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When using NDSolve to solve 2 pdes with different version of Mathematica, I obtained totally different results. The code is as follows.

L = 2*π; c = 1/20; ϵ = 35/100; α = 54/10; β = 35/100; (*parameters*)
(*define functions*)
cF0[z_, t_] := (8 - q[z, t]*S[z, t]*Log[S[z, t]/β])/(15*Log[S[z, t]/β] - 8*Log[S[z, t]/α]);
cA0[z_, t_] := (15 - q[z, t]*S[z, t]*Log[S[z, t]/α])/(15*Log[S[z, t]/β] - 8*Log[S[z, t]/α]);
F[z_, t_] := (cA0[z, t]*D[S[z, t], z])/S[z, t] + D[cA0[z, t], z]*Log[S[z, t]/β];
p0[z_, t_] := -(1/S[z, t]) + ϵ^2*D[S[z, t], {z, 2}] + 30/(2*S[z, t]^2)*(15*cF0[z, t]^2 - 8*cA0[z, t]^2);

(*pdes*)
eqS = 8*D[S[z, t]^2, t] == D[(1 - ϵ*D[p0[z, t], z])*(α^2 - S[z, t]^2)^2 - 
 2*S[z, t]*((1 - ϵ*D[p0[z, t], z])*S[z, t] + 60*ϵ*q[z, t]*F[z, t])*(α^2 - S[z, t]^2 + 
    2*S[z, t]^2*Log[S[z, t]/α]), z];

eqQ = D[q[z, t], t] + (1/4*(1 - ϵ*D[p0[z, t], z])*(α^2 - S[z, t]^2 + 
      2*S[z, t]^2*Log[S[z, t]/α]) + ϵ*30*q[z, t]*S[z, t]*F[z, t]*Log[S[z, t]/α])*D[q[z, t], z] + 
q[z, t]*(1/8*(1 - α^2/S[z, t]^2 + 2*Log[S[z, t]/α])*D[(1 - ϵ*D[p0[z, t], z])*S[z, t]^2 + 60*ϵ*q[z, t]*S[z, t]*F[z, t], z] - 
   1/16*ϵ*D[p0[z, t], {z, 2}]*(2*α^2 - 3*S[z, t]^2 + α^4/S[z, t]^2) + ϵ*30*q[z, t]*F[z, t]*D[S[z, t], z]) == 1/S[z, t]*(16*cF0[z, t] - 7*cA0[z, t]);

(*initial conditions*)
SeedRandom[1]
iniS[z_] = 22/5 + c*BSplineFunction[RandomReal[{-1, 1}, 50], SplineClosed -> True][z/L];
SeedRandom[2]
iniq[z_] = 1/8 + c*BSplineFunction[RandomReal[{-1, 1}, 50], SplineClosed -> True][z/L];

endtime = 400 (*600*);  nZ = 1001; xdifforder = 6;
eqns = {eqS, eqQ, S[z, 0] == iniS[z], q[z, 0] == iniq[z],
S[0, t] == S[L, t],
Derivative[1, 0][S][0, t] == Derivative[1, 0][S][L, t],
Derivative[2, 0][S][0, t] == Derivative[2, 0][S][L, t],
Derivative[3, 0][S][0, t] == Derivative[3, 0][S][L, t],
q[0, t] == q[L, t], Derivative[1, 0][q][0, t] == Derivative[1, 0][q][L, t]};

{solnS, solnQ} = NDSolveValue[eqns, {S, q}, {z, 0, L}, {t, 0, endtime}, MaxSteps -> ∞, Method -> {"MethodOfLines",
"Method" -> "LSODA", "TemporalVariable" -> t, "SpatialDiscretization" -> {"TensorProductGrid", 
  "MinPoints" -> nZ, "MaxPoints" -> nZ, "DifferenceOrder" -> xdifforder}}]

When I run above code in v9.0 and v11.3, it gives different result when plotting with

Splot = Plot[solnS[z, 400], {z, 0, L}, PlotRange -> {{0, L},All}, Axes -> False, 
Frame -> True, AspectRatio -> 0.3, ImageSize -> 600, PlotStyle -> Blue]

enter image description here

My findings:

  1. v9 gives a plausible result and runs faster than v11.3 in general;

  2. v9 yields a larger data than v11.3 with the same code;

  3. Since I am suspicious of the results, I found the fix function by @xzczd here, which allows us to have more control into the black box. So, I tried it as well (note fix function can only be used in v9), but the result is also very different from that of v9 with non-fixed difference order (6th-order) when running up to a longer time, for example, endtime=600.

     {solnS, solnQ} = fix[endtime, xdifforder]@NDSolveValue[eqns, {S, q}, {z, 0, L}, {t, 0, endtime},
     MaxSteps -> ∞, Method -> {"MethodOfLines", "Method" -> "LSODA", 
     "TemporalVariable" -> t, "SpatialDiscretization" -> {"TensorProductGrid", "MinPoints" -> nZ, "MaxPoints" -> nZ, "DifferenceOrder" -> xdifforder}}]
    
  4. As required in this question, I check the conservation of volume with

     volume = ListPlot[Table[Quiet@NIntegrate[π*solnS[z, t]^2, {z, 0, L}, Method -> {Automatic, "SymbolicProcessing" -> False}], {t, 0, tmax, 1}], PlotRange -> All]
    

With the fix function, I get more than 5% error between the initial volume iniVol = N[π*(α - 1)^2*L] and the finial one. This may be an example to explain why NDSolve chooses different difference orders for spatial derivatives.

In principle, I could reduce the error by in increasing mesh. However, even increasing to nZ = 2001 (note: do not run with nZ = 2001 because it ran > 20hrs on my desktop and the data become too larger to save), both v9 and v11 give a warning : >> estimated initial error on the specified spatial grid in the direction of independent variable z exceeds prescribed error tolerance. Btw, with nZ = 1801 the code will run for 12+ hrs and the data have a reasonable size to be saved.

  1. I also tried "DifferenceOrder"->"Pseudospectral", which gave an obviously wrong result (noise and zig-zag)

Dose anyone have any suggestions? My objective is to have the volume conserved to within about 0.01%. Thank you very much in advance.

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  • 1
    $\begingroup$ Versions you testing are too old. We now testing v.12.3, and they promise us v.13 soon. $\endgroup$ Dec 12, 2021 at 6:00
  • $\begingroup$ @AlexTrounev if my problem can be solved in these new versions, I am happy with it. $\endgroup$
    – Nobody
    Dec 12, 2021 at 6:51
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    $\begingroup$ @AlexTrounev Version 9 isn't bad for the problem. See my answer below. $\endgroup$
    – xzczd
    Dec 12, 2021 at 8:02
  • $\begingroup$ This may be a complete shot in the dark, but have you tested whether SeedRandom[1]; RandomReal[{-1, 1}] gives the same result in both versions? $\endgroup$
    – Domen
    Dec 12, 2021 at 8:05
  • $\begingroup$ @Domen Thanks! I just tested, SeedRandom[1]; RandomReal[{-1, 1}] gives completely the same result in v9.0 and v11.3 at least. $\endgroup$
    – Nobody
    Dec 12, 2021 at 8:10

1 Answer 1

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Based on the experience obtained in these posts, we know if conservation law involves in a PDE/PDE system, it's better not to expand the equation, and this is true not only for the spatial derivative, but also for the temporal derivative.

If I read it right, S[z, t] is always larger than 0, so let's introduce a new variable S2[z, t] that satisfies S2[z, t] == S[z, t]^2 to avoid the expansion of D[S[z, t]^2, t]:

$Version
(* "9.0 for Microsoft Windows (64-bit) (February 13, 2013)" *)

(* Code that's not modified is omitted here *)
srule = S -> (Sqrt[S2[#, #2]] &);

eqS = 8*D[S2[z, t], t] == 
   D[(1 - ϵ*D[p0[z, t], z])*(α^2 - S[z, t]^2)^2 - 
     2*S[z, t]*((1 - ϵ*D[p0[z, t], z])*S[z, t] + 
     60*ϵ*q[z, t]*F[z, t])*(α^2 - S[z, t]^2 + 2*S[z, t]^2*Log[S[z, t]/α]), z];

eqns = {eqS, eqQ,
        S2[z, 0] == iniS[z]^2, q[z, 0] == iniq[z],
        S2[0, t] == S2[L, t], q[0, t] == q[L, t]} /. srule;

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

{solnS2, solnQ} = 
   NDSolveValue[eqns, {S2, q}, {z, 0, L}, {t, 0, endtime}, MaxSteps -> ∞, 
    Method -> mol[100, 4]]; // AbsoluteTiming
(* {14.470040, Null} *)

S2plot = Plot[solnS2[z, 400], {z, 0, L}, PlotRange -> {{0, L}, All}, Axes -> False, 
  Frame -> True, AspectRatio -> 0.3, ImageSize -> 600]

enter image description here

Remark

I've simplified the code for the periodic b.c.s a bit, but it's equivalent to the original. Actually according to the observation in this post, even a single q[0, t] == q[L, t] is enough to specify periodic b.c.s for all the variables.

Increasing spatial grid points to 200 visually doesn't change the solution. I've also tested in v12.3, the solution looks the same. Finally, the solution obeys the conservation law well:

ListPlot[Table[
  Quiet@NIntegrate[solnS2[z, t], {z, 0, L}, 
    Method -> {Automatic, "SymbolicProcessing" -> 0}], {t, 0, endtime, 10}], 
 PlotRange -> All]

enter image description here

So I think it's safe to say the solution is reliable.

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  • $\begingroup$ thank you, but your bcs appear to be different from mine. $\endgroup$
    – Nobody
    Dec 12, 2021 at 8:14
  • $\begingroup$ @Nobody It's the same, actually even a single q[0, t] == q[L, t] is enough to specify periodic b.c. for all the variables. This is observed in the following post: mathematica.stackexchange.com/a/182050/1871 $\endgroup$
    – xzczd
    Dec 12, 2021 at 8:16
  • $\begingroup$ It looks like in v.12.3 we have similar results with your code, nice approach (+1). $\endgroup$ Dec 12, 2021 at 8:36
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    $\begingroup$ eerri is merely a warning. (For example, it never disappears for a unsmooth initial condition. ) As mentioned in the answer, with different grid size the solution is virtually the same, so the solution is probably reliable. @nobody $\endgroup$
    – xzczd
    Dec 12, 2021 at 9:13
  • 1
    $\begingroup$ @Nobody 1. I'm sorry, but $S2[z,t]$ is just terrible in my eyes. I've clarified my answer a bit, hopefully it's better now. 2. Yes, this example shows the design of "DifferenceOrder" may lead to better result compared to a fixed order difference discretization in certain cases. $\endgroup$
    – xzczd
    Dec 12, 2021 at 11:16

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