What are the equilibrium solutions in a differential equation?

[sara_87]

Homework Statement

what is the meaning of the coefficients 8 and 80 in the following differential equation?

Homework Equations

dP/dt=2*(10^-3)(P-8)(80-P)

The Attempt at a Solution

I know that they are equilibrium solutions but i have to explain.

Thank you

 

[Dick]

Figure out where dP/dt is positive, negative and zero as a function of P. If dP/dt=0 then you have an equilibrium value. Now explain why each equilibrium is stable or unstable.

 

[sara_87]

yeah i did all that but i just have to explain in words the meaning of the coeff. 8 and 80.

 

[sutupidmath]

Well, to do that i think we have to know what is the original problem, i mean to know what P stands for, is it something about a population model or?? If you want to explain the meaning of those constants in the context that the problem is stated!

 

[sara_87]

oh right, yes it is a population model. P is the population and t is time

 

[sutupidmath]

oh right, yes it is a population model. P is the population and t is time

Well i think that in this case, an explanation would be like this:
I am not sure whether you got this part right, is it (P-8) or 8-P ?
When the initial size of the population you are dealing with is greater than 80 whatever the units are, it means that that population model will start decreasing, and it will reach, converge somehow to 80, it means that the most it can decrease is 80.
If 8<Po<80, it means that the most that population size can grow is about 80, so it means that the population starts grwoing but the largest it can get is 80(million, hundred, thousand or whatever). I am not sure about the last part, when the Po<8. It looks like if the initial population has a size of smaller than 8 for some reason it will start decreasing, but it looks weird though!
I am sure Dick will give a much more accurate and decent explanation, or someone else, though i myself have just started to make a start on diff. eq.

 

[sara_87]

thank you. that does help a lot. now, i can think outside the box.

 

[sara_87]

if the equation is altered to:
dP/dt=2*(10^-3)(P-8)(80-P)-H
Where P is the population of fish, and
where H is rate per month. If the initial population is 16, then what would you recommend for a limit to the monthly catch of this fish population?
(we don't want the fish to be extinct and we don't want to starve).

 

[Dick]

You want to fix the population at P=16? Put dP/dt=0 (since you want a stable population), P=16 and solve for H.

 

[sara_87]

ah ok.
I found H to be:
less than or equal to 1.024
above this value would give below the solution.
is that it?
dont i have to consider that we don't want to starve?

 

[Dick]

P=16, H=1.024, gives you dP/dt=0 So the solution is P(t)=16 for all t. It's not a stable solution. If the population is a little above P=16 it will start to increase and if it drops a little below it will keep dropping. Do you need a stable solution at some value of P besides 16?

 

[sara_87]

i need a value for H so that the fish doesn't become extinct and at the same time...we don't starve
or at least some explanation about this case.

 

[Dick]

You already did. dP/dt=0 means P is CONSTANT at P=16 and H=1.024. That's neither increasing nor decreasing.