Cloaking using complementary media for the Helmholtz equation and a three spheres inequality for second order elliptic equations

Abstract

Cloaking using complementary media was suggested by Lai et al. in 2009. This was proved by H.-M. Nguyen (2015) in the quasistatic regime. One of the difficulties in the study of this problem is the appearance of the localized resonance, i.e., the fields blow up in some regions and remain bounded in some others as the loss goes to 0. To this end, H.-M. Nguyen introduced the technique of removing localized singularity and used a standard three spheres inequality. The method used also works for the Helmholtz equation. However, it requires small size of the cloaked region for large frequency due to the use of the (standard) three spheres inequality. In this paper, we give a proof of cloaking using complementary media in the finite frequency regime without imposing any condition on the cloaked region; the cloak works for an arbitrary fixed frequency provided that the loss is sufficiently small. To successfully apply the above approach of Nguyen, we establish a new three spheres inequality. A modification of the cloaking setting to obtain illusion optics is also discussed.

1. Introduction

Negative index materials (NIMs) were investigated theoretically by Veselago in 36. The existence of such materials was confirmed by Shelby, Smith, and Schultz in 35. The study of NIMs has attracted a lot of attention in the scientific community thanks to their interesting properties and applications. An appealing one is cloaking using complementary media.

Cloaking using NIMs or more precisely cloaking using complementary media was suggested by Lai et al. in 11. Their work was inspired by the notion of complementary media suggested by Pendry and Ramakrishna in 32. Cloaking using complementary media was established in 21 in the quasistatic regime using slightly different schemes from 11. Two difficulties in the study of cloaking using complementary media are as follows. Firstly, this problem is unstable since the equations describing the phenomenon have sign changing coefficients, hence the ellipticity and the compactness are lost. Secondly, the localized resonance, i.e., the field blows up in some regions and remains bounded in some others, might appear. To handle these difficulties, in 21 the author introduced the removing localized singularity technique and used a standard three spheres inequality. The approach in 21 also involved the reflecting technique introduced in 18. The method in 21 also works for the Helmholtz equation; however since the largest radius in the (standard) three spheres inequality is small as frequency is large (see Section 2 for further discussion), the size of the cloaked region is required to be small for large frequency.

In this paper, we present a proof of cloaking using complementary media in the finite frequency regime. Our goal is to not impose any condition on the size of the cloaked region (Theorem 1); the cloak works for an arbitrary fixed frequency as long as the loss is sufficiently small. To successfully apply the approach in 21, we establish a new three spheres inequality for the second order elliptic equations which holds for arbitrary radius (Theorem 2 in Section 2). This inequality is inspired from the unique continuation principle and its proof is in the spirit of Protter in 34. A modification of the cloaking setting to obtain illusion optics is discussed in Section 4 (Theorem 3). This involves the idea of superlensing in 19. Cloaking using complementary media for electromagnetic waves is investigated in 24.

In addition to cloaking using complementary media, other application of NIMs are superlensing using complementary media as suggested in 293033 (see also 28) and confirmed in 1922, and cloaking via anomalous localized resonance 15 (see also 31020). Complementary media were studied in a general setting in 1822 and played an important role in these applications; see 171920212225.

Let us describe the problem more precisely. Assume that the cloaked region is the annulus $B_{\gamma r_2} \setminus B_{r_2}$ for some $r_{2}> 0$ and $1 < \gamma < 2$ in which the medium is characterized by a matrix $a$ and a function $\sigma$. The assumption on the cloaked region by all means imposes no restriction since any bounded set is a subset of such a region provided that the radius and the origin are appropriately chosen. The idea suggested by Lai et al. in 11 in two dimensions is to construct its complementary medium in $B_{r_2} \setminus B_{r_1}$ for some $0 < r_1 < r_2$.

In this paper, instead of taking the schemes of Lai et al., we use a scheme from 21 which is inspired but different from the ones from 11. Following 21, the cloak contains two parts. The first one, in $B_{r_2} \setminus B_{r_1}$, makes use of complementary media to cancel the effect of the cloaked region, and the second one, in $B_{r_1}$, is to fill the space which “disappears” from the cancellation by the homogeneous media. Concerning the first part, instead of $B_{\gamma r_2} \setminus B_{r_2}$, we consider $B_{r_3} \setminus B_{r_2}$ with $r_3 = 2 r_2$ (the constant $2$ considered here is just a matter of simple representation) as the cloaked region in which the medium is given by

$$\begin{equation*} \hat{a}, \hat{\sigma }= \left\{ \begin{array}{cl} a, \sigma & \text{ in } B_{\gamma r_2} \setminus B_{r_2}, \\[6.0pt] I, 1 & \text{ in } B_{r_3} \setminus B_{\gamma r_2}. \end{array} \right. \end{equation*}$$

The complementary medium in $B_{r_2} \setminus B_{r_1}$ is given by

$$\begin{equation*} - F^{-1}_*\hat{a}, - F^{-1}_*\hat{\sigma }, \end{equation*}$$

where $F: B_{r_2} \setminus \bar{B}_{r_1} \to B_{r_3} \setminus \bar{B}_{r_2}$ is the Kelvin transform with respect to $\partial B_{r_2}$, i.e.,

$$\begin{equation} F(x) = \frac{r_2^2}{|x|^2} x. {\tag{1.1}} \end{equation}$$

Here

$$\begin{equation*} T_*\hat{a} (y) = \frac{D T (x) \hat{a}(x) D T (x)^T}{J(x)} \quad \text{ and } \quad T_* \hat{\sigma }(y) = \frac{\hat{\sigma }(x)}{J(x)}, \end{equation*}$$

where $x =T^{-1}(y)$ and $J(x) = |\det D T(x)|$ for a diffeomorphism $T$. It follows that

$$\begin{equation} r_1 = r_2^2/ r_3. {\tag{1.2}} \end{equation}$$

Concerning the second part, the medium in $B_{r_1}$ is given by

$$\begin{equation} \Big (r_3^2/r_2^2 \Big )^{d-2} I, \Big (r_3^2/r_2^2 \Big )^{d}. {\tag{1.3}} \end{equation}$$

The reason for this choice will be explained later.

With the loss, the medium is characterized by $s_\delta A, s_0 \Sigma$ ($\delta > 0$), where

$$\begin{equation} A, \Sigma = \left\{ \begin{array}{cl} \hat{a}, \hat{\sigma }& \text{ in } B_{r_3} \setminus B_{r_2}, \\[6.0pt] F^{-1}_*\hat{a}, F^{-1}_* \hat{\sigma }& \text{ in } B_{r_2} \setminus B_{r_1}, \\[6.0pt] \Big (r_3^2/r_2^2 \Big )^{d-2} I, \Big (r_3^2/r_2^2 \Big )^{d} & \text{ in } B_{r_1},\\[6.0pt] I, 1 & \text{ otherwise}, \end{array} \right. {\tag{1.4}} \end{equation}$$

and, for $\delta \ge 0$,

$$\begin{equation} s_\delta : = \left\{ \begin{array}{cl} -1 + i \delta & \text{ in } B_{r_2} \setminus B_{r_1}, \\[6.0pt] 1 & \text{ otherwise}. \end{array} \right. {\tag{1.5}} \end{equation}$$

Physically, the imaginary part of $s_\delta A$ is the loss of the medium (more precisely the loss of the medium in $B_{r_2} \setminus B_{r_1}$). Here and in what follows, we assume that

$$\begin{equation} \frac{1}{\Lambda } |\xi |^{2} \le a (x) \xi \cdot \xi \le \Lambda |\xi |^{2} \quad \forall \, \xi \in \mathbb{R}^{d}, \text{ for a.e. } x \in B_{\gamma r_{2}} \setminus B_{r_2}, {\tag{1.6}} \end{equation}$$

for some $\Lambda \ge 1$. In what follows, we assume in addition that

$$\begin{equation} \hat{a} \, \text{is Lipschitz in}\, B_{r_3} \setminus B_{r_2}. \tag{1.7} \end{equation}$$

One can verify that medium $(s_0 A,s_0\Sigma )$ is of reflecting complementary property, a concept introduced in 18, Definition 1, by considering the diffeomorphism $G: \mathbb{R}^{d} \setminus \bar{B}_{r_{3}} \to B_{r_{3}} \setminus \{0\}$ which is the Kelvin transform with respect to $\partial B_{r_{3}}$, i.e.,

$$\begin{equation} G(x) = r_{3}^{2} x/ |x|^{2}. {\tag{1.8}} \end{equation}$$

It is important to note that

$$\begin{equation} G_{*} F_{*} A = I \ \text{and } G_*F_*1=1\text{ in } B_{r_{3}}, {\tag{1.9}} \end{equation}$$

since $G\circ F (x) = (r_3^2/ r_2^2)x$. This is the reason for choosing $(A,\Sigma )$ in 1.3.

Let $\Omega$ be a smooth open subset of $\mathbb{R}^d$ ($d=2, \, 3$) such that $B_{r_3} \subset \subset \Omega$. Given $f \in L^{2}(\Omega )$, let $u_\delta , \, u \in H^1_{0}(\Omega )$ be respectively the unique solution to

$$\begin{equation} {\mathrm{div}}(s_\delta A \nabla u_\delta ) + s_0 k^2 \Sigma u_\delta = f \text{ in } \Omega , {\tag{1.10}} \end{equation}$$

and

$$\begin{equation} \Delta u + k^2 u = f \text{ in } \Omega . {\tag{1.11}} \end{equation}$$

As in 18, we assume that

$$\begin{equation} \text{equation \xhref[disp-formula]{#eq-uu}{1.11} with $f = 0$ has only a zero solution in $H^1_0(\Omega )$}. \tag{1.12} \end{equation}$$

Our result on cloaking using complementary media is:

Theorem 1.

Let $d=2, \, 3$, $f \in L^{2}(\Omega )$ with $\operatorname {supp}f \subset \Omega \setminus B_{r_{3}}$ and let $u$ and $u_\delta$ in $H^1_{0}(\Omega )$ be the unique solution to 1.10 and 1.11, resp. There exists $\gamma _0> 1$, depending only on $\Lambda$ and the Lipschitz constant of $\hat{a}$, such that if $1 < \gamma < \gamma _0$, then

$$\begin{equation} u_{\delta } \to u \text{ weakly in } H^{1} (\Omega \setminus B_{r_3}) \text{ as } \delta \to 0. {\tag{1.13}} \end{equation}$$

For an observer outside $B_{r_3}$, the medium in $B_{r_3}$ looks like the homogeneous one by 1.13 (and also 1.11): one has cloaking.

Remark 1.

Since $\Delta (u_\delta - u) + k^2 (u_\delta - u) = 0$ in $\Omega \setminus B_{r_3}$, it follows from Theorem 1 that $u_\delta \to u$ in $C^m_{\mathrm{loc}}(\Omega \setminus \bar{B}_{r_3})$ for $m \in \mathbb{N}$. A discussion on the rate of the convergence is given in Remark 4 after the proof of Theorem 1.

Remark 2.

The constant $\gamma _0$ in Theorem 1 depends only on $\Lambda$ and the Lipschitz constant of $\hat{a}$. Hence, by choosing $r_2$ large enough and $\gamma = \gamma _0/2$, the cloaked region $B_{\gamma r_2} \setminus B_{r_2}$ can be arbitrarily large.

Remark 3.

The case $k=0$ was established in 21. The proof of Theorem 1 has its root from there.

The proof of Theorem 1 is given in Section 3. It is based on the removing localized singularity technique introduced in 21 and uses a new three sphere inequality (Theorem 2) discussed in the next section. The discussion on illusion optics is given in Section 4.

2. Three spheres inequalities

Let $v$ be a holomorphic function defined in $B_{R_3}$. Hadamard in 8 proved the following famous three spheres inequality:

$$\begin{equation} \|v\|_{L^{\infty }(\partial B_{R_2})} \leq \|v\|_{L^{\infty }(\partial B_{R_1})}^{\alpha } \|v\|_{L^{\infty }(\partial B_{r_3})}^{1 - \alpha }, {\tag{2.1}} \end{equation}$$

for all $0 < R_1 < R_2 < R_3$, where

$$\begin{equation*} \displaystyle \alpha = \log \Big ( \frac{R_3}{R_2} \Big ) \Big / \log \Big (\frac{R_3}{R_1} \Big ). \end{equation*}$$

A three spheres inequality for general elliptic equations was proved by Landis 13 using Carleman type estimates. Landis proved 13, Theorem 2.1 that⁠¹ if $v$ is a solution to

¹

In fact, 13, Theorem 2.1 deals with the non-divergent form; however since $M$ is assumed $C^2$, the two forms are equivalent.

✖

$$\begin{equation} \operatorname {div}(M \nabla v) + \vec{b} \cdot \nabla v + c v = 0 \text{ in } B_{R_3}, {\tag{2.2}} \end{equation}$$

where $M$ is elliptic, symmetric, and of class $C^2$, $\vec{b}, c \in C^1$, and $c \le 0$, then there is a constant $C>0$ such that

$$\begin{equation} \|v\|_{L^{\infty }(\partial B_{R_2})} \leq C \|v\|_{L^{\infty }(\partial B_{R_1})}^{\alpha } \|v\|_{L^{\infty }(\partial B_{r_3})}^{1 - \alpha }, {\tag{2.3}} \end{equation}$$

for some $\alpha \in (0, 1)$ depending only on $R_2/R_1, R_2/R_3$, the ellipticity constant of $M$, and the regularity constants of $M$, $\vec{b}$, and $c$. The assumption $c \le 0$ is crucial and this is discussed in the next paragraph. Another proof was obtained by Agmon 1 in which he used the logarithmic convexity. Garofalo and Lin in 6 established similar results where the $L^\infty$-norm is replaced by the $L^2$-norm, $M$ is of class $C^1$, and $\vec{b}$ and $c$ are in $L^\infty$:

$$\begin{equation} \|v\|_{L^{2}(\partial B_{R_2})} \leq C \|v\|_{L^{2}(\partial B_{R_1})}^{\alpha } \|v\|_{L^2(\partial B_{r_3})}^{1 - \alpha } {\tag{2.4}} \end{equation}$$

using the frequency function.

A typical example of 2.2 when $c>0$ is the Helmholtz equation:

$$\begin{equation} \Delta v + k^2 v = 0 \text{ in } B_{R_3}. {\tag{2.5}} \end{equation}$$

Given $k> 0$, neither 2.4 nor 2.3 holds for all $R_1 < R_2 < R_3$. Indeed, consider first the case $d=2$. It is clear that for $n \in \mathbb{Z}\setminus \{0\}$, the function $J_n(k r ) e^{i n \theta }$ is a solution to 2.5 in $\mathbb{R}^2 \setminus \{0\}$, where $J_n$ is the Bessel function of order $n$. By taking $R_1$, $R_2$, and $R_3$ such that $J_n(kR_1) = 0 \neq J_n(kR_2)$, one reaches the fact that neither 2.4 nor 2.3 is valid. The same conclusion holds in the higher dimensional case by similar arguments. In the case $c > 0$, 2.4 holds under the smallness of $R_3$ (see, e.g., 2, Theorem 4.1); this condition is equivalent to the smallness of $c$ for a fixed $R_3$ by a scaling argument.

In this paper, we establish a new type of three spheres inequalities without imposing the smallness condition on $R_3$. This inequality will play an important role in the proof of Theorem 1. Define

$$\begin{equation} \| v \|_{{\mathbf{H}}(\partial B_r)} = \| v\|_{H^{1/2}(\partial B_r)} + \| M \nabla v \cdot \nu \|_{H^{-1/2}(\partial B_r)}. {\tag{2.6}} \end{equation}$$

Here and in what follows, $\nu$ denotes the outward normal vector on a sphere.

Our result on three spheres inequalities is:

Theorem 2.

Let $d \ge 2$, $c_1, c_2 >0$, $0 < R_* < R_1 < R_2 < R_3 < R^*$, and let $M$ be a Lipschitz uniformly elliptic symmetric matrix-valued function defined in $B_{R^*}$. Assume $v \in H^1(B_{{R_3}} \setminus \overline{B}_{{R_1}})$ satisfies

$$\begin{equation} |{\mathrm{div}}(M \nabla v)| \le c_1 |\nabla v| + c_2|v|, \quad \text{ in } {B_{R_3}\setminus \overline{B}_{R_1}}. {\tag{2.7}} \end{equation}$$

There exists a constant $q \ge 1$, depending only on $d$ and the elliptic and the Lipschitz constants of $M$, such that, for any $\lambda _0 > 1$ and $R_2 \in (\lambda _ 0 R_1, R_3/\lambda _0)$, we have

$$\begin{equation} \| v\|_{{\mathbf{H}}(\partial B_{R_2})} \le C\| v\|_{{\mathbf{H}}(\partial B_{R_1})} ^{\alpha } \| v\|_{{\mathbf{H}}(\partial B_{R_3})}^{1-\alpha }, \quad \text{ where } \quad \alpha := \frac{R_2^{-q} - R_3^{-q}}{R_1^{-q} - R_{3}^{-q}}. {\tag{2.8}} \end{equation}$$

Here $C$ is a positive constant depending on the elliptic and the Lipschitz constants of $M$, $c_1,$ $c_2,$ $R_*, R^*, d$, and $\lambda _0$ but independent of $v$.

In Theorem 2, one does not impose any smallness condition on $R_1, R_2, R_3$ and the exponent $\alpha$ is independent of $c_1$ and $c_2$. The proof of Theorem 2 is inspired by the approach of Protter in 34. Nevertheless, different test functions are used. The ones in 34 are too concentrated at 0 and not suitable for our purpose. The connection between three spheres inequalities and the unique continuation principle, and the application of three spheres inequalities for the stability of Cauchy problems can be found in 2.

The rest of this section contains two subsections. In the first one, we present some lemmas used in the proof of Theorem 2. The proof of Theorem 2 is given in the second subsection.

2.1. Preliminaries

This section contains several lemmas used in the proof of Theorem 2. These lemmas are in the spirit of 34. Nevertheless, the test functions used here are different from there. Let $0 < R_1 < R_3 < + \infty$. In this section, we assume that $M$ is a Lipschitz symmetric matrix-valued function defined in $\overline{B}_{R_3} \setminus B_{R_1}$ and satisfies

$$\begin{equation*} \frac{1}{\Lambda } |\xi |^2 \le M(x) \xi \cdot \xi \le \Lambda |\xi |^2 \quad \forall \, \xi \in \mathbb{R}^d, \end{equation*}$$

for a.e. $x \in \overline{B}_{R_3} \setminus B_{R_1}$, for some $\Lambda \ge 1$. Set

$$\begin{equation} L : = \| M \|_{L^\infty } + R_3\|\nabla M \|_{L^\infty }. {\tag{2.9}} \end{equation}$$

All functions considered in this section are assumed to be real.

The first lemma is:

Lemma 1.

Let $d \ge 2$ and $z \in H^2({B_{R_3}\setminus \overline{B}_{R_1}})$. We have

$$\begin{equation*} \int _{B_{R_3}\setminus \overline{B}_{R_1}}(x \cdot M \nabla z) \, {\mathrm{div}}(M\nabla z) \ge - \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} C L^2|\nabla z|^2 - \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} C L^2 r |\nabla z|^2, \end{equation*}$$

for some positive constant $C$ depending only on $d$.

Proof.

An integration by parts gives

$$\begin{multline} \int _{B_{R_3}\setminus \overline{B}_{R_1}}(x \cdot M \nabla z) \, {\mathrm{div}}(M\nabla z) = - \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} \nabla ( x \cdot M \nabla z) \cdot M\nabla z \\ + \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} (x \cdot M\nabla z) \, M \nabla z \cdot \nu . {\tag{2.10}} \end{multline}$$

Using the symmetry of $M$, we have⁠²

²

In what follows, the repeated summation is used.

✖

$$\begin{equation} \begin{split} \frac{\partial }{\partial x_i}(x \cdot M\nabla z) &= \frac{\partial }{\partial x_i} \Big ( M_{k j} x_j \frac{\partial z}{\partial x_k} \Big )\\ & = M_{k j} x_j \frac{\partial ^2 z}{\partial x_i \partial x_k} + M_{k i } \frac{\partial z}{\partial x_k} + x _j \frac{\partial M_{k j}}{\partial x_i} \frac{\partial z}{\partial x_k} \end{split}{\tag{2.11}} \end{equation}$$

and

$$\begin{multline} -\int _{{B_{R_3}\setminus \overline{B}_{R_1}}} 2 x_j M_{kj} \frac{\partial ^2 z}{\partial x_i \partial x_k}M_{il} \frac{\partial z}{\partial x_l} = - \int _{B_{R_3}\setminus \overline{B}_{R_1}}x_j M_{kj} M_{il} \frac{\partial }{\partial x_k}\left(\frac{\partial z}{\partial x_i} \frac{\partial z}{\partial x_l}\right) \\ = \int _{B_{R_3}\setminus \overline{B}_{R_1}}\frac{\partial (x_j M_{kj} M_{il})}{\partial x_k}\frac{\partial z}{\partial x_i} \frac{\partial z}{\partial x_l} - \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} x_j M_{kj} M_{il}\frac{\partial z}{\partial x_i} \frac{\partial z}{\partial x_l} \nu _k. {\tag{2.12}} \end{multline}$$

We derive from 2.11 and 2.12 that

$$\begin{equation} - \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} \nabla ( x \cdot M\nabla z) \cdot M\nabla z \ge - \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} C L^2|\nabla z|^2 - \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} C L^2 r|\nabla z|^2. {\tag{2.13}} \end{equation}$$

The conclusion now follows from 2.10 and 2.13.

■

The second lemma is

Lemma 2.

Let $d \ge 2$, $\beta \in \mathbb{R}$, and $z \in H^2({B_{R_3}\setminus \overline{B}_{R_1}})$. There exists $p_{\Lambda , L} \geq 1$ such that if $p \ge p_{\Lambda , L}$ and $|\beta | R_3^{-p} \ge 2$, then

$$\begin{multline*} \int _{B_{R_3}\setminus \overline{B}_{R_1}}e^{\beta r^{-p}} (M x \cdot \nabla |z|^2) \, {\mathrm{div}}(M\nabla e^{-\beta r^{-p}}) + \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} C L^2 p^2 \beta ^2 r^{- 2 p-1} |z|^2 \\[6.0pt] \ge \int _{B_{R_3}\setminus \overline{B}_{R_1}}\frac{1}{2}\Lambda ^{-2} p^3 \beta ^2 r^{-2p - 2}|z|^2 - \int _{B_{R_3}\setminus \overline{B}_{R_1}}C L^2 |\nabla z|^2, \end{multline*}$$

for some positive constant $C$ depending only on $d$.

Proof.

A computation yields

$$\begin{align*} {\mathrm{div}}(M\nabla e^{-\beta r^{-p}})& = p \beta e^{-\beta r^{-p}}\big [p \beta r^{-2 p - 4} - ( p + 2) r^{-p - 4}\big ] x\\ &\quad \cdot Mx +p \beta r^{-p - 2} e^{-\beta r^{-p}}\, {\mathrm{div}}(Mx). \end{align*}$$

An integration by parts gives

$$\begin{equation} \int _{B_{R_3}\setminus \overline{B}_{R_1}}e^{\beta r^{-p}} (M x \cdot \nabla |z|^2) \, {\mathrm{div}}(M\nabla e^{-\beta r^{-p}}) = P + Q. {\tag{2.14}} \end{equation}$$

Here

$$\begin{equation} P = P_1 + P_2 + P_3 \end{equation}$$

with

$$\begin{equation*} \left\{ \begin{array}{rl}\displaystyle P_1 &= \displaystyle - \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} p^2 \beta ^2 |z|^2 {\mathrm{div}}\big [ r^{- 2 p - 4} (x \cdot M x) M x \big ], \\[6.0pt] \displaystyle P_2 &= \displaystyle \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} p \beta (p+2) |z|^2 {\mathrm{div}}\big [ r^{- p - 4} (x \cdot M x) M x \big ], \\[6.0pt] \displaystyle P_3 &= \displaystyle \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} 2 p \beta r^{-p -2} {\mathrm{div}}(M x) z \nabla z \cdot Mx, \end{array}\right. \end{equation*}$$

and

$$\begin{equation*} Q = \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} p \beta |z|^2 \Big ( \big [p \beta r^{- 2p - 4} -( p + 2) r^{-p - 4} \big ] x \cdot M x \Big ) M x \cdot \nu . \end{equation*}$$

We next estimate $P$ and $Q$. A computation yields

$$\begin{equation*} - {\mathrm{div}}\big [ r^{- 2 p - 4} (x \cdot M x) M x \big ] = (2p + 4) (x \cdot Mx)^2 r^{-2p - 6} - r^{-2p - 4} {\mathrm{div}}\big [ (x \cdot Mx) Mx\big ]. \end{equation*}$$

This implies

$$\begin{equation} P_1 \ge \int _{B_{R_3}\setminus \overline{B}_{R_1}}p^2 \beta ^2 r^{-2p - 2}|z|^2 \big [ (2p + 4)\Lambda ^{-2} - C L^2 \big ] . {\tag{2.15}} \end{equation}$$

Similarly,

$$\begin{equation} P_2 \ge - \int _{B_{R_3}\setminus \overline{B}_{R_1}}(p + 2) p |\beta | r^{-p - 2}|z|^2 \big [ (p + 4)\Lambda ^{-2} + C L^2 \big ]. {\tag{2.16}} \end{equation}$$

A combination of 2.15 and 2.16 yields

$$\begin{equation} P_1 + P_2 \ge \int _{B_{R_3}\setminus \overline{B}_{R_1}}\Lambda ^{-2} p^3 \beta ^2 r^{-2p - 2}|z|^2. {\tag{2.17}} \end{equation}$$

Here we used the fact that $p \ge p_{\Lambda , L}$ and $|\beta | R_3^{-p} \ge 2$. On the other hand, using Cauchy’s inequality, we have

$$\begin{align*} |P_3| \leq \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} p^2 \beta ^2 r^{-2p - 2} L^2 |z|^2 + \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} C L^2 |\nabla z|^2. \end{align*}$$

It follows from 2.17 that

$$\begin{align} P &\geq \int _{B_{R_3}\setminus \overline{B}_{R_1}}\frac{1}{2} p^3\beta ^2 \Lambda ^{-2} r^{-2p - 2}|z|^2 - \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} C L^2 |\nabla z|^2, \tag{2.18} \end{align}$$

provided that $p \geq 2\Lambda ^{2} L^2.$ Since

$$\begin{equation*} |Q| \le \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} 2 \Lambda ^2 p^2 \beta ^2 r^{-2p -1 } |z|^2, \end{equation*}$$

the conclusion follows.

■

Using Lemmas 1 and 2, we can prove the following result.

Lemma 3.

Let $d \ge 2$, $\beta \in \mathbb{R}$, and $v \in H^2({B_{R_3}\setminus \overline{B}_{R_1}})$. There exists a positive constant $p_{\Lambda , L} \geq 1$ such that if $p \ge p_{\Lambda , L}$ and $|\beta | R_3^{-p} \ge 2$, then

$$\begin{multline*} \int _{B_{R_3}\setminus \overline{B}_{R_1}}\frac{r^{p + 2}e^{2\beta r^{-p}}}{2p |\beta |}\big [{\mathrm{div}}(M\nabla v)\big ]^2 + \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} C L^2 e^{2 \beta r^{-p}} |\nabla v|^2 \\ + \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} CL^2 p^2 \beta ^2 r^{-2p-1} e^{2 \beta r^{-p}}|v|^2 + \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} C L^2 r e^{2 \beta r^{-p}}|\nabla v|^2 \\ \ge \int _{B_{R_3}\setminus \overline{B}_{R_1}}\frac{1}{2} \Lambda ^{-2} p^3 \beta ^2 r^{-2p - 2} e^{2 \beta r^{-p}} |v|^2, \end{multline*}$$

for some positive constant $C$ depending only on $d$.

Proof.

Set

$$\begin{equation*} z = e^{\beta r^{-p}} v \quad \text{ equivalently } v = e^{-\beta r^{-p}} z. \end{equation*}$$

Since ${\mathrm{div}}\big (M \nabla (g h) \big ) = 2 \nabla h \cdot M\nabla g + h {\mathrm{div}}(M \nabla g) + g {\mathrm{div}}(M \nabla h)$ ($M$ is symmetric), it follows that

$$\begin{align*} {\mathrm{div}}(M\nabla v) = 2\beta p r^{-p - 2} e^{-\beta r^{-p}} x \cdot M\nabla z + e^{-\beta r^{-p}} {\mathrm{div}}(M\nabla z) + z {\mathrm{div}}(M\nabla e^{-\beta r^{-p}}). \end{align*}$$

Using the inequality $(a + b + c)^2 \geq 2 a (b + c)$, we obtain

$$\begin{align*} &\frac{1}{2 }\big [ {\mathrm{div}}(M\nabla v) \big ]^2\\ &\qquad \ge 2|\beta | p r^{-p - 2} e^{-\beta r^{-p}} (x \cdot M\nabla z) \Big ( e^{-\beta r^{-p}} {\mathrm{div}}(M\nabla z) + z {\mathrm{div}}( M\nabla e^{-\beta r^{-p}})\Big ). \end{align*}$$

This implies

$$\begin{multline*} \int _{B_{R_3}\setminus \overline{B}_{R_1}}\frac{r^{p + 2}e^{2\beta r^{-p}}}{2p|\beta |}\big [{\mathrm{div}}(M\nabla v)\big ]^2 \geq \int _{B_{R_3}\setminus \overline{B}_{R_1}}2 (x \cdot M\nabla z ) \, {\mathrm{div}}(M\nabla z) \\ + \int _{B_{R_3}\setminus \overline{B}_{R_1}}e^{\beta r^{-p}} (Mx \cdot \nabla |z|^2) \, {\mathrm{div}}(M\nabla e^{-\beta r^{-p}}). \end{multline*}$$

Applying Lemmas 1 and 2, we have

$$\begin{multline} \int _{B_{R_3}\setminus \overline{B}_{R_1}}\frac{r^{p + 2}e^{2\beta r^{-p}}}{2p|\beta |}\big [{\mathrm{div}}(M\nabla v)\big ]^2\\ \ge \int _{B_{R_3}\setminus \overline{B}_{R_1}}\Big ( \Lambda ^{-2} p^3 \beta ^2 r^{-2p - 2}|z|^2 - C L^2 |\nabla z|^2 \Big ) \\[6.0pt] - \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} \Big ( C L^2 p^2 \beta ^2 r^{-2 p-1} |z|^2 + C L^2 r |\nabla z|^2 \Big ). {\tag{2.19}} \end{multline}$$

Since $z = e^{\beta r^{-p}} v$,

$$\begin{equation} |\nabla z|^2 \leq 2 e^{2 \beta r^{-p}} (|\nabla v|^2 + p^2\beta ^2 r^{-2 p - 2}|v|^2). {\tag{2.20}} \end{equation}$$

A combination of 2.19 and 2.20 yields, since $p \ge p_{\Lambda , L}$,

$$\begin{multline*} \int _{B_{R_3}\setminus \overline{B}_{R_1}}\frac{r^{p + 2}e^{2\beta r^{-p}}}{2p|\beta |}\big [{\mathrm{div}}(M\nabla v)\big ]^2\\ \ge \int _{B_{R_3}\setminus \overline{B}_{R_1}}e^{2 \beta r^{-p}} \left( \frac{1}{2} \Lambda ^{-2} p^3 \beta ^2 r^{-2p - 2} |v|^2 - C L^2 |\nabla v|^2 \right) \\[6.0pt] - \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} e^{2 \beta r^{-p}} \left( C L^2 p^2 \beta ^2 r^{-2p-1} |v|^2 + C L^2 r |\nabla v|^2 \right). \end{multline*}$$

The conclusion follows.

■

We also have

Lemma 4.

Let $d \ge 2$, $\beta \in \mathbb{R}$, and $v \in H^2({B_{R_3}\setminus \overline{B}_{R_1}})$. There exists a positive constant $p_{\Lambda , L} \geq 1$ such that if $p \ge p_{\Lambda , L}$ and $|\beta | R_3^{-p} \ge 2$, then

$$\begin{multline*} \int _{B_{R_3}\setminus \overline{B}_{R_1}}e^{2\beta r^{-p}} v \, {\mathrm{div}}(M\nabla v) + \int _{B_{R_3}\setminus \overline{B}_{R_1}}e^{2 \beta r^{-p}} |\nabla v|^2 \\ \le \int _{B_{R_3}\setminus \overline{B}_{R_1}}C \beta ^2 p^2 r^{-2p - 2} e^{2 \beta r^{-p}} |v|^2+ \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} C e^{2\beta r^{-p}} (r|\nabla v|^2 + r^{-1}|v|^2), \end{multline*}$$

for some positive constant $C$ depending only on $d$, $\Lambda$, and $L$.

Proof.

We have

$$\begin{equation} \begin{split} &- \int _{B_{R_3}\setminus \overline{B}_{R_1}}e^{2\beta r^{-p}} v \, {\mathrm{div}}(M\nabla v)\\ &\qquad = \int _{B_{R_3}\setminus \overline{B}_{R_1}}M\nabla v \cdot \nabla (e^{2\beta r^{-p}} v) - \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} e^{2\beta r^{-p}} v M \nabla v \cdot \nu . \end{split}{\tag{2.21}} \end{equation}$$

On the other hand,

$$\begin{equation} \begin{split} & \int _{B_{R_3}\setminus \overline{B}_{R_1}}M\nabla v \cdot \nabla (e^{2\beta r^{-p}} v)\\ &\qquad =\int _{B_{R_3}\setminus \overline{B}_{R_1}}\Big (e^{2\beta r^{-p}} M\nabla v \cdot \nabla v - 2 \beta p r^{-p - 2}e^{2\beta r^{-p}} v M\nabla v \cdot x \Big ) \end{split}{\tag{2.22}} \end{equation}$$

and

$$\begin{equation} \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} e^{2\beta r^{-p}} v M \nabla v \cdot \nu \le \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} e^{2\beta r^{-p}} (r|\nabla v|^2 + L^2 r^{-1}|v|^2). {\tag{2.23}} \end{equation}$$

Since

$$\begin{equation*} 2 \beta p r^{-p - 2}v M\nabla v \cdot x \le \frac{1}{2} \Lambda ^{-1} |\nabla v|^2 + 8 \beta ^2 p^2 L^2 \Lambda r^{-2p - 2} |v|^2, \end{equation*}$$

we derive from 2.21, 2.22, and 2.23 that

$$\begin{multline*} \int _{B_{R_3}\setminus \overline{B}_{R_1}}e^{2\beta r^{-p}} v \, {\mathrm{div}}(M\nabla v) + \int _{B_{R_3}\setminus \overline{B}_{R_1}}\frac{1}{2} \Lambda ^{-1} e^{2 \beta r^{-p}} |\nabla v|^2 \\ \le \int _{B_{R_3}\setminus \overline{B}_{R_1}}C \beta ^2 p^2 r^{-2p - 2} e^{2 \beta r^{-p}} |v|^2+ \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} C e^{2\beta r^{-p}} \big (r|\nabla v|^2 + r^{-1}|v|^2 \big ). \end{multline*}$$

The conclusion follows.

■

Combining the inequalities in Lemmas 3 and 4, we obtain

Lemma 5.

Let $d \ge 2$, $\beta \in \mathbb{R}$, and $v \in H^2({B_{R_3}\setminus \overline{B}_{R_1}})$. There exists a positive constant $p_{\Lambda , L} \geq 1$ such that if $p \ge p_{\Lambda , L}$ and $|\beta | \geq 2 R_3^{-p}$, then

$$\begin{multline} \int _{{B_{R_3}\setminus \overline{B}_{R_1}}} e^{2\beta r^{-p}}|\beta | p \Big (p^3 \beta ^2 r^{-2p - 2} e^{2 \beta r^{-p}} |v|^2 + |\nabla v|^2 \Big )\\ \leq C \int _{B_{R_3}\setminus \overline{B}_{R_1}}r^{p+2} e^{2\beta r^{-p}}|{\mathrm{div}}(M\nabla v)|^2 \\ + C \int _{\partial ({B_{R_3}\setminus \overline{B}_{R_1}})} |\beta | p e^{2\beta r^{-p}} \big (r|\nabla v|^2 + p^2 \beta ^2 r^{-2p -1}|v|^2 \big ), {\tag{2.24}} \end{multline}$$

for some positive constant $C$ depending only on $d$, $\Lambda$, and $L$.

Proof.

Note that

$$\begin{align*} |v \operatorname {div}(M \nabla v)| \le p|\beta |^2 |v|^2 r^{- 2p -2 } + \frac{4 }{p |\beta |} |\operatorname {div}(M \nabla v)|^2 r^{p+2}. \end{align*}$$

The conclusion now follows from Lemmas 3 and 4. The details are left to the reader.

■

2.2. Proof of Theorem 2

Let

$$\begin{equation*} 1 < \lambda < \lambda _0 \end{equation*}$$

(which will be defined later) and set

$$\begin{equation*} D = B_{{\lambda R_3}} \setminus \overline{B}_{{R_1/\lambda }}. \end{equation*}$$

Let $u_1 \in H^1(D \setminus \partial B_{R_1})$ and $u_3 \in H^1(D \setminus \partial B_{R_3})$ be respectively the unique solution to

$$\begin{equation*} \left\{ \begin{array}{cl} {\mathrm{div}}(M\nabla u_1) = 0 &\text{in } D \setminus \partial B_{R_1},\\[6.0pt] \displaystyle [u_1] = v; \; \displaystyle [M\nabla u_1 \cdot \nu ] = M\nabla v \cdot \nu &\text{on } \partial B_{R_1},\\[6.0pt] u_1 = 0 &\text{on } \partial D, \end{array} \right. \end{equation*}$$

and

$$\begin{equation*} \left\{ \begin{array}{cl} {\mathrm{div}}(M\nabla u_3) = 0 &\text{in } D \setminus B_{R_3},\\[6.0pt] \displaystyle [u_3] = v; \; \displaystyle [M\nabla u_3 \cdot \nu ] = M\nabla v \cdot \nu &\text{on } \partial B_{R_3},\\[6.0pt] u_3 = 0 &\text{on } \partial D. \end{array} \right. \end{equation*}$$

Here and in what follows, $[\cdot ]$ denotes the jump across a sphere and $\nu$ denotes the unit outward normal vector on a sphere. It follows that

$$\begin{equation} \| u_1 \|_{H^1(D\setminus \partial B_{R_1} )} \le C \|v \|_{{\mathbf{H}}(\partial B_{R_1})}, \quad \|u_1\|_{H^{3/2}(\partial B_{R_1/\gamma })} \leq C\|v\|_{{\mathbf{H}}(\partial B_{R_1})} {\tag{2.25}} \end{equation}$$

and

$$\begin{equation} \| u_3 \|_{H^1(D \setminus \partial B_{R_3})} \le C \|v \|_{{\mathbf{H}}(\partial B_{R_3})}, \quad \|u_3\|_{H^{3/2}(\partial B_{\gamma R_3})} \leq C\|v\|_{{\mathbf{H}}(\partial B_{R_3})}. {\tag{2.26}} \end{equation}$$

Here and in what follows in this proof, $C$ denotes a positive constant depending only on the elliptic and the Lipschitz constant of $M$, $c_1$, $c_2$, $\lambda _0$, $R_*$, $R^*$, and $d$. Set

$$\begin{equation*} d_1 = (\lambda -1)R_1 \quad \text{ and } \quad d_3 = (\lambda - 1)R_3/\lambda . \end{equation*}$$

Let $\varphi _1, \; \varphi _3 \in C^2_{c}(\mathbb{R}^d)$ be such that

$$\begin{equation*} \varphi _1 = \left\{ \begin{array}{ll} 1 & \text{ in } B_{R_1 + d_1/3} \setminus B_{R_1}, \\[6.0pt] 0 &\text{in } \mathbb{R}^d \setminus (B_{R_1 + d_1/2} \setminus B_{R_1/\lambda }), \end{array}\right. \end{equation*}$$

and

$$\begin{equation*} \varphi _3 = \left\{ \begin{array}{ll} 1 &\text{in } B_{R_3} \setminus B_{R_3-d_3/3},\\[6.0pt] 0 &\text{in } \mathbb{R}^d \setminus (B_{\lambda R_3} \setminus B_{R_3 - d_3/2}). \end{array} \right. \end{equation*}$$

Define

$$\begin{equation} V = \left\{\begin{array}{cl} v- \varphi _1 u_1 - \varphi _3 u_3 & \text{ in } {B_{R_3}\setminus \overline{B}_{R_1}}, \\[6.0pt] - \varphi _1 u_1 - \varphi _3 u_3 & \text{ in } D \setminus ({B_{R_3}\setminus \overline{B}_{R_1}}). \end{array}\right. {\tag{2.27}} \end{equation}$$

Applying Lemma 5, we obtain, for $|\beta | > 2(\gamma R_{3})^{p}$,

$$\begin{equation} \begin{split} & C \int _{D} e^{2\beta r^{-p}}\beta (\beta ^2 |V|^2 + |\nabla V|^2)\\ &\qquad \leq \int _{D} e^{2\beta r^{-p}}|{\mathrm{div}}(M\nabla V)|^2 + \int _{\partial D} |\beta | e^{2\beta r^{-p}} (|\nabla V|^2 + \beta ^2 |V|^2). \end{split}{\tag{2.28}} \end{equation}$$

The proof is now quite standard and divided into two cases.

Case 1.

$\| v\|_{{\mathbf{H}}(\partial B_{R_1})} \le \| v\|_{{\mathbf{H}}(\partial B_{R_3})}$. We deduce from 2.28 that for $\beta \ge \beta _0 := \max \{1, 2 (\gamma _0 R^*)^p\}$,

$$\begin{equation} C \int _{D} e^{2\beta r^{-p}}(|V|^2 + |\nabla V|^2) \leq \beta ^2 e^{2\beta \hat{R}_{3}^{-p}} \| v\|_{{\mathbf{H}}(\partial B_{R_3})}^2 + \beta ^2 e^{2\beta \hat{R}_{1}^{-p}} \|v\|_{{\mathbf{H}}(\partial B_{R_1})}^2, {\tag{2.29}} \end{equation}$$

where

$$\begin{equation*} \hat{R}_3 = R_3/ \lambda \quad \text{ and } \quad \hat{R}_1 = R_1 / \lambda . \end{equation*}$$

This implies

$$\begin{equation} C \|v \|_{{\mathbf{H}}(\partial B_{R_2})} \leq \beta e^{\beta (\hat{R}_3^{-p} - R_2^{-p})} \| v\|_{{\mathbf{H}}(\partial B_{R_3})} + \beta e^{\beta (\hat{R}_1^{-p} - R_2^{-p})} \|v\|_{{\mathbf{H}}(\partial B_{R_1})}. {\tag{2.30}} \end{equation}$$

Define $\alpha ' \in (0, 1)$ and $\beta > 0$ as follows:⁠³

³

Here we assume that $\| v\|_{{\mathbf{H}}(\partial B_{R_1})} \neq 0$ since otherwise $v= 0$. This fact is a consequence of the unique continuation principle and can be obtained from 2.30 by letting $\beta \to \infty$.

✖

$$\begin{equation*} \alpha ' = \frac{R_2^{-p} - \hat{R}_3^{-p}}{\hat{R}_1^{-p} - \hat{R}_3^{-p}} \quad \text{ and }\quad \beta (R_2^{-p} - \hat{R}_1^{-p}) = (1 - \alpha ') \ln \Big (\|v \|_{{\mathbf{H}}(\partial B_{R_3})} \big /\| v\|_{{\mathbf{H}}(\partial B_{R_1})} \Big ). \end{equation*}$$

Note that $0 < \alpha ' < 1$ since $R_2 < R_3/ \gamma$. We assume that $\|v \|_{{\mathbf{H}}(\partial B_{R_3})}> C \| v\|_{{\mathbf{H}}(\partial B_{R_1})}$ for some large $C$ such that $\beta \!\ge \! \max \{2R_3^{-p}, 2, \beta _0\}$ since if $\|v \|_{{\mathbf{H}}(\partial B_{R_3})} \!<\! C \| v\|_{{\mathbf{H}}(\partial B_{R_1})}$, the conclusion holds for any $\alpha \in (0, 1)$ by taking $\beta = \max \{2 R_3^{-p}, 2, \beta _0\}$ in 2.30. It follows from 2.30 and the choice of $\alpha '$ and $\beta$ that

$$\begin{equation} \| v \|_{{\mathbf{H}}(\partial B_{R_2})} \le C \beta \|v \|_{{\mathbf{H}}(\partial B_{R_1})}^{\alpha '}\| v\|_{{\mathbf{H}}(\partial B_{R_3})}^{1-\alpha '}. {\tag{2.31}} \end{equation}$$

Define

$$\begin{equation} \alpha : = \frac{R_2^{-2p} - R_3^{-2p}}{R_1^{-2p} - R_3^{-2p}}. {\tag{2.32}} \end{equation}$$

It is clear that $\alpha < \frac{R_2^{-p} - R_3^{-p}}{R_1^{-p} - R_3^{-p}}$. Hence, by choosing $\lambda$ close to 1,

$$\begin{equation} \alpha < \alpha ' \quad \forall R_2 \in (\gamma _0 R_1, R_3/ \gamma _0). {\tag{2.33}} \end{equation}$$

A combination of 2.31 and 2.33 implies

$$\begin{equation} \| v \|_{{\mathbf{H}}(\partial B_{R_2})} \le C \|v \|_{{\mathbf{H}}(\partial B_{R_1})}^{\alpha }\| v\|_{H(\partial B_{R_3})}^{1-\alpha }. {\tag{2.34}} \end{equation}$$

Case 2.

$\| v\|_{{\mathbf{H}}(\partial B_{R_1})} \ge \| v\|_{{\mathbf{H}}(\partial B_{R_3})}$. The proof is similar to the previous case by considering $\beta < - 2 (\gamma R_{3})^{-p}$. The details are left to the reader.■

3. Cloaking using complementary media. Proof of Theorem 1

This section containing two subsections is devoted to the proof of Theorem 1. In the first subsection, we present two useful lemmas. The proof of Theorem 1 is given in the second subsection.

3.1. Preliminaries

In this section, we present two lemmas which will be used in the proof of Theorems 1 and 3. The first lemma is on a change of variables and follows from 18, Lemma 1.

Lemma 6.

Let $d \ge 2$, $k>0$, and $0 < R_1 < R_2 < R_3$ with $R_3 = R_2^2/ R_1$. Let $a \in [L^\infty (B_{R_3} \setminus \overline{B}_{R_2})]^{d \times d}$ be a matrix-valued function, $\sigma \in L^\infty (B_{R_3} \setminus \overline{B}_{R_2})$ a complex function, and $K:B_{R_2} \setminus \bar{B}_{R_1} \to B_{R_3} \setminus \bar{B}_{R_2}$ the Kelvin transform with respect to $\partial B_{R_2}$, i.e.,

$$\begin{equation*} K(x) = R_2^2 x/ |x|^2. \end{equation*}$$

For $v \in H^1(B_{{R_3}} \setminus \overline{B}_{{R_2}})$, define $w = v \circ K^{-1}$. Then

$$\begin{equation*} \operatorname {div}(a \nabla v) + k^2 \sigma v = 0 \text{ in } B_{{R_3}} \setminus \overline{B}_{{R_2}} \end{equation*}$$

if and only if

$$\begin{equation*} \operatorname {div}(K_*a \nabla w) + k^2 K_* \sigma w = 0 \text{ in } B_{{R_2}} \setminus \overline{B}_{{R_1}}. \end{equation*}$$

Moreover,

$$\begin{equation*} w = v \quad \text{ and } \quad K_*a \nabla w \cdot \nu = - a \nabla v \cdot \nu \text{ on } \partial B_{R_2}. \end{equation*}$$

The second lemma is a stability estimate for solutions of 1.10.

Lemma 7.

Let $0 < \delta < 1$ and $f \in L^2(\Omega )$, and let $A \in [L^\infty (\Omega )]^{d \times d}$ and $\Sigma \in L^\infty (\Omega , \mathbb{C})$ be such that $A$ is Lipschitz and uniformly elliptic, $\Im (\Sigma ) \ge 0$, and $\Re (\Sigma ) \ge \lambda > 0$ for some $\lambda$. There exists a unique solution $u_\delta \in H^1_0(\Omega )$ of 1.10. Moreover,

$$\begin{equation} \| u_\delta \|_{H^1(\Omega )}^2 \le C \left(\delta ^{-1} \|f\|_{L^2(\Omega )}\|u_{\delta }\|_{L^2({\operatorname {supp}} f)} + \|f\|_{L^2(\Omega )}^2\right), {\tag{3.1}} \end{equation}$$

for some positive constant $C$ independent of $\delta$ and $f$.

Lemma 7 is a variant of 18, Lemma 1. The case $k=0$ and its variant in the case $k>0$ were considered in 21 and 19, respectively. The proof is similar to the one of 18, Lemma 1. For the convenience of the reader, we present the proof.

Proof.

The existence and uniqueness of $u_\delta$ are given in 18. We only establish 3.1 by contradiction. Assume that 3.1 is not true. Then there exist $\delta _n \to 0$ and $(f_n) \subset L^2(\Omega )$ such that

$$\begin{equation} \| u_n\|_{H^1(\Omega )} =1 \text{ and } \frac{1}{\delta _n} \|f_n\|_{L^2(\Omega )} \| u_n\|_{L^2({\operatorname {supp}} f_n)} + \| f_n\|_{L^2(\Omega )}^2 \to 0, {\tag{3.2}} \end{equation}$$

as $n \to \infty$, where $u_n \in H^1_0(\Omega )$ is the unique solution to

$$\begin{equation} \operatorname {div}(s_{\delta _n} A \nabla u_n) + k^2 s_0 \Sigma u_n= f_n \text{ in } \Omega . {\tag{3.3}} \end{equation}$$

Without loss of generality, one may assume that $u_n \to u$ weakly in $H^1(\Omega )$ and strongly in $L^2(\Omega )$; moreover, $u \in H^1_0(\Omega )$ and $u$ satisfies

$$\begin{equation} \operatorname {div}(s_0 A \nabla u) + k^2 s_0 \Sigma u = 0 \text{ in } \Omega . {\tag{3.4}} \end{equation}$$

Multiplying equation 3.3 by $\bar{u}_n$ (the conjugate of $u_n$) and integrating on $\Omega$, we have

$$\begin{equation*} \int _{\Omega } s_{\delta _n} A \nabla u_n \cdot \nabla \bar{u}_n \, dx - \int _{\Omega } k^2 s_0 \Sigma |u_n|^2 \, dx = - \int _\Omega f_n \bar{u}_n \, dx. \end{equation*}$$

Considering the imaginary part and using the fact that

$$\begin{equation*} \frac{1}{\delta _n}\Big | \int _\Omega f_n \bar{u}_n \Big | \le \frac{1}{\delta }\|f_n \|_{L^2(\Omega )} \| u_n\|_{L^2({\operatorname {supp}} f_n)} \to 0 \text{ as $n \to \infty $ by \xhref[disp-formula]{#contradict-assumption}{3.2}}, \end{equation*}$$

we obtain, by 1.6,

$$\begin{equation} \| \nabla u_n\|_{L^2(B_{{r_2}} \setminus \overline{B}_{{r_1}})} \to 0 \text{ as } n \to \infty . {\tag{3.5}} \end{equation}$$

Since $\operatorname {div}(A \nabla u_n) + k^{2} \Sigma u_n = f_n$ in $B_{r_2} \setminus B_{r_1}$ and $f_n \to 0$ in $L^2(\Omega )$, it follows from 3.5 that $u_n \to 0$ in the distributional sense. This in turn implies

$$\begin{equation} \| u_n \|_{L^{2}(B_{r_2} \setminus B_{r_1})} \to 0 \text{ as } n \to \infty . {\tag{3.6}} \end{equation}$$

A combination of 3.5 and 3.6 yields

$$\begin{equation} \| u_n \|_{H^{1}(B_{r_2} \setminus B_{r_1})} \to 0 \text{ as } n \to \infty . {\tag{3.7}} \end{equation}$$

Hence

$$\begin{equation*} u= 0 \text{ in } B_{r_2} \setminus B_{r_1} \end{equation*}$$

and

$$\begin{equation} \begin{split} \| u_n \|_{H^{1/2}(\partial B_{r_2})} & + \| u_n \|_{H^{1/2}(\partial B_{r_1})} + \| A \nabla u_n \cdot \nu \|_{H^{-1/2}(\partial B_{r_2})}\\ & + \| A \nabla u_n \cdot \nu \|_{H^{-1/2}(\partial B_{r_1})} \to 0 \text{ as }n \to \infty . \end{split}{\tag{3.8}} \end{equation}$$

Since $u=0$ in $B_{r_2} \setminus B_{r_1}$ and $u$ satisfies 3.4, it follows from the unique continuation principle that $u = 0$ in $\Omega$. Hence, since $u_n \to u$ in $L^2(\Omega )$,

$$\begin{equation} u_n \to 0 \text{ in } L^2(\Omega ) \text{ as } n \to \infty . {\tag{3.9}} \end{equation}$$

Multiplying 3.3 by $\bar{u}_n$ and integrating on $\Omega \setminus B_{r_2}$, we have

$$\begin{equation*} \int _{\Omega \setminus B_{r_2}} A \nabla u_n \cdot \nabla \bar{u}_n \, dx - \int _{\Omega \setminus B_{r_2}} k^2 s_0 \Sigma |u_n|^2 \, dx = - \int _\Omega f_n \bar{u}_n \, dx + \int _{\partial B_{r_2}} A \nabla u_n \cdot \nu \; \bar{u}_n. \end{equation*}$$

Using 3.8 and 3.9, we obtain

$$\begin{equation} \|\nabla u_n \|_{L^2(\Omega \setminus B_{r_2})} \to 0 \text{ as } n \to \infty . {\tag{3.10}} \end{equation}$$

Similarly,

$$\begin{equation} \|\nabla u_n \|_{L^2(B_{r_1})} \to 0 \text{ as } n \to \infty . {\tag{3.11}} \end{equation}$$

Combining 3.7, 3.9, 3.10, and 3.11, we obtain

$$\begin{equation*} \|u_n \|_{H^1(\Omega )} \to 0 \text{ as } n \to \infty , \end{equation*}$$

which contradicts 3.2. The proof is complete.

■

3.2. Proof of Theorem 1

We use the approach in 21 with some modifications from 19 so that the same proof also gives the result on illusion optics (Theorem 3 in Section 4). However, instead of applying the standard three sphere inequality as in 21, we use Theorem 2.

We have, by Lemma 7,

$$\begin{equation} \| u_{\delta } \|_{H^{1}(\Omega )}^2 \le C \Big ( \delta ^{-1} \| f\|_{L^2(\Omega )} \|u_{\delta } \|_{L^2(\Omega \setminus B_{r_3})} + \| f\|_{L^2(\Omega )}^2 \Big ). {\tag{3.12}} \end{equation}$$

As in 21, let $u_{1, \delta }$ be the reflection of $u_{\delta }$ through $\partial B_{r_{2}}$ by $F$, i.e.,

$$\begin{equation} u_{1, \delta } = u_{\delta } \circ F^{-1} \text{ in } \mathbb{R}^{d} \setminus \bar{B}_{r_{2}}, {\tag{3.13}} \end{equation}$$

and let $u_{2, \delta }$ be the reflection of $u_{1, \delta }$ through $\partial B_{r_2}$ by $G$, i.e.,

$$\begin{equation} u_{2, \delta } = u_{1, \delta } \circ G^{-1} \text{ in } B_{r_{3}}. {\tag{3.14}} \end{equation}$$

By Lemma 6,

$$\begin{equation} \operatorname {div}(A \nabla u_{1, \delta }) + \frac{1}{1- i \delta } k^2 \Sigma u_{1, \delta } = 0 \text{ in } B_{r_3} \setminus B_{r_2}, {\tag{3.15}} \end{equation}$$

$$\begin{equation} \Delta u_{2, \delta } + k^2 u_{2, \delta } = 0 \text{ in } B_{r_3}. {\tag{3.16}} \end{equation}$$ Applying Lemma 6 again and using the fact that $F_*A = A$ in $B_{r_3} \setminus B_{r_2}$, we have

$$\begin{equation} u_{1, \delta } = u_{\delta } \Big |_{+} \text{ on } \partial B_{r_{2}} \quad \text{ and } \quad (1 - i \delta ) A \nabla u_{1, \delta } \cdot \nu = A \nabla u_{ \delta } \cdot \nu \Big |_{+} \text{ on } \partial B_{r_{2}}. {\tag{3.17}} \end{equation}$$

Let $V_{1, \delta } \in H^1(B_{r_3} \setminus B_{r_2})$ be the unique solution to

$$\begin{equation} \left\{\begin{array}{cl} {\mathrm{div}}(A \nabla V_{1, \delta }) + k^2 \Sigma V_{1, \delta } = \displaystyle - \frac{i \delta }{1 - i \delta } k^2 \Sigma u_{1, \delta } \text{ in } & B_{r_3} \setminus B_{r_2}, \\[6.0pt] A \nabla V_{1, \delta } \cdot \nu - i k V_{1, \delta } = 0 & \text{ on } \partial B_{r_2}, \\[6.0pt] V_{1, \delta } = 0 & \text{ on } \partial B_{r_3}. \end{array}\right. {\tag{3.18}} \end{equation}$$

By Fredholm’s theory,

$$\begin{equation} \| V_{1, \delta }\|_{H^1(B_{r_3} \setminus B_{r_2})} \le C \delta \| u_\delta \|_{H^1(\Omega )}. {\tag{3.19}} \end{equation}$$

Define $U_{1, \delta }$ in $B_{r_{3}} \setminus B_{r_2}$ as

$$\begin{equation} U_{1, \delta } =u_{\delta } - u_{1, \delta } - V_{1, \delta }. {\tag{3.20}} \end{equation}$$

Then $U_{1, \delta } \in H^1(B_{r_3} \setminus B_{r_2})$ and $U_{1, \delta }$ satisfies

$$\begin{equation*} {\mathrm{div}}(A \nabla U_{1, \delta }) + k^2 \Sigma U_{1, \delta } = 0 \text{ in } B_{r_3} \setminus B_{r_2}, \end{equation*}$$

$$\begin{equation*} \| U_{1, \delta }\|_{H^{1/2}(\partial B_{r_2})} + \| A \nabla U_{1, \delta } \cdot \nu \|_{H^{1/2}(\partial B_{r_2})} \le C \delta \| u_\delta \|_{H^1(\Omega )}, \end{equation*}$$ and

$$\begin{equation*} \| U_{1, \delta }\|_{H^{1/2}(\partial B_{r_3})} + \| A \nabla U_{1, \delta } \cdot \nu \|_{H^{1/2}(\partial B_{r_3})} \le C \| u_\delta \|_{H^1(\Omega )}. \end{equation*}$$

Applying Theorem 2, we have

$$\begin{equation} \| U_{1, \delta }\|_{H^{1/2}(\partial B_{\gamma r_2})} + \| A \nabla U_{1, \delta } \cdot \nu \|_{H^{1/2}(\partial B_{\gamma r_2})} \le C \delta ^{\alpha } \| u_\delta \|_{H^1(\Omega )}, {\tag{3.21}} \end{equation}$$

where $\alpha$ is given in 2.8 with $R_1 = r_2$, $R_2 = \gamma r_2$, $R_3 = r_3$. By choosing $\gamma _0$ close enough to 1, from 2.8, we can assume that

$$\begin{equation} \alpha > 1/2. {\tag{3.22}} \end{equation}$$

Here is the place where the condition $\gamma < \gamma _0$ is required. A combination of 3.19 and 3.21 yields

$$\begin{equation} \| u_\delta - u_{1, \delta } \|_{H^{1/2}(\partial B_{\gamma r_2})} + \|A \nabla (u_\delta - u_{1, \delta } ) \cdot \nu \|_{H^{-1/2}(\partial B_{\gamma r_2})} \le C \delta ^{\alpha } \| u_\delta \|_{H^1(\Omega )}. {\tag{3.23}} \end{equation}$$

In what follows, we assume that $k=1$ for notational ease. Define $U_{2, \delta }$ in $B_{r_3} \setminus B_{\gamma r_2}$ as

$$\begin{equation*} U_{2, \delta } = u_{1, \delta } - u_{2, \delta } + V_{1, \delta }. \end{equation*}$$

Then

$$\begin{equation} \Delta U_{2, \delta } + U_{2, \delta } = 0 \quad \text{ in } B_{r_3} \setminus B_{\gamma r_2} {\tag{3.24}} \end{equation}$$

and

$$\begin{equation} U_{2, \delta } = 0 \quad \text{and} \quad \partial _r U_{2, \delta } = -\frac{i\delta }{1 - i\delta } \partial _r u_{1, \delta } + \partial _r V_{1, \delta } \quad \text{ on } \partial B_{r_3}. {\tag{3.25}} \end{equation}$$

Case 1.

$d=2$. As in 19, define

$$\begin{equation*} \hat{J}_n(t) = 2^n n! J_n(t) \quad \text{ and } \quad \hat{Y}_n(t) = - \frac{\pi }{2^{n} (n-1)!} Y_n(t), \end{equation*}$$

where $J_n$ and $Y_n$ are the Bessel and Neumann functions of order $n$. It follows from 5, (3.80) and (3.81) that

$$\begin{equation} \hat{J}_n (t) = t^{n}\big [1 + o(1) \big ] {\tag{3.26}} \end{equation}$$

and

$$\begin{equation} \hat{Y}_n (t) = t^{-n} \big [1 + o(1) \big ], {\tag{3.27}} \end{equation}$$

as $n \to + \infty$.

From 3.24 one can represent $U_{2, \delta }$ as

$$\begin{equation} \begin{split} U_{2, \delta } &= a_0 \hat{J}_0(|x|) + b_0 \hat{Y}_0(|x|)\\ &\quad + \sum _{n=1}^\infty \sum _{\pm } \big [a_{n, \pm } \hat{J}_{n}(|x|) + b_{n, \pm } \hat{Y}_{n}(|x|) \big ] e^{\pm i n \theta } \quad \text{ in } B_{r_3} \setminus B_{\gamma r_2}, \end{split}{\tag{3.28}} \end{equation}$$

for $a_0, b_0, a_{n, \pm }, b_{n, \pm } \in \mathbb{C}$ ($n \ge 1$). Assume that

$$\begin{equation*} \partial _r U_{2, \delta } = c_0 + \sum _{n \ge 1} \sum _{\pm } c_{n, \pm } e^{\pm i n \theta } \quad \text{ on } \partial B_{r_3}. \end{equation*}$$

Then, by 3.18, 3.19, and 3.25,

$$\begin{equation} |c_0|^2 + \sum _{n \ge 1} \sum _{\pm } n^{-1} |c_{n , \pm }|^2 \sim \|\partial _r U_{2, \delta } \|_{H^{-1/2}(\partial B_{r_3})}^2 \le C \delta ^2 \| u \|_{H^{1}(\Omega )}^2. {\tag{3.29}} \end{equation}$$

Using 3.25 again, we have

$$\begin{equation*} \left\{\begin{array}{c} a_{n, \pm } \hat{J}_n (r_3) + b_{n, \pm } \hat{Y}_n (r_3) = 0, \\[6.0pt] a_{n, \pm } \hat{J}_n' (r_3) + b_{n, \pm } \hat{Y}_n' (r_3) = c_{n, \pm }, \end{array}\right. \quad \text{ for } n \ge 0. \end{equation*}$$

Here we denote $a_{0, \pm } = a_0/2$, $b_{0, \pm } = b_0/2$, and $c_{0, \pm } = c_0/ 2$. It follows that

$$\begin{equation} \left\{ \begin{array}{l} \displaystyle a_{n, \pm } = c_{n, \pm } AC_{n}, \\[6.0pt] \displaystyle b_{n, \pm } = c_{n, \pm } BC_{n}, \end{array} \right. \quad \text{ for } n \ge 0, {\tag{3.30}} \end{equation}$$

where

$$\begin{equation*} AC_{n} = - \frac{\hat{Y}_n}{\hat{J}_n \hat{Y}_n' - \hat{J}_n' \hat{Y}_n}(r_3) \quad \text{ and } \quad BC_{n} = - \frac{\hat{J}_n}{\hat{Y}_n \hat{J}_n' - \hat{Y}_n' \hat{J}_n}(r_3). \end{equation*}$$

Using 3.26 and 3.27, we derive that

$$\begin{equation*} AC_{n} = - \frac{1}{2n} r_3^{1-n} \Big ( 1 + o(1) \Big ) \quad \text{ and } BC_{n} = \frac{1}{2n} r_3^{1 + n} \Big ( 1 + o(1) \Big ). \end{equation*}$$

We now make use of the removing of localized singularity technique introduced in 1921. Set

$$\begin{equation*} \hat{u}_\delta (x)= \sum _{n=1}^\infty \sum _{\pm } b_{n, \pm } \hat{Y}_{n}(|x|) e^{\pm i n \theta } \quad \text{ in } B_{r_3} \setminus B_{\gamma r_2}. \end{equation*}$$

We claim that, for $\gamma r_2 \le r \le r_3$,

$$\begin{equation} \|U_{2, \delta } - \hat{u}_\delta \|_{H^{1/2}(\partial B_{r})} + \|\partial _r U_{2, \delta } - \partial _r \hat{u}_\delta \|_{H^{-1/2}(\partial B_{r})} \le C \delta \|u_\delta \|_{H^1(\Omega )}. {\tag{3.31}} \end{equation}$$

Indeed, for $\gamma r_2 \le r \le r_3$,

$$\begin{align*} \|U_{2, \delta } - \hat{u}_\delta \|_{H^{1/2}(\partial B_{r})}^2 & = \|\sum _{n \ge 0} \sum _{\pm } a_{n, \pm } \hat{J}_n(|x|) e^{i n \theta } \|_{H^{1/2}(\partial B_{r})}^2\\[6.0pt] & \sim \sum _{n \ge 0} \sum _{\pm } (n+1)|a_{n, \pm }|^2 |\hat{J}_n(|x|)|^2 \\[6.0pt] &\sim \sum _{n \ge 0} \sum _{\pm } (n+1)|c_{n, \pm } AC_n|^2 |\hat{J}_n(|x|)|^2\\[6.0pt] & \le C \sum _{n \ge 0} \sum _{\pm } (n+1)^{-1}| c_{n, \pm }|^2 (r/ r_3)^{2n}. \end{align*}$$

It follows from 3.29 that

$$\begin{equation*} \|U_{2, \delta } - \hat{u}_\delta \|_{H^{1/2}(\partial B_{r})} \le C \delta \| u_\delta \|_{H^1(\Omega )}, \end{equation*}$$

for $\gamma r_2 \le r \le r_3$. Similarly,

$$\begin{equation*} \|\partial _r U_{2, \delta } - \partial _r \hat{u}_\delta \|_{H^{-1/2}(\partial B_{r})} \le C \delta \| u_\delta \|_{H^1(\Omega )}, \end{equation*}$$

for $\gamma r_2 \le r \le r_3$. As a consequence of 3.19 and 3.31, we obtain for $\gamma r_2 \le r \le r_3$,

$$\begin{equation} \|u_{1, \delta } - u_{2, \delta } - \hat{u}_\delta \|_{H^{1/2}(\partial B_{r})} + \|\partial _r u_{1, \delta } - \partial u_{2, \delta } - \partial _r \hat{u}_\delta \|_{H^{-1/2}(\partial B_{r})} \le C \delta \|u_\delta \|_{H^1(\Omega )}. {\tag{3.32}} \end{equation}$$

Define

$$\begin{equation*} U_\delta = \left\{\begin{array}{cl} u_\delta & \text{ in } \Omega \setminus B_{r_3}, \\[6.0pt] u_\delta - \hat{u}_\delta & \text{ if } x \in B_{r_3} \setminus B_{\gamma r_2}, \\[6.0pt] u_{2, \delta } & \text{ if } x \in B_{\gamma r_2}. \end{array} \right. \end{equation*}$$

We have

$$\begin{equation*} {\mathrm{div}}(A \nabla U_\delta ) + k^2 \Sigma U_{\delta } = f \text{ in } \Omega \setminus (\partial B_{r_3} \cup \partial B_{\gamma r_2}). \end{equation*}$$

On the other hand, from 3.23 and 3.32, we obtain

$$\begin{equation} \|[U_\delta ] \|_{H^{1/2}(\partial B_{\gamma r_2})} + \|[\partial _r U_\delta \cdot \nu ] \|_{H^{-1/2}(\partial B_{\gamma r_2})} \le C \delta ^{\alpha } \| u_\delta \|_{H^1(\Omega )} {\tag{3.33}} \end{equation}$$

and

$$\begin{equation} \|[U_\delta ] \|_{H^{1/2}(\partial B_{r_3})} + \|[\partial _r U_\delta \cdot \nu ] \|_{H^{-1/2}(\partial B_{r_3})} \le C \delta ^{\alpha } \| u_\delta \|_{H^1(\Omega )}. {\tag{3.34}} \end{equation}$$

Using 3.12, we derive that

$$\begin{align*} &\| U_\delta \|_{H^1(\Omega \setminus (\partial B_{r_3} \cup \partial B_{\gamma r_2}) )}\\ &\qquad \le C \delta ^{\alpha } \Big ( \delta ^{-1/2}\| U_\delta \|_{L^2(\Omega \setminus B_{r_3})}^{1/2} \|f \|_{L^2(\Omega )}^{1/2} + \| f\|_{L^2(\Omega )}\Big ) + C \|f \|_{L^2(\Omega )}. \end{align*}$$

Since $\alpha > 1/2$, it follows that $U_\delta$ is bounded in $H^1\big (\Omega \setminus (\partial B_{r_3} \cup \partial B_{\gamma r_2}) \big )$. Without loss of generality, one may assume that $U_{\delta } \to U$ weakly in $H^1\big (\Omega \setminus (\partial B_{r_3} \cup \partial B_{\gamma r_2}) \big )$ as $\delta \to 0$; moreover, $U \in H^1(\Omega )$ and

$$\begin{equation*} \Delta U + k^2 U= f \text{ in } \Omega \text{ and } U = 0 \text{ on } \partial \Omega . \end{equation*}$$

Hence $U = u$. Since the limit is unique, we have the convergence for the family $(U_{\delta })$ as $\delta \to 0$.

Case 2.

$d=3$. Define

$$\begin{equation*} \hat{j}_n(t) =1 \cdot 3 \cdots (2n + 1) j_n(t) \quad \text{ and } \quad \hat{y}_n = - \frac{y_n(t)}{1 \cdot 3 \cdots (2n-1)} , \end{equation*}$$

where $j_n$ and $y_n$ are the spherical Bessel and Neumann functions of order $n$. Then, for $n$ large enough (see, e.g., 5, (2.38) and (2.39)),

$$\begin{equation} \hat{j}_n(t) = t^n \big (1 + O(1/n) \big ) \quad \text{ and } \quad \hat{y}_n(r) = t^{-n-1} \big (1 + O(1/n) \big ). {\tag{3.35}} \end{equation}$$

Thus one can represent $U_{2, \delta }$ of the form

$$\begin{equation} U_{2, \delta } = \sum _{n=1}^\infty \sum _{m = -n }^n \big [a^m_n \hat{j}_{n}(|x|) + b^m_n \hat{y}_{n}(|x|) \big ] Y^m_n(\hat{x} ) \quad \text{ in } B_{r_3} \setminus B_{r_0}, {\tag{3.36}} \end{equation}$$

for $a^n_m, b^n_m \in \mathbb{C}$ and $\hat{x} = x/ |x|$, where $Y^m_n$ is the spherical function of degree $n$ and of order $m$. The proof now follows similarly as in the case $d=2$. The details are left to the reader. ■

Remark 4.

Define $V_\delta = U_\delta - u$ in $\Omega$. Then $V_\delta \in H^1\big (\Omega \setminus (\partial B_{\gamma r_2} \cup \partial B_{r_3}) \big )$,

$$\begin{equation*} \Delta V_\delta + k^2 V_\delta = 0 \text{ in } \Omega \setminus (\partial B_{\gamma r_2} \cup \partial B_{r_3}), \quad V_\delta = 0 \text{ on } \partial \Omega , \end{equation*}$$

and, from 3.33 and 3.34,

$$\begin{equation*} \|[V_\delta ] \|_{H^{1/2}(\partial B_{\gamma r_2} \cup \partial B_{r_3})} + \|[\partial _r V_\delta \cdot \nu ] \|_{H^{-1/2}(\partial B_{\gamma r_2} \cup \partial B_{r_3})} \le C \delta ^{\alpha - 1/2} \| f \|_{L^2(\Omega )}. \end{equation*}$$

It follows that $\|V_\delta \|_{H^1\big (\Omega \setminus (\partial B_{\gamma r_2} \cup \partial B_{r_3}) \big )} \le C \delta ^{\alpha - 1/ 2} \|f \|_{L^2}$. This implies that $\|u_\delta - u \|_{H^1(\Omega \setminus \bar{B}_{r_3})} \le C \delta ^{\alpha - 1/ 2} \|f \|_{L^2}$. Note that $\alpha$ can be close to $1$ if $\gamma$ is sufficiently close to 1 (in order to keep the size of the cloaked object unchanged, one needs to have large $r_2$; see also Remark 2).

Remark 5.

In the proof, we use essentially the fact $(A, \Sigma ) = (I, 1)$ in $B_{r_3} \setminus B_{\gamma r_2}$ to use separation of variables in this region. In fact, this condition is not necessary by using the technique of separation of variables for a general structure in 20.

Remark 6.

The construction of the cloak given by 1.4 is not restricted to the Kelvin transforms $F$ (and $G$). In fact, one can extend this construction to a general class of reflections considered in 18.

Remark 7.

The condition $(F_*A, F_*\Sigma ) = (A, \Sigma )$ in $B_{r_3} \setminus B_{r_2}$ is necessary to ensure that cloaking can be achieved and the localized resonance might take place; see 23 (see also 4 for related results).

Remark 8.

Cloaking can also be achieved via schemes generated by changes of variables 71431. Resonance might also appear in this context but for specific frequencies see 916. It was shown in 16 that in the resonance case cloaking might not be achieved and the field inside the cloaked region can depend on the field outside. Cloaking can also be achieved in the time regime via change of variables 2627.

4. Illusion optics using complementary media

We next discuss briefly how to obtain illusion optics in the spirit of Lai et al. in 12. The scheme used here is a combination of the ones used for cloaking and superlensing in 1921 and is slightly different from 12. More precisely, set

$$\begin{equation*} m = r_3^2/ r_2^2. \end{equation*}$$

Let $a_c \in [L^\infty (B_{r_2/m})]^{d \times d}$ be elliptic and $\sigma _c \in L^{\infty }(B_{r_2^2/r_3^2}, \mathbb{C})$ with $\Im (\sigma _c) \geq 0$. Define

$$\begin{equation} A_1, \Sigma _1 = \left\{ \begin{array}{cl} A, \Sigma & \text{ in } \Omega \setminus B_{r_2/ m}, \\[6.0pt] a_c, \sigma _c & \text{ in } B_{r_2/ m}, \end{array} \right. {\tag{4.1}} \end{equation}$$

and

$$\begin{equation} \hat{A}_1, \hat{\Sigma }_1 = \left\{ \begin{array}{cl} I, 1 & \text{ in } \Omega \setminus B_{r_2}, \\[6.0pt] (r_3/ r_2)^{2-d} a_c(x/ m), (r_3/ r_2)^{-d}\sigma _c(x/m) & \text{ in } B_{r_2}. \end{array} \right. {\tag{4.2}} \end{equation}$$

Recall that $(A, \Sigma )$ is defined in 1.4. We assume that the following equation has only a zero solution in $H^1_0(\Omega )$:

$$\begin{equation} {\mathrm{div}}(A_1\nabla v) + k^2 \Sigma _1 v = 0 \text{ in } \Omega . {\tag{4.3}} \end{equation}$$

We obtain the following result on illusion optics:

Theorem 3.

Let $d=2, \, 3$ and $f \in L^{2}(\Omega )$ with $\operatorname {supp}f \subset \Omega \setminus B_{r_{3}}$, and let $u$ and $u_\delta$ in $H^1_{0}(\Omega )$ be respectively the unique solution of

$$\begin{equation*} \operatorname {div}(s_\delta A_1 \nabla u_\delta ) + k^2 s_0 \Sigma _1 u_\delta = f \quad \text{in } \Omega \end{equation*}$$

and

$$\begin{equation*} \operatorname {div}(\hat{A}_1 \nabla u) + k^2 \hat{\Sigma }_1 u= f \quad \text{in } \Omega . \end{equation*}$$

There exists $\gamma _0> 1$, depending only on $\Lambda$ and the Lipschitz constant of $\hat{a}$ such that if $1 < \gamma < \gamma _0$, then

$$\begin{equation} u_{\delta } \to u \text{ weakly in } H^{1} (\Omega \setminus B_{r_3}) \text{ as } \delta \to 0. {\tag{4.4}} \end{equation}$$

For an observer outside $B_{r_3}$, the medium in $B_{r_3}$ looks like $(\hat{A}_1, \hat{\Sigma }_1)$: one has illusion optics.

Proof.

The proof is similar to the one of Theorem 1. Note that in the proof of Theorem 1, we do not use the information of the medium inside $B_{r_2/m}$. The details are left to the reader.

■