2. Application to the optical imaging using plasmonic contrast agents

We develop a quantitative imaging procedure that can reconstruct the permittivity function $\epsilon _{0}({\cdot})$ within the bounded medium $\Omega$ , to be imaged, using the expansion given in Theorem1.1, formula (1.14). To do this, we start by recalling (1.14) that

\begin{align*} & \langle ( E^{\infty } - V^{\infty } )({-} \theta , \theta , q, \omega ) , ( \theta \times q ) \rangle \\ &\quad = - \frac {\mu \, a^{3}}{4 \, \pi } \, \sum _{j=1}^{\aleph } \, \frac {\omega ^{2}_{P_{\ell },n_{0},j} \, \epsilon _{0}(z_{j}) \, \big(\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j} \big) \big)}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \, \left ( \left \langle V (z_{j}, \theta , q ), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \right )^{2} \\ &\qquad + \mathcal{O}\big( a^{\min \left ( (3-s), \left(6-3t-2h-\frac {3s}{2}\right), (4 - h - s) \right )}\big), \end{align*}

which, by keeping only its dominant term, becomes

(2.1) \begin{equation} \left \langle \left ( E^{\infty } - V^{\infty }\right )({-} \theta , \theta , q, \omega ) , \left ( \theta \times q \right ) \right \rangle \simeq - \frac {\mu \, a^{3}}{4 \, \pi \, \lambda _{n_{0}}^{(3)}(B) } \, \sum _{j=1}^{\aleph } \, \mathcal{J}(\omega , z_{j}), \end{equation}

where the imaging functional $\mathcal{J}\left ({\cdot} , \cdot \right )$ , depending on both the frequency parameter $\omega$ and the location points $\{ z_{j} \}_{j=1}^{\aleph }$ , is given by

(2.2) \begin{equation} \mathcal{J}(\omega , z_{j}) \,:\!= \frac {\omega ^{2}_{P_{\ell },n_{0},j} \, \epsilon _{0}(z_{j}) \, \big(\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) \big)}{ \Lambda _{n_{0},j} ( \omega )} \, \left ( \left \langle V (z_{j}, \theta , q ), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \right )^{2}, \end{equation}

with $\Lambda _{n_{0},j} ( \cdot )$ being the dispersion equation given by (1.16). It is evident from the imaging functional expression, see (2.2), that the reconstruction of the permittivity function $\epsilon _{0}({\cdot})$ on the location points, i.e. $\{ \epsilon _{0}(z_{j}) \}_{j=1}^{\aleph }$ , can be derived from the reconstruction of the imaging functional on the same location points, i.e. $\{ \mathcal{J}(\omega , z_{j}) \}^{\aleph }_{j=1}$ , with

\begin{equation*} \omega \in \mathbf{I} \,:\!= \left ( \omega _{0}\, , \, \sqrt {\omega _{0}^{2} + \frac {\omega _{p}^{2}}{\lambda ^{(3)}_{n_{0}}(B)}} \right )\!. \end{equation*}

More details will be provided later in Subsection 3.3. Furthermore, the contrast between the two far fields $V^{\infty }({\cdot})$ and $E^{\infty }({\cdot})$ , given by (2.1), measured in the back-scattered direction, i.e. $\hat {x} = - \, \theta$ , provides us with

(2.3) \begin{equation} \mathcal{F}(\omega , \aleph ) \,:\!= \sum _{j=1}^{\aleph } \, \mathcal{J}(\omega , z_{j}), \end{equation}

up to a known multiplicative constant given by $-\dfrac {\mu \, a^{3}}{4 \, \pi \, \lambda _{n_{0}}^{(3)}(B) }$ . Unfortunately, we cannot determine the $\aleph$ imaging functional $\mathcal{J}(\omega , z_{j})$ , for $1 \leq j \leq \aleph$ , from the mean (sum) of the imaging functional $\mathcal{F}(\omega , \aleph )$ , given by (2.3), as the problem is not uniquely solvable. Hence, the reconstruction of the permittivity function $\epsilon _{0}({\cdot})$ on the location points $\{ z_{j} \}_{j=1}^{\aleph }$ cannot be done, in a straightforward manner, as explained above. In order to solve this issue, we propose employing an iterative method that requires knowledge of $\left \{ \mathcal{F}(\omega , \ell ) \right \}_{\ell = 1}^{\aleph }$ , given by (2.3). In this case, each imaging functional $\mathcal{J}(\omega , z_{j})$ can be obtained as

\begin{align*} \mathcal{J}\left (\omega , z_{1} \right ) &= \mathcal{F}(\omega , 1) = \text{Initial Reconstructed Term}, \\ \mathcal{J}(\omega , z_{j}) &= \mathcal{F}(\omega , j) - \mathcal{F}(\omega , j-1), \quad \text{for} \quad 2 \leq j \leq \aleph . \end{align*}

Based on the explanations mentioned above, our proposed imaging procedures follow as below.

1. (1) Step 1). Reconstructing $\mathcal{J}\left (\omega , z_{1} \right )$ .

   1. (a) Collect the far field before injecting any plasmonic nanoparticle inside $\Omega$ , in the back-scattered direction at a single incident wave $\theta$ , i.e. $V^{\infty }({-} \theta , \theta , q, \omega )$ , where $\omega \in \mathbf{I}$ .

   2. (b) Collect the far field after injecting the first plasmonic nanoparticle $D_{1}$ , in the back-scattered direction at a single incident wave $\theta$ , i.e. $E^{\infty }({-} \theta , \theta , q, \omega )$ , where $\omega \in \mathbf{I}$ .

   At this stage, since we have inserted only one single plasmonic nanoparticle inside $\Omega$ , the equation (2.1) will collapse to the following single term

   \begin{equation*} \left \langle \left ( E^{\infty } - V^{\infty }\right )({-} \theta , \theta , q, \omega ) , \left ( \theta \times q \right ) \right \rangle \simeq -\frac {\mu \, a^{3}}{4 \, \pi \, \lambda _{n_{0}}^{(3)}(B) } \, \mathcal{J}(\omega , z_{1} ), \end{equation*}
   where the left-hand side is a known (measured) term, see for instance (1a) and (1b). In addition, as stated previously, the constant $-\dfrac {\mu \, a^{3}}{4 \, \pi \, \lambda _{n_{0}}^{(3)}(B) }$ appearing on the right-hand side is known. Consequently, we deduce the reconstruction of the first imaging functional
   (2.4) \begin{equation} \omega \longrightarrow \mathcal{J}(\omega , z_{1}), \quad \text{with} \quad \omega \in \mathbf{I}. \end{equation}

2. (2) Step 2). Reconstructing $\epsilon _{0}(z_{1})$ .

   As we have reconstructed the first imaging functional $\mathcal{J}\left (\omega , z_{1} \right )$ , see for instance (2.4), and by recalling its analytical expression given by (2.2),

   \begin{equation*} \mathcal{J}\left (\omega , z_{1} \right ) = \frac {\omega ^{2}_{P_{\ell },n_{0},1} \, \epsilon _{0}(z_{1}) \, \big(\epsilon _{0}(z_{1}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},1}\big) \big) \, \left ( \left \langle V(z_{1}, \theta , q ), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \right )^{2}}{ \Lambda _{n_{0},1} ( \omega )}, \end{equation*}
   which can be rewritten, by introducing the proportional symbol $\propto$ , like
   (2.5) \begin{equation} \mathcal{J} (\omega , z_{1} ) \propto \frac {1}{ \Lambda _{n_{0},1} ( \omega )}, \quad \omega \in \mathbf{I}, \end{equation}
   since, with respect to the parameter $a$ , we know that
   \begin{equation*} \omega ^{2}_{P_{\ell },n_{0},1} \, \epsilon _{0}(z_{1}) \, \big(\epsilon _{0}(z_{1}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},1}\big) \big) \, \left( \left\langle V(z_{1}, \theta , q ), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right\rangle \right )^{2} \sim 1 . \end{equation*}
   Clearly, from (2.5), we observe that the argmax (i.e. the input point at which a function’s output value is maximized) of the function $\mathcal{J}\left ({\cdot} , z_{1} \right )$ is exactly the “quasi-root” of the first dispersion equation $\Lambda _{n_{0},1}\left ( \cdot \right )$ , given by (1.16), and vice versa.Footnote ¹ Here, “quasi-root” is referred to as the root of the real part of the disperison equation $\Lambda _{n_{0},1}\left ( \cdot \right )$ , noting that it contains a small imaginary part. Thus, we can estimate the first plasmonic resonance $\omega _{P_{\ell },n_{0},1}$ . Moreover, by solving (1.16), we deduce
   \begin{equation*} \epsilon _{0}( z_{1}) = \epsilon _{p}\big( \omega _{P_{\ell },n_{0},1}\big) \, \frac {\lambda _{n_{0}}^{(3)}\left (B\right )}{\left ( \lambda _{n_{0}}^{(3)}\left (B\right ) - 1 \right )}. \end{equation*}
   This justifies the reconstruction of $\epsilon _{0}( z_{1})$ .

3. (3) Step 3). Reconstructing $\mathcal{J}(\omega , z_{j})$ and $\epsilon _{0}( z_{j})$ , for $2 \leq j \leq \aleph$ .

   As explained through Step 1, by injecting the first plasmonic nanoparticle $D_{1}$ and measuring the contrast between the far fields that we obtain before injecting $D_{1}$ and after injecting $D_{1}$ , we can reconstruct the first imaging functional $\mathcal{J}\left ({\cdot} , z_{1} \right )$ , see for instance (2.4). After that, we inject the second plasmonic nanoparticle $D_{2}$ and we return to (2.1), to obtain

   \begin{equation*} \langle ( E^{\infty } - V^{\infty })({-} \theta , \theta , q, \omega ) , ( \theta \times q ) \rangle \simeq - \frac {\mu \, a^{3}}{4 \, \pi \, \lambda _{n_{0}}^{(3)}(B) } \, \sum _{j=1}^{2} \, \mathcal{J}(\omega , z_{j}), \end{equation*}
   or, equivalently,
   \begin{equation*} \mathcal{J}\left (\omega , z_{2} \right ) \simeq - \mathcal{J}\left (\omega , z_{1} \right ) - \frac {4 \, \pi \, \lambda _{n_{0}}^{(3)}(B)}{\mu \, a^{3}} \, \left \langle \left ( E^{\infty } - V^{\infty }\right )({-} \theta , \theta , q, \omega ) , \left ( \theta \times q \right ) \right \rangle . \end{equation*}
   The right-hand side of the formula above is already known (measured). Therefore, we deduce the reconstruction of the second imaging functional, i.e.
   (2.6) \begin{equation} \omega \longrightarrow \mathcal{J}(\omega , z_{2} ), \quad \text{with} \quad \omega \in \mathbf{I}. \end{equation}
   Additionally, by using the same arguments as those used in Step (2), we can determine the reconstruction of $\epsilon _{0}(z_{2})$ from $\mathcal{J}\left ({\cdot} , z_{2} \right )$ , see for instance (2.6). Hence, we have reconstructed
   \begin{equation*} (\mathcal{J}({\cdot} , z_{2} );\ \epsilon _{0}(z_{2}) ). \end{equation*}
   Finally, through an iterative process, as explained in Steps (1–3), we can reconstruct
   \begin{equation*} (\mathcal{J}({\cdot} , z_{j} );\ \epsilon _{0}(z_{j}) )_{j=2}^{\aleph }. \end{equation*}

4. (4) Step 4). Reconstructing $\epsilon _{0}({\cdot})$ within $\Omega$ .

   To reconstruct the permittivity function $\epsilon _{0}({\cdot})$ , inside $\Omega$ , or at least approximate it, from its already reconstructed point-wise values $\{ \epsilon _{0}( z_{j} ) \}_{j=1}^{\aleph }$ , we can use the Dual Reciprocity Method. Other interpolation methods can also be used. For a more detailed study of the Dual Reciprocity Method and its application, [Reference Cruse16, Reference Ghassemi, Fazelifar and Nadery31] and [Reference Partridge, Brebbia and Wrobel48] are recommended. In what follows, we provide a detailed explanation for Step 4) to accomplish the imaging procedures.

   We start by assuming that $\epsilon _{0}\left ({\cdot} \right )$ is smooth enough, and we approximate it by a linear combination of a finite number of basis functions, i.e.

   (2.7) \begin{equation} \epsilon _{0}\left ( z \right ) \simeq \sum _{k=1}^{\aleph } \beta _{k} \, f_{k}(z),\quad z \in \Omega , \end{equation}
   with $\left (\beta _{1}, \cdots , \beta _{\aleph } \right ) \in \mathbb{C}^{\aleph }$ being a vector to be determined, and $\left (f_{1}({\cdot}), \cdots , f_{\aleph }({\cdot})\right )$ are suitably chosen basis functions. The accuracy of the approximation (2.7) depends on the choice of $\left \{ f_{k}({\cdot}) \right \}_{k=1}^{\aleph }$ and the number $\aleph$ . The Dual Reciprocity Method has been applied by using various types of basis functions in different literature. Among them, without being exhaustive, we can list

   1. (a) A linear basis function given by $f_{k}({\cdot}) = 1 + \left \vert \cdot - x_{k} \right \vert$ ;

   2. (b) A Gaussian basis function given by $f_{k}({\cdot}) = e^{\left \vert \cdot - x_{k} \right \vert ^{2}}$ ;

   3. (c) A thin plate spline basis function given by $f_{k}({\cdot}) = \left \vert \cdot - x_{k} \right \vert ^{2}\ln \left (\left \vert \cdot - x_{k} \right \vert \right )$ ,

   where the set of points $\left \{ x_{k} \right \}_{k=1}^{\aleph }$ , called collocation points, is generated randomly inside $\Omega$ . Thus, we know the set of points $\left \{ x_{k} \right \}_{k=1}^{\aleph }$ .

   Now, by choosing a type for the basis functions, generating randomly a set of collocations points $\left \{ x_{k} \right \}_{k=1}^{\aleph }$ inside $\Omega$ and evaluating both sides of (2.7) at the a priori assumed known location $\{ z_{j} \}_{j=1}^{\aleph }$ , we obtain the following algebraic system,

   (2.8) \begin{equation} \begin{pmatrix} f_{1}(z_{1}) & \quad \cdots & \quad f_{\aleph }(z_{1}) \\ \vdots & \quad \ddots & \quad \vdots \\ f_{1}(z_{\aleph }) & \quad \cdots & \quad f_{\aleph }(z_{\aleph }) \end{pmatrix} \cdot \begin{pmatrix} \beta _{1} \\ \vdots \\ \beta _{\aleph } \end{pmatrix} = \begin{pmatrix} \epsilon _{0}(z_{1}) \\ \vdots \\ \epsilon _{0}(z_{\aleph }) \end{pmatrix}. \end{equation}
   Thanks to Steps (1–3), the right-hand side of (2.8) is a known vector. Moreover, by its construction, the matrix appearing on the left-hand side of (2.8) can be computed explicitly. Then, by solving (2.8), we can determine the unknown vector $\left (\beta _{1}, \cdots , \beta _{\aleph } \right )$ . The final stage involves inserting the obtained vector into (2.7), to come up with an approximation of the permittivity function $\epsilon _{0}({\cdot})$ inside $\Omega$ .

3. Proof of the main result, i.e. Theorem1.1

We divide this section into four subsections. In the first subsection, we will focus on deriving the algebraic system related to the solution of the electromagnetic problem (1.2). In the second subsection, we derive the dominant term related to the optical electric field and show its dependence on the eigen-system related to the Magnetization operator, the permittivity of the medium to be imaged, i.e $\epsilon _{0}({\cdot})$ and the permittivity of the used plasmonic nanoparticles, i.e. $\epsilon _{p}(\omega )$ . In the third subsection, we shall use the results from the first and the second subsections to estimate the scattered fields connected to the electric field. In the last subsection, we will use the derived result on the scattered fields to estimate the corresponding far-fields. To clear the structure, before moving forward to the main proof, we shall briefly summarize its key steps as follows.

• • Step 1: We derive the algebraic system satisfied by the solution of (1.2). The expression of the algebraic system is given by (3.15).

• • Step 2: On basis of Step 1, under appropriate conditions for the used parameters, we deduce the dominant term related to the solution of (1.2). The formula is given by (3.29).

• • Step 3: We provide the expression of the contrast between the scattered fields (solution of (1.2)), before and after inserting the plasmonic nanoparticles, by using the dominant terms obtained in Step 2. See the expression (3.40).

• • Step 4: From the precise form of the contrast between the scattered fields which has been deduced in Step 3, we formulate the corresponding far-fields contrast representation. This will be achieved after proving the Mixed Electromagnetic Reciprocity Relation, see Proposition 3.1.

3.1. Deriving the algebraic system satisfied by the solution of (1.2)

As shown in [Reference Ghandriche and Sini25, Section 3.2], the solution of (1.2) can be written as the solution to the following Lippmann–Schwinger system of equations

(3.1) \begin{equation} u_{1}(x) + \omega ^{2} \, \int _{D} G_{k}(x,y) \cdot u_{1}(y) \, \big(\boldsymbol{n}_{0}^{2}(y) - \boldsymbol{n}^{2}(y) \big) \, dy = u_{0}(x), \quad x \in \mathbb{R}^{3}, \end{equation}

where $u_0({\cdot}) {\,:\!=V}$ (respectively. $u_1({\cdot}) {\,:\!=E}$ ) denotes the electromagnetic field before (respectively. after) injecting the nanoparticles, $D = \overset {\aleph }{\underset {j=1}\cup } D_{j}$ , with $\aleph \sim a^{-s}$ , $s \gt 0$ , and $G_{k}({\cdot} ,\cdot )$ is solution to (1.10). To give sense of the integral appearing on the left-hand side of (3.1), i.e.

(3.2) \begin{equation} Int(x) \,:\!= \int _{D} G_{k}({x},y) \cdot u_{1}(y) \, \big(\boldsymbol{n}_{0}^{2}(y) - \boldsymbol{n}^{2}(y) \big) \, dy, \quad x \in \mathbb{R}^{3}, \end{equation}

we start by recalling that

(3.3) \begin{equation} G_{k}(x, \cdot ) = \Pi _{k}(x, \cdot ) + {\Gamma _{1}}(x, \cdot ), \end{equation}

see [Reference Ghandriche and Sini25, Subsection 3.2], where

(3.4) \begin{equation} \Pi _{k}(x,y) \,:\!= \frac {1}{k^{2}} \, \underset {y}{\nabla } \, \underset {y}{\nabla } \, \Phi _{k}(x,y) + \Phi _{k}(x,y) \, I_{3}, \end{equation}

and $\Gamma _{1}(x, \cdot ) \in \mathbb{L}^{\frac {3}{2} - \delta }\left ( \Omega \right )$ stands for the remainder term of the Green’s kernel, with $\delta$ being a small positive real number. Then, using (3.3), (3.4) and the definition of both the Magnetization operator and the Newtonian operator in (1.7), (3.2) should be understood as

(3.5) \begin{align} Int(x) &= - \frac {1}{k^{2}} \, \nabla M^{k}_{D}\left ( u_{1}({\cdot}) \, \big(\boldsymbol{n}_{0}^{2}({\cdot}) - \boldsymbol{n}^{2}({\cdot}) \big)\right )(x) + N^{k}_{D}\big( u_{1}({\cdot}) \, \big(\boldsymbol{n}_{0}^{2}({\cdot}) - \boldsymbol{n}^{2}({\cdot}) \big)\big)(x) \nonumber\\ &\quad +\int _{D} {\Gamma _{1}}(x,y) \cdot u_{1}(y) \, \big(\boldsymbol{n}_{0}^{2}(y) - \boldsymbol{n}^{2}(y) \big) \, dy, \quad x \in \mathbb{R}^{3}. \end{align}

It is clear that the first term and the second term on the right-hand side of (3.5) are well defined. In addition, the existence of the third term on the right-hand side of (3.5) can be deduced by using the fact that ${\Gamma _{1}}(x, \cdot ) \in \mathbb{L}^{\frac {3}{2} - \delta }\left ( \Omega \right )$ . This provides an explanation of how (3.2) can be interpreted. Therefore, from now on whenever we use the notation (3.2), it should be understood as (3.5).

Recalling the definition of the index of refraction $\boldsymbol{n}({\cdot})$ , see (1.5), we rewrite (3.1) as

(3.6) \begin{equation} u_{1}(x) + \omega ^{2} \, \mu \, \int _{D} G_{k}(x,y) \cdot u_{1}(y) \, (\epsilon _{0}(y)-\epsilon _{p}(\omega )) \, dy = u_{0}(x), \quad x \in \mathbb{R}^{3}, \end{equation}

where $\epsilon _{p}({\cdot})$ is the given permittivity of the injected nanoparticles, through the Lorentz model, by (1.1). Then, by restricting $x \in D_{m}$ , we rewrite (3.6) as

\begin{align*} u_{1}(x) & + \omega ^{2} \, \mu \, \int _{D_{m}} G_{k}(x,y) \cdot u_{1}(y) \, (\epsilon _{0}(y)-\epsilon _{p}(\omega )) \, dy + \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \int _{D_{j}} G_{k}(x,y) \cdot u_{1}(y) \, (\epsilon _{0}(y)-\epsilon _{p}(\omega )) \, dy \\&\quad = u_{0}(x). \end{align*}

By expanding the function $(\epsilon _{0}({\cdot}) - \epsilon _{p}(\omega ))$ near the centres, we obtain

(3.7) \begin{align} u_{1}(x) &+ \omega ^{2} \, \mu \, (\epsilon _{0}(z_{m})-\epsilon _{p}(\omega )) \, \int _{D_{m}} G_{k}(x,y) \cdot u_{1}(y) \, dy + \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \, (\epsilon _{0}(z_{j})-\epsilon _{p}(\omega )) \, \int _{D_{j}} G_{k}(x,y) \cdot u_{1}(y) \, dy \nonumber\\ &\quad = u_{0}(x) - \omega ^{2} \, \mu \, \int _{D_{m}} G_{k}(x,y) \cdot u_{1}(y) \, \int _{0}^{1} \nabla \epsilon _{0}(z_{m}+t(y-z_{m})) \cdot (y - z_{m}) \, dt \, dy \nonumber \\ &\qquad - \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \, \int _{D_{j}} G_{k}(x,y) \cdot u_{1}(y) \, \int _{0}^{1} \nabla \epsilon _{0}(z_{j}+t(y-z_{j})) \cdot (y - z_{j}) \, dt \, dy. \end{align}

Thanks to [Reference Ghandriche and Sini25, Theorem 2.1, Formula (2.3)], regarding the Green’s kernel $G_{k}({\cdot} ,\cdot )$ , we have the following expansion

(3.8) \begin{equation} G_{k}(x,y) = \frac {1}{\omega ^{2} \, \mu \, \epsilon _{0}(y)} \, \underset {x}{\nabla } \, \underset {x}{\nabla } \Phi _{0}(x,y) + {\Gamma _{2}}(x,y), \quad x \neq y, \end{equation}

where the term ${\Gamma _{2}}({\cdot} , \cdot )$ Footnote ² is given by

(3.9) \begin{equation} {\Gamma _{2}}(x,y) = \frac {- \, 1}{\omega ^{2} \, \mu \, \left ( \epsilon _{0}(y) \right )^{2}} \underset {x}{\nabla } \underset {x}{\nabla } M\left ( \Phi _{0}({\cdot} , y) \nabla \epsilon _{0}(y) \right )(x) + \Xi (x,z) , \end{equation}

with $\nabla M({\cdot})$ being the Magnetization operator defined by (1.9) and $\Xi ({\cdot} ,z) \in \mathbb{L}^{3-\delta }(D)$ , where $\delta$ is a small positive parameter. The decomposition (3.8) here can be utilized to comprehend (3.2). Then, by denoting

\begin{equation*}\tau _{m} \,:\!= \frac {(\epsilon _{0}(z_{m}) - \epsilon _{p}(\omega ))}{\epsilon _{0}(z_{m})},\end{equation*}

using (3.8) and (3.9), we derive from (3.7) the following equation

(3.10) \begin{equation} \left [ I - \tau _{m} \, \nabla M_{D_{m}} \right ](u_{1})(x) + \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \tau _{j} \, \epsilon _{0}(z_{j}) \, \int _{D_{j}} G_{k}(x,y) \cdot u_{1}(y) \, dy = u_{0}(x) + Err_{0}(x), \end{equation}

where

(3.11) \begin{align} Err_{0}(x) &\, :\!= - \, \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \int _{D_{j}} G_{k}(x,y) \cdot u_{1}(y) \int _{0}^{1} \nabla (\epsilon _{0})(z_{j}+t(y-z_{j})) \cdot (y - z_{j}) \, dt \, dy \nonumber\\ &\quad - \omega ^{2} \, \mu \, \int _{D_{m}} G_{k}(x,y) \cdot u_{1}(y) \int _{0}^{1} \nabla (\epsilon _{0})(z_{m}+t(y-z_{m})) \cdot (y - z_{j}) \, dt \, dy \nonumber \\ &\quad - \omega ^{2} \, \mu \, ( \epsilon _{0}(z_{m}) - \epsilon _{p}(\omega ) ) \, \int _{D_{m}} {\Gamma _{2}}(x,y) \cdot u_{1}(y) \, dy \nonumber \\ &\quad + (\epsilon _{0}(z_{m}) - \epsilon _{p}(\omega )) \, \nabla M_{D_{m}} \left (u_{1}({\cdot}) \, \int _{0}^{1} \nabla \big(\epsilon _{0}^{-1}\big)(z_{m}+t({\cdot} - z_{m})) \cdot ({\cdot} - z_{m}) \, dt \right )(x). \end{align}

We have the invertibility of the operator $[ I - \tau _{m} \, \nabla M_{D_{m}} ]$ due to the fact that $\tau ^{-1}_{m} \notin \sigma \left ( \nabla M_{D_{m}} \right ) \cup \left \{ 1 \right \}$ , where $\sigma \left ({\cdot} \right )$ stands for the spectrum set. Successively, by taking the inverse of the operator $[ I - \tau _{m} \, \nabla M_{D_{m}} ]$ on the both sides of the equation (3.10), integrating over the domain $D_{m}$ and using the definition

\begin{equation*} W_{m}({\cdot}) \,:\!= \left [ I - \tau _{m} \, \nabla M_{D_{m}} \right ]^{-1}(I)({\cdot}), \quad \text{in} \; D_{m}, \end{equation*}

we obtain

(3.12) \begin{align} \int _{D_{m}} u_{1}(x) \, dx &+ \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \tau _{j} \, \epsilon _{0}(z_{j}) \, \int _{D_{m}} W_{m}(x) \cdot \int _{D_{j}} G_{k}(x,y) \cdot u_{1}(y) \, dy \, dx \nonumber\\[3pt] &= \int _{D_{m}} W_{m}(x) \cdot u_{0}(x) \, dx + \int _{D_{m}} W_m(x) \cdot Err_{0}(x) \, dx. \end{align}

Next, by setting

\begin{equation*} \mathcal{C}_{m} \,:\!= \int _{D_{m}} W_{m}(x) \, dx, \end{equation*}

and expanding both the Green’s kernel $G_{k}({\cdot} , \cdot )$ and the vector field $u_{0}({\cdot})$ near the centres, we derive from (3.12) the following algebraic system

(3.13) \begin{equation} \mathcal{C}_{m}^{-1} \cdot \int _{D_{m}} u_{1}(x) \, dx + \omega ^{2} \, \mu \, \sum _{j = 1 \atop j \neq m}^{\aleph } \tau _{j} \, \epsilon _{0}(z_{j}) \, G_{k}(z_{m}, z_{j}) \cdot \mathcal{C}_{j} \cdot \mathcal{C}_{j}^{-1} \cdot \int _{D_{j}} u_{1}(y) \, dy = u_{0}(z_{m}) + \mathcal{C}_{m}^{-1} \cdot Error_{1,m}, \end{equation}

where

(3.14) \begin{align} Error_{1,m}\, &:\!= - \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \tau _{j} \, \epsilon _{0}(z_{j}) \, \int _{D_{m}} W_{m}(x) \cdot \int _{D_{j}} \int _{0}^{1} \nabla G_{k}(x,z_{j}+t(y-z_{j})) \cdot (y - z_{j}) \, dt \cdot u_{1}(y) \, dy \, dx\nonumber \\[2pt] &\quad - \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \tau _{j} \, \epsilon _{0}(z_{j}) \, \int _{D_{m}} W_{m}(x) \cdot \int _{0}^{1} \nabla G_{k}(z_{m}+t(x-z_{m}),z_{j}) \cdot (x - z_{m}) \, dt \, dx \cdot \int _{D_{j}} u_{1}(y) \, dy \nonumber \\[2pt] &\quad + \int _{D_{m}} W_{m}(x) \cdot \int _{0}^{1} \nabla u_{0}(z_{m}+t(x-z_{m})) \cdot (x - z_{m}) \, dt \, dx + \int _{D_{m}} W_{m}(x) \cdot Err_{0}(x) \, dx. \end{align}

Moreover, for (3.13), we correspond to the following matrix form,

(3.15) \begin{equation} \begin{pmatrix} \mathcal{Q}_{1} \\ \mathcal{Q}_{2} \\ \vdots \\ \mathcal{Q}_{\aleph } \end{pmatrix} + \omega ^{2} \, \mu \, \mathcal{M} \cdot \begin{pmatrix} \mathcal{Q}_{1} \\ \mathcal{Q}_{2} \\ \vdots \\ \mathcal{Q}_{\aleph } \end{pmatrix} = \begin{pmatrix} u_{0}(z_{1}) \\ u_{0}(z_{2}) \\ \vdots \\ u_{0}(z_{\aleph }) \end{pmatrix} + \begin{pmatrix} \mathcal{C}_{1}^{-1} \cdot Error_{1,1} \\ \mathcal{C}_{2}^{-1} \cdot Error_{1,2} \\ \vdots \\ \mathcal{C}_{\aleph }^{-1} \cdot Error_{1,\aleph } \end{pmatrix}, \end{equation}

where $\mathcal{M}$ is the matrix given by

\begin{equation*} \mathcal{M} \,:\!= \begin{pmatrix} 0 & \quad G_{k}(z_{1},z_{2}) \cdot \tau _{2} \epsilon _{0}(z_{2}) \, \mathcal{C}_{2} & \quad \cdots & \quad G_{k}(z_{1},z_{\aleph }) \cdot \tau _{\aleph } \epsilon _{0}(z_{\aleph }) \, \mathcal{C}_{\aleph } \\[4pt] G_{k}(z_{2},z_{1}) \cdot \tau _{1} \, \epsilon _{0}(z_{1}) \mathcal{C}_{1} & \quad 0 & \quad \cdots & \quad G_{k}(z_{2},z_{\aleph }) \cdot \tau _{\aleph } \epsilon _{0}(z_{\aleph }) \, \mathcal{C}_{\aleph } \\[4pt] \vdots & \quad \vdots & \quad \ddots & \quad \vdots \\[4pt] G_{k}(z_{\aleph },z_{1}) \cdot \tau _{1} \epsilon _{0}(z_{1})\, \mathcal{C}_{1} & \quad G_{k}(z_{\aleph },z_{2}) \cdot \tau _{2} \epsilon _{0}(z_{2})\, \mathcal{C}_{2} & \quad \cdots & \quad 0 \\ \end{pmatrix}, \end{equation*}

$Error_{1,m}$ is given by (3.14), and

(3.16) \begin{equation} \mathcal{Q}_{m} \,:\!= \mathcal{C}_{m}^{-1} \cdot \int _{D_{m}} u_{1}(x) \, dx, \quad 1 \leq m \leq \aleph . \end{equation}

3.2. Deriving the dominant term of the optical field

To extract the dominant term related to the optical field, we start by proving the invertibility of the algebraic system (3.15). To achieve this, it is sufficient to derive a condition such that the matrix appearing on the L.H.S of (3.15) is a diagonal dominant block matrix, i.e.

(3.17) \begin{equation} \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \left \vert G_{k}(z_{m},z_{j}) \cdot \tau _{j} \epsilon _{0}(z_{j}) \, \mathcal{C}_{j} \right \vert \lt \left \vert I_{3} \right \vert . \end{equation}

To do this, we have

\begin{equation*} \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \left \vert G_{k}(z_{m},z_{j}) \cdot \tau _{j} \epsilon _{0}(z_{j})\, \mathcal{C}_{j} \right \vert \lesssim \sum _{j=1 \atop j \neq m}^{\aleph } \left \vert G_{k}(z_{m},z_{j}) \right \vert \, \left \vert \tau _{j} \right \vert \, \left \vert \epsilon _{0}(z_{j}) \right \vert \, \left \vert \mathcal{C}_{j} \right \vert \lesssim \sum _{j=1 \atop j \neq m}^{\aleph } \left \vert G_{k}(z_{m},z_{j}) \right \vert \, \left \vert \mathcal{C}_{j} \right \vert . \end{equation*}

In addition, thanks to [Reference Ghandriche and Sini25, Proposition 2.2], we know that

(3.18) \begin{equation} \mathcal{C}_{j} = \mathcal{O}( a^{3-h} ), \end{equation}

and this implies

\begin{equation*} \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq m}^{\aleph } \left \vert G_{k}(z_{m},z_{j}) \cdot \tau _{j} \epsilon _{0}(z_{j}) \, \mathcal{C}_{j} \right \vert \lesssim a^{3-h} \, \sum _{j=1 \atop j \neq m}^{\aleph } \frac {1}{d^{3}_{mj}} = \mathcal{O}( a^{3-h} \, d^{-3} \, \aleph ) = \mathcal{O}( a^{3 - h - 3 t - s} ). \end{equation*}

Hence, under the condition

(3.19) \begin{equation} 3 - h - 3 t - s \, \gt \, 0, \end{equation}

the condition (3.17) will be satisfied. By using Born series and keeping the dominant terms, we get

(3.20) \begin{equation} \begin{pmatrix} \mathcal{Q}_{1} \\ \mathcal{Q}_{2} \\ \vdots \\ \mathcal{Q}_{\aleph } \end{pmatrix} = \begin{pmatrix} u_{0}(z_{1}) \\ u_{0}(z_{2}) \\ \vdots \\ u_{0}(z_{\aleph }) \end{pmatrix} + \begin{pmatrix} Error_{2,1} \\ Error_{2,2} \\ \vdots \\ Error_{2,\aleph } \end{pmatrix}, \end{equation}

where

\begin{equation*} \begin{pmatrix} Error_{2,1} \\ Error_{2,2} \\ \vdots \\ Error_{2,\aleph } \end{pmatrix} :\!= \begin{pmatrix} \mathcal{C}_{1}^{-1} \cdot Error_{1,1} \\ \mathcal{C}_{2}^{-1} \cdot Error_{1,2} \\ \vdots \\ \mathcal{C}_{\aleph }^{-1} \cdot Error_{1,\aleph } \end{pmatrix} - \omega ^{2} \, \mu \, \mathcal{M} \cdot \begin{pmatrix} \mathcal{Q}_{1} \\ \mathcal{Q}_{2} \\ \vdots \\ \mathcal{Q}_{\aleph } \end{pmatrix}, \end{equation*}

which can be rewritten as

\begin{equation*} \begin{pmatrix} Error_{2,1} \\ Error_{2,2} \\ \vdots \\ Error_{2,\aleph } \end{pmatrix} = \begin{pmatrix} \mathcal{C}_{1}^{-1} \cdot Error_{1,1} - \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq 1}^{\aleph }G_{k}(z_{1},z_{j}) \cdot \tau _{j} \epsilon _{0}(z_{j}) \, \mathcal{C}_{j} \, \mathcal{Q}_{j} \\[8pt] \mathcal{C}_{2}^{-1} \cdot Error_{1,2} - \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq 2}^{\aleph }G_{k}(z_{2},z_{j}) \cdot \tau _{j} \epsilon _{0}(z_{j}) \, \mathcal{C}_{j} \, \mathcal{Q}_{j} \\[4pt] \vdots \\[4pt] \mathcal{C}_{\aleph }^{-1} \cdot Error_{1,\aleph } - \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq \aleph }^{\aleph }G_{k}(z_{\aleph },z_{j}) \cdot \tau _{j} \epsilon _{0}(z_{j}) \, \mathcal{C}_{j} \, \mathcal{Q}_{j} \end{pmatrix}. \end{equation*}

Next, we need to estimate the term $Error_{2,k}$ , for $k = 1, \cdots , \aleph$ . To do this, we recall that

\begin{align*} Error_{2,k} \, &:\!= \mathcal{C}_{1}^{-1} \cdot Error_{1,k} - \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq k}^{\aleph }G_{k}(z_{k},z_{j}) \cdot \tau _{j} \epsilon _{0}(z_{j}) \, \mathcal{C}_{j} \, \mathcal{Q}_{j} \\ Error_{2,k} &\!\overset {(3.16)}{=} \mathcal{C}_{1}^{-1} \cdot Error_{1,k} - \omega ^{2} \, \mu \, \sum _{j=1 \atop j \neq k}^{\aleph } \tau _{j} \, \epsilon _{0}(z_{j}) \, G_{k}(z_{k},z_{j}) \cdot \int _{D_{j}} u_{1}(x) \, dx \\ \left \vert Error_{2,k} \right \vert \; & \lesssim \left \vert \mathcal{C}_{1}^{-1} \right \vert \, \left \vert Error_{1,k} \right \vert + \sum _{j=1 \atop j \neq k}^{\aleph } \, \left \vert G_{k}(z_{k},z_{j}) \right \vert \, \left \Vert 1 \right \Vert _{\mathbb{L}^{2}(D_{j})} \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{j})} \\ &\!\! \overset {(3.18)}{\lesssim } a^{h-3} \, \left \vert Error_{1,k} \right \vert + a^{\frac {3}{2}} \, \sum _{j=1 \atop j \neq k}^{\aleph } \, \frac {1}{d^{3}_{kj}} \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{j})} \\ & \lesssim a^{h-3} \, \left \vert Error_{1,k} \right \vert + a^{\frac {3}{2}} \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D)} \, \left ( \sum _{j=1 \atop j \neq k}^{\aleph } \, \frac {1}{d^{6}_{kj}} \right )^{\frac {1}{2}} \\ & \lesssim a^{h-3} \, \left \vert Error_{1,k} \right \vert + a^{\frac {3}{2}} \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D)} \, d^{-3} \, \aleph ^{\frac {1}{2}}. \end{align*}

Knowing that

(3.21) \begin{equation} \Vert u_{1} \Vert _{\mathbb{L}^{2}(D)} = \mathcal{O}\big( a^{\frac {3}{2}-h-\frac {s}{2}} \big), \end{equation}

see Proposition A.1 for its justification, we deduce that

(3.22) \begin{equation} \left \vert Error_{2,k} \right \vert \; \lesssim \; a^{h-3} \, \left \vert Error_{1,k} \right \vert + a^{3-h-3t-{s}}. \end{equation}

To accomplish the estimation of (3.22), we need to estimate $\left \vert Error_{1,k} \right \vert$ . For (3.14), we take the modulus on its both sides to get

\begin{align*} \vert Error_{1,m} \vert & \lesssim \Vert W_{m} \Vert _{\mathbb{L}^{2}(D_{m})} \left [ \left \Vert 1 \right \Vert _{\mathbb{L}^{2}(D_{m})} \sum _{j=1 \atop j \neq m}^{\aleph } \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{j})} \frac {1}{d^{4}_{mj}} \left [ \int _{D_{j}} \vert y - z_{j} \vert ^{2} dy \right ]^{\frac {1}{2}}\right. \\ &\quad+ \left. \left \Vert 1 \right \Vert _{\mathbb{L}^{2}(D_{m})}\, \left [ \int _{D_{m}} \left \vert x - z_{m} \right \vert ^{2} dx \right ]^{\frac {1}{2}} \sum _{j=1 \atop j \neq m}^{\aleph } \frac {1}{d^{4}_{mj}} \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{j})} + \left [ \int _{D_{m}} \left \vert x - z_{m} \right \vert ^{2} \, dx \right ]^{\frac {1}{2}} + \left \Vert Err_{0} \right \Vert _{\mathbb{L}^{2}(D_{m})}\right], \end{align*}

where we have used the fact that $u_{0}({\cdot})\in \mathcal{C}^1$ and the singularity of the Green’s kernel $G_{k}({\cdot} , \cdot )$ is of order 3. Besides, by the use of

\begin{equation*} \int _{D_{j}} \vert x - z_{j} \vert ^{2} \, dx = \mathcal{O}( a^{5}), \quad 1 \leq j \leq \aleph , \end{equation*}

we obtain

(3.23) \begin{align} \vert Error_{1,m} \vert & \lesssim \left \Vert W_{m} \right \Vert _{\mathbb{L}^{2}(D_{m})} \, \left [ a^{4} \, \sum _{j=1 \atop j \neq m}^{\aleph } \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{j})} \frac {1}{d^{4}_{mj}} \, + a^{\frac {5}{2}} + \left \Vert Err_{0} \right \Vert _{\mathbb{L}^{2}(D_{m})} \right ] \nonumber\\[3pt] & \lesssim \left \Vert W_{m} \right \Vert _{\mathbb{L}^{2}(D_{m})} \, \big[ a^{4} \, d^{-4} \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D)} \, \aleph ^{\frac {1}{2}} + a^{\frac {5}{2}} + \left \Vert Err_{0} \right \Vert _{\mathbb{L}^{2}(D_{m})} \big] \nonumber \\[3pt] &\!\! \overset {(3.21)}{\lesssim } \left \Vert W_{m} \right \Vert _{\mathbb{L}^{2}(D_{m})} \, \big[ a^{\frac {11}{2}-h-4t-s} + a^{\frac {5}{2}} + \left \Vert Err_{0} \right \Vert _{\mathbb{L}^{2}(D_{m})} \big]. \end{align}

To estimate of $\left \Vert Err_{0} \right \Vert _{\mathbb{L}^{2}(D_{m})}$ , we go back to (3.11) and take the $\left \Vert \cdot \right \Vert _{\mathbb{L}^{2}(D_{m})}$ -norm on the both sides of the equation to obtain

\begin{align*} \left \Vert Err_{0} \right \Vert _{\mathbb{L}^{2}(D_{m})} & \lesssim \sum _{j=1 \atop j \neq m}^{\aleph } \left \Vert \int _{D_{j}} G_{k}({\cdot} ,y) \cdot u_{1}(y) \int _{0}^{1} \nabla (\epsilon _{0})(z_{j}+t(y-z_{j})) \cdot (y - z_{j}) \, dt \, dy \right \Vert _{\mathbb{L}^{2}(D_{m})} \\ &\quad + \left \Vert \int _{D_{m}} G_{k}({\cdot} ,y) \cdot u_{1}(y) \int _{0}^{1} \nabla (\epsilon _{0})(z_{m}+t(y-z_{m})) \cdot (y - z_{j}) \, dt \, dy \right \Vert _{\mathbb{L}^{2}(D_{m})} \\ &\quad + \left \Vert \int _{D_{m}}\!\! {\Gamma _{2}}({\cdot} , y) \cdot u_{1}(y)\, dy \right \Vert _{\mathbb{L}^{2}(D_{m})} \!+ \left \Vert \nabla M_{D_{m}} \!\left (u_{1}\! \int _{0}^{1}\!\! \nabla (\epsilon _{0}^{-1})(z_{m}\!+t({\cdot} - z_{m})) \cdot ({\cdot} - z_{m})\, dt \right )\right \Vert _{\mathbb{L}^{2}(D_{m})}\!, \end{align*}

and, by using the smoothness of the function $\epsilon _{0}({\cdot})$ , the dominant term related to the Green’s kernel, see (3.8), and the dominant term related to the kernel ${\Gamma _{2}}({\cdot} , \cdot )$ , see (3.9), we reduce the previous estimation to

\begin{align*} \left \Vert Err_{0} \right \Vert _{\mathbb{L}^{2}(D_{m})} & \lesssim a \, \sum _{j=1 \atop j \neq m}^{\aleph } \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{j})} \, \left [ \int _{D_{m}} \, \int _{D_{j}} \vert G_{k}(x,y) \vert ^{2} \, dy \, dx \right ]^{\frac {1}{2}} \\ &\quad + \left \Vert \nabla M_{D_{m}}\left ( \frac {u_{1}({\cdot})}{\epsilon _{0}({\cdot})} \int _{0}^{1} \nabla (\epsilon _{0})(z_{m}+t({\cdot} -z_{m})) \cdot ({\cdot} - z_{j}) \, dt \right ) \right \Vert _{\mathbb{L}^{2}(D_{m})} \\ &\quad + \left \Vert \int _{D_{m}} \nabla \nabla M \big ( \Phi _{0}({\cdot} ,y) \nabla \epsilon _{0}^{-1}(y) \big )({\cdot}) \cdot u_{1}(y) \, dy \right \Vert _{\mathbb{L}^{2}(D_{m})} \\ &\quad + \left \Vert \nabla M_{D_{m}} \left (u_{1}({\cdot}) \, \int _{0}^{1} \nabla \big(\epsilon _{0}^{-1}\big)(z_{m}+t({\cdot} - z_{m})) \cdot ({\cdot} - z_{m}) \, dt \right )\right \Vert _{\mathbb{L}^{2}(D_{m})}. \end{align*}

Moreover, since $\left \Vert \nabla M_{D_{m}} \right \Vert _{\mathcal{L}\left (\mathbb{L}^{2}(D_{m});\mathbb{L}^{2}(D_{m})\right )} = 1$ , see [Reference Ghandriche and Sini25, Lemma 5.5], and using the fact that from singularity point of view we have

\begin{equation*} \nabla \nabla M \left ( \Phi _{0}({\cdot} ,y) \nabla \epsilon _{0}^{-1}(y) \right )({\cdot}) \sim \nabla \left ( \Phi _{0}({\cdot} , y) \, \nabla \epsilon _{0}^{-1}(y) \right )\!, \end{equation*}

we deduce the following estimation

\begin{align*} \left \Vert Err_{0} \right \Vert _{\mathbb{L}^{2}(D_{m})} & \lesssim a \, \sum _{j=1 \atop j \neq m}^{\aleph } \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{j})} \, \left [ \int _{D_{m}} \, \int _{D_{j}} \vert G_{k}(x,y) \vert ^{2} \, dy \, dx \right ]^{\frac {1}{2}} + a \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{m})} \\ &\quad + \left \Vert \int _{D_{m}} \nabla \big ( \Phi _{0}({\cdot} ,y) \nabla \epsilon _{0}^{-1}(y) \big ) \cdot u_{1}(y) \, dy \right \Vert _{\mathbb{L}^{2}(D_{m})} \, + a \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{m})}. \end{align*}

On the R.H.S, to evaluate the first term we expand the Green’s kernel $G_{k}({\cdot} ,\cdot )$ near the centres, together with the explicit computation for the third term, we end up with the following estimation

\begin{equation*} \Vert Err_{0} \Vert _{\mathbb{L}^{2}(D_{m})} \lesssim a^{4} \, \sum _{j=1 \atop j \neq m}^{\aleph } \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{j})} \, \vert G_{k}(z_{m},z_{j}) \vert + a \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{m})} + \big\Vert \nabla N_{D_{m}}\big( \nabla \epsilon _{0}^{-1} \cdot u_{1} \big) \big\Vert _{\mathbb{L}^{2}(D_{m})}. \end{equation*}

Now, by using the fact that

\begin{equation*} \big \Vert \nabla N_{D_{m}} \big \Vert _{\mathcal{L}\big (\mathbb{L}^{2}(D_{m}), \mathbb{L}^{2}(D_{m}) \big )} = \mathcal{O} ( a ) \quad \text{and} \quad \vert G_{k}(z_{m},z_{j}) \vert \sim \frac {1}{d^{3}_{mj}}, \end{equation*}

we obtain

(3.24) \begin{align} \left \Vert Err_{0} \right \Vert _{\mathbb{L}^{2}(D_{m})} & \lesssim a^{4} \, \sum _{j=1 \atop j \neq m}^{\aleph } \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{j})} \, \frac {1}{d^{3}_{mj}} + a \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{m})} \nonumber\\ & \lesssim a^{4} \, \left ( \sum _{j=1 \atop j \neq m}^{\aleph } \Vert u_{1} \Vert ^{2}_{\mathbb{L}^{2}(D_{j})} \right )^{\frac {1}{2}} \, \left ( \sum _{j=1 \atop j \neq m}^{\aleph } \frac {1}{d^{6}_{mj}} \right )^{\frac {1}{2}} + a \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D_{m})} \nonumber \\ & \lesssim \big( a^{4} \, d^{-3} \, \aleph ^{\frac {1}{2}} + \, a \big) \, \Vert u_{1} \Vert _{\mathbb{L}^{2}(D)} \nonumber \\ &\!\! \overset {(3.21)}{=} \mathcal{O}\big( a^{\frac {5}{2}-h-\frac {s}{2}} \big) + \mathcal{O}\big( a^{\frac {11}{2}-3t-s-h}\big) \overset {(3.19)}{=} \mathcal{O}\big( a^{\frac {5}{2}-h-\frac {s}{2}} \big). \end{align}

By returning to (3.23) and using (3.24), we deduce that

\begin{equation*} \left \vert Error_{1,m} \right \vert \lesssim \left \Vert W_{m} \right \Vert _{\mathbb{L}^{2}(D_{m})} \, a^{\frac {5}{2}-h + \min \left (3 - 4 t - s; -\frac {s}2 \right )}. \end{equation*}

In addition, thanks to [Reference Ghandriche and Sini25, Proposition 2.2], we know that $\left \Vert W_{m} \right \Vert _{\mathbb{L}^{2}(D_{m})} = \mathcal{O} ( a^{\frac {3}{2}-h} )$ , and this implies

\begin{equation*} \left \vert Error_{1,m} \right \vert = \mathcal{O}\big( a^{(4 - 2 h) + \min \left (3 - 4 t - s; -\frac {s}2 \right )} \big). \end{equation*}

Finally, by plugging the previous estimation into (3.22), we obtain

(3.25) \begin{equation} \left \vert Error_{2,m} \right \vert = \mathcal{O}\left ( a^{\min ((3 - 3t - h - s);(1 - h-\frac {s}{2}))} \right )\!. \end{equation}

Hence, by gathering (3.20) and (3.25), we conclude that

\begin{equation*} \mathcal{Q}_{m} = u_{0}(z_{m}) + \mathcal{O}\left ( a^{\min ((3 - 3t - h - s);(1 - h - \frac {s}{2}))} \right )\!. \end{equation*}

Consequently, by using (3.16) and (3.18), we obtain

\begin{equation*} \int _{D_{m}} u_{1}(x) \, dx = \mathcal{C}_{m} \cdot u_{0}(z_{m}) + \mathcal{O}\left ( a^{\min ((6 - 3t - 2 h - s);(4 - 2 h -\frac {s}{2}))} \right )\!. \end{equation*}

Furthermore, from [Reference Ghandriche and Sini25, Proposition 2.2], we know that

(3.26) \begin{equation} \mathcal{C}_{m} = \frac {a^{3} \, \epsilon _{0}(z_{m})}{\big (\epsilon _{0}(z_{m}) - \lambda _{n_{0}}^{(3)}(B) \, (\epsilon _{0}(z_{m}) - \epsilon _{p}(\omega ) ) \big )} \, \int _{B} e_{n_{0}}^{(3)}(x) \, dx \otimes \int _{B} e_{n_{0}}^{(3)}(x) \, dx + \mathcal{O}(a^{3}). \end{equation}

This implies,

(3.27) \begin{equation} \int _{D_{m}} u_{1}(x)\, dx = \frac {a^{3} \; \epsilon _{0}(z_{m}) \, \left \langle u_{0}(z_{m}) , \int _{B} e^{(3)}_{n_{0}}(x) \, dx \right \rangle }{(\epsilon _{0}(z_{m}) - (\epsilon _{0}(z_{m}) - \epsilon _{p}(\omega ) ) \, \lambda ^{(3)}_{n_{0}}(B) )} \, \int _{B} e^{(3)}_{n_{0}}(x) \, dx + \, \mathcal{O}\left ( a^{\min (3;(6 - 3t - 2 h - s);(4 - 2 h - \frac {s}{2}))} \right )\!, \end{equation}

where $h$ is taken to be

(3.28) \begin{equation} h \lt \min \left ( 1 - \frac {s}{2}; - s + 3 (1-t) \right )\!. \end{equation}

It is important to note that if $\omega ^{2} - \omega ^{2}_{p_{\ell }, n_{0}, m} \sim a^{h}$ is satisfied, we get $(\epsilon _{0}(z_{m}) - (\epsilon _{0}(z_{m}) - \epsilon _{p}(\omega ) ) \lambda ^{(3)}_{n_{0}}(B) ) \sim a^{h}$ , see (3.39), thus the first term on the R.H.S of (3.27) will be of order $ \sim a^{3-h}$ . Furthermore, the condition (3.28) is added to ensure that the R.H.S of (3.27) is well defined and, of course, to also satisfy (3.19).

In the subsequent discussions, we will use the notation $V \,:\!= u_{0}$ , respectively $E \,:\!= u_{1}$ for the solutions to (1.2) signifying the electric field respectively before and after injecting the nanoparticles $D$ inside $\Omega$ . Taking into account the introduced notation, we rewrite (3.27) as

(3.29) \begin{equation} \int _{D_{m}} E(x)\, dx = \frac {a^{3} \; \epsilon _{0}(z_{m}) \, \big\langle V ( z_{m}, \theta , q ) , \int _{B} e^{(3)}_{n_{0}}(x) \, dx \big\rangle }{\big(\epsilon _{0}(z_{m}) - (\epsilon _{0}(z_{m}) - \epsilon _{p}(\omega ) ) \, \lambda ^{(3)}_{n_{0}}(B) \big )} \, \int _{B} e^{(3)}_{n_{0}}(x) \, dx + \, \mathcal{O}\big ( a^{\min (3;(6 - 3t - 2 h - s);(4 - 2 h - \frac {s}{2}))} \big ). \end{equation}

3.3. Estimation of the scattered fields

To derive the estimate of the scattered field, we take $x$ away from $D$ in (3.6),

\begin{equation*} E(x) + \omega ^{2} \, \mu \, \int _{D} G_{k}(x,y) \cdot E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy = V(x). \end{equation*}

In addition, as $E = E^{I} + E^{s}$ and $V = V^{I} + V^{s}$ , with $V^{I} = E^{I}$ , we obtain

(3.30) \begin{align} E^{s}(x) - V^{s}(x) &= - \omega ^{2} \, \mu \, \int _{D} G_{k}(x,y) \cdot E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy \nonumber\\ &= - \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } \int _{D_{j}} G_{k}(x,y) \cdot E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy, \end{align}

which, by Taylor expansion for the permittivity function $\epsilon _{0}({\cdot})$ , gives us

\begin{equation*} E^{s}(x) - V^{s}(x) = - \, \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } ( \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) ) \, \int _{D_{j}} G_{k}(x,y) \cdot E(y) \, dy \; - \; T_{1}(x), \end{equation*}

where

\begin{equation*} T_{1}(x) \,:\!= \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } \int _{D_{j}} G_{k}(x,y) \cdot E(y) \, \int _{0}^{1} \, \nabla \epsilon _{0}(z_{j}+t(y-z_{j})) \cdot (y-z_{j}) \, dt \, dy. \end{equation*}

We need to estimate the term $T_{1}({\cdot})$ . We have

\begin{align*} \left \vert T_{1}(x) \right \vert & \lesssim \sum _{j=1}^{\aleph } \, \int _{D_{j}} \left \vert G_{k}(x,y) \cdot E(y) \, \int _{0}^{1} \, \nabla \epsilon _{0}(z_{j}+t(y-z_{j})) \cdot (y-z_{j}) \, dt \, \right \vert \, dy \\ & \lesssim \sum _{j=1}^{\aleph } \, \left \Vert G_{k}(x,\cdot ) \right \Vert _{\mathbb{L}^{2}(D_{j})} \, \left \Vert E \right \Vert _{\mathbb{L}^{2}(D_{j})} \, \left \Vert \int _{0}^{1} \, \nabla \epsilon _{0}(z_{j}+t({\cdot} -z_{j})) \cdot ({\cdot} -z_{j}) \, dt \, \right \Vert _{\mathbb{L}^{\infty }(D_{j})} \\ & \lesssim a \, \sum _{j=1}^{\aleph } \, \left \Vert G_{k}(x,\cdot ) \right \Vert _{\mathbb{L}^{2}(D_{j})} \, \left \Vert E \right \Vert _{\mathbb{L}^{2}(D_{j})} \lesssim a \, \left ( \sum _{j=1}^{\aleph } \, \left \Vert G_{k}(x,\cdot ) \right \Vert ^{2}_{\mathbb{L}^{2}(D_{j})} \right )^{\frac {1}{2}} \, \left \Vert E \right \Vert _{\mathbb{L}^{2}(D)}\!, \end{align*}

which, by knowing that $x$ is away from $D$ , such that the Green’s kernel $G_{k}(x,\cdot )$ is smooth, allows us to get

\begin{equation*} \left \vert T_{1}(x) \right \vert = \mathcal{O}\big( a^{\frac {5-s}{2}} \; \left \Vert E \right \Vert _{\mathbb{L}^{2}(D)}\big). \end{equation*}

Then,

\begin{equation*} E^{s}(x) - V^{s}(x) = - \, \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } ( \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) ) \, \int _{D_{j}} G_{k}(x,y) \cdot E(y) \, dy + \mathcal{O}\big( a^{\frac {5-s}{2}} \; \left \Vert E \right \Vert _{\mathbb{L}^{2}(D)}\big ). \end{equation*}

In addition, by Taylor expansion for $G_{k}(x,\cdot )$ , we obtain

\begin{align*} E^{s}(x) - V^{s}(x) &= - \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } ( \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) ) \, G_{k}(x,z_{j}) \cdot \int _{D_{j}} E(y) \, dy \\ &\quad - T_{2}(x) + \mathcal{O}\big( a^{\frac {5-s}{2}} \; \left \Vert E \right \Vert _{\mathbb{L}^{2}(D)}\big), \end{align*}

where

\begin{equation*} T_{2}(x) \,:\!= \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } \, ( \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) ) \, \int _{D_{j}} \int _{0}^{1} \nabla G_{k}(x,z_{j}+t(y-z_{j})) \cdot (y-z_{j}) \, dt \cdot E(y) \, dy. \end{equation*}

We estimate the above term as

\begin{equation*} \left \vert T_{2}(x) \right \vert \lesssim \sum _{j=1}^{\aleph } \vert \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) \vert \, \left \vert \int _{D_{j}} \int _{0}^{1} \nabla G_{k}(x,z_{j}+t(y-z_{j})) \cdot (y-z_{j}) \, dt \cdot E(y) \, dy \right \vert , \end{equation*}

which, by using the fact that $\left \vert \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) \right \vert \sim 1$ and the Hölder inequality, gives us

\begin{align*} \left \vert T_{2}(x) \right \vert & \lesssim \sum _{j=1}^{\aleph } \left \Vert \int _{0}^{1} \nabla G_{k}(x,z_{j}+t({\cdot} -z_{j})) \cdot ({\cdot} -z_{j}) \, dt \right \Vert _{\mathbb{L}^{2}(D_{j})} \, \left \Vert E \right \Vert _{\mathbb{L}^{2}(D_{j})} \\ & \lesssim \left ( \sum _{j=1}^{\aleph } \left \Vert \int _{0}^{1} \nabla G_{k}(x,z_{j}+t({\cdot} -z_{j})) \cdot ({\cdot} -z_{j}) \, dt \right \Vert ^{2}_{\mathbb{L}^{2}(D_{j})} \right )^{\frac {1}{2}} \, \left \Vert E \right \Vert _{\mathbb{L}^{2}(D)}. \end{align*}

Again, using the smoothness of $\nabla G_{k}(x, \cdot )$ , for $x$ away from $D$ , allows us to deduce that

\begin{equation*} \left \vert T_{2}(x) \right \vert = \mathcal{O}\big( a^{\frac {5-s}{2}} \; \left \Vert E \right \Vert _{\mathbb{L}^{2}(D)}\big). \end{equation*}

Then,

(3.31) \begin{align} E^{s}(x) - V^{s}(x) &= - \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } ( \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) ) \, G_{k}(x,z_{j}) \cdot \int _{D_{j}} E(y) \, dy + \mathcal{O}\big ( a^{\frac {5-s}{2}} \; \left \Vert E \right \Vert _{\mathbb{L}^{2}(D)}\big ) \nonumber\\ &\!\!\overset {(3.21)}{=} - \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } ( \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) ) \, G_{k}(x,z_{j}) \cdot \int _{D_{j}} E(y) \, dy + \mathcal{O}( a^{4 -h -s} ). \end{align}

In addition, by plugging (3.29) into (3.31), we derive that

(3.32) \begin{align} & E^{s}(x) - V^{s}(x) \nonumber\\&\quad = - \, \mu \, a^{3} \, \omega ^{2} \, \sum _{j=1}^{\aleph } \frac {\epsilon _{0}(z_{j}) \, ( \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) )}{\Lambda _{n_{0},j}(\omega )} \, \left \langle V (z_{j}, \theta , q ), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \, G_{k}(x,z_{j}) \cdot \int _{B} e_{n_{0}}^{(3)}(y) \, dy \nonumber\\ &\qquad + \mathcal{O}\big( a^{\min \left ((3-s);\left(6-3t-2h-\frac {3s}{2}\right)\right )}\big), \end{align}

where $\Lambda _{n_{0},j}(\omega )$ is a family of dispersion equations given by

(3.33) \begin{equation} \Lambda _{n_{0},j}(\omega ) \,:\!= \big ( \epsilon _{0}(z_{j}) - \lambda _{n_{0}}^{(3)}(B) \, ( \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega )) \big). \end{equation}

Next, we solve the real part of the above dispersion equation. More precisely, we set $\omega _{P_{\ell },n_{0},j}$ to be solution to

(3.34) \begin{equation} Re\big(\Lambda _{n_{0},j}\big(\omega _{P_{\ell },n_{0},j}\big)\big) = \big ( Re(\epsilon _{0}(z_{j})) - \lambda _{n_{0}}^{(3)}(B) \, \big ( Re(\epsilon _{0}(z_{j})) - Re\big (\epsilon _{p}\big(\omega _{P_{\ell },n_{0},j}\big)\big ) \big ) \big ) = 0, \end{equation}

which gives us

(3.35) \begin{equation} Re\big(\epsilon _{p}\big(\omega _{P_{\ell },n_{0},j}\big)\big) = Re(\epsilon _{0}(z_{j})) \, \frac {\big(\lambda _{n_{0}}^{(3)}(B) - 1 \big)}{\lambda _{n_{0}}^{(3)}(B)}. \end{equation}

The existence and uniqueness of the solution $ \omega _{P_{\ell },n_{0},j}$ to (3.34), or equivalently to (3.35), have already been discussed in [Reference Ghandriche and Sini25, Lemma 5.7]. We have,

(3.36) \begin{equation} \omega ^{2}_{P_{\ell },n_{0},j} = \frac {1}{2} \, \left [ 2 \, \omega ^{2}_{0} - \gamma ^{2} + \frac {\omega ^{2}_{p} \, \lambda ^{(3)}_{n_{0}}(B) \, \epsilon _{\infty }}{ \left [ {{\mathrm{Re\,}}}(\epsilon _{0}(z_{j})) \, \big(1 - \lambda ^{(3)}_{n_{0}}(B) \big ) + \lambda ^{(3)}_{n_{0}}(B) \, \epsilon _{\infty } \right ]} + \sqrt {\Delta } \right ], \end{equation}

where $\Delta$ is such that

\begin{align*} \Delta &= \left ( \frac {\omega ^{2}_{p} \, \lambda ^{(3)}_{n_{0}}(B) \, \epsilon _{\infty }}{ \left [ {{\mathrm{Re\,}}}(\epsilon _{0}(z_{j})) \, \left (1 - \lambda ^{(3)}_{n_{0}}(B) \right ) + \lambda ^{(3)}_{n_{0}}(B) \, \epsilon _{\infty } \right ]} \right )^{2} \\[5pt] &\quad - \gamma ^{2} \left ( 4 \, \omega ^{2}_{0} - \gamma ^{2} + \frac {2 \, \omega ^{2}_{p} \, \epsilon _{\infty } \, \lambda ^{(3)}_{n_{0}}(B)}{\left [ {{\mathrm{Re\,}}}(\epsilon _{0}(z_{j})) \left (1 - \lambda ^{(3)}_{n_{0}}(B) \right ) + \lambda ^{(3)}_{n_{0}}(B) \, \epsilon _{\infty } \right ]} \right )\!. \end{align*}

Furthermore, the frequency $\omega _{P_{\ell },n_{0},j}$ satisfies

\begin{align*} \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) &= Im ( \epsilon _{0}(z_{j}) ) \, \big( 1 - \lambda ^{(3)}_{n_{0}}(B) \big) + \lambda ^{(3)}_{n_{0}}(B) \, Im\left ( \epsilon _{p}\big(\omega _{P_{\ell },n_{0},j}\big) \right ) \\[3pt] \left \vert \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) \right \vert & \lesssim \left \Vert Im\left ( \epsilon _{0} \right ) \right \Vert _{\mathbb{L}^{\infty }(\Omega )} + \left \vert Im\left ( \epsilon _{p}\big(\omega _{P_{\ell },n_{0},j}\big) \right ) \right \vert \overset {(1.1)}{\underset {(1.4)}{=}} \mathcal{O}( \gamma ), \end{align*}

Consequently, if we assume that

(3.37) \begin{equation} \gamma = \mathcal{O}( a^{h}), \end{equation}

we derive from (3.37) the coming estimation

\begin{equation*} \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) = \mathcal{O}( a^{h}). \end{equation*}

In addition, if we let $\omega$ such that

(3.38) \begin{equation} \omega ^{2} = \omega ^{2}_{P_{\ell },n_{0},j} \, \pm \, a^{h}, \end{equation}

we get

\begin{align*} \Lambda _{n_{0},j} ( \omega ) & \overset {(3.33)}{=} \epsilon _{0}(z_{j}) - \lambda _{n_{0}}^{(3)}(B) \, ( \epsilon _{0}(z_{j}) - \epsilon _{p}(\omega ) ) \\[2pt] &\ = Re ( \epsilon _{0}(z_{j}) ) - \lambda _{n_{0}}^{(3)}(B) \, Re\left ( \epsilon _{0}(z_{j}) - \epsilon _{p}\big(\omega _{P_{\ell },n_{0},j}\big) \right ) - \lambda _{n_{0}}^{(3)}(B) \, Re\left ( \epsilon _{p}\big(\omega _{P_{\ell },n_{0},j}\big) - \epsilon _{p}(\omega ) \right ) \\[2pt] &\quad + i \, Im(\epsilon _{0}(z_{j})) - \lambda _{n_{0}}^{(3)}(B) \, i \, ( Im ( \epsilon _{0}(z_{j}) ) - Im ( \epsilon _{p}(\omega )))\\[2pt] &\!\overset {(3.34)}{=} \lambda _{n_{0}}^{(3)}(B) \, Re\left (\epsilon _{p}(\omega ) - \epsilon _{p}\big(\omega _{P_{\ell },n_{0},j}\big) \right ) + i \, Im(\epsilon _{0}(z_{j})) \left ( 1 - \lambda _{n_{0}}^{(3)}(B) \right ) + i \, \lambda _{n_{0}}^{(3)}(B) \, \, Im (\epsilon _{p}(\omega )). \end{align*}

Hence, by taking the modulus, we obtain

\begin{align*} \left \vert \Lambda _{n_{0},j} ( \omega ) \right \vert &\lesssim \left \vert Re\left (\epsilon _{p}(\omega ) - \epsilon _{p}\big(\omega _{P_{\ell },n_{0},j}\big) \right ) \right \vert + \left \Vert Im\left (\epsilon _{0} \right ) \right \Vert _{\mathbb{L}^{\infty }(\Omega )} + \vert Im (\epsilon _{p}(\omega )) \vert \\[4pt] & \overset {(1.1)}{\underset {(1.4) } \lesssim } , \left \vert Re\left (\epsilon _{p}(\omega ) - \epsilon _{p}\big(\omega _{P_{\ell },n_{0},j}\big) \right ) \right \vert + \gamma \, \overset {(1.1)}{\lesssim } \left \vert \omega ^{2} - \omega ^{2}_{P_{\ell },n_{0},j} \right \vert + \gamma , \end{align*}

and, by using (3.37) and (3.38), we deduce that

(3.39) \begin{equation} \Lambda _{n_{0},j} ( \omega ) = \mathcal{O}( a^{h}), \end{equation}

with $\omega$ satisfying (3.38). Now, by returning to (3.32) and using (3.33), we derive the following relation on the scattered fields

(3.40) \begin{align} & E^{s}(x) - V^{s}(x)\nonumber\\ & \quad = - \, \mu \, a^{3} \, \omega ^{2} \, \sum _{j=1}^{\aleph } \, \frac {\epsilon _{0}(z_{j}) \left (\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j} ( \omega ) \right )}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \, \left \langle V(z_{j}, \theta , q) , \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \, G_{k}(x,z_{j}) \cdot \int _{B} e_{n_{0}}^{(3)}(y) \, dy \nonumber\\ &\qquad + \mathcal{O}\big( a^{\min \left ( (3-s), \left(6-3t-2h-\frac {3s}{2}\right) \right )}\big). \end{align}

This proves (1.12).

3.4. Estimation of the far fields

For the L.H.S of (3.40), in order to get the corresponding far-field expressions, we return to (3.30) that

(3.41) \begin{equation} E^{s}(x) - V^{s}(x) = - \, \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } \int _{D_{j}} G_{k}(x,y) \cdot E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy. \end{equation}

Furthermore, thanks to [Reference Colton and Kress15, Theorem 6.9], we know that the following asymptotic relation holds

(3.42) \begin{equation} G_{k}(x,y) = \frac {e^{i \, k \, \vert x \vert }}{ \vert x \vert } \, \left ( G_{k}^{\infty }(\hat {x},y) + \mathcal{O}\left ( \frac {1}{ \vert x \vert } \right ) \right )\!, \quad \vert x \vert \rightarrow + \infty . \end{equation}

Then, by plugging (3.42) into (3.41), we obtain

\begin{align*} E^{s}(x) - V^{s}(x) &= \frac {e^{i \, k \, \vert x \vert }}{ \vert x \vert } \, \Bigg [ {-} \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } \int _{D_{j}} G_{k}^{\infty }(\hat {x},y) \cdot E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy \\ &\quad + \mathcal{O}\left ( \frac {1}{ \vert x \vert } \sum _{j=1}^{\aleph } \int _{D_{j}} E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy \right ) \Bigg ]. \end{align*}

By estimating the second term on the R.H.S, we obtain

\begin{align*} T_{0}(x) \, &:\!= \frac {1}{ \vert x \vert } \sum _{j=1}^{\aleph } \int _{D_{j}} E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy \\ \left \vert T_{0}(x) \right \vert & \leq \frac {1}{ \vert x \vert } \sum _{j=1}^{\aleph } \left \vert \int _{D_{j}} E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy \right \vert \\ & \leq \frac {1}{ \vert x \vert } \sum _{j=1}^{\aleph } \left \Vert E \right \Vert _{\mathbb{L}^{2}(D_{j})} \; \Vert \epsilon _{0}({\cdot}) - \epsilon _{p}(\omega ) \Vert _{\mathbb{L}^{\infty }(D_{j})} \, \vert D_{j} \vert ^{\frac {1}{2}}, \end{align*}

which, by knowing that $\left \Vert \epsilon _{0}({\cdot}) - \epsilon _{p}(\omega ) \right \Vert _{\mathbb{L}^{\infty }(D_{j})} \sim 1$ , gives

\begin{equation*} \left \vert T_{0}(x) \right \vert \lesssim \frac {1}{ \vert x \vert } \sum _{j=1}^{\aleph } \left \Vert E \right \Vert _{\mathbb{L}^{2}(D_{j})} \; \vert D_{j} \vert ^{\frac {1}{2}} \lesssim \frac {1}{ \vert x \vert } \, \left \Vert E \right \Vert _{\mathbb{L}^{2}(D)} \; \left \vert D \right \vert ^{\frac {1}{2}} \overset {(3.21)}{=} \mathcal{O}\left ( \frac {a^{3-h-s}}{ \vert x \vert } \right )\!. \end{equation*}

Hence,

\begin{equation*} E^{s}(x) - V^{s}(x) = \frac {e^{i \, k \, \vert x \vert }}{ \vert x \vert } \, \left [ - \, \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } \int _{D_{j}} G_{k}^{\infty }(\hat {x},y) \cdot E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy + \mathcal{O}\left ( \frac {a^{3-h-s}}{ \vert x \vert } \right ) \right ]\!. \end{equation*}

This implies,

\begin{equation*} E^{\infty }(\hat {x}) - V^{\infty }(\hat {x}) = - \, \omega ^{2} \, \mu \, \sum _{j=1}^{\aleph } \int _{D_{j}} G_{k}^{\infty }(\hat {x},y) \cdot E(y) \, ( \epsilon _{0}(y) - \epsilon _{p}(\omega ) ) \, dy. \end{equation*}

Furthermore, by repeating the same arguments used to derive (3.40) and letting $ \vert x \vert \gg 1$ in (3.40), we obtain

\begin{align*} & E^{\infty }(\hat {x}) - V^{\infty }(\hat {x}) \\ & \quad = - \mu \, a^{3} \, \omega ^{2} \, \sum _{j=1}^{\aleph } \, \frac {\epsilon _{0}(z_{j}) \, \big (\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j} ( \omega ) \big )}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \, \left \langle V(z_{j}, \theta , q ), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \, G^{\infty }_{k}(\hat {x},z_{j}) \cdot \int _{B} e_{n_{0}}^{(3)}(y) \, dy \\ & \qquad + \mathcal{O}\big( a^{\min \left ( (3-s), \left(6-3t-2h-\frac {3s}{2}\right) \right )}\big) , \end{align*}

which by using (3.38) further indicates,

\begin{align*} & E^{\infty }(\hat {x}) - V^{\infty }(\hat {x}) \\ &\quad = - \mu \, a^{3} \, \sum _{j=1}^{\aleph } \frac { \omega ^{2}_{P_{\ell },n_{0},j} \, \epsilon _{0}(z_{j}) \, \left (\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j} ( \omega ) \right )}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \, \left \langle V(z_{j}, \theta , q), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \, G^{\infty }_{k}(\hat {x},z_{j}) \cdot \int _{B} e_{n_{0}}^{(3)}(y) \, dy \\ &\qquad + \mathcal{O}\big( a^{\min \left ( (3-s), \left(6-3t-2h-\frac {3s}{2}\right) \right )}\big), \end{align*}

with $\hat {x} \in \mathbb{S}^{2}$ and $G^{\infty }_{k}({\cdot} ,z)$ being the far field associated to the Green’s kernel solution to (1.10). Besides, we have

\begin{align*} \frac {\epsilon _{0}(z_{j}) \, \left (\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j} ( \omega ) \right )}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} &= \frac {\epsilon _{0}(z_{j}) \, \big(\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) \big)}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} + \frac {\epsilon _{0}(z_{j}) \, \big(\Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) - \Lambda _{n_{0},j} ( \omega ) \big)}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \\[5pt] &= \frac {\epsilon _{0}(z_{j}) \, \big(\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) \big)}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} + \mathcal{O}\left ( 1 \right )\!. \end{align*}

Then,

(3.43) \begin{align} & E^{\infty }(\hat {x}) - V^{\infty }(\hat {x}) \nonumber\\ &\quad = - \mu \, a^{3} \, \sum _{j=1}^{\aleph } \, \frac {\omega ^{2}_{P_{\ell },n_{0},j} \, \epsilon _{0}(z_{j}) \, \big(\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) \big)}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \, \left \langle V(z_{j}, \theta , q), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \, G^{\infty }_{k}(\hat {x},z_{j}) \cdot \int _{B} e_{n_{0}}^{(3)}(y) \, dy \nonumber \\ &\qquad + \mathcal{O}\big( a^{\min \left ( (3-s), \left(6-3t-2h-\frac {3s}{2}\right) \right )}\big).\end{align}

We will delay the proof of Theorem1.1 until we announce and comment on a technical result named Mixed Electromagnetic Reciprocity Relation. Because the kernel $G^{\infty }_{k}({\cdot} , \cdot )$ is unknown on the right-hand side of (3.43), since the kernel $G_{k}({\cdot} ,\cdot )$ is unknown, we need to substitute it with a known (measured) term. To achieve this, we propose setting the following result.

Proposition 3.1. The following Mixed Electromagnetic Reciprocity Relation holds,

(3.44) \begin{equation} \left (G_{k}^{\infty }\right )^{\top }( \hat {x},z) \cdot ( \hat {x} \times q ) = -\frac {1}{4 \, \pi } \, V(z, - \, \hat {x},q), \quad z \in \mathbb{R}^{3} \quad \text{and} \quad \hat {x}, q \in \mathbb{S}^{2}. \end{equation}

Proof. See Subsection A.1. To avoid confusion, let us recall that the electromagnetic reciprocity relation refers to the relationship between two existing electromagnetic wave fields in terms of their incident, propagation and polarization directions. As per the proposition above, there is a correlation between the total field $V({\cdot} ,\cdot ,\cdot )$ and the far field corresponding to the Green’s kernel $G_{k}({\cdot} ,\cdot )$ , solution of (1.10) for the heterogeneous medium, and it is named the Mixed Electromagnetic Reciprocity Relation. It’s worth to emphasize that for the case of a homogeneous medium a reciprocity relation has been proved in [Reference Nédélec46, Section 5.1], and for the particular case of an exterior domain with specific boundary conditions, the mixed reciprocity relation has already been proved in [Reference Potthast49, Theorem 5]. Here, we extend this mixed reciprocity relation to our transmission problem.

By taking the inner product on each side of the equation given by (3.43) with the vector $ ( \hat {x} \times q )$ , we obtain

\begin{align*} & \langle ( E^{\infty } - V^{\infty })(\hat {x}) , ( \hat {x} \times q ) \rangle \\[2pt] &\quad = - \mu \, a^{3} \, \sum _{j=1}^{\aleph } \, \frac {\omega ^{2}_{P_{\ell },n_{0},j} \, \epsilon _{0}(z_{j}) \, \big(\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) \big)}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \, \left \langle V(z_{j}, \theta , q), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \\[2pt] &\qquad\cdot \left \langle G^{\infty }_{k}(\hat {x},z_{j}) \cdot \int _{B} e_{n_{0}}^{(3)}(y) \, dy , ( \hat {x} \times q ) \right \rangle + \mathcal{O}\big( a^{\min \left ( (3-s), \left(6-3t-2h-\frac {3s}{2}\right) \right )}\big), \end{align*}

or, equivalently

\begin{align*} & \left \langle \left ( E^{\infty } - V^{\infty }\right )(\hat {x}) , ( \hat {x} \times q ) \right \rangle \\ &\quad = - \mu \, a^{3} \, \sum _{j=1}^{\aleph } \, \frac {\omega ^{2}_{P_{\ell },n_{0},j} \, \epsilon _{0}(z_{j}) \, \big(\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) \big)}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \, \left \langle V(z_{j}, \theta , q), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \\ &\qquad\cdot \left \langle \int _{B} e_{n_{0}}^{(3)}(y) \, dy , \left ( G^{\infty }_{k}\right )^{\top }(\hat {x},z_{j}) \cdot ( \hat {x} \times q ) \right \rangle +\mathcal{O}\big( a^{\min \left ( (3-s), \left(6-3t-2h-\frac {3s}{2}\right) \right )}\big). \end{align*}

Thanks to the Mixed Electromagnetic Reciprocity Relation in Proposition 3.1, formula (3.44), the above equation becomes

\begin{align*} & \langle ( E^{\infty } - V^{\infty } )(\hat {x}) , ( \hat {x} \times q ) \rangle \\ &\quad = \frac { \mu \, a^{3}}{4 \, \pi } \, \sum _{j=1}^{\aleph } \, \frac {\omega ^{2}_{P_{\ell },n_{0},j} \, \epsilon _{0}(z_{j}) \, \big(\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) \big)}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \, \left \langle V(z_{j}, \theta , q), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle\\ &\qquad \cdot\left \langle V(z_{j}, - \hat {x}, q ), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle + \mathcal{O}\big( a^{\min \left ( (3-s), \left(6-3t-2h-\frac {3s}{2}\right) \right )}\big), \end{align*}

which proves (1.13). In the particular case for the back-scattered direction, i.e. $ \hat {x} = - \, \theta$ , we deduce

\begin{align*} & \left \langle \left ( E^{\infty } - V^{\infty }\right )({-} \theta ) , \left ( \theta \times q \right ) \right \rangle \\ &\quad= - \frac {\mu \, a^{3}}{4 \, \pi } \, \sum _{j=1}^{\aleph } \, \frac {\omega ^{2}_{P_{\ell },n_{0},j} \, \epsilon _{0}(z_{j}) \, \big (\epsilon _{0}(z_{j}) - \Lambda _{n_{0},j}\big( \omega _{P_{\ell },n_{0},j}\big) \big)}{\lambda _{n_{0}}^{(3)}(B) \, \Lambda _{n_{0},j} ( \omega )} \, \left ( \left \langle V (z_{j}, \theta , q ), \int _{B} e_{n_{0}}^{(3)}(y) \, dy \right \rangle \right )^{2} \\ &\qquad+ \mathcal{O}\big( a^{\min \left ( (3-s), \left(6-3t-2h-\frac {3s}{2}\right) \right )}\big). \end{align*}

This ends the proof Theorem1.1.

4. Conclusion

In the recent years, there has been a growing interest, in the engineering community, in using contrast agents for different imaging modalities as ultrasound, optic tomography, elastography and different hybridized modalities. Such contrast agents are usually given by micro-scaled bubbles, nano-scaled particles or micro-mechanical inclusions. In this work, we described and analysed a quantitative optical imaging modality using plasmonic nanoparticles as contrast agents. We provided with a novel method to reconstruct the permittivity distribution, of an object to image, from the remotely measured electromagnetic fields.

The key principle is that these nanoparticles enjoy resonant effects and hence have the potential in enhancing the applied incident fields, while excited at certain particular frequencies called plasmonic resonances. We characterized these resonances, at the subwavelenght regime, as related to the eigenvalues of the Magnetization operator and then derived the generated electromagnetic fields in a heterogeneous medium containing multiple of such plasmonic nanoparticles. Based on these expansions, we designed an imaging functional, built-up from the remotely measured electric fields, that depends solely on the used incident frequencies. We showed that such a functional reaches its maximum values, in terms of the used incident frenquencies, only on the related plasmonic resonances. This allowed us to recover these resonances from which we extract the values of the electric permittivity distribution.

The proposed idea looks to be original and promising as it is fundamentally based on the resonant character of the contrast agents. As next steps, we need to confirm this approach computationally by providing numerical tests. In addition, we need to improve it in a number of directions. For example, we plan to study this approach while injecting the nanoparticles all-at-once and not one-by-one as it is proposed currently. To tackle this, we propose first to consider the case when these contrast agents are distributed regularly following a given density of distribution. In this case, we expect deterministic estimates. The ultimate step is to handle the case where such contrast agents are sent randomly and then derive probabilistic estimates.