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\begin{document}

\title{Stat 992: Lecture 21\\
Diffusion smoothing on sphere.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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{\bf Problem 16.} What is the heat kernel on a unit sphere ? {\em
Solution.} Read S. Rosenberg's The Laplacian on a Riemannian
Manifold (1997 Cambridge University Press), I Chavel's Eigenvalues
in Riemannian geometry (Academic Press, 1984), E.B. Davies' Heat
kernels and spectral theory (Cambridge University Press, 1989). P.
Olver and C. Shakiban's Fundamentals of Applied Mathematics,
Prentice-Hall, in preparation.


\begin{enumerate}
\item {\em 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
spherical coordinate system: $$x=\sin \theta \cos \psi, \; y=\sin
\theta \sin \psi, \; z = \cos \theta.$$ On a unit sphere
$p=(x,y,z) \in S^2$, heat kernel $K_t$ must satisfy isotropic heat
equation
$$\frac{\partial K_t}{\partial t} = \Delta_{S^2} K_t.$$
where $$\Delta_{S^2}= \frac{1}{\sin \theta}\frac{\partial
}{\partial \theta}\Big(\sin \theta \frac{\partial}{\partial
 \theta}\Big) +\frac{1}{\sin^2 \theta} \frac{\partial^2 }
 {\partial^2 \psi}.$$
It should satisfy the following three conditions
$K_t(p,q)=K_t(q,p)$, $\lim_{t \to 0} K_t(p,q) = \delta(p-q)$ and
$K_t(p,q)=\int_M K_s(p,r)K_{t-s}(r,q)\; dr$. Then the following
PDE $$\frac{\partial f}{\partial t} = \Delta_{S^2} f$$ with
initial condition $f(0,p) = Y(p)$ has a unique solution
$$f(t,p) = \int_{S^2} K_t(p,q)Y(p) \;d\mu(p).$$
In lecture 20, we showed how to solve it numerically.
 \item
{\em Spherical Helmholtz equation.} We will follow lecture 11. In
the previous lecture, we showed the metric to be $g_{11}=1,
g_{12}=g_{21}=0, g_{22}=\sin^2 \theta.$ Define $L^2(S^2,g)$ with
respect to the Riemannian metric $g$ with inner product
$$\langle f,g \rangle = \int_{S^2} f(p)g(p) \;d\mu(p).$$
where the Lebesgue measure on a sphere is given by
$$d\mu = \sqrt{\det g} \; d\theta d\psi = \sin \theta d\theta d\psi.$$
This is the area element you must have seen it in vector calculus
course. Note that $\int_{S^2} \; d\mu = 4 \pi$, the area of the
unit sphere.

The gradient $\nabla: C^{\infty}(S^2) \to T_p(S^2)$ in local
coordinates is given by \bqn \nabla f &=& \sum_{i,j}g^{ij}
\frac{\partial
f}{\partial u^j} \frac{\partial p}{\partial u^i}\\
&=& \frac{\partial f}{\partial \theta} \frac{\partial p}{\partial
\theta} + \frac{1}{\sin^2 \theta}\frac{\partial f}{\partial \psi}
\frac{\partial p}{\partial \psi}.\eqn Then it can be shown that
$$\langle \Delta_{S^2}f,g\rangle = - \langle \nabla f, \nabla g
\rangle = \langle f, \Delta_{S^2} g \rangle$$ So the spherical
Laplacian is compact self-adjoint operator and it should have
eigenvalues $\lambda_j$ and eigenfunctions $H_j$ such that
$$\Delta_{S^2} H_j(p) = \lambda_j H_j(p)$$
and $0=|\lambda_0| \leq |\lambda_1| \leq |\lambda_2| \cdots.$ To
solve for eigenvalues and eigenfunctions of Laplacian, we need to
solve the {\em spherical Helmholtz equation} given by
$$\frac{1}{\sin \theta}\frac{\partial }{\partial \theta}\Big(\sin
\theta \frac{\partial H}{\partial
 \theta}\Big) +\frac{1}{\sin^2 \theta} \frac{\partial^2 H}
 {\partial^2 \psi} - \lambda H =0$$

Let us use the separation of variable technique: $H(\theta,\psi) =
\alpha(\theta)\beta(\psi)$. Then substituting the term, we get
$$ \sin^2 \theta \frac{\alpha''}{\alpha} + \cos \theta \sin \theta
\frac{\alpha'}{\alpha} -\sin^2 \theta \lambda =
-\frac{\beta''}{\beta}= c$$ Note that the left side is a function
of $\theta$ while the right side is a function of $\psi$. The only
way both side equal is when they are constant $c$.

The right hand side is trivial to solve:
$$ \beta'' + c \beta = 0$$
Since $\beta(\psi)$ must be a periodic function with period
$2\pi$, the solutions are
$$\beta(\psi) = \cos (m \psi),\; \sin (m \psi)$$
where $m= \sqrt{c}= 0, 1, 2,\cdots$.

The left hand side is not so obvious to solve. We will change
variable $t = \cos \theta$ and let $$\alpha(\theta) =
\alpha(cos^{-1} t) = P(t).$$ From the chain rule,
$$\alpha' = -\sqrt{1-t^2}P',$$
$$\alpha''=(1-t^2)P'' - t P'.$$
Substituting these into the left side, we get
$$(1-t^2)^2P'' - 2t(1-t^2)P' - [\lambda(1-t^2) +m^2]P=0$$
which is the {\em Legendre equation}. This is a well known
ordinary differential equation (ODE) and its solution is the {\em
Legendre functions}. Note that the solution is defined in the
domain $-1 \leq t=\cos \theta \leq 1$ and we give boundary
condition $|P(-1)| < \infty, |P(1)| < \infty$. For $m=0$,
$P_n(t)=\frac{1}{2^n n!}\frac{d^n}{dt^n}(t^2-1)^n, n
=0,1,2,\cdots$ are the Legendre polynomials. For general $m >0$,
the Legendre functions are constructed in the following way
$$P_n^m(t) = (1-t^2)^{m/2}\frac{d^m}{dt^m}P_n(t),\; n=m,m+1,\cdots$$
Even ordered Legendre functions are polynomials but for odd
ordered functions there is an additional factor $\sqrt{1-t^2}$. In
summarizing the result, the complete set of solutions for the left
hand side is
$$\alpha(\theta) = P_n^m(\cos \theta), n=m,m+1,\cdots.$$
For instance $P_0^0=1,P_1^0=\cos\theta,P_1^1=-\sin \theta,
P_2^0=\frac{1}{4}+\frac{3}{4}\cos 2\theta,\cdots.$

\item {\em Spherical harmonics.} By combining the solution from
both ODEs, we have the complete set of solution to the spherical
Helmholtz equation gives spherical harmonics:
$$H_n^m(\theta,\psi) = \cos (m\psi)P_n^m(\cos \theta),$$
$$\widetilde{H}_n^m(\theta,\psi) = \sin (m\psi) P_n^m(\cos
\theta)$$ where $n=0,1,2,\cdots, m=0,1,\cdots,n.$ The spherical
harmonics satisfy
$$\Delta_{S^2} H_n^m + n(n+1)Y_n^m = \Delta_{S^2} \widetilde{H}_n^m +
n(n+1)Y_n^m=0.$$ Note that $\widetilde{H}_n^0=0$. The
corresponding eigenvalues are $\lambda_n = -n(n+1)$. These
harmonics are orthogonal but not normalized. It can be shown that
$$\|H_n^0\|^2 = \frac{4\pi}{2n+1}, \|H_n^m\|^2 =
\|\widetilde{H}_n^m\|^2 = \frac{2\pi (n+m)!}{(2n+1)(n-m)!}.$$ We
will assume $H_n^m$ and $\widetilde{H}_n^m$ are normalize by
dividing by the norms. Then the spherical harmonics can span any
function $f \in L^2(S^2)$ so that
$$f(\theta,\psi) = \sum_{n = 0}^{\infty} \sum_{m =0}^n c_{m,n} \big[H_n^m +
\widetilde{c}_{m,n}\widetilde{H}_n^m \big]$$ where $c_{m,n} =
\langle f, H_n^m \rangle$, $\widetilde{c}_{m,n} = \langle f,
\widetilde{H}_n^m \rangle$. For simplicity let us order spherical
harmonics $H_n^m$ and $\widetilde{H}_n^m$ by $H_j$. Then it can be
shown that the spherical heat kernel can be extended as
$$K_t(p,q) = \sum_{j=0}^{\infty} e^{-\lambda_j t}H_j(p)H_j(q).$$
Hence solution to diffusion smoothing on the unit sphere is
exactly expressed as $$f(t,q) = \sum_{j=0}^{\infty}e^{-\lambda_j
t}H_j(q) \int_0^{2\pi}\int_0^{\pi}  H_j(p)Y(p) \sin\theta \;
d\theta d\psi.$$
\end{enumerate}

{\bf 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.


{\bf Problem 33.} Following above derivation, solve heat equation
on a unit circle $S^1$. Show every detail.
\end{document}