Question

Q. How to control size of h in Runge Kutta Fehlberg Method?

Answer 1

Answer on Question #76732 – Math – Quantitative Methods

Question

How to control size of $h$ 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 $y_{n+1}^p$ and $y_{n+1}^{p+1}$ be the approximations of $y_{n+1}$ computed using the methods of order $p$ and $p+1$ respectively. The local truncation error in these two methods is given by

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

The error between two solutions is

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

If the two answers are in close agreement ($\varepsilon_{n+1} \leq \varepsilon$), the approximation is accepted. If the two answers do not agree to a specified accuracy ($\varepsilon$), the step size is reduced. If the answers agree to more significant digits than required, the step size is increased. Our goal is to determine how to modify $h$. Because $\varepsilon_{n+1}$ is the error of a method that is $p$-th order accurate, then if we replace $h$ by $\delta \cdot h$, the error is multiplied by $\delta^p$. To calculate the new step, we must solve the inequality:

$$\left| \delta^p \frac{y_{n+1}^{p+1} - y_{n+1}^p}{h} \right| < \varepsilon$$

Solving for $\delta$:

$$\delta < \left(\frac{\varepsilon \cdot h}{|y_{n+1}^{p+1} - y_{n+1}^p|}\right)^{1/p} = \left(\frac{\varepsilon}{\varepsilon_{n+1}}\right)^{1/p}$$

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.

$$\varepsilon_ {n + 1} = \frac {1}{h} \left| y _ {k + 1} ^ {4} - y _ {k + 1} ^ {5} \right|$$ $$\delta < \left(\frac {\varepsilon}{\varepsilon_ {n + 1}}\right) ^ {1 / 4}$$

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