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\Jlpa1t is a solution of the diffusion equation and satisfies the boundary conditions for an unbounded domain.

The functionpx, t) defined in Eq. (2.53) is called the Gaussian or normal distribution with mean vt and variance o2t. If you did the exercise, you know that it satisfies the boundary conditions and the differential equation. But how do we know that it satisfies the normalization condition Eq. (2.52)? The following exercise helps with that.

First show that setting u — (x — v^/Vo"2^ means that the normalization condition is equivalent to showing that J1 1/\/2re exp(—u2/2)du — 1 Second, to show that this is true, consider the double integral HH exp(—u2/2) exp(—w2/2)dudw, switch to polar coordinates in which r — Vu2 + w2 and evaluate the resulting integral to show that the integral is 2p.

But we are not done with the solution given by Eq. (2.53). In Figure 2.18, I have plotted four Gaussian distributions with mean 0 (i.e. v — 0 so that the original walk is unbiased in either direction) and — 0.1, 0.5, 1, or 3. As the variance decreases, the curves become more peaked and centered around the origin. Now the area under each of this curves is 1 (because of the normalization constant). Let us think about the limit ofpx, t), for the case in which v — 0, as t approaches 0. The function we are considering is thus (1/V2po2t) exp(—x2/2o2t). Now, if x — 0, as t approaches 0, the reciprocal of the square root goes to infinity, but the exponential function goes to 0 and since exponentials a Figure 2.18. Gaussian distributions with mean 0 and o2t — 0.1,0.5, 1, or 3.

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