SUMMARY OF DOCS IN THIS FOLDER

Summary of docs in this folder PhL 1.15.11 As of Sunday 1.16.11 Jim has not responded to either of the last two docs. I think I wore him out. 0. My 1st response to Jims 1st proposed L .doc ( 1 page, 1.18.11) 1 1. Jim delta 1_11_11.doc (1.5 pages, 1.11.11) 1 2. Answer to question 2.doc (11 pages, 1.12.11: this review is a 7 page rewrite!) 2 Part 1: The Answer to Question #2 (this is a summary of things shown in later parts) 2 Part 2: Proof of Equation (1.1) 2 Part 3: Jim's second conjecture 2 Part 4: Email to Jim ( a section of the doc) 4 3. Jim delta 1_13_11.doc (2.5 pages, 1.13.11) 8 4. Jim delta 1.13.11 5PM.doc (8 pages, 1.17.11) 8 5. Jim delta 1_14_11 11AM.doc (4 pages, 1.14.11)) 8 6. Stakgold on the delta theorem.doc (7 pages, 1.14.11) 9 7. Jim delta 1_18_11.doc (2 pages, 1.18.11) 9 ________________________________________________________________________________ 0. My 1st response to Jims 1st proposed L .doc ( 1 page, 1.18.11) On 1/4/11 I presented my Lg=δ problem to Jim on the phone and he said he would think about it. A few days later he called back with what I call "Jim's first proposal". I then sent this doc as an email in which I noted that the dimensions of his proposal were wrong, AND that I did not know how he showed that it was in fact a solution. Jim had mentioned on the phone some sort of square root regulation method which turned out to be exactly the right method. In retrospect, he probably came up with his proposal starting with the 3D known fact, just as I did in my 1.18 wrap-up doc to him. I wish he would have explained this. 1. Jim delta 1_11_11.doc (1.5 pages, 1.11.11) "Let's start all over". Question #2: What Cartesian 2D differential operator Lxy makes this equation be true: Lxy ( 1/ ) = δ(x-x')δ(y-y') Answer: I don't know the answer. It might be that no operator Lxy exists which solves Problem #2. Or maybe it is some horrible thing like this, with horrible coefficient functions, Lxy = a(x,y)∂x + a(y,x)∂y + b(x,y) ∂x2 + b(y,x) ∂y2 + c(x,y)∂x∂y + lots of higher terms. I am pretty sure this is NOT the answer Lxy = ∂x2 + ∂y2 ________________________________________________________________________________ 2. Answer to question 2.doc (11 pages, 1.12.11: this review is a 7 page rewrite!) (dated 1.12.11, but added to since email was sent) I found many confusions and various errors in this doc, so basically what follows is a rewrite of it: Part 1: The Answer to Question #2 (this is a summary of things shown in later parts) I agree finally that the following is true (I derive it in Part 2 below, perhaps just as you did) - limε→0 ∂r2(r[1 /]) = δ(r)/r = 2π δ(2)(r) = 2πδ(x)δ(y) (1.1) and this is then our regulated definition of the meaning of - ∂r2( r [1/r]) . Let's then suppose that my Question #2 was this: " Find Lxy such that Lxy(1/) = 2πδ(x)δ(y) " Then the answer seems to be: Lxy  is a differential operator which, when expressed in polar coordinates, has the form Lf = - ∂r2(r f) + (1/r) ( some angular part that involves ∂θ derivatives) f I then convert this to Cartesian coordinates adding in no angular part and I get Lxy(1) ≡ - ∂r2(r ...) = - (1/[ x2 ∂x2 + y2∂y2 +2xy∂x∂y + 2x ∂x + 2y∂y ] (1.2) Part 2: Proof of Equation (1.1) I prove equation (1.1) above using the Stakgold theorem for n = 2. That is, - ∂r2 ( r [1/r]) = δ(r)/r . - limε→0 ∂r2(r[1 /]) = δ(r)/r = 2π δ(2)(r) = 2πδ(x)δ(y) (2.1) Part 3: Jim's second conjecture Based on dimensions, I assumed this new "second conjecture" was (here f is any function f) - Lxy(2)f ≡ (∂x2+∂y2)(f) = 22D(rf) => - Lxy(2) = 22D(r ...) (3.1) We know that 2D2 = ∂x2+∂y2 = r-1∂r(r ∂r ) + r-2∂φ2  = 2D,radial2 + r-2∂φ2 (3.2) where 2D,radial2 = r-1∂r(r ∂r ) = r-1[r ∂r2 + ∂r] = [∂r2+ (1/r) ∂r] (3.3) Therefore we can write - Lxy(2)f = 22D(rf) = [2D,radial2 + r-2∂φ2](rf) = [∂r2+ (1/r) ∂r] (rf) + r-1∂φ2f (3.4) which we can apply to this special radial function case, ` - Lxy(2) [1/r] = (∂r2(r [1/r]) + (1/r)∂r(r [1/r]) = - Lxy(1) [1/r] + Lxy(3) [1/r] (3.5) where Lxy(1)was defined in Part 1 above, and I here define the second term to be Lxy(3) ≡ (1/r)∂r(r ...) = (1/r)[r ∂r + 1 ] = ∂r + (1/r) (3.6) Note carefully that we do not have - Lxy(2) = - Lxy(1) + Lxy(3) in general. This is only true when applied to some F(r) ! The full connection between these three operators can be computed as follows: - Lxy(2) ≡ 22D(r ...) = { r-1∂r(r ∂r ) + r-2∂φ2} (r ...) = ∂r2(r...) + (1/r)∂r(r...) + r-1∂φ2 (...) (3.7) - Lxy(1) ≡ ∂r2(r ...) (3.8) = (1.2) Lxy(3) ≡ (1/r)∂r(r ...) (3.6) Therefore we find that - Lxy(2) = - Lxy(1) + Lxy(3) + r-1∂φ2 or Lxy(1) – Lxy(2) = Lxy(3) + r-1∂φ2 (3.9) I then study this second term Lxy(3)[1/r] and apply to it the Stakgold n = 2 analysis to get Lxy(3) [1/r] = limε→0 (1/r)∂r(r [1/]) = + 2π δ(2)(r) (3.10) Lxy(1) [1/r] = - limε→0 ∂r2(r[1 /]) = + 2π δ(2)(r) (3.11) = (2.1) where on the second line I show my Part 2 result, so now I have two different candidate solutions to my Question #2! Since above we have - Lxy(2)[1/r] = - Lxy(1)[1/r] + Lxy(3)[1/r], (3.11a) an immediate implication of this result is this: for the operator which was "Jim's second conjecture", we get - Lxy(2) [1/r] = 22D(r[1/r]) = - Lxy(1)[1/r] + Lxy(3)[1/r] = - 2π δ(2)(r) + 2π δ(2)(r) = 0 (3.12) Since this is a full 2 operator, I am able to apply the 2D divergence theorem, and this then provides independent confirmation that in fact - Lxy(2) [1/r] = 22D(r[1/r]) = 0 ! I then convert Jim's Lxy(2) to Cartesian coordinates and I get Lxy(2) = – 22D(r..) = – (1/)[ x2(∂x2+∂y2) + y2(∂x2+∂y2) + 2x∂x + 2y∂y + 1] (3.13) At this point we then have these Cartesian results ( I do Lxy(3) below ) Lxy(1) ≡ – ∂r2(r ...) = – (1/[ x2 ∂x2 + y2∂y2 +2xy∂x∂y + 2x ∂x + 2y∂y ] (3.14) Lxy(2) = – 22D(r..) = – (1/)[ x2(∂x2+∂y2) + y2(∂x2+∂y2) + 2x∂x + 2y∂y + 1] (3.13) Part 4: Email to Jim ( a section of the doc) I quote the results shown earlier in this manner (1/r)∂r(r [1/r]) = limε→0 {(1/r)∂r(r [1/]) } = + 2π δ(2)(r) (4.1) = (3.10) ∂r2(r[1/r]) = limε→0 {∂r2(r[1 /])} = – 2π δ(2)(r) (4.2) = (2.1) and I show that these add up to 22D(r [1/r]) and the fact that they are equal and opposite is verified by the 2D divergence theorem. Using these facts, x = rcosφ ∂φx = -rsinφ = -y ∂rx = cosφ = (x/r) y = rsinφ ∂φy = rcosφ = x ∂ry = sinφ = (y/r) ∂φ = (∂φy)∂y + (∂φx)∂x = (x∂y -y∂x) ∂r = (∂ry)∂y + (∂rx)∂x = (1/r)(y∂y +x∂x) we get ∂φ = (x∂y -y∂x) (4.3) r∂r = (y∂y +x∂x) (4.4) I then compute from the expressions shown above, Lxy(1) - Lxy(2) = (1/)[ x2∂y2 + y2∂x2 – 2xy∂x∂y + 1] (4.5) = (1/) [ (x∂y-y∂x)2 +(y∂y+x∂x) + 1] (4.6) = (1/r) ∂φ2 + ∂r + (1/r) (4.7) According to (3.9) above we should have Lxy(1) – Lxy(2) = Lxy(3) + r-1∂φ2 (4.8) = (3.9) Comparing these two results requires that Lxy(3) = [ ∂r + (1/r)] (4.9) and this agrees with (3.6). Comment: We have shown that both these guys solve our little equation Lxy[1/r] = 2π δ(2)(r). Lxy(3)[1/r] = limε→0 (1/r)∂r(r [1/]) = + 2π δ(2)(r) (3.10) Lxy(1) [1/r] = - limε→0 ∂r2(r[1 /]) = + 2π δ(2)(r) (3.11) = (2.1) The difference of these two solutions is given by Lxy(3) – Lxy(1) = – Lxy(2) - r-1∂φ2 = 22D(r..) - r-1∂φ2 . The difference must satisfy the homogeneous equation Lxy[1/r]= 0. We have show already that the first term solves this equation (as we prove using the divergence theorem for example), and it is obvious that the second term also solves the homo equation. As a further check, let's express Lxy(3) in Cartesians: Lxy(3) ≡ (1/r)∂r(r ...) = (1/r)2(x∂x+y∂y) (r ...) since ∂r = (1/r)(x∂x+y∂y) Here is what Maple says: so we conclude that Lxy(3) = (1/) [ x∂x + y∂y + 1] = (1/r) [ r∂r + 1 ] = ∂r + (1/r) (4.10) which also agrees with (3.6). At this point I want to make this comment. We have found two solutions to Lxy([1/r]) = 2π δ(2)(r) Those solutions are Lxy(1) and Lxy(3) as shown in (3.10) and (3.11) above. If we take the difference of these solutions, we should find that Lxy(1) [1/r] – Lxy(3) [1/r] = 0 From (3.11a) above we see that this implies that Lxy(2) [1/r] = 0 From (3.1) we can express this as 22D(r [1/r]) = 0 and this agrees with (3.12). Once again, here are the two solutions I have found to the equation Lxy([1/r]) = 2π δ(2)(r) : Lxy(3) = (1/) [ x∂x + y∂y + 1] = [ ∂r + (1/r) ] = (1/r)∂r(r...) (4.11) Lxy(1) ≡ – (1/) [ x2 ∂x2 + y2∂y2 +2xy∂x∂y + 2x ∂x + 2y∂y ] = – ∂r2(r ...) Certainly the first operator has the simpler form when written out in Cartesians. The conclusion is that (1/r)∂r(r [1/r] ) = 2π δ(2)(r) = Lxy(3)([1/r] ) (4.12) or (1/) [ x∂x + y∂y + 1] () [1/] = 2π δ(x)δ(y) (*) (4.13) where we think of [1/] = limε→0 (1/) Notice that the work done here does not seem to solve this problem Lxy (1/) = 2π δ(x-x')δ(y-y') (4.14) and THIS is the thing I was really trying to solve. Suppose we define x = X-X' ∂x = ∂X y = Y-Y' ∂y = ∂Y Then (*) above says (1/)[(X-X')∂X+(Y-Y')∂Y+1] { [1/] } = 2π δ(X-X')δ(Y-Y') = 2π δ(2)(R - R') Having derived this, let's now change everything to lower case (1/)[(x-x')∂x+(y-y')∂y+1] { [1/] } = 2π δ(x-x')δ(y-y') = 2π δ(2)(x-x') (4.15) This is of the form Lxy,x'y' ( [1/] ) = 2π δ(2)(x-x') (4.16) but the form I am always looking for is this Lxy( [1/] ) = 2π δ(2)(x-x') (4.17) Therefore, my solving this problem Lxy([1/r]) = 2π δ(x)δ(y) has not helped one bit ! Many hours down the drain, but interesting results found. This probably means my Richard Price email has problems. // This evening 1.15.11 I updated my Richard email so at least the above error is removed from it. ______________________________________________________________________________ 3. Jim delta 1_13_11.doc (2.5 pages, 1.13.11) Here I respond to Jim's suggestion that maybe this is true Lxy([1/r]) = (∂x2+∂y2)( ([1/r]) = 2(r[1/r]) = 2π δ(2)(r) // not true! I give two arguments why I think in fact 2(r[1/r]) = 0 rather than a delta function. On argument is the divergence theorem which I present in some but not total detail. The other is that I claim I have shown limε→0 ∂r2(r[1 /]) = – 2π δ(2)(r) limε→0 {(1/r)∂r(r [1/]) } = + 2π δ(2)(r) and if you add these up, you get 2(r[1/r]) = 0. But I don't back up these two claims. ______________________________________________________________________________ 4. Jim delta 1.13.11 5PM.doc (8 pages, 1.17.11) This is a longer doc where I back up my various claims. In Section 1 I quote Stak's theorem. In Sections 2 and 3 I apply it for two different sequences. In Section 3 I show that these results are consistent with the 2D divergence theorem which says the results of Sections 1 and 2 must add up to 0. In Section 4 I show why the 2D and 3D divergence theorems give results of a different nature (0 and non 0) for our problems of interest. In Section 5 I show how to apply the Stak n=3 theorem to the usual 3D case we know and love. In Section 6, added just today, I show that, while the regulation method [1/r] ≡ limε→0 (1/) works for the 3D case treated in Section 5, the regulation method [1/r] ≡ limε→0 (1/(r+ε)) fails! Here is the TOC for this doc: ______________________________________________________________________________ 5. Jim delta 1_14_11 11AM.doc (4 pages, 1.14.11)) In the previous doc in item 3 I did the n=2 Stakgold proof that limε→0 [(1/r)∂r(r/) ] = δ(r)/r. I realized this ought to work if you multiply both sides by r to get an n=1 Stakgold theorem situation. This is my first n = 1 derivation and two special issues come up which I explain. First is the factor of 1/2 that you must add to the theorem due to "half a delta function", and second is the way to understand why boundary integral is 2. I put this stuff into a little Appendix. After Jim read this, he said he didn't believe any of this delta stuff, so I sent the next item. ______________________________________________________________________________ 6. Stakgold on the delta theorem.doc (7 pages, 1.14.11) I took photos of two Stakgold pages, then pasted pieces of these pages into a doc interleaved with my comments. Luckily Stak included an exercise which was exactly a case Jim was challenging. I want to put Jim into battle with Stakgold rather than me. ______________________________________________________________________________ 7. Jim delta 1_18_11.doc (2 pages, 1.18.11) This is a little wrap-up and thank-you doc sent to Jim on this day. In it I derive both his solutions without any need for the Stakgold theorem math, and I acknowledge his contributions. ______________________________________________________________________________