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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.





\*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]\), \(x\)]\((
\*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],
share|improve this question
You've already seen this? – J. M. Jun 21 '12 at 0:54
@J.M. Yes I have. What If I don't want to use stiffness switching? Plus I already use the BDF method which is useful for stiff equations.. but only up to a point. – drN Jun 21 '12 at 1:52
Just making sure. I'm not sure if it's straightforward to hijack the process to quit if stiffness is seen, but I'll look into it. – J. M. Jun 21 '12 at 1:56
@J.M. I was reading about stiffness and obv. I came across the Jacobian and such. I was wondering if there is anyway of checking the jacobian periodically to detect stiffness.... – drN Jun 21 '12 at 1:59
Have you looked into tutorial/NDSolveStiffnessTest? It looks like there are some options for the stiffness test and I could imagine that you can set them so that it won't switch to the stiff method. There is also an example which uses an option "MethodMonitor" which you might be able to use to stop the integration (there are examples for the event locator method which show how to stop). I haven't tried any of the mentioned options to achieve what you want, though. If you provide a self contained example probably someone might start experimenting... – Albert Retey Jun 21 '12 at 7:52
up vote 6 down vote accepted

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]\), \(x\)]\((
\*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;
 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])}}

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;
  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]
share|improve this answer
Thats very interesting! I didn't think about using MaxSteps. I'll give that a shot as you have demonstrated for a bunch of other problems. Thanks! – drN Jun 23 '12 at 14:07

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