Is Strang Splitting Second-Order Accurate for Time-Dependent Operators?

[Dustinsfl]

How do I show Strang splitting has second order accuracy?

For the the following equation
\[
u_t = (L + M)u,
\]
where \(L\) and \(M\) are linear time-dependent operators, I am trying to show that the Strang splitting scheme
\[
u_{n + 1} = e^{LH / 2}e^{MH}e^{LH / 2}u_n
\]
has second-order accuracy in time (here \(h\) is the time step).

I am not sure how to do that though.

 

[Aaron Bex]

To prove that Strang splitting has second-order accuracy in time, we need to show that the error of the scheme is \(O(h^2)\) as \(h \rightarrow 0\). We can do this by Taylor expanding the exact solution \(u\) and the approximate solution \(u_{n+1}\) about the point \(nh\).

For the exact solution, we have:
\begin{align*}
u(t_{n+1}) &= u(nh + h) \\
&= u(nh) + hu'(nh) + \frac{h^2}{2}u''(nh) + O(h^3)
\end{align*}

For the approximate solution, we have:
\begin{align*}
u_{n+1} &= e^{LH/2}e^{MH}e^{LH/2}u_n \\
&= u_n + \frac{h^2}{4}[(L + M)^2 + (L + M)(LH/2) + (LH/2)(L + M)]u_n + O(h^3)
\end{align*}

Comparing the two expressions for the exact and approximate solutions, we can see that the error between them is:
\begin{align