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9.2 Runge-Kutta Methods: Class Notes

9.2

A generalization of Euler’s formula is $y_{n+1}=y_n+h(w_1{k}_1+w_2{k}_2+\cdots +w_m{k}_m)$, where the quantity in parentheses is a weighted average of slopes.

For example, $k_1$ can be the slope $y\prime =f(x,y)$ at the current point, and $k_2$ can be the predicted slope $f(x_{n+1},y_{n+1}^{*})$ from the improved Euler’s method.

Notes: $\sum _{i=1}^m{w}_i=1$, $k_i$ are recursively defined, and $m$ is the order of the method.

Definition

For $y_{n+1}=y_n+hf(x_n,y_n)$, we have $m=1$, $w_1=1$, and $k_1=f(x_n,y_n)$. Thus Euler’s method is a first-order Runge-Kutta method.

For reference: A second-order Runge-Kutta method has the form $y_{n+1}=y_n+h(w_1{k}_1+w_2{k}_2)$, where $k_1=f(x_n,y_n)$ and $k_2=f(x_n+\alpha h,y_n+\beta h{k}_1)$. When $w_1+w_2=1$, $w_2\alpha =1/2$, and $w_2\beta =1/2$, the results agree with a Taylor polynomial of degree 2. (Note that this is only one choice for the constants; there are others, though we won’t be exploring that.) If we choose $w_2=1/2$, then $w_1=1/2$, and $\alpha =\beta =1$. The result agrees with the improved Euler’s method, thus making it a second-order Runge-Kutta method.

A fourth-order Runge-Kutta method (RK4)

An RK4 method uses a weighted average of four slopes to approximate $y$-values. The basis of the technique is to find parameters so the formula

$$y_{n+1}=y_n+h(w_1{k}_1+w_2{k}_2+w_3{k}_3+w_4{k}_4),\text{where}$$

$$\begin{array}{cc}\hfill {\displaystyle k_1}& {\displaystyle =f(x_n,y_n)}\hfill \\ \hfill {\displaystyle k_2}& {\displaystyle =f(x_n+{\alpha }_{1}h,y_n+{\beta }_{1}h{k}_1)}\hfill \\ \hfill {\displaystyle k_3}& {\displaystyle =f(x_n+{\alpha }_{2}h,y_n+{\beta }_{2}h{k}_1+{\beta }_{3}h{k}_2)}\hfill \\ \hfill {\displaystyle k_4}& {\displaystyle =f(x_n+{\alpha }_{3}h,y_n+{\beta }_{4}h{k}_1+{\beta }_{5}h{k}_2+{\beta }_{6}h{k}_3)}\hfill \end{array}$$

agrees with a 4th-degree Taylor polynomial. The resulting system has 11 equations in 13 unknowns, and common choices for the parameters yields

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y_{n+1}}& {\displaystyle =y_n+\frac{h}{6}(k_1+2k_2+2k_3+k_4)}\hfill \\ \hfill {\displaystyle k_1}& {\displaystyle =f(x_n,y_n)}\hfill \\ \hfill {\displaystyle k_2}& {\displaystyle =f(x_n+\frac{1}{2}h,y_n+\frac{1}{2}h{k}_1)}\hfill \\ \hfill {\displaystyle k_3}& {\displaystyle =f(x_n+\frac{1}{2}h,y_n+\frac{1}{2}h{k}_2)}\hfill \\ \hfill {\displaystyle k_4}& {\displaystyle =f(x_n+h,y_n+h{k}_3)}\hfill \end{array}}$$

In the general form above, each slope is built from the ones before it: $k_1$ is used in $k_2$, $k_3$, and $k_4$; $k_2$ in $k_3$ and $k_4$; and $k_3$ in $k_4$.

Local and global truncation errors using RK4 are $O(h^5)$ and $O(h^4)$, respectively.

For comparison: Euler’s method has error $O(h)$, and the improved Euler’s method has error $O(h^2)$.

Example

Use the RK4 method to obtain a four-decimal approximation of the indicated value. First use $h=0.1$ and then use $h=0.05$.

$$y\prime =4x-2y,y(0)=2;y(0.5)$$

Solution

The work is done in a spreadsheet, one row per step:

At $x=0.5$ the relative error is $0.001576\%$, compared with $0.78\%$ for the improved Euler’s method with the same $h=0.1$ (Section 9.1).

Note that $h\to \frac{1}{2}h$ leads to an error about $\frac{1}{16}$ as large: ${\displaystyle \frac{0.001576}{0.000091}\approx 17}$, close to $16$. The error is $O(h^4)$.