Logistic map (discrete dynamical system) vs logistic differential equation
$begingroup$
I have to roughly illustrate the logistic discrete dynamical system (as a model for population growth) to some non mathematics students. I'm not an analyst or an expert of dynamical systems.
Looking things up in the internet, I find the logistic map
$$x_{n+1}=rx_n(1-x_n)$$
with initial condition $x_0in [0,1]$, and where $rin[0,4]$ is a parameter (the condition $0leq r leq 4$ guarantees that $x_n$ doesn't escape the unit interval $[0,1]$ throughout the evolution of the system). Here $x_n$ represents the ratio between the population at time $n$ and the total population the environment is able to support.
I find also different behaviors according to the value of the parameter $rin[0,4]$. For example,
- For $0<rleq 1$ there's extinction of the population.
- For $1<rleq 3$ the sequence tends to a stable equilibrium $x_infty:=1-1/r$.
- For $3<rleq 1+sqrt{6}$ there's convergence to a period-$2$ cycle.
- For $1+sqrt{6}<rleq r^*$ (where $r^*$ is a certain constant) several bifurcations occur with limit a cycle of period that doubles as $r$ traverses that range.
- For $r>r^*$ there's chaotic behavior.
I would expect that a similar span of different behaviors also happens for the logistic differential equation
$$dot{x}(t)=rx(t)(1-x(t))$$
upon varying the parameter $r$. But on the internet I found no reference to anything like this. On the contrary, many pages care to solve the differential equation explicitely and illustrate the solution, which is the famous logistic function: the S-shaped increasing curve (depending on $r$) with a horizontal asymptote and an inflection point. It seems this solution is obtainable no matter what $r$ is. This looks only like case number 2. of the discrete dynamical system above.
So where are the analogous to cases 1.,3.,4. and 5. where the time is continuous??
Or am I misunderstanding some aspects of how a continuous-time dynamical system gets discretized?
Also,
Which correspondence is there between the $r$ of the discrete version and the $r$ of the continuous version?
ordinary-differential-equations recurrence-relations dynamical-systems nonlinear-system
$endgroup$
add a comment |
$begingroup$
I have to roughly illustrate the logistic discrete dynamical system (as a model for population growth) to some non mathematics students. I'm not an analyst or an expert of dynamical systems.
Looking things up in the internet, I find the logistic map
$$x_{n+1}=rx_n(1-x_n)$$
with initial condition $x_0in [0,1]$, and where $rin[0,4]$ is a parameter (the condition $0leq r leq 4$ guarantees that $x_n$ doesn't escape the unit interval $[0,1]$ throughout the evolution of the system). Here $x_n$ represents the ratio between the population at time $n$ and the total population the environment is able to support.
I find also different behaviors according to the value of the parameter $rin[0,4]$. For example,
- For $0<rleq 1$ there's extinction of the population.
- For $1<rleq 3$ the sequence tends to a stable equilibrium $x_infty:=1-1/r$.
- For $3<rleq 1+sqrt{6}$ there's convergence to a period-$2$ cycle.
- For $1+sqrt{6}<rleq r^*$ (where $r^*$ is a certain constant) several bifurcations occur with limit a cycle of period that doubles as $r$ traverses that range.
- For $r>r^*$ there's chaotic behavior.
I would expect that a similar span of different behaviors also happens for the logistic differential equation
$$dot{x}(t)=rx(t)(1-x(t))$$
upon varying the parameter $r$. But on the internet I found no reference to anything like this. On the contrary, many pages care to solve the differential equation explicitely and illustrate the solution, which is the famous logistic function: the S-shaped increasing curve (depending on $r$) with a horizontal asymptote and an inflection point. It seems this solution is obtainable no matter what $r$ is. This looks only like case number 2. of the discrete dynamical system above.
So where are the analogous to cases 1.,3.,4. and 5. where the time is continuous??
Or am I misunderstanding some aspects of how a continuous-time dynamical system gets discretized?
Also,
Which correspondence is there between the $r$ of the discrete version and the $r$ of the continuous version?
ordinary-differential-equations recurrence-relations dynamical-systems nonlinear-system
$endgroup$
add a comment |
$begingroup$
I have to roughly illustrate the logistic discrete dynamical system (as a model for population growth) to some non mathematics students. I'm not an analyst or an expert of dynamical systems.
Looking things up in the internet, I find the logistic map
$$x_{n+1}=rx_n(1-x_n)$$
with initial condition $x_0in [0,1]$, and where $rin[0,4]$ is a parameter (the condition $0leq r leq 4$ guarantees that $x_n$ doesn't escape the unit interval $[0,1]$ throughout the evolution of the system). Here $x_n$ represents the ratio between the population at time $n$ and the total population the environment is able to support.
I find also different behaviors according to the value of the parameter $rin[0,4]$. For example,
- For $0<rleq 1$ there's extinction of the population.
- For $1<rleq 3$ the sequence tends to a stable equilibrium $x_infty:=1-1/r$.
- For $3<rleq 1+sqrt{6}$ there's convergence to a period-$2$ cycle.
- For $1+sqrt{6}<rleq r^*$ (where $r^*$ is a certain constant) several bifurcations occur with limit a cycle of period that doubles as $r$ traverses that range.
- For $r>r^*$ there's chaotic behavior.
I would expect that a similar span of different behaviors also happens for the logistic differential equation
$$dot{x}(t)=rx(t)(1-x(t))$$
upon varying the parameter $r$. But on the internet I found no reference to anything like this. On the contrary, many pages care to solve the differential equation explicitely and illustrate the solution, which is the famous logistic function: the S-shaped increasing curve (depending on $r$) with a horizontal asymptote and an inflection point. It seems this solution is obtainable no matter what $r$ is. This looks only like case number 2. of the discrete dynamical system above.
So where are the analogous to cases 1.,3.,4. and 5. where the time is continuous??
Or am I misunderstanding some aspects of how a continuous-time dynamical system gets discretized?
Also,
Which correspondence is there between the $r$ of the discrete version and the $r$ of the continuous version?
ordinary-differential-equations recurrence-relations dynamical-systems nonlinear-system
$endgroup$
I have to roughly illustrate the logistic discrete dynamical system (as a model for population growth) to some non mathematics students. I'm not an analyst or an expert of dynamical systems.
Looking things up in the internet, I find the logistic map
$$x_{n+1}=rx_n(1-x_n)$$
with initial condition $x_0in [0,1]$, and where $rin[0,4]$ is a parameter (the condition $0leq r leq 4$ guarantees that $x_n$ doesn't escape the unit interval $[0,1]$ throughout the evolution of the system). Here $x_n$ represents the ratio between the population at time $n$ and the total population the environment is able to support.
I find also different behaviors according to the value of the parameter $rin[0,4]$. For example,
- For $0<rleq 1$ there's extinction of the population.
- For $1<rleq 3$ the sequence tends to a stable equilibrium $x_infty:=1-1/r$.
- For $3<rleq 1+sqrt{6}$ there's convergence to a period-$2$ cycle.
- For $1+sqrt{6}<rleq r^*$ (where $r^*$ is a certain constant) several bifurcations occur with limit a cycle of period that doubles as $r$ traverses that range.
- For $r>r^*$ there's chaotic behavior.
I would expect that a similar span of different behaviors also happens for the logistic differential equation
$$dot{x}(t)=rx(t)(1-x(t))$$
upon varying the parameter $r$. But on the internet I found no reference to anything like this. On the contrary, many pages care to solve the differential equation explicitely and illustrate the solution, which is the famous logistic function: the S-shaped increasing curve (depending on $r$) with a horizontal asymptote and an inflection point. It seems this solution is obtainable no matter what $r$ is. This looks only like case number 2. of the discrete dynamical system above.
So where are the analogous to cases 1.,3.,4. and 5. where the time is continuous??
Or am I misunderstanding some aspects of how a continuous-time dynamical system gets discretized?
Also,
Which correspondence is there between the $r$ of the discrete version and the $r$ of the continuous version?
ordinary-differential-equations recurrence-relations dynamical-systems nonlinear-system
ordinary-differential-equations recurrence-relations dynamical-systems nonlinear-system
edited Dec 9 '18 at 19:59
Qfwfq
asked Dec 9 '18 at 19:52
QfwfqQfwfq
319112
319112
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add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Only some elements
A differential equation and it associated discretized problem may have different stability behaviors depending on the way you perform the discretization. See for example Wikipedia - discretization, paragraph Approximations.
Nevertheless, approximating $dot{x}(t) = rx(t)(1-x(t))$ by $x_{n+1}=r x_n(1-x_n)$ is rather strange. It should be better discretized by $x_{n+1} -x_n =r x_n(1-x_n)$. Because usually
$$dot{x}(t) = frac{x_{n+1}-x_n}{t_{n+1}-t_n} =x_{n+1}-x_n$$ if you consider equal discretization of the time.
$endgroup$
$begingroup$
I exactly thought the same thing: the ODE should be discretized by $x_{n+1}-x_n=rx_n(1-x_n)$ because the finite difference ratio $(x_{n+1}-x_n)/1$ plays the role of $dx/dt$. But I don't know...
$endgroup$
– Qfwfq
Dec 9 '18 at 20:22
$begingroup$
In the paragraph you cite (of the wikipedia link) it seems they're considering linear ODEs, while the logistic is nonlinear - not sure how this affects the continuous/discrete relationship.
$endgroup$
– Qfwfq
Dec 9 '18 at 20:25
$begingroup$
It is even worse... if You already have issue with linear problems!
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:26
add a comment |
$begingroup$
In the logistic equation you get transforming it as Bernoulli equation
$$
frac{d}{dt}x(t)^{-1}=r(1-x(t)^{-1})implies x(t)^{-1}=1+ce^{-rt}.
$$
Now compare the values for $t$ and $t+h$,
$$
x(t+h)^{-1}-1 = ce^{-rt-rh}=e^{-rh}(x(t)^{-1}-1).
$$
For the sequence $x_k=x(kh)$ you get thus the recursion formula
$$
x_{k+1}=frac{x_k}{1-e^{-rh}+e^{-rh}x_k}
$$
which looks rather different than the logistic map.
Any sane discretization using the Euler method or similar would use a sufficiently small $h$, so that in the Euler forward discretization $$x_{k+1}=x_k(1+hr)-hrx_k^2.$$ To get to the normal form one would have to rescale $y_k=ax_k$, so that then $$y_{k+1}=y_k(1+hr)-frac{hr}ay_k^2=(1+hr)y_kleft(1-frac{hr}{a(1+hr)}y_kright)$$ giving $a=frac{hr}{1+hr}$. This is indeed case $2$ in the list, with equilibrium at $y_infty=1-frac1{1+hr}=frac{hr}{1+hr}=a$ or translated back at $x_infty=1$.
$endgroup$
$begingroup$
Ok, this is to say the recursion you get depends heavily on how you choose to discretize, I assume? But is it relevant to the question? Which is the discretization method in the OP case, and does it lead to different phenomena (differences between the continuum and the discrete)?
$endgroup$
– Qfwfq
Dec 9 '18 at 21:32
1
$begingroup$
Too long for a comment, added remark on discretization to the answer. One can conclude that the interesting phenomena happen when $h$ is so large that the discretization error can no longer be classified as $O(h^2)$ as the higher order terms are equally large or even dominate.
$endgroup$
– LutzL
Dec 9 '18 at 21:50
$begingroup$
So it's true that chaotic features are not present in the continuous model but appear only once I discretize in that specific way?
$endgroup$
– Qfwfq
Dec 9 '18 at 22:19
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Only some elements
A differential equation and it associated discretized problem may have different stability behaviors depending on the way you perform the discretization. See for example Wikipedia - discretization, paragraph Approximations.
Nevertheless, approximating $dot{x}(t) = rx(t)(1-x(t))$ by $x_{n+1}=r x_n(1-x_n)$ is rather strange. It should be better discretized by $x_{n+1} -x_n =r x_n(1-x_n)$. Because usually
$$dot{x}(t) = frac{x_{n+1}-x_n}{t_{n+1}-t_n} =x_{n+1}-x_n$$ if you consider equal discretization of the time.
$endgroup$
$begingroup$
I exactly thought the same thing: the ODE should be discretized by $x_{n+1}-x_n=rx_n(1-x_n)$ because the finite difference ratio $(x_{n+1}-x_n)/1$ plays the role of $dx/dt$. But I don't know...
$endgroup$
– Qfwfq
Dec 9 '18 at 20:22
$begingroup$
In the paragraph you cite (of the wikipedia link) it seems they're considering linear ODEs, while the logistic is nonlinear - not sure how this affects the continuous/discrete relationship.
$endgroup$
– Qfwfq
Dec 9 '18 at 20:25
$begingroup$
It is even worse... if You already have issue with linear problems!
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:26
add a comment |
$begingroup$
Only some elements
A differential equation and it associated discretized problem may have different stability behaviors depending on the way you perform the discretization. See for example Wikipedia - discretization, paragraph Approximations.
Nevertheless, approximating $dot{x}(t) = rx(t)(1-x(t))$ by $x_{n+1}=r x_n(1-x_n)$ is rather strange. It should be better discretized by $x_{n+1} -x_n =r x_n(1-x_n)$. Because usually
$$dot{x}(t) = frac{x_{n+1}-x_n}{t_{n+1}-t_n} =x_{n+1}-x_n$$ if you consider equal discretization of the time.
$endgroup$
$begingroup$
I exactly thought the same thing: the ODE should be discretized by $x_{n+1}-x_n=rx_n(1-x_n)$ because the finite difference ratio $(x_{n+1}-x_n)/1$ plays the role of $dx/dt$. But I don't know...
$endgroup$
– Qfwfq
Dec 9 '18 at 20:22
$begingroup$
In the paragraph you cite (of the wikipedia link) it seems they're considering linear ODEs, while the logistic is nonlinear - not sure how this affects the continuous/discrete relationship.
$endgroup$
– Qfwfq
Dec 9 '18 at 20:25
$begingroup$
It is even worse... if You already have issue with linear problems!
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:26
add a comment |
$begingroup$
Only some elements
A differential equation and it associated discretized problem may have different stability behaviors depending on the way you perform the discretization. See for example Wikipedia - discretization, paragraph Approximations.
Nevertheless, approximating $dot{x}(t) = rx(t)(1-x(t))$ by $x_{n+1}=r x_n(1-x_n)$ is rather strange. It should be better discretized by $x_{n+1} -x_n =r x_n(1-x_n)$. Because usually
$$dot{x}(t) = frac{x_{n+1}-x_n}{t_{n+1}-t_n} =x_{n+1}-x_n$$ if you consider equal discretization of the time.
$endgroup$
Only some elements
A differential equation and it associated discretized problem may have different stability behaviors depending on the way you perform the discretization. See for example Wikipedia - discretization, paragraph Approximations.
Nevertheless, approximating $dot{x}(t) = rx(t)(1-x(t))$ by $x_{n+1}=r x_n(1-x_n)$ is rather strange. It should be better discretized by $x_{n+1} -x_n =r x_n(1-x_n)$. Because usually
$$dot{x}(t) = frac{x_{n+1}-x_n}{t_{n+1}-t_n} =x_{n+1}-x_n$$ if you consider equal discretization of the time.
answered Dec 9 '18 at 20:08
mathcounterexamples.netmathcounterexamples.net
26.2k21955
26.2k21955
$begingroup$
I exactly thought the same thing: the ODE should be discretized by $x_{n+1}-x_n=rx_n(1-x_n)$ because the finite difference ratio $(x_{n+1}-x_n)/1$ plays the role of $dx/dt$. But I don't know...
$endgroup$
– Qfwfq
Dec 9 '18 at 20:22
$begingroup$
In the paragraph you cite (of the wikipedia link) it seems they're considering linear ODEs, while the logistic is nonlinear - not sure how this affects the continuous/discrete relationship.
$endgroup$
– Qfwfq
Dec 9 '18 at 20:25
$begingroup$
It is even worse... if You already have issue with linear problems!
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:26
add a comment |
$begingroup$
I exactly thought the same thing: the ODE should be discretized by $x_{n+1}-x_n=rx_n(1-x_n)$ because the finite difference ratio $(x_{n+1}-x_n)/1$ plays the role of $dx/dt$. But I don't know...
$endgroup$
– Qfwfq
Dec 9 '18 at 20:22
$begingroup$
In the paragraph you cite (of the wikipedia link) it seems they're considering linear ODEs, while the logistic is nonlinear - not sure how this affects the continuous/discrete relationship.
$endgroup$
– Qfwfq
Dec 9 '18 at 20:25
$begingroup$
It is even worse... if You already have issue with linear problems!
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:26
$begingroup$
I exactly thought the same thing: the ODE should be discretized by $x_{n+1}-x_n=rx_n(1-x_n)$ because the finite difference ratio $(x_{n+1}-x_n)/1$ plays the role of $dx/dt$. But I don't know...
$endgroup$
– Qfwfq
Dec 9 '18 at 20:22
$begingroup$
I exactly thought the same thing: the ODE should be discretized by $x_{n+1}-x_n=rx_n(1-x_n)$ because the finite difference ratio $(x_{n+1}-x_n)/1$ plays the role of $dx/dt$. But I don't know...
$endgroup$
– Qfwfq
Dec 9 '18 at 20:22
$begingroup$
In the paragraph you cite (of the wikipedia link) it seems they're considering linear ODEs, while the logistic is nonlinear - not sure how this affects the continuous/discrete relationship.
$endgroup$
– Qfwfq
Dec 9 '18 at 20:25
$begingroup$
In the paragraph you cite (of the wikipedia link) it seems they're considering linear ODEs, while the logistic is nonlinear - not sure how this affects the continuous/discrete relationship.
$endgroup$
– Qfwfq
Dec 9 '18 at 20:25
$begingroup$
It is even worse... if You already have issue with linear problems!
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:26
$begingroup$
It is even worse... if You already have issue with linear problems!
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:26
add a comment |
$begingroup$
In the logistic equation you get transforming it as Bernoulli equation
$$
frac{d}{dt}x(t)^{-1}=r(1-x(t)^{-1})implies x(t)^{-1}=1+ce^{-rt}.
$$
Now compare the values for $t$ and $t+h$,
$$
x(t+h)^{-1}-1 = ce^{-rt-rh}=e^{-rh}(x(t)^{-1}-1).
$$
For the sequence $x_k=x(kh)$ you get thus the recursion formula
$$
x_{k+1}=frac{x_k}{1-e^{-rh}+e^{-rh}x_k}
$$
which looks rather different than the logistic map.
Any sane discretization using the Euler method or similar would use a sufficiently small $h$, so that in the Euler forward discretization $$x_{k+1}=x_k(1+hr)-hrx_k^2.$$ To get to the normal form one would have to rescale $y_k=ax_k$, so that then $$y_{k+1}=y_k(1+hr)-frac{hr}ay_k^2=(1+hr)y_kleft(1-frac{hr}{a(1+hr)}y_kright)$$ giving $a=frac{hr}{1+hr}$. This is indeed case $2$ in the list, with equilibrium at $y_infty=1-frac1{1+hr}=frac{hr}{1+hr}=a$ or translated back at $x_infty=1$.
$endgroup$
$begingroup$
Ok, this is to say the recursion you get depends heavily on how you choose to discretize, I assume? But is it relevant to the question? Which is the discretization method in the OP case, and does it lead to different phenomena (differences between the continuum and the discrete)?
$endgroup$
– Qfwfq
Dec 9 '18 at 21:32
1
$begingroup$
Too long for a comment, added remark on discretization to the answer. One can conclude that the interesting phenomena happen when $h$ is so large that the discretization error can no longer be classified as $O(h^2)$ as the higher order terms are equally large or even dominate.
$endgroup$
– LutzL
Dec 9 '18 at 21:50
$begingroup$
So it's true that chaotic features are not present in the continuous model but appear only once I discretize in that specific way?
$endgroup$
– Qfwfq
Dec 9 '18 at 22:19
add a comment |
$begingroup$
In the logistic equation you get transforming it as Bernoulli equation
$$
frac{d}{dt}x(t)^{-1}=r(1-x(t)^{-1})implies x(t)^{-1}=1+ce^{-rt}.
$$
Now compare the values for $t$ and $t+h$,
$$
x(t+h)^{-1}-1 = ce^{-rt-rh}=e^{-rh}(x(t)^{-1}-1).
$$
For the sequence $x_k=x(kh)$ you get thus the recursion formula
$$
x_{k+1}=frac{x_k}{1-e^{-rh}+e^{-rh}x_k}
$$
which looks rather different than the logistic map.
Any sane discretization using the Euler method or similar would use a sufficiently small $h$, so that in the Euler forward discretization $$x_{k+1}=x_k(1+hr)-hrx_k^2.$$ To get to the normal form one would have to rescale $y_k=ax_k$, so that then $$y_{k+1}=y_k(1+hr)-frac{hr}ay_k^2=(1+hr)y_kleft(1-frac{hr}{a(1+hr)}y_kright)$$ giving $a=frac{hr}{1+hr}$. This is indeed case $2$ in the list, with equilibrium at $y_infty=1-frac1{1+hr}=frac{hr}{1+hr}=a$ or translated back at $x_infty=1$.
$endgroup$
$begingroup$
Ok, this is to say the recursion you get depends heavily on how you choose to discretize, I assume? But is it relevant to the question? Which is the discretization method in the OP case, and does it lead to different phenomena (differences between the continuum and the discrete)?
$endgroup$
– Qfwfq
Dec 9 '18 at 21:32
1
$begingroup$
Too long for a comment, added remark on discretization to the answer. One can conclude that the interesting phenomena happen when $h$ is so large that the discretization error can no longer be classified as $O(h^2)$ as the higher order terms are equally large or even dominate.
$endgroup$
– LutzL
Dec 9 '18 at 21:50
$begingroup$
So it's true that chaotic features are not present in the continuous model but appear only once I discretize in that specific way?
$endgroup$
– Qfwfq
Dec 9 '18 at 22:19
add a comment |
$begingroup$
In the logistic equation you get transforming it as Bernoulli equation
$$
frac{d}{dt}x(t)^{-1}=r(1-x(t)^{-1})implies x(t)^{-1}=1+ce^{-rt}.
$$
Now compare the values for $t$ and $t+h$,
$$
x(t+h)^{-1}-1 = ce^{-rt-rh}=e^{-rh}(x(t)^{-1}-1).
$$
For the sequence $x_k=x(kh)$ you get thus the recursion formula
$$
x_{k+1}=frac{x_k}{1-e^{-rh}+e^{-rh}x_k}
$$
which looks rather different than the logistic map.
Any sane discretization using the Euler method or similar would use a sufficiently small $h$, so that in the Euler forward discretization $$x_{k+1}=x_k(1+hr)-hrx_k^2.$$ To get to the normal form one would have to rescale $y_k=ax_k$, so that then $$y_{k+1}=y_k(1+hr)-frac{hr}ay_k^2=(1+hr)y_kleft(1-frac{hr}{a(1+hr)}y_kright)$$ giving $a=frac{hr}{1+hr}$. This is indeed case $2$ in the list, with equilibrium at $y_infty=1-frac1{1+hr}=frac{hr}{1+hr}=a$ or translated back at $x_infty=1$.
$endgroup$
In the logistic equation you get transforming it as Bernoulli equation
$$
frac{d}{dt}x(t)^{-1}=r(1-x(t)^{-1})implies x(t)^{-1}=1+ce^{-rt}.
$$
Now compare the values for $t$ and $t+h$,
$$
x(t+h)^{-1}-1 = ce^{-rt-rh}=e^{-rh}(x(t)^{-1}-1).
$$
For the sequence $x_k=x(kh)$ you get thus the recursion formula
$$
x_{k+1}=frac{x_k}{1-e^{-rh}+e^{-rh}x_k}
$$
which looks rather different than the logistic map.
Any sane discretization using the Euler method or similar would use a sufficiently small $h$, so that in the Euler forward discretization $$x_{k+1}=x_k(1+hr)-hrx_k^2.$$ To get to the normal form one would have to rescale $y_k=ax_k$, so that then $$y_{k+1}=y_k(1+hr)-frac{hr}ay_k^2=(1+hr)y_kleft(1-frac{hr}{a(1+hr)}y_kright)$$ giving $a=frac{hr}{1+hr}$. This is indeed case $2$ in the list, with equilibrium at $y_infty=1-frac1{1+hr}=frac{hr}{1+hr}=a$ or translated back at $x_infty=1$.
edited Dec 9 '18 at 21:47
answered Dec 9 '18 at 21:01
LutzLLutzL
57.7k42054
57.7k42054
$begingroup$
Ok, this is to say the recursion you get depends heavily on how you choose to discretize, I assume? But is it relevant to the question? Which is the discretization method in the OP case, and does it lead to different phenomena (differences between the continuum and the discrete)?
$endgroup$
– Qfwfq
Dec 9 '18 at 21:32
1
$begingroup$
Too long for a comment, added remark on discretization to the answer. One can conclude that the interesting phenomena happen when $h$ is so large that the discretization error can no longer be classified as $O(h^2)$ as the higher order terms are equally large or even dominate.
$endgroup$
– LutzL
Dec 9 '18 at 21:50
$begingroup$
So it's true that chaotic features are not present in the continuous model but appear only once I discretize in that specific way?
$endgroup$
– Qfwfq
Dec 9 '18 at 22:19
add a comment |
$begingroup$
Ok, this is to say the recursion you get depends heavily on how you choose to discretize, I assume? But is it relevant to the question? Which is the discretization method in the OP case, and does it lead to different phenomena (differences between the continuum and the discrete)?
$endgroup$
– Qfwfq
Dec 9 '18 at 21:32
1
$begingroup$
Too long for a comment, added remark on discretization to the answer. One can conclude that the interesting phenomena happen when $h$ is so large that the discretization error can no longer be classified as $O(h^2)$ as the higher order terms are equally large or even dominate.
$endgroup$
– LutzL
Dec 9 '18 at 21:50
$begingroup$
So it's true that chaotic features are not present in the continuous model but appear only once I discretize in that specific way?
$endgroup$
– Qfwfq
Dec 9 '18 at 22:19
$begingroup$
Ok, this is to say the recursion you get depends heavily on how you choose to discretize, I assume? But is it relevant to the question? Which is the discretization method in the OP case, and does it lead to different phenomena (differences between the continuum and the discrete)?
$endgroup$
– Qfwfq
Dec 9 '18 at 21:32
$begingroup$
Ok, this is to say the recursion you get depends heavily on how you choose to discretize, I assume? But is it relevant to the question? Which is the discretization method in the OP case, and does it lead to different phenomena (differences between the continuum and the discrete)?
$endgroup$
– Qfwfq
Dec 9 '18 at 21:32
1
1
$begingroup$
Too long for a comment, added remark on discretization to the answer. One can conclude that the interesting phenomena happen when $h$ is so large that the discretization error can no longer be classified as $O(h^2)$ as the higher order terms are equally large or even dominate.
$endgroup$
– LutzL
Dec 9 '18 at 21:50
$begingroup$
Too long for a comment, added remark on discretization to the answer. One can conclude that the interesting phenomena happen when $h$ is so large that the discretization error can no longer be classified as $O(h^2)$ as the higher order terms are equally large or even dominate.
$endgroup$
– LutzL
Dec 9 '18 at 21:50
$begingroup$
So it's true that chaotic features are not present in the continuous model but appear only once I discretize in that specific way?
$endgroup$
– Qfwfq
Dec 9 '18 at 22:19
$begingroup$
So it's true that chaotic features are not present in the continuous model but appear only once I discretize in that specific way?
$endgroup$
– Qfwfq
Dec 9 '18 at 22:19
add a comment |
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