Runge-Kutta Methods | Sahithyan's Notes

Runge-Kutta methods approximate IVP solutions using higher-order Taylor expansions, without evaluating derivatives of $f$ .

The order- $n$ Taylor expansion of $y$ needs derivatives of $f$ up to order $n-1$ . These are costly to compute and may not exist everywhere. Runge-Kutta methods replace each derivative term with values of $f$ at selected $(t, y)$ points.

Mesh points are $t_i = a + ih$ with $h = \dfrac{b - a}{N}$ , as in Euler’s method.

Midpoint Method

Second-order Taylor, with the derivative term replaced using

f(t, y) + \frac{h}{2} f'(t, y) = f\left( t + \frac{h}{2},\ y + \frac{h}{2} f(t, y) \right) + O(h^2)

\begin{aligned} w_0 &= \alpha \\ w_{i+1} &= w_i + h\, f\left( t_i + \frac{h}{2},\ w_i + \frac{h}{2} f(t_i, w_i) \right) \end{aligned}

Local error $O(h^2)$ .

Modified Euler Method

Third-order Taylor, with the derivative term replaced by an average of the endpoint slopes.

\begin{aligned} w_0 &= \alpha \\ w_{i+1} &= w_i + \frac{h}{2}\left[ f(t_i, w_i) + f\big(t_{i+1},\ w_i + h f(t_i, w_i)\big) \right] \end{aligned}

Local error $O(h^3)$ .

Order 4 Runge-Kutta

\begin{aligned} w_0 &= \alpha \\ k_1 &= h\, f(t_i, w_i) \\ k_2 &= h\, f\left( t_i + \tfrac{h}{2},\ w_i + \tfrac{1}{2} k_1 \right) \\ k_3 &= h\, f\left( t_i + \tfrac{h}{2},\ w_i + \tfrac{1}{2} k_2 \right) \\ k_4 &= h\, f\left( t_{i+1},\ w_i + k_3 \right) \\ w_{i+1} &= w_i + \tfrac{1}{6}\left( k_1 + 2k_2 + 2k_3 + k_4 \right) \end{aligned}

for $i = 0, 1, \ldots, N-1$ . Local error $O(h^4)$ .

Error Orders

Method	Asymptotic error bound
Euler	$O(h)$
Midpoint	$O(h^2)$
Modified Euler	$O(h^3)$
Order 4 Runge-Kutta	$O(h^4)$

Higher order gives a better approximation for the same $h$ .

Memorization

All four have the form $w_{i+1} = w_i + h \times (\text{slope estimate})$ . They differ only in how the slope is estimated.

Euler
Slope at the left end, $f(t_i, w_i)$ . Nothing else.
Midpoint
Half-step to the middle with Euler, then take the slope there. One evaluation at $\left(t_i + \tfrac{h}{2},\ w_i + \tfrac{h}{2} f(t_i, w_i)\right)$ .
Modified Euler
Average of the two end slopes. Left slope $f(t_i, w_i)$ , right slope at the full Euler-predicted point $\left(t_{i+1},\ w_i + h f(t_i, w_i)\right)$ , weight $\tfrac{1}{2}$ each.
Order 4 Runge-Kutta
Weighted average of four slopes with weights $1, 2, 2, 1$ over $6$ (Simpson’s rule). Sample points step through $t_i,\ t_i + \tfrac{h}{2},\ t_i + \tfrac{h}{2},\ t_{i+1}$ , and each $k_j$ feeds the $y$ -argument of the next: $k_1$ into $k_2$ , $k_2$ into $k_3$ , $k_3$ into $k_4$ , all with factor $\tfrac{1}{2}$ except $k_3 \to k_4$ which uses the full $k_3$ .

Worked Example

IVP $y'(t) = y - t^2 + 1$ , $0 \leq t \leq 2$ , $y(0) = 0.5$ , with $N = 4$ , so $h = 0.5$ and $f(t, y) = y - t^2 + 1$ . Exact solution $y(t) = (t+1)^2 - 0.5 e^t$ , giving $y(2) \approx 5.3055$ .

Midpoint

\begin{aligned} w_1 &= w_0 + 0.5\, f\big(0.25,\ 0.5 + 0.25 f(0, 0.5)\big) = 0.5 + 0.5\, f(0.25,\ 0.875) = 1.4063 \\ w_2 &= 2.5977, \quad w_3 = 3.9399, \quad w_4 = 5.2149 \end{aligned}

Modified Euler

\begin{aligned} w_1 &= w_0 + 0.25\left[ f(0, 0.5) + f\big(0.5,\ 0.5 + 0.5 f(0, 0.5)\big) \right] = 0.5 + 0.25\,[1.5 + 2.0] = 1.375 \\ w_2 &= 2.5156, \quad w_3 = 3.7754, \quad w_4 = 4.9163 \end{aligned}

Order 4 Runge-Kutta

First step, $t_0 = 0$ , $w_0 = 0.5$

\begin{aligned} k_1 &= 0.5\, f(0,\ 0.5) = 0.75 \\ k_2 &= 0.5\, f(0.25,\ 0.5 + 0.375) = 0.9063 \\ k_3 &= 0.5\, f(0.25,\ 0.5 + 0.4531) = 0.9453 \\ k_4 &= 0.5\, f(0.5,\ 0.5 + 0.9453) = 1.0977 \\ w_1 &= 0.5 + \tfrac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) = 1.4251 \end{aligned}

Continuing, $w_2 = 2.6396$ , $w_3 = 4.0068$ , $w_4 = 5.3016$ .

Method	$w_4 \approx y(2)$	Error
Euler	$4.4375$	$0.868$
Midpoint	$5.2149$	$0.091$
Modified Euler	$4.9163$	$0.389$
Order 4 Runge-Kutta	$5.3016$	$0.004$