I need to solve using first order Euler method some simple equation $ \frac{dc}{dt} +3 c = f(t) $

where $f(t)$ is random delta-correlated force: $<f(t_1)f(t_2)>=2A \delta(t_1-t_2)$

Searching for similar problems I found that:

My code:

 MyEuler[start_, end_, initialvalue_, nrOfsteps_] :=
    Module[{a = start, b = end, j, m = nrOfsteps}, h = (b - a)/m; 
    T = Table[a + (j - 1) h, {j, 1, m + 1}];
    rnd = RandomVariate[NormalDistribution[0, 1]];
    Y = Table[initialvalue, {j, 1, m + 1}];
    For[j = 1, j <= m, j++, 
    Y[[j + 1]] = Y[[j]] + h f[T[[j]], Y[[j]]]] + (2 rnd)/Sqrt[h];
    Transpose@{T, Y}]
    f[t_, x_] = -3 x;(*rhs of ODE without thermal noise*)
    pts = MyEuler[0.0, 3.0, 10.0, 20];
    ListLinePlot[pts, Mesh -> All, MeshStyle -> Red, Frame -> True]

But the results is just decaying amplitude without any random force indications in it. How should I modify this code to get correct answer? Namely $c(t)$ should relax not to zero value but to some level given by a random force amplitude I believe.

  • $\begingroup$ Not sure about the meaning of $<f(t_1)f(t_2)>=2A \delta(t_1-t_2)$ but 1. + (2 rnd)/Sqrt[h] should be inside For; 2. rnd = should probably be rnd := $\endgroup$
    – xzczd
    Dec 22, 2017 at 13:35
  • $\begingroup$ Thx for your suggestions. Puting random part inside of for helps, but resulting function is random, but not what I expect - just random walks around 0 value and not relaxing into some value $\endgroup$
    – denkorw
    Dec 22, 2017 at 13:51
  • 3
    $\begingroup$ Have a look at ItoProcess in order to get a solution which might be a useful guide to your approach. $\endgroup$ Dec 22, 2017 at 14:22
  • 1
    $\begingroup$ Someone can correct me if I'm wrong, but isn't this an Ornstein-Uhlenbeck Process? If so, there's a builtin OrnsteinUhlenbeckProcess function. Otherwise, as @b.gatessucks says, ItoProcess can handle more general stochastic differential equations. $\endgroup$
    – Chris K
    Dec 22, 2017 at 14:32
  • 3
    $\begingroup$ If I understand the question right, you want to solve the ode c'[t]+3 c[t]==2 A DiracDelta[t-t1]? To use Euler etc. is your solution idea and not a requirement? If so, Mathematica is able to solve the problem c'[t]+3 c[t]==DiracDelta[t-t1] analytically. The solution is greensfunction which can be used to describe the general solution of your ode... $\endgroup$ Dec 22, 2017 at 19:54

1 Answer 1


Taking into account the comments made xzczd and rewriting your code to both simplify it and make more robust, I come up with this.

myEuler[func_, start_, end_, initialvalue_, nrOfsteps_] :=
  Module[{h, T, rnd, Y, j, m = nrOfsteps + 1},
    rnd := RandomVariate[NormalDistribution[0, 1]];
    h = (end - start)/nrOfsteps;
    T = start + h (Range[m] - 1);
    Y = ConstantArray[initialvalue, m];
    For[j = 1, j < m, j++,
      Y[[j + 1]] = Y[[j]] + h (func[T[[j]], Y[[j]]] + (2 rnd)/Sqrt[h])];
    Transpose[{T, Y}]]

f[t_, x_] := -3 x + t

pts = myEuler[f, 0.0, 3.0, 10.0, 50];
  PlotRange -> All,
  Mesh -> All,
  MeshStyle -> Red,
  Frame -> True]



Here is a functional programming solution that greatly simplifies the function myEuler. It will produce exactly same result as the implementation using For. It may be a little harder to understand, but it is worth studying because it is very concise and much more efficient than any code building arrays and processing them with For.

myEuler[func_, start_, end_, initialvalue_, nrOfsteps_] :=
  Module[{rnd, h},
    rnd := RandomVariate[NormalDistribution[0, 1]];
    h = (end - start)/nrOfsteps;
      {{#2, #1}, #1 + h (func[#2, #1] + (2 rnd)/Sqrt[h])} &, 
      start + h Range[0, nrOfsteps]]]
  • $\begingroup$ this class of problem needs specialized methods - not my area of expertise for sure, but I do know enough that comments in the question itself are on the mark (Ito process, Ornstein-Uhlenbeck process). $\endgroup$
    – Paul_A
    Dec 24, 2017 at 2:14
  • $\begingroup$ wolfram.com/broadcast/… for example (I am not affiliated with Wolfram or Pavlyk) $\endgroup$
    – Paul_A
    Dec 24, 2017 at 2:26
  • $\begingroup$ @Paul_A. I am not addressing the underlying issue of whether the OP can hope to get reasonable results by employing Euler's method. I am only trying to address the programming issues his question raises. $\endgroup$
    – m_goldberg
    Dec 24, 2017 at 3:33
  • $\begingroup$ I missed that, my apologies. I read many of your polished and detailed MMA programming solutions - am an admirer of yours in that regard. $\endgroup$
    – Paul_A
    Dec 24, 2017 at 4:54
  • $\begingroup$ Thx for your answer, but why f[t_, x_] := -3 x + t and not just f[x]=-3x? Also from previous comments I'm still not sure how noise should be included: as h (2 rnd)/Sqrt[h] or h (2 rnd)/Sqrt[h]) $\endgroup$
    – denkorw
    Dec 24, 2017 at 12:55

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