2-Dimensional Sloshing Flowsfor the

Shallow Water Equations

J.M. GreenbergProfessor Emeritus

Department of Mathematical SciencesCarnegie Mellon University

Pittsburgh, PA 15213email: greenber@andrew.cmu.edu

Introduction

In this talk I’ll discuss “sloshing” 2-Dimensional flows forthe “shallow-water” equations. The model describes themotion of a finite volume of viscous fluid taking placein container whose bottom is described by a paraboloidallike surface of the form:

z = (αx2 + βy2)/2, α > 0, β > 0

or more generally

z = a(x, y)

where a → ∞ as (x2 + y2) → ∞. The model includesgravity, coriolis, and viscous forces.

Euler Description

∂h

∂t+

∂

∂x(hu) +

∂

∂y(hv) = 0 continuity

∂

∂t(hu) +

∂

∂x(hu2) +

∂

∂y(huv) +

g

2

∂

∂x(h2) + gh

∂a

∂x=

−bhu + fhv +

(

∂

∂x(ǫhpσ) +

∂

∂y(ǫhpτ)

)

x mom

∂

∂t(hv) +

∂

∂x(huv) +

∂

∂y(hv2) +

g

2

∂

∂y(h2) + gh

∂a

∂y=

−bhv − fhu +

(

∂

∂x(ǫhpτ) −

∂

∂y(ǫhpσ)

)

y mom

σ =∂u

∂x−

∂v

∂yand τ =

∂u

∂y+

∂v

∂x

The stress tensor,

σ , τ

τ , −σ

, is trace free

and

trace

σ , τ

τ , −σ

∂u

∂x

∂u

∂y

∂v

∂x

∂v

∂y

=

(

∂u

∂x−

∂v

∂y

)2

+

(

∂u

∂y+

∂v

∂x

)2

≥ 0.

h(x, y, t) height of the water above z = a(x, y)

u(x, y, t) x component of the velocity at the point (x, y) at time t

v(x, y, t) y component of the velocity at the point (x, y) at time t

g the gravational constant

f is the rotation rate of the earth and b is the bottom drag coefficient

c = (u, v,0) × (0,0, f) = det

i j k

u v 00 0 f

= f(vi − uj)

is the coriolis acceleration

p..an integer . . . typically 1 or 2

The equations hold in

Ω(t) = (x, y)|h(x, y, t) > 0

and this region, and its boundary,

∂Ω(t) = (x, y)|h(x, y, t) = 0

must be determined as part of the solution.

Alternate-Semi Lagrangian Form of x and y Momentum Equations

h

(

∂u

∂t+ u

∂u

∂x+ v

∂u

∂y+ g

∂a

∂x− fv + bu

)

+g

2

∂

∂x(h2)

=

(

∂

∂x(ǫhpσ) +

∂

∂y(ǫhpτ)

)

x mom

h

(

∂v

∂t+ u

∂v

∂x+ v

∂v

∂y+ g

∂a

∂y+ fu + bv

)

+g

2

∂

∂y(h2)

=

(

∂

∂x(ǫhpτ) −

∂

∂y(ǫhpσ)

)

y mom

Lagrangian Description

These are solutions of

∂X

∂t(X, Y, t) = u(X(X, Y, t),Y(X, Y, t), t)

∂Y

∂t(X, Y, t) = v(X(X, Y, t),Y(X, Y, t), t)

X(X, Y,0) = X and Y(X, Y, t) = Y.

It is possible to transform the Euler Description to equa-tions in the variables X and Y . The derivation of theseequations is straightforward though rather tedious. Welet

h(X, Y, t) = h(X(X, Y, t),Y(X, Y, t), t)u(X, Y, t) = u(X(X, Y, t),Y(X, Y, t), t)v(X, Y, t) = v(X(X, Y, t),Y(X, Y, t), t).

These satisfy X,t = u and Y,t = v trajectories

h(X, Y, t)D(X, Y, t) = h0(X, Y ) continuity

where h0(X, Y ) is the initial data for h and

D(X, Y, t) =(

X,XY,Y −X,Y Y,X

)

(X, Y, t) > 0

is the jacobian of (X, Y ) → (X(X, Y, t),Y(X, Y, t)).

h0u,t +g

2

(

(

Y,Y h2)

,X−(

Y,Xh2)

,Y

)

+ gh0a,x(X ,Y) = −bh0u+

fh0v +

(

(

ǫhp(

Y,Y σ − X,Y τ))

,X+(

ǫhp(

X,Xτ − Y,Xσ))

,Y

)

and

h0v,t +g

2

(

−(

X,Y h2)

,X+(

X,Xh2)

,Y

)

+ gh0a,y(X ,Y) = −bh0v

−fh0u +

((

ǫhp(

Y,Y τ + X,Y σ)

,X−(

ǫhp(

Y,Xτ + X,Xσ))

,Y

)

where

σ = u,x− v,y = ((Y,Y u,X +X,Y v,X)− (Y,Xu,Y +X,Xv,Y ))/D

and

τ = u,y+v,x =((

Y,Y v,X − X,Y u,X

)

+(

X,Xu,Y − Y,Xv,Y

))

/D.

Outline of the talk

• Center of Mass Equations when the bottom is givenby

z = (αx2 + βy2)/2

• Interesting Change of Variables Factoring out the Cen-ter of Mass Motion

• Crucial Energy Identity

• Some insightful exact solutions

• Lagrangian Formulation

– discussion of potential pluses and minuses

– some interesting simulations which show robust-ness of this approach.

Center of Mass Equations

Recall

Ω(t) = (x, y)|h(x, y, t) > 0 and ∂Ω(t) = (x, y)|h(x, y, t) = 0 .

A Leibnitz (Leibniz) Formula

∫∫

Ω(t)

h(x, y, t)dxdy =

∫∫

Ω(0)

(hD)(X, Y, t)dXdY =

∫∫

Ω(0)

h0(X, Y )dXdY ≡ m.

Using the above identity, and the fact that points (X, Y )go into (X(X, Y, t),Y(X, Y, t)) we find that for any smoothfunction f

d

dt

∫∫

Ω(t)

h(x, y, t)f(x, y, t)dxdy =

∫∫

Ω(t)

h(x, y, t)(

f,t + uf,x + vf,y

)

(x, y, t)dxdy.

We now suppose a(x, y) = (αx2 + βy2)/2 and thereforea,x = αx and a,y = βy.

We let

mxc ≡∫∫

Ω(t)

h(x, y, t)x dxdy and

myc ≡∫∫

Ω(t)

h(x, y, t)y dxdy.

The Leibnitz Formula implies

mdxc

dt=

∫∫

Ω(t)

h(x, y, t)u(x, y, t)dxdy ≡ muc

and

mdyc

dt=

∫∫

Ω(t)

h(x, y, t)v(x, y, t)dxdy ≡ mvc

Moreover,

mduc

dt=

∫∫

Ω(t)

h(x, y, t)(u,t+uu,x+vu,y)(x, y, t) = fmvc−gαmxc−bmuc

and

mdvc

dt= −fmuc − gβmyc − bmvc

Change of Variable factoring out the motion of the Center of MassSuppose a = (αx2+βy2)/2 and that h, u, and v satisfy thecontinuity and momentum equations and xc, yc, uc, and vcsatisfy the center of mass equations.

Then, if we let

x1 = x − xc(t)y1 = y − yc(t)h1(x1, y1, t) = h(x1 + xc(t), y1 + yc(t), t)u1(x1, y1, t) = u(x1 + xc(t), y1 + yc(t), t) − uc(t)v1(x1, y1, t) = v(x1 + xc(t), y1 + yc(t), t) − vc(t),

the functions h1, u1 and v1 satisfy the continuity and mo-mentum equations in the subscripted variables in

Ω1(t) = (x1, y1) | (x1 + xc(t), y1 + yc(t)) ∈ Ω(t)

The functions h1, u1, and v1 further satisfy∫∫

Ω1(t)

h1(x1, y1, t) (1, x1, y1, u1(x1, y1, t), v1(x1, y1, t)) dx1dy1 ≡ (m,0,0,0,0)

Basic Energy Identity for Arbitrary Bottom z = a(x, y)

h∂t

(

u2 + v2

2+ g(a + h/2)

)

+hu∂x

(

u2 + v2

2+ g(a + h/2)

)

+ ∂x(gh2u/2)

+hv∂y

(

u2 + v2

2+ g(a + h/2)

)

+ ∂y

(

gh2v/2)

= ∂x ((ǫhpu (u,x − v,y)) + (ǫhpv (v,x + u,y)))

+∂y ((ǫhpu (u,y + v,x)) − (ǫhpv (u,x − v,y)))

−ǫhp(

(u,y + v,x)2 + (u,x − v,y)

2)

− bh(u2 + v2).

The last identity implies that

d

dt

∫∫

Ω(t)

h

(

u2 + v2

2+ g

(

a +h

2

)

)

(x, y, t) dxdy =

−∫∫

Ω(t)

(

ǫhp(

(u,y + v,x)2 + (u,x − v,y)

2)

+ bh(

u2 + v2))

(x, y, t)dxdy ≤ 0.

When a =(

αx2 + βy2)

/2 the energy and center of mass

identities also imply that

d

dt

1

2

∫∫

Ω(t)

h(

(u − uc)2 + (v − vc)

2 + g(

α (x − xc)2 + β (y − yc)

2 + h))

dxdy

= −∫∫

Ω(t)

ǫhp(

(u,y + v,x)2 + (u,x − v,y)

2)

dxdy

−b∫∫

Ω(t)

h(

(u − uc)2 + (v − vc)

2)

dxdy

Remark The last identity is crucial. It implies that

0 < D(t) =

∫∫

Ω(t)

ǫhp(

(u,y + v,x)2 + (u,x − v,y)

2)

(x, y, t)dxdy

is L1(0,∞) and leads one to suspect that

(i) limt→∞

D(t) = 0

and

(ii) that limt→∞

(u,x−v,y)(x, y, t) = 0 and limt→∞

(u,y+v,x)(x, y, t) = 0.

These suspicions are true.

Special Solutions Satisfying

(0,0,0,0) =

∫∫

Ω(t)

h(x, y, t)(x, y, u(x, y, t), v(x, y, t))dxdy

and

m ≡∫∫

Ω(t)

h(x, y, t)dxdy.

We seek solutions of the form

u = ((d + σ)x + (w + τ)y)/2

v = ((τ − w)x + (d − σ)y)/2

h = h0 −1

2(h11x2 + 2h12xy + h22y2)

where

d, w, σ, τ, h0, h11, h12, and h22

are functions t only. When p = 2 these solutions are exactand when p = 1 they are approximate.

We want

h0 > 0 , h11 > 0 , h22 > 0 and h11h22 − h212 > 0 and

Ω(t) =

(x, y)|√

x2 + y2 = r ≤

√

4h0(h11+h22)+(h11−h22) cos 2θ+2h12 sin 2θ

Insertion of the ansatz into the Euler Equations yieldsh0 = −dh0β = −2dβ + (h12w − lσ)

h12 = −2dh12 − (lτ + βω)and

l = −2dl − (βσ + h12τ)where β = (h11 − h22)/2 and l = (h11 + h22)/2.

w = −d(w − f) − 4ǫ(h12σ − βτ)σ = −dσ + fτ + 2gβ − 4ǫlστ = −dτ − fσ + 2gh12 − 4ǫlτ

d = 2g(l − α) − fw + (w2 − d2 − σ2 − τ2)/2 − 4ǫ(βσ + h12τ)

Remark

(β, h12, σ, τ) ≡ (0,0,0,0) is invariant

Integration Scheme

d = 2A/A , h0 = 1/A2

w = (w − f)

w = Ω/A2 , σ = Σ/A2 , τ = T /A2

l = L/A4 , β = B/A4 , h12 = H12/A4

Ω = −4ǫ(H12Σ − BT )/A4

L = −(BΣ + H12T )/A2

B = −

(

LΣ

A2−

(

Ω

A2+ f

)

H12

)

H12 = −

(

LT

A2+

(

Ω

A2+ f

)

B

)

Σ = −fT +2gB

A2−

4ǫLΣ

A4

T = fΣ +2gH12

A2−

4ǫLT

A4

2A +

(

2gα +f

2

2)

A −

(

Ω2

2+ 2gL

)

/A3

= −(

Σ2 + T 2)

/2A3 − 4ǫ(BΣ + H12T )/A5

Theorems

• L2 − B2 − H212 ≡ L2(0) − B2(0) − H2

12(0) ≡ L2∞ > 0

•

4g√

L2∞ + B2 + H2

12 + 8ǫ∫ t

0

LA4

(

Σ2 + T 2)

(s)ds +(

Σ2 + T 2)

≡ 4gL(0) +(

Σ2(0) + T 2(0))

≡ 4gLmax

• (B, H12, T ,Σ) → 0,Ω → Ω∞ and L → L∞ as t → ∞.

A asymptotically satisfies

2A +

(

2gα + f2

2)

A −

(

Ω2∞2 + 2L∞g

)

/A3 = 0

(A)2 +

(

gα + f4

2)

A2 +

(

Ω2∞4 + L∞g

)

1A2 = E2

∞

E2∞ ≥ 2

(

Ω2∞4 + L∞g

)1/2 (f2

4 + gα

)1/2

Aeq =

(

Ω2∞4 + L∞g

)1/4/

(

f2

4 + gα

)1/4

Lagrangian Description

These are solutions of

∂X

∂t(X, Y, t) = u(X(X, Y, t),Y(X, Y, t), t)

∂Y

∂t(X, Y, t) = v(X(X, Y, t),Y(X, Y, t), t)

X(X, Y,0) = X and Y(X, Y, t) = Y.

It is possible to transform the Euler Description to equa-tions in the variables X and Y . The derivation of theseequations is straightforward though rather tedious. Welet

h(X, Y, t) = h(X(X, Y, t),Y(X, Y, t), t)u(X, Y, t) = u(X(X, Y, t),Y(X, Y, t), t)v(X, Y, t) = v(X(X, Y, t),Y(X, Y, t), t).

These satisfy X,t = u and Y,t = v trajectories

h(X, Y, t)D(X, Y, t) = h0(X, Y ) continuity

where h0(X, Y ) is the initial data for h and

D(X, Y, t) =(

X,XY,Y −X,Y Y,X

)

(X, Y, t) > 0

is the jacobian of (X, Y ) → (X(X, Y, t),Y(X, Y, t)).

h0u,t +g

2

(

(

Y,Y h2)

,X−(

Y,Xh2)

,Y

)

+ gh0a,x(X ,Y) = −bh0u+

fh0v +

(

(

ǫhp(

Y,Y σ − X,Y τ))

,X+(

ǫhp(

X,Xτ − Y,Xσ))

,Y

)

and

h0v,t +g

2

(

−(

X,Y h2)

,X+(

X,Xh2)

,Y

)

+ gh0a,y(X ,Y) = −bh0v

−fh0u +

((

ǫhp(

Y,Y τ + X,Y σ)

,X−(

ǫhp(

Y,Xτ + X,Xσ))

,Y

)

where

σ = u,x− v,y = ((Y,Y u,X +X,Y v,X)− (Y,Xu,Y +X,Xv,Y ))/D

and

τ = u,y+v,x =((

Y,Y v,X − X,Y u,X

)

+(

X,Xu,Y − Y,Xv,Y

))

/D.

Comments

• The independent variables, X and Y , represent theinitial coordinates of trajectories and this points outone of the potential benefits of using the LagrangianFormulation; we only need to describe the initial do-main occupied by the fluid; i.e. work with a singlefixed grid.

• With the Euler Description we have a continuouslychanging computational domain for h, u, and v whoseboundary we have to track.

• One can encounter problems with the Lagrangian For-mulation with large, non-decaying stresses; in ourproblem

σ = u,x − v,y and τ = u,y + v,x.

These can produce severe “distortion” of the grid.Specificlly, regions which were triangular at t = 0 withno small angels can go into triangular regions wherethe triangles have “small”acute angels. This does nothappen in our case since σ and τ decay.

For N = 1,2, . . ., we’ll work with grids with 2N + 1 hor-izontal and vertical lines. Grids when N = 1 are shownin Figures 1 and 2. To get grids when N = 2 we cutand paste 3 additional copies of the N = 1 grid to theoriginal-one to the right and two above-and then rescaleto [0,1]× [0,1]. We call this process “squaring” the first

grid. The same procedure allows us to go from the N th

to the (N + 1)st grid. The control volumes of the (i, j)th

mesh points are the rectangular regions bordered by thered lines. We obtain a discretization of the Lagrangianequations in each “control-volume”assuming that the mo-tion in each deformed triangle is piecewise linear. The La-grangian continuity equation, hD = h0, implies that themass in each deformed triangle is spatially constant. Ad-ditionally, the mass in each deformed “control-volume” isspatially constant. I’ll spare the audience the gory detailsof the discrete system.

I note that when displaying results, we convert to an EulerRepresentation of all fields; that is we display quantitiesagainst (xi,j, yi,j), the current locations of (Xij, Yi,j) rather

than against (Xi,j, Yi,j).

http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/pw/crls.rxml

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