# What is $\frac {\partial \Gamma} {\partial \nu}$ on $\partial B_{\rho} (y)$?

I am studying Laplace's equation from the book "Elliptic partial differential equations of second order" written by Gilbarg and Trudinger. Here I am struggling to grasp a concept regarding the fundamental solution of Laplace's equation. Let $n \geq 3$ then the fundamental solution of Laplace's equation at a point $y \in \Omega$ is given by

$$\Gamma (x-y) = \Gamma (|x-y|) = \frac {1} { n (2-n)\omega_n} |x-y|^{2-n},$$ where $x \in \Omega \setminus \{y \}$.Let $B_{\rho} (y)$ denote an open ball centered at $y$ having some small radius $\rho$ . Then this book claims that $\frac {\partial \Gamma} {\partial \nu} = -\Gamma'(\rho)$ on $\partial B_{\rho} (y)$ (where $\nu$ is the unit outward normal to $\partial (\Omega-B_{\rho}))$ just before the equation $(2.16)$ but I couldn't figure out why it should be so.

• So you are the same person posting this question? – user99914 Apr 15 '18 at 21:22
• No he is my friend.I also don't understand this part. So I opted to post it separately. – Dbchatto67 Apr 15 '18 at 21:24

At $x\in \partial B_\rho (y)$, the vector $v_x$ is given by

$$v_x =- \frac{ x-y}{|x-y|}.$$

Thus

$$\frac{\partial \Gamma}{\partial v} = v \cdot \nabla \Gamma =-\frac{x-y}{|x-y|} \cdot \left( \frac{n}{\omega_n} |x-y|^{1-n} \frac{x-y}{|x-y|}\right) = -\frac{n}{\omega_n} |x-y|^{1-n}.$$

On the other hand, writing $\rho = |x-y|$, so

$$\Gamma'(\rho) = \left( \frac{n}{(2-n)\omega_n} \rho^{2-n}\right)' = \frac{n}{\omega_n} \rho^{1-n}.$$

So $\frac{\partial \Gamma}{\partial v} = - \Gamma'(\rho)$.