We first look at the 2D Laplace equation , say on the upper half plane: $$\Delta u=0,\quad -\infty<x<\infty, y>0$$ $$u(x,0)=g(x),$$ where $g\in L^p(\mathbb{R})$ for some $1\leq p<\infty$. Then the general solution can be represented using the Poisson kernel $$P_y(x)=\frac{y}{\pi(y^2+x^2)},$$ with $$u(x,y)=(P_y*g)(x)=\frac{1}{\pi}\int_{-\infty}^\infty \frac{y}{y^2+(x-t)^2}g(t)dt.$$ Now if we define the following linear operator on $L^p(\mathbb{R})$: $$T_yg(x)=(P_y*g)(x).$$ Then we can verify that the family $\{T_y\}_{y\geq 0}$, satisfies the semi-group properties:
- $T_0=\mathrm{id}$, i.e. $T_0$ is the identity operator;
- $T_{y+s}=T_yT_s$ for any $y,s\geq 0$.
Thus we see that we can study solutions of the Laplace equation from the view of semi-group theory.
Here is my question: Can we perform similar analysis to the Poission equation? i.e. consider the solutions of the poisson equation from the view of semi-group theory? The Poisson equation is basically the laplace equation with a source term $$-\Delta u=f(x,y),\quad -\infty<x<\infty, y>0$$ $$u(x,0)=g(x),$$ here we use the same domain as above. In this case the general solution can be represented by using the Green's function: $$G(x,y)=\frac{1}{2\pi}\ln\sqrt{x^2+y^2},$$ with $$u(x,y)=\int_{\mathbb{R}\times\mathbb{R}^+}G(x-x',y-y')f(x',y')dx'dy'+\int_{\{y=0\}}g(x')\frac{\partial G}{\partial\mathbf{n}}(x-x',y-y')dS,$$ where in the second integral above $\mathbf{n}$ is the normal vector of $\{y=0\}$ pointing ourwards the domain $\mathbb{R}\times\mathbb{R}^+$. If we want to view the solution from semi-group theory, then we need to find a suitable Banach space $X$ and a family of bounded linear operators $\{T_t\}_{t\geq 0}$ on $X$ which form a semi-group. But I'm not sure whether this can be done. Any ideas on this question are greatly appreciated.