24 1. Random Walk and Discrete Heat Equation

This nice form leads us to try to find eigenvalues and eigenfunctions

of Q, i.e., to find λ, φ such that

(1.16) Qφ(x) = λ φ(x),

with φ ≡ 0 on ∂A.

♦

The “algorithmic” way to find the eigenvalues and eigenvectors for

a matrix Q is first to find the eigenvalues as the roots of the characteristic

polynomial and then to find the corresponding eigenvector for each eigenvalue.

Sometimes we can avoid this if we can make good guesses for the eigenvectors.

This is what we will do here.

The sum rule for sine,

sin((x ± 1)θ) = sin(xθ) cos(θ) ± cos(xθ) sin(θ),

tells us that

Q{sin(θx)} = λθ {sin(θx)}, λθ = cos θ,

where {sin(θx)} denotes the vector whose component associated to

x ∈ A is sin(θx). If we choose θj = πj/N, then φj(x) = sin(πjx/N),

which satisfies the boundary condition φj(0) = φj(N) = 0. Since

these are eigenvectors with different eigenvalues for a symmetric ma-

trix Q, we know that they are orthogonal, and hence linearly inde-

pendent. Hence every function f on A can be written in a unique

way as

(1.17) f(x) =

N−1

j=1

cj sin

πjx

N

.

This sum in terms of trigonometric functions is called a finite Fourier

series. The solution to the heat equation with initial condition f is

pn(y) =

N−1

j=1

cj cos

jπ

N

n

φj(y).

Orthogonality of eigenvectors tells us that

N−1

x=1

sin

πjx

N

sin

πkx

N

= 0 if j = k.