Download Population models

Dynamical Systems
A dynamical system is a mathematical model of a portion of the real world that is changing with time. For
example,
1.
a moving object or a collection of moving objects such as the sun and planets,
2.
an electric circuit,
3.
a chemical reaction,
4.
the concentration of a drug in different parts of the body,
5.
the numbers of different species of animals in the environment.
One feature of the model is whether we are describing the real world for all times or just for a set of discrete
times. In the first case we usually use differential equations while in the second we use difference
equations. We begin with models involving a single differential equation.
1
Single Differential Equations
In mathematics books single differential equations are often written as
(1)
dx
= f(x, t)
dt
Here
t = time measured from some starting time measured in some convenient units
x = x(t) = amount of some physical quantity which is changing with time
dx
= rate of change of the physical quantity with respect to time
dt
f(x, t) = some formula that expresses how the rate of change is related to the amount and
time
In Example 1 below the equation (1) is
dx
= 2x
dt
with x being the number of halibut in a region and t is time in months measured from some convenient
starting time. In this case f(x, t) = 2x. We begin with models of animal populations.
1.1 - 1
1.1 Population Modeling.
1.1.1 Basic Concepts.
Here we are trying to describe how the number of some species of animal changes with time.
Example 1. Let
x = x(t) = number of halibut in the North Pacific near the coast of North America
t = time measured in months from some starting time
dx
= rate of change of population
dt
Assume x is measure in units of 100,000's of halibut so that x = 2 would mean 200,000 halibut. This
dx
dx
implies the units of
is 100,000's of halibut per month so that = - 0.1 would mean the halibut
dt
dt
population is decreasing at a rate of 10,000 per month.
We assume that the change in the population is due to four things
1.
births
2.
deaths
3.
immigration (halibut entering the North Pacific from some outside source)
4.
emigration (halibut leaving the North Pacific to some outside source)
Immigration could include stocking from a fish farm. Emigration could include halibut caught by
fishermen. In particular,
(1)
dx
= (birth rate) - (death rate) + (immigration rate) - (emigration rate)
dt
In the simplest models we assume there is no immigration or emigration and the birth and death rates are
proportional to the population size, i.e.
(2)
immigration rate = 0
(3)
emigration rate = 0
(4)
birth rate = bx
(5)
death rate = dx
where
1.1.1 - 2
b = specific birth rate (or per capita birth rate)
d = specific death rate
Often books omit "specific" and simply refer to b and d as the birth and death rates. For example, if b = 4
and x = 2, then halibut are being born at a rate of 800,000 per month.
Combining (1) – (5) we get
dx
= bx – dx = (b – d)x
dt
or
(6)
dx
= rx
dt
where
r = b – d = specific growth rate
For example, if b = 4 and d = 2, then r = 2 and (6) becomes
(7)
dx
= 2x
dt
One popular method to solve the differential equation (7), or more generally equations of the form
as follows.
Bring the term rx onto the left side so it is
dx
- rx = 0
dt
Multiply by e-rt so it is
e-rt
dx
- e-rtrx = 0
dt
Note that - e-rtr =
e-rt
d -rt
[e ], so we have
dt
dx
d
+ x [e-rt] = 0
dt
dt
By the product rule for derivatives, the right side is
d -rt
[e x], so we have
dt
d -rt
[e x] = 0
dt
1.1.1 - 3
dx
= rx is
dt
e-rt x = C
where C is a constant. Therefore,
x = Cert
The constant C can be found from some other piece of information. For example, if we denote x(0) by x0,
then we have
x0 = Ce(r)(0)
which implies C = x0. So the solution to (6)
x = x0ert
For example, the solution to
dx
= 2x along with the initial condition
dt
x(0) = 3 is
150
100
x = 3e2t
50
The graph is at the right. If we vary the initial conditions, then the
0.5
1.0
graphs of several solutions with different initial conditions is at the
15
right.
10
1.5
2.0
1
2
5
Only x0  0 is physically meaningful, but mathematically it could be
2
any number.
1
5
10
Note that if x0 = 0 then x(t) = 0 for all t. Physically, if we start with
15
no halibut, then it stays this way for all t. x(t) = 0 is called an equilibrium solution or a steady state
solution, and the number zero is called an equilibrium point.
In general, if we have a differential equation, then an equilibrium solution or a steady state solution is one
that is constant in time, i.e x(t) = x0 for all t where x0 is a constant which is called the equilibrium point.
Physically this corresponds to a situation where things are not changing with time. Some examples are the
population size is constant, an object is not moving or the current in an electric circuit is not changing. One
of the main things we are interested in this course is equilibrium solutions and whether other solutions
approach an equilibrium point as t gets large.
In our example
dx
= 2x the solutions x = x0e2t move away from the equilibrium solution x(t) = 0 as t
dt
increases. We say x = 0 is a source or x = 0 is unstable or x = 0 is a repellor.
Suppose, on the other hand the death rate exceeds the birth rate. For example, suppose b = 4 and d = 6 so
that r = b – d = 4 – 6 = - 2. Then (6) becomes
1.1.1 - 4
(8)
dx
= - 2x
dt
15
10
The solutions are now x(t) = x0e-2t. x(t) = 0 is still an equilibrium point.
5
However, now the solutions move toward the equilibrium point as t
2
1
1
2
5
increases. See the graph at the right.
We say x = 0 is a sink or x = 0 is
10
asymptotically stable or x = 0 is a attractor. Often people omit the word
15
asymptotically and simply say x = 0 is stable. However, there is a technical distinction between stable and
asymptotically stable that we shall discuss later.
Example 2. Let's modify Example 1 by assuming that halibut are stocked from a fish farm at a rate of
50,000 halibut per month. Let's also assume the net growth rate due to births and deaths is - 2x. Then
equation (1) becomes
(9)
dx
1
= - 2x +
dt
2
1
1
dx dy
so that x = y + and = . Substituting
4
4
dt dt
dy
1 1
dy
into (9) we get = - 2(y + ) + . This simplifies to the simpler equation = - 2y. From our work above
dt
4 2
dt
1
1
1
1
1
1
this implies y = y0e-2t. Since y = x - and y0 = x0 - we get x - = (x0 - )e-2t or x = + (x0 - )e-2t.
4
4
4
4
4
4
One way to solve this is to make the change of variables y = x -
Note that now x =
1
1
is an equilibrium point in the sense that if the halibut start out at then they remain that
4
4
way for all time. This equilibrium point for x corresponded to the equilibrium point y = 0 for y.
1
Furthermore, x = is a sink since if the halibut start out at any other
2.0
4
1.5
1
value then they approach as t  . At the equilibrium point the net
1.0
4
0.5
decrease in the population due to births and deaths is balance by the
increase due to stocking from the fish farm. The graph of some
solutions is at the right.
1.0
0.5
0.5
1.0
0.5
1.0
1.5
Another way to solve (9) is to proceed as follows
dx
1
dx
1
dx
1 2t
d 2t
1 2t
= - 2x +

+ 2x =
 e2t
+ e2t 2x =
e 
[e x] =
e
dt
2
dt
2
dt
2
dt
2
 e2t x =
1
1
1
1
2t
2t
-2t
-(2)(0)

 2 e dt = 4 e + C  x = 4 + C e  x(0) = 4 + C e
 C = x0 -
1
1
1
 x =
+ (x0 - ) e-2t
4
4
4
1.1.1 - 5
1.5
2.0