# Green’s Function and Its Properties

In this section, we would like to discuss Dirichlet boundary-value problem. The method is to use Green’s identity and Green’s second formula to transform the problem to another specialized Dirichlet boundary-value problem. In the process, we naturally derive Green’s function.

Generally speaking, There are two class of Dirichlet boundary-value problem for elliptic partial equations, i.e., the Dirichlet problem of harmonic equation

\begin{cases}
\Delta u=0, &\text{in }\Omega\\
u=\phi,&\text{on }\pt\Omega,
\end{cases}

and the Dirichlet problem of Poisson equation
\label{eq:2}
\begin{cases}
\Delta u=f, &\text{in }\Omega\\
u=\phi,&\text{on }\pt\Omega,
\end{cases}

where $f$ is a continuous function in $\bar\Omega$ and $\phi$ is a continuous function on $\pt\Omega$. Of course, the first problem can be viewed as a special case of the second one.

The elementary mathematics concern about the qualitative theory, i.e., the existence and uniqueness of the solution; whereas the applied mathematics concern more about the property of solution.

The elliptic equation always have classic solutions (compared with weak solution), we can say that it is complete. Thus, the elliptic equation is quite important in the constructing of general theory of PDE.

Let $u\in C^1(\bar\Omega)\cap C^2(\Omega)$ be a solution of \eqref{eq:2}, by Green’s representation
\label{eq:3}
u(x)=\int_\Omega\Gamma\Delta u\rd y
-\int_{\pt\Omega}\left(
\Gamma\frac{\pt u}{\pt\n}-u\frac{\pt\Gamma}{\pt\n}
\right)\rd S
thus, the only unknown term is $\frac{\pt u}{\pt\n}$ on $\pt\Omega$, we should note that we cannot prescribe $u$ and $\frac{\pt u}{\pt\n}$ on $\pt\Omega$ simultaneously. In fact, we will eliminate this term by adjusting $\Gamma$.

Our observation is that the Green’s representation is mostly similar to the Green’s second formula. Consider a harmonic function $\Phi\in C^1(\bar\Omega)\cap C^2(\Omega)$, by Green’s second formula, we have
$\int_\Omega \Phi\Delta u-u \Delta\Phi\rd y =\int_{\pt\Omega}\left(\Phi\frac{\pt u}{\pt\n}-u \frac{\pt\Phi}{\pt\n}\right)\rd S,$
i.e.,
\label{eq:4}
0=\int_\Omega\Phi\Delta u\rd y-\int_{\pt\Omega}\left(\Phi\frac{\pt u}{\pt\n}-u \frac{\pt\Phi}{\pt\n}\right)\rd S.

Set $G=\Gamma+\Phi$, and if we can find $\Phi$ (adjusting!), such that $G\equiv0$ on $\pt\Omega$, then, by \eqref{eq:3} and \eqref{eq:4}
\label{eq:5_1}
u(x)=\int_\Omega G\Delta u\rd y
+\int_{\pt\Omega}u\frac{\pt G}{\pt\n}\rd S.

Thus, we obtain the solution for the Dirichlet problem \eqref{eq:2}
\label{eq:5}
u(x)=\int_\Omega G(x,y) f(y)\rd y
+\int_{\pt\Omega}\phi(y)\frac{\pt G}{\pt\n_y}(x,y)\rd S.

Definition 1. The function $G(x,y)\eqdef \Gamma(x-y)+\Phi(x,y)$ defined as a function of $y\in\bar\Omega\setminus\set{x}$ for each fixed $x\in\Omega$ is called Green’s function of the domain $\Omega$. \eqref{eq:6} is called Green’s representation with Green’s function.

From the above discussion, we have transform the problem of solve \eqref{eq:2} to the following Dirichlet problem
\label{eq:6}
\begin{cases}
\Delta_y\Phi(x,y)=0, &y\in\Omega\\
\Phi(x,y)=-\Gamma(x-y),&y\in\pt\Omega,
\end{cases}

where $x\in\Omega$ is a fixed point, and $\Phi(x,\cdot)\in C^1(\bar\Omega)\cap C^2(\Omega)$.
Remark 1. Although we have transform the original Dirichlet problem to the special one \eqref{eq:7}, for a general domain, the Dirichlet problem is as hard as the original one. In despite of this, the Green’s function method has its own significance:
• If we obtain the Green’s function for some domain $\Omega$, then the existence of Dirichlet problem \eqref{eq:2} have solved and its solution can be represented by \eqref{eq:6};
• Use the solution \eqref{eq:6} we can discussion the property of solution;
• For some special domain, such as ball, half-space, the first octant, we can find Green’s function by elementary method, and the Dirichlet problem of these domain are important in the theory of PDE.

Now we turn to the properties of Green’s function.
Proposition 2. Suppose $\Gamma(x-y)$ is the fundamental solution of Laplace equation and $G(x,y)$ is the Green’s function on a domain $\Omega\subset\R^n$. Then for any fixed $x\in\Omega$, we have
1. $\Gamma(x-y) < G(x,y) < 0$, for any $y\in\Omega$ and $y\neq x$;
2. $G(x,y)$ as a function of $y$ is a harmonic function on $\Omega\setminus\set{x}$, and $\lim\limits_{y\to x}G(x,y)=\infty$, furthermore, if $n\geq3$, then $G(x,y)=O(1/|x-y|^{n-2})$;
3. $G(x,y)=G(y,x)$, for any $x,y\in\Omega$ and $x\neq y$;
4. $\int_{\pt\Omega}\frac{\pt G}{\pt\n}\rd S=1.$
Proof . we only proof (3) and leave others as an excise. Fix $x_1,x_2\in\Omega$ with $x_1\neq x_2$, we will prove that $G(x_1,x_2)=G(x_2,x_1)$. Set $G_i=G_i(y)=G(x_i,y)$, and $\Gamma_i=\Gamma_i(y)=\Gamma(x_i-y)$, $\Phi_i=\Phi_i(y)=\Phi(x_i,y)$, then $G_i=\Phi_i+\Gamma_i$, $i=1,2$. Apply Green's second formula in $\Omega\setminus(B_r(x_1)\cup B_r(x_2))$, $r < |x_1-x_2|/2$, we have $\begin{split} \int_{\Omega\setminus(B_r(x_1)\cup B_r(x_2))} &(G_1\Delta G_2-G_2\Delta G_1)\rd y =\int_{\pt\left(\Omega\setminus(B_r(x_1)\cup B_r(x_2))\right)} \left(G_1\frac{\pt G_2}{\pt\n}-G_2\frac{\pt G_1}{\pt\n}\right)\rd S\\ &=\left\{\int_{\pt\Omega}+\int_{\pt B_r(x_1)}+\int_{\pt B_r(x_2)}\right\} \left(G_1\frac{\pt G_2}{\pt\n}-G_2\frac{\pt G_1}{\pt\n}\right)\rd S \end{split}$ Note that $G_i$ is harmonic on $\Omega\setminus(B_r(x_1)\cup B_r(x_2))$ and vanished on $\pt\Omega$, thus $0=\int_{\pt B_r(x_1)}\left(G_1\frac{\pt G_2}{\pt\n}-G_2\frac{\pt G_1}{\pt\n}\right)\rd S +\int_{\pt B_r(x_2)}\left(G_1\frac{\pt G_2}{\pt\n}-G_2\frac{\pt G_1}{\pt\n}\right)\rd S.$ Now, since $\Phi_1=G_1-\Gamma_1$ and $G_2\in C^2(\bar B_r(x_1))$ are harmonic in $B_r(x_1)$, by Green's second formula $\int_{\pt B_r(x_1)}\left((G_1-\Gamma_1)\frac{\pt G_2}{\pt\n} -G_2\frac{\pt(G_1-\Gamma_1)}{\pt\n}\right)\rd S=0,$ similarly, $\int_{\pt B_r(x_2)}\left(G_1\frac{\pt (G_2-\Gamma_2)}{\pt\n} -(G_2-\Gamma_2)\frac{\pt G_1}{\pt\n}\right)\rd S=0.$ Thus, $0=\int_{\pt B_r(x_1)}\left(\Gamma_1\frac{\pt G_2}{\pt\n}-G_2\frac{\pt\Gamma_1}{\pt\n}\right)\rd S +\int_{\pt B_r(x_2)}\left(G_1\frac{\pt\Gamma_2}{\pt\n}-\Gamma_2\frac{\pt G_1}{\pt\n}\right)\rd S.$ Use the same skill as in the proof of Green's identity, we have, $\lim_{r\to0}\int_{\pt B_r(x_1)}\Gamma_1\frac{\pt G_2}{\pt\n} =\lim_{r\to0}\int_{\pt B_r(x_2)}\Gamma_2\frac{\pt G_1}{\pt\n}=0,$ and $\lim_{r\to0}\int_{B_r(x_1)}G_2\frac{\pt\Gamma_1}{\pt\n}=G_2(x_1), \quad\lim_{r\to0}\int_{B_r(x_2)}G_1\frac{\pt\Gamma_2}{\pt\n}=G_1(x_2),$ the result has obtained since $G(x_1,x_2)=G_1(x_2)=G_2(x_1)=G(x_2,x_1)$.
Problem 1. Complete the proof of Proposition prop:2.