Domain Optimization for an Acoustic Waveguide Scattering Problem

Abstract

We consider a domain optimization problem of an unbounded domain, which models scattering of a time-harmonic acoustic wave at the junction of two closed waveguides. Solutions of our problem fulfill the Helmholtz equation with a real wavenumber, a modal radiation condition and homogeneous Dirichlet boundary conditions. We derive an a-priori bound for the solution on a certain class of domains (which is compact in the Hausdorff metric topology) and show that within this class the solution depends $H^1$-continuously on the domain. Furthermore, we show some numerical examples to illustrate our results, which were calculated using the domain (or shape) derivative of our problem.

Appendices

Appendix 1: On the Trace Spaces

This section will deal briefly with our trace spaces $V^{1/2}(S_\pm )$, and prove the boundedness of the corresponding trace-operators, that is, we want to prove

Theorem 2.3

The trace is a surjective bounded operator between the spaces

$$\begin{aligned} T_{\Gamma _\pm }: \mathcal H\rightarrow V^{1/2}(S_\pm ), \end{aligned}$$

where $(T_{\Gamma _\pm } \psi )(z):=\psi (\pm d, z)$ for $z\in S_\pm$ and $\psi \in \mathcal D_0(\Omega )$.

Proof

We will only consider the cylindric domains of the form $\Omega =(0,1)\times S$, and thus omit the explicit denotation of “$\pm$”. Note that each waveguide junction contains such domains close to $\Gamma$ by property (2.1). Without loss of generality we only consider $T_{\{0\}\times S}:\mathcal H((0,1)\times S)\rightarrow V^{1/2}(S)$:

• (1) We start by showing an equivalent norm on $\mathcal H$: The functions

  $$\begin{aligned} \psi _{nm}(x,z):=\frac{2}{\pi }\cos (mx)\theta _n(z)\, \, n\in {\mathbb {N}},m\in {\mathbb {N}}_0 \end{aligned}$$

  form a complete orthonormal system for $L^2((0,\pi )\times S)=L^2(0,\pi )\otimes L^2(S)$. Note that we used Neumann eigenfunctions for $x$ and Dirichlet eigenfunctions for $z$; By partial integration we have for $\phi \in \mathcal D_0(\Omega )$

  $$\begin{aligned} \langle \phi ,\psi _{nm}\rangle _{\mathcal H}&=k^2 \int _\Omega \phi \overline{\psi }_{nm}~dr+\int _\Omega \nabla \phi \cdot \nabla \overline{\psi }_{nm}~dr\\&=k^2\langle \phi ,\psi _{nm}\rangle _\Omega +\int _\Omega \phi (-\Delta \overline{\psi }_{nm})~dr+\int _{\partial \Omega } \phi \partial _\nu \overline{\psi }_{nm} ~do\\&=(k^2+m^2+\mu _n)\langle \phi ,\psi _{nm}\rangle _{\Omega }. \end{aligned}$$

  where we used $-\Delta \psi _{nm}=(m^2+\mu _n)\psi _{nm}$ and that the boundary integral vanished, since either $\partial _\nu \psi _{nm}=0$ or $\phi =0$ holds. By density and continuity, this extends to all $\phi \in \mathcal H$, which shows that the $(\psi _{nm})_{n\in {\mathbb {N}}, m\in {\mathbb {N}}_0}$ form a complete normal system of $\mathcal H$. Accordingly we obtain for $\phi \in \mathcal H$:

  $$\begin{aligned} \Vert \phi \Vert _{\mathcal H}^2=\sum _{n\in {\mathbb {N}}, m\in {\mathbb {N}}_0}(k^2+m^2+\mu _n)\vert \langle \phi ,\psi _{nm}\rangle _\Omega \vert ^2.\qquad \end{aligned}$$

  (5.1)

• (2) Next comes a technical estimate. Note that for $\alpha >0$ the following formula holds

  $$\begin{aligned} \sum _{m\in {\mathbb {N}}_0}\frac{1}{\alpha ^2 + m^2}=\frac{\pi \alpha \coth (\pi \alpha )+1}{2\alpha ^2}, \end{aligned}$$

  and we obtain for all $n\in {\mathbb {N}}$ that

  $$\begin{aligned} (\mu _n)^{1/2}\sum _{m\in {\mathbb {N}}_0} (k^2+m^2+\mu _n)^{-1}&=(\mu _n)^{1/2}\frac{\pi \sqrt{k^2+\mu _n} \coth \left( \pi \sqrt{k^2+\mu _n}\right) +1}{2 (\sqrt{k^2+\mu _n})^2}\nonumber \\&\le \frac{\pi }{2} \coth \left( \pi \sqrt{k^2+\mu _1}\right) +\frac{1}{2\sqrt{k^2+\mu _1}}\nonumber \\&=: C, \end{aligned}$$

  (5.2)

  where we used that $\coth$ is monotonously decreasing on $(0,\infty )$ while $\mu _n$ is monotonously increasing.

• (3) Let us get back to our actual trace operator: Let $\phi =\sum _{n,m\in {\mathbb {N}}} \langle \phi ,\psi _{nm}\rangle _\Omega \psi _{nm}\in \mathcal H$ be some function; Then its trace $T_{\{0\}\times S}\phi$ is given by

  $$\begin{aligned} T_{\{0\}\times S}\phi&=\sum _{n\in {\mathbb {N}}, m\in {\mathbb {N}}_0}\langle \phi ,\psi _{nm}\rangle _\Omega \cos (0) ~\theta _n\\&=\sum _{n\in {\mathbb {N}}}\left( \sum _{m\in {\mathbb {N}}}\langle \phi ,\psi _{nm}\rangle _\Omega \right) \theta _n. \end{aligned}$$

  Let us show that $T_{\{0\}\times S}\phi \in V{^{1/2}(S)}$: By definition and the Cauchy-Schwarz inequality we have

  $$\begin{aligned} \Vert T_{\{0\}\times S}\phi \Vert _{V^{1/2}(S)}^2&=\sum _{n\in {\mathbb {N}}} (\mu _n)^{1/2} \left| \sum _{m\in {\mathbb {N}}_0}\langle \phi ,\psi _{nm}\rangle _\Omega \right| ^2\\&\le \sum _{n\in {\mathbb {N}}}(\mu _n)^{1/2}\left( \sum _{m\in {\mathbb {N}}_0}(k^2+m^2+\mu _n)^{-1}\right) \\&\qquad \cdot \left( \sum _{m\in {\mathbb {N}}_0}(k^2+m^2+\mu _n)\left| \langle \phi ,\psi _{nm}\rangle _\Omega \right| ^2\right) \\&\le C\sum _{n\in {\mathbb {N}}} \sum _{m\in {\mathbb {N}}_0}(k^2+m^2+\mu _n)\left| \langle \phi ,\psi _{nm}\rangle _\Omega \right| ^2\\&=C \Vert \phi \Vert _{\mathcal H}^2, \end{aligned}$$

  where we used inequality (5.2) for the second estimate and (5.1) for the last statement. This finishes the proof of the boundedness.

For the surjectivity, we just remark that the mapping $R:V^{1/2}(S)\rightarrow \mathcal H$ defined by

$$\begin{aligned} R\phi (x,z):=\sum _{n\in {\mathbb {N}}}e^{-\sqrt{\mu _n}x} \theta _n(z) \langle \phi ,\theta _n\rangle _{S} \end{aligned}$$

is a bounded linear right-inverse of $T_{\{0\}\times S}$ (which can be verified easily by direct calculation).

Let us remark that $V^{1/2}(S)$ is exactly the space of functions in $H^{1/2}(S)$, which can be extended by $0$ to get some function in $H^{1/2}(S)$: This follows from the commutativity of the diagram

$$\begin{aligned} \begin{array}{cccc} \mathcal H\quad &{}\xrightarrow {T_{\{0\}\times S}} &{}V^{1/2}(S)\\ \downarrow E_0 &{} &{}\downarrow \tilde{E}_0 \\ H^1((a,b)\times {\mathbb {R}}^{N-1})&{}\xrightarrow {T_{\{0\}\times {\mathbb {R}}^{N-1}}}&{}H^{1/2} ({\mathbb {R}}^{N-1}) \end{array} \end{aligned}$$

where $E_0:\mathcal H\rightarrow H^1((a,b)\times R^{N-1})$ and $\tilde{E}_0: V^{1/2}(S)\rightarrow H^{1/2}({\mathbb {R}}^{N-1})$ denote the extensions by $0$. The boundedness of $\tilde{E}_0$ can be seen by using the bounded right-inverse $R$ of $T_{\{0\}\times S}$ from the previous proof and going “backwards” in the diagram. Precisely this extension operator is discontinuous if $V^{1/2}(S)$ is replaced by $H^{1/2}(S)$ (see Theorem 1 in section 4.3.2 of [15]).

Appendix 2: An A-Priori Bound for Solutions

The proof is a combination of arguments of the two papers [4] and [12], where the latter discusses a situation very similar to ours. Most of the difficulties of our situation has already been tackled successfully there.

Lemma 2.14

Assume $k^2\ne \mu _n^\pm ~\text{ for } \text{ all } n\in {\mathbb {N}}$, $d(\mu _{\hat{N}_\pm +1}-k^2)^{-1/2}>2$ and let the cladding $\Gamma$ be at least $C^{1,1}$ smooth. Further assume that $\Gamma$ suffices

$$\begin{aligned} x \nu _x\le 0 \text{ on } \Gamma , \end{aligned}$$

(2.21)

where $\nu _x$ denotes the $x-$component of the outward normal $\nu$. Then for all $g\in L^2(\Omega )$, each solution $u\in \mathcal H$ of

$$\begin{aligned} b(u,\psi )=-\langle g,\psi \rangle _\Omega ~~~\text{ for } \text{ all } \psi \in \mathcal H \end{aligned}$$

(2.22)

suffices

$$\begin{aligned} \Vert u \Vert _\mathcal H\le C\Vert g\Vert _\Omega , \end{aligned}$$

(2.23)

where the constant $C>0$ only depends on $S_\pm , d,\delta$ and $k$.

Proof

The proof is relatively long and technical, which is why we split into a number of parts. First fix some $u\in \mathcal H, g\in L^2(\Omega )$ sufficing (2.22).

• (1) We start by some regularity arguments: First note that we can extend $u$ to $\Omega _\pm$ as in Lemma 2.12 by the modal expansion (where $u_{inc}=0$) to obtain some $\hat{u}\in H^1_{loc}(\hat{\Omega })$ solution of $\Delta \hat{u}+k^2\hat{u}=g$ in $\hat{\Omega }$, where we extended $g$ by $0$ outside of $\Omega$. Note that $\partial \hat{\Omega }$ is $C^{1,1}$, and we obtain by standard regularity results (compare for example [11]) that $\hat{u}\in H^2(\hat{\Omega }\cap B_R(0))$ for all $R>0$, and thus also $u\in H^2(\Omega )$, and we have $\Delta u+k^2u=g$ almost everywhere.

• (2) We show an estimate for certain boundary terms: We claim that

  $$\begin{aligned} \Vert \partial _x u\Vert ^2_{S_\pm }-\Vert \nabla _z u\Vert ^2_{S_\pm }+k^2\Vert u\Vert _{S_\pm }^2\le 2 k \mathfrak {I}(\Lambda _\pm u(u)). \end{aligned}$$

  (6.1)

  (Note that each quantity is well-defined since $u\in H^2(\Omega )$). Denoting $u_n^\pm :=\langle u,\theta _n^\pm \rangle _{S_\pm }$, we re-calculate the left hand side of (6.1) using the orthogonality of the $\theta _n^\pm$ and $\nabla \theta _n^\pm$:

  $$\begin{aligned} \Vert \partial _x u\Vert ^2_{S_\pm }\!-\!\Vert \nabla _z u\Vert ^2_{S_\pm }\!+\!k^2\Vert u\Vert _{S_\pm }^2&\!=\! \sum _{n\in \mathbb N}\Vert \lambda _n^\pm u_n^\pm \theta _n^\pm \Vert ^2_{S_\pm }\!-\! \Vert u_n^\pm \nabla \theta _n^\pm \Vert ^2_{S_\pm }\!+\! k^2\Vert u_n^\pm \theta _n^\pm \Vert ^2_{S_\pm }\\&=\sum _{n\in \mathbb N} \left( \vert \lambda _n^\pm \vert ^2-\mu _n^\pm +k^2\right) \vert u_n^\pm \vert ^2. \end{aligned}$$

  By (2.8) we have

  $$\begin{aligned} \vert \lambda _n^\pm \vert ^2=\left\{ \begin{matrix} k^2-\mu _n^\pm &{} \text{ if } n\le \hat{N}_\pm \\ \mu _n^\pm -k^2 &{} \text{ if } n> \hat{N}_\pm \end{matrix}\right. \end{aligned}$$

  and accordingly

  $$\begin{aligned} \Vert \partial _x u\Vert ^2_{S_\pm }-\Vert \nabla u\Vert ^2_{S_\pm }+k^2\Vert u\Vert _{S_\pm }^2&=\sum _{n=1}^{\hat{N}_\pm }2\vert \lambda _n^\pm \vert ^2\vert u_n^\pm \vert ^2\\&\le \sum _{n=1}^{\hat{N}_\pm }2k\vert \lambda _n^\pm \vert \vert u_n^\pm \vert ^2= 2k \mathfrak {I}(\Lambda _\pm u(u)).\\ \end{aligned}$$

• (3) For $u\in \mathcal H$, the following Poincaré-type inequality holds

  $$\begin{aligned} \Vert u \Vert ^2_\Omega \le 2d\left( \sum _\pm \Vert u \Vert _{S_\pm }^2\right) +d^2\Vert \partial _x u \Vert ^2_\Omega . \end{aligned}$$

  (6.2)

  Let us prove this: We denote $\Omega _{x<0}:=\{(x,z)\in \Omega ,x<0\}$. Let $\psi \in \mathcal D_0(\Omega )$. Then we have

  $$\begin{aligned} \psi (x,z)=\psi (-d,0)+\int _{-d}^x\partial _x\psi (s,z)~ds \end{aligned}$$

  and thus by the Chauchy-Schwarz-inequality:

  $$\begin{aligned} \int _{-d}^0 \vert \psi (x,z)\vert ^2~dx&\le 2\int _{-d}^0\left( \vert \psi (-d,z)\vert ^2+\left| \int _{-d}^x \partial _x \psi (s,z)~ds\right| ^2 \right) ~dx\\&\le 2d \vert \psi (-d,z)\vert ^2 +2\int _{-d}^0 \int _{-d}^x 1^2 ds \int _{-d}^x \vert \partial _x\psi (s,z)\vert ^2 ~ds ~dx\\&\le 2d \vert \psi (-d,z)\vert ^2 +d^2 \int _{-d}^0 \vert \partial _x\psi (s,z)\vert ^2 ~ds. \end{aligned}$$

  Integrating the last statement over $S_-$ then yields

  $$\begin{aligned} \int _{\Omega _{x<0}}\vert \psi \vert ^2~dr\le 2d\int _{S_-} \vert \psi \vert ^2 dz +d^2 \int _{\Omega _{x<0}} \vert \partial _x u\vert ^2 ~dr \end{aligned}$$

  with an analogous statement for the integrals over $\Omega _{x>0}$. Summing both statements up we obtain

  $$\begin{aligned} \Vert \psi \Vert _\Omega ^2\le 2d\left( \sum _\pm \Vert \psi \Vert _{S_\pm }^2\right) + d^2\Vert \partial _x \psi \Vert ^2_\Omega , \end{aligned}$$

  which can be extended from $\mathcal D_0(\Omega )$ to $\mathcal H(\Omega )$ by density and continuity.

• (4) We now prove a critical Rellich-type identity: Since $(\Delta +k^2)u=g$ almost everywhere, we have

  $$\begin{aligned} 2\mathfrak {R}\int _\Omega xg\partial _x\overline{u} ~dr&=2\mathfrak {R}\int _\Omega x(\Delta u+k^2u)\partial _x\overline{u} ~dr \nonumber \\&=\int _\Omega 2 \mathfrak {R}\mathrm{div }(x(\partial _x\overline{u})\nabla u)\!-\! \vert \partial _x u\vert ^2\!-\!x\partial _x (\vert \nabla u \vert ^2)\!+\!k^2 x\partial _x(\vert u \vert ^2)~dr \end{aligned}$$

  (6.3)

  where the last equality can be easily verified by direct calculation. Using Gauss’ theorem on the divergence term and partial integration on the last two terms yields

  $$\begin{aligned} 2\mathfrak {R}\int _\Omega xg\partial _x\overline{u} ~dr&=\int _\Omega -\vert \partial _x u\vert ^2+\vert \nabla u\vert ^2- k^2 \vert u \vert ^2 ~dr\nonumber \\&\quad +\int _{\partial \Omega } 2\mathfrak {R}(x(\partial _x\overline{u})(\partial _\nu u)) - \nu _x x\vert \nabla u\vert ^2 + k^2\nu _xx\vert u\vert ^2 ~do. \end{aligned}$$

  (6.4)

  Let us split the boundary $\partial \Omega =\Gamma _-\cup \Gamma \cup \Gamma _+$. On $\Gamma$ we have $u=0$ and thus $\nabla u=\nu \partial _\nu u$,

  $$\begin{aligned} \partial _x u= e_x\cdot \nabla u = e_x\cdot \nu \partial _\nu u=\nu _x\partial _\nu u. \end{aligned}$$

  This yields for the corresponding boundary term of (6.4)

  $$\begin{aligned} \int _{\Gamma } 2\mathfrak {R}(x(\partial _x\overline{u})(\partial _\nu u)) - \nu _x x\vert \nabla u\vert ^2 + k^2\nu _xx\vert u\vert ^2 ~do=\int _{\Gamma }x\nu _x \vert \partial _\nu u\vert ^2 do. \end{aligned}$$

  On $\Gamma _-$ we have $x=-d$ and $\nu =-e_x$, and thus $x\partial _xu=d \partial _\nu u$, yielding

  $$\begin{aligned}&\int _{\Gamma _-} 2\mathfrak {R}(x(\partial _x\overline{u})(\partial _\nu u)) - \nu _x x\vert \nabla u\vert ^2 + k^2\nu _xx\vert u\vert ^2 ~do\\&\qquad =\int _{\Gamma _-}2d\vert \partial _x u\vert ^2-d\vert \nabla u \vert ^2 +k^2d\vert u\vert ^2~dr\\&\qquad =d(\Vert \partial _x u\Vert ^2_{S_-}-\Vert \nabla _zu\Vert ^2_{S_-}+k^2\Vert u\Vert ^2_{S_-}). \end{aligned}$$

  with an analogous equation for the boundary terms on $\Gamma _+$. Using these relations, we can rewrite (6.4) as follows:

  $$\begin{aligned} 2\mathfrak {R}\int _\Omega xg\partial _x\overline{u} ~dr&=-\Vert \partial _x u\Vert ^2_\Omega +\Vert \nabla u\Vert ^2_\Omega -k^2\Vert u\Vert ^2_\Omega +\int _\Gamma x\nu _x \vert \partial _\nu u\vert ^2 ~do\nonumber \\&\quad +d\sum _\pm (\Vert \partial _x u\Vert ^2_{S_\pm }-\Vert \nabla _zu\Vert ^2_{S_\pm }+k^2\Vert u\Vert ^2_{S_\pm }), \end{aligned}$$

  (6.5)

• (5) We will now derive a bound for $\Vert \partial _xu\Vert ^2_\Omega$: By our geometric assumption, we have $\nu _x x\le 0$, and thus with our Rellich type identity (6.5):

  $$\begin{aligned} \Vert \partial _xu\Vert _\Omega ^2&\le - \int _\Gamma x\nu _x \vert \partial _\nu x\vert ^2~do+\Vert \partial _x u\Vert ^2_{\Omega }\nonumber \\ {}&=d\sum _\pm \left( \Vert \partial _x u\Vert ^2_{S_\pm }-\Vert \nabla _z u \Vert ^2_{S_\pm }+k^2 \Vert u \Vert ^2_{S_\pm }\right) \nonumber \\&\quad +\Vert \nabla u\Vert ^2_\Omega -k^2\Vert u\Vert ^2_\Omega - 2\mathfrak {R}\int _\Omega x g\partial _x \overline{u} ~dx. \end{aligned}$$

  (6.6)

  Furthermore, by taking the real part of $b(u,u)=-\langle g,u\rangle _\Omega$, we obtain

  $$\begin{aligned} \Vert \nabla u\Vert ^2_\Omega -k^2\Vert u\Vert ^2_\Omega +\sum _\pm \mathfrak {R}(\Lambda _\pm u (u))=-\mathfrak {R}\langle g,u\rangle _\Omega . \end{aligned}$$

  (6.7)

  Using inequality (6.1) for the boundary related terms of (6.6) and the last equation for the $L^2(\Omega )$ norms, we obtain

  $$\begin{aligned} \Vert \partial _xu\Vert _\Omega ^2&\le d\sum _{\pm }2k\mathfrak {I}(\Lambda _\pm u(u))\nonumber \\&\quad \ -\left( \sum _{\pm }\mathfrak {R}(\Lambda _\pm u(u))\right) - \mathfrak {R}\langle g,u\rangle _\Omega -2\mathfrak {R}\langle xg,\partial _xu\rangle _\Omega . \end{aligned}$$

  (6.8)

• (6) Reformulating (6.7) yields by the Poincaré type inequality (6.2) yields

  $$\begin{aligned} \Vert u\Vert _{\mathcal H}^2&=2k^2\Vert u\Vert _\Omega -\left( \sum _\pm \mathfrak {R}(\Lambda _\pm u (u))\right) -\mathfrak {R}\langle g,u\rangle _\Omega \\&\le 4k^2d\left( \sum _\pm \Vert u\Vert ^2_{S_\pm }\right) +2k^2d^2 \Vert \partial _x u\Vert _\Omega ^2-\mathfrak {R}\langle g,u\rangle _\Omega , \end{aligned}$$

  where we used that $\mathfrak {R}(\Lambda _\pm u (u))\ge 0$. Applying our estimate for $\Vert \partial _x u\Vert _\Omega$ of inequality (6.8), we obtain:

  $$\begin{aligned} \Vert u \Vert ^2_{\mathcal H}&\le 2k^2d\left( \sum _\pm 2\Vert u\Vert _{S_\pm } ^2 - d\mathfrak {R}(\Lambda _\pm u(u))+kd^2\mathfrak {I}(\Lambda _\pm u(u)) \right) \nonumber \\&\quad - \left( (1+k^2d^2)\mathfrak {R}\langle g,u\rangle _\Omega +2 k^2d^2\mathfrak {R}\langle xg, \partial _xu\rangle _\Omega \right) . \end{aligned}$$

  (6.9)

  We now want to estimate the boundary terms appearing in the last equation: There we have (denoting $u_n^\pm :=\langle u, \theta _n^\pm \rangle _{S_\pm }$):

  $$\begin{aligned} 2\Vert u\Vert _{S_\pm } ^2&- d \mathfrak {R}(\Lambda _\pm u(u))+kd^2\mathfrak {I}(\Lambda _\pm u(u))\nonumber \\&= \sum _{n=1}^{\hat{N}_\pm }(2+kd^2\vert \lambda _n^\pm \vert )\vert u_n^\pm \vert ^2+\sum _{\hat{N}_\pm +1}^\infty (2- d\lambda _n^\pm ) \vert u_n^\pm \vert ^2 \nonumber \\&\le \sum _{n=1}^{\hat{N}_\pm }(2+kd^2\vert \lambda _n^\pm \vert )\vert u_n^\pm \vert ^2 \end{aligned}$$

  (6.10)

  where we used that by assumption (b) we have $d\lambda _n^\pm \ge d\lambda _{\hat{N}_\pm +1}^\pm \ge 2$ for all $n\ge \hat{N}_\pm +1$. On the other hand, we have for $n\le \hat{N}_\pm$ that $\vert \lambda _n^\pm \vert \ge \vert \lambda _{\hat{N}_\pm }^\pm \vert$, and thus

  $$\begin{aligned} \sum _\pm 2\sum _{n=1}^{\hat{N}_\pm }\vert u_n^\pm \vert ^2&\le 2\max \left( \vert \lambda _{\hat{N}_+} \vert ^{-1},\vert \lambda _{\hat{N}_-} \vert ^{-1}\right) \sum _\pm \sum _{n=1}^{\hat{N}_\pm } \vert \lambda _n^\pm \vert \vert u_n^\pm \vert ^2\\&=2\max \left( \vert \lambda _{\hat{N}_+} \vert ^{-1},\vert \lambda _{\hat{N}_-} \vert ^{-1}\right) \sum _\pm \mathfrak {I}(\Lambda _\pm u (u))\\&=-2\max \left( \vert \lambda _{\hat{N}_+} \vert ^{-1},\vert \lambda _{\hat{N}_-} \vert ^{-1}\right) \mathfrak {I}\langle g, u\rangle _\Omega \end{aligned}$$

  where we used that $\mathfrak {I}b(u,u)=\sum _\pm \mathfrak {I}(\Lambda _\pm u (u))=-\mathfrak {I}\langle g,u\rangle _\Omega$ for the last equality. Let us denote

  $$\begin{aligned} C(k,d,S_\pm ):=2\max \left( \vert \lambda _{\hat{N}_+} \vert ^{-1},\vert \lambda _{\hat{N}_-} \vert ^{-1}\right) +kd^2. \end{aligned}$$

  (6.11)

  Then we obtain from (6.10) and the previous discussion that

  $$\begin{aligned} \sum _\pm 2\Vert u\Vert _{S_\pm } ^2 - d \mathfrak {R}(\Lambda _\pm u(u))+kd^2\mathfrak {I}(\Lambda _\pm u(u)) \le -C(k,d,S_\pm )\mathfrak {I}\langle g,u\rangle _\Omega . \end{aligned}$$

• (7) Let us finish the proof: We can now rewrite (6.9) with the previous estimate to obtain

  $$\begin{aligned} \Vert u\Vert _\mathcal H^2\!{\le } \!-k^2 d~ 2C(k,d,S_\pm ) \mathfrak {I}\langle g, u\rangle _\Omega \!-\! \left( (1\!{+}\!k^2d^2)\mathfrak {R}\langle g,u\rangle _\Omega \!+\!2 k^2d^2\mathfrak {R}\langle xg, \partial _xu\rangle _\Omega \right) \end{aligned}$$

  and since $\Vert x g \Vert _\Omega \le d\Vert g \Vert _\Omega$ as well as $k\Vert u\Vert _\Omega \le \Vert u \Vert _\mathcal H$, we obtain by the Cauchy-Schwarz inequality

  $$\begin{aligned} \Vert u\Vert _\mathcal H^2&\le 2k^2 d~ C(k,d,S_\pm ) \Vert u \Vert _\Omega \Vert g\Vert _\Omega \!+\! (1+k^2d^2)\Vert u \Vert _\Omega \Vert g\Vert _\Omega \!+\!2k^2d^3 \Vert u\Vert _\mathcal H \Vert g\Vert _\Omega \\&\le (2kb C(k,S_\pm ) + 1+kd^2+2k^2d^3)\Vert u\Vert _\mathcal H \Vert g\Vert _\Omega \end{aligned}$$

  that is, we have shown

  $$\begin{aligned} \Vert u \Vert _\mathcal H\le (2kd C(k,d,S_\pm ) + 1+kd^2+2k^2d^3)\Vert g\Vert _\Omega \end{aligned}$$

  which is the statement of the theorem.

Appendix 3: A Domain Approximation Lemma

To extend the estimate of the previous section to Lipschitz domains, we used the following domain-approximation Lemma.

Lemma 3.5

Let $\Omega \in \mathcal P_\delta$ be some junction and let $\psi \in \mathcal D_0(\Omega )$ be some testfunction. Then there exists some $\Omega '\in \mathcal P_{\delta /2}$, which is $C^{1,1}$ smooth and contains the support of $\psi$ (Fig. 7). $\square$

Fig. 7

Transformations in the proof of Lemma 3.5: a $\hat{\Omega }\rightarrow \Omega ^*$: Cutting out a small piece near $x=0$ b $\Omega ^*\rightarrow \tilde{\Omega }$: Shrinking and smoothing of the domain. c $\tilde{\Omega }\rightarrow \Omega '$: Reconnecting to the waveguides

Sketch of Proof

Denote $K=\overline{ \mathrm{supp }(\psi )}$ and consider the whole waveguide-junction $\hat{\Omega }=\Omega \cup \Omega _-\cup \Omega _+$. Note that the defining property (a) of $P_\delta$ is equivalent to

$$\begin{aligned}&\text{ If } (x,z)\in \hat{\Omega } \text{ and } x\ge 0, \text{ then } (x,+\infty )\times \{z\}\subset \hat{\Omega }\\&\text{ and } \text{ if } (x,z)\in \hat{\Omega } \text{ and } x\le 0, \text{ then } (-\infty ,x)\times \{z\}\subset \hat{\Omega }\end{aligned}$$

(a'')

In this section, for some set $A\subset {\mathbb {R}}^M$, we denote by $\complement A:={\mathbb {R}}^M\setminus A$ its complement in ${\mathbb {R}}^M$ (and not, as in section $3$, its complement in $D$). Now note that some open $A\subset {\mathbb {R}}^N$ fulfills property $(a'')$ if and only if the distance function of its complement suffices

$$\begin{aligned} \begin{array}{ll} &{}d_{\complement A}(x_1,z)\ge d_{\complement A}(x_2,z) \text{ for } \text{ all } x_1\le x_2\le 0, z\in {\mathbb {R}}^{N-1} \\ &{}d_{\complement A}(x_1,z)\le d_{\complement A}(x_2,z) \text{ for } \text{ all } 0\le x_1\le x_2, z\in {\mathbb {R}}^{N-1}, \end{array} \end{aligned}$$

(a''')

that is, if the distance function is decreasing in $x$ for $x\le 0$ and increasing in $x$ if $x\ge 0$.

The construction of the junction $\Omega '$ will be split into two steps: First, we construct a smoothed, monotone version of $\Omega$ (see Fig. 7a, b), and connect it in a sufficiently smooth manner to the waveguides $\Omega _\pm$ in the second step (Fig. 7c). Denote $\rho :=d_H(K,\complement \hat{\Omega })>0$. Now, we choose some $\epsilon >0$ such that (a) $\epsilon <\rho /4$, (b) $\epsilon <\delta /12$ and (c) $\epsilon$ fulfills additional restriction depending on the two domains $S_\pm$, which we will specify later on.

Denote by $Q\!:=\!\{z:(0,z)\!\in \!\Omega \}$ the intersection of $\Omega$ with the plane $\{x\!=\!0\}$. First put

$$\begin{aligned} \Omega ^*:=\hat{\Omega }\setminus ( [-\epsilon ,\epsilon ] \times ({\mathbb {R}}^{N-1}\setminus Q)) \end{aligned}$$

Note that $\Omega ^*$ fulfills $(a'')$ and $d_H(K,\complement \Omega ^*)>\rho -\epsilon >3\epsilon$. Further note that for $(x_1,z),(x_2,z)\in \Omega ^*$, with $x_1,x_2\in (-\epsilon ,\epsilon )$ we have

$$\begin{aligned} d_{\complement \Omega ^*}(x_1,z)=d_{\complement \Omega ^*}(x_2,z). \end{aligned}$$

(7.1)

Let $\phi _1:{\mathbb {R}}\rightarrow {\mathbb {R}}$ and $\phi _2:{\mathbb {R}}^{N-1}\rightarrow {\mathbb {R}}$ be two smooth functions such that

$$\begin{aligned} \begin{array}{ll} &{}\phi _1(x)=0 \text{ for } \vert x\vert >\epsilon /2,~ \phi _1(x)\ge 0 \text{ for } x\in {\mathbb {R}} \text{ and } \int _{\mathbb {R}}\phi _1 dx=1,\\ &{}\phi _2(z)=0 \text{ for } \vert z\vert >\epsilon /2,~ \phi _2(z)\ge 0 \text{ for } z\in {\mathbb {R}}^{N-1} \text{ and } \int _{{\mathbb {R}}^{N-1}} \phi _2 dz=1. \end{array} \end{aligned}$$

Furthermore, let $\phi (x,z):=\phi _1(x)\phi _2(z)$ (thus $\phi (r)=0$ for $\vert r\vert >\epsilon$ and $\int \phi dr=1$). Define the smooth function $f$ by

$$\begin{aligned} f(x,z):=(\phi *d_{\complement \Omega ^*})(x,z):=\int _{{\mathbb {R}}^N}\phi (\xi ,\zeta )d_{\complement \Omega ^*}(x-\xi ,z-\zeta )~d\xi d\zeta . \end{aligned}$$

Denote $X:=\{r\in {\mathbb {R}}^N:\nabla f(r)=0\}$. By the lemma of Sard, the set $f(X)$ has Lebesgue measure $0$, and we can find some $\eta \in (\epsilon , 2\epsilon )$ such that $\eta \notin f(X)$. This implies that the domain

$$\begin{aligned} \tilde{\Omega }:= \{(x,z)\in {\mathbb {R}}^N:f(x,z)>\eta \} \end{aligned}$$

has a smooth boundary by the implicit function theorem and the smoothness of $f$. We claim that (i) $\tilde{\Omega }$ fulfills $(a'')$, (ii) $K\subset \tilde{\Omega }$ and (iii) $\tilde{\Omega }\subset \hat{\Omega }$. Let us sketch how each is shown:

• (i) First note that $(a''')$ for $\Omega ^*$ and the property (7.1) imply that if $(x_1,z),(x_2,z)\in {\Omega ^*}$ and $x_1\le x_2\le +\epsilon$, then $d_{\complement \Omega ^*}(x_1,z)\ge d_{\complement \Omega ^*}(x_2,z)$ (For $\hat{\Omega }$, this property holds only for $x_1\le x_2\le 0$). Let $(\tilde{x}, z)\in \tilde{\Omega }$ and $\hat{x}\le \tilde{x}\le 0$. Then we have

  $$\begin{aligned} \begin{array}{ll} f(\hat{x}, z)&{}=\displaystyle \int _{{\mathbb {R}}^N}\phi (\xi ,\zeta )d_{\complement \Omega ^*}(\hat{x}-\xi ,z-\zeta )~d\xi d\zeta \\ &{}=\displaystyle \int _{\{\vert (\xi ,\zeta )\vert <\epsilon \}}\phi (\xi ,\zeta )d_{\complement \Omega ^*}(\hat{x}-\xi ,z-\zeta )~d\xi d\zeta \\ &{}\ge \displaystyle \int _{\{\vert (\xi ,\zeta )\vert <\epsilon \}}\phi (\xi ,\zeta )d_{\complement \Omega ^*}(\tilde{x}-\xi ,z-\zeta )~d\xi d\zeta \\ &{}=f(\tilde{x},z)>\eta \end{array} \end{aligned}$$

  and therefore $(\hat{x}, z)\in \tilde{\Omega }$. The corresponding statement for $0\le \tilde{x}\le \hat{x}$ is proved analogously. This implies that $\tilde{\Omega }$ fulfills $(a''')$, which is equivalent to $(a'')$.

• (ii) Note that $K\subset \{r\in \Omega ^*: d_{\complement \Omega ^*}(r)>3\epsilon \}$. Take some $(x,z)\in \{r\in \Omega ^*: d_{\complement \Omega ^*}(r)>3\epsilon \}$. We show that $(x,z)\in \tilde{\Omega }$: First note for all $A\subset {\mathbb {R}}^N$ the distance function $d_{A}$ is Lipschitz, that is for $r,r'\in {\mathbb {R}}^N$ we have

  $$\begin{aligned} \vert d_A(r)-d_A(r')\vert \le \vert r-r'\vert . \end{aligned}$$

  This implies for $(\xi ,\zeta )\in {\mathbb {R}}^N$ such that $\vert (\xi ,\zeta )\vert <\epsilon$

  $$\begin{aligned} d_{\complement \Omega ^*}(x-\xi ,z-\zeta )\ge d_{\complement \Omega ^*}(x,z)-\epsilon , \end{aligned}$$

  and accordingly

  $$\begin{aligned} \begin{array}{ll} f(x,z)&{}=\displaystyle \int _{\{\vert (\xi ,\zeta )\vert <\epsilon \}}\phi (\xi ,\zeta )d_{\complement \Omega ^*}(\hat{x}-\xi ,z-\zeta )~d\xi d\zeta \\ &{}\displaystyle \ge \int _{\{\vert (\xi ,\zeta )\vert <\epsilon \}}\phi (\xi ,\zeta )(d_{\complement \Omega ^*}(\hat{x},z)-\epsilon )~d\xi d\zeta \\ &{}=d_{\complement \Omega ^*}(\hat{x},z)-\epsilon \\ &{}>2\epsilon >\eta . \end{array} \end{aligned}$$

  Which shows that $(x,z)\in \tilde{\Omega }$, proving $K\subset \tilde{\Omega }$.

• (iii) To see that $\tilde{\Omega }\subset \hat{\Omega }$, note that if $(x,z)\in \tilde{\Omega }$, we have again by the Lipschitz continuity for $\vert (\xi ,\zeta )\vert <\epsilon$ that

  $$\begin{aligned} d_{\complement \Omega ^*}(x,z)\ge d_{\complement \Omega ^*}(x-\xi ,z-\zeta )-\epsilon , \end{aligned}$$

  which implies by a calculation similar to step (ii)

  $$\begin{aligned} d_{\complement \Omega ^*}(x,z)\ge f(x,z)-\delta \ge \eta -\epsilon >0, \end{aligned}$$

  proving that $(x,z)\in \Omega ^*\subset \hat{\Omega }$.

Hence our domain $\tilde{\Omega }$ fulfills all properties we demand from $\Omega '$, except that it does not connect to the two waveguides $\Omega _\pm$. The rest of the proof will be concerned with the construction of a sufficiently smooth transition to the waveguide $\Omega _-$ in the slab $\{-d+\delta /2<x<-d+\delta \}$. The construction of the the transition to $\Omega _+$ then is analogous.

First denote $S_-^c:={\mathbb {R}}^{N-1}\setminus S_-$. By $b_{S_-^c}:{\mathbb {R}}^{N-1}\rightarrow {\mathbb {R}}$ we denote the oriented distance function of $S_-^c$, defined by

$$\begin{aligned} b_{S_-^c}(z)=\left\{ \begin{array}{ll} -d_{S_-}(z) &{} \text{ if } z\in S_-^c\\ d_{S_-^c}(z) &{} \text{ if } z\in S_-. \end{array}\right. \end{aligned}$$

We start by noting that since $S_-$ is $C^{1,1}$, there exists some $\epsilon _1^->0$ such that $d_{S_-^c}$ is in $C^{1,1}(\overline{W})$, where $W:=\{z\in {\mathbb {R}}^{N-1}: d_{\partial S_-}(z)<\epsilon _1\}$. This is an immediate consequence of Theorem 4.3 in [5] and the compactness of $\partial S_-$. Thus $\nabla b_{S_-^c}$ is Lipschitz continuous, and we can find some constant $\epsilon _2^-$ such that

$$\begin{aligned} \text{ if } z_1,z_2\in \overline{W} \text{ and } \vert z_1-z_2\vert <\epsilon _2^-, \text{ then } \vert \nabla b_{S_-^c}(z_1)- \nabla b_{S_-^c}(z_2)\vert <\frac{1}{2}. \end{aligned}$$

(7.2)

We now specify our last assumption for $\epsilon$ and demand that (c) $\epsilon <\epsilon _2^\pm$ and $\epsilon <4\epsilon ^\pm _1$. First note that for $\xi <-d+\delta -4\epsilon$ and $0\le b_{S_-^c}(\zeta )<4\epsilon$, we have that $d_{\complement \Omega ^*}(\xi ,\zeta )=b_{S_-^c}(\zeta )$, which implies for $(x,z)$ such that $\epsilon < b_{S_-^c}(z)<3\epsilon$ and $x<\delta -5\epsilon$ by the definition of $f$ that

$$\begin{aligned} \begin{array}{ll} f(x,z)&{}=\displaystyle \int _{\vert \xi \vert <\epsilon /2,\vert \zeta \vert <\epsilon /2}\phi _1(\xi )\phi _2(\zeta )d_{\complement \Omega ^*}(x-\xi ,z-\zeta )~d\xi d\zeta \\ &{}=\displaystyle \int _{\vert \xi \vert <\epsilon /2,\vert \zeta \vert <\epsilon /2}\phi _1(\xi )\phi _2(\zeta )b_{S_-^c}(z-\zeta )~d\xi d\zeta \\ &{}=\displaystyle \int _{{\mathbb {R}}^{N-1}}\phi _2(\zeta )b_{S_-^c}(z-\zeta )~d\zeta \\ &{}=:\tilde{f}(z). \end{array} \end{aligned}$$

We obtain that $S^*_-:=\{z\in {\mathbb {R}}^{N-1}:(-d+\delta -\epsilon /5,z)\in \Omega ^*\}=\tilde{f}^{-1}((\eta ,+\infty ))$ and $\partial S^*_-=\tilde{f}^{-1}(\{\eta \})$. We now define for $t\in [0,1]$ the function $H_t:{\mathbb {R}}^{N-1}\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} H_t(z):=(1-t)(\tilde{f}(z)-\eta ) + t b_{S_-^c}(z), \end{aligned}$$

and define the set $S_t:=H_t^{-1}((0,+\infty ))$. Let us show some properties of these set:

• (i) If $0\le t_1\le t_2\le 1$, then $S_{t_1}\subset S_{t_2}$. This obviously follows if for $z\in {\mathbb {R}}^{N-1}$ the function $t\rightarrow H_t(z)$ is monotonously increasing. This is the case if $b_{S_-^c}(z)\ge \tilde{f}(z)-\eta$, which can be seen to hold by an easy calculation using the Lipschitz continuity of $b_{S_-^c}$.

• (ii) $S_0=S^*_-$, $S_1=S_-$, which is clear by definition.

• (iii) For all $t\in [0,1]$, $S_t$ is a $C^{1,1}$ domain. First note that by (ii), we have that $\partial S_t\subset \{z\in {\mathbb {R}}^{N-1}, d_{\partial S_-}(z)<3\epsilon \}=:W_{3\epsilon }$. Since $H_t$ is $C^{1,1}$ on $W_{3\epsilon }$ by assumption (c), it is sufficient to show that $H_t$ has no critical points on $W_{3\epsilon }$ by the implicit function theorem. Take $z\in W_{3\epsilon }$, then we have

  $$\begin{aligned} \begin{array}{ll} \nabla H_t(z)&{}=\displaystyle \int _{{\mathbb {R}}^{N-1}}\left( (1-t)\nabla b_{S_-^c}(z-\zeta )+ t\nabla b_{S_-^c}(z)\right) \phi _2(\zeta ) ~d\zeta \\ &{}=\displaystyle \int _{{\mathbb {R}}^{N-1}}\left( (1-t)(\nabla b_{S_-^c}(z)+F(\zeta ))+ t\nabla b_{S_-^c}(z)\right) \phi _2(\zeta ) ~d\zeta \\ &{}\displaystyle =\nabla b_{S_-^c}(z)+\int _{\vert \zeta \vert <\epsilon }(1-t)F(\zeta )\phi _2(\zeta ) ~d\zeta \end{array} \end{aligned}$$

  where $\vert F(\zeta )\vert <1/2$ for $\vert \zeta \vert <\epsilon$ by (7.2). Thus

  $$\begin{aligned} \vert \nabla H_t(z)\vert >\vert \nabla b_{S_-^c}(z) \vert -\int _{{\mathbb {R}}^{N-1}}(1-t)\frac{1}{2}\phi _2(\zeta )~d\zeta \ge 1-1/2=1/2, \end{aligned}$$

  which shows that $H_t$ has no critical point in $W_{3\epsilon }$.

Let $\psi :(-\infty ,\infty )\rightarrow [0,1]$ be some smooth function such that $\psi =1$ on $(-\infty ,d+\delta /2]\cup [d+\delta /2,+\infty )$ and $\psi =0$ on $[-d+\delta -\epsilon /5,+d-\delta +\epsilon /5]$. We now define the function $g$ on ${\mathbb {R}}^{N}$ by

$$\begin{aligned} g(x,z):=\left\{ \begin{array}{ll} (1-\psi (x)) (f (x,z)-\eta )+\psi (x)b_{S_-^c}(z)&{} \text{ if } x\le 0\\ (1-\psi (x)) (f (x,z)-\eta )+\psi (x)b_{S_+^c}(z)&{} \text{ if } x>0. \end{array}\right. \end{aligned}$$

And put $\hat{\Omega }':=g^{-1}((0,+\infty ))$. We claim that $\Omega ':=\{(x,z)\in \hat{\Omega }': -d<x<d\}$ is in $\mathcal P_{\delta /2}$, has a $C^{1,1}$ boundary and suffices $K\subset \Omega '\subset \Omega$: First note that for all $x_0\in {\mathbb {R}}$ we have

$$\begin{aligned} (\tilde{\Omega }\cap \{x=x_0\})\subset (\Omega '\cap \{x=x_0\})\subset (\Omega \cap \{x=x_0\}), \end{aligned}$$

which is clear for $x_0\in [-d+\delta -6\epsilon ,+d-\delta +6\epsilon ]$ and follows by the properties (i) and (ii) of $S_t$ else wise. Thus $K\subset \tilde{\Omega }\subset \Omega '\subset \Omega$. Likewise, the monotony property follows from the monotony of $\Omega ^*$ and property (i) of $S_t$. Finally, by property (iii) of $S_t$ and the choice of $\eta$, we see that $g$ is $C^{1,1}$ in some neighborhood of $\partial \Omega '=D^{-1}(\{0\})$ and has no critical values, which implies by the implicit function theorem that $\partial \Omega '$ is $C^{1,1}$. Also note that $\Omega '$ is connected to the waveguides, since for $x_0\in (-d,-d+\delta /2)$ we have $D(x_0,z)=b_{S_-^c}(z)$, and thus $\{(x_0,z)\in \Omega '\}=\{x_0\}\times \complement S_-^c=\{x_0\}\times S_-$. $\square$

Appendix 4: The Domain Derivative

Domain derivatives are well know for scattering problems, and the proof we show is oriented at the paper [10].

Theorem 4.2

Under the assumptions of Definition 4.1, assume additionally that the reference junction $\Omega$ has a $C^{1,1}$ cladding. Then $\mathcal T_\pm$ is Fréchet differentiable at $0$, and its derivative in direction $a \in \mathrm{Var }(\Omega )$ is given by

$$\begin{aligned} D\mathcal T_\pm (0)a=T_{\Gamma _\pm }v_a, \end{aligned}$$

(4.5)

where $v_a\in H^1_{loc}(\hat{\Omega })$ is the unique variational solution to the problem

$$\begin{aligned} \left\{ \begin{array}{rl} \Delta v_a+k^2 v_a=0 &{}\quad \text{ in } \hat{\Omega }\\ v_a=-a^\top \nu ~\partial _\nu u_\beta &{}\quad \text{ on } \Gamma \\ v_a=0 &{}\quad \text{ on } \partial \hat{\Omega }\setminus \Gamma \\ v_a&{}\quad \text{ is } \text{ outgoing } \,\,\text{ on } \Omega _\pm , \end{array}\right. \end{aligned}$$

(4.6)

where $\nu$ denotes the outward normal on $\Omega$ and $u_\beta =\mathcal B_0^{-1}G_\beta$ the solution on $\Omega$ to the scattering problem with incident wave $\beta$.

Proof

The proof will be again a long one, and we structure it in multiple steps. First note, that we drop the $\beta$-index of $u=u_\beta$. In the following, for some $a\in \mathrm{Var }_\epsilon (\Omega )$, we will denote by $u_a=\mathcal B_a^{-1} G_\beta$ the solution to the scattering problem on the domain $\Omega _a$ with incident wave $\beta$. The basic idea of proof is the following: We pull back the solution $u_a$ on $\Omega _a$ by the diffeomorphism $f_a$ to obtain $\tilde{u}_a:=u_a\circ f_a$. Then we essentially show that $\Vert \epsilon a\Vert _{C^1(\Omega )}^{-1} (\tilde{u}_{\epsilon a}-u)\rightarrow v_a+a^\top \nabla u=:w_a$ as $\epsilon \rightarrow 0$, where the convergence takes place in $\mathcal H(\Omega )$. Since $T_{\Gamma _\pm } v_a=T_{\Gamma _\pm } (v_a+a^\top \nabla u)$, this will imply (by the continuity of the traces) the differentiability of $T_{\Gamma _\pm } u_a$ with the representation of above. However, since we show Frechét differentiability, the proof will get more technical than in this outline.

• (1) We start by having a closer look at the function $w_a:=v_a+a^\top \nabla u$: First note that $w_a\in \mathcal H(\Omega )$, since $T_\Gamma w_a=T_\Gamma (v_a+a^\top \nabla u)=-a^\top \nu \partial _\nu u+a^\top \nu \partial _\nu u=0$, where we used that $\nabla u=\nu \partial _\nu u$ on $\Gamma$, since $u$ obeys Dirichlet boundary conditions there. We will now show that

  $$\begin{aligned} b(w_a,\psi )&= \int _\Omega \nabla u^\top (-\mathrm{div }(a)I+da+da^\top )\nabla \overline{\psi }-k^2 \mathrm{div }(a)u\overline{\psi }~dr~~~\nonumber \\&\quad \text{ for } \text{ all } \psi \in \mathcal H(\Omega ), \end{aligned}$$

  (8.1)

  where $\mathrm{div }$ denotes the divergence, $da$ the Jacobian of $a$ and $I$ the $3\times 3$ identity matrix. Let us denote the right hand side by $J$, and choose some $\psi \in \mathcal H(\Omega )$. Without loss of generality, we assume that $\psi \in C^2(\Omega )$. First note that

  $$\begin{aligned} \begin{array}{ll} \nabla u^\top &{}\left( -\mathrm{div }(a) I + da^\top +da \right) \nabla \overline{\psi }\\ &{}=\mathrm{div }\left[ (a^\top \nabla u) \nabla \overline{\psi }+(a^\top \nabla \overline{\psi }) \nabla u - (\nabla \overline{\psi }^\top \nabla u) a \right] -(a^\top \nabla u) \Delta \overline{\psi }\\ &{}\qquad -(a^\top \nabla \overline{\psi }) \Delta u. \end{array} \end{aligned}$$

  Applying Gauss’ theorem to the divergence term yields

  $$\begin{aligned} \begin{array}{ll} J= &{}\displaystyle \int _{\Omega } -(a^\top \nabla u) \Delta \overline{\psi }-(a^\top \nabla \overline{\psi })\Delta u - k^2 \mathrm{div }(a) u\overline{\psi }~ dr \\ &{}\displaystyle +\int _{\Gamma } \left[ (a^\top \nabla u)\nabla \overline{\psi }+ (a^\top \nabla \overline{\psi }) \nabla u - (\nabla \overline{\psi }^\top \nabla u) a \right] ^\top \nu ~ds, \end{array} \end{aligned}$$

  where the boundary integrals over $\Gamma _\pm$ vanished since $a=0$ there. Since $u$ and $\psi$ obey homogeneous Dirichlet boundary condition on $\Gamma$, we can rewrite the gradients on the boundary by $\nabla u= \nu \partial _\nu u$ and $\nabla \psi = \nu \partial _\nu \psi$ and hence

  $$\begin{aligned} (a^\top \nabla \overline{\psi }) \nabla u ^\top \nu = \left( a^\top \nu \frac{\partial \overline{\psi }}{\partial \nu } \right) \nu ^\top \frac{\partial u}{\partial \nu } \nu = \left( \nabla \overline{\psi }^\top \nabla u\right) a^\top \nu ~, \end{aligned}$$

  which implies that the last two terms in the boundary integral cancel out. Using Greens’ Theorem on the part containing $\Delta \overline{\psi }$ and the Helmholtz equation on the part containing $\Delta u$ gives

  $$\begin{aligned} J&= \int _{\Omega } \nabla ( a^\top \nabla u ) \nabla \overline{\psi }+ k^2 \left[ (a^\top \nabla \overline{\psi }) u + u \overline{\psi }\mathrm{div }(a) + (a^\top \nabla u) \nabla \overline{\psi }\right] \\&\quad -k^2 (a^\top \nabla u) \overline{\psi }~dr~. \end{aligned}$$

  The middle term vanishes by Gauss’ theorem since

  $$\begin{aligned} \mathrm{div }( u \overline{\psi }a)=(a^\top \nabla \overline{\psi }) u + u \overline{\psi }\mathrm{div }(a) + (a^\top \nabla u) \nabla \overline{\psi }~. \end{aligned}$$

  Thus we get

  $$\begin{aligned} J=\int _{\Omega } \nabla ( a^\top \nabla u ) \nabla \overline{\psi }-k^2 (a^\top \nabla u) \overline{\psi }~dr ~. \end{aligned}$$

  This gives the following statement (which is valid for $\psi \in \hat{H}^1_0(\Omega )$ by its continuity with respect to $H^1$ convergence):

  $$\begin{aligned} \begin{array}{ll} b(w_a,\psi )\!-\!J&{}=\displaystyle \int _{\Omega } \nabla (w_a\!-\!a^\top \nabla u)^\top \nabla \overline{\psi }\!-\! k^2 (w_a\!-\!a^\top \nabla u) \overline{\psi }~dr \!-\! \sum _{\pm }\Lambda _\pm w_a(\psi )\\ &{} = b(v_a,\psi )~, \end{array} \end{aligned}$$

  where we used that $T_{\Gamma _\pm }w_a=T_{\Gamma _\pm }v_a$ since $a^\top \nabla u=0$ on $\Gamma _{\pm }$. Recall that the variational formulation of $(\Delta +k^2) v_a=0$ in $\Omega$, $v_a$ outgoing on $\Omega _\pm$ is given by:

  $$\begin{aligned} b(v_a,\psi )=0 \text{ for } \text{ all } \psi \in \mathcal H(\Omega ), \end{aligned}$$

  which implies $b(w_a,\psi )=J$, showing that (8.1) holds.

• (2) For $a\in \mathrm{Var }_\epsilon (\Omega )$, we now look at the (unique) solution $u_a\in \mathcal H(\Omega _a)$ of $b(u_a,.)=G_\beta (.)$. It fulfills

  $$\begin{aligned} \int _{\Omega _a}\nabla u_a^\top \nabla \overline{\psi }- k^2 u_a \overline{\psi }~dr - \sum _{\pm }\Lambda _\pm u_a(\psi )= G_\beta (\psi ) \text{ for } \text{ all } \psi \in \mathcal H(\Omega _a).\nonumber \\ \end{aligned}$$

  (8.2)

  Using the transformation theorem with the diffeomorphism $f_a:\Omega \rightarrow \Omega _a$, the integral can be transformed to get

  $$\begin{aligned} \begin{array}{ll} \displaystyle \int _{\Omega ^J_a}\nabla _r u_a^\top \nabla _r \overline{\psi }- k^2 u_a \overline{\psi }~dr&{}=\displaystyle \int _{\Omega ^J}\left[ \left( df_a^{-1} \nabla _\rho (u_a\circ f_a)\right) ^\top \left( df_a^{-1} \nabla _\rho (\overline{\psi }\circ f_a) \right) \right. \\ &{}\quad \displaystyle \left. - k^2 (u_a\circ f_a) (\overline{\psi }\circ f_a) \right] \det (df_a) ~ d\rho . \end{array} \end{aligned}$$

  Since $\mathcal H(\Omega )\rightarrow \mathcal H(\Omega _a),\psi \mapsto \psi \circ f_a$ is bijective, the function $\tilde{u} _a := u_a\circ f_a\in \mathcal H(\Omega )$ satisfies for all $\psi \in \mathcal H(\Omega )$

  $$\begin{aligned} \begin{array}{ll} b_a(\tilde{u}_a, \psi )&{}:= \displaystyle \int _ {\Omega ^J} \nabla \tilde{u}_a^\top \left( \det (df_a) df_a^{-\top } df_a^{-1} \right) \nabla \overline{\psi }- k^2 \det (df_a) \tilde{u}_a \overline{\psi }~dr\\ &{}\quad \displaystyle + \sum _{\pm } \Lambda _\pm \tilde{u}_a(\psi ) \\ &{}= G_\beta (\psi ), \end{array} \end{aligned}$$

  where we used that $a=0$ on $\Gamma _\pm$(i.e. the traces remain unchanged).

• (3) We will prove that $\tilde{u}_a\rightarrow u$ in $\hat{H}^1_0(\Omega ^J)$ for $a\rightarrow 0$. Take some arbitrary $\psi \in \hat{H}^1_0(\Omega ^J)$, then we have

  $$\begin{aligned} \begin{array}{ll} &{}\vert b(u-\tilde{u}_a, \psi )\vert \\ &{}\quad =\vert b(u,\psi )-b(\tilde{u}_a, \psi ) \vert \\ &{}\quad = \vert b_a(\tilde{u}_a,\psi )-b(\tilde{u}_a, \psi )\vert \\ &{}\quad = \left| \displaystyle \int _ {\Omega ^J} \nabla \tilde{u}_a^T \left( \det (df_a) df_a^{-T} df_a^{-1} \!-\!I \right) \nabla \overline{\psi }\!-\! k^2 (\det (df_a) \!-\!1) \tilde{u}_a \overline{\psi }~dr \right| \quad \\ &{}\quad \le \Vert \tilde{u}_a \Vert _{\mathcal H}\left( \Vert \det (df_a) df_a^{-T} df_a^{-1} -I\Vert _{C^0}+\Vert \det (df_a) -1 \Vert _{C^0}\right) \Vert \psi \Vert _\mathcal H\\ &{}\quad \le 2\Vert \tilde{u}_a \Vert _{\mathcal H} \Vert a \Vert _{C^1}\Vert \psi \Vert _{\mathcal H} , \end{array} \end{aligned}$$

  (8.3)

  where we used that $b(u,\psi )=b_a(\tilde{u}_a,\psi )$ by (8.2). For the last inequality, we refer to step (5), where we show calculations that imply the estimate there. In view of the uniqueness on $\Omega$ and the Fredholm alternative, we have a boundedly invertable and bounded operator $B: \mathcal H(\Omega )\rightarrow \mathcal H(\Omega )$ such that for all $g\in \mathcal H(\Omega )$

  $$\begin{aligned} b(B g, \psi )=\langle g, \psi \rangle _{\mathcal H} \text{ for } \text{ all } \psi \in \mathcal H(\Omega ). \end{aligned}$$

  By the boundedness of $B$ and by Eq. (8.3) we obtain

  $$\begin{aligned} \begin{array}{ll} \Vert u-\tilde{u}_a \Vert _{\mathcal H}^2 &{}\le \Vert B\Vert ^2 \Vert B^{-1}(u-\tilde{u}_a) \Vert _{\mathcal H}^2\\ &{}=\Vert B \Vert ^2 \langle ~ B^{-1}( u-\tilde{u}_a), B^{-1}(u-\tilde{u}_a) ~\rangle _{\mathcal H}\\ &{}= \Vert B \Vert ^2 \vert ~ b(u-\tilde{u}_a, B^{-1}( u-\tilde{u}_a)) ~\vert \\ &{}\le 2\Vert B\Vert ^2 \Vert \tilde{u}_a \Vert _{\mathcal H} \Vert a \Vert _{C^1}\Vert B^{-1}( u-\tilde{u}_a) \Vert _{\mathcal H} \\ &{}\le 2 \Vert B\Vert ^2\Vert B^{-1}\Vert \Vert \tilde{u}_a \Vert _{\mathcal H} \Vert a \Vert _{C^1}\Vert u-\tilde{u}_a \Vert _{\mathcal H} \\ \end{array} \end{aligned}$$

  where we used the boundedness of $B^{-1}$ for the last step. And thus

  $$\begin{aligned} \frac{\Vert u-\tilde{u}_a \Vert _{\mathcal H}}{\Vert \tilde{u}_a \Vert _{\mathcal H}} \le 2\Vert B\Vert ^2\Vert B^{-1}\Vert \Vert a \Vert _{C^1}. \end{aligned}$$

  (8.4)

  One easily sees that this estimate implies that $\Vert \tilde{u}_a \Vert _{\mathcal H}$ is bounded for $\Vert a \Vert _{C^1}\rightarrow 0$ (assume it is not: Then there exists a subsequence such that the left hand side of (8.4) converges to $1$). On the other hand, by rewriting (8.4)

  $$\begin{aligned} \Vert u-\tilde{u}_a \Vert _{\mathcal H} \le 2\Vert B\Vert ^2\Vert B^{-1}\Vert \Vert a \Vert _{C^1}\Vert \tilde{u}_a \Vert _{\mathcal H} \end{aligned}$$

  we immediately see by the boundedness of $\Vert \tilde{u}_a \Vert _{\mathcal H}$ that $a\rightarrow 0$ implies $\tilde{u}_a\rightarrow u$ in $\mathcal H(\Omega )$.

• (4) We now state a perturbation argument, which we will need in the next step: Let $(u_n)\subset \mathcal H(\Omega )$, $(A_n),(\tilde{A}_n)\subset C^0(\Omega ,{\mathbb {R}}^{N\times N})$, $(B_n),(\tilde{B}_n)\subset C^0(\Omega , {\mathbb {R}})$ be sequences such that

  • $u_n \rightarrow u$ in $\mathcal H(\Omega )$,

  • $(A_n),(\tilde{A}_n)$ are bounded in $C^0(\Omega ,{\mathbb {R}}^{N\times N})$ and $A_n-\tilde{A}_n\rightarrow 0$ in $C^0(\Omega ,{\mathbb {R}}^{N\times N})$,

  • $(B_n),(\tilde{B}_n)$ are bounded in $C^0(\Omega ,{\mathbb {R}}^{N\times N})$ and $B_n-\tilde{B}_n\rightarrow 0$ in $C^0(\Omega ,{\mathbb {R}})$.

  Then there exists some sequence $(c_n)\subset (0,\infty ), c_n\rightarrow 0$ such that for all $\psi \in \mathcal H(\Omega )$ and all $n\in {\mathbb {N}}$ there holds

  $$\begin{aligned} \int _{\Omega }\nabla u_n^\top \tilde{A}_n \nabla \overline{\psi }- B_n u\overline{\psi }- \left( \nabla u_n^\top \tilde{A}_n \nabla \overline{\psi }- \tilde{B}_n u\overline{\psi }\right) ~dr \le c_n \Vert \psi \Vert _\mathcal H. \end{aligned}$$

  (8.5)

  The proof is just a multiple application of Hölder’s inequality after rearranging the terms under the integral suitably in differences that converge to $0$.

• (5) By the definition of the Fréchet derivative we have to prove (where the traces are taken on $\Gamma _\pm$)

  $$\begin{aligned} \frac{\Vert T u_a - T u - T v_a \Vert _{V^{1/2}(S_\pm )} }{ \Vert a \Vert _{C^1}} \rightarrow 0 \text{ for } \Vert a \Vert _{C^1}\rightarrow 0~. \end{aligned}$$

  However, since $a$ vanishes on $\Gamma _\pm$, we have that $T u_a=T\tilde{u}_a$ as well as $T v_a= T w_a$. Thus, by the continuity of $T$, it suffices to show

  $$\begin{aligned} \gamma _a:= \frac{1}{\Vert a \Vert _{C^1}} \left( \tilde{u}_a - u - w_a \right) \rightarrow 0 \text{ in } \mathcal H(\Omega ) \text{ for } \Vert a \Vert _{C^1}\rightarrow 0. \end{aligned}$$

  Let’s apply the sesquilinear form $b(.,.)$ to $\gamma _a$ with some $\psi \in \mathcal H(\Omega )$

  $$\begin{aligned} \begin{array}{ll} b(\gamma _a,\psi ) &{}= \frac{1}{\Vert a \Vert _{C^1}} \left( b( \tilde{u}_a,\psi )-b(u, \psi ) - b(w_a,\psi ) \right) \\ &{}= \frac{1}{\Vert a \Vert _{C^1}} \left( b( \tilde{u}_a,\psi )-b_a(\tilde{u}_a, \psi ) - b(w_a,\psi ) \right) \end{array} \end{aligned}$$

  where Eq. (8.2) was used to get that $b_a(\tilde{u}_a, \psi )=b(u, \psi )$. Now, using step (1) for the last term as well as the definitions of $b$ and $b_a$, we get

  $$\begin{aligned} \begin{array}{ll} b(\gamma _a,\psi )= &{} \displaystyle \int _{\Omega } \nabla \tilde{u}^T_a \underbrace{\left( \frac{I\!-\!\det (df_a) df^{-T}_a df^{-1}_a }{\Vert a\Vert _{C^1}} \right) }_{=:\tilde{A}_{a}} \nabla \overline{\psi }\!+\! \underbrace{\frac{k^2(1\!-\!\det (df_a))}{\Vert a\Vert _{C^1}}}_{=:\tilde{B}_a}\tilde{u}_a\overline{\psi }~ dr \\ &{}\displaystyle -\int _{\Omega }\nabla u ^T \underbrace{\left( \frac{-\mathrm{div }(a) I + da +da^T}{\Vert a\Vert _{C^1}}\right) }_{=:A_a} \nabla \overline{\psi }- \underbrace{\frac{k^2 \mathrm{div }(a) }{\Vert a\Vert _{C^1}}}_{=:B_a} u\overline{\psi }~dr. \end{array} \end{aligned}$$

  We want to apply step (4). Using the geometric series for matrices we obtain

  $$\begin{aligned} df_a^{-1}=(1+da)^{-1}=\sum _{l=0}^\infty -(da)^l = 1-da+O(\Vert a \Vert _{C^1}^2), \end{aligned}$$

  and similarly by elementary calculation

  $$\begin{aligned} \det (df_a)=1+\mathrm{div }(a)+O(\Vert a \Vert _{C^1}^2). \end{aligned}$$

  Using these relations and elementary calculation yields

  $$\begin{aligned} \begin{array}{ll} \tilde{A}_a - A_a = O(\Vert a \Vert _{C^1})\\ \tilde{B}_a - B_a = O(\Vert a \Vert _{C^1}) \end{array} \end{aligned}$$

  as well as $\tilde{A}_a, A_a, \tilde{B}_a, B_a = O(1)$, that is their boundedness. Now step (4) yields some $c_a>0$ with $c_a\rightarrow 0$ such that

  $$\begin{aligned} \vert b(\gamma _a, \psi )\vert \le c_a \Vert \psi \Vert _{\mathcal H}. \end{aligned}$$

  (8.6)

  Inserting $\psi =B^{-1}\gamma _a$ into (8.6) yields

  $$\begin{aligned} \Vert B^{-1}\gamma _a \Vert ^2_{\mathcal H} = (B^{-1}\gamma _a,B^{-1}\gamma _a)_{\mathcal H} = \vert b(\gamma _a , B^{-1}\gamma _a ) \vert \le c_a \Vert B^{-1} \gamma _a \Vert _{\mathcal H}. \end{aligned}$$

  And by the boundedness of $B$

  $$\begin{aligned} \Vert \gamma _a \Vert _{\mathcal H} \le \Vert B \Vert \Vert B^{-1}\gamma _a \Vert _{\mathcal H} \le \Vert B \Vert c_a. \end{aligned}$$

  This implies $\gamma _a\rightarrow 0$ in $\mathcal H(\Omega )$ for $a\rightarrow 0$ in $C^1$, which ends the proof. $\square$