Heat Equation and Brownian Motion — Mathematics Knowledge System

Overview

Brownian motion and the heat equation are deeply connected: the transition density of Brownian motion satisfies the heat equation, and solutions to the heat equation have probabilistic representations via Brownian motion (the Feynman-Kac formula). This connection unifies PDE theory and probability, providing both analytical tools for studying diffusions and probabilistic methods for solving PDEs. The Dirichlet problem, the connection to harmonic functions, and the Feynman-Kac formula are central results.

Intuition

Heat diffuses through a material just as the probability distribution of a Brownian particle spreads over time. Starting from a point heat source, the temperature profile is a Gaussian -- exactly the density of Brownian motion. This is not a coincidence: the operator (1/2)d^2/dx^2 governing heat diffusion is the generator of Brownian motion. Conversely, to solve a PDE, you can run Brownian motion and average the boundary values it hits.

Formal Definition

Definition

The transition density p(t, x, y) = P(B_t = y | B_0 = x) of standard Brownian motion is the heat kernel. It satisfies the heat equation in both variables. For the Dirichlet problem, the Feynman-Kac formula gives a probabilistic solution.

p(t,x,y) = \frac{1}{\sqrt{2\pi t}}\exp\!\left(-\frac{(y-x)^2}{2t}\right)

Heat kernel (Brownian transition density)

\frac{\partial p}{\partial t} = \frac{1}{2}\frac{\partial^2 p}{\partial x^2} \quad \text{(heat equation)}

Heat equation for transition density

u(x) = \mathbb{E}^x[g(B_\tau)] \text{ solves } \frac{1}{2}u'' = 0 \text{ on } D, \; u|_{\partial D} = g

Harmonic functions / Dirichlet problem

u(t,x) = \mathbb{E}^x\!\left[\exp\!\left(\int_0^t V(B_s)\,ds\right)g(B_t)\right] \text{ solves } \frac{\partial u}{\partial t} = \frac{1}{2}u_{xx} + Vu

Feynman-Kac formula

Notation

Notation	Meaning
$p(t,x,y)$	Heat kernel: probability density of B_t = y given B_0 = x
$\mathbb{E}^x$	Expectation under Brownian motion started at x
$\tau$	First exit time from domain D: tau = inf{t > 0 : B_t not in D}

Theorems

Theorem 1: Theorem 1 (Brownian Motion and the Heat Equation)

\text{The function } u(t,x) = \mathbb{E}^x[f(B_t)] = \int_{-\infty}^\infty f(y) p(t,x,y)\,dy \text{ satisfies the heat equation } \partial_t u = \frac{1}{2}\partial_{xx} u \text{ with } u(0,x) = f(x).

Theorem 2: Theorem 2 (Feynman-Kac Formula)

\text{Let } V : \mathbb{R} \to \mathbb{R} \text{ be bounded. The unique bounded solution to } \partial_t u = \frac{1}{2}\partial_{xx} u + V u, \; u(0,x) = g(x), \text{ is } u(t,x) = \mathbb{E}^x\!\left[g(B_t)\exp\!\left(\int_0^t V(B_s)\,ds\right)\right].

Worked Examples

1
E^x[B_t^2] = E[(x + B_t - B_0)^2] where B starts at x. Since B_t - B_0 ~ N(0,t), E^x[B_t^2] = x^2 + t.
$u(t,x) = \mathbb{E}^x[B_t^2] = x^2 + t$
2
Compute partial derivatives: u_t = 1, u_x = 2x, u_{xx} = 2.
$u_t = 1, \quad \frac{1}{2}u_{xx} = \frac{1}{2} \cdot 2 = 1$
3
Check: u_t = 1 = (1/2) u_{xx}. The heat equation is satisfied.
$u_t = \frac{1}{2}u_{xx} \checkmark$

✓ Answer

u(t,x) = x^2 + t satisfies the heat equation with initial condition u(0,x) = x^2.

Practice Problems

Mediumfree response

The heat kernel p(t, x, y) satisfies the heat equation in x. Verify directly that partial_t p = (1/2) partial_{xx} p.

Hardfree response

State the Feynman-Kac formula and explain how it connects PDEs and Brownian motion.

Common Mistakes

Common Mistake

Thinking the heat equation generator is d^2/dx^2 (without the 1/2)

The generator of standard Brownian motion is (1/2) d^2/dx^2. If you use dB_t = sqrt(2) dW_t, the heat equation becomes partial_t u = partial_{xx} u.

Common Mistake

Confusing forward and backward Kolmogorov equations

The forward (Fokker-Planck) equation governs the density p(t,x,y) as a function of y. The backward equation governs it as a function of x. Both are heat equations but in different variables.

Quiz

Summary

The Brownian transition density p(t,x,y) = (2pi t)^{-1/2} exp(-(y-x)^2/(2t)) satisfies the heat equation.
The generator of Brownian motion is (1/2) d^2/dx^2 -- the Laplacian scaled by 1/2.
Dirichlet problem: u(x) = E^x[g(B_tau)] is the unique harmonic function with boundary data g.
Feynman-Kac: u(t,x) = E^x[g(B_t) exp(int V(B_s) ds)] solves partial_t u = (1/2) u_{xx} + Vu.

References

BookDurrett, R. -- Brownian Motion and Martingales in Analysis
BookKaratzas, I. and Shreve, S. -- Brownian Motion and Stochastic Calculus, Chapter 4
WebsiteWikipedia -- Feynman-Kac formula