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?

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    $\begingroup$ It is impossible to have zero gradient if $f$ is of constant sign (Hopf's Lemma). $\endgroup$ Mar 3, 2022 at 15:16
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    $\begingroup$ Moreover, you at the very least need $\int_\Omega f = 0$ as testing the equation with $1$ shows. $\endgroup$
    – Keba
    Mar 3, 2022 at 15:22

1 Answer 1


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 $\left (P(iD)\hat f\right )(0)=0$.


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