Semi-group theory and Poisson equation on the upper half plane

Question

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:

1. $T_0=\mathrm{id}$, i.e. $T_0$ is the identity operator;
2. $T_{y+s}=T_yT_s$ 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.

Answer 1

I suspect that the answer is no, at least it is if we're trying to do it in the exact same way as for the Laplace equation. Here's why:

The semigroup associated to a differential equation usually takes the "initial condition" (in the Laplace case, the boundary condition) as input and then returns the solution to the equation at some "time" $t$ ($y$, in the above case). So from our point of view, I would presume that we see the right-hand side $f$ as fixed and define $T_{y}g(x)$ to be the solution $u(x,y)$ of

$$-\Delta u=f(x,y),\quad -\infty<x<\infty, y>0,$$ $$ u(x,0)=g(x),$$ Then, $(T_y)_{y\geq 0}$ is a family of bounded linear operators. Do they form a semigroup? We have $$\Delta(T_{y}g)(x) = f(x,y),\quad y > 0.$$ Let $s > 0$. Defining $\hat g := T_{s}g$, we find that $$\Delta(T_{y+s}g)(x) = f(x,y+s),$$ but $$\Delta(T_{y}T_{s}g)(x) = \Delta(T_{y}\hat g)(x) = f(x,y).$$ So, for this approach to work, $f$ should be constant in $y$.

This is of course not a definitive answer to your question, but it seems to suggest to me that it is not possible to apply semigroup theory here in a meaningful way.

Answer 2

The answer is no, i.e. the solution operator of the boundary problem value of every non-homogeneous linear equation is not a semigroup of operators respect to any of the space or time variables involved. I will show this in two steps: I'll construct the general solution operator $T$ for the Dirichlet problem for the Poisson equation in the upper half plane and then show that this cannot be a semigroup.

First Step. Let's precisely define the Green function for a non homogeneous boundary value problem and explicitly calculate it for the Dirichlet problem in the upper half-plane for the Poisson equation as required by the OP, i.e. $$ \begin{cases} \Delta u(x,y)=f(x,y),\quad (x,y)\in\mathbb{R}\times\mathbb{R}^+\\ u(x,0)=g(x). \end{cases}\tag{1}\label{1} $$ Definition. The Green function of the (linear) boundary value problem $$ \begin{cases} P(x,D) u(x)=f(x),\quad x\in G\subset\mathbb{R}^n\\ Bu(x)=g(x),\quad x\in\partial G \end{cases}\tag{2}\label{2} $$ for the linear partial differential operator $P(x,D)$ is the distribution solution of the following associated boundary problem $$ \begin{cases} P(x,D) \mathscr{G}(x,t)=\delta(x-t),\quad x,t\in G\subset\mathbb{R}^n\\ B\mathscr{G}(x,t)=0,\quad x\in\partial G \end{cases}\tag{3}\label{3} $$ where

• $G$ is a domain with a "sufficiently regular" boundary $\partial G$,
• $B$ is a linear boundary operator defined on $\partial G$.

Notes.

• Clearly the Dirichlet problem \eqref{1} is of type \eqref{2} since $B u(x,y)=u|_{\partial G}=u(x,0)$ is a linear boundary operator, and
• a solution of problem \eqref{3} for \eqref{1} exists under reasonable conditions on $\partial G$ and on $P(x,D)$ (Vladimirov (1983) §29.1, p. 369).

The Green's function $\mathscr{G}((x,y),(t,s))$ for \eqref{1} is the solution of the following Dirichlet boundary value problem: $$ \begin{cases} \Delta \mathscr{G}((x,y),(t,s))=\delta((x,y)-(t,s)),\quad (x,y),(t,s)\in \mathbb{R}\times\mathbb{R}^+\\ \mathscr{G}((x,0),(t,s))=0. \end{cases}\tag{3'}\label{3'} $$ Note. The vector $(t,s)\in \mathbb{R}\times\mathbb{R}^+$ is a parameter which has an interesting physical interpretation in the theory of the static electric field (see Vladimirov (1983) §29.1 for the details)

The solution of problem \eqref{3'} has the form (Vladimirov (1983) §29.1, p. 368) $$ \mathscr{G}((x,y),(t,s))=\frac{1}{2\pi}\ln|(x,y)-(t,s)|+g((x,y),(t,s)) $$ where the first term on the right side is the fundamental solution of the laplacian while the second one is a function harmonic in the whole $\mathbb{R}\times\mathbb{R}^+$. In the problem posed by the OP, $$ \mathscr{G}((x,y),(t,s))=\frac{1}{2\pi}\big(\ln|(x,y)-(t,s)|-\ln|(x,y)-(t,-s)|\big).\tag{4}\label{4} $$ Now, noting that $\frac{\partial\mathscr{G}}{\partial\mathbf{n}}=\frac{\partial\mathscr{G}}{\partial t}$ on the boundary $\{y=0\}$ of the upper half plane, by Green's formula we obtain the formula for the general solution of \eqref{1}: $$ \begin{split} u(x,y)&=\int\limits_{\mathbb{R}\times\mathbb{R}^+}\mathscr{G}((x,y),(t,s))f(t,s)\mathrm{d}t\mathrm{d}s+\int\limits_{\{s=0\}}g(t)\frac{\partial\mathscr{G}}{\partial t}((x,y),(t,0))\mathrm{d}t\\ &=\int\limits_{\mathbb{R}\times\mathbb{R}^+}\mathscr{G}((x,y),(t,s))f(t,s)\mathrm{d}t\mathrm{d}s+\frac{1}{\pi}\int\limits_{-\infty}^\infty \frac{y}{y^2+(x-t)^2}g(t)\mathrm{d}t\\ &\overset{\mathrm{def}}{=} T(g;f)(x,y) \end{split}\tag{5}\label{5} $$ Second step. Now note that the operator $T$ defined by \eqref{5} is a semigroup only if $f\equiv 0$. To see this, suppose that we have homogeneous boundary conditions, i.e. $g\equiv 0$ and assume $y$ as the parameter of the hypothetical semigroup, i.e $T(g;f)(x,y)=T_y(g;f)(x)$: for $y=0$ we have $$ T_0(0;f)(x)\equiv 0 \quad \forall x\in\mathbb{R}\text{ by equation \eqref{4} } $$ If $T_y$ defined by \eqref{5} would be a semigroup, the equation above would imply that $$ T_y(0,f)(x)=T_{y+0}(0,f)=T_{y}T_{0}(0,f)(x)\equiv 0 $$ by the second property of the semigroup and by the linearity of $T$, and this is clearly false since the first term of \eqref{5} is not necessarily $\equiv0$. If instead of $y$ we try to assume $x$ as the parameter of the hypothetical semigroup by posing $T(g;f)(x,y)=T_x(g;f)(y)$, it is simple to see that $T_0\neq \mathrm{id}$.

Last notes

• This proposition holds for general boundary problems \eqref{2} since it can be proved that, for the general Green's function defined by \eqref{3} $$ B\mathscr{G}(x,t)=0,\quad t\in\partial G. $$ See again Vladimirov (1983) §29.1 for the details when $P(x,D)=\Delta$.
• Alberto Cialdea showed me the main argument of the second step in the proof above, and I would like to thank him publicly.

[1] Vladimirov, V. S. (1983)[1970], Equations of mathematical physics, Moscow: Mir Publishers, 2nd ed., pp. 464, MR0764399, Zbl 0207.09101 (the Zbmath review refers to the first English edition).