00:01
Here i'll be going over the logistic differential equation.
00:05
I tend to like to talk about this system.
00:09
But we're going to look at the general form of the equation, what the general form of the solutions look like, and then solve for a particular example and see how that works.
00:22
So usually the logistic equation governs a population with limited resources, so there is a term in the differential equation in the rates on the right -hand side that looks very much like exponential growth, but population subject to resource limitations.
00:52
So that could be food, it could be water, it could be land.
00:59
Oftentimes we don't really say too much about what those limitations are.
01:04
But the first term, the r times p, if that was the only term in the equation, what would happen is you would have exponential growth, where r is a growth rate in a number of organisms per unit time.
01:31
So it is a growth rate, an inverse time, more or less.
01:39
The second term is where the limitations come in.
01:44
And what's important to note is that that term can be either positive or negative.
01:51
So the resource limitation term comes from the second factor.
01:57
And in particular, the thing to note is k is what's called the carrying capacity.
02:05
We're going to take a look at the situation with that caring capacity term.
02:15
So if all you had was the first factor, the r times p, you'd have exponential growth.
02:21
It's that second term that's going to govern what really happens to that population over time.
02:28
And the first term is always positive.
02:35
The second term can be either positive or negative.
02:40
And what happens, there's an attractor in the system.
02:51
So if we look at population over time, what happens is if the population starts out less than the caring capacity.
03:00
So there's our resource limitation.
03:04
If the population starts out less than the caring capacity, what will happen is there will be some growth.
03:17
And then that growth will slow down as the population reaches that carrying capacity.
03:27
And so if that initial value is less than the k, you approach the tractor from below and asymptotically approach it.
03:38
If you start off, though, with a population bigger than the carrying capacity, well, guess what happens there? again, you are going to be attracted to k.
03:50
Sorry, i should label that k, carrying capacity.
03:57
So that is what is sometimes called an attractor because all solutions lead to that final answer at infinity.
04:10
But if you start with population of zero bigger than the carrying capacity, what will happen? stuff will die off.
04:19
Bad, mad, mad, mad, bad.
04:20
Stuff's going to die.
04:22
Okay, bad, mad, mad, mad, mad.
04:25
Sad.
04:26
Yeah, let's make a crying population.
04:30
So the population will die off and approach that carrying capacity from above.
04:36
So that's what we mean by an attractor is that all solutions approach that.
04:42
So let's take a look at a particular example of logistic equation, dp by d t and i'll go ahead and write that out as two factors 0 .5 p times 1 minus 0 .825 p sorry i said those zero's a little funny um and so our rate constant is 0 .5 that's a fairly rapid growth uh our carrying capacity 1 over k is equal to 0 .0 .05.
05:27
I think i said that a little bit better than oat.
05:30
And if we invert that, of course, we get the carrying capacity of 200.
05:37
Ok, so there's our tractor.
05:42
And it turns out there's a way to solve this differential equation by separating and integrating.
05:51
And the other math tool that you use is something called the method of partial fraction let me say that right.
06:00
So solve by, first of all, separating, and then integrating.
06:17
And the method that you're going to use to integrate is called partial fractions.
06:23
And this is where you need to know what your initial condition is.
06:28
And i'll talk a little bit about that.
06:31
When we get into the math a little bit, there's kind of a little trick that you have to pull.
06:36
Pool depending on whether there's going to be a great die -off or an exponential growth, depending on what side of the carrying capacity you start from.
06:48
Okay, but anyway, we can do the little stuff that you usually do, d .p, and then we take all the p terms together.
06:58
Actually, we'll put the half with the dt.
07:01
So we'll divide by p on both sides, and then we'll divide by 1 minus that.
07:08
Carrying capacity term times p.
07:12
So both factors are in the denominator, and we'll keep the rate, the rate constant with the dt, because that is units of one over time.
07:27
It should go with a time.
07:30
And now we'll integrate.
07:34
Don't forget that when you do that, you have to add an arbitrary constant onto the other side.
07:40
That comes about due to the initial condition on the population.
07:48
So that's your arbitrary constant.
07:58
Okay, now, what about the method of partial fractions? what you basically do is you say that the product of those two factors is a sum of their individual factors in the denominator, and you give arbitrary coefficients, to those two terms.
08:27
And then you do a little bit of fractions in order, fractions in common denominator in order to find out what a and b are.
08:36
So what you're going to be doing is matching a and b to the appropriate term in the numerator on the left -hand side.
08:47
So that's done after you find the common denominator.
08:55
So if you hated adding fractions.
08:58
This would probably give you a little creepy creeps.
09:02
But you basically multiply the first term by the factor that's missing in there, and the second term by the factor that's missing in its denominator.
09:15
And then the denominator is the common p times 1 minus point aught 0 .15 p.
09:25
And that's all added in the numerator.
09:29
And then you expand the numerator.
09:32
So we're just going to take a look at the numerator.
09:36
Drops numerator.
09:38
So we have 1 plus 0 times p times p multiplier there.
09:49
So it doesn't look like a decimal is equal to a minus, oops, i have to p out of there.
09:58
Yeah, there we go.
10:00
I'll get that right.
10:01
A minus a times point a ought ought 5 p, expanding out that first term, and then plus b p.
10:15
Okay, and then we see that the constant on the right -hand term side is just a, and that's got a match up with one.
10:24
So we can see immediately that a is equal to one.
10:29
And if we have to match the other terms with the p coefficients, so we have 0 is equal to minus point aught 5a plus b.
10:47
And since a is one, we can put that in, and so that tells us that b is point aught 5.
10:58
So now we can simply take that integral and break it into two pieces on the left -hand side.
11:04
So we're going to break up left -hand side into two integrals.
11:20
So we're going to have dp over p.
11:30
Let's see, and then we have b plus point aught 5 over the integral of d p, 1 minus point a watt 5p is equal to 0 .5 t plus constant.
11:56
And at this point, it really makes sense to think about your initial condition.
12:03
If p is bigger than the carrying capacity, some point of time in reference, say time t equals zero, then that second integral is going to have a negative denominator.
12:22
And so both forms are a logarithm of some sort.
12:26
Okay, one over something.
12:29
And if your p is bigger than the carrying capacity, you're going to have issues with this term because then you'll have the algorithm of a negative number, which you cannot do.
12:53
That is undefined.
12:56
Whereas if p, and i shouldn't say it has issues, it is negative, and it is positive, and it is positive, if p starts off less than.
13:16
And i really should say that...