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An oriented Nedelec edge circulation on one tetrahedron

Example

On the reference tetrahedron with vertices x0=0, x1=e1, x2=e2, x3=e3, take the Whitney edge field ϕ01=λ0∇λ1−λ1∇λ0 and orient the face [0,1,2] by the vertex order shown. Compute the circulation on [0,1] and the curl flux through [0,1,2].

Facts & Assumptions

Given: The local edge basis and oriented edge and face moments of Lowest-order Whitney and Nedelec de Rham spaces, their unisolvence from Lowest-order Nedelec edge circulation unisolvence, and Stokes commutation from Whitney edge, face, and cell moments commute with grad, curl, and div.

[F1]

The barycentric edge field is ϕ01=λ0∇λ1−λ1∇λ0, and its edge circulation is the canonical degree of freedom. (Lowest-order Whitney and Nedelec de Rham spaces)

[F2]

The six local edge circulations form the dual basis for the Whitney edge fields. (Lowest-order Nedelec edge circulation unisolvence)

[F3]

On the oriented face [0,1,2], the curl flux equals the signed sum of its three boundary edge circulations. (Whitney edge, face, and cell moments commute with grad, curl, and div)

Verification

1.1F1constructalgebra

In the reference coordinates, λ1=x1, λ0=1−x1−x2−x3, so ϕ01=(1−x2−x3,x1,x1). Along [0,1], parameterized by x(t)=(t,0,0), this field has tangent component 1; therefore its oriented circulation is ∫011 dt=1.

2.1F1F2F3step 1.1algebra∎

Differentiating the displayed polynomial gives curl⁡ϕ01=(0,−2,2). The oriented face [0,1,2] lies in x3=0 with normal e3 and area 1/2, so its curl flux is 2⋅(1/2)=1. Its oriented boundary is [1,2]−[0,2]+[0,1]; by [F2], the first two circulations vanish and the last is one, agreeing with the flux by [F3].

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