Integral Equations In Electromagnetics

Integral Equations in ElectromagneticsMassachusetts Institute of Technology6.635 lecture notesMost integral equations do not have a closed form solution. However, they can often bediscretized and solved on a digital computer.Proof of the existence of the solution to an integral equation by discretization was firstpresented by Fredholm in 1903.In general, integral equations can be divided into two families:1. When the unknown is in the integral only, the integral equation is called of the first kind.2. When the unknown is both inside and outside the integral, the integral equation is calledof the second kind.For electromagnetic applications, we can have both scalar and vector integral equations.1Scalar integral equationsLet us consider the situation of Fig. 1: two regions are defined in space, region 2 bounded bythe closed surface S and region 1 being all the remaining space, bounded by S and S , in whichsources are located.#1Jn̂S#2S Figure 1: Definition of the geometry of the problem.It is already known that we can write:( 2 k12 ) Φ1 (r̄) J(r̄) ,(1a)( 2 k22 ) Φ2 (r̄) 0 ,(1b)1

2Section 1. Scalar integral equationsand similarly for the Green’s functions( 2 k12 ) g1 (r̄, r̄0 ) δ(r̄ r̄ 0 )2( k22 ) g2 (r̄, r̄0 )0 δ(r̄ r̄ )in region 1,(2a)in region 2.(2b)Upon performing (Eq. (1a)g1 (r̄, r̄0 ) Eq. (2b)Φ1 (r̄, r̄0 )), we get:[g1 (r̄, r̄0 ) 2 Φ1 (r̄) Φ1 (r̄) 2 g1 (r̄, r̄0 )] J(r̄)g1 (r̄, r̄0 ) δ(r̄ r̄ 0 )Φ1 (r̄) .(3)Upon integrating Eq. (3) over the entire volume and using the identity · (g Φ Φ g) Φ 2 g, we get:ZZ00dvJ(r̄) g(r̄, r̄ 0 ) Φ1 (r̄0 ) .(4)dv · [g1 (r̄, r̄ )Φ1 (r̄) Φ1 (r̄) g1 (r̄, r̄ )] g 2 ΦVVBy Gauss theorem, we reduce the left-hand side integral to a surface integral. AlsoZdvJ(r̄)g1 (r̄, r̄0 ) Φinc (r̄0 ) .(5)VWe therefore obtainZds n̂ · [g1 (r̄, r̄0 ) Φ1 (r̄) Φ1 (r̄) g1 (r̄, r̄0 )] Φinc (r̄) Φ1 (r̄0 ) , r̄0 V .(6)S S By invoking the radiation condition, the integral over S vanishes, leaving (and exchangingr̄ and r̄0 so that primed coordinates correspond to sources and unprimed ones to observation):Φ1 (r̄) Φinc (r̄) Zds0 n̂ · [g1 (r̄, r̄0 ) 0 Φ1 (r̄0 ) Φ1 (r̄0 ) 0 g1 (r̄, r̄0 )]r̄ V1 .(7)SFor r̄ V2 , the wave equation has no source and therefore the integration of the deltafunction yields a zero value. Performing the same steps for this second case, we get the genericrelation: Φ (r̄) r̄ V11ds0 n̂ · [g1 (r̄, r̄0 ) 0 Φ1 (r̄0 ) Φ1 (r̄0 ) 0 g1 (r̄, r̄0 )] Φinc (r̄) 0Sr̄ V2Z(8)This is directly evocative of Huygens’ principle: For r̄ V1 : the total field Φ1 (r̄) is the sum of the incident field plus the field due to thesurface currents on the surface S. For r̄ V2 : the surface source on S produces a field that exactly opposes Φinc , yieldingthe extinction theorem.Applying the same reasoning to region 2, we write (where there is no incident field):Zds0 n̂ · [g2 (r̄, r̄0 ) 0 Φ2 (r̄0 ) Φ2 (r̄0 ) 0 g2 (r̄, r̄0 )] S 0r̄ V1 Φ2 (r̄) r̄ V2(9)

3Note that the sign reversal is due to the definition of the normal vector n̂ which has to pointoutward from the surface. Here, since we use the same n̂ as before, we have to take it as beingnegative.Eqs. (8) and (9) have four independent unknowns:Φ1 , Φ2 ; n̂ · Φ1 , n̂ · Φ2 ,(10)which can be related by the boundary conditions. Here, g(r̄, r̄ 0 ) and n̂ · g(r̄, r̄ 0 ) are the kernelof the integral equation.2Vector integral equationFor the sake of completeness, we shall write the vector wave equation as well, although we willnot use it here directly.Considering the same situation as above, we know that the fields have to satisfy: , Ē1 (r̄) ω 2 ²1 µ1 Ē1 (r̄) iωµ1 J(r̄)(11a) Ē2 (r̄) ω 2 ²2 µ2 Ē2 (r̄) 0 ,(11b)and the Green’s functions: G1 (r̄, r̄0 ) ω 2 ²1 µ1 G1 (r̄, r̄0 ) Iδ(r̄ r̄ 0 ) ,(12a) G2 (r̄, r̄0 ) ω 2 ²2 µ2 G2 (r̄, r̄0 ) Iδ(r̄ r̄ 0 ) .(12b)RBy the same technique as before ( V dv [Eq. (11a) · G1 (r̄, r̄0 ) Eq. (12a) · Ē1 (r̄)]dv) ·ZZ00 · G1 (r̄, r̄0 ) Ē1 (r̄) .dv J(r̄)dv Ē1 (r̄) · G1 (r̄, r̄ ) Ē1 (r̄) · G1 (r̄, r̄ ) iωµ1VV(13)The left-hand side can be transformed into a surface integral (left as exercise) and the right-handside written in terms of incident field, yielding:½¾Z0000ds n̂· [ Ē1 (r̄)] G1 (r̄, r̄ ) Ē1 (r̄) G1 (r̄, r̄ ) , r̄0 V1 . (14)Ē1 (r̄ ) Ēinc (r̄ ) S S By reciprocity of the Green’s functions, we can transformn̂ · [ Ē1 (r̄)] G1 (r̄, r̄0 ) n̂ [ Ē1 (r̄)] · G1 (r̄, r̄0 ) iωµ1 G1 (r̄, r̄0 ) · n̂ H̄1 (r̄) .(15)In addition, for an unbounded homogeneous medium ( G(r̄, r̄ 0 ) is reciprocal):n̂ · Ē1 (r̄) G1 (r̄, r̄0 ) n̂ Ē1 (r̄) · G1 (r̄, r̄0 ) [ G1 (r̄, r̄0 )] · n̂ Ē1 (r̄) .(16)

4Section 3. Problem with the internal resonanceFor Green’s functions that satisfy the radiation condition, Eq. (14) becomes:½¾Z0000ds n̂ · iωµ1 G1 (r̄, r̄ ) · n̂ H̄1 (r̄) [ G1 (r̄, r̄ )] · n̂ Ē1 (r̄) .Ē1 (r̄ ) Ēinc (r̄ ) (17)SWe perform the same steps for the other region to eventually obtain (and again interchanging r̄and r̄0 ): ¾ ½Ē1 (r̄) r̄ V1Ēinc (r̄) ds0 iωµ1 G1 (r̄, r̄0 ) · n̂ H̄1 (r̄0 ) [ 0 G1 (r̄, r̄0 )] · n̂ Ē1 (r̄0 ) 0Sr̄ V2Z(18a) Z¾½00000ds iωµ2 G2 (r̄, r̄ ) · n̂ H̄2 (r̄ ) [ G2 (r̄, r̄ )] · n̂ Ē2 (r̄ ) 0S 0r̄ V1 Ē2 (r̄) r̄ V2(18b)Together with the boundary conditionsn̂ H̄1 (r̄) n̂ H̄2 (r̄) ,(19a)n̂ Ē1 (r̄) n̂ Ē2 (r̄) ,(19b)this system can be solved for n̂ Ē1 (r̄) and n̂ H̄1 (r̄).Note also that Eqs. (18) can be written in terms of electric and magnetic currents, andmagnetic Green’s functions:J eq (r̄0 ) n̂ H̄1 (r̄0 ) ; M̄eq (r̄0 ) n̂ Ē1 (r̄0 ) ,(20a)so that for example: ½¾ Ē1 (r̄) r̄ V1 0 ) Gm (r̄, r̄0 ) · M̄ (r̄0 ) Ēinc (r̄) ds0 iωµ1 Ge1 (r̄, r̄0 ) · J(r̄1 0Sr̄ V2Z3(21)Problem with the internal resonanceA question arises: is the integral equation equivalent to Maxwell’s equations? Or asked differently, if we solve the integral equation and Maxwell’s equations, do we get the same solution?The answer is actually “no”, that is they are not always equivalent to each-other. Theproblem comes from spurious solutions at frequencies corresponding to the eigenfrequencies ofthe cavity enclosed by the surface S. This problem is generally referred to as the “internalresonance of the integral equation”.However, this lack of complete equivalence between the physical problem and its definingintegral equation is rather minor and infrequent phenomenon, and is therefore often toleratedin practice.

54Scattering by a rough surfaceLet us consider the 2D problem (for which we shall use a scalar integral equation) depicted inFig. 2.z²0f (x)PSfrag replacementsx²1Figure 2: Rough surface S separating two media.The integral equation in scalar form is given by:Φinc (r̄) 4.1Zds0 n̂ · [Φ(r̄ 0 ) g(r̄, r̄ 0 ) g(r̄, r̄ 0 ) Φ(r̄0 )] SDirichlet boundary conditions: EFIE Φ(r̄) 0r̄ V0r̄ V1(22)For a TE wave (Ē E ŷ) and PEC surface, the boundary condition isΦ(r̄) 0 ,for r̄ S .(23)The integral equation becomes, for r̄ S , r̄ 0 S:Φinc (r̄) Zds0 g(r̄, r̄0 ) n̂ · Φ(r̄ 0 ) S Φ(r̄)(24) 0This equation, in which Φ represent the electric field, is referred to as the electric field integralequation (EFIE).Note that as r̄ gets closer to the surface, Φ(r̄) 0 (from the boundary condition) so thatwe do not need to distinguish between approaching the surface from one side or the other. Infact, we can unify the equations and write:ZΦinc (r̄) ds0 g(r̄, r̄0 ) n̂ · Φ(r̄ 0 ) 0 ,r̄ S , r̄ 0 S .(25)Sg(r̄, r̄ 0 )In addition,has an integrable singularity as r̄ r̄ 0 . Let us consider the surfacedepicted in Fig. 2, with z f (x):sµ ¶2pdf22ds dx dz dx 1 ,(26)dxsuch that the integral equation becomes:sµ ¶2Zdfdx0 1 Φinc (r̄) g(x, f (x), x0 , f (x0 )) (n̂ · Φ(r̄ 0 )) ,dx xat z 0 f (x0 ) ,(27)

64.2Neumann boundary conditions: MFIEwhere we can limit x to [ L/2, L/2].By lettingsµ ¶2df(n̂ · Φ(r̄ 0 )) z 0 f (x0 ) U (x0 ) ,1 dx(28a)Φinc (x, f (x)) b(x) ,(28b)K(x, x0 ) g(x, f (x), x0 , f (x0 )) ,(28c)we can rewrite the integral equation asZ L/2dx0 K(x, x0 ) U (x0 ) b(x) ,(29) L/2which has to be solved numerically.Before doing this, we shall write the integral equation with Neumann boundary conditions.4.2Neumann boundary conditions: MFIEThis corresponds to a TM case with H̄ H ŷ with a PEC surface. In this case, the boundarycondition isn̂ · Φ(r̄) 0 ,(30)and the integral equation becomes Φ(r̄) r̄ V0Φinc (r̄) ds0 Φ(r̄) n̂ · g(r̄, r̄ 0 ) 0Sr̄ V1Z(31)In this case, it makes a difference if we approach the surface from the top or from the bottom.In factZΦinc (r̄ ) ds0 Φ(r̄0 ) n̂ · g(r̄ , r̄0 ) Φ(r̄ ) ,(32a)SZΦinc (r̄ ) ds0 Φ(r̄0 ) n̂ · g(r̄ , r̄0 ) 0 .(32b)SThese two equations seems inconsistent with one another. We shall show that in fact, theyare actually consistent with each-other, due to the singularity of the Green’s functions.We shall examine what happens when we let r̄ approach the surface. Fig. 3 is an illustrationof the situation at the immediate vicinity of the surface. The integral part of the equation canbe written as:ZZZ000000ds0 Φ(r̄0 ) n̂ · g(r̄, r̄ 0 ) ,(33)ds Φ(r̄ ) n̂ · g(r̄, r̄ ) ds Φ(r̄ ) n̂ · g(r̄, r̄ ) P VSpieceSwhere P V denotes the principal value and ‘piece’ refers to the integration over the local domainshown in Fig. 3. For this integral, we use the local coordinates:ds0 dX 0 ,n̂ Ẑ 0 , etc(34)

7r̄Z0PSfrag replacements ar̄ (0, Z)a X0Figure 3: Zoom on the rough surface.The integral becomes:ZZ000ds Φ(r̄ ) n̂ · g(r̄, r̄ ) lim lim Φ(r̄)a 0 Z 0pieceadX 0 a g(r̄, r̄0 ) . Z 0(35)Over the small piece, we have: r̄ r̄0 pX 02 (Z Z 0 )2 ,(36a)1i (1) r̄ g H0 ( r̄ r̄ 0 ) ln(2π2µ 4¶Z g. Z 0 Z 0 0 2π(X 02 Z 2 )Thus, the integral becomesZZ lim lim Φ(r̄)piecear̄0 ),1Z02a 0 Z 02π X Z 2 a· 0 aΦ(r̄) 1 X lim limtana 0 Z 0 2πZ a 1 Φ(r̄)for Z 0 2 1 Φ(r̄) for Z 0(36b)(36c)dX 0(37)2The two parts of the integral then become:ZΦinc (r̄) P Vds0 Φ(r̄0 ) n̂ · g(r̄, r̄ 0 ) SZds0 Φ(r̄0 ) n̂ · g(r̄, r̄ 0 ) Φinc (r̄) P VS1Φ(r̄) Φ(r̄) ,21Φ(r̄) 0 ,2(38a)(38b)(38c)which is cast into one equation:Φinc (r̄) P VZ1ds0 Φ(r̄0 ) n̂ · g(r̄, r̄ 0 ) Φ(r̄) .2S(39)Eq. (39) is called the magnetic field integral equation (since Φ represents the magnetic field)and is an integral equation of the second kind.

8Section 5. Solving the EFIEIn the same way as before, we can define the kernel as0K(x, x ) s1 µdfdx¶2n̂ · g(r̄, r̄ 0 ) z f (x),z 0 f (x0 ) ,(40)and write the MFIE asΦinc (r̄) P V5Z dx0 Φ(x0 ) K(x, x0 ) 1Φ(x) .2(41)Solving the EFIEUpon using the notation introduced in Section 4.1, the problem is reduced to solving the followingintegral equation:Z L/2dx0 K(x, x0 ) U (x0 ) b(x) .(42) L/2Let us subdivide the integration domain into N small elements, each of length L/N ,and centered at xm (m [1, N ]). Thus, constraining the observation at these discrete locations,the integral equation becomesZL/2dx0 K(xm , x0 ) U (x0 ) b(xm ) .(43) L/2Next, if we suppose that U (x) is constant in each interval, we replace the integral by a sum overall segments, excluding the singular term:ZX xK(xm , xn ) U (xn ) U (xm )dx0 K(xm , x0 ) b(xm ) .(44)mn 1n6 mNote that we have to single out the singularity, i.e. the mth interval because K(xm , x0 ) issingular at x0 xm . This part is known as the self-patch contribution.For a 2D problem, we have:K(xm , x0 ) i (1) pH (k (xm x0 )2 (f (xm ) f (x0 ))2 ) .4 0(45)Upon approximating f (x0 ) f (xm ) ' f 0 (xm )(x0 xm ), we write:K(xm , x0 ) (1)pi (1)H0 (k x0 xm 1 f 0 (xm )2 ) .4(46)For small argument: H0 (α) 1 i π2 ln(αγ/2) where γ is the Euler constant (γ ' 1.78).Therefore:µ·¶ pγi10002k(x xm ) 1 f (xm ) .K(xm , x ) 1 i ln(47)4π2