Appendix I v 2 REVIEWED

11.14.13 This is where I developed Appendix I which is now installed in lines doc. I split the 3D and 2D stuff into two separate appendices. This is "version 2" of an earlier draft of this appendix, don't know what was wrong with the first version. Appendix I : The Poisson and Helmholtz Free-Space Propagators in 2D Note: Appendix H deals with these propagators in 3D rather than 2D. Sections I.1 and I.2 below are basically "cut, paste and edit" versions of Sections H.1 and H.2, and we have made equation numbers match. However, Section I.3 is something new since it involves a "special function". __________________________________________________________________________________ I.1 Overview and Meaning of Free-Space Propagators This appendix proves two Facts: (H0(1) is a Hankel function ) Fact 1 : -2[ln(1/r)/2π] = δ(r) (I.2.1) (I.1.1) Fact 2 : - (2+k2) [(j/4) H0(1)(kr)] = δ(r) (I.3.1) (I.1.2) Throughout this Appendix, 2 is the usual 2D Laplacian operator, 2 = ∂x2 + ∂y2. and δ(r) = δ(x) δ(y) . (I.1.3) In the two Facts above, if one replaces r → r-r' (a simple translational shift of origin) ones finds -2[ln(1/R)/2π] = δ(r-r') R = | r - r' | (I.1.4) - (2+k2) [(j/4) H0(1)(kR)] = δ(r-r') δ(r-r') = δ(x-x') δ(y-y') (I.1.5) The quantities in brackets are known as free-space Green's Functions (Green Functions) or propagators, or as "fundamental solutions" : ln(1/R) = the Poisson 2D free-space propagator (I.1.6) (j/4) H0(1)(kR) = 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' [ln(1/R)/2π] s(x') + homogeneous solutions The Poisson Equation (I.1.8) - (2+k2) f(x) = s(x) => f(x) = ∫d2x' [(j/4) H0(1)(kR)] 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 curves), 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' [ln(1/R)/2π] s(x') + fh(x) -2 f(x) = ∫d2x' (-2 [ln(1/R)/2π] ) 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' [ln(1/R)/2π] 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. __________________________________________________________________________________ I.2 Fact 1: 2[ln(1/r)] = - 2πδ(r) (I.2.1) Proof: Let area A be all of 2D space. Cut out from A a small spherical hole of radius a centered at r = 0. If we call this spherical area Aa and then A' = A - Aa is the original area with the spherical hole cut out: In order to show that some function g(r) = δ(r), one has to show that lima→0 ∫A'dA g(r) = 0 (I.2.2a) lima→0 ∫AdA 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) for any r0. 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 without the z) 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 A' for any a > 0. It remains to verify condition (I.2.2b). Consider the integral appearing in the left side of (I.2.2b) ∫AdA g(r) = - (1/4π) ∫AdA 2[1/r] = - (1/4π) ∫AdA [1/r] . (I.2.6) The divergence Fact in 2D says ∫A dA div F = C ds F (I.2.7) where A is any closed area whose bounding curve is C, and where ds = ds where is normal to C at any given point on C. Notice that this closed area is necessarily planar since everything is 2D here. Using A = Aa = disk of radius a and F = [ln(1/r)] = ∂r(ln(1/r)) = - ∂r(lnr) = [ -r-1] we find that LHS (I.2.7) = ∫AdA div [ln(1/r)] = ∫AdA 2[ln(1/r)] = ∫AdA [-2πg(r)] = -2π ∫AdA dA g(r) RHS (I.2.7) = ∫C ds [ ln(1/r)] = ∫ [adθ ] [ ln(1/r)]|r=a = ∫dθ [a ] [-a-1] = -2π which tells us that ∫Aa dA g(r) = 1 for any a. Thus, lima→0 ∫Aa dA 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 __________________________________________________________________________________ I.3 Fact 2: - (2+k2) [(j/4) H0(1)(kr)] = δ(r) (I.3.1) We seek the solution E(r) of this equation - (2+k2 ) E(r) = δ(r) where E(r→∞) = 0 (I.3.2) which we write as 2E+ k2E = - δ(r). Using polar coordinates this says r-1∂r(r∂rE) + k2E = - δ(r) or E" + r-1E' + k2E = - δ(r) or r2E"(r) + rE'(r) + r2k2E(r) = - δ(r) . (I.3.3) Writing E(r) = F(kr) we get r2k2F"(kr) + rk F'(kr) + r2k2F(kr) = - δ(r) or (rk)2F"(kr) + (rk) F'(kr) +(rk)2F(kr) = - δ(r) or z2F"(z) + z F'(z) + z2F(z) = - δ(r) where z = kr . (I.3.4) Away from r = z = 0, this is Bessel's equation of index 0 (A&S 10.2.1) so solutions are Bessel functions like these F(z) = J0(z), Y0(z), H0(1)(z), H0(2)(z). z = kr (I.3.5) which are Bessel functions of the first, second and third kind. The third kind functions (the H's) are called Hankel Functions. If we assume that k has a tiny positive imaginary part (see Comments later), then of all the functions just listed, only H0(1)(kr) has decaying behavior for large r (A&S 10.2.5). We therefore put forward the following candidate for a delta function g(r) = - (2+k2 ) C H0(1)(kr) . (I.3.6) Recall from Section I.2 that a successful δ(r) candidate must satisfy these two conditions lima→0 ∫A'dA g(r) = 0 (I.2.2a) lima→0 ∫AdA g(r) = 1 (I.2.2b) Our candidate g(r) vanishes within any region A' no matter how small the hole because g(r) = 0 for any r> 0, so the first condition is already met. It remains only to show that the second condition is also met. We must then show that lima→0 ∫A dA {- (2+k2 ) C H0(1)(kr)} = 1 . (I.3.7) Since Aa is a very small disk as we approach the limit, we may use the small argument behavior of our candidate g(r) in studying the situation. We know that H0(1)(kr) ≈ (2j/π) ln(kr) // A&S 10.7.2 (I.3.8) so what we need to show is that lima→0 ∫AdA {- (2+k2) C (2j/π) ln(kr)} = 1 or - C (2j/π) lima→0 ∫AdA { (2+k2) ln(kr)} = 1 or - C (2j/π)2π lima→0 !Syntax Error, Irdr{ (2+k2) ln(kr)} = 1 // ∫dθ = 2π or C (4/j) lima→0 !Syntax Error, Irdr{ (2+k2) ln(kr)} = 1 . (I.3.9) Now consider : lima→0 [!Syntax Error, Irdr ln(kr)] = lima→0 [(1/4)a2{2ln(ka)-1}] = 0 . (I.3.10) Thus, the k2 ln(kr) term in (I.3.9) makes no contribution in the limit, so we then have to show that C (4/j) lima→0 !Syntax Error, Irdr 2 [ ln(kr)] = 1 . (I.3.11) But (I.2.9) says that 2[ln(r)] = 2πδ(r) . (I.2.9) Now δ(r) = δ(x)δ(y) = δ(r)/2πr (I.3.12) since 1 = ∫∫dxdy δ(x)δ(y) = ∫rdr∫dθ δ(r)/2πr = 2π∫rdr δ(r)/2πr = ∫dr δ(r) = 1 . Therefore 2[ln(r)] = δ(r)/r (I.3.13) and then 2[ln(kr)] = 2[ln(k) + ln(r)] = 2[ln(r)] = δ(r)/r . (I.3.14) Inserting this last result into (I.3.11) then gives C (4/j) lima→0 !Syntax Error, Irdr 2 [ ln(kr)] = 1 C (4/j) lima→0 !Syntax Error, Irdr δ(r)/r = 1 C (4/j) lima→0 !Syntax Error, Idr δ(r) = 1 C (4/j) lima→0 1 = 1 C (4/j) = 1 . Thus, we have a solution if we select constant C = (j/4). Therefore, the solution to (I.3.2) is E(r) = C H0(1)(kr) = (j/4) H0(1)(kr) . (I.3.15) Stakgold Vol II page 55 (5.120) confirms this result where = k. Therefore we have shown that - (2+k2) [(j/4) H0(1)(kr)] = δ(r) (I.3.16) which is the Fact stated as (I.3.1). QED On page 54 Stakgold gives the solution to - (2+k2 ) E(r) = δ(r) for n≥2 dimensions as (5.118): Comments: 1. Complex Helmholtz Parameter. Stakgold considers the Helmholtz operator to be λ which is our k2. He regards λ as a complex variable which can lie anywhere in the complex λ plane. If we consider the function k(λ) = λ1/2, we find that it has a branch point at λ = 0. If we take the branch cut to the right, then one of the two Riemann sheets in λ-space for this function maps to the upper half k-plane as shown. This is the branch of λ1/2 that Stakgold selects and that is why we think of k and therefore k2 as having a tiny positive imaginary part. The point is that we approach the positive real axis from above, not from below. It is this assumption that causes the large-r-decaying solution to our problem to be H0(1)(kr) instead of H0(2)(kr) . 2. Helmholtz morphs into Poisson. We have shown that - (2+k2) [(j/4) H0(1)(kr)] = δ(r) . (I.3.16) In the limit that k << 1, we showed above that H0(1)(kr) ≈ (2j/π) ln(kr) // A&S 10.7.2 (I.3.8) In this limit we then have - (2+k2) [(j/4)) (2j/π) ln(kr) ] = δ(r) or - (2) [(1/2π) ln(kr) ] = δ(r) and this is in agreement with the Poisson result (I.1.1). So as the Helmholtz equation morphs into the Poisson equation as k → 0, the Helmholtz propagator morphs into the Poisson propagator.