Vector ParametricNDSolve and FindRoot interaction

Question

This question came out of this question.

I have a set of differential equations, written in vector form. I'm only interested in the value of these at the endpoint, and so I use ParametricNDSolve, asking it to only return that function of the vectors. This works fine on its own, and is slightly quicker than asking for the whole solution to be returned:

Clear[test];
A = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {q, 0, 0, 0}}; test = 
 ParametricNDSolveValue[{Y'[x] == A.Y[x], Y[0] == Table[1, 4]}, 
  Y[4].Y[4], {x, 0, 4}, q];
First@AbsoluteTiming[test /@ Range[0, 10, 0.1];]
(* 0.037914 *)

However, if I try to use this same function in FindRoot, it now takes much longer to evaluate at the same points afterwards:

Clear[test];
A = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {q, 0, 0, 0}}; test = 
 ParametricNDSolveValue[{Y'[x] == A.Y[x], Y[0] == Table[1, 4]}, 
  Y[4].Y[4], {x, 0, 4}, q];
Quiet[FindRoot[test[q], {q, 3}]];
First@AbsoluteTiming[test /@ Range[0, 10, 0.1];]
(* 0.24924 *)

Which is 6 times longer than it took to do exactly the same calculation. Note that the function definition is identical, just the use of the function in the FindRoot has changed (which is also much slower than just getting the entire interpolation functions out and then calculating only the part I need).

Can anyone explain what is going on? I get the same timings on 11.3 and 12.0 on my mac.

Answer 1

We can add a method, then the time is reduced by an order. In this example, the test-1000 has a root

Clear[test];
A = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {q, 0, 0, 0}}; test = 
 ParametricNDSolveValue[{Y'[x] == A.Y[x], Y[0] == Table[1, 4]}, 
  Y[4].Y[4], {x, 0, 4}, q];
Quiet[FindRoot[test[q] - 1000, {q,0,1}, Method -> "Secant"]];
First@AbsoluteTiming[test /@ Range[0, 10, 0.1];]
(*0.0341673*)

Compare without method and without FindRoot[]

 Clear[test];
A = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {q, 0, 0, 0}}; test = 
 ParametricNDSolveValue[{Y'[x] == A.Y[x], Y[0] == Table[1, 4]}, 
  Y[4].Y[4], {x, 0, 4}, q];
Quiet[FindRoot[test[q] - 1000, {q, 3}]];
First@AbsoluteTiming[test /@ Range[0, 10, 0.1];]

(*0.271661*)

 Clear[test];
A = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {q, 0, 0, 0}}; test = 
 ParametricNDSolveValue[{Y'[x] == A.Y[x], Y[0] == Table[1, 4]}, 
  Y[4].Y[4], {x, 0, 4}, q];
First@AbsoluteTiming[test /@ Range[0, 10, 0.1];]

(* 0.0395219 *)

With the option Method -> "Secant" code works even faster than without FindRoot[].If we use the option Method -> "AffineCovariantNewton", then the time increases:

 Clear[test];
A = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {q, 0, 0, 0}}; test = 
 ParametricNDSolveValue[{Y'[x] == A.Y[x], Y[0] == Table[1, 4]}, 
  Y[4].Y[4], {x, 0, 4}, q];
Quiet[FindRoot[test[q] - 1000, {q, 1}, 
   Method -> "AffineCovariantNewton"]];
First@AbsoluteTiming[test /@ Range[0, 10, 0.1];]

(* 0.298559 *)

Consequently, Newton's method (the default method) can slow down the code in this combination.