# 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?

• It's hard to see how, from your differential equation, $c_{intra}$ at equilibrium could be $cextra0/2$. At equilibrium, your equation tells us that, at equilibrium, $c_{intra}=cextra0$. I suspect an issue with the mathematics. – Chet Miller Dec 14 '15 at 15:49
• @ChesterMiller I never had a course on differential equations, so this might well be. My train of thought: Forward and reverse reaction rate are the same, and $c_{intra0}=0$, hence the equilibrium of $c_{extra}/2$. This is also the result when simulating it in MATLAB – Dahlai Dec 14 '15 at 15:55
• This is not the solution if $c_{extra}$ does not change from its initial value of $c_{extra0}$ – Chet Miller Dec 14 '15 at 16:10
• Yes. You need an entirely different approach for that. You want to specify $c_{extra}(t)$ (i.e., as a function of time t) and you want of determine the response $c_{intra}(t)$ as a function of time t, correct? We can provide the general solution for that if this is what you want. – Chet Miller Dec 14 '15 at 21:34
• @ChesterMiller Yes, exactly – Dahlai Dec 15 '15 at 11:07

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).

• Agree. OP: use Bernoulli method to integrate your eq. analytically and obtain (1). At $t\rightarrow \infty$, $c_{intra}=c_{extra}=c_{extra0}$. The second eq. above is mole balance (no change of moles, because no reaction, change in T and P). – John_West Dec 14 '15 at 16:34
• I think I didn't explain myself precisely enough. I don't want $c_{extra}$ to be constrained to its initial value. I will update the original post so that I can hopefully clarify my question. Sorry for the inconvenience. – Dahlai Dec 14 '15 at 18:17
• Did you see the addition I edited into my answer for the case in which the external concentration is not constant? – Chet Miller Dec 16 '15 at 15:34