In this introductory course on Ordinary Differential Equations, we first provide basic terminologies on the theory of differential equations and then proceed to methods of solving various types of ordinary differential equations. We handle first order differential equations and then second order linear differential equations. We also discuss some related concrete mathematical modeling problems, which can be handled by the methods introduced in this course. The lecture is self contained. However, if necessary, you may consult any introductory level text on ordinary differential equations. For example, "Elementary Differential Equations and Boundary Value Problems by W. E. Boyce and R. C. DiPrima from John Wiley & Sons" is a good source for further study on the subject. The course is mainly delivered through video lectures. At the end of each module, there will be a quiz consisting of several problems related to the lecture of the week.
4-4 Population Dynamics III
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來自 Korea Advanced Institute of Science and Technology 的課程
Introduction to Ordinary Differential Equations
63 個評分
Korea Advanced Institute of Science and Technology
63 個評分
從本節課中
Mathematical Modeling and Applications
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Kwon, Kil HyunProfessor
Department of Mathematical Sciences
More precisely, you can make the following guess.
Logistic curve for the logistic equation p prime is equal
to r minus ap times p. Now,
this horizontal axis is the time axis and this vertical axis is
the population axis p of t. Using this table,
you can make the following guess.
If your initial population, that is Po,
when t is equal to zero,
you are starting from Po.
So if the initial population is greater than r over a,
then the information we have in the data says,
the graph of p of t is decreasing and the concave up.
So, the graph should look like this red curves.
And p is equal to identical r over a.
This horizontal blue line,
this is a solution called the equilibrium solution.
That's a solution too.
On the other hand,
if we use initial population Po is between zero and r over a,
the information we get from the table tells us the graph of the P of t looks like.
We have two different cases.
If your initial population is between 0 and r over 2a.
And if the initial population is between r over 2a and r over a.
In this second interval for the whole interval the graph must be always increasing.
But depending on this small interval or that small interval, concavity may changes.
If your initial population is between these two,
then it should be always a concave down and the increasing.
So you get graph of this type;
Increasing and the concave graph.
If your initial population is between 0 and r over 2a,
then up to a certain time
your graph is always increasing but your concavity may changes, somewhere in between.
So, increasing concave up and the changes concavity,
still increasing by concave down.
That's the shape of the general logistic curve.
You can read out some others things too from this graph.
For example, any logistic curves starting from the initial point
Po which is greater than zero approaches the equilibrium solution why?
the P is equal r over a but it
never reaches the value r over a at the finite time by Picard's theorem.
Can you figure it out?
why can touch the line P is equal to r over a In any finite time?
Here I claim the following.
If we started from any point,
starting from any point
which Po is a positive and look at the red curves and blue curves,
both of them they approach is to this equilibrium position,
equilibrium state as t turns to infinity.
But I claim that,
it never reaches the value r of a by the Picard's theorem.
Why is it then? Lets assume that for example,
here is the P axis and here is the time axis.
Here's the equilibrium solution,
which is the line p is equal to r over a.
That's the equilibrium solution.
Let's take any one of them for example this one.
Lets assume that solutions starting from somewhere from there,
this is a Po.
If it touches the value r over a as a finite time at a finite time say To.
If it can touch it and keep going away,
consider the following the initial value problem,
p prime is equal to (r-nP)P and p of
To that is equal to r over a. P of To is equal to exactly this equilibrium value.
Consider this initial value problem.
It satisfies all the conditions required by the Picard's theorem. What does that mean?
By the Picard's Theorem,
this initial value problem should have
a unique solution defined [inaudible].
There should be only one solution satisfying this initial value problem.
But as I assume if you have a solution starting from
this point and it touches the point To,
r over a at the finite time To then,
In any n value for To you have two distant solution.
This is the solution,
equilibrium solution, another solution given by this curve.
Can you see it right?
You have at least two different solution.
They cannot coincide because one
represents a straight line and the other one is a genuine curve.
That's a contradiction, because
for the given initial value problem you have two distinct solution,
which is impossible by the Picard's Theorem which says,
certain initial value problem should have one should have only one solution.
That's why there is such a solution starting from any positive initial value
approaches to this equilibrium solution as t times to infinity and never touches it.
Moreover, if your initial point Po
is between 0 and r over
a, so moving here.
If your initial population is between these two,
then your solution, your integral curve is these blue curves.
Their approach is to the value r over a as to test infinity but never touches
it and always should lie below this equilibrium line.
This horizontal line.
So, this value is P is equal to r over a is
a kind of upper limit for all those solutions.
So we call this number r over a.
It's very reasonable to call this number r over
a the carrying capacity or the saturation level of this environment.
So far we get all those informations about
the logistic curves without solving the given logistic differential equation explicitly.
We get all those informations on the logistic curves by the qualitative analysis.
But in fact in this case,
it's easy enough to solve this logistic equation.
The logistic equations say p prime is equal to r minus ap
times p. Because as I said before this is a separable differential equation.
So, why not separate the variables?
Divide it through by r minus ap times p and the multiply dt on both sides,
you will get this one.
One over one minus ap times the pdp that is equal to dt.
Let's take a partial fraction decomposition, one over one mines ap, times p,
which is one over r over p,
plus a over r over r minus ap.
That's a partial fraction decomposition of this quotient here.
We know what to the next right?
Take integration of both sides,
you are going to get,
from the left you will get,
one of r times log absolute value of p,
plus one of r times log,
negative one over r times
the log absolute value of a over r minus ap.
So by the log rule,
you are going to have the left hand side,
one over r log of the absolute value of p over minus aP,
that should be equal to integral of dt.
That is the T plus some integral constants of one.
So, we can get to the general solution of
the given logistic differential equation very simply through these manipulations.
The general solution of this logistic equation is given by p of t is equal to r times
C over AC plus exponential negative r of t. C is arbitrary constant now.
The initial condition P of zero is called a Po,
which I assume to be non-negative.
You can compute this C in terms of Po,
that is C is equal to Po over r minus a times Po.
Plugin this quantity for C into this expression and simplify,
then you will get P of t is equal to r times Po over a times Po,
plus r minus a times Po,
times e to the negative rt.
This is the general solution of this logistic differential equation.
The function given in here.
The equation number five,
we called it as the logistic function.
Using this explicit expression of p of t you can
confirm our previous consequences obtained by
qualitative analysis of the differential equation easily.
You can check all the things that we get through the table,
by using this expressive form of the solution.
