Euler method

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Task
Euler method
You are encouraged to solve this task according to the task description, using any language you may know.

Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in the wikipedia page. The ODE has to be provided in the following form:

\frac{dy(t)}{dt} = f(t,y(t))

with an initial value

y(t0) = y0

To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:

\frac{dy(t)}{dt}  \approx \frac{y(t+h)-y(t)}{h}

then solve for y(t + h):

y(t+h) \approx y(t) + h \, \frac{dy(t)}{dt}

which is the same as

y(t+h) \approx y(t) + h \, f(t,y(t))

The iterative solution rule is then:

y_{n+1} = y_n + h \, f(t_n, y_n)

h is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.

Example: Newton's Cooling Law

Newton's cooling law describes how an object of initial temperature T(t0) = T0 cools down in an environment of temperature TR:

\frac{dT(t)}{dt} = -k \, \Delta T

or

\frac{dT(t)}{dt} = -k \, (T(t) - T_R)

It says that the cooling rate \frac{dT(t)}{dt} of the object is proportional to the current temperature difference ΔT = (T(t) − TR) to the surrounding environment.

The analytical solution, which we will compare to the numerical approximation, is

T(t) = T_R + (T_0 - T_R) \; e^{-k t}

Task

The task is to implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of 2 s, 5 s and 10 s and to compare with the analytical solution. The initial temperature T0 shall be 100 °C, the room temperature TR 20 °C, and the cooling constant k 0.07. The time interval to calculate shall be from 0 s to 100 s.

A reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.

Euler Method Newton Cooling.png

Contents

[edit] Ada

The solution is generic, usable for any floating point type. The package specification:

 
generic
type Number is digits <>;
package Euler is
type Waveform is array (Integer range <>) of Number;
function Solve
( F  : not null access function (T, Y : Number) return Number;
Y0  : Number;
T0, T1 : Number;
N  : Positive
) return Waveform;
end Euler;
 

The function Solve returns the solution of the differential equation for each of N+1 points, starting from the point T0. The implementation:

 
package body Euler is
function Solve
( F  : not null access function (T, Y : Number) return Number;
Y0  : Number;
T0, T1 : Number;
N  : Positive
) return Waveform is
dT : constant Number := (T1 - T0) / Number (N);
begin
return Y : Waveform (0..N) do
Y (0) := Y0;
for I in 1..Y'Last loop
Y (I) := Y (I - 1) + dT * F (T0 + dT * Number (I - 1), Y (I - 1));
end loop;
end return;
end Solve;
end Euler;
 

The test program:

 
with Ada.Text_IO; use Ada.Text_IO;
with Euler;
 
procedure Test_Euler_Method is
package Float_Euler is new Euler (Float);
use Float_Euler;
 
function Newton_Cooling_Law (T, Y : Float) return Float is
begin
return -0.07 * (Y - 20.0);
end Newton_Cooling_Law;
 
Y : Waveform := Solve (Newton_Cooling_Law'Access, 100.0, 0.0, 100.0, 10);
begin
for I in Y'Range loop
Put_Line (Integer'Image (10 * I) & ":" & Float'Image (Y (I)));
end loop;
end Test_Euler_Method;
 

Sample output:

 0: 1.00000E+02
 10: 4.40000E+01
 20: 2.72000E+01
 30: 2.21600E+01
 40: 2.06480E+01
 50: 2.01944E+01
 60: 2.00583E+01
 70: 2.00175E+01
 80: 2.00052E+01
 90: 2.00016E+01
 100: 2.00005E+01

[edit] ALGOL 68

Translation of: D
Note: This specimen retains the original D coding style.
Works with: ALGOL 68 version Revision 1 - no extensions to language used.
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny.
#
Approximates y(t) in y'(t)=f(t,y) with y(a)=y0 and
t=a..b and the step size h.
#

PROC euler = (PROC(REAL,REAL)REAL f, REAL y0, a, b, h)REAL: (
REAL y := y0,
t := a;
WHILE t < b DO
printf(($g(-6,3)": "g(-7,3)l$, t, y));
y +:= h * f(t, y);
t +:= h
OD;
printf($"done"l$);
y
);
 
# Example: Newton's cooling law #
PROC newton cooling law = (REAL time, t)REAL: (
-0.07 * (t - 20)
);
 
main: (
euler(newton cooling law, 100, 0, 100, 10)
)

Ouput:

 0.000: 100.000
10.000:  44.000
20.000:  27.200
30.000:  22.160
40.000:  20.648
50.000:  20.194
60.000:  20.058
70.000:  20.017
80.000:  20.005
90.000:  20.002
done

[edit] BBC BASIC

      PROCeuler("-0.07*(y-20)", 100, 0, 100, 2)
PROCeuler("-0.07*(y-20)", 100, 0, 100, 5)
PROCeuler("-0.07*(y-20)", 100, 0, 100, 10)
END
 
DEF PROCeuler(df$, y, a, b, s)
LOCAL t, @%
@% = &2030A
t = a
WHILE t <= b
PRINT t, y
y += s * EVAL(df$)
t += s
ENDWHILE
ENDPROC

Output:

     0.000   100.000
     2.000    88.800
     4.000    79.168
     6.000    70.884
     8.000    63.761
    10.000    57.634
...
     0.000   100.000
    10.000    44.000
    20.000    27.200
    30.000    22.160
    40.000    20.648
    50.000    20.194
    60.000    20.058
    70.000    20.017
    80.000    20.005
    90.000    20.002
   100.000    20.000

[edit] C

#include <stdio.h>
#include <math.h>
 
typedef double (*deriv_f)(double, double);
#define FMT " %7.3f"
 
void ivp_euler(deriv_f f, double y, int step, int end_t)
{
int t = 0;
 
printf(" Step %2d: ", (int)step);
do {
if (t % 10 == 0) printf(FMT, y);
y += step * f(t, y);
} while ((t += step) <= end_t);
printf("\n");
}
 
void analytic()
{
double t;
printf(" Time: ");
for (t = 0; t <= 100; t += 10) printf(" %7g", t);
printf("\nAnalytic: ");
 
for (t = 0; t <= 100; t += 10)
printf(FMT, 20 + 80 * exp(-0.07 * t));
printf("\n");
}
 
double cooling(double t, double temp)
{
return -0.07 * (temp - 20);
}
 
int main()
{
analytic();
ivp_euler(cooling, 100, 2, 100);
ivp_euler(cooling, 100, 5, 100);
ivp_euler(cooling, 100, 10, 100);
 
return 0;
}
output
    Time:        0      10      20      30      40      50      60      70      80      90     100
Analytic: 100.000 59.727 39.728 29.797 24.865 22.416 21.200 20.596 20.296 20.147 20.073
Step 2: 100.000 57.634 37.704 28.328 23.918 21.843 20.867 20.408 20.192 20.090 20.042
Step 5: 100.000 53.800 34.280 26.034 22.549 21.077 20.455 20.192 20.081 20.034 20.014
Step 10: 100.000 44.000 27.200 22.160 20.648 20.194 20.058 20.017 20.005 20.002 20.000

[edit] C++

Translation of: D
#include <iomanip>
#include <iostream>
 
typedef double F(double,double);
 
/*
Approximates y(t) in y'(t)=f(t,y) with y(a)=y0 and
t=a..b and the step size h.
*/

void euler(F f, double y0, double a, double b, double h)
{
double y = y0;
for (double t = a; t < b; t += h)
{
std::cout << std::fixed << std::setprecision(3) << t << " " << y << "\n";
y += h * f(t, y);
}
std::cout << "done\n";
}
 
// Example: Newton's cooling law
double newtonCoolingLaw(double, double t)
{
return -0.07 * (t - 20);
}
 
int main()
{
euler(newtonCoolingLaw, 100, 0, 100, 2);
euler(newtonCoolingLaw, 100, 0, 100, 5);
euler(newtonCoolingLaw, 100, 0, 100, 10);
}

Last part of output:

...
0.000 100.000
10.000 44.000
20.000 27.200
30.000 22.160
40.000 20.648
50.000 20.194
60.000 20.058
70.000 20.017
80.000 20.005
90.000 20.002
done

[edit] C#

using System;
 
namespace prog
{
class MainClass
{
const float T0 = 100f;
const float TR = 20f;
const float k = 0.07f;
readonly static float[] delta_t = {2.0f,5.0f,10.0f};
const int n = 100;
 
public delegate float func(float t);
static float NewtonCooling(float t)
{
return -k * (t-TR);
}
 
public static void Main (string[] args)
{
func f = new func(NewtonCooling);
for(int i=0; i<delta_t.Length; i++)
{
Console.WriteLine("delta_t = " + delta_t[i]);
Euler(f,T0,n,delta_t[i]);
}
}
 
public static void Euler(func f, float y, int n, float h)
{
for(float x=0; x<=n; x+=h)
{
Console.WriteLine("\t" + x + "\t" + y);
y += h * f(y);
}
}
}
}

[edit] Clay

 
import printer.formatter as pf;
 
euler(f, y, a, b, h) {
while (a < b) {
println(pf.rightAligned(2, a), " ", y);
a += h;
y += h * f(y);
}
}
 
main() {
for (i in [2.0, 5.0, 10.0]) {
println("\nFor delta = ", i, ":");
euler((temp) => -0.07 * (temp - 20), 100.0, 0.0, 100.0, i);
}
}
 

Example output:

For delta = 10:
 0 100
10 43.99999999999999
20 27.2
30 22.16
40 20.648
50 20.1944
60 20.05832
70 20.017496
80 20.0052488
90 20.00157464

[edit] COBOL

Translation of: C#
Works with: Visual COBOL

The following is in the Managed COBOL dialect:

       DELEGATE-ID func.
PROCEDURE DIVISION USING VALUE t AS FLOAT-LONG
RETURNING ret AS FLOAT-LONG.
END DELEGATE.
 
CLASS-ID. MainClass.
 
78 T0 VALUE 100.0.
78 TR VALUE 20.0.
78 k VALUE 0.07.
 
01 delta-t INITIALIZE ONLY STATIC
FLOAT-LONG OCCURS 3 VALUES 2.0, 5.0, 10.0.
 
78 n VALUE 100.
 
METHOD-ID NewtonCooling STATIC.
PROCEDURE DIVISION USING VALUE t AS FLOAT-LONG
RETURNING ret AS FLOAT-LONG.
COMPUTE ret = - k * (t - TR)
END METHOD.
 
METHOD-ID Main STATIC.
DECLARE f AS TYPE func
SET f TO METHOD self::NewtonCooling
 
DECLARE delta-t-len AS BINARY-LONG
MOVE delta-t::Length TO delta-t-len
PERFORM VARYING i AS BINARY-LONG FROM 1 BY 1
UNTIL i > delta-t-len
DECLARE elt AS FLOAT-LONG = delta-t (i)
INVOKE TYPE Console::WriteLine("delta-t = {0:F4}", elt)
INVOKE self::Euler(f, T0, n, elt)
END-PERFORM
END METHOD.
 
METHOD-ID Euler STATIC.
PROCEDURE DIVISION USING VALUE f AS TYPE func, y AS FLOAT-LONG,
n AS BINARY-LONG, h AS FLOAT-LONG.
PERFORM VARYING x AS BINARY-LONG FROM 0 BY h UNTIL x >= n
INVOKE TYPE Console::WriteLine("x = {0:F4}, y = {1:F4}", x, y)
COMPUTE y = y + h * RUN f(y)
END-PERFORM
END METHOD.
END CLASS.

Example output:

delta-t = 10.0000
x = 0.0000, y = 100.0000
x = 10.0000, y = 44.0000
x = 20.0000, y = 27.2000
x = 30.0000, y = 22.1600
x = 40.0000, y = 20.6480
x = 50.0000, y = 20.1944
x = 60.0000, y = 20.0583
x = 70.0000, y = 20.0175
x = 80.0000, y = 20.0052
x = 90.0000, y = 20.0016

[edit] Common Lisp

;; 't' usually means "true" in CL, but we need 't' here for time/temperature.
(defconstant true 'cl:t)
(shadow 't)
 
 
;; Approximates y(t) in y'(t)=f(t,y) with y(a)=y0 and t=a..b and the step size h.
(defun euler (f y0 a b h)
 
;; Set the initial values and increments of the iteration variables.
(do ((t a (incf t h))
(y y0 (incf y (* h (funcall f t y)))))
 
;; End the iteration when t reaches the end b of the time interval.
((>= t b) 'DONE)
 
;; Print t and y(t) at every step of the do loop.
(format true "~6,3F ~6,3F~%" t y)))
 
 
;; Example: Newton's cooling law, f(t,T) = -0.07*(T-20)
(defun newton-cooling (time T) (* -0.07 (- T 20)))
 
;; Generate the data for all three step sizes (2,5 and 10).
(euler #'newton-cooling 100 0 100 2)
(euler #'newton-cooling 100 0 100 5)
(euler #'newton-cooling 100 0 100 10)
Example output:

 0.000  100.000
10.000  44.000
20.000  27.200
30.000  22.160
40.000  20.648
50.000  20.194
60.000  20.058
70.000  20.017
80.000  20.005
90.000  20.002

[edit] D

import std.stdio, std.range;
 
/**
Approximates y(t) in y'(t)=f(t,y) with y(a)=y0 and
t=a..b and the step size h.
*/

void euler(F)(in F f, in double y0,
in double a, in double b, in double h) {
double y = y0;
foreach (t; iota(a, b, h)) {
writefln("%.3f  %.3f", t, y);
y += h * f(t, y);
}
writeln("done");
}
 
void main() {
/// Example: Newton's cooling law
static newtonCoolingLaw(in double time, in double t) {
return -0.07 * (t - 20);
}
 
euler(&newtonCoolingLaw, 100, 0, 100, 2);
euler(&newtonCoolingLaw, 100, 0, 100, 5);
euler(&newtonCoolingLaw, 100, 0, 100, 10);
}

Last part of the output:

...
0.000  100.000
10.000  44.000
20.000  27.200
30.000  22.160
40.000  20.648
50.000  20.194
60.000  20.058
70.000  20.017
80.000  20.005
90.000  20.002
done

[edit] Euler Math Toolbox

 
>function dgleuler (f,x,y0) ...
$ y=zeros(size(x)); y[1]=y0;
$ for i=2 to cols(y);
$ y[i]=y[i-1]+f(x[i-1],y[i-1])*(x[i]-x[i-1]);
$ end;
$ return y;
$endfunction
>function f(x,y) := -k*(y-TR)
>k=0.07; TR=20; TS=100;
>x=0:1:100; dgleuler("f",x,TS)[-1]
20.0564137335
>x=0:2:100; dgleuler("f",x,TS)[-1]
20.0424631834
>TR+(TS-TR)*exp(-k*TS)
20.0729505572
>x=0:5:100; plot2d(x,dgleuler("f",x,TS)); ...
> plot2d(x,TR+(TS-TR)*exp(-k*x),>add,color=red);
>ode("f",x,TS)[-1] // Euler default solver LSODA
20.0729505568
>adaptiverunge("f",x,TS)[-1] // Adaptive Runge Method
20.0729505572
 

[edit] F#

let euler f (h : float) t0 y0 =
(t0, y0)
|> Seq.unfold (fun (t, y) -> Some((t,y), ((t + h), (y + h * (f t y)))))
 
let newtonCoolíng _ y = -0.07 * (y - 20.0)
 
[<EntryPoint>]
let main argv =
let f = newtonCoolíng
let a = 0.0
let y0 = 100.0
let b = 100.0
let h = 10.0
(euler newtonCoolíng h a y0)
|> Seq.takeWhile (fun (t,_) -> t <= b)
|> Seq.iter (printfn "%A")
0

Output for the above (step size 10)

(0.0, 100.0)
(10.0, 44.0)
(20.0, 27.2)
(30.0, 22.16)
(40.0, 20.648)
(50.0, 20.1944)
(60.0, 20.05832)
(70.0, 20.017496)
(80.0, 20.0052488)
(90.0, 20.00157464)
(100.0, 20.00047239)

[edit] Forth

: newton-cooling-law ( f: temp -- f: temp' )
20e f- -0.07e f* ;
 
: euler ( f: y0 xt step end -- )
1+ 0 do
cr i . fdup f.
fdup over execute
dup s>f f* f+
dup +loop
2drop fdrop ;
 
100e ' newton-cooling-law 2 100 euler cr
100e ' newton-cooling-law 5 100 euler cr
100e ' newton-cooling-law 10 100 euler cr

[edit] Fortran

Works with: Fortran version 2008
program euler_method
use iso_fortran_env, only: real64
implicit none
 
abstract interface
! a derivative dy/dt as function of y and t
function derivative(y, t)
use iso_fortran_env, only: real64
real(real64) :: derivative
real(real64), intent(in) :: t, y
end function
end interface
 
real(real64), parameter :: T_0 = 100, T_room = 20, k = 0.07, a = 0, b = 100, &
h(3) = [2.0, 5.0, 10.0]
 
integer :: i
 
! loop over all step sizes
do i = 1, 3
call euler(newton_cooling, T_0, a, b, h(i))
end do
 
contains
 
! Approximates y(t) in y'(t) = f(y, t) with y(a) = y0 and t = a..b and the
! step size h.
subroutine euler(f, y0, a, b, h)
procedure(derivative) :: f
real(real64), intent(in) :: y0, a, b, h
real(real64) :: t, y
 
if (a > b) return
if (h <= 0) stop "negative step size"
 
print '("# h = ", F0.3)', h
 
y = y0
t = a
 
do
print *, t, y
t = t + h
if (t > b) return
y = y + h * f(y, t)
end do
end subroutine
 
 
! Example: Newton's cooling law, f(T, _) = -k*(T - T_room)
function newton_cooling(T, unused) result(dTdt)
real(real64) :: dTdt
real(real64), intent(in) :: T, unused
dTdt = -k * (T - T_room)
end function
 
end program

Output for h = 10:

# h = 10.000
   0.0000000000000000        100.00000000000000     
   10.000000000000000        43.999999761581421     
   20.000000000000000        27.199999856948853     
   30.000000000000000        22.159999935626985     
   40.000000000000000        20.647999974250794     
   50.000000000000000        20.194399990344049     
   60.000000000000000        20.058319996523856     
   70.000000000000000        20.017495998783350     
   80.000000000000000        20.005248799582862     
   90.000000000000000        20.001574639859214     
   100.00000000000000        20.000472391953071  

[edit] freebasic

 
 
'Freebasic .9
 
'Custom rounding
#define round(x,N) Rtrim(Rtrim(Left(Str((x)+(.5*Sgn((x)))/(10^(N))),Instr(Str((x)+(.5*Sgn((x)))/(10^(N))),".")+(N)),"0"),".")
 
#macro Euler(fn,_y,min,max,h,printoption)
Print "Step ";#h;":":Print
Print "time","Euler"," Analytic"
If printoption<>"print" Then Print "Data omitted ..."
Scope
Dim As Double temp=(min),y=(_y)
Do
If printoption="print" Then Print temp,round(y,3),20+80*Exp(-0.07*temp)
y=y+(h)*(fn)
temp=temp+(h)
Loop Until temp>(max)
Print"________________"
Print
End Scope
#endmacro
 
Euler(-.07*(y-20),100,0,100,2,"don't print")
Euler(-.07*(y-20),100,0,100,5,"print")
Euler(-.07*(y-20),100,0,100,10,"print")
Sleep
 

outputs (steps 5 and 10)

Step 2:

time          Euler          Analytic
Data omitted ...
________________

Step 5:

time          Euler          Analytic
 0            100            100
 5            72             76.37504717749707
 10           53.8           59.72682430331276
 15           41.97          47.99501992889243
 20           34.281         39.72775711532852
 25           29.282         33.90191547603561
 30           26.034         29.79651426023855
 35           23.922         26.90348691994964
 40           22.549         24.86480501001743
 45           21.657         23.42817014936322
 50           21.077         22.41579067378548
 55           20.7           21.70237891507017
 60           20.455         21.19964614563822
 65           20.296         20.84537635070821
 70           20.192         20.59572664567395
 75           20.125         20.41980147193451
 80           20.081         20.29582909731863
 85           20.053         20.20846724147268
 90           20.034         20.14690438216231
 95           20.022         20.10352176843727
 100          20.014         20.07295055724436
________________

Step 10:

time          Euler          Analytic
 0            100            100
 10           44             59.72682430331276
 20           27.2           39.72775711532852
 30           22.16          29.79651426023855
 40           20.648         24.86480501001743
 50           20.194         22.41579067378548
 60           20.058         21.19964614563822
 70           20.017         20.59572664567395
 80           20.005         20.29582909731863
 90           20.002         20.14690438216231
 100          20             20.07295055724436
________________

[edit] Go

package main
 
import (
"fmt"
"math"
)
 
// fdy is a type for function f used in Euler's method.
type fdy func(float64, float64) float64
 
// eulerStep computes a single new value using Euler's method.
// Note that step size h is a parameter, so a variable step size
// could be used.
func eulerStep(f fdy, x, y, h float64) float64 {
return y + h*f(x, y)
}
 
// Definition of cooling rate. Note that this has general utility and
// is not specific to use in Euler's method.
 
// newCoolingRate returns a function that computes cooling rate
// for a given cooling rate constant k.
func newCoolingRate(k float64) func(float64) float64 {
return func(deltaTemp float64) float64 {
return -k * deltaTemp
}
}
 
// newTempFunc returns a function that computes the analytical solution
// of cooling rate integrated over time.
func newTempFunc(k, ambientTemp, initialTemp float64) func(float64) float64 {
return func(time float64) float64 {
return ambientTemp + (initialTemp-ambientTemp)*math.Exp(-k*time)
}
}
 
// newCoolingRateDy returns a function of the kind needed for Euler's method.
// That is, a function representing dy(x, y(x)).
//
// Parameters to newCoolingRateDy are cooling constant k and ambient
// temperature.
func newCoolingRateDy(k, ambientTemp float64) fdy {
crf := newCoolingRate(k)
// note that result is dependent only on the object temperature.
// there are no additional dependencies on time, so the x parameter
// provided by eulerStep is unused.
return func(_, objectTemp float64) float64 {
return crf(objectTemp - ambientTemp)
}
}
 
func main() {
k := .07
tempRoom := 20.
tempObject := 100.
fcr := newCoolingRateDy(k, tempRoom)
analytic := newTempFunc(k, tempRoom, tempObject)
for _, deltaTime := range []float64{2, 5, 10} {
fmt.Printf("Step size = %.1f\n", deltaTime)
fmt.Println(" Time Euler's Analytic")
temp := tempObject
for time := 0.; time <= 100; time += deltaTime {
fmt.Printf("%5.1f %7.3f %7.3f\n", time, temp, analytic(time))
temp = eulerStep(fcr, time, temp, deltaTime)
}
fmt.Println()
}
}

Output, truncated:

...
 85.0  20.053  20.208
 90.0  20.034  20.147
 95.0  20.022  20.104
100.0  20.014  20.073

Step size = 10.0
 Time Euler's Analytic
  0.0 100.000 100.000
 10.0  44.000  59.727
 20.0  27.200  39.728
 30.0  22.160  29.797
 40.0  20.648  24.865
 50.0  20.194  22.416
 60.0  20.058  21.200
 70.0  20.017  20.596
 80.0  20.005  20.296
 90.0  20.002  20.147
100.0  20.000  20.073

[edit] Groovy

Generic Euler Method Solution

The following is a general solution for using the Euler method to produce a finite discrete sequence of points approximating the ODE solution for y as a function of x.


In the eulerStep closure argument list: xn and yn together are the previous point in the sequence. h is the step distance to the next point's x value. dydx is a closure representing the derivative of y as a function of x, itself expressed (as required by the method) as a function of x and y(x).


The eulerMapping closure produces an entire approximating sequence, expressed as a Map object. Here, x0 and y0 together are the first point in the sequence, the ODE initial conditions. h and dydx are again the step distance and the derivative closure. stopCond is a closure representing a "stop condition" that causes the the eulerMapping closure to stop after a finite number of steps; the given default value causes eulerMapping to stop after 100 steps.

def eulerStep = { xn, yn, h, dydx ->
(yn + h * dydx(xn, yn)) as BigDecimal
}
 
Map eulerMapping = { x0, y0, h, dydx, stopCond = { xx, yy, hh, xx0 -> abs(xx - xx0) > (hh * 100) }.rcurry(h, x0) ->
Map yMap = [:]
yMap[x0] = y0 as BigDecimal
def x = x0
while (!stopCond(x, yMap[x])) {
yMap[x + h] = eulerStep(x, yMap[x], h, dydx)
x += h
}
yMap
}
assert eulerMapping.maximumNumberOfParameters == 5


Specific Euler Method Solution for the "Temperature Diffusion" Problem (with Newton's derivative formula and constants for environment temperature and object conductivity given)

def dtdsNewton = { s, t, tR, k ->  k * (tR - t) }
assert dtdsNewton.maximumNumberOfParameters == 4
 
def dtds = dtdsNewton.rcurry(20, 0.07)
assert dtds.maximumNumberOfParameters == 2
 
def tEulerH = eulerMapping.rcurry(dtds) { s, t -> s >= 100 }
assert tEulerH.maximumNumberOfParameters == 3


Newton's Analytic Temperature Diffusion Solution (for comparison)

def tNewton = { s, s0, t0, tR, k ->
tR + (t0 - tR) * Math.exp(k * (s0 - s))
}
assert tNewton.maximumNumberOfParameters == 5
 
def tAnalytic = tNewton.rcurry(0, 100, 20, 0.07)
assert tAnalytic.maximumNumberOfParameters == 1


Specific solutions for 3 step sizes (and initial time and temperature)

[10, 5, 2].each { h -> 
def tEuler = tEulerH.rcurry(h)
assert tEuler.maximumNumberOfParameters == 2
println """
STEP SIZE == ${h}
time analytic euler relative
(seconds) (°C) (°C) error
-------- -------- -------- ---------"""

tEuler(0, 100).each { BigDecimal s, tE ->
def tA = tAnalytic(s)
def relError = ((tE - tA)/(tA - 20)).abs()
printf('%5.0f  %8.4f %8.4f %9.6f\n', s, tA, tE, relError)
}
}


Selected output

STEP SIZE == 10
  time   analytic   euler   relative
(seconds)  (°C)     (°C)     error
-------- -------- -------- ---------
    0    100.0000 100.0000  0.000000
   10     59.7268  44.0000  0.395874
   20     39.7278  27.2000  0.635032
   30     29.7965  22.1600  0.779513
   40     24.8648  20.6480  0.866798
   50     22.4158  20.1944  0.919529
   60     21.1996  20.0583  0.951386
   70     20.5957  20.0175  0.970631
   80     20.2958  20.0052  0.982257
   90     20.1469  20.0016  0.989281
  100     20.0730  20.0005  0.993524

STEP SIZE == 5
  time   analytic   euler   relative
(seconds)  (°C)     (°C)     error
-------- -------- -------- ---------
    0    100.0000 100.0000  0.000000
     ... yada, yada, yada ...
  100     20.0730  20.0145  0.801240

STEP SIZE == 2
  time   analytic   euler   relative
(seconds)  (°C)     (°C)     error
-------- -------- -------- ---------
    0    100.0000 100.0000  0.000000
     ... yada, yada, yada ...
  100     20.0730  20.0425  0.417918

Notice how the relative error in the Euler method sequences increases over time in spite of the fact that all three the Euler approximations and the analytic solution are approaching the same asymptotic limit of 20°C.


Notice also how smaller step size reduces the relative error in the approximation.

[edit] Haskell

import Text.Printf
 
euler :: (Num a, Ord a) => (a -> a -> a) -> a -> a -> a -> a -> [(a,a)]
euler f y0 a b h =
(a, y0) :
if a < b
then euler f (y0 + (f a y0) * h) (a + h) b h
else []
 
newtonCooling :: Double -> Double -> Double
newtonCooling _ t = -0.07 * (t - 20)
 
main = do
mapM_ (uncurry $ printf "%6.3f %6.3f\n") $ euler newtonCooling 100 0 100 10
putStrLn "DONE"

Output:

 0.000 100.000
10.000 44.000
20.000 27.200
30.000 22.160
40.000 20.648
50.000 20.194
60.000 20.058
70.000 20.017
80.000 20.005
90.000 20.002
100.000 20.000
DONE

[edit] Icon and Unicon

Translation of: Common Lisp

This solution works in both Icon and Unicon. It takes advantage of the proc procedure, which converts a string naming a procedure into a call to that procedure.

 
invocable "newton_cooling" # needed to use the 'proc' procedure
 
procedure euler (f, y0, a, b, h)
t := a
y := y0
until (t >= b) do {
write (right(t, 4) || " " || left(y, 7))
t +:= h
y +:= h * (proc(f) (t, y)) # 'proc' applies procedure named in f to (t, y)
}
write ("DONE")
end
 
procedure newton_cooling (time, T)
return -0.07 * (T - 20)
end
 
procedure main ()
# generate data for all three step sizes [2, 5, 10]
every (step_size := ![2,5,10]) do
euler ("newton_cooling", 100, 0, 100, step_size)
end
 

Sample output:

   0 100    
  10 44.0   
  20 27.2   
  30 22.16  
  40 20.648 
  50 20.1944
  60 20.0583
  70 20.0174
  80 20.0052
  90 20.0015
DONE

[edit] J

Solution:

NB.*euler a Approximates Y(t) in Y'(t)=f(t,Y) with Y(a)=Y0 and t=a..b and step size h.
euler=: adverb define
'Y0 a b h'=. 4{. y
t=. i.@>:&.(%&h) b - a
Y=. (+ h * u)^:(<#t) Y0
t,.Y
)
 
ncl=: _0.07 * -&20 NB. Newton's Cooling Law
 

Example:

   ncl euler 100 0 100 2
... NB. output redacted for brevity
ncl euler 100 0 100 5
... NB. output redacted for brevity
ncl euler 100 0 100 10
0 100
10 44
20 27.2
30 22.16
40 20.648
50 20.1944
60 20.0583
70 20.0175
80 20.0052
90 20.0016
100 20.0005

[edit] Java

 
public class Euler {
private static void euler (Callable f, double y0, int a, int b, int h) {
int t = a;
double y = y0;
while (t < b) {
System.out.println ("" + t + " " + y);
t += h;
y += h * f.compute (t, y);
}
System.out.println ("DONE");
}
 
public static void main (String[] args) {
Callable cooling = new Cooling ();
int[] steps = {2, 5, 10};
for (int stepSize : steps) {
System.out.println ("Step size: " + stepSize);
euler (cooling, 100.0, 0, 100, stepSize);
}
}
}
 
// interface used so we can plug in alternative functions to Euler
interface Callable {
public double compute (int time, double t);
}
 
// class to implement the newton cooling equation
class Cooling implements Callable {
public double compute (int time, double t) {
return -0.07 * (t - 20);
}
}
 

Output for step = 10;

Step size: 10
0 100.0
10 43.99999999999999
20 27.199999999999996
30 22.159999999999997
40 20.648
50 20.194399999999998
60 20.05832
70 20.017496
80 20.0052488
90 20.00157464
DONE

[edit] Lua

T0 = 100
TR = 20
k = 0.07
delta_t = { 2, 5, 10 }
n = 100
 
NewtonCooling = function( t ) return -k * ( t - TR ) end
 
 
function Euler( f, y0, n, h )
local y = y0
for x = 0, n, h do
print( "", x, y )
y = y + h * f( y )
end
end
 
 
for i = 1, #delta_t do
print( "delta_t = ", delta_t[i] )
Euler( NewtonCooling, T0, n, delta_t[i] )
end
 

[edit] Mathematica

Better methods for differential equation solving are built into Mathematica, so the typical user would omit the Method and StartingStepSize options in the code below. However since the task requests Eulers method, here is the bad solution...

 
euler[step_, val_] := NDSolve[{T'[t] == -0.07 (T[t] - 20), T[0] == 100}, T, {t, 0, 100}, Method -> "ExplicitEuler", StartingStepSize -> step][[1, 1, 2]][val]
 
Output:
euler[2, 100]
20.0425

euler[5, 100]
20.0145

euler[10, 100]
20.0005

[edit] Maxima

euler_method(f, y0, a, b, h):= block(
[t: a, y: y0, tg: [a], yg: [y0]],
unless t>=b do (
t: t + h,
y: y + f(t, y)*h,
tg: endcons(t, tg),
yg: endcons(y, yg)
),
[tg, yg]
);
 
/* initial temperature */
T0: 100;
 
/* environment of temperature */
Tr: 20;
 
/* the cooling constant */
k: 0.07;
 
/* end of integration */
tmax: 100;
 
/* analytical solution */
Tref(t):= Tr + (T0 - Tr)*exp(-k*t);
 
/* cooling rate */
dT(t, T):= -k*(T-Tr);
 
/* get numerical solution */
h: 10;
[tg, yg]: euler_method('dT, T0, 0, tmax, h);
 
/* plot analytical and numerical solution */
plot2d([Tref, [discrete, tg, yg]], ['t, 0, tmax],
[legend, "analytical", concat("h = ", h)],
[xlabel, "t / seconds"],
[ylabel, "Temperature / C"]);
 

[edit] МК-61/52

П2	С/П	П3	С/П	П4	ПП	19	ИП3	*	ИП4
+ П4 С/П ИП2 ИП3 + П2 БП 05 ...
... ... ... ... ... ... ... ... ... В/О

Instead of dots typed calculation program equation f(u, t), where the arguments are t = Р2, u = Р4.

Input: Initial time С/П Time step С/П Initial value С/П.

The result is displayed on the indicator.

[edit] Objeck

 
class EulerMethod {
T0 : static : Float;
TR : static : Float;
k : static : Float;
delta_t : static : Float[];
n : static : Float;
 
function : Main(args : String[]) ~ Nil {
T0 := 100;
TR := 20;
k := 0.07;
delta_t := [2.0, 5.0, 10.0];
n := 100;
 
f := NewtonCooling(Float) ~ Float;
for(i := 0; i < delta_t->Size(); i+=1;) {
IO.Console->Print("delta_t = ")->PrintLine(delta_t[i]);
Euler(f, T0, n->As(Int), delta_t[i]);
};
}
 
function : native : NewtonCooling(t : Float) ~ Float {
return -1 * k * (t-TR);
}
 
function : native : Euler(f : (Float) ~ Float, y : Float, n : Int, h : Float) ~ Nil {
for(x := 0; x<=n; x+=h;) {
IO.Console->Print("\t")->Print(x)->Print("\t")->PrintLine(y);
y += h * f(y);
};
}
}
 

Output:

delta_t = 2
        0       100
        2       88.8
        4       79.168
        6       70.88448
        ...
delta_t = 10
        0       100
        10      44
        20      27.2
        30      22.16
        40      20.648

[edit] OCaml

(* 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)

Using the above functions to produce the task results:

(* 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.)

Example output:

# results (cool ~step:10.);;
	  time	 euler	analytic
	  0.000	100.000	100.00000
	 10.000	 44.000	 59.72682
	 20.000	 27.200	 39.72776
	 30.000	 22.160	 29.79651
	 40.000	 20.648	 24.86481
	 50.000	 20.194	 22.41579
	 60.000	 20.058	 21.19965
	 70.000	 20.017	 20.59573
	 80.000	 20.005	 20.29583
	 90.000	 20.002	 20.14690
	100.000	 20.000	 20.07295
- : unit = ()

[edit] Pascal

Translation of: C

Euler code for Free Pascal - Delphi mode. Apart from the function-pointer calling convention for the NewtonCooling method, this example is ISO-7185 standard Pascal.

 
 
{$mode delphi}
PROGRAM Euler;
 
TYPE TNewtonCooling = FUNCTION (t: REAL) : REAL;
 
CONST T0 : REAL = 100.0;
CONST TR : REAL = 20.0;
CONST k : REAL = 0.07;
CONST time : INTEGER = 100;
CONST step : INTEGER = 10;
CONST dt : ARRAY[0..3] of REAL = (1.0,2.0,5.0,10.0);
 
VAR i : INTEGER;
 
FUNCTION NewtonCooling(t: REAL) : REAL;
BEGIN
NewtonCooling := -k * (t-TR);
END;
 
PROCEDURE Euler(F: TNewtonCooling; y, h : REAL; n: INTEGER);
VAR i: INTEGER = 0;
BEGIN
WRITE('dt=',trunc(h):2,':');
REPEAT
IF (i mod 10 = 0) THEN WRITE(' ',y:2:3);
INC(i,trunc(h));
y := y + h * F(y);
UNTIL (i >= n);
WRITELN;
END;
 
PROCEDURE Sigma;
VAR t: INTEGER = 0;
BEGIN
WRITE('Sigma:');
REPEAT
WRITE(' ',(20 + 80 * exp(-0.07 * t)):2:3);
INC(t,step);
UNTIL (t>=time);
WRITELN;
END;
 
BEGIN
WRITELN('Newton cooling function: Analytic solution (Sigma) with 3 Euler approximations.');
WRITELN('Time: ',0:7,10:7,20:7,30:7,40:7,50:7,60:7,70:7,80:7,90:7);
Sigma;
FOR i := 1 to 3 DO
Euler(NewtonCooling,T0,dt[i],time);
END.
 
 


Output:

Newton cooling function: Analytic solution (Sigma) with 3 Euler approximations.
Time:       0     10     20     30     40     50     60     70     80     90
Sigma: 100.000 59.727 39.728 29.797 24.865 22.416 21.200 20.596 20.296 20.147
dt= 2: 100.000 57.634 37.704 28.328 23.918 21.843 20.867 20.408 20.192 20.090
dt= 5: 100.000 53.800 34.280 26.034 22.549 21.077 20.455 20.192 20.081 20.034
dt=10: 100.000 44.000 27.200 22.160 20.648 20.194 20.058 20.017 20.005 20.002

[edit] Perl

sub euler_method {
my ($t0, $t1, $k, $step_size) = @_;
my @results = ( [0, $t0] );
 
for (my $s = $step_size; $s <= 100; $s += $step_size) {
$t0 -= ($t0 - $t1) * $k * $step_size;
push @results, [$s, $t0];
}
 
return @results;
}
 
sub analytical {
my ($t0, $t1, $k, $time) = @_;
return ($t0 - $t1) * exp(-$time * $k) + $t1
}
 
my ($T0, $T1, $k) = (100, 20, .07);
my @r2 = grep { $_->[0] % 10 == 0 } euler_method($T0, $T1, $k, 2);
my @r5 = grep { $_->[0] % 10 == 0 } euler_method($T0, $T1, $k, 5);
my @r10 = grep { $_->[0] % 10 == 0 } euler_method($T0, $T1, $k, 10);
 
print "Time\t 2 err(%) 5 err(%) 10 err(%) Analytic\n", "-" x 76, "\n";
for (0 .. $#r2) {
my $an = analytical($T0, $T1, $k, $r2[$_][0]);
printf "%4d\t".("%9.3f" x 7)."\n",
$r2 [$_][0],
$r2 [$_][1], ($r2 [$_][1] / $an) * 100 - 100,
$r5 [$_][1], ($r5 [$_][1] / $an) * 100 - 100,
$r10[$_][1], ($r10[$_][1] / $an) * 100 - 100,
$an;
}
 
Output:
Time          2     err(%)      5     err(%)    10      err(%)  Analytic
----------------------------------------------------------------------------
   0      100.000    0.000  100.000    0.000  100.000    0.000  100.000
  10       57.634   -3.504   53.800   -9.923   44.000  -26.331   59.727
  20       37.704   -5.094   34.280  -13.711   27.200  -31.534   39.728
  30       28.328   -4.927   26.034  -12.629   22.160  -25.629   29.797
  40       23.918   -3.808   22.549   -9.313   20.648  -16.959   24.865
  50       21.843   -2.555   21.077   -5.972   20.194   -9.910   22.416
  60       20.867   -1.569   20.455   -3.512   20.058   -5.384   21.200
  70       20.408   -0.912   20.192   -1.959   20.017   -2.808   20.596
  80       20.192   -0.512   20.081   -1.057   20.005   -1.432   20.296
  90       20.090   -0.281   20.034   -0.559   20.002   -0.721   20.147
 100       20.042   -0.152   20.014   -0.291   20.000   -0.361   20.073

[edit] Perl 6

sub euler ( &f, $y0, $a, $b, $h ) {
my $y = $y0;
my @t_y;
for $a, * + $h ... * > $b -> $t {
@t_y[$t] = $y;
$y += $h * f( $t, $y );
}
return @t_y;
}
 
constant COOLING_RATE = 0.07;
constant AMBIENT_TEMP = 20;
constant INITIAL_TEMP = 100;
constant INITIAL_TIME = 0;
constant FINAL_TIME = 100;
 
sub f ( $time, $temp ) {
return -COOLING_RATE * ( $temp - AMBIENT_TEMP );
}
 
my @e;
@e[$_] = euler( &f, INITIAL_TEMP, INITIAL_TIME, FINAL_TIME, $_ ) for 2, 5, 10;
 
say 'Time Analytic Step2 Step5 Step10 Err2 Err5 Err10';
 
for INITIAL_TIME, * + 10 ... * >= FINAL_TIME -> $t {
 
my $exact = AMBIENT_TEMP + (INITIAL_TEMP - AMBIENT_TEMP)
* (-COOLING_RATE * $t).exp;
 
my $err = sub { @^a.map: { 100 * abs( $_ - $exact ) / $exact } }
 
my ( $a, $b, $c ) = map { @e[$_][$t] }, 2, 5, 10;
 
say $t.fmt('%4d '), ( $exact, $a, $b, $c )».fmt(' %7.3f'),
$err.([$a, $b, $c])».fmt(' %7.3f%%');
}
Output:
Time Analytic   Step2   Step5  Step10     Err2     Err5    Err10
   0  100.000 100.000 100.000 100.000   0.000%   0.000%   0.000%
  10   59.727  57.634  53.800  44.000   3.504%   9.923%  26.331%
  20   39.728  37.704  34.281  27.200   5.094%  13.711%  31.534%
  30   29.797  28.328  26.034  22.160   4.927%  12.629%  25.629%
  40   24.865  23.918  22.549  20.648   3.808%   9.313%  16.959%
  50   22.416  21.843  21.077  20.194   2.555%   5.972%   9.910%
  60   21.200  20.867  20.455  20.058   1.569%   3.512%   5.384%
  70   20.596  20.408  20.192  20.017   0.912%   1.959%   2.808%
  80   20.296  20.192  20.081  20.005   0.512%   1.057%   1.432%
  90   20.147  20.090  20.034  20.002   0.281%   0.559%   0.721%
 100   20.073  20.042  20.014  20.000   0.152%   0.291%   0.361%

[edit] PicoLisp

(load "@lib/math.l")
 
(de euler (F Y A B H)
(while (> B A)
(prinl (round A) " " (round Y))
(inc 'Y (*/ H (F A Y) 1.0))
(inc 'A H) ) )
 
(de newtonCoolingLaw (A B)
(*/ -0.07 (- B 20.) 1.0) )
 
(euler newtonCoolingLaw 100.0 0 100.0 2.0)
(euler newtonCoolingLaw 100.0 0 100.0 5.0)
(euler newtonCoolingLaw 100.0 0 100.0 10.0)

Output:

...
0.000 100.000
10.000 44.000
20.000 27.200
30.000 22.160
40.000 20.648
50.000 20.194
60.000 20.058
70.000 20.018
80.000 20.005
90.000 20.002

[edit] PL/I

test: procedure options (main); /* 3 December 2012 */
 
declare (x, y, z) float;
declare (T0 initial (100), Tr initial (20)) float;
declare k float initial (0.07);
declare t fixed binary;
declare h fixed binary;
 
x, y, z = T0;
/* Step size is 2 seconds */
h = 2;
put skip data (h);
put skip list (' t By formula', 'By Euler');
do t = 0 to 100 by 2;
put skip edit (t, Tr + (T0 - Tr)/exp(k*t), x) (f(3), 2 f(17,10));
x = x + h*f(t, x);
end;
 
/* Step size is 5 seconds */
h = 5;
put skip data (h);
put skip list (' t By formula', 'By Euler');
do t = 0 to 100 by 5;
put skip edit ( t, Tr + (T0 - Tr)/exp(k*t), y) (f(3), 2 f(17,10));
y = y + h*f(t, y);
end;
 
/* Step size is 10 seconds */
h = 10;
put skip data (h);
put skip list (' t By formula', 'By Euler');
do t = 0 to 100 by 10;
put skip edit (t, Tr + (T0 - Tr)/exp(k*t), z) (f(3), 2 f(17,10));
z = z + h*f(t, z);
end;
 
f: procedure (dummy, T) returns (float);
declare dummy fixed binary;
declare T float;
 
return ( -k*(T - Tr) );
end f;
 
end test;

Only the final two outputs are shown, for brevity.

H=        5;
  t    By formula       By Euler 
  0   100.0000000000   100.0000000000
  5    76.3750457764    72.0000000000
 10    59.7268257141    53.7999992371
 15    47.9950218201    41.9700012207
 20    39.7277565002    34.2805023193
 25    33.9019165039    29.2823257446
 30    29.7965145111    26.0335121155
 35    26.9034862518    23.9217834473
 40    24.8648052216    22.5491600037
 45    23.4281692505    21.6569538116
 50    22.4157905579    21.0770206451
 55    21.7023792267    20.7000637054
 60    21.1996459961    20.4550418854
 65    20.8453769684    20.2957763672
 70    20.5957260132    20.1922550201
 75    20.4198017120    20.1249656677
 80    20.2958297729    20.0812282562
 85    20.2084674835    20.0527992249
 90    20.1469039917    20.0343189240
 95    20.1035213470    20.0223064423
100    20.0729503632    20.0144996643
H=       10;
  t    By formula       By Euler 
  0   100.0000000000   100.0000000000
 10    59.7268257141    44.0000000000
 20    39.7277565002    27.2000007629
 30    29.7965145111    22.1599998474
 40    24.8648052216    20.6480007172
 50    22.4157905579    20.1944007874
 60    21.1996459961    20.0583209991
 70    20.5957260132    20.0174961090
 80    20.2958297729    20.0052490234
 90    20.1469039917    20.0015754700
100    20.0729503632    20.0004730225

[edit] PureBasic

Define.d
Prototype.d Func(Time, t)
 
Procedure.d Euler(*F.Func, y0, a, b, h)
Protected y=y0, t=a
While t<=b
PrintN(RSet(StrF(t,3),7)+" "+RSet(StrF(y,3),7))
y + h * *F(t,y)
t + h
Wend
EndProcedure
 
Procedure.d newtonCoolingLaw(Time, t)
ProcedureReturn -0.07*(t-20)
EndProcedure
 
 
If OpenConsole()
Euler(@newtonCoolingLaw(), 100, 0, 100, 2)
Euler(@newtonCoolingLaw(), 100, 0, 100, 5)
Euler(@newtonCoolingLaw(), 100, 0, 100,10)
 
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf
...
 85.000  20.053
 90.000  20.034
 95.000  20.022
100.000  20.014
  0.000 100.000
 10.000  44.000
 20.000  27.200
 30.000  22.160
 40.000  20.648
 50.000  20.194
 60.000  20.058
 70.000  20.017
 80.000  20.005
 90.000  20.002
100.000  20.000

[edit] Python

Translation of: Common Lisp
def euler(f,y0,a,b,h):
t,y = a,y0
while t < b:
print "%6.3f %6.3f" % (t,y)
t += h
y += h * f(t,y)
 
def newtoncooling(time, temp):
return -0.07 * (temp - 20)
 
euler(newtoncooling,100,0,100,10)
 

Output:

 0.000 100.000
10.000 44.000
20.000 27.200
30.000 22.160
40.000 20.648
50.000 20.194
60.000 20.058
70.000 20.017
80.000 20.005
90.000 20.002


[edit] Racket

The ODE solver:

 
(define (ODE-solve f init
#:x-max x-max
#:step h
#:method (method euler))
(reverse
(iterate-while (λ (x . y) (<= x x-max)) (method f h) init)))
 

It uses the default integration method euler, defined separately.

 
(define (euler F h)
(λ (x y) (list (+ x h) (+ y (* h (F x y))))))
 

A general-purpose procedure which evalutes a given function f repeatedly starting with argument x, while all results satisfy a predicate test. Returns a list of iterations.

 
(define (iterate-while test f x)
(let next ([result x]
[list-of-results '()])
(if (apply test result)
(next (apply f result) (cons result list-of-results))
list-of-results)))
 

Textual output:

 
> (define (newton-cooling t T)
(* -0.07 (- T 20)))
> (ODE-solve newton-cooling '(0 100) #:x-max 100 #:step 10)
'((0 100)
(10 44.)
(20 27.2)
(30 22.16)
(40 20.648)
(50 20.1944)
(60 20.05832)
(70 20.017496)
(80 20.0052488)
(90 20.00157464)
(100 20.000472392))
 

Plotting results:

 
> (require plot)
> (plot
(map (λ (h c)
(lines
(ODE-solve newton-cooling '(0 100) #:x-max 100 #:step h)
#:color c #:label (format "h=~a" h)))
'(10 5 1)
'(red blue black))
#:legend-anchor 'top-right)
 

Euler1.jpg

High modularity of the program allows to implement very different solution metods. For example 2-nd order Runge-Kutta method:

 
(define (RK2 F h)
(λ (x y)
(list (+ x h) (+ y (* h (F (+ x (* 1/2 h))
(+ y (* 1/2 h (F x y)))))))))
 

Two-step Adams–Bashforth method

 
(define (adams F h)
(case-lambda
 ; first step using Runge-Kutta method
[(x y) (append ((RK2 F h) x y) (list (F x y)))]
[(x y f′)
(let ([f (F x y)])
(list (+ x h) (+ y (* 3/2 h f) (* -1/2 h f′)) f))]))
 

Adaptive one-step method modifier using absolute accuracy ε

 
(define ((adaptive method ε) F h0)
(case-lambda
[(x y) (((adaptive method ε) F h0) x y h0)]
[(x y h)
(match-let* ([(list x0 y0) ((method F h) x y)]
[(list x1 y1) ((method F (/ h 2)) x y)]
[(list x1 y1) ((method F (/ h 2)) x1 y1)]
[τ (abs (- y1 y0))]
[h′ (if (< τ ε) (min h h0) (* 0.9 h (/ ε τ)))])
(list x1 (+ y1 τ) (* 2 h′)))]))
 
 

Comparison of different integration methods

 
> (define (solve-newton-cooling-by m)
(ODE-solve newton-cooling '(0 100)
#:x-max 100 #:step 10 #:method m))
> (plot
(list
(function (λ (t) (+ 20 (* 80 (exp (* -0.07 t))))) 0 100
#:color 'black #:label "analytical")
(lines (solve-newton-cooling-by euler)
#:color 'red #:label "Euler")
(lines (solve-newton-cooling-by RK2)
#:color 'blue #:label "Runge-Kutta")
(lines (solve-newton-cooling-by adams)
#:color 'purple #:label "Adams")
(points (solve-newton-cooling-by (adaptive euler 0.5))
#:color 'red #:label "Adaptive Euler")
(points (solve-newton-cooling-by (adaptive RK2 0.5))
#:color 'blue #:label "Adaptive Runge-Kutta"))
#:legend-anchor 'top-right)
 

Euler2.jpg

See also Runge-Kutta method#Racket

[edit] REXX

Translation of: PLI
/* REXX ***************************************************************
* 24.05.2013 Walter Pachl translated from PL/I
**********************************************************************/

Numeric Digits 100
T0=100
Tr=20
k=0.07
 
h=2
x=t0
Call head
do t=0 to 100 by 2
Select
When t<=4 | t>=96 Then
call o x
When t=8 Then
Say '...'
Otherwise
Nop
End
x=x+h*f(x)
end
 
h=5
y=t0
Call head
do t=0 to 100 by 5
call o y
y=y+h*f(y)
end
 
h=10
z=t0
Call head
do t=0 to 100 by 10
call o z
z=z+h*f(z)
end
Exit
 
f: procedure Expose k Tr
Parse Arg t
return -k*(T-Tr)
 
head:
Say 'h='h
Say ' t By formula By Euler'
Return
 
o:
Parse Arg v
Say right(t,3) format(Tr+(T0-Tr)/exp(k*t),5,10) format(v,5,10)
Return

Output:

h=2
  t    By formula       By Euler
  0   100.0000000000   100.0000000000
  2    89.5486587628    88.8000000000
  4    80.4626994233    79.1680000000
...
 96    20.0965230572    20.0574137147
 98    20.0839131147    20.0493757946
100    20.0729505571    20.0424631834
h=5
  t    By formula       By Euler
  0   100.0000000000   100.0000000000
  5    76.3750471216    72.0000000000
 10    59.7268242534    53.8000000000
 15    47.9950199099    41.9700000000
 20    39.7277571000    34.2805000000
 25    33.9019154664    29.2823250000
 30    29.7965142633    26.0335112500
 35    26.9034869314    23.9217823125
 40    24.8648050015    22.5491585031
 45    23.4281701466    21.6569530270
 50    22.4157906708    21.0770194676
 55    21.7023789162    20.7000626539
 60    21.1996461464    20.4550407250
 65    20.8453763508    20.2957764713
 70    20.5957266443    20.1922547063
 75    20.4198014729    20.1249655591
 80    20.2958290978    20.0812276134
 85    20.2084672415    20.0527979487
 90    20.1469043822    20.0343186667
 95    20.1035217684    20.0223071333
100    20.0729505571    20.0144996367
h=10
  t    By formula       By Euler
  0   100.0000000000   100.0000000000
 10    59.7268242534    44.0000000000
 20    39.7277571000    27.2000000000
 30    29.7965142633    22.1600000000
 40    24.8648050015    20.6480000000
 50    22.4157906708    20.1944000000
 60    21.1996461464    20.0583200000
 70    20.5957266443    20.0174960000
 80    20.2958290978    20.0052488000
 90    20.1469043822    20.0015746400
100    20.0729505571    20.0004723920

[edit] Ruby

Translation of: Python
def euler(y, a, b, h)
a.step(b,h) do |t|
puts "%7.3f %7.3f" % [t,y]
y += h * yield(t,y)
end
end
 
[10, 5, 2].each do |step|
puts "Step = #{step}"
euler(100,0,100,step) {|time, temp| -0.07 * (temp - 20) }
puts
end
Output:
Step = 10
  0.000 100.000
 10.000  44.000
 20.000  27.200
 30.000  22.160
 40.000  20.648
 50.000  20.194
 60.000  20.058
 70.000  20.017
 80.000  20.005
 90.000  20.002
100.000  20.000

Step = 5
  0.000 100.000
  5.000  72.000
 10.000  53.800
 15.000  41.970
 20.000  34.280
 25.000  29.282
 30.000  26.034
 35.000  23.922
 40.000  22.549
 45.000  21.657
 50.000  21.077
 55.000  20.700
 60.000  20.455
 65.000  20.296
 70.000  20.192
 75.000  20.125
 80.000  20.081
 85.000  20.053
 90.000  20.034
 95.000  20.022
100.000  20.014

Step = 2
  0.000 100.000
  2.000  88.800
  4.000  79.168
  6.000  70.884
  8.000  63.761
 10.000  57.634
 12.000  52.365
 14.000  47.834
 16.000  43.937
 18.000  40.586
 20.000  37.704
 22.000  35.226
 24.000  33.094
 26.000  31.261
 28.000  29.684
 30.000  28.328
 32.000  27.163
 34.000  26.160
 36.000  25.297
 38.000  24.556
 40.000  23.918
 42.000  23.369
 44.000  22.898
 46.000  22.492
 48.000  22.143
 50.000  21.843
 52.000  21.585
 54.000  21.363
 56.000  21.172
 58.000  21.008
 60.000  20.867
 62.000  20.746
 64.000  20.641
 66.000  20.551
 68.000  20.474
 70.000  20.408
 72.000  20.351
 74.000  20.302
 76.000  20.259
 78.000  20.223
 80.000  20.192
 82.000  20.165
 84.000  20.142
 86.000  20.122
 88.000  20.105
 90.000  20.090
 92.000  20.078
 94.000  20.067
 96.000  20.057
 98.000  20.049
100.000  20.042

[edit] Run BASIC

x = euler(-0.07,-20, 100, 0, 100, 2)
x = euler-0.07,-20, 100, 0, 100, 5)
x = euler(-0.07,-20, 100, 0, 100, 10)
end
 
FUNCTION euler(da,db, y, a, b, s)
print "===== da:";da;" db:";db;" y:";y;" a:";a;" b:";b;" s:";s;" ==================="
t = a
WHILE t <= b
PRINT t;chr$(9);y
y = y + s * (da * (y + db))
t = t + s
WEND
END FUNCTION
===== da:-0.07 db:-20 y:100 a:0 b:100 s:2 ===================
0	100
2	88.8
4	79.168
6	70.88448
8	63.7606528
10	57.6341614
12	52.3653788
14	47.8342258
......
===== da:-0.07 db:-20 y:100 a:0 b:100 s:10 ===================
0	100
10	44.0
20	27.2
30	22.16
40	20.648
50	20.1944
60	20.05832
70	20.017496
80	20.0052488

[edit] Scala

 
object App{
 
def main(args : Array[String]) = {
 
def cooling( step : Int ) = {
eulerStep( (step , y) => {-0.07 * (y - 20)} ,
100.0,0,100,step)
}
cooling(10)
cooling(5)
cooling(2)
}
def eulerStep( func : (Int,Double) => Double,y0 : Double,
begin : Int, end : Int , step : Int) = {
 
println("Step size: %s".format(step))
 
var current : Int = begin
var y : Double = y0
while( current <= end){
println( "%d %.5f".format(current,y))
current += step
y += step * func(current,y)
}
 
println("DONE")
}
 
}
 

Output for step = 10;

Step size: 10
0 100.00000
10 44.00000
20 27.20000
30 22.16000
40 20.64800
50 20.19440
60 20.05832
70 20.01750
80 20.00525
90 20.00157
DONE

[edit] Smalltalk

ODESolver>>eulerOf: f init: y0 from: a to: b step: h
| t y |
t := a.
y := y0.
[ t < b ]
whileTrue: [
Transcript
show: t asString, ' ' , (y printShowingDecimalPlaces: 3);
cr.
t := t + h.
y := y + (h * (f value: t value: y)) ]
 
ODESolver new eulerOf: [:time :temp| -0.07 * (temp - 20)] init: 100 from: 0 to: 100 step: 10
 

Transcript:

0 100.000
10 44.000
20 27.200
30 22.160
40 20.648
50 20.194
60 20.058
70 20.017
80 20.005
90 20.002

[edit] Tcl

Translation of: C++
proc euler {f y0 a b h} {
puts "computing $f over \[$a..$b\], step $h"
set y [expr {double($y0)}]
for {set t [expr {double($a)}]} {$t < $b} {set t [expr {$t + $h}]} {
puts [format "%.3f\t%.3f" $t $y]
set y [expr {$y + $h * double([$f $t $y])}]
}
puts "done"
}

Demonstration with the Newton Cooling Law:

proc newtonCoolingLaw {time temp} {
expr {-0.07 * ($temp - 20)}
}
 
euler newtonCoolingLaw 100 0 100 2
euler newtonCoolingLaw 100 0 100 5
euler newtonCoolingLaw 100 0 100 10

End of output:

...
computing newtonCoolingLaw over [0..100], step 10
0.000	100.000
10.000	44.000
20.000	27.200
30.000	22.160
40.000	20.648
50.000	20.194
60.000	20.058
70.000	20.017
80.000	20.005
90.000	20.002
done

[edit] XPL0

include c:\cxpl\codes;  \intrinsic 'code' declarations
 
proc Euler(Step); \Display cooling temperatures using Euler's method
int Step;
int Time; real Temp;
[Text(0, "Step "); IntOut(0, Step); Text(0, " ");
Time:= 0; Temp:= 100.0;
repeat if rem(Time/10) = 0 then RlOut(0, Temp);
Temp:= Temp + float(Step) * (-0.07*(Temp-20.0));
Time:= Time + Step;
until Time > 100;
CrLf(0);
];
 
real Time, Temp;
[Format(6,0); \display time heading
Text(0, "Time ");
Time:= 0.0;
while Time <= 100.1 do \(.1 avoids possible rounding error)
[RlOut(0, Time);
Time:= Time + 10.0;
];
CrLf(0);
 
Format(3,2); \display cooling temps using differential eqn.
Text(0, "Dif eq "); \ dTemp(time)/dtime = -k*Temp
Time:= 0.0;
while Time <= 100.1 do
[Temp:= 20.0 + (100.0-20.0) * Exp(-0.07*Time);
RlOut(0, Temp);
Time:= Time + 10.0;
];
CrLf(0);
 
Euler(2); \display cooling temps for various time steps
Euler(5);
Euler(10);
]

Output:

Time         0    10    20    30    40    50    60    70    80    90   100
Dif eq  100.00 59.73 39.73 29.80 24.86 22.42 21.20 20.60 20.30 20.15 20.07
Step 2  100.00 57.63 37.70 28.33 23.92 21.84 20.87 20.41 20.19 20.09 20.04
Step 5  100.00 53.80 34.28 26.03 22.55 21.08 20.46 20.19 20.08 20.03 20.01
Step 10 100.00 44.00 27.20 22.16 20.65 20.19 20.06 20.02 20.01 20.00 20.00
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