R x r y r x y similarly since x r cos r sin y r

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Unformatted text preview: ∂x + + ∂η ∂y 2 ∂η ∂x D ∂u ∂y ∂η ∂y dxdy 2 dxdy. If we neglect the last term in this expression (which is justified if η and its derivatives are assumed to be small), we find that New potential energy − Old potential energy = T ∇u · ∇ηdxdy. D It follows from the divergence theorem in the plane and the fact that η vanishes on the boundary ∂D that T ∇u · ∇ηdxdy + T η ∇ · ∇udxdy = D D T η ∇u · Nds = 0, ∂D and hence Change in potential = − η (x, y )T (∇ · ∇u)(x, y )dxdy. (5.15) D The work performed must be minus the change in potential energy, so it follows from (5.14) and (5.15) that η (x, y )ρ D ∂2u (x, y )dxdy = ∂t2 η (x, y )T (∇ · ∇u)(x, y )dxdy. D Since this equation holds for all choices of η , it follows that ρ ∂2u = T ∇ · ∇u, ∂t2 which simplifies to ∂2u = c2 ∇ · ∇u, ∂t2 where c2 = ∂2u ∂2u +2 ∂x2 ∂y T , ρ . or equivalently ∂2u = c2 ∂t2 This is just the wave equation. The equations of a perfect gas and sound waves. Next, we will describe Euler’s equations for a perfect gas in (...
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This document was uploaded on 01/12/2014.

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