# Can you deduce Neumann boundary data from Dirichlet boundary data?

Say for the following problem, suppose boundary of $\Omega$ is $C^{1,1}$: \left\{ \begin{aligned} -\Delta \phi &= \mathrm{div} \,\vec{u}\quad \text{ in } \Omega \\ \phi&=0 \quad \text{ on }\partial \Omega \end{aligned} \right. Could we deduce $\nabla \phi\cdot \vec{n}$ on $\partial \Omega$ by the following reasoning? Multiply a test function $v\in H^1(\Omega)$ and by doing integration by parts we have: $$\int_{\Omega} \nabla \phi\cdot \nabla v -\int_{\partial \Omega} (\nabla \phi\cdot\vec{n})\,v = -\int_{\Omega} \vec{u}\cdot \nabla v +\int_{\partial \Omega} (\vec{u}\cdot\vec{n})\,v$$ Could we say that by the arbitrariness of $v$ that $\nabla \phi\cdot \vec{n} = -\vec{u}\cdot \vec{n}$?

No, you can't deduce this from the arbitrariness of $v$, because $\nabla v$ isn't independent of $v$. Also, you didn't, as the title suggests, deduce the Neumann boundary data from the Dirichlet boundary data; you tried to deduce them from the differential equation (at least I don't see where you used $\phi=0$). That can't work, since the boundary data are not determined by the differential equation.
Another way to see that this can't be right is that the solenoidal (divergence-free) part of $\vec u$ doesn't enter into the equation at all, whereas it does enter into the boundary condition you deduced.