1.3. Heat equation 27

In fact, the φj can be chosen to be orthonormal,

φj,φk :=

x∈A

φj(x) φk(x) = δ(k − j).

♦

Here we have introduced the delta function notation, δ(z) = 1 if z = 0

and δ(z) = 0 if z = 0.

Since pn(x) → 0 as n → ∞, we know that the eigenvalues have

absolute value strictly less than one. We can order the eigenvalues

1 λ1 ≥ λ2 ≥ · · · ≥ λN −1.

We will write p(x, y; A) to be the solution of the heat equation with

initial condition equal to one at x and 0 otherwise. In other words,

pn(x, y; A) = P{Sn = y, TA n | S0 = x}, x, y ∈ A.

Then if #(A) = N,

pn(x, y; A) =

N

j=1

cj(x) λj

n

φj(y),

where cj(x) have been chosen so that

N

j=1

cj(x)φj(y) = δ(y − x).

In fact, this tells us that cj(x) = φj(x). Hence,

pn(x, y; A) =

N

j=1

λj

n

φj(x) φj(y).

Note that the quantity on the right is symmetric in x, y. One can

check that the symmetry also follows from the definition of pn(x, y; A).

The largest eigenvalue λ1 is often denoted λA. We can give a

“variational” definition of λA as follows. This is really just a theorem

about the largest eigenvalue of symmetric matrices.

Theorem 1.8. If A is a finite subset of Zd, then λA is given by

λA = sup

f

Qf, f

f, f

,