I am trying to solve a system of ODEs with the Magnus expansion. Unfortunately, even though I know the solution exists (and I also exactly know it), it seems the Magnus expansion is not converging. My system of ODEs is a bit too complicated to write here, but a toy model that exhibits the same problem is the following: $$ \Delta'=B \Delta, ~~~~~~~~~~~~~~~~~(1) $$ with $\Delta(x)=(y(x), z(x))$ and $$ B=\left( \begin{array}{cc} 1 & \frac{1}{x} \\ 1 & \frac{1}{x} \\ \end{array} \right). ~~~~~(2) $$ I set the initial conditions at $x_i=10^{-2}$ and I'm only interested in solving in the range $x\in[x_i,1]$. In this case, the system can actually be solved analytically and the solution is given in terms of the exponential integral function $Ei(z)$ (it's too long so, I won't give it here).
Trying to solve Eq. (1) with the Magnus expansion fails, as it seems the series doesn't converge since the integral of the norm $\int_0^T ||B(x)||_2 dx $ doesn't converge for any $T$.
My question is: Is there any way to fix this?
What I have tried:
1) Change of variables (from $x$ to $a$):
This doesn't work as $\Delta'(x)=\frac{d\Delta}{dx}=\frac{d\Delta}{da}\frac{da}{dx}$ and Eq (1) becomes:
$$ \Delta'(a)=B \frac{dx}{da} \Delta(a), \\ =\tilde{B}\Delta(a) $$ In this case the hope is that $\tilde{B}$ will not be singular, however using the properties of the norm $\int_0^T ||\tilde{B}(a)||_2 da=\int_0^T ||B(x)||_2 \frac{dx}{da} da= \int_0^T ||B(x)||_2 dx$, which again doesn't converge.
2) Redefining $\Delta$:
Assume that $\Delta(x)=\Gamma(x) \tilde{\Delta}(x)$, then the system of Eq.(1) can be written as $$ \tilde{\Delta}'=\Gamma^{-1}(B\Gamma-\Gamma') \tilde{\Delta}, ~~~~~~~~~~~~~~~~~(3) $$
One solution would then be to demand that $\Gamma^{-1}(B\Gamma-\Gamma')$ is the 2x2 Identity matrix $I_2$, ie we can rewrite (3) as $$ \tilde{\Delta}'= \tilde{\Delta} $$ but then this means that $\Gamma$ satisfies $\Gamma'=(B-I_2)\Gamma$, which brings us back to square one.
Any help is appreciated.
PS The Matrix to the full problem (whose analytical solution is also known) is given by $$ B=\left( \begin{array}{cc} \frac{27}{20 x^2} & \frac{243 + 10 (27 - 20 x) x}{ 20 \sqrt{30} x^{\frac{5}{2}}} \\ -\frac{3 \sqrt{\frac{3}{10}}}{2 x^{\frac{3}{2}}} & -\frac{ 27 + 20 x}{20 x^2} \\ \end{array} \right), ~~~~~(4) $$ while the initial conditions are $\Delta_0=(\frac{91}{100}, -\frac{\sqrt{\frac{3}{10}}}{10}$) at $x_0=10^{-2}$.