a review of the generic singular BC problem

A review of the generic singular BC problem PhL 8.10.11 Begin scan of the "regular" doc and look for things that are now different. 1 Digression to review of my Stak Meta Notes on this subject 1 Resume Scan of the Regular doc, looking for differences 3 The Eigenfunction expansion for g 3 Completeness of the eigenfunctions 5 The integral of g formula 6 Begin scan of the "regular" doc and look for things that are now different. I will now just naively read through the regular BC doc and look for things that might be a problem in the case that b= ∞. I think formal self-adjointness is OK, since we ignore parts. Lemma 1 is OK. Lemma 2 is OK. However, when we integrate Lemma 2 we get !Syntax Error, Idx { Lu - u} = p(∞) W[u,; ∞] - p(a) W[u,; a] where recall u and v are ANY functions. We are concerned about those ∞ appearances in the above! Digression to review of my Stak Meta Notes on this subject Now we have to pay attention to the meta notes on this subject. The trick is to realize that there must exist two linearly independent functions φ and ψ of Lλu=0 which have certain unmixed BC's at x=a. Each has a different BC there, but our only interest is in having two lin indep solutions of the ODE (without regard to the original problem BC's) with both p(a)W(φ,; a) = 0 and p(a)W(ψ,; a) = 0. Stak shows that p(a)W(φ,ψ; a) = - (α12 + α22) ≠ 0 and that is why they are lin indep. The BC's at a for φ and ψ are carefully selected to make this these two pW objects vanish and W(φ,ψ) not vanish. Our concern is the integrability of a candidate solution at the b=∞ endpoint. We try u = φ + mψ as our candidate for a function which is finite-s-norm at the upper endpoint, and of course u and ψ are lin indep. We back off the limit b0→ b = ∞ and we select β1,2 in order to get p(bo) W( u, ; b0) = 0 from a right-end unmixed BC. Imposing this BC puts a condition on our parameter m. Since everything is a function of λ (things like φ,ψ and u, because they are all solutions of Lλf=0), we find that m is also a function of λ, and we call it m(λ). We define h = β1/β2 from our imposed BC. As we let this ratio vary all it wants, the solution value for m(λ) lies on a circular locus in the m=plane. Here we are varying our b0 BC in every possible way before taking the limit b0→b = ∞. We want to see every possible way that we can obtain our desired result p(bo) W( u, ; b0) = 0 at the right endpoint. Stak gives expressions for the center A and radius r of this circle. If λ = real the circle happens to be the real axis, so assume λ not real. So, m lies on a certain circle in the complex m plane of some radius which is a function of b0. As we take b0→ b, the limiting circle might have a finite radius (limit circle case) or a zero radius (limit point case). In the first case, it turns out that ψ is not finite-s-norm at the upper endpoint, but u is. In the circle case, both u and ψ are finite s norm, and we have two possible solutions at the upper endpoint. Now in either case, our one or two viable solutions satisfy p(b) W( u, ; b) = 0. Now we must keep in mind that everything hinges on our choice for parameter λ. For one λ we might in theory get limit circle, and for some other λ we might get limit point. But it turns out (Weyl's theorem) that if you are limit circle for one λ, then you are limit circle for all λ! So you are motivated to pick a simple λ to see which case you are in. So how does this help us? Here is one payoff. We selected some arbitrary λ and said Lλf= 0 for all three functions φ,ψ and u= φ + mψ . Since we are ignoring our original problem BC's, λ can just be any complex number. But equation 4.80a is true for any solution of Lλf= 0. We have just shown that for either our one limit-point solution u, or for our two limit circle solutions u and ψ, pW = 0 both at x=a and at x=b. Therefore 4.80a tells us that Im(λ) = 0. So we recover "eigenvalues must be real". Question: Suppose there is only one limit-point solution u = φ+mψ for some m. Can we argue that the BC at x=a on u is arbitrary? Lets try to write Au(a) + Bu'(a) = 0 A[ φ(a) + mψ(a)] + B[ φ'(a) + mψ'(a)] = 0 A[ -α2 + mα1] + B[ α1/p(a) + mα2/p(a)] = 0 α1[mA + B/p(a)] + α2[ -A +Bm/p(a)] = 0 α1/α2 = [ A - Bm/p(a)]/[ mA + B/p(a)] = [ (A/B) - m/p(a)]/[ m(A/B) + 1/p(a)] For a given m on the solution circle or point, the only problem we might have is if the ratio of A and B are such that m(A/B)= -1/p(a). If we assume that our BC avoids this particular ratio, then we can just pick any α1 and α2 such that α1/α2 = the RHS shown above. If we hit the bad ratio, we just select α1= 0 and α2 = any non-zero value. Remember the condition is that you cannot have both α1 and α2 = 0 at the same time. So, I think we can claim that our homo unmixed BC at the left end is in fact arbitrary. So we now look at our original singular problem. We have some arbitrary unmixed BC at the left end a. We know there are either 1 or 2 solutions of Lλf= 0 that are integrable at the high end b = ∞. So consider the two cases: Limit Point Case: solution = u(x) meets arbitrary BC on the left, if finite s norm on the right In this case, we know there IS a solution on our interval (a,∞) that meets our left end BC whatever (almost) it is. In effect, our right end BC is just that the function be finite s norm there, which surely means u→0 at the high end. Limit Circle Case: solutions = u(x), ψ(x) In this case, we can write a candidate general solution as w(x) = c u(x)+d ψ(x). We are OK at the right end (we meet the effective BC there) for and c,d choice. Our left end BC is this Aw(a) + Bw'(a) = 0 A[ c u(a) + dψ(a)] + B[ c u'(a) + dψ'(a)] = 0 c [A u(a) + B u'(a) ] + d [ Aψ(a) + Bψ'(a)] = 0 (c/d) = - [ Aψ(a) + Bψ'(a)]/ [A u(a) + B u'(a) ] = - [ (A/B)ψ(a) + ψ'(a)]/ [( A/B) u(a) + u'(a) ] In general for arbitrary A and B, it is clear that we can find a ratio (c/d) that works! If the denominator happens to vanish then we just set c = 0 and d = anything. Notice that in all cases, the overall scale of any solution of Lλf=0 is arbitrary. Conclusion: In either case, circle or point, if we specify an unmixed BC at the low end a, there will be some unique solution to our singular BV problem for interval (a,∞) for given real λ unless λ happens to not be in the spectrum, in which case there will be no solution. If we ignore our low end BC, we conclude that for general λ, there will be one or two viable solutions (finite s norm) based on limit or circle. But if λ = real, there might be no solutions! Although Stak uses a low end BC in his Weyl proofs, he is mostly interested in deciding whether there are 1 or 2 finite-s-norm solutions at a given singular endpoint, without regard to a BC at the other endpoint. Indeed, you might have both endpoints be singular. Resume Scan of the Regular doc, looking for differences The Three Basic Theorems seem still valid, since I did not assume discrete spectrum there. And g is still symmetric for the same reason. Solution u in terms of g stays the same. The Eigenfunction expansion for g Here we do have to make changes. We know that the continuous part of any singular spectrum comes from coalescence of discrete points as b0→ b = ∞. So in the limit we have to included potentially both a discrete and continuous contribution to an expansion, so we now write u(x) = Σλ=λn uλnφλn(x) + ∫dλ' uλ' φλ'(x) F(x) = Σλ=λn Fλnφλn(x) + ∫dλ' Fλ' φλ'(x) This is what a complete expansion now looks like in the general case. So here we go Lu-λsu = f = sF (L-λs) [Σλ=λn uλnφλn(x) + ∫dλ' uλ' φλ'(x)] = s [Σλ=λn Fλnφλn(x) + ∫dλ' Fλ' φλ'(x)] [Σλ=λn uλn(L-λs)φλn(x) + ∫dλ' uλ' (L-λs)φλ'(x)] = s [Σλ=λn Fλnφλn(x) + ∫dλ' Fλ' φλ'(x)] [Σλ=λn uλn(λns-λs)φλn(x) + ∫dλ' uλ' (λ's-λs)φλ'(x)] = s [Σλ=λn Fλnφλn(x) + ∫dλ' Fλ' φλ'(x)] [Σλ=λn uλn(λn-λ)φλn(x) + ∫dλ' uλ' (λ'-λ)φλ'(x)] = [Σλ=λn Fλnφλn(x) + ∫dλ' Fλ' φλ'(x)] Σλ=λn [ uλn(λn-λ) - Fλn] φλn(x) + ∫dλ' [ uλ' (λ'-λ) - Fλ']φλ'(x) = 0 Now in the usual way we appeal to completeness in each EF in our set and we conclude uλn(λn-λ) - Fλn = 0 uλ' (λ'-λ) - Fλ' = 0 uλn(λn-λ) = Fλn uλ' (λ'-λ) = Fλ' uλn = Fλn/ (λn-λ) uλ' = Fλ'/(λ'-λ) Remember that λ is our thing in Lλ. Our solution is therefore u(x) = Σλ=λn uλnφλn(x) + ∫dλ' uλ' φλ'(x) = Σλ=λn [Fλn/ (λn-λ)]φλn(x) + ∫dλ'[Fλ'/(λ'-λ)] φλ'(x) Now let's try our special case of f = δ as we did before. F(x) = δ(x-ξ)/s(x) Fλn = <φλn, sF> = < φλn,sδ/s> = < φλn n,δ> = φλn (ξ)* Fλ = <φλ, sF> = <φλ,sδ/s> = <φλ,δ> = φλ(ξ)* We have then shown that g(x) = g(x|ξ;λ) = Σλ=λn [Fλn/ (λn-λ)]φλn(x) + ∫dλ'[Fλ'/(λ'-λ)] φλ'(x) = Σλ=λn [φλn (ξ)*/ (λn-λ)]φλn(x) + ∫dλ'[φλ(ξ)*/(λ'-λ)] φλ'(x) = Σλ=λn φλn (ξ)* φλn(x)/ (λn-λ) + ∫dλ'φλ(ξ)* φλ'(x)/(λ'-λ)] or if we throw in the symmetry of g, we get g(x|ξ;λ) = Σλ=λn φλn (x)* φλn(ξ)/ (λn-λ) + ∫dλ'φλ(x)* φλ'(ξ)/(λ'-λ)] I did all this work just to show that there are no weird extra weight functions appearing in the integral. Completeness of the eigenfunctions Here we need some work as well. I think we can normalize the φλn (x) and the φλ(x) [ remember that these are in theory unrelated functions! ] we will get <φλm,sφλn> = δn,m <φλ,sφλ'> = δ(λ-λ') <φλ,sφλn> = 0 The last item is the most mysterious, but with the notion of the cut as limit of poles, it ought to work. This tells us that the completeness relation must have this form δ(x-ξ)/s(x) = Σλn φλn*(x)φλn(ξ) + ∫dλ'φλ'(x)* φλ'(ξ) Keep in mind that we have Lλ φλn = s λn φλn for λn in the point spectrum Lλ φλ = s λ φλ for λ in the continuous spectrum so we never want to write the dummy integration variable as λ ! To verify this completeness result, we take the weighed inner product against each eigenfunction of our complete basis φλm(ξ) = <φλm, s(x)δ(x-ξ)/s(x)> = <φλm, s(x) [Σλn φλn*(x)φλn(ξ) + ∫dλ'φλ(x)* φλ'(ξ)]> = <φλm, s(x) [Σλn φλn*(x)φλn(ξ)] = Σλn φλn(ξ) δm,n = φλm(ξ) and similarly for the other case we will verify that φλ(ξ) = φλ(ξ) . Notice that the third result above is essential. The integral of g formula We have already shown that (our expansion formula with a mixed spectrum) g(x|ξ;λ) = Σλ=λn φλn (x)* φλn(ξ)/ (λn-λ) + ∫dλ'φλ(x)* φλ'(ξ)/(λ'-λ)] where the λn and λ' lie on the real axis. If we use a CCW great circle contour, it certainly will include that real axis, and we can then say (1/2πi) ∫C dλ g(x|ξ;λ) = Σλn φλn(x)*φλn(ξ) (1/2πi) ∫C dλ/(λn-λ) + ∫dλ'φλ(x)* φλ'(ξ) (1/2πi) ∫C dλ/(λ'-λ) = - Σλn φλn(x)*φλn(ξ) - ∫dλ'φλ(x)* φλ'(ξ) The contour is in the correct direction around each pole. The minus sign arise because the residue formula requires 1/(λ-λn) and (λ-λ') on the bottom! We can now combine our integral of g formula with our completeness formula to say - (1/2πi) ∫C dλ g(x|ξ;λ) = Σλn φλn*(x)φλn(ξ) + ∫dλ'φλ(x)* φλ'(ξ) = δ(x-ξ)/s(x) We have now proven this to be true for any singular BV problem! A common method we use to find the eigenfunctions is to compute g using standard methods, and try to get it into a Σn + ∫ form with a symmetric summand, and then we can just "read off" the φn(x) and φλ(x)normalized eigenfunctions! I have never seen this done in the mixed case, but it ought to work. I like writing the integration variable as λ' because it reminds me that it really is the eigenvalue parameter itself, not something related to it. That is to say, φλ'(x) satisfies Lλ' φλ'(x) = 0. Sometimes people replace λ' by ν, but then this leads to confusion if we happen to have ν with meaning λ = ν2 say. The "measure" in the integral part above is just dλ', not something else. Summary of the Above Assume a mixed spectrum, and assume orthonormal eigenfunctions, which means this <φλm,sφλn> = δn,m <φλ,sφλ'> = δ(λ-λ') <φλ,sφλn> = 0 Notice that the inner product associated with our Hilbert Space contains the weight function s ! These eigenfunctions are eigenfunctions of self-adjoint operator Lλ = L - λs(x) // as on p 268 so Lλ φλn = s λn φλn for λn in the point spectrum Lλ φλ = s λ φλ for λ in the continuous spectrum Theorem 1: Assuming the usual stuff, this set of eigenfunctions is complete, and the proper way to express this fact is ( a one line proof is given above) δ(x-ξ)/s(x) = Σλn φλn*(x)φλn(ξ) + ∫dλ'φλ'(x)* φλ'(ξ) Theorem 2: Bilinear expansion of g Consider this Fred 2 inhomo Lλu = Lu-λsu = f = sF We show above that the solution to this problem is u(x) = Σλ=λn [Fλn/ (λn-λ)]φλn(x) + ∫dλ'[Fλ'/(λ'-λ)] φλ'(x) where F(x) = Σλ=λn Fλn φλn(x) + ∫dλ' Fλ' φλ'(x) We then apply this to the case f = δ and find F = δ/s so that Fλn = φλn (ξ)* etc and we find: Lλg = Lg-λsg = δ g = g(x|ξ) = Σλ=λn φλn (ξ)* φλn(x)/ (λn-λ) + ∫dλ'φλ(ξ)* φλ'(x)/(λ'-λ)] This is the bilinear expansion for g. Theorem 3: Integral of g relation. - (1/2πi) ∫C dλ g(x|ξ;λ) = Σλn φλn*(x)φλn(ξ) + ∫dλ'φλ'(x)* φλ'(ξ) Combined Results We can combine all these results into the following dense form: - (1/2πi) ∫C dλ g(x|ξ;λ) = Σλn φλn*(x)φλn(ξ) + ∫dλ'φλ'(x)* φλ'(ξ) = δ(x-ξ)/s(x) <φλm,sφλn> = δn,m <φλ,sφλ'> = δ(λ-λ') <φλ,sφλn> = 0 Notice carefully that in the integral, the integration variable is the actual eigenvalue variable. We can state this symbolically where we include the integral with the sum - (1/2πi) ∫C dλ g(x|ξ;λ) = Σλn φλn*(x)φλn(ξ) = δ(x-ξ)/s(x) <φλm,sφλn> = δn,m Notice there is no mysterious factor Kn as I have in some of my earlier quotes of this result. Example 1: the KL transform business We start with -(xg')' +μxg -λx-1g = δ(x-ξ) Lλ = L - λs s = x-1 L = -(xg')' +μxg We show (see KL doc) that the free-space solution g is not unique, but that this is a good one to use: g(x|ξ;λ) = I-iγ(kx<) K-iγ(kx>) where γ ≡ , μ = k2 Since this is a purely continuous spectrum problem, we know that - (1/2πi) ∫C dλ g(x|ξ;λ) = ∫dλ'φλ(x)* φλ'(ξ) We show in that same KL doc that - dλ g(x|ξ; λ) = !Syntax Error, Idλ' (2/π)sin(πα)Kα(kx)Kα(kξ) where α = i To be λ purist, we rewrite Kα(x) = Ki(x) sin(πα) = sin(πi) = i sh(π) so the above becomes - dλ g(x|ξ; λ) = !Syntax Error, Idλ' (2i/π) sh(π) Ki(kx) Ki(kξ) - (1/2πi) dλ g(x|ξ; λ) = (1/π2)!Syntax Error, Idλ' sh(π) Ki(kx) Ki(kξ) Therefore we may directly identify !Syntax Error, Idλ'φλ'(x)* φλ'(ξ) = (1/π2)!Syntax Error, Idλ' sh(π) Ki(kx) Ki(kξ) The integrands must be equal, so we get φλ'(x)* φλ'(ξ) = (1/π2) sh(π) Ki(kx) Ki(kξ) We then have this completely unambiguous expression for the eigenfunction (remember the order symmetry of the K, and remember that Ki(kx) is real. φλ(x) = (1/π) Ki(kx) Now at this point we can use γ ≡ or λ = γ2, and choose to label φ with γ, so we have φγ(x) = (1/π) Kiγ(kx) Then we have from above λ = γ2 dλ = 2γdγ - (1/2πi) dλ g(x|ξ; λ) = (1/π2)!Syntax Error, Idλ' sh(π) Ki(kx) Ki(kξ) = (1/π2)!Syntax Error, I[ 2γdγ] sh(πγ) Kiγ(kx) Kiγ(kξ) = (2/π2)!Syntax Error, Idγ γ sh(πγ) Kiγ(kx) Kiγ(kξ) = δ(x-ξ)/s(x) = x δ(x-ξ) // agrees with 4.137 Stak Vol I Now we can look at orthogonality: <φλ,sφλ'> = δ(λ-λ') !Syntax Error, Idx s(x) (1/π) Ki(kx) (1/π) Ki(kx) = δ(λ-λ') or (1/π2) !Syntax Error, Idx s(x) Ki(kx) Ki(kx) = δ(λ-λ') or !Syntax Error, Idx x-1 Ki(x) Ki(x) = π2 δ(λ-λ')' / sh(π) If we now again write λ = γ2 we have δ(λ-λ') = δ(γ2 - γ'2 ) =δ(γ-γ')/|2γ| = δ(γ-γ')/(2γ) Then orthogonality can be written, since γ = , !Syntax Error, Idx x-1 Kiγ(kx) Kiγ'(kx) = π2 { δ(γ-γ')/(2γ) } /sh(πγ) !Syntax Error, Idx x-1 Kiγ(kx) Kiγ'(kx) = δ(γ-γ') π2/[2γ sh(πγ)] Where can I verify this? You can see why the RHS is independent of k, just change variables on the LHS to y = kx and it then looks the same with k =1. I don't think Stak writes this down anywhere. Amazingly I find this in a 2010 short journal article which I have saved as bielski.pdf and there is my formula!!!! The two referenced papers are 2006 and 2009. In this paper the authors more or less do what I just did above and in fact quote Stakgold 1998. So there is my verification! Now here is the transform taken from my KL doc f(x) = (2/π2) !Syntax Error, I dγ γ sinh(πγ) Kiγ(kx) F(ν) which is 4.139. F(γ) ≡ !Syntax Error, Idx f(x) Kiγ(kx) /x Finally I am going to attempt to gather the basic four results in one place and then try to convert it all to generic variable names. In this Example we have shown that -(xg')' +μxg -λx-1g = δ(x-ξ) where γ ≡ , μ = k2 Lλ = L - λs s = x-1 L = -(xg')' +μxg φλ(x) = (1/π) Ki(kx) φγ(x) = (1/π) Kiγ(kx) γ = x δ(x-ξ) = (1/π2)!Syntax Error, Idλ' sh(π) Ki(kx) Ki(kξ) = (2/π2)!Syntax Error, Idγ γ sh(πγ) Kiγ(kx) Kiγ(kξ) !Syntax Error, Idx x-1 Kiγ(kx) Kiγ'(kx) = δ(γ-γ') π2/[2γ sh(πγ)] f(x) = (2/π2) !Syntax Error, I dγ γ sinh(πγ) Kiγ(kx) F(ν) which is 4.139. F(γ) ≡ !Syntax Error, Idx f(x) Kiγ(kx) /x So one way to do all this stuff is to write μ = k2 in the ODE and this remove μ completely: -(xg')' +k2xg -λx-1g = δ(x-ξ) where γ ≡ , μ = k2 Lλ = L - λs s = x-1 L = -(xg')' +k2xg φλ(x) = (1/π) Ki(kx) φγ(x) = (1/π) Kiγ(kx) γ = x δ(x-ξ) = (1/π2)!Syntax Error, Idλ' sh(π) Ki(kx) Ki(kξ) = (2/π2)!Syntax Error, Idγ γ sh(πγ) Kiγ(kx) Kiγ(kξ) !Syntax Error, Idx x-1 Kiγ(kx) Kiγ'(kx) = δ(γ-γ') π2/[2γ sh(πγ)] f(x) = (2/π2) !Syntax Error, I dγ γ sinh(πγ) Kiγ(kx) F(γ) which is 4.139. F(γ) ≡ !Syntax Error, Idx f(x) Kiγ(kx) /x Now why not just set λ = γ2 in the ODE and then be rid of λ (now that we are done using it) -(xg')' +k2xg -γ2x-1g = δ(x-ξ) Lλ = L - γ2s s = x-1 L = -(xg')' +k2xg φγ(x) = (1/π) Kiγ(kx) x δ(x-ξ) = (2/π2)!Syntax Error, Idγ γ sh(πγ) Kiγ(kx) Kiγ(kξ) !Syntax Error, Idx x-1 Kiγ(kx) Kiγ'(kx) = δ(γ-γ') π2/[2γ sh(πγ)] f(x) = (2/π2) !Syntax Error, I dγ γ sinh(πγ) Kiγ(kx) F(γ) F(γ) ≡ !Syntax Error, Idx f(x) Kiγ(kx) /x Now make replacements k2 = α2 and γ2 = β2 -(xu')' + α2xu -β2x-1u = 0 Lλu = Lu - β2s u = -(xu')' + α2xu - β2x-1 u = 0 Lλ = L - β2s s = x-1 L = -(xu')' + α2xu φβ(x) = (1/π) Kiβ(αx) L φβ(x) = β2(x-1) φβ(x) x δ(x-ξ) = (2/π2)!Syntax Error, Idβ β sh(πβ) Kiβ(αx) Kiβ(αξ) !Syntax Error, Idx x-1 Kiβ(αx) Kiβ'(αx) = δ(β-β') π2/[2β sh(πβ)] f(x) = (2/π2) !Syntax Error, I dβ β sinh(πβ) Kiβ(αx) F(β) F(β) ≡ !Syntax Error, Idx f(x) Kiβ(αx) /x Then maybe reorder this into my usual ordering -(xu')' + α2xu -β2x-1u = 0 or Lu = β2x-1u where L = -(xu')' + α2xu or Lλu = 0 where Lλ = L - β2x-1 λ = β2 s(x) = x-1 φβ(x) = (1/π) Kiβ(αx) f(x) = (2/π2) !Syntax Error, I dβ β sh(πβ) Kiβ(αx) Fα(β) // expansion Fα(β) ≡ !Syntax Error, Idx x-1 f(x) Kiβ(αx) // projection !Syntax Error, Idx x-1 Kiβ(αx) Kiβ'(αx) = δ(β-β') π2/[2βsh(πβ)] // orthogonality x δ(x-x') = (2/π2)!Syntax Error, Idβ β sh(πβ) Kiβ(αx) Kiβ(αx') // completeness Example 2: the Hankel transform business We start with (see Hankel doc) -(xw')' + ν2 w/x - λxw = 0 λ = k2 k = Lg - λs(x)g = δ(x-ξ) Lg = -(xg')' + ν2 g/x s(x) = x x in (0,∞) We write the Green's Function problem and then solve it to get g(x|ξ; λ) = (iπ/2) Jν(x<) H ν(1)( x>). I then show that – (1/2πi) ∫C dλ g(x|ξ; λ) = (1/2) !Syntax Error, Idλ' Jν(x<)Jν(x>) Direct comparison to - (1/2πi) ∫C dλ g(x|ξ;λ) = ∫dλ'φλ'(x)* φλ'(ξ)= δ(x-ξ)/s(x) tells us two things. First, (1/2) !Syntax Error, Idλ' Jν(x<)Jν(x>) = δ(x-ξ)/x Second ∫dλ'φλ'(x)* φλ'(ξ) = (1/2) !Syntax Error, Idλ' Jν(x<)Jν(x>) so we conclude from this that φλ(x) = (1/) Jν(x) We can now relable this orthonormal eigenfunction with k = k2 = λ 2kdk = dλ φk(x) = (1/) Jν(kx) Then completeness above becomes (1/2) !Syntax Error, I[2k'dk'] Jν(k'x)Jν(k'ξ) = δ(x-ξ)/x or !Syntax Error, Idk' k' Jν(k'x)Jν(k'ξ) = δ(x-ξ)/x or !Syntax Error, Idk k Jν(kx)Jν('ξ) = δ(x-ξ)/x This equation I know is part of this data set f(x) = !Syntax Error, Idk k Jν(kx) Fν(k) // expansion Fν(k) = !Syntax Error, Idx x Jν(kx) f(x) // projection !Syntax Error, Idx x Jν(kx) Jν(k'x) = δ(k-k')/k // orthogonality !Syntax Error, Idk k Jν(kx) Jν(kx') = δ(x-x')/x // completeness So now I will build the larger data set similar to the above -(xu')' + ν2 u/x - k2xu = 0 or Lu = λxu where L = -(xu')' + ν2 u/x or Lλu = 0 where Lλ = L - λx λ = k2 s(x) = x φk(x) = (1/) Jν(kx) f(x) = !Syntax Error, Idk k Jν(kx) Fν(k) // expansion Fν(k) = !Syntax Error, Idx x Jν(kx) f(x) // projection !Syntax Error, Idx x Jν(kx) Jν(k'x) = δ(k-k')/k // orthogonality !Syntax Error, Idk k Jν(kx) Jν(kx') = δ(x-x')/x // completeness Now replace ν2 = b2 k2 = a2 and rewrite it all -(xu')' + b2 x-1u - a2xu = 0 or Lu = a2xu where L = -(xu')' + b2 u/x or Lλu = 0 where Lλ = L - a2x λ = a2 s(x) = x φa(x) = (1/) Jb(ax) // normalized eigenfunction f(x) = !Syntax Error, Ida a Jb(ax) Fb(a) // expansion Fb(a) = !Syntax Error, Idx x Jb(ax) f(x) // projection !Syntax Error, Idx x Jb(ax) Jb(a'x) = δ(a-a')/k // orthogonality !Syntax Error, Idk k Jb(ax) Jb(ax') = δ(x-x')/x // completeness and right here I will replicate the above KL block for comparison -(xu')' + α2xu -β2x-1u = 0 or Lu = β2x-1u where L = -(xu')' + α2xu or Lλu = 0 where Lλ = L - β2x-1 λ = β2 s(x) = x-1 φβ(x) = (1/π) Kiβ(αx) // normalized eigenfunction f(x) = (2/π2) !Syntax Error, I dβ β sh(πβ) Kiβ(αx) Fα(β) // expansion Fα(β) ≡ !Syntax Error, Idx x-1 f(x) Kiβ(αx) // projection !Syntax Error, Idx x-1 Kiβ(αx) Kiβ'(αx) = δ(β-β') π2/[2βsh(πβ)] // orthogonality x δ(x-x') = (2/π2)!Syntax Error, Idβ β sh(πβ) Kiβ(αx) Kiβ(αx') // completeness