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?


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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Chet Miller Dec 14 '15 at 15:49
  • $\begingroup$ @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 $\endgroup$ – Dahlai Dec 14 '15 at 15:55
  • $\begingroup$ This is not the solution if $c_{extra}$ does not change from its initial value of $c_{extra0}$ $\endgroup$ – Chet Miller Dec 14 '15 at 16:10
  • $\begingroup$ 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. $\endgroup$ – Chet Miller Dec 14 '15 at 21:34
  • $\begingroup$ @ChesterMiller Yes, exactly $\endgroup$ – 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.


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).

  • $\begingroup$ 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). $\endgroup$ – John_West Dec 14 '15 at 16:34
  • $\begingroup$ 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. $\endgroup$ – Dahlai Dec 14 '15 at 18:17
  • $\begingroup$ Did you see the addition I edited into my answer for the case in which the external concentration is not constant? $\endgroup$ – Chet Miller Dec 16 '15 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.