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Now we let r ! 0, D ! 0 and assume that as this happens (m1D) / ->v, (m2D2)/r ! a2, o(D2)/r! 0, which is the definition of the quantities v and a2. When we do this, the resulting equation is

With this equation, we have much to talk about (but little to reminisce, as we will by Chapters 7 and 8). First, let us consider these new parameters. Since D has units of distance, r has units of time, and both m1 and m2 are pure numbers (averages of the jump size), we see that v has units of distance/time - it is a velocity. On the other hand, a has units of (length)2/time; these units make it a diffusion coefficient. Hence, Eq. (2.51) is called a diffusion equation.

We need to discuss one subtlety of this limit. In particular, how is P(x, t) interpreted now that a set of discrete sites has become a continuum (as D ! 0)? Recall, that we earlier agreed to think of px, t) as the probability that the walker was within ±D/2 of the spatial point x (site i) at time t. In the limit, we interpretpx, t)dx as the probability that the walker is within dx of the spatial point x at time t. Since the walker must be some place, we obtain a normalization condition px, t)dx = 1 (2.52)

Equation (2.51) also involves one time derivative and two spatial derivatives. This means that there are three conditions needed to completely specify the solution. One is an initial condition: we specify px, 0) - the chance of initially finding the walker at spatial point x. The choices about boundary conditions depend upon the nature of the spatial region.

First, suppose that this region is unbounded. It is reasonable to expect, then, that the chance that the walker can reach ±1 in any finite time is 0. We then specify thatpx, t)! 0 as |x| !±i. Sometimes we can get a solution of this equation virtually for free, as in the following exercise.

Exercise 2.13 (M)

Show that

0 0