green's functions

2Green's Functions

In 1828 George Green wrote an essay entitled "On the application of mathematical analysisto the theories of electricity and magnetism" in which he developed a method for obtainingsolutions to Poisson's equation in potential theory. The method, which makes use of a potentialfunction that is the potential from a point or line source of unit strength, has been expandedto deal with a large number of different partial differential equations. The technique alwaysmakes use of a function that is the field from a source of unit strength; this function is called theGreen's function for the problem. In the early applications of Green's method no mathematicalexpression was introduced to represent a unit source. The Green's function was defined insteadthrough its singular behavior in the vicinity of the unit-strength source.

In 1927 the delta function was introduced by Dirac as the limit of a sequence of functions,i.e.,

o(X -x') == lim Un(x -x')n~oo

where Un(x - x') can be, for example,

These functions have the property that

orsin n(x - x')

1r(x - x') ·

lim jf(xI)Un(X - x') dx' == f(x).n~oo

The area under the curve of Un(x - x') is unity for all n. It is now common practice to usethe delta function to represent a unit source in one dimension and to use a product of deltafunctions to represent a unit source in a multidimensional space. The functions Un(X -x') arelegitimate functions, but once the limit has been taken the result is no longer a function in theclassical sense and o(x - x') is called, for this reason, a symbolic function. It can be definedfor most purposes by the operational property given by (96) in Chapter 1. The delta functiondid not come into widespread use before around 1940. Even Baker and Copson in their 1950book [1.30] did not make use of the delta function. In 1950 Laurent Schwartz developed thedistribution theory which provided a rigorous nonoperational method of handling entities suchas the delta function. A short introduction to distribution theory is given in Section 2.21.

The first two sections of this chapter present the general Green's function method for solvingPoisson's equation. This is done to provide the reader with an introduction to the subject. Anumber of the following sections take up the problem of how to construct Green's functions fordifferential equations of the Sturm-Liouville type. The construction ofGreen's functions formultidimensional problems and techniques for obtaining alternative representations are thengiven. The chapter includes a treatment of dyadic Green's functions which are needed for

55

56 FIELD THEORY OF GUIDED WAVES

the vector problem. Integral equations for scattering are also discussed. In Section 2.20 wepresent a brief discussion of procedures that must be followed to construct Green's functionsfor non-self-adjoint differential equations.

The field point is described in terms of the coordinates x, y, Z or the position vectorr. The source point is described in terms of the coordinates xo, Yo, Zo or x', y', z' withcorresponding position vector ro or r'. In general, we use the xo, Yo, Zo notation when anumber of differentiations with respect to xo, Yo, Zo are involved. The distance Ir - ro I isrepresented by the symbol R.

2.1. GREEN'S fuNCTIONS FOR POISSON'S EQUATION

The scalar potential in electrostatics is a solution of Poisson's equation

( 1)

where p is the density of the charge distribution which gives rise to the scalar potential <1>, and€ is assumed constant. In the absence of any boundaries the solution is given by the integral

<1>( ) == fff p(xo, Yo, zo) dVx, y, z JJJ 47r€R 0

v

(2)

where R == [(x - XO)2 + (y - YO)2 + (z - ZO)2]1/2 and is the distance from the source pointto the point in space where the potential is evaluated. A source of unit strength located atthe point (xo, Yo, zo) can be conveniently represented by a three-dimensional delta functiono(x - xo)o(y - yo)o(z - zo), where the delta function is zero at all points except Xo == x,Yo ==y, Zo ==Z.

If P(xo, Yo, zo) is an arbitrary vector function which is continuous at (x, y, z), then [see(96) in Chapter 1]

IIIP(xo,· · .)o(x - xo)o(y - Yo)o(z - zo) d Vi,v

{O'

P(x, y, z),

(x, y, z) not in V

(x, y, z) in V.(3)

A similar result holds for a scalar function. Our discussion of the solution of the inhomoge-neous Helmholtz equation in Section 1.9 showed that - \72(47rR)-1 has the properties givenby (3) above. The three-dimensional solution of the following inhomogeneous differentialequation defines the free-space three-dimensional Green's function for Poisson's equation:

\72G(X, y, ZIxo, Yo, zo) == -o(x - xo)o(y - yo)o(z - zo). (4)

The Green's function G is a function of both the point of observation or field point and thesource point. From the known properties of the function (47rR)-1 we see at once that this isthe solution of (4), and hence

G(x, y, zlxo, Yo, zo) == 1/47rR. (5)

GREEN'S FUNCTIONS

Fig. 2.1. Illustration for Green's theorem.

57

The Green's function is symmetrical in the variables xo, Yo, Zo and x, y, z. This propertywill be found to be true in most cases that we will encounter, and is equivalent to the fieldobeying a reciprocity principle. Once the solution for a unit source is known, the solution forany given arbitrary source distribution follows at once by superposition, i.e.,

(x,. · .) = ~IIIG(x, y, zlxo, Yo, zo)p(xo. Yo, zo)dVo

v

(6)

which, with the aid of (5), is certainly recognizable as the standard solution of Poisson'sequation.

In addition to the notation used above, it will be convenient, in what follows, to use thefollowing abbreviated notation as well:

(r) = ~fflG(r, ro)p(ro)dVo.

v

When G is symmetrical, G(r, ro) == G(ro, r).Let cP and 1/; be two scalar functions of position which are continuous with continuous partial

derivatives. Let the region of space V contain conducting bodies with surfaces 8 1,82, ... ,8n ,

as illustrated in Fig. 2.1. By introducing suitable cuts, the region of interest can be made simplyconnected. Let 8 be a closed surface consisting of the surfaces 8 1, ••. ,8n» the surfaces alongthe cuts, and the surface of an infinite sphere which surrounds all the conducting bodies.

Applying Green's second identity! to the functions <P and 1/; gives

where n is the normal to S and is directed into the volume V. We now let <P be the requiredsolution of Poisson's equation, and let 1/; be the Green's function as given by (5). Replacing

1Section A.lc.

\75q> by - pj€ and \75G by the negative delta function, we find that

- fff(ro)o(r - ro)dVo+ ~ fffp(ro)G(r, ro)dVov v

fj aq>(ro) fj aG(r, ro)== {jflG(r, ro)dSo - <p(ro) an dSo

s s

where

(7a)

aG- == (\7oG)-oan

and \70 designates differentiation with respect to xo, Yo, zoo It is important to note that, ifxo, Yo, Zo are used as the variables of integration in Green's identity, then all derivatives offunctions in the integrands must also be with respect to xo, Yo, Zoo Using the property of thedelta function as given by (3) shows that the required solution for q> is

(r) = ~ fffp(ro)G(r, ro)dVo + fj (~~ - G~:) as; (7b)v s

The surface integrals along the cuts do not contribute since they are traversed twice in theopposite sense and with.the normal 0 oppositely directed. The three terms contributing to q>are the volume distribution of charge p, a surface distribution of charge - €(a<p jan), and asurface distribution of electric dipoles of dipole moment € <P per unit area [2.1].

In a two-dimensional space the Green's function is defined by the differential equation

(8a)

The source function in this case is an infinite line source of unit density. The solution for 0is readily obtained by rewriting (8a) in cylindrical coordinates r, 0 as follows:

1 a oc 1 a20--r- +-- == -o(r)r ar ar r 2 ao2 (8b)

where the origin for the r, 0 coordinate system has been chosen coincident with the linecharge, i.e., at x 0, Yo. From symmetry considerations we see that G is not a function of 0,and hence depends only on the radial distance r from the line source. If we surround the linesource by a cylindrical surface S as illustrated in Fig. 2.2 and integrate (8b) throughout thevolume enclosed by a unit length of S, we get

fffv-vo av = fj vo.as = - fffo(r)dV =-1

v s v

by virtue of the properties of the delta function. Since G is a function of r only, we get(aG jar)21rr == -1, and a further integration gives the desired result, which is

G = - 1;: . (9a)

GREEN'S FUNCTIONS 59

..z

s-r--Ir--l-I -------+1---...I II V I

_L .J_

'-x'r/ "'I \

S I 8 \'Xo,Yo I\ I, /

" '"~-...",

Fig. 2.2. Surface S surrounding a line source.

Transforming back to our X, Y coordinate system, we have

(9b)

2.2. MODIFIED GREEN'S FUNCTIONS

The solution for 4> given by (7b) is not too useful in practice since usually we know 4>on part of the boundary and a4>jan on the remainder, but rarely do we know both on thecomplete surface S. However, by a suitable modification of our Green's function it is possible,in principle at least, to obtain a solution for 4> from a knowledge of either 4> or a4> jan on theboundary. If the required boundary conditions specify the value of 4> on the complete boundary,the potential 4> is said to satisfy Dirichlet boundary conditions. If the normal derivative a4> janis specified on the whole boundary, the corresponding boundary conditions are referred to asNeumann boundary conditions. Finally, if 4> is specified on part of the boundary and a4> janon the remainder, we have mixed boundary conditions.

If 4> is known on the complete boundary, we see from an examination of the solution (7b)that the offending term involving a4> jan can be eliminated by making the Green's functionvanish on the complete boundary, i.e., on the surface S. Thus we define the Green's functionof the first kind as a solution to the equation

(lOa)

and obeying the boundary condition

on S. (lOb)

The solution for the potential 4> is thus given by

(r) = JJJ;01 dVo + fj a~1 as;v s

(11)

When a4> jan is given everywhere on S, we modify our Green's function to make ao janvanish on the boundary. The Green's function of the second kind is defined as a solution of

(I2a)

subject to the boundary condition

a02 == 0an

onS. (I2b)

60

The solution for ~ in this case is given by

FIELD THEORY OF GUIDED WAVES

(r) = ff! ;02av; - If ~:02 as;V s

(13)

When we have mixed boundary conditions, it is desirable, if possible, to find a Green'sfunction which vanishes on that part of S for which we know ~, and which has a normalderivative equal to zero over that part of S for which we know 8~/8n. From (11) and (13)we see that the Green's functions are useful for solving boundary-value problems, when thevolume distribution of charge p is zero, as well. Needless to say, quite often in practice itis about as difficult to find a solution for the Green's functions as it is to solve the originalboundary-value problem. The use of Green's functions converts a differential equation togetherwith the boundary conditions to an integral equation, which in many cases is more readilyattacked by approximate techniques.

We will now show that the Green's function is symmetric in the variables x, Y, z andXo, Yo, Zo, that is,

G(r, ro) == G(ro, r). (14)

This property might have been anticipated since the delta function is symmetrical, i.e., o(r -ro) = o(ro -r). This reciprocity principle states that the potential at (x, Y, z) from a unit chargeat (xo, Yo, zo) is the same as the potential at (xo, Yo, zo) due to a unit charge at (x, Y, z).Let G 1(r, rOl) be the Green's function of the first kind for a point charge at (X01, Y01, ZOI),and let G 1(r, r02) be the Green's function for a point charge at (X02, Y02, Z02). If these twofunctions are used in Green's second identity, we get

!!!£o\(r, rOl)V'20\(r, r02) -O\(r, r02)V'20\(r, rOl)]dVv

__If [G ( )8G1(r, r02) _ G ( )8G 1(r, r01)] dS- 1 r, rOI 8n 1 r, r02 an ·

s

Since G 1 vanishes on Sand

\72G1(r, rOl) = -o(r - rOl)

\72G1(r, r02) == -o(r - r02)

we obtain the following result from the volume integral:

(15)

Since (X01, Y01, ZOI) and (X02, Y02, Z02) are arbitrary points, the stated reciprocity principleis proved. A similar proof holds for the Green's function of the second kind since aG2/8nequals zero in this case and the surface integral again vanishes.

GREEN'S FUNCTIONS

2.3. STURM-LIOUVILLE EQUATION

61

Many of the boundary-value problems encountered in connection with the study of guidedelectromagnetic waves lead to the Sturm-Liouville equation. In this section we will review thebasic properties of the Sturm-Liouville system but without detailed derivation.

The Sturm-Liouville equation is of the form

:XP(X)d~~) + [q(x) + AI1(X)]lf(X) = 0 (16)

where A is a separation constant. In practice both p and a are usually positive and continuousfunctions of x. We will consider the properties of this equation when the interval in x is finite,say 0 ::;x ::; a.

The solutions to (16) are fixed only if certain boundary conditions are first specified. Thethree common boundary conditions are 1/; == 0, d1/;[dx == 0 and 1/;+K(d1/;/dx) == 0 at x == 0, a,where K is a suitable constant. Once a set of boundary conditions has been specified (onecondition at x == 0 and one at x == a) then one finds an infinite set of solutions 1/;n to (16)corresponding to particular values of A, say An. The functions 1/;n are called eigenfunctionsand the An are called eigenvalues. In all cases, the 1/;n form an orthogonal set of functionsover the interval 0 ::; x ::; a. The orthogonality of the functions may be proved as follows:Multiplying the equation for 1/;n by 1/;m and vice versa, subtracting and integrating gives

fa ( d d1/;m d d1/;n) faJo lfndx P dx -lfm dx P dx dx = Jo (An - Am)l1lfnlfm dx

since the term involving q vanishes. Integrating the left-hand side by parts gives

(lfndlfm -lfm dlfn ) p 1° - r P (d1/;m d1/;n _ d1/;n d1/;m) dx

dx dx 0 io dx dx dx dx

(d1/;m d1/;n) ·I

afa

= P lfn dx -lfm dx 0 = (An - Am)Jo I1lfnlfm dx.

When 1/In and 1/Im satisfy the same boundary conditions, of the type given earlier, the integratedterms vanish. For example, if 1/1; +K(d1/l;/dx) == °at x == 0, a, with i == n, m, then we canwrite

which clearly vanishes at x == 0, a. When An 1= Am we conclude that

(17)

i.e., the 1/;n form an orthogonal set with respect to the weighting function u(x). If An == Amwe have a degeneracy but suitable combinations of 1/;n and 1/;m can be chosen to yield anorthogonal set of functions.

For example, if 1/;1 and 1/;2 are eigenfunctions with degenerate eigenvalues Al == A2 andJ: u1/;I1/;2 dx =1= 0, then we can choose new eigenfunctions ~1 == 1/;1 and ~2 == 1/11 + C1/;2


(18)n =1= m.

n==m

10 {I,u1/;n1/;m dx ==

o 0,

and force these to be orthogonal. This means that the constant C is chosen such thatJ; u(1/;i +C1/;I1/;2)dx == O. The new eigenfunctions are now orthogonal but have the commoneigenvalue AI. This procedure may be extended to the case where there are N degenerateeigenvalues with corresponding eigenfunctions. The procedure just described is an exampleof the Gram-Schmidt orthogonalization procedure.

We will now assume that the 1/;n have been normalized so that they form an orthonormalset, i.e.,

The eigenfunctions form a complete set and may be used to expand an arbitrary piecewisecontinuous function f(x) into a Fourier-type series. Thus let

00

f(x) == L an1/;n(X).n=l

(19)

To find the unknown coefficients multiply both sides by u1/;m, integrate over 0 to a, and use(18); thus

10 00 1°uf1/;m dx == Lan u1/;n1/;m dx == am·o n=l 0

(20)

Our main interest in the orthogonality property of the eigenfunctions is for the purpose offinding the coefficients in Fourier series-type expansions.

The notion:of completeness for the space of functions 1/;n defined on the interval 0 ::; x ::;ainvolves the following: Let f(x) be a piecewise continuous function on 0 ::; x ::;a that isquadratically integrable with u(x) as a weighting function, i.e.,

We assume that u(x) is always positive.Consider now the approximation

to f(x). If the limit as N ~ ()() of the integrated square of the error tends to zero, then the1/;n form a complete set. Completeness thus implies that

a N 2

lim r f(x) - LCn1Pn(X) u(x)dx = o.N~ooJo n=l

(21)

This equation becomes, in the limit,

(22)


which is Parseval's theorem. The space of functions 1/In is said to be complete when (21) istrue. The functions 1/In then span the space or domain of functions that the Sturm-Liouvilledifferential operator operates on. The functions are said to form a basis for this domain, whichmeans that they may be used to expand functions in the domain into Fourier-like series. It isimportant to note that the functions 1/In are not a basis for expanding functions that do not liein the domain of the Sturm-Liouville differential operator. For example, they cannot be usedto give an expansion of the function defined on the interval 0 ::;x ::; b with b > a.

A method of showing that the eigenfunctions for the Sturm-Liouville differential operatorform a complete set is given in Chapter 6 and is based on an associated variational principlefor the eigenvalues. The eigenfunctions form an orthogonal set when the differential operatoris self-adjoint. When the operator is not self-adjoint it is necessary to find the eigenfunctionsof the adjoint operator in order to have an orthogonality principle that can be used to developeigenfunction expansions. Non-self-adjoint operators are discussed in Section 2.20.

The normalized eigenfunctions correspond to unit vectors in a linear vector function space.When the eigenfunctions are orthogonal they correspond to orthogonal unit vectors. Thesymmetrical scalar product

(23)

is analogous to finding the projection of the vector f(x) onto the unit vector 1/In(x). Whenthe operator is not self-adjoint the eigenfunctions are not orthogonal and we then need tointroduce a reciprocal set of unit vectors in order to find the components of f. The normalizedeigenfunctions of the adjoint operator correspond to the reciprocal set of unit vectors in anonorthogonal system.

2.4. GREEN'S FUNCTION G(x, X')

The Green's function is the response of a linear system to a point source of unit strength.A source of unit strength, at the position x' can be conveniently represented by the Diracdelta function o(x - x'). From a heuristic viewpoint we can think of o(x - x') as a narrowrectangular pulse of width ~x and height 1/ ~x, and thus'of unit area, in the limit as ~x ~ 0as shown in Fig. 2.3(a). The Green's function G(x, x') is the solution of the Sturm-Liouvilleequation when the source term or forcing function is a point source of unit strength. TheGreen's function satisfies the equation,

d dG I-p- + (q + "Au)G == -o(x - x ).dx dx

(24)

The Green's function must also satisfy appropriate boundary conditions at x == 0, a. We willdiscuss two basic solutions to (24).

Method I

We may expand G in terms of the eigenfunctions 1/In, with eigenvalues "An, that satisfythe same boundary conditions as G. Thus let G(x, x') == E~lan1/ln(X). Substituting in theequation for G gives E~lan("A - "An)u1/ln == -o(x - x') since 1/In is a solution of (16) for


L1x

--lr1

1L1x

--+-------...&...--"""------ Xx'

(a)

---I~-------.L..--------x

G

~III

x'

(b)

pC::;

--+-----+------~ x

Fig. 2.3. (a) One-dimensional delta function. (b) Behavior of G and its derivative near x',

A == An. If we now multiply both sides by 1/;m, integrate over 0 to a, and use (18) we obtain

upon using (3) to evaluate the last integral. Using this solution for am we obtain

G(x, x') = _fVtn~!!~(X').n=l n

Note that G has poles at the points A == An. We will make use of this property later on.

Method II

(25)

We note that at all points x =1= x' the Green's function is a solution of the homogeneousequation (d /dx)p(dG /dX)+(q+AU)G == O. At x' G must be continuous but p(dG /dx) mustbe discontinuous by a unit amount [see Fig. 2.3(b)]. The second derivative (d /dx)p(dG /dx)then yields a delta function singularity. If G was not continuous at x' the resulting singularitywould be of too high an order.

Let A q,1 (x) be a solution of the homogeneous equation that satisfies the boundary conditionat x == O. Similarly, let Bq,2(X) be a solution, linearly independent of q,1, that satisfies theboundary condition at x == a. Thus let G == Aq,l(X), x 5:. x', and G == Bq,2(X), x ? x'.Continuity of G at x == x' requires that A q,l(x') == Bq,2(X'). If 'Ye integrate the equation forG from x' - T to x' + T and let T ~ 0 and note that limT--+o Jx~~; (q + AU)Gdx == 0 since

q, o , and G are all continuous at x', then we obtain

lx ' +7 d dG dG IX~ lX'+7

lim -p- dx == p - == - o(x -x')dx == -1.7~O X' -7 dx dx dx x': X'-7

In other words, p dG jdx evaluated between adjacent sides of the point x' must undergo astep change as shown in Fig. 2.3(b). Hence we have

x'dGI + 1 " "- ==---, ==BiP2(x)-AiPI(x)dx s': p(x)

where iP' stands for diPjdx. The solution for A and B may be found from the above twoconditions and is A == -iP2(x')/p(x')W(x'), b == -iPI(x')/p(x')W(x') where the Wronskiandeterminant W(x') == iPl (x')iP~(x') - iP~ (x')iP2(x') does not equal zero since iPl and iP2 arelinearly independent solutions.

Our solution for the Green's function is thus

(26)

x'<» <«.

O:::;x <x'iPl (x)iP2(x')

{

- p(x')W(x') ,G(x, x') ==

iPl (x')iP2(x)- p(x')W(x') ,

The Wronskian determinant W will be a function of the parameter A and will vanish wheneverA == An so G will have the poles that are explicitly exhibited by the first solution (25). Although(25) and (26) are different in form they are equal.

It is convenient to define the following:

x., == the greater of x or x'

x< == the lesser of x or x',

We can then express the solution for G(x, x') in the condensed form

G(x, x') ==<PI(x <)<P2(X»

p(x')W(x')(27)

which includes both forms given by (26).A useful property of the Sturm-Liouville equation is that the product of p(x)W(x) is a

constant. We can prove this result as follows: We form the product of iPl (x) with the equationfor iP2(x) and subtract from this the corresponding result with iPl and iP2 interchanged; thus

d diP2 d diPl<PIdxP dx + (q + Au)iPl iP2 - iP2dxP dx - (q + Au)iPl iP2

d diP2 d diPl

== iPl dxP dx - iP2 dx P dx

66

From the last result we conclude that


where the constant C is a function of A.

(28)

Example

We wish to find the eigenfunctions and eigenvalues for the equation d2t/;/ dx? + At/; == 0with boundary conditions t/; == 0 at x == 0, a. This is a special case with P == a == 1 andq == O. A general solution for t/; is A sin J>..x + B cos VAx. To make t/; == 0 at x == 0we must choose B == O. To make t/; vanish at x == a we must have sin ~a == O. Thus~a == nx, n == 1,2,3, ... or An == (n7r/a)2. Hence the eigenfunctions are sin nxx ]a,Since J; sin2(nxx fa) dx == a /2 the normalized eigenfunctions are JfTO sin (nxx fa) == t/;n.

We now wish to solve d2G [dx? + AG == -5(x - x'). According to Method I we letG == E~lanJfTO sin (nxx ia). By substituting in the equation for G we find E~lan(A-n27r2/ a2)JfTO sin (nxx fa) == -5(x - x'). We now multiply both sides byJfTO sin (mxx fa), integrate from 0 to a, and obtain

(m27r2) ~la mxx ~ mxx'am A--- ==- - 5(x-x')sin--dx==- -sin--.

a2 a 0 a. a a

Thus

~. nxx~ . nxx'- SIn - - sIn --

00 a a a aG(x, X') = - L 2 2 / 2 •

n=1 A - n 7r a(29a)

We may also use Method II. We let <1>1 == sin J>..x and <1>2 == sin ~(a - x). Notethat <1>1 and <1>2 are linearly independent and satisfy the boundary conditions at x ==0, a, respectively. The Wronskian determinant W(x') == <1>1 (x') <1>2 (x') - <1>i (X')<1>2(X') ==(sin J>..x')[-~ cos ~(a - x')] - ~ cos J>..x' sin ~(a - x') == -~ sin ~a. From(27) we get

G( ') == sin vfXx<. sin /X(a -x»x,x A A ·

VA sin vAa(29b)

Note that sin /Xa == 0 whenever A == (n7r/a)2 so that this second solution for G does havethe poles at A == n27r2/a2 that are shown explicitly in the first solution. A Fourier seriesexpansion of the second solution would yield the first solution.

2.5. SOLUTION OF BOUNDARy-VALUE PROBLEMS

Let it be required to find a solution to the equation

d dt/;dx P dx +(q + AlJ)t/; == -f(x) (30)


with either 1/;, d1/;/dx, or 1/; +K(d1/;/dx) specified at x == 0, Q. In (30) f(x) is a distributedforcing function. Let G(x, x') be a Green's function that is a solution of

d dG I-p- + (q + Aa)G == -o(x -x).dx dx

(31)

(32)

We now replace x by x' for convenience, multiply the equation for 1/; by G and the equationfor G by 1/;, subtract, and integrate to obtain

10 [G( I )~ ( l)d1/;(X') __"( I)~ ( l)dG(X', X)] d Ix ,x d ,P X d I 'Y X d ,P X d I Xo X x x x

=1° -f(x')G(x', x)dx' +1°1/!(X')o(x' -x)dx'

since the terms involving the factor q + Aa cancel. After an integration by parts we obtain

1/!(x) = 1°G(x', x)f(x') dx' + [G(X', x)p(x') d~~')

_ 1/!(x')p(x')dG~:: x) ] I: .If 1/; satisfies one of the boundary conditions 1/; == 0, d1/; / dx' == 0, 1/; +K (d1/; / dx') == 0, at eachend x == 0 or Q, and we choose G to satisfy the same boundary conditions, then the boundaryterms vanish and the solution for 1/; is simply

1/!(x) = 1°G(x', x)f(x')dx'. (33)

This solution is a superposition of the response from each differential element of force f dx'with G being the response of the system due to a point source of unit strength. It is theability to represent the solution to a linear system driven by an arbitrary forcing function f(x)by a superposition integral such as (33) that makes the theory of Green's functions of greatimportance in practice.

Sometimes the boundary conditions on 1/; may be of the form K t1/; +K2(d1/;/dx) == K 3 (K 1

or K 2 might be zero and the K, may be different at each end). In this case we choose theboundary conditions on G in such a way that we can evaluate the boundary terms in (32) interms of known quantities. If 1/; is given on the boundary (K2 == 0) we choose G == 0 on theboundary; if d1/;/dx is given (K 1 == 0) we choose dG /dx == 0; and finally if K 11/;+K2(d1/;/dx)

is given we choose K IG +K 2(dG /dx) == 0 on the boundary. We will illustrate the last caseexplicitly. The boundary terms in (32) may be written as

which by virtue of the boundary conditions on 1/; and G becomes the known quantity(K3/K2)p(x')G(x ', x)lg·

At this point we digress in order to introduce some mathematical concepts that will be more

c£ + O"A)l/I

Sourcespace S

CR

Fig. 2.4. Mapping of functions from the domain ~ of an operator into its range space CR. Thesource function f(x) from the source space S should lie in the range space for a solution of thedifferential equation to exist. Functions in the null space are not mapped into the range space.

important in connection with dyadic Green's functions but which can be illustrated in a simplesetting here.

For convenience we repeat (30) in the form

£'1/; + ACJ1/; == - f(x) (34)

where £, denotes the operator (d /dx)p(x)(d /dx) +q(x). This operator operates on functions1/; that are in the space of functions called the domain ~ of the operator. This space is thespace of functions that are twice differentiable, are quadratically integrable as defined in Section2.3, exist on the interval 0 ::;x ::; a, and satisfy the boundary conditions at x == 0, a. Theboundary conditions are an important part of the specification of the domain of the operator.The operator £, +0"Acan be viewed as operating on functions 1/; in the domain ~ and producingnew functions that lie in a space of functions called the range space <R of the operator. Thisconcept is illustrated in Fig. 2.4. It should be obvious that (34) does not have a solution if thesource function f(x) is not in the range space <R of the operator. Hence a necessary conditionfor (34) to have a solution is that the source f(x) be in the range space, i.e., f(x) E <R whereE means that f(x) is in <R.

The solution for 1/; in (34) when f(x) E <R is unique only if there are no solutions exceptthe trivial one u == 0 to the homogeneous equation [solutions to the homogeneous equationare not related to the source f (x)]

£'U + CJAU == O.

Any nonzero solution U to this homogeneous equation is said to be in the null space of theoperator and is not mapped into the range space. For A == An, an eigenvalue, U == 1/;n is anonzero solution and is in the null space of (£, + CJAn ) . Hence for a unique solution to (34) Amust not be equal to an eigenvalue. All values of A that correspond to the discrete eigenvaluesAn make up the point spectrum of the operator. All values of Afor which (34) has a bounded(finite) solution belong to the resolvent set. Later on we will encounter problems where A hasa continuous range of values forming a continuous spectrum. The continuous spectrum is notpart of the resolvent set. When A is in the resolvent set (34) can be solved (resolved).

The completeness of the eigenfunction set enables us to replace the operator £ +CJAoperatingon 1/; by

(£, + O"A)1/; == LCJ(A - An)Cn1/;nn

GREEN'S FUNCTIONS

where

We can represent the operator cC + (J A as

cC + (JA == 2:(A - An)(J1/;n(x)1/;n(x')n

69

where the action of the operator on a function 1/;(x') is obtained by multiplying by 1/;(x') andintegrating over x'. This representation of the operator is called the spectral representation. Ifwe define the scalar product ((J1/;n, 1/;) to be

((l1/;n(X'), 1/J(x')) =1°a(x')1/Jn (x')1/J(x') dx'

then symbolically we can write

(cC + (JA, 1/;) == 2:(A - An)1/;n(x) ((J1/;n , 1/;)n

as describing the action of the operator on 1/;.When we substitute the series expansion for 1/; into (34) we obtain

2:Cn(A - An)(J1/;n(x) == -f(x).n

By means of the standard procedure

C -1 ra

f I I d In = A_ An Jo (x )1/Jn(x) X

and

1/J(x) = - f:~~ r!(X')1/Jn(X')dx'.n=1 n io

We now rewrite this solution in the form

which allows us to identify the Green's function operator as

G(x', x) = - f 1/Jn~~~(X) .n=1 n

(35a)

(35b)

(36)

This is the spectral representation of the inverse operator (cC + (J A)-1. It is irrelevant whetheror not this series converges since it should be viewed as an operator that is to be used in a term

(37)

by term integration so as to generate the series expansion for the solution 1/; as in (35a). Forone-dimensional problems the Green's function series does converge to an ordinary function,but in two- and three-dimensional problems the Green's function series is not convergent whenthe field point and source point coincide. For this reason it is not meaningful to always considersumming the series, which cannot be done in general. The concept of the Green's function asan operator as used in (35a) is, however, valid.

In the space of functions 5) we have, for a function h(x),

h(x) = 'fOnt{;n(X), On = 1°h(x')(J(X')t{;n(x')dx'n=l

and hence

h(x) = 'ft{;n(X)1°h(x')(J(x')t{;n(x')dx'n=l 0

= [h(X') [~(J(x')t{;n(x')t{;n(x)]dx'.

The series in (37) is seen to have the same operational property as the symbolic functiono(x - x'). Hence the series, which is not a convergent one for all x, x', is a representationfor o(x - x') in the space 5), Le.,

00

o(x - x') == LU(X')1/;n(X')1/;n(X).n=l

(38)

This representation for o(x - x') is also to be viewed as an operator, not as a function, andbe used in a term by term integration to yield h(x) as in (37). It is important to note thatthis eigenfunction representation for o(x - x') is valid only for functions in the domain 5)

for which the eigenfunctions form a complete set of functions, i.e., span the domain 5). Withthese qualifications the relationship exhibited in (38) is said to be the completeness relationfor the eigenfunction set.

Consider the equation (cC + UAk)1/; == - f for a given specified eigenvalue Ak. The eigen-function 1/;k is in the null space of the operator cC + UAk since (cC + UAk )1/;k == o. Clearly thespace of functions in the range space of cC + UAk does not contain the function 1/;k. Conse-quently, the equation has a solution only if the source function f is orthogonal to 1/;k. We seethat the range space is orthogonal to the null space for a self-adjoint operator.

Not all differential operators are self-adjoint operators. Non-self-adjoint operators are de-scribed in Section 2.20. In this section we will give an example of a non-self-adjoint operatorand its implication. Consider a differential operator of the form

.£t{; = d2t{;

_ .dt{; + At{; =0dx' J dx

with boundary conditions 1/; == 0 at x == 0, a. Let this equation be multiplied by a function and integrate by parts so as to transform the differential operators onto ; thus the symmetricproduct

(<1>, .£t{;) = r<1>d2~ dx - j r<1> ddt{; dx + A r<1>t{;dx = 0io dx io x io

GREEN'S FUNCTIONS

becomes

71

d1/; la d4J la 1ad

24J la4J- -1/;- + 1/; -2 dx -j4J1/;dx 0 dx 0 0 dx 0

fa d4J fa+ j Jo I/; dx dx + AJo (pI/;dx = O.

If we wish to make the integral of the product 4J1/; equal to zero, then 4J must be a solution of

and. in order to make the integrated terms vanish 4J == 0 at x == 0, a. The new differentialoperator shown above is called the adjoint operator. It differs from the original operator byhaving j instead of - j as a multiplier in front of the first derivative. With the conditionsabove imposed, we write

(39)

which is the formal definition for the adjoint operator. If an operator is not self-adjoint, then itis the eigenfunctions, say 4Jn , of the adjoint operator that are orthogonal to the eigenfunctions,say 1/;m, of the original operator. A different expansion procedure is required to obtain a spectralexpansion of the solution for the inhomogeneous equation. The details for the modificationsrequired are described in Section 2.20.

When the eigenfunctions are complex we can choose to define the scalar product in thefollowing way:

where 1/;* is the complex conjugate of 1/;. If we choose this as our scalar product then a similaranalysis would show that cCa == cC so that the operator is now self-adjoint. Thus the choice ofscalar product dictates in many instances whether or not an operator is self-adjoint. SometimescCa == cC but the boundary conditions for 4J are different from those for 1/;, in which case theoperator is classified as non-self-adjoint. An operator is not fully defined until the boundaryconditions are also given since this determines the domain of the operator. Generally, whenwe have complex eigenfunctions it is preferable to use the scalar product involving 4J1/;*. Themathematical theory for differential and integral operators that involve complex functions isbased on the use of the scalar product involving 4J1/;* because this allows the concept of length(norm) to be defined as a positive real quantity, i.e., the length of a function 4J would be givenby

II (Pll = ((P, (p)1/2 = [1 0

(P(P*'". 1/2

which is analogous to the length of a Euclidean vector A with real components. The latter isgiven by

72 FIELD THEORY OF GUIDED WAYES

(40)

and is always a real positive quantity. These concepts are described more completely in Section2.20.

2.6. MULTIDIMENSIONAL GREEN'S FUNCTIONS AND ALTERNATIVE REPRESENTATIONS

[2.10], [2.13], [2.14]

We have noted that there is more than one representation for a one-dimensional Green'sfunction. Also we have noted that the Green's function has poles for certain values of theseparation constant X. These poles are called the spectrum of the Green's function. For finite-range problems the spectrum is discrete, but for infinite-range problems the spectrum iscontinuous and the Green's function will have branch point singularities instead of poles.

In two and three dimensions there are many different possible representations for the Green'sfunctions. For this reason we wish to present a unified approach to multidimensional Green'sfunctions.

First we will present a theorem that is useful in constructing two- and three-dimensionalGreen's functions from Green's functions for one-dimensional problems.

Theorem I:1 f I o(x -x')

-2. G(x, x , "A)d"A = - (/1r) C U X )

where C is a closed contour in the complex X plane that encloses all the singularities ofG(x, x', X).

Proof: Use the form

G( I '\.) == _~ 1/;n(x)1/;n(x')X,X,/\ LJ X-X

n=l n

and Cauchy's theorem to give

To complete the proof we show that this equals the right-hand side of (40) by expandingo(x - x') in terms of a series in the 1/;n. Let o(x - x') == E~lCn1/;n(x). We multiply bothsides by u(x)1/;m(x) and use (18) to obtain Cm == J; o(x - x ')u(x)1/;m(x) dx == u(x l)1/;m(x

').Hence

which completes the proof.

o(x - x') _ ~." ( ).1, ( ')( ') - LJ'Yn X vn X

U X n=l(41)

Example

Consider d 2G /dx2 +XG == -o(x -x'), G == 0 at x == 0, a. By using Method II we readilyfind that

sin VXx < sin JX(a - x ,G(x, x', X) == ----------

JX sin JXa

GREEN'S FUNCTIONS

A plane

c

Fig. 2.5. Contour C enclosing the poles of the Green's function.

73

Since G is an even function of ~ it has no branch points; its only singularities are poles at~a == ns: or A == (n« /a)2, n == 1, 2, 3, .... Theorem I gives

1 f I-. GdA == -o(x -x)21f'j c

where C is the contour shown in Fig. 2.5. Near An == (n1f' /a)2 we have sin ~a == sin a(~A+A) == sin[(~-A)a+n1f'] == cos nr sin(~-A)a~ (cos n1f')(~-A)a ==cos n1f'[(A - An) /(~ + A)]a. Thus the residue from the term 1/(~ sin ~a) at the nthpole is (2A)/(aA cos n1f') == (-I)n(2/a). The residue expansion of the integral maynow readily be found:

1 f G ~ Loo

n 2 . nxx; . nt:-. d/\ == (-1) - SIn -- SIn -(a - x»21f'j c a a an=l

00 2 . n1f'X . nsx'== - L - SIn -- sIn --

n=l a a a

since

. n1f'() . nxx-,SIn - a -x> == -cos n1f' sIn --.a a

The latter series is easily verified to be the expansion of - o(x - x') in conformity withTheorem I.

We will use the above theorem to show that multidimensional Green's functions can beconstructed in the form of contour integrals taken over the spectrum of a product of associ-ated one-dimensional Green's functions. The theory will be developed by means of suitableexamples.

Green's Function for Two-Dimensional Laplace's Equation

Consider

G==Oatx==O,a; y==O,b. (42)


y

b~-----

.X', y'

--+-------...---~ xa

Fig. 2.6. Line charge along z.

Apart from a factor l/eo the function G represents the electric potential from a z-directed linecharge inside a conducting rectangular tube as shown in Fig. 2.6.

Let £ == 82 j8x2 + 82 j8y 2 and £x == 82 j8x2, £ y == 82 j8y 2 . The equation £1/; == 0

separates into two one-dimensional equations £x1/;x(x) + Ax1/;x == 0 and £y1/;y(y) + Ay1/;y == 0with Ax +Ay == 0 provided we assume a solution in product form, i.e., 1/;(x, y) == 1/;x(x)1/;y(y).Corresponding to each of the separated equations there are associated one-dimensional Green'sfunction problems of the form

Gx == 0 at x == 0, a

Gy==Oaty==O,b. (43)

We will now show that the solution to the original problem (42) can be expressed in termsof the solutions to the simpler one-dimensional problems (43). Specifically, we will show that

G(x, x', y, y') == 2- ~f Gx(x, x', Ax)Gy(y, v', -Ax)dAx7rJ C X

-If I I==~ Gx(x, x , -Ay)Gy(y, y , Ay)dAy7r} Cy

(44)

where Cx encloses the singularities of Gx and excludes those of G y and similarly C y enclosesonly the singularities of Gy. The formula (44) synthesizes an appropriate two-dimensionalGreen's function from the associated one-dimensional Green's functions.

To prove (44) we only need to show that it is a solution of (42). Clearly G satisfies the correctboundary conditions in view of the conditions imposed on Gx and Gy at the boundaries. Wealso have

£G == (£x + Ax +£y + Ay)G

- 1 f 0 I I- --2' (a1Jx + Ax +£y + Ay)Gx(X, x , Ax)Gy(y, y , Ay)dAx7r) Cx

==-21 .f [-o(x-x')Gy(y,y', -AX)-GX(X,X', Ax)O(y_y')]dAx7r} c.

upon using (43) to replace (£x +Ax)Gx by -o(x -x') and (£y +Ay)Gy by -o(y - y'). SinceC; excludes the singularities of G y we obtain £G == (lj27rj)fc

xGx(x, x', Ax)dAx o(y -

y') == -o(X - X')o(y - y') upon using Theorem I (note that a == 1 in this problem). In otherwords, the contour integral of Gy is zero because no singularities are enclosed by virtue ofthe manner in which Cx was chosen. A similar proof holds if we use the form involvingintegration around the contour C y in (44).

We will now illustrate the above procedure by constructing G directly and then show thatthe same solution is obtained from using (44) and solutions to (43).

The normalized eigenfunctions of the equation d2Gx j dx? + AxGx == 0 which vanish atx == 0, a, are yff1Q sin (nxx Ia), n == 1,2 ... , and Ax == (n1rja)2. These form a completeset so we may assume that

00

G(x, x', y, y') == L:an(y)<Pn(x),n=I

n,. ~. nxx'J!n == - SIn --.a a

When we substitute into the equation for G, i.e., into (42), we obtain

00 [ 2 ()2 ]d an(y) nr8 dy

2 - a an (y) q)n(X) = -o(x - x')o(y - y').

We now multiply both sides by <Pm(x) and integrate to get

d2 ()2am m1r I I--2 - - am == -<Pm(x )o(y - y )dy a

since the functions <Pn are orthogonal. We can readily solve this equation for am(y) accordingto Method II given earlier. We find that

n,. ( ') inh mx inh mx b'J!m X SI -y< SI -( - y»a a

am(y) == -----------mx inh mxb-SI --a a

upon choosing <PI == sinh (mxy jb), q>2 == sinh [(m1r ja)(b - y)]. Our solution for G is thus

o n1rX 0 nxx' 0 nt: 0 nt:00 2 SIn -- sIn -- SInh-y< SInh-(b - y»

G == L:- a a a a

1 nt: inh nxbn= SI --a

(45)

We could equally well have made the first expansion with respect to y and we would thenhave found that

(46)

We can find yet another form for G by assuming that it can be expanded as a double Fourier

76

series. Thus let


Fig. 2.7. The proper contour ex.

00 00

G ,",,~C . nxx . mxy== L.-t nm SIn -- SIn --.

a bn=lm=l

If we substitute this solution into (42) we obtain

~~ [(n7r)2 (m7r)2] . nxx . mxy "- ~~Cnm a + b sm a sm -b- == -o(x -x )o(y - y).

The coefficient C nm may be found by multiplying both sides by sin (n7rxfa) sin (mxy /b) andintegrating over x and y. The final result is

. nxx . nxx' . mxy . mxy'00 00 SIn -- sIn -- sIn -- sIn --

G =~~a~ a a b b

n=lm=l ( na1rr+ (~1rr (47)

(48a)

(48b)

We will now show how the different solutions (45)-(47) are all contained in the generalformulation (44). We first need the solutions to the one-dimensional Green's function problemsgiven in (43). By using Method II these are readily found in the forms

sin J}:;x< sin J}:;(a - x ,Gx == -----------

J}:; sin J}:;a

sin j>:;y< sin j>:;(b - y»Gy == ---=----_...:.....--_---

j>:; sin j>:;bWe note that Gx(x, x', Ax) has poles at Ax == int:/a)2, n == 1, 2, ... ,and that Gy(y, y', -Ax)has poles when sin J-Axb == 0 or at Ax == -(n7r/b)2. Thus we choose the contour C, asshown in Fig. 2.7. According to (44) we now have

GREEN'S FUNCTIONS

'.::::------..---............ -1... ....

.....,<,

Fig. 2.8. Deformation of contour ex into C y.

77

The residue expansion? of this integral in terms of the residues at the poles of Gx(Ax) givesthe following solution for G:

. n7f' . n7f' ( ) . . nt: ..nt: b00 SIn -X< sIn - a -x> sIn J-Y< sIn J-( - Y»

G = -2: a a a a.nx a . .nsb

n=l J-- cos nt: sin J--a 2 a

. n7f'X . n7f'X'. n7f' n7f'00 2 SIn -- SIn -- SInh -Y< sinh -(b - Y»=2: a a a a

n= l inh nxbn7f'SI --a

which is the same as (45).Since the integral over any portion of a circle with infinite radius is zero, the contour Cx

may be deformed into the contour Cy as shown in Fig. 2.8. If this is done the integral may beevaluated in terms of the residues at the poles of Gy ( -Ax) which occur at Ax == -(n7f' /b)2.This evaluation would give the same solution as (46).

If we construct the solution for G y according to Method I we would obtain [see (29a)]

2 . macy . mxy'00 - SIn -- SIn --

O( 'A) __"b b by Y,y, y - ~ Xy -(m7rjbi (49)

If we use (48a) for Ox and (49) for Oy in the formula (44) we would obtain the Green'sfunction solution given by (47) provided that we integrate around a contour C; enclosing thesingularities of Ox(x, x', Ax). We leave the details as Problem 2.6 to be carried out by thereader.

The eigenfunctions l/!nm(x, y) of the two-dimensional Laplacian operator are solutions of

(50)

2The residues may be found by evaluating the derivative of the denominator with respect to Ax at the poles.


y

bt------......

~--------~ xx' a

_______I_Ig_e_iw

_

t ----~ zz'

Fig. 2.9. A line source in a rectangular waveguide.

where 'if;nm == 0 on the boundary shown in Fig. 2.6. The normalized eigenfunctions are

~. nxx . mxy

- SIn -- SIn --ab a b

and Anm == (n7r/ a)2 + imt:/ b)2. When these eigenfunctions are used to solve (42) the solutionis that given by (47). Thus (47) represents the complete eigenfunction expansion for theGreen's function. The other solutions given by (45) and (46) are eigenfunction expansionswith respect to x or y and closed-form solutions with respect to y or x, respectively.

2.7. GREEN'S fuNCTION FOR A LINE SOURCE IN A RECTANGULAR WAVEGUIDE

Figure 2.9 shows a uniform current filament directed along y, at x', z', in a rectangularwaveguide. The current has a time dependence ej wt and is not a function of y. This currentsource will excite TEno modes only. The vector potential is a solution of (A == A yay == 'if;ay)

(82 8

22) .t I I

8x2 + 8z2 + ko 'II = -/LoIgO(X - x )o(z - z ) (51)

with 'if; == 0 at x == 0, a and 'if; representing outward-propagating waves at Iz I approachinginfinity. If we solve the Green's function problem

(52)G == 0; x == 0, a(82 8

22) I I- + - +ko G == -o(x -x )o(z -z),

8x 2 8z2

then 'if; == p,oIgG and E, == -jwp,oIgG. In (51) and (52) kij ==w2p,O€O ==w2/C2.Consider now the homogeneous equation (82/8x2 + 82/8z2 + kij)G == O. If we assume

G(x, x', Z, Zl) == Gx(x, x')GZ(z, z') we obtain

so we must have

and

(53)

From these we can identify the associated one-dimensional Green's function problems to be

Gx == 0; x == 0, a (54a)

(54b)

Gz is an outward-propagating wave as lz] -+ 00.

We may readily solve (54a) by Method I or II. If we use Method I we obtain [see (29a)]

. nxx . nxx'00 SIn -- sIn --G -_~~ a a

x - LJ a A _ n27r2ja2 ·n=l x

Since the range in z is infinite we use Fourier transforms to solve (54b). We will defineGz({3 ) to be the transform of Gz(z), Le.,

(55a)

Then

(55b)

If we take the Fourier transform of (54b) and note that the transform of 82Gz/8z

2 is - (32Gz(obtained through two integrations by parts) we find that

(56)

(57)

We wish to consider Az as a complex variable and so we will arbitrarily choose that branchof the two-valued function A which has Imag. A < O. In the complex (3 plane o, thenhas two poles located as shown in Fig. 2.10. The solution for Gz is given by

1 /00 e-j {3 (z - z ' )

G z == - f\ d(3.27r -00 «(3 - V Az)«(3 + A)

When z > z' the factor e- j{3(z-z') becomes exponentially small (we write (3 as (3' + j(3") inthe lower half plane [proportional to e{3"(z-z') with (3/1 < 0]. Hence we can close the contourby a semicircle of infinite radius in the lower half plane and evaluate the integral in terms ofthe residue at the pole at (3 == A. For Z < z' we can close the contour in the upper half planeand evaluate the integral in terms of the residue at - A. Note that there is no contribution tothe integral from the semicircles and that a negative sign enters when the contour is traversedin a clockwise sense. We are thus led to the following solution for Gz :

{-2!;>:;e-h / >;; (Z- ZI ) , z > z'

Gz == .__J_ejy'}\;(Z-Z') z < z'2A '

-.JI;•

80

»>:-:/

//

II


j/3"

----<,"'\ for z < z'

\\\

•-n;----+--------+----------t--~ /3'

I/

/AforZ>Z'

/"--./'"

Fig. 2.10. Pole locations in the complex {3 plane.

which may be also expressed as

Gz

== __J_'_e-j~lz-z'l

2A all z. (58)

Note that with Imag. A < 0 chosen as the branch of the function A that Gz will tendtoward zero as Iz I~ 00 for complex Az . Also note that Gz does not have any poles but insteadhas a branch point at Az == O. If we wish to always remain on the branch for which A has anegative imaginary part, then the angle of Az in the complex plane must be restricted to - 211"to O. This may be accomplished by placing a branch line (or cut) along the positive real axisbut an infinitesimal distance above it as shown in Fig. 2.11. This forces us to measure theangle of Xz from the positive real axis in a clockwise or negative sense and hence restrictsthe angle to the range 0 to - 211". The spectrum of Gz is the continuous range of values ofAz along the branch cut (singular line, which if crossed changes the sign of A). This is inaccord with our earlier statement that for infinite-interval problems the Green's function hasa continuous spectrum.

We now apply (44) to obtain G in the form

G(x, x', Z, ZI) == -21 .f Gx(x, x', Ax)Gz(Z, Z', k6 - Ax)dAx7r} c.

== -21 .f Gx(x, x', k6 - Az)Gz(Z, Z', Az)dAz· (59)7r} c.

Note that when we integrate over Ax we use the condition (53) to express Az in terms of Axand vice versa. In the Ax plane the poles of Gx are at Ax == (n1l" /a)2 and the branch linefor Gz becomes the line along which Imag. Jk5 - Ax == O. This is the line extending fromAx == k5 (branch point) along the negative axis. The contour ex is chosen to enclose the polesbut not the branch point at k5 as shown in Fig. 2.12(a). On the other hand, if we choose tointegrate over Az, then the contour C z is chosen to enclose the branch cut for Gz but suchthat the poles of Ox, which now occur when k5 - Az == tn« /a)2 or Az == k5 - (n« /a)2, are

GREEN'S FUNCTIONS

Az plane

Sr. point

/

Sr. cut

81

L Az is negative

Fig. 2.11. Branch cut for Gz(z, z', Az)·

Sr. cut - 00 to k~

(a)

k§ - (~rPoles

(b)

Sr. cut 0 to 00

Fig. 2.12. Proper choice of the contours Cx and Cz .

excluded [Fig. 2.12(b)]. If only the TE10 mode propagates, then k o > x[a but k o < nr]afor n > 1. The contour C z can be deformed to cross the branch cut and thus exclude the poleat Az == k5 - (1r /2)2 without changing the value of the integral, as long as the branch point isenclosed.

Either of the two prescriptions given by (59) may be used for determining G. If we use thefirst one then the integral may be evaluated in terms of the residues at the poles. We have

.n7rX . nrx'. -jvk5-Axlz-z'l 00 SIn a sm -a-

G == --1-1 Je L 2 dAx21rj ex aJk'fi - Ax n=l Ax - (n1r/a)

1~ 1 . nxx . nxx' -rnlz-z'l== - L....J - SIn -- SIn --ea n=l rn a a

(60)

where rn == J(n7r/a)2 -k5 and Imag. rn > O. If k o < nx]« for n > 1, then the solutionconsists of the propagating dominant mode for which r I == j Jk5 - 1r2 / a2 plus an infinitenumber of evanescent or nonpropagating TEno modes. These modes exist on both sides of thesource point at z'. If x' == a /2, then only the modes for n == 1, 3, 5, ... , etc. are excited.

If the prescription involving the contour C z in (59) is used, then the solution is expressedsimply as a branch cut integral which may be interpreted as a continuous sum (integral) over the


continuous spectrum of Oz in place of the sum over the discrete spectrum of Ox as exhibitedin (60).

We could equally well have chosen the branch Imag. A > 0 and would have arrived at thesame solution. The reader may find it of interest to make this choice and to show that (60) isstill obtained for the final result. If we make the change of variable w2 == kij - Ax the branchcut integral in (60) becomes an inverse Fourier transform.

At this point it will be instructive to show the relationship of this problem to the conceptsintroduced in Section 2.5. The equation of interest is

(61)

with 1/; == 0 at x == 0, a; 1/; bounded at z == ± 00; and f(x, z) being a source function. Wewish to solve this equation using the eigenfunctions of

Since - 00 ::; z ::; 00 we first Fourier transform this equation with respect to Z to obtain

(822 + A - W 2) Joo if;(x, z)ejwzdz = 0

ax -00

or

(a2 2) A

ax2+A-W 1/;(x,W) ==0.

For the x dependence we choose sin nxx[a and we then find that a solution exists if

)

22 n7r

A=W +(0 ·Corresponding to this eigenvalue is the unnormalized eigenfunction

_ -jwz . nxx1/Inw(x, z) - e SIn --.

a

The normalization factor is obtained from the integral

JOO 1° aJooif;nw(x, z)if;;w'(x, z)dxdz = "2 e-j(w-W')Z dz-00 0 -00

a ,== 27r-o(w - W )

2

upon using the Fourier transform property

(62)

(63)


This normalization procedure is analogous to setting the integral of the product of the one-dimensional eigenfunction sin tnxx fa) with itself equal to unity. The normalization factor andorthogonality properties are expressed through the integral

]

00 1° nxx m1rx· ,sin -- sin __ej(w-w )Z dx dz-00 0 a a

== 1raOnmo(w - w') (64)

where the Kronecker delta onm == 0 if n =1= m and. onn 1. When a continuous spectrumis involved the normalization-orthogonality integral results in a Dirac delta function. Thenormalized eigenfunctions are seen to be

1 . nxx -jwz-- SIn --e .Vira a

We can expand the solution of (61) in terms of these as follows: Let

00 ]00 mxx ., dw'1/;(x, z) == L Cm(w') sin __e-jWz__

m=l -00 a Vira

(65)

which is equivalent to solving (61) using a combination of Fourier series and a Fourier trans-form. The sum over the continuous spectrum of eigenfunctions is an integral over w'. We nowsubstitute into (61) to obtain

00]00 1 mxx .,L c;Cm(w')[k~ -~] sin --e-jWZdw' == -f(x, z)m=l -00 Y 1ra a

where ~ == wa + imt: / a)2. In order to find C n (w) we multiply by the complex conjugatedeigenfunction (1/Vira) sin (n1rx/a)e jwz, integrate over x and z, and use (64) to obtain

00 ]00L (k~ - ~)Cm(w')onmo(w - w')dw'm=l -00

== (k~ - ~)Cn(w)

1 ]001° nxx .== -. c; f(x, z) sin -ejWZdxdz.y1ra -00 0 a

The expansion coefficients are now known and thus the solution for 1/; is readily constructed:

~jOO 1 . nxx .1/;(x, z) == L...J ---- SIn __e-jWZn=l -00 1ra(~ - k~) a

]

00 1° nxx' . ,x f(x', z") sin __ejWZ dx' dz' dw.-00 0 a(66)

We can rewrite this solution in a form that will enable us to identify the Green's function

84

operator. We have

t/;(x, z) == joo t"f(x', z')-ooJo


(68)

. ntoc' . nxx00 00 SIn -- SIn --

x L:j a 2 a e-jW(z-z') dw dx'dz' (67)n=l -00 1ra("A - ko)

which shows that

. nxx' . nxx00 00 SIn -- SIn --

G(x', z'; x, z) == L:j a 2 a e-jW(z-z')dwn=l -00 1ra("A - k o)

where "A == w2 + (n1r/a)2. This Green's function is expressed as a complete eigenfunctionexpansion in terms of the eigenfunctions of the operator given after (61). Its use is accordingto the sequence of operations shown in (66) so as to generate the eigenfunction expansionof t/;(x, z). If the integral over w in (68) is carried out, then the eigenfunction expansion ofthe Green's function gets converted over to the expansion in terms of the waveguide modesas given by (60). Note that the expansion of G in terms of modes is not the same as theeigenfunction expansion even though they are equivalent. The integrand in (68) has poles oforder one at w == ± jfn so the mode expansion is readily found from an evaluation of theintegral in terms of the residues at the poles.

When the order of integration over x', z' and the integration over wand summation overn are interchanged from that shown in (66) the interchange must, in general, be justified withregard to its validity. When the integration over w leads to well-defined functions, which itdoes for the present problem, it can be carried out first.

The series that occurs in the mode expansion of G is convergent for all values of x =1= x' ,even if z == z' so that the exponential decay of the individual terms vanishes. For large valuesof n we have Tn ~ n1r / a so the behavior of G is governed by the dominant series

00 1 'S L: . nxx . nxx -n7rlz-z'l/a== - SIn -- sIn --e

ns: a an=l

L:oo

1 [n1r(X-X') n1r(x+x')] -n7rlz-z'l/a== -- cos - cos en=12n1r a a

The sum of series of this type is given in the Mathematical Appendix. Thus we find that Gbehaves like the function S given by

1S == -- In

41r

h2 1r(z - z') 2 1r(X - x')cos 2a - cos 2a

h2 1r(z - z') 2 1r(X+x')cos 2a - cos 2a

(69)

GREEN'S FUNCTIONS

When z - z' and x - x' are both very small we readily find that 8 has the limiting form

85

18 "J - - In

41r

(x - X')2 + (z - Z')2

(')2 2 (2a) 2 . 2 1r (x +x')z -z + - sIn

7r 2a(70)

which shows that G has the expected logarithmic singularity near the line source. For aninfinitely long line source in free space the static vector potential is given by (I is the currentin the line source)

A = _/L01

In[(x -x'i +(z -z'ily 47r

which is similar to the expression in (70).In practice it is often useful to extract the dominant part (singular part) of the Green's

function as a closed-form expression since the remainder is a rapidly converging series; i.e.,we express G in the form

G ==8 +(G -8)

where (G - 8) converges rapidly because the terms in 8 become essentially equal to thosein G for large n. The singular behavior of the Green's function is the same as that for staticconditions (ko == 0).

We can use the eigenfunctions of the differential operator in (61) to also obtain a represen-tation of the delta function operator 5(x - x')5(z - z') that is valid for functions in the domainof the operator. A function htx, z) in the domain of the operator can be expanded as follows:

~jOO 1. nxx "hex, z) == Z:: An(w).~ sm _e-j WZ dw.

n=l -00 Y 7ra a

By following the standard procedure we find that

/00 10 nxx' " ,

v!1WAn (w ) == h(x', z") sin __eJ WZ dx'dz'-00 0 a

and hence

~jOO 1 . nxx "h(x, z) ==~ - sm __e-j WZ

n=l -00 7ra a

j 00 1° n x'x hex', z') sin ~ejwZ' dx' dz' dw.-00 0 a

We now rewrite this last expression as

hex, z) == JOO t"h(x', z')-ooJo

[~jOO 1 . nsx' . nxx -j"W(Z-Z') d ] d ' d 'x ~ - sm -- sm -- e w x Zn=l -00 7ra a a

(71)

86

from which we can identify the delta function operator as


00]00 1 nxx' nxx . ,o(x -x')o(Z -Z') == L - sin -- sin __e-Jw(z-z )dw.

n=l -00 7ra a a(72)

This result can also be obtained by carrying out a formal expansion of the two-dimensionaldelta function in terms of a Fourier series in x and a Fourier integral in z.

From the results given above it can be correctly inferred that the Green's function operatorand the delta function operator or unit source function can be represented in terms of theeigenfunctions of the differential operator in a multidimensional space. When one or more ofthe coordinate variables extends over an infinite interval (or semi-infinite interval) the relatedeigenvalue spectrum is a continuous one rather than a discrete one. For a continuous spectrumthe sum over eigenfunctions becomes an integral and the normalization integral results in adelta function replacing the Kronecker delta onm. In rectangular coordinates the continuousspectrum is described by a Fourier transform. In cylindrical and spherical coordinates we willfind that the continuous spectrum is described in terms of the Hankel transforms.

2.8. THREE-DIMENSIONAL GREEN'S FUNCTIONS

The theory presented in the previous sections may be applied to three-dimensional problemsalso. We will illustrate the case of the scalar Helmholtz equation in rectangular coordinates.We have

(\72 + k5)G == -o(x - x')o(y - y')o(z - z'). (73)

The equation (\72 + k5)t/; == 0 separates into three one-dimensional equations of the form82t/;x j8x2 + Axt/;x == 0, 82t/;y j8y 2 + Ayt/;y == 0, 82t/;z j8z2 + Azt/;z == 0 with

Ax +Ay +Az ==k5·

Consequently, the associated one-dimensional Green's function problems are

d2Gx ,

--2 + AxGx == -o(x - x )dx

(74)

(75a)

d2Gy ,

--2 + AyGy == -o(y - y ) (75b)dy

d2G

z , 75dz2 + AzGz = -o(z - z ) ( c)

along with appropriate boundary conditions.When (75a)-(75c) have been solved then the solution to (73) is given by

2

G(X,y'Z'X"Y"Z')==(2~~) f f Gx(Ax)Gy(Ay)Gz(k6-AX-Ay)dAXdAyJ c, c,

== (2~~)2f rGx(Ax)Gy(k6-i; -Az)Gz(Az)dAxdAzJ cxJcz

= (2~~rfc f c Gx(k6 - i; - Az)Gy(Ay)Gz(Az) dAy dAz,y z (76)


Note that there is an integration over two of the Ai and that the third A is expressed in termsof the two being integrated over by means of the condition (74) on the separation constants.The contours C i enclose only the singularities of the G i with i == x, Y, z. The proof is readilydeveloped by noting that

(V2 +k~)G = [(::2 + AX) + (:;2 +Ay ) + (::2 +k~ -Ax - Ay ) ] G

==(2~~)2f f [-Gy(Ay)Gz(k6-AX-Ay)O(X-x')J c, c,

- Gx(Ax)Gz(k6 - Ax - Ay)O(Y - Y') - Gx(Ax)Gy(Ay)O(Z - z')] dAx dAy.

The integral around ex of the first term vanishes while the integral around Cy of the secondterm vanishes because no singularities of Gz are included. The integral of the third term gives- o(x - x')o(y - y')o(z - z') by application of Theorem I twice.

The above technique is applicable in other coordinate systems also. Some applications aregiven in papers by Marcuvitz [2.13] and Felsen [2.14]. The method can be applied in a veryeffective way to obtain the Green's function in the form of a very rapidly converging seriesfor the problem of diffraction by a very large cylinder or sphere [2.13]-[2.15].

2.9. GREEN'S fuNCTION AS A MULTIPLE-REFLECTED WAVE SERIES

The Green's function can be constructed in terms of the free-space waves radiated by thesource and with these waves undergoing multiple reflections at the boundaries. This techniqueis useful in a number of different problems for the purpose of generating a series representationof the Green's function that may be more rapidly converging. The method of images is a specialcase of the method of multiple-reflected waves. We will illustrate it by considering a uniformline source that is parallel to the z axis and located between two concentric cylinders of radiia and b as illustrated in Fig. 2.13. The mathematical statement of the problem is

o(r --; r') 0(8),r

l/; == 0 at r == a, b. (77)

We can expand l/;(r, 0) into a Fourier series in 0; thus

ex)

l/;(r, 0) == Lfn(r) cos nO.n=O

(78)

We now substitute this expansion into (77), multiply by cos nO, integrate over 0 from 0 to21(', and then obtain

1 8 8fn n22 EOn ,

-8-r-8 - 2 f n +kOfn == --2,o(r -r)r r r r l('r (79)

where EOn == 1 if n == 0 and equals 2 for n > O. In this equation the source function can beviewed as a uniform cylindrical sheet of current at r'. When r i= r' the homogeneous equationis Bessel's differential equation for which the solutions are Bessel functions of order n, i.e. ,In(kor) and Y n(kor) or combinations of these functions in the form of Hankel functions

y

r

....)I'---+-.........-+-~--~x

lI/=O

lI/=OFig. 2.13. A line source radiating between two cylinders.

H~(kor) and H~(kor) of the first and second kinds where

H~(kor) == J n(kor) + jYn(kor)

H~(kor) == J n(kor) - jYn(kor).

When the argument kor is large the asymptotic forms for the Hankel functions are

These asymptotic forms show that the Hankel function of the first kind represents an inward-propagating cylindrical wave, while the Hankel function of the second kind represents anoutward-propagating cylindrical wave. Bessel's differential equation is a Sturm-Liouville equa-tion with p(r) == r, q(r) == -n2 [r , and (J(r) == r, and k5 can be regarded as the eigenvalueparameter A.

Initially we will ignore the boundaries at r == a, b. We can then choose

r < r'

r > r'

as the solution for In(r). At r == r' the solution must be continuous so

We can integrate the differential equation about a vanishingly small interval centered on r' toobtain

r 2 1

r din I + == - €On == kor' (Bn dHn -An dHn ) I

dr" 21r dkor dkor"


which fixes the step change in the first derivative across the point r'. These two source-relatedboundary conditions allow us to find the amplitude coefficients An and B n- The solutions forAn and B; involve the Wronskian determinant

W =H2dH~ _HI dH~n dkor n dkor

evaluated at r == r '. We can use the property that p(r)W(r) is a constant and then use theasymptotic expressions given earlier to evaluate the constant since W(r') is readily found fromthe asymptotic forms. Since p(r) == r we find that

I 4jW(r) == -kI·

7r or

After solving for An and B; we obtain

(80)

We now take the effect of the boundaries into account in the following way. The inward-propagating wave, apart from a multiplying factor, is H~(kor) and is reflected at r = a toproduce an outward-propagating wave H~(kor). At r = a the reflected wave must cancel theincident wave so the reflection coefficient r a is given by the condition

H~(koa) + fa H~(koa) == o.

Thus the reflected wave is

H~(koa)H 2(k r).H~(koa) n 0

This reflected outward-propagating wave is in turn reflected at r == b and generates a newinward-propagating wave f b H~(kor) where

This new wave is again reflected at r == a. The sum of all of the multiple-reflected wavesgenerated by the primary inward-propagating wave is

raH~(kor) + rarbH~(kor) + r~rbH~(kor) + r~riH~(kor) + ...00

= L [r~rb-lH~(kor) + r~rbH~(kor)].m=l

(81)

The initially outward-propagating primary wave H~(kor) is reflected at r == b and subse-quently undergoes multiple reflections thus creating the wave series

fbH~(kor) + fafbH~(kor) + faf~H~(kor) + ...00

= L [rbr~-lH~(kor) +r~rbH~(kor)].m=l

(82)

The solution for the function f n(r) is a superposition of the primary solution given by (80)and the two wave series (81) and (82) multiplied respectively by the factors - jEonH~(kor')/8 and-jEofl~(kor')/8. Hence a solution for in is

f JEOn HI k 2 kn(r) == --8- n( or<)Hn( or»

_jEoni (rafb)m-I{[H~(kor) +r Jlf~(kor)]r"H~(kor') + [H~(kor)+r "H~(kor)]rJlf~(kor')}

8 m=l

We can use this series representation in (79) to complete the solution for the Green's function.The method just described is quite general and can be applied to problems with other

boundary conditions by using the appropriate expressions for the reflection coefficients. If themethod is applied to the problem of a line source in a rectangular waveguide as analyzedearlier, the x-dependent part of the Green's function is the image series.

For the rectangular waveguide problem the Green's function can be expressed as an inverseFourier transform in the form

1 100

. ( ')G(x, x'; z, Z') == 2 f(x, x', w)e-JW z-z dw7C' -00

where f is a solution of

(d2 2) f I- +1' == -o(x -x)

dx 2

(84)

(85)

(86)

with 1'2 == k5 - w2 and f == 0 at x == 0, a. One solution for this Green's function is givenby (60) and also by (68). If we ignore the boundary conditions at x == 0, a, then the primaryfield that is a solution of (85) can be found using Method II and is

f = -l./-hlx-x/,.

The primary field ei"(x-x') incident on the reflecting boundary at x == 0 produces the multiple-reflected wave series

00

== L(fnf~-I e-j"(x + fnf~ ej"(X)e-j"(x'

n=l

(87)

where I' == -1 and fa == _e-2j "(a . In a similar way the primary wave e-j"(X-x') incident onthe boundary at x == a produces the multiple-reflected wave series

ej"(x' [fa ej"(x + I'Ta e - jvx + ff~ ej"(x + ...J00

= L(r~rn-l eh x + ~rn e-hX)eh x'.

n=l

(88)

By combining the two wave series and the primary field we find, after simplification, that the

GREEN'S FUNCTIONS

solution for f is

• [ 00 00 ]It», x', w) = - f", e-h1x-x'l + n~~e-hlx-x'-2nal - n~ooe-hlx+xl+2nal

91

(89)

where the prime on the summation sign signifies that the n == 0 term is omitted. This seriescan be used in (84) to obtain the Green's function G. The series represents waves generatedby the images of the line source.

The integral in (84) over w can be evaluated by using the Fourier transform relation

(90)

By using this result the Green's function can be expressed as a series of cylindrical wavesradiated from the original line source and its images in the two side walls of the waveguide.This representation is

• 00J I- 4: L [H~(koJ(x -x' -2na)2 +(z -Z')2) -H~(koJ(x+x' +2na)2 +(z -Z')2)].

n=-oo(91)

The expansion of G in terms of waveguide modes can be obtained by using the Poissonsummation formula to transform the series in (91) into series in terms of the modes (see theMathematical Appendix).

It should be apparent by now that there are many equivalent expressions for the Green'sfunction for a given problem. Which form is best depends on the physical nature of theproblem to be solved. As a general rule, if one series representation is very slowly convergingan alternative series representation usually converges more rapidly. For example, in (91) onlythe first term is singular at x == x', z == Z' and since Hij rv - (j /7r) In [(x - X')2 + (z - ZI)2]

for small arguments the singular part of the Green's function is known explicitly.

2.10. FREE-SPACE GREEN'S DYADIC FUNCTION

The vector potential A is a solution of the inhomogeneous vector Helmholtz equation

(92)

(93)

In Section 1.9 it was shown that a particular solution is

fff -jklr-rll

A(r) = :7rjjjJ(rl)elr_rll dV1•

vIf p,J is replaced by a unit vector source (ax +ay +az)o(rl -ro), we find at once by substitutinginto (93) that the solution is

(94)


and this may be considered as a vector Green's function. For a general current distributionJ, the solution would be obtained by a superposition integral provided we make the rule thatthe x component of the current Jx is to be associated with the unit vector ax in the Green'sfunction, and similarly for J y and J z- This rule is readily taken account of by introducing thedyadic Green's function, which is defined as a solution of the equation

(95)

where I is the unit dyadic or idemfactor axax +ayay +azaz. Since multiplying the source byI simply multiplies the solution by I also, we have the result

-jklr-rol- -eG(r, ro) = I I l47r r - ro

(96)

Each component of the current gives rise to a vector field, and hence the general linear relationbetween the current and the field is a dyadic relation. Therefore, for vector fields the Green'sfunction is a dyadic quantity. Equation (96) is the free-space dyadic Green's function for thevector Helmholtz equation. For an arbitrary current distribution the solution for the vectorpotential is

A(r) = IJ.jjjG(r, ro)·J(ro)dVo

v

(97)

since the scalar product of J with G associates Jx with ax, etc. The above dyadic Green'sfunction is a symmetrical dyadic so that J. G == G·J. It is also symmetrical in the variablesrand roo The use of dyadic Green's functions for the vector Helmholtz equation results in aconsiderable advantage in notation.

2.11. MODIFIED DYADIC GREEN'S FUNCTIONS

In the presence of boundaries on which the electromagnetic field must satisfy prescribedboundary conditions, the free-space dyadic Green's function suffers from the same short-comings that the free-space Green's function for Poisson's equation does In the solution ofelectrostatic-potential boundary-value problems. The dyadic Green's functions are, therefore,modified so as to facilitate the solution of the given boundary-value problem. The sort ofmodification we should make is best determined by an examination of the type of solution weget by using the free-space dyadic Green's function.

Consider a region of space V bounded by a closed surface S, within which a currentdistribution J exists. Let n be the unit inward normal to S as illustrated in Fig. 2.14. Considerthe following dyadic quantities:

V'o·[A X (Y'o X B)]

V'o·[(V'o X A) X B]

V'o·[(V'o·A)B]

V'o·[A(V'o·B)]

(98a)

(98b)

(98c)

(98d)

GREEN'S FUNCTIONS

s

Fig. 2.14. Region V with source function J.

93

where B is a symmetric dyadic. By expanding the above expressions, we shall be able toobtain a combined vector and dyadic equivalent of Green's second identity. The dyadic Bwill be replaced by the dyadic Green's function later on. Although B is a symmetric dyadic,\70 X B is not symmetric, and special care must be taken in expanding the above expressions,and also in the order in which the terms are written. Perhaps the easiest way to proceed is towrite the dyadic B as the sum of three vectors associated with the unit vectors ax, ay , az , thatis,3

Thus, for (98a) we have

\7o-[A X (\70 X B)] == \7o-(A X \70 X B1)ax

+ Vo-(A X \70 X B2)ay + \7o-(A X \70 X B3)az •

Each term may now be expanded according to the rules of vector calculus to obtain

\7o-(A X \70 X B1) == (\70 X A)-(\70 X B1) - (\70 X \70 X B1)-A

== (\70 X A)-(\70 X B1) - A- \70\70-B 1 + A- \7~Bl (99)

and similarly for the other two terms. By arranging the terms in the expansion so that thevector B1 always appears on the right-hand side of each term, we may again associate B1 withax, and hence we can combine the end results for all three associated vectors into a dyadicagain. Thus the expansion of (98a) gives

\7o-[A X (\70 X B)] == (\70 X A)-(\7ox B) -A-(\7o\7o-B) +A-(\7~B). (l00a)

Expanding (98b)-(98d) in a similar fashion, we obtain

- - 2 - -\70-[(\70 X A) X B] == (\7o\7o-A)-B - (\7oA)-B - (\70 X A)-(\70 X B)

\70- [(\7o-A)B] == (\7o-A)(\7o-B) + (\70 \7o-A)-B

\70- [A(\7o-B)] == (\7o-A)(\7o-B) + A- (\70 \7o-B).

Adding (l00a), (l00b), and (lOOd) and subtracting (lOOc) from the sum gives

(100b)

(100c)

(l00d)

\70- [A X (\70 X B) + (\70 X A) X B + A(\7o-B) - (\7o-A)B]2 - 2-== A-(\7oB) - (\7oA)-B (101)

3Section A.2.


which may now be used in the divergence theorem to give us the equivalent of Green'ssecond identity. Integrating (101) over the volume V and converting the volume integral of thedivergence to a surface integral gives

iii[A. V'~B - (V'~A).B]dVov

= -IfO.[A X (V'o x B)

s

+ (Vo X A) x B +A(Vo-B) - (Vo-A)B] dSo (102a)

when n is the unit inward normal to S. A term such as VijA-B may be written as B- VijAsince B is a symmetric dyadic. The first term in the surface integral in (102a) may be writtenas

(0 X A)-(Vo x B),

the second term as

[0 X (Vo X A)]-B,

and the third and fourth terms as

n-AVo-B n-BVo-A.

Hence, we obtain the following final result, which is the basic starting point for the applicationof dyadic Green's functions to the solution of the vector Helmholtz equation:

iii[(V'~A).B - A· V'~B]av;v

= ff [(0 X A)·(V'o X B)

s

+ (0 X Vo X A)-B + o-AVo-B - o-BVo-A] dSo• (102b)

This result is also valid when B is not a symmetric dyadic.If the dyadic Bis now replaced by a dyadic Green's function which is a solution of (95) and A

is the vector potential satisfying (92), then VijB and VijA may be replaced by -k2B-I5(r-ro)

and -k2A - ILJ, respectively, in (102b). The terms in k 2 cancel, and, by using the property ofthe three-dimensional delta function, we obtain the following solution for the vector potentialA:

A(r) = iff jLJ(ro)·G(ro, r)dVo+If [(0 X A)·(V'o X G)

v s

+ (0 X Vo X A)-G + n-AVo-G - o-GVo-A] d'S«. (103)

GREEN'S FUNCTIONS

The electric field is given by

VV·AE == -jwA + -.-- == -jwA - V~

}W€J.L

while the magnetic field is given by

The scalar potential ~ is related to the vector potential A by the Lorentz condition

- jW€J.L~ == V·A.

95

(l04a)

(l04b)

(105)

If the boundary conditions specify that the tangential component of E vanishes on S, thenn X E == 0 on S, and hence from (l04a) we see that we must have n X A == 0 on S,n X V~ == 0 on S. The condition n X V~ == 0 on S means that the derivatives of ~ on thesurface S are zero, and hence ~ is constant, which may be taken as zero, on the surface S. If~ vanishes on S, then V0 •A == 0 on S also by virtue of (105).

An examination of (103) shows that in this case the first and last terms in the surface integralvanish. If we are able to modify our Green's function so that n X G and V0 •G vanish on thesurface S, then the whole surface integral vanishes, and the solution for A is given by thevolume integral of J.LJ. G only.

Sometimes it is more convenient to deal directly with the equation

(106)

which the electric field E satisfies. In this case the Green's function is taken as a solution ofthe corresponding equation

(107)

The appropriate Green's identity to use in this case is obtained by adding (l00a) and (l00b)and converting the volume integral of the divergence to a surface integral again to get

JJJ[('\i1o X V'o X A).B - A·(V'o X V'o X B))dVov

= - fj {[o X (V'o X A)).B + (0 X A).{V'o X B)}dSoo (l08)

s

Replacing A by E and B by G and using (106) and (107), the following solution for E isobtained:

E(r) = -jwp.JJJJ(ro)·G(ro, r)dVov

+ff [(0 X V'o X E).G + (0 X E). V'o X G)as;s

(109)


Since \7 X E == - jwp,H, the first term in the surface integral may be interpreted as a con-tribution from a surface current of density D X H on S. The second term may be considereda contribution from a magnetic current sheet of density E X D. If the tangential componentsof E, that is, n X E, are known or specified on S, then, if possible, we should modify ourGreen's function so that D X Gvanishes on S. Thus only the second term, the one which weknow, contributes to the surface integral. In practice, the particular boundary values we aregiven dictate the choice of which Green's function to use.

2.12. SOLUTION FOR ELECTRIC FIELD DYADIC GREEN'S FUNCTION

In this section we will derive a solution for the electric field dyadic Green's function infree space. This Green's function is a solution of (107) and will now be represented by thesymbol Ge to emphasize that it is the dyadic Green's function for (106). One way to obtain thesolution is to make use of the relationship between the electric field and the vector potentialin the Lorentz gauge. Since

E == -jwA - j(w/k~)\7\7.A

we readily see that for a current element J(ro) and using the vector potential Green's dyadicgiven by (96) that

E(r) == - jwp,oJ(ro)·Ge(ro, r)

= -jWJLO [J(ro)oG(ro, r) + :~ V'V'oG(r, ro)oJ(ro)]

[_e- j k OR 1 e-j k OR

]

= - jWJLoJ(ro) 0 I 41fR + k~ V'V' 41fR

where R == [r - roI. Note that the operation \7\7. G is equal to \7\7. 1 == \7\7 operating on thescalar Green's function. From the above relation we find that

_ (_ 1 ) e-j k oR

Ge(ro, r) = I + k~ V'V' 41fR' (110)

Another way to derive (110) is a generalization of the method used in Section 1.9 to obtaina solution for the vector potential. The scalar Green's function g(ro, r) == e-j k oR /47rR is asolution of the equation

(111)

so the left-hand side can be viewed as a representation for the delta function. The divergenceof (107) gives -k5\7·Ge == \7.lo(r-ro) == \7o(r-ro). If we expand the \7 X \7 X operatorand use the above result we obtain

or equivalently

2 2 - (- 1 )(V' + ko)Ge = - I + k~ V'V' o(r - ro).

GREEN'S FUNCTIONS

We now use (111) to replace the unit source function; thus

2 2 - (- 1 ) 2 2(V' + ko)Ge = I + k~ V'V' (V' + ko)g

or

2 2 [- (- 1 )](V' + ko) Ge - I + k~ V'V' g = O.

A particular solution to this equation is

97

(112)- (- 1 )Ge = I + k~ V'V' g ,

If necessary, solutions to the homogeneous equation can be added to take care of boundaryconditions. In free space Ge and g both satisfy the radiation condition at infinity; i.e., theyrepresent outward-propagating waves so only the particular solution is needed. Hence we arriveat the solution given earlier by (110).

The above method of derivation was first used by Levine and Schwinger [2.16]. We canapply it also to the case where g is represented in terms of cylindrical waves instead of sphericalwaves. Consider a nonuniform line source I(z) located along the Z axis and the scalar fieldwhich is a solution of

(113)

and represents outward-propagating waves. By means of a Fourier transform with respect toZ we obtain (1'2 == k5 - w2)

For r > 0 the solution to Bessel's equation that represents outward-propagating cylindricalwaves is ~(r, w) == CHij('Yr) where C is a constant to be determined. If we apply thedivergence theorem in two-dimensional form to the equation

(V'2 + 'Y2)~ = _1o(r)t 27rf

we obtain, for a circular disk of radius a,

1°127r

(V't· V't~ + 'Y2~)r ae dr

= _.l r t"o(r)dOdr27r io io

=127r

V't~·aradO +1°12

1<'Y2~rdOdr = -1,

When we make the radius a vanishingly small, the integral of 1'2~ vanishes since for r very


small H5('Yr) f'V - j(2/1f) In r. We also can use V'tCH5('Yr) == -j(2/1f)Ca,/r for r verysmall and we then obtain j4C == i. The solution for l/; is now seen to be

1 foo ji(w) .l/; == - - --H5('Yr)e-J WZ dw

21f -00 4

1 foo [fOO - . ]= - I(Z') -.lH~(rJk~ - w2)e-j W(Z- z ' ) dw dz'21f -00 -00 4

from which we can identify the scalar Green's function, expressed in cylindrical coordinates,as

1 foo .g == - - J H2(rJk2 - w2)e- j W(Z - Z' ) dw21f 4 0 0 •

-00

If the unit source is located at x', y' then we replace r == (X2+y2)1/2 by [(X-X')2+(y-y')2]1/2

to obtain

g(r, zlr', z') == -8 j foo H~( Ir - r'IJk'ij - w2 )e- j w(z - Z' ) dw (114)1f -00

where r == xa, + yay and r' == x'ax + y'ay. We can apply the operation shown in (112) tothis function so as to obtain the dyadic Green's function for the electric field as a function ofcylindrical coordinates.

The solutions for the dyadic Green's functions are the solutions outside the unit dyadic pointsource. These solutions are the limit of the solutions for a uniform-density source contained,for example, within a small spherical volume of radius a as the volume shrinks to zero butthe total source strength is kept equal to unity. The limiting procedure is that described inSection 1.9. It is important to keep this limiting aspect in mind in order to obtain a correctinterpretation of the electric field solution given by (109).

For currents contained in a finite volume V in free space the surface integral in (109) istaken over a sphere of infinite radius. When r is much larger than all ro in V the asymptoticform of the radiated electric field is

e-j k o'E(r) f'V u». cP)-4-

1fr

where f is a vector function of the polar angle 0 and azimuth angle cP and has ao and a,components only, i.e., is an outward-propagating spherical TEM wave. We can approximateR == Ir - roI by r - a,· ro upon using the binomial expansion and we then find that theasymptotic form for the dyadic Green's function is

at infinity. For an outward-propagating spherical wave at infinity the magnetic field is relatedto the electric field by the equation ZoH == a, x E. Since \7 X E == - jWJ.toH == - j koZoHthe radiation condition at infinity can be stated in the form

lim r(\7 X E + jkoa, X E) == 0'---tOO

(115a)

GREEN'S FUNCTIONS

and for the Green's dyadic function

lim ,(~ X Ge + jk08 , X Ge) == O.'----+00

99

(115b)

These relations are valid in any isotropic medium as well provided ko is replaced by k. Withthese radiation conditions the reader can readily verify that to terms of order 1/,2 the integrandin the surface integral in (109) vanishes. Hence in free space the solution for the electric fieldis given by

E(r) = -jwlloJJJJ(ro)·Ge(ro, r)dVo.

v

(116)

If the asymptotic form (115) is used for Ge the electric field clearly has the form given earlierfor, » '0.

For observation points r that are outside V there is no difficulty in evaluating the integral in(116) since Ge is not singular within V. However, if r is located in V then rand ro can coincideand Ge will become singular. The most singular part of Ge is the term ~~(e-jkoR /47rkijR).For Ir - roI very small the exponential function can be replaced by unity so the dominantsingular part is

111 1411"k5\l\lIi = 411"k5\lo\loIi'

One way to overcome the difficulty in integrating this singular term is to reformulate theproblem using the finite unit source discussed in Section 1.9. Thus we consider a scalarGreen's function that is a solution of

{-b

(~~ + k~)g == 47ra'0,

R == [r - roI <5: a

R == Ir-rol >a

where a is the radius of a small spherical volume Vo centered on r. Outside the source regionthe solution for g is still given by Ae-j koR /47rR since this is a solution of the homogeneousequation. In this solution A is a constant to be determined. The Green's function g that satisfiesthe above equation is finite everywhere so in the region R ~ a we can choose a power series

for the solution of the Helmholtz equation in spherical coordinates,

R 5:.a.

There is no dependence on the angles () and cP because of the spherical symmetry. When wesubstitute the series expansion into this equation we obtain


By equating the coefficients of like powers of R it is readily found that for the particularsolution Co == C 1 == C 3 == Cs == ... == 0

1C2 == ---3'

811'"0n == 4,6, ....

When the series is written out it can be recognized as the solution

R :S o.3 [ sin koR ]g\ = 41l"03k~ koR - 1 ,

A solution to the homogeneous equation that is finite at R == 0 is

_ B sin koRgo - koR' R :S a,

We now make g and 8g j8R continuous at R == a, This allows the two constants A and B tobe determined. The final solution obtained for g is

{

3 ( sin koR _ 1) + 3 sin koR [(1 -vi« ) -jkoa - 1] R :S 0411'"03k6 koR 47rk603R J 00 e ,

g == (117)3 e-j koR

k603(sin koa - koa cos koa) 41l"R ' R ~ a.

If we choose koo to be very small, then koR is also small in the volume Vo and we canapproximate g as follows:

(118)

R ~Q

R ? Q.

(3 R

2)

g ~ { 811'"0 - 87r03

'

e-j koR

411'"R 'This approximation is obtained by using the small argument approximation for the trigono-metric functions. The approximation for g in R :S 0 is the static field approximation used inSection 1.9.

In (116) we now split the volume into a small spherical volume Vo plus the remainderV - Vo. The integral over V - Vo can be carried out since Ge is not singular in this region.We can choose Vo so small that J(ro) can be approximated by its value at the center. Inaddition, the only term that will remain as we make Vo shrink to zero is the term V'0 V'og andwe evaluate the integral of this term using the approximation given in (118). We now have

. rrr (_R2

) . f aR \7oR2

1- jW/loJ(r)· JJJ V'o V'o k581l"a3 d V« = jW/loJ(r)· k581l"a3 R= dSoVo 0

upon converting the volume integral of the gradient to a surface integral using a vector identity(see the Mathematical Appendix). At this point we note that V'OR 2 == 2RaR and then byexpressing aR aR in rectangular coordinates we readily find that the integral gives a finiteresult independent of the radius of the small sphere. Hence we find that the contribution tothe electric field from the currents in Vo is given by

!J~if- I.J(r)

lim - jWjlo J(ro)·Ge(ro, ri d V« == jWJ.t0--2-.a~O 3ko

Vo

(119)


At the end of Section 2.15 we will show that the above result can also be obtained by usingthe eigenfunction expansion of the dyadic Green's function.

In classical potential theory where a scalar Green's function behaving like 1/R is involvedthe integral in (109) can be successfully defined as the limit of an improper integral. Thisapproach has also been applied to (109) [2.20], [2.21], but as shown by the author [2.17] itis not an entirely satisfactory method.

The classical approach is to surround the singular point r by a small volume Vo, applyGreen's theorem to the region V - Yo, and take the limit as Vo tends to zero. For the scalarHelmholtz equation one term obtained from the limiting procedure is the scalar field at thesingular point. If we apply (109) with the point r excluded we obtain

- jwp,o lim rrrJ(ro)·Ge(ro, r)dVovo~oJJJ

V-Yo

+ lim 1[(0 X \70 X E)eGe +(0 X E)e\70 X Ge]dSo ==0. (120)vo~o

o

It can be shown that both limits depend on the shape of the excluded volume [2.17], [2.21].For a spherical volume the limit as Vo goes to zero gives

2 1- -E(r) - -\7 X \7 X E(r)

3 3k6

- jwp,o lim rrrJ·Ge dVo = 0vo~oJJJ

V-Yo

(121)

which is not an explicit solution for the electric field [2.17]. However, if we assume that theelectric field in (121) does obey the original equation (106) then we obtain

E . j.J(r) . 1· ~~ifJ G- dV(r) == jwP.0--2- - jWp.o im · e 03ko Vo~O

V-Yo

(122)

which is the same result as that given by (119) when the contribution from the volume V - Vois added. Since the principal-volume approach does not yield a direct solution for the electricfield the dyadic Green's function does not serve as the inverse operator for the electric fielddifferential equation (106). For this reason we do not regard the principal-volume procedure asthe appropriate procedure for obtaining an inverse operator that will solve (106). The limitingprocedure using a finite unit source as given earlier does permit the integration to be carriedout over the whole volume containing J even when the field point is located in that volume.As we will see later, the eigenfunction expansion of Ge also enables the integral to be carriedout.

Thus (116) can be evaluated and clearly Ge then functions as an inverse operator that solves(106), whereasthe ptincipal-volume approach results in a new differential equation for E(r)as given by (121).

The use of the principal-volume result (122) to evaluate the electric field also presentsdifficulties that often mitigate against its use. In a cavity or a waveguide the satisfactionof the boundary conditions is most easily accomplished by using an eigenfunction or modeexpansion of the dyadic Green's function. The use of (122) then considerably complicates


the solution procedure since the eigenfunctions are not orthogonal over the volume V - Vo.As a consequence, infinite series occur that must be handled carefully in the limiting processsince some of these series become divergent. When an eigenfunction expansion of the dyadicGreen's function is used there is no need to introduce an exclusion volume since no singularintegrals occur.

2.13. RECIPROCITY RELATION FOR DYADIC GREEN'S FUNCTIONS

In free space, the dyadic Green's function for the vector Helmholtz equation is a symmetricaldyadic. In the presence of boundaries on which G satisfies prescribed boundary conditions,the dyadic Green's function is not a symmetrical dyadic in general. The nonsymmetry isbrought about by the boundary conditions that G must satisfy. Any number of examples maybe constructed to demonstrate the validity of the above statements. With reference to Fig.2.15, let G be the Green's function for the half space z > 0 and subject to the boundarycondition n X G == 0 on the perfectly conducting plane at z == O. The Green's dyadic is afunction of the coordinates of the field point, the coordinates xo, Yo, zo of the source point,and the coordinates xo, Yo, -Zo of the image source. It may be constructed from the free-spaceGreen's dyadic by using image theory. However, we will not require the analytical form of Gto demonstrate the nonsymmetry of the dyadic. A current element axJx produces an electricfield with a Z component given by J xGxz, while a current element azJz produces a field withan x component given by JzGzx. If Gxz == Gzx, that is, if G is a symmetric dyadic, thesecomponents of the field would be equal when J x == J z- For the problem under consideration,the current axJx produces a field with a nonzero Z component, while azJz produces a fieldwith an x component which vanishes on the conducting plane. Therefore, Gxz =I- Gzx, and theGreen's dyadic is not a symmetrical dyadic in general.

Even though G is not a symmetric dyadic in general, a reciprocity principle does hold forG. Let G(rl, r) be the Green's dyadic such that a current J1(rl) at rl produces an electricfield EI (r2) == J I (rl)· G(rl, r2) at the point r2. Similarly, a current element J 2(r2) produces afield E2(r l) == J2(r2)· G(r2, r.) at the point rl. According to the Lorentz reciprocity theorem(Section 1.10 and assuming that the surface integral is zero),

and hence

(123)

Since J I and J 2 are arbitrary, we must have for an arbitrary current element

(124)

where Gt is the transposed dyadic. Thus the Green's dyadic with the variables rand rointerchanged is equal to the transpose of the original dyadic. If J 1 == ax and J2 == ay, weobtain from (123) the result Gxy(rl, r2) == Gyx(r2, rj), which is a special case of (124). Aformal proof of (124) may be obtained by using the vector form of Green's identity instead ofthe Lorentz reciprocity theorem. The details are left as a problem at the end of the chapter.

GREEN'S FUNCTIONS

z-o

103

,~ Image'f source

n

~-~V J

Fig. 2.15. Boundary-value problem for which G is not a symmetric dyadic.

2.14. EIGENFUNCTION EXPANSIONS OF DYADIC GREEN'S FUNCTIONS

In order to find the spectral representation of the differential operator in the-vector Helmholtzequation and the vector wave equation we need to construct the vector eigenfunctions for thetwo equations

(\72 + X)1/J == 0

(\7 x \7 X - X)1/J == 0

(125a)

(125b)

where 1/J is a vector eigenfunction and Xis the eigenvalue parameter. The vector eigenfunctionscan be separated into solenoidal or transverse eigenfunctions F that have the properties

\7·F == 0, \7xF#O

and irrotational or longitudinal eigenfunctions L with the properties

\7 X L == 0, \7·L # O.

If we expand the curl curl operator in (125b) to get \7\7. - \72, then since \7. F == 0 we seethat

so the solenoidal functions can be eigenfunctions of both operators shown in (125). If wesubstitute the longitudinal functions L into (125b) we see that because \7 X L == 0 thesenonzero functions lie in the null space of the curl curl operator and the eigenvalue X must bezero. This is an important point that we will return to when we consider the eigenfunctionexpansion for the electric field dyadic Green's function.

Our ability to construct vector eigenfunctions is limited to a relatively small number ofcoordinate systems. For spherical (and conical) coordinate systems Hansen [2.23] showed thatthe solenoidal functions could be obtained from the scalar functions of the scalar Helmholtzequation


The F functions consist of two sets called the M and N functions which are given by

M == \7 X (a,rt/;)

~N == \7 x \7 x (a,rt/;).

The longitudinal functions can be generated from the gradient of t/;, i.e.,

L == \7t/;.

(126a)

(126b)

(126c)

The reader can readily verify that in spherical coordinates the M and N functions satisfy bothequations in (125) and that the L functions satisfy (125a). For the free-space problem thesame scalar function t/; generates all three sets of vector functions .. In cavity and waveguideproblems the scalar functions that generate the M, N, and L functions are generally differentbecause they are required to satisfy different boundary conditions.

In cylindrical coordinate systems such as circular, elliptical, parabolic, and rectangularcylindrical coordinates the vector functions can be found from the operations

M == \7 x azt/; == -az x \7t/;

r: at/; 2Y AN == \7 x \7 x azt/; == \7az - az\7 t/;

L == \7t/;

(127a)

(127b)

(127c)

where again t/; is a solution of the scalar Helmholtz equation. In rectangular coordinates theunit vector az can be replaced by ax or a, also.

Vector Eigenfunctions in Spherical Coordinates

The scalar Helmholtz equation in spherical coordinates is

where 0 is the polar angle relative to the z axis and 4> is the azimuth angle measured fromthe x axis. We have also replaced A by k 2 as the eigenvalue parameter. If we assume aproduct solution of the form t/; == f(r)g(O)h( 4», then the Helmholtz equation separates intothe following three ordinary differential equations:

d . 0dg A(j> ~. 0 0dO SIn dO - sin 0g + 1\, SIn g ==

(129a)

(129b)

(129c)

where A(j> and A, are separation constants. The above equations are all of the Sturm-Liouville


type. For the free-space problem the angle cP covers the whole interval 0 to 211'" so h(cP) mustbe periodic in cP and hence Act> == m2 where m is an integer 0, 1,2, 3, .... The solutions forh are cos mcP and sin mo.

The equation for g(O) is Legendre's equation. There are two independent solutions and bothare singular on the Z axis unless Ar is chosen equal to n(n+1) where n is an integer. In this caseone solution is finite everywhere and is called the associated Legendre polynomial designatedby the symbol P':(cos 0). When m == 0 the solutions are called Legendre polynomials. Thesesolutions can be obtained using Rodrigues' formula

(130)

For m > n the solutions are zero. The orthogonality of the different solutions for differentvalues of n is with respect to sin 0 as a weighting factor. For the same n but different m thesolutions are orthogonal with respect to 1/ sin 0 as a weighting factor. These orthogonality re-lations follow directly from the properties of the Sturm-Liouville equation. A number of usefulformulas, including recurrence relations, are summarized in the Mathematical Appendix.

The equation for the function j(r) is a modified form of Bessel's differential equation. Thesolutions are called spherical Bessel functions and are related to the Bessel cylinder functionsas follows:

jn(kr) = J2;r J n+1/2(kr)

Yn(kr) = J2;r y n+l/2(kr)

h~(kr) == jn(kr) + jYn(kr)

h~(kr) == jn(kr) - jYn(kr).

(131a)

(131b)

(131c)

(131d)

The basic properties and recurrence relations for the spherical Bessel functions are also sum-marized in the Mathematical Appendix.

In rectangular coordinates the continuous spectrum of eigenfunctions can be representedby a Fourier integral. In spherical coordinates the radial eigenfunction can be chosen as thej n (kr) function and its completeness properties are expressed through the Hankel transforms

j(r) ={f100

F(k)jn(kr)k2dk

F(k) ={f100

j(r)jn(kr)r2dr

and the completeness relations

(132a)

(132b)

(133a)

(133b)


In view of these relations the eigenfunctions for the scalar Helmholtz equation, for free space,can be seen to be

cos1/;~~ == P'[tco« O)jn(kr) mcjJ

sin(134)

where e stands for the even function cos mcjJ and 0 identifies the odd function sin mo. Thesefunctions remain finite at the origin and vanish at infinity.

From the scalar function given in (134) we can now construct the vector eigenfunctionsusing the operations shown in (126). Thus we have

cos d'Le,o == V_'Ie, 0 == pm m,l,.k~a

nm 'Ynm n. w d(kr) rSIn

. cos 1 dP'::+ In . mcjJ--dO ao

SIn r

m sin=f-'-OP'::jn mcjJalj>

r SIn cos

k sinM e,° - n k _'le,O _ m m .nm - v X ra''Ynm -=f-'-OPnJn mcjJao

SIn cos

1 1Ne,° == - V X V X kra .te.o == - V X Me,°nm k r'rnm k nm

n(n + 1) m : cos dP':: cos 1 d(krjn)P nIn . mo», + -dO . mcjJ- d(k ) ao

r SIn sm r r

m m sin 1 d(krjn)=f-·-O

P n mcP- d(k ) a",.SIn cos r r

(135a)

(135b)

(135c)

Note that we have included a factor k in the definition of the solenoidal functions.The vector eigenfunctions are orthogonal among themselves as well as with respect to

each other when integrated over all values of r, 0, and cjJ. Consider the product L- M ==kV1/;- V X (r1/;a,) == kV -[1/;V X (r1/;a,)] -1/;V-V X (r1/;a,). Hence we have, upon using thedivergence theorem,

111L·MdV = If l#M.ar dS = 0v s

because M does not have a radial component. In a similar way we find that

111L·N dV =If l#N.ar dS = 0v s


because -.f; decreases like r-1, N· a, decreases like r- 2 , and the surface area of a sphereincreases only as r 2 so that for the surface of the sphere with a radius that becomes infinitethe surface integral vanishes. We have thus shown that the L functions are orthogonal to theM and N functions.

Consider next the relation \7. (kr-.f;a,) X N == \7 X (kr-.f;a,).N - kr-.f;a,. \7 X N == M·Nsince \7 X N == kM and does not have a radial component. We now use the divergencetheorem to get

because a, X N·a, == (a, X a.j-N == O. Consequently, we find that the M and N functionsare also orthogonal. Note that from (135c) we have k\7 X N == \7 X \7 X M == k2M whichis the relationship we used above.

The orthogonality of the vector eigenfunctions among themselves for pairs of values nmkand n'm'k' follows in a straightforward way from the orthogonality properties of the P,:/,cos mo, sin mo, and the jn(kr) functions, the latter being expressed by (133a).

The normalization integrals for the M and N functions are the same. We can prove this bydirect evaluation. The normalization integral for the N functions is not straightforward so wewill consider it in detail and follow the procedure given by Tai [2.12].

By using (135c) and integrating over c/J we readily find that

111N~~({', cP, kr).N~~(O, cP, k'r)dVv

2 r roo { [(dPm )2 (mpm )2]= fo:10 10 [n(n+l)P~fjn(kr)jn(k'r)+ dOn + sin2nO

X [drj;;kr)drj~~klr)]} sin OdrdO.

We now use the known integrals

r [P':(cos O)f sin 0 dO = _2_ (n + m)!Jo 2n + 1 (n - m)!

I' [(dP': ) 2 + (mp~i] sin 0dO == 2n(n + 1) (n + m)!Jo dO sirr' 0 2n + 1 (n - m)!

to obtain

211" 2n(n + 1) (n + m)!100{ ..,

- 2 + 1 ( )' n(n + l)Jn(kr)Jn(k r)EOm n n - m. 0

drjn(kr)drjn(k'r)} d+ dr dr r.

(136a)

(136b)

In order to proceed further we simplify the integrand by using the following two relations for

108

the spherical Bessel functions:


d. U 1 . .du [uJn(u)] == 2n + 1[(n + )In-l - nJn+l]

jn(u) = 2nU+

1Un-l + jn+1)'

The use of these relations enables us to obtain

(137a)

(137b)

:rrjn(kr) ~rjn(k'r)

= kk'r2

[(n + l)jn-l(kr)jn-l(k'r) + njn+l(kr)jn+l(k'r)] - n(n + l)jn(kr)jn(k'r)2n + 1

by eliminating the cross-product terms jn-ljn+l through use of the expansion of jn (kri]; (k'r)and (137b). By employing the above relation the integrand reduces to

The integration may be completed by using (133a) to give the final result, which is

rr{Ne,O(O q, kr).Ne,O(O q, k'r)dV= 7r22n(n + l )(n + m)!O(k _ k ' ) (138a)JJJ nm " nm v:» fom(2n + 1)(n - m)! ·

v

The other two normalization integrals are straightforward to obtain and are

fffMe,o«(} ,k kr).Me,O«(} ,k k'r)dV= 7r22n(n + l)(n + m)!O(k _ k ' ).JJJ nm ,0/, nm v:> 0/, fom (2n + 1)(n - m)!

v

fff e ° 0 k e ° 0 k' d 271"2(n + m)! ~ k k'JJJ Ln~ ( ,q" r)· Ln~ ( ,q" r) V = fOm(2n + l)(n _ m)! u( - ).v

(138b)

(138c)

In evaluating the normalization integral (138c) the following relationship obtained from (137)was used:

udjn(u) U. .du == 2n + 1 [nJn-l(U) - (n + I)Jn+l(U)].

For later use we will define the normalization factor Qnm by the relation

271"2(n + m)!Qnm == ·

fom(2n + 1)(n - m)!(139)

Now that the complete set of vector eigenfunctions in spherical coordinates has been obtainedwe are in a position to consider the eigenfunction expansion of the solutions for the vectorHelmholtz and vector wave equations. This topic is taken up next for the vector and scalar

potentials in the Coulomb gauge. The results are then used to obtain the complete eigenfunctionexpansion of the electric field.

Eigenfunction Expansion of Vector and Scalar Potentials: Coulomb Gauge

In Section 1.7 it was shown that in the Coulomb gauge where \7. A == 0 the vector andscalar potentials are solutions of the equations

\7 X \7 X A - k5A == J.toJ(r) - jW€oJ.to\7ip(r) == S(r)

\72ip == _ p(r) .€o

(140a)

(140b)

The source function S(r) for A(r) has zero divergence since \7.J == -jwp so that \7·S ==-jwJ.toP -jwJ.tO€0\72ip == 0 when ip satisfies (140b). The L functions lie in the null spaceof the curl curl operator but since both A and S are orthogonal to this space of functionsthe solution of (140a) can be constructed using only the M and N functions. We assume thatthe current and charge are contained in a bounded volume Vo in free space. We will orderthe eigenfunction labels n, m, e, and 0 in such a way that a single subscript j will denote aparticular combination. We can easily show that the projection of the source function S ontothe space of Lj functions is zero by showing that

for all j. We note that Lj .S == \71/;j. S == \7. (1/;jS) -1/;j \7. S == \7. (1/;jS). Hence by using thedivergence theorem we have

IIILj.SdV = Ill/;js.ndsv s

where S is the surface at infinity. On So, the boundary of the source volume, we have S· n ==p,oJ·n - jWfoJ.ton· \7q,. If n-J is not zero on So, then a compensating layer of surface chargewith density

+ n·JPs == -fo\7·nl_ == -.-

}w

exists on So where \7ip. n I! is the discontinuity in the potential gradient across the charge layer.The two discontinuities, namely, that in n-J and that in n- \7ip, cancel so n-S is continuousacross So. For this reason we could apply the divergence theorem to the total volume V. Since1/;j and S vanish at infinity faster than r-2 the surface integral vanishes thus proving that S hasno projection on the functions Lj ,

We can let A be expanded as follows:

A(r) = 2;100

[Bj(k)Mj(O, €j), kr) +Cj(k)Nj(O, €j), kr)] dk.J

We will also express Mj(8, cP, kr) in the form Mj(r, k) and similarly for the Nj function.

110

When we substitute this expression into (140a) we obtain


We can get a solution for Bj(k) by scalar multiplying both sides by Mj(O, cP, k'r) and inte-grating over V. By virtue of the orthogonal properties of the vector eigenfunctions and usingthe normalization integral (138b) we obtain

tk'? - k~)Bj(k')n(n + l)Qnm = 111S(ro)·Mj(ro, k')dVo

v

since the volume integral, as (138b) shows, gives a delta function factor o(k - k') and theintegral over k then selects the value k == k'. The coefficient C j(k') is obtained in a similarway. We easily find that

A(r) =~100

n(n + 1)Q:m(k2 _ k~) [Mj(r, k) 111Mj(ro, k)·S(ro)dVoJ v

+ Nj(r, k). 1[1Nj(ro, k).S(rO)dvo] dk (141)

which is the eigenfunction expansion of the solution to (140a). The solution for A(r) can berewritten in the form

From this expression we can identify the free-space dyadic Green's function for the vectorpotential in the Coulomb gauge; thus

We emphasize that the proper interpretation of this eigenfunction expansion of the dyadicGreen's function is that it is to be used in a term by term integration so as to yield the samesolution as that given by (141).

For functions F that have V'.F == 0 the vector eigenfunctions Mj and Nj form a completeset in the domain of the operator. Consequently, in this space of functions the unit sourcefunction has the representation

100 dkIo(r - ro) = L ( 1)Q [Mj(ro, k)Mj(r, k) + Niro, k)Nir, k)].

. 0 n n + nmJ

(144)

This result can be obtained by a formal expansion of the left-hand side or by beginning withthe eigenfunction expansion of F which is

F(r) = L roo dk [Mj(r, k) (vir{Mj(ro, k).F(ro)dVo. Jo n(n + 1)Qnm JJJ

J

+ Nj(r, k)J[JNj(ro, k).F(rO)dVo] (145)

and rewriting this in a form that will lead to the identification of the delta function operator.Note that even though the current J and charge p are zero outside Vo the source function

S has a nonzero term proportional to \7<P outside yo. However, \7<P and the lamellar part J,of J have a zero projection on the functions M j and Nj . Consider, for example,

Now Mj"odS.

s

This integral vanishes when S is a spherical surface since Mj-ar == O. A similar result holdsfor the Nj functions because <pNj-ar varies like r-3 at infinity. It now follows that S can bereplaced by JLoJ in the solution given by (141) or (142).

The solution for the scalar potential can be expanded in terms of the scalar eigenfunctionsof the Laplacian operator. This eigenfunction set is the set 1/;j == 1/;~~. Thus we let

<l>(r) = ~100

o,o», (r, k)dk.J

The normalization integral for the 1/;j functions is

({{I/;·(r, k)I/;·(r, k')dV = 211"2(n +m)! o(k <k'). (146)JJJ J J Eomk 2(2n + 1)(n - m)!v

By following the standard procedure we obtain

We now use \7-J == -jwp, n-J == jiao, on So, and \7-(1/;jJ) == 1/;j\7-J + \71/;j-J to expressthe volume integral in the form

(148)

If n-J does not equal zero on So then the volume integral in (147) must also include thesurface integral

where jwps == n-J and Ps is the charge density on So. This integral needs to be added to(148); when this is done we obtain

We can use this result to express the solution for <l> in the form

1 100

27l"2(n + m)! ~~if<l>(r) == -.-2:: (2 + 1)( )' if;j(r, k) Lj(ro, k)JWf:o . 0 f:Om n n - m .

J V

.J(ro)dVodk. (150)

The electric field is given by E(r) == - jwA(r) - V<l>(r). By using the solutions obtainedfor the potentials we can easily write down the eigenfunction expansion for the electric field.We will do this in a form that will allow us to identify the Green's function operator Ge forthe electric field. From (142) and (150) we obtain

Clearly the electric field dyadic Green's function operator is

- roo dk { 1Ge(ro, r) = ~10 Qnm n(n + 1)(k2 _ k~) [Mj(ro, k)Mj(r, k)

J

+ Nj(ro, k)Nj(r, k)] - -;Lj(ro, k)Lj(r, k)} . (152)k o

The expansion of the electric field requires both the transverse and longitudinal vector eigen-functions and hence the dyadic Green's function must also have both sets of eigenfunctions inits expansion. Even though V· E == 0 outside the source region the scalar potential <l> is notzero and hence E has a lamellar part given by - V<l> outside the source region. We point outonce more that (152) is to be used in a term by term integration to yield the eigenfunctionexpansion of the electric field.

If we want to obtain an eigenfunction expansion of the electric field directly we considerthe equation

V x \7 x E - k6E == -jwp-oJ. (153)

Since the L j functions lie in the null space of the curl curl operator and hence are notmapped into the range space of the operator, (153) does not have a solution in terms of theeigenfunctions of the differential operator unless J lies in the range space. But, in general,V·J i=- 0 so J is not always in the range space of the operator. The range space only containsfunctions with nonzero curl. The part of J that does not lie in the range space <R lies in theorthogonal complement <R.l of the range space. This is the space containing the L, functionsand coincides with the null space of the curl curl operator since the latter is a self-adjointoperator. We can overcome the difficulty in solving (153) by the simple expedient of findingthe projection of both sides on the space of L j functions and removing this part from theequation. Thus we split E into its longitudinal and transverse parts E, +E, and let

Similarly, we let the longitudinal part of J be represented by

From (153) we see that

v X V X E, - k5Et - k5E/ == -jwp.oJt - jwp.oJ,

and thus we must have k6Cj == jwp.oDj . By following the procedure that by now should beclear we readily find that

When we remove the longitudinal part from (153) the equation that remains can be solved interms of the Mj and Nj functions which are the eigenfunctions of the curl curl operator. Theresult is, of course, the same solution as that given earlier by (151).

Another way to approach (153) is to take the divergence of both sides to obtain

Now let E/ == - V'~ and use the continuity equation to get

V'2ep = _ jW~o V'.J = _!!....ko €o

The solution for the scalar potential can be constructed in terms of the t/;j functions and simplyreturns us to our earlier solution (150) . Note that even though J may be zero outside a finitevolume Vo the separate parts J/ and Jt are generally not zero outside Yo. However, outsideVo the transverse part Jt cancels the longitudinal part J/.

The eigenfunction expansion of the scalar Green's function for the Helmholtz equation

(154)


can be identified from the solution (147) for 4>(r) by inserting a factor k2 - kij into thedenominator and keeping the k 2 factor in the normalization constant. The solution for g is

For the scalar Green's function j includes the n == m == 0 terms.

2.15. EXPANSION OF THE ELECTRIC FIELD IN SPHERICAL MODES [2.18]

In this section we will show that the eigenfunction expansion (151) for the electric field canbe converted into an expansion in terms of spherical standing waves and outward-propagatingspherical waves. The mode expansion outside the source region involves only M and N func-tions that are derived from the scalar wave functions

cos1/;j == P':(cos fJ) mc/Jjn(kor)

sin

cos1/;1 ==P':(cos fJ) mc/Jh~(kor).

sin

The N function spectrum of eigenfunctions contributes a zero frequency or "static-like" setof modes that cancels the longitudinal eigenfunction contributions outside the source region.Inside the source region the cancellation is not complete but the remainder can be expressedas a delta function contribution. The mode expansion is obtained by evaluating the integralover k in the eigenfunction expansion of the dyadic Green's function given by (152).

We will consider the mode expansion of the part of (143) involving the Nj functions first.If we examine (135c) for the N j functions we see that the basic integral over k that has to beevaluated is

(156)

Actually, various derivatives with respect to rand ro are also involved. According to thesequence of operations specified in (141) the derivatives should be taken before the integrationover k is carried out. However, we can keep the derivative operations outside the integralprovided due care is taken in differentiation of the resultant functions of rand ro, some ofwhich have discontinuous derivatives at the point r == roo

In order to evaluate the integral we convert it to a contour integral by using the substitution

when r > roo We note that when kr is very small j n(kr) behaves like (2kr)nn! /(2n + I)! whileh~,2(kr) behaves like =r:-j2(2n)!/n!(2kr)n+l. Thus the product jn(kro)h~,2(kr) is asymptoticto =r:-jr8/(2n + 1)rn+1k and has a first-order pole at k == O. In the sum h~ +h~ the pole term


is canceled and indeed the integrand in (156) is regular at k = o. We will return to this pointshortly.

The circuit relations

can be used to replace the integral of jn(kro)h~(kr) from 11 to infinity by an integral fromminus infinity to - 11, i.e. ,

100 jn(kro)h~(kr)dk _ J-oo jn( -kro)h~(-kr)( -dk)k 2 k2 - k 2 k 2

11 - 0 -11 - 0

-J-l1jn(kro)h~(kr) dk- 2 2 •

-00 k - ko

Later on we will take the limit as 11 approaches zero. When we split the integral ofj n(k r0)[h ~ (k r) + h~ (k r)] in the manner indicated we must take the resultant integral as aCauchy principal-value integral in order to obtain the required pole cancellation at k = O.With these considerations the integral in (156) can be replaced by

I = !pJoo jn(kro)h~(kr) dk2 -00 k 2 - k5

= limJ-'1 jn(kro)h~(kr)dk + lim100

jn(kro)h~(kr)dk. (157)11~0 -00 2(k2 - k5) 11~0 11 2(k2 - k5)

We can convert the principal-value integral to a contour integral along either of the two contoursshown in Fig. 2.16, but we must then add or subtract, respectively, 21rj times one-half of theresidue at the pole k = O. This last term is ± 1rj(jr8/(2n + l)rn+1

) = =f1rr8/ (2n + l)rn+1•

We can initially consider the medium to have a small loss so that k o = kb - j kg. Thus thepole at k = ko lies above the contour. The solution must be an analytic function of the lossparameters so consequently when we reduce the loss to zero the contour must be indentedabove the pole at k o and below the pole at - k o as shown in Fig. 2.16. We cannot allow thepole to move across the contour since this will create a discontinuity in the function equal to± 21rj times the residue at the pole.

When kr is very large the spherical Bessel functions have the following asymptotic behavior:

. 1 ( n+l)In(kr) rv kr cos kr - -2-1r

Consequently, in the lower half of the complex k plane the product jn(kro)h~(kr) becomesexponentially small for r > roo We can, therefore, close the contour by a semicircle in thelower half plane; when the radius of this circle becomes infinite we do not get any contributionto the integral from this part of the contour. We can now evaluate the contour integral usingresidue theory. We get residue contributions from the two poles at k = 0 and k = ko whenwe use the contour shown in Fig. 2.16(a). With this choice of contour we must add to the

116

(a)


(b)

Fig. 2.16. Contour for converting the eigenfunction expansion into a mode expansion. The twoalternative semi-circle contours about the point at the origin converts the principal value integral toa contour integral.

contour integral one-half of the pole contribution at k == o. Thus we find that the integral [has the value

It. ) - 2 .jn(koro)h~(kor)r, ro - - 7r} 4k

o(158)

The result is valid for r == ro also since the contribution to the integral from the semicircle isstill zero when r == ro because the integrand decreases like k :" while the circumference ofthe circle increases only like Ik I.

When r < ro we use the substitution 2jn(kro) == h~(kro) + h~(kro). A similar analysisthen shows that the value of the integral is given by (158) but with the roles of rand rointerchanged. For all values of r, ro we have

(159)

The contribution to Ge from the NjNj terms in (152) can now be expressed in the form

_ 1 [ cos ]Gel == 2: \70 X \70 X a'orOP':/(COS ( 0 ) mc/JO

. Qnmn(n + 1) sinJ

[

COS ]x \7 X \7 X a,rP':(cos 0) . mc/J [(ro, r)

SIn

(160)

where [(ro, r) is given by (159). For convenience we will define the following functions:

cosNj(r) == \7 X \7 X a,rjn(kor)P':(cos 0) mc/J

sin

cosNj(r) == \7 X \7 X a,rh~(kor)P,:/(cos 0) mc/J

sin

(161a)

(161b)

r < r»

r > ro(161c)

GREEN'S FUNCTIONS

ro < r

ro > r.

117

(161d)

The subscript (J is an indicator that tells which of r, ro is to be used according to the rulesgiven in (161c) and (161d).

We also need symbols for the static-like modes generated from the other term in I(ro, r).For these we will use the symbols Njo(r), Njo(r) , and Njao(r). The function Njo(r) is thesame as that in (161a) but with jn(kor) replaced by r", The function N)o(r) is the same asthat in (161b) but with h~(kor) replaced by r:":', The (J indicator is applied according tothe rules already given.

The function I(ro, r) is continuous at r == ro, but the derivative with respect to r or ro hasa step change across the point r == roo The operations in (161) require the derivative operation

in order to evaluate the aoo ao and act>o act> contributions to GeI. This operation produces deltafunction terms. The delta function terms come from the second derivative of 1 with respect toro and r. The step change in dl ldr is

where the prime means a derivative with respect to koro. By using the Wronskian relation

we obtain

,+dl I 0 7r 7r

dr TO = - 2k~r~ + 2k~r6

so the two step changes are equal in value but of opposite sign. When dl ldr is regarded asa function of r, the two terms undergo the above step changes as r crosses the point ro fromthe region r < ro to the region r > roo But when dl ldri, is viewed as a function of ro, thestep changes are the negative of the above amounts as the point ro crosses the point r fromthe region ro < r to the region ro > r. Clearly, the delta functions generated from the secondderivative cancel. If the derivative operations are retained in the integrand, then the integrandcan be rewritten in a form that will give the same cancellation of terms.

With all of the above definitions and canceling delta function terms taken into account wecan express Gel in the concise form

(162)


/('0, ,) ==

For the L, function contribution to the Green's dyadic function the integral to be evaluatedis

I = roo j n(krO)j;(kr) dk.io -ko

By means of the same substitutions as used before, the integral can be transformed into thesame form as in (157) except that the denominator term is simply -kij. The only pole presentis that at k == 0 so the value of the integral is given by the second term only in (159), i.e.,

2k~(2n + l)r;+1 ·

The contribution to Ge from the L j Lj terms is

1r [ cos cos ],nGe2 == - L 2 \70\7 P'[tco« Oo)P':(cos 0) mcPo mcP n~1 •

. 2ko(2n + I)Qnm sin sin r>J

By evaluating the gradients we find that

1 1r m m cos cos 0(r - '0)- 2:L -2QP n (cos Oo)Pn (cos 0) . mcPo . mcP 2 a.a,; (163)

k o · nm SIn SIn '0J

The derivative of a unit step function, which gives a delta function, is sometimes called ageneralized derivative. The sum of the series is equal to 0(0 - Oo)O( cP - cPo) / sin O. Thiscan be verified by expanding the delta functions in terms of the Legendre functions and thetrigonometric functions. By comparing (163) and (162) we see that the LjLj terms cancel thestatic-like modes that are a part of the Nj spectrum of modes outside the source region. Insidethe source region the cancellation is not complete and the residual part can be expressed bythe delta function term

(164)

The last contribution to Ge is that from the MjMj terms. The integral over k that must beevaluated for these terms is

Since the numerator has a factor k 2 , there is no pole at k == O. The evaluation of the integralis carried out using the same procedure as before and it has the value

GREEN'S FUNCTIONS

At this point we introduce the functions

cosMj(r) == \7 X 8,korP'::(cos 8) mf/>jn(kor)

sin

cosMj(r) == \7 X 8,korP':(cos 8) . mf/>h~(kor)

SIn

119

(165a)

(165b)

r < ro

r >'0(165c)

with similar definitions when rand ro are interchanged. The mode expansion for the electricfield dyadic Green's function in free space can now be expressed in the form

- '"'" j 7r 8,8,Ge(ro, r) = - ~2k ( l)Q [Mj(J(ro)Mj(J(r) + Nj(J (ro)Nj(J (r)] - -k2 o(r - ro).. on n + nm 0

J

(166)

This mode expansion for the dyadic Green's function was first obtained by Tai [2.24].The eigenfunction expansion of Ge as given by (152) clearly shows that the dyadic Green's

function has a nonzero lamellar part outside the source region. The mode expansion given by(166) on the other hand involves only transverse wave functions outside the source region.However, these transverse wave functions have components that are not all continuous withcontinuous derivatives in the source region, and therefore do not represent purely transverseor solenoidal fields. Consequently, Ge as given by (166) should not be interpreted to meanthat Ge does not have a lamellar part outside the source region.

The mode expansion for Ge can be effectively employed to evaluate the electric field in theinterior of a small spherical volume V0 of radius a for a current density J (ro) that we willassume can be considered as constant and in the z direction. The source point ro and the fieldpoint r both lie within the small volume Vo but are otherwise arbitrary points. In view of thesymmetry about the z axis, all of the terms m > 0 in 8z •Ge when integrated over the angle~ give zero. Consequently, the Mj functions do not contribute to the field. By referring to(135c) it can be seen that the scalar product 8z •Nj gives rise to two integrals over (Jo that aregiven by

{'If dPn(cos (Jo) . 2 8 d8- Jo d80

SIn 0 o·

Since cos 80 == PI (COS ( 0) the first integral is zero if n i- 1 because of the orthogonal propertyof the Legendre polynomials. For n == 1 the value of the first integral is, from (136a), ~. Thesecond integral can be integrated by parts to give

The integrated term vanishes and the integral that remains has a nonzero value of 1only forn == 1. In view of these results we can conclude that only the N, mode for m == 0, n == 1 isneeded to describe the electric field. This result is solely dependent on the fact that the currentis constant and in the Z direction and is true for any spherical volume Vo.

Inside the volume Vo the electric field is given by

upon using (151) and (166). After performing the integration over 80 and cPo we obtain

jZoJ cos 8 {+ r [ . d.]E(r) = ko

ar-ZOJ Nl(r)}o rO}l(koro)+rodr/O}l(koro) dr«

+ N l (r)10

[rohi(koro)+ rod~/ohi(koro)] dro}.

The spherical Bessel functions of order 1 are given by

. (k ) sin koro cos koroJl oro == 2 -

(koro) koro

From these expressions we find-that the two integrands reduce to

ro sin koro and

The integration over ro is easily done to give

jZoJ cos 8 ZoJ.E(r) == k a, - -2[(SIn kor - kor COS kor)Nt(r)

o k o

+«j - koa)e-jkoa - (j - kor)e-jkO')Nl(r)],

where Nt and N1 are given by

2. 1. drjl(kor)Nl == - cos 8Jl(kor)a, - - SIn 8 d ao

r r r

+ 2 2 1. drhi(kor)N l == - cos 8h l(kor)a, - - SIn 8 d ao·

r r r

r<a (167)

When kor is small sin kor -kor cos kor ~ (k or)3/3 so the term involving Nt remains finiteat r == o. If we set r == 0 then the electric field at the center of the spherical volume is found

GREEN'S FUNCTIONS

to be equal to

E(O) == jZoJ _ jZoJ (k )23ko az 3ko 0

0 az

121

(168)

(169)

when k00 is also small. The limit of (168) as the radius 0 tends to zero is thus a constant andfinite value as found earlier and given by (119). Outside the sphere the electric field is givenby

ZoJ . +E(r) == --2(sin koo - koo cos koo)N 1 (r).

ko

For koo very small (169) reduces to - (ZokoI /41r)Nt (r) where I == 41r03J /3. For suitablevalues of k-a , which are apparent from (169), the radiated field vanishes.

2.16. DYADIC GREEN'S FUNCTION EXPANSION IN CYLINDRICAL COORDINATES

In cylindrical coordinates r, et>, z the solutions to the scalar Helmholtz equation that arefinite at the origin are

where

cosl/;~' 0 == In(kr)e- j wz net>

sin(170)

(171)

and k 2 == A - w2 • For this problem there are two continuous eigenvalue parameters k and wsince the interval in z is infinite and that for r is semi-infinite. A general solution to the scalarHelmholtz equation is of the form

100 roo 00

-oolo ~[Cn(W, k) cos ncf> +Dn(w, k) sin ncf>]Jn(kr)e-jwz

dk dw

where en and Dnare expansion coefficients depending on wand k. The orthogonality andnormalization properties of the eigenfunctions e-j wz are expressed by the Fourier transformproperty

I:e-jwz+jw'z dz = 21l"o(w - w'). (172)

The corresponding property for the eigenfunctions J n (kr) is given by the Hankel transformrelations for the cylinder Bessel functions:

100

F(k)Jn(kr)kdk =j(r)

100

j(r)Jn(kr)r dr = F(k)

(173a)

(173b)


roo fJ(k k')Jo In(kr)Jn(k'r)rdr = ;

rooJ n(kr)Jn(kr')k dk = o(r - r') .io r

The vector eigenfunctions are given by U stands for a particular n, e, 0 combination)

L, == V'1/;j

M j == V' X az1/;j == -az X V'1/;j

pNj == V' X V' X az1/;j == V'V'-az1/;j - azV'21/;j

== p2az1/;j - jwV'1/;j == azp21/;j - jwLj

(173c)

(173d)

(174a)

(174b)

(174c)

where p2 == A.In order to derive the normalization integrals the following two Bessel function differential

and recurrence relations are used:

(175a)

(175b)

The explicit forms for the vector eigenfunctions are

dJn(kr) -jwz cos,l.. nJ

(k ) -jwz sin ,I..

L, == a, dr e no/ =f= aq,- n r e no/sin r cos

cos- azjwJn(kr)e-jwz ncP

sin

N- iw dJn(kr) -jwz cos ,I.. ± jw n J (k )

j - -ar - d e no/ ae/> - - n rp r sin p r

. sin k 2 . cosx e-JWZ ncP + az-Jn(kr)e-JWZ ne,

cos P sin

(176a)

(176b)

(176c)

Normalization

In this section we will use 1/;j with k, wand 1/;j with k', w' as the eigenvalue parameters.By using (175) it is readily found that the scalar product L j - L j, after integration over cP,equals

r21r{kk'Jo Lj"Lj deb = T[Jn+l(kr)Jn+l(k'r) + In-l(kr)Jn-I(kr')]

+ ww'J n(kr)Jn(k'r)}(21r/f:on)e-j(W-W')z.

GREEN'S FUNCTIONS

The integral over all values of rand z, upon using (172) and (173c), is given by

]00 1001211" 411"2 o(k - k')

Lj.LjrdcPdrdz == -(kk' +ww')o(w -w') k'-00 0 0 EOn

4 2 k2 2~ + w o(w - w')o(k - k').EOn k

Next consider M j • Mj and carry out the integration over cP to obtain

[27<Mj"Mj deb = 211" [kk'J~(kr)J~(k'r)+ n: J n(kr)Jn(k'r)] e-j(W-W')z.Jo EOn r

123

(177a)

By using (175) to simplify the integrand, the result is readily integrated over rand z to yield

rrr * 411"2 , k k'JJJ Mj"Mj dr = EOn ko(w - w )o( - ).

V

In a very similar way one obtains

IIINj"Nj dr = ~:: ko(w - w')o(k - k').v

Orthogonality

(177b)

(177c)

A volume integral of Nn·M~" Nn·L~" or Mn·L~, is clearly zero if n i= n' and w i= w'because of the orthogonal property of the cos ncP and sin ncP functions and the Fourier integralrelation (172). Hence, it suffices to consider only the case n' == nand w' == w, k i= k'.

First consider

1211"] <X> 411"2 ( - jW) [ 12 ,Lj.NjdzdcP==- --,- k JnCkr)JnCkr)

o -00 E~ P

- kk'J~(kr)J~(k'r) - ;:In(kr)Jn(k'r)] o(w -w').

When we compare the second term on the right with the normalization integral for M j . Mjand also use (173c) we see that the integral over r gives a factor

(k,2 _ t2)o(k - k') = 0k

so L, and Nj are orthogonal.After integrating over cP and z the other two scalar products, namely, Ll: Mj and Nj • Mj ,

are both found to have the factor

[kJ~(kr)Jn(k'r) + k'Jn(kr)J~(k'r)]r-l.

124

The use of (175) allows this factor to be expressed as


for which the integral over r equals zero. Thus the orthogonality of the vector eigenfunctionset is established.

Dyadic Green's Function

The dyadic Green's function Ge is a solution of the equation

2 - - ,(V X V X - ko)Ge == Io(r - r ). (178)

The fact that the L j functions are in the null space of the curl curl operator does not causeany difficulty if we simply include the L j functions in the expansion of Ge since this willautomatically make the projection of - k6Ge and I5(r - r') onto this space of functions equal.Hence we let

where Aj, Bj, and C, are vector expansion coefficients. It is a straightforward matter to usethe orthogonality and normalization properties of the eigenfunctions to solve for the expansioncoefficients after the assumed solution for Ge is substituted into (178). We will omit the detailsand only give the final result which is

- , 1 ~Ge(r, r) == -2 LJ€On

47f' n=O

1OO J OO [Mj(k' w, r)Mj(k, w, r') + Nj(k, w, r)Nj(k, w, r')

x 2 2 2o -00 k(k + w - ko)

k Lj(k, w, r)Lj(k, w, r')]- - dwdk.

k5 (k 2 + w2)

(179)

The vector eigenfunctions are given explicitly by (176).The eigenfunction expansion of Ge can be converted into a mode expansion. We can carry

out the integral over w to obtain a mode expansion along z while leaving a spectral integralover k. Alternatively, we can carry out the integration over k and then be left with an inverseFourier transform to be carried out to obtain the explicit dependence on z. Unfortunately, ineither case it is only possible to evaluate the first integral using residue theory. We will evaluatethe integral over k and leave it to the reader to evaluate the alternative solution obtained byintegrating over w first.

The integral over k will be converted to an integral from - 00 to 00 by using

In(kr) == ![H~(kr) +H~(kr)]

H~( -kr) == -e-jn7rH~(kr)

(180a)

(180b)

GREEN'S FUNCTIONS

Fig. 2.17. Contour for obtaining the mode expansion.

125

(180c)

The integrand in (179) does not have any singularities at k == 0 since the M j and Nj functionsvanish for k == O. When the Hankel functions are introduced the integral from - 00 to 00

must be chosen as a principal-value integral so as to avoid introducing any pole singularityat k == O. The Y n function which is part of the Hankel function has a part that contains alogarithmic factor In k r and thus a branch point at the origin is introduced along with theHankel functions. The branch point, however, does not cause any difficulty in the evaluationof the integrals. The Bessel function circuit relations given in (180) are valid when the phaseangle of kr is between - 7r and 7r. A branch cut along the negative real axis is introducedand the contour of integration is chosen as in Fig. 2.17.

A typical integral to be evaluated is of the form

100

f(k)kJn(kr)Jn(kr') dk

where f(k) is an even function of k. For r > r' we use (180a) to replace In(kr) and thenupon using (180b) and (180c) we have

lim100

f(k)kJn(kr)Jn(kr') dk'YI~O 'YI

foo l= p -00 2f(k)kH~(kr)Jn(kr') dk

by following the same steps used to obtain (157) for the mode expansion in spherical coordi-nates. The derivative operations with respect to rand r' can be brought outside the integraland applied later with due care being taken in differentiating functions that have a discontin-uous first derivative when r == r'. The principal-value integral can be expressed as a contourintegral along the path shown in Fig. 2.17 provided «i times the residue at any pole at k == 0is subtracted. The contour can be closed by a semicircle in the lower half of the complex kplane since the asymptotic behavior of the Bessel functions makes the product J n(kr')H~(kr)behave like (l/k)e- j k (r - r ' ) which is exponentially decaying in the lower half plane whenr > r'. The integral over the semicircle of infinite radius does not contribute as long as theconditions for Jordan's lemma hold. This requires that when r > r' the integrand behaves like

- n2kJ n(kr)Jn(kr')

(k2 + w2)r r '

Ikl T with T < 0, apart from the exponential factor. When r == r' the integrand must decreaseat least as fast as Ikl-1+T with T < O. The only term for which these conditions are violatedis the arar, component of the LjLj contribution. For this term the integrand is of the form

and cannot be evaluated using residue theory. However, when the derivatives with respect tokr and kr' are brought outside the integral, a k 2 factor is eliminated and residue theory can beused. The resultant function of rand r' has a discontinuous derivative at r == r' so the resultof applying the operator d2/dr dr' is a delta function term in addition to the usual terms thatarise. An alternative procedure is to express k2/(k2+w2) as l-w2/(k2+w2). The integrandfor the arar, part of the LjLj contribution is

k dJn(kr) dJn(kr')

k 2 + »;2 dr dr'

We can use (175) and the splitting of k 2/(k2 + w2) just noted to rewrite the integrand in theform

kw?2 2 [Jn+l (kr)Jn+l (kr')

2(/( + w )

+ I n--1(kr)Jn-1(kr')] + ~[Jn+l(kr)Jn+l(kr')

+J n--l (kr)Jn-l (kr')].

The integral over k of the last term gives o(r - r')/r as reference to (173d) shows. Theremaining terms can be integrated using standard residue theory after the replacement ofone of the Bessel functions by a Hankel function. In general, we can justify the formalprocedure of differentiating discontinuous functions to obtain delta functions by extracting thedelta function contributions directly from the integrals by rewriting the integrands in such aform that (173d) can be used. Of course (173d) should not be viewed as a classical function;it is an eigenfunction representation of the delta function operator.

When r < r' the Bessel function J n(kr') is expressed in terms of the Hankel functions anda similar evaluation is carried out. Note that the conjugation operation shown in (179) appliesonly to the jw factor. The complex conjugate of the Hankel functions is not taken since theseare replacements for the real J n functions.

For the MjMj contribution in (179) the denominator is k(k2 + W 2 -kij) so poles at k == 0 andk == (kij _W2)1/2 occur. We will show that the residue contribution at the k == 0 pole is canceledby a corresponding contribution from the NjNj terms. For the NjNj terms the denominatoris k(k2 + w2)(k 2 + w2 - kij) because of the extra factor l/p2 that occurs. The residue atthe pole k == - jw in the lower half plane will be shown to be canceled by a correspondingcontribution from the LjLj term. In fact, these residue terms completely cancel the LjLjcontributions outside, as well as inside, the source region. The only remaining contributionfrom the LjLj terms is a delta function term coming from the arar, part as described earlier.Thus, apart from the delta function term, the mode expansion is made up solely from theresidue contributions at the pole k == (kij - W2)1/2 that is present in the integral for the M j

and Nj contributions.

At this point we must verify the residue cancellations described above and also evaluate theresidual delta function term that is left over as part of the contribution from the longitudinaleigenfunctions. In order to evaluate the singular behavior at k == 0, in the various integralswe have to deal with, the following small argument approximations for the Bessel functionsare used:

, 1 (kr,)n 1 (kr,)n+2In(kr) '" n! 2 - (n + I)! 2

[ .2( kr)] j (2)nH~(kr) '" 1 - J;;: 'Y + In 2 Jn(kr) + ;;:(n - I)! kr

2 .2 ( kr)Ho(kr) rv -J; l -l-In 2

(181a)

(181b)

(18Ic)

where l == 0.5772 is Euler's constant. From these relations it is clear that the productJn(kr')H~(kr) is finite at k == 0 and In (kr /2) is multiplied by a k" factor. Only the n == 0term exhibits a logarithmic singularity. We will use the notation that Lt ' Mt ' and Nt denotethe functions given in (176) but with J n(kr) replaced by H~(kr). The az component of theNj functions is multiplied by k 2 so this component vanishes at k == 0 and does not have aresidue contribution. Apart from the z- and et>-dependent factors, the parts of the MjMj andNjNj terms that contribute nonzero residues at k == 0 are (note that these are the factorsmultiplying the same sin net> or cos net> terms)

[_ dH~(kr) ± ~H2(k)] [_ dJn(kr') ± !!'-J (k ')]a, d a, n r a<t>, dr' a", n rr r r r

and

Upon using the small argument approximations these expressions become

w 2 n r'":'2 -1i+1( ± a, + a,)(=t=a<t>' + a,,).p 1r r

At the pole k == 0 the factor w2/ p2 equals unity since p2 == k 2 + w2. Thus, as is readily

apparent, the above two factors will cancel so the residues at the k == 0 pole cancel.Next we consider the residues at the pole k == - jw that occurs in connection with the

LjLj and NtNj terms. The integrand for the NtNj term is proportional to the expression

[az(k2 + w2)H~(kr) - jwi;l[az(k

2 + w2)J n(kr') + jwi;l

as can be seen by referring to (174c). The caret above the Lj functions signifies that thecommon functions of et>, et>', z, and z' are not included. The residue contributions come only

128

from the part


for the Nj functions. For the Lj functions the residues come from the factor

,,+,,*k LjLj

- k5 k 2 + w2 •

When k == - jw the two multiplying factors are w2/ jwk~ and jw /k~, respectively, and hence

the residue contributions are of opposite sign and cancel.As noted earlier, the integral of the ftrBr' part of the LjLj contribution has to be evaluated

either by extracting the delta function part or before the operation d2/dr dr' is carried out.This is not necessary for the NjNj terms. The result is that we obtain a delta functioncontribution which is given by the second derivative at r == r', i.e.,

jeo foe . , cos cos d2 Ia,.ar·2: ~ _",,e-]W(Z-Z). nq, . nq/ dr dr H~( -jwr»Jn ( -jwrd dw.n 81T'4) SIn SIn > < r=r'

The step change in the first derivative with respect to r at r == r' is

J ( _ . ') dH~(-jwr)1 -H2(_. ') dJn(-jwr) I

n jwr d n jwr dr r' r r'

2j- 'If'r'

upon using the Wronskian relationship for the Bessel functions involved. The step change,when the derivative is viewed as a function of r', is the negative of this so the delta functionterm that is produced is (2j j1rr')o(r -r'). We now note that the integral over w can be carriedout to give 2'1f'o(z - z'); hence we get

o(r - r'v- (X) EOn ..- arar,o(z - z') 2 ~-2(cos ncP cos ncP' + SIn ncP SIn ncP')

kor :1=0 1r

1 o(r - r') ~(,I,. ,I,.')~( ') _ a.a, ~( ')== -2 U 0/ - 0/ U Z - Z BrBr - --2-u r - r~ r ~

since the series is the expansion of o( cP - cP').The final expression for the mode expansion of the electric field dyadic Green's function in

cylindrical coordinates is

, j(X) ~jEOn , , dw a.a, ,Ge(r, r) == ~-8-[Mju(r)Mjer(r) + Nju(r)Nju(r)] 2 _ k2 - k2o(r - r) (182)

-(X) j 'If' woo

where the mode functions are defined by

cosMj(r) == \7 x BzJn('Yr)e-jwz ncP

sin(183a)

GREEN'S FUNCTIONS

cosMj(r) == \7 X 8zH~('Yr)e-jwZ . net>

SIn

coskoNj(r) == \7 X \7 X 8zJ n('Yr)e-jwz . net>

SIn

COS

koNj(r) == \7 X \7 X 8zH~('Yr)e-jWZ . net>SIn

129

(183b)

(183c)

(183d)

r < r'

r > r'

r < r'

r > r'

(183e)

(1830

with 'Y == (kij - W 2)1/2. The functions of r' are defined the same way except that e-jwz isreplaced by e jwz' . The (J indicator is also applied the same way but with the roles of rand r'interchanged. The mode expansion including the delta function term given in (182) was alsoderived first by Tai [2.24].

If the integral over w is carried out first then there will be residue terms at w == (kij - k 2) 1/2

for the MjMj and NjNj terms and these produce a mode expansion along z. There areresidues from poles at w == -jk for the LjLj and NjNj terms also. However, these residueterms cancel with the exception of a delta function term that arises from the 8z8z part of theLjLj contribution. In evaluating the integrals over w the - jw factors that appear in theexpressions for the MI» NI» and Lj functions in (176) are brought outside the integral asa d [dz operator so as to allow evaluation of the integrals using residue theory. The resultof such an evaluation gives the following alternative mode expansion for the dyadic Green'sfunction:

c,o. r') = _LL ['XJ Mju (r)Mju(r') + Nju(r)Nju(r') dk - az~z o(r - r') (184)

471" j Jo kJkfi - k2 ko

where now the mode functions are given by

(185a)

(185b)

{

Mj(r),Mj(J(r) == _

M j (r),

z >z'

z <z'(185c)

coskoNj(r) == \7 X \7 X 8zJ n(kr)e-j-yz . net>

SIn

coskoNj(r) == \7 X \7 X 8zJ n(kr)e

j-yz . net>sm

(185d)

(185e)

{

Nj(r), z > z'koN j (1 (r) == (185t)

Nj(r), z < z'

where v == (kij - k2)1/2.The functions of r' are defined the same way but with the roles of rand r' interchanged.

The a indicator specifies the functions with superscript + or - in accordance with z' > z orz' < z, respectively.

2.17. ALTERNATIVE REPRESENTATIONS FOR DYADIC GREEN'S FUNCTIONS

For certain types of scattering and diffraction problems it is useful to have other represen-tations for the dyadic Green's functions that may exhibit a more rapid convergence than theseries forms given in the two previous sections. This is very much the case for scatteringand diffraction by very large cylinders and spheres for which the standard mode expansionsconverge very slowly.

One way to obtain an alternative representation is to synthesize an alternative form for thescalar Green's function g that is a solution of the Helmholtz equation and obtain the Green'sdyadic function by means of the operations shown in (112). We will illustrate the method inconnection with cylindrical coordinates.

Associated with the three-dimensional scalar Green's function problem

(186)

are three one-dimensional characteristic Green's function problems which can be identifiedfrom the separated partial differential equation. These one-dimensional Green's function prob-lems are

Id2gt/> ,dqi + "A",g", = -o(cP - cP )

d2gz ,dz2 + Azgz == -o(z - z )

1 d dg, p2 o(r - r')--dr-

d+ Argr - 2 gr ==

r r r r r

(187a)

(187b)

(187c)

cP < cP'

where At/> == p2, Ar == kij - Az , in order that gr(r), gt/>(cP), gz(z) satisfy the homogeneousHelmholtz equation when r i= r'.

For a specific problem to solve we choose the scattering of the field from an infinitelylong cylinder of radius a as illustrated in Fig. 2.18. On the cylinder surface we will imposeDirichlet boundary conditions which require g == 0 at r == a. In addition, g must representoutward-propagating waves at infinity.

The function g must be periodic in cP with a period of 21r. A useful solution can be obtainedby considering cP as a rectangular coordinate that varies from - 00 to 00. The periodicityrequirement can be met by placing unit sources at cP == cP' +2n1r, n == ± 1, ± 2, .... For thesingle source at cP' we can solve (187a) by choosing

{

Cle-jPcP,

gt/> ==C2e j pcP ,


z

•/T',!/J',Z'a

IIIIIII

"....J.- - - -t---~ Y./

x

Fig. 2.18. A cylindrical scattering problem.

where v = y'Y\;. At e/> = e/>' we require gcP to be continuous and in addition

dgcP Iq,~ =-1.

de/> cP'-

By imposing these source boundary conditions we find that a solution for g c/> is

This wave solution is called a creeping wave since it represents a field that propagates orcreeps around the cylinder. We can now superimpose the solutions from all the other sourcesto obtain the Green's function that has the required periodicity. Thus a valid Green's functionis

00

j "'" - iv Ic/>-c/>' -2n1r1gc/> = - 2v LJ e ·

n=-oo

(188)

A similar solution exists for gz so we choose

(189)

For the function g r a suitable solution is

Or = {Cl[H;(Ar)Jp(Aa) -H;(Aa)Jp(Ar)],

C2H;(Ar),

a ~ r ~ r'

r ~ r'.

This solution has the desired properties of vanishing at r = a and corresponding to outward-propagating waves for large r. We require gr to be continuous at r = r' and also

d

Ir~gr

r dr r'. =-1.

.-Vm

.-V2

Fig. 2.19. Location of the complex roots JIm.

When we impose the source boundary conditions and use the Wronskian relation

J ( I}: l)dH;(~rl) _H2( I}: l)dJv(~rl) == _ 2}v V I\,r d I v V I\,r d I Ir r ~r

we readily find that

The spectra associated with g</> and gz arise from the branch points at the origin. In order thatthe wave solutions do not grow exponentially the branches are chosen such that

Imag. A == Imag. v :::; 0

Imag. A <o.(191a)

(191b)

The function H;(Aa) has an infinite number of zeroes for complex values of v . Henceg, has a discrete spectrum. The zeroes are located in the second and fourth quadrants of thecomplex v plane as shown in Fig. 2.19. The two sets of zeroes are the negatives of each other.In the A</> plane only the zeroes with Imag. v < 0 lie on the Riemann sheet of interest in viewof the manner in which the desired branch was specified by (191a).

According to the synthesis method presented in Section 2.8 the three-dimensional Green'sfunction is given by

(192)

where we have chosen to integrate around the branch cuts in the A(j) and Az planes. These cutsare placed along, but just above, the real axis so that the angle of A(j) and Az ranges from 0to - 2~. This ensures that the conditions given by (191) will hold. It is convenient to changevariables according to

dAz == 2wdw

dA(j) == z» a».

GREEN'S FUNCTIONS

Az plane

133

-------4Gc::::=~=== Br. cut

c,

(a)

v plane

•• Vm

••

(c)

____-+~======Br.cut

(b)

w plane

(d)

Fig. 2.20. Contours of integration for synthesizing the three-dimensional Green's function. (a)Branch cut and contour in AcP plane. (b) Branch cut and contour in Az plane. (c) Location of polesin v plane. (d) The contour Cz mapped into the w plane.

The new contours now become the whole real axis in the wand v planes. The various contoursare shown in Fig. 2.20.

We can express g in the following form after the above change of variables has been made:

x e-jwlz-z'l dw dv, (193)

For a line source the solution is obtained by integrating over Z' from - 00 to 00 to give211"o(w). The integral over w then selects the value w == o. We can close the contour by asemicircle in the lower half of the complex u plane and evaluate the integral over v in termsof the residues at the poles v == Vm == (Jm - ln«. If we denote the residue associated withH;( Jk5 - w2a) by

1R

Jlm== --------..,......-

ddH; <Jkfi -W20)1v JIm

our solution for g can be expressed as

1 00 00•

g(r r') == - '""" '"""e-(jum+ 7Jm) lcP - Q>' -2n1r1 J1I" R H 2 ('\If ), 211"· L....J L....J 2 JIm JIm I >

J n=-oom=l


s

n

v

(195)

Fig. 2.21. Illustration for a field equivalence principle.

where v == (kfi - W2)I/2. The significant advantage obtained with this solution is that when 'YOis large the imaginary parts 11m of Pm are large so that only a few terms in the sum over nandm need be retained because of the rapid decay of the cP function. The evaluation of the rootsand the residues is quite complex and the reader is referred to the literature for the details[2.15], [2.25].

We can construct a dyadic Green's function from (194) but this function will, in general,not satisfy the boundary conditions D X Ge == 0 on the cylinder surface. However, if we havea z-directed electric current element the function

-,logGe(r, r ). 8z == 8zg + 2" \7!}

ko uz

will have zero tangential components on the cylinder surface and the problem of scatteringof electromagnetic waves, radiated by a z-directed electric current element, from a perfectlyconducting cylinder is solved. For the more general case it would be necessary to add solutionsto the homogeneous vector wave equation to take care of boundary conditions.

2.18. DYADIC GREEN'S FUNCTIONS AND FIELD EQUIVALENCE PRINCIPLES

Let the sources for an electromagnetic field be contained in a volume V I bounded by asmooth closed surface S as shown in Fig. 2.21. The unit normal directed out from VI is D.

The electric field outside S is a solution of the source-free equation

\7 X \7 X E - k6E == O.

Consider also a Green's dyadic function that is a solution of

- 2 - -\7 X \7 X G(r, ro) - koG == Io(r - ro)

and satisfies a radiation condition at infinity. We now form the scalar products

E. \7 X \7 X G - \7 X \7 X E.G == E(r)o(r - ro).

The left-hand side can be written as

- \7. [E X \7 X G + \7 X E X G]

(196)

(197)


as direct expansion of the divergence will verify. We now integrate over the volume V outsideS and use the divergence theorem to obtain

If {E(ro),(0 X E· \7 X G + 0 X \7 X E.G)dS ==

8 0,

ro E V(198)

where the integral over a spherical surface at infinity has been set to zero since it vanishesbecause both G and E satisfy a radiation condition. From Maxwell's equations \7 X E ==- jW/LoH, so for ro in V

E(ro) = If n X E· \7 X CdS - jwp,oIf n X H·CdS

8 8

(199)

which gives the field in V in terms of boundary values on S. This is the mathematical statementof Love's field equivalence theorem discussed in Section 1.8. The boundary values can beinterpreted as equivalent surface currents Jms == -0 X E and Jes == 0 X H. The dyadicGreen's function can be chosen as the one for unbounded free space.

A similar solution for the magnetic field can be constructed since H also satisfies (196).Hence

H(ro) =If n X H· \7 X CdS + jWfoIf n X E·CdS.

8 8

(200)

The solutions given by (199) and (200) have the expected property that the functions of rogiven by the integrals involving \7 X G have tangential components that undergo a step changeequal to the boundary value 0 X E or 0 X H as the observation point ro passes through thesurface S from the exterior side V to the interior side VI. We can readily prove this property.

Let So be a very small circular disk that is a part of S as shown in Fig. 2.22. The surfaceintegral in (199) is split into an integral over So and an integral over S -So. The latter integralis a continuous function of ro since r =1= ro and hence does not have a discontinuous behavioracross S. For the integral over So we choose So so small that 0 X E can be considered to beconstant on So. If we use the free-space dyadic Green's function given by (112), then

_ _e-j koR 1 t\7 X G == \7 X 1-- ~ - \7 X -

41rR 41r R

since the exponential factor can be approximated by unity over the small disk and in the regionimmediately surrounding it. The integral of interest is thus approximated as follows:

1 li r 1 li 1--0 X E· \7 X - dS == -0 X E· \7- dS X I.

41r R 41r R80 80

From Fig. 2.22 we see that \7(I/R) can be expressed as

1 aR 0 ar .\7- == - - == - cos () - - SIn ()

R R2 R2 R2

where 8R == (r - ro)/R.


Fig. 2.22. A small circular area So on S.

From symmetry considerations the integral of the component along 8, over the circular diskis zero. The remaining integral

nff C;2 () dS = nff dn = nn

So So

where 0 is the solid angle subtended by the disk at the observation point. Our final result forthe integral is thus

-0 0 0n X Evn X 1- == (n X E) X n- == -n X (n X E)-4 . (201)

411'" 41r 11'"

The solid angle 0 increases up to a value of 211'" as ro approaches S- and then jumps to -211'"as ro passes through S. Since n X (n X E) == (n-Ejn - (n·n)E == -Etan , we see that (201)shows that the tangential value of E given by

lim {'in X E· \7 X GdSso-+oJJ

So

approaches a value equal to one-half of the boundary value obtained from n X E as ro ap-proaches S from the exterior region. The other half of the total tangential electric field comesfrom the sources on S - So. Note from (201) that the integral does not contribute a normalcomponent when So is a circular disk.

Let the z axis be oriented along n from the center of the circular disk assumed to have aradius a. We now consider

ffnx HoGdS~nx Hoff (1+ :~vv) 4:R dSSo So

where we have again approximated e-j koR by one. If ro lies on the z axis at Zo and p is aradial coordinate on the disk, then the first integral is given by

121l"l Q

d dcPJP = 21r[Ja2 +zfi -Izoll·o 0 p2 + z5

This function is continuous as the point zo passes through the surface and vanishes as aapproaches zero.


In order to evaluate the second integral we will orient the x axis parallel to n X H == Jesand first find the x component of the contribution to the field. This is given by (the integral isevaluated first and then Zo can be set equal to zero)

jW/Lo 1211'1a a21- --2Jes -2 odo do

411"ko 0 0 ax VP2 +z~

where p2 == x 2 + y2. The integral can be expressed as

. 1a1~a2 1jW/Lo- --2Jes4 -2 dxdy

411"kO 0 0 ax .JX2 + y2 + Z~

Ja2 _ y 2

= -jW/l~Jes4 r~ 1 dy411"ko Jo ax .Jx2 + y2 +z~ 0

. 1a_/ 2 2

- jW/LO J 4 v a - y dy- 411"k5 es 0 (a2 + Z5)3/2

== jW/LOJes [YVa2 _ y2 +a2 sin-1 ~] la211"k5(a2 + Z5)3/2 a 0

• 2_ jW/Lo J a- 4k5 es(a2 + Z5)3/2 ·

This contribution is also a continuous function of Zo and vanishes when a approaches zerofor all finite values of zo. For Zo == 0 the tangential electric field increases proportional toa-I. This singular behavior is an artificial one due to the line charge Pi == T·Jes / .jt» on theboundary C on So when the current disk is isolated. The integral over S - So produces acompensating singularity. For numerical evaluation the patch size would be kept finite. Froma theoretical point of view the integral can be evaluated by an integration by parts to showexplicitly the line charge contribution that produces the singular behavior, as we will showlater. If we excluded the line charge contribution from both the integral over So and that overS - So both integrals would have much better convergence properties.

By means of a similar analysis, we find that the contribution to the tangential component ofthe electric field that is perpendicular to Jes is zero.

The normal component of the electric field at the center of the disk cannot be found usingthe assumption that the current Jes on the disk is constant since this is equivalent to assumingthat the charge density given by jwps == - \7s ·Jes is zero. In order to evaluate the normalcomponent we consider (the limit as So approaches zero is implied)

-~:~~n.ff (vvk) ·Jes dSSo

= ~:~~n.ff (Vovk) ·Jes dSSo

jW/Lo ({ [ (Jes) 1 ]= 411"k~n'JJ Vo V· If - RV •J es dS.So


When So approaches zero we can replace V'-Jes by - jwps outside the integral. Hence weobtain, upon using the divergence theorem,

jWILof 1 Ps li 1--2 T-Jeso(ro)-V'0R- .u - 4--0 - V'0R- dS411"ko 1I"€0

c ~

where T is a unit outward normal to the boundary of So and lies on S.We can choose ro to coincide with the center of So. On So we can expand Jes(r) in a

Taylor series about roo The dominant contribution to the contour integral comes from theJes(ro) term. For this term T-Jes(ro) == \Jes(ro)I cos cP where cP is the angle between T and theconstant vector Jes(ro). When ro lies on So we have o(ro)- V'o(I/R) == cos (} /R2 where (} isthe angle between o(ro) and aR. On C the distance R is never zero and since R is a constantalong the contour C of the circular disk the contour integral is zero because the integral ofcos cP over the range 0 to 211" is zero. The second integral equals ps(ro)O/411"€0 which is theexpected value since for Zo == 0 this makes 0 -E == Ps/2€0 at the center of the disk. There is anequal and opposite directed electric field normal to the disk on the interior side. The sourceson S - So produce a field that will cancel the interior field, which, as (198) shows, is zero.

In principle we can construct dyadic Green's functions that satisfy one of the followingboundary conditions on S:

nxG==O

oxV'xG==O

In this instance the solutions for E( ro) reduce to

on S

onS.

E(ro) = fJn X E·V X GdSs

and

E(ro) = -jWllofJn X H·GdS.s

(202a)

(202b)

These two solutions correspond to Schelkunoff's field equivalence principles.The two Green's functions appearing in (202) are functions of rand ro but cannot be

continued as a function of ro into the region VI. When boundary conditions are imposed ona Green's function its region of validity becomes restricted to the region for which it wasconstructed. Nevertheless, the function of ro in (202a) will exhibit the same discontinuousbehavior in its tangential value in that as ro passes through the surface S the tangential valuegiven by 0 X E on the exterior side is reduced to zero on the interior side. This property stemsfrom the boundary condition 0 X G == 0 on S. The integral in (202b) has a discontinuous curl.The discontinuous behavior exhibited by (202a) and (202b) is consistent with the interpretationof - 0 X E in (202a) being an equivalent magnetic surface current on a perfect electricconducting surface and that of 0 X H in (202b) being an equivalent electric current on aperfect magnetic conducting surface.

In practice there are relatively few problems for which dyadic Green's functions satisfyingeither of the two above boundary conditions can be constructed. For those problems for which


the Green's functions can be constructed the discontinuous behavior described above can bereadily verified.

2.19. INTEGRAL EQUATIONS FOR SCATTERING

The scattering of an incident electromagnetic field by an obstacle may be formulated as anintegral equation to be solved numerically rather than by solving the partial differential equationwith boundary conditions. In order to formulate the basic integral equations for scattering,consider a current source Ja outside a perfectly conducting surface S (see Fig. 2.21). In placeof (199) one readily finds that the total electric field is given by

E(ro) = fin X E·V7 x CdS -jw/Lofln X H·GdS -jW/Lo!!!Jo.GdS.s s v

The last integral gives the incident electric field E; in the absence of the obstacle when G isthe free-space dyadic Green's function. The induced current Jes on S is given by n X H. Theintegral equation for the unknown current Jes is obtained by setting the tangential componentof the total electric field equal to zero on the perfectly conducting surface S; thus

n X E(ro) = jW/Lo fI n X H.C X n(ro)dS + n X Ej = 0

s

or equivalently

jW/LO fI Jes(r).C(r, ro) X n(ro)dS = -n(ro) X Ej(ro),

sro on S (203)

which is the electric field integral equation (EFIE). 4 This integral equation has a unique solutionprovided there are no current distributions on S for which the integral on the left gives a valueof zero. In order that a solution exists the source function - n X E; must be orthogonalto the null space of the integral operator; Le., it must lie completely in the range space ofthe operator. Unfortunately, at discrete values of k o corresponding to the internal resonantfrequencies of the cavity formed by S the currents Jes that correspond to those associated withthe cavity resonant modes are in the null space of the integral operator. When the scatteringobject is large in terms of wavelength the resonant frequencies are closely spaced so thenumerical solution of (203) is not feasible without some modification of the integral equation.A suitable augmentation will be described later.

The proof that the current on S that corresponds to a resonant cavity mode is in the nullspace of the operator is straightforward. The application of Green's theorem in the region VIshows that

E(ro) =jW/LoflJes.GdS - fin X E·V7 X CdS,s s

where n is still directed out from V I .

The eigenvalue equation for Jes for a resonant mode is obtained by equating n X E to zero

4The integral must be evaluated using an appropriate limiting procedure as described in the previous section.

140

on S. This gives


jWlLoIf Jes(r).G(r, ro) X n(ro)dS = 0,s

ro on S.

But this equation is the same as the homogeneous equation obtained from (203) when n X E,is set equal to zero. Hence, resonant mode surface currents are in the null space of the integraloperator. The eigenvalue equation has solutions at the resonant wavenumbers for the cavity.

An eigenmode current Jes does not produce a radiated field outside S since it gives n X E ==oon S. We can show that this current is orthogonal to the incident field Ej • For this purposewe need to choose an appropriate scalar product. We will choose as the scalar product theintegral of E, -Jes over the surface S for the simple reason that the results we wish to establishlater can then be obtained without introducing an adjoint integral operator. The same resultscan be obtained by using Ej-J;s in the integral over S.

From a physical point of view, the resonant cavity mode current distribution on S supportsthe nonzero interior cavity resonant mode field. This field has a zero tangential electric fieldon the closed surface S, a finite tangential magnetic field on the interior side of S, and a zerofield on the exterior side of S.

When Ja is the source for the incident field outside S the reciprocity theorem gives

JJJE; ·Jes dV = JJJE·Ja dV = 0v v

where E, the field radiated by Jes , is zero outside S. The volume integral on the left reducesto the surface integral

and involves only the tangential part of Ej. Since the integral is zero we conclude that Ej,tan isorthogonal to the eigenmode current. For this reason a solution to the integral equation exists.In principle, the solution can be made unique by removing the eigenmode currents from anycurrent distribution Jes on S. In practice, it is not easy to accomplish this goal since neitherthe eigenfrequencies nor the eigenmode currents are known.

An approximation numerical solution of the EFIE (203) can be achieved using the methodof moments [2.26]. The current Jes may be expanded in terms of a set of vector basis functionsdefined on S, i.e.,

N

Jes(r) == LInJn(r)n=l

where the In are unknown amplitudes and the In(r) are basis functions in the domain of theoperator. These should be part of a complete set of functions on S. When this expansion isused in (203) and the integration over r is carried out we obtain

N

jwp,oLInGn(ro) ~ - o(ro) X Ej(ro)n=l

(204)

GREEN'S FUNCTIONS

where

Gn(ro) = if In(r)· G(r, ro) X n(ro) as.s

141

In evaluating the Gn{ro) an appropriate limiting procedure must be used when the two pointsrand ro can coincide. For example, a small circular disk area can be isolated in the mannerdescribed earlier so that the integral is split into an integral over So and one over S - So.When only a finite number of basis functions are used the integral equation can hold in anapproximate sense only; this is reflected in (204).

A system of N algebraic equations for determining the I n can be obtained by introducing Nweighting functions Wm{ro) that are part ofa complete set in the range space of the operator,and equating the N integrated weights of the two sides of (204) to each other. Thus

where

N

jWJ1.oLlnGnm ==,Sm,n=l

M==1,2, ... ,N (205)

Gnm = ifGn(ro)·Wm(ro) dSos

s; = -if n(ro) X Ej(ro)· Wm(ro) as«s

The reciprocity theorem shows that In(r).G(r, ro)·Jm(ro) == Jm(ro)·G(ro, r)·Jn(r) and sincethe free-space dyadic Green's function is symmetrical in r, ro, i.e., G(r, ro) == G(ro, r),we see that if we choose Wm == J m the matrix with elements Gnm will be symmetrical.The above system of equations can be solved provided the matrix with elements Gnm is notsingular. However, when ko is equal to an interior cavity resonant wavenumber the matrixhas a zero (or near zero, for finite N) determinant because an eigenvector exists at eachresonant frequency and the eigenvalue of the operator is zero. Even if k o is only close to aresonant wavenumber the inversion of (205) is numerically difficult because the matrix is illconditioned, i.e., almost singular. A solution can be obtained using generalized eigenvectors[2.9] but it is preferable to find some way to avoid a singular matrix altogether.

An integral equation involving the magnetic field can also be formulated. For a sourcecurrent Ja (r) the magnetic field is a solution of

This equation can be solved using the same free-space dyadic Green's function. The solutionis given by

H(ro) = ifn X H· V' X GdS + jWEoif n X E·GdS +IIIV' X Jo·GdS.s s v

On the surface S we have n x E == 0, n X H == Jes , so upon replacing the last integral by the

142

incident magnetic field Hi we obtain


H(ro) = If Jes'V' X GdS + Hj.S

(206)

The surface integral gives the scattered magnetic field H,;In the formulation of the electric field integral equation (203) the observation point ro

could be placed on the surface S since the tangential component of the integral on the leftis continuous across S. In (206) the tangential component of the integral is a discontinuousfunction of ro and changes discontinuously as ro crosses the surface S [see (201) for a similarcase]. For the magnetic field we have

(207)

where S+ denotes the exterior side of S.If the integral in (206) is split into an integral over S - So and an integral over So where

So is a small circular disk, then

-oxH(ro)== lim [fr{Jes.V' X aas « 0] + lim {{Jes·V'xGdSxo-oxHj.so~o } so~o }}

S -so fO~s+ So

The first integral is called the principal-value integral even though the term principal value asused here is not the same as the Cauchy principal value of an integral of a complex function. 5

The second integral has the value (H X 0)/2 == -Jes /2as may be deduced from (201) whichwas obtained for a similar integral. Since 0 X H(ro) == Jes we finally obtain

Jes(ro) + lim [jr f Jes(r). \7 X G(r, ro) dS X o(ro)] = o(ro) X ";(ro),2 So~o J

S-So

which is Maue's magnetic field integral equation (MFIE).If the derivation of (201) is examined it can be seen that

lim 0 {{ CO~ () dS = lim 0 {{ dO = 211"0zo~o}} R Zo~O }}

So So

ro on S

(208a)

is independent of the shape of So. Thus the factor of ~ multiplying Jes(ro) in (208a), whichstems from this integral, is independent of the shape of So.

If the unit normal o(ro) is brought inside the integral then the integrand behaves like (Ir -ro)I- 1 as ro approaches r and it is not necessary to introduce a limiting procedure. A common

5Sancer has pointed out that the use of the phrase "principal value" implies that the integral converges only if Sois a symmetric area centered on the point fo such that in the integration negative and positive diverging contributionscancel. In the MFIE canceling contributions are not required to obtain a converging improper integral [2.2]. Theintegral, without the limiting designation, is often written If as a shorthand notation to mean that it is evaluated ina "principal-value sense."


s

Fig. 2.23. Illustration for the magnetic field integral equation.

form of the MFIE often quoted in the literature is

Jes(ro) If-2- +. [Jes(r) X \7g(r, ro)] X o(ro)dS == o(ro) X ";(ro)s

(208b)

where g is the free-space scalar Green's function. The validity of (208b) is readily established.We will not give a detailed proof but will show in a geometrical way how the proof can beconstructed.

With reference to Fig. 2.23 we note that \7g is a vector along the line joining the two pointsrand ro, which lie on the surface S. We now pass a plane through the surface S which cutsthe surface along the curve joining the points r and rOe The current Jes may be expressed asa component Jp along the curve and a component J t perpendicular to Ja- As the point roapproaches r the angle cP approaches 0 and the cross product Jp X \7g has a magnitude thatapproaches

. 1 [r - rollV'glJp SID () = I 12J p- 2-

411'" r - ro p

where p is the radius of curvature of the surface, in the chosen plane, at the point r. Theother vector component J t X \7g approaches o(r) in direction as ro approaches r such thatI(Jt X \7g) X o(ro) 1is also proportional to sin O. Thus the integrand in (208b) has only theintegrable singularity Ir - ro1-1 as ro approaches r and a limiting formulation is not neededwhen o(ro) is brought under the integral sign." If the cross product with o(ro) is not takenfirst, then Jt X \7g will have a Ir - ro1-2 singularity.

The magnetic field integral equation is a statement of a necessary boundary condition. Ingeneral, it is not a sufficient condition that will guarantee a unique solution for the scatteredmagnetic field. At those frequencies corresponding to the resonant frequencies of the internalcavity modes, the solution is not unique. An example of the failure of the MFIE to providea unique solution at a resonant frequency will be given later for the case of scattering by aperfectly conducting sphere.

6For this reason the integral operator in (208b) is a compact operator (see [2.3]).


The introduction of a limiting procedure in (208a) is only for the purpose of providinga method of evaluating the surface integral. An equivalent expression for the magnetic fieldintegral equation that comes directly from (207) is

Jes + lim IfJes - \7 X GdS X 0 == 0 X Hi.ro----'S+

S

If the homogeneous equation, obtained by setting Hi == 0, has a nonzero solution it followsthat the integral equation does not have a unique solution.

Equation (208), which is a Fredholm integral equation of the second kind since Jes alsooccurs outside the integral, has solutions that make the left-hand side vanish when ko is equalto anyone of the resonant wavenumbers for the internal cavity bounded by S. A proof for thisis developed below.

For the interior problem Green's theorem gives

H(ro) = -Ifn x H· \7 x CdS - jWfoIf n x E.CdS, ro E VI·

S S

For a cavity with a perfectly conducting surface 0 X E == 0 on S. We again introduce a limitingprocedure and let ro approach S-, Le., S from the interior side. If we then use the result in(201) and the fact that now 0 X H == -Jes , since 0 is still directed out of VI, we obtain

(209)

for the interior eigenvalue problem. The eigenmode current that satisfies this equation is not thesolution of the homogeneous equation obtained from (208a) even though the integral operatorsin the two equations are the same. However, as we will show, the two solutions are closelyrelated.

In order to conveniently relate the two solutions we will recast the interior and exterioreigenvalue problems in a different but equivalent form. In place of Jes in (209) write - 0 X Hand bring o(ro) inside the integral; thus

If H· [n(r) x \7 x C(r, ro) X n(ro)] dS - ~H(ro)· [n(ro) X I] = 0

S

(210a)

where we have used 0 X H == -H X 0 == -H-I X 0 == -H-(o X I). The homogeneousequation for the exterior problem can be written in a similar form, namely,

If H· [n(r) X \7 X C X n(ro)] dS + ~H(ro)· [n(ro) X I] = o.S

(210b)

We can show that the scalar product H 1(r)-[0(r) X \7 X G(r, ro) X 0(ro)]-H2(ro) is sym-metric for any two vectors HI and H 2 that are tangential to S. Let J I(r) == o(r) X HI (r)and J2(rO) == o(ro) X H 2(ro). The scalar product becomes - JI(r)- \7 X G-J2(rO) ==-JI(r)-(\7g X 1)-J2(ro) == -JI(r)- v« X J2(rO) == J 2(ro)- \7g X JI(r) where we have used

GREEN'S FUNCTIONS

v X G == V X Ig == V'g X I and g is the scalar Green's function

e- jko Ir-roIg(r, ro) == 4 I I-

1r r - ro

If we interchange the labels on rand ro and note that V'g == - Vog, then we obtain

J2(r)- V'og X J t(ro) == -J2(r)- Vg X J 1(ro)

== -J2(r)- V' X G-Jt(ro)

== H2(r)-[0(r) X V X G X o(ro)]-HI(ro)

145

which proves that the scalar product is symmetric.The other scalar product HI-(o X I)-H2 is antisymmetric since H 1-(0 X 1)-H2

HI-o X H2 == H2-H1 X 0 == -H2-0 X HI == -H2-(0 X I)-HI. Note that 0 X I is a rota-tion operator. It rotates a vector by 90°.

Let the solution for H in (210a) be expanded in terms of a complete set of vector basisfunctions In(r) defined on S; thus

H(r) == LI~Jn(r)-n

We use this expansion in (210a), integrate over r, then scalar multiply by Jm(ro) and integrateover ro. This leads to the following system of equations:

where

m==I,2, ... (211a)

Gnm = fj fj In(r)· [n(r) X V' X G X n(ro)]·Jm(ro) dS as;s s

T nm = fjJn(roHn(ro) X I]·Jm(ro)dSo.s

Note that Gnm == Gmn, T mn == -Tnm» and T nn == 0 because of the symmetrical and antisym-metrical properties of the scalar products. In (21Ia) the I~ are the unknown amplitudes.

For the solution to (210b) we assume an expansion of the form

H(r) == LI~Jn(r).n

By a similar procedure we obtain the system of equations

m == 1, 2, ... _ (211b)

For a numerical solution only a finite number of basis functions can be used.Solutions for I~ and I~ exist only if the determinant of the system of equations vanishes.

146

The transpose of the matrixin (211b) has elements


which are the elements of the matrix in (211a). Both systems have the same determinant.Hence, if there is an eigenvector for the interior problem there will also be an eigenvector forthe exterior problem. We can express (211a) and (211b) in matrix form as

[0 - ~T] [il = 0

[0 + ~T] Wl = 0 = Wlt [0 - ~T] .By subtracting the two equations we get

(212a)

(212b)

If we assume that [Ii] == [Ie], then since [1] is a nonzero matrix we would get [Ii] == [Ie] == o.Hence we conclude that [Ii] =I [Ie]. We see that if [Ii] is a right eigenvector of [0 - !T] then[Ie] is a left eigenvector. Thus the interior and exterior eigenvalue problems have solutions atthe resonant wavenumbers for the internal cavity but the two current distributions on S are notthe same. This result is a very important one since it allows us to combine the electric fieldand magnetic field integral equations into a combined field integral equation that does not haveany nonzero solutions in the null space.

Yaghjian has given a physical description of the current which is a solution of the homo-geneous MFIE [2.29]. We will present the same results from a somewhat different approach.Let Fn(r) be a solution inside S of \7 X \7 X Fn - k5F n == 0 with the boundary conditionn X Fn == 0 on S. Thus Fn corresponds to the electric field of an interior cavity resonantmode. The corresponding magnetic field and current on S are

jH= koZ

ov x r,

jJes = Jen = -0 X H = - koZ

o0 X V' X F,

where the unit normal n is directed outward from S. By duality, a solution of Maxwell'sequations with H == Fn also exists and represents the resonant mode in a cavity with perfectlyconducting magnetic walls. The electric field of this mode has a tangential value on S givenby

j j ~on X E == -k y n X \7 X H == -k y n X \7 X r, == -Jen •

o 0 0 0 EO

The tangential component of H is zero on S. We can construct a unique solution for theelectric field outside S such that this electric field has tangential components that match thoseof the interior electric field on S. From this electric field we can find the magnetic field using


Maxwell's equations. We will call the external field Es, Us. In order to support the externalmagnetic field a current 0 X Us must be placed on S. This current is also given by

jJes = 0 X H, = koZ

o0 X V X E,

and is different from Jen. We will now show that the current Jes as specified is a solution ofthe homogeneous MFIE.

By means of an application of Green's theorem, the magnetic field exterior to S is given interms of boundary values on S by (200) and is

n, = fjOX Hs'Vx GdS +jkoYofjO x Es·GdS.s s

On the surface S we have

n X Us + lim fjo X Us· \7 X GdS X o(ro) == -jkoYofjO X Es·GdS X o(ro).ro~S+

S S

The integral on the right side of this equation equals zero because on S we have n X E, ==(JJ.o / fO)Jen so the integral gives the tangential value of the electric field radiated by the resonantcavity mode current Jen and the latter is zero. If we now replace 0 X Us by Jes we obtain thehomogeneous MFIE.

In order to be a valid solution of Maxwell's equations, the solution of the homogeneousMFIE would require that an actual physical current Jes be placed on S. In a scattering problemsuch a current distribution does not exist on S. Hence the solution for the homogeneous MFIEis spurious. It corresponds to an undamped external oscillation; such undamped oscillationscan only be maintained by an impressed source since energy is lost by radiation. In addition,this homogeneous solution has an internal cavity electromagnetic field associated with it.

As we have noted, the solutions to the homogeneous EFIE and MFIE represent differentcurrent distributions on the surface of the scatterer, but at the same frequency. Neither currentdistribution can satisfy the homogeneous equation that results from the combination of the twofield integral equations.

The combined field integral equation is, upon combining (203) and (208) [2.27]:

. fj - 1'Jes (ro) # G- dS-]koZo Jes·Gxn(ro)dS+ 2 +1' Jes·\7x xo(ro)

S S

(213)

where l' is an arbitrary constant. Since IZoD; I is equal to lEd for an incident plane wave, l'is commonly chosen equal to Zoo

Another procedure that can be used to obtain an integral equation that does not have aresonant mode solution gives the combined source integral equation [2.28]. In this formulationthe scattered electric field is expressed as a field produced by a system of electric and magneticcurrents on S. This integral equation is obtained from Love's field equivalence theorem as


stated by (199). The combined source integral equation is

n X E, == -0 X Ei == lim 0 X fjJmse \7 X GdS - jkoZoo X fjJeseGdSro~8+

S 8or

J;s + s~i~on X 11Jms• \7 X GdS - jkoZon X fj Jes·GdS = -n X E; on S.

8-80 8(214)

In obtaining (214) we replaced 0 X E by - Jms and 0 X H by Jes in (199). Mautz andHarrington chose Jms == Zoo X Jes and then showed that the homogeneous equation obtainedfrom (214) does not have a solution [2.27]. The two sources in (214) are equivalent sourcesonly. The actual current induced on S has to be determined by finding the total tangentialmagnetic field at the surface S.

Yaghjian has reviewed the various integral equation formulations and the reader is referredto [2.29] for more details on methods that can be used to circumvent the problem associatedwith the cavity resonances. Yaghjian has also shown that the electric field and magnetic fieldintegral equations can be made unique by imposing a condition on the normal components ofthe field at the surface S [2.29]. This latter method has the advantage of keeping the resultantintegral equations relatively simple, whereas the combined field and combined source equationsare quite complex. The resonant mode currents can also be eliminated by requiring that thetotal field inside S vanish at a number of selected points. A review of the literature using thisapproach has also been given by Yaghjian [2.29].

The solution of the integral equations can be developed in terms of the eigenvectors of thematrix operators shown in (205) and (211b). These eigenvectors are called the characteristicmodes and can be used as a basis to expand the current or tangential magnetic field respectivelyon the surface S. A treatment of the characteristic modes associated with scattering can befound in [2.30].

The magnetic field integral equation cannot be used to solve the scattering problem involvingan open surface where the surface consists of an infinitely thin perfectly conducting metalsurface. The reason is that the incident magnetic field has a tangential component with equalvalues on the two sides of the surface and hence the boundary condition 0 X H;- - 0 X H; ==Jes does not involve Hi.

In recent years a large amount of work has been done on transient or pulsed electromagneticfield scattering from perfectly conducting obstacles using Baum's singularity expansion method(SEM). An extensive bibliography of the relevant literature is given in a recent volume of theElectromagnetics Journal [2.4]. A conducting body supports an infinite number of externalnatural damped oscillations corresponding to complex resonant frequencies. The scatteredfield may be described in terms of these natural modes [2.31]. The spurious solution that isobtained for the homogeneous MFIE with steady-state sinusoidal excitation is not part of thespectrum of natural modes because it is an undamped solution. Undamped natural oscillationsexterior to a scattering body do not exist because the oscillating modes must lose energy byradiation. In the exterior transient problem the pure imaginary js» axis poles do not contributewhen the Laplace transform solution is inverted so the spurious modes are eliminated." Sancer

7In actuality the j w axis poles in the eigenfunction expansion are canceled by numerator mode coupling coefficientsthat vanish at the same frequencies.


et ale have shown that current distributions on the scatterer that arise at the interior cavityresonant frequencies are not excited [2.32]. Thus the transient scattering problem is a betterposed problem than that of scattering by a time harmonic incident field.

Example: Scattering from a Perfectly Conducting Sphere

The problem of scattering of an incident field by a perfectly conducting sphere of radius acan be solved exactly. Thus this problem provides an instructive example of the application ofthe EFIE and MFIE and insight into the effects of the homogeneous integral equation solutionson the determination of unique solutions for the scattered fields.

We will assume that the source for the incident field is located at a distance r > R > a awayfrom the sphere. In the region r :S R the incident field can then be expanded in terms of theMnm and Nnm functions given by (135) with k == k«. We will assume that the incident fieldcan be represented by a finite number of modes; thus"

N M

E; == L L (AnmM~m +BnmN~m)n=l m=ON N

H; == LLjYo(AnmN~m +BnmM~m)·n=l m=O

(215a)

(215b)

The magnetic field is obtained by using V X E == -jwp,oH, V X Mnm koNnm, andV' X Nnm == koM nm. For simplicity we choose the cos mcP and sin mcP functions in theexpressions for Mnm and Nnm given by (135) so that E;8 is an even function about cP == O. Forthe scattering problem we also need the outward-propagating modes M~;,t and N~~+. Thesemode functions are also given by (135) with k == k o and jn(kor) replaced by h~(kor).

The classical method to solve the scattering problem is to let the scattered field be describedin terms of a series of outward-propagating modes in the form

N M

Es == L L (CnmM~;"+ +DnmN~~+).n=l m=O

(216)

The coefficients Cnm» D nm are found by requiring that ar X Es == -ar X E; at r == a. Thisgives, in view of the orthogonal property of the modes,

DnmN~~+ x ar == -BnmN~m x a,

at r == a

at r == a. (217)

We will now assume that j 1(koa) == 0 so that M~l' M~o, and Mf1 represent degenerateresonant modes inside the sphere. Thus, part of the incident field will have an electric fieldwith a zero tangential component on the surface of the sphere. The sphere will not scatter theM~i + mode as is quite clear from the solutions for C nm and Dnm; i.e., C 11 == 0 because theM~m functions have a jn(kor) radial dependence and jl(koa) == O.

In order to implement the EFIE we use the dyadic Green's function given by (166) and note

8For an incident plane wave an infinite number of terms is needed in the expansion.


(218)

that the field point ro approaches the spherical surface from the exterior. Thus the EFIE is

cx:>nfj { ZE j X aro = ~~ s Jes(r). 2n(n: ~)Qnm [M~m(r)M~;,t(ro)+ M~m(r)M~~+(ro)

+ N~m(r)N~~+(ro) + N~m(r)N~~+(ro)]} X aro dS

where , = '0 = a .At this point we will introduce the following surface vector functions:

mpm sin dpm cosSe, 0 _ n A. n A.nm - =f-'-(J mo/ao - -d(J mo/a(j>

SIn cos sin

dpm cos mpm sin~~ = d(Jn mcPao=f~(J mcPa(j>.

sin SIn cos

(219a)

(219b)

Note that S~~ = T~~ X a, and T~~ = -S~~ X a,. These vector functions are all mutuallyorthogonal with respect to integration over the surface of a sphere. They have the samenormalization values which can be found using (136b); thus

121("1 1(" Te,o. Te,0 sin (J d(J dA. = 1

21("11("se,«. se, 0 sin (J d(J dA.nm nm 0/ nm nm 0/

o 0 0 0

= 2n(n + l)(n + m)! 21r = 2n(n + 1)Qnm(2n + 1)(n - m)! €om 1r

(219c)

where Qnm is given by (139). These vector functions may be used as a basis set to expand thecurrent on the sphere. Note that the M nm functions are equal to Snm multiplied by a functionof r and that the () and cP components of Nnm are given by T nm multiplied by a radial function.The vector functions given by (219) are the characteristic modes for a sphere. Note that theSlo and Tlomodes do not exist.

The current on the sphere can be expanded in the form

Jes = L(I~mS~m +I~mS~m +J~mT~m +J~mT~m)·n. m

(220)

The I~;; and J~ii: are unknown amplitude constants. When we use this expansion in (218) andthen scalar multiply (218) by the basis functions used to expand Jes it is clear that the matrixmultiplying the I nm and J nm will be diagonal and the matrix elements corresponding to theIll' I~o, and I~1 terms will be zero. Thus the determinant of the matrix is zero. However,the term E; X a,o when scalar multiplied by S11 and integrated over S is also zero.

When the expansion (220) is used in the EFIE (218) the following system of equations isobtained:


m=0,1,2,···,n; n = 1, 2, ... , 00.

From these equations we readily see that I~o and I~l are unspecified when i, (koa) = O. Thusthe currents I~oS~o and I~1S~1 are solutions of the homogeneous EFIE. 9 All the other I~m

and J~m coefficients are zero (except at the resonant frequencies where jn(koa) or [koajnl'vanish). The solutions for I~m and J~m for n = 1, ... ,N, m = 1, ... ,M, are

provided we cancel the common factors on both sides of the equations for I~m and J~m'

All of the other I~m and J~m are zero. If we do not cancel the common factor jl(koa)then the equation for IYl is indeterminant. Consequently, we do not obtain a unique so-lution for IYl; this is consistent with the fact that Sfl is also a solution of the homo-geneous EFIE. Apart from the lack of uniqueness for the solutions for IYl' I~o' and I~l theresults agree with those obtained by the classical method. The current modes Sfl' S~o, andS~1 do not radiate because they do not couple to the mode spectrum outside the sphere sincethe M~l' M~o, and M~1 factors in the Green's function are zero at r = a.

If the surface integrals were evaluated on a computer, then because of the finite precisionof the computer the matrix element and the coupling of the incident field to the S~1 currentmode would be small but not zero. As a consequence, the Mfi + mode would appear as partof the scattered field. If the current were expanded in terms of a set of basis functions that donot lead to a diagonal matrix we would have an ill-conditioned (almost singular) matrix andwould not be able to easily identify the current that is a solution of the homogeneous integralequation. The same difficulties would occur if j 1(koa) were very small but not equal to zero.

From an analytical point of view the EFIE gives the correct solution for the sphere problem,but the solution is not unique because the homogeneous integral equation has a nontrivialsolution.

If we examine the expression for the incident magnetic field as given by (215) we find thatthe modes jYOBlOM~o and jYOBllM~l equal zero at r = a because i, (koa) = O. From theclassical solution to the scattering problem as given by (217) we see that the scattered magneticfield modes jYODlOM~(t and jYODllM~i+ do not have zero amplitudes. Consequently, wecan anticipate that it might be difficult to determine the scattered magnetic field modes fromthe MFIE.

The MFIE based on (207) can be expressed in the form

a., X Hi = Jes + f t fjJese [2 ( - jl~Q ] [N~m(r)M~~+(ro) + N~m(r)M~;,,+(ro)n=l m=O S n n + nm

+ M~m(r)N~~+(ro) + M~m(r)N~;,,+(ro)l dS X 8 ro• (221)

A complete expansion for the current is given by (220). We can simplify (221) by introducing

9The origin for 4> can always be chosen so that the incident field does not have the M~ 1 mode (Problem 2.33).


the surface vector functions and using (219). Thus we find that (221) can be expressed as

N M

L L [-jYo(koajn)'AnmS~m + jYo(koajn)BnmT~m]n=1 m=O

00 n

== «s; - jaL L [(koajn)(koah~)'(/~mS~m +l~mS~m)n=1 m=O

where the prime denotes a derivative with respect to koa. We now scalar multiply theseequations by the various basis functions and integrate over () and cP. This procedure yields theequations

m ==0,1,2,· ··,n; n == 1, 2, ... , 00.

We make note of the following Wronskian relationship:

d[aj t(koa)]h2(k ) _ [dahf(koa)] . (k ) == d[a j 1(koa)]h2(k ) == Ld 1 oa d J 1 oa d 1 oa ksa'a a a oa

The formal solutions for I~m and J~m are

10 == - jYoAnm(koajn)' A nmnm a[1 - j(koajn)(koah~)'] - aZo(koah~)

e jYoBnm(koajn) BnmJnm = a[1 +j(koajn)/(koah~)] - aZo(koah~)'

(222)

upon using (222). These solutions are the same as those obtained using the EFIE. The solutionsremain finite even at the resonant frequencies. However, if we had not used the Wronskianrelation (222) the solutions for J~m would have taken the indeterminate form of zero dividedby zero and would have required a limiting procedure for evaluation. From a numerical pointof view indeterminant forms cannot be easily evaluated on a computer.

The solutions to the homogeneous MFIE are the current modes JY1T~l' J10T~o, and J11 T~ 1

at the frequency where jl(koa) == O. These solutions are different from those for the homo-geneous EFIE at the same resonant frequency. The coefficient al'[, == -jYO(koajl)'A ll andthe current IY1SYl supports the incident jYOA llNYl == jYoAll(koajl)'a-1TYl mode. All ofthe coefficients I~m' J~m for n up to Nand m up to M can be solved for, but the solutions for


J~o and J~l are not unique because the equations for them have a zero factor on both sides.It is seen that the MFIE also gives a correct solution for the sphere problem but the solution

is not unique at the resonant frequencies. Furthermore, the solutions to the homogeneousMFIE give rise to spurious radiated modes since the N~l' N~o, and N~l mode functions arenot zero at j 1(koa) == O.

As a final comment we point out that if the scattered magnetic field inside the sphereis desired then in the Green's dyadic function the modal functions representing outward-propagating waves will appear on the left. By using the Wronskian determinant one can thenreadily show that

fj sJes • V' X CdS X a,

has the correct discontinuous behavior as the observation point fO passes through S fromthe exterior side to the interior side. Of course, this is not a surprising outcome since thecorrect discontinuous behavior is a built-in feature of the eigenfunction expansion of thedyadic Green's function. One also readily finds that for the interior homogeneous MFIE thesolution is I~1s~1 + IY1s~1 + I~oS~o which is the current associated with the M~ l' M~o, andM~l electric field modes.

2.20. NON-SELF-ADJOINT SYSTEMS

Let cC denote a differential operator and consider the equation

cCt/! + At/! == f(x). (223)

The variable is x where 0 :S x :S a. The domain ~ of the operator cC is the domain of functionst/! that satisfy certain continuity conditions and specified boundary conditions. We can definea scalar product in more than one way- for example,

or

or

(cP, 1/;) = 1a

cP(x)1/;(x)dx

(cP, 1/;) =1a

cP(x)1/;(x)u(x) dx

(cP, 1/;) = 1a

cP(x)1/;*(x)dx

(224a)

(224b)

(224c)

etc. The definition (224c) is used in quantum mechanics (Hilbert space), while (224b) is usedif (223) has the function a as a factor with A.

Once the scalar product has been defined the adjoint operator cCo is defined to be thatoperator which satisfies

(225)


If cC == cCo and the domain ~o of cCo coincides with the domain ~ of cC, then cC is calleda self-adjoint operator. Sometimes ~ and 5)0 differ even though cC == cCo and in this case cCis only formally self-adjoint. In practice cCo is determined by integration by parts to transferdifferentiations on t/; to differentiations on c/>.

Example

x==O,a.

t/; may be expressed in terms of the eigenfunctions of d2t/; /dx? + Xt/; == O. It is easily verifiedthat t/;n == sin (n-xxfa) - j(n1r fa) cos (nsx fa). If we choose (224a) for the scalar productthen

n =1= m.

The domain of cC is the domain of functions that satisfy t/; + j(dt/;/dx) == 0 at x == 0, a.With (224a) as the scalar product we have

The integrated terms vanish if c/> + j (d c/> /dx) == 0 at x == 0, a. Hence cCo == cC; since 5) == 5)0

also, the operator is self-adjoint.However, if we choose (224c) as the scalar product then

The boundary terms which can be written as

[(cJ> _ j dcJ» dt/;* _ (l/;*_jdl/;*) dcPJl

o

dx dx dx dx 0

vanish if c/> - j(dc/>/dx) == 0 at x == 0, a. In this case cCo == cC but 5)0 =1= ~ so the operator isnot self-adjoint. Note that the adjointness properties of an operator are dependent on the choiceof scalar product. When an operator is not self-adjoint the eigenfunctions are not orthogonalwith respect to the scalar product used.

With the scalar product (224a) cC is self-adjoint and J; t/;nt/;m dx == 0, n =1= m. With thescalar product (224c) cC is not self-adjoint and this implies J; t/;n t/;;" dx =1= 0 for n =1= m. In anon-self-adjoint system the appropriate orthogonality principle is instead

n =1= m. (226)


We may show this as follows: Let cPn, J.tn be the eigenfunctions and eigenvalues of £0; then£ocPn + J.tncPn == O. Since £1/;m + 'Am1/;m == 0 also, we have (cPn, £1/;m) == -'A;"(cPn, 1/;m) ==(£ocPn, 1/;m) == -J.tn (cPn, 1/;m) if (224c) is used for the scalar product [if (224a) is used, replace'A;" by 'Am]. If 'A;" i= J.tn, then (cPn, 1/;m) == o.

The eigenvalues of the adjoint operator are related to those of the original operator. Wehave (£0 + J.tn)cPn, 1/;) == 0 for all 1/; in ~. But this also equals (£ocPn, 1/;) + J.tn (cPn, 1/;) ==(cPn, £1/;) + (cPn, J.t~1/;) == (cPn, (£ + J.t~)1/;) == O. Since cPn is not zero (£ + J.t~)1/; == 0 so J.t~ isan eigenvalue of £ when (224c) is used for the scalar product. When (224a) is used J.tn is aneigenvalue of both £0 and £. With proper indexing J.tn == 'A~ (or J.tn == 'An).

In our example if we use (224c) for the scalar product then cPn == 1/;; and the orthogonalityrelation is (1/;n, cPm) == 0, n i= m. But this equals J; 1/;ncP;" dx == J; 1/;n1/;m dx == 0 in agreementwith the results obtained if (224a) is used for the scalar product.

For the Sturm-Liouville equation we choose £ == (l/a)(d/dx)p(d/dx) + q/a so theequation becomes £1/; + 'A1/; == f [o . We now use (224b) for the scalar product. For (cP, cC1/;)we get

t' ucP [!!!..-Pdy; + ~y;] dx == [PcP dy; _ py; dcP] 1° + t"uy; [!!!..-PdcP + ~cP] dxio a dx dx a dx dx 0 io a dx dx a

upon integrating by parts twice. If K I1/; +K 2 d1/;/ dx == 0 on the boundary then the integratedterms vanish if K1cP +K2dcP/dx == 0 on the boundary. Hence £ == £0 and f) == f)o so theSturm-Liouville equation is self-adjoint.

Many people, outside the field of quantum mechanics, use (224a) for the scalar product.We can use this for the Sturm-Liouville equation also if we choose (d jdx)p(d jdx) + q asthe operator cC. The orthogonality condition is then expressed as (cPn, acPm) == 0, n i= m. Inthe next section on Green's function we will follow this convention.

Green's Function

Consider the Green's function for a self-adjoint system with

£G(x, x') + 'AG == -o(x - x'),

We then have

(227)

(G(x, x~), £G(x, x~)) - (G(x, x~), cCG(x, x~))

+ 'A (G(x , x~), G(x, x~)) - 'A (G(x , x~), G(x, x~))

== -(G(x, x~), o(x -x~)) + (G(x, x~), o(x -x~)).

But (G(x, x~), £G(x, x~)) == (cCG(x, x~), G(x, x~)) for a self-adjoint system so the left-hand side vanishes. Thus (G(x, x~), o(x -x~)) == (G(x, x~), o(x -x~)) or

G(x~, x~) == G(x~, x~)

i.e., the Green's function is symmetrical in x~, x~.

For a non-self-adjoint system we consider (227) along with

cCoGo(x, x') + 'AGo == -o(x -x').

(228)

(229)

156

We can now write


(G(x, x~), £oGo(x, x~)) + A(G, Go) - (Go(x, x~), £G(x, x~)) - A(Go, G)

== -(G(x, x~), o(x -x~)) + (Go(x, x~), o(x -x~)) == 0

since (G, £oGo) == (£G, Go) == (Go, £G). Hence

For a non-self-adjoint system the Green's function is not symmetric.Consider now a linear system

£l/; +Al/; ==1·

(230)

(231)

If we solve (227) and make G satisfy the same boundary conditions as l/; does, then bysuperposition the solution for l/; is

!/;(x) = -LaG(x, x')f(x') dx' (232)

since G(x, x') is the field at x due to a unit source at x'. From (230) we see that this solutioncan also be written as

!/;(x) = -LaGa(x', x)f(x')dx'. (233)

If l/;n, An are the eigenfunctions and eigenvalues of £, and cPn, An, those of £0' then

and

G( ') == _~ l/;n (x)cPn (x')x,x c: A-A

n n

G ( ') == _~ l/;n(x')cPn(x)a x,x L-t A-A ·

n n

(234a)

(234b)

For the purpose of obtaining eigenfunction expansions and expressions for Green's functionsthe choice of scalar (inner) product to use is often dictated by what is most convenient forthe problem at hand. For many problems in electromagnetic theory the use of (224a) as thescalar product makes it unnecessary to introduce an adjoint operator. The standard reciprocitytheorems are based on (224a) as the scalar product, e.g.,

On the other hand, if we wish to investigate convergence properties of series expansionsthat involve complex functions the scalar product shown in (224c) is used. In a linear vectorspace of complex functions the concept of length or norm of a function is a positive quantity.Thus if l/;n(x) is a complex eigenfunction and we want its length to be unity we normalize this

GREEN'S FUNCTIONS

function by dividing it by the scalar quantity

157

(235)

A complete linear vector space of complex functions with the metric (measure of distance)defined by (224c) as the scalar product is called a Hilbert space. In this space the distancebetween two functions 1/; and tP is given by

[fa ] 1/2

(111P - <1>11>1/2 = (1P - <1>, 1P - <1» 1/2 = io (1P - <1»(1P - <1»* dx

when the functions are defined on the interval 0 ~ x ~ a. If 1/; is a sequence of functions 1/;N,which might be generated by a Fourier series,

N

1/;N == LCntPnn=1

then we say that 1/;N converges to tP as N tends to infinity if the distance between 1/;N and tPvanishes as N tends to infinity.

The use of the scalar product (224c) is important if we want to make use of the well-developed mathematical theory of linear operators in a Hilbert space [2.9], [2.11].

The condition for a system such as (231) to have a solution is that the source function bein the range of the operator. For a unique solution the equation should not have any solutionsfor the homogeneous equation obtained by setting! equal to zero. If cC is not a self-adjointoperator the solvability condition can also be stated as the requirement that! be orthogonal tothe functions in the null space of the adjoint operator [2.11].

2.21. DISTRIBUTION THEORY

In this section we will give a brief introduction to distribution theory for the purpose ofshowing how this theory gives a mathematical structure within which the delta function canbe treated in a rigorous manner.

In distribution theory a function !(x) is characterized by a functional mapping. As a func-tion, f(x) is characterized by the scalar numbers y == f(x) for each assigned value of x. Inthe Fourier integral of !(x), i.e.,

F(w) =f:ej wx

!(x) dx

the function !(x) is characterized by the scalars F(w) for each value of w. This is an exampleof a functional mapping and !(x) is viewed as a functional on the space of ej wx functions.

Consider now a space of continuous functions cP(w, x) with continuous derivatives of allorders and with bounded support [all cP(w, x) vanish for IxIgreater than some finite number].The members of the set of functions cP(w, x) are identified by the parameter w. The cP(w, x)are called testing functions. An example of a testing function that vanishes for IxI 2: 1 is

1

{exp~, Ixl < 1

cP(x) == x-I0, Ixl 2: 1.

This function has continuous derivatives of all orders and they all vanish at x == ± 1.

(236)

If I (x) is an integrable function it can be characterized in terms of the distribution F (w)defined by

F(w) = i:cf>(w, x)!(x)dx. (237)

It is common practice to simply write I(x) for the distribution F(w), it being understood thatI(x) viewed as a distribution is given by (237). We will deviate from this practice and usea subscript d to denote the distribution given by (237), i.e., Id(X) = F(w). The functionalmapping given by (237) is unique. All functions that have or generate the same distributionare identical with the exception of possible values assigned at a number of discrete points. Forexample, the two functions

o:::;x < 1

O:::;x<!, !<x:::;1

_1x- 2

have the same distribution. For physical problems functions such as 12 do not arise. Thedistribution I d(X) is not a function of x. The notation is merely a reminder that the distributionwas generated from a function I(x) of x. The definition (237) characterizes I(x) as a functionalon the space of all testing functions. The derivative of a distribution Id(X) is defined as

/

00 dl /00 dePew, x)!d(X) = -00cf>(w, x) dx dx = !(x)cf>(w, x) I~oo - _oo!(x) dx dx

= _/00 !(x) dcf>~, x) dx. (238)-00 x

The integrated part vanishes since eP has a bounded support.Since eP(w, x) has a bounded support the Heaviside unit step function hex) given by

{

O,h(x) =

1,

can be characterized as a distribution

x <0

x >0

hd(x) = /00 h(x)cf>(w, x)dx = roocf>dx.-00 io (239a)

By means of the definition (238) we can also define the derivative of hex) in a distributionalsense; i.e., we write

h~ = dhdd(X) = _/00 h(X)dcf>~W, x) dx = _ (OOddcf> dx = cf>(w, 0).X -00 x io x (239b)

The derivative does not exist in the classical sense but does exist as a distribution since- deP/dx is also a testing function, as is any derivative of eP. In (239b) a knowledge of all

dePew, x)/dx defines the functional hd.The delta function distribution Od(X-x') is defined as follows (the integral is only a formal

GREEN'S FUNCTIONS

intermediate step):

159

(240)Od(X - x') = i:o(x - X')c/J(W , x) dx = c/J(w, x'),

Consequently, the distribution Od(X) = cJ>(w, 0). But this is equal to (239b) so in a distributionalsense the derivative of the Heaviside unit step distribution is the delta function distribution.

Consider the potential 1/I(r) from a unit point charge. It is given by 1/41rr and is a solutionof Laplace's equation for r > 0

~r2~_I_ =0,ar ar 41rr

r #0.

The derivative of 1/ r is not defined at r = 0 in a classical sense. From a distributional pointof view

(a 2 a 1) roo ( 2 8 1) 8cJ>arr ar 47rr d = - 10 r ar 47rr ar dr

= _ ( r 28cJ» _1_1 00

+ roo_l_~r28cJ> dr8r 41rr 0 io 41rr8r 8r

= {00_1_~r28cJ> dr.io 41rrar Br

We can rewrite this integral in the form

.100

1 8 28cJ>I=hm --r -dr0-+0 0 41rrar 8r

since the derivatives of cJ> exist and are bounded at r = O. Thus the limit exists. We nowintegrate by parts to reverse our equation back to the original form, Le.,

I = lim [~ acJ> 1

00 -100

r2acJ>~ (_1) dr]0-+0 41r ar 0 0 ar Br 41["r

• [ 2 8 ( 1)1

00 100

8 2 8 ( 1) ]= lim -cJ>r - - + cJ>-r - - dr .0-+0 8r 41rr 0 ·0 8r 8r 41rr

The first integrated part vanishes as a ---+ 0 and for r = 00. The last integral vanishes sincefor all finite r, (8/8r)r2(81/1/8r) = O. Also 8(I/r)/8r = -1/r2 so

I = lim [cJ>(W, r) 1

00

] = _ cJ>(w, 0) = _ od(r)0-+0 41[" 0 41[" 41r

by comparison with (240). Hence in a distributional sense

8 281/1 o(r)-r -=--.ar ar 41r

The potential 1/1 is now viewed as a distribution. The source for 1/Id can be regarded as adelta function distribution. The factor of 41[" in the denominator arises because we used a

one-dimensional form of Laplace's equation. The total source strength is unity since

11r12 1r1OO

o(r)--2r2 sin 8drdcPd8 == 1.o 0 0 47rr

The above examples clearly show that distribution theory does provide a mathematical frame-work in which entities such as the delta function and the derivative of a step function can bedefined in a rigorous manner.

When we represent o(x - x') in terms of an infinite series of eigenfunctions that seriesshould be viewed as a distribution. The series may be differentiated and integrated term byterm in a distributional sense.

Consider the space of testing functions with support 0 ::; x ::; a. The distribution

00 10 2 nxx' nxxOd(X -x') == L .. - sin -- sin -cP(w, x)dx

o a a an=l

(241)

is a distribution depending on x' as a parameter and represents the delta function as a distri-butional series. A function j(x) has the distribution

h(x) =1Q

f(x')~(w. x') dx'

when j(O) == j(a) == O. We can also express this in the form

1

0 00 2 n7rx'l° nxxj(x')L - sin -- cP(w, x) sin - dx dx'

o n=l a a 0 a

(242a)

(242b)

upon introducing a Fourier series representation for cP(w, x'), which always exists. By com-parison with (241) we see that the distribution fd(X) is also given by

where Od(X - x') is regarded as a distribution. We can also write

1

0 [00 2 n7rx1° n7rx']fd(X) == cP(w, x) L - sin - f(x ') sin -- dx' dxo n=l a a 0 a

(242c)

(242d)

which is clearly the distributional definition of the Fourier series representation of f (x).Note that in (242a) to (242d) fd(X) as a distribution is not a function of x. The conventionof representing the function and its distribution by the same notation f(x) requires carefulattention to the context. To assist the reader in this regard we have attached the subscript d tothe distribution.

As a final example consider the Fourier series for the triangle function

{

2x /a ,f(x) ==

2(I-x/a),

o<x <a /2

a/2 ::; x ::;a

GREEN'S FUNCTIONS

which is

f L:oo 8 . nx . nxx

(x) == -- SIn - SIn --.n=l (n1r)2 2 a

161

(243)

This series is uniformly convergent. The derivative of f(x) is the piecewise continuous function2/a for 0 ~ x < a /2 and - 2/a for a /2 < x ~ a. The second and higher order derivativesof f(x) do not exist at x == a /2 in the classical sense. In a distributional sense derivatives off(x) of any order have a meaning. The distribution for d2f[dx? is

i " 1a~ 8 . nt: . nxx 4 /d(X) == LJ - 2" SIn -2 SIn -cP(w, x)dx == --Od(X - a 2)o n=l a a a

(244)

by comparison with (241). We note that the step discontinuity at x == a/2 is - 4/a soformally we would expressf"(x) as -4/a multiplied by the delta function. The distributionalrepresentation of the third derivative is

4 dOd(X - a/2) f"'() 1a~ 8 . nr . nxx dcP d- - == d X == LJ - SIn - SIn ---- x

a dx 0 n=l a2 2 a dx

fa~ 8n1r . nt: nxx= -}o LJ 7 sin 2 cos a¢(w, x)dx

o n=l

(245)

with the latter obtained by an integration by parts. This example shows that distribution theorygives a meaning to highly divergent series when these are viewed as distributions.

It is not necessary to know the explicit form of the testing functions cP(w, x). It is sufficientto know that they exist in order to apply the theory.

The above discussion of distribution theory is only a sketchy outline and is presented mainlyfor the purpose of showing in a heuristic way how distribution theory deals with delta functionsand divergent series that occur in boundary-value problems and the eigenfunction representa-tion of Green's functions. The reader should consult the references for a rigorous developmentof the theory. For most problems of engineering importance one can use the operational ap-proach followed in this chapter with the assurance that a rigorous approach exists that justifiesthe formal procedures.

REFERENCES AND BIBLIOGRAPHY

[2.1] J. A. Stratton, Electromagnetic Theory. New York, NY: McGraw-Hill Book Company, Inc., 1941, ch.3.

[2.2] M. I. Sancer, private communication. (See also: M. I. Sancer, "The magnetic field integral equation singu-larity and its numerical consequences," Tech. Dig., URSI National Radio Science Meeting, Boulder, CO,Jan. 1981.)

[2.3] D. S. Jones, Methods in Electromagnetic Wave Propagation. Oxford: Clarendon Press, 1979, sect.6.16.

[2.4] Electromagnetics, vol. 1, no. 4, 1981.

General Mathematical Theory

[2.5] P. M. Morse and H. Feshbach, Methods of Theoretical Physics, parts I and II. New York, NY: McGraw-Hill Book Company, Inc., 1953.

[2.6] A. G. Webster, Partial Differential Equations of Mathematical Physics. New York, NY: Hafner Pub-lishing Company, 1950.


[2.7] R. Courant and D. Hilbert, Methods of Mathematical Physics, vol. 1, English ed. New York, NY:Interscience Publishers, Inc., 1953.

[2.8] A. Sommerfeld, Partial Differential Equations in Physics. New York, NY: Academic Press, Inc., 1949.[2.9] B. Friedman, Principles and Techniques of Applied Mathematics. New York, NY: John Wiley & Sons,

Inc., 1956.[2.10] E. C. Titchmarsh, Eigenfunction Expansions Associated with Second Order Differential Equations.

Oxford: Clarendon Press, 1946.[2.11] I. Stakgold, Green's Functions and Boundary Value Problems. New York, NY: John Wiley & Sons,

Inc., 1979.[2.12] C. T. Tai, Dyadic Green's Functions in Electromagnetic Theory. Scranton, PA: Intext, 1971.

Special Topics

[2.13] N. Marcuvitz, "Field representations in spherically stratified regions," Commun. Pure Appl. Math., vol.4, pp. 263-315, sect. 5, Aug. 1951.

[2.14] L. Felson, "Alternative field representations in regions bounded by spheres, cones, and planes," IRE Trans.Antennas Propagat., vol. AP-5, pp. 109-121, Jan. 1957.

[2.15] S. Sensiper, "Cylindrical radio waves," IRE Trans. Antennas Propagat., vol. AP-5, pp. 56-70, 1957.[2.16] H. Levine and J. Schwinger, "On the theory of electromagnetic wave diffraction by an aperture," Commun.

Pure Appl. Math., vol. 3, pp. 355-391, 1950.[2.17] R. E. Collin, "The dyadic Green's function as an inverse operator," Radio Sci., vol. 21, pp. 883-890,

1986.[2.18] R. E. Collin, "Dyadic Green's function expansions in spherical coordinates," Electromagnetics J., vol. 6,

pp. 183-207, 1986.[2.19] L. Wilson Pearson, "On the spectral expansion of the electric and magnetic dyadic Green's functions in

cylindrical coordinates," Radio Sci., vol. 18, pp. 166-174, 1983.[2.20] J. Van Bladel, "Some remarks on Green's dyadic for infinite space," IEEE Trans. Antennas Propagat.,

vol. AP-9, pp. 563-566, 1961.[2.21] A. D. Yaghjian, "Electric dyadic Green's functions in source regions," Proc. IEEE, vol. 68, pp. 248-263,

1980. (See also: A. D. Yaghjian, "Maxwellian and cavity fields within continuous sources," Amer. J. Phys.,vol. 53, pp. 859-863, 1985.)

[2.22] W. A. Johnson, Q. Howard, and D. G. Dudley, "On the irrotational component of the electric Green'sdyadic," Radio Sci., vol. 14, pp. 961-967, 1979.

[2.23] W. W. Hansen, "A new type of expansion in radiation problems," Phys. Rev., vol. 47, pp. 139-143,1935.[2.24] C. T. Tai, "Singular terms in the eigenfunction expansion of the electric dyadic Green's functions," Tech.

Rep. RL-750, Rad. Lab., The University of Michigan, Mar. 1980.[2.25] J. R. Wait, Electromagnetic Radiation from Cylindrical Structures. New York, NY: Pergamon Press,

1959.[2.26] R. F. Harrington, Field Computation by Moment Methods. Malabar, FL: Kreiger Publishing Co., Inc.,

1968.[2.27] J. R. Mautz and R. F. Harrington, "H-field, E-field, and combined field solutions for conducting bodies of

revolution," A.E.U. (Germany), vol. 32, pp. 157-164, 1978.[2.28] J. R. Mautz and R. F. Harrington, "A combined-source solution for radiation from a perfectly conducting

body," IEEE Trans. Antennas Propagat., vol. AP-27, pp. 445-454, 1979.[2.29] A. D. Yaghjian, "Augmented electric and magnetic field integral equations," Radio Sci., vol. 16, pp.

987-1001, 1981.[2.30] R. F. Harrington and J. R. Mautz, "Theory of characteristic modes for conducting bodies," IEEE Trans.

Antennas Propagat., vol. AP-19, pp. 622-628, 1971.[2.31] L. Marin, "Natural-mode representation of transient scattered fields," IEEE Trans. Antennas Propagat.,

vol. AP-21, pp. 809-818, 1973.[2.32] M. Sancer, A. D. Varvatsis, and S. Siegel, "Pseudosymmetric eigenmode expansion for the magnetic field

integral equation and SEM consequences," Interaction Note 355, Air Force Weapons Laboratory, KirtlandAir Force Base, NM, 87117, 1978.

[2.33] J. Van Bladel, Singularities of the Electromagnetic Field. London: Oxford University Press. To bepublished.

PROBLEMS

2.1. Make a Fourier series expansion of (29b) in terms of the eigenfunctions # sin (ns»: fa) and show that(29a) is the result obtained.

2.2. Consider the equation d2l/;/dx2 + Xl/; = 0 with boundary conditions l/; + 2dl/;/dx = 0 at x = 0 and l/; = 0at x = a, Show that the eigenvalues are determined by the transcendental equation tan ~o = 2~ and that theeigenfunctions (unnormalized) are sin ';>;:;'x - 2';>;:;' cos .;>;:;,x, with Xn the nth eigenvalue.

2.3. Consider the lossless transmission-line circuit illustrated in Fig. P2.3. The line has inductance L and capac-itance C per meter. At the end Z = 0 the line is terminated in a reactance jX, while at Z = 0 it is short-circuited.At z = z' the line is excited by a time harmonic current generator with output I g el'", Because of the current sourceat Z' the current on the line undergoes a step change I g at Z' as shown in the figure. Consequently, 8I/8z has adelta function change I gO(z - z') at z'. The voltage is continuous at Z' but its slope, proportional to I, has a stepchange as shown. The equations describing the line are 8V/8z = - jial.L, 8I/8z = - jwCV + I gO(z - z'). Showthat 82 V /8z2 + w2LCV = - jtal.IgO(z - z'). Find two solutions for V (Methods I and II). Note that at Z = 0,V = 0 and at z = 0, V = jXI = -(X/wL)(8V /8z) or V + (X /wL) (8V /8z) = O. Note that the poles determinethe resonant frequencies.

Answers:Method I.

00~ jwLIg sin knz< sin knz>

V= L..J_ (X _ k 2 ) (~ _ sin 2k nO )

n-l n 0 2 4kn

where Xn = k~ and tan kno = -knX/wL.

Method II.

V = jZcI sin koz<[XYc cos ko(o - z» + sin ko(o - z»]g sin koo +XYc cos koo

where k5 = w2LC and Y~ = C /L.

Resonant frequencies are given by tan k;« = -XYc where k n = W n VEe.

z'

z=a

~jX

_+-- .........1- .. Z

z'-+-------&.------~z

I(z) .rII

z'

dIliZ

JCa function

II

V(z)

--+-----""-----~z

z'

Fig. P2.3.

dVdz

-+-----+------~z

2.4. For the above problem let X = w2LC. The boundary condition is then (tan ~a)/~ = -X /wL =- XY c /~ at Z = 0 and the eigenvalue is now part of the boundary condition. This is no longer a standard boundary-

value problem. A solution can be obtained by extending the definition of the operator and that of its domain. Theprocedure to follow is given below.

Let U be the two-element column matrix

U == [W(Z)](z)

and define .cU by

[

d(z)/dz ].cU-

-dw(z)/dz

Show that the system

w(O) == 0 dw(a) == _ koZcw(a), dz X

is equivalent to

[d/dz ] [W].cU - -ko

-dw/dz 

with boundary conditions w(a) = XYc(a) and '11(0) = d(O)/dz == O.Consider the eigenvalue problem

Show that for two solutions U, and Um

so the eigenfunctions are orthogonal. Choose wn == sin knz and n = -cos knz. You will need to use tan kno =-XYc.

Show that (Un, Un) == a which is the normalization constant.Show that if V satisfies the same boundary conditions as U, (V, cCU) = (U, cCV) so the operator cC is self-adjoint.Show that

is equivalent to

Let

[V] == Len [ sinknz ]

 -cos knz

GREEN'S FUNCTIONS

and use the orthogonality of the eigenfunctions to show that a third solution for V(z) is

165

Note that this solution involves different eigenvalues and eigenfunctions than the solutions for Problem 2.3. (See B.Friedman, Principles and Techniques ofApplied Mathematics. New York, NY: John Wiley & Sons, Inc., 1956,p. 205.)

2.5. Verify the equivalence of the solutions obtained in Problems 2.3 and 2.4. Evaluate the following integralusing residue theory

-1-1 jZJgwVLCsinwz< [~cosw(a-z»+sinW(a-z»]dw

21rj 2 2 ( • wX )c (w -w LC) SIn wa + wL cos wa

and show that the residue series from the poles at w = ± v1£w gives the Method II solution in Problem 2.3 andthe residue series from the poles at tan wna = -w,x/wL, i.e., w = ± vXfi, gives the negative of the Method Isolution. The residues can be found by evaluating the derivative of the denominator at the poles. You will need touse the eigenvalue equation to simplify the results. Show that on circle C with infinite radius, with w = u + jv theintegrand behaves like {exp [-IvI(z> - Z<)] } [w and hence the integral is zero and solutions 1 and 2 in Problem 2.3are equivalent. Consider next the integral

1 1 jZCIg sin wz<[XYc cos w(a -z» + sin w(a -z»]-- dw21rj C (w - wVLC)(sin wa +XYc cos wa)

and show that the residue evaluation demonstrates that solutions 1 and 3 are equivalent. Thus all three solutions areequivalent.

2.6. Use (48a) and (49) in (44) and integrate around a contour C; enclosing the poles of Ox and show that (47)is the solution obtained for O.

2.7. Consider an infinitely long transmission line excited by a current source I ge':" at Z' (see Fig. P2. 7). Solved2V/dz 2 + kk5 V = - jol.IgO(z - z') by means of a Fourier transform.

HINT: Assume that there is a small shunt conductance so that k5 = w2LC(1 - jO /wC) and k o = k~ - jk~'. Thiswill displace the poles away from the real axis. Note that for Z > z' the inversion contour can be closed in the lowerhalf plane, and for Z < z' it can be closed in the upper half plane. Thus the inverse transform can be evaluated interms of residues.

., z

z'

Fig. P2.7.

2.8. Consider an axial line source inside a rectangular pipe as shown in Fig. P2.8. Find the Green's functionwhich satisfies

o = 0 on boundary

in terms of a double Fourier series of eigenfunctions for the equation

1/;nm = 0 on boundary.


y

Linesource•

x', y'

bt------------.

L----------I~--____i~Xa

Fig. P2.8.

2.9. Find G for Problem 2.8 in terms of an eigenfunction expansion along x but as a closed form in the y variable,Le., in the form

2.10. Find the Green's function for Problem 2.8 in terms of a contour integral of Gx(Ax)Gy(Ay) where

a2Gy ,-2- + AyG y = -o(y - y ).ay

Verify your result with those obtained in Problems 2.8 and 2.9.2.11. Determine the Green's function of the first kind (G = 0 at r = 0) for Poisson's equation for a line source

parallel and outside a conducting cylinder of radius a. The source is at r' > a, Use (a) the method of images and(b) expansion in a suitable set of eigenfunctions.

2.12. For the line source in a rectangular waveguide use Method II to show that an equivalent solution for Gx is

Use this solution in (59) and make the variable change Az = w 2 to show that

where 'Y = Jk5 - w2 • Evaluate the integral by closing the contour in the lower half plane and using residue theory.Note that the contour runs above the pole at w = 'Y and below the pole at w = -'Y.

2.13. By using Laplace transforms, find the Green's function for the following differential equation:

(d

2G R dG 1) ,

L - + - - + -G = 0(1 - t )dt 2 L dt LC

G = dG =0dt

at t = O.

This Green's function represents the charge flowing in a series RLC circuit with a unit voltage impulse input at timet = t',

Answer:

1 I

G(t It') = -e-a(t+t ) sin w(t - t'), t > t'wL

GREEN'S FUNCTIONS

where

2.14. For Problem 2.13 show that

(2) 1/2

W = (LC)-lj2 1 - ~f ·

167

gives the usual steady-state solution for the charge in the circuit due to an impressed voltage source ej wot•

2.15. Let G(rlro) be a solution of \7 X \7 X G - k 2G = Io(r - ro) in a volume V. On the surface S surroundingV let G satisfy one of the boundary conditions, n X G or n X \7 X G = O. Use the vector form of (108) to show

that

HINT: Replace the dyadic jj in (108) by a vector function B. Next choose for A and B the following:

A = G(r, rl)·JI(r.)

B = G(r, r2)·J2(r2).

2.16. Derive the normalization integrals (138b), (138c).2.17. Derive the solution for the scalar Green's function g given by (155).2.18. Evaluate the integral over k in (155) and show that

'k 00 ne-J oR • ",,,,eom(2n + 1)(n - m)! m m' , . h2 k

g = 41rR = -JkoL....tL....t 41r(n + m)! P n (cos 8)Pn (cos 8 ) cos m«(j> - (j> )In(ko'<) n( 0'»'

n=O m=O

2.19. Show that the Green's function for Poisson's equation in free space, which is a solution of

0(' - ")0(8 - 8')0( (j> - (j>')

,2 sin 8

is given by

00 n 100

g = 4 lR =LL fom(2n

2+ l)(n - m)! P:;'(cos 8)P:;'(cos 8') cos m(t/> - cJ/)jn(kr)jn(kr') dk .

1r n=O m=O 0 21r (n + m)!

HINT: Expand the solution in terms of the eigenfunctions of the Laplacian operator.2.20. Evaluate the integral over k in Problem 2.19 and show that another solution is

00 n

1 LLeOm(n - m)! m 8 m' , ,~g = 4-R = 4 ( )' P n (cos )Pn (cos 8 ) cos m«(j> - (j> )--n+f'

1r 1r n + m . '>n=O m=O

Take the limit of the solution in Problem 2.18 as ko approaches zero and show that the same result is obtained.2.21. Find a solution for the Green's function in Problem 2.19 using the following method: Expand g in the form

00 n

g = LLfn(')PZ'(COS 8) cos m«(j> - (j>').

n=O m=O

Substitute this into the differential equation and make use of the orthogonal properties of the P,:/ and cos m( c/> - c/>')functions to show that

€Om(2n + 1)(n - m)! pm(cos 8) o(r - r') .41r(n+ m)! n r2

Use Method I and a Hankel transform to find a solution for In; i.e., let

Also use Method II and let

r :::; r'

r 2:: r'

to find a solution. Verify your solutions by comparison with those given in Problems 2.19 and 2.20.2.22. In cylindrical coordinates show that a solution of

o(r - r')o( c/> - c/>')o(z - z')r

for the free-space scalar Green's function is

~j~l~ ,= '"' ~ (,I,. _ ,1,.') -jw(z-z') kJn(kr)Jn(kr ) dk dg ~ 2 cos n 0/ 0/ e 2 2 2 W.

n=O -~ 0 41r k + w - ko

Evaluate the integral over k and show that

where)' = Jk5 - w2 . Show that when r' = 0 the solution reduces to

where R = Jr 2 + (z - Z')2. Note that a useful inverse Fourier transform has been evaluated.2.23. For Problem 2.22 expand g in the form

~j~g = L In(r) cos n(c/> - c/>')e- jwz dw

n=O -~

and by Fourier analysis show that

Assume that In = CnJn()'r<)H~()'r» and use the Wronskian relation W = JnH~' - J~H~ = -2j /1r)'r' to showthat Cn = -j(€on/81r)ejWZ' and that the solution given in Problem 2.22 is obtained.

GREEN'S FUNCTIONS

2.24. Consider the scalar Green's function within a sphere of radius a that is a solution of

169

g(a) = O.

Show that two solutions for g are

where k, are the roots of jn(ka) = O. The Wronskian determinant is W = jn(u)y~(u) - j~(u)Yn(u) = l/u2• Show

that when ko approaches zero the limiting solution is

-:<g = (2n + l)an

The series that represents this function can be found by putting k 0 = 0 in the above series.2.25. For Problem 2.24 let a become infinite and use Hankel transforms to show that g is given by

= 3.1CX) jn (kr)jn (kr') k 2 dk.g 7r k2 - k 2

o 0

Evaluate the integral over k to show that

2.26. Show that when the boundary condition in Problem 2.24 is changed to d(rg)/dr = 0 at r = a the twosolutions for g are

where l, are the roots of d[rj n(kr)]/dr = 0 at r = a. Show that the limiting form as ko tends to zero is

r~ [(n + l)an + nr~ ] .

g = (n + 1)(2n + 1) r~+l an+1

Derive a series representation for this function.2.27. Show that the two solutions for g given in Problem 2.24 have the same residues at ko = k..HINT: Expand jn(~a) in a Taylor series in k~ about the point k~ = kr. Use the Wronskian determinant

to evaluate Yn(kia) to obtain Yn = -(k;a2j~)-l. Show that g vanishes when k o becomes infinite. Note that twofunctions with the same pole singularities in the complex plane and that vanish at infinity must be identical. Hencethe two solutions represent the same function.

2.28. A point source at r' > a radiates in the presence of a sphere of radius a. Find the solution for the scalarGreen's function that satisfies the Dirichlet boundary condition g = 0 at r = a.

HINT: Use the mode solution in Problem 2.18 and superimpose outward-propagating waves that cancel the incidentfield at r = a.

2.29. In the solution given in Problem 2.18 let r' tend to infinity, use the asymptotic value of h~(kor), and derivethe following expansion for a plane wave:

CX) n

ejkor.a,o = L Lj" fom(2~n+;~~!- m)! P::'(cos O)P::'(cos 0') cos m(</J - </J')jn(kor)

n=O m=O

for r finite.

170

HINT: Express e-j koR /41rR in the form


e - jko Ir-ro I ejkoro.-~-- rv __ elkor-sro41r [r - roI 41rro

If the point source is located on the Z axis only the terms for m = 0 are needed.2.30. A dyadic Green's function for the electric field may be obtained from (155) by applying the operator shown

in (112). Comment on this solution with respect to the decomposition of the dyadic Green's function into transversespherical TE and TM waves, longitudinal waves, and orthogonality properties.

2.31. Take a three-dimensional Fourier transform of

(V x V x - k5)G = o(r - r')1

to obtain

To invert the dyadic assume that

0-1 = [(k2 -k5)1 - kk]-1 =Akk +BI

and use O·(A kk + BI) = I to find A and B. Thus show that

00

G(r r') = _1_ fff k~I - kk e-jk-(r-r') dk., 81r3 JJJ k~(k2 - k~)

-00

In order to evaluate the integral, change to spherical coordinates in k space. Choose the polar axis to lie along r - t' .

Then

The integration over () and cP is readily carried out after bringing kk outside the integral as the operator - VV. Thefinal integral over k can be expressed as an integral from - 00 to 00 and evaluated by residues. Complete the detailsof this evaluation and show that the solution given by (112) is obtained.

The solution for G given above is an eigenfunction expansion in terms of plane waves_ It can be split intolongitudinal and transverse waves. The longitudinal waves can be obtained by taking a scalar product with k / k. Thusshow that

The first integral is well behaved and can be evaluated by bringing k 21 - kk outside the integral sign as the operator(VV· - V2)1. Show that the contribution to the Green's dyadic is

If in the second term kk is brought outside the integral sign as the operator - VV show that the contribution is

However, the second integral is not a convergent one so replacing kk by - \7\7 cannot be justified. What this meansis simply that the eigenfunction solution for the Green's dyadic "function" cannot generally be summed to give a truefunction. However, the above solution can be used in the usual way by scalar multiplying with the source functionJ(r/) and integrating over t' before completing the integration over k.

A mode expansion in terms of waves propagating along z can also be obtained by integrating over k z first. Showthat in the expansion of G the longitudinal modes are canceled by contributions from the transverse modes apart froma residual delta function term - azazo(r - r/)/k5' Show that the remaining transverse mode contribution is given by

+00

- j {{ [Ik5- (axkx + ayky + azkz sg (z - Z/»2] e-jk/.(r-r')-jkz Iz-z' Idk811'2 )} k5kz t

-00

where kt == axkx + ayky and k z == Jk5 - k;. ,Note that the longitudinal modes have the factor e-k t Iz-z I in the integrand and are nonpropagating modes along

z. An often overlooked point in connection with the above Fourier transform solution for G is the fact that G in theform

cannot be Fourier transformed if the differentiations are carried out because of the 11R3 singularity. Hence thesolution given above should be interpreted as the plane wave eigenfunction expansion of the Green's dyadic operatorand integration over the source coordinates is to be performed first before inverting the Fourier transform.

2.32. Consider the one-dimensional Green's function problem for the interior of a cylinder

Show that two solutions are

! ~ r dg + k 2g == o(r - r ') ,r dr dr r

dgdr == 0 at r == a.

00 I

L 2JO(qmr la)Jo(qm r la) 2 11'g == 2 2 2 1 2 + 22 == 2J (k )Jo(kr<)[J I (ka)Yo(kr» - Jo(kr»Y I(ka)]

a Jo(qm)[k -(qm a)] a k I am=1

where qm are the roots of dJo(kr)/dr == 0 at r == a. Note that because of the Neumann boundary condition at r == athere is an m == 0 eigenfunction with a zero eigenvalue.

2.33. Let the incident electric field given by (215a) contain the additional mode CllM~I' Show that the follow-ing current is induced on the sphere (assume that the common factor jl(koa) is canceled): (AIIS~1 + CllS~I)1

[aZo(koahi)]· Let All == JAil + cil cos a and C ll = JAil + Cil sin a and show that the induced current is

proportional to JA iI +Ci I S~I (cP + a). Show that S~ I (cP + a) is an orthogonal current distribution and a solutionof the homogeneous integral equation in the limit as i, (koa) approaches zero.

2.34. Consider the radiation condition given by (115) and the solution (109) for the electric field where S ischosen as a sphere of infinite radius. Use (115) to eliminate lim, ---+00 r \7 X E and lim, ---+00 r \7 x G and show thatthe integrand in (109) vanishes when the radiation condition holds.

2.35. Let E, H be the difference field in V I as discussed in Section 1.8 where the sources for the field arecontained in V2. Then on S and the sphere of infinite radius

fjE X H·odS + lim fjE X H·odS == 0

'---+00S Soo

since 0 X (E I -~) == 0 X E == 0 and 0 X (HI - H2) == 0 X H == 0 on S and the difference field satisfies theradiation condition.

Consider a field E/, H' that is a solution of Maxwell's equations due to an arbitrary source J in VI and whichsatisfies the radiation condition. Then

fj (E X H' -E' X H)·DdS =0

S


and according to an extension of the results of Problem 1.17 J (r /). E(r') == 0 for all t' in V I. Hence E == 0 in V I

and therefore the two solutions E I and E2 are equal throughout V I. Maxwell's equations then require that 8 I == 8 2

and hence the solution to the external radiation problem is unique if the radiation condition is satisfied and n X E I

and n X HI are given on the surface S. If only n X E I is given on S then E/, H' can be chosen so that n X E' = 0on S and the reciprocity theorem as used above will again show that E is zero throughout V I because the surfaceintegrals over S and the sphere at infinity are zero.

For scattering from a perfectly conducting object the two candidate solutions EI and E2 are required to satisfy theboundary condition 0 X E I == 0 X E2 == -0 X E; where E; is the incident field. Thus on S we have 0 X (E I - E2) ==n X E == 0 but 0 X H is left unspecified. As noted above, the solution is unique. When the scattering problem issolved using either the EFIE or the MFIE the lack of uniqueness is a fault of the method of solution and is not due toan intrinsic nonuniqueness in the problem. (A much longer and detailed proof of uniqueness based on the properties ofthe spherical vector wave functions may be found in: C. Muller, Mathematical Theory ofElectromagnetic Waves.New York, NY: Springer-Verlag, 1969.)

green's functions

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