$ \frac{\partial}{ \partial t} = \frac 1 2 \Delta \implies $ Brownian Motion

Question

Apparently, the diff equation : $ \frac{\partial}{ \partial t} = \frac 1 2 \Delta \implies $ Brownian Motion. I.E. all solutions of the equations are brownian motions.

How ?

Answer 1

$\newcommand{\D}{\mathit{\Delta}}$ If you discretize the equation to $$ u(x,t+\D t)-u(x,t)=\frac{\D t}{2\D x^2}[u(x+\D x,t)-2u(x,t)+u(x-\D x,t)] $$ and then choose the step sizes so that $\D t=\D x^2$, the equation simplifies to $$ u(x,t+\D t)=\frac12[u(x+\D x,t)+u(x-\D x,t)] $$ and then \begin{align} u(x,t+2\D t)&=\frac14[u(x+2\D x,t)+2u(x,t)+u(x-2\D x,t)] \\ u(x,t+3\D t)&=\frac14[u(x+3\D x,t)+3u(x+\D x,t)+3u(x-\D x,t)+u(x-3\D x,t)] \end{align} etc. down the Pascal's triangle. This can be interpreted as summing over all random walks from $t$ to $t+N\D t$ moving up or down $\D x=\sqrt{\D t}$ with equal probability $\frac12$ in every time step. The limit of these random walks is the Brownian motion.