Let $\Delta u = 0$ in the upper half space $z > 0$ in $\mathbb{R^3}$ with $\frac{\partial u}{\partial z} = f$ on $z = 0.$ Find the kernel $P(x,y)$ such that for $f$ continuous and vanishing for $|x|$ large, $$ u(x) = \int_{\mathbb{R^2}} P(x,y)f(y) \; dy $$ gives a solution to this problem which converges to zero as $|x| \to \infty.$

I tried to solve this using a Green's formula to provide a representation of $u$ at any point in the upper half space. Letting the upper half space be denoted $U,$ $$ \int_U (u\Delta w - w \Delta u) dy = \int_{\partial U} u \frac{\partial w}{\partial n} - w \frac{\partial u}{\partial n} \; dS, $$ with $w = \frac{1}{|x-y|}$ gives us $$ u(x) = \frac{1}{4\pi} \int_{\partial U} \bigg[\frac{1}{|x-y|} \frac{\partial u(y)}{\partial n} - u(y) \frac{\partial }{\partial n} \frac{1}{|x-y|}\bigg] \; dS - \frac{1}{4\pi} \int_{U} \frac{\Delta u(y)}{|x-y|} dy $$

Then because $\Delta u = 0$ in $U,$ and using the Neumann condition, I got $$ u(x) = \frac{1}{4\pi} \int_{\mathbb{R^2}} \bigg[\frac{f(y)}{|x-y|} - u(y) \frac{\partial }{\partial n} \frac{1}{|x-y|}\bigg] \; dy $$

Not really sure what to do with $u(y)$ restricted to $\partial U$ in the integral. It doesn't feel right. There is also that even in the statement of the problem, I am not sure if $x \in \mathbb{R}$ or $x \in \mathbb{R^3}.$ Anyone have any suggestions, solutions, comments? Thanks in advance!

  • $\begingroup$ Yes that is right, thanks for the catch. $\endgroup$ – Merkh Jun 1 '17 at 21:02
  • $\begingroup$ The discussion of Green's theorem here may be of interest: farside.ph.utexas.edu/teaching/jk1/lectures/node22.html. In particular, one should note the appearance of the surface dipole distribution $D(x)$ which is exactly the $u|_{\partial U}$ factor of interest. (In your case you've got $\rho=0$ in $U$ so therefore one of their terms doesn't appear in your equations.) The next page may also be illuminating. $\endgroup$ – Semiclassical Jun 2 '17 at 17:26

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