Comparison of KL and Hankel transforms

Comparison of KL and Hankel transforms PhL 7.18.11 I have updated this comparison better in " a review of generic singular BC...". as of 8.29.11 I will try a side-by-side comparison here: (note that the interval in both cases is 0,∞ ) The KL case The Hankel case ODE: -(xg')' +ν2xg -λg/x = δ(x-ξ) -(xg')' + ν2 g/x - λxg = δ(x-ξ) s Green's: I-ik(νx<) K-ik(νx>), k ≡ (iπ/2) Jν(kx<) H ν(1)(kx>), k = I-i(νx<) K-i(νx>) (iπ/2) Jν(x<) H ν(1)(x>) Lg: Lg = -(xg')' +ν2xg Lg = -(xg')' + ν2g/x p(x) = x , q(x) = ν2x p(x) = x , q(x) = ν2/x [L-λs(x)]g: -(xg')' +ν2xg - λg/x s(x)=1/x -(xg')' + ν2g/x - λxg s(x)=x completeness: x δ(x-ξ) δ(x-ξ)/x = (2/π2) !Syntax Error, Idk k sinh(πk)Kik(νx) Kik(νξ) = !Syntax Error, Idk k Jν(kx)Jν(kξ) eigenfunctions: φk(x) = ( /π) Kik(νx) //wrong φk(x) = Jν(kx) //wrong Note that K-ν(z) = Kν(z) so K-ik(νx) OK. This last we know as follows. Looking at vol I p 304 we have (for a general spectrum situation) – (1/2πi) dλ g(x|ξ; λ) = δ(x-ξ)/s(x) = Σn φn(x) n(ξ) + ∫dα φα(x) α(ξ) Suppose there is only some continuous spectrum we label by α say in (0,∞). Then we have δ(x-ξ) = s(x) !Syntax Error, Idα φα(ξ) α(x) Apply ∫ dx φα'(x) to both sides to get ∫ dx φα'(x) δ(x-ξ) = ∫ dx s(x) φα'(x) !Syntax Error, Idα φα(ξ) α(x) => φα'(ξ) = !Syntax Error, Idα φα(ξ) ∫ dx s(x) φα'(x) α(x) This is consistent with this normalization: ∫ dx s(x) φα'(x) α(x) = δ(α-α') Now, in the KL case if we think of α = k as our spectrum label, then we can write δ(x-ξ) = s(x) !Syntax Error, Idα φα(ξ) α(x) = (1/x) !Syntax Error, Idk φk(ξ) k(x) = (1/x) (2/π2) !Syntax Error, Idk k sinh(πk)Kik(νx) Kik(νξ) from which we conclude that ( remember K order symmetry!) φk(x) = ( /π) Kik(νx) // wrong! with normalization given by ∫ dx (1/x) φk'(x) k(x) = δ(k-k') In the Hankel case we again think of α = k and we have δ(x-ξ) = s(x) !Syntax Error, Idα φα(ξ) α(x) = x !Syntax Error, Idk φk(ξ) k(x) = x !Syntax Error, Idk k Jν(kx)Jν(kξ) from which we conclude that φk(x) = Jν(kx) // wrong with normalization given by ∫ dx (x) φk'(x) k(x) = δ(k-k') or ∫ dx (x) Jν(k'x) Jν(kx) = δ(k-k') or ∫ dx x Jν(k'x) Jν(kx) = δ(k-k')/ = δ(k-k')/k which agrees with my transforms.doc grouping which I quote here: φk(x; ν) = Jν(kx) // complete orthonormal set f(ρ) = !Syntax Error, Idk k Jν(kρ) Fν(k) // expansion Fν(k) = !Syntax Error, Idρ ρ Jν(kρ) f(ρ) // projection !Syntax Error, Idρ ρ Jν(kρ) Jν(k'ρ) = δ(k-k')/k // orthogonality !Syntax Error, Idk k Jν(kρ) Jν(kρ') = δ(ρ-ρ')/ρ // completeness Statement of the KL transform. We have from above that φk(x; ν) = ( /π) Kik(νx) and then we can write orthogonality this way ∫ dx (x) φk'(x) k(x) = δ(k-k') (2/π2) ∫ dx x Kik(νx) Kik(νx) = δ(k-k') Now in the KL doc we wrote -(xg')' +μxg -λx-1g = δ(x-ξ) λ = γ2 , μ = k2 f(x) = (2/π2) !Syntax Error, I κdκ sinh(πκ) Kiκ(x) F(κ) which is 4.139. F(κ) ≡ !Syntax Error, Idx f(x) Kiκ(x) /x which we can translate to say -(xg')' +μxg -λx-1g = δ(x-ξ) λ = γ2 , μ = k2 f(x) = (2/π2) !Syntax Error, I κdκ sinh(πκ) Kiκ(x) F(κ) which is 4.139. F(κ) ≡ !Syntax Error, Idx f(x) Kiκ(x) /x So in our current document we will then have μ → ν2 and -(xg')' +μxg -λx-1g = δ(x-ξ) f(x) = (2/π2) !Syntax Error, I κdκ sinh(πκ) Kiκ(x) F(κ) which is 4.139. F(κ) ≡ !Syntax Error, Idx f(x) Kiκ(x) /x And here are the corresponding K-L results f(x) = (2/π2) !Syntax Error, Idκ κ sinh(πκ) Kiκ(x) F(κ) // expansion F(κ) ≡ !Syntax Error, Idx f(x) Kiκ(x) /x // projection (2/π2)∫ dx (1/x) Kik(νx) Kik'(νx) = δ(k-k') // orthogonality (2/π2) !Syntax Error, Idk k sinh(πk)Kik(νx) Kik(νx') = xδ(x-x') // completeness Comments: The ODE's are the same if we just swap ν2 ↔ -λ . This rule converts either into the other. If we make this swap, either Green's function becomes the other Green's function (as must be the case) I-i(νx<) K-i(νx>) = Iν(-ix<)Kν(-ix>) = i-ν Jν(i * -ix<) (π/2) iν+1 Hν(1)(i*-ix>) = (iπ/2) Jν(x<) Hν(1) (x>) This swap does NOT convert one completeness relation into the other. One has an integration of a factor in the Bessel function order, the other an integration of a factor in the Bessel function argument. The underlying L operators are different. The s(x) and q(x) functions are different. Both transforms are associated with a Bessel equation with two parameters I have called ν2 and λ. And both transforms involve the interval (0,∞). Both SL problems have a λ spectrum of the positive real axis. The two associated eigenvalue problems are different even though the spectrum is the same: Lν uν = λ s(x) uν : KL: Lνu = -(xu')' +ν2xu s(x) = 1/x Hankel: Lνu = -(xu')' +ν2(1/x)u s(x) = x Question: would you refer to either of these L's as a Bessel Operator? I guess they both are such, in the sense that if you consider the EV problem associated with that L, and if you pick the right s(x), you arrive at a 2-parameter Bessel ODE in either case. It is not the same ODE, however, since -λ and ν2 are swapped. The official Bessel's equation [ Schaum p 136 with ν=n ] has one numerator parameter ν2 which is the order. But we can write a "scaled Bessel's equation" which has another numerator parameter k2. Then we go on two different paths if we take one or the other of these two parameters to be the λ eigenvalue parameter of a SL problem. Each path has its own q(x) and s(x). Each path is a different SL problem and each path results in a different transform. In either path, the functions involved are always something in the "family of Bessel functions".