I need your help to understand the following formula:
Let be $\Gamma$ a regular curve boundary of $\Omega \subset \mathbb{R}^2$. The boundary $\Gamma$ can be parametrized in terms of the curvilinear abscissa $s$ as $s \to x_\Gamma(s)$, $s \in [0,L)$, with $\|\frac{dx_\Gamma(s)}{ds}\| = 1$, where $L$ is the length of $\Gamma$. We assume that this parametrization defines a clockwise orientation. Let $n(s)$ be the unitary normal vector at $x_\Gamma(s)$ directed to the exterior of $\Omega$ and set $\tau(s) = \frac{dx_\Gamma(s)}{ds}$ which is a unitary vector tangential to $\Gamma$ at $x_\Gamma(s)$. The curvature $c$ can be defined by
$$c(s) := \tau(s) \cdot \frac{dn(s)}{ds}$$
Omitting some details, we can descompose any $x$ outside $\Omega$ as $x = x_\Gamma(s) + \nu n(s)$. ($x_\Gamma$ is the orthogonal projection of $x$ on $\Gamma$), then for a function $u$ defined outside $\Omega$ we can write
$$\tilde{u}(s,\nu) := u(x)$$
Then, show that
$$\nabla u(x) = \frac{1}{1 + \nu c} \partial_s \tilde{u} \tau + \partial_\nu \tilde{u}n$$
I can't arrive to the $\frac{1}{1 + \nu c} \partial_s \tilde{u} \tau$ expression (calculating $h_s = \|\frac{\partial}{\partial s} (x_\Gamma(s) + \nu n(s))\|$). I would appreciate your help, suggestion or ideas.
Thanks in advance!
if $x_\Gamma(s)$ -is the curve $\Gamma$ whith curvature c(s) then $$x(s,\nu)=x_\Gamma(s)+\nu n(s)$$ is a curve equidistant (parallel) to curve $\Gamma$, where $\nu$ is a distanse between the two curves and we have $$dx(s,\nu)=dx_\Gamma(s)+d \nu n(s)+ \nu dn(s)=\tau ds+d\nu n(s)-\nu c(s) \tau (s)ds=(1-\nu c)ds \tau + d\nu n$$ so metric is $$ (dx)^2=(1-\nu c)^2 ds^2+d\nu^2$$
$c=1/r$ where r is curvature radius , so $$1-\nu c=(r-\nu)/r$$. if you change $r$ to $r+\nu$ then will be $$ r/(r+\nu)=1/(1+\nu c)$$