Sneddon Parallel Treatment of the two dual integral equation J0 cases
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Working notes by Phil dated 7.25.10 that solve two dual integral equations with a J0 kernel, one with weight ξ^-1 and one with ξ^+1, using parallel steps. The steps are the ansatz for A(ξ), the discontinuous J0 sine and cosine integrals, and Abel inversion. In the second pair the parts form and non-parts form must be swapped at certain steps. He concludes the result matches Sneddon but disagrees with Polyanin. Equations are lost in the extracted text.
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Sneddon Parallel Treatment of the two dual integral equation J0 cases PhL 7.25.10
Things are so slippery here, that I want to try an exact parallel development so I can see exactly how the two cases are different. It is like handling mercury, you simply cannot track it.
Treatment of the First Pair
1. Statement of the Pair:
!Syntax Error, Idξ ξ-1J0(ξρ)A(ξ) = f(ρ) ρ < 1 3.2.1
!Syntax Error, Idξ J0(ξρ)A(ξ) = 0 ρ > 1 3.2.2
2. Definition of the two functions
F(ρ) ≡ !Syntax Error, Idξ ξ-1J0(ξρ)A(ξ) all ρ
G(ρ) ≡ !Syntax Error, Idξ J0(ξρ)A(ξ) all ρ
3. Ansatz for the Form of A(ξ) and its parts form
A(ξ) = ξ !Syntax Error, Idt φ(t) cos(ξt) 3.5.1
A(ξ) = φ(1)sinξ - !Syntax Error, Idt φ'(t)sin(ξt) 3.5.2
4. Proof that this Form solves the second pair member. This requires showing that G(ρ>1) = 0 if we use the ansatz form. We install the parts form of A(ξ) into G to get:
G(ρ) = !Syntax Error, Idξ J0(ξρ)A(ξ) = !Syntax Error, Idξ J0(ξρ){ φ(1)sinξ - !Syntax Error, Idt φ'(t)sin(ξt)}
= φ(1) !Syntax Error, Idξ J0(ξρ) sinξ - !Syntax Error, Idt φ'(t) !Syntax Error, Idξ J0(ξρ) sin(ξt) p 75 A
The discontinuous integral of interest here is this
!Syntax Error, Idx J0(ρx) sin(ax) = 0 a < ρ // which are 2.1.14
= 1/ a > ρ
We have ρ > 1 for first integral, and we have ρ > 1 > t for second integral, so both are 0. QED.
5. Now, install the non-parts form of A(ξ) into the first of the dual pair F(ρ), for which ρ < 1
!Syntax Error, Idξ ξ-1J0(ξρ)A(ξ) = !Syntax Error, Idξ ξ-1J0(ξρ) [ξ !Syntax Error, Idt φ(t) cos(ξt)]
= !Syntax Error, Idt φ(t) !Syntax Error, Idξ J0(ξρ) cos(ξt) (*)
= [ !Syntax Error, Idt φ(t) + !Syntax Error, Idt φ(t) ] !Syntax Error, Idξ J0(ξρ) cos(ξt) // since ρ < 1, can do this
Now we need the other discontinuous integral which is this:
!Syntax Error, Idx J0(ρx) cos(ax) = 1/ a <ρ // which are 2.1.13
= 0 a > ρ
In the second integral, ρ < t so that term vanishes. In the first integral, t > ρ and we are left with
F(ρ) ≡ !Syntax Error, Idξ ξ-1J0(ξρ)A(ξ) = !Syntax Error, Idt φ(t)/ ρ < 1 3.5.6
In the other case that ρ > 1, we have ρ > t in (*) and we don't split into two integrals and we have
F(ρ) ≡ !Syntax Error, Idξ ξ-1J0(ξρ)A(ξ) = !Syntax Error, Idt φ(t)/ ρ > 1 3.5.7
6. Use the Abel inversion formula. First, quote the inversion theorem of interest from Chap 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 }
and now make edits for our current situation
S1: !Syntax Error, Idt φ(t) / = f(ρ) interval for x,t is [a,b]
=> φ(t) = (2/π)∂t { !Syntax Error, Idρ ρ f(ρ) / }
Applying this to 3.5.6 we get this inversion
φ(t) = (2/π)∂t { !Syntax Error, Idρ ρ f(ρ) / } 3.5.9
We install this into the non-parts Form for A(ξ) and get our result
A(ξ) = ξ !Syntax Error, Idt φ(t) cos(ξt) = ξ !Syntax Error, Idt [(2/π)∂t { !Syntax Error, Idρ ρ f(ρ) / } ] cos(ξt)
= (2/π) ξ !Syntax Error, Idt cos(ξt) ∂t { !Syntax Error, Idρ ρ f(ρ) / }
and this agrees with the claim of Polyanin.
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Now I am going to copy and paste the entire section above, and just edit in changes as needed.
Treatment of the Second Pair
1. Statement of the Pair:
!Syntax Error, Idξ ξ+1J0(ξρ)A(ξ) = f(ρ) ρ < 1 3.5.15
!Syntax Error, Idξ J0(ξρ)A(ξ) = 0 ρ > 1 3.5.16
2. Definition of the two functions
F(ρ) ≡ !Syntax Error, Idξ ξ+1J0(ξρ)A(ξ) all ρ
G(ρ) ≡ !Syntax Error, Idξ J0(ξρ)A(ξ) all ρ
3. Ansatz for the Form of A(ξ) and its parts form
A(ξ) = 1 !Syntax Error, Idt φ(t) sin(ξt) φ(0) = 0 3.5.17
A(ξ) = ξ-1{- φ(1)cosξ + !Syntax Error, Idt φ'(t) cos(ξt)} p 77 A; we used φ(0) = 0 !
4. Proof that this Form solves the second pair member. This requires showing that G(ρ>1) = 0 if we use the ansatz form. We install the parts form of A(ξ) into G to get:
G(ρ) = !Syntax Error, Idξ J0(ξρ)A(ξ) = !Syntax Error, Idξ J0(ξρ) ξ-1{- φ(1)cosξ + !Syntax Error, Idt φ'(t) cos(ξt)}
= - φ(1) !Syntax Error, Idξ ξ-1 J0(ξρ) cosξ - !Syntax Error, Idt φ'(t) !Syntax Error, Idξ ξ-1J0(ξρ) cos (ξt)
But this does not work now because the two integrals diverge at the low end! So we have to change the thread right at this point. So let's try instead installing the non-parts form of A(ξ) into G:
G(ρ) = !Syntax Error, Idξ J0(ξρ)A(ξ) =!Syntax Error, Idξ J0(ξρ) 1 !Syntax Error, Idt φ(t) sin(ξt)
= !Syntax Error, Idt φ(t) !Syntax Error, Idξ J0(ξρ) sin(ξt)
Comments: (1) since we used the non-parts form, we have the sin trig function here, which is then the same as we had for the first dual pair; (2) since we used the non-parts form, there is only one of these J0 sin integrals, not two as we had before; (3) since we used the non-parts form, we have φ(t) appearing instead of φ'(t) as last time !!
The discontinuous integral of interest here is this (same as for first pair)
!Syntax Error, Idx J0(ρx) sin(ax) = 0 a < ρ // which are 2.1.14
= 1/ a > ρ
Since our second dual equation is for ρ > 1, we have ρ > 1 > t for our integral above, so the integral is 0. Thus, we find the same fact we found last time, which is that G(ρ>1) = 0. I think this is WHY we used a sin in the ansatz form instead of a cos for this case!
5. Now, install the non-parts form of A(ξ) into the first of the dual pair F(ρ), for which ρ < 1
!Syntax Error, Idξ ξ+1J0(ξρ)A(ξ) = !Syntax Error, Idξ ξ+1J0(ξρ) [1 !Syntax Error, Idt φ(t) sin(ξt) ]
= !Syntax Error, Idt φ(t) !Syntax Error, Idξ ξ+1J0(ξρ) cos(ξt) (*)
But again we have a problem: this integral now diverges at the high end. So we make our second change in the thread at this point, and we instead install the parts form into the first of the dual pair:
!Syntax Error, Idξ ξ+1J0(ξρ)A(ξ) = !Syntax Error, Idξ ξ+1J0(ξρ)[ ξ-1{- φ(1)cosξ + !Syntax Error, Idt φ'(t) cos(ξt)} ] p 77A
= !Syntax Error, Idξ J0(ξρ) {- φ(1)cosξ + !Syntax Error, Idt φ'(t) cos(ξt)}
= - φ(1) !Syntax Error, Idξ J0(ξρ) cosξ + !Syntax Error, Idt φ'(t) !Syntax Error, Idξ J0(ξρ) cos(ξt) p 77 B
Now we need the other discontinuous integral which is this: (same as for first pair)
!Syntax Error, Idx J0(ρx) cos(ax) = 1/ a <ρ // which are 2.1.13
= 0 a > ρ
We are doing ρ < 1 now for first of the dual pair, so the first integral vanishes and we then have
!Syntax Error, Idξ ξ+1J0(ξρ)A(ξ) = !Syntax Error, Idt φ'(t) !Syntax Error, Idξ J0(ξρ) cos(ξt)
= [ !Syntax Error, Idt φ'(t) + !Syntax Error, Idt φ'(t) ] !Syntax Error, Idξ J0(ξρ) cos(ξt) // since ρ < 1, can do this
This is exactly the situation we had for the first pair, except here we have φ'(t) instead of φ(t), so the same conclusions will occur: In the second integral, ρ < t so that term vanishes. In the first integral, t > ρ and we are left with
F(ρ) ≡ !Syntax Error, Idξ ξ+1J0(ξρ)A(ξ) = !Syntax Error, Idt φ'(t)/ ρ < 1 p 77 C
In the other case that ρ > 1, we have ρ > t in (*) and we don't split into two integrals and we have
F(ρ) ≡ !Syntax Error, Idξ ξ+1J0(ξρ)A(ξ) = !Syntax Error, Idt φ'(t)/ ρ > 1 not shown
6. Use the Abel inversion formula. First, quote the inversion theorem of interest from Chap 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 }
and now make edits for our current situation
S1: !Syntax Error, Idt φ'(t) / = f(ρ) interval for x,t is [a,b]
=> φ'(t) = (2/π)∂t { !Syntax Error, Idρ ρ f(ρ) / }
Applying this to p77C above and we get this inversion
φ'(t) = (2/π)∂t { !Syntax Error, Idρ ρ f(ρ) / } p 77 D
We install this into the non-parts Form for A(ξ) and get our result -- but no can do! The non-parts form does not have φ'(t) in it. So instead let's try installing this into the parts form:
A(ξ) = ξ-1{- φ(1)cosξ + !Syntax Error, Idt φ'(t) cos(ξt)}
= - φ(1) ξ-1 cosξ + ξ-1!Syntax Error, Idt cos(ξt) φ'(t)
= - φ(1) ξ-1 cosξ + ξ-1!Syntax Error, Idt cos(ξt) { (2/π)∂t { !Syntax Error, Idρ ρ f(ρ) / }}
But this is not a form anybody claims, and it has that φ(1), a non-starter.
So instead, let's go back to
φ'(t) = (2/π)∂t { !Syntax Error, Idρ ρ f(ρ) / } p 77 D
and integrate this from 0 to t. The LHS gives just φ(t) since φ(0) = 0. The RHS is the RHS without the ∂t,
φ(t) = (2/π) !Syntax Error, Idρ ρ f(ρ) / 3.5.18
Now install this into the non-parts Form to get
A(ξ) = 1 !Syntax Error, Idt φ(t) sin(ξt) = !Syntax Error, Idt sin(ξt) {(2/π) !Syntax Error, Idρ ρ f(ρ) / }
= (2/π) !Syntax Error, Idt sin(ξt) !Syntax Error, Idρ ρ f(ρ) / 3.5.19
and this agrees with Sneddon but disagrees with the claim of Polyanin! I now think Polyanin has an error here! Amazing, since this is one of the simples canonical cases!