GEOPHYSICAL FLUID DYNAMICS:Understanding (almost) everything with

rotating shallow water models.

Hints and solutions to the problems

Vladimir Zeitlin

Laboratory of Dynamical Meteorology,Sorbonne University and Ecole Normale Superieure

Paris, France

1

Solutions to Problems of Chapter 2

1.1 Problem 1

To prove thatd

dt=

∂

∂t+ v · ∇v, (1.1)

as applied to any dynamical variable F (x, t), which is dened in the domain of the ow,we use the Euler-Lagrange duality and consider the space coordinates x as position ofa uid parcel passing through this point, and the Eulerian velocity v as the velocityof the parcel. We recall that the Lagrangian derivative is to be calculated along theparcel's trajectory, that is the change of x is ∆x = v∆t , and hence

∆F∆t

=F (x + v∆t, t+ ∆t)− F (x, t)

∆t, (1.2)

which, in the limit ∆t→ 0 gives (1.1).

1.2 Problem 2

The demonstration follows by direct calculation of the Lagrangian derivative, and the

use of the Lagrangian expression of divergence ∇ · v = ∂X∂X + ∂Y

∂Y + ∂Z∂Z :

d

dt(ρJ ) =

dρ

dtJ + ρ

dJdt

=dρidt

= 0, (1.3)

dJdt

=∂(X, Y, Z)∂(x, y, z)

+∂(X, Y , Z)∂(x, y, z)

+∂(X,Y, Z)∂(x, y, z)

=

(∂(X, Y, Z)∂(X,Y, Z)

+ ...

)J =

(∂X

∂X+∂Y

∂Y+∂Z

∂Z

)J ⇒

dρ

dt+ ρ∇ · v = 0↔ ∂ρ

∂t+∇ · (ρv) = 0. (1.4)

1.3 Problem 3

In the direct calculation of the time-derivative, the fact that the contour is Lagrangian,that is its elements are moving with the uid, should not be forgotten, which means

that for an element of the contour l = dx = x2 − x1 , where x1,2 are positions ofneighbouring uid parcels on the contour, d

dt l = v2 − v1 = dv

dγ

dt=∫

Γ

(dl · dv

dt+ v · d

dtl)

= −∫

Γ

∇Pρ· dl +

∫Γ

v · dv. (1.5)

The last integral in (1.5) is an integral of the full dierential dv2

2 over the closedcontour, and is therefore zero. In the case of barotropic uid the integrand in the rstintegral becomes the dierential of enthalpy, and the integral vanishes by the samereason.

1.4 Problem 4

The variation of the action with Lagrangian (2.26) gives:

δS =∫dt

∫d3a

[dXdt· δX − P

(∂(δX, Y, Z)∂(a, b, c)

+∂(X, δY, Z)∂(a, b, c)

+∂(X,Y, δZ)∂(a, b, c)

)− δP

(∂(X,Y, Z)∂(a, b, c)

− 1)]

. (1.6)

Vanishing δS for independent variations of δX, δY , δZ, and δP gives the equationsof motion. The latter variation straightforwardly gives incompressibility condition inLagrangian form:

∂(X,Y, Z)∂(a, b, c)

= 1. (1.7)

In order to make appear variations of δX, δY , δZ instead of their derivatives, integra-tions by parts in t, a, b, and c in (1.6) under condition of vanishing variations at theboundaries are necessary.

In this way we get

d2X

dt2= −∂(P, Y, Z)

∂(a, b, c)= − ∂(P, Y, Z)

∂(X,Y, Z)= − ∂P

∂X, (1.8)

where (1.7) was used, and similarly for Y and Z. Replacing dXdt by v, and using Euler-

Lagrange duality, we see that these equations are Euler equations for incompressibleuid with constant density.

1.5 Problem 5

In many textbooks, e.g. Landau & Lifshitz (1975), the Navier-Stokes equations arewritten in the coordinates using co-latitude θ instead of latitude φ. The coordinatesare thus (0 ≤ r ≤ ∞, 0 ≤ θ ≤ π, 0 ≤ λ < 2π), and we will use them below. Theformulas in coordinates (0 ≤ r ≤ ∞, 0 ≤ λ < 2π,−π2 ≤ θ ≤ π

2 ) are obtained bysubstitution θ = π

2 − φ, and the change of sign of the corresponding component ofvelocity due to dierent orientation of the coordinate basis, see below. The right or-thonormal coordinate basis is then (r, θ, λ), where the unit vectors are pointed in the

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sense of increase of corresponding coordinates. The main technical diculty of usingcurvilinear coordinates, with respect to Cartesian ones, is that the orientation of thebasis depends on position. The position of a point is given by the radius-vector r. Aninnitesimal change of position is given by

dr = drr + rdθθ + r sin θdλλ. (1.9)

By denition, the velocity of the point is

v =dr

dtr + r

dθ

dtθ + r sin θ

dλ

dtλ. (1.10)

The expressions in front of the unit vectors are respective components of the velocity

vr =dr

dt, vθ = r

dθ

dt, vλ = r sinλ

dλ

dt. (1.11)

When calculating the acceleration dvdt the changes of basis vectors with the change of

position, i.e. in time, should be taken into account. As the length of the unit vectorsdoes not change, the change of basis consists in a solid rotation. It is easy to checkthat it is given then by the following formulas:

drdt = vθ

r θ + vλr λ,

dθdt = −vθr θ + cot θ vλr λ,dλdt = −vθr θ + cot θ vλr λ,

(1.12)

and hence, the vector of acceleration takes the form

dv

dt=dvrdt−(v2λ + v2

θ

r

)r

=(dvθdt

+vrvθ − v2

λ cotλr

)θ

=(dvλdt

+vrvλ + vθvλ cot θ

r

)λ. (1.13)

This determines the acceleration terms in Euler equations. The pressure gradient termsare obtained with the help of expression of the operator nabla in spherical coordinates:

∇ = r∂

∂r+ θ

∂

r ∂θ+ λ

∂

r sin θ ∂λ, (1.14)

which is also to be used in full (Lagrangian) time-derivatives in (1.13): ddt = ∂

∂t +v ·∇.The Coriolis acceleration is calculated from the decomposition of the angular velocityvector in the spherical coordinates basis: Ω = Ω cos θr − Ω sin θθ, and the denitionof the vector product in a right orthonormal basis:

Ω ∧ v =

∣∣∣∣∣∣r θ λ

Ω cos θ −Ω sin θ 0vr vθ vλ

∣∣∣∣∣∣ . (1.15)

All ingredients of the Euler equations are thus expressed in spherical coordinates.To obtain the continuity equation, the divergence of velocity should be expressed in

3

spherical coordinates, taking into account that dierentiations with respect to thecoordinates are now applied not only to the components of velocity, but also to thebasis vectors themselves. The formulas for the dierentials of the basis vectors can bereconstructed from (1.12), by re-multiplying them by dt and recalling the denitions(2.15). This gives

∇ · v =1r2

∂(r2vr)∂r

+1

r sin θ

(∂(sin θvθ)

∂θ+∂vλ∂λ

). (1.16)

1.6 Problem 6

The demonstration is a straightforward calculation, taking the curl of the momentumequations, using the hydrodynamic identity, and retaining the dominant, in the domainof parameters of the primitive equations, terms in the curl.

1.7 Problem 7

By denition q = ~ζa · ∇ρ, and it is sucient to prove that ∂t~ζa · ∇ρ = −~v ·(~ζa · ∇ρ

).

The left-hand side of the last equality is calculated using the equations of motion for~ζa and ρ, the fact that ∇ · ~ζa = ∇ · ~v = 0, and the following identities of the vectoranalysis:

∇A ·(∇∧ ~B

)= −∇ ·

(∇A ∧ ~B

), (1.17)

~A ∧(~B ∧ ~C

)= ~B

(~A · ~C

)− ~C

(~A · ~B

), (1.18)

Proof

∂t

(~ζa · ∇ρ

)= (∂t~ζa) · ∇ρ+ ~ζa · ∇(∂tρ)

= ∇ρ ·(∇∧

(~v ∧ ~ζa

))− ~ζa · ∇ (~v · ∇ρ)

= −∇ ·(∇ρ ∧

(~v ∧ ~ζa

))− ~ζa · ∇ (~v · ∇ρ)

= −∇ ·(~v(~ζa · ∇ρ

))+∇ ·

(~ζa (~v · ∇ρ)

)− ~ζa · ∇ (~v · ∇ρ) = −~v · ∇

(~ζa · ∇ρ

). (1.19)

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1.8 Problem 8

For the internal waves of the form

(u, v, φ) =(u, v, φ

)ei(kx+ly+mw−ωt) + c.c., (1.20)

with the dispersion relation:

4

ω2 =(N2 k

2 + l2

m2+ f2

), (1.21)

polarisation relations are obtained by solving the inhomogeneous algebraic system oftwo Fourier-transformed momentum equations at a given amplitude of geopotentialperturbation, which gives:

u =ωk + ilf

ω2 − f2φ (1.22)

v =ωl − ikfω2 − f2

φ (1.23)

Perturbations of density and vertical velocity in terms of the perturbation of geopoten-tial follow straightforwardly from the hydrostatic and linearised continuity equations.

Polarisation relation for the root ω = 0 follows from the Fourier-transformed equa-tion (2.80).

1.9 Problem 9

The fast way to show that inertia-gravity waves have no PV anomaly is to use theFourier-transformed equation (2.80). The time-derivative in this equation becomesmultiplication by ω, and the expression in brackets is the linearised PV anomaly. Soif ω 6= 0, which is the case for inertia-gravity waves, it is PV anomaly which shouldvanish. This can be alternatively shown by a direct calculations using polarisationrelations of Problem 8.

1.10 Problem 10

As the horizontal momentum equations in (2.82) are the same as in (2.74), with thereplacement P → φ the polarisation relations for u and v are the same. On the contrary,the third polarisation relation, between w and P is dierent, and easy to nd fromthe third equation in (2.80). Hence relations between the components of the horizontalvelocity and the vertical velocity are dierent.

5

2

Solutions to Problems of Chapter 3

2.1 Problem 1

We give the derivation of the two-layer model with a free surface. The one-layer modelis directly recovered from the equations for the upper layer below.

Conventions:Upper layer - layer 1, position of the upper surface - z2, position of the interface

- z1, position of the bottom, which is taken to be at - z0 = const, pressure over theupper surface P2 = const, pressure at the interface - P1, pressure at the bottom - P0.Corresponding densities: ρ1,2 = const

2.1.1 Upper layer

Master equation:

ρ1(z2 − z1)(dv1

dt+ f z ∧ v1

)=

−∇[−ρ1g

(z2 − z1)2

2+ (z2 − z1)P1

]−∇z1P1 +∇z2P2 (2.1)

After substitution of the hydrostatic relation

P1 = P2 + ρ1g(z2 − z1), (2.2)

the right-hand side of (2.1) becomes, recalling that P2 is constant:

ρ1g(z2−z1)∇(z2−z1)−P2∇(z2−z1)−2ρ1g(z2−z1)∇(z2−z1)+P2∇(z2−z1)−ρ1g(z2−z1)∇z1 = ρ1g(z2−z1)∇z2

Recalling that z2−z1 = h1, and z1−z0 = h2, where h1,2 are respective thicknessesof the layers, we thus get

dv1

dt+ f z ∧ v1 = −g∇(h1 + h2) (2.3)

The equations for h1,2 follow straightforwardly from the mass conservation in thelayers, which becomes volume conservation at constant densities.

2.1.2 Lower layer

Master equation:

ρ2(z1 − z0)(dv2

dt+ f z ∧ v2

)=

−∇[−ρ2g

(z1 − z0)2

2+ (z1 − z0)P0

]−∇z0P0 +∇z1P1 (2.4)

After substitution of the hydrostatic relation P0 = P2 +ρ2g(z1−z0) the right-handside of (2.1) becomes, recalling that z0 is constant:

ρ2g(z1−z0)∇(z1−z0)−P1∇z1+P1∇z1−(z1−z0)∇P1 = −ρ2g(z1−z0)∇(z1−z0)−ρ1g(z1−z0)∇(z2−z1),

where we used (2.2) and condition P2 = const. Recalling that z2 − z1 = h1, andz1 − z0 = h2, where h1,2 are respective thicknesses of the layers, we thus get

dv2

dt+ f z ∧ v2 = −g∇

(h2 +

ρ1

ρ2h1

)(2.5)

2.2 Problem 2

2.2.1 Master equation and boundary conditions

Before solving this problem, the analogue of master equations should be derived for theprimitive equations in pseudo-height coordinates for the atmosphere. The derivationgoes along the same lines as in the oceanic case, but the incompressibility condition∇ · v + ∂zw = 0, where v is the horizontal velocity, should be averaged over the layer(z2, z1), giving, for the averaged horizontal velocity, without change of notation:

∂t(z2 − z1) +∇ · ((z2 − z1)v) = 0 (2.6)

Averaging of the momentum equation over the layer gives:

(z2 − z1)(dv1

dt+ f z ∧ v1

)=

−∇[−g θ

θ0

(z2 − z1)2

2+ (z2 − z1) φ|z1

]−∇z1P1 +∇z2 φ|z2 − φ|z1 ∇z1, (2.7)

where φ is geopotential, θ = const is the mean potential temperature in the layer, andθ0 is a reference value.

As in the Problem 1 above, the master equation (2.7) is applied to the lower (1)and upper (2) layers. The dierence, however, is that boundary conditions are not thesame. The upper boundary z2 is considered to be at, while the lower boundary z0 isfree, but the geopotential at z = z0 is constant.

7

2.2.2 Lower layer

The hydrostatic equation for the lower layer (layer 1) is

φ1 = φ0 + gθ1

θ0(z1 − z0), (2.8)

where the subscript in geopotential indicates the corresponding vertical position, asfor the pressure in Problem 1. Using this formula in the master equation applied tothe lower layer we get

dv1

dt+ f z ∧ v1 = +g

θ1

θ0∇z0 = −g θ1

θ0∇(h1 + h2), (2.9)

where we used that z2 = h1 + h2 + z0 = const.

2.2.3 Upper layer

The hydrostatic equation for the upper layer (layer 2) is

φ2 = φ1 + gθ2

θ0(z2 − z1). (2.10)

Applying (2.7) we get:

ρ1(z2 − z1)(dv2

dt+ f z ∧ v2

)=

−∇[−g θ2

θ0

(z2 − z1)2

2+ (z2 − z1)φ1

]−∇z1φ1 +∇z2φ2, (2.11)

and using the boundary conditions and (2.8) we obtain

dv2

dt+ f z ∧ v1 = −g θ2

θ0∇h2 − g

θ1

θ0∇h1. (2.12)

We can either consider now that the mean potential temperature in the lower layerhas the reference value: θ1 = θ0, or introduce the reduced gravity g′ = g θ1θ0 . In both

cases the stratication parameter α = θ2θ1> 1 appears in the equation for the upper

layer.The continuity equations for thicknesses h1,2 of both layers follow directly from

mass conservation.

2.3 Problem 3

As there is no dependence on the vertical coordinate in the RSW equations, we onlyneed the polar coordinates 0 ≤ r < ∞, 0 ≤ θ < 2π in the plane.The corresponding

8

orthogonal unit vectors are r, θ. Similarly to Problem 5 in Chapter 1 of this manual,the innitesimal change of the radius vector r = rr is

dr = drr + rdθθ, (2.13)

and velocity is

v =dr

dtr + r

dθ

dtθ, (2.14)

The expressions in front of the unit vectors are respective components of the velocity

vr =dr

dt, vθ = r

dθ

dt. (2.15)

While calculating the derivative of the velocity, that is acceleration, a change in ori-entation of the basis with the change of position should be taken into account. It iseasy to see that it is only the change of polar angle θ that changes orientation, and

dr = dθθ, dθ = −dθr (2.16)

Hence

dv

dt=dvrdt−(v2θ

r

)r

=(dvθdt

+vrvθr

)θ. (2.17)

The operator nabla in polar coordinates takes the form:

∇ = r∂

∂r+ θ

∂

r ∂θ, (2.18)

which is to be used in the gradient and divergence operators. The divergence of velocitythus becomes

∇ · v =1r

∂(rvr)∂r

+1r

∂vθ∂θ

. (2.19)

Finally the components of the Coriolis acceleration in the cylindrical basis r, θ, z are

2Ω ∧ v =

∣∣∣∣∣∣r θ z0 0 2Ωvr vθ 0

∣∣∣∣∣∣ . (2.20)

Injecting all these ingredients in the RSW equations gives these equations in polarcoordinates, which were used in applications to vortex stability in the book.

2.4 Problem 4

The demonstration is provided by a straightforward calculation of the time-derivativeof the energy density using the expressions for time-derivative of height and velocityfollowing from the equations of motion:

∂te = ∂th(vkvk

2+ gh

)+ hvk∂tvk = −∂i(hvi)

(vkvk2

+ gh)

+ (hvi)∂i(vkvk

2+ gh

),

(2.21)where we used tensor notation ∂k = ∂

∂xk, x1 = x, x2 = y, the corresponding ex-

pression for advective derivative ∂t + vk∂k, and Einstein's rule of summation over

9

repeting indices. It is easy to see that the left-hand side of (2.21) is a divergence of

hv(

v2

2 + gh).

2.5 Problem 5

The vorticity equation is derived by cross-dierentiation of the the equations for u andv, respectively by y and x, which eliminates h, and rearranging the remaining terms.

2.6 Problem 6

As the integration is over the xed domain of the ow, the time-derivative of CF isequal to the integral of the time-derivative of the integrand, which is

∂t (hF(q)) = ∂thF(q) + hF ′(q)∂tq = −∇ · (hv)F(q)− hv · ∇qF ′(q) = −∇ · (hvF(q)) .(2.22)

Using the Gauss theorem the integral over the domain of the divergence of the vectoreld hvF is transformed into the total ux through the boundary, which is zero forisolated system.

2.7 Problems 7 and 8

These problems parallel the corresponding problems for the primitive equations, andthe methods of solution are the same, as well as the polarisation relations, with thereplacement φ→ gh.

2.8 Problem 9

The equations are straightforwardly obtained by taking weighted sums and dierencesof the corresponding equations in (3.43).

2.9 Problem 10

The invariance of the action with respect to time-shifts implies that the expression(3.57) vanishes. This expression is the well-known form of energy conservation in eldtheory. The rst term represents the time invariance of energy, as the time-derivative,with a xed boundary, can be taken out of the integral, and if no-ux boundaryconditions are imposed, the other term vanish, afer applying Gauss theorem

d

dt

∫dxdy

(δLδX

X +δLδY

Y − L)

= 0. (2.23)

The Lagrangian energy thus is

E =∫dxdy

(δLδX

X +δLδY

Y − L)

=∫dxdy

(X2 + Y 2

2+ g

h(X,Y )2

). (2.24)

In Eulerian description, the coordinates are X, Y themselves, and the correspond-ing change of integration variables should be made in (2.25): (x, y)→ (X,Y ) with the

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Jacobian which is hH0

, as the Lagrangian Labels can be supposed to correspond to theat surface h = H0, without loss of generality. Thus

E =1H0

∫dXdY h(X,Y )

(X2 + Y 2

2+ g

h(X,Y )2

), (2.25)

and we recognise in the integrand in new variables the Eulerian energy density of(3.27), (3.28). The other terms in (3.57) can be transformed similarly and, togetherwith replacing the Lagrangian time derivative by its Eulerian expression, give (3.28).

2.10 Problem 11

The Lagrangian (3.52) is obviously invariant with respect to translations X → X +∆X , Y → Y + ∆Y with constant ∆X,Y , as it depends only on derivatives of X andY . Let us consider, for simplicity, the one-dimensional RSW and translations in x-direction: X → X+∆. X ′ = Xx will denote the x-derivative of X. The demonstrationin the two-dimensional case is the same, with more cumbersome formulae. The formalcalculation of the variation of the action gives (cf. the derivation of the equations ofmotion in the main text):

δS = ρ

∫dt

∫dx δL =

ρ

∫dt

∫dx

[(δLδX− d

dt

δLδX− ∂

∂x

δLδX ′

)∆ +

d

dt

(δLδX

∆)

+∂

∂x

(δLδX ′

∆)]

,(2.26)

and the expression in the right-hand side should vanish. The rst bracket has equationsof motion inside, and vanishes. hence the invariance of action at any ∆ implies

d

dt

(δLδX

)+

∂

∂x

(δLδX ′

)= 0, (2.27)

which gives

X + g∂h

∂X= 0, (2.28)

following the same manipulation with the space derivative of h as those used in (3.54),(3.55).

11

3

Solutions to Problems of Chapter 4

3.1 Problem 1

The demonstration follows the same lines as in Problem 8 to Chapter 3: the linearisedpotential vorticity has the same form in the presence of the boundary, and non-zerowave-frequency implies vanishing of the anomaly of PV. A straightforward calculationwith the polarisation relation for Kelvin waves gives the same.

3.2 Problem 2

See Chapter 12.2. These are baroclinic and barotropic Kelvin waves, and to obtaincorresponding solutions in the two-layer model, it is sucient to apply the procedureused in the one-layer model to (3.45) and (3.46).

3.3 Problem 3

We take f = 1 to simplify the formulae. Solution for the Kelvin wave is obtained atn = 0 and l < 0. From (4.30) we get

p+12−√l2 − l

ω+

14, (3.1)

where p is given by (4.28). Exponentiating (3.1) twice, we get

(ω2 − 1)(ω4 − l2 − 2l) = 0. (3.2)

The roots ω2 = 1 should be discarded, as in order to eliminate u0 and v0 in favour ofη0 in (4.15) we have to suppose ω

2 6= 1. The positive root at negative l gives the Kelvin-wave branch of Fig. 4.3. The solution for η0(x) readily follows from (4.32) and consistsof a single exponent. The corresponding polarisation relations are straightforwardlyobtained from (4.15). An important dierence with the abrupts shelf is that in spitethat the other characteristics of the wave are similar, the across-coast component ofvelocity is not zero. This can be checked directly in (4.15): if a solution is sought withu0 ≡ 0 a contradiction arises, as the rst two equations lead to a solution for η0(x) ina form of an exponential, the last equation does not.

3.4 Problem 4

To nd the wave solutions over escarpment in form of the step function, which we willorient along the x-axis in the f -plane and place at y = 0, without loss of generality,

we linearise the RSW equations at each side of the escarpment, and eliminate allvariables in favour of the deviation of the free surface η. The only dierence between theequations in the lower and upper half-planes is in the mean depth: h = H± respectively.By looking for solutions in the form (u±, v±, η±) = (u±(y), v±(y), η±(y))ei(kx−ωt) wethus get

gH±η′′± + (ω2 − f2 − gH±k2)η′′± = 0, (3.3)

with(ω2 − f2)v± + iωgη′± + ikfgη± = 0, (3.4)

where prime denotes y- dierentiation. For decaying o the escarpment solutions η± =A±e

∓a± , a± > 0 we get

ω2 − f2 − gH±k2 + gH±a2± = 0. (3.5)

It remains to match the solutions in the lower and upper half-planes using boundaryconditions at the escarpment, that is at y = 0. They are 1) continuity of pressure,η+ = η−, and 2) continuity of the transverse mass ux H+v+ = H−v−. The rstimmediately gives A+ = A−, and the second, with the help of (3.4), gives

ω(a+H+ + a−H−) = kf(H+ −H−). (3.6)

with a± > 0 satisfying (3.5). This is the dispersion relation in question. The phasevelocity c = ω

k is dermined by the orientation of the step, such as the waves always runleaving shallow water on their right, as coastal Kelvin waves. The dispersion relationsimplies in the case of long waves k → 0, and becomes

ω = fk

|k|H+ −H−H+ +H−

(3.7)

3.5 Problem 5

Erratum: The second inequality in the upper line of (4.48) should be opposite. The one-

dimensional Green's functions (4.46), (4.47) are calculated by the method of residuesin both cases. We recall that the method is based on the Cauchy formula for anyfunction f(z) of the complex variable, which is analytic inside the contour C on thecomplex plane, and a point a inside the contour:∮

C

dzf(z)z − a

= 2πif(a) (3.8)

If a lies outside the contour, the integral is zero.

The integrand in (4.45) is represented as eik(x−x′) 12

√1U

(1

k−√

1U− 1

k+√

1U

), and

the contour is chosen to consist of the real axis and demi-circle of radius tendingto innity either in upper, or in lower half-plane, depending on the sign of x − x′,to render the integrand in the Cauchy formula non-singular inside the contour. Theapplication of (3.8) gives the result (4.46) directly for negative U , as the poles of the

13

integrand lie on the imaginary axis of the complex plane. For positive U the poles ofthe integrand lie on the real axis, and additional hypotheses are needed to prescribehow to circumvent them. It is clear that the result of the integration, whatever the rulegives either zero or oscillating in x−x′, that is non-decaying, function. As usual in suchcases the boundary conditions of causality should be imposed. Causality means thatthere is no reaction before action, and that the response of the ow to the mountain isdownstream, not upstream. This is achieved by chifting the integration contour alongthe real axis downwards, i.e. adding −iε, ε→ 0 in the denominator of the integrand,which leads to (4.47).

The two-dimensional Green's function (4.48) is obtained by Fourier-transformingthe equation, renormalising one of spatial variables with

√F 2 − 1 which leads to the

integral of the form ∫dkdl

ei(kx+ly)

k2 − l2 − const

The integration over k goes as in the one-dimensional example above, and conditionx > 0 arises as in (4.47). The remaining integration in l is not elementary and theintegral can be found in tables of integrals.

3.6 Problem 6

Ordinary dierential equation in (4.56) is inhomogeneous, with constant coecients,if PV is constant, Q(y) = Q0. According to standard rules, its general solution is a

general solution of the corresponding homogeneous equation H = Ae√Q0y +Be−

√Q0y

plus a particular solution of the inhomogeneous equation H = 1Q0

. The constants A, Bare determined from the boundary conditions.

3.7 Problem 7

Although the problem (4.64) is linear, the presence of y in the coecients leads totechnical diculties while consecutively eliminating variables to obtain a single equa-tion. The safe way of doing it is 1) to express ux in terms of other variables, anddierentiate the two remaining equations in x, in order to use the expression for ux.The x-momentum equation gives (∂2

xx − ∂2tt)η as a function of various derivatives of v

and y. Acting by the operator ∂2xx−∂2

tt upon the remaining equations gives the result,up to an extra dierentiation by x in the left-hand side, which can be removed withoutloss of generality for wave solutions.

3.8 Problem 8

Solution is obtained by straightforward calculations. For Kelvin waves it is given inthe text. For Yanai waves u = h by construction, and the dispersion relation followsfrom (4.45). For other waves, once a solution for v is found in terms of Gauss-Hermitefunctions, the zonal momentum and continuity equations, after Fourier-transformationin x and t give a pair of algebraic equations for u and η. For example, for Rossby waveswe get:

(u, v, η) = (U(y), V (y), H(y)) ei(kx−ωt), (3.9)

14

with

V (y) = φn(y), U(y) = −ikφ′n(y)− ωnyφn(y)

ω2n − k2

,

H(y) = iωnφ

′n(y)− kyφn(y)ω2n − k2

. (3.10)

3.9 Problem 9

Erratum: nabla is missing in the last term in the rst line of eq. (4.90), this termshould read −qv · ∇v

The derivation is a direct computations, taking rst the weighted sum and thedierence of the momentum equations in the layers, in order to obtain the momentumequations for barotropic and baroclinic components, respectively, cross-dierentiatingthe equations for barotropic u and barotropic v to get an equation for ψ, and expressingvelocity components in nonlinear terms through the inverse transformation (4.86), and(4.88) - see also solution of Problem 7 to Chapter 5.

3.10 Problem 10

We rst rewrite the Eulerian RSW equations in cylindrical coordinates (r, θ) andassume exact axial symmetry:

(∂t + ur ∂r)ur − uθ(f +

uθr

)+ g∂rh = 0 ,

(∂t + ur ∂r)uθ + ur

(f +

uθr

)= 0 , (3.11)

∂th+1r∂r(r ur h) = 0 .

Here ur, uθ are the radial and azimuthal components of velocity. Note that the ad-justed stationary state changes character as compared to the rectilinear case: it veriesconditions of the cyclo-geostrophic balance, and not of the purely geostrophic one:

uθ

(f +

uθr

)= ∂rh, ur = 0. (3.12)

Multiplying the second equation in (3.11) by r, we recover the conservation of angularmomentum:

(∂t + ur ∂r)(ruθ + f

r2

2

)= 0 , (3.13)

which replaces the conservation of geostrophic momentum in the plane-parallel case.Equations (3.11) can be rewritten using the Lagrangian coordinate R(r, t). Integrating(3.13) gives:

R(r, t)uθ(r, t) + fR2(r, t)

2= r uθI(r) + f

r2

2≡ G(r) , (3.14)

15

where uθI is the initial azimuthal velocity prole. Using the above expression we get:

uθ

(f +

uθR

)=

1R

(G− f R

2

2

)(f +

G

R2− f

2

)=

1R3

(G2 − f2R4

4

). (3.15)

The mass conservation is expressed by the following relation:

h(r, t)R(r, t) dR = hI(r) r dr . (3.16)

With the help of (3.15), (3.16), and the denition R(r, t) = ur(r, t) the radial momen-tum equation becomes:

R+f2

4R− 1

R3G2 +

1∂rR

∂r

(r hIR∂rR

)= 0 . (3.17)

to be solved with initial conditions R(r, 0) = r, R(r, 0) = urI .A direct check shows that

hI(r) =h0

2

(1− r2

L2

), R(r, t) = r, uθI(r) = rΩ, (3.18)

where Ω is determined by cyclo-geostrophic balance

Ω(Ω + f)r − g h0

L2r = 0, (3.19)

is an exact solution of (3.17). We could linearise the (3.17) about this solution to ndsmall oscillations, but it is possible to do better and nd nonlinear oscillation whichgive an exact solution (1). Indeed, if we make the same ansatz with arbitrary constantΩ, and time-dependent radial position

hI(r) =h0

2

(1− r2

L2

), R(r, t) = rφ(t), uθI(r) = rΩ, Ω = const, (3.20)

non-dimensionalise the system in the obvious way, with L as the length scale, f−1 asthe time scale, introduce the Burger number γ, and denote M = 1

2 + Ωf we get:

φ+φ

4− M2

φ3− γ

φ3= 0, (3.21)

to be solved with initial conditions φ(0) = 1, φ(0) = urI . A drastic simplication ofthis equation is provided by the substitution φ2 = χ which immediately gives theequation of the harmonic oscillator with shifted equilibrium position:

χ+ χ− 4E = 0, E =u2rI

2+

18

+M2 + γ

2> 0. (3.22)

A "pulson" solution satisfying the initial conditions χ(0) = 1, χ(0) = 2urI is given by

χ(t) = 4E + (1− 4E) cos t+ (2urI + 1− 4E) sin t. (3.23)

and represents nonlinear inertial oscillations, as its frequency is always one (in non-dimensional terms). A small-amplitude version of this solution is obtained by straight-forward linearisation.

16

The details of solution can be found in (Zeitlin 2010).

17

4

Solutions to Problems of Chapter 5

4.1 Problem 1

Introducing parameters, as in the main text, we write down the denition of PV:

q =vx − uy + f0 + βy

H0(1 + λη). (4.1)

The QG scaling is: U for both components of velocity, L for both coordinates, and allnondimensional parameters small and of the same order: λ ∼ βL ∼ ε = U

f0L. The PV

thus becomes

q =f0

H0

1 + ε(vx − uy + y

1 + εη. (4.2)

Expanding the denominator in Taylor series, and retaining the leading terms gives

q =f0

H0(1 + ε(vx − uy + y − η)) , (4.3)

where the leading-order in ε expressions u = −ηy, v = ηx should be used.With the QG scaling for time: T = L/U , the leading-order in ε term in the advec-

tion operator is ddt = ∂t + ηx∂y − ηy∂x, and applying it to (4.3) we recover the QG

equation.

4.2 Problem 2

We take as dispersion relation ω = −β kk2+l2 , where the wave vector is k = (k, l). The

presence of an additional constant in the denominator of the dispersion relation doesnot change the analysis of phase and group velocities. The expression for the phasevelocity is

c =ωkk2

=(−β k2

(k2 + l2)2, 2β

kl

(k2 + l2)2

), k = (k, l). (4.4)

Its remarkable property is that the zonal phase velocity is negative-denite, i.e. thewaves propagate always to the West.

The expression for the group velocity is

cg = (∂kω, ∂lω) =(−β k2 − l2

(k2 + l2)2,−β kl

(k2 + l2)2

). (4.5)

The zonal group velocity is not sign-denite and, at a given l, is positive for short waves(large enough k), and negative for long waves (small enough k). So although wavesare always propagating westward, the modulations (as well as energy, momentum, etc)can propagate both ways, depending on the wavelength.

4.3 Problem 3

Following the same lines as in the problem 1 above, we get for the PV

q =f0

H0

1 + ε(vx − uy + y

h, (4.6)

as in the frontal regime it does not make sense to split h into the mean and a per-turbation. Hence, unlike the QG regime the zero-order PV is not constant. However,it is not advected by the geostrophic velocity, as J (h, 1

h ) ≡ 0. So the rst-slow-timederivative of h is zero, and we need to introduce a slower time τ = εt. By the samereason we have to keep the rst ageostrophic correction to advection, which is givenby the equation (5.14) in the main text, where the time-derivative ∂t should be omit-ted by the just explained reason. Retaining the leading -order terms in the resultingexpression for the Lagrangian conservation of PV:

(u(0)∂x + v(0)∂y + ε(∂τ + u(1)∂x + v(1)∂y)

) 1 + ε(∂xv(0) − ∂yu(0) + y)h

= 0, (4.7)

with

v(0) = hx, u(0) = −hy, v(1) = −J (h, hy)− yhx, u(1) = −J (h, hx) + yhy, (4.8)

and calculating derivatives leads to the FG equation, up to multiplication by 1h2 .

4.4 Problem 4

It is easy to see that the linearisation over the state of rest with constant thicknessh = H + η leads to a simple-wave equation resulting from the rst and the last termsin (5.38):

ηt −Hηx = 0. (4.9)

It describes non-dispersive waves propagating westward. This is a long-wave limitof the Rossby-wave equation, which is consistent with the nature of the FG regimeapplicable to the scales which are much larger than the deformation radius.

4.5 Problem 5

Erratum: in equation (5.49) a plus/minus sign between the two brackets inside thesquare brackets is missing in the rst print.

The polarisation relation is given by any of the two equations in (5.47) after Fourier-transformation in space and time. After substitution of the expressions for ω , it givesthat π1 = π2 for ω = ωbt. This means that geostrophic velocities are the same in thelayers, which is the essence of the barotropic motion. For ω = ωbc the pressures andthe geostrophic velocities are opposite.

19

4.6 Problem 6

The most unstable mode corresponds to the maximum of the growth rate, i.e. of theimaginary part of the eigenfrequency corresponding to (5.54). It is clear from (5.54)that the maximum of the growth rate is achieved at the minimal possible meridionalwavenumber ky. Although the boundary conditions were not discussed in Chapter 5.3.3of the main text, it is clear that we need some, at least in y. As the interface betweenthe layers has a constant inclination, the available potential energy of the system isinnite, if there are no lateral boundaries. A natural conguration is a zonal channel,cf. Chapter 10 of the main text, where the baroclinkc instability is treated in moredetail. So the meridional component of velocity v should vanish at the boundary, whichmeans, in turn, that the Fourier-transforms of π1,2 in x and t vanish. This implies thatzero-mode with constant in y-direction π1,2 is to be discarded, and the meridionalwavenumber ky is discrete and non-zero. Let the minimal value of ky be lmin. We givea solution of the problem in the case U1 = −U2 = U/2, and F1 = F2 = F (layersof the same depth in the reference frame moving with the mean zonal velocity of thesystem). The expression for the growth rate then is

σ(k, l) = kU

2

√(k2 + l2)2 − 4F 2

k2 + l2 + F 2, (4.10)

where we denote kx by k.Dierentiating this expression with respect to k gives

4F 4 − (k2 + l2)3 − F 2(3k4 + l2(−4 + l2) + 4k2(1 + l2))(F 2 + k2 + l2)2

√4F 2 − (k2 + l2)2

, (4.11)

and zeroes of the numerator give extrema of the growth rate. As is easy to see, e.g. byplotting σ, the growth rate has a single maximum. The corresponding value of k = k0

is given by the real root of the cubic equation, which can be obtained by using Cardanoformulas, or numerically. For example, for lmin = 1, F = 1, the zone of unstable k isthe interval[0, 1], with k monotonously increasing from zero to k0 ≈ 0.65, and thendecreasing to zero, with σmax = 0.19U in non-dimensional terms.

4.7 Problem 7

By introducing baroclinic and barotropic components of velocity

vbt =1H

(h1v1 + h2v2), vbc = v1 − v2, (4.12)

where the subscripts 1(2) refer to upper (lower) layer and H = const is the totalthickness of the system, it is easy to see that vbt is divergenceless, and its componentscan be expressed in terms of stream-function ψ, as usual. Taking the curl of thebarotropic momentum equation on the f - plane gives

∂t∇2ψ + J (ψ,∇2ψ) + (∂2xx − ∂2

yy)(h(H − h)

H2uv

)− ∂2

xy

(h(H − h)

H2(u2 − v2)

)= 0,

(4.13)

20

where h ≡ h1, and we omitted the subscript bc at the components of the baroclinicvelocity u, v, for which we have:

∂tu+J (ψ, u)−(u∂2

xy + v∂2y2

)ψ+

H − 2hH

(u∂x + v∂y)u− u

H(u∂x + v∂y)h−fv+g∂xh = 0,

(4.14)

∂tv+J (ψ, v)−(u∂2

x2 + v∂2xy

)ψ+

H − 2hH

(u∂x + v∂y) v− v

H(u∂x + v∂y)h+fu+g∂yh = 0.

(4.15)The mass conservation in the upper layer takes the form:

∂th+ J (ψ, h) + ∂x

(h(H − h)

Hu

)+ ∂y

(h(H − h)

Hv

)= 0. (4.16)

Dening the baroclinic Rossby number ε, as usual, with the help of the typical baro-clinic velocity and horizontal scale, choosing the aspect ratio and the barotropic Rossbynumber both to be of the order ε2, and the time-scale to be of the order of ε2, as well,we get from (4.14) and (4.15) the expressions for the u and v in terms of h and ψ,as an expansion in ε about the geostrophic wind. By injecting them into (4.13) and(4.16) we get a system of coupled equations for h and ψ, where h can be expressed interms of deviation of the interface η as h = 1− η.

A detailed derivation can be found in (Benilov & Reznik 1996).

4.8 Problem 8

Erratum: h(0) should be understood as its value at the coast, that is a vertical barwith subscript x = 0 should be inserted before the equality sign. Equation (5.68)

is obtained by straightforward substitution of expressions for zero-order elds, i.e.geostrophic balance equations, in the right-hand side of (5.67).

Equation (5.72), where h(0) should be understood as its value at the coast, isobtained similarly, with substitution of the zero-order expression for v(0), and u(0) = 0from (5.65) and elementary integrations, with integrations by parts when necessary.

4.9 Problem 9

Imposing a rigid lid in shallow water equations means neglecting free-surface elevationin the mass conservation equation, and replacing it by barotropic pressure in themomentum equations. The continuity equation becomes a time-independent constraint

(Hu)x + (Hv)y = 0, (4.17)

and can be explicitly resolved with the help of stream-function ψ:

Hu = −ψy, Hv = +ψx (4.18)

As suggested in the Problem, we will work with one-dimensional topographyH = H(x)and get the following expression for the divergence:

21

H = H(x)⇒ ux + vy = −H′(x)

H(x)u = D. (4.19)

The relative orticity becomes

ζ = vx − uy =(

ψxH(x)

)x

+ψyyH(x)

, (4.20)

and vorticity equation ddt (ζ+f)+(ζ+f)D = 0 becomes an equation for ψ. Its inearised

vertion on the f - plane is:

ζt + fD = 0 ↔[(

ψxH(x)

)x

+ψyyH(x)

]t

+ fH ′(x)H2(x)

ψy = 0. (4.21)

Looking for wave solutions ψ = φ(x)ei(ωt−ly) + c.c., we get:(φ′

H

)′− l2

Hφ− f

c

H ′

H2φ = 0, c =

ω

l. (4.22)

which is a Sturm - Liouville problem for eigenfunctions φn and eigenvalues cn(l). It canbe shown that for escarpment-like topography the spectrum is discrete and correspondsto trapped waves with a number of nodes n = 0, 1, 2, ....

4.10 Problem 10

Let us consider the Korteweg-deVries equation in a most general form:

∂tu+ c0∂xu+ αu∂xu+ β∂3xxxu = 0. (4.23)

We are looking for solutions propagating without change of form:

u = u(x− V t) ⇒

(c0 − V )u′ + αuu′ + βu′′′ = 0, (4.24)

which are localised (solitary waves, solitons). With this hypothesis we integrate (4.24)once:

(c0 − V )u+ αu2

2+ βu′′ = 0, (4.25)

then multiply by u′ and integrate once more:

(c0 − V )u2

2+ α

u3

6+ β

u′2

2= 0. (4.26)

This equation can be interpreted as an equation of a material point moving in apotential, given by the rst two terms in (4.26), with zero total energy, and integratedby separation of variables giving the solution u(x− V t) = 3

αV−c0

cosh2qV−c0

4β (x−V t), where

V = c0 + α3 umax, and thus the speed of solution depends on amplitude, which is a

purely nonlinear eect.

22

5

Solutions to Problems of Chapter 6

5.1 Problem 1

The demonstration is given by calculating the circulation∫Cudx+ vdy of the velocity

eld corresponding to the Green's function of the source situated at the origin: x′ = 0:

v =x

2π(x2 + y2), u = − y

2π(x2 + y2)

over a circle C, by passing to polar coordinates r, θ with the help of relations dy =cos θdθ, dx = − sin θdθ. The circulation thus is equal to one. By Stokes theorem, thisis equal to the integral of corresponding vorticity over the area of the circle. As iseasy to see, the vorticity is zero everywhere except the origin, and thus is given by thedelta-function.

5.2 Problem 2

The demonstration is the same as in Problem 6 in Chapter 2

5.3 Problem 3

Erratum: An overall minus sign is missing in the expression of energy (6.20).

The conservation of energy follows directly from its conservation in the continu-ous system. The demonstration for two other supposed integrals of motion consists intaking the time derivative of each of them, in substituting the expression for time-derivatives (6.19), and using relabelling of indices and symmetry properties, e. g.∑ij κiκjxij =

∑ji κjκixji = −

∑ij κiκjxij , and hence

∑ij κiκjxij = 0.

5.4 Problem 4

The solution of this problem can be found in the main text: equations (6.84) and(6.23), and can be obtained by straightforward application of (6.19) to the two-vortexsystem.

5.5 Problem 5

The solution of this problem is geometric and does not necessitate any calculations: itis sucient to plot the vectors of velocity eld due to two other vortices at the locationof a given vortex, that is at the vertex of the triangle, sum them up and show that theresulting vector is perpendicular to the radius-vector giving the position of the vortex.

5.6 Problem 6

The solution of the problem readily follows from the analysis of equations (6.23) and(6.24): if a wall is put at the axis, we get a solution of the Euler equations in a half-plane, as the boundary condition of zero normal velocity is identically satised. Thus,a vortex near the wall forms a dipole with its image, and moves along the wall.

5.7 Problem 7

Erratum: There is an overall minus sign missing in the rst print in the expression ofvorticity in (6.14).

First of all we notice that the Jacobian of the complex Lagrangian mapping:

Z = αξeiωt + βξ∗, Z∗ = α∗ξ∗eiωt + β∗ξ, (5.1)

is |α|2 − |β|2 6= 1, and this should be taken into account while calculating variousphysical quantities, like vorticity.

It is easy to see that (5.1) satises (6.13) identically, as in the complex form theequations of motion contain Jacobians of Z and Z∗, and its complex conjugate, andtheir respective contributions cancel each other, as Z contains only ξ, and no ξ∗.

By changing variables (X,Y ) → (Z,Z∗) in (6.14), and renormalising with theabove Jacobian gives

ζ =2ω|α|2

|α|2 − |β|2= const (5.2)

The proposed solution of 2D Euler equatons, thus, has uniform vorticity inside theunit circle in the ξ - plane.

Taking the real and imaginary part of Z given by (5.1) at ξ = eiθ gives a parametricequations of the ellipse, whose orientation is dened by t, at xed ω;

Matching with the outer solution at the unit circle is straightforward as at the unitcircle ξ∗ = ξ−1. Note that in order to ensure matching the complex velocity shouldbe understood as V = X + iY = u + iv here. The proposed velocity eld has zerovorticity out of the unit circle in the ξ- plane.

5.8 Problem 8

Erratum: The sign in the equation (5.36) in the main text was erroneous in the rstprint. The factor 2π in (5.35), (5.36) can be removed by renormalisation, to get thestandard expressions for Green's functions.

We are looking for steady solutions ψ = ψ(x − Ut), moving with constant zonalvelocity U , of the barotropic beta-plane equation

∂t∇2ψ + J (ψ,∇2ψ) + β∂xψ = 0, (5.3)

which can be, thus rewritten as

−U∂x′(∇2ψ − β

Uψ

)− J

(ψ,∇2ψ − β

Uψ

)= 0, (5.4)

wherre x′ = x− Ut.

24

Equation (5.4) can be interpreted as vorticity equation for steady moving congu-rations, with a new "vorticity" ∇2ψ − β

Uψ. For a given "vorticity" conguration, thestreamfunction, and thus velocity, is given by the Green's function, obeying

∇2G(|x− x′|)− β

Uψ = δ(|x− x′|) (5.5)

Depending on the sign of U , this is either −K0( |x−x′|R ) (negative U), or 1

4Y0( |x−x′|R )

(positive U). We here introduced the notation R =√|U |β . In both cases, for the vortex

dipole of intensity κ situated at x = 0, y = ±d2 we get

ψ(x, y) = κ

[G(x, y − d

2)−G(x, y +

d

2)]. (5.6)

The dipole is a solution of (5.3) if the velocity of its displacement, induced by thevelocity induced by a partner vortex at the location of a given vortex coincides withU , which gives the following relation:

U =κ

RG′(d

R

)→ U

32 = κ

√βG′

(d

√β

U

). (5.7)

This equation can be solved, qualitatively, by looking for intersection of the curvesrepresenting the left- and the right-hand sides. Already this analysis shows a qualitativedierence between the cases positive (eastward propagation) and negative (westwardpropagation) U : there are only two solutions in the rst case, and an innite seriesof solutions in the second. The structure of solutions, given by modied and ordinaryBessel functions, respectively, is totally dierent: exponentially decaying in the rstcase, and having an weakly decaying oscillating tail in the second. The solutions areanalogs of the modon solution described in the main text, with an addition of theRossby-wave tail in the case of westward propagation. Exact solutions can be foundby solving (5.7)

This solution was found in (Gryanik 1986)

5.9 Problem 9

Conservation of energy and enstrophy within the triad means that for variations ofenergy of each component of the triad δEi, i = 1, 2, 3 we get a system of two equations

δE1 + δE2 + δE3 = 0, k21δE1 + k2

2δE2 + k23δE3 = 0. (5.8)

If the variation of the median component δE2 is given, (5.8) becomes an inhomogeneoussystem of two linear equations for δEi, i = 1, 2. To make things transparent, let us takea = 2, b = 3. The solution of (5.8) then is δE1 = − 5

8δE2, δE3 = − 38δE2, that is, more

energy goes to smaller k. At the same time, k21δE1 = − 5

32k22δE2, k

23δE3 = − 27

32k22δE2,

that is more enstrophy goes to higher k.

25

6

Solutions to Problems of Chapter 7

6.1 Problem 1

Erratum To be compatible with the denitions given in the main text, the inequalityin the announcement should be inverted: [h] > 0

Let us consider the RH conditions for one-dimensional shallow water in the absolutereference frame, that is with U = 0, denote the quantities behind (in front of) theshock as u+(−), h+(−), and suppose that the shock is moving from left to right, thatis u+(−) > 0 (this means that the uid goes through the shock from right to left). Thethicknesses h± are positive, by denition. The RH conditions become:

u+h+ = u−h− (6.1)

u2+h+ +

gh2+

2= u2

−h− +gh2−

2(6.2)

u3+h+ + gh2

+u+ ≤ u3−h− + gh2

−u− (6.3)

Substituting u+ = h−h+u−, as follows from (6.1), in (6.3), and dividing by h−u−, which

is positive, as supposed, we get

g(h+ − h−) ≤(

1−h2−h2

+

u2−

)(6.4)

Substituting u+ = h−h+u− in (6.2) gives the expression of u2

− in terms of h±:

u2− = g

h+ + h−2

h+

h−(6.5)

Substituting this expression in (6.4) gives quadratic inequality for h+

g(h+ − h−) ≤ g (h+ + h−)2(h+ − h−)2h+h−

, (6.6)

which can be only satised for h+ > h−.

6.2 Problem 2

The demonstration follows by using denitions and straightforward algebra.

6.3 Problem 3

Adding viscous dissipation to the right-hand sides of the momentum equations in (7.1)in the main text, that is, respectively ν∇2u and ν∇2v, and then dierentiating the v- equation with respect to x, we get the vorticity equation in the presence of viscosity:

d

dt(vx + f) + (vx + f)ux = ν∇2 (vx + f) . (6.7)

combining this equation with the mass-conservation equation written in the form

d

dth+ hux = 0, (6.8)

we getd

dt

vx + f

h= ν∇2 (vx + f)

h. (6.9)

6.4 Problem 4

The 2-layer 1d RSW model with a rigid lid read :

∂tu1 + u1∂xu1 + ρ−11 ∂xπ1 =0 ,

∂tu2 + u2∂xu2 + ρ−12 ∂xπ1 + g′∂xη =0 ,

∂t(H1 − η) + ∂x((H1 − η)u1) =0 ,∂t(H2 + η) + ∂x((H2 + η)u2) =0 ,

Their linearisation about a solution with constant pressure and constant velocities inthe layers u1,2 = U1,2 gives

D1u1 + ρ−11 ∂xπ1 =0 ,

D2u2 + ρ−12 ∂xπ1 + g′∂xη =0 ,−D1η +H1∂xu1 =0 ,D2η +H2∂xu2 =0 ,

where D1,2 = ∂t+U1,2∂x. By eliminating subsequently u1,2 and then π1, with the helpof extra dierentiations, and recalling the denition of reduced gravity, we arrive to asingle equation for η:

ρ1

H1D2

1η +ρ2

H2D2

2η − g∆ρ∂2xxη = 0. (6.12)

After Fourier - transformation η = ηei(ωt−kx) (6.12) gives a quadratic dispersion rela-tion with the discriminant

D = g∆ρ(ρ1

H1+ρ2

H2

)− (U1 − U2)2 ρ1

H1

ρ2

H2, (6.13)

which, if negative, corresponds to instability. This gives the required condition of KHinstability.

27

In the presence of rotation, we can not work in the one-dimensional framework anymore, as the background ow should be in geostrophic equilibrium, which necessitatesacross-ow gradients of pressure. The linearised about the ow with layer-wise constantcomponent u of velocity, as above, equations on the f -plane read in this case:

D1u1 − fv1 + ρ−11 ∂xπ1 =0 ,

D1v1 + fu1 + ρ−11 ∂yπ1 =0D2u2 − fv2 + ρ−1

2 ∂xπ1 + g′∂xη = 0 ,

D2v2 + fu2ρ−12 ∂yπ1 + g′∂yη =0−D1η +H1(∂xu1 + ∂yv1) = 0 ,

D2η +H2(∂xu2 + ∂yv2) =0 .

These equations can be processed by introducing relative vorticity ζ = ∂xv1,2−∂yu1,2,and divergence χ = ∂yv1,2 + ∂xu1,2, cross dierentiating momentum equations to getthe equations for vorticity D1,2ζ1,2 + fχ1,2 = 0, allowing to express divergence interms of vorticity, which allows, in turn, to express vorticity in each layer in termsof η from the mass conservation equations. Finally, as in the case without rotation,the Laplacian of the barotropic pressure, entering the remaining divergence equationslayer-wise, can be eliminated, which leads to the equation for η, which is similar to(6.12), but contains a correction due to rotation:

ρ1

H1D2

1η +ρ2

H2D2

2η − f2

(ρ1

H1+ρ2

H2

)− g∆ρ∂2

xxη = 0. (6.15)

The discriminant of the dispersion relation resulting after Fourier-transformation is

D = k2

[g∆ρ

(ρ1

H1+ρ2

H2

)− (U1 − U2)2 ρ1

H1

ρ2

H2

]+ f2

(ρ1

H1+ρ2

H2

)2

. (6.16)

As is easy to see, the new term increases the threshold for the instability, by anamount which is negligible at high k (short-wave limit), and is substantial for small k(long-wave limit). Hence, rotation tends to stabilise the ow, especially for long-waveperturbations.

6.5 Problem 5

We start with 2-layer 1.5d RSW model with a rigid lid:

∂tu1 + u1∂xu1 − fv1 + ρ−11 ∂xπ1 =0 ,

∂tv1 + u1(f + ∂xv1) =0 ,

∂tu2 + u2∂xu2 − fv2 + ρ−12 ∂xπ1 + g′∂xη =0 ,

∂tv2 + u2(f + ∂xv2) =0 ,∂t(H1 − η) + ∂x((H1 − η)u1) =0 ,∂t(H2 + η) + ∂x((H2 + η)u2) =0 ,

where (u1, v1), (u2, v2) are velocities in the upper/lower layer; π2 = π1+g(ρ1h1+ρ2h2), η - interface displacement,H1 andH2 are the heights of two layers at rest, g

′ is reduced

28

gravity: g′ = g(ρ2− ρ1)/ρ2. We will be using the full layers' heights h1,2 = H1,2∓ η asdynamical variables.

By eliminating the barotropic pressure and imposing condition of zero overall trans-verse momentum, we get 4 equations for 4 variables u2, h2, v2, v1:

du2

dt− fv2 +

ρ1

ρ2h1 + ρ1h2

(f(h1v1 + h2v2)

− ∂x(h1u

21 + h2u

22

)+g∆ρρ1

h1∂xh2

)= 0,

dh2

dt+ h2∂xu2 = 0,

dv2

dt+ fu2 = 0,

dv1

dt+ (u1 − u2)∂xv1 + fu1 = 0,

u1 =h2u2

h2 −H, h1 = H − h2.

Here ddt = ∂t + u2∂x is Lagrangian derivative, as usual.

Further simplications are achieved by using the mass Lagrangian variable a: h2 =∂a/∂x):

du2

dt− fv2 +

ρ1

ρ2h1 + ρ1h2

(f(h1v1 + h2v2)

− h2∂

∂a

(h1u

21 + h2u

22

)+g∆ρρ1

h1h2∂ah2

)= 0,

dh2

dt+ h2∂au2 = 0,

dv2

dt+ fu2 = 0,

dv1

dt+ (u1 − u2)h2∂av1 + fu1 = 0.

These equations can be rewritten in matrix form:

d

dt

v2

v1

h2

u2

+

0 0 0 00 T 0 00 0 0 h2

2

0 0 M 2N

∂

∂a

v2

v1

h2

u2

=

−fu2

−fu1

0fh1

ρ2v2−ρ1v1ρ2h1+ρ1h2

,

where we introduce the notation:

T = (u1 − u2)h2 =Hu2

H − h2h2,

M =ρ1h2

ρ2h1 + ρ1h2

(g

∆ρρ1

(H − h2)− H2u22

(H − h2)2

),

29

N = − ρ1h2

ρ2h1 + ρ1h2

Hh2

H − h2u2,

and are, thus, quasi-linear. For nontrivial eigenvalues (the trivial ones are 0 and T ),we get:

det

(−λ h2

2

M 2N − λ

)= λ2 − 2Nλ− h2

2M = 0 (6.18)

with the solution

λ± = N ±√N2 +Mh2

2. (6.19)

The discriminant of this equation is

D =ρ1ρ2h1h

32

(ρ2h1 + ρ1h2)2

g∆ρ

(h1

ρ1+h2

ρ2

)− H2u2

2

h21

. (6.20)

The eigenvalues (6.18) are real and, hence, the system is hyperbolic when D is positive.On the contrary, if

(u2 − u1)2 =H2u2

2

h21

> g∆ρ(h1

ρ1+h2

ρ2

)(6.21)

then D < 0 and the system loses hyperbolicity. The former condition may be rewrittenas a condition on u2

u22 > g∆ρ

(h1

ρ1+h2

ρ2

)h2

1

H2.

We, thus, see that, unlike its one-layer counterpart, the 2-layer RSW changes type ifthe vertical shear of the transverse velocity is too strong. as h1 = H−h2, the conditionof KH instability of the previous Problem is sucent for hyperbolicity loss.

30

References

Zeitlin, V., 2010, "Lagrangian dynamics of fronts, vortices and waves: understandingthe (semi-) geostrophic adjustment", Lecture Notes in Physics, 805, 109137.

Benilov, E.S., and Reznik, G.M., 1996, "The complete classication of large-amplitudegeostrophic ows in a two-layer uid", Geophys. Astrophys. Fluid Dyn., 82, 122.

Gryanik, V. M. , 1986, "The singular geostrophic vortices on the beta-plane as amodel for synoptic vortices, Okeanologia, 26, 174179 (in Russian)