First order differential equation for transmembrane diffusion

I want to model transmembrane diffusion using a first order reversible differential equation. The diffusion rate is known from literature. There are two volumes $V\_{\mathrm{intra}}$ and $V\_{\mathrm{extra}}$. The former is $10^{12}$ times smaller than the latter. $c\_{\mathrm{intra0}}=0~\mathrm{mol~L^{-1}}$, and $c\_{\mathrm{extra0}}=x~\mathrm{mol~L^{-1}}$, with $x>0$.


Thus far I am using this:
$$\frac{\mathrm{d}c\_{\mathrm{intra}}}{\mathrm{d}t}=k\_{\mathrm{diff}}\cdot c\_{\mathrm{extra}} - k\_{\mathrm{diff}} \cdot c\_{\mathrm{intra}}$$


However, this does not account for the fact, that $V\_{\mathrm{extra}}$ is much larger than $V\_{\mathrm{intra}}$. I expect $c\_{\mathrm{extra}}$ to be virtually constant. So far the the simulation results in the equilibrium being at $c\_{\mathrm{extra0}}/2$.


Can someone suggest a method to account for the volume discrepancy?




---


**Edit:**


From the outlined scenario above, I expect in REALITY, my $c\_{\mathrm{extra}}$ to not noticeably change from its initial value due to the large difference between $V\_{\mathrm{intra}}$ and $V\_{\mathrm{extra}}$.


However, in the simulation I don't want to set $c\_{\mathrm{extra}}$ as constant, because I want to apply a 'extracellular' decay rate later on. Basically, I want to be able to modulate the extracellular concentration over time and see how the concentration within the cell changes.


This is why I want to account for the volume. Maybe, one needs an entirely different approach for that?




The solution to your differential equation, subject to the prescribed initial conditions is $$c\_{intra}=c\_{extra0}(1-e^{-k\_{diff}t})\tag{1}$$ under the constraint that $c\_{extra}$ does not change from its initial value of $c\_{extra0}$. The true material balance constraint is $$V\_{intra}c\_{intra}+V\_{extra}c\_{extra}=V\_{extra}c\_{extra0}$$
which would yield a result virtually indistinguishable from Eqn. 1.


**CASE WHERE $c\_{extra}$ IS NOT CONSTANT**


OK. Here is the solution for the case where $c\_{extra}$ is a function of time $c\_{extra}(t)$, and $c\_{intra}=c\_{intra}(0)$ at time t = 0:


$$c\_{intra}(t)=c\_{intra}(0)e^{-k\_{diff}t}+k\_{diff}\int\_0^t{e^{-k\_{diff}\tau}c\_{extra}(t-\tau)d\tau}\tag{2}$$
where $\tau$ is a dummy variable of integration and where $c\_{extra}(t-\tau)$ represents the value of the concentration $c\_{extra}$ at time $t-\tau$. This solution reduces to the previous result if $c\_{intra}(0)=0$ and $c\_{extra}$ is constant. 


So, to find the value of $c\_{intra}$ at time t, you need to use Eqn. 2 and perform the indicated integration (usually numerically).


 1