Coverage for pySDC/implementations/problem_classes/Quench.py: 77%

152 statements  

« prev     ^ index     » next       coverage.py v7.6.7, created at 2024-11-16 14:51 +0000

1import numpy as np 

2import scipy.sparse as sp 

3from scipy.sparse.linalg import spsolve, gmres 

4from scipy.linalg import inv 

5 

6from pySDC.core.errors import ProblemError 

7from pySDC.core.problem import Problem, WorkCounter 

8from pySDC.helpers import problem_helper 

9from pySDC.implementations.datatype_classes.mesh import mesh, imex_mesh 

10 

11 

12# noinspection PyUnusedLocal 

13class Quench(Problem): 

14 """ 

15 This is a toy problem [1]_ to emulate a magnet that has been cooled to temperatures where superconductivity is possible. 

16 However, there is a leak! Some point in the domain is constantly heated and when this has heated up its environment 

17 sufficiently, there will be a runaway effect heating up the entire magnet. 

18 This effect has actually lead to huge magnets being destroyed at CERN in the past and hence warrants investigation. 

19 

20 The model we use is a 1d heat equation with Neumann-zero boundary conditions, meaning this magnet is totally 

21 insulated from its environment except for the leak. 

22 We add a non-linear term that heats parts of the domain that exceed a certain temperature threshold as well as the 

23 leak itself. 

24 

25 The problem is discretised with finite difference in space and treated *fully-implicitly*. 

26 

27 Parameters 

28 ---------- 

29 Cv : float, optional 

30 Volumetric heat capacity. 

31 K : float, optional 

32 Thermal conductivity. 

33 u_thresh : float, optional 

34 Threshold for temperature. 

35 u_max : float, optional 

36 Maximum temperature. 

37 Q_max : float, optional 

38 Maximum heat source power density. 

39 leak_range : tuple of float 

40 Range of the leak. 

41 leak_type : str, optional 

42 Type of leak, choose between ``'linear'`` or ``'exponential'``. 

43 leak_transition : str, optional 

44 Indicates how the heat in the leak propagates, choose between ``'step'`` and ``'Gaussian'``. 

45 order : int, optional 

46 Order of the finite difference discretization. 

47 stencil_type : str, optional 

48 Type of stencil for finite differences. 

49 bc : str, optional 

50 Type of boundary conditions. Default is ``'neumann-zero'``. 

51 nvars : int, optional 

52 Spatial resolution. 

53 newton_tol : float, optional 

54 Tolerance for Newton to terminate. 

55 newton_maxiter : int, optional 

56 Maximum number of Newton iterations to be done. 

57 lintol : float, optional 

58 Tolerance for linear solver to be done. 

59 liniter : int, optional 

60 Maximum number of linear iterations inside the Newton solver. 

61 direct_solver : bool, optional 

62 Indicates if a direct solver should be used. 

63 inexact_linear_ratio : float, optional 

64 Ratio of tolerance of linear solver to the Newton residual, overrides `lintol` 

65 min_lintol : float, optional 

66 Minimal tolerance for the linear solver 

67 reference_sol_type : str, optional 

68 Indicates which method should be used to compute a reference solution. 

69 Choose between ``'scipy'``, ``'SDC'``, or ``'DIRK'``. 

70 

71 Attributes 

72 ---------- 

73 A : sparse matrix (CSC) 

74 FD discretization matrix of the ND grad operator. 

75 Id : sparse matrix (CSC) 

76 Identity matrix of the same dimension as A. 

77 dx : float 

78 Distance between two spatial nodes. 

79 xv : np.1darray 

80 Spatial grid values. 

81 leak : np.1darray of bool 

82 Indicates the leak. 

83 

84 References 

85 ---------- 

86 .. [1] Thermal thin shell approximation towards finite element quench simulation. E. Schnaubelt, M. Wozniak, S. Schöps. 

87 Supercond. Sci. Technol. 36 044004. DOI 10.1088/1361-6668/acbeea 

88 """ 

89 

90 dtype_u = mesh 

91 dtype_f = mesh 

92 

93 def __init__( 

94 self, 

95 Cv=1000.0, 

96 K=1000.0, 

97 u_thresh=3e-2, 

98 u_max=6e-2, 

99 Q_max=1.0, 

100 leak_range=(0.45, 0.55), 

101 leak_type='linear', 

102 leak_transition='step', 

103 order=2, 

104 stencil_type='center', 

105 bc='neumann-zero', 

106 nvars=2**7, 

107 newton_tol=1e-8, 

108 newton_maxiter=99, 

109 lintol=1e-8, 

110 liniter=99, 

111 direct_solver=True, 

112 inexact_linear_ratio=None, 

113 min_lintol=1e-12, 

114 reference_sol_type='scipy', 

115 ): 

116 """ 

117 Initialization routine 

118 

119 Args: 

120 problem_params (dict): custom parameters for the example 

121 dtype_u: mesh data type (will be passed parent class) 

122 dtype_f: mesh data type (will be passed parent class) 

123 """ 

124 # invoke super init, passing number of dofs, dtype_u and dtype_f 

125 super().__init__(init=(nvars, None, np.dtype('float64'))) 

126 self._makeAttributeAndRegister( 

127 'Cv', 

128 'K', 

129 'u_thresh', 

130 'u_max', 

131 'Q_max', 

132 'leak_range', 

133 'leak_type', 

134 'leak_transition', 

135 'order', 

136 'stencil_type', 

137 'bc', 

138 'nvars', 

139 'direct_solver', 

140 'reference_sol_type', 

141 localVars=locals(), 

142 readOnly=True, 

143 ) 

144 self._makeAttributeAndRegister( 

145 'newton_tol', 

146 'newton_maxiter', 

147 'lintol', 

148 'liniter', 

149 'inexact_linear_ratio', 

150 'min_lintol', 

151 localVars=locals(), 

152 readOnly=False, 

153 ) 

154 

155 self._makeAttributeAndRegister( 

156 'newton_tol', 

157 'newton_maxiter', 

158 'lintol', 

159 'liniter', 

160 'direct_solver', 

161 localVars=locals(), 

162 readOnly=False, 

163 ) 

164 

165 # setup finite difference discretization from problem helper 

166 self.dx, xvalues = problem_helper.get_1d_grid(size=self.nvars, bc=self.bc) 

167 

168 self.A, self.b = problem_helper.get_finite_difference_matrix( 

169 derivative=2, 

170 order=self.order, 

171 stencil_type=self.stencil_type, 

172 dx=self.dx, 

173 size=self.nvars, 

174 dim=1, 

175 bc=self.bc, 

176 ) 

177 self.A *= self.K / self.Cv 

178 

179 self.xv = xvalues 

180 self.Id = sp.eye(np.prod(self.nvars), format='csc') 

181 

182 self.leak = np.logical_and(self.xv > self.leak_range[0], self.xv < self.leak_range[1]) 

183 

184 self.work_counters['newton'] = WorkCounter() 

185 self.work_counters['rhs'] = WorkCounter() 

186 if not self.direct_solver: 

187 self.work_counters['linear'] = WorkCounter() 

188 

189 def eval_f_non_linear(self, u, t): 

190 """ 

191 Get the non-linear part of f. 

192 

193 Parameters 

194 ---------- 

195 u : dtype_u 

196 Current values of the numerical solution: 

197 t : float 

198 Current time at which the numerical solution is computed. 

199 

200 Returns 

201 ------- 

202 me : dtype_u 

203 The non-linear part of the right-hand side. 

204 """ 

205 u_thresh = self.u_thresh 

206 u_max = self.u_max 

207 Q_max = self.Q_max 

208 me = self.dtype_u(self.init) 

209 

210 if self.leak_type == 'linear': 

211 me[:] = (u - u_thresh) / (u_max - u_thresh) * Q_max 

212 elif self.leak_type == 'exponential': 

213 me[:] = Q_max * (np.exp(u) - np.exp(u_thresh)) / (np.exp(u_max) - np.exp(u_thresh)) 

214 else: 

215 raise NotImplementedError(f'Leak type \"{self.leak_type}\" not implemented!') 

216 

217 me[u < u_thresh] = 0 

218 if self.leak_transition == 'step': 

219 me[self.leak] = Q_max 

220 elif self.leak_transition == 'Gaussian': 

221 me[:] = np.max([me, Q_max * np.exp(-((self.xv - 0.5) ** 2) / 3e-2)], axis=0) 

222 else: 

223 raise NotImplementedError(f'Leak transition \"{self.leak_transition}\" not implemented!') 

224 

225 me[u >= u_max] = Q_max 

226 

227 me[:] /= self.Cv 

228 

229 return me 

230 

231 def eval_f(self, u, t): 

232 """ 

233 Evaluate the full right-hand side. 

234 

235 Parameters 

236 ---------- 

237 u : dtype_u 

238 Current values of the numerical solution. 

239 t : float 

240 Current time at which the numerical solution is computed. 

241 

242 Returns 

243 ------- 

244 f : dtype_f 

245 The right-hand side of the problem. 

246 """ 

247 f = self.dtype_f(self.init) 

248 f[:] = self.A.dot(u.flatten()).reshape(self.nvars) + self.b + self.eval_f_non_linear(u, t) 

249 

250 self.work_counters['rhs']() 

251 return f 

252 

253 def get_non_linear_Jacobian(self, u): 

254 """ 

255 Evaluate the non-linear part of the Jacobian only. 

256 

257 Parameters 

258 ---------- 

259 u : dtype_u 

260 Current values of the numerical solution. 

261 

262 Returns 

263 ------- 

264 scipy.sparse.csc 

265 The derivative of the non-linear part of the solution w.r.t. to the solution. 

266 """ 

267 u_thresh = self.u_thresh 

268 u_max = self.u_max 

269 Q_max = self.Q_max 

270 me = self.dtype_u(self.init) 

271 

272 if self.leak_type == 'linear': 

273 me[:] = Q_max / (u_max - u_thresh) 

274 elif self.leak_type == 'exponential': 

275 me[:] = Q_max * np.exp(u) / (np.exp(u_max) - np.exp(u_thresh)) 

276 else: 

277 raise NotImplementedError(f'Leak type {self.leak_type} not implemented!') 

278 

279 me[u < u_thresh] = 0 

280 if self.leak_transition == 'step': 

281 me[self.leak] = 0 

282 elif self.leak_transition == 'Gaussian': 

283 me[self.leak] = 0 

284 me[self.leak][u[self.leak] > Q_max * np.exp(-((self.xv[self.leak] - 0.5) ** 2) / 3e-2)] = 1 

285 else: 

286 raise NotImplementedError(f'Leak transition \"{self.leak_transition}\" not implemented!') 

287 me[u > u_max] = 0 

288 

289 me[:] /= self.Cv 

290 

291 return sp.diags(me, format='csc') 

292 

293 def solve_system(self, rhs, factor, u0, t): 

294 r""" 

295 Simple Newton solver for :math:`(I - factor \cdot f)(\vec{u}) = \vec{rhs}`. 

296 

297 Parameters 

298 ---------- 

299 rhs : dtype_f 

300 Right-hand side. 

301 factor : float 

302 Abbrev. for the local stepsize (or any other factor required). 

303 u0 : dtype_u 

304 Initial guess for the iterative solver. 

305 t : float 

306 Current time (e.g. for time-dependent BCs). 

307 

308 Returns 

309 ------- 

310 u : dtype_u 

311 The solution as mesh. 

312 """ 

313 u = self.dtype_u(u0) 

314 res = np.inf 

315 delta = self.dtype_u(self.init, val=0.0) 

316 

317 # construct a preconditioner for the space solver 

318 if not self.direct_solver: 

319 M = inv((self.Id - factor * self.A).toarray()) 

320 zero = self.dtype_u(self.init, val=0.0) 

321 

322 for n in range(0, self.newton_maxiter): 

323 # assemble G such that G(u) = 0 at the solution of the step 

324 G = u - factor * self.eval_f(u, t) - rhs 

325 self.work_counters[ 

326 'rhs' 

327 ].decrement() # Work regarding construction of the Jacobian etc. should count into the Newton iterations only 

328 

329 res = np.linalg.norm(G, np.inf) 

330 if res <= self.newton_tol and n > 0: # we want to make at least one Newton iteration 

331 break 

332 

333 if self.inexact_linear_ratio: 

334 self.lintol = max([res * self.inexact_linear_ratio, self.min_lintol]) 

335 

336 # assemble Jacobian J of G 

337 J = self.Id - factor * (self.A + self.get_non_linear_Jacobian(u)) 

338 

339 # solve the linear system 

340 if self.direct_solver: 

341 delta = spsolve(J, G) 

342 else: 

343 delta, info = gmres( 

344 J, 

345 G, 

346 x0=zero, 

347 M=M, 

348 rtol=self.lintol, 

349 maxiter=self.liniter, 

350 atol=0, 

351 callback=self.work_counters['linear'], 

352 ) 

353 

354 if not np.isfinite(delta).all(): 

355 break 

356 

357 # update solution 

358 u = u - delta 

359 

360 self.work_counters['newton']() 

361 

362 return u 

363 

364 def u_exact(self, t, u_init=None, t_init=None): 

365 r""" 

366 Routine to compute the exact solution at time :math:`t`. 

367 

368 Parameters 

369 ---------- 

370 t : float 

371 Time of the exact solution. 

372 

373 Returns 

374 ------- 

375 me : dtype_u 

376 The exact solution. 

377 """ 

378 

379 me = self.dtype_u(self.init, val=0.0) 

380 

381 if t > 0: 

382 if self.reference_sol_type == 'scipy': 

383 

384 def jac(t, u): 

385 """ 

386 Get the Jacobian for the implicit BDF method to use in `scipy.solve_ivp` 

387 

388 Parameters 

389 ---------- 

390 t : float 

391 The current time. 

392 u : dtype_u 

393 Current solution. 

394 

395 Returns 

396 ------- 

397 scipy.sparse.csc 

398 The derivative of the non-linear part of the solution w.r.t. to the solution. 

399 """ 

400 return self.A + self.get_non_linear_Jacobian(u) 

401 

402 def eval_rhs(t, u): 

403 """ 

404 Function to pass to `scipy.solve_ivp` to evaluate the full right-hand side. 

405 

406 Parameters 

407 ---------- 

408 t : float 

409 Current time. 

410 u : numpy.1darray 

411 Current solution. 

412 

413 Returns 

414 ------- 

415 numpy.1darray 

416 Right-hand side. 

417 """ 

418 return self.eval_f(u.reshape(self.init[0]), t).flatten() 

419 

420 me[:] = self.generate_scipy_reference_solution(eval_rhs, t, u_init, t_init, method='BDF', jac=jac) 

421 

422 elif self.reference_sol_type in ['DIRK', 'SDC']: 

423 from pySDC.implementations.controller_classes.controller_nonMPI import controller_nonMPI 

424 from pySDC.implementations.hooks.log_solution import LogSolution 

425 from pySDC.helpers.stats_helper import get_sorted 

426 

427 description = {} 

428 description['problem_class'] = Quench 

429 description['problem_params'] = { 

430 'newton_tol': 1e-10, 

431 'newton_maxiter': 99, 

432 'nvars': 2**10, 

433 **self.params, 

434 } 

435 

436 if self.reference_sol_type == 'DIRK': 

437 from pySDC.implementations.sweeper_classes.Runge_Kutta import DIRK43 

438 from pySDC.implementations.convergence_controller_classes.adaptivity import AdaptivityRK 

439 

440 description['sweeper_class'] = DIRK43 

441 description['sweeper_params'] = {} 

442 description['step_params'] = {'maxiter': 1} 

443 description['level_params'] = {'dt': 1e-4} 

444 description['convergence_controllers'] = {AdaptivityRK: {'e_tol': 1e-9, 'update_order': 4}} 

445 elif self.reference_sol_type == 'SDC': 

446 from pySDC.implementations.sweeper_classes.generic_implicit import generic_implicit 

447 

448 description['sweeper_class'] = generic_implicit 

449 description['sweeper_params'] = {'num_nodes': 3, 'QI': 'IE', 'quad_type': 'RADAU-RIGHT'} 

450 description['step_params'] = {'maxiter': 99} 

451 description['level_params'] = {'dt': 0.5, 'restol': 1e-10} 

452 

453 controller_params = {'hook_class': LogSolution, 'mssdc_jac': False, 'logger_level': 99} 

454 

455 controller = controller_nonMPI( 

456 description=description, controller_params=controller_params, num_procs=1 

457 ) 

458 

459 uend, stats = controller.run( 

460 u0=u_init if u_init is not None else self.u_exact(t=0.0), 

461 t0=t_init if t_init is not None else 0, 

462 Tend=t, 

463 ) 

464 

465 u_last = get_sorted(stats, type='u', recomputed=False)[-1] 

466 

467 if abs(u_last[0] - t) > 1e-2: 

468 self.logger.warning( 

469 f'Time difference between reference solution and requested time is {abs(u_last[0]-t):.2e}!' 

470 ) 

471 

472 me[:] = u_last[1] 

473 

474 return me 

475 

476 

477class QuenchIMEX(Quench): 

478 """ 

479 This is a toy problem [1]_ to emulate a magnet that has been cooled to temperatures where superconductivity is possible. 

480 However, there is a leak! Some point in the domain is constantly heated and when this has heated up its environment 

481 sufficiently, there will be a runaway effect heating up the entire magnet. 

482 This effect has actually lead to huge magnets being destroyed at CERN in the past and hence warrants investigation. 

483 

484 The model we use is a 1d heat equation with Neumann-zero boundary conditions, meaning this magnet is totally 

485 insulated from its environment except for the leak. 

486 We add a non-linear term that heats parts of the domain that exceed a certain temperature threshold as well as the 

487 leak itself. 

488 

489 The problem is discretised with finite difference in space and treated *semi-implicitly*. 

490 """ 

491 

492 dtype_f = imex_mesh 

493 

494 def eval_f(self, u, t): 

495 """ 

496 Routine to evaluate the right-hand side of the problem. 

497 

498 Parameters 

499 ---------- 

500 u : dtype_u 

501 Current values of the numerical solution. 

502 t : float 

503 Current time of the numerical solution is computed. 

504 

505 Returns 

506 ------- 

507 f : dtype_f 

508 The right-hand side of the problem. 

509 """ 

510 

511 f = self.dtype_f(self.init) 

512 f.impl[:] = self.A.dot(u.flatten()).reshape(self.nvars) 

513 f.expl[:] = self.eval_f_non_linear(u, t) + self.b 

514 

515 self.work_counters['rhs']() 

516 return f 

517 

518 def solve_system(self, rhs, factor, u0, t): 

519 r""" 

520 Simple linear solver for :math:`(I - factor \cdot f_{expl})(\vec{u}) = \vec{rhs}`. 

521 

522 Parameters 

523 ---------- 

524 rhs : dtype_f 

525 Right-hand side for the linear system. 

526 factor : float 

527 Abbrev. for the local stepsize (or any other factor required). 

528 u0 : dtype_u 

529 Initial guess for the iterative solver. 

530 t : float 

531 Current time (e.g. for time-dependent BCs). 

532 

533 Returns 

534 ------- 

535 me : dtype_u 

536 The solution as mesh. 

537 """ 

538 

539 me = self.dtype_u(self.init) 

540 me[:] = spsolve(self.Id - factor * self.A, rhs.flatten()).reshape(self.nvars) 

541 return me 

542 

543 def u_exact(self, t, u_init=None, t_init=None): 

544 r""" 

545 Routine to compute the exact solution at time :math:`t`. 

546 

547 Parameters 

548 ---------- 

549 t : float 

550 Time of the exact solution. 

551 

552 Returns 

553 ------- 

554 me : dtype_u 

555 The exact solution. 

556 """ 

557 me = self.dtype_u(self.init, val=0.0) 

558 

559 if t == 0: 

560 me[:] = super().u_exact(t, u_init, t_init) 

561 

562 if t > 0: 

563 

564 def jac(t, u): 

565 """ 

566 Get the Jacobian for the implicit BDF method to use in `scipy.solve_ivp`. 

567 

568 Parameters 

569 ---------- 

570 t : float 

571 Current time. 

572 u : dtype_u 

573 Current solution. 

574 

575 Returns 

576 ------- 

577 scipy.sparse.csc 

578 The derivative of the non-linear part of the solution w.r.t. to the solution. 

579 """ 

580 return self.A 

581 

582 def eval_rhs(t, u): 

583 """ 

584 Function to pass to `scipy.solve_ivp` to evaluate the full right-hand side. 

585 

586 Parameters 

587 ---------- 

588 t : float 

589 Current time 

590 u : numpy.1darray 

591 Current solution 

592 

593 Returns 

594 ------- 

595 numpy.1darray 

596 The right-hand side. 

597 """ 

598 f = self.eval_f(u.reshape(self.init[0]), t) 

599 return (f.impl + f.expl).flatten() 

600 

601 me[:] = self.generate_scipy_reference_solution(eval_rhs, t, u_init, t_init, method='BDF', jac=jac) 

602 return me