Logistic map (discrete dynamical system) vs logistic differential equation












5












$begingroup$


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,




  1. For $0<rleq 1$ there's extinction of the population.

  2. For $1<rleq 3$ the sequence tends to a stable equilibrium $x_infty:=1-1/r$.

  3. For $3<rleq 1+sqrt{6}$ there's convergence to a period-$2$ cycle.

  4. 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.

  5. 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?











share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    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,




    1. For $0<rleq 1$ there's extinction of the population.

    2. For $1<rleq 3$ the sequence tends to a stable equilibrium $x_infty:=1-1/r$.

    3. For $3<rleq 1+sqrt{6}$ there's convergence to a period-$2$ cycle.

    4. 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.

    5. 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?











    share|cite|improve this question











    $endgroup$















      5












      5








      5


      3



      $begingroup$


      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,




      1. For $0<rleq 1$ there's extinction of the population.

      2. For $1<rleq 3$ the sequence tends to a stable equilibrium $x_infty:=1-1/r$.

      3. For $3<rleq 1+sqrt{6}$ there's convergence to a period-$2$ cycle.

      4. 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.

      5. 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?











      share|cite|improve this question











      $endgroup$




      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,




      1. For $0<rleq 1$ there's extinction of the population.

      2. For $1<rleq 3$ the sequence tends to a stable equilibrium $x_infty:=1-1/r$.

      3. For $3<rleq 1+sqrt{6}$ there's convergence to a period-$2$ cycle.

      4. 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.

      5. 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






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 19:59







      Qfwfq

















      asked Dec 9 '18 at 19:52









      QfwfqQfwfq

      324112




      324112






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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



















          1












          $begingroup$

          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 $x_infty=1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 20:08









          mathcounterexamples.netmathcounterexamples.net

          27k22158




          27k22158












          • $begingroup$
            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










          • $begingroup$
            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











          1












          $begingroup$

          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 $x_infty=1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          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 $x_infty=1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          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 $x_infty=1$.






          share|cite|improve this answer











          $endgroup$



          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 $x_infty=1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 21:47

























          answered Dec 9 '18 at 21:01









          LutzLLutzL

          59.7k42057




          59.7k42057












          • $begingroup$
            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






          • 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


















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