Integrate Diffusion Equation: $\frac{dG(t)}{dt} = k(\frac{a}{b+G(t)} - \frac{c}{d-G(t)})$

Question

Background

I am trying to determine the total diffusion of water across a cell membrane after some time $t$. The experiment is set up by placing potato cores in sucrose solutions of different concentrations and measuring weight change of the potato cores after a constant time. (I know this sounds like biology but this question is strictly mathematical).

My Attempt

Basically, from Fick's Second Law of Diffusion, we know that the rate of diffusion is proportional to the concentration gradient. In my case, this means:

$$J = k \Delta C$$

where $J$ is the rate of diffusion (I'll use L/s for simplicity), $k$ is a diffusion constant and $\Delta C$ is the concentration gradient ($C_{out}-C_{in}$ where $C$ is solute concentration in M).

I can rewrite the equation to be more explicit like so:

$$J = k(\frac{a}{b}-\frac{c}{d})$$

where $a$ and $c$ are moles of solute in the potato core and solution respectively and $b$ and $d$ are liters of water in each respectively. As the water diffuses, $a$ and $c$ remain constant.

Therefore, I can determine $J(t)$ as the following:

$$J(t) = k(\frac{a}{b+\int_{0}^{t}J dt} - \frac{c}{d - \int_{0}^{t}Jdt})$$

Or perhaps simpler, given $G(t) = \int_{0}^{t}Jdt$, (the total amount of water diffused by time $t$),

$$\frac{dG(t)}{dt} = k(\frac{a}{b+G(t)} - \frac{c}{d-G(t)})$$

(after time $t$, a certain amount of water $G(t)$ has crossed the membrane and increased the volume in the potato cores while decreasing the volume in the solution)

At this point I am stuck. I have not been able to figure out a general solution for this differential equation. I feel like this will end up being some exponential solution but it is beyond my capabilities to solve. Could someone with more advanced skills help me out?

Goal

I would like to obtain $G$, the total water diffused, as a function of $t$, the elapsed time, given $a$, $b$, $c$, $d$, and $k$ as constants.

Answer 1

I do not know whether your physics is right or not. Here I just talk abut the integration.

As you pointed that \begin{equation} \frac{dG(t)}{dt} = k \left( \frac{a}{b+G} -\frac{c}{d-G}\right) \end{equation}

Then you can rewrite it into \begin{equation} \frac{dG(t)}{dt} = k\left( \frac{(a-c)G+ad-bc}{(b+G)(d-G)} \right) \end{equation} This leads to \begin{equation} \int \frac{(b+G)(d-G)}{(a-c)G+ad-bc}dG = \int kdt \end{equation} This further leads to \begin{equation} -\frac{1}{a-c}\int \left(G+\frac{ad-bc}{a-c} + \frac{\frac{(a-c)(b-d)-2(ab-bc)}{a-c}G-\left(\frac{ad-bc}{a-c} \right)^2-bd}{G+\frac{ad-bc}{a-c}}\right)dG = \int kdt \end{equation}

The second integration term could be written to more general form as $\frac{k_1G+k_2}{k_3G+k_4}$. Now considering a general form integration of function $\frac{ax+b}{cx+d}$ whose integration result is as \begin{equation} \int \frac{ax+b}{cx+d} dx = a\left( \frac{x}{c}-\frac{d}{c^2}\ln(cx+d)\right) + b\ln(cx+d) \end{equation}

So using this general integration result you would be able to get the integration with respect to G

Answer 2

It is just a bit of linear algebra. The process looks like this

\begin{equation} \frac{ (b+G)(d-G)}{(a-c)G+ad-bc} \end{equation} It could be rewritten as \begin{equation} \frac{-1}{a-c} \frac{(G+b)(G-d)}{G-\frac{ad-bc}{a-c}} \end{equation} Further goes to \begin{equation} \frac{-1}{a-c} \frac{\left(G+\frac{ad-bc}{a-c}\right)^2+\frac{(a-c)(b-d)-2(ab-bc)}{a-c}G-\left(\frac{ad-bc}{a-c} \right)^2-bd}{G-\frac{ad-bc}{a-c}} \end{equation} This goes to the integration form as in the first answer.