a new lagrangian coastal circulation model
TRANSCRIPT
A New Lagrangian Coastal Circulation Model
J.M. Greenberg†
Prof. Emeritus, Mathematical SciencesCarnegie Mellon University
Pittsburgh, PA 15213and
Visiting Research ProfessorUnited States Naval Academy
Annapolis, MD 21402
and
Ensign Daniel HartigUnited States Naval Academy
Annapolis, MD 21402
†The research of the first author was partially supported by the U.S. Office of Naval Research.
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1 Introduction.
To mankind, the most important areas of the ocean are those closest to us. Thelittoral regions of the world include the gulfs and straits and seas that are most heavilyfished and trafficked by merchant vessels. The majority of the earth’s population andconflict areas both reside in these coastal regions. The deadliest natural disasters–hurricanes and tsunamis and the flooding they induce–occur at the interface of landand sea.
The currents, wave heights and surges in these littoral environments differ sig-nificantly from circulation in the open ocean. In deep seas, currents are largelypredictable. The different seasons’ prevailing currents have been charted since thedays of the earliest oceanic explorers. However, currents and wave heights in coastalregions are heavily affected by local winds and bathymetry. In the open ocean, thesea floor is of little consequence to all but the deepest-diving submarines while incoastal regions the ocean bottom threatens ships of all sizes. Navies, shippers, in-surers and governmental planners and policy makers need good predictive modelswhich can update, over a reasonable time horizon, key hydrodynamical quantitiesgiven a knowledge of the current state of the system, local bathymetry and externalforces acting on the system. We present a model that will create these forecasts ofhydrodynamic properties.
In section 2 we review the relevant equations describing an incompressible oceansystem. These equations take into account Coriolis, gravitational and dissipative“turbulent” or viscous like internal forces acting on the ocean as well as externalwind and frictional forces. We present a “closed” system which accounts for themovement of a finite volume of fluid. The basic difficulty we are confronted withis the free-boundary nature of our problem; we have to solve the governing flowequations in the time-varying region where the water column height is positive. Thefree-boundary where the water column height vanishes is not known a-priori and mustbe determined as part of the solution.
In section 3 we “scale” the model developed in section 2 and recast it in dimension-less form. The scalings reflect the fact that we want a simpler model valid in shallowwater regions. In this section we obtain our first essential simplification, namely thehydrostatic approximation to the pressure.
In section 4 we take the dimensionless system developed in section 3 which holdsin a time varying domain of variable thickness and introduce the “s-coordinate”transformation used by Mellor et. al. [1]-[4] to map the dimensionless system into anequivalent system which holds on a domain of constant thickness. We also developequations for the depth-averaged flow velocities and velocity residuals. Exploiting thesmallness of certain dimensionless parameters, we show that—to within error termswe are prepared to tolerate—the velocity residuals may be computed in terms ofdepth-averaged flow velocities and water column height but have no influence on howthese latter quantities evolve; this is a key result.
In section 5 we derive a set of a-priori estimates and other key facts about thesolution of the closed system developed in section 4 which governs the evolution ofthe depth-averaged flow velocities and water column height. We also present an
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interesting exact Rotating-Pulsating solution to this system.In section 6 we look at a Lagrangian reformulation of the depth-averaged system
developed at the end of section 4. This new system holds on a fixed spatial domainwhich corresponds to the region in space which is “wet” at time t = 0. With thisreformulation we avoid having to explicitly track the unknown free-boundary wherethe height of the water column vanishes. From the Lagrangian model we derive ourcomputational model.
Section 7 is devoted to a discussion of the parallel implementation of the computa-tional model. We include a thorough discussion of the justification for partitioning our“square-domain” into a rectangular rather than a “lattice-like” sub-square partition.The principal advantage of this partitioning is a simpler communication protocol.We present performance results and timing comparisons for the parallel implementa-tion of the computational model; our speedups are impressive. We end the sectionwith the presentation of two simulations which show that Lagrangian computationalmodel produces steady-state or long-time solutions like the exact solutions producedin section 5. These results validate the effectiveness of the Lagrangian reformulationof the problem.
2 Basic Hydrodynamic Model.
Our starting point is the equations of motion for an incompressible fluid of constantdensity ρ0. Specifically, we are interested in solutions of the system:
ux + vy + wz = 0, (C − 2)
ut +(
u2)
x+ (uv)y + (uw)z + px =
fv + (σ11)x + (σ21)y + (σ31)z , (Mx − 2)
vt + (uv)x +(
v2)
y+ (vw)z + py =
−f u + (σ12)x + (σ22)y + (σ32)z , (My − 2)
and
wt + (uw)x + (vw)y +(
w2)
z+ pz =
−g + (σ13)x + (σ23)y + (σ33)z . (Mz − 2)(1)
(1)Equation (C-2) implies thatut +
(
u2)
x+ (uv)y + (uw)z = ut + uux + vuy + wuz,
vt + (uv)x +(
v2)
y+ (vw)z = vt + uvx + vvy + wvz, and
wt + (uw)x + (vw)y +(
w2)
z= wt + uwx + vwy + wwz.
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Here, (x, y, z) are the Cartesian components of a point in the fluid, t is time,(u, v, w)
(
x, y, z, t)
are the Cartesian components of the Eulerian velocity field at
(x, y, z) and time t, p is the ratio of the hydrostatic pressure P to the constant massdensity ρ0, and σij = σji, 1 ≤ i, j ≤ 3, are the components of the “Reynolds” or“Turbulent” stress tensor. We assume that
σ11 = 2Eux , σ22 = 2Evy, and σ33 = 2Ewz (2.1)
and that
σ12 = σ21 = E (uy + vx) , (2.2)
σ13 = σ31 = E (uz + wx) , and (2.3)
σ23 = σ32 = E (vz + wy) . (2.4)
E is typically referred to as the “Eddy Viscosity” and is assumed to be a function of(x, y) and t; i.e. E is independent of the vertical coordinate, z, and satisfies
Ez = 0. (2.5)
E is not a fundamental physical quantity but rather a mathematical construct we arefree to choose (for details see Section 6 and in particular (6.30) and (6.31)). We dothis in such a way that the spurious oscillations are removed from our solutions. Oneshould think that our real interest is in the “perfect” model when the “Turbulent”stresses are set to zero. The “Turbulent” stresses are included as regularizing termsin our equations.
The terms (fv) and (−fu) in (Mx−2) and (My−2) are the Coriolis accelerationsand are present because we are operating in a coordinate system which is rotatingabout the z-axis with angular velocity 2f . The term (−g) in (Mz− 2) represents thegravitational acceleration in the vertical direction; thus ours is a flat earth model.
Equation (C-2) is referred to as the continuity equation and the equations(Mx − 2), (My − 2), and (Mz − 2) represent the balance of momentum in the x, y,and z directions. These equations hold in the time varying domain
D(t) =
(x, y, z)∣
∣
∣a(x, y) < z < a(x, y) + h(x, y, t) and h(x, y, t) > 0
. (2.6)
The bottom surface of the fluid, namely
z = a(x, y) (2.7)
is given data but the upper surface of the fluid, namely
z = a(x, y) + h(x, y, t) (2.8)
must be determined as part of the solution. We must also determine
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Ω(t) =
(x, y)∣
∣
∣h(x, y, t) > 0
(2.9)
and
∂Ω(t) =
(x, y)∣
∣
∣h(x, y, t) = 0
. (2.10)
The unknown h(x, y, t) has the interpretation as the height of the water column abovethe bottom surface.
If the velocity field is known, the particle trajectories may be computed by solving
d
dt
(
X(t), Y (t), Z(t))
= (u, v, w)(
X(t), Y (t), Z(t), t)
(2.11)
and the positions of particles at time t which started at (X0, Y0, Z0) at t = 0 will bedenoted by (X, Y , Z)(X0, Y0, Z0, t).
We conclude this section with a brief discussion of boundary conditions for thesystem (C-2), (Mx − 2), (My − 2) and (Mz − 2).
Kinematic Boundary Conditions
Our kinematic boundary conditions are that
w(x, y, a(x, y), t) = u(x, y, a(x, y), t)ax(x, y) + v(x, y, a(x, y), t)ay(x, y),
for z = a(x, y) and (x, y) ∈ Ω(t) (BC − a)
and
w(x, y, (a + h)(x, y, t), t) = u(x, y, (a + h)(x, y, t), t)(ax + hx)(x, y, t)
+v(x, y, (a + h)(x, y, t), t)(ay + hy)(x, y, t) + ht(x, y, t),
for z = a(x, y) + h(x, y, t) and (x, y) ∈ Ω(t). (BC−(a + h))
These conditions imply that particles which start in the surfaces z = a(x, y) andz = a(z, y) + h(z, y, t) and flow with the velocity field (u, v, w) stay in these surfacesfor all t.
Force Boundary Conditions
At the upper surface z = a(x, y) + h(x, y, t) we assume that
p(x, y, a(x, y) + h(x, y, t), t) = 0. (BC − p)
On the surfaces z = a(x, y) and z = a(x, y) + h(x, y, t) we also impose boundaryconditions for the “surface tractions” in the x and y directions.
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We note that the “normals” to the surfaces z = a(x, y) and z = a(x, y)+ h(x, y, t).are given by:
na = (−ax, −ay, 1)T /
√
1 + a2x + a2
y (2.12)
and
na+h =
(
−(ax + hx), −(ay + hy), 1)T
/
√
1 + (ax + hx)2 + (ay + hy)2. (2.13)
Recalling that the “surface tractions” to any surface with unit normal
N = (N1, N2, N3)T /
√
N21 + N2
2 + N23 (2.14)
and are evaluated by computing
σ11 σ12 σ13
σ12 σ22 σ23
σ13 σ23 σ23
N =
σ11N1 + σ12N2 + σ13N3
σ12N1 + σ22N2 + σ23N3
σ13N1 + σ23N2 + σ33N3
/√
N21 + N2
2 + N23 (2.15)
and noting that the x and y components of the “surface traction” are given by
Tx = (σ11N1 + σ12N2 + σ13N3) /√
N21 + N2
2 + N23 (2.16)
and
Ty = (σ12N1 + σ22N2 + σ13N3) /√
N21 + N2
2 + N23 (2.17)
we arrive at our remaining boundary conditions on the surfaces z = a(x, y) andz = a(x, y) + h(x, y, t). They are
T ax = (−axσ11 − ayσ12 + σ13) /
√
1 + a2x + a2
y =
Kfrhu(
x, y, a(x, y), t)
/√
1 + a2x + a2
y, (2.18)
T ay = (−axσ12 − ayσ22 + σ23) /
√
1 + a2x + a2
y =
Kfrhv(
x, y, a(x, y), t)
/√
1 + a2x + a2
y, (2.19)
T(a+h)x =
(
−(ax + hx)σ11 − (ay + hy)σ12 + σ13
)
/
√
1 + (ax + hx)2 + (ay + hy)2 =
Kwind
(
uwind − u(x, y, a(x, y) + h(x, y, t), t))
/
√
1 + (ax + hx)2 + (ay + hy)2
(2.20)
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and
T(a+h)y =
(
−(ax + hx)σ12 − (ay + hy)σ22 + σ23
)
/
√
1 + (ax + hx)2 + (ay + hy)2 =
Kwind
(
vwind − v(x, y, a(x, y) + h(x, y, t), t))
/
√
1 + (ax + hx)2 + (ay + hy)2. (2.21)
The right hand sides of (2.18) and (2.19) represent the x and y components of thefrictional forces on the surface z = a(x, y) and Kfr has the dimension of 1/Time.Similarly, the right hand sides of (2.20) and (2.21) represent the x and y componentsof the wind forces on the surface z = a(x, y) + h(x, y, t), Kwind has the dimension ofLength/Time, and uwind and vwind are the prescribed wind velocities in the x and ydirections.
3 Shallow Water Scalings
Our task in this section is to cast our fluid model in dimensionless form. The scal-ings we employ reflect the fact that the underlying flows are taking place in shallowdomains. We let
x = Lx , y = Ly , z = Hz and t =L
Ut; (3.1)
a(x) = Ha(x) and h(x, y, t) = Hh(x, y, t); (3.2)
u(x, y, z, t) = Uu(x, y, z, t) , v(x, y, z, t) = Uv(x, y, z, t) and
w(x, y, z, t) =UH
Lw(x, y, z, t); (3.3)
and
p(x, y, z, t) = gHp(x, y, z, t). (3.4)
We also introduce the distinguished dimensionless scalars λ, ǫ,D, C2, and F by
λ =H
L, ǫ =
E
ULD =
ǫ
λ2, C2 =
gH
U2, and F =
fL
U. (3.5)
λ is referred to as the aspect ratio, ǫ is the inverse lateral Reynolds number andfunctions as the dimensionless lateral viscosity, D is the effective vertical viscosity,C
2 is the inverse Froude number, and F is the Rossby number. These parameterswill appear as coefficients in our dimensionless equations. In shallow water problems,the aspect ratio, λ, satisfies
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0 < λ ≪ 1 (3.6)
and this reflects the fact H ≪ L. For the oceanographic problems of interest to ushere, ǫ and C
2 satisfy
0 < ǫ ≪ 1 and C2 > 1. (3.7)
The definition of D guarantees that D ≫ ǫ. We shall in fact assume that
D ≫ 1. (3.8)
We note that our hypothesis (2.5) on E implies that both ǫ and D satisfy
ǫz = Dz = 0. (3.9)
If we now substitute (3.1) - (3.4) into (C − 2), (Mx− 2), (My − 2) and (Mz − 2)and (3.9) we arrive at the following dimensionless system:
ux + vy + wz = 0, (C − 3)
ut + (u2)x + (uv)y + (uw)z + C2px =
F v + (ǫ(ux − vy))x + (ǫ(uy + vx))y + Duzz + (ǫ(ux + vy + wz))x − ǫxwz.(Mx − 3)
If we exploit (C-3), we see that the 5th term on the right hand side of (Mx − 3) iszero and thus (Mx − 3) reduces to
ut + (u2)x + (uv)y + (uw)z + C2px =
F v + (ǫ(ux − vy))x + (ǫ(uy + vx))y + Duzz − ǫxwz.(Mx − 3)
(My − 2) similarly transforms to
vt + (uv)x + (v2)y + (wv)z + C2py =
−Fu + (ǫ(uy + vx))x + (ǫ(vy − ux))y + Dvzz − ǫywz
(My − 3)
while (Mz − 2) becomes
λ2(wt + (uw)x + (vw)y + (w2)z) + C2(pz + 1) =
(ǫ(uz + λ2wx))x + (ǫ(vz + λ2wy))y + 2ǫwzz.(Mz − 3)
We now exploit C2 > 1, λ ≪ 1, and ǫ ≪ 1 and replace (Mz−3) by the equilibrium
approximation
pz + 1 = 0. (Mz − 3)
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These equations hold in
D(t) =
(x, y, z)∣
∣a(x, y) < z < a(x, y) + h(x, y, t) and h(x, y, t) > 0
. (3.10)
The domains Ω(t) and ∂Ω(t) defined in (2.9) and (2.10) transform to
Ω(t) =
(x, y∣
∣h(x, y, t) > 0
(3.11)
and
∂Ω(t) =
(x, y)∣
∣h(x, y, t) = 0
. (3.12)
The boundary conditions (BC − a), (BC − (a + h)) and (BC − p) transform to
w(x, y, a(x, y), t) = u(x, y, a(x, y), t)ax(x, y) + v(x, y, a(x, y), t)ay(x, y),
for z = a(x, y) and (x, y) ∈ Ω(t), (BC − a)
w(x, y, (a + h)(x, y, t), t) = u(x, y, (a + h)(x, y, t), t)(ax + hx)(x, y, t)
+v(x, y, (a + h)(x, y, t), t)(ay + hy)(x, y, t) + ht(x, y, t),
for z = a(x, y) + h(x, y, t) and (x, y) ∈ Ω(t). (BC − (a + h))
and
p(x, y, (a + h)(x, y, t), t) = 0. (BC − p)
Moreover, if we combine (Mz − 3) and (BC − p), we obtain the hydrostatic approxi-mation to the pressure:
p(x, y, t) = a(x, y) + h(x, y, t) − z. (H − 3)
When solving (Mx − 3) and (My − 3) we shall use this relation for the pressure.Finally, if we exploit (3.1)-(3.4) and equations (2.1)-(2.5) relating the σij’s to the
velocity gradients we arrive at the dimensionless version of our traction boundaryconditions (2.18)-(2.21). They are
ǫ (−2axux − ay(uy + vx) + wx) (x, y, a+(x, y), t) + Duz(x, y, a+(x, y), t)
= kfrhu(x, y, a+(x, y), t), (3.13)
ǫ (−ax(uy + vx) − 2ayvy + wy) (x, y, a+(x, y), t) + Dvz(x, y, a+(x, y), t)
= kfrhv(x, y, a+(x, y), t) (3.14)
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ǫ(
−2(ax + hx)ux − (ay + hy)(uy + vx) + wx
)
(
x, y, (a + h)−
(x, y, t), t))
+Duz(x, y, (a + h)−
(x, y, t), t) = kwind(uwind − u)(x, y, (a + h)−
(x, y, t), t)
(3.15)
and
ǫ(
−(ax + hx)(uy + vx) − 2(ay + hy)vy + wy
)
(x, y, (a + h)−
(x, y, t), t)
+Dvz(x, y, (a + h)−
(x, y, t), t) = kwind(vwind − v)(x, y, (a + h)−
(x, y, t), t).
(3.16)
The parameters kfr and kwind are related to Kfr and Kwind by
Kfr =U
Lkfr and Kwind = kwindUλ (3.17)
and (uwind, vwind) is related to (uwind, uwind) via
(uwind, vwind) = U(uwind, vwind). (3.18)
Finally, for any function (x, y, z, t) → f(x, y, z, t) defined ina(x, y) < z < a(x, y) + h(x, y, t) we have adopted the notation
f(x, y, a+(x, y), t) = limδ>0δ→0
f(x, y, a(x, y) + δ, t) (3.19)
and
f(x, y, (a + h)−
(x, y, t), t) = limδ>0δ→0
f(x, y, (a + h)(x, y, t) − δ, t). (3.20)
We conclude this section with some observations about the tractions defined bythe left hand side of (3.13)-(3.16). We note that
−2axux + wx = −ax(ux − vy) − ax(ux + vy) + wx = −ax(ux − vy) + wx + axwz
and thus the left hand side of (3.13) may be rewritten as
ǫ (−ax(ux − vy) − ay(uy + vx)) (x, y, a+(x, y), t)) + ǫ ∂∂x
(w(x, y, a(x, y), t)
+Duz(x, y, a+(x, y), t).
Similar formulas hold for the left hand sides of (3.14)-(3.16).
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4 s-Coordinate Transformation
In this section, we introduce the s-coordinate transformation which takes our original,variable thickness, domain
D(t) =
(x, y, z, t)∣
∣a(x, y) < z < a(x, y) + h(x, y, t) and h(x, y, t) > 0
(4.1)
into one of constant thickness
D(t) =
(x, y, s, t)∣
∣0 < s < 1 and (x, y) ∈ Ω(t) = (x, y)| h(x, y, t) > 0
. (4.2)
This transformation was used by Mellor [1], Mellor and Blumberg [2] and [3] andMellor and Yamada [4]. The transformation is given by
x = x, y = y, t = t and z = a(x, y) + sh(x, y, t) for 0 < s < 1. (4.3)
We also let
a(x, y) = a(x, y),
h(x, y, t) = h(x, y, t),
(u, v, w)(x, y, s, t) = (u, v, w)(x, y, a(x, y) + sh(x, y, t), t),
p(x, y, s, t) = p(x, y, a(x, y) + sh(x, y, t), t),
(4.4)
and introduce
φ(x, y, s, t)def= w − (ax + shx)u − (ay + shy)v − sht. (4.5)
We note that the boundary conditions (BC − a) and (BC − (a + h)) of section 3imply that
φ(x, y, 0+, t) = φ(x, y, 1−, t) = 0. (4.6)
We record next how the differential operators ∂x, ∂y, ∂z, and ∂t transform. The resultsare
∂x = ∂x −(ax + shx)
h∂s =
1
h(∂x(h·) − ∂s ((ax + shx)·)) , (4.7)
∂y = ∂y −(ay + shy)
h∂s =
1
h(∂y(h·) − ∂s((ay + shy)·)) , (4.8)
∂z =1
h∂s, (4.9)
and
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∂t = ∂t −sht
h∂s =
1
h(∂t(h·) − ∂s((sht)·)) . (4.10)
Our next task is to transform the equations (C−3), (Mx−3) and (My−3) whenp is given by the hydrostatic approximation
p(x, y, z, t) = a(x, y) + h(x, y, t) − z. (H − 3)
The latter relation and (4.3) imply that
p(x, y, s, t) = (1 − s)h(x, y, t). (H − 4)
Exploiting (4.7)2, (4.8)2, (4.9) and (4.5) we find the (C − 3) becomes
ht + (hu)x + (hv)y + φs = 0 (C − 4)
where once again φ is defined in (4.5) and satisfies (4.6). Similarly, (4.7)2, (4.8)2,(4.9), (4.5), (H − 3) and (H − 4) imply that (Mx − 3) transforms to
1
h
(
(hu)t + (hu2)x + (huv)y + (φu)s
)
+ C2(ax + hx) =
Fv +1
h((hǫ(ux − vy))x + (hǫ(uy + vx))y)
−1
h(ǫ(ax + shx)(ux − vy) + ǫ(ay + shy)(uy + vx))s
+D
h2uss −
ǫxws
h(Mx − 4)
and (My − 3) transforms to
1
h
(
(hv)t + (huv)x + (hv2)y + (φv)s
)
+ C2(ay + hy) =
−Fu +1
h((hǫ(uy + vx))x + (hǫ(vy − ux))y)
−1
h(ǫ(ax + shx)(uy + vx) + ǫ(ay + shy)(vy − ux))s
+D
h2vss −
ǫyws
h. (My − 4)
Finally we note that
ux − vy = (ux − vy) −(ax + shx)
hus +
(ay + shy)
hvs (4.11)
12
and
uy + vx = (uy + vx) −(ay + shy)
hus −
(ax + shx)
hvs (4.12)
and our traction boundary conditions (3.13)-(3.16) become:
−ǫ (ax(ux − vy) + ay(uy + vx)) (x, y, 0, t)
+D(
1 + λ2(a2x + a2
y)) us(x, y, 0, t)
h+ ǫwx(x, y, 0, t) = kfrhu(x, y, 0, t), (4.13)
−ǫ (ax(uy + vx) + ay(vy − ux)) (x, y, 0, t)
+D(
1 + λ2(
a2x + a2
y
)) vs
h(x, y, 0, t) + ǫwy(x, y, 0, t) = kfrhv(x, y, 0, t), (4.14)
−ǫ ((ax + hx)(ux − vy) + (ay + hy)(uy + vx)) (x, y, 1, t)
+D(
1 + λ2(
(ax + hx)2 + (ay + hy)
2)) us
h(x, y, 1, t) + ǫwx(x, y, 1, t)
= kwind (uwind − u) (x, y, 1, t) (4.15)
and
−ǫ ((ax + hx)(uy + vx) + (ay + hy)(vy − ux)) (x, y, 1, t)
+D(
1 + λ2(
(ax + hx)2 + (ay + hy)
2)) vs
h(x, y, 1, t) + ǫwy(x, y, 1, t)
= kwind(vwind − v)(x, y, 1, t). (4.16)
Velocity Decompositions
Our strategy now is to write the lateral velocities as
(u, v)(x, y, s, t) = (¯u, ¯v) (x, y, t) + (u, v)(x, y, s, t) (4.17)
where the depth-averaged mean velocities are
(¯u, ¯v) (x, y, t) =
∫ 1
0
(u, v)(x, y, s, t)ds (4.18)
and the residual components are
13
∫ 1
0
(u, v) (x, y, s, t)ds = (0, 0). (4.19)
Our goal is to use (C − 4), (Mx − 4) and (My − 4) to obtain equations for h, (¯u, ¯v)and (u, v). Equation (C − 4) implies that
ht + (h¯u)x + (h¯v)y + (hu)x + (hv)y + φs = 0
and integrating this identity over 0 ≤ s ≤ 1 and exploiting (4.19) and (4.6) yields
ht + (h¯u)x + (h¯v)y = 0 (C − 4av)
and
φ(x, y, s, t) = −
(
h
∫ s
0
u(x, y, η, t)dη
)
x
−
(
h
∫ s
0
v(x, y, η, t)dη
)
y
(4.20)
for 0 ≤ s ≤ 1.In the remainder of this section we work towards obtaining a significant
model reduction and ultimately obtaining a closed system for h, ¯u and ¯v.This reduction exploits the hypotheses (3.6)-(3.8). The accepting readeris advised to skip immediately to the conclusions portion of this section.
Multiplying (Mx − 4) by h, integrating the resulting expression over 0 ≤ s ≤ 1and exploiting the boundary conditions (4.13) and (4.15) yields
(h¯u)t +
(
h
(
¯u2 +
∫ 1
0
u2(x, y, s, t)ds
))
x
+
(
h
(
¯u¯v +
∫ 1
0
(uv) (x, y, s, t)ds
))
y
+C2hax + C
2(h2/2)x = Fh¯v + (hǫ (¯ux − ¯vy))x + (hǫ (¯uy + ¯vx))y
+kwind (uwind − ¯u − u(x, y, 1, t)) − kfrh (¯u + u(x, y, 0, t))
−
(
ǫ
∫ 1
0
((ax + shx)us(x, y, s, t) − (ay + shy)vs(x, y, s, t)) ds
)
x
−
(
ǫ
∫ 1
0
((ay + shy)us(x, y, s, t) + (ax + shx)vs(x, y, s, t)) ds
)
y
− (ǫ(w(x, y, 1, t) − w(x, y, 0, t)))x. (4.21)
If we now let ∂ be one of the operators ∂x or ∂y and exploit the identity
14
∫ 1
0
(∂a + s∂h) (u, v)s (x, y, s, t)ds =
(∂a + ∂h)(u, v)(x, y, 1, t) − ∂a(u, v)(x, y, 0, t)
−∂h
∫ 1
0
(u, v) (x, y, s, t)ds =
(∂a + ∂h) (u, v) (x, y, 1, t) − ∂a(u, v)(x, y, 0, t) (4.22)
we find that (4.21) reduces to
(h¯u)t +
(
h
(
¯u2 +
∫ 1
0
u2(x, y, s, t)ds
))
x
+
(
h
(
¯u¯v +
∫ 1
0
(uv)(x, y, s, t)ds
))
y
C2hax + C
2(h2/2)x =
Fh¯v + (hǫ (¯ux − ¯vy))x + (hǫ (¯uy + ¯vx))y
+kwind (uwind − ¯u − u(x, y, 1, t)) − kfrh (¯u + u(x, y, 0, t))
− (ǫ ((ax + hx)u(x, y, 1, t) − axu(x, y, 0, t) − (ay + hy)v(x, y, 1, t) + ayv(x, y, 0, t)))x
− (ǫ ((ay + hy)u(x, y, 1, t) − ayu(x, y, 0, t) + (ax + hx)v(x, y, 1, t) − axv(x, y, 0, t)))y
− (ǫ (w(x, y, 1, t) − w(x, y, 0, t)))x . (Mx − 4av)
Similar arguments applied to (My − 4) yield
(h¯v)t +
(
h
(
¯u¯v +
∫ 1
0
(uv)(x, y, s, t)ds
))
x
+
(
h
(
¯v2 +
∫ 1
0
v2(x, y, s, t)ds
))
y
+C2hay + C
2(h2/2)y =
−Fh¯u + (hǫ (¯uy + ¯vx))x + (hǫ (¯vy − ¯ux))y
+kwind (vwind − ¯v − v(x, y, 1, t)) − kfrh (¯v + v(x, y, 0, t))
− (ǫ ((ay + hy)u(x, y, 1, t) − ayu(x, y, 0, t) + (ax + hx)v(x, y, 1, t) − axv(x, y, 0, t)))x
− (ǫ ((ay + hy)v(x, y, 1, t) − ayv(x, y, 0, t) − (ax + hx)u(x, y, 1, t) + axu(x, y, 0, t)))y
− (ǫ (w(x, y, 1, t) − w(x, y, 0, t)))y . (My − 4av)
15
To obtain the evolution equations for u and v we subtract (Mx − 4av) fromh × (Mx − 4) and (My − 4av) from h × (My − 4). This process yields:
hut = G1(x, y, s, t) + D((
1 + λ2((ax + shx)2 + (ay + shy)
2))
us
)
s/h
− (ǫ((ax + shx) ((¯ux + ux) − (¯vy + vy)) + (ay + shy) ((¯uy + uy) + (¯vx + vx))))s
− (kwind (uwind − ¯u − u(x, y, 1, t)) − kfr (¯u + u(x, y, 0, t)))
+ (ǫ (w(x, y, 1, t) − w(x, y, 0, t)))x − ǫxws (4.23)
and
hvt = G2(x, y, s, t) + D((
1 + λ2(
(ax + shx)2 + (ay + shy)
2))
us
)
s/h
− (ǫ ((ax + shx) ((uy + uy) + (¯vx + vx)) + (ay + shy) ((¯vy + vy) − (¯ux + ux))))s
− (kwind (vwind − ¯v − v(x, y, 1, t)) − kfr (¯v + v(x, y, 0, t)))
+(ǫ(w(x, y, 1, t) − w(x, y, 0, t))y − ǫyws, (4.24)
and u and v satisfy the traction boundary conditions
D(
1 + λ2(a2x + a2
y)) us(x, y, 0, t)
h+ ǫwx(x, y, 0, t)
−ǫ(ax((¯ux + ux) − (¯vy + vy)) + ay(x, y)((¯uy − uy) + (¯vx + vx))(x, y, 0, t)
= kfrh (¯u + u(x, y, 0, t)) (4.25)
D(1 + λ2(a2x + a2
y))vs
h(x, y, 0, t) + ǫwy(x, y, 0, t)
−ǫ(ax((¯uy + uy) + (¯vx + vx)) + ay((¯vy + vy) − (¯ux + ux)))(x, y, 0, t)
= kfrh (¯v + v(x, y, 0, t)) , (4.26)
16
D(
1 + λ2(
(ax + hx)2 + (ay + hy)
2)) us
h(x, y, 1, t) + ǫwx(x, y, 1, t)
−ǫ ((ax + hx) ((¯ux + ux) − (¯vy + vy)) + (ay + hy) ((¯uy + uy) + (¯vx + vx))) (x, y, 1, t)
= kwind (uwind − ¯u − u(x, y, 1, t)) (4.27)
and
D(
1 + λ2(
(ax + hx)2 + (ay + hy)
2)) vs(x, y, 1, t)
h+ ǫwy(x, y, 1, t)
−ǫ ((ax + hx) ((¯uu + uy) + (¯vx + vx)) + (ay + hy) ((¯vy + vy) − (¯ux + ux))) (x, y, 1, t)
= kwind (vwind − ¯v − v(x, y, 1, t)) . (4.28)
The functions G1 and G2 are defined by
G1def= −htu −
(
h
(
2¯uu + u2 −
∫ 1
0
u2(x, y, η, t)dη
))
x
−
(
h
(
¯uv + ¯vu + uv −
∫ 1
e
(uv)
)
(x, y, η, t)dη
)
y
− ¯uφs − (φu)s
+Fhv + (hǫ (ux − vy))x + (hǫ (uy + vx))y + (ǫ (hxu − hyv))x
− (ǫ ((ax + shx)u − (ay + shy)v))xs
+(ǫ((ax + hx)u(x, y, 1, t) − axu(x, y, 0, t) − (ay + hy)v(x, y, 1, t)
+ayv(x, y, 0, t)))x + (ǫ(hyu + hxv))y − (ǫ((ay + shy)u + (ax + shx)v))ys
+ (ǫ((ay + hy)u(x, y, 1, t) − ayu(x, y, 0, t) + (ax + hx)v(x, y, 1, t) − axv(x, y, 0, t)))y
(4.29)
and
17
G2def= −htv −
(
h
(
¯uv + ¯vu + uv −
∫ 1
0
(uv)(x, y, η, t)dη
))
x
−
(
h
(
2¯vv + v2 −
∫ 1
0
v2(x, y, η, t)dη
))
y
− ¯vφs − (φv)s
−Fhu + (hǫ (uy + vx))x + (hǫ (vy − ux))y + (ǫ (hxv + hyu))x
− (ǫ((ax + shx)v + (ay + shy)u))xs
+ (ǫ((ax + hx)v(x, y, 1, t) − axv(x, y, 0, t) + (ay + hy)u(x, y, 1, t) − axu(x, y, 0, t)))x
+ (ǫ (hyv − hxu))y − (ǫ ((ay + shy)v − (ax + shx)u))ys
+ (ǫ((ay + hy)v(x, y, 1, t) − ayv(x, y, 0, t) − (ax + hx)u(x, y, 1, t) + ayu(x, y, 0, t)))y
(4.30)
and both satisfy
∫ 1
0
(G1, G2)(x, y, s, t)ds = (0, 0). (4.31)
We are now going to exploit the fact that
0 < ǫ = Dλ2 ≪ 1 ≪ D (4.32)
and replace (4.23) and (4.24) by the approximating equations:
hut = G1 +D
huss
−kwind (uwind − ¯u − u(x, y, 1, t)) + kfrh (¯u + u(x, y, 0, t)) (4.33)
and
hvt = G2 +D
huss
−kwind (vwind − ¯v − v(x, y, 1, t)) + kfrh (¯v + v(x, y, 0, t)) , (4.34)
and (4.25)-(4.28) by the approximate boundary conditions:
18
Dus(x, y, 0, t)
h= kfrh (¯u + u(x, y, 0, t)) , (4.35)
Dus(x, y, 1, t)
h= kwind (uwind − ¯u − u(x, y, 1, t)) , (4.36)
Dvs
h(x, y, 0, t) = kfrh (¯v + v(x, y, 0, t)) , (4.37)
andDvs
h(x, y, 1, t) = kwind (vwind − ¯v − v(x, y, 1, t)) . (4.38)
We attempt to write the solutions of (4.33)-(4.38) as
u = U + u1 (4.39)
and
v = V + v1 (4.40)
where U and V satisfy the “approximate” steady state problems:
Uss =(
kwindh (uwind − ¯u − U(1)) − kfrh2 (¯u + U(0))
)
/D (4.41)
and
Vss =(
kwindh (vwind − ¯v − V (1)) − kfrh2 (¯v + V (0))
)
/D (4.42)
and the boundary conditions
Us(0) =kfrh
2(¯u + U(0))
Dand Us(1) =
kwindh(uwind − ¯u − U(1))
D(4.43)
and
Vs(0) =kfrh
2(¯v + V (0))
Dand Vs(1) =
kwindh(¯vwind − ¯v − V (1))
D(4.44)
and the transients u1 and v1 satisfy
hu1t = G1 +D
hu1ss + kwindu1(x, y, 1, t) + kfrhu1(x, y, 0, t) (4.45)
19
and
hv1t = G2 +D
hv1ss + kwindv1(x, y, 1, t) + kfrhv1(x, y, 0, t) (4.46)(2)
and the boundary conditions
D
hu1s(x, y, 0, t) = kfrhu1(x, y, 0, t), (4.47)
D
hu1s(x, y, 1, t) = −kwindu1(x, y, 1, t), (4.48)
D
hv1s(x, y, 0, t) = kfrhv1(x, y, 0, t), and (4.49)
D
hv1s(x, y, 1, t) = −kwindv1(x, y, 1, t). (4.50)
We note that the approximate steady state solutions admit the representations
U = U(0) +kfrh
2(¯u + U(0))s
D+
(kwindh(uwind − ¯u − U(1)) − kfrh2(¯u + U(0)))s2
2D(4.51)
and
V = V (0) +kfrh
2(¯v + V (0))s
D+
(kwindh(vwind − ¯v − V (1)) − kfrh2(¯v + V (0)))s2
2D(4.52)
for 0 ≤ s ≤ 1. Moreover, the boundary conditions (4.43)2, (4.44)2 and the“zero-mean” condition
∫ 1
0
(U, V )(s)ds = (0, 0) (4.53)
imply that
(U, V )(0) = −kwindh((uwind − ¯u), (vwind − ¯v))
6DB−
(1 + kwindh/4D)kfrh2(¯u, ¯v)
6DB(4.54)
and
(U, V )(1) =(1/3 + kfrh
2/12D)kwindh((uwind − ¯u), (vwind − ¯v))
DB+
kfrh2(¯u, ¯v)
6DB(4.55)
(2)In deriving these equations we have neglected the fact that U and V depend on x, y and t through
the dependence of h, ¯u, ¯v, and D on these quantities.
20
where
B = 1 +(kwh + kfrh
2)
3D+
kwindkfrh3
12D2. (4.56)
The important point about these formulas is that U(0), U(1), and s → U(s),0 ≤ s ≤ 1 are linear in (uwind − ¯u) and ¯u and are O(1/D) ≪ 1. The latter conclusionfollows from the hypothesis (3.8) that D ≫ 1. Of course a similar statement holdsfor V (0), V (1), and s → V (s), 0 ≤ s ≤ 1 except that the factors (uwind − ¯u) and ¯uare replaced by (uwind − ¯v) and ¯v.
We note that if Dt/D = O(1) then (U, V )t, (U, V )x, and (U, V )y are also O(1/D),a desirable property. We further observe that if
udef= U + u1 and v
def= V + v1 (4.57)
are also O(1/D), then the terms G1 and G2 defined in (4.29) and (4.30) satisfy
G1 and G2 = O(
max(
1/D, 1/D2,ǫ
D= λ2
))
≪ 1 (4.58)
and thus approximations to u1 and v1 may be obtained by solving (4.45)-(4.50) withG1 and G2 identically zero. Of course this system is solved with u1 and v1 satisfyingthe initial conditions
u1(x, y, s, 0+) = u(x, y, s, 0) − U(x, y, s, 0+) = O(1/D) (4.59)
and
v1(x, y, s, 0+) = v(x, y, s, 0) − V (x, y, s, 0+) = O(1/D). (4.60)
Noting that solutions of (4.45)-(4.50) with G1 = G2 = 0 satisfy the estimates
∫ 1
0
(u21, v
21)(x, y, s, t)ds ≤
∫ 1
0
(u21, v
21)(x, y, s, 0)ds
we find that evolving u1 and v1 via (4.45) and (4.46) produces consistent results withthe hypothesis that u and v are O(1/D). We emphasize that this scale for (u, v) wasnot arbitrary, it was dictated by the size of U and V .
Conclusions
The upshot of the previous calculations is that to within terms which are at worst
O
(
max
(
1
D,
1
D2,
ǫ
D= λ2
))
the governing equations for h, ¯u, and ¯v is given by
ht + (h¯u)x + (h¯v)y = 0, (C − 4av)
(h¯u)t + (h¯u2)x + (h¯u¯v)y + C2hax + C
2(h2/2)x =
Fh¯v + (hǫ(¯ux − ¯vy))x + (hǫ(¯uy + ¯vx))y + kwind(uwind − ¯u) − kfrh¯u,(Mx − 4av)
21
and
(h¯v)t + (h¯u¯v)x + (h¯v2)y + C2hax + C
2(h2/2)y =
−Fh¯u + (hǫ(¯uy + ¯vx))x + (hǫ(¯vy − ¯ux))y + kwind(vwind − ¯v) − kfrh¯v.(My − 4av)(3)
This is an important simplification since it implies that with our scalings thefields (u, v)(x, y, s, t) do not influence the depth-averaged fields ¯u and ¯v. Of coursethe converse statement is not true; clearly h, ¯u, and ¯v drive u, v, and φ. A differentapproach which leads to a similar depth-averaged system may be found in Gerbeauand Perthame[5].
We conclude this section by once again noting that the equations (C − 4av),(Mx − 4av), and (My − 4av) hold in the time varying domain
Ω(t) = (x, y)|h(x, y, t) > 0 (4.61)
whose boundary
∂Ω(t) = (x, y)|h(x, y, t) = 0 (4.62)
must be determined as part of the solution. At time t = 0 we require initial data for(h, ¯u, ¯v) in Ω(0).
5 A Priori Estimates and Exact Solutions
In this section we seek qualitative/quantitative a-priori information about solutionsof the depth-averaged shallow water system developed at the end of the last section.We also present a simple rotating-pulsating solution to this system. Our reason forpresenting this rather ”academic” example is it will form a baseline by which we canassess the performance of the computational model we shall develop in section 6.
In this section we neglect the “wind” and “bottom frictional” forces and set thecoefficients kwind and kfr to zero. Thus, we are left with the system
ht + (h¯u)x + (h¯v)y = 0, (C)
(3)Equation (C − 4av) implies that(h¯u)t + (h¯u2)x + (h¯u¯v)y = h(¯ut + ¯u¯ux + ¯v ¯uy)(h¯v)t + (h¯u¯v)x + (h¯v2)y = h(¯vt + ¯u¯vx + ¯u¯vy)
22
(h¯u)t + (h¯u2)x + (h¯u¯v)y + C2hax + C
2(h2/2)x =
Fh¯v + (hpǫ(¯ux − ¯vy))x + (hpǫ(¯uy + ¯vx))y , (Mx)
and
(h¯v)t + (h¯u¯v)x + (h¯v2)y + C2hay + C
2(h2/2)y =
−Fh¯u + (hpǫ(¯uy + ¯vx))x + (hpǫ(¯vy − ¯ux))y. (My)
The observant reader will note that we have added a power p to h in the “turbulent”stress terms in (Mx) and (My). We have in mind the situation where p = 1 or p = 2.The latter case can be obtained from the former by reinterpreting ǫ.
The type of solutions we have in mind contain a finite amount of fluid and occupythe time varying region
Ω(t) = (x, y)|h(x, y, t) > 0 (5.1)
whose boundary
∂Ω(t) = (x, y)|h(x, y, t) = 0 (5.2)
must be determined as part of the solution.For the moment, we assume we have a solution to (C), (Mx) and (My) and we
introduce particle trajectories (X ,Y)(X,Y, t) as the unique solutions of
Xt = ¯u(X ,Y , t) and Yt = ¯v(X ,Y , t) (5.3)
taking on the initial condition
X (X,Y, 0) = X and Y(X,Y, 0) = Y (5.4)
where (X,Y ) ∈ Ω(0). We assume that for each t > 0 that the mapping(X,Y ) → (X ,Y)(X,Y, t) is one to one and that the Jacobian determinant
D(X,Y, t) = (XXYY −XY YX)(X,Y, t) > 0. (5.5)
For future reference, we note that Ω(t) is the image of Ω(0) under the mapping(X,Y ) → (X ,Y)(X,Y, t).
Though we defer until the next section the complete Lagrangian formulation of thesystem (C), (Mx) and (My) we record here the Lagrangian version of the continuityequation, namely the identity
h(X,Y, t)D(X,Y, t) = h0(X,Y ), (X,Y ) ∈ Ω(0). (5.6)
23
Once again, we are assuming that (h, ¯u, ¯v) is the solution to (C), (Mx) and (My),that
h(X,Y, t)def= h(X (X,Y, t),Y(X,Y, t), t), (X,Y ) ∈ Ω(0) (5.7)
and that
h0(X,Y ) = h(X,Y, 0), (X,Y ) ∈ Ω(0). (5.8)
One simple consequence of the change of variable formula for integrals and (5.6)is the identity that for all t > 0
mdef=
∫∫
Ω(0)
h0(X,Y )dXdY =
∫∫
Ω(t)
h(x, y, t)dxdy. (5.9)
Another is that for any smooth function (x, y, t) → f(x, y, t)
d
dt
∫∫
Ω(t)
h(x, y, t)f(x, y, t)dxdy
=
∫∫
Ω(t)
h(x, y, t) (ft + ¯ufx + ¯vfy) (x, y, t)dxdy. (5.10)
The identify (5.10), with judicious choices of f , yields some of the a-priori infor-mation we seek.
We define the center of mass of the system by
m(xc, yc)(t)def=
∫∫
Ω(t)
h(x, y, t)(x, y)dxdy. (5.11)
Then, (5.10) yields
m
(
dxc
dt,dyc
dt
)
(t) =
∫∫
Ω(t)
h(¯u, ¯v)(x, y, t)dxdydef= m(uc, vc)(t) (5.12)
and (5.12) implies that
m
(
duc
dt,dvc
dt
)
(t) =
∫∫
Ω(t)
h(x, y, t)(¯ut+¯u¯ux+¯v ¯uy, ¯vt+¯u¯vx+¯v ¯vy)(x, y, t)dxdy. (5.13)
If we now exploit the fact that (Mx) and (My) may be rewritten as
h(¯ut + ¯u¯ux + ¯v ¯uy) =
−C2hax − C
2(h2/2)x + Fh¯v + (hpǫ(¯ux − ¯vy))x + (hpǫ(¯uy + ¯vx))y
and
24
h(¯vt + ¯u¯vx + ¯v ¯vy) =
−C2hay − C
2(h2/2)y − Fh¯u + (hpǫ(¯uy + ¯vx))x + (hpǫ(¯vy − ¯ux))y
we see that if the bottom surface z = a(x, y) is a paraboloid; i.e.
a(x, y) = (αx2 + βy2)/2
with α and β positive, then (5.13), the divergence theorem, and the fact that h = 0for (x, y) ∈ ∂Ω(t) implies
m
(
duc
dt,dvc
dt
)
(t) = m(
−αC2xc + Fvc,−βC
2yc − Fuc
)
(t). (5.14)
The system (5.12), and (5.14) are nothing more than a coupled pair of harmonicoscillators.
The system has the energy integral:
(u2c + v2
c + C2(αx2
c + βy2c )) ≡ E1 (5.15)
and when β = α > 0 it is endowed with the additional angular momentum integral:
2(xcvc − ycuc)(t) + E2 =F
αC2(u2
c + v2c )(t). (5.16)
In this latter case, we also have that Rdef= (x2
c + y2c )(t) satisfies
d2R
dt2+ (F 2 + 4αC
2)R = H (5.17)
where
H =
(
2 +F 2
αC2
)
((uoc)
2 + (voc)
2 + αC2((xo
c)2 + (yo
c)2))
−F
(
F
αC2((uo
c)2 + (vo
c)2) − 2(xo
cvoc − yo
cuoc)
)
= 2αC2((xo
c)2 + (yo
c)2) + (Fxo
c + voc)
2 + (Fyoc − uo
c)2 + ((uo
c)2 + (vo
c)2) > 0 (5.18)
and
(xoc, y
oc , u
oc, v
oc) = (xc, yc, uc, vc)(0). (5.19)
We leave it to the reader to verify the following theorem.Theorem 1 Suppose the bottom surface of the fluid is the paraboloid
z = (αx2 + βy2)/2, (5.20)
25
that (h, ¯u, ¯v) is the solution of (C), (Mx) and (My), and that (xc, yc, uc, vc) is thesolution of the center of mass system (5.12) and (5.14). Then, if we let
t1 = t, x1 = x − xc(t), and y1 = y − yc(t) (5.21)
h1(x1, y1, t1) = h(x1 + xc(t1), y1 + yc(t1), t1), (5.22)
ǫ1(x1, y1, t1) = ǫ(x1 + xc(t1), y1 + yc(t1), t1) (5.23)
u1(x1, y1, t1) = ¯u(x1 + xc(t1), y1 + yc(t1), t1) − uc(t1), (5.24)
and
v1(x1, y1, t1) = ¯v(x1 + xc(t1), y1 + yc(t1), t1) − vc(t1) (5.25)
they satisfy
h1t1
+ (h1u1)x1 + (h1v1)y1 = 0, (C)
(h1u1)t1 + (h1(u1)2)x1 + (h1u1v1)y1 + C2αh1x1 + C
2((h1)2/2)x1 =
Fh1v1 + ((h1)pǫ1(u1x1
− v1y1
))x1 + ((h1)pǫ1(u1y1
+ v1x1
))y1 , (Mx1)
and
(h1v1)t1 + (h1u1v1)x1 + (h1(v1)2)y1 + C2βh1y1 + C
2((h1)2/2)y1 =
−Fh1u1 + ((h1)pǫ1(u1y1
+ v1x1
))x1 + ((h1)pǫ1(v1y1− u1
x1))y1 (My1)
in
Ω1(t1) = (x1, y1)|(x1 + xc(t1), y1 + yc(t1)) ∈ Ω(t1). (5.26)
Moreover, these functions further satisfy
∫∫
Ω1(t1)
h1(x1, y1, t1)(1, x1, y1, u1(x1, y1, t1), v
1(x1, y1, t1))dx1dy1 = (m, 0, 0, 0, 0).
(5.27)The center of mass results and the results of Theorem 1 are well known for the
system (C), (Mx) and (My) when ǫ ≡ 0 (for details see [6-9]). The same is alsotrue for the rotating-pulsating solutions we shall construct at the end of this section.Before turning to that solution we record two additional “energy” type identities validfor solutions of (C), (Mx) and (My).
26
We first note that if (h, ¯u, ¯v) is a solution of (C), (Mx) and (My) then the followingidentity holds for all (x, y) in Ω(t):
h
(
(¯u2 + ¯v2)
2+ C
2
(
a +h
2
))
t
+h¯u
(
(¯u2 + ¯v2)
2+ C
2
(
a +h
2
))
x
+h¯v
(
(¯u2 + ¯v2)
2+ C
2
(
a +h
2
))
y
+(
h2 ¯u/2)x + C2(h2 ¯v/2
)
y=
(hpǫ(¯u(¯ux − ¯vy) + ¯v(¯vx + ¯uy)))x
+(hpǫ(¯u(¯uy + ¯vx) + ¯v(¯vy − ¯ux)))y
−hpǫ((¯uy + ¯vx)2 + (¯ux − ¯vy)
2).
(5.28)
This identity, when combined with (5.10) and the fact that h = 0 for points(x, y) ∈ ∂Ω(t) yields:
d
dt
∫∫
Ω(t)
h((¯u2 + ¯v2) + C2(2a + h))(x, y, t)dxdy =
−2
∫∫
Ω(t)
(hpǫ((¯uy + ¯vx)2 + (¯ux − ¯vy)
2))(x, y, t)dxdy.
(5.29)
In the situation when a = (αx2 + βy2)/2, (5.29) and the center of mass identities(5.12), (5.14) and (5.15) further imply that
d
dt
∫∫
Ω(t)
h((¯u − uc)2 + (¯v − vc)
2 + C2(α(x − xc)
2 + β(y − yc)2 + h))(x, y, t)dxdy
= −2
∫∫
Ω(t)
(hpǫ((¯uy + ¯vx)2 + (¯ux − ¯vy)
2))(x, y, t)dxdy. (5.30)
The identity (5.29) or its companion (5.30) are crucial; they imply that
D(t)def=
∫∫
Ω(t)
(hpǫ((¯uy + ¯vx)2 + (¯ux − ¯vy)
2))(x, y, t)dxdy (5.31)
27
is L1(0,∞) and lead one to suspect that
limt→∞
D(t) = 0 (5.32)
or, better yet, that for (x, y) ∈ Ω(t)
limt→∞
((¯ux − ¯vy), (¯uy + ¯vx))(x, y, t) = (0, 0). (5.33)
We shall show that these suspicions are in fact true for the rotating-pulsating solutionswe construct next.
Rotating-Pulsating Solutions
We assume that the bottom surface is given by
a(x, y) = α(x2 + y2)/2 (5.34)
with α > 0 and our solutions will be of the form
h = h0 − (h11x2 + 2h12xy + h22y
2)/2, (5.35)
¯u = ((d + σ)x + (τ + w)y)/2, (5.36)
and
¯v = ((τ − w)x + (d − σ)y)/2 (5.37)
where h0, h11, h12, h22, d, w, σ and τ are function of t only. When p = 2 thesesolutions are exact and when p = 1 they are approximate. For definiteness we assumethat ǫ > 0 is a constant.
We note that
¯ux + ¯vy = d,
¯ux − ¯vy = σ,
¯uy − ¯vx = w,
and
¯uy + ¯vx = τ
and thus the turbulent stresses are given by
hpǫ(¯ux − ¯vy) = hpǫσ (5.38)
and
28
hpǫ(¯uy + ¯vx) = hpǫτ. (5.39)
We want h0 > 0, h11 > 0, h22 > 0 and h11h22 − h212 > 0. For such a solution
Ω(t) =
(x, y)|√
x2 + y2 = r ≤
√
4h0
(h11 + h22) + (h11 − h22) cos 2θ + 2h12 sin 2θ
(5.40)and
∫∫
Ω(t)
(h(x, y, ¯u, ¯v))(x, y, t)dxdy = (0, 0, 0, 0). (5.41)
Insertion of (5.35)-(5.37) into (C), (Mx) and (My) when p = 2 yields
•
h0 = −dh0 (5.42)
•
l = −2dl − (βσ + h12τ), (5.43)
•
β = −2dβ + (h12w − lσ), (5.44)
•
h12 = −2dh12 − (lτ + βw), (5.45)
•w = −d(w − F ) − 4ǫ(h12σ − βτ), (5.46)
•σ = −dσ + Fτ + 2C
2β − 4ǫlσ, (5.47)
•τ = −dτ − Fσ + 2C
2h12 − 4ǫlτ, (5.48)
and
•
d = 2C2(l − α) − Fw +
(w2 − d2 − σ2 − τ 2)
2− 4ǫ(βσ − h12τ)). (5.49)
The quantities l and β are related to h11 and h22 by
l =h11 + h22
2and β =
h11 − h22
2. (5.50)
and for any function g,•g = dg/dt.
Our first observation about the system (5.42)-(5.49) is that the manifold
(β, h12, σ, τ) ≡ (0, 0, 0, 0) (5.51)
29
is invariant. To integrate (5.42)-(5.49) we introduce an auxiliary unknown, A > 0,related to d by
2A
A= d. (5.52)
Once we determine A then h0 is recovered by
h0 = 1/A2. (5.53)
If we now let
Ω = A2(w − f), Σ = A2σ and T = A2τ (5.54)
and
L = A4l, B = A4β and H12 = A4h12, (5.55)
then it is easily checked that these new quantities satisfy
•
L = −(BΣ + H12T )/A2, (5.56)
•
B = −(LΣ − (Ω + FA2)H12)/A2, (5.57)
•
H12 = −(LT + (Ω + FA2)B)/A2, (5.58)
•
Ω = −4ǫ(H12Σ − BT )/A2, (5.59)
•
Σ = +FT +2C
2B
A2−
4ǫLΣ
A4, (5.60)
•
T = −FΣ +2C
2H12
A2−
4ǫLT
A4, (5.61)
and
2••
A +
(
2C2α +
F 2
2
)
A −
(
Ω2
2+ 2C
2L
)
/A3
= −(Σ2 + T 2)/2A3 − 4ǫ(BΣ + H12T )/A5. (5.62)
This latter system has the invariant manifold
(B,H12, Σ, T ) = (0, 0, 0, 0) (5.63)
and on this manifold the system reduces to
•
L =•
Ω = 0 (5.64)
30
or
(L, Ω) ≡ (L∞ > 0, Ω∞) (5.65)
and
2••
A +
(
2C2α +
F
2
2)
A −
(
Ω2∞
2+ 2C
2L∞
)
/A3 = 0. (5.66)
The last equation satisfies the “energy” identity
(•
A)2 +
(
C2α +
F
4
2)
A2 +
(
Ω2∞
4+ C
2L∞
)
/A2 = E2∞ (5.67)
and the total energy, E2∞, satisfies the inequality
E2∞ ≥ 2
(
Ω2∞
4+ C
2L∞
)1/2 (
F 2
4+ C
2α
)1/2
. (5.68)
If E2∞ is equal to the lower bound, then
A(t) ≡
(
Ω2∞
4+ C
2L∞
)1/4
/
(
F 2
4+ C
2α
)1/4
(5.69)
which is the unique positive equilibrium solution of (5.66).The following Theorem applies to solutions of the full system (5.56)-(5.62) taking
on initial data
(L,B,H12, Ω, Σ, T )(0) = (L0, B0, H012, Ω
0, Σ0, T 0). (5.70)
We shall assume that
(L0)2 − (B0)2 − (H012)
2 > 0. (5.71)
This latter assumption is equivalent to
(h11(0))(h22(0)) − (h12(0))2 > 0. (5.72)
Theorem 2. The solution of (5.56)-(5.62) taking on the initial data (5.70) whichsatisfies (5.71) satisfies
31
· (L2 − B2 − H212)(t) ≡ (L0)2 − (B0)2 − (H0
12)2 def
= L2∞, (5.73)
· 4C2√
L2∞ + (B2 + H2
12)(t) + 8ǫ
∫ t
0
L(Σ2 + T 2)(s)ds
A4
+(Σ2 + T 2)(t) ≡ 4C2L0 + (Σ0)2 + (T 0)2 def
= 4C2Lmax, (5.74)
· (B,H12, Σ, T )(t) → (0, 0, 0, 0), Ω(t) → Ω∞ and L(t) → L∞ as t → ∞,
and finally
· as t → ∞, A asymptotically solves(5.66) and (5.67).
Proof. To establish (5.73) we multiply (5.56) by L, (5.57) by (−B) and (5.58) by(−H12) and add. To obtain (5.74) we multiply (5.60) by (2Σ), (5.61) by (2T ), andadd and make use of (5.56) and the fact that (5.73) implies that
L(t) ≡√
L2∞ + (B2 + H2
12)(t) ≥ L∞. (5.75)
The third assertion follows from similar arguments and relies on the identity
d
dt(BΣ + H12T )
= −4ǫL
A4(BΣ + H12T ) + F (H12Σ − BT )
+2C
2(B2 + H212)
A2−
L(Σ2 + T 2)
A2. (5.76)
If A(·) satisfies 0 < Amin ≤ A(t) ≤ Amax, then (5.76) may be used to establish that(B2 + H2
12)(·) is in L2(0,∞) and satisfies
∫ ∞
0
(B2 + H212)(t)dt ≤ K1
∫ ∞
0
(Σ2 + T 2)(t)dt. (5.77)
Equations (5.73), (5.74) and (5.77) then imply that (B,H12, Σ, T )(·) are in L∞(0,∞)∩
L2(0,∞) while (5.56)-(5.61) imply that (•
B,•
H12,•
Σ,•
T )(·) are also in L∞(0,∞) andthese facts suffice to establish that
limt→∞
(B,H12, Σ, T )(t) = (0, 0, 0, 0). (5.78)
The existence of constants Amin and Amax so that
0 < Amin ≤ A(t) ≤ Amax
32
follows from the energy identity
dE
dt=
Ω•
Ω
2+ C
2•
L −(Σ2 + T 2)
4
•
/A2 − ǫ(BΣ + H12T )•/A4, (5.79)
where
E(t) = (•
A)2 +
(
C2α +
F 2
4
)
A2 +
(
Ω2
4+ C
2L + (Σ2 + T 2)
)
/A2− ǫ(BΣ+H12T )/A4.
(5.80)We note that if A(0) ≥ Amin > 0 and if
((Σ0)2 + (T 0)2)(1 + ǫ/(Amin)2) + ǫ((B0)2 + (H0
12)2)/(Amin)
2 ≤C
2L∞
2(5.81)
then
E(0) ≥ (•
A(0))2 +
(
C2α +
F 2
4
)
A2(0) +
(
Ω20
4+
C2L∞
2
)
/(A(0))2 def= P 2
0 > 0. (5.82)
Moreover, (5.74) implies that
L∞ ≤ L(t) ≤ Lmax (5.83)
and
(B2 + H212)
2L∞
(t) + (Σ2 + T 2)(t) ≤(B0)2 + (H0
12)2
2L∞
+ ((Σ0)2 + (T 0)2) (5.84)
and thus by further constraining the data (B0, H012, Σ
0, T 0) we can be assured thatso long as A(t) ≥ Amin > 0 the inequality
(Σ2 + T 2)(1 + ǫ/(Amin)2) + ǫ(B2 + H2
12)/(Amin)2 ≤
C2L∞
2(5.85)
will obtain and thus E(t) will satisfy
E(t) ≥ (•
A)2 +
(
C2α +
F 2
4
)
A2 +
(
Ω2
4+
C2L∞
2
)
/A2. (5.86)
The observation that the right hand side of (5.79) is L1(0,∞) and satisfies
∫ ∞
0
|rhs of (5.79)|dt ≤K2
(Amin)6((B0)2 + (H0
12)2 + (Σ0)2 + (T 0)2) (5.87)
33
then implies that by possibly constraining the data (B0, H012, Σ
0, T 0) further we obtainthe inequality that
(•
A)2 +
(
C2α +
F 2
4
)
A2 +
(
Ω2
4+
C2L∞
2
)
/A2 ≤ E(t) ≤ 4K23P
20 (5.88)
where P 20 is defined in (5.82) and K2
3 > 0 is such that E(0) ≤ K23P
20 . The latter
inequality implies we may choose
A2min =
C2L∞
8K23P
20
and A2max =
4K23P
20
(C2α + F 2
4)
. (5.89)
The remaining assertions are left to the reader. This completes section 5.
6 The Lagrangian Formulation and Computational
Model
The principal difficulty in solving the initial value problem for (C−4av), (Mx−4av)and (My − 4av) is that the domain of the solution is time varying and not a-prioriknown; that is the sets
Ω(t) = (x, y)|h(x, y, t) > 0 (6.1)
and
∂Ω(t) = (x, y)|h(x, y, t) = 0 (6.2)
must be determined as part of the solution. The Lagrangian formulation of ourunderlying system and the computational model associated with this formulationrepresent a system of equations which hold in the fixed domain
Ω(0) = (X,Y )|h(X,Y, 0) = h0(X,Y ) > 0 (6.3)
which is known. This is an essential simplification.The basic unknowns of the Lagrangian formulation are (X ,Y , u, v, h) which are
defined for (X,Y ) ∈ Ω(0) and t ≥ 0. These are related to the unknowns of the Eulerformulation, namely (¯u, ¯v, h), by
Xt(X,Y, t) = u(X,Y, t) = ¯u(X (X,Y, t),Y(X,Y, t), t), (6.4)
Yt(X,Y, t) = v(X,Y, t) = ¯v(X (X,Y, t),Y(X,Y, t), t) (6.5)
and
(X (X,Y, 0),Y(X,Y, 0)) = (X,Y ). (6.6)
h is related to h by
34
h(X,Y, t) = h(X (X,Y, t),Y(X,Y, t), t) (6.7)
ǫ to ǫ by
ǫ(X,Y, t) = ǫ(X (X,Y, t),Y(X,Y, t), t). (6.8)
Our first task is the development of the Lagrangian form of the continuity equation(C − 4av). We first note that (C − 4av) may be rewritten as
(ht + ¯uhx + ¯vhy + h(¯ux + ¯vy)) (X (X,Y, t),Y(X,Y, t), t) = 0 (6.9)
and this identity, when combined with (6.4)-(6.7) and
∂x =1
D(YY ∂X − YX∂Y ) =
1
D((YY •)X − (YX•)Y ), (6.10)
∂y =1
D(XX∂Y −XY ∂X) =
1
D((XX•)Y − (XY •)X), (6.11)
D(X,Y, t) = (XXYY −XY YX)(X,Y, t) > 0, (6.12)
and
D(X,Y, 0) = 1 (6.13)
implies that
ht(X,Y, t) = (ht + ¯uhx + ¯vhy)(X (X,Y, t),Y(X,Y, t), t), (6.14)
and
(¯ux + ¯vy)(X (X,Y, t),Y(X,Y, t), t) =
1
D(YY uX + XX vY − YX uY −XY vX)(X,Y, t) =
1
D(YY XXt + XXYY t − YXXY t −XY YXt) =
Dt
D(6.15)
and thus (6.9) becomes
ht +hDt
D= 0, (X,Y ) ∈ Ω(0) (6.16)
or equivalently
(hD)t = 0 , (X,Y ) ∈ Ω(0). (6.17)
Equations (6.13) and (6.17) then yield
35
(hD)(X,Y, t) = h0(X,Y ) , (X,Y ) ∈ Ω(0) (C − Lag)
where again h0 is the initial data for h.Similar arguments applied to (Mx − 4av) and (My − 4av) yield
h0ut + C2h0ax(X ,Y) +
C2
2((YY h2)X − (YX h2)Y ) =
Fh0v + (hpǫ(YY σ −XY τ))X + (hpǫ(XX τ − YX σ))Y
+kwindD(uwind(X ,Y) − u) − kfrh0u (Mx − Lag)
and
h0vt + C2h0ay(X ,Y) +
C2
2((XX h2)Y − (XY h2)X) =
−Fh0u + (hpǫ(YY τ + XY σ))X − (hpǫ(YX τ + XX σ))Y
+kwindD(vwind(X ,Y) − v) − kfrh0v (My − Lag)
where
σ = (¯ux − ¯vy) = (YY uX + XY vX − YX uY −XX vY )/D (6.18)
and
τ = (¯uy + ¯vx) = (YY vX −XY uX + XX uY − YX VY )/D. (6.19)
Again, the exponent p is 1 or 2.
The Computational ModelGeometric Preliminaries
Our task now is the development of a computational model for the system (6.4),(6.5), (6.6), (6.12), (6.13), (C − Lag), (Mx − Lag) and (My − Lag). In addition tothe initial data for X ,Y and h we require data for u and v for points (X,Y ) ∈ Ω(0).
For simplicity we assume that Ω(0) is the unit square
Ω(0) = (X,Y )|L ≤ X ≤ L + 1 and B ≤ Y ≤ B + 1. (6.20)
On Ω(0) we lay in an odd number of vertical grid lines Xi2N+1i=1 and horizontal grid
lines Yj2N+1j=1 where
Xi = L+(i − 1)
2Nfor i = 1 : 2N +1 and Yj = B+
(j − 1)
2Nfor j = 1 : 2N +1 (6.21)
36
Figure 1: The Region Ω(0)
37
and the (i, j)th grid point is that point with coordinates (Xi, Yj).On this grid we overlay a rather elaborate tiling by triangles. Edges of these tri-
angles will include the vertical grid lines Xi2N+1i=1 , the horizontal grid lines Yj
2N+1j=1
and a collection of straight lines with slopes (+1) and (−1). The lines with slope(+1) start at (X1, Y2k−1)
Nk=1 and (X2k−1, Y1)
Nk=2 and terminate respectively at
(X2(N−k)+3, Y2N+1)Nk=1 and (X2N+1, Y2(N−k)+3)
Nk=2. The lines with slope (−1) start
at (X2k+1, Y1)Nk=1 and (X2N+1, Y2k−1)
Nk=2 and terminate respectively at (X1, Y2k+1)
Nk=1
and (X2k−1, Y2N+1)Nk=2.
If (Xi, Yj), 2 ≤ i, j ≤ 2N is one of the interior grid points and if (i + j) is even,then the neighborhood structure of (Xi, Yj) is as shown in Figure 2. Similarly, if (i+j)is odd then the neighborhood structure of the point (Xi, Yj, )is as shown in Figure3. In each figure the smaller square centered at (Xi, Yj) is the computational cellwe shall associate to the point (Xi, Yj). In Figure 2 that cell is made up of portionsof eight distinct triangles while in Figure 3 that cell is made up of portions of fourdistinct triangles.
Figure 2: (i + j) is even
38
Figure 3: (i + j) is odd
39
Boundary points (X1, Yj)2Nj=2, (X2N+1, Yj)
2Nj=2, (Xi, Y1)
2Ni=2 and (Xi, Y2N+1)
2Ni=2 and
the four corners (X1, Y1), (X1, Y2N+1), (X2N+1, Y1) and (X2N+1, Y2N+1) are handledsimilarly. For these points we add appropriate “ghost” points and tile the augmenteddomain consistently. In the portion of augmented domain which lies outside of Ω(0)we assign h to be zero.
In what follows we let (Xi,j,Yi,j, ui,j, vi,j)(t) denote the approximate value of(X ,Y , u, v)(Xi, Yj, t) for 1 ≤ i, j ≤ 2N + 1 and insist that for 1 ≤ i, j ≤ 2N + 1
•
X i,j = ui,j and•
Y i,j = vi,j (6.22)
and
Xi,j(0) = Xi and Yi,j(0) = Yj. (6.23)
In the sequel we let
•
f = ft. (6.24)
In our computational model we assume that the fields(X,Y ) → (X ,Y , u, v)(X,Y, t) are piecewise linear in each of the triangles of our tiling.Thus, if (i + j) is even (see Figure 4) we have
X (X,Y, t) = Xi,j +2N(Xi+1,j −Xi,j)(X −Xi)+2N(Xi+1,j+1−Xi+1,j)(Y −Yj) (6.25)
for Yj ≤ Y ≤ X − Xi + Yj and Xi ≤ X ≤ Xi+1 in the bottom triangle and
X (X,Y, t) = Xi,j +2N(Xi+1,j+1−Xi,j+1)(X −Xi)+2N(Xi,j+1−Xi,j)(Y −Yj) (6.26)
for Xi ≤ X ≤ Y +Xi −Yj and Yj ≤ Y ≤ Yj+1 in the top triangle. The same formulasapply to the field Y , u and v if we simply replace the Xα,β’s by Yα,β, uα,β or vα,β asappropriate. On the other hand, if (i + j) is odd then Figure 4 is replaced by Figure5 and the piecewise linear approximation to X becomes
X (X,Y, t) = Xi,j + 2N(Xi+1,j −Xi,j)(X − Xi) + 2N(Xi,j+1 −Xi,j)(Y − Yj) (6.27)
for Yj ≤ Y ≤ Xi+1 + Yj − X and Xi ≤ X ≤ Xi+1 in the bottom triangle and
X (X,Y, t) = Xi,j+1 + 2N(Xi+1,j+1 −Xi,j+1)(X − Xi) + 2N(Xi+1,j+1 −Xi+1,j)(Y − Yj)
(6.28)
40
Figure 4: (i + j) is even
41
Figure 5: (i + j) is odd
42
for Xi+1 + Yj − X ≤ Y ≤ Yj+1 and Xi ≤ X ≤ Xi+1 in the top triangle. Once again,the same formulas apply to the fields Y , u and v if we replace the Xα,β’s by Yα,β, uα,β
or vα,β as appropriate. The reader should note there are a number of equivalent waysto write the formulas (6.25)-(6.28).
The basic kinematic unknowns of our computational model are functions(Xi,j,Yi,j, ui,j, vi,j)(t) for 1 ≤ i, j ≤ 2N +1. A knowledge of these quantities allows usto construct the approximate, piecewise linear fields (X ,Y , u, v)(X,Y, t) using (6.25)-(6.28) and their extension to the fields Y , u and v. The reader will note that (6.25)-(6.28) and their extensions are endowed with the following properties:
(i) the X and Y derivatives of X ,Y , u and v are constant in the interior of each ofthe triangles of our tiling, and
(ii) the Jacobian, D = XXYY −XY YX , is also constant in each of these triangles asare the quantities σ and τ defined in (6.18)-(6.19).
If, at t = 0, we assume that the initial data h0(·, ·) is constant in the interior ofthe triangles of our tiling, then (C −Lag) may be used to compute the current valueof h; specifically, in the interior of each of the triangles of our tiling
h = h0/D. (6.29)
Up to this point, we have said virtually nothing about the functional form of the“Eddy” viscosity E or its dimensionless forms ǫ and ǫ. As we said previously weregard ǫ > 0 as a mathematical construct that we are free to choose. We choose it toreduce spurious high frequency oscillations from our solutions; there is no fundamentalphysical law giving us E. For definiteness, we assume that ǫ is spatially constant in theinterior of each of the triangles of our tiling. In the case where p = 1 (see (Mx−Lag)and (My − Lag)) we assume that in the interior each triangle
ǫ = ǫ1def= µ1
D(∆X)2
∆t, µ1 a constant, (6.30)
whereas when p = 2 we assume that in the interior each triangle
ǫ = ǫ2def= µ2
D2(∆X)2
∆t, µ2 a constant. (6.31)
In formulas (6.30) and (6.31), ∆X =1
2Nand ∆t will be the time step in our numerical
integration scheme and again D is the Jacobian defined in (6.12).To obtain our computational model we integrate the equations (Mx − Lag) and
(My − Lag) over the computational cells which are interior to the red boxes shownin Figures 2 and 3. To keep things simple we assume that
h0(X,Y ) ≡
h0 > 0, (X,Y ) ∈ Ω(0)
0, (X,Y ) ∈ Ω(0)c
(6.32)
43
where once again Ω(0) is the unit square defined (6.20) and
Ω(0)c = (X,Y )|(X,Y ) /∈ Ω(0). (6.33)
The spatially discrete-continuous time approximation to (Mx − Lag) and(My − Lag) follow easily from some integral identities we present next.
We let
Xi−1/2 = (Xi−1 + Xi)/2 and Xi+1/2 = (Xi + Xi+1)/2 (6.34)
and
Yj−1/2 = (Yj−1 + Yj)/2 and Yj+1/2 = (Yj + Yj+1)/2 (6.35)
and
Ω(i,j) = (X,Y )|Xi−1/2 < X < Xi+1/2 and Yj−1/2 < Y < Yj+1/2 (6.36)
is the computational cell shown in Figures 2 and 3. Given (6.32) we note that for2 ≤ i, j ≤ 2N
Mi,j =
∫∫
Ω(i,j)
h0dXdY = h0(∆X)2 =h0
4N2(6.37)
and for the boundary points (X1, Yj)2Nj=2
Ω(1,j) = (X,Y )|X1 ≤ X ≤ X3/2 and Yj−1/2 ≤ Y ≤ Yj+1/2 (6.38)
M1,j =
∫∫
Ω(1,j)
h0dXdY =h0
8N2. (6.39)
The other boundary points (X2N+1, Yj)2Nj=2, (Xi, Y1)
2Ni=2 and (Xi, Y2N+1)
2Ni=2 are handled
similarly. For the corner (X1, Y1)
Ω(1,1) = (X,Y )|X1 ≤ X ≤ X3/2 and Y1 ≤ Y ≤ Y3/2 (6.40)
and
M1,1 =
∫∫
Ω(1,1)
h0dXdY =h0
16N2. (6.41)
The other corners are handled similarly and we obtain
M2N+1,1 = M1,2N+1 = M2N+1,2N+1 =h0
16N2. (6.42)
44
If f is any continuous (or smoother) function on Ω(0) we approximate
∫∫
Ω(i,j)
h0fdXdY by Mi,jf(Xi, Yj) (6.43)(4)
for 1 ≤ i, j ≤ 2N + 1. Some may find these approximations questionable at theboundary and corner points.
Finally, for 2 ≤ i, j ≤ 2N we replace
∫∫
Ω(i,j)
∂f
∂X(X,Y, t)dXdY and
∫∫
Ω(i,j)
∂g
∂Y(X,Y, t)dY dX by boundary integrals; specifically
∫∫
Ω(i,j)
∂f
∂X(X,Y, t)dXdY =
∫ Yj+1/2
Yj−1/2
(f(Xi+1/2, Y, t) − f(Xi−1/2, Y, t))dY (6.44)
and
∫∫
Ω(i,j)
∂g
∂Y(X,Y, t)dY dX =
∫ Xi+1/2
Xi−1/2
(g(X,Yj+1/2, t) − g(X,Yj−1/2, t))dX. (6.45)
For i = 1 and j = 2 : 2N we use formula (6.44) with f(X−1/2, Y, t) = 0 while fori = 2N + 1 and j = 2 : 2N we use (6.44) with f(X2N+3/2, Y, t) = 0. These extensionsreflect the fact that terms which are differentiated with respect to X in (Mx − Lag)and (My − Lag) are all multiplied by some power of h and these terms vanish forpoints in Ωc(0). Similar formulas hold for i = 2 : 2N and j = 1 or 2N + 1.
The corner points (X1, Y1), (X2N+1, Y1), (X1, Y2N+1) and (X2N+1, Y2N+1) are han-
dled analogously. Specifically we replace
∫∫
Ω(1,1)
∂f
∂X(X,Y, t)dXdY with
∫ Y3/2
Y1
f(X3/2, Y, t)dY . Analogous formulas hold for the other three corners.
(4)In the case where f is piecewise-linear in each triangle of our tiling, the integrals
∫∫
Ω(i,j)
h0fdXdY
may be computed exactly. Specifically, if (i + j) is even (see Figure 2)
∫∫
Ω(i,j)
h0fdXdY =mfi,j
3+
m
12(fi+1,j + fi,j+1 + fi−1,j + fi,j−1)
+m
12(fi+1,j+1 + fi−1,j+1 + fi−1,j−1 + fi+1,j−1)
whereas if (i + j) is odd (see Figure 3)
∫∫
Ω(i,j)
h0fdXdY =m
2fi,j +
m
8(fi+1,j + fi,j+1 + fi−1,j + fi,j−1)
In each of the above, m = h0
(
1
2N
)2
. We elected not to use these exact formulas since their
use would have presented us with a large linear algebra problem at each step of our “leap-frog”
algorithm (for details see (6.60) and (6.61)).
45
A careful look at (Mx−Lag) and (My−Lag), recalling the properties (i) and (ii) ofour approximate fields and (6.29), yields that all of the functions being differentiatedwith respect to X have the property that
f(Xi+1/2, Y, t) =
f−(Xi+1/2, t), Yj−1/2 < Y < Yj
f+(Xi+1/2, t), Yj < Y < Yj+1/2
(6.46)
while the functions g being differentiated with respect to Y satisfy
g(X,Yj+1/2, t) =
g−(Yj+1/2, t), Xi−1/2 < X < Xi
g+(Yj+1/2, t), Xi < X < Xi+1/2.(6.47)
For definiteness, we shall explicitly carry out the computation when f = YY h2. If(i + j) is even (see Figure 2), then
(XX)+(Xi+1/2, t) = 2N(Xi+1,j −Xi,j),
(XY )+(Xi+1/2, t) = 2N(Xi+1,j+1 −Xi+1,j),
(YX)+(Xi+1/2, t) = 2N(Yi+1,j − Yi,j),
(YY )+(Xi+1/2, t) = 2N(Yi+1,j+1 − Yi+1,j),
D+(Xi+1/2, t) = ((XX)+(YY )+ − (XY )+(YX)+)(Xi+1/2, t) (6.48)
h+(Xi+1/2, t) = h0/D+(Xi+1/2, t), (6.49)
(XX)−(Xi+1/2, t) = (XX)+(Xi−1/2, t),
(XY )−(Xi+1/2, t) = 2N(Xi+1,j −Xi+1,j−1),
(YX)−(Xi+1/2, t) = (YX)+(Xi+1/2, t),
(YY )−(Xi+1/2, t) = 2N(Yi+1,j − Yi+1,j−1),
D−(Xi+1/2, t) = ((XX)−(YY )−) − (XY )−(YX)−)(Xi+1/2, t), (6.50)
and
h−(Xi+1/2, t) = h0/D−(Xi+1/2, t). (6.51)
When (i + j) is odd (see Figure 3) we obtain
46
(XX)+(Xi+1/2, t) = (XX)−(Xi+1/2, t) = 2N(Xi+1,j −Xi,j),
(XY )+(Xi+1/2, t) = 2N(Xi,j+1 −Xi,j),
(YX)+(Xi+1/2, t) = (YX)−(Xi+1/2, t) = 2N(Yi+1,j − Yi,j), p
(YY )+(Xi+1/2, t) = 2N(Yi+1,j − Yi,j),
(XY )−(Xi+1/2, t) = 2N(Xi,j −Xi,j−1),and
(YY )−(Xi+1/2, t) = 2N(Yi,j − Yi,j−1).
From these latter quantities we compute D+, D−, h+, and h− using (6.48), (6.50),(6.49) and (6.51).
The Approximating ODE Model
Using the tools developed in the sub-section Geometric Preliminaries our tasknow is to translate (Mx − Lag) and (My − Lag) into an approximating system ofordinary differential equations for the matrices ui,j(t)
2N+1i,j=1 and vi,j(t)
2N+1i,j=1 . These
equations take the form
M ∗•u = G(X ,Y , u, v) (6.52)
and
M ∗•v = H(X ,Y , u, v). (6.53)
Here M is the (2N + 1)× (2N + 1) square matrix whose (i, j)th entries are defined in
(6.37), (6.38), (6.40) and (6.41) and the (i, j)th components of the products (M ∗•u)i,j
and (M ∗•v)i,j are give by
(M ∗•u)i,j = Mi,j
•ui,j (6.54)
and
(M ∗•v)i,j = Mi,j
•vi,j. (6.55)
G and H are (2N +1)× (2N +1) matrix valued functions of the (2N +1)× (2N +1)matrices Xi,j
2N+1i,j=1 , Yi,j
2N+1i,j=1 , ui,j
2N+1i,j=1 and vi,j
2N+1i,j=1 . G has a contribution due
to
1. the potential term −C2h0ax(X ,Y) whose (i, j)th component is evaluated using
(E6.43) and is given by
Gpotentiali,j = −C
2Mi,jax(Xi,j,Yi,j); (6.56)
47
2. the Coriolis term Fh0v and bottom friction term −kfrh0u whose (i, j)th com-ponents are evaluated using (6.43) and are given by
GCoriolisi,j = FMi,j vi,j and Gfriction
i,j = −kfrMi,jui,j; (6.57)
3. the wind term kwindD(uwind(X ,Y) − u) whose (i, j)th component is evaluatedusing (6.43) and is given by (see Figures 2 and 3)
Gwindi,j =
kwind
32N2
(
Σ8α=1D
αi,j
)
(uwind(Xi,j,Yi,j) − ui,j) , (i + j) even
kwind
16N2
(
Σ4α=1D
αi,j
)
(uwind(Xi,j,Yi,j) − ui,j), (i + j) odd
(6.58)
when 2 ≤ i, j ≤ 2N ;(5)
4. the pressure term −C2
2((YY h2)X−(YX h2)Y ) whose (i, j)th component is denoted
by Gpressrei,j and is evaluated using (6.44)-(6.47); and finally
5. the turbulent stress term
(hpǫ(YY σ −XY τ))X + (hpǫ(XX τ − YX σ))Y
whose (i, j)th component is denoted by GT.stressi,j and is evaluated using (6.44)-
(6.47). The function H is evaluated using similar techniques.(6)
Concluding Remarks
Our final system is (6.22), (6.52) and (6.53). We update these via a “leap-frog”algorithm. We let ∆t be our time-step parameter and assume that at
tn = n∆t n = 0, 1, . . . (6.59)
we know the matrices
X ni,j
2N+1
i,j=1,
Yni,j
2N+1
i,j=1,
uni,j
2N+1
i,j=1and
vni,j
2N+1
i,j=1.
These represent the approximate values of our solution at tn. Given these data andthe tiling of Ω(0) we can compute the quantities Dn, hn and ǫn in the interior of eachtriangle of our tiling at tn = n∆t.
(5)Analogous formulas hold for the boundary and corner points.(6)The details of these terms may be found by examining our Matlab codes SLOSHP1.m and
SLOSHP2.m which may be found at www.math.cmu.edu/people/fac/greenberg.html.
48
We use (6.52) and (6.53) to obtain
un+1i,j = un
i,j +∆t
Mi,j
Gi,j(Xn,Yn, un, vn) (6.60)
and
vn+1i,j = vn
i,j +∆t
Mi,j
Hi,j(Xn,Yn, un, vn) (6.61)
and then use (6.22) to obtain
X n+1i,j = X n
i,j + ∆t un+1i,j (6.62)
and
Yn+1i,j = Yn
i,j + ∆t vn+1i,j . (6.63)
This is our “leap-frog” scheme. The details of the serial implementation may be foundin at www.math.cmu.edu/people/fac/greenberg.html. The codes with p = 1 and 2correspond to the listings SLOSHP1.m and SLOSHP2.m respectively.
We note that these codes not only work on a square grid with (2N +1)× (2N +1)grid points but they work equally well in rectangular grids with (2K + 1)× (2N + 1)
points so long as ∆X = ∆Y =1
4N2.
7 Implementation in Parallel
To implement this “leap-frog” scheme in parallel, work must be divided equally amongseveral processors. A simple way to do this is to divide the junction points of thegrid evenly between the number of available processors. At each grid point we arerequired to solve a separate set of differential equations–specifically (6.22), (6.52)and (6.53). A communication problem arises when not all data required to updateinformation at a specific grid point is assigned to the same processor. Overcomingthis problem requires a scheme for passing data between the processors, and thisrequirement drives the system by which the sub-domains are constructed from thewhole two-dimensional grid and assigned to processors.
Communication is, in general, time consuming. To optimize a parallel implemen-tation, communication must be avoided as much as possible; therefore, the processfor splitting the two-dimensional grid is designed to keep the amount of necessarycommunication small. To balance the parallel processors’ loads, each should be as-signed an approximately equal share of the computational work, because every time aprocessor pauses to communicate, it must wait for all processors to complete their as-signed computational tasks. Assuming that the processors are of roughly equal speed,each processor should therefore be assigned approximately equal sized sub-domainsand equally sized sub-sets of the differential equations.
49
Figure 6: Two-dimensional grid divided into square sub-domains
Figure 7: Two-Dimensional grid divided into rectangular sub-domains
50
Two of the easiest ways to partition the two-dimensional square grid into sub-domains are by square sub-domains as depicted in (Figure 6) or by rectangular sub-domains as in (Figure 7). If we have P 2 processors, P = 2, 3, . . . and a squaregrid with n2 points, then for P = 2 the square and rectangular partitioning requireessentially the same amount of communication while for P ≥ 3 the partitioning by
squares requires less communication than partitioning by rectangles;4(n)
Pv.s. 2(n).
Despite requiring more communication, the rectangular partition has several ad-vantages. The first is that we partition in only one coordinate direction; this allowseach processor to evaluate over the interval 1 : n in the unpartitioned direction. Thisnot only simplifies writing our numerical algorithm but greatly simplifies messagepassing. Another advantage of the rectangular method is in the pattern of commu-nication. In a computing cluster using the square method for creating sub-domains,each processor may need to communicate with as many as eight other processors.With the rectangular method, each processor is guaranteed to have at most two com-munication partners. Depending on the configuration of the hardware of the cluster,this can be a great advantage. For clusters organized in a lattice or a ring, the neces-sity to only communicate with two adjacent processors eliminates the need for anycommunication to pass through multiple processors in order to get from its source toits destination.
We will use a rectangular partition of our original square domain; to implementthis certain technical constraints on this partition must be met. In what follows weshall assume that our square-grid has (2M + 1) × (2M + 1), M = 1, 2, . . ., points.Since we are electing to use a rectangular partition we no longer need require thatthe number of processors be a perfect square; instead we assume that
P = 2p p = 1, 2, . . . , < M.
Thus the number of grid points on each processor will be approximately2M−p× (2M +1). If F is any of our (2M +1)× (2M +1) data matrices the preliminaryassignment of F to the individual processors would be
F(k) = F ((k − 1)2M−p + 1 : k2M−p, :) for 1 ≤ k ≤ 2p−1
and
F(2p) = F ((2p − 1)2M−p + 1 : 2M + 1, :) for k = 2p.
Figure 7 shows this preliminary assignment when M = 4, p = 2 and P = 4.This preliminary data assignment is only part of the story. We next observe that
our basic serial code runs on any rectangular grid with (2K + 1) × (2M + 1) pointsand this code invokes our implementation of the h = 0 boundary condition on theouter edges and corner points of the grid.
Our desire is to choose K and an appropriate augmentation of the matricesF(k), 1 ≤ k ≤ 2p so that during each update we can run the identical (2K+1)×(2M+1)serial code on each processor and be guaranteed that if, after doing the update, we
51
reform the F matrices we have the same answer as obtained via serial code on theoriginal (2M + 1) × (2M + 1) square grid.
We let K = 2M−p−1 + 1 and note that 2K + 1 = 2M−p + 3. The final assignmentof F to the individual processors is
F(1) = F (1 : 2M−p + 3, :)
F(k) = F ((k − 1)2M−p − 1 : k2M−p + 1, :) for 2 ≤ k ≤ 2p−1
F(2p) = F ((k − 1)2M−p − 1 : 2M + 1, :).
All of these matrices are of the same size, namely (2M−p + 3) × (2M + 1). This steprepresents how we pass to the initial data the individual processors.
After a one-step update the F(k)’s are modified via:
F(1) = [F(1)(1 : 2M−p, :), F(2)(3 : 5, :)]
F(k) = [F(k−1)(2M−p + 1 : 2M−p + 2, :), F(k)(3 : 2M−p + 2, :), F(k+1)(3, :)]
for 2 ≤ k ≤ 2p−1
and
F(2p) = [F(2p−1)(2M−p + 1 : 2M−p + 2, :), F(2p)(3 : 2M−p + 3, :)]
These formulas are our message passing protocol. Of course F may be recovered fromthe F(k)’s via
F = [F(1)(1 : 2M−p, :), F(k)(3 : 2M−p + 2, :), F(2p)(3 : 2M−P + 3)] for 2 ≤ k ≤ 2p−1
Results of Test Data
Table 1: Results of TestingDomain Size Serial 4 Proc. 4 Processor 8 Proc. 8 Processor 16 Proc. 16 Processor
Rect. Parallel Rect. Parallel Rect. Parallel
65 x 65 0.7953 0.2839 0.3766 0.1939 0.2976 0.1461 0.3158129 x 129 4.1803 0.8650 0.9338 0.4878 0.5743 0.3205 0.5210257 x 257 24.5459 4.7274 5.0877 1.8032 1.9852 0.9316 1.1136
The data in (Table 1) was generated through test runs on the U.S. NavalAcademy’s Beowulf cluster with the MATLAB Distributed Computing Toolbox. The
52
Figure 8: Sub-domain scheme with adjacent columns as used in three-dimensionalmodel
tests are run on square two-dimensional grids whose sides are of length 2M +1. Threegrid sizes were tested: M = 6, M = 7, and M = 8. The code for both the serial andparallel versions were modified to track runtime. Note that the times in (Table 1)are not processor times but “real” times. The code starts tracking time for each testafter the initialization of data is complete and before the “leap-frog” loop begins. Inthe parallel version, this means that the initial grid and data are distributed to allprocessors before the timing starts.
The domain size is the number of grid points along each edge of the square gridcreated for the test. The column of data labeled “Serial” is the runtime for thatdomain size in the serial version. The data labeled “Rect.” represents the runtime for a
single processor running on a grid of size
(
2M
P+ 3
)
×(
2M + 1)
. Note that this is the
size of the grids used in our parallel scheme and represents the theoretical maximumspeed of the model without any communication. The data labeled “Parallel” is theruntime for P parallel processors. The data in (Table 2) is calculated from the data in(Table 1) and shows the speedup factor for each domain size and number of processors,as well as the percentage of time spent communicating.
The results of these tests show that the implementation of the Lagrangian com-putational model in parallel was successful. For a sufficiently large grid, increasingthe number of processors increases the computational speed of the model but there isa reduction in efficiency. This loss of efficiency is the result of increased communica-
53
Table 2: Speedup and Communication TimeDomain Size 4 Processor 4 Processor 8 Processor 8 Processor 16 Processor 16 Processor
Speedup Comm. Time Speedup Comm. Time Speedup Comm. Time
65 x 65 2.11 24.6% 2.67 34.8% 2.52 53.7%129 x 129 4.48 7.4% 7.28 15.1% 8.02 38.5%257 x 257 4.82 7.1% 12.36 9.2% 22.04 16.4%
tion. However, this increased communication time is counterbalanced by a reductionin runtime for smaller domains. The ratio of run times between larger and smallerdomains is greater than the ratio between the number of nodes in each domain. Thismeans that while one domain may be half the size of another, the processing timefor the smaller domain will be less than half of the time for the larger domain. Wesuspect that this is because larger domains require more memory usage and makesproportionately less use of caching.
We see that the speedup on larger domains can exceed the number of processors.This can be seen in the tests on the grid of size M = 8, where the speedup with 8processors is over 12, and the speedup with 16 processors is over 22. However, oncethe number of processors becomes too large, the losses introduced by communicationovercome the advantages gained by using smaller domains. In the grid of sizeM = 6, 16 processors actually run slower than 8 processors.
The results of these tests confirm that the methodology used here is viable. Asmore complexity is added to the model the parallelization strategy should remainvalid.
Two Simulations
We conclude by showing how our computational model captures the “Rotating-Pulsating” solution discussed in section 5.
For definiteness, we take the bottom surface to be the paraboloid
a(x, y) =1
2(x2 + y2). (7.1)
For this surface
(ax, ay) = (x, y). (7.2)
We also take
(C2, F ) = (1, 1.5) (7.3)
and kwind and kfr are both chosen to be zero.The initial wet region is
Ω(0) =
(x, y)| − .5 + (.5)1/2 ≤ x ≤ .5 + (.5)1/2 and − .5 ≤ y ≤ .5
(7.4)
and
54
h(x, y, 0) =
1 if (x, y) ∈ Ω(0)and0 otherwise .
(7.5)
The initial values for the velocities u and v are zero. A quick computation shows thatat t = 0, the coordinates of the center of mass are given by
(xc, yc)(0) = ((.5)1/2, 0). (7.6)
The reader will note that we have chosen data so that the center of mass satisfies
•xc = uc and
•yc = vc, (7.7)
•uc = −xc +
3vc
2and
•vc = −yc −
3uc
2(7.8)
and
(xc, yc)(0) = ((.5)1/2, 0) and (uc, vc)(0) = (0, 0). (7.9)
The system (7.7)-(7.9) has the two first integrals
(x2c + y2
c + u2c + v2
c )(t) ≡1
2(7.10)
and
(
3
2(u2
c + v2c ) − 2(xcvc − ycuc)
)
(t) ≡ 0 (7.11)
and
R(t)def= (x2
c + y2c )(t) (7.12)
satisfies
••
R +25
4R =
17
8(7.13)
and the initial conditions
R(0) =1
2and
•
R(0) = 0. (7.14)
The solution to (7.13) and (7.14) is given by
R(t) = .34 + .16 cos5t
2(7.15)
and this latter function is periodic with period T given by
T =4π
5= 2.51327 . . . . (7.16)
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The simulations we show in the next two figures were obtained running the codesSLOSHP1.m and SLOSHP2.m. For both codes we used a 65 by 65 grid with ∆X =∆Y = 1/64 and ∆t = .025∆X. In the former simulation we used the “Eddy”viscosity given by (6.30) with µ1 = .1 while in the latter simulation the viscosity isgiven by (6.31) with µ2 = .1. The time-span of both simulations was 50.3 which isapproximately 20 periods of the function R(·).
Given the raw outputs of these code we empirically compute the center of massby
(xc, yc) =1
Ma(Σi,jMi,jXi,j, Σi,jMi,jYi,j) (7.17)
and (uc, vc) by
(uc, vc) =1
Ma(Σi,jMi,jui,j, Σi,jMi,j vi,j) . (7.18)
The numbers Mi,j are the point mass given by (6.37), (6.39), (6.41) and (6.42) andMa = Σi,jMi,j. The empirical value of R is obtained using (7.12) and (7.17).
In both figures we show a surface plot of the water column height as a functionof x− xc and y − yc in the top of the first column. In the bottom of the first column
we show both the total water height h +(x2 + y2)
2and the visible portion of the
paraboloid(x2 + y2)
2which is wet.
In the top picture in the second column we show plots of u − uc and v − vc asfunctions of x − xc and y − yc. The u − uc field is the one with embedded blackcontours lines.
In the middle picture we show the empirical trajectory t → (xc, yc)(t) over aninterval slightly larger than 5T . The black + denotes the current position of (xc, yc).
In the bottom picture we show a graph of the empirical function R over a windowof length 10.
After 20 periods of R it is clear that h has gone to the desired paraboloidal shapeand u − u0 and v − vc have converged to linear functions of x − xc and y − yc.
If the reader has access to Matlab we suggest they run our serial code SLOSHP1.mand SLOSHP2.m in order to get a better feel for the dynamics of this model.
Conclusions
The simulations just presented demonstrate that for the type of free-boundaryproblem we have been focusing on the Lagrangian reformulation and associated com-putational model offer a viable solution alternative to an explicit tracking methodthat attempts to capture the curve where h = 0 superimposed on a viscous solverapplied to the Eulerian version of the equations of motion. Our formulation of the“Eddy” viscosities also seems simpler than ones based on turbulence closure models;for example see [3] and [4]. Our particular parallel implementation of spreading theproblem across multiple processors and our simple communication protocol seemspromising and allows us to reuse our serial code with virtually no modifications. Thisis one of the few examples we know of where a slightly modified version of one’s“dusty deck” yields quite acceptable parallel results.
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Figure 9
Figure 10
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[8] Thacker, W.C., Some exact solutions to the nonlinear sallow-water wave equa-tions, J. Fluid Mech., 113 (1981), 499-508.
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