Prove that the nonhomogeneous Neumann problem has a solution only if $\int_Ωf = 0$.

Question

Let $Ω ⊂ \mathbb{R}^n$ be a bounded domain with smooth boundary $∂Ω$, and denote by $\vec{n}$ the outer unit normal to $∂Ω$. Prove that the nonhomogeneous Neumann problem

$$ \begin{cases} ∆u = f & \text{ in }Ω \\ \langle∇u ,\vec{n}\rangle = 0 & \text{ on }∂Ω \\ \end{cases} $$

has a solution only if $\int_Ωf = 0$.

How would I go about doing this? Can anyone help to me understand where to start better?

Necessary condition: by divergence theorem. The Laplacian is the divergence of gradient. — , Nov 24 '15 at 02:45
Also of existence of the solution of Neumann problem in $ \mathbb{R}^3$ if you are only interested in the necessity part. — , Nov 24 '15 at 02:50

score 2 · Answer 1 · answered Nov 24 '15 at 02:47

2

Assuming the correct regularity on $f$, this is an application of the divergence theorem. Indeed notice that $$\int_{\Omega}f\,dx = \int_{\Omega}\Delta u\, dx = \int_{\partial \Omega}\nabla u \cdot n\, d\mathcal{H}^{N-1} = 0,$$ where the last equality follows from the Neumann condition. This is usually referred to as a compatibility condition.

answered Nov 24 '15 at 02:47

Giovanni

6,321

Could you expand just a bit on that last equality and why it follows from the Neumann condition? – Jimm Nov 24 '15 at 02:50
The Neumann condition says that the integrand in last integral is constant and equal to $0$. (I've used a different notation for the inner product, maybe that caused confusion..) – Giovanni Nov 24 '15 at 02:53
Ah ok I can follow that now. Thank you – Jimm Nov 24 '15 at 02:55

Prove that the nonhomogeneous Neumann problem has a solution only if $\int_Ωf = 0$.

1 Answers1