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:
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:
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
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 shall be 100 °C, the room temperature 20 °C, and the cooling constant 0.07. The time interval to calculate shall be from 0 s to 100 s.
An reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
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
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
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 (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)</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 DONE
D
<lang 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)(F f, double y0, double a, double b, double h) {
double y = y0; foreach (t; iota(a, b, h)) { writefln("%.3f %.3f", t, y); y += h * f(t, y); } writeln("done");
}
/// Example: Newton's cooling law auto newtonCoolingLaw(double time, double t) {
return -0.07 * (t - 20);
}
void main() {
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
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
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
'Y a b h'=. 4{. y t=. a while. b >: tincr=. {:t + h do. Y=. Y, (+ h * u) {:Y t=. t, tincr end. 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
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>
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
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
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