Epidemiology

Abstract

This chapter gives an introduction into the most basic models for the dynamics of ecosystems. Starting with the simplest case of one single homogeneous population which inhabits an ecosystem, we proceed to two-population models. They are prototypes for the different interaction patterns between species: one species may predate upon the other, the two species may compete for common resources, or they may support each other (mutualism). Typical questions arising are: Why do predators stabilise an ecosystem, or why do similar species coexist though there is the “principle of competitive exclusion”. These ideas are also transferred to the case of three species, leading towards the theory of food chains and food webs. Finally, we develop some methods to understand the complexity of even more complex food webs via the question in how many different ways a system can have developed.

Appendix: Epidemiology

4.1.1 1 Solutions

4.1 We find that S + I = N is constant; thus, the equation for I can be written in the form

$$\displaystyle{I^{{\prime}} =\beta \frac{(N - I)I} {N} -\alpha I = (\beta -\alpha )I(1 - I/K)}$$

where

$$\displaystyle{K = N \frac{\beta -\alpha } {\beta }.}$$

• (a) The stationary points read

  $$\displaystyle{I_{1} = 0,\quad I_{2} = K = N(1 -\alpha /\beta ).}$$

  If we define R ₀ = β∕α (is this reasonable in the view of Definition 4.1) we have simply I ₂ = N(1 − 1∕R ₀). As the r.h.s. of I ^′ is a quadratic polynomial where the leading order has a negative coefficient − N∕β, the larger stationary point is always linearly stable I ₂ ≠ 0. This is, for β < α, the trivial solution I ₁ is linearly stable, for β > α the solution I ₂ is linearly stable. As we have a non-degenerated, one-dimensional system for β ≠ α, the linear stability and the local asymptotic stability are identical.

• (b) The differential equation for I alone given above is a logistic equation, with the solution

  $$\displaystyle{I(t) = \frac{I_{0}} {e^{-(\beta -\alpha )t} + I_{0}(1 - e^{-(\beta -\alpha )t})/K}.}$$

4.2

• (a) The meat industry controls the cattle population according to their requirements, and not according to some natural population dynamics. This is, if μ is appropriately constant and I rather small in comparison with S, there is a constant outflow μ S that is balanced by a constant inflow b. I.e., b ≈ μ S.

• (b) Mass action law implies that the contact rate increases proportional to the population. We may assume that the fraction of (undetected) infected slaughtered cows is proportional to the fraction of infected cows. This is, I∕(I + S) will be the fraction of diseased cows that are used in animal food production. The material of these cows, however, are mixed into a large amount of food. This is, the larger the animal population, the larger the number of healthy cows that will feed on this product. Therefore, the contact rate is approximately proportional to the complete population, and we find the incidence function

  $$\displaystyle{\mathop{\underbrace{(\beta (S + I))}}\limits _{contact\,\,rate}\,\,I/(I + S)\,\,\,S =\beta SI.}$$

• (c) Stationary points:

  Case 1::

  I ₁ = 0. Then, S ₁ = b∕μ.

  Case 2::

  I ₂ > 0. Then (from I ^′ = 0) we find $S_{2} =\tilde{\mu } /\beta$, and hence

  $$\displaystyle{I_{2} = \frac{b -\mu S_{2}} {\beta S_{2}} = \frac{b} {\beta } \,\, \frac{1} {S_{2}} -\frac{\mu } {\beta } = \frac{b} {\tilde{\mu }} -\frac{\mu } {\beta } = \frac{b\beta -\mu \tilde{\mu }} {\tilde{\mu }\beta }.}$$

  The Jacobian read

  $$\displaystyle{J = \left (\begin{array}{cc} -\beta I-\mu &-\beta S\\ \beta I & \beta S-\tilde{\mu }\end{array} \right ).}$$

  Case 1::

  (S ₁, I ₁) = (b∕μ, 0). Then,

  $$\displaystyle{J = \left (\begin{array}{cc} -\mu &-\beta b/\mu \\ 0 & \beta b/\mu -\tilde{\mu } \end{array} \right ).}$$

  The first eigenvalue, λ ₁ = −μ < 0, is always negative. The second eigenvalue, $\lambda _{2} =\beta b/\mu -\tilde{\mu }$ changes sign at $\beta =\beta _{0}:=\tilde{\mu }\mu /b$. This is, for β < β ₀, (S ₁, I ₁) is linearly stable (a stable node), for β > β ₀ linearly unstable (it is a saddle).

  Case 2::

  $(S_{2},I_{2}) = (\tilde{\mu }/\beta,(b\beta -\mu \tilde{\mu })/\tilde{\mu }\beta )$.

  $$\displaystyle{J = \left (\begin{array}{cc} (-b\beta +\mu \tilde{\mu })/\tilde{\mu }-\mu &-\beta \tilde{\mu } /\beta \\ (b\beta -\mu \tilde{\mu })/\tilde{\mu } & \beta \tilde{\mu }/\beta -\tilde{\mu } \end{array} \right ) = \left (\begin{array}{cc} - b\beta /\tilde{\mu }&-\tilde{\mu }\\ b\beta /\tilde{\mu }-\mu & 0 \end{array} \right ).}$$

  This is, $\mbox{ tr}(J) = -b\beta /\tilde{\mu } <0$, and

  $$\displaystyle{\mbox{ det}(J) = b\beta -\mu \tilde{\mu }.}$$

  Therefore, this stationary point is linearly stable for β > β ₀, and linearly unstable for β < β ₀. As the two stationary points meet for β = β ₀ and exchange their stability, we find a transcritical bifurcation.

4.3 If we multiply the vector field by 1∕(I ₁(N ₁ − I ₁)I ₂(N ₂ − I ₂)), we obtain

$$\displaystyle\begin{array}{rcl} & & \partial _{I_{1}}\left ( \frac{1} {I_{1}(N_{1} - I_{1})I_{2}(N_{2} - I_{2})}\left [(N_{1} - I_{1})(\beta _{1,1}I_{1} +\beta _{1,2}I_{2}) -\alpha I_{1}\right ]\right ) {}\\ & & +\partial _{I_{2}}\left ( \frac{1} {I_{1}(N_{1} - I_{1})I_{2}(N_{2} - I_{2})}\left [(N_{1} - I_{2})(\beta _{2,1}I_{1} +\beta _{2,2}I_{2}) -\alpha I_{2}\right ]\right ) {}\\ & =& \frac{\partial _{I_{1}}\left (\beta _{1,1} +\beta _{1,2}\frac{I_{2}} {I_{1}} - \frac{\alpha } {N_{ 1}-I_{1}} \right )} {I_{2}(N_{2} - I_{2})} + \frac{\partial _{I_{2}}\left (\beta _{2,1}\frac{I_{1}} {I_{2}} +\beta _{2,2} - \frac{\alpha } {N_{ 2}-I_{2}} \right )} {I_{2}(N_{2} - I_{2})} {}\\ & =& \frac{-\beta _{1,2}I_{2}/I_{1}^{2} -\alpha /(N_{1} - I_{1})^{2}} {I_{2}(N_{2} - I_{2})} + \frac{-\beta _{2,1}I_{1}/I_{2}^{2} -\alpha /(N_{2} - I_{2})^{2}} {I_{1}(N_{1} - I_{1})} <0 {}\\ \end{array}$$

4.4 The data about the inter-generation interval are approximately normally distributed, with (Fig. 4.15)

$$\displaystyle{T_{a} = 11\ \mbox{ days},\quad \sigma = 2.5\ \mbox{ days}.}$$

A linear fit yields the slope in the onset of the epidemic

$$\displaystyle{r = 0.5/\mbox{ month} \approx 0.0161/\mbox{ day}.}$$

Therefore, using the formula M(−r) = 1∕R _v where M(. ) is the moment generating functions of the normal distribution, we find the effective reproduction number as

$$\displaystyle{R_{v} \approx e^{rT_{a}-\sigma ^{2}r^{2}/2 } = 1.2.}$$

The epidemic happened in a partially immunised community. Therefore the effective R _v is considerably smaller than R ₀ (the reproduction number of measles is often estimated to range between 10 and 12). The article [38] mentions that the vaccination coverage has been around 70 %. This is, we compute

$$\displaystyle{R_{0} = R_{v}/(1 - 0.7) = 4.}$$

This estimation is larger than expected. One reason could be that T _a is not estimated appropriately.

Fig. 4.15

Left: inter-generation time. Right: fit in the exponential growing phase (see Exercise 4.4)