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AerE 546 Lecture 18 Hyperbolic Equations

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Hyperbolic Equations Physics: Convection and wave propagation; sound in compressible flow. ∂tq = -∇·Fq . Now consider convective (dominated) flux uq. A. Examples: Linearized compressible potential flow (1-M2) ∂x2ϕ + ∂y2ϕ = 0 M1 is hyperbolic: recall asymptotes=radiation # # # # # c.f. Mach waves Shallow water waves (long waves; non-dispersive): k = 2π/λ; ω = 2π/T ∂t 2h - gH∂x2h = 0 (or ∇2h) ; a = √gH = |ω/k| ; ω = ±k√gH Pictures: expansions and shocks; .gif animations. Will encounter in 1-D model equation and Euler equations. Physics - numerics. B. Hyperbolic conservation laws: now sound, later 1-D Euler 1. Mass conservation control volume sketch, mass in-out = rate of change ∂tρ + ∇·(uρ) = 0 ; or ∂tρ + (u · ∇)ρ + ρ ∇ ·u = 0 ; or Dtρ + ρ ∇ ·u = 0 conservation form! ! ! ! ! ! convective form 2. Momentum conservation: F/vol = ma/vol = ρDtu. F = pressure force: ( A Pleft -A Pright) /Vol = -∆P/∆x - -∇P. Inviscid momentum equation ρ Dtu = -∇P (convective form) Conservation form for momentum: ρ(∂tu + (u · ∇)u) = -∇P u(∂tρ + ∇·(uρ) = 0) → ∂tρu + ∇·(u⊗uρ) = -∇P ; ∂tρui + ∂j(ujuiρ) = -∂iP Note ∇P = ∇·(I P): ∂tρu + ∇·(uuρ+I P) = 0: Hyperbolic conservation law ∂tρu + ∇·Fu = 0, Fu = uuρ+I P is flux function (tensor) ∂tρ + ∇· Fρ = 0, Fρ = uρ will encounter later 3. Total enthalpy (or thermodynamic variable. Euler equations later). For now: isentropic - sound speed given a2 = ∂P/∂ρ |s A

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AERE 546

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AerE 546 # Lecture 18



Hyperbolic Equations

Physics: Convection and wave propagation; sound in compressible flow.

∂tq = -∇·Fq . Now consider convective (dominated) flux uq.

A. Examples:
Linearized compressible potential flow
(1-M2) ∂x2ϕ + ∂y2ϕ = 0 M>1 is hyperbolic: recall asymptotes=radiation
# # # # # c.f. Mach waves
Shallow water waves (long waves; non-dispersive): k = 2π/λ; ω = 2π/T
∂t2h - gH∂x2h = 0 (or ∇2h) ; a = √gH = |ω/k| ; ω = ±k√gH


Pictures: expansions and shocks; .gif animations. Will encounter in 1-D model equation and
Euler equations. Physics <-> numerics.


B. Hyperbolic conservation laws: now sound, later 1-D Euler
1. Mass conservation control volume sketch, mass in-out = rate of change
∂tρ + ∇·(uρ) = 0 ; or ∂tρ + (u · ∇)ρ + ρ ∇ ·u = 0 ; or Dtρ + ρ ∇ ·u = 0
! conservation form! ! ! ! ! ! convective form

2. Momentum conservation: F/vol = ma/vol = ρD t u. F = pressure force:
( A Pleft -A Pright) /Vol = -∆P/∆x -> -∇P. Inviscid momentum equation
ρ Dtu = -∇P (convective form)
Conservation form for momentum:
ρ(∂tu + (u · ∇)u) = -∇P
u(∂tρ + ∇·(uρ) = 0)
# → ∂tρu + ∇·(u⊗uρ) = -∇P ; ∂tρui + ∂j(ujuiρ) = -∂iP

Note ∇P = ∇·(I P): ∂tρu + ∇·(uuρ+I P) = 0: Hyperbolic conservation law

∂tρu + ∇·Fu = 0, Fu = uuρ+I P is flux function (tensor)
∂tρ + ∇· Fρ = 0, Fρ = uρ will encounter later

3. Total enthalpy (or thermodynamic variable. Euler equations later). For now: isentropic
<-> sound speed given
# a2 = ∂P/∂ρ |s




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, AerE 546 # Lecture 18




C. Compressible flow contains sound waves. Sound waves are implicit in equations, affect
numerics They are small disturbances.
Linearize about uniform flow U = (U,0,0), ρ̄ where U and ρ̅ are constant. u = U + u’(x,t).
ρ = ρ̄ + ρ’ ; small perturbations |u|<<U etc. .
u·∇ u = (U + u’)·∇ (U + u’) = (U + u’)·∇ u’ ≈ U∂xu’


ρ̄ ∂tu’ + ρ̄U∂xu’ = -∇P’
∂tρ’ + U∂xρ‘ + ρ̄(∂xu’ + ∂yv’) = 0


2 equations in 3 unknowns. Additional piece of info: a2 = ∂P/∂ρ measures
compressibility of a barotropic fluid: will show that a is sound speed; in an isentropic gas
a2 =γRT. Use dP’ = a2 dρ’ in first equation.

∇·{ ρ̄ Dtu’ = -a2 ∇ρ’ } → ρ̄ Dt ∇·u’ + a2 ∇2ρ’ = 0
Dt { Dtρ‘ +ρ̅∇·u= 0 } → D2tρ‘ +ρ̅ Dt ∇·u = 0
Give
Dt2 ρ‘ − a2 ∇ 2 ρ’ = 0
This is the (linear) equation of sound propagation on a mean flow; with Dt = ∂t + U∂x
#
Mach
waves
D. Case A. Steady (drop primes)
U2∂x2ρ − a2 (∂x2ρ+∂y2ρ) = 0
or
characteristics
or ( M2 - 1 )∂x2ρ - ∂y2ρ = 0

where M=U/a is the Mach number. Slender body equation. Hyperbolic
in super-sonic case.

Solution: try ρ = f(x-αy) : η = x-αy; ∂xρ = f’(η) ; ∂yρ = -αf’(η) etc. Substitute and find
that α = √ M2 - 1.
ρ is constant along Mach lines: dρ= dx ∂xρ + dy ∂yρ→
# dy/dx = - ∂xρ/∂yρ = 1/α = 1/√ M2 - 1
Mach angle: tanθ = 1/√ M2 -1 or sinθ = 1/M




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