The output of NDSolve is a (list of)InterpolatingFunction(s). In standard output format the ‎elements of interpolation are not printed explicitly. I want to know if there is any way to access these ‎values?‎
I need to obtain the value of the solution of an ODE system, at the endpoint of the interval, let's say‎t=‎3‎. I use NDSolve as follows: ‎ enter image description here
The output is a vector of InterpolatingFunctions. When I put the answers vector in Psi[t](a vector of functions of t) and ‎then replace the variable t by 3, it seems that Mathematica starts interpolating all functions and ‎then replaces t=‎‏3‎.‎ But for some large systems (more than ‎‏30000‏ x ‏30000‏‎). this procedure becomes unbearably time ‎consuming or even impossible for a PC (Corei‏5‏ & ‎‏4‏GB of RAM) to perform. Thus, if both of the ‎interpolating points and the corresponding values be available (at least at boundary points), they ‎can be used instead, much more simply.‎


1 Answer 1


In version 9 you can use Part to access the parts of an InterpolatingFunction:

points = {{0, 0}, {1, 1}, {2, 3}, {3, 4}, {4, 3}, {5, 0}};
ifun = Interpolation[points]
(* InterpolatingFunction[{{0,5}},<>] *)

{ifun[[3]], ifun[[4]]}
(* {{{0,1,2,3,4,5}},{{0},{1},{3},{4},{3},{0}}} *)

You can also access Properties of ifun using (not ifun["Properties"] as one would expect) ifun["Methods"]:

(* {"Coordinates", "DerivativeOrder", "Domain", "ElementMesh", 
     "Evaluate", "Grid", "InterpolationOrder", "MethodInformation", 
     "Methods", "Properties", "ValuesOnGrid"} *)

{ifun["Coordinates"][[1]], ifun["ValuesOnGrid"]}
(* {{0,1,2,3,4,5}, {0,1,3,4,3,0}} *)

Using the above with NDSolve

s = NDSolve[{y'[x] == y[x] Cos[x + y[x]], y[0] == 1}, y, {x, 0, 30}]
(* {{y -> InterpolatingFunction[{{0.,30.}},<>]}} *)
if1 = s[[1, All, -1]]
(* InterpolatingFunction[{{0.,30.}},<>] *)

{if1["Coordinates"][[1]], if1["ValuesOnGrid"]} // Short
(* {{0.,0.00017069,0.00034138,<<332>>,29.7515,29.8757,30.},{<<1>>}}*)

coords = Transpose[{if1["Coordinates"][[1]], if1["ValuesOnGrid"]}];

enter image description here

Related Q/As:

How to splice together several instances of InterpolatingFunction

Incompatible InterpolatingFunction between V9 and v10

  • $\begingroup$ Thanks for your attention. I've tried it, but even for a partly small example that I've solved before in less than 5 seconds, this syntax caused to an immediate memory overflow! A note I should stated before is that I'm using Mathematica 8.0 . $\endgroup$
    – Toughee
    Sep 25, 2014 at 2:56
  • $\begingroup$ Would you explain that why you've used the term : if1 = s[[1, -1, -1]]? I think I'm misusing this part of syntax. $\endgroup$
    – Toughee
    Sep 25, 2014 at 3:03
  • $\begingroup$ @Toughee, despite the typo (it was meant to be s[[1,1,-1]]) this syntax happens to work fine with a single function y. However, the correct form that works in general is s[[1,All,-1]]. It is an alternative to {x,y,z}/. s //First to get the right-hand-sides of Rules inside s. $\endgroup$
    – kglr
    Sep 25, 2014 at 3:17
  • $\begingroup$ By v11.3, ifun["Methods"] has grown to {"Coordinates", "DerivativeOrder", "Domain", "ElementMesh", "Evaluate", "GetPolynomial", "Grid", "InterpolationMethod", "InterpolationOrder", "MethodInformation", "Methods", "OutputDimensions", "Periodicity", "PlottableQ", "Properties", "QuantityUnits", "Unpack", "ValuesOnGrid"}. $\endgroup$
    – bbgodfrey
    May 13, 2018 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.