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Consider

$$\begin{aligned} \partial_t v + b\cdot \nabla \phi &=0 \\ \partial_t \phi + b\cdot \nabla v &= 0 \end{aligned}$$

for $v:(x,y)\mapsto \mathbb{R}$, unknown and time-dependent ($\phi$ similary), and $b\in \mathbb{R}^2$ given, and divergence-free. A weak form in $H^1$ is

$$\begin{aligned} \int \partial_t v w -\int \phi b\cdot \nabla w &= -\int_\Gamma \phi_\Gamma w b \cdot e_n \quad \forall w\in H^1\\ \int \partial_t \phi \psi -\int v b\cdot \nabla \psi &= -\int_\Gamma V_\Gamma \psi b\cdot e_n \quad \forall \psi\in H^1 \end{aligned}$$

where $\Gamma$ denotes the domain boundary, and $\int$ denotes a (non-parametrized) integral over the domain.

Am I right to assume, that a Neumann boundary condition, like ($n$ denoting the outward-pointing, unit normal vector)

$$n\cdot \nabla v \equiv g$$

can not be enforced in a "natural" way, i.e. by manipulating the above weak form?

If so, does this imply that the BC has to be enforced by posing the problem on an adapted function space, which includes the condition in its definition, e.g.

$$H^1_\nabla \equiv \left \{ u\in H^1 : (n\cdot \nabla u)|_{(x,y)\in \Gamma} \equiv g \right \} ?$$


Background info: I would like to solve the problem using FEM.

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  • $\begingroup$ I don't think you need to rewrite $\int w b\cdot\nabla \phi$ using the divergence theorem, and similarly for the other one, it doesn't gain you anything as far as I can tell. As far as the BC goes I think you would need to prescribe it explicitly in some manner. $\endgroup$
    – lightxbulb
    Nov 7, 2023 at 18:45
  • $\begingroup$ The application of the divergence theorem does not allow the enforcement of the stated BC. It does, however, allow for less regular solutions, or more concretely: solutions that are only weakly differentiable in the spatial coordinates. Please correct me, if my reasoning is incorrect. $\endgroup$ Nov 7, 2023 at 20:23
  • $\begingroup$ With "prescribe it explicitly in some manner", do you mean the way I have proposed in the question, i.e. to include the condition in the function space definition? $\endgroup$ Nov 7, 2023 at 20:25
  • $\begingroup$ The divergence theorem here really doesn't seem to do anything except move the derivative to the test function. If you are considering Petrov-Galerkin then it may make sense, however if the solution and test functions space is the same there's no difference except you now also have to deal with a boundary term. As far as prescribe it explicitly, I mean something like that yes. Once you discretise you will have to enforce this as an additional linear constraint acting on your node values. $\endgroup$
    – lightxbulb
    Nov 7, 2023 at 20:31
  • $\begingroup$ Thank you for clarifying. $\endgroup$ Nov 7, 2023 at 20:56

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