Master equation -> diffusion equation

[Mikhail_MR]

Homework Statement

I am trying to understand the derivation of the diffusion equation from the Master equation for a 1D chain. We have an endless 1D discrete chain. State from $n$ can jump to $n+1$ and $n-1$ with equal probabilities. The distance between chain links is $a$.

Homework Equations

Master equation: $$\frac{d\rho_n}{dt} = W\left(-2\rho_n + \rho_{n+1} + \rho_{n-1}\right)$$

The Attempt at a Solution

With $\Delta n = 1$ we have $$\frac{d\rho_n}{dt}=Wa^2 \frac{(\rho_{n-1}-\rho_n) + (\rho_{n+1}-\rho_n)}{(\Delta n)^2 a^2},$$ where we could write $(\Delta n)^2 a^2 = (\Delta x)^2$, since $na=x$. In limit $\Delta x \rightarrow 0$ we get $$\frac{d\rho_n}{dt} = Wa^2 \frac{\partial^2 \rho_n}{\partial x^2}$$.
Unfortunately, I can not understand the last step. The definition of a derivative is $f(x_0+\Delta x) - f(x_0) /\Delta x$, but I do not understand how it is used here.

Any help would be appreciated.

 

[mjc123]

Using your definition (with brackets inserted correctly), what is the expression for dρ/dx at x₀ = n-1? What is the expression at x₀ = n? So what is d(dρ/dx)/dx ?

 

[Mikhail_MR]

$d\rho/dx$ would be $\frac{\rho_n - \rho_{n-1}}{\Delta x}$ at $x_0=n-1$ and $\frac{\rho_{n+1} - \rho_{n}}{\Delta x}$ at $x_0=n$.
Then: $$d(d\rho/dx)dx= [(d\rho/dx)_{x_0=n} - (d\rho/dx)_{x_0=n-1}] / \Delta x$$. Oh, now I see it. Thank you!

 

[Chestermiller]

A better way of doing this would be in terms of central differences: $d\rho/dx$ would be $\frac{\rho_n - \rho_{n-1}}{\Delta x}$ at $x_0=n-1/2$ and $\frac{\rho_{n+1} - \rho_{n}}{\Delta x}$ at $x_0=n+1/2$