Stochastic Processes and Financial Mathematics
(part two)

11.3 Brownian motion and the heat equation \(\offsyl \)

Brownian motion lies at the heart of many modern models of the physical world. Before we study Brownian motion in its own right, let us give one example of a model with close connection to Brownian motion, namely heat diffusion. Note that this section is marked with a \(\offsyl \), meaning that it is off syllabus.

Consider a long thin metal rod. If, initially, some parts of the rod are hot and some are cold, then as time passes heat will diffuse through the rod: the differences in temperature slowly average out. Suppose that the temperature of the metal in the rod at position \(x\) at time \(t\) is given by \(u(t,x)\), where \(t\in [0,\infty )\) represents time and \(x\in \R \) represents space. Suppose that, at time \(0\), the temperature at the point \(x\) is \(f(x)=u(0,x)\).

Then (as you may have seen from e.g. MAS222), it is well known that the heat equation

\begin{equation} \label {eq:heat} \frac {\p u}{\p t}=\frac 12\frac {\p ^2 u}{\p x^2} \end{equation}

with the initial condition

\begin{equation} \label {eq:heat_init} u(0,x)=f(x) \end{equation}

describes how the temperature \(u(t,x)\) changes with time, from an initial temperature of \(f(x)\) at site \(x\). This equation has a close connection to Brownian motion, which we now explore.

If we start Brownian motion from \(x\in \R \), then \(B_t\sim x+N(0,t)\sim N(x,t)\), so we can write down its probability density function

\[\phi _{t,x}(y)=\frac {1}{\sqrt {2\pi t}}\exp \l (-\frac {(y-x)^2}{2t}\r ).\]

Proof: Note that if any function \(u(t,x)\) satisfies the heat equation, so does \(u(t,x-y)\) for any value of \(y\). So, we can assume \(y=0\) and need to show that

\[\phi _{t,x}(0)=\frac {1}{\sqrt {2\pi t}}\exp \l (-\frac {x^2}{2t}\r )\]

satisfies (11.3). This is an exercise in partial differentiation. Using the chain and product rules:

\begin{align*} \frac {\p \phi }{\p t}&=\frac {\frac {-1}{2}}{\sqrt {2\pi t^3}}\exp \l (-\frac {x^2}{2t}\r )+\frac {1}{\sqrt {2\pi t}}\frac {-x^2(-1)}{2t^2}\exp \l (-\frac {x^2}{2t}\r )\\ &=\frac {1}{\sqrt {2\pi }}\l (\frac {x^2}{2t^{5/2}}-\frac {1}{2t^{3/2}}\r )\exp \l (-\frac {x^2}{2t}\r ) \end{align*} and

\begin{align*} \frac {\p \phi }{\p x} &=\frac {1}{\sqrt {2\pi t}}\frac {-2x}{2t}\exp \l (-\frac {x^2}{2t}\r )\\ &=\frac {-x}{\sqrt {2\pi t^3}}\exp \l (-\frac {x^2}{2t}\r ) \end{align*} so that

\begin{align*} \frac {\p ^2\phi }{\p x^2} &=\frac {-1}{\sqrt {2\pi t^3}}\exp \l (-\frac {x^2}{2t}\r )+\frac {-x}{\sqrt {2\pi t^3}}\frac {-2x}{2t}\exp \l (-\frac {x^2}{2t}\r )\\ &=\frac {1}{\sqrt {2\pi }}\l (\frac {x^2}{t^{5/2}}-\frac {1}{t^{3/2}}\r )\exp \l (-\frac {x^2}{2t}\r ). \end{align*} Hence, \(\frac {\p \phi }{\p t}=\frac 12\frac {\p ^2\phi }{\p x^2}\).   ∎

We can use Lemma 11.3.2 to give a physical explanation of the connection between Brownian motion and heat diffusion. We define

\begin{align} \label {eq:heat_duality} w(t,x)&=\E _x[f(B_t)] \end{align} That is, to get \(w(t,x)\), we start a particle at location \(x\), let it perform Brownian motion for time \(t\), and then take the expected value of \(f(B_t)\).

Before we give the proof, let us discuss the physical interpretation of this result. Within our metal rod, the metal atoms have fixed positions. But atoms that are next to each other transfer heat between each other, in random directions. If we could pick on an individual ‘piece’ of heat and watch it move, it would move like a Brownian motion. Since there are lots of little pieces of heat moving around, and they are very small, when we measure temperature we only see the average effect of all the little pieces, corresponding to \(\E [\ldots ]\).

We should think of the Brownian motion in (11.5) as running in reverse time, so as it tracks (backwards in time) the path through space that a typical piece of heat has followed. Then, after running for time \(t\), it looks at the initial condition to find out how much heat there was initially that its eventual location.

Proof: We have \(B_0=x\), so

\[w(0,x)=\E _x[f(B_0)]=\E _x[f(x)]=f(x).\]

Hence the initial condition (11.4) is satisfied. We still need to check (11.3). To do so we will allow ourselves to swap \(\int \)s and partial derivatives². We have

\begin{align*} w(t,x)&=\E _x[f(B_t)]\\ &=\int _{-\infty }^\infty f(y)\phi _{t,x}(y)\,dy \end{align*} so, by Lemma 11.3.2,

\begin{align*} \frac {\p w}{\p t} &=\frac {\p }{\p t}\int _{-\infty }^\infty f(y)\phi _{t,x}(y)\,dy\\ &=\int _{-\infty }^\infty f(y)\frac {\p }{\p t}\phi _{t,x}(y)\,dy\\ &=\int _{-\infty }^\infty f(y)\frac 12\frac {\p ^2}{\p x^2}\phi _{t,x}(y)\,dy\\ &=\frac 12\frac {\p ^2}{\p x^2}\int _{-\infty }^\infty f(y)\phi _{t,x}(y)\,dy\\ &=\frac 12\frac {\p ^2 w}{\p x^2}\\ \end{align*} as required.   ∎

There are many similar ways to connect various stochastic processes to PDEs. This kind of connection can be very useful, because it allows us to transfer knowledge about stochastic process to (and from) knowledge about PDEs. We’ll see a more sophisticated example in Section 14.1.