Let $R\in SO(3)$, and $\widehat \Omega \in \mathfrak{se}(3)$, where $SO(3)$ is the special orthogonal Lie group, and $\mathfrak{se}(3)$ the corresponding Lie algebra. The "hat" operator $\widehat{\cdot}: \mathbb{R}^3 \rightarrow \mathfrak{se}(3)$. Let $u\in\mathbb{R}^3$ be a known control input. I want to simulate the following system in some software environment such as Matlab: $$ \dot R(t) = R(t)\widehat{\Omega}(t)\\ \dot \Omega(t) = u(t). $$ Clearly, the first equation is not defined on $\mathbb{R}^n$ and we cannot use naive integration techniques such as Newton's forward or backward method to integrate the given differential equations, i.e., $R(t+1) = R(t) + dt(R(t)\widehat{\Omega}(t))$, for some small discretization step $dt$. However, we know that the solution of the first differential equation in closed form is $R(t+1) = R(t)exp(dt\widehat{\Omega}(t))$.
Here is my question: can I use the following two equations to integrate the above system without going into the details of integration on manifolds, such as the one mentioned in [1]? $$ R(t+1) = R(t)exp(dt\widehat{\Omega}(t))\\ \Omega(t+1) = \Omega(t) + dt(u(t)). $$
Thanks!
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