Neumann Boundary Condition around a square

Question

Let's say there's a Poisson equation to solve. The domain is a unit square $\Omega=(0,1)\times(0,1)$. It is subject to Neumann boundary condition $\partial_nu=g(x,y)$, for $(x,y)\in\partial\Omega$, where $\partial_nu$ denotes differentiation in the direction of the outward normal to the boundary.

In the text it says the boundary condition implies $-u_x=g(0,y)$ in $(0<y<1)$, i.e., along $x=0$ and $u_y=g(x,1)$ in $(0<x<1)$, i.e., along $y=1$. How so?

Answer 1

For instance, on the left boundary ($x = 0, \, 0\leq y \leq 1$) the outward normal is $(-1,0)$. Assuming differentiability, since $\partial_n u = \partial_x u \, n_x + \partial_y u \, n_y$, you get the condition $$ \underbrace{\partial_n u (0,y)}_{=g(0,y)}=\partial_x u(0,y)\times (-1)+\partial_x u(0,y)\times 0 = -\partial_x u(0,y). $$

So, in fact, it is true that $-\partial_x u(0,y) = g(0,y)$. Similar considerations can be performed for the other sections of the boundary:

1. $\partial_x u (1,y) = g(1,y)$
2. $\partial_y u (x,1) = g(x,1)$
3. $-\partial_y u(x,0) = g(x,0)$