stakgold chap 7 exercise 7_34

Exercise 7.34: The 2D Exterior Helmholtz Green's Function for a circle. Exercise 7.34: The 2D Exterior Helmholtz Green's Function for a circle. 1 Part (a) : Use the Addition Theorem to get g. 1 Part (b): Use the 1D Green's method to get g. 3 Take the electrostatics λ→0 limit of this g. 5 _______________________________________________________________________________ Overview from Meta Notes. Exercise 7.34 ( p 290) The problem is to find the 2D Helmholtz exterior Green's function outside a circle on which g=0, and we solve the problem by two different methods. The first method is my "Green's Function by Dirichlet" method where we then get a homo PDE with a certain BC for v where g = E2 + v. We solve the homo by first selecting einθ and having unknown vn(r). The Green's ODE then for vn(r) is something we already figured out near page 269B with homo solution Zn(r), but in this simple exterior case we must have Zn(r) = H(1)n(r) so now vn(r) = Vn H(1)n(r) and we have only to find the constants Vn. The trick is that we can look at r=a where we know vn(a) = -E2 and this (with an addition theorem for E2) then tells us the Vn and we get our solution which is g(r,θ|r0,θ0) = (i/4) Σn ein(θ-θ0) * r< = min(r,r0) { Jn(r<)H(1)n(a) - Jn(a) H(1)n(r<) } [H(1)n(r>) / H(1)n(a)] The second method is more conventional. We start with the δδ Green's, select the same einθ eigenfunctions, and get then our usual 1D ODE Green's in r which we then just solve in the usual manner and the same result is obtained as shown above. Then just for fun I took the "electrostatics limit" λ→0 of the above result and got agreement with 6.100 which was the solution we got eons ago back in potential theory. __________________________________________________________________________________ Consider a circle of radius a. We have BC that u=0 on the boundary of the circle, and we want a 2D Helmholtz Green's Function for a point source outside this circle. We surely want u=0 at r=∞ as well. Just out of interest, what problems like this were done in the text? Stak first listed off the fundies in this section. He then did some weird expansions for E2 and then did the infinite wedge problem. Then one of the E2 expansions by another method. Then came the "ring source" example. Here the ring source is the driver of the H equation, so it is "like" a Green's Function problem, but instead of a point source, you have this delta ring thing. The main idea is that you have a distribution as your driving function. Stak solves this two ways, doing partial eigenfunctions in r and in z. Then he is off into the half plane stuff, another H Green's problem in 2D. Then finally he pulsed source at t=0 in this half-plane problem. That is it. So no, he did not do this exterior to disk H Green's Function problem. So here is the problem. We have in 2D (-2 -λ)g(r,θ) = δ(2)(r-r0) = δ(r-r0)/r * δ(θ-θ0) // Green's point charge at r0,θ0 g(a,θ) = 0 g(∞,θ) = 0 // boundary conditions Part (a) : Use the Addition Theorem to get g. His suggestion first off is to write g = E2+v, one of my standard methods ("the Dirichlet method of computing Green's Functions"). Then since E2 satisfies the PDE, we will get homo for v, but with different BC's: (-2 -λ)v(r,θ) = 0 v(a,θ) = -E2(a,θ|r0,θ0) v(∞,θ) = 0 // boundary conditions We did something like this back on page 274-5. That problem was a 2D Green's for the half-line boundary, and he used the KL transform to replace r by γ and got a simple equation in θ. Hold on that. Now, what can we say about the atoms for this Helmholtz problem? Looking back at p 268 where we considered this same PDE (but driven by a point source with no BC's), we assume v(r,θ) = Σn vn(r)einθ and we can steal from p 269B to get the ODE for vn(r) except we have no driver. On that page we saw that the form of the homo solution is Jn(r). For our exterior problem, we can only have H(1)n(r), so we know that vn(r) = constant * H(1)n(r) . If we call this constant vn, we get v(r,θ) = Σn vn(r)einθ = Σn vn H(1)n(r) einθ where we have all integer sum, and thus, we have done part (a) of this problem! Now how can we obtain the vn values? At r = a we do have this fact v(a,θ) = Σn vn H(1)n(a) einθ = -E2(a,θ|r0,θ0) (*) But now we can write E2 using the expansion 7.163 (his "addition theorem" or "sum rule"): E2(r,θ|r0,θ0) = (i/4) Σn Jn(r<) H(1)n(r>)ein(θ-θ0) where the < > refer to a,r0. But since r0 > a for sure, we then have E2(a,θ|r0,θ0) = (i/4) Σn Jn(a) H(1)n(r0)ein(θ-θ0) Therefore, installing this in (*) above we find Σn vn H(1)n(a) einθ = - (i/4) Σn Jn(a) H(1)n(r0)ein(θ-θ0) Since the einθ form a complete set, we may identify vn H(1)n(a) = - (i/4) Jn(a) H(1)n(r0) e-inθ0 and then we have our vn vn = - (i/4) [Jn(a)/ H(1)n(a) ] H(1)n(r0) e-inθ0 and so this coefficient is really vn( a, λ, r0, θ0) as you would expect. This concludes part a. Our solution of course is then this v(r,θ|r0,θ0) = Σn vn H(1)n(r) einθ = - (i/4) Σn [Jn(a)/ H(1)n(a) ] H(1)n(r0) H(1)n(r) ein(θ-θ0) and we see our expected r,r0 symmetry. Now if we want g, we have to do g = E2+v so g(r,θ|r0,θ0) = - (i/4) Σn [Jn(a)/ H(1)n(a) ] H(1)n(r0) H(1)n(r) ein(θ-θ0) + (i/4) Σn Jn(r<) H(1)n(r>)ein(θ-θ0) To keep this result general, we cannot do much with the r< type variables, so all we can do is this: g(r,θ|r0,θ0) = (i/4) Σn ein(θ-θ0) * r< = min(r,r0) { Jn(r<) H(1)n(r>) - [Jn(a)/ H(1)n(a) ] H(1)n(r0) H(1)n(r) } If we set r = a, then we know that r< = a and r> = r0 and this becomes g(a,θ|r0,θ0) = (i/4) Σn ein(θ-θ0) * { Jn(a) H(1)n(r0) - [Jn(a)/ H(1)n(a) ] H(1)n(r0) H(1)n(a) } = (i/4) Σn ein(θ-θ0) * { Jn(a) H(1)n(r0) - Jn(a) H(1)n(r0) } = 0 so we verify that g=0 on the circle at r=a. For large r we get the desired decay in r. Let's go for a cleaner way to write g. Just process it a bit g(r,θ|r0,θ0) = (i/4) Σn ein(θ-θ0) * r< = min(r,r0) { Jn(r<) H(1)n(r>) - [Jn(a)/ H(1)n(a) ] H(1)n(r0) H(1)n(r) } g(r,θ|r0,θ0) = (i/4) Σn ein(θ-θ0) * r< = min(r,r0) { Jn(r<) H(1)n(r>) H(1)n(a) - Jn(a) H(1)n(r0) H(1)n(r) }/ H(1)n(a) g(r,θ|r0,θ0) = (i/4) Σn ein(θ-θ0) * r< = min(r,r0) { Jn(r<) H(1)n(r>) H(1)n(a) - Jn(a) H(1)n(r<) H(1)n(r>) }/ H(1)n(a) g(r,θ|r0,θ0) = (i/4) Σn ein(θ-θ0) * r< = min(r,r0) { Jn(r<)H(1)n(a) - Jn(a) H(1)n(r<) } [H(1)n(r>) / H(1)n(a)] Now when r< = a, we see at once that g = 0, so this is a nicer form. Below I give this alternate form g(r,θ|r0,θ0) = (i/4) Σn εn cos(n[θ-θ0]) * { Jn(r<)Hn(1)(a) - Jn(a) Hn(1)(r<) } [ Hn(1)(r>) /Hn(1)(a)] where we have just folded things over. So these are the results for this problem! Part (b): Use the 1D Green's method to get g. Now Stak tells us to go solve this little ODE p 269 B in its homo form. Well, this will be very much like my Exercise 7.33 piece of work. The difference is that here we have a 2D 2 so the ODE. Our starting point here is this (-2 -λ)g(r,θ) = δ(2)(r-r0) = δ(r-r0)/r * δ(θ) // Green's point charge at r0,0 (θ0= 0 now) g(a,θ) = 0 g(∞,θ) = 0 // boundary conditions so the PDE says (-r-1∂r(r∂r..) -r-2∂θ2 -λ)g(r,θ) = δ(r-r0)/r * δ(θ) (-∂r(r∂r..) -r-1∂θ2 -rλ)g(r,θ) = δ(r-r0) δ(θ) -∂r(r∂rg) -r-1∂θ2g -rλ g = δ(r-r0) δ(θ) - (rg')' -r-1∂θ2g -rλ g = δ(r-r0) δ(θ) // where prime means ∂r This is equation 7.162 page 269 and we know that if we expand g = Σneinθan(r) we get p 269 B which is -(ra'n)' - λran + n2an/r = δ(r-r0)/2π So this is our 1D Green's problem, and we already know that solutions form is Jn(r). So, we know the form for our Green's function, and it is exactly like what we did in Exercise 7.33 with these two changes: (1) we have general n instead of n = 1/2 (2) we have no 1/ factors as part of the general solution form. So I can just read off various results making these changes. For example U(r,s) = C (rr0)-1/2 [ H1/2(1)(a)J1/2(r<) - J1/2(a) H1/2(1)(r<)] H1/2(1)(r>) becomes for us here an(r; a,r0,λ) = C [ Hn(1)(a)Jn(r<) - Jn(a) Hn(1)(r<)] Hn(1)(r>) If we compute the jump in this thing at r = r0 we got before that jump = C (2i/πr02) H1/2(1)(a) Since we are missing the (rr0)-1/2 factor, our result here will be jump = C (2i/πr0) Hn(1)(a) Now going back to our ODE -(ra'n)' - λran + n2an/r = δ(r-r0)/2π which we write as -ra"n - a'n - λran + n2an/r = δ(r-r0)/2π from which we see that our famous jump "a0" (overloaded!) is -r0 so jump = 1/a0 = -1/r0 here. But we have the extra 1/2π, so we have jump = -1/(2πr0) whereas in our Ex 7.33 we had jump = -1/(4πr02). So we set things equal here to get C (2i/πr0) Hn(1)(a) = -1/(2πr0) C (4i) Hn(1)(a) = -1 C = -(1/4i) 1/ Hn(1)(a) = (i/4) [ Hn(1)(a)]-1 which we can compare to C = (i/8) [H1/2(1)(a)]-1 of our Ex 7.33, pretty close! So we now have our Green's Function for the current 1D problem an(r; a,r0,λ) = C [ Hn(1)(a)Jn(r<) - Jn(a) Hn(1)(r<)] Hn(1)(r>) = (i/4) [ Hn(1)(a)]-1 [ Hn(1)(a)Jn(r<) - Jn(a) Hn(1)(r<)] Hn(1)(r>) = (i/4) [ Hn(1)(a)Jn(r<) - Jn(a) Hn(1)(r<)] [ Hn(1)(r>) /Hn(1)(a)] Now we install this into our expansion and we get our solution g(a,θ|r0,0) = Σneinθan(r) = (i/4) Σneinθ [ Hn(1)(a)Jn(r<) - Jn(a) Hn(1)(r<)] [ Hn(1)(r>) /Hn(1)(a)] This agrees exactly with the result obtained in part (a). Take the electrostatics λ→0 limit of this g. Now I know that if we take λ→0, somehow this has to give the Green's function for a grounded disk where the point charge is in the plane of the disk?? NO, that is a 3D problem, here we have a 2D problem, so this limit does not give that result. The limiting electrostatics problem is a line charge external to a grounded infinite cylinder, or the 2D Green's function of for a circle. Here are small z limits So you see that Jn(z) Hn(1)(z') = (z/2)n / Γ(n+1) * (1/iπ) Γ(n) (z'/2)-n = (z/z')n / n* (1/iπ) = (1/iπn) (z/z')n Therefore we have Jn(r) Hn(1)( r') = (1/iπn) (r/r')n and λ goes away! Similarly. Hn(1) (r) / Hn(1)( r') = (r/r')-n Thus we can just write our electrostatics limit of the above Green's Function g(r,θ|r0,0) = Σneinθan(r) = (i/4) Σneinθ [ Hn(1)(a)Jn(r<) - Jn(a) Hn(1)(r<)] [ Hn(1)(r>) /Hn(1)(a)] = (i/4) Σneinθ [ (1/iπn) (r</a)n - (1/iπn) (a/r<)n] [(r>/a)-n] = (i/4) Σneinθ (1/iπn) [ (r</a)n - (a/r<)n] (r>/a)-n = (1/4π) Σneinθ (1/n) [ (r</a)n - (a/r<)n] (r>/a)-n = (1/4π) Σneinθ (1/n) [ (r</a)n(r>/a)-n - (a/r<)n(r>/a)-n] = (1/4π) Σneinθ (1/n) [ (r</r>)n - (a2/r<r>)n] = (1/4π) Σneinθ (1/n) [ (r</r>)n - (a2/r<r>)n] // wrong, corrected below and probably this is a problem we did in Chapter 6 somewhere. This problem is addressed on page 148, but I don't think a solution in the above form was given. You can sort of see an image charge here and I could figure it out but won't do that right now. Something looks wrong here because the negative n terms will diverge at r=∞. The reason is that the AS limit for Hn(1) does not apply for negative n, so you have to mess with connection formulas first, that is For ν = integer, each of these formulas just says you pick up (-1)n when you negate order. So a product JK or a ratio H/H will be symmetric in the order. Then we can say, g(a,θ|r0,0) = Σneinθan(r) = (i/4) Σneinθ [ Hn(1)(a)Jn(r<) - Jn(a) Hn(1)(r<)] [ Hn(1)(r>) /Hn(1)(a)] = (i/4) Σneinθ fn = (i/4) [ f0 + Σn=1∞einθ fn + Σn=1∞e-inθ fn ] since f-n = fn = (i/4) [ f0 + 2Σn=1∞cos(nθ) fn ] = (i/4) Σn=0∞cos(nθ) εnfn = (i/4) Σn=0∞εn cos(nθ) [ Hn(1)(a)Jn(r<) - Jn(a) Hn(1)(r<)] [ Hn(1)(r>) /Hn(1)(a)] This is a good form for the result of this problem, by the way. NOW we can install our λ→0 limits as per above to get g(r,θ|r0,0) = (1/4π) Σn=0∞εncos(nθ) (1/n) [ (r</r>)n - (a2/r<r>)n] and now we are good as r→∞. The n=0 term is a little unusual and gives f(n)/g(n) = 0/0 so [ (r</r>)n - (a2/r<r>)n] / n = { exp[x ln(r</r>)] - exp[xln (a2/r<r>)] }/x as x→ 0 { exp[αx] - exp[βx] }/x = (α-β) = ln(r</r>) - ln (a2/r<r>) = ln[(r</r>) /(a2/r<r>) ] = ln[(r<2) /(a2) ] = 2 ln(r</a) If this is right, we then have g(r,θ|r0,0) = (1/4π){ 2 ln(r</a) + 2 Σn=1∞cos(nθ) (1/n) [ (r</r>)n - (a2/r<r>)n] } Now suppose we can apply the famous sum rule (6.22) to each term! Σn=1∞cos(nθ) (1/n) [ (r</r>)n] = -(1/2) ln(1 + (r</r>)2 - 2(r</r>)cosθ) Σn=1∞cos(nθ) (1/n) [(a2/r<r>)n] = -(1/2) ln(1 + (a2/r<r>)2 - 2(a2/r<r>)cosθ) Then we have some horrible log answer: g(r,θ|r0,0) = (1/2π) ln(r</a) + (1/2π)[ -(1/2) ln(1 + (r</r>)2 - 2(r</r>)cosθ)] - (1/2π)[ -(1/2) ln(1 + (a2/r<r>)2 - 2(a2/r<r>)cosθ ] g(r,θ|r0,0) = (1/2π) ln(r</a) - (1/4π) ln(1 + (r</r>)2 - 2(r</r>) cosθ) +(1/4π) ln(1 + (a2/r<r>)2 - 2(a2/r<r>)cosθ) Now suppose we define a certain image distance r<* ≡ a2/r< Then we have g(r,θ|r0,0) = (1/2π) ln(r</a) - (1/4π) ln(1 + (r</r>)2 - 2(r</r>) cosθ) + (1/4π) ln(1 + (r<*/r>)2 - 2(r<*r>)cosθ) Suppose we add these terms, making no difference in so doing, - (1/4π) ln(r>2) + (1/4π) ln(r>2) and we absorb the first into the first log, and the second into the second, getting g(r,θ|r0,0) = (1/2π) ln(r</a) - (1/4π) ln(r>2 + r<2 - 2r<r> cosθ) +(1/4π) ln(r>2 + r<*2 - 2r<*r> cosθ) Now let's just imagine that the log arguments are |r>- r<|2 and |r>- r<*|2 so we then have g(r,θ|r0,0) = (1/2π) ln(r</a) - (1/2π) ln |r>- r<| +(1/2π) ln |r>- r<*| = (1/2π) ln(r</a) + E2(r>| r<) - E2(r>| r<*) Now suppose I assume that a = 1 and r > r0. Then this result becomes = (1/2π) ln(r0) + E2(r| r0) - E2(r| r0*) and this agrees exactly with 6.100. Very good my friend.