# Strang splitting

I have recently come across the Strang splitting and have some questions. For the differential equation of the form

$$dy/dt = (L_1 + L_2)y$$

Strang splitting implement the time splitting as

$$\begin{eqnarray*} \tilde y_1 & = & e^{L_1 \Delta t/2} y_0, & & \bar y_1 = e^{L_2 \Delta t} \tilde y_1, & & y_1 = e^{L_1 \Delta t/2} \bar y_1 \\ \tilde y_2 & = & e^{L_1 \Delta t/2} y_1, & & \bar y_2 = e^{L_2 \Delta t} \tilde y_2, & & y_2 = e^{L_1 \Delta t/2} \bar y_2 \\ ... \\ \tilde y_n & = & e^{L_1 \Delta t/2} y_{n-1}, & & \bar y_n = e^{L_2 \Delta t} \tilde y_n, & & y_n = e^{L_1 \Delta t/2} \bar y_n \\ \end{eqnarray*}$$

However, from the equations above, it is obviously that the half time step with $y_1$ and $\tilde y_2$ can be combined into a single time step, so it is equivalent to:

$$\begin{eqnarray*} \tilde y_1 & = & e^{L_1 \Delta t/2} y_0, & & \bar y_1 = e^{L_2 \Delta t} \tilde y_1 \\ \tilde y_2 & = & e^{L_1 \Delta t} \bar y_1, & & \bar y_2 = e^{L_2 \Delta t} \tilde y_2 \\ ... \\ \tilde y_n & = & e^{L_1 \Delta t} \bar y_{n-1}, & & \bar y_n = e^{L_2 \Delta t} \tilde y_n \\ y_n & = & e^{L_1 \Delta t/2} \bar y_n \end{eqnarray*}$$

This look like that it is exactly the same as the first order time splitting scheme except the first and last half time step, and the computation is faster with the reduction. Am I missing something here?

Also, what is the time splitting method with four order? How does it look like explicitly? I am looking for some numerical algorithms for solving nonlinear Schroedinger equation.

• You're not missing anything, I think: the first-order splitting scheme shadows the true solutions to second-order accuracy—it gives a second-order accurate solution to a modified problem where initial values are modified slightly (by $O(\Delta t)$), and your analysis proves as much. It even says so in epubs.siam.org/doi/abs/10.1137/0705041 on p.511. – Kirill Jul 27 '16 at 23:25

This look like that it is exactly the same as the first order time splitting scheme except the first and last half time step, and the computation is faster with the reduction. Am I missing something here?

You're exactly correct, as others have already said.

Also, what is the time splitting method with four order?

There are many higher-order splitting schemes, some developed very recently. This is the most comprehensive list I know of and includes various methods of order 4.

• Thanks. It seems that a good choice is a splitting with real coefficient and least number of step $s$. Can you explain a little bit why there are many difference method? I expect there is a unique BCH expansion for a given order. What are the advantage of complex coefficient over the real coefficient? – unsym Jul 29 '16 at 0:30
• @hwlau Please go ahead and post any questions you have as new questions. It's not recommended to post additional questions as comments. – David Ketcheson Jul 29 '16 at 10:38

You can combine those two steps since both of them involve the same operator ($L_1$).

For the fourth-order method I recommend that you check out my answer to a similar question on Stack Overflow and also Section 4 in the following paper (see also the discussion below):

Yoshida, H. (1990). Construction of higher order symplectic integrators. Physics Letters A, 150(5–7), 262–268. http://doi.org/10.1016/0375-9601(90)90092-3

Let $a = L_1$ and $b = L_2$. The analysis of this type of schemes is done using the Baker-Campbell-Hausdorff (BCH) formula, which tells us that the expression for $c$ in $$\exp\{t a\} \exp\{t b\} = \exp\{t c\},$$ is given in terms of Lie brackets involving $a$ and $b$ (recall that $[a, b] = ab - ba$): $$c = a + b + \frac{t}{2} \left[a, b\right] + \frac{t^2}{12} \left( \left[a, \left[a, b\right]\right] + \left[b, \left[b, a\right]\right] \right) + \frac{t^3}{24} \left[a, \left[b, \left[b, a\right]\right]\right] + \dotsb$$ The BCH formula for the particular case of the Strang splitting is given by $$\exp\left\{\tfrac{t}{2} a\right\} \exp\left\{t b\right\} \exp\left\{\tfrac{t}{2} a\right\} = \exp\left\{t c_1 + t^3 c_3 + t^5 c_5 + \dotsb \right\},$$ where \begin{align*} c_1 &= a + b, \\ c_3 &= -\frac{1}{24} \left[a, \left[a, b\right]\right] + \frac{1}{12} \left[b, \left[b, a\right]\right], \\ &\dotsb \end{align*} In general, $c_3$ does not vanish and this implies that the scheme is second order. In order to construct higher order methods, one can choose a splitting such that the appropriate terms in the BCH formula vanish (the paper by Yoshida cited above discusses this in depth).

Particular choices of $a$ and $b$ will yield symplectic methods but from the discussion above we see that the order of the scheme does not necessarily have to do with it being symplectic.

• It is still quite vague. In my understading, we can only have symplectic integrator when the operator splitting takes the form $L_1(q)$ and $L_2(p)$ which is not always the case, right? – unsym Jul 29 '16 at 0:40
• That is one way of obtaining a symplectic integrator. The reason why that is the case is because the systems $dq/dt = L_1(q), dp/dt = 0$ and $dq/dt = 0$, $dp/dt = L_2(p)$ can be solved exactly, and the solutions are of course a symplectic mappings. Therefore, the composition of $\exp{t L_1}$ and $\exp{t L_2}$ will be symplectic.One could devise other symplectic integrators, though (e.g.: exploiting particular features of the vector fields under consideration). – Juan M. Bello-Rivas Jul 29 '16 at 9:50