Find L for Ls(x,y)=d(x-y) in 1D

How to find L in 1D Lxs(x,y)=δ(x-y) PhL 1.9.11 (See Summary in separate summary doc. ) This entire doc considers only 1D situations of Lx s(x,y) = δ(x-y) and special case Lxr(x) = δ(x). When I wrote this doc, I did not realize that Lxg(x)=δ(x) Lxg(x-x')=δ(x-x') (except for constant coefficient Lx), but I have reread all and have written an Overview, and this does not invalidate much that I wrote, and I comment in the appropriate places. All this thrashing around boils down to this question stated in section 5 below: "Given a function r(x), is it, or is it not, a Green's Function of some self-adjoint differential operator? " I am unable to answer this question, but my gut feeling is that if you select some r(x) at random, such as r(x) = 1/x4, you will find that there is no Lx that will work. I could be wrong. Assuming I am right, one could extend the conclusion to higher dimensions. The gut feeling then is that Green's Functions are a very narrow class of functions in the universe of all functions. In Section 6 at the very end I think I have proven this theorem: "If the kernel is of the r(x) form and is even in x and has a Fourier Integral Transform, then Lxr(x) = δ(x) has no solution Lx which is a differential operator having constant coefficients. " But in this case I cannot rule out Lx having some other form. The Love integral equation has a kernel that fits this theorem. Overview (2 pages, written 1.16.11) 1 1. s(x,y) kernel which has single Fourier Int expansion and general an(x) coefficients in Lx. 3 2. s(x,y) kernel which has double Fourier Int expansion and constant an coefficients in Lx. 9 3. r(x-y) kernel which has Fourier Int expansion and general an(x) coefficients in Lx. 11 4. Known r(x-y) example does not have a Fourier Transform! 13 5. Expand r(x) in a complete set of basis functions 15 6. r(x-y) kernel which has Fourier Int expansion and constant an coefficients in Lx. 17 __________________________________________________________________________________ Overview (2 pages, written 1.16.11) In Section 1 (8 pages) I consider the problem of finding Lx for Lx s(x,y) = δ(x-y). I Fourier-Integral expand s(x,y) into S(k,y) and expand the δ as well. I first assume in parts (a)→(d) a general form Lx = Σn=0∞ an(x) ∂xn. I Fourier expand the an(x) as well (which might be a bad idea for a function like x2! ) and I end up with this very ugly and intractable equation which has to somehow be valid for all real y in our interval, and the qn are known, Σn=0∞ !Syntax Error, I dk qn(k'-k; y)An(k) = - (1/2π)eik'y Then in part (e) I start over assuming Lx = Σn=0∞an∂xn with constant coefficients. Our only chance here is if s(x,y) = r(x-y), but then our hopes are dashed by Theorem 2 and 2A which show that, if r(x) is even, which is always the case for me, then R(k) = F(|k|) and is not anlytic in k at k=0 and therefore cannot be matched with a power series in k, and therefore Lx = Σn=0∞an∂xn fails, and this is true on (a,b) or (-∞,∞). In the latter case, I can at least diagonalize our integral equation g(x) = (1/2π) !Syntax Error, Idy 2πr(x-y)f(y) into G(k) = 2πR(k)F(k) and then R(k) = F(|k|) is not a problem and I can solve for f using F(k). This assumes of course that g(x) has a FI transform G(k). In Section 2 (2 pages) I redo Section 1 but I do a double Fourier Integral transform of s(x,y) into S(k,k') and assume Lx has constant coefficients. This leads to this strange requirement [Σn=0∞ an (-ik)n] S(k,k') = (1/2π)δ(k+k') [Σn=0∞ an ∂xn ] s(x,y) = δ(x-y) where I put the starting equation below the result. I don't know how to make a series add up to a delta function. Probably if I assume the r(x-y) situation, S(k,k') will cancel that delta, but I have really conclusively treated this situation in Section 1, so Section 2 brings nothing useful to the party. I was just blindly "trying things" at the time. In Section 3 (3 pages) I assume that s(x,y) = r(x-y). I assume general function coefficients an(x) for Lx and FI transform them into An(k) as I did in Section 1. This leads to Σn=0∞!Syntax Error, Idk un(k'-k) An(k) = -(1/2π) where un(k) = Σn=0∞ (-ikn } R(k) which is similar to the first equation in Section 1, where here it has to somehow be true for all k' As in Section 1, I regard this as intractable. In Section 4 (2 pages) I try the example r(x-y) = |x-y| which appears in Stakgold. But I quickly realize that r(x) has no R(k) FI transform. The lesson here is this: when considering Lx r(x-y) = δ(x-y), if you want to transform both sides of this equation, you better use a convergent transform for r(x). Find some complete set of functions other than eikx on which to expand both sides! In Section 5 (3 pages) I follow the suggestion of Section 4. I assume some complete set φn(x) and I define Ln'n = <φn'| Lx | φn> and end up with this infinite matrix equation: Σn'=0∞ Lnn' rn' = φn(0), which is what has become of Lx r(x) = δ(x) [ I did not realize at this time that solving this does not solve the equation Lx r(x-y) = δ(x-y), so this whole section is wrongly aimed.] rn are of course the coefficients of r(x) = Σn=0∞ rnφn(x). The ∞ matrix equation is Lr = φ(0) which maybe you could solve numerically. If I redo this sectino properly for Lx r(x-y) = δ(x-y), I end up with this Σn=0∞φn(x-y) Σn'=0∞ rn' Lnn' = Σn=0∞ φn(x) φn(y) and I cannot then appeal to completeness of the φn(x) to get a simple result, end of the line. The only way out of course is to use φn(x) = exp(iknx) on the interval (a,b) so φn(x-y) = φn(x) φn(y) and then it seems you end up with infinite matrix equation Lr = 1 which you might solve for some working Ln'n. I then make the "radical step" of assuming that φn(x) are eigenfunctions of our unknown Lx. This leads to being able to write r(x-y) = Σn=0∞ φn(x)φn(y)/λn which is the usual Green's Function form, but this is really useless since I don't know Lx, so I don't know λn or φn(x) ! It's just a circle. At this point I sent off my first Richard Price email. In Section 6 (2 pages) I repeat Section 3 r(x-y), but with Lx having constant coefficients. I end up with Σn=0∞ bn kn = (1/2π)(1/R(k)) which is what I found at the end of Section 1, and again I consider r(x) = (d/π) / (d2 +x2) and R(k) = e-d|k| /2d and how then no series exists. And I conclude again this situation arises for any function r(x) which is even in x. _______________________________________________________________________________ 1. s(x,y) kernel which has single Fourier Int expansion and general an(x) coefficients in Lx. As Shankar suggested, I think the easiest way is to expand f and the delta function into a Fourier Transform in the appropriate number of dimensions. (a) First, let's consider the action of a differential operator Lx on an integral equation in 1D. I assume I can move Lx through the integral, which needs to be validated in a rigorous analysis, but not here: g(x) = ∫dy s(x,y)f(y) (1.1) Lx g(x) = ∫dy Lx s(x,y)f(y) (b) In order for this integral equation to be converted into a differential equation, I have to find Lx which has this property Lx s(x,y) = δ(x-y) (2.1) and if I can find such an Lx then acting on our integral equation results in this differential equation: Lx g(x) = f(x) Thus, a differential equation corresponding to our integral equation only exists if Lx exists such that Lx s(x,y) = δ(x-y). If Lx s(x,y) produces just a function h(x,y), then the integral remains and I have then an integro-differential equation which is not what I want. Notice that y is a dummy integration variable in (1.1). I certainly cannot have Lx being a function of y! That is why I write it as Lx. Apart from that, I are not narrowing our world of possible Lx operators at this time, everything is possible. (c) So how can I determine whether such an Lx exists for a given kernel s(x,y) ? A first idea is to treat y as a bystander parameter and try to find an expansion of this form: s(x,y) = !Syntax Error, Idk e-ikx S(k, y) => S(k, y) = (1/2π) !Syntax Error, Idx e+ikx s(x, y) I am using capital letters for momentum-space objects. I can at the same time expand 2πδ(x) = !Syntax Error, Idk e-ikx Then our equation (2.1) becomes Lx !Syntax Error, Idk e-ikx S(k, y) = (1/2π) !Syntax Error, Idk e-ik(x-y) Again we assume we can move Lx through the integral (and rewrite things on both sides a bit) to get !Syntax Error, Idk [ Lx e-ikx ] S(k, y) = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] (d) Now we shall make an assumption about Lx, namely, Lx = Σn=0∞ an(x) ∂xn Each ∂x on e-ikx brings down a factor if (-ik), so we then have !Syntax Error, Idk S(k, y) [Σn=0∞ an(x) (-ik)n ] e-ikx = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] Next, I suppose we have to Fourier expand each of our coefficient functions an(x) = !Syntax Error, Idk' e-ik'x An(k') => An(k) = (1/2π) !Syntax Error, Idx e+ikx an(x) Then we have !Syntax Error, Idk S(k, y) [Σn=0∞ an(x) (-ik)n ] e-ikx = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] !Syntax Error, Idk S(k, y) [Σn=0∞ {!Syntax Error, Idk' e-ik'x An(k')} (-ik)n ] e-ikx = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] !Syntax Error, Idk S(k, y)!Syntax Error, Idk' e-i(k+k')x [Σn=0∞ { An(k')} (-ik)n ] = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] Now replace k' by k" = k+k' on the LHS to get !Syntax Error, Idk S(k, y)!Syntax Error, Idk" e-ik"x [Σn=0∞ An(k"-k) (-ik)n ] = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] Now swap variable names k↔k" on the LHS !Syntax Error, Idk" S(k", y)!Syntax Error, Idk e-ikx [Σn=0∞ An(k-k") (-ik")n ] = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] !Syntax Error, Idk e-ikx !Syntax Error, Idk" S(k", y) [Σn=0∞ An(k-k") (-ik")n ] = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] Now, finally, all the x dependence is in the e-ikx factors and we can claim these are a complete set and we then can equate the integrands. This gives !Syntax Error, Idk" S(k", y) [Σn=0∞ An(k-k") (-ik")n ] = (1/2π)eiky Now let k - k" = k' on the LHS - !Syntax Error, Idk' S(k-k', y) [Σn=0∞ An(k') (-i)n(k-k')n ] = (1/2π)eiky and now swap k ← k' !Syntax Error, Idk S(k'-k, y) [Σn=0∞ An(k) (-i)n(k'-k)n ] = - (1/2π)eik'y Σn=0∞ !Syntax Error, Idk An(k) { (-i)n (k'-k)n S(k'-k, y) } = - (1/2π)eik'y Σn=0∞ !Syntax Error, Iqn(k'-k; y)An(k) dk = - (1/2π)eik'y where qn(k'-k; y)= (-i)n (k'-k)n S(k'-k, y) qn(k; y)= (-i)n kn S(k, y) What we have here is a single equation with an infinite series on the LHS. Each term in this series is an integral of an Lx Fourier coefficient against a known kernel. But this equation must be true for each value of y, so it is then a continuous infinite system of equations with an infinite discrete number of unknown An(k) functions, Σn=0∞ Qn(y) An = - (1/2π)eik'y y = all values in (-∞,∞) The kernels at least are of the difference form. We could go on here, but it seems fairly intractable. There could in theory be solutions An(k), I guess we cannot rule them out. Or solutions might not exist. (e) Suppose we assumed constant coefficients instead? Back up a bit: !Syntax Error, Idk [ Lx e-ikx ] S(k, y) = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] Lx = Σn=0∞ an ∂xn !Syntax Error, Idk [Σn=0∞ an(-ik)n e-ikx ] S(k, y) = !Syntax Error, Idk e-ikx [ (1/2π)eiky ] [Σn=0∞ an(-ik)n ] S(k, y) = (1/2π)eiky (*) which compare to starting equation [ Σn=0∞ an ∂xn] s(x,y) = δ(x-y) We could write (*) this way with bn = an(-i)n [Σn=0∞ bn kn ] = (1/2π)[ eiky / S(k, y)] The equation has to be true for all y. So the only chance you would have for a solution here was if it happened that eiky / S(k, y)] = 1/R(k) = independent of y => S(k, y) = R(k) eiky Therefore, for some general s(x,y) you come up with (which has a single Fourier Transform on the first variable), it is pretty unlikely you will get a constant coefficient Lx that works. In fact, you would have to have S(k, y) = eiky R(k) as the form. What would that imply for the kernel? s(x,y) = !Syntax Error, Idk e-ikx S(k, y) = !Syntax Error, Idk e-ikx eiky R(k) = !Syntax Error, Idk e-ik(x-y) R(k) = g(x-y) Interesting. If the kernel is of "convolution form" then we stand a chance. So let's assume that s(x,y) = r(x-y) Then from this kernel we compute R(k) = (1/2π) !Syntax Error, Idx e+ikx r(x) Then we need to find the bn coefficients such that [Σn=0∞ bn kn ] = (1/2π)[ eiky / S(k, y)] = (1/2π) ( 1/R(k) ) Assuming we can find the bn, our solution is then Lx = [ Σn=0∞ an ∂xn] where an = bn(i)n Wow, we have a solution! Let's state this as a Theorem: Theorem 1. Our integral equation of interest is as follows, and we seek Lx to do as shown g(x) = !Syntax Error, Idy s(x,y)f(y). Lx g(x) = !Syntax Error, Idy Lx s(x,y)f(y) = !Syntax Error, Idy δ(x-y)f(y) = f(x) Lx s(x,y) = δ(x-y) We consider only the special case where s(x,y) = r(x-y). We assume that r(x) has a Fourier Transform R(k) = (1/2π) !Syntax Error, Idx e+ikx r(x) We assume we can find coefficients bn such that (ie, we can expand 1/R(k) in a power series in k) Σn=0∞ bn kn = 1/[2πR(k)] Then the solution Lx is this: (that is, we have constant coefficients) Lx = Σn=0∞ an ∂xn where an = (i)nbn Is there an easier way to understand what is happening here? We have g(x) = (1/2π) !Syntax Error, Idy 2πh(x-y)f(y). Lx g(x) = !Syntax Error, Idy Lx r(x-y)f(y) = !Syntax Error, Idy δ(x-y)f(y) = f(x) Lx h(x-y) = δ(x-y) The first equation is of convolution form with the correct group range. So it will diagonalize into G(k) = 2πR(k)F(k) // see "Diag. of Convolution...doc" Then we have F(k) = G(k) / [2πR(k)] So we know how to solve the first equation without even using the Lx h(x-y) = δ(x-y) idea. But OK, let's now start with just Lx h(x-y) = δ(x-y). We Fourier Expand the h and the δ Lx !Syntax Error, Idk e-ik(x-y) R(k) = !Syntax Error, Idk e-ik(x-y)(1/2π) Assume Lx = Σn=0∞ an ∂xn and move it in to get !Syntax Error, Idk [ Σn=0∞ an (-ik)n ] e-ik(x-y) R(k) = !Syntax Error, Idk e-ik(x-y)(1/2π) Now claim completenss to say [ Σn=0∞ an (-ik)n ] R(k) = (1/2π) Therefore we have Σn=0∞ an (-ik)n = 1/ [2πR(k)] which replicates the claim of Theorem 1. Now how is this different with finite endpoints? g(x) = (1/2π) !Syntax Error, Idy 2πr(x-y)f(y). Lx g(x) = !Syntax Error, Idy Lx r(x-y)f(y) = !Syntax Error, Idy δ(x-y)f(y) = f(x)θ(x>a)θ(x<b) Lx r(x-y) = δ(x-y) The result now has theta functions. But if we can still say this: Lx g(x) = f(x) valid for a ≤ x ≤b so we still seem to have a solution if we can find Lx such that Lx r(x-y) = δ(x-y). In the original integral equation, r(x-y) has an argument that runs from a-b up to b-a which is a range of 2(b-a) = 2L. In other words, the function h(x) need only be defined on the range (-L,L) centered on the origin. In order to do our Fourier Integral expansion idea, we would need to somehow "extend" the function h(x) out beyond this range to both endpoints. For the moment, perhaps r(x) = 1/x2 and this naturally extends itself to the two limits. In fact, let's do exactly this case. Then R(k) = (1/2π) !Syntax Error, Idx e+ikx [ 1/x2] = (1/2π) !Syntax Error, Idx cos(kx) [ 1/x2] = (1/π) !Syntax Error, Idx cos(kx)/x2 But this integral diverges at the origin. So instead let's do our Love case of r(x) = 1/(d2 +x2) and then we know R(k) =(1/π) !Syntax Error, Idx cos(kx)/ (d2 +x2) = e-d|k| /2d Then we are stuck with this series problem: Σn=0∞ an (-ik)n = 1/ [2πR(k)] = (1/2π) 2d e+d|k| = (1/πd) e+d|k| and we have the famous problem that our function f(k) = e+d|k| is not analytic at k = 0. In fact we know that there is a slope discontinuity there since f '(0+) = d and f '(0-) = -d . Thus, you cannot have a power series around k = 0. Thus, there is no solution Lx of this form. In fact, for any even r(x), we know that if it exists we will get R(k) = F(|k|). Since |k| is not analytic, even if F is continuous, F(|k|) won't be analytic and there will be no possible series. Most kernels I am familiar with are in fact even. Theorem 2: Consider this situation g(x) = (1/2π) !Syntax Error, Idy 2πr(x-y)f(y). Lx g(x) = !Syntax Error, Idy Lx r(x-y)f(y) = !Syntax Error, Idy δ(x-y)f(y) = f(x)θ(x>a)θ(x<b) Lx r(x-y) = δ(x-y) and assume that r(x) is even and can be naturally extended to (-∞,∞). And assume that the Fourier Transform of r(x) exists, which we call R(k). Then our theorem states that R(k) has the form F(|k|), and therefore Theorem 1 fails because we cannot find coefficients bn , and therefore we cannot find Lx of the form Lx = Σn=0∞ an ∂xn with constant coefficients. This theorem applies for any fixed endpoints a,b includiong -∞ and +∞. Let's restate this result more compactly: Theorem 2A. Consider the problem Lx r(x-y) = δ(x-y) on an interval (a,b) where we seek Lx given r. Then if r(x) is even, and if when extended to (-∞,∞) it has a Fourier transform R(k), then there is no solution Lx which has constant coefficients. 2. s(x,y) kernel which has double Fourier Int expansion and constant an coefficients in Lx. (a) Let's back up and take a different path just to see where it leads. Go all the way back to here: Lx s(x,y) = δ(x-y) (2.1) Now let's assume we can do a double Fourier of s(x,y) and see what happens: s(x,y) = !Syntax Error, Idk e-ikx!Syntax Error, Idk' e-ik'y S(k,k') S(k,k') = (1/2π)2!Syntax Error, Idx e+ikx!Syntax Error, Idy e+ik'y s(x,y) Then we have Lx s(x,y) = δ(x-y) Lx !Syntax Error, Idk e-ikx!Syntax Error, Idk' e-ik'y S(k,k') = (1/2π) !Syntax Error, Idk e-ik(x-y) Move Lx inside, !Syntax Error, Idk [Lx e-ikx] !Syntax Error, Idk' e-ik'y S(k,k') = (1/2π) !Syntax Error, Idk e-ik(x-y) Negate k on the RHS only !Syntax Error, Idk [Lx e-ikx] !Syntax Error, Idk' e-ik'y S(k,k') = (1/2π) !Syntax Error, Idk e+ik(x-y) Swap dummy variable names k ↔ k' on LHS only !Syntax Error, Idk' [Lx e-ik'x] !Syntax Error, Idk e-iky S(k',k) = (1/2π) !Syntax Error, Idk e+ikx e-iky Rewrite !Syntax Error, Idk e-iky !Syntax Error, Idk' [Lx e-ik'x] S(k',k) = (1/2π) !Syntax Error, Idk e-iky e+ikx Now use completeness of e-iky to claim that !Syntax Error, Idk' [Lx e-ik'x] S(k',k) = (1/2π) e+ikx Now swap k ↔ k' again !Syntax Error, Idk [Lx e-ikx] S(k,k') = (1/2π) e+ik'x Now assume for the moment the simple case that Lx has constant coefficients Lx = Σn=0∞ an ∂xn Then we have !Syntax Error, Idk [Σn=0∞ an (-ik)n] e-ikx S(k,k') = (1/2π) e+ik'x This seems to require that [Σn=0∞ an (-ik)n] S(k,k') = (1/2π)δ(k+k') (*) which compare to starting point [Σn=0∞ an ∂xn ] s(x,y) = δ(x-y) I have no idea how to deal with an equation like (*), so I now let this path come to rest. I think the single Fourier analysis of the last section is better, the double is unnecessary for this application. 3. r(x-y) kernel which has Fourier Int expansion and general an(x) coefficients in Lx. Now, finally, let's make the Big Assumption that our original kernel has the form s(x,y) = r(x-y) so our integral equation is then g(x) = ∫dy s(x,y)f(y) = ∫dy r(x-y)f(y) and we then go back to Lx s(x,y) = δ(x-y) (2.1) Lx r(x-y) = δ(x-y) (2.1) Notice that this is the case for the Love integral equation, and for my 2D 1/r problem. Since this must be true for all y, it is true for y = 0 and this must be true Lx r(x) = δ(x) Now we construct Fourier Integral expansions for r(x) and δ(x) r(x) = !Syntax Error, Idk e-ikx R(k) => R(k) = (1/2π) !Syntax Error, Idx e+ikx r(x) Then we have Lx !Syntax Error, Idk e-ikx R(k) = (1/2π)!Syntax Error, Idk e-ikx !Syntax Error, Idk Lx e-ikx R(k) = (1/2π)!Syntax Error, Idk e-ikx (6.1) Now let's first try the most general case that an(x) = !Syntax Error, Idk' e-ik'x An(k') => An(k) = (1/2π) !Syntax Error, Idx e+ikx an(x) Lx = Σn=0∞ an(x) ∂xn = Σn=0∞ !Syntax Error, Idk' e-ik'x An(k') ∂xn => Lx e-ikx = Σn=0∞ !Syntax Error, Idk' e-ik'x An(k') ∂xn e-ikx Lx e-ikx = Σn=0∞ !Syntax Error, Idk' e-ik'x An(k') (-ik)n e-ikx Then (6.1) becomes !Syntax Error, Idk { Σn=0∞ !Syntax Error, Idk' e-ik'x An(k') (-ik)n e-ikx } R(k) = (1/2π)!Syntax Error, Idk e-ikx !Syntax Error, Idk { Σn=0∞ !Syntax Error, Idk' e-i(k+k')x An(k') (-ik)n } R(k) = (1/2π)!Syntax Error, Idk e-ikx Now we shall follow the same steps followed earlier in the more general case. Let k" = k+k' on LHS !Syntax Error, Idk { Σn=0∞ !Syntax Error, Idk" e-ik"x An(k"-k) (-ik)n } R(k) = (1/2π)!Syntax Error, Idk e-ikx And now swap k and k" on the LHS only !Syntax Error, Idk" { Σn=0∞ !Syntax Error, Idk e-ikx An(k-k") (-ik")n } R(k") = (1/2π)!Syntax Error, Idk e-ikx !Syntax Error, Idk e-ikx !Syntax Error, Idk" { Σn=0∞ An(k-k") (-ik")n } R(k") = (1/2π)!Syntax Error, Idk e-ikx Now use completeness of the e-ikx to claim !Syntax Error, Idk" { Σn=0∞ An(k-k") (-ik")n } R(k") = (1/2π) Now replace k" so that k-k" = k' and keep k where it was !Syntax Error, Idk' { Σn=0∞ An(k') (-i[k-k')n } R(k-k') = (1/2π) for all k Now swap k and k' !Syntax Error, Idk { Σn=0∞ An(k) (-i[k'-k)n } R(k'-k) = (1/2π) for all k' Now we identify a kernel here un(k'-k) = Σn=0∞ (-i[k'-k)n } R(k'-k) un(k) = Σn=0∞ (-ikn } R(k) and we then have Σn=0∞!Syntax Error, Idk un(k'-k) An(k) = -(1/2π) And so we arrive at a conclusion very similar to what we got before. There are an infinite discrete number of unknown functions An(k), and the equation must be true for an infinite continuum of values of k', so it is an infinite system of equations with an infinite number of unknowns. It happens that the RHS is independent of k' in this case. Again, this is pretty hopeless. There may or may not be solutions. 4. Known r(x-y) example does not have a Fourier Transform! 6A. I happen to know that this is true (Stak II p51 top) ∂x2[ (1/2)|x-y| ] = δ(x-y) s(x,y) = |x-y|/2 r(x) = |x|/2 Since this is a case that we know has a solution, we want to just see "how it works out". We had r(x) = !Syntax Error, Idk e-ikx R(k) => R(k) = (1/2π) !Syntax Error, Idx e+ikx r(x) So we then have R(k) = (1/2π) !Syntax Error, Idx e+ikx |x|/2 = (1/4π) !Syntax Error, Idx e+ikx |x| Since |x| is even in x, only the cosine term survives and we have = (1/4π) !Syntax Error, Idx cos(kx) |x| = (1/2π) !Syntax Error, Idx cos(kx) |x| = (1/2π) !Syntax Error, Idx cos(kx) x But we are suddenly dead in the water. The Fourier Transform only exists in general for functions which are square or abs integrable, We can see that !Syntax Error, Idx cos(kx) x diverges and is meaningless. So our whole method here breaks down completely. One could add that only integrable an(x) are allowed in what we did above. Can this example be worked using some other transform? Let's go back to : ∂x2[ (1/2)|x| ] = δ(x) Well, I know we can prove this doing a finite integral !Syntax Error, I∂x2[ (1/2)|x| ] = 1 LHS = ∂x [ (1/2)|x| ] |ε-ε = (1/2) - (-1/2) = 1 which is just the usual jump business in 1D. I suppose we could try expanding |x| and δ(x) on some orthogonal set on some interval like -1 to 1, perhaps the Pn(x). The lesson here is this: when you consider a priori finding Lx such that Lx r(x) = δ(x) and if you are going to try to solve this by doing some transform expansion, you better use transform such that r(x) has an expansion in that transform! In all my stuff above, I was always using the Fourier Transform. 5. Expand r(x) in a complete set of basis functions Here we generalize from Fourier Transform with basis functions e-ikx to an arbitrary φn(x) basis. Let's assume we have some orthonormal basis functions φn(x) on some interval surrounding x = 0. And assume that r(x) is expandable in this basis. Then we have r(x) = Σn=0∞ rnφn(x) δ(x-x') = Σn=0∞ φn(x) φn(x') So then we have for Lx r(x) = δ(x), Lx[Σn=0∞ rnφn(x)] = Σn=0∞ φn(x) φn(0) [Σn=0∞ rn Lx φn(x)] = Σn=0∞ φn(x) φn(0) Now since φn(x) is a complete set, we can write Lx φn(x) = Σn'=0∞ Ln'n φn'(x) Ln'n = <φn'| Lx | φn> Then we have Σn=0∞ rn{ Σn'=0∞ Ln'n φn'(x)} = Σn=0∞ φn(x) φn(0) Σn=0∞ Σn'=0∞ φn'(x) rn Ln'n = Σn=0∞ φn(x) φn(0) Now swap n ↔ n' on the LHS Σn=0∞ Σn'=0∞ φn(x) rn' Lnn' = Σn=0∞ φn(x) φn(0) Σn=0∞φn(x) Σn'=0∞ rn' Lnn' = Σn=0∞ φn(x) φn(0) and then appeal to completeness to claim Σn'=0∞ Lnn' rn' = φn(0) L R = φ(0) where we have an infinite matrix equation. Now let's take a radical next step. Let's assume our φn(x) are not just ANY orthonormal functions, but they are that particular set of orthonormal functions that diagonalize Lx. We know this set is the set of eigenfunctions of Lx (assuming it has eigenfunctions as an operator). So suppose Lx φn(x) = λn φn(x) Ln'n = <φn'| Lx | φn> = λn <φn'| φn> = λn δn,n' Then we go back to our last equation above Σn'=0∞ Lnn' rn' = φn(0) Σn'=0∞ λn δn,n' rn' = φn(0) λn rn = φn(0) We know rn , but we don't know anything about Lx so we don't know the φn or the λn. This seems to me to be a pretty stringent requirement. For example, suppose we were to just try as candidate Lx = the Legendre operator, and we know that Lx φn(x) = n(n+1)φn(x) = λn φn(x). We then have our candidate set of eigenfunctions and eigenvalues. The equation Lx r(x) = δ(x) only has a solution if it happens that rn = φn(0)/λn. This says r(x) has to be the Green's function of Lx. For Legendre, we know that Σn=0∞(1/Kn0) Pn(x) Pn(0) = δ(x) that is Σn=0∞ φn(x)φn(0) = δ(x) Then we consider Lx r(x) = δ(x) Lx r(x) = Σn=0∞ φn(x)φn(0) Lx Σn=0∞ rnφn(x) = Σn=0∞ φn(x)φn(0) Σn=0∞ rnλn φn(x) = Σn=0∞ φn(x)φn(0) rn = φn(0)/λn Then we have r(x) = Σn=0∞ rnφn(x) = Σn=0∞ φn(0)/λn φn(x) and this is just the famous result r(x;0) = Σn=0∞ φn(x)φn(0)/λn = g(x; 0) I guess I am just unable to answer this question: Question: Given a function r(x), is it, or is it not, a Green's Function of some self-adjoint differential operator? I have only scratched the surface of this question. I know how to show that r(x) is or is not the Green's Function of a differential operator with constant coefficients. There are an infinite number of self-adjoint differential operators, so there are an infinite number of Green's functions. But, I think if you pick a typical function like r(x) = 1/x3 , you will find that it is not a Green's function of any self-adjoint differential operator. I just don't know how to prove whether it is or is not. I just send off an email to Richard Price, maybe he will have something to add. 6. r(x-y) kernel which has Fourier Int expansion and constant an coefficients in Lx. Now let's maintain our Big Assumption of section 6 that s(x,y) = r(x-y), and this time let's try to search for an Lx with constant coefficients. We assume that r(x) has a Fourier Transform. We then have g(x) = ∫dy r(x-y)f(y) (1.1) // this is just the associated integral equation Lx r(x) = δ(x) r(x) = !Syntax Error, Idk e-ikx R(k) => R(k) = (1/2π) !Syntax Error, Idx e+ikx r(x) !Syntax Error, Idk Lx e-ikx R(k) = (1/2π)!Syntax Error, Idk e-ikx (6.1) Lx = Σn=0∞ an ∂xn Lx e-ikx = Σn=0∞ an (-ik)n e-ikx Then we have from 6.1, !Syntax Error, Idk { Σn=0∞ an (-ik)n e-ikx } R(k) = (1/2π)!Syntax Error, Idk e-ikx !Syntax Error, Idk e-ikx { Σn=0∞ an (-ik)n } R(k) = (1/2π)!Syntax Error, Idk e-ikx and completeness right away tells us that { Σn=0∞ an (-ik)n } R(k) = (1/2π) Let's define bn = (-i)n an then we have (this must be valid for all k in the range (-∞,∞) ! ) Σn=0∞ bn kn = (1/2π)(1/R(k)) If we can express 1/R(k) as a power series in k, then we have our solution! In the Love example, treated in a separate doc, we find r(x) = (d/π) / (d2 +x2) R(k) = e-d|k| /2d and the absolute value makes things not work and there is no solution Lx with constant coefficients. I think if r(x) is any even function, as in the Love example, which has a Fourier Transform, you are going to find that R(k) is even in k, and is therefore R(|k|) and you will be stuck with Σn=0∞ bn kn = (1/2π)(1/R(|k|)) = f( |k| ) and then there will be no solution for Lx which has constant coefficients.