1.3. Heat equation 23
If x A and the initial condition is f(x) = 1 and f(z) = 0 for z = x,
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
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) =
If pn is of this form, then
∂npn(x) =
φ(x) = 1)
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