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Again I have question I hope someone can answer. I made a very simple example illustrating my more difficult problem.

I want to solve a differential equation numerically with a conditional in the equation depending on a dataset. Mathematica does not give an error but never evaluates my conditional to true. When I try the same thing with a function in the conditional instead of an InterpolatingFunction it does work. What am I doing wrong?

datafunc = Interpolation[{{0, 0}, {5, 0}, {6, 2}, {10, 0}, {15, 1}, {21, 1}}, InterpolationOrder -> 0];
Plot[datafunc[t], {t, 0, 21}]

enter image description here

solvefunc = NDSolve[{D[f[t], t] == Piecewise[{{7, datafunc[t] > 0.}}, -5], f[0] == 0}, f[t], {t, 0, 20}];
output[t_] := Evaluate[f[t] /. solvefunc];

Plot[output[t], {t, 0, 20}]

enter image description here

As you see the result only takes evaluates the -5 part.

However when I take a function instead of the Interpolating function, it does seem to work. 

testfunc[t_] := Sin[t]
solvetest = NDSolve[{D[f[t], t] == Piecewise[{{7, testfunc[t] > 0.}}, -5], f[0] == 0}, f[t], {t, 0, 20}];
outputtest[t_] := Evaluate[f[t] /. solvetest];
Plot[outputtest[t], {t, 0, 20}]

enter image description here

Can someone explain me the difference between the two. And why the first example does not work?

Edit: I noticed now that when I start the interpolating function with a number >0. It does work for the first switch. But then when it changes back to 0 again it does not.

datafunc1 = Interpolation[{{0, 1}, {5, 1}, {6, 2}, {10, 0}, {15, 0}, {21, 1}}, InterpolationOrder -> 0]; 
solvefunc1 = NDSolve[{D[f[t], t] == Piecewise[{{7, datafunc1[t] > 0.}}, -5], f[0] == 0}, f[t], {t, 0, 20}] 
output1[t_] := Evaluate[f[t] /. solvefunc1]; 
Plot[output1[t], {t, 0, 20}]

enter image description here

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  • $\begingroup$ Very odd, considering that directly plugging in the equivalent Piecewise[{{2, 5 < x <= 6}, {1, 10 < x <= 21}}] works. But here is something interesting: if you do datafunc[x_?NumericQ] := Piecewise[{{2, 5 < x <= 6}, {1, 10 < x <= 21}}], then you get the same result as using Interpolation[]. I guess you might need to reformulate this in terms of WhenEvent[]. $\endgroup$
    – J. M.'s eventual burnout ♦
    Mar 17, 2016 at 9:46
  • $\begingroup$ Thanks for your idea. The problem is that for my "real" problem, the interpolating function is a list of daily data over a year. That's why I change it into an interpolating function instead of plugging in piecewise directly. Do you have an idea how to work around that? $\endgroup$
    – Wiebe
    Mar 17, 2016 at 10:26

1 Answer 1

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Encapsulating the right-hand side in a NumericQ-protected function and setting the MaxStepSize to 1 seems to work for me:

df[t_?NumericQ] := Piecewise[{{7, datafunc[t] > 0.}, {-5, datafunc[t] <= 0.}}];
solvefunc = NDSolve[{D[f[t], t] == df[t], f[0] == 0}, f[t], {t, 0, 20}, MaxStepSize -> 1];
output[t_] := Evaluate[f[t] /. solvefunc];
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  • $\begingroup$ This seems to work, indeed. Still do not get why my code did not work. But I am happy with this solution! Thanks $\endgroup$
    – Wiebe
    Mar 17, 2016 at 16:32
  • $\begingroup$ Glad it helps. We'll need a real M guru to weigh in on the why & why not question! $\endgroup$
    – Chris K
    Mar 17, 2016 at 17:18

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