Practice DE Modeling for Rhinoceros Population

[HeLiXe]

This is not a homework question or anything assigned, but I would like some practice / assistance with modeling as we have not covered that in my DE class.

Homework Statement

The rhinoceros is now extremely rare. Suppose 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. However if the population is too small, the fertile adults will have difficulty finding each other when it is time to mate. Write a differential equation that models the rhinoceros population based on these assumptions. (Note that there is more than one reasonable model that fits these assumptions.)

Homework Equations

The Attempt at a Solution

Ok so if I were to model this--I would say that the rate of change in the rhinoceros population with respect to time is proportional to the size of the population times the parameter which would be the growth coefficient...and the growth coefficient would be affected if the population is small. There are no limiting resources or predators...

Independent variable = time (t)
Dependent variable = population of rhinoceroses (P)
Parameter = (k)

I've never used LaTex before -_-
so it would be basic unlimited population growth model to start with
dP/dt = kP
but I think there would need to be another parameter (g) the smallest possible population for which the adults can still meet. So if P < g then dP/dt < 0...and I'm not sure how to express that in the model .

 

[I like Serena]

Good! You defined your variables and constants.
And you set up a proper reasoning how the population expansion would be proportional to the population size.

dP/dt = kP
but I think there would need to be another parameter (g) the smallest possible population for which the adults can still meet. So if P < g then dP/dt < 0...and I'm not sure how to express that in the model .

I think you just did!

Aren't you saying something like:
\frac {dP} {dt} = <br /> \left\{ \begin{array}{ll}<br /> &lt; 0, &amp; \mbox{ if } P &lt; g \\<br /> k P, &amp; \mbox{ otherwise}<br /> \end{array} \right.

Perhaps you might consider graphing a function that describes what you need.
What kind of graph would match the function just defined?

 

[elegysix]

If rhino's being to spread apart affects your model, try relating your parameter to a density function. (rhinos/area) good luck!

 

[HeLiXe]

Good! You defined your variables and constants.
And you set up a proper reasoning how the population expansion would be proportional to the population size.

Yayyy! My programming teacher done taught me well

I think you just did!

Aren't you saying something like:
\frac {dP} {dt} = <br /> \left\{ \begin{array}{ll}<br /> &lt; 0, &amp; \mbox{ if } P &lt; g \\<br /> k P, &amp; \mbox{ otherwise}<br /> \end{array} \right.

Perhaps you might consider graphing a function that describes what you need.
What kind of graph would match the function just defined?

Yes this is what I am saying. I will think about the graph and mess around in Maplesoft a bit and come back with my findings :) Thanks I Like Serena!

If rhino's being to spread apart affects your model, try relating your parameter to a density function. (rhinos/area) good luck!

This is very interesting, I did not think of this at all...I will see how I can implement this. I have to go, but when I come back I will post what I've come up with. Thanks elegysix.

 

[I like Serena]

Well, here's a link to WolframAlpha, and the corresponding picture.
Somehow, I don't think this is what you had in mind!

I have set g=2, k=1/2, and the negative part to -1.
http://www.wolframalpha.com/input/?i=Piecewise[{{-1,+P+<+2}},+(1/2)+P

Can you pinpoint what is wrong with this graph for being a model to your problem?

 

[HeLiXe]

Well, here's a link to WolframAlpha, and the corresponding picture.
Somehow, I don't think this is what you had in mind!

I have set g=2, k=1/2, and the negative part to -1.
http://www.wolframalpha.com/input/?i=Piecewise[{{-1,+P+<+2}},+(1/2)+P

Can you pinpoint what is wrong with this graph for being a model to your problem?

Sorry I like Serena...I'm studying for a chemistry exam I have on Monday. I have a slope field that resembles what I have in mind

 

[I like Serena]

Sorry I like Serena...I'm studying for a chemistry exam I have on Monday. I have a slope field that resembles what I have in mind

No hurry, finish your chemistry exam first.
I have to admit that I was wondering if you had already lost interest, but for now I'll assume you just had pressing business elsewhere. :)

Now, let's take a look at your attempt.

It seems you've tried to set up the model, calculate, and show the results all in one go.
(My graph was only of the growth rate dP/dt versus population P.)
I presume your field is population P versus time t?

Assuming it is, there are a few things wrong with it.
I'll mention one: apparently the population growth is near infinite at any given time when there is a positive population!
That can't be right.
Can you point out a few other flaws?

Basically that's what modelling is about. You try to find a simple model to fit the facts.
Work it out. See if your model fits. And if it's not complete, add something to your model.
Until you cannot solve it anymore. Then you simplify the model again...

 

[HeLiXe]

No hurry, finish your chemistry exam first.
I have to admit that I was wondering if you had already lost interest, but for now I'll assume you just had pressing business elsewhere. :)

I never lose interest in math :) And modeling is something I have not figured out so I am FAR from losing interest in it . I will examine these things and will get back to you after my exam. We have to turn in our homework on our exam date and my prof gives 100+ problems for every chapter and they either require a lot of steps or are so simple and repetitious that it becomes a pain doing them -_-


Yes the field was dP/dt but I have to confess I did it very quickly just to give an impression of what I was idealising and I was trying to relate it to the figures you were using lol.
I definitely do not want an infinite growth at all times, I was idealising two points of equilibrium, exponential growth above the highest point and a curve declining between the highest and lowest equilibrium points and ending at the lowest point. This leads me to believe there has to be a fraction in the model with at least one variable (numerically speaking...it could be a fraction of a parameter over a variable) in the numerator and the denominator--and I am considering that along with elegysix's hint about relating the parameter to a density function, but I have not given this the attention I want to yet.

 

[I like Serena]

So we're agreed that we'll first try to find a function that properly describes the relation between current population P and growth rate dP/dt?
(Note that this is a function and not a vector field?)

I'll leave you to it for now, and I'll look forward to your response in which you may show some incredible insight!
(Or just show you want to learn more... whatever.)

Btw, will you let us know how your chem test went?

 

[HeLiXe]

So we're agreed that we'll first try to find a function that properly describes the relation between current population P and growth rate dP/dt?
(Note that this is a function and not a vector field?)

I'll leave you to it for now, and I'll look forward to your response in which you may show some incredible insight!
(Or just show you want to learn more... whatever.)

Btw, will you let us know how your chem test went?

omg lol yes I know this is a function we are trying to define. I really should not have responded until I have given it proper thought. And I will definitely show you that I need to learn more lol this is my second attempt at modeling ever :) Will certainly let you know about the chem exam also.

And I should note I am notorious for making the simple complicated and doing things backwards lol

 

[HeLiXe]

So we're agreed that we'll first try to find a function that properly describes the relation between current population P and growth rate dP/dt?
(Note that this is a function and not a vector field?)

I wonder now if there is something fundamentally wrong with my reasoning. Isn't a vector field defined by a function? Like the divergence of a vector field is a function? Sometimes it is easier for me to visualise the vector field if I am not sure of the model for the function or sure how the equation would look precisely in space...the vector field helps me to visualise this. I have not learned about vector fields in calculus yet, but "slope fields" in differential equations which I understand to be the same thing.

 

[I like Serena]

In a literal sense you are right. :)

A vector field is indeed a function. A special one, where a vector is assigned to each point in the plane.
I was talking more of a regular simple function, where a scalar value (population growth) is assigned to a scalar value (population size).

So we have a simple function f(P) that describes the population growth as a (simple) function of population.

I guess you can also visualize this as a vector field, based on population P and time t.
That is, a vector field that for each time t and population P shows what the growth rate of P versus time is.

 

[HeLiXe]

So we have a simple function f(P) that describes the population growth as a (simple) function of population.

-_- The "S" word again lol

I guess you can also visualize this as a vector field, based on population P and time t.
That is, a vector field that for each time t and population P shows what the growth rate of P versus time is.

Yes this is what I was doing. I will come back to this on Tuesday or Monday evening :) Thanks for all of your help so far I like Serena ! And thanks for the explanation.

 

[HeLiXe]

dP/dt = kP((P/g)-1)?
Because of this "However if the population is too small, the fertile adults will have difficulty finding each other when it is time to mate"
so P/g provides a ratio of the current population to the smallest possible population for which the adults can still meet. P < g then dP/dt < 0 if kP is multiplied by (P/g)-1 because if the population is greater than the smallest possible population for which adults can still meet then P/g should be > 1; if g>P then P/g<1 and if we subtract 1 from P/g when g>p then (P/g) - 1 < 0 and dP/dt<0.

I hope this makes sense I am severely lacking sleep and I feel I may be over looking something.

 

[HeLiXe]

It seems like it could be a possible model...

 

[I like Serena]

Looks good.
Could you make a graph?
Since I like pictures.

And can you solve the DE?

 

[HeLiXe]

Looks good.

Really? lol Yayyyyy

Could you make a graph?
Since I like pictures.

And can you solve the DE?

Will do as soon as I finish writing up this lab :)

 

[I like Serena]

And? Did you finish your lab? :)

Btw, you may have found it's rather hard to solve the DE.

 

[HeLiXe]

Patience is a virtue ILS Actually I did not finish the lab lol the Prof extended the due date because we have 4 lab reports due at the same time, so I have been working on all of them plus an extra credit assignment I got my DE lab back and got 100% though :) I have not even attempted to solve this equation, but have pondered it and I cannot think of how to solve it readily. Please do not give me any hints or anything yet because I have not seriously considered solving it yet. Sorry that I am taking long to get back to you with this model, but my chemistry class is coming to an end and the prof postponed some things because of the holidays, so now she is trying to cram everything in before the class ends.

 

[I like Serena]

Whaaaat? I think you have it backward lol.
Patience is not a virtue, impatience is!
See for instance here: http://patrickg.net/the-3-virtues-of-a-programmer/

(Weren't you interested in learning how to program too? )

 

[HeLiXe]

I promise I am not being lazy this time >_>

I am trying to get everything finished by Saturday and then I will be completely devoted to DE Although... I have been trying to sneak some of my DE skills into my chem lab reports

 

[Redbelly98]

I'm just now joining in this interesting thread.

dP/dt = kP((P/g)-1)

I was going to chime in with a somewhat lengthy explanation of a suitable equation, then realized that it was equivalent in form to this one. I'll just add that I would take g=1, the reasoning is that if there are even 2 rhinos then there is a nonzero probability that they could produce offspring together.

I have not even attempted to solve this equation, but have pondered it and I cannot think of how to solve it readily. Please do not give me any hints or anything yet because I have not seriously considered solving it yet.

I'll just mention, be prepared that you may have to solve this numerically.

EDIT: ... or go to wolfram alpha to see the solution. (The solution surprised me, but I won't spoil it for HeLiXe.)

 

[I like Serena]

Hey RB, welcome to this thread!
(The more the merrier.)

the reasoning is that if there are even 2 rhinos then there is a nonzero probability that they could produce offspring together.

Yes, but since the chance that they would meet is supposedly low, wouldn't the expectancy for the number of rhino's in the next generation be a decreasing number?

 

[HeLiXe]

I'm just now joining in this interesting thread.

I was going to chime in with a somewhat lengthy explanation of a suitable equation, then realized that it was equivalent in form to this one. I'll just add that I would take g=1, the reasoning is that if there are even 2 rhinos then there is a nonzero probability that they could produce offspring together.

Yes I'll take this into consideration. All lengthy explanations are welcome

(The solution surprised me, but I won't spoil it for HeLiXe.)

Thank you Redbelly98! Please no one tell me -_- i will avoid Wolfram Alpha like the plague! Thanks for putting up with me taking long to respond I am going through my Chem work at a good pace and should be finished tomorrow. As soon as I am done, I will be immersed in DE until Wednesday of next week =)

 

[Redbelly98]

Hey RB, welcome to this thread!
(The more the merrier.)

Thanks!

Yes, but since the chance that they would meet is supposedly low, wouldn't the expectancy for the number of rhino's in the next generation be a decreasing number?

I had the impression that we were ignoring rhinos dying, in which case the expectancy is still a slightly larger number. If we want to include dying off in the model, I am thinking we would add a term -k_2 \ p to the differential equation:

\frac{dp}{dt} = k_1 p (p/g - 1) -k_2 \ p

Oh, hey, wait a minute...

\begin{equation*} <br /> \begin{split} <br /> &amp; = k_1 \ p \ [p/g - (1+k_2/k_1)] \\<br /> &amp; = k_1 \ (1+k_2/k_1) \ p \ [\frac{p}{g (1+k_2/k_1)} - 1] \\<br /> &amp; = k \ p \ [\frac{p}{g&#039;} - 1]<br /> \end{split} \end{equation*} <br />
So never mind -- we get the same equation again. Mortality can be accounted for by choosing a larger value for g.

All lengthy explanations are welcome

Okay, here is what I was thinking:

1. The probability that a single rhino meets another rhino and reproduces (in some time interval) is proportional to the density of rhinos (or to the number of rhinos, if we take the land area to be a fixed parameter).

2. The rate of population growth is proportional to the number of rhinos, and to the probability that a single rhino reproduces, i.e. \frac{dp}{dt}=kp^2

3. If we modify the probability that a single rhino reproduces to include a threshold, or include a mortality rate, we get \frac{dp}{dt}=kp(p/g - 1) instead.

 

[HeLiXe]

Thanks!I had the impression that we were ignoring rhinos dying, in which case the expectancy is still a slightly larger number. If we want to include dying off in the model, I am thinking we would add a term -k_2 \ p to the differential equation:

\frac{dp}{dt} = k_1 p (p/g - 1) -k_2 \ p

Oh, hey, wait a minute...

\begin{equation*} <br /> \begin{split} <br /> &amp; = k_1 \ p \ [p/g - (1+k_2/k_1)] \\<br /> &amp; = k_1 \ (1+k_2/k_1) \ p \ [\frac{p}{g (1+k_2/k_1)} - 1] \\<br /> &amp; = k \ p \ [\frac{p}{g&#039;} - 1]<br /> \end{split} \end{equation*} <br />
So never mind -- we get the same equation again. Mortality can be accounted for by choosing a larger value for g.

How did you get g' ?

Okay, here is what I was thinking:

1. The probability that a single rhino meets another rhino and reproduces (in some time interval) is proportional to the density of rhinos (or to the number of rhinos, if we take the land area to be a fixed parameter).

2. The rate of population growth is proportional to the number of rhinos, and to the probability that a single rhino reproduces, i.e. \frac{dp}{dt}=kp^2

3. If we modify the probability that a single rhino reproduces to include a threshold, or include a mortality rate, we get \frac{dp}{dt}=kp(p/g - 1) instead.

Ahh so this is what elegysix meant by the density function. Coincidentally I was looking over this last night and noticed it is an odd numbered problem and the answer given by the text is the model you described in #3. They also indicate that there are many more correct models for this scenario...maybe this one is correcter

Still I am having trouble with land being a fixed parameter here ... how is it indicated in the model? Or better stated how is it defined? I have never encountered the term "fixed parameter." I did some searching last night and can only find it related to algorithms in programming--if I try to translate what I read about fixed parameters in programming to this model I would think that land as a fixed parameter is indicated by the value of g in the function...(g-1).

 

[I like Serena]

Hi HeLiXe!

How did you get g' ?

RB defined g&#039; = g(1+k2/k1).

g' is a new constant (not a derivative), which is made up of the 3 original constants (also called parameters).

it is an odd numbered problem

What do you mean by "an odd numbered problem"?

Still I am having trouble with land being a fixed parameter here ... how is it indicated in the model? Or better stated how is it defined? I have never encountered the term "fixed parameter." I did some searching last night and can only find it related to algorithms in programming--if I try to translate what I read about fixed parameters in programming to this model I would think that land as a fixed parameter is indicated by the value of g in the function...(g-1).

A fixed parameter for the model would be any constant related to the problem.

In this case the problem states "However if the population is too small, the fertile adults will have difficulty finding each other when it is time to mate. "

But what is "too small"?
This sentence hides a number of parameters.
One is land size, which is supposedly "large enough".

This is indeed captured in your parameter "g", but it probably is also part of your parameter "k".
Somehow g will be dependent on land size.
Because if the land size increases, logically g will increase as well.
However, there is no need (at this stage) to specify how g depends on land size.

RB is calling it a "fixed" parameter to emphasize that the parameter "land size" won't change, and that there is no need to define it as a separate parameter to the model.

 

[Redbelly98]

How did you get g' ?

Just combining constants from the previous equation:
g&#039; = g(1+\frac{k_2}{k_1})

Ahh so this is what elegysix meant by the density function. Coincidentally I was looking over this last night and noticed it is an odd numbered problem and the answer given by the text is the model you described in #3. They also indicate that there are many more correct models for this scenario...maybe this one is correcter

Still I am having trouble with land being a fixed parameter here ... how is it indicated in the model? Or better stated how is it defined? I have never encountered the term "fixed parameter." I did some searching last night and can only find it related to algorithms in programming--if I try to translate what I read about fixed parameters in programming to this model I would think that land as a fixed parameter is indicated by the value of g in the function...(g-1).

"Fixed parameter" is redundant, I guess, since parameters are normally held fixed anyway while the variables, uh, vary. 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&#039; - 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.

 

[Redbelly98]

What do you mean by "an odd numbered problem"?

It is a problem from a textbook. It's customary (in the USA, at least) for textbooks to have answers to odd-numbered problems, but not even-numbered ones, in the back of the book. Then professors have a choice of assigning problems where students either can or cannot check their answers before turning in a homework assignment.

 

[HeLiXe]

Hi HeLiXe!



RB defined g&#039; = g(1+k2/k1).

g' is a new constant (not a derivative), which is made up the 3 original constants (also called parameters).

Ahhhhh okay I totally missed that! I was thinking g prime as in a derivative.

What do you mean by "an odd numbered problem"?

The starting rhinoceros scenario is an exercise from my textbook and the answers to the odd numbered exercises are in the back of the book, this one has just the equation as redbelly had in #3. I thought this was an even numbered problem.

A fixed parameter for the model would be any constant related to the problem.

In this case the problem states "However if the population is too small, the fertile adults will have difficulty finding each other when it is time to mate. "

But what is "too small"?
This sentence hides a number of parameters.
One is land size, which is supposedly "large enough".

This is indeed captured in your parameter "g", but it probably is also part of your parameter "k".
Somehow g will be dependent on land size.
Because if the land size increases, logically g will increase as well.
However, there is no need (at this stage) to specify how g depends on land size.

RB is calling it a "fixed" parameter to emphasize that the parameter "land size" won't change, and that there is no need to define it as a separate parameter to the model.

OK I totally get it...I was actually thinking along these lines at first since I did not see a parameter added, but then I looked it up and started to make my own definition lol Thanks for explaining this for me!

 

[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&#039; - 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