In discrete growth, cell division occurs at fixed time intervals.

The number of cells (the abundance \(N\)) at a future time \(t+1\) is the abundance of cells at the current time \(t\) plus the number of new cells added \(b\cdot N_{t}\). Here, \(b\) is the per-capita reproduction rate (birth rate), the proportion of cells that divide.

The abundance \(N\) is either dimensionless and means “number of individuals,” but it can also be specified with a unit of measurement, e.g., individuals per square meter or per liter. A concentration value is also possible, e.g., mg/L biomass or mg/L carbon.

Instead of the birth rate \(b\), the symbol \(r\) is often used, representing the net reproduction rate. This consists of two parts: the birth rate \(b\) and the death rate \(d\). Depending on whether the difference \(r=b-d\) is positive or negative, the population grows or declines.

A birth rate \(r=1\) means that for every existing individual, a new individual is added per time step, resulting in a doubling of the population.

When considering a very large number of individuals that reproduce continuously—that is, not at fixed intervals—we refer to this as exponential growth. The equation above becomes an exponential function:

This function describes the development of the number of individuals (abundance, \(N\)) at time \(t\) as a function of the net reproduction rate \(r\) and the initial abundance \(N_0\) at time \(t=0\).

For animals that reproduce only once a year, discrete growth models are typically used in practice, while continuous models are typically used for microorganisms.

A so-called system diagram can be used for illustration purposes:

• In this diagram, a thick-bordered rectangle represents a state variable, e.g., the abundance of a population,
• thick arrows represent a flow of matter or energy, and
• thin arrows represent an interaction.
• Metabolic processes, e.g., growth, are symbolized by a hexagon in this example,
• while fixed parameters, metabolic rates, and auxiliary variables are represented by circles.

Positive (enhancing) feedback can be symbolized by a (+), and negative feedback by a (-).

1. Given a growth rate of \(r=1\) d^-1, how many cells will develop from a single cell in one day:

1. in discrete growth at fixed time intervals?
2. with continuous (exponential) growth?

2. A green algae population starts with 10 cells per liter and a growth rate of \(r=1\) d^-1. How many cells per liter will there be after 7 days of exponential growth?

3. A cyanobacteria population (blue-green algae) grows at a rate of \(r=0.5\) d^-1. After approximately how many days will the abundance have increased tenfold under exponential growth?

4. A clear-water situation has developed in a lake. The Daphnia population, with a concentration of 50 individuals per liter, is suffering from acute food shortage and is no longer reproducing. However, due to predation by fish, the exponential mortality rate is \(d=0.1\) d^-1. After how many days will there be only one Daphnia left in one liter of water?

5. A thought experiment: A beaker containing 1 L of nutrient solution holds algae with a fresh mass of 1 mg L^-1, and the growth rate is 1 d^-1. How long would it take until the beaker is completely filled with algal biomass if the algae could grow without limitation?

• If the numbers get too large, you can shorten the simulation period.
• You can read the numbers directly from the graph using the mouse pointer.
• You can save the results graph by clicking the camera icon.
• Instead of a simulation, a pen and paper or a calculator can sometimes be useful—and thinking things through is always a good idea.

• If a population is dying off, i.e., \(b < d\), then the reproduction rate \(r\) is negative.
• The density of algae is only slightly greater than that of water. We can practically assume a density of 1 g cm^-3.

The introduction assumed a doubling rate, but for continuous growth, an exponential function is used. Why an exponential function and not a power law??

In reality, not all algae cells divide at the same time. Some divide earlier, others later, and the ones that divide earlier may already be dividing for the second time. As a result, we do not obtain a geometric sequence, but rather exponential growth.

From the step-by-step (iterative) equation:

\[ N_{t+1} = N_{t} + r \cdot N_{t} \] we obtain, by rearranging, a difference equation:

\[ \frac{N_{t+1} - N_t}{\Delta t} = r \cdot N_t \] or:

If we consider a very large number of cells and short time intervals, we obtain a differential equation:

This equation can be analytically integrated using the so-called variable separation method to yield the exponential formula:

The exponential formula has the advantage that one does not have to iterate step by step, but can calculate the values for all time steps directly.

In logistic growth, the growth rate \(r\) is no longer constant but is a function of abundance:

Here, the actual growth rate \(r\) is given by the product of the species-specific (intrinsic) reproduction rate \(r_{max}\) and a dimensionless term \(\left(1 - \frac{N}{K}\right)\), where \(r\) is the realized reproduction rate and \(K\) is the carrying capacity.

The growth equation can then be written as:

This looks quite complicated, but it is easy to calculate. In reality, this formula is the solution to a differential equation:

This equation describes the change in population per unit of time, exactly as shown in the system diagram.

Growth continues until the carrying capacity \(K\) is reached. At low population density, i.e., when \(N \ll K\), the term in parentheses is approximately equal to 1 and growth is at its maximum. The product \(r \cdot \left(1 - \frac{N}{K}\right)\) is approximately \(r\).

The closer the abundance \(N\) approaches the carrying capacity, the smaller \(r \cdot \left(1 - \frac{N}{K}\right)\) becomes, converging to zero as \(N \rightarrow K\).

1. Set the logistic growth rate to the smallest possible value (\(r_{max}=0.02\)) and the carrying capacity \(K=500\):

   • What shape does the growth curve take?

2. Now adjust the \(K\) slider.

   • Why does increasing \(K\) have only a minor effect?

   • Why does decreasing \(K\) have a greater effect?

3. Now set \(r_{max}=0.1\) and \(K=100\) again.

   • What happens if you change the initial abundance \(N_0\)?

   • Now set the initial abundance to a larger value, e.g., 120. What happens?

4. In the following, we will examine different types of equilibrium:

   • \(N_0=0, r_{max}=0.1, K=100\)
   • \(N_0=1, r_{max}=0, K=100\)
   • \(N_0=100, r_{max}=0.1, K=100\)

1. First, explain by interpretation how the different equilibrium states arise.

2. Try to understand why equilibrium is reached by substituting the values into the equations. For this, the derivative equation \(dN/dt= ...\) is easier to understand than the integral form.

• When the initial abundance \(N_0\) is small compared to \(K\), the growth curve initially approaches exponential growth. The abundance is far from \(K\), so its increase no longer has any effect.

• When \(K\) is reduced to a value close to the abundance, the limitation becomes clearly visible and the growth curve flattens out.

• The value of \(N_0\) only influences how quickly the value of \(K\) is reached. It has no influence on the final value.

• If the initial abundance exceeds the carrying capacity, the population decreases until \(K\) is reached.

In the experiment, we can identify three different types of equilibria. If \(N_0=0\), there are no organisms present that can grow. We have, so to speak, a “sterile system.” The abundance remains at 0.

If \(r_{max}=0\) and \(N_0 > 0\), then organisms are present, but they do not reproduce. It is also possible that they reproduce, but reproduction and death are exactly balanced.

Both equilibria are “labile” (or unstable). If even a small value is added to \(N_0 = 0\), population growth begins, the equilibrium is left, and the abundance moves toward \(K\). The same phenomenon can be observed in the second case when \(r_{max} > 0\).

In the third case (at \(N = K\)), the equilibrium is stable. It is reached when \(N\) is less than \(K\) but also when \(N\) is greater than \(K\). Even if the abundance temporarily deviates from \(K\), its value us quickly restored. This is called “resilience.”

The logistic model provides the theoretical basis for distinguishing between so-called r- and K-strategists. An r-strategist refers to species that reproduce rapidly, “at any cost”. K-strategists, on the other hand, are species that can achieve a high abundance (a high K). According to a more modern interpretation, these are organisms that can survive well and minimize losses even during resource shortages.

See also r/K selection theory on Wikipedia.

The logistic model and its equilibria are a textbook example from chaos theory. The experiments described under “Exercises” assume that population growth occurs continuously, i.e., can be described mathematically by a differential equation.

If population growth occurs in the form of generations, e.g., in wild animals that reproduce only once a year, then the time step is “discrete”. Mathematically, this is a difference equation.

In this case, the carrying capacity may be temporarily exceeded. This is then followed by a partial collapse of the population, e.g., due to a lack of food. In the following time step, the population can recover again, and so on. This creates a cycle and, under certain conditions, a chaotic pattern.

More information on this can be found under [Logistic Map]https://en.wikipedia.org/wiki/Logistic_map#Discretization_of_the_logistic_equation) on Wikipedia.

If you want to try it out yourself, you can send an email to the app’s developers. Perhaps they will then incorporate the chaos model into the app.

In this example, we will describe the population growth of phytoplankton as a function of phosphorus (\(P\) in mg m^-3). To avoid duplicate letters, we will refer to the phytoplankton as “algae” \(A\). As a unit of measurement, we will use the biovolume or the carbon contained in the organisms in mg m^-3. For time, we will use measurement hours (h).

In more modern models, biomass is often expressed in units of carbon. Carbon is the “element of life” and accounts for approximately 50% of the dry weight in most organisms.

The first equation \(dA/dt\) is, in principle, the exponential growth equation again; however, the growth rate \(r\) is now a function of \(P\), i.e., \(r=f(P)\). When the nutrient \(P\) is depleted, the growth rate \(r(P)\) approaches zero.

The first part of the phosphorus equation \(dP/dt\) corresponds to the algae equation, but with a negative sign. When algae grow, phosphorus is consumed. The conversion factor \(1/Y\) is used to convert the phosphorus and carbon content of the algae stoichiometrically. Here, one typically uses the reciprocal of the yield coefficient \(Y\) (“Yield”). A value of \(Y=41\) means that for every 1 mg of phosphorus, 41 mg of carbon is bound in the biomass.

The function \(r(P)\) describes the dependence of the growth rate \(r\) on phosphorus as so-called saturation kinetics. As the phosphorus concentration increases, the growth rate initially rises steeply. When there is a nutrient surplus, the curve flattens out and approaches a maximum value \(r_{max}\). The value \(k_P\) is the nutrient concentration at which 50% of the maximum growth rate is reached. The smaller \(k_P\) is, the better an algal species is adapted to low nutrient concentrations.

1. Compare resource-limited growth with logistic growth.

   1. What is the advantage of resource-limited growth over the logistic growth function?

   2. Are there any disadvantages?

2. Examine the influence of \(r\), \(k_P\), and \(Y\) on growth.

   1. At what values of \(r\) is growth fastest?

   2. At what values of \(k_P\) is growth fastest and nutrient utilization best?

   3. What determines the steepness of the transition to the limiting phase?

   4. Which parameter and which initial value determine the maximum phytoplankton abundance reached? This value corresponds to the carrying capacity \(K\) in the logistic model.

The relationship between the growth rate \(r\) and the nutrient concentration is described by what is known as kinetics. This refers to how fast the algae grow; this is also called a functional response. In contrast, the actual growth models (e.g., the logistic and resource-limited models) describe how many algae can grow in total. This is called a numerical response.

For the dependence of the growth rate on a resource (e.g., phosphorus), saturation kinetics are used. The more nutrients are available, the faster the growth. Above a certain nutrient concentration, saturation is reached, meaning that even more nutrients do not lead to further growth.

There are various ways to describe this mathematically. In the Blackman model, the growth rate initially increases proportionally with the nutrient concentration. Here, \(k_b\) is the proportionality factor. When the nutrient concentration (\(P\)) exceeds the maximum possible growth rate (\(r_{max}\)), the growth rate \(r\) remains constant at \(r_{max}\):

\[ r = r_{max} \cdot \min(k_b \cdot P, 1) \] The Monod model assumes a gradual transition from the growth phase to saturation. In the equation used for this purpose:

the limitation is achieved through a ratio in which the denominator is a so-called half-saturation rate \(k_P\), which has the same unit of measurement as \(P\). This makes this term dimensionless. If the phosphorus concentration and the half-saturation rate are equal (i.e., \(P=k_P\)), then the ratio \(P/(P + P) = 0.5\). However, if \(P\) is very large (\(P >> k_P\)), then the ratio converges to 1 and the growth rate \(r\) converges to \(r_{max}\).

Fast-growing phytoplankton species have a high maximum growth rate, while species that can survive on few nutrients have a low semi-saturation constant.

The use of fertilizers in agriculture is very important for our food supply, but it is crucial to avoid overfertilization. Excess phosphorus and nitrogen cannot be absorbed by plants and ends up in our rivers and lakes, and ultimately in the sea.

In these bodies of water, increased phosphorus and nitrogen inputs lead to increased phytoplankton growth, and the water becomes turbid. In deep water, the decay of algae then leads to oxygen depletion. The increased growth of plankton and aquatic plants and the resulting deterioration of water quality is called eutrophication.

Further information on nutrient input and eutrophication can be found, for example, on the website of the German Federal Environment Agency.

Unlike in logistic or resource-limited growth, stable equilibria are rarely observed in nature; instead, population abundance fluctuates. There are various reasons for this, such as the reproductive cycle or variations in food availability due to the seasons and weather.

However, population fluctuations can also result from interactions between populations.

Such interactions were investigated and mathematically described in the early 20th century by the Italian mathematician and physicist Vito Volterra and the Austrian-American chemist and actuary Alfred Lotka.

Their model consists of two equations, one for the prey population (e.g., algae) and one for the predator population (e.g., daphnia). As a modification to the notation of the equations originally used by Volterra (1926), many scientists and textbooks use different symbols and variants. In the following, we use a simplified notation with \(B\) for the prey population and \(R\) for the predator population. For the model parameters, we use consecutive letters \(a, b, c, d\), which can be interpreted as follows:

\(a\): Reproduction rate of phytoplankton (prey population)

\(b\): Loss of phytoplankton (prey) due to predation by daphnia (predators)

\(c\): Growth efficiency of daphnia (predators) due to food intake

\(d\): Mortality rate of daphnia (predators)

The model essentially consists of three processes. The growth of the prey population (e.g., algae) and the decline of the predator population (e.g., Daphnia) each correspond to an exponential growth model with a positive or negative rate of change, respectively.

Exponential growth of the prey population:

Exponentielles Absterben der Räuberpopulation:

The third process describes the interaction between predators and prey. Here, the rate of decline in the prey population \((b \cdot R)\) depends on the abundance of the predator population, and the growth rate of the predator population depends on the abundance of the prey population \((c \cdot B)\). By choosing different values for \(b\) and \(c\), one can account for trophic efficiency (\(c/b\)), e.g., that only 10% of the food is used for reproduction.

This results in the following system of equations, in which, for clarity, the loss rate of the prey population due to predation and the growth rate of the predator population are denoted by additional parentheses:

The parentheses are for illustrative purposes only and are usually omitted.

The initial standard parameters and starting values are as follows:

\(a=b=c=d=0.1\), \(B_0=1\), \(R_0=0.5\).

1. Change the initial value for the predator (\(R_0\)) to 2 or 1.

• What happens?

2. Set \(R_0=2\) again. Now try to find a prey reproduction rate (\(a\)) at which an equilibrium is reached again.

• You can find the desired value either by trial and error or by rearranging the system of equations.

First, restore the default parameters, e.g., by reloading the app.

3. By changing the parameters \(a, b, c, d\), you can alter the period. For simplicity, let all parameter values be equal at first (\(a = b = c = d\)). Now find values such that the period:

   1. shortens
   2. lengthens

   3. Why do the parameters affect the period in the observed direction?

Note: Checking the “Scan values” box activates additional guide lines. You can read specific values using the mouse.

4. Now set the parameter values back to their defaults.

   1. Double the prey reproduction rate \(a\) to 0.2. How can the result be explained?

   2. Double the predator mortality rate \(d\) to 0.2. How can the result be explained?

   3. What conclusions can be drawn regarding turnover and population density in real-world systems?

5. Double and halve the turnover between \(B\) and \(R\). Use parameters in the range: \(b = c = 0.05 \dots 0.2\).

• How can the result be explained? What information is needed to assess whether the result is realistic?

6. Rearrange the system of equations and determine the equilibrium state.

   1. At equilibrium, the first derivatives of both equations are equal to 0. Return to Problem (2) and find the value of the parameter a by rearranging and solving the system of equations.

   2. Using the “Read Values” checkbox, you can estimate the mean values of prey and predator for each cycle from the displayed points. Here, a cycle always extends from one maximum to the next.

   3. Compare the equilibrium state (horizontal dotted lines) with the mean values estimated in (b). Consider or research how the mean value can be obtained directly from the initial values and the model parameters.

We are usually taught that coordinate axes should always be labeled with units of measurement. This is very important, but there are exceptions. In theoretical models, sometimes only the variable name – without a unit of measurement – is written on the axis, e.g., “time,” “frequency,” or “abundance.”

Physicists go even further and scale all axes to an interval from zero to one. In our apps, we have chosen not to do this, so that users can visualize what the values represent. This allows us to specify a unit of measurement depending on the application, e.g., “hours” or “days” for fast-growing algae and Daphnia, or “years” for snowshoe hares and lynx.

When it comes to units of measurement for abundance and biomass, it is important to note that algae and Daphnia vary in size and have different body masses. If we omit the unit of measurement, the values on the y-axis are merely relative quantities that cannot be compared with one another.

When considering biomass, we can use a trick to establish approximately comparable ratios. As a rule of thumb, about 10% of the energy is transferred from one trophic level to the next in a food chain.

We can account for this if we set the parameter \(c\) (“growth efficiency”) to one-tenth of \(b\) (“feeding intensity”), i.e., for example, \(b=0.2\) and \(c=0.02\). In this case, the initial values must also be adjusted accordingly, e.g., \(B_0=1\) and \(R_0=0.05\).

By changing the parameters, combinations can also be set that appear nonsensical or unrealistic. This is partly because the parameter values are specified relative to time, e.g., growth per day or per hour. Therefore, it is sometimes necessary to adjust the time unit as well and subject the values and units of measurement to a thorough reality check.

To make the results of the app easier to understand, we can disregard the units of measurement and consider the results only qualitatively, i.e., small-large, cycle–equilibrium, larger amplitude–smaller amplitude, longer cycle–shorter cycle.

The example of snowshoe hares and lynx is often cited in connection with the predator-prey model. Both species live in North America, primarily in Canada.

It has been observed that hunters’ fur sales fluctuate in a cycle of 9 to 11 years. It seemed likely that this was a Lotka–Volterra cycle. Upon closer examination, however, inconsistencies became apparent, and various other explanations were proposed. One explanation was that the fluctuating fur sales were due to the influence of hunters. Another explanation is that the predator-prey cycle results from interactions at a lower trophic level—that is, not as an interaction between snowshoe hares and lynx, but between vegetation and the snowshoe hares. The lynx merely follows the cycle. More details can be found in the Wikipedia article Lotka–Volterra equations.

Historical dataset of lynx catches in Canada, from Brockwell & Davis (1991)

The algae-Daphnia system is also used as a textbook example in theoretical ecology. In real water bodies, however, only short section of these dynamics can be observed. An annual spring peak in algae is immediately followed by a distinct clear-water phase. The continuation of these oscillations is then prevented by damping mechanisms such as resource limitations (particularly phosphorus) and integration into a more complex food web (Sommer et al., 2012).

Another prominent example is the interrelationship between moose and wolves. Research on this topic was conducted in the Isle Royale National Park, an island in one of the Great Lakes in North America. It is the longest-running study of predator-prey systems in the world. With the exception of researchers and summer visitors, no humans live on the island.

Lotka–Volterra-type interactions were indeed observed between the moose and the wolves. However, these occurred only during brief phases of a cycle. In reality, additional factors also come into play here, such as parasites that weaken the moose. Conversely, the wolves were temporarily infected by a viral disease and nearly went extinct.

During a particularly cold winter, wolves were observed traveling a distance of 22 km across the frozen lake to reach the island. This confirmed a long-standing assumption that the wolf population on the island is not completely isolated. Interaction with the mainland is apparently vital for the wolf population’s survival, but due to global warming, such ice bridges no longer form every 3 to 4 years, but only once every 10 years.

More information can be found on the National Park’s website, on the research project’s website (https://isleroyalewolf.org/), and in the book by Vucetich (2024). There are also articles on Wikipedia on this topic. The videos about the project on YouTube are highly recommended, e.g., https://youtu.be/DS-4IsDg7mA. Additional videos can be found by searching for the keywords: wolves isle royale.

The examples show that predator-prey interactions are far more complex in practice than the Lotka-Volterra model. The model is not suitable for making predictions, but, as in the app’s simulation experiments, one can derive fundamental insights into the emergence of population fluctuations or the establishment of equilibria.

In practical models, various model components are combined: exponential growth, logistic growth, multiple resources, and predator-prey interactions. Additionally, the models incorporate current data on nutirients, weather, climate, and human influence.

Brockwell, P. J. and Davis, R. A. (1991). Time Series and Forecasting Methods. Second edition. Springer. Series G (p. 557).

Sommer, U., Adrian, R., De Senerpont Domis, L., Elser, J. J., Gaedke, U., Ibelings, B., Jeppesen, E., Lürling, M., Molinero, J. C., Mooij, W. M., van Donk, E., & Winder, M. (2012). Beyond the Plankton Ecology Group (PEG) Model: Mechanisms Driving Plankton Succession. Annual Review of Ecology, Evolution, and Systematics, 43, 429–448. doi:10.1146/annurev-ecolsys-110411-160251

Vucetich, J.A. (2021) Restoring the Balance: What Wolves Tell Us about Our Relationship with Nature. Johns Hopkins University Press.