10 Introducing models of cancer dynamics

Introducing cancer models

More than one in three people will develop some form of cancer during their lifetime. Given this, it is no surprise that cancer research is a highly active field, and, relevant to this course, that there are a great many researchers interested in modelling the dynamics of cancer growth and its treatments. It can also mean it is a difficult subject for some to think about – I have personally lost much-loved family to cancer, and it is only recently I have felt able to start teaching content in this area – but the more research we can do, the sooner better treatments and even cures will come.

In this chapter we will examine a number of different model forms for thinking about how cancer tumours – which we essentially treat as localised populations of cells – grow or proliferate over time (see chapter references). For the most part we will not do any in-depth analysis of the models here, but we will look at one specific model in more detail in the next chapter.

Single variable models

From a mathematical viewpoint we can essentially think about the growth of a population of cancer cells in much the same way as we would the growth of any population. We have seen some of these model structures already in this textbook, but will cover them again here for completeness.

Linear growth (with or without mortality)

The most basic model for growth of a population of cancer cells, [latex]c[/latex] – even more basic model than we have examined before – would be for linear growth, given by,

[latex]\begin{equation} \frac{dc}{dt} = r. \end{equation}[/latex]

While you may spot a few flaws in such a model (for example we have positive proliferation even with zero cancer cells), it has been used to describe the dynamics of certain cancers. There is also no mortality of cells here, or shrinkage of the tumour. Such a term is readily added to give,

[latex]\begin{equation} \frac{dc}{dt} = r-kc. \end{equation}[/latex]

We can solve this model either by separation of variables or an integrating factor to give,

[latex]\begin{equation} c(t) =\frac{r}{k}+\left(c(0)-\frac{r}{k}\right)e^{-kt}. \end{equation}[/latex]

This suggests the tumour will tend towards an intermediate size of [latex]c=r/k[/latex] (which we can also see by just looking for the equilibrium from the ODE).

Exponential growth (with or without mortality)

The more classic example of population growth we saw at the very start of this resource was that of exponential growth, where the population growth depends on the current density, that is,

[latex]\begin{equation} \frac{dc}{dt} = rc. \end{equation}[/latex]

There might then be a question of whether [latex]r[/latex] is purely the proliferation rate or, as we assumed earlier, the difference between proliferation and shrinkage. Let’s say that we do include a separate shrinkage term, the equation can still be solved using separation of variables,

[latex]\begin{align*} \frac{dc}{dt} &= rc-kc,\\ \implies c(t)&=c(0)e^{(r-k)t}, \end{align*}[/latex]

giving either exponential growth or decay of the tumour depending on whether proliferation or shrinkage is greater.

Logistic growth

Again, as we saw back in chapter 1, there is a realism problem with exponential growth in that it predicts growth to infinite numbers of cells. We introduced one approach to deal with this which is to assume a linear decline in the growth rate, [latex]r[/latex], as the population density increases, leading to a carrying capacity at size [latex]K[/latex] after which the population declines. This is given by the ODE,

[latex]\begin{equation} \frac{dc}{dt} = r_0c\left(1-\frac{c}{K}\right). \end{equation}[/latex]

We solved the non-dimensionalised version of this earlier, with the full solution here being given by,

[latex]\begin{equation} c(t) = \frac{Kc(0)}{c(0)+(K-c(0))e^{-r_0t}}. \end{equation}[/latex]

Gompertz growth

The logistic equation assumes the growth rate decreases linearly with the population density, but experimental studies have indicated that the decrease in cell proliferation is often closer to exponential. We can readily make our density-dependent growth rate take a different functional form to represent different scenarios. One classic example that has been used to good effect in cancer modelling is the Gompertz model, with the ODE given as,

[latex]\begin{equation} \frac{dc}{dt} = r_0c\ln\left(\frac{K}{c}\right). \end{equation}[/latex]

Now the individual-level growth rate (I’d like to say per-capita, but it has been pointed out to me that cells do not have heads) is [latex]r_0\ln(K/c)[/latex]. This equals 0 when [latex]c=K[/latex], retaining the meaning of [latex]K[/latex] as a carrying capacity. Note we do have an issue that we cannot let [latex]c=0[/latex]. What is the solution to this?


Using the substitution [latex]u=\ln\left(\frac{K}{c}\right)[/latex], show that the solution to the Gompertz model is,

[latex]\begin{equation} c(t)=K\left(\frac{c(0)}{K}\right)^{e^{-rt}}. \end{equation}[/latex]

Click for solution

If we’re given a substitution we can assume we want to replace all instances of [latex]c[/latex] with some function of [latex]u[/latex]. If [latex]u=\ln(K/c)[/latex], we can find that [latex]du/dc=-1/c[/latex]. Putting this all together we can rewrite the ODE as follows,

[latex]\begin{align*} \frac{1}{c\ln(K/c)}\frac{dc}{dt}&=r\\ \implies -\frac{1}{u}\frac{du}{dc}\frac{dc}{dt}&=r\\ \implies \frac{1}{u}\frac{du}{dt}&=-r. \end{align*}[/latex]

At this point we can now use separation of variables to find the solution for [latex]u[/latex],

[latex]\begin{align*} \ln(u)&=-rt+A\\ \implies u&=A_1e^{-rt}\\ \implies \ln\left(\frac{K}{c}\right)&=A_1e^{-rt} \end{align*}[/latex]

where [latex]A[/latex] is the constant of integration and [latex]A_1=e^A[/latex]. If we have density [latex]c(0)[/latex] at [latex]t=0[/latex] we can find that [latex]A_1=\ln(K/c(0))[/latex]. Then if we remember that [latex]\ln(A/B)=\ln(A)-\ln(B)[/latex], this means that [latex]-\ln(A/B)=\ln(B/A)[/latex]. Using this fact we can then say,

[latex]\begin{align*} \ln\left(\frac{K}{c}\right)&=\ln\left(\frac{K}{c(0)}\right)e^{-rt}\\ \implies \ln\left(\frac{c}{K}\right)&=\ln\left(\frac{c(0)}{K}\right)e^{-rt}\\ \implies \frac{c}{K}&=e^\left[\ln\left(\frac{c(0)}{K}\right)e^{-rt}\right]\\ \implies \frac{c}{K}&=\left(\frac{c(0)}{K}\right)^{e^{-rt}}. \end{align*}[/latex]

We then just multiply the $K$ to the other side to reach the required solution.

This results in a rather more complex solution that we saw for logistic growth, but at the benefit of a curve that often fits real data much better.

Volume-based growth

Let’s return to the exponential growth model for a moment. A more general form for this is given by,

[latex]\begin{equation} \frac{dc}{dt}=rc^b. \end{equation}[/latex]

It has been suggested that the best choice for the power is not [latex]b=1[/latex] (as we implicitly assumed in our approach to the exponential model) but [latex]b=2/3[/latex]. Why might that be? Unlike an ecological population, a cancer tumour forms as a roughly spherical object. If it has volume [latex]V[/latex], then its surface area scales with [latex]V^{2/3}[/latex]. If we assume that all resources that the tumour needs must enter through the outer edge of the tumour, then its growth rate will be dependent on its surface area rather than its volume. Note that this need not be limited to the exponential growth model, and may equally well form the growth rate in more complex models.

Two variable models

All of the model forms we have looked at so far assume that there is only one variable of interest – the density (or maybe volume) of a cancer tumour. However, there are many reasons why we may wish to explore models with two or more variables. We will explore some of these here.

Proliferating and quiescent cells

Not all cancer cells in a tumour are growing all the time, and this may have impacts on the overall dynamics. We might therefore choose to separate out the cells into those that are proliferating and those that are not, which we call quiescent. A simple model structure for this case would be,

[latex]\begin{align} \frac{dP}{dt}&=f(P)-m_1P+m_2Q\\ \frac{dQ}{dt}&=m_1P-m_2Q. \end{align}[/latex]

The function [latex]f(P)[/latex] describes the growth dynamics of the tumour, likely using one of the model forms we saw earlier. There is then a simple linear transfer of cells between the proliferative and quiescent states. The ability to solve this system will depend on the nature of [latex]f(P)[/latex]. If it is linear we will be able to solve the system explicitly. Otherwise we would use our qualitative approaches to find the long-term behaviour. Let’s look at a quick example here.


Find the possible equilibria for [latex]P[/latex] and [latex]Q[/latex] for the system,

[latex]\begin{align*} \frac{dP}{dt}&=f(P)-m_1P+m_2Q\\ \frac{dQ}{dt}&=m_1P-m_2Q. \end{align*}[/latex]

when there is logistic growth of proliferating cells with basic growth rate [latex]r_0[/latex] and the carrying capacity for all cells is [latex]K[/latex].

Click for solution

If we have logistic growth with the carrying capacity determined for all cells we can write our system as,

[latex]\begin{align*} \frac{dP}{dt}&=r_0P\left(1-\frac{P+Q}{K}\right)-m_1P+m_2Q\\ \frac{dQ}{dt}&=m_1P-m_2Q. \end{align*}[/latex]

Since this is non-linear it looks like we will not be able to solve it explicitly. Instead, let us determine the equilibria and their stability. If we set the second ODE to 0 we find $Q=Pm_1/m_2$. If we substitute this into the first ODE and set it to 0 we obtain,

[latex]\begin{equation*} r_0P\left(1-\frac{P(1+m_1/m_2)}{K}\right)=0. \end{equation*}[/latex]

The long-term equilibria are therefore either,

  • [latex]P^*=0[/latex], meaning [latex]Q^*=0[/latex];
  • [latex]P^*=\frac{Km_2}{m_1+m_2}[/latex], meaning [latex]Q^*=\frac{Km_1}{m_1+m_2}[/latex].

In the first case the tumour is absent, while at the second the tumour is present, and its total size is at its carrying capacity, but only a proportion of those cells are proliferating.

We can check the stability of these equilibria by writing out the Jacobian,

[latex]\begin{align*} J=&\left( \begin{array}{cc} r_0-2\frac{r_0P^*}{K}-\frac{r_0Q^*}{K}-m_1 & -\frac{r_0P^*}{K}+m_2 \\ m_1 & -m_2 \end{array} \right). \end{align*}[/latex]

At the no tumour equilibrium this reduces to,

[latex]\begin{align*} J=&\left( \begin{array}{cc} r_0-m_1 & m_2 \\ m_1 & -m_2 \end{array} \right), \end{align*}[/latex]

which gives [latex]tr=r_0-m_1-m_2[/latex] and [latex]\det=-r_0m_2[/latex]. Since the determinant is negative, this equilibrium is always a saddle. For the equilibrium where the tumour is present, if we substitute in the equilibria and then do some cancelling we have,

[latex]\begin{align*} J=&\left( \begin{array}{cc} -\frac{r_0m_2}{m_1+m_2}-m_1 & -\frac{r_0m_2}{m_1+m_2}+m_2 \\ m_1 & -m_2 \end{array} \right). \end{align*}[/latex]

Here we have [latex]tr=-\frac{r_0m_2}{m_1+m_2}-m_1-m_20[/latex] and [latex]\det=r_0m_2>0[/latex], so it is definitely stable. Substituting in some values, it is quite easy to find examples where [latex]tr^2-4\det\lt 0[/latex], meaning we can have a stable spiral into the equilibrium. Note that this is different to if we just had proliferating cells with logistic growth, where no such damped oscillations would be possible.

Drug-resistant and drug-sensitive cells

An important question when thinking about treatment strategies for tumours is whether they can develop drug resistance. If so, we might divide our population into two compartments: sensitive and resistant. We would assume drug-sensitive cells have their proliferation rate reduced through treatment but drug-resistant cells do not. Different assumptions might then be made over whether resistance is a pre-existing trait or if it can be acquired. Similarly, we might explore whether cells can switch back and forth between being sensitive and resistant. Such a model might be represented as,

[latex]\begin{align} \frac{dS}{dt}&=f_S(S)-m_1S+m_2R\\ \frac{dR}{dt}&=f_R(R)+m_1S-m_2R. \end{align}[/latex]

This looks similar to our previous model for proliferating and quiescent cells, except here both types of cell can proliferate, just at different rates. We might also think about whether the transition rates, [latex]m_1[/latex] and [latex]m_2[/latex], depend on the level of treatment.


We have seen just a few examples of model structures here, and there are many more we haven’t covered, for example directly including the effects of treatment or the immune system. In the next chapter we will see another specific example where we think about how a tumour both grows and impacts its own carrying capacity through angiogenesis.

It is worth stressing that these different model forms are not necessarily limited to cancer modelling, and you may find that some of these different scenarios are more or less suited to the system you are interested in.

Key Takeaways

  • We can model growth of tumours in many different ways.
  • Single variable models can often be solved explicitly, and include examples such as logistic growth and Gompertz growth.
  • Two variable models allow us to explore cell dynamics in more biological detail.

Chapter references


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