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 [Reference Monk11]), 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 [Reference Monk11, 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2], 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 [Reference Monk11, 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 [Reference Ammari, Buffa and Nédélec1] 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 [Reference Schmidt, Sterz and Hiptmair12] 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2, Reference Ledger and Lionheart5, Reference Ledger and Lionheart8], 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2, Reference Ledger and Lionheart5, Reference Ledger and Lionheart8], 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2] 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2], 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2] 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, [Reference Monk11, 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2], $\nabla \cdot{\boldsymbol{E}}_\alpha = \nabla \cdot{\boldsymbol{E}_0} =0$ 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2] 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2], 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2], 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2], 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 [Reference Ledger and Lionheart6]:

(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 [Reference Ledger and Lionheart5]. Further simplifications are also possible if the eddy current approximation applies [Reference Ledger and Lionheart5], 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 [Reference Ammari, Chen, Chen, Garnier and Volkov2], 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 [Reference Ledger and Lionheart5] 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.

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:

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 $\hat{{\boldsymbol{r}}} \times (\hat{{\boldsymbol{r}}} \times{\boldsymbol{a}} ) =\hat{{\boldsymbol{r}}} (\hat{{\boldsymbol{r}}} \cdot{\boldsymbol{a}}) -{\boldsymbol{a}} (\hat{{\boldsymbol{r}}} \cdot \hat{{\boldsymbol{r}}} ) = \hat{{\boldsymbol{r}}} (\hat{{\boldsymbol{r}}} \cdot{\boldsymbol{a}}) -{\boldsymbol{a}}$ .

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 [Reference Ledger and Lionheart5, Reference Ledger and Lionheart9] 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 [Reference Ledger and Lionheart5, Reference Ledger and Lionheart9]. 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 [Reference Wait13] 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 [Reference Ledger and Lionheart7].

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 [Reference Ledger and Lionheart9] and Lemma 3 of [Reference Ledger and Lionheart7]. Note that Lemma 3 of [Reference Ledger and Lionheart7] 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. [Reference Ledger and Lionheart6]).

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 [Reference Ammari, Vogelius and Volkov3, Reference Ammari and Volkov4].

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 $\sigma _{\ast}=0$ .