Euler method

Method description

Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is a explicit method for solving initial value problems (IVPs), as described in Euler method

The ODE has to be provided in the folloving form:

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

with an initial value

y(t_{0})=y_{0}

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 $T(t_{0})=T_{0}$ cools down in an environment of temperature $T_{R}$ :

{\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 $\Delta T=(T(t)-T_{R})$ 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^{-kt}

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 $T_{0}$ shall be 100 °C, the room temperature $T_{R}$ 20 °C, and the cooling constant $k$ 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.

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.

<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++

Translation of: D

<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

'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

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

  ncl euler 100 0 100 5

... NB. output redacted

  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>

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