Nondimensionalizing DE: Plankton Herbivore Model

  • Thread starter fauboca
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In summary: You have 3 variables including time so you can make 3 substitutions. You have 6 constants. You can not therefore make them all 1 for instance. You have a choice - there is not one unique 'correct' substitution. Your p = P/C makes sense -just do it, i.e. wherever you see P write Cp. But be careful to remember you have P in the left hand side too in eq. 1. Then your K = kC is reasonable too. Do it and you see you can take your C out of the bracket and it cancels nicely the C you just introduced on the LHS. Then you can change the
  • #1
fauboca
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This is a plankton herbivore model.

The dimensionalized model is

[tex]\displaystyle
\frac{dP}{dt} = rP\left[(K-P)-\frac{BH}{C+P}\right], \quad \frac{dH}{dt} = DH\left[\frac{P}{C+P} - AH\right][/tex]where r, K, A, B, C, and H are positive constants.

The dimensions of K, P, B, H, C have to be population (that is the only way I can see it to make since) then we have pop^2 - pop^2.

Then D or A has to be (pop)^{-1}.

I am trying to nondimensionalize to

[tex]\displaystyle
\frac{dp}{d\tau} = p\left[(k-p) - \frac{h}{1+p}\right], \quad \frac{dh}{d\tau} = dh\left[\frac{p}{1+p} -ah\right][/tex]

I have that [itex]p=\dfrac{P}{C}[/itex] but I can't figure out any others.

I think that [itex]k=\dfrac{K}{C}[/itex] and [itex]\tau = tr[/itex]

Are those correct? Even if they are, I can't figure out what H will be.
 
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  • #2
fauboca said:
This is a plankton herbivore model.

The dimensionalized model is

[tex]\displaystyle
\frac{dP}{dt} = rP\left[(K-P)-\frac{BH}{C+P}\right], \quad \frac{dH}{dt} = DH\left[\frac{P}{C+P} - AH\right][/tex]


where r, K, A, B, C, and H are positive constants.

The dimensions of K, P, B, H, C have to be population (that is the only way I can see it to make since) then we have pop^2 - pop^2.

Then D or A has to be (pop)^{-1}.

I am trying to nondimensionalize to

[tex]\displaystyle
\frac{dp}{d\tau} = p\left[(k-p) - \frac{h}{1+p}\right], \quad \frac{dh}{d\tau} = dh\left[\frac{p}{1+p} -ah\right][/tex]

I have that [itex]p=\dfrac{P}{C}[/itex] but I can't figure out any others.

I think that [itex]k=\dfrac{K}{C}[/itex] and [itex]\tau = tr[/itex]

Are those correct? Even if they are, I can't figure out what H will be.

Is anyone familiar with nondimensionalizing?
 
  • #3
Never mind nondimensionalized it.

How can I find the a,k parameter plane so I can answer:
Hence show that in the a,k parameter plane a necessary condition for a periodic solution to exist is that a,k lie in the domain bounded by a = 0 and [itex]a=4(k-1)/(k+1)^3[/itex]. Hence show that if a < 4/27 there is a window of values of k where periodic solutions are possible.
 
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  • #4
Sure you have never had any lessons and have no books nor examples that say a thing about this? (Though I recognise there are not a lot of suitable books and your teacher may be making up his own course).

I have played from time to time with this sort of equation. What I would ask myself, given that you are alerted to the existence of a centre is - where is it? Would have to find the stationary point(s). Which would be about the first thing you do in a problem like this anyway, centres or no centres. Then I would see what it looks like with equations rewritten with a s.p as origin.

Edit: There is one obvious stationary point - at H=0, P=0. There are probably others.
You have to linearise equations about a stationary point and analyse the stability of the linearised equation which is the local stability in the overall system. I think that says it all but if not the books surely explain it.
 
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  • #5
epenguin said:
Sure you have never had any lessons and have no books nor examples that say a thing about this? (Though I recognise there are not a lot of suitable books and your teacher may be making up his own course).

I have played from time to time with this sort of equation. What I would ask myself, given that you are alerted to the existence of a centre is - where is it? Would have to find the stationary point(s). Which would be about the first thing you do in a problem like this anyway, centres or no centres. Then I would see what it looks like with equations rewritten with a s.p as origin.

Have you ever used J.D. Murray's book? This is a challenging book. The problem I am having is find the a,k parameter plane. If I could get that, I could probably do the rest.
 
  • #6
You have 3 variables including time so you can make 3 substitutions. You have 6 constants. You can not therefore make them all 1 for instance. You have a choice - there is not one unique 'correct' substitution.

Your p = P/C makes sense -just do it, i.e. wherever you see P write Cp. But be careful to remember you have P in the left hand side too in eq. 1. Then your K = kC is reasonable too. Do it and you see you can take your C out of the bracket and it cancels nicely the C you just introduced on the LHS. Then you can change the whole outside multiplying constant to 1 if you like (but see below) by changing t (in both equations!). Inside your first square bracket you've got a combination BH/C. You can chose to define BH/C = h and make that substitution also in the second eq.

Whenever you have a function of constants only, you can replace it by defining a single new constant, so alternatively you could call AH h (h = H/A) and then in the first eq. you've got I think BH/C which becomes Bh/CA and then you define B/CA as E or some other letter.

However I think it is better to concentrate the changes in one of the equations. And I think it is most helpful to do it in the simplest equation so as to have one thing as simple as possible, so I would change the second equation rather than the first.

Finally for some purposes a different substitution may be convenient, that is you may at some point be analysing not the time derivative but dP/dH so you may make a choice that gives you unit coefficient for that.

Yes I have Murray's book and I met him once or twice long ago. I can't say I've read or studied it but I get his general ideas. Things like this a probably not real math for him but obvious preliminaries.
 
  • #7
fauboca said:
Never mind nondimensionalized it.

I have already nondimensionalized the problem.

I am trying to find the a,k parameter plane now.
 
  • #8
It would help to see what your dimensionless equations are.
 
  • #9
vela said:
It would help to see what your dimensionless equations are.

That is in the first post.
 
  • #10
I see the problem is an exercise from the book and all the principles/methods required are explained there and other examples given.

When you have found the domain it would be interesting to see your answers and computations showing the cycles.
 
  • #11
epenguin said:
I see the problem is an exercise from the book and all the principles/methods required are explained there and other examples given.

When you have found the domain it would be interesting to see your answers and computations showing the cycles.

The book doesn't explain how to find the a,k parameter plane for a system of ODEs. It shows how to parametrize two variables of a single equation only.
 
  • #12
fauboca said:
The book doesn't explain how to find the a,k parameter plane for a system of ODEs. It shows how to parametrize two variables of a single equation only.

Sorry, what you say you have just done is parametrise the variables which are in two equations. Moreover he has given you the result (one of the possible ones). (So that one way of understanding it would have been to work backwards from this result.)

This is an exercise where he is pushing you along the path in an application of the chapter where there are other illustrations, essentially same ideas different equations, so are like worked examples.

He asks you to show existence of a positive steady state. Convince yourself without formulae or calculations! Curve sketching. (Like for our other thread).

Then I would jump a bit and see what the nullclines are suggesting about the behaviour, again just qualitatively, I think you can see his behaviour description is at least plausible.

Then do what he asks - you have to rework the equation to make the nonzero s.p. your new origin as I mentioned at the start. Then go through what he calls the community matrix (I think proper mathematicians call it the Jacobian) of which he gives other examples.
 
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FAQ: Nondimensionalizing DE: Plankton Herbivore Model

What is the purpose of nondimensionalizing a differential equation?

Nondimensionalization is a process used in mathematical modeling to simplify and streamline a differential equation by removing any units and scaling the variables to more manageable values. This allows for easier interpretation and analysis of the equation.

How is nondimensionalization achieved in the Plankton Herbivore Model?

In the Plankton Herbivore Model, nondimensionalization is achieved by scaling the variables using characteristic values, such as the initial values of the variables or maximum values of their rates of change. This results in dimensionless equations that can be more easily studied and compared.

What are the benefits of nondimensionalizing a differential equation?

Nondimensionalization allows for the identification of key parameters and relationships between variables in a differential equation. It also simplifies the equation, making it easier to analyze and solve. Additionally, it can provide insights into the behavior and stability of the system being modeled.

How does nondimensionalization affect the interpretation of the Plankton Herbivore Model?

By removing units and scaling the variables, the Plankton Herbivore Model becomes more abstract and general, allowing for a broader understanding of the relationships between the variables and their behavior. This can aid in the prediction and control of the system being modeled.

Are there any limitations to nondimensionalization in the Plankton Herbivore Model?

While nondimensionalization can greatly simplify and enhance the understanding of the Plankton Herbivore Model, it does have its limitations. It may not accurately represent the true behavior of the system, as some important parameters or relationships may be lost in the scaling process. It is important to carefully consider the chosen characteristic values and their impact on the model before nondimensionalizing.

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