 Solution

Put $\delta =\Delta t$ and $L=A+B$, keeping the spatial grid fixed for the temporal error assertion. The matrix exponential expansions giveHence the Lie-Trotter splitting commutator error isThe 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:For $n\delta \le T$, this givesThus 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.