Eigenfunctions on the surface of a sphere In spherical coordinates, the Laplacian is u = u rr + 2 r u r + 1 r 2 [ uφφ sin 2 (θ) + 1 sin θ (sin θ u θ) θ ].
Eigenfunctions on the surface of a sphere In spherical coordinates, the Laplacian is u = u rr + 2 r u r + 1 r 2 [ uφφ sin 2 (θ) + 1 sin θ (sin θ u θ) θ Separating out the r variable, left with the eigenvalue problem for v(φ, θ) ]. s v + λv = 0, s v v φφ sin 2 (θ) + 1 sin θ (sin θ v θ) θ.
Eigenfunctions on the surface of a sphere In spherical coordinates, the Laplacian is u = u rr + 2 r u r + 1 r 2 [ uφφ sin 2 (θ) + 1 sin θ (sin θ u θ) θ Separating out the r variable, left with the eigenvalue problem for v(φ, θ) ]. s v + λv = 0, s v v φφ sin 2 (θ) + 1 sin θ (sin θ v θ) θ. Let v = p(θ)q(φ) and separate variables: q q + sin θ(sin θ p ) + λ sin 2 θ = 0. p
Eigenfunctions on the surface of a sphere In spherical coordinates, the Laplacian is u = u rr + 2 r u r + 1 r 2 [ uφφ sin 2 (θ) + 1 sin θ (sin θ u θ) θ Separating out the r variable, left with the eigenvalue problem for v(φ, θ) ]. s v + λv = 0, s v v φφ sin 2 (θ) + 1 sin θ (sin θ v θ) θ. Let v = p(θ)q(φ) and separate variables: q q + sin θ(sin θ p ) + λ sin 2 θ = 0. p The problem for q is familiar: q /q = constant with periodic boundary conditions gives q = cos(mφ), sin(mφ), m = 0, 1, 2,..., (complex form: q = exp(imφ), m = 0, ±1, ±2,...)
Eigenfunctions on the surface of a sphere, cont. With q q = m2, and equation for p(θ) is 1 sin θ (sin θ p ) + (λ m2 sin 2 )p = 0. θ
Eigenfunctions on the surface of a sphere, cont. With q q = m2, and equation for p(θ) is 1 sin θ (sin θ p ) + (λ Lucky change of variables s = cos θ gives m2 sin 2 )p = 0. θ [(1 s 2 )p ] +[λ m 2 /(1 s 2 )]p = 0 (associated Legendre s equation)
Eigenfunctions on the surface of a sphere, cont. With q q = m2, and equation for p(θ) is 1 sin θ (sin θ p ) + (λ Lucky change of variables s = cos θ gives m2 sin 2 )p = 0. θ [(1 s 2 )p ] +[λ m 2 /(1 s 2 )]p = 0 (associated Legendre s equation) Boundary conditions on p: insist solution is bounded at θ = 0, π. In the s variable, this implies p(s = ±1) is bounded.
Solution of the associated Legendre s equation With m = 0, look for a solution of the form p(s) = k=0 a ks k.
Solution of the associated Legendre s equation With m = 0, look for a solution of the form p(s) = k=0 a ks k. Substituting into the equation (k + 2)(k + 1)a k+2 s k (k 2 + k λ)a k s k = 0. k=2 k=0
Solution of the associated Legendre s equation With m = 0, look for a solution of the form p(s) = k=0 a ks k. Substituting into the equation (k + 2)(k + 1)a k+2 s k (k 2 + k λ)a k s k = 0. k=2 k=0 Equating coefficients of s k leads to the recursion relation k(k + 1) λ a k+2 = a k (k + 2)(k + 1).
Solution of the associated Legendre s equation With m = 0, look for a solution of the form p(s) = k=0 a ks k. Substituting into the equation (k + 2)(k + 1)a k+2 s k (k 2 + k λ)a k s k = 0. k=2 k=0 Equating coefficients of s k leads to the recursion relation k(k + 1) λ a k+2 = a k (k + 2)(k + 1). If λ = l(l + 1) for l = 0, 1, 2, 3,... series solution has zero coefficients for k > l; these are famous Legendre polynomials P l (s)
Solution of the associated Legendre s equation With m = 0, look for a solution of the form p(s) = k=0 a ks k. Substituting into the equation (k + 2)(k + 1)a k+2 s k (k 2 + k λ)a k s k = 0. k=2 k=0 Equating coefficients of s k leads to the recursion relation k(k + 1) λ a k+2 = a k (k + 2)(k + 1). If λ = l(l + 1) for l = 0, 1, 2, 3,... series solution has zero coefficients for k > l; these are famous Legendre polynomials P l (s) If λ l(l + 1) for positive integer l, power series has one as radius of convergence, and thus series diverges at s = ±1.
Solution of the associated Legendre s equation With m = 0, look for a solution of the form p(s) = k=0 a ks k. Substituting into the equation (k + 2)(k + 1)a k+2 s k (k 2 + k λ)a k s k = 0. k=2 k=0 Equating coefficients of s k leads to the recursion relation k(k + 1) λ a k+2 = a k (k + 2)(k + 1). If λ = l(l + 1) for l = 0, 1, 2, 3,... series solution has zero coefficients for k > l; these are famous Legendre polynomials P l (s) If λ l(l + 1) for positive integer l, power series has one as radius of convergence, and thus series diverges at s = ±1. For m > 0: remarkable formula p(s) = P m l (s) (1 s 2 ) m/2 d m This requires m l to get nonzero answer. ds m P l(s).
Eigenfunctions on the surface of a sphere, summary Complete set of eigenfunctions is cos(mφ)p m l (cos θ), sin(mφ)p m l (cos θ), l = 0, 1, 2,... m = 0, 1, 2,.. with corresponding eigenvalues λ = l(l + 1).
Eigenfunctions on the surface of a sphere, summary Complete set of eigenfunctions is cos(mφ)p m l (cos θ), sin(mφ)p m l (cos θ), l = 0, 1, 2,... m = 0, 1, 2,.. with corresponding eigenvalues λ = l(l + 1). Complex form, which gives Y m l (φ, θ) = e imφ P m l (cos θ), l = 0, 1, 2,... m = 0, ±1, ±2,..., ±l. These are the famous spherical harmonics.
Eigenfunctions on the surface of a sphere, summary Complete set of eigenfunctions is cos(mφ)p m l (cos θ), sin(mφ)p m l (cos θ), l = 0, 1, 2,... m = 0, 1, 2,.. with corresponding eigenvalues λ = l(l + 1). Complex form, which gives Y m l (φ, θ) = e imφ P m l (cos θ), l = 0, 1, 2,... m = 0, ±1, ±2,..., ±l. These are the famous spherical harmonics. s is self-adjoint with the inner product v 1, v 2 = π 2π 0 0 v 1 (φ, θ)v 2 (φ, θ) sin θdφdθ.
Eigenfunctions on the surface of a sphere, summary Complete set of eigenfunctions is cos(mφ)p m l (cos θ), sin(mφ)p m l (cos θ), l = 0, 1, 2,... m = 0, 1, 2,.. with corresponding eigenvalues λ = l(l + 1). Complex form, which gives Y m l (φ, θ) = e imφ P m l (cos θ), l = 0, 1, 2,... m = 0, ±1, ±2,..., ±l. These are the famous spherical harmonics. s is self-adjoint with the inner product v 1, v 2 = π 2π 0 0 v 1 (φ, θ)v 2 (φ, θ) sin θdφdθ. Spherical harmonics are therefore orthogonal: < Yl m, Yl m >= 0 unless l = l and m = m
Spherical harmonics in pictures
Example: solving Laplace equation inside sphere Problem in spherical coordinates is u = u rr + 2 r u r + 1 r 2 su = 0, u(φ, θ, a) = f (φ, θ)
Example: solving Laplace equation inside sphere Problem in spherical coordinates is u = u rr + 2 r u r + 1 r 2 su = 0, u(φ, θ, a) = f (φ, θ) Separating variables u = R(r)v(φ, θ) gives thus s v + λv = 0. r 2 R 2rR R = sv v = λ
Example: solving Laplace equation inside sphere Problem in spherical coordinates is u = u rr + 2 r u r + 1 r 2 su = 0, u(φ, θ, a) = f (φ, θ) Separating variables u = R(r)v(φ, θ) gives thus s v + λv = 0. r 2 R 2rR R = sv v = λ Good news! We have solved the eigenvalue problem v = Y m l (φ, θ), λ = l(l+1), l = 0, 1, 2,... m = 0, ±1, ±2,..., ±l.
Example: solving Laplace equation inside sphere, cont. Now for R equation r 2 R + 2rR l(l + 1)R = 0, which is an Euler equation with solutions R = r α.
Example: solving Laplace equation inside sphere, cont. Now for R equation r 2 R + 2rR l(l + 1)R = 0, which is an Euler equation with solutions R = r α. Substituting gives α(α 1) + 2α l(l + 1) = (α l)(α + l + 1) = 0 Thus α = l or α = l 1. (reject latter)
Example: solving Laplace equation inside sphere, cont. Now for R equation r 2 R + 2rR l(l + 1)R = 0, which is an Euler equation with solutions R = r α. Substituting gives α(α 1) + 2α l(l + 1) = (α l)(α + l + 1) = 0 Thus α = l or α = l 1. (reject latter) Now put it all together as superposition: l u = A lm r l Yl m (φ, θ), A lm are potentially complex. l=0 m= l
Example: solving Laplace equation inside sphere, cont. Now for R equation r 2 R + 2rR l(l + 1)R = 0, which is an Euler equation with solutions R = r α. Substituting gives α(α 1) + 2α l(l + 1) = (α l)(α + l + 1) = 0 Thus α = l or α = l 1. (reject latter) Now put it all together as superposition: l u = A lm r l Yl m (φ, θ), A lm are potentially complex. l=0 m= l Use orthogonality to find coefficients: l f (φ, θ) = so that A lm = 1 a l f, Y m Y m l l l=0 m= l, Yl m, v 1, v 2 = A lm a l Y m l (φ, θ) π 2π 0 0 v 1 (φ, θ)v 2 (θ, φ) sin θdφdθ.
Spherical coordinates in Cartesian coordinates Consider a separated solution P(x) = r l Yl m (φ, θ) of Laplace s equation.
Spherical coordinates in Cartesian coordinates Consider a separated solution P(x) = r l Yl m (φ, θ) of Laplace s equation. The function P(x) is homogeneous of degree l, which means P(αx) = α l P(x), α > 0.
Spherical coordinates in Cartesian coordinates Consider a separated solution P(x) = r l Yl m (φ, θ) of Laplace s equation. The function P(x) is homogeneous of degree l, which means P(αx) = α l P(x), α > 0. Theorem about homogeneous functions: If P(x) : R n R is continuous and homogeneous, of degree l, then P(x) is a polynomial.
Spherical coordinates in Cartesian coordinates Consider a separated solution P(x) = r l Yl m (φ, θ) of Laplace s equation. The function P(x) is homogeneous of degree l, which means P(αx) = α l P(x), α > 0. Theorem about homogeneous functions: If P(x) : R n R is continuous and homogeneous, of degree l, then P(x) is a polynomial. Therefore Yl m = P(x)/r l, where P(x) is a homogeneous polynomial of degree l which solves Laplace s equation.
Spherical coordinates in Cartesian coordinates Consider a separated solution P(x) = r l Yl m (φ, θ) of Laplace s equation. The function P(x) is homogeneous of degree l, which means P(αx) = α l P(x), α > 0. Theorem about homogeneous functions: If P(x) : R n R is continuous and homogeneous, of degree l, then P(x) is a polynomial. Therefore Yl m = P(x)/r l, where P(x) is a homogeneous polynomial of degree l which solves Laplace s equation. For l = 0, P = 1 and get Y 0 0 = 1. For l = 1, P = x, y, z gives z/r = Y 0 1, and x/r,y/r which are the real and imaginary parts of Y 1 1.