d dt ∫ f(x, t)

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)