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By Frederick S. Woods

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The Kutta–Joukowski theorem gives the force exerted on B. Kutta–Joukowski Theorem Consider incompressible potential flow exterior to a region B. Let the velocity field approach the constant value (U, V ) = U at infinity. 10) where ΓC is the circulation around B and n is a unit vector orthogonal to U. Proof By assumption, the complex velocity F is an analytic function outside the body B. It may therefore, be expanded in a Laurent series. Because F tends to a constant U at infinity, no positive powers of z can occur in the expansion.

Is symmetric. 7 Since σ is symmetric, if follows from properties 1 and 2 that σ can depend only on the symmetric part of ∇u; that is, on the deformation D. Because σ is a linear function of D, σ and D commute and so can be simultaneously diagonalized. Thus, the eigenvalues of σ are linear functions of those of D. By property 2, they must also be symmetric because we can choose U to permute two eigenvalues of D (by rotating through an angle π/2 about an eigenvector), and this must permute the corresponding eigenvalues of σ.

The reason is that basically C ∪ C forms the boundary of a surface Σ in D. Stokes’ theorem then gives ξ · dA = Σ u · ds − C u · ds = ΓC − ΓC C and because ξ = 0 in D, we get ΓC = ΓC . 2, the circulation around a curve is constant in time. Thus, the circulation around an obstacle in the plane is well-defined and is constant in time. Next, consider incompressible potential flow in a simply connected domain D. From u = grad ϕ and div u = 0, we have ∆ϕ = 0. 1. The circulations about C and C are equal if the flow is potential in Σ.

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A course in mathematics,: For students of engineering and applied science, by Frederick S. Woods

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