PERTEMUAN VIII
(Aplikasi Persamaan Differensial )
Population Dynamics:
Example: Let P(t) be the population of a certain animal species. Assume that
P(t) satisfies the logistic growth equation
1. Is the above differential equation separable?
2. Is the differential equation autonomous?
3. Is the differential equation linear?
4. Without solving the differential equation, give a sketch of the graph of P(t).
5. What is the long-term behavior of the population P(t)?
6. Show that the solution is of the form
Find A and B.
Hint: Use the initial condition and the result of 5.
7. Where is the solution's inflection point?
Hint: This can be done without using the answer of 6.
8. What is special about the growth rate of the population P(t) at the
inflection point (found in 7)?
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Answer:
1. This differential equation is autonomous, i.e. the variable t is missing.
Therefore, this equation is indeed separable.
2. The answer is Yes! (see 1.) Every autonomous differential equation is
separable.
3. The equation is not linear because of the presence of
.
3. The graph of the solution P(t) is as follows:
5. Clearly, because of the initial condition (see the graph of the solution below),
we have
,
200 being the carrying capacity.
6. Let us solve this equation (use the technique for solving separable
equations). First, we look for the constant solutions (equilibrium points or
critical points). We have
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Then, the non-constant solutions can be generated by separating the
variables
,
and the integration
.
Next, the left hand-side can be handled by using the technique of
integration of rational functions. We get
,
which gives
Hence, we have
.
Easy algebraic manipulations give
,
where
.
Therefore, all the solutions are
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where C is a constant parameter.
Remark: We may rewrite the non-constant solutions as
,
where a and B are two parameters. If we use the conditions
we will be able to get the desired solution. Indeed, we have
Thus,
and consequently
.
7. From the graph (see 4.), the graph of P(t) has an inflection point between 0
and 200. Let us find it. Since we must have P''(t) = 0, we get
,
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where we used the chain rule and the fact that
.
Since our solution is not one of the two constant solutions we are only left
with the equation
This simplifies to
,
which gives P=100 (half-way between 0 and 200).
Remark: You still need to convince yourself that t=100 is indeed the
inflection point, that is the second derivative changes sign at that
point.
8. The growth rate at t=100 is maximal.
Example:
SOLUTIONS TO SOME OF THE APPLICATION PROBLEMS
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Example:
Example:
A tank holds 100 gallons of liquid. The tank is half full with a salt water
solution that contains 0.1 lb of salt per gallon. Pure water is added to the
container at the rate of 2 gallons per minute and at the same time one gallon
of salt water per minute is removed from the tank. Assume that the
concentration of salt in the water in the tank remains uniform
throughout. When the tank becomes full it begins to overflow and at that
time a total of 2 gallons per minute of salt water will be leaving the
tank. Construct a piecewise continuous function S(t) that gives the amount
of salt in the tank as a function of time t where t = 0 represents the time
when the 2 gallons per minute of pure water began being added to the tank
and the 1 gallon per minute of "well-stirred, uniform" mixture began exiting
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the tank. The picture at the right shows
the amount of salt in the tank as a
function of time.
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Example:
Here is an example of the logistic equation which describes growth with a
natural population ceiling:
The solutions of this logistic equation have the following form:
VII-8
Equilibria of Autonomous Equations
Example. Draw some solutions for the equation
Answer. Note that this equation models the logistic growth with threshold.
The equilibria or constant solutions are given by
or y = 0, y = 2, and y = 5. The graph of the function f(y) = .2y(5-y)(y-2) is
given in the next picture
So we have:
the solution which satisfies the initial condition
is increasing. Moreover we have
, with
,
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the solution which satisfies the initial condition
decreasing. Moreover we have
, with
, is
the solution which satisfies the initial condition
increasing. Moreover we have
, with
, is
the solution which satisfies the initial condition
decreasing. Moreover we have
, with
, is
In the picture below we draw some solutions
Phase Line
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Let us reconsider the above example. Let us focus our attention on the
solutions which satisfy the initial condition
already know that
and
. We
It is somehow amazing that this conclusion is valid regardless whether
is
close to 5 or close to 2. In this case we say that the equilibrium point 5 is
attractive from below. In fact, in this example, this equilibrium point is
attractive from below and above. On the other hand, the solutions get away
from the equilibrium point 2 from below and above. We will say that the
equilibrium point 2 is repelling. Note that there is no general agreement on
the words used to describe these phenomena.
Example. Classify the equilibrium points of the equation
as source, sink, or node.
Answer. The equilibrium points are 0, 2, and 5. Using the above results, we
see that 0 and 5 are sinks while 2 is a source.
Example. Find and classify the equilibrium points of the equation
as source, sink, or node.
Answer. It is easy to see that the only equilibrium point is 0. Since
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is always positive, then the solutions will always be increasing.
Using previous results, we conclude:
if y(t) is a solution satisfying y(0) < 0, then we have
if y(t) is a solution satisfying y(0) > 0, then we have
Hence, the equilibrium point 0 is a node. In the picture below we draw some
solutions.
There is a very nice way to put all this information together. Indeed, draw a
vertical line (where the variable describing it is y) and start by marking the
equilibrium points of the equation
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on the y-axis.
Then using the sign of f(y), we draw arrows pointing upward in a region where
f(y) is positive, and downward in a region where f(y) is negative. This vertical
line is called the phase line of the equation.
An equilibrium point is a sink, if the arrows on both sides point towards the
equilibrium point, and it is a source, if both arrows point away from it.
TERIMA KASIH
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