Welcome to our community

Be a part of something great, join today!

Steady State Question Understanding

mt91

New member
Jul 30, 2020
15
1596323380616.png

I've got 2 questions here. I was able to work out question 5 and calculate the steady states. However for question 6 I've got no idea with the wording of the equation and where you would start, so any sort of help would be really helpful, cheers
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
Okay, so for #5 you saw that the "steady state" solutions satisfy u(1- u)(1+u)= Eu. An obvious solution is u=0. If u is not 0, we can divide both sides by u and have $(1- u)(1+ u)= 1- u^2= E$ so $u^2= 1- E$ and the two other steady state solutions are $u= \sqrt{1- E}$ and $u= -\sqrt{1- E}$ which, of course, are only defined for E< 1. u= 0 is obviously a stable solution. The other two are unstable. You haven't said what "u" represents so I have no idea what "biologically relevant" could mean here. (If u is the population of some species then "$u= -\sqrt{1- E}$" is obviously NOT "biologically relevant" since u cannot be negative.)

For #6, if y= Eu(E) then y'= u(E)+ Eu'(E)= 0 at a maximum. Obviously "u(x)= 0" will NOT give "maximum yield". With $u(E)= \sqrt{1- E}= (1- E)^{1/2}$, $u'= -\frac{1}{2}(1- E)^{-1/2}$ and $y'= u(E)+ Eu'(E)= (1- E)^{1/2}-\frac{1}{2}E(1- E)^{-1/2}= 0$ so $(1- E)^{1/2}= \frac{1}{2}E(1- E)^{-1/2}$ and $2(1- E)= E$, $2- 2E= E$, $2= 3E$, and $E= \frac{2}{3}$.