Worked Example: A Three-Stage Maturation Model

Building a Three-Stage Maturation Model

A cell population moving through a sequence of maturation stages can be represented by linked compartments. Cells enter the first stage, may be lost from any stage, and otherwise progress forward. This structure distinguishes the number of cells present in each stage from the flux of cells leaving the final stage as mature cells.

Let \(X_1(t)\), \(X_2(t)\), and \(X_3(t)\) denote the cell counts in the three maturation stages at time \(t\). A constant precursor input \(u\) enters stage 1, measured in cells per unit time. Cells transit from stage 1 to stage 2 at per-cell rate \(k_1\), from stage 2 to stage 3 at rate \(k_2\), and leave stage 3 as mature cells at rate \(k_3\). The stage-specific loss rates are \(\delta_1\), \(\delta_2\), and \(\delta_3\). All rates have units of inverse time.

The resulting model is

The model assumes constant input and rates, well-mixed compartments, and first-order exits: over a short interval, the expected number leaving a stage is proportional to the number currently in it. Transit and loss are competing exits, and cells move only forward through the chain. The formulation has no feedback, division, or return transitions. Assume \(u\geq 0\), \(k_i\geq 0\), \(\delta_i\geq 0\), and \(a_i>0\), so each stage has a positive total exit rate.

Steady State and Output

At steady state, each stage count is constant, so its time derivative is zero. Solving the equations in sequence gives the stage 1 count first, then the stage 2 count, then the stage 3 count.

The mature-cell output is a flux, not a cell count:

This expression shows that the steady output depends on the fraction of cells progressing rather than being lost at each stage. In stage \(i\), the forward fraction is \(k_i/a_i\). Thus \(J_{\mathrm{mature}}^*/u=(k_1/a_1)(k_2/a_2)(k_3/a_3)\). The ratio is dimensionless; it describes the fraction of input ultimately leaving through the mature-cell route under this model.

Transient Response to a Constant Input

Steady-state calculations do not describe how quickly the counts approach equilibrium or how maturation delays output after an input begins. For constant \(u\), the first stage has a direct solution. The later stages are driven by the preceding stage, so their solutions can be written as integrals of the upstream counts.

These integral forms follow by multiplying each linear stage equation by its integrating factor. For example, \(dX_2/dt+a_2X_2=k_1X_1\); multiplying by \(e^{a_2t}\) makes the left side \(d(e^{a_2t}X_2)/dt\). Integrating from 0 to \(t\) and rearranging gives the displayed expression for \(X_2(t)\). The same argument gives the stage 3 expression.

A useful closed form occurs when all total exit rates are equal, \(a_1=a_2=a_3=a\), and the chain starts empty. Define \(z=at\). Substitution through the three equations gives

For stage 2, the integral of the stage 1 response introduces the factor \(1-(1+z)e^{-z}\). Integrating that response once more gives the stage 3 factor \(1-(1+z+z^2/2)e^{-z}\). These factors are initially zero and approach one as \(t\) increases, so the counts rise from zero toward their corresponding steady states. The progressively higher powers of \(z\) also show why downstream stages respond more gradually.

Parameter Changes, Checks, and Limitations

Several checks help detect errors. The equations preserve nonnegative counts from nonnegative initial counts and input: each stage receives nonnegative influx and has an outflow proportional to its own count. Also, summing the three equations gives

The internal transit fluxes \(k_1X_1\) and \(k_2X_2\) cancel in the total, as they should. At steady state, input must equal total losses plus mature output. If a calculation violates this balance, check whether a transit term was omitted, counted as a loss, or assigned the wrong units.

The system is triangular: each stage is driven by the stage before it, and its own deviation from equilibrium decays at rate \(a_i\), with upstream deviations also contributing to downstream transients. Because every \(a_i>0\), those exponential components decay, and the steady state is stable. Equal exit rates can produce polynomial factors multiplying an exponential, as in the step-response formulas, but those terms still decay to zero.

A three-stage chain is more flexible than a single compartment because it represents a spread of maturation times rather than one simple exponential waiting-time profile. A fixed-delay model is another possible representation, but it treats maturation as occurring after a specified delay rather than through these continuously exiting compartments. Neither choice is automatically correct: stage counts or time-course data should support the selected structure. The present model is limited by constant rates, constant input in the solutions above, and the assumption that cells within a stage share the same exit rates. It also treats mature-cell output as a flux; modelling a mature-cell count requires an additional compartment with its own balance equation.

A common mistake is to interpret \(X_3\) itself as the number of mature cells produced per day. \(X_3\) is a count, measured in cells; production is the flux \(k_3X_3\), measured in cells per day. Another mistake is to use \(k_3\), rather than \(k_3+\delta_3\), as the total stage 3 exit rate when calculating its steady count. Both exits remove cells from stage 3, even though only one contributes to mature output.