def main(nelems:int, etype:str, btype:str, degree:int, poisson:float, angle:float, restol:float, trim:bool):
'''
Deformed hyperelastic plate.
.. arguments::
nelems [10]
Number of elements along edge.
etype [square]
Type of elements (square/triangle/mixed).
btype [std]
Type of basis function (std/spline).
degree [1]
Polynomial degree.
poisson [.25]
Poisson's ratio, nonnegative and stricly smaller than 1/2.
angle [20]
Rotation angle for right clamp (degrees).
restol [1e-10]
Newton tolerance.
trim [no]
Create circular-shaped hole.
'''
domain, geom = mesh.unitsquare(nelems, etype)
if trim:
domain = domain.trim(function.norm2(geom-.5)-.2, maxrefine=2)
bezier = domain.sample('bezier', 5)
ns = function.Namespace(fallback_length=domain.ndims)
ns.x = geom
ns.angle = angle * numpy.pi / 180
ns.lmbda = 2 * poisson
ns.mu = 1 - 2 * poisson
ns.ubasis = domain.basis(btype, degree=degree)
ns.u_i = 'ubasis_k ?lhs_ki'
ns.X_i = 'x_i + u_i'
ns.strain_ij = '.5 (d(u_i, x_j) + d(u_j, x_i))'
ns.energy = 'lmbda strain_ii strain_jj + 2 mu strain_ij strain_ij'
sqr = domain.boundary['left'].integral('u_k u_k J(x)' @ ns, degree=degree*2)
sqr += domain.boundary['right'].integral('((u_0 - x_1 sin(2 angle) - cos(angle) + 1)^2 + (u_1 - x_1 (cos(2 angle) - 1) + sin(angle))^2) J(x)' @ ns, degree=degree*2)
cons = solver.optimize('lhs', sqr, droptol=1e-15)
energy = domain.integral('energy J(x)' @ ns, degree=degree*2)
lhs0 = solver.optimize('lhs', energy, constrain=cons)
X, energy = bezier.eval(['X', 'energy'] @ ns, lhs=lhs0)
export.triplot('linear.png', X, energy, tri=bezier.tri, hull=bezier.hull)
ns.strain_ij = '.5 (d(u_i, x_j) + d(u_j, x_i) + d(u_k, x_i) d(u_k, x_j))'
ns.energy = 'lmbda strain_ii strain_jj + 2 mu strain_ij strain_ij'
energy = domain.integral('energy J(x)' @ ns, degree=degree*2)
lhs1 = solver.minimize('lhs', energy, arguments=dict(lhs=lhs0), constrain=cons).solve(restol)
X, energy = bezier.eval(['X', 'energy'] @ ns, lhs=lhs1)
export.triplot('nonlinear.png', X, energy, tri=bezier.tri, hull=bezier.hull)
return lhs0, lhs1
def main(nelems: int, etype: str, btype: str, degree: int, poisson: float):
'''
Horizontally loaded linear elastic plate.
.. arguments::
nelems [10]
Number of elements along edge.
etype [square]
Type of elements (square/triangle/mixed).
btype [std]
Type of basis function (std/spline), with availability depending on the
configured element type.
degree [1]
Polynomial degree.
poisson [.25]
Poisson's ratio, nonnegative and strictly smaller than 1/2.
'''
domain, geom = mesh.unitsquare(nelems, etype)
ns = function.Namespace()
ns.x = geom
ns.basis = domain.basis(btype, degree=degree).vector(2)
ns.u_i = 'basis_ni ?lhs_n'
ns.X_i = 'x_i + u_i'
ns.lmbda = 2 * poisson
ns.mu = 1 - 2 * poisson
ns.strain_ij = '(d(u_i, x_j) + d(u_j, x_i)) / 2'
ns.stress_ij = 'lmbda strain_kk δ_ij + 2 mu strain_ij'
sqr = domain.boundary['left'].integral('u_k u_k J(x)' @ ns,
degree=degree * 2)
sqr += domain.boundary['right'].integral('(u_0 - .5)^2 J(x)' @ ns,
degree=degree * 2)
cons = solver.optimize('lhs', sqr, droptol=1e-15)
res = domain.integral('d(basis_ni, x_j) stress_ij J(x)' @ ns,
degree=degree * 2)
lhs = solver.solve_linear('lhs', res, constrain=cons)
bezier = domain.sample('bezier', 5)
X, sxy = bezier.eval(['X', 'stress_01'] @ ns, lhs=lhs)
export.triplot('shear.png', X, sxy, tri=bezier.tri, hull=bezier.hull)
return cons, lhs
def main(nelems: int, etype: str, btype: str, degree: int, traction: float,
maxrefine: int, radius: float, poisson: float):
'''
Horizontally loaded linear elastic plate with FCM hole.
.. arguments::
nelems [9]
Number of elements along edge.
etype [square]
Type of elements (square/triangle/mixed).
btype [std]
Type of basis function (std/spline), with availability depending on the
selected element type.
degree [2]
Polynomial degree.
traction [.1]
Far field traction (relative to Young's modulus).
maxrefine [2]
Number or refinement levels used for the finite cell method.
radius [.5]
Cut-out radius.
poisson [.3]
Poisson's ratio, nonnegative and strictly smaller than 1/2.
'''
domain0, geom = mesh.unitsquare(nelems, etype)
domain = domain0.trim(function.norm2(geom) - radius, maxrefine=maxrefine)
ns = function.Namespace()
ns.x = geom
ns.lmbda = 2 * poisson
ns.mu = 1 - poisson
ns.ubasis = domain.basis(btype, degree=degree).vector(2)
ns.u_i = 'ubasis_ni ?lhs_n'
ns.X_i = 'x_i + u_i'
ns.strain_ij = '(d(u_i, x_j) + d(u_j, x_i)) / 2'
ns.stress_ij = 'lmbda strain_kk δ_ij + 2 mu strain_ij'
ns.r2 = 'x_k x_k'
ns.R2 = radius**2 / ns.r2
ns.k = (3 - poisson) / (1 + poisson) # plane stress parameter
ns.scale = traction * (1 + poisson) / 2
ns.uexact_i = 'scale (x_i ((k + 1) (0.5 + R2) + (1 - R2) R2 (x_0^2 - 3 x_1^2) / r2) - 2 δ_i1 x_1 (1 + (k - 1 + R2) R2))'
ns.du_i = 'u_i - uexact_i'
sqr = domain.boundary['left,bottom'].integral('(u_i n_i)^2 J(x)' @ ns,
degree=degree * 2)
cons = solver.optimize('lhs', sqr, droptol=1e-15)
sqr = domain.boundary['top,right'].integral('du_k du_k J(x)' @ ns,
degree=20)
cons = solver.optimize('lhs', sqr, droptol=1e-15, constrain=cons)
res = domain.integral('d(ubasis_ni, x_j) stress_ij J(x)' @ ns,
degree=degree * 2)
lhs = solver.solve_linear('lhs', res, constrain=cons)
bezier = domain.sample('bezier', 5)
X, stressxx = bezier.eval(['X', 'stress_00'] @ ns, lhs=lhs)
export.triplot('stressxx.png',
X,
stressxx,
tri=bezier.tri,
hull=bezier.hull)
err = domain.integral('<du_k du_k, sum:ij(d(du_i, x_j)^2)>_n J(x)' @ ns,
degree=max(degree, 3) * 2).eval(lhs=lhs)**.5
treelog.user('errors: L2={:.2e}, H1={:.2e}'.format(*err))
return err, cons, lhs
def main(nelems: int, ndims: int, degree: int, timescale: float,
newtontol: float, endtime: float):
'''
Burgers equation on a 1D or 2D periodic domain.
.. arguments::
nelems [20]
Number of elements along a single dimension.
ndims [1]
Number of spatial dimensions.
degree [1]
Polynomial degree for discontinuous basis functions.
timescale [.5]
Fraction of timestep and element size: timestep=timescale/nelems.
newtontol [1e-5]
Newton tolerance.
endtime [inf]
Stopping time.
'''
domain, geom = mesh.rectilinear([numpy.linspace(0, 1, nelems + 1)] * ndims,
periodic=range(ndims))
ns = function.Namespace()
ns.x = geom
ns.basis = domain.basis('discont', degree=degree)
ns.u = 'basis_n ?lhs_n'
ns.f = '.5 u^2'
ns.C = 1
res = domain.integral('-basis_n,0 f d:x' @ ns, degree=5)
res += domain.interfaces.integral(
'-[basis_n] n_0 ({f} - .5 C [u] n_0) d:x' @ ns, degree=degree * 2)
inertia = domain.integral('basis_n u d:x' @ ns, degree=5)
sqr = domain.integral(
'(u - exp(-?y_i ?y_i)(y_i = 5 (x_i - 0.5_i)))^2 d:x' @ ns, degree=5)
lhs0 = solver.optimize('lhs', sqr)
timestep = timescale / nelems
bezier = domain.sample('bezier', 7)
with treelog.iter.plain(
'timestep',
solver.impliciteuler('lhs',
res,
inertia,
timestep=timestep,
lhs0=lhs0,
newtontol=newtontol)) as steps:
for itime, lhs in enumerate(steps):
x, u = bezier.eval(['x_i', 'u'] @ ns, lhs=lhs)
export.triplot('solution.png',
x,
u,
tri=bezier.tri,
hull=bezier.hull,
clim=(0, 1))
if itime * timestep >= endtime:
break
return lhs
def main(nelems: int, etype: str, btype: str, degree: int):
'''
Laplace problem on a unit square.
.. arguments::
nelems [10]
Number of elements along edge.
etype [square]
Type of elements (square/triangle/mixed).
btype [std]
Type of basis function (std/spline), availability depending on the
selected element type.
degree [1]
Polynomial degree.
'''
# A unit square domain is created by calling the
# :func:`nutils.mesh.unitsquare` mesh generator, with the number of elements
# along an edge as the first argument, and the type of elements ("square",
# "triangle", or "mixed") as the second. The result is a topology object
# ``domain`` and a vectored valued geometry function ``geom``.
domain, geom = mesh.unitsquare(nelems, etype)
# To be able to write index based tensor contractions, we need to bundle all
# relevant functions together in a namespace. Here we add the geometry ``x``,
# a scalar ``basis``, and the solution ``u``. The latter is formed by
# contracting the basis with a to-be-determined solution vector ``?lhs``.
ns = function.Namespace()
ns.x = geom
ns.basis = domain.basis(btype, degree=degree)
ns.u = 'basis_n ?lhs_n'
# We are now ready to implement the Laplace equation. In weak form, the
# solution is a scalar field :math:`u` for which:
#
# .. math:: ∀ v: ∫_Ω \frac{dv}{dx_i} \frac{du}{dx_i} - ∫_{Γ_n} v f = 0.
#
# By linearity the test function :math:`v` can be replaced by the basis that
# spans its space. The result is an integral ``res`` that evaluates to a
# vector matching the size of the function space.
res = domain.integral('d(basis_n, x_i) d(u, x_i) J(x)' @ ns,
degree=degree * 2)
res -= domain.boundary['right'].integral(
'basis_n cos(1) cosh(x_1) J(x)' @ ns, degree=degree * 2)
# The Dirichlet constraints are set by finding the coefficients that minimize
# the error:
#
# .. math:: \min_u ∫_{\Gamma_d} (u - u_d)^2
#
# The resulting ``cons`` array holds numerical values for all the entries of
# ``?lhs`` that contribute (up to ``droptol``) to the minimization problem.
# All remaining entries are set to ``NaN``, signifying that these degrees of
# freedom are unconstrained.
sqr = domain.boundary['left'].integral('u^2 J(x)' @ ns, degree=degree * 2)
sqr += domain.boundary['top'].integral(
'(u - cosh(1) sin(x_0))^2 J(x)' @ ns, degree=degree * 2)
cons = solver.optimize('lhs', sqr, droptol=1e-15)
# The unconstrained entries of ``?lhs`` are to be determined such that the
# residual vector evaluates to zero in the corresponding entries. This step
# involves a linearization of ``res``, resulting in a jacobian matrix and
# right hand side vector that are subsequently assembled and solved. The
# resulting ``lhs`` array matches ``cons`` in the constrained entries.
lhs = solver.solve_linear('lhs', res, constrain=cons)
# Once all entries of ``?lhs`` are establised, the corresponding solution can
# be vizualised by sampling values of ``ns.u`` along with physical
# coordinates ``ns.x``, with the solution vector provided via the
# ``arguments`` dictionary. The sample members ``tri`` and ``hull`` provide
# additional inter-point information required for drawing the mesh and
# element outlines.
bezier = domain.sample('bezier', 9)
x, u = bezier.eval(['x', 'u'] @ ns, lhs=lhs)
export.triplot('solution.png', x, u, tri=bezier.tri, hull=bezier.hull)
# To confirm that our computation is correct, we use our knowledge of the
# analytical solution to evaluate the L2-error of the discrete result.
err = domain.integral('(u - sin(x_0) cosh(x_1))^2 J(x)' @ ns,
degree=degree * 2).eval(lhs=lhs)**.5
treelog.user('L2 error: {:.2e}'.format(err))
return cons, lhs, err
def main(etype: str, btype: str, degree: int, nrefine: int):
'''
Adaptively refined Laplace problem on an L-shaped domain.
.. arguments::
etype [square]
Type of elements (square/triangle/mixed).
btype [h-std]
Type of basis function (h/th-std/spline), with availability depending on
the configured element type.
degree [2]
Polynomial degree
nrefine [5]
Number of refinement steps to perform.
'''
domain, geom = mesh.unitsquare(2, etype)
x, y = geom - .5
exact = (x**2 + y**2)**(1 / 3) * function.cos(
function.arctan2(y + x, y - x) * (2 / 3))
domain = domain.trim(exact - 1e-15, maxrefine=0)
linreg = util.linear_regressor()
with treelog.iter.fraction('level', range(nrefine + 1)) as lrange:
for irefine in lrange:
if irefine:
refdom = domain.refined
ns.refbasis = refdom.basis(btype, degree=degree)
indicator = refdom.integral(
'd(refbasis_n, x_k) d(u, x_k) J(x)' @ ns,
degree=degree * 2).eval(lhs=lhs)
indicator -= refdom.boundary.integral(
'refbasis_n d(u, x_k) n(x_k) J(x)' @ ns,
degree=degree * 2).eval(lhs=lhs)
supp = ns.refbasis.get_support(
indicator**2 > numpy.mean(indicator**2))
domain = domain.refined_by(refdom.transforms[supp])
ns = function.Namespace()
ns.x = geom
ns.basis = domain.basis(btype, degree=degree)
ns.u = 'basis_n ?lhs_n'
ns.du = ns.u - exact
sqr = domain.boundary['trimmed'].integral('u^2 J(x)' @ ns,
degree=degree * 2)
cons = solver.optimize('lhs', sqr, droptol=1e-15)
sqr = domain.boundary.integral('du^2 J(x)' @ ns, degree=7)
cons = solver.optimize('lhs', sqr, droptol=1e-15, constrain=cons)
res = domain.integral('d(basis_n, x_k) d(u, x_k) J(x)' @ ns,
degree=degree * 2)
lhs = solver.solve_linear('lhs', res, constrain=cons)
ndofs = len(ns.basis)
error = domain.integral('<du^2, sum:k(d(du, x_k)^2)>_i J(x)' @ ns,
degree=7).eval(lhs=lhs)**.5
rate, offset = linreg.add(numpy.log(len(ns.basis)),
numpy.log(error))
treelog.user(
'ndofs: {ndofs}, L2 error: {error[0]:.2e} ({rate[0]:.2f}), H1 error: {error[1]:.2e} ({rate[1]:.2f})'
.format(ndofs=len(ns.basis), error=error, rate=rate))
bezier = domain.sample('bezier', 9)
x, u, du = bezier.eval(['x', 'u', 'du'] @ ns, lhs=lhs)
export.triplot('sol.png', x, u, tri=bezier.tri, hull=bezier.hull)
export.triplot('err.png', x, du, tri=bezier.tri, hull=bezier.hull)
return ndofs, error, lhs
def main(nrefine: int, traction: float, radius: float, poisson: float):
'''
Horizontally loaded linear elastic plate with IGA hole.
.. arguments::
nrefine [2]
Number of uniform refinements starting from 1x2 base mesh.
traction [.1]
Far field traction (relative to Young's modulus).
radius [.5]
Cut-out radius.
poisson [.3]
Poisson's ratio, nonnegative and strictly smaller than 1/2.
'''
# create the coarsest level parameter domain
domain, geom0 = mesh.rectilinear([1, 2])
bsplinebasis = domain.basis('spline', degree=2)
controlweights = numpy.ones(12)
controlweights[1:3] = .5 + .25 * numpy.sqrt(2)
weightfunc = bsplinebasis.dot(controlweights)
nurbsbasis = bsplinebasis * controlweights / weightfunc
# create geometry function
indices = [0, 2], [1, 2], [2, 1], [2, 0]
controlpoints = numpy.concatenate([
numpy.take([0, 2**.5 - 1, 1], indices) * radius,
numpy.take([0, .3, 1], indices) * (radius + 1) / 2,
numpy.take([0, 1, 1], indices)
])
geom = (nurbsbasis[:, numpy.newaxis] * controlpoints).sum(0)
radiuserr = domain.boundary['left'].integral(
(function.norm2(geom) - radius)**2 * function.J(geom0),
degree=9).eval()**.5
treelog.info('hole radius exact up to L2 error {:.2e}'.format(radiuserr))
# refine domain
if nrefine:
domain = domain.refine(nrefine)
bsplinebasis = domain.basis('spline', degree=2)
controlweights = domain.project(weightfunc,
onto=bsplinebasis,
geometry=geom0,
ischeme='gauss9')
nurbsbasis = bsplinebasis * controlweights / weightfunc
ns = function.Namespace()
ns.x = geom
ns.lmbda = 2 * poisson
ns.mu = 1 - poisson
ns.ubasis = nurbsbasis.vector(2)
ns.u_i = 'ubasis_ni ?lhs_n'
ns.X_i = 'x_i + u_i'
ns.strain_ij = '(d(u_i, x_j) + d(u_j, x_i)) / 2'
ns.stress_ij = 'lmbda strain_kk δ_ij + 2 mu strain_ij'
ns.r2 = 'x_k x_k'
ns.R2 = radius**2 / ns.r2
ns.k = (3 - poisson) / (1 + poisson) # plane stress parameter
ns.scale = traction * (1 + poisson) / 2
ns.uexact_i = 'scale (x_i ((k + 1) (0.5 + R2) + (1 - R2) R2 (x_0^2 - 3 x_1^2) / r2) - 2 δ_i1 x_1 (1 + (k - 1 + R2) R2))'
ns.du_i = 'u_i - uexact_i'
sqr = domain.boundary['top,bottom'].integral('(u_i n_i)^2 J(x)' @ ns,
degree=9)
cons = solver.optimize('lhs', sqr, droptol=1e-15)
sqr = domain.boundary['right'].integral('du_k du_k J(x)' @ ns, degree=20)
cons = solver.optimize('lhs', sqr, droptol=1e-15, constrain=cons)
# construct residual
res = domain.integral('d(ubasis_ni, x_j) stress_ij J(x)' @ ns, degree=9)
# solve system
lhs = solver.solve_linear('lhs', res, constrain=cons)
# vizualize result
bezier = domain.sample('bezier', 9)
X, stressxx = bezier.eval(['X', 'stress_00'] @ ns, lhs=lhs)
export.triplot('stressxx.png',
X,
stressxx,
tri=bezier.tri,
hull=bezier.hull,
clim=(numpy.nanmin(stressxx), numpy.nanmax(stressxx)))
# evaluate error
err = domain.integral('<du_k du_k, sum:ij(d(du_i, x_j)^2)>_n J(x)' @ ns,
degree=9).eval(lhs=lhs)**.5
treelog.user('errors: L2={:.2e}, H1={:.2e}'.format(*err))
return err, cons, lhs
def main(nelems: int, etype: str, btype: str, degree: int,
epsilon: typing.Optional[float], contactangle: float, timestep: float,
mtol: float, seed: int, circle: bool, stab: stab):
'''
Cahn-Hilliard equation on a unit square/circle.
.. arguments::
nelems [20]
Number of elements along domain edge.
etype [square]
Type of elements (square/triangle/mixed).
btype [std]
Type of basis function (std/spline), with availability depending on the
configured element type.
degree [2]
Polynomial degree.
epsilon []
Interface thickness; defaults to an automatic value based on the
configured mesh density if left unspecified.
contactangle [90]
Wall contact angle in degrees.
timestep [.01]
Time step.
mtol [.01]
Threshold value for chemical potential peak to peak difference, used as
a stop criterion.
seed [0]
Random seed for the initial condition.
circle [no]
Select circular domain as opposed to a unit square.
stab [linear]
Stabilization method (linear/optimal/none).
'''
mineps = 1. / nelems
if epsilon is None:
treelog.info('setting epsilon={}'.format(mineps))
epsilon = mineps
elif epsilon < mineps:
treelog.warning('epsilon under crititical threshold: {} < {}'.format(
epsilon, mineps))
domain, geom = mesh.unitsquare(nelems, etype)
bezier = domain.sample('bezier', 5) # sample for plotting
ns = function.Namespace()
if not circle:
ns.x = geom
else:
angle = (geom - .5) * (numpy.pi / 2)
ns.x = function.sin(angle) * function.cos(angle)[[1, 0
]] / numpy.sqrt(2)
ns.epsilon = epsilon
ns.ewall = .5 * numpy.cos(contactangle * numpy.pi / 180)
ns.cbasis = ns.mbasis = domain.basis('std', degree=degree)
ns.c = 'cbasis_n ?c_n'
ns.dc = 'cbasis_n (?c_n - ?c0_n)'
ns.m = 'mbasis_n ?m_n'
ns.F = '.5 (c^2 - 1)^2 / epsilon^2'
ns.dF = stab.value
ns.dt = timestep
nrg_mix = domain.integral('F J(x)' @ ns, degree=7)
nrg_iface = domain.integral('.5 sum:k(d(c, x_k)^2) J(x)' @ ns, degree=7)
nrg_wall = domain.boundary.integral('(abs(ewall) + c ewall) J(x)' @ ns,
degree=7)
nrg = nrg_mix + nrg_iface + nrg_wall + domain.integral(
'(dF - m dc - .5 dt epsilon^2 sum:k(d(m, x_k)^2)) J(x)' @ ns, degree=7)
numpy.random.seed(seed)
state = dict(c=numpy.random.normal(0, .5, ns.cbasis.shape),
m=numpy.random.normal(0, .5,
ns.mbasis.shape)) # initial condition
with treelog.iter.plain('timestep', itertools.count()) as steps:
for istep in steps:
E = sample.eval_integrals(nrg_mix, nrg_iface, nrg_wall, **state)
treelog.user(
'energy: {0:.3f} ({1[0]:.0f}% mixture, {1[1]:.0f}% interface, {1[2]:.0f}% wall)'
.format(sum(E), 100 * numpy.array(E) / sum(E)))
x, c, m = bezier.eval(['x', 'c', 'm'] @ ns, **state)
export.triplot('phase.png', x, c, tri=bezier.tri, clim=(-1, 1))
export.triplot('chempot.png', x, m, tri=bezier.tri)
if numpy.ptp(m) < mtol:
break
state['c0'] = state['c']
state = solver.optimize(['c', 'm'],
nrg,
arguments=state,
tol=1e-10)
return state
