Explicit counter-example for $\exp(\int A)' = A \exp(\int A)$

Question

If $A \in \mathbb R^{d \times d}$ is a matrix it is known for the purpose of ODE theory that $$ \frac{d}{dt}e^{tA} = A e^{tA}. $$ It is also known that this does not easily generalizes to non autonomous case: $$ \exp\left( \int_0^t A(s)ds \right) X_0 $$ does not give necessarily the solution of the Cauchy problem $$ \dot X (t) = A(t) X(t),\quad X(0)=X_0. $$ I am wondering if there are easy and explicit situations where this phenomenon could be observed, that is $A : \mathbb R \rightarrow \mathbb R^{d \times d}$ smooth but $$ \frac{d}{dt} \exp\left( \int_0^t A(s)ds \right) \neq A(t) \exp\left( \int_0^t A(s)ds \right). $$ Actually any discussion on the differentiation of the above quantity is of interest.

Answer 1

Try $A(t) = \begin{bmatrix} 1 & t \\ 0 & 0 \end{bmatrix}$. Then $B(t)=\int_0^t A(s)ds = \begin{bmatrix} t & {1 \over 2} t^2 \\ 0 & 0 \end{bmatrix}$, and $B^n(t) = t^n A({1 \over 2} t)$ for $n \ge 1$. Hence $\Phi(t)=e^{B(t)} = I + (e^t-1) A({ 1\over 2} t) = \begin{bmatrix} e^t & {1 \over 2} t (e^t -1 ) \\ 0 & 1 \end{bmatrix}$.

Then $\dot{\Phi}(t)= \begin{bmatrix} e^t & {1 \over 2} (t e^t+e^t -1 ) \\ 0 & 0 \end{bmatrix}$ and $A(t)\Phi(t) = \begin{bmatrix} e^t & {1 \over 2} (t e^t+t ) \\ 0 & 0 \end{bmatrix}$.

Assuming I have made no mistake in computation, $\dot{\Phi}(t) \ne A(t)\Phi(t)$ for $t \neq 0$.

Answer 2

This is not a full answer, but it could lead to constructing counterexamples:

Theorem. If $A:I\to \mathbb R^{d\times d}$, continuous, where $I$ interval, and $$ A(s)A(t)=A(t)A(s), \quad \text{for all $s,t\in I$}, $$ then $$ \frac{d}{dt}\exp\Big(\int_\tau^t A(s)\,ds\Big)=A(t)\exp\Big(\int_\tau^t A(s)\,ds\Big). $$