Accessing array of unknown dofs

[TomvanSluijs]

Hi,

We want to take the time derivative of a residual by using a chain rule. As shown in the mwe below, we want to take the derivative to the dofs of variable u and we still need to multiply by the dofs of the time derivative of u, which is a separate field called dudt. However, we are not able to do this multiplication. Can you help us with the syntax of the line where dres_dt is defined?

Best,
Tom

from nutils import mesh, function, solver, export, testing
from nutils.expression_v2 import Namespace

nelems: int = 10
etype: str = 'square'
btype: str = 'std'
degree: int = 1

domain, geom = mesh.unitsquare(nelems, etype)

ns = Namespace()
ns.x = geom
ns.define_for('x', gradient='∇', normal='n', jacobians=('dV', 'dS'))
ns.basis = domain.basis(btype, degree=degree)
setattr(ns, 'u', function.dotarg('udofs',ns.basis))
setattr(ns, 'v', function.dotarg('vdofs',ns.basis))
setattr(ns, 'dudt', function.dotarg('dudtdofs',ns.basis))

res = domain.integral('∇_i(v) ∇_i(u) dV' @ ns, degree=degree*2)
res -= domain.boundary['right'].integral('v cos(1) cosh(x_1) dS' @ ns, degree=degree*2)

# Here we want to create the time derivative of the residual
dres_dt = res.derivative("udofs") # How do we multiply this by the vector of dudt degrees of freedom?

[joostvanzwieten]

This should work

dres_dt = res.derivative("udofs") @ function.Argument('dudtdofs', ns.basis.shape)

Alternatively you can define u as u0 + t dudt and take the derivative to t:

ns = Namespace()
ns.x = geom
ns.define_for('x', gradient='∇', normal='n', jacobians=('dV', 'dS'))
ns.basis = domain.basis(btype, degree=degree)
ns.add_field(('u0', 'dudt', 'v'), ns.basis)
ns.add_field('t')
ns.u = 'u0 + t dudt'

res = domain.integral('∇_i(v) ∇_i(u) dV' @ ns, degree=degree*2)
res -= domain.boundary['right'].integral('v cos(1) cosh(x_1) dS' @ ns, degree=degree*2)

dres_dt = res.derivative('t')
'''