Euler method: Difference between revisions
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=={{header|Maple}}== |
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Build-in function Euler: |
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<lang Maple> |
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with(Student[NumericalAnalysis]); |
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k := 0.07: |
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TR := 20: |
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Euler(diff(T(t), t) = -k*(T(t) - TR), T(0) = 100, t = 100, numsteps = 50); |
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Euler(diff(T(t), t) = -k*(T(t) - TR), T(0) = 100, t = 100, numsteps = 20); |
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Euler(diff(T(t), t) = -k*(T(t) - TR), T(0) = 100, t = 100, numsteps = 10);</lang> |
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{{out}} |
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<pre> |
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20.04 |
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20.01 |
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20.00 |
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</pre> |
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=={{header|Mathematica}} / {{header|Wolfram Language}}== |
=={{header|Mathematica}} / {{header|Wolfram Language}}== |
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Revision as of 18:52, 5 January 2021
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:
with an initial value
To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
then solve for :
which is the same as
The iterative solution rule is then:
where 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 cools down in an environment of temperature :
or
It says that the cooling rate of the object is proportional to the current temperature difference to the surrounding environment.
The analytical solution, which we will compare to the numerical approximation, is
- Task
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.
- Initial values
-
- initial temperature shall be 100 °C
- room temperature shall be 20 °C
- cooling constant shall be 0.07
- time interval to calculate shall be from 0 s ──► 100 s
A reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
11l
<lang 11l>F euler(f, y0, a, b, h)
V t = a
V y = y0
L t <= b
print(‘#2.3 #2.3’.format(t, y))
t += h
y += h * f(t, y)
V newtoncooling = (time, temp) -> -0.07 * (temp - 20)
euler(newtoncooling, 100.0, 0.0, 100.0, 10.0)</lang>
- 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
Ada
The solution is generic, usable for any floating point type. The package specification: <lang Ada> 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; </lang> The function Solve returns the solution of the differential equation for each of N+1 points, starting from the point T0. The implementation: <lang Ada> 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; </lang> The test program: <lang Ada> 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; </lang> 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
ALGOL 68
Note: This specimen retains the original D coding style.
<lang algol68># 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)
)</lang> 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
BASIC
BBC BASIC
<lang bbcbasic> 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</lang>
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
FreeBASIC
<lang 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 </lang> 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 ________________
Run BASIC
<lang rinbasic>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</lang>
===== 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
C
<lang 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; }</lang>output<lang> 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</lang>
C#
<lang csharp>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); } } } }</lang>
C++
<lang cpp>#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);
}</lang> 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
Clay
<lang 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);
}
} </lang>
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
Clojure
<lang lisp>(ns newton-cooling
(:gen-class))
(defn euler [f y0 a b h]
"Euler's Method.
Approximates y(time) in y'(time)=f(time,y) with y(a)=y0 and t=a..b and the step size h."
(loop [t a
y y0
result []]
(if (<= t b)
(recur (+ t h) (+ y (* (f (+ t h) y) h)) (conj result [(double t) (double y)]))
result)))
(defn newton-coolling [t temp]
"Newton's cooling law, f(t,T) = -0.07*(T-20)" (* -0.07 (- temp 20)))
- Run for case h = 10
(println "Example output") (doseq [q (euler newton-coolling 100 0 100 10)]
(println (apply format "%.3f %.3f" q)))
</lang>
- Output:
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 20.002 100.000 20.000
COBOL
The following is in the Managed COBOL dialect: <lang cobol> 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.</lang>
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
Common Lisp
<lang 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 (+ t h))
(y y0 (+ 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)</lang>
<lang lisp>;; slightly more idiomatic Common Lisp version
(defun newton-cooling (time temperature)
"Newton's cooling law, f(t,T) = -0.07*(T-20)" (declare (ignore time)) (* -0.07 (- temperature 20)))
(defun euler (f y0 a b h)
"Euler's Method.
Approximates y(time) in y'(time)=f(time,y) with y(a)=y0 and t=a..b and the step size h."
(loop for time from a below b by h
for y = y0 then (+ y (* h (funcall f time y)))
do (format t "~6,3F ~6,3F~%" time y)))</lang>
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
D
<lang d>import std.stdio, std.range, std.traits;
/// 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) @safe if (isCallable!F && __traits(compiles, { real r = f(0.0, 0.0); })) {
double y = y0;
foreach (immutable t; iota(a, b, h)) {
writefln("%.3f %.3f", t, y);
y += h * f(t, y);
}
"done".writeln;
}
void main() {
/// Example: Newton's cooling law.
enum newtonCoolingLaw = (in double time, in double t)
pure nothrow @safe @nogc => -0.07 * (t - 20);
euler(newtonCoolingLaw, 100, 0, 100, 2); euler(newtonCoolingLaw, 100, 0, 100, 5); euler(newtonCoolingLaw, 100, 0, 100, 10);
}</lang> 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
Delphi
Elixir
<lang elixir>defmodule Euler do
def method(_, _, t, b, _) when t>b, do: :ok def method(f, y, t, b, h) do :io.format "~7.3f ~7.3f~n", [t,y] method(f, y + h * f.(t,y), t + h, b, h) end
end
f = fn _time, temp -> -0.07 * (temp - 20) end Enum.each([10, 5, 2], fn step ->
IO.puts "\nStep = #{step}"
Euler.method(f, 100.0, 0.0, 100.0, step)
end)</lang>
- 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
Erlang
<lang erlang> -module(euler). -export([main/0, euler/5]).
cooling(_Time, Temperature) -> (-0.07)*(Temperature-20).
euler(_, Y, T, _, End) when End == T -> io:fwrite("\n"), Y;
euler(Func, Y, T, Step, End) -> if T rem 10 == 0 -> io:fwrite("~.3f ",[float(Y)]); true -> ok end, euler(Func, Y + Step * Func(T, Y), T + Step, Step, End).
analytic(T, End) when T == End -> io:fwrite("\n"), T;
analytic(T, End) -> Y = (20 + 80 * math:exp(-0.07 * T)), io:fwrite("~.3f ", [Y]), analytic(T+10, End).
main() -> io:fwrite("Analytic:\n"), analytic(0, 100), io:fwrite("Step 2:\n"), euler(fun cooling/2, 100, 0, 2, 100), io:fwrite("Step 5:\n"), euler(fun cooling/2, 100, 0, 5, 100), io:fwrite("Step 10:\n"), euler(fun cooling/2, 100, 0, 10, 100), ok. </lang>
- Output:
Analytic: 100.000 59.727 39.728 29.797 24.865 22.416 21.200 20.596 20.296 20.147 Step 2: 100.000 57.634 37.704 28.328 23.918 21.843 20.867 20.408 20.192 20.090 Step 5: 100.000 53.800 34.280 26.034 22.549 21.077 20.455 20.192 20.081 20.034 Step 10: 100.000 44.000 27.200 22.160 20.648 20.194 20.058 20.017 20.005 20.002 ok
Euler Math Toolbox
<lang 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
</lang>
F#
<lang fsharp>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</lang>
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)
Factor
<lang factor>USING: formatting fry io kernel locals math math.ranges sequences ; IN: rosetta-code.euler-method
- euler ( quot y! a b h -- )
a b h <range> [
:> t
t y "%7.3f %7.3f\n" printf
t y quot call h * y + y!
] each ; inline
- cooling ( t y -- x ) nip 20 - -0.07 * ;
- euler-method-demo ( -- )
2 5 10 [ '[ [ cooling ] 100 0 100 _ euler ] call nl ] tri@ ;
MAIN: euler-method-demo</lang>
- 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
Forth
<lang 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</lang>
Fortran
<lang fortran>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</lang>
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
Futhark
Specialised to the cooling function. We produce an array of the temperature at each step subtracted from the analytically determined temperature (so we are computing the error).
<lang Futhark> let analytic(t0: f64) (time: f64): f64 =
20.0 + (t0 - 20.0) * f64.exp(-0.07*time)
let cooling(_time: f64) (temperature: f64): f64 =
-0.07 * (temperature-20.0)
let main(t0: f64) (a: f64) (b: f64) (h: f64): []f64 =
let steps = i32.f64 ((b-a)/h)
let temps = replicate steps 0.0
let (_,temps) = loop (t,temps)=(t0,temps) for i < steps do
let x = a + f64.i32 i * h
let temps[i] = f64.abs(t-analytic t0 x)
in (t + h * cooling x t,
temps)
in temps
</lang>
Go
<lang 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()
}
}</lang> 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
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.
<lang groovy>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</lang>
Specific Euler Method Solution for the "Temperature Diffusion" Problem (with Newton's derivative formula and constants for environment temperature and object conductivity given)
<lang groovy>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</lang>
Newton's Analytic Temperature Diffusion Solution (for comparison)
<lang groovy>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</lang>
Specific solutions for 3 step sizes (and initial time and temperature)
<lang groovy>[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)
}
}</lang>
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.
Haskell
Modular solution which separates the solver and a method. Moreover it works on a given mesh which can be irregular. <lang Haskell>-- the solver dsolveBy _ _ [] _ = error "empty solution interval" dsolveBy method f mesh x0 = zip mesh results
where results = scanl (method f) x0 intervals
intervals = zip mesh (tail mesh)</lang>
It is better to use strict Data.List.scanl' in the solver but avoiding highlighting problems we leave lazy scanl function.
Some possible methods:
<lang haskell>-- 1-st order Euler euler f x (t1,t2) = x + (t2 - t1) * f t1 x
-- 2-nd order Runge-Kutta rk2 f x (t1,t2) = x + h * f (t1 + h/2) (x + h/2*f t1 x)
where h = t2 - t1
-- 4-th order Runge-Kutta rk4 f x (t1,t2) = x + h/6 * (k1 + 2*k2 + 2*k3 + k4)
where k1 = f t1 x
k2 = f (t1 + h/2) (x + h/2*k1)
k3 = f (t1 + h/2) (x + h/2*k2)
k4 = f (t1 + h) (x + h*k3)
h = t2 - t1</lang>
Graphical output, using EasyPlot:
<lang haskell>import Graphics.EasyPlot
newton t temp = -0.07 * (temp - 20)
exactSolution t = 80*exp(-0.07*t)+20
test1 = plot (PNG "euler1.png")
[ Data2D [Title "Step 10", Style Lines] [] sol1
, Data2D [Title "Step 5", Style Lines] [] sol2
, Data2D [Title "Step 1", Style Lines] [] sol3
, Function2D [Title "exact solution"] [Range 0 100] exactSolution ]
where sol1 = dsolveBy euler newton [0,10..100] 100
sol2 = dsolveBy euler newton [0,5..100] 100
sol3 = dsolveBy euler newton [0,1..100] 100
test2 = plot (PNG "euler2.png")
[ Data2D [Title "Euler"] [] sol1
, Data2D [Title "RK2"] [] sol2
, Data2D [Title "RK4"] [] sol3
, Function2D [Title "exact solution"] [Range 0 100] exactSolution ]
where sol1 = dsolveBy euler newton [0,10..100] 100
sol2 = dsolveBy rk2 newton [0,10..100] 100
sol3 = dsolveBy rk4 newton [0,10..100] 100</lang>
Icon and Unicon
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.
<lang Icon> 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 </lang>
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
J
Solution: <lang j>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 </lang> Example: <lang j> 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</lang>
Java
<lang 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);
}
} </lang>
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
JavaScript
<lang javascript> // Function that takes differential-equation, initial condition, // ending x, and step size as parameters function eulersMethod(f, x1, y1, x2, h) { // Header console.log("\tX\t|\tY\t"); console.log("------------------------------------");
// Initial Variables var x=x1, y=y1;
// While we're not done yet // Both sides of the OR let you do Euler's Method backwards while ((x<x2 && x1<x2) || (x>x2 && x1>x2)) { // Print what we have console.log("\t" + x + "\t|\t" + y);
// Calculate the next values y += h*f(x, y) x += h; }
return y; }
function cooling(x, y) { return -0.07 * (y-20); }
eulersMethod(cooling, 0, 100, 100, 10); </lang>
jq
<lang jq># euler_method takes a filter (df), initial condition
- (x1,y1), ending x (x2), and step size as parameters;
- it emits the y values at each iteration.
- df must take [x,y] as its input.
def euler_method(df; x1; y1; x2; h):
h as $h
| [x1, y1]
| recurse( if ((.[0] < x2 and x1 < x2) or
(.[0] > x2 and x1 > x2)) then
[ (.[0] + $h), (.[1] + $h*df) ]
else empty
end )
| .[1] ;
- We could now solve the task by writing for each step-size, $h
- euler_method(-0.07 * (.[1]-20); 0; 100; 100; $h)
- but for clarity, we shall define a function named "cooling":
- [x,y] is input
def cooling: -0.07 * (.[1]-20);
- The following solves the task:
- (2,5,10) | [., [ euler_method(cooling; 0; 100; 100; .) ] ]
</lang> For brevity, we modify euler_method so that it only shows the final value of y: <lang jq> def euler_solution(df; x1; y1; x2; h):
def recursion(exp): reduce recurse(exp) as $x (.; $x);
h as $h
| [x1, y1]
| recursion( if ((.[0] < x2 and x1 < x2) or
(.[0] > x2 and x1 > x2)) then
[ (.[0] + $h), (.[1] + $h*df) ]
else empty
end )
| .[1] ;</lang>
Example: <lang jq>(1,2,5,10,20) | [., [ euler_solution(cooling; 0; 100; 100; .) ] ]</lang>
- Output:
<lang sh>$ jq -M -n -c -f Euler_method.jq [1,[20.05641373347389]] [2,[20.0424631833732]] [5,[20.01449963666907]] [10,[20.000472392]] [20,[19.180799999999998]]</lang>
Julia
<lang julia>euler(f::Function, T::Number, t0::Int, t1::Int, h::Int) = collect(begin T += h * f(T); T end for t in t0:h:t1)
- Prints a series of arbitrary values in a tabular form, left aligned in cells with a given width
tabular(width, cells...) = println(join(map(s -> rpad(s, width), cells)))
- prints the table according to the task description for h=5 and 10 sec
for h in (5, 10)
print("Step $h:\n\n")
tabular(15, "Time", "Euler", "Analytic")
t = 0
for T in euler(y -> -0.07 * (y - 20.0), 100.0, 0, 100, h)
tabular(15, t, round(T,digits=6), round(20.0 + 80.0 * exp(-0.07t), digits=6))
t += h
end
println()
end</lang>
- Output:
Step 5: Time Euler Analytic 0 72.0 100.0 5 53.8 76.375047 10 41.97 59.726824 15 34.2805 47.99502 20 29.282325 39.727757 25 26.033511 33.901915 30 23.921782 29.796514 35 22.549159 26.903487 40 21.656953 24.864805 45 21.077019 23.42817 50 20.700063 22.415791 55 20.455041 21.702379 60 20.295776 21.199646 65 20.192255 20.845376 70 20.124966 20.595727 75 20.081228 20.419801 80 20.052798 20.295829 85 20.034319 20.208467 90 20.022307 20.146904 95 20.0145 20.103522 100 20.009425 20.072951 Step 10: Time Euler Analytic 0 44.0 100.0 10 27.2 59.726824 20 22.16 39.727757 30 20.648 29.796514 40 20.1944 24.864805 50 20.05832 22.415791 60 20.017496 21.199646 70 20.005249 20.595727 80 20.001575 20.295829 90 20.000472 20.146904 100 20.000142 20.072951
Kotlin
<lang scala>// version 1.1.2
typealias Deriv = (Double) -> Double // only one parameter needed here
const val FMT = " %7.3f"
fun euler(f: Deriv, y: Double, step: Int, end: Int) {
var yy = y
print(" Step %2d: ".format(step))
for (t in 0..end step step) {
if (t % 10 == 0) print(FMT.format(yy))
yy += step * f(yy)
}
println()
}
fun analytic() {
print(" Time: ")
for (t in 0..100 step 10) print(" %7d".format(t))
print("\nAnalytic: ")
for (t in 0..100 step 10)
print(FMT.format(20.0 + 80.0 * Math.exp(-0.07 * t)))
println()
}
fun cooling(temp: Double) = -0.07 * (temp - 20.0)
fun main(args: Array<String>) {
analytic()
for (i in listOf(2, 5, 10))
euler(::cooling, 100.0, i, 100)
}</lang>
- 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
Lambdatalk
Translation of Python <lang scheme> {def eulersMethod
{def eulersMethod.r
{lambda {:f :b :h :t :y}
{if {<= :t :b}
then {tr {td :t} {td {/ {round {* :y 1000}} 1000}}}
{eulersMethod.r :f :b :h {+ :t :h} {+ :y {* :h {:f :t :y}}}}
else}}}
{lambda {:f :y0 :a :b :h}
{table {eulersMethod.r :f :b :h :a :y0}}}}
{def cooling
{lambda {:time :temp}
{* -0.07 {- :temp 20}}}}
{eulersMethod cooling 100 0 100 10} ->
0 100 10 44 20 27.2 30 22.16 40 20.648 50 20.194 60 20.058 70 20.017 80 20.005 90 20.002 100 20
</lang>
Lua
<lang 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 </lang>
Maple
Build-in function Euler: <lang Maple> with(Student[NumericalAnalysis]); k := 0.07: TR := 20: Euler(diff(T(t), t) = -k*(T(t) - TR), T(0) = 100, t = 100, numsteps = 50); Euler(diff(T(t), t) = -k*(T(t) - TR), T(0) = 100, t = 100, numsteps = 20); Euler(diff(T(t), t) = -k*(T(t) - TR), T(0) = 100, t = 100, numsteps = 10);</lang>
- Output:
20.04
20.01
20.00
Mathematica / Wolfram Language
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... <lang Mathematica> 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] </lang>
- Output:
euler[2, 100] 20.0425 euler[5, 100] 20.0145 euler[10, 100] 20.0005
Maxima
<lang 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"]);
</lang>
МК-61/52
<lang>П2 С/П П3 С/П П4 ПП 19 ИП3 * ИП4 + П4 С/П ИП2 ИП3 + П2 БП 05 ... ... ... ... ... ... ... ... ... ... В/О</lang>
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.
Nim
<lang nim>import strutils
proc euler(f: proc (x,y: float): float; y0, a, b, h: float) =
var (t,y) = (a,y0) while t < b: echo formatFloat(t, ffDecimal, 3), " ", formatFloat(y, ffDecimal, 3) t += h y += h * f(t,y)
proc newtoncooling(time, temp: float): float =
-0.07 * (temp - 20)
euler(newtoncooling, 100.0, 0.0, 100.0, 10.0)</lang>
- 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
Objeck
<lang 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);
};
}
} </lang>
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
OCaml
<lang 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)</lang>
Using the above functions to produce the task results: <lang OCaml>(* 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.)</lang>
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 = ()
Oforth
<lang oforth>: euler(f, y, a, b, h) | t |
a b h step: t [
System.Out t <<wjp(6, JUSTIFY_RIGHT, 3) " : " << y << cr
t y f perform h * y + ->y
] ;</lang>
Usage :
<lang oforth>: newtonCoolingLaw(t, y)
y 20 - -0.07 * ;
- test
euler(#newtonCoolingLaw, 100.0, 0.0, 100.0, 2) euler(#newtonCoolingLaw, 100.0, 0.0, 100.0, 5) euler(#newtonCoolingLaw, 100.0, 0.0, 100.0, 10) ;</lang>
- Output:
....
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
Pascal
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.
<lang 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.
</lang>
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
Perl
<lang 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;
} </lang>
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
Phix
<lang Phix>constant FMT = " %7.3f"
procedure ivp_euler(integer f, atom y, integer step, integer end_t) integer t = 0;
printf(1, " Step %2d: ", step);
while t<=end_t do
if remainder(t,10)==0 then printf(1, FMT, y) end if
y += step * call_func(f,{t, y});
t += step
end while
printf(1, "\n");
end procedure
procedure analytic()
printf(1, " Time: ");
for t = 0 to 100 by 10 do printf(1," %7g", t) end for
printf(1, "\nAnalytic: ");
for t = 0 to 100 by 10 do
printf(1, FMT, 20 + 80 * exp(-0.07 * t))
end for
printf(1,"\n");
end procedure
function cooling(atom /*t*/, atom temp)
return -0.07 * (temp - 20);
end function constant r_cooling = routine_id("cooling")
analytic(); ivp_euler(r_cooling, 100, 2, 100); ivp_euler(r_cooling, 100, 5, 100); ivp_euler(r_cooling, 100, 10, 100);</lang>
- 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
PicoLisp
<lang 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)</lang> 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
PL/I
<lang pli>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;</lang>
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
PowerShell
<lang PowerShell> function euler (${f}, ${y}, $y0, $t0, $tEnd) {
function f-euler ($tn, $yn, $h) {
$yn + $h*(f $tn $yn)
}
function time ($t0, $h, $tEnd) {
$end = [MATH]::Floor(($tEnd - $t0)/$h)
foreach ($_ in 0..$end) { $_*$h + $t0 }
}
$time = time $t0 10 $tEnd
$time5 = time $t0 5 $tEnd
$time2 = time $t0 2 $tEnd
$yn10 = $yn5 = $yn2 = $y0
$i2 = $i5 = 0
foreach ($tn10 in $time) {
while($time2[$i2] -ne $tn10) {
$i2++
$yn2 = (f-euler $time2[$i2] $yn2 2)
}
while($time5[$i5] -ne $tn10) {
$i5++
$yn5 = (f-euler $time5[$i5] $yn5 5)
}
[pscustomobject]@{
t = "$tn10"
Analytical = "$("{0:N5}" -f (y $tn10))"
"Euler h = 2" = "$("{0:N5}" -f $yn2)"
"Euler h = 5" = "$("{0:N5}" -f $yn5)"
"Euler h = 10" = "$("{0:N5}" -f $yn10)"
"Error h = 2" = "$("{0:N5}" -f [MATH]::abs($yn2 - (y $tn10)))"
"Error h = 5" = "$("{0:N5}" -f [MATH]::abs($yn5 - (y $tn10)))"
"Error h = 10" = "$("{0:N5}" -f [MATH]::abs($yn10 - (y $tn10)))"
}
$yn10 = (f-euler $tn10 $yn10 10)
}
} $k, $yr, $y0, $t0, $tEnd = 0.07, 20, 100, 0, 100 function f ($t, $y) {
-$k *($y - $yr)
} function y ($t) {
$yr + ($y0 - $yr)*[MATH]::Exp(-$k*$t)
} euler f y $y0 $t0 $tEnd | Format-Table -AutoSize </lang> Output:
t Analytical Euler h = 2 Euler h = 5 Euler h = 10 Error h = 2 Error h = 5 Error h = 10 - ---------- ----------- ----------- ------------ ----------- ----------- ------------ 0 100.00000 100.00000 100.00000 100.00000 0.00000 0.00000 0.00000 10 59.72682 57.63416 53.80000 44.00000 2.09266 5.92682 15.72682 20 39.72776 37.70413 34.28050 27.20000 2.02363 5.44726 12.52776 30 29.79651 28.32850 26.03351 22.16000 1.46801 3.76300 7.63651 40 24.86481 23.91795 22.54916 20.64800 0.94685 2.31565 4.21681 50 22.41579 21.84311 21.07702 20.19440 0.57268 1.33877 2.22139 60 21.19965 20.86705 20.45504 20.05832 0.33260 0.74461 1.14133 70 20.59573 20.40788 20.19225 20.01750 0.18784 0.40347 0.57823 80 20.29583 20.19188 20.08123 20.00525 0.10395 0.21460 0.29058 90 20.14690 20.09027 20.03432 20.00157 0.05664 0.11259 0.14533 100 20.07295 20.04246 20.01450 20.00047 0.03049 0.05845 0.07248
PureBasic
<lang PureBasic>Define.d Prototype.d Func(Time, t)
Procedure.d Euler(*F.Func, y0, a, b, h)
Protected y=y0, t=a While t<=b PrintN(RSet(StrF(t,3),7)+" "+RSet(StrF(y,3),7)) y + h * *F(t,y) t + h Wend
EndProcedure
Procedure.d newtonCoolingLaw(Time, t)
ProcedureReturn -0.07*(t-20)
EndProcedure
If OpenConsole()
Euler(@newtonCoolingLaw(), 100, 0, 100, 2) Euler(@newtonCoolingLaw(), 100, 0, 100, 5) Euler(@newtonCoolingLaw(), 100, 0, 100,10) Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input() CloseConsole()
EndIf</lang>
... 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
Python
<lang python>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) </lang> 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
R
<lang rsplus>euler <- function(f, y0, a, b, h) {
t <- a
y <- y0
while (t < b)
{
cat(sprintf("%6.3f %6.3f\n", t, y))
t <- t + h
y <- y + h*f(t, y)
}
}
newtoncooling <- function(time, temp)
return(-0.07*(temp-20))
euler(newtoncooling, 100, 0, 100, 10)</lang> 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
Racket
The ODE solver: <lang racket> (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)))
</lang>
It uses the default integration method euler, defined separately.
<lang racket> (define (euler F h)
(λ (x y) (list (+ x h) (+ y (* h (F x y))))))
</lang>
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.
<lang racket> (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)))
</lang>
Textual output: <lang racket> > (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))
</lang>
Plotting results: <lang racket> > (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)
</lang>
High modularity of the program allows to implement very different solution metods. For example 2-nd order Runge-Kutta method:
<lang racket> (define (RK2 F h)
(λ (x y)
(list (+ x h) (+ y (* h (F (+ x (* 1/2 h))
(+ y (* 1/2 h (F x y)))))))))
</lang>
Two-step Adams–Bashforth method <lang racket> (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))]))
</lang>
Adaptive one-step method modifier using absolute accuracy ε <lang racket> (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′)))]))
</lang>
Comparison of different integration methods <lang racket> > (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)
</lang>
See also Runge-Kutta method#Racket
Raku
(formerly Perl 6) <lang perl6>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%%');
}</lang>
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%
REXX
version 1
<lang rexx>/* 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
exp: Procedure
Parse Arg x,prec If prec<9 Then prec=9 Numeric Digits (2*prec) Numeric Fuzz 3 o=1 u=1 r=1 Do i=1 By 1 ra=r o=o*x u=u*i r=r+(o/u) If r=ra Then Leave End Numeric Digits (prec) r=r+0 Return r
</lang> 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
version 2
This REXX version allows values to be specified via the command line (CL).
It also shows the percentage difference (analytic vs. Euler's method) for each calculation.
<lang rexx>/*REXX pgm solves example of Newton's cooling law via Euler's method (diff. step sizes).*/
e=2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138
numeric digits length(e) - length(.) /*use the number of decimal digits in E*/
parse arg Ti Tr cc tt ss /*obtain optional arguments from the CL*/
if Ti= | Ti="," then Ti= 100 /*given? Default: initial temp in ºC.*/
if Tr= | Tr="," then Tr= 20 /* " " room " " " */
if cc= | cc="," then cc= 0.07 /* " " cooling constant. */
if tt= | tt="," then tt= 100 /* " " total time seconds. */
if ss= | ss="," then ss= 2 5 10 /* " " the step sizes. */
@= '═' /*the character used in title separator*/
do sSize=1 for words(ss); say; say; say center('time in' , 11)
say center('seconds' , 11, @) center('Euler method', 16, @) ,
center('analytic', 18, @) center('difference' , 14, @)
$=Ti; inc= word(ss, sSize) /*the 1st value; obtain the increment.*/
do t=0 to Ti by inc /*step through calculations by the inc.*/
a= format(Tr + (Ti-Tr)/exp(cc*t),6,10) /*calculate the analytic (exact) value.*/
say center(t,11) format($,6,3) 'ºC ' a "ºC" format(abs(a-$)/a*100,6,2) '%'
$= $ + inc * cc * (Tr-$) /*calc. next value via Euler's method. */
end /*t*/
end /*sSize*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ exp: procedure expose e; arg x; ix= x%1; if abs(x-ix)>.5 then ix=ix+sign(x); x= x-ix; z=1
_=1; w=1; do j=1; _= _*x/j; z= (z+_)/1; if z==w then leave; w=z
end /*j*/; if z\==0 then z= e**ix * z; return z</lang>
- output when using the default inputs:
time in
══seconds══ ══Euler method══ ═════analytic═════ ══difference══
0 100.000 ºC 100.0000000000 ºC 0.00 %
2 88.800 ºC 89.5486588319 ºC 0.84 %
4 79.168 ºC 80.4626993165 ºC 1.61 %
6 70.884 ºC 72.5637455852 ºC 2.31 %
8 63.761 ºC 65.6967251079 ºC 2.95 %
10 57.634 ºC 59.7268243033 ºC 3.50 %
12 52.365 ºC 54.5368418743 ºC 3.98 %
14 47.834 ºC 50.0248879081 ºC 4.38 %
16 43.937 ºC 46.1023835698 ºC 4.70 %
18 40.586 ºC 42.6923221200 ºC 4.93 %
20 37.704 ºC 39.7277571153 ºC 5.09 %
22 35.226 ºC 37.1504881142 ºC 5.18 %
24 33.094 ºC 34.9099180832 ºC 5.20 %
26 31.261 ºC 32.9620600747 ºC 5.16 %
28 29.684 ºC 31.2686736737 ºC 5.07 %
30 28.328 ºC 29.7965142602 ºC 4.93 %
32 27.163 ºC 28.5166803503 ºC 4.75 %
34 26.160 ºC 27.4040462008 ºC 4.54 %
36 25.297 ºC 26.4367685400 ºC 4.31 %
38 24.556 ºC 25.5958577396 ºC 4.06 %
40 23.918 ºC 24.8648050100 ºC 3.81 %
42 23.369 ºC 24.2292582991 ºC 3.55 %
44 22.898 ºC 23.6767405319 ºC 3.29 %
46 22.492 ºC 23.1964046609 ºC 3.04 %
48 22.143 ºC 22.7788207156 ºC 2.79 %
50 21.843 ºC 22.4157906738 ºC 2.55 %
52 21.585 ºC 22.1001875173 ºC 2.33 %
54 21.363 ºC 21.8258153140 ºC 2.12 %
56 21.172 ºC 21.5872875795 ºC 1.92 %
58 21.008 ºC 21.3799215292 ºC 1.74 %
60 20.867 ºC 21.1996461456 ºC 1.57 %
62 20.746 ºC 21.0429222563 ºC 1.41 %
64 20.641 ºC 20.9066730524 ºC 1.27 %
66 20.551 ºC 20.7882236849 ºC 1.14 %
68 20.474 ºC 20.6852487518 ºC 1.02 %
70 20.408 ºC 20.5957266457 ºC 0.91 %
72 20.351 ºC 20.5178998655 ºC 0.81 %
74 20.302 ºC 20.4502405132 ºC 0.73 %
76 20.259 ºC 20.3914202980 ºC 0.65 %
78 20.223 ºC 20.3402844596 ºC 0.58 %
80 20.192 ºC 20.2958290973 ºC 0.51 %
82 20.165 ºC 20.2571814620 ºC 0.45 %
84 20.142 ºC 20.2235828220 ºC 0.40 %
86 20.122 ºC 20.1943735676 ºC 0.36 %
88 20.105 ºC 20.1689802617 ºC 0.32 %
90 20.090 ºC 20.1469043822 ºC 0.28 %
92 20.078 ºC 20.1277125344 ºC 0.25 %
94 20.067 ºC 20.1110279436 ºC 0.22 %
96 20.057 ºC 20.0965230571 ºC 0.19 %
98 20.049 ºC 20.0839131146 ºC 0.17 %
100 20.042 ºC 20.0729505572 ºC 0.15 %
time in
══seconds══ ══Euler method══ ═════analytic═════ ══difference══
0 100.000 ºC 100.0000000000 ºC 0.00 %
5 72.000 ºC 76.3750471775 ºC 5.73 %
10 53.800 ºC 59.7268243033 ºC 9.92 %
15 41.970 ºC 47.9950199289 ºC 12.55 %
20 34.281 ºC 39.7277571153 ºC 13.71 %
25 29.282 ºC 33.9019154760 ºC 13.63 %
30 26.034 ºC 29.7965142602 ºC 12.63 %
35 23.922 ºC 26.9034869199 ºC 11.08 %
40 22.549 ºC 24.8648050100 ºC 9.31 %
45 21.657 ºC 23.4281701494 ºC 7.56 %
50 21.077 ºC 22.4157906738 ºC 5.97 %
55 20.700 ºC 21.7023789151 ºC 4.62 %
60 20.455 ºC 21.1996461456 ºC 3.51 %
65 20.296 ºC 20.8453763507 ºC 2.64 %
70 20.192 ºC 20.5957266457 ºC 1.96 %
75 20.125 ºC 20.4198014719 ºC 1.44 %
80 20.081 ºC 20.2958290973 ºC 1.06 %
85 20.053 ºC 20.2084672415 ºC 0.77 %
90 20.034 ºC 20.1469043822 ºC 0.56 %
95 20.022 ºC 20.1035217684 ºC 0.40 %
100 20.014 ºC 20.0729505572 ºC 0.29 %
time in
══seconds══ ══Euler method══ ═════analytic═════ ══difference══
0 100.000 ºC 100.0000000000 ºC 0.00 %
10 44.000 ºC 59.7268243033 ºC 26.33 %
20 27.200 ºC 39.7277571153 ºC 31.53 %
30 22.160 ºC 29.7965142602 ºC 25.63 %
40 20.648 ºC 24.8648050100 ºC 16.96 %
50 20.194 ºC 22.4157906738 ºC 9.91 %
60 20.058 ºC 21.1996461456 ºC 5.38 %
70 20.017 ºC 20.5957266457 ºC 2.81 %
80 20.005 ºC 20.2958290973 ºC 1.43 %
90 20.002 ºC 20.1469043822 ºC 0.72 %
100 20.000 ºC 20.0729505572 ºC 0.36 %
Ring
<lang ring> decimals(3) see euler("return -0.07*(y-20)", 100, 0, 100, 2) + nl see euler("return -0.07*(y-20)", 100, 0, 100, 5) + nl see euler("return -0.07*(y-20)", 100, 0, 100, 10) + nl
func euler df, y, a, b, s
t = a
while t <= b
see "" + t + " " + y + nl
y += s * eval(df)
t += s
end
return y</lang>
Output:
0 100 2 88.800 4 79.168 6 70.884 8 63.761 10 57.634
Ruby
<lang ruby>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</lang>
- 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
Rust
<lang Rust>fn header() {
print!(" Time: ");
for t in (0..100).step_by(10) {
print!(" {:7}", t);
}
println!();
}
fn analytic() {
print!("Analytic: ");
for t in (0..=100).step_by(10) {
print!(" {:7.3}", 20.0 + 80.0 * (-0.07 * f64::from(t)).exp());
}
println!();
}
fn euler<F: Fn(f64) -> f64>(f: F, mut y: f64, step: usize, end: usize) {
print!(" Step {:2}: ", step);
for t in (0..=end).step_by(step) {
if t % 10 == 0 {
print!(" {:7.3}", y);
}
y += step as f64 * f(y);
}
println!();
}
fn main() {
header();
analytic();
for &i in &[2, 5, 10] {
euler(|temp| -0.07 * (temp - 20.0), 100.0, i, 100);
}
}</lang>
- Output:
Time: 0 10 20 30 40 50 60 70 80 90 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
Scala
<lang 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")
}
} </lang>
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
SequenceL
<lang sequencel>import <Utilities/Conversion.sl>; import <Utilities/Sequence.sl>;
T0 := 100.0; TR := 20.0; k := 0.07;
main(args(2)) := let results[i] := euler(newtonCooling, T0, 100, stringToInt(args[i]), 0, "delta_t = " ++ args[i]); in delimit(results, '\n');
newtonCooling(t) := -k * (t - TR);
euler: (float -> float) * float * int * int * int * char(1) -> char(1); euler(f, y, n, h, x, output(1)) := let newOutput := output ++ "\n\t" ++ intToString(x) ++ "\t" ++ floatToString(y, 3); newY := y + h * f(y); newX := x + h; in output when x > n else euler(f, newY, n, h, newX, newOutput);</lang> Based on C# version [1] but using tail recursion instead of looping.
- Output:
For step size 10:
main.exe 10
"delta_t = 10
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
100 20.000"
Sidef
<lang ruby>func euler_method(t0, t1, k, step_size) {
var results = 0, t0 for s in (step_size..100 -> by(step_size)) { t0 -= ((t0 - t1) * k * step_size) results << [s, t0] } return results;
}
func analytical(t0, t1, k, time) {
(t0 - t1) * exp(-time * k) + t1
}
var (T0, T1, k) = (100, 20, .07) var r2 = euler_method(T0, T1, k, 2).grep { _[0] %% 10 } var r5 = euler_method(T0, T1, k, 5).grep { _[0] %% 10 } var r10 = euler_method(T0, T1, k, 10).grep { _[0] %% 10 }
say "Time\t 2 err(%) 5 err(%) 10 err(%) Analytic" say "-"*76
r2.range.each { |i|
var an = analytical(T0, T1, k, r2[i][0])
printf("%4d\t#{'%9.3f' * 7}\n",
r2[i][0],
r2[i][1], ( r2[i][1] / an) * 100 - 100,
r5[i][1], ( r5[i][1] / an) * 100 - 100,
r10[i][1], (r10[i][1] / an) * 100 - 100,
an)
}</lang>
- 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 34.281 -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
Smalltalk
<lang 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 </lang> 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
Tcl
<lang tcl>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 $y])}]
} puts "done"
}</lang> Demonstration with the Newton Cooling Law: <lang tcl>proc newtonCoolingLaw {time temp} {
expr {-0.07 * ($temp - 20)}
}
euler newtonCoolingLaw 100 0 100 2 euler newtonCoolingLaw 100 0 100 5 euler newtonCoolingLaw 100 0 100 10</lang> 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
VBA
<lang vb>Private Sub ivp_euler(f As String, y As Double, step As Integer, end_t As Integer)
Dim t As Integer
Debug.Print " Step "; step; ": ",
Do While t <= end_t
If t Mod 10 = 0 Then Debug.Print Format(y, "0.000"),
y = y + step * Application.Run(f, y)
t = t + step
Loop
Debug.Print
End Sub
Sub analytic()
Debug.Print " Time: ",
For t = 0 To 100 Step 10
Debug.Print " "; t,
Next t
Debug.Print
Debug.Print "Analytic: ",
For t = 0 To 100 Step 10
Debug.Print Format(20 + 80 * Exp(-0.07 * t), "0.000"),
Next t
Debug.Print
End Sub
Private Function cooling(temp As Double) As Double
cooling = -0.07 * (temp - 20)
End Function
Public Sub euler_method()
Dim r_cooling As String r_cooling = "cooling" analytic ivp_euler r_cooling, 100, 2, 100 ivp_euler r_cooling, 100, 5, 100 ivp_euler r_cooling, 100, 10, 100
End Sub</lang>
- 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 34,281 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
XPL0
<lang 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); ]</lang>
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
Wren
<lang ecmascript>import "/fmt" for Fmt import "/trait" for Stepped import "/math" for Math
var euler = Fn.new { |f, y, step, end|
Fmt.write(" Step $2d: ", step)
for (t in Stepped.new(0..end, step)) {
if (t%10 == 0) Fmt.write(" $7.3f", y)
y = y + step * f.call(y)
}
System.print()
}
var analytic = Fn.new {
System.write(" Time: ")
for (t in Stepped.new(0..100, 10)) Fmt.write(" $7d", t)
System.write("\nAnalytic: ")
for (t in Stepped.new(0..100, 10)) {
Fmt.write(" $7.3f", 20 + 80 * Math.exp(-0.07*t))
}
System.print()
} var cooling = Fn.new { |temp| -0.07 * (temp - 20) }
analytic.call() for (i in [2, 5, 10]) euler.call(cooling, 100, i, 100)</lang>
- Output:
Time: 0 10 20 30 40 50 60 70 80 90 100 Analytic: 100.000 59.727 39.728 29.797 24.865 22.416 21.200 20.596 20.296 20.147 20.073 Step 2: 100.000 57.634 37.704 28.328 23.918 21.843 20.867 20.408 20.192 20.090 20.042 Step 5: 100.000 53.800 34.280 26.034 22.549 21.077 20.455 20.192 20.081 20.034 20.014 Step 10: 100.000 44.000 27.200 22.160 20.648 20.194 20.058 20.017 20.005 20.002 20.000
zkl
<lang zkl>const FMT=" %7.3f";
fcn ivp_euler(f,y,step,end_t){
print(" Step %2d: ".fmt(step));
foreach t in ([0..end_t,step]){
if (t % 10 == 0) print(FMT.fmt(y));
y += f(t,y) * step;
}
println();
}
fcn analytic{
print(" Time: ");
foreach t in ([0..100,10]){ print(" %7g".fmt(t)) }
print("\nAnalytic: ");
foreach t in ([0..100,10]){ print(FMT.fmt(20.0 + 80.0 * (-0.07 * t).exp())) }
println();
}
fcn cooling(_,temp){ return(-0.07 * (temp - 20)) }
analytic(); ivp_euler(cooling, 100.0, 2, 100); ivp_euler(cooling, 100.0, 5, 100); ivp_euler(cooling, 100.0, 10, 100);</lang>
- Output:
Time: 0 10 20 30 40 50 60 70 80 90 100 Analytic: 100.000 59.727 39.728 29.797 24.865 22.416 21.200 20.596 20.296 20.147 20.073 Step 2: 100.000 57.634 37.704 28.328 23.918 21.843 20.867 20.408 20.192 20.090 20.042 Step 5: 100.000 53.800 34.280 26.034 22.549 21.077 20.455 20.192 20.081 20.034 20.014 Step 10: 100.000 44.000 27.200 22.160 20.648 20.194 20.058 20.017 20.005 20.002 20.000
ZX Spectrum Basic
<lang zxbasic>10 LET d$="-0.07*(y-20)": LET y=100: LET a=0: LET b=100: LET s=10 20 LET t=a 30 IF t<=b THEN PRINT t;TAB 10;y: LET y=y+s*VAL d$: LET t=t+s: GO TO 30</lang>
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