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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?

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1 Answer 1

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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 $\bar n(y_1)=(0,\pm1)$.

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