Driving a system of differential equations with an AR1-Process

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I have the following system of differential equations: 



v[t_] := RandomVariate[NormalDistribution];
sol = NDSolve[x1'[t] == -0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 
 1.22 x4[t] + v[t], 
 x2'[t] == 
 0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] + 0.28 x4[t] + v[t], 
 x3'[t] == 
 0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t] + v[t], 
 x4'[t] == 
 0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t] + v[t], 
 x1[0] == 0, x2[0] == 0, x3[0] == 0, x4[0] == 0, x1, x2, x3, 
 x4, t, 0, 60];
graphAll = 
 Plot[Evaluate[x1[t], x2[t], x3[t], x4[t] /. sol], t, 0, 60, 
 PlotRange -> All]


I want to drive the system with an AR1-Process. So far I was only able to specify some random variate from a normal distribution, but it does not seem to work properly.



Any ideas how to do this?






differential-equations random-process







share|improve this question

























    up vote
    2
    down vote

    favorite












    I have the following system of differential equations: 



    v[t_] := RandomVariate[NormalDistribution];
    sol = NDSolve[x1'[t] == -0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 
     1.22 x4[t] + v[t], 
     x2'[t] == 
     0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] + 0.28 x4[t] + v[t], 
     x3'[t] == 
     0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t] + v[t], 
     x4'[t] == 
     0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t] + v[t], 
     x1[0] == 0, x2[0] == 0, x3[0] == 0, x4[0] == 0, x1, x2, x3, 
     x4, t, 0, 60];
    graphAll = 
     Plot[Evaluate[x1[t], x2[t], x3[t], x4[t] /. sol], t, 0, 60, 
     PlotRange -> All]


    I want to drive the system with an AR1-Process. So far I was only able to specify some random variate from a normal distribution, but it does not seem to work properly.



    Any ideas how to do this?






    differential-equations random-process







    share|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I have the following system of differential equations: 



      v[t_] := RandomVariate[NormalDistribution];
      sol = NDSolve[x1'[t] == -0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 
       1.22 x4[t] + v[t], 
       x2'[t] == 
       0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] + 0.28 x4[t] + v[t], 
       x3'[t] == 
       0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t] + v[t], 
       x4'[t] == 
       0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t] + v[t], 
       x1[0] == 0, x2[0] == 0, x3[0] == 0, x4[0] == 0, x1, x2, x3, 
       x4, t, 0, 60];
      graphAll = 
       Plot[Evaluate[x1[t], x2[t], x3[t], x4[t] /. sol], t, 0, 60, 
       PlotRange -> All]


      I want to drive the system with an AR1-Process. So far I was only able to specify some random variate from a normal distribution, but it does not seem to work properly.



      Any ideas how to do this?






      differential-equations random-process







      share|improve this question













      I have the following system of differential equations: 



      v[t_] := RandomVariate[NormalDistribution];
      sol = NDSolve[x1'[t] == -0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 
       1.22 x4[t] + v[t], 
       x2'[t] == 
       0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] + 0.28 x4[t] + v[t], 
       x3'[t] == 
       0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t] + v[t], 
       x4'[t] == 
       0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t] + v[t], 
       x1[0] == 0, x2[0] == 0, x3[0] == 0, x4[0] == 0, x1, x2, x3, 
       x4, t, 0, 60];
      graphAll = 
       Plot[Evaluate[x1[t], x2[t], x3[t], x4[t] /. sol], t, 0, 60, 
       PlotRange -> All]


      I want to drive the system with an AR1-Process. So far I was only able to specify some random variate from a normal distribution, but it does not seem to work properly.



      Any ideas how to do this?






      differential-equations random-process




      differential-equations random-process






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 46 mins ago









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          1 Answer
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          You need to use RandomFunction[ItoProcess] instead of NDSolve. The syntax is a frustratingly a little different than NDSolve. Driven by a WienerProcess as in your example:



          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == [DifferentialD]v[t] + 
           (-0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == [DifferentialD]v[t] + 
           (0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] +0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == [DifferentialD]v[t] +
           (0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == [DifferentialD]v[t] +
           (0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           x1[t], x2[t], x3[t], x4[t], x1, x2, x3, x4, 0, 0, 0, 0, t,
           v [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol, PlotRange -> All]


          Mathematica graphics



          For the AR-1 part, I think you need to add a first-order decay equation for v[t] driven by a WienerProcess like:



          ϕ = 1;
          σ = 1;
          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == (v[t] - 0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == (v[t] + 0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] + 0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == (v[t] + 0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == (v[t] + 0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           [DifferentialD]v[t] == -ϕ v[t] [DifferentialD]t + σ [DifferentialD]W[t],
           x1[t], x2[t], x3[t], x4[t], v[t], x1, x2, x3, x4, v, 0, 0, 0, 0, 0, t,
           W [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol]


          Mathematica graphics






          share|improve this answer






















          • @MariuszIwaniuk Yeah I made a few mistakes in the AR(1) part and also missed a couple commas -- could you try again?
            – Chris K
            12 mins ago











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          1 Answer
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          active

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          up vote
          3
          down vote













          You need to use RandomFunction[ItoProcess] instead of NDSolve. The syntax is a frustratingly a little different than NDSolve. Driven by a WienerProcess as in your example:



          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == [DifferentialD]v[t] + 
           (-0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == [DifferentialD]v[t] + 
           (0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] +0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == [DifferentialD]v[t] +
           (0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == [DifferentialD]v[t] +
           (0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           x1[t], x2[t], x3[t], x4[t], x1, x2, x3, x4, 0, 0, 0, 0, t,
           v [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol, PlotRange -> All]


          Mathematica graphics



          For the AR-1 part, I think you need to add a first-order decay equation for v[t] driven by a WienerProcess like:



          ϕ = 1;
          σ = 1;
          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == (v[t] - 0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == (v[t] + 0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] + 0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == (v[t] + 0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == (v[t] + 0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           [DifferentialD]v[t] == -ϕ v[t] [DifferentialD]t + σ [DifferentialD]W[t],
           x1[t], x2[t], x3[t], x4[t], v[t], x1, x2, x3, x4, v, 0, 0, 0, 0, 0, t,
           W [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol]


          Mathematica graphics






          share|improve this answer






















          • @MariuszIwaniuk Yeah I made a few mistakes in the AR(1) part and also missed a couple commas -- could you try again?
            – Chris K
            12 mins ago















          up vote
          3
          down vote













          You need to use RandomFunction[ItoProcess] instead of NDSolve. The syntax is a frustratingly a little different than NDSolve. Driven by a WienerProcess as in your example:



          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == [DifferentialD]v[t] + 
           (-0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == [DifferentialD]v[t] + 
           (0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] +0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == [DifferentialD]v[t] +
           (0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == [DifferentialD]v[t] +
           (0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           x1[t], x2[t], x3[t], x4[t], x1, x2, x3, x4, 0, 0, 0, 0, t,
           v [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol, PlotRange -> All]


          Mathematica graphics



          For the AR-1 part, I think you need to add a first-order decay equation for v[t] driven by a WienerProcess like:



          ϕ = 1;
          σ = 1;
          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == (v[t] - 0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == (v[t] + 0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] + 0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == (v[t] + 0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == (v[t] + 0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           [DifferentialD]v[t] == -ϕ v[t] [DifferentialD]t + σ [DifferentialD]W[t],
           x1[t], x2[t], x3[t], x4[t], v[t], x1, x2, x3, x4, v, 0, 0, 0, 0, 0, t,
           W [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol]


          Mathematica graphics






          share|improve this answer






















          • @MariuszIwaniuk Yeah I made a few mistakes in the AR(1) part and also missed a couple commas -- could you try again?
            – Chris K
            12 mins ago













          up vote
          3
          down vote










          up vote
          3
          down vote









          You need to use RandomFunction[ItoProcess] instead of NDSolve. The syntax is a frustratingly a little different than NDSolve. Driven by a WienerProcess as in your example:



          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == [DifferentialD]v[t] + 
           (-0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == [DifferentialD]v[t] + 
           (0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] +0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == [DifferentialD]v[t] +
           (0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == [DifferentialD]v[t] +
           (0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           x1[t], x2[t], x3[t], x4[t], x1, x2, x3, x4, 0, 0, 0, 0, t,
           v [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol, PlotRange -> All]


          Mathematica graphics



          For the AR-1 part, I think you need to add a first-order decay equation for v[t] driven by a WienerProcess like:



          ϕ = 1;
          σ = 1;
          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == (v[t] - 0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == (v[t] + 0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] + 0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == (v[t] + 0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == (v[t] + 0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           [DifferentialD]v[t] == -ϕ v[t] [DifferentialD]t + σ [DifferentialD]W[t],
           x1[t], x2[t], x3[t], x4[t], v[t], x1, x2, x3, x4, v, 0, 0, 0, 0, 0, t,
           W [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol]


          Mathematica graphics






          share|improve this answer














          You need to use RandomFunction[ItoProcess] instead of NDSolve. The syntax is a frustratingly a little different than NDSolve. Driven by a WienerProcess as in your example:



          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == [DifferentialD]v[t] + 
           (-0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == [DifferentialD]v[t] + 
           (0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] +0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == [DifferentialD]v[t] +
           (0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == [DifferentialD]v[t] +
           (0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           x1[t], x2[t], x3[t], x4[t], x1, x2, x3, x4, 0, 0, 0, 0, t,
           v [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol, PlotRange -> All]


          Mathematica graphics



          For the AR-1 part, I think you need to add a first-order decay equation for v[t] driven by a WienerProcess like:



          ϕ = 1;
          σ = 1;
          sol = RandomFunction[ItoProcess[
           [DifferentialD]x1[t] == (v[t] - 0.33 x1[t] - 1.13 x2[t] - 1.84 x3[t] - 1.22 x4[t]) [DifferentialD]t,
           [DifferentialD]x2[t] == (v[t] + 0.15 x1[t] - 0.57 x2[t] + 0.29 x3[t] + 0.28 x4[t]) [DifferentialD]t,
           [DifferentialD]x3[t] == (v[t] + 0.24 x1[t] + 0.34 x2[t] - 0.48 x3[t] + 0.38 x4[t]) [DifferentialD]t,
           [DifferentialD]x4[t] == (v[t] + 0.17 x1[t] + 0.18 x2[t] + 0.32 x3[t] - 0.56 x4[t]) [DifferentialD]t,
           [DifferentialD]v[t] == -ϕ v[t] [DifferentialD]t + σ [DifferentialD]W[t],
           x1[t], x2[t], x3[t], x4[t], v[t], x1, x2, x3, x4, v, 0, 0, 0, 0, 0, t,
           W [Distributed] WienerProcess[0, 1]], 0, 100, 0.01];
          ListLinePlot[sol]


          Mathematica graphics







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 12 mins ago

























          answered 29 mins ago









          Chris K

          5,82221738




          5,82221738











          • @MariuszIwaniuk Yeah I made a few mistakes in the AR(1) part and also missed a couple commas -- could you try again?
            – Chris K
            12 mins ago

















          • @MariuszIwaniuk Yeah I made a few mistakes in the AR(1) part and also missed a couple commas -- could you try again?
            – Chris K
            12 mins ago
















          @MariuszIwaniuk Yeah I made a few mistakes in the AR(1) part and also missed a couple commas -- could you try again?
          – Chris K
          12 mins ago





          @MariuszIwaniuk Yeah I made a few mistakes in the AR(1) part and also missed a couple commas -- could you try again?
          – Chris K
          12 mins ago


















           

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          Confectionery

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          Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of ...
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