scalar coefficient in a vector form

[gdmcbain]

Say τ is ElementTriRT0 and φ is ElementTriP0 and I want to define the linear form (φ, div τ). How do I interpolate the scalar coefficient φ onto the common quadrature points?

Motivation: Thinking about the (weak) gradient of a piecewise-constant function like an indicator function over in #821 on Friday, I suggested :

Ah, the characteristic function might need to be continuous, like ElementTriP1.

But seeing as we might not want that for the purposes of defining a discontinous conductivity as in ex17 (and while we're at it just use the conductivity instead of a zero–one characteristic function), what about a Raviart–Thomas gradient of the P0 conductivity/characteristic function?

My thinking was that a weak definition of σ = grad φ would be (σ, τ) = (grad φ, τ) for all τ and integrating the right-hand side by parts would give (grad φ, τ) = (n φ, τ)_∂Ω − (φ, div τ).

Then to begin with I thought I'd make my life easier by considering only the indicator function of an internal subdomain so that φ would vanish on ∂Ω and I'd only have to solve (σ, τ) = − (φ, div τ).

The LHS is a vector-mass matrix, but what about the RHS?

I had a quick go at this with:

from skfem import *
from skfem.helpers import div, dot

import ex17


@BilinearForm
def vector_mass(sigma, tau, w):
    return dot(sigma, tau)


@LinearForm
def linf(tau, w):
    return - w["phi"] * div(tau)


mesh, conductivity = ex17.basis0.mesh.to_meshtri(ex17.conductivity)
basis = CellBasis(mesh, ElementTriRT0())

M = vector_mass.assemble(basis)
rhs = linf.assemble(basis, w=conductivity)

but of course this raises

ValueError: Input array has wrong size.

This (viz., have a scalar coefficient in a vectorial form) seems like an obvious kind of thing to want to do; am I overlooking something simple?

[gdmcbain]

rhs = linf.assemble(basis, phi=basis.with_element(ElementTriP0()).interpolate(indicator))

[gdmcbain]

Sorry, that is fairly obvious, isn't it. Now: how to render Raviart–Thomas vector-fields, but that's a separate question.