Time-Varying Proliferation and Death Rates

Why Rates That Change Over Time Matter

A cell population’s count can change because cells proliferate, because cells die, or both. In a controlled experiment or treatment scenario, either process may vary over time: a drug may alter cell loss after dosing, or proliferation may change as cells respond to a stimulus. A single constant net rate can obscure the timing and duration of these changes. A time-varying model keeps that schedule explicit and links it to the entire count trajectory.

The earlier material on per-capita rates and exponential cell growth provides the constant-rate case. Here the rates may vary continuously or switch at specified times. The central calculation is the accumulated net rate, not just the rate at the final time.

Definition: Let $N(t)$ be the cell count at time $t$, $p(t)$ the per-capita rate of cell-number gain from proliferation, and $d(t)$ the per-capita death rate. Both rates have units of inverse time. Under a well-mixed, deterministic count model with no immigration or other cell flows, the net rate is $r(t)=p(t)-d(t)$, and the governing equation is $$ \frac{dN(t)}{dt}=[p(t)-d(t)]N(t)=r(t)N(t). $$ For binary division in which one parent cell is replaced by two daughters, the event produces a net gain of one cell; if the division-event hazard is $p(t)$, it is also the per-capita cell-number gain rate under this accounting.

The model assumes that each cell at time $t$ has the same instantaneous per-capita proliferation and death rates, so total gain and loss are proportional to the current count. The rates may be specified functions of time, including a scheduled change, but are not functions of $N(t)$ in this model. There is no instantaneous addition or removal of cells at a rate switch. Thus a rate may jump at a switch time, while the count itself remains continuous. The equation describes a deterministic population-scale trajectory; it is not a claim that individual cells divide fractionally.

A positive $r(t)$ means the count is increasing at that instant; a negative $r(t)$ means it is decreasing. Either sign is possible even though $p(t)$ and $d(t)$ themselves are nonnegative rates. If a model uses a different convention for what “proliferation rate” counts, its cell-number gain term must be defined accordingly before applying this equation.

From Rates to a Count Trajectory

For an initial count $N(0)=N_0>0$, divide the governing equation by $N(t)$ and integrate. The rate functions must be integrable over the time interval; piecewise-continuous rates, including finite scheduled jumps, satisfy this condition. The solution is:

\frac{1}{N(t)}\frac{dN(t)}{dt}=r(t) \quad\Longrightarrow\quad \ln\!\left(\frac{N(t)}{N_0}\right)=\int_0^t r(u)\,du \quad\Longrightarrow\quad N(t)=N_0\exp\!\left(\int_0^t [p(u)-d(u)]\,du\right).

The integral is dimensionless: a rate in inverse days integrated over days gives a dimensionless exponent. It is the cumulative net rate that determines the multiplicative change in count. If $p$ and $d$ are constant, the integral becomes $(p-d)t$, recovering the exponential-growth or exponential-decline formula. For a rate that varies over time, using only its value at the endpoint generally gives the wrong trajectory.

The same result follows by an integrating factor. Write $N'(t)-r(t)N(t)=0$, and multiply by $\exp[-\int_0^t r(u)\,du]$. The left side becomes the derivative of $N(t)\exp[-\int_0^t r(u)\,du]$, which is zero. This product is therefore constant and equals $N_0$ at $t=0$; rearranging gives the stated solution. This also shows why a finite jump in $r(t)$ does not cause a jump in $N(t)$: the integral of a bounded rate over a zero-length interval is zero.

Worked Example: Rates That Vary Continuously

Suppose a hypothetical culture begins with $N_0=12{,}000$ cells. Over the first 5 days, let proliferation increase linearly as $p(t)=0.030+0.002t$ per day, where $t$ is measured in days, while death remains $d(t)=0.010$ per day. Calculate the count at day 5.

The net rate is $r(t)=0.020+0.002t$ per day. The coefficient $0.002$ has units of inverse days squared, so $0.002t$ has units of inverse days. The rates are nonnegative over this interval, and the model assumes they apply uniformly to the cells present.

The cumulative net rate is: $$ \int_0^5 (0.020+0.002t)\,dt =0.020(5)+0.001(5^2) =0.100+0.025 =0.125. $$ Thus: $$ N(5)=12{,}000e^{0.125} =12{,}000(1.133148\ldots) =13{,}597.8\ldots $$ The predicted count is approximately $13{,}598$ cells at day 5. The count is larger than the starting count because the cumulative net rate is positive.

Piecewise-Constant Rates and Scheduled Changes

When rates are constant within intervals but change at known times, integrate each interval separately and add the interval contributions. Suppose the rate on interval $j$ is $r_j$, the interval lasts $\Delta t_j$, and $N_j$ is the count at its start. Then $N_{j+1}=N_j e^{r_j\Delta t_j}$. Applying this successively is equivalent to using the total cumulative net rate across all elapsed intervals.

N(t_k)=N_0\exp\!\left(\sum_{j=1}^{k} r_j\Delta t_j\right), \qquad r_j=p_j-d_j.

This formula assumes each interval’s rates are constant throughout that interval and that a scheduled switch changes only the rates, not the count. If a procedure instead causes an instantaneous cell infusion, removal, or other count jump, that event requires a separate update to $N$; it is not represented by changing $p$ or $d$ alone.

Write the schedule.
List the time intervals and specify $p_j$ and $d_j$ on each one, with units.

Find each net rate.
Calculate $r_j=p_j-d_j$, keeping the sign and units explicit.

Accumulate the rate.
Multiply each $r_j$ by its interval duration and sum the products up to the requested time.

Calculate and check the count.
Multiply the initial count by the exponential of the cumulative net rate. Counts must agree from either side of a rate switch.

Worked Example: Checking Counts at Scheduled Switches

Consider a hypothetical population with $N(0)=50{,}000$ cells. The rates are constant on each interval: from day 0 to day 2, $p=0.18$ per day and $d=0.08$ per day; from day 2 to day 5, $p=0.06$ per day and $d=0.10$ per day; from day 5 to day 7, $p=0.09$ per day and $d=0.07$ per day. Calculate the count at each switch and at day 7.

The net rates are $0.10$, $-0.04$, and $0.02$ per day. These remain constant over their respective intervals, and all specified proliferation and death rates are nonnegative. At day 2: $$ N(2)=50{,}000e^{(0.10)(2)} =50{,}000e^{0.20} =61{,}070.1\ldots \approx 61{,}070\text{ cells}. $$

From day 2 to day 5 the cumulative net contribution is $(-0.04)(3)=-0.12$. Therefore: $$ N(5)=50{,}000e^{0.20-0.12} =50{,}000e^{0.08} =54{,}164.4\ldots \approx 54{,}164\text{ cells}. $$

From day 5 to day 7 the contribution is $(0.02)(2)=0.04$. Retaining the unrounded cumulative calculation gives: $$ N(7)=50{,}000e^{0.20-0.12+0.04} =50{,}000e^{0.12} =56{,}374.8\ldots \approx 56{,}375\text{ cells}. $$ The count decreases during days 2–5 and rises during days 5–7, consistent with the signs of the net rates. For clarity, applying the final interval to the rounded day-5 count gives $54{,}164e^{0.04}=56{,}374.5\ldots$, which rounds to $56{,}374$ cells. The small difference arises from rounding the intermediate count; retaining unrounded values gives the cumulative result of approximately $56{,}375$ cells.

Worked Example: Comparing Two Scheduled Rate Profiles

A hypothetical culture starts with $8{,}000$ cells. In one schedule, the net rate is $0.05$ per day from day 0 to day 3, $-0.03$ per day from day 3 to day 6, and $0.02$ per day from day 6 to day 10. Find the count at days 3, 6, and 10. Assume no instantaneous count changes at the scheduled switches.

Each interval rate is the difference between nonnegative proliferation and death rates: for example, $0.05$ could result from $p=0.10$ and $d=0.05$ per day, and $-0.03$ from $p=0.04$ and $d=0.07$ per day. The count at day 3 is: $$ N(3)=8{,}000e^{(0.05)(3)} =8{,}000e^{0.15} =9{,}294.7\ldots \approx 9{,}295\text{ cells}. $$

At day 6, include both the first and second intervals: $$ N(6)=8{,}000e^{(0.05)(3)+(-0.03)(3)} =8{,}000e^{0.06} =8{,}494.7\ldots \approx 8{,}495\text{ cells}. $$ At day 10, the final interval contributes $(0.02)(4)=0.08$: $$ N(10)=8{,}000e^{0.15-0.09+0.08} =8{,}000e^{0.14} =9{,}202.2\ldots \approx 9{,}202\text{ cells}. $$ The final count exceeds the initial count because the cumulative net rate, $0.14$, is positive, even though the count declined during the middle interval.

Choosing a Calculation and Checking Its Limits

For rates that are constant on known intervals, the interval-by-interval exponential calculation is direct and transparent. For smoothly varying rates with an analytically tractable integral, the integral solution gives the trajectory directly. If the rate functions are specified numerically or the integral has no convenient closed form, a numerical integration method can approximate the ODE solution. In all cases, the model equation and rate schedule should be the same; the methods differ in how the cumulative change is calculated.

Replacing time-varying rates with one constant rate is an approximation, not a general identity. Over a fixed interval of length $T$, choosing the constant rate equal to the time-average net rate, $\bar r=T^{-1}\int_0^T r(u)\,du$, reproduces the model’s endpoint count because $\bar rT=\int_0^T r(u)\,du$. It does not reproduce the within-interval trajectory unless the original rate is constant. An unweighted average of rates across intervals is not appropriate when the intervals have different durations.

Useful checks include confirming that each rate has units of inverse time, that every exponent is dimensionless, and that the count stays continuous at a rate switch. The direction of change should also agree with the sign of $p(t)-d(t)$ in each interval. If the computed count jumps at a switch, the interval calculation or the model description needs review. If it rises during an interval where $p<d$, check the sign and the subtraction order.

A common mistake is to use the net rate from the final interval for the whole observation period. For example, in the scheduled-switch example, using $0.02$ per day for all 7 days would give $50{,}000e^{0.14}$, rather than the correct $50{,}000e^{0.12}$. The correction is to account for the duration and net rate of every interval. Another pitfall is rounding a count at each switch before continuing; keep full precision internally and round only the reported results.

This formulation has important limits. It assumes per-capita rates that are shared by the modeled population and specified independently of its count. It does not represent density-dependent rates, variation among cell subpopulations, delays in cell-cycle progression, or random variation in small populations. If those mechanisms materially affect the question, the model needs additional structure rather than a different arithmetic shortcut.

Check Your Understanding

Try each problem first, then compare your reasoning with the solutions below.

A population has $N_0=2{,}000$ cells and a constant proliferation rate of $0.12$ per day and death rate of $0.07$ per day for 4 days. What is the predicted count?
For the first 2 days, a population has net rate $0.04$ per day; for the next 3 days, its net rate is $-0.02$ per day. What is the cumulative net rate through day 5, and how does it determine the endpoint count relative to $N_0$?
At a scheduled switch, $p$ changes from $0.08$ to $0.03$ per day and $d$ remains $0.05$ per day. Does the count jump at the switch, and what are the net rates before and after it?
Why might a constant rate equal to the time-average net rate reproduce an endpoint but not the intermediate counts?

Solutions. (1) The net rate is $0.12-0.07=0.05$ per day, so $N(4)=2{,}000e^{(0.05)(4)}=2{,}000e^{0.20}=2{,}442.8\ldots$, or approximately $2{,}443$ cells. (2) The cumulative net rate is $(0.04)(2)+(-0.02)(3)=0.08-0.06=0.02$. Thus $N(5)=N_0e^{0.02}$, about $1.0202N_0$; the endpoint is about 2.02% above the initial count. (3) The net rate changes from $0.08-0.05=0.03$ per day to $0.03-0.05=-0.02$ per day. The rate changes, but the count does not jump in this model. (4) The time-average net rate preserves the total integral through the endpoint, so its exponential gives the same endpoint multiplier. It generally distributes that cumulative change differently over time, so intermediate counts can differ.

Time-Varying Proliferation and Death Rates

What You'll Learn