For the <Strang splitting>
$$
F(t)=e^{tA/2}e^{tB}e^{tA/2},
$$
reversing $t$ reverses all three factors, so
$$
F(-t)=e^{-tA/2}e^{-tB}e^{-tA/2}=F(t)^{-1}.
$$
It is therefore a <time-symmetric numerical method>. Multiplication of the three <matrix exponential>[exponential] series shows agreement with $e^{t(A+B)}$ through degree two. Equivalently, the <Baker--Campbell--Hausdorff formula> gives an odd modified generator
$$
\log F(t)=t(A+B)+t^3D+O(t^5)
$$
for a matrix $D$ made from nested <commutator>[commutators]. Exponentiating gives
$$
\boxed{F(t)=e^{t(A+B)}+Ct^3+O(t^4)}
$$
for a matrix $C$ depending on $A$ and $B$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-341/section-a/5/b/solution