How to apply improved Euler method to a systems of differential equation?












0












$begingroup$


I have the following problem (linearized pendulum problem):



$x_1'=x_2$



$x_2'=-frac{g}{L}*sin(x_1)$



with the following initial condition:



$x_1(t_0)=0$



$x_2(t_0)=0$



Improved Euler Method says that:



$Y_{k+1}=Y_k+frac{h}{2}*[f(t_k, Y_k)+f(t_{k+1}, Y_k+h*f(t_k, Y_k))]$



In this case I have done the following:



$x_{1, k+1}=x_{1, k}+frac{h}{2}*[x_{1, k}+(x_{1, k}+h*x_{1, k})]$



$x_{2, k+1}=x_{2, k}+frac{h}{2}*[-frac{g}{L}*sin(x_{1, k})+-frac{g}{L}*sin(x_{1, k}+h*(-frac{g}{L}*sin(x_{1, k}))))]$



Where is my mistake?






numerical-methods







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  • $begingroup$
    Why do you think there's a mistake?
    $endgroup$
    – caverac
    Jan 12 at 11:35










  • $begingroup$
    @caverac: I've been using it in a MATLAB code and it differs very much from the exact solution. That's why I have supposed that the method I've written is mistaken.
    $endgroup$
    – user2235427
    Jan 12 at 11:39










  • $begingroup$
    How could you notice a difference? It should be all zero with these initial values.
    $endgroup$
    – LutzL
    Jan 12 at 15:27
















0












$begingroup$


I have the following problem (linearized pendulum problem):



$x_1'=x_2$



$x_2'=-frac{g}{L}*sin(x_1)$



with the following initial condition:



$x_1(t_0)=0$



$x_2(t_0)=0$



Improved Euler Method says that:



$Y_{k+1}=Y_k+frac{h}{2}*[f(t_k, Y_k)+f(t_{k+1}, Y_k+h*f(t_k, Y_k))]$



In this case I have done the following:



$x_{1, k+1}=x_{1, k}+frac{h}{2}*[x_{1, k}+(x_{1, k}+h*x_{1, k})]$



$x_{2, k+1}=x_{2, k}+frac{h}{2}*[-frac{g}{L}*sin(x_{1, k})+-frac{g}{L}*sin(x_{1, k}+h*(-frac{g}{L}*sin(x_{1, k}))))]$



Where is my mistake?






numerical-methods







share|cite|improve this question









$endgroup$












  • $begingroup$
    Why do you think there's a mistake?
    $endgroup$
    – caverac
    Jan 12 at 11:35










  • $begingroup$
    @caverac: I've been using it in a MATLAB code and it differs very much from the exact solution. That's why I have supposed that the method I've written is mistaken.
    $endgroup$
    – user2235427
    Jan 12 at 11:39










  • $begingroup$
    How could you notice a difference? It should be all zero with these initial values.
    $endgroup$
    – LutzL
    Jan 12 at 15:27














0












0








0


1



$begingroup$


I have the following problem (linearized pendulum problem):



$x_1'=x_2$



$x_2'=-frac{g}{L}*sin(x_1)$



with the following initial condition:



$x_1(t_0)=0$



$x_2(t_0)=0$



Improved Euler Method says that:



$Y_{k+1}=Y_k+frac{h}{2}*[f(t_k, Y_k)+f(t_{k+1}, Y_k+h*f(t_k, Y_k))]$



In this case I have done the following:



$x_{1, k+1}=x_{1, k}+frac{h}{2}*[x_{1, k}+(x_{1, k}+h*x_{1, k})]$



$x_{2, k+1}=x_{2, k}+frac{h}{2}*[-frac{g}{L}*sin(x_{1, k})+-frac{g}{L}*sin(x_{1, k}+h*(-frac{g}{L}*sin(x_{1, k}))))]$



Where is my mistake?






numerical-methods







share|cite|improve this question









$endgroup$




I have the following problem (linearized pendulum problem):



$x_1'=x_2$



$x_2'=-frac{g}{L}*sin(x_1)$



with the following initial condition:



$x_1(t_0)=0$



$x_2(t_0)=0$



Improved Euler Method says that:



$Y_{k+1}=Y_k+frac{h}{2}*[f(t_k, Y_k)+f(t_{k+1}, Y_k+h*f(t_k, Y_k))]$



In this case I have done the following:



$x_{1, k+1}=x_{1, k}+frac{h}{2}*[x_{1, k}+(x_{1, k}+h*x_{1, k})]$



$x_{2, k+1}=x_{2, k}+frac{h}{2}*[-frac{g}{L}*sin(x_{1, k})+-frac{g}{L}*sin(x_{1, k}+h*(-frac{g}{L}*sin(x_{1, k}))))]$



Where is my mistake?






numerical-methods




numerical-methods






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 11:33









user2235427user2235427

1083




1083












  • $begingroup$
    Why do you think there's a mistake?
    $endgroup$
    – caverac
    Jan 12 at 11:35










  • $begingroup$
    @caverac: I've been using it in a MATLAB code and it differs very much from the exact solution. That's why I have supposed that the method I've written is mistaken.
    $endgroup$
    – user2235427
    Jan 12 at 11:39










  • $begingroup$
    How could you notice a difference? It should be all zero with these initial values.
    $endgroup$
    – LutzL
    Jan 12 at 15:27


















  • $begingroup$
    Why do you think there's a mistake?
    $endgroup$
    – caverac
    Jan 12 at 11:35










  • $begingroup$
    @caverac: I've been using it in a MATLAB code and it differs very much from the exact solution. That's why I have supposed that the method I've written is mistaken.
    $endgroup$
    – user2235427
    Jan 12 at 11:39










  • $begingroup$
    How could you notice a difference? It should be all zero with these initial values.
    $endgroup$
    – LutzL
    Jan 12 at 15:27
















$begingroup$
Why do you think there's a mistake?
$endgroup$
– caverac
Jan 12 at 11:35




$begingroup$
Why do you think there's a mistake?
$endgroup$
– caverac
Jan 12 at 11:35












$begingroup$
@caverac: I've been using it in a MATLAB code and it differs very much from the exact solution. That's why I have supposed that the method I've written is mistaken.
$endgroup$
– user2235427
Jan 12 at 11:39




$begingroup$
@caverac: I've been using it in a MATLAB code and it differs very much from the exact solution. That's why I have supposed that the method I've written is mistaken.
$endgroup$
– user2235427
Jan 12 at 11:39












$begingroup$
How could you notice a difference? It should be all zero with these initial values.
$endgroup$
– LutzL
Jan 12 at 15:27




$begingroup$
How could you notice a difference? It should be all zero with these initial values.
$endgroup$
– LutzL
Jan 12 at 15:27










2 Answers
2






active

oldest

votes


















1












$begingroup$

One of the most common problems in numerical integration is the time step. Make sure it is small enough so the system does not explode. Beyond that here are a couple of tips




  1. The system has a fixed point at $(x_1, x_2) = 0$. That means that if your initial condition starts there, the system will not move


  2. Define a function that takes x (array) and returns dx (array) with dx[1] = x[2] and dx[2] = - g * sin(x[1]) / l. This way is simpler to evaluate the system, and your code is easier to debug


  3. Here's a solution using $h = 0.01$ in python, just for reference



enter image description here






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You have to use the correct right side for the increments of the variables. If
    $$
    x_1'=f(x_2)\
    x_2'=g(x_1)
    $$
    then the method step reads
    $$
    x_{1,k+1}=x_{1,k}+frac h2[f(x_{2,k})+f(x_{2,k}+hg(x_{1,k}))]\
    x_{2,k+1}=x_{2,k}+frac h2[g(x_{1,k})+g(x_{1,k}+hf(x_{1,k}))]\
    $$
    To keep track of these mixings you can either use a vector-based code or compute the steps like in the standard formulation of Runge-Kutta methods,
    begin{align}
    Δ_1x_1 &= f(x_{2,k})Δt,& Δ_1x_2 &= g(x_{1,k})Δt,\
    Δ_2x_1 &= f(x_{2,k}+Δ_1x_2)Δt,& Δ_2x_2 &= g(x_{1,k}+Δ_1x_1)Δt,\[2em]hline
    x_{1,k+1}&=x_{1,k}+frac{Δ_1x_1+Δ_2x_1}2,&x_{2,k+1}&=x_{2,k}+frac{Δ_1x_2+Δ_2x_2}2.
    end{align}






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      One of the most common problems in numerical integration is the time step. Make sure it is small enough so the system does not explode. Beyond that here are a couple of tips




      1. The system has a fixed point at $(x_1, x_2) = 0$. That means that if your initial condition starts there, the system will not move


      2. Define a function that takes x (array) and returns dx (array) with dx[1] = x[2] and dx[2] = - g * sin(x[1]) / l. This way is simpler to evaluate the system, and your code is easier to debug


      3. Here's a solution using $h = 0.01$ in python, just for reference



      enter image description here






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        One of the most common problems in numerical integration is the time step. Make sure it is small enough so the system does not explode. Beyond that here are a couple of tips




        1. The system has a fixed point at $(x_1, x_2) = 0$. That means that if your initial condition starts there, the system will not move


        2. Define a function that takes x (array) and returns dx (array) with dx[1] = x[2] and dx[2] = - g * sin(x[1]) / l. This way is simpler to evaluate the system, and your code is easier to debug


        3. Here's a solution using $h = 0.01$ in python, just for reference



        enter image description here






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          One of the most common problems in numerical integration is the time step. Make sure it is small enough so the system does not explode. Beyond that here are a couple of tips




          1. The system has a fixed point at $(x_1, x_2) = 0$. That means that if your initial condition starts there, the system will not move


          2. Define a function that takes x (array) and returns dx (array) with dx[1] = x[2] and dx[2] = - g * sin(x[1]) / l. This way is simpler to evaluate the system, and your code is easier to debug


          3. Here's a solution using $h = 0.01$ in python, just for reference



          enter image description here






          share|cite|improve this answer









          $endgroup$



          One of the most common problems in numerical integration is the time step. Make sure it is small enough so the system does not explode. Beyond that here are a couple of tips




          1. The system has a fixed point at $(x_1, x_2) = 0$. That means that if your initial condition starts there, the system will not move


          2. Define a function that takes x (array) and returns dx (array) with dx[1] = x[2] and dx[2] = - g * sin(x[1]) / l. This way is simpler to evaluate the system, and your code is easier to debug


          3. Here's a solution using $h = 0.01$ in python, just for reference



          enter image description here







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 12 at 12:18









          caveraccaverac

          14.8k31130




          14.8k31130























              0












              $begingroup$

              You have to use the correct right side for the increments of the variables. If
              $$
              x_1'=f(x_2)\
              x_2'=g(x_1)
              $$
              then the method step reads
              $$
              x_{1,k+1}=x_{1,k}+frac h2[f(x_{2,k})+f(x_{2,k}+hg(x_{1,k}))]\
              x_{2,k+1}=x_{2,k}+frac h2[g(x_{1,k})+g(x_{1,k}+hf(x_{1,k}))]\
              $$
              To keep track of these mixings you can either use a vector-based code or compute the steps like in the standard formulation of Runge-Kutta methods,
              begin{align}
              Δ_1x_1 &= f(x_{2,k})Δt,& Δ_1x_2 &= g(x_{1,k})Δt,\
              Δ_2x_1 &= f(x_{2,k}+Δ_1x_2)Δt,& Δ_2x_2 &= g(x_{1,k}+Δ_1x_1)Δt,\[2em]hline
              x_{1,k+1}&=x_{1,k}+frac{Δ_1x_1+Δ_2x_1}2,&x_{2,k+1}&=x_{2,k}+frac{Δ_1x_2+Δ_2x_2}2.
              end{align}






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You have to use the correct right side for the increments of the variables. If
                $$
                x_1'=f(x_2)\
                x_2'=g(x_1)
                $$
                then the method step reads
                $$
                x_{1,k+1}=x_{1,k}+frac h2[f(x_{2,k})+f(x_{2,k}+hg(x_{1,k}))]\
                x_{2,k+1}=x_{2,k}+frac h2[g(x_{1,k})+g(x_{1,k}+hf(x_{1,k}))]\
                $$
                To keep track of these mixings you can either use a vector-based code or compute the steps like in the standard formulation of Runge-Kutta methods,
                begin{align}
                Δ_1x_1 &= f(x_{2,k})Δt,& Δ_1x_2 &= g(x_{1,k})Δt,\
                Δ_2x_1 &= f(x_{2,k}+Δ_1x_2)Δt,& Δ_2x_2 &= g(x_{1,k}+Δ_1x_1)Δt,\[2em]hline
                x_{1,k+1}&=x_{1,k}+frac{Δ_1x_1+Δ_2x_1}2,&x_{2,k+1}&=x_{2,k}+frac{Δ_1x_2+Δ_2x_2}2.
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You have to use the correct right side for the increments of the variables. If
                  $$
                  x_1'=f(x_2)\
                  x_2'=g(x_1)
                  $$
                  then the method step reads
                  $$
                  x_{1,k+1}=x_{1,k}+frac h2[f(x_{2,k})+f(x_{2,k}+hg(x_{1,k}))]\
                  x_{2,k+1}=x_{2,k}+frac h2[g(x_{1,k})+g(x_{1,k}+hf(x_{1,k}))]\
                  $$
                  To keep track of these mixings you can either use a vector-based code or compute the steps like in the standard formulation of Runge-Kutta methods,
                  begin{align}
                  Δ_1x_1 &= f(x_{2,k})Δt,& Δ_1x_2 &= g(x_{1,k})Δt,\
                  Δ_2x_1 &= f(x_{2,k}+Δ_1x_2)Δt,& Δ_2x_2 &= g(x_{1,k}+Δ_1x_1)Δt,\[2em]hline
                  x_{1,k+1}&=x_{1,k}+frac{Δ_1x_1+Δ_2x_1}2,&x_{2,k+1}&=x_{2,k}+frac{Δ_1x_2+Δ_2x_2}2.
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  You have to use the correct right side for the increments of the variables. If
                  $$
                  x_1'=f(x_2)\
                  x_2'=g(x_1)
                  $$
                  then the method step reads
                  $$
                  x_{1,k+1}=x_{1,k}+frac h2[f(x_{2,k})+f(x_{2,k}+hg(x_{1,k}))]\
                  x_{2,k+1}=x_{2,k}+frac h2[g(x_{1,k})+g(x_{1,k}+hf(x_{1,k}))]\
                  $$
                  To keep track of these mixings you can either use a vector-based code or compute the steps like in the standard formulation of Runge-Kutta methods,
                  begin{align}
                  Δ_1x_1 &= f(x_{2,k})Δt,& Δ_1x_2 &= g(x_{1,k})Δt,\
                  Δ_2x_1 &= f(x_{2,k}+Δ_1x_2)Δt,& Δ_2x_2 &= g(x_{1,k}+Δ_1x_1)Δt,\[2em]hline
                  x_{1,k+1}&=x_{1,k}+frac{Δ_1x_1+Δ_2x_1}2,&x_{2,k+1}&=x_{2,k}+frac{Δ_1x_2+Δ_2x_2}2.
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 12 at 15:25









                  LutzLLutzL

                  59.6k42057




                  59.6k42057






























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