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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Briggs–Bers criterion ： ウィキペディア英語版 Briggs–Bers criterion In stability theory, the Briggs–Bers criterion is a criterion for determining whether the trivial solution to a linear partial differential equation with constant coefficients is stable, convectively unstable or absolutely unstable. This is often useful in applied mathematics, especially in fluid dynamics, because linear PDEs often govern small perturbations to a system, and we are interested in whether such perturbations grow or decay. The Briggs–Bers criterion is named after R. J. Briggs and A. Bers. Suppose that the PDE is of the form $Ly=0$, where $y\; =\; y(x,t)$ is a function of space $x$ and time $t$. The partial differential operator $L=L(\backslash partial\_x,\backslash partial\_t)$ has constant coefficients, which do not depend on $x$ and $t$. Then a suitable ansatz for $y$ is the normal mode solution :$y=\backslash hat\; \backslash exp(ikx-i\backslash omega\; t).$ Making this ansatz is equivalent to considering the problem in Fourier space – the solution may be decomposed into its Fourier components in space and time. Making this ansatz, the equation becomes :$L(ik,-i\backslash omega)\backslash hat\; \backslash exp(ikx-i\backslash omega\; t)\; =\; 0;$ or, more simply, :$L(ik,-i\backslash omega)\; =\; 0.$ This is a dispersion relation between $k$ and $\backslash omega$, and tells us how each Fourier component evolves in time. In general, the dispersion relation may be very complicated, and there may be multiple $\backslash omega$ which satisfy the relation for a given value of $k$, or vice-versa. The solutions to the dispersion relation may be complex-valued.〔 Now, an initial condition $y(x,0)$ can be written as a superposition of Fourier modes of the form $\backslash exp(ikx)$. In practice, the initial condition will have components of all frequencies. Each of these components evolves according to the dispersion relation, and therefore the solution at a later time $y(x,t)$ may be obtained by Fourier inversion. In the simple case where $L$ is first-order in time, the dispersion relation determines a unique value of $\backslash omega(k)$ for each given value of $k$, and so :$y(x,t)\; =\; \backslash frac\; \backslash int\; \backslash hat(k)\; \backslash exp(ikx-i\backslash omega(k)t)\; \backslash ,\; dk$ where :$\backslash hat(k)\; =\; \backslash int\; y(x,0)\; \backslash exp(-ikx)\; \backslash ,\; dx$ is the Fourier transform of the initial condition. In the more general case, the Fourier inversion must be performed by contour integration in the complex $k$ and $\backslash omega$ planes.〔 While it may not be possible to evaluate the integrals explicitly, asymptotic properties of $y(x,t)$ as $t\backslash rightarrow\backslash infty$ may be obtained from the integral expression, using methods such as the method of stationary phase or the method of steepest descent. In particular, we can determine whether $y(x,t)$ decays or grows exponentially in time, by considering the largest value that $\backslash Im\backslash omega$ may take. If the dispersion relation is such that always, then any solution will decay as $t\backslash rightarrow\backslash infty$, and the trivial solution $y=0$ is stable. If there is some mode with $\backslash Im\backslash omega>0$, then that mode grows exponentially in time. By considering modes with zero group velocity and determining whether they grow or decay, we can determine whether an initial condition which is localised around $x=0$ moves away from $x=0$ as it grows, with $y(0,t)\backslash rightarrow\; 0$ (convective instability); or whether $y(0,t)\backslash rightarrow\backslash infty$ (absolute instability). ==Transient growth== Suppose the PDE is of the form :$y\_t\; =\; A\; y$ where $A$ is a linear differential operator in $x$. In general, $A$ is not a normal operator. While the large-time behaviour of $y$ is still determined by the eigenvalues of $A$, the behaviour which takes place before this large-time behaviour may be dramatically different. In particular, while the eigenvalues of $A$ may all have negative real part, which would predict that $y$ decays exponentially at large times and that the trivial state $y=0$ is stable, it is possible for $y$ to grow transiently and become large before decaying.〔 In practice, the linear equations that we work with are linearisations of more complicated governing equations such as the Navier–Stokes equations about some base state, with the linearisations carried out under the assumption that the perturbation quantity $y$ is small. Transient growth may violate this assumption. When nonlinear effects are considered, then a system may be unstable even if the linearised system is 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「Briggs–Bers criterion」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.