 Euler-Heun embedded pair

The Euler-Heun 1(2) pair combines the explicit Euler method and Heun's method. Here 1(2) names their respective order of a numerical method; the accepted endpoint in this article is the second-order one. Its shared Runge-Kutta stages and endpoints areThe first output row in this Butcher tableau is Heun's method; the second is the explicit Euler method. Both outputs cost two evaluations of $f$ together, rather than one for Euler method plus two for a separately computed Heun's method.

For the initial value problem $y^{\prime }=y$, $y(0)=1$, take $h=0.2$. The first slope is $k_1=1$; the stage state is $1+0.2k_1=1.2$, giving $k_2=1.2$. ThereforeThe exact exponential function gives signed local defects approximately $0.02140275816$ for the explicit Euler method and $0.00140275816$ for Heun's method. Their difference is exactly $0.02$. Thus the pair difference estimates the larger Euler defect; calling it the actual Heun defect would miss by a factor about fourteen here.

More generally, a step from $y_n$ on $y^{\prime }=y$ hasThese original Taylor expansions separate three quantities: the accepted Heun's method has local truncation error $O(h^3)$ and global error $O(h^2)$ under suitable numerical stability, while the tested local error estimator is $O(h^2)$. For $E=|d|/s$ with $s$ held approximately fixed, use $h_{\mathrm{new}}=\eta h{E}^{-1/2}$. A cube-root update would model a different quantity, the actual higher-order local defect.