By A. J. Chorin, J. E. Marsden
The objective of this article is to offer a number of the uncomplicated rules of fluid mechanics in a mathematically beautiful demeanour, to provide the actual history and motivation for a few buildings which have been utilized in contemporary mathematical and numerical paintings at the Navier-Stokes equations and on hyperbolic platforms and to curiosity many of the scholars during this appealing and hard topic. The 3rd version has included a few updates and revisions, however the spirit and scope of the unique publication are unaltered.
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Extra resources for A mathematical introduction to fluid mechanics, Second Edition
A vortex tube with nonzero strength cannot “end” in the interior of the ﬂuid. It either forms a ring (such as the smoke from a cigarette), extends to inﬁnity, or is attached to a solid boundary. The usual argument supporting this statement goes like this: suppose the tube ended at a certain cross section S, inside the ﬂuid. Because the tube cannot be extended, we must have ξ = 0 on C1 . Thus, the strength is zero—a contradiction. This “proof” is hopelessly incomplete. First of all, why should a vortex tube end in a nice regular way on a surface?
2 2 L ∂x L2 ∂x ∂y ∂z 2 Similar equations hold for the y and z components. 5) where p = p/(ρ0 U 2 ). Incompressibility still reads div u = 0. 5) are the Navier–Stokes equations in dimensionless variables. We deﬁne the Reynolds number R to be the dimensionless number LU R= . ν For example, consider two ﬂows past two spheres centered at the origin but with diﬀering radii, one with a ﬂuid where U∞ = 10 km/hr past a sphere of radius 10 m and the other with the same ﬂuid but with U∞ = 100 km/hr and radius = 1 m.
Then a parameterization of Ct is ϕ(x(s), t), 0 ≤ s ≤ 1. Thus, by deﬁnition of the line integral and the material derivative, d dt u · ds = Ct 1 d dt u(ϕ(x(s), t), t) · 0 1 = 0 ∂ ϕ(x(s), t) ds ∂s Du ∂ (ϕ(x(s), t), t) · ϕ(x(s), t) ds Dt ∂s 1 u(ϕ(x(s), t), t) · + 0 ∂ ∂ ϕ(x(s), t) ds. ∂t ∂s 22 1 The Equations of Motion Because ∂ϕ/∂t = u, the second term equals 1 u(ϕ(x(s), t), t) · 0 ∂ u(ϕ(x(s), t), t) ds ∂s = 1 2 1 0 ∂ (u · u)(ϕ(x(s), t), t) ds = 0 ∂s (since Ct is closed). The ﬁrst term equals Ct Du ds, Dt so the lemma is proved.