The Mellin Transform

The Mellin Transform PhL 8.10.11 The Mellin Sine Transform Consider this ODE on the interval (0,a) -(ru')' = λ (1/r)u s(r) = 1/r Lu = -(ru')' -(ru')' - (λ/r)u = 0 -(ru')' - (ν2/r)u = 0 This is the limit of a Bessel equation when k→0, -(xu')' + ν2u/x - k2xu = 0 Zν(kx) (*) -(xu')' – ν2u/x - k2xu = 0 Z±iν(kx) (*) -(xu')' – ν2u/x + k2xu = 0 K±iν(kx) (*) where in the last form (which will match ours for k=0) we see the real-valued KL function solutions. Our limiting situation does not give of course a Bessel function. To avoid the need for ugly factors in exponents and elsewhere, let's use λ = ν2 as it appears in this last Bessel equation. The solutions are then of the form r±iν as we now confirm u = r±iν u' = (±i)ν r±iν-1 (ru') = (±i)ν r±iν (ru')' = (±i)ν (±i)ν r±iν-1 = - ν2 r±iν/r = -ν2u/r => - (ru')' = + ν2u/r QED However, if ν = 0, our solutions are r±i0 = 1, and then there must be a second independent solution which is ln(r), so in this case ν = 0 we get the famous A + B ln(r) solution. A linear combination which vanishes at r=a is given by uλ(r) = (1/2)[ (r/a)iν - (r/a)-iν] = sin[ ν ln(r/a) ] ν = and again if λ=ν=0 we would use u0(r)= ln(r/a) to replace the above, which is in agreement with the limit of the sin function for small ν, ignoring the constant ν. Probably r=0 is a limit-point situation [yes, p 316] and the limit of uλ(r=0) only exists in some distributional sense. But we know that our solution has finite s norm at r = 0 with our s(r) = 1/r because {∫ dr (1/r) sin[ b ln(r/a) ] }|r=0 = { cos(b ln(r/a))/b }|r=0 which is no doubt 0 as a distribution, and in any event is bounded above and below by 1 and -1, so one cannot say that this endpoint diverges. The functions uλ(r) = sin[ ln(r/a) ] form a complete set on (0,a) as λ sweeps the + real axis. We know this because these are the EF's of a SL problem. It turns out that we don't have to worry about the λ=0 case in this transform, but I am not sure why. This is at the very end of the "cut" in the λ plane for the g integral. I will just ignore this issue for now. We can establish normalization by requiring [ remember that s(r) = 1/r ] uλ(r) = C sin[ ln(r/a) ] !Syntax Error, Idr(1/r) C2 sin[ ln(r/a) ] sin[ ln(r/a) ] = δ(λ-λ') Change variables to x = r/a to get C2!Syntax Error, Idx(1/x) sin[ ln(x) ] sin[ ln(x) ] = δ(λ-λ') Let = k ( called ν above), change to y = lnx to get [ dy = dx/x ] C2!Syntax Error, Idy sin(ky)sin(k'y) = δ(λ-λ') C2!Syntax Error, Idy sin(ky)sin(k'y) = δ(λ-λ') (*) But we know from our Fourier Sine Transform work that !Syntax Error, Idy sin(ky)sin(k'y) = (π/2)δ(k-k') And we also know that δ(λ-λ') = δ(k2-k'2) = (1/2|k|)δ(k-k') = (2)-1 δ(k-k') // k and k' both > 0 so that δ(k-k') = (2) δ(λ-λ') and putting this in (*) above we have C2 (π/2)δ(k-k') = δ(λ-λ') = (2)-1 δ(k-k') C2 π = (1/) C2 = (π)-1 C = (π)-1/2 Thus, our properly normalized eigenfunctions are uλ(r) = (π)-1/2 sin[ ln(r/a) ] **** We can then call upon our Big Result from "review of generic singular...doc" which says - (1/2πi) ∫C dλ g(x|ξ;λ) = Σλn φλn*(x)φλn(ξ) + ∫dλ'φλ(x)* φλ'(ξ) = δ(x-ξ)/s(x) and since we have only a continuous spectrum - (1/2πi) ∫C dλ g(x|ξ;λ) = ∫dλ'φλ'(x)* φλ'(ξ) = δ(x-ξ)/s(x) So we then find that δ(r-r')r = ∫dλ' (π)-1 sin[ ln(r/a) ] sin[ ln(r'/a) ] δ(r-r')r = (1/π) ∫dλ' (1/) sin[ ln(r/a) ] sin[ ln(r'/a) ] If we now change to = k' then dk' = (1/2) λ'-1/2dλ' = (2)-1dλ' then dλ' = (2) dk' so that dλ' (1/) = 2 dk' and we find δ(r-r')r = (2/π)!Syntax Error, Idk' sin[ k' ln(r/a) ] sin[ k' ln(r'/a) ] and this agrees with Stak 4.134 on page 316 vol I. I will rewrite this using k δ(r-r')r = (2/π)!Syntax Error, Idk sin[ k ln(r/a) ] sin[ k ln(r'/a) ] **** Then suppose we try this for a transform F(k) = !Syntax Error, Idr sin[ k ln(r/a) ] f(r) **** f(r) = (2/π) !Syntax Error, Idk sin[ k ln(r/a) ] F(k) Stak calls this the Mellin Sine Transform "for want of a better term". And the conjugate to r is here called k, but we know it is really k = where λ is the true continuous eigenvalue. On page 316 Stak asks the reader to obtain the completeness statement using the other method of first finding the Green's Function on (0,a) and then stuffing that into the g integral formula. You see that his Green's has a λ cut, and this creates a discontinuity which then gives the formula. Notice that his Green's function vanishes at r = 0 cleanly if you set λ to λ +iε, take the limit r→0, then set ε = 0. This is his method of limited absorption. I never did this Exercise 4.28 in my raw Chapter 4 notes. Stak's method shows that there is no "special case" to worry about at λ = 0 If you replace the BC that u(a) = 0 with u'(a) = 0, you get the Mellin Cosine Transform as commented on in Exercise 4.29. Summary of the Mellin Sine Transform uλ(r) = (π)-1/2 sin[ ln(r/a) ] k = // eigenfunctions !Syntax Error, Idr sin(kr)sin(k'r) = (π/2)δ(k-k') // orthogonality !Syntax Error, Idr(1/r) uλ(r) uλ'(r) = δ(λ-λ') s(r) = 1/r (2/π)!Syntax Error, Idk sin[ k ln(r/a) ] sin[ k ln(r'/a) ] = δ(r-r')r // completeness !Syntax Error, Idλ φλ(r)* φλ(r') = δ(r-r') r F(k) = !Syntax Error, Idr sin[ k ln(r/a) ] f(r) // projection f(r) = (2/π) !Syntax Error, Idk sin[ k ln(r/a) ] F(k) // expansion