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Master equation for reproduction and mutual annihilation process

Physics Asked on April 30, 2021

I was solving some exercises regarding the Master Equation and couldn’t solve the following problem.

Consider a population with individuals $A$. This population can suffer the following processes:

i) Reproduction: $A xrightarrow{text{$sigma$}} A + A$

ii) Mutual annihilation: $A + A xrightarrow{text{$lambda$}} 0 $

where $sigma$ and $lambda$ are rates of evolution.

My attempt

I started by writing the transition rates $omega_{n,n’}$ such as described in Reichl – A modern course in statistical physics, page 260. Hence, for process i) I got

$$omega_{n,n+1} Delta t = sigma n Delta t = sigma_n Delta t$$

while for the second

$$omega_{n,n-2} Delta t = frac{1}{2} n(n-1) lambda Delta t = lambda_n Delta t $$

which yielded the master equation

$$partial_{t}P(n,t) = -(lambda_n+sigma_n)P(n,t) + sigma_{n-1}P(n-1,t) + lambda_{n+2}P(n+2,t) $$

I was wondering: is this correct? I am unable to check this result.

One Answer

After searching in several books, I found out that the expression is, indeed, correct. However, some authors prefer to absorve the $frac{1}{2}$ in the definition of $lambda$.

For future reference, after writing the Generating Function and simplifying, one should obtain

$$partial_t langle n(t) rangle = langle n(t) rangle (sigma + lambda) - lambda langle n^2(t) rangle$$

However, if absorving the factor as said above,

$$partial_t langle n(t) rangle = langle n(t) rangle (sigma + 2lambda) - 2 lambda langle n^2(t) rangle$$

This is practically the result we get by a qualitative analysis, in first order!

Correct answer by miniplanck on April 30, 2021

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