Now I am trying to solve the following 2D+1 type of PDE:

$\partial_t u(t,x,y)=-y\partial_{x}u+\partial_{y}\left[a y+b sin(x)u+c\partial_{y}u\right]$

with $u(0,x,y)=\frac{1}{2\pi}e^{-((x-\pi/4)^2+y^2)/2}$

and periodic boundary condition:


In $y$-direction, it is unbounded.

Here is the code:

a = 1;
T = 1;
ωcb = -50;
ωct = 50;
ωb = -5;
ωt = 5;
A = 10;
γ = 0.1;
kT = 0.1;
ufun = NDSolveValue[{D[u[t, θ, ω], t] == -D[ω u[t, θ, ω], θ] - 
       D[-A Sin[θ] u[t, θ, ω] - γ kT D[u[t, θ, ω], ω], ω] + 0.1 D[ω u[t, θ, ω], ω], 
    u[0, θ, ω] == 
     1/(2 a π)
       E^(-((θ - π/4)^2/(2 a^2)) - (ω)^2/(2 a^2)),
    u[t, -π, ω] == u[t, π, ω], 
    u[t, θ, ωcb] == u[t, θ, ωct]}, 
   u, {t, 0, 
    T}, {θ, -π, π}, {ω, ωcb, ωct},
    Method -> {"MethodOfLines", 
     "SpatialDiscretization" -> {"TensorProductGrid", 
       "MinPoints" -> 200, "MaxPoints" -> 1000}}];
plots = Table[
     ufun[t, θ, ω]], {θ, -π, π}, {ω, ωb, ωt}, PlotRange -> All, 
    ColorFunction -> "LakeColors"], {t, 0, T, .1}];

The solution is a mess. I believe there are some problem on mesh?

enter image description here

Also there are some message

NDSolveValue::mxsst: Using maximum number of grid points 1000 allowed by the MaxPoints or MinStepSize options for independent variable $\theta$.

NDSolveValue::ibcinc: Warning: boundary and initial conditions are inconsistent.

NDSolveValue::eerr: Warning: scaled local spatial error estimate of 62.663959713694915at t = 1. in the direction of independent variable θ is much greater than the prescribed error tolerance. Grid spacing with 1001 points may be too large to achieve the desired accuracy or precision. A singularity may have formed or a smaller grid spacing can be specified using the MaxStepSize or MinPoints method options.

But when I try

Method -> {"MethodOfLines", 
         "SpatialDiscretization" -> {"TensorProductGrid", 
           "MinPoints" -> 200, "MaxPoints" -> 10000}}

It cannot manage to finish the program. What's going on ? Thanks for any suggestion!

Reference: enter image description here

  • $\begingroup$ Is this equation taken from literature or derived by yourself? Can you add some background information? $\endgroup$
    – xzczd
    Aug 16 '18 at 9:06
  • $\begingroup$ Yes, I derive it by myself. It's the Klein-Kramers equation or 2D Fokker-Planck equation in rotational coordinate with sinusoidal potential. $\endgroup$
    – Bob Lin
    Aug 16 '18 at 10:09
  • $\begingroup$ The initial data is not symmetric, so the periodic boundary conditions for the variable θ are inconsistent with the initial data. $\endgroup$ Aug 16 '18 at 11:25
  • 1
    $\begingroup$ We must pay attention to the code or the equations that are declared by the author? The equations in the code and in the beginning of the topic do not coincide. $\endgroup$ Aug 16 '18 at 12:00
  • $\begingroup$ @AlexTrounev I guess OP's equation in the code is a variant of the one in the reference. (He mentioned that "I derive it by myself" in his first comment. ) Of course this is something that OP should clarify. $\endgroup$
    – xzczd
    Aug 16 '18 at 12:12

You need the magic of "Pseudospectral":

mol[n_Integer, o_:"Pseudospectral"] := {"MethodOfLines", 
  "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, 
    "MinPoints" -> n, "DifferenceOrder" -> o}}

With[{u = u[t, θ, ω]}, 
  eq = D[u, t] == -D[ω u, θ] - D[-A Sin[θ] u, ω] - γ kT D[u, ω] + 1/10 D[ω u, ω];

  ic = u == E^(-((θ - π/4)^2/(2 a^2)) - ω^2/(2 a^2))/(2 a π) /. t -> 0];

ufun = NDSolveValue[{eq, ic, u[t, -π, ω] == u[t, π, ω], u[t, θ, ωcb] == u[t, θ, ωct]}, 
    u, {t, 0, T}, {θ, -π, π}, {ω, ωcb, ωct}, 
    Method -> mol[81]]; // AbsoluteTiming
(* {24.137131, Null} *)

plots = Table[
   Plot3D[Abs[ufun[t, θ, ω]], {θ, -π, π}, {ω, ωb, ωt}, 
     PlotRange -> All, AxesLabel -> Automatic, PlotPoints -> 50, 
    BoxRatios -> {Pi, ωb, 1}], {t, 0, T, .04}];

enter image description here

Update: A faster approach

Your equation turns out to be another example where the somewhat strange difference strategy discussed in this post causes trouble, so the problem can be solved With the fix function in that post:

(* The fix function isn't included in this post, 
   please find it in the link above. *)
ufun = fix[T, 4]@
    NDSolveValue[{eq, ic, u[t, -π, ω] == u[t, π, ω], 
      u[t, θ, ωcb] == u[t, θ, ωct]}, 
     u, {t, 0, T}, {θ, -π, π}, {ω, ωcb, ωct}, 
     Method -> mol[81]]; // AbsoluteTiming
(* {13.415239, Null} *)

The resulting animation is similar so I'd like to omit it here.

  • $\begingroup$ The method is good, but the problem is not correct, so we have a message on the output NDSolveValue::ibcinc: Warning: boundary and initial conditions are inconsistent. $\endgroup$ Aug 16 '18 at 11:34
  • $\begingroup$ @Alex ibcinc warning pops up because the i.c. isn't periodic technically, but it's close to constant near $\theta=\pm \pi$ i.e. approximately periodic, so the warning isn't a big deal, I think. One can eliminate the warning by e.g. modifying i.c. to With[{u = u[t, θ, ω]}, ic = u == Piecewise[{{E^(-((θ - π/4)^2/(2 a^2)) - ω^2/(2 a^2))/( 2 a π), Abs@θ < 0.9 Pi}}] /. t -> 0], and the solution won't change much. Of course when the b.c. involves derivative, ibcinc often means fatal error, see this post for more information. $\endgroup$
    – xzczd
    Aug 16 '18 at 11:57
  • $\begingroup$ Note that the equations in the code and in the beginning of the topic do not coincide. $\endgroup$ Aug 16 '18 at 12:08
  • $\begingroup$ Thanks for pointing out the problem. Indeed, they looks different but essentially the same. $\endgroup$
    – Bob Lin
    Aug 16 '18 at 13:41
  • $\begingroup$ Could someone explain what's going on? I am confused with the method. What's the problem with my original code? Anyway, thanks! $\endgroup$
    – Bob Lin
    Aug 16 '18 at 13:43

I took the equations and the statement of the problem at the beginning of the topic and wrote my code. I got the expected rotation of the wave as a result.

{a, b, c, k, L, T} = {1, 10, 1, 1, 5, 2};
    u0[x_, y_] := Exp[-(x^2 + y^2)/2]/(2*Pi)
eq = D[u[t, x, y], t] == -k*y*D[u[t, x, y], x] + 
    D[a*y + b*Sin[x]*u[t, x, y] + c*D[u[t, x, y], y], y];
ic = u[0, x, y] == u0[x, y];
bc = {u[t, x, -L] == u[t, x, L], u[t, -Pi, y] == u[t, Pi, y]};
   sol = NDSolveValue[{eq, ic, bc}, 
  u, {t, 0, T}, {x, -Pi, Pi}, {y, -L, L}, 
  Method -> {"MethodOfLines", 
    "SpatialDiscretization" -> {"TensorProductGrid", 
      "MinPoints" -> 40, "MaxPoints" -> 100, 
      "DifferenceOrder" -> "Pseudospectral"}}, MaxSteps -> 10^6]
plots = Table[
   Plot3D[Abs[sol[t, x, y]], {x, -Pi, Pi}, {y, -L, L}, 
    PlotRange -> All, ColorFunction -> "LakeColors", 
    Mesh -> None], {t, 0, T, .05*T}];


  • $\begingroup$ The solution is perfect! Many thanks. Do you have any opinion on the mesh you use? I am not familiar with specifying these things. $\endgroup$
    – Bob Lin
    Aug 16 '18 at 14:38
  • $\begingroup$ I used a very simple option Mesh -> None $\endgroup$ Aug 16 '18 at 14:46
  • $\begingroup$ Sorry I must misunderstand. What I want to ask is Method -> {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MinPoints" -> 40, "MaxPoints" -> 100, "DifferenceOrder" -> "Pseudospectral"}. How did you find those setup? $\endgroup$
    – Bob Lin
    Aug 16 '18 at 14:50
  • $\begingroup$ I solved a similar equation, only non-linear in the complex plane. I picked up the parameters experimentally. $\endgroup$ Aug 16 '18 at 15:19
  • $\begingroup$ Sorry.. I copy your code but I cannot reproduce your solution. If your ic is centered at origin, I guess it won't rotate, right? $\endgroup$
    – Bob Lin
    Aug 16 '18 at 17:33

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