|
| 1 | +import pytest |
| 2 | +import numpy as np |
| 3 | +import sympy |
| 4 | +from sympy import lambdify, simplify, diff |
| 5 | +import firedrake |
| 6 | +from firedrake import ( |
| 7 | + as_vector, |
| 8 | + max_value, |
| 9 | + Constant, |
| 10 | + Function, |
| 11 | + derivative, |
| 12 | + NonlinearVariationalProblem, |
| 13 | + NonlinearVariationalSolver, |
| 14 | + DirichletBC, |
| 15 | +) |
| 16 | +from icepack2 import constants |
| 17 | +from icepack2.model.minimization import ( |
| 18 | + viscous_power, |
| 19 | + friction_power, |
| 20 | + calving_terminus, |
| 21 | + momentum_balance, |
| 22 | + ice_shelf_momentum_balance, |
| 23 | +) |
| 24 | + |
| 25 | + |
| 26 | +sparams = { |
| 27 | + "solver_parameters": { |
| 28 | + "snes_type": "newtonls", |
| 29 | + "snes_max_it": 200, |
| 30 | + "snes_linesearch_type": "nleqerr", |
| 31 | + "ksp_type": "gmres", |
| 32 | + "pc_type": "lu", |
| 33 | + "pc_factor_mat_solver_type": "mumps", |
| 34 | + }, |
| 35 | +} |
| 36 | + |
| 37 | + |
| 38 | +@pytest.mark.parametrize("degree", (1, 2)) |
| 39 | +def test_composite_rheology_floating(degree): |
| 40 | + lx, ly = 20e3, 20e3 |
| 41 | + Lx, Ly = Constant(lx), Constant(ly) |
| 42 | + h0, dh = Constant(500.0), Constant(100.0) |
| 43 | + u_inflow = Constant(100.0) |
| 44 | + |
| 45 | + ρ_I = Constant(constants.ice_density) |
| 46 | + ρ_W = Constant(constants.water_density) |
| 47 | + g = Constant(constants.gravity) |
| 48 | + |
| 49 | + # Use a combination of two different rheology exponents. |
| 50 | + n_1 = firedrake.Constant(1.0) |
| 51 | + n_3 = firedrake.Constant(constants.glen_flow_law) |
| 52 | + |
| 53 | + τ_c = Constant(0.1) |
| 54 | + ε_c = Constant(0.01) |
| 55 | + A_1 = ε_c / τ_c ** n_1 |
| 56 | + A_3 = ε_c / τ_c ** n_3 |
| 57 | + |
| 58 | + # This is an exact solution for the velocity of a floating ice shelf with |
| 59 | + # constant temperature and linearly decreasing thickness. See Greve and |
| 60 | + # Blatter for the derivation. |
| 61 | + def exact_δu(x, n, A): |
| 62 | + ρ = ρ_I * (1 - ρ_I / ρ_W) |
| 63 | + h = h0 - dh * x / Lx |
| 64 | + P = ρ * g * h / 4 |
| 65 | + P_0 = ρ * g * h0 / 4 |
| 66 | + δP = ρ * g * dh / 4 |
| 67 | + return Lx * A * (P_0 ** (n + 1) - P ** (n + 1)) / ((n + 1) * δP) |
| 68 | + |
| 69 | + errors, mesh_sizes = [], [] |
| 70 | + k_min, k_max, num_steps = 5 - degree, 8 - degree, 9 |
| 71 | + for nx in np.logspace(k_min, k_max, num_steps, base=2, dtype=int): |
| 72 | + mesh = firedrake.RectangleMesh(nx, nx, lx, ly, diagonal="crossed") |
| 73 | + x, y = firedrake.SpatialCoordinate(mesh) |
| 74 | + |
| 75 | + # Make some function spaces. |
| 76 | + cg = firedrake.FiniteElement("CG", "triangle", degree) |
| 77 | + dg = firedrake.FiniteElement("DG", "triangle", degree - 1) |
| 78 | + Q = firedrake.FunctionSpace(mesh, cg) |
| 79 | + V = firedrake.VectorFunctionSpace(mesh, cg) |
| 80 | + Σ = firedrake.TensorFunctionSpace(mesh, dg, symmetry=True) |
| 81 | + Z = V * Σ |
| 82 | + z = Function(Z) |
| 83 | + z.sub(0).assign(Constant(u_inflow, 0)) |
| 84 | + |
| 85 | + expr = u_inflow + exact_δu(x, n_1, A_1) + exact_δu(x, n_3, A_3) |
| 86 | + u_exact = Function(V).interpolate(as_vector((expr, 0))) |
| 87 | + |
| 88 | + h = Function(Q).interpolate(h0 - dh * x / Lx) |
| 89 | + |
| 90 | + # Make the boundary conditions. |
| 91 | + inflow_ids = (1,) |
| 92 | + outflow_ids = (2,) |
| 93 | + side_wall_ids = (3, 4) |
| 94 | + |
| 95 | + inflow_bc = DirichletBC(Z.sub(0), Constant((u_inflow, 0)), inflow_ids) |
| 96 | + side_wall_bc = DirichletBC(Z.sub(0).sub(1), 0, side_wall_ids) |
| 97 | + bcs = [inflow_bc, side_wall_bc] |
| 98 | + |
| 99 | + # Make the model specification and solver. |
| 100 | + u, M = firedrake.split(z) |
| 101 | + fields = {"velocity": u, "membrane_stress": M, "thickness": h} |
| 102 | + boundary_ids = {"outflow_ids": outflow_ids} |
| 103 | + |
| 104 | + L = ( |
| 105 | + viscous_power(**fields, flow_law_exponent=n_1, flow_law_coefficient=A_1) + |
| 106 | + viscous_power(**fields, flow_law_exponent=n_3, flow_law_coefficient=A_3) + |
| 107 | + ice_shelf_momentum_balance(**fields, **boundary_ids) |
| 108 | + ) |
| 109 | + |
| 110 | + F = derivative(L, z) |
| 111 | + qdegree = max(8, degree ** constants.glen_flow_law) |
| 112 | + pparams = {"form_compiler_parameters": {"quadrature_degree": qdegree}} |
| 113 | + problem = NonlinearVariationalProblem(F, z, bcs, **pparams) |
| 114 | + solver = NonlinearVariationalSolver(problem, **sparams) |
| 115 | + |
| 116 | + # Solve the problem using a continuation method -- step the flow law |
| 117 | + # exponent from 1 to 3 in small intervals. |
| 118 | + num_steps = 5 |
| 119 | + for exponent in np.linspace(1.0, constants.glen_flow_law, num_steps): |
| 120 | + n_3.assign(exponent) |
| 121 | + solver.solve() |
| 122 | + |
| 123 | + u, M = z.subfunctions |
| 124 | + error = firedrake.norm(u - u_exact) / firedrake.norm(u_exact) |
| 125 | + δx = mesh.cell_sizes.dat.data_ro.min() |
| 126 | + mesh_sizes.append(δx) |
| 127 | + errors.append(error) |
| 128 | + print(".", end="", flush=True) |
| 129 | + |
| 130 | + log_mesh_sizes = np.log2(np.array(mesh_sizes)) |
| 131 | + log_errors = np.log2(np.array(errors)) |
| 132 | + slope, intercept = np.polyfit(log_mesh_sizes, log_errors, 1) |
| 133 | + print(f"degree {degree}: log(error) ~= {slope:g} * log(dx) {intercept:+g}") |
| 134 | + assert slope > degree + 0.9 |
| 135 | + |
| 136 | + |
| 137 | +def test_manufactured_solution(): |
| 138 | + L, ρ_I, ρ_W, g = sympy.symbols("L ρ_I ρ_W g", real=True, positive=True) |
| 139 | + A_1, A_3 = sympy.symbols("A_1 A_3", real=True, positive=True) |
| 140 | + n_1, n_3 = sympy.symbols("n_1 n_3", integer=True, positive=True) |
| 141 | + |
| 142 | + def strain_rate(M, A_1, A_3, n_1, n_3): |
| 143 | + return A_1 * (M / 2)**n_1 + A_3 * (M / 2)**n_3 |
| 144 | + |
| 145 | + def sliding_velocity(τ, K_1, K_3, m_1, m_3): |
| 146 | + return -K_1 * τ**m_1 - K_3 * τ**m_3 |
| 147 | + |
| 148 | + def stress_balance(x, h, s, M, τ): |
| 149 | + return diff(h * M, x) + τ - ρ_I * g * h * diff(s, x) |
| 150 | + |
| 151 | + def boundary_condition(x, u, h, s, M): |
| 152 | + d = (s - h).subs(x, L) |
| 153 | + τ_c = (ρ_I * g * h**2 - ρ_W * g * d**2) / 2 |
| 154 | + return (h * M - τ_c).subs(x, L) |
| 155 | + |
| 156 | + # Create the thickness and surface elevation |
| 157 | + x = sympy.symbols("x", real=True) |
| 158 | + h0, δh = sympy.symbols("h0 δh", real=True, positive=True) |
| 159 | + h = h0 - δh * x / L |
| 160 | + |
| 161 | + s0, δs = sympy.symbols("s0 δs", real=True, positive=True) |
| 162 | + s = s0 - δs * x / L |
| 163 | + |
| 164 | + # The variable `β` is the fraction of the driving stress that membrane |
| 165 | + # stress divergence will take up; this is chosen to eliminate some extra |
| 166 | + # terms and make the rest of the algebra simpler. See the website also. |
| 167 | + h_L = h0 - δh |
| 168 | + s_L = s0 - δs |
| 169 | + β = δh / δs * (ρ_I * h_L ** 2 - ρ_W * (s_L - h_L) ** 2) / (ρ_I * h_L**2) |
| 170 | + |
| 171 | + ρ = β * ρ_I * δs / δh |
| 172 | + P = ρ * g * h / 4 |
| 173 | + δP = ρ * g * δh / 4 |
| 174 | + P0 = ρ * g * h0 / 4 |
| 175 | + |
| 176 | + M = ρ * g * h / 2 |
| 177 | + τ = -(1 - β) * ρ_I * g * h * δs / L |
| 178 | + |
| 179 | + # The exact velocity field looks similar to the floating case by using a |
| 180 | + # careful choice of `β`. |
| 181 | + u_0 = sympy.symbols("u_0", real=True, positive=True) |
| 182 | + δu_1 = L * A_1 * (P0 ** (n_1 + 1) - P ** (n_1 + 1)) / ((n_1 + 1) * δP) |
| 183 | + δu_3 = L * A_3 * (P0 ** (n_3 + 1) - P ** (n_3 + 1)) / ((n_3 + 1) * δP) |
| 184 | + u = u_0 + δu_1 + δu_3 |
| 185 | + |
| 186 | + # Make the linear part of the rheology a much smaller effect than the |
| 187 | + # nonlinear part -- at a stress of 100kPa, the nonlinear rheology gives |
| 188 | + # a strain rate of 10 m/yr/km, while the linear part only 1 m/yr/km. |
| 189 | + ε_1 = 0.001 |
| 190 | + ε_3 = 0.01 |
| 191 | + τ_c = 0.1 |
| 192 | + |
| 193 | + values = { |
| 194 | + u_0: 100, |
| 195 | + h0: 500, |
| 196 | + δh: 100, |
| 197 | + s0: 150, |
| 198 | + δs: 90, |
| 199 | + L: 20e3, |
| 200 | + A_1: ε_1 / τ_c, |
| 201 | + A_3: ε_3 / τ_c ** constants.glen_flow_law, |
| 202 | + ρ_I: constants.ice_density, |
| 203 | + ρ_W: constants.water_density, |
| 204 | + n_1: 1, |
| 205 | + n_3: constants.glen_flow_law, |
| 206 | + g: constants.gravity, |
| 207 | + } |
| 208 | + |
| 209 | + # Check the momentum conservation law and boundary condition. |
| 210 | + τ_b = lambdify(x, τ.subs(values), "numpy") |
| 211 | + τ_d = lambdify(x, (-ρ_I * g * h * diff(s, x)).subs(values), "numpy") |
| 212 | + τ_m = lambdify(x, simplify(diff(h * M, x)).subs(values), "numpy") |
| 213 | + xs = np.linspace(0, values[L], 21) |
| 214 | + |
| 215 | + tolerance = 1e-8 |
| 216 | + assert abs(boundary_condition(x, u, h, s, M).subs(values)) < tolerance |
| 217 | + |
| 218 | + stress_norm = np.max(np.abs(τ_d(xs))) |
| 219 | + assert np.max(np.abs(τ_m(xs) + τ_b(xs) + τ_d(xs))) < tolerance * stress_norm |
| 220 | + |
| 221 | + # Check the constitutive relation. |
| 222 | + ε_u = lambdify(x, simplify(diff(u, x).subs(values)), "numpy") |
| 223 | + ε_M = lambdify( |
| 224 | + x, simplify(strain_rate(M, A_1, A_3, n_1, n_3).subs(values)), "numpy" |
| 225 | + ) |
| 226 | + assert np.max(np.abs(ε_u(xs) - ε_M(xs))) < tolerance * np.max(np.abs(ε_u(xs))) |
| 227 | + |
| 228 | + # Check the sliding law. |
| 229 | + α = 0.1 |
| 230 | + m_1 = 1 |
| 231 | + m_3 = constants.glen_flow_law |
| 232 | + K_1 = α * u / abs(τ) ** m_1 |
| 233 | + K_3 = (u - K_1 * abs(τ) ** m_1) / abs(τ)**m_3 |
| 234 | + |
| 235 | + U = lambdify(x, simplify(u).subs(values), "numpy") |
| 236 | + U_b = lambdify( |
| 237 | + x, simplify(sliding_velocity(τ, K_1, K_3, m_1, m_3).subs(values)), "numpy" |
| 238 | + ) |
| 239 | + assert np.max(np.abs(U(xs) - U_b(xs))) < tolerance * np.max(np.abs(U(xs))) |
| 240 | + |
| 241 | + |
| 242 | +@pytest.mark.parametrize("degree", (1, 2)) |
| 243 | +def test_composite_rheology_grounded(degree): |
| 244 | + lx, ly = 20e3, 20e3 |
| 245 | + Lx, Ly = Constant(lx), Constant(ly) |
| 246 | + h0, dh = Constant(500.0), Constant(100.0) |
| 247 | + s0, ds = Constant(150.0), Constant(90.0) |
| 248 | + u_inflow = Constant(100.0) |
| 249 | + |
| 250 | + # See the previous test for explanations of all the setup. |
| 251 | + ρ_I = Constant(constants.ice_density) |
| 252 | + ρ_W = Constant(constants.water_density) |
| 253 | + g = Constant(constants.gravity) |
| 254 | + |
| 255 | + n_1 = firedrake.Constant(1.0) |
| 256 | + n_3 = firedrake.Constant(constants.glen_flow_law) |
| 257 | + |
| 258 | + m_1 = firedrake.Constant(1.0) |
| 259 | + m_3 = firedrake.Constant(constants.weertman_sliding_law) |
| 260 | + |
| 261 | + τ_c = Constant(0.1) |
| 262 | + ε_1 = Constant(0.001) |
| 263 | + ε_3 = Constant(0.01) |
| 264 | + A_1 = ε_1 / τ_c ** n_1 |
| 265 | + A_3 = ε_3 / τ_c ** n_3 |
| 266 | + |
| 267 | + h_L = h0 - dh |
| 268 | + s_L = s0 - ds |
| 269 | + β = dh / ds * (ρ_I * h_L**2 - ρ_W * (s_L - h_L)**2) / (ρ_I * h_L**2) |
| 270 | + |
| 271 | + def exact_δu(x, n, A): |
| 272 | + ρ = β * ρ_I * ds / dh |
| 273 | + h = h0 - dh * x / Lx |
| 274 | + P = ρ * g * h / 4 |
| 275 | + dP = ρ * g * dh / 4 |
| 276 | + P0 = ρ * g * h0 / 4 |
| 277 | + du = Lx * A * (P0 ** (n + 1) - P ** (n + 1)) / ((n + 1) * dP) |
| 278 | + return du |
| 279 | + |
| 280 | + errors, mesh_sizes = [], [] |
| 281 | + k_min, k_max, num_steps = 5 - degree, 8 - degree, 9 |
| 282 | + for nx in np.logspace(k_min, k_max, num_steps, base=2, dtype=int): |
| 283 | + mesh = firedrake.RectangleMesh(nx, nx, lx, ly, diagonal="crossed") |
| 284 | + x, y = firedrake.SpatialCoordinate(mesh) |
| 285 | + |
| 286 | + # Make some function spaces. |
| 287 | + cg = firedrake.FiniteElement("CG", "triangle", degree) |
| 288 | + dg = firedrake.FiniteElement("DG", "triangle", degree - 1) |
| 289 | + Q = firedrake.FunctionSpace(mesh, cg) |
| 290 | + V = firedrake.VectorFunctionSpace(mesh, cg) |
| 291 | + Σ = firedrake.TensorFunctionSpace(mesh, dg, symmetry=True) |
| 292 | + T = firedrake.VectorFunctionSpace(mesh, dg) |
| 293 | + Z = V * Σ * T |
| 294 | + z = Function(Z) |
| 295 | + z.sub(0).assign(Constant((u_inflow, 0))) |
| 296 | + |
| 297 | + # Make the exact velocity, thickness, surface elevation, and sliding |
| 298 | + # coefficients. |
| 299 | + u_expr = u_inflow + exact_δu(x, n_1, A_1) + exact_δu(x, n_3, A_3) |
| 300 | + u_exact = Function(V).interpolate(as_vector((u_expr, 0))) |
| 301 | + |
| 302 | + h = Function(Q).interpolate(h0 - dh * x / Lx) |
| 303 | + s = Function(Q).interpolate(s0 - ds * x / Lx) |
| 304 | + |
| 305 | + α = Constant(0.1) |
| 306 | + τ_expr = (1 - β) * ρ_I * g * h * ds / Lx |
| 307 | + K_1 = α * u_expr / τ_expr ** m_1 |
| 308 | + K_3 = (u_expr - K_1 * τ_expr ** m_1) / τ_expr ** m_3 |
| 309 | + |
| 310 | + # Create the boundary conditions. |
| 311 | + inflow_ids = (1,) |
| 312 | + outflow_ids = (2,) |
| 313 | + side_wall_ids = (3, 4) |
| 314 | + |
| 315 | + inflow_bc = DirichletBC(Z.sub(0), Constant((u_inflow, 0)), inflow_ids) |
| 316 | + side_wall_bc = DirichletBC(Z.sub(0).sub(1), 0, side_wall_ids) |
| 317 | + bcs = [inflow_bc, side_wall_bc] |
| 318 | + |
| 319 | + # Create the model specification and solvers. |
| 320 | + u, M, τ = firedrake.split(z) |
| 321 | + fields = { |
| 322 | + "velocity": u, |
| 323 | + "membrane_stress": M, |
| 324 | + "basal_stress": τ, |
| 325 | + "thickness": h, |
| 326 | + "surface": s, |
| 327 | + } |
| 328 | + |
| 329 | + rheology_1 = { |
| 330 | + "flow_law_exponent": n_1, |
| 331 | + "flow_law_coefficient": A_1, |
| 332 | + "sliding_exponent": m_1, |
| 333 | + "sliding_coefficient": K_1, |
| 334 | + } |
| 335 | + |
| 336 | + rheology_3 = { |
| 337 | + "flow_law_exponent": n_3, |
| 338 | + "flow_law_coefficient": A_3, |
| 339 | + "sliding_exponent": m_3, |
| 340 | + "sliding_coefficient": K_3, |
| 341 | + } |
| 342 | + |
| 343 | + boundary_ids = {"outflow_ids": outflow_ids} |
| 344 | + L = ( |
| 345 | + viscous_power(**fields, **rheology_1) + |
| 346 | + viscous_power(**fields, **rheology_3) + |
| 347 | + friction_power(**fields, **rheology_1) + |
| 348 | + friction_power(**fields, **rheology_3) + |
| 349 | + calving_terminus(**fields, **boundary_ids) + |
| 350 | + momentum_balance(**fields) |
| 351 | + ) |
| 352 | + |
| 353 | + F = derivative(L, z) |
| 354 | + qdegree = max(8, degree ** constants.glen_flow_law) |
| 355 | + pparams = {"form_compiler_parameters": {"quadrature_degree": qdegree}} |
| 356 | + problem = NonlinearVariationalProblem(F, z, bcs, **pparams) |
| 357 | + solver = NonlinearVariationalSolver(problem, **sparams) |
| 358 | + |
| 359 | + # Solve the problem using a continuation method. |
| 360 | + num_steps = 5 |
| 361 | + for exponent in np.linspace(1.0, constants.glen_flow_law, num_steps): |
| 362 | + n_3.assign(exponent) |
| 363 | + m_3.assign(exponent) |
| 364 | + solver.solve() |
| 365 | + |
| 366 | + u, M, τ = z.subfunctions |
| 367 | + error = firedrake.norm(u - u_exact) / firedrake.norm(u_exact) |
| 368 | + δx = mesh.cell_sizes.dat.data_ro.min() |
| 369 | + mesh_sizes.append(δx) |
| 370 | + errors.append(error) |
| 371 | + print(".", end="", flush=True) |
| 372 | + |
| 373 | + log_mesh_sizes = np.log2(np.array(mesh_sizes)) |
| 374 | + log_errors = np.log2(np.array(errors)) |
| 375 | + slope, intercept = np.polyfit(log_mesh_sizes, log_errors, 1) |
| 376 | + print(f"degree {degree}: log(error) ~= {slope:g} * log(dx) {intercept:+g}") |
| 377 | + assert slope > degree + 0.9 |
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