Value of p for dp/dt to be maximum

[songoku]

Homework Statement

Let p(t) denote the number of bacteria at time t measured so that the carrying capacity is reached when p = 1. The model of p(t) is:
$$\frac{dp}{dt}=\frac{k}{c} p(t)(1-p(t)^{c})$$
where k and c > 0

What is the value $p_{max}$ of $p$ does $\frac{dp}{dt}$ is maximum? What does changing the value of c do to the $p_{max}$?

Relevant Equations

Integration and Differentiation

To find maximum of $\frac{dp}{dt} \rightarrow \frac{d^{2} p}{dt^{2}}=0$
$$\frac{k}{c}p'(t)(1-p(t)^{c})-\frac{k}{c}p(t)(c.p(t)^{c-1}).p'(t)=0$$
$$p'(t)=0~\text{or}~p(t)=\left(\frac{1}{c+1}\right)^{\frac{1}{c}}$$

For p'(t) = 0 → p(t) = 0 or p(t) = 1

So I have 3 values of p(t), how to know which one will result in maximum $\frac{dp}{dt}$?

I tried using sign diagram for 2nd derivative but the term $\frac{k}{c}p'(t)(1-p(t)^{c})$ and $\frac{k}{c}p(t)(c p(t)^{c-1}).p'(t)$ both are always positive so I can't determine the sign of their subtraction

Thanks

 

[Mark44]

Homework Statement:: Let p(t) denote the number of bacteria at time t measured so that the carrying capacity is reached when p = 1. The model of p(t) is:
$$\frac{dp}{dt}=\frac{k}{c} p(t)(1-p(t)^{c})$$
where k and c > 0

What is the value $p_{max}$ of $p$ does $\frac{dp}{dt}$ is maximum? What does changing the value of c do to the $p_{max}$?
Relevant Equations:: Integration and Differentiation

To find maximum of $\frac{dp}{dt} \rightarrow \frac{d^{2} p}{dt^{2}}=0$
$$\frac{k}{c}p'(t)(1-p(t)^{c})-\frac{k}{c}p(t)(c.p(t)^{c-1}).p'(t)=0$$
$$p'(t)=0~\text{or}~p(t)=\left(\frac{1}{c+1}\right)^{\frac{1}{c}}$$

The above is easier to read if you use prime notation omit the "function of t" notation.
Find the maximum of p' by setting p'' = 0.
$$p' = \frac k c[p(1 - p^c)]$$
$$\Rightarrow p'' = \frac k c[p'(1 - p^c) + p(-cp^{c - 1})]$$
$$= \frac k c[p' - p'p^c - cp^c] = \frac k c[p'(1 - p^c) - cp^c]$$
$$p'' = 0 \Rightarrow p'(1 - p^c) = cp^c$$

I get p'(t) = 0, but I get something else for the other value you show for p'(t).

For p'(t) = 0 → p(t) = 0 or p(t) = 1

So I have 3 values of p(t), how to know which one will result in maximum $\frac{dp}{dt}$?

I tried using sign diagram for 2nd derivative but the term $\frac{k}{c}p'(t)(1-p(t)^{c})$ and $\frac{k}{c}p(t)(c p(t)^{c-1}).p'(t)$ both are always positive so I can't determine the sign of their subtraction

Thanks

 

[Orodruin]

$$\Rightarrow p'' = \frac k c[p'(1 - p^c + p(-cp^{c - 1}]$$
$$= \frac k c[p' - p'p^c - cp^c] = \frac k c[p'(1 - p^c) - cp^c]$$
$$p'' = 0 \Rightarrow p'(1 - p^c) = cp^c$$

I get p'(t) = 0, but I get something else for the other value you show for p'(t).

You are missing a $p’$ on the third term in the second line here. Perhaps the result of the unmatched parentheses on line 1? This propagates to the end result (to which $p’ = 0$ is also not a solution).

 

[Mark44]

Perhaps the result of the unmatched parentheses on line 1?

Well, there are the unmatched parens, but I also neglected to use the chain rule when I differentiated $p^c$. Doh!
Thanks for pointing it out! I'm about to head out the door for an overnight trip, so I won't be able to make a complete correction.

 

[Orodruin]

So I have 3 values of p(t), how to know which one will result in maximum dpdt?

Two of those correspond to $p’ = 0$ (which are also stationary points…) and the third is positive. Which of those is larger?

 

[songoku]

Two of those correspond to $p’ = 0$ (which are also stationary points…) and the third is positive. Which of those is larger?

I should take the positive one.

Thank you very much Mark44 and Orodruin

 

[epenguin]

Please check and report exact wording of the question.

 

[songoku]

Please check and report exact wording of the question.

I posted the exact wording of the question

 

[epenguin]

OK I was confused by your answer, not seeing where your three solutions come from so as we have often expended efforts on what turned out to be not the real question - the phrase "What is the value pmax of p does dpdt is maximum?" does not quite make sense- I asked.

Maybe you are mixing up $dp/dt=0$ with $d(dp/dt)/dp=0$, which latter has only one* solution?

*Practically - you might also say multiple coincident.

 

[Orodruin]

OK I was confused by your answer, not seeing where your three solutions come from

$p’ = 0$ corresponds to two solutions for $p$ since $p’ = p(1-p^c)$, the solutions being $p=0$ and $p=1$. These are both stationary solutions to the differential equation (one stable and one unstable) since $p’ =0$.

Now, $dp’/dt = (dp’/dp) p’$ so the zero of $dp’/dp$ will of course coincide with $dp’/dt=0$ if $p’ \neq 0$ so the result is the same.

 

[epenguin]

I get that these are solutions of the d.e. but that was not what the question was askin.

 

[Orodruin]

I get that these are solutions of the d.e. but that was not what the question was askin.

You misread me. They are also solutions to where $p’$ has time derivative zero and therefore has a possible max if you just consider $dp’/dt$. Further investigation however results in those $p’$ being constant and zero, leaving the remaining solution.

As I stated, you can use either $dp’/dt$ or $dp’/do$. The result is the same.

 

[Mark44]

As I stated, you can use either $dp’/dt$ or $dp’/do$. The result is the same.

Did you mean $dp'/dp$ for the latter?

 

[Orodruin]

Did you mean $dp'/dp$ for the latter?

Yes. New phone and fat fingers.

 

[Mark44]

New phone and fat fingers.

That's a good one! I'll have to remember that when I make a typo!

 

[epenguin]

Well if I may say so there seem to me to have been fat fingers in the thread from the start.
. I presume the question really was intended to be:
"What is the value pmax of p where dp/dt is maximum?
What does changing the value of c do to the pmax?"

And I found calling that quantity "pmax" not overly helpful.

There is only one such maximum. Then, yes, the stationary points, not asked about, need be brought into a total account of the system.

It would be good if the student as well as completing the answer to the question stated what behaviour/s the model predicts which is the point of the exercise.

 

[Prof B]

Multiply out the right side before taking the derivative. It will be easier.

 

[Prof B]

Yes. New phone and fat fingers.

If you differentiate with respect to p then you don't have to write p' at all and you don't get the spurious solution p'=0, so it's easier. It's like differentiating a polynomial.

 

[Orodruin]

If you differentiate with respect to p then you don't have to write p' at all and you don't get the spurious solution p'=0, so it's easier. It's like differentiating a polynomial.

Sure, but that is not what OP did. I was explaining why the answer was the same.

 

[epenguin]

Multiply out the right side before taking the derivative. It will be easier.

That's exactly what I did originally, hence my initial puzzlement at the other solutions.

Doing just that gives you the answer to the question (which the OP did give).
The question and answer however are only useful as part of describing the behaviour of the system as a whole - which I think does not have a useful analytical solution.

This has not been done here.

 

[Orodruin]

The question and answer however are only useful as part of describing the behaviour of the system as a whole - which I think does not have a useful analytical solution.

The solution satisfies
$$
\frac{p}{(1-p^c)^{1/c}} = e^{t-t_0}
$$
This leads to
$$
p^c = (1-p^c) e^{c(t-t_0)},\qquad
p^c (1+e^{c(t-t_0)}) = e^{c(t-t_0)} \qquad
p = \left(\frac{1}{1+e^{-c(t-t_0)}}\right)^{1/c}
$$

… unless I made arithmetic errors along the way …

I might have put k=c but I really don’t want to trace it at the moment. Probably changes c to k in the exponent …

 

[epenguin]

The solution satisfies
$$
\frac{p}{(1-p^c)^{1/c}} = e^{t-t_0}
$$
This leads to
$$
p^c = (1-p^c) e^{c(t-t_0)},\qquad
p^c (1+e^{c(t-t_0)}) = e^{c(t-t_0)} \qquad
p = \left(\frac{1}{1+e^{-c(t-t_0)}}\right)^{1/c}
$$

… unless I made arithmetic errors along the way …

I might have put k=c but I really don’t want to trace it at the moment. Probably changes c to k in the exponent …

I'm sure you'll get it right. Is that standard theory?
Could you set out the solution more completely?

All I was saying above is that one ought to set out a brief statement in words of the behaviour or the population predicted by this model, for which an analytical solution is not necessary.

 

[Orodruin]

Could you set out the solution more completely?

You have $dp/dt = f(p)$. This is separable:
$$
\frac{dp}{f(p)} = dt
$$
so integrate both sides and solve for $p(t)$.

 

[epenguin]

Aaargh that's what I thought several times - and then concluded somehow it was wrong... have not been feeling too well of recent.

Perhaps Songaku will want to complete the analytical and qualitative analysis of this system.

 

[Orodruin]

I mean, even if it is separable that is no guarantee that you can write $p(t)$ on a simple form. It just so happens to be the case here.

 

[epenguin]

Quite, and I for some reason had taken a standard solvable type for the other type.