Derivative with respect to the normal?

Question

I am trying to use greens theorem to show the following: $$\int \int (f_{xx}+f_{yy}) \, dx \, dy=\int\frac{\partial f}{\partial n} \, ds$$ I am not completely sure how to treat the $d/dn$. I have simplified to a point where I have $$\int f_x \, dy-f_y \, dx=\int \int (f_{xx}+f_{yy}) \, dx \, dy$$ So I need to go from the left side of the last equation to the right side of the first equation. Any intuition on what the derivative implies?

Answer 1

The derivative with respect to the normal is the rate of change in the direction of the normal: $$ \partial_n f=n\cdot\nabla f $$ where $n$ is the unit normal.

Along a curve piece $(\mathrm{d}x,\mathrm{d}y)$, the unit outward normal to a counterclockwise curve times the length of the curve piece is $(\mathrm{d}y,-\mathrm{d}x)=n\,\mathrm{d}s$. Thus, we get $$ \begin{align} \int n\cdot\nabla f\,\mathrm{d}s &=\int \nabla f\cdot(\mathrm{d}y,-\mathrm{d}x)\\ &=\int f_x\,\mathrm{d}y-f_y\,\mathrm{d}x \end{align} $$