2-Dimensional Sloshing Flows for the Shallow Water Equations · Introduction In this talk I’ll...
Transcript of 2-Dimensional Sloshing Flows for the Shallow Water Equations · Introduction In this talk I’ll...
2-Dimensional Sloshing Flowsfor the
Shallow Water Equations
J.M. GreenbergProfessor Emeritus
Department of Mathematical SciencesCarnegie Mellon University
Pittsburgh, PA 15213email: [email protected]
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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