stakgold chap 7 exercise 7_49

Stakgold Exercise 7.49 (p 327) PhL 12.31.11 This is the 3D problem of a point charge off the edge of a half plane. Charge is at x = -a as usual. Helmholtz of course is this, where right off the bat he uses λ = -k2, (-∂x2-∂y2-∂z2+k2)v = δ(x+a)δ(y)δ(z) (*) We are Dirichlet BC so u = 0 on the half plane. This half plane emerges from the plane of paper with z pointing to the viewer, so the two sides of the half plane are at y = ±ε . That is to say, y is up in our little 2D projected picture with the half plane appearing as a half line on the right x ≥ 0. I therefore agree with all parts of p 327C for the problem setup. We are instructed to do a z FT with α conjugate to z. I will use the Stak p 23 FT form v^(x,y; α) = !Syntax Error, Idz eiαz v(x,y,z) v(x,y,z) = (1/2π) !Syntax Error, Idα e-iαz v^(x,y; α) So apply the projection operator !Syntax Error, Idz eiαz to both sides of (*) to get !Syntax Error, Idz eiαz (-∂x2-∂y2-∂z2+k2) v(x,y,z) = !Syntax Error, Idz eiαz δ(x+a)δ(y)δ(z) On the LHS we do double parts with the ∂z2 term so that -∂z2 → - (iα)2 = α2 and we through the actual parts away at z = ± ∞. We then have (-∂x2-∂y2+α2+k2) !Syntax Error, Idz eiαz v(x,y,z) = δ(x+a)δ(y) (-∂x2-∂y2+α2+k2) v^(x,y; α) = δ(x+a)δ(y) (-∂x2-∂y2+κ2) v^(x,y; α) = δ(x+a)δ(y) κ2 = α2+k2 This is a 2D Green's function problem, but what are the BC's? Well v^(x,±ε; α) = !Syntax Error, Idz eiαz v(x,±ε,z) = !Syntax Error, Idz eiαz 0 = 0 where I don't think there are technical problems with the integral. So our 2D Green's problem is: (-∂x2-∂y2+κ2) v^(x,y; α) = δ(x+a)δ(y) κ2 = α2+k2 v^(x,±ε; α) = 0 As Stak points out, this is the same as 7.295 (321) but with κ for k and with v^ for w. This was the setup for the 2D half line problem, or the 3D problem with the full line source. This is in the Weiner-Hopf section of the text I suddenly realize! This in fact is WH example 2, and in ω-space we get to (7.301) which is the ω-space equation we need to "segregate". This required lots of work. Remember we are doing the "integral equation method" with WH. When the dust all settles, we find the current on the wire is 7.308 on p 326. WE can use instead 7.307R if we want the result in k instead of λ. Although Stak did not do the last step, that step would be to insert this current I(ξ) into 7.249 (301) to get us and we would call this v^s(x,y; α). We would then do inverse FT for vs(x,y,z) and then since vi = E3(source at point x = -a) we would know the solution v. That outlines the method. Stak is only asking us here about the current, so we can forget the rest of the outline above. We therefore know this fact: (this is the total current on both sides of half-plane I^(x,0; α) = a1/2 exp(-(k2+α2)1/2(x+a) / [ π x1/2(x+a) ] But in our current problem, this is not the current, it is the z FT of the current, so we now have to do this: I(x,0,z) = (1/2π) !Syntax Error, Idα e-iαz I^(x,0; α) = (1/2π) !Syntax Error, Idα e-iαz a1/2 exp(-(k2+α2)1/2(x+a)) / [ π x1/2(x+a) ] = a1/2 (1/2π2) [ x1/2(x+a) ]-1 !Syntax Error, Idα e-iαz exp[-(k2+α2)1/2(x+a)] We now need this integral I = !Syntax Error, Idα e-iαz exp[-(k2+α2)1/2(x+a)] Let x+a = c so this says I = !Syntax Error, Idα e-iαz exp[-c(α2+k2)1/2] with c > 0. The integrand is even in α, so this integral is twice the Fourier Cosine Transform of the function exp[-c(α2+k2)1/2]. This does not appear in my simple Schaum table. Let's try Bateman IT 1 from which I now quote: So first I rewrite my integral this way I = !Syntax Error, Idx e-ixz exp[-c(x2+k2)1/2] = 2 !Syntax Error, Idx cos(xz) exp[-c(x2+k2)1/2] = 2 g(z) So set β = c and α = k and then we seem to have I = 2 kc (z2+c2)-1/2 K1[ k (z2+c2)1/2] Now I can replace c = x+a to get I = 2 k(x+a) (z2+(x+a)2)-1/2 K1[ k (z2+(x+a)2)1/2] And finally we look at K properties from AS2010 p 251 so that I then have I = - 2 k(x+a) (z2+(x+a)2)-1/2 K0'[ k (z2+(x+a)2)1/2] MY answer is then I(x,0,z) = a1/2 (1/2π2) [ x1/2(x+a) ]-1 I = - a1/2 (1/2π2) [ x1/2(x+a) ]-1 2 k(x+a) (z2+(x+a)2)-1/2 K0'[ k (z2+(x+a)2)1/2] = - (k/π2) (1/) (z2+(x+a)2)-1/2 K0'[ k (z2+(x+a)2)1/2] and this agrees exactly with p 328A, hurray! What is this result saying? as you approach the edge, you get the usual 1/ for any z! But surely the other factors show decay in the z direction. Here is K1 so as |z| increases, the K0' function decays very fast and is helped by the other z factor in this decay. Now what is the DC limit? k = 0, so need small arg behavior of K1 Therefore K1(z) ~ (1/2) (z/2)-1 -K0'[ k (z2+(x+a)2)1/2] ~ (1/2) [(k/2) (z2+(x+a)2)1/2]-1 = (1/2) (2/k) (z2+(x+a)2)-1/2 I(x,0,z) =- (k/π2) (1/) (z2+(x+a)2)-1/2 K0'[ k (z2+(x+a)2)1/2] = (k/π2) (1/) (z2+(x+a)2)-1/2 (1/k) (z2+(x+a)2)-1/2 = (/π2) (1/) (z2+(x+a)2)-1 Now this is I and Stak says that σ = -I. I will verify this momentarily against my own notes, but first let's assume it is true, then my result says σ = - (/π2) (1/) (z2+(x+a)2)-1 and this then is the charge induced on a half plane due to a point charge of +1 sitting off at x = -a. So I have exposed another Stak typo!! He has k in the numerator but of course k = 0 in our limit so that should not be there. Was this corrected in Stak's later edition? No it was not! I just updated my errata doc with the last two errors I found ( p 312 and 328, both not fixed in 2000) and added a comment about our email exchange. Regarding I, this is introduced by Stak on page 301. From my memory of Ch6 and more directly from the minus sign in 7.249 if is clear that I → -σinduced in the DC limit! So we have this result: σ(x,z) = - (/π2) (1/) (z2+(x+a)2)-1 The following we know must be true !Syntax Error, Idz !Syntax Error, Idx (/π2) (1/) (z2+(x+a)2)-1 = +1 Stak asks us to do the integral and verify this fact. So we need to show that K ≡ !Syntax Error, Idx (1/) !Syntax Error, Idz (z2+(x+a)2)-1 = π2/ where I will try to do the z integral first. Let (x+a)2 = b2 so we write K = !Syntax Error, Idx (1/) I I = !Syntax Error, Idz 1/ (z2+b2) = !Syntax Error, Idz 1/ [(z+ib)(z-ib)] and I will close up and assume b > 0 to get = 2πi Res{1/ [(z+ib)(z-ib)}z=+ib = 2πi {1/ [(ib+ib) } = 2πi/2ib = π/b which agrees with Maple. So we are then faced with K = !Syntax Error, Idx (1/) π/(x+a) = π !Syntax Error, Idx x-1/2 / (x+a) For a > 0, Maple says this integral is π/ ( we do not run through the pole), so then K = π π/ = π2/ and that is what we wanted to show! Summary: This is probably a fairly hard 3D problem. In the above method, we take advantage of our Weiner-Hopf solution of the line-charge problem (really a 2D problem) and with a z Fourier Transform, we are able to come up with the solution current on the half plane for the full 3D problem. This problem got me back into serveral things of Olde: Bateman ET lookup Maple integrals AS2010 lookups on function properties residue integrals use of the FT