Characterising small objects in the regime between the eddy current model and wave propagation

Abstract

Being able to characterise objects at low frequencies, but in situations where the modelling error in the eddy current approximation of the Maxwell system becomes large, is important for improving current metal detection technologies. Importantly, the modelling error becomes large as the frequency increases, but the accuracy of the eddy current model also depends on the object topology and on its materials, with the error being much larger for certain geometries compared to others of the same size and materials. Additionally, the eddy current model breaks down at much smaller frequencies for highly magnetic conducting materials compared to non-permeable objects (with similar conductivities, sizes and shapes) and, hence, characterising small magnetic objects made of permeable materials using the eddy current at typical frequencies of operation for a metal detector is not always possible. To address this, we derive a new asymptotic expansion for permeable highly conducting objects that is valid for small objects and holds not only for frequencies where the eddy current model is valid but also for situations where the eddy current modelling error becomes large and applying the eddy approximation would be invalid. The leading-order term we derive leads to new forms of object characterisations in terms of polarizability tensor object descriptions where the coefficients can be obtained from solving vectorial transmission problems. We expect these new characterisations to be important when considering objects at greater stand-off distance from the coils, which is important for safety critical applications, such as the identification of landmines, unexploded ordnance and concealed weapons. We also expect our results to be important when characterising artefacts of archaeological and forensic significance at greater depths than the eddy current model allows and to have further applications parking sensors and improving the detection of hidden, out-of-sight, metallic objects.

1. Introduction

Improving the characterisation of metal objects from electromagnetic field perturbations, when their location is not in the immediate proximity of the exciting and sensing coils, is important for a range of applications. For example, while traditional walk-through metal detectors are used in transport hubs, court rooms, museums, galleries and at concerts and public events for identifying potential threat objects in close proximity (typically much less than a metre) of the coils, there is also considerable interest from security and police forces in being able to identify potential terrorist threat objects, including guns, knives and improvised weapons, at greater stand-off distances (in the order of several metres). The ability to characterise objects at greater stand-off distances could also bring benefits to other metal detection applications including improving safety for anti-personal landmine disposal teams, by identifying unexploded ordnance and landmines at greater distances, improving the detection of smaller, and deeper buried artefacts of archaeological significance and improving the detection of items in forensic searches, which are beyond the reach of current metal detection technologies. There are further applications in parking sensors and in the detection of out-of-vision objects, such as cyclists, for large vehicles, such as lorries and buses.

The metal detection problem is usually described by the eddy current approximation of the Maxwell system, since the conductivities of the objects of interest are high and the frequencies of approximation are low. The eddy current approximation neglects the displacement currents in the Maxwell system, which is responsible for wave propagation and is often applied if the wavelength is large compared to the object size. Ammari, Buffa and Nédélec [1] show, depending on the topology of the object and its materials, that the eddy current approximation is either first- or second-order accurate when considering an asymptotic expansion of the electric and magnetic fields as the excitation frequency tends to zero. In practice, this means that the eddy current model will break down at a lower frequency for a horseshoe-shaped conductor compared to a similar sized solid sphere made of the same material. Since the wavelength needs to be large compared to the object size, this also means that the eddy current model breaks down at a lower frequency compared to non-magnetic objects of the same shape, size and conductivity. The modelling error in the eddy current approximation has been considered by Schmidt, Hiptmair and Sterz [12] who find that the approximation is accurate provided that

(1.1) \begin{align} C_1 \epsilon _{\ast} \mu _{\ast} \omega ^2 \alpha ^2 \ll 1, \qquad C_2 \frac{\omega \epsilon _{\ast}}{ \sigma _{\ast}} \ll 1, \end{align}

where $\epsilon _{\ast}$ , $\mu _{\ast}$ and $\sigma _{\ast}$ denote the conducting object’s permittivity, permeability and conductivity, respectively, $\omega$ is the angular frequency and $\alpha$ is a measure of the object size. The constants $C_1$ and $C_2$ depend on the topology of the object and, if they are of a moderate size, (1.1) reduces approximately down to the condition that the wavelength needs to be large compared to the object size (quasi-static limit) and $\epsilon _{\ast} \omega \ll \sigma _{\ast}$ (conductivities are large), respectively. They also provide a procedure for estimating $C_1$ and $C_2$ .

For the eddy current problem, Ammari, Chen, Chen, Volkov and Garnier [2] have derived the leading-order term in an asymptotic expansion for the perturbation of the magnetic field caused by the presence of a permeable conducting object as it size tends to zero for the eddy current problem. There are conditions imposed on the material parameters, which allows this result to be applied to the case of constant $\sigma _{\ast}$ and $\mu _{\ast}$ as $\alpha \to 0$ . We have shown, [5], that the leading-order term they have been derived can be simplified and provides an object characterisation, in terms of a rank 2 magnetic polarizability tensor (MPT) that depends on the object shape, object size, its material parameters and the frequency of excitation but is independent of its position. The object location is separated from the other object characterisation information with the background field being evaluated at the object position. In [8], we have extended the leading-order term obtained by Ammari et al. in [2] to a complete expansion of the perturbed magnetic field and introduced the concept of generalised MPTs, which provide improved object characterisation. We have derived alternative forms of the MPT and considered spectral behaviour in [9]. Together with Wilson, we have proposed an efficient computational procedure for the computation of the MPT spectral signatures [14] and shown how it can be used to characterise threat and non-threat objects of relevance to security screening [10] and proposed an approach to classification using machine learning [15].

For the full Maxwell system, Ammari, Vogelius and Volkov [3] have derived the leading-order term in an asymptotic expansion of the perturbed electric and magnetic fields caused by the presence of an object as its size tends to zero. In this case, there is no simplification in the Maxwell system and wave propagation effects are included. The resulting expansions shows that leading-order provides an object characterisation in terms of the simpler Póyla–Szegö tensor parameterised by contrasts in (complex) permittivity and permeability, which can be computed from a scalar transmission problem. Unlike the eddy current case, conditions are not imposed on the material parameters and, hence, $\epsilon _{\ast}, \sigma _{\ast}$ and $\mu _{\ast}$ are allowed to change as $\alpha \to 0$ . In the limiting case of magnetostatics, it can be shown that the MPT simplifies to the simpler Póyla–Szegö tensor description [9], whose coefficients are obtained by solving scalar transmission problems, but, in the eddy current regime, the calculation of the MPT coefficients requires the solution of vectorial curl–curl transmission problems. However, it was not clear what form the characterisation takes as the modelling error in the eddy current model becomes large and wave propagation effects become important.

Our present work contributes to understanding the regime where frequencies are small, but the displacement currents remain important through the following novelties:

1. 1. We derive a new asymptotic formula for the magnetic field perturbation caused by the presence of a conducting permeable object as its size tends to zero. We establish the leading order for the full Maxwell system with specific constraints applied to the object’s materials and the frequency of excitation.

2. 2. We obtain new forms of polarizability tensors for characterising objects, whose coefficients are obtained by post-processing solutions to vectorial transmission problems.

3. 3. We show that our expansion reduces to the result obtained by Ammari et al. [2] if the eddy current approximation is made. Our new polarizability tensors reduce to an object characterisation using the MPT in this case.

4. 4. Additionally, we obtain an expansion of the form obtained by Ammari et al. [3] if the conditions we impose are relaxed.

The material proceeds as follows: In Section 2, we introduce some notation that will aid the presentation of the material. Then, in Section 3, we introduce the mathematical model that will be of interest in this work. Section 4 presents our main result, the proof of which will be established subsequent sections. Section 5 establishes some energy estimates, Section 6 establishes an integral representation formula and Section 7 the remaining parts of the proof. Then, Section 8 presents alternative forms of our main result and presents simplifications and relaxation of assumptions that make comparisons with previous object characterisations possible.

2. Notation

For what follows it is beneficial to introduce the following notation: we will use boldface for vector quantities (e.g. $\boldsymbol{u}$ ) and denote by ${\boldsymbol{e}}_j$ , $j=1,2,3$ the units vectors associated with an orthonormal coordinate system. We denote the $j$ -th component of a vector ${\boldsymbol{u}}$ in this coordinate system by $({\boldsymbol{u}} )_j ={\boldsymbol{u}} \cdot{\boldsymbol{e}}_j = u_j$ . We will use calligraphic symbols to denote rank 2 tensors, for example, ${\mathcal N} ={\mathcal N}_{ij}{\boldsymbol{e}}_i \otimes{\boldsymbol{e}}_j$ , where Einstein summation convention is implied, and denote their coefficients by ${\mathcal N}_{ij}$ .

We recall that for $0\leqslant \ell \lt \infty$ , $0\leqslant p\lt \infty$ ,

\begin{equation*} \|{\boldsymbol{u}} \|_{W^{\ell,p}(B_\alpha )} \;:\!=\; \!\left ( \sum _{j=0}^\ell \int _{B_\alpha } |{\boldsymbol{D}}^j({\boldsymbol{u}} ({\boldsymbol{x}} ))|^p \mathrm{d}{\boldsymbol{x}} \right )^{1/p}, \end{equation*}

where the derivatives are defined in a weak sense and

\begin{equation*} \|{\boldsymbol{u}} \|_{W^{\ell,\infty }(B_\alpha )} \;:\!=\;\mathop{\text{ess sup}}_{{\boldsymbol{x}} \in B_\alpha } \sum _{j=0}^\ell |{\boldsymbol{D}}^j({\boldsymbol{u}} ({\boldsymbol{x}} ))|. \end{equation*}

3. Mathematical model

3.1 Governing equations

For linear materials, the time harmonic Maxwell equations are as follows:

(3.1a) \begin{align} \nabla \times\boldsymbol{\mathcal{E}} & ={\mathrm{i}} \omega \mu _0\boldsymbol{\mathcal{H}}, \end{align}

(3.1b) \begin{align} \nabla \times\boldsymbol{\mathcal{H}} &= \sigma\boldsymbol{\mathcal{E}} +\boldsymbol{\mathcal{J}}_0 -{\mathrm{i}} \omega \epsilon\boldsymbol{\mathcal{E}}, \end{align}

(3.1c) \begin{align} \nabla \cdot ( \epsilon\boldsymbol{\mathcal{E}} ) & = \frac{1}{{\mathrm{i}} \omega } \nabla \cdot ( \sigma\boldsymbol{\mathcal{E}} ), \end{align}

(3.1d) \begin{align} \nabla \cdot ( \mu\boldsymbol{\mathcal{H}} ) & = 0, \end{align}

where $\boldsymbol{\mathcal{E}}$ and $\boldsymbol{\mathcal{H}}$ denote the complex amplitudes of the electric and magnetic field intensity vectors, respectively, for an assumed $e^{-{\mathrm{i}} \omega t}$ time variation with angular frequency $\omega \gt 0$ and ${\mathrm{i}} \;:\!=\;\sqrt{-1}$ . In addition, $\boldsymbol{\mathcal{J}}_0$ denotes the complex amplitude of an external solenoidal current source, and the parameters $\epsilon, \mu$ and $\sigma$ denote the permittivity, permeability and conductivity, respectively, and satisfy

(3.2) \begin{align} 0 \lt \epsilon ^{\text{min}} \leqslant \epsilon \leqslant \epsilon ^{\text{max}} \lt \infty, \qquad 0 \lt \mu ^{\text{min}} \leqslant \mu \leqslant \mu ^{\text{max}} \lt \infty, \qquad 0 \leqslant \sigma \leqslant \sigma ^{\text{max}} \lt \infty. \end{align}

As standard (e.g. Monk [11]), the scaled fields are introduced as ${\boldsymbol{E}}_\alpha = \epsilon _0^{1/2}\boldsymbol{\mathcal{E}}$ , ${\boldsymbol{H}}_\alpha = \mu _0^{1/2}\boldsymbol{\mathcal{H}}$ and ${\boldsymbol{J}}_0 = \mu _0^{1/2}\boldsymbol{\mathcal{J}}_0$ , where $\epsilon _0 = 8.854 \times 10^{-12}$ F/m and $\mu _0 = 4\pi \times 10^{-7}$ H/m are the free space values of the permittivity and permeability, respectively, leading to

(3.3a) \begin{align} \nabla \times{\boldsymbol{E}}_\alpha & ={\mathrm{i}} k \tilde{\mu }_r{\boldsymbol{H}}_\alpha, \end{align}

(3.3b) \begin{align} \nabla \times{\boldsymbol{H}}_\alpha & ={\boldsymbol{J}}_0 -{\mathrm{i}} k \tilde{\epsilon }_r{\boldsymbol{E}}_\alpha, \end{align}

(3.3c) \begin{align} \nabla \cdot ( \tilde{\epsilon }_r{\boldsymbol{E}}_\alpha ) & = 0, \end{align}

(3.3d) \begin{align} \nabla \cdot ( \tilde{\mu }_r{\boldsymbol{H}}_\alpha ) & = 0, \end{align}

where

\begin{align*} \tilde{\epsilon }_r\;:\!=\; \frac{1}{\epsilon _0} \!\left ( \epsilon + \frac{{\mathrm{i}} \sigma }{\omega } \right ), \qquad \tilde{\mu }_r \;:\!=\; \frac{\mu }{\mu _0}, \qquad k\;:\!=\; \omega ( \epsilon _0 \mu _0) ^{1/2}, \end{align*}

and $\tilde{\epsilon }_r, \tilde{\mu }_r$ are, in general, functions of position.

3.2 Perturbed field formulation

We describe a single connected inclusion by $B_\alpha \;:\!=\; \alpha B +{\boldsymbol{z}}$ , which means that it could be thought of a unit-sized object $B$ located at the origin, scaled by $\alpha$ and translated by $\boldsymbol{z}$ . Its boundary, $\Gamma _\alpha \;:\!=\;\partial B_\alpha$ , is equipped with unit outward normal $\boldsymbol{n}$ , and we assume that the object is homogeneous with material coefficients $\epsilon _{\ast}, \mu _{\ast}$ and $\sigma _{\ast}$ . The object is surrounded by an unbounded region of free space $B_\alpha ^c \;:\!=\;{\mathbb R}^3 \setminus \overline{B_\alpha }$ with material coefficients $\epsilon _0, \mu _0$ and $\sigma =0$ so that, henceforth,

\begin{align*} \tilde{\epsilon }_r ({\boldsymbol{x}}) \;:\!=\; \left \{ \begin{array}{l@{\quad}l} \epsilon _r \;:\!=\; \dfrac{1}{\epsilon _0} \!\left ( \epsilon _{\ast} + \dfrac{{\mathrm{i}} \sigma _{\ast}}{\omega } \right ) &{\boldsymbol{x}} \in B_\alpha \\[9pt] 1 &{\boldsymbol{x}}\in B_\alpha ^c \end{array} \right ., \qquad \tilde{\mu }_r \;:\!=\;\left \{ \begin{array}{l@{\quad}l} \mu _r \;:\!=\; \dfrac{\mu _{\ast}}{\mu _0} &{\boldsymbol{x}} \in B_\alpha \\[5pt] 1 &{\boldsymbol{x}} \in B_\alpha ^c \end{array} \right. . \end{align*}

Furthermore, we assume that ${\boldsymbol{J}}_0$ has support only in $B_\alpha ^c$ and for it to be located away from $B_\alpha$ . The electric and magnetic fields obey the standard transmission conditions:

(3.4a) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{E}}_\alpha ]_{\Gamma _\alpha } & =\textbf{0}, \end{align}

(3.4b) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{H}}_\alpha ]_{\Gamma _\alpha } & =\textbf{0}, \end{align}

(3.4c) \begin{align} [{\boldsymbol{n}} \cdot \tilde{\epsilon }_r{\boldsymbol{E}}_\alpha ]_{\Gamma _\alpha } & = 0, \end{align}

(3.4d) \begin{align} [{\boldsymbol{n}} \cdot \tilde{\mu }_r{\boldsymbol{H}}_\alpha ]_{\Gamma _\alpha } & = 0, \end{align}

on $\Gamma _\alpha$ where $[\!\cdot\!]_{\Gamma _\alpha } = \cdot |_+ -\cdot |_-$ denotes the jump and subscript $+/-$ denotes the evaluation just outside/inside of $B_\alpha$ , respectively. By ${\boldsymbol{n}}^+$ and ${\boldsymbol{n}}^-$ , we denote the unit outward normal pointing from the $+/-$ side of $\Gamma _\alpha$ into $B_\alpha$ and $B_\alpha ^c$ , respectively, and with similar meanings on other interfaces. In absence of $B_\alpha$ , the background (incident) fields $\boldsymbol{E}_0$ and $\boldsymbol{H}_0$ satisfy (3.3) with $\tilde{\epsilon }_r = \tilde{\mu }_r=1$ in ${\mathbb R}^3$ . The fields ${\boldsymbol{E}}_\Delta \;:\!=\;{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0}$ and ${\boldsymbol{H}}_\Delta \;:\!=\;{\boldsymbol{H}}_\alpha -{\boldsymbol{H}_0}$ , which represent the perturbation in the electric and magnetic fields due to the presence of $B_\alpha$ , respectively, satisfy the radiation conditions:

\begin{align*} \lim _{x\to \infty } x \!\left ( ( \nabla \times{\boldsymbol{E}}_\Delta ) \times \hat{{\boldsymbol{x}}} -{\mathrm{i}} k{\boldsymbol{E}}_\Delta \right ) = \textbf{0}, \\[5pt] \lim _{x\to \infty } x\!\left ( ( \nabla \times{\boldsymbol{H}}_\Delta ) \times \hat{{\boldsymbol{x}}} -{\mathrm{i}} k{\boldsymbol{H}}_\Delta \right ) = \textbf{0}, \end{align*}

where $x=|{\boldsymbol{x}}|$ and $\hat{{\boldsymbol{x}}} ={\boldsymbol{x}}/ x$ . By eliminating ${\boldsymbol{H}}_\alpha$ , we arrive at the following transmission problem for ${\boldsymbol{E}}_\Delta$ :

(3.5a) \begin{align} \nabla \times{\mu }_r^{-1} \nabla \times{\boldsymbol{E}}_\Delta -k^2 \epsilon _r{\boldsymbol{E}}_\Delta &= \nabla \times ( 1- \mu _r^{-1}) \nabla \times{\boldsymbol{E}_0} -k^2(1-\epsilon _r ){\boldsymbol{E}_0} &&{} \text{in $B_\alpha $}, \end{align}

(3.5b) \begin{align} \nabla \times \nabla \times{\boldsymbol{E}}_\Delta -k^2{\boldsymbol{E}}_\Delta &=\textbf{0} &&\text{in $B_\alpha ^c$}, \end{align}

(3.5c) \begin{align} \nabla \cdot ( \tilde{\epsilon }_r{\boldsymbol{E}}_\Delta ) & = 0 && \text{in ${\mathbb R}^3$}, \end{align}

(3.5d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{E}}_\Delta ]_{\Gamma _\alpha } & =\textbf{0} &&\text{on $\Gamma _\alpha $}, \end{align}

(3.5e) \begin{align} [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla \times{\boldsymbol{E}}_\Delta ]_{\Gamma _\alpha } & = -[\tilde{\mu }_r^{-1}]_{\Gamma _\alpha }{\boldsymbol{n}} \times \nabla \times{\boldsymbol{E}_0} &&\text{on $\Gamma _\alpha $}, \end{align}

(3.5f) \begin{align} \lim _{x\to \infty } x \!\left ( ( \nabla \times{\boldsymbol{E}}_\Delta ) \times \hat{{\boldsymbol{x}}} -{\mathrm{i}} k{\boldsymbol{E}}_\Delta \right ) &=\textbf{0}, \end{align}

for ${\boldsymbol{E}}_\Delta$ , while an analogous transmission problem could also be derived for ${\boldsymbol{H}}_\Delta$ . Given the solenoidal behaviour of $\boldsymbol{E}_0$ , the condition $\nabla \cdot ( \tilde{\epsilon }_r{\boldsymbol{E}}_\Delta ) = 0$ in ${\mathbb R}^3$ can be disregarded provided that $k^2\gt 0$ is not an eigenvalue of (3.5), which we shall assume throughout.

The system (3.5) is set on an unbounded domain, and we follow the approach of Monk [11, Chapter 10, Chapter 11] to reduce this to a problem on a bounded domain. Compared to the static decay of the fields considered for the eddy current problem in [2], a different treatment is required before the weak variational statement can be introduced due to the presence of the radiation condition in (3.5f). We assume that there is a radius $R_0$ , centred on the origin, such that $B_\alpha \subset S_{R_0}$ with $S_{R_0}$ being a solid sphere of radius $R_0$ . Next, we introduce a solid sphere $S_R$ , with radius $R\gt R_0$ , and let $\Omega \;:\!=\; S_R$ , which is a bounded domain. We denote the boundary of $\Omega$ by $\Sigma$ and note that $\Gamma _\alpha \bigcap \Sigma = \emptyset$ . This leads to the alternative transmission problem for ${\boldsymbol{E}}_\Delta$ :

(3.6a) \begin{align} \nabla \times{\mu }_r^{-1} \nabla \times{\boldsymbol{E}}_\Delta -k^2 \epsilon _r{\boldsymbol{E}}_\Delta &= \nabla \times ( 1- \mu _r^{-1}) \nabla \times{\boldsymbol{E}_0} -k^2(1-\epsilon _r ){\boldsymbol{E}_0} &&{} \text{in $B_\alpha $}, \end{align}

(3.6b) \begin{align} \nabla \times \nabla \times{\boldsymbol{E}}_\Delta -k^2{\boldsymbol{E}}_\Delta &=\textbf{0} &&\text{in $\Omega \setminus \overline{B_\alpha }$}, \end{align}

(3.6c) \begin{align} \nabla \times \nabla \times{\boldsymbol{E}}_\Delta -k^2{\boldsymbol{E}}_\Delta &=\textbf{0} &&\text{in ${\mathbb R}^3 \setminus \overline{\Omega }$}, \end{align}

(3.6d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{E}}_\Delta ]_{\Gamma _\alpha } & =\textbf{0} &&\text{on $\Gamma _\alpha $}, \end{align}

(3.6e) \begin{align} [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla \times{\boldsymbol{E}}_\Delta ]_{\Gamma _\alpha } & = -[\tilde{\mu }_r^{-1}]_{\Gamma _\alpha }{\boldsymbol{n}} \times \nabla \times{\boldsymbol{E}_0} &&\text{on $\Gamma _\alpha $}, \end{align}

(3.6f) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{E}}_\Delta ]_{\Sigma } & =\textbf{0} &&\text{on $\Sigma $}, \end{align}

(3.6g) \begin{align} [{\boldsymbol{n}} \times \nabla \times{\boldsymbol{E}}_\Delta ]_{\Sigma } & ={\boldsymbol 0} &&\text{on $\Sigma $}, \end{align}

(3.6h) \begin{align} \lim _{x\to \infty } x \!\left ( ( \nabla \times{\boldsymbol{E}}_\Delta ) \times \hat{{\boldsymbol{x}}} -{\mathrm{i}} k{\boldsymbol{E}}_\Delta \right ) &=\textbf{0}. \end{align}

By integrating over $\Omega$ , we have

\begin{align*} \int _{\Omega \setminus \overline{B_\alpha }} \!\left ( \nabla \times{\boldsymbol{E}}_\Delta \cdot \nabla \times \overline{\boldsymbol{v}} - k^2{\boldsymbol{E}}_\Delta \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} &+ \int _{B_\alpha } \!\left ( \mu _r^{-1} \nabla \times{\boldsymbol{E}}_\Delta \cdot \nabla \times \overline{\boldsymbol{v}} -k^2 \epsilon _r{\boldsymbol{E}}_\Delta \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \\[5pt] & + \int _{\Sigma }{\boldsymbol{n}}^- \times \nabla \times{\boldsymbol{E}}_\Delta \cdot ({\boldsymbol{n}}^- \times \overline{\boldsymbol{v}} ) \times{\boldsymbol{n}} ^- \mathrm{d}{\boldsymbol{x}} \\[5pt] & = \int _{B_\alpha } \!\left ( (1- \mu _r^{-1}) \nabla \times{\boldsymbol{E}_0} \cdot \nabla \times \overline{\boldsymbol{v}} -k^2 (1- \epsilon _r){\boldsymbol{E}_0}\cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}}, \end{align*}

and, by applying the transmission condition in (3.6e), followed by the Dirichlet to Neumann map $ \Lambda _e^{(k)} ({\boldsymbol{n}} \times{\boldsymbol{E}}_\Delta ) \;:\!=\;{\boldsymbol{n}} \times \frac{1}{{\mathrm{i}} k} \nabla \times{\boldsymbol{E}}_\Delta$ , in a similar manner to Monk [11, Equation (10.2), see also Chapter 11] (who instead of $\Lambda _e^{(k)}$ denotes the map by $G_e$ ), leads to the weak statement. Find ${\boldsymbol{E}}_\Delta \in H(\text{curl})$

(3.7) \begin{align} \int _{\Omega \setminus \overline{B_\alpha }} \!\left ( \nabla \times{\boldsymbol{E}}_\Delta \cdot \nabla \times \overline{\boldsymbol{v}} - k^2{\boldsymbol{E}}_\Delta \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} &+ \int _{B_\alpha } \!\left ( \mu _r^{-1} \nabla \times{\boldsymbol{E}}_\Delta \cdot \nabla \times \overline{\boldsymbol{v}} -k^2 \epsilon _r{\boldsymbol{E}}_\Delta \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \nonumber \\[5pt] &+ {\mathrm{i}} k \int _{\Sigma }{\Lambda }_e^{(k)} ({\boldsymbol{n}}^- \times{\boldsymbol{E}}_\Delta ) \cdot ({\boldsymbol{n}}^- \times \overline{\boldsymbol{v}} ) \times{\boldsymbol{n}}^- \mathrm{d}{\boldsymbol{x}} \nonumber \\[5pt] & = \int _{B_\alpha } \!\left ( (1- \mu _r^{-1}) \nabla \times{\boldsymbol{E}_0} \cdot \nabla \times \overline{\boldsymbol{v}} -k^2 (1- \epsilon _r){\boldsymbol{E}_0}\cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}}, \end{align}

for all ${\boldsymbol{v}} \in H(\text{curl})$ .

We also recall that

\begin{equation*} G_k({\boldsymbol{x}},{\boldsymbol{y}}) \;:\!=\; \frac{e^{{\mathrm{i}} k |{\boldsymbol{x}}-{\boldsymbol{y}}|}}{4\pi |{\boldsymbol{x}}-{\boldsymbol{y}}|}, \end{equation*}

is the Helmholtz’s free space Green’s function, which becomes the Laplace free space Green’s function if $k=0$ .

4. Main result

As mentioned in the introduction, Ammari, Buffa and Nédélec [1] have established that the eddy current model is a low-frequency approximation to the Maxwell system, which is first- or second-order accurate depending on the topology of the object and its materials. Furthermore, the conditions stated in (1.1) for the validity of the eddy current model depend on the $B_\alpha$ topology-dependent constants $C_1$ and $C_2$ , and Schmidt, Hiptmair and Sterz [12] have proposed a methodology for estimating their values. Nonetheless, the modelling error in the eddy current model increases as $\omega$ increases and will be large, and the eddy current model is invalid, as $C_1 \epsilon _{\ast} \mu _{\ast} \omega ^2 \alpha ^2 \to 1$ . In this case, the formulae for computing the MPT coefficients [2, 5, 8], which is based on the eddy current model, will also need to be replaced.

Our goal is to derive an asymptotic formula for ${\boldsymbol{H}}_\Delta ({\boldsymbol{x}}) = ({\boldsymbol{H}}_\alpha -{\boldsymbol{H}_0})({\boldsymbol{x}})$ for $\boldsymbol{x}$ away from $B_\alpha$ as $\alpha \to 0$ leading to a characterisation of $B_\alpha$ , which includes the situation where the modelling error in the eddy current model becomes large. To investigate this, we will require that $C(B) \epsilon _{\ast} \mu _{\ast} \omega _2 \alpha ^2 \lt 1$ and $ C(\Omega ) \epsilon _0\mu _0 \omega ^2 R^2 \lt 1$ , but, importantly, we will retain displacement current terms in order to investigate the form the characterisation takes when the eddy current model becomes invalid. Additionally, we introduce $ \nu = O(1)$ as $\alpha \to 0$ where

\begin{equation*} \nu =\alpha ^2 k^2 (\epsilon _r-1) = \nu _{\textrm{r}} +{\mathrm{i}} \nu _{\mathrm{i}},\qquad \nu _{\textrm{r}} =(\epsilon _{\ast}- \epsilon _0) \mu _0 \omega ^2 \alpha ^2, \qquad \nu _{\mathrm{i}} = \sigma _{\ast}\mu _0 \omega \alpha ^2, \end{equation*}

and $\mu _r =O(1)$ . Our treatment will allow an extension of the results described in [2, 5, 8], which considered the eddy current model, where the quasi-static assumption (i.e. $\alpha \ll \lambda _0= 2\pi/ k$ ) and large conductivities ( $\sigma _{\ast} \gg \epsilon _{\ast} \omega$ ) were assumed, and, instead, considered the regime where $ \nu _{\mathrm{i}} =O(1)$ and $\mu _r =O(1)$ as $\alpha \to 0$ . Fixing $\nu =O(1)$ , rather than $ \nu _{\mathrm{i}} =O(1)$ , means that, in addition to including the case where $\sigma _{\ast}$ and $\omega$ are constant, we include the situation where $\epsilon _{\ast}$ is constant. These quantities are also allowed to decrease with $\alpha$ , but not faster than $1/\alpha ^2$ .

We focus on ${\boldsymbol{H}}_\Delta ({\boldsymbol{x}}) = ({\boldsymbol{H}}_\alpha -{\boldsymbol{H}_0})({\boldsymbol{x}})$ to allow ease of comparison with these earlier results. Our main result is as follows:

Theorem 4.1. For $\boldsymbol{x}$ away from $B_\alpha = \alpha B +{\boldsymbol{z}}$ , and the conditions stated in Section 4 , the following expansion of ${\boldsymbol{H}}_\Delta ({\boldsymbol{x}}) = ({\boldsymbol{H}}_\alpha -{\boldsymbol{H}_0})({\boldsymbol{x}})$ holds

(4.1) \begin{align} ({\boldsymbol{H}}_\Delta ({\boldsymbol{x}}))_j = & -{\mathrm{i}} k (\nabla _x G_k({\boldsymbol{x}},{\boldsymbol{z}}))_p \varepsilon _{jpr} ({\mathcal A}_{ri} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +{\mathcal B}_{ri}({\boldsymbol{E}_0}({\boldsymbol{z}}))_i )\nonumber \\[5pt] & + ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}))_{\ell m} \varepsilon _{j\ell s}{\mathcal C}_{msi} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i \nonumber \\[5pt] & +( ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}))_{j r}+ k^2 \delta _{jr} G_k({\boldsymbol{x}},{\boldsymbol{z}})){\mathcal N}_{ri} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i + (\boldsymbol{R}({\boldsymbol{x}}))_j, \end{align}

where ${\mathcal A},{\mathcal B},{\mathcal C}$ and $\mathcal N$ are polarizability tensors with coefficients:

(4.2a) \begin{align} {\mathcal A}_{ri} \;:\!=\; & \frac{{\mathrm{i}} k \alpha ^4 (\epsilon _r -1)}{2}{\boldsymbol{e}}_r \cdot \int _{B} \!\left ({\boldsymbol{e}}_i \times{\boldsymbol{\xi }} +{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }}, \end{align}

(4.2b) \begin{align} {\mathcal B}_{ri} \;:\!=\; & \alpha ^3 (\epsilon _r -1){\boldsymbol{e}}_r\cdot \int _B \!\left ({\boldsymbol{e}}_i +{\boldsymbol{\phi }}_i \right ) \mathrm{d}{\boldsymbol{\xi }}, \end{align}

(4.2c) \begin{align} {\mathcal C}_{msi} \;:\!=\;& -\frac{k^2 \alpha ^5 (\epsilon _r -1)}{2}{\boldsymbol{e}}_s \cdot \int _B \xi _m \!\left ({\boldsymbol{e}}_i \times{\boldsymbol{\xi }} +{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }}, \end{align}

(4.2d) \begin{align} {\mathcal N}_{ri} \;:\!=\; & \alpha ^3 (1-\mu _r^{-1}){\boldsymbol{e}}_r \cdot \int _B \!\left ({\boldsymbol{e}}_i + \frac{1}{2} \nabla \times{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }}, \end{align}

that depend on the solution to the transmission problems:

(4.3a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i -k^2 \alpha ^2 \epsilon _r{\boldsymbol{\theta }}_i & = -k^2\alpha ^2 (1-\epsilon _r){\boldsymbol{e}}_i\times{\boldsymbol{\xi }} && \text{in}\, B, \end{align}

(4.3b) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{\theta }}_i -k^2\alpha ^2{\boldsymbol{\theta }}_i & = \boldsymbol{0} && \text{in}\, B^c\;:\!=\;{\mathbb R}^3 \setminus \overline{B}, \end{align}

(4.3c) \begin{align} \nabla _\xi \cdot{\boldsymbol{\theta }}_i & = 0 && \text{in}\,{\mathbb R}^3, \end{align}

(4.3d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{\theta }}_i ]_\Gamma =\boldsymbol{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i ]_\Gamma & = -2 (1-\mu _r^{-1}){\boldsymbol{n}} \times{\boldsymbol{e}}_i && \text{on}\; \Gamma \;:\!=\; \partial B, \end{align}

(4.3e) \begin{align} \lim _{\xi \to \infty }\xi \!\left ( ( \nabla _\xi \times{\boldsymbol{\theta }}_i ) \times \hat{{\boldsymbol{\xi }}} -{\mathrm{i}} k \alpha{\boldsymbol{\theta }}_i \right ) & = \boldsymbol{0}, \end{align}

and

(4.4a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\phi }}_{i} -k^2 \alpha ^2 \epsilon _r{\boldsymbol{\phi }}_{i} & = -k^2\alpha ^2 (1-\epsilon _r){\boldsymbol{e}}_i && \text{in}\; B, \end{align}

(4.4b) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{\phi }}_{i} -k^2\alpha ^2{\boldsymbol{\phi }}_{i} & = \boldsymbol{0} && \text{in}\; B^c, \end{align}

(4.4c) \begin{align} \nabla _\xi \cdot{\boldsymbol{\phi }}_{i} & = 0 && \text{in}\,{\mathbb R}^3, \end{align}

(4.4d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{\phi }}_{i} ]_\Gamma =\boldsymbol{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times{\boldsymbol{\phi }}_{i} ]_\Gamma & = \boldsymbol{0}&& \text{on}\, \Gamma, \end{align}

(4.4e) \begin{align} \lim _{\xi \to \infty } \xi \!\left ( ( \nabla _\xi \times{\boldsymbol{\phi }}_{i} ) \times \hat{{\boldsymbol{\xi }}} -{\mathrm{i}} k \alpha{\boldsymbol{\phi }}_{i} \right ) & = \boldsymbol{0}. \end{align}

The residual $\boldsymbol{R}({\boldsymbol{x}})$ satisfies $ |\boldsymbol{R}({\boldsymbol{x}})| \leqslant C \!\left ( \alpha ^4 \|{\boldsymbol{H}_0} \|_{W^{2,\infty }(B_\alpha )} + \alpha ^4 k \!\left ( |\epsilon _r-1| + \alpha k^2 \!\left |1 -\frac{1}{\epsilon _r} \right | \right ) \|{\boldsymbol{E}_0} \|_{W^{1,\infty }(B_\alpha )} \right )$ .

The proof of this theorem follows in Section 7. Beforehand, we derive results that our proof will draw on.

5. Energy estimates

To arrive at an energy estimate for an eddy current problem, Ammari, Chen, Chen, Garnier and Volkov [2] have introduced the following equations:

(5.1) \begin{align}\boldsymbol{F}({\boldsymbol{x}}) & \;:\!=\; \frac{1}{2} (\nabla _z \times{\boldsymbol{E}_0}({\boldsymbol{z}}))\times ({\boldsymbol{x}}-{\boldsymbol{z}}) + \frac{1}{3}{\boldsymbol{D}}_z (\nabla _z \times{\boldsymbol{E}_0}({\boldsymbol{z}}))({\boldsymbol{x}}-{\boldsymbol{z}}) \times ({\boldsymbol{x}}-{\boldsymbol{z}}) \nonumber \\[5pt] & = \frac{{\mathrm{i}} k }{2} \sum _{i=1}^3 ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i{\boldsymbol{e}}_i \times ({\boldsymbol{x}}-{\boldsymbol{z}}) + \frac{{\mathrm{i}} k}{3} \sum _{i,j=1}^3{\boldsymbol{D}}_z ({\boldsymbol{H}_0}({\boldsymbol{z}}) )_{ij} ({\boldsymbol{x}}-{\boldsymbol{z}})_j{\boldsymbol{e}}_i \times ({\boldsymbol{x}}-{\boldsymbol{z}}), \end{align}

which has the curl:

(5.2) \begin{align} \nabla _x \times\boldsymbol{F} & = \nabla _z \times{\boldsymbol{E}_0}({\boldsymbol{z}}) +{\boldsymbol{D}}_z (\nabla _z \times{\boldsymbol{E}_0}({\boldsymbol{z}}))({\boldsymbol{x}}-{\boldsymbol{z}}) \nonumber \\[5pt] & = {\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{z}}) +{\mathrm{i}} k \sum _{i,j=1}^3{\boldsymbol{D}}_z ({\boldsymbol{H}_0}({\boldsymbol{z}}))_{ij} ({\boldsymbol{x}} -{\boldsymbol{z}} )_j{\boldsymbol{e}}_i. \end{align}

In the above equation, $\nabla _x \times\boldsymbol{F}$ corresponds to the first two terms in a Taylor’s series expansion of ${\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{x}})$ about $\boldsymbol{z}$ as $|{\boldsymbol{x}}-{\boldsymbol{z}}| \to 0$ . We adopt a different form for $\boldsymbol{F}$ as follows:

(5.3) \begin{align}\boldsymbol{F}({\boldsymbol{x}}) & \;:\!=\; \frac{1}{2} (\nabla _z \times{\boldsymbol{E}_0}({\boldsymbol{z}}))\times ({\boldsymbol{x}}-{\boldsymbol{z}}) + \frac{1}{3}{\boldsymbol{D}}_z (\nabla _z \times{\boldsymbol{E}_0}({\boldsymbol{z}}))({\boldsymbol{x}}-{\boldsymbol{z}}) \times ({\boldsymbol{x}}-{\boldsymbol{z}}) +{\boldsymbol{E}_0}({\boldsymbol{z}}) \nonumber \\[5pt] & = \frac{{\mathrm{i}} k }{2} \sum _{i=1}^3 ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i{\boldsymbol{e}}_i \times ({\boldsymbol{x}}-{\boldsymbol{z}}) + \frac{{\mathrm{i}} k}{3} \sum _{i,j=1}^3{\boldsymbol{D}}_z ({\boldsymbol{H}_0}({\boldsymbol{z}}) )_{ij} ({\boldsymbol{x}}-{\boldsymbol{z}})_j{\boldsymbol{e}}_i \times ({\boldsymbol{x}}-{\boldsymbol{z}}) +{\boldsymbol{E}_0}({\boldsymbol{z}}), \end{align}

so that

(5.4) \begin{align} \nabla _x \times\boldsymbol{F} = {\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{z}}) +{\mathrm{i}} k \sum _{i,j=1}^3{\boldsymbol{D}}_z ({\boldsymbol{H}_0}({\boldsymbol{z}}))_{ij} ({\boldsymbol{x}} -{\boldsymbol{z}} )_j{\boldsymbol{e}}_i, \end{align}

is still the first two terms in a Taylor’s series expansion of ${\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{x}})$ about $\boldsymbol{z}$ as $|{\boldsymbol{x}}-{\boldsymbol{z}}| \to 0$ . Note that $\boldsymbol{F}({\boldsymbol{z}}) ={\boldsymbol{E}}_0({\boldsymbol{z}})$ and $(\nabla _x \times\boldsymbol{F})({\boldsymbol{z}}) ={\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{z}})$ . We can also determine

(5.5) \begin{align} \nabla _x \cdot\boldsymbol{F} &= - \frac{{\mathrm{i}} k}{3} \sum _{i=1}^3 ( \nabla _z \times{\boldsymbol{H}_0}({\boldsymbol{z}}))_{i}{\boldsymbol{e}}_i \cdot ({\boldsymbol{x}} -{\boldsymbol{z}} ) \nonumber \\[5pt] &= \frac{ k^2}{3} \sum _{i=1}^3 ({\boldsymbol{E}_0}({\boldsymbol{z}}))_{i}{\boldsymbol{e}}_i \cdot ({\boldsymbol{x}} -{\boldsymbol{z}} ) \nonumber \\[5pt] &= \nabla ^2 \phi _0, \end{align}

where, up to a constant,

(5.6) \begin{align} \phi _0 = \frac{k^2}{18} \!\left ( ({\boldsymbol{E}_0}({\boldsymbol{z}}))_1 (x_1-z_1)^3 + ({\boldsymbol{E}_0}({\boldsymbol{z}}))_2 (x_2-z_2)^3 + ({\boldsymbol{E}_0}({\boldsymbol{z}}))_3 (x_3-z_3)^3 \right ). \end{align}

Related to (3.7) in [2], and following a similar treatment to (3.7), we introduce $\boldsymbol{w}$ as the unique solution to the weak problem. Find ${\boldsymbol{w}} \in H(\text{curl})$ s.t.

(5.7) \begin{align} & \int _{\Omega \setminus \overline{B_\alpha }} \!\left ( \nabla \times{\boldsymbol{w}} \cdot \nabla \times \overline{\boldsymbol{v}} - k^2{\boldsymbol{w}}\cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} + \int _{B_\alpha } \!\left ( \mu _r^{-1} \nabla \times{\boldsymbol{w}} \cdot \nabla \times \overline{\boldsymbol{v}} -k^2 \epsilon _r{\boldsymbol{w}} \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \nonumber \\[5pt] & \quad + {\mathrm{i}} k \int _{\Sigma }{\Lambda }_e^{(k)} ({\boldsymbol{n}}^- \times{\boldsymbol{w}} ) \cdot ({\boldsymbol{n}}^- \times \overline{\boldsymbol{v}} ) \times{\boldsymbol{n}}^- \mathrm{d}{\boldsymbol{x}}\nonumber \\[5pt] & \qquad= \int _{B_\alpha } \!\left ( (1- \mu _r^{-1}) \nabla \times\boldsymbol{F} \cdot \nabla \times \overline{\boldsymbol{v}} -k^2 (1- \epsilon _r)\boldsymbol{F} \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}}, \end{align}

for all $\boldsymbol{v} \in H(\text{curl})$ .

The following updated form of Lemma 3.2 from [2] relies on a Friedrichs’ type inequality that will allow us to estimate a divergence-free field ${\boldsymbol{u}}\in H(\text{curl})\cap H(\text{div})$ in a bounded Lipschitz domain $B_\alpha$ in terms of its curl and ${\boldsymbol{n}} \cdot{\boldsymbol{u}}$ on $\Gamma _\alpha$ , for example, [11, Corollary 3.52, pg 72]:

(5.8) \begin{align} \|{\boldsymbol{u}} \|_{L^2(B_\alpha )} \leqslant C \!\left ( \| \nabla \times{\boldsymbol{u}} \|_{L^2(B_\alpha )} + \|{\boldsymbol{n}} \cdot{\boldsymbol{u}} \|_{L^2(\Gamma _\alpha )} \right ), \end{align}

which we assume holds with $\alpha =1$ while, in general,

(5.9) \begin{align} \|{\boldsymbol{u}} \|_{L^2(B_\alpha )} \leqslant C \alpha \!\left ( \| \nabla \times{\boldsymbol{u}} \|_{L^2(B_\alpha )} + \|{\boldsymbol{n}} \cdot{\boldsymbol{u}} \|_{L^2(\Gamma _\alpha )} \right ), \end{align}

leading to:

Lemma 5.1. Let $\boldsymbol{w}$ be defined by ( 5.7 ). For $|\epsilon _r|$ such that $ \frac{1}{|\epsilon _r|} \|{\boldsymbol{n}} \cdot ({\boldsymbol{E}}_\alpha |_+-{\boldsymbol{E}_0} -{\boldsymbol{w}}|_+)\|_{L^2( \Gamma _\alpha ) } \leqslant \| \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0}-{\boldsymbol{w}}) \|_{L^2(B_\alpha )}$ , and the conditions stated in Section 4 , there exists a constant $C$ such that

\begin{align*} \| \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} ) \|_{L^2(B_\alpha )} & \leqslant C \alpha ^{7/2} \mu _r \!\left (|1 - \mu _r^{-1}| + | \nu | \right ) \| \nabla \times{\boldsymbol{E}_0}\|_{W^{2,\infty }(B_\alpha )}, \\[5pt] \left \|{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} - \frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right \|_{L^2(B_\alpha )} & \leqslant C \alpha ^{9/2} \mu _r \!\left (|1 - \mu _r^{-1}| + | \nu | \right ) \| \nabla \times{\boldsymbol{E}_0}\|_{W^{2,\infty }(B_\alpha )}. \end{align*}

Proof. For the eddy current model considered by [2], $\nabla \cdot{\boldsymbol{E}}_\alpha = \nabla \cdot{\boldsymbol{E}_0} =0$ in $B_\alpha$ and ${\boldsymbol{n}} \cdot{\boldsymbol{E}}_\alpha |_-=0$ on $\Gamma _\alpha$ . Also, they have $\nabla \cdot{\boldsymbol{w}} = - \nabla \cdot\boldsymbol{F}$ in $B_\alpha$ and ${\boldsymbol{n}} \cdot{\boldsymbol{w}}|_- = -{\boldsymbol{n}} \cdot\boldsymbol{F}$ on $\Gamma _\alpha$ . Our situation is different, however, so that choosing $\boldsymbol{v}= \nabla \vartheta$ for some appropriate $\vartheta$ in (5.7) we find that

\begin{align*} \nabla \cdot (\epsilon _r{\boldsymbol{w}} ) = \nabla \cdot ((1-\epsilon _r)\boldsymbol{F}) \qquad \text{in $B_\alpha $}, \qquad{\boldsymbol{n}} \cdot{\boldsymbol{w}}|_+ -{\boldsymbol{n}} \cdot \epsilon _r{\boldsymbol{w}}|_- = - (1-\epsilon _r){\boldsymbol{n}} \cdot\boldsymbol{F} \qquad \text{on} \,\Gamma _\alpha. \end{align*}

We also know that for the problem we are considering

\begin{align*} \nabla \cdot (\epsilon _r{\boldsymbol{E}}_\alpha ) = 0 \qquad \text{in $B_\alpha $}, \qquad{\boldsymbol{n}} \cdot{\boldsymbol{E}}_\alpha |_+ -{\boldsymbol{n}} \cdot \epsilon _r{\boldsymbol{E}}_\alpha |_- = 0 \qquad \text{on $\Gamma _\alpha $}. \end{align*}

We follow [2] and introduce $\phi _0$ in a similar way such that $\nabla \cdot ({\boldsymbol{E}_0} + \nabla \phi _0) = \nabla \cdot\boldsymbol{F}$ in $B_\alpha$ with ${\boldsymbol{n}} \cdot ({\boldsymbol{E}_0} +\nabla \phi _0) |_- ={\boldsymbol{n}} \cdot\boldsymbol{F}$ on $\Gamma _\alpha$ , which is justified based on (5.6) and knowing that $\nabla \cdot{\boldsymbol{E}_0}=0$ in $B_\alpha$ . Since $\nabla \cdot (\boldsymbol{F} - \nabla \phi _0) =0$ in $B_\alpha$ , then $\nabla \cdot ((1-\epsilon _r ) (\boldsymbol{F} - \nabla \phi _0)) = \nabla \cdot (\epsilon _r{\boldsymbol{w}} - (1-\epsilon _r) \nabla \phi _0) =0$ and $\nabla \cdot ({\boldsymbol{w}} -\frac{1-\epsilon _r}{\epsilon _r} \nabla \phi _0) =0$ in $B_\alpha$ . Then,

(5.10) \begin{align} \nabla \cdot \left ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} - \frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right ) =0 \qquad \text{in $B_\alpha $}, \end{align}

and, by combining the above results, we also have

(5.11) \begin{align}{\boldsymbol{n}} \cdot \left ({\boldsymbol{E}}_\alpha |_- -{\boldsymbol{w}}|_- -{\boldsymbol{E}_0} - \frac{\epsilon _r-1}{\epsilon _r}\nabla \phi _0 \right ) -\frac{1}{\epsilon _r}{\boldsymbol{n}} \cdot \left ({\boldsymbol{E}}_\alpha |_+ -{\boldsymbol{w}}|_+ -{\boldsymbol{E}_0} \right ) =0 \qquad \text{on $\Gamma _\alpha $}. \end{align}

For $|\epsilon _r| \to \infty$ , then $ \frac{1}{\epsilon _r}{\boldsymbol{n}} \cdot \left ({\boldsymbol{E}}_\alpha |_+ -{\boldsymbol{w}}|_+ -{\boldsymbol{E}_0} \right )= \frac{1}{\epsilon _r}{\boldsymbol{n}} \cdot \left ({\boldsymbol{E}}_\Delta |_+ -{\boldsymbol{w}}|_+ \right ) \to 0$ on $\Gamma _\alpha$ and, the presence of this term, means that rather than

(5.12) \begin{align} \left \|{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0}-{\boldsymbol{w}} -\frac{\epsilon _r-1}{\epsilon _r }\nabla \phi _0 \right \|_{L^2(B_\alpha )} \leqslant C \alpha \| \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0}-{\boldsymbol{w}}) \|_{L^2(B_\alpha )}, \end{align}

we should use (5.8) leading to an estimate of the form:

(5.13) \begin{align} \left \|{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0}-{\boldsymbol{w}} -\frac{\epsilon _r-1}{\epsilon _r }\nabla \phi _0 \right \|_{L^2(B_\alpha )} & \leqslant C \alpha \!\left ( \| \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0}-{\boldsymbol{w}}) \|_{L^2(B_\alpha )}\right. \nonumber \\[5pt] &\left. \qquad+\; \frac{1}{|\epsilon _r|} \|{\boldsymbol{n}} \cdot ({\boldsymbol{E}}_\alpha |_+-{\boldsymbol{E}_0}-{\boldsymbol{w}}|_+)\|_{L^2( \Gamma _\alpha ) } \right ) \!. \end{align}

Next, we construct

(5.14) \begin{align} \int _{\Omega \setminus \overline{B_\alpha }} & \left ( \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} ) \cdot \nabla \times \overline{\boldsymbol{v}} - k^2 ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} )\cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \nonumber \\[5pt] &+ \int _{B_\alpha } \!\left ( \mu _r^{-1} \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} ) \cdot \nabla \times \overline{\boldsymbol{v}} - k^2 \epsilon _r \!\left ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}}-\frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right )\cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \nonumber \\[5pt] &+{\mathrm{i}} k \int _{\Sigma }{\Lambda }_e^{(k)} ({\boldsymbol{n}}^- \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}}) ) \cdot ({\boldsymbol{n}}^- \times \overline{\boldsymbol{v}})\times{\boldsymbol{n}} ^- \mathrm{d}{\boldsymbol{x}} \nonumber \\[5pt] & \qquad= \int _{B_\alpha } \!\left ( (1-\mu _r^{-1})\nabla \times ({\boldsymbol{E}_0}-\boldsymbol{F}) \cdot \nabla \times \overline{\boldsymbol{v}} - k^2 (1-\epsilon _r) ({\boldsymbol{E}_0}+ \nabla \phi _0 -\boldsymbol{F}) \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}}. \end{align}

By noting the curl of the gradient of a scalar vanishes, we can write

(5.15) \begin{align} \int _{\Omega \setminus \overline{B_\alpha } } & \left ( \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} ) \cdot \nabla \times \overline{\boldsymbol{v}} - k^2 ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} )\cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \nonumber \\[5pt] &+ \int _{B_\alpha } \!\left ( \mu _r^{-1} \nabla \times \left ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} - \frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right ) \cdot \nabla \times \overline{\boldsymbol{v}} - k^2 \epsilon _r \!\left ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}}-\frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \!\right )\cdot \overline{\boldsymbol{v}} \!\right )\! \mathrm{d}{\boldsymbol{x}} \nonumber \\[5pt] &+{\mathrm{i}} k \int _{\Sigma } \Lambda _e^{(k)} ({\boldsymbol{n}}^- \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}}) ) \cdot ({\boldsymbol{n}}^- \times \overline{\boldsymbol{v}})\times{\boldsymbol{n}}^- \mathrm{d}{\boldsymbol{x}} \nonumber \\[5pt] & \qquad= \int _{B_\alpha } \!\left ( (1-\mu _r^{-1})\nabla \times ({\boldsymbol{E}_0}-\boldsymbol{F}) \cdot \nabla \times \overline{\boldsymbol{v}} - k^2 (1-\epsilon _r) ({\boldsymbol{E}_0}+ \nabla \phi _0 -\boldsymbol{F}) \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}}. \end{align}

Choosing $\boldsymbol{v} = \left \{{\begin{array}{l@{\quad}l}{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} - \frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 & \text{in}\, B_\alpha \\[5pt] {\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} & \text{in}\, \Omega \setminus \overline{B_\alpha } \end{array}} \right .$ and taking modulus of both sides of (5.15), we observe that the resulting left-hand side has the form $|Z| = (Z_{\textrm{r}}^2+Z_{\mathrm{i}}^2)^{1/2}$ where

\begin{align*} Z_{\textrm{r}}\;:\!=\; & \left \| \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} ) \right \|_{L^2(\Omega \setminus \overline{B_\alpha })}^2 - k^2 \!\left \|{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} \right \|_{L^2(\Omega \setminus \overline{B_\alpha })}^2 \\[5pt] &+ \mu _r^{-1}\!\left \| \nabla \times \left ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} - \frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right ) \right \|_{L^2(B_\alpha )}^2 - k^2 \frac{\epsilon _{\ast}}{\epsilon _0} \!\left \|{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}}-\frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right \|_{L^2(B_\alpha )}^2 \\[5pt] &+ k \textrm{Im}\!\left ( \int _{\Sigma } \Lambda _e^{(k)} ({\boldsymbol{n}}^- \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}}) ) \cdot ({\boldsymbol{n}}^- \times \overline{\boldsymbol{v}})\times{\boldsymbol{n}}^- \mathrm{d}{\boldsymbol{x}}\right )\!, \\[5pt] Z_{\mathrm{i}} \;:\!=\; &- k^2 \frac{\sigma _{\ast}}{\omega \epsilon _0} \!\left \|{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}}-\frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right \|_{L^2(B_\alpha )}^2 \\[5pt] &+ k \textrm{Re}\!\left ( \int _{\Sigma } \Lambda _e^{(k)} ({\boldsymbol{n}}^- \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}}) ) \cdot ({\boldsymbol{n}}^- \times \overline{\boldsymbol{v}})\times{\boldsymbol{n}}^- \mathrm{d}{\boldsymbol{x}}\right )\!, \end{align*}

and clearly $|Z_{\textrm{r}}| \lt |Z|$ . Considering the conditions needed for

\begin{align*} \left \| \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} ) \right \|_{L^2(\Omega \setminus \overline{B_\alpha })}^2 - k^2 \!\left \|{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} \right \|_{L^2(\Omega \setminus \overline{B_\alpha })}^2 & \gt 0, \\[5pt] \mu _r^{-1}\!\left \| \nabla \times \left ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} - \frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right ) \right \|_{L^2(B_\alpha )}^2 - k^2 \frac{\epsilon _{\ast}}{\epsilon _0} \!\left \|{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}}-\frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right \|_{L^2(B_\alpha )}^2 & \gt 0, \end{align*}

leads to $C(\Omega ) \epsilon _0 \mu _0 \omega ^2 R^2\lt 1$ and $C(B) \epsilon _{\ast} \mu _{\ast} \omega ^2 \alpha ^2\lt 1$ (assuming boundary terms can be absorbed) and, hence, the criteria in Section 4. In this case, we have

(5.16) \begin{align} \mu _r^{-1} &\left \| \nabla \times \left ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} - \frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right ) \right \|_{L^2(B_\alpha )} \nonumber \\[5pt] & \leqslant |Z| = \left | \int _{B_\alpha } \!\left ( (1-\mu _r^{-1})\nabla \times ({\boldsymbol{E}_0}-\boldsymbol{F}) \cdot \nabla \times \overline{\boldsymbol{v}} - k^2 (1-\epsilon _r) ({\boldsymbol{E}_0}+\nabla \phi _0 -\boldsymbol{F} ) \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \right |. \end{align}

Also,

(5.17) \begin{align} \left | \int _{B_\alpha } (1-\mu _r^{-1}) \!\left ( \nabla \times ({\boldsymbol{E}_0}-\boldsymbol{F}) \cdot \nabla \times \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \right | \leqslant C | 1-\mu _r^{-1} | \alpha ^{7/2}\| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty }(B_\alpha )} \| \nabla \times\boldsymbol{v} \|_{L^2(B_\alpha )}, \end{align}

and

(5.18) \begin{align} \left | k^2 \int _{B_\alpha } (1-\epsilon _r) ({\boldsymbol{E}_0}+\nabla \phi _0 -\boldsymbol{F} )\cdot \overline{\boldsymbol{v}} \mathrm{d}{\boldsymbol{x}} \right | & \leqslant k^2 |1-\epsilon _r | \|{\boldsymbol{E}_0}+ \nabla \phi _0 -\boldsymbol{F} \|_{L^2(B_\alpha )} \|\boldsymbol{v} \|_{L^2(B_\alpha )} \nonumber \\[5pt] & \leqslant C\alpha ^2 k^2 |1-\epsilon _r | \| \nabla \times ({\boldsymbol{E}_0}-\boldsymbol{F} ) \|_{L^2(B_\alpha )} ( \| \nabla \times\boldsymbol{v} \|_{L^2(B_\alpha )} +\|{\boldsymbol{n}} \cdot\boldsymbol{v} \|_{L^2(\Gamma _\alpha )} ) \nonumber \\[5pt] & \leqslant C \alpha ^{7/2} | \nu | \| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty } (B_\alpha )}( \| \nabla \times\boldsymbol{v} \|_{L^2(B_\alpha )} +\|{\boldsymbol{n}} \cdot\boldsymbol{v} \|_{L^2(\Gamma _\alpha )}), \end{align}

which follows since ${\boldsymbol{E}_0}+\nabla \phi _0-\boldsymbol{F}$ is divergence-free in $B_\alpha$ and has vanishing ${\boldsymbol{n}} \cdot ({\boldsymbol{E}_0}+\nabla \phi _0-\boldsymbol{F})$ on $\Gamma _\alpha$ and so

(5.19) \begin{align} \|{\boldsymbol{E}_0}+\nabla \phi _0 -\boldsymbol{F} \|_{L^2(B_\alpha )} \leqslant C\alpha \| \nabla \times ({\boldsymbol{E}_0}-\boldsymbol{F} ) \|_{L^2(B_\alpha )} & \leqslant C\alpha \alpha ^{3/2} \alpha ^2 \| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty }(B_\alpha )}\nonumber \\[5pt] &= C\alpha ^{9/2} \| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty }(B_\alpha )}. \end{align}

Also, since $\boldsymbol{v}$ is divergence-free, $\|\boldsymbol{v} \|_{L^2(B_\alpha )} \leqslant C \alpha \!\left ( \| \nabla \times\boldsymbol{v} \|_{L^2(B_\alpha )} + \|{\boldsymbol{n}} \cdot\boldsymbol{v} \|_{L^2(\Gamma _\alpha )} \right )$ . Using (5.17) and (5.18) in (5.16), then

(5.20) \begin{align} & \mu _r^{-1} \!\left \| \nabla \times \left ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0}-\frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 -{\boldsymbol{w}} \right ) \right \|_{L^2(B_\alpha )}^2 \leqslant C | 1-\mu _r^{-1} | \alpha ^{7/2}\| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty }(B_\alpha )} \| \nabla \times\boldsymbol{v} \|_{L^2(B_\alpha )} \nonumber \\[5pt] &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad + C \alpha ^{7/2} | \nu | \| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty } (B_\alpha )} (\| \nabla \times\boldsymbol{v} \|_{L^2(B_\alpha )}+\|{\boldsymbol{n}} \cdot\boldsymbol{v} \|_{L^2(\Gamma _\alpha )}), \end{align}

and for $|\epsilon _r|$ such that $\|{\boldsymbol{n}} \cdot\boldsymbol{v} \|_{L^2(\Gamma _\alpha )} = \frac{1}{|\epsilon _r|} \|{\boldsymbol{n}} \cdot ({\boldsymbol{E}}_\Delta |_+-{\boldsymbol{w}}|_+)\|_{L^2( \Gamma _\alpha ) } \leqslant \| \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0}-{\boldsymbol{w}}) \|_{L^2(B_\alpha )}$ , then

(5.21) \begin{align} \| \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} -{\boldsymbol{w}} )\|_{L^2(B_\alpha )} \leqslant C \mu _r \alpha ^{7/2}\!\left ( | 1-\mu _r^{-1} | + | \nu | \right ) \| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty }(B_\alpha )}. \end{align}

By additionally combining this with (5.13) completes the proof.

Unlike the eddy current problem considered in [2], we are not guaranteed to have $\nabla _x \times{\boldsymbol{H}_0}({\boldsymbol{x}})=\textbf{0}$ in $B_\alpha$ and so an integration by parts yields

(5.22) \begin{align} \int _{B_\alpha } (1-\mu _r^{-1}) \nabla \times\boldsymbol{F}\cdot \nabla \times \overline{\boldsymbol{v}} \mathrm{d}{\boldsymbol{x}} = & \int _{\Gamma _\alpha } [ \tilde{\mu }_r^{-1} \nabla \times\boldsymbol{F} \times{\boldsymbol{n}}^-]_{\Gamma _\alpha } \cdot \overline{\boldsymbol{v}} \mathrm{d}{\boldsymbol{x}} + \int _{B_{\alpha }} (1-\mu _r^{-1}) \overline{\boldsymbol{v}} \cdot \nabla \times \nabla \times\boldsymbol{F} \mathrm{d}{\boldsymbol{x}} \end{align}

and, hence, the weak problem for $\boldsymbol{w}$ becomes. Find ${\boldsymbol{w}} \in H(\text{curl})$ such that

(5.23) \begin{align} &\int _{\Omega \setminus \overline{B_\alpha }} \!\left ( \nabla \times{\boldsymbol{w}} \cdot \nabla \times \overline{\boldsymbol{v}} - k^2{\boldsymbol{w}}\cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} + \int _{B_\alpha } \!\left ( \mu _r^{-1} \nabla \times{\boldsymbol{w}} \cdot \nabla \times \overline{\boldsymbol{v}} -k^2 \epsilon _r{\boldsymbol{w}} \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \nonumber \\[2pt] & \quad+{\mathrm{i}} k \int _{\Sigma } \Lambda _e^{(k)} ({\boldsymbol{n}}^- \times{\boldsymbol{w}} ) \cdot ({\boldsymbol{n}}^- \times \overline{\boldsymbol{v}})\times{\boldsymbol{n} }^- \mathrm{d}{\boldsymbol{x}} \nonumber \\[2pt] & \qquad= \int _{\Gamma _\alpha } [ \tilde{\mu }_r^{-1} \nabla \times\boldsymbol{F} \times{\boldsymbol{n}}^-]_{\Gamma _\alpha } \cdot \overline{\boldsymbol{v}} \mathrm{d}{\boldsymbol{x}} -k^2\int _{B_\alpha } \!\left ( (1- \epsilon _r)\boldsymbol{F} \cdot \overline{\boldsymbol{v}} \right )\! \mathrm{d}{\boldsymbol{x}} \nonumber \\[2pt] &\qquad \quad + \int _{B_{\alpha }} (1-\mu _r^{-1}) \overline{\boldsymbol{v}} \cdot \nabla \times \nabla \times\boldsymbol{F} \mathrm{d}{\boldsymbol{x}}, \end{align}

for all $\boldsymbol{v} \in H(\text{curl})$ . This, in turn, motivates the strong form for $\boldsymbol{w}$ as:

(5.24a) \begin{align} \nabla \times \mu _r^{-1} \nabla \times{\boldsymbol{w}} -k^2 \epsilon _r{\boldsymbol{w}} & = (1-\mu _r^{-1}) \nabla \times \nabla \times\boldsymbol{F} -k^2(1-\epsilon _r)\boldsymbol{F} && \text{in $B_\alpha $}, \end{align}

(5.24b) \begin{align} \nabla \times \nabla \times{\boldsymbol{w}} -k^2{\boldsymbol{w}} & = \textbf{0} && \text{in $\Omega \setminus \overline{B_\alpha } $}, \end{align}

(5.24c) \begin{align}\nabla \cdot{\boldsymbol{w}}& = 0 && \text{in $\Omega $}, \end{align}

(5.24d) \begin{align} \nabla \times \nabla \times{\boldsymbol{w}} -k^2{\boldsymbol{w}} & = \textbf{0} && \text{in ${\mathbb R}^3 \setminus \overline{\Omega }$}, \end{align}

(5.24e) \begin{align} \nabla \cdot{\boldsymbol{w}} & = 0 && \text{in ${\mathbb R}^3 \setminus \overline{\Omega } $}, \end{align}

(5.24f) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{w}}]_{\Gamma _\alpha } =\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu _r}^{-1} \nabla \times{\boldsymbol{w}}]_{\Gamma _\alpha } & = -(1-\mu _r^{-1}){\boldsymbol{n}} \times \nabla \times\boldsymbol{F} && \text{on $\Gamma _\alpha $}, \end{align}

(5.24g) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{w}}]_{\Sigma } =\textbf{0}, \qquad [{\boldsymbol{n}} \times \nabla \times{\boldsymbol{w}}]_{\Sigma } & ={\boldsymbol 0} && \text{on $\Sigma $}, \end{align}

(5.24h) \begin{align} \lim _{x\to \infty } x \!\left ( ( \nabla \times{\boldsymbol{w}}) \times \hat{{\boldsymbol{x}}}-{\mathrm{i}} k{\boldsymbol{w}} \right ) & = \textbf{0}. \end{align}

which can also be expressed as:

(5.25a) \begin{align} \nabla \times \mu _r^{-1} \nabla \times{\boldsymbol{w}} -k^2 \epsilon _r{\boldsymbol{w}} &= (1-\mu _r^{-1}) \nabla \times \nabla \times\boldsymbol{F} -k^2(1-\epsilon _r)\boldsymbol{F} && \text{in $B_\alpha $}, \end{align}

(5.25b) \begin{align} \nabla \times \nabla \times{\boldsymbol{w}} -k^2{\boldsymbol{w}} &= \textbf{0} && \text{in $B_\alpha ^c$}, \end{align}

(5.25c) \begin{align} \nabla \cdot{\boldsymbol{w}} &= 0 && \text{in ${\mathbb R}^3$}, \end{align}

(5.25d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{w}}]_{\Gamma _\alpha }=\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu _r}^{-1} \nabla \times{\boldsymbol{w}}]_{\Gamma _\alpha } &= -(1-\mu _r^{-1}){\boldsymbol{n}} \times \nabla \times\boldsymbol{F} && \text{on $\Gamma _\alpha $}, \end{align}

(5.25e) \begin{align} \lim _{x\to \infty } x \!\left ( ( \nabla \times{\boldsymbol{w}}) \times \hat{{\boldsymbol{x}}}-{\mathrm{i}} k{\boldsymbol{w}} \right ) &= \textbf{0}. \end{align}

By introducing ${\boldsymbol{w}} ({\boldsymbol{x}}) = \alpha{\boldsymbol{w}}_0\!\left ( \frac{{\boldsymbol{x}}-{\boldsymbol{z}}}{\alpha } \right ) = \alpha{\boldsymbol{w}}_0({\boldsymbol{\xi }})$ , we see ${\boldsymbol{w}}_0({\boldsymbol{\xi }})$ satisfies

(5.26a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{w}}_0 -k^2 \epsilon _r\alpha ^2{\boldsymbol{w}}_0 &= (1-\mu _r^{-1}) \nabla \times \nabla \times [\alpha\boldsymbol{F}] -k^2\alpha ^2 (1-\epsilon _r) [\alpha ^{-1}\boldsymbol{F}] && \text{in $B$}, \end{align}

(5.26b) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{w}}_0 -k^2 \alpha ^2{\boldsymbol{w}}_0 &= \textbf{0} && \text{in $B^c$}, \end{align}

(5.26c) \begin{align} \nabla _\xi \cdot{\boldsymbol{w}}_0 &= 0 && \text{in ${\mathbb R}^3$}, \end{align}

(5.26d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{w}}_0]_\Gamma =\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times{\boldsymbol{w}}_0]_\Gamma &= -(1-\mu _r^{-1}){\boldsymbol{n}} \times \nabla \times [\alpha ^{-1}\boldsymbol{F}] && \text{on $\Gamma $}, \end{align}

(5.26e) \begin{align} \lim _{\xi \to \infty } \xi \!\left ( ( \nabla _\xi \times{\boldsymbol{w}}_0) \times \hat{{\boldsymbol{\xi }}} -{\mathrm{i}} k \alpha{\boldsymbol{w}}_0 \right ) &= \textbf{0}, \end{align}

and, by introducing a bounded domain $\Omega _0$ with boundary $\Sigma _0 \;:\!=\; \partial \Omega _0$ the weak solution of (5.26) can be expressed in a similar way to (5.7).

Hence, Theorem 3.1 in [2], which follows directly from Lemma 5.1, is replaced by:

Theorem 5.2. For $|\epsilon _r|$ such that $ \frac{1}{|\epsilon _r|} \|{\boldsymbol{n}} \cdot ({\boldsymbol{E}}_\Delta |_+-{\boldsymbol{w}}|_+)\|_{L^2( \Gamma _\alpha ) } \leqslant \| \nabla \times ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0}-{\boldsymbol{w}}) \|_{L^2(B_\alpha )}$ , and the conditions stated in Section 4 , there exists a constant $C$ such that

\begin{align*} \left \| \nabla \times \left ({\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} - \alpha{\boldsymbol{w}}_0 \!\left ( \frac{{\boldsymbol{x}}-{\boldsymbol{z}}}{\alpha } \right ) \right ) \right \|_{L^2(B_\alpha )} & \leqslant C \alpha ^{7/2} \mu _r \!\left (|1 - \mu _r^{-1}| + | \nu | \right ) \| \nabla \times{\boldsymbol{E}_0}\|_{W^{2,\infty }(B_\alpha )}, \\[5pt] \left \|{\boldsymbol{E}}_\alpha -{\boldsymbol{E}_0} - \alpha{\boldsymbol{w}}_0 \!\left ( \frac{{\boldsymbol{x}}-{\boldsymbol{z}}}{\alpha } \right ) - \frac{\epsilon _r-1}{\epsilon _r} \nabla \phi _0 \right \|_{L^2(B_\alpha )} & \leqslant C \alpha ^{9/2} \mu _r \!\left (|1 - \mu _r^{-1}| + | \nu | \right ) \| \nabla \times{\boldsymbol{E}_0}\|_{W^{2,\infty }(B_\alpha )}. \end{align*}

From (5.3), we have

\begin{align*} \alpha ^{-1}\boldsymbol{F} ( \alpha \xi +{\boldsymbol{z}} ) & = \frac{{\mathrm{i}} k }{2}\sum _{i=1}^3 ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i{\boldsymbol{e}}_i \times{\boldsymbol{\xi }} + \frac{{\mathrm{i}} k \alpha }{3} \sum _{i,j=1}^3 ({\boldsymbol{D}}_z{\boldsymbol{H}_0}({\boldsymbol{z}}))_{ij} \xi _j{\boldsymbol{e}}_i \times{\boldsymbol{\xi }} + \alpha ^{-1} \sum _{i=1}^3 ({\boldsymbol{E}_0}({\boldsymbol{z}}))_i{\boldsymbol{e}}_i, \end{align*}

and similarly

(5.27) \begin{align}{\boldsymbol{w}}_0 ({\boldsymbol{\xi }} ) = \frac{{\mathrm{i}} k}{2} \sum _{i=1}^3 ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i{\boldsymbol{\theta }}_i({\boldsymbol{\xi }}) + \frac{{\mathrm{i}} k \alpha }{3}\sum _{i=1,j}^3 ({\boldsymbol{D}}_z{\boldsymbol{H}_0}({\boldsymbol{z}}))_{ij}{\boldsymbol{\psi }}_{ij} ({\boldsymbol{\xi }}) + \alpha ^{-1} \sum _{i=1}^3 ({\boldsymbol{E}_0}({\boldsymbol{z}}))_i{\boldsymbol{\phi }}_i ({\boldsymbol{\xi }}). \end{align}

In the above equation, ${\boldsymbol{\theta }}_i({\boldsymbol{\xi }})$ solves

(5.28a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i -k^2 \alpha ^2 \epsilon _r{\boldsymbol{\theta }}_i &= -k^2\alpha ^2 (1-\epsilon _r){\boldsymbol{e}}_i\times{\boldsymbol{\xi }} && \text{in $B$}, \end{align}

(5.28b) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{\theta }}_i -k^2\alpha ^2{\boldsymbol{\theta }}_i & = \textbf{0} && \text{in $B^c$}, \end{align}

(5.28c) \begin{align}\nabla _\xi \cdot{\boldsymbol{\theta }}_i & = 0 && \text{in ${\mathbb R}^3$}, \end{align}

(5.28d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{\theta }}_i ]_\Gamma =\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i ]_\Gamma & = -2 (1-\mu _r^{-1}){\boldsymbol{n}} \times{\boldsymbol{e}}_i && \text{on $\Gamma $}, \end{align}

(5.28e) \begin{align} \lim _{\xi \to \infty } \xi \!\left ( ( \nabla _\xi \times{\boldsymbol{\theta }}_i ) \times \hat{{\boldsymbol{\xi }}} -{\mathrm{i}} k \alpha{\boldsymbol{\theta }}_i \right ) & = \textbf{0}, \end{align}

${\boldsymbol{\psi }}_{ij} ({\boldsymbol{\xi }})$ solves

(5.29a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\psi }}_{ij} -k^2 \alpha ^2 \epsilon _r{\boldsymbol{\psi }}_{ij} & = (1-\mu _r^{-1}){\boldsymbol{e}}_j \times{\boldsymbol{e}}_i -k^2\alpha ^2 (1-\epsilon _r) \xi _j{\boldsymbol{e}}_i\times{\boldsymbol{\xi }} && \text{in $B$}, \end{align}

(5.29b) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{\psi }}_{ij} -k^2\alpha ^2{\boldsymbol{\psi }}_{ij} & = \textbf{0} && \text{in $B^c$}, \end{align}

(5.29c) \begin{align} \nabla _\xi \cdot{\boldsymbol{\psi }}_{ij} & = 0 && \text{in ${\mathbb R}^3$}, \end{align}

(5.29d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{\psi }}_{ij} ]_\Gamma =\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i ]_\Gamma & = -3 (1-\mu _r^{-1}){\boldsymbol{n}} \times \xi _j{\boldsymbol{e}}_i && \text{on $\Gamma $}, \end{align}

(5.29e) \begin{align} \lim _{\xi \to \infty } \xi \!\left ( ( \nabla _\xi \times{\boldsymbol{\psi }}_{ij} ) \times \hat{{\boldsymbol{\xi }}} -{\mathrm{i}} k \alpha{\boldsymbol{\psi }}_{ij} \right ) & = \textbf{0}, \end{align}

and ${\boldsymbol{\phi }}_{i} ({\boldsymbol{\xi }})$ solves

(5.30a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\phi }}_{i} -k^2 \alpha ^2 \epsilon _r{\boldsymbol{\phi }}_{i} & = -k^2\alpha ^2 (1-\epsilon _r){\boldsymbol{e}}_i && \text{in $B$}, \end{align}

(5.30b) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{\phi }}_{i} -k^2\alpha ^2{\boldsymbol{\phi }}_{i} & = \textbf{0} && \text{in $B^c$}, \end{align}

(5.30c) \begin{align} \nabla _\xi \cdot{\boldsymbol{\phi }}_{i} & = 0 && \text{in ${\mathbb R}^3$}, \end{align}

(5.30d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{\phi }}_{i} ]_\Gamma =\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times{\boldsymbol{\phi }}_{i} ]_\Gamma & =\textbf{0} && \text{on $\Gamma $}, \end{align}

(5.30e) \begin{align} \lim _{\xi \to \infty } \xi \!\left ( ( \nabla _\xi \times{\boldsymbol{\phi }}_{i} ) \times \hat{{\boldsymbol{\xi }}} -{\mathrm{i}} k \alpha{\boldsymbol{\phi }}_{i} \right ) & = \textbf{0}. \end{align}

Furthermore, given that the only source term is the gradient of a scalar, the solution to (5.30) can be expressed in terms of ${\boldsymbol{\phi }}_i = (\epsilon _r-1) \nabla \vartheta _i$ where $\vartheta _i$ solves

(5.31a) \begin{align} \nabla _\xi \cdot \epsilon _r\nabla _\xi \vartheta _{i} & = 0 && \text{in $B$}, \end{align}

(5.31b) \begin{align}\nabla _\xi \cdot \nabla _\xi \vartheta _{i} & = 0 && \text{in $B^c$}, \end{align}

(5.31c) \begin{align} [\vartheta _i]_\Gamma = 0 \qquad{\boldsymbol{n}} \cdot \nabla _\xi \vartheta _i|_+ -{\boldsymbol{n}} \cdot \epsilon _r \nabla _\xi \vartheta _i |_- & = {\boldsymbol{n}} \cdot{\boldsymbol{e}}_i && \text{on $\Gamma $}, \end{align}

(5.31d) \begin{align} \vartheta _i & \to 0 && \text{as $|{\boldsymbol{\xi }}|\to \infty $}. \end{align}

6. Integral representation formulae

A different integral representation formula for ${\boldsymbol{H}}_\Delta ({\boldsymbol{x}} ) = ({\boldsymbol{H}}_\alpha -{\boldsymbol{H}_0})({\boldsymbol{x}})$ for $\boldsymbol{x}$ away from $B_\alpha$ is required compared to that used in [2], since the eddy current approximation is no longer applied. The following is appropriate for describing the perturbed magnetic field outside of $B_\alpha$ and is the same as used in [6]:

(6.1) \begin{equation}{\boldsymbol{H}}_\Delta ({\boldsymbol{x}}) =\nabla _x \times \left ( \int _{\Gamma _\alpha } G_k({\boldsymbol{x}},{\boldsymbol{y}}){\boldsymbol{n}} \times{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) |_+ \mathrm{d}{\boldsymbol{y}} \right ) - \frac{{\mathrm{i}}}{k} \nabla _x \times \left (\nabla _x \times \int _{\Gamma _\alpha } G_k({\boldsymbol{x}},{\boldsymbol{y}}){\boldsymbol{n}} \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) |_+ \mathrm{d}{\boldsymbol{y}} \right ). \end{equation}

We transform this result to be expressed in terms of volume integrals over $B_\alpha$ and express the result in the lemma below.

Lemma 6.1. An integral representation formula for ${\boldsymbol{H}}_\Delta ({\boldsymbol{x}} ) = ({\boldsymbol{H}}_\alpha -{\boldsymbol{H}_0})({\boldsymbol{x}})$ for $\boldsymbol{x}$ away from $B_\alpha$ expressed in terms of volume integrals over $B_\alpha$ is

(6.2) \begin{align}{\boldsymbol{H}}_\Delta ({\boldsymbol{x}}) &= -{\mathrm{i}} k(\epsilon _r -1) \int _{B_\alpha } \nabla _x G_k ({\boldsymbol{x}},{\boldsymbol{y}}) \times{\boldsymbol{E}}_\alpha ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} + k^2 (\mu _r -1) \int _{B_\alpha } G_k({\boldsymbol{x}},{\boldsymbol{y}}){\boldsymbol{H}}_\alpha ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} \nonumber \\[5pt] &\qquad + (\mu _r-1) \int _{B_\alpha }{\boldsymbol{D}}_y^2 G_k({\boldsymbol{x}},{\boldsymbol{y}}){\boldsymbol{H}}_\alpha ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}}. \end{align}

Proof. Using the transmission conditions (3.4), transforming the first term of (6.1) to a volume integral and then applying $\nabla \times ({\boldsymbol{a}} \times{\boldsymbol{b}}) ={\boldsymbol{a}} \nabla \cdot{\boldsymbol{b}} -{\boldsymbol{b}} \nabla \cdot{\boldsymbol{a}} + ({\boldsymbol{b}} \cdot \nabla ){\boldsymbol{a}} - ({\boldsymbol{a}} \cdot \nabla ){\boldsymbol{b}}$ gives

(6.3) \begin{align} \nabla _x \times & \left ( \int _{\Gamma _\alpha } G_k({\boldsymbol{x}},{\boldsymbol{y}}){\boldsymbol{n}} \times{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) |_+ \mathrm{d}{\boldsymbol{y}} \right ) = \nabla _x \times \left ( \int _{B_\alpha } \nabla _y \times (G_k({\boldsymbol{x}},{\boldsymbol{y}}){\boldsymbol{H}}_\Delta ({\boldsymbol{y}})) \mathrm{d}{\boldsymbol{y}} \right ) \nonumber \\[5pt] &= \nabla _x \times \left ( \int _{B_\alpha } \nabla _y G_k({\boldsymbol{x}},{\boldsymbol{y}}) \times{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) + G_k({\boldsymbol{x}},{\boldsymbol{y}}) \nabla _y \times{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} \right ) \nonumber \\[5pt] &= \int _{B_\alpha } \!\left ( \nabla _y G_k({\boldsymbol{x}},{\boldsymbol{y}}) (\nabla _x \cdot{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) )-{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) \nabla _x\cdot \nabla _y G_k({\boldsymbol{x}},{\boldsymbol{y}}) \right. \nonumber \\[5pt] & \qquad \left .+ ({\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) \cdot \nabla _x) (\nabla _y G_k({\boldsymbol{x}},{\boldsymbol{y}}) )- (\nabla _y G_k({\boldsymbol{x}},{\boldsymbol{y}}) \cdot \nabla _x)({\boldsymbol{H}}_\Delta ({\boldsymbol{y}})) + \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}}) \times \nabla _y \times{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) \right )\! \mathrm{d}{\boldsymbol{y}} \nonumber \\[5pt] &= \int _{B_\alpha } \!\left (-k^2{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) G_k({\boldsymbol{x}},{\boldsymbol{y}}) -{\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{y}}) ({\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) )+ \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}}) \times \nabla _y \times{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) \right )\! \mathrm{d}{\boldsymbol{y}}, \end{align}

where $\nabla _x\cdot \nabla _y G_k({\boldsymbol{x}},{\boldsymbol{y}}) = -\nabla _x\cdot \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}}) = k^2 G_k({\boldsymbol{x}},{\boldsymbol{y}})$ and $\nabla _x \nabla _y G_k({\boldsymbol{x}},{\boldsymbol{y}}) = -{\boldsymbol{D}}_x^2 G_k ({\boldsymbol{x}},{\boldsymbol{y}})$ have been applied. Considering the second term in (6.1), and applying similar ideas, gives

(6.4) \begin{align} \nabla _x &\times \left (\!\nabla _x \times \int _{\Gamma _\alpha } G_k({\boldsymbol{x}},{\boldsymbol{y}}){\boldsymbol{n}} \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) _+ \mathrm{d}{\boldsymbol{y}} \!\right ) = \nabla _x \times \left (\!\nabla _x \times \int _{B_\alpha } \nabla _y G_k({\boldsymbol{x}},{\boldsymbol{y}}) \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) + G_k({\boldsymbol{x}},{\boldsymbol{y}}) \nabla _y \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} \!\right ) \nonumber \\[5pt] & = \nabla _x \times \int _{B_\alpha } - k^2{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) G_k({\boldsymbol{x}},{\boldsymbol{y}}) -{\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{y}}){\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) + \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}}) \times \nabla _y \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} \nonumber \\[5pt] & = \int _{B_\alpha } \!\left ( - k^2\nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}}) \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) - \nabla _y \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) (\nabla _x \cdot \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}})) + (\nabla _y \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) \cdot \nabla _x ) \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}}) \right )\! \mathrm{d}{\boldsymbol{y}} \nonumber \\[5pt] & = \int _{B_\alpha } \!\left ( - k^2\nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}}) \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) +k^2 \nabla _y \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) G_k({\boldsymbol{x}},{\boldsymbol{y}}) +{\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{y}}) (\nabla _y \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) ) \right )\! \mathrm{d}{\boldsymbol{y}}. \end{align}

Using (6.3) and (6.4) in (6.1) gives

(6.5) \begin{align}{\boldsymbol{H}}_\Delta ({\boldsymbol{x}}) & = -k^2 \int _{B_\alpha }{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) G_k({\boldsymbol{x}},{\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} - \int _{B_\alpha }{\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{y}}) ({\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) )\!\mathrm{d}{\boldsymbol{y}} + \int _{B_\alpha } \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}}) \times \nabla _y \times{\boldsymbol{H}}_\Delta ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} \nonumber \\[5pt] & \qquad -{\mathrm{i}} k \int _{B_\alpha } \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{y}}) \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} -{\mathrm{i}} k \int _{B_\alpha } \nabla _y \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) G_k({\boldsymbol{x}},{\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} \nonumber \\[5pt] & \qquad - \frac{{\mathrm{i}}}{k} \int _{B_\alpha }{\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{y}}) (\nabla _y \times{\boldsymbol{E}}_\Delta ({\boldsymbol{y}}) ) \mathrm{d}{\boldsymbol{y}}. \end{align}

Then, using (3.3), we have

(6.6a) \begin{align} \nabla \times{\boldsymbol{E}}_\Delta & ={\mathrm{i}} k \tilde{\mu }_r{\boldsymbol{H}}_\Delta +{\mathrm{i}} k (\tilde{\mu _r}-1){\boldsymbol{H}_0}, \end{align}

(6.6b) \begin{align} \nabla \times{\boldsymbol{H}}_\Delta & = -{\mathrm{i}} k \tilde{\epsilon }_r{\boldsymbol{E}}_\Delta -{\mathrm{i}} k (\tilde{\epsilon }_r-1){\boldsymbol{E}_0}, \end{align}

and inserting these into (6.5) completes the proof.

7. Proof of Theorem 4.1

We write (6.2) as the sum:

(7.1) \begin{align}{\boldsymbol{H}}_\Delta ({\boldsymbol{x}}) &= -{\mathrm{i}} k(\epsilon _r -1) \int _{B_\alpha } \nabla _x G_k ({\boldsymbol{x}},{\boldsymbol{y}}) \times{\boldsymbol{E}}_\alpha ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} + (\mu _r-1) \int _{B_\alpha }({\boldsymbol{D}}_y^2 G_k({\boldsymbol{x}},{\boldsymbol{y}})+k^2G_k({\boldsymbol{x}},{\boldsymbol{y}}){\mathbb I}){\boldsymbol{H}}_\alpha ({\boldsymbol{y}}) \mathrm{d}{\boldsymbol{y}} \nonumber \\[5pt] & ={\boldsymbol{T}}_1 +{\boldsymbol{T}}_2. \end{align}

Approximation of ${\boldsymbol{T}}_1$

Let ${\boldsymbol{T}}_1 ={\boldsymbol{T}}_1^a +{\boldsymbol{T}}_1^b +{\boldsymbol{T}}_1^c +{\boldsymbol{T}}_1^d$ where

\begin{align*}{\boldsymbol{T}}_1^a & \;:\!=\; -{\mathrm{i}} k(\epsilon _r -1) \int _{B_\alpha } \nabla _x G_k ({\boldsymbol{x}},{\boldsymbol{y}}) \times \left ({\boldsymbol{E}}_\alpha ({\boldsymbol{y}}) -{\boldsymbol{E}_0}({\boldsymbol{y}}) - \alpha{\boldsymbol{w}}_0 \!\left ( \frac{{\boldsymbol{y}} -{\boldsymbol{z}}}{\alpha } \right ) - \frac{\epsilon _r-1}{\epsilon _r }\nabla \phi _0 \right )\!\mathrm{d}{\boldsymbol{y}}, \\[5pt] {\boldsymbol{T}}_1^b & \;:\!=\; -{\mathrm{i}} k(\epsilon _r -1) \int _{B_\alpha } \nabla _x G_k ({\boldsymbol{x}},{\boldsymbol{y}}) \times \left ({\boldsymbol{E}_0}({\boldsymbol{y}})+\nabla \phi _0 -\boldsymbol{F}({\boldsymbol{y}}) \right )\! \mathrm{d}{\boldsymbol{y}} -{\mathrm{i}} k\frac{(\epsilon _r -1)}{\epsilon _r} \int _{B_\alpha } \nabla _x G_k ({\boldsymbol{x}},{\boldsymbol{y}}) \times \nabla \phi _0 \mathrm{d}{\boldsymbol{y}}, \\[5pt] {\boldsymbol{T}}_1^c & \;:\!=\; -{\mathrm{i}} k(\epsilon _r -1) \int _{B_\alpha } \!\left ( \nabla _x G_k ({\boldsymbol{x}},{\boldsymbol{y}}) - \nabla _x G_k ({\boldsymbol{x}},{\boldsymbol{z}}) +{\boldsymbol{D}}_x^2 G_k ({\boldsymbol{x}},{\boldsymbol{z}}) ({\boldsymbol{y}}-{\boldsymbol{z}}) \right )\times \left (\boldsymbol{F}({\boldsymbol{y}}) + \alpha{\boldsymbol{w}}_0 \!\left ( \frac{{\boldsymbol{y}} -{\boldsymbol{z}}}{\alpha } \right ) \right )\!\mathrm{d}{\boldsymbol{y}}, \\[5pt] {\boldsymbol{T}}_1^d & \;:\!=\; -{\mathrm{i}} k(\epsilon _r -1) \int _{B_\alpha } \!\left ( \nabla _x G_k ({\boldsymbol{x}},{\boldsymbol{z}}) -{\boldsymbol{D}}_x^2 G_k ({\boldsymbol{x}},{\boldsymbol{z}}) ({\boldsymbol{y}}-{\boldsymbol{z}}) \right )\times \left (\boldsymbol{F}({\boldsymbol{y}}) + \alpha{\boldsymbol{w}}_0 \!\left ( \frac{{\boldsymbol{y}} -{\boldsymbol{z}}}{\alpha } \right ) \right )\!\mathrm{d}{\boldsymbol{y}}. \end{align*}

Using Theorem 5.2, we estimate

\begin{align*} |{\boldsymbol{T}}_1^a| & \leqslant C k |\epsilon _r -1| \alpha ^{3/2} \alpha ^{9/2} \| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty }(B_\alpha )} \mu _r \!\left (|1- \mu _r^{-1} |+ | \nu | \right ) \\[5pt] & \leqslant C \alpha ^4 \|{\boldsymbol{H}_0} \|_{W^{2,\infty }(B_\alpha )}. \end{align*}

Next, using (5.19) and (5.6) we estimate

\begin{align*} |{\boldsymbol{T}}_1^b| \leqslant & C k |\epsilon _r -1| \alpha ^{3/2}\alpha ^{9/2} \| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty }(B_\alpha )}+ C k^3 \!\left |1 -\frac{1}{\epsilon _r} \right | \alpha ^{3/2}\alpha ^{3/2} \alpha ^2 \|{\boldsymbol{E}_0} \|_{W^{1,\infty }(B_\alpha )} \\[5pt] \leqslant & C\!\left ( \alpha ^4 \|{\boldsymbol{H}_0} \|_{W^{2,\infty }(B_\alpha )} + k^3 \alpha ^5 \!\left |1 -\frac{1}{\epsilon _r} \right | \|{\boldsymbol{E}_0} \|_{W^{1,\infty }(B_\alpha )} \right ). \end{align*}

Then, using (5.3) and (5.27),

\begin{align*} |{\boldsymbol{T}}_1^c| \leqslant & C k |\epsilon _r -1| \alpha ^{3/2}\alpha ^{2} \alpha ^{3/2} \alpha \| \nabla \times{\boldsymbol{E}_0} \|_{W^{2,\infty }(B_\alpha )} \\[5pt] \leqslant & C \alpha ^4 \|{\boldsymbol{H}_0} \|_{W^{2,\infty }(B_\alpha )}. \end{align*}

Similarly, ${\boldsymbol{T}}_1^{d}={\boldsymbol{T}}_1^{d,1} +{\boldsymbol{T}}_1^{d,2}+{\boldsymbol{T}}_1^{d,3}+{\boldsymbol{T}}_1^{d,4} +{\boldsymbol{T}}_1^{d,5} +{\boldsymbol{T}}_1^{d,6}$ where

\begin{align*}{\boldsymbol{T}}_1^{d,1}& \;:\!=\; -\frac{{\mathrm{i}} k \alpha ^4 (\epsilon _r -1)}{2} \sum _{i=1}^3 \nabla _x G_k ({\boldsymbol{x}},{\boldsymbol{z}})\times \int _{B} \!\left ({\boldsymbol{e}}_i \times{\boldsymbol{\xi }} +{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }} ({\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{z}}))_i, \\[5pt] {\boldsymbol{T}}_1^{d,2} &\;:\!=\; -{\mathrm{i}} k \alpha ^3 (\epsilon _r -1) \sum _{i=1}^3 \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{z}}) \times \int _B \!\left ({\boldsymbol{e}}_i +{\boldsymbol{\phi }}_i \right ) \mathrm{d}{\boldsymbol{\xi }} ({\boldsymbol{E}_0}({\boldsymbol{z}}))_i, \\[5pt] {\boldsymbol{T}}_1^{d,3} & \;:\!=\; - \frac{{\mathrm{i}} k \alpha ^5 (\epsilon _r -1)}{3} \sum _{i,j=1}^3 \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{z}}) \times \int _B \!\left ( \xi _j{\boldsymbol{e}}_i \times{\boldsymbol{\xi }} +{\boldsymbol{\psi }}_{ij} \right ) \mathrm{d}{\boldsymbol{\xi }} ({\boldsymbol{D}}_z({\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{z}})))_{ij}, \\[5pt] {\boldsymbol{T}}_1^{d,4} & \;:\!=\; \frac{{\mathrm{i}} k \alpha ^5 (\epsilon _r -1)}{2} \sum _{i=1}^3 \int _B{\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}){\boldsymbol{\xi }} \times \left ({\boldsymbol{e}}_i \times{\boldsymbol{\xi }} +{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }} ({\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{z}}))_i, \\[5pt] {\boldsymbol{T}}_1^{d,5} & \;:\!=\; -{\mathrm{i}} k \alpha ^4 (\epsilon _r -1) \sum _{i=1}^3 \int _B{\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}){\boldsymbol{\xi }} \times \left ({\boldsymbol{e}}_i +{\boldsymbol{\phi }}_i \right ) \mathrm{d}{\boldsymbol{\xi }} ({\boldsymbol{E}_0}({\boldsymbol{z}}))_i, \\[5pt] {\boldsymbol{T}}_1^{d,6} & \;:\!=\; - \frac{{\mathrm{i}} k \alpha ^6 (\epsilon _r -1)}{3} \sum _{i,j=1}^3 \int _B{\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}){\boldsymbol{\xi }} \times \left (\xi _j{\boldsymbol{e}}_i \times{\boldsymbol{\xi }} +{\boldsymbol{\psi }}_{ij} \right ) \mathrm{d}{\boldsymbol{\xi }} ({\boldsymbol{D}}_z({\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{z}})))_{ij}, \end{align*}

Defining $\boldsymbol{R}({\boldsymbol{x}})$ such that $ |\boldsymbol{R}({\boldsymbol{x}})| \leqslant C \!\left ( \alpha ^4 \|{\boldsymbol{H}_0} \|_{W^{2,\infty }(B_\alpha )} + \alpha ^4 k \!\left ( |\epsilon _r-1| + k^2 \alpha \!\left |1 -\frac{1}{\epsilon _r} \right | \right ) \|{\boldsymbol{E}_0} \|_{W^{1,\infty }(B_\alpha )} \right ),$ we see that ${\boldsymbol{T}}_1^{a},{\boldsymbol{T}}_1^{b},{\boldsymbol{T}}_1^{c}$ and ${\boldsymbol{T}}_1^{d,5}$ are elements of $\boldsymbol{R}({\boldsymbol{x}})$ and, since ${\boldsymbol{D}}_z({\mathrm{i}} k{\boldsymbol{H}_0}({\boldsymbol{z}}))$ is related to $\boldsymbol{E}_0$ , ${\boldsymbol{T}}_1^{d,3}$ and ${\boldsymbol{T}}_1^{d,6}$ are also elements of $\boldsymbol{R}({\boldsymbol{x}})$ .

Remark 7.1. In the case of the eddy current regime, where (5.28) and (5.30) simplify to

(7.2a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i -\nu _{\mathrm{i}}{\boldsymbol{\theta }}_i & = \nu _{\mathrm{i}}{\boldsymbol{e}}_i\times{\boldsymbol{\xi }} && \text{in $B$}, \end{align}

(7.2b) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{\theta }}_i & = \textbf{0} && \text{in $B^c$}, \end{align}

(7.2c) \begin{align} \nabla _\xi \cdot{\boldsymbol{\theta }}_i & = 0 && \text{in ${\mathbb R}^3$}, \end{align}

(7.2d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{\theta }}_i ]_\Gamma =\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i ]_\Gamma & = -2 (1-\mu _r^{-1}){\boldsymbol{n}} \times{\boldsymbol{e}}_i && \text{on $\Gamma $}, \end{align}

(7.2e) \begin{align} {\boldsymbol{\theta }}_i & = O(|{\boldsymbol{\xi }}|^{-1}) && \text{as $|{\boldsymbol{\xi }}| \to \infty $}, \end{align}

and

(7.3a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\phi }}_{i} -\nu _{\mathrm{i}}{\boldsymbol{\phi }}_{i} & = \nu _{\mathrm{i}}{\boldsymbol{e}}_i && \text{in $B$}, \end{align}

(7.3b) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{\phi }}_{i} & = \textbf{0} && \text{in $B^c$} \end{align}

(7.3c) \begin{align} \nabla _\xi \cdot{\boldsymbol{\phi }}_{i} & = 0 && \text{in ${\mathbb R}^3$}, \end{align}

(7.3d) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{\phi }}_{i} ]_\Gamma =\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times{\boldsymbol{\phi }}_{i} ]_\Gamma & =\textbf{0} && \text{on $\Gamma $}, \end{align}

(7.3e) \begin{align} {\boldsymbol{\phi }}_i & = O(|{\boldsymbol{\xi }}|^{-1}) && \text{as $|{\boldsymbol{\xi }}| \to \infty $}, \end{align}

respectively, we find that ${\boldsymbol{T}}_1^{d,1}={\boldsymbol{T}}_1^{d,2}=\textbf{0}$ by using the above transmission problems and integration by parts. Similarly, it can be shown that ${\boldsymbol{T}}_1^{d,3}=\textbf{0}$ by simplifying the transmission problem for ${\boldsymbol{\psi }}_{ij}$ in a similar way to the above.

In general, we write

(7.4) \begin{align}{\boldsymbol{T}}_1 & = -{\mathrm{i}} k{\boldsymbol{e}}_j (\nabla _x G_k({\boldsymbol{x}},{\boldsymbol{z}}))_p \varepsilon _{jpr} ({\mathcal A}_{ri} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +{\mathcal B}_{ri}({\boldsymbol{E}_0}({\boldsymbol{z}}))_i ) \nonumber \\[5pt] &\qquad +{\boldsymbol{e}}_j (({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}})) _{\ell m}{\mathcal P}_{\ell m ji } ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +\boldsymbol{R}({\boldsymbol{x}}), \end{align}

where

(7.5a) \begin{align} {\mathcal A}_{ri} & \;:\!=\; \frac{{\mathrm{i}} k \alpha ^4 (\epsilon _r -1)}{2}{\boldsymbol{e}}_r \cdot \int _{B} \!\left ({\boldsymbol{e}}_i \times{\boldsymbol{\xi }} +{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }}, \end{align}

(7.5b) \begin{align} {\mathcal B}_{ri} &\;:\!=\; \alpha ^3 (\epsilon _r -1){\boldsymbol{e}}_r\cdot \int _B \!\left ({\boldsymbol{e}}_i +{\boldsymbol{\phi }}_i \right ) \mathrm{d}{\boldsymbol{\xi }}, \end{align}

(7.5c) \begin{align} {\mathcal P}_{\ell m ji} & \;:\!=\; -\frac{ k^2 \alpha ^5 (\epsilon _r -1)}{2}{\boldsymbol{e}}_j \cdot \int _B{\boldsymbol{e}}_{\ell } \times (\xi _m \!\left ({\boldsymbol{e}}_i \times{\boldsymbol{\xi }} +{\boldsymbol{\theta }}_i \right )) \mathrm{d}{\boldsymbol{\xi }} \nonumber \\[5pt] & \ = \varepsilon _{j \ell s}{\mathcal C}_{m si}, \end{align}

(7.5d) \begin{align} {\mathcal C}_{msi}& \;:\!=\; \frac{1}{2} \varepsilon _{sj\ell }{\mathcal P}_{\ell m ji} =-\frac{k^2 \alpha ^5 (\epsilon _r -1)}{2}{\boldsymbol{e}}_s \cdot \int _B \xi _m \!\left ({\boldsymbol{e}}_i \times{\boldsymbol{\xi }} +{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }}, \end{align}

and the latter follows by considering the skew symmetry of $\mathcal P$ w.r.t indices $j$ and $\ell$ and applying similar arguments to [5]. Further simplifications are also possible if the eddy current approximation applies [5], and we shall consider these later in Section 8.

Approximation of ${\boldsymbol{T}}_2$

Let ${\boldsymbol{T}}_2 = (\mu _r -1)({\boldsymbol{T}}_2^a +{\boldsymbol{T}}_2^b +{\boldsymbol{T}}_2^c+{\boldsymbol{T}}_2^d)$ where

\begin{align*}{\boldsymbol{T}}_2^a & \;:\!=\; \int _{B_\alpha } ({\boldsymbol{D}}_x^2 G_k ({\boldsymbol{x}},{\boldsymbol{y}})+k^2 G_k({\boldsymbol{x}},{\boldsymbol{y}}){\mathbb I}) \!\left ({\boldsymbol{H}}_\alpha ({\boldsymbol{y}}) - \mu _r^{-1}{\boldsymbol{H}_0}({\boldsymbol{y}}) - \mu _r^{-1}{\boldsymbol{H}_0}^* \!\left ( \frac{{\boldsymbol{y}} -{\boldsymbol{z}}}{\alpha }\right ) \right )\! \mathrm{d}{\boldsymbol{y}}, \\[5pt] {\boldsymbol{T}}_2^b & \;:\!=\; \mu _r^{-1} \int _{B_\alpha } ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{y}}) -{\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}} )+k^2(G_k({\boldsymbol{x}},{\boldsymbol{y}})-G_k({\boldsymbol{x}},{\boldsymbol{z}})){\mathbb I}) \!\left ({\boldsymbol{H}_0}({\boldsymbol{y}}) +{\boldsymbol{H}_0}^* \!\left ( \frac{{\boldsymbol{y}} -{\boldsymbol{z}}}{\alpha }\right ) \right )\!\mathrm{d}{\boldsymbol{y}}, \\[5pt] {\boldsymbol{T}}_2^c & \;:\!=\; \mu _r^{-1} \int _{B_\alpha } ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) +k^2G_k( \boldsymbol{v},\boldsymbol{z}){\mathbb I} )\!\left ({\boldsymbol{H}_0}({\boldsymbol{y}}) -{\boldsymbol{H}_0} ({\boldsymbol{z}} ) \right )\! \mathrm{d}{\boldsymbol{y}}, \\[5pt] {\boldsymbol{T}}_2^d & \;:\!=\; \mu _r^{-1} \int _{B_\alpha } ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}})+k^2G_k({\boldsymbol{x}},{\boldsymbol{z}}){\mathbb I}) \!\left ({\boldsymbol{H}_0}({\boldsymbol{z}}) +{\boldsymbol{H}_0}^* \!\left ( \frac{{\boldsymbol{y}} -{\boldsymbol{z}}}{\alpha }\right ) \right )\! \mathrm{d}{\boldsymbol{y}}, \end{align*}

and ${\boldsymbol{H}_0}^*({\boldsymbol{\xi }}) = \frac{1}{{\mathrm{i}} k} \nabla \times{\boldsymbol{w}}_0$ . Following similar arguments to the above, and in [2], we have

\begin{align*} |{\boldsymbol{T}}_2^a | & \leqslant C \alpha ^4 \|{\boldsymbol{H}_0} \|_{W^{2,\infty }(B_\alpha )}, \\[5pt] |{\boldsymbol{T}}_2^b | & \leqslant C \alpha ^4 \|{\boldsymbol{H}_0} \|_{W^{2,\infty }(B_\alpha )}, \\[5pt] |{\boldsymbol{T}}_2^c | & \leqslant C \alpha ^4 \|{\boldsymbol{H}_0} \|_{W^{2,\infty }(B_\alpha )}, \end{align*}

and we can express ${\boldsymbol{T}}_2^d$ as:

\begin{align*}{\boldsymbol{T}}_2^d & = \mu _r^{-1}\alpha ^3 \!\left ( \sum _{i=1}^3 ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) +k^2G_k({\boldsymbol{x}},{\boldsymbol{z}}){\mathbb I})\int _{B} \!\left ({\boldsymbol{e}}_i + \frac{1}{2} \nabla \times{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i \right ) \\[5pt] &\qquad + \mu _r^{-1}\alpha ^4 \!\left ( \sum _{i,j=1}^3 ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) +k^2G_k({\boldsymbol{x}},{\boldsymbol{z}}){\mathbb I})\int _{B} \!\left (\xi _j{\boldsymbol{e}}_i + \frac{1}{3} \nabla \times{\boldsymbol{\psi }}_{ij} \right ) \mathrm{d}{\boldsymbol{\xi }} ({\boldsymbol{D}}_z({\boldsymbol{H}_0}({\boldsymbol{z}})))_{ij} \right ) \\[5pt] & = \mu _r^{-1}\alpha ^3 \sum _{i=1}^3 ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) +k^2G_k({\boldsymbol{x}},{\boldsymbol{z}}){\mathbb I})\int _{B} \!\left ({\boldsymbol{e}}_i + \frac{1}{2} \nabla \times{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +\boldsymbol{R}({\boldsymbol{x}}). \end{align*}

So that

(7.6) \begin{align}{\boldsymbol{T}}_2 = ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) +k^2G_k({\boldsymbol{x}},{\boldsymbol{z}}) {\mathbb I}){\boldsymbol{e}}_j{\mathcal N}_{ji} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +\boldsymbol{R}({\boldsymbol{x}}), \end{align}

where

(7.7) \begin{align}{\mathcal N}_{ji} \;:\!=\; \alpha ^3 (1-\mu _r^{-1}){\boldsymbol{e}}_j \cdot \int _B \!\left ({\boldsymbol{e}}_i + \frac{1}{2} \nabla \times{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }}. \end{align}

8. Simplifications and alternative forms

Lemma 8.1. Using ${\boldsymbol{\phi }}_i = (\epsilon _r-1) \nabla \vartheta _i$ where $\vartheta _i$ solves ( 5.31 ) then ${\mathcal B}_{ri}$ becomes

(8.1) \begin{align}{\mathcal B}_{ri} = \alpha ^3 \!\left ( (\epsilon _r-1) |B| \delta _{ri} + (\epsilon _r-1)^2 \int _B{\boldsymbol{e}}_r \cdot \nabla \vartheta _i \mathrm{d}{\boldsymbol{\xi }} \right ), \end{align}

which are also the coefficients of the symmetric Póyla–Szegö tensor ${\mathcal T}[\alpha B, \epsilon _r]$ of an object $B_\alpha$ for a contrast $\epsilon _r$ .

Proof. The proof directly follows from (7.5b) and (5.31).

Lemma 8.2. Up to a residual term, ${\mathcal C}_{msi}$ is skew-symmetric w.r.t. $m$ and $i$ and hence

(8.2) \begin{align}{\mathcal C}_{msi} = \varepsilon _{msr} \check{{\mathcal C}}_{ri}+ R_{msi}, \qquad \check{{\mathcal C}}_{ri}\;:\!=\;- \frac{ \nu \alpha ^3}{4}{\boldsymbol{e}}_r \cdot \int _B{\boldsymbol{\xi }} \times ({\boldsymbol{\theta }}_i +{\boldsymbol{e}}_i \times{\boldsymbol{\xi }} ) \mathrm{d}{\boldsymbol{\xi }}, \end{align}

where $R_{msi}=O(\alpha ^4 )$ for fixed $k$ as $\alpha \to 0$ .

Proof. We first write the transmission problem for ${\boldsymbol \theta }_i$ in an alternative form (5.28):

(8.3a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i -k^2 \alpha ^2 \epsilon _r{\boldsymbol{\theta }}_i & = -k^2\alpha ^2 (1-\epsilon _r){\boldsymbol{e}}_i\times{\boldsymbol{\xi }} && \text{in $B$}, \end{align}

(8.3b) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{\theta }}_i -k^2\alpha ^2{\boldsymbol{\theta }}_i & = \textbf{0} && \text{in $\Omega _0 \setminus \overline{B}$}, \end{align}

(8.3c) \begin{align} \nabla _\xi \cdot{\boldsymbol{\theta }}_i & = 0 && \text{in $\Omega _0$}, \end{align}

(8.3d) \begin{align} \nabla _\xi \times \nabla _\xi \times{\boldsymbol{\theta }}_i -k^2\alpha ^2{\boldsymbol{\theta }}_i & = \textbf{0} && \text{in ${\mathbb R}^3 \setminus \overline{\Omega _0}$}, \end{align}

(8.3e) \begin{align} \nabla _\xi \cdot{\boldsymbol{\theta }}_i & = 0 && \text{in ${\mathbb R}^3 \setminus \overline{\Omega _0}$}, \end{align}

(8.3f) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{\theta }}_i ]_\Gamma =\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i ]_\Gamma & = -2 (1-\mu _r^{-1}){\boldsymbol{n}} \times{\boldsymbol{e}}_i && \text{on $\Gamma $}, \end{align}

(8.3g) \begin{align} [{\boldsymbol{n}} \times{\boldsymbol{\theta }}_i ]_{\Sigma _0} =\textbf{0}, \qquad [{\boldsymbol{n}} \times \nabla _\xi \times{\boldsymbol{\theta }}_i ]_{\Sigma _0} & = {\boldsymbol 0} && \text{on $\Sigma _0$}, \end{align}

(8.3h) \begin{align} \lim _{\xi \to \infty } \xi \!\left ( ( \nabla _\xi \times{\boldsymbol{\theta }}_i ) \times \hat{{\boldsymbol{\xi }}} -{\mathrm{i}} k \alpha{\boldsymbol{\theta }}_i \right )& = \textbf{0}, \end{align}

where $\Omega _0$ and $\Sigma _0$ are introduced in a similar way to $\Omega$ and $\Sigma$ in Section 3.2. Then, an integration by parts can be applied to (7.5) in a similar manner to the proof of Lemma 4.2 in [5] leading to

(8.4) \begin{align} -\frac{2C_{msi}}{\alpha ^3} & = \int _B{\boldsymbol{e}}_m \times{\boldsymbol{e}}_s \cdot \mu _r^{-1}\nabla \times{\boldsymbol \theta }_i \mathrm{d}{\boldsymbol \xi } - \alpha ^2 k^2 \int _B{\boldsymbol \theta }_i \cdot{\boldsymbol{e}}_s \xi _m \mathrm{d}{\boldsymbol \xi } \nonumber \\[5pt] &\qquad -2 [\tilde{\mu _r}^{-1} ]_\Gamma \int _B{\boldsymbol{e}}_i \cdot{\boldsymbol{e}}_m \times{\boldsymbol{e}}_s \mathrm{d}{\boldsymbol \xi } - \int _\Gamma{\boldsymbol{n}}^+ \cdot \nabla \times{\boldsymbol \theta }_i \times ( \xi _m{\boldsymbol{e}}_s )|_+ \mathrm{d}{\boldsymbol \xi }. \end{align}

Also,

(8.5) \begin{align} \int _\Gamma{\boldsymbol{n}}^+ \cdot \nabla \times{\boldsymbol \theta }_i \times ( \xi _m{\boldsymbol{e}}_s ) |_+ \mathrm{d}{\boldsymbol \xi } & = - \int _{\Sigma _0} ({\boldsymbol{n}}^- \times ( \xi _m{\boldsymbol{e}}_s )) \times{\boldsymbol{n}}^- \cdot{\boldsymbol{n}}^- \times \nabla \times{\boldsymbol \theta }_i |_- \mathrm{d}{\boldsymbol \xi } \nonumber \\[5pt] & \qquad + \int _{\Omega _0 \setminus \overline{B}} \!\left ( \alpha ^2 k^2{\boldsymbol \theta }_i \cdot{\boldsymbol{e}}_s \xi _m -\nabla \times{\boldsymbol \theta }_i \cdot \nabla \times (\xi _m{\boldsymbol{e}}_s ) \right ) \mathrm{d}{\boldsymbol \xi } \nonumber \\[5pt] & = -{\mathrm{i}} k \alpha \int _{\Sigma _0} ({\boldsymbol{n}}^- \times ( \xi _m{\boldsymbol{e}}_s )) \times{\boldsymbol{n}}^- \cdot \Lambda _e^{( \alpha k)} ({\boldsymbol{n}}^- \times{\boldsymbol \theta }_i) \mathrm{d}{\boldsymbol \xi } \nonumber \\[5pt] & \qquad + \int _{\Omega _0 \setminus \overline{B}} \!\left ( \alpha ^2 k^2{\boldsymbol \theta }_i \cdot{\boldsymbol{e}}_s \xi _m \mathrm{d}{\boldsymbol \xi } -\nabla \times{\boldsymbol \theta }_i \cdot \nabla \times (\xi _m{\boldsymbol{e}}_s ) \right ) \mathrm{d}{\boldsymbol \xi }. \end{align}

Thus, we find (8.2) holds with

(8.6) \begin{equation} R_{msi} = \frac{{\mathrm{i}} \alpha ^4 k}{2} \int _{\Sigma _0} ({\boldsymbol{n}}^- \times ( \xi _m{\boldsymbol{e}}_s )) \times{\boldsymbol{n}}^- \cdot \Lambda _e^{( \alpha k)} ({\boldsymbol{n}}^- \times{\boldsymbol \theta }_i) \mathrm{d}{\boldsymbol \xi } - \frac{\alpha ^5 k^2}{2}{\boldsymbol{e}}_s \cdot \int _{\Omega _0} \xi _m{\boldsymbol{\theta }}_i \mathrm{d}{\boldsymbol \xi }, \end{equation}

which is not skew-symmetric with respect to $m$ and $i$ , and consequently we see the estimate holds.

Corollary 8.3. If $\nu$ reduces to $\nu _{\mathrm{i}}$ , the transmission problem for ${\boldsymbol{\theta }}_i({\boldsymbol{\xi }})$ provided in ( 5.28 ) reduces to ( 7.2 ) and $R_{msi}=0$ , since this is now identical to the case considered in Lemma 4.2 of [5].

By differentiation, it is easily established that

\begin{align*} \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{z}}) & = \frac{({\boldsymbol{x}}-{\boldsymbol{z}})}{4\pi |{\boldsymbol{x}} -{\boldsymbol{z}}|^2 } \!\left ({\mathrm{i}} k - \frac{1}{ |{\boldsymbol{x}} -{\boldsymbol{z}}| } \right ) e^{{\mathrm{i}} k |{\boldsymbol{x}} -{\boldsymbol{z}}|}, \\[5pt] {\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) & = \frac{1}{4\pi } \!\left ( \frac{1}{ |{\boldsymbol{x}} -{\boldsymbol{z}}| ^3} \!\left (\frac{3 ({\boldsymbol{x}}-{\boldsymbol{z}}) \otimes ({\boldsymbol{x}}-{\boldsymbol{z}}) }{|{\boldsymbol{x}} -{\boldsymbol{z}}|^2 }- \mathbb{I} \right ) -\frac{{\mathrm{i}} k}{ |{\boldsymbol{x}} -{\boldsymbol{z}}| ^2 } \!\left ( \frac{3 ({\boldsymbol{x}}-{\boldsymbol{z}}) \otimes ({\boldsymbol{x}}-{\boldsymbol{z}})}{ |{\boldsymbol{x}} -{\boldsymbol{z}}| ^2} -{\mathbb I}\right ) \right. \\[5pt] &\qquad \left. -\frac{k^2}{ |{\boldsymbol{x}} -{\boldsymbol{z}}|^3 } ({\boldsymbol{x}}-{\boldsymbol{z}}) \otimes ({\boldsymbol{x}}-{\boldsymbol{z}}) \right ) e^{{\mathrm{i}} k |{\boldsymbol{x}} -{\boldsymbol{z}}|}, \end{align*}

and, by introducing ${\boldsymbol{r}}\;:\!=\;{\boldsymbol{x}} -{\boldsymbol{z}}$ , $r=|{\boldsymbol{r}}|$ and $\hat{{\boldsymbol{r}}}={\boldsymbol{r}}/r$ , these derivatives can be expressed as:

\begin{align*} \nabla _x G_k({\boldsymbol{x}},{\boldsymbol{z}}) & = \frac{\hat{{\boldsymbol{r}}}}{4\pi r } \!\left ({\mathrm{i}} k - \frac{1}{ r} \right ) e^{{\mathrm{i}} k r}, \\[5pt] {\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) & = \frac{1}{4\pi } \!\left ( \frac{1}{r^3 } (3 \hat{{\boldsymbol{r}}} \otimes \hat{{\boldsymbol{r}}} - \mathbb{I} ) - \frac{{\mathrm{i}} k}{ r^2} ( 3 \hat{{\boldsymbol{r}}} \otimes \hat{{\boldsymbol{r}}} -{\mathbb I} ) -\frac{k^2}{r} \hat{{\boldsymbol{r}}} \otimes \hat{{\boldsymbol{r}}} \right ) e^{{\mathrm{i}} k r}. \end{align*}

Noting that ${\mathcal C}_{msi} = \varepsilon _{msr} \check{{\mathcal C}}_{ri} +R_{msi}$ by Lemma 8.2, we get

\begin{align*} ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) )_{\ell m } \varepsilon _{j\ell s} \varepsilon _{msr} \check{{\mathcal C}}_{ri} & = - ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) )_{\ell m } \varepsilon _{s j\ell } \varepsilon _{smr} \check{{\mathcal C}}_{ri} \\[5pt] & = - ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) )_{\ell m } ( \delta _{jm} \delta _{\ell r} - \delta _{jr} \delta _{\ell m} ) \check{{\mathcal C}}_{ri} \\[5pt] &= (- ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) )_{r j } + ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) )_{m m }\delta _{jr}) \check{{\mathcal C}}_{ri} \\[5pt] &= -( ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) )_{jr } +k^2 \delta _{jr}G_k({\boldsymbol{x}},{\boldsymbol{z}})) \check{{\mathcal C}}_{ri}. \end{align*}

Furthermore, we introduce the rank 2 tensor $\mathcal M$ with coefficients ${\mathcal M}_{ri} \;:\!=\;{\mathcal N}_{ri } - \check{{\mathcal C}}_{ri}$ , which is symmetric as the following lemma shows:

Lemma 8.4. The tensor ${\mathcal M}={\mathcal N}-\check{\mathcal C}$ is complex symmetric with coefficients satisfying ${\mathcal M}_{ri} ={\mathcal M}_{ir}$ .

Proof. By an application of an integration by parts to $\check{C}_{ri}$ , we find

\begin{align*} \frac{-\check{{\mathcal C}}_{ri}}{\alpha ^3} =& \frac{1}{4 \nu }\int _B \nabla \times \mu _r^{-1} \nabla \times{\boldsymbol{\theta }}_i \cdot \nabla \times \mu _r^{-1} \nabla \times{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} \\[5pt] &-\frac{1}{4} \!\left (\frac{\epsilon _r}{\epsilon _r-1} -1 \right ) \!\left ( \int _{\Omega _0 } \tilde{\mu }_r^{-1} \nabla \times{\boldsymbol{\theta }}_i \cdot \nabla \times{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} -\int _{\Omega _0 \setminus \overline{B}} \alpha ^2 k^2{\boldsymbol{\theta }}_i \cdot{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} - 2[\tilde{\mu }_r^{-1}]\int _B{\boldsymbol{e}}_r \cdot \nabla \times{\boldsymbol{\theta }}_i \mathrm{d}{\boldsymbol{\xi }} \right. \\[5pt] &\left. - \int _{\Sigma _0}{\boldsymbol \theta }_r \cdot{\boldsymbol{n}}^- \times \nabla \times{\boldsymbol \theta }_i |_-\mathrm{d}{\boldsymbol \xi } \right ) \\[5pt] &-\frac{1}{4} \!\left (\frac{\epsilon _r}{\epsilon _r-1} \right ) \!\left ( \int _{\Omega _0 } \tilde{\mu }_r^{-1} \nabla \times{\boldsymbol{\theta }}_i \cdot \nabla \times{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} -\int _{\Omega _0 \setminus \overline{B}} \alpha ^2 k^2{\boldsymbol{\theta }}_i \cdot{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} - 2[\tilde{\mu }_r^{-1}]\int _B{\boldsymbol{e}}_i \cdot \nabla \times{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} \right. \\[5pt] &\left. - \int _{\Sigma _0}{\boldsymbol \theta }_i \cdot{\boldsymbol{n}}^- \times \nabla \times{\boldsymbol \theta }_r |_-\mathrm{d}{\boldsymbol \xi } \right ) \\[5pt] &+\frac{1}{4} \!\left (\frac{\epsilon _r}{\epsilon _r-1} -1 \right ) \!\left ( \int _{B} \alpha ^2 k^2 \epsilon _r{\boldsymbol{\theta }}_i \cdot{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} \right ), \end{align*}

so that

(8.7) \begin{align} &\frac{{\mathcal N}_{ri} -\check{{\mathcal C}}_{r i}}{\alpha ^3} = \frac{1}{4 \nu }\int _B \nabla \times \mu _r^{-1} \nabla \times{\boldsymbol{\theta }}_i \cdot \nabla \times \mu _r^{-1} \nabla \times{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} + \nonumber \\[5pt] & \frac{1}{4} \int _{\Omega _0} \tilde{\mu }_r^{-1} \nabla \times{\boldsymbol{\theta }}_i \cdot \nabla \times{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} + [\tilde{\mu }_r^{-1}]_\Gamma \int _B \delta _{r i} \mathrm{d}{\boldsymbol{\xi }} - \frac{\alpha ^2 k^2}{4} \int _{\Omega _0 }\tilde{\epsilon }_r{\boldsymbol{\theta }}_i \cdot{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} - \frac{1}{4} \int _{\Sigma _0}{\boldsymbol \theta }_i \cdot{\boldsymbol{n}}^- \times \nabla \times{\boldsymbol \theta }_r |_- \mathrm{d}{\boldsymbol \xi } \nonumber \\[5pt] &-\frac{1}{2} \!\left (\frac{\epsilon _r}{\epsilon _r-1} \right ) \!\left ( \int _{\Omega _0 } \tilde{\mu }_r^{-1} \nabla \times{\boldsymbol{\theta }}_i \cdot \nabla \times{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} -\int _{\Omega _0 \setminus \overline{B}} \alpha ^2 k^2{\boldsymbol{\theta }}_i \cdot{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} - [\tilde{\mu }_r^{-1}]_\Gamma \int _B ({\boldsymbol{e}}_r \cdot \nabla \times{\boldsymbol{\theta }}_i +{\boldsymbol{e}}_i \cdot \nabla \times{\boldsymbol{\theta }}_r) \mathrm{d}{\boldsymbol{\xi }} \right. \nonumber \\[5pt] & \left. - \int _{\Sigma _0}{\boldsymbol \theta }_r \cdot{\boldsymbol{n}}^- \times \nabla \times{\boldsymbol \theta }_i |_- \mathrm{d}{\boldsymbol \xi } - \int _{\Sigma _0}{\boldsymbol \theta }_i \cdot{\boldsymbol{n}}^- \times \nabla \times{\boldsymbol \theta }_r |_- \mathrm{d}{\boldsymbol \xi } \right ) \nonumber \\[5pt] &+\frac{1}{4} \!\left (\frac{\epsilon _r}{\epsilon _r-1} \right ) \!\left ( \int _{B} \alpha ^2 k^2 \epsilon _r{\boldsymbol{\theta }}_i \cdot{\boldsymbol{\theta }}_r \mathrm{d}{\boldsymbol{\xi }} \right ), \end{align}

and, hence, ${\mathcal N}_{ri} -\check{{\mathcal C}}_{ri}$ is symmetric in $r$ and $i$ up to the term $\frac{\alpha ^3 }{4} \int _{\Sigma _0}{\boldsymbol \theta }_i \cdot{\boldsymbol{n}}^- \times \nabla \times{\boldsymbol \theta }_r |_- \mathrm{d}{\boldsymbol \xi } = \frac{{\mathrm{i}} \alpha ^4 k }{4} \int _{\Sigma _0} ({\boldsymbol{n}}^- \times{\boldsymbol \theta }_i) \times{\boldsymbol{n}}^- \cdot \Lambda _e^{(\alpha k)} ({\boldsymbol{n}}^- \times{\boldsymbol \theta }_r) \mathrm{d}{\boldsymbol \xi }$ ; however, we note this remaining term can be absorbed in to $R_{msi}$ .

It follows that an alternative form to (4.1) is

(8.8) \begin{align} ({\boldsymbol{H}}_\Delta ({\boldsymbol{x}})) & = -{\mathrm{i}} k{\boldsymbol{e}}_j (\nabla _x G_k({\boldsymbol{x}},{\boldsymbol{z}}))_p \varepsilon _{jpr} ({\mathcal A}_{ri} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +{\mathcal B}_{ri}({\boldsymbol{E}_0}({\boldsymbol{z}}))_i )\nonumber \\[5pt] & \qquad +{\boldsymbol{e}}_j ( ({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}))_{j r}+ k^2 \delta _{jr} G_k({\boldsymbol{x}},{\boldsymbol{z}})){\mathcal M}_{ri} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +\boldsymbol{R}({\boldsymbol{x}}) \nonumber \\[5pt] & = \frac{e^{{\mathrm{i}} k r}}{4\pi } \!\left \{- \left ( - \frac{k^2}{ r} \hat{{\boldsymbol{r}}} \times ({\mathcal A}{\boldsymbol{H}_0}({\boldsymbol{z}}) +{\mathcal B}{\boldsymbol{E}_0}({\boldsymbol{z}})) -\frac{{\mathrm{i}} k}{ r^2 } \hat{{\boldsymbol{r}}} \times ({\mathcal A}{\boldsymbol{H}_0}({\boldsymbol{z}}) +{\mathcal B}{\boldsymbol{E}_0}({\boldsymbol{z}})) \right ) \right. \nonumber \\[5pt] & \qquad + \frac{1}{r^3} \!\left ( 3 \hat{{\boldsymbol{r}}} \cdot ({\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) ) \hat{{\boldsymbol{r}}} -{\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) \right ) - \frac{{\mathrm{i}} k}{r^2} \!\left ( 3 \hat{{\boldsymbol{r}}} \cdot ({\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) ) \hat{{\boldsymbol{r}}} -{\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) \right ) \nonumber \\[5pt] & \qquad \left. - \frac{k^2}{r} \!\left ( \hat{{\boldsymbol{r}}} \cdot{\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) -{\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) \right ) \right \} +\boldsymbol{R}({\boldsymbol{x}}) \nonumber \\[5pt] &= \frac{e^{{\mathrm{i}} k r}}{4\pi } \!\left \{ \frac{1}{r^3} \!\left ( 3 \hat{{\boldsymbol{r}}} \cdot ({\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) ) \hat{{\boldsymbol{r}}} -{\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) \right ) \right. \nonumber \\[5pt] & \qquad -\frac{{\mathrm{i}} k}{ r^2 } \!\left ( 3 \hat{{\boldsymbol{r}}} \cdot ({\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) ) \hat{{\boldsymbol{r}}} -{\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) - \hat{{\boldsymbol{r}}} \times ({\mathcal A}{\boldsymbol{H}_0}({\boldsymbol{z}}) +{\mathcal B}{\boldsymbol{E}_0}({\boldsymbol{z}})) \right ) \nonumber \\[5pt] & \qquad \left. - \frac{k^2}{r} \!\left ( \hat{{\boldsymbol{r}}} \times \hat{{\boldsymbol{r}} } \times ({\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) ) - \hat{{\boldsymbol{r}}} \times ({\mathcal A}{\boldsymbol{H}_0}({\boldsymbol{z}}) +{\mathcal B}{\boldsymbol{E}_0}({\boldsymbol{z}})) \right ) \right \} +\boldsymbol{R}({\boldsymbol{x}}), \end{align}

where we have assumed that $({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) )_{\ell m } \varepsilon _{j\ell s} R_{msi} ({\boldsymbol{H}}_0 ({\boldsymbol{z}}))_i$ can be grouped with $\boldsymbol{R}({\boldsymbol{x}})$ and have applied $\hat{{\boldsymbol{r}}} \times (\hat{{\boldsymbol{r}}} \times{\boldsymbol{a}} ) =\hat{{\boldsymbol{r}}} (\hat{{\boldsymbol{r}}} \cdot{\boldsymbol{a}}) -{\boldsymbol{a}} (\hat{{\boldsymbol{r}}} \cdot \hat{{\boldsymbol{r}}} ) = \hat{{\boldsymbol{r}}} (\hat{{\boldsymbol{r}}} \cdot{\boldsymbol{a}}) -{\boldsymbol{a}}$ .

Remark 8.5. Expression (8.8) bears close resemblance to the result in Theorem 4.1 of [6], which describes the field perturbation for wave-dominated problems, where, $\mathcal B$ is the Póyla–Szegö tensor parameterised by $\epsilon _r$ , as expected, but, instead of a Póyla–Szegö tensor parameterised by $\mu _r$ we have the complex symmetric rank 2 tensor $\mathcal M$ , which depends on $\mu _r$ and $\nu$ . Furthermore, we have the additional term ${\mathcal A}{\boldsymbol{H}_0}({\boldsymbol{z}})$ . We explore this connection further in later sections.

8.1 Quasi-static regime

Assuming that $\alpha \ll \lambda _0$ and noting that $({\boldsymbol{D}}_x^2 G_k({\boldsymbol{x}},{\boldsymbol{z}}) )_{jr } +k^2 \delta _{jr }G_k({\boldsymbol{x}},{\boldsymbol{z}}) = ({\boldsymbol{D}}_x^2 G_0({\boldsymbol{x}},{\boldsymbol{z}}) )_{jr }+O(k)$ and $\nabla _x G_k({\boldsymbol{x}},{\boldsymbol{z}}) = \nabla _x G_0({\boldsymbol{x}},{\boldsymbol{z}}) +O(k)$ as $k\to 0$ then expression (8.8) reduces to

\begin{align*} ({\boldsymbol{H}}_\Delta ({\boldsymbol{x}})) &= -{\mathrm{i}} k{\boldsymbol{e}}_j (\nabla _x G_0({\boldsymbol{x}},{\boldsymbol{z}}))_p \varepsilon _{jpr} ({\mathcal A}_{ri} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +{\mathcal B}_{ri}({\boldsymbol{E}_0}({\boldsymbol{z}}))_i )\\[5pt] &\qquad +{\boldsymbol{e}}_j ({\boldsymbol{D}}_x^2 G_0({\boldsymbol{x}},{\boldsymbol{z}}))_{j r}{\mathcal M}_{ri} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +\boldsymbol{R}({\boldsymbol{x}}), \end{align*}

and, in the near field, the dominant response is

(8.9) \begin{align} ({\boldsymbol{H}}_\Delta ({\boldsymbol{x}})) & \approx {\boldsymbol{e}}_j ({\boldsymbol{D}}_x^2 G_0({\boldsymbol{x}},{\boldsymbol{z}}))_{j r}{\mathcal M}_{ri } ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i, \end{align}

where the coefficients of ${\mathcal M}={\mathcal N}-\check{\mathcal C}$ depend on $\epsilon _{\ast}$ , $\mu _{\ast}$ , $\sigma _{\ast}$ , $\alpha$ and $\omega$ and are obtained from Lemma 8.2 and (7.7) using (5.28). The expression (8.9) describes the magnetic field perturbation in terms of a complex symmetric rank 2 tensor $\mathcal M$ , extending the MPT characterisation of objects considered in [5, 9] for the eddy current problem, to the regime where $\alpha \ll \lambda _0$ , but without requiring $\sigma _{\ast}\gg \epsilon _{\ast} \omega$ .

8.2 Eddy current regime

The eddy current regime is a low-frequency approximation of the Maxwell system where, in addition to the quasi-static approximation, we have $\sigma _{\ast}\gg \epsilon _{\ast} \omega$ and $\epsilon _{\ast}= \epsilon _0$ so that $\nu _{\mathrm{i}} = O(1)$ and displacement currents are neglected. Hence, (5.28) reduces to (7.2) and (5.30) to (7.3). Then, by an application of an integration by parts on ${\mathcal A}_{ri}$ and ${\mathcal B}_{ri}$ in (7.5a) and (7.5b) gives ${\mathcal A}_{ri}=0$ and ${\mathcal B}_{ri}=0$ . Furthermore, by Corollary 8.3, ${\mathcal C}_{msi} = \varepsilon _{msr} \check{{\mathcal C}}_{ri}$ with $R_{msi}=0$ in this case and ${\mathcal M}={\mathcal N}-\check{\mathcal C}$ becomes the MPT discussed in [5, 9]. Hence,

(8.10) \begin{align} ({\boldsymbol{H}}_\Delta ({\boldsymbol{x}})) &= {\boldsymbol{e}}_j ({\boldsymbol{D}}_x^2 G_0({\boldsymbol{x}},{\boldsymbol{z}}) )_{jr }{\mathcal M}_{ri} ({\boldsymbol{H}_0}({\boldsymbol{z}}))_i +\boldsymbol{R}({\boldsymbol{x}}) \nonumber \\[5pt] &= \frac{1}{4 \pi r^3} \!\left ( 3 \hat{{\boldsymbol{r}}} \cdot ({\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) ) \hat{{\boldsymbol{r}}} -{\mathcal M}{\boldsymbol{H}_0}({\boldsymbol{z}}) \right ) +\boldsymbol{R}({\boldsymbol{x}}), \end{align}

where the coefficients of $\mathcal M$ no longer depend on $\epsilon _{\ast}$ and are obtained using (7.2) instead of (5.28).

Remark 8.6. In practice, all metals have $\sigma _{\ast}\gg \epsilon _0 \omega$ over all the frequencies for which $\sigma _{\ast}$ can be regarded as a constant so that $\epsilon _r \approx \epsilon _r-1 \approx \sigma _{\ast}/({\mathrm{i}} \epsilon _0 \omega )$ holds over these frequencies and, for sufficiently small objects, these also extend to all the frequencies for which the quasi-static approximation holds. This means that the coefficients of $\mathcal M$ obtained for the eddy current model using (7.2) are almost indistinguishable from those obtained using (5.28) provided that $\sigma _{\ast}\gg \epsilon _0 \omega$ and the object is sufficiently small. As an example, we compare the diagonal coefficients of $\mathcal M$ obtained using the two models using the analytical solution of Wait [13] in Figure 1. Of course, by making the eddy current approximation, and using (7.2), then ${\mathcal A}_{ri}=0$ and, using (7.3), ${\mathcal B}_{ri}=0$ indicating that (8.9) is indeed the dominant response for the quasi-static regime for metallic objects, which holds not only in the near field.

8.3 Regime with non-constant parameters

In this regime, we no longer assume that the parameters are constant and relax the assumption of $\nu =O(1)$ . First, we consider the case of $\sigma _{\ast}=0$ and require the other parameters ( $\epsilon _r=\epsilon _{\ast}/\epsilon _0$ , $\mu _r$ ) to change so that $k \to 0$ as $\alpha \to 0$ . Second, we require the parameters ( $\epsilon _r=1/\epsilon _0 ( \epsilon _{\ast} + \sigma _{\ast}/({\mathrm{i}} \omega ))$ , $\mu _r$ and $k$ ) to change as $\alpha \to 0$ .

Figure 1. Comparison of the diagonal coefficients of $\mathcal M$ for a conducting sphere with $\mu _r=100$ , $\sigma _{\ast}=1\times 10^6$ S/m and radius $\alpha =0.01$ m for frequencies $1\times 10^1 \leqslant \omega \leqslant 1 \times 10^9$ rad/s for results obtained using the quasi-static model and eddy current approximation.

8.3.1 Small $k$ and $\sigma _{\ast} =0$

We consider small $k$ and $\sigma _{\ast}=0$ so that $\epsilon _r=\epsilon _{\ast}/\epsilon _0$ . In this case, we have ${\boldsymbol{\theta }}_i= \tilde{{\boldsymbol{\theta }}}_i^{(0)} + O(k)$ as $k \to 0$ where $\tilde{{\boldsymbol{\theta }}}_i^{(0)}$ solves

(8.11a) \begin{align} \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times \tilde{{\boldsymbol{\theta }}}_i^{(0)} &= \textbf{0} && \;\text{in $B$}, \end{align}

(8.11b) \begin{align} \nabla _\xi \times \nabla _\xi \times \tilde{{\boldsymbol{\theta }}}_i^{(0)} &= \textbf{0} && \;\text{in $B^c$}, \end{align}

(8.11c) \begin{align} \nabla _\xi \cdot \tilde{{\boldsymbol{\theta }}}_i^{(0)} &= 0 && \;\text{in ${\mathbb R}^3$}, \end{align}

(8.11d) \begin{align} [{\boldsymbol{n}} \times \tilde{{\boldsymbol{\theta }}}_i^{(0)} ]_\Gamma =\textbf{0}, \qquad [{\boldsymbol{n}} \times \tilde{\mu }_r^{-1} \nabla _\xi \times \tilde{{\boldsymbol{\theta }}}_i^{(0)} ]_\Gamma &= -2 (1-\mu _r^{-1}){\boldsymbol{n}} \times{\boldsymbol{e}}_i && \;\text{on $\Gamma $}, \end{align}

(8.11e) \begin{align} \tilde{{\boldsymbol{\theta }}}_i^{(0)} &= O(|{\boldsymbol{\xi }}|^{-1}|) && \;\text{as $|{\boldsymbol{\xi }}| \to \infty $}, \end{align}

which allow us to deduce the following about the coefficients of $\mathcal M$ :

Lemma 8.7. For small $k$ and $\sigma _{\ast}= 0$ , ${\boldsymbol{\theta }}_i = \tilde{{\boldsymbol{\theta }}}_i^{(0)} + O(k)$ as $k \to 0$ and ${\mathcal M}_{ri} ={\mathcal N}_{ri} -\check{{\mathcal C}}_{ri}$ reduces to ${\mathcal M}_{ri} ={\mathcal N}^{(0)}_{ri} +O(k)$ where

(8.12) \begin{align}{\mathcal N}^{(0)}_{r i} = \frac{\alpha ^3 }{4} \int _{B\cup B^c} \tilde{\mu }_r^{-1} \nabla \times \tilde{{\boldsymbol{\theta }}}_i^{(0)} \cdot \nabla \times \tilde{{\boldsymbol{\theta }}}_r^{(0)} \mathrm{d}{\boldsymbol{\xi }} + [\tilde{\mu }_r^{-1}] \int _B \delta _{r i} \mathrm{d}{\boldsymbol{\xi }}, \end{align}

which are also the coefficients of the Póyla–Szegö tensor ${\mathcal T}[\alpha B,\mu _r]$ of an object $B_\alpha$ for a contrast $\mu _r$ as defined, for example, in Lemma 3 of [7].

Proof. For the transmission problem (8.11), we can show that

(8.13) \begin{align} \int _{B\cup B^c} \tilde{\mu }_r^{-1} \nabla \times \tilde{{\boldsymbol{\theta }}}_i^{(0)} \cdot \nabla \times \tilde{{\boldsymbol{\theta }}}_r^{(0)} \mathrm{d}{\boldsymbol{\xi }} = [\tilde{\mu }_r^{-1}]\int _B ({\boldsymbol{e}}_r \cdot \nabla \times \tilde{{\boldsymbol{\theta }}}_i^{(0)} +{\boldsymbol{e}}_i \cdot \nabla \times \tilde{{\boldsymbol{\theta }}}_r^{(0)}) \mathrm{d}{\boldsymbol{\xi }}, \end{align}

which is obtained by integrating by parts $\int _{B\cup B^c} \!\left ( \nabla \times{\tilde{\mu }}_r^{-1} \nabla \times \tilde{{\boldsymbol{\theta }}}_i^{(0)} \cdot \tilde{{\boldsymbol{\theta }}}_r^{(0)} + \nabla \times \tilde{\mu }_r^{-1} \nabla \times \tilde{{\boldsymbol{\theta }}}_r^{(0)} \cdot \tilde{{\boldsymbol{\theta }}}_i^{(0)} \right ) \mathrm{d}{\boldsymbol{\xi }} =0$ . Then, for small $k$ , we have ${\boldsymbol{\theta }}_i= \tilde{{\boldsymbol{\theta }}}_i^{(0)} + O(k)$ and, by letting the solid sphere $\Omega _0$ have radius that tends towards infinity, considering the far field decay of $\tilde{{\boldsymbol{\theta }}}_i^{(0)}$ and using (8.13) in (8.7), we find that ${\mathcal N}_{ri} -\check{{\mathcal C}}_{ri}$ reduces to ${\mathcal N}^{(0)}_{ri}$ , and this is also equivalent to the coefficients of the Póyla–Szegö tensor ${\mathcal T}[\alpha B, \mu _r]$ of an object $B$ for a contrast ${\mu }_r$ , as shown by Theorem 3.2 of [9] and Lemma 3 of [7]. Note that Lemma 3 of [7] considered the question of connectedness of $B$ and showed that this reduction holds independent of the first Betti number of $B$ .

By an integration by parts and then using ${\boldsymbol{\theta }}_i= \tilde{{\boldsymbol{\theta }}}_i^{(0)} + O(k)$ , we get

(8.14) \begin{align}{\mathcal A}_{ri}& \;:\!=\; \frac{{\mathrm{i}} \alpha ^3 }{2\alpha k}{\boldsymbol{e}}_r \cdot \int _{B} \!\left ( \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i - k^2 \alpha ^2{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }} \nonumber \\[5pt] &\ ={-\frac{{\mathrm{i}} \alpha ^4 k }{2}{\boldsymbol{e}}_r \cdot \int _{\Omega _0}{\boldsymbol{\theta }}_i \mathrm{d}{\boldsymbol{\xi }} -\frac{\alpha ^3}{2}{\boldsymbol{e}}_r\cdot \int _{\Sigma _0} \Lambda _{e}^{(\alpha k)} ({\boldsymbol{n}}^- \times{\boldsymbol \theta }_i) \mathrm{d}{\boldsymbol{\xi }} }= O(k), \end{align}

as $k \to 0$ by letting the solid sphere $\Omega _0$ have radius that tends towards infinity and considering the far field decay of $\tilde{{\boldsymbol{\theta }}}_i^{(0)}$ .

Using the above results and Lemma 8.1 means that (8.8) now becomes

(8.15) \begin{align} ({\boldsymbol{H}}_\Delta ({\boldsymbol{x}})) &= \frac{e^{{\mathrm{i}} k r}}{4\pi } \!\left \{ \frac{1}{r^3} \!\left ( 3 \hat{{\boldsymbol{r}}} \cdot ({\mathcal T} [\alpha B, \mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) ) \hat{{\boldsymbol{r}}} -{\mathcal T} [\alpha B, \mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) \right ) \right. \nonumber \\[5pt] & \qquad -\frac{{\mathrm{i}} k}{ r^2 } \!\left ( 3 \hat{{\boldsymbol{r}}} \cdot ({\mathcal T} [\alpha B, \mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) ) \hat{{\boldsymbol{r}}} -{\mathcal T} [\alpha B, \mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) - \hat{{\boldsymbol{r}}} \times \left ({\mathcal T} [\alpha B, \epsilon _r]{\boldsymbol{E}_0}({\boldsymbol{z}}) \right) \right ) \nonumber \\[5pt] & \qquad \left. - \frac{k^2}{r} \!\left ( \hat{{\boldsymbol{r}}} \times \hat{{\boldsymbol{r}} } \times ({\mathcal T} [\alpha B, \mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) ) - \hat{{\boldsymbol{r}}} \times ({\mathcal T} [\alpha B, \epsilon _r]{\boldsymbol{E}_0}({\boldsymbol{z}})) \right ) \right \} +\boldsymbol{R}({\boldsymbol{x}}), \end{align}

which agrees with the known results for low-frequency scattering from dielectric and permeable bodies (e.g. [6]).

8.3.2 Small $\alpha$ and small $\nu$

We now consider the case of small $\alpha$ and require $k,\epsilon _r,\mu _r$ to change with $\alpha$ such that $\nu$ is also small. For small $\alpha, \nu$ then ${\boldsymbol{\theta }}_i= \tilde{{\boldsymbol{\theta }}}_i^{(0)} + O(\alpha )$ as $\alpha \to 0$ where $\tilde{{\boldsymbol{\theta }}}_i^{(0)}$ solves (8.11), which allow us to show the following about the coefficients of $\mathcal M$ :

Lemma 8.8. For small $\alpha$ and $\nu$ , ${\boldsymbol{\theta }}_i={\boldsymbol{\theta }}_i^{(0)} +O(\alpha )$ as $\alpha \to 0$ and ${\mathcal M} _{ri} ={\mathcal N}_{ri} -\check{\mathcal C}_{ri }$ reduces to ${\mathcal M}_{ri} ={\mathcal N}^{(0)}_{ri} +O(\alpha ^5)$ where

(8.16) \begin{align}{\mathcal N} ^{(0)}_{ri} = \frac{\alpha ^3 }{4} \int _{B\cup B^c} \tilde{\mu }_r^{-1} \nabla \times \tilde{{\boldsymbol{\theta }}}_i^{(0)} \cdot \nabla \times \tilde{{\boldsymbol{\theta }}}_r^{(0)} \mathrm{d}{\boldsymbol{\xi }} + [\tilde{\mu }_r^{-1}] \int _B \delta _{ri} \mathrm{d}{\boldsymbol{\xi }}, \end{align}

which are also the coefficients of the Póyla–Szegö tensor ${\mathcal T} [\alpha B, \mu _r]$ of an object $B_\alpha$ for a contrast $\mu _r$ .

Proof. The proof is similar to the proof of Lemma (8.7) and again uses (8.13). For small $\alpha, \nu$ , we have ${\boldsymbol{\theta }}_i= \tilde{{\boldsymbol{\theta }}}_i^{(0)} + O(\alpha )$ as $\alpha \to 0$ and find that ${\mathcal N}_{ri} -\check{{\mathcal C}}_{ri}$ reduces to ${\mathcal N}^{(0)}_{ri}$ .

Similarly, by an integration by parts and then using ${\boldsymbol{\theta }}_i= \tilde{{\boldsymbol{\theta }}}_i^{(0)} + O(\alpha )$ , we get

(8.17) \begin{align}{\mathcal A}_{ri} & \;:\!=\; \frac{{\mathrm{i}} \alpha ^3 }{2\alpha k}{\boldsymbol{e}}_r \cdot \int _{B} \!\left ( \nabla _\xi \times \mu _r^{-1} \nabla _\xi \times{\boldsymbol{\theta }}_i - k^2 \alpha ^2{\boldsymbol{\theta }}_i \right ) \mathrm{d}{\boldsymbol{\xi }} \nonumber \\[5pt] &\ ={-\frac{{\mathrm{i}} \alpha ^4 k }{2}{\boldsymbol{e}}_r \cdot \int _{\Omega _0}{\boldsymbol{\theta }}_i \mathrm{d}{\boldsymbol{\xi }} -\frac{\alpha ^3}{2}{\boldsymbol{e}}_r\cdot \int _{\Sigma _0} \Lambda _{e}^{(\alpha k)} ({\boldsymbol{n}}^- \times{\boldsymbol \theta }_i) \mathrm{d}{\boldsymbol{\xi }} } = O(\alpha ^4), \end{align}

as $\alpha \to 0$ by letting the solid sphere $\Omega _0$ have radius that tends towards infinity and considering the far field decay of $\tilde{{\boldsymbol{\theta }}}_i^{(0)}$ .

Using the above results and Lemma 8.1 means that (8.8) now becomes

(8.18) \begin{align} ({\boldsymbol{H}}_\Delta ({\boldsymbol{x}})) &= \frac{e^{{\mathrm{i}} k r}}{4\pi } \!\left \{ \frac{1}{r^3} \!\left ( 3 \hat{{\boldsymbol{r}}} \cdot ({\mathcal T}[\alpha B, \mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) ) \hat{{\boldsymbol{r}}} -{\mathcal T} [\alpha B,\mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) \right ) \right. \nonumber \\[5pt] & \qquad -\frac{{\mathrm{i}} k}{ r^2 } \!\left ( 3 \hat{{\boldsymbol{r}}} \cdot ({\mathcal T} [\alpha B,\mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) ) \hat{{\boldsymbol{r}}} -{\mathcal T} [\alpha B,\mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) - \hat{{\boldsymbol{r}}} \times ({\mathcal T} [\alpha B,\epsilon _r]{\boldsymbol{E}_0}({\boldsymbol{z}})) \right ) \nonumber \\[5pt] & \qquad \left. - \frac{k^2}{r} \!\left ( \hat{{\boldsymbol{r}}} \times \hat{{\boldsymbol{r}} } \times ({\mathcal T} [\alpha B,\mu _r]{\boldsymbol{H}_0}({\boldsymbol{z}}) ) - \hat{{\boldsymbol{r}}} \times ({\mathcal T} [\alpha B,\epsilon _r]{\boldsymbol{E}_0}({\boldsymbol{z}})) \right ) \right \} +\boldsymbol{R}({\boldsymbol{x}}), \end{align}

which agrees with the known results for scattering from small bodies [3, 4].

Remark 8.9. While the forms for ${\boldsymbol{H}}_\Delta ({\boldsymbol{x}})$ derived in (8.8) and (8.18) look very similar, the former holds for small $k$ and requires $\sigma _{\ast}=0$ and the latter is for the case of small $\alpha$ and small $ \nu$ but does not require $\sigma _{\ast}=0$ .