University:
Widener University
Course:
ESSC 103 | Intriduction To Earth Science
Academic year:

2023
Views:

40

Pages:

3
Author:

fedello3d4

PV fronts: dispersion relations Consider a line Y = η(x, t) separating two regions of constant PV. South of the front, we have q = q0 while to the north q = q0 + ∆ q0 + ∆ q 0 Front separating two constant PV regions. The dashed lines show the anomaly in ψ. The interface between the two regions is a material surface; ﬂuid parcels cannot move from one side to the other, since they would have to change their PV which is prohibited. Therefore, we have ∂ ∂ η + u(x, η, t) η = v(x, η.t) ∂t ∂x The PV inversion formula gives 2 2 ∇ ψ−γ ψ = q0 + ∆ y > η y<η q0 with γ = 0 (BT) or 1/Rd (BC). We’ll discuss two ways of solving the linearized problem where η = η0 exp(ıkx) 1 Matching We can solve the PV equation using a Fourier expansion in x; for the linearized problem we only need the x-independent and exp(ıkx) modes. We’ll talk about the barotropic case:  y2   (q0 + ∆) + A exp(ıkx − ky) y > η 2 ψ 2 y   q0 + B exp(ıkx + ky) y<η 2 ∂ ψ (or u) at the perturbed boundary keeping terms of order Now we match ψ (or v) and ∂y η, A, B but not higher order. Matching ψ gives A exp(ıkx) = B exp(ıkx) Matching ∂ ∂y ψ ⇒ A=B gives (q0 + ∆)η0 exp(ıkx) − kA exp(ıkx) = q0 η0 exp(ıkx) + kA exp(ıkx) ⇒ A= 1 ∆η0 2k The linearized kinematic condition for the interface tells us we need u to order one and v to order η. These are just 0 and v(x, η, t) ∆ ∂ η 2k ∂x respectively. Therefore, the interface evolves according to ∂ ∆ ∂ η η= ∂t 2k ∂x or ω=− ∆ 2 , 2 c=− ∆ 2k Greens function Alternatively, we split ψ into a background and a ﬂuctuation part (∇2 − γ 2 )ψ = q0 + ∆H(y) (∇2 − γ 2 )ψ � = ∆[H(y − η) + H(y)] where H is the Heaviside step function. Diﬀerentiating the ﬁrst equation by y gives ( ∂2 − γ 2 )u = −∆δ(y) ∂y 2 Linearizing the second by Taylor-expanding H(y − η) H(y) − ηδ(y)... gives (∇2 − γ 2 )ψ � = −∆ηδ(y) For a single wave, the last equation becomes ( ∂2 − k 2 − γ 2 )ψ � = −∆ηδ(y) ∂y 2 Deﬁning the Greens function ( ∂2 − α2 )Gα (y − y � ) = δ(y − y � ) ∂y 2 gives with K = u = −∆Gγ (y) ψ � = −∆η GK (y) k 2 + γ 2 and our interface satisﬁes ∂ ∂ ∂ η = ∆Gγ (0) η − ∆GK (0) η ∂x ∂t ∂x so that c = ∆GK (0) − ∆Gγ (0) For the barotropic problem Gk (0) = − 1 2k , G0 (0) = 0 3 ⇒ c=− ∆ 2k