coupling between ice sheets movement and subglacial hydrology · 2017-07-25 · coupling between...
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Coupling Between Ice Sheets Movement and Subglacial Hydrology
Luca Bertagna1, Max Gunzburger2, Mauro Perego1, Konstantin Pieper2
1Sandia National Laboratories, 2Florida State University
Atlanta, March 2nd 2017
Sandia National Laboratories is a multi-mission laboratory managed and operated by Sandia Corporation, a wholly owned subsidiary of Lockheed Martin
Corporation, for the U.S. Department of Energy’s National Nuclear Security Administration under contract DEAC0494AL85000.
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 1 / 21
SAND2017-2256C
Introduction
Introduction
Ice dynamics is complex:
nonlinear rheology
coupling with other components(atmosphere, ocean, land)
long time scales compared tosurrounding air/water
complex boundary conditions atbedrock interface
Our focus: boundary condition at the bedrock interface.
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 2 / 21
Introduction
The governing equations
The velocity and pressure of the ice are widely assumed to satisfy the Stokes problem
−∇ ·[2µ(u)ε(u)
]+∇p = ρig
∇ · u = 0
with g = (0, 0,−g). The nonlinear viscosity is given by Glen’s law
µ(u) =1
2A−
1n (T )ε
1n−1
e
where, usually, n = 3, and
εij =1
2
(∂ui∂xj
+∂uj∂xi
), εe =
√1
2
∑ij
εijεij .
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 3 / 21
Introduction
The First Order (FO) model
Under certain assumptions, a simplified model (Blatter 1995, Pattyn 2003) can bederived for the horizontal velocity u = (ux , uy ). In particular, the assumptions are
negligible bridging effects
vertical derivatives of horizontal velocities much larger than horizontal derivatives ofvertical velocity
Under these assumptions, εxz ' 12∂ux/∂z , and εyz ' 1
2∂uy/∂z , and the Stokes problem is
simplified to
−∇ · σ = −ρig∇s (1)
where s is the surface elevation, and
σ = 2µ(u)
[2εxx + εyy εxy εxz
εxy εxx + 2εyy εxz
]The constitutive law is still Glen’s law, where εe can be written as
2ε2e = ε2
xx + ε2yy + εxxεyy + ε2
xy + ε2xz + ε2
yz
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 4 / 21
Introduction
Boundary conditions
For ice-air interface, we impose a stress-free condition
σn = 0.
For the ice-bedrock interface, we consider the ice-bedrock sliding condition
σn + βu = 0.
where β > 0 is a friction coefficient.
Obvious question: how to choose β?
Constant field β = β(x)
pro: linear BC
con: 2D field to estimate,problem dependent, does notevolve in time
Functional form β = β(u, p)
pro: potentially general (evolvesin time with u)
con: nonlinear BC, needs goodfunctional form guess
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 5 / 21
Introduction
Sliding laws
One reasonable class of β is (Schoof 2004)
β = τc f (u)1
u(2)
where u = |u|. Some requirements on f :
continuous
non-decreasing
f (0) = 0
limu→∞ f (u) = 1
τc is the yield stress. Typically, τc = µfN, where N = ρigH + ρwgzb − pw is the effectivepressure and µf a scalar friction coefficient.
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 6 / 21
Introduction
Examples:
f (u) = uqN r f (u) =
(u
u + λANn
)q
with q, r > 0, A as in Glen’s law, and λ geometric parameter for bed roughness (ratiobetween maximum bed bump length and maximum bed bump slope).
We consider the latter, which gives the boundary condition
σn + µfN
(u
u + λANn
)qu
u= 0 (3)
Note: Schoof suggests q = 1/n, with n as in Glen’s law.
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 7 / 21
Introduction
Note: the field N (or, equivalently, pw ) is still unknown. Two possibilities:
Couple ice problem with a 2D subglacial hydrlogy model for N
Use a surrogate of field N.
For the latter, one can use the approximation (Bueler et al., 2008)
pw = αρigHh
hmax
where h is the thickness of water film between ice and bedrock.
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Introduction
Figure: Basal friction coefficient: 2D field estimation (left), parameter fitting (right).
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Distributed subglacial hydrology
Subglacial hydrology
In order to have a more realistic N, one needs to consider subglacial hydrology, and add a2D problem on Γb. For instance (Hewitt, Schoof, Werder, 2012)
∂h
∂t+∇ · q = ω
∂h
∂t=
hr − h
lr|ub| − AhNn
q = −k0h3
µw∇pw
pw = ρigH + ρwgzb − N := p0 − N
with h water layer thickness, k0 transmissivity constant, hr , lr bed bumps characteristiclength/height, ω water source, and n = 3 (as in Glen’s law).Assuming h known, one can derive an equation for N
−∇ ·(k0h
3
µw∇(N − p0)
)+ AhNn =
hr − h
lr|ub| − ω (4)
This is endowed with zero porewater pressure boundary condition, hence N = p0.
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 10 / 21
Distributed subglacial hydrology
The well posedness of the hydrology follows from convex optimization. Define
J(N) =1
2
∫k0h
3
µw|∇N|2 +
1
4
∫AhN4 −
∫k0h
3
µw∇p0 · ∇N −
∫ (hr − h
lr|ub| − ω
)N.
Note: since dim(Γb) = 2, if we ask N ∈ H1(Γb), we have N ∈ Lq(Γb), ∀q ∈ [1,∞)(Sobolev embedding theorem), so all the integrals are well defined, provided h ∈ L∞(Γb),ω ∈ L2(Γb).
J is differentiable, so N is a weak solution iff ∂J(N) = 0. Since J is coercive and strictlyconvex, such point (the minimizer) exists and is unique, provided that h ≥ h0 > 0.Moreover,
‖N2 − N1‖H1 ≤ C(data)‖u1 − u2‖L
43
(5)
Note: With a little more work, using Stampacchia’s method, we can even show thatN ∈ L∞(Γb).
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 11 / 21
Distributed subglacial hydrology
The coupled problem
The coupled ice-hydrology problem reads−∇ · σ = −ρig∇s
σn + µfN
(u
u + λANn
)qu
u= 0
−∇ ·(k0h
3
µw∇(N − p0)
)+ AhNn =
hr − h
lr|ub|+ ω
in Ω
on Γb
in Γb
(6)
existence of a solution?
uniqueness?
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 12 / 21
Distributed subglacial hydrology
Existence/Uniqueness
Lemma
There are A,B,C > 0 independent on N, u such that
‖u‖W
1, 43 (Ω)≤ A
‖u1 − u2‖W
1, 43 (Ω)≤ B‖N1 − N2‖H1(Γb)
‖N2 − N1‖H1(Γb) ≤ C‖u1 − u2‖L
43 (Γb)
LetG : N → u(N), H : u→ N(u), T = G H.
The inequalities in the lemma ensure that
G is continuous
H is continuous
T is continuous (by composition)
v = T (u) is uniformly bounded
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 13 / 21
Distributed subglacial hydrology
To use Schaefer’s theorem and prove existence, we need T to be compact. If
un u in W 1, 43 (Ω), letting vn, v be the trace of un, u on Γb, we have
un u in W 1, 43 (Ω)⇒ vn vn in W
14, 4
3 (Γb)⇒ vn → vn in L43 (Γb)
where the last implication follows from Sobolev’s embedding theorem. By virtue of the
previous lemma, we conclude T (un)→ T (u) in W 1, 43 (Ω). Therefore, using Schaefer’s
theorem,
Theorem
There is at least one solution to the coupled problem (6)
Combining the bounds in the previous lemma, one gets
‖T (u1)− T (u2)‖ ≤ γ‖u1 − u2‖.
If γ < 1, T is a contraction, so we have uniqueness.
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 14 / 21
Distributed subglacial hydrology
Unsteady
We add the cavities equation to evolve h:
−∇ · σ = −ρig∇s
σn + µfN
(u
u + λANn
)qu
u= 0
−∇ ·(k0h
3
µw∇(N − p0)
)+ AhNn =
hr − h
lr|ub|+ ω
∂h
∂t=
hr − h
lr|ub| − AhNn
(7)
Using Stampacchia’s method, we can show that u and N are bounded in L∞. In particular
‖u‖∞ ≤ C1, ‖N‖∞ ≤ C2|h|h3
0
‖u‖∞.
so that |∂th| ≤ Q(|h|, h0), with Q rational function. If h0 + ε < h(x , 0) < M − ε, then,for t ∈ (0, δ) we have |h| ∈ (h0,M).
Question: how big can δ be?
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 15 / 21
Distributed subglacial hydrology
A model with basal melting
We may consider also water formation due to melting:
m =1
L(G − τb · u) (8)
with L latent heat, G (net) geothermal flux and τb ice basal stress. Using sliding law,τb = −βu. The hydrology problem becomes
∂h
∂t+∇ · q = ω + 1
ρw L(G − β|ub|2)
∂h
∂t= 1
ρiL(G − β|ub|2) +
hr − h
lr|ub| − AhNn
q = −k0h3
µw∇pw = −k0h
3
µw∇(p0 − N)
The equation for N becomes
−∇ ·(k0h
3
µw∇(N − p0)
)+ AhNn − ρ
Lβ(N)|u|2 =
ρ
LG +
hr − h
lr|ub| − ω (9)
where ρ = 1/ρi − 1/ρw .
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 16 / 21
Distributed subglacial hydrology
We can define the functional
J(N) =1
2
∫k0h
3
µw|∇N|2dx +
1
4
∫AhN4dx − ρ
L
∫ (∫ N
0
β(s)ds
)|u|2dx
−∫
k0h3
µw∇p0 · ∇Ndx −
∫ (ρ
LG +
hr − h
lr|ub| − ω
)Ndx .
Problem: J is not convex. The non convex term is continuous, and∣∣∣∣∫ (∫ N
0
β(s)ds
)|u|2dx
∣∣∣∣ ≤ ‖u‖L 43 (Γb)‖N‖2
L8(Γb) ≤ C‖u‖L
43 (Γb)‖N‖2
L2(Γb).
Therefore, J is still coercive, so that
Existence
The hydrology problem (9) has at least one solution
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 17 / 21
Distributed subglacial hydrology
The lack of convexity does not allow to immediately get uniqueness. However, we canshow that any two solutions N1,N2 must satisfy
Ah0‖N2 − N1‖2L4(Γb) ≤
ρ
L‖u‖
L43 (Γb)
. (10)
Convexity is recovered for small data. In particular, we still have convexity and continuityof solution w.r.t. data if
k0θ3
µw≥ (1 + c2
p )C8,4ρ
L‖u‖
L43 (Γb)
which however is too strict of an assumption for real data.
Without well posedness (in particular, continuity of N w.r.t u), we cannot prove existenceof solutions for the coupled problem.
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 18 / 21
Distributed subglacial hydrology
Steady hydrology
Recall the hydrology problem
∂h
∂t+∇ · q = ω +
m
ρw∂h
∂t=
m
ρi+
hr − h
lr|ub| − AhNn
q = −k0h3
µw∇pw
pw = ρigH − N := pi − N
If we assume ∂h∂t
= 0, then we have
h =hr/lr + G/(Lρi ) + β(N)|u|2/(Lρi )
|ub|/lr + ANn
And the equation for N becomes
−∇ ·(k0h(N)3
µw∇(N − pi )
)+β(N)|u|2
ρwL= −ω − G
ρwL(11)
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 19 / 21
Distributed subglacial hydrology
pro: no need to assume h known (although we assumed ∂h/∂t = 0)
con: diffusion coefficient not guaranteed to be positive
Idea: add box constraint for N (Schoof, Hewitt, Werder, 2012). Physical argumentssuggest 0 ≤ N ≤ p0. We then turn (11) into a variational inequality. Wherever N = 0 orN = p0, (11) is regarded as an equation for h (not interesting for ice coupling).
Other issue: h = h(N, |ub|) is singular if N → 0 and |ub| → 0. However, if we assume|ub| ≥ um > 0, and we use box constraints for N, then the steady hydrology problem iswell posed.
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 20 / 21
Conclusions
Looking ahead
sliding laws (e.g., regularized Coulomb) are appealing since they can evolve with thesolution
need to identify the parameters in the sliding law
regularized Coulomb requires to model effective pressure N
proved well posedness of quasi-static coupled FO-hydrology problem (at least forsmall data)
todo extend well-posedness do time-dependent. Under what conditions we have wellposedness for arbitrary large times?
todo Numerically solve the coupled FO-hydrology problem (Albany, ongoing).
doubt Is the case with melting intrinsically ill-posed?
Luca Bertagna (SNL) Ice-hydrology coupling Atlanta, March 2nd 2017 21 / 21