old section G_4 retired REVIEWED

Retiring old section G.4 PhL 2.7.14 Here I had my "double" evaluation of the integral here called I1(r,r') which I realized is the same as Q from Appendix B. So here is the old section just for the record. G.4 Solution for Az using the 2D Helmholtz Integral This method is technically more difficult than the first two methods shown in Section G.2 (solving the ODE and adding a homogeneous solution to meet the boundary conditions) and Section G.3 (instant Ampere's Law solution), so we show a lot of detail. The method is important because our entire Chapter 4 is based on using Helmholtz integrals to develop the theory of transmission line parameters, and this is one Helmholtz integral that we can actually compute without too much effort. An important result we find is that, when μ1≠μ2, the Helmholtz integral by itself does not supply the complete solution, and one must add in some amount of homogeneous solution of 22D Az(r) = 0 in order to meet the required boundary conditions at r = a. We really have a Poisson integral since β12 = 0, but the full Helmholtz integral works the same way so we keep referring to it as a Helmholtz integral. Recall from above: 22D Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθ with Bθ = - ∂rAz . (G.1.7) Using the 2D free-space Green's function (propagator) as reviewed in Appendix I equation (I.1.6), g(x|x') = (1/2π) ln(1/R) = - (1/4π) ln(R2) R = R = |x-x'| (G.4.1) we may write the particular solution to (G.1.7) as the following "Helmholtz" integral [see (I.1.8)] AzH(r) = - ∫dA' [(1/4π) ln(R2) ] [Iμ2/(πa2)] θ(r≤a) = -!Syntax Error, Ir' dr'!Syntax Error, Idθ' [(1/4π) ln(r2+ r'2- 2rr'cos(θ-θ'))] [Iμ2/(πa2)] = - (1/4π) [Iμ2/(πa2)] !Syntax Error, Ir' dr' [ !Syntax Error, Idθ' ln(r2+ r'2- 2rr'cos(θ-θ')) ] = - (1/4π) [Iμ2/(πa2)] !Syntax Error, Ir' dr' [ 2 I1(r,r') ] (G.4.2) where I1(r,r') ≡ (1/2) !Syntax Error, Idθ' ln(r2+ r'2- 2rr'cos(θ-θ')) = (1/2) !Syntax Error, Idθ" ln(r2+ r'2- 2rr'cos(θ")) = !Syntax Error, Idx ln(r2+ r'2- 2rr'cosx) . (G.4.3) Since this integral I1 is quite important, we evaluate it two independent ways: _______________________________________________________________________________ Method 1: I1 = !Syntax Error, Idx ln[r2(1+ r'2/r2- 2r'/rcosx) ] = !Syntax Error, Idx ln[r2(1+ α2- 2αcosx) ] α ≡ r'/r = !Syntax Error, Idx ln[r2] + !Syntax Error, Idx ln (1+ α2- 2α cosx) = π ln(r2) + I2 . (G.4.4) We then use this easily derived expansion Stakgold Vol II p 104 (6.22), ln(1 + α2 - 2α cosx ) = -2 Σn=1∞ (αn/n)cos(nx) |α| < 1 (G.4.5) which is just a Fourier Cosine Series expansion of ln(1 + α2 - 2α cosx ). Then, I2 = !Syntax Error, Idx ln (1+ α2- 2αcosx) = -2 Σn=1∞(αn/n)!Syntax Error, Idx cos(nx) = -2 Σn=1∞(αn/n) [ (1/n)sin(nx)]|π0 = -2 Σn=1∞(αn/n2) [sin(nπ) - sin(0)] = 0 and then I1 = π ln (r2) + I2 = π ln (r2) + 0 = 2π ln(r) α < 1 r'<r Since I1(r,r') is symmetric under r↔r' we must have I1 = 2π = 2π . (G.4.6) Method 2: I1 = !Syntax Error, Idx ln(r2+ r'2- 2rr'cosx) = !Syntax Error, Idx ln (a + b cosx) . This integral may be evaluated using GR7 p 531 4.224, with a = r'2 +r2 and b = -2rr' and a2-b2 = (r'2-r2)2 so that = | r'2-r2 |. The condition a > |b| > 0 is met since (r±r')2 > 0 => r2+r'2 > ±2rr' which says a > ±b so a > |b|. Thus, I1 = !Syntax Error, Idx ln [r'2 +r2-2rr' cosx)] = π ln [ ] = = 2π (G.4.7) which agrees with the result of method 1. ___________________________________________________________________________________ Using the notation θ(x >a) ≡ θ(x-a) [ Heaviside step ] we write (G.4.7) as I1 = 2π { ln(r)θ(r'<r) + ln(r') θ(r'>r) } . (G.4.8) Note: We use notation θ(x >a) in place of the more correct θ(x-a) because it requires "less thinking" to interpret. Our particular "Helmholtz" integral (G.4.2) is then AzH(r) = - (1/4π) Iμ2/(πa2) !Syntax Error, Ir' dr' [ 2 I1(r,r') ] = - (1/4π) Iμ2/(πa2) !Syntax Error, Ir' dr' 4π { ln(r)θ(r'<r) + ln(r') θ(r'>r) } = - Iμ2/(πa2) !Syntax Error, Ir' dr' { ln(r)θ(r'<r) + ln(r') θ(r'>r) } = - Iμ2/(πa2) { ln(r)!Syntax Error, Ir' dr' θ(r'<r) + !Syntax Error, Ir' dr' ln(r') θ(r'>r) } . (G.4.9) The two integrals are !Syntax Error, Ir' dr' θ(r'<r) = !Syntax Error, Ir' dr' = (1/2) [min(a,r)]2 (G.4.10) !Syntax Error, Ir' dr' ln(r') θ(r'>r) = θ(r<a) !Syntax Error, Ir' dr' ln(r') = θ(r<a) !Syntax Error, Ix lnx dx = θ(r<a) [(1/4) (r2-a2) + (1/2) a2lna - (1/2) r2lnr ] . (G.4.11) Then AzH(r) = - Iμ2/(πa2) { ln(r)!Syntax Error, Ir' dr' θ(r'<r) + !Syntax Error, Ir' dr' ln(r') θ(r'>r) } = - Iμ2/(πa2) { ln(r) (1/2) [min(a,r)]2 + θ(r<a) [(1/4) (r2-a2) + (1/2) a2lna - (1/2) r2lnr ] } . For r > a we get AzH(r) = - Iμ2/(πa2) { ln(r) (1/2)a2 } = - Iμ2/(2π) ln(r) (G.4.12) ∂rAzH(r) = - Iμ2/(2π) (1/r) . For r < a the result is AzH(r) = - Iμ2/(πa2) { ln(r) (1/2)r2 + (1/4) (r2-a2) + (1/2) a2lna - (1/2) r2lnr } = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } (G.4.13) ∂rAzH(r) = - Iμ2/(πa2) (r/2) . Recall the two boundary conditions [Az(a)]1 = [Az(a)]2 . (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 . (G.2.6) For our particular Helmholtz integral AzH(r) we evaluate these boundary conditions to find - Iμ2/(2π) ln(a) = - Iμ2/(πa2) (1/2) a2lna (1/μ1) [- Iμ2/(2π) (1/a)] = (1/μ2) [- Iμ2/(πa2) (a/2)] or 1 = 1 (μ2/μ1) = 1 Thus, only in the case μ1 = μ2 does the Helmholtz particular solution meet both boundary conditions. If μ1≠ μ2, we must add to the particular solution some amount of homogeneous solution of 22D Az(r,θ) = 0. So we then write generally, Az(r) = AzH(r) + Azhomo(r) (G.4.14) where we know that Azhomo(r) can only have terms α + β ln r. We then write for the two regions Az(r) = - Iμ2/(2π) ln(r) + α + β lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + α ' + β' lnr r<a As earlier, we choose the zero point for Az(r) by requiring that the large r behavior be K ln(r) without a constant added, which then means α = 0. And for r<a we must have β' = 0 to be finite at r = 0. So Az(r) = [- Iμ2/(2π) + β] lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + α ' r<a (G.4.15) -∂rAz(r) = [Iμ2/(2π) - β](1/r) r>a -∂rAz(r) = Iμ2/(πa2) (2r) r<a The boundary conditions are then [Az(a)]1 = [Az(a)]2 (1/μ1) [(-∂rAz)(a)]1 = (1/μ2) [(-∂rAz)(a)]2 // (G.2.6) repeated or [- Iμ2/(2π) + β] lna = - Iμ2/(πa2) { (1/4) (a2-a2) + (1/2) a2lna } + α ' (1/μ1)[ Iμ2/(2π) - β](1/a) = (1/μ2) Iμ2/(πa2) (2a) // insert expressions, set r = a or [- Iμ2/(2π) + β] lna = - Iμ2/(2π) lna + α ' (1/μ1)[ Iμ2/(2π) - β](1/a) = I/(2πa) // simplify or β lna = α ' Iμ2/(2π) - β = μ1I/(2π) // simplify some more so we find that β = I/(2π) (μ2-μ1) α' = I/(2π) (μ2-μ1) lna (G.4.16) The full solution is then Az(r) = [- Iμ2/(2π) + β] lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + α ' r<a or Az(r) = [- Iμ2/(2π) + {I/(2π) ( μ2-μ1)}] lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + { I/(2π) ( μ2-μ1) lna } r<a or Az(r) = I/(2π) [- μ2 + ( μ2-μ1)] lnr r>a Az(r) = - Iμ2/(πa2) (1/4) (r2-a2) - Iμ2/(πa2) (1/2) a2lna + I/(2π) ( μ2-μ1) lna r<a or Az(r) = - (I/(2π) [μ1] lnr r>a Az(r) = - Iμ2/(πa2) (1/4) (r2-a2) - Iμ1/(π) (1/2) lna r<a or Az(r) = - [Iμ1/2π] lnr r>a Az(r) = - [Iμ2/(4πa2)] (r2-a2) - [Iμ1/2π] lna r<a (G.4.17) This result matches the results (G.2.7) and (G.3.3) of the previous two methods and then results in the B field solution Bθ = - ∂rAz = [Iμ1/2π]∂rln(r) = [Iμ1/2π](1/r) r > a region 1 Bθ = - ∂rAz = [Iμ2/(2πa2)] r = [Iμ2/2π](r/a2) r < a region 2 (G.2.8)