stak chap 6 scraps

Chapter 6 scraps PhL 7.21.09 hn ≡ !Syntax Error, I dφ f(φ) sin(πnφ/α) Cn bπn/α = (2/α) hn Cn = (2b-nπ/α/α) hn u(r,φ) = Σn=1∞ Cn rπn/α sin(πnφ/α) = Σn=1∞ (2b-nπ/α/α) hn rπn/α sin(πnφ/α) so we have u(r,φ) = Σn=1∞ bn (r/b)nπ/α sin(πnφ/α) bn ≡ (2/α)!Syntax Error, I dφ f(φ) sin(πnφ/α) where here bn is the "official" Fourier sine coefficient, and this "b" is unrelated to radius b. First our "periodic range" is 2L = α so that L = α/2. This is because f(φ) is only defined on (0,φ). The series we see on page 131 is then this: f(φ) = ao/2 + Σn=1∞ ( an cos(2nπφ/α) + bn sin(2nπφ/α) ) but this does not agree with what we have in our current problem. So define 2n = m and write the above Fourier form as f(φ) = ao/2 + Σm=2,4,..∞ ( am cos(mπφ/α) + bm sin(mπφ/α) ) = Σm=2,4,..∞ bm sin(mπφ/α) and this does match our form, but is then missing half the terms! Notice that this is a partial Fourier Series. The cos terms are missing because u = 0 on φ = 0, and this statement includes the constant term. If we look at Schaum p 131, we see the right form for a FS and we can read of the projection formula Cn bπn/α = (1/α) !Syntax Error, I dφ f(φ) sin(πnφ/α) n = 1,2,3... HOWEVER, something is not quite right here because f(φ) is only defined in the range (0,α). The condition that R(0) = 0 requires that C0 = 0, so we end up with these as the basis solutions to our problem: Φσ(φ) = sin(πnφ/α) Rσ(r) = rπn/α Our general solution is then u = Σσ Rσ(r) Φσ(φ) = Σn=1∞ An rπn/α sin(πnφ/α) -- there is no n=0 contribution for the reason just noted. Actually, n = 0 kills Solve Laplace's equation for a disk wedge where we are given certain values all around the wedge boundary. I screwed this up and only got cleared up with Plan G, so you should skip all the other stuff and jump down to that point. I keep the raw screw up notes anyway. Plan A. So here is one approach: u(r,φ) = Σ fn r|n|einφ // wrong form for these BC's !!!! 1. u(r,0) = Σ fn r|n| = 0 for any r 2. u(r,α) = Σ fn r|n|einα = 0 for any r 3. u(1,φ) = Σ fn einφ = f(φ) for φ = 0 to α Question: the first of these conditions would seem to be a different equation for each value or r, so it is really an infinite number of equations. But an is an infinite number of coefficients. Not sure how to proceed here. Plan B Wedge Superposition? Call the wedge solution u(r,φ). Suppose there were N wedges around the circle. Then the BC's superpose OK so we could say this: u(r,φ) + u(r,φ+(2π/N)) + u(r,φ+2(2π/N)) + ... + u(r,φ+(N-1)(2π/N)) = 6.11 where we set f(ψ) = f(ψ+i(2π/N)) and so is periodic. i = 0,1,2...N-1 But this is not right, because we have extra slit BC's where u = 0 is being forced, so this superposition does not work right. Plan C. Real and Imaginary with original idea of Plan A (which is wrong!): Let fn = an+ i bn 1. Σ an r|n| + Σ i bn r|n| = 0 => Σ an r|n| = 0 AND Σ bn r|n| = 0 2. Σ (an + i bn) r|n|( cos(nα) + isin(nα)) = 0 => Σ r|n| [ancos(nα) - bn sin(nα)] = 0 => Σ r|n| [bn cos(nα) + an sin(nα)] = 0 3. Σ (an + i bn) ( cos(nφ) + isin(nφ)) = f(φ) assume real => Σ [ancos(nφ) - bn sin(nφ)] = f(φ) // both these on just 0 to α => Σ [bncos(nφ) + an sin(nφ)] = 0 So now I seem to have 6 equations to play with: I assume that f(φ) = 0 at its two endpoints. Σ an r|n| = 0 // r in 0,1 Σ bn r|n| = 0 Σ r|n| [ancos(nα) - bn sin(nα)] = 0 // r in 0,1 Σ r|n| [bn cos(nα) + an sin(nα)] = 0 Σ [ancos(nφ) - bn sin(nφ)] = f(φ) // both these on just 0 to α Σ [bncos(nφ) + an sin(nφ)] = 0 These should be valid with r = 1 I would guess, so have 1a Σ an = 0 1b Σ bn = 0 2a Σ [ancos(nα) - bn sin(nα)] = 0 2b Σ [bn cos(nα) + an sin(nα)] = 0 3a Σ [ancos(nφ) - bn sin(nφ)] = f(φ) // both these on just 0 to α 3b Σ [bncos(nφ) + an sin(nφ)] = 0 But if I combine 2a with 3a, I get f(φ) = 0 which is nonsense, so I guess r = 1 was no go. Let's try r = 1/2 instead: Σ an (1/2)|n| = 0 Σ bn (1/2)|n| = 0 Σ (1/2)|n| [ancos(nα) - bn sin(nα)] = 0 Σ (1/2)|n| [bn cos(nα) + an sin(nα)] = 0 Σ [ancos(nφ) - bn sin(nφ)] = f(φ) // both these on just 0 to α Σ [bncos(nφ) + an sin(nφ)] = 0 Well this is taking me nowhere and just seems all wrong somehow. Plan D. Can I write a Fourier transform on a wedge instead of a circle. Here is the usual FT for a periodic function: f(φ) = Σnfn einφ fn = (1/2π) !Syntax Error, I f(φ) e-inφ dφ Now change variables to ψ so that integral goes from -α/2 to +α/2. Thus have ψ = φ * (α/2π). Then I guess we have: (page 94) f(φ(ψ)) = Σnfn ein(2πψ/α) = Σnfn ei(2πn/α)ψ fn = (1/2π) !Syntax Error, I f(φ(ψ)) e-i(2πn/α)ψ dψ (2π/α) = (1/α) !Syntax Error, I f(φ(ψ)) e-i(2πn/α)ψ dψ So let's call f(φ(ψ)) = F(ψ) and it is defined only from -α/2 to +α/2. Our Fourier Series is this: F(ψ) = Σnfn ei(2πn/α)ψ fn = (1/α) !Syntax Error, I dψ F(ψ) e-i(2πn/α)ψ Now on our wedge we are given some F(ψ) specified on the rim of the wedge. We are at r = 1. Our general model for a harmonic solution is this: u(r,φ) = Σn=-∞∞ an r|n|einφ // wrong! or u(r,φ(ψ)) = Σ an r|n| ei(2πn/α)ψ . Then on our wedge edge we have u(1,φ(ψ)) = Σ an ei(2πn/α)ψ = f(ψ) Using our adjusted wedge FS, we then have an = (1/α) !Syntax Error, I dψ f(ψ) e-i(2πn/α)ψ which in turn implies a-n = (1/α) !Syntax Error, I dψ f(ψ) e+i(2πn/α)ψ Now let's rewrite the series with r = 1 like this: u(r,φ) = Σ an r|n|einφ = a0 + Σn=1∞ rn an einφ + Σn=1∞ rn a-n e-inφ = a0 + Σn=1∞ rn { (an + a-n) cos(nφ) + i (an – a-n) sin(nφ) } Now we have an + a-n = (1/α) !Syntax Error, I dψ f(ψ) 2 cos(2πnψ/α) an – a-n = - (i/α) !Syntax Error, I dψ f(ψ) 2 sin(2πnψ/α) So this leaves us with u(r,φ) = ao + Σn=1∞ rn (1/α) !Syntax Error, I dψ f(ψ) 2 { cos(2πnψ/α) cos(nφ) + sin(2πnψ/α) sin(nφ) } The trig rule becomes just cos(2πnψ/α - nφ), so I then get u(r,φ) = ao + Σn=1∞ rn (2/α) !Syntax Error, I dψ f(ψ) cos(2πnψ/α - nφ) So I have something with the right kind of sum, but I am not getting the power rnπ/α , and other things are wrong as well. Plan E. Go way back to top of page 92. Let's try changing variables at this point to ψ = φ * (α/2π) so that dψ = dφ * (α/2π) and ∂ψ= ∂φ (2π/α). Then our first equation there becomes -∂ψ2Φ[φ(ψ)] = -(2π/α)2∂φ2Φ[φ] = (2π/α)2 λ Φ[φ(ψ)] which we might write as -∂ψ2Ψ(ψ) = (2π/α)2 λ Ψ(ψ) But now I don't have any requirement at boundaries on Ψ(ψ) so nowhere to go. [ but that is the whole point, the wedge provides such BC's, see G below) Plan F. Same point. What happens if you impose "periodic BC's" such that [ but what justification is there for doing this? The wedge does not have periodic BC's! ] Φ(φ) = Φ(φ + α) and same for derivative Then if we try Φ = eiσφ so that Φ' = iσ Φ and Φ" = -σ2Φ, so what values can λ = σ2 have? eiσφ = eiσ(φ+α) => eiσα = 1 => σα = n(2π) => σ = n (2π/α) Then λ = n2 (2π/α)2. I think the r solutions are then r2πn/α which reduces to rn for α = 2π. But Stak does not have that factor of 2 in his exponent, so this path is not going to the right place either. Plan G. The correct solution (with a little help from the web) http://quantumrelativity.calsci.com/Physics/EandM6.html Look at Stak page 92 but, as above, write λ = σ2 and we try Φ = eiσφ so that Φ' = iσ Φ and Φ" = -σ2Φ, so what values can λ = σ2 have? Let's DEFER this question for a moment and just think of σ or λ as a set of eigenvalues to be determined later. Then we know that our general solution is this u(r,φ) = Σσ rσ ( Aσ cos(σφ) + Bσ sin(σφ) ) where we sum over reasonable eigenvalues. An eigenvalue is going to be a value of σ2 = λ which meets our boundary conditions, but now we are talking the wedge, not an open periodic function around the circle. So what do our BC conditions tell us: 1. u(r,0) = 0 for any r 2. u(r,α) = 0 for any r 3. u(1,φ) = f(φ) for φ = 0 to α The first condition says u(r,0) = Σσ rσ ( Aσ 1) ) = 0 => Aσ = 0 Second condition says u(r,α) = Σσ rσ ( Bσ sin (σα) ) = 0 => σα = nπ => σ = nπ/α So NOW we know that our eigenvalues are σ = nπ/α where n = 0,1,2... We reject negative powers because they make us blow up at r = 0, So we then have u(r,φ) = Σn=0∞ rnπ/α ( bn sin (nπφ/α) ) But the n=0 term gives nothing due to the sine, so we have u(r,φ) = Σn=1∞ rnπ/α ( bn sin (nπφ/α) ) // agrees with p 107 A At r = 1 we then have u(1,φ) = Σn=1∞ bn sin (nπφ/α) = f(φ) which is condition 3. But this stuff is only defined on (0,α) so I don't think we can use the normal FS inversion. Let's define a new variable x = (π/α)φ. Then an integral dφ from 0 to α would run x = 0 to π. So multiply both sides of the above by sin (mπφ/α) and integrate dφ from 0 to α, !Syntax Error, I dφ sin (mπφ/α) f(φ) = !Syntax Error, I dφ sin (mπφ/α) Σn=1∞ bn sin (nπφ/α) = Σn=1∞ bn !Syntax Error, I dφ sin (mπφ/α) sin (nπφ/α) = Σn=1∞ bn !Syntax Error, I (α/π) dx sin (mx) sin (nx) = Σn=1∞ bn (π/2) (α/π) δn,m Schaum p 96 = (α/2) bm So our projection formula must be bm = (2/α) !Syntax Error, I sin (mπφ/α) f(φ) // agrees with p 107 B ___________________________________________ annulus scraps __________________________________________ Here we do have the normal "n" boundary conditions, so general solution is this: u(r,φ) = Σn=1∞ (An(r/a)n + Bn(r/a)-n)( cos(φn) + Cnsin(φn) ) + A0 + B0 ln(r/a) where I want a < b. Our boundary conditions might be taken as u=f on the outside, u=0 on the inside: u(b,φ) = Σn=1∞ (An(b/a)n + Bn(b/a)-n)( cos(φn) + Cnsin(φn) ) + A0 + B0 ln(b/a) = f(φ) u(a,φ) = Σn=0∞ (An + Bn)( cos(φn) + Cnsin(φn) ) + A0 + B0 ln(1) = 0 This last would seem to require that Bn = - An for n > 1, A0 = 0, an B0 undetermined. From that condition only we learn that u(r,φ) = Σn=1∞ An ( (r/a)n –(r/a)-n)( cos(φn) + Cnsin(φn) ) + B0 ln(r/a) The first condition then says u(b,φ) = Σn=1∞ An ( (b/a)n –(b/a)-n)( cos(φn) + Cnsin(φn) ) + B0 ln(b/a) = f(φ) Let's define some new constants: An ( (b/a)n –(b/a)-n) = an An ( (b/a)n –(b/a)-n) Cn = bn Then we have u(b,φ) = Σn=1∞ { an cos(φn) + bn sin(φn) } + B0 ln(b/a) = f(φ) where φ ranges 0 to 2π. We set L = π on Schaum page 131 and we can then read off the coefficients, where we make the association that B0 ln(b/a) = a0/2 : an = (1/π) !Syntax Error, Idφ f(φ) cos(nφ) n = 0,1,2... => a0 = 2B0ln(b/a) = (1/π) !Syntax Error, Idφ f(φ) bn = (1/π) !Syntax Error, I dφ f(φ) sin(nφ) n = 1,2,3... Notice that the first equation with n=0 determines the constant B0 Our complete solution must then be u(r,φ) = Σn=1∞ An ( (r/a)n – (r/a)-n)( cos(φn) + Cnsin(φn) ) + B0 ln(r/a) where An = an/ [(b/a)n –(b/a)-n] and AnCn = bn/ [(b/a)n –(b/a)-n]. So install this to get u(r,φ) = Σn=1∞ [ (r/a)n – (r/a)-n] / [(b/a)n –(b/a)-n] ( an cos(φn) + bn sin(φn) ) + B0 ln(r/a) The last term is B0 ln(r/a) = a0/2 * 1/ln(b/a)} * ln(r/a) = (a0/2) ln(r/a)/ ln(b/a) We could now install the known coefficients and see what happens: u(r,φ) = Σn=1∞ [ (r/a)n – (r/a)-n] / [(b/a)n –(b/a)-n] + B0 ln(r/a) { (1/π) !Syntax Error, Idφ' f(φ') cos(nφ') cos(φn) +(1/π) !Syntax Error, Idφ' f(φ') sin(nφ') sin(φn) ) } = (1/π) Σn=1∞ [ (r/a)n – (r/a)-n] / [(b/a)n –(b/a)-n] !Syntax Error, Idφ' f(φ') cos(n[φ-φ']) + B0 ln(r/a) To verify, at r = a we get 0. At r = b we get u(b,φ) = (1/π) Σn=1∞ !Syntax Error, Idφ' f(φ') cos(n[φ-φ']) + B0 ln(b/a) = (1/π) !Syntax Error, I dφ' f(φ') Σn=1∞ cos(n[φ-φ']) + B0 ln(b/a) where B0 ln(b/a) = (a0/2) = (1/2π) !Syntax Error, Idφ f(φ) so write as = (1/π) !Syntax Error, I dφ' f(φ') Σn=1∞ cos(n[φ-φ']) + (1/2π) !Syntax Error, Idφ' f(φ') = (1/2π) !Syntax Error, I dφ' f(φ') { 2 Σn=1∞ cos(n[φ-φ']) + 1 } so it must be that 1 + 2 Σn=1∞ cos(nθ) = 2π δ(θ) We did once have this on the real axis x from volume I page 45 Σint k δ(x-k) = 1 + 2 Σn=1∞ cos(2nπx) Take this last and change variables to 2πx = θ. We then get 1 + 2 Σn=1∞ cos(nθ) = Σint k δ(x-k) = Σint k 2π δ(2πx-2πk) = Σint k 2π δ(θ-2πk) = 2π δ(θ) QED since only one δ hit in range 0 ≤ θ < 2π, k = 0 Now I don't see a simple way to "do" the sum as we did in the disk case. The [(b/a)n –(b/a)-n] factor confounds things. Cannot write things as a power of some z as done page 94. So leave as is. Here is my final result: u(a,φ) = 0 u(b,φ) = f(φ) b > a u(r,φ) = Σn=1∞ [ (r/a)n – (r/a)-n] / [(b/a)n –(b/a)-n] ( fn cos(φn) + gn sin(φn) ) + (f0/2) ln(r/a)/ ln(b/a) fn = (1/π) !Syntax Error, Idφ f(φ) cos(nφ) gn = (1/π) !Syntax Error, Idφ f(φ) sin(nφ) Now, suppose we want to solve for an annulus which has g on the inner and 0 on the outer. Hopefully you just swap the role of a and b and the answer is this: u(b,φ) = 0 u(a,φ) = g(φ) b > a u(r,φ) = Σn=1∞ [ (r/b)n – (r/b)-n]/[(a/b)n – (a/b)-n] ( Fn cos(φn) + Gn sin(φn) ) + (F0/2) ln(r/b)/ ln(a/b) Fn = (1/π) !Syntax Error, Ig(φ) cos(nφ) dφ Gn = (1/π) !Syntax Error, Ig(φ) sin(nφ) dφ Now finally if we want f on the outer and g on the inner we can superpose our two solutions. This does the right things for BC's since we were careful to put 0 on "the other" circle. So here is the general solution: u(b,φ) = f(φ) u(a,φ) = g(φ) b > a u(r,φ) = Σn=1∞ [ (r/a)n – (r/a)-n] / [(b/a)n –(b/a)-n] ( fn cos(φn) + gn sin(φn) ) + (f0/2) ln(r/a)/ ln(b/a) + Σn=1∞ [ (r/b)n – (r/b)-n] / [(a/b)n –(a/b)-n] (Fn cos(φn) + Gn sin(φn) ) + (F0/2) ln(r/b)/ ln(a/b) fn = (1/π) !Syntax Error, If(φ) cos(nφ) dφ gn = (1/π) !Syntax Error, If(φ) sin(nφ) dφ Fn = (1/π) !Syntax Error, Ig(φ) cos(nφ) dφ Gn = (1/π) !Syntax Error, Ig(φ) sin(nφ) dφ You do wonder what happens as you take a limit a → 0 say of the first solution f outside, 0 inside. The limit is some situation that has u = 0 at the center and so is not the correct MVT situation. So I don't see this limit as being relevant. Recall from above (where I just now made up the second line) u(b,φ) = Σn=1∞ { fn cos(φn) + gn sin(φn) } + (f0/2) = f(φ) u(a,φ) = Σn=1∞ { Fn cos(φn) + Gn sin(φn) } + (F0/2) = g(φ) This problem is treated in a book on the web here http://books.google.com/books?id=QGw2EkI3WNIC&pg=PA7&lpg=PA7&dq=laplace+annulus&source=bl&ots=rvzNcD0nPJ&sig=tH6U4g9r57j6GcunK2_Ve9FZ50I&hl=en&ei=VhNlSqq2N4ygsgOQpIHqDg&sa=X&oi=book_result&ct=result&resnum=9 and I just copy some of the results: This author did not use ratios the way I did, maybe that makes the results a lot simpler. The various coefficients in the above he then gives as where f1 and f2 are the boundary conditions on the two circles. He uses s in place of φ. So yes, his result is simpler without using ratios. But the general nature of this web author's solution confirms that I did things pretty much on the right track. Were there some compact formula result, he would surely have stated it in his book, which, by the way, is this: )________________________________________________________ The first problem here is to the previous problem as the disk was to the wedge. The start for either of the two parts here is this: Write down the general form of the solution u(r,φ). That general form is this u(r,φ) = Σσ { Aσ rσ + Bσ r-σ ) cos(σφ) + Σσ { Cσ rσ + Dσ r-σ ) sin(σφ) + A0 + B0 ln(r) where the eigenvalues in the sum are σ = nπ/α , so write as u(r,φ) = Σn=1∞ { An rnπ/α + Bn r-nπ/α ) cos(nπφ/α) + A0 + B0 ln(r) + Σn=1∞ { Cn rnπ/α + Dn r-nπ/α ) sin(nπφ/α) At this point, why not just treat r as a constant, and do a Fourier Series processing in φ. Use Schaum page 131 with these interpretations: L = α an(r) = An rnπ/α + Bn r-nπ/α n = 1,2,3... bn(r) = Cn rnπ/α + Dn r-nπ/α n = 1,2,3... (a0(r)/2) = A0 + B0 ln(r) The coefficients are given by an(r) = (1/α) ∫02α dφ u(r,φ) cos(nπφ/α) bn(r) = (1/α) ∫02α dφ u(r,φ) sin(nπφ/α) Part (a) Put u = 0 on the inner radial part. We then get 0 = Σn=1∞ { An anπ/α + Bn a-nπ/α ) cos(nπφ/α) + A0 + B0 ln(a) + Σn=1∞ { Cn anπ/α + Dn a-nπ/α ) sin(nπφ/α) This tells me that An anπ/α + Bn a-nπ/α = 0 Cn anπ/α + Dn a-nπ/α = 0 A0 + B0 ln(a) = 0 because we have to have 0 coefficient on each φ dependent basis function. We can thus replace: Bn = – An a2nπ/α Dn = – Cn a2nπ/α B0 = – A0/ln(a) So general solution with this BC is now given by u(r,φ) = Σn=1∞ { An rnπ/α + Bn r-nπ/α ) cos(nπφ/α) + A0 + B0 ln(r) + Σn=1∞ { Cn rnπ/α + Dn r-nπ/α ) sin(nπφ/α) = Σn=1∞ An { rnπ/α + Bn/An r-nπ/α ) cos(nπφ/α) + A0[1 + B0/A0 ln(r) ] + Σn=1∞ Cn { rnπ/α + Dn/Cn r-nπ/α ) sin(nπφ/α) = Σn=1∞ An { rnπ/α – a2nπ/α r-nπ/α ) cos(nπφ/α) + A0[1 – ln(r)/ln(a) ] + Σn=1∞ Cn { rnπ/α – a2nπ/α r-nπ/α ) sin(nπφ/α) Now put u = f on the outer radial part u(r,φ) = Σn=1∞ An { rnπ/α – a2nπ/α r-nπ/α ) cos(nπφ/α) + A0[1 – ln(r)/ln(a) ] + Σn=1∞ Cn { rnπ/α – a2nπ/α r-nπ/α ) sin(nπφ/α) f(φ) = u(b,φ) = Σn=1∞ An { bnπ/α – a2nπ/α b-nπ/α ) cos(nπφ/α) + A0[1 – ln(b)/ln(a) ] + Σn=1∞ Cn { bnπ/α – a2nπ/α b-nπ/α ) sin(nπφ/α) But I think we have what we need already earlier, an(b) = (1/α) ∫02α dφ u(b,φ) cos(nπφ/α) = (1/α) ∫02α dφ f(φ) cos(nπφ/α) bn(b) = (1/α) ∫02α dφ u(b,φ) sin(nπφ/α) = (1/α) ∫02α dφ f(φ) sin(nπφ/α) an(b) = An bnπ/α + Bn b-nπ/α n = 1,2,3... bn(b) = Cn bnπ/α + Dn b-nπ/α n = 1,2,3... (a0(b)/2) = A0 + B0 ln(b) Install our results from above that Bn = – An a2nπ/α Dn = – Cn a2nπ/α B0 = – A0/ln(a) Then we have an(b) = An bnπ/α + Bn b-nπ/α = An ( bnπ/α – a2nπ/α b-nπ/α) = An bnπ/α [ 1 - (a/b)2nπ/α ] bn(b) = Cn bnπ/α [ 1 - (a/b)2nπ/α ] a0(b) = 2 A0[ 1 – ln(b)/ln(a) ] Thus, we do the dφ integrals shown above to get an(b) and bn(b) and that tells us An and Cn and A0. We then have the complete solution to our problem: u(r,φ) = Σn=1∞ An { rnπ/α – a2nπ/α r-nπ/α ) cos(nπφ/α) + A0[1 – ln(r)/ln(a) ] + Σn=1∞ Cn { rnπ/α – a2nπ/α r-nπ/α ) sin(nπφ/α) _______________________________________________ _________________________________________- For the case n = 0 and σ = 0, we do get that extra piece with the ln(r). So I guess this is our general solution that meets the radial BC's u(r,φ) = Σσ { Aσ rσ + Bσ r-σ ) cos(σφ) + Σσ { Cσ rσ + Dσ r-σ ) sin(σφ) + A0 + B0 ln(r) = Σσ Aσ { rσ – a2σ r-σ ) cos(σφ) + Σσ Cσ { rσ – a2σ r-σ ) sin(σφ) + A0 + B0 ln(r) The log term looks fishy. We need A0 + B0 ln(r) to be 0 at both r = a and r = b. A0 + B0 ln(a) = 0 A0 + B0 ln(b) = 0 Subtract to get B0 ln(a/b) = 0 so B0 = 0 and then we need A0 = 0. So here is our solution now: u(r,φ) = Σσ Aσ { rσ – a2σ r-σ ) cos(σφ) + Σσ Cσ { rσ – a2σ r-σ ) sin(σφ) where σ = n π/ ln(b/a) with n = ± 1, ± 2, ... . Lets define 1/ln(b/a) = c. Then we have σ = nπc u(r,φ) = Σ'n=-∞∞ {An ( rnπc – a2πnc r-nπc ) cos(σφ) + Cn ( rnπc – a2πnc r-nπc ) sin(σφ) } = Σ'n=-∞∞ ( rnπc – a2πnc r-nπc ) { An cos(σφ) + Cn sin(σφ) } Now at last we can impose our condition at φ = 0 which says u = 0 u(r,0) = Σ'n=-∞∞ ( rnπc – a2πnc r-nπc ) { An } = 0 I think this is saying An = 0, so then our general solution has this form: u(r,φ) = Σ'n=-∞∞ ( rnπc – a2πnc r-nπc ) Cn sin(nπcφ) and then our condition at φ = α is this: u(r,α) = Σ'n=-∞∞ ( rnπc – a2πnc r-nπc ) Cn sin(nπcα) = f(r) Now I am dead in the water because I don't know how to solve this for Cn. We really need some kind of orthonormal set on the interval (a,b) in terms of these powers, like Legendres. So I think I have done enough on this particular problem. Let's move on! Suppose we start off by imposing the condition at r = a where u = 0, as we did above, leaving us with this general form: u(r,φ) = Σn=1∞ An { rnπ/α – a2nπ/α r-nπ/α ) cos(nπφ/α) + A0[1 – ln(r)/ln(a) ] + Σn=1∞ Cn { rnπ/α – a2nπ/α r-nπ/α ) sin(nπφ/α) Now let's impose that this be 0 at r = b: u(b,φ) = Σn=1∞ An { bnπ/α – a2nπ/α b-nπ/α ) cos(nπφ/α) + A0[1 – ln(b)/ln(a) ] + Σn=1∞ Cn { bnπ/α – a2nπ/α b-nπ/α ) sin(nπφ/α) = 0 But this seems to force An = 0 and Cn = 0 and A0 = 0, which is obviously wrong. __________________________________ ************************************************************************* Meanwhile, our equation 6.55 has this form u(r) = –(1/8π) ∫dΩ' b(r') / |r - r'| // WRONG! where r is inside the sphere, and r' is on the sphere. We can project this onto Y's analogously to how we did it above, but now |r| = r and not 1, so there is a slight difference. Let's review our projection above: f(r) = -b(r)/2 - 1/8π∫b(r')dΩ' / |r-r'| R = |r-r'| fmn = -bmn/2 + (1/8π) N0n bmn u(r) = –(1/8π) ∫dΩ' b(r') / |r - r'| umn = –(1/8π) N0n bmn rn = - (1/2) bmn rn/(2n+1) // my guessed result, confirmed below = - (1/2) {– (2n+1)/(n+1) * fmn } rn/(2n+1) = +(1/2(n+1)) * fmn rn Then our u(r) is given by u(r) = ΣnmYnm(Ω) umn = ΣnmYnm(Ω) (+(1/2(n+1)) * fmn rn} = + Σnm rn (1/2(n+1)) fmn Ynm(Ω) Now if we want to have this be an integral over angles, we could do this fmn = (1/Nnm) ∫dΩ' Ynm*(Ω') f(Ω') Then our final result would be u(r) = Σnm rn (1/2(n+1)) Ynm(Ω){ (1/Nnm) ∫dΩ' Ynm*(Ω') f(Ω')} = ∫dΩ' f(Ω') { Σnm rn (1/2(n+1)) Ynm(Ω) Ynm*(Ω')/Nnm } If we compare this to 6.26, we would have to conclude that { Σnm rn (1/2(n+1)) Ynm(Ω) Ynm*(Ω')/Nnm } = - (r2-1)/4π * 1/(1+r2-2rcosγ)3/2 But from A.14 we do know this: (true for any angle, so replace θ with γ) (r2-1) /(1+r2-2rcosγ)3/2 = – Σn(2n+1) rnPn(cosγ) so our task would then be to show that { Σnm rn (1/2(n+1)) Ynm(Ω) Ynm*(Ω')/Nnm } = (1/4π) Σn(2n+1) rnPn(cosγ) or for each rn term that Σm (1/2(n+1)) Ynm(Ω) Ynm*(Ω')/Nnm = (1/4π) (2n+1) Pn(cosγ) (*) But from A.11 or p 126 A we do know that Pn(cosγ) = N0nΣm Ynm(Ω) Ynm*(Ω')/Nnm so the RHS of (*) becomes (1/4π) (2n+1) N0nΣm Ynm(Ω) Ynm*(Ω')/Nnm = Σm {(1/4π) N0n (2n+1)} Ynm(Ω) Ynm*(Ω')/Nnm We would then have our desired equality if it were true that (1/2(n+1)) = (1/4π) N0n (2n+1) But N0n = 4π/(2n+1) so this would say (1/2(n+1)) = 1 so we are close but no cigar. Somewhere I have made an error. I have already found several errors but some still remain somewhere! **********************************************8 How are we going to handle this 1/R3 factor ?? I can set cosγ = P1(cosγ) and expand that on Y's, and the same for b(r'), but we still have to deal with 1/R3. Here is an idea. Let's consider A.8 which says: (1/4π)R-1 = Σnm rn Ynm(Ω)Y*nm(Ω')/ [ (2n+1)Nmn] Differentiate this with respect to r to get - (1/4π) R-2 ∂rR = Σnm n rn-1 Ynm(Ω)Y*nm(Ω')/ [ (2n+1)Nmn] But R = [1 + r2-2r cosγ]1/2 so ∂rR = 1/2 [1 + r2-2r cosγ]-1/2 (2r - 2cosγ) = (r-cosγ)/R which tells us then that - (1/4π) (r-cosγ)/R3 = Σnm n rn-1 Ynm(Ω)Y*nm(Ω')/ [ (2n+1)Nmn] which says: 1/4πR3 = – (r-cosγ)-1 Σnm n rn-1 Ynm(Ω)Y*nm(Ω')/ [ (2n+1)Nmn] and then we have u(r) = - ∫ dΩ' (r cos γ - 1)/(r - cosγ) b(r') Σnm n rn-1 Ynm(Ω)Y*nm(Ω')/ [ (2n+1)Nmn] But this is a huge mess that I don't like! So I am now blocked. Anther approach is geometry which shows this: cosψ = (r2- 1 - R2)/2R so I would then have u(r) = ∫cos(r-r',n at r')b(r')dSr' / 4π R2 R = |r - r'| = ∫dΩ' b(r') (r2- 1 - R2) * 1/8π R3 but then what?? So time to activate "the Master Debugger". We need to produce a factor of (n+1) somewhere, or show that this (n+1) should really be 1.2. (1) Let's check on "my guessed result" and see if somehow that is where the error lies: u(r) = –(1/8π) ∫dΩ' b(r') / |r - r'| (1/Nmn)∫dΩ Y*nm(Ω) to both sides, giving umn = –(1/8π) (1/Nmn)∫dΩ Y*nm(Ω) ∫dΩ' b(r') / |r - r'| = –(1/2) ∫ b(r') dΩ'(1/Nmn) ∫dΩ Y*nm(Ω) / 4π|r - r'| Now use A.8 with r< = r and r> = 1 to get 1/4π|r - r'| = Σn'm' rn' Yn'm'(Ω) Y*n'm'(Ω') / { (2n'+1)Nm'n' } Then we have umn == –(1/2) ∫ b(r') dΩ'(1/Nmn) ∫dΩ Y*nm(Ω) { Σn'm' rn' Yn'm'(Ω) Y*n'm'(Ω') / [ (2n'+1)Nm'n' ] } = –(1/2) ∫ b(r') dΩ'(1/Nmn) { Σn'm' rn' Y*n'm'(Ω') / [ (2n'+1)Nm'n' ] ∫dΩ Y*nm(Ω)Yn'm'(Ω) } = –(1/2) ∫ b(r') dΩ'(1/Nmn) { Σn'm' rn' Y*n'm'(Ω') / [ (2n'+1)Nm'n' ] δnn'δmm'Nnm } = –(1/2) ∫ b(r') dΩ'(1/Nmn) { rn Y*nm(Ω') / [ (2n+1) ] } = –(1/2) * rn/ [ (2n+1) ] (1/Nmn) ∫ b(r') dΩ' Y*nm(Ω') } = –(1/2) * rn/ [ (2n+1) ] * bnm // which agrees with "my guessed result". So this did not produce my error. 2. Is my result for bmn correct. In this book,