RL-452 ITERATIVE SOLUTIONS OF MAXWELL'S EQUATIONS by John Stelios Asvestas A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in The University of Michigan 1968& Doctoral Committee: Doctor Ralph E. Kleinman, Co-chairman Professor Chen-To Tai, Co-chairman Professor Chiao-Min Chu Doctor Raymond F. Goodrich Doctor Thomas B. A. Senior i

ITERATIVE SOLUTIONS OF MAXWELL'S EQUATIONS by John Stelios Asvestas Co-Chairmen: Ralph E. Kleinman Chen-To Tai The problem of scattering of electromagnetic waves by a closed, bounded, smooth, perfectly conducting surface immersed in vacuum is considered and a method for determining the scattered electric and magnetic field vectors o (solutions of the homogeneous Maxwell equations satisfying well known boundary conditions on the surface and the Silver-MUller radiation condition at infinity) everywhere exterior to the surface is presented. Specifically, two integral equations are derived, one for each scattered field vector. These equations are coupled. The kernels of the equations are dyadic functions of position and can be derived from the solutions of standard interior and exterior potential problems. Once these dyadic kernels are determined for a particular surface geometry the integral equations can be solved by iteration for the wave number k being sufficiently small. Alternatively, the scattered fields in the integral equations may be expanded in a power series of the wave number k and recursion formulas be found for the unknown coefficients in the expansions by equating equal powers of k. As a check, the method is applied to the problem of scattering of a plane electromagnetic wave by a perfectly conducting sphere. The first two terms in the low frequency expansions of the electric and magnetic scattered fields are found and are shown to be in complete agreement with known results.

A CKNOWLEDGEMENTS The author wishes to thank Dr. R. E. Kleinman for suggesting the problem and for his continuous guidance and encouragement. The author wishes to thank the members of his committee for their support during the execution of this work. In particular, he wishes to express his appreciation to Prof. C-T Tai for his interest in the problem and for his inspired teaching. ii

TABLE OF CONTENTS INTRODUCTION 1 I THE DERIVATION OF TWO INTEGRAL EQUATIONS 9 II THE FIELDS OF INFINITESIMAL DIPOLES AND THE SOLUTION OF THE TWO DYADIC PARTIAL DIFFERENTIAL EQUATIONS 18 III INTEGRAL REPRESENTATIONS OF THE ELECTROMAGNETIC SCATTERED FIELDS 34 IV AN EXAMPLE: THE SPHERE. 47 V CONCLUSIONS. 74 REFERENCES 75 APPENDIX A: THE BEHAVIOR AT INFINITY OF THE =(l) =(1) DYADICS H AND E 76 e m iii

INTRODUCTION The subject of low frequency electromagnetic scattering dates back to Lord Rayleigh (1897). In his well-known paper Lord Rayleigh examined the scattering of both acoustical and electromagnetic waves by two-dimensional as well as three-dimensional bodies and he showed that in the limit as the wave number k tends to zero the electric and magnetic scattered vectors in the near field can be expressed in terms of solutions of standard potential problems. Furthermore, he was able to continue these solutions into the far field region and arrive at his famous fourth power of frequency law for the scattering cross section of objects whose characteristic dimension is small compared with the wavelength. Since that time considerable work has been done in this direction. An extensive bibliography for both acoustical and electromagnetic low frequency scattering is given by Kleinman (1965a). Much of the work in deriving higher order terms in the low frequency expansion,however, has depended explicitly on a particular geometry for the scatterer and on a particular type of source with a restricted direction of incidence. Stevenson (1953) overcame these limitations by means of the Stratton-Chu integral representation of the scattered fields. He showed that the scattered fields for a sufficiently smooth three-dimensional scatterer and for arbitrary excitation can be written in the form E =F + V m m -S - 5 H = G +V,m m m m where E and H are the coefficients of k in the k-series expansions of the m m electric and magnetic scattered fields, respectively. The vectors F and - S -.ft - S G are known in terms of the previous coefficients E,...,E and. -bS s s H..., H The sealar functions 0 and pi are solutions of Laplace's 0 m-1 m m equation satisfying known conditions on the scatterer arising from the properties of the electromagnetic field. They also satisfy the Kellogg regularity 1

2 conditions at infinity. The vectors E and H m can be continued into the far field by substituting them in the Stratton- Chu equations and using the far field approximation for the free space Green's function which is involved in these equations. Kleinman (1965b) showed, however, that Stevenson's method leads to incorrect results after the first few terms and proposed an alternate scheme largely based on Stevenson's. We will undertake this point in Ch. Im. In either case, however, the labor for obtaining higher order terms becomes prohibitive at a very early stage. Inherent to all three-dimensional low frequency techniques is the assumption Q that the scattered electric and magnetic fields can be written in a power series of the wave number k. Werner (1963) put the whole subject on a rigorous mathematical basis by showing that this assumption was correct. Specifically, he proved that the scattered electric field E tends, as k ->- 0, analytically to a corresponding electrostatic field. In the present work we propose an alternate low frequency scheme by means of which one can obtain as many terms as desired in the expansions of the scattered fields by operating on potential functions (solutions of Laplace's equation) satisfying certain boundary conditions on the scatterer and the Kellogg regularity conditions at infinity. The advantage of the present method over Stevenson's is that it does away with the determination of 2m potential functions (O and /s ). That is, for every E and H we wish to determine we do not m m m 5 5 have to solve two boundary value problems to determine 0 and (t. Moreover, m in we believe (though we do not prove) that the resulting series expansions through the present method represent the scattered fields not only in the near region but everywhere in space. Its disadvantage is that it applies only to perfectly conducting scatterers while Stevenson's applies to scatterers of finite or zero conductivity also. In order to facilitate reading of this work and to clarify our approach we include an introductory section where the problem, its motivation, and the main results are presented.

3 The Problem, its Motivation, and the Principal Results The problem under consideratioh is the following: In the three-dimensional frbe space (vacuum) we have a closed, bounded, perfectly conducting surface S which separates the whole space into two regions: the finite region Vi enclosed by S and the rest of the space V. The surface S is sufficiently smooth to guarantee the existence of a normal at all of its points. A time harmonic source of electromagnetic waves is located in V and its electric and magnetic fields are denoted by E and H, respectively. The time dependence e is omitted. The presence of the perfectly conducting surface S gives rise to an electromagnetic wave whose electric and magnetic vectors we denote by E and H, respectively. These two vectors satisfy 1) Maxwell's equations VxE =ikZH,VxH =-ikYE, (1) Z = 1/Y, the free space characteristic impedance, 2) the homogeneous vector wave equation Vx Vx s -k2 s =0 in V, (2) a consequence of Maxwell's equations, 3) the boundary conditions nxE =-nxE, n- H =-n- H on S, (3) A where n is the unit normal on S directed out of V and into Vi, and 4) the radiation conditions R) R[Rx( VxE )+ik]E = 0, AA li S. uniformly in R, (4) RRx( VxH )+ikH.: =O. R being the radial unit vector and R the distance from the origin of a coordinate system to a point in V.

4 Our intention is to determine E and H for k "sufficiently" small. The plan is as follows: First, we express the scattered fields in terms of two coupled integral equations. The kernels of these equations are dyadic functions of position which are derivable from solutions of Laplace 's equation. The equations are coupled in the sense that both E and H appear in each of them. Secondly, for k "sufficiently" small, we iterate these equations in an alternating manner to produce a Neumann series for each of the scattered fields. The motivation for such an approach to the problem is a paper by Kleinman (1965c) entitled "The Dirichlet Problem for the Helmholtz Equation. " In it the author derives a new integral equation for the regular part of the Dirichlet Green's function for the Helmholtz equation. The "kernel" of this equation is not, as is commonly the case, the free space Green's function but the Dirichlet Green's function for the Laplace equation. The equation can be solved by iteration as a Neumann series to produce the regular part of the Dirichlet Green's function for the Helmholtz equation for the absolute value of k sufficiently small. The practical value of this method lies in the fact that it employs the static Dirichlet Green's function which is known for most coordinate systems of interest and which usually involves special functions whose properties have been studied extensively. What is more important, however, is its conceptual value: for the first time it was rigorously demonstrated that the slowly varying dynamic Green's function can be obtained by suitably perturbing the corresponding static function. That this could be true had long been 2 felt among workers in the field. The same feeling certainly existed regarding the electromagnetic problem and now Kleinman's method suggested a way of attacking it. In connection with this work it should be mentioned that the Neumann problem for the Helmholtz equation has been treated in a manner analogous to the Dirichlet by Ar and Kleinman (1966). Cf. M. M. Schiffer's work in 'Lecture Series on Partial Differential Equations, " The University of Kansas Press, 1957.

5 The equivalent concept of a scalar Green's function in the vector case is the dyadic Green's function. After Levine and Schwinger (1950), we define it as follows: 1) VxVxG(RIRt)-k2G(RIR')=16(RR') inV (5) 2) Either A I= nxG = 0 on S (6) or nxVxG=0 on S (7) 3) liro R ( VxG-ikRxG) = 0. (8) We have in reality defined two dyadic Green's function depending on whether we choose the boundary condition (6) or the boundary condition (7). (That the Green's function must be a dyadic instead of, say, a vector is necessitated by the fact that we wish to obtain a linear relation between the field vector in V and the field vectors on the scatterer, and the most general linear relation between two vectors is a dyadic.) Using these dyadics (one at a time) in the dyadic form of Green's theorem we can find two sets of two integral equations each for the scattered fields E and H. The suitable form of Green's theorem in this case is f [VVXVQ) PQ. (VxVx P]dV= fn LQx(VX P)+(VX )X P dS, V S+S'+S+ (9) where S' is the surface of a small sphere centered at the singularity atR't, S is the surface of a sphere with infinite radius, and n is the unit normal always directed out of V. The vector Q stands for either of the scattered fields while the dyadic P usually stands for one of the dyadics defined above. In our case, however, we wish to use dyadics which Arrows (-5) over letters denote vectors while double bars (-) denote dyadics. Carets (^) over letters denote unit vectors.

6 are derivable from potential functions. At this point one must exercise care in defining these dyadics. They must have an appropriate singularity to make (9) give a desirable result (i. e. evaluate the field at the singularity) and they must satisfy appropriate conditions on S so that together with the natural boundary conditions of the electromagnetic fields they will make the integral over the scattering surface S in (9) a known term. The boundary conditions on S are readily determined by inspection of the surface integral in (9). The appropriate singularity was found by noticing _k2 -h that vector wave functions for the equation VxVxA -k A = 0 are formed by letting A = Vx (ci/), c being a constant unit vector and p a solution of the Helmholtz equation. In our case we let > be the free space Green's function for the Laplace equation. From now on the road is open and we can reach the following result (with respect to the geometry defined at the outset). If: = Vx MI) 1) H Vx - I+He, (10) 4,f f R -R E r L- 4wR-R 1RI r =(1) H e 2) VxVx = O, in V (12) =() r 3) nxE O, nxVxH, on S (13) m e 4) R 2(RXA)I < oo R 3 Vx A < oo, as R - o, (14) where A stands for either H( or E ( e m 5) E and H regular in V (15)

7 6) lvx(RE) <co, IVx (RH)<o, asR-.w, (16) then V'xE(R') =- (Vx VxE). dV+ (nx) (VxE (1))dS, (17) and -~ -~^xx ~ -* =1) dS (18) V'xH (R')=- (VxVxH) H dV+f nx(Vx H] dS. (18) V S This is the principal result of Ch. I. (The notation used for the two dyadics will become clear |when we recognize their physical significance.) Wilcox' (1956) expansion theorem makes it obvious that the scattered fields defined in (1) - (4) satisfy condition (16) and we can therefore write E (R,)=-ikZ H H dV- (nxE) H dS (19) e e V S and ikZH S (R E dV- (nxE. (Vx E ) dS. (20) Jm m V S The last equation can be written in a better form. To do this we need the explicit representation of E ) in terms of potential functions which leads us to the question of how to determine, the dyadics defined by (10) - (14). This problem may be attacked from a purely mathematical point of view. We had the feeling, however, that these functions should be related to the electrostatic and magnetostatic fields of infinitesimal electric and magnetic dipoles. Indeed, we can show (for our geometry) that H is the coefficient of k in e the low frequency expansion of the total magnetic field of three orthogonally crossed infinitesimal electric dipoles at the point R' of V. Similarly, E( is the coefficient of k in the low frequency expansion of the total electric field of three orthogonally crossed infinitesimal magnetic dipoles at the point

8 R' of V. After this the determination of the two dyadics becomes an easy matter. In Chapter II we derive these dyadics in their explicit form. In Chapter HI we start with Eqs. (19) and (20) and modify them in a way that will render them amenable to iteration. Finally, in Chapter IV we apply our method to the sphere.

Chapter I THE DERIVATION OF TWO INTEGRAL EQUATIONS In this chapter we proceed to derive in detail the integral equations (17) and (18) of the Introduction. To do this we need the divergence theorem in its dyadic form. 1 If V is a volume bounded by a regular surface S (S e L2, where L is P the class of surfaces whose equations have continuous derivatives up to and including ptth order and whose p'th derivatives satisfy a Holder condition2) and if A is regular in V and on S (A E C(, where C) is the class of functions with continuous derivatives up to and including the n'th order), then J V A dV= n. AdS, (1.1) V S A where n is the unit normal directed out of V. The proof of this theorem follows immediately from the corresponding theorem for vector functions. Attention should be drawn to the fact that the dot product in the surface, integral is not commutative. By writing A in the form A = Q x (VxP) +(VxQ) xP,(1.2) we obtain the followinglGreen's identity f EVXVxQ)P-Q. (VxVxP)]dV=f LQx(VxP)+(VxQ)xP]dS, (1.3) All dyadic identities that will be employed subsequently may be found in Van Bladel (1964). 2 - - A function f(R) is said to satisfy a Holder condition at R if there are three positive constants A, B, C such that If(R)-f(Ro)l < AIR-RoI for all points R for which j R-R < C. 9

10 where, in arriving at (1. 2), we used the dyadic identity V. (axb)=(Vxa) - b-a (Vxb). (1.4) Equation (1. 3) is the form of the divergence theorem that we will use to derive the two integral equations. The First Integral Equation Let S be a closed, bounded, regular surface. This surface separates the whole space into two regions: the finite region Vi enclosed by S and the rest of the space V. Let E( be a function of position defined in V and on S by (1) = Vx + m 'tl5 E4v I] + (1.5) -(1) where Em is regular in V and on S and satisfies mr =(1) VxVxE =, in V and on S. (1.6) Moreover, A (1) n x E =, onS, (1.7) and RR (x Em) < oo and R3 Vx | < o, as R oo (1 8) A where R is the radial unit vector and R the distance from the origin of a rectangular coordinate system to a point in V. For convenience, the origin of the coordinate system is located in Vi. Let, finally, E be a regular A, - = function of position in V and apply (1. 3) in V with Q substituted by E and P by E m(1) Since the dyadic E is singular at R', we exclude a small sphere from V centered at R' with radius r and surface S'. Eq. (1. 3) then becomes,

11 j (Vx VX E) E dV dSn - x(Vx E )+(Vx)xE, (1. 9) V S+S'+S ao where S is the surface of a large sphere, with center at the origin, bounding the volume V at infinity (see Fig. 1). We now proceed to examine the surface integrals one by one. First we examine the integral over S. Using the dyadic identity, A - = *.,. --. A,.- = (axb) A= a (bxA) = -b (axA) (1.10) and the boundary condition (1. 7), we have that Is= (n(x E) - (Vx E )dS. (1.11) s m S - Next we examine the integral over S', where we take S' to be the surface of a sphere of radius r centered at R'. We intend first to evaluate the integral and then let the radius r of the sphere go to zero. In the process, =(1) the integral involving the regular part of E ( will go to zero and we are left with the following expression =lim A- xVxVx1 s' r-o VxVx (Vx E)xVx. Jt L 47r IR-R' E 47 1-R1; (1.12) A A Since on S' n= -r and since Vx vx - -- =vv ~ - V2 - 1 - =VV - IV2 ( ) -z R-ZR I~ [R-R

12 S, n v i FIG. 1: GEOMETRY FOR THE APPLICATION OF GREEN'S IDENTITY. -IDENTcITY.

13 we have that li=r o (-r [ExWv(- V +(VxE)xVx(- 4r )] dS, (1. 13) S' where r =IR-R'I, (1.14) Now Vx (b)=(Vx)b - a x Vb. (1.15) Substitution of this relation in (1.13) leads to r E) V(- Vrr ) - Vx(EV(- 4r) + S' + (VxE)xVx(- 4Ar dS. (1.16) A By Stokes' theorem the part of the integral involving r Vx vanishes and by (1. 10) r -V E)V(- 4r)+(Vx E)x Vx (- Jr = =(r Vx E) V(- r)-(Vx E) Vx ( — r= =(r. V x E) V( )-(Vx E)- rx - )x] = =(r Vx E)V(- r)-(VxE) - V(- )( )- r V(- = =(r. Vx&V(-42T)+ 1E):~ V(-x)]r+r. V(-4-r)(VxE) = =(Vx.E)x )xr] +r* V(- )(Vx E )r V( ) (VxE) (1.17) where, above, we made use of the dyadic identity, ax, -b A.. - - = b.1 a x (bxA)=b(a.-A)-A(a b). (1. 18)

14 Equation (1.16) then becomes =I r-o 4-rr )(Vx E) dS -lim f (Vx E)I = dS ( = - V'xE (R') (1.19) r-* o dS 2 4' r 4rr We are now left with the evaluation of the integral over S. To do this 00 we draw a sphere of radius R and center at the origin of the coordinate system. We then let R go to infinity and we require the integral to vanish in the A limit. If R is the unit vector in the radial direction we have that (see Eq. 1. 9) I- = R - x(Vx E1 )+ (Vx E)x E dS 00 J m Sco By (1. 10) 7r 27 IS = I R2sinOeded RxE)- (VxE( ) )-(VxE). (Rx E( S f R m m( from which we can write 7T 27r lim f f sind {R2(xE (VxE1)) + R-* oD ISO I< R-)PO |D sinO dO do lR RxE),, EM) + 0 O '2.,. A =1) | + R2 (VxE) (RxE) ) I (1.20) In order that this integral vanish in the limit we must have R- I im R2(xE (VxE( ) ) =0 and lim iR2(Vx ^ R E)- (Rx E) 0 o

15 which, together with (1. 8), imply that RxE1<CD andIRVxEI <o as R oo. (1.21) Since Vx(RE) = RxE +R VxE Eq. (1.21) can be written jVx(RE) <oo asR- co. (1.22) This is the regularity condition on E if the integral (1.20) is to vanish. Collecting our results from (1. 9), (1. 11) and (1. 19), we can then state the following theorem: Theorem A: If V is the volume exterior to a closed, bounded, regular.surface S and E E is a regular function of position in V and on S satisfying the regularity condition IVx (RE)k oo, as R -* co, then E satisfies the integral equation vf(vXvX%. - (l)dv+f -(I) Vx E(R) (m dV ^x E) (Vx E ) dS, (1.23) V where E is defined in (1. 5) - (1. 8), and n is the unit normal on S directed m out of V and into the interior of S. The Second Integral Equation The derivation of the second integral equation follows closely that of the first and for this reason we shall be brief. The geometry of the problem remains the same. We define the dyadic =(1) H as follows: e H + He (1.24) 47r IR-R,' e r

16 =(1) where He is regular in V and on S and satisfies er VxVx H) =, in V and on S. (1.25) er r Moreover, A(1) nxVxHe =0, onS, (1.26) and IR2(RXH () | < co and jR3 Vx H() < oo, as R o. (1.27) = (1) A.Letting in (1. 3), P be HI and Q be H (H regular in V) we have that e e e ', I (Vx Vx H) H)dV = x(Vx ) + (Vx H) x H dS. (128) V S+S'+So By (1.26) and the dyadic identity (1. 10), the integral over S becomes IS nx(VxH) He d S. (1.29) S The integral over S' is evaluated as in the previous section yielding IS =-V'xH (R). (1.30) Similarly, the integral over Sio vanishes in the limit as R * oo provided |Vx (RH) I < C as R -n. (1.31) Collecting these results we can state the following theorem: Theorem B. If V is the volume exterior ito a closed, bounded, regular surface S and H is a regular function of osition in V and on S satisfying the regularity condition |Vx (RH ) | < co, as R -oo, then H satisfies the integral equation

17 V'xH (R)= - ((VxVxH) H * H)d x(x) dS (1.32) S where H(e is defined in (1. 24) - (1. 27), and n is the unit normal on S directed out of V and into the interior of S.. At this point we conclude Chapter I, the main results being Theorems A and B. The integral equations (1.23) and (1. 32) will be subsequently employed to find integral representations for the scattered field defined in (1) - (4) of the Introduction. Our immediate concern, however, is the explicit representation of H and E), the dyadic kernels of these equations, in terms of e m potential functions. This we proceed to do in the next chapter.

Chapter I THE FIELDS OF INFINITESIMAL DIPOLES AND THE SOLUTION OF THE TWO DYADIC PARTIAL DIFFERENTIAL EQUATIONS As mentioned on p. 7 of the Introduction, the problem of finding explicit solutions in terms of potential functions for the dyadic kernels H and E e m can be dealt with either from a mathematical point of view (i. e. without taking recourse to the physical significance of the dyadics) or by recognizing the relation between these dyadics and the fields of static electric and magnetic dipoles and proceed to determine them by utilizing the available knowledge on potential theory. We chose the second course of action. Let S be a closed, bounded, perfectly conducting surface immersed in vacuum. This surface separates the whole three dimensional space into two regions: the finite region Vi enclosed by S and the rest of the space V. It is, moreover, regular in the sense that it satisfies the requirements of Green's theorem: S e L2, where Lp is the class of surface whose equations have continuous derivatives up to and including pth order and whose pth derivatives satisfy a Holder condition. For later purposes, erect a rectangular coordinate system x, y, z with origin in Vi. Let J be the volume dyadic current density -iwt in a finite region of V. We let J have a harmonic time variation e which we suppress throughout this work. Then the electromagnetic fields in dyadic form satisfy 1) the dyadic Maxwell's equations Vx E = ikZH (2.1) Vx H = J -ikYE (2.2) Z = 1/Y, the free space characteristic impedance, 2) the dyadic wave equations Vx Vx E -k2 E = ikZ J (2.3) Vx Vx H -k2H = Vx, (2.4) 18

19 3) the boundary conditions nxE =0, n.H = OonS, (2.5) the unit normal on S directed out of V and into Vi. Moreover, these fields satisfy a radiation condition in which we are not presently interested. Except for the harmonic time variation we have left the current distribution J completely unspecified. We now turn our attention to two types of it, namely: J =-iki 6 (R R), (2.6) and J =-YVx (R R') (2.7) where I is the identity dyadic defined by =A A A ^ 1 I = aa a2 2+a3a3 (2.8) The current distribution in (2. 6) is that of three harmonically oscillating infinitesimal electric dipoles situated, each along one coordinate direction, at the point R' of V and of dipole moment & 1^ Pei= a, i=1,2, 3, (2.9) ei c 1 c being the speed of light in vacuum. Similarly, the current distribution in (2. 7) is that of three harmonically oscillating infinitesimal magnetic dipoles situated, each along one coordinate direction, at the point R' of V and of dipole moment Pmi= -Y a., i=1, 2, 3.(2.10) 1 Let now E and H be the fields due to the current distribution (2. 6) and e e expand them in a power series of k: 0o E ==(n) n =(n E = (ik) E, H = y He. (2.11) e e ' e / e n=o n=o 1 AA4 AA A For convenience we use al, a2, a3 instead of x,y, z for the rectangular unit vectors.

20 Substitution of these equations together with (2. 6) in (2. 1) - (2. 5) leads to the following relations =(O) VxE 0= (2. 12a) e (n+l) =(n) Vx E =-ZH n=, 1,2,.. (2. 12b) e e VxH(O) = (2. 13a) Ml) Af A =O VxH = -I6(RRI') -YE (2. 13b) vxH(+)= -YE n= 1.,2, 3,.. (2. 13c) e e =(0) VXVXE =0 (2. 14a) e VxoxVE - 0 (2. 14b) e =(2vx ) =(2).(0) =. VxVxE ~E =IZ6(RRIi) (2. 14c) e e =(n+2) (n) Vx Vx E +E =0, n=1, 2, 3,... (2. 14d) e e =(l) VxVxH = L R (2. 15a) Vx Vx He -vx 6 (R Rii (2. 15b) Vx Vx Af(n+2) (nL=0 jn=0, 1, 2,... (2. 15c) e e A (n) A (n nXE =( n O., n=Orl, 2,..., on S. (2.16) e e Repeating the procedure for the fields (~Kn=(n) =(n) m 7 H 4(ik) H (2.17) n=o of the current distribution (2. 7) we obtain =(O) Vx E 0 (2. 18a) m (n+1) =(n) Vx E =_ZH n = 0,1, 2... (2. 18b) m m

21 Vx H = -YVx 6(R I R' (2.19a) m Vx l)= Y E() n = 0, 1, 2,... (2.19b) mn m VxVxE(0) =0 (2.20a) m VxVxEm) =- vx 6(RR'1 (2. 20b).VXVXE Vx 6 (2. 20b) (n+2) (n) VxVx E +E = O, n=0,1, 2,... (2.20c) m m (1) VxVx 0 -Y VVx Vx 6 (Rl R1 (2. 21a) Vx VxH( = 0 (2.21b) m =(n) =(n) _ VxVx H + = 0, n=0, 1, 2,... (2. 21c) A =(n) (n) nxE =0, n H = 0, n=0,1, 2,..., on S. (2.22) m m At this point we offer some relief to the reader by saying that, of all this multitude of equations, we are only interested in those involving He and E. e mn These two dyadics are directly related (if not identifiable) with those defined in Eqs. (10) - (14) of the Introduction: By (2. 15b), H can be written in the form (10); by Eq. (2.13b) and the boundary condition (2. 16) on E, it satisfies (13). (1) e Similarly, by (2. 20b), E can be written in the form (11); by (2. 22), it m -'1) =(l) satisfies (13). Our next step is to find explicit forms for H() and E() in e m terms of potential functions. Subsequently, we will verify that these solutions satisfy the regularity condition (14) of the Introduction and, therefore, they qualify as kernels of the integral equations (17) and (18). In effect, we will have shown that the kernels of the integral equations can be obtained from the solutions of electrostatic and magnetostatic problems.

22 The Dyadic H in Terms of Potential Functions e Let H-(1) - — H(1) (2.23) e | irrR-R'|u r where VxVxHe =0. (2.24) r =(1) The dyadic H ) satisfies Eq. (2.15b), namely VxVxH Vx 6 (R\R') (2.25) and is related to E() by (2.13b). Equation (2.12a) permits us to write E ) e e in the form 3 E(0) (a~)i. (2.26) i=1 Substituting this expression in (2. i3b) and taking the divergence of the resulting equation we have 3 V2 a = * V 6 (RiR'). (2.27) el 1 i=1 We wish then to find solutions to the problem V 0ei = a.-V 6 (RlR', i=1,2,3. (2.28) By virtue of (2. 16), these scalar functions satisfy the boundary condition n (0)2. 29) nxVO. = O onS, i=1,2,3. (2.29)

23 For the total field (0) we can write ei (0) A 1 (o) ei av -i(RIR'), (2.30) el _ eir-' where v i (RIR') =. (2.31) r Stokes' theorem together with the boundary condition (2. 29) implies that (ei) is a constant on the surface S. From (2. 30) we can then write (0) A - - __ _i (Rs ) VE 4| i sVR-i)2+C R E S, (2.32) r 47r JRs-R where C,i a constant, is the value of (ei on S. We now employ Green's theorem and write (0). (0) e) eir ei (Rs R) an G (R, R ) dS, (2. 33) (e, where G(e) is the exterior Dirichlet Green's function for S: (e) M - G (e) i +G (RIR'), (2.34) 47r R-R' with V2G( )(RIR') =0, in V (2.35) and RsG( R')I= 0, Rs ES. (2.36) Substituting.(2. 32) in (2. 33) we have

24 ~(0) elB,)(ai V ') I, + f Ge(R)slR) dS + i(RR =(iV ) L 47lR R J anS eiei +Ci a G (R slR)dS =-(ai V')G( (RI R')+Cei an G (RsR) dS (2.37) S By a simple application of the divergence theorem we find that 1 a L 1 dS = 0; (2. 38) therefore, i (RR -(ai V')G((RI R')+ Cei n Gr ((Rs R) dS (2.39) ei i r Ci r r r S S Substituting (2.39) in (2.30) and taking (2. 34) into account we have (0) (aV,)G(e)(R )+c F1 G(e)(RR)dS (2.40 -()G Rl R+C G R) dS (2. 40) ei (ieiJ an r In order to determine the constant C'. we employ the relation el fnsVs ( (RR') dS = 0. (2.41) eirrs S This is a consequence of (2.13b) and Stokes' theorem. It is a mathematical statement of the physical fact that the total induced static charge on the perfectly conducting surface must be zero. Substituting (2. 39) in this expression we have that

25 A a ( e) -" —. (a - V)J G (Rsl R)d S ____an______r C e el = = (ai V) D(R') (2.42) fdT - S ds an (e)(. -I ) fdS G (RI S s. where an Ge) (Rs R') dS De(R') = fdT fdS Ge)(R T) anT r s is r (2.43) The electric field dyadic (2. 26) can then be written 3 Ee(RR')=Z OV V)G)(RR-C dS a- G(e)(R e ei an r s i=l S S or by (2. 42) e a n r r and if we let 0)(RR) = G(e)(RlR) -D (R') dS G e(R R), 1 an r s[R) then E(RlR')= Z V V, (o) (R R') e (-e e A ai, (2.44) (2.45) (2.46) (2. 47) In passing we note that this is the electric field due to three orthogonally crossed electric dipoles with moments defined by (2. 9). Now that we have determined E(0) we return to (2.13b) and substitute our result in it: e Vx() - I 6 (R|R') - VV (~) (2.48) e e

26 Substituting (2. 23) and (2. 46) in this expression and employing the identity dyadic VxVx (I ) =VVq -I V2 p (2.49) we obtain wvv E i =IV E i+vxH2 = e r 4v|R-R L 47rIR-Ri' + rVx = -I 6(R R')- VV'G ( RIR)-D (R')dS - G (RSI R. (2.50) (e) By taking into account the definition (2. 34) of G and that vE 2 -- l = 6(RlR) (2.51) 47r I R-R' and v II 4I-R]E ].VI (2.52) 47r I R-RI ' 4lT1R-R.' Eq. (2. 50) reduces to VxH()- v () (R R') -D (R') JdS an G (R, (2. 53) S (0) (2.54) r where er (RIR) is the regular part of I( 0(R R'). This dyadic equation can be broken into three vector equations of the form VxHe i = - (ai. V) V e (2.55) r r From (2. 40),'(2. 42) and (2. 46) we can relate 0(0) to () as follows: ''ei e (0) (0) (2.56) ei= VI V)e 2.6

27 and consequently, (0) A (0) ei (ai.l) (2. 57 r r Equation (2. 55) can then be written -(1) (0) VxH. = V0. (2. 58) ei ei r r Stevenson (1954) has shown that the necessary and sufficient conditions for (2. 58) to have a solution are V2(O = inV (2.59) ei and (0) fi Vs dS. (2. 60) sr S (0) But ) was constructed to satisfy Laplace's equation and (2. 60) is automatier cally satisfied if Cei is chosen according to (2. 42). According to Kleinman (1965b), a particular solution, He ii of (2. 60) may be cast in either of the eir following forms: -)1p,i ei Hl)P(I')= VxdV + r 47r J R 1 & - VSei V. 1 1a where Vi is the volume interior to S, or (-0) () Ne ( Ni)- 1 SA (~p (R I H& 1 e(R I R') -N ei(R) n 1 1' L eir s ei n s HeiRR = V. (2.62) r I R|-Rs

28 (i) A The function N. (R), R interior to S, is a potential function satisfying ex V2N( (R)= 0, R interior to S (2. 63) ei with A (i) A (0) --- n.V N(R )= n V (R I R'), Rs E S. (2. 64) s s ei s s s eir This is a standard interior Neumann problem and has a solution provided that A (i) n V N (Rs) dS = 0, (2. 65) S which is satisfied by virtue of (2. 64) and (2. 60) The complete solution of H can then be written: H()= Vx E -L + VNei ai (2.66) e e el 1 47r[ R-R' r i=l (e) where VN(e is a solution of the homogeneous part of (2. 58). The functions (e) ei Ne. are exterior Neumann functions and can be uniquely determined from the boundary condition (2.16) on H 1): e V2N(e) = 0 inV Vel (e) N(. regular at infinity (2. 67) ei A Vn -(e xA ___ _ _. on S s S ei S - s e S

29 The Dyadic E in Terms of Potential Functions Let E EVx- I +E, (2.68) with VxVxE() = 0. (2.69) m r The dyadic E satisfies (2. 20b), namely VxVxE )=-VxE 6(RlR, - (2.70) and is related to H ) by (2.18b). From (2.19a) and (2. 21a) we can write =(0) m H in either of two forms: Either m H()= -Y I 6 (R | R') + H() (2. 71) or m m or = YVxVx -E —1 +H0, (2.72) 4r |R-R~- " r where Vx0H =0 andVxVxH(0) = 0. (2.73) r r From (2. 18b) we see, however, that if we are to be consistent with (2. 68) we must choose H( ) according to (2. 72). Thus substituting (2. 68) and (2. 73) in (2.18b) we obtain =(i) -(0) Vx E = ZH. (2.74) The first of (2. 73) permits us to write H( in the form H Mr = -Y ( V mir) ai (2. 75) i=l 1 and by taking the divergence of (2. 74) we see that V20 = 0, i=1,2, 3. (2.76) ~mir(.6

30 From the boundary condition (2. 22) on H(0) and (2. 72) we can write mn A F I A Y. VxvSx - - ] +hs 4T IRs-R' I M() H(m= 0, onS. r (2.77) But VxVx -,I IVV4 1 = -v2 1, 47rR-R L 47r R'I]- 4v [ i-R] By (2. 75) and (2. 78) the boundary condition (2. 77) becomes (2. 78) A n 4 4VV 4|Rg-R -I 3 -ns' Vs ( 0. ai i0 i=mi i=1 (2. 79) or A 1 n V V s s s R\ -R L \ 4 R S-R'| S ~S =0, (2.80) or $ n.V s ^ V) (- )i -0 aiO 0 (2. 81) The last relation gives us a ans ) (RS R') mi r a 1 an _ (_ (a, i=1, 2,3. s 47rw Rg-RI/I (2. 82) We now employ the scalar Green's theorem and write (0m) (IR') m r '0 -% r (e) (Rs R) r I, an5 dS s (2.83) where N() (RR') is the exterior Neumann Green's function for S: N(e)(R ') - 1 + N(e) (R|RI) 4r + R- IR') 4wIR-R'I (2. 84)

31 with V2 N(e)(R R) = 0, in V, (2.85) and an N (RSI R) =0, Rs e S. (2.86) Substituting (2. 82) in (2. 83) we have (0)- (e) M(i 1V s (RF. raa. i- ldS r,gL 4~R-R 11 S = V') N(e)(RRI '). (2) (2.87) 1 r Equation (2. 73) then becomes 3 =() '0) A (e) VX = - VPmir) ai=- W N (R R'). (2.88) The necessary and sufficient conditions (2. 59) and (2. 60) for the equation Vx E( = -_ V (2.89) mi mi r r (0) to have a solution can be seen to be readily satisifed: 0mir satisfies Laplace's equation (2. 76) and it also satisfies the condition n V5 dS = (2.90) S by virtue of (2. 82) and (2. 38). According to Kleinman (1965b) a particular L( l)p solution, E. ( is ml r It is interesting to note that substitution of this expression in (2. 76) and then in (2. 71) gives the total magnetic field of three orthogonally crossed static magnetic dipoles of moments defined by (2.10) as H=( -Y VV'N(e)-YI6(RI R ).

32 [r() (R I R) n (i) )-] The function N.(R), R interior to S, is an interior Neumann function satismi VN (R) = 0 R interior to S (2.92) with N ( 0) (0) nsVsNmi(Rs)n8 (R s R), Rs S. (2. 93) r This is a standard interior Neumann problem and has a solution provided nVsNi (Rs) dS.0 (2.94) a condition guaranteed by (2. 90). The complete solution of E can then be written m where V G. is a solution of the homogeneous part of (2. 89). The functions (e) ml (e) mi (1) boundary condition (2 22) on E I mi mi m i i=2,. (2.97)in (e) G' regular at infinity (2. 96) n xVG =-ni X V xa-i --- -- -n x E on S S s Mi S S 47|R-R1/ OnS mr (e) To completely determine G. we employ the additional condition ns [S + G. d = 0 S, =1.1.2,3. (2. 97) S

33 This condition arises from (2. 19b) and Stokes' theorem over a closed surface. Since Emir is the curl of another vector (see 2. 91), by Stokes' theorem again, its normal component integrated over a closed surface is zero. Equation (2. 97) then becomes AI n V G(e dS = 0, i=1,2, 3. (2. 98) S At this point we conclude Chapter II. In summary, we have derived explicit expressions for the dyadics H() and E, defined at the beginning of the chapter, in terms of potential functions. These dyadics satisfy Eqs. (10) - (13) of the Introduction. They also satisfy the regularity condition (14) as shown in Appendix A. It is of interest to note that in order to determine the two dyadics we employed more boundary conditions than those specified in Eq. (13). Specifically, for H( we used (2. 16) (n H 1)= 0 ) and (2. 41); for = ) A =used (2. 22)(n H(0)e E we used (2. 22) (n H = ) and (2. 98). The question then arises m m whether the boundary conditions (13) together with the regularity condition (14) of the Introduction determine uniquely the two dyadics, as specified there. If the dyadics are determined uniquely, one should be able to show that the two boundary conditions, together with the regularity conditions, imply the additional ones mentioned above. This question has not been answered as yet. The only statement we can make (from Appendix A) is that the boundary condition (2. 41) implies the regularity condition |R3 (VxH1) <oo asR-oo

Chapter III INTEGRAL REPRESENTATIONS OF THE ELECTROMAGNETIC SCATTERED FIELDS In this chapter we employ theorems A and B of Chapter I to find integral representations for the electric and magnetic fields scattered by a perfectly conducting surface. We start by defining the geometry of the problem and the properties of the scattered fields. In the three-dimensional free space (vacuum) we have a closed, bounded, perfectly conducting, regular surface S which separates the whole space into two regions: the finite region Vi enclosed by S and the rest of space V. A time harmonic source of electromagnetic waves is located in V and its electric and magnetic fields are denoted by E1 and H, respectively. The time dependence e-iwt is omitted. The presence of the perfectly conducting surface S gives rise to an electromagnetic wave whose electric and magnetic vectors we denote by E and H respectively. These vectors satisfy 1) Maxwell's equations VxE= ikZH, VxH= -ik YES (3.1) Z = Y, the free space characteristic impedance, 2) the homogeneous vector wave equation Vx Vx[ } -k2 s -= 0 inV, (3.2) a consequence of Maxwell's equations, 3) the boundary conditions Aix5= -nxE, n* = -no H on S (3.3) where Ai is the unit normal on S directed out of V and into Vi, and 34

35 4) the radiation conditions lirn -Nx s -% R- x(VxE )+]ikE, uniformly in R, (3.4) lir a RS S - RLx(Vx H )+ikH = 0, R being the radial unit vector and R the distance from the origin of a rectangular coordinate system with origin in Vi to a point in V. Our intention is to substitute E and H in the integral equations (1. 23) and (1. 32) of Theorems A and B, respectively. In order to do this, however, we must show that JVx(RE ) < o and |Vx(RH) < oo, as R -> oD. This follows from the following expansion theorem by Wilcox (1956): Theorem C Let A(R) be a vector radiation function for a region R > c where (R, 0, 0) - -& are spherical coordinates. Then A(R) has an expansion ikR A (0O, A(R)) = n (3.5) R. R n=o which is valid for R > c and which converges absolutely and uniformly in the parameters R, 0, 0 in any region R > c + e > c. The series can be differentiated term by term with respect to R, 0, and 0 any number of times and the resulting series all converge absolutely and uniformly. It immediately follows from this theorem that Vx(REs) < o and |Vx(RH) |I < oD, as R w o. (3. 6) Letting then E and H stand for E and H of Eqs. (1.23) and (1. 32), respectively, and at the same time, employing (3. 1) - (3. 3), we obtain A vector radiation function is one that satisfies Eqs. (3. 2) and (3. 4).

36 -is 2CS (l)r A L =(l) ikZH (R')=-k E E ) dV- (nx E) (Vx E) dS, (3.7) V S and E (R') =-ikZ H( dV- xE H( d S. (3.8) e e V Equation (3. 7) can be written in a better form. By Eq. (2. 68) Vx E VV)+Vxm, R R', (3.9) mR-Rl r and by (2. 81) Vx - (- 41-1') V- N- ('r1) 4 R-RI = - W'Ne(RIR') ', R f R ', (3.10) where N( is the exterior Neumann's function for S defined in (2. 84). Employing the identities (a x ) * A = * (b x A) and Vx (b) = (Vx a)b -ax Vb we have (nxE ) (Vx E() )=.Ex (VxE = -n * x VV' N(( =- [(Vx EV' Ne)-Vx(EV' N(e) (3.11) The second term of this last expression vanishes by Stokes' theorem when integrated over the closed surface S. Employing (3. 1) in the first part we have that S (nSxE) (VxE( ))dS= -ikZ (n. (12) S S

37 Substituting this result in (3. 7) we have that HRI) =ikY E f (1)3. 13) s(R kYf. 1 dV+Vj (n. H)N (Rsj R') dS. (3.13) V S We can then state the following theorem. Theorem D The fields E and H defined in (3. 1) - (3. 4) satisfy the integral equations H-(R') = ikY E a E+ dV + V' H )N() (Rs R') dS (3.14) V S and E (R')= -ikZi H (V- (n (x). H(1) dS (3. 15) e e V S where E is defined by (2.95), H(1) by (2. 66) and N() by (2. 84)-(2.86). m e Equations (3. 14) and (3. 15) constitute a system of two coupled integral equations for the scattered fields E and. They can be written in operator iorra by deiinirg L1=ikYfE ]'E ()(RIR')dV, L2= -ikZf '] H()(RR') dV V V (3.16) and (n H (s| RI)dS, E(o) - (nxE) H (R|R') dS (o) s(o) j e d S (3.17) With these definitions, (3. 14) and (3. 15) becomes S -.s -.s H =L E +H o (3. 18) and E =L2H +E(o) (3. 19)

33 For the wave number k sufficiently small we can solve these two equations by the method of successive approximations. We let H(0) and E( be the first approximation to H and E, respectively. The first correction to this solution is (1) 1 E(0) +H(0) (3. 20) E5 =LH +E5 (3. 21) (1) 2 (0) +(0) and the second one (2) 1 (1)+H (0) 1 2 (0) 1E (0) +H(0) (3.22) E =LH +E =L L E +L H O+E? (3.23) (2) 2 (1) (0) 2 1 (0) 2 (0) (0) (3. 23) and so on. In this manner we generate two sequences of functions HN) and { E(N) }, which we must show to be convergent for a certain range of values of k and, also, that they converge to the desired solutions, i. e. S lim fS -s- lim S (3.24) HN-o (N) E -N-o)o E(N) (3.24) We strongly suspect, however, that this approach would lead to divergent volume integrals quite early in the process. The reason for this is the following: The incident fields of the zeroth order iterates (3. 17), be they dipole fields or plane waves, are independent of the primed coordinates. This fact together with our knowledge of the nature of N) and H from Chapter II e 'S -&.S -l bl leads us to the conclusion that H(0) (R') and E(0) (R1) do not contain the exponential ei. In fact, by a simple inductive argument it can be shown that none of the iterates contain the eikR' as a factor. From (3. 5), however, we know that the scattered fields should contain this factor. We are then led to the conclusion that eikR' appears in the iterates expanded in a power series in k. As the iteration proceeds there will come a point when positive powers of R will appear in the volume integrals of the operator and these integrals will diverge and this is precisely what happens when applying Stevenson's method.

39 To avoid this difficulty, we must remove the troublesome eikR factor from E and H Accordingly, we propose to use the following vector functions in the integral equations -ikR -'s - -ikR -s e =e E and h= e H (3.25) The motivation for doing so lies in (3. 5) of Wilcox' expansion theorem. From this equation we see that, at least in the region where the expansion holds, the ~ -ikR new fields e and h do not contain the troublesome factor e. Our next step is to rewrite the integral equations (1. 23) and (1. 32) in terms of e and h. That these two fields satisfy the regularity conditions | Vx(Re) <co and jVx(Rh) | <o, as R - o, (3.26) so as to be admissible in the integral equations, is obvious from their definition and Eq. (3.5). In terms of e and h Eqs. (1.23) and (1. 32) are written tdV + x~~ (Vx`E(1) V'xe(R') = - (VxVx)e E(1 dV +(txe) (VxE~ )dS, (3.27) j m m V S and V'xh(R')= - (VxVxh) H)dV+ x(Vx H d S (3. 28) e e V S A. From the definition (3. 25) the functions e and h satisfy: By (3. 1) Vx = ik (Zh -Rxe), (3.29) Vxh = -ik (Ye+Rxh); (3. 30) and by (3.2) VxVx = k2(e+ ZRxh) -ikVx(Rx), (3. 31) VxVxh2 YRx ikVxRxh. (3. 32) VxVxh= k (h-YRx e) -ikVx (Rxh ). (3.32) Substituting these expression in the integral equations we obtain

40 ik h(R')-R'x e(R') -k +ZRxh) E ) dV + V +ik x(Rxe * () dV+(nxe)* (VxE() ) dS, (3.33) J m J V S and r A - 2 A (1) -i e(R')+Rxh(R' = -k | (h-YRxe)* H' dV + J e V +ik [x(Rxhi H dV-ik x(Ye+Rxh He dS (3.34) V S The second of the volume integrals in each of these expressions may be written in a different form. We start with that of (3. 33): By the dyadic identity V- (a xA)=(Vx a) ~ A-a- (VxA) (3.35) we have AdV+ R A (2) f [vx(RX3] Et dV=f V [RX)x E( )JdV+ (Rxel) (VxEM) )dV. V V V V (3.36) By (2. 66) and (2. 88) and application of the divergence theorem to the first integral on the right, (3. 36) becomes: J[V RXj E )dV= [(nRxe)xE( dS-fI (Rxe )' W N(e)dV. (3. 37) V S+S' V A (1) Using the boundary condition n x E = 0 on S and the identity V- ( b) = (V. a)b+ -Vb (3. 38) ) In the discussion that follows we will use the notation of Chapter I: S' denotes a spherical surface over the singularity at R' and the volume enclosed by this surface is excluded from V. Integrals over the surface at infinity (Sw), will be omitted if it is clear that they vanish. A knowledge of the properties of the dyadics and the notation of Chapter I will also be assumed.

41 we can write (3. 37) as follows: f [Vx (xRxj E dV=f n. [(Rx^xEldS- VRxe)Vt dV+V (RxV dV= V St V V =n. (xe)xE dS- f. Rxe)V'N( dS+ [e Vx R-R Vx eV'N dV -ikJ R.L zhReV'N() dV (3.9) V where, above, we made use of (3. 29). Notice now that the part of the integral over S' involving the regular parts of E() and N() will vanish in the limit. We then write - (Rxe) V rR d + I xe) (d - S-1S VA A (e) -ik zf h -Rx hVN' dV.(3.4) V omitwit. The form of (3. 40) with the integral over S ' evaluated then is rr(l). _A r FA(1) -n(e) *' V J LVx(Ii e m t dvh - re xe(Rtp)+Rrt(so ndVN dS-wikZ ai in h VlNemdS.W V S V (3.41) (3. 41)b

42 Substituting this expression in (3. 33) we have that ikZh(R')= -k (e+ZRxh) * E() dV+k2 ZV R hN(edV + V V +ikI Re nxe)V'edS+ (nxe) * Vx E()dS. (3.42) S S According to (3.11) the last integral above can be written (nx) * Vx E(dS = - (n Vx e) V'N(e)dS (3.43) S S and by (3 29) (nxe') Vx E( dS= -ik n (Zh-Rxe) V N(S = -ikZ n. h) V'N dS - SS S -ik R* [(nxe) V' N( dS. (3.44) Substituting this last expression in (3. 42) we have that h(R')I=ik (Ye+Rxh) Em dV-ikV' R. hN(dV - Vf. h)N(e dS. V V S (3.45) A. This is the first integral equation for h and e We now proceed in the same manner as above to modify the second volume integral of (3. 34). The result is e - ] e VS r+ikY f(Rt) j O)dV, (3.)46) +ikY VI (R'e) edV, (3.46) V

43 where qp( is defined in (2. 46). Substituting this expression in.(3. 34) we have -ikYe(R') -k I (h-YRxe) He dV-kyV' (R V) dV - V V -ikV V n (Rxh)) dS-ikY ( anxe) * H(1) dS. (3. 47) S S But on the surface S (0)= -D (R') so that e e A 0) d eA A,) A A (Rxh)/ dS=-D (RI) nV (Rxh)dS=-D (R') VO (Rxh)dV = ' e e -S S S A rA A A = D(R')I R VxhdV = -ikYD (R') | R edV.(3.48) V v Substituting this result in (3. 47) we have e(R) =-ik (Z-Rxe) +D HdV-ik + (nxeo ' 1 dS e e e (3.49) This is the second integral equation for e and h. Employing then the relations in (3. 25) we can state the following theorem: Theorem E The scattered fields H and E defined in (3. 1)-(3. 4) satisfy the integral equations r Hs(R,)=ikYeikR -ikE R). E( (RR)dV+ m tikekR e LH R) E (R R) - H ( vN (R R dV V r -ikR + ikR' V'e n H(R se) R') dS, (3. 50) S

44 and E (R)=-ikZeR e- H (R)H (RlR) dV + + ikeR ei EtlSRx(} H(e)(RR )-R E s(R)] vi(LI + e J e-1 ~l =>i --,r(o ) -* ( -eikR e x Ei(Rs)] H(1 (R R') dS, (3. 51) S, (0)by where E() is defined by (2.93), (1) by(2 64), N(e) by (2. 84)-(2. 86), e) by m e e (2. 46) and D by (2. 43). e Though these equations give directly the scattered fields it may prove more convenient to work with Eqs. (3. 45) and (3. 46) for the vectors h and e rather than the above. These equations may be solved in either of two ways for small values of k. We can either iterate the equations to form two sequences of functions which, hopefully, converge to the actual functions e and J. h or we can expand these fields in powers of k of the form ek A. n A(ik) h= (ik) h (3.52) en n n=o n=o and then substitute them in (3. 45) and (3. 46) to find a recursion formula for the coefficients. That e and h can be written in the form (3. 52) follows from (3. 25) and a result by Werner (1963), who showed that the scattered electric field E tendS, as k -* 0, analytically to a corresponding electrostatic field. When we use iteration to find e and h we let h(n=- V' ) Ne dS, (3.53) S S

45 be the first approximation to h and e, respectively. The first correction to this solution is. = () (e) h() =ik (Ye(0)+Rxh(0))E dV-ikV' h) N dV+h(0 (3. 54) ) j (O) (0) m J (0) 0 )' ) V V and the second h = +ARx =(l\ eA) h(2) = L)Rxhl) E( )h NV dV-+h( (3. 56) V V AW A Al)( A (0) e(2) =-ik(Zhl) -Rx ))H dV-ik V (R )e +DdV+e(0) V V (3.57) and so on. In this way we generate two sequences of functions, 1 hL and e(N) which, we hope, for a certain range of values of k, converge to A., h and e respectively as N-* o. On the other hand, when we use the low frequency expansions (3. 52) we proceed as follows: First we expand the known surface integrals in power series of k - V (n.h)N dS= (ik)n ( () dS= ik) (3 58) v ( ) j) m n 1 Jne )dS / n l S n=o S n=o and then substitute (3. 52) in (3. 45) and (3. 46) and collect coefficients of equal powers of k to obtain the result hv(R') = v, (3.59) A. -n i Ay A ga stwo A su (e AI h (R Ye +xh E dV-V Rh N(e)dV+f n= ri f (3.60) n Jn n Em n nl 1,. V V (3. 60)

46 e (RHt=.,, (3. 61).0 e 1(RI) = -f Zh -Rxe )h"H("dV-V f(R9 ) +D2 dV+g~1 V V n=O, 11,..a. (3. 62)

Chapter IV AN EXAMPLE: THE SPHERE In this chapter we apply our results to the problem of scattering of a plane electromagnetic wave by a perfectly conducting sphere. First we will employ the results of Chapter II to determine the dyadics H and E and e m then the results of Chapter II to determine the first two terms in the low frequency expansion of the scattered fields. The sphere is of radius a and its center coincides with the origin of a rectangular coordinate system (x, y, z) (see Fig. 2). Using the notation of the previous chapters, Vi denotes the volume of the sphere while V the rest of space. The surface of the sphere is denoted by S and the unit normal,, on it is directed out of V and into Vi. The plane wave propagates along the negative z-axis with its electric vector polarized along the positive x-axis. We shall use both the above rectangular coordinate system and its related spherical coordinate system (R, 0, O). z H E V y x FIG. 2: GEOMETRY FOR THE SPHERE PROBLEM. 47

48 We start with some expressions we shall be using rather often: The expansion of the free space static Grecen's function in spherical harmonics is 1 1 (n-m)! mn< - -R - - 2E f Pn (cose 1)Pncos)cosm(-') 447 m (n+m)! n n n+I n=o m=o R (4. 1) where R<=min (R, R'), R>=max(R, R'), and m is the Neumann factor: EO=l, e =2 for m=1, 2,.... The functions P are the associated Legendre functions m n defined by (n (n )(l-x2) m 2Fl(l+m+n, m-n;l+m; -x n m m! (n-m)! 21: 2 -1 <x<1. (4.2) This definition is according to Magnus et al (1966) and all the contiguous relations for these functions that we shall subsequently use can be found there (p. 171). The regular part of the exterior static Dirichlet Green's function for the sphere as defined in (2. 34) - (2. 36) is given by (e) 1 n M2n+1 (e)_ 1 ~-' (n-i)! m m a ) n P(cose) Pr c~s cosm(- ) R-n-in - r 47r c m (n+m)! n n (os)cos( n+l n+ R' R n=o m=o (4. 3) while the regular part of the corresponding Neumann function as defined in (2. 84)-(2. 86) is given by 0o n 2n+l (e) 1 - '(nn-m)P n __m p na N - P (cos0')p (cs 0)cosm (-0' ) r 47r m i n+l (+m)! n n c i RPRn+1 n=o m=o (4.4) We are now ready to proceed with the determination of the two dyadics.

49 (1) The Explicit Form of H( e First we determine the constant De(R') of (2. 43). From (4. 3) - 1 a (e) G an G (RJs R)dS= -fTR G( (RSI R) dS = S S 2 Z (n+1) (n+-)! Pn (co01) a| a2sinOsdOsd2s x 4n+ jnJ(n-m)! s s n=o m=o O 0 m a x P (cosOs)cosm(0 s-0)= R' (4 5) a (e) (s T. JdT a dS an G (RR = 47 a (4.6) S S Substitutingthese two results in (2. 43) we have that L 1 D(R )4+= (4.7) By (4. 5) the regular part of q0) as defined in (2. 46) is e r e,iO(lt) = GJ)RIR a ' (R|RI)-Gr (R|RI), (4.8) 47R'R (. r (e) where G) is defined in (4. 3). By (2. 57) -ei V R - 41 2R'(4. r A A A A A A where al, a2, a3 stand for the rectangular unit vectors x, y, z. In order to determine the particular solution (2. 62) of (2. 58) we need to determine first the interior Neumann function defined in (2. 63) - (2. 65).

50 By (2. 64) A (i).ah a (0) A nVN R) --- =-ai V') s ei s aRei ai 47r R=a n= 1 n Ze (n+l) x M=m m=o x (nm)) n-1 x (n-m)! pm(cose')pm(cose)cosm(o-0')., i = 1, 2, 3. (n+m)! n n (4.10) The interior Neumann function that satisfies this boundary condition is n(- 1 (- m n+1 (n-m)! os)cosm R ei R)-r ai. / m n (n+m)! n n+1 ei 4RI 1 n Lmn n=l m=o RE Vi, i=l, 2, 3, (4. 11) where v. is an arbitrary constant. We now form the difference el ( ) (R N(iR )= — (a V') e 2n+l (n-m)! os) x 'eid ei s 47r ai. v (n+m)l n n=l m=o m an x P (cosO)cosm(-') an+l- ei, i = 1,2, 3. nR' (4. 12) Before proceeding to evaluate (2. 62) we note that the fact that v.ei has not been determied is not distrbing. Since, by Stokes' eorem, determined is not disturbing. Since, by Stokes' theorem, n ff Vxns 1 Vx dS - = SV, x n = 0 J s 's (4.13) we see that the part of (2. 62) involving the constant vei vanishes. In writing (2. 62), therefore, we shall omit v. From (4. 1), (4. 12) and (2. 62) we have tei that

51 1)] ei! Am= (2n+l (n-nm)! (I-t)! n=l m=o It=o t=o n+> 7r 27r n+e x P (cos t')Ptcos) s) R + a2 sinOedOsds R P (coss) x n R R,+lRx+l s s s s n 2 x Pt(cose )cosm( 0') cost (0 -0), i=l, 2, 3 (4 xP; ( 4.14) In order to evaluate the integral we write the radial unit vector on the surface A of the sphere, R, in terms of its rectangular components, A A A A R =x sinO cos0 +y sinO sino +z cos0, S 5 5 5 5 5 (4.15) and then we employ the orthogonality properties of the trigonometric and Legendre functions involved in the integration. The final result is rather simple in form: HI1P(RIRI)= - - (aiS Vt) 0 2 L n (n+m) (coso) x n=l m=o 2n 9. x To Pn (cosO)cosm(_-0') Rn+ln- J i= 1, 2, 3 (4.16) We notice here that this vector is transverse to the radial direction. The labor involved in obtaining this result is substantial and it is rather fortunate that the operations in (4. 14) have to be performed only once and not three times (one for each i). We now turn to the determination of the exterior Neumann functions defined in (2. 67). From the nature of the boundary condition these functions have to be found one at a time. We will show how to find the first one and will just give the results for the other two. From (2. 67) we write

52 -nVN=e LV 4xr soS ocSRsoSs H 1s) Ln-R-RI 03 r =R* V -h. x(RsonO') cosO )csinS _OiO (nnm) n) + +, (~) n n Labl m sn oTj n-(4.17) (e) F nml) m(CSI Fromthi bondar coditonI cn be deeminedeeryher inVete byiseto or byCorause ofi Gren' theorem Nn)R~)- N seRIRIR) (e) el n. dS. (4.18) + - P (C~~On8 P (OO)S

CD ~ a) 'O) FLA C) CD CD~e CD If (D CD 0 z CD ' 'I x 0: I 0 01 -0 02 '-A +, z x __ o o 02 ii + _________ w! + Me 4 bA +1 -4-A 1+ 0 0 02 S __ x 0 0 02 x p '1 CD CD x . 02 I-i. + - + - +1 + 0 0 02 6 0 0 02 x Iti 0 0 H CD CD 02 Uly C,43 oll"% -1. tplw t-'s C, I.-I P-a CD0

54 (The Explct Form of E1) The Explicit Form of Em First we determine the interior Neumann functions defined in (2. 92) - (2. 94). By (2. 93) and (2. 87) and (4. 4) we have 88 Nmi( = - aR Pmir(R ) R=a 1 - = -4 — (ai V') 4 ri n0 n=o E n (nm) P (cose')P (cos)cosm-') x m=o n m) m=o n —1 a n+l RI, i=l,2,3, (4.22) from which we can write (i)() 4 Ai ) mi (aVi c n (n-r)! pm(cOsO,)Pm em (n+n! n ns(cos0)cosm(0-0') m=o n R -+ x + v,REV. n+1 mi' Vil RI x i=l, 2, 3, (4.23) where v. is an arbitrary constant. From Eqs. (2. 87), (4. 4) and the above mn result we form the difference 0n 1co 1 A '. Z - 4T (ai' o nto n. 2n+l M n+l m=o (n-m)! Pn(cos')Pm(oses) X (n+m)' n x cosm(Os-0') a - RI i = 1, 2, 3. (4.24) Substituting this result together with; (4. 1) in (2. 91) we have that

55 miQ (RIR))= - - a.V) Vx r (47t) Z n=o n+. m t a(cos) +R+ x P (cosO ')Pi(CosO) RIR n Go I 2n+l (n-m)! (Q-t)! x:m t n+l (n-mn)! (l+t)! m=o Q=o t=o 7r 27r J a2sinOsdO sd R P (cosOs) x s s s s n x. cosm(0S-0')cos t(os-0), i=l, 2, 3. (4.25) In this expression we have omitted the constant vmi of (4.24) since, as we explained through (4. 13),it does not contribute to the integration. Equation (4.25) is evaluated in the same manner as (4. 14) yielding the result R n R 47r i sinO e r n=l m=l m (n-m)! mos) x n+1 (n+m)! n ' 2n+1 o i a A 7 1 x P (coso)sinm(0-0') n+ + 0 e x R' R n m=o n+ 0 2n+l x (nm)! P (cosOe) o. P (cose)cos m(_-0') Rn+i n+l (n+m) n d i=' i = 1,2,3. I (4.26) Note the similarity between this expression and (4. 16). We now turn to the determination of the exterior Dirichlet function defined by (2. 96). These functions have to be found one at a time. From the boundary condition in (2. 96) and the preceding results we have that on the surface of the sphere e

56 A VG(e) A - 5 xerni -ns ( AF =- 4r Rx0I 1 I A A x V x x -n x 8 s 47r R - %Now 8 r n-I (- ) rMn-m)(n+rn). n (Csln (Cs LrrJm+10 M=O n-i a x l Rnl 1 co (n-rn)' m(0a)n-1 m(n+rn-1)! n n s s~ni~ryJ n+1 - 2 (0.VI) sinO S n m (n-rn)! n anir(~-' (n-rn)! ~lIfl(0?),nn( 8R n + + 0~nL m~o rn (n-rn)! Prn(cosO1) Pn-1 (CosO5)Cosj rn+l)05-rnoj.-nj1 + +1 2 n-i (n-rn),' mrn 0!mr (cs (okr-) -rnl (n+rn-1)!- n n-i RSJ Rn+i A o n-i 1 (n-rn)' m ~ d Pr(00)csr(-' -(x~~~~ ~ ~ ~ ~ an] csl)-P(oe)CS(-1 A teoernbyKelog (153 p.l 143m) alwsu toxpnd G s)i nasre rnl of shercal arronic ofthe orr 7) (e) 1 o G 1)E rn1l47r ~b m.r n=o rn=o ~~(oe A cosrn0+B sinrn0]. R~+ n.,rn n.rn (4.28)

57 from which we write n VG = -— Rx 'FE ~e P (cosO ) A cosm 3B sinm0i s ml 47r m de sn n+2 n,m n,m n= m=o s a s n= m= a A 2 IPm(c1os0) n+2 [A sinmna- fsB cos (4.29) We now equate corresponding vector components in (4. 27) and (4. 29), thus obtaining two expressions for the uiknown coefficients A and B n, m n, m Using the orthogonality properties of the trigonometric functions involved there we obtain the following. 2n+l A (n-m)' m+l a A = P (cosel) sin(m+l) -f n, m 2n (n+m)! n R' +l s 2n+l 1 (+! (cos) a sin(m-1)', n> 1, 0 < m<n, (4.30) 2n (n+m-1)! n Rn+l 2n+l B -1 (n-m)' pm- l(os) cos (m+l)' - n, m 2n (n+m)' n Rn+l 2n+l 1 (n+m+l)! pm- (cose') a + cos(m-1)p', n > 1 1 m < n. (4. 31) R' These expressions determine all the constants except A0 0. In order to determine this constant we employ the condition (2. 98). Transforming the integral over the surface to an integral over a spherical surface at infinity (by the divergence theorem) we have lim Ja (e) R- l -G dS=O, i=l, 2,3. (4. 32) R-Ioo E-R m i

58 Substituting (4.28), in this expression we find that if the integral is to vanish., we must have A0 =0. (4. 33) From this result and (4. 30) and (4. 31)., Eq. (4. 28) can be put in the form (e) 1(n-m m! m+l G RlRI).= -P (CosO')P (cs) x ml rnlmon n+m)! n n (OO 2n+1 x sin[(m+1)0.-mO?1a n+1 + 11 (n-m+l).'m 1 2n+1 ) x sin m-1)0-mO' n1 (4. 34) R'. R In a similar manner we find (e) — 1A 1 (n-in)! jn m+l G (RRIR)= ~ -2I a ii( Pm! (coset)P (COSO) m2I4In n (m- )! n (cs) n(oO 2n+1 x Cos Lm+1) 0-MO,1 a n+(435 P (cosO')P ((ose) si (-' m3n (~&s~n~m (n)! n n n~l 2n+1 aa n+1o n+10-o (4. 36) RI Noie) th iiaiybten these! the exrsin n h orsodn ones (4. 19) - (4. 21)

59 Equations (4. 1), (4. 26) and (4. 34) - (4. 36) completely determine the dyadic E defined in (2. 95). We are now ready to proceed with the scattering problem. m Scattering of a Plane Wave by a Perfectly Conducting Sphere at Low Frequencies As shown in Fig. 2, the incident plane wave propagates along the negative z-axis with the electric field vector polarized along the positive x-axis. Taking the amplitude of the electric field to be equal to one we write Ei A -ikz -.i ^ -ikz. E e ikz H -yYe (4.37) where Y is the free space characteristic admittance. We shall first determine the zeroth order iterates given by (3. 53). By (3. 25) and the boundary conditions. (3. 3) we have that on the surface of the sphere A ^- -ikaAi -i A - ikaAbi(J -( Substitution of these boundary conditions in (3. 53) leads to the following expressions for the zeroth order iterates (0) (R and S -ika - ' (RI)= -e (nx )e H (R R) dS. (4.40) X -ikaa(cos (0) e s By (4. 1), (4. 4) and (4. 37), the integral of (4. 39) can be written (R e ' dS 1 OD n 2n+l (n-i)! m an+ S 4 () - e P (cos(1) - no mNo" m n+l (n+m)! n nn+l? 27r r 2: -ika coseO x sine d sdose Pn (cos0 )sin cos m(0s-) O O(4 (4.41)

60 The indicated integrations are performed by using the expansion (Magnus, et al, 1966, p. 108) ikacos (-i) (2m+1)J 1 (ka)Pm(cos0), (4.42) m=o m+ - 2 and the orthogonality properties of the trigonometric and Legendre functions. The function J is a Bessel function of half order. The resulting expression is m+l/2 i A' n 2n+1 ) ( dS -i) -n+l J (ka) P (cos0') x f n=l en+ S2 x sin p1 a+2. (4.43) Performing the remaining operation in (4. 39) we obtain iYe -ika A- Co n+2 h -i)n 2a -R (2n+l)J (ka)Pn(Cos)sinn + + (0) ka VF n=1 n+jk)n Rn+2 n+2 0 (-) J p(coso)sin~ + 2n+2 sin (i) n+ laPn (cos0)cos n n+2 (4.44) n=l n+, R We now turn to (4. 40) which, by (4. 37), can be written e(0) (') e e + 0 sinic )-H) (Rh R')dS. (4. 45) S In order to perform this integration we split the vector integrand into its rectangular components and integrate component by component. The dyadic H is defined in (2. 66) and, for the sphere problem, it can be written in e

61 terms of (4. 1), (4. 16) and (4. 19) - (4. 21). Performing the dot product indicated in (4.45) results in expressions similar to those in (4.41) which can be integrated readily. We omit all these operations because of their great length and we only give the resulting expressions; thus, 00 n+1 ka ( A nT (ka)P (cos a + (0) 2k 1 n e+l n=l n OD n+1 /I.0.0 a J n(n- )J (ka) P (cos)+ 2ka n2 n- - R + 2 OD in+1 n 2 a 2 (i) j 1(ka)P n(cosO)cos2 n+1 n=2 n+- R F n+n+ 1 (ka)P (cose)cos2an+l 2ka 2 1 n i + n= n- n R-+ 2n+2 00 1 n+2 +i j (ka)Pn(cosO)cosa 12ka crx n 1 n n=1 n- R I OD n+2 ik ia~~T 3' (.)n~n 1 j k)2aa (_,)n n (ka)Pc(coss)cos On (4.46) (0) 22ka +x1 n~ = n+ 2 ika A L 1 2 a e e(0) (R) ' y= 2 2 1 j (ka)P n(coSO)sina~ nT+n l OD (1nc n1 n+2 a i (ka)Pi 1(co)sin )o n+1 n=1n+ n- n 23 -+i _i)Z i (ka) P(coso)cosO ~ 2ka o-yn 1. n -ka, ka 1(kI ~n, 1Oa) n ese+ 1 (4tn+(.47) n=1 n+ Z'

62 oD n+1. e ne (-i)n. z -1+' (ka)P1 (cosO)cosO an n= n+- R n1 1 n' a + - / (-i) nJ 1(ka)P (cos0)cos at + ka v2ka 1 n n+l n=l n- R o n+1 + F(_-)nj (ka) (coso) coso a + J2ka 3 1. n+1 n=l n- R Cr t_ Zi n (ka)P (cos)cos n+a a n=l n- R - a s -i)j l (ka)P (cose) coso a (4.48) (4. 48) n=1 n+ R Certain simplifications take place in the above three expressions, when the indicated differentiations are performed, by employing the properties of the Bessel functions involved. Nothing nearly as simple as (4.44) results, however, either in rectangular or spherical coordinates. From the zeroth order iterates h(0) and e(o) we can obtain the zeroth order coefficients h and e,respectively, of the low frequency expansions (3. 52). By (3. 53) and (3. 58) fo(R)e R) - k o h(l')(R) (4.5 o o k —o (0 ' ( k- o (4.50) go(R) ~o(R) k-o e(o)(R) In order to calculate these limits we expand the Bessel functions involved in (4. 44) and (4. 46) - (4. 48) in a power series of ka of the form (Magnus, et al, 1966, p. 65) (ka) J (ka) 2 3 ( -. (4.51) + =o I ' 2 '(n+o+ 1 )

63 Proceeding next to the limit we find that 3 3 3 A - A 3 A a3 a h (R) = R sisino-sn- cosOsin — 3 coso (4.52) 0 R 2R3 2R3. A 2a3 A 3 a3 e (R) =Ra sincos cos cos sin. (4. 53) 0 R R3 R By (3. 25) these are also the zeroth order terms in the low frequency expansions LS -JS of the scattered fields H and E, respectively, and are in complete agreement with Kleinman's (1965b) results as derived using the modified Stevenson method. They can also be obtained from Rayleigh's (1897) theory. We now turn to the calculation of the first order terms in the low frequency expansions for h and e. From (3. 60). -L -it A A (e) Ab A h(R') (Ye +Rxh E dV - V (R h )N )dV+f (R') (4. 54) V V By (4. 52) and (4. 53) a A af1 A 0Y +Rx R, smincos+e-(2 -cosO)cosA 0- 3(1-cos)sin. Ye+ 2 Aa3 1 R3 -cosO 4 4.55) By (2. 59) the dyadic E) can be written V(1) v - x+ +G. a. (4.56) m m 4 1 m 1 We now define the vector A as follows A=(Ye+Rxho) 4 V)(- xi =(Yeo+Rxh) xV (- - R-R' 4r ]R-R'] (4. 57) and we proceed to integrate it over V as indicated in (4. 55). In doing so we use the expansion (4. 1). The integration is performed in a standard manner: we split A into its rectangular components and we integrate using the orthogonality properties of the trigonometric and Legendre functions. We

64 will just give the results for the x and y components of A, while we show part of the integration for the z component: (a, R'-a (A. dV=- 12 P2(cose'sin2 -, (4.58) RI V f(A^y)dV= -Ya + Y3 P.(cose')-+ Ya- P (cose')R + 6 6 R'2 2 R, 3 Ve a R'+ Ya P2(cosO')cos2' - (4.59) 12 2 3 After performing the angular integrations for the z component of A we are left with (the Legendre functions are functions of cos0') 3 R T) 3 aR 2 Ya 31 dR a Ya 1 2i, dR a J )dV= TP sin'+ R2 P22sin' R < Ya 1sin dR < Ya 3' 3 R Pssinn f dRR+ 3 P ins J 4R2 2(4.60) aR' a integrals of the equations (3. 60) and (3. 62) involving the dad E and 3 ) R ri Ya3 1. -dR,2 Ya3l dR i Yan3 4dR 2 Ya3.. 3 dR 2 2 RH'3 0 R5 behave in a very undesirable manner, namely log R'/R'2, and that these terms ionvegrals of the sm e will converge for all n. Terms of the same kind also appeared in the integration

65 of the y component of A and they also cancelled out. Performing the indicated integrations in (4. 60) we obtain 3 3 (A. z)dV= - P(cosO')sin- P1 s)sin (4. 61) (2 6 1) R R' o Collecting our results from (4. 58), (4. 59) and (4. 61) we have that A -~ Ya 2 12 3 | (Ye+Rxh). (.-fr)xc dV= - P(coss ' + 3 +P(os sn' x Ya 2 1R'-a 2 RIa + +4P (cos') +2P (cosO) -+P (cose')cos2O' -?+ 12 R2 1 2 +2P(cs ') 3 2 3 ~+ P2 -+ - 2p(cos0')sinm + P(cosO)sin0. (4. 62) 6 1 23 R' R We now turn to the second term of (4.56). By (4. 26) we have that (Ye +Rxh ) =)p V ' +Rxho ) (4. 63) r where 6 is the vector of (4. 26), i. e. O n 2n+l m m m a - P (cos)P c,+-, sinO _ iL n+l (n-m)! n n (osn+l n+l1 n=1 R' R 'n 2n+J1 0 o 1 (n-m)! m d a + Zm Pn (cos0') o P (cos0)cosm (-, --- n1 mo n+l (n+m)! n dO n n n+1 J (4.64) From this expression and (4. 55) we form the dot product indicated in (4. 63) and then we integrate over V to find (YI A Ya. 1 1 L (Ye+Rxh) * 2dV= -12 P(cos0')sin '. (4.65) V

66 Employing now (4. 63) we have that A 1)p Ya (____ (Yo +Rxh o)E dV= - ho(R'), (4. 66) 'm 6 12 R'2 6 0 r / where h is given by (4. 52). 0 We now turn to the last terms of the dyadic (4. 56). The exterior Dirichlet (e) functions G. are given by (4. 34) - (4. 36). With relatively little labor we find min that3 3 3 mi 3 1 2 1 V This calculation completes the evaluation of the first integral of (4. 54). To evaluate the second integral we employ the definition of N(e) in (2. 84) - (2. 86) and the expansions (4. 1) and (4. 4). The resulting expression is 3 4 (A e) Ya 1 1 Y 1 1 (R-h )NN dV=2 P(cos0')sin't R 4 - l(cos0t)sin. (4.68) V RI Taking the gradient of this expression we obtain Ya3 A A A - V'f (R h )N )dV= 2(Rsin'sin'-'sn cos0'sin'-0'cos 0)- ho(R') V R(4. 69) To conclude the evaluation of h in (4. 54) we need f. From (3. 58) and iY AA 1 a A5 d 1 (4. 44) we have that (R) R)+ sin5J(a)P2(cos0)sin'- -0 — P (cos0)sin h - 5 k R R 2 R A5 1 co,nScos1(a4 - J5(ka)P2(cos 0)cos

67 From (4.51) J (ka) = (2 LS + (k2. (4.71) Substituting this expression in (4. 70) we find that Ya (Ya A 3 A f (R)=-ah (R-)+a (-Rsin2Osin0+Ocos20sin+4cosOcos). (4. 72) 1 3R4 2s Before collecting our results, we write (4. 62) and (4. 67) in spherical coordinates: A1 f (e) t o omi r(Yeo' +Rxho) [ Vt' 1.. dV+ Cai (Ye -+Rxh ) VGe) dV V V Ya A ^A a 2(4. 73) R,2 {-R'sin0'sin0'+0'2sin5'+~'2cos0'cos0 + - h (R'). (4. 73) 2R' Collecting our results from (4. 66), (4. 69), (4. 71) and (4. 73) and substituting them in (4. 54) we obtain the following expression for hi: f r3A A hl(R)= -YaR2sin0sin0+0(2+cos0)sino+0(2cos0+l)cosJ + 1 2R2 ( Ya5 (A 3 A A + -4-R2sin2Osin0 +0cos20sinO+Ocos0cos. (4. 74) 3R Our next step is to determine el. From (3. 62) o o e L'e e(Rf) -f(Zh -Rxe) HedV-Vf (R 6) eVi)+D dV+g1(R) (4. 75) V V The procedure for finding el is analogous to that followed for h1 of (4. 54). For this reason we will be brief. From (4. 52) and (4. 53) A 3{ 1 A 1 Rx -Zh - Rsinsin+0 ( cos-l)sin+ ( -coscos (4. 76) R3L 2 o~~4 76),u H ' )

68 The dyadic H(1) is given by (2. 66): H e (- -6.* I p E ei 9i (4. 77) Proceeding as in (4. 5 7) -(4. 6 1) we find that O~x - h )& Vx IdV= - ----- +- -P1(cosO',). + OL\4J7r R 1R3R'2R a3 vR1-a +.-p (cos6'-..-_3 2 3 3 a 2 Rt-a IA P RcsI~ 2 '3....I + a3 2 RI-a], + --- P2 (CosO'~sin2At 3..~ + a 1(cosO') cosot 3~L z 32 ~RI A a3 1 coO)osIk LTp(OO)OO + (4. 78) We mention that in deriving this expression undesirable terms (log R'/ RI2) arose but cancelled out. The dyadic H is given by (4 16) and if we define er 1Fr0sinO n- n(n+m)! Pn (otpn 9CS)shim(0-0')a+ A 1D n 1 (n-m)! m d m 2n+l m n (n-un)! n 'dTo n n+nF m=o RIn lRn I, we readily find that A A 4; dV pia L'*LOU)O0 (4. 79) (4.80) from which rA 4 V ('tIx' -..Zh ) J~dV =a~- V, Joo 3 V (sinO5cost)= -a. * Rf2 5~ eo(RI) (4.81)

69 The exterior Neumann functions, N(e) of (4. 77) are given by (4. 19) - ei ' (4.21). Performing the integration we find that A Af( R4 Zh. )VN(e) dV a 1 A e e-Z -VN xdV- -P(cos0') - f (A 0 e 6t V3 -- P l(cosO')cos 2 z. (4.82) From (4. 78) and (4. 82) we find that in spherical coordinates (Rxe o-Zh J' 4V 1 _ I xf dV+ 4aiJ (RXo-Zho) N(edV = 47r R-RI V - V =-Ra3sin1cos0-2 + 0 eo (R'). (4.83) Proceeding to the nextvolume integral in (4. 75) we have from (4. 7), (4. 8) and (4. 53) r(Ro e o)s 0Ge )+DedV2a3 dV 2a4 f sin0osdV jR 4r R' R3 f2a3 sincos0 V ' + (4.84) Clearly, the last integral vanishes due to the cosp term. Employing (4. 1) and (4. 3) for G(e) we readily find (Re eo) V () +D dV= a3 1cose')cosp' R -a4Pl(coso')cos~' 1 -JR1 9v (4.85) from which (A ) (0 d a A A A & V' (RO o) q +D dV = (rsine'cos,'-'cosOcost,+',sin0,)-aeo(R,). V (4.86)

70 We are now left with the determination of gl. From (3. 58) and (4. 46) - (4. 48) we have that 3 a 3 a.-i_.A, 3 ikg ) =-i J3(ka)P1(cose) + J (ka) P3(cose) a - R R 2 a4 2 TIkaJ5a)P (COS 2ka 12 J5(ka)P (cose)sin2O4y +:2 2 2 4 ka (a)P )cos (ka)P (cos). (4.87) 2 By (4. 51) 1 \ - s2 cos co-cosssin. (4. 9) 2R 2 2 R 3 Jc(ka) os(cs-,c+Oska2 ). (4.89) rectanar to spherical coordinat (4. 75) we obtain 2A3 A A + |-R- sin2( +cos2-cosa-0cosssin. (4. 90) gl(R) =.-ag (ecoso- O coseo + 22- (-R s sinco)s~ 1 2R22R4, R 2 sin2e cosO+e cos2e cost-O cose sino (4.90) 2 g

71 A. b In order to check the correctness of our expressions for h and e we go back to (3. 25) and expand the scattered fields in power series of ik. The first order terms are given by - S. -b -&S Ih =Rh +h El= Re + (4. 91) Substitution of (4. 52), (4. 53), (4. 75) and (4. 90) in these two expressions leads to -s A.C. AYa A3 A H1(R)= -2-(0sin0+ cosOcos)+ - (-R-sin2Osin+e0cos20sini + RH 3R4 + 0cosecos), (4.92) 3 A A 5 A 3 A E (R) a=- ( coso- 0 cos sin).+ -4(-R sin20 cos+ 0 cos20 coso - 1 2R 27r -Ocos0sino). (4.93) Both expressions agree with Kleinman's (1965b) results as derived using the modified Stevenson method. We will conclude the example by finding the terms of t2 and e5 that behave as 1/R. From these terms we can find the first term in the low frequency expansion of the far field for both the electric and the magnetic scattered vectors. By (3. 60) " ^" ^- VV' fR(e)) h2( dR')) (4.94) niV V Employing (4. 74) and (4. 90) we have that

72 A A2Ya3 Ya5 I 3 Ye +Rxh1 -R 2 s + sin2cos++ R2 RI A 1 A 1 +0(cos2- 3 cos0)cos+ (- cos20-cosO)sin. (4.95) Performing the first volume integration in (4. 94) as we did in the previous cases we find that (Yel+Rxh)E dV=Y R Pl(cos0') - a Pl(cose0)sin0'+O(- 2) I 1 1 m RI R' 1 (4. 96) Similarly, A Ya Ya R 1 isin - Ya sin sosi sin0 (4.97) and A L(e) Ya 1 N Rh1N(e)dV= -2 Pl(cos0')sin'+O(R,), (4.98) hdV =- -2 P1R(4.98) V from which we have that v AfR. 1N(e)d Ya3 1 V' Rh1 N(dV= 2 V (sinO'sin0')+O (-) = V ya3 A A 1 2R (0' coso' sin' + 'cos ') + O ( (4.99) From Eq. (4. 44) it is clear that none of the f 's contribute a 1/R term. n Transforming then (4. 96) into spherical coordinates and substituting the result together with (4. 99) in (4. 94) we have that Ya A A A h (R) =- (Osina+cosocos^)- 2 (0 cos0 sin0+Ocos()+ 0 (2) = R Rso - Yaf3 A 1 A1 (1+ (+cos0)sin+ (-+cos0) cos+) + (4. 100) R. I2 2 ^ R

73 In order to determine the corresponding terms of e2 we substitute (3. 52) in (3. 30) to obtain A. ^ A Vx h1= -Y e -Rxh n = 0, 1, 2,.. (4. 101) n+l n n But, by (4. 100), Vxh = O(2 ). (4.102) 2 R Substituting then (4. 100) in (4. 101) we have that 3 e2(R)= { (1+ cosO)sin 0-(O +cosO) cos 0 (R2). (4.103) To check the results we obtained for e and h2 we substitute them in (3. 25): H (R)= e (ik)2 h2(R)+0 (k3) = = -- (ka) 30(1 + - cosO)sin +cosO)cos + (k4) +0 (); (4.104) Similarly, E (R)=e (ik) e (R)+o(k3)l= JI A A (4.105) These two results for the scattered fields are in complete agreement with Lord Rayleights (1897) results.

v CONCLUSIONS In this work we have developed a technique for determining the electromagnetic fields scattered by a perfectly conducting surface in three space when the characteristic dimension of the scatterer is small compared with the wavelength of the excitation fields. Though the method appears to work well there are two significant questions that were left unanswered: first the identification of the dyadic kernels of Chapter I with the dyadic dipole fields of Chapter II; second, the convergence of the volume integrals in the higher order approximations and in the higher order terms of the low frequency expansions for e' and h of Chapter III. These two questions will be part of the work we plan to do in the near future. This work will also involve the question of convergence of the sequence of the iterates when we solve the Ah. integral equations for e and h by the method of successive approximations as well as an example more complicated than the sphere. The most probable candidate is the prolate spheroid which has only two degrees of symmetry. From the results for the prolate spheroid one can also obtain the corresponding results for the sphere, the oblate spheroid and the disc by simple transformations. 74

REFERENCES Ar, Ergun and R. E. Kleinman (1966), "The Exterior Neumann Problem for the Three-dimensional Helmholtz Equation, " Arch. Rational Mech. Analysis, 23 (218-236). Kellogg, O. D. (1953), Foundations of Potential Theory, Dover Publications, New York. Kleinman, R. E. (1965a), "The Rayleigh Region, " Proc. IEEE, 53 (848-856). Kleinman, R. E. (1965b), "Low Frequency Solutions of Three-dimensional Scattering Problems, " The University of Michigan Radiation Laboratory Report 7133-4-T, AD 623 852. Kleinman, R. E. (1965c), "The Dirichlet Problem for the Helmholtz Equation," Arch. Rational Mech., Analysis, 18 (205-229). Levine H. and J. Schwinger (1950), "On the Theory of Electromagnetic Wave Diffraction by an Aperture in an Infinite Conducting Screen, "Comm. Pure and Appl. Math., 3 (355-391). Lord Rayleigh (1897), "On the Incidence of Aerial and Electric Waves upon Small Obstacles in the Form of Ellipsoids or Elliptic Cylinders and on the Passage of Electric Waves through a Circular Aperture in a Conducting Screen, " Philo. Mag., XLIV (28-52). Magnus, D., F. Oberhettinger and R. P. Soni (1966), Formulas and Theorems for the Special Functions of Mathematical Physics, Springer-Verlag, Inc. New York. Stevenson, A. F. (1953), "Solution of Electromagnetic Scattering Problems as a Power Series in the Ratio (dimensionof scatterer)/Wavelength, " J.Appl.Phys., 24 (1134-1142). Stevenson, A. F' (1954), "Note on the Existence and Determination of a Vector Potential," Q. Appl. Math., XI (194-197). Van Bladel, J. (1964) Electromagnetic Fields, McGraw-Hill Book Company, New York. Werner, P. (1963), "On the Exterior Boundary Value Problem of Perfect Reflection for Stationary Electromagnetic Wave Fields," J. Math. Anal. and Applic., 7 (348-396). Wilcox, C. H. (1956), "An Expansion Theorem for Electromagnetic Fields," Comm. Pure and Appl. Math., IX (115-134). 75

APPENDIX A THE BEHAVIOR AT INFINITY OF THE DYADICS H(1) AND E( e m In this appendix we prove that the dyadics H )and E(1) as defined by Eqs. e m (2. 66) and (2. 95), respectively, satisfy the regularity conditions R2 (RxA)l < o and | R x A < oj, as R -, (A. 1) where A stands for either H or E ) e mI The Regularity of He The expression for He is given by (2. 66) which we repeat here for convenience He ' I+er + VNei (A. 2) i=l We will now examine the behavior of each of the three terms for large R. First we expand the distance function 1/ R-R'I in spherical harmonics for R> R'oD n |R-R n m(nm) n n n+losm (A. 3) where e is the Neumann factor; em = 1, for m=1, e = 2 for m = 2, 3.... The gradient of this expression is 76

77 =%-0 A -R n-o (l)(n-rn)! In, ROPkcs') nI... n n R+ mn(n)m)! Pno+ o d ll cosO" cos )Cfl__ ----?NcsO)RI' o m(O-.0I') - m n m)! dO n n R 2 m=o -O sinO A 2 00 m=l m(n-rn)! pm cs)m ()R"' (n-~m)! n R+ (A. 4) At large distances (R-> co) this expression becomes vi. R-R A R R2 R R -~ co. (A. 5) We now turn to the second term in (A. 2) and write it in terms of its vector components: H(~ = er i=1 He a. % r, (A. 6) By (2. 62) -.&()P,.L.-A. 1 el(R R)= vxfdsRLeiN ei Rsj x dS 5' A 5 1 (0) (iJ ___ A ( R )-N.(R )i 47r Oei 5 h ei -U& xnS rU IRE-RI For R large we can employ (A. 5) in this expression to get (A. 7) Hei -- 2 r 47rR A f-l(O) -3 1 R x CA O (R )- '(i ) dS+0( ),R-oD.A. Jn[ei )-NeRjso( Rw(A8 r

78 In order to evaluate the last term in (A. 2) for large R we employ a theorem by Kellogg (1953, p. 143) which says that if a function satisfies Laplace's equation then it can be written in a series of spherical harmonics of the form Ne Yin(O., (e) n+l Ne = Rn) (A. 9) n=o where Yin is a nth order spherical harmonic: n = A. P (cos )eim. (A. 10) n mnn m=-n The series (A. 9) is uniformly and absolutely convergent outside a sphere enclosing all sources. Clearly, ^ A VN = -R - + ( ), R-o0. (A. 11) ei R R3 R Collecting our results we have that =(l) 1 A 1A 0i~) (i -d1 A He 4R2 RX 4 ei (Rs)-N i(R dS a. - i=l 2 S 2 A 3 A - Ai ai R o (A. 12) 2, 00' i=l From this last expression we conclude that R (Rx ) | < OD J as R o. (A. 13) In order to prove the second statement in (A. 1) we turn to Eq. (2. 23) and write

79 Vx H V ( 1- +Vx R R'. (A. 14) 4 R-R r Splitting the dyadics into their vector components and employing (2. 58) we have that x (1) A V Vx He = -(a V') V ( )V i=1, 2, 3. (A. 15) \ 47r R-Ri r We now examine (A. 4) and we see that the n=0 term of the R-component.(which is the only term that behaves as 1/R2) is independent of the primed coordinates. The rest of the terms in (A. 4) behave at least as 1/R3. We then conclude that (aiV)v(-)=o( ), R-o. (A. 16) 4 7R-Rt' R To show that the second term of (A. 15) behaves similarly we employ (2.41): A ^~(0) 'A 17 n AV ei (R R')dS =0 4A 17) Sr This condition together with the divergence theorem imply that lim A (0) R-o R Vei (RiR') dS = (A. 18) Co or (0) ei lim 2 r lira J dQR2 R =0 (A. 19) R-*co 3aR S (0) But eir can be written in the form (A. 9). If condition (A. 19) is to hold, however, the constant Aio in (A. 10) must be zero. We therefore have (0eir = Y ' (A. 20) e l n-Il n=l

80 which implies that VO~ = 10(-~), asR-*o. (A. 21) er R asR From (A. 16) and (A. 21) we can readily conclude that | R3 VxH | < o, as R -. (A. 22) The Regularity of Em To prove thatR2(RxEm ) | < o, as R -> co, we start with (2. 95), which reads =m - -R XI+EM + VG a,.,.(m)=rV 1 xt+2.()p+,V ai, (A.23) 4\ R-R' i=l and we proceed in exactly the same manner as we did above for 1. The behavior of the functions involved being the same as those for i1), we omit e the proof. To prove that R 3VxE () <co, as R -> o, we start with (2. 68): MVx E = (- ) R R'. (A. 24) \ R-R Employing (2. 88) in this expression we get Vx E = - (- )- VV'N( (R|R'), R R' (A. 25) 47r R-R' or, in terms of the vector component of the dyadics, Vx E - (aiV')V (- ) - (NOV?)Ne) (R|RI') \ 4;] R-R'Ii=l, 2,3. (A. 26) The function N(e, however, is the regular part of the exterior Neumann Green's function for the surface and it, therefore, has the property that

81 ns VNr )dS = 0. (A. 27) S Transforming this integral by the divergence theorem to an integral over a (e) surface at infinity and writing N in a series of spherical harmonics we r conclude that (e) 1(e) 1 N = O(- ) VN ( ---3, as R -c. (A. 28) This result together with (A. 16) leads us to the conclusion that R3 VxEm1)I <oo asR- >o. (A. 29) m I