= Solution
Put $\delta=\Delta t$ and $L=A+B$, keeping the spatial grid fixed for the temporal error assertion. The <matrix exponential> expansions give
$$
S_\delta:=e^{\delta A}e^{\delta B}
=I+\delta L+\delta^2(\tfrac12A^2+AB+\tfrac12B^2)+O(\delta^3),
$$
$$
E_\delta:=e^{\delta L}=I+\delta L+\tfrac12\delta^2(A^2+AB+BA+B^2)+O(\delta^3).
$$
Hence the <Lie-Trotter splitting commutator error> is
$$
S_\delta-E_\delta=\tfrac12\delta^2[A,B]+O(\delta^3).
$$
The energy estimate from part 1 makes $e^{tA}$, $e^{tB}$ and $e^{tL}$ contractions, so both $S_\delta$ and $E_\delta$ have norm at most one. Telescope their $n$th powers:
$$
S_\delta^n-E_\delta^n
=\sum_{j=0}^{n-1}S_\delta^{n-1-j}(S_\delta-E_\delta)E_\delta^j.
$$
For $n\delta\le T$, this gives
$$
\boxed{\|S_\delta^nu^0-e^{n\delta L}u^0\|_h
\le nC_h\delta^2\|u^0\|_h\le C_hT\delta\|u^0\|_h.}
$$
Thus the <Lie-Trotter splitting> has global temporal error $O(\Delta t)$. The constant here may depend on the fixed spatial operators; a mesh-uniform joint-limit error estimate needs additional regularity and <commutator> bounds. If $A$ and $B$ commute, this particular exponential splitting is exact, but variable diffusion coefficients generally remove that commutativity.
Back to article page
