Second normal derivative on a curved boundary The motivation is that I am trying to understand the equation $$\tag{1}\label{eq:1}u_{nn} - \Delta u = -(\kappa u_n + u_{tt})$$
on the boundary $\partial\Omega$ of some bounded open set $\Omega\subset\mathbb R^2$ ($\kappa$ is the signed curvature of the boundary). Basically, if $\partial \Omega$ is piecewise straight, then $\kappa=0$ and the tangential $t$ and normal $n$ simply form a local coordinate system by rotating the standard coordinates and hence \eqref{eq:1} is clear since the Laplacian $\Delta u := D^2u(x_1,x_1) + D^2u(x_2,x_2) = D^2u(t,t) + D^2u(n,n)$ is invariant under such rotation.
In the case of arbitrary, say piecewise $C^2$, boundary I am a bit confused. What does $u_{tt}$ and $u_{nn}$ even stand for, it cannot simply be $D^2u(t,t)$ and $D^2u(n,n)$ since then there would be no curvature term in \eqref{eq:1}.

*

*My intuition for $u_{tt}$: The tangential derivative $u_t=\frac{\partial}{\partial t} u=\nabla u\cdot t$ can be written as $\frac{d}{ds}u(\gamma(s))$ for some arc-length parametrisation $\gamma$ of the boundary. Hence $u_{tt} = \frac{d^2}{ds^2}u(\gamma(s)) = D^2u(\gamma'(s), \gamma'(s)) + \nabla u\cdot \gamma''(s)= D^2u(t(s), t(s)) - \kappa u_n$ since $\gamma'(s)= t(s), \gamma''(s) = -\kappa n(s)$.


*For $u_{nn}$: The normal derivative $\frac{\partial}{\partial n} u_n$ as a directional derivative is the limit when approaching the boundary normally, but $u_n$ is only defined on the boundary. So does this mean that for any extension $n^*$ of the normal field $n$, that$\frac{\partial}{\partial n} u_n = \frac{\partial}{\partial n} u_{n^*}$ is equal? And how to see this? Is there a better way of interpreting $u_{nn}$ in the first place?
I guess I am making this way to complicated - thanks for your help!
 A: I don't think it is wise to think of these derivatives on the boundary in terms of derivatives w.r.t. the Euclidean coordinates.
A more natural way is to define suitable normal
coordinates, so that a neighborhood of a boundary point looks
like a neighborhood  $U$ of the origin in the half-plane
${\Bbb R}_- \times {\Bbb R}$. We write $(x_\alpha)_{\alpha=1,2}$
for the coordinates in $\Omega\supset \Omega$.
Let $\gamma(s)$, $s\in ]-\delta,\delta[$
be a piece of the arc-length parametrized boundary with
$p=\gamma(0) \in \partial \Omega$. Let $t(s)=\gamma'(s)$ be
the corresponding unit tangent vector and $n(s)$ an outward unit normal
at $\gamma(s)$. We then have $t'(s)=-k n(s)$ and consequently
$n'(s)=k t(s)$.
An explicit way to construct normal coordinates is now
to pose $\xi(r,s):=\gamma(s) + r n(s)$
in a neighborhood of the origin
$(r,s)=(0,0)$.
The map verifies for $r=0$ (Fact 2):
$$   \partial_s \xi_{|r=0} = t(s), \ \
   \partial_r \xi_{|r=0}  = n(s),  \ \
   \partial^2_s \xi_{|r=0} = -k n(s),  \ \
   \partial^2_r \xi_{|r=0} = 0. 
   $$
In particular, we see that
the jacobian matrix at a boundary point is the orthogonal matrix
$J=J_{|r=0}=\begin{pmatrix} n(s) & t(s) \end{pmatrix} $. Note that from $JJ^T=I$
we get (Fact 1) for
$\alpha,\beta=1,2$: $n_\alpha n_\beta+t_\alpha t_\beta = \delta_{\alpha,\beta}$
(Kronecker-delta). As $J$ is invertible, $\xi$ defines a local diffeomorphism
$$ \xi : U \cap ({\Bbb R}_- \times {\Bbb R}) \to V \cap \Omega$$
from a neighborhood $U$ of $(0,0)$ to a neighborhood $V$
of $p\in \Omega$.
Under this coordinate transformation
we get the following derivatives (using the Einstein convention of summing
over repeated indices and natural abbreviations):
$$ \partial_r =
\frac{\partial}{\partial r} = 
\frac{\partial x_\alpha}{\partial r} 
\frac{\partial }{\partial x_\alpha} 
= n_\alpha(s) \partial_\alpha
$$
$$ \partial_s =
\frac{\partial}{\partial s} = 
\frac{\partial x_\alpha}{\partial s} 
\frac{\partial }{\partial x_\alpha} 
= (1+kr) t_\alpha(s)  \partial_\alpha
$$
$$ \partial_r^2 
= n_\alpha n \partial_\beta
\partial_\alpha \partial_\beta
$$
$$ \partial_s^2 =
 (1+kr) (-k n_\alpha)  \partial_\alpha + (1+kr)^2 t_\alpha t_\beta
\partial_\alpha \partial_\beta
$$
Finally, restricting to the boundary, i.e. $r=0$
and using Fact 1 for the equality in the middle:
$$ \left(\partial_r^2  + 
 \partial_s^2 \right)_{|r=0}
= 
 -k n_\alpha  \partial_\alpha + 
( n_\alpha n_\beta +
 t_\alpha t_\beta)
\partial_\alpha \partial_\beta =
-k \partial_r + \partial_\alpha \partial_\alpha
=
-k \partial_r + \Delta ,
$$
as we wanted.
A priori the above is defined only in the interior of $U$ (or $\Omega$).
In the $(r,s)$ coordinates, however, $C^2$-smoothness with
suitable boundary behavior is easy to define
(one advantage of the half-plane) and on the boundary you may identify the
$r,s$ derivatives with the $n,t$ derivatives which you mention. Regarding generality, any smooth map $\xi$ verifying Fact 2 will do.
A: It is probably difficult to guess the symbols' meaning without the context of what the mathematics is used to describe.  There is a similarity in symbols and forms to this question.  Note the answer added in 2018 by @armando.sano.  The original reference is not free.  But there is a lecture note from Purdue on the same topic.
From that answer, the equation looks like:
$$
\begin{align}
\nabla_s^2 f = \nabla f^2 - \kappa \frac{\partial f}{\partial n} - \pmb{n}^T H(f) \pmb{n}
\end{align}
$$
But that solution is about the boundary of a soap bubble in 3 or more dimensions, whereas this question is about the boundary of a 2-d function. I think the theory still applies, and I am guessing that $u_{nn}=D_n u_n$ is the Hessian in 1-d, and $u_{tt}$ is the surface Laplacian in 1-d, the "surface" being the 1-d boundary.
