153 lines
		
	
	
		
			3.9 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
		
		
			
		
	
	
			153 lines
		
	
	
		
			3.9 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
|   | --- | ||
|  | id: 59880443fb36441083c6c20e | ||
|  | title: Euler method | ||
|  | challengeType: 5 | ||
|  | forumTopicId: 302258 | ||
|  | dashedName: euler-method | ||
|  | --- | ||
|  | 
 | ||
|  | # --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](https://en.wikipedia.org/wiki/Euler method "wp: Euler method"). | ||
|  | 
 | ||
|  | The ODE has to be provided in the following form: | ||
|  | 
 | ||
|  | <ul style='list-style: none;'> | ||
|  |   <li><big>$\frac{dy(t)}{dt} = f(t,y(t))$</big></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | with an initial value | ||
|  | 
 | ||
|  | <ul style='list-style: none;'> | ||
|  |   <li><big>$y(t_0) = y_0$</big></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation: | ||
|  | 
 | ||
|  | <ul style='list-style: none;'> | ||
|  |   <li><big>$\frac{dy(t)}{dt}  \approx \frac{y(t+h)-y(t)}{h}$</big></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | then solve for $y(t+h)$: | ||
|  | 
 | ||
|  | <ul style='list-style: none;'> | ||
|  |   <li><big>$y(t+h) \approx y(t) + h \, \frac{dy(t)}{dt}$</big></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | which is the same as | ||
|  | 
 | ||
|  | <ul style='list-style: none;'> | ||
|  |   <li><big>$y(t+h) \approx y(t) + h \, f(t,y(t))$</big></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | The iterative solution rule is then: | ||
|  | 
 | ||
|  | <ul style='list-style: none;'> | ||
|  |   <li><big>$y_{n+1} = y_n + h \, f(t_n, y_n)$</big></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | 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 $T_R$: | ||
|  | 
 | ||
|  | <ul style='list-style: none;'> | ||
|  |   <li><big>$\frac{dT(t)}{dt} = -k \, \Delta T$</big></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | or | ||
|  | 
 | ||
|  | <ul style='list-style: none;'> | ||
|  |   <li><big>$\frac{dT(t)}{dt} = -k \, (T(t) - T_R)$</big></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | 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 | ||
|  | 
 | ||
|  | <ul style='list-style: none;'> | ||
|  |   <li><big>$T(t) = T_R + (T_0 - T_R) \; e^{-k t}$</big></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | # --instructions--
 | ||
|  | 
 | ||
|  | Implement a routine of Euler's method and then use it to solve the given example of Newton's cooling law for three different step sizes of: | ||
|  | 
 | ||
|  | <ul> | ||
|  |   <li><code>2 s</code></li> | ||
|  |   <li><code>5 s</code> and</li> | ||
|  |   <li><code>10 s</code></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | and compare with the analytical solution. | ||
|  | 
 | ||
|  | **Initial values:** | ||
|  | 
 | ||
|  | <ul> | ||
|  |   <li>initial temperature <big>$T_0$</big> shall be <code>100 °C</code></li> | ||
|  |   <li>room temperature <big>$T_R$</big> shall be <code>20 °C</code></li> | ||
|  |   <li>cooling constant <big>$k$</big> shall be <code>0.07</code></li> | ||
|  |   <li>time interval to calculate shall be from <code>0 s</code> to <code>100 s</code></li> | ||
|  | </ul> | ||
|  | 
 | ||
|  | First parameter to the function is initial time, second parameter is initial temperature, third parameter is elapsed time and fourth parameter is step size. | ||
|  | 
 | ||
|  | # --hints--
 | ||
|  | 
 | ||
|  | `eulersMethod` should be a function. | ||
|  | 
 | ||
|  | ```js | ||
|  | assert(typeof eulersMethod === 'function'); | ||
|  | ``` | ||
|  | 
 | ||
|  | `eulersMethod(0, 100, 100, 2)` should return a number. | ||
|  | 
 | ||
|  | ```js | ||
|  | assert(typeof eulersMethod(0, 100, 100, 2) === 'number'); | ||
|  | ``` | ||
|  | 
 | ||
|  | `eulersMethod(0, 100, 100, 2)` should return 20.0424631833732. | ||
|  | 
 | ||
|  | ```js | ||
|  | assert.equal(eulersMethod(0, 100, 100, 2), 20.0424631833732); | ||
|  | ``` | ||
|  | 
 | ||
|  | `eulersMethod(0, 100, 100, 5)` should return 20.01449963666907. | ||
|  | 
 | ||
|  | ```js | ||
|  | assert.equal(eulersMethod(0, 100, 100, 5), 20.01449963666907); | ||
|  | ``` | ||
|  | 
 | ||
|  | `eulersMethod(0, 100, 100, 10)` should return 20.000472392. | ||
|  | 
 | ||
|  | ```js | ||
|  | assert.equal(eulersMethod(0, 100, 100, 10), 20.000472392); | ||
|  | ``` | ||
|  | 
 | ||
|  | # --seed--
 | ||
|  | 
 | ||
|  | ## --seed-contents--
 | ||
|  | 
 | ||
|  | ```js | ||
|  | function eulersMethod(x1, y1, x2, h) { | ||
|  | 
 | ||
|  | } | ||
|  | ``` | ||
|  | 
 | ||
|  | # --solutions--
 | ||
|  | 
 | ||
|  | ```js | ||
|  | 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; | ||
|  | } | ||
|  | ``` |