翻訳と辞書
Words near each other
・ "O" Is for Outlaw
・ "O"-Jung.Ban.Hap.
・ "Ode-to-Napoleon" hexachord
・ "Oh Yeah!" Live
・ "Our Contemporary" regional art exhibition (Leningrad, 1975)
・ "P" Is for Peril
・ "Pimpernel" Smith
・ "Polish death camp" controversy
・ "Pro knigi" ("About books")
・ "Prosopa" Greek Television Awards
・ "Pussy Cats" Starring the Walkmen
・ "Q" Is for Quarry
・ "R" Is for Ricochet
・ "R" The King (2016 film)
・ "Rags" Ragland
・ ! (album)
・ ! (disambiguation)
・ !!
・ !!!
・ !!! (album)
・ !!Destroy-Oh-Boy!!
・ !Action Pact!
・ !Arriba! La Pachanga
・ !Hero
・ !Hero (album)
・ !Kung language
・ !Oka Tokat
・ !PAUS3
・ !T.O.O.H.!
・ !Women Art Revolution


Dictionary Lists
翻訳と辞書 辞書検索 [ 開発暫定版 ]
スポンサード リンク

linear stability : ウィキペディア英語版
linear stability
In mathematics, in the theory of differential equations and dynamical systems, a particular stationary or quasistationary solution to a nonlinear system is called linearly unstable if the linearization of the equation at this solution has the form \frac=A r, where ''A'' is a linear operator whose spectrum contains eigenvalues with ''positive'' real part. If all the eigenvalues have ''negative'' real part, then the solution is called linearly stable. Other names for linear stability include exponential stability or stability in terms of first approximation.〔V.I. Arnold, Ordinary Differential Equations. MIT Press, Cambridge, MA (1973)〕〔P. Glendinning, Stability, instability and chaos: an introduction to the theory of nonlinear differential equations. Cambridge university press, 1994.〕 If there exist an eigenvalue with ''zero'' real part then the question about stability cannot be solved on the basis of the first approximation and we approach the so-called "centre and focus problem".〔V.V. Nemytskii, V.V. Stepanov, "Qualitative theory of differential equations", Princeton Univ. Press (1960)〕
==Example 1: ODE==

The differential equation
:\frac=x-x^2
has two stationary (time-independent) solutions: ''x'' = 0 and ''x'' = 1.
The linearization at ''x'' = 0 has the form
\frac=x. The linearized operator is ''A''0 = 1. The only eigenvalue is \lambda=1. The solutions to this equation grow exponentially;
the stationary point ''x'' = 0 is linearly unstable.
To derive the linearizaton at ''x'' = 1, one writes
\frac=(1+r)-(1+r)^2=-r-r^2, where ''r'' = ''x'' − 1. The linearized equation is then \frac=-r; the linearized operator is ''A''1 = −1, the only eigenvalue is \lambda=-1, hence this stationary point is linearly stable.

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
ウィキペディアで「linear stability」の詳細全文を読む



スポンサード リンク
翻訳と辞書 : 翻訳のためのインターネットリソース

Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.