Vrftjkry

The population growth can be modelled by the function ${dPover dt}=rP(1-{Pover k})$ and $P$ will go from $P_0$ (the initial value) to $k$.

But, I am trying to understand the behaviour of the term $f(P)=P(1-{Pover k})$. So, I solved the ODE equation ( ${dPover dt}$) over time (using Matlab ode 45) and stored the value of $P$ at each time point. Then using these $P$ values I plotted $P(1-{Pover k})$. This is the code and the result of $f(P)$ in log scale.

p=1000;

time=100;

[t,y]=ode45(@(t,y)LogisticEq(t,y),0:time,p);



k=10^5;

figure

plot(t/24,y(:,1).*(1-(y(:,1)/k)))



function s= LogisticEq(~,y)



r1=0.5;

k=10^5;

s=zeros(1,1);



s(1)=r1*y(1)*(1-(y(1)/k));



end

In here, it can be seen that $f(P)$ is negative for some values. Is it possible for $f(P)$ to be negative? $f(P)=0$ for $P=0$ and $P=k$ and as the maximum value of $P$ is $k$, why does $f(P)$ become negative?

What does it mean for $f(P)$ to be negative?Does it mean that over time the population decreases?

Is there a condition that can be posed to have only positive values for $f(P)$?

With $f(P)$ given by
$$
f(P)=Pleft(1-frac{P}{k}right)
$$
we have $f(P)<0$ if $P<0$ (which is not reasonable) or if $P>k$. It means that the system cannot support a population larger than $k$; in other words, $k$ is "so-called carrying capacity (i.e., the maximum sustainable population)" (according to Wolfram MathWorld). If we are searching for a steady state solution we must set $dP/dt=0$. Therefore, we have $P=0$ or $P=k$, which means that, if the population is zero or if the population achieved the 'carrying capacity', the population won't change anymore.

Since $P=k$ is a fixed point of the system, we can also see the behavior of $P$ when it's close to $k$. If $P=k+Delta P$, we have
$$
frac{dP}{dt} = r f(k+Delta P) = r left( (k+Delta P) left( 1 - frac{k+Delta P}{k} right) right) = -r left( Delta P + frac{Delta P^2}{k}right) approx -rDelta P,
$$
where I dropped out $Delta P^2$ because I assumed $Delta P$ very small. If $P$ is close to $k$, but a bit larger, we have $P=k+Delta P$ with positive $Delta P$. Since close to $P=k$ we have $dP/dt propto -Delta P$, $P$ decreases until reaching $P=k$. On the other hand, if $P$ is slightly smaller than $k$, $P$ will increase until reaching $P=k$. However, since the 'velocity' with which $P$ changes is proportional to the distance to $k$ (remember that $dP/dt propto Delta P$) $P$ never actually reaches $k$, but tends asymptotically to it. You can use
$$
frac{dP}{dt} approx r(k-P)
$$
to show that $P to k+C exp (-rt)$ as $Pto k$.

Of course, all of this is useless, since we know the analytical solution for $P$, which is
$$
P(t) = k P(0) frac{exp(rt)}{k+P(0)(exp(rt)-1)},
$$
as Claude Leibovici stated in the comments. I only did all this asymptotic stuff to show, without resorting to plottings, that $f(P)$ can be negative, but cannot change its sign. We saw that $f(P)$ can only change its sign if $P$ 'cross' the line $P=k$ (or $P=0$, which is out of question). However, we also saw that $P$ always approaches $P=k$ asymptotically. Therefore, $f(P)$ cannot change its sign.

I tried to replicate your results, without success. I can't tell exactly is wrong with your code, but I'm not sure if you should specify the time array that way; when I used MATLAB I usually specified only the initial and final time (like timespan = [t0 tend]) and left to the integrator to choose the inner points.