ordered exponential of unbounded operators - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:14:02Z http://mathoverflow.net/feeds/question/109319 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109319/ordered-exponential-of-unbounded-operators ordered exponential of unbounded operators André Henriques 2012-10-10T19:02:25Z 2012-10-10T20:15:54Z <p>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 <I><a href="http://en.wikipedia.org/wiki/Ordered_exponential" rel="nofollow">ordered exponential</a></i> of the family $A_t$.</p> <p>If $A_t=A$ is independ of $t$, then the solution of the <i>ODE</i> is the usual exponential $E_t=e^{-tA}$.<br> Note that the above operators $E_t$ are <b><i>bounded</i></b>.</p> <p>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 $\le 1$.</p> <p><b>Q:</b> Which properties should I impose on the family $A_t$ in order for the solution of (1) to be well defined and bounded?</p> http://mathoverflow.net/questions/109319/ordered-exponential-of-unbounded-operators/109324#109324 Answer by András Bátkai for ordered exponential of unbounded operators András Bátkai 2012-10-10T19:32:54Z 2012-10-10T20:15:54Z <p>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. </p> <p>I would say that strong resolvent continuity and a sufficiently big common domain is sufficient in your case. See Section 5.3 in <a href="http://books.google.hu/books/about/Semigroups_of_Linear_Operators_and_Appli.html?id=sIAyOgM4R3kC&amp;redir_esc=y" rel="nofollow">Pazy</a>.</p> <p>I can also give a related <a href="http://arxiv.org/abs/1105.6372" rel="nofollow">self-reference</a>, where also the investigation of this product appears and some of the ideas are explained in a simpler situation.</p> <p><strong>ADDED:</strong> 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.</p> http://mathoverflow.net/questions/109319/ordered-exponential-of-unbounded-operators/109325#109325 Answer by Delio Mugnolo for ordered exponential of unbounded operators Delio Mugnolo 2012-10-10T19:38:52Z 2012-10-10T19:44:29Z <p>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.</p> <p>In the "variationaL" setting you consider, it is probably most efficient to introduce a weak formulation based on a family of quadratic forms. </p> <p>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.</p> http://mathoverflow.net/questions/109319/ordered-exponential-of-unbounded-operators/109326#109326 Answer by Bazin for ordered exponential of unbounded operators Bazin 2012-10-10T19:54:00Z 2012-10-10T19:54:00Z <p>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 $$\Vert E(t)u\Vert^2\le e^{-2c_0t}\Vert E(0)u\Vert^2.$$</p>