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#!/usr/bin/env python
"""
A simple WaveSolver class for evolving the wave equation in 2D.
This works in parallel by using a RectPartitioner object.
Authors
-------
* Xing Cai
* Min Ragan-Kelley
"""
import time
from numpy import arange, newaxis, sqrt, zeros
def iseq(start=0, stop=None, inc=1):
"""
Generate integers from start to (and including!) stop,
with increment of inc. Alternative to range/xrange.
"""
if stop is None: # allow isequence(3) to be 0, 1, 2, 3
# take 1st arg as stop, start as 0, and inc=1
stop = start
start = 0
inc = 1
return arange(start, stop + inc, inc)
class WaveSolver:
"""
Solve the 2D wave equation u_tt = u_xx + u_yy + f(x,y,t) with
u = bc(x,y,t) on the boundary and initial condition du/dt = 0.
Parallelization by using a RectPartitioner object 'partitioner'
nx and ny are the total number of global grid cells in the x and y
directions. The global grid points are numbered as (0,0), (1,0), (2,0),
..., (nx,0), (0,1), (1,1), ..., (nx, ny).
dt is the time step. If dt<=0, an optimal time step is used.
tstop is the stop time for the simulation.
I, f are functions: I(x,y), f(x,y,t)
user_action: function of (u, x, y, t) called at each time
level (x and y are one-dimensional coordinate vectors).
This function allows the calling code to plot the solution,
compute errors, etc.
implementation: a dictionary specifying how the initial
condition ('ic'), the scheme over inner points ('inner'),
and the boundary conditions ('bc') are to be implemented.
Normally, values are legal: 'scalar' or 'vectorized'.
'scalar' means straight loops over grid points, while
'vectorized' means special NumPy vectorized operations.
If a key in the implementation dictionary is missing, it
defaults in this function to 'scalar' (the safest strategy).
Note that if 'vectorized' is specified, the functions I, f,
and bc must work in vectorized mode. It is always recommended
to first run the 'scalar' mode and then compare 'vectorized'
results with the 'scalar' results to check that I, f, and bc
work.
verbose: true if a message at each time step is written,
false implies no output during the simulation.
final_test: true means the discrete L2-norm of the final solution is
to be computed.
"""
def __init__(
self,
I,
f,
c,
bc,
Lx,
Ly,
partitioner=None,
dt=-1,
user_action=None,
implementation={
'ic': 'vectorized', # or 'scalar'
'inner': 'vectorized',
'bc': 'vectorized',
},
):
nx = partitioner.global_num_cells[0] # number of global cells in x dir
ny = partitioner.global_num_cells[1] # number of global cells in y dir
dx = Lx / float(nx)
dy = Ly / float(ny)
loc_nx, loc_ny = partitioner.get_num_loc_cells()
nx = loc_nx
ny = loc_ny # now use loc_nx and loc_ny instead
lo_ix0 = partitioner.subd_lo_ix[0]
lo_ix1 = partitioner.subd_lo_ix[1]
hi_ix0 = partitioner.subd_hi_ix[0]
hi_ix1 = partitioner.subd_hi_ix[1]
x = iseq(dx * lo_ix0, dx * hi_ix0, dx) # local grid points in x dir
y = iseq(dy * lo_ix1, dy * hi_ix1, dy) # local grid points in y dir
self.x = x
self.y = y
xv = x[:, newaxis] # for vectorized expressions with f(xv,yv)
yv = y[newaxis, :] # -- " --
if dt <= 0:
dt = (1 / float(c)) * (1 / sqrt(1 / dx**2 + 1 / dy**2)) # max time step
Cx2 = (c * dt / dx) ** 2
Cy2 = (c * dt / dy) ** 2
dt2 = dt**2 # help variables
u = zeros((nx + 1, ny + 1)) # solution array
u_1 = u.copy() # solution at t-dt
u_2 = u.copy() # solution at t-2*dt
# preserve for self.solve
implementation = dict(implementation) # copy
if 'ic' not in implementation:
implementation['ic'] = 'scalar'
if 'bc' not in implementation:
implementation['bc'] = 'scalar'
if 'inner' not in implementation:
implementation['inner'] = 'scalar'
self.implementation = implementation
self.Lx = Lx
self.Ly = Ly
self.I = I
self.f = f
self.c = c
self.bc = bc
self.user_action = user_action
self.partitioner = partitioner
# set initial condition (pointwise - allows straight if-tests in I(x,y)):
t = 0.0
if implementation['ic'] == 'scalar':
for i in range(0, nx + 1):
for j in range(0, ny + 1):
u_1[i, j] = I(x[i], y[j])
for i in range(1, nx):
for j in range(1, ny):
u_2[i, j] = (
u_1[i, j]
+ 0.5 * Cx2 * (u_1[i - 1, j] - 2 * u_1[i, j] + u_1[i + 1, j])
+ 0.5 * Cy2 * (u_1[i, j - 1] - 2 * u_1[i, j] + u_1[i, j + 1])
+ dt2 * f(x[i], y[j], 0.0)
)
# boundary values of u_2 (equals u(t=dt) due to du/dt=0)
i = 0
for j in range(0, ny + 1):
u_2[i, j] = bc(x[i], y[j], t + dt)
j = 0
for i in range(0, nx + 1):
u_2[i, j] = bc(x[i], y[j], t + dt)
i = nx
for j in range(0, ny + 1):
u_2[i, j] = bc(x[i], y[j], t + dt)
j = ny
for i in range(0, nx + 1):
u_2[i, j] = bc(x[i], y[j], t + dt)
elif implementation['ic'] == 'vectorized':
u_1 = I(xv, yv)
u_2[1:nx, 1:ny] = (
u_1[1:nx, 1:ny]
+ 0.5
* Cx2
* (u_1[0 : nx - 1, 1:ny] - 2 * u_1[1:nx, 1:ny] + u_1[2 : nx + 1, 1:ny])
+ 0.5
* Cy2
* (u_1[1:nx, 0 : ny - 1] - 2 * u_1[1:nx, 1:ny] + u_1[1:nx, 2 : ny + 1])
+ dt2 * (f(xv[1:nx, 1:ny], yv[1:nx, 1:ny], 0.0))
)
# boundary values (t=dt):
i = 0
u_2[i, :] = bc(x[i], y, t + dt)
j = 0
u_2[:, j] = bc(x, y[j], t + dt)
i = nx
u_2[i, :] = bc(x[i], y, t + dt)
j = ny
u_2[:, j] = bc(x, y[j], t + dt)
if user_action is not None:
user_action(u_1, x, y, t) # allow user to plot etc.
# print(list(self.us[2][2]))
self.us = (u, u_1, u_2)
def solve(self, tstop, dt=-1, user_action=None, verbose=False, final_test=False):
t0 = time.time()
f = self.f
c = self.c
bc = self.bc
partitioner = self.partitioner
implementation = self.implementation
nx = partitioner.global_num_cells[0] # number of global cells in x dir
ny = partitioner.global_num_cells[1] # number of global cells in y dir
dx = self.Lx / float(nx)
dy = self.Ly / float(ny)
loc_nx, loc_ny = partitioner.get_num_loc_cells()
nx = loc_nx
ny = loc_ny # now use loc_nx and loc_ny instead
x = self.x
y = self.y
xv = x[:, newaxis] # for vectorized expressions with f(xv,yv)
yv = y[newaxis, :] # -- " --
if dt <= 0:
dt = (1 / float(c)) * (1 / sqrt(1 / dx**2 + 1 / dy**2)) # max time step
Cx2 = (c * dt / dx) ** 2
Cy2 = (c * dt / dy) ** 2
dt2 = dt**2 # help variables
# id for the four possible neighbor subdomains
lower_x_neigh = partitioner.lower_neighbors[0]
upper_x_neigh = partitioner.upper_neighbors[0]
lower_y_neigh = partitioner.lower_neighbors[1]
upper_y_neigh = partitioner.upper_neighbors[1]
u, u_1, u_2 = self.us
# u_1 = self.u_1
t = 0.0
while t <= tstop:
t_old = t
t += dt
if verbose:
print(
'solving ({} version) at t={:g}'.format(implementation['inner'], t)
)
# update all inner points:
if implementation['inner'] == 'scalar':
for i in range(1, nx):
for j in range(1, ny):
u[i, j] = (
-u_2[i, j]
+ 2 * u_1[i, j]
+ Cx2 * (u_1[i - 1, j] - 2 * u_1[i, j] + u_1[i + 1, j])
+ Cy2 * (u_1[i, j - 1] - 2 * u_1[i, j] + u_1[i, j + 1])
+ dt2 * f(x[i], y[j], t_old)
)
elif implementation['inner'] == 'vectorized':
u[1:nx, 1:ny] = (
-u_2[1:nx, 1:ny]
+ 2 * u_1[1:nx, 1:ny]
+ Cx2
* (
u_1[0 : nx - 1, 1:ny]
- 2 * u_1[1:nx, 1:ny]
+ u_1[2 : nx + 1, 1:ny]
)
+ Cy2
* (
u_1[1:nx, 0 : ny - 1]
- 2 * u_1[1:nx, 1:ny]
+ u_1[1:nx, 2 : ny + 1]
)
+ dt2 * f(xv[1:nx, 1:ny], yv[1:nx, 1:ny], t_old)
)
# insert boundary conditions (if there's no neighbor):
if lower_x_neigh < 0:
if implementation['bc'] == 'scalar':
i = 0
for j in range(0, ny + 1):
u[i, j] = bc(x[i], y[j], t)
elif implementation['bc'] == 'vectorized':
u[0, :] = bc(x[0], y, t)
if upper_x_neigh < 0:
if implementation['bc'] == 'scalar':
i = nx
for j in range(0, ny + 1):
u[i, j] = bc(x[i], y[j], t)
elif implementation['bc'] == 'vectorized':
u[nx, :] = bc(x[nx], y, t)
if lower_y_neigh < 0:
if implementation['bc'] == 'scalar':
j = 0
for i in range(0, nx + 1):
u[i, j] = bc(x[i], y[j], t)
elif implementation['bc'] == 'vectorized':
u[:, 0] = bc(x, y[0], t)
if upper_y_neigh < 0:
if implementation['bc'] == 'scalar':
j = ny
for i in range(0, nx + 1):
u[i, j] = bc(x[i], y[j], t)
elif implementation['bc'] == 'vectorized':
u[:, ny] = bc(x, y[ny], t)
# communication
partitioner.update_internal_boundary(u)
if user_action is not None:
user_action(u, x, y, t)
# update data structures for next step
u_2, u_1, u = u_1, u, u_2
t1 = time.time()
print(
'my_id=%2d, dt=%g, %s version, slice_copy=%s, net Wtime=%g'
% (
partitioner.my_id,
dt,
implementation['inner'],
partitioner.slice_copy,
t1 - t0,
)
)
# save the us
self.us = u, u_1, u_2
# check final results; compute discrete L2-norm of the solution
if final_test:
loc_res = 0.0
for i in iseq(start=1, stop=nx - 1):
for j in iseq(start=1, stop=ny - 1):
loc_res += u_1[i, j] ** 2
return loc_res
return dt
|
