Frobenius Method Applied to the Associated Legendre Equation

Frobenius Method Applied to the Associated Legendre Equation PhL 7.26.09 I have run into a lot of confusion and contradiction here, so we now have a separate document to do battle with the subject. First, here is the ODE of interest: (1-z2) u" - 2zu' + [ λ - m2/(1-z2) ] u = 0 // this form appears Schaum p 149 Expand around the point z = 1 To this end, I want to cast this equation into "standard Frobenius form" with respect to z = 1. This will take a few processing steps: (1-z2) u" - 2zu' + [ λ - m2/(1-z2) ] u = 0 // as above (1+z) (1-z) u" - 2zu' + [ λ - m2/{(1+z) (1-z)} ] u = 0 // trivial (1-z) u" - 2z/(1+z)u' + [ λ/(1+z) - m2/{(1+z)2 (1-z)} ] u = 0 // divide through by (1+z) (1-z)2 u" - 2z(1-z)/(1+z)u' + [ λ(1-z)/(1+z) - m2/{(1+z)2} ] u = 0 // mult thru by (1-z) (z-1)2 u" + 2z(z-1)/(1+z)u' + [ -λ(z-1)/(1+z) - m2/{(1+z)2} ] u = 0 // change some signs (z-1)2 u" + (z-1) {2z /(1+z) }u' + { -λ(z-1)/(1+z) - m2/(1+z)2 } u = 0 // std form We can now read off our standard form which is this for general point z0 (z-z0)2u"(z) + (z-z0) P(z)u'(z) + Q(z)u(z) We therefore find that P(z) = {2z /(1+z) } Q(z) = { -λ(z-1)/(1+z) - m2/(1+z)2 } P(1) = 1 Q(1) = -m2/4 The indicial equation with reference to z = 1 expansion has this form r(r-1) + rP(1) + Q(1) = 0 r(r-1) + r -m2/4 = 0 r2 = m2/4 r = ± m/2 Thus, we can take r1 = |m|/2 r2 = - |m|/2 Case 3: r1 = r2 + s s = integer > 0 w1(z) = (z-1)|m|/2 [ 1 + Σn=1∞ An(1) (z-1)n] w2(z) = (z-1)-|m|/2 [ - 1/s + Σn=1∞ hn (z-1)n ] + gs w1(z) ln(z-1) w2(z) = (z-1)-|m|/2 [ 1 + Σn=1∞ (-shn) (z-1)n ] +(-sgs) w1(z) ln(z-1) taken from my "regular singular points" doc. Now for m = positive integer, we know we can write solutions in this manner: (AS p 334) Pνm(z) = (z-1)m/2 {(z+1)m/2 ∂zm Pν(z) } where ν(ν+1) = λ = arbitrary value This is consistent with the w1(z) solution shown above, that is my main point. The series in w1 must be an infinite series in (z-1) which adds up to (z+1)m/2 ∂zm Pν(z) . More detail added 10.8.10. For example, consider f(z) = (z+1)a where we have in mind a = m/2. Expand this in a Taylor series Σn=0∞ bn (z-1)n about the point z = 1. We have f(z) = (z+1)a = 2a f'(z) = a(z+1)a-1 = a 2a-1 f(2)(z) = a(a-1)(z+1)a-2 = a(a-1) 2a-2 etc (z+1)a = 2a + (z-1) a 2a-1 + (z-1)2/2! * a(a-1) 2a-2 + ... If you have doubts about this series, try it for a = 2 so f = (z+1)2. This is recovered by the series for which only the first three terms are non-vanishing. So, my point is that, in this same manner, you could expand the quantity {(z+1)m/2 ∂zm Pν(z) } in a power series about (z-1) by doing a similar Taylor series and relying on the fact that the P functions derivatives are continuous at z = 1. So when you see the total form of the result Pνm(z) = (z-1)m/2 {(z+1)m/2 ∂zm Pν(z) }, if you are thinking about the regular point z = 1 and m>0, it really is showing here as the form w1(z) = (z-1)|m|/2 [ Σn=0∞ An(1) (z-1)n]. The factor (z+1)m/2 is analytic at z=1 so there is no problem doing a power series about z = 1, as shown above. we know that we can expand: (z+1)m/2 = Expand around the point z = -1 Start in the same place: (1-z2) u" - 2zu' + [ λ - m2/(1-z2) ] u = 0 // as above (1+z) (1-z) u" - 2zu' + [ λ - m2/{(1+z) (1-z)} ] u = 0 // trivial (1+z) u" - 2z/(1-z)u' + [ λ/(1-z) - m2/{(1-z)2 (1+z)} ] u = 0 // divide through by (1-z) (1+z)2 u" - 2z(1+z)/(1-z)u' + [ λ(1+z)/(1-z) - m2/{(1-z)2} ] u = 0 // mult thru by (1+z) (z+1)2 u" - 2z(z+1)/(1-z)u' + [ λ(z+1)/(1-z) - m2/{(1-z)2} ] u = 0 // replace 1+z = z+1 (z+1)2 u" + (z+1){-2z/(1-z)}u' + { λ(z+1)/(1-z) - m2/{(1-z)2} u = 0 // standard form P(z) = {-2z /(1-z) } Q(z) = { λ(z+1)/(1-z) - m2/{(1-z)2} P(-1) = 1 Q(-1) = -m2/4 The indicial equation with reference to z = -1 expansion has this form r(r-1) + rP(-1) + Q(-1) = 0 r(r-1) + r -m2/4 = 0 r2 = m2/4 r = ± m/2 So we have the same two exponents that we had at the z = 1 point. Case 3: r1 = r2 + s s = integer > 0 w1(z) = (z+1)|m|/2 [ 1 + Σn=1∞ An(1) (z+1)n] w2(z) = (z+1)-|m|/2 [ - 1/s + Σn=1∞ hn (z+1)n ] + gs w1(z) ln(z+1) w2(z) = (z+1)-|m|/2 [ 1 + Σn=1∞ (-shn) (z+1)n ] +(-sgs) w1(z) ln(z+1) Now for m = positive integer, we know we can write solutions in this manner: (AS p 334) Pνm(z) = (z+1)m/2 {(z-1)m/2 ∂zm Pν(z) } This is consistent with the w1(z) solution shown above, that is my main point. The series in w1 must be an infinite series which adds up to (z-1)m/2 ∂zm Pν(z) . Conclusion: For positive integer m and general ν (λ), our known form for the associated Legendres Pνm(z) = (z2- 1)m/2 ∂zm Pν(z) is consistent with our Frobenius analysis at both z = 1 and z = -1. By the way, if z = ±1, then if m is any positive integer, we get Pνm(1) = 0 (if ν = n). Then Jackson p 65 3.51 makes this true also for m being any negative integer. If m = 0, we get Pν0(1) = Pν(1) = 1.