I am trying to solve a system of differential equations which model reaction kinetics using NDSolve. I would like to generate a surface-plot of the function value as a function of time and the initial value of one of the functions. For example, suppose I want to solve

NDSolve[{x'[t] == y[t], y'[t] == -x[t], x[0] == 1, y[0] == c}, {x, y}, {t, 0, 10}]

where c is a parameter that I vary. In my case, the solution can only be found numerically; no analytical solutions exist. I want a surface plot of y(t,c). Does anyone know of an easy way to do this? I haven't found much help online.

  • $\begingroup$ Have you seen Plot3D[]? $\endgroup$ Mar 9, 2016 at 2:54
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    – Michael E2
    Mar 9, 2016 at 3:20
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    $\begingroup$ ParametricNDSolve. $\endgroup$
    – march
    Mar 9, 2016 at 3:21
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    $\begingroup$ To be clear; my system can not be solved analytically -- I gave an example to clarify my meaning, but it is not the system I am working with, and I don't think the output from NDSolve is compatible with Plot3D ParametricNDSolve is more along the lines of what I am looking for; is there a version that lets you create a surface plot? $\endgroup$ Mar 9, 2016 at 4:09
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    $\begingroup$ Well, you could use a slight change of march's suggestion; that is, look up ParametricNDSolveValue[]; you should be able to use that with Plot3D[]. $\endgroup$ Mar 9, 2016 at 4:10

1 Answer 1


I guess I can post an answer at this point (with a hat tip to march):

pf = ParametricNDSolveValue[{x'[t] == y[t], y'[t] == -x[t], x[0] == 1, y[0] == c},
                            y, {t, 0, 10}, c];

Plot3D[pf[c][t], {t, 0, 10}, {c, 0, 4}]

plot of a parametrized numerical solution

  • $\begingroup$ Does ParametricNDSolveValue still solve for both x and y under the hood, just returns only the solution for y? $\endgroup$
    – LLlAMnYP
    Mar 9, 2016 at 13:17
  • $\begingroup$ I believe so (it's a coupled DE system after all); since only y was needed, this is reflected in the second argument of ParametricNDSolveValue[]. $\endgroup$ Mar 9, 2016 at 13:28
  • $\begingroup$ It seems to give identical results either way, though I wouldn't trust MMA to know that the list of functions is {x,y} without explicitly being told so. $\endgroup$
    – LLlAMnYP
    Mar 9, 2016 at 13:31

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