Coverage for pySDC/implementations/problem_classes/AllenCahn_2D_FFT.py: 21%

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1import numpy as np 

2 

3from pySDC.core.errors import ProblemError 

4from pySDC.core.problem import Problem 

5from pySDC.implementations.datatype_classes.mesh import mesh, imex_mesh 

6 

7 

8# noinspection PyUnusedLocal 

9class allencahn2d_imex(Problem): 

10 r""" 

11 Example implementing the two-dimensional Allen-Cahn equation with periodic boundary conditions :math:`u \in [-1, 1]^2` 

12 

13 .. math:: 

14 \frac{\partial u}{\partial t} = \Delta u + \frac{1}{\varepsilon^2} u (1 - u^\nu) 

15 

16 on a spatial domain :math:`[-\frac{L}{2}, \frac{L}{2}]^2`, and constant parameter :math:`\nu`. Different initial conditions 

17 can be used, for example, circles of the form 

18 

19 .. math:: 

20 u({\bf x}, 0) = \tanh\left(\frac{r - \sqrt{x_i^2 + y_j^2}}{\sqrt{2}\varepsilon}\right), 

21 

22 or *checker-board* 

23 

24 .. math:: 

25 u({\bf x}, 0) = \sin(2 \pi x_i) \sin(2 \pi y_j), 

26 

27 or uniform distributed random numbers in :math:`[-1, 1]` for :math:`i, j=0,..,N-1`, where :math:`N` is the number of 

28 spatial grid points. For time-stepping, the problem is treated *semi-implicitly*, i.e., the diffusion part is solved by 

29 Fast-Fourier Transform (FFT) and the nonlinear term is treated explicitly. 

30 

31 An exact solution is not known, but instead the numerical solution can be compared via a generated reference solution computed 

32 by a ``SciPy`` routine. 

33 

34 Parameters 

35 ---------- 

36 nvars : List of int tuples, optional 

37 Number of unknowns in the problem, e.g. ``nvars=[(128, 128), (128, 128)]``. 

38 nu : float, optional 

39 Problem parameter :math:`\nu`. 

40 eps : float, optional 

41 Scaling parameter :math:`\varepsilon`. 

42 radius : float, optional 

43 Radius of the circles. 

44 L : float, optional 

45 Denotes the period of the function to be approximated for the Fourier transform. 

46 init_type : str, optional 

47 Indicates which type of initial condition is used. 

48 

49 Attributes 

50 ---------- 

51 xvalues : np.1darray 

52 Grid points in space. 

53 dx : float 

54 Mesh width. 

55 lap : np.1darray 

56 Spectral operator for Laplacian. 

57 """ 

58 

59 dtype_u = mesh 

60 dtype_f = imex_mesh 

61 

62 def __init__( 

63 self, 

64 nvars=None, 

65 nu=2, 

66 eps=0.04, 

67 radius=0.25, 

68 L=1.0, 

69 init_type='circle', 

70 ): 

71 """Initialization routine""" 

72 

73 if nvars is None: 

74 nvars = (128, 128) 

75 

76 # we assert that nvars looks very particular here.. this will be necessary for coarsening in space later on 

77 if len(nvars) != 2: 

78 raise ProblemError('this is a 2d example, got %s' % nvars) 

79 if nvars[0] != nvars[1]: 

80 raise ProblemError('need a square domain, got %s' % nvars) 

81 if nvars[0] % 2 != 0: 

82 raise ProblemError('the setup requires nvars = 2^p per dimension') 

83 

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

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

86 self._makeAttributeAndRegister( 

87 'nvars', 'nu', 'eps', 'radius', 'L', 'init_type', localVars=locals(), readOnly=True 

88 ) 

89 

90 self.dx = self.L / self.nvars[0] # could be useful for hooks, too. 

91 self.xvalues = np.array([i * self.dx - self.L / 2.0 for i in range(self.nvars[0])]) 

92 

93 kx = np.zeros(self.init[0][0]) 

94 ky = np.zeros(self.init[0][1] // 2 + 1) 

95 

96 kx[: int(self.init[0][0] / 2) + 1] = 2 * np.pi / self.L * np.arange(0, int(self.init[0][0] / 2) + 1) 

97 kx[int(self.init[0][0] / 2) + 1 :] = ( 

98 2 * np.pi / self.L * np.arange(int(self.init[0][0] / 2) + 1 - self.init[0][0], 0) 

99 ) 

100 ky[:] = 2 * np.pi / self.L * np.arange(0, self.init[0][1] // 2 + 1) 

101 

102 xv, yv = np.meshgrid(kx, ky, indexing='ij') 

103 self.lap = -(xv**2) - yv**2 

104 

105 def eval_f(self, u, t): 

106 """ 

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

108 

109 Parameters 

110 ---------- 

111 u : dtype_u 

112 Current values of the numerical solution. 

113 t : float 

114 Current time of the numerical solution is computed. 

115 

116 Returns 

117 ------- 

118 f : dtype_f 

119 The right-hand side of the problem. 

120 """ 

121 

122 f = self.dtype_f(self.init) 

123 tmp = self.lap * np.fft.rfft2(u) 

124 f.impl[:] = np.fft.irfft2(tmp) 

125 if self.eps > 0: 

126 f.expl[:] = 1.0 / self.eps**2 * u * (1.0 - u**self.nu) 

127 return f 

128 

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

130 """ 

131 Simple FFT solver for the diffusion part. 

132 

133 Parameters 

134 ---------- 

135 rhs : dtype_f 

136 Right-hand side for the linear system. 

137 factor : float 

138 Abbrev. for the node-to-node stepsize (or any other factor required). 

139 u0 : dtype_u 

140 Initial guess for the iterative solver (not used here so far). 

141 t : float 

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

143 

144 Returns 

145 ------- 

146 me : dtype_u 

147 The solution as mesh. 

148 """ 

149 

150 me = self.dtype_u(self.init) 

151 

152 tmp = np.fft.rfft2(rhs) / (1.0 - factor * self.lap) 

153 me[:] = np.fft.irfft2(tmp) 

154 

155 return me 

156 

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

158 r""" 

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

160 

161 Parameters 

162 ---------- 

163 t : float 

164 Time of the exact solution. 

165 u_init : pySDC.implementations.problem_classes.allencahn2d_imex.dtype_u 

166 Initial conditions for getting the exact solution. 

167 t_init : float 

168 The starting time. 

169 

170 Returns 

171 ------- 

172 me : dtype_u 

173 The exact solution. 

174 """ 

175 

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

177 

178 if t == 0: 

179 if self.init_type == 'circle': 

180 xv, yv = np.meshgrid(self.xvalues, self.xvalues, indexing='ij') 

181 me[:, :] = np.tanh((self.radius - np.sqrt(xv**2 + yv**2)) / (np.sqrt(2) * self.eps)) 

182 elif self.init_type == 'checkerboard': 

183 xv, yv = np.meshgrid(self.xvalues, self.xvalues) 

184 me[:, :] = np.sin(2.0 * np.pi * xv) * np.sin(2.0 * np.pi * yv) 

185 elif self.init_type == 'random': 

186 me[:, :] = np.random.uniform(-1, 1, self.init) 

187 else: 

188 raise NotImplementedError('type of initial value not implemented, got %s' % self.init_type) 

189 else: 

190 

191 def eval_rhs(t, u): 

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

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

194 

195 me[:, :] = self.generate_scipy_reference_solution(eval_rhs, t, u_init, t_init) 

196 

197 return me 

198 

199 

200class allencahn2d_imex_stab(allencahn2d_imex): 

201 r""" 

202 This implements the two-dimensional Allen-Cahn equation with periodic boundary conditions :math:`u \in [-0.5, 0.5]^2` 

203 with stabilized splitting 

204 

205 .. math:: 

206 \frac{\partial u}{\partial t} = \Delta u + \frac{1}{\varepsilon^2} u (1 - u^\nu) + \frac{2}{\varepsilon^2}u 

207 

208 on a spatial domain :math:`[-\frac{L}{2}, \frac{L}{2}]^2`, and constant parameter :math:`\nu`. Different initial conditions 

209 can be used here, for example, circles of the form 

210 

211 .. math:: 

212 u({\bf x}, 0) = \tanh\left(\frac{r - \sqrt{x_i^2 + y_j^2}}{\sqrt{2}\varepsilon}\right), 

213 

214 or *checker-board* 

215 

216 .. math:: 

217 u({\bf x}, 0) = \sin(2 \pi x_i) \sin(2 \pi y_j), 

218 

219 or uniform distributed random numbers in :math:`[-1, 1]` for :math:`i, j=0,..,N-1`, where :math:`N` is the number of 

220 spatial grid points. For time-stepping, the problem is treated *semi-implicitly*, i.e., the diffusion part is solved with 

221 Fast-Fourier Transform (FFT) and the nonlinear term is treated explicitly. 

222 

223 An exact solution is not known, but instead the numerical solution can be compared via a generated reference solution computed 

224 by a ``SciPy`` routine. 

225 

226 Parameters 

227 ---------- 

228 nvars : List of int tuples, optional 

229 Number of unknowns in the problem, e.g. ``nvars=[(128, 128), (128, 128)]``. 

230 nu : float, optional 

231 Problem parameter :math:`\nu`. 

232 eps : float, optional 

233 Scaling parameter :math:`\varepsilon`. 

234 radius : float, optional 

235 Radius of the circles. 

236 L : float, optional 

237 Denotes the period of the function to be approximated for the Fourier transform. 

238 init_type : str, optional 

239 Indicates which type of initial condition is used. 

240 

241 Attributes 

242 ---------- 

243 xvalues : np.1darray 

244 Grid points in space. 

245 dx : float 

246 Mesh width. 

247 lap : np.1darray 

248 Spectral operator for Laplacian. 

249 """ 

250 

251 def __init__(self, nvars=None, nu=2, eps=0.04, radius=0.25, L=1.0, init_type='circle'): 

252 """Initialization routine""" 

253 

254 if nvars is None: 

255 nvars = [(256, 256), (64, 64)] 

256 

257 super().__init__(nvars, nu, eps, radius, L, init_type) 

258 self.lap -= 2.0 / self.eps**2 

259 

260 def eval_f(self, u, t): 

261 """ 

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

263 

264 Parameters 

265 ---------- 

266 u : dtype_u 

267 Current values of the numerical solution. 

268 t : float 

269 Current time of the numerical solution is computed. 

270 

271 Returns 

272 ------- 

273 f : dtype_f 

274 The right-hand side of the problem. 

275 """ 

276 

277 f = self.dtype_f(self.init) 

278 tmp = self.lap * np.fft.rfft2(u) 

279 f.impl[:] = np.fft.irfft2(tmp) 

280 if self.eps > 0: 

281 f.expl[:] = 1.0 / self.eps**2 * u * (1.0 - u**self.nu) + 2.0 / self.eps**2 * u 

282 return f 

283 

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

285 """ 

286 Simple FFT solver for the diffusion part. 

287 

288 Parameters 

289 ---------- 

290 rhs : dtype_f 

291 Right-hand side for the linear system. 

292 factor : float 

293 Abbrev. for the node-to-node stepsize (or any other factor required). 

294 u0 : dtype_u 

295 Initial guess for the iterative solver (not used here so far). 

296 t : float 

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

298 

299 Returns 

300 ------- 

301 me : dtype_u 

302 The solution as mesh. 

303 """ 

304 

305 me = self.dtype_u(self.init) 

306 

307 tmp = np.fft.rfft2(rhs) / (1.0 - factor * self.lap) 

308 me[:] = np.fft.irfft2(tmp) 

309 

310 return me