If a vector field with zero divergence vanishes on a flat portion of boundary, its normal derivative is zero

Question

Let's consider a vector field $v$ in a bounded region $R$ of the space; assume that $\operatorname{div}v=0$ and $v=0\,\,\text{on}\,\, \partial R$; I have to prove that $$(\nabla v)^Tn=0$$ where $n$ is the outward unit normal to $\partial R$, for a plane portion of boundary. I have evaluated $(\nabla v)^Tn$ and I get $\nabla v_3$ ($v=(v_1,v_2,v_3)$). I don't succeed in proving it is zero; any suggestion? I have also tried to use the divergence theorem but then I have problems to localize the result.

Answer 1

Let's say the plane is $x_3=0$; this appears to be your assumption too. For the points in the plane, $$\frac{\partial v_i}{\partial x_j} =0,\qquad i=1,2,3,\quad j=1,2$$ because all components are zero along the $x_1$ and $x_2$ directions.

Hence, the divergence of $v$ (which is zero by assumption) is $$\sum_{i=1}^3 \frac{\partial v_i}{\partial x_i} = \frac{\partial v_3}{\partial x_3}$$ This completes the proof of $\nabla v_3=0$.