In many papers about Magnus integrators, like this one at slide
7, one truncates the Magnus expansion and apply a quadrature rule to the integrals.
My question is about the expression one gets if uses just the first two terms and apply two-points Gauss quadrature rule with nodes $c_{1,2}=\frac{1}{2} \mp \frac{\sqrt{3}}{6}$.
The term to approximate is $$\int_0^h A(t_n + \tau) d \tau – \frac{1}{2}\int_0^h [\int_0^{\tau}A(t_n+ \sigma) d\sigma,A(t_n+\tau)] d \tau$$
where $$[\cdot,\cdot]$$ is the classical commutator.
For the first integral it's easy to see that applying Gauss quadrature rule with those $c_1,c_2$ its approximation is $$ \frac{h}{2}(A(t_n + c_1 h) + A(t_n + c_2h))$$
My question is about the other one: I'm having serious problem in finding the term in the literature
What I have to approximate is $$\int_0^h [\int_0^{\tau}A(t_n+ \sigma) d\sigma,A(t_n+\tau)] d \tau$$
The first thing that I had in mind is to see the inner part of the integral as a function of $\tau$,i.e. $$f(\tau)=[\int_0^{\tau}A_n(\sigma) d \sigma,A_n(\tau)]$$
and hence I see my integral as $I = \int_0^h f(\tau) d \tau$
Now, applying Gauss 2-point quadrature I get
$$ I \approx \frac{h}{2}(f(c_1 h)+f(c_2 h))$$
The problem is that now I don't know how to see that it is equal to $\frac{\sqrt{3}}{12}h^2[A_n(c_2h),A_n(c_1h)]$
Best Answer
A journal reference with explanation and analysis is
Iserles, A.; Nørsett, S. P., On the solution of lienar differential equations in Lie groups, Philos. Trans. R. Soc. Lond., Ser. A, Math. Phys. Eng. Sci. 357, No. 1754, 983-1019 (1999). ZBL0958.65080.
See the discussion after Corollary 3.3 in particular. The idea is to approximate $A$ by an interpolating polynomial, in this case
$$A(x)\approx A(c_1h)\frac{x-c_2h}{c_1h-c_2h} + A(c_2h)\frac{x-c_1h}{c_2h-c_1h}$$
I'll set $t_n=0$ and $h=1.$ Then \begin{align*} &\int_0^h \int_0^\kappa [A(\xi),A(\kappa)] \;d\xi \;d\kappa\\ &\approx[A(c_1),A(c_2)]\int_0^1 \int_0^\kappa \frac{\xi-c_2}{c_1-c_2}\frac{\kappa-c_1}{c_2-c_1}-\frac{\kappa-c_2}{c_1-c_2}\frac{\xi-c_1}{c_2-c_1} \;d\xi \;d\kappa\\ &=[A(c_1),A(c_2)]\int_0^1 \int_0^\kappa \frac{\kappa-\xi}{c_2-c_1} \;d\xi \;d\kappa\\ &=[A(c_1),A(c_2)]\int_0^1 \frac{\tfrac12\kappa^2}{c_2-c_1} \;d\kappa\\ &=[A(c_1),A(c_2)]\frac{\sqrt{3}}{6} \end{align*}
Going back to the original question, the term you're asking about is
$$ - \frac{1}{2}\int_0^h [\int_0^{\tau}A(t_n+ \sigma) d\sigma,A(t_n+\tau)] \;d \tau$$ Changing variable names (to match the Iserles and Nørsett reference), setting $t_n=0,$ and pulling out the integral, this term is \begin{align*} &- \frac{1}{2}\int_0^h \int_0^{\kappa}[A(\xi),A(\kappa)] \;d\xi \;d \kappa\\ &=-\frac12 [A(c_1),A(c_2)]\frac{\sqrt{3}}{6}\qquad\text{ by the previous calculation}\\ &=\frac12 [A(c_2),A(c_1)]\frac{\sqrt{3}}{6}\qquad\text{because $[a,b]=-[b,a]$}\\ &=[A(c_2),A(c_1)]\frac{\sqrt{3}}{12} \end{align*}