SCATTERING OF ELECTROMAGNETIC WAVES BY THIN HIGH CONTRAST DIELECTRICS: EFFECTS OF THE OBJECT BOUNDARY

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1 COMMUN. MATH. SCI. Vol. 11, No. 1, pp c 2013 International Press SCATTERING OF ELECTROMAGNETIC WAVES BY THIN HIGH CONTRAST DIELECTRICS: EFFECTS OF THE OBJECT BOUNDARY DAVID M. AMBROSE AND SHARI MOSKOW Abstract. We study te scattered field from a tin ig contrast dielectric volume of finite extent. Te waves are modeled by te full tree dimensional time-armonic Maxwell equations wile accounting for material boundaries. We derive a formulation of Lippmann-Scwinger type for a dielectric scatterer; tis formulation as an additional surface term to account for te material discontinuities. Te layer potential operator resulting from tis surface term is sown to converge in a weak sense to an explicitly computable limit as te tickness of te domain approaces zero. By properly accounting for te boundary effects, we sow two results about te tin ig contrast limit: First, te normal component of te electric field s interior trace on te lateral boundary approaces zero. Second, te tird component of te electric field (wic corresponds to te direction perpendicular to te slab) goes to zero inside te slab. We propose a new two-dimensional limiting equation as a first-order computational tecnique. Key words. Scattering, Maxwell s equation, asymptotics, tin dielectric. AMS subject classifications. 78A45, 45E99, 34E Introduction In electromagnetic scattering teory, volume integral formulations for Maxwell s equations are frequently very elpful for studying certain asymptotic regimes or inverse problems. However, tese formulations can lead to difficulties dealing wit material jump discontinuities, and a typical approac is to smoot out suc discontinuities. Here we are motivated by te study of tin potonic band gap materials, in wic te magnitude of suc a jump may not lend itself to regularization. Furtermore, even wen regularization is suitable, it can be quite difficult to take te limit as te regularization vanises. Here we propose instead to start wit a formulation for te full Maxwell s equations wic takes te jumps into account. We ten perform an asymptotic analysis wic does not require regularization. In te paper [9], Santosa, Zang, and te second autor proposed an approximate metod to compute scattered fields from tin ig contrast dielectric structures, using te Helmoltz equation. In tis model, te squared index of refraction is assumed to be on te order of te reciprocal of te tickness of te structure, represented by a small parameter. Te metod starts wit te time-armonic wave equations, and applies a perturbation approac based on expansions wit respect to tis small. (We remind te reader tat if te index of refraction were bounded te object would disappear in te limit.) Tis asymptotic approac was extended to Maxwell s equations in te presence of smootly varying dielectrics in [1]. Te advantage of tese metods is tat tey reduce te complexity of te computation by one dimension (i.e., a treedimensional volume integral equation reduces to a set of two-dimensional integral equations). Tis leads to a igly efficient computational metod for obtaining te scattered field anywere in R 3 : one can first inexpensively solve for te field using te Received: February 14, 2012; accepted (in revised form): May 9, Communicated by Ricard Tsai. Department of Matematics, Drexel University, Piladelpia, PA 19104, USA (ambrose@ mat.drexel.edu). Department of Matematics, Drexel University, Piladelpia, PA 19104, USA (moskow@ mat.drexel.edu). 293

2 294 SCATTERING BY THIN DIELECTRICS dimensionally reduced system inside te scatterer, and ten use tis approximation as input into a Lippmann-Scwinger form. In te current paper, we study te same scattering problem, using te full tree-dimensional Maxwell time-armonic system, allowing for jumps in te material wit respect to a constant background. We find tat surface terms arising from material jumps do indeed impact te resulting system and approximations. In our first teorem, we derive a generalized Lippmann-Scwinger equation for a Maxwell scatterer in te presence of jump discontinuities. Te proof uses te Maxwell scattering formulations from [7] and [8], originally derived for factorization metods in inverse problems. In te absence of discontinuities, tis formulation developed in te current work reduces to te Lippmann-Scwinger form studied in [1]. Te generalized version as a surface integral on te jumps, unlike te original equation wic as only volume integrals. Te discontinuities tat tis formulation allows for can be from te external boundary of te scattering object and te background, or could just as well be from internal discontinuities suc as a periodic array of air oles. Te new surface term corresponds to an integral operator tat depends on our small parameter. Correct asymptotic analysis will require tat we understand te limiting beavior of tis operator as te tickness of te domain vanises. We compute te limit of tis operator viewed on a fixed, scaled domain. Te operators are of layer potential type, but converge to one tat is instead algebraic in nature. Tis is wat we sow in our second teorem. We restrict te electric field to its interior limit up to te object boundary using te integral formulation. In te ig contrast case, we again take te limit as te tickness is going to zero, now wit te dielectric constant simultaneously going to infinity. Under certain uniform regularity assumptions on te field, we sow tat te normal component of te field on te boundary of te object goes to zero. Tat is, in te limiting sense, te wave is trapped in te object. Next, we sow tat te tird component of te electric field, wic corresponds to te direction perpendicular to te scatterer, goes to zero inside of te scatterer. Tese two results about tis ig contrast media model are new, since tis beavior was not observed in te models in [9] or [1]. Oter previous work on tin dielectrics can be found in [2] and [10]. Tis paper is organized as follows. In Section 2, we give te precise setup for te tin ig contrast dielectric scattering problem. We ten derive te Lippmann- Scwinger form wic we will use to analyze te problem, and prove tat any regular enoug solution to te differential equation satisfies it. Section 3 contains a study of te boundary integral operator wic arises from te surface term on te jumps. Te teorems about te trapping of te wave and te convergence of te perpendicular component are in Section 4. Finally, in Section 5 we discuss te implications of tis work for asymptotic expansions and te related computational tecnique. 2. Problem formulation 2.1. Derivation. Let E R 3 denote te electric field, wic is governed by Maxwell s equations wit a variable index of refraction and wit te magnetic permeability, µ, given by µ=1. After eliminating te magnetic field tis yields te following equation for E: E k 2 n 2 (x)e=0, (2.1) were k>0 is te normalized frequency and n 2 (x) is te squared index of refraction. We assume tat we ave a cylindrical structure of small tickness on wic te

3 D.M. AMBROSE AND S. MOSKOW 295 index of refraction is large. We incorporate tis tin ig contrast structure into te definition of n 2 (x): 1 for x 3 >/2, n 2 (x)= ε 0 (x 1,x 2 )/ for x 3 </2, (x 1,x 2 ), (2.2) 1 for x 3 </2, (x 1,x 2 ), were is a bounded domain in R 2 wit smoot boundary. We denote te region of te dielectric object by = ( /2,/2). We assume tat ε 0 is smoot inside of, but ere n 2 (x) as a jump discontinuity across. Note tat nowere do we assume is convex, and terefore te structure may contain oles wose boundaries are part of. Te total field, E, can be written as E=E i +E s, were E i is a given incident wave satisfying te background equation E i k 2 E i =0, (2.3) and E s is te scattered field wic satisfies te Sommerfeld radiation condition. Te 3 3 tensor G(x,y)=φ(x,y)I+ 1 k 2 ydiv y (φ(x,y)i) (2.4) is te free space fundamental solution for te operator k 2 on R 3, or te so-called dyadic Green s function. Here φ is te free space Helmoltz fundamental solution: φ(x,y)= 1 e ik x y 4π x y. (2.5) Wit tis dyadic Green s function we ave te following Lippmann-Scwinger form for te equation for E: were E(x)=E i (x)+k 2 G(x,y)q(y)E(y)dy, (2.6) q(y)=n 2 (y) 1= ε 0(y 1,y 2 ) 1. Note tat tis form is a 3 3 system of integral equations for te tree components of E. Plugging in te formula for G, we find E(x)=E i (x)+k 2 φ(x,y)q(y)e(y)dy+ y div y (φ(x,y)i)q(y)e(y)dy. (2.7) Let us assume, just for te time being, tat te above form makes perfect sense despite te singular integral in te last term. Using te equation (2.1) for E, we can write E as being proportional to a curl: E= E k 2 (1+q(y)). (2.8)

4 296 SCATTERING BY THIN DIELECTRICS Using (2.8) and integrating by parts yields y div y (φ(x,y)i)q(y)e(y)dy = 1 q(y) k 2 y div y (φ(x,y)i) 1+q(y) E(y)dy = 1 ( ) q(y) k 2 div y (φ(x,y)i) E(y)dy 1+q(y) + 1 q(y) k 2 div y (φ(x,y)i) 1+q(y) ( E(y) ν) dσ y ( ) q(y) = div y (φ(x,y)i) (1+q(y))E(y)dy 1+q(y) + div y (φ(x,y)i)q(y)(e(y) ν) dσ y, were we ave used te fact tat a curl of a gradient is zero. Here ν is te outward unit normal to. A simple calculation sows tat ( ) q (1+q) = q 1+q (1+q). (2.9) Applying tis in (2.7) yields E(x)=E i (x)+k 2 φ(x,y)q(y)e(y)dy Since we ave div y (φ(x,y)i) q(y) 1+q(y) E(y)dy + div y (φ(x,y)i)q(y)(e(y) ν) dσ y. (2.10) div y (φ(x,y)i)= y φ(x,y)= x φ(x,y), E(x)=E i (x)+k 2 φ(x,y)q(y)e(y)dy + x φ(x,y) q(y) 1+q(y) E(y)dy+ y φ(x,y)q(y)(e(y) ν) dσ y. (2.11) Note tat witout te last boundary term, tis is precisely te formulation used in [1] for te case of smoot n 2 (x), wic is derived from te Stratton-Cu formulas [3]. Te last term appears ere because tere is a jump between te tin dielectric object and te omogeneous background. Replacing q we ave ( E(x)=E i (x)+k 2 ε0 (y) φ(x,y) )E(y)dy+ 1 x φ(x,y) ε 0(y) ε 0 (y) E(y)dy ( ) ε0 (y) + y φ(x,y) 1 (E(y) ν) dσ y. (2.12)

5 D.M. AMBROSE AND S. MOSKOW 297 Of te tree integral kernel terms in te rigt and side of te formulation for E above, te first is te same tat occurs for te special case of te Helmoltz equation, and its asymptotics wit respect to were first analyzed in [9]. Te second, as was previously mentioned, occurs in te formulation for Maxwell s equations for smoot n R 3, and was first analyzed in [1]. Here we now take into account te jump between te scatterer and te background, and tis leads to te final boundary integral term above. It is te analysis of tis term and te resulting limiting equations for E as 0 tat are te primary subject of te remaining sections of tis paper Justification of integral formulation. Tanks to te statement of Teorem 9 in [8] (wic was proved in [6]), we ave tat if E H loc (curl,r 3 ) is a solution of (2.1) wit te radiation conditions, ten its restriction to satisfies E(x)=E i (x)+k 2 φ(x,y)q(y)e(y)dy+ x div x φ(x,y)q(y)e(y)dy. (2.13) In fact, we also ave from [6] tat tis integral equation as a unique solution in H(curl, ), and tis solution can be extended by te rigt and side to a radiating solution of (2.1) in all of R 3. Now, let us assume for simplicity tat tis solution E is also smoot inside and up to te boundary from te interior (but not across ). Consider div x φ(x,y)q(y)e(y) dy= x φ(x,y) E(y)q(y) dy, (2.14) were we ave intercanged integral and derivative since te kernel is only weakly singular. Since we integrate by parts to get x φ(x,y)= y φ(x,y), div x φ(x,y)q(y)e(y)dy = y φ(x,y) E(y)q(y)dy = φ(x,y)div(e(y)q(y))dy φ(x,y)q(y)(e ν) dσ y, (2.15) were (E ν) represents te limit on te boundary from te interior of. Using (2.8) and (2.9) and te fact te divergence of a curl is zero, and ence x div x φ(x,y)q(y)e(y)dy = x div(e(y)q(y))= q(y) 1+q(y) E(y), φ(x,y) q(y) 1+q(y) E(y)dy x φ(x,y)q(y)(e ν) dσ y

6 298 SCATTERING BY THIN DIELECTRICS = x φ(x,y) q(y) 1+q(y) E(y)dy+ y φ(x,y)q(y)(e ν) dσ y, were we ave again intercanged derivative and integral and replaced te x derivative wittenegativeoftederivativewitrespecttoy. Notetatforanyxinteinterior of te boundary integral is not singular and te intercange is completely justified. We ave sown te following: Teorem 2.1. Let E H loc (curl,r 3 ) be te unique radiating solution to (2.1). Assume tat we additionally ave E C 0 ( ) H 1 ( ). Ten for interior x, E(x) satisfies te integral equation E(x)=E i (x)+k 2 φ(x,y) ( n 2 (y) 1 ) E(y)dy + x φ(x,y) n2 (y) n 2 (y) E(y)dy+ y φ(x,y) ( n 2 (y) 1 ) (E(y) νy ) dσ y. (2.16) Additionally, let ν be te outward unit normal to, and (E(x) ν) be te limit on te boundary from te interior of te normal component of E. Ten, for x, (E(x) ν) satisfies (E(x) ν) =E i (x) ν+k 2 ν φ(x,y) ( n 2 (y) 1 ) E(y)dy +ν x φ(x,y) n2 (y) n 2 (y) E(y)dy+1 2 (n2 (x) 1) (E(x) ν) νx φ(x,y) ( n 2 (y) 1 ) (E(y) νy ) dσ y. (2.17) Te statement (2.17) follows from te fact tat te normal component of te last term in (2.16) is te negative of a normal derivative of a single layer potential, and ence as a jump across te boundary of ; see, for example, Teorem 3.28 of [4]. We also note tat te representation (2.16) extends Teorem 9.1 of [3] to te case were n 2 (x) is piecewise smoot; te last term accounts for te jump in n 2 across. Remark 2.1. Note tat in te above teorem we need to assume tat E C 0 ( ). We expect tat tis will be te case for any fixed >0, altoug we are not aware of a specific result to tis effect. In future work te autors intend to make a detailed study of te regularity of solutions of Maxwell s equations for tis geometry. Tis future study may also explore te extent to wic we are able to relax assumptions on te domain, suc as te smootness of te boundary. We also remark tat te specific cylindrical geometry is not necessary for te above teorem; te teorem olds as long as E satisfies te required regularity assumption inside te scatterer. 3. Te boundary layer potential operator and its limit Crucial to te above formulation (2.16) for Maxwell s equations is te last term on te rigt and side, wic involves E ν on te boundary of te scatterer. In order to correctly understand tis term, we need te boundary formulation (2.17), wic contains te surface operator νx φ(x,y) ( n 2 (y) 1 ) (E(y) νy ) dσ y.

7 D.M. AMBROSE AND S. MOSKOW 299 We note tat tis term plays a crucial role for any scatterer, not just te ig contrast model presented ere. It as a nontrivial limit for tin domains, wic we analyze in tis section. Consider te linear operator defined by T :C 0 ( ) C 0 ( ) (Tf)(x)= νx φ(x,y)f(y) dσ y. (3.1) Tis is te dual of a double layer potential operator. We can sow tat wen viewed on an appropriately scaled domain, te operators T converge in a weak sense to an operator T0. To be precise, define te scaled domain S= ( 1/2,1/2) associated wit te cange of variables to ỹ given by and for any given f(y), let y=(ỹ 1,ỹ 2,ỹ 3 ), (3.2) f(ỹ)=f(y). In wat follows we will not want to require tat functions are continuous at te corner curves of te pillbox, so we define te space CP 0 =C 0( { 1 C 2}) 0( { 1 }) C 0( [ 1 ]) 2 2,1 (3.3) 2 of piecewise continuous functions on S. Tat is, CP 0 is te space of functions wic are separately continuous on te closures of te top, bottom, and lateral side of te scaled domain. Ten define by T :C 0 P C 0 P T f=t f. Note tat te elements of CP 0 are not actually functions on S since tey could be multiply defined on te corner curves {1/2} and { 1/2}. To be precise, any f C P 0 is a triple f=(f1,f2,f3) were te first two components are functions on (x 1,x 2 ) (representing te top and bottom respectively), and te tird component is a function on (x 1,x 2, x 3 ) [ 1/2,1/2] (representing te lateral side). For te sake of clarity of exposition, owever, in wat follows we sometimes abuse notation and identify elements of C 0 P wit functions on S. Given tis notation, we can write

8 300 SCATTERING BY THIN DIELECTRICS ( T f)( x)= νx φ(x,y)f 1 (ỹ)dỹ+ νx φ(x,y)f 2 (ỹ)dỹ { 1 2 } { 1 2 } + νx φ(x,y)f 3 (ỹ)dσỹ, (3.4) ( 1 2,1 2 ) were we note tat te area element dỹ=dy 1 dy 2 in te top and bottom pieces is uncangedintenewvariables. Inordertocomputetelimitof{ T }asapproaces 0, it is more straigtforward to compute te limit of its L 2 -based adjoint. Te adjoint of T is te complex conjugate of te usual double layer potential operator (T f)(x)= νy φ(y,x)f(y)dσ y, (3.5) wic we also rescale to get ( T f)( x)= y3 φ(y,x)f 1 (ỹ) dỹ y3 φ(y,x)f 2 (ỹ) dỹ { 12 } { 1 2 } + νy φ(y,x)f 3 (ỹ) dσỹ. (3.6) ( 1 2,1 2 ) Wile te true Banac space adjoint is defined on te larger dual space (C 0 P ), we consider its natural restriction to te piecewise continuous functions in C 0 P and compute its weak limit. Teorem 3.1. Te scaled double layer potential operators (3.6) defined on piecewise continuous functions on te boundary of te scaled domain S, T :C 0 P C 0 P, for C 0 P defined by (3.3), converge pointwise weakly to a bounded operator T 0 :C 0 P C 0 P, in te sense tat T f,g T 0 f,g for any f,g CP 0, were, is te L2 ( S) inner product. For a given f CP 0, T 0 f is defined by f 1 0 1/2 0 f 1 T 0 f 2 = 1/2 0 0 f 2, 1/4R 1/4R 0 were te operator f 3 f 3 R:C 0 () C 0 ( ( 1/2,1/2)) is a restriction operator to followed by a constant extension in te x 3 direction. Proof. We will complete te proof in two steps. First, we will demonstrate tat T f as te desired limit, pointwise in x. Second, we will demonstrate te weak convergence.

9 D.M. AMBROSE AND S. MOSKOW 301 Step 1: Pointwise convergence of T f: Let us assume first tat (x 1,x 2 ) \ and x 3 =/2, ( x 3 =1/2). Tat is, we will first find te limit for ( T f) restricted to te interior top portion of te scaled domain. In wat follows we may use eiter x 3 or /2, but tey are always equal in tis part of te calculation. We begin wit (3.6). Since (x 1,x 2 ) is away from te lateral side, te last term is not singular and goes to zero O(). Hence ( T f)( x)= (y 1,y 2,/2,x 1,x 2,/2)f 1 (y) dy 1 dy 2 y 3 (y 1,y 2, /2,x 1,x 2,/2)f 2 (y) dy 1 dy 2 +O(). (3.7) y 3 Notice tat for x 3 =y 3, we ave φ y 3 (x,y)=0; tis implies tat te first integral is identically zero. Wit te approximation f 2 (y 1,y 2 ) f 2 (x 1,x 2 ) in mind, we define te remainder term R 1 by te following equation: ( T f)( x)= f 2 (x 1,x 2 ) (y 1,y 2, /2,x 1,x 2,/2) dσ y +R 1. (3.8) y 3 To see tat te remainder R 1 goes to zero wit for continuous f 2, consider Since we ave tat = 1 y 3 4π y 3 (y,x) = 1 4π y 3 e ik x y ( (x 1 y 1 ) 2 +(x 2 y 2 ) 2 + ( 2 ỹ 3 e ik x y 1 ik x y, x 3 ỹ 3 ( (x 1 y 1 ) 2 +(x 2 y 2 ) 2 + ( 2 ỹ 3 ) 2 ) 3/2 +O ) 2 ) 1/2. (3.9) ( ) x3 ỹ 3 x y 2. Since x 3 =/2, te remainder term will yield factors of multiplying kernels wic are integrable in te limit as 0. So, wen we make te replacement f 2 (y 1,y 2 ) f 2 (x 1,x 2 ), te remaining term is of te form 1 4π (f 2 (x 1,x 2 ) f 2 (y 1,y 2 )) ((x 1 y 1 ) 2 +(x 2 y 2 ) ) 3/2 dy 1dy 2. (3.10) Now, if f 2 is a C 1 function, tis integral clearly goes to zero wit since in tis case f 2 (x 1,x 2 ) f 2 (y 1,y 2 ) =O( x y ) and te integrand is only weakly singular in te limit as 0. So, let us approximate f 2 by a sequence of C 1 functions, {f2 n }, tat will converge to f 2 in te uniform norm as n. Ten, we write te following: (f 2 (x) f 2 (y)) ((x 1 y 1 ) 2 +(x 2 y 2 ) ) dy 1dy 3/2 2

10 302 SCATTERING BY THIN DIELECTRICS =(f 2 (x) f2 n (x)) + + ((x 1 y 1 ) 2 +(x 2 y 2 ) ) 3/2dy 1dy 2 (f2 n (x) f2 n (y)) ((x 1 y 1 ) 2 +(x 2 y 2 ) ) 3/2dy 1dy 2 (f2 n (y) f 2 (y)) ((x 1 y 1 ) 2 +(x 2 y 2 ) ) 3/2dy 1dy 2 =I+II+III. (3.11) One can ceck tat by integrating around a small ball, te integral ((x 1 y 1 ) 2 +(x 2 y 2 ) ) 3/2dy 1dy 2 is bounded as 0. So, for any ǫ>0 we can coose n large enoug so tat terms I and III are smaller tan ǫ/3 for any (0,1]. Ten, since f2 n is C 1, we can coose small enoug tat II is also less tan ǫ/3, so tat te wole integral in (3.11) is less tan ǫ. Tis establises tat R 1 does go to zero wen goes to zero, as claimed; furtermore, tis convergence is uniform in x. We ten ave ( T f)( x)= 1 4π f 2(x 1,x 2 ) x 3 +/2 ( (x1 y 1 ) 2 +(x 2 y 2 ) 2 +( )2) 3/2 dy 1dy 2 +R 2, were again R 2 goes to zero wit. Since (x 1,x 2 ) \, tere exists δ>0 suc tat te ball of radius δ centered at (x 1,x 2 ), B δ \. We notice tat outside of tis ball, te integral will go to zero wit. We are ten left wit ( T f)( x)= 1 4π f 2(x 1,x 2 ) B δ ((x 1 y 1 ) 2 +(x 2 y 2 ) ) 3/2dy 1dy 2 +R 3, were R 3 goes to zero wit. Tis integral can be calculated explicitly wit polar co-ordinates. Te result is ( T f)( x)=f2 (x 1,x 2 )( /2)(r ) 1/2 r=δ r=0 +R 3. We can evaluate tis and include te r=δ contribution into te remainder, finding f 2 (x 1,x 2 ) ( T f)( x)= +R 4, 2 were R 4 goes to zero wit. A similar calculation yields tat for x on te bottom in te interior of, (x 1,x 2 ) \,x 3 = /2, f 1 (x 1,x 2 ) ( T f)( x)= +R 5, 2 were R 5 goes to zero wit. Let us next consider x ( 1/2,1/2). Altoug te last term in (3.4) as a singularity, it still goes to zero because of te factor of in front, and te fact tat if x and y are bot on te lateral boundary, (x y) ν y =O( x y 2 ),

11 D.M. AMBROSE AND S. MOSKOW 303 since x y is close to tangential wen x y is small [4]. However, now neiter te integral over te top nor te bottom is identically zero and bot will contribute to te limit. Hence, we can write ( T f)( x) = 1 4π f x 3 /2 1(x 1,x 2 ) ((x 1 y 1 ) 2 +(x 2 y 2 ) 2 +(x 3 /2) 2) dy 3/2 1dy π f x 3 +/2 2(x 1,x 2 ) ((x 1 y 1 ) 2 +(x 2 y 2 ) 2 +(x 3 +/2) 2) dy 3/2 1dy 2 +R 6. Now, since (x 1,x 2 ), we can no longer take a ball contained in about tis point. However, for smoot, we can take a alf-ball of radius δ about (x 1,x 2 ) wit flat side tangential to ; call tis H δ. Introduction of tis alf-ball does introduce an error term unless te boundary is actually flat at (x 1,x 2 ); owever, tis term can be made arbitrarily small by coosing δ small, since is smoot. Te remainder of bot of tese integrals will go to zero wit : ( T f)( x) = 1 4π f ( x 3 1/2) 1(x 1,x 2 ) H δ ((x 1 y 1 ) 2 +(x 2 y 2 ) 2 +(( x 3 1/2)) 2) dy 3/2 1dy π f ( x 3 +1/2) 2(x 1,x 2 ) H δ ((x 1 y 1 ) 2 +(x 2 y 2 ) 2 +(( x 3 +1/2)) 2) dy 3/2 1dy 2 +R 7. Tis yields ( T f)( x)=(f 1 (x 1,x 2 ))(/4)( x 3 1/2)(r 2 +(( x 3 1/2)) 2 ) 1/2 r=δ r=0 (f 2 (x 1,x 2 ))(/4)( x 3 +1/2)(r 2 +(( x 3 +1/2)) 2 ) 1/2 r=δ r=0 +R 7. (3.12) Again, te r=δ term goes in wit te remainder, so as 0, tis converges to 1/4f 1 +1/4f 2. Step 2: Convergence of T f,g : Looking at te formula (3.6), we introduce te auxiliary operators S 1,, S 2,, and S 3,, so tat T =S 1, +S 2, +S 3,. Here, S 1, is te integral over te top surface, S 2, is te integral over te bottom surface, and S 3, is te integral over te lateral surface. For j {1,2,3}, we will sow tat S j, f,g converges to S j,0 f,g, were S j,0 are te relevant pieces of T 0. We begin wit S 3,, since tis is te most intricate piece of te weak convergence argument. We need to prove g(x) f 3 (y) νy φ(x,y) dσy dx 0 as 0. [ 1 2,1 2 ]

12 304 SCATTERING BY THIN DIELECTRICS We will consider tis first for x { 1 2 }, and ten we will consider x on te oter pieces of te boundary. Towards tis end, we consider ere ( ) I= g 1 (x) f 3 (y) νy φ(x,y) dy dx. { 1 2 } [ 1 2,1 2 ] We bound f 3 and g 1 by teir infinity norms, and we cange te order of integration: I g 1 f 3 νy φ(x,y) dxdy. { 1 2 } [ 1 2,1 2 ] We rewrite te integral over te lateral boundary as being an arclengt integral over and an integral wit respect to y 3 : I g 1 f { 1 2 } νy φ(x,(y1 (s),y 2 (s),ỹ 3 )) dxdsdỹ 3. On te lateral boundary, te y 3 direction is tangential, so te normal vector is only in te y 1 and y 2 directions. Terefore, φ νy φ(x,y) (x,y) y 1 + φ (x,y) y 2. Based on tis, we write I=I 1 +I 2, so tat I j is te contribution from φ y j. We can calculate φ (x,y)= 1 y 1 4π e ik (x 1 y 1) 2 +(x 2 y 2) ( x 3 ỹ 3) 2 ((x 1 y 1 ) 2 +(x 2 y 2 ) ( x 3 ỹ 3 ) 2 ) 3/2(x 1 y 1 ) ik e ik (x 1 y 1) 2 +(x 2 y 2) ( x 3 ỹ 3) 2 4π (x 1 y 1 ) 2 +(x 2 y 2 ) ( x 3 ỹ 3 ) 2(x 1 y 1 ). (3.13) For a fixed s and fixed ỹ 3, we use polar coordinates for x. We ave x 1 y 1 (s)=rcos(θ) and x 2 y 2 (s)=rsin(θ). We ten ave te following inequality: φ (x,y) y 1 c r (r ( x 3 ỹ 3 ) 2 ) +c r 3/2 r ( x 3 ỹ 3 ) 2. If we coose R to be te diameter of, ten is a subset of te (two-dimensional) ball centered at (y 1 (s),y 2 (s)) wit radius R; terefore, 1 2 I 1 c g 1 f π R 0 0 r 2 (r ( x 3 ỹ 3) 2 ) 3/2 + r 2 r ( x 3 ỹ 3) 2 drdθdsdỹ3. In tis last integral, notice tat te integrand does not depend on s or θ. Since s and θ are bot taken from a bounded interval, we can bound tese integrals by a constant times te remaining integrals. We get I 1 c g 1 f R 0 r 2 (r ( x 3 ỹ 3 ) 2 ) 3/2 + r 2 r ( x 3 ỹ 3 ) 2 drdỹ 3.

13 We ave te following indefinite integral: Using tis antiderivative, we ave D.M. AMBROSE AND S. MOSKOW 305 r 2 ( (r 2 +α 2 ) dr=ln r+ r 2 +α ) 2 3/2 I 1 c f 3 g r r2 +α +c ln(( x 3 ỹ 3 )) dỹ 3. Since te natural logaritm is integrable, we conclude tat I 1 c( ln() +1) f 3 g 1. Of course, we ave te same bound for I 2. Tis implies I c( ln() +1) f 3 g 1. Clearly, ten, I goes to zero as goes to zero. If we ad integrated over te bottom surface instead, we would find te same conclusion. Also, wen x ( 1 2, 1 2 ), we see from te pointwise convergence argument tat S 3, f(x) converges uniformly to zero. Taking all of tis togeter, we ave S 3, f,g converges to zero as 0. We turn now to considering S 1,. First, notice tat if x { 1 2 }, ten S 1,f(x)= 0. Let K be a compact subset of, suc tat K =. Ten, by examining te pointwise convergence argument, we can see tat S 1, f(x) converges uniformly for x K { 1 2 }. More generally, for x { 1 2 }, we ave te following uniform bound: S 1 f(x) C y3 φ(x,y) f1 (ỹ) dỹ (3.14) { 1 2 } C f 1 ((x 1 y 1 ) 2 +(x 2 y 2 ) ) dy 1dy 3/2 2 (3.15) C f 1. (3.16) Tis uniform bound implies tat te integral g 2 (x) y3 φ(x,y)f1 (y) dydx (\K) { 1 2 } { 1 2 } can be made small by coosing K appropriately. Furtermore, for x ( 1 2, 1 2 ), we see from te pointwise convergence argument tat te convergence of S 1, f(x) is uniform. We ave now establised te convergence of S 1, f,g as goes to zero. Te weak convergence argument for S 2, is te same as for S 1,. Tis completes te proof. Remark 3.1. Te weak pointwise convergence of T to T 0 automatically implies te weak pointwise convergence of te adjoints, tat is, T f,g T 0f,g for any f,g C 0 P, were T 0 is te adjoint of T 0.

14 306 SCATTERING BY THIN DIELECTRICS In te boundary equation (2.17), te last two terms become ( ) 1 ((n 2 I T 2 1) (E ν) ). Tis motivates te following lemma, wic will be useful in te next section. Lemma 3.2. Let T be given by (3.1) and let T 0 be its scaled pointwise limit as given in Teorem 3.1. For any fixed, assume tat k 2 is not a Neumann eigenvalue for te Laplacian on. Ten te operator ( 1 ) 2 I+T is invertible. However, te scaled limiting operator ( 1 ) 2 I+ T 0 is not invertible, and it as a null space consisting of all functions on S wic ave equal values on te top and bottom and are zero on te lateral boundary. Proof. Recall tat since for fixed, T is te dual of a double layer potential, it is well known tat 1 2 I+T is Fredolm. Terefore, it is invertible if it as trivial null space. Suppose β C 0 ( ) suc tat 1 2 β+t β=0. Define w to be te single layer potential wit moment β, w(x)= φ(x,y)β(y)dσ y, defined for any x R 3. By standard potential teory we ave tat te interior limit for te normal derivative of w satisfies tat w satisfies te Helmoltz equation ( ν w) = 1 2 β+t β=0, (3.17) w+k 2 w=0 bot in te interior and te exterior R 3 \, and we ave tat te jumps across te boundary of are given by and [w] = 0 (3.18) [ w ] =β. (3.19) ν Assuming k 2 is not a Neumann eigenvalue for te Laplacian on, since w satisfies (3.17), it must be identically zero inside of. Since te jump (3.18) is zero, tis means tat on te exterior, w satisfies te Diriclet problem for te Helmoltz

15 D.M. AMBROSE AND S. MOSKOW 307 equation wit zero boundary data. So, w must be identically zero on te exterior by Teorem 3.7 of [3]. Hence w is identically zero on R 3. Hence its normal derivative jump across must be zero, so from (3.19) we ave β=0. Terefore we ave tat te null space is trivial and te operator 1 2 I+T is invertible. For te limiting operator, te proof is simple algebra. Using te notation in te previous teorem, we calculate ( 1 ) 2 I+ T 0 f= R R were R is te adjoint of te restriction and extension operator, and its range is in te dual space to continuous functions on. One can calculate tat for continuous f 3, ( 1/2 R f 3 = f 3 d x 3 )δ, 1/2 a Dirac mass on te boundary multiplied by te x 3 average of f 3. We see directly tat te null space consists of all f suc tat f 1 =f 2 and f 3 =0. 4. Limiting beavior of E Teorem 2.1 gives us a Lippman-Scwinger type of formulation for any Maxwell scatterer wit regular enoug boundary. In particular, wen te index of refraction is of te specific form (2.2), it is useful for analyzing te beavior of te electric field as 0. In tis section we sow two teorems about te asymptotic beavior of E in tis case. Te first is tat, on te lateral boundary of te domain, te interior limit of te normal component of te electric field, (E ν), goes to zero. Te second result is tat wit some assumptions of uniform boundedness, te tird component of te electric field, E 3, goes to zero wit. Tese two results togeter sow tat te ig contrast structure does indeed trap te wave, in te sense tat te flow out of te scatterer is small wit respect to. Before stating tese teorems, let us define some operators to more compactly describe te system (2.16). Set, for f [ L 2 ( ) ] 3, ( ) K ε0 (y) f= φ(x,y) 1 f(y) dy, (4.1) S f= x φ(x,y) ε 0(y) ε 0 (y) f 1 f 2 f 3, f(y) dy, (4.2) ( ) L ε0 (y) f= y φ(x,y) ( /2,/2) 1 f(y) ν dσ y, (4.3) ( ) Q ε0 (y) f= y φ(x,y) {/2} 1 { /2} f 3 (y) dy 1 dy 2 ( ) ε0 (y) y φ(x,y) 1 f 3 (y) dy 1 dy 2, (4.4)

16 308 SCATTERING BY THIN DIELECTRICS were ν=(ν 1,ν 2,0) is te outward unit normal to te lateral boundary L= ( /2,/2), and f 3 (y) is te tird component of te vector-valued f. Wit tis notation, E satisfies E=E i +k 2 K E+S E+L E+Q E, (4.5) were K is te operator coming from te Helmoltz-like term, S is te operator coming from te smoot variation in n in te interior, L is te operator resulting from te lateral boundary of te object, and finally Q is te operator coming from te top and bottom boundary of te object (combined). From [9], we know tat wile from [1] we ave K f C f, (4.6) S f C log f, (4.7) were in bot cases C is independent of. Similar to CP 0, define te piecewise Cα space on te scaled boundary: C α P =C α( { 1 2}) C α( { 1 }) C α( [ 1 ]) 2 2,1. 2 Also, we represent all functions in te scaled variables (3.2) wit a tilde, tat is, Ẽ(ỹ)=E(y). Teorem 4.1. Let E solve (2.16) wit n 2 (x) defined by (2.2), and let (E ν) be te interior limit of E ν on te boundary. Assume tat ε 0 C 0 ( ) is suc tat ε 0 is bounded away from zero. Assume tat E is bounded in C 0 ( ) independently of, and tat (Ẽ ν) is bounded in CP α independently of, for some α>0. Ten we ave tat (Ẽ ν) 0, (4.8) uniformly on te lateral boundary ( 1/2,1/2), as 0. Proof. Consider (2.17) wit n 2 (x)=ε 0 (x)/ in te interior of. Multiplying everyting by and collecting wat appear to be O(1) terms, we ave tat 1 2 ε0(e ν) νx φε 0(E ν y) dσ 3 y=(e i ν 2 (E ν) +k 2 (K E) ν+(s E) ν). (4.9) Note tat by te uniform boundedness assumption on E and bounds (4.6) and (4.7), te rigt and side is continuous on and O() in L ( ). Consider, for x, ψ (x)=ε 0 (x)(e ν) (x). Ten ψ (x) satisfies te Fredolm equation 1 2 ψ +T ψ =e, (4.10) were T is te operator (3.1), te unscaled version of te dual of operator from Teorem 3.1. Also, e is te negative of te rigt and side of (4.9) above. Rescaling in x 3, we ave ( 1 2 I+ T ) ψ =ẽ. (4.11)

17 D.M. AMBROSE AND S. MOSKOW 309 Note tat by our assumptions, te { ψ } are uniformly bounded in C α separately on eac piece of S. By te Arzela-Ascoli Teorem, tere exists a subsequence wit a uniform limit; call tis limit ψ 0. For any f C 0 P, ( 1 2 I+ T ) ψ,f = ẽ,f (4.12) were te inner product is te usual one for L 2 ( S). By taking te dual of te operator, tis gives ψ,( 1 2 I+ T ) f = ẽ,f, (4.13) were T is te dual to T, as in Teorem 3.1. Also from tis teorem we know tat T converges pointwise weakly to T 0. By adding and subtracting we ave ψ,( 1 2 I+ T ) 0 f = ψ,( T 0 T )f + ẽ,f (4.14) = ψ ψ 0,( T 0 T )f + ψ 0,( T 0 T )f + ẽ,f. (4.15) Notice tat since T f converges weakly, te Principle of Uniform Boundedness implies tat it is bounded uniformly wit respect to. Tis can also be seen directly by te weak convergence proof of te operators. Terefore, if we let 0, everyting on te rigt-and side goes to zero and we ave ψ0,( 1 2 I+ T ) 0 f =0, (4.16) wic implies tat ( 1 2 I+ T 0) ψ0,f =0 (4.17) for any f CP 0. Hence ψ 0 lies in te null space of ( 1 2 I+ T 0). From Lemma 3.1 tis implies tat ψ 0 is zero on te lateral boundary (and equal on te top and bottom). Terefore any sequence of {Ẽ ν} as a subsequence wic converges uniformly to zero on te lateral boundary; tis implies te conclusion of te teorem. Teorem 4.2. Let E 3 C α ( ) for some 0<α 1 be te solution to (2.16) suc tat E 3 Cα ( ) M α, were M α is independent of. Ten, pointwise for x, tere exists C depending on α and possibly x suc tat, for α<1, we ave and for α=1 we ave were C is independent of. E 3 (x) CM α α, E 3 (x) CM 1 log,

18 310 SCATTERING BY THIN DIELECTRICS Proof. We will examine te term in (4.5) wic we expect will blow up and multiply it by. We first fix x 3 =0 and compute tat (Q E) 3 = (x,y)(n 0 (y) )E 3 (y) y3=/2 dy 1 dy 2 y 3 (x,y)(n 0 (y) )E 3 (y) y3= /2 dy 1 dy 2 y 3 ( =(n 0 (x) )E 3 (x) (x,y) ) y3=/2 (x,y) y3= /2 dy 1 dy 2 y 3 y 3 + (x,y)[(n 0 (y) )E 3 (y) (n 0 (x) )E 3 (x)] y3=/2 dy 1 dy 2 y 3 + (x,y)[(n 0 (y) )E 3 (y) (n 0 (x) )E 3 (x)] y3= /2 dy 1 dy 2. y 3 From te Hölder condition on E and te smootness assumption on n 0, (n 0 (y) )E 3 (y) (n 0 (x) )E 3 (x) C α x y α. Tis implies (x,y) y3=/2[(n 0 (y) )E 3 (y) (n 0 (x) )E 3 (x)] dy 1 dy 2 y 3 C (x,y) y3=/2 y 3 x y α dy 1 dy 2. If we define r to be te two-dimensional distance r= (x 1,x 2 ) (y 1,y 2 ), (4.18) we see tat (x,y) y 3 y3=/2 C 1 (r ) 3/2, so for some C independent of, we ave (letting R be sufficiently large) (x,y) y3=/2 r α y 3 x y α dy 1 dy 2 C (r ) 3/2dy 1dy 2 A computation yields tat for 0<α<1, and for α=1, R 0 R 0 C R r 1+α (r ) 3/2dr Cα 1 +O(1), r 2 (r ) 3/2dr C log +O(1). 0 r 1+α (r ) 3/2dr. (4.19)

19 D.M. AMBROSE AND S. MOSKOW 311 Te same bound also olds for te last term in te rigt and side of (4.18). Tis ten gives tat, for 0<α<1, ( (Q E) 3=(n 0(x) )E 3(x) (x,y) y3 =/2 ) (x,y) y3 = /2 dy 1dy 2+O( α ). y 3 y 3 Set ( I = (x,y) ) y3=/2 (x,y) y3= /2 dy 1 dy 2. y 3 y 3 Using te above in te tird component of (4.5) we ave (n 0 (x) )E 3 (x)i =E 3 (E i ) 3 k 2 (T E) 3 (S E) 3 +(L E)+O( α ). (4.20) Now, we will sow tat I is actually bounded from below away from zero. We recall te definition of φ: φ(x,y)= 1 e ik x y 4π x y. We take te derivative of tis wit respect to y 3 to get Next, we set x 3 =0 so tat (x,y) y 3 (x,y) = 1 e ik x y y 3 4π x y 3(x 3 y 3 ) ik e ik x y 4π x y 2(x 3 y 3 ). x3=0 = y 3 4π e ik x y x y 3 x3=0 + iky 3 4π e ik x y x y 2 x3=0 We let r= (x 1 y 1 ) 2 +(x 2 y 2 ) 2, plug in y 3 =/2 and y 3 = /2, and subtract to obtain (x,y) (x,y) = e ik(r2 + 2 /4) e ik(r2 + 2 /4) +ik y 3 y 3 4π (r /4) 3/2 4π (r /4). x3=0, y 3=/2 x3=0, y 3= /2 Rewriting tis and adding and subtracting 1 from te first exponential we ave (x,y) (x,y) y 3 y 3 x3=0, y 3=/2 = 1 4π (r /4) +ik 3/2 4π x3=0, y 3= /2 e ik(r2 + 2 /4) (r /4) 4π. (e ik(r2 + 2 /4) 1). (4.21) (r /4) 3/2 Since (x 1,x 2 ) is in te interior of, tere exists δ>0 suc tat B δ (x 1,x 2 ). We terefore write I (x 1,x 2,0)=A 1+A 2+A 3+ (x,y) (x,y) dy 1dy 2; y 3 y 3 \B δ (x 1,x 2 ) x3 =0, y 3 =/2 x3 =0, y 3 = /2 we write it in tis fasion since we ave tree terms on te rigt-and side of (4.21), and we write te region of integration as =B δ (x 1,x 2 ) (\B δ (x 1,x 2 )).

20 312 SCATTERING BY THIN DIELECTRICS Using (4.21), we can see tat te integral over te complement of te ball goes to zero wit, tat is (x,y) (x,y) dy 1 dy 2 =O(). y 3 y 3 \B δ (x 1,x 2) x3=0, y 3=/2 x3=0, y 3= /2 We can bound A 1 by switcing to polar coordinates as follows: A 1 = B δ (x 1,x 2) 4π 1 (r /4) dy 1dy 3/2 2 = δ 2 0 For A 2 we start by stating its definition: A 2 = B δ (x 1,x 2) r dr (r /4) 3/2 = δ 1 = 1+O(). (4.22) 2 r2 + 2 /4 ike ik(r2 + 2 /4) 4π r /4 dy 1dy 2. We switc to polar coordinates, and we bound te absolute value of tis as A 2 k 2 δ 0 r r /4 dr= k δ 4 ln(r2 + 2 /4) =O(ln()). We need finally to bound A 3. Te definition of A 3 is (e ik(r2 + 2 /4) 1) A 3 = dy B δ (x 1,x 2) 4π (r /4) 3/2 1 dy 2. By eiter te Caucy or Lagrange form of te remainder in Taylor s Teorem, tere exists a constant, C, suc tat e ik(r2 + 2 /4) 1 C(r /4). We again use polar coordinates. We estimate te absolute value of A 3 as A 3 C δ r dr=o(). 2 0 (r ) 1/2 Taking tese calculations togeter, we ave sown tat I (x 1,x 2,0)= 1+O(ln()). Hence from (4.20) we ave tat for 0<α<1 and for α=1 n 0 (x)e 3 (x 1,x 2,0)=O( α ), n 0 (x)e 3 (x 1,x 2,0)=O(log). Using te C α boundedness of E 3, te same bounds old on all of. 0 0

21 D.M. AMBROSE AND S. MOSKOW Discussion We ave derived and used te formulation in Teorem 2.1 to analyze te beavior of a tin ig contrast dielectric. Tis formulation treats any jumps in te material wit a surface integral, and we ave sown tat te surface terms force te normal component of te electric field to go to zero on te boundary of te tin ig contrast domain, so tey are indeed playing a crucial role. Here we discuss te implications for te limiting equations and furter asymptotic expansions in. If we assume tat E(y)=E (0) (y 1,y 2 )+E (1) (y)+o( 2 ), and we plug tis into (2.16) and let 0, ten from [9] we know tat te first integral term becomes k 2 φ((x 1,x 2,0),(y 1,y 2,0))ε 0 (y 1,y 2 )E (0) (y 1,y 2 )dy 1 dy 2, and from [1] we know tat te second integral term is O(log), so it disappears in te igest order limit. For te surface integral term, ( ) ε0 (y) y φ(x,y) 1 (E(y) ν) dσ y, (5.1) te situation is more complicated. Tis term as of course tree vector components, and in eac of tese components we ave terms corresponding to te top, bottom, and lateral sides of te scatterer. In te first two components, te integral over te top and bottom parts of te boundary cancel wit eac oter in te limit (tis will be described in detail in a fortcoming work). Te limit of te first two components of (5.1) terefore correspond to an integral over te lateral side, wic becomes te curve integral y φ(x,y)ε 0 E (0) νds y. (5.2) If x is in te interior of, tis term as no singularity, and by Teorem 4.1 it sould disappear to first order, yielding te same limiting equations as in te Helmoltz and smoot cases [1], [9]. Te limit in te first two components became nice because tere was cancellation between te top and bottom parts of te surface integral as tey get close to eac oter. However, for te tird component, te top and bottom terms are te same sign and do not cancel in te limit, and get more singular as tey approac; it is on te order of E 3 /. Tis is anoter way to see tat E 3 goes to zero in te interior of te scatterer, since it allows te equations to balance. Hence te analysis (formally) yields te limiting system E (0) 1 (x 1,x 2 )=(E i ) 1 (x 1,x 2,0)+k 2 E (0) 2 (x 1,x 2 )=(E i ) 2 (x 1,x 2,0)+k 2 φε 0 E (0) 1, (5.3) φε 0 E (0) 2, (5.4) E (0) 3 (x 2,x 2 )=0. (5.5) We propose te following first-order approximation to te field anywere in R 3 : solve te above two dimensional system for E (0), and insert tis solution E (0) into te rigt and side of (2.6) to compute te field elsewere. Tis sould be very efficient, and

22 314 SCATTERING BY THIN DIELECTRICS muc more accurate tan te commonly used Born approximation of just inserting te incident wave E i. We ave indeed sown tat (under certain regularity assumptions on E), te perpendicular component of E to te slab, E 3, goes to zero inside te slab. Tis dimensionally reduced system will also be useful for approximating resonances as in [5], and for inversion metods suc as in [11]. We ave also sown ere tat on te lateral boundary to te slab, te interior normal component (E ν) 0 in Teorem 4.1. Tis normal component to now involves a combination of E 1 and E 2. However, te solution to (5.3),(5.4) is unique (see [9]), and cannot in general satisfy tat te normal component on is zero for arbitrary E i. Tis implies tat any convergence of E E (0) can not be uniform in general: some sort of boundary layer must exist near. Tis makes sense, since te kernel in (5.2) becomes singular as x approaces te boundary of, and tis term can no longer be discarded. We now comment on ow te results ere relate to te previous results in [9] and [1]. Te former only considered a scalar Helmoltz system, and tere te first-order limiting system for te polarized field is te same as (5.3), and te convergence is uniform and O(). In [1], full Maxwell is considered, but tere te squared index of refraction is assumed to be everywere C 2. In particular, te effects of material jumps between and te background are not considered. Our analysis ere includes jumps, suc as oles in, and no smooting at te edge is required. It implies tat, up to first order in te interior of, one gets te same limiting system for te first two components, but te surface term does force a different limit (namely zero) for te tird component E 3. Te surface term also plays a role in boundary beavior and will be crucial for deriving accurate corrections. Furter asymptotic analysis of te boundary and te next order corrections will be te subject of a fortcoming paper. Acknowledgment. D.M. Ambrose was supported by te National Science Foundation under grants DMS and DMS S. Moskow was supported by te National Science Foundation under grant DMS REFERENCES [1] H. Ammari, H. Kang, and F. Santosa, Scattering of electromagnetic waves by tin dielectric planar structures, SIAM J. Mat. Anal., 38, , [2] G. Boucitté, Analyse limite de la diffraction d ondes électromagnétiques par une structure mince, C.R. Acad. Sci. Paris Sér. II Méc. Pys. Cim. Sci. Univers Sci. Terre, 311, 51 56, [3] D. Colton and R. Kress, Inverse Acoustic and Electromagnetic Scattering Teory, Appl. Mat. Sci., Springer, 93, [4] G.B. Folland, Introduction to Partial Differential Equations, Princeton University Press, Princeton, N.J., [5] J. Gopalakrisnan, S. Moskow, and F. Santosa, Asymptotic and numerical tecniques for resonances of tin potonic structures, SIAM J. Appl. Mat., 69, 37 63, [6] A. Kirsc, An integral equation for Maxwell s equations in a layered medium wit an application to te factorization metod, J. Int. Eqs. Appl., 19, , [7] A. Kirsc and N. Grinberg, Te Factorization Metod for Inverse Problems, Oxford Lecture Series in Matematics and its Applications, Oxford University Press, Oxford, 36, [8] A. Kirsc and A. Lecleiter, Te operator equations of Lippmann-Scwinger type for acoustic and electromagnetic scattering problems in L 2, Appl. Anal., 88, , [9] S. Moskow, F. Santosa, and J. Zang, An approximate metod for scattering by tin structures, SIAM J. Appl. Mat., 66, , [10] J.H. Ricmond, Scattering by tin dielectric strips, IEEE Trans. Ant. Prop., 33, 64 68, [11] N. Zeev and F. Cakoni, Te identification of tin dielectric objects from far field or near field scattering data, SIAM J. Appl. Mat., 69, , 2009.