# Anomalous Localized Resonance Phenomena in the Nonmagnetic, Finite-Frequency Regime

Authors: Daniel Onofrei; Andrew E. Thaler
Journal: Advances in Mathematical Physics
 (2016)

---

## Abstract

The phenomenon of anomalous localized resonance (ALR) is observed at the interface between materials with positive and negative material parameters and is characterized by the fact that when a given source is placed near the interface, the electric and magnetic fields start to have very fast and large oscillations around the interface as the absorption in the materials becomes very small while they remain smooth and regular away from the interface. In this paper, we discuss the phenomenon of anomalous localized resonance (ALR) in the context of an infinite slab of homogeneous, nonmagnetic material (μ=1) with permittivity ϵs=-1-iδ for some small loss δ≪1 surrounded by positive, nonmagnetic, homogeneous media. We explicitly characterize the limit value of the product between frequency and the width of slab beyond which the ALR phenomenon does not occur and analyze the situation when the phenomenon is observed. In addition, we also construct sources for which the ALR phenomenon never appears.


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## Body

## 1. Introduction

In the following, we discuss the anomalous localized resonance phenomenon (ALR) appearing at the interface between materials with positive and negative material parameters in the finite-frequency regime. We consider the particular slab geometry described by (see Figure1)(1)C≡x,y∈R2:x<0;S≡x,y∈R2:0<x<a;M≡x,y∈R2:x>a,where a>0 denotes the width of the slab and the sets C, S, and M represent the regions to the left of the slab, within the slab, and to the right of the slab, respectively. We also define (2)d0≡min⁡x:x,y∈supp⁡f,d1≡max⁡x:x,y∈supp⁡f.Figure 1
In this figure, we illustrate the geometry of the problem we consider in this paper.In this geometry, we assume that all materials are homogeneous and nonmagnetic (i.e., with magnetic permeabilityμ=1); the electrical permittivity is given by(3)ε≡1forx<0,-1-iδfor0<x<a,1forx>afor some δ∈(0,1). We consider the following partial differential equation (PDE) in 2D:(4)∇·1ε∇V+k02+iζV=-finR2,where ζ≥0, k0>0, f∈L2(M) with compact support in M, and ε is given in (3) (see Section A.5 for a derivation of (4) from the Maxwell equations).For convenience, we define(5)Vc≡VC;Vs≡VS;Vm≡VM.We assume the solution V also satisfies the following continuity conditions across the boundaries at x=0 and x=a for almost every y∈R:(6)Vc0,y=Vs0,y;∂Vc∂x0,y=1-1-iδ∂Vs∂x0,y,Vsa,y=Vma,y;1-1-iδ∂Vs∂xa,y=∂Vm∂xa,y.In what follows we assume that the parameters and data are such that problem (4) and (6) admit a unique solution V∈Lloc2(R2) with V(x,·)∈H1(R) and ∂V/∂x(x,·)∈L2(R) for almost every x∈R.Remark 1.
Note that, in the case whenζ>0, the unique solution of the problem will have the property that V(x,y)→0 as x→∞ for almost every y∈R; for ζ≪1, this solution will be well approximated by the solution in the case ζ=0.We say anomalous localized resonance (ALR) occurs if the following two properties hold asδ→0+ [1]:(1)
V→∞ in certain localized regions with boundaries that are not defined by discontinuities in the relative permittivity.(2)
V approaches a smooth limit outside these localized regions.In [1], Milton et al. showed that if f is a dipole and εc=εm=1, then ALR occurs if a<d0<2a, where d0 is the location of the dipole. In this case, there are two locally resonant strips—one centered on each face of the slab. As the loss parameter (represented by δ) tends to zero, the potential diverges and oscillates wildly in these resonant regions. Outside these regions, the potential converges to a smooth function. Also, if the source is far enough away from the slab, that is, if d0>2a, then there is no resonance and again the potential converges to a smooth function.Applications of ALR to superlensing were first discussed by Nicorovici et al. in [2] and were analyzed in more depth in [1] (see also the works by Yan et al. [3], Bergman [4], Nguyen [5], Pendry [6], and Pendry and Ramakrishna [7] for a description of superlensing phenomena).Applications of ALR to cloaking in the quasistatic regime were first analyzed Milton and Nicorovici [8]; they showed that if εc=εm=1 and a fixed field is applied to the system (e.g., a uniform field at infinity), then a polarizable dipole located in the region a<d0<3a/2 causes anomalous localized resonance and is cloaked in the limit δ→0+. Cloaking due to anomalous localized resonance (CALR) in the quasistatic regime was further discussed in [9–17]. CALR in the long-time limit regime was discussed in [16, 18] (see also [19]).In [20], Nicorovici et al. studied CALR for the circular cylindrical superlens in the finite-frequency case; they showed that for small values of δ the cloaking device (the superlens) can effectively cloak a tiny cylindrical inclusion located within the cloaking region but that the superlens does not necessarily cloak itself—they deemed this phenomenon to be the “ostrich effect.” The finite-frequency case was further discussed by Kettunen et al. [21] and Nguyen [22].In the present report we prove, analytically and numerically, the existence of a limit valueγ∗, such that, for k0 with k0a>γ∗, ALR does not occur regardless of the position of the source with respect to the slab interface. Under suitable conditions on the source, we present numerical evidence for the occurrence of ALR in the regime k0a<γ∗ when the source is close enough to the material interface, and we discuss some characteristics of the phenomenon in this frequency regime as well. In the end we present two examples of sources f which do not generate ALR regardless of the frequency regime and their relative position with respect to the material interface.The paper is organized as follows: in Section1.1 we present highlights of the derivation of the unique solution in the Fourier domain while in Section 1.2 we describe the energy around the right interface of the slab. In Section 3, we show the absence of ALR phenomena for large enough values of k0a while in Section 3.2 we present an interesting side effect of the nonmagnetic case, namely, the shielding effect of the slab which behaves as an almost perfect reflector. Next, for suitable conditions on the source, in Section 4.1 we present numerical evidence for the ALR phenomenon in the case of small enough values of k0a. In Section 4.2, we construct two examples of possible sources for which there is no ALR phenomenon regardless of the range of k0a or the relative position of the source with respect to the slab interface. The Appendix contains the technical proofs and derivations which were not included in the main text.


### 1.1. Solution in Fourier Domain

Due to our well-posedness assumption, it follows that our problem will admit a unique solution after applying the Fourier transform with respect to they variable. Recall that, for a given function h(x,·)∈L2(R) for some x∈R, the Fourier transform of h with respect to y is(7)h^x,q=∫-∞∞hx,ye-iqydy.We will study the Fourier domain solution in each of the relevant subdomains defined in (1).


#### 1.1.1. The Solution inC

In the regionC, the relevant equation is(8)∂2Vc∂x2+∂2Vc∂y2+k02Vc=0.Taking the Fourier transform of (8) with respect to y, we find that V^c(x,q) satisfies(9)∂2V^c∂x2-k02νc2V^c=0,whereνc2≡q2k02-1.Remark 2.
Here and throughout the paper, we take the principal square root of complex numbers; that is, for a complex numberz=z′+iz′′=zeiθ, where θ∈(-π,π], we take(10)z=z1/2eiθ/2, where θ/2∈(-π/2,π/2]. In particular, this implies Rez≥0.Remark2 implies(11)
ν
c
=
i
q
2
k
0
2
-
1
i
f
q
2
k
0
2
<
1
,
q
2
k
0
2
-
1
i
f
q
2
k
0
2
≥
1
.Then the general solution to (9) is(12)V^cx,q=Aqek0νcx+Bqe-k0νcxfor coefficients Aq and Bq that are independent of x.Ifq2/k02<1, then νc is purely imaginary. Because Vc should be outgoing (i.e., left-going) as x→-∞ and we are considering eiωt time dependence (see Section A.5), we should have(13)V^c~eik0q2/k02-1xasx⟶-∞.From (11) and (12), we see that we can ensure this by taking Bq=0.On the other hand, ifq2/k02>1, then νc>0. Thus we take Bq=0 in this case to ensure that V^c(x,q)→0 as x→∞. Finally, without loss of generality we may also take Bq=0 for q2/k02=1. Therefore,(14)V^cx,q=Aqek0νcx.


#### 1.1.2. The Solution inS

In the regionS, the Fourier transform of Vs satisfies(15)∂2V^s∂x2-k02νs2V^s=0,whereνs2≡q2k02+1+iδ.The general solution is(16)V^sx,q=Cqek0νsx+Dqe-k0νsx;the coefficients Cq and Dq may be found by using the continuity conditions across x=0 from (6). In particular, we find(17)V^sx,q=Aqα+12αek0νsx1+Re-2k0νsx,where(18)α≡νs-1-iδνc,R≡α-1α+1=νs+1+iδνcνs-1+iδνc.Although one can observe thatα degenerates for q2=k02 we will see in (25) and (27) that Aq/α is well defined in the limit when q2=k02.


#### 1.1.3. The Solution inM

In the regionM, the Fourier transform of Vm satisfies(19)∂2V^m∂x2-k02νm2V^m=-f^x,q,whereνm2≡q2k02-1.If q2/k02≠1, then the general solution to (19) can be found using the Laplace transform and the continuity conditions across x=a from (6) [23, 24]; we have(20)
V
^
m
x
,
q
=
e
k
0
ν
m
x
2
A
q
e
-
k
0
ν
m
a
ψ
q
+
+
ψ
q
-
ν
m
-
1
k
0
ν
m
∫
d
0
x
e
-
k
0
ν
m
s
f
^
s
,
q
d
s
+
e
-
k
0
ν
m
x
2
A
q
e
k
0
ν
m
a
ψ
q
+
+
ψ
q
-
ν
m
+
1
k
0
ν
m
∫
d
0
x
e
k
0
ν
m
s
f
^
s
,
q
d
s
,where(21)ψq+≡1AqV^sa,q=α+12αek0νsa1+Re-2k0νsa;ψq-≡1k0Aq1-1-iδ∂V^s∂xa,q=νs-1-iδα+12αek0νsa1-Re-2k0νsa.Ifq2/k02<1, then νm is purely imaginary. Because Vm should be outgoing (i.e., rightgoing) as x→∞ and we are considering eiωt time dependence, we should have(22)V^m~e-ik0q2/k02-1xasx⟶∞. To ensure this, we take the first expression in brackets in (20) to be zero and find that(23)Aq≡Iqek0νmak0νmψq++ψq-,where(24)Iq≡∫d0d1f^s,qe-k0νmsds.Ifq2/k02>1, then νm>0; to ensure that V^m(x,q)→0 as x→∞, we again take Aq as in (23).Finally, ifq2/k02=1, then we can use the Laplace transform and the continuity conditions across x=a to find that(25)V^mx,±k0=A±k0ψ±k0+-aψ±k0-+∫d0xsf^s,±k0ds+xk0A±k0ψ±k0--∫d0xf^s,±k0ds,where(26)ψ±k0+=1+e-2k0νsa;ψ±k0-=νs-1-iδ1-e-2k0νsawith νs defined at (15) being computed for q=±k0 and where again we take A±k0 so that we ensure V^m is outgoing as x→∞; in this case(27)A±k0=1k0ψ±k0-∫d0d1f^s,±k0ds.


### 1.2. Energy Discussion

For0<ξ≤a, we define the strip(28)Sξ≡x,y∈R2:a-ξ<x<a.Then, due to the Plancherel theorem and properties of Fourier transforms, we have (29)∇VL2Sξ2=∫a-ξa∫-∞∞∇Vsx,y2dydx=∫a-ξa∫-∞∞∂2Vs∂x22+∂2Vs∂y22dydx=12π∫a-ξa∫-∞∞∂2V^s∂x22+q2V^s2dqdx. Using (17)-(18) and (21)–(24) in this expression, switching the order of integration, computing the integral with respect to x, using the fact that ∇Vs2 is an even function of q if f is real-valued, making the change of variables p=q/k0, and simplifying the resulting expression, we obtain(30)
E
δ
ξ
≡
∇
V
L
2
S
ξ
2
=
1
+
δ
2
π
∫
0
∞
I
p
2
e
2
k
0
ν
m
′
a
ν
s
-
1
+
i
δ
ν
m
2
g
δ
p
;
γ
2
·
ν
s
2
+
p
2
1
-
e
-
2
k
0
ν
s
′
ξ
2
ν
s
′
+
R
2
e
-
4
k
0
ν
s
′
a
-
ξ
/
2
1
-
e
-
2
k
0
ν
s
′
ξ
2
ν
s
′
+
2
-
ν
s
2
+
p
2
e
-
2
k
0
ν
s
′
a
Im
⁡
R
¯
e
2
i
k
0
ν
s
′
′
a
1
-
e
-
2
i
k
0
ν
s
′
′
ξ
2
ν
s
′
′
d
p
,where(31)gδp;γ≡νs-1+iδνm2-νs+1+iδνm2e-2γνs,we have used the fact that νc=νm (see (11) and (19)), and we have replaced q by k0p throughout the integrand (e.g., we have νm=p2-1).Similarly, we have(32)VL2Sξ2=1+δ2π∫0∞Ip2e2k0νm′aνs-1+iδνm2gδp;γ2·1-e-2k0νs′ξ2νs′+R2e-4k0νs′a-ξ/21-e-2k0νs′ξ2νs′+2e-2k0νs′aIm⁡R¯e2ik0νs′′a1-e-2ik0νs′′ξ2νs′′dp.Remark 3.
One of the quantities we are most interested in studying in this paper is(33)VH1Sξ2=VL2Sξ2+∇VL2Sξ2.Due to the similarity between the expressions in (30) and (32), without loss of generality we focus on ∇VL2(Sξ)2. In particular, our arguments depend heavily on the exponential terms in the integrands in (30) and (32), so the additional terms νs2 and q2 in (30) will have no bearing on our results.


## 1.1. Solution in Fourier Domain

Due to our well-posedness assumption, it follows that our problem will admit a unique solution after applying the Fourier transform with respect to they variable. Recall that, for a given function h(x,·)∈L2(R) for some x∈R, the Fourier transform of h with respect to y is(7)h^x,q=∫-∞∞hx,ye-iqydy.We will study the Fourier domain solution in each of the relevant subdomains defined in (1).


### 1.1.1. The Solution inC

In the regionC, the relevant equation is(8)∂2Vc∂x2+∂2Vc∂y2+k02Vc=0.Taking the Fourier transform of (8) with respect to y, we find that V^c(x,q) satisfies(9)∂2V^c∂x2-k02νc2V^c=0,whereνc2≡q2k02-1.Remark 2.
Here and throughout the paper, we take the principal square root of complex numbers; that is, for a complex numberz=z′+iz′′=zeiθ, where θ∈(-π,π], we take(10)z=z1/2eiθ/2, where θ/2∈(-π/2,π/2]. In particular, this implies Rez≥0.Remark2 implies(11)
ν
c
=
i
q
2
k
0
2
-
1
i
f
q
2
k
0
2
<
1
,
q
2
k
0
2
-
1
i
f
q
2
k
0
2
≥
1
.Then the general solution to (9) is(12)V^cx,q=Aqek0νcx+Bqe-k0νcxfor coefficients Aq and Bq that are independent of x.Ifq2/k02<1, then νc is purely imaginary. Because Vc should be outgoing (i.e., left-going) as x→-∞ and we are considering eiωt time dependence (see Section A.5), we should have(13)V^c~eik0q2/k02-1xasx⟶-∞.From (11) and (12), we see that we can ensure this by taking Bq=0.On the other hand, ifq2/k02>1, then νc>0. Thus we take Bq=0 in this case to ensure that V^c(x,q)→0 as x→∞. Finally, without loss of generality we may also take Bq=0 for q2/k02=1. Therefore,(14)V^cx,q=Aqek0νcx.


### 1.1.2. The Solution inS

In the regionS, the Fourier transform of Vs satisfies(15)∂2V^s∂x2-k02νs2V^s=0,whereνs2≡q2k02+1+iδ.The general solution is(16)V^sx,q=Cqek0νsx+Dqe-k0νsx;the coefficients Cq and Dq may be found by using the continuity conditions across x=0 from (6). In particular, we find(17)V^sx,q=Aqα+12αek0νsx1+Re-2k0νsx,where(18)α≡νs-1-iδνc,R≡α-1α+1=νs+1+iδνcνs-1+iδνc.Although one can observe thatα degenerates for q2=k02 we will see in (25) and (27) that Aq/α is well defined in the limit when q2=k02.


### 1.1.3. The Solution inM

In the regionM, the Fourier transform of Vm satisfies(19)∂2V^m∂x2-k02νm2V^m=-f^x,q,whereνm2≡q2k02-1.If q2/k02≠1, then the general solution to (19) can be found using the Laplace transform and the continuity conditions across x=a from (6) [23, 24]; we have(20)
V
^
m
x
,
q
=
e
k
0
ν
m
x
2
A
q
e
-
k
0
ν
m
a
ψ
q
+
+
ψ
q
-
ν
m
-
1
k
0
ν
m
∫
d
0
x
e
-
k
0
ν
m
s
f
^
s
,
q
d
s
+
e
-
k
0
ν
m
x
2
A
q
e
k
0
ν
m
a
ψ
q
+
+
ψ
q
-
ν
m
+
1
k
0
ν
m
∫
d
0
x
e
k
0
ν
m
s
f
^
s
,
q
d
s
,where(21)ψq+≡1AqV^sa,q=α+12αek0νsa1+Re-2k0νsa;ψq-≡1k0Aq1-1-iδ∂V^s∂xa,q=νs-1-iδα+12αek0νsa1-Re-2k0νsa.Ifq2/k02<1, then νm is purely imaginary. Because Vm should be outgoing (i.e., rightgoing) as x→∞ and we are considering eiωt time dependence, we should have(22)V^m~e-ik0q2/k02-1xasx⟶∞. To ensure this, we take the first expression in brackets in (20) to be zero and find that(23)Aq≡Iqek0νmak0νmψq++ψq-,where(24)Iq≡∫d0d1f^s,qe-k0νmsds.Ifq2/k02>1, then νm>0; to ensure that V^m(x,q)→0 as x→∞, we again take Aq as in (23).Finally, ifq2/k02=1, then we can use the Laplace transform and the continuity conditions across x=a to find that(25)V^mx,±k0=A±k0ψ±k0+-aψ±k0-+∫d0xsf^s,±k0ds+xk0A±k0ψ±k0--∫d0xf^s,±k0ds,where(26)ψ±k0+=1+e-2k0νsa;ψ±k0-=νs-1-iδ1-e-2k0νsawith νs defined at (15) being computed for q=±k0 and where again we take A±k0 so that we ensure V^m is outgoing as x→∞; in this case(27)A±k0=1k0ψ±k0-∫d0d1f^s,±k0ds.


## 1.1.1. The Solution inC

In the regionC, the relevant equation is(8)∂2Vc∂x2+∂2Vc∂y2+k02Vc=0.Taking the Fourier transform of (8) with respect to y, we find that V^c(x,q) satisfies(9)∂2V^c∂x2-k02νc2V^c=0,whereνc2≡q2k02-1.Remark 2.
Here and throughout the paper, we take the principal square root of complex numbers; that is, for a complex numberz=z′+iz′′=zeiθ, where θ∈(-π,π], we take(10)z=z1/2eiθ/2, where θ/2∈(-π/2,π/2]. In particular, this implies Rez≥0.Remark2 implies(11)
ν
c
=
i
q
2
k
0
2
-
1
i
f
q
2
k
0
2
<
1
,
q
2
k
0
2
-
1
i
f
q
2
k
0
2
≥
1
.Then the general solution to (9) is(12)V^cx,q=Aqek0νcx+Bqe-k0νcxfor coefficients Aq and Bq that are independent of x.Ifq2/k02<1, then νc is purely imaginary. Because Vc should be outgoing (i.e., left-going) as x→-∞ and we are considering eiωt time dependence (see Section A.5), we should have(13)V^c~eik0q2/k02-1xasx⟶-∞.From (11) and (12), we see that we can ensure this by taking Bq=0.On the other hand, ifq2/k02>1, then νc>0. Thus we take Bq=0 in this case to ensure that V^c(x,q)→0 as x→∞. Finally, without loss of generality we may also take Bq=0 for q2/k02=1. Therefore,(14)V^cx,q=Aqek0νcx.


## 1.1.2. The Solution inS

In the regionS, the Fourier transform of Vs satisfies(15)∂2V^s∂x2-k02νs2V^s=0,whereνs2≡q2k02+1+iδ.The general solution is(16)V^sx,q=Cqek0νsx+Dqe-k0νsx;the coefficients Cq and Dq may be found by using the continuity conditions across x=0 from (6). In particular, we find(17)V^sx,q=Aqα+12αek0νsx1+Re-2k0νsx,where(18)α≡νs-1-iδνc,R≡α-1α+1=νs+1+iδνcνs-1+iδνc.Although one can observe thatα degenerates for q2=k02 we will see in (25) and (27) that Aq/α is well defined in the limit when q2=k02.


## 1.1.3. The Solution inM

In the regionM, the Fourier transform of Vm satisfies(19)∂2V^m∂x2-k02νm2V^m=-f^x,q,whereνm2≡q2k02-1.If q2/k02≠1, then the general solution to (19) can be found using the Laplace transform and the continuity conditions across x=a from (6) [23, 24]; we have(20)
V
^
m
x
,
q
=
e
k
0
ν
m
x
2
A
q
e
-
k
0
ν
m
a
ψ
q
+
+
ψ
q
-
ν
m
-
1
k
0
ν
m
∫
d
0
x
e
-
k
0
ν
m
s
f
^
s
,
q
d
s
+
e
-
k
0
ν
m
x
2
A
q
e
k
0
ν
m
a
ψ
q
+
+
ψ
q
-
ν
m
+
1
k
0
ν
m
∫
d
0
x
e
k
0
ν
m
s
f
^
s
,
q
d
s
,where(21)ψq+≡1AqV^sa,q=α+12αek0νsa1+Re-2k0νsa;ψq-≡1k0Aq1-1-iδ∂V^s∂xa,q=νs-1-iδα+12αek0νsa1-Re-2k0νsa.Ifq2/k02<1, then νm is purely imaginary. Because Vm should be outgoing (i.e., rightgoing) as x→∞ and we are considering eiωt time dependence, we should have(22)V^m~e-ik0q2/k02-1xasx⟶∞. To ensure this, we take the first expression in brackets in (20) to be zero and find that(23)Aq≡Iqek0νmak0νmψq++ψq-,where(24)Iq≡∫d0d1f^s,qe-k0νmsds.Ifq2/k02>1, then νm>0; to ensure that V^m(x,q)→0 as x→∞, we again take Aq as in (23).Finally, ifq2/k02=1, then we can use the Laplace transform and the continuity conditions across x=a to find that(25)V^mx,±k0=A±k0ψ±k0+-aψ±k0-+∫d0xsf^s,±k0ds+xk0A±k0ψ±k0--∫d0xf^s,±k0ds,where(26)ψ±k0+=1+e-2k0νsa;ψ±k0-=νs-1-iδ1-e-2k0νsawith νs defined at (15) being computed for q=±k0 and where again we take A±k0 so that we ensure V^m is outgoing as x→∞; in this case(27)A±k0=1k0ψ±k0-∫d0d1f^s,±k0ds.


## 1.2. Energy Discussion

For0<ξ≤a, we define the strip(28)Sξ≡x,y∈R2:a-ξ<x<a.Then, due to the Plancherel theorem and properties of Fourier transforms, we have (29)∇VL2Sξ2=∫a-ξa∫-∞∞∇Vsx,y2dydx=∫a-ξa∫-∞∞∂2Vs∂x22+∂2Vs∂y22dydx=12π∫a-ξa∫-∞∞∂2V^s∂x22+q2V^s2dqdx. Using (17)-(18) and (21)–(24) in this expression, switching the order of integration, computing the integral with respect to x, using the fact that ∇Vs2 is an even function of q if f is real-valued, making the change of variables p=q/k0, and simplifying the resulting expression, we obtain(30)
E
δ
ξ
≡
∇
V
L
2
S
ξ
2
=
1
+
δ
2
π
∫
0
∞
I
p
2
e
2
k
0
ν
m
′
a
ν
s
-
1
+
i
δ
ν
m
2
g
δ
p
;
γ
2
·
ν
s
2
+
p
2
1
-
e
-
2
k
0
ν
s
′
ξ
2
ν
s
′
+
R
2
e
-
4
k
0
ν
s
′
a
-
ξ
/
2
1
-
e
-
2
k
0
ν
s
′
ξ
2
ν
s
′
+
2
-
ν
s
2
+
p
2
e
-
2
k
0
ν
s
′
a
Im
⁡
R
¯
e
2
i
k
0
ν
s
′
′
a
1
-
e
-
2
i
k
0
ν
s
′
′
ξ
2
ν
s
′
′
d
p
,where(31)gδp;γ≡νs-1+iδνm2-νs+1+iδνm2e-2γνs,we have used the fact that νc=νm (see (11) and (19)), and we have replaced q by k0p throughout the integrand (e.g., we have νm=p2-1).Similarly, we have(32)VL2Sξ2=1+δ2π∫0∞Ip2e2k0νm′aνs-1+iδνm2gδp;γ2·1-e-2k0νs′ξ2νs′+R2e-4k0νs′a-ξ/21-e-2k0νs′ξ2νs′+2e-2k0νs′aIm⁡R¯e2ik0νs′′a1-e-2ik0νs′′ξ2νs′′dp.Remark 3.
One of the quantities we are most interested in studying in this paper is(33)VH1Sξ2=VL2Sξ2+∇VL2Sξ2.Due to the similarity between the expressions in (30) and (32), without loss of generality we focus on ∇VL2(Sξ)2. In particular, our arguments depend heavily on the exponential terms in the integrands in (30) and (32), so the additional terms νs2 and q2 in (30) will have no bearing on our results.


## 2. Properties ofgδ(p;γ)

In this section, we collect some essential properties about the denominatorgδ2 in (30). As we will see, the parameter(34)γ≡k0aplays a crucial role in the behavior of the solution V and Eδ(a) in the limit δ→0+.Lemma 4.
Supposegδ is defined as in (31). Then for p≥0 and γ>0 one has(35)limδ→0+gδp;γ=g0p;γ≡p2+1-p2-12-p2+1+p2-12e-2γp2+1.Proof.
The result follows from direct calculations sincegδ is a continuous function of δ.The next lemma plays an essential role in the following discussion.Lemma 5.
Supposeg0(p;γ) is defined as in (35) for p≥0 and γ>0. Then there is γ∗≈0.9373 such that (1)
if0<γ<γ∗, then g0(p;γ) has two distinct real roots of order 1, namely, 1<pγ1<pγ2;(2)
ifγ>γ∗, then g0(p;γ) has no real roots.We note thatγ∗ can be computed as the solution of an optimization problem; more importantly, we emphasize that Lemmas 4 and 5 areindependent of the source term f in (4). We will see later that the roots of g0(p;γ) are indicative of anomalous localized resonance. For brevity, we defer the proof of Lemma 5 to the Appendix.


## 3. Short Wavelength/High Frequency Regime (γ>γ∗)

In this section, we prove that, forγ>γ∗ (where γ was introduced at (34)), Ea(δ) remains bounded as δ→0+ for all sources f∈L2(M) with bounded support in M, regardless of how close the source is to the slab. In addition, we also prove that the slab lens behaves as a “shield” in the sense that the solution to the left of the lens, that is, Vc, is vanishingly small in the limit δ→0+.


### 3.1.Eδ(a) for γ>γ∗

From (30), we have(36)Eδa=∫0∞Lδp;γdp,where, for δ>0, p≥0, and γ>0,(37)
L
δ
p
;
γ
≡
I
p
2
e
2
γ
ν
m
′
g
δ
p
;
γ
2
M
δ
p
;
γ
,
(38)
M
δ
p
;
γ
≡
1
+
δ
2
π
ν
s
-
1
+
i
δ
ν
m
2
ν
s
2
+
p
2
1
-
e
-
2
γ
ν
s
′
2
ν
s
′
+
R
2
e
-
2
γ
ν
s
′
1
-
e
-
2
γ
ν
s
′
2
ν
s
′
+
2
-
ν
s
2
+
p
2
e
-
2
γ
ν
s
′
Im
⁡
R
¯
e
2
i
γ
ν
s
′
′
1
-
e
-
2
i
γ
ν
s
′
′
2
ν
s
′
′
.We now state the main theorem from this section.Theorem 6.
Supposeγ>γ∗ (where γ∗ is introduced in Lemma 5). If there is a constant C>0 such that(39)Ip≤Cfor0≤p≤1,Ce-γd0/ap2-1for1≤p<∞,then there is a constant Cγ>0 and δγ>0 such that VH1(Sa)≤Cγ as for all δ≤δγ.The proof of this theorem is somewhat tedious and may be found in the Appendix—although we only prove the theorem for∇VL2(Sa)2, Remark 3 implies that it holds for VL2(Sa) as well. In the next lemma, we show that bound (39) holds for very general sources f.Lemma 7.
Supposef∈L2(M) with compact support; then (39) holds.Proof.
For0≤p≤1, recall from (24) that(40)Ip=∫d0d1f^s,k0pe-ik01-p2sds.Then the triangle, Cauchy-Schwarz, and Jensen inequalities imply that(41)Ip≤∫d0d1f^s,k0pds≤d1-d01/2∫d0d1f^s,k0p2ds1/2=d1-d01/2∫d0d1∫-∞∞fs,ye-ik0pydy2ds1/2≤d1-d01/2∫d0d1∫-∞∞fs,y2dyds1/2=d1-d01/2fL2M.
Similarly, forp≥1, recall from (24) that(42)Ip=∫d0d1f^s,k0pe-k0p2-1sds.Then (43)Ip≤∫d0d1f^s,k0pe-k0p2-1sds≤d1-d01/2∫d0d1f^s,k0p2e-2k0p2-1sds1/2≤d1-d01/2fL2Me-k0d0p2-1=d1-d01/2fL2Me-γd0/ap2-1.
To complete the proof, we defineC≡(d1-d0)1/2fL2(M).


### 3.2. Shielding Effect for Largeγ

It turns out that the slab lens behaves as a shield and acts as an almost perfect reflector. This fact was also observed in [21] where it was explained based on the fact that, at least in the lossless nonmagnetic case ϵ=-1, μ=1 will give a purely imaginary wavenumber inside the slab and thus no propagation beyond the slab in region C. We have the following.Theorem 8.
Supposeγ≥2γ∗, Ip satisfies (39), and choose 0<η<1; then there is a constant Cη>0 such that(44)Vcx,y≤Cηe-ηγ∀x,y∈C.In particular,(45)Vcx,y⟶0ask0⟶∞∀x,y∈C.Remark 9.
Lemma7 implies that Theorems 6 and 8 hold for all sources f∈L2(M) with compact support. However, the bound in (39) is stronger than we need. For example, suppose there is a positive, real-valued function B(p;γ) that is continuous for 0≤p<∞ and γ∗≤γ<∞. In addition, for every ϵ>0, suppose that(46)limp→∞Bp;γe-ϵγp2-1=0∀γ≥γ∗,limγ→∞Bp;γe-ϵγp2-1=0∀p≥1.For example, if B(p;γ) is a continuous function of p and γ that is of polynomial order for p→∞ and γ→∞, it will satisfy (46). Finally, suppose(47)Ip≤Bp;γfor0≤p≤1,Bp;γe-γd0/ap2-1for1≤p<∞.Then, by appropriately modifying (A.72)–(A.75), one can prove that the result of Theorem 6 will hold for sources satisfying (47). Similarly, by appropriately modifying (A.82)–(A.86), one can show that Theorem 8 also holds for sources satisfying (47) as long as we replace (44) by(48)Vcx,y≤Cηe-η-ϵγ, where 0<ϵ<η.
In particular, certain distributional sources such as dipoles and quadrupoles satisfy (47)—see Section A.2 for more details.In Figure2, we plot the solution V to (4) in the case where f is a dipole with dipole moment [1,0]T, δ=10-12, and γ=4γ∗ (we take a=1 in all figures throughout the paper). In Figures 2(a) and 2(b), the dipole is located at the point (d0,0)=(4a,0); in Figures 2(c) and 2(d), the dipole is located closer to the slab at the point (d0,0)=(1.2a,0). The solution V is smooth throughout the domain; in addition, we observe the “shielding effect” from Theorem 8 in the region to the left of the lens.Figure 2
This is a plot ofV, the solution to (4), when f is a dipole and γ=2γ∗: (a) Re(V) and (b) Im(V) for d0=4a; (c) Re(V) and (d) Im(V) for d0=1.2a. To make the behavior of V clearer, we clipped the maximum and minimum values in each plot to 0.2 (yellow) and -0.2 (blue), respectively.
(a) (b) (c) (d)In Figure3, we plot Eδ(a) as a function of various parameters for a dipole source f. The parameters we used are in the ranges 10-12≤δ≤10-10, 1.01γ∗≤γ≤2γ∗, and 1.2a≤d0≤2a. We note that Eδ(a) depends strongly on δ, γ, and d0, but, because γ>γ∗, Eδ(a) is quite small.Figure 3
These are plots ofEδ(a) as a function of (a) δ and γ (d0=1.2a); (b) δ and γ (d0=4a); (c) δ and d0 (γ=1.01γ∗); (d) δ and d0 (γ=2γ∗); (e) γ and d0 (δ=10-10); (f) γ and d0 (δ=10-12).
(a) (b) (c) (d) (e) (f)Figure4 is similar to Figure 2, except in Figure 4 we take(49)fx,y=C2d1-d0x-d0-132d1-d0x-d0-1-13·2h1-h0y-h0-132h1-h0y-d0-1-13ford0≤x≤d1,h0≤y≤h10otherwise.Although this source has compact support, in contrast to the dipolar source considered above it is in L2(M) and is twice continuously differentiable; we chose this source to emphasize that the ideas presented in this paper do not rely on the extreme nature of distributional sources such as dipoles.Figure 4
This is a plot ofV, the solution to (4), when f is the function in (49) and γ=2γ∗: (a) Re(V) and (b) Im⁡(V) for d0=4a; (c) Re(V) and (d) Im⁡(V) for d0=1.2a. To make the behavior of V clearer, we clipped the maximum and minimum values in each plot to 0.2 (yellow) and -0.2 (blue), respectively.
(a) (b) (c) (d)To construct the plots, we have takenC=104,h1=-h0=1, and d1=d0+2. The solution V is smooth throughout the domain and very small in the region to the left of the slab.


## 3.1.Eδ(a) for γ>γ∗

From (30), we have(36)Eδa=∫0∞Lδp;γdp,where, for δ>0, p≥0, and γ>0,(37)
L
δ
p
;
γ
≡
I
p
2
e
2
γ
ν
m
′
g
δ
p
;
γ
2
M
δ
p
;
γ
,
(38)
M
δ
p
;
γ
≡
1
+
δ
2
π
ν
s
-
1
+
i
δ
ν
m
2
ν
s
2
+
p
2
1
-
e
-
2
γ
ν
s
′
2
ν
s
′
+
R
2
e
-
2
γ
ν
s
′
1
-
e
-
2
γ
ν
s
′
2
ν
s
′
+
2
-
ν
s
2
+
p
2
e
-
2
γ
ν
s
′
Im
⁡
R
¯
e
2
i
γ
ν
s
′
′
1
-
e
-
2
i
γ
ν
s
′
′
2
ν
s
′
′
.We now state the main theorem from this section.Theorem 6.
Supposeγ>γ∗ (where γ∗ is introduced in Lemma 5). If there is a constant C>0 such that(39)Ip≤Cfor0≤p≤1,Ce-γd0/ap2-1for1≤p<∞,then there is a constant Cγ>0 and δγ>0 such that VH1(Sa)≤Cγ as for all δ≤δγ.The proof of this theorem is somewhat tedious and may be found in the Appendix—although we only prove the theorem for∇VL2(Sa)2, Remark 3 implies that it holds for VL2(Sa) as well. In the next lemma, we show that bound (39) holds for very general sources f.Lemma 7.
Supposef∈L2(M) with compact support; then (39) holds.Proof.
For0≤p≤1, recall from (24) that(40)Ip=∫d0d1f^s,k0pe-ik01-p2sds.Then the triangle, Cauchy-Schwarz, and Jensen inequalities imply that(41)Ip≤∫d0d1f^s,k0pds≤d1-d01/2∫d0d1f^s,k0p2ds1/2=d1-d01/2∫d0d1∫-∞∞fs,ye-ik0pydy2ds1/2≤d1-d01/2∫d0d1∫-∞∞fs,y2dyds1/2=d1-d01/2fL2M.
Similarly, forp≥1, recall from (24) that(42)Ip=∫d0d1f^s,k0pe-k0p2-1sds.Then (43)Ip≤∫d0d1f^s,k0pe-k0p2-1sds≤d1-d01/2∫d0d1f^s,k0p2e-2k0p2-1sds1/2≤d1-d01/2fL2Me-k0d0p2-1=d1-d01/2fL2Me-γd0/ap2-1.
To complete the proof, we defineC≡(d1-d0)1/2fL2(M).


## 3.2. Shielding Effect for Largeγ

It turns out that the slab lens behaves as a shield and acts as an almost perfect reflector. This fact was also observed in [21] where it was explained based on the fact that, at least in the lossless nonmagnetic case ϵ=-1, μ=1 will give a purely imaginary wavenumber inside the slab and thus no propagation beyond the slab in region C. We have the following.Theorem 8.
Supposeγ≥2γ∗, Ip satisfies (39), and choose 0<η<1; then there is a constant Cη>0 such that(44)Vcx,y≤Cηe-ηγ∀x,y∈C.In particular,(45)Vcx,y⟶0ask0⟶∞∀x,y∈C.Remark 9.
Lemma7 implies that Theorems 6 and 8 hold for all sources f∈L2(M) with compact support. However, the bound in (39) is stronger than we need. For example, suppose there is a positive, real-valued function B(p;γ) that is continuous for 0≤p<∞ and γ∗≤γ<∞. In addition, for every ϵ>0, suppose that(46)limp→∞Bp;γe-ϵγp2-1=0∀γ≥γ∗,limγ→∞Bp;γe-ϵγp2-1=0∀p≥1.For example, if B(p;γ) is a continuous function of p and γ that is of polynomial order for p→∞ and γ→∞, it will satisfy (46). Finally, suppose(47)Ip≤Bp;γfor0≤p≤1,Bp;γe-γd0/ap2-1for1≤p<∞.Then, by appropriately modifying (A.72)–(A.75), one can prove that the result of Theorem 6 will hold for sources satisfying (47). Similarly, by appropriately modifying (A.82)–(A.86), one can show that Theorem 8 also holds for sources satisfying (47) as long as we replace (44) by(48)Vcx,y≤Cηe-η-ϵγ, where 0<ϵ<η.
In particular, certain distributional sources such as dipoles and quadrupoles satisfy (47)—see Section A.2 for more details.In Figure2, we plot the solution V to (4) in the case where f is a dipole with dipole moment [1,0]T, δ=10-12, and γ=4γ∗ (we take a=1 in all figures throughout the paper). In Figures 2(a) and 2(b), the dipole is located at the point (d0,0)=(4a,0); in Figures 2(c) and 2(d), the dipole is located closer to the slab at the point (d0,0)=(1.2a,0). The solution V is smooth throughout the domain; in addition, we observe the “shielding effect” from Theorem 8 in the region to the left of the lens.Figure 2
This is a plot ofV, the solution to (4), when f is a dipole and γ=2γ∗: (a) Re(V) and (b) Im(V) for d0=4a; (c) Re(V) and (d) Im(V) for d0=1.2a. To make the behavior of V clearer, we clipped the maximum and minimum values in each plot to 0.2 (yellow) and -0.2 (blue), respectively.
(a) (b) (c) (d)In Figure3, we plot Eδ(a) as a function of various parameters for a dipole source f. The parameters we used are in the ranges 10-12≤δ≤10-10, 1.01γ∗≤γ≤2γ∗, and 1.2a≤d0≤2a. We note that Eδ(a) depends strongly on δ, γ, and d0, but, because γ>γ∗, Eδ(a) is quite small.Figure 3
These are plots ofEδ(a) as a function of (a) δ and γ (d0=1.2a); (b) δ and γ (d0=4a); (c) δ and d0 (γ=1.01γ∗); (d) δ and d0 (γ=2γ∗); (e) γ and d0 (δ=10-10); (f) γ and d0 (δ=10-12).
(a) (b) (c) (d) (e) (f)Figure4 is similar to Figure 2, except in Figure 4 we take(49)fx,y=C2d1-d0x-d0-132d1-d0x-d0-1-13·2h1-h0y-h0-132h1-h0y-d0-1-13ford0≤x≤d1,h0≤y≤h10otherwise.Although this source has compact support, in contrast to the dipolar source considered above it is in L2(M) and is twice continuously differentiable; we chose this source to emphasize that the ideas presented in this paper do not rely on the extreme nature of distributional sources such as dipoles.Figure 4
This is a plot ofV, the solution to (4), when f is the function in (49) and γ=2γ∗: (a) Re(V) and (b) Im⁡(V) for d0=4a; (c) Re(V) and (d) Im⁡(V) for d0=1.2a. To make the behavior of V clearer, we clipped the maximum and minimum values in each plot to 0.2 (yellow) and -0.2 (blue), respectively.
(a) (b) (c) (d)To construct the plots, we have takenC=104,h1=-h0=1, and d1=d0+2. The solution V is smooth throughout the domain and very small in the region to the left of the slab.


## 4. Long-Wavelength/Low Frequency Regime (γ<γ∗)

Unfortunately, the complicated nature of expression (30) has thus far prevented us from deriving lower bounds on Eδ(a) that would allow us to prove that Eδ(a)→∞ as δ→0+. Undaunted, in this section we present an heuristic argument, coupled with numerical experiments, to illustrate why we believe the slab lens under consideration exhibits ALR in the long-wavelength regime.


### 4.1. Blow-Up ofEδ(a)

The key result of this section is Lemma5: g0p;γ has two real roots when γ<γ∗; namely, 1<pγ1<pγ2. Because both roots are larger than 1, the main contribution to the blow-up of Eδ(a) comes from the integral over the interval 1≤p<∞. Indeed, the following lemma shows that we do not need to worry about the integral over the interval 0≤p≤1.Lemma 10.
Suppose0<γ≤γ∗ and f∈L2(M) with compact support. Then there is a positive constant Cγ and δγ>0 such that(50)∫01Lδp;γdp=∫01Ip2gδ2Mδp;γdp≤Cγfor all 0<δ≤δγ.Remark 11.
We emphasize that Lemma10 also holds for those sources for which the bound in (47) holds (e.g., dipole sources)—see Remark 9.Proof.
First, we note thatMδ(p;γ) is continuous for δ∈[0,1], p∈[0,1], and γ∈[0,γ∗], so it is bounded by a constant independent of δ, p, and γ. Additionally, Ip2 is also bounded by a constant, thanks to Lemma 7. All that remains for us to show is that gδp;γ is bounded away from 0.
We define the function(51)Ξδγ≡maxp∈0,1gδp;γ-g0p;γ.Because gδp;γ and g0p;γ are both continuous for 0≤p≤1, the above maximum is attained, say at p=pδ∗(γ). This means that(52)Ξδγ=gpδ∗γ;γ-g0pδ∗γ;γ.
Now let{δn}n=1∞ be a sequence converging to 0 as n→∞. Because pδn∗(γ) is a bounded sequence, it has a convergent subsequence pδnk∗(γ). Along this subsequence,(53)Ξδnkγ=gδnkpδnk∗γ;γ-g0pδnk∗γ;γ⟶0⁡ask⟶∞ by Lemma 4. In other words, every sequence Ξδn(γ) has a subsequence that converges to 0, which implies that every sequence Ξδn converges to 0. Because the original sequence δn was arbitrary, this implies that(54)limδ→0+Ξδγ=0. In combination with (51), this implies that gδp;γ converges to g0p;γuniformly in p for 0≤p≤1. Thus for every ϵ>0 there is δγ>0 such that(55)gδp;γ≥g0p;γ-ϵ for all p∈[0,1] and all 0<δ≤δγ. If we take(56)ϵ=12min0≤p≤1g0p;γ, then(57)gδp;γ≥12g0p;γ≥Cγ>0for all p∈[0,1] (the last two inequalities hold because the roots of g0 are larger than 1 by Lemma 5). Combining this result with the first paragraph of the proof gives us the bound(58)∫01Ip2gδp;γ2Mδp;γdp≤C∫011g0p;γ2dp≤Cγ for some constant Cγ>0.The preceding lemma proves that we only need to study the integral in (30) over the interval 1≤p<∞. Because gδp;γ→g0p;γ as δ→0+, it should be the case that gδp;γ≈0 near the roots of g0p;γ. Inspired by our earlier work in the quasistatic regime, we conjecture that gδpγ1;γ and gδpγ2;γ are on the order of δ as δ→0+.Conjecture 12.
Suppose0<γ<γ∗, and let 1<pγ1<pγ2 be the roots of g0(p;γ). Then there is δγ>0 such that gδp;γ≠0 for all 1≤p<∞ and all 0<δ≤δγ; however, gδpγ1;γ=O(δ) and gδpγ2;γ=O(δ) as δ→0+.One way to prove this conjecture would be to expandgδpγj;γ (for j=1, 2) in Taylor series around δ=0 and then prove that ∂gδpγj;γ/∂δ is uniformly bounded for p∈[1,∞) and δ small enough. Unfortunately, these derivatives are quite complicated; moreover, numerical experiments indicate that they become unbounded as p→∞, so it is unlikely that this technique would work even if the expressions were suitable for analytic study. To provide some justification for Conjecture 12, in Figures 5(a) and 5(b) we plot(59)gδpγ1;γδ,gδpγ1;γδas functions of δ and γ over the ranges 10-12≤δ≤10-10 and 0.1γ∗≤γ≤0.99γ∗ (we believe the functions in (59) remain bounded as δ→0 for all 0<γ<γ∗; however, pγ2→∞ as γ→0, so the numerical computation of the roots becomes more difficult as γ gets closer to 0. Similarly, pγ∗1=pγ∗2, so as γ gets close to γ∗ it becomes difficult to distinguish the roots). For each γ, we see that the functions in (59) remain bounded as δ gets close to 0, which seems to indicate that gδpγ1;γ=O(δ) and gδpγ2;γ=O(δ) as δ→0. Curiously, both functions in (59) seem to depend very weakly on δ.Figure 5
In this figure, we plot (a)gδpγ1;γ/δ and (b) gδpγ2;γ/δ over the range 10-12≤δ≤10-10 and 0.1γ∗≤γ≤0.99γ∗.
(a) (b)Next, we conjecture that theO(δ) behavior of gδp;γ near pγ1 and pγ2 is not canceled by the term Mδ(p;γ) in the numerator.Conjecture 13.
Suppose0<γ<γ∗, and define Mδ(p;γ) as in (38). Then there exist positive constants δγ and Cγ such that Mδ(p;γ)≥Cγ near pγ1 and pγ2 for all 0<δ≤δγ.If Conjectures12 and 13 are true, then (36)-(37) imply that the part of the integrand Lδ(p;γ) that is independent of the source f, namely,(60)e2γνm′Mδp;γgδp;γ2, is on the order of δ-2 near pγ1 and pγ2 as δ→0+. If Ip2 is also bounded away from 0 near pγ1 and pγ2, the entire integrand Lδ(p;γ) will have values on the order of δ-2 near pγ1 and pγ2.To provide some justification for Conjecture13, in Figures 6(a) and 6(b) we plot(61)Mδpγ1;γ,Mδpγ2;γ as functions of δ and γ over the same intervals as in Figure 5. In particular, we note that Mδ(pγ1;γ) and Mδ(pγ2;γ) are both bounded away from 0 and seem to depend quite weakly on δ.Figure 6
In this figure, we plot (a)Mδ(pγ1;γ) and (b) Mδ(pγ2;γ) over the range 10-12≤δ≤10-10 and 0.1γ∗≤γ≤0.99γ∗.
(a) (b)Finally, to obtain a blow-up inEδ(a), it should be the case that Ip does not conquer the small values of gδ near pγ1 and pγ2. Heuristically, there will be no blow-up if Ip≈0 near pγ1 and pγ2. In the next section, we present numerical evidence that suggests that sources with Ipγ1=Ipγ2=0 do not lead to ALR.On the other hand, recall from (24) that (62)Ip=∫d0d1f^x,k0pe-k0p2-1sds. Again we take our inspiration from the quasistatic case [23, 24]. If d0≫a, then the exponential in the above integrand will be extremely small (especially because pγ1 and pγ2 are both greater than 1). In particular, the exponential may be small enough so that it cancels out the effect of the denominator near pγ1 and pγ2. We emphasize that this is not rigorous, but we hope that it may provide a starting point for future investigations.Conjecture 14.
Suppose0<γ<γ∗. Then there exist sources f∈L2(M) with compact support or distributional sources such as dipoles such that, for any 0<ξ≤a, Eδ(ξ)→∞ if d0 is “close enough” to a and Eδ(ξ)≤Cγ for some positive constant Cγ if d0 is “far enough away” from a. This critical distance may depend on γ.
Moreover, there are positive constantsbγ, Cγ, and δγ such that, for all 0<δ≤δγ,(63)Vx,y≤Cγ for all (x,y)∈C∪M with x>bγ.Remark 15.
If it is only the case that(64)limsupδ→0+Eδξ=∞, then we say thatweak ALR occurs. Because Eδ(ξ) is difficult to deal with analytically, we cannot say much more on this. It is difficult to determine whether(65)limsupδ→0+Eδξ=∞orlimδ→0+Eδξ=∞ using only numerical techniques. In particular, if the limit supremum of Eδ(ξ) is ∞, there is at least one sequence δn→0+ along which Eδn(ξ)→∞; however, it may be the case that Eδn(ξ)→∞ for all sequences δn→0+ except a few very special sequences that would be extremely difficult to find via numerical experiments alone.Figures7 and 8 are exactly the same as Figures 2 and 4 except γ=0.5γ∗ in Figures 7 and 8. In Figures 7(a), 7(b), 8(a), and 8(b), the sources (a dipole in Figure 7 and the source f from (49) in Figure 8) are located at d0=4a, and the solution V appears to be smooth throughout the domain. As the sources move closer to the slab, resonant regions appear around both boundaries of the slab at x=0 and x=a. Figures 7(c), 7(d), 8(c), and 8(d) contain plots of V when d0=1.2a. From these figures we see that the extreme oscillations of V are contained near the boundaries of the slab and that the boundaries between the resonant and nonresonant regions are sharp and not defined by the boundaries of the slab; away from the slab, V is smooth and bounded. This is highly characteristic of ALR (see, e.g., [1, 23] and the references therein). Moreover, Figures 7 and 8 indicate that an image of (part of) the solution V is focused in the region to the left of the lens (outside of the resonant region); this is in stark contrast to the high frequency regime illustrated in Figures 2 and 4, in which the solution V in the region to the left of the slab is barely noticeable. Indeed, in the quasistatic regime, ALR is closely associated with this so-called superlensing [1]; since ALR does not occur for γ>γ∗ (see Theorem 6), we do not expect to see the superlensing effect in this regime (see Theorem 8).Figure 7
This is a plot ofV, the solution to (4), when f is a dipole and γ=0.5γ∗: (a) Re(V) and (b) Im(V) for d0=4a; (c) Re(V) and (d) Im(V) for d0=1.2a. To make the behavior of V clearer, we clipped the maximum and minimum values in each plot to 0.2 (yellow) and -0.2 (blue), respectively. The vertical red line in (c) extends a distance of 2λγ, where λγ is defined in (66).
(a) (b) (c) (d)Figure 8
This is a plot ofV, the solution to (4), when f is the function in (49) and γ=0.5γ∗: (a) Re(V) and (b) Im(V) for d0=4a; (c) Re(V) and (d) Im(V) for d0=1.2a. To make the behavior of Re(V) clearer, in (a) and (c) we clipped the maximum and minimum values in each plot to 0.1 (yellow) and -0.1 (blue), respectively. The vertical red line in (c) extends a distance of 2λγ, where λγ is defined in (66).
(a) (b) (c) (d)Figures7(c) and 8(c) provide an additional insight into Conjecture 14. In general, for q≈k0pγ2 (where pγ2 is the larger root of g0(p;γ)), the coefficient Aq from (23) becomes very large since its denominator is proportional to gδ(p;γ) and gδ(pγ2;γ)≈g0(pγ2;γ)=0 for δ small enough. Recalling that the Fourier transform variable q=k0p represents a wavenumber in the y-direction with corresponding wavelength λ=2π/q, this implies that the solution V should exhibit prominent oscillations with wavelength on the order of(66)λγ=2πk0pγ2.In Figures 7(c) and 8(c), we have drawn a vertical red line of length 2λγ. This red line covers approximately 2 wavelengths of oscillation in the resonant region, which seems to indicate that at least one of the zeros of g0, namely, pγ2, is responsible for ALR. Because pγ2 is independent of f, the above argument also suggests that the wavelength of the resonant oscillations of V is also independent of the source f. We emphasize that this is speculative at best, but it would be interesting to investigate further.To illustrate how drastically different the behavior ofV is for γ>γ∗ and γ<γ∗, in Figure 9 we plotted V corresponding to a dipole source located at d0=1.2a for two different values of γ. In Figures 9(a) and 9(b), we took γ=1.01γ∗ while in Figures 9(c) and 9(d) we took γ=0.99γ∗. The ALR is present when γ<γ∗ in Figure 9(c); on the other hand, in Figure 9(a) there are a few oscillations near the x-axis, but they quickly die out as y grows.Figure 9
This is a plot ofV, the solution to (4), when f is a dipole located at d0=1.2a: (a) Re(V) and (b) Im⁡(V) for γ=1.01γ∗; (c) Re(V) and (d) Im⁡(V) for γ=0.99γ∗. To make the behavior of V clearer, in (a), (c), and (d) we clipped the maximum and minimum values in each plot to 0.1 (yellow) and -0.1 (blue), respectively.
(a) (b) (c) (d)Unfortunately, we cannot provide a figure analogous to Figure3 for Eδ(a) when f is a dipole source—MATLAB is unable to accurately compute the integral(67)∫1∞Lδp;γdp because gδp;γ is very close to 0 near the roots of g0(p;δ) for small values of δ (see Conjecture 12). However, to get a sense of what is going on, we plotted Lδ(p;γ) on a logarithmic scale for a dipole source f with γ=0.99γ∗ in Figures 10(a) and 10(b) and γ=1.01γ∗ in Figures 10(c) and 10(d). Each curve is Lδ(p;γ) as a function of p for various values of δ. In Figures 10(a) and 10(b), where γ<γ∗, we see that Lδ(p;γ) is quite large near the poles of g0(p;γ), even if δ=10-4. Additionally, on comparing the y-axis scales in Figures 10(a) and 10(b), we note that the poles seem somewhat less severe in Figure 10(a) than in Figure 10(b), which, in combination with results from the quasistatic regime [24], lends credence to our conjecture (Conjecture 14) that ALR may be present only if the source is located close enough to the lens. On the other hand, in Figures 10(c) and 10(d), γ>γ∗ and we see that Lδ(p;γ) remains bounded regardless of d0 (in Figure 10, all of the functions rapidly tend to 0 for larger values of p (not shown in the figure)).Figure 10
A plot of the integrandLδ(p;γ) from (36) for several parameter values. The separate curves in each plot represent different values of δ, indicated in the legend: (a) γ=0.99γ∗ and d0=1.2a; (b) γ=0.99γ∗ and d0=4a; (c) γ=1.01γ∗ and d0=1.2a; (d) γ=1.01γ∗ and d0=4a.
(a) (b) (c) (d)


### 4.2. Sources for Which ALR Does Not Occur

When0<γ<γ∗, the conjectures from the previous section suggest that the zeros of g0(p;γ) are responsible for forcing Eδ(a) to blow up in the limit as δ→0+. This begs the question of whether one can design a (realistic) source in the finite-frequency regime (with 0<γ<γ∗) that effectively cancels the poles that show up in the limit δ→0+. In other words, we would like to design a source such that Ip=0 exactly at the zeros of g0(p;γ); heuristically, in the limit as δ→0+, this would force the integrand in (30) to remain bounded at the zeros of g0(p;γ) and annihilate the anomalous localized resonance that occurs in this limit. Recall from (24) that(68)Ip=∫d0d1f^s,k0pe-k0νmsds.Lemma 5 implies that g0(p;γ) has two roots 1<pγ1<pγ2. Using this and (68), we see that an “ALR-busting” source f(x,y) can be constructed by choosing f such that f^(s,k0pγ1)=f^(s,k0pγ2)=0 for all s∈[d0,d1] (which implies Ipγ1=Ipγ2=0). We do not want to just choose any f^ satisfying this property; however, we restrict ourselves to those sources f∈L2(M) with compact support. In summary, we make the following conjecture.Conjecture 16.
Supposef∈L2(M) has compact support and(69)f^x,k0pγ1=f^x,k0pγ2=0, where 1<pγ1<pγ2 are the zeros of g0(p;γ) from Lemma 5 and k0pγj are zeros of order at least 1 for f^(x,k0p). Then there is δγ>0 and a constant Cγ>0 such that Eδ(a)≤Cγ for all 0<δ≤δγ.There are many sources that satisfy the hypotheses of this theorem. We will present2 examples here. First, consider(70)f^x,q=-iχd0,d1xsinc⁡α1qsin⁡α2q,where sinc⁡(x)=sin⁡(x)/x,(71)χd0,d1x=1ford0<x<d1,0otherwise,α1≡πk0pγ1,α2≡πk0pγ2.Then f^(x,·)∈L2(R) and, hence, f(x,·)∈L2(R) by the Plancherel theorem; moreover, f^(x,k0pγ1)=f^(x,k0pγ2)=0, where the zeros are order 1. Finally, by direct calculations we have(72)fx,y=χd0,d1x·14α1H-y-α1-α2-H-y-α1+α2+H-y+α1+α2-H-y-α2+α1,where H(z) is the Heaviside step function; this f∈L2(M) has compact support and thus satisfies the hypotheses of Conjecture 16. We may also take(73)f^x,q=χd0,d1xJ0β0qJ1β1q,where J0 and J1 are the Bessel functions of the first kind of orders 0 and 1, respectively, and β0 and β1 are such that J0(β0k0pγ1)=J1(β1k0pγ2)=0 (we note that these zeros are also of order 1). Because the Bessel functions of the first kind are O(q-1/2) as q→∞ [25], we have f^(x,·)∈L2(R). By the convolution theorem for Fourier transforms,(74)fx,y=χd0,d1xf0∗f1y,where ∗ denotes convolution and f0 and f1 are the inverse Fourier transforms of J0(β0q) and J1(β1q), respectively; in particular, we obtain(75)f0y=1πβ02-y2χ-β0,β0y,f1y=-yβ1πβ12-y2χ-β1,β1y.Although the convolution in (74) is difficult to compute analytically, since f0 and f1 both have compact support, the convolution of f0 with f1 will as well. Thus f as defined in (74) is in L2(M) and has compact support.In Figures11(a) and 11(b) we plot Re(V) and Im⁡(V), respectively, for the source from (70) (equivalently, (72)); in Figures 11(c) and 11(d), we plot Re(V) and Im(V), respectively, for the source from (73) (equivalently, (74)). We take the same parameters that we used in Figures 7(c), 7(d), 8(c), and 8(d), namely, d0=1.2a and γ=0.5γ∗. In stark contrast with those figures, the solution V is well-behaved in Figure 11.Figure 11
This is a plot ofV, the solution to (4), for two different sources with d0=1.2a and γ=0.5γ∗: (a) Re(V) and (b) Im(V) for f as in (70) (equivalently, (72)); (c) Re(V) and (d) Im(V) for f as in (73) (equivalently (74)).
(a) (b) (c) (d)


#### 4.2.1. Current Sources for Which ALR Does Not Occur

In the monochromatic electromagnetic setting,f must satisfy additional restrictions for it to represent a realistic (divergence-free) current source—see Section A.5. In particular, the function f should be in L2(M) with compact support and be of the form(76)f=μ0∂J~y∂x-∂J~x∂yfor a current(77)J~=J~xx,yex+J~yx,yeysatisfying the continuity equation(78)∂J~x∂x+∂J~y∂y=0.We now construct a sourcef satisfying the hypotheses of Conjecture 16 that is of the form (76)–(78). For simplicity, we assume that the current from (77) has the form(79)J~xx,y=r1xt1y,J~yx,y=r2xt2y.with r1, r2, t1, t2 smooth enough. Then continuity equation (78) becomes(80)r1′xt1y+r2xt2′y=0. Taking the Fourier transform of this equation with respect to y gives(81)r1′xt^1q+r2xiqt^2q=0.We further simplify the problem by taking(82)t^1q=iqt^2q.Then (81) becomes(83)r1′x+r2xt^1q=0, which is satisfied for all x and q if(84)r2x=-r1′x.Then (76), (79), (82), and (84) imply that(85)f^x,q=μ0r2′xt^2q-r1xiqt1q=μ0-r1′′xt^2q+r1xq2t^2q.At this point, f satisfies (76)–(78).By the Plancherel theorem and (85), f∈L2(M) if and only if t^2(q)∈L2(R) and q2t^2(q)∈L2(R). We must also be careful to also choose t^2(q) in such a way that t^2(k0pγ1)=t^2(k0pγ2)=0 and f(x,y)=μ0-r1′′(x)t2(y)-r1xt2′′(y) has compact support in y.There are many examples of functions that accomplish these tasks. Unfortunately, the functions in (72) and (74) lead to current sources that are discontinuous and, hence, not divergence-free, so we need to be a bit more careful. To find a smooth current with compact support satisfying our requirements, we take(86)t^2q=sinc3α1qsinc2α2q,α1≡πk0pγ1,α2≡πk0pγ2.Then t^2(q)∈L2(R), q2t^2(q)∈L2(R), and f^(x,k0pγ1)=f^(x,k0pγ2)=0 (here the zeros are of orders 3 and 2, resp., so they are stronger than what we need according to Conjecture 16; we take these higher-order zeros to ensure that t2(y) and t2′′(y) are both continuous; there may be other choices of continuous functions t2(y) and t2′′(y) such that t^2(q)∈L2(R) has zeros of order 1 at q=k0p1 and q=k0p2).One possible choice ofr1(x) from (85) is(87)r1x=C2d1-d0x-d0-132d1-d0x-d0-1-13ford0≤x≤d1,0otherwise,where C is a nonzero constant. The function r1(x) is twice continuously differentiable and has compact support, so r1′′(x) is continuous with compact support.Finally,f^(x,q) may be computed via (85), (86), and (87). We note that the inverse Fourier transform of f^(x,q) can be computed analytically; for the benefit of the reader, we avoid writing out the expression. Importantly, f is continuous with compact support.The current source corresponding to thisf may be computed via (77), (79), (82), (84), (86), and (87). We emphasize that both J~x(x,y) and J~y(x,y) are continuously differentiable functions with compact support in M.In Figure12, we plot the source f defined by (85)–(87) with C=103μ0-1. In Figures 13(a) and 13(b), we plot Re(V) and Im⁡(V), respectively, corresponding to this source. As expected, we see that there is no resonant region near the slab even though the source is quite close to the slab (d0=1.2a), γ=0.5γ∗, and δ=10-12.Figure 12
This is the sourcef defined by (85)–(87) with the parameters d0=1.2a and d1=d0+2.Figure 13
In this figure, we plot (a)Re(V) and (b) Im⁡(V), where V is the solution to (4) corresponding to the source f defined through (85)–(87) with the parameters d0=1.2a, γ=0.5γ∗, and δ=10-12. To make the behavior of V clearer, we clipped the maximum and minimum values in each plot to 0.2 (yellow) and -0.2 (blue), respectively.
(a) (b)


## 4.1. Blow-Up ofEδ(a)

The key result of this section is Lemma5: g0p;γ has two real roots when γ<γ∗; namely, 1<pγ1<pγ2. Because both roots are larger than 1, the main contribution to the blow-up of Eδ(a) comes from the integral over the interval 1≤p<∞. Indeed, the following lemma shows that we do not need to worry about the integral over the interval 0≤p≤1.Lemma 10.
Suppose0<γ≤γ∗ and f∈L2(M) with compact support. Then there is a positive constant Cγ and δγ>0 such that(50)∫01Lδp;γdp=∫01Ip2gδ2Mδp;γdp≤Cγfor all 0<δ≤δγ.Remark 11.
We emphasize that Lemma10 also holds for those sources for which the bound in (47) holds (e.g., dipole sources)—see Remark 9.Proof.
First, we note thatMδ(p;γ) is continuous for δ∈[0,1], p∈[0,1], and γ∈[0,γ∗], so it is bounded by a constant independent of δ, p, and γ. Additionally, Ip2 is also bounded by a constant, thanks to Lemma 7. All that remains for us to show is that gδp;γ is bounded away from 0.
We define the function(51)Ξδγ≡maxp∈0,1gδp;γ-g0p;γ.Because gδp;γ and g0p;γ are both continuous for 0≤p≤1, the above maximum is attained, say at p=pδ∗(γ). This means that(52)Ξδγ=gpδ∗γ;γ-g0pδ∗γ;γ.
Now let{δn}n=1∞ be a sequence converging to 0 as n→∞. Because pδn∗(γ) is a bounded sequence, it has a convergent subsequence pδnk∗(γ). Along this subsequence,(53)Ξδnkγ=gδnkpδnk∗γ;γ-g0pδnk∗γ;γ⟶0⁡ask⟶∞ by Lemma 4. In other words, every sequence Ξδn(γ) has a subsequence that converges to 0, which implies that every sequence Ξδn converges to 0. Because the original sequence δn was arbitrary, this implies that(54)limδ→0+Ξδγ=0. In combination with (51), this implies that gδp;γ converges to g0p;γuniformly in p for 0≤p≤1. Thus for every ϵ>0 there is δγ>0 such that(55)gδp;γ≥g0p;γ-ϵ for all p∈[0,1] and all 0<δ≤δγ. If we take(56)ϵ=12min0≤p≤1g0p;γ, then(57)gδp;γ≥12g0p;γ≥Cγ>0for all p∈[0,1] (the last two inequalities hold because the roots of g0 are larger than 1 by Lemma 5). Combining this result with the first paragraph of the proof gives us the bound(58)∫01Ip2gδp;γ2Mδp;γdp≤C∫011g0p;γ2dp≤Cγ for some constant Cγ>0.The preceding lemma proves that we only need to study the integral in (30) over the interval 1≤p<∞. Because gδp;γ→g0p;γ as δ→0+, it should be the case that gδp;γ≈0 near the roots of g0p;γ. Inspired by our earlier work in the quasistatic regime, we conjecture that gδpγ1;γ and gδpγ2;γ are on the order of δ as δ→0+.Conjecture 12.
Suppose0<γ<γ∗, and let 1<pγ1<pγ2 be the roots of g0(p;γ). Then there is δγ>0 such that gδp;γ≠0 for all 1≤p<∞ and all 0<δ≤δγ; however, gδpγ1;γ=O(δ) and gδpγ2;γ=O(δ) as δ→0+.One way to prove this conjecture would be to expandgδpγj;γ (for j=1, 2) in Taylor series around δ=0 and then prove that ∂gδpγj;γ/∂δ is uniformly bounded for p∈[1,∞) and δ small enough. Unfortunately, these derivatives are quite complicated; moreover, numerical experiments indicate that they become unbounded as p→∞, so it is unlikely that this technique would work even if the expressions were suitable for analytic study. To provide some justification for Conjecture 12, in Figures 5(a) and 5(b) we plot(59)gδpγ1;γδ,gδpγ1;γδas functions of δ and γ over the ranges 10-12≤δ≤10-10 and 0.1γ∗≤γ≤0.99γ∗ (we believe the functions in (59) remain bounded as δ→0 for all 0<γ<γ∗; however, pγ2→∞ as γ→0, so the numerical computation of the roots becomes more difficult as γ gets closer to 0. Similarly, pγ∗1=pγ∗2, so as γ gets close to γ∗ it becomes difficult to distinguish the roots). For each γ, we see that the functions in (59) remain bounded as δ gets close to 0, which seems to indicate that gδpγ1;γ=O(δ) and gδpγ2;γ=O(δ) as δ→0. Curiously, both functions in (59) seem to depend very weakly on δ.Figure 5
In this figure, we plot (a)gδpγ1;γ/δ and (b) gδpγ2;γ/δ over the range 10-12≤δ≤10-10 and 0.1γ∗≤γ≤0.99γ∗.
(a) (b)Next, we conjecture that theO(δ) behavior of gδp;γ near pγ1 and pγ2 is not canceled by the term Mδ(p;γ) in the numerator.Conjecture 13.
Suppose0<γ<γ∗, and define Mδ(p;γ) as in (38). Then there exist positive constants δγ and Cγ such that Mδ(p;γ)≥Cγ near pγ1 and pγ2 for all 0<δ≤δγ.If Conjectures12 and 13 are true, then (36)-(37) imply that the part of the integrand Lδ(p;γ) that is independent of the source f, namely,(60)e2γνm′Mδp;γgδp;γ2, is on the order of δ-2 near pγ1 and pγ2 as δ→0+. If Ip2 is also bounded away from 0 near pγ1 and pγ2, the entire integrand Lδ(p;γ) will have values on the order of δ-2 near pγ1 and pγ2.To provide some justification for Conjecture13, in Figures 6(a) and 6(b) we plot(61)Mδpγ1;γ,Mδpγ2;γ as functions of δ and γ over the same intervals as in Figure 5. In particular, we note that Mδ(pγ1;γ) and Mδ(pγ2;γ) are both bounded away from 0 and seem to depend quite weakly on δ.Figure 6
In this figure, we plot (a)Mδ(pγ1;γ) and (b) Mδ(pγ2;γ) over the range 10-12≤δ≤10-10 and 0.1γ∗≤γ≤0.99γ∗.
(a) (b)Finally, to obtain a blow-up inEδ(a), it should be the case that Ip does not conquer the small values of gδ near pγ1 and pγ2. Heuristically, there will be no blow-up if Ip≈0 near pγ1 and pγ2. In the next section, we present numerical evidence that suggests that sources with Ipγ1=Ipγ2=0 do not lead to ALR.On the other hand, recall from (24) that (62)Ip=∫d0d1f^x,k0pe-k0p2-1sds. Again we take our inspiration from the quasistatic case [23, 24]. If d0≫a, then the exponential in the above integrand will be extremely small (especially because pγ1 and pγ2 are both greater than 1). In particular, the exponential may be small enough so that it cancels out the effect of the denominator near pγ1 and pγ2. We emphasize that this is not rigorous, but we hope that it may provide a starting point for future investigations.Conjecture 14.
Suppose0<γ<γ∗. Then there exist sources f∈L2(M) with compact support or distributional sources such as dipoles such that, for any 0<ξ≤a, Eδ(ξ)→∞ if d0 is “close enough” to a and Eδ(ξ)≤Cγ for some positive constant Cγ if d0 is “far enough away” from a. This critical distance may depend on γ.
Moreover, there are positive constantsbγ, Cγ, and δγ such that, for all 0<δ≤δγ,(63)Vx,y≤Cγ for all (x,y)∈C∪M with x>bγ.Remark 15.
If it is only the case that(64)limsupδ→0+Eδξ=∞, then we say thatweak ALR occurs. Because Eδ(ξ) is difficult to deal with analytically, we cannot say much more on this. It is difficult to determine whether(65)limsupδ→0+Eδξ=∞orlimδ→0+Eδξ=∞ using only numerical techniques. In particular, if the limit supremum of Eδ(ξ) is ∞, there is at least one sequence δn→0+ along which Eδn(ξ)→∞; however, it may be the case that Eδn(ξ)→∞ for all sequences δn→0+ except a few very special sequences that would be extremely difficult to find via numerical experiments alone.Figures7 and 8 are exactly the same as Figures 2 and 4 except γ=0.5γ∗ in Figures 7 and 8. In Figures 7(a), 7(b), 8(a), and 8(b), the sources (a dipole in Figure 7 and the source f from (49) in Figure 8) are located at d0=4a, and the solution V appears to be smooth throughout the domain. As the sources move closer to the slab, resonant regions appear around both boundaries of the slab at x=0 and x=a. Figures 7(c), 7(d), 8(c), and 8(d) contain plots of V when d0=1.2a. From these figures we see that the extreme oscillations of V are contained near the boundaries of the slab and that the boundaries between the resonant and nonresonant regions are sharp and not defined by the boundaries of the slab; away from the slab, V is smooth and bounded. This is highly characteristic of ALR (see, e.g., [1, 23] and the references therein). Moreover, Figures 7 and 8 indicate that an image of (part of) the solution V is focused in the region to the left of the lens (outside of the resonant region); this is in stark contrast to the high frequency regime illustrated in Figures 2 and 4, in which the solution V in the region to the left of the slab is barely noticeable. Indeed, in the quasistatic regime, ALR is closely associated with this so-called superlensing [1]; since ALR does not occur for γ>γ∗ (see Theorem 6), we do not expect to see the superlensing effect in this regime (see Theorem 8).Figure 7
This is a plot ofV, the solution to (4), when f is a dipole and γ=0.5γ∗: (a) Re(V) and (b) Im(V) for d0=4a; (c) Re(V) and (d) Im(V) for d0=1.2a. To make the behavior of V clearer, we clipped the maximum and minimum values in each plot to 0.2 (yellow) and -0.2 (blue), respectively. The vertical red line in (c) extends a distance of 2λγ, where λγ is defined in (66).
(a) (b) (c) (d)Figure 8
This is a plot ofV, the solution to (4), when f is the function in (49) and γ=0.5γ∗: (a) Re(V) and (b) Im(V) for d0=4a; (c) Re(V) and (d) Im(V) for d0=1.2a. To make the behavior of Re(V) clearer, in (a) and (c) we clipped the maximum and minimum values in each plot to 0.1 (yellow) and -0.1 (blue), respectively. The vertical red line in (c) extends a distance of 2λγ, where λγ is defined in (66).
(a) (b) (c) (d)Figures7(c) and 8(c) provide an additional insight into Conjecture 14. In general, for q≈k0pγ2 (where pγ2 is the larger root of g0(p;γ)), the coefficient Aq from (23) becomes very large since its denominator is proportional to gδ(p;γ) and gδ(pγ2;γ)≈g0(pγ2;γ)=0 for δ small enough. Recalling that the Fourier transform variable q=k0p represents a wavenumber in the y-direction with corresponding wavelength λ=2π/q, this implies that the solution V should exhibit prominent oscillations with wavelength on the order of(66)λγ=2πk0pγ2.In Figures 7(c) and 8(c), we have drawn a vertical red line of length 2λγ. This red line covers approximately 2 wavelengths of oscillation in the resonant region, which seems to indicate that at least one of the zeros of g0, namely, pγ2, is responsible for ALR. Because pγ2 is independent of f, the above argument also suggests that the wavelength of the resonant oscillations of V is also independent of the source f. We emphasize that this is speculative at best, but it would be interesting to investigate further.To illustrate how drastically different the behavior ofV is for γ>γ∗ and γ<γ∗, in Figure 9 we plotted V corresponding to a dipole source located at d0=1.2a for two different values of γ. In Figures 9(a) and 9(b), we took γ=1.01γ∗ while in Figures 9(c) and 9(d) we took γ=0.99γ∗. The ALR is present when γ<γ∗ in Figure 9(c); on the other hand, in Figure 9(a) there are a few oscillations near the x-axis, but they quickly die out as y grows.Figure 9
This is a plot ofV, the solution to (4), when f is a dipole located at d0=1.2a: (a) Re(V) and (b) Im⁡(V) for γ=1.01γ∗; (c) Re(V) and (d) Im⁡(V) for γ=0.99γ∗. To make the behavior of V clearer, in (a), (c), and (d) we clipped the maximum and minimum values in each plot to 0.1 (yellow) and -0.1 (blue), respectively.
(a) (b) (c) (d)Unfortunately, we cannot provide a figure analogous to Figure3 for Eδ(a) when f is a dipole source—MATLAB is unable to accurately compute the integral(67)∫1∞Lδp;γdp because gδp;γ is very close to 0 near the roots of g0(p;δ) for small values of δ (see Conjecture 12). However, to get a sense of what is going on, we plotted Lδ(p;γ) on a logarithmic scale for a dipole source f with γ=0.99γ∗ in Figures 10(a) and 10(b) and γ=1.01γ∗ in Figures 10(c) and 10(d). Each curve is Lδ(p;γ) as a function of p for various values of δ. In Figures 10(a) and 10(b), where γ<γ∗, we see that Lδ(p;γ) is quite large near the poles of g0(p;γ), even if δ=10-4. Additionally, on comparing the y-axis scales in Figures 10(a) and 10(b), we note that the poles seem somewhat less severe in Figure 10(a) than in Figure 10(b), which, in combination with results from the quasistatic regime [24], lends credence to our conjecture (Conjecture 14) that ALR may be present only if the source is located close enough to the lens. On the other hand, in Figures 10(c) and 10(d), γ>γ∗ and we see that Lδ(p;γ) remains bounded regardless of d0 (in Figure 10, all of the functions rapidly tend to 0 for larger values of p (not shown in the figure)).Figure 10
A plot of the integrandLδ(p;γ) from (36) for several parameter values. The separate curves in each plot represent different values of δ, indicated in the legend: (a) γ=0.99γ∗ and d0=1.2a; (b) γ=0.99γ∗ and d0=4a; (c) γ=1.01γ∗ and d0=1.2a; (d) γ=1.01γ∗ and d0=4a.
(a) (b) (c) (d)


## 4.2. Sources for Which ALR Does Not Occur

When0<γ<γ∗, the conjectures from the previous section suggest that the zeros of g0(p;γ) are responsible for forcing Eδ(a) to blow up in the limit as δ→0+. This begs the question of whether one can design a (realistic) source in the finite-frequency regime (with 0<γ<γ∗) that effectively cancels the poles that show up in the limit δ→0+. In other words, we would like to design a source such that Ip=0 exactly at the zeros of g0(p;γ); heuristically, in the limit as δ→0+, this would force the integrand in (30) to remain bounded at the zeros of g0(p;γ) and annihilate the anomalous localized resonance that occurs in this limit. Recall from (24) that(68)Ip=∫d0d1f^s,k0pe-k0νmsds.Lemma 5 implies that g0(p;γ) has two roots 1<pγ1<pγ2. Using this and (68), we see that an “ALR-busting” source f(x,y) can be constructed by choosing f such that f^(s,k0pγ1)=f^(s,k0pγ2)=0 for all s∈[d0,d1] (which implies Ipγ1=Ipγ2=0). We do not want to just choose any f^ satisfying this property; however, we restrict ourselves to those sources f∈L2(M) with compact support. In summary, we make the following conjecture.Conjecture 16.
Supposef∈L2(M) has compact support and(69)f^x,k0pγ1=f^x,k0pγ2=0, where 1<pγ1<pγ2 are the zeros of g0(p;γ) from Lemma 5 and k0pγj are zeros of order at least 1 for f^(x,k0p). Then there is δγ>0 and a constant Cγ>0 such that Eδ(a)≤Cγ for all 0<δ≤δγ.There are many sources that satisfy the hypotheses of this theorem. We will present2 examples here. First, consider(70)f^x,q=-iχd0,d1xsinc⁡α1qsin⁡α2q,where sinc⁡(x)=sin⁡(x)/x,(71)χd0,d1x=1ford0<x<d1,0otherwise,α1≡πk0pγ1,α2≡πk0pγ2.Then f^(x,·)∈L2(R) and, hence, f(x,·)∈L2(R) by the Plancherel theorem; moreover, f^(x,k0pγ1)=f^(x,k0pγ2)=0, where the zeros are order 1. Finally, by direct calculations we have(72)fx,y=χd0,d1x·14α1H-y-α1-α2-H-y-α1+α2+H-y+α1+α2-H-y-α2+α1,where H(z) is the Heaviside step function; this f∈L2(M) has compact support and thus satisfies the hypotheses of Conjecture 16. We may also take(73)f^x,q=χd0,d1xJ0β0qJ1β1q,where J0 and J1 are the Bessel functions of the first kind of orders 0 and 1, respectively, and β0 and β1 are such that J0(β0k0pγ1)=J1(β1k0pγ2)=0 (we note that these zeros are also of order 1). Because the Bessel functions of the first kind are O(q-1/2) as q→∞ [25], we have f^(x,·)∈L2(R). By the convolution theorem for Fourier transforms,(74)fx,y=χd0,d1xf0∗f1y,where ∗ denotes convolution and f0 and f1 are the inverse Fourier transforms of J0(β0q) and J1(β1q), respectively; in particular, we obtain(75)f0y=1πβ02-y2χ-β0,β0y,f1y=-yβ1πβ12-y2χ-β1,β1y.Although the convolution in (74) is difficult to compute analytically, since f0 and f1 both have compact support, the convolution of f0 with f1 will as well. Thus f as defined in (74) is in L2(M) and has compact support.In Figures11(a) and 11(b) we plot Re(V) and Im⁡(V), respectively, for the source from (70) (equivalently, (72)); in Figures 11(c) and 11(d), we plot Re(V) and Im(V), respectively, for the source from (73) (equivalently, (74)). We take the same parameters that we used in Figures 7(c), 7(d), 8(c), and 8(d), namely, d0=1.2a and γ=0.5γ∗. In stark contrast with those figures, the solution V is well-behaved in Figure 11.Figure 11
This is a plot ofV, the solution to (4), for two different sources with d0=1.2a and γ=0.5γ∗: (a) Re(V) and (b) Im(V) for f as in (70) (equivalently, (72)); (c) Re(V) and (d) Im(V) for f as in (73) (equivalently (74)).
(a) (b) (c) (d)


### 4.2.1. Current Sources for Which ALR Does Not Occur

In the monochromatic electromagnetic setting,f must satisfy additional restrictions for it to represent a realistic (divergence-free) current source—see Section A.5. In particular, the function f should be in L2(M) with compact support and be of the form(76)f=μ0∂J~y∂x-∂J~x∂yfor a current(77)J~=J~xx,yex+J~yx,yeysatisfying the continuity equation(78)∂J~x∂x+∂J~y∂y=0.We now construct a sourcef satisfying the hypotheses of Conjecture 16 that is of the form (76)–(78). For simplicity, we assume that the current from (77) has the form(79)J~xx,y=r1xt1y,J~yx,y=r2xt2y.with r1, r2, t1, t2 smooth enough. Then continuity equation (78) becomes(80)r1′xt1y+r2xt2′y=0. Taking the Fourier transform of this equation with respect to y gives(81)r1′xt^1q+r2xiqt^2q=0.We further simplify the problem by taking(82)t^1q=iqt^2q.Then (81) becomes(83)r1′x+r2xt^1q=0, which is satisfied for all x and q if(84)r2x=-r1′x.Then (76), (79), (82), and (84) imply that(85)f^x,q=μ0r2′xt^2q-r1xiqt1q=μ0-r1′′xt^2q+r1xq2t^2q.At this point, f satisfies (76)–(78).By the Plancherel theorem and (85), f∈L2(M) if and only if t^2(q)∈L2(R) and q2t^2(q)∈L2(R). We must also be careful to also choose t^2(q) in such a way that t^2(k0pγ1)=t^2(k0pγ2)=0 and f(x,y)=μ0-r1′′(x)t2(y)-r1xt2′′(y) has compact support in y.There are many examples of functions that accomplish these tasks. Unfortunately, the functions in (72) and (74) lead to current sources that are discontinuous and, hence, not divergence-free, so we need to be a bit more careful. To find a smooth current with compact support satisfying our requirements, we take(86)t^2q=sinc3α1qsinc2α2q,α1≡πk0pγ1,α2≡πk0pγ2.Then t^2(q)∈L2(R), q2t^2(q)∈L2(R), and f^(x,k0pγ1)=f^(x,k0pγ2)=0 (here the zeros are of orders 3 and 2, resp., so they are stronger than what we need according to Conjecture 16; we take these higher-order zeros to ensure that t2(y) and t2′′(y) are both continuous; there may be other choices of continuous functions t2(y) and t2′′(y) such that t^2(q)∈L2(R) has zeros of order 1 at q=k0p1 and q=k0p2).One possible choice ofr1(x) from (85) is(87)r1x=C2d1-d0x-d0-132d1-d0x-d0-1-13ford0≤x≤d1,0otherwise,where C is a nonzero constant. The function r1(x) is twice continuously differentiable and has compact support, so r1′′(x) is continuous with compact support.Finally,f^(x,q) may be computed via (85), (86), and (87). We note that the inverse Fourier transform of f^(x,q) can be computed analytically; for the benefit of the reader, we avoid writing out the expression. Importantly, f is continuous with compact support.The current source corresponding to thisf may be computed via (77), (79), (82), (84), (86), and (87). We emphasize that both J~x(x,y) and J~y(x,y) are continuously differentiable functions with compact support in M.In Figure12, we plot the source f defined by (85)–(87) with C=103μ0-1. In Figures 13(a) and 13(b), we plot Re(V) and Im⁡(V), respectively, corresponding to this source. As expected, we see that there is no resonant region near the slab even though the source is quite close to the slab (d0=1.2a), γ=0.5γ∗, and δ=10-12.Figure 12
This is the sourcef defined by (85)–(87) with the parameters d0=1.2a and d1=d0+2.Figure 13
In this figure, we plot (a)Re(V) and (b) Im⁡(V), where V is the solution to (4) corresponding to the source f defined through (85)–(87) with the parameters d0=1.2a, γ=0.5γ∗, and δ=10-12. To make the behavior of V clearer, we clipped the maximum and minimum values in each plot to 0.2 (yellow) and -0.2 (blue), respectively.
(a) (b)


## 4.2.1. Current Sources for Which ALR Does Not Occur

In the monochromatic electromagnetic setting,f must satisfy additional restrictions for it to represent a realistic (divergence-free) current source—see Section A.5. In particular, the function f should be in L2(M) with compact support and be of the form(76)f=μ0∂J~y∂x-∂J~x∂yfor a current(77)J~=J~xx,yex+J~yx,yeysatisfying the continuity equation(78)∂J~x∂x+∂J~y∂y=0.We now construct a sourcef satisfying the hypotheses of Conjecture 16 that is of the form (76)–(78). For simplicity, we assume that the current from (77) has the form(79)J~xx,y=r1xt1y,J~yx,y=r2xt2y.with r1, r2, t1, t2 smooth enough. Then continuity equation (78) becomes(80)r1′xt1y+r2xt2′y=0. Taking the Fourier transform of this equation with respect to y gives(81)r1′xt^1q+r2xiqt^2q=0.We further simplify the problem by taking(82)t^1q=iqt^2q.Then (81) becomes(83)r1′x+r2xt^1q=0, which is satisfied for all x and q if(84)r2x=-r1′x.Then (76), (79), (82), and (84) imply that(85)f^x,q=μ0r2′xt^2q-r1xiqt1q=μ0-r1′′xt^2q+r1xq2t^2q.At this point, f satisfies (76)–(78).By the Plancherel theorem and (85), f∈L2(M) if and only if t^2(q)∈L2(R) and q2t^2(q)∈L2(R). We must also be careful to also choose t^2(q) in such a way that t^2(k0pγ1)=t^2(k0pγ2)=0 and f(x,y)=μ0-r1′′(x)t2(y)-r1xt2′′(y) has compact support in y.There are many examples of functions that accomplish these tasks. Unfortunately, the functions in (72) and (74) lead to current sources that are discontinuous and, hence, not divergence-free, so we need to be a bit more careful. To find a smooth current with compact support satisfying our requirements, we take(86)t^2q=sinc3α1qsinc2α2q,α1≡πk0pγ1,α2≡πk0pγ2.Then t^2(q)∈L2(R), q2t^2(q)∈L2(R), and f^(x,k0pγ1)=f^(x,k0pγ2)=0 (here the zeros are of orders 3 and 2, resp., so they are stronger than what we need according to Conjecture 16; we take these higher-order zeros to ensure that t2(y) and t2′′(y) are both continuous; there may be other choices of continuous functions t2(y) and t2′′(y) such that t^2(q)∈L2(R) has zeros of order 1 at q=k0p1 and q=k0p2).One possible choice ofr1(x) from (85) is(87)r1x=C2d1-d0x-d0-132d1-d0x-d0-1-13ford0≤x≤d1,0otherwise,where C is a nonzero constant. The function r1(x) is twice continuously differentiable and has compact support, so r1′′(x) is continuous with compact support.Finally,f^(x,q) may be computed via (85), (86), and (87). We note that the inverse Fourier transform of f^(x,q) can be computed analytically; for the benefit of the reader, we avoid writing out the expression. Importantly, f is continuous with compact support.The current source corresponding to thisf may be computed via (77), (79), (82), (84), (86), and (87). We emphasize that both J~x(x,y) and J~y(x,y) are continuously differentiable functions with compact support in M.In Figure12, we plot the source f defined by (85)–(87) with C=103μ0-1. In Figures 13(a) and 13(b), we plot Re(V) and Im⁡(V), respectively, corresponding to this source. As expected, we see that there is no resonant region near the slab even though the source is quite close to the slab (d0=1.2a), γ=0.5γ∗, and δ=10-12.Figure 12
This is the sourcef defined by (85)–(87) with the parameters d0=1.2a and d1=d0+2.Figure 13
In this figure, we plot (a)Re(V) and (b) Im⁡(V), where V is the solution to (4) corresponding to the source f defined through (85)–(87) with the parameters d0=1.2a, γ=0.5γ∗, and δ=10-12. To make the behavior of V clearer, we clipped the maximum and minimum values in each plot to 0.2 (yellow) and -0.2 (blue), respectively.
(a) (b)



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*Source: 4156072-2016-10-11.xml*