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A simple counterexample related to the Lie-Trotter product formula. (English) Zbl 1256.47026

The “addition formula” \[ e^{A + B} = \lim_{n \to \infty} \Big( e^{A /n} e^{B/n}\Big)^n \] for matrix exponentials is due to Sophus Lie. Its generalization to semigroups \[ U(t)u = \lim_{n \to \infty} \Big( S \Big( {t \over n} \Big) T \Big( {t \over n} \Big) \Big)^n u \eqno(1) \] was given by H. F. Trotter [Proc. Am. Math. Soc. 10, 545–551 (1959; Zbl 0099.10401)]; here, \(A\) (resp., \(B)\) is the infinitesimal generator of the strongly continuous semigroup \(S(t)\) (resp., \(T(t))\) and \(U(t)\) is the semigroup generated by \(\overline{A + B}\). In the language of differential equations in Banach spaces, formula (1) is the basis of the splitting method of assembling the solution \(u'(t) = (A + B)u(t)\) from the solutions of \(u'(t) = Au(t)\) and \(u'(t) = Bu(t).\)
The validity of Trotter’s formula (1) requires additional assumptions. In a slightly generalized version, it is assumed that \[ \Big \| \Big(S\Big( {t \over n} \Big) T \Big( {t \over n} \Big) \Big)^n \Big \| \leq Me^{\omega t} \quad (t \geq 0, \;n = 1, 2, \dots) \eqno(2) \] and that \(\overline{A + B}\) is an infinitesimal generator; in another version, this last requirement follows from other hypotheses.
The question arises: does the sole assumption that \(\overline{A + B}\) is a semigroup generator guarantee (1)? The answer is known to be “no” through a counterexample of F. Kühnemund and M. Wacker [Semigroup Forum 60, No. 3, 478–485 (2000; Zbl 0976.47021)]. The authors provide a simpler counterexample using operator matrices; the limit in (1) is zero, while \(U(t)u \neq 0\).

MSC:

47D06 One-parameter semigroups and linear evolution equations
34G10 Linear differential equations in abstract spaces
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References:

[1] Bressan, A.: Hyperbolic Systems of Conservation Laws. Oxford Lecture Series in Mathematics and Its Applications, vol. 20. Oxford University Press, Oxford (2000). The one-dimensional Cauchy problem · Zbl 0997.35002
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[5] Kühnemund, F., Wacker, M.: The Lie-Trotter product formula does not hold for arbitrary sums of generators. Semigroup Forum 60(3), 478–485 (2000) · Zbl 0976.47021 · doi:10.1007/s002339910039
[6] Kühnemund, F., Wacker, M.: Commutator conditions implying the convergence of the Lie-Trotter products. Proc. Am. Math. Soc. 129(12), 3569–3582 (2001) · Zbl 0987.34061 · doi:10.1090/S0002-9939-01-06034-8
[7] Kurtz, T.G., Pierre, M.: A counterexample for the Trotter product formula. J. Differ. Equ. 52(3), 407–414 (1984) · Zbl 0533.47029 · doi:10.1016/0022-0396(84)90170-0
[8] Trotter, H.F.: On the product of semi-groups of operators. Proc. Am. Math. Soc. 10, 545–551 (1959) · Zbl 0099.10401 · doi:10.1090/S0002-9939-1959-0108732-6
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