Mixing
Understanding logistic growth
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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 $f(P)$?
calculus differential-equations
asked 2 days ago
sam_rox
4852919
|
show 3 more comments
up vote
0
down vote
favorite
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 $f(P)$?
calculus differential-equations
asked 2 days ago
sam_rox
4852919
I can't see any negative values. The lowest term of your y-axis is $10^{-1} >0$ and x-axis starts from $0$. In order, thought, for someone to properly understand the problem, I think you should give a rigorous explanation of every variable/function and their given domains-restrictions.
– Rebellos
2 days ago
@Rebellos The "blank" spaces in the plot correspond to negative values of $f(P)$, since what's being plotted is $log f(P)$.
– rafa11111
2 days ago
@rafa11111 The function seems rather continuous so it doesn't make much sense what's happening if we can't see domains and restrictions and not properly understand what each thing is.
– Rebellos
2 days ago
1
@Rebellos I agree that the question is far from being clear. I just wanted to point out that the graph has logarithmic $y$ axis.
– rafa11111
2 days ago
As t is indepent of the right hand side, P is a linear function.
– William Elliot
2 days ago
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 $f(P)$?
calculus differential-equations
asked 2 days ago
sam_rox
4852919
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 $f(P)$?
calculus differential-equations
calculus differential-equations
asked 2 days ago
sam_rox
4852919
asked 2 days ago
sam_rox
4852919
asked 2 days ago
sam_rox
4852919
asked 2 days ago
sam_rox
4852919
asked 2 days ago
sam_rox
4852919
4852919
I can't see any negative values. The lowest term of your y-axis is $10^{-1} >0$ and x-axis starts from $0$. In order, thought, for someone to properly understand the problem, I think you should give a rigorous explanation of every variable/function and their given domains-restrictions.
– Rebellos
2 days ago
@Rebellos The "blank" spaces in the plot correspond to negative values of $f(P)$, since what's being plotted is $log f(P)$.
– rafa11111
2 days ago
@rafa11111 The function seems rather continuous so it doesn't make much sense what's happening if we can't see domains and restrictions and not properly understand what each thing is.
– Rebellos
2 days ago
1
@Rebellos I agree that the question is far from being clear. I just wanted to point out that the graph has logarithmic $y$ axis.
– rafa11111
2 days ago
As t is indepent of the right hand side, P is a linear function.
– William Elliot
2 days ago
|
show 3 more comments
I can't see any negative values. The lowest term of your y-axis is $10^{-1} >0$ and x-axis starts from $0$. In order, thought, for someone to properly understand the problem, I think you should give a rigorous explanation of every variable/function and their given domains-restrictions.
– Rebellos
2 days ago
@Rebellos The "blank" spaces in the plot correspond to negative values of $f(P)$, since what's being plotted is $log f(P)$.
– rafa11111
2 days ago
@rafa11111 The function seems rather continuous so it doesn't make much sense what's happening if we can't see domains and restrictions and not properly understand what each thing is.
– Rebellos
2 days ago
1
@Rebellos I agree that the question is far from being clear. I just wanted to point out that the graph has logarithmic $y$ axis.
– rafa11111
2 days ago
As t is indepent of the right hand side, P is a linear function.
– William Elliot
2 days ago
I can't see any negative values. The lowest term of your y-axis is $10^{-1} >0$ and x-axis starts from $0$. In order, thought, for someone to properly understand the problem, I think you should give a rigorous explanation of every variable/function and their given domains-restrictions.
– Rebellos
2 days ago
I can't see any negative values. The lowest term of your y-axis is $10^{-1} >0$ and x-axis starts from $0$. In order, thought, for someone to properly understand the problem, I think you should give a rigorous explanation of every variable/function and their given domains-restrictions.
– Rebellos
2 days ago
@Rebellos The "blank" spaces in the plot correspond to negative values of $f(P)$, since what's being plotted is $log f(P)$.
– rafa11111
2 days ago
@Rebellos The "blank" spaces in the plot correspond to negative values of $f(P)$, since what's being plotted is $log f(P)$.
– rafa11111
2 days ago
@rafa11111 The function seems rather continuous so it doesn't make much sense what's happening if we can't see domains and restrictions and not properly understand what each thing is.
– Rebellos
2 days ago
@rafa11111 The function seems rather continuous so it doesn't make much sense what's happening if we can't see domains and restrictions and not properly understand what each thing is.
– Rebellos
2 days ago
1
1
@Rebellos I agree that the question is far from being clear. I just wanted to point out that the graph has logarithmic $y$ axis.
– rafa11111
2 days ago
@Rebellos I agree that the question is far from being clear. I just wanted to point out that the graph has logarithmic $y$ axis.
– rafa11111
2 days ago
As t is indepent of the right hand side, P is a linear function.
– William Elliot
2 days ago
As t is indepent of the right hand side, P is a linear function.
– William Elliot
2 days ago
|
show 3 more comments
1 Answer
1
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up vote
0
down vote
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.
answered yesterday
rafa11111
748315
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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.
answered yesterday
rafa11111
748315
add a comment |
up vote
0
down vote
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.
answered yesterday
rafa11111
748315
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
answered yesterday
rafa11111
748315
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.
answered yesterday
rafa11111
748315
answered yesterday
rafa11111
748315
answered yesterday
rafa11111
748315
answered yesterday
rafa11111
748315
748315
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I can't see any negative values. The lowest term of your y-axis is $10^{-1} >0$ and x-axis starts from $0$. In order, thought, for someone to properly understand the problem, I think you should give a rigorous explanation of every variable/function and their given domains-restrictions.
– Rebellos
2 days ago
@Rebellos The "blank" spaces in the plot correspond to negative values of $f(P)$, since what's being plotted is $log f(P)$.
– rafa11111
2 days ago
@rafa11111 The function seems rather continuous so it doesn't make much sense what's happening if we can't see domains and restrictions and not properly understand what each thing is.
– Rebellos
2 days ago
1
@Rebellos I agree that the question is far from being clear. I just wanted to point out that the graph has logarithmic $y$ axis.
– rafa11111
2 days ago
As t is indepent of the right hand side, P is a linear function.
– William Elliot
2 days ago