MATH 311Topics in Applied MathematicsLecture 25:Bessel functions (continued).Bessel’s differential equation of order m ≥ 0:z2d2fdz2+ zdfdz+ (z2− m2)f = 0The equation is considered on the interval (0, ∞).Solutions are called Bessel functions of order m.Jm(z): Bessel function of the first kind,Ym(z): Bessel function of the second kind.The general Bessel function of order m isf (z) = c1Jm(z) + c2Ym(z), where c1, c2areconstants.Bessel functions of the 1st and 2nd kindAsymptotics at the originJm(z) is regular while Ym(z) has a singularity at 0.As z → 0, we have for any integer m > 0Jm(z) ∼12mm!zm, Ym(z) ∼ −2m(m − 1)!πz−m.Also, J0(z) ∼ 1, Y0(z) ∼2πlog z.To get the asymptotics for a noninteger m, wereplace m! by Γ(m + 1) and (m − 1)! by Γ(m).Jm(z) is uniquely determined by this asymptoticswhile Ym(z) is not.Asymptotics at infinityAs z → ∞, we haveJm(z) =r2πzcosz −π4−mπ2+ O(z−1),Ym(z) =r2πzsinz −π4−mπ2+ O(z−1).Both Jm(z) and Ym(z) are uniquely determined bythis asymptotics.For m = 1/2, these are exact formulas.Original definition by Bessel (only for integer m):Jm(z) =1πZπ0cos(z sin τ − mτ) dτ=1πZπ0cos(z sin τ) cos(mτ) dτ +1πZπ0sin(z sin τ) sin(mτ) dτ.The first integral is 0 for any odd m while thesecond integral is 0 for any even m. It follows thatcos(z sin τ) = J0(z) + 2X∞n=1J2n(z) cos(2nτ),sin(z sin τ) = 2X∞n=1J2n−1(z) sin(2n − 1)τ.Zeros of Bessel functionsLet 0 < jm,1< jm,2< . . . be zeros of Jm(z) and0 < ym,1< ym,2< . . . be zeros of Ym(z).Let 0 ≤ j′m,1< j′m,2< . . . be zeros of J′m(z) and0 < y′m,1< y′m,2< . . . be zeros of Y′m(z).(We let j′0,1= 0 while j′m,1> 0 if m > 0.)Then the zeros are interlaced:m ≤ j′m,1< ym,1< y′m,1< jm,1<< j′m,2< ym,2< y′m,2< jm,2< . . .Asymptotics of the nth zeros as n → ∞:j′m,n≈ ym,n∼ (n +12m −34)π,y′m,n≈ jm,n∼ (n +12m −14)π.Dirichlet Laplacian in a circleEigenvalue problem:∇2φ + λφ = 0 in D = {(x, y ) : x2+ y2≤ R2},φ|∂D= 0.Separation of variables in polar coordinates:φ(r, θ) = f (r)h(θ). Reduces the problem to twoone-dimensional eigenvalue problems:r2f′′+ rf′+ (λr2− µ)f = 0, f (R) = 0, |f (0)| < ∞;h′′= −µh, h(−π) = h(π), h′(−π) = h′(π).The latter problem has eigenvalues µm= m2,m = 0, 1, 2, . . . , and eigenfunctions h0= 1,hm(θ) = cos mθ,˜hm(θ) = sin mθ, m ≥ 1.The 1st intermediate eigenvalue problem:r2f′′+ rf′+ (λr2− m2)f = 0, f (R) = 0, |f (0)| < ∞.New variable z =√λ · r reduces the equation toBessel’s equation of order m. Hence the generalsolution is f (r) = c1Jm(√λ r) + c2Ym(√λ r),where c1, c2are constants.Singular condition |f (0)| < ∞ holds if c2= 0.Nonzero solution exists if Jm(√λ R) = 0.Thus there are infinitely many eigenvalues λm,1, λm,2, . . . ,wherepλm,nR = jm,n, i.e., λm,n= (jm,n/R)2.Associated eigenfunctions: fm,n(r) = Jm(jm,nr/R).The eigenfunctions fm,n(r) = Jm(jm,nr/R) areorthogonal relative to the inner producthf , gir=ZR0f (r) g (r ) r dr.Any piecewise continuous function g on [0, R] isexpanded into a Fourier-Bessel seriesg(r) =X∞n=1cnJmjm,nrR, cn=hg, fm,nirhfm,n, fm,nir,that converges in the mean (with weight r).If g is piecewise smooth, then the series convergesat its points of continuity.Eigenvalue problem:∇2φ + λφ = 0 in D = {(x, y ) : x2+ y2≤ R2},φ|∂D= 0.Eigenvalues: λm,n= (jm,n/R)2, wherem = 0, 1, 2, . . . , n = 1, 2, . . . , and jm,nis the nthpositive zero of the Bessel function Jm.Eigenfunctions: φ0,n(r, θ) = J0(j0,nr/R).For m ≥ 1, φm,n(r, θ) = Jm(jm,nr/R) cos mθ and˜φm,n(r, θ) = Jm(jm,nr/R) sin mθ.Neumann Laplacian in a circleEigenvalue problem:∇2φ + λφ = 0 in D = {(x, y ) : x2+ y2≤ R2},∂φ∂n∂D= 0.Again, separation of variables in polar coordinates,φ(r, θ) = f (r)h(θ), reduces the problem to twoone-dimensional eigenvalue problems:r2f′′+ rf′+ (λr2− µ)f = 0, f′(R) = 0, |f (0)| < ∞;h′′= −µh, h(−π) = h(π), h′(−π) = h′(π).The 2nd problem has eigenvalues µm= m2,m = 0, 1, 2, . . . , and eigenfunctions h0= 1,hm(θ) = cos mθ,˜hm(θ) = sin mθ, m ≥ 1.The 1st one-dimensional eigenvalue problem:r2f′′+ rf′+ (λr2− m2)f = 0, f′(R) = 0, |f (0)| < ∞.For λ > 0, the general solution of the equation isf (r) = c1Jm(√λ r) + c2Ym(√λ r), where c1, c2areconstants.Singular condition |f (0)| < ∞ holds if c2= 0.Nonzero solution exists if J′m(√λ R) = 0.Thus there are infinitely many eigenvalues λm,1, λm,2, . . . ,wherepλm,nR = j′m,n, i.e., λm,n= (j′m,n/R)2.Associated eigenfunctions: fm,n(r) = Jm(j′m,nr/R).λ = 0 is an eigenvalue only for m = 0.Eigenvalue problem:∇2φ + λφ = 0 in D = {(x, y ) : x2+ y2≤ R2},∂φ∂n∂D= 0.Eigenvalues: λm,n= (j′m,n/R)2, wherem = 0, 1, 2, . . . , n = 1, 2, . . . , and j′m,nis the nthpositive zero of J′m(exception: j′0,1= 0).Eigenfunctions: φ0,n(r, θ) = J0(j′0,nr/R).In particular, φ0,1= 1.For m ≥ 1, φm,n(r, θ) = Jm(j′m,nr/R) cos mθ and˜φm,n(r, θ) = Jm(j′m,nr/R) sin mθ.Laplacian in a circular sectorEigenvalue problem:∇2φ + λφ = 0 in D = {(r, θ) : r < R, 0 < θ < L},φ|∂D= 0.Again, separation of variables in polar coordinates,φ(r, θ) = f (r)h(θ), reduces the problem to twoone-dimensional eigenvalue problems:r2f′′+ rf′+ (λr2− µ)f = 0, f (0) = f (R) = 0;h′′= −µh, h(0) = h(L) = 0.The 2nd problem has eigenvalues µm= (mπL)2,m = 1, 2, . . . , and eigenfunctions hm(θ) = sinmπθL.The 1st one-dimensional eigenvalue problem:r2f′′+ rf′+ (λr2− ν2)f = 0, f (0) = f (R) = 0.Here ν2= µm. We may assume that λ > 0.The general solution of the equation isf (r) = c1Jν(√λ r) + c2Yν(√λ r), where c1, c2areconstants.Boundary condition f (0) = 0 holds if c2= 0.Nonzero solution exists if Jν(√λ R) = 0.Thus there are infinitely many eigenvalues λm,1, λm,2, . . . ,wherepλm,nR = jν,n, i.e., λm,n= (jν,n/R)2.Associated eigenfunctions: fm,n(r) = Jν(jν,nr/R).Note that ν = mπ/L.Eigenvalue problem:∇2φ + λφ = 0 in D = {(r, θ) : r < R, 0 < θ < L},φ|∂D= 0.Eigenvalues: λm,n= (jmπL,n/R)2, wherem = 1, 2, . . . , n = 1, 2, . . . , and jmπL,nis the nthpositive zero of the Bessel function JmπL.Eigenfunctions:φm,n(r, θ) = JmπL(jmπL,n· r/R)