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Ingy döt Net 2023-07-01 11:58:00 -04:00
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(* Euler integration by recurrence relation.
* Given a function, and stepsize, provides a function of (t,y) which
* returns the next step: (t',y'). *)
let euler f ~step (t,y) = ( t+.step, y +. step *. f t y )
(* newton_cooling doesn't use time parameter, so _ is a placeholder *)
let newton_cooling ~k ~tr _ y = -.k *. (y -. tr)
(* analytic solution for Newton cooling *)
let analytic_solution ~k ~tr ~t0 t = tr +. (t0 -. tr) *. exp (-.k *. t)

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(* Wrapping up the parameters in a "cool" function: *)
let cool = euler (newton_cooling ~k:0.07 ~tr:20.)
(* Similarly for the analytic solution: *)
let analytic = analytic_solution ~k:0.07 ~tr:20. ~t0:100.
(* (Just a loop) Apply recurrence function on state, until some condition *)
let recur ~until f state =
let rec loop s =
if until s then ()
else loop (f s)
in loop state
(* 'results' generates the specified output starting from initial values t=0, temp=100C; ending at t=100s *)
let results fn =
Printf.printf "\t time\t euler\tanalytic\n%!";
let until (t,y) =
Printf.printf "\t%7.3f\t%7.3f\t%9.5f\n%!" t y (analytic t);
t >= 100.
in recur ~until fn (0.,100.)
results (cool ~step:10.)
results (cool ~step:5.)
results (cool ~step:2.)