Practice DE Modeling for Rhinoceros Population

[I like Serena]

And?
Did you try to solve the DE?

As it is you have a model that satisfies the constraints of the problem, but which is hard to solve.

An alternative road is to simplify the model.
Any ideas on that?

(Btw, there's nothing wrong with doing both! )

 

[HeLiXe]

Hi redbelly,
I was a bit impatient and asked ILS to answer me lol. I was trying to be patient, but I really wanted to know what a fixed parameter is.

"Fixed parameter" is redundant, I guess, since parameters are normally held fixed anyway while the variables, uh, vary.

lolololl

But land area is not one of the explicit parameters in this problem -- k and g are. Somehow the area is incorporated in one or both of them.

Thinking about the "growth rate" for a single rhino (in a population of many), we might expect it depends on two things:

1. The probability of meeting another rhino and mating successfully, which is proportional to population density p/A. (Here is where the land area explicitly comes into play.) Call this growth rate (or probability) kp/A.
2. The probability that the rhino dies, which we can take to be constant for simplicity. Call this g.

So we get $k \frac{p}{A} - g$ as the growth "per rhino", per unit time. Multiply that by p, the total number of rhinos, to get the growth rate of the entire population:

$$\frac{dp}{dt}=p(kp/A - g)$$

we can rearrange things and combine the constants, so that A gets incorporated into k, in which case it won't explicitly appear in the equation.

Also, it's instructive to have the factor in parentheses in the form $(p/g' - 1)$. Then we see immediately see that p=g' represents a "break even" point for the population, where births and deaths exactly balance out. Of course, g' here is the g you had written earlier, and it can depend on things like mortality rate and land area.

Hope that helps. By now in this thread we've seen several different forms for the dp/dt equation, but they're essentially the same, just with different definitions of the constants.

Excellent this is a very good explanation.

When you say that we can rearrange things and combine A into k...how is this done? I am not asking for you to work it out for me, just an indication of what you mean exactly.

 

[I like Serena]

And since I like pictures, here's a graph of our current model (k=2, g=10).
It's a graph with P on the horizontal axis, and dP/dt on the vertical axis.
I like to think of dP/dt as the change in population in one generation.

http://www.wolframalpha.com/input/?i=Plot[2+p+(p+/+10+-+1),+{p,+-10,+30}]

I'll just point out a few things to note.

1. At P=0, the population remains constant at zero, which is good.

2. At P=10, the population remains constant as well, which is our break even point (g).

3. At P=5, the population halves in the next generation, which sounds about right.

4. At P=20, the population triples in one generation, I think I overdid the parameter k.

5. At P=30, the population is quintupled in one generation, which is too much I think.
And at higher population sizes, this "problem" becomes unrealistically worse.

6. For negative P, the population increases, which is not right.
So we should set the model for instance to zero for negative P.

 

[HeLiXe]

And?
Did you try to solve the DE?

Well ...you may be partially pleased to know that I have finally started working on it lol

As it is you have a model that satisfies the constraints of the problem, but which is hard to solve.

An alternative road is to simplify the model.
Any ideas on that?

(Btw, there's nothing wrong with doing both! )

I will give it some thought and continue with trying to work this one out since I just started with it.

 

[ehild]

Should not this be a two variable problem? If I remember well, there has to be a male and a female to produce a baby, but only females can give birth. So the growth rates of both male and female rhinos are proportional to the probability that a female rhino meets a male during the mating period, and the number of female rhinos.
The baby can be a male or female, say, with equal probability and you have to subtract the dying rate. But it can be even more complicated than that.

ehild

 

[Redbelly98]

When you say that we can rearrange things and combine A into k...how is this done? I am not asking for you to work it out for me, just an indication of what you mean exactly.

I mean, when you can identify an expression that contains only constants, then that expression is also a constant. This is what I did in Post #25 when I came up with the new constant g'. The previous equation contained the expression $g(1+k2/k1)$ with all constants. So that expression is a constant too. All those constants get combined into a single constant g'.

In Post #28, I had written
$$\frac{dp}{dt}=p(kp/A−g)$$
Focus on the term kp/A, which contains two constants. If we define a new constant k'=___(?), then we can reduce that to one constant.

Should not this be a two variable problem? If I remember well, there has to be a male and a female to produce a baby, but only females can give birth. So the growth rates of both male and female rhinos are proportional to the probability that a female rhino meets a male during the mating period, and the number of female rhinos.
The baby can be a male or female, say, with equal probability and you have to subtract the dying rate. But it can be even more complicated than that.

ehild

Yes, strickly speaking you're right. We could have even more than two variables to model different age groups, since newborns and extremely aged rhinos have a different mortality rate than adult males.

But for an introductory problem on population models, it's probably best to keep things simple. If this is the first time you're modelling a population, just model it as a single population to familiarize yourself with the concepts.

 

[ehild]

We could have even more than two variables to model different age groups, since newborns and extremely aged rhinos have a different mortality rate than adult males.

But for an introductory problem on population models, it's probably best to keep things simple. If this is the first time you're modelling a population, just model it as a single population to familiarize yourself with the concepts.

You are right, and your model dp/dt=p(kp/A−g) looks quite appropriate. Is g the dying rate?
The solution is rather strange, however.

I just wonder how is it possible to include the condition that the male rhinos do not find a female if the population density is too low.

ehild

 

[Redbelly98]

Hey, I am really liking this whole discussion.

5. At P=30, the population is quintupled in one generation, which is too much I think.
And at higher population sizes, this "problem" becomes unrealistically worse.

An underlying assumption here is that the population is sufficiently small. From the OP:

"...enough game preserve land is set aside so that there is sufficient room for many more rhinoceros territories than there are rhinoceroses. Consequently there will be no danger of overcrowding."​

So unrealistic behavior at higher population levels is not surprising. The probability that an individual produces offspring in one breeding period increases linearly with population density in our model, but we know it can never exceed 100% in reality. To alter the model, this probability should have a maximum level. But that is something to try after HeLiXe has solved the problem at hand first. One step at a time.

6. For negative P, the population increases, which is not right.
So we should set the model for instance to zero for negative P.

What would constitute "right" behavior for a negative population? Negative populations are unphysical to begin with, so do we really care what the math says about them? It's neither right nor wrong, as you could never do an experimental test of the model for negative p.

(In case it's not evident, I am an experimental physicist by training, not a mathematician.)

 

[Redbelly98]

You are right, and your model dp/dt=p(kp/A−g) looks quite appropriate. Is g the dying rate?
The solution is rather strange, however.

Yes, in that form of the equation g is the death rate. In HeLiXe's dpt/dt=kp(p/g-1), the death rate is k. k and g mean different things in the two equations.

The solution does get strange as the population gets larger. But the model is not valid for large populations -- see my previous post.

I just wonder how is it possible to include the condition that the male rhinos do not find a female if the population density is too low.

ehild

As long as there are two rhinos, there is >0 probability they will meet, and that they are of opposite genders. To include the possibility that it doesn't happen (i.e., that either they don't meet, or they are the same gender), you could add a random number generator to a computer simulation of the problem. But so far, we are essentially taking expectation values, and treating population as a continuous variable -- which breaks down when the population is just 2.

 

[ehild]

I imagine that each rhino occupies a territory, and these territories or cells are regular hexagons of the same size. Each such cell has 6 neighbours. Imagine the central cell is occupied by a female, she is waiting for a male. She can meet a male if a neighbouring cell is occupied by one.
There is lot of land, so the probability that a cell is occupied can be considered constant. But the probability that an occupied cell has a male or female inhabitant is proportional to the ratio of males (females) to the whole population.
It is needed to estimate the probability that no females are in the neighbourhood and at least 1 cell is occupied by a male. If yes, there can be a baby. The number of babies is proportional to this probability and the number of females. The babies can be boys and girls with equal probability. And there is a death rate, equal to both sex.

The problem can be simplified by assuming nearly equal number of males and females, and solving for small deviations, (linearising the problem).

ehild

 

[HeLiXe]

Still working on it everyone...no hints yet please! lol I think this is an autonomous nonlinear equation because it is similar to the logistic differential equation. I'm trying to solve it analytically using separation of variables but I am getting stuck, so I might try the numerical deal redbelly mentioned

 

[elegysix]

I'll just point out a few things to note.

1. At P=0, the population remains constant at zero, which is good.

2. At P=10, the population remains constant as well, which is our break even point (g).

3. At P=5, the population halves in the next generation, which sounds about right.

4. At P=20, the population triples in one generation, I think I overdid the parameter k.

5. At P=30, the population is quintupled in one generation, which is too much I think.
And at higher population sizes, this "problem" becomes unrealistically worse.

6. For negative P, the population increases, which is not right.
So we should set the model for instance to zero for negative P.

Have you considered trying dP/dt = const*(P - 10) ? This would have the same intercept as your current one, where growth is 0 at P = 10.


Letting Po be initial population, growth would be negative for 0<Po<10, and for Po>10 it would not grow as quickly as your current model. It is also an easy DE to solve, and P will be an exponential function, which I think is the normally used to model population growth.

P would eventually be negative for initial P<10 , but I think its reasonable to assume that a negative population is just 0.

 

[I like Serena]

Hey, I am really liking this whole discussion.

An underlying assumption here is that the population is sufficiently small. From the OP:

"...enough game preserve land is set aside so that there is sufficient room for many more rhinoceros territories than there are rhinoceroses. Consequently there will be no danger of overcrowding."​

So unrealistic behavior at higher population levels is not surprising. The probability that an individual produces offspring in one breeding period increases linearly with population density in our model, but we know it can never exceed 100% in reality. To alter the model, this probability should have a maximum level. But that is something to try after HeLiXe has solved the problem at hand first. One step at a time.

Just on what feels right, I'd like the model to behave more or less up to 1000 pop I guess.

What would constitute "right" behavior for a negative population? Negative populations are unphysical to begin with, so do we really care what the math says about them? It's neither right nor wrong, as you could never do an experimental test of the model for negative p.

(In case it's not evident, I am an experimental physicist by training, not a mathematician.)

The mathematician in me says the domain should be limited to non-negative populations.
So the model should be undefined for negative P.

As a computer scientist I'd be using numerical methods to solve the model.
When it oversteps into negative P, I don't want the solution to spin out of control, but preferably kind of "freeze".
Setting it to zero will do that. :shy:

As a physicist I don't care what happens, I just want to see results that match with experimental data.

(In case it's not evident, I had difficulty choosing, although I did specialize in software engineering.)

I imagine that each rhino occupies a territory, and these territories or cells are regular hexagons of the same size. Each such cell has 6 neighbours. Imagine the central cell is occupied by a female, she is waiting for a male. She can meet a male if a neighbouring cell is occupied by one.

Hmm, nice idea!
But... I don't see how we could define and solve that as a DE...

Still working on it everyone...no hints yet please! lol I think this is an autonomous nonlinear equation because it is similar to the logistic differential equation. I'm trying to solve it analytically using separation of variables but I am getting stuck, so I might try the numerical deal redbelly mentioned

Go go go!

(I didn't think I could solve the DE until I saw the solution at WolframAlpha. )

Have you considered trying dP/dt = const*(P - 10) ? This would have the same intercept as your current one, where growth is 0 at P = 10.


Letting Po be initial population, growth would be negative for 0<Po<10, and for Po>10 it would not grow as quickly as your current model. It is also an easy DE to solve, and P will be an exponential function, which I think is the normally used to model population growth.

P would eventually be negative for initial P<10 , but I think its reasonable to assume that a negative population is just 0.

Very good!

I think this would be a pretty good model.
And simple!
I like simple (until it's not good enough to do what you want it to do).

Do you have a picture?
(I like pictures! )

 

[HeLiXe]

Do you have a picture?
(I like pictures! )

Here is a really cool picture:

I refuse to check wolfram alpha but I may ask for help. I will post my progress in a few hours--I might even try to do it in laTEX :D.

 

[I like Serena]

Here is a really cool picture:

I refuse to check wolfram alpha but I may ask for help. I will post my progress in a few hours--I might even try to do it in laTEX :D.

Cool!

Did you see how the horns are shaped like different models to the problem?

 

[HeLiXe]

Cool!

Did you see how the horns are shaped like different models to the problem?

 

[HeLiXe]

I guess a few hours has turned into a few days + more

I am sorry everyone and I really appreciate your patience with me. I have another chem exam this week, a final exam next week and a DE exam tomorrow so my attention has not been with this model. I really feel bad about not getting back to this sooner but it is really a hectic time in the semester -_- + my brain is partially fried so I'm operating at like 60% efficiency lol

 

[HeLiXe]

OK so it is good I took a break from this lol. I just did it in 30 mins and I have a solution...which may not be the correct solution but it is a solution. I want to try LaTex, but I think it would be easier for me to just take a pic of my work and post it lol. I got ce^rt/[(P/g)-1]=P? If this is it I can understand why you were saying it is surprising lol I did all of these algebraic gymnastics and looked down on my paper and saw P((P/g)-1) again lol.

btw I was working on the model I came up with, not the "correcter" one

 

[Redbelly98]

I got ce^rt/[(P/g)-1]=P?

Cool, let's think about this...

You have P on the left-hand-side there. Did you not get P as an explicit function of t?

It would be good to verify that this solution is consistent with the differential equation. To do that, I'd multiply through by [(P/g)-1] to get rid of the fraction, then try implicit differentiation.

EDIT: just thought I'd go ahead and LaTeX this thing:

$$\frac{c e^{rt}}{\frac{P}{g}-1} = P$$

 

[HeLiXe]

Cool, let's think about this...

You have P on the left-hand-side there. Did you not get P as an explicit function of t?

It would be good to verify that this solution is consistent with the differential equation. To do that, I'd multiply through by [(P/g)-1] to get rid of the fraction, then try implicit differentiation.

EDIT: just thought I'd go ahead and LaTeX this thing:

$$\frac{c e^{rt}}{\frac{P}{g}-1} = P$$

ok first I had it as

$$c e^{rt} = {P}({\frac{P}{g}-1})$$
Then I was thinking I have to solve for P and divided by
$${\frac{P}{g}-1}$$
I did not like that P on the left side but was unsure what to do.

This LaTex thing isn't so bad

 

[Redbelly98]

Well, you have a quadratic expression in P: ${P}({\frac{P}{g}-1}) = P^2/g - P$. That's solvable, but kinda messy, so-o-o-o...

... before we go solving for P, I recommend confirming whether it satisfies the differential equation first. (If it doesn't, then there's no need to spend time on solving for P ) So, implicit differentiation and $c e^{rt} = {P}({\frac{P}{g}-1})$ ...

 

[I like Serena]

OK so it is good I took a break from this lol. I just did it in 30 mins and I have a solution...which may not be the correct solution but it is a solution. I want to try LaTex, but I think it would be easier for me to just take a pic of my work and post it lol. I got ce^rt/[(P/g)-1]=P? If this is it I can understand why you were saying it is surprising lol I did all of these algebraic gymnastics and looked down on my paper and saw P((P/g)-1) again lol.

btw I was working on the model I came up with, not the "correcter" one

I'm glad you decided to stay and solve it. :)

But I don't understand how you got to your solution.
What did you do to arrive at it?

 

[HeLiXe]

Yes ILS I thought you were losing faith in me

I solved it analytically by separation of variables. I think I will just post a picture of the paper I worked on...well...I will rewrite it and post it because that paper is a bit wild. And it is ce^kt not rt...we were working with r as a constant on my last DE exam and I arbitrarily posted it here.

 

[HeLiXe]

Well, you have a quadratic expression in P: ${P}({\frac{P}{g}-1}) = P^2/g - P$. That's solvable, but kinda messy, so-o-o-o...

... before we go solving for P, I recommend confirming whether it satisfies the differential equation first. (If it doesn't, then there's no need to spend time on solving for P ) So, implicit differentiation and $c e^{rt} = {P}({\frac{P}{g}-1})$ ...

Ok redbelly...I think I see where you are going with this. I am still foggy about DE because we are just solving everything in class, but I am not really understanding the application of them or why they are being used (which is why I started the modeling practice). So I could be missing some fundamental things and I appreciate if anyone points them out to me!

I will post my work for this solution.

 

[HeLiXe]

Well in writing it over I noticed that I made a mistake when I was doing the partial fraction decomposition..."A" was supposed to = -1 and I wrote it as 1 when I was putting the value of A in place of A, so that changes the answer to

$${\frac{\frac{P}{g}-1}{P}} = c e^{kt}$$

Here is my work
[PLAIN]http://img810.imageshack.us/img810/4249/rhincalc.png
the "g" got cut off at the bottom there :D

 

[Redbelly98]

Well in writing it over I noticed that I made a mistake when I was doing the partial fraction decomposition..."A" was supposed to = -1 and I wrote it as 1 when I was putting the value of A in place of A, so that changes the answer to

$${\frac{\frac{P}{g}-1}{P}} = c e^{kt}$$

And the best part is, that can be solved for P explicitly, no sweat

(And then you can get dP/dt, and confirm whether the original differential equation is satisfied)

 

[HeLiXe]

cool

 

[I like Serena]

Well in writing it over I noticed that I made a mistake when I was doing the partial fraction decomposition..."A" was supposed to = -1 and I wrote it as 1 when I was putting the value of A in place of A, so that changes the answer to

$${\frac{\frac{P}{g}-1}{P}} = c e^{kt}$$

Very good!

Sorry I tried to get you to do the simpler case.
Apparently that would have been no challenge at all!

But, uhh, what is P explicitly?

And I want a picture!

 

[HeLiXe]

Very good!

Sorry I tried to get you to do the simpler case.
Apparently that would have been no challenge at all!

I think it would have been for me -_- The things others usually refer to as "simple" are usually very complicated for me lol

But, uhh, what is P explicitly?

And I want a picture!

Is not the picture of my work good enough??
I have some errands to run and stuff but when I get a chance I will do the explicit thing and post a picture of it

 

[I like Serena]

I think it would have been for me -_- The things others usually refer to as "simple" are usually very complicated for me lol

Aaaah, so that's what the remark about the S-word was about!
So S-problems are more of a challenge for you?

Is not the picture of my work good enough??
I have some errands to run and stuff but when I get a chance I will do the explicit thing and post a picture of it

Well... errr... yes... the picture of your work was quite good... but... I can only say..

More! More! More!