Let $H$ be a Hilbert space, and let $A_t$ be a family of unbounded positive (self-adjoint) operators on $H$ parametrized by $\mathbb t\in R_{\ge 0}$. Consider the ordinary differential equation $$ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\frac{d}{dt} E_t = -A_tE_t \quad\qquad\qquad\qquad\qquad\qquad\qquad(1) $$ that defines the ordered exponential of the family $A_t$.

If $A_t=A$ is independ of $t$, then the solution of the ODE is the usual exponential $E_t=e^{-tA}$.
Note that the above operators $E_t$ are bounded.

I suspect that, if I put appropriate hypotheses on $A_t$, (such as having a common dense domain, depending continuously on $t$, whatever that might mean, etc.) the solution of (1) will also be bounded. Intuitively, it's kind of clear: $$ E_t = \lim_{N\to\infty} \Big(e^{-\frac t N A_t} \cdot e^{-\frac t N A_{t(1-1 /N)}} \cdot e^{-\frac t N A_{t(1-2 /N)}}\cdots \cdot e^{-\frac t N A_{t(3 /N)}} \cdot e^{-\frac t N A_{t(2 /N)}} \cdot e^{-\frac t N A_{t(1 /N)}}\Big) $$ and each little exponential in the above product has norm $\le 1$.

Q: Which properties should I impose on the family $A_t$ in order for the solution of (1) to be well defined and bounded?

  • $\begingroup$ It would interest me if you have made any progress. $\endgroup$ – András Bátkai Jan 9 '13 at 10:23
  • $\begingroup$ Don't you need $A_t$ bounded from below to start with? $\endgroup$ – lcv Dec 6 '16 at 2:43
  • $\begingroup$ Opps.. has the word 'positive' been there all along? Anyway there is a fair amount of literature for the somewhat more general case where $-A_t$ is a generator of a contraction semigroup for all (fixed) $t\ge0$. You can look for example at the first lemmas in arxiv.org/abs/1106.4661 $\endgroup$ – lcv Dec 6 '16 at 6:51

This is usually called the magnus expansion method and has a nice literature in numerical analysis. Kato also used this method to show the existence of solutions in the hyperbolic case.

I would say that strong resolvent continuity and a sufficiently big common domain is sufficient in your case. See Section 5.3 in Pazy.

I can also give a related self-reference, where also the investigation of this product appears and some of the ideas are explained in a simpler situation.

ADDED: My answer concentrates on the method you propose to converge to the solution. To make the content of the references short: yes. A common dense domain and continuity of hte maps $t\mapsto A_tx$ implies the convergence of the product to the solution of the differential equation.

  • $\begingroup$ I don't want to assume that the resolvents $A_t^{-1}$ are bounded. How should I interpret "strong resolvent continuity" then? $\endgroup$ – André Henriques Oct 10 '12 at 19:48
  • $\begingroup$ What about $(1+A_t)^{-1}$? $\endgroup$ – András Bátkai Oct 10 '12 at 19:59
  • $\begingroup$ This is usually satisfied if you have a large common domain ant the maps $t\mapsto A_t x$ are continuous. $\endgroup$ – András Bátkai Oct 10 '12 at 20:00
  • $\begingroup$ @André Henriques: What do you mean then by the statement you do not want to assume boundedness of resolvents? By selfadjointness, you have planty of resolvent points... $\endgroup$ – András Bátkai Oct 11 '12 at 23:11
  • $\begingroup$ I mistakenly thought that "resolvent" means the same thing as "inverse". $\endgroup$ – André Henriques Oct 12 '12 at 15:55

There are zillions of articles and books on this topic, for more and more general families of operators, where usually you can choose a specific trade-off between regularity of solutions, variability of domains of the operators, and smoothness of the dependence on time.

In the "variational" setting you consider, it is probably most efficient to introduce a weak formulation based on a family of quadratic forms.

In this case, there is a classical theory by Lions (see e.g. Chapt. 3 of Equations Differentielles Opérationelles et Problèmes aux Limites, Springer 1961), which essentially says that if the family of forms is equi-continuous and equi-coercive, and the dependence on time is merely measurable, then you have well-posedness (in a certain weak sense) of your equation (1) (even if you add an inhomogeneous term in the equation), as well as boundedness. Also Chapt. 4 of Tanabe's Equations of Evolutions (Pitman 1979) is a good reference.

  • $\begingroup$ May I ask: what does "equi-coercive" mean? $\endgroup$ – André Henriques Oct 10 '12 at 19:52
  • $\begingroup$ Take the quadratic form associated with the operator $A_t$, say $a_t$. This is defined on a Hilbert space, say $V_t$, but as I wrote above in Lions' scenario one is assuming that $V_t\equiv V$. For a certain fixed $t$, coercivity of $a_t$ means that $$a_t(u)\ge \alpha_t \|u\|_V$$ for some $\alpha_t>0$ and all $u\in V$. Equi-coercivity of $(a_t)$ means that the same estimate holds for all $t$ for some constant $\alpha$ not depending on $t$, i.e., $$a_t(u)\ge \alpha \|u\|_V$$ for some $\alpha>0$ and all $t$. $\endgroup$ – Delio Mugnolo Oct 10 '12 at 20:04
  • $\begingroup$ (and all $u\in V$, of course). $\endgroup$ – Delio Mugnolo Oct 10 '12 at 20:04

Certainly, one should pay attention to the domain of the operator. However, the following argument should survive a reasonable assumption. We have $$ \frac{d}{dt}(E^\ast(t)E(t))=-2 E^\ast(t) A(t) E(t) $$ and thus for a fixed $u$ $$ \frac{d}{dt}(\Vert E(t)u\Vert^2)=-2\langle A(t)E(t)u, E(t)u(t)\rangle\le 0\Longrightarrow \Vert E(t)u\Vert^2\le \Vert E(0)u\Vert^2, $$ implying boundedness of $E(t)$, provided $E(0)$ is bounded. Assuming $A(t)\le -c_0Id$, the same inequality along with Gronwall gives $$ \Vert E(t)u\Vert^2\le e^{-2c_0t}\Vert E(0)u\Vert^2. $$

  • $\begingroup$ Yes, this is pretty much what's behind the above mentioned Lions' result. $\endgroup$ – Delio Mugnolo Oct 10 '12 at 20:05
  • $\begingroup$ According to the OPs comment to my answer, he wants to have $c_0=0$. $\endgroup$ – András Bátkai Oct 11 '12 at 15:39
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    $\begingroup$ I may miss something, but how does this imply the existence of the family $E(t)$? $\endgroup$ – András Bátkai Oct 11 '12 at 15:41

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