Bacterial Population Growth

No panic and do not run away because of seeing some mathematical formula : maths are fun and some parts are understandable by everyone ! Morever, we have tried to illustrate our work with figures which could be understood for themselves, without knowing the method to get it. We will just add one or two -for the warrior- star(s) to parts which need more mathematical background in order to don't disgust not-mathematical readers from mathematics.

Introduction

This part aims to predict the bacterial population growth on an ellipsoidal object - a fake lemon in practice - over time.

Our work will focus on two different aspects of the bacterial population growth : first, we will study the overall growth, exprimed in concentration of bacteria, then we will have a look on the growth of a single bacterial colony, coming from one bacterium.

Overall growth

We are here considering a bacterial population uniformly spread on a surface in the euclidian space, in crowd-free conditions and with unlimited food resource.

Pure birth process

In this part, we assume that :

• bacteria do not die,
• they develop without interacting with each other,
• the birth rate, $\lambda$, is the same for all the organisms, regardless of their age and does not change with time.

Determininistic model

Let $N(t)$ denote the population size at time $t$.

Then in the subsequent small time interval of length $h$ the increase in the population due to a single organism is $\lambda\times h$ - i.e the rate $\times$ the time - so the increase in size due to all $N(t)$ organisms is $\lambda\times h\times N(t)$. Thus \[ N(t+h) = N(t) + \lambda h N(t) \] which on dividing both sides by h gives \[ \frac{N(t+h)-N(t)}{h} = \lambda N(t) \] Letting $h$ approach zero then yields the differential equation \[ \frac{dN(t)}{dt} = \lambda N(t) \] which integrates to give \[ N(t) = N_0 \exp{\lambda t} \] where $N_0$ denotes the initial population size at time $t=0$. This form for $N(t)$ is known as the Malthusian expression for population development, and shows that the simple rules we used rises to exponential growth.

Stochastic model

A deterministic model only give us an average solution of the problem. In order to take into account the unpredictability of biology, we need a stochastic approach. Usualy, the stochastic model converge to a limit which is similar to the deterministic model. A stochastic model is all the more pertinent as the initial bacterial population is small. It give us an array of probabilities, describing each possible state for the population at time t.

As in the deterministic model, we call $\lambda$ the birth rate : in a short time interval of length $h$ the probability that any particular cell will divide is $\lambda h$. Then for the population to be of the size $N$ at the time $t+h$, either it is of the size $n$ at time $t$ and no birth occurs in the subsequent short interval $(t,t+h)$, or else it is of size $N-1$ at time $t$ and exactly one birth occurs in $(t,t+h)$. In fact, by choosing $h$ sufficiently small we may ensure that the probability of more than one birth occuring is negligible.

Since the probability of $N$ increasing to $N+1$ in $(t,t+h)$ is $\lambda h N$, it follows that the probability of no increase in $(t,t+h)$ is $1-\lambda h N$. Similarly, the probability of $N-1$ increasing to $N$ in $(t,t+h)$ is $\lambda(N-1)h$. Thus on denoting \[ p_N(t)(t) = \mathbb{P}(population~is~of~size~N~at~time~t) \] we have \[ p_N(t+h) = p_N(t)\times\mathbb{P}(no~birth~in~(t,t+h)) + p_{N-1}(t)\times\mathbb{P}(one~birth~in~(t,t+h)) \] i.e \[ p_N(t+h) = p_N(t)\times(1-\lambda N h)+p_{N-1}(t)\times\lambda(N-1)h \] On dividing both sides by $h$ \[ \frac{p_N(t+h)-p_N(t)}{h} = - \lambda N p_N(t)+\lambda(N-1)p_{N-1}(t) \] and as it approaches zero this becomes \[ \frac{dp_N(t)}{dt} = \lambda(N-1)p_{N-1}(t)-\lambda p_N(t) \] for $N=N_0,~N_0+1,...$.

The solution of the above give equation is \[ p_N(t) = \left( \begin{array}{c} N-1 \\ N_0-1 \end{array} \right) e^{-\lambda N_0 t}(1-e^{-\lambda t})^{N-N_0} \quad;\quad N=N_0, N_0+1,... \] which is the negative binomial distribution where, for conveniance, we have written $N(0)$ as $N_0$. In a pratical way, this differential equation can be solved by a variety of theoretical techniques. While we are more interested here with the result than the mathematical formulae, we let the readers interested in the proof consult Bailey's book The elements of stochastic processes [Bai], for instance.

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As our solution is of standard negative binomial form, the mean -i.e the average value- and the variance -i.e the average way we move away from the mean- are given by : \[ m(t)=N_0e^{\lambda t}\quad and\quad V(t)=N_0e^{\lambda t}(e^{\lambda t}-1) \]

Especially, we see that the mean of the stochastic model is exactly what we have found by having a deterministic approach.

Simple birth-death process

In fact, the model we exposed in the previous part was not realistic and we have to consider that our bacteria could also die. We introduce here the death rate $\mu$ which is also supposed be the same for all bacteria, regardless of their age and don't change with time, and we still asume that bacteria develop whihout interacting with each other.

Determininistic model

We still have the same notation and $N(t)$ denote the population size at time $t$.

We proceed in a same way than in the pure birth process but, this time, death will lead to a decrease of the population, that's why the $\mu$ is preceded by a minus : \[ \frac{dN(t)}{dt} = (\lambda-\mu) N(t) \] which integrates to give \[ N(t) = N_0 \exp{\lambda t} \] where $N_0$ denotes the initial population size at time $t=0$.

We still find an exponential growth, but the coefficient $(\lambda-\mu)$ can now be either positive or either negative in function of the values of $\lambda$ and $\mu$.

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Stochastic model

Establishment of the model

Analysis of the stochastic behaviour follows along exactly the same lines as for the pure birth process, except that in the sort time $(t,t+h)$, there is now a probability $\lambda h$ that a particular bacterium gives birth \emph{and} a probability $\mu h$ that it dies. With a population of size $N(t)$ at sime $t$, the probability that no events ocures is therefore $1-\lambda Nh-\mu Nh$, since $h$ is assumed to be sufficiently smal to ensure that the probability of more than one event occuring in $(t,t+h)$ is negligible.

As state $N$ can be reached from states $N-1$ (by a birth), $N+1$ (by a death) or $N$ (either), we find : \[ p_N(t+h) = p_N(t)\times(1-(\lambda+\mu)N h)+p_{N-1}(t)\times\lambda(N-1)h+p_{N+1}(t)\times\mu(N+1)h+o(h) \] Dividing by $h$ and letting $h$ approach zero then yields the set of equations \[ \left\{ \begin{array}{l} {\displaystyle \frac{dp_N(t)}{dt} = \lambda(N-1)p_{N-1}(t)-(\lambda+\mu)N p_N(t)+\mu(N+1)p_{N+1}(t)} \\ {\displaystyle p_N(0)=\delta_{N,N_0}} \end{array}\right. \] over $N=0,1,2\ldots$ and $t\geqslant0$ on which, for $N=0$, $p_{-1}(t)$ is identically zero.

Generating function *

For the following part, the reader could refer to the book of Cox and Miller, The Theory of Stochastic Processes [Mil]. Let \[ G(z,t) = \sum_{n=0}{\infty}p_n(t)z^n \] thus, $P_N(t)$ will be the coefficient before $z^N$ on $G(z,t)$ We multiply the previous equation by $(z^0,z^1,z^2,\ldots)$ and add to obtain : \[ \begin{array}{rcl} {\displaystyle \sum_{n=0}^{\infty} \frac{dp_n(t)}{dt} z^n} &=& {\displaystyle \sum_{n=0}^{\infty} (\lambda(n-1)p_{n-1}(t)-(\lambda+\mu)p_n(t)+\mu(n+1)p_{n+1}(t))z^n} \\ &=& {\displaystyle \lambda \sum_{n=0}^{\infty}p_{n-1}(t)(n-1)z^n - (\lambda+\mu)\sum_{n=0}^{\infty}p_n(t)nz^n + \mu\sum_{n=0}^{\infty}p_{n+1}(n+1)z^n} \end{array} \] and while we know that $p_{-1}=0$ \[ \begin{array}{rcl} {\displaystyle \sum_{n=0}^{\infty} \frac{dp_n(t)}{dt} z^n} &=& {\displaystyle \lambda \sum_{n=0}^{\infty}p_{n}(t)(n)z^{n+1} - (\lambda+\mu)\sum_{n=0}^{\infty}p_n(t)nz^n + \mu\sum_{n=0}^{\infty}p_{n}(n)z^{n-1}} \\ &=& {\displaystyle \lambda z^2\sum_{n=0}^{\infty}p_{n}(t)(n)z^{n-1} - (\lambda+\mu)z\sum_{n=0}^{\infty}p_n(t)nz^{n-1} + \mu\sum_{n=0}^{\infty}p_{n}(n)z^{n-1}} \\ &=& {\displaystyle (\lambda z^2-(\lambda+\mu)z+\mu) \sum_{n=0}^{\infty}p_{n}(t)(z^n)'} \end{array} \] and finally \[ \begin{array}{rcl} {\displaystyle \frac{\partial G(z,t)}{\partial t}} &=& {\displaystyle (\lambda z^2-(\lambda+\mu)z+\mu) \frac{\partial G(z,t)}{\partial z}(z,t)} \\ &=& {\displaystyle (\lambda z-\mu)(z-1)\frac{\partial G(z,t)}{\partial z}(z,t)} \end{array} \] which can be written as \[ \partial_tG(z,t)+f(z)\partial_zG(z,t) = 0\quad where\quad f(z) := -(\lambda z-\mu)(z-1) \]

Lax-Milgram theorem **

We will try to apply the Lax-Milgram theorem in the following paragraph. Readers which are not familiar with the theory of distribution and how to use it in order to solve partial derivative equation (PDE) could consult the notes of N.Burq and P.Gerard Contrôle optimal des équations aux dérivees partielles [Bur].

As from now, we work in the distributions space, $\mathcal{D}(\mathbb{R^2})$. To all agree with the notation, we remind (or not) the definition of a Sobolev space. First we define \[ \mathcal{H}^1(\mathbb{R^2})=\{\phi\in\mathbb{L}_2(\mathbb{R^2})~|~\triangledown\phi\in\mathbb{L}_2(\mathbb{R}) \] In particular, we have \[ \mathcal{H}^1(\mathbb{R^2})\subset\mathcal{C}^{\infty}_0=\{\phi\in\mathcal{C}^\infty(\mathbb{R}^2)~|~Supp(\phi)~is~compact\} \] thus justify to set down \[ \mathcal{H}^1_0(\mathbb{R^2}) = \overline{\mathcal{C}^{\infty}_0}^{\mathcal{H}^1}\subset\mathcal{H}^1 \] where $\overline{\ldots}^{\mathcal{H}^1}$ is a notation to say adherance in $\mathcal{H}^1$. $\mathcal{H}^1_0$ is an exemple of a Sobolev space. So, like the whole Sobolev space, it is an Hilbert space and we note $<~|~>$ his scalar product. \[ \forall u,v\in\mathcal{H}^1_0,\quad < u~|~v>~=~< u~|~v>_{\mathcal{H}^1_0}~:=~< u~|~v>_{\mathbb{L}^2}+< \triangledown u~|~\triangledown v>_{\mathbb{L}^2} \]

\[ \begin{array}{l} {\displaystyle \forall(z,t)\in\mathbb{R}^2,\quad\partial_tG(z,t)+f(z)\partial_zG(z,t) = 0\quad} \\ {\displaystyle \quad\quad ssi \quad\forall\phi\in\mathcal{C}^{\infty}_0(\mathbb{R}^2),\quad< \partial_tG+f\partial_zG~|~\phi> =0} \\ {\displaystyle \quad\quad ssi \quad\forall\phi\in\mathcal{C}^{\infty}_0(\mathbb{R}^2),\quad \int{\partial_tG\phi+f\partial_zG\phi} = 0} \\ {\displaystyle \quad\quad ssi \quad\forall\phi\in\mathcal{C}^{\infty}_0(\mathbb{R}^2),\quad a(G,\phi) = 0\quad where\quad a(G,\phi)=\int{\partial_tG\phi+f\partial_zG\phi}} \end{array} \]

Because of the lenearity of the integral, $a$ is a form bilinear. So we just have to proof that $a$ is continue ant coercive.

• $a$ is continue :

Let $u$ and $v$ in $\mathcal{H}^1_0$. Because of the density of $\mathcal{C}^{\infty}_0$ in $\mathcal{H}^1_0$, it exists a sequence $(\phi_n)_{n\in\mathbb{N}}$ of $\mathcal{C}^{\infty}_0$ which converge on $v$. Let $n\in\mathbb{N}$ \[ \begin{array}{rcl} {\displaystyle |~a(u,\phi_n)~|=|~\int_{\mathbb{R}^2}{\partial_tu\phi_n+f\partial_zu\phi_n}~|} &=& {\displaystyle |~\int_{Supp(\phi_n)}{\partial_tu\phi_n+f\partial_zu\phi_n}~|} \\ &\leqslant& {\displaystyle \int_{Supp(\phi_n)}{|~\partial_tu\phi_n+f\partial_zu\phi_n~|}} \\ &\leqslant& {\displaystyle (1+\int_{Supp(\phi_n)}{f})||\triangledown u||_{\mathbb{L}^2}||\triangledown\phi_n||_{\mathbb{L}^2}} \\ &\leqslant& {\displaystyle (1+\underset{x\in Supp(\phi_n)}{\sup}f)||u||_{\mathcal{H}^1_0}||\phi_n||_{\mathcal{H}^1_0}} \end{array} \] Let ${\displaystyle M= \underset{n\in\mathbb{N}}{\sup}\underset{x\in Supp(\phi_n)}{\sup}f)}$. And the Heine theorem assure that $M$ is finit. Actually, $f$ is continue on the compact $Supp(\phi_n)$ for all $n$ in $\mathbb{N}$ so is bound on it. By passing to the limit, we find \[ {\displaystyle |~a(u,v)~|\leqslant (1+M)||u||_{\mathcal{H}^1_0}||\phi_n||_{\mathcal{H}^1_0}} \] which is what we search.

• $a$ is coercive :

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So, according to the Lax-Milgram theorem \[ \exists ! G\in\mathcal{H}^1_0,\quad\forall\phi\in\mathcal{H}^1_0,\quad a(G,\phi)=0 \] With other words, it means that, in the distribution space, if we can exhibit a solution, it is the only solution of this problem. And the solution exists. If the solution we find is sufficiently regular (in a way to define more precisly), general theorems assure that the solution we find in not only a solution in the distribution space but is also a solution in "the real life".

Precisely, we can verify with a calcul that \[ \gamma(z,t)~:=~\left(\frac{\mu(1-z)-(\mu-\lambda z)e^{-(\lambda-\mu)t}}{\lambda(1-z)-(\mu-\lambda z)e^{-(\lambda-\mu)t}}\right)^{N_0} \] is solution to the equation, i.e that \[ \forall(z,t)\in\mathbb{R}^2, \partial t\gamma(z,t)+f(z,t)\partial z\gamma(z,t)=0 \] The calcul is verry long and not really interessant (it suffices to derive $\gamma$ and substitue in the equation) that's why we will not detail it in this text...

So, while the solution exists and is unique according th the Lax-Milgram theorem, we have found it and \[ \forall(z,t)\in\mathbb{R}^2,~G(z,t)~=~\left(\frac{\mu(1-z)-(\mu-\lambda z)e^{-(\lambda-\mu)t}}{\lambda(1-z)-(\mu-\lambda z)e^{-(\lambda-\mu)t}}\right)^{N_0} \] and for all time $t$, $P_N(t)$ is the coefficient before $z^N$ in the previous expression.

Coefficient of Variation

In this small parts, we will just try to have an idea of the influence of the value of $\lambda$ and $\mu$ on the probability. To do that, we calcul the mean and the variance of population size. Actually, while we consider a discret probability, \[ m(t) = \sum_{n=0}^{\infty}{n p_n(t)} = \ldots = N_0 e^{(\lambda-\mu)t} \] and by a similar calcul, if $\lambda\neq\mu$, \[ V(t) = N_0\frac{\lambda+\mu}{\lambda-\mu}e^{(\lambda-\mu)t}(e^{(\lambda-\mu)t}-1) \]

Unlike $m(t)$ and $V(t)$ depend not only on the difference between the birth and the death rates, but also on their absolute magnetudes. This is what we should expect, because predictions about the future size of a population will be less precise if birth and death occur in rapid succesion than if they occur only occasionally.

We defin the coefficient of variation $\displaystyle CV(t)~:=~\frac{\sqrt{V(t)}}{m(t)}$ which qualify the variation of the system and we will study the effect of the relative values of $\lambda$ and $\mu$. \[ \begin{array}{rcl} \forall\lambda,\mu,~\forall t, CV(t) &=& {\displaystyle\sqrt{N_0\frac{\lambda+\mu}{\lambda-\mu}e^{(\lambda-\mu)t}(e^{(\lambda-\mu)t}-1)}\times\frac{1}{N_0}e^{-(\lambda-\mu)t}} \\ &=& {\displaystyle \sqrt{\frac{\lambda+\mu}{N_0(\lambda-\mu)}}\times\sqrt{e^{(\lambda-\mu)t}-1}\times e^{-\frac{1}{2}(\lambda-\mu)t}}

\end{array} \]

If $\lambda\gg\mu$, \[ CV(t)~\sim~\sqrt{\frac{\lambda+\mu}{N_0(\lambda-\mu)}} \] On the contrary, if $\lambda\ll\mu$, \[ CV(t)~\sim~\sqrt{\frac{\lambda+\mu}{N_0(\lambda-\mu)}}e^{\frac{1}{2}(\lambda-\mu)t} \]

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At this point it is worth reflecting on a a criticism sometimes levelled against this model of exponential population growth, namely that if $\lambda$ exceeds $\mu$ then it ultimatly leads to populations so large that their existence is physically impossible.

A far more serious, but often neglected, question is How far into the future is ecological prediction, based on simple models, feasible ? The answer must depend on the situation being considered, since it is influenced to a large extent by biological factors such as over what length of time the (actual) birth and death rates can be expected to remain reasonably constant and how large $N(t)$ may become before organisms can no longer be assumed to develop independently of each other. Clearly, whenever a biological process, no matter how innocently simple it may first apprear, is being modelled, the underlying assumption used in the contruction of the model must be constantly questioned. The feasibly predictable future may well be disapointingly short.

Probability of extinction

References

[Bai] Norman T.J Bailey, The Elements of Stochastic Processes with Applications to the Natural Sciences, New York, Wiley (1964).

[Bur] Nicolas Burq & Patrick Gérard, Contrôle optimal des équations aux dérivées partielles, Ecole polytechnique (2002)

[Mil] D.R. Cox & H.D. Miller, The Theory of Stochastic Processes, London : Methuen (1965)

[Ren] Eric Renshaw, Modelling Biological Populations in Space and Times, Cambridge university press (1991).