The Heat Equation, Semigroups, and Brownian Motion

Abstract

We first want to reinterpret some of our results about the heat equation. For that purpose, we again consider the heat kernel of ${\mathbb{R}}^{d}$, which we now denote by p(x, y, t),

$$p(x,y,t) = \frac{1} {{(4\pi t)}^{\frac{d} {2} }}{ \mathrm{e}}^{-\frac{{\left \vert x-y\right \vert }^{2}} {4t} }.$$

(8.1.1)

For a continuous and bounded function $f : {\mathbb{R}}^{d} \rightarrow \mathbb{R}$, by Lemma 5.2.1,

$$u(x,t) =\int\limits_{{\mathbb{R}}^{d}}p(x,y,t)f(y)\mathrm{d}y$$

(8.1.2)

then solves the heat equation

$$\Delta u(x,t) - {u}_{t}(x,t) = 0.$$

(8.1.3)

For t > 0, and letting C _b ⁰ denote the class of bounded continuous functions, we define the operator

$${P}_{t} : {C}_{b}^{0}({\mathbb{R}}^{d}) \rightarrow {C}_{ b}^{0}({\mathbb{R}}^{d})$$

via

$$({P}_{t}f)(x) = u(x,t),$$

(8.1.4)

with u from (8.1.2). By Lemma 5.2.2

$${P}_{0}f :{=\lim }_{t\rightarrow 0}{P}_{t}f = f;$$

(8.1.5)

i.e., P ₀ is the identity operator. The crucial point is that we have for any ${t}_{1},{t}_{2} \geq 0$,

$${P}_{{t}_{1}+{t}_{2}} = {P}_{{t}_{2}} \circ {P}_{{t}_{1}}.$$

(8.1.6)

Written out, this means that for all $f \in {C}_{b}^{0}({\mathbb{R}}^{d})$,

$$\begin{array}{rcl} & & \int\limits_{{\mathbb{R}}^{d}} \frac{1} {{\left (4\pi \left ({t}_{1} + {t}_{2}\right )\right )}^{\frac{d} {2} }}{ \mathrm{e}}^{- \frac{{\left \vert x-y\right \vert }^{2}} {4({t}_{1}+{t}_{2})} }f(y)\,\mathrm{d}y \\ & & \quad =\int\limits_{{\mathbb{R}}^{d}} \frac{1} {{(4\pi {t}_{2})}^{\frac{d} {2} }}{ \mathrm{e}}^{-\frac{{\left \vert x-z\right \vert }^{2}} {4{t}_{2}} }\int\limits_{{\mathbb{R}}^{d}} \frac{1} {{(4\pi {t}_{1})}^{\frac{d} {2} }}{ \mathrm{e}}^{-\frac{{\left \vert z-y\right \vert }^{2}} {4{t}_{1}} }f(y)\,\mathrm{d}y\,\mathrm{d}z.\end{array}$$

(8.1.7)

This follows from the formula

$$\frac{1} {{\left (4\pi \left ({t}_{1} + {t}_{2}\right )\right )}^{\frac{d} {2} }}{ \mathrm{e}}^{- \frac{{\left \vert x-y\right \vert }^{2}} {4({t}_{1}+{t}_{2})} } = \frac{1} {{(4\pi {t}_{2})}^{\frac{d} {2} }} \frac{1} {{(4\pi {t}_{1})}^{\frac{d} {2} }} \int\limits_{{\mathbb{R}}^{d}}{\mathrm{e}}^{-\frac{{\left \vert x-z\right \vert }^{2}} {4{t}_{2}} }{\mathrm{e}}^{-\frac{{\left \vert z-y\right \vert }^{2}} {4{t}_{1}} }\,\mathrm{d}z,$$

(8.1.8)

which can be verified by direct computation (cf. also Exercise 5.3).