I have the following problem
$$
\dot{R}(t)=R(t)
\begin{pmatrix}
0 & 0 & -\cos{t} \\
0 & 0 & -\sin{t} \\
\cos{t} & \sin t & 0
\end{pmatrix}
\equiv R(t) A(t), \quad R(0)=\mathbb{1}.
$$
Basicly, I need to compute time orderd exponential. I was thinking of formula
$$ R(t) = e^{\int_0^{t}A(\tau)d\tau},$$
but it turned out that $[A(t), A(t')]\neq 0$. So, the only way is to use general formula
$$
\begin{equation}
R(t) = \mathcal{T}\{ e^{\int_0^{t} A(\tau) d\tau}\},
\end{equation}
$$
where $\mathcal{T}$ denotes time-ordering,
$$\mathcal{T}\{ e^{\int_0^{t} A(\tau) d\tau}\} \equiv \sum\limits_{n=0}^{\infty} \frac{1}{n!}\int_0^t \ldots \int_0^t \mathcal{T}\{A(\tau_1) \ldots A(\tau_n) \}.$$
I wonder, is there any way to do it in this general case? i also have the following information about this system
- The eigen values of $A(t)$ are $i, -i, 0$ and corresponding eigen vectors are
$$
\begin{pmatrix}
i\cos{t} \\
i\sin t \\
1
\end{pmatrix},
\begin{pmatrix}
-i\cos{t} \\
-i\sin t \\
1
\end{pmatrix},
\begin{pmatrix}
i\tan{t} \\
1\\
0
\end{pmatrix}
$$ - $A(t) = \cos{t} A_1 + \sin t A_2$, where
$$
A_1 =
\begin{pmatrix}
0 & 0 & -1 \\
0 & 0 & 0\\
1 & 0 & 0
\end{pmatrix}, \quad A_2 =
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0
\end{pmatrix}, \quad A_3 =
\begin{pmatrix}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
$$
are generators of Lie algebra $\mathfrak{so}(3)$.
Best Answer
$ \def\z{\zeta} \def\e{{\cal E}} \def\a{\alpha} \def\b{\beta} \def\l{\lambda} \def\o{{\tt1}} \def\p{\partial} \def\BR#1{\left[#1\right]} \def\LR#1{\left(#1\right)} \def\unskew#1{\operatorname{unskew}\LR{#1}} \def\skew#1{\operatorname{skew}\LR{#1}} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\qif{\quad\iff\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\mc#1{\left[\begin{array}{c}#1\end{array}\right]} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} \def\frob#1{\left\| #1 \right\|} $It is well known that a rotation matrix can be constructed from a skew-symmetric matrix which in turn can be constructed from a vector $$\eqalign{ R &= \exp(B),\quad B = \m{ 0 & -b_3 & b_2 \\ b_3 & 0 & -b_1 \\ -b_2 & b_1 & 0 \\ },\quad b = \m{b_1\\b_2\\b_3} \\ B &= \skew b = -\e\cdot b \\ b &= \unskew B = -\tfrac12\e:B \\ b &= 0 \qiq B=0 \qiq R=I \\ }$$ ${\large[}\,\sf NB\!:\;\e$ is the Levi-Civita tensor$\large\,]$
This is pretty standard stuff, but this paper contains an amazing formula $$\eqalign{ Y &= \fracLR{bb^T+(R^T-I)B}{\frob{b}^2} \qiq \grad Rb = -R\cdot\e\cdot Y \\ }$$ which can be applied to your problem $$\eqalign{ \dot R = \gradLR Rb\cdot\dot b = R\cdot\CLR{-\e\cdot Y\cdot\dot b} = R\:\c{\skew{Y\dot b}} \,\equiv\, R\c{A} \\ }$$ This reduces your problem to $$\eqalign{ &{Y\dot b} = \unskew A = a \;\equiv\; \m{\sin t\\-\cos t\\0\;\;\;} \\ &\dot b = Y^{-1} a,\qquad b(0) = 0 \\ }$$ Your original problem is a matrix-valued ODE. Worse, it requires special processing to maintain the orthogonality of $R$ after every time-step.
The reduced equation is a simple (non-stiff) vector ODE with no orthogonality constraints.
Better yet, an explicit formula exists for the matrix inverse $$\eqalign{ &Y^{-1} = I \;+\; \tfrac12 B \;+\; \z B^2 \\ &\b = \frob b,\qquad \z = \fracLR{{\o-\cos\b}-\tfrac12\b\sin\b}{{\b^2-\b^2\cos\b}} \\ \\ &\lim_{\b\to0}\: \z B^2 \,=\; \lim_{\b\to0}\, B = 0 \qiq \lim_{\b\to0}\: Y^{-1} = I \\ }$$ The formula for $Y^{-1}$ has a nice limit at $\b=0,$ however $Y$ is singular at other points where $\b$ is a multiple of $2\pi.$
Similar formulas exist for $\LR{R,Y}$ in terms of $B$ $$\eqalign{ &R = \LR{I+\l B+\mu B^2},\qquad\quad Y = \LR{I-\mu B+\nu B^2} \\ &\l = \fracLR{\sin\b}{\b},\quad \mu=\fracLR{\o-\cos\b}{\b^2},\quad \nu=\fracLR{\o-\l}{\b^2} \\ &\lim_{\b\to0}\: \left\{\l,\,\mu,\,\nu,\,\z\right\} \,=\; \Big\{\o,\,\tfrac12,\,\tfrac16,\,\tfrac1{12}\Big\} \\ }$$ and these are stable for all values of $\b$