Let us consider the interior Robin problem for the Helmholtz equation in a starlike annulus , whose boundaries {\partial}_{\pm}\mathcal{A} are described by the polar equations r={R}_{\pm}(\vartheta ) respectively:

\{\begin{array}{cc}\mathrm{\Delta}v(x,y)+{k}^{2}v(x,y)=0,\hfill & (x,y)\in \stackrel{\u02da}{\mathcal{A}},\hfill \\ {\lambda}_{\pm}v(x,y)+{\gamma}_{\pm}\frac{\partial v}{\partial \nu}(x,y)={f}_{\pm}(x,y),\hfill & (x,y)\in {\partial}_{\pm}\mathcal{A},\hfill \end{array}

(12)

where k>0 denotes the propagation constant, {\stackrel{\u02c6}{\nu}}_{\pm}={\stackrel{\u02c6}{\nu}}_{\pm}(\vartheta ) are the outward-pointing normal unit vectors to the domain boundaries {\partial}_{\pm}\mathcal{A}, respectively, and {\lambda}_{\pm}, {\gamma}_{\pm} are given regular weighting coefficients.

Under the mentioned assumptions, one can prove the following theorem.

**Theorem**
*Let*

{\psi}_{\pm}(\vartheta )=\frac{d}{d\vartheta}ln{R}_{\pm}(\vartheta )=\frac{{\dot{R}}_{\pm}(\vartheta )}{{R}_{\pm}(\vartheta )},

(13)

*and*

\begin{array}{rl}{f}_{\pm}({R}_{\pm}(\vartheta )cos\vartheta ,{R}_{\pm}(\vartheta )sin\vartheta )& ={F}_{\pm}(\vartheta )\\ =\frac{1}{\sqrt{1+{\psi}_{\pm}{(\vartheta )}^{2}}}\sum _{m=0}^{+\mathrm{\infty}}({\alpha}_{m}^{(\pm )}cosm\vartheta +{\beta}_{m}^{(\pm )}sinm\vartheta ),\end{array}

(14)

*where*

\left\{\begin{array}{c}{\alpha}_{m}^{(\pm )}\\ {\beta}_{m}^{(\pm )}\end{array}\right\}=\frac{{\u03f5}_{m}}{2\pi}{\int}_{0}^{2\pi}{F}_{\pm}(\vartheta )\sqrt{1+{\psi}_{\pm}{(\vartheta )}^{2}}\left\{\begin{array}{c}cosm\vartheta \\ sinm\vartheta \end{array}\right\}\phantom{\rule{0.2em}{0ex}}d\vartheta ,

(15)

{\u03f5}_{m} *being the usual Neumann symbol*. *Then boundary value problem* (12) *for the Helmholtz equation admits a classical solution* v(x,y)\in {L}^{2}(\mathcal{A}) *such that the following Fourier*-*Hankel series expansion holds true*:

\begin{array}{c}v(\frac{(b-\varrho ){R}_{-}(\vartheta )-(a-\varrho ){R}_{+}(\vartheta )}{b-a}cos\vartheta ,\frac{(b-\varrho ){R}_{-}(\vartheta )-(a-\varrho ){R}_{+}(\vartheta )}{b-a}sin\vartheta )\hfill \\ \phantom{\rule{1em}{0ex}}=U(\varrho ,\vartheta )\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{+\mathrm{\infty}}[{H}_{m}^{(1)}\left(k\frac{(b-\varrho ){R}_{-}(\vartheta )-(a-\varrho ){R}_{+}(\vartheta )}{b-a}\right)({A}_{1,m}cosm\vartheta +{B}_{1,m}sinm\vartheta )\hfill \\ \phantom{\rule{2em}{0ex}}+{H}_{m}^{(2)}\left(k\frac{(b-\varrho ){R}_{-}(\vartheta )-(a-\varrho ){R}_{+}(\vartheta )}{b-a}\right)({A}_{2,m}cosm\vartheta +{B}_{2,m}sinm\vartheta )].\hfill \end{array}

(16)

*For each index* *m*, *define*

\begin{array}{rl}\left[\begin{array}{c}{\xi}_{p,m}^{(\pm )}(\vartheta )\\ {\eta}_{p,m}^{(\pm )}(\vartheta )\end{array}\right]=& \left[\begin{array}{cc}cosm\vartheta & -sinm\vartheta \\ sinm\vartheta & cosm\vartheta \end{array}\right]\\ \cdot \left[\begin{array}{c}{\lambda}_{\pm}{H}_{m}^{(p)}(k{R}_{\pm}(\vartheta ))\sqrt{1+{\psi}_{\pm}{(\vartheta )}^{2}}\pm k{\gamma}_{\pm}{\dot{H}}_{m}^{(p)}(k{R}_{\pm}(\vartheta ))\\ \mp {\gamma}_{\pm}\frac{m}{{R}_{\pm}(\vartheta )}{H}_{m}^{(p)}(k{R}_{\pm}(\vartheta )){\psi}_{\pm}(\vartheta )\end{array}\right],\end{array}

(17)

*with* {H}_{m}^{(p)}(\cdot ) *denoting the Hankel function of kind* p=1,2 *and order* *m*. *Hence*, *the coefficients* {A}_{p,m}, {B}_{p,m} *in* (16) *can be determined by solving the infinite linear system*

\sum _{m=0}^{+\mathrm{\infty}}\left[\begin{array}{cccc}{\mathrm{X}}_{1,1,n,m}^{(-)}& {\mathrm{Y}}_{1,1,n,m}^{(-)}& {\mathrm{X}}_{1,2,n,m}^{(-)}& {\mathrm{Y}}_{1,2,n,m}^{(-)}\\ {\mathrm{X}}_{2,1,n,m}^{(-)}& {\mathrm{Y}}_{2,1,n,m}^{(-)}& {\mathrm{X}}_{2,2,n,m}^{(-)}& {\mathrm{Y}}_{2,2,n,m}^{(-)}\\ {\mathrm{X}}_{1,1,n,m}^{(+)}& {\mathrm{Y}}_{1,1,n,m}^{(+)}& {\mathrm{X}}_{1,2,n,m}^{(+)}& {\mathrm{Y}}_{1,2,n,m}^{(+)}\\ {\mathrm{X}}_{2,1,n,m}^{(+)}& {\mathrm{Y}}_{2,1,n,m}^{(+)}& {\mathrm{X}}_{2,2,n,m}^{(+)}& {\mathrm{Y}}_{2,2,n,m}^{(+)}\end{array}\right]\cdot \left[\begin{array}{c}{A}_{1,m}\\ {B}_{1,m}\\ {A}_{2,m}\\ {B}_{2,m}\end{array}\right]=\left[\begin{array}{c}{\alpha}_{n}^{(-)}\\ {\beta}_{n}^{(-)}\\ {\alpha}_{n}^{(+)}\\ {\beta}_{n}^{(+)}\end{array}\right],

(18)

*where*

{\mathrm{X}}_{\left\{\begin{array}{c}1\\ 2\end{array}\right\},p,n,m}^{(\pm )}=\frac{{\u03f5}_{n}}{2\pi}{\int}_{0}^{2\pi}{\xi}_{p,m}^{(\pm )}(\vartheta )\left\{\begin{array}{c}cosn\vartheta \\ sinn\vartheta \end{array}\right\}\phantom{\rule{0.2em}{0ex}}d\vartheta ,

(19)

{\mathrm{Y}}_{\left\{\begin{array}{c}1\\ 2\end{array}\right\},p,n,m}^{(\pm )}=\frac{{\u03f5}_{n}}{2\pi}{\int}_{0}^{2\pi}{\eta}_{p,m}^{(\pm )}(\vartheta )\left\{\begin{array}{c}cosn\vartheta \\ sinn\vartheta \end{array}\right\}\phantom{\rule{0.2em}{0ex}}d\vartheta ,

(20)

*with* p=1,2 *and* m,n\in {\mathbb{N}}_{0}:=\mathbb{N}\cup \{0\}.

*Proof* Upon noting that in the stretched coordinate system *ϱ*, *ϑ* introduced in the *x*, *y* plane, the considered domain turns into the circular annulus of radii *a* and *b*, one can readily adopt the usual eigenfunction method [10] in combination with the separation of variables (with respect to *r* and *ϑ*). As a consequence, elementary solutions of the problem can be searched in the form

u(r,\vartheta )=U(\frac{b[r-{R}_{-}(\vartheta )]-a[r-{R}_{+}(\vartheta )]}{{R}_{+}(\vartheta )-{R}_{-}(\vartheta )},\vartheta )=\mathrm{P}(r)\mathrm{\Theta}(\vartheta ).

(21)

Substituting into the Helmholtz equation, one easily finds that the functions \mathrm{P}(\cdot ), \mathrm{\Theta}(\cdot ) must satisfy the ordinary differential equations

\frac{{d}^{2}\mathrm{\Theta}(\vartheta )}{d{\vartheta}^{2}}+{\mu}^{2}\mathrm{\Theta}(\vartheta )=0,

(22)

{r}^{2}\frac{{d}^{2}\mathrm{P}(r)}{d{r}^{2}}+r\frac{d\mathrm{P}(r)}{dr}+({k}^{2}{r}^{2}-{\mu}^{2})\mathrm{P}(r)=0,

(23)

respectively. The parameter *μ* is a separation constant whose choice is governed by the physical requirement that at any fixed point in the real plane the scalar field u(r,\vartheta ) must be single-valued. So, by setting \mu =m\in {\mathbb{N}}_{0}, one can easily find

\mathrm{\Theta}(\vartheta )={a}_{m}cosm\vartheta +{b}_{m}sinm\vartheta ,

(24)

where {a}_{m},{b}_{m}\in \mathbb{C} denote arbitrary constants. The radial function \mathrm{P}(\cdot ) satisfying (23) can be readily expressed as follows:

\mathrm{P}(r)={c}_{m}{H}_{m}^{(1)}(kr)+{d}_{m}{H}_{m}^{(2)}(kr),

(25)

with {c}_{m},{d}_{m}\in \mathbb{C}. Therefore, the general solution of Robin problem (12) can be searched in the form

\begin{array}{rl}u(r,\vartheta )=& \sum _{m=0}^{+\mathrm{\infty}}[{H}_{m}^{(1)}(kr)({A}_{1,m}cosm\vartheta +{B}_{1,m}sinm\vartheta )\\ +{H}_{m}^{(2)}(kr)({A}_{2,m}cosm\vartheta +{B}_{2,m}sinm\vartheta )].\end{array}

(26)

Enforcing the Robin boundary condition yields

\begin{array}{rl}{F}_{\pm}(\vartheta )& ={\lambda}_{\pm}u({R}_{\pm}(\vartheta ),\vartheta )+{\gamma}_{\pm}\frac{\partial u}{\partial \nu}({R}_{\pm}(\vartheta ),\vartheta )\\ ={\lambda}_{\pm}u({R}_{\pm}(\vartheta ),\vartheta )+{\gamma}_{\pm}\mathrm{\nabla}u({R}_{\pm}(\vartheta ),\vartheta )\cdot {\stackrel{\u02c6}{\nu}}_{\pm}(\vartheta ),\end{array}

(27)

where

\mathrm{\nabla}u(r,\vartheta )=\stackrel{\u02c6}{r}\frac{\partial u(r,\vartheta )}{\partial r}+\stackrel{\u02c6}{\vartheta}\frac{1}{r}\frac{\partial u(r,\vartheta )}{\partial \vartheta},

(28)

and

{\stackrel{\u02c6}{\nu}}_{\pm}(\vartheta )=\pm \frac{\stackrel{\u02c6}{r}-{\psi}_{\pm}(\vartheta )\stackrel{\u02c6}{\vartheta}}{\sqrt{1+{\psi}_{\pm}{(\vartheta )}^{2}}}.

(29)

Hence, combining equations above and using the classical Fourier projection method, equations (17)-(20) follow after some algebraic manipulations. □

It is worth noting that the derived expressions still hold under the assumption that {R}_{\pm}(\vartheta ) are piecewise continuous functions, and the boundary values are described by square integrable, not necessarily continuous, functions, so that the relevant Fourier coefficients {\alpha}_{m}^{(\pm )}, {\beta}_{m}^{(\pm )} in equation (14) are finite quantities.