Numerical modeling of ice-sheet dynamics supervizor: Prof. RNDr. Zdeněk Martinec, DrSc. Department of Geophysics, Faculty of Mathematics and Physics, Charles University in Prague Research Institute of Geodesy, Topography and Cartography, Zdiby.6.
Overview Ice-sheets and their dynamics The Shallow Ice approximation The SIA-I iterative algorithm Numerical benchmarks ISMIP-HOM A exp. - steady-state ISMIP-HOM F exp. - prognostic EISMINT - Greenland Ice Sheet models
Scheme of an ice sheet
Transport processes in an ice sheet
Simplified ice sheet model z Atmosphere Cold ice F (x,t) = s Temp. ice F (x,t) = CTS F (x,t) = b Lithosphere x,y Cold-ice zone (pure ice only) Temperate-ice zone (liquid water present) Free surface and glacier base represented by differentiable surfaces
Modelled problem - cold ice zone Stokes flow problem divτ +ρ g = Rheology τ ij = pδ ij +σ ij Equation of continuity (Constant homogeneous ice density) div v = Energy balance Heat transport equation ρc(t)ṫ = div(κ(t)gradt)+τ. ε σ ij = η ε ij ε ij = ( vi + v ) j x j x i η = A(T ) 3 ε 3 II ( ) A(T Q ) = A exp R(T + T ) εkl ε kl ε II =
Boundary conditions Free surface Kinematic boundary condition fs + v gradfs = a s t Dynamic boundary condition Traction-free τ n s = Surface temperature T = T s ( x, t) Accumulation-ablation function Glacier base Kinematic boundary condition f ( x, t) given b Dynamic boundary conditions Frozen-bed conditions, T < T M : v = Sliding law T = T M : v = g(τ n) Geothermal heat flux q + = q geo ( x, t) a s = a s ( x, t)
Scaling Approximation plausibility and motivation Glaciers and ice sheets are typically very flat features, with the aspect (height:horizontal) ratio less than for small valley glaciers and one order less for big ice sheets ( )
Scaling Approximation plausibility and motivation Natural scaling may be introduced: kinematic quantities (x, x, x 3) = ([L] x,[l] x,[h] x 3) (v, v, v 3) = ([V ]ṽ,[v ]ṽ,[v ]ṽ 3) (f s(x, x ), f b (x, x )) = [H]( f s( x, x ), f b ( x, x )) ε = [H] [L] = [V ] [V ] dynamic quantities - p = ρg[h] p (σ 3,σ 3) = ρg[h]( σ 3, σ 3) (σ,σ,σ ) = ρg[h]( σ, σ, σ )
Scaling Approximation plausibility and motivation Natural scaling may be introduced: kinematic quantities (x, x, x 3) = ([L] x,[l] x,[h] x 3) (v, v, v 3) = ([V ]ṽ,[v ]ṽ,[v ]ṽ 3) (f s(x, x ), f b (x, x )) = [H]( f s( x, x ), f b ( x, x )) ε = [H] [L] = [V ] [V ] dynamic quantities - Shallow Ice Approximation SIA p = ρg[h] p (σ 3,σ 3) = ερg[h]( σ 3, σ 3) (σ,σ,σ ) = ε ρg[h]( σ, σ, σ )
Scaling Approximation plausibility and motivation Standard scaling perturbation series constructed: all field unknowns (ṽ i, σ ij, f s, f b ) are expressed by a power series in the aspect ratio ε q = q () +ε q () +ε q () +... and inserting these expressions into governing equations (eq. of motion, continuity, rheology) leads to separation of these equations according to the order of ε. (Implicit assumption: not only q is now scaled to unity but also its gradient (???))
Scaling Approximation plausibility and motivation Keeping only lowest-order terms, we arrive at the Shallow Ice Approximation, which can be explicitly solved where ( ) (x, x ) p () (, x 3) = f () s ( ) x 3 σ () 3 σ () 3 (, x3) = f () s ( ) x (f () s ( ) x 3) (, x3) = f () s ( ) x (f () s ( ) x 3)
Scaling Approximation plausibility and motivation Also the velocities may be expressed semi-analytically as x3 v () (, x3) = v() ( ) sl + A(T (, x 3)) f () b ( ) v () (, x3) = v() ( ) sl + v 3 from equation of continuity x3 f () b ( ) A(T (, x 3)) ( x3 v () 3 (, x3) = v() 3 ( ) sl f () b ( ) v () ( ( σ () 3 σ () 3 () ) +σ 3 σ () 3 (, x 3)dx 3 () ) +σ 3 σ () 3 (, x 3)dx 3 ) (, x 3)+ v() (, x 3) x x dx 3
Problems of SIA Zeroth order model looses validity in regions where higher-order terms become important: Regions of high curvature of the surface
Problems of SIA Zeroth order model looses validity in regions where higher-order terms become important: Regions of high curvature of the surface Regions where a-priori dynamic scaling assumptions are violated (floating ice SSA, ice streams)
Problems of SIA Zeroth order model looses validity in regions where higher-order terms become important: Regions of high curvature of the surface Regions where a-priori dynamic scaling assumptions are violated (floating ice SSA, ice streams) Figure: RADARSAT Antarctic Mapping Project
Solution? Higher-order models continuation of the expansion procedure and solving for higher-order corrections (Pattyn F., Rybak O.) Full Stokes Solvers Finite Elements Methods (Elmer - Gagliardini O.), Spectral methods (Hindmarsch R.) PROBLEM is the speed of full-stokes and higher-order techniques, the computational demands disable usage of these techniques in large-scale evolutionary models
Any ideas? Aim: Find an "intermediate" technique that would exploit the scaling assumptions of flatness of ice but would provide "better" solution than SIA: KEY IDEA: Apply the scaling assumptions of smallness not on the particular stress components but only on their deviations from the full-stokes solution
SIA-I u n δ u n+ u n+ = u n +ǫ δu n+ u n+ = ( ǫ ) u n+ +ǫ u n+ u n+ v n+
Ice-Sheet Model Intercomparison Project - Higher Order Models (ISMIP-HOM) - experiment A f s(x, x ) = x tanα, α =.5, f b (x, x ) = f s(x, x ) + 5 sin(ωx ) sin(ωx ) Ice considered isothermal with ω = π. [L] Boundary conditions Free surface Frozen bed At the sides, periodic boundary conditions are prescribed: v(x,, x 3) = v(x,[l], x 3) v(, x, x 3) = v([l], x, x 3)
ISMIP - HOM experiment A 8 9 v x (m a - ) at f s 6 4 σ xz [kpa] at f b 8 7 6 5 4..4.6.8 3..4.6.8 x x Figure: Pattyn F., Ondřej ISMIP-HOM Souček Ph.D. results defense preliminary report
ISMIP - HOM experiment A 8 9 v x (m a - ) at f s 6 4 σ xz [kpa] at f b 8 7 6 5 4..4.6.8 3..4.6.8 x x Figure: Pattyn F., ISMIP-HOM results preliminary report
ISMIP - HOM experiment A Computational details Resolution: 4 4 4 Number of iterations: ε = 4 iter., ε = iter. Each iter. step took approximately. s at Pentium 4, 3..GHz For comparison Elmer (Gagliardini) Full-Stokes solver 4 CPU s
ISMIP-HOM experiment F Experiment description - ice slab flowing downslope (3 ) over a Gaussian bump Linear rheology! Output: Steady state free surface profile and velocities Comparison with ISMIP-HOM full-stokes finite-element solution (Olivier Gagliardini computing by Elmer)
ISMIP-HOM experiment F Free surface (left - Elmer, right- SIA-I)
ISMIP-HOM experiment F Numerical performance Surface velocities Figure: Gagliardini & Zwinger, 8
ISMIP-HOM experiment F Surface velocities (left-elmer, right-sia-i) Numerical performance CPU-time/time-step [s] 8 6 4 time_demands.dat 4 6 8 4 6 8 Iteration
Extension for non-linear rheology We take setting from ISMIP-HOM experiment A (L=8) - flow of an inclined ice slab over a sinusoidal bump, periodically elongated Impossible to compare with evolution in Elmer (too CPU time demanding) Time demands of the SIA-I algorithm practically the same as for the linear case!!! Let s compare only velocities at particular time instants Evolution of free surface, steady state
Extension for non-linear rheology t = 5a (left - Elmer, right - SIA-I) t = a (left - Elmer, right - SIA-I)
EISMINT benchmark - Greenland Ice Sheet Models Paleoclimatic simulation Prognostic experiment - global warming scenario
EISMINT Greenland - Paleoclimatic experiment Two glacial cycles (cca 5 ka) Temperature + sea-level forcing based on ice-core δ 8 O isotope record T (K) 6 4 - -4-6 -8 - - -4-5 - -5 - -5 time (ka) Sea level (m) - -4-6 -8 - - -4-5 - -5 - -5 time (ka)
3 EISMINT Greenland - Paleoclimatic experiment Surface topography (km), age = 5 ka.5. y ( 3 km).5 3..5 3...5..5 x (Ph.D. 3 km) defense
3 EISMINT Greenland - Paleoclimatic experiment Surface topography [km], age = 35 ka.5. y [ 3 km].5 3..5 3...5..5 x [Ph.D. 3 km] defense
EISMINT Greenland - Paleoclimatic experiment Surface topography (km), age = 5 ka.5. y ( 3 km).5 3..5...5..5 x (Ph.D. 3 km) defense
EISMINT Greenland - Paleoclimatic experiment Surface topography (km), age = 5 ka.5. y ( 3 km).5 3..5...5..5 x (Ph.D. 3 km) defense
3 EISMINT Greenland - Paleoclimatic experiment Surface topography (km), age = 75 ka.5. 3 y ( 3 km).5..5 3...5..5 x (Ph.D. 3 km) defense
3 EISMINT Greenland - Paleoclimatic experiment Surface topography (km), age = a.5. 3 y ( 3 km).5..5 3...5..5 x (Ph.D. 3 km) defense
EISMINT Greenland - Paleoclimatic experiment Glaciated area ( 6 km ).6.4..8.6.4. -5 - -5 - -5 time (ka) Ice-sheet volume ( 6 km 3 ) 5 4.5 4 3.5 3.5.5-5 - -5 - -5 time (ka)
EISMINT Greenland - Paleoclimatic experiment (Huybrechts, 998) Glaciated area ( 6 km ).6.4..8.6.4. -5 - -5 - -5 time (ka)
Prognostic experiment Model response to an artificial warming scenario temperature increase by.35 C per year for the first 8 years (total.8 C increase) by.7 per year C for the remaining 4 years (.74 C) In total temperature increase of 3.54 C within 5 years
3 Prognostic experiment Surface topography (km), t = a.5. y ( 3 km).5..5...5..5 x ( 3 km)
3 Prognostic experiment Surface topography (km), t = a Initial accumulation ablation function (m a ).6.5.5.8.4..8.4.4.4.4.6.8. 3..8.4.4.8..6.6..8.. 3.6 3..8.4.6. 4 3.6.4.8 3...6.4.8.4 y ( 3 km).5. y ( 3 km).5..8 4.8 4 5. 8 8.4 9.6 3.6 9. 8.8 3. 4.4.4 7.6.4.8.4 5. 6.8 4.4 4.8 7. 6.8 3.6 4 6.4 5.6 3..4.4.6..8 4 4.4 5.6 3.6 3. 3. 3.6 4.4 4.4.8.6.8.4.8..6.4. 6.8.4 5. 4.8 6 5. 5.6 4.8.5.5 5.6 6.8 6 6.4.6 7.. 8.8 8.4 9.6 7.6 8 9. 4.8 4.4 5. 4.8.8.4 8.4 7.6 7. 6.8 6.4 8 5. 5.6 6 4.8.4 3.6 3..4.4 4.4.8 3. 3.6 4...5..5 x ( 3 km).8.4.8.8..8.6..6...5..5 x ( 3 km)
3 3 Prognostic experiment Surface topography (km), t = a Surface topography (km), t = 5 a Initial accumulation ablation function (m a ).6.5.5.5.4.4.6.8..8.4..8.4.4 3..8.4.4.8..6.6..8... 3.6 3..8.4.6. 4 3.6.4.8 3...6.4.8.4 y ( 3 km).5. y ( 3 km).5. y ( 3 km).5..8 4.8 4 5. 8 8.4 9.6 3.6 9. 8.8 3. 4.4.4 7.6.4.8.4 5. 6.8 4.4 4.8 7. 6.8 3.6 4 6.4 5.6 3..4.4.6..8 4 4.4 5.6 3.6 3. 3. 3.6 4.4 4.4.8.6.8.4.8..6.4. 6.8.4 5. 4.8 6 5. 5.6 4.8.5.5.5 5.6 6.8 6 6.4.6 7.. 8.8 8.4 9.6 7.6 8 9. 4.8 4.4 5. 4.8.8.4 8.4 7.6 7. 6.8 6.4 8 5. 5.6 6 4.8.4 3.6 3..4.4 4.4.8 3. 3.6 4...5..5 x ( 3 km)...5..5 x ( 3 km).8.4.8.8..8.6..6...5..5 x ( 3 km)
3 3 Prognostic experiment Surface topography (km), t = a Surface topography (km), t = 5 a Topography difference (m) y ( 3 km).5..5..5...5..5 y ( 3 km).5..5..5...5..5 y ( 3 km).5..5..5...5..5 6 6 4 8 7 6 5 4 3 3 4 5 6 7 8 9 x ( 3 km) x ( 3 km) x ( 3 km)
Thank you for your attention!