Iris Green's by partial eigenfunctions attempt

Iris Green's by partial eigenfunctions (cylindrical) PhL 4.30.10 Summary of this doc: In Section 1 I use Stakgold's "partial eigenfunction method" to derive (full detail) an expression for the Green's function of a point charge with no boundary conditions in cylindrical coordinates. This leads to an expression for 1/R which is different from the one quoted by Jackson in his (3.149) page 86. Here is the comparison [ Green's point charge is at (ρ',φ'=0,z'), observation point is (ρ,φ,z) ] 1/R = (2/π) Σm=0∞ εm cos(mφ)!Syntax Error, Idk cos[k(z-d)] Im(kρ<) Km(kρ>) // Jackson 1/R = Σm=0∞ εm cos(mφ)!Syntax Error, Idk Jm(kρ) Jm(kρ') e-k|z-z'| // me Both expressions are correct and I verified "mine" against various external sources. It is a matter of which coordinate you take to be non-oscillatory. For Jackson it is ρ, for me it is z. In Section 2 I try to use the same method for the iris Green's function, without success. It seems that the boundary conditions entangle both the ρ and z coordinates and this makes the method not work. 1. The Free Space Green's Function in Cylindricals with z as the non-osc. 2 (a) Processing the φ dimension. 2 (b) Processing the ρ dimension. 3 (c) Solve the 1D Green's function problem in z 6 (d) Assemble the complete Green's function solution 7 (e) Comparison to Jackson and others. 7 2. The Iris Green's Function in Cylindricals with z as the non-osc. 8 (a) Plan A: try expanding on Jm(kρ) 9 (b) Plan B: try expanding on cos[k(z-z')]. 10 (c) Conclusions: 13 I don't expect to get far, but let's at least give this a shot. The method is discussed in Stak p 154, but he only does the 2D examples of a rectangle and a unit circle where he knocks out one variable using an expansion. In 3D you have to knock out two variables with such expansions, so how might this go? My Wilcox guy below refers to Stak' "method of partial eigenfunctions" as "the reduced Green's function technique". Rather than try the iris, I am going to first to the free space Green's function, which Stak always calls the fundamental solution. I use Jackson 4π conventions. 1. The Free Space Green's Function in Cylindricals with z as the non-osc. The first order of business is to actually write down the Green's ODE, something I rarely do, and I want to do it in cylindricals. Here I imply that 2 is wrt unprimed variables, and primed variables are parameters marking the Green's charge location. 2g(ρ,φ,z; ρ',φ',z') = - 4π δ(z-z') δ(φ-φ') δ(ρ-ρ')/ρ from d3r = dz ρdρdφ where I am staring at Jackson page 69 for conventions. This is Poisson's equation Jackson p 13 in the units system where V = q/R for a point charge. (a) Processing the φ dimension. Let's first try g(ρ,φ,z; ρ',φ',z') = Σm=0∞ gm(ρ,z; ρ',z') cos[m(φ-φ')] We have now to write out, using Jackson page 69 (agrees with my TK page of course) 2 = ∂ρ2 + (1/ρ)∂ρ + (1/ρ2)∂φ2+ ∂z2 = ρ2 + z2+ φ2 = ρz2 + φ2 where I am just making up names for the various pieces of 2. Note that ρ2 = ∂ρ2 + (1/ρ)∂ρ = (1/ρ)∂ρ(ρ∂ρ) Meanwhile, from my new transforms doc I know that Σm=0+∞εm cos[m(φ-φ')] = 2π δ(φ-φ') // completeness Now inserting our expansion we get 2π 2{ Σm=0∞ gm(ρ,z; ρ',z') cos[m(φ-φ')]} = - 4π δ(z-z') { Σm=1+∞εm cos[m(φ-φ')]}δ(ρ-ρ')/ρ We now have to figure out 2 { gm(ρ,z; ρ',z') cos[m(φ-φ')] } = (ρz2 + φ2){ gm(ρ,z; ρ',z') cos[m(φ-φ')] } = cos[m(φ-φ')] ρz2 gm(ρ,z; ρ',z') + gm(ρ,z; ρ',z') φ2 cos[m(φ-φ')] But of course φ2 cos[m(φ-φ')] = -(m2/ρ2) cos[m(φ-φ')] so we have = cos[m(φ-φ')] ρz2 gm(ρ,z; ρ',z') - (m2/ρ2) gm(ρ,z; ρ',z') cos[m(φ-φ')] = [ ρz2 gm(ρ,z; ρ',z') - (m2/ρ2) gm(ρ,z; ρ',z') ] cos[m(φ-φ')] So our equation above becomes: 2π Σm=0∞[ ρz2 gm(ρ,z; ρ',z') - (m2/ρ2) gm(ρ,z; ρ',z') ] cos[m(φ-φ')] = - 4π δ(z-z') { Σm=0+∞εm cos[m(φ-φ')]}δ(ρ-ρ')/ρ = 2π Σm=0+∞[ - (2εm) δ(z-z') δ(ρ-ρ')/(ρ) ] cos[m(φ-φ')] (*) Aside: Now if someone tells you that Σm=0∞ am cos[m(φ-φ')] = Σm=0∞ bm cos[m(φ-φ')] where you know that cos[m(φ-φ')] is a "complete set" for expansion Σm=0+∞εm cos[m(φ-φ')] = 2π δ(φ-φ') you must conclude that am = bm. We could of course prove this by setting φ' = 0 which would say Σm=0∞ am cos[m(φ)] = Σm=0∞ bm cos[m(φ)] and then we could apply !Syntax Error, I dφ cos(m'φ) to both sides and use orthogonality !Syntax Error, I dφ cos(mφ)cos(m'φ) = (π/εm) δmm' and we would conclude that am = bm. So, comparison of the two sides of (*) in terms of the complete set of functions shown tells us that ρz2 gm(ρ,z; ρ',z') - (m2/ρ2) gm(ρ,z; ρ',z') = - (2εm) δ(z-z') δ(ρ-ρ')/(ρ) [ρz2 - (m2/ρ2)] gm(ρ,z; ρ',z') = -(2εm) δ(z-z') δ(ρ-ρ')/ρ and we have now successfully reduced our Green's problem by one variable, having replaced φ with m. (b) Processing the ρ dimension. I choose ρ because Jm(ρ) is a complete set in the general atomic forms I want to use. We know that !Syntax Error, Idρ ρ Jm(kρ) Jm(k'ρ) = δ(k-k')/k // orthogonality !Syntax Error, Idk k Jm(kρ) Jm(kρ') = δ(ρ-ρ')/ρ // completeness which is the Hankel transform (again, from transforms.doc). So let's try this: gm(ρ,z; ρ',z') = !Syntax Error, Idk k gmk(z; z') Jm(kρ) Jm(kρ') Here I want to just expand on the Jm(kρ), but I know that a Green's function has to be symmetric. This is Theorem 2 of Stakgold section 6.6 (see meta meta Chap 6), and I think this is a very major step! The final function gmk(z; z') will also be symmetric in its arguments. I threw in a k above since I know it is the required weight function, just part of the definition then of gmk. We now need to compute [ρz2 - (m2/ρ2)] gm(ρ,z; ρ',z') = [z2 + ρ2 -(m2/ρ2)] gm(ρ,z; ρ',z') = [ρ2 + z2 - (m2/ρ2)] !Syntax Error, Idk k gmk(z; z') Jm(kρ) Jm(kρ') !Syntax Error, Idk k Jm(kρ) Jm(kρ') [z2 - (m2/ρ2)]gmk(z; z') + !Syntax Error, Idk k gmk(z; z') Jm(kρ') [ρ2 Jm(kρ)] We now turn to our Jackson cylindricals doc from which we quote: [ x∂x(x∂x) + (a2x2-n2)] Jn(ax) = 0 and translate to [ ρ∂ρ(ρ∂ρ) + (k2ρ2-m2)] Jm(kρ) = 0 [ (1/ρ)∂ρ(ρ∂ρ) + (k2-m2/ρ2)] Jm(kρ) = 0 [ρ2 + (k2-m2/ρ2)] Jm(kρ) = 0 [ρ2 Jm(kρ)] = (m2/ρ2- k2) Jm(kρ) We can then install this to get [ρz2 - (m2/ρ2)] gm(ρ,z; ρ',z') = = !Syntax Error, Idk k Jm(kρ) Jm(kρ') [z2 - (m2/ρ2)]gmk(z; z') + !Syntax Error, Idk k gmk(z; z') Jm(kρ') [ρ2 Jm(kρ)] = !Syntax Error, Idk k Jm(kρ) Jm(kρ') [z2 - (m2/ρ2)]gmk(z; z') + !Syntax Error, Idk k gmk(z; z') Jm(kρ') [(m2/ρ2- k2) Jm(kρ)] = !Syntax Error, Idk k Jm(kρ) Jm(kρ') [z2 - (m2/ρ2)]gmk(z; z') + !Syntax Error, Idk k Jm(kρ) Jm(kρ') [(m2/ρ2- k2)] gmk(z; z') = !Syntax Error, Idk k Jm(kρ) Jm(kρ') { z2 - (m2/ρ2) + (m2/ρ2- k2) }gmk(z; z') = !Syntax Error, Idk k Jm(kρ) Jm(kρ') { z2 - k2} gmk(z; z') = - (εm/2π) δ(z-z') δ(ρ-ρ')/ρ // the RHS of our 2D Green's equation found above = - (εm/2π) δ(z-z') !Syntax Error, Idk k Jm(kρ) Jm(kρ') // installing Bessel completeness So we now have !Syntax Error, Idk k Jm(kρ) Jm(kρ') { z2 - k2} gmk(z; z') = !Syntax Error, Idk k Jm(kρ) Jm(kρ') { - (2εm) δ(z-z') } (*) Aside: Now if someone tells you that !Syntax Error, Idk k Jm(kρ) Jm(kρ') a(k) = !Syntax Error, Idk k Jm(kρ) Jm(kρ') b(k) (**) where you know that Jm(kρ) Jm(kρ') is a "complete set" for expansion !Syntax Error, Idk k Jm(kρ) Jm(kρ') = δ(ρ-ρ')/ρ you must conclude that a(k) = b(k). We could of course prove this by applying !Syntax Error, Idρ ρ to both sides of (**) , !Syntax Error, Idk k {!Syntax Error, Idρ ρ Jm(kρ) Jm(kρ') }a(k) = !Syntax Error, Idk k {!Syntax Error, Idρ ρ Jm(kρ) Jm(kρ') }b(k) and we then use orthogonality !Syntax Error, Idρ ρ Jm(kρ) Jm(k'ρ) = δ(k-k')/k to get !Syntax Error, Idk k { δ(k-k')/k }a(k) = !Syntax Error, Idk k { δ(k-k')/k }b(k) a(k') = b(k') Therefore, looking at equation (*) above we may conclude that { z2 - k2} gmk(z; z') = - (2εm) δ(z-z') (c) Solve the 1D Green's function problem in z We have this problem now { z2 - k2} gmk(z; z') = - (2εm) δ(z-z') which is a 1D Helmholtz problem. Up to this point, I have said nothing about boundary conditions! The Green's solution for our iris problem has to vanish on the iris which is z = 0 but only when ρ ≥ a. But I have ignored this fact totally, and I think we are "too late". So let's just find the Green's function in the case that there are NO boundary conditions and see what we get. We solve the above like any other Green's problem, our interval is now (-∞,∞) for z. We start with the obvious homogeneous solutions g = Ae-k(z-z') z > z' g = Be+k(z-z') z < z' Our "pillbox" condition is found by integrating the ODE from z'-ε to z'+ε and that gives ∂zg(z = z'+ε) - ∂zg(z = z'-ε) = - (2εm) which is the famous "jump condition" in the 1D world. So: ∂z g = A (-k)e-k(z-z') z > z' ∂z g = B(+k)e+k(z-z') z < z' A (-k) - B(+k) = - (2εm) A (k) + B(+k) = (2εm) => (A+B)k = (2εm) => A+B = (2εm/k) But of course the solution also has to be continuous at z = z' which says A = B. So 2A = (2εm/k) => A = B = (εm/k) and our solution is then gmk(z; z') = (εm/k) e-k|z-z'| which is noted to be symmetric in z. (d) Assemble the complete Green's function solution g(ρ,φ,z; ρ',φ',z') = Σm=0∞ gm(ρ,z; ρ',z') cos[m(φ-φ')] gm(ρ,z; ρ',z') = !Syntax Error, Idk k gmk(z; z') Jm(kρ) Jm(kρ') gmk(z; z') = (εm/k) e-k|z-z'| Therefore g(ρ,φ,z; ρ',φ',z') = Σm=0∞ εm !Syntax Error, Idk Jm(kρ) Jm(kρ') e-k|z-z'| cos[m(φ-φ')] This can be interpreted as the potential of a unit positive point charge located at (ρ',φ',z'), because our starting equation was -2g = 4πδ(r-r') which is just Poisson's equation Jackson p 13 (1.28) without the 4π. We have V = q/R for a point charge. This, we have also shown that 1/R = 1/|r - r'| = Σm=0∞ εm !Syntax Error, Idk Jm(kρ) Jm(kρ') e-k|z-z'| cos[m(φ-φ')] (e) Comparison to Jackson and others. Jackson green (same in red) on page 84 derives a different form for 1/R. He starts with the same Green's equation (3.138) on page 84 (we ignore the 4π) , but he assumes oscillatory in the z direction, not the ρ direction, so his result is (3.149) which involves the I and K functions instead of the J functions. (Wilcox) Make SURE you look soon at this web page: http://en.wikipedia.org/wiki/Green%27s_function_for_the_three-variable_Laplace_equation But for the moment we continue our search. I have a nice downloaded book chapter 3 on doing cylindrical electrostatics which claims: (the sum is on m) First off, if we expand eim(φ-φ') = cos[m(φ-φ')] + i sin[m(φ-φ')], the sine term is odd in m for non-zero m, and is zero for m=0, so makes no contribution at all. If we then fold the m<0 terms over, we get a doubling for m > 0 but not for m = 0, and this makes our εm factor. So, this is full confirmation! I have not made any errors in the work above. The above comes from Chapter 3 (spherical and cylindrical electrostatics) of a web book written by Walter Wilcox of the Bayler Physics department, http://bearspace.baylor.edu/Walter_Wilcox/open_text/em3.html " Baylor University in Waco, Texas, is a private Baptist university, and a nationally ranked liberal arts institution. Chartered in 1845 by the Republic of Texas, Baylor is the oldest, continually operating university in the state. Though 80 percent of our students come from within Texas, we are home to students from all 50 states, and 80 countries." I downloaded the 2 part contents of his Jackson-like book, and a the first few chapters. Smythe has Jacksons I K form of 1/R but not my form. M&F vol 2 on page 1262 develop my result in an unusual manner, and here is their result: and I note that M&F use Neumann's factor εm. It probably appears in their Fourier section. It is good to compare this result to Jackson p 69 (see also p 65) 1/R = 4π Σlm (2l+1)-1 r<l r>-l-1 Y*Y Y = ((2l+1)/4π)1/2 f(l,-m)1/2 Plm(z)eimφ when we insert this Y twice, the obvious factors cancel and we get these to compare: 1/R = Σlm eim(φ-φ') f(l,-m) r<l r>-l-1 Plm(z) Plm(z') 1/R Σm=0∞ !Syntax Error, Idk cos[m(φ-φ')] εm e-k|z-z'| Jm(kρ) Jm(kρ') I suppose there is some general way to write this in any coordinate system. But you have to decide which is your non-oscillatory coordinate, then you have to do the separation of variables, and so on. 2. The Iris Green's Function in Cylindricals with z as the non-osc. OK, so how does the above presentation change, and does it become impossible? Our starting point is now this ODE system ( I don't bother to write g = 0 at ∞) 2g(ρ,φ,z; ρ',φ',z') = - 4π δ(z-z') δ(φ-φ') δ(ρ-ρ')/ρ g(z=0) = 0 for ρ > a ∂zg(z=0+) = ∂zg(z=0-) for ρ < a The continuity condition just says there is no surface charge in the hole of the iris. Were we to select z' = 0 for the Green's charge location, we could replace the second condition with ∂zg(z=0) = 0, but let's try to stay general. Let's now imagine we can carry out the φ processing as was done above, so that we arrive at this first intermediate point: g(ρ,φ,z; ρ',φ',z') = Σm=0∞ gm(ρ,z; ρ',z') cos[m(φ-φ')] [ρz2 - (m2/ρ2)] gm(ρ,z; ρ',z') = -(2εm) δ(z-z') δ(ρ-ρ')/ρ where we now apply our two BC's to the function gm(ρ,z; ρ',z'). So we have this new PDE system: [ρz2 - (m2/ρ2)] gm(ρ,z; ρ',z') = -(2εm) δ(z-z') δ(ρ-ρ')/ρ gm(z=0) = 0 for ρ > a ∂zgm(z=0+) = ∂zgm(z=0-) for ρ < a limz→±∞ gm(z) = 0 (a) Plan A: try expanding on Jm(kρ) The next step is not very clear to me. Suppose we try the same expansion as above: gm(ρ,z; ρ',z') = !Syntax Error, Idk k gmk(z; z') Jm(kρ) Jm(kρ') Justification might be that we know that gm is going to be at least a continuous function of ρ on (0,∞), so we ought to be able to expand it on the interval (0,∞) in this manner, since the Jm(kρ) form a complete orthogonal set on that interval. And I think we are still symmetric in the arguments as before. Barring order-interchange issues, we can trace this down to the next conclusion which is this: { z2 - k2} gmk(z; z') = - (2εm) δ(z-z') But something must be wrong at this point because ρ no longer appears anywhere, so how are you going to impose your required boundary conditions? This then suggests to me that the "form" shown above is not correct. One fact is that if we try the above form, applying ∂z does not help us much. We get the following "system" { z2 - k2} gmk(z; z') = - (2εm) δ(z-z') !Syntax Error, Idk k gmk(0; z') Jm(kρ) Jm(kρ') = 0 ρ > a !Syntax Error, Idk k [∂zgmk(0+; z') – ∂zgmk(0+; z') ] Jm(kρ) Jm(kρ') = 0 ρ < a I don't think this is a well-defined system, but it might be. What happens now if we apply !Syntax Error, Idρ' ρ' Jm(k'ρ') to these two boundary conditions? In each case, we think of ρ as fixed somewhere in its appropriate range, and we then get !Syntax Error, Idk k gmk(0; z') Jm(kρ) !Syntax Error, Idρ' ρ' Jm(k'ρ')Jm(kρ') = 0 ρ > a !Syntax Error, Idk k [∂zgmk(0+; z') – ∂zgmk(0+; z') ] Jm(kρ) !Syntax Error, Idρ' ρ' Jm(k'ρ') Jm(kρ') = 0 ρ < a or !Syntax Error, Idk k gmk(0; z') Jm(kρ) δ(k-k')/k = 0 ρ > a !Syntax Error, Idk k [∂zgmk(0+; z') – ∂zgmk(0+; z') ] Jm(kρ) δ(k-k')/k = 0 ρ < a or gmk'(0; z') Jm(k'ρ) = 0 ρ > a [∂zgmk'(0+; z') – ∂zgmk'(0+; z') ] Jm(k'ρ) ρ < a which seems to imply that gmk(0; z') = 0 [∂zgmk'(0+; z') – ∂zgmk'(0+; z') ] = 0 so maybe we just have this system as our result: { z2 - k2} gmk(z; z') = - (2εm) δ(z-z') gmk(0; z') = 0 [∂zgmk'(0+; z') – ∂zgmk'(0+; z') ] = 0 limz→±∞ gmk(z; z') = 0 But this is an overdetermined system. Instead of the two homo endpoint conditions, we have two BC's at z = 0, one at z = +∞ and one at z = - ∞. There is no solution to this system, so I guess the form must be no good, but I don't really know what is going wrong here, I cannot point to the step that is bad. (b) Plan B: try expanding on cos[k(z-z')]. Here is where we were: [ρz2 - (m2/ρ2)] gm(ρ,z; ρ',z') = -(2εm) δ(z-z') δ(ρ-ρ')/ρ gm(z=0) = 0 for ρ > a ∂zgm(z=0+) = ∂zgm(z=0-) for ρ < a limz→±∞ gm(z) = 0 Suppose we sort of ignore the last condition and hope it works out and we then try this next step: gm(ρ,z; ρ',z') = !Syntax Error, Idk gmk(ρ;ρ') cos[k(z-z')] The justification here would be based on the completeness of the z functions, !Syntax Error, I dz cos[(k-k')z] = 2π δ(k-k') // orthogonality !Syntax Error, I dk cos[k(z-z')] = 2π δ(z-z') // completeness The form is symmetric in z ↔ z'. Our first boundary conditions would then be gm(ρ,0; ρ',z') = !Syntax Error, Idk gmk(ρ;ρ') cos[kz'] = 0 ρ > a To get the other BC, we would say ∂zgm(ρ,z; ρ',z') = -!Syntax Error, Idk k gmk(ρ;ρ') sin[k(z-z')] ∂zgm(ρ,0±; ρ',z') = -!Syntax Error, Idk k gmk(ρ;ρ') sin[kz'] The second BC would be met no matter what since 0± gives the same result. So we then have only that first BE which says: !Syntax Error, Idk gmk(ρ;ρ') cos[kz'] = 0 ρ > a // and = unknown for ρ < a So OK, maybe we can live with this "integral boundary condition". Let's now try to continue forward. gm(ρ,z; ρ',z') = !Syntax Error, Idk gmk(ρ;ρ') cos[k(z-z')] We then need to compute [ρz2 - (m2/ρ2)] gm(ρ,z; ρ',z') = [z2 +ρ2 - (m2/ρ2)] gm(ρ,z; ρ',z') = !Syntax Error, Idk { gmk(ρ;ρ') z2 cos[k(z-z')] + cos[k(z-z')] [ρ2 - (m2/ρ2)] gmk(ρ;ρ')} = !Syntax Error, Idk { gmk(ρ;ρ') (-k2) cos[k(z-z')] + cos[k(z-z')] [ρ2 - (m2/ρ2)] gmk(ρ;ρ')} = !Syntax Error, Idk cos[k(z-z')] [ρ2 - (m2/ρ2) - k2] gmk(ρ;ρ') Our starting position was this: [ρz2 - (m2/ρ2)] gm(ρ,z; ρ',z') = -(2εm) δ(z-z') δ(ρ-ρ')/ρ so we replace δ(z-z') = (1/2π) !Syntax Error, I dk cos[k(z-z')] and we end up with !Syntax Error, Idk cos[k(z-z')] [ρ2 - (m2/ρ2) - k2] gmk(ρ;ρ') = !Syntax Error, I dk cos[k(z-z')] (1/2π) -(2εm) δ(ρ-ρ')/ρ We then invoke our completeness of the cos[k(z-z')] to conclude that [ρ2 - (m2/ρ2) - k2] gmk(ρ;ρ') = -(εm/π) δ(ρ-ρ')/ρ We then end up with this system: [ρ2 - (m2/ρ2) - k2] gmk(ρ;ρ') = -(εm/π) δ(ρ-ρ')/ρ !Syntax Error, Idk gmk(ρ;ρ') cos[kz'] = 0 ρ > a // and = unknown for ρ < a But this is another of our "ill-defined looking" systems. The solution of the homo ODE is this: Amk I(kρ) + Bmk K(kρ) We will get some solution that looks like this: gmk(ρ;ρ') = cmk Im(kρ<) Km(kρ>) where cmk is fully determined. As usual, our pillbox integration tells us that ∂ρgmk(ρ;ρ')+ – ∂ρgmk(ρ;ρ')- = -(εm/π)/ρ' ∂ρgmk(ρ;ρ')+ = ∂ρ { cmk Im(kρ<) Km(kρ>)} = ∂ρ { cmk Im(kρ') Km(kρ)} = cmk Im(kρ') k Km'(kρ') ∂ρgmk(ρ;ρ')- = ∂ρ { cmk Im(kρ<) Km(kρ>)} = ∂ρ { cmk Im(kρ) Km(kρ')} = cmk Km(kρ') k Im'(kρ') ∂ρgmk(ρ;ρ')+ – ∂ρgmk(ρ;ρ')- = k cmk W[ Im(kρ'), Km(kρ') = k cmk (-1/[kρ']) = - cmk/ρ' Therefore we find that cmk = (εm/π) and gmk(ρ;ρ') = (εm/π) Im(kρ<) Km(kρ>) Now what do I have to say about our "boundary condition" !Syntax Error, Idk gmk(ρ;ρ') cos[kz'] = 0 !Syntax Error, Idk Im(kρ<) Km(kρ>)cos[kz'] = 0 ρ > a // and = unknown for ρ < a But this is an integral I ran into doing the Iris Attempt #3, !Syntax Error, Idk cos[kd] Im(kc) Km(kρ) = (1/2) 1/ Qm-1/2 [(ρ2+ c2+ d2)/(2cρ)] Re(ρ) > |Re(c)| d>0 Re(m) > -1/2 which I translate to be !Syntax Error, Idk cos[kz'] Im(kρ') Km(kρ) = (1/2) 1/ Qm-1/2 [(ρ2+ ρ'2+ z'2)/(2ρ'ρ)] Re(ρ) > |Re(ρ')| z'>0 Re(m) > -1/2 So this result seems valid for ρ > ρ' . So I can certainly find a value of ρ which is larger than both a and ρ'. This integral is NOT zero, so our BC is NOT going to be met. So again, things just "don't work". Some earlier assumption is wrong. (c) Conclusions: The problem might have to do with Stakgold type limits where you have order interchange, as we saw in his boundary layers section. We do have some ∂z action going on here, after all. But just in general, this "method" seems to have trouble when you have boundary conditions that involve two of the three variables. In this problem the BC's involve both ρ and z. 2g(ρ,φ,z; ρ',φ',z') = - 4π δ(z-z') δ(φ-φ') δ(ρ-ρ')/ρ g(z=0) = 0 for ρ > a ∂zg(z=0+) = ∂zg(z=0-) for ρ < a I guess I can say that I have done "due diligence" in that I have at least tried to find the iris Green's function by the "partial eigenfunction expansion method" of Stakgold.