Consider the following homogeneous IVP:

$$\begin{cases} \dot{u}(t)+A(t)u(t)=0 \\ u(0)=u_0 \end{cases}$$

for $u:[0,1]\to \mathbb{R}^n$ (some interval to some finite dimensional Hilbert space, let's use this special case for simplicity) and $u_0\in\mathbb{R}^n$.

If $A$ is continuous for $t\in [0,1]$, then this system has a unique solution (Picard–Lindelöf) and one can define the evolution operator $U(t,s)$ such that $u(t)=U(t,0)u_0$. This is all standard. In particular, we could just as well consider continuous and bounded $A$ on $\mathbb{R}$ and extend the solution $u$ to $\mathbb{R}$.

Now suppose $A$ is only (Lebesgue)-ingegrable. Since $u(t)=u_0+\int_0^t A(s)ds$ is absolutely continuous and therefore differentiable almost everywhere, my expectation would be that everything above is still true (in particular, the evolution operator exists and fulfills the same differential equation as $u$) with the only exception that the differential equation is only satisfied "almost everywhere".

This seems to be implied here, too ("$A(t)$ is a bounded operator [...] (or, more generally, are measurable and integrable on every finite interval)"), but there is no proof.

For the ODE, one could use Carathéodory's existence theorem ($A$ is supposed to be bounded for all $t$) and then work some more for uniqueness (theorem 2.1 of Chapter 2 in Coddington/Levinson: Theory of ODEs should do), and then one has to do similar things for the evolution operator, but this seems to me too complicated for the case at hand:

Q: Can I not just use the proof for continuous $A$ (e.g. Picard–Lindelöf and the link above) and the definition of the evolution operator for the linear ODE, and whenever derivatives are involved add "almost everywhere"?

It seems to me that the continuity of $A$ is needed nowhere except to make sure that $u$ and $U$ are differentiable everywhere. However, I have never seen this done (then again, I have found it difficult to find suitable references for linear ODEs with measurable/integrable $A$. Do you have any suggestion?)

  • $\begingroup$ $A$ has to be integrable, not only measurable. $\endgroup$ Dec 8, 2015 at 17:58
  • $\begingroup$ That is indeed true. By now I found a suitable reference that answers all of those questions. $\endgroup$
    – Martin
    Dec 8, 2015 at 17:59
  • $\begingroup$ But it is best if you edit the question to be correct! The question may have stopped being of use to you, but others will come, and surely you see that it is best for it to use the correct hypothesis. $\endgroup$ Dec 8, 2015 at 18:02
  • $\begingroup$ @MarianoSuárez-Alvarez: Done. I had already corrected this in the answer I gave and thought this to be sufficient (I had simply not thought about it - the particular example I'm interested in is trivially integrable), but I fixed it in the question, too. $\endgroup$
    – Martin
    Dec 8, 2015 at 18:21

1 Answer 1


The answer to the question should be yes, if $A$ is integrable.

In any case, another proof that this is indeed true is given in the book "Mathematical Control Theory" by Eduardo D. Sontag (online version available) in Theorem 55 in appendix C. The question about the propagator is also answered in that appendix.


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