YAHWT: Non-linear systems with Piece-wise Naka-Rushton component

[Stiganator]

This is a homework problem, but I have been trying to find the equilibrium of hours now.

dE(1)/dt= 1/10(-E(1)+S(80-5E(2)-5E(3))

dE(2)/dt= 1/10(-E(2)+S(80-5E(1)-5E(3))

dE(3)/dt= 1/10(-E(3)+S(80-5E(1)-5E(2))

S(x)= 100(x)^2/(40^2+x^2) when x>=0
0, when x<0


The problem is to find all the equilibrium and describe the stability.

So, I figured there were 3 classes (which my professor confirmed) E(1)=E(2)=E(3), two equal and one zero, and two equal to zero (winner take all).

The case where they are equal should be easy right.

E=100(80-10E)^2/(40^2+(80-10E)^2)

Expanding it and putting it in matlab roots command yields 100,8,8 as roots. Funny thing is, none of those work.

 

[CycloWizard]

The steady state point where they are equal should be solved as 0=..., since dE/dt=0 at steady state. It's not clear to me why there is an x dependence - should that be a t dependence (i.e. S(t)=...)? It could be that there are both spatial and temporal variations in the population, but most such models assume spatial homogeneity. If it is S(t), it's much easier to solve, but if not, then I'll have to go back to the book and see if I can remember how to do it.

 

[toslat]

Ur systems is in equilibrium when E1'=E2'=E3'=0, hence ur diff eqn reduces to a set of polynomials of the form X(40^2+(80-5Y-5Z)^2) = 100(80-5Y-5Z)^2 where {X,Y,Z}={E(1), E(2), E(3)}.

For X=Y=Z we have 100X^3 -11600X^2 +168000X-640000=0 (expanded form of ur expression above with X=E(1)) which gives X=6.9102 or 9.2793 or 99.8105.

For using MatLab, expression shld be 'roots([100 -11600 168000 -640000])'

 

[HardcoreRomantic]

I figured it out this morning, forgot the 40^2 in the expansion. Thanks for the help. It's always the little things.

 

[CycloWizard]

Originally posted by: Stiganator
This is a homework problem, but I have been trying to find the equilibrium of hours now.

dE(1)/dt= 1/10(-E(1)+S(80-5E(2)-5E(3))

dE(2)/dt= 1/10(-E(2)+S(80-5E(1)-5E(3))

dE(3)/dt= 1/10(-E(3)+S(80-5E(1)-5E(2))

S(x)= 100(x)^2/(40^2+x^2) when x>=0
0, when x<0


The problem is to find all the equilibrium and describe the stability.

So, I figured there were 3 classes (which my professor confirmed) E(1)=E(2)=E(3), two equal and one zero, and two equal to zero (winner take all).

The case where they are equal should be easy right.

E=100(80-10E)^2/(40^2+(80-10E)^2)

Expanding it and putting it in matlab roots command yields 100,8,8 as roots. Funny thing is, none of those work.

Originally posted by: HardcoreRomantic
I figured it out this morning, forgot the 40^2 in the expansion. Thanks for the help. It's always the little things.

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