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 , where r is the perturbation to the steady state, 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. 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". The differential equation has two stationary (time-independent) solutions: x = 0 and x = 1. The linearization at x = 0 has the form The linearized operator is A0 = 1. The only eigenvalue is . The solutions to this equation grow exponentially; the stationary point x = 0 is linearly unstable. To derive the linearization at x = 1, one writes where r = x − 1. The linearized equation is then ; the linearized operator is A1 = −1, the only eigenvalue is , hence this stationary point is linearly stable. The nonlinear Schrödinger equation where u(x,t) ∈ C and k > 0, has solitary wave solutions of the form . To derive the linearization at a solitary wave, one considers the solution in the form The linearized equation on is given by where with and the differential operators. According to Vakhitov–Kolokolov stability criterion, when k > 2, the spectrum of A has positive point eigenvalues, so that the linearized equation is linearly (exponentially) unstable; for 0 < k ≤ 2, the spectrum of A is purely imaginary, so that the corresponding solitary waves are linearly stable. It should be mentioned that linear stability does not automatically imply stability; in particular, when k = 2, the solitary waves are unstable.

À propos de ce résultat
Cette page est générée automatiquement et peut contenir des informations qui ne sont pas correctes, complètes, à jour ou pertinentes par rapport à votre recherche. Il en va de même pour toutes les autres pages de ce site. Veillez à vérifier les informations auprès des sources officielles de l'EPFL.

Graph Chatbot

Chattez avec Graph Search

Posez n’importe quelle question sur les cours, conférences, exercices, recherches, actualités, etc. de l’EPFL ou essayez les exemples de questions ci-dessous.

AVERTISSEMENT : Le chatbot Graph n'est pas programmé pour fournir des réponses explicites ou catégoriques à vos questions. Il transforme plutôt vos questions en demandes API qui sont distribuées aux différents services informatiques officiellement administrés par l'EPFL. Son but est uniquement de collecter et de recommander des références pertinentes à des contenus que vous pouvez explorer pour vous aider à répondre à vos questions.