Let $\mathbb R^2_+=\{x=(x_1,x_2)\in \mathbb R^2 : x_2 >0 \}$ and let $H(x,y)=\frac{1}{2\pi}\ln|x-y|+\frac{1}{2\pi}\ln|x-\bar y|$ where if $y=(y_1,y_2)$ then $\bar y =(y_1,-y_2)$.
I need to calculate the directional derivative $\frac{\partial H(x,y)}{\partial \nu_x}$ where $\nu_x$ is the exterior normal at $x$ where $x\in \partial \mathbb R^2_+$.
I am supposed the get that $\frac{\partial H(x,y)}{\partial \nu_x}$ is constant for $x\in \partial \mathbb R^2_+$. But I am unsure what the exterior normal is. Is it the vector $(0,-1)$? If so then $\frac{\partial H(x,y)}{\partial \nu_x}=\nabla_x H(x,y) \cdot (0,-1) $
But when calculating $\frac{\partial H(x,y)}{\partial x_k}$ for $k \in \{1,2\}$ I wasn't able to show that $\frac{\partial H(x,y)}{\partial \nu_x}$ was constant in $x$.
The derivatives I got were $\frac{\partial H(x,y)}{\partial x_k}=\dfrac{x_k-y_k}{2\pi|x-y|^2}+\dfrac{x_k+y_k(-1)^k}{2\pi|x-\bar y |^2}$ for $k\in \{1,2\}$.
Any help is appreciated.
I believe I answered it. When $x$ is in the boundary of $\{x=(x_1,x_2) \in \mathbb R^2 : x_2>0\}$ then $x=(x_1,0)$. If we plug this $x$ into $\dfrac{\partial H(x,y)}{\partial \nu_x)}$ and take $\nu_x = (0,-1)$ then we see that this derivative is $0$.