I am trying to plot the Logistic system using ode45 solver in matlab, removing the initial transient, then re-plotting the result again.
The system is:
dx/dt = N0*x*(1-x/k)
Where:
N0= Growth rate
k= Carrying capacity
This system has two equilibrium point 0 and k.
At 0 the system is unstable and at k the system is stable.
This is my code:
function RunLogistic
N0=2;
K=10;
tspan=[0:0.1:200];
x0=.1;
[t,x]=ode45(@logistic511,tspan,x0,[],N0,K);
figure(1)
plot(t,x)
figure(2),
xx=x(65:2001);
tt=t(65:2001);
plot(tt,xx);
xlabel('time')
ylabel('population Density')
1;
%function dxdt=logistic511(t,x,N0,K)
% dxdt= N0*x*(1-x/K);
%end
Note: I am using the default tolerances for relTol and relAbs.
An ordinary differential equation in one independent variable doesn't have chaotic behavior, so that's not an issue. The ode solver, on the other hand, replaces the differential equation by a recursion relation, a map on an interval-and that *can* show chaotic behavior. Since this depends on the step size, which is an unphysical parameter, for the differential equation, such chaotic behavior is irrelevant to the properties of the functions described by the equation itself and is, just, an artifact of the numerical method, that should be avoided by choosing the step size appropriately. For this equation, of course, this is the object of the exercise, since the equation can be solved in closed form, that's easy to handle analytically and numerically. It shows that the differential equation can have ``trivial'' behavior, while the difference equation can have very rich behavior.
According to the manual on the method-which is useful to read, it's not a good idea to use a routine like a black box-this is a 4th or 5th (depending on some parameter) order Runge-Kutta, so, it does give rise to a quadratic recursion relation that, for a step size chosen (in)appropriately, gives rise to a sequence that can have nothing to do with the solution of the differential equation. This isn't a surprise, every numerical method of integrating a differential equation is subject to stability issues. So write down the recursion relation and study the map. You'll end up with x(n+1)=F(x(n)), from [0,k]->[0,k], only if the step size satisfies some condition. For the map to describe an approximation to the differential equation (that can be solved analytically), it shouldn't show periodic behavior, but converge to a fixed point. And that condition will give you the condition on the stability interval for th step size. So the reason for the oscillatory behavior is an inappropriate choice for the value of the step size: since 0
Hi Avan
The supposed 'oscillation' is caused by numerical instability. Partially possibly caused by the numerical solutions 'too rough' estimates at proximity of equilibrium point x=10.
If the numerical solution gets some rough estimates around constant.sols/equilibria its probably an issue as Stam Nicolis pointed out and explained.
If I did not copy wrong the exact solution should be x(t)=10/(1+9exp(-2t)).
Since x=10 is stable (sink) the solution with ic. x(0)=1 should tend quite fast to
x(t)--> 10 as t grows and the solution with this ic can not exceed 10. Well this is quite obvious from the exact solution also. (Well all sols. with ic 0
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