5
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

6
$\begingroup$

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 $$ \begin{align} \int n\cdot\nabla f\,\mathrm{d}s &=\int \nabla f\cdot(\mathrm{d}y,-\mathrm{d}x)\\ &=\int f_x\,\mathrm{d}y-f_y\,\mathrm{d}x \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.