Incompatible right-hand side in Neumann demo

[Anders Logg (Chalmers)]

The demo neumann-poisson uses a right-hand side which seems to be incompatible with the Neumann boundary condition (we need \int_{\Omega} f \dx = - \int_{\partial\Omega} g \ds, which does not seem to be the case. The Lagrange multiplier takes care of this but a comment/explanation should be added.

[Jan Blechta]

Testing by $v=1$ you actually get, $$ c |\Omega| = \int_\Omega f \dx + \int_{\partial\Omega} g \ds. $$ Unknown $c$ therefore takes care of this incompatibility and a necessary condition given above does not apply.

[Matthias Liertzer]

Issue #142 was marked as a duplicate of this issue.

[Matthias Liertzer]

I am sorry that I reported a duplicate of this issue. I originally searched for demo 13 instead of poisson neumann.

However, the Lagrange multiplier doesn't take care of the issue that the problem, as it is stated, is ill-defined. Using fenics one can actually show this easily. After the Poisson equation is solved the term \nabla^2 u + f should by definition be zero on the domain \Omega.

So let's plot this term using fenics: (Note, in order for the projection to work, one first needs to increase the polynomial degree of the function space to >= 2)

plot(project(nabla_div(nabla_grad(u)), V) + f, interactive=True)

The term, however, is not zero and therefore the Poisson equation is not fulfilled. In a next step one can fix f to fulfill the compatibility condition:

fV = interpolate(f, V)
f = f - assemble(fV*dx + g*ds))

and if one plots \nabla^2 u + f again, one can see that this term is now indeed zero on \Omega.

Note, that the result u itself is the same as before. The Lagrange multiplier only takes care of the problem in the sense, that it "redefines" f such that it fulfills the compatibility condition \int_{\Omega} f \dx = - \int_{\partial\Omega} g \ds, but the original problem is still ill-defined and therefore doesn't have a solution.

Note, that "Demo 13" is actually the first link that one gets when searching for "pure neumann condition" on google, so I would fix this rather sooner than later...

[Jan Blechta]

In some sense you're correct that the equation - \nabla^2 u = f is not fulfilled. But the problem which is being solved in the demo, i.e. the weak problem with the Lagrange multiplier, is definitely well-posed. Your interpretation that the Lagrange multiplier redefines f is correct.

[Anders Logg (Chalmers)]

Fix issue 117: Incompatible right-hand side in Neumann demo

→ <<cset 2e82a0ed195c>>

[Anders Logg (Chalmers)]

Resolved by explaining in the documentation how the incompatible right-hand side is handled by the variational formulation.

[Prof Garth Wells]

Removing milestone: 1.3 (automated comment)