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Unformatted text preview: ∂x +
∂y 2 ∂η
∂x D ∂u
∂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
(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
= T ∇ · ∇u,
∂t2 which simpliﬁes to
= c2 ∇ · ∇u,
∂t2 where c2 = ∂2u ∂2u
ρ . or equivalently
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.
- Winter '14