Homogenous Matrix-valued ODE with left and right multiplication

Question

Considering the differential equation

$X'=X \cdot A + B \cdot X$,

where $A, B, X: I \rightarrow \mathbb{R}^{n \times n}$ are matrix-valued functions. What can be said about the solutions for such a differential equation? Can it be reduced to a simpler case? Does it have a name?

To give a bit of context: When looking for canonical curves in the context of Cartan geometries, I came upon this differential equation. In my case $B$ happens to be constant and $X$ is required to be of a relatively special form, namely

$X = \begin{pmatrix} a^{-1} & v & -\frac{1}{2} a \langle v,v \rangle_{p,q} \\ 0 & A & -aAv \\ 0 & 0 & a \end{pmatrix}$

for some $a: I \rightarrow \mathbb{R}$ a real valued function, $v:I \rightarrow \mathbb{R}^n$ a vector valued function, $A: I \rightarrow \mathcal{O}(p,q)$ an orthogonal matrix at each point in time. In particular $X$ is an upper triangular matrix with determinant $1$.

It looks very similar to the case $X'=X \cdot A$, where a solution is given by the matrix exponential, but I did not find that to be of any help so far.

Answer 1

If $A,B$ are constant matrices, then the solution s.t. $X(0)=X_0$ is $X(t)=e^{tB}X_0e^{tA}$.

Answer 2

As stated in a comment before: An ODE of the form $X'(t)=A(t)X(t)+X(t)B(t)$ is called Sylvester ODE. It is treated in "Sylvester matrix Differential Equations: Analytical and Numerical Solutions" by Laurene V. Fausett.

I will quickly copy the general solution here: Let $Y$ be a solution of $T'=AT$ and $Z$ be a solution of $T'=TB$. Let $C$ be any constant square matrix. Then $X(t):=Y(t) C Z(t)$ satisfies:

$X'(t) =Y'(t) C Z(t)+ Y(t) C Z'(t) =A(t) Y(t) C Z(t) + Y(t) C Z(t) B(t) =A(t) X(t) + X(t) B(t)$,

i.e. is a solution to the above mentioned ODE. Given any initial data $X(0)=X_0$ we can choose $C$ to satisfy there data. By the Picard-Lindelöf theorem this solution to the IVP is unique.