A FEniCS Tutorial - FEniCS Project

implies a boundary integral over the complete boundary, or in FEniCS terms,
an integral over all exterior facets. However, the contributions from the parts
of the boundary where we have Dirichlet conditions are erased when the linear
system is modified by the Dirichlet conditions. We would like, from an efficiency
point of view, to integrate v*g*ds only over the parts of the boundary where we
actually have Neumann conditions. And more importantly, in other problems
one may have different Neumann conditions or other conditions like the Robin
type condition. With the mesh function concept we can mark different parts
of the boundary and integrate over specific parts. The same concept can also
be used to treat multiple Dirichlet conditions. The forthcoming text illustrates
how this is done.
Essentially, we still stick to the model problem from Section 1.14, but replace
the Neumann condition at y = 0 by a Robin condition:
− ∂u
∂n = p(u−q),
where p and q are specified functions. The Robin condition is most often used
to model heat transfer to the surroundings and arise naturally from Newton’s
cooling law.
Since we have prescribed a simple solution in our model problem, u = 1+
x 2 +2y 2 , we adjust p and q such that the condition holds at y = 0. This implies
that q = 1+x 2 +2y 2 and p can be arbitrary (the normal derivative at y = 0:
∂u/∂n = −∂u/∂y = −4y = 0).
Now we have four parts of the boundary: Γ N which corresponds to the upper
side y = 1, Γ R which corresponds to the lower part y = 0, Γ 0 which corresponds
to the left part x = 0, and Γ 1 which corresponds to the right part x = 1. The
complete boundary-value problem reads
−∇ 2 u = −6 in Ω, (90)
u = u L on Γ 0 , (91)
u = u R on Γ 1 , (92)
− ∂u
∂n = p(u−q) on Γ R, (93)
− ∂u
∂n = g on Γ N . (94)
Theinvolvedprescribedfunctionsareu L = 1+2y 2 ,u R = 2+2y 2 ,q = 1+x 2 +2y 2 ,
p is arbitrary, and g = −4y.
Integration by parts of − ∫ Ω v∇2 udx becomes as usual
∫
−
Ω
∫
v∇ 2 udx =
Ω
∫
∇u·∇vdx−
∂Ω
∂u
∂n vds.
The boundary integral vanishes on Γ 0 ∪Γ 1 , and we split the parts over Γ N and
Γ R since we have different conditions at those parts:
∫
− v ∂u ∫
∂n ds = − v ∂u ∫
Γ N
∂n ds− v ∂u ∫ ∫
Γ R
∂n ds = vgds+ vp(u−q)ds.
Γ N Γ R
∂Ω
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