appendix I REVIEWED

11.14.13 This is the "version 1" draft of Appendix I, it got replaced right away by Version 2 for some reason. In version 2 theorems become facts, and there is some reorganization. Appendix I : The Poisson and Helmholtz Free-Space Propagators in 2D Note: This Appendix was constructed from Appendix H using the time-honored method of "copy, paste and edit". I.1 Overview and Meaning of Free-Space Propagators This appendix contains two theorems and an application of the second theorem: Theorem : -2[1/4πr] = δ(r) (I.2.1) (I.1.1) Theorem : -2[h(r)/r] = 4π h(0) δ(r) - h"(r)/ r (I.3.1) (I.1.2) Application of Theorem B : - (2+k2) (e-jkr/4πr) = δ(r) (I.3.5) (I.1.3) Throughout, 2 is the usual 2D Laplacian operator 2 = ∂x2 + ∂y2. In the first and last results above, if one replaces r → r-r' (a simple translational shift of origin) ones finds -2[1/4πR] = δ(r-r') R = | r - r' | (I.1.4) - (2+k2) [e-jkR/4πR] = δ(r-r') δ(r-r') = δ(x-x') δ(y-y') δ(z-z') (I.1.5) The quantities in brackets are known as free-space Green's Functions (Green Functions) or propagators, or as "fundamental solutions": 1/4πR = the Poisson 2D free-space propagator (I.1.6) e-jkR/4πR = the Helmholtz 2D free-space propagator (I.1.7) The significance of these propagators is the following: -2 f(x) = s(x) => f(x) = ∫d2x' [1/4πR] s(x') + homogeneous solutions The Poisson Equation (I.1.8) - (2+k2) f(x) = s(x) => f(x) = ∫d2x' [e-jkR/4πR] s(x') + homogeneous solutions The Helmholtz Equation (I.1.9) The equations on the left are inhomogeneous partial differential equations driven by source function s(x). If one is careful to include in s(x) all source contributions (such as those on boundary surfaces), one generally does not have to add any homogeneous solutions on the right. A homogeneous solution refers to -2 fh(x) = 0, for example. The solutions shown on the right above can be instantly verified as follows: f(x) = ∫d2x' [1/4πR] s(x') + fh(x) -2 f(x) = ∫d2x' (-2 [1/4πR] ) s(x') -2 fh(x) = ∫d3x' δ(r-r') s(x') - 0 = s(x) (I.1.10) and similarly for - (2+k2) f = g. A "free space" Green's Function gF in general is a solution of D gF(r, r') = δ(r-r'), gF(r, r') → 0 as r → ∞ (I.1.11) where D is some differential operator. The condition on the right says gF must vanish on the Great Circle. More generally one can write D g(r, r') = δ(r-r'), g(r, r') = 0 for r on some closed curve enclosing a region of interest (I.1.12) In this second form, the ∫d2x' is over the area inside that closed surface. We shall not make use of this more general form in this document. George Green (1793-1841), by the way, was an English grain miller. Looking at f(x) = ∫d2x' [1/4πR] s(x') = ∫ gF(x,x') [s(x') d2x'], one can say that the kernel Green's Function gF(x,x') "propagates" a tiny piece of "source" [s(x')d2x'] from location x' to location x so that the solution f(x) is then a sum of all such propagated contributions as the source ranges over the entire volume of interest, which for us is all 2D space where the source is non-vanishing. H.2 Theorem: -2[1/r] = 4πδ(r) (I.2.1) Proof: Let volume V be all of 2D space. Carve out from V a small spherical hole of radius a centered at r = 0. If we call this spherical area Va and then V' = V - Va is the original area with the spherical hole carved out: In order to show that some function g(r) = δ(r), one has to show that lima→0 ∫V' dV g(r) = 0 (I.2.2a) lima→0 ∫Va dV g(r) = 1 (I.2.2b) This is basically the definition of δ(r). Since δ(r) has units L-2, g(r) has units L-2. Comment: When any differential operator like or 2 is applied to ln(r0/r), the result is independent of r0 so we can always take r0 = 1. For example, ∂x [ln(r0/r)] = ∂x [ lnr0 + ln(1/r)] = ∂x ln(1/r). In what follows, ln(r) and ln(1/r) are always acted upon by differential operators, so we can interpret these objects as dimensionless quantities ln(r/r0) and ln(r0/r). Then it is clear below that dim [g(r)] = L-2. Our candidate function of interest is g(r) = - (1/2π) 2[ln(1/r)] = +(1/2π) 2 [ ln(r) ] . (I.2.3) Using 2 in polar (cylindrical) coordinates acting on a function of r, one finds that, since ∂r(1) = 0, 2[ln(r)] = (1/r)∂r(r∂r) [ln(r)] = 0 r > 0 (I.2.4) so that g(r) = - (1/2π) 2[ln(1/r)] = 0 r > 0 . (I.2.5) Thus, condition (I.2.2a) is trivially satisfied since r > 0 everywhere in area V'. It remains to verify condition (I.2.2b). Consider the integral appearing in the left side of (I.2.2b) ∫Va dV g(r) = - (1/4π) ∫Va dV 2[1/r] = - (1/4π) ∫Va dV [1/r] . (I.2.6) The divergence theorem says ∫V dV div F = ∫S dS F (I.2.7) where V is any closed area whose bounding curve is S, and dS points out. Using V = Va and F = [ln(1/r)] = ∂r(ln(1/r)) = - ∂r(lnr) = [ -r-1] we find that LHS (I.2.7) = ∫Va dV div [ln(1/r)] = ∫Va dV 2[ln(1/r)] = ∫Va dV [-2πg(r)] = -2π ∫Va dV g(r) RHS (I.2.7) = ∫S dS [ ln(1/r)] = ∫dθ [a ] [ ln(1/r)]|r=a = ∫dθ [a ] [-a-1] = -2π which tells us that ∫Va dV g(r) = 1 for any a. Thus, lima→0 ∫Va dV g(r) = 1 and we have then verified (I.2.2b). Therefore we conclude that the candidate g(r) of (I.2.3) is in fact the same as δ(r) so - (1/2π)2[ln(1/r)] = δ(r) (I.2.8) or 2[ln(1/r)] = - 2πδ(r) (I.2.9) which is (I.2.1). QED H.3 Theorem: 2[h(r)/r] = - 4π h(0) δ(r) + h"(r)/ r (I.3.1) Proof: Start with this vector identity, which is valid in 2D as well as 3D, 2(φψ) = φ2ψ + ψ2φ + 2 φ ψ (I.3.2) This identity is valid in any number of dimensions (implied sum on i from 1 to N) , ∂i2(φψ)= ∂i[ (∂iφ)ψ + ψ(∂iφ)] = (∂i2φ)ψ + (∂iφ) (∂iψ) + φ(∂i2ψ) + (∂iφ) (∂iψ) . We insert this small fact to be used below (cylindrical coordinates) ln(1/r) = [- ln(r)] = ∂r[- ln(r)] = - (1/r) So apply (I.3.2) to the case φ = h and ψ = ln(1/r), 2(h ln(1/r)) = h2(ln(1/r)) + ln(1/r) 2h + 2 h ln(1/r) = - h 2π δ(r) + ln(1/r)2h + 2 [ h' (-r-1) ] // using (I.2.9) and fact just above = - 2π h(0) δ(r) + ln(1/r)2h - 2 r-1 h'(r) . (I.3.3) Algebra shows that, using polar (cylindrical) coordinates, 2h = (1/r)∂r(r∂r)h(r) = h"(r) + r-1 h'(r) (I.3.4) so then 2(h ln(1/r)) = - 2π h(0) δ(r) + ln(1/r) [h"(r) + r-1h'(r) ] - 2 r-1 h'(r) = - 2π h(0) δ(r) + ln(1/r)h"(r) + [ ln(1/r) - 2] r-1 h'(r) which is the claim of (I.3.1). QED Application: Let h(r) = e-jkr h = e-jkr h(0) = 1 h' = -jk e-jkr h" = -k2 e-jkr 2[e-jkr ln(1/r) ] = - 2π δ(r) + ln(1/r) [-k2 e-jkr] + [ ln(1/r) - 2] r-1 [-jk e-jkr] = - 2π δ(r) + e-jkr { -k2 ln(1/r) -jk r-1 [ ln(1/r) - 2] } Thus, ( 2+k2) [e-jkr ln(1/r) ] = - 2πδ(r) -jk r-1 [ ln(1/r) - 2] e-jkr I was not expecting this mess!! What does Stak say? or - (2+k2) (e-jkr/4πr) = δ(r) (I.3.5) __________________________________________________________________________________ Theorem 3: 2[ln(1/r)] = - 2πδ(r) in two dimensions (I.1)' Our 2D proof here will be an "copy, paste and edit" version of the 3D proof of Theorem 1. We shall use the word "volume" with symbol V to denote area. A "spherical cavity" now means a circular hole. It should be understood that 2 = ∂x2+∂y2 and δ(r)= δ(x)δ(y). We indicate all edits in red and equation numbers are the same as for Theorem 1 with a prime added. Proof: Let volume V be all of 2D space. Carve out from V a small spherical cavity of radius a centered at r = 0. If we call this spherical volume Va and then V' = V - Va is the original volume with the spherical cavity. In order to show that some function g(r) = δ(r), one has to show that lima→0 ∫V' dV g(r) = 0 (I.2a)' lima→0 ∫Va dV g(r) = 1 (I.2b)' This is basically the definition of δ(r). Since δ(r) has units L-2, g(r) has units L-2. Comment: When any differential operator like or 2 is applied to ln(r0/r), the result is independent of r0 so we can always take r0 = 1. For example, ∂x [ln(r0/r)] = ∂x [ lnr0 + ln(1/r)] = ∂x ln(1/r). In what follows, ln(r) and ln(1/r) are always acted upon by differential operators, so we can interpret these objects as dimensionless quantities ln(r/r0) and ln(r0/r). Then it is clear below that dim [g(r)] = L-2. Our candidate function of interest is g(r) = - (1/2π) 2[ln(1/r)] = +(1/2π) 2 [ ln(r)] (I.3)' Using 2 in polar coordinates acting on a function of r, one finds that, since ∂r(1) = 0, 2[ln(r)] = (1/r)∂r(r∂r) [ln(r)] = 0 r > 0 (I.4)' so that g(r) = - (1/2π) 2[ln(1/r)] = 0 r > 0 . (I.5)' Thus, condition (I.2a)' is trivially satisfied since r > 0 everywhere in volume V'. It remains to verify condition (I.2b)'. Consider the integral appearing in the left side of (I.2b)' ∫Va dV g(r) = - (1/2π) ∫Va dV 2[ln(1/r)] = - (1/2π) ∫Va dV [ln(1/r)] . (I.6)' The divergence theorem says ∫V dV div F = ∫S dS F (I.7)' where V is any closed volume whose surface is S, and dS points out. Using V = Va and F = [ln(1/r)] = ∂r(ln(1/r)) = - ∂r(lnr) = [ -r-1] we find that [ here dS = dS where dS = adθ = a piece of circumference of the circle bounding Va ] LHS (I.7) = ∫Va dV div [ln(1/r)] = ∫Va dV 2[ln(1/r)] = ∫Va dV [-2πg(r)] = -2π ∫Va dV g(r) RHS (I.7) = ∫S dS [ ln(1/r)] = ∫dθ [a ] [ ln(1/r)]|r=a = ∫dθ [a ] [-a-1] = -2π which tells us that ∫Va dV g(r) = 1 for any a. Thus, lima→0 ∫Va dV g(r) = 1 and we have then verified (I.2b)'. Therefore we conclude that - (1/2π)2[ln(1/r)] = δ(r) (I.8)' or 2[ln(1/r)] = - 2πδ(r) which is (I.1)'. QED