Question

using the classical R-K method of O(h4) calculate approximate solution of the IVP y'=1-x+4y, y(0) =1 at x=0.6, taking h=0.1 and 0.2. use extrapolation technique to improve the accuracy.

Answer 1

Question #74129 - Math - Quantitative Methods

using the classical R-K method of O(h4) calculate approximate solution of the IVP $y' = 1 - x + 4y$ , $y(0) = 1$ at $x = 0.6$ , taking $h = 0.1$ and 0.2. use extrapolation technique to improve the accuracy.

Solution: initial value problem

$$y ^ {\prime} = f (x, y) = 1 - x + 4 y, \quad y (x _ {0}) = y _ {0} = 1, \quad x _ {0} = 0. 6$$

In general a Runge-Kutta method of order 4 can be written as:

$$y _ {n + 1} = y _ {n} + \frac {h}{6} (k _ {1} + 2 k _ {2} + 2 k _ {3} + k _ {4})$$

where

$$k _ {1} = f (x _ {n}, y _ {n}) = 1 - x _ {n} + 4 y _ {n}$$ $$\begin{array}{l} k _ {2} = f \left(x _ {n} + \frac {h}{2}, y _ {n} + \frac {h}{2} k _ {1}\right) = 1 - x _ {n} - \frac {h}{2} + 4 \left(y _ {n} + \frac {h}{2} (1 - x _ {n} + 4 y _ {n})\right) \\ = 1 + \frac {3}{2} h - x _ {n} (1 + 2 h) + 4 y _ {n} (1 + 2 h) = 1 + \frac {3}{2} h + (1 + 2 h) (4 y _ {n} - x _ {n}) \\ \end{array}$$ $$\begin{array}{l} k _ {3} = f \left(x _ {n} + \frac {h}{2}, y _ {n} + \frac {h}{2} k _ {2}\right) = 1 - x _ {n} - \frac {h}{2} + 4 \left(y _ {n} + \frac {h}{2} \left(1 + \frac {3}{2} h + (1 + 2 h) (4 y _ {n} - x _ {n})\right)\right) \\ = 1 + \frac {3}{2} h + 3 h ^ {2} + (4 y _ {n} - x _ {n}) (1 + 2 h + 4 h ^ {2}) \\ \end{array}$$ $$\begin{array}{l} k _ {4} = f (x _ {n} + h, y _ {n} + h k _ {3}) \\ = 1 - x _ {n} - h + 4 \left(y _ {n} + h \left(1 + \frac {3}{2} h + 3 h ^ {2} + (4 y _ {n} - x _ {n}) (1 + 2 h + 4 h ^ {2})\right)\right) \\ = 1 + 3 h + 6 h ^ {2} + 1 2 h ^ {3} + (4 y _ {n} - x _ {n}) (1 + 4 h + 8 h ^ {2} + 1 6 h ^ {3}) \\ \end{array}$$ $$\begin{array}{l} k _ {1} + 2 k _ {2} + 2 k _ {3} + k _ {4} \\ = 1 + 2 \left(1 + \frac {3}{2} h\right) + 2 \left(1 + \frac {3}{2} h + 3 h ^ {2}\right) + 1 + 3 h + 6 h ^ {2} + 1 2 h ^ {3} \\ + (4 y _ {n} - x _ {n}) (1 + 2 (1 + 2 h) + 2 (1 + 2 h + 4 h ^ {2}) + (1 + 4 h + 8 h ^ {2} \\ \left. + 1 6 h ^ {3}\right)) = 6 + 9 h + 1 2 h ^ {2} + 1 2 h ^ {3} + (4 y _ {n} - x _ {n}) (6 + 1 2 h + 1 6 h ^ {2} + 1 6 h ^ {3}) \\ = (4 y _ {n} - x _ {n} + 1) (6 + 9 h + 1 2 h ^ {2} + 1 2 h ^ {3}) + (4 y _ {n} - x _ {n}) (3 h + 4 h ^ {2} + 4 h ^ {3}) \\ \end{array}$$

Let

$$g (h, x _ {n}, y _ {n}) = h (k _ {1} + 2 k _ {2} + 2 k _ {3} + k _ {4})$$

For step h1=0.1

$$g (0. 1, x _ {n}, y _ {n}) = 0. 7 0 3 2 - 0. 7 3 7 6 x _ {n} + 2. 9 5 0 4 y _ {n}$$ $$y _ {n + 1} (0. 1) = y _ {n} (0. 1) + g (0. 1, x _ {n}, y _ {n}) / 6$$

For step h2=0.2

$$g (0. 2, x _ {n}, y _ {n}) = 1. 6 7 5 2 - 1. 8 3 3 6 x _ {n} + 7. 3 3 4 4 y _ {n}$$ $$y _ {n + 1} (0. 2) = y _ {n} (0. 2) + g (0. 2, x _ {n}, y _ {n}) / 6$$

To calculate the error of the solution, we use the analytic solution of IVP (y(x)-exact value):

$$y (x) = \frac {4 x - 3 + 1 9 * e ^ {4 x}}{1 6}$$ $$e r r _ {n + 1} (0. 1) = y \left(x _ {n + 1}\right) - y _ {n + 1} (0. 1)$$ $$e r r _ {n + 1} (0. 2) = y \left(x _ {n + 1}\right) - y _ {n + 1} (0. 2)$$ $$e r r _ {n + 1} = y \left(x _ {n + 1}\right) - y _ {n + 1}$$

Where $y_{n+1}$ approximate solution of the IVP with use extrapolation technique on the step $n+1$ . Richardson extrapolation technique (h1=h2/2):

$$y _ {n + 1} = y _ {n + 1} (0. 1) + \frac {y _ {n + 1} (0 . 1) - y _ {(n + 1) * 2} (0 . 2)}{2 ^ {4} - 1}$$

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