Derive the solution of the diffusion equation from the solution of a random walk

Question

Summary

The probability distribution (pdf) of a random walk in 1 dimension is represented by a Bessel function. On the other hand, the pdf of a Brownian motion in free space is represented by a Gaussian distribution. While it is possible to derive the diffusion equation from the master equation of a random walk by taking the limit of small jumps, is it possible to directly derive the Gaussian distribution as some asymptotic form of the Bessel function? In other words, can I derive the pdf of a Brownian motion directly from the pdf of a random walk, rather than from the equation of the random walk?

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Consider a continuous time random walk on a neareast-neighbour 1d lattice, where each jump is of size $d$. The probability of being at site $n$, i.e. being at a distance $nd$ of the origin is described by: $$\dot{P}_{n}(t)=\frac{1}{2}\left(P_{n-1}(t)+P_{n+1}(t)\right)-P_{n}(t)$$

The solution is $P_n(t)=I_n(t)e^{-t}$, where $I_n(t)$ is the Bessel function.

When we extend the random walk on the real line, i.e $d\to 0$, the probability is described by the diffusion equation:

$$\frac{\partial p(x,t)}{\partial t}=D\frac{\partial^2p(x,t)}{\partial x^2}$$

The solution is $p(x,t)=\frac{1}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}$.

Is it possible to derive the expression of $p(x,t)$ from our expression of $P_n(t)=I_n(t)e^{-t}$, instead of starting from the master equation? I.e. how does the Bessel function relate to the Gaussian distribution in the limit $d\to0$?

If the question is not clear, or you would like more details, please let me know and I will edit the question.

Answer 1

To carry out the limit, it helps to start from an integral representation of the Bessel function, $$P_n(T)=e^{-T}I_n(T)=\frac{1}{2\pi}\int_{-\pi}^\pi \exp [i k n+T \cos k-T]\,dk.$$ For $T\gg 1$ this may be approximated by expansion of the exponent to second order in $k$, $$P_n(T)\approx\frac{1}{2\pi}\int_{-\infty}^\infty \exp[ikn-\tfrac{1}{2}Tk^2]\,dk=(2\pi T)^{-1/2}e^{-n^2/2T}.$$

Define $x=nd$, $t=\tau T$, $D=\tfrac{1}{2}d^2/\tau$ to arrive at $$d^{-1}P_n(T)\approx \frac{1}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}\equiv p(x,t). $$

Here is a plot for $d=0.1$, $D=1$, $t=1$; blue is $p(x,t)$, red is $d^{-1}P_n(T)$, the two curves are indistinguishable.