“linear theory of equilibria”的概念、定义、翻译、参考文献-科学参考

linear theory of equilibria

Mathematics

For small perturbations about an equilibrium, terms of higher order than linear may be considered negligible (see linearization). Small angles θ‎, that a simple pendulum makes about the vertical, satisfy $\ddot{θ} = - (g / l) θ$ ; this is SHM, which has small bounded trigonometric solutions, and so θ‎ = 0 is a stable equilibrium. For θ‎ = π‎ + ε‎, that is, small perturbations about the upward vertical, we have

$\ddot{ε} = \ddot{θ} = - \frac{g}{l} sin θ = - \frac{g}{l} sin (π + ε) = \frac{g}{l} sin ε \approx \frac{g}{l} ε,$

which has unbounded exponential solutions, and so θ‎ = π‎ is an unstable equilibrium.

Linearizing about (0,0,0), the three equations governing the Lorenz attractor give

$\frac{d x}{d t} = α (y - x), \frac{d y}{d t} = β x - y, \frac{d z}{d t} = - γ z,$

where α‎,β‎,γ‎ > 0. So r(t) = (x(t), y(t),z(t))^T satisfies

$\frac{d r}{d t} = (\begin{matrix} - α & α & 0 \\ β & - 1 & 0 \\ 0 & 0 & - γ \end{matrix}) r .$

If the matrix has distinct real eigenvalues λ‎₁,λ‎₂,λ‎₃ with eigenvectors v₁,v₂,v₃, then the solution is

$r (t) = A_{1} e^{λ_{1} t} v_{1} + A_{2} e^{λ_{2} t} v_{2} + A_{3} e^{λ_{3} t} v_{3} .$

This is a small bounded solution if λ‎₁,λ‎₂,λ‎₃ are all negative; so the origin is stable when β‎<1.

单词	linear theory of equilibria
释义	linear theory of equilibria Mathematics For small perturbations about an equilibrium, terms of higher order than linear may be considered negligible (see linearization). Small angles θ‎, that a simple pendulum makes about the vertical, satisfy $\ddot{θ} = - (g / l) θ$ ; this is SHM, which has small bounded trigonometric solutions, and so θ‎ = 0 is a stable equilibrium. For θ‎ = π‎ + ε‎, that is, small perturbations about the upward vertical, we have $\ddot{ε} = \ddot{θ} = - \frac{g}{l} sin θ = - \frac{g}{l} sin (π + ε) = \frac{g}{l} sin ε \approx \frac{g}{l} ε,$ which has unbounded exponential solutions, and so θ‎ = π‎ is an unstable equilibrium. Linearizing about (0,0,0), the three equations governing the Lorenz attractor give $\frac{d x}{d t} = α (y - x), \frac{d y}{d t} = β x - y, \frac{d z}{d t} = - γ z,$ where α‎,β‎,γ‎ > 0. So r(t) = (x(t), y(t),z(t))^T satisfies $\frac{d r}{d t} = (\begin{matrix} - α & α & 0 \\ β & - 1 & 0 \\ 0 & 0 & - γ \end{matrix}) r .$ If the matrix has distinct real eigenvalues λ‎₁,λ‎₂,λ‎₃ with eigenvectors v₁,v₂,v₃, then the solution is $r (t) = A_{1} e^{λ_{1} t} v_{1} + A_{2} e^{λ_{2} t} v_{2} + A_{3} e^{λ_{3} t} v_{3} .$ This is a small bounded solution if λ‎₁,λ‎₂,λ‎₃ are all negative; so the origin is stable when β‎<1.
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