At $x_0 \in \partial \Omega$, $\text{Hess}u(x_0) = \Delta u(x_0) (\nu \otimes \nu)$, $\nu$ outer unit normal

Question

Let $x_0 \in \partial \Omega$ and $\nu(x_{0})$ is the exterior unit normal at $x_0$, with $\nu_1(x_0) > 0$, where $u \in C^{2}(\overline{\Omega_{\epsilon}})$ and $\Omega_{\epsilon}=\Omega \cap \{|x-x_0|<\epsilon\}$, $u>0$ in $\Omega$ and $u=0$ on $\partial \Omega \cap \{|x-x_0|<\epsilon\}$. We know that $u_{x_1}(x_0) = 0$. Why does the following hold?

$$ \Delta u = - f(0) > 0 \implies u_{x_{i},x_{j}}=-f(0)\nu_{i}\nu_{j} \quad \text{ at } x_0 $$

Moreover, in their paper, Gidas, Ni and Nirenberg claim, right before what inspires the question, that $\nabla u(x_0) = 0$. It's a mistake, isn't it? Up to my understanding, we only know that $u_1(x_0) = 0$.

This expression appears specifically in the last part of the demonstration of Lemma 2.1 of this article https://projecteuclid.org/download/pdf_1/euclid.cmp/1103905359 of Gidas-Ni-Nirenberg's and I think that it is related to the second directional derivative and Hessian matrix.

Answer 1

It’s not a mistake. Since $u=0$ on the boundary, the tangential derivative of $u$ is zero on the boundary. If $t_1$, $\ldots$, $t_{n-1}$ are the tangent vectors, you can write $$e_1=\sum_{i=1}^{n-1}c_it_i+c_n\nu$$ Taking the inner product with $\nu$ you get that $c_n=\nu_1>0$. Hence $0=u_1=\nabla u\cdot e_1=c_nu_\nu$ at $x_0$ and so also the normal derivative is zero. Hence the gradient is zero. For the Laplacian you have to do something similar. All the second derivatives in the tangential directions will be zero.

Answer 2

The following argument is due to Adam Rudnik and Vitor Borges, making use of the hint by Gio67

Since the Laplace operator is invariant by rotations and translations, and since every tangential derivative is $0$ we can write it with respect to the basis $\{t_1, \ldots, t_{N - 1}, \nu\}$ as $$ \Delta u(x_0) = u_{\nu \nu}. $$ But $$ u_{ij} = [u_{ij}](e_i, e_j) = [u_{ij}] (\nu_i \nu, \nu_j \nu) = \nu_i \nu_j [u_{ij}](\nu, \nu) = \nu_i \nu_j u_{\nu \nu} = \nu_i \nu_j \Delta u(x_0), $$