Strang splitting

Remarks

For clause 1, expand both sides in powers of $h$: the terms of order $0$, $1$ and $2$ agree, because the symmetric product gives $\frac{h^2}{2}\left(\frac{A^2}{4} + \frac{A^2}{4} + B^2 + AB + BA + \frac{A^2}{2}\right) = \frac{h^2}{2}(A + B)^2$. For clause 2, write the difference of $n$-th powers as a telescoping sum of $n$ terms. Each term is a local error $O(n^{-3})$ multiplied by matrices of norm at most $e^{t(\norm{A} + \norm{B})}$. The Lie–Trotter product formula uses $e^{tA/n} e^{tB/n}$ instead, and its error is only $O(n^{-1})$; the symmetric arrangement gains one order. If $AB = BA$, both splittings are exact. In applications, $e^{hA}$ and $e^{hB}$ are replaced by the flows $\varphi^A$ and $\varphi^B$ of $u' = A(u)$ and $u' = B(u)$ for nonlinear operators. The composition $\varphi^A_{h/2} \circ \varphi^B_h \circ \varphi^A_{h/2}$ is still a method of order $2$ for smooth problems.

Going deeper

For a student

Strang splitting is used for reaction-diffusion equations, for the Schrödinger equation (kinetic and potential parts, solved by FFT and pointwise), and for advection in several dimensions. Consecutive half steps of $A$ merge, so the cost is nearly that of the first-order Lie–Trotter splitting. The leapfrog method for Hamiltonian systems is the Strang splitting of the kinetic and potential flows.