Solve $\dot{S}(t)=S(t)A+A^TS(t)$.

Question

I am faced with the matrix differential equation $\dot{S}(t)=S(t)A+A^TS(t)$ for which I have to find a solution. I remember I learned that the matrix differential equation $\dot{S}(t) = AS(t)$ has the solution $S = S(0)e^{At}$ but I am not sure this helps me here.

Answer 1

One approach is as follows. Let $s(t) = \operatorname{vec}(S(t))$, where vec denotes the vectorization operator. Vectorizing both sides of the differential equation gives us $$ \dot s(t) = (A^T \otimes I + I \otimes A^T) s(t), $$ where $\otimes$ denotes a Kronecker product. From there, you could apply your solution. If you prefer, we could rewrite this in terms of the Kronecker sum (see the same page) $$ \dot s(t) = (A^T \oplus A^T) s(t). $$

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On the other hand, here is a much fancier solution. Rearrange the equation to get $$ \dot S(t) - S(t) A - A^TS(t) = 0 \implies\\ [e^{-tA^T}\dot S(t) e^{-tA} - A^Te^{-tA^T}S(t)e^{-tA} - e^{-tA^T}S(t) Ae^{tA}] = 0 \implies\\ \frac d{dt} [e^{-tA^T}S(t)e^{-tA}] = 0 \implies\\ e^{-tA^T}S(t)e^{-tA} = S(0) \implies\\ S(t) = e^{tA^T} S(0) e^{tA}. $$ We could have arrived at this conclusion using the first method if we used the identity $$ \exp(t A \oplus tB) = \exp(tA) \otimes \exp(tB). $$