A Breakthrough in Resonance Analysis for Complex Multilayered Structures in Small, High-Contrast Nonlinear Media with Kerr Effect

1. Resonances in Small, Doublelayered, High-Contrast, Nonlinear Scatterers with Arbitrary Geometry

1.1. The Original Work

In this section we present the results in [4], as we use aspects of the solution method to compute the new formulae. The reader can refer to [4] for more details and examples.

Consider a small volume high contrast nonlinear medium exhibiting Kerr effect of arbitrary geometry. The governing equation is

$$\Delta u+k^2(1+\eta \left(x\right))u+k^2\beta \left(x\right){\left|u\right|}^{2}u=0$$

subject to Sommerfeld radiation condition at infinity.

where k is the wave number, $\eta$ is the medium susceptibility coefficient, and u is the scalar field. Here, $\lambda =k^2$ is the spectral parameter.

Lippmann-Schwinger Integral solution is given by:

$$\begin{array}{cc}\hfill u\left(x\right)& =u_i\left(x\right)+k^2{\int }_{hB}{\displaystyle \frac{{\eta }_0\left(y\right)}{h^2}}G(x,y)u\left(y\right)dy\hfill \\ \hfill & +k^2{\int }_{hB}{\displaystyle \frac{{\beta }_0\left(y\right)}{h^2}}G(x,y){\left|u\left(y\right)\right|}^{2}u\left(y\right)dy,\hfill \end{array}$$

where $u_i$ is the incident field and G in 3D is defined as:

$$G(x,y)={\displaystyle \frac{1}{4\pi }}{\displaystyle \frac{exp(i\sqrt{\lambda }|x-y|)}{|x-y|}}$$

Setting $u_i=0$, we obtain the integral eigenvalue problem that is nonlinear in both $\lambda$ and u:

$$\lambda T\left(\lambda \right)u=u$$

where

$$T\left(\lambda \right)\left(u\right)\left(x\right)={\displaystyle \frac{1}{4\pi }}{\int }_{hB}\eta \left(y\right){\displaystyle \frac{exp(i\sqrt{\lambda }|x-y|)}{|x-y|}}u\left(y\right)dy$$

$$+{\displaystyle \frac{1}{4\pi }}{\int }_{hB}\beta \left(y\right){\displaystyle \frac{exp(i\sqrt{\lambda }|x-y|)}{|x-y|}}{\left|u\left(y\right)\right|}^{2}u\left(y\right)dy$$

By scaling the spatial variables and the media properties as follows:

$$\eta ={\chi }_{hB}{\displaystyle \frac{{\eta }_0}{h^2}},\beta ={\chi }_{hB}{\displaystyle \frac{{\beta }_0}{h^2}},x=h\tilde{x},y=h\tilde{y}$$

we obtain:

$${\lambda }_h{T}_h\left({\lambda }_h\right)u_h=u_h$$

where

$$T_h\left(\lambda \right)\left(u\right)\left(\tilde{x}\right)={\displaystyle \frac{1}{4\pi }}{\int }_B{\eta }_0\left(h\tilde{y}\right){\displaystyle \frac{exp(i\sqrt{\lambda }h|\tilde{x}-\tilde{y}\left|\right)}{|\tilde{x}-\tilde{y}|}}u\left(\tilde{y}\right)d\tilde{y}$$

(1)

$$+{\displaystyle \frac{1}{4\pi }}{\int }_B{\beta }_0\left(h\tilde{y}\right){\displaystyle \frac{exp(i\sqrt{\lambda }h|\tilde{x}-\tilde{y}\left|\right)}{|\tilde{x}-\tilde{y}|}}u\left(\tilde{y}\right){\left|u\left(\tilde{y}\right)\right|}^{2}d\tilde{y}$$

The associated limiting eigenvalue problem, as $h\to 0$ is:

$${\lambda }_0{T}_0\left({\lambda }_0\right)u_0=u_0,$$

where

$$T_0\left(u\right)\left(\tilde{x}\right)={\displaystyle \frac{1}{4\pi }}{\int }_B{\eta }_0\left(0\right){\displaystyle \frac{u\left(\tilde{y}\right)}{|\tilde{x}-\tilde{y}|}}d\tilde{y}+{\displaystyle \frac{1}{4\pi }}{\int }_B{\beta }_0\left(0\right){\displaystyle \frac{u\left(\tilde{y}\right)}{|\tilde{x}-\tilde{y}|}}{\left|u\left(\tilde{y}\right)\right|}^{2}d\tilde{y}$$

Now that $T_h$ and $T_0$ are well defined, we can compute the asymptotic formula by executing the inner product in the formula

$${\lambda }_h={\lambda }_0+{{\lambda }_0}^2〈(T_0-T_h\left({\lambda }_0\right))u_0,u_0〉+\mathcal{O}\left(h^2\right)$$

$T_0$ is a self-adjoint compact operator with positive eigenvalues [7]. Performing Taylor exapnsion on the functions

$$h⟼{\eta }_0\left(h\tilde{y}\right){\displaystyle \frac{exp(i\sqrt{\lambda }h|\tilde{x}-\tilde{y}\left|\right)}{|\tilde{x}-\tilde{y}|}}\mathrm{and}h⟼{\beta }_0\left(h\tilde{y}\right){\displaystyle \frac{exp(i\sqrt{\lambda }h|\tilde{x}-\tilde{y}\left|\right)}{|\tilde{x}-\tilde{y}|}}$$

gives a first order approximation of the resonances as follows. See [4] for more details. Let

$$U_0={\int }_B{u}_0\left(x\right)dx,\mathrm{and}{u}_h=u_0+u_{1}h+o\left(h\right)$$

The value ${\lambda }_1$ in ${\lambda }_h={\lambda }_0+{\lambda }_{1}h+\mathcal{O}\left(h^2\right)$ is given by:

$$\begin{array}{c}-4\pi {\lambda }_1=2{\beta }_0\left(0\right){\lambda }_0^2{\int }_B{\int }_B{\displaystyle \frac{u_0\left(y\right)Re\left(u_1\left(y\right)\right)}{|x-y|}}{u}_0\left(x\right)dydx\hfill \\ +{\lambda }_0^2{\int }_B{\int }_B{\displaystyle \frac{\nabla {\eta }_0\left(0\right)·y{u}_0\left(y\right)}{|x-y|}}{u}_0\left(x\right)dydx+i{\lambda }_0^{{\displaystyle \frac{5}{2}}}{\eta }_0\left(0\right)U_0^2\\ \hfill +{\lambda }_0^2{\int }_B{\int }_B{\displaystyle \frac{\nabla {\beta }_0\left(0\right)·y{u}_0^3\left(y\right)}{|x-y|}}{u}_0\left(x\right)dydx+i{\lambda }_0^{{\displaystyle \frac{5}{2}}}{\beta }_0\left(0\right)U_0{\int }_B{u}_0^3\left(y\right)dy\end{array}$$

(2)

$$\begin{array}{c}\mathrm{Re}(u_1\left)\mathrm{obeys}\text{:}4\pi Re\right(u_1\left(x\right))=\hfill \\ {\lambda }_0{\int }_B{\displaystyle \frac{\nabla {\eta }_0·y{u}_0+\nabla {\beta }_0·y{u}_0^3}{|x-y|}}dy\\ +{\lambda }_0{\int }_B{\displaystyle \frac{Re\left(u_1\right)}{|x-y|}}({\eta }_0+3{\beta }_0{u}_0^2)dy\\ -{\displaystyle \frac{{\lambda }_0}{4\pi }}{u}_0\left(x\right)(2{\beta }_0\left(0\right){\int }_B{\int }_B{\displaystyle \frac{u_0\left(y\right)Re\left(u_1\left(y\right)\right)}{|x-y|}}{u}_0\left(x\right)dydx\\ \hfill +{\int }_B{\int }_B{\displaystyle \frac{\nabla {\eta }_0\left(0\right)·y{u}_0\left(y\right)}{|x-y|}}{u}_0\left(x\right)dydx+{\int }_B{\int }_B{\displaystyle \frac{\nabla {\beta }_0\left(0\right)·y{u}_0^3\left(y\right)}{|x-y|}}dydx)\end{array}$$

In practice, we will show that few terms in formula (2) are zero. For example, when ${\eta }_0$ and ${\beta }_0$ are constants, corollary 3.1 shows that:

$$\begin{array}{cc}\hfill -4\pi {\lambda }_1=2{\beta }_0{\lambda }_0^2{\int }_B{\int }_B{\displaystyle \frac{u_0\left(y\right)Re\left(u_1\right)}{|x-y|}}{u}_0\left(x\right)dydx+& i{\lambda }_0^{{\displaystyle \frac{5}{2}}}{\eta }_0{U}_0^2+i{\lambda }_0^{{\displaystyle \frac{5}{2}}}{\beta }_0{U}_0{U}_1\hfill \end{array}$$

(3)

where

$$U_0={\int }_B{u}_0\left(x\right)dx,U_1={\int }_B{u}_0{\left(x\right)}^{3}dx$$

and $Re\left(u_1\right)$ solves the integral equation:

$$\begin{array}{cc}\hfill 4\pi Re\left(u_1\left(x\right)\right)=& {\lambda }_0{\int }_B{\displaystyle \frac{Re\left(u_1\right)}{|x-y|}}({\eta }_0+3{\beta }_0{u}_0^2)dy\hfill \\ \hfill & -{\displaystyle \frac{{\lambda }_0}{4\pi }}2{\beta }_0{u}_0\left(x\right){\int }_B{\int }_B{\displaystyle \frac{u_0\left(y\right)Re\left(u_1\right)}{|x-y|}}{u}_0\left(x\right)dydx.\hfill \end{array}$$

(4)

Furthermore, in the specific case of a spherical scatterer, we found that $R\left(u_1\right)$ vanishes, i.e., $R\left(u_1\right)=0$.

While the general derivation presented in equation (2) may appear complex and potentially challenging to implement due to the non-constant nature of $\eta$ and $\beta$, and the arbitrary geometry involved, it often simplifies significantly in practical physical scenarios. This apparent inconvenience actually proves to be a major advantage, a point this paper will thoroughly illustrate.

1.2. Case 1: Single-Layer Kerr Effect

Consider the scattering problem of high contrast small volume, $hB$. The small scatterer has two concentric inner and outer layers, with the latter exhibiting nonlinear Kerr effect.

$$hB=h{B}_{\mathrm{in}}\cup h{B}_{\mathrm{out}}$$

$$\eta \left(x\right)={\chi }_{h{B}_{\mathrm{in}}}\left(x\right){\eta }_{\mathrm{in}}+{\chi }_{h{B}_{\mathrm{out}}}\left(x\right){\eta }_{\mathrm{out}}={\chi }_{h{B}_{\mathrm{in}}}{\displaystyle \frac{{\eta }_{\mathrm{in}}}{h^2}}+{\chi }_{h{B}_{\mathrm{out}}}{\displaystyle \frac{{\eta }_{\mathrm{out}}}{h^2}}$$

$$\beta \left(x\right)={\chi }_{h{B}_{out}}{\displaystyle \frac{{\beta }_0}{h^2}}$$

for constant ${\eta }_0$ and ${\beta }_0$.

Within the context of this new configuration, the operators $T_h$, previously defined by (1), can be expressed as:

$$\begin{array}{cc}\hfill T_h\left(\lambda \right)\left(u\right)\left(x\right)& ={\displaystyle \frac{{\eta }_{in}^0}{4\pi }}{\int }_{B_{in}}{\displaystyle \frac{exp(i\sqrt{\lambda }h|x-y|)}{|x-y|}}u\left(y\right)dy\hfill \\ \hfill & +{\displaystyle \frac{{\eta }_{out}^0}{4\pi }}{\int }_{B_{out}}{\displaystyle \frac{exp(i\sqrt{\lambda }h|x-y|)}{|x-y|}}u\left(y\right)dy\hfill \\ \hfill & +{\displaystyle \frac{{\beta }_0}{4\pi }}{\int }_{B_{out}}{\displaystyle \frac{exp(i\sqrt{\lambda }h|x-y|)}{|x-y|}}u\left(y\right){\left|u\left(y\right)\right|}^{2}dy\hfill \end{array}$$

Hence we define the limiting operator, $T_0$, as $h\to 0$ as:

$$\begin{array}{cc}\hfill T_0\left(\lambda \right)\left(u\right)\left(x\right)& ={\displaystyle \frac{{\eta }_{in}^0}{4\pi }}{\int }_{B_{in}}{\displaystyle \frac{u\left(y\right)}{|x-y|}}dy\hfill \\ \hfill & +{\displaystyle \frac{{\eta }_{out}^0}{4\pi }}{\int }_{B_{out}}{\displaystyle \frac{u\left(y\right)}{|x-y|}}dy\hfill \\ \hfill & +{\displaystyle \frac{{\beta }_0}{4\pi }}{\int }_{B_{out}}{\displaystyle \frac{u\left(y\right)}{|x-y|}}{\left|u\left(y\right)\right|}^{2}dy\hfill \end{array}$$

As it was introduced in section 1.1, we are solving for the resonances, ${\lambda }_h$, the eigenvalue problem

$${\lambda }_h{T}_h{u}_h=u_h$$

via the limiting eigenvalue problem

$${\lambda }_0{T}_0{u}_0=u_0$$

In other words, we first solve the easier latter equation for $u_0$ and ${\lambda }_0$. Now let us expand $T_h$ by performing Taylor expansion of $h⟼exp\left(i\sqrt{\lambda }h\right|x-y\left|\right)$:

$$\begin{array}{cc}\hfill & T_h\left(\lambda \right)\left(u\right)\left(x\right)=T_0\left(\lambda \right)\left(u\right)\left(x\right)+{\displaystyle \frac{i\sqrt{\lambda }h}{4\pi }}\left({\eta }_{in}^0{\int }_{B_{in}}u\left(y\right)dy\right.\hfill \\ \hfill & \left.+{\eta }_{out}^0{\int }_{B_{out}}u\left(y\right)dy+{\beta }_0{\int }_{B_{out}}u\left(y\right){\left|u\left(y\right)\right|}^{2}dy\right)+\mathcal{O}\left(h^2\right)\hfill \end{array}$$

Evaluating at $u_0$ and ${\lambda }_0$ we obtain:

$$\begin{array}{cc}\hfill & (T_0-T_h\left({\lambda }_0\right))u_0=-{\displaystyle \frac{i\sqrt{{\lambda }_0}h}{4\pi }}\left({\eta }_{in}^0{\int }_{B_{in}}{u}_0\left(y\right)dy\right.\hfill \\ \hfill & \left.+{\eta }_{out}^0{\int }_{B_{out}}{u}_0\left(y\right)dy+{\beta }_0{\int }_{B_{out}}{u}_0^3\left(y\right)dy\right)+\mathcal{O}\left(h^2\right)\hfill \end{array}$$

or

$$(T_0-T_h\left({\lambda }_0\right))u_0=-{\displaystyle \frac{i\sqrt{{\lambda }_0}h}{4\pi }}\left({\eta }_{in}^0{U}_{in}+{\eta }_{out}^0{U}_{out}+{\beta }_0{U}_{out\beta }\right)+\mathcal{O}\left(h^2\right)$$

where

$$U_{in}={\int }_{B_{in}}{u}_0\left(y\right)dy,U_{out}={\int }_{B_{out}}{u}_0\left(y\right)dy$$

and

$$U_{out\beta }={\int }_{B_{out}}{u}_0^3\left(y\right)dy$$

At this point, we are ready to compute ${\lambda }_1$ by executing the expansion asymptotic formula:

$${\lambda }_h={\lambda }_0+{\lambda }_0^2\langle (T_0-T_h\left({\lambda }_0\right))u_0,u_0\rangle +\mathcal{O}\left(h^2\right)$$

(5)

Let

$${\lambda }_h={\lambda }_0+{\lambda }_{1}h+\mathcal{O}\left(h^2\right)$$

It follows that the correction ${\lambda }_1$ is

$${\lambda }_{\mathbf{1}}=-{\displaystyle \frac{\mathbf{i}{\lambda }_{\mathbf{0}}^{\frac{\mathbf{5}}{\mathbf{2}}}}{\mathbf{4}\pi }}\left({\eta }_{\mathbf{in}}^{\mathbf{0}}{\mathbf{U}}_{\mathbf{in}}+{\eta }_{\mathbf{out}}^{\mathbf{0}}{\mathbf{U}}_{\mathbf{out}}+{\beta }_{\mathbf{0}}{\mathbf{U}}_{\mathbf{out}\beta }\right){\mathbf{U}}_{\mathbf{0}}$$

where

$$U_0={\int }_B{u}_0\left(y\right)dy$$

1.3. Case 2: Dual-Layer Kerr Effect

This configuration offers greater flexibility in designing the 3D scatterer, allowing for scenarios with either distinct or identical susceptibility coefficients (${\eta }_{\mathrm{in}}\ne {\eta }_{\mathrm{out}}$ or ${\eta }_{\mathrm{in}}={\eta }_{\mathrm{out}}$) and Kerr coefficients (${\beta }_{\mathrm{in}}\ne {\beta }_{\mathrm{out}}$ or ${\beta }_{\mathrm{in}}={\beta }_{\mathrm{out}}$). All coefficients are assumed to be constant.

$$\beta \left(x\right)={\chi }_{h{B}_{in}}{\displaystyle \frac{{\beta }_{in}^0}{h^2}}+{\chi }_{h{B}_{out}}{\displaystyle \frac{{\beta }_{out}^0}{h^2}}$$

In this case $T_h$ and $T_0$ write:

$$\begin{array}{cc}\hfill T_h\left(\lambda \right)\left(u\right)\left(x\right)& ={\displaystyle \frac{{\eta }_{in}^0}{4\pi }}{\int }_{B_{in}}{\displaystyle \frac{exp(i\sqrt{\lambda }h|x-y|)}{|x-y|}}u\left(y\right)dy\hfill \\ \hfill & +{\displaystyle \frac{{\eta }_{out}^0}{4\pi }}{\int }_{B_{out}}{\displaystyle \frac{exp(i\sqrt{\lambda }h|x-y|)}{|x-y|}}u\left(y\right)dy\hfill \\ \hfill & +{\displaystyle \frac{{\beta }_{in}^0}{4\pi }}{\int }_{B_{in}}{\displaystyle \frac{exp(i\sqrt{\lambda }h|x-y|)}{|x-y|}}u\left(y\right){\left|u\left(y\right)\right|}^{2}dy\hfill \\ \hfill & +{\displaystyle \frac{{\beta }_{out}^0}{4\pi }}{\int }_{B_{out}}{\displaystyle \frac{exp(i\sqrt{\lambda }h|x-y|)}{|x-y|}}u\left(y\right){\left|u\left(y\right)\right|}^{2}dy\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill T_0\left(\lambda \right)\left(u\right)\left(x\right)& ={\displaystyle \frac{{\eta }_{in}^0}{4\pi }}{\int }_{B_{in}}{\displaystyle \frac{u\left(y\right)}{|x-y|}}dy\hfill \\ \hfill & +{\displaystyle \frac{{\eta }_{out}^0}{4\pi }}{\int }_{B_{out}}{\displaystyle \frac{u\left(y\right)}{|x-y|}}dy\hfill \\ \hfill & +{\displaystyle \frac{{\beta }_{in}^0}{4\pi }}{\int }_{B_{in}}{\displaystyle \frac{u\left(y\right)}{|x-y|}}{\left|u\left(y\right)\right|}^{2}dy\hfill \\ \hfill & +{\displaystyle \frac{{\beta }_{out}^0}{4\pi }}{\int }_{B_{out}}{\displaystyle \frac{u\left(y\right)}{|x-y|}}{\left|u\left(y\right)\right|}^{2}dy\hfill \end{array}$$

Similar to case 1, we perform a Taylor expansion of $h⟼exp\left(i\sqrt{\lambda }h\right|x-y\left|\right)$, yielding:

$$\begin{array}{cc}\hfill & T_h\left(\lambda \right)\left(u\right)\left(x\right)=T_0\left(\lambda \right)\left(u\right)\left(x\right)+{\displaystyle \frac{i\sqrt{\lambda }h}{4\pi }}\left({\eta }_{in}^0{\int }_{B_{in}}u\left(y\right)dy\right.\hfill \\ \hfill & \left.+{\eta }_{out}^0{\int }_{B_{out}}u\left(y\right)dy+{\beta }_{in}^0{\int }_{B_{in}}{u\left(y\right)\left|u\left(y\right)\right|}^{2}dy+{\beta }_{out}^0{\int }_{B_{out}}u\left(y\right){\left|u\left(y\right)\right|}^{2}dy\right)+\mathcal{O}\left(h^2\right)\hfill \end{array}$$

Subsequently, the correction ${\lambda }_1$ is derived by evaluating the inner product in equation (5).

$${\lambda }_{\mathbf{1}}=-{\displaystyle \frac{\mathbf{i}{\lambda }_{\mathbf{0}}^{\frac{\mathbf{5}}{\mathbf{2}}}}{\mathbf{4}\pi }}\left({\eta }_{\mathbf{in}}^{\mathbf{0}}{\mathbf{U}}_{\mathbf{in}}+{\eta }_{\mathbf{out}}^{\mathbf{0}}{\mathbf{U}}_{\mathbf{out}}+{\beta }_{\mathbf{in}}^{\mathbf{0}}{\mathbf{U}}_{\mathbf{in}\beta }+{\beta }_{\mathbf{out}}^{\mathbf{0}}{\mathbf{U}}_{\mathbf{out}\beta }\right){\mathbf{U}}_{\mathbf{0}}$$

where

$$U_{in\beta }={\int }_{B_{in}}{u}_0^3\left(y\right)dy,U_{out\beta }={\int }_{B_{out}}{u}_0^3\left(y\right)dy$$

$$U_{in}={\int }_{B_{in}}{u}_0\left(y\right)dy,U_{out}={\int }_{B_{out}}{u}_0\left(y\right)dy,\mathrm{and}{U}_0={\int }_B{u}_0\left(y\right)dy$$

Note that we recover ${\lambda }_1$ in case 1 by setting ${\beta }_{in}^0=0$. Furthermore, setting ${\beta }_{out}^0=0$ provides the formula for ${\lambda }_1$ when only the inner layer exhibits the Kerr effect.

2. Resonances in Small, Multilayered, High-Contrast, Nonlinear Scatterers with Arbitrary Geometry

Suppose $B_h$ is composite and optically inhomogenous with n layers of concentric high contrast media $\{B_i{\}}_{1\le i\le n}$, such that

$$hB={\cup }_{i=1}^{i=n}h{B}_i,$$

$$\eta \left(x\right)=\sum _{i=1}^{i=n}{\chi }_{h{B}_i}\left(x\right){\eta }^i=\sum _{i=1}^{i=n}{\chi }_{h{B}_i}\left(x\right){\displaystyle \frac{{\eta }_0^i}{h^2}},$$

$$\beta \left(x\right)=\sum _{i=1}^{i=n}{\chi }_{h{B}_i}\left(x\right){\beta }^i=\sum _{i=1}^{i=n}{\chi }_{h{B}_i}\left(x\right){\displaystyle \frac{{\beta }_0^i}{h^2}},$$

$${\eta }_0\left(x\right)=\sum _{i=1}^{i=n}{\chi }_{B_i}\left(x\right){\eta }_0^i,{\beta }_0\left(x\right)=\sum _{i=1}^{i=n}{\chi }_{B_i}\left(x\right){\beta }_0^i$$

Similar to the previous cases, the operators $T_h$ and $T_0$ write:

$$\begin{array}{c}{T}_h\left(u\right)\left(x\right)\left(\lambda \right)=\sum _{k=1}^{k=n}{\displaystyle \frac{{\eta }_0^k}{4\pi }}{\int }_{B_k}{\displaystyle \frac{exp\left(i\sqrt{\lambda }h\right|x-y|)}{|x-y|}}u\left(y\right)dy\hfill \\ \hfill +{\displaystyle \frac{{\beta }_0^k}{4\pi }}{\int }_{B_k}{\displaystyle \frac{exp\left(i\sqrt{\lambda }h\right|x-y|)}{|x-y|}}u\left(y\right){\left|u\left(y\right)\right|}^{2}dy\end{array}$$

(6)

$$T_0\left(u\right)\left(x\right)=\sum _{k=1}^{k=n}{\displaystyle \frac{{\eta }_0^k}{4\pi }}{\int }_{B_k}{\displaystyle \frac{1}{|x-y|}}u\left(y\right)dy+{\displaystyle \frac{{\beta }_0^k}{4\pi }}{\int }_{B_k}{\displaystyle \frac{1}{|x-y|}}u\left(y\right){\left|u\left(y\right)\right|}^{2}dy$$

(7)

We perform a Taylor expansion of $h⟼exp\left(i\sqrt{\lambda }h\right|x-y\left|\right)$ again, yielding:

$$T_h\left(\lambda \right)\left(u\right)\left(x\right)=T_0\left(\lambda \right)\left(u\right)\left(x\right)+{\displaystyle \frac{i\sqrt{\lambda }h}{4\pi }}\left(\sum _{k=1}^n{\eta }_0^k{\int }_{B_k}u\left(y\right)dy+{\beta }_0^k{\int }_{B_k}u\left(y\right){\left|u\left(y\right)\right|}^{2}dy+\right)+\mathcal{O}\left(h^2\right)$$

We can now evaluate (5) at $u_0$ and ${\lambda }_0$, which gives:

$${\lambda }_{\mathbf{h}}={\lambda }_{\mathbf{0}}-\mathbf{i}{\displaystyle \frac{{{\lambda }_{\mathbf{0}}}^{\frac{\mathbf{5}}{\mathbf{2}}}}{\mathbf{4}\pi }}{\mathbf{U}}_{\mathbf{0}}\left[\sum _{\mathbf{k}=\mathbf{1}}^{\mathbf{n}}{\eta }_{\mathbf{0}}^{\mathbf{k}}{\mathbf{U}}_{\mathbf{k}}+{\beta }_{\mathbf{0}}^{\mathbf{k}}{\mathbf{U}}_{\mathbf{k}}^{\beta }\right]\mathbf{h}+\mathcal{O}\left({\mathbf{h}}^{\mathbf{2}}\right)$$

where

$$U_0={\int }_B{u}_0\left(x\right)dx,$$

and

$$U_k={\int }_{B_k}{u}_0\left(x\right)dx,U_k^{\beta }={\int }_{B_k}{u}_0^3\left(x\right)dx,1\le k\le n.$$

In the future, we aim to investigate the higher-order terms, such as the second-order component of ${\lambda }_h$, to gain better control over the physical parameters contributing to the scattering phenomenon in small, high-contrast nonlinear media. A similar investigation was conducted for linear media in [18].