Question

Q. How to control error in Runge Kutta Fehlberg Method?

Answer 1

Answer on Question #76731 – Math – Quantitative Methods

Question

How to control error in Runge Kutta Fehlberg Method?

Solution

To approximate the solution to the 1st order IVP:

$$y' = f(x, y), \qquad y(x_0) = y_0$$

we seek:

$$y_{n+1} = y_n + h \sum_{i=1}^{s} b_i k_i + O(h^{s+1})$$

The adaptive method is designed to produce an estimate of the local truncation error of a single Runge-Kutta step. Let $\gamma^{p_{n+1}}$ and $\gamma^{q_{n+1}}$ be the approximations of $\gamma\{\mathbf{x}_{n+1}\}$ computed using the methods of order $p$ and $q$ respectively. The local truncation error in these two methods is given by

$$\varepsilon_{n+1}^p = \frac{y_{n+1} - y_{n+1}^p}{h} = O(h^{p+1}), \qquad \varepsilon_{n+1}^q = \frac{y_{n+1} - y_{n+1}^q}{h} = O(h^{q+1})$$

The error between two solutions is

$$\varepsilon_{n+1} = \frac{|y_{n+1}^q - y_{n+1}^p|}{h}$$

If they differ by no more than $\varepsilon$ - the required error, then the approximation is accepted. This method estimates the error of the lower order scheme.

The Runge-Kutta-Fehlberg method is a one-step method with the approximations calculated using the Runge-Kutta method of order 4 and 5. For this method each step requires the use of the following six values:

$$k_1 = h \cdot f(x_k, y_k)$$ $$k_2 = h \cdot f\left(x_k + \frac{1}{4} h, y_k + \frac{1}{4} k_1\right)$$ $$k_3 = h \cdot f\left(x_k + \frac{3}{8} h, y_k + \frac{3}{32} k_1 + \frac{9}{32} k_2\right)$$ $$k_4 = h \cdot f\left(x_k + \frac{12}{13} h, y_k + \frac{1932}{2197} k_1 - \frac{7200}{2197} k_2 + \frac{7296}{2197} k_3\right)$$ $$k_5 = h \cdot f\left(x_k + h, y_k + \frac{439}{216} k_1 - 8 k_2 + \frac{3680}{513} k_3 - \frac{845}{4104} k_4\right)$$ $$k_6 = h \cdot f\left(x_k + \frac{1}{2} h, y_k - \frac{8}{27} k_1 + 2 k_2 - \frac{3544}{2565} k_3 + \frac{1859}{4104} k_4 - \frac{11}{40} k_5\right)$$

Then we calculate the approximation to the solution with the help of the method of the fourth order:

$$y _ {k + 1} ^ {4} = y _ {k} + \frac {25}{216} k _ {1} + \frac {1408}{2565} k _ {3} + \frac {2197}{4101} k _ {4} - \frac {1}{5} k _ {5}, \quad \text{Error} = O (h ^ {4})$$

And the approximation to the solution with the help of the method of the 5th order:

$$y _ {k + 1} ^ {5} = y _ {k} + \frac {16}{135} k _ {1} + \frac {6656}{12825} k _ {3} + \frac {28561}{56430} k _ {4} - \frac {9}{50} k _ {5} + \frac {2}{55} k _ {6}, \quad \text{Error} = O (h ^ {5})$$

At each step, two different approximations for the solution are made and compared.

$$\varepsilon_ {n + 1} = \frac {1}{h} \left| y _ {k + 1} ^ {4} - y _ {k + 1} ^ {5} \right|$$

Answer provided by https://www.AssignmentExpert.com