Heun's Method

Heun’s Method improves the estimation of the slope done by Euler’s Method by calculating an average slope over the interval $h$. This is a predictor/corrector method.

Starting on the left side, we can evaluate the slope using the equation and our initial condition (the predictor).

$$y_i^{\prime }=f(x_i,y_i)$$

The right side (the corrector) has slope:

$$y_{i+1}^{\prime }=f(x_{i+1},y_{i+1}^0)$$

This is more difficult since we don’t know $y_{i+1}$ to begin. We can estimate $y_{i+1}$ to begin, using the same extrapolation as the Euler’s method to get $y_{i+1}^0$, an intermediate value.

$$y_{i+1}^0=y_i+f(x_i,y_i)h$$

Then, we find the average slope over the interval:

$${\bar{y}}^{\prime }=\frac{y_i^{\prime }+y_{i+1}^{\prime }}{2}=\frac{f(x_i+y_i)+f(x_{i+1},y_{i+1}^0)}{2}$$

This average slope is then used to extrapolate to the next point:

$$y_{i+1}=y_i+{\bar{y}}^{\prime }h$$

Example

Let’s say we have $\frac{dy}{dx}=\frac{x^3+1}{y}$, $0\le x\le 10$, $h=0.5$ with an I.C. of $y(0)=2$.

The left slope (predictor) would be:

$$y^{\prime }=f(x_0,y_0)=\left(\frac{0^3+1}{2}\right)=\frac{1}{2}$$

Then:

$$y_{i+1}^0=y_i+f(x_i,y_i)h=2+\left(\frac{1}{2}\right)(0.5)=2.25$$

Then, the slope at $y_{i+1}$ is $f(x_{i+1},y_{i+1}^0)$:

$$y_{i+1}^0=\frac{0.5^3+1}{2.25}=\frac{1}{2}$$

Then:

$$\bar{y}=\frac{1/2+1/2}{2}=\frac{1}{2}$$

and

$$y=2+\left(\frac{1}{2}\right)0.5=2.25$$