# Normal derivative of the autonomous single layer potential

Let $G$ be the fundamental solution of the Laplace equation. Let $\Omega$ be an open and bounded subset of $\mathbb{R}^n$ regular enough. It is known that:

$$\int_{\partial \Omega}\frac{\partial G(x-y)}{\partial \nu(y)}d\sigma(y)= \begin{cases} -1 \mbox{ if } x\in \Omega,\\ -1/2 \mbox{ if } x\in \partial\Omega,\\ 0 \mbox{ if } x\in \mathbb{R}^n\setminus\bar\Omega.\\ \end{cases}$$ Where $\nu(y)$ is the unit outer normal to $\partial \Omega$ in $y$. In other words this is the jump formula for the double layer potential when the density is constant.

How about the value of the normal derivative of the single layer potential? Let $x_0\in\partial\Omega$. I want to compute: $$\int_{\partial \Omega}\frac{\partial G(x-y)}{\partial \nu(x_0)}d\sigma(y)$$

• This is covered in Taylors second PDE book, proposition 11.3, page 37. It involves the normal derivative of the single layer potential. If you can access that book, I suggest it, otherwise I can write down the formula for it. – mickep Feb 15 '15 at 18:52
• Thank you. Now I can't access that book, but I will in these days. If you can, could you only tell me the values of the integral, without the proof? – foo90 Feb 15 '15 at 19:21

OK, here is the formula (slightly different notation than yours): Let $$\mathcal{Sl}f(x)=\int_{\partial\Omega}f(y)E(x,y)\,dS(y).$$ Then $$\frac{\partial}{\partial \nu}\mathcal{Sl}f_{\pm}(x)=\frac{1}{2}\bigl(\mp f+N^{\#}f\bigr),$$ where $$N^{\#}f(x)=2\int_{\partial\Omega}f(y)\frac{\partial E}{\partial \nu_x}(x,y)\,dS(y).$$
• This is the jump formula for the single layer potential, right? I need, with this notation, to compute explicitly $$N^\#1(x)\int_{\partial \Omega}\frac{\partial E}{\partial \nu_x}dS(y).$$ It follows from the jump relation? – foo90 Feb 15 '15 at 19:51
• I have missed a $=$ – foo90 Feb 15 '15 at 20:02
• Yes, then you have to compute $\frac{\partial}{\partial\nu}\mathcal{Sl}1(x)$. – mickep Feb 16 '15 at 7:24