stakgold chap 7 exercise 7_36

Stakgold chap 7 exercise 7_36 PhL 8.8.11 We have here a slab that runs from z=0 to h, but is infinite in the x and y directions. On each of the two faces, we want no heat loss, so to speak, so we want ∂zu(x,y,z) = 0 at both z=0 and z=h. We have a driving Helmholtz source in the form of a ring around the z axis, radius r0, and at z = z0. I presume he means that this ring lies inside the slab so that 0 < z0 < h. Notice that the entire "boundary" consists of the two surfaces and perhaps some surface at infinity. This is a 3D Neumann Helmholtstatics problem with azimuthal symmetry. Here is our PDE in cylindrical coordinates: (-2-λ)u(r,z) = δ(r-r0)δ(z-z0)/2πr dim(λ) = L-2 dim(u) = L-1 dim(RHS)=L-3 where I assume the total source charge is 1 unit -- if you integrate the above source over dV = dz rdrdθ you get 1. -(1/r)∂r(r∂ru) - ∂z2u - λu(r,z) = δ(r-r0)δ(z-z0)/2πr -∂r(r∂ru) - r ∂z2u - λru = δ(r-r0)δ(z-z0)/2π ∂zu(r,0) = ∂zu(r,h) = 0 z in (0,h) r in (0,∞) We are going to do this in "both ways" , r eigenfunctions and z eigenfunctions. r means ρ of course. 1. Atoms which appear in the homo solution Our usual first order of business is to separate the homo equation so we can know the nature of solutions away from the distributional source. So try u = R(r)Z(z) : -∂r(r∂ru) - r ∂z2u - λru = 0 -∂r(r∂rRZ) - r ∂z2RZ - λrRZ = 0 dim(R) = dim(Z) = L-1 Z [-∂r(r∂rR)] - rR ∂z2Z - λrRZ = 0 [-∂r(r∂rR)]/R - r ∂z2Z/Z - λr = 0 (1/r)[-∂r(r∂rR)]/R - ∂z2Z/Z - λ = 0 (1/r)[-∂r(r∂rR)]/R = ∂z2Z/Z + λ ≡ ν, our separation constant (1/r)[-∂r(r∂rR)]/R = ν r in (0,∞) ∂z2Z/Z = ν-λ z in (0,h) -∂r(r∂rR)/R = νr r in (0,∞) ∂z2Z/Z = ν-λ z in (0,h) -(rR')' = νrR r in (0,∞) L-2 - L-2 = L-2 so OK Z" = (ν-λ)Z z in (0,h) -(rR')' -νrR = 0 r in (0,∞) Z" = (ν-λ)Z z in (0,h) where ' and " have the obvious derivative meanings. We now focus on the R equation only. We now want to refer to ν as λ in this equation, so we write -(rR')' -λrR = 0 r in (0,∞) λ = ν Z" = (λ-λH)Z z in (0,h) The R equation is of the Hankel form with Bessel order = 0. If we let λ = κ2, solutions are J0(κr). The corresponding transform is the Hankel with order = 0, which is to say Rh(k) = !Syntax Error, Idr r J0(kr) R(r) R(r) = !Syntax Error, Idk k J0(kr) Rh(k) The second just says that the J0(kr) for positive real k form a complete set for functions or r. Question: if we transform the R equation from r-space to k-space, what do we expect to see in k-space? If we define LR = -(rR')' Then our eigenvalue problem is (L-rλ)uλ = 0 or (L-rκ2)uκ or Luκ = rκ2uκ for example LJ0(κr) = rκ2 J0(κr) So we can write our r-equation like this -(rR')' -λrR = 0 LR -λrR = 0 Now let's install a Hankel expansion for R to get L !Syntax Error, Idk k J0(kr) Rh(k) -λr!Syntax Error, Idk k J0(kr) Rh(k) = 0 Move in the L and use LJ0(kr) = rk2 J0(kr) and we then have !Syntax Error, Idk k J0(kr) rk2 Rh(k) = !Syntax Error, Idk k J0(kr) λr Rh(k) = 0 Since the J0(kr) are a complete set, we conclude that rk2 Rh(k) = λr Rh(k) λ = k2 I think I just drove off the road. The radial r equation IS the EV equation. All we know is that we can write the solutions of this homo equation in the form J0(kr) where k = . [ You are still on the road. The J0(κr) ARE the eigenfunctions, you have found them. So this is the TYPE of homo R equation solution away from the ring source. ] Back up. Here where we were: -(rR')' -λrR = 0 r in (0,∞) λ = ν Z" = (λ-λH)Z z in (0,h) We were looking for homo solution forms away from the source. The Z solutions are of the form Z(z) = exp( ± z) So we know a little know about the "atoms" that will appear in the solution to this problem. 2. Pause for Comment on Approach At this point all I have done is write down the atomic forms for the homo Helmholtz equation. But I am not sure that is relevant. It would be if I were doing a "Smythian form" solution attempt. In this method, you would then divide the slab region into perhaps 4 regions: inside and outside the ring, above and below the ring. But I don't want to try a Smythian form solution here, I want to exercise the methods taught in this chapter! So time for a little review. In Ex 7.31 we had a "similar" problem with a point source involving two variables on the RHS. In THAT problem, what did I do? (a) In one method, I noted the homo θ eigenfunctions (Legendre) and applied the Legendre projector. After MUCH fiddling, this led to a 1D Green's equation in r with just a single δ driving that equation. The projector trivially removes the other δ from the full RHS. (b) In a second method, I instead expanded on these same Legendres and ended up with the same 1D Greens. In both these methods, I essentially had u(r,z) = Σn=0∞ an(r) Pn(z) and it was for an(r) that I had a 1D Green's equation. I found that an(r) = C [r<-1/2 Jn+1/2(r<)][r>-1/2H(1)n+1/2(r>)] and I went on to find the constant C, which was then C = i [ (2n+1)/8 ]. I then "put all the pieces together" to get u(r,z) = Σn=0∞ an(r) Pn(z) = Σn=0∞ i [ (2n+1)/8 ] [r<-1/2 Jn+1/2(r<)][r>-1/2H(1)n+1/2(r>)] Pn(z) So basically this is the old "partial EF method" of the Laplace world. Then in Ex 7.32, we had a ring source lying on the cone instead of a point charge on the z axis. In this problem, the LHS is exactly the same as in Ex 7.31, so I applied that same Legendre projector to both sides and just reused the LHS result and got this for the 1D Green's function problem: - (r2an'(r))' – [λr2– n(n+1)] an(r) = δ(r-r0) (2n+1) Pn(cosθ0)/4π Using the same Legendre expansion, the solution was an(r) = i [ (2n+1) Pn(cosθ0)/[8] Jn+1/2(r<) H(1)n+1/2(r>) so u(r,θ;λ) = (i/[8]) Σn=0∞ Pn(cosθ) Pn(cosθ0) (2n+1) Jn+1/2(r<) H(1)n+1/2(r>) So how does that relate to the current problem? Well, the problem is this: (-2-λ)u(r,z) = δ(r-r0)δ(z-z0)/2πr = ring source where now z = rcosθ. Since we are now in cylindrical coordinates, only single r in denominator. So this is going to be a different problem in this sense: since we are no longer in spherical coordinates, we can no longer use the Legendre projector. But that is fine, I see that there are two ways to proceed. Method A. Learn the r-space atoms (they are J0(kr) with the Hankel projector), then apply the Hankel projector to both sides, and you will end up with a z-space 1D Green's problem which you then have to go and solve. Method B. Learn the z-space atoms (they are exp( ± z), but probably now we have to do something like λ = -κ2 so we have a transform in z) , then apply this transform projector to both sides, and you should end up with an r-space 1D Green's problem. Stak in fact asks for a solution by both methods! Method A. I will do expansion instead of projector, either is OK. Now maybe we are supposed to just realize this fact, and go back to the inhomo equation (this is cylindrical coordinates!!) -∂r(r∂ru) - r ∂z2u - λH ru = δ(r-r0)δ(z-z0)/2π Lu - r ∂z2u - λH ru = δ(r-r0)δ(z-z0)/2π Lu = -∂r(r∂ru) Now expand on the known radial homo atoms, u(r,z) = !Syntax Error, Idk J0(kr) U(k,z) Lu = !Syntax Error, Idk LJ0(kr) U(k,z) = !Syntax Error, Idk rk2 J0(kr) U(k,z) Then our PDE becomes !Syntax Error, Idk rk2 J0(kr) U(k,z) - r!Syntax Error, Idk J0(kr) ∂z2U(k,z) - λH r!Syntax Error, Idk J0(kr) U(k,z) = δ(z-z0)/2π * r!Syntax Error, Idk k J0(kr)J0(kr0) ` // δ(r-r0) from "transforms" We cancel all the r's, !Syntax Error, Idk k2 J0(kr) U(k,z) - !Syntax Error, Idk J0(kr) ∂z2U(k,z) - !Syntax Error, Idk J0(kr) λH U(k,z) = !Syntax Error, Idk k J0(kr)J0(kr0) δ(z-z0)/2π Now we appeal to completeness and conclude that k2 U(k,z) -∂z2U(k,z) - λU U(k,z) = J0(kr0) δ(z-z0)/2π -∂z2U(k,z) + (k2- λH) U(k,z) = [J0(kr0)/2π] δ(z-z0) So if we start with our PDE and do a Hankel transform from u(r,z) to U(k,z), we end up with a 1D Green's function problem in variable z. This is what happens if you start with the correct "r-eigenfunctions". This is what happens if you transform the PDE with the Hankel transform of order 0. So our next business is to identify homo solutions of this 1D thing, and then find the Green's function. Suppose we define κ2 = - [k2 –λH] Then homo solutions are e±iκz or cos(κz) and sin(κz) It seems then that our Green's might have this form U(k,z) = [J0(kr0)/2π]cos(κz<) [ Asin(κz>) + B cos(κz>) ] I suspect that A and B will be determined by the usual Green's 1D jump condition and by the BC at z = h. We have already met the BC at z = 0 with our form above. So then the final solution will be u(r,z) = !Syntax Error, Idk J0(kr) U(k,z) = !Syntax Error, Idk J0(kr) [J0(kr0)/2π]cos(κz<) [ Asin(κz>) + B cos(κz>) ] κ2 = - [k2 –λH] So this is just an outline of how Method A would go. Method B. Start again with the full PDF -∂r(r∂ru) - r ∂z2u - λH ru = δ(r-r0)δ(z-z0)/2π Lu - r ∂z2u - λH ru = δ(r-r0)δ(z-z0)/2π Lu = -∂r(r∂ru) Now we do a different starting point expansion. But first it seems a good idea to select z atoms that meet the z BC's. How about fk(z) = cos(kz) f 'k(z) = -ksin(kz) Thus meets the BC at z = 0 for any k. But we also need f 'k(h) = -ksin(kh) = 0 => kh = nπ k = nπ/h So we have a discrete spectrum here and our guys are then fn(z) = cos(knz) kn = nπ/h This sounds like a Fourier Cosine Series Transform situation. Let's define an angle θ which is NOT the spherical polar angle, it is a special thing for this problem. I make it range θ = 0 to π. θ = π (z/h) kn z = nπ/h * hθ/π = nθ I am not saying anything is even here, our only z range is 0 to h. I think all is OK. So here are the EF's of interest fn(θ) = cos(nθ) Now we call upon transforms.doc which says f(θ) = a0/2 + Σn=1∞ an cos(nθ) = (1/2) Σn=0∞ εn an cos(nθ) // expansion an = (2/π) !Syntax Error, Idθ f(θ) cos(nθ) where εn = 2-δn,0 // projection !Syntax Error, I dθ cos(nθ)cos(n'θ) = (π/εn) δnn' // orthogonality Σn=0+∞εn cos[n(θ-θ')] = 2π δ(θ-θ') // completeness Then we have δ(z-z0) = δ( hθ/π- hθ0/π) = (π/h) δ(θ-θ0). ∂z2 = ? dz = (h/π) dθ dz2 = (h/π)2 dθ2 ∂z2 = (π/h)2∂θ2 I think it might be best to apply "the projector" in this problem rather than the expansion. But first, here is our equation transformed from z to θ, and u = u(r,θ) here: Lu - r (π/h)2∂θ2u - λH ru = (π/h) δ(r-r0)δ(θ-θ0)/2π = (1/2h) δ(r-r0)δ(θ-θ0) Lu = -∂r(r∂ru) Now apply (2/π) !Syntax Error, Idθ cos(nθ) to all terms. Acting on u this gives un(r) we can call it LHS = Lun - r (π/h)2(2/π) !Syntax Error, Idθ cos(nθ) ∂θ2u - λH run RHS = (2/π) !Syntax Error, Idθ cos(nθ) (1/2h) δ(r-r0)δ(θ-θ0) = (2/π) (1/2h) δ(r-r0) cos(nθ0) = (1/πh) δ(r-r0) cos(nθ0) Now we have to do the parts twice in the usual manner. First parts !Syntax Error, Idθ cos(nθ) ∂θ2u = – !Syntax Error, Idθ ∂θ cos(nθ) ∂θu + [ cos(nθ) ∂θu] | π0 But ∂θu ~ ∂zu = 0 at both these endpoints from our BC's, so no parts. so !Syntax Error, Idθ cos(nθ) ∂θ2u = – !Syntax Error, Idθ ∂θ cos(nθ) ∂θu Now second parts on the RHS: = – { – !Syntax Error, Idθ ∂2θ cos(nθ) u + [∂θcos(nθ) u] | π0 } But this time the parts vanish because sin(nθ) = 0 at both endpoints. So as expected, !Syntax Error, Idθ cos(nθ) ∂θ2u = + !Syntax Error, Idθ ∂2θ cos(nθ) u = -n2 !Syntax Error, Idθ cos(nθ) u and then (2/π) !Syntax Error, Idθ cos(nθ) ∂θ2u = -n2 (2/π) !Syntax Error, Idθ cos(nθ) u = -(2n2/π) un So here then is our projected equation, Lun - r (π/h)2[-(2n2/π)] un - λH run = (1/πh) δ(r-r0) cos(nθ0) Lun + r (2n2/h2) un - λH run = (1/πh) cos(nθ0)δ(r-r0) and this then is our Method B 1D Green's equation in r, where r is really cylindrical ρ. So write it that way -∂ρ(ρ∂ρun) + ρ (2n2/h2) un - λH ρun = (1/πh) cos(nθ0)δ(ρ-ρ0) We then consult our Bessel doc to find the homo solutions of this thing: -(ρun')' + [(2n2/h2) –λH]ρ un = (1/πh) cos(nθ0)δ(ρ-ρ0) We should define κ2 = - [(2n2/h2) –λH] kn = nπ/h so we have -(ρun')' -κ2 ρ un = (1/πh) cos(nθ0)δ(ρ-ρ0) and our doc says -(xu')' + ν2u/x - k2xu = 0 Zν(kx) so I conclude that ν = 0 and our solutions are of the form Z0(κρ). So in the usual way, I would conclude that un = an(ρ) = C (1/πh) cos(nθ0) J0(κρ<) H(1)0(κρ>) and assume that we trivially find C, skip for now. Then our solution becomes u(z,ρ) = (1/2) C (1/πh) Σn=0∞ εn cos(nθ) cos(nθ0) J0(κρ<) H(1)0(κρ>) = (1/2) C (1/πh) Σn=0∞ εn cos(nπz/h) cos(nπz0/h) J0(κρ<) H(1)0(κρ>) OK, I don't have to do any more detail here. The result has the correct symmetry. The ring is at z0 and some ρ0 . and we will have to include how our BC's which sort of pass through I think. I need to pause and write down the general theory here, though Stak did that in his "partial eigenfunction method", but I think this deserves its own doc.