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.

Please help me in this regard. Then it will be really helpful for me.

Thank you in advance.

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

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

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


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

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


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.