22 Jun 2025
Euler and Fourth-Order Runge-Kutta Methods
Consider the initial-value problem
\[\frac{dy}{dx}=f(x,y),\qquad y(x_0)=y_0.\]A step method constructs $y_{n+1}$ at $x_{n+1}=x_n+h$ from information near $(x_n,y_n)$.
The figure is reproducible from its TikZ source.
Euler’s method
Taylor expansion gives
\[y(x_n+h)=y(x_n)+hy'(x_n)+\frac{h^2}{2}y''(\xi_n).\]Using $y’=f(x,y)$ and dropping terms of order $h^2$ produces
\[\boxed{y_{n+1}=y_n+h f(x_n,y_n)}.\]Thus Euler’s method follows the tangent at the start of each interval. Its local truncation error is $O(h^2)$; after roughly $(b-a)/h$ steps, these accumulated errors give global error $O(h)$.
Fourth-order Runge-Kutta method
RK4 samples four slopes:
\[k_1=h f(x_n,y_n),\] \[k_2=h f\left(x_n+\frac h2,y_n+\frac{k_1}{2}\right),\] \[k_3=h f\left(x_n+\frac h2,y_n+\frac{k_2}{2}\right),\] \[k_4=h f(x_n+h,y_n+k_3).\]Their weighted combination is
\[\boxed{y_{n+1}=y_n+\frac{k_1+2k_2+2k_3+k_4}{6}}.\]The weights are chosen to match the Taylor expansion through $h^4$. This can be seen directly for $y’=y$:
\[k_1=hy,\] \[k_2=hy\left(1+\frac h2\right),\] \[k_3=hy\left(1+\frac h2+\frac{h^2}{4}\right),\] \[k_4=hy\left(1+h+\frac{h^2}{2}+\frac{h^3}{4}\right).\]Substitution in the weighted update gives
\[y_{n+1}=y_n\left( 1+h+\frac{h^2}{2}+\frac{h^3}{6}+\frac{h^4}{24} \right),\]which is the fourth-degree Taylor polynomial of $e^h$. For a smooth general equation, RK4 has local error $O(h^5)$ and global error $O(h^4)$.
One step for $y’=y$
Let $y(0)=1$ and $h=0.1$. Euler gives
\[y_1^{\rm Euler}=1+0.1(1)=1.1.\]For RK4,
\[k_1=0.1,\qquad k_2=0.105,\qquad k_3=0.10525,\qquad k_4=0.110525,\]so
\[y_1^{\rm RK4} =1+\frac{0.1+2(0.105)+2(0.10525)+0.110525}{6} =1.105170833\ldots.\]The exact value is $e^{0.1}=1.105170918\ldots$. The same step size therefore gives a visibly smaller RK4 error.
The symbolic expansion in the Maxima worksheet returns
rk4_residual = 0
where the residual is RK4 minus the fourth-degree Taylor update for $y’=y$.
Discussion