The Runge–Kutta method · Differential Equations

Intuition

If two slopes averaged give order two, more slopes, well placed, give more. The classical Runge–Kutta method reads the slope four times in each step: at the start, twice in the middle, and at the end, each reading taken at a point reached with the one before. Their weighted average, with the middle counted twice, is accurate to order four: halve the step and the error falls sixteenfold. For decades it was the method, and it is still the first one to reach for.

Surveying the slope of a field with four readings, two of them checks in the middle, instead of one at the gate: a little more walking, and a far better map.

plane diagram: three-methods-one-step-size

Three methods with the same step $h=0.5$ on $y'=y$ , $y(0)=1$ . At $x=2$ Euler’s method reaches $5.06$ , Heun’s $6.97$ , and Runge–Kutta, the dots, $7.384$ — against $e^{2}\approx7.389$ .

Four slopes, weighted

The classical Runge–Kutta method, often called RK4, takes each step with four slopes. $k_{1}$ is the slope at the start; $k_{2}$ the slope in the middle of the step, at the point reached with $k_{1}$ ; the slope in the middle again, at the point reached with ; and the slope at the end, at the point reached with . The step uses their weighted average, with weights over . It evaluates four times per step and is of order four.

\begin{aligned}k_{1}&=f(x_{n},y_{n})\\k_{2}&=f\bigl(x_{n}+\tfrac h2,\ y_{n}+\tfrac h2k_{1}\bigr)\\k_{3}&=f\bigl(x_{n}+\tfrac h2,\ y_{n}+\tfrac h2k_{2}\bigr)\\k_{4}&=f(x_{n}+h,\ y_{n}+hk_{3})\\y_{n+1}&=y_{n}+\tfrac h6\bigl(k_{1}+2k_{2}+2k_{3}+k_{4}\bigr)\end{aligned}

What the four slopes buy

For $y'=f(x)$ , with no $y$ on the right, $k_{2}=k_{3}$ and the step is Simpson’s rule for .

Runge–Kutta matches the exponential to fourth order

\begin{aligned}&y'=\lambda y,\ z=h\lambda\ \Longrightarrow\\&y_{n+1}=\Bigl(1+z+\frac{z^{2}}{2}+\frac{z^{3}}{6}+\frac{z^{4}}{24}\Bigr)y_{n}\end{aligned}

On y′ = λy every slope is λ times the height it is read at, so each one times h is z = hλ times y_n times a polynomial in z. The second slope is read at (1 + z/2) y_n, the third at (1 + z/2 + z²/4) y_n, the fourth at (1 + z + z²/2 + z³/4) y_n. Weighting them 1, 2, 2, 1 over 6 and stepping multiplies y_n by the first five terms of the series of e^z, which is what the exact solution multiplies by. The first term missed is z⁵/120: a local error of order h⁵, and a global error of order h⁴.

Proof steps

hk_{1}=z\,y_{n}

The slope at the start, times h.

hk_{2}=z\Bigl(1+\tfrac z2\Bigr)y_{n}

Read at $y_{n}+\tfrac h2k_{1}=\bigl(1+\tfrac z2\bigr)y_{n}$ .

hk_{3}=z\Bigl(1+\tfrac z2+\tfrac{z^{2}}{4}\Bigr)y_{n}

Read at $y_{n}+\tfrac h2k_{2}$ .

hk_{4}=z\Bigl(1+z+\tfrac{z^{2}}{2}+\tfrac{z^{3}}{4}\Bigr)y_{n}

Read at $y_{n}+hk_{3}$ , at the end of the step.

y_{n+1}=\Bigl(1+z+\tfrac{z^{2}}{2}+\tfrac{z^{3}}{6}+\tfrac{z^{4}}{24}\Bigr)y_{n}

Weight $1,2,2,1$ over $6$ and collect the powers of $z$ : the series of $e^{z}$ up to .

Applications

Practice

Four Slopes

Start, middle, middle again, end: each slope is read at a point reached with the one before, and the middle two count double.

y_{n+1}=y_{n}+\tfrac h6\bigl(k_{1}+2k_{2}+2k_{3}+k_{4}\bigr)

Try it

A Runge–Kutta step with $h=1$ on $y'=y$ , $y(0)=1$ has $k_{1}=1$ , $k_{2}=1.5$ , $k_{3}=1.75$ and $k_{4}=2.75$ . What is $y_{1}$ ? Give four decimal places.