[SOLVED]Nondimensional

dwsmith

Well-known member
This a plankton herbivore model.

The dimensionalized model is

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

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

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

I am not sure what is a good starting point. I need a hint on one dimensionless unit.

Jester

Well-known member
MHB Math Helper
Start with $p = \dfrac{P}{C}.$

dwsmith

Well-known member
Start with $p = \dfrac{P}{C}.$
How were you able to identify that as a substitution?

$k = \frac{K}{C}$ then correct?

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Jester

Well-known member
MHB Math Helper
Well, $P$ and $C$ have the same dimenson, right? So let $P = Cp$ so that $p$ is dimensionless.

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dwsmith

Well-known member
Well, $p$ and $C$ have the same dimenson, right? So let $P = pC$ so that $p$ is dimensionless.
How I see the problem I believe they do unless I am wrong. I wasn't told any dimensions but I know P and H are population and t is time.

Jester

Well-known member
MHB Math Helper
How I see the problem I believe they do unless I am wrong.
You might want to rephrase this statement - it makes no sense!

dwsmith

Well-known member
You might want to rephrase this statement - it makes no sense!
How I see the problem, I believe P and C have the same dimensions unless I am wrong.

HallsofIvy

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MHB Math Helper
You said they were both "population". What different dimensions could there be? I suppose P could be measured in "people" and C in "thousands of people" but the ratio would still be dimensionless- if P is 120000 people and C is 400 "thousands of people", P/C= 12/3 (people/1000 people)= 4/1000.

dwsmith

Well-known member
Are these substitutions correct?

$k=\dfrac{K}{C}$ and $\tau=tr$.

I can't figure out $r,H,B$ in the first DE as well and $A$ and $D$ in the second.

dwsmith

Well-known member
So skipping the nondimensionalizing, next I am trying to show that for 0 < k < 1 the positive steady state is stable by noting the signs $\frac{dp}{d\tau}$ and $\frac{dh}{d\tau}$.

So the derivatives are

$$k - 2p - \dfrac{h - p}{(1 + p)^2} \ \text{and} \ \dfrac{dp}{1 + p} - 2dah$$

I don't get how the derivatives are going to help answer this question. Could it make $k - 2p - \dfrac{h - p}{(1 + p)^2} < 0$?

dwsmith

Well-known member
The dimensions I know are:

$$p =\frac{P}{C}, \ k=\frac{K}{C}, \ h=\frac{HB}{C^2}, \ \tau=Crt$$

That means
$$a = \frac{AC^2}{B} \ \text{and} \ d = \frac{D}{Cr}$$

But those wouldn't be dimensionless.

I found the problem. A has to be 1 over population. I am still not sure of D though.

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dwsmith

Well-known member
This a plankton herbivore model.

The dimensionalized model is

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

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

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

I am not sure what is a good starting point. I need a hint on one dimensionless unit.
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 $a=4(k-1)/(k+1)^3$. Hence show that if a < 4/27 there is a window of values of k where periodic solutions are possible.