Rotation & nonlinear effects in shallow water:the Rossby adjustment problem

Ross Tulloch

April 29, 2004

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1 Shallow water equations

Consider, as in [6], a homogeneous incompressible 2D fluid in a channelwith negligible vertical velocity, roughly constant horizontal velocity for eachvertical column and height h(x, t) shallow relative to typical wavelengths.The total mass in [x1, x2] at time t is

total mass in [x1, x2] =

∫ x2

x1

ρh(x, t)dx,

where ρ is constant. The momentum at (x, t) is ρu(x, t), so integratingvertically gives the mass flux ρu(x, t)h(x, t) and therefore conservation ofmass gives:

∂

∂t

∫ x2

x1

ρhdx = ρuh(x1, t)− ρuh(x2, t) ⇒ ht + (uh)x = 0. (1)

Conservation of momentum is the same as in the Euler equations

(ρhu)t + (ρhu2 + p)x = 0 (2)

and the pressure can be eliminated using the hydrostatic balance approxi-mation, integrating vertically from z = 0 (flat bottom) to z = h(x, t) gives

p(z) = ρgz ⇒ p =1

2ρgh2, (3)

so the momentum equation becomes

(hu)t +

(hu2 +

1

2gh2

)x

= 0 ⇒ ut +

(u2

2+ gh

)x

= 0 (4)

when equation (1) is used to cancel the ht term.If we now allow for flow v in the y direction (into the plane), but enforce

no y dependence we have the following nonlinear shallow water PDE system

ht + uhx + hux = 0

ut + uux + ghx = 0

vt + uvx = 0. (5)

The shallow water equations describe the motion of a barotropic fluid withfree surface and depth much smaller than the horizontal length scale. This

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system of equations is a simplified and tractable version of the primitiveequations (Navier-Stokes with rotation and stratification) and is a very usefulmodel in geophysical fluid dynamics. One reason the shallow water modelis so important is because there is an analogy between it and descriptionsusing isopycnal or isentropic co-ordinates, see [11] for more details.

1.1 Adding rotation

Now if the channel is also rotating, Coriolis forcing adds important sourceterms to (5) as follows:

ht + uhx + hux = 0

ut + uux + ghx = fv

vt + uvx = −fu, (6)

where f is twice the angular velocity.

1.2 Linear shallow water: “gravity wave equations”

One always gains insight into a nonlinear problem by first linearizing abouta base state and solving the linearized version of the problem. In this case,the velocity and height fields are linearized about (u, v) = (U, 0) and h = H.So decompose the fields into ‘base’ and ‘perturbed’ variables

u(x, t) = U + u′(x, t), v(x, t) = v′(x, t), h(x, t) = H + h′(x, t), (7)

then substitute these into (6) and cancel all quadratic terms

h′t + (U + u′)h′x + (H + h′)u′x = 0 ⇒ h′t + Uh′x +Hu′x = 0

u′t + (U + u′)u′x + gh′x = fv′ ⇒ u′t + Uu′x + gh′x = fv′

v′t + (U + u′)v′x = −fu′ ⇒ v′t + Uv′x = −fu′. (8)

Finally, setting U = 0 (no mean wind) and dropping primes gives the equa-tion set for the Rossby adjustment problem described in [2] and [3], calledthe “gravity wave equations”:

∂u

∂t− fv + g

∂h

∂x= 0,

∂v

∂t+ fu = 0,

∂h

∂t+H

∂u

∂x= 0. (9)

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1.3 Project objectives and outline

The goal of this project will be to numerically analyze the time dependentevolution of shallow water equations in various regimes given a particularinitial condition. Specifically, the effects of rotation and nonlinearity ongeostrophic adjustment (described in the next section) are examined as in[5]. The four regimes are: (a) nonrotating-linear, (b) rotating-linear, (c)nonrotating-nonlinear and (d) rotating-nonlinear. Case (a) is just the waveequation (utt = gHuxx), which we can write down the solution for immedi-ately using d’Alembert’s formula. Case (b) is solved using the stencils shownin Problem 12 of Chapter 3 in [2], ie. using leapfrog on an unstaggeredmesh as well as using forward-backward time-differencing on a staggeredmesh. An attempt is made to reproduce the cover art on Durran’s textbook,which shows the emission of gravity waves during the adjustment process.Case (c) is the classic dam break problem which is solved using conserva-tion law methods, as described in [6], [7]. Case (d) is the most difficultbecause it involves a nonlinear system with source terms (coriolis forces).In [5] the emphasis is on analyzing the transient behaviour and not numer-ical method, so they used Leveque’s CLAWPACK conservation law package(amath.washington.edu/ claw) to solve (d). While CLAWPACK would be auseful tool to learn in its own right, it is not clear to me that it will handlethe source terms properly. In the steady-state, the source terms should beexactly balanced by the flux gradients. To achieve this I will experimentwith a “central-upwind” scheme adapted from [4], which examined the shal-low water system with bottom topography as the source term (instead ofrotation).

2 The Rossby adjustment problem

Rossby was the first to completely discuss the question of how a rotatingfluid with an initial perturbation adjusts under gravity. The geostrophicequilibrium that is achieved cannot be found by solving the steady-stateequations because they are degenerate in that any solution of the momentumequations satisfies the continuity equation exactly [3]. So the equilibriumstate depends on the initial state.

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2.1 Initial conditions

The problem discussed by Gill involves the linear evolution of an initial stepin the height field, with the rotating fluid initially at rest, so at t = 0

u(x, t = 0) = 0,

v(x, t = 0) = 0,

h(x, t = 0) =

{hl ≡ η0, x < 0hr ≡ −η0, x > 0

(10)

2.2 Steady-state analytical solution of the linear prob-lem

The analytical solution of (9) is found by first noting that the potentialvorticity is conserved, which in this case means

Qt =∂

∂t

(vx

f− h

H

)= 0 (11)

and (10) gives Q(t = 0) = (η0/H) sgn(x). Manipulating (9) gives a singlesecond order equation

htt +Hfvx − gHhxx = 0 (12)

which in steady state, after substituting Q(t = 0) and h for v, becomes

−gHhxx + f 2h = −f 2η0sgn(x) (13)

and the solution is

h

η0

=

{−1 + e−x/a

1− ex/a

for x > 0for x < 0,

v = −gη0

fae−|x|/a (14)

where a =√gH/f is called the Rossby radius of deformation. Figure 1, as

in [3], illustrates the initial and steady states.

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Figure 1: Initial height field (top) and geostrophic equilibrium height andmeridional velocity (below). The horizontal is scaled by the Rossby radiusa =

√gH/f and the vertical is scaled by the magnitude of the initial height

perturbation η0. Parameter values are g = 10m/s2, H = 10m and f =10−4s−1.

2.3 Transients in the linear problem

The homogenous problem for the height field, upon substituting Q as in (11)for vx in (40), is

htt − gHhxx + f 2h = 0 (15)

with initial condition

h(t = 0) = −η0sgn(x)− hsteady = −η0e−|x|/asgn(x) = −2η0

π

∫ ∞

0

ksinkx

k2 + a−2dk,

where k is the horizontal wave number assuming wavelike solutions of theform h ∝ ei(kx−ωt). Preserving antisymmetry in h, the transient height and

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zonal (u) velocity field are

h = −2η0

π

∫ ∞

0

ksinkxcosωt

k2 + a−2dk,

∂u

∂x= − 1

H

∂h

∂t⇒ u =

2η0

π

√g

H

∫ ∞

0

sinωtcoskx√k2 + a−2

dk,

∂v

∂t= −fu⇒ v =

2fη0

πH

∫ ∞

0

cosωtcoskx

k2 + a−2dk (16)

where the dispersion relation from (15) is given by ω2 = f 2 + k2gH. Anequivalent expression for u in terms of Bessel functions is

u =

{η0

π

√gHJ0

(f√t2 − x2

gH

)0

for | x |<√gHt

for | x |>√gHt,

, (17)

and is plotted for various times in Figure 2. Notice the emission of gravitywaves from the initial discontinuity. In this study, the nature of gravity waveemission will be analyzed by comparing transients in the zonal velocity uand height h for the 4 regimes discussed in Section 1.3. Also, the adjusted(steady state) height h and meridional velocity v profiles will be comparedfor the four regimes. Note that the final value of u will be uniformly zerowhen there’s not dependence in the meridional (y) direction.

The best possible approximation to the initial conditions in (10) for eachnumerical scheme will be used. That is, standard finite difference approxima-tions to the linear equations linear equations may require smoothing of thediscontinuity over a few grid cells because features of order 2∆x and smallerare poorly resolved [2]. In Section 4.1 smoothed-out step is used as describedin [2] and in Section 4.4 the initial height field has an arctangent profile inorder to mimic the cover art of [2]. To compare transients no boundary con-ditions need be specified, a large periodic domain is sufficient. Nonreflecting(also known as “open”, “radiating”, or “outflow”) boundary conditions areapplied in order to run to steady state.

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Figure 2: Transient profiles of u scaled by 1/η0 for the initial conditions inequation (10). Parameters are g = 10m/s2, H = 10m, f = 10−4s−1.

3 Case (a): Solution of the nonrotating-linear

problem

In the nonrotating-nonlinear problem there is no meridional velocity (v) andequations (9) reduce to

∂u

∂t+ g

∂h

∂x= 0,

∂h

∂t+H

∂u

∂x= 0 (18)

and decouple as

∂2u

∂t2− gH

∂2u

∂x2= 0,

∂2h

∂t2− gH

∂2h

∂x2= 0 (19)

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With initial conditions in (10), the solution is

h(x, t) =h(x−

√gHt, 0) + h(x+

√gHt, 0)

2,

u(x, t) = − 1

2√gH

∫ x+√

gHt

x−√

gHt

∂

∂sh(s, 0)ds

=h(x−

√gHt)− h(x+

√gHt)

2√gH

, (20)

which is plotted in Figure 3.

Figure 3: Transient profiles of h (left) and u (right) for nonrotating-linearadjustment, both scaled by 1/η0 and plotted against x/a. Parameters areg = 10m/s2, H = 10m, f = 10−4s−1.

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4 Case (b): Numerical Solution of the rotating-

linear problem

4.1 Problem 12, Chapter 3: Durran’s textbook

The inspiration for this project comes from an exercise in [2] which asks tocompare solutions of the linearized one-dimensional Rossby adjustment prob-lem obtained using leapfrog on an unstaggered grid versus forward-backwardtime differencing on a staggered grid. The problem is posed with typical pa-rameter values (f = 10−4s−1, g = 10m/s2, H = 10m) and initial conditions(step function as in Figure 1). Note that the initial conditions are smoothedout with three iterative applications of the filter

φfj =

1

4(φj+1 + 2φj + φj−1). (21)

Two common schemes are used to solve this problem, the first is leapfrog onan unstaggered mesh

δ2tu− fv + gδ2xh = 0,

δ2tv + fu = 0,

δ2th+Hδ2xu = 0. (22)

and the second is forward-backward time-differencing on a staggered mesh

δtun+1/2j − fvn

j + gδxhnj = 0,

δtvn+1/2j + fun+1

j = 0,

δthn+1/2j+1/2 +Hδxu

n+1j+1/2 = 0. (23)

The results of the two schemes are shown in Figure 4, forward-backward dif-ferencing appears to do better than leapfrog for both large and small Courantnumbers (see Section 4.3 for explanation of the relevance of Courant num-bers for these schemes). The plots show nondimensionalized height and bothvelocity fields at t = 2 in nondimensional time as question 12 in [2] asks for.See Section 4.5 for transient and steady state plots. Note that both schemesexhibit dispersion due to the sharp gradient (≈ 3∆x) in the initial condition.

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(a) Leapfrog, CFL=0.9 (b) Leapfrog, CFL=0.1

(c) FBS, CFL=0.9 (d) FBS, CFL=0.1

Figure 4: Plots of h, u, and v fields for leapfrog (top) and forward-backwarddifferencing (bottom) with Courant numbers 0.9 (left) and 0.1 (right) at timet = 21000s ≈ ft = 2 on domain x = 600km to x = 1400km. Parameters areg = 10m/s2, H = 10m, f = 10−4s−1, ∆x = 10

3km.

4.2 Consistency

Consistency is quite easy to show for the above schemes but stability issomewhat more difficult. The unstaggered leapfrog is second order accurate

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and therefore pointwise consistent

un+1j − un−1

j

2∆t− fun

j + ghn

j+1 − hnj−1

2∆x=

∆t2

6

∂3u

∂t3+g∆x2

6

∂3h

∂x3+ · · ·

vn+1j − vn−1

j

2∆t+ fun

j =∆t2

6

∂3v

∂t3+ · · ·

hn+1j − hn−1

j

2∆t+H

unj+1 − un

j−1

2∆x=

∆t2

6

∂3h

∂t3+H∆x2

6

∂3u

∂x3+ · · · . (24)

The forward-backward staggered time differencing is first order accurate intime and second order accurate in space

un+1j − un

j

∆t− fun

j + ghn

j+1/2 − hnj−1/2

∆x=

∆t

2

∂2u

∂t2+g∆x2

24

∂3h

∂x3+ · · ·

vn+1j − vn

j

∆t+ fun

j =∆t

2

∂2v

∂t2+ · · ·

hn+1j+1/2 − hn

j+1/2

∆t+H

un+1j+1 − un+1

j

∆x=

∆t

2

∂2h

∂t2+H∆x2

24

∂3u

∂x3+ · · · . (25)

4.3 Stability

Since this is a system of equations stability is ensured (sufficient) when thenorm of the amplification matrix (~vn

k = Ank~v

0n) is

‖Ak‖ ≤ 1. (26)

A necessary condition for stability is that the spectral radius is less or equalto unity, that is ρ(Ak) ≤ 1 (which is sufficient in the case that A is diagonal-izable. My attempt to show stability in this way was somewhat unsuccessful.Defining φ, ψ, and χ as past values of u, v, and h respectively gives theamplification matrix for the leapfrog scheme

uvhφψχ

n+1

j

=

0 2∆tf i2g∆t sin k∆x

∆x1 0 0

−2∆tf 0 0 0 1 0i2H∆t sin k∆x

∆x0 0 0 0 1

1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 0

uvhφψχ

n

j

(27)

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To find the norm ‖Ak‖, take the square root of the largest eigenvalue ofB = A∗

kAk. After a lengthy hand calculation that Maple appeared unwillingto perform, the characteristic equation of B appears to be

(1− λ)2[(1− λ)2 − λ(g2α2 + F 2)

] [(1− λ)2 − λ(H2α2 + F 2)

]= 0

⇒ λ2 − (2 + g2α2 + F 2)λ+ 1 = 0,

λ2 − (2 +H2α2 + F 2)λ+ 1 = 0 (28)

where α = (2∆t/∆x) sin k∆x and F = 2∆tf . Apparently ρ(B) > 1 so thiscondition is not satisfied, but that does not mean the scheme is necessarilyunstable. Progress can be made by taking a dispersion relation approach toderive a necessary condition for stability. Seek wave like solutions of the form

unj = u0e

i(kj∆x−ωn∆t), vnj = v0e

i(kj∆x−ωn∆t), hnj = h0e

i(kj∆x−ωn∆t). (29)

In matrix form the unstaggered leapfrog equations become − sin ω∆t∆t

−if g sin k∆x∆x

if − sin ω∆t∆t

0H sin k∆x

∆x0 − sin ω∆t

∆t

u0

v0

h0

= 0 (30)

which have nontrivial solutions when the determinant is zero, so the disper-sion relation is

sinω∆t

∆t= ±

[f 2 + gH

(sin k∆x

∆x

)2]1/2

(31)

and ω = 0 is another solution. So a necessary condition for stability in theunstaggered leapfrog scheme is∣∣∣∣∣∆t

√f 2 +

gH

∆x2

∣∣∣∣∣ ≤ 1. (32)

Similar analysis on the forward-backward staggered scheme gives the disper-sion relation

sinω∆t

∆t/2= ±

[f 2 + gH

(sin k∆x/2

∆x/2

)2]1/2

(33)

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so a necessary condition for stability is then∣∣∣∣∣∆t√f 2

4+gH

∆x2

∣∣∣∣∣ ≤ 1. (34)

Given the parameters used by Durran in [2] (gH = 100, f = 10−4, ∆x =104/3) it is easy to see that rotation has almost no effect on this condition,which is approximately the same as the condition from the one way waveequation, that is require

√gH∆t/∆x ≤ 1.

4.4 Durran’s cover art

A side objective of this project is to reproduce, as exactly as possible, thecover art on [2] which shows the emission of gravity waves in the transientprofile of the zonal velocity u, see Figure 5. After some tuning, I obtained a

Figure 5: Cover art of Dale Durran’s textbook: Numerical Methods for WaveEquations in Geophysical Fluid Dynamics [2] showing the time evolutionof u(x, t) at t = 943s, t = 1222s, and t = 1501s with initial conditionh(x, t = 0) =arctan(x/20km). Parameters are g = 10m/s2, H = 10m,f = 10−4s−1, ∆x = 10

3km.

very similar plot (Figure 6) using forward-backward time-differencing. Thepreface of [2] specifies all the parameters used to obtain the plots in Figure 5

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Figure 6: Transient profiles of u for rotating-linear adjustment at times t =1062∆t, t = 1375∆t, and t = 1689∆t where ∆t = 200s. Parameters areg = 10m/s2, H = 10m, f = 10−4s−1, ∆x = 3km.

except for the time step ∆t. Apparently the times quoted in [2] to produceFigure 5 are actually time step numbers n (unless the parameters of theproblem are different), in which case the time step was ∆t ≈ 224s.

4.5 Transient and steady-state solutions

Adding rotation to the linear equations causes gravity waves to disperse out-wards as in Figure 7 and allows for a nonzero steady state to develop, whichis shown in Figure 9. The agreement between the analytical solution de-scribed in Section 2.3 and the numerical solution is shown in Figure 8. Thenumerical scheme does a poor job capturing the front as expected becauseit is a small scale O(∆x) feature. Also notice that some dispersion is nearthe front. Otherwise the numerical scheme matches the analytical predic-tion quite well. To evolve the system to steady state, “outflow” boundary

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Figure 7: Transient profiles, using forward-backward time-differencing, of h(left) and u (right) for rotating-linear adjustment, both scaled by 1/η0 andplotted against x/a. Parameters are g = 10m/s2, H = 10m, f = 10−4s−1,∆x = 3km, ∆t = 200s.

conditions were applied to the forward-backward scheme

un+1N = un

N −∆t√gH

unN − un

N−1

∆x

un+11 = un

1 + ∆t√gH

un2 − un

1

∆x

hn+1N−1/2 = hn

N−1/2 −∆t√gH

hnN−1/2 − hn

N−3/2

∆x

hn+11/2 = hn

1/2 + ∆t√gH

hn3/2 − hn

1/2

∆x. (35)

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Figure 8: Numerical (solid) versus theoretical transient (dashed) profiles ofu at t = 12/f for rotating-linear adjustment. The initial height perturbationwas smoothed with 10 iterations of equation (21).

The steady height profile is identical to the predicted height profile in Figure 1while the velocity v has a slightly rounded vertex at x = 0 because the initialconditions were smoothed.

5 Case (c): Solution of the nonrotating-nonlinear

problem

5.1 Dam break problem

The nonrotating-nonlinear shallow water equations (5) with initial conditions

u(x, t = 0) = 0,

v(x, t = 0) = 0,

h(x, t = 0) =

{hl ≡ η0 +H, x < 0hr ≡ −η0 +H, x > 0

(36)

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Figure 9: Initial height field (top) and geostrophic equilibrium height andmeridional velocity (below) for the rotating-linear equations.

is known as the dam break problem, which is solved semi-analytically in [9]using conservation law theory. In this configuration, a 1-rarefaction wave anda 2-shock form, as illustrated in Figure 10. One can solve for the intermediatestate (numerically) by connecting the rightward Riemann invariant with theintermediate state (u∗, h∗)

u∗ − ul = 2(√

ghl −√gh∗

)(37)

and using the Rankine-Hugoniot jump conditions on the shock

u∗ − ur = (h∗ − hr)

√g

2

(h∗ + hr

h∗hr

). (38)

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One can solve for the rarefaction state (u(x′, t′), h(x′, t′) by matching therightward Riemann invariant with the rarefaction state

ul + 2√ghl = u(x′, t′) + 2

√gh(x′, t′) (39)

and noting that the leftwards characteristic is valid in the rarefaction

u(x′, t′)−√gh(x′, t′) =

x′

t′, (40)

giving the solution

u(x′, t′) =2

3

(√ghl +

x′

t′

)h(x′, t′) =

1

9g

(2√ghl −

x′

t′

)2

. (41)

The total solution is

h(x, t) =

hl, x <

√ghlt

19g

(2√ghl − x

t

)2, −

√ghlt < x <

(u∗ −

√gh∗

)t

h∗,(u∗ −

√gh∗

)t < x < ct

hr, x > ct

(42)

u(x, t) =

ul, x <

√ghlt

23

(√ghl + x

t

), −

√ghlt < x <

(u∗ −

√gh∗

)t

u∗,(u∗ −

√gh∗

)t < x < ct

ur, x > ct

(43)

where c = (hrur − h∗u∗)/(ur − u∗) is the speed of the shock. With initialconditions (36), ul = ur = 0 so c = h∗. The solution for hl = 11 and hr = 9corresponding to η0 = 1 andH = 10 is plotted in Figure 10. The intermediatestate is (u∗, h∗) = (1.00175, 9.97444), obtained from equations (37) and (38).

5.2 Finite volume methods

Typically, as in [5] and [7], the dam break problem is solved numericallyusing finite volume methods. Specifically, equation (5) is integrated usingGodunov’s method with Roe linearization to solve approximate Riemann

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Figure 10: Transient profiles of h (left) and u (right) for nonrotating-nonlinear adjustment, with η0 = 1 and H = 10. Other parameters areg = 10m/s2, f = 10−4s−1.

problems. Then to improve accuracy “flux” (or “slope”) limiters are addedto the flux term based on local slopes and an “entropy fix” is applied becausethe linear equations cannot capture rarefaction waves.

The idea behind Godunov’s method is to solve piecewise constant localRiemann problems “exactly”. The exact solution qn(x, t) can be determinedby piecing together the solutions of the Riemann problems at each cell inter-face if the time step ∆t is sufficiently small. The new solution is then definedas the average of the exact solutions over each cell

Qn+1j =

1

∆x

∫ xj+1/2

xj−1/2

qn(x, tn+1)dx (44)

which does not need to be explicitly integrated since the conservation law

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Qt + F (Q)x = 0 implies

Qn+1j = Qn

j −∆t

∆x(F n

j+1/2 − F nj−1/2) (45)

whereF n

j−1/2 = f(q∗(Qnj−1, Q

nj ) (46)

and q∗(ql, qr) is the solution to the Riemann problem between states ql andqr.

Solving the Riemann problems exactly at each cell is numerically ex-pensive so typically the nonlinear problem qt + f(q)x = 0 is replaced by alocally defined linear problem q + Aj−1/2qx = 0 where A is diagonalizablewith real eigenvalues and is consistent with the original conservation law:Aj−1/2 → f ′(q)asQj−1, Qj → q. For A to be Roe approximate then it must

also satisfy Aj−1/2(Qj − Qj−1) = f(Qj) = f(Qj−1). The details of the Roesolver, entropy fix, and flux limiters become quite technical. Since we alreadyhave the solution to the dam break problem in Figure 10 let us proceed tothe nonlinear-rotating problem and take a look at central upwind schemes.

6 Case (d): Solution of the rotating-nonlinear

problem

6.1 Central upwind schemes

The idea to solve the rotating-nonlinear problem using a central upwindscheme comes from [4], which analyzed the nonrotating-nonlinear problemwith bottom topography

ht + (hu)x = 0

(hu)t +

(hu2 +

1

2gh2

)x

= −ghBx, (47)

where B(x) represents the bottom elevation (topography). The right handside of (47) acts as a source term analogously to the coriolis terms in (6).taken for such terms to ensure that they exactly the balance the appropriateflux gradients to ensure the correct steady state develops. The authors of [4]claim that standard numerical schemes typically fail to preserve the balancebetween the fluxes and the source terms.

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Central upwind schemes are supposed to be simple, efficient and robust.They are based on Godunov’s method in that they are based on an exactevolution of a approximate piecewise polynomial reconstruction and do notrequire any Riemann solvers. However the major drawback to of centralschemes is the large numerical dissipation (O(∆t)2r−1) where r is a “formalorder of accuracy” [4]. Apparently this dissipation also affects long timesimulations such as steady-state computations (as Figure 12 attests).

The central upwind scheme for a one dimensional problem given in [5]is as follows. Evolve the cell averaged conservation law with source termS(u(x, t), x, t):

d

dtuj(t) +

f(u(xj+1/2, t)

)− f

(u(xj−1/2, t)

)∆x

=1

∆x

∫ xj+1/2

xj−1/2

Sdx (48)

and the cell average u is defined as

uj(t) =1

∆x

∫ xj+1/2

xj−1/2

u(x, t)dx. (49)

Substituting numerical fluxes in (48) gives

d

dtuj(t) = −

Hj+1/2(t)−Hj−1/2(t)

∆x+ Sj(t), (50)

where the fluxes Hj+1/2 are given by

Hj+1/2(t) =a+

j+1/2f(u−j+1/2)− a−j+1/2f(u+j+1/2)

a+j+1/2 − a−j+1/2

+a+

j+1/2a−j+1/2

a+j+1/2 − a−j+1/2

[u+

j+1/2 − u−j+1/2

].

Note that u+j+1/2 = pj+1(xj+1/2, t) is the right value of a (conservative, non-

oscillatory) piecewise polynomial interpolant at x = xj+1/2 and u−j+1/2 =

pj+1(xj+1/2, t) is the corresponding left value at x = xj+1/2. The details ofnon-oscillatory piecewise quadratic reconstructions are given in the appendixof [4]. Also, the local speeds of propagation a±j+1/2 are determined by theeigenvalues of the system

a+j+1/2 = max

{λN

(∂f

∂u(u−j+1/2)

), λN

(∂f

∂u(u+

j+1/2)

), 0

},

a−j+1/2 = min

{λ1

(∂f

∂u(u−j+1/2)

), λ1

(∂f

∂u(u+

j+1/2)

), 0

}, (51)

22

given λ1 < · · · < λN .In the Rossby adjustment problem (6) the system eigenvalues are u and

u ±√gH, obtained from the Jacobian in conservation form ~pt + f(~p)x = 0

(ignoring source terms)

~p =

hhuhv

=

hrs

⇒ f =

rr2/h+ gh2/2

rs/h

(52)

⇒ J =∂fi

∂pj

=

0 1 0gh− u2 2u 0−uv v u

. (53)

In steady state of the Rossby adjustment problem must support geostrophicbalance, which in the one dimensional system (6) with no y-dependence is

u = 0, v =g

f

∂h

∂x. (54)

The conservation form of (6) is

ht + (uh)x = 0

(hu)t +

(hu2 +

1

2gh2

)x

= fvh

(hv)t + (huv)x = −fuh, (55)

so in the numerical discretization we want to balance

g

2(h2)x ≈

g

2

h2j+1/2 − h2

j−1/2

∆x≈ fvj

hj+1/2 + hj−1/2

2≈ fvh (56)

⇒ let vj =g

f

hj+1/2 − hj−1/2

∆x. (57)

6.2 Transient and steady state solutions

Nonlinearity has a few obvious effects on geostrophic adjustment as illus-trated in the transient profiles of h and u in Figure 11. First note thatthe right-left symmetry of the rotating-linear problem in Figure 7 is broken.One would expect this considering the rarefaction wave that developed in

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Figure 11: Transient profiles of h (left) and u (right) for rotating-nonlinearadjustment, with η0 = 1 and H = 10. Other parameters are g = 10m/s2,f = 10−4s−1.

the break problem (Figure 10). Notice also that the initial leftward rarefac-tion wave is overtaken by dispersive wave, which should form a shock butapparently my code is quite dissipative. Numerical dissipation is even moreevident in Figure 12 which shows the “steady” profiles of h and v (after along simulation time). Part of the dissipation could be because I had troublewith quadratic polynomial interpolants and am currently using only linearinterpolants.

For more complicated theoretical considerations of the Rossby adjustmentproblem see [1], [10], and [12]. Also see [5] for a proper numerical study and [8]for a recent experimental study which claims to have quantitatively measuredpotential vorticity and the flow balance after a geostrophic adjustment for

24

Figure 12: Initial height field (top) and geostrophic equilibrium height andmeridional velocity (below) for the rotating-nonlinear equations.

the first time by taking simultaneous measurements the horizontal velocityfield and the vertical density field.

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References

[1] Bouchet, F., Submitted 2003: Frontal geostrophic adjustment and non-linear wave phenomena in one dimensional rotating shallow water. Part2: high-resolution numerical solutions. Journal of Fluid Mechanics, Cam-bridge.

[2] Durran, D., 1998: Numerical Methods for Wave Equations in GeophysicalFluid Dynamics. Spring-Verglag, New York.

[3] Gill, A., 1982: Atmosphere-Ocean Dynamics. International geophysicsseries, Academic Press, San Diego.

[4] Kurganov, A., D. Levy, 2002: Central-Upwind Schemes for the Saint-Venant System. Mathematical Modelling and Numerical Analysis, 36,397-425.

[5] Kuo, A., L. Polvani: Time-Dependent Fully Nonlinear Geostrophic Ad-justment. Journal of Physical Oceanography, 27, 1614-1634.

[6] LeVeque, R., 1990: Numerical Methods for Conservation Laws.Birkhauser, Boston.

[7] LeVeque, R., 2002: Finite Volume Methods for Hyperbolic Problems.Cambridge University Press, New York.

[8] Stegner, A., P. Bouruet-Aubertot, T. Pichon, 2004: Nonlinear adjust-ment of density fronts. Part 1. The Rossby scenario and the experimentalreality. Journal of Fluid Mechanics, 502, 335-360.

[9] Stoker, J., 1958: Water Waves. John Wiley, New York.

[10] Reznik, G., V. Zeitlin, M. Ben Jelloul, 2001: Nonlinear theory ofgeostrophic adjustment. Part I : Rotating shallow water model. Journalof Fluid Mechanics, 481, 269-290.

[11] Vallis, G., 2003: Atmospheric and Ocean Fluid Dynamics. Online text:http://www.princeton.edu/ gkv/aofd.

[12] Zeitlin, V., S. Medvedev, R. Plougonven, 2003: Frontal geostrophicadjustment, slow manifold and nonlinear wave phenomena in 1d rotatingshallow water. Part 1. Theory. Journal of Fluid Mechanics, 445, 93-120.

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