2013
DOI: 10.1002/mana.201200095
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Scattering for strictly hyperbolic systems with time‐dependent coefficients

Abstract: The present paper is devoted to finding conditions on the occurrence of scattering for strictly hyperbolic systems with time-dependent coefficients whose time-derivatives are in L 1 in time. More precisely, it will be shown that the solutions are asymptotically free if the coefficients are stable in the sense that their improper Riemann integrals converge as t → ±∞, while each nontrivial solution with radially symmetric data is never asymptotically free provided that the coefficients are not stable as t → ±∞. … Show more

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Cited by 5 publications
(7 citation statements)
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“…Let us first analyse certain basic properties of characteristic roots ϕ k (t, ξ). The next proposition is established in [17].…”
Section: Asymptotic Integrations For Linear Hyperbolic Systemmentioning
confidence: 97%
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“…Let us first analyse certain basic properties of characteristic roots ϕ k (t, ξ). The next proposition is established in [17].…”
Section: Asymptotic Integrations For Linear Hyperbolic Systemmentioning
confidence: 97%
“…In this section we shall derive asymptotic integrations for linear hyperbolic systems with time-dependent coefficients, a kind of representation formula for their solutions. In fact, we have discussed such arguments in our recent paper [17] in the context of the scattering problems. To make the argument self-contained, we must give the proof completely, because the Fourier integral form of solutions U to Kirchhoff system (1.2) will be obtained by a careful observation of the construction of asymptotic integrations for linear systems.…”
Section: Asymptotic Integrations For Linear Hyperbolic Systemmentioning
confidence: 99%
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“…(1.3) with time-dependent coefficients. There are a number of results concerning (1.2)-(1.3) (see, for example, [1,8,10,11,12,6], [14, Section 2] and references therein).…”
Section: Introductionmentioning
confidence: 99%
“…10) for every 0 ≤ s ≤ t. From (3.5), it follows thatlim k→∞ y k (t) − y(t) H×H = 0 for every t ∈ (0, ∞) \ N 1 , and therefore lim k→∞ (E(λ)y k )(t) − (E(λ)y)(t) H×H = 0…”
mentioning
confidence: 98%