It is already formulated in the title. NDSolve takes sometimes a considerable piece of time. It would be very practical to have some information on how long it is still to wait. So, any ideas?

To be specific, here is the equation along with the boundary and initial conditions:

     Clear[s, bc1, bc2, ic, Lx, Ly, eq];
Lx = 90;
Ly = 90;
bc1 = z[t, x, -Ly] == z[t, x, Ly] == 0;
bc2 = z[t, -Lx, y] == z[t, Lx, y] == 0;
ic = z[0, x, y] == 0.5*Exp[-(x^2 + y^2)/10];
U[x_, y_] := -((
   y (-2 x + Sqrt[x^2 + y^2]) Sqrt[x + Sqrt[x^2 + y^2]])/(
   2 Sqrt[2] (x^2 + y^2 + 0.00001)^(3/2)));

mol = "MethodOfLines";

eq = D[z[t, x, y], t] == 
   D[z[t, x, y], {x, 2}] + 
    D[z[t, x, y], {y, 2}] - (0.03 - U[x, y])*z[t, x, y] - z[t, x, y]^3;

s = NDSolve[{eq, bc1, bc2, ic}, 
   z, {t, 0, 100}, {x, -Lx, Lx}, {y, -Ly, Ly}, Method -> mol][[1, 1]]

This equation solves nicely and rather fast. However, the maximum time is here chosen to be 100. It may, however, be necessary to solve it with the max time 10 to 30 fold greater. In this case I would like to have a visual indication of how far is the progress.

  • 1
    $\begingroup$ I presume you have already looked at StepMonitor? Are you prepared to explicitly provide Method options? $\endgroup$
    – Mr.Wizard
    Commented Jan 5, 2017 at 10:14
  • $\begingroup$ No, I missed this. I am just looking at it, but so far do not see a reasonable visual implementation like the ProgressIndicator. By Methosdo you mean the method used in NDSolve? If yes, it is the MethodOfLines. $\endgroup$ Commented Jan 5, 2017 at 10:36
  • $\begingroup$ I don't see a simple way to use StepMonitor to find out how much of the evaluation is left to do. I suspect that to use it one will need to know what is being done internally and for that that one will need to specify a Method for NDSolve. $\endgroup$
    – Mr.Wizard
    Commented Jan 5, 2017 at 10:45

3 Answers 3


Here are a few versions I use:


showStatus[status_] := 
clearStatus[] := showStatus[""];

NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0, 
  u[t, 0] == Sin[t], u[t, 5] == 0}, u, {t, 0, 10}, {x, 0, 5}, 
 EvaluationMonitor :> showStatus["t = " <> ToString[CForm[t]]]];


The status is reported in the lower left corner of the notebook.


tEnd = 100;
ProgressIndicator[Dynamic[currentTime], {0, tEnd}]

currentTime = 0;
NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0, 
  u[t, 0] == Sin[t], u[t, 5] == 0}, u, {t, 0, tEnd}, {x, 0, 5}, 
 EvaluationMonitor :> (currentTime = t;)]


Monitor[NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0, 
   u[t, 0] == Sin[t], u[t, 5] == 0}, u, {t, 0, 100}, {x, 0, 5}, 
  MaxStepSize -> 0.01, 
  EvaluationMonitor :> (monitor = Row[{"t = ", CForm[t]}])], monitor]

I hope you find something to your liking.

Monitor[NDSolve[{eq, bc1, bc2, ic}, 
  z, {t, 0, 100}, {x, -Lx, Lx}, {y, -Ly, Ly}, Method -> mol, 
  StepMonitor :> (time = t)], 

enter image description here

This doesn't appear to slow down the evaluation in any significant way:

  NDSolve[{eq, bc1, bc2, ic}, 
   z, {t, 0, 100}, {x, -Lx, Lx}, {y, -Ly, Ly}, Method -> mol], 60]

 Monitor[NDSolve[{eq, bc1, bc2, ic}, 
   z, {t, 0, 100}, {x, -Lx, Lx}, {y, -Ly, Ly}, Method -> mol, 
   StepMonitor :> (time = t)], ProgressIndicator[time/100]], 60]

 Monitor[NDSolve[{eq, bc1, bc2, ic}, 
   z, {t, 0, 100}, {x, -Lx, Lx}, {y, -Ly, Ly}, Method -> mol, 
   EvaluationMonitor :> (time = t)], ProgressIndicator[time/100]], 60]




  • 2
    $\begingroup$ Simpler than I expected. +1 :-) $\endgroup$
    – Mr.Wizard
    Commented Jan 5, 2017 at 11:03

You should be aware of the fact that using either StepMonitor or EvaluationMonitor could considerably slow down the execution of NDSolve, so the progress indication might come at a price. As there might be several evaluations per step, StepMonitor will be excecuted less often than EvaluationMonitor so might be a better choice in most cases. It also is one of those cases where restricting the update of the progress indiciation might make sense. Of course, the more expensive a single step/evaluation within NDSolve is, the less will the percentual effort for the updates. Here are some example with which you can experiment, of course timings might be very different for different problems...

A simple example from the documentation:

 p = 20; tmax = 100;
 runNDSolve[opts___] := NDSolve[
   {x''[t] + x'[t]/10 + x[t]/(1 + x[t]^2) == Sin[t], x[0] == 1, 
   x'[0] == 0}, x, {t, 0, tmax},
   PrecisionGoal -> p, AccuracyGoal -> p, WorkingPrecision -> 2*p,
   MaxSteps -> Infinity, 
   Method -> {"ExplicitRungeKutta", "DifferenceOrder" -> 8},

you might adjust the precision to adjust runtimes to a few seconds so you can see what actually happens. Here is a run with no monitoring at all as a reference:


Compare EvaluationMonitor to StepMonitor:

 AbsoluteTiming[runNDSolve[EvaluationMonitor :> (tt = t)];]
 AbsoluteTiming[runNDSolve[StepMonitor :> (tt = t)];]

For this example on my computer the restriction of dynamic updating doesn't make much difference, but I have seen examples where doing this was crucial to not slow down the evaluation unacceptably, which can easily happen if you do something more expensive than update a progressbar, like e.g. show a plot.

 tt = 0;
 PrintTemporary@ProgressIndicator[Dynamic[tt], {0, tmax}];
 AbsoluteTiming[runNDSolve[StepMonitor :> (tt = t)];]

 tt = 0;
    Refresh[tt, UpdateInterval -> 1, TrackedSymbols :> {}]
 ], {0, tmax}];
 AbsoluteTiming[runNDSolve[StepMonitor :> (tt = t)];]

Here is a version of Feyre's elegant solution using Monitor which restricts the update to a given interval. AFAIK this is not really documented but doesn't come as a surprise when assuming that the second argument will be wrapped in a Dynamic at some point:

   Monitor[runNDSolve[StepMonitor :> (tt = t)], 
     Refresh[ProgressIndicator[tt/tmax], UpdateInterval -> 1, 
       TrackedSymbols :> {}]];]
  • $\begingroup$ Happy New Year! Hope to see you at some conference ;-) $\endgroup$
    – user21
    Commented Jan 5, 2017 at 11:49
  • $\begingroup$ Following your answer I decided to run a benchmark. Granted I have a superior set up to most people, but I didn't run into any serious slow down, and had no difference between StepMonitor and EvaluationMonitor $\endgroup$
    – Feyre
    Commented Jan 5, 2017 at 19:50
  • 1
    $\begingroup$ @Feyre: I hope my answer made it clear that it very much depends on the problem you are solving. For many problems including the OPs example the differences might not be relevant. If you try my example you should see a larger difference, even with a better setup than mine. I just wanted to direct the attention of readers to the potential problem. $\endgroup$ Commented Jan 6, 2017 at 1:30
  • $\begingroup$ You might want to change "will" in the first sentence to "could". :) Your answer adds a lot of useful information, but I thought it relevant to differentiate OP's situation here. $\endgroup$
    – Feyre
    Commented Jan 6, 2017 at 11:47
  • $\begingroup$ @Feyre: good point, just did that. The answer does now much more reflect my initial intend... $\endgroup$ Commented Jan 6, 2017 at 17:04

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