The Abel Transform

The Abel Transform PhL 2.28.11 The work has been done in my Sneddon Chapter 2 raw notes, here I just want to "report out" the results in their various forms. 1. The trig forms. 1 2. The quadratic forms. 2 3. The linear forms. 3 Appendix A: Derivation of the linear forms from the quadratic forms. 3 Part (a) : obtaining the linear forms 3 Part (b): Doing the derivatives 5 Appendix B: Show that Polyanin's S1 linear form results agree: 7 1. The trig forms. Sneddon gives a very general monotonic Srivastav transform form (page 40-41). The theorem is limited to the range 0 < α < 1 where α is the power appearing in the denominator fo 2.3.1 or 2.3.3. He then applies this theorem to some special cases, one of which I shall call "the trig case". Notice we have a separate transform for for the two possible Volterra endpoint positions. We are thinking about a real interval (a,b): S1: !Syntax Error, Idt f(t) / [cos(t) - cos(x)]α = g(x) => f(t) = π-1sin(πα) ∂t { !Syntax Error, Idu sin(u)g(u) / [cos(u) - cos(t)]1-α } S2: !Syntax Error, Idt f(t) / [cos(x) - cos(t)]α = g(x) => f(t) = – π-1sin(πα) ∂t { !Syntax Error, Idu sin(u)g(u) / [cos(t) - cos(u)]1-α } We can specialize these for α = 1/2 which is the "non-generalized" Abel Transform S1: !Syntax Error, Idt f(t) / [cos(t) - cos(x)]1/2 = g(x) => f(t) = π-1 ∂t { !Syntax Error, Idu sin(u)g(u) / [cos(u) - cos(t)]1/2 } // which are 2.3.5 S2: !Syntax Error, Idt f(t) / [cos(x) - cos(t)]1/2 = g(x) => f(t) = – π-1 ∂t { !Syntax Error, Idu sin(u)g(u) / [cos(t) - cos(u)]1/2 } //which are 2.3.6 which we might write one more time as S1: !Syntax Error, Idt f(t) / = g(x) => f(t) = π-1 ∂t { !Syntax Error, Idu sinu g(u) / } // agrees Sned 2.3.5 S2: !Syntax Error, Idt f(t) / = g(x) => f(t) = – π-1 ∂t { !Syntax Error, Idu sinu g(u) / } // agrees Sned 2.3.6 2. The quadratic forms. Similar to the above, we just quote this different special case of Srivastav : S1: !Syntax Error, Idt f(t) / [x2 - t2]α = g(x) interval for x,t is [a,b] => f(t) = (2/π)sin(πα) ∂t { !Syntax Error, Idu u g(u) / [t2- u2]1-α } // which are 2.3.7 S2: !Syntax Error, Idt f(t) / [t2 - x2]α = g(x) interval for x,t is [a,b] => f(t) = – (2/π) sin(πα) ∂t { !Syntax Error, Idu u g(u) / [u2- t2]1-α } // which are 2.3.8 and again specialized to α = 1/2 S1: !Syntax Error, Idt f(t) / [x2 - t2]1/2 = g(x) interval for x,t is [a,b] => f(t) = (2/π)∂t { !Syntax Error, Idu u g(u) / [t2- u2]1/2 } S2: !Syntax Error, Idt f(t) / [t2 - x2]1/2 = g(x) interval for x,t is [a,b] => f(t) = – (2/π) ∂t { !Syntax Error, Idu u g(u) / [u2- t2]1/2 } which we might rewrite as S1: !Syntax Error, Idt f(t) / = g(x) => f(t) = (2/π)∂t { !Syntax Error, Idu u g(u) / } S2: !Syntax Error, Idt f(t) / = g(x) => f(t) = – (2/π) ∂t { !Syntax Error, Idu u g(u) / } 3. The linear forms. I did a little bit in my Sneddon Chap 2 raw notes, but I complete the job in Appendix A below and quote the final results here, where now I do the ∂t derivatives to get alternate answers: S1: !Syntax Error, I dt f(t) / [x - t]α = g(x) => f(t) = (1/π)sin(πα) ∂t { !Syntax Error, Idu g(u) / [t - u]1-α } = (1/π)sin(πα) { g(a) / [t -a]1-α + !Syntax Error, Idu g'(u) / [t - u]1-α } S2: !Syntax Error, I dt f(t)/ [t - x]α = g(x) => f(t) = – (1/π) sin(πα) ∂t { !Syntax Error, I du g(u) / [u- t]1-α } = (1/π) sin(πα) { g(b) / [b - t]1-α – !Syntax Error, Idu g'(u) / [u - t]1-α } We can specialize to the case α = 1/2 to get S1: !Syntax Error, I dt f(t) / = g(x) => f(t) = (1/π)∂t { !Syntax Error, Idu g(u) / } = (1/π) { g(a) / + !Syntax Error, Idu g'(u) / } S2: !Syntax Error, I dt f(t)/ = g(x) => f(t) = – (1/π) ∂t { !Syntax Error, I du g(u) / } = (1/π) { g(b) / – !Syntax Error, Idu g'(u) / } Appendix A: Derivation of the linear forms from the quadratic forms. Part (a) : obtaining the linear forms We start with A statement of the quadratic forms S1: !Syntax Error, Idt f(t) / [x2 - t2]α = g(x) interval for x,t is [a,b] => f(t) = (2/π)sin(πα) ∂t { !Syntax Error, Idu u g(u) / [t2- u2]1-α } // which are 2.3.7 S2: !Syntax Error, Idt f(t) / [t2 - x2]α = g(x) interval for x,t is [a,b] => f(t) = – (2/π) sin(πα) ∂t { !Syntax Error, Idu u g(u) / [u2- t2]1-α } // which are 2.3.8 Let χ = x2 and τ = t2 and μ = u2 and A = a2 and B = b2 and use capital letters in this sense: G(μ) = g(u). We can then edit all of the above: dμ = 2u du => du u = dμ (1/2) dτ = 2tdt = 2dt dt = dτ/(2) ∂t = (2)∂τ S1: !Syntax Error, I dτ/(2) F(τ) / [χ - τ]α = G(χ) => F(τ) = (2/π)sin(πα) (2)∂τ { !Syntax Error, Idμ(1/2) G(μ) / [τ - μ]1-α } S2: !Syntax Error, I dτ/(2) F(τ) / [τ - χ]α = G(χ) => F(τ) = – (2/π) sin(πα) (2)∂τ { !Syntax Error, I dμ(1/2) G(μ) / [μ- τ]1-α } To get these into standard forms, we now define F(τ) = F(τ)/ (2) Then we have S1: !Syntax Error, I dτ F(τ) / [χ - τ]α = G(χ) => F(τ) = (2/π)sin(πα) ∂τ { !Syntax Error, Idμ(1/2) G(μ) / [τ - μ]1-α } S2: !Syntax Error, I dτ F(τ) / [τ - χ]α = G(χ) => F(τ) = – (2/π) sin(πα) ∂τ { !Syntax Error, I dμ(1/2) G(μ) / [μ- τ]1-α } Now that we have all our results, we first change τ → t μ → u χ → x A → a B → b F → f G → g τ to t everywhere, μ to u everywhere, and write 2(1/2) = 1, S1: !Syntax Error, I dt f(t) / [x - t]α = g(x) => f(t) = (1/π)sin(πα) ∂t { !Syntax Error, Idu g(u) / [t - u]1-α } S2: !Syntax Error, I dt f(t)/ [t - x]α = g(x) => f(t) = – (1/π) sin(πα) ∂t { !Syntax Error, I du g(u) / [u- t]1-α } Part (b): Doing the derivatives We now refer to Appendix A of our Sneddon Chapter 2 raw notes where we show that g(x) = !Syntax Error, Idt k(x-t)f(t) => ∂x g(x) = f(a) k(x-a) + !Syntax Error, Idt k(x-t)f '(t) g(x) = !Syntax Error, Idt k(t-x)f(t) => ∂x g(x) = – f(b) k(b-x) + !Syntax Error, Idt k(t-x) f '(t) We of course have k(z) = 1/zα so we rewrite these lines as g(x) = !Syntax Error, Idt f(t)/ [x - t]α => ∂x g(x) = f(a) / [x -a]α + !Syntax Error, Idt f '(t) / [x - t]α g(x) = !Syntax Error, Idt f(t)/ [t - x]α => ∂x g(x) = – f(b) / [b - x]α + !Syntax Error, Idt f '(t) / [t - x]α Now on the RHS's replace t→ u everywhere THEN replace x → t everywhere g(t) = !Syntax Error, Idu f(u)/ [t - u]α => ∂t g(t) = f(a) / [t -a]α + !Syntax Error, Idu f '(u) / [t - u]α g(t) = !Syntax Error, Idu f(u)/ [u - t]α => ∂t g(t) = – f(b) / [b - t]α + !Syntax Error, Idu f '(u) / [u - t]α Now let g→h and THEN let f→g h(t) = !Syntax Error, Idu g(u)/ [t - u]α => ∂t h(t) = g(a) / [t -a]α + !Syntax Error, Idu g'(u) / [t - u]α h(t) = !Syntax Error, Idu g(u)/ [u - t]α => ∂t h(t) = – g(b) / [b - t]α + !Syntax Error, Idu g'(u) / [u - t]α We can then make these replacements ( we take α → 1-α to match forms above) ∂t { !Syntax Error, Idu g(u) / [t - u]1-α } = g(a) / [t -a]1-α + !Syntax Error, Idu g'(u) / [t - u]1-α ∂t { !Syntax Error, I du g(u) / [u- t]1-α } = – g(b) / [b - t]1-α + !Syntax Error, Idu g'(u) / [u - t]1-α Then we can restate our results from the end of part (a) S1: !Syntax Error, I dt f(t) / [x - t]α = g(x) => f(t) = (1/π)sin(πα) ∂t { !Syntax Error, Idu g(u) / [t - u]1-α } = (1/π)sin(πα) [g(a) / [t -a]1-α + !Syntax Error, Idu g'(u) / [t - u]1-α ] S2: !Syntax Error, I dt f(t)/ [t - x]α = g(x) => f(t) = – (1/π) sin(πα) ∂t { !Syntax Error, I du g(u) / [u- t]1-α } = – (1/π) sin(πα) [– g(b) / [b - t]1-α + !Syntax Error, Idu g'(u) / [u - t]1-α ] And let's restate this one more time with some sign changes S1: !Syntax Error, I dt f(t) / [x - t]α = g(x) => f(t) = (1/π)sin(πα) ∂t { !Syntax Error, Idu g(u) / [t - u]1-α } = (1/π)sin(πα) { g(a) / [t -a]1-α + !Syntax Error, Idu g'(u) / [t - u]1-α } S2: !Syntax Error, I dt f(t)/ [t - x]α = g(x) => f(t) = – (1/π) sin(πα) ∂t { !Syntax Error, I du g(u) / [u- t]1-α } = (1/π) sin(πα) { g(b) / [b - t]1-α – !Syntax Error, Idu g'(u) / [u - t]1-α } We can then specilialized these last results to α = 1/2 and sin(π/2) = 1 S1: !Syntax Error, I dt f(t) / [x - t]1/2 = g(x) => f(t) = (1/π)∂t { !Syntax Error, Idu g(u) / [t - u]1/2 } = (1/π) { g(a) / [t -a]1/2 + !Syntax Error, Idu g'(u) / [t - u]1/2 } S2: !Syntax Error, I dt f(t)/ [t - x]1/2 = g(x) => f(t) = – (1/π) ∂t { !Syntax Error, I du g(u) / [u- t]1/2 } = (1/π) { g(b) / [b - t]1/2 – !Syntax Error, Idu g'(u) / [u - t]1/2 } And one more time with radicals S1: !Syntax Error, I dt f(t) / = g(x) => f(t) = (1/π)∂t { !Syntax Error, Idu g(u) / } = (1/π) { g(a) / + !Syntax Error, Idu g'(u) / } S2: !Syntax Error, I dt f(t)/ = g(x) => f(t) = – (1/π) ∂t { !Syntax Error, I du g(u) / } = (1/π) { g(b) / – !Syntax Error, Idu g'(u) / } Appendix B: Show that Polyanin's S1 linear form results agree: Our linear results from above were: S1: !Syntax Error, I dt f(t) / [x - t]α = g(x) => f(t) = (1/π)sin(πα) ∂t { !Syntax Error, Idu g(u) / [t - u]1-α } = (1/π)sin(πα) { g(a) / [t -a]1-α + !Syntax Error, Idu g'(u) / [t - u]1-α } S2: !Syntax Error, I dt f(t)/ [t - x]α = g(x) => f(t) = – (1/π) sin(πα) ∂t { !Syntax Error, I du g(u) / [u- t]1-α } = (1/π) sin(πα) { g(b) / [b - t]1-α – !Syntax Error, Idu g'(u) / [u - t]1-α } We shall now make a series of changes to get things into Polyanin form. First, f→y S1: !Syntax Error, I dt y(t) / [x - t]α = g(x) => y(t) = (1/π)sin(πα) ∂t { !Syntax Error, Idu g(u) / [t - u]1-α } = (1/π)sin(πα) { g(a) / [t -a]1-α + !Syntax Error, Idu g'(u) / [t - u]1-α } S2: !Syntax Error, I dt y(t)/ [t - x]α = g(x) => y(t) = – (1/π) sin(πα) ∂t { !Syntax Error, I du g(u) / [u- t]1-α } = (1/π) sin(πα) { g(b) / [b - t]1-α – !Syntax Error, Idu g'(u) / [u - t]1-α } Next, g → f and at the same time lets do α → λ S1: !Syntax Error, I dt y(t) / [x - t]λ = f(x) => y(t) = (1/π)sin(πλ) ∂t { !Syntax Error, Idu f(u) / [t - u]1-λ } = (1/π)sin(πλ) { f(a) / [t -a]1-λ + !Syntax Error, Idu f '(u) / [t - u]1-λ } S2: !Syntax Error, I dt y(t)/ [t - x]λ = f(x) => y(t) = – (1/π) sin(πλ) ∂t { !Syntax Error, I du f(u) / [u- t]1-λ } = (1/π) sin(πλ) { f(b) / [b - t]1-λ – !Syntax Error, Idu f '(u) / [u - t]1-λ } Now in the y(t) = equations of each set, replace t by x S1: !Syntax Error, I dt y(t) / [x - t]λ = f(x) => y(x) = (1/π)sin(πλ) ∂x { !Syntax Error, Idu f(u) / [x - u]1-λ } = (1/π)sin(πλ) { f(a) / [x -a]1-λ + !Syntax Error, Idu f '(u) / [x - u]1-λ } S2: !Syntax Error, I dt y(t)/ [t - x]λ = f(x) => y(x) = – (1/π) sin(πλ) ∂x { !Syntax Error, I du f(u) / [u- x]1-λ } = (1/π) sin(πλ) { f(b) / [b - x]1-λ – !Syntax Error, Idu f '(u) / [u - x]1-λ } Now in the y(x) equations, change integration varaible from u to t S1: !Syntax Error, I dt y(t) / [x - t]λ = f(x) => y(x) = (1/π)sin(πλ) ∂x { !Syntax Error, Idt f(t) / [x - t]1-λ } = (1/π)sin(πλ) { f(a) / [x -a]1-λ + !Syntax Error, Idt f '(t) / [x - t]1-λ } S2: !Syntax Error, I dt y(t)/ [t - x]λ = f(x) => y(x) = – (1/π) sin(πλ) ∂x { !Syntax Error, I dt f(t) / [t- x]1-λ } = (1/π) sin(πλ) { f(b) / [b - x]1-λ – !Syntax Error, Idt f '(t) / [t - x]1-λ } Polyanin gives the form S1, but I could not find the form S2 in his book. The S1 agrees. S1: !Syntax Error, I dt y(t) / [x - t]λ = f(x) => y(x) = (1/π)sin(πλ) ∂x { !Syntax Error, Idt f(t) / [x - t]1-λ } = (1/π)sin(πλ) { f(a) / [x -a]1-λ + !Syntax Error, Idt f '(t) / [x - t]1-λ } Appendix C: is there a SL problem associated with this transform? I don't remember Sneddon ever asking this question. We are looking for eigenfunctions of this form, where x is a continuous parameter: φx(t) =1/ = (x-t)-1/2 φx'(t) =1/ = (1/2)(x-t)-3/2 φx"(t) =1/ = (3/4)(x-t)-5/2 So maybe this ODE, where x is just a parameter (4/3) (x-t)2 φx"(t) + φx(t) = 0 L = (4/3) (x-t)2 D2 + 1 Maybe our interval is t in (a,x) . L looks self adjoint to me. But φx(t) blows up at the x endpoint. Consider then !Syntax Error, Idt | φx(t) |2 = !Syntax Error, Idt 1/(x-t) = log divergent at the upper endpoint so this is not an L2 situation, but it is OK as L1. Can you have SL with L1 norm? When you talk about the Laplace Transform, you don't normally think of eigenfunctions but they are there. Our candidate eigenfunctions are φx(t) =1/ but these are pretty clearly not oscillatory, so how can you ever have orthogonality? You would want something like this !Syntax Error, Idt (1/)(1/) ~ δ(x-y) which seems pretty unlikely. I think the answer is that SL theory does not apply. I will put this question aside for rainy day 4,234,567.