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Master equation -> diffusion equation

  1. Jun 7, 2017 #1
    1. The problem statement, all variables and given/known data
    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##.

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

    3. 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.
     
  2. jcsd
  3. Jun 7, 2017 #2
    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 ?
     
  4. Jun 7, 2017 #3
    ##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!
     
  5. Jun 8, 2017 #4
    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##
     
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