MHB Steady State Question Understanding

mt91
Messages
13
Reaction score
0
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
 
Physics news on Phys.org
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 $E= \frac{2}{3}$.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top