Heun’s method · Differential Equations

Intuition

Euler’s method uses the slope at the start of a step as if it held all the way across, and it does not: by the end of the step the slope has changed. Heun’s idea is to take the Euler step only as a trial, read the slope where it lands, and then take the real step with the average of the two slopes. The average is much closer to the true mean slope over the step, and the error falls with the square of the step: halve h and the error is a quarter.

Estimating the average speed of a drive from the speedometer at the start alone, or from the start and the finish. Two readings averaged are far closer to the truth than one, for the price of looking twice.

One step of Heun’s method with $h=1$ on $y'=y$ from $(0,1)$ . The trial Euler step lands at $2$ , where the slope is $2$ ; averaged with the slope $1$ at the start it gives $1.5$ , and the step with that slope ends at $2.5$ — against Euler’s $2$ and the solution’s $e\approx2.72$ .

Predict, then correct

Heun’s method, also called the improved Euler method: take a trial Euler step, then step again from $(x_{n},y_{n})$ with the average of the slope at the start and the slope at the trial point. In the integral form, it replaces the integral of the slope over the step by the trapezium rule. Each step evaluates $f$ twice, and the method is of order two: its global error is at most a constant times .

\begin{aligned}\tilde y&=y_{n}+h\,f(x_{n},y_{n})\\y_{n+1}&=y_{n}+\tfrac{h}{2}\bigl[f(x_{n},y_{n})+f(x_{n+1},\tilde y)\bigr]\end{aligned}

Two slopes instead of one

The trial value $\tilde y$ is the predictor and the averaged step the corrector; only the corrected value is kept.
Writing $k_{1}=f(x_{n},y_{n})$ and , the step is .

Heun’s step matches the exponential to second order

y'=\lambda y\ \Longrightarrow\ y_{n+1}=\Bigl(1+h\lambda+\frac{(h\lambda)^{2}}{2}\Bigr)y_{n}

On y′ = λy the two slopes can be written out. The slope at the start is λ times y_n; the trial point is (1 + hλ) times y_n, where the slope is λ(1 + hλ) times y_n. Averaging and stepping multiplies y_n by 1 + hλ + (hλ)²/2 — the first three terms of the series of e^{hλ}, which is what the exact solution multiplies by over one step. The first term missed is (hλ)³/6, so the local error is of order h³, and over a number of steps proportional to 1/h the global error is of order h².

Proof steps

k_{1}=\lambda y_{n}

The slope at the start of the step.

\tilde y=y_{n}+h\lambda y_{n}=(1+h\lambda)\,y_{n}

The trial Euler step.

k_{2}=\lambda\tilde y=\lambda(1+h\lambda)\,y_{n}

The slope at the trial point.

\begin{aligned}y_{n+1}&=y_{n}+\tfrac h2(k_{1}+k_{2})\\&=\Bigl(1+h\lambda+\tfrac{(h\lambda)^{2}}{2}\Bigr)y_{n}\end{aligned}

Average the two slopes and step.

e^{h\lambda}=1+h\lambda+\tfrac{(h\lambda)^{2}}{2}+\tfrac{(h\lambda)^{3}}{6}+\cdots

The exact solution multiplies by $e^{h\lambda}$ : the step agrees up to the $h^{2}$ term, so the local error is of order $h^{3}$ .

Applications

Practice

Two Slopes, Averaged

Take a trial Euler step, read the slope where it lands, and step again from the start with the average of the two slopes.

y_{n+1}=y_{n}+\tfrac{h}{2}\bigl(k_{1}+k_{2}\bigr)

Try it

One step of Heun’s method with $h=1$ on $y'=y$ , $y(0)=1$ : what is $y_{1}$ ?