Comparing Tumor Growth Laws Across Size Ranges

Why Compare Growth Laws Across Size Ranges?

A growth law can match tumor measurements over one interval and still make very different predictions outside it. The reason is that the laws encode different relationships between tumor size and growth rate. Exponential growth assigns the same proportional growth rate at every size; logistic growth reduces that rate as the tumor approaches a carrying capacity; and Gompertz growth makes the proportional rate decline with the logarithm of size relative to that capacity.

The comparison here concerns untreated, deterministic trajectories. Let \(N(t)\) denote tumor burden at time \(t\), measured in \(\text{mm}^3\). The equations describe the burden directly; interpreting measured volume as viable tumor-cell burden requires a separate measurement model. Assume throughout that \(N(0)>0\), time is measured in days, and any carrying capacity \(K\) is positive and constant. These are simplified, autonomous growth models: they omit treatment, changes in the tumor environment, and variation among tumors.

Definition: The per-capita growth rate is the instantaneous rate of change divided by current burden, \(g(N)=(1/N)(dN/dt)\). The absolute growth rate is \(dN/dt\), the change in burden per unit time. Per-capita rates have units of \(\text{day}^{-1}\); with burden measured in \(\text{mm}^3\), absolute rates have units of \(\text{mm}^3/\text{day}\).

Three Growth Laws and Their Assumptions

In the exponential model, the per-capita rate is a positive constant \(r\). Its defining equation and trajectory are:

\frac{dN}{dt}=rN,\qquad N(t)=N(0)e^{rt},\qquad r>0.

This law assumes that the proportional growth rate does not depend on tumor burden. Thus its absolute growth rate \(rN\) increases in direct proportion to size. There is no built-in upper limit: for \(r>0\), burden grows without bound as time increases.

The logistic model adds a constant carrying capacity \(K\). Its per-capita rate decreases linearly with burden:

\frac{dN}{dt}=rN\left(1-\frac{N}{K}\right),\qquad N(t)=\frac{K}{1+\left(K/N(0)-1\right)e^{-rt}}.

Here \(r>0\) is the low-burden limiting per-capita growth rate. For \(0<N<K\), burden increases and approaches \(K\); at \(N=K\), growth is zero. The model represents a density-dependent brake that becomes stronger as the burden approaches \(K\). It does not specify the biological mechanism producing that brake.

The Gompertz model also has an upper limiting burden \(K\), but its per-capita rate declines with the logarithm of the ratio between that limit and current burden:

\frac{dN}{dt}=\beta N\ln\left(\frac{K}{N}\right),\qquad N(t)=K\exp\left[-\ln\left(\frac{K}{N(0)}\right)e^{-\beta t}\right], \qquad \beta>0.

For \(0<N<K\), the Gompertz burden increases toward \(K\). The parameter \(\beta\) has units of \(\text{day}^{-1}\), as does \(r\). Unlike the logistic law, the Gompertz per-capita rate does not approach a finite constant as \(N\) becomes very small: \(\beta\ln(K/N)\) increases without bound as \(N\) tends to zero. This mathematical feature should not be read as evidence that arbitrarily small tumors have an infinite biological growth rate.

How the Rate Laws Differ

The per-capita rates make the main contrast explicit. Exponential growth has \(g(N)=r\); logistic growth has \(g(N)=r(1-N/K)\); and Gompertz growth has \(g(N)=\beta\ln(K/N)\). If the logistic and Gompertz models share the same \(K\), both rates fall to zero as \(N\) approaches \(K\), but they fall according to different functions. Exponential growth has no such decline.

Absolute rates also matter because they describe the actual burden change per day. Logistic absolute growth is \(rN(1-N/K)\), which reaches its maximum at \(N=K/2\). Gompertz absolute growth is \(\beta N\ln(K/N)\), which reaches its maximum at \(N=K/e\), where \(e\) is the base of natural logarithms. Both absolute rates tend to zero as \(N\) approaches \(K\), though their trajectories approach the limit differently.

Justification: For the logistic law, write the absolute rate as \(rN-rN^2/K\). Its derivative with respect to \(N\) is \(r(1-2N/K)\), which is zero at \(N=K/2\), positive below that point, and negative above it. For the Gompertz law, differentiate \(\beta N\ln(K/N)\) with respect to \(N\): the derivative is \(\beta[\ln(K/N)-1]\), zero at \(N=K/e\), positive below it, and negative above it. These stationary points are maxima because the derivatives change from positive to negative.

At small positive burdens, the logistic per-capita rate is close to \(r\), since \(1-N/K\) is close to one. The Gompertz per-capita rate can be substantially larger, depending on \(K/N\). However, a high per-capita rate does not necessarily mean a large absolute increase: \(\beta N\ln(K/N)\) tends to zero as \(N\) tends to zero, because \(N\ln(K/N)\) tends to zero. At large burdens near \(K\), both capacity-limited models predict slowing growth; the exponential model continues to accelerate in absolute terms.

Worked Example: Comparing Ten-Day Predictions From a Small Burden

Consider hypothetical untreated tumors beginning at \(N(0)=50\ \text{mm}^3\), with \(K=1{,}000\ \text{mm}^3\) for the logistic and Gompertz models. Set the exponential per-capita rate to \(0.200\ \text{day}^{-1}\). For an informative comparison at the starting burden, calibrate the other models to have the same initial per-capita rate, \(0.200\ \text{day}^{-1}\).

For the logistic model, \(g(50)=r(1-50/1{,}000)\). Therefore \(r(0.95)=0.200\), so \(r=0.200/0.95=0.2105\ \text{day}^{-1}\). For the Gompertz model, \(\beta\ln(1{,}000/50)=0.200\), giving:

\beta=\frac{0.200}{\ln(1{,}000/50)} =\frac{0.200}{\ln(20)} \approx 0.06676\ \text{day}^{-1}.

After 10 days, the exponential prediction is:

N(10)=50e^{0.200(10)} =50e^2 \approx 369.5\ \text{mm}^3.

For the logistic model, substitute \(r=0.200/0.95\) and \(K/N(0)-1=19\):

N(10)=\frac{1{,}000}{1+19e^{-(0.200/0.95)(10)}} \approx \frac{1{,}000}{1+19(0.1218)} \approx 301.7\ \text{mm}^3.

For the Gompertz model, use the unrounded expression for \(\beta\) in the substitution:

N(10)=1{,}000\exp\left\{\ln(0.05)e^{-[0.200/\ln(20)](10)}\right\} \approx 215.1\ \text{mm}^3.

The models were calibrated to have the same instantaneous proportional growth rate at \(50\ \text{mm}^3\), yet their ten-day predictions differ. The exponential trajectory rises fastest over this interval, while the Gompertz trajectory rises least. That ordering is a result of these specified equations and parameters, not a universal ranking for every possible calibration or time interval.

Predictions at Larger Burdens

The models’ differences are especially consequential when extrapolating toward large burdens. With a shared \(K\), logistic and Gompertz predictions remain below that limit when starting below it and approach it over time. Their slopes, however, need not be similar at a given burden. Exponential predictions have no \(K\), and eventually exceed any fixed capacity.

Worked Example: Comparing Instantaneous Rates Across Size

Use the parameters from the preceding example: exponential \(r=0.200\ \text{day}^{-1}\), logistic \(r=0.2105\ \text{day}^{-1}\), Gompertz \(\beta=0.200/\ln(20)\ \text{day}^{-1}\), and \(K=1{,}000\ \text{mm}^3\) for the capacity-limited models. Compare the instantaneous rates at \(N=500\ \text{mm}^3\).

For exponential growth, \(g=0.200\ \text{day}^{-1}\), so the absolute rate is \(0.200(500)=100.0\ \text{mm}^3/\text{day}\). For logistic growth:

g=0.2105(1-500/1{,}000)\approx 0.1053\ \text{day}^{-1}, \qquad \frac{dN}{dt}\approx 500(0.1053)=52.63\ \text{mm}^3/\text{day}.

For Gompertz growth:

g=\frac{0.200}{\ln(20)}\ln(1{,}000/500) \approx 0.04628\ \text{day}^{-1}, \qquad \frac{dN}{dt}\approx 500(0.04628)=23.14\ \text{mm}^3/\text{day}.

At this burden, exponential growth has the largest absolute rate, followed by logistic and then Gompertz growth. The values illustrate why the two rate definitions should not be conflated: a per-capita rate is not itself a burden increment.

Worked Example: Forecasting From a Larger Starting Burden

Now suppose the hypothetical starting burden is \(500\ \text{mm}^3\), while the model parameters remain as above. This deliberately changes the initial size without recalibrating the rates. At day 10, exponential growth predicts:

N(10)=500e^2\approx 3{,}695\ \text{mm}^3.

For the logistic model, \(K/N(0)-1=1\), so:

N(10)=\frac{1{,}000}{1+e^{-(0.200/0.95)(10)}} \approx \frac{1{,}000}{1+0.1218} \approx 891.4\ \text{mm}^3.

For Gompertz growth, \(\ln(K/N(0))=\ln 2\):

N(10)=1{,}000\exp\left[-\ln(2)e^{-[0.200/\ln(20)](10)}\right] \approx 1{,}000e^{-0.3555} \approx 700.8\ \text{mm}^3.

These are model-based predictions, not observations. The exponential result exceeds the capacity used in the other two models because exponential growth has no upper limit. The logistic and Gompertz predictions differ despite sharing the same \(K\), because their size-dependent rates differ. Since this example does not recalibrate each model at \(500\ \text{mm}^3\), their initial proportional rates are not equal there.

Choosing and Diagnosing a Growth Law

There is no universally best growth law independent of the question, observed size range, and intended prediction. Exponential growth can be a useful local approximation when burden is far below any relevant capacity and proportional growth is roughly stable. Logistic growth is a reasonable choice when a finite capacity and a rate that declines approximately linearly with burden on a proportional scale are scientifically useful. Gompertz growth represents a logarithmically declining per-capita rate and may be useful when observed trajectories show strong slowing across size.

Feature	Exponential	Logistic	Gompertz
Per-capita rate	Constant \(r\)	\(r(1-N/K)\)	\(\beta\ln(K/N)\)
Upper limit in the model	None	\(K\)	\(K\)
Small-burden rate behavior	Constant proportional rate	Per-capita rate approaches \(r\)	Per-capita rate increases as \(N\) decreases
Absolute-rate maximum below \(K\)	No finite maximum	At \(K/2\)	At \(K/e\)

Diagnostics should focus on the implications of a fitted model, not only on whether its curve passes near observed points. Plot predicted trajectories and their rates across the full size range relevant to the scientific question. Look for systematic residual patterns, implausible extrapolation, or a fitted \(K\) that is unsupported by data because observations never show slowing. If observations cover only a narrow range far below \(K\), logistic and Gompertz trajectories may look similar to each other or to exponential growth over that interval. A good local fit then does not establish which law is appropriate for long-range prediction.

The main alternative to choosing a single law is to compare candidate laws using the same observations and a clearly stated prediction objective. Such comparisons should account for their different parameter meanings and complexity; matching parameter symbols or fixing the same numerical rate across models does not generally ensure equal initial growth rates. If the scientific purpose is prediction at large burdens, assess performance in that range rather than relying only on fit near the observed starting sizes. No curve comparison removes the need to question whether a constant environment and a single burden variable are adequate assumptions.

Common Pitfalls and Limitations

Comparing parameters as if they meant the same thing. The exponential and logistic parameter \(r\) is not generally the Gompertz parameter \(\beta\). Compare implied rates at specified burdens, or explicitly calibrate models to a common rate at a stated burden.
Confusing proportional and absolute growth. A high per-capita rate at small \(N\) does not imply a large absolute increase. Compute \(N g(N)\) when the question concerns burden change per day.
Treating \(K\) as a measured biological constant by definition. In these equations, \(K\) is a model parameter representing a limiting burden. Its interpretation and support depend on the data and modeling assumptions.
Extrapolating a local fit without checking its consequences. A trajectory that fits a short interval can diverge substantially from other laws outside the observed size range.

All three laws omit potentially important features, including changing resources, spatial structure, measurement error, and tumor heterogeneity. The Gompertz per-capita rate’s unbounded increase as \(N\) tends to zero is a particularly clear reminder that a mathematical law should not be extrapolated beyond its scientifically plausible range. These untreated growth laws also do not specify how therapy changes tumor burden.

Key takeaway: Compare growth laws through their size-dependent rates and the predictions those rates imply. Exponential growth keeps a constant proportional rate; logistic growth declines linearly with \(N/K\); and Gompertz growth declines with \(\ln(K/N)\). Similar fits over a limited range can conceal very different predictions elsewhere.

Check Your Understanding

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

For a logistic model with \(r=0.30\ \text{day}^{-1}\), \(K=800\ \text{mm}^3\), and \(N=200\ \text{mm}^3\), calculate the per-capita and absolute growth rates.
At what burden is Gompertz absolute growth maximal, and why is that not the same as the burden where Gompertz per-capita growth is maximal?
Why can two capacity-limited models fit observations at small burdens yet make different large-burden predictions?
An exponential model predicts burden above \(K\) at a future time, while a logistic model approaches \(K\). Is that a mathematical error? Explain.

Solutions. (1) The logistic per-capita rate is \(0.30(1-200/800)=0.225\ \text{day}^{-1}\). The absolute rate is \(200(0.225)=45.0\ \text{mm}^3/\text{day}\). (2) The Gompertz absolute rate is maximal at \(K/e\), as follows by differentiating \(\beta N\ln(K/N)\). Its per-capita rate \(\beta\ln(K/N)\) decreases as \(N\) increases, so it has no interior maximum over \(0<N<K\); it is largest toward the small-burden boundary. (3) Over a narrow small-burden range, the data may not reveal enough curvature to distinguish their different rate functions. Their different size-dependent rates then produce divergent predictions as burden approaches \(K\). (4) No. Exponential growth has no carrying capacity, so it may exceed the \(K\) used by a different model. That discrepancy is a consequence of different assumptions, not an algebraic error.