I have this simple system :
dx/dt = x(1-x)
I am trying to solve this system analytically and numerically using ode45 in MATLAB, then trying to find the Fisher Information according to this formula :
I = integral ( (1- 2*x)^2 / (x-x^2)^2) .
I wrote three different codes to find that but I am not feeling confident with the all three for many reasons: I am getting different results for the numerical and analytical, My system is stable at the carrying capacity but what I am get is oscillation or an increasing curve and infinite when I am using different time vector.
Can anyone give me any idea about how to write a code in Matlab to find that?
dx/dt = x(1-x). Thus the slope (responsiveness of x to t) is small when x is close to zero or 1. The point of inflection can be solved for either by plotting the values of the derivative and finding the values of x and t at which dx/dt reaches a maximum,
Models like this are great for logistic growth (S-curves). There can be additional parameters, such as dx/dt = b x (K - x ), without changing too much, if b > 0:
See application in attached paper (p 39).
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The analytical solution of your simple differential equation is:
x(t) = 1/(1+C*exp(-t))
and it is a sigmoid curve.
If you have x(0)=x0, then the solution is:
x(t) = 1/(1-exp(-t)*(x0-1)/x0)
Dear Avan Al-Saffar,
A hint to get the analytical solution:
change of unknown: X(t) = Y(t) exp(t). Then you get an equation for Y:
dY/dt = - Y^2 exp(t), or equivalently if Y is different from 0:
- (dY/dt)/(y^2) = exp(t), and the left hand-side integrates to 1/Y and the right-hand side to exp(t). Changing back to X(t) gives you the solution pointed out by Demetris.
Then you can prove that if you start from x0 in [0,1] then x(t) remains in [0,1] and tends to 1 as t grows.
Concerning your integral, its integrand tends to infinity when t grows because x(t) tends to 1, thus (x-x^2)^2 tends to 0, while the numerator tends to 1.
If you want to solve your EDO on your own using a software like scilab, octave or matlab, I suggest you write your own explicit Euler code:
( X(n+1) - X(n) )/dt = X(n) - ( X(n) )^2
which gives you X(n+1) = ( 1 + dt*( 1 - X(n) ) ) * X(n)
Then you can prove that if 0 < X(n) < 1 then you'll get 0 < X(n+1) < 1 provided that dt < 1/X(n). Since X(n) remains then in [0,1], it is sufficient to take dt < 1.
Hope this helps,
Pascal
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