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In this second interval for the whole interval the graph must be always increasing.

Â But depending on this small interval or that small interval, concavity may changes.

Â If your initial population is between these two,

Â then it should be always a concave down and the increasing.

Â So you get graph of this type;

Â Increasing and the concave graph.

Â If your initial population is between 0 and r over 2a,

Â then up to a certain time

Â your graph is always increasing but your concavity may changes, somewhere in between.

Â So, increasing concave up and the changes concavity,

Â still increasing by concave down.

Â That's the shape of the general logistic curve.

Â You can read out some others things too from this graph.

Â For example, any logistic curves starting from the initial point

Â Po which is greater than zero approaches the equilibrium solution why?

Â the P is equal r over a but it

Â never reaches the value r over a at the finite time by Picard's theorem.

Â Can you figure it out?

Â why can touch the line P is equal to r over a In any finite time?

Â Here I claim the following.

Â If we started from any point,

Â starting from any point

Â which Po is a positive and look at the red curves and blue curves,

Â both of them they approach is to this equilibrium position,

Â equilibrium state as t turns to infinity.

Â But I claim that,

Â it never reaches the value r of a by the Picard's theorem.

Â Why is it then? Lets assume that for example,

Â here is the P axis and here is the time axis.

Â Here's the equilibrium solution,

Â which is the line p is equal to r over a.

Â That's the equilibrium solution.

Â Let's take any one of them for example this one.

Â Lets assume that solutions starting from somewhere from there,

Â this is a Po.

Â If it touches the value r over a as a finite time at a finite time say To.

Â If it can touch it and keep going away,

Â consider the following the initial value problem,

Â p prime is equal to (r-nP)P and p of

Â To that is equal to r over a. P of To is equal to exactly this equilibrium value.

Â Consider this initial value problem.

Â It satisfies all the conditions required by the Picard's theorem. What does that mean?

Â By the Picard's Theorem,

Â this initial value problem should have

Â a unique solution defined [inaudible].

Â There should be only one solution satisfying this initial value problem.

Â But as I assume if you have a solution starting from

Â this point and it touches the point To,

Â r over a at the finite time To then,

Â In any n value for To you have two distant solution.

Â This is the solution,

Â equilibrium solution, another solution given by this curve.

Â Can you see it right?

Â You have at least two different solution.

Â They cannot coincide because one

Â represents a straight line and the other one is a genuine curve.

Â That's a contradiction, because

Â for the given initial value problem you have two distinct solution,

Â which is impossible by the Picard's Theorem which says,

Â certain initial value problem should have one should have only one solution.

Â That's why there is such a solution starting from any positive initial value

Â approaches to this equilibrium solution as t times to infinity and never touches it.

Â Moreover, if your initial point Po

Â is between 0 and r over

Â a, so moving here.

Â If your initial population is between these two,

Â then your solution, your integral curve is these blue curves.

Â Their approach is to the value r over a as to test infinity but never touches

Â it and always should lie below this equilibrium line.

Â This horizontal line.

Â So, this value is P is equal to r over a is

Â a kind of upper limit for all those solutions.

Â So we call this number r over a.

Â It's very reasonable to call this number r over

Â a the carrying capacity or the saturation level of this environment.

Â So far we get all those informations about

Â the logistic curves without solving the given logistic differential equation explicitly.

Â We get all those informations on the logistic curves by the qualitative analysis.

Â But in fact in this case,

Â it's easy enough to solve this logistic equation.

Â The logistic equations say p prime is equal to r minus ap

Â times p. Because as I said before this is a separable differential equation.

Â So, why not separate the variables?

Â Divide it through by r minus ap times p and the multiply dt on both sides,

Â you will get this one.

Â One over one minus ap times the pdp that is equal to dt.

Â Let's take a partial fraction decomposition, one over one mines ap, times p,

Â which is one over r over p,

Â plus a over r over r minus ap.

Â That's a partial fraction decomposition of this quotient here.

Â We know what to the next right?

Â Take integration of both sides,

Â you are going to get,

Â from the left you will get,

Â one of r times log absolute value of p,

Â plus one of r times log,

Â negative one over r times

Â the log absolute value of a over r minus ap.

Â So by the log rule,

Â you are going to have the left hand side,

Â one over r log of the absolute value of p over minus aP,

Â that should be equal to integral of dt.

Â That is the T plus some integral constants of one.

Â So, we can get to the general solution of

Â the given logistic differential equation very simply through these manipulations.

Â The general solution of this logistic equation is given by p of t is equal to r times

Â C over AC plus exponential negative r of t. C is arbitrary constant now.

Â The initial condition P of zero is called a Po,

Â which I assume to be non-negative.

Â You can compute this C in terms of Po,

Â that is C is equal to Po over r minus a times Po.

Â Plugin this quantity for C into this expression and simplify,

Â then you will get P of t is equal to r times Po over a times Po,

Â plus r minus a times Po,

Â times e to the negative rt.

Â This is the general solution of this logistic differential equation.

Â The function given in here.

Â The equation number five,

Â we called it as the logistic function.

Â Using this explicit expression of p of t you can

Â confirm our previous consequences obtained by

Â qualitative analysis of the differential equation easily.

Â You can check all the things that we get through the table,

Â by using this expressive form of the solution.

Â