3.3 Establishing species

Let’s look at another example where from we will determine a differential equation model from a context:

A newly introduced plant species is introduced to a region. It competes with another established species for nutrients (and is a better competitor). However, the growth rate of the new species is proportional to the difference between the current number of established species and the number of new species. You may assume that the number of established species is a constant E.

For this problem we will start by naming our variables. Let \(N\) represent number of new species and \(E\) the number of established species. We will break this down accordingly:

  • “the growth rate of the new species” means \(\displaystyle \frac{dN}{dt}\).
  • “is proportional to the difference between the current number of established species and the number of new species” means \(\displaystyle k \cdot (E-N)\), where \(k\) is the proportionality constant. Including this parameter helps to avoid assuming we have a 1:1 correspondence between the growth rate of the new species and the population difference.
  • “and is a better competitor” helps to explain why the term is \(\displaystyle k \cdot (E-N)\) insteady of \(\displaystyle k \cdot (N-E)\). We know that the newly established species will start out in much smaller numbers than \(N\). But since it is a better competitor, we would expect its rate to increase initially. So \(\displaystyle \frac{dN}{dt}\) should be positive rather than negative. Assuming \(N < E\), then \(E-N > 0\), which guarantees that the new species will grow.

So this description could be modeled with Equation (3.6): \[\begin{equation} \frac{dN}{dt} = k(E-N) \tag{3.6} \end{equation}\]

Does Equation (3.6) seem familiar to you? It is similar to Equation (1.4) in Section 1 for the spread of Ebola! While this may seem surprising, it is often the case that similar equations appear in different contexts. So do you have to memorize every possible model for every possible context? The answer is emphatically NO! Rather it is more advantageous to learn how to analyze models qualitatively rather than memorize several different types of models and not see the connections between them.

An interesting solution to a differential equation is the steady state or equilibrium solution. We find this where the rate equals zero. Let’s take a look how to do that for our establishing plant model

Example 3.1 What is the steady state value for the differential equation \(\displaystyle \frac{dE}{dt} = k(E-N)\)? (That is solve for \(E\) when \(\displaystyle \frac{dE}{dt} = 0\).)
Solution. Let’s solve \(\displaystyle \frac{dE}{dt} = k(E-N) = 0\). For this equation we want to express the right hand side in terms of \(E\). The parameter \(k\) is a constant \(k > 0\), so really the steady state occurs when \(E-N = 0\), or when \(N = E\).

What this model tells us that eventually the new species will overtake the established species \(E\).