Reynolds transport theorem: Z Z ∂f d f (x, t) dΩ = (x, t) + ∇ · (f ~u ) dΩ dt ∂t Ω(t) Preliminaries Ω(t) Let Ω0 ⊂ Rn and let φ : Ω0 × [0, ∞) → Ω(t) ⊂ Rn be the transformation defined by (x, t) 7→ y := φ(x, t). ⇒ d d y =: u(y, t) = u(φ(x, t), t) = φ(x, t) =: φt (x, t) dt dt Calcuation of the Jacobian-matrix: S.v.S Dx φt (x, t) =: φxt (x, t) = φtx (x, t) = ux (φ(x, t), t) · φx (x, t) Proof: For a more detailed proof we refer to the lecture notes of Prof. Dr. J. Lorenz, “Die Naviers-Stokes Gleichung ..”, RWTH-Aachen, 1994. Z Z d 1 d f (y, t) dΩ = f (φ(x, t), t) |det (φx (x, t))| dΩ0 dt dt Ω0 Ω(t) Z d d = [f (φ(x, t), t)] |det (φx (x, t))| + f (φ(x, t), t) |det (φx (x, t))| dΩ0 dt dt Ω0 Z ∂f 2 (φ(x, t), t) + ∇f (φ(x, t), t) · u(φ(x, t), t) |det (φx (x, t))| = ∂t Ω0 + f (φ(x, t), t) (∇ · u)(φ(x, t), t) |det (φx (x, t))|} dΩ0 Z ∂f (φ(x, t), t) + ∇f (φ(x, t), t) · u(φ(x, t), t) |det (φx (x, t))| = ∂t Ω0 + f (φ(x, t), t) (∇ · u)(φ(x, t), t)} |det (φx (x, t))| dΩ0 Z ∂f 1 (y, t) + ∇ · [f (y, t) u(y, t)] dΩ = ∂t Ω(t) 1. Transformation theorem with respect to φ(t, ·) : Ω0 → Ω(t) 2. Derivation rule for Wronski determinants: (würde hier das ganze Lemma hinschreiben sonst ist nicht d klar was Y (t) bzw. A(t) überhaupt ist.) dt detY (t) = trA(t) · detY (t) d |det (φx (x, t))| = trux (φ(x, t), t) |det (φx (x, t))| dt and trux (φ(x, t), t) = (∇ · u)(φ(x, t), t)
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