Product integrals of continuous resolvents: existence and nonexistence
type: Item
instance of: scholarly article
title: Product integrals of continuous resolvents: existence and nonexistence
zbMATH Open document ID: 0537.34060
DOI: 10.1007/BF02760628
author: M. A. Freedman
published in: Israel Journal of Mathematics
publication date: 1983-01-01T00:00:00Z
review text: Of concern is the initial value problem (*) \(dy(t)/dt=A(t)y(t)\), 0\(\leq t\leq T\), \(y(0)=y_ 0\). Assume, for simplicity, that each A(t) is densely defined and m-dissipative on a Banach space X. If A is independent of t, then existence and uniqueness holds for (*), whether A is linear or not. When A depends on t nicely, then a solution can be constructed by a product integral formula, namely \[ y(t)=\lim_{n\to \infty}J_{t/n}(((n-1)/n)t)...J_{t/n}((2/n)t)J_{t/n}((1/n)t)y_ 0 \] where \(J_{\lambda}(s)=(I-\lambda A(s))^{-1}\). The usual smoothness condition in t is that for \(y_ 0\in X,\) \[ (+)\quad \| J_{\lambda}(t)y_ 0-J_{\lambda}(s)y_ 0\| \leq \lambda \| f(t)-f(s)\| L(\| y_ 0\|) \] where L is increasing and \(f:[0,T]\to X\) is continuous. Condition \((+)\) can be weakened slightly, but the author shows that replacing \((+)\) by the condition that \((\lambda,t,y_ 0)\to J_{\lambda}(t)y_ 0\) is continuous on \((0,\infty)\times [0,T]\times X\) is not adequate to make the product integral converge or (*) have a solution, even when each A(t) is a bounded linear operator. The construction of the counterexample is quite clever.
Mathematics Subject Classification ID: 34G10
Mathematics Subject Classification ID: 47D03
Mathematics Subject Classification ID: 47B44
zbMATH DE Number: 3853556
zbMATH Keywords: product integrals
zbMATH Keywords: time dependent evolution equations
reviewed by: Jerome A. Goldstein
MaRDI profile type: Publication
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full work available at URL: https://doi.org/10.1007/bf02760628
OpenAlex ID: W2036949628
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description: scientific article; zbMATH DE number 3853556