A normal derivative on a given contour $\Gamma$ of a function $u$ expressed in Cartesian coordinates $u=u(x,y)$, can be calculated as:
$$\frac{\partial u}{\partial n} = \nabla u \cdot n =
\frac{\partial u}{\partial x} n_x +
\frac{\partial u}{\partial y} n_y
$$
where symbols are depicted in the figure.
When a segment of a curve is parametrized by normalized variable $\xi \in [-1,1]$ we can describe its geometry as:
$x=x(\xi)= \sum_i x_i N_i(\xi)$ and
$y=y(\xi)= \sum_i y_i N_i(\xi)$
where $(x_i, y_i)$ are nodal points, and $N_i$ – interpolation polynomials, e.g. Lagrange polynomials.
This leads to the derivatives:
$\frac{dx}{d\xi}= \sum_i x_i \frac{dN_i}{d\xi}$ and
$\frac{dy}{d\xi}= \sum_i y_i \frac{dN_i}{d\xi}$
The Jacobian of the transformation between local and global coordinates can be calculated as
$$d\Gamma = \sqrt{dx^2+dy^2} $$
and the components of the normal vector are equal
$n_x = \cos \alpha = \frac{dy}{d\Gamma}$
$n_y = – \sin \alpha = -\frac{dx}{d\Gamma}$
In this way one can describe components of the normal vector in terms of Cartesian coordinates and consequently in terms of parameter $\xi$.
Now what if the function $u$ and a curve $\Gamma$ are expressed in polar coordinates
$r = r(\theta)$, $\theta = [0,2\pi]$, $u=u(r,\theta)$? Recently, some paper that I found provided the following formula: $$\frac{\partial u}{\partial n} = \frac{\partial u}{\partial r} – \frac{r'}{r^2} \frac{\partial u}{\partial \theta}.$$
According to the theory the normal derivative in polar coordinates is given by:
$$\frac{\partial u}{\partial n} = \nabla u \cdot n =
\frac{\partial u}{\partial r} n_r +
\frac{1}{r} \frac{\partial u}{\partial \theta} n_\theta
$$
and what I need is to express the radial and angular components of the normal vector in terms of polar coordinates, or in fact in terms of parameter $\theta$ because $r=r(\theta)$.
Taking into account that
$x=r(\theta) \cos \theta$,
$y=r(\theta) \sin \theta$ and
$d\Gamma = \sqrt{dx^2+dy^2} $
it can be obtained that
$$d\Gamma = \sqrt{r'^2 + r^2} $$
but I can't find any trigonometric relations between $dr$, $d\theta$ and $n_r$, $n_\theta$ as it can be found for $dx$, $dy$, $n_x$, $n_y$. So my question is: Could anyone provide formulae for $n_r$ and $n_\theta$ in terms of polar coordinates?
EDIT:
I found another paper which contains formula that I was looking for. It is different from the formula from formerly cited paper by the same author. What is more, it is a bit different from what Ted Schifrin derived.
$$\frac{\partial u}{\partial n} =
\frac{r}{\sqrt{r'^2 + r^2}}
\left[ \frac{\partial u}{\partial r} – \frac{r'}{r^2}
\frac{\partial u}{\partial \theta} \right]
$$
The appendix of the paper gives the derivation of the formula. Now everything seems OK to me.
Best Answer
The contour at hand is assumed to be given by $r=r(\theta)$.
EDIT: OK, I've tried to decipher what's going on here. I presume that the article you cite is referring to the normal derivative along a level (contour) curve of $u$. The formula does not seem to be correct with my guess that the author is parametrizing the curve by $r=r(\theta)$. With that assumption, I get the formula $$\frac{\partial u}{\partial n} = \frac{r\frac{\partial u}{\partial r}-r'\frac{\partial u}{\partial\theta}}{\sqrt{r^2 + r'^2}}=\frac{\frac{\partial u}{\partial r}-\frac{r'}r\frac{\partial u}{\partial\theta}}{\sqrt{1 + \big(\frac{r'}r\big)^2}}.$$