Prove that divergence and curl free vector field is a constant vector field

Question

I need to prove the fact that a vector field $\vec{B}$ that is divergence and curl free, is a constant vector field.

I have attempted to prove this by referring to the divergence, but realized that this will not work since the fields of the vector field not necessary must be all constants to obtain a divergence of 0. I believe that both properties (divergence free and curl free) should be used to prove it.

Answer 1

The result is not true. If you take any harmonic function $u \colon \Omega \rightarrow \mathbb{R}^2$ and let $\vec{B} = \mathrm{grad}(u)$ then $\mathrm{curl}(\vec{B}) = \mathrm{curl}(\mathrm{grad}(u)) = 0$ and $\mathrm{div}(\vec{B}) = \mathrm{div}(\mathrm{grad}(u)) = \Delta u = 0$. For most harmonic functions, $\vec{B}$ won't be a constant vector field.

Answer 2

By specifying the vector field's divergence and curl you still have ambiguity in its normal component at the boundary. If you require that the normal component at the boundary is a constant vector then the vector field throughout the volume is a constant one. Similarly if you require that the normal component to the boundary vanishes then the vector field is identically zero.