Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A / 5

Near $t=0$, take $\Omega (t)=\log F(t)$ to be the continuous local matrix logarithm with $\Omega (0)=0$. The symmetry identity gives $F(-t)=F(t{)}^{-1}$, and uniqueness of this logarithm yieldsThus $\Omega$ is an odd function. The local-logarithm qualification is necessary because the matrix exponential is not globally injective; the statement is naturally understood either near $t=0$ or as an identity of formal power series.

For the Strang splittingreversing $t$ reverses all three factors, soIt is therefore a time-symmetric numerical method. Multiplication of the three exponential series shows agreement with $e^{t(A+B)}$ through degree two. Equivalently, the Baker--Campbell--Hausdorff formula gives an odd modified generatorfor a matrix $D$ made from nested commutators. Exponentiating givesfor a matrix $C$ depending on $A$ and $B$.

The three substeps have total length $2\alpha +(1-2\alpha )=1$. Their leading cubic defects add, so cancellation requiresTaking the real cube root and solving givesThis is the coefficient in the higher-order composition of a symmetric splitting; the middle substep $1-2\alpha$ is negative.

The composition is palindromic, and henceThus $G_{\alpha }$ is symmetric. Part c cancels the cubic term in its odd formal logarithm; symmetry forbids a fourth-degree term, so the next possible defect has degree five. Thereforewhich makes the composition a fourth-order method.