}$, where $\mathrm{\nabla }E:\overline{\mathrm{\nabla }v}:={\sum }_{i,m}{\mathrm{\partial }}_i{E}_m\overline{{\mathrm{\partial }}_i{v}_m}$ and $⟨T_{R}E,v{⟩}_{S_R}:={\sum }_{m=1}^3⟨T_R{E}_m,v_m⟩$ uses the scalar DtN pairing of [L1] on the components. Then $\mathrm{\mid }a(E,v)\mathrm{\mid }\le C\mathrm{\parallel }E{\mathrm{\parallel }}_{H^1(A)}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1(A)}$: the first two terms are bounded by Cauchy-Schwarz [L19]; the boundary term is bounded by $\mathrm{\parallel }\kappa {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }E{\mathrm{\parallel }}_{L^2(\mathrm{\partial }O)}\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2(\mathrm{\partial }O)}\le C^{\mathrm{\prime }}\mathrm{\parallel }E{\mathrm{\parallel }}_{H^1}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$ by [L10]; and the last by the boundedness of $T_R:H^{1\mathrm{/}2}(S_R)\to H^{-1\mathrm{/}2}(S_R)$ [L1] combined with the exact fractional trace bound [L21] for $H^1(A)$ functions on the smooth sphere $S_R$.

Collar trace inequality on the inner boundary. For every $\delta >0$ there is $C_{\delta }$ such that every $u\in H^1(A;{\mathbb{C}}^3)$ satisfies $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(\mathrm{\partial }O)}^2\le \delta \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{L^2(A)}^2+C_{\delta }\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(A)}^2$. Indeed, cover $\mathrm{\partial }O$ by finitely many flattening charts of the smooth boundary (Bounded C^k domains and boundary charts) and let $\chi$ be a smooth cutoff, equal to $1$ near $\mathrm{\partial }O$ and supported in the resulting collar. On the collar, in normal coordinates $x=\mathrm{\Phi }(y,t)$ with $t$ proportional to the distance to $\mathrm{\partial }O$, the strip estimate [L12] with $p=2$, divided by $h$ and integrated over $y$ with the bounded chart Jacobians, gives $\mathrm{\parallel }u\chi {\mathrm{\parallel }}_{L^2(\mathrm{\partial }O)}^2\le C(h^{-1}\mathrm{\parallel }u\chi {\mathrm{\parallel }}_{L^2(A)}^2+h\mathrm{\parallel }\mathrm{\nabla }(u\chi ){\mathrm{\parallel }}_{L^2(A)}^2)$ for every smooth $u$; choosing $h$ so small that $2Ch\le \delta$, expanding $\mathrm{\nabla }(u\chi )=\chi \mathrm{\nabla }u+u\mathrm{\nabla }\chi$ and absorbing the cutoff terms into $C_{\delta }\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(A)}^2$ yields the claim for smooth $u$, and both sides are continuous in the $H^1(A)$ norm (boundary term by [L10]), while products of global $C_c^{\mathrm{\infty }}$ approximations with a fixed cutoff are dense in the $H^1$ classes supported in the collar by Compactly supported smooth functions are dense in W^{k,p}(R^n) and [L13]; the inequality therefore extends to all $u\in H^1(A;{\mathbb{C}}^3)$.

Smooth componentwise lift of the tangential datum. For $j=1,2,3$ let $(e_T{)}_j\in C^{\mathrm{\infty }}(\mathrm{\partial }O)$ be the Cartesian components of $e_T$ and extend them by zero to $\mathrm{\partial }A=\mathrm{\partial }O\cup S_R$, obtaining smooth functions $g_j\in C^{\mathrm{\infty }}(\mathrm{\partial }A)$. The domain $A\subset {\mathbb{R}}^3$ is bounded and smooth, and $\epsilon =I_3$ is real symmetric with $\alpha =\beta =1$. Thus [L3] with $\rho =0$, applied separately to the real and imaginary parts, there is a unique $G_j\in H^1(A)$ with trace $g_j$ and ${\int }_A\mathrm{\nabla }\bar{v}\cdot \mathrm{\nabla }{G}_j=0$ for all $v\in H_0^1(A)$; by [L4] with $\rho =0$ and the smooth Dirichlet data $g_j$, each $G_j$ is $C^{\mathrm{\infty }}$ on $\bar{A}$, is harmonic on $A$, and $G_j{\mathrm{\mid }}_{\mathrm{\partial }A}=g_j$. Hence $G:=(G_1,G_2,G_3)$ is $C^{\mathrm{\infty }}$ on $\bar{A}$, satisfies $\mathrm{\Delta }G=0$ componentwise, $G{\mathrm{\mid }}_{\mathrm{\partial }O}=e_T$ and $G{\mathrm{\mid }}_{S_R}=0$, and is tangential on $\mathrm{\partial }O$.

Coercivity of the shifted form. With $\lambda :=\mathrm{\parallel }\kappa {\mathrm{\parallel }}_{\mathrm{\infty }}(1+C_1)+k^2+1$, where $C_1$ is the constant of step 1.2 for $\delta :=\frac{1}{2}(\mathrm{\parallel }\kappa {\mathrm{\parallel }}_{\mathrm{\infty }}+1{)}^{-1}$, one has $\mathrm{Re}(a(E,E)+\lambda \mathrm{\parallel }E{\mathrm{\parallel }}_{L^2(A)}^2)\ge c\mathrm{\parallel }E{\mathrm{\parallel }}_{H^1(A)}^2$ for every $E\in X$, with $c=\frac{1}{2}$. Indeed $\mathrm{Re}a(E,E)=\mathrm{\parallel }\mathrm{\nabla }E{\mathrm{\parallel }}_2^2-k^2\mathrm{\parallel }E{\mathrm{\parallel }}_2^2+{\int }_{\mathrm{\partial }O}\kappa \mathrm{\mid }{E}_n{\mathrm{\mid }}^2-\mathrm{Re}⟨T_{R}E,E⟩$; the last term is nonnegative by $\mathrm{Re}⟨T_{R}f,f⟩\le 0$ of [L1] applied componentwise to $f=E{\mathrm{\mid }}_{S_R}$; and the boundary term obeys $\mathrm{\mid }{\int }_{\mathrm{\partial }O}\kappa \mathrm{\mid }{E}_n{\mathrm{\mid }}^2\mathrm{\mid }\le \mathrm{\parallel }\kappa {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }E{\mathrm{\parallel }}_{L^2(\mathrm{\partial }O)}^2\le \frac{1}{2}\mathrm{\parallel }\mathrm{\nabla }E{\mathrm{\parallel }}_2^2+\mathrm{\parallel }\kappa {\mathrm{\parallel }}_{\mathrm{\infty }}{C}_1\mathrm{\parallel }E{\mathrm{\parallel }}_2^2$ by step 1.2 and [L19].

The shifted form is boundedly invertible. By the Riesz representation theorem [L8], applied to the bounded conjugate-linear functional $v\mapsto a(E,v)+\lambda (E,v{)}_{L^2}$ on the complete space $X$, the shifted form is represented by a bounded linear operator $S:X\to X$ with $\mathrm{Re}⟨Sx,x⟩\ge c\mathrm{\parallel }x{\mathrm{\parallel }}^2$ and $\mathrm{\parallel }S\mathrm{\parallel }\le M$ for constants $c,M>0$ (step 2.1 and step 1.1). Choose $0<t<2c\mathrm{/}{M}^2$ and put $q:=\sqrt{1-2tc+t^2{M}^2}<1$. For $x,y\in X$, $\mathrm{\parallel }(I-tS)(x-y){\mathrm{\parallel }}^2\le q^2\mathrm{\parallel }x-y{\mathrm{\parallel }}^2$ by expanding the square and using [L19]; hence for every $f\in X$ the iteration $x_{n+1}:=(I-tS)x_n+tf$ is a strict contraction of the complete space $X$, so $\mathrm{\parallel }{x}_{n+1}-x_n\mathrm{\parallel }\le q^n\mathrm{\parallel }{x}_1-x_0\mathrm{\parallel }$. The geometric tail estimate makes $(x_n)$ Cauchy; completeness supplies its limit $x$, continuity gives $x=(I-tS)x+tf$ and $Sx=f$, and the same contraction inequality makes the fixed point unique. If $Sx=0$ then $c\mathrm{\parallel }x{\mathrm{\parallel }}^2\le \mathrm{Re}⟨Sx,x⟩=0$, so $x=0$; thus $S$ is boundedly invertible with $\mathrm{\parallel }{S}^{-1}\mathrm{\parallel }\le c^{-1}$.

The kernel of the form is trivial. Let $E\in X$ satisfy $a(E,v)=0$ for every $v\in X$. Taking $v=E$ and imaginary parts in step 2.1 gives $0=\mathrm{Im}a(E,E)=-\mathrm{Im}⟨T_{R}E,E⟩$, where $f=E{\mathrm{\mid }}_{S_R}$; by [L1] the imaginary pairing equals ${\sum }_m{\sum }_{l,m^{\mathrm{\prime }}}(\mathrm{Im}{t}_l)\mathrm{\mid }(f_m{)}_{l{m}^{\mathrm{\prime }}}{\mathrm{\mid }}^2$ with $\mathrm{Im}{t}_l>0$, so every coefficient vanishes, $f=0$ on $S_R$, and consequently $T_{R}f=0$. For every $v\in X$ supported away from $\mathrm{\partial }O$ the identity $a(E,v)=0$ has no inner-boundary contribution and reads ${\int }_A(\mathrm{\nabla }E:\overline{\mathrm{\nabla }v}-k^{2}E\cdot \bar{v})dx=⟨T_{R}E,v{⟩}_{S_R}=⟨T_{R}f,v{\mathrm{\mid }}_{S_R}⟩=0$; in particular every $\phi \in C_c^{\mathrm{\infty }}(D)$, whose restriction to $A$ is an admissible test vanishing near $\mathrm{\partial }O$, satisfies ${\int }_A\mathrm{\nabla }E:\overline{\mathrm{\nabla }\phi }dx=k^2{\int }_{A}E\cdot \bar{\phi }dx$, and taking $v$ compactly supported in the interior of $A$ gives $-\mathrm{\Delta }E=k^{2}E$ weakly on $A$. Define $\tilde{E}:=E$ on $A$ and $\tilde{E}:=0$ on $\{\mathrm{\mid }x\mathrm{\mid }>R\}$; assign any value on the null sphere $S_R$, so $\tilde{E}$ defines an $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2(D)$ class, and its only interface is the sphere $S_R$, across which the full trace of the field is $f=0$. For $\phi \in C_c^{\mathrm{\infty }}(D)$ the Sobolev-Gauss-Green formula [L11] on the bounded $C^1$ domain $A$, applied componentwise with $u=E_m\in W^{1,2}(A)$ and $v=\phi {\mathrm{\mid }}_A$, gives ${\int }_A{E}_m\overline{{\mathrm{\partial }}_i\phi }dx=-{\int }_A({\mathrm{\partial }}_i{E}_m)\bar{\phi }dx+{\int }_{\mathrm{\partial }A}(T{E}_m)(T\phi ){\nu }_{i}dS$; the boundary term over $\mathrm{\partial }O$ vanishes because $\phi$ is compactly supported in $D$, and the term over $S_R$ vanishes because $T{E}_m=f_m=0$ there; hence ${\int }_D{\tilde{E}}_m\overline{{\mathrm{\partial }}_i\phi }dx=-{\int }_A({\mathrm{\partial }}_i{E}_m)\bar{\phi }dx$, so the distributional gradient of $\tilde{E}$ is the field $(\mathrm{\nabla }E){\mathbf{1}}_A\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2(D)$, $\tilde{E}\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(D)$, and the weak identity above gives ${\int }_D\mathrm{\nabla }\tilde{E}:\overline{\mathrm{\nabla }\phi }dx=k^2{\int }_D\tilde{E}\cdot \bar{\phi }dx$. By [L6] each real and imaginary part of each component of $\tilde{E}$ lies in $H_{\mathrm{l}\mathrm{o}\mathrm{c}}^2(D)$ and satisfies $\mathrm{\Delta }w=-k^{2}w$ distributionally; $D$ is nonempty, open and connected and $\tilde{E}=0$ on the nonempty open set $D\cap \{\mathrm{\mid }x\mathrm{\mid }>R\}$, so [L5] gives $\tilde{E}=0$ on $D$, in particular $E=0$ on $A$.

Compactness of the $L^2$ pairing and the Fredholm reduction. Use the Riesz identification fixed in step 3.1 and let $J:X\to X$ be the represented $L^2$ pairing, $⟨Jx,v{⟩}_X=(x,v{)}_{L^2(A)}$. A conjugate-linear functional in $v$ is represented here by applying [L8] to its complex conjugate and then conjugating the pairing, so this identification is linear in $x$. The inclusion $X\hookrightarrow H^1(A;{\mathbb{C}}^3)$ is continuous and $A$ is a bounded $W^{1,2}$-extension domain [L13], so by [L14] with $p=2<n=3$ every bounded sequence in $H^1(A)$ has an $L^2$-convergent subsequence; by [L16] the Countable and Dependent Choice assumptions of the sequential criterion [L15] are available, so the inclusion $H^1(A)\hookrightarrow L^2(A)$ is compact, and so is its restriction $X\hookrightarrow L^2(A)$, hence also the composition $J$ of that restriction with the bounded represented pairing $L^2(A)\to X$. Therefore $K:=S^{-1}J:X\to X$ is compact, and the operator $A_{\mathrm{o}\mathrm{p}}:=S-\lambda J$ represented by $a$ satisfies $S^{-1}{A}_{\mathrm{o}\mathrm{p}}=I-\lambda K$; by the Fredholm alternative [L9] the operator $S^{-1}{A}_{\mathrm{o}\mathrm{p}}$ is boundedly invertible as soon as its kernel is trivial, and its kernel consists exactly of the $E\in X$ with $a(E,v)=0$ for all $v\in X$.

Bounded invertibility of $a$. By step 3.2 the kernel of $S^{-1}{A}_{\mathrm{o}\mathrm{p}}=I-\lambda K$ is trivial, so the Fredholm alternative [L9] makes it boundedly invertible; hence $A_{\mathrm{o}\mathrm{p}}=S(I-\lambda K)$ and the form $a$ are boundedly invertible: for every bounded conjugate-linear functional $F$ on $X$ there is exactly one $E\in X$ with $a(E,v)=F(v)$ for all $v\in X$, and $\mathrm{\parallel }E{\mathrm{\parallel }}_{H^1(A)}\le C\mathrm{\parallel }F{\mathrm{\parallel }}_{X^{\ast }}$.

The weak solution with prescribed tangential trace. Define on $X$ the functional $F(v):=-a(G,v)-{\int }_{\mathrm{\partial }O}({\mathrm{div}}_{\mathrm{\partial }O}{e}_T)\overline{v_n}dS$. Both summands are bounded in $v$: the first by step 1.1, the second because ${\mathrm{div}}_{\mathrm{\partial }O}{e}_T\in C^{\mathrm{\infty }}(\mathrm{\partial }O)$ and $v\mapsto v_n{\mathrm{\mid }}_{\mathrm{\partial }O}$ is trace-continuous into $L^2(\mathrm{\partial }O)$ by [L10], applied in the boundary charts. By step 5.1 there is exactly one $u\in X$ with $a(u,v)=F(v)$ for all $v\in X$; set $E:=G+u\in H^1(A;{\mathbb{C}}^3)$. Then $E_{\text{tangential}}=e_T$ on $\mathrm{\partial }O$ because $u\in X$ and $G$ is tangential there, and testing the defining identity with $v$ compactly supported in the interior of $A$ gives $-\mathrm{\Delta }E=k^{2}E$ weakly on $A$. Moreover, for every $v\in X$ supported away from $\mathrm{\partial }O$ one has $a(E,v)=a(G,v)+a(u,v)=-{\int }_{\mathrm{\partial }O}({\mathrm{div}}_{\mathrm{\partial }O}{e}_T)\overline{v_n}dS=0$, so the same integration-by-parts computation as in step 3.2 shows that the normal trace of $E$ on $S_R$ equals $T_{R}f$ in the weak trace sense, where $f=E{\mathrm{\mid }}_{S_R}=u{\mathrm{\mid }}_{S_R}\in H^{1\mathrm{/}2}(S_R;{\mathbb{C}}^3)$.

Gluing across the artificial sphere and smoothness on the open set $D$. Let $E_{\mathrm{o}\mathrm{u}\mathrm{t}}$ be the outgoing series of [L1] built from the trace $f=E{\mathrm{\mid }}_{S_R}$, defined on $\{\mathrm{\mid }x\mathrm{\mid }>R\}$; by [L1] it is a finite-local-energy solution there with normal derivative $T_{R}f$, and it is smooth away from the sphere. Define $\tilde{E}:=E$ on $A$ and $\tilde{E}:=E_{\mathrm{o}\mathrm{u}\mathrm{t}}$ on $\{\mathrm{\mid }x\mathrm{\mid }>R\}$. The two pieces have the same trace $f$ on $S_R$; applying the Sobolev-Gauss-Green formula [L11] on $A$ and on the exterior annulus $B_M\setminus {\bar{B}}_R$ (with the outward normals $+n$ and $-n$ on $S_R$) and adding, the two interface pairings cancel, so $\tilde{E}\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(D)$ with weak gradient the two-sided gradient. For every test $\phi \in C_c^{\mathrm{\infty }}(D)$ the weak equations on both sides then give ${\int }_D\mathrm{\nabla }\tilde{E}:\overline{\mathrm{\nabla }\phi }=k^2{\int }_D\tilde{E}\cdot \bar{\phi }$: on $A$ this is the weak identity of step 6.1, on the exterior it is the weak Helmholtz equation satisfied by the outgoing series, and the interface terms have cancelled. Since $k^2\tilde{E}\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^k(D)$ for every $k$ as soon as $\tilde{E}\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^k(D)$, repeated application of the interior gain-two estimate [L7] starting from $H_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ gives $\tilde{E}\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^m(D)$ for every $m$, hence a $C^{\mathrm{\infty }}(D)$ representative satisfying the classical Helmholtz equation on $D$.

Homogeneous tangential traces before boundary localization. Put $u:=E-G$, with the smooth harmonic lift $G$ of step 1.3; by step 6.1, $u\in X$, so its tangential trace on $\mathrm{\partial }O$ is zero. Fix a smooth normal collar chart with coordinates $(y,s)$, $s<0$ in $A$, and a smooth orthonormal frame $Q=({\tau }_1,{\tau }_2,n)$ extended constantly along its normal lines. Write $u=QU$; thus $U_1=U_2=0$ on the flat boundary. The weak identity for $u$ is $a(u,v)=-a(G,v)-{\int }_{\mathrm{\partial }O}({\mathrm{div}}_{\mathrm{\partial }O}{e}_T)\overline{v_n}$. Since $\mathrm{\Delta }G=0$, integrating the smooth lift term by parts in a chart away from $S_R$ gives the smooth volume datum $k^{2}G$ and normal boundary datum $-{\mathrm{div}}_{\mathrm{\partial }O}{e}_T-{\mathrm{\partial }}_n{G}_n$ (the trace $G_n$ is zero and ${\mathrm{\partial }}_{n}n=0$). Transporting and localizing this identity by a cutoff $\eta$ gives, for every compactly localized $H^1$ test $V$ with $V_1=V_2=0$ on the flat boundary, ${\sum }_m\int a^{ij}{\mathrm{\partial }}_j{U}_m\overline{{\mathrm{\partial }}_i{V}_m}+{\int }_B\beta U_3\overline{V_3}=\int f\cdot \bar{V}+{\sum }_i\int G_i\cdot \overline{{\mathrm{\partial }}_{i}V}+{\int }_{B}h\overline{V_3}\mathrm{.}$ The principal matrix $a^{ij}$ is smooth, uniformly positive and scalar in $m$, since $Q^{\mathsf{T}}Q=I$; $\beta =\kappa J_B$ is smooth and possibly signed; $f=k^{2}J(U+Q^{\mathsf{T}}G)+(\text{smooth})U+(\text{smooth})\mathrm{\partial }U\in L^2$; $G_i=(\text{smooth})U$ plus smooth lift/cutoff terms belongs to $H^1$; and $h=-({\mathrm{div}}_{\mathrm{\partial }O}{e}_T+{\mathrm{\partial }}_n{G}_n)\circ \psi J_B$ times the localized cutoff is smooth. Tests are admissible: their first two components have zero trace, so odd reflection of those components and even reflection of the third, followed by parity-preserving mollification and cutoff, approximate them in $H^1$; pulling back the smooth frame and chart and extending by zero produces fields in $X$. The difference-quotient tests below therefore act on $U$, whose first two traces actually vanish, rather than on $Q^{\mathsf{T}}E$ with its prescribed nonzero traces.

Tangential second derivatives and recovery of the normal derivatives. Let $k\in \{1,2\}$ be a tangential direction, put $W:={\delta }_h^{k}U=(U(y+h{e}_k,s)-U(y,s))\mathrm{/}h$ and test the system of step 7.2 with $V:=-{\delta }_{-h}^k({\eta }^{2}W)$ on a nested pair of half-boxes; this test has $V_1=V_2=0$ on $B$ and compact support, so it is admissible. The discrete integration by parts of [L17] and the shifted product rule give a principal term bounded below by $\alpha \mathrm{\parallel }\eta \mathrm{\nabla }W{\mathrm{\parallel }}_{L^2}^2$ with $\alpha >0$ the ellipticity constant of $a^{ij}$, while all remaining terms involve ${\delta }_{h}a$, $\mathrm{\nabla }\eta$, $W$, $\mathrm{\nabla }U$, $f$ or the transferred quotients of $G$, and are bounded by $ϵ\mathrm{\parallel }\eta \mathrm{\nabla }W{\mathrm{\parallel }}_2^2+C_{ϵ}(\mathrm{\parallel }U{\mathrm{\parallel }}_{H^1}^2+\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}^2+\mathrm{\parallel }G{\mathrm{\parallel }}_{H^1}^2)$ using [L17], the increment estimate $\mathrm{\parallel }{\delta }_h^{k}U{\mathrm{\parallel }}_{L^2}\le \mathrm{\parallel }{\mathrm{\partial }}_{k}U{\mathrm{\parallel }}_{L^2}$, and [L19]. On $B$ the Robin term gives ${\int }_B\beta (y+h{e}_k){\eta }^2\mathrm{\mid }{W}_3{\mathrm{\mid }}^2+{\int }_B({\delta }_h^k\beta )U_3{\eta }^2\overline{W_3}$; the trace estimate $\mathrm{\parallel }Tz{\mathrm{\parallel }}_{L^2(B)}^2\le ϵ\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_{L^2}^2+C_{ϵ}\mathrm{\parallel }z{\mathrm{\parallel }}_{L^2}^2$ of step 1.2 (with $B$ the flattened $\mathrm{\partial }O$) absorbs the first term for either sign of $\beta$, and bounded ${\delta }_h\beta$, the trace bound for $U_3$ and [L19] bound the second; the smooth datum $h$ contributes ${\int }_B({\delta }_h^{k}h)\overline{W_3}$, bounded by $\mathrm{\parallel }h{\mathrm{\parallel }}_{H^1(B)}\mathrm{\parallel }T({\eta }^2{W}_3){\mathrm{\parallel }}_{L^2(B)}$ and absorbed in the same way. Choosing the finitely many $ϵ$ small gives $\mathrm{\parallel }\eta \mathrm{\nabla }{\delta }_h^{k}U{\mathrm{\parallel }}_{L^2}\le C(\mathrm{\parallel }U{\mathrm{\parallel }}_{H^1}+\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}+\mathrm{\parallel }G{\mathrm{\parallel }}_{H^1}+\mathrm{\parallel }h{\mathrm{\parallel }}_{H^1(B)})$ uniformly in $h$, hence every tangential second derivative of $U$ lies in $L^2$ after the limit argument in which the bounded functional $\psi \mapsto {\lim }_h\int ({\mathrm{\partial }}_i{U}_m){\delta }_{-h}^k({\eta }^2\psi )$ is represented in $L^2$ by [L8]. From the interior equation $-{\mathrm{\partial }}_i(a^{ij}{\mathrm{\partial }}_j{U}_m)=f_m-{\mathrm{\partial }}_i{G}_{i,m}$ one then recovers $a^{33}{\mathrm{\partial }}_{ss}{U}_m={\mathrm{\partial }}_i{G}_{i,m}-f_m-{\sum }_{(i,j)\mathrm{\ne }(3,3)}{a}^{ij}{\mathrm{\partial }}_{ij}{U}_m-{\sum }_{i,j}({\mathrm{\partial }}_i{a}^{ij}){\mathrm{\partial }}_j{U}_m$, whose right side is now $L^2$ and whose coefficient satisfies $a^{33}\ge \alpha >0$; so all second derivatives of $U$ lie in $L^2$ on the smaller half-box, and a finite cover of the compact boundary gives $u\in H^2$ up to $\mathrm{\partial }O$ and hence $E=u+G\in H^2$, since $G$ is smooth.

Higher-order bootstrap and the classical boundary conclusion. Suppose $U\in H^{r+1}$ on the larger half-box for some $r\ge 1$; the data of the local system then satisfy $f\in H^r$, each divergence datum $G_i\in H^{r+1}$ and the lift-adjusted boundary datum $h$ is smooth because all coefficients are smooth. For a tangential multi-index $\alpha$ of order $r$ put $Z:=D_y^{\alpha }U$; differentiating the system of step 7.2 by $D_y^{\alpha }$ keeps the principal matrix $a^{ij}$ and the Robin coefficient $\beta$, and replaces the data by $f_{\alpha }\in L^2$, $G_{\alpha }\in H^1$ and $h_{\alpha }\in H^1(B)$ built from $D_y^{\alpha }f$, $D_y^{\alpha }G$ and the finitely many commutator terms $D_y^{\gamma }(\text{coefficient})D_y^{\alpha -\gamma }U$ with $0<\gamma \le \alpha$, all known by the induction hypothesis; the tangential traces of $Z$ are again zero in the first two components and free in the third, and localized tests for $Z$ are admissible by the test approximation of step 7.2; the differentiated Robin commutators contain traces of tangential derivatives of $U$ of order at most $r-1$, their tangential derivatives of one further order have $L^2$ traces because the corresponding interior derivatives of order at most $r$ belong to $H^1$; trace continuity and smooth approximation therefore place those commutators in $H^1(B)$ by [L10]. Applying step 8.1 to $Z$ gains two further tangential or mixed derivatives, and differentiating the normal-recovery equation of step 8.1 and inducting on the number of normal derivatives recovers the remaining ones; nested half-boxes and a finite boundary cover give $U\in H^m$ on each smaller patch for every $m$, with constants depending on the data, $\kappa$, the chart and the fixed nested covers. By the Sobolev embedding [L18] on a rounded bounded smooth subdomain of the chart that agrees with $s=0$ near the smaller patch and lies inside the larger half-box (hence an extension domain by [L13]) each $H^m$ class with $m>j+\frac{3}{2}$ has a $C^j$ representative on the closed patch, compatible for different $j$; hence $u$, and therefore $E=u+G$, has all derivatives continuous up to $\mathrm{\partial }O$; combining with the smooth gluing at $S_R$, [L22] on the bounded annulus $A$ supplies one smooth ambient representative near its closure, and the classical integration by parts in the local frame (equivalently [L11] with classical traces) turns $a(E,v)=-{\int }_{\mathrm{\partial }O}({\mathrm{div}}_{\mathrm{\partial }O}{e}_T)\overline{v_n}$ for admissible inner-collar tests into the boundary identities ${\mathrm{\partial }}_n{E}_n+\kappa E_n=-{\mathrm{div}}_{\mathrm{\partial }O}{e}_T$ and $E_{\text{tangential}}=e_T$ on $\mathrm{\partial }O$.

The divergence constraint. Let $q:=\mathrm{div}E$ on $D$. By step 9.1 and step 7.1, $E\in C^{\mathrm{\infty }}(\bar{D};{\mathbb{C}}^3)$ once the two-sided solution is matched at $S_R$; the mixed partials commute [L20], so $\mathrm{\Delta }q=\mathrm{div}\mathrm{\Delta }E=-k^2\mathrm{div}E=-k^{2}q$ on $D$. On $\mathrm{\partial }O$ the exact boundary identity $\mathrm{div}E={\mathrm{\partial }}_n{E}_n+(\mathrm{div}n)E_n+{\mathrm{div}}_{\mathrm{\partial }O}{E}_{\text{tangential}}$ holds in the classical sense with the chosen normal extension, and its first two terms sum to $-{\mathrm{div}}_{\mathrm{\partial }O}{e}_T$ by step 9.1 while the third equals ${\mathrm{div}}_{\mathrm{\partial }O}{e}_T$; hence $q=0$ on $\mathrm{\partial }O$. To check outgoingness without assuming it for derivatives, let $H_l(x)=\mathrm{\mid }x{\mathrm{\mid }}^l{Y}_{lm}(x\mathrm{/}\mathrm{\mid }x\mathrm{\mid })$ be the homogeneous harmonic polynomial of a single angular mode and set $J_i(x)=x_i{H}_l(x)-\mathrm{\mid }x{\mathrm{\mid }}^2{\mathrm{\partial }}_i{H}_l(x)\mathrm{/}(2l+1)$. Product differentiation gives $\mathrm{\Delta }(x_i{H}_l)=2{\mathrm{\partial }}_i{H}_l$ and $\mathrm{\Delta }(\mathrm{\mid }x{\mathrm{\mid }}^2{\mathrm{\partial }}_i{H}_l)=(4l+2){\mathrm{\partial }}_i{H}_l$, so $J_i$ is harmonic of degree $l+1$, while ${\mathrm{\partial }}_i{H}_l$ is harmonic of degree $l-1$ (zero when $l=0$), as in [L24]. Direct differentiation of the finite polynomial formula [L23] gives $h_l^{\mathrm{\prime }}=l{h}_l\mathrm{/}z-h_{l+1}$ for $l\ge 0$ and $h_l^{\mathrm{\prime }}=h_{l-1}-(l+1)h_l\mathrm{/}z$ for $l\ge 1$. The Cartesian derivative therefore is ${\mathrm{\partial }}_i[h_l(kr)Y_{lm}(n)]=-k{h}_{l+1}(kr)J_i(n)+k{h}_{l-1}(kr){\mathrm{\partial }}_i{H}_l(n)\mathrm{/}(2l+1)$ for $l\ge 1$; for $l=0$ only the first term occurs, since $h_0^{\mathrm{\prime }}=-h_1$ and ${\mathrm{\partial }}_i{H}_0=0$. Expand these two harmonic polynomials in the finite bases of their degrees [L24]. The all-derivative convergence of [L1] justifies differentiation and angular projection on every annulus strictly outside $S_R$; a fixed output degree receives terms only from the two adjacent input degrees, so each radial coefficient of $q$ is a multiple of its outgoing $h_j(kr)$, with no incoming partner. Since $q$ is smooth across $S_R$, its coefficients and radial derivatives extend continuously to $R$; nonvanishing of $h_j(kR)$ [L23] identifies those coefficients with the outgoing series for $q{\mathrm{\mid }}_{S_R}$ and gives ${\mathrm{\partial }}_{r}q=T_R(q{\mathrm{\mid }}_{S_R})$. Thus $q$ belongs to the exact outgoing class of [L1]. The exterior scalar uniqueness lemma [L2] applies to $q$ on the connected $D$ and gives $q=0$ on $D$, that is, the divergence constraint.