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title	id	challengeType
Euler method	59880443fb36441083c6c20e	5

Description

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

The ODE has to be provided in the following form:

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

with an initial value

:: $y(t_0) = y_0$

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:

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

where $h$ 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 $T_R$:

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

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

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

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

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 $T_0$ shall be 100 °C

::* room temperature $T_R$ shall be 20 °C

::* cooling constant $k$ shall be 0.07

::* time interval to calculate shall be from 0 s ──► 100 s

Instructions

Tests

tests:
  - text: <code>eulersMethod</code> is a function.
    testString: 'assert(typeof eulersMethod === "function", "<code>eulersMethod</code> is a function.");'
  - text: '<code>eulersMethod(0, 100, 100, 10)</code> should return a number.'
    testString: 'assert(typeof eulersMethod(0, 100, 100, 10) === "number", "<code>eulersMethod(0, 100, 100, 10)</code> should return a number.");'
  - text: '<code>eulersMethod(0, 100, 100, 10)</code> should return 20.0424631833732.'
    testString: 'assert.equal(eulersMethod(0, 100, 100, 2), 20.0424631833732, "<code>eulersMethod(0, 100, 100, 10)</code> should return 20.0424631833732.");'
  - text: '<code>eulersMethod(0, 100, 100, 10)</code> should return 20.01449963666907.'
    testString: 'assert.equal(eulersMethod(0, 100, 100, 5), 20.01449963666907, "<code>eulersMethod(0, 100, 100, 10)</code> should return 20.01449963666907.");'
  - text: '<code>eulersMethod(0, 100, 100, 10)</code> should return 20.000472392.'
    testString: 'assert.equal(eulersMethod(0, 100, 100, 10), 20.000472392, "<code>eulersMethod(0, 100, 100, 10)</code> should return 20.000472392.");'

Challenge Seed

function eulersMethod (x1, y1, x2, h) {
  // Good luck!
}

Solution

function eulersMethod(x1, y1, x2, h) {
  let x = x1;
  let y = y1;

  while ((x < x2 && x1 < x2) || (x > x2 && x1 > x2)) {
    y += h * (-0.07 * (y - 20));
    x += h;
  }

  return y;
}

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