Proving Mass Conservation In Coupled Tissue Compartments

Why Sum the Compartment Balances?

A multi-compartment tissue model represents drug in distinct spaces so that transport between them can occur at finite rates. Those exchanges can make each compartment’s amount rise or fall, but an exchange across an internal membrane does not, by itself, add drug to or remove drug from the tissue as a whole. This distinction is a basic structural check on the model: internal transport redistributes amount; only terms crossing the boundary of the tracked tissue system change its total amount.

The conservation argument does not depend on the particular permeability equation used for a membrane flux. It depends on writing the same flux once as an amount leaving its donor compartment and once as an equal amount entering its receiver compartment. The earlier material on extracellular–intracellular exchange and endosomal exchange supplies the compartment structure and membrane-transfer interpretation; here the aim is to establish what follows when all three balances are added.

Definition: Let $A_e(t)$, $A_i(t)$, and $A_{\mathrm{end}}(t)$ denote the drug amounts in the extracellular, intracellular, and endosomal compartments of a tissue at time $t$. The total amount in the tracked tissue compartments is $$ A_{\mathrm{tissue}}(t)=A_e(t)+A_i(t)+A_{\mathrm{end}}(t). $$ An internal flux transfers amount between two compartments included in this total. An external input or external output transfers amount across the boundary of the tracked system.

Write Each Internal Flux Once, With a Direction

Use a signed flux for each membrane connection. Define $J_{ei}$ as positive for transfer from extracellular to intracellular space, and $J_{i,\mathrm{end}}$ as positive for transfer from intracellular space to the endosome. A negative value means that the actual transfer is in the opposite direction. Let $I_k$ and $O_k$ be the nonnegative external input and output rates, respectively, for compartment $k$, where $k$ is $e$, $i$, or $\mathrm{end}$. An external rate may represent, for example, delivery from blood, return to blood, or a loss from the tracked drug amount. The balances are

\begin{aligned} \frac{dA_e}{dt} &= I_e-O_e-J_{ei},\\ \frac{dA_i}{dt} &= I_i-O_i+J_{ei}-J_{i,\mathrm{end}},\\ \frac{dA_{\mathrm{end}}}{dt} &= I_{\mathrm{end}}-O_{\mathrm{end}}+J_{i,\mathrm{end}}. \end{aligned}

The same $J_{ei}$ appears with opposite signs because it leaves one compartment and enters the other. The same is true of $J_{i,\mathrm{end}}$. These signs are required whether a flux is calculated from permeability, a concentration difference, or another transport law. The law determines the value of the flux; it does not change how that transfer enters the two amount balances.

This formulation assumes that each internal transfer is represented as a single, matched rate in the connected compartments, and that the listed compartments constitute the system whose total is being counted. External inputs and outputs may vary with time. Internal fluxes may also vary with time or depend nonlinearly on the compartment states; neither possibility changes the cancellation. The conservation result is a statement about the accounting structure, not an assumption that concentrations or fluxes are constant.

Deriving the Total-Amount Balance

Differentiate the definition of $A_{\mathrm{tissue}}$ and substitute the three compartment equations. Group the external terms and the membrane terms separately:

\begin{aligned} \frac{dA_{\mathrm{tissue}}}{dt} &=\frac{dA_e}{dt}+\frac{dA_i}{dt}+\frac{dA_{\mathrm{end}}}{dt}\\ &=(I_e-O_e-J_{ei}) +(I_i-O_i+J_{ei}-J_{i,\mathrm{end}})\\ &\qquad +(I_{\mathrm{end}}-O_{\mathrm{end}}+J_{i,\mathrm{end}})\\ &=(I_e+I_i+I_{\mathrm{end}}) -(O_e+O_i+O_{\mathrm{end}}). \end{aligned}

The two internal contributions vanish because $-J_{ei}+J_{ei}=0$ and $-J_{i,\mathrm{end}}+J_{i,\mathrm{end}}=0$. Thus the total amount changes only through the sum of external inputs minus the sum of external outputs.

Theorem: For the three-compartment model above, if each internal flux enters its donor and receiver balances with equal magnitude and opposite signs, then $$ \frac{dA_{\mathrm{tissue}}}{dt} =\sum_{k} I_k-\sum_{k} O_k, $$ where both sums are over $k=e,i,\mathrm{end}$. Internal membrane exchange has no direct effect on the total amount in the tracked tissue compartments.

Integrating this rate balance from time $t_0$ to $t_1$ gives the corresponding finite-interval statement:

A_{\mathrm{tissue}}(t_1)-A_{\mathrm{tissue}}(t_0) =\int_{t_0}^{t_1}\left(\sum_k I_k(t)-\sum_k O_k(t)\right)\,dt.

This follows by integrating the derivative of $A_{\mathrm{tissue}}$ over the interval. The integral of each internal flux cancels with its opposite-signed occurrence, even if the flux changes direction during the interval. If there are no external inputs or outputs, the total amount is constant, although the amount in each individual compartment can still change.

Worked Examples

Worked Example: Checking Instantaneous Conservation

Consider a hypothetical moment at which the extracellular compartment receives an external input of $1.20$ nmol/min and has an external output of $0.40$ nmol/min. The endosome has an external output of $0.10$ nmol/min; there are no other external terms. At this moment, $J_{ei}=0.80$ nmol/min and $J_{i,\mathrm{end}}=0.30$ nmol/min. Find each compartment’s amount rate of change and verify the total balance.

For the extracellular amount, $$ \frac{dA_e}{dt}=1.20-0.40-0.80=0.00\ \text{nmol/min}. $$ For the intracellular amount, $$ \frac{dA_i}{dt}=0+0+0.80-0.30=0.50\ \text{nmol/min}. $$ For the endosomal amount, $$ \frac{dA_{\mathrm{end}}}{dt}=0-0.10+0.30=0.20\ \text{nmol/min}. $$

The sum of the compartment rates is $0.00+0.50+0.20=0.70$ nmol/min. The external net rate is $1.20-0.40-0.10=0.70$ nmol/min. The results agree: the internal fluxes redistribute amount, while the net external input raises the tissue total at $0.70$ nmol/min.

Worked Example: Verifying Conservation Over a Time Interval

In a hypothetical two-hour interval, the initial amounts are $A_e=5.00$ nmol, $A_i=2.00$ nmol, and $A_{\mathrm{end}}=1.00$ nmol. Over the interval, $3.00$ nmol enters the extracellular compartment externally, $1.00$ nmol leaves it externally, and $0.40$ nmol leaves the endosome externally. The integrated internal transfers are $2.50$ nmol from extracellular to intracellular space and $0.70$ nmol from intracellular space to the endosome. Calculate the final compartment amounts and compare the total change with net external transfer.

Apply each transfer with a negative sign in its donor balance and a positive sign in its receiver balance: $$ \begin{aligned} A_e(t_1)&=5.00+3.00-1.00-2.50=4.50\ \text{nmol},\\ A_i(t_1)&=2.00+2.50-0.70=3.80\ \text{nmol},\\ A_{\mathrm{end}}(t_1)&=1.00+0.70-0.40=1.30\ \text{nmol}. \end{aligned} $$

The initial total is $5.00+2.00+1.00=8.00$ nmol, and the final total is $4.50+3.80+1.30=9.60$ nmol. Its increase is $9.60-8.00=1.60$ nmol. Independently, the net external transfer is $3.00-1.00-0.40=1.60$ nmol. The internal transfers do not appear in that total because each is counted once as a loss and once as a gain.

Worked Example: Finding a Sign Error by Summing

Suppose a hypothetical implementation uses the extracellular equation $\frac{dA_e}{dt}=I_e-O_e-J_{ei}$ but incorrectly writes the intracellular equation as $\frac{dA_i}{dt}=I_i-O_i-J_{ei}-J_{i,\mathrm{end}}$. The endosomal equation is written correctly as $\frac{dA_{\mathrm{end}}}{dt}=I_{\mathrm{end}}-O_{\mathrm{end}}+J_{i,\mathrm{end}}$. At an instant when the net external rate is $0.20$ nmol/min, $J_{ei}=0.40$ nmol/min, and $J_{i,\mathrm{end}}=0.10$ nmol/min, determine what the sum reveals.

Adding the implemented equations leaves the external net rate and both contributions from $J_{i,\mathrm{end}}$ cancelling, but $J_{ei}$ appears twice with a negative sign: $$ \frac{dA_{\mathrm{tissue}}}{dt} =0.20-2(0.40)=-0.60\ \text{nmol/min}. $$

That result is inconsistent with the external net rate of $0.20$ nmol/min. The intracellular equation treats transfer from extracellular space as another loss, rather than as an intracellular gain. Replacing $-J_{ei}$ in that equation with $+J_{ei}$ makes the paired terms cancel and restores the required total rate of $0.20$ nmol/min.

Practical Checks, Alternatives, and Limitations

Summing the amount equations is a direct structural check during model formulation. A useful procedure is to label the donor and receiver of every internal flux, confirm that its signs are opposite in those two equations, and then add the equations symbolically. A remaining internal flux in the total balance signals a possible sign error, a missing receiving or donor term, or a transfer that has been included in only one compartment. A flux that is intended to cross the tissue boundary should instead appear as an external output or input.

A second check uses integrated changes: compare the simulated change in $A_{\mathrm{tissue}}$ with the accumulated external inputs minus outputs over the same time interval. Small differences can arise from numerical integration tolerances, but a discrepancy that persists or grows as tolerances are tightened warrants inspection of the equations and implementation. This check does not identify the exact error by itself; it shows that the bookkeeping or numerical calculation does not match the stated balance.

The amount formulation matters. If the state variables are concentrations $C_k$, the conserved total is generally the sum of amounts, not the unweighted sum of concentrations. For fixed compartment volumes $V_k$, that total is $\sum_k V_k C_k$. If a volume changes with time, differentiating $A_k=V_kC_k$ gives $\frac{dA_k}{dt}=V_k\frac{dC_k}{dt}+C_k\frac{dV_k}{dt}$; omitting the volume-change contribution can make a concentration-based accounting appear inconsistent.

One alternative is to inspect each compartment equation separately and track every term. That is useful for finding where an implementation is wrong, but summing the equations is the most direct way to establish the total balance. Another alternative is to simulate the compartments and check their total numerically; this is a diagnostic, not a substitute for proving the symbolic cancellation. Neither approach requires the internal fluxes to be at equilibrium or instantaneous.

The conclusion applies to the amount being tracked. If a term represents metabolism of the parent drug, it is an output from the parent-drug balance unless the metabolite amount is also included in the tracked system with an appropriate production term. Likewise, any compartment omitted from the total can behave as an external source or sink relative to that reduced system. Conservation does not mean that every chemical species is conserved separately, nor does it guarantee that the model’s flux magnitudes or other physiological assumptions are correct.

Common Pitfall

Wrong reasoning: “The membrane flux is present in two equations, so it doubles the tissue’s transport rate.” A paired appearance is not a pair of losses or gains; one term removes amount from the donor while the other adds the same amount to the receiver.
Correction: Fix a direction for each flux and use one signed value with opposite signs in the two connected amount balances. When those balances are added, the internal terms must cancel. If they do not, check the signs, the affected compartments, and whether the transfer actually crosses the boundary of the tracked system.

Key takeaway: For extracellular, intracellular, and endosomal amount balances, every internal membrane transfer must appear once as a loss and once as an equal gain. Therefore the tissue total changes only by net external input and output.

Check Your Understanding

Try each question, then compare your reasoning with the solutions.

If there are no external inputs or outputs, what does the summed balance imply about $A_{\mathrm{tissue}}$? Can individual compartment amounts still change?
A flux $J_{ei}$ is defined as positive from extracellular to intracellular space. What signs must it have in the extracellular and intracellular amount equations?
Over an interval, the tissue total rises by $0.90$ nmol. External input is $1.40$ nmol and external output is $0.50$ nmol. What is the net internal contribution to the total change?
Why is $C_e+C_i+C_{\mathrm{end}}$ generally not the conserved total when the compartments have different volumes?

Solutions. (1) The total remains constant because its derivative is zero; individual amounts can change as internal fluxes redistribute drug. (2) The term is $-J_{ei}$ in the extracellular balance and $+J_{ei}$ in the intracellular balance. (3) Net external transfer is $1.40-0.50=0.90$ nmol, equal to the observed total increase, so the net internal contribution is zero. (4) Concentrations are amount per volume, so they cannot generally be added as amounts; for fixed volumes the total is $V_eC_e+V_iC_i+V_{\mathrm{end}}C_{\mathrm{end}}$.