By A. J. Chorin, J. E. Marsden

The target of this article is to offer the various simple principles of fluid mechanics in a mathematically appealing demeanour, to provide the actual heritage and motivation for a few structures which were utilized in contemporary mathematical and numerical paintings at the Navier-Stokes equations and on hyperbolic structures and to curiosity a number of the scholars during this attractive and hard topic. The 3rd variation has integrated a few updates and revisions, however the spirit and scope of the unique ebook are unaltered.

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**Extra resources for A mathematical introduction to fluid mechanics, Second Edition **

**Sample text**

We use this remark to prove existence. Indeed, given w, let p be deﬁned by the solution to the Neumann problem ∆p = div w in D, with ∂p = w · n on ∂D. ∂n It is known10 that the solution to this problem exists and is unique up to the addition of a constant to p. With this choice of p, deﬁne u = w−grad p. Then, clearly u has the desired properties div u = 0, and also u · n = 0 by construction of p. 2. 2. Decomposing a vector ﬁeld into a divergence-free and gradient part. It is natural to introduce the operator P, an orthogonal projection operator, which maps w onto its divergence-free part u.

Hilbert [1953], Methods of Mathematical Physics, Wiley. The equation ∆p = f, ∂p/∂n = g has a solution unique up to a constant if and only if D f dV = ∂D g dA. The divergence theorem ensures that this condition is satisﬁed in our case. 38 1 The Equations of Motion provided div u = 0 and u · n = 0, and that P(grad p) = 0. 5). If we apply the operator P to both sides, we obtain P(∂t u + grad p) = P −(u · ∇)u + 1 ∆u . R Because u is divergence-free and vanishes on the boundary, the same is true of ∂t u (if u is smooth enough).

In fact, if one just assumes the force is a continuous function of n, then, using balance of momentum, one can prove it is linear in n. 6 Our assumptions on σ are as follows: 1. σ depends linearly on the velocity gradients ∇u that is, σ is related to ∇u by some linear transformation at each point. 2. σ is invariant under rigid body rotations, that is, if U is an orthogonal matrix, σ(U · ∇u · U−1 ) = U · σ(∇u) · U−1 . This is reasonable, because when a ﬂuid undergoes a rigid body rotation, there should be no diﬀusion of momentum.