Solving Diffusion Equations: F'(t) / [1 - F(t)] = p + q F(t)

[Enuma_Elish]

The diff. eq. F'(t) / [1 - F(t)] = p + q F(t) or F'(t) = [p + q F(t)][1 - F(t)] has the solution F(t) = [1 - exp(-bt)]/[1 + a exp(-bt)] where a = q / p and b = p + q.

How can I write the system of equations with the solution {F1(t), F2(t)} = { [1 - exp(-b*t)]/[1 + a exp(-b*t)] , [1 - exp(-b*(t-T))]/[1 + a2 exp(-b*(t-T))] } in a similar way, with F1' and F2' on the left-hand side, and p and q (or equivalently a and b) on the right-hand side?

I tried defining L1(t) = F1(t)[1 - F2(t - T)], L2(t) = F2(t - T)[1 + F1(t)] and differentiating, but that did not seem to get me to L1' / (1 - L1) = p + q*L1, L2' / (1 - L2) = p + q*L2.

Thanks for any thoughts and ideas.

 

[coomast]

Hello Enuma_Elish,

I looked into your equation and came up with the following conclusions. The solution to the DE is given by you without any boundary condition. Now solving the DE gives as solution the following with K as integration constant:

$$F(t)=\frac{K-e^{-(p+q)t}}{K+\frac{p}{q}e^{-(p+q)t}}$$

Assuming now the following boundary condition, $F(0)=0$ holds, it is easily written as your solution. Indeed, we have:

$$F(0)=0=\frac{K-1}{K+\frac{p}{q}}$$

From which, $K=1$ and $\frac{p}{q} \neq -1$. This gives thus:

$$F(t)=\frac{1-e^{-(p+q)t}}{1+\frac{p}{q}e^{-(p+q)t}}$$

However notice the assumption on the ratio of p and q, if this is not fullfilled you will have to integrate the DE again and you will obtain a different solution. Try it.
The second part is to write it as a system of equations, but I don't think it is necessary, you can obtain the other solution by setting the boundary condition to $F(T)=0$. This gives then $K=e^{-(p+q)T}$, from which:

$$F(t)=\frac{1-e^{-(p+q)(t-T)}}{1+\frac{p}{q}e^{-(p+q)(t-T)}}$$

I don't see any reason for setting a system of DEs up for two different particular solutions of the same DE. Hope this helps.

coomast