0
$\begingroup$

It is known than for a second order elliptic boundary-value problem with pure Neumann conditions, a certain compatibility condition between the data must be satisfied. For example, in the case of the Poisson equation with pure Neumann data: $$-\Delta u = f\quad\text{in }\Omega, $$ $$\frac{\partial u}{\partial \nu} = g\quad\text{on }\partial\Omega, $$ the following must hold: $$\int_\Omega f\,dx + \int_{\partial\Omega}g\,ds = 0. $$

My question is how we can generalise the compatibility condition for a general problem, i.e. $$-\sum_{i, k = 1}^{n} \partial_i(a_{ik}\partial_k u) + a_0u = f\quad \text{on }\Omega, $$ $$\sum_{i, k}\nu_i a_{ik} \partial_k u = h \quad \text{on } \partial\Omega. $$ where the differential operator given in divergence form is elliptic, and $\nu_i$ denotes the $i$-th component of the outward-pointing normal.

$\endgroup$

1 Answer 1

2
$\begingroup$

The compatibility condition comes from integrating the equation on both sides. In your case you have

$$-\text{div}(A\nabla u) + a_0 u = f.$$

Integrate both sides over $\Omega$ and integrate by parts (i.e., use the divergence theorem) to find that

$$-\int_{\partial \Omega} \nu \cdot A \nabla u \, dS + a_0 \int_\Omega u\, dx = \int_\Omega f\, dx.$$

The first term is exactly your boundary condition (with a negative sign), so we obtain

$$\int_\Omega f\, dx + \int_{\partial \Omega} h \, dS = a_0 \int_\Omega u\, dx.$$

When $a_0\neq 0$, this is not so much a compatibility condition as it is just a condition on the solution $u$. When $a_0=0$ it is the same compatibility condition as in your simple example.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .