Illustrate my problem with figure:
For a given curve (solid line '-'). $\phi$ is the signed distance function of curve, $n$ is the normal vector at $\phi^{-1} = 0$, the dashed line '--' is the level curve with distance $\eta$.
Define $\kappa$ is the curvature of level curve, I want find the relation between curvature of $\kappa(x)$ and $\kappa(y)$ where $y = x + \eta n$. More detailed, I want express the relation by the form of $ \kappa(y) = a \kappa(x) + O(\eta^2) $ where $a$ is constant or geometry on given curve.
For a simple example. Let curve $\phi^{-1}=0$ is a circle with radius of $r$. Then we have $ \kappa(y) = (1 + \eta \kappa(x) )^{-1} \kappa(x)$ since $\kappa(x) = \frac{1}{r}$, and $\kappa(y) = \kappa(x+\eta n) = \frac{1}{r+\eta}$.
I have try to expansion by Taylor expansion, $\kappa(y) = \kappa(x) + \eta n \cdot \nabla \kappa + O(\eta^2)$. But I want more simple formula since $\nabla \kappa$ is complex to calculate in my problem.
Is there some books or reference recommend for this kind problem? Thanks
This is only intuition, no prove: I'll write $\kappa_\eta(t)$ to denote the curvature of $t\mapsto c(t)+\eta n(t)$ at $t$, where $c$ is the initial curve. By using \begin{equation}|\kappa|=R^{-1}, \end{equation} where $R$ is the radius of the best fitting circle, we get (always for $\eta$ small enough) \begin{equation}\kappa_\eta=-(R+\eta)^{-1}=\frac{-|\kappa|}{1+\eta |\kappa|}=\frac{\kappa}{1-\eta \kappa} \end{equation} for points with initial negative curvature $\kappa$ and \begin{equation}\kappa_\eta=(R-\eta)^{-1}=\frac{|\kappa|}{1-\eta |\kappa|}=\frac{\kappa}{1-\eta \kappa} \end{equation} for points with initial positive curvature $\kappa.$ So in general (for $\eta$ small enough) \begin{equation}\kappa_\eta=\frac{\kappa}{1-\eta \kappa}=\kappa\sum_{i=0}^\infty (\kappa \eta)^i=\kappa+\kappa^2\eta+\kappa^3\eta^2+... \end{equation}