Population Dynamics

7048 /www/nrich/html/content/id/7048/ 1 2011-11-28T14:45:50 <mdoxml version="1.0"><h3>The Logistic Equation</h3> Unlimited growth is generally impossible, because birth and death rates are affected by factors such as lack of food, water and land. We now incorporate the effect of the environment on population size into the exponential <a href="http://nrich.maths.org/7104?part=index">model</a>: $\frac {\mathrm{d}N}{\mathrm{d}t} =rN(t)$ Below are two derivations of this new model, called the logistic equation. It was originally devised by the mathematician <a href="http://desip.igc.org/malthus/">Thomas Malthus</a>. Notes on the discrete form of this equation can be found <a href="http://nrich.maths.org/7253?part=index">here</a>. <h3> </h3> <h3>First Derivation</h3> The carrying capacity, K, is the largest population that can be supported indefinitely, given the resources available in the environment. When the population size is far below K, its growth is exponential. As the population approaches K, it begins to be affected by the reduced ability of the environment to provide necessary resources. In order to have this effect, we consider the term $\frac {K-N}{K}$ . Note that when $N< < K$ then $\frac {K-N}{K}\approx 1$ and when $N \rightarrow K$ then $\frac {K-N}{K} \rightarrow 0$. Including the above term, we now get the equation: $$\frac {\mathrm{d}N}{\mathrm{d}t}=r \frac{K-N}{K} N(t)= rN\left(1-\frac{N}{K}\right)$$Note that each individual added to the population reduces the rate of increase of the whole population. <h3>Second Derivation</h3> Consider a population of lions of size y, and begin by assuming the simple exponential equation $\frac {\mathrm{d}y}{\mathrm{d}t}=r y$ , where r is the intrinsic growth rate. <mdo:image alt="" src="male_lions-188.jpg" style="width: 277px; height: 191px;"></mdo:image> If the probability of food being found by an individual lion is y, then the probability of some food being found by two lions is proportional to $y^2$. Fighting will occur within the species for these limited resources, so the death rate due to fighting can be represented by $\gamma y^2$ . Our equation then becomes: $$\frac {\mathrm{d}y}{\mathrm{d}t}=ry-\gamma y^2=ry\left(1-\frac{y}{Y}\right)$$where $Y=\frac{r}{\gamma}$. Again this is the logistic equation. Question: Graph the logistic equation to find the equilibrium points. Then solve the equation using standard integrals, showing that the solution is given by $N(t)=\frac {K N_0 e^{rt}}{K+N_0 (e^{rt}-1)}$ Graphically we describe this kind of population growth by a sigmoid, or S-shaped growth curve: <mdo:image alt="" src="Sigmoid%20Num%202.jpg" style="width: 536px; height: 302px;"></mdo:image> By looking at the blue curve, we see the size of the population begins to level off and reaches a stable value, which must be less than the carrying capacity of the environment (marked in red). Mathematically, we can show this by looking at the above solution for $N(t)$ and showing $\lim_{t\to\infty} N(t)=K$ Question: A small lake has a carrying capacity of 100 geese. Starting with a pair of geese, how would the population change over 70 years if $r=0.1$? Draw a sigmoid graph of this change. <mdo:image alt="" src="flock-geese.jpg" style="width: 272px; height: 181px;"></mdo:image> Recall that r is the intrinsic growth rate. How will this affect the time taken for the number of geese to reach the carrying capacity of the lake? Perhaps start by drawing graphs with different r-values.</mdoxml> <mdoxml version="1.0"> <mdo:image alt="" src="Logistic%20phase%20portrait2.jpg" style="width: 429px; height: 243px;"></mdo:image> We first solve this equation: $$\begin{align*} \frac{\mathrm{d}N}{\mathrm{d}t}&=rN(1-\frac{N}{K}) \\ \frac{1}{K} \frac{\mathrm{d}N}{\mathrm{d}t} &= \frac{N}{K} r(1-\frac{N}{K}) \end{align*}$$ To simplify this, let $x=\frac{N}{K}$ $$\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t}&=xr(1-x) \\ \frac{\mathrm{d}x}{x(1-x)}&=r\mathrm{d}t \\ \Big( \frac{1}{x}+\frac{1}{1-x} \Big) \mathrm{d}x&=r\mathrm{d}t \\ ln\Big(\frac{x}{1-x}\Big)&=rt+c \\ \frac{x}{1-x}&=X_0e^{rt} \\ x&=X_0e^{rt}-x X_0e^{rt} \\ x&=\frac{X_0e^{rt}}{1+X_0e^{rt}} \end{align*}$$ Putting N back in: $$N(t)=\frac {K N_0 e^{rt}}{K+N_0 (e^{rt}-1)}$$As expected, over time the population tends to the maximum allowed by the environment: $$\lim_{t\to\infty} N(t)=K$$ <mdo:image alt="" src="Sigmoids%20different%20r.JPG" style="width: 488px; height: 258px;"></mdo:image> </mdoxml> <mdoxml version="1.0"> Another Question: The logistic equation above, models the r/K selection theory which is the reproductive strategy of organisms. In stable environments K-selection predominates, as the populations are generally close to the carrying capacity. Characteristics of K-selection organisms are: <ul> <li>longer maturity time and lifespan</li> <li>fewer offspring with high level of parental care</li> <li>low mortality rate and high offspring survival rate</li> </ul> <mdo:image alt="" src="elephant-baby_1004458i.jpg" style="width: 320px; height: 206px;"></mdo:image> Elephants are an example of K-selection species. How do you think we could model a population of elephants in a stable environment? What adaptions to the logistic equation would we need to make? In unstable environments, r-selection predominates. Look up examples of r-selection organisms and find their characteristics. Think about individual size, number of children and life span. </mdoxml> 2 4 0 0 0 0 1 Population Dynamics - part 3 Third in our series of problems on population dynamics for advanced students. Applications biology Using, Applying and Reasoning about Mathematics Mathematical modelling Pre-Calculus and Calculus Calculus generally Collections Population Dynamics