δ u = - iω ξ

. The continuity and adiabatic equations give the Lagrangian perturbations

\frac{Δ _{L} ρ}{ρ} = - Δ, \frac{Δ _{L} p}{p} = - γ Δ, Δ = i k_{x} ξ_{x} + \frac{d ξ _{z}}{d z} .

p \propto ρ^{γ}

, so the displacement terms cancel and

\frac{δ p}{p} = γ \frac{δ ρ}{ρ} .

Using

d p / d z = - ρ g

then gives

Ψ = \frac{δ p}{ρ} = g ξ_{z} - v_{s}^{2} Δ, v_{s}^{2} = \frac{γ p}{ρ} .

Eliminating

ξ_{y}

from the two horizontal momentum equations and combining gravity with the vertical pressure force yields

- (ω^{2} - 4 Ω^{2}) ξ_{x} = - i k_{x} Ψ, - ω^{2} ξ_{z} = - \frac{d Ψ}{d z} .

Thus

ξ_{x} = \frac{i k _{x} Ψ}{ω ^{2} - 4 Ω ^{2}},

and the requested coupled first-order system is

\frac{d Ψ}{d z} = ω^{2} ξ_{z}, \frac{d ξ _{z}}{d z} = \frac{g}{v _{s}^{2}} ξ_{z} + (\frac{k _{x}^{2}}{ω ^{2} - 4 Ω ^{2}} - \frac{1}{v _{s}^{2}}) Ψ .

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