disk green attempt 3

Disk Green Attempt #3 PhL 4.22.10 This is another cos(kz) attempt. 1. Preliminaries concerning Region 1. 1 2. A Smythian three-region form 2 3. The pillbox analysis at ρ = b: determination of bm(k) . 3 4. How do we find the remaining cm(k) coefficients? 4 5. The "pillbox" analysis at the 1/2 region boundary: 5 (5a) stay away from z = 0, it tries to make cm(k) = 0 5 (5b) work at z = 0, it tries to make cm(k) = – (qεm/π2ε)/ Km(ka) 7 Summary of Section 5: 10 6. Try to obtain a dual integral equation. 10 1. Preliminaries concerning Region 1. First we have a picture showing the geometry and the three regions: One idea for a region 1 expansion is this: V1(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz)Am(k) Im(kρ) (1.1) A requirement on this "form" is that the potential vanish on the disk, so we must have Σm=0∞ cos(mφ) !Syntax Error, Idk Am(k) Im(kρ) = 0 which in each partial wave becomes !Syntax Error, Idk Am(k) Im(kρ) = 0 0 ≤ ρ ≤ a (1.2) This looks a little like Jackson p 91 3.173, so we don't give up at once. We selected Im go get finite behavior at ρ = 0. It seems at least possible that as z increases, cos(kz) moves faster and chops up the integral causing V1 → 0. I think this is a viable Smythian form for the central cylinder region. The only other choice would be this: ( I throw in a factor of k for reasons seen below, just definition of B) U1(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk k e-k|z|Bm(k) Jm(kρ) (1.3) and as with the above, this would lead to the requirement that !Syntax Error, Idk k Bm(k) Jm(kρ) = 0 0 ≤ ρ ≤ a (*) (1.4) I cannot think of any other possible "forms" for this region 1. The only other choice would be that you cannot represent the potential in cylindrical atoms, as in the spherical case, but I think you can. The reason is that, unlike powers rn, the functions Jm(kρ) may form an appropriate orthogonal set. We would like to use this Hankel Transform: !Syntax Error, Idρ ρJm(k'ρ) Jm(kρ) = δ(k-k')/k orthogonality !Syntax Error, Idk kJm(kρ') Jm(kρ) = δ(ρ-ρ')/ρ completeness (1.5) We would like to take (*) above and apply !Syntax Error, Idρ ρJm(k'ρ) to both sides. IF (*) were valid on ρ in (0,∞), this is what would happen: 0 = !Syntax Error, Idρ ρJm(k'ρ) !Syntax Error, Idk k Bm(k) Jm(kρ) = !Syntax Error, Idk k Bm(k) δ(k-k')/k = Bm(k') and we would conclude that our coefficient was 0. Although perhaps bad news, at least this is a well defined answer. Our problem of course is that we DON'T have (*) being true for all ρ. The upshot is that equation (*) could be true, but we don't have enough information to be able to invert it to find Bm(k). So we have to regard (*) as a condition on Bm(k), a boundary condition perhaps, but (*) does not fully determine Bm(k). We cannot solve this problem, then, without information on what is happening in the other regions, I think that is the point here. 2. A Smythian three-region form I argue elsewhere for the following form, using the cos(kz) model as in (1.1) above, V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka)Km(kb) Im(kρ) + cm(k) Km(ka)Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { bm(k) Im(ka) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } (2.1) Although these things look a little complicated, it is really just this, V1(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz)Am(k) Im(kρ) V2(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) [ Bm(k) Im(kρ) + Cm(k) Km(kρ)] V3(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz) Dm(k) Km(kρ) (2.2) but we have built in the requirements that the potentials match at the 1/2 and 2/3 cylinder boundaries. This matching is "manifestly clear" looking at the first three lines above. The 1/2 boundary is ρ = a, and the 2/3 boundary is ρ = b. So this form really is Smythian in its nature. The forms (2.2) are forced on us by the requirements on the Bessel functions at ρ = 0 and ρ = ∞. 3. The pillbox analysis at ρ = b: determination of bm(k) . I did all the details in my "other doc". I set ∂ρV2 - ∂ρV3 = (q/ε)(1/b) δ(φ)δ(z) and compute the LHS using the forms above. The result is this: (q/ε)δ(φ)δ(z) = Σm=0∞Cmφ!Syntax Error, Idk Ckz bm(k) Im(ka) I then apply !Syntax Error, Idφ Cm'φ to both sides and this gives: (q/ε)δ(z) = 2π/(2-δm,0) !Syntax Error, Idk cos(kz) bm(k) Im(ka) εm = (2-δm,0) At this point, I would like to use the Fourier Cosine Transform outlined in "other doc". This analysis starts off dealing with z in range (-∞,∞) but then we restrict to functions f(z) which are even in z and we can then fold the integral. The completeness relation is written this way: !Syntax Error, Idk cos(kz) cos(kz') = (π/2)δ(z-z') // completeness and if we set z' = 0 this says !Syntax Error, Idk cos(kz) = (π/2)δ(z) So, let us use this expansion on the left of the above to get (q/ε) (2/π) !Syntax Error, Idk cos(kz) = 2π/(2-δm,0) !Syntax Error, Idk cos(kz) bm(k) Im(ka) We would argue that each "k spectral component" must match, so we get (q/ε) (2/π) = 2π/(2-δm,0) bm(k) Im(ka) = (2π/εm) bm(k) Im(ka) bm(k) = (q/ε) (2/π) (εm/2π) / Im(ka) = (qεm/π2ε)/ Im(ka) Therefore, it does seem that our "usual pillbox analysis" as used in Smythian form work has yielded up one of our two coefficients. The other coefficient could still be anything because it cancels out in this pillbox analysis, a standard fact we have seen many times before. 4. How do we find the remaining cm(k) coefficients? Our Smythian forms with the bm(k) installed now look like this (no red or blue this time) V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ) } V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } We should point out that these forms have "manifest" matching of the potentials at the 1/2 and 2/3 boundaries. But what about matching the electric fields at these boundaries? We know that anywhere on these two cylinder boundaries, at least away from the grounded disk, the E field has to be continuous at the boundary in all its components, since there is no charge floating around out there. One would think this requirement would have some significant effect on the above situation. And we also have our condition for potential on the plate, which is this: V1(0,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk {εm (q/π2ε) Km(kb) + cm(k) Km(ka) Km(kb)} Im(kρ) = 0 ρ≤a where we now see that {εm (q/π2ε) Km(kb) + cm(k) Km(ka) Km(kb)} = Am(k) of our opening discussion above. We can write (1.2) in this way: !Syntax Error, Idk {εm (q/π2ε) Km(kb) + cm(k) Km(ka) Km(kb)} Im(kρ) = 0 (*) As already noted in our first section, this condition is not enough to determine in this case cm(k), but it is a condition we have to make sure is satisfied. So, what about the electric field matching stuff? We can think of the E field as having components related to ∂φV, ∂zV and ∂ρV with some scale factors involved. If we just stare at the forms above, and in particular, at the Cmφ and Ckz factors, we can see what will happen if we require continuity of ∂φV or ∂zV: the condition of continuity will be the same as the condition for V to be continuous! Thus, continuity of the E field in these two directions is already built in. So we only need worry about ∂ρV. But at the 2/3 boundary, we already "worried about ∂ρV" when we did our pillbox analysis. So what remains is the 1/2 boundary. We must take care, because the metal disk rim lies on this boundary, and we expect this rim to have perhaps an infinite charge density. Our hope then is that we can study this 1/2 ∂ρV matching condition and maybe it will yield a condition which, combined with (*) above, will give us perhaps some "dual integral equations" which then will fully determine the cm(k) , sort of in analogy with Jackson's charge disk problem. So let's look into the 1/2 matching condition for ∂ρV . 5. The "pillbox" analysis at the 1/2 region boundary: (5a) stay away from z = 0, it tries to make cm(k) = 0 As long as we stay away from z=0 where the disk is located, we expect this to be true: ∂ρV1 - ∂ρV2 = 0 z ≠ 0 at ρ = a From above we have V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } ∂ρV1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { (qεm/π2ε) Km(kb) Im'(kρ) + cm(k) Km(ka) Km(kb) Im'(kρ)} ∂ρV2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { (qεm/π2ε) Km(kb) Im'(kρ) + cm(k) Km'(kρ) Km(kb) Im(ka) } ∂ρV1(z,a,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km(ka) Km(kb) Im'(ka)} ∂ρV2(z,a,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km'(ka) Km(kb) Im(ka) } So we want to set the above two lines equal, at least for z ≠ 0. Certainly the Cmφ functions are orthogonal so we can get it quickly to this point: !Syntax Error, Idk k Ckz { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km(ka) Km(kb) Im'(ka)} = !Syntax Error, Idk k Ckz { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km'(ka) Km(kb) Im(ka) } (**) At this point, we are tempted to use our Fourier Cosine Transform, f(z) = !Syntax Error, Idk cos(kz) G(k) // expansion G(k) = !Syntax Error, Idz cos(kz) f(z) // projection !Syntax Error, Idz cos(kz) cos(k'z) = (π/2) δ(k-k') // orthogonality !Syntax Error, Idk cos(kz) cos(kz') = (π/2)δ(z-z') // completeness but I am worried that this is not valid because our f(z) is singular at z = 0. Aside on this issue: What can I say about ∂ρV(0,ρ,φ) as we approach the disk edge from region 2 ? If I sit just outside the disk edge in the z = 0 plane, I certainly expect to see some radial electric field component. In the charged disk problem, the potential in the z = 0 plane is given by V(ρ) ~ sin-1(1/ρ) in region 2 if a=1 so ∂ρV(ρ) ~ - 1/[ρ sqrt(ρ2-1)] As we approach ρ = 1 from the outside, we do in fact get ∂ρV(ρ) = ∞ so Eρ = ∞ just outside the surface. I expect this might happen as well in our current problem. The point is that I expect ∂ρV2(z,ρ,φ) to be singular, but only at the location ρ = a and z = 0. This is very reminiscent of Stakgold's surface layers discussion. Meanwhile, it seems pretty clear that ∂ρV(0,ρ,φ) = 0 if we approach the disk edge from the inside, since this is in fact true for all ρ ≤ a. Since ∂ρV2(z,ρ,φ) is singular at z = 0 which makes up part of my 1/2 boundary, it seems likely that this singularity will find its way into some of not all the partial waves (maybe only m = 0 as in the charged disk problem). Therefore, when I end up with the condition f(z) = !Syntax Error, Idk k cos(kz) {stuff(k)} = 0 z ≠ 0 I am hesitant to conclude that stuff(k) = 0. For the moment, let's ignore this "cause for concern" and use our orthogonality anyway to conclude the next step: { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km(ka) Km(kb) Im'(ka)} = { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km'(ka) Km(kb) Im(ka) } The first terms cancel so this says cm(k) Km(ka) Km(kb) Im'(ka) = cm(k) Km'(ka) Km(kb) Im(ka) cm(k) Km(ka) Im'(ka) = cm(k) Km'(ka) Im(ka) cm(k) W[Km(ka), Im(ka) ] = 0 From our "other doc" we have W[ Km(ka), Im(ka)] = 1/(ka) so we get cm(k)/(ka) = 0 This is trying to fool us into accepting the solution cm(k) = 0 which of course makes ∂ρV1 - ∂ρV2 = 0, but we know this is wrong for the following reason. If cm(k) = 0, then V1(z,ρ,φ) = V2(z,ρ,φ) identically everywhere. In other words, we would have: V12(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) } This says that the expression on the right gives the potential for 0 ≤ ρ < b . If we set z = 0 we get V12(0,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk { (qεm/π2ε) Km(kb) Im(kρ) } We know this has to be 0 for ρ ≤ a, so we have 0 = Σm=0∞Cmφ!Syntax Error, Idk { (qεm/π2ε) Km(kb) Im(kρ) } ρ ≤ a Orthogonality of the Cmφ then tells us that !Syntax Error, Idk Km(kb) Im(kρ) = 0 ρ ≤ a (*) But how can this be true where b is a free parameter with b > a ? I think in fact this integral diverges at the high end. From Jackson page 75 we have Km(kb) → (kb)-1/2 exp(-kb) Im(kρ) → (kρ)-1/2 exp(+kρ) Km(kb) Im(kρ) → (1/k) exp[-k(b-ρ)] → 0 so I was wrong, the integral does converge. I cannot "disprove" (*) above because this integral does not appear in GR. Well, here is an easy disproof. A&S page 374 plot Io and Ko and both are positive for all real positive argument. They state this to be true for both Im and Km for any m > -1. Therefore, my integrand above is always positive, so how can the integral of a positive definite integrand be 0! QED. So this confirms that the cm(k)= 0 solution is incorrect, as suspected. (5b) work at z = 0, it tries to make cm(k) = – (qεm/π2ε)/ Km(ka) The ρ-radial E field will be 0 along the disk as you approach the edge (otherwise current would flow). So ∂ρV = 0 as you approach the edge at z = 0 from the inside, which is to say, in region 1. What does this fact tell us in terms of our form in region 1? V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} ∂ρV1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { (qεm/π2ε) Km(kb) Im'(kρ) + cm(k) Km(ka) Km(kb) Im'(kρ)} ∂ρV1(z,a,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km(ka) Km(kb) Im'(ka)} ∂ρV1(0,a,φ) = Σm=0∞Cmφ!Syntax Error, Idk k { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km(ka) Km(kb) Im'(ka)} Our "fact" is that ∂ρV1(0,a,φ) = 0. So we have: Σm=0∞Cmφ!Syntax Error, Idk k { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km(ka) Km(kb) Im'(ka)} = 0 As usual, we can apply the φ orthogonality and get !Syntax Error, Idk k { (qεm/π2ε) Km(kb) Im'(ka) + cm(k) Km(ka) Km(kb) Im'(ka)} = 0 !Syntax Error, Idk k { (qεm/π2ε) + cm(k) Km(ka) } Km(kb) Im'(ka) = 0 This is a new condition I have not seen before. One way to make it be true is this: { (qεm/π2ε) + cm(k) Km(ka) } = 0 cm(k) = – (qεm/π2ε)/ Km(ka) But I am suspicious of this solution as well. If we install this into our region 1 potential we get V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) - (qεm/π2ε) Km(kb) Im(kρ)} ≡ 0 But we know that the potential does not vanish everywhere in region 1, so this is a bad "way" to make the above condition be true. This is the same "solution" for cm(k) that makes Am(k) ≡ 0 and we say this same problem arise in our preliminary section above. So we seek some OTHER solution cm(k) of our conditions: !Syntax Error, Idk {εm (q/π2ε) Km(kb) + cm(k) Km(ka) Km(kb)} Im(kρ) = 0 // V=0 on disk !Syntax Error, Idk k { (qεm/π2ε) + cm(k) Km(ka) } Km(kb) Im'(ka) = 0 // ∂ρV(ρ=a-) = 0 which we can rewrite as !Syntax Error, Idk Km(kb){ (q εm /π2ε) + cm(k) Km(ka) } Im(kρ) = 0 // V=0 on disk ρ ≤ a !Syntax Error, Idk Km(kb) { (qεm/π2ε) + cm(k) Km(ka) } k Im'(ka) = 0 // ∂ρV(ρ=a-) = 0 Notice that the same {} factor appears on both lines. Recall now our earlier definition: Km(kb){ (q εm /π2ε) + cm(k) Km(ka) } = Am(k) so our two conditions are really these: !Syntax Error, Idk Am(k) Im(kρ) = 0 // V=0 on disk !Syntax Error, Idk Am(k) k Im'(ka) = 0 // ∂ρV(ρ=a-) = 0 I think I can make a more general statement of the second line here, but I need to back up and redo things: The radial E field is not just 0 at the edge of the disk, it is zero for all ρ ≤ a in region 1. So we have V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ)} ∂ρV1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk k Ckz { (qεm/π2ε) Km(kb) Im'(kρ) + cm(k) Km(ka) Km(kb) Im'(kρ)} ∂ρV1(0,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk k { (qεm/π2ε) Km(kb) Im'(kρ) + cm(k) Km(ka) Km(kb) Im'(kρ)} So our statement really is this: Σm=0∞Cmφ!Syntax Error, Idk k { (qεm/π2ε) Km(kb) Im'(kρ) + cm(k) Km(ka) Km(kb) Im'(kρ)} = 0 ρ ≤ a which becomes !Syntax Error, Idk Km(kb) { (qεm/π2ε) + cm(k) Km(ka) } k Im'(kρ) = 0 ρ ≤ a !Syntax Error, Idk Am(k) k Im'(kρ) = 0 ρ ≤ a So now we can restate our two conditions: !Syntax Error, Idk Am(k) Im(kρ) = 0 / V=0 on disk ρ ≤ a !Syntax Error, Idk Am(k) k Im'(kρ) = 0 // ∂ρV = 0 on disk ρ ≤ a These conditions bear at least some resemblance to Jackson 3.173 p 91, but they are on the same range of ρ, a big difference. So these two integral equations do not fall into the "dual integral equation" class as discussed in Bateman at the end of the Bessel section. Suppose we go way back and just look again at our region 1 all by itself: V1(z,ρ,φ) = Σm=0∞ cos(mφ) !Syntax Error, Idk cos(kz)Am(k) Im(kρ) (1.1) Once we postulate this "form" for region 1, we arrive immediately at our two conditions above, there is no need to think about any other "regions". Imagine these "vectors", where I suppress the bystander label m: Ak = Am(k) Bk = Im(kρ) Ck = k Im'(kρ) Then we have our two conditions as A B = 0 A C = 0 We are saying that, in our infinite dimensional space of k, the vector A is perpendicular to both B and C. But I would think there would be lots of vectors A that satisfy these conditions. If N = 3, then maybe only one solution, but not if N = 4. Summary of Section 5: I am unable to come up with a believable result for coefficient cm(k). If I look at the 1/2 boundary away from z = 0, and I do "bad math", exact matching on the cylinder could be achieved with cm(k) = 0. But this says that regions 1 and 1 have the exact same "form" and this quickly leads to a contradiction, a positive integral being equal to 0. On the other hand, if I look at z = 0 and do another form of "bad math", I say that Am(k) = 0 ( by saying cm(k) = – (qεm/π2ε)/ Km(ka) ) and this then says that the potential is identically 0 in all of region 1, another nonsense result. Doing "good math", I am able to write two integral conditions for Am(k), but I don't think these are sufficient to determine Am(k), and they do not form a dual integral equation. I note in passing that in all my Smythe problems, I never separated out the potential of the induced charge from that of the point charge. They were always treated together, as I have done here. 6. Try to obtain a dual integral equation. First, here is some quotation from our attempt #1 doc: "If we lower our point r onto the z = 0 plane outside the disk, we know that the potential will have this shape as we go up and down in z: We don't know it is cupping down as I have drawn it, but we know that (1) V(z=0) = finite; (2) V(z) is symmetrical about z = 0. Therefore, we know that ∂zV = 0 at z = 0, just as in Jackson's problem. But on the z = 0 plane, the quantity ∂zV is proportional to the charge density there which is of course zero (except near the Green's charge). So just as in Jackson's problem, we have these conditions: V(ρ,φ,0) = V1 ρ ≤ a ∂zV(ρ,φ,0) = 0 ρ > a but away from the point charge " The idea here is that we know from symmetry alone that ∂zV = 0 in the z = 0 plane outside the disk, except for the Green's charge, and this is very similar to what Jackson did. Later we might move the Green's charge off the z = 0 plane just to make this claim stronger. So in our triple model, we have to say something like this: V1(0,ρ,φ) = 0 ρ ≤ a Dirichlet ∂zV2(0,ρ,φ) = 0 a < ρ ≤ b Neumann ∂zV3(0,ρ,φ) = 0 b < ρ < ∞ Neumann, except for the Green's charge With our forms above, where does this take us? V1(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) + cm(k) Km(ka) Km(kb) Im(kρ) } V2(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kb) Im(kρ) + cm(k) Km(kρ) Km(kb) Im(ka) } V3(z,ρ,φ) = Σm=0∞Cmφ!Syntax Error, Idk Ckz { (qεm/π2ε) Km(kρ) Im(kb) + cm(k) Km(kρ) Km(kb) Im(ka) } We know that ∂zCkz= -kSkz so we find that ∂zV2(0,ρ,φ) = 0 automatically as part of our form, and the same for V3(z,ρ,φ). That is to say, Skz = 0 at z = 0. So we don't GET any "second" dual integral equation. This fact might be pointing out the "flaw" in our taking a cos(kz) approach in the first place.