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Stat 992: Lecture 21
Diffusion smoothing on sphere.
December 8, 2003
Problem 16. What is the heat kernel on a unit sphere? Solution. Read S. Rosenberg’s The Laplacian on a Rie- mannian Manifold (1997 Cambridge University Press), I Chavel’s Eigenvalues in Riemannian geometry (Aca- demic Press, 1984), E.B. Davies’ Heat kernels and spec- tral theory (Cambridge University Press, 1989). P. Olver and C. Shakiban’s Fundamentals of Applied Mathemat- ics, Prentice-Hall, in preparation.
- Diffusion on sphere. Following lecture 7 and 20, we show the construction of heat kernel on a unit sphere. We have already showed how to construct it in implicit numerical scheme using the iterated kernel smoothing on manifolds. We may use spher- ical coordinate system:
x = sin θ cos ψ, y = sin θ sin ψ, z = cos θ.
On a unit sphere p = (x, y, z) ∈ S^2 , heat kernel Kt must satisfy isotropic heat equation
∂Kt ∂t
= ∆S 2 Kt.
where
∆S 2 =
sin θ
∂θ
sin θ
∂θ
sin^2 θ
∂^2
∂^2 ψ
It should satisfy the following three conditions Kt(p, q) = Kt(q, p), limt→ 0 Kt(p, q) = δ(p − q) and Kt(p, q) =
M Ks(p, r)Kt−s(r, q)^ dr. Then the following PDE
∂f ∂t
= ∆S 2 f
with initial condition f (0, p) = Y (p) has a unique solution
f (t, p) =
S^2
Kt(p, q)Y (p) dμ(p).
In lecture 20, we showed how to solve it numeri- cally.
- Spherical Helmholtz equation. We will follow lec- ture 11. In the previous lecture, we showed the met- ric to be g 11 = 1, g 12 = g 21 = 0, g 22 = sin^2 θ. Define L^2 (S^2 , g) with respect to the Riemannian metric g with inner product
〈f, g〉 =
S^2
f (p)g(p) dμ(p).
where the Lebesgue measure on a sphere is given by dμ =
det g dθdψ = sin θdθdψ. This is the area element you must have seen it in vector calculus course. Note that
S^2 dμ^ = 4π, the area of the unit sphere. The gradient ∇ : C∞(S^2 ) → Tp(S^2 ) in local coor- dinates is given by
∇f =
i,j
gij^ ∂f ∂uj
∂p ∂ui
= ∂f ∂θ
∂p ∂θ
sin^2 θ
∂f ∂ψ
∂p ∂ψ
Then it can be shown that
〈∆S 2 f, g〉 = −〈∇f, ∇g〉 = 〈f, ∆S 2 g〉
So the spherical Laplacian is compact self-adjoint operator and it should have eigenvalues λj and eigenfunctions Hj such that
∆S 2 Hj (p) = λj Hj (p)
and 0 = |λ 0 | ≤ |λ 1 | ≤ |λ 2 | · · ·. To solve for eigen- values and eigenfunctions of Laplacian, we need to solve the spherical Helmholtz equation given by 1 sin θ
∂θ
sin θ
∂H
∂θ
sin^2 θ
∂^2 H
∂^2 ψ
− λH = 0
Let us use the separation of variable technique: H(θ, ψ) = α(θ)β(ψ). Then substituting the term, we get
sin^2 θ α′′ α
− sin^2 θλ = − β′′ β
= c
Note that the left side is a function of θ while the right side is a function of ψ. The only way both side equal is when they are constant c. The right hand side is trivial to solve:
β′′^ + cβ = 0
Since β(ψ) must be a periodic function with period 2 π, the solutions are
β(ψ) = cos(mψ), sin(mψ)
where m =
c = 0, 1 , 2 , · · ·. The left hand side is not so obvious to solve. We will change variable t = cos θ and let
α(θ) = α(cos−^1 t) = P (t).
From the chain rule,
α′^ = −
1 − t^2 P ′,
α′′^ = (1 − t^2 )P ′′^ − tP ′. Substituting these into the left side, we get
(1−t^2 )^2 P ′′^ − 2 t(1−t^2 )P ′^ −[λ(1−t^2 )+m^2 ]P = 0
which is the Legendre equation. This is a well known ordinary differential equation (ODE) and its solution is the Legendre functions. Note that the so- lution is defined in the domain − 1 ≤ t = cos θ ≤ 1 and we give boundary condition |P (−1)| < ∞, |P (1)| < ∞. For m = 0, Pn(t) = (^2) n^1 n!^ d n dtn^ (t
1)n, n = 0, 1 , 2 , · · · are the Legendre polynomials. For general m > 0 , the Legendre functions are con- structed in the following way
P (^) nm (t) = (1−t^2 )m/^2 dm dtm^
Pn(t), n = m, m+1, · · ·
Even ordered Legendre functions are polynomials but for odd ordered functions there is an additional factor
1 − t^2. In summarizing the result, the com- plete set of solutions for the left hand side is
α(θ) = P (^) nm (cos θ), n = m, m + 1, · · ·.
For instance P 00 = 1 , P 10 = cos θ, P 11 = − sin θ, P 20 = 14 + 34 cos 2θ, · · ·.
- Spherical harmonics. By combining the solution from both ODEs, we have the complete set of so- lution to the spherical Helmholtz equation gives spherical harmonics:
Hnm (θ, ψ) = cos(mψ)P (^) nm (cos θ),
H^ ˜nm (θ, ψ) = sin(mψ)P (^) nm (cos θ) where n = 0 , 1 , 2 , · · · , m = 0 , 1 , · · · , n. The spherical harmonics satisfy
∆S 2 Hnm +n(n+1)Y (^) nm = ∆S 2 H˜nm +n(n+1)Y (^) nm = 0.
Note that H˜ n^0 = 0. The corresponding eigenvalues are λn = −n(n + 1). These harmonics are orthog- onal but not normalized. It can be shown that
‖H^0 n‖^2 = 4 π 2 n + 1 , ‖Hnm ‖^2 = ‖ H˜mn ‖^2 = 2 π(n + m)! (2n + 1)(n − m)!
We will assume Hnm and H˜nm are normalize by di- viding by the norms. Then the spherical harmonics can span any function f ∈ L^2 (S^2 ) so that
f (θ, ψ) =
∑^ ∞
n=
∑^ n
m=
cm,n
[
Hmn + ˜cm,n H˜nm
]
where cm,n = 〈f, Hnm 〉, ˜cm,n = 〈f, H˜nm 〉. For simplicity let us order spherical harmonics Hnm and H^ ˜nm by Hj. Then it can be shown that the spherical heat kernel can be extended as
Kt(p, q) =
∑^ ∞
j=
e−λj^ tHj (p)Hj (q).
Hence solution to diffusion smoothing on the unit sphere is exactly expressed as
f (t, q) =
∑^ ∞
j=
e−λj^ tHj (q)
∫ (^2) π
0
∫ (^) π
0
Hj (p)Y (p) sin θ dθdψ.
Problem 32. Simplify the above expression. Please make sure your simplified expression and the above derivation is correct by directly showing it satisfies the heat equation. Problem 33. Following above derivation, solve heat equation on a unit circle S^1. Show every detail.
