logistic differential equation?

Question

For the logisitc differential equation dA/dt= .8A(1- A/600) where t is the time in years and A(0) = 150, how do you find the limit of A(t) as t approaches infinity. You do do 150 plus the carrying capacity of 600? or is it just 600 regardless of the initial value? how do you figure that out?

Answer 1

If dA/dt = cA+kA^2 and A(0)=150, then
A(t)=c/((|(c+k*150)/150)|*e^(-c*t)-k).
With c=.8 and k=-.8/600,
A(t)=.8/((|.8+.8*150/600)/150)|*e^(-.8*t)-.8/600)
As t goes to infinity, A(t) goes to -c/k = -.8/(-.8/600) = 600

Answer 2

Here's a BASIC program for you to play with.

100 DT = 0.001
110 T = 0
120 A = 150
130 for i = 1 to 100000
140 DADT = 0.8 * A * (1-A/600)
150 print T,A,DADT
160 T = T + DT
170 A = A + DT*DADT
180 next i

Your logistic equation has horizontal slopes at A=0 and at A=600.

You're given the initial condition A(t=0) = 150.

As t increases, A will approach the limit of 600 horizontally asymptotically. (And dA/dt will approach zero.)

Answer 3

A constant A0 is a stationary solution of the equation dA/dt = F(A,t) if F(A0,t) = 0 for all t. In this case, your F(A,t) is just the right side of your equation. What constant values of A cause it to vanish? That will give you an idea of what the limit should be.

See the source below for more info.