L
Lv 7
L 發問時間: 科學數學 · 1 0 年前

[動態系統] 一題關於 Lorenz system 解的問題

Lorenz system

dx/dt = -σx + σy

dy/dt = rx - y - xz

dz/dt = -bz + xy

設 X(t) = (x(t),y(t),z(t))^t 是 Lorenz system 的一個解, 且 X(0) 落在 z 軸上,

證明 X(t) 的整個軌跡都在 z 軸上且 X(t) -> 0 as t -> oo

像這種我們不知道解的公式是什麼, 通常有哪些方法去看解在時間無窮大時的行為了 QQ 這是我書上後面的習題, 只剩這題不會做.剛接觸動態系統才發現高速電腦和程式真的很重要.

1 個解答

評分
  • Eric
    Lv 6
    1 0 年前
    最佳解答

    Consider the initial value problem determined by the ODEdx/dt = -σx + σydy/dt = rx - y - xzdz/dt = -bz + xy and initial conditionsx(0) = y(0) = 0, z(0) = z0. Lemma.  If (x(t),y(t),z(t))T is a solution to the initial value problem, then so is (-x(t),-y(t),z(t))T.Proof.  This is a simple verification.d(-x)/dt = -(-σx + σy) = -σ(-x) + σ(-y)d(-y)/dt = -(rx - y - xz) = -r(-x) - (-y) - (-x)zdz/dt = -bz + xy = -bz + (-x)(-y)-x(0) = -y(0) = 0, z(0) = z0. ∎Now, assume that X(t) = (x(t),y(t),z(t))T (t ≥ 0) is a solution the the initial-value problem.Claim. X(t) stays on the z-axis for all t > 0.Proof. Say t* > 0. By uniqueness properties (the right-hand side of the ODE is Lipschitz on any closed and bounded rectangle centered at (t*,X(t*))T, so X(t) is the unique solution of the ODE passing through (t*,X(t*))T in some neighborhood of t*),and from the above lemma, we have for all t in a neighborhood of t* that x(t) = -x(t) and y(t) = -y(t), and hence x(t*) = y(t*) = 0. ∎Claim. X(t) → 0 as t → ∞.Proof. From the previous result, we know thatx(t) = y(t) = 0for all t ≥ 0. Hence, z(t) satisfies the initial value problemdz/dt = -bz,z(0) = z0whose unique solution isz(t) = z0e-btwhich clearly tends to zero as t → ∞. ∎微分方程理論非常深奧,沒有單一的解決方法。

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