discretize a BilinearForm with spaces of different degree

[FrederikSchnack]

This PR deals with issue #515 .

By exposing the pads keyword to discretize_space in psydac/api/discretize.py, we can manually set its padding to match with the degree of the other space. This resolves the assertion error mentioned in the issue and seems to allow discretizing BilinearForms with spaces of different degree.

This change was proposed by @vcarlier in the branch maxwell_ssc.

A MWE is given by:

from sympde.topology import Square
from sympde.topology.analytical_mapping import IdentityMapping
from sympde.topology import Derham
from sympde.topology  import element_of

from sympde.expr.expr import BilinearForm
from sympde.expr.expr import integral

from psydac.api.discretization import discretize

ncells = [8, 8]
degree = [3, 3]
d_degree = [d - 1 for d in degree]

log_domain = Square('Omega', bounds1=(0, 1), bounds2=(0, 1))
mapping = IdentityMapping('M1', dim=2)
domain = mapping(log_domain)
derham = Derham(domain, sequence=['h1', 'hcurl', 'l2'])

domain_h = discretize(domain, ncells=ncells)

p_derham_h = discretize(derham, domain_h, degree=degree)
d_derham_h = discretize(derham, domain_h, degree=d_degree, pads=degree)

pV0h, pV1h, pV2h = p_derham_h.spaces
dV0h, dV1h, dV2h = d_derham_h.spaces

pV = pV0h.symbolic_space
dV = dV0h.symbolic_space

u = element_of(pV, name='u')
v = element_of(dV, name='v')

a = BilinearForm((u, v), integral(domain, u*v))
ah = discretize(a, domain_h, [pV0h, dV0h])
A = ah.assemble()

That this produces the correct solution has to be verified!

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