Introduction to Population Models and Logistic Equation (Differential Equations 31)

How differential equations can be applied to population models. We also explore the Logistic Equation, Population Explosion, and Population Extinction from a mathematical perspective involving limits.

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[wisamdoghoz]

You guys I have noticed someting off about the explosion condition. It does explode when p0 > m but only at a specific point in the futrue when mtk=ln(P0/m) which gives us zero in the denominator and in return either plus or minus infinity depending if you are approaching the point from the left which gives us plus infinity i.e. the explosion we are talking about. However, if you move a tiny bit in the future you get minus infinity and after a little while it reaches zero and stays there. It was pretty confusing to me at the beginning and needed to spend some time to get it right.

[nathantraylor]

Thank you for everything. It's funny how things never make sense until they do.

[TimothyShevgun]

Great video! Great job!
Logistic equation is a hardcore thing indeed. Much more difficult than one may guess, especially when it comes to the general case.
This should be the best explanation on the topic I have ever seen so far.

However, I must notice there is a sort of EXPLANATORY MISTAKE where it comes to 'explosion/extinct' version of the model at 59:30. No formal mistakes are though.

Part 1. Dynamics analysis.
Let's look at the equation dP/dt = k*P*(P-M), where k>0 and M>0.
We can identify 3 cases.
1) If current value P>M (for instance, as P0>M), then factor (P-M)>0 and the right-hand side (RHS) of the equation gives us dP/dt>0. It means that P(t)->+inf.
2) If current value 0<P<M (such as 0<P0<M), then (P-M)<0 and the RHS of the equation gives us dP/dt<0. It drives P(t)->0 (0+ to be a bit more precise).
3) If P<0 (just for the purpose of generality), then both (P-M) and p are less than 0 and the RHS gives us dP/dt>0. It again moves P(t)->0 (but it is 0- in this case).
Since we deal with the differential equation with continous RHS, no possible way seems to exist for jumping between regions described in cases 1, 2 and 3. Actually it is a debatable question, especially about the regions in case....1 and 3, we may see it a bit later, at the second glance, that they turn out to make a joint region.

Some more details:
Equ. made for RHS k*P*(P-M)=0 gives us two roots P=0 and P=M. The first is a point of stable equilibrium, while the second (P=M) is non-stable equilibrium. This can simply be determined by the sign of d[k*P*(P-M)]/dP evaluated at those points.
[Cf.: the logistic version has the same equilibrium points but of vise versa stability].

Part 2. The analysis using the formal expression.

This expression is the easiest to understand in case 2, if 0<P0<M. So, -(P0-M)>0, and we may see positive denominator, exponentially increasing with t->+inf.
In case 3, where P0<0, it can be shown that the denominator remains positive at least for all t>=0. Indeed, abs(P0-M)>abs(P0), since P0<0, M>0.
But if we really want to clarify case 1 (P0>M), we should rewrite our expression in the following form:
P(t)=M/[1-exp(k*M*(t-tau))],
where tau = -1/(kM)*ln(1-M/P0). Notice tau>0 since M<P0 and ln(1-M/P0)<0
Only now can we see what is going on! If P0 > M we will see an explosive growth toward +inf up to the moment t = tau (0<t<tau). There, at t=tau, we get so called ‘singularity’ point. One may roughly say the infinite jump happens. However, just after it, where t>tau (starting from t=tau+0), we will get the negative value for P(t) by means of our expression. If we go to t->+inf after it (t>tau), we will see P(t) move from -inf to 0-.

Final comment.
The solution for ‘logistic’ version of the equation will have the same ‘feature’ if P0>M, but tau<0 in that version. However, as long as t>0 is region of interest, this issue stays in the shade.

[ayushishrivastava]

Sir your videos are amazingly helpful, after watching your videos I feel more confident with topics related to differential equations..
Kindly consider partial differential equations next.
Thank-you.

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[benradick]

Sir I think there is a mistake around 58:43. Isn't the limit of P(t) in this case going to be 0, no matter the starting condition?

For M>Po, the equation is a scaled version of:

P=1/(1-e^t)

And for M<Po:

P=1/(1+e^t)

The limit of these as t--> infinity is 0.


Perhaps I have made a mistake?

[doomerman]

Hey professor please do Laplace transforms soon!

[hitm]

1:00:00 if the second term in the denominator is getting larger and larger and we subtract it from Po, wouldn't that mean the denominator approaches negative infinity and therefore P(t) -> 0? The conclusion of P(t) -> infinity makes more sense if the second term in the denominator approaches Po such that the denominator gets smaller and smaller and therefore P(t) larger and larger. Not sure if that is what was meant here.

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[nicholi]

If you had been my professor I may have learned something in dif eq. Though I seem to know more than I thought, this was still good to go over. I know in chemistry reactions can reach an equilibria, is this not a term used for your first case?

[carolinemitchell]

Holy cow. Fantastic. Thank you so much!

[aysezulalyilmaz]

Can a threshold value of a population be a rational number, or do I need to approximate the number to the nearest integer while making calculations?