MHB How Do Steady State Equations Predict Animal Population Dynamics?

[mt91]

Anyone able to calculate the steady states?

 

[HOI]

Do you not know what a "steady state" IS? A steady state is a constant solution to the differential equation. And if it is constant, $\frac{du*}{dt}= 0$
So $\frac{du*}{dt}= 0= u*(1- u*)(1+ u*)- Eu*$.

Solve that third degree equation for u*.

 

[mt91]

Do you not know what a "steady state" IS? A steady state is a constant solution to the differential equation. And if it is constant, $\frac{du*}{dt}= 0$
So $\frac{du*}{dt}= 0= u*(1- u*)(1+ u*)- Eu*$.

Solve that third degree equation for u*.

Cheers, it was the wording of the question I struggled with, because it said as a function of E. So this is what I did:

\[ Eu*=u*(1-u*)(1+u*) \]
\[ E=(1-u*)(1+u*) \]
\[ ∴u*=1-E, u*=E-1 \]

I think that u*=0 would be a solution, but I'm not sure with the wording of "function of E" what it wants me to do, cheers

 

[HOI]

Then you should have said, in your first post, that your difficulty was with solving the cubic equation, not with the "steady state" of the differential equation! I am concerned that, in a problem involving differential equations, your difficulty is with solving a quadratic equation! In fact, your difficulty is with solving an equation of the form "$x^2= a$".

Yes, Eu*= u*(1- u*)(1+ u*). u*= 0 is a solution. If u* is not 0 we can divide by u* to get E= (1- u*)(1+ u*).

But that does NOT give u*= 1- E or u*= 1+ E as solutions! You would have seen that if you had checked: If u*= 1- E then 1- u*= 1-1+ E= E and 1+ E= 2- E. $(1- u*)(1+ u*)= E(2- E)= 2E- E^2$.

From $E= (1- u*)(1+ u*)= 1- u*^2$ we get $u*^2= 1- E$ so that $u*= \pm\sqrt{1- E}$.

 

[mt91]

Then you should have said, in your first post, that your difficulty was with solving the cubic equation, not with the "steady state" of the differential equation! I am concerned that, in a problem involving differential equations, your difficulty is with solving a quadratic equation! In fact, your difficulty is with solving an equation of the form "$x^2= a$".

Yes, Eu*= u*(1- u*)(1+ u*). u*= 0 is a solution. If u* is not 0 we can divide by u* to get E= (1- u*)(1+ u*).

But that does NOT give u*= 1- E or u*= 1+ E as solutions! You would have seen that if you had checked: If u*= 1- E then 1- u*= 1-1+ E= E and 1+ E= 2- E. $(1- u*)(1+ u*)= E(2- E)= 2E- E^2$.

From $E= (1- u*)(1+ u*)= 1- u*^2$ we get $u*^2= 1- E$ so that $u*= \pm\sqrt{1- E}$.

Meaning therefore we have 3 steady state solutions?
\[ u*=0 \]
\[ u*=sqrt(1-E) \]
\[ u*=-sqrt(1-E) \]

Thanks for your help by the way

 

[HOI]

Yes.

 

[mt91]

Yes.

Thanks, do you know much about how you would interpret relevant steady states? So the steady state equation was about an animal population. If a steady state is 0 is this indicating that there's no population? and the negative square root steady state is indicating a decline in the population?

 

[HOI]

Thanks, do you know much about how you would interpret relevant steady states? So the steady state equation was about an animal population. If a steady state is 0 is this indicating that there's no population?

The population is u(t). If for any t, u(t)= 0, there is no population so, of course, it cannot decrease and is not going to increase so stays 0. 0 is always a "steady state" for a population. (Unless you allow "spontaneous generation"!)

and the negative square root steady state is indicating a decline in the population?

No, the whole point of "steady state" is that it is steady, it is NOT changing, neither decreasing nor increasing. A negative value here is a mathematically a solution to the equation but is not a realistic solution- the number of animals cannot be negative.