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My Problem is this: given Initial Value Problem

$$y^{\prime}=\frac{3x-2y}{x}\quad y(1)=0$$

There is a given Interval of $[1,2]$ and a given step size $h$ of $h=0.1$

After using Euler's and Heun's method I am looking for a way to solve it with the Runge-Kutta method for differential equations / IV-Problems of first order.

My Approach: To understand the differences between these mentioned methods it seems important for me to calculate at least one example. We already solved this Problem with Euler's and Heun's.

Additionally we have determined the exact solution: $$y(x) = \dfrac{x^3-1}{x^2}$$ at $x=2$ it is: $$y(2) = \dfrac{7}{4} = 1.75$$

But what is the way to solve it with the Runge-Kutta method? I am stuck in applying it on the solution. But I guess the solution is even more accurate than the solution we got from the Heun's method.

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You're writing code to do this in some language like Matlab right? You wouldn't want to do this by hand. It should be straightforward to modify your Heun's method code to use the RK4 recursion instead. –  littleO Jul 9 '13 at 8:59
    
No, at the Moment i just want to get known to the different methods for solving the Initial Value Problem. No Matlab in sight :/ –  Toralf Westström Jul 9 '13 at 9:07
1  
Ah, well I think it'll be much more enlightening for you to just code them up. –  littleO Jul 9 '13 at 9:14
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@ToralfWestström: Did the answer below make sense? Regards –  Amzoti Jul 10 '13 at 3:23
    
yes it does, thanks alot. –  Toralf Westström Sep 19 '13 at 7:49
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1 Answer

up vote 2 down vote accepted

To be complete, it is worth noting that there are many variants of Runge-Kutta (RK), including some new ones. These include, but are not limited to:

  • Generalized RK, RK1, RK2, RK3, RK4, Simple RK, Euler-Cauchy, Optimal Method, RK5, New Variant RK, RK-Merson (RKM), RK-Fehlberg (RKF), Cash-Karp-RK (CKRK), Implicit RK (IRK) ...

For this problem, I will assume you want to typical fourth order RK (RK4).

Initialization

  • $h = \dfrac{b-a}{N} = .1 = \dfrac{2-1}{N} \rightarrow N = 10$
  • $x_0 = a = 1, y_0 = y(a) = y(1) = \alpha = 0 \rightarrow a = 1, \alpha = 0$

Iteration using RK4

  • $y'(x) = f(x, y) = \dfrac{3 x -2 y}{x}, y(1) = 0$
  • $k_1 = h f(x_i, y_i)$
  • $k_2 = h f(x_i + h/2, y_i + k_1/2)$
  • $k_3 = h f(x_i + h/2, y_i + k_2/2)$
  • $k_4 = h f(x_i + h, y_i + k_3)$
  • $y_{i + 1} = y_i + \dfrac{1}{6}\left( k_1+2k_2+2k_3+k_4 \right)$
  • $x_{i + 1} = x_i + h = 1 + 0.1 i$

Results

  • For $i= 1$, we have:
  • $x_0 = 1, y_0 = 0$
  • $k_1 = h f(x_i, y_i) = \dfrac{1}{10} \left(\dfrac{3 x_{i} - 2 y_{i}}{x_{i}}\right) = \dfrac{1}{10}\left(\dfrac{3 x(0) - 2y(0)}{x(0)}\right) = \dfrac{1}{10}\left(\dfrac{3 (1) - 2(0)}{1}\right) = 0.3$
  • $k_2 = h f(x_i + h/2, y_i + k_1/2) = \dfrac{1}{10} \left(\dfrac{3 (x_{i}+(1/20)) - 2(y_{i} + (0.3/2))}{x_{i}+(1/20)}\right) = \dfrac{1}{10} \left(\dfrac{3 (1+(1/20)) - 2(0 + (0.3/2))}{1+(1/20)}\right) = 0.271428571428571428571428571428571428571428571428571428571428$
  • $k_3 = h f(x_i + h/2, y_i + k_2/2) = \dfrac{1}{10} \left(\dfrac{3 (x_{i}+(1/20)) - 2(y_{i} + (0.27142857/2))}{x_{i}+(1/20)}\right) = \dfrac{1}{10} \left(\dfrac{3 (1+(1/20)) - 2(0 + (0.27142857/2))}{1+(1/20)}\right) = 0.274149659863945578231292517006802721088435374149659863945578$
  • $k_4 = h f(x_i + h, y_i + k_3) = \dfrac{1}{10} \left(\dfrac{3 (x_{i}+(1/10)) - 2(y_{i} + 0.2741496598)}{x_{i}+(1/10)}\right) = \dfrac{1}{10} \left(\dfrac{3 (1+(1/10)) - 2(0 + 0.2741496598)}{1+(1/10)}\right)= 0.250154607297464440321583178726035868893011750154607297464440$
  • $y_{i + 1} = y_i + \dfrac{1}{6}\left( k_1+2k_2+2k_3+k_4 \right) = 0.273551844980416408987837559266130694702123273551844980090476 \approx 0.273552$
  • $x_{i + 1} = x_i + h = 1 + 0.1 i = 1.1$

Continuing with this itearation, we generate the following table.

$~~~~~\text{Step} ~~|~~~~ x ~~~|~~ y $

  • $~~00 ~~|~~ 1.0 ~~|~~ 0. 00000$
  • $~~01 ~~|~~ 1.1 ~~|~~ 0.273552$
  • $~~02 ~~|~~ 1.2 ~~|~~ 0.505553$
  • $~~03 ~~|~~ 1.3 ~~|~~ 0.708281$
  • $~~04 ~~|~~ 1.4 ~~|~~ 0.889793$
  • $~~05 ~~|~~ 1.5 ~~|~~ 1.05555$
  • $~~06 ~~|~~ 1.6 ~~|~~ 1.20937$
  • $~~07 ~~|~~ 1.7 ~~|~~ 1.35398$
  • $~~08 ~~|~~ 1.8 ~~|~~ 1.49136$
  • $~~09 ~~|~~ 1.9 ~~|~~ 1.62299$
  • $~~10 ~~|~~ 2.0 ~~|~~ 1.75$

The exact solution is given by:

$$y(x) = \dfrac{x^3-1}{x^2}$$

At $x=2$, we have: $y(2) = \dfrac{7}{4} = 1.75$

Compare that to Runge-Kutta's-4 method, which has $1.75$.

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