One-Variable Systems¶
A Simple Population Model¶
Think of an animal population, say $X = $ the number of animals.
What changes $X$?
One thing that changes $X$ is animal births, and another is animal deaths.
There are other flows too, animals from other population may join, may leave.
But let's keep it simple and say:
$$X' = \text{birth rate} − \text{death rate}$$
All models make assumptions, and it is critical to be able to state what they are for a given model.
The validity of a model depends strongly on its assumptions.
How do we represent the birth rate? Let’s make some assumptions:
(1) all animals are capable and have he same likelihood of giving birth,
(2) an animal’s ability to give birth is constant over its lifetime from birth to death.
Overall, we are saying that there is a single constant rate $b$ at which that animal gives birth, let’s say $b = 0.5$ babies per year (i.e. one baby every two years).
Then we say that the per capita birth rate is: $b = 0.5$ per year
Another way of saying this is that the birth rate is proportional to the population with the proportional constant the per capita birth rate.
$$ X' = bX = 0.5X$$
Likewise, one can image something like a per capita death rate $d$ (per $yr$) for the population.
Hence, the change equation would look like
$$ X' = bX -dX = (b-d)X = rX$$
where $r$ is the net per capita growth rate.
To summarize, a model of a process is a change equation, in which the changes in a system depend on the current states. We write
$$X' = f(X)$$
Exercise 1.4.10 Write change equations for the following situations. You can use $X$ or anything you prefer for your state variable.
a) A population has a per capita birth rate of $0.3$.
b) A population has a per capita death rate of $0.4$.
c) A population has a per capita birth rate of $0.25$ and a per capita death rate of $0.15$.
d) A population has a per capita birth rate of $0.1$ and a per capita death rate of $0.2$.
Exercise 1.4.11 In each of the cases in the previous exercise, is the population growing or shrinking?
1.4.10, 11
a) $X' = 0.3X$ (growing)
b) $X' = -0.4X$ (skrinking)
c) $X' = 0.25X - 0.15X = 0.1X$ (growing)
d) $X' = 0.1X - 0.2X = -0.1X$ (shrinking)
A Population Model With Crowding¶
The model of rabbit population $X' = bX$ is pretty unrealistic if we take it too seriously at large values of $X$.
What is a much more reasonable model would be have a "crowding factor" which will be some number $\le 1$ and the change equation would look like:
$$X' = bX(\text{crowding factor})$$
A reasonable assumption that leads to a crowding factor would be an assumption on the environment carrying capacity.
Let's assume the environment has a capacity of carrying $k$ animals.
Then $X/k$ would represent the fraction of the carrying capacity that is already being used by the present population $X$, which leaves $(1 − X/k )$ as the fraction of resources that are currently unused and therefore available.
So our new change equation of the population would looks like:
$$ X' = bX\left( 1 - \frac{X}{k} \right) $$
where $k$ is called the carrying capacity of the environment and $(1-X/k)$ is called the crowding factor.
There is another approach to same equation. Think about the crowding factor is casued by competition between animals.
How often would two rabbits landing on one small lettuce patch?
It turns out that this chance to proportional to the square of the rabbit population, i.e. $cX^2$ for some constant $c >0$.
Thus, the change equation for the population would be
$$ X' =bX - cX^2$$
An equation of this form is called a logistic equation.
Note also that by factoring out $bX$ we arrive to the same equation
$$ X' = bX\left(1-\frac{c}{b}X\right) = bX(1-\frac{X}{k})$$
where $k = b/c$. So we see that logistic questions are the same simple growth equations with crowding factors.
It is often important to understand the sign of $X'$.
The case $X' > 0$ means $X$ is increasing (as time) while $X' < 0$ indicates a decrease of $X$ over time.
Let's again consider the logistic equation. Since $X$, the population and $b$ the per capita birth rate are always positive (the model is trivial when either of them is 0), so the sign of $X'$ is the same as the sign of the crowding factor $1-X/k$.
Since $X < k$ means $X/k < 1$, i.e. $1-X/k > 0$. Similarly, $X > k$, means $1-X/k < 0$. Thus,
\begin{cases} \text{population is growing} &\text{if } X < k \\ \text{population is shrinking} &\text{if } X > k. \end{cases}
Two-Variable Systems¶
Romeo & Juliet¶
Let $R$ represent Romeo’s feelings for Juliet, and $J$ represent Juliet’s feelings for Romeo. Positive values represent love and negative values represent hatred.
The state space for the Romeo–Juliet system is the 2-dimensional space $(R, J)$.
What changes $R$ and $J$? We need some assumptions to build a model.
- Let’s assume that the changes in Juliet’s love do not depend on her own feelings, but are purely a reflection of Romeo’s love for her. If his love is positive, hers grows, and if he hates her, her love will decrease, possibly even into hate. In other words, let's assume $J'$ is proportional to $R$ and for simplicity with proportional constant $1$.
That is, we assume $$ J' = R$$
- Romeo also does not care about his own feelings and only reacts to Juliet, but in his case, the reaction is negative. If Juliet loves him, his love declines, and if she hates him, his love will increase.
$$R' = -J$$
Exercise 1.4.13 Suppose that in addition to being turned off by Juliet’s love, Romeo is turned off by his own love for her.
Specifically, Romeo’s love declines at a rate proportional to itself with proportionality constant $k$. Write a model for the Romeo–Juliet system that adds in this assumption.
$J' = R$, $R' = - J -kR$
Springs¶
State Variables: $X$ (displacement of mass in m), $V$ (velocity in m/s)
State Space: $\mathbb{R}^2$
Change equations:
$X' = V$ (definition of velocity)
$V' = F/m$ (Newton's 2nd Law, $m$ is the mass of the object, a parameter of this system)
And $F = -kX$ (Hooke's Law) where $k$ is called the stiffness (or spring constant) of the spring. which is another parameter of this system.
So $V' = -(k/m)X$. The negative sign indicates that the force is of opposite direction to the displacement of the object.
Exercise 1.4.14 Write an expression for friction. You can make up parameters as necessary.
Exercise 1.4.15 Write the model for the spring with friction.
1.4.14 $F_f = -cV$ where $c > 0$ is the fictional constant.
(Issue of graph on p.34)
1.4.15 \begin{align*} X' &= V,\\ V' &= -(k/m)X - cV \end{align*}
Sharks and Tuna (Lotka-Volterra predator-prey model)¶
State variables: $S$ (shark population) and $T$ (Tuna population)
State space: $[0,\infty) \times [0,\infty)$
Change equations
$S' = $ shark birth rate - shark death rate
Assumptions:
Sharks die at a constant per capita rate $d$, so shark death rate = $dS$
Sharks birth rate is proportional (with $m$ as the proportional constant) to the amount of food available,
And the amount of food available is proportional to the tuna population
(let $\beta$ be the proportional constant which can be interpreted as the success rate of a shark catching a tuna) so
$$ \text{Shark Birth rate} = m \beta ST $$
Putting these together
$$ S' = m\beta ST - dS \qquad \text{individual/unit time}$$
Similar reasoning tells us that
$$T' = bT-\beta ST \qquad \text{individual/unit time}$$
where $b$ is the birth rate per capita for Tuna. In this model tuna is the food for (eaten by) sharks which is decreased at the rate $\beta ST$.
Exercise 1.4.16 Describe this reasoning. Where does each term in the $T'$ equation come from? What assumptions do we need to make to derive it?
Assumptions
Assume sharks only die of natural causes. There could be other circumstances that cause shark death.
Assuming only one tuna predator. There could be many others.
Disregarding tuna natural death.
Assume no migration/emigration of shark/tuna species from/to other geographical regions.
By choosing and adjusting the model parameters suitably, one can gain a better qualitative view of the model from the simplied system of equations:
\begin{align*} S' & = ST - S \\ T' &= -ST + T \end{align*}