Deriving Gompertzian Tumor Growth From a Declining Growth Rate

From Declining Per-Capita Growth to a Tumor Trajectory

A tumor can continue to grow while its relative growth rate slows. The distinction matters: the per-capita growth rate describes change relative to current burden, whereas the absolute rate of increase also depends on how much burden is already present. The Gompertz model gives a precise trajectory when per-capita growth declines in proportion to the logarithm of tumor burden.

The model is deterministic and assumes a single positive burden $N(t)$, measured in a consistent unit such as viable cells. It treats the tumor as a homogeneous burden and assumes that its per-capita rate depends on current burden through a linear relation with log burden. In the derivation below, the decline parameter is constant in time; treatment effects, changing environments, and random variation are not included.

Definition: Let $g(N)$ be the per-capita growth rate, so that $dN/dt=N g(N)$. The Gompertz assumption is that $g(N)$ declines linearly with log burden: $$ g(N)=g_0-\beta\ln\left(\frac{N}{N_0}\right), \qquad \beta>0. $$ Here $N_0=N(0)>0$ is initial burden, $g_0=g(N_0)$ is the initial per-capita growth rate, and $\beta$ is the rate at which per-capita growth declines as log burden increases. Both $g_0$ and $\beta$ have units of inverse time.

The ratio $N/N_0$ makes the logarithm dimensionless. If $g_0>0$, burden initially grows. As $N$ rises, $\ln(N/N_0)$ rises, reducing $g(N)$. The rate reaches zero at an implied asymptotic burden, not at a finite time.

Deriving the Closed-Form Trajectory

The differential equation is $dN/dt=N g(N)$. To solve it, track log burden relative to its initial value: let $x(t)=\ln(N(t)/N_0)$. By differentiation, $dx/dt=(1/N)dN/dt=g(N)$. Substituting the assumed rate gives a linear differential equation in $x$:

\frac{dx}{dt}=g_0-\beta x,\qquad x(0)=0.

An integrating factor gives the solution directly. Multiply both sides by $e^{\beta t}$; the left side becomes the derivative of $e^{\beta t}x(t)$. Integrating from $0$ to $t$ and using $x(0)=0$ yields:

\frac{d}{dt}\left(e^{\beta t}x(t)\right)=g_0e^{\beta t}, \qquad e^{\beta t}x(t)=\frac{g_0}{\beta}\left(e^{\beta t}-1\right), \qquad x(t)=\frac{g_0}{\beta}\left(1-e^{-\beta t}\right).

Since $x(t)=\ln(N(t)/N_0)$, exponentiating gives the burden trajectory. The same solution can be written using $K$, the burden at which the model's per-capita rate is zero. Solving $g(K)=0$ gives $K=N_0e^{g_0/\beta}$. Substituting this identity produces the usual Gompertz form:

N(t)=N_0\exp\left[\frac{g_0}{\beta}\left(1-e^{-\beta t}\right)\right] =K\exp\left[-\ln\left(\frac{K}{N_0}\right)e^{-\beta t}\right], \qquad K=N_0e^{g_0/\beta}.

For $g_0>0$, $K>N_0$, and $N(t)$ increases toward $K$ as $t$ tends to infinity. The parameter $g_0$ is the initial per-capita growth rate because $N'(0)/N_0=g_0$. The parameter $\beta$ controls how quickly that rate declines: differentiating the model's per-capita rate along the trajectory gives $g'(t)=-\beta g(t)$, and hence $g(t)=g_0e^{-\beta t}$. The quantity $K$ is the model's asymptotic burden; it is not automatically a measured biological carrying capacity.

Key result: With $N_0>0$, $\beta>0$, and $g_0>0$, the Gompertz trajectory is positive, grows monotonically, and approaches $K=N_0e^{g_0/\beta}$. Its per-capita growth rate is $g_0e^{-\beta t}$, so it is positive at every finite time and tends to zero asymptotically.

Curvature: Increasing Burden Does Not Always Mean Deceleration

A common oversimplification is to assume that a tumor following the Gompertz model must be concave downward at every time. The burden increases for $g_0>0$, but its curvature depends on the size of the current per-capita rate relative to $\beta$. Since $N'=Ng$ and $g'=-\beta g$, differentiating once more gives:

N''(t)=N(t)\left[g(t)^2+g'(t)\right] =N(t)g(t)\left[g(t)-\beta\right].

For an initially growing tumor, $N(t)>0$ and $g(t)>0$. Thus $N''(t)>0$ when $g(t)>\beta$, and $N''(t)<0$ when $g(t)<\beta$. If $g_0>\beta$, the trajectory initially accelerates and changes to concave downward when $g(t)=\beta$, at $t=\ln(g_0/\beta)/\beta$. If $g_0\leq\beta$, it is concave downward from the start, except that $N''(0)=0$ when $g_0=\beta$.

Worked Example: Calculate Burden and Locate the Inflection

Suppose a hypothetical tumor has $N_0=1.00\times10^8$ viable cells, an initial per-capita growth rate $g_0=0.0800\ \text{day}^{-1}$, and $\beta=0.0400\ \text{day}^{-1}$. Calculate its burden after 20 days and determine whether it is accelerating or decelerating then.

First calculate the asymptotic burden: $$ K=N_0e^{g_0/\beta} =(1.00\times10^8\ \text{cells})e^{0.0800/0.0400} =(1.00\times10^8)e^2 =7.39\times10^8\ \text{cells}. $$

At $t=20$ days, the exponent in the trajectory is: $$ \frac{g_0}{\beta}\left(1-e^{-\beta t}\right) =2\left(1-e^{-0.800}\right) =2(1-0.4493) =1.101. $$ Therefore, $$ N(20)=(1.00\times10^8)e^{1.101} \approx3.01\times10^8\ \text{cells}. $$

The per-capita rate at 20 days is $g(20)=0.0800e^{-0.800}=0.0359\ \text{day}^{-1}$, which is less than $\beta=0.0400\ \text{day}^{-1}$. The tumor is still growing, because $g(20)>0$, but its trajectory is concave downward at that time. Initially $g_0>\beta$, so it was concave upward; the inflection occurred at $\ln(0.0800/0.0400)/0.0400=17.3$ days.

Using the Parameters in Either Direction

The model can be expressed in terms of $N_0$, $g_0$, and $\beta$, or in terms of $N_0$, $K$, and $\beta$. Their relationship is $g_0=\beta\ln(K/N_0)$. This identity helps interpret parameters and check whether a proposed parameterization is internally consistent. In particular, $K>N_0$ corresponds to initial growth, $K=N_0$ to zero initial growth, and $K<N_0$ to initial decline under this model.

Worked Example: Recover the Decline Parameter from Initial Growth and Asymptote

A hypothetical model starts at $N_0=2.00\times10^8$ cells, has initial per-capita growth rate $g_0=0.100\ \text{day}^{-1}$, and is specified to approach $K=1.00\times10^9$ cells. Find $\beta$ and calculate burden at 14 days.

From $K=N_0e^{g_0/\beta}$, take logarithms and solve for $\beta$: $$ \beta=\frac{g_0}{\ln(K/N_0)} =\frac{0.100\ \text{day}^{-1}}{\ln(1.00\times10^9/2.00\times10^8)} =\frac{0.100}{\ln 5} =0.0621\ \text{day}^{-1}. $$

The exponential decline factor at 14 days is $e^{-\beta t}=e^{-(0.0621)(14)}\approx0.419$. Using the $K$-parameter form, $$ N(14)=K\exp\left[-\ln(K/N_0)e^{-\beta t}\right] =(1.00\times10^9)\exp[-(1.609)(0.419)] \approx5.09\times10^8\ \text{cells}. $$

This predicted burden is between $N_0$ and $K$, as expected for a growing tumor at a finite time. The calculation also shows that a larger $\beta$ means a faster decline in per-capita growth, holding the other parameterized quantities under consideration fixed.

Worked Example: Find the Time to Half the Asymptotic Burden

For a hypothetical tumor, let $N_0=1.00\times10^8$ cells, $K=8.00\times10^8$ cells, and $\beta=0.0500\ \text{day}^{-1}$. How long does the model take to reach $4.00\times10^8$ cells?

Starting from $N(t)=K\exp[-\ln(K/N_0)e^{-\beta t}]$, take logarithms and rearrange: $$ \ln\left(\frac{K}{N(t)}\right) =\ln\left(\frac{K}{N_0}\right)e^{-\beta t}, \qquad t=\frac{1}{\beta}\ln\left[ \frac{\ln(K/N_0)}{\ln(K/N(t))} \right]. $$

Here $\ln(K/N_0)=\ln 8=2.079$, and $\ln(K/N(t))=\ln 2=0.6931$. Thus $$ t=\frac{1}{0.0500\ \text{day}^{-1}} \ln\left(\frac{2.079}{0.6931}\right) =\frac{\ln 3}{0.0500\ \text{day}^{-1}} =22.0\ \text{days}. $$

The target is half of $K$, but the tumor does not reach $K$ at this or any other finite time under the growing Gompertz model. $K$ is a limit, while $4.00\times10^8$ cells is a finite target below it.

Assumptions, Diagnostics, and Alternatives

The derivation is exact for the stated differential equation, but the equation is a biological modeling assumption, not a universal law. It presumes that per-capita growth is a linear function of log burden and that the slope $-\beta$ is constant over the period being modeled. The predicted asymptote also assumes that the same relationship continues indefinitely. A fit to a limited observation period may not establish that long-run limit reliably.

When evaluating whether the assumption is plausible, check whether observed growth slows as burden rises and whether a linear relation is reasonable between estimated per-capita growth and log burden. Be cautious: estimating per-capita rates from differences in noisy burden measurements can amplify measurement error. A trajectory that repeatedly accelerates at larger burdens, changes slope markedly over time, or shows a sustained decline while the model predicts growth may indicate that the simple model is inadequate. Such patterns call for reassessing the model and data assumptions, rather than interpreting $K$ literally.

Approach	Per-capita growth assumption	Useful distinction
Exponential growth	Constant over time and burden	Does not represent a declining per-capita rate.
Gompertz growth	Declines linearly with log burden	Decline slows as the rate approaches zero; burden tends toward $K$.
Logistic growth	Declines linearly with burden under its standard form	Uses a different relationship between burden and per-capita growth.

These models can give similar predictions over a restricted range of observed burdens, so a good fit in that range alone does not establish which growth law is appropriate. The Gompertz model also has no explicit cell compartments, spatial structure, or mechanism for treatment response. Its $K$ is an implication of the assumed rate relationship, not proof of a fixed physical limit in the tumor microenvironment.

Common Mistake

Pitfall: “The per-capita rate declines, so the total burden must be concave downward throughout.” This does not follow. The total rate is $N'=Ng$, and its derivative is $N''=Ng(g-\beta)$. When $g>\beta$, the tumor burden is increasing and concave upward even though its per-capita growth rate is already declining. Check the sign of $g-\beta$ before describing curvature.

Check Your Understanding

Try each question, then check the solutions below.

For $N_0=5.00\times10^7$ cells, $g_0=0.0600\ \text{day}^{-1}$, and $\beta=0.0300\ \text{day}^{-1}$, calculate $K$.
For the parameters in question 1, is the burden initially concave upward or downward? When does its curvature change sign?
If $K=N_0$, what is $g_0$ under the Gompertz parameter relationship?
For $N_0=1.00\times10^8$ cells, $K=4.00\times10^8$ cells, and $\beta=0.0500\ \text{day}^{-1}$, calculate the time to reach $2.00\times10^8$ cells.
Why does a declining per-capita growth rate not imply that tumor burden is declining?

Solutions. (1) $K=N_0e^{g_0/\beta}=(5.00\times10^7)e^2=3.69\times10^8$ cells. (2) Since $g_0=0.0600>\beta=0.0300\ \text{day}^{-1}$, burden is initially concave upward; the sign changes at $t=\ln(2)/0.0300=23.1$ days. (3) $g_0=\beta\ln(K/N_0)=\beta\ln 1=0$. (4) The target is half of $K$, so $t=\ln[\ln 4/\ln 2]/0.0500=\ln 2/0.0500=13.9$ days. (5) Burden increases whenever its per-capita rate remains positive, since $N'=Ng>0$ for $N>0$ and $g>0$; a declining positive rate is still positive.