# Finding expression for poisson kernel of upper half plane

For the upper half plane, we have that the green function is

$$G(x,y) = -\frac{1}{2\pi}(\log|y-(x_1,x_2)| -\log|y-(x_1,-x_2)|)$$

The solution $$u(x)$$ of the dirichlet problem

$$-\Delta u = f\\u = g$$

is given by

$$u(x) = -\int_{\partial\Omega}\frac{\partial G(x,y)}{\partial n}\ dS(y) + \int_{\Omega} f(y)G(x,y)\ dy$$

For the poisson kernel we consider $$-\Delta u = f = 0$$ because $$u$$ is harmonic and then we consider only the integral

$$u(x) = -\int_{\partial\Omega}\frac{\partial G(x,y)}{\partial n}\ dS(y)$$

So I need to find $$\frac{\partial G}{\partial n}$$ where $$n$$ is the normal vector on the boundary of the half space, that is, on the region $$\{(x_1,x_2)| x_1\in\mathbb{R},x_2 = 0\}$$.

For me it looks like the expression for such unit normal is

$$n(y) = n((y_1,y_2)) = \left(0,\frac{-\mbox{sign}(y_1)y_1}{|y|}\right)$$

I just reflected the point $$y = (y_1,0)$$ to $$(0,-y_1)$$ in the case $$y_1>0$$ and to $$(0,y_1)$$ in the case $$y_1<0$$. And of course, divided everything by $$|y|$$ to make the norm $$1$$.

We need to take $$\frac{\partial G(x,y)}{\partial n} = \nabla G(x,y)\cdot n = \left(-\frac{1}{2\pi}\frac{1}{|y-x|}\frac{y-x}{|y-x|}\right)\cdot \left(0,\frac{-\mbox{sign}(y_1)y_1}{|y|}\right)$$

But this doesn't looks like it will give the formula Integral of Poisson Kernel for upper halfspace

What is wrong?

The normal $$\bar n$$ is parallel to $$Ox_2$$ axe, so the answer for the Poisson kernel is $$\pm\frac{\partial G(x,y)}{\partial y_2}|_{y_2=0}$$, the sign depending upon the direction of $$\bar n$$. And $$\bar n(y_1)=(0,\pm1)$$.