# The behavior of $\nabla u$ on the boundary for Poisson equations

Let $$\Omega$$ be a bounded domain with smooth boundary. Consider the Poisson equation $$\begin{eqnarray} -\Delta u&=&f\text{ in }\Omega\\ u&=&0\text{ on }\partial\Omega \end{eqnarray}$$ where $$f\in C_0^{\infty}(\Omega)$$. By using the Lax-Milgram theorem, we can find the solution in $$H_0^1(\Omega)$$ and then enhance the regularity for $$u$$. I want to ask what condition should I assume for $$f$$ can I obatin that $$\nabla u=0$$ on $$\partial\Omega$$. More generaly we can replace the operator $$-\Delta$$ by other elliptic operators. Can you give me some referances or hints?

• It is impossible to have zero gradient if $f$ is of constant sign (Hopf's Lemma). Mar 3 at 15:16
• Moreover, you at the very least need $\int_\Omega f = 0$ as testing the equation with $1$ shows.
– Keba
Mar 3 at 15:22

The first observation is that the $$u$$ above satisfies $$\nabla u=0$$ on $$\partial \Omega$$ if and only if $$f$$ is orthogonal to all harmonic functions $$v$$ in $$\Omega$$, continuous up the the boundary. In fact, $$\int_{\Omega} fv=\int_{\Omega} (\Delta u) v=\int_{\Omega} u \Delta v=0$$, by the boundary conditions. Conversely, if this holds for $$f$$, then, since $$u=0$$ on $$\partial \Omega$$, $$0=\int_{\Omega} (\Delta u) v=\int_{\partial \Omega} v \frac{\partial u}{\partial n}$$ for every harmonic $$v$$. This gives $$\nabla u=0$$ at the boundary, since $$v$$ is arbitrary on $$\partial \Omega$$ and the tangential derivatives of $$u$$ are zero by the boundary conditions.
The second remark is that such $$u$$ vanishes in a neighborhood of the boundary where $$f$$ is zero (hence $$u$$ is harmonic). In fact, the equation and the boundary conditions imply that all the derivatives up to the second order vanish on the boundary. Continuing $$u$$ by zero across the boundary we obtain an harmonic function vanishing in an open set and hence in any connected set containing the boundary where $$f=0$$.
This observation allows to change $$\Omega$$ to a ball $$B_R$$ containing it. In fact, if $$u$$ solves the problem above in $$\Omega$$, since it is zero in a neighborhood of the boundary, it solves the same probelm in $$B_R$$ and conversely if it solves in $$B_R$$ then it is 0 in $$B_R \setminus \Omega$$, by the above argument and $$u, \nabla u=0$$ at $$\partial \Omega$$.
Let us take therefore $$\Omega=B_R$$. By the above discussion and by density, the problem has a solution iff $$f$$ is orthogonal to all harmonic polynomials $$P$$, that is $$\int_{\mathbb R^n} P(x)f(x)=0$$ or $$\left (P(iD)\hat f\right )(0)=0$$.