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If $A$ is a symmetric matrix satisfying

$$ \dot A = AB^T + B A, \qquad A(0) = A_0, $$

for a square matrix $B$ and a symmetric matrix $A_0$, then

$$ A(t) = e^{Bt} \, A_0 \, e^{B^T t}. $$

Now consider that $B$ depends on time, i.e. $B = B(t)$, and define $\Pi(\cdot, \cdot)$ as the matrix function such that, for any $s, t \in \mathbb R$ and any $x_0$, $x(t) = \Pi(s, t) x_0$ satisfies:

$$ \dot x(t) = B(t) \, x(t), \qquad x(s) = x_0. $$

Clearly, $\Pi(s, s) = I$ for all $s$, and $\partial_t \Pi(s, t) = B \, \Pi(s, t)$.

Does a formula similar to the one above for $A(t)$ exist in this case, possibly using $\Pi(\cdot, \cdot)$ instead of the matrix exponential? The naive extension

$$ A(t) = \Pi(0, t) \, A_0 \, \Pi(0, t)^T $$

does not work, because $B(t)$ and $\Pi(0, t)$ do not necessarily commute.

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  • $\begingroup$ Never mind, the naive extension actually works. :) $\endgroup$ – Roberto Rastapopoulos Jun 12 at 13:45

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