Parametric plot from the results of NDSolve

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I'm having a bit of trouble making a ParametricPlot from two curves, At fisrt I was having difficulty solving my systems of autonomous ODEs, but then finally got a way to solve it. However, I am now not able to figure out how to get the parametric plot.



solution[t_] = 
With[n = 1.5,
NDSolve[
q'[t] ==
((q[t]^2)/3)*(3 - n)*x[t]^2 - ((q[t]^2)/3)*n*(y[t] - 1) - ((q[t]^2)/3) + (q[t]*x[t]/3)*(3 - n)*x[t]^2 - (q[t]*x[t]/3)*n*(y[t] - 1) + (q[t]*x[t]/3) + (1/3)*x[t]^2 - (1/3) + (1/3)*((n*y[t]/(n - 1))),
y'[t] ==
((2*y[t]*x[t]^2)/3)*(3 - n)*(x[t] + q[t]) + (2*x[t]*y[t]/3)*((n^2 - 2*n + 2)/(n - 1)) - (2*x[t]*y[t]/3)*n*y[t] + (2*q[t]*n*y[t]*((1 - y[t])/3)),
x'[t] ==
((x[t]^3)/3)*(3 - n)*(q[t] + x[t]) + ((x[t]^2)/3)*n*(2 - y[t]) - 5*((x[t]^2)/3) + (q[t]*x[t]/3)* n*(1 - y[t]) - (q[t]*x[t]) + (1/3)*((n*(n - 2)*y[t])/(n - 1)) - (1/3)*n + 2/3,
q[0] == -1.33, y[0] == 0.88, x[0] == 0.33,
q[t], y[t], x[t], t, 0, 10]][[1, All, 2]]

Plot[solution[t], t, 0, 10]


I want to get a parametric plot of q[t] and y[t]. And eventually a 3D plot of q[t], y[t], x[t].










share|improve this question









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    up vote
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    I'm having a bit of trouble making a ParametricPlot from two curves, At fisrt I was having difficulty solving my systems of autonomous ODEs, but then finally got a way to solve it. However, I am now not able to figure out how to get the parametric plot.



    solution[t_] = 
    With[n = 1.5,
    NDSolve[
    q'[t] ==
    ((q[t]^2)/3)*(3 - n)*x[t]^2 - ((q[t]^2)/3)*n*(y[t] - 1) - ((q[t]^2)/3) + (q[t]*x[t]/3)*(3 - n)*x[t]^2 - (q[t]*x[t]/3)*n*(y[t] - 1) + (q[t]*x[t]/3) + (1/3)*x[t]^2 - (1/3) + (1/3)*((n*y[t]/(n - 1))),
    y'[t] ==
    ((2*y[t]*x[t]^2)/3)*(3 - n)*(x[t] + q[t]) + (2*x[t]*y[t]/3)*((n^2 - 2*n + 2)/(n - 1)) - (2*x[t]*y[t]/3)*n*y[t] + (2*q[t]*n*y[t]*((1 - y[t])/3)),
    x'[t] ==
    ((x[t]^3)/3)*(3 - n)*(q[t] + x[t]) + ((x[t]^2)/3)*n*(2 - y[t]) - 5*((x[t]^2)/3) + (q[t]*x[t]/3)* n*(1 - y[t]) - (q[t]*x[t]) + (1/3)*((n*(n - 2)*y[t])/(n - 1)) - (1/3)*n + 2/3,
    q[0] == -1.33, y[0] == 0.88, x[0] == 0.33,
    q[t], y[t], x[t], t, 0, 10]][[1, All, 2]]

    Plot[solution[t], t, 0, 10]


    I want to get a parametric plot of q[t] and y[t]. And eventually a 3D plot of q[t], y[t], x[t].










    share|improve this question









    New contributor




    Logan Jacobs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I'm having a bit of trouble making a ParametricPlot from two curves, At fisrt I was having difficulty solving my systems of autonomous ODEs, but then finally got a way to solve it. However, I am now not able to figure out how to get the parametric plot.



      solution[t_] = 
      With[n = 1.5,
      NDSolve[
      q'[t] ==
      ((q[t]^2)/3)*(3 - n)*x[t]^2 - ((q[t]^2)/3)*n*(y[t] - 1) - ((q[t]^2)/3) + (q[t]*x[t]/3)*(3 - n)*x[t]^2 - (q[t]*x[t]/3)*n*(y[t] - 1) + (q[t]*x[t]/3) + (1/3)*x[t]^2 - (1/3) + (1/3)*((n*y[t]/(n - 1))),
      y'[t] ==
      ((2*y[t]*x[t]^2)/3)*(3 - n)*(x[t] + q[t]) + (2*x[t]*y[t]/3)*((n^2 - 2*n + 2)/(n - 1)) - (2*x[t]*y[t]/3)*n*y[t] + (2*q[t]*n*y[t]*((1 - y[t])/3)),
      x'[t] ==
      ((x[t]^3)/3)*(3 - n)*(q[t] + x[t]) + ((x[t]^2)/3)*n*(2 - y[t]) - 5*((x[t]^2)/3) + (q[t]*x[t]/3)* n*(1 - y[t]) - (q[t]*x[t]) + (1/3)*((n*(n - 2)*y[t])/(n - 1)) - (1/3)*n + 2/3,
      q[0] == -1.33, y[0] == 0.88, x[0] == 0.33,
      q[t], y[t], x[t], t, 0, 10]][[1, All, 2]]

      Plot[solution[t], t, 0, 10]


      I want to get a parametric plot of q[t] and y[t]. And eventually a 3D plot of q[t], y[t], x[t].










      share|improve this question









      New contributor




      Logan Jacobs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I'm having a bit of trouble making a ParametricPlot from two curves, At fisrt I was having difficulty solving my systems of autonomous ODEs, but then finally got a way to solve it. However, I am now not able to figure out how to get the parametric plot.



      solution[t_] = 
      With[n = 1.5,
      NDSolve[
      q'[t] ==
      ((q[t]^2)/3)*(3 - n)*x[t]^2 - ((q[t]^2)/3)*n*(y[t] - 1) - ((q[t]^2)/3) + (q[t]*x[t]/3)*(3 - n)*x[t]^2 - (q[t]*x[t]/3)*n*(y[t] - 1) + (q[t]*x[t]/3) + (1/3)*x[t]^2 - (1/3) + (1/3)*((n*y[t]/(n - 1))),
      y'[t] ==
      ((2*y[t]*x[t]^2)/3)*(3 - n)*(x[t] + q[t]) + (2*x[t]*y[t]/3)*((n^2 - 2*n + 2)/(n - 1)) - (2*x[t]*y[t]/3)*n*y[t] + (2*q[t]*n*y[t]*((1 - y[t])/3)),
      x'[t] ==
      ((x[t]^3)/3)*(3 - n)*(q[t] + x[t]) + ((x[t]^2)/3)*n*(2 - y[t]) - 5*((x[t]^2)/3) + (q[t]*x[t]/3)* n*(1 - y[t]) - (q[t]*x[t]) + (1/3)*((n*(n - 2)*y[t])/(n - 1)) - (1/3)*n + 2/3,
      q[0] == -1.33, y[0] == 0.88, x[0] == 0.33,
      q[t], y[t], x[t], t, 0, 10]][[1, All, 2]]

      Plot[solution[t], t, 0, 10]


      I want to get a parametric plot of q[t] and y[t]. And eventually a 3D plot of q[t], y[t], x[t].







      differential-equations






      share|improve this question









      New contributor




      Logan Jacobs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Logan Jacobs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited yesterday









      m_goldberg

      82.6k869190




      82.6k869190






      New contributor




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      asked yesterday









      Logan Jacobs

      161




      161




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      New contributor





      Logan Jacobs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.




















          2 Answers
          2






          active

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          up vote
          3
          down vote













          The issue you are facing is to correctly call out the desired dependent variables. So, you need to specify the position of the output. For example, you want to have q[x] then you need to use solution[t][[1]].



          For parametric plot, try this out,



          ParametricPlot[solution[t][[1]], solution[t][[2]], t, 0, 10]


          and then for 3D



          ParametricPlot3D[solution[t][[1]], solution[t][[2]], solution[t][[3]], t, 0, 10]





          share|improve this answer





























            up vote
            2
            down vote













            Redefine your solution a little bit to



            solution := 
            With[n = 1.5,
            NDSolve[q'[t] == ((q[t]^2)/3)*(3 - n)*x[t]^2 - ((q[t]^2)/3)*
            n*(y[t] - 1) - ((q[t]^2)/3) + (q[t]*x[t]/3)*(3 - n)*
            x[t]^2 - (q[t]*x[t]/3)*n*(y[t] - 1) + (q[t]*x[t]/3) + (1/3)*
            x[t]^2 - (1/3) + (1/3)*((n*y[t]/(n - 1))),
            y'[t] == ((2*y[t]*x[t]^2)/3)*(3 - n)*(x[t] + q[t]) + (2*x[t]*
            y[t]/3)*((n^2 - 2*n + 2)/(n - 1)) - (2*x[t]*y[t]/3)*n*
            y[t] + (2*q[t]*n*y[t]*((1 - y[t])/3)),
            x'[t] == ((x[t]^3)/3)*(3 - n)*(q[t] + x[t]) + ((x[t]^2)/3)*
            n*(2 - y[t]) -
            5*((x[t]^2)/3) + (q[t]*x[t]/3)*
            n*(1 - y[t]) - (q[t]*
            x[t]) + (1/3)*((n*(n - 2)*y[t])/(n - 1)) - (1/3)*n + 2/3,
            q[0] == -1.33, y[0] == 0.88, x[0] == 0.33, q , y , x , t, 0,
            10]] [[1]]


            Now you can plot your results



            Plot[Evaluate[ q[t], y[t] /. solution ], t, 0, 10]


            enter image description here






            share|improve this answer




















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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              3
              down vote













              The issue you are facing is to correctly call out the desired dependent variables. So, you need to specify the position of the output. For example, you want to have q[x] then you need to use solution[t][[1]].



              For parametric plot, try this out,



              ParametricPlot[solution[t][[1]], solution[t][[2]], t, 0, 10]


              and then for 3D



              ParametricPlot3D[solution[t][[1]], solution[t][[2]], solution[t][[3]], t, 0, 10]





              share|improve this answer


























                up vote
                3
                down vote













                The issue you are facing is to correctly call out the desired dependent variables. So, you need to specify the position of the output. For example, you want to have q[x] then you need to use solution[t][[1]].



                For parametric plot, try this out,



                ParametricPlot[solution[t][[1]], solution[t][[2]], t, 0, 10]


                and then for 3D



                ParametricPlot3D[solution[t][[1]], solution[t][[2]], solution[t][[3]], t, 0, 10]





                share|improve this answer
























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  The issue you are facing is to correctly call out the desired dependent variables. So, you need to specify the position of the output. For example, you want to have q[x] then you need to use solution[t][[1]].



                  For parametric plot, try this out,



                  ParametricPlot[solution[t][[1]], solution[t][[2]], t, 0, 10]


                  and then for 3D



                  ParametricPlot3D[solution[t][[1]], solution[t][[2]], solution[t][[3]], t, 0, 10]





                  share|improve this answer














                  The issue you are facing is to correctly call out the desired dependent variables. So, you need to specify the position of the output. For example, you want to have q[x] then you need to use solution[t][[1]].



                  For parametric plot, try this out,



                  ParametricPlot[solution[t][[1]], solution[t][[2]], t, 0, 10]


                  and then for 3D



                  ParametricPlot3D[solution[t][[1]], solution[t][[2]], solution[t][[3]], t, 0, 10]






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited yesterday

























                  answered yesterday









                  zhk

                  8,50411433




                  8,50411433




















                      up vote
                      2
                      down vote













                      Redefine your solution a little bit to



                      solution := 
                      With[n = 1.5,
                      NDSolve[q'[t] == ((q[t]^2)/3)*(3 - n)*x[t]^2 - ((q[t]^2)/3)*
                      n*(y[t] - 1) - ((q[t]^2)/3) + (q[t]*x[t]/3)*(3 - n)*
                      x[t]^2 - (q[t]*x[t]/3)*n*(y[t] - 1) + (q[t]*x[t]/3) + (1/3)*
                      x[t]^2 - (1/3) + (1/3)*((n*y[t]/(n - 1))),
                      y'[t] == ((2*y[t]*x[t]^2)/3)*(3 - n)*(x[t] + q[t]) + (2*x[t]*
                      y[t]/3)*((n^2 - 2*n + 2)/(n - 1)) - (2*x[t]*y[t]/3)*n*
                      y[t] + (2*q[t]*n*y[t]*((1 - y[t])/3)),
                      x'[t] == ((x[t]^3)/3)*(3 - n)*(q[t] + x[t]) + ((x[t]^2)/3)*
                      n*(2 - y[t]) -
                      5*((x[t]^2)/3) + (q[t]*x[t]/3)*
                      n*(1 - y[t]) - (q[t]*
                      x[t]) + (1/3)*((n*(n - 2)*y[t])/(n - 1)) - (1/3)*n + 2/3,
                      q[0] == -1.33, y[0] == 0.88, x[0] == 0.33, q , y , x , t, 0,
                      10]] [[1]]


                      Now you can plot your results



                      Plot[Evaluate[ q[t], y[t] /. solution ], t, 0, 10]


                      enter image description here






                      share|improve this answer
























                        up vote
                        2
                        down vote













                        Redefine your solution a little bit to



                        solution := 
                        With[n = 1.5,
                        NDSolve[q'[t] == ((q[t]^2)/3)*(3 - n)*x[t]^2 - ((q[t]^2)/3)*
                        n*(y[t] - 1) - ((q[t]^2)/3) + (q[t]*x[t]/3)*(3 - n)*
                        x[t]^2 - (q[t]*x[t]/3)*n*(y[t] - 1) + (q[t]*x[t]/3) + (1/3)*
                        x[t]^2 - (1/3) + (1/3)*((n*y[t]/(n - 1))),
                        y'[t] == ((2*y[t]*x[t]^2)/3)*(3 - n)*(x[t] + q[t]) + (2*x[t]*
                        y[t]/3)*((n^2 - 2*n + 2)/(n - 1)) - (2*x[t]*y[t]/3)*n*
                        y[t] + (2*q[t]*n*y[t]*((1 - y[t])/3)),
                        x'[t] == ((x[t]^3)/3)*(3 - n)*(q[t] + x[t]) + ((x[t]^2)/3)*
                        n*(2 - y[t]) -
                        5*((x[t]^2)/3) + (q[t]*x[t]/3)*
                        n*(1 - y[t]) - (q[t]*
                        x[t]) + (1/3)*((n*(n - 2)*y[t])/(n - 1)) - (1/3)*n + 2/3,
                        q[0] == -1.33, y[0] == 0.88, x[0] == 0.33, q , y , x , t, 0,
                        10]] [[1]]


                        Now you can plot your results



                        Plot[Evaluate[ q[t], y[t] /. solution ], t, 0, 10]


                        enter image description here






                        share|improve this answer






















                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          Redefine your solution a little bit to



                          solution := 
                          With[n = 1.5,
                          NDSolve[q'[t] == ((q[t]^2)/3)*(3 - n)*x[t]^2 - ((q[t]^2)/3)*
                          n*(y[t] - 1) - ((q[t]^2)/3) + (q[t]*x[t]/3)*(3 - n)*
                          x[t]^2 - (q[t]*x[t]/3)*n*(y[t] - 1) + (q[t]*x[t]/3) + (1/3)*
                          x[t]^2 - (1/3) + (1/3)*((n*y[t]/(n - 1))),
                          y'[t] == ((2*y[t]*x[t]^2)/3)*(3 - n)*(x[t] + q[t]) + (2*x[t]*
                          y[t]/3)*((n^2 - 2*n + 2)/(n - 1)) - (2*x[t]*y[t]/3)*n*
                          y[t] + (2*q[t]*n*y[t]*((1 - y[t])/3)),
                          x'[t] == ((x[t]^3)/3)*(3 - n)*(q[t] + x[t]) + ((x[t]^2)/3)*
                          n*(2 - y[t]) -
                          5*((x[t]^2)/3) + (q[t]*x[t]/3)*
                          n*(1 - y[t]) - (q[t]*
                          x[t]) + (1/3)*((n*(n - 2)*y[t])/(n - 1)) - (1/3)*n + 2/3,
                          q[0] == -1.33, y[0] == 0.88, x[0] == 0.33, q , y , x , t, 0,
                          10]] [[1]]


                          Now you can plot your results



                          Plot[Evaluate[ q[t], y[t] /. solution ], t, 0, 10]


                          enter image description here






                          share|improve this answer












                          Redefine your solution a little bit to



                          solution := 
                          With[n = 1.5,
                          NDSolve[q'[t] == ((q[t]^2)/3)*(3 - n)*x[t]^2 - ((q[t]^2)/3)*
                          n*(y[t] - 1) - ((q[t]^2)/3) + (q[t]*x[t]/3)*(3 - n)*
                          x[t]^2 - (q[t]*x[t]/3)*n*(y[t] - 1) + (q[t]*x[t]/3) + (1/3)*
                          x[t]^2 - (1/3) + (1/3)*((n*y[t]/(n - 1))),
                          y'[t] == ((2*y[t]*x[t]^2)/3)*(3 - n)*(x[t] + q[t]) + (2*x[t]*
                          y[t]/3)*((n^2 - 2*n + 2)/(n - 1)) - (2*x[t]*y[t]/3)*n*
                          y[t] + (2*q[t]*n*y[t]*((1 - y[t])/3)),
                          x'[t] == ((x[t]^3)/3)*(3 - n)*(q[t] + x[t]) + ((x[t]^2)/3)*
                          n*(2 - y[t]) -
                          5*((x[t]^2)/3) + (q[t]*x[t]/3)*
                          n*(1 - y[t]) - (q[t]*
                          x[t]) + (1/3)*((n*(n - 2)*y[t])/(n - 1)) - (1/3)*n + 2/3,
                          q[0] == -1.33, y[0] == 0.88, x[0] == 0.33, q , y , x , t, 0,
                          10]] [[1]]


                          Now you can plot your results



                          Plot[Evaluate[ q[t], y[t] /. solution ], t, 0, 10]


                          enter image description here







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered yesterday









                          Ulrich Neumann

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