Skip to main content

differential equations - Return partial result when MemoryConstrained aborts NDSolve


I use NDSolve to solve a large set (~400) of coupled ODEs. Sometimes, the memory (~4GB) gets filled up, and my computer becomes impossible to work with, because it spends too much time writing to swap and the process can only be killed violently by the OS.



I circumvent this by using MemoryConstrained, but when the solver reaches the memory limit it is simply aborted and does not return the solution it obtained so far. Is there a way to obtain this solution (much like what happens when the solver encounters a singularity or reaches MaxSteps)?


Note: using a hack of the form


 StepMonitor :> If[MemoryInUse[]>...,...]

results in serious computational overhead.



Answer



Borrowing from an example of WhenEvent from the documentation in which a Button is used to stop the integration, I came up with this.


ClearAll[ndsolveMemConstrained];
SetAttributes[ndsolveMemConstrained, HoldFirst];
ndsolveMemConstrained::mlim = "Memory used `` exceeded limit ``.";

ndsolveMemConstrained[(nd_: NDSolve | NDSolveValue)[eqns_, rest___], bytes_] :=
Module[{sol, stop, task, mstart},
mstart = MemoryInUse[];
stop = False;
task = RunScheduledTask[
stop = (ndsolve`mem = MemoryInUse[] - mstart) > bytes,
0.2];
sol = nd[Append[eqns,
WhenEvent[stop,
Message[ndsolveMemConstrained::mlim, ndsolve`mem, bytes];

"StopIntegration"]],
rest];
RemoveScheduledTask[task];
sol]

As a baseline, here is an example DE from the documentation:


NDSolveValue[{D[u[t, x], t, t] == D[u[t, x], x, x], 
u[0, x] == Exp[-10 x^2], Derivative[1, 0][u][0, x] == 0,
u[t, -10] == u[t, 10]}, u, {t, 0, 100}, {x, -10, 10},
Method -> "StiffnessSwitching"] // AbsoluteTiming


(* {10.969550,InterpolatingFunction[{{0.,100.},{\[Ellipsis],-10.,10.,\[Ellipsis]}},<>]} *)

When the memory is not exceeded, it takes about the same amount of time:


ndsolveMemConstrained[
NDSolveValue[{
D[u[t, x], t, t] == D[u[t, x], x, x], u[0, x] == Exp[-10 x^2],
Derivative[1, 0][u][0, x] == 0, u[t, -10] == u[t, 10]},
u, {t, 0, 100}, {x, -10, 10}, Method -> "StiffnessSwitching"],
8000000] // AbsoluteTiming

ndsolve`mem

(* {10.962278, InterpolatingFunction[{{0., 100.}, {..., -10., 10.,...}}, <>]} *)
(* 6992160 *)

When the memory limit is exceeded, there is frequently an extra warning message generated. I assume it has to do with where the solver is when stop is checked. (It's odd that it doesn't always produce the convergence warning.)


ndsolveMemConstrained[
NDSolveValue[{
D[u[t, x], t, t] == D[u[t, x], x, x], u[0, x] == Exp[-10 x^2],
Derivative[1, 0][u][0, x] == 0, u[t, -10] == u[t, 10]},

u, {t, 0, 100}, {x, -10, 10}, Method -> "StiffnessSwitching"],
4000000] // AbsoluteTiming
ndsolve`mem


NDSolveValue::evcvmit: Event location failed to converge to the requested accuracy or precision within 100 iterations between t = 56.32617731294334and t = 56.50060870314276. >>


ndsolveMemConstrained::mlim: Memory used 4158544 exceeded limit 4000000.



(* {5.595887, InterpolatingFunction[{{0., 56.3262}, {..., -10., 10.,...}}, <>]} *)
(* 4047584 *)


You can also monitor memory usage if the following is executed before ndsolveMemConstrained.


Dynamic @ ndsolve`mem

(* 6992160 *)

Comments

Popular posts from this blog

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}]

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}]

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....