The Runge-Kutta method

The picture shows the flow lines for a differential equation of the form $\dot{x}=f(t,x)$. To solve the equation, we need to follow the flow lines. However, we do not initially know a formula for the flow lines. We only know how to evaluate the function $f(t,x)$, i.e. how to determine the slope of the flow lines at any given point $(t,x)$.

Consider a point $(t_0,x_0)$. We want to follow the black flow line from there to determine the value of $x$ at time $t_0+h$.

As a first attempt, we draw a line from $(t_0,x_0)$ with slope $m_0=f(t_0,x_0)$, so the line is tangent to the flow line at $(t_0,x_0)$. In Euler's method, we would just follow this line to time $t_0+h$. Because the flow lines are strongly curved, this would be a poor approximation.

In the Runge-Kutta method, we instead follow this line only to $t_1=t_0+h/2$ and call the resulting point $(t_1,x_1)$.

We then put $m_1=f(t_1,x_1)$, which is the slope of the flow line at $(t_1,x_1)$.

We draw a line of slope $m_1$ through $(t_0,x_0)$ and follow this line to time $t_2=t_1=t_0+h/2$. This gives us a new point $(t_2,x_2)$.

We then put $m_2=f(t_2,x_2)$, which is the slope of the flow line at $(t_2,x_2)$.

We draw a line of slope $m_2$ through $(t_0,x_0)$ and follow this line to time $t_3=t_0+h$. This gives us a new point $(t_3,x_3)$.

We then put $m_3=f(t_3,x_3)$, which is the slope of the flow line at $(t_3,x_3)$.

We now have four different slopes $m_0,m_1,m_2,m_3$ which are potentially relevant.

We take the weighted average $m_4=(m_0+2m_1+2m_2+m_3)/6$, draw a line of slope $m_4$ through $(t_0,x_0)$ and follow this line to time $t_4=t_0+h$. This gives us a new point $(t_4,x_4)$. You can see that this is very close to the black flow line, as desired.