A mathematical introduction to fluid mechanics, Second by A. J. Chorin, J. E. Marsden

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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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.

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