Deriving Net Tumor Dynamics Under Constant Drug Kill

From Growth and Killing to Net Tumor Dynamics

A tumor can contain cells that are dividing while treatment is removing cells. Under a simple model, both processes act continuously and in proportion to the current viable tumor burden. The key consequence is that the two per-capita rates combine into one net rate: its sign determines whether the modeled burden grows or shrinks.

Here, “constant drug kill” means a constant per-capita rate of cell loss over the interval being modeled. It does not mean that the drug removes a fixed number of cells per day. If the burden is twice as large, a constant per-capita kill rate removes twice as many cells per unit time.

Definition: Let $N(t)$ be viable tumor cell burden at time $t$, with $N(0)=N_0>0$. Let $r$ be the constant per-capita tumor growth rate and $k$ the constant per-capita drug-kill rate, both in inverse time units. The net dynamics model is $$ \frac{dN}{dt}=rN-kN=(r-k)N. $$ The model assumes exponential untreated growth, a constant kill rate, and a single well-mixed burden whose cells share these rates. It also assumes no delay, changing sensitivity, or treatment interruption during the modeled interval.

The quantities $rN$ and $kN$ are rates of change in cells per unit time; $r$ and $k$ themselves have units such as day$^{-1}$. For $N>0$, $r$ is positive growth and $k$ is a nonnegative loss rate. In this model, treatment changes the rate of change rather than imposing a fixed ceiling on tumor size.

Deriving the Trajectory and the Threshold

Write $a=r-k$, which is constant under the assumptions above. Separating variables and integrating from time 0 to time $t$ gives a complete solution:

\begin{aligned} \frac{1}{N}\frac{dN}{dt} &= a,\\ \int_{N_0}^{N(t)} \frac{1}{u}\,du &= \int_0^t a\,ds,\\ \ln\!\left(\frac{N(t)}{N_0}\right) &= at,\\ N(t) &= N_0 e^{(r-k)t}. \end{aligned}

The division by $N$ is valid because a positive initial burden stays positive at every finite time in this solution. The result also verifies the initial condition: setting $t=0$ gives $N(0)=N_0$. Differentiating the expression gives $dN/dt=(r-k)N(t)$, so it satisfies the stated differential equation.

Theorem: For $N_0>0$ and constant $r$ and $k$, the sign of $r-k$ determines the direction of change at every time. If $k<r$, burden grows exponentially; if $k=r$, burden is constant; if $k>r$, burden declines exponentially. Thus $k=r$ is the threshold separating growth from shrinkage.

When $r=k$, the solution is $N(t)=N_0$: this is a balance of the two modeled rates, not a stable equilibrium that restores burden after a perturbation. When $k>r$, the solution approaches zero asymptotically but does not reach zero at a finite time. A model prediction of a fractional cell count near zero should therefore not be read as a literal description of extinction; the continuous-burden approximation has limits at very small counts.

If untreated exponential growth is described by a doubling time $T_d$, then $r=\ln(2)/T_d$. Substitution gives a threshold kill rate $k_{\text{crit}}=\ln(2)/T_d$. A kill rate greater than this value produces net shrinkage in the model; a lower rate still permits net growth, even though drug killing is occurring.

Worked Example: Calculating Net Shrinkage

Suppose a hypothetical tumor has untreated doubling time $T_d=30$ days, initial viable burden $N_0=2.00\times10^9$ cells, and a constant per-capita kill rate $k=0.0300$ day$^{-1}$. Calculate the predicted burden after 28 days.

The exponential-growth assumption gives $r=\ln(2)/T_d$, so

r=\frac{\ln(2)}{30\ \text{days}} =0.0231049\ \text{day}^{-1}.

The drug kill rate is constant and is expressed in the same inverse-time units, so the net rate is

r-k=0.0231049-0.0300 =-0.0068951\ \text{day}^{-1}.

Because $k>r$, the model predicts shrinkage. Applying the derived solution,

\begin{aligned} N(28) &=2.00\times10^9 \exp\!\left[(-0.0068951\ \text{day}^{-1})(28\ \text{days})\right]\\ &=2.00\times10^9 e^{-0.193063}\\ &\approx 1.6489\times10^9\ \text{cells}. \end{aligned}

Thus the predicted burden is about $1.65\times10^9$ viable cells after 28 days, a decrease of about $3.51\times10^8$ cells from baseline. This is a model prediction under constant rates, not evidence that a real tumor must follow this trajectory.

Reading the Net Rate in Practice

The net rate $r-k$ is the slope of log burden against time in this model. Taking logarithms of the solution gives $\ln N(t)=\ln N_0+(r-k)t$. Consequently, a constant negative slope on a log-burden scale corresponds to exponential decline, while a positive slope corresponds to exponential growth. The magnitude also matters: a net rate close to zero implies comparatively little change over a given interval.

Worked Example: A Kill Rate Just Below the Threshold

Consider a hypothetical tumor with untreated doubling time 20 days, initial burden $N_0=5.00\times10^8$ cells, and constant kill rate $k=0.0340$ day$^{-1}$. Predict the burden after 60 days and classify the response.

First calculate the growth rate and compare it with the kill rate:

r=\frac{\ln(2)}{20}=0.0346574\ \text{day}^{-1}, \qquad r-k=0.0346574-0.0340 =0.0006574\ \text{day}^{-1}.

The kill rate is below the threshold, so the model predicts growth, despite ongoing drug killing. Over 60 days,

\begin{aligned} N(60) &=5.00\times10^8 \exp\!\left[(0.0006574\ \text{day}^{-1})(60\ \text{days})\right]\\ &=5.00\times10^8 e^{0.039444}\\ &\approx 5.20\times10^8\ \text{cells}. \end{aligned}

The increase is modest because the net rate is small. This example shows why observing a positive kill rate alone does not establish shrinkage: it must exceed the growth rate in this model.

The model also gives a convenient time calculation when net shrinkage is predicted. If $0<f<1$ is the desired fraction of initial burden, solving $f=N(t)/N_0=e^{(r-k)t}$ yields $t=\ln(f)/(r-k)$, which is positive when $k>r$. In particular, the modeled half-time is $\ln(2)/(k-r)$. This is a half-time for the net decline, not the untreated doubling time.

Worked Example: Time to Half the Initial Burden

Suppose a hypothetical tumor has untreated doubling time 24 days and constant per-capita kill rate $k=0.0480$ day$^{-1}$. Its initial burden is $1.20\times10^9$ cells. Find the net shrinkage half-time and the predicted burden after 40 days.

The untreated growth rate and net rate are

r=\frac{\ln(2)}{24}=0.0288811\ \text{day}^{-1}, \qquad r-k=0.0288811-0.0480 =-0.0191189\ \text{day}^{-1}.

Since $k>r$, the half-time is

t_{1/2}=\frac{\ln(2)}{k-r} =\frac{0.693147}{0.0480-0.0288811} \approx 36.25\ \text{days}.

For the 40-day prediction, use the same net rate and the initial burden:

\begin{aligned} N(40) &=1.20\times10^9 \exp\!\left[(-0.0191189\ \text{day}^{-1})(40\ \text{days})\right]\\ &=1.20\times10^9 e^{-0.764756}\\ &\approx 5.59\times10^8\ \text{cells}. \end{aligned}

The 40-day burden is below half of baseline because 40 days exceeds the calculated half-time of about 36.25 days. The model predicts a continuous exponential decline, not a sudden halving at that time.

Assumptions, Diagnostics, and Alternatives

The solution is useful only when its assumptions are reasonable over the interval of interest. It treats untreated growth as exponential, assumes the effective kill rate does not change with time, and applies the same rates to all cells represented by $N(t)$. It also treats burden as a continuous quantity. If the observed outcome is tumor volume rather than viable cell count, interpretation requires care: volume can include components that are not represented by viable-cell burden.

A practical diagnostic is to plot the logarithm of repeated positive burden measurements against time. A roughly straight trend is compatible with a constant net rate over that interval. Systematic curvature, a change in slope, or a trend that alternates between growth and decline signals that a single constant net rate may not describe the data. Such patterns can arise from changing growth or kill rates, measurement features, or processes omitted from the model; the plot alone does not identify which explanation is correct.

A further limitation is that a trajectory identifies the net rate, not necessarily its two components. A measured log-burden slope of $-0.01$ day$^{-1}$, for example, implies $r-k=-0.01$ day$^{-1}$ under this model. Without separate information about untreated growth or kill, many pairs $(r,k)$ produce that same slope. A good fit to the trajectory therefore does not by itself establish that either rate has been estimated independently.

This formulation differs from cytostasis: cytostasis reduces or halts the growth term, whereas the present model explicitly includes a loss term $kN$. It also differs from logistic or Gompertz growth, whose growth rate changes with burden rather than remaining constant. These alternatives can be appropriate when their assumptions better match the biological question and data, but they do not change the threshold result derived here for constant exponential growth opposed by constant per-capita killing.

Common pitfall: “The drug kills cells, so the tumor must shrink.” This conclusion ignores untreated growth. In the stated model, the correct comparison is $k$ versus $r$: shrinkage occurs only when $k>r$; when $k<r$, the burden still grows. Also, a constant per-capita kill rate is not a fixed number of cells removed each day.

Check Your Understanding

Answer each question using the constant per-capita kill model. Brief solutions follow the questions.

For $r=0.025$ day$^{-1}$ and $k=0.025$ day$^{-1}$, what does the model predict for a positive initial burden?
A hypothetical tumor doubles every 40 days. What is the threshold kill rate, and would $k=0.020$ day$^{-1}$ predict growth or shrinkage?
For $r=0.03$ day$^{-1}$, $k=0.05$ day$^{-1}$, and $N_0=100$ million cells, calculate $N(10)$.
Repeated measurements appear curved when log burden is plotted against time. What assumption should be questioned, and what does the curvature not establish on its own?

Solutions. (1) Since $r-k=0$, $N(t)=N_0$: the modeled burden remains constant. (2) The threshold is $k_{\text{crit}}=\ln(2)/40=0.01733$ day$^{-1}$. Since $0.020>0.01733$, the model predicts shrinkage. (3) The net rate is $0.03-0.05=-0.02$ day$^{-1}$, so $N(10)=100e^{-0.2}\approx81.87$ million cells. (4) The assumption of a constant net rate, and hence a straight line on the log-burden scale, should be questioned. Curvature alone does not identify its cause or establish which model component changed.

Deriving Net Tumor Dynamics Under Constant Drug Kill

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