Two-Dimensional Elastic Scattering Coefficients and Enhancement of Nearly Elastic Cloaking

Abstract

The concept of scattering coefficients has played a pivotal role in a broad range of inverse scattering and imaging problems in acoustic, and electromagnetic media. In view of their promising applications in inverse problems related to mathematical imaging and elastic cloaking, the notion of elastic scattering coefficients of an inclusion is presented from the perspective of boundary layer potentials and a few properties are discussed. A reconstruction algorithm is developed and analyzed for extracting the elastic scattering coefficients from multi-static response measurements of the scattered field in order to cater to inverse scattering problems. The decay rate, stability and error analyses, and the estimate of maximal resolving order in terms of the signal-to-noise ratio are discussed. Moreover, scattering-coefficients-vanishing structures are designed and their utility for enhancement of nearly elastic cloaking is elucidated.

Appendices

Appendix A: Multipolar Expansion of Elastodynamic Fundamental Solution

Note that, by Helmholtz decomposition, $\boldsymbol {\Gamma }^{\omega}({\mathbf {x}},{\mathbf{y}})\mathbf{p}$ can be decomposed for any constant vector $\mathbf{p}\in \mathbb {R}^{2}$ and ${\mathbf{x}}\neq {\mathbf{y}}$ as (see, for instance, [1, 27])

$$\begin{aligned} \boldsymbol {\Gamma }^{\omega}({\mathbf{x}},{\mathbf{y}})\mathbf{p}= \nabla_{{\mathbf {x}}}G_{P} ({\mathbf{x}},{\mathbf{y}}) +\vec{ \nabla}_{\perp,{\mathbf {x}}}\times G_{S} ({\mathbf{x}},{\mathbf{y}}), \end{aligned}$$

(A.1)

with

$$G_{P}({\mathbf{x}},{\mathbf{y}}):=-\frac{1}{{\kappa }_{P}^{2}}\nabla_{{\mathbf {x}}} \cdot\bigl(\boldsymbol {\Gamma }^{\omega}({\mathbf{x}},{\mathbf{y}})\mathbf{p}\bigr) \quad\text{and} \quad G_{S}({\mathbf{x}},{\mathbf{y}}):=\frac{1}{{\kappa }_{S}^{2}} { \nabla}_{\perp ,{\mathbf{x}}}\times\bigl(\boldsymbol {\Gamma }^{\omega}({\mathbf{x}},{\mathbf{y}}) \mathbf{p}\bigr). $$

By (2.2), one can easily show that

$$\begin{aligned} & G_{P}({\mathbf{x}},{\mathbf{y}})=-\frac{1}{\rho_{0}\omega^{2}} \nabla _{{\mathbf{x}}}g({\mathbf{x}}-{\mathbf{y}},{\kappa }_{P})\cdot\mathbf{p} = \frac{1}{\rho_{0}\omega^{2}} \nabla_{{\mathbf{y}}}g({\mathbf{x}}-{\mathbf{y}},{\kappa }_{P}) \cdot\mathbf{p}, \end{aligned}$$

(A.2)

$$\begin{aligned} & G_{S}({\mathbf{x}},{\mathbf{y}})=-\frac{1}{\rho_{0}\omega^{2}} \vec{ \nabla }_{\perp,{\mathbf{x}}}\times g({\mathbf{x}}-{\mathbf{y}},{\kappa }_{S})\cdot \mathbf{p} =\frac{1}{\rho_{0}\omega^{2}} \vec{\nabla}_{\perp, {\mathbf{y}}}\times g({\mathbf{x}}- {\mathbf{y}},{\kappa }_{S})\cdot\mathbf{p}, \end{aligned}$$

(A.3)

where the reciprocity relations

$$g({\mathbf{x}}-{\mathbf{y}},{\kappa }_{\alpha})= g({\mathbf{y}}-{\mathbf{x}},{\kappa }_{\alpha}) \quad\text{and}\quad \nabla_{{\mathbf{x}}}g({\mathbf{x}}-{\mathbf{y}},{\kappa }_{\alpha})= -\nabla _{{\mathbf{y}}}g({\mathbf{x}}-{\mathbf{y}},{\kappa }_{\alpha}), $$

have been used. Recall that, by Graf’s addition formula (see, for example, [38, Formula 10.23.7]), we have

$$H_{0}^{(1)}({\kappa }|{\mathbf{x}}-{\mathbf{y}}|)=\sum _{n\in} H_{n}^{(1)}({\kappa }|{ \mathbf{x}}|)e^{in\theta_{{\mathbf{x}}}} \overline{J_{n}({\kappa }|{\mathbf{y}}|)e^{in\theta_{{\mathbf{y}}}}}. $$

Consequently, it follows from (A.2), (A.3), and (2.3) that

$$\begin{aligned} G_{P}({\mathbf{x}},{\mathbf{y}})&=\frac{i}{4\rho_{0}\omega^{2}}\sum _{n\in } H_{n}^{(1)}({\kappa }_{P}|{ \mathbf{x}}|)e^{in\theta_{{\mathbf{x}}}}\, \overline{\nabla\bigl[ J_{n}( {\kappa }_{P}|{\mathbf{y}}|)e^{in\theta_{{\mathbf{y}}}}\bigr]\cdot \mathbf{p}}, \\ G_{S}({\mathbf{x}},{\mathbf{y}})&=\frac{i}{4\rho_{0}\omega^{2}}\sum _{n\in } H_{n}^{(1)}({\kappa }_{S}|{ \mathbf{x}}|)e^{in\theta_{{\mathbf{x}}}} \overline{\vec{\nabla}_{\perp}\times\bigl[ J_{n}({\kappa }_{S}|{\mathbf{y}}|)e^{in\theta_{{\mathbf{y}}}}\bigr]\cdot \mathbf{p}}. \end{aligned}$$

The identity (3.11) follows by substituting the values of $G_{\alpha}$ in the decomposition (A.1) and using the definition of $\mathbf {J}^{\alpha}$ and $\mathbf {H}^{\alpha}$.

Appendix B: Proof of Lemma 3.5

Proof

Since our formulation here is based on an integral representation in terms of the densities ${\boldsymbol{\varphi}}$ and ${\boldsymbol{\psi}}$ satisfying (2.8), we take a different route than those already discussed in [46, 49] without directly invoking the argument of reciprocity.

Let us first fix some notation. For any ${\mathbf{v}},{\mathbf{w}}\in H^{3/2}(D)^{2}$ and $a,b\in \mathbb {R}_{+}$, define the quadratic form

$$\begin{aligned} \langle{\mathbf{v}},{\mathbf{w}}\rangle^{a,b}_{D}:= \int_{D} \biggl[ a(\nabla\cdot{\mathbf{v}}) (\nabla\cdot{ \mathbf{w}})+\frac {b}{2} \bigl( \nabla{\mathbf{v}}+\nabla{ \mathbf{v}}^{\top} \bigr) :\bigl(\nabla{\mathbf{w}}+\nabla{ \mathbf{w}}^{\top}\bigr) \biggr] d{\mathbf{x}}, \end{aligned}$$

where double dot ‘ : ’ denotes the matrix contraction operator defined for two matrices $\mathbf{A}=(a_{ij})$ and $\mathbf {B}=(b_{ij})$ by $\mathbf{A}:\mathbf{B}:=\displaystyle \sum_{i,j}a_{ij}b_{ij}$. It is easy to get from the definition of $\langle \cdot{,}\cdot\rangle^{a,b}_{D}$ that

$$\begin{aligned} \int_{\partial D}{\mathbf{v}}\cdot\frac{\partial{\mathbf {w}}}{\partial\nu}d\sigma({ \mathbf{x}})= \int_{D}{\mathbf{v}}\cdot \mathcal {L}_{a,b}[{\mathbf{w}}]d{ \mathbf{x}}+\langle{\mathbf{v}},{\mathbf {w}}\rangle^{a,b}_{D}. \end{aligned}$$

(B.1)

Note that if ${\mathbf{w}}$ is a solution of the Lamé equation $\mathcal {L}_{a,b}[{\mathbf{w}}]+c\omega^{2}{\mathbf{w}}=\mathbf{0}$ then

$$\begin{aligned} \int_{\partial D}{\mathbf{v}}\cdot\frac{\partial{\mathbf {w}}}{\partial\nu}d\sigma({ \mathbf{x}})=-c \omega^{2} \int _{D}{\mathbf{v}}\cdot{\mathbf{w}}d{\mathbf{x}}+\langle{ \mathbf {v}},{\mathbf{w}}\rangle^{a,b}_{D}, \end{aligned}$$

and consequently from (B.1)

$$\begin{aligned} \int_{\partial D}{\mathbf{v}}\cdot\frac{\partial{\mathbf {w}}}{\partial\nu}d\sigma({ \mathbf{x}})= \int_{\partial D} \frac {\partial{\mathbf{v}}}{\partial\nu}\cdot{\mathbf{w}}d\sigma ({ \mathbf{x}})-c\omega^{2} \int_{D}{\mathbf{v}}\cdot{\mathbf {w}}d{\mathbf{x}}- \int_{D}\mathcal {L}_{a,b}[{\mathbf{v}}]\cdot{\mathbf {w}}d{ \mathbf{x}}. \end{aligned}$$

(B.2)

Moreover, if ${\mathbf{v}}$ solves $\mathcal {L}_{a,b}[{\mathbf{v}}]+c\omega ^{2}{\mathbf{v}}=\mathbf{0}$ then

$$\begin{aligned} \int_{\partial D}{\mathbf{v}}\cdot\frac{\partial{\mathbf {w}}}{\partial\nu}d\sigma({ \mathbf{x}})= \int_{\partial D}\frac {\partial{\mathbf{v}}}{\partial\nu}\cdot{\mathbf{w}}d\sigma ({ \mathbf{x}}). \end{aligned}$$

(B.3)

We will also require the constants

$$\begin{aligned} \eta_{P}&:=\frac{\mu_{0}}{\mu_{1}-\mu_{0}}, \\ \widetilde{\eta}_{P}&:=\frac{\mu_{1}}{\mu_{1}-\mu_{0}}, \\ \eta_{S}&:=\frac{\lambda_{0}+\mu_{0}}{(\lambda_{1}-\lambda _{0})+(\mu_{1}-\mu_{0})}, \\ \widetilde{\eta}_{S}&:=\frac{\lambda_{1}+\mu_{1}}{(\lambda _{1}-\lambda_{0})+(\mu_{1}-\mu_{0})}. \end{aligned}$$

Let $({\boldsymbol{\varphi}}_{n}^{\alpha}, {\boldsymbol{\psi}}_{n}^{\alpha})$ and $({\boldsymbol{\varphi}}_{n}^{\beta}, {\boldsymbol{\psi}}_{m}^{\beta})$ be the solutions of (2.8) with ${\mathbf{u}}^{\mathrm{inc}}=\mathbf {J}^{\alpha}$ and ${\mathbf{u}}^{\mathrm{inc}}=\mathbf {J}^{\beta}$ respectively, i.e.

$$\begin{aligned} &\widetilde{\mathcal{S}}^{\omega}_{D}{\boldsymbol{\varphi}}_{n}^{\alpha}- \mathcal {S}^{\omega}_{D}{\boldsymbol{\psi}}_{n}^{\alpha} = \mathbf {J}^{\alpha}_{n} \big|_{\partial D}, \end{aligned}$$

(B.4)

$$\begin{aligned} &\frac{\partial}{\partial\widetilde{\nu}}\widetilde{\mathcal {S}}^{\omega}_{D} {\boldsymbol{\varphi}}_{n}^{\alpha} \bigg|_{-} -\frac{\partial}{\partial\nu} \mathcal{S}^{\omega}_{D}{\boldsymbol{\psi}}_{n}^{\alpha} \bigg|_{+} =\displaystyle \frac{\partial \mathbf {J}^{\alpha}_{n}}{\partial\nu} \bigg|_{\partial D}, \end{aligned}$$

(B.5)

and

$$\begin{aligned} &\widetilde{\mathcal{S}}^{\omega}_{D}{\boldsymbol{\varphi}}_{m}^{\beta}- \mathcal {S}^{\omega}_{D}{\boldsymbol{\psi}}_{m}^{\beta}= \mathbf {J}^{\beta}_{m} \big|_{\partial D}, \end{aligned}$$

(B.6)

$$\begin{aligned} &\frac{\partial}{\partial\widetilde{\nu}}\widetilde{\mathcal {S}}^{\omega}_{D} {\boldsymbol{\varphi}}_{m}^{\beta} \bigg|_{-} -\frac{\partial}{\partial\nu} \mathcal{S}^{\omega}_{D}{\boldsymbol{\psi}}_{m}^{\beta} \bigg|_{+}=\displaystyle \frac{\partial \mathbf {J}^{\beta}_{m}}{\partial \nu} \bigg|_{\partial D}. \end{aligned}$$

(B.7)

Then, by making use of the jump conditions (2.4), $W^{\alpha ,\beta}_{m,n}$ can be expressed as

$$\begin{aligned} W^{\alpha,\beta}_{m,n}=\displaystyle \int_{\partial D}\overline{\mathbf {J}_{n}^{\alpha}}\cdot {\boldsymbol{\psi}}_{m}^{\beta} d\sigma({\mathbf{x}})= \int _{\partial D}\overline{\mathbf {J}_{n}^{\alpha}}\cdot \biggl[ \frac {\partial}{\partial\nu}\mathcal{S}^{\omega}_{D}\bigl[ {\boldsymbol{\psi}}^{\beta }_{m}\bigr] \big|_{+}-\frac{\partial}{\partial\nu} \mathcal{S}^{\omega }_{D}\bigl[{\boldsymbol{\psi}}^{\beta}_{m} \bigr] \big|_{-} \biggr] d\sigma({\mathbf{x}}). \end{aligned}$$

Further, by invoking (B.7) and subsequently using (B.2) and (B.3), one gets the expression

$$\begin{aligned} W^{\alpha,\beta}_{m,n}&=- \int_{\partial D}\overline{\mathbf {J}_{n}^{\alpha}}\cdot \frac{\partial \mathbf {J}^{\beta}_{m}}{\partial\nu }d\sigma({\mathbf{x}})+ \int_{\partial D}\overline{\mathbf {J}_{n}^{\alpha }}\cdot \biggl[ \frac{\partial}{\partial\widetilde{\nu}}\widetilde {\mathcal{S}}^{\omega}_{D} \bigl[{\boldsymbol{\varphi}}^{\beta}_{m}\bigr] \big|_{-} - \frac{\partial}{\partial\nu}\mathcal{S}^{\omega}_{D}\bigl[{\boldsymbol{\psi}}^{\beta}_{m}\bigr] \big|_{-} \biggr] d\sigma({ \mathbf{x}}) \\ &=- \int_{\partial D}\overline{\mathbf {J}_{n}^{\alpha}}\cdot \frac {\partial \mathbf {J}^{\beta}_{m}}{\partial\nu}d\sigma({\mathbf{x}})+ \int _{\partial D} \biggl[ \frac{\partial\overline{\mathbf {J}_{n}^{\alpha }}}{\partial\widetilde{\nu}}\cdot\widetilde{ \mathcal{S}}^{\omega }_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr] -\frac{\partial\overline{\mathbf {J}_{n}^{\alpha}}}{\partial\nu}\cdot \mathcal{S}^{\omega}_{D}\bigl[ {\boldsymbol{\psi}}^{\beta}_{m}\bigr] \biggr] d\sigma ({\mathbf{x}}) \\ &\quad{}-\rho_{1}\omega^{2} \int_{D}\overline{\mathbf {J}^{\alpha}_{n}}\cdot \widetilde{\mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d{\mathbf {x}}- \int_{D}\mathcal {L}_{\lambda_{1},\mu_{1}}\bigl[\overline{ \mathbf {J}^{\alpha }_{n}}\bigr]\cdot\widetilde{\mathcal{S}}^{\omega}_{D} \bigl[{\boldsymbol{\varphi}}^{\beta }_{m}\bigr]d{\mathbf{x}}. \end{aligned}$$

This, together with (B.6), leads to

$$\begin{aligned} W^{\alpha,\beta}_{m,n}&=- \int_{\partial D}\overline{\mathbf {J}_{n}^{\alpha}}\cdot \frac{\partial \mathbf {J}^{\beta}_{m}}{\partial\nu }d\sigma({\mathbf{x}}) + \int_{\partial D}\frac{\partial\overline{\mathbf {J}_{n}^{\alpha }}}{\partial\widetilde{\nu}}\cdot\widetilde{ \mathcal{S}}^{\omega }_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr] d\sigma({\mathbf{x}}) - \int_{\partial D}\frac{\partial\overline{\mathbf {J}_{n}^{\alpha }}}{\partial\nu}\cdot\widetilde{ \mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d\sigma({\mathbf{x}}) \\ &\quad{}+ \int_{\partial D}\frac{\partial\overline{\mathbf {J}}_{n}^{\alpha }}{\partial\nu}\cdot \mathbf {J}_{m}^{\beta} \cdot d\sigma({\mathbf{x}}) -\rho_{1}\omega^{2} \int_{D}\overline{\mathbf {J}^{\alpha}_{n}}\cdot \widetilde{\mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d{\mathbf {x}}- \int_{D}\mathcal {L}_{\lambda_{1},\mu_{1}}\bigl[\overline{ \mathbf {J}^{\alpha }_{n}}\bigr]\cdot\widetilde{\mathcal{S}}^{\omega}_{D} \bigl[{\boldsymbol{\varphi}}^{\beta }_{m}\bigr]d{\mathbf{x}}. \end{aligned}$$

It is easy to see that the first and the fourth terms cancel out each other thanks to (B.3). Therefore,

$$\begin{aligned} W^{\alpha,\beta}_{m,n}&= \int_{\partial D} \biggl[ \frac{\partial\overline{\mathbf {J}_{n}^{\alpha }}}{\partial\widetilde{\nu}}-\frac{\partial\overline{\mathbf {J}_{n}^{\alpha}}}{\partial\nu} \biggr] \cdot\widetilde{\mathcal {S}}^{\omega}_{D}\bigl[ {\boldsymbol{\varphi}}^{\beta}_{m}\bigr]d\sigma({\mathbf{x}})-\rho _{1} \omega^{2} \int_{D}\overline{\mathbf {J}^{\alpha}_{n}}\cdot \widetilde {\mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d{\mathbf{x}} \\ &\quad{}- \int_{D}\mathcal {L}_{\lambda_{1},\mu_{1}}\bigl[\overline{ \mathbf {J}^{\alpha }_{n}}\bigr]\cdot\widetilde{\mathcal{S}}^{\omega}_{D} \bigl[{\boldsymbol{\varphi}}^{\beta }_{m}\bigr]d{\mathbf{x}}. \end{aligned}$$

(B.8)

Remark that $\nabla\cdot \mathbf {J}^{S}_{n}=0=\nabla\times \mathbf {J}^{P}_{n}$. Therefore, it is easy to verify by definition of the surface traction operator that

$$\begin{aligned} \frac{\partial\overline{\mathbf {J}_{n}^{\alpha}}}{\partial\widetilde {\nu}}-\frac{\partial\overline{\mathbf {J}_{n}^{\alpha}}}{\partial\nu }=\frac{1}{\eta_{\alpha}}\frac{\partial\overline{\mathbf {J}_{n}^{\alpha }}}{\partial\nu} = \frac{1}{\widetilde{\eta}_{\alpha}}\frac {\partial\overline{\mathbf {J}_{n}^{\alpha}}}{\partial\widetilde{\nu }}. \end{aligned}$$

(B.9)

Thus, using right most quantity of (B.9) in (B.8) and subsequently invoking identity (B.2), one gets

$$\begin{aligned} \widetilde{\eta}_{\alpha} W^{\alpha,\beta}_{m,n} &= \int_{\partial D}\frac{\partial\overline{\mathbf {J}^{\alpha }_{n}}}{\partial\widetilde{\nu}}\cdot\widetilde{\mathcal {S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d\sigma({\mathbf {x}})-\widetilde{\eta}_{\alpha}\rho_{1} \omega^{2} \int _{D}\overline{\mathbf {J}^{\alpha}_{n}}\cdot \widetilde{\mathcal {S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d{\mathbf{x}} \\ &\quad{}-\widetilde{\eta}_{\alpha} \int_{D}\mathcal {L}_{\lambda_{1},\mu_{1}} \bigl[\overline{ \mathbf {J}^{\alpha}_{n}} \bigr]\cdot\widetilde{\mathcal {S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d{\mathbf{x}} \\ &= \int_{\partial D}\overline{\mathbf {J}^{\alpha}_{n}}\cdot \frac{\partial }{\partial\widetilde{\nu}}\widetilde{\mathcal{S}}^{\omega }_{D}\bigl[ {\boldsymbol{\varphi}}^{\beta}_{m}\bigr] \big|_{-}d\sigma({\mathbf{x}}) +(1-\widetilde{\eta}_{\alpha})\rho_{1}\omega^{2} \int_{D}\overline {\mathbf {J}^{\alpha}_{n}}\cdot \widetilde{\mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m}\bigr]d{\mathbf{x}} \\ &\quad{}+(1-\widetilde{\eta}_{\alpha}) \int_{D}\mathcal {L}_{\lambda_{1},\mu_{1}} \bigl[\overline{ \mathbf {J}^{\alpha}_{n}}\bigr]\cdot\widetilde{\mathcal{S}}^{\omega }_{D} \bigl[{\boldsymbol{\varphi}}^{\beta}_{m}\bigr]d{\mathbf{x}}. \end{aligned}$$

This, together with (B.4) and (B.7), provides

$$\begin{aligned} \widetilde{\eta}_{\alpha} W^{\alpha,\beta}_{m,n} &= \int_{\partial D}\overline{\widetilde{\mathcal{S}}^{\omega }_{D} \bigl[{\boldsymbol{\varphi}}^{\alpha}_{n}\bigr]}\cdot\frac{\partial}{\partial\widetilde {\nu}} \widetilde{\mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr] \big|_{-}d\sigma({\mathbf{x}}) - \int_{\partial D}\overline{\mathcal{S}^{\omega}_{D} \bigl[{\boldsymbol{\psi}}^{\alpha }_{n}\bigr]}\cdot\frac{\partial}{\partial\nu} \mathcal{S}^{\omega }_{D}\bigl[{\boldsymbol{\psi}}^{\beta}_{m} \bigr] \big|_{+}d\sigma({\mathbf{x}}) \\ &\quad{}- \int_{\partial D}\overline{\mathcal{S}_{D}^{\omega} \bigl[{\boldsymbol{\psi}}^{\alpha}_{n}\bigr]}\cdot\frac{\partial \mathbf {J}^{\beta}_{m}}{\partial\nu }d\sigma({ \mathbf{x}})+(1-\widetilde{\eta}_{\alpha})\rho _{1} \omega^{2} \int_{D}\overline{\mathbf {J}^{\alpha}_{n}}\cdot \widetilde {\mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d{\mathbf{x}} \\ &\quad{}+(1-\widetilde{\eta}_{\alpha}) \int_{D}\mathcal {L}_{\lambda_{1},\mu _{1}}\bigl[\overline{ \mathbf {J}^{\alpha}_{n}}\bigr]\cdot\widetilde{\mathcal {S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d{\mathbf{x}}. \end{aligned}$$

(B.10)

Similarly, substituting the first relation of (B.9) back in (B.8) and invoking (B.6), one obtains

$$\begin{aligned} \eta_{\alpha} W^{\alpha,\beta}_{m,n}&= \int_{\partial D}\frac {\partial\overline{\mathbf {J}^{\alpha}_{n}}}{\partial\nu}\cdot \widetilde{ \mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr]d\sigma ({\mathbf{x}})-\eta_{\alpha}\rho_{1} \omega^{2} \int_{D}\overline {\mathbf {J}^{\alpha}_{n}}\cdot \widetilde{\mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m}\bigr]d{\mathbf{x}} \\ &\quad{}-\eta_{\alpha} \int_{D}\mathcal {L}_{\lambda_{1},\mu_{1}}\bigl[\overline{\mathbf {J}^{\alpha}_{n}}\bigr]\cdot\widetilde{\mathcal{S}}^{\omega}_{D} \bigl[{\boldsymbol{\varphi}}^{\beta}_{m}\bigr]d{\mathbf{x}} \\ &= \int_{\partial D}\frac{\partial\overline{\mathbf {J}^{\alpha }_{n}}}{\partial\nu}\cdot\mathcal{S}^{\omega}_{D} \bigl[{\boldsymbol{\psi}}^{\beta }_{m}\bigr]d\sigma({\mathbf{x}})+ \int_{\partial D}\frac{\partial \overline{\mathbf {J}^{\alpha}_{n}}}{\partial\nu}\cdot \mathbf {J}^{\beta}_{m} d\sigma({\mathbf{x}}) \\ &\quad{}-\eta_{\alpha}\rho_{1}\omega^{2} \int_{D}\overline{\mathbf {J}^{\alpha }_{n}}\cdot \widetilde{\mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta }_{m} \bigr]d{\mathbf{x}} -\eta_{\alpha} \int_{D}\mathcal {L}_{\lambda_{1},\mu_{1}}\bigl[\overline{\mathbf {J}^{\alpha}_{n}}\bigr]\cdot\widetilde{\mathcal{S}}^{\omega}_{D} \bigl[{\boldsymbol{\varphi}}^{\beta}_{m}\bigr]d{\mathbf{x}}. \end{aligned}$$

(B.11)

Finally, subtracting (B.11) from (B.10) and noting that $\widetilde{\eta}_{\alpha}-\eta_{\alpha}=1$, one finds out that

$$\begin{aligned} W^{\alpha,\beta}_{m,n}&= \int_{\partial D}\overline{\widetilde{\mathcal{S}}^{\omega}\bigl[ {\boldsymbol{\varphi}}^{\alpha}_{n}\bigr]}\cdot\frac{\partial}{\partial\widetilde{\nu }}\widetilde{ \mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\beta}_{m} \bigr] \big|_{-}d\sigma({\mathbf{x}}) - \int_{\partial D}\overline{\mathcal{S}^{\omega}_{D} \bigl[{\boldsymbol{\psi}}^{\alpha }_{n}\bigr]}\cdot\frac{\partial}{\partial\nu} \mathcal{S}^{\omega }_{D}\bigl[{\boldsymbol{\psi}}^{\beta}_{m} \bigr] \big|_{+}d\sigma({\mathbf{x}}) \\ &\quad{}- \int_{\partial D}\overline{\mathcal{S}_{D}^{\omega} \bigl[{\boldsymbol{\psi}}^{\alpha}_{n}\bigr]}\cdot\frac{\partial \mathbf {J}^{\beta}_{m}}{\partial\nu }d\sigma({ \mathbf{x}}) - \int_{\partial D}\frac{\partial\overline{\mathbf {J}^{\alpha }_{n}}}{\partial\nu}\cdot\mathcal{S}^{\omega}_{D} \bigl[{\boldsymbol{\psi}}^{\beta }_{m}\bigr]d\sigma({\mathbf{x}}) - \int_{\partial D}\frac{\partial\overline{\mathbf {J}^{\alpha }_{n}}}{\partial\nu}\cdot \mathbf {J}^{\beta}_{m} d\sigma({\mathbf{x}}). \end{aligned}$$

(B.12)

Similarly, we have

$$\begin{aligned} W^{\beta,\alpha}_{n,m}&= \int_{\partial D}\overline{\widetilde{\mathcal{S}}^{\omega}\bigl[ {\boldsymbol{\varphi}}^{\beta}_{m}\bigr]}\cdot\frac{\partial}{\partial\widetilde{\nu }}\widetilde{ \mathcal{S}}^{\omega}_{D}\bigl[{\boldsymbol{\varphi}}^{\alpha}_{n} \bigr] \big|_{-}d\sigma({\mathbf{x}}) - \int_{\partial D}\overline{\mathcal{S}^{\omega}_{D} \bigl[{\boldsymbol{\psi}}^{\beta }_{m}\bigr]}\cdot\frac{\partial}{\partial\nu} \mathcal{S}^{\omega }_{D}\bigl[{\boldsymbol{\psi}}^{\alpha}_{n} \bigr] \big|_{+}d\sigma({\mathbf{x}}) \\ &\quad{} - \int_{\partial D}\overline{\mathcal{S}_{D}^{\omega} \bigl[{\boldsymbol{\psi}}^{\beta }_{m}\bigr]}\cdot\frac{\partial \mathbf {J}^{\alpha}_{n}}{\partial\nu}d\sigma ({ \mathbf{x}}) - \int_{\partial D}\frac{\partial\overline{\mathbf {J}^{\beta }_{m}}}{\partial\nu}\cdot\mathcal{S}^{\omega}_{D} \bigl[{\boldsymbol{\psi}}^{\alpha }_{n}\bigr]d\sigma({\mathbf{x}}) - \int_{\partial D}\frac{\partial \overline{\mathbf {J}^{\beta}_{m}}}{\partial\nu}\cdot \mathbf {J}^{\alpha}_{n} d\sigma({\mathbf{x}}) \\ &= \int_{\partial D}\frac{\partial}{\partial\widetilde{\nu }}\overline{\widetilde{ \mathcal{S}}^{\omega}\bigl[{\boldsymbol{\varphi}}^{\beta }_{m}\bigr]} \big|_{-} \cdot\widetilde{\mathcal{S}}^{\omega}_{D} \bigl[{\boldsymbol{\varphi}}^{\alpha }_{n}\bigr]d\sigma({\mathbf{x}}) - \int_{\partial D}\frac{\partial}{\partial\nu}\overline{\mathcal {S}^{\omega}_{D}\bigl[{\boldsymbol{\psi}}^{\beta}_{m}\bigr]} \big|_{+}\cdot\mathcal {S}^{\omega}_{D}\bigl[ {\boldsymbol{\psi}}^{\alpha}_{n}\bigr]d\sigma({\mathbf{x}}) \\ &\quad{} - \int_{\partial D}\overline{\mathcal{S}_{D}^{\omega} \bigl[{\boldsymbol{\psi}}^{\beta }_{m}\bigr]}\cdot\frac{\partial \mathbf {J}^{\alpha}_{n}}{\partial\nu}d\sigma ({ \mathbf{x}}) - \int_{\partial D}\frac{\partial\overline{\mathbf {J}^{\beta }_{m}}}{\partial\nu}\cdot\mathcal{S}^{\omega}_{D} \bigl[{\boldsymbol{\psi}}^{\alpha }_{n}\bigr]d\sigma({\mathbf{x}}) - \int_{\partial D}\overline{\mathbf {J}^{\beta}_{m}} \cdot \frac{\partial \mathbf {J}^{\alpha}_{n}}{\partial\nu}d\sigma({\mathbf{x}}). \end{aligned}$$

(B.13)

The proof is completed by taking complex conjugate of expression (B.13) and comparing the result with equation (B.12). □

Appendix C: Proof of Theorem 3.7

In order to prove identity (3.23), we follow the approach taken by [46]. Since $\mathbf{W}_{\infty}$ is independent of the choice of incident field, we consider the case when the plane waves

$${\mathbf{u}}^{\mathrm{inc}}_{P}({\mathbf{x}}):= \nabla e^{i{\kappa }_{P} {\mathbf {x}}\cdot {\mathbf{d}}} \quad \text{and}\quad {\mathbf{u}}^{\mathrm{inc}}_{S}({\mathbf{x}}):= \vec{ \nabla}_{\perp}\times e^{i{\kappa }_{S} {\mathbf{x}}\cdot {\mathbf{d}}}, $$

are incident simultaneously and use the superposition principle for the optical theorem thanks to the linearity of the RHS of identity (3.22). Note that the coefficients $a^{\alpha}_{m} ({\mathbf{u}}^{\mathrm{inc}}_{\alpha})$ and $\gamma^{\alpha }_{n}$ in this case are given by

$$\begin{aligned} a^{\alpha}_{m}\bigl({\mathbf{u}}^{\mathrm{inc}}\bigr)= e^{im(\pi/2-\theta_{{\mathbf{d}}})} \quad\text{and}\quad \gamma^{\alpha}_{n}= \frac{i}{4\rho_{0}\omega^{2}}\sum _{m\in \mathbb {Z}} \bigl[ a^{P}_{m} W^{\alpha,P}_{m,n}+a^{S}_{m} W^{\alpha ,S}_{m,n} \bigr] . \end{aligned}$$

To facilitate ensuing discussion, let us define

$$\mathbf{A}:= \begin{pmatrix} \mathbf{A}_{P}\\\mathbf{A}_{S} \end{pmatrix} \quad \text{and} \quad {\boldsymbol{ \gamma}}:= \begin{pmatrix} {\boldsymbol{\gamma}}_{P}\\{\boldsymbol{\gamma}}_{S} \end{pmatrix} , \quad\text{with } ( \mathbf{A}_{\alpha} ) _{m}:= a^{\alpha}_{m} \bigl({\mathbf{u}}^{\mathrm{inc}}\bigr) \text{ and } ( {\boldsymbol{ \gamma}}_{\alpha} ) _{m}:=\gamma^{\alpha}_{m},\ \forall m\in \mathbb {Z}. $$

It can be easily seen, by the definitions of $\mathbf{A}$ and ${\boldsymbol{\gamma}}$, and the fact that $\mathbf{W}_{\infty}$ is Hermitian, that

$$\begin{aligned} {\boldsymbol{\gamma}}= \frac{i}{4\rho_{0}\omega^{2}} \mathbf {A}^{\top} \mathbf{W}_{\infty} \quad\text{and}\quad {\boldsymbol{\gamma}}\cdot \overline{{\boldsymbol{\gamma}}}= \frac {1}{(4\rho_{0}\omega^{2})^{2}}\mathbf{A}^{T} \mathbf{W}_{\infty }\overline{\mathbf{W}_{\infty}}\overline{ \mathbf{A}}. \end{aligned}$$

On the other hand, using the orthogonality relations (3.1)–(3.2) of the cylindrical surface vector potentials and fairly easy manipulations, we have

$$\begin{aligned} \int_{0}^{2\pi} \biggl( \frac{1}{{\kappa }_{P}}\bigl\vert {\mathbf{u}}^{\infty }_{P}({\hat{{\mathbf{x}}}};\hat {{\mathbf{d}}})\bigr\vert ^{2}+ \frac{1}{{\kappa }_{S}}\bigl\vert {\mathbf{u}}^{\infty }_{S}({\hat{{\mathbf{x}}}};\hat {{\mathbf{d}}})\bigr\vert ^{2} \biggr) d\theta = 4 {\boldsymbol{\gamma}}\cdot\overline{{ \boldsymbol{\gamma}}}= \frac{4}{(4\rho_{0}\omega^{2})^{2}}\mathbf{A}^{T}\mathbf {W}_{\infty}\overline{\mathbf{W}_{\infty}}\overline{\mathbf{A}}. \end{aligned}$$

(C.1)

Similarly, by virtue of superposition principle, the RHS of the identity (3.22) can be written as

$$\begin{aligned} & 2 \biggl[\sqrt{\frac{2\pi}{ {\kappa }_{P}}} \Im m \bigl\{ \sqrt{i} {\mathbf{u}}^{\infty}_{P}( \hat {{\mathbf{d}}};\hat {{\mathbf{d}}}, P)\cdot\hat {{{\boldsymbol {e}}}}_{r} \bigr\} - \sqrt{ \frac{2\pi}{ {\kappa }_{S}}}\Im m \bigl\{ \sqrt{i} {\mathbf{u}}^{\infty}_{S}( \hat {{\mathbf{d}}};\hat {{\mathbf{d}}}, S)\cdot\hat {{{\boldsymbol {e}}}}_{\theta} \bigr\} \biggr] \\ &\quad =\frac{4}{4\rho_{0}\omega^{2}}\Im m \bigl\{ \mathbf{A}^{\top } \mathbf{W}_{\infty} \overline{\mathbf{A}} \bigr\} . \end{aligned}$$

(C.2)

Substituting (C.1) and (C.2) in (3.22), one gets

$$\begin{aligned} \frac{1}{4\rho_{0}\omega^{2}} \mathbf{A}^{T}\mathbf{W}_{\infty }\overline{ \mathbf{W}_{\infty}}\overline{\mathbf{A}}=\Im m \bigl\{ \mathbf{A}^{\top}\mathbf{W}_{\infty} \overline{\mathbf{A}} \bigr\} . \end{aligned}$$

(C.3)

Finally, note that

$$\begin{aligned} \Im m \bigl\{ \mathbf{A}^{\top}\mathbf{W}_{\infty} \overline { \mathbf{A}} \bigr\} &= \Re e \bigl\{ \mathbf{A}^{\top} \bigr\} \Im m \{ \mathbf{W}_{\infty} \} \Re e \{ \mathbf {A} \} -\Re e \bigl\{ \mathbf{A}^{\top} \bigr\} \Re e \{ \mathbf{W}_{\infty} \} \Im m \{ \mathbf{A} \} \\ &\quad{}+\Im m \bigl\{ \mathbf{A}^{\top} \bigr\} \Re e \{ \mathbf {W}_{\infty} \} \Re e \{ \mathbf{A} \} +\Im m \bigl\{ \mathbf{A}^{\top} \bigr\} \Im m \{ \mathbf {W}_{\infty} \} \Im m \{ \mathbf{A} \} . \end{aligned}$$

Recall that each term on the RHS of the above equation is a scalar and the matrix $\mathbf{W}_{\infty}$ is Hermitian. Thus, the second term cancels out the third one on transposition. Finally, the first and the fourth terms can be combined to yield

$$\begin{aligned} \Im m \bigl\{ \mathbf{A}^{\top}\mathbf{W}_{\infty} \overline { \mathbf{A}} \bigr\} = \mathbf{A}^{\top}\Im m \{ \mathbf {W}_{\infty} \} \overline{\mathbf{A}}. \end{aligned}$$

(C.4)

The relation (3.23) follows by substituting (C.4) back in (C.3). This completes the proof.

Appendix D: Proof of Theorem 5.2

Recall that, for $t\to0$,

$$\begin{aligned} &J_{n}(t)=\phantom{-}\frac{t^{n}}{2^{n}\varGamma(n+1)}+O\bigl(t^{n+1} \bigr), \\ &J'_{n}(t)=\frac{nt^{n-1}}{2^{n}\varGamma(n+1)}+O \bigl(t^{n}\bigr), \\ & H_{n}^{(1)}(t)=-i\frac{2^{n}\varGamma(n)}{\pi t^{n}}+O \bigl(t^{-n+1}\bigr), \\ &\bigl(H_{n}^{(1)}\bigr)'(t)=i \frac{2^{n}\varGamma(n+1)}{\pi t^{n+1}}+O\bigl(t^{-n}\bigr). \end{aligned}$$

Hence, by the definition of $B^{\alpha}_{n}(t,\lambda,\mu)$, $C^{\alpha}_{n}(t,\lambda,\mu)$, $\widehat{B}^{\alpha }_{n}(t,\lambda,\mu)$ and $\widehat{C}^{\alpha}_{n}(t,\lambda,\mu )$, we have

$$\begin{aligned} &B^{P}_{n}(t,\lambda,\mu)=-\frac{i\mu2^{n+1}\varGamma(n+1)}{\pi t^{n}}+O \bigl(t^{-n+1}\bigr), \\ &C^{S}_{n}(t,\mu)=\frac{i\mu2^{n+1}\varGamma(n+1)}{\pi t^{n}}+O \bigl(t^{-n+1}\bigr), \\ &C^{P}_{n}(t,\mu)=-B^{S}_{n}(t, \lambda,\mu) =-\frac{\mu\,2^{n+1}\varGamma(n+1)\,(n+1)}{\pi t^{n}}+O\bigl(t^{-n+1}\bigr), \\ &\widehat{B}^{P}_{n}(t,\lambda,\mu) =- \frac{\mu t^{n}}{2^{n-1}\varGamma(n)}+O\bigl(t^{n+1}\bigr), \\ &\widehat{C}^{S}_{n}(t,\lambda,\mu) = \frac{\mu t^{n}}{2^{n-1}\varGamma(n)}+O\bigl(t^{n+1}\bigr), \\ &\widehat{C}^{P}_{n}(t,\lambda,\mu)=- \widehat{B}^{S}_{n}(t,\lambda ,\mu) =i\frac{\mu(n-1) t^{n}}{2^{n-1}\varGamma(n)}+O \bigl(t^{n+1}\bigr), \end{aligned}$$

as $t\to0$. Inserting the previous asymptotic behavior into the expression of $\mathbf{M}_{n,j}$, we get

$$\begin{aligned} \mathbf{M}_{n,j}= \begin{pmatrix} \mathbf{A}_{11} & \mathbf{A}_{12} \\ \mathbf{A}_{21} & \mathbf{A}_{22} \end{pmatrix} , \end{aligned}$$

where

$$\begin{aligned} \mathbf{A}_{11} &= \frac{n}{2^{n}\varGamma(n+1)} \begin{pmatrix} t^{n}_{j,P} & i t^{n}_{j,S} \\ i t^{n}_{j,P} & -t^{n}_{j,S} \end{pmatrix} +O\bigl(\omega^{n+1}\bigr), \\ \mathbf{A}_{12} &= \frac{2^{n}\varGamma(n+1)}{\pi} \begin{pmatrix} it^{-n}_{j,P} & t^{-n}_{j,S} \\ t^{-n}_{j,P} & -it^{-n}_{j,S} \end{pmatrix} +O\bigl(\omega^{-n+1}\bigr), \\ \mathbf{A}_{21} &= -\frac{\mu}{2^{n-1}\varGamma(n)} \begin{pmatrix} -t^{n}_{j,P} & -i(n-1) t^{n}_{j,S} \\ i(n-1) t^{n}_{j,P} & t^{n}_{j,S} \end{pmatrix} +O\bigl(\omega^{n+1}\bigr), \\ \mathbf{A}_{22} &= -\frac{2^{n+1}\mu\varGamma(n+1)}{\pi} \begin{pmatrix} it^{-n}_{j,P} & -(n+1)t^{-n}_{j,S} \\ (n+1)t^{-n}_{j,P} & -it^{-n}_{j,S} \end{pmatrix} +O\bigl(\omega^{-n+1}\bigr). \end{aligned}$$

It implies that

$$\begin{aligned} \mathbf{M}_{n,j} &= \begin{pmatrix} O(\omega^{n+1}) & O(\omega^{-n+1}) \\O(\omega^{n}) & O(\omega^{-n}) \end{pmatrix} , \quad j=1,\ldots, L, \end{aligned}$$

(D.1)

$$\begin{aligned} \mathbf{M}_{n,L} &= \begin{pmatrix} 0 & 0 \\O(\omega^{n}) & O(\omega^{-n}) \end{pmatrix} , \quad\text{as} \;\; \omega\to0. \end{aligned}$$

(D.2)

Moreover, the inverse of $\mathbf{M}_{n,j}$ can be expressed as

$$\begin{aligned} \mathbf{M}_{n,j}^{-1} &= \begin{pmatrix} \mathbf{A}_{11}^{-1}+\mathbf{A}_{11}^{-1}\mathbf{A}_{12}\mathbf {B}^{-1} \mathbf{A}_{21}\mathbf{A}^{-1}_{11} & -\mathbf{A}_{11}^{-1}\mathbf{A}_{12} \mathbf{B}^{-1} \\ -{\mathbf{B}}^{-1}\mathbf{A}_{21}\mathbf{A}_{11}^{-1} & \mathbf{B}^{-1} \end{pmatrix} , \end{aligned}$$

where $\mathbf{B}$ is the Schur’s complement of $\mathbf{A}_{22}$, that is,

$$\mathbf{B}:=\mathbf{A}_{22}-\mathbf{A}_{21}\mathbf {A}_{11}^{-1}\mathbf{A}_{12}. $$

Since

$$\begin{aligned} \mathbf{A}_{11}^{-1}=O\bigl(\omega^{-n-1}\bigr), \quad \mathbf{A}_{11}^{-1}\mathbf{A}_{12}=O\bigl( \omega^{-2n}\bigr), \quad \mathbf{A}_{21} \mathbf{A}_{11}^{-1}=O\bigl(\omega^{-1}\bigr), \quad\text{and}\quad \mathbf{B}^{-1}=O\bigl(\omega^{n} \bigr), \end{aligned}$$

it follows that

$$\begin{aligned} \mathbf{M}^{-1}_{n,j}= \begin{pmatrix} O(\omega^{-n-1}) & O(\omega^{-n}) \\O(\omega^{n-1}) & O(\omega^{n}) \end{pmatrix} , \quad\text{as }\omega\to0. \end{aligned}$$

(D.3)

Inserting (D.1), (D.2) and (D.3) into the expression (5.7) of $\mathbf{Q}^{(n)}$ and then making use of the series expansions of $J_{n}$, $Y_{n}$, $J'_{n}$ and $Y_{n}'$, we find out that

$$\begin{aligned} \mathbf{Q}^{(n)}_{21}\bigl(\lambda,\mu,\rho \omega^{2}\bigr) &=\omega^{n} \Biggl( \mathbf {G}_{n,0}( \lambda,\mu,\rho)+\sum_{l=1}^{N-n}\sum _{j=0}^{L+1} \mathbf {G}_{n,l}^{(j)}( \lambda,\mu,\rho)\omega^{2l}(\ln\omega)^{j} + o \bigl( \omega^{2(N-n)} \bigr) \Biggr) , \\ \mathbf{Q}^{(n)}_{22}\bigl(\lambda,\mu,\rho \omega^{2}\bigr) &=\omega^{-n} \Biggl( \mathbf {H}_{n,0}( \lambda,\mu,\rho)+\sum_{l=1}^{N-n}\sum _{j=0}^{L+1} \mathbf {H}_{n,l}^{(j)}( \lambda,\mu,\rho)\omega^{2l}(\ln\omega)^{j} + o \bigl( \omega^{2(N-n)} \bigr) \Biggr) , \end{aligned}$$

for some functions $\mathbf{G}_{n,0}$, $\mathbf{G}_{n,l}^{(j)}$, $\mathbf{H}_{n,0}$, $\mathbf{H}_{n,l}^{(j)}$. This together with (5.8) yields (5.9). Here, the remaining terms $o(\omega^{2(N-n)})$ are understood element-wise for the matrices.