Skip to main content

performance tuning - How to tell mathematica not to resolve stiffness issues


Very often I solve partial differential equations that are nonlinear and could be up to 4th order. In these cases, it is usual for the solution determined by NDSolve to be stiff during a later stage. What I suspect NDSolve does in this case is to resolve the stiffness until the error/local accuracy is very poor. That is when it quits the problem and gives you an Interpolating function polynomial.


Whilst using the BDF method to MaxOrder of 1 for instance, is there someway to tell Mathematica to quit as soon as stiffness is encountered in the solution so that I save time? I don't want to resolve the stiff portion and just stop my solution just as it gets stiff.


The below example looks like a mess in plain text but it copies fine. It gets stiff at t=4806. However, is lots of problems, NDSolve lingers at the time at which stiffness is achieved to try and resolve the features that I would like to circumvent completely.


I will obv. look into the stiffness switching stuff again.


Example


{xMin,xMax}={-4\[Pi]/0.0677,4\[Pi]/0.0677};

k=0.0677/4;


TMax=5000;

uSolpbc[t_,x_]=u[t,x]/.NDSolve[{\!\(
\*SubscriptBox[\(\[PartialD]\), \(t\)]\(u[t, x]\)\)==-100\!\(
\*SubscriptBox[\(\[PartialD]\), \(x\)]\((
\*SuperscriptBox[\(u[t, x]\), \(3\)]\ 
\*SubscriptBox[\(\[PartialD]\), \(x, x, x\)]u[t, x])\)\)+1/3 \!\(
\*SubscriptBox[\(\[PartialD]\), \(x\)]\((
\*SuperscriptBox[\(u[t, x]\), \(3\)]\ 

\*SubscriptBox[\(\[PartialD]\), \(x\)]u[t, x])\)\)-5 \!\(
\*SubscriptBox[\(\[PartialD]\), \(x\)]\((
\*SuperscriptBox[\((
\*FractionBox[\(u[t, x]\), \(1 + u[t, x]\)])\), \(2\)]\ 
\*SubscriptBox[\(\[PartialD]\), \(x\)]u[t, x])\)\),u[0,x]==1-0.1 Cos[k*x],
u[t,xMin]== u[t,xMax],
Derivative[0,1]u[t,xMin]==Derivative[0,1]u[t,xMax],
Derivative[0,2]u[t,xMin]==Derivative[0,2]u[t,xMax],
Derivative[0,3]u[t,xMin]==Derivative[0,3]u[t,xMax]},
u,

{t,0,TMax},
{x,xMin,xMax},
MaxStepFraction->1/150][[1]]

Answer



Having played around with your example I don't think that there is any method switching involved at all, as that only seems to be the case when Method is explicitly set to "StiffnessSwitching", which you didn't do (you also haven't specified "BDF" and I'm not sure what NDSolve actually chose...). What you see is that NDSolve just makes the step size smaller and smaller because the errors get worse and worse. As you vary MaxStepFraction you will find that the point where it complains about an effectively zero step size will change. This I think you already have found yourself and thus I agree now that my comment about stiffness switching wasn't very useful.


You could stop the integration when the step size gets smaller than a reasonable amount, e.g. with something like this (there might be better ways to achieve the same thing):


pde = {\!\(
\*SubscriptBox[\(\[PartialD]\), \(t\)]\(u[t, x]\)\) == -100 \!\(
\*SubscriptBox[\(\[PartialD]\), \(x\)]\((
\*SuperscriptBox[\(u[t, x]\), \(3\)] 

\*SubscriptBox[\(\[PartialD]\), \(x, x, x\)]u[t, x])\)\) + 1/3 \!\(
\*SubscriptBox[\(\[PartialD]\), \(x\)]\((
\*SuperscriptBox[\(u[t, x]\), \(3\)] 
\*SubscriptBox[\(\[PartialD]\), \(x\)]u[t, x])\)\) - 5 \!\(
\*SubscriptBox[\(\[PartialD]\), \(x\)]\((
\*SuperscriptBox[\((
\*FractionBox[\(u[t, x]\), \(1 + u[t, x]\)])\), \(2\)] 
\*SubscriptBox[\(\[PartialD]\), \(x\)]u[t, x])\)\), 
   u[0, x] == 1 - 0.1 Cos[k*x], u[t, xMin] == u[t, xMax], 
   Derivative[0, 1][u][t, xMin] == Derivative[0, 1][u][t, xMax], 

   Derivative[0, 2][u][t, xMin] == Derivative[0, 2][u][t, xMax], 
   Derivative[0, 3][u][t, xMin] == Derivative[0, 3][u][t, xMax]};

{xMin, xMax} = {-4 \[Pi]/0.0677, 4 \[Pi]/0.0677};
k = 0.0677/4;
TMax = 5000;
thisstep = 0;
laststep = 0;
Timing[
 uSolpbc = u /. NDSolve[pde, u, {t, 0, TMax}, {x, xMin, xMax},

     MaxStepFraction -> 1/150,
     StepMonitor :> (
       laststep = thisstep; thisstep = t; 
       stepsize = thisstep - laststep;
       ),
     Method -> {"MethodOfLines", 
       Method -> {"EventLocator", 
         "Event" :> (If[stepsize < 10^-4, 0, 1])}}
     ][[1]]
 ]


unfortunately for the given problem this will be even slower than letting NDSolve run into "effectively zero stepsize". It might be different for other problems, but using event locators is in general rather slow, so I wouldn't have much hope to achieve any speedup this way except when the time for a single step is much larger than in this example. You could use something like If[t > 4500, Print[t -> stepsize]] within the StepMonitor to see what happens and check that it does what it is supposed to do.


A much easier approach is to just limit the maximal numbers of steps to be used which will also limit the maximal runtime. It is of course depending on the system you solve how far that number of steps will take you. So it probably needs some estimation about what a good value would be, for your example a value of 200 already seems to show the characteristics of the solution and is about twice as fast:


{xMin, xMax} = {-4 \[Pi]/0.0677, 4 \[Pi]/0.0677};
k = 0.0677/4;
TMax = 5000;
thisstep = 0;
laststep = 0;
Timing[Quiet[
  uSolpbc = u /. NDSolve[pde, u, {t, 0, TMax}, {x, xMin, xMax},

      MaxStepFraction -> 1/150,
      MaxSteps -> 200
      ][[1]], NDSolve::mxst]]

Plot3D[uSolpbc[t, x], {t, 
  0, (uSolpbc@"Domain")[[1, 2]]}, {x, -185.618, 185.618}, 
 PlotRange -> All]

Comments

Post a Comment

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.
Post a Comment
Read more