It comes from the steady-heat equation in the upper half-plane $\Bbb R_+^2$ .The solution of this question is to use Fourier transofrm so that you can transform this equation to a new one and the last term $e^{-2\pi |\xi| y}$ is the Fourier transform of the Poisson kernel for $\Bbb R_+^2$
However,I don’t understand why this kernel corresponds to $\Bbb R_+^2$ .In other words,if we exchange $\Bbb R_+^2$ to other planes,can we apply this method similarly?(I understand the accessibility of this method of Fourier transform requires “rapidly decreasing”but it doesn’t explain why it is feasible for $\Bbb R_+^2$ )
It occurs to me when I first meet the steady-state heat equation for a disc,the Poisson kernel for a disc which we use polar coordinates to represent it corresponds to this situation so naturally because the plane is a disc.But this time,I don’t get any links between Poisson kernel for $\Bbb R_+^2$ and the upper half plane?
What am I missing?
The Poisson problem for the upper half-plane $\mathbb{R}_{+}$ is $$ \nabla^2 f(x,y)=0\\ -\infty < x < \infty,\; 0 < y < \infty, \\ f(x,0) = g(x). $$ You could assume $g\in L^{\infty}$, and solve for a bounded solution $f$. Or you could assume $g\in L^p$ for some $1 \le p < \infty$ and require that $\sup_{y > 0} \int_{-\infty}^{\infty}|f(x,y)|^p dx < \infty$. The fundamental solution is $$ (Pg)(x,y)= \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y g(x')}{(x-x')^2+y^2}dx'. $$