Properties of Bessel Functions with Green's Function Problem Examples

Properties of Bessel Functions with Green's Function Problem Examples PhL 8.4.11 1. Selected formulas 1 2. Forms of the homogeneous Bessel Equation and their solutions 2 3. The 3D and 2D Helmholtz problems. 3 4. How are the ODE equations different for the I and K functions? 5 5. Examples of Problems involving Bessel Functions 5 Example 1: The 3D free-space Helmholtz radial exterior Green's Function problem. 5 Example 2: The 2D free-space Helmholtz radial exterior Green's Function problem. 7 Example 1A: The 3D Helmholtz radial exterior-to-sphere Green's Function problem. 8 Example 2A: The 2D Helmholtz radial exterior-to-circle Green's Function problem. 10 1. Selected formulas These are taken from AS2010 and verified against green Jackson. The Bessel Functions: Jν(z) ≡ (z/2)ν Σk=0∞(-z2/4)k/[k! Γ(ν + 1 + k)] cuts all run (0,-∞) Yν(z) ≡ [ cos(πν)Jν(z) – J-ν(z)] /sin(πν) cos(πν)Jν(z) – J-ν(z) = sin(πν) Yν(z) Hν(1)(z) ≡ Jν(z) + i Yν(z) Hν(2)(z) ≡ Jν(z) - i Yν(z) The Modified Bessel Functions: Iν(z) ≡ i-ν Jν(iz) cuts run (0,+i∞) Kν(z) ≡ (iπ/2) iν Hν(1)(iz) = "MacDonald function" Iν(z) Kν(z) = (iπ/2) Jν(iz) Hν(1)(iz) Negative order: J-ν(z) = cos(πν)Jν(z) - sin(πν) Yν(z) // Maple confirmed Y-ν(z) = sin(πν)Jν(z) + cos(πν) Yν(z) // Maple confirmed H-ν(1)(z) = e+iπν Hν(1)(z) // AS H-ν(2)(z) = e-iπν Hν(1)(z) // AS K-ν(z) = Kν(z) I-ν(z) = Iν(z) + (2/π)sin(πν) Kν(z) Small z: Jν(z) ≈ (z/2)ν / Γ(ν+1) ν ≠ -1,-2... Yν(z) ≈ – (Γ(ν)/π) (z/2)-ν Re(ν) > 0 Hν(1)(z) ≈ – i(Γ(ν)/π) (z/2)-ν Re(ν) > 0 Hν(2)(z) ≈ + i(Γ(ν)/π) (z/2)-ν Re(ν) > 0 Iν(z) ≈ (z/2)ν / Γ(ν+1) Kν(z) ≈ (z/2)-ν Γ(ν)/2 Re(ν)>0 K0(z) ≈ -ln(z) - 0.5772 Large z: Jν(z) ≈ cos(z - νπ/2 -π/4) Yν(z) ≈ sin(z - νπ/2 -π/4) Hν(1)(z) ≈ e+i(z-νπ/2-π/4) Hν(2)(z) ≈ e-i(z-νπ/2-π/4) Iν(z) ≈ ez / arg(z) in (-π/2,π/2) Kν(z) ≈ e-z arg(z) in (-3π/2,3π/2) Negative argument: Jν(-z) = e∓iπν Jν(z) Im(z) 0 other cases can be obtained from this case, see GR4 p 968 ODE Forms (see below) [ If k2 terms have opposite sign, then replace Z by I/K. ] -(xu')' + ν2u/x - k2xu = 0 Zν(kx) (*) - (z2u')' - k2z2u + (ν2-1/4) u=0 Zν(kz)/ (**) Wronskians: W[Jν(z), Yν(z)] = 2/(πz) W[Jν(z), J-ν(z)] = -sin(πν) 2/(πz) W[Jν(z), Hν(1,2)(z)] = ± i 2/(πz) W[Hν(1)(z), Hν(2)(z)] = -2i 2/(πz) W[Kν(z), Iν(z)] = 1/z W[Iν(z), I-ν(z)] = -sin(πν) 2/(πz) Note on Wronskians : Wz[Iν(z), I-ν(z)] = Wz[Jν(iz), J-ν(iz)] = W-iξ[Jν(ξ), J-ν(ξ)] = i Wξ[Jν(ξ), J-ν(ξ)] = i { -sin(πν) 2/(πξ)} = i { -sin(πν) 2/(πiz)} = -sin(πν) 2/(πz) = W[Jν(z), J-ν(z)] . So you don't just naively replace z→iz here; you have to consider the meaning of the derivative in W. On the other hand, we can say that Wz[u(z),v(z)] = f(z) => Wkz[u(kz),v(kz)] = f(kz) and here z→kz works fine. In other words, the meaning of W[u(kz),v(kz)] is really Wkz[u(kz),v(kz)]. All these functions are analytic in ν. Cuts in z as specified above. 2. Forms of the homogeneous Bessel Equation and their solutions (a) The basic single-parameter equation and solution is this: (Schaum p 136 for example) x2y" +xy' + (x2- ν2)y = 0 y = Jν(x) xy" +y' + xy - ν2y/x = 0 (xy')' + xy - ν2y/x = 0 -(xy')' + ν2y/x - xy = 0 (b) Suppose y(x) solves the above ODE. Then what equation does w(x) ≡ y(kx) solve? To answer this, we can define x' = kx and then we have [ we are just doing this case as an example ] w(x) = y(kx) w'(x) = ky'(kx) w"(x) = k2y"(kx) dx' = kdx w(x) = y(x') w'(x) = ky'(x') w"(x) = k2y"(x') ∂x'= k-1∂x We know from (a) that xy" +y' + xy - ν2y/x = 0 // quoted from part (a) x∂x2y(x) +∂xy(x) + xy(x) - ν2y(x)/x = 0 // wrote out derivatives x'∂x'2y(x') +∂x'y(x') + x'y(x') - ν2y(x')/x' = 0 // changed x to x' everywhere (kx)k-2∂x2y(kx) +k-1∂xy(kx) +kxy(kx) - ν2y(kx)/(kx) = 0 // installed above facts x ∂x2y(kx) +∂xy(kx) +k2xy(kx) - ν2y(kx)/x = 0 // multiplied by k x ∂x2 w(x) +∂x w(x) +k2x w(x) - ν2 w(x)/x = 0 // replaced y with w x w" +∂x w' +k2x w - ν2 w/x = 0 // deleted obvious arguments x w" +∂x w' +k2x w - ν2 w/x = 0 // removed blanks from line (xw')' +k2x w - ν2 w/x = 0 // combine first two terms - (xw')' - k2x w + ν2 w/x = 0 // change signs. Therefore, we know that Jν(kx) solves this last equation, and notice that k appears in only one term and it is squared. So we have established -(xu')' + ν2u/x - xu = 0 Zν(x) Z = J, Y or an H -(xu')' + ν2u/x - k2xu = 0 Zν(kx) Huge lists of other ODE forms appear GR4 p 971, GR7 p 931. Here is just one example If we set α = -1/2 and γ = 1 and β = k this becomes u" +(2/z)u' + [ k2 + ((1/4) - ν2)/z2 ] u=0 u = Zν(kz)/ z2u" +2zu' + [ k2z2 - (ν+1/2)(ν-1/2) ] u=0 u = Zν(kz)/ (z2u')' + [ k2z2 - (ν+1/2)(ν-1/2) ] u=0 u = Zν(kz)/ - (z2u')' - k2z2u + (ν+1/2)(ν-1/2) u=0 u = Zν(kz)/ - (z2u')' - k2z2u + (ν2-1/4) u=0 u = Zν(kz)/ (**) - (z2u')' - k2z2u + n(n+1) u=0 u = Zn+1/2(kz)/ So let's now restate our two principle results to this point: -(xu')' + ν2u/x - k2xu = 0 Zν(kx) (*) - (z2u')' - k2z2u + n(n+1) u=0 Zn+1/2(kz)/ (**) 3. The 3D and 2D Helmholtz problems. In this section, I will temporarily use LHelm = -2 -k2 with k>0, so in Stak terms λ = +k2. You always have to think separately about the two possible signs of the H parameter λ, and here we assume +. In a wave situation where we have e±iωt, we know ∂t2→ -k2and we arrive at our situation here for the H equation arising from the wave equation. In a heat situation where we have e-at, we have ∂t = -a and λ = a > 0. But if we just think of a "Hemholtstatics problem", λ could have either sign. Equation (**) is the homogeneous radial equation for (-2 -k2)u(r,θ,φ) = 0. To see why, we write (-2 -k2)u(r,θ,φ) = (-r2 -(1/r2)L2 -k2)Ynm(θ,φ)R(r) = 0 r2u = (1/r2)(r2u')' L2 Ynm = n(n+1) Ynm (-r2 +(1/r2)L2 -k2)Ynm(θ,φ)R(r) = 0 (-r2 +(1/r2)n(n+1) -k2)Ynm(θ,φ)R(r) = 0 (-r2 +(1/r2)n(n+1) -k2) R(r) = 0 -(1/r2)(r2R')' +(1/r2)n(n+1)R -k2R = 0 - (r2R')' +n(n+1)R -k2r2R = 0 which does agree with the above. The radial functions are then R = Jn+1/2(kr)/. One often uses the functions jn(r) ≡ Jn+1/2(r) in this case, which are called the spherical Bessel functions, and more generally jn(kr) ≡ Jn+1/2(kr) . See green Jackson p 539. Here is the corresponding 2D Helmholtz problem (-2 -k2)u(r,φ) = (-r2 -(1/r2)∂φ2 -k2)einφ R(r) = 0 r2u = (1/r)(ru')' ∂φ2 einφ = -n2 einφ (-r2 -(1/r2)∂φ2 -k2)einφ R(r) = 0 (-r2 + (1/r2)n2 -k2)einφ R(r) = 0 -(1/r)(rR')' + (1/r2)n2R -k2R = 0 - (rR')' + n2R/r -k2rR = 0 and this matches our Jn(kr) form of part (b) above, so in this 2D problem the radial functions are just Jn(kr). All these functions of course have 2nd and 3rd kind analogs, I am just using J here to represent the class. 4. How are the ODE equations different for the I and K functions? For example, consider these two: -(xu')' + ν2u/x - k2xu = 0 Zν(kx) - (z2u')' - k2z2u + n(n+1) u=0 Zn+1/2(kz)/ All that happens is the k2 term changes sign, so we have -(xu')' + ν2u/x + k2xu = 0 Kν(kx) or Iν - (z2u')' + k2z2u + n(n+1) u=0 Kn+1/2(kz)/ 5. Examples of Problems involving Bessel Functions Example 1: The 3D free-space Helmholtz radial Green's Function problem. If we define the Helmholtz operator as -2-λ , and do separation for a φ azisym problem like the free-space situation, and if we assume w = Σn=0∞ un(r) Pn(cosθ), and if we put our unit point charge on the +z axis at z = r0, we arrive at this radial Green's Function equation (as I show in Stak Ex 7.31) (I suppress the n index on un) : - (r2u')' – [λr2– n(n+1)] u = δ(r-r0) (2n+1)/4π u(∞) = 0 u(0) = finite If we then set λ = -k2 as Stak always does, we get - (r2u')' +k2r2u+ n(n+1) u = δ(r-r0) (2n+1)/4π Rather than find u directly for this equation, let's instead treat this more generic equation (ν = n+1/2) - (r2g')' +k2r2g + (ν2-1/4) g = δ(r-r0) => dim(g) = 1/L We recognize this as (**) but with k2→-k2. Thus, we expect our free-space Green's Function to have the form g = C [ Iv(kr<)/] [ Kv(kr>)/] = (C/) Iv(kr<) Kv(kr>) => dim(C) = 1 We like this form on (0,∞) due to the small-r limit of I, and the large r limit of K: In+1/2(kr)/ → r-1/2 (kr/2)ν / Γ(ν+1) = r-1/2 (kr/2)n+1/2 /Γ(n+3/2) ~ rn = finite r→0 Kn+1/2(kr)/ → r-1/2e-kr ~ e-kr /r → 0 r→∞ We want to compute C. We know staring at our ODE above that jump = -1/r02. But we know jump = C[ Iv(kr0)/] [ Kv(kr)/] ' – C[ Iv(kr)/] ' [ Kv(kr0)/] The two terms involving ∂r(1/) cancel when r=r0, so we can simplify the above to jump = (C/r0) {[ Iv(kr0)] [ Kv(kr)] ' – [ Iv(kr)] ' [ Kv(kr0)]} = (Ck/r0)W[Iv(kr0), Kv(kr0)] = (Ck/r0)(-1/kr0) = -C/r02 Comparison tells us that C = 1. Thus we have obtained our generic Green's function - (r2g')' +kr2g+ (ν2-1/4) g = δ(r-r0) => g = (1/) Iv(kr<) Kv(kr>) **** If we now set ν = n+1/2 and add factor (2n+1)/4π to the RHS, we get - (r2g')' +kr2g+ n(n+1) g = δ(r-r0) (2n+1)/4π => g = [(2n+1)/4π] (1/) In+1/2(kr<) Kn+1/2(kr>) Using Iν(z) Kν(z) = (iπ/2) Jν(iz) Hν(1)(iz) we could write this as g = (iπ/2) [(2n+1)/4π] (1/) Jn+1/2(ikr<) Hν(1)n+1/2(ikr>) = (i/8) (2n+1) Jn+1/2(ikr<) Hν(1)n+1/2(ikr>) and this is the form I obtained in Exercise 7.31. We can then insert this into an expansion for w we find w = Σn=0∞ un(r) Pn(cosθ) = = Σn=0∞ Pn(cosθ) [(2n+1)/4π] (1/) In+1/2(kr<) Kn+1/2(kr>) = E3(λ = -k2) But we know that (see Stak p 55 5.119 and 5.121) E3(λ=-k2) = (1/2π) (k/2πR)1/2K1/2(kR) = e-kR/(4πR) so we then have this little addition theorem e-kR/(4πR) = (1/4π)(1/) Σn=0∞ Pn(cosθ) (2n+1) In+1/2(kr<) Kn+1/2(kr>) **** R2 = r2+ r02 - 2rr0cosθ which we see in 7.225. I cannot find this in Bateman. But Bateman does have p 43 (3) which we can apply with ν = 1/2 to get R-1/2K1/2(R) = (rr0/2)-1/2Γ(1/2) x Σn=0∞ (1/2 +n)C(1/2)n(cosθ) In+1/2(r<) Kn+1/2(r>) But we know from Bateman p 179 (3) that C(1/2)n(cosθ) = Pn(cosθ), so we then have R-1/2K1/2(R) = (rr0/2)-1/2 (1/2) Σn=0∞ (2n+1) Pn(cosθ) In+1/2(r<) Kn+1/2(r>) = (1/2) 1/ Σn=0∞ (2n+1) Pn(cosθ) In+1/2(r<) Kn+1/2(r>) So Bateman's result gives K1/2(R)/ = (1/2) 1/ Σn=0∞ (2n+1) Pn(cosθ) In+1/2(r<) Kn+1/2(r>) But from above we have (1/2π) (k/2πr)1/2K1/2(kr) = e-kr/(4πr) (1/2π) (1/2πR)1/2K1/2(R) = e-kR/(4πR) K1/2(R)/ = 2π e-kr/(4πR) = e-kR/2R So Bateman's result becomes e-kR/(4πR) = (1/4π) 1/ Σn=0∞ (2n+1) Pn(cosθ) In+1/2(r<) Kn+1/2(r>) which agrees with the Stak result. Example 2: The 2D free-space Helmholtz radial Green's Function problem. If we define the Helmholtz operator as -2-λ , and if we assume w = Σn un(r) einθ, and if we put our unit point charge on the +x axis (θ=0) at x = r0, we arrive at this radial Green's Function equation (as I show in Stak Ex 7.34 part (b), although there I was doing range (a,∞), whereas here I am doing (0,∞))(I suppress the n index on un) : -(ru')' - λru + n2u/r = δ(r-r0)/2π u(∞) = 0 u(0) = finite If we then set λ = -k2 as Stak always does, we get -(ru')' +k2ru + n2u/r = δ(r-r0)/2π Rather than find u directly for this equation, let's instead treat this more generic equation -(ru')' +k2ru + ν2u/r = δ(r-r0) dim(u) = 1 We recognize this as (*) but with k2→-k2. Thus, we expect our free-space Green's Function to have this form, g = C [ Iv(kr<)] [ Kv(kr>)] => dim(C) = 1 We like this form on (0,∞) due to the small-r limit of I, and the large r limit of K: Iν(kr) → (kr/2)ν / Γ(ν+1) ~ rν = finite r→0 ν = 0,1,2... Kν(kr) → e-kr ~ e-kr /r → 0 r→∞ We want to compute C. We know staring at our ODE above that jump = -1/r0. But we know jump = C[ Iv(kr0)] [ Kv(kr)] ' – C[ Iv(kr)] ' [ Kv(kr0)] = CkW[Iv(kr0), Kv(kr0)] = Ck(-1/kr0) = -C/r0 Comparison tells us that C = 1. Thus we have obtained our generic Green's function -(ru')' +k2ru + ν2u/r = δ(r-r0) => g = Iv(kr<) Kv(kr>) **** If we now set ν = n and add factor 1/2π to the RHS, we get - (r2g')' +kr2g+ n(n+1) g = δ(r-r0) (2n+1)/4π => g = (1/2π) In(kr<) Kn(kr>) We can then insert this into an expansion for w we find w = (1/2π) Σn In(kr<) Kn(kr>) einθ = E2(λ=-k2) But we know that (see Stak p 266 76.154) E2(λ=-k2) = (i/4)H0(1)(ikR) = (i/4)(2/iπ)K0(kR) = (1/2π)K0(kR) // as in 7.155 so we then have this little addition theorem K0(kR) = Σn In(kr<) Kn(kr>) einθ R2 = r2+ r02 - 2rr0cosθ **** which we see sort of in 7.166, and in Bateman p 44 top, Example 1A: The 3D Helmholtz radial exterior-to-sphere Green's Function problem. We follow Example 1 down to the generic equation (a is radius of sphere, r0 radius of point source) - (r2g')' +k2r2g + (ν2-1/4) g = δ(r-r0) => dim(g) = 1/L but now we have to take g to have the form g = C [ (Iv(kr<)/) (Kv(ka)/) – (Kv(kr<)/) (Iv(ka)/)] [ Kv(kr>)/] which is a Smythian form making exterior g(r=a) = 0, our sphere BC. As before, we know jump = -1/r02. When we compute the jump from the above form for g, the second term will make no contribution to the jump since W[K,K] = 0. The remaining term is just what we had in Example 1 except we have an extra factor of (Kv(ka)/). Thus, we shall obtain C (Kv(ka)/) = 1 instead of C = 1, and g will be g = [ (Iv(kr<)/) (Kv(ka)/) – (Kv(kr<)/) (Iv(ka)/)] [ Kv(kr>)// Kv(ka)/] Each term here has / = = in the denominator, so simplify to g = (1/)[ Iv(kr<) Kv(ka)) – Kv(kr<) Iv(ka)] [ Kv(kr>)/ Kv(ka)] Thus we have this result - (r2g')' +k2r2g + (ν2-1/4) g = δ(r-r0) g(a) = g(∞) = 0 g = (1/)[ Iv(kr<) Kv(ka)) – Kv(kr<) Iv(ka)] [ Kv(kr>)/ Kv(ka)] **** Now moving back from the generic to the Helmholtz problem, we have (ν = n+1/2) - (r2u')' +k2r2u+ n(n+1) u = δ(r-r0) (2n+1)/4π g = (2n+1) (1/4π)[ In+1/2(kr<) Kn+1/2(ka)) – Kn+1/2 (kr<) In+1/2(ka)] [Kn+1/2(kr>)/ Kn+1/2(ka)] and then the solution to our exterior-to-grounded-sphere Helmholtstatics 3D Green's Function problem is (λ = -k2) w(r,θ) =(1/4π) Σn=0∞ (2n+1)Pn(cosθ) * [ In+1/2(kr<) Kn+1/2(ka)) – Kn+1/2 (kr<) In+1/2(ka)] [Kn+1/2(kr>)/ Kn+1/2(ka)] If we change to J and H functions, we will pick up a factor of (iπ/2) and arguments gain factor i. The leading factor is then (1/4π)(iπ/2) = (i/8). This makes the above result for w agree with the result I got for U(r,s) in the Ex 7.33 doc, which I now quote ( and in which we have n = 0) U(r,s) = = (i/8) [ H1/2(1)(a)J1/2(r<) - J1/2(a) H1/2(1)(r<)] [H1/2(1)(r>)/ H1/2(1)(a)] where the connection is = ik = i. The reason we only got n=0 there was due to the spherical symmetry of the problem posed in Ex 7.33. In contrast, our Helmholtstatics problem is only azisym. Example 2A: The 2D Helmholtz radial exterior-to-circle Green's Function problem. We follow Example 2 down to the generic equation -(ru')' +k2ru + ν2u/r = δ(r-r0) dim(u) = 1 but now we have to take g to have the form g = C [ Iv(kr<)Kv(ka) – Kv(kr<) Iv(ka)] [ Kv(kr>)] which is a Smythian form making exterior g(r=a) = 0, our circle BC. As before, we know jump = -1/r0. When we compute the jump from the above form for g, the second term will make no contribution to the jump since W[K,K] = 0. The remaining term is just what we had in Example 2 except we have an extra factor of Kv(ka). Thus, we shall obtain C Kv(ka) = 1 instead of C = 1, and g will be g = [ Iv(kr<)Kv(ka) – Kv(kr<) Iv(ka)] [ Kv(kr>)/ Kv(ka) ] Thus we have this result -(ru')' +k2ru + ν2u/r = δ(r-r0) g(a) = g(∞) = 0 g = [ Iv(kr<)Kv(ka) – Kv(kr<) Iv(ka)] [ Kv(kr>)/ Kv(ka) ] **** For the actual Helmholtz problem instead of the generic, we add factor 1/2π and set ν = n. Then -(ru')' +k2ru + n2u/r = δ(r-r0)/2π g = (1/2π) [ In(kr<)Kn(ka) – Kn(kr<) In(ka)] [ Kn(kr>)/ Kn(ka) ] and then we have our full solution, w = Σn un(r) einθ = (1/2π) Σn einθ[ In(kr<)Kn(ka) – Kn(kr<) In(ka)] [ Kn(kr>)/ Kn(ka) ] Then using Iν(z) Kν(z) = (iπ/2) Jν(iz) Hν(1)(iz) and Kn(x)/ Kn(y) = Hn(1)(ix)/ Hn(1)(iy), we can rewrite this, noting that (1/2π) (iπ/2) = (i/4), w(r,θ) = (i/4) Σn einθ[Jn(ikr<) Hn(1)(ika) – Hn(1)(ikr<) Jν(ika)] [Hn(1) (ikr>)/ Hn(1) (ika) ] which agrees with the end of part (b) of Exercise 7.34. Comment: Notice that, in our 1A and 2A examples above, if we take a→0, we do NOT obtain the solutions to the free-space problems 1 and 2. The reason of course is that the free-space problems do not have w=0 at the origin, whereas the limit of the a problems does.