Advances in Mathematical Physics
Volume 2016 (2016), Article ID 4156072, 28 pages
http://dx.doi.org/10.1155/2016/4156072

Research Article

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

Daniel Onofrei¹ and Andrew E. Thaler²

¹Department of Mathematics, University of Houston, Houston, TX 77204, USA
²Institute for Mathematics and Its Applications, University of Minnesota, College of Science and Engineering, Minneapolis, MN 55455, USA

Received 23 May 2016; Accepted 7 August 2016

Academic Editor: Jacopo Bellazzini

Copyright © 2016 Daniel Onofrei and Andrew E. Thaler. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 () with permittivity for some small loss 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.

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 Figure 1)

where denotes the width of the slab and the sets , , and represent the regions to the left of the slab, within the slab, and to the right of the slab, respectively. We also define

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 permeability ); the electrical permittivity is given by

for some . We consider the following partial differential equation (PDE) in 2D:

where , , with compact support in , and is given in (3) (see Section A.5 for a derivation of (4) from the Maxwell equations).

For convenience, we define

We assume the solution also satisfies the following continuity conditions across the boundaries at and for almost every :

In what follows we assume that the parameters and data are such that problem (4) and (6) admit a unique solution with and for almost every .

Remark 1. Note that, in the case when , the unique solution of the problem will have the property that as for almost every ; for , this solution will be well approximated by the solution in the case .

We say anomalous localized resonance (ALR) occurs if the following two properties hold as [1]:

(1) in certain localized regions with boundaries that are not defined by discontinuities in the relative permittivity.(2) approaches a smooth limit outside these localized regions.

In [1], Milton et al. showed that if is a dipole and , then ALR occurs if , where 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 ) 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 , 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 and a fixed field is applied to the system (e.g., a uniform field at infinity), then a polarizable dipole located in the region causes anomalous localized resonance and is cloaked in the limit . 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 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 value , such that, for with , 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 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 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 Section 1.1 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 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 . In Section 4.2, we construct two examples of possible sources for which there is no ALR phenomenon regardless of the range of 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 the variable. Recall that, for a given function for some , the Fourier transform of with respect to is

We will study the Fourier domain solution in each of the relevant subdomains defined in (1).

1.1.1. The Solution in

In the region , the relevant equation is

Taking the Fourier transform of (8) with respect to , we find that satisfies

Remark 2. Here and throughout the paper, we take the principal square root of complex numbers; that is, for a complex number , where , we take

where . In particular, this implies .

Remark 2 implies

Then the general solution to (9) is

for coefficients and that are independent of .

If , then is purely imaginary. Because should be outgoing (i.e., left-going) as and we are considering time dependence (see Section A.5), we should have

From (11) and (12), we see that we can ensure this by taking .

On the other hand, if , then . Thus we take in this case to ensure that as . Finally, without loss of generality we may also take for . Therefore,

1.1.2. The Solution in

In the region , the Fourier transform of satisfies

The general solution is

the coefficients and may be found by using the continuity conditions across from (6). In particular, we find

where

Although one can observe that degenerates for we will see in (25) and (27) that is well defined in the limit when .

1.1.3. The Solution in

In the region , the Fourier transform of satisfies

If , then the general solution to (19) can be found using the Laplace transform and the continuity conditions across from (6) [23, 24]; we have

where

If , then is purely imaginary. Because should be outgoing (i.e., rightgoing) as and we are considering time dependence, we should have

To ensure this, we take the first expression in brackets in (20) to be zero and find that

where

If , then ; to ensure that as , we again take as in (23).

Finally, if , then we can use the Laplace transform and the continuity conditions across to find that

where

with defined at (15) being computed for and where again we take so that we ensure is outgoing as ; in this case

1.2. Energy Discussion

For , we define the strip

Then, due to the Plancherel theorem and properties of Fourier transforms, we have

Using (17)-(18) and (21)–(24) in this expression, switching the order of integration, computing the integral with respect to , using the fact that is an even function of if is real-valued, making the change of variables , and simplifying the resulting expression, we obtain

where

we have used the fact that (see (11) and (19)), and we have replaced by throughout the integrand (e.g., we have ).

Similarly, we have

Remark 3. One of the quantities we are most interested in studying in this paper is

Due to the similarity between the expressions in (30) and (32), without loss of generality we focus on . In particular, our arguments depend heavily on the exponential terms in the integrands in (30) and (32), so the additional terms and in (30) will have no bearing on our results.

2. Properties of

In this section, we collect some essential properties about the denominator in (30). As we will see, the parameter

plays a crucial role in the behavior of the solution and in the limit .

Lemma 4. Suppose is defined as in (31). Then for and one has

Proof. The result follows from direct calculations since is a continuous function of .

The next lemma plays an essential role in the following discussion.

Lemma 5. Suppose is defined as in (35) for and . Then there is such that

(1)if , then has two distinct real roots of order , namely, ;(2)if , then has no real roots.

We note that can be computed as the solution of an optimization problem; more importantly, we emphasize that Lemmas 4 and 5 are independent of the source term in (4). We will see later that the roots of are indicative of anomalous localized resonance. For brevity, we defer the proof of Lemma 5 to the Appendix.

3. Short Wavelength/High Frequency Regime ()

In this section, we prove that, for (where was introduced at (34)), remains bounded as for all sources with bounded support in , 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, , is vanishingly small in the limit .

3.1. for

From (30), we have

where, for , , and ,

We now state the main theorem from this section.

Theorem 6. Suppose (where is introduced in Lemma 5). If there is a constant such that

then there is a constant and such that as for all .

The proof of this theorem is somewhat tedious and may be found in the Appendix—although we only prove the theorem for , Remark 3 implies that it holds for as well. In the next lemma, we show that bound (39) holds for very general sources .

Lemma 7. Suppose with compact support; then (39) holds.

Proof. For , recall from (24) that

Then the triangle, Cauchy-Schwarz, and Jensen inequalities imply that

Similarly, for , recall from (24) that

Then

To complete the proof, we define .

3.2. Shielding Effect for Large

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 , will give a purely imaginary wavenumber inside the slab and thus no propagation beyond the slab in region . We have the following.

Theorem 8. Suppose , satisfies (39), and choose ; then there is a constant such that

In particular,

Remark 9. Lemma 7 implies that Theorems 6 and 8 hold for all sources with compact support. However, the bound in (39) is stronger than we need. For example, suppose there is a positive, real-valued function that is continuous for and . In addition, for every , suppose that

For example, if is a continuous function of and that is of polynomial order for and , it will satisfy (46). Finally, suppose

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

where .
In particular, certain distributional sources such as dipoles and quadrupoles satisfy (47)—see Section A.2 for more details.

In Figure 2, we plot the solution to (4) in the case where is a dipole with dipole moment , , and (we take in all figures throughout the paper). In Figures 2(a) and 2(b), the dipole is located at the point ; in Figures 2(c) and 2(d), the dipole is located closer to the slab at the point . The solution is smooth throughout the domain; in addition, we observe the “shielding effect” from Theorem 8 in the region to the left of the lens.

(a)

(b)

(c)

(d)

Figure 2: This is a plot of , the solution to (4), when is a dipole and : (a) and (b) for ; (c) and (d) for . To make the behavior of clearer, we clipped the maximum and minimum values in each plot to (yellow) and (blue), respectively.

In Figure 3, we plot as a function of various parameters for a dipole source . The parameters we used are in the ranges , , and . We note that depends strongly on , , and , but, because , is quite small.

(a)

(b)

(c)

(d)

(e)

(f)

Figure 3: These are plots of as a function of (a) and (); (b) and (); (c) and (); (d) and (); (e) and (); (f) and ().

Figure 4 is similar to Figure 2, except in Figure 4 we take

Although this source has compact support, in contrast to the dipolar source considered above it is in 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.

(a)

(b)

(c)

(d)

Figure 4: This is a plot of , the solution to (4), when is the function in (49) and : (a) and (b) for ; (c) and (d) for . To make the behavior of clearer, we clipped the maximum and minimum values in each plot to (yellow) and (blue), respectively.

To construct the plots, we have taken , and . The solution is smooth throughout the domain and very small in the region to the left of the slab.

4. Long-Wavelength/Low Frequency Regime ()

Unfortunately, the complicated nature of expression (30) has thus far prevented us from deriving lower bounds on that would allow us to prove that as . 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 of

The key result of this section is Lemma 5: has two real roots when ; namely, . Because both roots are larger than , the main contribution to the blow-up of comes from the integral over the interval . Indeed, the following lemma shows that we do not need to worry about the integral over the interval .

Lemma 10. Suppose and with compact support. Then there is a positive constant and such that

for all .

Remark 11. We emphasize that Lemma 10 also holds for those sources for which the bound in (47) holds (e.g., dipole sources)—see Remark 9.

Proof. First, we note that is continuous for , , and , so it is bounded by a constant independent of , , and . Additionally, is also bounded by a constant, thanks to Lemma 7. All that remains for us to show is that is bounded away from .
We define the function

Because and are both continuous for , the above maximum is attained, say at . This means that

Now let be a sequence converging to as . Because is a bounded sequence, it has a convergent subsequence . Along this subsequence,

by Lemma 4. In other words, every sequence has a subsequence that converges to , which implies that every sequence converges to . Because the original sequence was arbitrary, this implies that

In combination with (51), this implies that converges to uniformly in for . Thus for every there is such that

for all and all . If we take

then

for all (the last two inequalities hold because the roots of are larger than by Lemma 5). Combining this result with the first paragraph of the proof gives us the bound

for some constant .

The preceding lemma proves that we only need to study the integral in (30) over the interval . Because as , it should be the case that near the roots of . Inspired by our earlier work in the quasistatic regime, we conjecture that and are on the order of as .

Conjecture 12. Suppose , and let be the roots of . Then there is such that for all and all ; however, and as .

One way to prove this conjecture would be to expand (for , ) in Taylor series around and then prove that is uniformly bounded for and small enough. Unfortunately, these derivatives are quite complicated; moreover, numerical experiments indicate that they become unbounded as , 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

as functions of and over the ranges and (we believe the functions in (59) remain bounded as for all ; however, as , so the numerical computation of the roots becomes more difficult as gets closer to . Similarly, , so as gets close to it becomes difficult to distinguish the roots). For each , we see that the functions in (59) remain bounded as gets close to , which seems to indicate that and as . Curiously, both functions in (59) seem to depend very weakly on .

(a)

(b)

Figure 5: In this figure, we plot (a) and (b) over the range and .

Next, we conjecture that the behavior of near and is not canceled by the term in the numerator.

Conjecture 13. Suppose , and define as in (38). Then there exist positive constants and such that near and for all .

If Conjectures 12 and 13 are true, then (36)-(37) imply that the part of the integrand that is independent of the source , namely,

is on the order of near and as . If is also bounded away from near and , the entire integrand will have values on the order of near and .

To provide some justification for Conjecture 13, in Figures 6(a) and 6(b) we plot

as functions of and over the same intervals as in Figure 5. In particular, we note that and are both bounded away from and seem to depend quite weakly on .

(a)

(b)

Figure 6: In this figure, we plot (a) and (b) over the range and .

Finally, to obtain a blow-up in , it should be the case that does not conquer the small values of near and . Heuristically, there will be no blow-up if near and . In the next section, we present numerical evidence that suggests that sources with do not lead to ALR.

On the other hand, recall from (24) that

Again we take our inspiration from the quasistatic case [23, 24]. If , then the exponential in the above integrand will be extremely small (especially because and are both greater than ). In particular, the exponential may be small enough so that it cancels out the effect of the denominator near and . We emphasize that this is not rigorous, but we hope that it may provide a starting point for future investigations.

Conjecture 14. Suppose . Then there exist sources with compact support or distributional sources such as dipoles such that, for any , if is “close enough” to and for some positive constant if is “far enough away” from . This critical distance may depend on .
Moreover, there are positive constants , , and such that, for all ,

for all with .

Remark 15. If it is only the case that

then we say that weak ALR occurs. Because is difficult to deal with analytically, we cannot say much more on this. It is difficult to determine whether

using only numerical techniques. In particular, if the limit supremum of is , there is at least one sequence along which ; however, it may be the case that for all sequences except a few very special sequences that would be extremely difficult to find via numerical experiments alone.

Figures 7 and 8 are exactly the same as Figures 2 and 4 except 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 from (49) in Figure 8) are located at , and the solution 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 and . Figures 7(c), 7(d), 8(c), and 8(d) contain plots of when . From these figures we see that the extreme oscillations of 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, 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 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 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 (see Theorem 6), we do not expect to see the superlensing effect in this regime (see Theorem 8).

(a)

(b)

(c)

(d)

Figure 7: This is a plot of , the solution to (4), when is a dipole and : (a) and (b) for ; (c) and (d) for . To make the behavior of clearer, we clipped the maximum and minimum values in each plot to (yellow) and (blue), respectively. The vertical red line in (c) extends a distance of , where is defined in (66).

(a)

(b)

(c)

(d)

Figure 8: This is a plot of , the solution to (4), when is the function in (49) and : (a) and (b) for ; (c) and (d) for . To make the behavior of clearer, in (a) and (c) we clipped the maximum and minimum values in each plot to (yellow) and (blue), respectively. The vertical red line in (c) extends a distance of , where is defined in (66).

Figures 7(c) and 8(c) provide an additional insight into Conjecture 14. In general, for (where is the larger root of ), the coefficient from (23) becomes very large since its denominator is proportional to and for small enough. Recalling that the Fourier transform variable represents a wavenumber in the -direction with corresponding wavelength , this implies that the solution should exhibit prominent oscillations with wavelength on the order of

In Figures 7(c) and 8(c), we have drawn a vertical red line of length . This red line covers approximately wavelengths of oscillation in the resonant region, which seems to indicate that at least one of the zeros of , namely, , is responsible for ALR. Because is independent of , the above argument also suggests that the wavelength of the resonant oscillations of is also independent of the source . We emphasize that this is speculative at best, but it would be interesting to investigate further.

To illustrate how drastically different the behavior of is for and , in Figure 9 we plotted corresponding to a dipole source located at for two different values of . In Figures 9(a) and 9(b), we took while in Figures 9(c) and 9(d) we took . The ALR is present when in Figure 9(c); on the other hand, in Figure 9(a) there are a few oscillations near the -axis, but they quickly die out as grows.

(a)

(b)

(c)

(d)

Figure 9: This is a plot of , the solution to (4), when is a dipole located at : (a) and (b) for ; (c) and (d) for . To make the behavior of clearer, in (a), (c), and (d) we clipped the maximum and minimum values in each plot to (yellow) and (blue), respectively.

Unfortunately, we cannot provide a figure analogous to Figure 3 for when is a dipole source—MATLAB is unable to accurately compute the integral

because is very close to near the roots of for small values of (see Conjecture 12). However, to get a sense of what is going on, we plotted on a logarithmic scale for a dipole source with in Figures 10(a) and 10(b) and in Figures 10(c) and 10(d). Each curve is as a function of for various values of . In Figures 10(a) and 10(b), where , we see that is quite large near the poles of , even if . Additionally, on comparing the -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), and we see that remains bounded regardless of (in Figure 10, all of the functions rapidly tend to for larger values of (not shown in the figure)).

(a)

(b)

(c)

(d)

Figure 10: A plot of the integrand from (36) for several parameter values. The separate curves in each plot represent different values of , indicated in the legend: (a) and ; (b) and ; (c) and ; (d) and .

4.2. Sources for Which ALR Does Not Occur

When , the conjectures from the previous section suggest that the zeros of are responsible for forcing to blow up in the limit as . This begs the question of whether one can design a (realistic) source in the finite-frequency regime (with ) that effectively cancels the poles that show up in the limit . In other words, we would like to design a source such that exactly at the zeros of ; heuristically, in the limit as , this would force the integrand in (30) to remain bounded at the zeros of and annihilate the anomalous localized resonance that occurs in this limit. Recall from (24) that

Lemma 5 implies that has two roots . Using this and (68), we see that an “ALR-busting” source can be constructed by choosing such that for all (which implies ). We do not want to just choose any satisfying this property; however, we restrict ourselves to those sources with compact support. In summary, we make the following conjecture.

Conjecture 16. Suppose has compact support and

where are the zeros of from Lemma 5 and are zeros of order at least for . Then there is and a constant such that for all .

There are many sources that satisfy the hypotheses of this theorem. We will present examples here. First, consider

where ,

Then and, hence, by the Plancherel theorem; moreover, , where the zeros are order . Finally, by direct calculations we have

where is the Heaviside step function; this has compact support and thus satisfies the hypotheses of Conjecture 16. We may also take

where and are the Bessel functions of the first kind of orders and , respectively, and and are such that (we note that these zeros are also of order ). Because the Bessel functions of the first kind are as [25], we have . By the convolution theorem for Fourier transforms,

where denotes convolution and and are the inverse Fourier transforms of and , respectively; in particular, we obtain

Although the convolution in (74) is difficult to compute analytically, since and both have compact support, the convolution of with will as well. Thus as defined in (74) is in and has compact support.

In Figures 11(a) and 11(b) we plot and , respectively, for the source from (70) (equivalently, (72)); in Figures 11(c) and 11(d), we plot and , 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, and . In stark contrast with those figures, the solution is well-behaved in Figure 11.

(a)

(b)

(c)

(d)

Figure 11: This is a plot of , the solution to (4), for two different sources with and : (a) and (b) for as in (70) (equivalently, (72)); (c) and (d) for as in (73) (equivalently (74)).

4.2.1. Current Sources for Which ALR Does Not Occur

In the monochromatic electromagnetic setting, must satisfy additional restrictions for it to represent a realistic (divergence-free) current source—see Section A.5. In particular, the function should be in with compact support and be of the form

for a current

satisfying the continuity equation

We now construct a source 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

with , , , smooth enough. Then continuity equation (78) becomes

Taking the Fourier transform of this equation with respect to gives

We further simplify the problem by taking

Then (81) becomes

which is satisfied for all and if

Then (76), (79), (82), and (84) imply that

At this point, satisfies (76)–(78).

By the Plancherel theorem and (85), if and only if and . We must also be careful to also choose in such a way that and has compact support in .

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

Then , , and (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 and are both continuous; there may be other choices of continuous functions and such that has zeros of order at and ).

One possible choice of from (85) is

where is a nonzero constant. The function is twice continuously differentiable and has compact support, so is continuous with compact support.

Finally, may be computed via (85), (86), and (87). We note that the inverse Fourier transform of can be computed analytically; for the benefit of the reader, we avoid writing out the expression. Importantly, is continuous with compact support.

The current source corresponding to this may be computed via (77), (79), (82), (84), (86), and (87). We emphasize that both and are continuously differentiable functions with compact support in .

In Figure 12, we plot the source defined by (85)–(87) with . In Figures 13(a) and 13(b), we plot and , 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 (), , and .

Figure 12: This is the source defined by (85)–(87) with the parameters and .

(a)

(b)

Figure 13: In this figure, we plot (a) and (b) , where is the solution to (4) corresponding to the source defined through (85)–(87) with the parameters , , and . To make the behavior of clearer, we clipped the maximum and minimum values in each plot to (yellow) and (blue), respectively.

Appendix

A. Proofs and Derivations Omitted in the Text

In this appendix, we provide detailed proofs we omitted in the main body of the paper.

A.1. Proof of Lemma 5

Setting , defining a new variable , and simplifying, we find that is equivalent to having

We define

Then if and only if . We will complete the proof of the lemma in several steps.

(1) Claim. for . Proof of Claim. For ,

 Since for , the only point at which could possibly be is . But, for ,

 In particular, this proves that if is a root of , then .(2) Claim. For , the function is concave for . Proof of Claim. For , we have

 We note that for and . This proves the claim for . Now consider

 We note that for , so for . Then, for , we have

 For , the left-hand side of the above inequality satisfies

 Using calculus and Maple, it can be shown that

 In combination with (A.7) and (A.8), this implies

 for as long as and verifies the claim.(3) Claim. For , the function has two real zeros . Proof of Claim. We begin by defining the functions

 for and . Then

 For we have

 Similarly, for we have

 Then for and for . Next, we note that

 In addition, we have

 Because is decreasing for and for , the above inequality implies

 In fact, increases enough on this interval to become positive; we have

 For increases without bound while will get arbitrarily close to ; the upshot of this is that becomes arbitrarily negative for large enough. In summary, we have ; the function increases at least until , where . Next, will continue increasing until becomes larger than ; then will decrease toward as approaches . Thus has real zeros for ; by item (1) above, both of these zeros must be larger than . Finally, since is strictly greater than , by continuity the roots cannot be equal. This also proves that both of the zeros are of order . This proves the claim.

We have shown that has two real roots provided . For , the function is concave for by item (2) above. Thus has a unique maximum and will have two real roots of order if the maximum is positive and no real roots if the maximum is negative.

Because the maximum of is positive for (see the proof of item (3) above) and decreases at an exponential rate as a function of , there exists such that

In particular, we have

MATLAB gives . This completes the proof.

A.2. for Dipole Sources

In this section, we derive an explicit formula for , defined in (24), when the source is a dipole. In particular, we consider a source of the form

here is the dipole moment. Then (7) gives

Next, (24) implies

If we take and (the typical case), this becomes

which, after the changes of variables and , becomes

From here the triangle inequality implies that satisfies bound (47) with

Finally, the same computations as those leading up to (A.25) imply that quadrupole, octopole, and higher-order distributional sources satisfy equations similar to (A.25) with higher-order powers of and . Therefore, as discussed in Remark 9, such sources satisfy Theorems 6 and 8.

A.3. Proof of Theorem 6

We will prove this theorem in several steps. Essentially, our goal is to bound the integrand from above by a function that is integrable and independent of . We begin by finding a lower bound on ; in particular, we note that (31) and the reverse triangle inequality imply that

In the next lemma, we provide a lower bound on the first term in (A.27).

Lemma A.1. Suppose and ; then

Proof. We have

If , then

If , then

Squaring both sides of (A.30) and (A.31) and utilizing (A.29) give us the desired result.

In the next lemma, we provide upper bounds on the second term in (A.27).

Lemma A.2. Suppose ; then there is a constant such that

Proof. The triangle inequality, the assumptions that and , and the bound

imply that

For and , the definitions of and from (15) and (19), respectively, imply that the above bound becomes

Therefore, if we take , we have

Now we consider the case . The triangle inequality, the bound in (A.33), and the definitions of and for (see (15) and (19), resp.), imply that

Applying the bound for and to the right-hand side of the above inequality implies

Squaring the term in brackets on right-hand side of the above expression gives

where

For and , we have the bound

Using this bound in (A.40) gives

Combining (A.38), (A.39), and (A.42) gives us the second bound in (A.32).

We are now ready to give lower bounds on .

Lemma A.3. Suppose ; then there exist satisfying and a positive constant such that, for ,

Proof. If and , then (A.27) and Lemmas A.1 and A.2 imply that there is a constant such that

This gives us first part of (A.43).
We now assume . First, note that for all because it has no zeros by Lemma 5 and . Then (35), (A.27), and Lemmas A.1 and A.2 imply that

for all . Therefore,

for all . From (A.47), for all , we have

Thanks to the exponential decay in the second term on the right-hand side of (A.48), there exists such that the expression on the right-hand side of (A.48) is strictly positive for all . Therefore,

Finally, we consider . Because both terms on the right-hand side of (A.46) are continuous, we may define

because by Lemma 5. Hence (A.46) becomes

which is nonnegative if we take

Therefore

We define ; then (A.49) and (A.53) imply that

This completes the proof.

Recalling that our ultimate goal is to prove Theorem 6, in the next lemma we derive upper bounds on , defined in (38).

Lemma A.4. Suppose and , where is defined in Lemma A.3. Then there exists a constant such that

for all and all .

Proof. By the triangle inequality, we may derive upper bounds on each term of individually.

(1)For the term outside of the braces in (38), we have

 Therefore, for all and all , we have

(2)For the first term in brackets in (38), we have, thanks to (A.33),

 Because the above function is continuous for and tends to as , it attains its maximum value on . Thus there is a constant such that

(3)Using our result from item (2) and (18), we find that the second term in brackets in (38) satisfies

 where is the constant from (A.59). Applying the bounds from (A.28) and (A.32) as well as the bounds and to the above expression gives

 The function on the right-hand side of the above inequality is continuous as a function of for and decays to as . Thus it attains its maximum value on (this maximum value is independent of ); this and (A.61) imply that there is a constant such that

 for all and all .(4)For the last term in (38), we have

 Arguments similar to those in item can be used to show that there is a positive constant such that

 Because

 the function

 is continuous for and (after modification at ). Moreover, this function goes to as , so it attains its maximum value. This implies that there is a constant such that

 Inserting (A.64) and (A.67) into (A.63) implies that there is a constant such that

 for all and all .

Using the bound

which holds for all and all , as well as the bounds (A.57), (A.59), (A.62), and (A.68) in (38) gives

for some positive constant .

We are finally ready to complete the proof of Theorem 6. First, we split the integral in (36) to obtain

We will focus on each integral separately. For , we use the bounds from Lemma A.3, (A.55), and (39) in (37) to obtain the bound

which holds for all and . The function on the right-hand side of the above inequality is continuous for , so it attains its maximum value. Thus there is a positive constant such that

Similarly, for , the bounds from Lemmas A.3, (A.55), and (39) imply that there is a positive constant such that

this bound holds for all and all . Note from (35) that as at an algebraic (i.e., nonexponential) rate. Because , the exponential term in the numerator in (A.74) defeats the nonexponential terms in the braces and in ; in other words, there is a positive constant such that

Using (A.73) and (A.75) in (36) gives

This completes the proof of the theorem.

A.4. Proof of Theorem 8

We begin by taking care of an important technicality. Recall from Section A.3 that was chosen so that the expression on the right-hand side of (A.48) is strictly positive for . Because of this choice, depends on in a nontrivial way—see (A.53). However, there is such that if , the right-hand side of (A.48) is positive for all (thanks to the exponential decay in of the last term in (A.48)). Thus, for , (A.48) immediately implies that there is , independent of , such that and

for all , all , and all . To visualize this, in Figure 14 we plot the expression on the right-hand side of (A.48) as a function of for different values of . Figure 14(a) is a plot over the interval while Figure 14(b) is a plot over the interval ; in addition, the blue dashed curve is for , the red dotted curve is for , and the yellow solid curve is for (the and curves overlap in Figure 14(b). In Figure 14(a), we note that the expression is negative for some values of if ; however, if or , the curve is always positive. Therefore, for the remainder of this proof, we will assume and .

(a)

(b)

Figure 14: This figure contains a plot of the expression on the right-hand side of (A.48). In both plots, the blue dashed curve corresponds to , the red dotted curve corresponds to , and the yellow solid curve corresponds to . In particular, we have plotted the expression in (A.48) over the intervals (a) and (b) .

The Fourier inversion theorem, the triangle inequality, and (14) imply that

Using (21)–(23) and (31) and making the change of variables in the above integral gives

Case 1 (). In this case, the integral in (A.79) (restricted to ) is

For and , (A.33) and Lemmas A.3 and 7 imply that there is a constant such that the above integral is less than or equal to

Because the integrand in (A.81) is continuous, it is bounded above by a constant (independent of , , and ). Thus (A.81) implies that there is such that

Case 2 (). In this case, the integral in (A.79) (restricted to ) is

For and , (A.33) and Lemmas A.3 and 7 imply that there is a constant such that the above integral is less than or equal to

From (35), the denominator of the above integrand is an increasing function of . Together with the fact that all of the exponential terms in the numerator attain their maximum values at , this implies, for , that the above integral is bounded above by

Because has no roots (by Lemma 5) and tends to as only algebraically, the above integral converges. This implies that

for some . Using (A.82) and (A.86) in (A.79) gives

this bound holds for all , , all , and all . Thus goes to exponentially as . This completes the proof of the theorem.

A.5. Derivation of Helmholtz Equation from Maxwell Equations

When no charge source is present, the Maxwell equations are

In linear media the relevant fields satisfy the constitutive relations

By taking the divergence of the Ampére law with the Maxwell correction (the fourth equation) and utilizing the Gauss law (the first equation), we find that the current must satisfy the continuity equation, namely,

We then take the curl of the Ampére law with the Maxwell correction and apply a vector identity to obtain

In an isotropic and homogeneous medium (where and are constant scalars), the above equation becomes

Utilizing the Faraday law in combination with the fact that is divergence-free, we obtain

Finally, we assume that all fields have harmonic time-dependence of the form ; in particular, we assume that and . We also define

Then, thanks to (A.93), we find that satisfies

where and . Thus each component of satisfies a 3D Helmholtz equation.

A.5.1. 2D Helmholtz Equation

We now assume that the current source is a line current of the form

where, because must satisfy (A.90),

By symmetry, none of the fields will depend on . In addition, we assume that we are dealing with nonmagnetic materials for which . Thus (A.95) and (A.96) imply that the -component of satisfies

Finally, the Maxwell equations can be used to show that and must be continuous across the boundaries of the slab at and [26]. Then (A.98) can be written in divergence form as

where

Competing Interests

The authors declare that there are no conflicts of interests regarding the publication of this article.

Acknowledgments

The research of Andrew E. Thaler was supported in part by the Institute for Mathematics and Its Applications with funds provided by the National Science Foundation through NSF Award 0931945 and in part by AFOSR under the 2013 YIP Award FA9550-13-1-0078.

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