In this case the solution is rr arn brn once again

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Unformatted text preview: ewrite this equation as ∂v 1 + (v · ∇)v = − ∇p. ∂t ρ (5.18) Note that this equation is nonlinear because of the term (v · ∇)v. To finish the Euler equations, we need an equation of state , which relates pressure and density. The equation of state could be determined by experiment, the simplest equation of state being p = a2 ργ , (5.19) where a2 and γ are constants. (An ideal monatomic gas has this equation of state with γ = 5/3.) The Euler equations (5.16), (5.18), and (5.19) are nonlinear, and hence quite difficult to solve. However, one explicit solution is the case where the fluid is motionless, ρ = ρ0 , p = p0 , v = 0, where ρ0 and p0 satisfy p0 = a2 ργ . 0 Linearizing Euler’s equations near this explicit solution gives rise to the linear differential equation which governs propagation of sound waves. Let us write ρ = ρ0 + ρ , p = p 0 + p , v = v , where ρ , p and v are so small that their squares can be ignored. Substitution into Euler’s equations yields ∂ρ + ρ0 ∇ · (v ) = 0, ∂t ∂v 1 = − ∇p , ∂t ρ0 and p = [a2 γ (ρ0 )(γ −1) ]ρ = c2 ρ , where c2 is a new constant. It follows from these three equations that ∂2ρ ∂v = −ρ0 ∇ · ∂ t2 ∂t = ∇...
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