stakgold chap 7 exercise 7_31,32

Exercises 7.31 Page 290, Second Chapter 7 Midterm Exam 8.2.11 It took 15 pages to do the first problem, so it went into a separate document rather than try to pack this into the raw3 notes. Exercise 7.31: An expansion for exp(iR)/4πR for r0 on the z axis (290) 1 Obtain the 1D Green's equation in r. 1 The Legendre Equation in θ. 5 Obtain the 1D Green's equation in r by a different method. 6 Compute the 1D Green's function for the ODE in variable r. 7 Compute the jump constant C 9 Take the λ→0 limit: 12 Write in the other form valid when λ= -k2. 13 Take the r0→ 0 limit of the first result 14 Exercise 7.32 Ring source expansion (290) 8.2.11 15 ____________________________________________________________________________________ Overview from Meta Notes: Exercise 7.31 ( p 290). This problem (and the next) are solved in a separate doc in the Ch 7 support section. We consider here the 3D Helmholtz Green's PDE with a point source on the + z axis, so there is φ azisym. Our task is to solve the PDE for u(r,θ), so we have a 2-variable problem. We select θ for eigenfunctions which are Pn(cosθ), and in the usual way this leads to a 1D Green's in variable r which we solve in terms of certain J and H Bessel functions, and the resulting u(r,θ) is the RHS of 7.224. But since there were no local BC's in this problem, that RHS must also be E3 which is then the LHS of 7.224. In the usual way we rewrite this equation for λ = -k2 and then the Bessel I and K functions appear in 7.225. Exercise 7.32 ( p 290). Same problem as above, but replace the z-axis point source with a ring source. It turns out that, when you write the δδ for this ring source, it is exactly the same as the δδ source for the previous problem except now we have δ(θ-θ0) instead of δ(θ). We are still azisym so still an r,θ problem. The exact same solution method then applies and you get the exact same result except where we used to have a 1, we now have Pn(cosθ0). __________________________________________________________________________________ Exercise 7.31: An expansion for exp(iR)/4πR for r0 on the z axis (290) We are given the Helmholtz equation in 3D driven by a point charge on the + z axis. I confirmed the distribution description of this point charge by integrating over a small curved frustum containing such a point. The meaning of δ(θ) is limε→0δ(θ+ε), by which I mean it is a "full" delta, not a half delta. This is always an issue for a δ at a finite range endpoint. Note that this is an azisym situation. We know the fundy solution of this problem from page 55 to be u = E(r|r0) = exp(iR)/4πR where R = |r-r0| This azisym problem involves coordinates r and θ only. A complete set of functions for the θ variable is the Legendre polys Pn(cosθ). So we construct u(r,z) = Σn=0∞ an(r) Pn(z). an(r) = (n+1/2)!Syntax Error, Idz Pn(z) u(r,z) where I use my result in transforms.doc with m=0 (any m is valid). Notice that this applies to θ in the full sphere range (0,π) which is appropriate for a fundy situation. Here is the full transforms.doc quote: g(z) = Σn=|m|∞gnm Pnm(z) // expansion gnm = (1/Knm) ∫dz Pnm(z) g(z) Knm = (n+1/2)-1 f(n,m) // projection Obtain the 1D Green's equation in r. Now the usual step is to apply (n+1/2)!Syntax Error, Idz Pn(z) to each term in our PDE, so we better state that PDE: (-2-λ)u(r,θ) = δ(r-r0)δ(θ)/[2πr2sinθ] // RHS is unit charge on +z axis ! -r-2∂r(r2∂ru) - r-2/sinθ * ∂θ(sinθ∂θu) - λu = δ(r-r0)δ(θ)/[2πr2sinθ] - ∂r(r2∂ru) - 1/sinθ * ∂θ(sinθ∂θu) - λr2u = δ(r-r0)δ(θ)/[2πsinθ] So let's stick to θ coordinates and apply (n+1/2)!Syntax Error, Idθ sinθ Pn(cosθ). There are three terms on the LHS and we will do them one at a time: (n+1/2)!Syntax Error, Idθ sinθ Pn(cosθ) { - ∂r(r2∂ru(r,θ)} = = - ∂r(r2∂r { (n+1/2)!Syntax Error, Idθ sinθ Pn(cosθ) u(r,θ) } = - ∂r(r2∂r { an(r) } // first term (n+1/2)!Syntax Error, Idθ sinθ Pn(cosθ) { - λr2u } = -λ r2an(r) // third term We saved the harder middle term for last: (n+1/2)!Syntax Error, Idθ sinθ Pn(cosθ) { - 1/sinθ * ∂θ(sinθ∂θu)} // middle term = - (n+1/2)!Syntax Error, Idθ Pn(cosθ) ∂θ(sinθ∂θu(r,θ)) ≡ - (n+1/2) qn(r) We are going to do parts integration twice. We need to know certain facts: Pn(cosθ)|θ=π = 1 Pνμ(1) = δ0μ μ=integer Pn(cosθ)|θ=-π = (-1)n Let's just work now with qn(r) ≡ !Syntax Error, Idθ Pn(cosθ) ∂θ(sinθ∂θu(r,θ)) Here is our first parts: = - !Syntax Error, Idθ ∂θPn(cosθ) (sinθ∂θu(r,θ)) + [ Pn(cosθ) (sinθ∂θu(r,θ))] |π-π The parts part involves sin(±π) at each endpoint times finite functions, so vanishes. We know ahead of time in any SL problem like this that the parts will always vanish since we have a self-adjoint L, but we will carry these parts and check them anyway, it does not hurt. So we have = - !Syntax Error, Idθ ∂θPn(cosθ) (sinθ∂θu(r,θ)) = - !Syntax Error, Idθ [ sinθ ∂θPn(cosθ)] ∂θu(r,θ) Now here comes our second parts: = + !Syntax Error, Idθ ∂θ[ sinθ ∂θPn(cosθ)] u(r,θ) - [ sinθ ∂θPn(cosθ) u(r,θ)] |π-π and the parts contributions vanish as before, leaving us with = + !Syntax Error, Idθ ∂θ [ sinθ ∂θPn(cosθ)] u(r,θ) At this point I am sure we need Legendre's equation (1-z2)Pn"(z) - 2zPn'(z) + n(n+1)Pn(z) = 0 z = cosθ dz = -sinθdθ ∂z = ∂/∂z = ∂θ/∂z ∂/∂θ = (-sinθ)-1∂θ Pn'(z) = ∂z Pn(z) = (-sinθ)-1∂θ Pn(z) Pn''(z) = ∂z Pn'(z) = (-sinθ)-1∂θ [(-sinθ)-1∂θ Pn(z)] So in terms of ugly θ, we have sin2θ { (sinθ)-1∂θ [(sinθ)-1∂θ Pn(z)] } + 2cosθ { (sinθ)-1∂θ Pn(z)} + n(n+1)Pn(z) = 0 { (sinθ) ∂θ [(sinθ)-1∂θ Pn(z)] } + 2cosθ { (sinθ)-1∂θ Pn(z)} + n(n+1)Pn(z) = 0 This does not seem to be helping me much! But persist a little longer. The above is T1+T2+T3=0. We have T1 = (sinθ) ∂θ [(sinθ)-1∂θ Pn(z)] = (sinθ) { (sinθ)-1∂θ2 Pn(z) - (sinθ)-2cosθ ∂θ Pn(z) } = ∂θ2 Pn(z) - (sinθ)-1cosθ ∂θ Pn(z) Then we have T1+T2 = ∂θ2 Pn(z) + (sinθ)-1cosθ ∂θ Pn(z) Now consider T4 ≡ ∂θ [ sinθ ∂θPn(cosθ)] = sinθ ∂θ2 Pn(z) + cosθ ∂θPn(z) (sinθ)-1T4 = ∂θ2 Pn(z) + (sinθ)-1cosθ ∂θPn(z) Thus I have proven that (sinθ)-1T4 = T1+T2 = -T3 = - n(n+1)Pn(z) which is to say (sinθ)-1∂θ [ sinθ ∂θPn(cosθ)] = - n(n+1)Pn(z) In retrospect, this just says -LYn0 = -n(n+1)Yn0 as I review below. Now recap what happened earlier. We showed that, after two parts integrations, qn(r) = + !Syntax Error, Idθ ∂θ [ sinθ ∂θPn(cosθ)] u(r,θ) And now we see that we must then have qn(r) = !Syntax Error, Idθ sinθ { (sinθ)-1∂θ [ sinθ ∂θPn(cosθ)]} u(r,θ) = !Syntax Error, Idθ sinθ { - n(n+1)Pn(cosθ)} u(r,θ) = - n(n+1) !Syntax Error, Idθ sinθ Pn(cosθ)} u(r,θ) = - n(n+1) {(n+1/2)-1} an(r) Now our "middle term" above was middle term = - (n+1/2) qn(r) = +n(n+1) an(r) and THAT certainly took a long time to obtain. So we add the three terms and we have for our transformed LHS of our original equation - ∂r(r2∂r { an(r) } + n(n+1) an(r) -λr2 an(r) The transformed RHS is this (n+1/2)!Syntax Error, Idz Pn(z) { δ(r-r0)δ(θ)/[2πsinθ] } = (n+1/2)!Syntax Error, Idθ sinθ Pn(cosθ) { δ(r-r0)δ(θ)/[2πsinθ] } = (n+1/2)!Syntax Error, Idθ Pn(cosθ) { δ(r-r0)δ(θ)/[2π] } = (n+1/2) Pn(1) { δ(r-r0) /[2π] } = { (n+1/2) δ(r-r0) /[2π] } = δ(r-r0) (n+1/2) /[2π] = δ(r-r0) (2n+1)/4π and then here is our transformed equation - ∂r(r2∂r { an(r) } + n(n+1) an(r) -λr2an(r) = δ(r-r0) (2n+1)/4π - (r2an'(r))' + n(n+1) an(r) - λr2an(r) = δ(r-r0) (2n+1)/4π - (r2an'(r))' – [λr2– n(n+1)] an(r) = δ(r-r0) (2n+1)/4π The Legendre Equation in θ. This really was the missing piece of information above. Nobody ever writes the Legendre equation in θ, it is always in z. For example Schaum p 146 which I quoted above, (1-z2)Pn"(z) - 2zPn'(z) + n(n+1)Pn(z) = 0 Rather than convert this from z =cosθ to θ, here is a faster way to get the same answer. From my phys/qm/angmom we know that L2 = (1/S) [ S ] (1/sin)2 2 and that 2 = r2 - r-2L2 We know that L2 Ylm(θ,φ) = l(l+1) Ylm(θ,φ) and in particular we know that L2 Yl0(θ,φ) = l(l+1) Yl0(θ,φ) Regardless of normalization of the Y, this does tell us that L2 Pl(cosθ) = l(l+1) Pl(cosθ) and thus we know that (1/S) ([ S ] Pl(cosθ)) = l(l+1) Pl(cosθ) and so this is the Legendre equation in θ coordinates. Rewrite as - (1/sinθ) ∂θ[ sinθ ∂θPn(cosθ)] = n(n+1) Pn(cosθ) or (1/sinθ) ∂θ[ sinθ ∂θPn(cosθ)] + n(n+1) Pn(cosθ) = 0 Obtain the 1D Green's equation in r by a different method. Go back to the full PDE, - ∂r(r2∂ru) - 1/sinθ * ∂θ(sinθ∂θu) - λr2u = δ(r-r0)δ(θ)/[2πsinθ] Insert the expansion for u u(r,θ) = Σn=0∞ an(r) Pn(cosθ) First term: - ∂r(r2∂ru) = Σn=0∞ Pn(cosθ) {- ∂r(r2∂r an(r))} Third term: - λr2u = Σn=0∞ Pn(cosθ) {- λr2an(r)} Second term: - 1/sinθ * ∂θ(sinθ∂θ{ Σn=0∞ an(r) Pn(cosθ)}) = Σn=0∞ an(r){- 1/sinθ * ∂θ(sinθ∂θ{ Pn(cosθ)})} Apply our Legendre equation in θ as just derived above to get = Σn=0∞ an(r){ n(n+1) Pn(cosθ)} = Σn=0∞ Pn(cosθ) n(n+1) an(r) Thus we can write our LHS in this way, including all three terms LHS = Σn=0∞ Pn(cosθ) { - ∂r(r2∂r an(r)) + n(n+1) an(r) - λr2an(r) } Now what about the RHS, which was this RHS = δ(r-r0)δ(θ)/[2πsinθ] From transforms.doc I quote Σn=|m|∞(1/Knm) Pnm(z') Pnm(z) = δ(z'-z) Knm = (n+1/2)-1 f(n,m) // completeness or Σn=0∞(1/Kn0) Pn(z') Pn(z) = δ(z'-z) or Σn=0∞(n+1/2) Pn(z') Pn(z) = δ(z'-z) Set z=1 and z'=cosθ to get Σn=0∞(n+1/2) Pn(cosθ) Pn(1) = δ(cosθ-1) or Σn=0∞(n+1/2) Pn(cosθ) = δ(θ)/ |sinθ| Since we are concerned only with (0,π), we can remove the ||. Then we have shown that RHS = δ(r-r0)δ(θ)/[2πsinθ] = δ(r-r0)(1/2π) Σn=0∞(n+1/2) Pn(cosθ) = Σn=0∞ Pn(cosθ) { δ(r-r0)(1/2π) (n+1/2)} = Σn=0∞ Pn(cosθ) { δ(r-r0)(1/4π) (2n+1)} Now using the fact that the Pn are a complete set, we set LHS = RHS to get - ∂r(r2∂r an(r)) + n(n+1) an(r) - λr2an(r) = δ(r-r0) (2n+1)/(4π) - (r2an'(r))' – [λr2– n(n+1)] an(r) = δ(r-r0) (2n+1)/4π and this agrees with our result obtained by the previous method. Compute the 1D Green's function for the ODE in variable r. Here is one of our Bessel equation forms which we then process (peeking at Stak's solution) w(x) = Jν(ρx) solves -(xw')' – ρ2xw + ν2 w/x = 0. w(x) = Jn+1/2(ρx) solves -(xw')' – ρ2xw + (n+1/2)2 w/x = 0. w(r) = Jn+1/2(r) solves -(rw')' – λrw + (n+1/2)2 w/r = 0. Our equation however has this form - (r2u')' – [λr2– n(n+1)] u = RHS so I will conjecture this connection: u = r-1/2w u' = r-1/2w' - (1/2)r-3/2 w r2u' = r3/2w' - (1/2)r1/2 w (r2u')' = r3/2w" + (3/2)r1/2 w' - (1/2) r1/2 w' - (1/2)2r-1/2w = r3/2w" + r1/2 w' - (1/4)r-1/2w But we know that (rw')' = rw" + w' so we have shown that (r2u')' = r1/2(rw" + w') - (1/4)r-1/2w or (r2u')' = r1/2(rw')' - (1/4)r-1/2w Therefore, our equation in u becomes - (r2u')' – [λr2– n(n+1)] u = RHS -r1/2(rw')' + (1/4)r-1/2w – [λr2– n(n+1)] r-1/2w = RHS ok -(rw')' + (1/4)r-1w – [λr2– n(n+1)]r-1w = r-1/2 RHS -(rw')' + (1/4)r-1w – [λr– n(n+1)r-1] w = r-1/2 RHS -(rw')' – [λr– {n(n+1)-1/4}r-1]w = r-1/2 RHS -(rw')' - [λr– (n+1/2)2 r-1]w = r-1/2 RHS -(rw')' - λrw + (n+1/2)2 w/r = r-1/2 RHS and, when RHS = 0, this matches our Bessel w equation above. Conclusion: We have this 1D Green's Function equation: - (r2an'(r))' – [λr2– n(n+1)] an(r) = δ(r-r0) (2n+1)/4π where we think of an(r) as our u(r) above. We then define u = r-1/2w an(r) = r-1/2 An(r) and our equation for An(r) is this -(r An ')' + (1/4) An - [λr2– (n+1/2)2] An = r1/2 δ(r-r0) (2n+1)/4π and our homo solutions are then of this form An(r) = Jn+1/2(r) an(r) = r-1/2 Jn+1/2(r) We know from the asymptotic behaviors that an(r) = C [r<-1/2 Jn+1/2(r<)][r>-1/2H(1)n+1/2(r>)] and our only task is to find C. Question: Why am I putting H(1) on the RHS here? What asymptotic behavior am I thinking of? Answer: Although I write all the time, I really have in mind that λ = -k2 where λ is "clean", and the we have in mind = +ik with k>0, so it is then H(1)ν(ikr) that is of interest, and this goes as ei(ikr)= e-kr and that is the asymptotic behavior I had in mind, so this is really the ~Kν(kr) decaying. Compute the jump constant C Let an(r) = u again and we have - (r2u')' – [λr2– n(n+1)] u = δ(r-r0) (2n+1)/4π -r2u" - 2ru' – [λr2– n(n+1)] u(r) = δ(r-r0) (2n+1)/4π We integrate this as usual from r0-ε to r0+ε0. -r2u" - 2ru' – [λr2– n(n+1)] u = δ(r-r0) (2n+1)/4π !Syntax Error, Idr { -r2du'/dr - 2ru' – [λr2– n(n+1)] u } = (2n+1)/4π -r02!Syntax Error, Idr du'/dr - 2r0!Syntax Error, Idr du/dr - [λ r02– n(n+1)] !Syntax Error, Idr u = (2n+1)/4π -r02!Syntax Error, Idu' - 2r0!Syntax Error, Idu - [λ r02– n(n+1)] !Syntax Error, Idr u = (2n+1)/4π -r02!Syntax Error, Idu' - 2r0!Syntax Error, Idu - [λ r02– n(n+1)] u(r0)!Syntax Error, Idr = (2n+1)/4π Since u is continuous, the third integral vanishes as shown, the second integral trivially vanishes, so -r02!Syntax Error, Idu' = (2n+1)/4π -r02[ u'(r)]|+- = (2n+1)/4π jump in u' = [ (2n+1)/4π ] /(-r02) where we recognize our familiar result that jump = 1/a0 but I always do it longhand every time. So u(r) = C [r<-1/2 Jn+1/2(r<)][r>-1/2H(1)n+1/2(r>)] u(r0+ε) = C [r0-1/2 Jn+1/2(r0)][r-1/2H(1)n+1/2(r)] u(r0- ε) = C [r-1/2 Jn+1/2(r)][r0-1/2H(1)n+1/2(r0)] u'(r0+ε) = C [r0-1/2 Jn+1/2(r0)] [r-1/2H(1)n+1/2(r)]' u'(r0- ε) = C [r-1/2 Jn+1/2(r)]' [r0-1/2H(1)n+1/2(r0)] u'(r0+ε) = C [r0-1/2 Jn+1/2(r0)] [r-1/2∂rH(1)n+1/2(r) -(1/2)r-3/2 H(1)n+1/2(r) ] u'(r0- ε) = C [r-1/2∂rJn+1/2(r) -(1/2)r-3/2 J(r) ] [r0-1/2H(1)n+1/2(r0)] Now use ∂rJn+1/2(r) = Jn+1/2'(r) and then we have u'(r0+ε) = C [r0-1/2 Jn+1/2(r0)] [r-1/2H(1)'n+1/2(r) -(1/2)r-3/2 H(1)n+1/2(r) ] u'(r0- ε) = C [r-1/2Jn+1/2'(r) -(1/2)r-3/2 J(r) ] [r0-1/2H(1)n+1/2(r0)] u'(r0+ε) = C/ [ Jn+1/2(r0)] [H(1)'n+1/2(r) -(1/2)r-1 H(1)n+1/2(r) ] u'(r0- ε) = C/ [Jn+1/2'(r) -(1/2)r-1 J(r) ] [H(1)n+1/2(r0)] Now we can take the difference u'(r0+ε)- u'(r0- ε) = C/ * { Jn+1/2(r0) [H(1)'n+1/2(r) - (1/2)r-1 H(1)n+1/2(r) ] H(1)n+1/2(r0) [Jn+1/2'(r) - (1/2)r-1 Jn+1/2(r) ] } But we want this difference evaluated at r = r0 so get [u'(r0+ε)- u'(r0- ε)]|r=r0 = (C/r0) * { Jn+1/2(r0) [H(1)'n+1/2(r0) - (1/2)r0-1 H(1)n+1/2(r0) ] H(1)n+1/2(r0) [Jn+1/2'(r0) - (1/2)r0-1 Jn+1/2(r0) ] } Now the two -(1/2) type terms cancel and we get jump in u' = (C/r0) { Jn+1/2(r0) H(1)'n+1/2(r0) - Jn+1/2'(r0) H(1)n+1/2(r0) } (C/r0) W[Jn+1/2(r0), H(1)n+1/2(r0)] and we go look up in AS2010 and conclude then that jump in u' = (C/r0) 2i/(πr0) = (C) 2i/(πr02) = C (2i/πr02) But from above we had jump in u' = [ (2n+1)/4π ] /(-r02) so comparison says C (2i/πr02) = [ (2n+1)/4π ] /(-r02) C (2i/π) = [ (2n+1)/4π ] /(-1) C (-2i) = [ (2n+1)/4 ] C = i [ (2n+1)/8 ] Then our 1D Green's function is this: an(r) = C [r<-1/2 Jn+1/2(r<)][r>-1/2H(1)n+1/2(r>)] = i [ (2n+1)/[8] Jn+1/2(r<) H(1)n+1/2(r>) Now we install this into our initial expansion to get u(r,z) = Σn=0∞ an(r) Pn(z) = (i/[8]) Σn=0∞ Pn(z) (2n+1) Jn+1/2(r<) H(1)n+1/2(r>) and we have therefore shown that exp(iR)/4πR = (i/[8]) Σn=0∞ Pn(cosθ) (2n+1) Jn+1/2(r<) H(1)n+1/2(r>) and this agrees with Stak's result 7.244 on page 290 ! Remember that this is the special case that r0 lies on the + z axis and r can be anywhere. Aside on Green's Functions. We just showed above that for - (r2an'(r))' – [λr2– n(n+1)] an(r) = δ(r-r0) (2n+1)/4π The Green's function on (a,∞) was this an(r) = i [ (2n+1)/[8] Jn+1/2(r<) H(1)n+1/2(r>) Therefore, if we were given this ODE - (r2g'(r))' – [λr2– n(n+1)] g(r) = δ(r-r0) we could conclude that g = 4π/(2n+1) * an(r) = 4π/(2n+1) *i [ (2n+1)/[8] Jn+1/2(r<) H(1)n+1/2(r>) = (iπ/2) Jn+1/2(r<) H(1)n+1/2(r>) ` Take the λ→0 limit: Get some z→0 limits from AS2010 Then Jn+1/2(r<) → (r</2)n+1/2 / Γ(n+3/2) H(1)n+1/2(r>) → (-i/π)Γ(n+1/2) (r>/2)-(n+1/2) Jn+1/2(r<) H(1)n+1/2(r>) → (-i/π) [Γ(n+1/2)/ Γ(n+3/2)] (r</2)n+1/2(r>/2)-(n+1/2) = (-i/π) [Γ(n+1/2)/ {(n+1/2)Γ(n+1/2)] (r</r>) n+1/2 = (-i/[π(n+1/2)]) (r</r>)n+1/2 Then we get 1/4πR = (i/[8]) Σn=0∞ Pn(cosθ) (2n+1) { (-i/[π(n+1/2)]) (r</r>)n+1/2} = r<-1/2 r>-1/2(i/8) Σn=0∞ Pn(cosθ) (2n+1) { (-i/[π(n+1/2)]) (r</r>)n+1/2} = r<-1/2 r>-1/2(i/8) Σn=0∞ Pn(cosθ) (2n+1) { (-2i/[π(2n+1)]) (r<)n+1/2 (r>)-n-1/2} = (i/8) Σn=0∞ Pn(cosθ) { (-2i/[π]) (r<)n (r>)-n-1} = (i/8) (-2i/[π]) Σn=0∞ Pn(cosθ) { (r<)n (r>)-n-1} = (1/4π) Σn=0∞ Pn(cosθ) { (r<)n (r>)-n-1} 1/R = Σn=0∞ Pn(cosθ) { (r<)n (r>)-n-1} Now from my curvilinear / spherical 1/R doc, I know that 1/R = Σn=0∞ Σm=0∞ εm f(n,-m) (r<)n(r>)-n-1 Pnm(z) Pnm(z0) cos[m(φ-φ0)] (**) when r0 is at (r0, θ0, φ0). If θ0 = 0 so that z0= 1, we know Pnm(1) = δm0 and ε0 = 1 so 1/R = Σn=0∞ (r<)n(r>)-n-1 Pn(z) and this agrees with our limit above. Write in the other form valid when λ= -k2. We are concerned about Jn+1/2(r<) when λ = -k2 and = +ik in which case we have Jn+1/2(ikr<). We call upon AS2010 This tells us that (assume z has positive imaginary part) Iν(z)Kν(z) = e-iπν/2 Jν(iz) (iπ/2)eiπν/2 H(1)(iz) = (iπ/2) Jν(iz) H(1)(iz) So we know that Jν(iz) H(1)(iz) = (2/iπ) Iν(z)Kν(z) and we let z = kr and we get Jν(ikr<) H(1)(ikr>) = (2/iπ) Iν(kr<)Kν(kr>) Then our expansion is exp(iR)/4πR = (i/[8]) Σn=0∞ Pn(cosθ) (2n+1) Jn+1/2(r<) H(1)n+1/2(r>) exp(-kR)/4πR = (i/[8]) Σn=0∞ Pn(cosθ) (2n+1) (2/iπ)In+1/2(kr<) Kn+1/2(kr>) = (i/[8]) (2/iπ) Σn=0∞ Pn(cosθ) (2n+1) (2/iπ)In+1/2(kr<) Kn+1/2(kr>) = (1/[4π]) Σn=0∞ Pn(cosθ) (2n+1) (2/iπ)In+1/2(kr<) Kn+1/2(kr>) and this agrees with (7.225). Take the r0→ 0 limit of the first result We move our point down the z axis to the origin. Let's use the first form exp(iR)/4πR = (i/[8]) Σn=0∞ Pn(cosθ) (2n+1) Jn+1/2(r<) H(1)n+1/2(r>) In this limit, r< is r0 so RHS = (i/[8]) Σn=0∞ Pn(cosθ) (2n+1) Jn+1/2(r0) H(1)n+1/2(r) I quoted the z→0 limit above so Jn+1/2(r0) → (r0/2)n+1/2/Γ(n+3/2) so the claim then is that exp(ir)/4πr = (i/[8]) Σn=0∞ Pn(cosθ) (2n+1) (r0/2)n+1/2/Γ(n+3/2) H(1)n+1/2(r) or exp(ir)/4πr = (i/[8]) Σn=0∞ Pn(cosθ) (2n+1) (r0/2)n+1/2/Γ(n+3/2) H(1)n+1/2(r) in the limit. The RHS has factor (r0)n so only the n=0 term survives and we claim then exp(ir)/4πr = (i/[8]) 1 (1) (r0/2)1/2/Γ(3/2) H(1)1/2(r) = (i/[8]) (/2)1/2/Γ(3/2) H(1)1/2(r) = (i/[8]) (/2)1/2 (2/) H(1)1/2(r) = i 2-3 r-1/2 λ1/4 2-1/2 2 π-1/2 H(1)1/2(r) = i 2-2 r-1/2 λ1/4 2-1/2 π-1/2 H(1)1/2(r) = (i/4) (/2πr)1/2 H(1)1/2(r) But according to p 54 5.118 for n=3 , this is exactly right for E3 with source at origin. Exercise 7.32 Ring source expansion (290) 8.2.11 How do we describe this little ring of charge at r = r0 and θ = θ0? I know how to do a point charge point at (r0,θ0, φ0) = δ(r-r0) = δ(r-r0)/r2 δ(θ-θ0)/sinθ δ(φ-φ0) Integrate this around the ring and get ring at (r0,θ0) = K δ(r-r0)/r2 δ(θ-θ0)/sinθ To get the constant, we want the integral of the above to be 1. So integrate over an upper cone of larger angle than θ0 and set to 1: !Syntax Error, Ir2dr !Syntax Error, Idθ sinθ !Syntax Error, Idφ { K δ(r-r0)/r2 δ(θ-θ0)/sinθ } = 1 !Syntax Error, Idr !Syntax Error, Idθ !Syntax Error, Idφ { K δ(r-r0) δ(θ-θ0) } = 1 K 2π = 1 K = 1/2π Just practicing this stuff. So our desired source is source = (1/2π) δ(r-r0)/r2 δ(θ-θ0)/sinθ But this is exactly the same source as in problem 7.31 but θ0 was 0 there. So we should be able to crib large parts of the previous exercise. In the first section, only the RHS is different under transformation. We get The transformed RHS is this (n+1/2)!Syntax Error, Idz Pn(z) { δ(r-r0) δ(θ-θ0)/[2πsinθ] } = (n+1/2)!Syntax Error, Idθ sinθ Pn(cosθ) { δ(r-r0) δ(θ-θ0)/[2πsinθ] } = (n+1/2)!Syntax Error, Idθ Pn(cosθ) { δ(r-r0) δ(θ-θ0)/[2π] } = (n+1/2) Pn(cosθ0) { δ(r-r0) /[2π] } = { (n+1/2) Pn(cosθ0) δ(r-r0) /[2π] } = δ(r-r0) Pn(cosθ0) (n+1/2) /[2π] = δ(r-r0) Pn(cosθ0) (2n+1)/4π so we just gain a new factor of Pn(cosθ0) on our RHS. Our transformed equation is now - (r2an'(r))' – [λr2– n(n+1)] an(r) = δ(r-r0) (2n+1) Pn(cosθ0)/4π In our "second method" this same factor appears when we write Σn=0∞(n+1/2) Pn(cosθ) Pn(cosθ0) = δ(cosθ-cosθ0) = δ(θ-θ0)/ |sinθ| Our 1D Green's function form is the same, C will be different. From our ODE we will get jump in u' = [ (2n+1) Pn(cosθ0)/4π ] /(-r02) and this will lead to C = i [ (2n+1) Pn(cosθ0)/8 ] and an(r) = i [ (2n+1) Pn(cosθ0)/[8] Jn+1/2(r<) H(1)n+1/2(r>) So our expansion will then be u(r,θ;λ) = (i/[8]) Σn=0∞ Pn(cosθ) Pn(cosθ0) (2n+1) Jn+1/2(r<) H(1)n+1/2(r>) and this shows the explicit symmetry were we to do r ↔ r0. And of course we will also get u(r,θ;λ=-k2) == (1/[4π]) Σn=0∞ Pn(cosθ) Pn(cosθ0) (2n+1) (2/iπ)In+1/2(kr<) Kn+1/2(kr> ) Of course since we have a ring of source, the LHS is no longer the E3 fundy exp(iR)/4πR . It is just the solution to this PDE in 3D: (-2-λ)u(r,θ) = δ(r-r0)δ(θ-θ0)/[2πr2sinθ] with BCs of being finite at r=0 and bounded at r=∞. It is a "fundy like" solution.