Order of accuracy of operator splitting Consider the initial value problem: $$\frac{\partial\phi}{\partial t}+A(\phi)=0 \qquad in \quad (0,T),$$ $$\phi(0)=\phi_0$$
Where $A$ is a linear operator (smooth enough) from a Hilbert space to itself. Suppose that $A$ has the decomposition: $$A=A_1+A_2$$
Then the solution of the IVP can be approximated in time by the solution of the following scheme . Let $\Delta t>0$ be a time discretization step such that $t_n =n\Delta t$ and let $\phi^n$ be an approximation of $\phi(t_n)$ then for $n \geq 0$ we compute $\phi^{n+1}$ by solving for $i=1,2$ $$\frac{\partial\phi}{\partial t}+A_i(\phi)=0 \qquad in \quad (t_n,t_{n+1}),$$ 
$$\phi(t_n)= \phi^{n+(i-1) / 2}$$
$$\phi(t_{n+1})= \phi^{n+i / 2}$$
How can we show that this method is first order accurate? i.e. $||\phi(t_n) - \phi^n|| =O(\Delta t)$ 
 A: The exact solution satisfies $\phi(t^{n+1}) = e^{\Delta t\, (A_2 + A_1)} \phi(t^{n})$ over one full step. Let us assume that each sub-step of the present splitting scheme (cf. the fractional step method in this article, or Godunov splitting) is solved exactly. Thus, we obtain on the first sub-step ($i=1$)
$$
\phi^{n+1/2} = e^{\Delta t\, A_1} \phi^n\, ,
$$
while on the second sub-step ($i=2$), we obtain
$$
\phi^{n+1} = e^{\Delta t\, A_2} \phi^{n+1/2} = e^{\Delta t\, A_2} e^{\Delta t\, A_1} \phi^n\, .
$$
By definition of the matrix exponential,
$$
e^{\Delta t\, (A_2 + A_1)} = I + \Delta t\, (A_2 + A_1) + \frac{{\Delta t}^2}{2} ({A_2}^2 + A_2A_1 + A_1A_2 + {A_1}^2) + \dots
$$
and
$$
e^{\Delta t\, A_2} e^{\Delta t\, A_1} = I + \Delta t\, (A_2 + A_1) + \frac{{\Delta t}^2}{2} ({A_2}^2 + 2A_2A_1 + {A_1}^2) + \dots
$$
Therefore, the Local Truncation Error (LTE) induced by the splitting is $O({\Delta t}^2)$:
$$
\phi^{n+1} - \phi(t^{n+1}) = \frac{{\Delta t}^2}{2}(A_2A_1 - A_1A_2)\phi^n + O({\Delta t}^3) \, .
$$
Finally, the global truncation error is $O(\Delta t)$.

Note that this splitting scheme is exact if $A_2$ and $A_1$ commute.
