def rhs_to_scipy_ode(rhs, t, x_vect, u_vect, constants, *args,
**lambdify_kwargs):
"""Convert rhs to lambda function and jacobian comaptible with scipy.
Given a state space description of a dynamic system, create
a lambda function with a signature compatible with
scipy.ode.integrate: f(t, x, u, *args).
Parameters
----------
rhs : sympy.Matrix
Column vector of expressions for
the derivative of x (continuous),
or for the change in x (discrete).
t : sympy.Symbol
The independent time-like variable.
x_vect : sympy.Matrix
Column vector of symbols in the x (state) vector.
u_vect : sympy.Matrix
Column vector of symbols in the u (input) vector.
Most controllers pass a vector of inputs and this
supports this structure.
constants: dict
Dictionary of constants to substitute into rhs.
*args: dict
Additioinal symbols in f_vect that are not states, inputs
or constants.
**lambdify_kwargs:
Additional arugments to pass to lambdify.
@see sympy.utilities.lambdify.lambdify
Returns
-------
f : function
A lambda function that compute the right hand side
of the ode: f(t, x, u, *f_args).
jac : function
A lambda function that computes the jacobian
of the right hand side with respect to the state x:
jac(t, x, u, *jac_args).
"""
if constants is not None:
rhs = rhs.subs(constants)
x = sympy.DeferredVector('x')
u = sympy.DeferredVector('u')
ss_subs = {x_vect[i]: x[i] for i in range(len(x_vect))}
ss_subs.update({u_vect[i]: u[i] for i in range(len(u_vect))})
if 'default_array' not in lambdify_kwargs.keys():
lambdify_kwargs['default_array'] = True
f = sympy.lambdify((t, x, u) + args, rhs.subs(ss_subs), **lambdify_kwargs)
jac_vect = rhs.jacobian(x_vect)
jac = sympy.lambdify((t, x, u) + args, jac_vect.subs(ss_subs),
**lambdify_kwargs)
return (f, jac)
def _integrate_exact(f, tetrahedron):
#
# Note that
#
# \int_T f(x) dx = \int_T0 |J(xi)| f(P(xi)) dxi
#
# with
#
# P(xi) = x0 * (1-xi[0]-xi[1]) + x1 * xi[0] + x2 * xi[1].
#
# and T0 being the reference tetrahedron [(0.0, 0.0), (1.0, 0.0), (0.0,
# 1.0)].
# The determinant of the transformation matrix J equals twice the volume of
# the tetrahedron. (See, e.g.,
# <http://math2.uncc.edu/~shaodeng/TEACHING/math5172/Lectures/Lect_15.PDF>).
#
xi = sympy.DeferredVector('xi')
x_xi = \
+ tetrahedron[0] * (1 - xi[0] - xi[1] - xi[2]) \
+ tetrahedron[1] * xi[0] \
+ tetrahedron[2] * xi[1] \
+ tetrahedron[3] * xi[2]
abs_det_J = 6 * quadpy.tetrahedron.volume(tetrahedron)
exact = sympy.integrate(
sympy.integrate(
sympy.integrate(abs_det_J * f(x_xi),
(xi[2], 0, 1 - xi[0] - xi[1])),
(xi[1], 0, 1 - xi[0])), (xi[0], 0, 1))
return float(exact)
def test_serendipity_derivatives():
cell = UFCQuadrilateral()
S = Serendipity(cell, 2)
x = sympy.DeferredVector("X")
X, Y = x[0], x[1]
basis_functions = [
(1 - X) * (1 - Y),
Y * (1 - X),
X * (1 - Y),
X * Y,
Y * (1 - X) * (Y - 1),
X * Y * (Y - 1),
X * (1 - Y) * (X - 1),
X * Y * (X - 1),
]
points = [[0.5, 0.5], [0.25, 0.75]]
for alpha, actual in S.tabulate(2, points).items():
expect = list(
sympy.diff(basis, *zip([X, Y], alpha))
for basis in basis_functions)
expect = list(
[basis.subs(dict(zip([X, Y], point))) for point in points]
for basis in expect)
assert actual.shape == (8, 2)
assert np.allclose(np.asarray(expect, dtype=float),
actual.reshape(8, 2))
def __init__(self):
from sympy import pi, sin, cos
x = sympy.DeferredVector('x')
u = (
+pi * 2 * sin(pi * x[1]) * cos(pi * x[1]) * sin(pi * x[0])**2,
-pi * 2 * sin(pi * x[0]) * cos(pi * x[0]) * sin(pi * x[1])**2,
)
p = cos(pi * x[0]) * sin(pi * x[1])
self.solution = {
'u': {
'value': u,
'degree': MAX_DEGREE
},
'p': {
'value': p,
'degree': MAX_DEGREE
},
}
self.mu = 1.0
self.f = {
'value': _get_stokes_rhs(u, p, self.mu),
'degree': MAX_DEGREE,
}
return
def to_riesz(self, poly_set):
es = poly_set.get_expansion_set()
ed = poly_set.get_embedded_degree()
result = numpy.zeros(es.get_num_members(ed))
sd = self.ref_el.get_spatial_dimension()
X = sympy.DeferredVector('x')
dX = numpy.asarray([X[i] for i in range(sd)])
# evaluate bfs symbolically
bfs = es.tabulate(ed, [dX])[:, 0]
n = self.n
qwts = self.Q.get_weights()
for i in range(len(result)):
thing = sympy.lambdify(
X,
sum([sympy.diff(bfs[i], dxi) * ni for dxi, ni in zip(dX, n)]))
for j, pt in enumerate(self.deriv_dict.keys()):
result[i] += qwts[j] * self.f_at_qpts[j] * thing(pt)
return result
def _setup(self):
self._F0_N_ssro, self._F1_N_ssro, self._F0_e_ssro, self._F1_e_ssro, self._F0_RO_pulse, self._F1_RO_pulse, self._P_min1, self._P_0 = \
sympy.symbols('F0_N_ssro F1_N_ssro F0_e_ssro F1_e_ssro F0_RO_pulse F1_RO_pulse P_min1 P_0')
self._p_correlations = sympy.DeferredVector('p_correlations')
# The total error matrix is the tensor product of the nitrogen and electron error matrices.
# The nitrogen error matrix is built up of three matrices RO_err, CNOT_err and Init_err.
# RO_err reflects fidelities of the electron RO,
# CNOT_err reflects the fidelities in the pi (F0) and 2pi (F1) parts of the pi-2pi pulse (as CNOT gate).
# Init_err includes populations in the three nitrogen lines after initialization in -1.
#The inverse matrices are calculated before the tensorproduct, since this is much faster.
self.error_matrix_N = (sympy.Matrix([[self._F0_N_ssro, 1.-self._F1_N_ssro],[1.-self._F0_N_ssro, self._F1_N_ssro]]) * \
sympy.Matrix([[self._F0_RO_pulse, 1.-self._F1_RO_pulse, 0.],[1.-self._F0_RO_pulse, self._F1_RO_pulse, 1.]]) * \
sympy.Matrix([[self._P_min1,self._P_0],[self._P_0,self._P_min1],[1.-self._P_0-self._P_min1,1.-self._P_0-self._P_min1]]))
self.error_matrix_e = (sympy.Matrix(
[[self._F0_e_ssro, 1. - self._F1_e_ssro],
[1. - self._F0_e_ssro, self._F1_e_ssro]]))
self.correction_matrix_N = self.error_matrix_N.inv()
self.correction_matrix_e = self.error_matrix_e.inv()
self.correction_matrix = TensorProduct(self.correction_matrix_N,
self.correction_matrix_e)
corr_vec = self.correction_matrix * \
sympy.Matrix([self._p_correlations[0], self._p_correlations[1], self._p_correlations[2], self._p_correlations[3]])
corr_p_correlations = corr_vec
self.p0_formula = error.Formula()
self.p0_formula.formula = corr_p_correlations
def problem_taylor():
'''Taylor--Green vortex, cf.
<http://en.wikipedia.org/wiki/Taylor%E2%80%93Green_vortex>.
'''
def mesh_generator(n):
return RectangleMesh(Point(0.0, 0.0), Point(2 * pi, 2 * pi), n, n,
'crossed')
mu = 1.0
rho = 1.0
cell_type = triangle
x = sympy.DeferredVector('x')
t = sympy.symbols('t')
x0 = x[0]
x1 = x[1]
# F = sympy.exp(-2*mu*t)
F = 1 - 2 * mu * t
u = (sympy.sin(x0) * sympy.cos(x1) * F, -sympy.cos(x0) * sympy.sin(x1) * F,
0)
p = rho / 4 * (sympy.cos(2 * x0) + sympy.cos(2 * x1)) * F**2
solution = {
'u': {
'value': u,
'degree': MAX_DEGREE
},
'p': {
'value': p,
'degree': MAX_DEGREE
},
}
f = {
'value': _get_navier_stokes_rhs(u, p),
'degree': MAX_DEGREE,
}
return mesh_generator, solution, f, mu, rho, cell_type
def problem_guermond2():
'''Problem 2 from section 3.7.2 in
An overview of projection methods for incompressible flows;
Guermond, Minev, Shen;
Comp. Meth. in Appl. Mech. and Eng., vol. 195, 44-47, pp. 6011-6045;
<http://www.sciencedirect.com/science/article/pii/S0045782505004640>.
'''
def mesh_generator(n):
return UnitSquareMesh(n, n, 'crossed')
cell_type = triangle
x = sympy.DeferredVector('x')
t = sympy.symbols('t')
u = (sympy.sin(x[0] + t) * sympy.sin(x[1] + t),
sympy.cos(x[0] + t) * sympy.cos(x[1] + t))
p = sympy.sin(x[0] - x[1] + t)
solution = {
'u': {
'value': u,
'degree': MAX_DEGREE
},
'p': {
'value': p,
'degree': MAX_DEGREE
},
}
f = {
'value': _get_navier_stokes_rhs(u, p),
'degree': MAX_DEGREE,
}
mu = 1.0
rho = 1.0
return mesh_generator, solution, f, mu, rho, cell_type
def problem_guermond1():
'''Problem 1 from section 3.7.1 in
An overview of projection methods for incompressible flows;
Guermond, Minev, Shen;
Comp. Meth. in Appl. Mech. and Eng., vol. 195, 44-47, pp. 6011-6045;
<http://www.sciencedirect.com/science/article/pii/S0045782505004640>.
'''
def mesh_generator(n):
return RectangleMesh(Point(-1, -1), Point(1, 1), n, n, 'crossed')
cell_type = triangle
x = sympy.DeferredVector('x')
t = sympy.symbols('t')
# m = sympy.exp(t) - 0.0
m = sympy.sin(t)
u = (+pi * m * 2 * sympy.sin(pi * x[1]) * sympy.cos(pi * x[1]) *
sympy.sin(pi * x[0])**2, -pi * m * 2 * sympy.sin(pi * x[0]) *
sympy.cos(pi * x[0]) * sympy.sin(pi * x[1])**2)
p = m * sympy.cos(pi * x[0]) * sympy.sin(pi * x[1])
solution = {
'u': {
'value': u,
'degree': MAX_DEGREE
},
'p': {
'value': p,
'degree': MAX_DEGREE
}
}
f = {'value': _get_navier_stokes_rhs(u, p), 'degree': MAX_DEGREE}
mu = 1.0
rho = 1.0
return mesh_generator, solution, f, mu, rho, cell_type
def __init__(self, f, past, helpers=(), control_pars=(), n_basic=None):
self.past = past
self.t, self.y, self.diff = self.past[-1]
self.n = len(self.y)
self.n_basic = n_basic or self.n
self.last_garbage = -1
self.old_new_y = None
self.parameters = []
from jitcdde._jitcdde import t, y, current_y, past_y, anchors
Y = sympy.DeferredVector("Y")
substitutions = list(helpers[::-1]) + [(y(i), Y[i])
for i in range(self.n)]
past_calls = 0
f_wc = []
for entry in f():
new_entry = entry.subs(substitutions).simplify(ratio=1.0)
past_calls += new_entry.count(anchors)
f_wc.append(new_entry)
F = sympy.lambdify([t, Y] + list(control_pars), f_wc,
[{
anchors.__name__: self.get_past_anchors,
past_y.__name__: self.get_past_value
}, "math"])
self.f = lambda *args: np.array(F(*args)).flatten()
self.anchor_mem = (len(past) - 1) * np.ones(past_calls, dtype=int)
def problem_whirl():
'''Example from Teodora I. Mitkova's text
"Finite-Elemente-Methoden fur die Stokes-Gleichungen".
'''
def mesh_generator(n):
return UnitSquareMesh(n, n, 'left/right')
cell_type = triangle
x = sympy.DeferredVector('x')
# t = sympy.symbols('t')
# Note that the exact solution is indeed div-free.
u = (x[0]**2 * (1 - x[0])**2 * 2 * x[1] * (1 - x[1]) * (2 * x[1] - 1),
x[1]**2 * (1 - x[1])**2 * 2 * x[0] * (1 - x[0]) * (1 - 2 * x[0]))
p = x[0] * (1 - x[0]) * x[1] * (1 - x[1])
solution = {
'u': {
'value': u,
'degree': 7
},
'p': {
'value': p,
'degree': 4
},
}
f = {
'value': _get_navier_stokes_rhs(u, p),
'degree': MAX_DEGREE,
}
mu = 1.0
rho = 1.0
return mesh_generator, solution, f, mu, rho, cell_type
def _lambdify(self, outputs):
# TODO : We could forgo this substitution for generation speed
# purposes and have lots of args for lambdify (like it used to be
# done) but there may be some limitations on number of args.
subs = {}
vec_inputs = []
if self.specifieds is None:
def_vecs = ['q', 'u', 'p']
else:
def_vecs = ['q', 'u', 'r', 'p']
for syms, vec_name in zip(self.inputs, def_vecs):
v = sm.DeferredVector(vec_name)
for i, sym in enumerate(syms):
subs[sym] = v[i]
vec_inputs.append(v)
try:
outputs = [me.msubs(output, subs) for output in outputs]
except AttributeError:
# msubs doesn't exist in SymPy < 0.7.6.
outputs = [output.subs(subs) for output in outputs]
modules = [{'ImmutableMatrix': np.array}, 'numpy']
return sm.lambdify(vec_inputs, outputs, modules=modules)
def _get_data(dim, n, point_fun, symbolic):
# points
idxs = numpy.array(get_all_exponents(dim + 1, n)[-1])
points = point_fun(idxs)
# weights
weights = numpy.empty(len(points))
kk = 0
for idx in idxs:
# Define the polynomial which to integrate over the simplex.
t = sympy.DeferredVector("t")
g = prod(
sympy.poly((t[i] - point_fun(k)) / (point_fun(m) - point_fun(k)))
for i, m in enumerate(idx) for k in range(m))
if isinstance(g, int):
exp = [0] * (dim + 1)
coeffs = g
else:
# make sure the exponents are all of the correct length
exp = [list(m) + [0] * (dim - len(m) + 1) for m in g.monoms()]
coeffs = g.coeffs()
weights[kk] = sum(c * integrate_bary(m, symbolic)
for c, m in zip(coeffs, exp))
kk += 1
return weights, points
def problem_taylor_cylindrical():
"""Taylor--Green vortex, cf.
<http://en.wikipedia.org/wiki/Taylor%E2%80%93Green_vortex>.
"""
alpha = 1.0
def mesh_generator(n):
return RectangleMesh(
Point(0.0 + alpha, 0.0), Point(2 * pi + alpha, 2 * pi), n, n, "crossed"
)
mu = 1.0
rho = 1.0
cell_type = triangle
x = sympy.DeferredVector("x")
t = sympy.symbols("t")
x0 = x[0] - alpha
x1 = x[1]
F = 1 - 2 * mu * t
u = (
+sympy.sin(x0) * sympy.cos(x1) * F / x[0],
-sympy.cos(x0) * sympy.sin(x1) * F / x[0],
0,
)
p = rho / 4 * (sympy.cos(2 * x0) + sympy.cos(2 * x1)) * F ** 2
solution = {
"u": {"value": u, "degree": numpy.infty},
"p": {"value": p, "degree": numpy.infty},
}
f = {"value": _get_navier_stokes_rhs_cylindrical(u, p), "degree": numpy.infty}
return mesh_generator, solution, f, mu, rho, cell_type
def problem_flat():
'''Nothing interesting happening in the domain.
'''
def mesh_generator(n):
return UnitSquareMesh(n, n, 'left/right')
cell_type = triangle
x = sympy.DeferredVector('x')
u = (0.0 * x[0], 0.0 * x[1])
# Only grad(p) is of physical significance, and the numerical procedure
# chooses p such that int(p) = 0. Also, numerical solutions have n.grad(p)
# = 0 for non-penetration boundaries. Make sure that the exact solution
# respects those things to make error analysis easier.
# p = sympy.cos(pi * x[1])
p = -x[1]
solution = {
'u': {
'value': u,
'degree': 1
},
'p': {
'value': p,
'degree': 1
},
}
f = {
'value': _get_navier_stokes_rhs(u, p),
'degree': MAX_DEGREE,
}
mu = 1.0
rho = 1.0
return mesh_generator, solution, f, mu, rho, cell_type
def __call__(self, fn):
"""Evaluate the functional on the function fn. Note that this depends
on sympy being able to differentiate fn."""
x = list(self.deriv_dict.keys())[0]
X = sympy.DeferredVector('x')
dX = numpy.asarray([X[i] for i in range(len(x))])
dvars = tuple(d for d, a in zip(dX, self.alpha) for count in range(a))
return sympy.diff(fn(X), *dvars).evalf(subs=dict(zip(dX, x)))
def _newton_cotes(n, point_fun):
'''
Construction after
P. Silvester,
Symmetric quadrature formulae for simplexes
Math. Comp., 24, 95-100 (1970),
<https://doi.org/10.1090/S0025-5718-1970-0258283-6>.
'''
degree = n
# points
idx = numpy.array([[i, j, k, n - i - j - k] for i in range(n + 1)
for j in range(n + 1 - i)
for k in range(n + 1 - i - j)])
bary = point_fun(idx, n)
points = bary[:, [1, 2, 3]]
# weights
if n == 0:
weights = numpy.ones(1)
return points, weights, degree
def get_poly(t, m, n):
return sympy.prod([
sympy.poly(
(t - point_fun(k, n)) / (point_fun(m, n) - point_fun(k, n)))
for k in range(m)
])
weights = numpy.empty(len(points))
idx = 0
for i in range(n + 1):
for j in range(n + 1 - i):
for k in range(n + 1 - i - j):
L = n - i - j - k
# Compute weight.
# Define the polynomial which to integrate over the
# tetrahedron.
t = sympy.DeferredVector('t')
g = get_poly(t[0], i, n) \
* get_poly(t[1], j, n) \
* get_poly(t[2], k, n) \
* get_poly(t[3], L, n)
# The integral of monomials over a tetrahedron are well-known,
# see Silvester.
weights[idx] = numpy.sum([
c * numpy.prod([math.factorial(k) for k in m]) * 6.0 /
math.factorial(numpy.sum(m) + 3)
for m, c in zip(g.monoms(), g.coeffs())
])
idx += 1
return points, weights, degree
def test_lambdify_matrix_vec_input():
X = sympy.DeferredVector('X')
M = Matrix([[X[0]**2, X[0] * X[1], X[0] * X[2]],
[X[1] * X[0], X[1]**2, X[1] * X[2]],
[X[2] * X[0], X[2] * X[1], X[2]**2]])
f = lambdify(X, M, [{'ImmutableMatrix': numpy.array}, "numpy"])
Xh = array([1.0, 2.0, 3.0])
expected = array([[Xh[0]**2, Xh[0] * Xh[1], Xh[0] * Xh[2]],
[Xh[1] * Xh[0], Xh[1]**2, Xh[1] * Xh[2]],
[Xh[2] * Xh[0], Xh[2] * Xh[1], Xh[2]**2]])
actual = f(Xh)
assert numpy.allclose(actual, expected)
def _integrate_exact(k, pyra):
def f(x):
return x[0] ** int(k[0]) * x[1] ** int(k[1]) * x[2] ** int(k[2])
# map the reference hexahedron [-1,1]^3 to the pyramid
xi = sympy.DeferredVector("xi")
pxi = (
+pyra[0] * (1 - xi[0]) * (1 - xi[1]) * (1 - xi[2]) / 8
+ pyra[1] * (1 + xi[0]) * (1 - xi[1]) * (1 - xi[2]) / 8
+ pyra[2] * (1 + xi[0]) * (1 + xi[1]) * (1 - xi[2]) / 8
+ pyra[3] * (1 - xi[0]) * (1 + xi[1]) * (1 - xi[2]) / 8
+ pyra[4] * (1 + xi[2]) / 2
)
pxi = [sympy.expand(pxi[0]), sympy.expand(pxi[1]), sympy.expand(pxi[2])]
# determinant of the transformation matrix
J = sympy.Matrix(
[
[
sympy.diff(pxi[0], xi[0]),
sympy.diff(pxi[0], xi[1]),
sympy.diff(pxi[0], xi[2]),
],
[
sympy.diff(pxi[1], xi[0]),
sympy.diff(pxi[1], xi[1]),
sympy.diff(pxi[1], xi[2]),
],
[
sympy.diff(pxi[2], xi[0]),
sympy.diff(pxi[2], xi[1]),
sympy.diff(pxi[2], xi[2]),
],
]
)
det_J = sympy.det(J)
# we cannot use abs(), see <https://github.com/sympy/sympy/issues/4212>.
# abs_det_J = sympy.Piecewise((det_J, det_J >= 0), (-det_J, det_J < 0))
# This is quite the leap of faith, but sympy will cowardly bail out
# otherwise.
abs_det_J = det_J
exact = sympy.integrate(
sympy.integrate(
sympy.integrate(abs_det_J * f(pxi), (xi[2], -1, 1)), (xi[1], -1, +1)
),
(xi[0], -1, +1),
)
return float(exact)
def _integrate_exact(f, hexa):
xi = sympy.DeferredVector('xi')
pxi = \
+ hexa[0] * 0.125*(1.0 - xi[0])*(1.0 - xi[1])*(1.0 - xi[2]) \
+ hexa[1] * 0.125*(1.0 + xi[0])*(1.0 - xi[1])*(1.0 - xi[2]) \
+ hexa[2] * 0.125*(1.0 + xi[0])*(1.0 + xi[1])*(1.0 - xi[2]) \
+ hexa[3] * 0.125*(1.0 - xi[0])*(1.0 + xi[1])*(1.0 - xi[2]) \
+ hexa[4] * 0.125*(1.0 - xi[0])*(1.0 - xi[1])*(1.0 + xi[2]) \
+ hexa[5] * 0.125*(1.0 + xi[0])*(1.0 - xi[1])*(1.0 + xi[2]) \
+ hexa[6] * 0.125*(1.0 + xi[0])*(1.0 + xi[1])*(1.0 + xi[2]) \
+ hexa[7] * 0.125*(1.0 - xi[0])*(1.0 + xi[1])*(1.0 + xi[2])
pxi = [
sympy.expand(pxi[0]),
sympy.expand(pxi[1]),
sympy.expand(pxi[2]),
]
# determinant of the transformation matrix
J = sympy.Matrix([
[
sympy.diff(pxi[0], xi[0]),
sympy.diff(pxi[0], xi[1]),
sympy.diff(pxi[0], xi[2])
],
[
sympy.diff(pxi[1], xi[0]),
sympy.diff(pxi[1], xi[1]),
sympy.diff(pxi[1], xi[2])
],
[
sympy.diff(pxi[2], xi[0]),
sympy.diff(pxi[2], xi[1]),
sympy.diff(pxi[2], xi[2])
],
])
det_J = sympy.det(J)
# we cannot use abs(), see <https://github.com/sympy/sympy/issues/4212>.
# pylint: disable=invalid-unary-operand-type
abs_det_J = sympy.Piecewise((det_J, det_J >= 0), (-det_J, det_J < 0))
g_xi = f(pxi)
exact = \
sympy.integrate(
sympy.integrate(
sympy.integrate(abs_det_J * g_xi, (xi[2], -1, 1)),
(xi[1], -1, 1)
),
(xi[0], -1, 1)
)
return float(exact)
def to_riesz(self, poly_set):
x = list(self.deriv_dict.keys())[0]
X = sympy.DeferredVector('x')
dx = numpy.asarray([X[i] for i in range(len(x))])
es = poly_set.get_expansion_set()
ed = poly_set.get_embedded_degree()
bfs = es.tabulate(ed, [dx])[:, 0]
# Expand the multi-index as a series of variables to
# differentiate with respect to.
dvars = tuple(d for d, a in zip(dx, self.alpha) for count in range(a))
return numpy.asarray(
[sympy.lambdify(X, sympy.diff(b, *dvars))(x) for b in bfs])
def problem_flat_cylindrical():
"""Nothing interesting happening in the domain.
"""
def mesh_generator(n):
return UnitSquareMesh(n, n, "left/right")
cell_type = triangle
# Coordinates ordered as (r, z, phi).
x = sympy.DeferredVector("x")
u = (0.0 * x[0], 0.0 * x[1], 0.0 * x[2])
p = -9.81 * x[1]
solution = {"u": {"value": u, "degree": 1}, "p": {"value": p, "degree": 1}}
f = {"value": _get_navier_stokes_rhs_cylindrical(u, p), "degree": numpy.infty}
mu = 1.0
rho = 1.0
return mesh_generator, solution, f, mu, rho, cell_type
def _get_stokes_rhs(u, p, mu):
'''Given a solution u of the Cartesian Navier-Stokes equations, return
a matching right-hand side f.
'''
x = sympy.DeferredVector('x')
# Make sure that the exact solution is indeed analytically div-free.
d = sympy.diff(u[0], x[0]) + sympy.diff(u[1], x[1])
d = sympy.simplify(d)
assert d == 0
f0 = (-mu * (sympy.diff(u[0], x[0], 2) + sympy.diff(u[0], x[1], 2)) +
sympy.diff(p, x[0]))
f1 = (-mu * (sympy.diff(u[1], x[0], 2) + sympy.diff(u[1], x[1], 2)) +
sympy.diff(p, x[1]))
f = (sympy.simplify(f0), sympy.simplify(f1))
return f
def to_riesz(self, poly_set):
x = list(self.deriv_dict.keys())[0]
X = sympy.DeferredVector('x')
dx = numpy.asarray([X[i] for i in range(len(x))])
es = poly_set.get_expansion_set()
ed = poly_set.get_embedded_degree()
bfs = es.tabulate(ed, [dx])[:, 0]
# We need the gradient dotted with the normal.
return numpy.asarray([
sympy.lambdify(
X,
sum([sympy.diff(b, dxi) * ni
for dxi, ni in zip(dx, self.n)]))(x) for b in bfs
])
def _integrate_exact(f, quadrilateral):
xi = sympy.DeferredVector("xi")
pxi = (quadrilateral[0] * 0.25 * (1.0 + xi[0]) * (1.0 + xi[1]) +
quadrilateral[1] * 0.25 * (1.0 - xi[0]) * (1.0 + xi[1]) +
quadrilateral[2] * 0.25 * (1.0 - xi[0]) * (1.0 - xi[1]) +
quadrilateral[3] * 0.25 * (1.0 + xi[0]) * (1.0 - xi[1]))
pxi = [sympy.expand(pxi[0]), sympy.expand(pxi[1])]
# determinant of the transformation matrix
det_J = +sympy.diff(pxi[0], xi[0]) * sympy.diff(
pxi[1], xi[1]) - sympy.diff(pxi[1], xi[0]) * sympy.diff(pxi[0], xi[1])
# we cannot use abs(), see <https://github.com/sympy/sympy/issues/4212>.
abs_det_J = sympy.Piecewise((det_J, det_J >= 0), (-det_J, det_J < 0))
g_xi = f(pxi)
exact = sympy.integrate(sympy.integrate(abs_det_J * g_xi, (xi[1], -1, 1)),
(xi[0], -1, 1))
return float(exact)
def problem_guermond1_cylindrical():
"""Cylindrical variation of Guermond's test problem.
"""
alpha = 1.5
def mesh_generator(n):
return RectangleMesh(
Point(-1 + alpha, -1), Point(1 + alpha, 1), n, n, "crossed"
)
cell_type = triangle
x = sympy.DeferredVector("x")
t = sympy.symbols("t")
x0 = x[0] - alpha
x1 = x[1]
# m = sympy.exp(t) - 0.0
m = sympy.sin(t) + 1.0
u = (
+pi
* m
* 2
* sympy.sin(pi * x1)
* sympy.cos(pi * x1)
* sympy.sin(pi * x0) ** 2
/ x[0],
-pi
* m
* 2
* sympy.sin(pi * x0)
* sympy.cos(pi * x0)
* sympy.sin(pi * x1) ** 2
/ x[0],
0,
)
p = m * sympy.cos(pi * x0) * sympy.sin(pi * x1)
solution = {
"u": {"value": u, "degree": numpy.infty},
"p": {"value": p, "degree": numpy.infty},
}
f = {"value": _get_navier_stokes_rhs_cylindrical(u, p), "degree": numpy.infty}
mu = 1.0
rho = 1.0
return mesh_generator, solution, f, mu, rho, cell_type
def _newton_cotes(n, point_fun):
degree = n
# points
idx = numpy.array([[i, j, k, n - i - j - k] for i in range(n + 1)
for j in range(n + 1 - i)
for k in range(n + 1 - i - j)])
points = point_fun(idx, n)
# weights
if n == 0:
weights = numpy.ones(1)
return weights, points, weights, degree
def get_poly(t, m, n):
return sympy.prod([
sympy.poly(
(t - point_fun(k, n)) / (point_fun(m, n) - point_fun(k, n)))
for k in range(m)
])
weights = numpy.empty(len(points))
idx = 0
for i in range(n + 1):
for j in range(n + 1 - i):
for k in range(n + 1 - i - j):
L = n - i - j - k
# Compute weight.
# Define the polynomial which to integrate over the
# tetrahedron.
t = sympy.DeferredVector("t")
g = (get_poly(t[0], i, n) * get_poly(t[1], j, n) *
get_poly(t[2], k, n) * get_poly(t[3], L, n))
# The integral of monomials over a tetrahedron are well-known,
# see Silvester.
weights[idx] = numpy.sum([
c * numpy.prod([math.factorial(k) for k in m]) * 6.0 /
math.factorial(numpy.sum(m) + 3)
for m, c in zip(g.monoms(), g.coeffs())
])
idx += 1
return weights, points, degree, citation
def __init__(self):
x = sympy.DeferredVector('x')
u = (0 * x[0], 0 * x[1])
p = -9.81 * x[1]
self.solution = {
'u': {
'value': u,
'degree': 1
},
'p': {
'value': p,
'degree': 1
},
}
self.mu = 1.0
self.f = {
'value': _get_stokes_rhs(u, p, self.mu),
'degree': MAX_DEGREE,
}
return
def problem_whirl_cylindrical():
"""Example from Teodora I. Mitkova's text
"Finite-Elemente-Methoden fur die Stokes-Gleichungen", adapted for
cylindrical Navier-Stokes.
"""
alpha = 1.0
def mesh_generator(n):
# return UnitSquareMesh(n, n, 'left/right')
return RectangleMesh(
Point(alpha, 0.0), Point(1.0 + alpha, 1.0), n, n, "left/right"
)
cell_type = triangle
x = sympy.DeferredVector("x")
# Note that the exact solution is indeed div-free.
x0 = x[0] - alpha
x1 = x[1]
u = (
x0 ** 2 * (1 - x0) ** 2 * 2 * x1 * (1 - x1) * (2 * x1 - 1) / x[0],
x1 ** 2 * (1 - x1) ** 2 * 2 * x0 * (1 - x0) * (1 - 2 * x0) / x[0],
0,
)
p = x0 * (1 - x0) * x1 * (1 - x1)
solution = {
"u": {"value": u, "degree": numpy.infty},
"p": {"value": p, "degree": 4},
}
plot_solution = False
if plot_solution:
sol_u = Expression(
(helpers.ccode(u[0]), helpers.ccode(u[1])),
degree=numpy.infty,
t=0.0,
cell=cell_type,
)
plot(sol_u, mesh=mesh_generator(20))
f = {"value": _get_navier_stokes_rhs_cylindrical(u, p), "degree": numpy.infty}
mu = 1.0
rho = 1.0
return mesh_generator, solution, f, mu, rho, cell_type
def _get_navier_stokes_rhs(u, p):
'''Given a solution u of the Cartesian Navier-Stokes equations, return
a matching right-hand side f.
'''
x = sympy.DeferredVector('x')
t, mu, rho = sympy.symbols('t, mu, rho')
# Make sure that the exact solution is indeed analytically div-free.
d = sympy.diff(u[0], x[0]) + sympy.diff(u[1], x[1])
d = sympy.simplify(d)
assert d == 0
# Get right-hand side associated with this solution, i.e., according
# the Navier-Stokes
#
# rho (du_x/dt + u_x du_x/dx + u_y du_x/dy)
# = - dp/dx + mu [d^2u_x/dx^2 + d^2u_x/dy^2] + f_x,
# rho (du_y/dt + u_x du_y/dx + u_y du_y/dy)
# = - dp/dx + mu [d^2u_y/dx^2 + d^2u_y/dy^2] + f_y,
# du_x/dx + du_y/dy = 0.
#
# rho (du/dt + (u.\nabla)u) = -\nabla p + mu [\div(\nabla u)] + f,
# div(u) = 0.
#
f0 = rho * (sympy.diff(u[0], t)
+ u[0] * sympy.diff(u[0], x[0])
+ u[1] * sympy.diff(u[0], x[1])
) \
+ sympy.diff(p, x[0]) \
- mu * (sympy.diff(u[0], x[0], 2) + sympy.diff(u[0], x[1], 2))
f1 = rho * (sympy.diff(u[1], t)
+ u[0] * sympy.diff(u[1], x[0])
+ u[1] * sympy.diff(u[1], x[1])
) \
+ sympy.diff(p, x[1]) \
- mu * (sympy.diff(u[1], x[0], 2) + sympy.diff(u[1], x[1], 2))
f = (sympy.simplify(f0), sympy.simplify(f1))
return f