1.3. Heat equation 23

If x ∈ A and the initial condition is f(x) = 1 and f(z) = 0 for z = x,

then

pn(y) = P{Sn∧TA = y | S0 = x}.

♦

The heat equation is a deterministic (i.e., without randomness) model

for heat flow. It can be studied without probability. However, probability adds

a layer of richness in terms of movements of individual random particles. This

extra view is often useful for understanding the equation.

Given any initial condition f, it is easy to see that there is

a unique function pn satisfying (1.12)–(1.14). Indeed, we just set:

pn(y) = 0 for all n ≥ 0 if y ∈ ∂A; p0(x) = f(x) if x ∈ A; and for

n 0, we define pn(x),x ∈ A recursively by (1.12). This tells us

that set of functions satisfying (1.12) and (1.14) is a vector space of

dimension #(A). In fact, {pn(x) : x ∈ A} is the vector

Qnf.

Once we have existence and uniqueness, the problem remains to

find the function. For a bounded set A, this is a problem in lin-

ear algebra and essentially becomes the question of diagonalizing the

matrix Q.

♦

Recall from linear algebra that if A is a k × k symmetric matrix with

real entries, then we can find k (not necessarily distinct) real eigenvalues

λk ≤ λk−1 ≤ · · · ≤ λ1,

and k orthogonal vectors v1, . . . , vk that are eigenvectors,

Avj = λj vj .

(If A is not symmetric, A might not have k linearly independent eigenvectors,

some eigenvalues might not be real, and eigenvectors for different eigenvalues

are not necessarily orthogonal.)

We will start by considering the case d = 1. Let us compute the

function pn for A = {1,...,N − 1}. We start by looking for functions

satisfying (1.12) of the form

(1.15) pn(x) =

λn

φ(x).

If pn is of this form, then

∂npn(x) =

λn+1

φ(x) −

λn

φ(x) = (λ − 1)

λn

φ(x).