Jims first proposed L

Jim's First Proposed L PhL 1.8.11 Overview (1 page, written 1.17.11) 1 1. Jim's First Proposal cuts the mustard. 2 2. How do you prove the usual 3D case without using the divergence theorem? 5 3. Stakgold on Delta Functions 6 __________________________________________________________________________________ Overview (1 page, written 1.17.11) To his credit, Jim was the first to propose any solution to my "Question #2" equation Lx[1/r] = 2πδ(2)(r). I had no solutions in mind at all when I asked him my Question #2. I was even confused about how to write δ(n)(r) in terms of δ(r) and he showed the correct way as noted below (and later confirmed in Stak). Jim was also the first to propose the right method of regulating [1/r] as limε→0 1/. Note that Section 1 was written AFTER the other sections of this doc!! In Section 1, I give an n=2 Stakgold Theorem proof of this equation, which form was Jim's "first proposal" - limε→0 ∂r2(r[1 /]) = δ(r)/r = 2π δ(2)(r) = 2πδ(x)δ(y) (1) I think this was my first use of Stak's theorem, my first proof of many to follow later on. In Section 2 I wonder just how one would prove the 3D equation 2[1/r] = -4π δ(3)(r) without using the divergence theorem. I wanted to know, because I don't have the divergence theorem available to prove equation (1) above. [ The answer I later found is: use the Stakgold theorem, OR, start with the 3D known fact and multiply by r.] In Section 3 I peruse Stak for the first time on this subject, and note that he talks about sequences of functions that become delta functions in a limit. I decide to search for a sequence that in some limit reproduces the result shown just above but for the 3D case, where RHS = -δ(r)/r2. My first attempt is sk(r) = ∂r(r2∂r (1/[r+k])) as k→0. This is the 3D radial part of 2 with the leading 1/r2 removed. Maple tells me that this function sk(r) happens to have 0 area on (0,∞) (it has a sign change), so that was discouraging no fly attempt. [ I later found out that I misquoted the Maple results here, and still later found out that this method of regulation in fact fails. ] I then note "Stak's Theorem" and ponder it a bit. My second attempt is to use - ∂r(r2∂r (1/r1+α)) where you regulate the exponent. Same operator, different regulation proposal. This fixes the "thing = 0 for r>0" issue, but in retrospect is not very good since 1/r1+α still diverges for small α > 0. I cannot make this form "match" the Stak theorem which requires limα→0 [ α-3 g(r/α) ]. At this point I give up. As noted above, the correct regulation method is 1/ which Jim came up with. My first attempt was 1/(r+ε) which I later showed fails to give a delta function. __________________________________________________________________________________ 1. Jim's First Proposal cuts the mustard. The problem was to find L such that L(1/r) = 4πδ(2)(r). Jim proposes Lf = r ∂r2(rf) which certainly meets the requirement that L(1/r) = 0 for r ≠ 0. Note Added 1.12.10: But as I note elsewhere, r ∂r2(r[1/r]) = 4πδ(2)(r) has the wrong dimensions, so perhaps we could amend Jim's proposal to say Lf = ∂r2(rf) so then ∂r2(r[1/r]) = 4πδ(2)(r). Meanwhile, from Stak II page 22 ( I just had to search for 8 minutes in Vol II to locate this thing!!! ) we do see Jim's claim that in 3D one gets δ(3)(r) = δ(r)/[4πr2] page 22 A though this result is concealed in other obscure notation. Just integrate both sides on a small sphere to see that this is true. It just then also follow that δ(2)(r) = δ(r)/[2πr] and here we would integrate this over a small disk. The powers of r cancel against the volume element, and the angle integral is done. Using this last result, Jim's proposal becomes k∂r2(r[1/r]) = 4πδ(2)(r) = 4π δ(r)/[2πr] = 2 δ(r)/r Again, where k is a constant TBD, ∂r2(r[1/r]) = (2/k) δ(r)/r So now the problem is entirely in one dimension, r that runs (0,∞). Jim proposes regulating the LHS this way limε→0 ∂r2(r[1 /]) = (2/k) δ(r)/r According to Maple, the LHS is this which is to say, we then have LHS = limε→0 ∂r2(r[1 /]) = -3rε / (r2+ε)5/2 so then the question is: can we show this: limε→0[-3rε / (r2+ε)5/2] = (2/k) δ(r)/r Let's rewrite the RHS as it used to be, so since δ(2)(r) = δ(r)/[2πr] (2/k) δ(r)/r = (2/k) 2π δ(2)(r) = (4π/k) δ(2)(r) Then we want to show that limε→0[-3rε / (r2+ε)5/2] = (4π/k) δ(2)(r) ? Let's right now replace ε → ε2 so ε and r have the same dimension. Then we have limε→0[-3rε2 / (r2+ε2)5/2] = (4π/k) δ(2)(r) ? But (r2+ε2)5/2 = ( ε2[(r/ε)2+1] )5/2 = ε5[(r/ε)2+1]5/2 dim = L5 rε2 = (r/ε)ε3 Then we want to show this limε→0[-3(r/ε)ε3 / { ε5[(r/ε)2+1]5/2] = (4π/k) δ(2)(r) ? limε→0[-3(r/ε)ε-2 / { [(r/ε)2+1]5/2] = (4π/k) δ(2)(r) ? limε→0[ε-2{-3(r/ε) / { [(r/ε)2+1]5/2}] = (4π/k) δ(2)(r) ? Suppose we now define h(r) = -3r / (r2+1)5/2 Then we want to ask it this is true: limε→0[ε-2 h(r/ε)] = (4π/k) δ(2)(r) This at least has a limit form as we see used in Stak Vol II Section 5.4 (page 13 bottom) where Stak says δ(n)(r) = limε→0 [ ε-ng(r/ε)] where the only condition on g(r) is that !Syntax Error, Idr rn-1 g(r) = 1/Sn(1) I think Jim has done it! If we go to n = 2 we get δ(2)(r) = limε→0 [ ε-2g(r/ε)] This says then that (4π/k) δ(2)(r) = limε→0 [ ε-2(4π/k)g(r/ε)] Comparing with the above, I have (4π/k)g(r/ε) = h(r/ε) => (4π/k)g(r) = h(r) => g(r) = (k/4π) h(r) = (k/4π)[ -3r / (r2+1)5/2] Now to get the normalization we need to get !Syntax Error, Idr r1 (k/4π)[ -3r / (r2+1)5/2] = 1/2π -(3k/4π)!Syntax Error, Idr r1 [ r / (r2+1)5/2] = 1/2π -(3k/2)!Syntax Error, Idr r1 [ r / (r2+1)5/2] = 1 -(3k/2)!Syntax Error, Idr r2 / (r2+1)5/2 = 1 Maple says the integral is 1/3 So we end up with -(3k/2)(1/3) = 1 -(k/2) = 1 => k = -2 Therefore I have shown that limε→0 ∂r2(r[1 /]) = (2/k) δ(r)/r = - δ(r)/r Again - limε→0 ∂r2(r[1 /]) = δ(r)/r = 2π δ(2)(r) = 2πδ(x)δ(y) So this is indeed a major step forward. So we then know that, in spherical coordinates, L = ∂r2(r ..) + angular part **************************************************************************** (these are older sections) 2. How do you prove the usual 3D case without using the divergence theorem? [ Note Added: In a later doc, I do prove the 3D case without using the divergence theorem! I use an n=3 application of Stakgold's Theorem. I sent this to Jim in an email, it was part of a large doc.] So now one needs to show this: ∫small disk dV L(1/r) = 4π By analogy, we know in 3D that the corresponding fact is this ∫small sphere dV [-2(1/r)] = 4π How would we prove this if we did not have the divergence theorem? I guess you install a test function: ∫small sphere dV φ(r) [-2(1/r)] = 4πφ(0) ? ∫small sphere dV φ(r) r-2∂r(r2∂r (1/r)) = - 4πφ(0) ? !Syntax Error, Ir2dr φ(r) r-2∂r(r2∂r (1/r)) = - φ(0) ? !Syntax Error, Idr φ(r) ∂r(r2∂r (1/r)) = - φ(0) ? Now what do you do, Mr. Stak? Cannot compute the derivative because it gives 0. Parts is pretty unclear. So I guess I have to go back and review "delta functions" in Stak as part of this mid term exam. 3. Stakgold on Delta Functions Vol I page 22. Here Stak considers the sequence sk(x) = k/[π(1+k2x2)] The claim is that this becomes a delta function as k → ∞. Plotting for k = 5 is promising: So on page 22-23 he throws in a test function φ(x) and he proves what he needs to prove, there is nothing too fancy in this proof. The area of sk(x) = 0. So can I come up with some kind of sequence for my 3D case above? !Syntax Error, Idr φ(r) ∂r(r2∂r (1/r)) = - φ(0) ? Perhaps I replace r by r+k and then later take the limit k → 0. When k > 0, I can compute stuff. Let sk(r) = ∂r(r2∂r (1/[r+k])) = -2(r+k)-2 + 4r(r+k)-3 = 2(r-k)(r+k)-3 according to Maple There are two problems. First, I am not sure how to deal with the fact that we are on (0,∞) so I somehow don't have access to the left half of the delta function! [ this is resolved elsewhere ] Second, Maple says that for any k, the area under my proposed sequence curve is 0. So I guess this is not a good "candidate" for the second reason. [ wrong, it turns out in fact to be infinite! ] ___________________________________________________________________________________ Note added 1.17.11: Somehow I entered things wrong, the Maple result above is wrong! This is one place where I failed to quote the Maple result, and naturally I got it wrong. I show in Section 6 of "jim delta 1.13.11 5PM.doc" that this method of regulation fails for our usual 3D problem being considered here. That is to say, sequence sε(x) = ∂r(r2∂r (1/(r+ε)) does not approach a delta function δ(r) in the limit ε→0. It approaches something bigger than a delta function! __________________________________________________________________________________ Regrettably, after studying Stak for a year, I have no idea what to do here. Where do I start reviewing? This is part of the game. You once read about a subject, and now you have a problem in that subject, so you have to review it because the attic is not large enough to retain it. I think Vol II Section 5.4 (page 10) might have what I need. I just reviewed Example 3 page 12 which is rather amazing and this is a deal in general Rn . Consider f(x) which has a unit volume integral in Rn. Then consider fα(x) = α-n f(x/α) where you scale each coordinate as in xi/α. The amazing claim is that in the limit that α → 0, this thing = δ(x) in n dimensions! The proof is pretty straightforward. Example 4 is even closer to our problem at hand! Suppose f(x) has the simple form g(r). Then of course we have fα(x) = α-n g(r/α). The integral requirement is that Sn(1)∫dr rn-1 g(r) = 1 where the S thing is the angular integral in n dimensions. A candidate for n=3 is this g(r) = π-3/2 exp(-r2) since Sn(1)∫dr r2 g(r) = 1 Then we find that limα→0 [ α-3 g(r/α) ] = δ3(r) So this seems to me a pretty powerful way to create delta functions. Is there some way I could use this idea to show that [-2(1/r)] = 4πδ(r) n = 3 - ∂r(r2∂r (1/r)) = 4πδ(r) This is a little different because there are derivatives on the LHS. I would have to write - ∂r(r2∂r (1/r)) = (1/4π) limα→0 [ α-3 g(r/α) ] so I would have to find some g(r) that makes this true. How do you do that? All you can do is try different things. For example - ∂r(r2∂r (1/r1+α)) = -α(α+1)/r1+α =?= (1/4π) α-3 g(r/α) => g(r/α) = 4πα3 [-α(α+1)/r1+α ] = -4πα4(α+1) (r/α)-1-α α-1-α = -4πα3-α(α+1) (r/α)-1-α => g(r) = -4πα3-α(α+1) (r)-1-α But g(r) is not allowed to depend on α, so this trial does not fly. Here is another trial - ∂r(r2∂r (1/(r+α))) = 2 rα (r+α)-3 = 2 rα (r/α+1)-3 α-3 = 2 rα-2 (r/α+1)-3 = 2α-1(r/α) (r/α+1)-3 =?= (1/4π) α-3 g(r/α) => g(r/α) = 4πα3 2α-1(r/α) (r/α+1)-3 = 8πα2(r/α) (r/α+1)-3 => g(r) = 8πα2 r(r+1)-3 This seems to be at least a better trial, but we still have α dependence, though in a simple way. Here is a third trial - ∂r(r2+β∂r (1/r1+α)) = -2(r+β)(β-α)/(r+α)3 = -2(r/α + β/α)α(β-α)/[ (r/α + 1)3 α3] =?= (1/4π) α-3 g(r/α) => g(r/α) = 4πα3 * -2(r/α + β/α)α(β-α)/[ (r/α + 1)3 α3] = -8π α(β-α) (r/α + β/α)/ (r/α + 1)3 But you are now stuck with α(β-α) which is α dependent. I think I could spend 100 hours here searching for a g(r). In his fundy section, Stak always calls upon the divergence theorem to get what he needs. What "handle" can one attach to Stak's discussion of g(r) ? I could scan some of my "famous math books" like Courant and Hilbert. Conclusion: I don't know how to prove the 3D case without the divergence theorem! Thus, I have no precedent on which to try to prove or disprove the 2D case. I don't really know how to do a web search on this subject. So Jim's suggestion is of no use because there is no way. [ But later a "way" was found! ]